1TECHNION - THE ISRAEL INSTITUTE OF TECHNOLOGYFACULTY OF INDUSTRIAL ENGINEERING & MANAGEMENTINFORMATION-BASED COMPLEXITYOF
CONVEX PROGRAMMING
A. Nemirovski

Fall Semester 1994/95

2
Information-Based Complexity of Convex Programming
Goals: given a class of Convex Optimization problems, one may look for an efficient algorithm for the class, i.e., an algorithm with a ”good” (best possible, polynomial time,...) theoretical worst-case efficiency estimate on the class. The goal of the course is to present a numberof efficient algorithms for several standard classes of Convex Optimization problems.
The course deals with the black-box setting of an optimization problem (all known in advanceis that the problem belongs to a given ”wide” class, say, is convex, convex of a given degree ofsmoothness, etc.; besides this a priory qualitative information, we have the possibility to ask an”oracle” for quantitative local information on the objective and the constraints, like their values
and derivatives at a point). We present results on the associated with this setting complexityof standard problem classes (i.e. the best possible worst-case # of oracle calls which allows tosolve any problem from the class to a given accuracy) and focus on the corresponding optimalalgorithms.
Duration: one semester
Prerequisites: knowledge of elementary Calculus, Linear Algebra and of the basic conceptsof Convex Analysis (like convexity of functions/sets and the notion of subgradient of a convex
function) is welcomed, although is not absolutely necessary.
Contents:
Introduction: problem complexity and method efficiency in optimization
Methods with linear dimension-dependent convergencefrom bisection to the cutting plane scheme
how to divide a n-dimensional pie: the Center-of-Gravity method
the Outer Ellipsoid method
polynomial solvability of Linear Programming
the Inner Ellipsoid method
convex-concave games and variational inequalities with monotone operators
Large-scale problems and methods with dimension-independent convergence
subradient and mirror descent methods for nonsmooth convex optimization
optimal methods for smooth convex minimizationstrongly convex unconstrained problems
How to solve a linear system: optimal iterative methods for unconstrained convex quadratic
minimization

3
About Exercises

The majority of Lectures are accompanied by the ”Exercise” sections. In several
cases, the exercises are devoted to the lecture where they are placed; sometimes they
prepare the reader to the next lecture.
The mark ∗ at the word ”Exercise” or at an item of an exercise means that you
may use hints given in Appendix ”Hints”. A hint, in turn, may refer you to thesolution of the exercise given in the Appendix ”Solutions”; this is denoted by themark + . Some exercises are marked by + rather than by ∗ ; this refers you directly
to the solution of an exercise.
Exercises marked by # are closely related to the lecture where they are placed;
it would be a good thing to solve such an exercise or at least to become acquainted
with its solution (if any is given).
Exercises which I find difficult are marked with > .
The exercises, usually, are not that simple. They in no sense are obligatory, andthe reader is not expected to solve all or even the majority of the exercises. Those
who would like to work on the solutions should take into account that the orderof exercises is important: a problem which could cause serious difficulties as it is
becomes much simpler in the context (at least I hope so).

4

Contents
1 Introduction: what the course is about
1.1 Example: one-dimensional convex problems . . . . . . . . . . . . . . . . . . . . .
1.2 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Exercises: Brunn, Minkowski and convex pie . . . . . . . . . . . . . . . . . . . .
1.3.1 Prerequisites . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3.2 Brunn, Minkowski and Convex Pie . . . . . . . . . . . . . . . . . . . . . .

9
9
17
17
17
18

2 Methods with linear convergence, I
2.1 Class of general convex problems: description and complexity . . . . . . . . . . .
2.2 Cutting Plane scheme and Center of Gravity Method . . . . . . . . . . . . . . . .
2.2.1 Case of problems without functional constraints . . . . . . . . . . . . . .
2.3 The general case: problems with functional constraints . . . . . . . . . . . . . . .
2.4 Exercises: Extremal Ellipsoids . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4.1 Tschebyshev-type results for sums of random vectors . . . . . . . . . . . .

29
29
31
31
36
40
46

3 Methods with linear convergence, II
3.1 Lower complexity bound . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 The Ellipsoid method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2.1 Ellipsoids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2.2 The Ellipsoid method . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3 Exercises: The Center of Gravity and the Ellipsoid methods . . . . . . . . . . . .
3.3.1 Is it actually difficult to find the center of gravity? . . . . . . . . . . . . .
3.3.2 Some extensions of the Cutting Plane scheme . . . . . . . . . . . . . . . .

51
51
56
57
59
61
61
65

4 Polynomial solvability of Linear Programming
4.1 Classes P and NP . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 Linear Programming . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2.1 Polynomial solvability of FLP . . . . . . . . . . . . . . . . . . . . . . . . .
4.2.2 From detecting feasibility to solving linear programs . . . . . . . . . . . .
4.3 Exercises: Around the Simplex method and other Simplices . . . . . . . . . . . .
4.3.1 Example of Klee and Minty . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3.2 The method of outer simplex . . . . . . . . . . . . . . . . . . . . . . . . .

75
75
78
80
82
84
84
85

5

6

CONTENTS

5 Linearly converging methods for games
5.1 Convex-concave games . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2 Cutting plane scheme for games: updating localizers . . . . . . . . . . . . . . . .
5.3 Cutting plane scheme for games: generating solutions . . . . . . . . . . . . . . .
5.4 Concluding remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.5 Exercises: Maximal Inscribed Ellipsoid . . . . . . . . . . . . . . . . . . . . . . . .

87
87
89
90
94
95

6 Variational inequalities with monotone operators
101
6.1 Variational inequalities with monotone operators . . . . . . . . . . . . . . . . . . 101
6.2 Cutting plane scheme for variational inequalities . . . . . . . . . . . . . . . . . . 108
6.3 Exercises: Around monotone operators . . . . . . . . . . . . . . . . . . . . . . . . 111
7 Large-scale optimization problems
115
7.1 Goals and motivations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
7.2 The main result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116
7.3 Upper complexity bound: the Gradient Descent . . . . . . . . . . . . . . . . . . . 117
7.4 The lower bound . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
7.5 Exercises: Around Subgradient Descent . . . . . . . . . . . . . . . . . . . . . . . 122
8 Subgradient Descent and Bundle methods
127
8.1 Subgradient Descent method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
8.2 Bundle methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132
8.2.1 The Level method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133
8.2.2 Concluding remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137
8.3 Exercises: Mirror Descent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138
9 Large-scale games and variational inequalities
143
9.1 Subrgadient Descent method for variational inequalities . . . . . . . . . . . . . . 144
9.2 Level method for variational inequalities and games . . . . . . . . . . . . . . . . . 147
9.2.1 Level method for games . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147
9.2.2 Level method for variational inequalities . . . . . . . . . . . . . . . . . . . 150
9.3 Exercises: Around Level . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152
9.3.1 ”Prox-Level” . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152
9.3.2 Level for constrained optimization . . . . . . . . . . . . . . . . . . . . . . 154
10 Smooth convex minimization problems
159
10.1 Traditional methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160
10.2 Complexity of classes Sn (L, R) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161
10.2.1 Upper complexity bound: Nesterov’s method . . . . . . . . . . . . . . . . 162
10.2.2 Lower bound . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166
10.2.3 Appendix: proof of Proposition 10.2.1 . . . . . . . . . . . . . . . . . . . . 169
11 Constrained smooth and strongly convex problems
171
11.1 Composite problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171
11.2 Gradient mapping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172
11.3 Nesterov’s method for composite problems . . . . . . . . . . . . . . . . . . . . . . 175

CONTENTS

7

11.4 Smooth strongly convex problems . . . . . . . . . . . . . . . . . . . . . . . . . . . 178
12 Unconstrained quadratic optimization
181
12.1 Complexity of quadratic problems: motivation . . . . . . . . . . . . . . . . . . . 181
12.2 Families of source-representable quadratic problems . . . . . . . . . . . . . . . . . 183
12.3 Lower complexity bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184
12.4 Complexity of linear operator equations . . . . . . . . . . . . . . . . . . . . . . . 187
12.5 Ill-posed problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192
12.6 Exercises: Around quadratic forms . . . . . . . . . . . . . . . . . . . . . . . . . . 192
13 Optimality of the Conjugate Gradient method
195
13.1 The Conjugate Gradient method . . . . . . . . . . . . . . . . . . . . . . . . . . . 196
13.2 Main result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197
13.3 Proof of the main result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198
13.3.1 CGM and orthogonal polynomials . . . . . . . . . . . . . . . . . . . . . . 198
13.3.2 Expression for inaccuracy . . . . . . . . . . . . . . . . . . . . . . . . . . . 201
13.3.3 Momentum inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201
13.3.4 Proof of (13.3.20) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202
13.3.5 Concluding the proof of Theorem 13.2.1 . . . . . . . . . . . . . . . . . . . 204
13.4 Exercises: Around Conjugate Gradient Method . . . . . . . . . . . . . . . . . . . 206
14 Convex Stochastic Programming
211
14.1 Stochastic Approximation: simple case . . . . . . . . . . . . . . . . . . . . . . . . 214
14.1.1 Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214
14.1.2 The Stochastic Approximation method . . . . . . . . . . . . . . . . . . . . 214
14.1.3 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216
14.2 MinMax Stochastic Programming problems . . . . . . . . . . . . . . . . . . . . . 218
Hints to exercises

221

Solutions to exercises

235

8

CONTENTSLecture 1

Introduction: what the course is
about
What we are interested in the course are theoretically efficient methods for convex optimization
problems. Almost each word in the previous sentence should be explained, and this explanation,that is, formulation of our goals, is the main thing I am going to speak about today. I believethat the best way to explain what we are about to do is to start with a simple example one-dimensional convex minimization - where everything is seen.

1.1

Example: one-dimensional convex problems

Consider one-dimensional convex problems
minimizef (x) s.t. x ∈ G = [a, b],

where [a, b] is a given finite segment on the axis. It is also known that our objective f is acontinuous convex function on G; for the sake of simplicity, assume that we know bounds, letthem be 0 and V , for the values of the objective on G. Thus, all we know about the objectiveis that it belongs to the family
P = {f : [a, b] → R | f is convex and continuous; 0 ≤ f (x) ≤ V, x ∈ [a, b]}.
And what we are asked to do is to find, for a given positive ε, an ε-solution to the problem, i.e.,a point x̄ ∈ G such thatf (x̄) − f ∗ ≡ f (x̄) − min f ≤ ε.
G

Of course, our a priori knowledge on the objective given by the inclusion f ∈ P, is, for smallε, far from being sufficient for finding an ε-solution, and we need some source of quantitative
information on the objective. The standard assumption here which comes from the optimization
practice is that we can compute the value and a subgradient of the objective at a point, i.e., wehave access to a subroutine, an oracle O, which gets, as an input, a point x from our segment
and returns the value f (x) and a subgradient f 0 (x) of the objective at the point.
We have subject the input to the subroutine to the restriction a < x < b, since the objective,
generally speaking, is not defined outside the segment [a, b], and its subgradient might be undefined at the endpoints of the segment as well. I should also add that the oracle is not uniquely
9

10

LECTURE 1. INTRODUCTION: WHAT THE COURSE IS ABOUT

defined by the above description; indeed, at some points f may have a ”massive” set of subgradients, not a single one, and we did not specify how the oracle at such a point chooses the
subgradient to be reported. We need exactly one hypothesis of this type, namely, we assume
the oracle to be local: the information on f reported at a point x must be uniquely defined by
the behaviour of f in a neighbourhood of x:{f, f¯ ∈ P, x ∈ int G, f ≡ f¯ in a neighbourhood of x } ⇒ O(f, x) = O(f¯, x).
What we should do is to find a method which, given on input the desired value of accuracyε, after a number of oracle calls produces an ε-solution to the problem. And what we areinterested in is the most efficient method of this type. Namely, given a method which solves
every problem from our family to the desired accuracy in finite number of oracle calls, let us
define the worst-case complexity N of the method as the maximum, over all problems from thefamily, of the number of calls; what we are looking for is exactly the method of the minimal
worst-case complexity. Thus, the question we are interested in is
Given- the family P of objectives f ,- a possibility to compute values and subgradients of f at a point of (a, b),- desired accuracy ε,what is the minimal #, Compl(ε), of computations of f and f 0 which is sufficient,
for all f ∈ P, to form an ε-minimizer of f ? What is the corresponding - i.e., theoptimal - minimization method?
Of course, to answer the question we should first specify the notion of a method. This isvery simple task. Indeed, let us think what a method, let it be called M, could be. It should
perform sequential calls for the oracle, at i-th step forwarding to it certain input xi ∈ (a, b), let
us call this input i-th search point. The very first input x1 is generated by the method whenthe method has no specific information on the particular objective f the method is applied to;thus, the first search point should be objective-independent:
x1 = S1M .

(1.1.1)

Now, the second search point is generated after the method knows the value and a subgradientof the objective at the first search point, and x2 should be certain function of this information:x2 = S2M (f (x1 ), f 0 (x1 )).

(1.1.2)

Similarly, i-th search point is generated by the method when it already knows the values andthe subgradients of f at the previous search points, and this is all the method knows about f
so far, so that i-th search point should be certain function of the values and the subgradients of
the objective at the previous search points:
xi = SiM (f (x1 ), f 0 (x1 ); ...; f (xi−1 ), f 0 (xi−1 )).

(1.1.3)

We conclude that the calls to the oracle are defined by certain recurrence of the type (1.1.3);
the rules governing this recurrence, i.e., the functions SiM (·), are specific for the method andform a part of its description.

1.1. EXAMPLE: ONE-DIMENSIONAL CONVEX PROBLEMS

11

I have said that the search rules SiM form only a part of the description of the method;indeed, the method should sometime stop and form the result. The moment of termination,same as the result found by the method, also should depend only on information accumulated
to the corresponding moment; we may assume that there is a sequence of termination tests functions
TiM (·) ∈ {STOP, CONTINUE}(1.1.4)taking values in the indicated two-element set, and the method terminates at the very first
moment i when it turns out that
TiM (f (x1 ), f 0 (x1 ); ...; f (xi ), f 0 (xi )) = STOP.
At this moment the method forms the result of its work:x̄(M, f ) = RiM (f (x1 ), f 0 (x1 ); ...; f (xi ), f 0 (xi ));(1.1.5)the termination tests TiM , same as the rules RiM for forming the result, also are a part of thedescription of the method.
Given the search rules, the termination tests and the rules for forming the result, we cancompletely define the behaviour of the method on every problem from our family, and we may
identify the method with the collection of these rules. Thus, by definitiona method is a collection of the search rules Si , the termination tests Ti and therules Ri for forming the result
In order to get the widest possible definition, we do not subject the rules comprising a method
to any further restrictions like ”computability in finitely many arithmetic operations”; the rules
might be arbitrary functions of the information on the problem accumulated to the step when
the rule should be used.
Now, the number of steps performed by a method M as applied to a problem f from ourfamily is called the complexity Compl(M, f ) of the method at f (this is a positive integer or +∞depending on whether the method stops at f or works at this problem forever). The complexityof the method on the whole family of problems is, by definition, the maximum of the complexity
at a problem f over all f :Compl(M) = max Compl(M, f ).
f ∈PThis is simply the worst-case # of oracle calls made by the method.
Similarly, we define the accuracy (it would be better to say inaccuracy) of the method M
at a problem f as
Accur(M, f ) = f (x̄(M, f )) − min f,Gi.e., as the residual, in the values of the objective, of the result obtained by the method asapplied to f , and define the accuracy of the method at the whole family looking at the worst
case:
Accur(M) = max Accur(M, f ).
f ∈PWith our now terminology, the problem of finding the optimal method is

12

LECTURE 1. INTRODUCTION: WHAT THE COURSE IS ABOUT
given the family
P = {f : G = [a, b] → R | f is convex and continuous on G , 0 ≤ f ≤ V }of problems and an ε > 0, find among the methods M with the accuracy on the family
not worse than ε that one with the smallest possible complexity on the family.

The complexity of the associated - optimal - method, i.e., the function
Compl(ε) = min{Compl(M) | Accur(M) ≤ ε}is called the complexity of the family. Thus, the question we are interested in is to find thecomplexity of the family P of one-dimensional convex minimization problems along with theassociated optimal method.
The answer to the question is given by the following statement.
Theorem 1.1.1 The complexity of the family in question satisfies the inequalities
1
VV
log2 ( ) − 1 < Compl(ε) ≤clog2 ( )b, 0 < ε < V.
5εε(1.1.6)The method associated with the upper bound (and thus optimal in complexity, up to an
absolute constant factor), is the usual bisection terminated after N =clog2 (V /ε)b steps.
Note that the range of values of ε in our statement is (0, V ), and this is quite natural: since allfunctions from the family take their values between 0 and V , any point of the segment solves
every problem from the family to the accuracy V , so that a nontrivial optimization problem
occurs only when ε < V .
Now let us prove the theorem.
10. Upper bound. I hope all of you remember what is the bisection method. In order tominimize f , we start with the midpoint of our segment, i.e., with
x1 =

a+b
,
2

and ask the oracle about the value and a subgradient of the objective at the point. If thesubgradient is zero, we are done - we have found an optimal solution. If the subgradient is
positive, then the function, due to convexity, to the right of x1 is greater than at the point, andwe may cut off the right half of our initial segment - the minimum for sure is localized in theremaining part. If the subgradient is negative, then we may cut off the left half of the initial
segment: Thus, we either terminate with an optimal solution, or find a new segment, twice
smaller than the initial domain, which for sure localizes the set of optimal solutions. In thislatter case we repeat the procedure, with the initial domain replaced by the new localizer, andso on. After we have performed the number of steps indicated in the formulation of the theorem
we terminate and form the result as the best - with the minimal value of f - of the search points
we have looked through:x̄ ∈ {x1 , ..., xN }; f (x̄) = min f (xi ).
1≤i≤N

Let me note that traditionally the approximate solution given by the bisection method is identified with the last search point (which is clearly at at the distance at most (b − a)2−N from

1.1. EXAMPLE: ONE-DIMENSIONAL CONVEX PROBLEMS

a)

b)

13

c)

The bisection step. a) f 0 (x1 ) > 0; b) f 0 (x1 ) = 0; c) f 0 (x1 ) < 0[ the dashed segment is the new localizer of the optimal set]

the optimal solution), rather than with the best point found so far. This traditional choice has
small in common with our accuracy measure (we are interested in small values of the objective
rather than in closeness to optimal solution) and is simply dangerous, as you can see from thefollowing example:

Here during the first N − 1 steps everything looks as if we were minimizing f (x) = x, so that the
N -th search point is xN = 2−N ; our experience is misleading, as you see from the picture, andthe relative accuracy of xN as an approximate solution to the problem is very bad, something
like V /2.
By the way, we see from this example that the evident convergence of the search points
to the optimal set at the rate at least 2−i does not imply automatically certain fixed rate ofconvergence in terms of the objective; it turns out, anyhow, that such a rate exists, but for thebest points found so far rather than for the search points themselves.
Let me present you an extremely simple and rather instructive proof of the rate of convergencein terms of the objective.
Let us start with the observation that if GN is the final localizer of optimum found during
the bisection, then outside the localizer the value of the objective is at least that one at the bestof the search points, i.e., at least the value at the approximate solution x̄ found by bisection:f (x) ≥ f (x̄) ≡ min f (xi ), x ∈ G\GN .
1≤i≤NIndeed, at a step of the method we cut off only those points of the domain G where f is atleast as large as at the current search point and is therefore ≥ its value at the best of the search
points, that is, to the value at x̄; this is exactly what was claimed.

14

LECTURE 1. INTRODUCTION: WHAT THE COURSE IS ABOUT

Now, let x∗ be an optimal solution to the problem, i.e., a minimizer of f ; as we know, such aminimizer does exist, since our continuous objective for sure attains its minimum over the finite
segment G. Let α be a real greater than 2−N and less than 1, and consider α-contraction of thesegment G to x∗ , i.e., the set
Gα = (1 − α)x∗ + αG ≡ {(1 − α)x∗ + αz | z ∈ G}.
This is a segment of the length α(b − a), and due to our choice of α the length is greater thanthat one of our final localizer GN . It follows that Gα cannot be inside the localizer, so that there
is a point, let it be y, which does not belong to the final localizer and belongs to Gα :∃y ∈ Gα : y 6∈ int GN .
Since y belongs to Gα , we havey = (1 − α)x∗ + αz
for some z ∈ G, and from convexity of f it follows thatf (y) ≤ (1 − α)f (x∗ ) + αf (z),which can be rewritten as
f (y) − f ∗ ≤ α (f (z) − f ∗ ) ≤ αV,
so that y is an αV -solution to the problem. On the other hand, y is outside GN , where, as wealready know, f is at least f (x̄):
f (x̄) ≤ f (y).
We conclude thatf (x̄) − f ∗ ≤ f (y) − f ∗ ≤ αV.
Since α can be arbitrary close to 2−N , we come tof (x̄) − f ∗ ≤ 2−N V = 2−clog2 (V /ε)b V ≤ ε.
Thus,
Accur(BisectionN ) ≤ ε.
The upper bound is justified.
20 . Lower bound. At the first glance, it is not clear where from could one get a lower boundfor the complexity of the family. Indeed, to find an upper bound means to invent a method
and to evaluate its behaviour on the family - this is what people in optimization are doing all
their life. In contrast to this, to find a lower bound means to say something definite about all
methods from an extremely wide class, not to investigate a particular method. Nevertheless, as
we shall see in a while, the task is quite tractable.
To simplify explanation, consider the case when the domain G of our problems is the segment[−1, 1] and the objectives vary from 0 to 1, i.e., let V be equal to 1; due to similarity reasons,with these assumptions we do not loose generality. Thus, given an ε ∈ (0, 1) we should provethat the complexity of our family for this value of ε is greater than11log2 ( ) − 1.
5ε1.1. EXAMPLE: ONE-DIMENSIONAL CONVEX PROBLEMS

15

In other words, we should prove that if M is a method which solves all problems from thefamily to the accuracy ε, then the complexity K 0 of the method on the family, i.e., the worst
case number of steps is at least the aforementioned quantity. Let M solve every problem tothe accuracy ε in no more than K 0 steps. By adding redundant steps, we may assume that M
performs exactly
K = K0 + 1
steps at every problem and that the result of M as applied to a problem is nothing but the last,
K-th search point.
Now let us prove the following basic lemma:
Lemma 1.1.1 For any i, 0 ≤ i ≤ K, our family contains an objective fi with the following two
properties:(1i ) : there is a segment ∆i ⊂ G = [−1, 1] - the active segment of fi - of the lengthδi = 21−2iwhere fi is a modulus-like function, namely,fi (x) = ai + 2−3i |x − ci |, x ∈ ∆i ,ci being the midpoint of ∆i ;(2i ) : The first i points x1 , ..., xi generated by M as applied to fi , are outside ∆i .
Proof will be given by induction on i.
Base i = 0 is immediate: we set
f0 (x) = |x|,∆0 = G = [−1, 1],

thus ensuring (10 ). Property (20 ) holds true by trivial reasons - when i = 0, then there are nosearch points to be looked at.
Step i ⇒ i + 1: Let fi be the objective given by our inductive hypothesis, let ∆i be the active
segment of this objective and let ci be the midpoint of the segment.
Let also x1 , ..., xi , xi+1 be the first i + 1 search points generated by M as applied to fi .
According to our inductive hypothesis, the first i of these points are outside the active segment.
In order to obtain fi+1 , we modify the function fi in its active segment and do not vary thefunction outside the segment. The way we modify fi in the active segment depends on whether
xi+1 is to the right of the midpoint ci of the segment (”right” modification), or this is not thecase and xi+1 either coincides with ci or is to the left of the point (”left” modification).
The ”right” modification is as follows: we replace the modulus-like in its active segment
function fi by a piecewise linear function with three linear pieces, as is shown on Figure 1.1.
Namely, we do not change the slope of the function in the initial 1/14 part of the segment, thenchange the slope from −2−3i to −2−3(i+1) and make a new breakpoint at the end ci+1 of thefirst quarter of the segment ∆i . Starting with this breakpoint and till the right endpoint of theactive segment, the slope of the modified function is 2−3(i+1) . It is easily seen that the modified
function at the right endpoint of ∆i comes to the same value as that one of fi and that themodified function is convex on the whole axis.

16

LECTURE 1. INTRODUCTION: WHAT THE COURSE IS ABOUT

Figure 1.1: ”Right” modification of fi . The slopes for the right modification, from left to right,
are −2−3i , −2−3(i+1) , 2−3(i+1) , the breakpoints u and ci+1 are at the distances |∆i |/14, |∆i |/4from the left endpoint of the active segment ∆i of fi . The bold segment on the axis is the active
segment of the modified function (right modification).
In the case of the ”left” modification, i.e., when xi+1 ≤ ci , we act in the ”symmetric” manner,13
so that the breakpoints of the modified function are at the distances 43 |∆i | and 14|∆i | from the−3(i+1)
left endpoint of ∆i , and the slopes of the function, from left to right, are −2, 2−3(i+1) and
2−3i .
Let us verify that the modified function fi+1 satisfies the requirements imposed by the lemma.
As we have mentioned, this is a convex continuous function; since we do not vary fi outside thesegment ∆i and do not decrease it inside the segment, the modified function takes its values in(0, 1) together with fi . It suffices to verify that fi+1 satisfies (1i+1 ) and (2i+1 ).
(1i+1 ) is evident by construction: the modified function indeed is modulus-like with required
slopes in a segment of a required length. What should be proved is (2i+1 ), the claim that themethod M as applied to fi+1 during the first i + 1 step does not visit the active segment of fi+1 .
To prove this, it suffices to prove that the first i + 1 search points generated by the method asapplied to fi+1 are exactly the search point generated by it when minimizing fi , i.e., they are
the points x1 , ..., xi+1 . Indeed, these latter points for sure are outside the new active segment the first i of them due to the fact that they even do not belong to the larger segment ∆i , andthe last point, xi+1 - by our construction, which ensures that the active segment of the modified
function and xi+1 are separated by the midpoint ci of the segment ∆i .
Thus, we come to the necessity to prove that x1 , ..., xi+1 are the first i + 1 points generated
by M as applied to fi+1 . This is evident: the points x1 , ..., xi are outside ∆i , where fi andfi+1 coincide; consequently, the information - the values and the subgradients - on the functions
along the sequence x1 , ..., xi also is the same for both of the functions. Now, by definition of amethod the information accumulated by it during the first i steps uniquely determines the firsti+1 search points; since fi and fi+1 are indistinguishable in a neighbourhood of the first i search
points generated by M as applied to fi , the initial (i + 1)-point segments of the trajectories of
M on fi and on fi+1 coincide with each other, as claimed.
Thus, we have justified the inductive step and therefore have proved the lemma.
It remains to derive from the lemma the desired lower complexity bound. This is immediate.

1.2. CONCLUSION

17

According to the lemma, there exists function fK in our family which is modulus-like in its
active segment and is such that the method during its first K steps does not visit this active
segment. But the K-th point xK of the trajectory of M on fK is exactly the result found by themethod as applied to the function; since fK is modulus-like in ∆K and is convex everywhere, it∗ at the midpoint c of the segment ∆ and outside ∆ is greater thanattains its minimum fKKKK∗ + 2−3K × 2−2k = 2−5k (the product here is half of the length of ∆ times the slope of f ).
fK
K
K
Thus,
∗
fK (x̄(M, fK )) − fK
> 2−5K .
On the other hand, M, by its origin, solves all problems from the family to the accuracy ε, andwe come to
2−5K < ε,i.e., to1
1
K ≡ K 0 + 1 < log2 ( ).
5εas required in our lower complexity bound.

1.2

Conclusion

The one-dimensional situation we have investigated is, of course, very simple; I spoke about it
only to give you an impression of what we are going to do. In the main body of the course weshall consider much more general classes of convex optimization problems - multidimensional
problems with functional constraints. Same as in our simple one-dimensional example, weshall ask ourselves what is the complexity of the classes and what are the corresponding optimal
methods. Let me stress that these are optimal methods we mainly shall focus on - it is much moreinteresting issue than the complexity itself, both from mathematical and practical viewpoint.
In this respect, one-dimensional situation is not typical - it is easy to guess that the bisection
should be optimal and to establish its rate of convergence. In several dimensions situation is far
from being so trivial and is incomparably more interesting.

1.3

Exercises: Brunn, Minkowski and convex pie

1.3.1

Prerequisites

Prerequisites for the exercises of this section are:the Separation Theorem for convex sets:
if G is a convex subset of Rn with a nonempty interior and x ∈ Rn does notbelong to the interior of G, then x can be separated from G by a linear functional,i.e., there exists a nonzero e ∈ Rn such thateT x ≥ eT y∀y ∈ G.

The Caratheodory Theorem. To formulate this theorem, let us start with the fundamental
notion of an extreme point of a convex set Q ⊂ Rn . Namely, a point x ∈ G is called an extreme

18

LECTURE 1. INTRODUCTION: WHAT THE COURSE IS ABOUT

point of G, if it cannot be represented as the midpoint of a nontrivial segment belonging to G,
i.e., if the relation
1
1
x = x0 + x00
22
with x0 and x00 belonging to G is possible only if x0 = x00 = x. For example, the extreme pointsof the segment [0, 1] ⊂ R are 0 and 1; the extreme points of a triangle in the plane are thevertices of the triangle; the extreme points of a circle in a plane are the points belonging to theboundary circumference.
The Caratheodory Theorem is as follows:
Let G be a(a) nonempty,(b) closed,(c) boundedand(d) convex
subset of Rn . Then the set Gext of extreme points of G is nonempty; moreover, every
point x ∈ G can be represented as a convex combination of at most n + 1 extreme
points of G: there exist x1 , ..., xn+1 ∈ Gext and nonnegative λ1 , ..., λn+1 with theunit sum such that
x=

n+1
X

λ i xi .

i=1

Corollary:Let x1 , ..., xm be certain points of Rn . Assume that a point x ∈ Rn can be
represented as a convex combination of x1 , ..., xm . Then x can be also represented
as a convex combination of at most n + 1 points from the set {x1 , ..., xm }.
Indeed, the convex hull G of {x1 , ..., xm }, i.e., the set of all convex combinations of x1 , ..., xm ,is a nonempty, closed and bounded convex subset of Rn (why?), and the extreme points of thisconvex hull clearly belong to {x1 , ..., xm } (why?), and it suffices to apply to G the Caratheodory
Theorem.

1.3.2

Brunn, Minkowski and Convex Pie

Consider the following game with two players - you and me. I am cooking a pie Q, which should
be a closed and bounded convex domain in the space Rn (the word ”domain” means that theinterior of Q is nonempty). After the pie is cooked, I show it to you and offer to divide it in thefollowing way: you choose a point x ∈ Rn ; given this point, I perform a cut passing through thepoint, i.e., choose a half-space Π with the boundary hyperplane passing through x:
Π = {y ∈ Rn | eT y > eT x}(e is a nonzero vector) and take, as my part, the intersection of Q with the half-space, and you
take the remaining part of the pie, i.e., the set
Qe = {y ∈ Q | eT y ≤ eT x},

1.3. EXERCISES: BRUNN, MINKOWSKI AND CONVEX PIE

19

x
e

Figure 1.2: Dividing a pie: you choose x, I choose Π and take the dashed part of the pie
see Figure 1.2. How should you choose the point x in order to get a ”substantial part” of thepie independently of how I have cooked the pie and perform the cut through the point chosen
by you, i.e., in order to ensure ”not too small” ratio
qe = Voln (Qe )/ Vol(Q)n(Voln stands for n-dimensional volume).
Some observations may be done immediately.
Exercise 1.3.1 # Prove that if x 6∈ int Q, then I can give you ”almost nothing”, i.e., I canchoose e resulting in Voln (Qe ) = 0.
Recall that a simplex in Rn is the convex hull of n + 1 points (vertices of the simplex) which areaffinely independent (the smallest affine subspace containing all the vertices is the whole Rn )).
Exercise 1.3.2 #∗ Prove that if I have cooked a simplex Q, then, independently of how you
choose x, I can leave you no more than an -th part of the pie, an = (n/(n + 1))n ≥ exp{−1},i.e., given x, I can choose e in such a way that
Voln (Qe ) ≤ an Voln (Q).
Prove that if you choose x not as the barycenter (the arithmetic mean of the vertices) of thesimplex, then I can make the above inequality strict.
A surprising fact is that the upper bound an on the worst-case value of your part is sharp:
given a closed and bounded convex domain Q ⊂ Rn , you can choose x ∈ int Q
in a way which ensures that

Voln {y ∈ Q | eT y ≤ eT x} ≥ an Vol(Q)nfor any nonzero e, where


an =

n
n+1

n

≥ exp{−1}.

(1.3.1)20

LECTURE 1. INTRODUCTION: WHAT THE COURSE IS ABOUT
In order to ensure this inequality, it suffices to choose x as the center of gravity of
Q, i.e., asZx = x∗ (Q) = Vol−1
n (Q)

xdx.
Q

This statement in its general (n-dimensional) form was first proved by B. Grunbaum, and wewill refer to it as to the Grunbaum Theorem.
We are about to prove the Grunbaum Theorem. To this end let us start with a simple
characterization of the center of gravity. Given a closed and bounded convex set G ∈ Rn andan affine functional f (x) on Rn , one can define the momentum of G with respect to f asZ

If (G) =f (x)dx;
Gin mechanical terms, If (G) is the momentum of the unform mass distribution on G with respect
to the hyperplane {f (x) = 0}.
Exercise 1.3.3 # Prove that the center of gravity x∗ (Q) of a closed and bounded convex domain
Q is uniquely defined by the relation
If (Q) = f (x∗ (Q))Voln (Q)for all affine functionals f .
In mechanical terms the statement of the latter exercise can be formulated as follows: themomentum of the uniform mass distribution on G with respect to any hyperplane is the sameas if all the mass of G would be located at the center of gravity; in particular, the momentumof this distribution with respect to any hyperplane passing through the center of gravity is 0.
Exercise 1.3.4 #∗ Prove that the center of gravity of a closed and bounded convex domain
Q ⊂ Rn is an interior point of Q. Show that the center of gravity is affine invariant: if
x 7→ A(x) = Ax + b is an invertible affine mapping of Rn onto itself and Q+ = A(Q) is theimage of Q under this mapping, thenx∗ (Q+ ) = A(x∗ (Q))
Exercise 1.3.5 #∗ Prove that the center of gravity of an n-dimensional simplex is the barycenter of the simplex, i.e., the arithmetic mean of the n + 1 vertices of the simplex. Is it true thatthe center of gravity of a polytope is the arithmetic mean of the vertices?
The statement given by the latter exercise can be extended as follows. Let Π be an affine
hyperplane in Rn , let B be a closed and bounded convex set in Π with a nonempty (n − 1)dimensional interior, and let a be a point not belonging to the hyperplane. Consider the convex
hull of a and B; this clearly is a closed and bounded convex domain in Rn . We shall say that
Q is a conic set with the base B and the vertex a, see Figure 1.3. E.g., a simplex is a conic set;
one can take any facet of the simplex as its base and the vertex not belonging to the base asvertex of this conic set.
What can be said about the center of gravity of a conic set?

1.3. EXERCISES: BRUNN, MINKOWSKI AND CONVEX PIE

21

a

B

Figure 1.3: A conic set: B-base, a - vertex.
Exercise 1.3.6 #+ Let Q be a conic set with base B and vertex a. The center of gravity of Q
belongs to the segment ∆ linking the vertex a and the center b of gravity of the base (regarded asa closed convex domain in (n − 1)-dimensional space) and divides ∆ in the ratio n : 1:x∗ (Q) = a/(n + 1) + nb/(n + 1).
To proceed with Grunbaum’s theorem, we need general and extremely important theorem
called the Brunn-Minkowski Symmetrization Principle.
Let Q be a closed and bounded convex domain in Rn , and let l be a straight line in Rn .
Given Q, one can construct another set, Ql , which is(a) symmetric with respect to l (i.e., its cross-section by an affine hyperplane orthogonal tol either is empty, or is a (n − 1)-dimensional Euclidean disk centered at the point where thehyperplane intersects l), and(b) has the same (n − 1)-volumes of cross-sections with orthogonal to l affine hyperplanes
as those for Q, i.e., an affine hyperplane Γ which is orthogonal to l intersects Q if and only if
it intersects Ql , the (n − 1)-dimensional volumes of the cross-sections of Q and Ql by Γ being
equal to each other.
Of course, properties (a) and (b) uniquely define Ql ; this set is called symmetrization of Q
with respect to the axis l.
Now, the Brunn-Minkowski Symmetrization Principle claims thatthe symmetrization of a closed convex domain Q with respect to an axis also is a closed
convex domain.
Now let us derive the Grunbaum Theorem from the Symmetrization Principle. Let Q be aclosed convex set, x∗ be the center of gravity of Q (your choice in the problem of dividing thepie Q recommended by Grunbaum’s theorem), and let e be a nonzero vector (my choice). Letl = x∗ + Re be chosen as the symmetrization axis, and let Ql be the symmetrization of Q withrespect to l. When dividing Q according to our choices, I take the part Q+ of Q to the right of
the hyperplane Γ passing through x∗ and orthogonal to l, and you take the part Q− of Q to theleft of Γ (l is oriented by the vector e). Let Ql+ and Ql− be similar parts of the ”symmetrizedpie” Ql , see Figure 1.4.

22

LECTURE 1. INTRODUCTION: WHAT THE COURSE IS ABOUT

x*e

Figure 1.4: Partitioning of a pie and the induced partitioning of the symmetrized pie.
Exercise 1.3.7 # Prove that x∗ is the center of gravity of Ql and that
Voln (Q± ) = Voln (Ql± ).
The latter exercise says that our parts in the pie Q given by the partitioning in question are thesame as if we were dividing a symmetric with respect to certain axis l pie Ql , x being chosen
as the center of gravity of Ql and e being the direction of the axis (let us call this the special
partitioning of Ql ). Note that the symmetrized pie Ql is convex - this crucial fact is given by theSymmetrization Principle. Thus, in order to establish the Grunbaum Theorem we may restrict
ourselves with axial-symmetric closed and bounded convex domains and their special partitions.
Of course, we are interested in the multi-dimensional case (since the Grunbaum theorem in the
one-dimensional case is evident).
Let Q be an axial-symmetric closed convex and bounded convex domain in Rn , n > 1.
Without loss of generality we may assume that the symmetry axis of Q is the last coordinate
axis and that the center of gravity of Q is at the origin. Denoting a point from Rn as x = (u, t)(u is (n − 1)-dimensional, t is scalar), we can represent Q as
Q = {(u, t) | u ∈ Rn−1 , t ∈ [−a, b], |u| ≤ φ(t)},(1.3.2)where a, b > 0 and φ is a concave on [a, b], positive on (a, b) (and, as it is easily seen, continuous)function on [a, b].
Exercise 1.3.8 # Let Q be the given by (1.3.2) axial-symmetric closed and bounded convex
domain with the center of gravity at the origin, and let Q− , Q+ be the parts of Q to the left andto the right of the hyperplane {t = 0}.
1)∗ Prove that the momentum I−t (Q− ) of Q− with respect to the functional (u, t) 7→ −t isequal to the momentum It (Q+ ) of Q+ with respect to the functional (u, t) 7→ t.
2)∗ Prove that there exist positive α and β such that the coneQ∗ = {(u, t) | −α ≤ t ≤ β, |u| ≤ ξ(t) ≡ α−1 (t + α)φ(0)}(i.e., the cone with the same symmetry axis and the same cross-section by the hyperplane {t = 0}
as those of Q) has volumes of the parts Q∗− , Q∗+ to the left and to the right of the hyperplane1.3. EXERCISES: BRUNN, MINKOWSKI AND CONVEX PIE

23

B

Q-

Q+

A

B’

A and Q∗ = ABB 0
Figure 1.5: Cone Q∗ : the volumes of the parts of the cone to the left and to the right of {t = 0}
are the same as the similar quantities for Q.
{t = 0} equal to the volumes of the similar parts Q− , Q+ of Q; in particular,Voln (Q∗ ) = Voln (Q).
3)∗ Prove that the momentum I−t (Q∗− ) is not less than the momentum I−t (Q− ), while themomentum It (Q∗+ ) is not greater than the momentum It (Q+ ).
4) Conclude from 1) and 3) that I−t (Q∗− ) ≥ It (Q∗+ ).
5) Conclude from 4) that the center of gravity of the cone Q∗ is to the left of the hyperplane{t = 0}.
6)∗ Conclude from 2) and 5) that
Voln (Q− ) = Voln (Q∗− ) ≥ (n/(n + 1))n Voln (Q∗ ) = (n/(n + 1))n Voln (Q),so that the Grunbaum theorem is valid for special partitions of axial-symmetric closed andbounded convex domains and therefore for all partitions of all closed and bounded convex domains.
To understand the power of Grunbaum’s theorem, let us think of whether the answer is”stable”, i.e., what may happen if you choose a point not in the center of gravity. Consider thesimple case when the pie is an n-dimensional Euclidean ball of the unit radius centered, say,at the origin. If you choose the origin, then, independently of the cut, you will get one half ofthe pie. Now, what happens when you choose a point x at a distance α, 0 < α < 1, from theorigin, say, at the distance 0.5, and my cutting plane is orthogonal to x (and, of course, I takethe larger part of the pie, i.e., that one which contains the origin)? The ”spherical hat” youget ”visually” looks large enough; I think you will be surprised that if n is large then you get”almost nothing”, namely, exponentially small part of the pie.
Exercise 1.3.9 + Prove that the volume of the ”spherical hat”
Vα = {x ∈ Rn | |x| ≤ 1, xn ≥ α}

24

LECTURE 1. INTRODUCTION: WHAT THE COURSE IS ABOUT(0 ≤ α ≤ 1) of the unit Euclidean ball V = {x ∈ Rn | |x| ≤ 1} satisfies the inequality as follows:
Voln (Vα )/Voln (V ) ≤ κ exp{−nα2 /2},
κ > 0 being a positive absolute constant. What is, numerically, the left hand side ratio when
α = 1/2 and n = 64?
The Grunbaum theorem says that the center of gravity of a closed and bounded convex
domain Q is, in certain sense, a ”far enough from the boundary” interior point of the domain.
There are other statements of this type, e.g., the following one:
Chord Partitioning Property:
let Q be a closed and bounded convex domain in Rn , let x∗ (Q) be the center ofgravity of Q and let δ be a chord of Q (i.e., a segment with the endpoints on theboundary of Q) passing through x∗ (Q). Then the ratio in which the center of gravity
divides δ is at most n : 1 (i.e., the larger part is at most n times the smaller one).
Exercise 1.3.10 ∗ Prove the Chord Partitioning Property.
The Grunbaum theorem says that if you have a unit mass uniformly distributed over a closed
and bounded convex domain, then there exists a point such that any half-space containing thepoint is of ”a reasonable”, not too small, mass. What can be said about an arbitrary distributed
unit mass? The answer is given by another theorem of Grunbaum which is as follows.
let µ be a probability distribution of mass on Rn . Then there exists a point x
such that, for any closed half-space Π containing x, the mass of Π is at least 1/(n+1).
It should be specified what is a ”probability distribution of mass”. Those familiar with themeasure theory should think about a Borel probability measure on Rn . A not so educatedreader may think about a finite set of points assigned nonnegative masses with the sum of themasses equal to 1; the mass of an arbitrary subset A of Rn is simply the sum of masses of thepoints belonging to A.
Note that the constant 1/(n+1) is sharp, as it is seen from the example when µ is concentrated
at vertices of a simplex, with the mass of each of the vertices equal to 1/(n + 1).
The second theorem of Grunbaum, to the best of my knowledge, has no applications in
Optimization, in contrast to the first one. I am citing it here because it gives me the possibility
to introduce to you a very important theorem of Helley (same as the first theorem of Grunbaum
gave us chance to become acquainted with the fundamental Symmetrization Principle). The
Helley theorem is as follows:let {Qα }α∈A be a family of closed convex subsets of Rn such that any n + 1 setfrom the family have a nonempty intersection. Assume, in addition, that either thefamily is finite, or the intersection of certain finite number of sets from the family isbounded. Then all the sets have a point in common.
Exercise 1.3.11 ∗ Derive the second theorem of Grunbaum from the Helley Theorem.
The actual challenge is, of course, to prove the Helley Theorem and the Symmetrization
Principle, and this is the goal of the remaining exercises. We start with the Helley theorem (aswith the simpler one).

1.3. EXERCISES: BRUNN, MINKOWSKI AND CONVEX PIE

25

Exercise 1.3.12 + Prove the following fundamental fact:
if f1 (x), ..., fm (x) are convex functions defined in a convex neighbourhood of a point x0 ∈ Rn
andf (x) = max fi (x),i=1,...,mthen the subgradient set ∂f (x0 ) can be found as follows: let us take those fi which are active at
x0 , i.e., fi (x0 ) = f (x0 ), and form the union U of the subgradient sets of these fi at x0 ; ∂f (x0 )is exactly the convex hull of U .
Conclude that if f attains its minimum at x0 , then one can select no more than n + 1functions from the family f1 , ..., fm in such a way that the selected functions are active at x0
and their maximum, same as f , attains its minimum at x0 .
Exercise 1.3.13 ∗ Let F = {Q1 , ..., Qk } be a finite family of closed and bounded convex setssatisfying the premise of the Helley theorem. Consider the functionr(x) = max dist(x, Qi ),
1≤i≤k

where
dist(x, Q) = min |x − y|y∈Qis the distance from x to Q. Prove that r(x) attains its minimum at certain x∗ and verify thatx∗ belongs to all Qi , i = 1, ..., k. Thus, the Helley theorem is valid for finite families of boundedsets.
Exercise 1.3.14 + Conclude from the statement of the latter exercise that the Helley theorem
is valid for finite families.
Exercise 1.3.15 + . Derive from the statement of the latter exercise the complete Helley theorem.
The Helley Theorem has a lot of applications in Convex Analysis. Let me present you thefirst which comes to my mind.
Let f be a continuous real-valued function defined on the finite segment [a, b] of the real axis
(a < b). There is a well-known problem of the finding the best polynomial approximation of f
in the uniform norm:
given a positive integer n, find a polynomial p of the degree < n which minimizes
k f − p k∞ ≡ max |f (t) − p(t)|t∈[a,b]over all polynomials of the degree < n.
The characterization of the solution is given by the famous Tschebyshev theorem:a polynomial p of a degree < n is the solution if and only if there exist n + 1
points t0 < ... < tn on the segment such thatk f − p k∞ = |f (ti ) − p(ti )|, i = 0, ..., n,
and neighbouring differences f (ti ) − p(ti ) have opposite signs.

26

LECTURE 1. INTRODUCTION: WHAT THE COURSE IS ABOUT

In fact this answer consists of two parts:(1) the claim that there exists a n + 1-point subset T of [a, b] which ”concentrates” the whole
difficulties of uniform approximating f , i.e., is such thatminp:deg p<nk f − p k∞ =min

p:deg p<nk f − p kT,∞(here k h kT,∞ = maxt∈T |h(t)|),and(2) characterization of the best uniform approximation of a function on a (n + 1)-point set
by a polynomial of a degree < n.
(2) heavily depends on the algebraic nature of polynomials; in contrast to this, (1) is a completely
general fact.
Exercise 1.3.16 + Prove the following generalization of (1):
Let X be a compact set and f , φ0 , ..., φn−1 be continuous mappings from X into an mdimensional linear space E provided with a norm k · k. Consider the problem of finding the best
uniform approximation of f by a linear combination of φi , i.e., of minimizingk f − p kX,∞ = max k f (t) − p(t) kt∈Xover p belonging to the linear span Φ of the functions φi , and let δ be the optimal value in the
problem.
Prove that there exists a finite subset T in X of cardinality at most n + 1 such thatδ = min k f − p kT,∞ .
p∈ΦNote that (1) is the particular case of this general statement associated with E = R, X = [a, b]
and φi (t) = ti .
Now let us come to the Symmetrization Principle. Thus, let Q be a closed and bounded convex domain in n dimensions; without loss of generality we may assume that the axis mentioned
in the principle is the last coordinate axis. It is convenient to denote the coordinate along thisaxis by t and the vector comprised of the first n − 1 coordinates by x. Let [a, b] be the image of
Q under the projection onto the axis, and let Qt = {x | (x, t) ∈ Q}, a ≤ t ≤ b, be the inverse
image of a point (0, ..., 0, t) under the projection. The Symmetrization Principle claims exactlythat the function
φ(t) = {Voln−1 (Qt )}1/(n−1)
is concave and continuous on [a, b] (this function is, up to a multiplicative constant, exactly theradius of the (n − 1)-dimensional disk of the same as Qt (n − 1)-dimensional volume).
It is quite straightforward to derive from the convexity and closedness of Q that φ is continuous; let us ignore this easy and not interesting part of the job (I hope you can do it without
my comments). The essence of the matter is, of course, to prove the concavity of φ. To this end
let us reformulate the task as follows.
Given two compact subsets H, H 0 in Rk , one can form their ”linear combination” with realcoefficients
λH + λ0 H 0 = {x = λx + λ0 x0 | x ∈ H, x0 ∈ H 0 },

1.3. EXERCISES: BRUNN, MINKOWSKI AND CONVEX PIE

27same as the ”product of H by a real λ”:
λH = {x = λx | x ∈ H}.
The results of these operations are, of course, again compact subsets in Rk (as images of thecompact sets H × H 0 , H under continuous mappings). It is clear that the introduced operations
share the following property of their arithmetic predecessors:
λH + λ0 H 0 = (λH) + (λ0 H 0 ).

(1.3.3)

Now, given a compact set H ⊂ Rk , one can define its ”average radius” rk (H) asrk (H) = (Volk (H))1/k(I use Volk as a synonym for the Lebesque measure; those not familiar with this notion, might
think of convex polytopes H and use the Riemann definition of volume). Note that rk (H) is, up
to a coefficient depending on k only, nothing but the radius of the Euclidean ball of the same
volume as that one of H.
It is clear thatrk (λH) = λrk (H), λ ≥ 0.
(1.3.4)
Now, our function φ(t) is nothing but rn−1 (Qt ). To prove that the function is concave means toprove that, for any given λ ∈ [0, 1] and all t0 , t00 ∈ [a, b] one has
rn−1 (Qλt0 +(1−λ)t00 ) ≥ λrn−1 (Qt0 ) + (1 − λ)rn−1 (Qt00 ).

(1.3.5)

Now, what do we know about Qλt0 +(1−λ)t00 is that this set contains the set λQt0 + (1 − λ)Qt00 ;this is an immediate consequence of the convexity of Q. Therefore the left hand side in (1.3.5)
isrn−1 (Qλt0 +(1−λ)t00 ) ≥ rn−1 (Q0 + Q00 ), Q0 = λQt0 , Q00 = (1 − λ)Qt00(we have used (1.3.3) and evident monotonicity of rk (·) with respect to inclusion), while the left
hand side of (1.3.5) is nothing but rn−1 (Q0 ) + rn−1 (Q00 ) (look at (1.3.4)). Thus, to prove theconcavity of φ, i.e., (1.3.5), it suffices to prove, for all k, the following statement:(SAk ): the function rk (H) is a super-additive function on the family of kdimensional compact sets, i.e.,rk (H + H 0 ) ≥ rk (H) + rk (H 0 ).

(1.3.6)

To prove the concavity of φ, it suffices to establish this statement only for convex compact sets,
but in fact it is true independently of the convexity assumptions.
The goal of the remaining exercises is to give an inductive in k proof of (SAk ). Unfortunately,
our scheme requires a little portion of the measure theory. Those familiar with this theory canomit the intuitive explanation of the required technique I start with.
Let H ⊂ Rk , k > 1, be a closed and bounded set.
As above, let (x, t) be the coordinates in the space (x is (k − 1)-dimensional, t is scalar), T
be the projection of H onto the t-axis and H t be the (k − 1)-dimensional inverse image of the

28

LECTURE 1. INTRODUCTION: WHAT THE COURSE IS ABOUT

point (0, ..., 0, t) under this projection. Then the k-dimensional volume of the set ”of course”
can be computed as follows:Z

Volk (H) =

T

t

Z

Volk−1 (H )dt =

T

k−1
rk−1
(H t )dt.

(the first equality, for the Lebesque integration theory, is the Fubini theorem; similar statement
exists in the Riemann theory of multidimensional integration, up to the fact that H should be
subject to additional restrictions which make all objects well-defined).
Now, for l ≥ 1 the integralZ
f l (t)dt

J=T

involving a bounded nonnegative function f defined on a bounded subset T of the axis can becomputed according to the following reasoning. Consider the flat region given by
R = {(t, s) ∈ R2 | t ∈ T, 0 ≤ s ≤ f (t)}and letZI=

sl−1 dtds;

R

integrating first in s and then in t, we come toI = l−1 J.
Integrating first in t and then in s, we come toZ ∞I=

sl−1 µ(s)ds,µ(s) = Vol1 {t ∈ T | f (t) ≤ s}0(µ(s) is the measure of the set of those t ∈ T for which (t, s) ∈ R). We took ∞ as the upper
integration limit in order to avoid new notation; in fact µ(s) is identically zero for all large
enough s, since R is bounded.
Thus, we come to the equality
Z

l

Z ∞

f (t)dt = lTsl−1 µ(s)ds, µ(s) = Vol1 {t ∈ T | f (t) ≥ s}

(1.3.7)0

Of course, to make this reasoning rigorous, we need some additional assumptions, depending on
what is the ”power” of the integration theory we use. Those who know the Lebesque integration
theory should realize that it for sure is sufficient to assume T and f to be measurable andbounded; in the case we are interested in (i.e., when T is the projection onto an axis of a closed
and bounded set H ∈ Rk and f (t) = rk−1 (Ht )) these assumptions are satisfied automatically(since T is a compact set and rk−1 (t), as it immediately follows from the compactness of H, isupper semicontinuous). Those who use a less general integration theory, are advised to trust in
the fact that there exists a theory of integration which makes the above ”evident” conclusions
rigorous.
With (1.3.7) in mind, we are enough equipped to prove (SAk ) by induction in k. Let us startwith the base:
Exercise 1.3.17 ∗ Prove (SA1 ).
The main effort, of course, is to make the inductive step.
Exercise 1.3.18 ∗> Given (SAk−1 ) for some k > 1 and (SA1 ), prove (SAk ) and thus provethat (SAk ) is valid for all k, which completes the proof of the symmetrization principle.

Lecture 2

Methods with linear convergence, I
In the introductory lecture we spoke about the complexity and optimal algorithm of minimizing
a convex function of one variable over a segment. From now on we shall study general typemultidimensional convex problems.

2.1

Class of general convex problems: description and complexity

A convex programming problem isminimize f (x) subject to gi (x) ≤ 0, i = 1, ..., m, x ∈ G ⊂ Rn .

(2.1.1)

Here the domain G of the problem is a closed convex set in Rn with a nonempty interior, theobjective f and the functional constraints gi , i = 1, ..., m, are convex continuous functions on
G. Some more related terminology: any point x ∈ G which satisfies the system
gi (x) ≤ 0, i = 1, ..., mof functional constraints is called a feasible solution to the problem, and the problem is calledfeasible, if it admits feasible solutions. The set of all feasible solutions is called the feasible setof the problem.
The lower boundf ∗ = inf{f (x) | x ∈ G, gi (x) ≤ 0, i = 1, ..., m}of the objective over the set of all feasible solutions is called the optimal value. For an infeasible
problem, by definition, the optimal value is +∞; this fits the standard convection that infimumof something over an empty set is = ∞ and sup of something over an empty set is −∞. For a
feasible problem, the optimal value is either a finite real (if the objective is below bounded on
the set of feasible solutions), or −∞.
A problem is called solvable, if it is feasible and the optimal value of the problem is attainedat certain feasible solution; every feasible solution with this latter property is called an optimal
solution to the problem, and the set of all optimal solutions in what follows will be denoted by
X ∗ . The optimal value of a solvable problem is, of course, finite.
29

30

LECTURE 2. METHODS WITH LINEAR CONVERGENCE, I

Following the same scheme as in the introductory lecture, let us fix a closed and bounded
convex domain G ⊂ Rn and the number m of functional constraints, and let P = Pm (G) be thefamily of all feasible convex problems with m functional constraints and the domain G. Notethat since the domain G is bounded and all problems from the family are feasible, all of them
are solvable, due to the standard compactness reasons.
In what follows we identify a problem instance from the family Pm (G) with a vector-valued
functionp = (f, g1 , ..., gm )comprised of the objective and the functional constraints.
What we shall be interested in for a long time are the efficient methods for solving problems
from the indicated very wide family. Similarly to the one-dimensional case, we assume that themethods have an access to the first order local oracle O which, given an input vector x ∈ int G,
returns the values and some subgradients of the objective and the functional constraints at x,
so that the oracle computes the mapping0x 7→ O(p, x) = (f (x)(, f 0 (x); g1 (x), g10 (x); ...; gm (x), gm(x)) : int G → R(m+1)×(n+1) .

The notions of a method and its complexity at a problem instance and on the whole family
are introduced exactly as it was done in the one-dimensional case, as a set of rules for forming
sequential search points, the moment of termination and the result as functions of the informationon the problem; this information is comprised by the answers of the oracle obtained to themoment when a rule is to be applied.
The accuracy of a method at a problem and on the family is defined in the way similar
to that one used in the one-dimensional case; the only question here is what to do with thefunctional constraints. A convenient way is to start with the vector of residuals of a point x ∈ Gregarded as an approximate solution to a problem instance p:
Residual(p, x) = (f (x) − f ∗ , (g1 (x))+ , ..., (gm (x))+ )which is comprised of the inaccuracy in the objective and the violations of functional constraints
at x. In order to get a convenient scalar accuracy measure, it is reasonable to pass from thisvector to the relative accuracyε(p, x) = max{

f (x) − f ∗(gm (x))+(g1 (x))+,, ...,};∗
maxG f − f (maxG g1 )+(maxG gm )+to get the relative accuracy, we normalize each of the components of the vector of residuals by
its maximal, over all x ∈ G, value and take the maximum of the resulting quantities. It is clearthat the relative accuracy takes its values in [0, 1] and is zero if and only if x is an optimal
solution to p, as it should be for a reasonable accuracy measure.
After we have agreed how to measure accuracy of tentative approximate solutions, we define
the accuracy of a method M at a problem instance as the accuracy of the approximate solution
found by the method when applied to the instance:
Accur(M, p) = ε(p, x̄(p, M)).

2.2. CUTTING PLANE SCHEME AND CENTER OF GRAVITY METHOD

31

The accuracy of the method on the family is its worse-case accuracy at the problems of thefamily:Accur(M) = sup Accur(M, p).
p∈Pm (G)Last, the complexity of the family is defined in the manner we already are acquainted with,
namely, as the best complexity of a method solving all problems from the family to a given
accuracy:
Compl(ε) = min{Compl(M) | Accur(M) ≤ ε}.
What we are about to do is to establish the following main result:
Theorem 2.1.1 The complexity Compl(ε) of the family Pm (G) of general-type convex problems
on an m-dimensional closed and bounded convex domain G satisfies the inequalitiesnbln( 1ε )1
c − 1 ≤ Compl(ε) ≤c2.181 n ln( )b.
6 ln 2ε(2.1.2)

Here the upper bound is valid for all ε < 1. The lower bound is valid for all ε < ε(G), whereε(G) ≥

1n3

for all G ⊂ Rn ; for an ellipsoid G one has
ε(G) =

1,n

and for a parallelotope G
ε(G) = 1.
Same as in the one-dimensional case, to prove the theorem means to establish the lower complexity bound and to present a method associated with the upper complexity bound (andthus optimal in complexity, up to an absolute constant factor, for small enough ε, namely,for 0 < ε < ε(G). We shall start with this latter task, i.e., with constructing an optimal method.

2.2

Cutting Plane scheme and Center of Gravity Method

The method we are about to present is based on a very natural extension of the bisection - thecutting plane scheme.

2.2.1

Case of problems without functional constraints

To explain the cutting plane scheme, let me start with the case when there are no functional
constraints at all, so that m = 0 and the family is comprised of problems
minimizef (x) s.t. x ∈ G(2.2.1)of minimizing convex continuous objectives over a given closed and bounded convex domain
G ⊂ Rn .

32

LECTURE 2. METHODS WITH LINEAR CONVERGENCE, I

To solve such a problem, we can use the same basic idea as in the one-dimensional bisection.
Namely, choosing somehow the first search point x1 , we get from the oracle the value f (x1 ) anda subgradient f 0 (x1 ) of f ; thus, we obtain a linear function
f1 (x) = f (x1 ) + (x − x1 )T f 0 (x1 )which, due to the convexity of f , underestimates f everywhere on G and coincides with f at x1 :
f1 (x) ≤ f (x), x ∈ G;

f1 (x1 ) = f (x1 ).

If the subgradient is zero, we are done - x1 is an optimal solution. Otherwise we can point out
a proper part of G which localizes the optimal set of the problem, namely, the set
G1 = {x ∈ G | (x − x1 )T f 0 (x1 ) ≤ 0};
indeed, outside this new localizer our linear lower bound f1 for the objective, and therefore theobjective itself, is greater than the value of the objective at x1 .
Now, our new localizer of the optimal set, i.e., G1 , is, same as G, a closed and bounded
convex domain, and we may iterate the process by choosing the second search point x2 inside
G1 and generating the next localizer
G2 = {x ∈ G1 | (x − x2 )T f 0 (x2 ) ≤ 0},
and so on. We come to the following generic cutting plane scheme:
starting with the localizer G0 ≡ G, choose xi in the interior of the current localizer
Gi−1 and check whether f 0 (xi ) = 0; if it is the case, terminate, xi being the result,
otherwise define the new localizer
Gi = {x ∈ Gi−1 | (x − xi )T f 0 (xi ) ≤ 0}(see Figure 2.1) and loop.
The approximate solution found after i steps of the routine is, by definition, thebest point found so far, i.e., the pointx̄i ∈ {x1 , ..., xi } such that f (x̄i ) = min f (xj ).
1≤j≤i

A cutting plane method, i.e., a method associated with the scheme, is governed by the rules
for choosing the sequential search points in the localizers. In the one-dimensional case there is,basically, the only natural possibility for this choice - the midpoint of the current localizer (thelocalizer always is a segment). This choice results exactly in the bisection and enforces the lengthsof the localizers to go to 0 at the rate 2−i , i being the step number. In the multidimensionalcase the situation is not so simple. Of course, we would like to decrease a reasonably definedsize of localizer at the highest possible rate; the problem is, anyhow, which size to choose andhow to ensure its decreasing. When choosing a size, we should take care of two things(1) we should have a possibility to conclude that if the size of a current localizer
Gi is small, then the inaccuracy of the current approximate solution also is small;(2) we should be able to decrease at certain rate the size of sequential localizers
by appropriate choice of the search points in the localizers.

2.2. CUTTING PLANE SCHEME AND CENTER OF GRAVITY METHOD

33

Figure 2.1: A step of a cutting plane method. Dashed region is the new localizer.
Let us start with a wide enough family of sizes which satisfy the first of our requirements.
Definition 2.2.1 A real-valued function Size(Q) defined on the family Q of all closed andbounded convex subsets Q ⊂ Rn with a nonempty interior is called a size, if it possesses thefollowing properties:(Size.1) Positivity: Size(Q) > 0 for any Q ∈ Q;
(Size.2) Monotonicity with respect to inclusion: Size(Q) ≥ Size(Q0 ) whenever Q0 ⊂ Q, Q,
Q0 ∈ Q;(Size.3) Homogeneity with respect to homotheties: if Q ∈ Q, λ > 0, a ∈ Rn and
Q0 = a + λ(Q − a) = {a + λ(x − a) | x ∈ Q}
is the image of Q under the homothety with the center at the point a and the coefficient λ, thenSize(Q0 ) = λ Size(Q).
Example 1. The diameterDiam(Q) = max{|x − x0 | | x, x0 ∈ Q}
is a size;
Example 2. The average diameter
AvDiam(Q) = (Voln (Q))1/n ,Voln being the usual n-dimensional volume, is a size.
To the moment these examples are sufficient for us.
Let us prove that any size of the indicated type satisfies requirement (1), i.e., if the size ofa localizer is small, then the problem is ”almost” solved.
Lemma 2.2.1 Let Size(·) be a size. Assume that we are solving a convex problem (2.2.1) by acutting plane method, and let Gi and x̄i be the localizers and approximate solutions generated by
the method. Then we have for all i ≥ 1ε(p, x̄i ) ≤ εi ≡p denoting the problem in question.

Size(Gi ),Size(G)(2.2.2)

34

LECTURE 2. METHODS WITH LINEAR CONVERGENCE, I

Proof looks completely similar to that one used for the bisection method. Indeed, let us fix
i ≥ 1. If εi ≥ 1, then (2.2.2) is evident - recall that the relative accuracy always is ≤ 1. Now
assume that εi < 1 for our i. Let us choose α ∈ (εi , 1], let x∗ be a minimizer of our objective fover G and let
Gα = x∗ + α(G − x∗ ).
ThenSize(Gα ) = α Size(G) > εi Size(G) = Size(Gi )(we have used the homogeneity of Size(·) with respect to homotheties). Thus, the size of Gα
is greater than that on of Gi , and therefore, due to the monotonicity of Size, Gα cannot be asubset of Gi . In other words, there exists a pointy ∈ Gα \Gi .
Since Gα clearly is contained in the domain of the problem and does not belong to the i-th
localizer Gi , we havef (y) > f (x̄i );indeed, at each step j, j ≤ i, of the method we remove from the previous localizer (which initiallyis the whole domain G of the problem) only those points where the objective is greater thanat the current search point xj and is therefore greater than at the best point x̄i found during
the first i steps; since y was removed at one of these steps, we conclude that f (y) > f (x̄i ), as
claimed.
On the other hand, y ∈ Gα , so thaty = (1 − α)x∗ + αz
with some z ∈ G. From convexity of f it follows thatf (y) ≤ (1 − α)f (x∗ ) + αf (z) = (1 − α) min f + α max f,GGwhencef (y) − min f ≤ α(max f − min f ).
GGG

As we know, f (y) > f (x̄i ), and we come tof (xi ) − min f ≤ α(max f − min f ),GGGor, which is exactly the same, toε(p, x̄i ) < α.
Since α is an arbitrary real > εi , we conclude that (2.2.2) holds.
Thus, we realize now what could be the sizes we are interested in, and the problem is how
to ensure certain rate of their decreasing along the sequence of localizers generated by a cutting
plane method. The difficulty here is that when choosing the next search point in the current
localizer, we do not know what will be the next cutting plane; the only thing we know is that
it will pass through the search point. Thus, we are interested in the choice of the search point
which guarantees certain reasonable, not too close to 1, ratio of the size of the new localizer to2.2. CUTTING PLANE SCHEME AND CENTER OF GRAVITY METHOD

35

that one of the previous localizer independently of what will be the cutting plane. Whether sucha choice of the search point is possible, it depends on the size we are using. For example, thediameter of a localizer, which is a very natural measure of it and which was successively used in
the one-dimensional case, would be a very bad choice in the multidimensional case. To realize
this, imagine that we are minimizing over the unit square on the two-dimensional plane, and our
objective in fact depends on the first coordinate only. All our cutting planes (in our example
they are lines) will be parallel to the second coordinate axis, and the localizers will be stripes ofcertain horizontal size (which we may enforce to tend to 0) and of some fixed vertical size (equal
to 1). The diameters of the localizers here although decrease but do not tend to zero. Thus, thefirst of the particular sizes we have looked at does not fit the second requirement. In contrast
to this, the second particular size, the average diameter AvDiam, is quite appropriate, due tothe following geometric fact (which was presented in the problems accompanying the previous
lecture):Proposition 2.2.1 (Grunbaum) Let Q be a closed and bounded convex domain in Rn , let
1
x (G) =
Voln (G)
∗

Z

xdx
G

be the center of gravity of Q, and let Γ be an affine hyperplane passing through the center ofgravity. Then the volumes of the parts Q0 , Q00 in which Q is partitioned by Γ satisfy the inequality
0



00

Voln (Q ), Voln (Q ) ≤ {1 −

n
n+1

1/n

}Voln (Q) ≤ exp{−κ}Voln (Q),

κ = ln(1 − 1/ e) = 0.45867...;in other words,

κ
AvDiam(Q0 ), AvDiam(Q00 ) ≤ exp{− } AvDiam(Q).
n(2.2.3)Note that the proposition states exactly that the smallest (in terms of the volume) fraction you
can cut off a n-dimensional convex body by a hyperplane passing through the center of gravityof the body is the fraction you get when the body is a simplex, the plane passes parallel to a
facet of the simplex and you cut off the part not containing the facet.
Corollary 2.2.1 Consider the Center of Gravity method, i.e., the cutting plane method withthe search points being the centers of gravity of the corresponding localizers:
xi = x∗ (Gi−1 ) ≡1Voln (Gi−1 )

Z

xdx.
Gi−1

For the method in question one hasκ
AvDiam(Gi ) ≤ exp{− } AvDiam(Gi−1 ), i ≥ 1;nconsequently (see Lemma 2.2.2) the relative accuracy of i-th approximate solution generated by
the method as applied to any problem p of minimizing a convex objective over G satisfies theinequalityκε(p, x̄i ) ≤ exp{− i}, i ≥ 1.
n

36

LECTURE 2. METHODS WITH LINEAR CONVERGENCE, I

In particular, to solve the problem within relative accuracy ε ∈ (0, 1) it suffices to perform no
more than
 
 
1
1
1
N =c n lnb≤c2.181n lnb(2.2.4)κεεsteps of the method.
Remark 2.2.1 The Center of Gravity method for convex problems without functional constraints was invented in 1965 independently by A.Yu.Levin in Russia and J. Newman in the
USA.

2.3

The general case: problems with functional constraints

The Center of Gravity method, as it was presented, results in the upper complexity boundstated in our Main Theorem, but only for problems without functional constraints. In order toestablish the upper bound in full generality, we should modify the cutting plane scheme in amanner which enables us to deal with these constraints. The very first idea is to act as follows:
after current search point is chosen and we have received the values and subgradients of theobjective and the constraints at the point, let us check whether the point is feasible; if it is thecase, then let us use the subgradient of the objective in order to cut off the part of the current
localizer where the objective is greater that at the current search point, same as it was donein the case when there were no functional constraints. Now, if there is a functional constraint
violated at the point, we can use its subgradient to cut off points where the constraint is forsure greater than it is at the search point; the removed points for sure are not feasible.
This straightforward approach cannot be used as it is, since the feasible set may have empty
interior, and in this case our process, normally, will never find a feasible search point and,
consequently, will never look at the objective. In this case the localizers will shrink to thefeasible set of the problem, and this is fine, but if the set is not a singleton, we would not have
any idea how to extract from our sequence of search points a point where the constraints are”almost satisfied” and at the same time the objective is close to the optimal value - recall that
we simply did not look at the objective when solving the problem!
There is, anyhow, a simple way to overcome the difficulty - we should use for the cut the
subgradient of the objective at the steps when the constraints are ”almost satisfied” at thecurrent search point, not only at the steps when the point is feasible. Namely, consider the
method as follows:
Cutting plane scheme for problems with functional constraints:
Given in advance the desired relative accuracy ε ∈ (0, 1), generate, starting with
G0 = G, the sequence of localizers Gi , as follows:
given Gi−1 (which is a closed and bounded convex subset of G with a nonempty
interior), act as follows:
1) choose i-th search pointxi ∈ int Gi−1
and ask the oracle about the valuesf (xi ), g1 (xi ), ..., gm (xi )2.3. THE GENERAL CASE: PROBLEMS WITH FUNCTIONAL CONSTRAINTS

37

and subgradients0f 0 (xi ), g10 (xi ), ..., gm(xi )of the objective and the constraints at xi .
2) form the affine lower bounds(i)gj (x) = gj (xi ) + (x − xi )T gj0 (xi )for the functional constraints (these actually are lower bounds since the constraints
are convex) and compute the quantities


(i)



gji,∗ = max gj (x)G, i = 1, ..., m.
+3) If, for all j = 1, ..., m,gj (xi ) ≤ εgj∗,i ,(2.3.1)claim that i is a productive step and go to 4.a), otherwise go to 4.b).
4.a) [productive step]If f 0 (xi ) 6= 0, define the new localizer Gi as
Gi = {x ∈ Gi−1 | (x − xi )T f 0 (xi ) ≤ 0}
and go to 5); otherwise terminate, xi being the result formed by the method.
4.b) [non-productive step]
Choose j such thatgj (xi ) > εgj∗,i ,define the new localizer Gi as
Gi = {x ∈ Gi−1 | (x − xi )T gj0 (xi ) ≤ 0}
and go to 5).
5) Define i-th approximate solution as the best (with the smallest value of the objective) of the search points xj corresponding to the productive steps j ≤ i. Replace
i by i + 1 and loop.
Note that the approximate solution x̄i is defined only if a productive step already has been
performed.
The rate of convergence of the above method is given by the following analogue of Lemma
2.2.2:
Proposition 2.3.1 Let Size be a size, and let a cutting plane method be applied to a convex
problem p with m functional constraints. Assume that for a given N the method either terminatesin course of the first N steps, or this is not the case and the relation
Size(GN )<εSize(G)(2.3.2)is satisfied. In the first case the result x̄ found by the method is a solution to the problem of
relative accuracy ≤ ε; in the second case the N -th approximate solution is well-defined and solves
the problem to the relative accuracy ε.

38

LECTURE 2. METHODS WITH LINEAR CONVERGENCE, I

Proof. Let us first note that for any i and j one has
gj∗,i ≤





max gj;G(2.3.3)+(i)

this is an immediate consequence of the fact that gj (x) is a lower bound for gj (x) (an immediate
consequence of the convexity of gj ), so that the maximum of this lower bound over x ∈ G, i.e.,gj∗,i , is at most the similar quantity for the constraint gj itself.
Now, assume that the method terminates at certain step i ≤ N . According to the descriptionof the method, it means that i is a productive step and 0 is a subgradient of the objective at xi ;the latter means that xi is a minimizer of f over the whole G, so thatf (xi ) ≤ f ∗ .
Besides this, i is a productive step, so that
gj (xi ) ≤ εgj∗,i ≤ ε max gjG, j = 1, ..., m+(we have used (2.3.3)); these inequalities, combined with the definition of the relative accuracy,state exactly that xi (i.e., the result obtained by the method in the case in question) solves the
problem within the relative accuracy ε, as claimed.
Now assume that the method does not terminate in course of the first N steps. In view ofour premise, here we have
Size(GN ) < ε Size(G).
(2.3.4)
Let x∗ be an optimal solution to the problem, and let
Gε = x∗ + ε(G − x∗ ).
Gε is a closed and bounded convex subset of G with a nonempty interior; due to homogeneityof Size with respect to homotheties, we have
Size(Gε ) = ε Size(G) > Size(Gε )(the second inequality here is (2.3.4)). From this inequality and the monotonicity of the size it
follows that Gε cannot be a subset of GN :∃y ∈ Gε \GN .
Now, y is a point of G (since the whole Gε is contained in G), and since it does not belong to
GN , it was cut off at some step of the method, i.e., there is an i ≤ N such thateTi (y − xi ) > 0,(2.3.5)ei being the linear functional defining i-th cut.
Note also that since y ∈ Gε , we have a representationy = (1 − ε)x∗ + εz

(2.3.6)

2.3. THE GENERAL CASE: PROBLEMS WITH FUNCTIONAL CONSTRAINTS

39

with certain z ∈ G.
Let us prove that in fact i-th step is productive. Indeed, assume it is not the case. Thenei = gj0 (xi )(2.3.7)for some j such that gj (xi ) > εgj∗,i . From this latter inequality combined with (2.3.7) and(2.3.5) it follows that(i)(i)
gj (y) > gj (xi ) = gj (xi ) > εgj∗,i .
(2.3.8)(i)

On the other hand, we havegj (y) = (1 − ε)gj (x∗ ) + εgj (z) ≤ εgj∗,i
(i)

(i)

(i)

(i)

(we have taken into account that gj is a lower bound for gj (·) and therefore this bound is
nonpositive at the optimal solution to the problem); the resulting inequality contradicts (2.3.8),
and thus the step i indeed is productive.
Now, since i is a productive step, we have ei = f 0 (xi ), and (2.3.5) implies therefore thatf (y) > f (xi );from this latter inequality and (2.3.6), exactly as in the case of problems with no functional
constraints, it follows thatf (xi ) − f ∗ < f (y) − f ∗ ≤ ε(max f − f ∗ ).
G

(2.3.9)Now let us summarize our considerations. We have proved that in the case in question (i.e.,
when the method does not terminate during first N steps and (2.3.2) is satisfied) there exist a
productive step i ≤ N such that (2.3.9) holds. Since the N -th approximate solution is the best(in terms of the values of the objective) of the search points generated at the productive steps
with step numbers ≤ N , it follows that x̄N is well-defined andf (x̄N ) − f ∗ ≤ f (xi ) − f ∗ ≤ ε(max f − f ∗ );G

(2.3.10)

since x̄N is, by construction, the search point generated at certain productive step i0 , we have
also


0
(i0 )
gj (x̄N ) = gj (xi0 ≤ εgj∗,i ≤ ε max gj, j = 1, ..., m;G+these inequalities combined with (2.3.10) results inε(p, x̄N ) ≤ ε,as claimed.
Combining Proposition 2.3.1 and the Grunbaum Theorem, we come to the Center of Gravity
method for problems with functional constraints. The method is obtained from our generalcutting plane scheme for constrained problems by the following specifications:first, we use, as a current search point, the center of gravity of the previous localizer:
xi =1Voln (Qi−1 )

Z

xdx;
Qi−1

40

LECTURE 2. METHODS WITH LINEAR CONVERGENCE, I
second, we terminate the method after N -th step, N being given by the relation
1
b.
ε

 

N =c2.181n ln

With these specifications the average diameter of i-th localizer at every step, due to the Grunbaum Theorem, decreases with i at least asκ
exp{− i} AvDiam(G),nκ = 0.45867...,and since κ1 < 2.181, we come to
AvDiam(GN ) < ε AvDiam(G);this latter inequality, in view of Proposition 2.3.1, implies that the method does find an ε-solution
to every problem from the family, thus justifying the upper complexity bound we are proving.

2.4

Exercises: Extremal Ellipsoids

There are two natural ways to define an ellipsoid W in Rn . The first is to represent W as theset defined by a convex quadratic constraint, namely, as
W = {x ∈ Rn | (x − c)T A(x − c) ≤ 1}

(2.4.1)A being a symmetric positive definite n × n matrix and c being a point in Rn (the center of theellipsoid).
The second way is to represent W as the image of the unit Euclidean ball under an affine
invertible mapping, i.e., as
W = {x = Bu + c | uT u ≤ 1},(2.4.2)where B is an n × n nonsingular matrix and c is a point from Rn .
Exercise 2.4.1 # Prove that the above definitions are equivalent: if W ⊂ Rn is given by (2.4.1),
then W can be represented by (2.4.2) with B chosen according to
A = (B −1 )T B −1
(e.g., with B chosen as A−1/2 ). Vice versa, if W is represented by (2.4.2), then W can be
represented by (2.4.1), where one should set
A = (B −1 )T B −1 .
Note that the (positive definite symmetric) matrix A involved into (2.4.1) is uniquely definedby W (why?); in contrast to this, a nonsingular matrix B involved into (2.4.2) is defined by W
up to a right orthogonal factor: the matrices B and B 0 define the same ellipsoid if and only if
B 0 = BU with an orthogonal n × n matrix U (why?)
From the second description of an ellipsoid it immediately follows that

2.4. EXERCISES: EXTREMAL ELLIPSOIDS

41

if
W = {x = Bu + c | u ∈ Rn , uT u ≤ 1}
is an ellipsoid andx 7→ p + B 0 x
is an invertible affine transformation of Rn (so that B 0 is a nonsingular n×n matrix),then the image of W under the transformation also is an ellipsoid.
Indeed, the image is nothing but
W 0 = {x = B 0 Bu + (p + B 0 c) | u ∈ Rn , uT u ≤ 1},
the matrix B 0 B being nonsingular along with B and B 0 . It is also worthy of note thatfor any ellipsoid
W = {x = Bu + c | u ∈ Rn , uT u ≤ 1}
there exists an invertible affine transformation of Rn , e.g., the transformation
x 7→ B −1 x − B −1 c,which transforms the ellipsoid exactly into the unit Euclidean ballV = {u ∈ Rn | uT u ≤ 1}.
In what follows we mainly focus on various volume-related issues; to avoid complicatedconstant factors, it is convenient to take, as the volume unit, the volume of the unit Euclideanball V in Rn rather than the volume of the unit cube. The volume of a body1 Q measured inthis unit, i.e., the ratio
Voln (Q),
Voln (V )Voln being the usual Lebesque volume in Rn , will be denoted voln (Q) (we omit the subscript n
if the value of n is clear from the context).
Exercise 2.4.2 # Prove that if W is an ellipsoid in Rn given by (2.4.2), thenvol W = | Det B|,(2.4.3)

vol W = | Det A|−1/2 .

(2.4.4)and if W is given by (2.4.1), then

Our local goal is to prove the following statement:
Let Q be a convex body in Rn (i.e., a closed and bounded convex set with a nonempty
interior). Then there exist ellipsoids containing Q, and among these ellipsoids there is one withthe smallest volume. This ellipsoid is unique; it is called the outer extremal ellipsoid associated1

in what follows ”body” means a set with a nonempty interior

42

LECTURE 2. METHODS WITH LINEAR CONVERGENCE, I

with Q. Similarly, there exist ellipsoids contained in Q, and among these ellipsoids there is
one with the largest volume. This ellipsoid is unique; it is called the inner extremal ellipsoidassociated with Q.
In fact we are not too interested in the uniqueness of the extremal ellipsoids (and you may
try to prove the uniqueness yourself); what actually is of interest is the existence and some
important properties of the extremal ellipsoids.
Exercise 2.4.3 # . Prove that if Q is a closed and bounded convex body in Rn , then there
exist ellipsoids containing Q and among these ellipsoids there is (at least) one with the smallest
volume.
Exercise 2.4.4 . Prove that if Q is a closed and bounded convex body in Rn , then there exist
ellipsoids contained in Q and among these ellipsoids there is (at least) one with the largest
volume.
Note that extremal ellipsoids associated with a closed and bounded convex body Q ”accompany Q under affine transformations”: if x 7→ Ax + b is an invertible affine transformation and
Q0 is the image of Q under this transformation, then the image W 0 of an extremal outer ellipsoid
W associated with Q (note the article: we has not proved the uniqueness!) is an extremal outerellipsoid associated with Q0 , and similarly for (an) extremal inner ellipsoid.
The indicated property is, of course, an immediate consequence of the facts that affine imagesof ellipsoids are again ellipsoids and that the ratio of volumes remains invariant under an affine
transformation of the space.
In what follows we focus on outer extremal ellipsoids. Useful information can be obtained
from investigating these ellipsoids for ”simple parts” of an Euclidean ball.
Exercise 2.4.5 #∗ Let n > 1,
V = {x ∈ Rn | |x|2 ≡

( n
X

)1/2

x2i

≤ 1}

i=1

be the unit Euclidean ball, let e be a unit vector in Rn and let
Vα = {x ∈ V | eT x ≥ α}, α ∈ [−1, 1](Vα is what is called a ”spherical hat”).
Prove that if
1< α < 1,nthen the set Vα can be covered by an ellipsoid W of the volume−n2voln (W ) ≤ { 2
}n/2
n −1

s

n−1
(1 − α2 )(n−1)/2 (1 − α) < 1 = voln (V );n+1

W is defined as
W = {x =
where
(

B=

n2
(1 − α2 ) 2n −1

1 + nαe + Bu | uT u ≤ 1},
n+1

)1/2

s




I − βeeT ,β =1−(1 − α)(n − 1)
(1 + α)(n + 1)

2.4. EXERCISES: EXTREMAL ELLIPSOIDS

43

In fact the ellipsoid given by the latter exercise is the extremal outer ellipsoid associated withVα .
Looking at the result stated by the latter exercise, one may make a number of useful conclusions.
1. When α = 0, i.e., when the spherical hat Vα is a half-ball, we have1voln (W ) = 1 + 2
n −1


n

≤ exp{1/(n2 − 1)}= exp{−

on/2

n/2 s

1−

2
≤
n−1

exp{−1/(n − 1)} =1
1n+2} < exp{−} = exp{−}voln (V );22(n − 1)2n − 22n − 2

thus, for the case of α = 0 (and, of course, for the case of α > 0) we may cover Vα byan ellipsoid with the volume 1 − O(1/n) times less than that one of V . In fact thesame conclusion (with another absolute constant factor O(1)) holds true when α isnegative (so that the spherical hat is greater than half-ball), but ”not too negative”,
1say, when α ≥ − 2n.
2. In order to cover Vα by an ellipsoid of absolute constant times less volume thanthat one of V we need α to be positive of order O(n−1/2 ) or greater. This fits ourobservation that the volume of Vα itself is at least absolute constant times less thanthat one of V only if α ≥ O(n−1/2 ) (exercise 1.9). Thus, whenever the volume of Vα
is absolute constant times less than that one of V , we can cover Vα by an ellipsoidof the volume also absolute constant times less than
√ that one of V ; this covering isalready given by the Euclidean ball of the radius 1 − α2 centered at the point αe(which, anyhow, is not the optimal covering presented in exercise 2.4.5).
Exercise 2.4.6 Let V be the unit Euclidean ball in Rn , e be a unit vector and let α ∈ (0, 1).
Consider the ”symmetric spherical stripe”
V α = {x ∈ V | −α ≤ eT x ≤ α}.
√
Prove that if 0 < α < 1/ n then V α can be covered by an ellipsoid W with the volume√voln (W ) ≤ α n(n(1 − α2 )n−1)(n−1)/2< 1 = voln (V ).

Find an explicit representation of the ellipsoid.
We see that in order to cover a symmetric spherical stripe of the unit Euclidean ball V by anellipsoid of volume less than that one of V , it suffices to have the ”half-thickness” α of the√
stripe to be < 1/ n, which again fits our observation (exercise 1.9) that basically all volume√of the unit n-dimensional Euclidean ball is concentrated in the O(1/ n) neighbourhood of its”equator” - the cross-section of the ball and a hyperplane passing through the center of the ball.
A useful exercise is to realize when a non-symmetric spherical stripe
V α,β = {x ∈ V | −α ≤ eT x ≤ β}

44

LECTURE 2. METHODS WITH LINEAR CONVERGENCE, I

of the (centered at the origin) unit Euclidean ball V can be covered by an ellipsoid of volume
less than that one of V .
The results of exercises 2.4.5 and 2.4.6 imply a number of important geometrical consequences.
Exercise 2.4.7 Prove the following theorem of Fritz John:
Let Q be a closed and bounded convex body in Rn . Then(i) Q can be covered by an ellipsoid W in such a way that the concentric to W n times
smaller ellipsoid11
W 0 = (1 − )c + Wnn(c is the center of W ) is contained in Q. One can choose as W the extremal outer ellipsoidassociated with Q.
(ii) If, in addition, Q is central-symmetric with respect to certain point c, then the above
result can be improved: Q can be covered by an ellipsoid W centered at c in such a way that the√concentric to W n times smaller ellipsoid1
1
W 00 = (1 − √ )c + √ Wnn
is contained in Q.
√
Note that the constants n and n in the Fritz John Theorem are sharp; an extremal example
for (i) is a simplex, and for (ii) - a cube.
Here are several nice geometrical consequences of the Fritz John Theorem:
Let Q be a closed and bounded convex body in Rn . Then
1. There exist a pair of concentric homothetic with respect to their common center
parallelotopes p, P with homothety coefficient equal to n−3/2 such that p ⊂ Q ⊂ P ;in other words, there exists an invertible affine transformation of the space such thatthe image Q0 of Q under this transformation satisfies the inclusions{x ∈ Rn | k x k∞ ≤1n3/2

} ⊂ Q0 ⊂ {x ∈ Rn | k x k∞ ≤ 1};here
k x k∞ = max |xi |1≤i≤nis the uniform norm of x.
Indeed, from the Fritz John Theorem it follows that there exists an invertible affine
transformation resulting in{x | |x|2 ≤ 1/n} ⊂ Q0 ⊂ {x | |x|2 ≤ 1},
Q0 being the image of Q under the transformation (it suffices to transform the outerextremal ellipsoid associated with Q into the unit Euclidean ball centered at theorigin). It remains to note that the smaller Euclidean ball in the above chain ofinclusions contains the cube {x | k x k∞ ≤ n−3/2 } and the larger one is contained in
the unit cube.

2.4. EXERCISES: EXTREMAL ELLIPSOIDS

45

2. If Q is central-symmetric, then the parallelotopes mentioned in 1. can bechosen to have the same center, and the homothety coefficient can be improved to
1/n; in other words, there exists an invertible affine transformation of the space
which makes the image Q0 of Q central symmetric with respect to the origin andensures the inclusions{x | k x k∞ ≤1
} ⊂ Q0 ⊂ {x | k x k∞ ≤ 1}.
n

The statement is given by the reasoning completely similar to that one used for 1.,
up to the fact that now we should refer to item (ii) of the Fritz John Theorem.
√
3. Any norm k · k on Rn can be approximated, within factor n, by a Euclidean
norm: given k · k, one can find a Euclidean norm|x|A = (xT Ax)1/2 ,A being a symmetric positive definite n × n matrix, in such a way that1
√ |x|A ≤k x k≤ |x|Anfor any x ∈ Rn .
Indeed, let B = {x | k x k≤ 1} be the unit ball with respect to the norm k · k; this isa closed and bounded convex body, which is central symmetric with respect to theorigin. By item (ii) of the Fritz John Theorem, there exists a centered at the origin
ellipsoid
W = {x | xT Ax ≤ n}
(A is an n × n symmetric positive definite matrix) which contains B, while theellipsoid{x | xT Ax ≤ 1}
is contained in B; this latter inclusion means exactly that|x|A ≤ 1 ⇒ x ∈ B ⇔k x k≤ 1,i.e., means that |x|A ≥k x k. The inclusion B ⊂ W , by similar reasons, implies thatk x k≥ n−1/2 |x|A .
Remark 2.4.1 The third of the indicated consequences says that any norm on Rn can be√approximated, within constant factor n, by an appropriately chosen Euclidean norm. It turns
out that the quality of approximation can be done much better, if we would be satisfied byapproximating the norm not at the whole space, but at a properly chosen subspace. Namely,
there exists a marvelous and important theorem of Dvoretsky which is as follows:
there exists a function m(n, ε) of positive integer n and positive real ε with the following
properties:
first,
lim m(n, ε) = +∞
n→∞

and,

46

LECTURE 2. METHODS WITH LINEAR CONVERGENCE, I

second, whenever k · k is a norm on Rn , one can indicate a m(n, ε)-dimensional subspaceE ⊂ Rn and a Euclidean norm | · |A on Rn such that | · |A approximates k · k on E within factor
1 + ε:(1 − ε)|x|A ≤k x k≤ (1 + ε)|x|A , x ∈ E.
In other words, the Euclidean norm is ”marked by God”: for any given integer k an arbitrary
normed linear space contains an ”almost Euclidean” k-dimensional subspace, provided that thedimension of the space is large enough.

2.4.1

Tschebyshev-type results for sums of random vectors

The series of exercises we are about to present deals with issues which, at the first glance,
are far from extremal ellipsoids and optimization - namely, with the Tschebyshev-type results
for sums of random vectors. These results (which are useful in various applications - nonparametric estimation, Stochastic Programming, etc.) are a surprising byproduct of our previous
considerations.
Recall that the standard Tschebyshev equality states that if {ξi }∞i=1 is a sequence of mutuallyindependent random reals such thatE{ξi } = 0, E{ξi2 } < ∞, i = 1, 2, ...,
(E stands for expectation), then
N
X

E{

!2

ξi

}≤

i=1

N
X

E{ξi2 }

i=1

(in fact, of course, there is equality rather than inequality). The proof is immediate: open theparentheses and take the termwise expectation; the expectations of ”cross-products” ξi ξj , i 6= j,
will disappear, since ξi and ξj are independent with zero means.
The issue we are interested in is: what can be said when the sequence is formed of random
vectors rather than random reals? Namely, let k · k be certain once for ever fixed norm on Rn ,nand let {ξi }∞i=1 be a sequence of mutually independent random vectors from R such thatE{ξi } = 0, E{k ξi k2 } ≡ s2i < ∞, i = 1, 2, ...
Now, let
ΞN =

N
X

ξi .

i=1

What can be said about the quantities
2
SN= E{k ΞN k2 } ?

First of all, if k · k is a Euclidean norm |x|A = (xT Ax)1/2 (A is an n × n symmetric positive
definite matrix), then we have exactly the same inequality as in the scalar case:
2
SN
≤

N
X
i=1

s2i

(2.4.5)

2.4. EXERCISES: EXTREMAL ELLIPSOIDS

47(in fact this is an equality). The proof remains the same:|ΞN |2A =NX

ξiT Aξj ;i,j=1

when taking expectation, cross-product terms disappear, and we come to (2.4.5). As we shall see
in a while, this is a ”bad” reasoning: in heavily exploits the algebraic structure of the Euclidean
norm and therefore cannot be extended onto more general norms.
The ”Euclidean” version of the Tschebyshev inequality immediately leads to the following
general statement: for any norm k · k on Rn one has
2
SN
≤n

N
X

s2i .

(2.4.6)i=1

Exercise 2.4.8 ∗ Prove (2.4.6).
Exercise 2.4.9 ∗ Prove that the factor n in (2.4.6) is sharp, as far as the order of dependence
on n is concerned: one can find a norm k · k on Rn and a sequence of iid (independent identically
distributed) random n-dimensional vectors {ξi } such thats2i ≡ E{k ξi k2 } = 1
and

S2
2
2
lim inf N →∞ SN/N ≡ lim inf N →∞ PNN 2 ≥ n.
π
i=1 si

Results stated in exercises 2.4.8 and 2.4.9 are quite satisfactory (and, let me say, not interestingat all) when the dimension n of our random vectors is regarded as something once for ever fixed;
in all applications of the ”vector Tschebyshev-type inequalities” known to me n is large, andwhat we actually are interested in is what is the (depending on the norm k · k) constant factor
γk·k in the estimate
2SN≤ γk·k

N
X

s2i .

i=1

As we just have seen, this factor always does not exceed n, and sometimes (i.e., for some particular norms on Rn and some random sequences) this value n cannot be significantly improved.
On the other hand, for some particular norms it can be improved (say, for a Euclidean norm it
can be set to 1 independently of the dimension). What are those norms which allow to reduce
the factor, as compared to its worst-case with respect to the norms value O(n)? The mostimportant specification of this question is as follows: what happens in the case of the uniform
norm|x|∞ = max |xi | ?
1≤i≤nThis norm arises in many applications, and, believe me, it is useful to know a ”good” Tschebyshevtype inequality for this norm.

48

LECTURE 2. METHODS WITH LINEAR CONVERGENCE, I
At the first glance it seems that the answer should be the same as that one for the 1-norm|x|1 =nX|xi |,

i=1

since the norms | · |1 and | · |∞ are dual to each other, so that the norms are, roughly speaking,of the same level of similarity (better to say, dissimilarity) to the Euclidean norm (for this latter
norm, γ = 1). Those who have solved exercise 2.4.9 know that for the 1-norm γ is as bad as it
could be, i.e., it is of order of n; therefore one could expect the uniform norm also to be bad.
Surprisingly, the latter norm is almost as good as the Euclidean one:γ|·|∞ = O(ln n).

(2.4.7)

To establish (2.4.7), this is the goal of the remaining exercises.
It is reasonable to deal not with the uniform norm directly, but with the family of thestandard norms ”linking” the 1-norm and the uniform one, i.e., with the family of the lp -norms
on Rn!
|x|p =nX

1/p|xi |p.

i=1Here 1 ≤ p ≤ ∞, and for p = ∞ the right hand side is, by definition, the uniform norm maxi |xi |.
Note that | · |2 is the standard Euclidean norm.
Let us start with the following simple observation:
Exercise 2.4.10 + One has
0n1/p −1/p |x|p ≥ |x|p0 ≥ |x|p , x ∈ Rn , ∞ ≥ p ≥ p0 ≥ 1;

(2.4.8)in particular, if n ≥ 3, then1
|x|2 ln n ≥ |x|∞ ≥ exp{− }|x|2 ln n .
2(2.4.9)To get an idea of what to do, let us look once again at the proof of the Tschebyshev inequality
for the case of the Euclidean norm | · |2 , but now let us pass to an ”equivalent” reasoning: rather
than to express |ΞN |22 in terms of pair inner products of the terms forming the sum ΞN , let us
use the recurrenceT|Ξk+1 |22 = |Ξk |22 + 2ξk+1
Ξk + |ξk+1 |22 ;(2.4.10)taking expectation, we come to2
Sk+1= Sk2 + s2k+1(2.4.11)T Ξ equals to zero, since the second factor depends(the expectation of the linear in ξk+1 term ξk+1
k
only on ξ1 , ..., ξk and is therefore independent of ξk+1 , while the expectation of the first factor
is zero).
Now, the Cosine Theorem|x + h|22 = |x|22 + 2hT x + |h|22

2.4. EXERCISES: EXTREMAL ELLIPSOIDS

49

underlying (2.4.10) is nothing but the second-order Taylor representation
1f (x + h) = f (x) + hT f 0 (x) + hT f 00 (x)h2of the quadratic function
f (x) = |x|22 ;for this particular function, the Taylor representation is an equality. Now note that if we, for agiven f (which should not necessarily be |x|22 ), start with an inequality of the typef (x + h) ≤ f (x) + hT f 0 (x) + r(h)(2.4.12)(i.e., with the first order Taylor expansion of f written as an inequality with an appropriate
remainder r(h) depending on h only and independent of x), we could evaluate from above theexpectations
σN ≡ E{f (ΞN )}acting exactly in the manner we just have used for the quadratic f :Tf (Ξk+1 ) ≤ f (Ξk ) + ξk+1
f 0 (Ξk ) + r(ξk+1 ) ⇒⇒ E{f (Ξk+1 )} ≤ E{f (Ξk )} + E{r(ξk+1 )}T f 0 (Ξ ) disappears since the first factor has zero(the expectation of the linear in ξk+1 term ξk+1
kmean and the second one is independent of the first). Thus, given (2.4.12), we may write therecurrenceE{f (Ξk+1 )} ≤ E{f (Ξk )} + E{r(ξk+1 )},(2.4.13)which allows to evaluate sequentially all the quantities E{f (ΞN )}, N = 1, 2, ..., in terms of si ,provided that we are able to bound from above, in terms of si , the expectations E{r(ξi )}. A
good news here is that our estimating abilities depend on the possibility to bound from above
the remainder r in the first order Taylor expansion of f , i.e., on smoothness of f , not on theparticular algebraic structure of f .
To summarize our considerations at this point, let us present the statement we just have
proved in the following, although slightly restricted, but quite sufficient for us form:
let f be a continuously differentiable function on Rn such that, for certain constant ωf ∈ [0, ∞) and all x, h ∈ Rn one hasf (x + h) ≤ f (x) + hT f 0 (x) + ωf f (h).

(2.4.14)

Let also {ξi } be a sequence of mutually independent random n-dimensional vectors
with zero means and finite quantities
(f )

σi

= E{f (ξi )}.

Then, for all N , one has
E{f (ΞN )} ≤ f (0) + ωf

N
X
(f )

σi ,

i=1

where, as always,
ΞN =

N
X
i=1

ξi .

(2.4.15)50

LECTURE 2. METHODS WITH LINEAR CONVERGENCE, I

The proof is an immediate consequence of recurrence (2.4.13).
In order to obtain a Tschebyshev-type inequality for the uniform norm, we could try to
use the latter statement, f (x) being |x|2∞ . An immediate difficulty is that the function is not
smooth, so that our reasoning cannot be applied to it directly. But this is a minor difficulty,
since we, due to the result given by exercise 2.4.10, can approximate the uniform norm by a
lp -norm with finite p > 2, the latter norm being smooth.
Exercise 2.4.11 + Let p ∈ (2, ∞), and let
f (x) = |x|2p : Rn → R.
Prove that this function satisfies (2.4.14) with
ωf = 2(2p−3)/(p−1) (p − 1) + 1 ≤ 4p − 3(2.4.16)and, consequently,
E{|ΞN |2p } ≤ (4p − 3)NX

E{|ξi |2p }(2.4.17)i=1

for every sequence {ξi } of mutually independent random n-dimensional vectors with zero mean
and finite variances E{|ξi |2p }.
Exercise 2.4.12 ∗ Prove the following Tschebyshev-type inequality for the uniform norm:
E{|ΞN |2∞ } ≤ (8 ln n − 3) exp{1}N
X

E{|ξi |2∞ }(2.4.18)

i=1

for any N , n ≥ 3 and any sequence {ξi } of random mutually independent n-dimensional vectors
with zero means and finite variances E{|xi |2∞ }.
Thus, the constant γ|·|∞ for the uniform norm on Rn can be chosen to be of order of ln n, i.e.,”almost independent of n”. You may try to prove that the constant should be of order of ln n,provided that no additional restrictions on {ξi } are imposed.

Lecture 3

Methods with linear convergence, II
3.1

Lower complexity boundTo complete the proof of Theorem 2.1.1, it remains to establish the lower complexity bound.
This is done, basically, in the same way as in the one-dimensional case. In order to avoid things
difficult for verbal presentation, I shall restrict myself with demonstrating a slightly worse lowerboundn ln( 1ε )
 , 0 < ε < ε∗ (G),Compl(ε) ≥ O(1) (3.1.1)1
ln n ln( ε )O(1) being a positive absolute constant and ε∗ (G) being certain positive quantity depending on
G only (we shall see that this quantity measures how G differs from a parallelotope).
The ”spoiled” bound (3.1.1) (which is by logarithmic denominator worse than the estimate
announced in the Theorem) is more or less immediate consequence of our one-dimensional considerations. Of course, it is sufficient to establish the lower bound for the case of problems
without functional constraints, since the constrained ones form a wider family (indeed, a problem without functional constraints can be thought of as a problem with a given number m oftrivial, identically zero functional constraints). Thus, in what follows the number of constraints
m is set to 0.
Let me start with the following simple observation. Let, for a given ε > 0 and a convex
objective f , the set Gε (f ) be comprised of all approximate solutions to f of relative accuracy
not worse than ε:Gε (f ) = {x ∈ G | f (x) − min f ≤ ε(max f − min f )}.
G

GGAssume that, for a given ε > 0, we are able to point out a finite set F of objectives with thefollowing two properties:(I) no different problems from F admit a common ε-solution:Gε (f ) ∩ Gε (f¯) = ∅whenever f, f¯ ∈ F and f 6= f¯;(II) given in advance that the problem in question belongs to F, one can compress ananswer of the first order local oracle to be a (log2 K)-bit word. It means the following. For
51

52

LECTURE 3. METHODS WITH LINEAR CONVERGENCE, II

certain positive integer K one can indicate a function I(f, x) taking values in a K-element set
and a function R(i, x) such that
O(f, x) = R(I(f, x), x),f ∈ F, x ∈ int G.

In other words, given in advance that the problem we are interested in belongs to F, a method canimitate the first-order oracle O via another oracle I which returns log2 K bits of information
rather than infinitely many bits contained in the answer of the first order oracle; given the”compressed” answer I(f, x), a method can substitute this answer, along with x itself, into auniversal (defined by F only) function in order to get the ”complete” first-order information on
the problem.
E.g., consider the family Fn comprised of 2n convex functions
f (x) = max i xi ,
i=1,...,nwhere all i are ±1. At every point x a function from the family admits a subgradient of theform I(f, x) = ±ei (ei are the orths of the axes), with i, same as the sign at ei , depending on
f and x. Assume that the first order oracle in question when asked about f ∈ Fn reports a
subgradient of exactly this form. Since all functions from the family are homogeneous, given x
and I(f, x) we know not only a subgradient of f at x, but also the value of f at the point:
f (x) = xT I(f, x).
Thus, our particular first-order oracle as restricted onto Fn can be compressed to log2 (2n) bits.
Now let us make the following observation:(*):
under assumptions (I) and (II) the ε-complexity of the family F, and thereforeof every larger family, is at leastlog2 |F|.
log2 K
Indeed, let M be a method which solves all problems from F within accuracy ε in no morethan N steps. We may assume (since informationally this is the same) that the method uses
the oracle I, rather than the first-order oracle. Now, the behaviour of the method is uniquely
defined by the sequence of answers of I in course of N steps; therefore there are at most K N
different sequences of answers and, consequently, no more than K N different trajectories of M.
In particular, the set X̄ formed by the results produced by M as applied to problems from F iscomprised of at most K N points. On the other hand, since M solves every of |F| problems of
the family within accuracy ε, and no two different problems from the family admit a common
ε-solution, X̄ should contain at least |F| points. Thus,
K N ≥ |F|,
as claimed.
As an immediate consequence of what was said, we come to the following result:the complexity of minimizing a convex function over an n-dimensional parallelotope G within
relative accuracy ε < 1/2 is at least n/(1 + log2 n).

3.1. LOWER COMPLEXITY BOUND

53

Indeed, all our problem classes and complexity-related notions are affine invariant, so that wealways may assume the parallelotope G mentioned in the assertion to be the unit cube
{x ∈ Rn | |x|∞ ≡ max |xi | ≤ 1}.
i

For any ε < 12 the aforementioned family
Fn = {f (x) = max i xi }iclearly possesses property (I) and, as we have seen, at least for certain first-order oracle possesses
also property (II) with K = 2n. We immediately conclude that the complexity of finding an
ε-minimizer, ε < 1/2, of a convex function over an n-dimensional parallelotope is, at least forsome first order oracle, no less thanlog2 |F|,log2 (2n)as claimed. In fact, of course, the complexity is at least n for any first order oracle, but to prove
the latter statement it requires more detailed considerations.
Now let us use the above scheme to derive the lower bound (3.1.1). Recall that when studying
the one-dimensional case, we have introduced certain family of univariate convex functions which
was as follows. The functions of the family form a tree, with the root (”generation 0”) being
the function
f root (x) = |x|;when subject to the left and to the right modifications, the function produces two ”children”, letthem be called fr and fl ; each of these functions, in turn, may be subject to the right and to theleft modification, producing two new functions, so that at the level of ”grandchildren” there are
four functions frr , frl , flr , fll , and so on. Now, every of the functions f· of a ”generation” k > 0
possesses its own active segment δ(f ) of the length 21−2k , and at this segment the function ismodulus-like:f· (x) = a(k) + 8−k |x − c(f )|,c(f ) being the midpoint of δ(f ). Note that a(k) depends only on the generation f· belongs to,
not on the particular representative of the generation; note also that the active segments ofthe 2k functions belonging to the generation k are mutually disjoint and that a function fromour ”population” coincides with its ”parent” outside the active segment of the parent. In what
follows it is convenient also to define the active segment of the root function f root as the wholeaxis.
Now, let Fk be the set of 2k functions comprising k-th generation in our population. Let
us demonstrate that any first order oracle, restricted onto this family of functions, admits compression to log2 (2k) bits. Indeed, it is clear from our construction that in order to restore, given
an x, f (x) and a subgradient f 0 (x), it suffices to trace the path of predecessors of f - its father,its grandfather, ... - and to find the ”youngest” of them, let it be f¯, such that x belongs to theactive segment of f¯ (let us call this predecessor the active at x predecessor of f ). The active at
x predecessor of f does exist, since the active segment of the ”common predecessor” f root is thewhole axis. Now, f is obtained from f¯ by a number of modifications; the first of them possibly
varies f¯ in a neighbourhood of x (x is in the active segment of f¯), but the subsequent modifications do not, since x is outside the corresponding active segments. Thus, in a neighbourhood

54

LECTURE 3. METHODS WITH LINEAR CONVERGENCE, IIof x f coincides with the function f˜ - the modification of f¯ which leads from f¯ to f . Now, toidentify the local behaviour of f˜ (i.e., that one of f ) at x, it suffices to indicate the ”age” of f˜,i.e., the number of the generation it belongs to, and the type of the modification - left or right- which transforms f¯ into f˜.
Indeed, given x and the age k̄ of f˜, we may uniquely identify the active segment of f¯ (since thesegments for different members of the same generation k̄ − 1 have no common points); given
the age of f¯, its active segment and the type of modification leading from f¯ to f˜, we, of course,know f˜ in a neighbourhood of the active segment of f¯ and consequently at a neighbourhood ofx.
Thus, to identify the behaviour of f at x and therefore to imitate the answer of any given
local oracle on the input x, it suffices to know the age k̄ of the active at x predecessor of f andthe type - left or right - of modification which ”moves” the predecessor towards f , i.e., to knowa point from certain (2k)-element set, as claimed.
Now let us act as follows. Let us start with the case when our domain G is a parallelotope;
due to affine invariance of our considerations, we may assume G to be the unit n-dimensionalcube:
G = {x ∈ Rn | |x|∞ ≤ 1}.
Given a positive integer k, consider the family F k comprised of the objectives
fi1 ,...,in (x) = max{fi1 (x1 ), ..., fin (xn )},fis ∈ Fk , s = 1, ..., n.

This family contains |F|n = 2nk objectives, all of them clearly being convex and Lipschitz
continuous with constant 1 with respect to the uniform norm |·|∞ . Let us demonstrate that there
exists a first order oracle such that the family, equipped with this oracle, possesses properties(I), and (II), where one should set
ε = 2−5k ,K = 2nk.

(3.1.2)

Indeed, a function fi1 ,...,in attains its minimum a(k) exactly at the point xi1 ,...,in with the coordinates comprised of the minimizers of fis (xs ). It is clear that within the cube
Cz 1 ,...,in = {x | |x − xi1 ,...,in |∞ ≤ 2 × 2−2k }(i.e., within the direct product of the active segments of fis , s = 1, ..., n) the function is simplya(k) + 8−k |x − xi1 ,...,in |∞ ,therefore outside this cube one has
fi1 ,...,in (x) − min fi1 ,...,in > 2−5k = ε.
GTaking into account that all our functions fi1 ,...,in , being restricted onto the unit cube G, taketheir values in [0, 1], so that for these functions absolute inaccuracy in terms of the objective ismajorated by the relative accuracy, we come toGε (fi1 ,...,in ) ⊂ Ci1 ,...,in .

3.1. LOWER COMPLEXITY BOUND55

It remains to note that the cubes C· corresponding to various functions from the family are mutually disjoint (since the active segments of different elements of the generation Fk are disjoint).
Thus, (I) is verified.
In order to establish (II), let us note that to find the value and a subgradient of fi1 ,...,in ata point x it suffices to know the value and a subgradient at xis of any function fis which is”active” at x, i.e., is ≥ all other functions participating in the expression for fi1 ,...,in . In turn,as we know, to indicate the value and a subgradient of fis it suffices to report a point from a(2k)-element set. Thus, one can imitate certain (not any) first order oracle for the family F k
via a ”compressed” oracle reporting log2 (2nk)-bit word (it suffices to indicate the number s,
1 ≤ s ≤ n of a component fis active at x and a point of a (2k)-element set to identify fis at xis ).
Thus, we may imitate certain first order oracle for the family F k (comprised of 2kn functions),
given a ”compressed” oracle with K = 2nk; it follows from (*) that the ε-complexity of F for
ε = 2−5k (see (3.1.2)) is at leastlog2 (2nk );log2 (2nk)
expressing k in terms of ε, we come toCompl(ε) ≥nlog2 ( 1ε )

,

5log2 nlog2 ( 2ε )ε = 2−5k , k = 1, 2, ...;

taking into account that Compl(ε) is a non-increasing function, we come to an estimate of the1
type (3.1.1) with ε∗ (G) = 32
(recall that we were dealing with the unit cube G).
Let me stress what was in fact was proved. We have demonstrated that it is possible toequip the family of all convex continuous functions on the unit cube with a first order oracle insuch a way that the complexity of the family would satisfy lower bound (3.1.1), provided that1
ε < 32.
Now, what happens in the case when G is not a parallelotope? In this case we can find apair of homothetic parallelotopes, p and P , such that p ⊂ G ⊂ P . Let us choose among thesepairs p, P that one with the smallest possible similarity ratio, let it be called α(G). Those whohave solved problems given at the previous lecture know, and other should believe thatα(G) ≤ n3/2
for any closed and bounded convex body G ⊂ Rn . After appropriate affine transformation of
the space (the transformation does not influence the complexity - once again, all our complexityrelated notions are affine invariant) we may assume that p is the unit cube. Thus, we come tothe situation as follows:Cn ≡ {x ∈ Rn | |x|∞ ≤ 1} ⊂ G ⊂ {x ∈ Rn | |x|∞ ≤ α(G)}.

(3.1.3)

Now let us look at the family of problems
minimize f (x) s.t. x ∈ G
associated with f ∈ F k . It is easily seen that all the objectives from the family attain theirglobal (i.e., over the whole space) minima within the unit cube and the sets of approximate

56

LECTURE 3. METHODS WITH LINEAR CONVERGENCE, II

solutions of absolute accuracy 2−5k to problems of the family are mutually disjoint (these setsare exactly the small cubes already known to us). Therefore our previous reasoning states that
it is possible to equip the family with a first order oracle in such a way that the worst-case, withrespect to the family F k , complexity of finding an approximate solution of absolute accuracy
2k
2−5k is at least lnn2ln(2nk) . On the other hand, the Lipschitz constant of every function from thefamily F k taken with respect to the uniform norm is, as we know, at most 1, so that the variation
max f − min fGGof such a function on the domain G is at most the diameter of G with respect to the uniform
norm; the latter diameter, due to (3.1.3), is at most 2α(G). It follows that any method whichsolves all problems from F k within relative accuracy 2−5k−1 /α(G) solves all these problems
within absolute accuracy 2−5k as well; thus, the complexity of minimizing convex function overn log2 k
G within relative accuracy 2−5k−1 /α(G) is at least log(2nk) :
2Compl(2−5k−1 /α(G)) ≥nlog2 k, k = 1, 2, ...
log2 (nk)This lower bound immediately implies thatnlog2

Compl(ε) ≥ O(1)

log2





1
α(G)ε


,

1
nlog2 ( α(G)ε)

α(G)ε <

1,64

whence, in turn,Compl(ε) ≥ O(1)n ln(1/ε),ln (n ln(1/ε))ε ≤ ε∗ (G) ≡

1
1
(≥
);
2
64α (G)
64n3

this is exactly what is required in (3.1.1).
Note that our reasoning results in a lower bound which is worse than that one indicatedin the Theorem not only by the logarithmic denominator, but also due to the fact that this isa lower bound for a particular first order oracle, not for an arbitrary one. In fact both theseshortcomings - I mean the presence of the denominator and the ”oracle-dependent” type of thelower bound - may be overcome by more careful reasoning, but this is what I am not going to
do here.

3.2

The Ellipsoid method

We have presented the Center of Gravity method which is optimal in complexity, up to an
absolute constant factor, at least when the required accuracy is small enough. The rate ofconvergence of the method, which is the best possible theoretically, looks fine also from thepractical viewpoint. The upper complexity bound O(1)n ln(1/ε) associated with the methodmeans that in order to improve inaccuracy by an absolute constant factor (say, reduce it byfactor 10) it suffices to perform O(1)n iterations more, which looks not too bad, taking into
account how wide is the family of problems the method can be applied to. Thus, one may ask:
what were the computational consequences of the method invented as early as in 1965? The

3.2. THE ELLIPSOID METHOD

57

answer is: no consequences at all, since the method cannot be used in practice, provided that thedimension of the problem is, say > 4. The reason is that the auxiliary problems arising at thesteps of the method, I mean those of finding centers of gravity, are computationally extremely
hard; the source of the difficulty is that the localizers arising in the method may be almost
arbitrary solids; indeed, all we can say is that if the domain G of the problem is a polytope
given by k0 linear inequalities, then the localizer Gi formed after i steps of the method is apolytope given by k0 + i linear inequalities; and all known deterministic methods for computing
the center of gravity of a polytope in Rn given by k > 2n linear inequalities take an exponential
in n time to find the center. In our definition of complexity we ignore the computational effort
required to implement the search rules; but in practice this effort, of course, should be taken into
account, this is why the Center of Gravity method, for which this latter effort is tremendous,
cannot be used at all.
The situation, nevertheless, is not as bad as one could think: to the moment we have notexploit all abilities of the cutting plane scheme.
Let us note that the cutting plane scheme can be ”spoiled” as follows. Given previous
localizer Gi−1 , we apply our basic scheme to produce a new localizer, Ḡi , but now this issomething intermediate, not the localizer we are forwarding to the next step (this is why wedenote by Ḡi the solid which in the basic scheme was designated Gi ); the localizer we do use
at the next step is certain larger solid Gi ⊃ Ḡi . Thus, at a step of the modified cutting plane
scheme we perform a cut, exactly as in the basic scheme, and enlarge the resulting localizer toobtain Gi .
At this point one could ask: what for should we add to an actual localizer something which forsure does not contain optimal solutions? The answer is: acting in this manner, we may stabilize
geometry of our localizers and enforce them to be convenient for numerical implementation of
the search rules. This is the idea underlying the Ellipsoid method I am about to present.

3.2.1

Ellipsoids

Recall that an ellipsoid in Rn is defined as a level set of a nondegenerate convex quadratic form,i.e., as a set of the type
W = {x ∈ Rn | (x − c)T A(x − c) ≤ 1},(3.2.1)

where A is an n × n symmetric positive definite matrix and c ∈ Rn is the center of the ellipsoid.
An equivalent definition, which will be more convenient for us, says that an ellipsoid is the imageof the unit Euclidean ball under a one-to-one affine transformation:
W = W (B, c) = {x = Bu + c | uT u ≤ 1},(3.2.2)where B is an n × n nonsingular matrix. It is immediately seen that one can pass from representation (3.2.2) to (3.2.1) by settingA = (B T )−1 B −1 ;(3.2.3)since any symmetric positive definite matrix A admits a representation of the type (3.2.3) (e.g.,with B = A−1/2 ), the above definitions indeed are equivalent.

58

LECTURE 3. METHODS WITH LINEAR CONVERGENCE, II
From (3.2.2) it follows immediately that
Voln (W (B, c)) = | Det B|Voln (V ),(3.2.4)

where V denotes the unit Euclidean ball in Rn .
Now, by compactness reasons it is easily seen that for any n-dimensional solid Q there exist
ellipsoids containing Q and among these ellipsoids there is at least one with the smallest volume;in fact this extremal outer ellipsoid of Q is uniquely defined, but we are not interested in theuniqueness issues. The average diameter of this extremal outer ellipsoid is certain function ofQ, let it be denoted by EllOut(Q) and called the outer ellipsoidal size:
EllOut(Q) = {min{Voln (W ) | W is an ellipsoid containing Q}}1/n .
It is immediately seen that the introduced function is a size, i.e., it is positive, monotone withrespect to inclusions and homogeneous with respect to similarity transformations of the homogeneity degree 1.
We need the following simple lemma
Lemma 3.2.1 Let n > 1, let
W = {x = Bu + c | uT u ≤ 1}
be an ellipsoid in Rn , and let
W̄ = {x ∈ W | (x − c)T q ≤ 0}
be a ”half-ellipsoid” - the intersection of W and a half-space with the boundary hyperplanepassing through the center of W (here q 6= 0). Then W̄ can be covered by an ellipsoid W + of
the volume
Voln (W + ) = κn (n)Voln (W ),n2
κn (n) = 2
n −1

s

n−1
1
≤ exp{−};n+1
2(n − 1)(3.2.5)in particular,
EllOut(W + ) ≤ κ(n) EllOut(W ) ≤ exp{−

1} EllOut(W ).
2n(n − 1)

The ellipsoid W + is given by
W + = {x = B + u + c+ | uT u ≤ 1},
B + = α(n)B − γ(n)(Bp)pT ,where(α(n) =n2n2 − 1)1/4

c+ = c −s,γ(n) = α(n)

n−1
,
n+1

1
Bp,
n+1

p= p

BT q
.
q T BB T q(3.2.6)

3.2. THE ELLIPSOID METHOD

59

To prove the lemma, it suffices to reduce the situation to the similar one with W being the unit
Euclidean ball V ; indeed, since W is the image of V under the affine transformation u 7→ Bu + c,
the half-ellipsoid W̄ is the image, under this transformation, of the half-ballV̄ = {u ∈ V | (B T q)T u ≤ 0} = {u ∈ V | pT u ≤ 0}.
Now, it is quite straightforward to verify that a half-ball indeed can be covered by an ellipsoid
V + with the volume being the required fraction of the volume of V ; to verify this, it was oneof the exercises distributed last time, and in the formulation of the exercise you were given
the explicit representation of V + . It remains to note that the image of V + under the affine
transformation which maps the unit ball V onto the ellipsoid W is an ellipsoid which clearly
contains the half-ellipsoid W̄ and is in the same ratio of volumes with respect to W as V +is with respect to the unit ball V (since the ratio of volumes remains invariant under affine
transformations). The ellipsoid W + given in formulation of the lemma is nothing but the imageof V + under our affine transformation.

3.2.2

The Ellipsoid method

Bearing in mind our ”spoiled” cutting plane scheme, we may interpret the statement of Lemma
3.2.1 as follows: assume that at certain step of a cutting plane method the localizer Gi−1 to be
updated is an ellipsoid. Let us choose the current search point as the center of the ellipsoid;then the cut will result in the intermediate localizer Ḡi which is a half-ellipsoid. Let us cover
this intermediate localizer by an ellipsoid given by our lemma and choose the latter ellipsoidas the new localizer Gi . Now we are in the same position as we were - the new localizer is anellipsoid, and we may proceed in the same manner. And due to our lemma, we do decrease incertain ratio certain size of localizers - namely, the outer ellipsoidal size EllOut.
Two issues should be thought of. First, how to initialize our procedure - i.e., how to enforce
the initial localizer to be an ellipsoid (recall that in our basic scheme the initial localizer was thedomain G of the problem). The answer is immediate - let us take as G0 an arbitrary ellipsoidcontaining G. The second difficulty is as follows: in our ”spoiled” cutting plane scheme thelocalizers, generally speaking, are not subsets of the domain of the problem; it may, consequently,
happen that the center of a localizer is outside the domain; how to perform a cut in the lattercase? Here again the difficulty can be immediately overcome. If the center xi of the current
localizer Gi−1 is outside the interior of G, then, due to the Separation Theorem for convex sets,
we may find a nonzero linear functional eT x which separates xi and int G:(x − xi )T e ≤ 0, x ∈ G.
Using e for the cut, i.e., settingḠi = {x ∈ Gi−1 | (x − xi )T e ≤ 0},
we remove from the previous localizer Gi−1 only those points which do not belong to the domainof the problem, so that Ḡi indeed can be thought of as a new intermediate localizer.
Thus, we come to the Ellipsoid method which, as applied to a convex programming problem
minimize f (x) s.t. gj (x) ≤ 0, j = 1, ..., m, x ∈ G ⊂ Rn
works as follows:

60

LECTURE 3. METHODS WITH LINEAR CONVERGENCE, IIInitialization. Choose n × n nonsingular matrix B0 and a point x1 such that theellipsoidG0 = {x = B0 u + x1 | uT u ≤ 1}
contains G. Choose β > 0 such thatβ≤

EllOut(G).
EllOut(G0 )i-th step, i ≥ 1. Given Bi−1 , xi , act as follows:
1) Check whether xi ∈ int G. If it is not the case, then call step i non-productive,
find a nonzero ei such that(x − xi )T ei ≤ 0 ∀x ∈ G
and go to 3), otherwise go to 2).
2) Call the oracle to compute the quantities
0f (xi ), f 0 (xi ), g1 (xi ), g10 (xi ), ..., gm (xi ), gm(xi ).

If one of the inequalities

gj (xi ) ≤ ε max{gj (xi ) + (x − xi )T gj0 (xi )}G, j = 1, ..., m(3.2.7)+is violated, say, k-th of them, call i-th step non-productive, set
ei = gk0 (xi )and go to 3).
If all inequalities (3.2.7) are satisfied, call i-th step productive and set
ei = f 0 (xi ).
If ei = 0, terminate, xi being the result found by the method, otherwise go to 3).
3) Set
p= q

T e
Bi−1
i
T e
eTi Bi−1 Bi−1i

, Bi = α(n)Bi−1 − γ(n)(Bi−1 p)pT , xi+1 = xi −1
Bi−1 p,n+1(3.2.8)α(n) and γ(n) being the quantities from Lemma 3.2.1.
If
κi (n) < εβ,(3.2.9)terminate with the result x̄ being the best, with respect to the objective, of thesearch points associated with the productive steps:x̄ ∈ Argmin{f (x) | x is one of xj with productive j ≤ i}otherwise go to the next step.

3.3. EXERCISES: THE CENTER OF GRAVITY AND THE ELLIPSOID METHODS

61

The main result on the Ellipsoid method is as follows:
Theorem 3.2.1 The associated with a given relative accuracy ε ∈ (0, 1) Ellipsoid method Ell(ε),
as applied to a problem instance p ∈ Pm (G), terminates in no more than
ln



1
βε



1
Compl(Ell(ε)) =cbb≤c2n(n − 1) ln
ln(1/κ(n))βε
steps and solves p within relative accuracy ε: the result x̄ is well defined andε(p, x̄) ≤ ε.
Given the direction ei defining i-th cut, it takes O(n2 ) arithmetic operations to update(Bi−1 , xi ) into (Bi , xi+1 ).
Proof. The complexity bound is an immediate corollary of the termination test (3.2.9). To
prove that the method solves p within relative accuracy ε, note that from Lemma 3.2.1 it follows
that
EllOut(Gi ) ≤ κi (n) EllOut(G0 ) ≤ κi (n)β −1 EllOut(G)(the latter inequality comes from the origin of β). It follows that if the method terminates at a
step N due to (3.2.9), then
EllOut(GN ) < ε EllOut(G).
Due to this latter inequality, we immediately obtain the accuracy estimate as a corollary of ourgeneral convergence statement on the cutting plane scheme (Proposition 2.3.1). Although thelatter statement was formulated and proved for the basic cutting plane scheme rather than forthe ”spoiled” one, the reasoning can be literally repeated in the case of the ”spoiled” scheme.
Note that the complexity of the Ellipsoid method depends on β, i.e., on how good is theinitial ellipsoidal localizer we start with. Theoretically, we could choose as G0 the ellipsoid of
the smallest volume containing the domain G of the problem, thus ensuring β = 1; for ”simple”
domains, like a box, a simplex or a Euclidean ball, we may start with this optimal ellipsoid not
only in theory, but also in practice. Even with this good start, the Ellipsoid method has O(n)times worse theoretical complexity than the Center of Gravity method (here it takes O(n2 ) steps
to improve inaccuracy by an absolute constant factor). As a compensation of this theoreticaldrawback, the Ellipsoid method is not only of theoretical interest, it can be used for practicalcomputations as well. Indeed, if G is a simple domain from the above list, then all actionsprescribed by rules 1)-3) cost only O(n(m + n)) arithmetic operations. Here the term mn comesfrom the necessity to check whether the current search point is in the interior of G and, if it is
not the case, to separate the point from G, and also from the necessity to maximize the linear
approximations of the constraints over G; the term n2 reflects complexity of updating Bi−1 7→ Bi
after ei is found. Thus, the arithmetic cost of a step is quite moderate, incomparably to thetremendous one for the Center of Gravity method.

3.3

Exercises: The Center of Gravity and the Ellipsoid methods

3.3.1

Is it actually difficult to find the center of gravity?

I cannot say that it is proved that, given a polytope in Rn defined by m linear inequalities, one
cannot find the center of gravity of the polytope at a ”moderate” (polynomial in n and m) # of

62

LECTURE 3. METHODS WITH LINEAR CONVERGENCE, II

arithmetic operations. What can be said is that efficient, in the indicated sense, deterministic
algorithms are unknown, and there are strong reasons to doubt whether such an algorithm
exists. Randomized algorithms, this is another story: it was proved recently that it is possible
to approximate the center of gravity of a polytope (within accuracy sufficient for our goals) by
certain Monte Carlo algorithm at a polynomial in n, m arithmetic cost. Nevertheless, I have
not heard much about applications of these new tools in optimization, since the theoreticalcomplexity of the algorithms, although polynomial, seems to be inappropriately large. There is,anyhow, hope that practical behaviour of the Monte Carlo routine will be much better than thatone prescribed by the theoretical analysis, so that the approach seems to be worthy of testingand you are welcome to do it. The idea of the Monte Carlo routine will be explained in a while,
but let me start with the deterministic approach.
The most natural (and actually the only one which comes to my mind) way to compute thecenter of gravity of a polytope P ⊂ Rn is to partition P into simplexes Si , i = 1, ..., n, with nosimplex in the partitioning covering an interior point of another simplex, and to compute thecenters of gravity xi and the volumes vi of the simplexes; as we already know (Exercise 1.3.5),
the center of gravity of a simplex
S = Conv{w0 , ..., wn }is the arithmetic mean of the vertices w0 , ..., wn , while the volume of S is1
| Det (w1 − w0 , w2 − w0 , ..., wn − w0 )|n!
(why?) Thus, given the partitioning, we can easily compute xi and vi and then setx∗ (P ) = {Xivi }−1Xvi xi .

(3.3.1)i

Exercise 3.3.1 # Prove (3.3.1).
Now, there is no theoretical difficulty to partition a polytope into simplexes. Indeed, let us use
induction in dimension: after all (n − 1)-dimensional facets of the polytope are partitioned into(n − 1)-dimensional simplexes, let us choose a vertex w in the interior of P ; the conic sets withthe bases being (n − 1)-dimensional simplexes coming from the facets and the vertex at w clearly
form a partitioning of P . Of course, this is a tremendous work (think of partitioning in thismanner an n-dimensional cube; you will get 2n n! simplexes!). But may be this is a bad way topartition a polytope and there are much better ones? The answer is negative:
Exercise 3.3.2 #∗ Let P ⊂ Rn be a polytope contained in the unit cube {|x|∞ ≤ 1} andcontaining, for some α > 0, a cube of the type {|x − a|∞ ≤ α}. Prove that any partitioning of
P into non-overlapping simplexes contains at least
√ !n
√
nα
(1 + o(1))
2πne√elements, with o(1) → 0 as n → ∞. Thus, if α n is not small, say, is ≥ 10, then the partitioning
contains at least exp{n} simplexes.

3.3. EXERCISES: THE CENTER OF GRAVITY AND THE ELLIPSOID METHODS

63

I have no idea whether the lower bound indicated in the latter exercise is ”tight”, but alreadythis bound says that simplicial partitions are of no use for computing the center of gravity of apolytope.
Now let me present you the Monte Carlo scheme. To compute the center of gravity means,by definition, to compute certain n-dimensional integral. It is quite traditional to compute an
integralZ
1
I=
φ(x)dxVol G G
via the Monte Carlo simulation: given a generator of a random vector ξ uniformly distributedin G, one can generate a sample ξ1 , ..., ξN of independent vectors with the indicated distribution
and approximate the integral (which is nothing but the expectation of the random variable φ(ξ))
by the empirical average
N
1 X
IN =
φ(ξi ).
N i=1
In the case we are interested in φ(x) = x is a vector-valued function; due to the Tschebyshev
equality for random vectors, the expected squared | · |2 -error of the approximation is given by
2
σN≡ E{|I − IN |22 } =where
σ 2 = E{|I − I1 |2 } =

1
Vol G

Z
G

σ2
,
N

|x − x∗ (G)|22 dx.

The rate of convergence seems to be quite reasonable: it takes O(σ 2 /ε2 ) calls for the generator
to approximate x∗ (G) within accuracy of order of ε. Of course, σ might be large (imagine that
G is ”long and thin”), but this is an illusion: an n-dimensional solid in Rn can be ”large andthin” only with respect to a badly chosen Euclidean metric; for a properly chosen metric, as weknow from Exercise 2.4.7, the ”asphericity” of G, i.e., the ratio of the radii of a pair of concentricballs, with the smaller one contained in G and the larger one containing G, does not exceed n.
It is easily seen that with respect to this metric one has σ 2 ≤ n2 , and, by the way, this is exactly
the metric in which we actually should measure inaccuracy when approximating the center ofgravity. Of course, the ”proper” metric is not given in advance, but we are not interested in
it: the Monte Carlo procedure is independent of any metric, the latter is involved only in theconvergence analysis.
Thus, everything looks fine - everything but one point: how to generate the uniformly
distributed in G random vector. An immediate idea is: let us embed G into a box G+ . Thereis no difficulty in generating a random vector uniformly distributed in a box (you should calln times the standard generator of a uniformly distributed in [0, 1] random variable and makethe outputs of the generator, after appropriate scalings, the coordinates of the random vector
we need). Now, when asked to generate a random vector from G, let us generate sequentiallyrandom vectors uniformly distributed in G+ until a vector belonging to G is met; this vector
will be the output of our new generator and will, of course, be distributed uniformly in G, as
required. Are we done? No! Indeed, let us look how long it takes to meet a vector from G whengenerating a sample of vectors uniformly distributed in the box G+ . The answer is immediate:
the expected # of elements in the sample is Vol(G+ )/ Vol(G). In other words, if the linear

64

LECTURE 3. METHODS WITH LINEAR CONVERGENCE, II

dimensions of G are something like 10% of those of G+ (which is not so small ratio), then thecomplexity of generating a point from G (# of calls for the inner generator) will be something
like 10n - and for n = 20 we shall for sure die before our generator produces a single answer.
Thus, we did not overcome what is called ”the damn of the dimension” - we simply havetransformed our initial question into that one of how to generate, at a moderate cost, a vector
uniformly distributed in G. The transformed question, anyhow, admits an affirmative answer
as follows.
Consider a discrete mesh formed by vectors with the coordinates being multiples of certain
small positive ν, and let Γ be the part of the mesh comprised of vectors belonging to G. Let
x0 ∈ Γ be once for ever fixed. Consider the random walk along Γ described as follows: themoving point, initially placed at x0 , at an integer moment t chooses randomly one of the 2ndirections ±ei , i = 1, ..., n (ei are the standard orths) and tries to make a step of the lengthν in the chosen direction. If the step keeps the point within G, then it actually is performed;
otherwise the point does not move at the moment t and repeats the attempt to move at thenext moment, t + 1, again choosing randomly the direction of the step.
It is clear that the described process (which can be easily simulated) moves the point along
the connectedness component, Γ0 , of the point x0 in the mesh Γ. This random walk is certain
Markov’s chain, and since the probability to come in a finitely many steps N ( N is equal to
the largest length of a simple path in Γ0 ) from any given point of Γ0 to any other point of Γ0 is
positive, there exists a unique steady-state probability distribution π of positions of the point,
an the distribution πt of the position of the point at a moment t converges, as t → ∞, to π.
Exercise 3.3.3 ∗ Prove that π is nothing but the uniform distribution on Γ0Thus, the steady-state distribution of the position of our walking point is uniform on Γ0 , andthe distributions πt , which we actually can simulate, for large t are almost uniform (since they
converge to π as t → ∞). Thus, choosing in advance certain ”large” T and simulating T -step
trajectory of the point, we can generate an ”almost uniformly distributed in Γ0 ” (and therefore”almost uniformly distributed in G”, provided that ν is small) vector.
Of course, the main issue here is how large should be our ”large” T - why could not we again
come to something like exp{O(n)}? It turns out, anyhow, that the things are not so bad: it
suffices to take T as a polynomial of n and 1/ν .This is a relatively recent and actually brilliant
result of Cobham, Dyer, Frieze; the proof heavily exploits, via the isoperimeter inequalities, theconvexity of G (as it should be: the same approach to generating vectors uniformly distributed
in a ”slightly nonconvex” G immediately results in an exponential in n value of T ). It remains
to note that it is not quite straightforward to apply this scheme to the problem of approximating
the center of gravity of a polytope; to this end one should relate the construction to a properlychosen Euclidean metric (since now a ”long and thin” G do causes troubles - it enforces ν to bevery small in order for Γ0 to be a ”good” discretization of G) and to choose small (although small”polynomially in n”, not exponentially) ν. As a result, T becomes a high-degree polynomial ofn, not speaking on preliminary computational effort required to choose a ”proper” Euclidean
metric underlying the mesh. Let me repeat, anyhow, that from the practical viewpoint these
difficulties might be partly imaginary and coming from rough estimates used in the proofs.
The scheme of the generator is simple, and it is worthy to test the approach numerically, notbothering much about ”bad” theoretical values of T and ν.

3.3. EXERCISES: THE CENTER OF GRAVITY AND THE ELLIPSOID METHODS

3.3.2

65

Some extensions of the Cutting Plane scheme

To the moment we have applied the Cutting Plane scheme to convex optimization problems inthe standard form
minimize f (x) s.t.

gj (x) ≤ 0, j = 1, ..., m, x ∈ G ⊂ Rn

(3.3.2)(G is a solid, i.e., a closed and bounded convex set with a nonempty interior, f and gj are convex
and continuous on G). In fact the scheme has a wider field of applications. Namely, consider a”generic” problem as follows:(f ) :

minimize f (x) s.t. x ∈ Gf ⊂ Rn ;(3.3.3)

here Gf is certain (specific for the problem instance) solid and f is a function taking values inthe extended real axis R ∪ {−∞} ∪ {+∞} and finite on the interior of G.
Let us make the following assumption on our abilities to get information on (f ):(A): we have an access to an oracle A which, given on input a point x ∈ Rn , informs us whether
x belongs to the interior of Gf ; if it is not the case, the oracle reports a nonzero functional ex
which separates x and Gf , i.e., is such that(y − x)T ex ≤ 0, y ∈ Gf ;if x ∈ int Gf , then the oracle reports f (x) and a functional ex such that the Lebesque set
L∗f (x) = {y ∈ Gf | f (y) < f (x)}is contained in the open half-space{y ∈ Rn | (y − x)T ex < 0}.
In the mean time we shall see that under assumption (A) we can efficiently solve (f ) by cutting
plane methods; but before coming to this main issue let me indicate some interesting particularcases of the ”generic” problem (f ).
Example 1. Convex optimization in the standard setting. Consider the usual convex
problem (3.3.2) and assume that the feasible set of the problem has a nonempty interior where
the constraints are negative. We can rewrite (3.3.2) as (3.3.3) by setting
Gf = {x ∈ G | gj (x) ≤ 0, j = 1, ..., m}.
Now, given in advance the domain G of the problem (3.3.2) and being allowed to use a first-orderoracle O for (3.3.2), we clearly can imitate the oracle A required by (A).
Example 1 is not so interesting - we simply have expressed something we already know in aslightly different way. The next examples are less trivial.
Example 2. Quasiconvex optimization. Assume that the objective f involved into
(3.3.2) and the functional constraints gj are quasiconvex rather than convex; recall that a function h defined on a convex set Q is called quasiconvex, if the sets
Lh (x) = {y ∈ Q | h(y) ≤ h(x)}66

LECTURE 3. METHODS WITH LINEAR CONVERGENCE, II

are convex whenever x ∈ Q (note the difference between Lh (x) and L∗h (x); the latter set isdefined via strict inequality <, the former one via ≤). Besides quasiconvexity, assume that thefunctions f , gj , j = 1, ..., m, are regular, i.e., continuous on G, differentiable on the interiorof G with a nonzero gradient at any point which is not a minimizer of the function over G.
Last, assume that the feasible set Gf of the problem possesses a nonempty interior and that theconstraints are negative on the interior of Gf .
If h is a regular quasiconvex function on G and x ∈ int G, then the set L∗h (x) = {y ∈ G |h(y) < h(x)} belongs to the half-spaceΠh (x) = {y ∈ G | (y − x)T h0 (x) < 0}.

Exercise 3.3.4 #+ Prove the latter statement.
Thus, in the case in question the sets {y ∈ G | f (y) < f (x)}, {y ∈ G | gj (y) < gj (x)} arecontained in the half-spaces {y | (y − x)T f 0 (x) < 0}, {y | (y − x)T gj0 (x) < 0}, respectively.
It follows that in the case in question, same as in the convex case, we, given an access to afirst-order oracle for (3.3.2), can imitate a required by (A) oracle A for the induced problem(3.3.3).
Example 3. Linear-fractional programming. Consider the problemaω (x)
minimize f (x) = maxω∈Ωbω (x)


s.t. gj (x) ≤ 0, j = 1, ..., m, bω (x) > 0, ω ∈ Ω, x ∈ G;

(3.3.4)

here G is a solid in Rn and Ω is a finite set of indices, aω and bω are affine functions and gj
are, say, convex and continuous on G. The problem is, as we see, to minimize the maximum ofratios of given linear forms over the convex set defined by the inclusion x ∈ G, convex functional
constraints gj (x) ≤ 0 and additional linear constraints expressing positivity of the denominators.
Let us set
Gf = {x ∈ G | gj (x) ≤ 0, bω (x) ≥ 0, ω ∈ Ω};
we assume that the (closed and convex) set Gf possesses a nonempty interior and that thefunctions gj are negative, while bω are positive on the interior of Gf .
By settingmaxω {aω (x)/bω (x)}x ∈ int Gf
f (x) =
(3.3.5)+∞,otherwise
we can rewrite our problem as
minimize f (x) s.t.

x ∈ Gf .

(3.3.6)

Now, assume that we are given G in advance and have an access to a first-order oracle O which,given on input a point x ∈ int G, reports the values and subgradients of functional constraints
at x, same as reports all aω (·), bω (·).
Under this assumptions we can imitate for (3.3.6) the oracle A required by the assumption(A). Indeed, given x ∈ Rn , we first check whether x ∈ int G, and if it is not the case, find anonzero functional ex which separates x and G (we can do it, since G is known in advance); ofcourse, this functional separates also x and Gf , as required in (A). Now, if x ∈ int G, we ask the3.3. EXERCISES: THE CENTER OF GRAVITY AND THE ELLIPSOID METHODS

67

first-order oracle O about the values and subgradients of gj , aω and bω at x and check whether
all gj are negative at x and all bω (x) are positive. If it is not the case and, say, gk (x) ≥ 0,
we claim that x 6∈ int Gf and set ex equal to gk0 (x); this functional is nonzero (since otherwise
gk would attain a nonnegative minimum at x, which contradicts our assumptions about the
problem) and clearly separates x and Gf (due to the convexity of gk ). Similarly, if one of thedenominators bω̄ is nonpositive at x, we claim that x 6∈ int Gf and set
ex = ∇bω̄ ;here again ex is nonzero and separates x and Gf (why?)
Last, if x ∈ int G, all gj are negative at x and all bω are positive at the point, we claim thatx ∈ int Gf (as it actually is the case), find the index ω = ω(x) associated with the largest at xof the fractions aω (·)/bω (·), compute the corresponding fraction at x (this is nothing but f (x))
and setex = ∇y aω(x) (y) − f (x)∇y bω (x)(y).
Since in the latter case we haveL∗f (x) ⊂ {y ∈ Gf |aω(x) (y)< f (x)} ⊂ {y ∈ Gf | (y − x)T ex < 0},bω(x) (y)we do fit the requirements imposed by (A).
Note that the problem of the type (3.3.4) arises, e.g., in the famous von Neumann Modelof Economy Growth which is as follows. Consider an economy where m kinds of goods are
circulating. The economy is described by a pair of m × n matrices A and B with positive
entries, where row index i stands for goods and column index j stands for ”processes”. A
process j takes, as input, aij units of good i and produces, as output, bij units of the same good,per year. Now, let xt be an n-dimensional vector with coordinates xtj being the ”intensities”
by which we let j-th process work in year t. Then the amount of goods consumed by all theprocesses run in year t is represented by the vector Axt , and the amount of goods produced in
the same year is given by the vector Bxt . If we have no external sources of goods, then the”trajectory” of our economy should satisfy the inequalities
Axt+1 ≤ Bxt , t = 0, 1, ...
(for t = 0 the right hand side should be replaced by a positive vector representing the ”starting”
amount of goods). Now, in the von Neumann Economic Growth problem it is asked what is thelargest growth factor, γ ∗ , for which there exists a ”semi-stationary growth trajectory”, i.e., a
trajectory of the type xt = γ t x0 . In other words, we should solve the problem
maximize γ s.t. γAx ≤ Bx for some nonzero x ≥ 0.
Without loss of generality, x in the above formulation can be taken as a point form the standard
simplexX
G = {x ∈ Rn | x ≥ 0,
xj = 1}j

68

LECTURE 3. METHODS WITH LINEAR CONVERGENCE, II(which should be regarded as a solid in its affine hull). It is clearly seen that the problem inquestion can be rewritten as follows:P

minimizeaij xjs.t.
j bij xjjmax P

i=1,...,mx ∈ G;(3.3.7)this is a problem of the type (3.3.6).
It is worthy to note that the von Neumann growth factor γ ∗ describes, in a sense, thehighest rate of growth of our economy (this is far from being clear in advance: why the ”Soviet”
proportional growth is the best one? Why could we not get something better along an oscillating
trajectory?) The exact statement on optimality of the von Neumann semi-stationary trajectory(or, better to say, the simplest one of these statements) is as follows:
Proposition 3.3.1 Let {xt }Tt=1 be a trajectory of our economy, so that xt are nonnegative,x0 6= 0 and
Axt+1 ≤ Bxt , t = 0, 1, ..., T − 1.
Assume that xT ≥ λT x0 for some positive λ (so that our trajectory results, for some T , in growthof the amount of goods in λT times in T years). Then λ ≤ γ ∗ .
Note that following the von Neumann trajectory
xt = (γ ∗ )T x0 ,x0 being the x-component of an optimal solution to (3.3.7), does ensure growth by factor (γ ∗ )Teach T years.
Exercise 3.3.5 ∗ Prove Proposition 3.3.1.
Example 4. Generalized Eigenvalue problem. Let us again look at problem (3.3.4).
What happens if the index set Ω becomes infinite? To avoid minor elaborations, let us assume(as it actually is the case in all applications known to me) that Ω is a compact set and thatthe functions aω (x), bω (x) are continuous in (x, ω) (and, as above, are affine in x). As far
as reformulation (3.3.5) - (3.3.6) is concerned, no difficulties occur. The possibility to imitate
the oracle A, this is another story (it hardly would be realistic to ask O to report infinitely
many numerators and denominators). Note, anyhow, that from the discussion accompanying
the previous example it is immediately seen that at a given x we are not interested in all aω ,bω ; what in fact we are interested in are the active at x numerator and denominator, i.e., either(any) bω (·) such that bω (x) ≤ 0, if such a denominator exists, or those aω (·), bω (·) with thelargest ratio at the point x, if all the denominators at x are positive. Assume therefore that weknow G in advance and have an access to an oracle O which, given on input x ∈ int G, reports
the values and subgradients of the functional constraints gj at x and, besides this, tells us (atleast in the case when all gj are negative at x) whether all the denominators bω (x), ω ∈ Ω,
are positive; if it is the case, then the oracle returns the numerator and the denominator withthe largest ratio at x, otherwise returns the denominator which is nonpositive at x. Looking at
the construction of A given in the previous example we immediately conclude that in our now
situation we again can imitate a compatible with (A) oracle A for problem (3.3.6).
In fact the ”semiinfinite fractional problem” we are discussing possesses interesting applications; let me introduce one of them which is of extreme importance for modern Control Theory3.3. EXERCISES: THE CENTER OF GRAVITY AND THE ELLIPSOID METHODS

69- the Generalized Eigenvalue problem. The problem is as follows: given two affine functions,A(x) and B(x), taking values in the space of symmetric m × m matrices (”affine” means thatthe entries of the matrices are affine functions of x), minimize, with respect to x, the Rayleighratio
ω T A(x)ω
maxω∈Rm \{0} ω T B(x)ω
of the quadratic forms associated with these matrices under the constraints that B(x) is positive
definite (and, possibly, under additional convex constraints on x). In other words, we are looking
for a pair (x, λ) satisfying the constraints
B(x) is positive definite , λB(x) − A(x) is positive semidefinite
and additional constraints
gj (x) ≤ 0, j = 1, ..., m, x ∈ G ⊂ Rn
(gj are convex and continuous on the solid G) and are interested in the pair of this type withthe smallest possible λ.
The Generalized Eigenvalue problem (the origin of the name is that in the particular casewhen B(x) ≡ I is the unit matrix we come to the problem of minimizing, with respect to x,
the largest eigenvalue of A(x)) can be immediately written down as a ”semiinfinite fractional
problem”
ω T A(x)ω
s.t. gj (x) ≤ 0, j = 1, ..., m, ω T B(x)ω > 0, ω ∈ Ω, x ∈ G;ω∈Ω ω T B(x)ω

minimize max(3.3.8)here Ω is the unit sphere in Rm . Note that the numerators and denominators in our ”objective
fractions” are affine in x, as required by our general assumptions on fractional problems.
Assume that we are given G in advance, same as the data identifying the affine in x matrixvalued functions A(x) and B(x), and let we have an access to a first-order oracle providing uswith local information on the ”general type” convex constraints gj . Then it is not difficult todecide, for a given x, whether B(x) is positive definite, and if it is not the case, to find ω ∈ Ωsuch that the denominator ω T B(x)ω is nonpositive at x. Indeed, it suffices to compute B(x)and to subject the matrix to the Choleski factorization (I hope you know what it means). If
factorization is successful, we find a lower-triangular matrix Q with nonzero diagonal such that
B(x) = QQT ,and B(x) is positive definite; if the factorization fails, then in course of it we automaticallymeet a unit vector ω which ”proves” that B(x) is not a positive semidefinite, i.e., is such that
ω T B(x)ω ≤ 0. Now, if B(x), for a given x, is positive semidefinite, then to find ω associatedwith the largest at x of the fractions
ω T A(x)ω
ω T B(x)ω
is the same as to find the eigenvector of the (symmetric) matrix Q−1 A(x)(QT )−1 associatedwith the largest eigenvalue of the matrix, Q being the above Choleski factor of B(x) (why?); tofind this eigenvector, this is a standard Linear Algebra routine.

70

LECTURE 3. METHODS WITH LINEAR CONVERGENCE, II

Thus, any technique which allows to solve (f ) under assumption (A) immediately implies anumerical method for solving the Generalized Eigenvalue problem.
It is worthy to explain what is the ”control source” of Generalized Eigenvalue problems. Let
me start with the well-known issue - stability of a linear differential equationz 0 (t) = αz(t)(z ∈ Rs ). As you for sure know, the maximal ”growth” of the trajectories of the equation as
t → ∞ is predetermined by the eigenvalue of α with the largest real part, let this part be λ;namely, all the trajectories admit, for any  > 0, the estimate|z(t)|2 ≤ C exp{(λ + )t}|z(0)|2 ,and vice versa: from the fact that all the trajectories admit an estimate|z(t)|2 ≤ C exp{at}|z(0)|2(3.3.9)it follows that a ≥ λ.
There are different ways to prove the above fundamental Lyapunov Theorem, and one of thesimplest is via quadratic Lyapunov functions. Let us say that a quadratic function z T Lz (L isa symmetric positive definite s × s matrix) proves that the decay rate of the trajectories is at
most a, if for any trajectory of the equation one hasz T (t)Lz 0 (t) ≤ az T (t)Lz(t), t ≥ 0;

(3.3.10)

if it is the case, then of coursed  Tln z (t)Lz(t) ≤ 2adt

and, consequently,1/2

z T (t)Lz(t)



1/2

≤ exp{at} z T (0)Lz(0),which immediately results in an estimate of the type (3.3.9). Thus, any positive definite symmetric matrix L which satisfies, for some a, relation (3.3.10) implies an upper bound (3.3.9) on
the trajectories of the equation, the upper bound involving just this a. Now, what does it meanthat L satisfies (3.3.10)? Since z 0 (t) = αz(t), it means exactly that1z T (t)Lαz(T ) ≡ z T (t)(Lα + αT L)z(t) ≤ az T (t)Lz(t)2for all t and all trajectories of the equation; since z(t) can be an arbitrary vector of Rs , thelatter inequality means that
2aL − (αT L + Lα) is positive semidefinite.

(3.3.11)Thus, any pair comprised of a real a and a positive definite symmetric L satisfying (3.3.11)
results in upper bound (3.3.9); the best (with the smallest possible a) bound (3.3.9) which canbe obtained on this way is given by the solution to the problem
minimize a s.t. L is positive definite, 2aL − (αT L − Lα) is positive definite;3.3. EXERCISES: THE CENTER OF GRAVITY AND THE ELLIPSOID METHODS

71

this is nothing but the Generalized Eigenvalue problem with B(L) = 2L, A(L) = αT L + Lα
and no additional constraints on x ≡ L. And it can be proved that the best a given by thisconstruction is nothing but the largest of the real parts of the eigenvalues of α, so that in the casein question the approach based on quadratic Lyapunov functions and Generalized Eigenvalue
problems results in complete description of the behaviour of the trajectories as t → ∞.
In fact, of course, what was said is of no ”literal” significance: what for should we solvea Generalized Eigenvalue problem in order to find something which can be found by a directcomputation of the eigenvalues of α? The indicated approach becomes meaningful when wecome from our simple case of a linear differential equation with constant coefficients to a much
more difficult (and more important for practice) case of a differential inclusion. Namely, assume
that we are given a multivalued mapping z 7→ Q(z) ⊂ Rs and are interested in bounding the
trajectories of the differential inclusionz 0 (t) ∈ Q(z(t)), t ≥ 0;

(3.3.12)such an inclusion may model, e.g., a time-varying dynamic systemz 0 (t) = α(t)z(t)with certain unknown α(·). Assume that we know finitely many matrices α1 ,...,αM such that
Q(z) = Conv{α1 z, ..., αM z}(e.g., we know bounds on entries of α(t) in the above time-varying system). In order to obtainan estimate of the type (3.3.9), we again may use a quadratic Lyapunov function z T Lz: if for
all trajectories of the inclusion one hasz(t)Lz 0 (t) ≤ az T (t)Lz(t),or, which is the same, if
z T Lz 0 ≤ az T Lz ∀(z ∈ Rs , z 0 ∈ Q(z))(3.3.13)then, same as above,(z T (t)Lz(t))1/2 ≤ exp{at}(z T (0)Lz(0))1/2
for all trajectories.
Now, the left hand side in (3.3.13) is linear in z 0 , and in order for the inequalityin (3.3.13)to

1
0
T
T
T
be satisfied for all z ∈ Q(z) it is necessary and sufficient to have z Lαi z ≡ 2 z (αi L + Lαi )z ≤z T Lz for all z and all i, i.e., to ensure positive semidefiniteness of the matrices 2aL−(αiT L+Lαi ),i = 1, ..., M . By setting
T
B(L) = Diag(2L, ..., 2L), A(L) = Diag(α1T L + Lα1 , ..., αM
L + LαM )we convert the problem of finding the best quadratic Lyapunov function (i.e., that one with thebest associated decay rate a) into the Generalized Eigenvalue problem
minimize a s.t. B(L) positive definite, aB(L) − A(L) positive semidefinite
with no additional constraints on x ≡ L.

72

LECTURE 3. METHODS WITH LINEAR CONVERGENCE, II

Note that in the case of differential inclusions (in contrast to that one of equations withconstant coefficients) the best decay rate which can be demonstrated by a quadratic Lyapunov
function is not necessarily the actually best decay rate possessed by the trajectories of the
inclusion; the trajectories may behave themselves better than it can be demonstrated by aquadratic Lyapunov function. This shortcoming, anyhow, is compensated by the fact that theindicated scheme is quite tractable computationally; this is why it becomes now one of thestandard tools for stability analysis and synthesis.
It is worthy to add that the von Neumann problem is a very specific case of a Generalized
Eigenvalue problem (make the entries of Ax and Bx the diagonal entries of diagonal matrices).
I have indicated a number of important applications of (f ); it is time now to think how tosolve the problem. There is no difficulty in applying to the problem the cutting plane scheme.
The Cutting Plane scheme for problem (f ):Initialization. Choose a solid G0 which covers Gf .
i-th step, i ≥ 1. Given a solid Gi−1 (the previous localizer), choosexi ∈ int Gi
and call the oracle A, xi being the input.
Given the answer ei ≡ exi of the oracle,- call step i productive if the oracle says that xi ∈ int Gf , and call the step nonproductive otherwise;- check whether ei = 0 (due to (A), this may happen only at a productive step); if
it is the case, terminate, xi being the result found by the method. Otherwise- define i-th approximate solution to (f ) as the best, in terms of the values of f ,of the search points xj associated with the productive steps j ≤ i;
- set
Ḡi = {x ∈ Gi−1 | (x − xi )T ei ≤ 0};- embed the intermediate localizer Ḡi into a solid Gi and loop.
The presented scheme defines, of course, a family of methods rather than a single method. The
basic implementation issues, as always, are how to choose xi in the interior of Gi−1 and how
to extend Ḡi to Gi ; here one may use the same tactics as in the Center of Gravity or in the
Ellipsoid methods. An additional problem is how to start the process (i.e., how to choose G0 );this issue heavily depends on a priori information on the problem, and here we hardly could do
any universal recommendations.
Now, what can be said about the rate of convergence of the method? First of all, we should
say how we measure inaccuracy. A convenient general approach here is as follows.
Let x ∈ Gf and let, for a given ε ∈ (0, 1),Gεf = x + ε(Gf − x) = {y = (1 − ε)x + z | z ∈ Gf }be the image of Gf under the similarity transformation which shrinks Gf to x in 1/ε times. Letfx (ε) = sup f (y).
y∈Gεf3.3. EXERCISES: THE CENTER OF GRAVITY AND THE ELLIPSOID METHODS73

The introduced quantity depends on x; let us take the infimum of it over x ∈ Gf :f ∗ (ε) = inf fx (ε).
x∈GfBy definition, an ε-solution to f is any point x ∈ Gf such thatf (x) ≤ f ∗ (ε).
Let us motivate the introduced notion. The actual motivation is, of course, that the notion
works, but let us start with a kind of speculation. Assume for a moment that the problem issolvable, and let x∗ be an optimal solution to it. One hardly could argue that a point x̄ ∈ Gf
which is at the distance of order of ε of x∗ is a natural candidate on the role of an ε-solution;
since all points from Gεx∗ are at the distance at most ε Diam(Gf ) from x∗ , all these points can be
regarded as ε-solutions, in particular, the worst of them (i.e., with the largest value of f ) pointx∗ (ε). Now, what we actually are interested in are the values of the objective; if we agree tothink of x∗ (ε) as of an ε-solution, we should agree that any point x ∈ Gf with f (x) ≤ f (x∗ (ε))
also is an ε-solution. But this latter property is shared by any point which is an ε-solution in
the sense of the above definition (look, f (x∗ (ε)) is nothing but fx∗ (ε)), and we are done - ourdefinition is justified!
Of course, this is nothing but a speculation. What might, and what might not be called
a good approximate solution, this cannot be decided in advance; the definition should comefrom the real-world interpretation of the problem, not from inside the Optimization Theory.
What could happen with our definition in the case of a ”bad” problem, it can be seen from thefollowing example:x
minimize, x ∈ Gf = [0, 1].
10−20 + x
Here in order to find a solution with the value of the objective better than, say, 1/2 (note thatthe optimal value is 0) we should be at the distance of order 10−20 of the exact solution x = 0.
For our toy problem it is immediate, of course, to indicate the solution exactly, but think what
happens if the same effect is met in the case of a multidimensional and nonpolyhedral Gf . We
should note, anyhow, that the problems like that one just presented are ”intrinsically bad”; in”good” situations our definition does work:
Exercise 3.3.6 # Prove that
1) if the function f involved into (f ) is convex and continuous on Gf , then any ε-solution x
to the problem satisfies the inequality
f (x) − min f ≤ ε(max f − min f ),
Gf

Gf

Gf

i.e., solves the problem within relative accuracy ε;2) if the function f involved into (f ) is Lipschitz continuous on Gf with respect to certain
norm | · |, Lf being the corresponding Lipschitz constant, and if x is a ε-solution to (f ), thenf (x) − min f ≤ Diam(Gf )Lf ε,
Gfwhere Diam is the diameter of Gf with respect to the norm in question.

74

LECTURE 3. METHODS WITH LINEAR CONVERGENCE, II

Now - the end of the story.
Exercise 3.3.7 #∗ Prove the following statement:
let a cutting plane method be applied to (f ), and let Size(·) be a size. Assume that for
certain N the method either terminates in course of the first N steps, or this is not the case, butSize(GN ) is smaller than Size(Gf ). In the first of the cases the result produced by the method is aminimizer of f over Gf , and in the second case the N -th approximate solution x̄N is well-defined
an is an ε0 -solution to (f ) for any
Size(GN )ε0 >.
Size(Gf )Exercise 3.3.8 # Write a code implementing the Ellipsoid version of the Cutting Plane schemefor (f ). Use the code to find the best decay rate for the differential inclusionz 0 (t) ∈ Q(z) ⊂ R3 ,where
Q(z) = Conv{α1 z, ..., αM z}
and αi , i = 1, 2, ..., M = 26 = 64, are the vertices of the polytope1P = { p21
p31


p12
1
p32

p13
p23  | |pij | ≤ 0.1}.
1Lecture 4

Polynomial solvability of Linear
Programming
This lecture is devoted to one of the most important theoretical applications of the Ellipsoid
method, namely, to the proof of polynomial solvability of Linear Programming. Let me first
explain what is the problem we are going to discuss.

4.1

Classes P and NP

Consider the following pair of problems:”Connectedness”:
Given a graph Γ, detect whether there exists a path leading from a given vertex a to anothergiven vertex b.
”Hamiltonian cycle”:
Given a graph Γ, detect whether there is a path leading from a given vertex a to anothergiven vertex b and visiting each of the remaining vertices exactly once.
Both of the problems belong to a very important world which is called the class of NP (Nondeterministic Polynomial) problems. The characteristic features of the citizens of this world are
as follows:
1) The data identifying a problem instance (in our example these are the graph Γ along withthe pair of vertices marked as a and b) are ”completely finite” and can be naturally represented
by a finite binary word x.
Indeed, the data form a finite sequence of integers - you should specify the number n of vertices
in the graph, the zero-one entries of the incidence matrix
(

Aij =

1, vertices i and j are linked by an arc
0, otherwise

to indicate what the arcs are, and the numbers ia and ib enumerating the vertices a and b. Anda finite sequence of integers can, of course, be coded by a finite binary word; it suffices to code
binary digits in the integers as 0 7→ 00, 1 7→ 11 and to use, say 01 to separate the subsequent
integers in the sequence.
75

76

LECTURE 4. POLYNOMIAL SOLVABILITY OF LINEAR PROGRAMMING

2) The problem itself is to detect whether in a given set of ”candidate” solutions there exists
one - the ”actual solution” - which satisfies a given condition P. Candidate solutions also admit
natural coding by finite binary words y.
Our problems ”Connectedness” and ”Hamiltonian cycle” are exactly of this type; here candidate
solutions are paths in a graph, i.e., finite sequences of integers, and such sequences, as we know,
can be naturally coded by finite binary words.
3) Given a binary word x specifying the data, i.e., the problem instance, and a binary word
y representing a candidate solution, one can easily check whether y actually is a solution to the
problem instance x. More precisely that means that there exists an algorithm M which, given
x and y, verifies efficiently whether y indeed satisfies the required condition P with respect tothe data x, i.e., computes the function(

P(x, y) =0, y is an actual solution to x
1, otherwise

in no more than polynomial p(l(x) + l(y)) number of elementary steps, where l(u) is the binarylength of a finite word u; a predicate P (x, y) possessing this property is called polynomiallycomputable.
In the above definition one can use any known from Mathematical Logic algorithmic models; allthese models lead to the same set of polynomially computable predicates.
Last characteristic property of problems from the class NP is as follows:4) When looking for a solution y to a problem instance x, one can in advance restrict himself
with ”not too long” candidate solutions: the binary length l(y) of any actual solution y to a
problem instance x is bounded from above by certain polynomial q(l(x)) of the binary lengthl(x) of the word x coding the instance.
You can easily verify that the above two problems - ”Connectedness” and ”Hamiltonian
cycle” - indeed belong to the class NP. This class contains also a lot of other extremely important
discrete problems, say, numerous problems involving graphs, Boolean programming, etc.
There is a straightforward way to solve a problem from the class NP. Indeed, given a problem
instance x, let us look through all binary words y with l(y) ≤ q(l(x)) - by definition, all actual
solutions to the problem, if any exists, belong to this finite set of binary words. After a candidate
solution y is chosen, let us test whether it is an actual solution; to this end it suffices to compute
P (x, y), and we know that it is possible to do in polynomial in l(x) + l(y) time. If y solves x,
we are done; if in course of our straightforward search no solution is found, we are in our rightto say that the problem instance is unsolvable. Since the set of candidate solutions is finite, thecomplete procedure will take finitely many elementary steps, and the problem instance will besolved in finite time. A bad news, of course, is that this ”finite time” for our brut force procedure
is extremely large, since the number of candidate solutions is exponential in the length l(x) of
the input data. And what we are interested in is a polynomial time solution algorithm, i.e., that
one with the running time bounded from above by a polynomial of the length l(x) of the input
data. In theoretical Computer Science polynomiality of an algorithm is regarded as a criterionof its effectiveness, and a problem is regarded as tractable if it admits a polynomial time solution
algorithm. The problems with this latter property are called polynomially solvable, and the partof the class NP formed by these ”tractable” problems is called P.
Thus, there exists a wide class of problems NP with a subclass of ”simple”, tractable, polynomially solvable problems P. I believe that everyone of you realizes that the first of our two

4.1. CLASSES P AND NP

77

problems, ”Connectedness”, belongs to this latter subclass, since one immediately can invent analgorithm which, as applied to an n-vertex graph, solves the problem in O(n3 ) elementary steps(this bound can be significantly improved, but for our now purposes it does not matter). In
contrast to the ”Connectedness” problem, nobody knows a polynomial algorithm for the problem of detecting whether there exists a Hamiltonian cycle in a graph; all we know is that this is
the most complicated, a so called NP-complete problem in the class NP. In fact the discovery of
NP-complete problems somewhere 25 years ago was a great event in Discrete and Combinatorial
Mathematics. Before this the situation with discrete problems was as follows. There were someof them which were known to be ”tractable” (as we say now, polynomially solvable), e.g., the”Connectedness” problem or a close to it problem of finding shortest paths in a graph with thearcs assigned positive integer lengths. There were numerous extremely important for applications problems, like Boolean Programming or the problem of finding a Hamiltonian cycle, forwhich efficient algorithms were unknown; every of these ”hard” problems had its own theory.
Along with introducing the notion of the class NP it was proved that there are ”most complicated” problems in the class, the NP-complete ones. The definition of a NP-complete problem
is as follows. Let P(·, ·) and P 0 (·, ·) be two problems from NP (it is very natural to identify
problems from NP with the related functions P(x, y)). We say that P is polynomially reducible
to P 0 , if there exists an algorithm N with the following properties:(1) given on input a binary word x representing an instance of the problem P, it transforms
it into a binary word N (x) which represents an instance of the problem P 0 , and the answer inthis new problem is ”yes” if and only if the answer in the original problem x is ”yes”; thus,
instead of solving an instance x of the problem P, it suffices to solve the instance N (x) of the
problem P 0 ;(2) the algorithm N computes N (x) in time polynomial in l(x).
It is clear that if P is polynomially reducible to P 0 and the latter problem is polynomiallysolvable, then the former one also polynomially solvable; to get a polynomial time algorithm
for P, it suffices to act on its data x by the algorithm N , which takes polynomial in l(x) timeand, consequently, produces the result N (x) of the binary length polynomial in l(x) (since thelength of the output cannot be essentially larger that the running time of N - you should atleast write down all binary digits of N (x), and it, for any model of algorithm, takes O(l(N (x)))
elementary steps). After N (x) is computed, let us apply to it the polynomial time algorithm for
P 0 , which, by assumption, do exists; the resulting answer ”yes/no” will be exactly the same as
for the original problem P (definition of reducibility), and the computation will take polynomialin l(N (x)) (and, consequently, in l(x)) time. Thus, the total time of solving x is bounded fromabove by a sum of two polynomials of l(x)i, and is therefore a polynomial in l(x).
Now, a problem P from NP is called NP-complete, if any other problem from NP is polynomially reducible to P. From the above discussion it is clear that polynomial solvability of anyof NP-complete problems would imply polynomial solvability of all problems from the class NP.
The latter conclusion is nothing but a tautology. The actual breakthrough was to provethat NP-complete problems do exist, which is by no means evident in advance. Moreover, thediscovery of NP-complete problems was not a pure existence theorem; several NP-complete
problems were indicated explicitly, e.g., the Boolean Programming problem or that one on
Hamiltonian cycle. Soon the situation became as follows: basically all important problems forwhich polynomial time algorithms remained unknown turned out to be NP-complete. This
was actually a great deed, since it became clear that when looking for theoretically efficient

78

LECTURE 4. POLYNOMIAL SOLVABILITY OF LINEAR PROGRAMMING

algorithms, you need not develop its own theory for every difficult problem; in a sense, we are
dealing with numerous equivalent formulations of a single difficult problem. Let me stress that
we still do not know whether the problem actually is difficult, whether the class NP is indeed
larger than the subclass of polynomially solvable problems P. The conjecture is that NP6=P- this conjecture simply summarizes the failure of at least 40-year intensive attempts to findefficient algorithms for Boolean Programming, Travelling Salesman problem and tens of others.
Nevertheless, the conjecture NP6=P, after 25 years, remains unproved; it is called the ”major
open question in Computer Science”.
I have said that soon after the notions of the class NP and of a NP-complete problem
were discovered in mid-sixties, it was found that almost all important problems for which no
polynomial time algorithms were known are NP-complete. There were, anyhow, two remarkable
exceptions: the Graph Isomorphism problem, the complexity status of this problem still being
unknown, and Linear Programming.

4.2

Linear Programming

A Linear Programming problem
minimize cT u s.t. P u ≤ q, x ∈ Rk

(4.2.1)(here A is an l × k matrix) is, in its nature, a problem of continuous optimization. Nevertheless,it can be easily included into the realm of problems from the class NP - it suffices to assume
the coefficients to be integer. Let me stress that we do not pass to Integer Programming - thecoefficients are integer, but the solution we are looking for may be real. To include the LP
problem with integer coefficients (for the time being, no other LP problems are considered) into
the class NP, it is convenient to write the problem down along with its dual, thus reducing the
problem to a system of linear equalities and inequalities with unknowns u ∈ Rk and v ∈ Rl :
P u ≤ q,




v≤0

P T v = c, Tc u − qT v = 0;(4.2.2)each linear equality can be represented by a pair of linear inequalities, and we conclude that LP
problems with integer data can be easily reduced to systems of linear inequalities, again withinteger data. Thus, we may speak about the generic problem of solving the system
Ax ≤ b,(4.2.3)A being an m × n matrix, b being an m-dimensional vector, all entries in A and b being integer.
The data in the problem form a finite sequence of integers and can be therefore coded by a finite
binary word. The length of this word is something like twice the total number L of bits in thebinary representation of all coefficients of the problem:
L=

m X
n
X

(clog2 (|Aij | + 1)b+1) +i=1 j=1mX
i=1(clog2 (|bi | + 1)b+1);

4.2. LINEAR PROGRAMMING

79

this quantity L is called the input length of the problem instance.
Now, the problem of solving a system of linear (or any other) inequalities is not of the typeallowed for NP problems; in these latter problems the question is whether a solution exists, notthe solution itself. Thus, for the time being let us speak about the reduced version of an LP
problem, namely, the LP Feasibility problem as follows:
LPF: given system of linear inequalities (4.2.3) with integer coefficients, detect whether thesystem is solvable
In this form the problem is of the type required by the definition of the class NP, which,anyhow, does not mean automatically that it is in NP - to verify this inclusion, we should verify
the remaining items of the definition of a NP problem. What about the second property of aproblem from the class NP - possibility to represent candidate solutions by finite binary words?
Although we are looking for solutions with real entries, it is well-known (and we shall prove it
in a while) that a solvable system of linear inequalities with integer coefficients always admits arational solution with the entries
x∗j =

pi
, |pi |, |qi | ≤ 2O(L) ;qi

it follows that without loss of generality we may look for a rational solution, i.e., for somethingwhich can be represented by a finite sequence of integers and therefore can be coded by afinite binary word. Moreover, from the above bounds on the magnitudes of the numerators andthe denominators in a solution it follows that the length of the word representing a solution
may be restricted to O(Ln). Since clearly n ≤ L, this upper bound on the output length ispolynomial in L, as required by the definition of the class NP. And, of course, given the data
and a candidate solution, we can check, simply by substitution, whether the candidate is asolution; this verification, of course, takes time polynomial in L and the binary length of thecandidate solution. Thus, all restrictions imposed by the definition of a problem from the class
NP are satisfied, and we see that the LP feasibility problem FLP with integer coefficients is inNP.
Now, what is the complexity status of the problem? Is it polynomially solvable or not? This
crucial question (which can be posed, of course, not only for the feasibility problem FLP, butalso for the standard Linear Programming problem with integer coefficients) remained open forsomething like 20 years. Of course, an LP program, even with real coefficients, can be solved
in finitely many arithmetic operations by the Simplex method, and this method was and is
extremely efficient in practice. Anyhow, is the Simplex method polynomial? The answer is”no”. In early sixties Klee and Minty have discovered a family of LP problems Pn with nvariables and 2n linear inequality constraints, n = 1, 2, ..., as follows: the input length of Pn isO(n), and for certain particular version of the Simplex method it takes 2n iterations to reach thesolution. Thus, at least this particular version of the method is not polynomial; similar ”bad”
examples were found also for all other basic versions of the method. Thus, the Simplex method,in spite of its excellent practical performance, is not polynomial, and therefore it does not allow
to understand whether LP is polynomially solvable.
Polynomial solvability of LP was established only in 1979 by Leonid Khachiyan, and thetool he used came from nonsmooth convex optimization - it was the Ellipsoid method invented
3 years earlier. Now we come to the Khachiyan Theorem on polynomial solvability of LP.

80

4.2.1

LECTURE 4. POLYNOMIAL SOLVABILITY OF LINEAR PROGRAMMING

Polynomial solvability of FLP

The key point of Khachiyan’s reasoning was to prove that the LP Feasibility problem is polynomially solvable. To this end Khachiyan acts as follows.
The first step. We may assume that the columns of the matrix A are linearly independent.
Indeed, we are asked to find a linear combination of the columns which does not exceed theright hand side vector; since the set of linear combinations of all columns is exactly the same as
the set of linear combinations of any column basis, we may from the very beginning look for a
combination of this latter type, thus setting to zero the variables associated with the non-basis
columns. In other words, we loose nothing when eliminate from our matrix A all ”redundant”
columns which are linear combinations of the remaining ones. Of course, to perform such anelimination, it requires to find a column basis, but this is a polynomial time linear algebra
routine. Thus, from now on let us assume that the columns in A are linearly independent.
The second step. Let us prove that if the system is solvable, then it admits a solution x∗
with not too large coordinates:x∗ ∈ B = {x ∈ Rn | |x|∞ ≤ 2L }.
We do not say that all solutions are inside the cube, we say that the solution set, if nonempty,
intersects the cube.
The proof is immediate. The set X ∗ of solutions, if nonempty, is a polyhedral set which
does not contain lines. Indeed, if it contains a line, that means that our linear inequalities are
satisfied along a line, which is possible, of course, only if the left hand sides of these inequalities
are zero at the direction of the line. But this means linear dependence of columns in A, andwe just have excluded this case. From the theory of Linear Programming it is known that any
polyhedral set defined by a system of linear inequalities and not containing lines has an extreme
point x∗ , and this extreme point, as substituted into the inequalities, makes equalities certain nlinearly independent inequalities of the system (and, possibly, some inequalities more). In otherwords, x∗ is a solution to a n × n linear subsystem
A0 x = b0
of our initial system, with the inequality signs replaced by the equality one, and n × n matrix
A0 here is nonsingular. Due to the Kramer rule, the entries of x∗ are ratios of the Kramer
determinants:∆j(x∗ )j =.
∆Now, the determinants here are the determinants of certain n × n submatrices of the integer
matrix [A, b]. According to the Hadamard inequality, the modulus of the determinant of an × n matrix with columns p1 , ..., pn does not exceed the product of the Euclidean norms of thecolumns:n| Det (p1 , ..., pn )| ≤Y

|pi |2i=1(”the volume of a parallelotope does not exceed the product of lengths of the vectors the parallelotope is spanned by”). We immediately conclude thatlog2 | Det (p1 , ..., pn )| ≤

n
X
i,j=1

log2 (1 + |(pi )j |)4.2. LINEAR PROGRAMMING

81(indeed, nj=1 (1+|aj |)2 ≥ nj=1 |aj |2 ). In our case, when ∆j and ∆ come from an integer matrix
with the input length L, we, by definition of the length, conclude thatQP|∆j |, |∆| ≤ 2L .

(4.2.4)Thus, the numerators in the Kramer formulae are of absolute value at most 2L ; the denominator
is a nonzero integer, and we conclude that|x∗ |j ≤ 2L , j = 1, ..., n,
as claimed.
Third step. Now let us reformulate our linear feasibility problem as a problem of nonsmoothconvex programming. To this end let us introduce the residual function
f (x) = max {aTi x − bi },
i=1,...,maTi being the rows of A. This is a piecewise linear convex function; it is nonpositive exactly
on the set of solutions to our system of inequalities. Thus, to detect whether the system has asolution in the box B (as we already know, this is exactly the same as to detect whether thesystem is solvable) is the same as to detect whether the optimal value f ∗ in the problem
minimize f (x) s.t. x ∈ B(4.2.5)is nonpositive or positive. The crucial point here is the following simple statement:
Lemma 4.2.1 Assume that the optimal value f ∗ in problem (4.2.5) is positive (so that system(4.2.3) is unfeasible). Then f ∗ is not too small:f ∗ ≥ 2−3L .

(4.2.6)

Proof. The optimal value f ∗ clearly is not less than the optimal value in the following Linear
Programming problem:(P ) : minimize t s.t. Ax ≤ b + te,e being the m-dimensional vector of ones. The problem is, of course, feasible; if f ∗ is positive,
then, as we know, the system Ax ≤ b is unsolvable, and consequently the optimal value t∗ in
our auxiliary LP problem is positive. Now, the rows in the matrix [A, −e] of problem (P ) arelinearly independent; indeed, the columns of A are independent by assumption, and if e were alinear combination of these rows:
e = Ax̄,then the feasible set of (P ) would contain, together with any point (x, t), the line (x + sx̄, t + s),s ∈ R, and the objective in (P ) would be below unbounded on the feasible set of the problem;
as we know, is not the case. Thus, the columns of [A, −e] are linearly independent, so thatthe feasible set of (P ) does not contain lines. From the theory of Linear Programming it now
follows that among optimal solutions to (P ) there is one, (x+ , t∗ ), which corresponds to an
extreme point of the feasible set, and this solutions makes equalities certain (n + 1) linearly
independent inequality constraints in (P ). Same as above, we conclude that the coordinates of

82

LECTURE 4. POLYNOMIAL SOLVABILITY OF LINEAR PROGRAMMING

this solutions are ratios of certain determinants coming from (n + 1) × (n + 1) submatrices of[A, −e]. In particular,
∆∗
t∗ =
∆
is the ratio of this type. The denominator here, exactly as above, does not exceed in absolutevalue 2(2m+L) (the binary length of the matrix [A, −e] is at most L + 2m), and the numerator isan integer. We conclude that if t∗ is positive (and we know that it is), then t∗ ≥ 2−2m−L ≥ 2−3L ,so that f ∗ ≥ t∗ ≥ 2−3L , as claimed.
Now we see that to find out whether (4.2.3) is solvable is the same as to evaluate, within
accuracy, say 2−3L /3, the optimal value to the nonsmooth convex optimization program (4.2.5).
If the approximate optimal value is less than 2−3L /2, we can say that the initial system issolvable, if it is not the case, the system is unsolvable. To solve the nonsmooth problem, one
can use the Ellipsoid method tuned to the relative accuracy likeε=2−3L
≤ 2−6L−33(2n2L × 2L )(the quantity in the parentheses in the denominator is a rough upper bound for the variationof the objective on the cube B; note also that n ≤ L). To solve the problem to the desired
accuracy, it takes O(1)n2 ln(1/ε) = O(1)n2 L steps of the method, with O(mn + n2 ) arithmetic
operations per step (it takes O(mn) operations to compute the value and a subgradient of theobjective at a point and O(n2 ) operations to update the current ellipsoid). Thus, the overall
effort is
N = O(1)n2 (n2 + mn)L ≤ O(L5 )arithmetic operations, which is a polynomial of L. Of course, in our basic representation of
the Ellipsoid method we dealt with operations of precise real arithmetic, which is not allowed
now: we should use completely finite model of computations and count bit operations. A
straightforward analysis demonstrates, anyhow, that we may replace the operations of the precise
real arithmetic with inexact operations which keep O(nL) digits of the result before and after
the dot, and this ”inexact” implementation of the method still evaluates the optimal value
within the desired accuracy. Thus, we may decide whether the initial system of inequalities issolvable or not in no more than O(L5 ) arithmetic operations with O(nL)-bit operands, whichtakes a polynomial in L number of elementary steps. We conclude that the feasibility problem
is polynomially solvable.

4.2.2

From detecting feasibility to solving linear programs

Recall that we have reduced the Linear Programming problem to that one of solving a systemof linear inequalities, but we still did not point out a polynomial time algorithm for this latterproblem; we know how to decide, in polynomial time, whether the system of inequalities is
feasible, but do not know how to find a solution if it exists. There are many ways to reduce
solving a system of linear inequalities to a ”small” series of feasibility problems, and the simplest
for explanation is as follows. Let us first check whether the system in question is solvable (wealready know how to do it in a polynomial time). If infeasibility is detected, we are done. Now
assume that we have found the system to be feasible. Let us prove that in this case it is possible

4.2. LINEAR PROGRAMMING

83

to delete some of inequalities of the system and to set the remaining inequalities to equalities insuch a way that the resulting system of linear equations would be solvable and any solution to it
would also be a solution to the initial system of inequalities. To this end consider the following
construction. We start with the initial system of inequalities, let it be called L0 , and at i-th
step of the construction have another system, Li−1 , which is obtained from the initial one bydeleting some inequalities and setting some inequalities to equalities in such a way that, first,
the number of inequalities in the updated system Li−1 is at most m + 1 − i,

Li−1 is solvable, and all solutions to the system are also solutions to the initial system L0 .
Note that for i = 1 these requirements are satisfied by trivial reasons.
Now assume that Li−1 includes at least one inequality. Let us choose one of the inequalitiesof the system, let us make it the equality and test whether the resulting system L0i is solvable.
If it is the case, then the system L0i clearly can be chosen as Li : it contains one inequality lessthan Li−1 , is solvable and all solutions to the system clearly are solutions to Li−1 and therefore,
by our inductive hypothesis, to L0 . Now assume that the system L0i is unfeasible. Since Li−1 issolvable and L0i is not, we conclude that the inequality we have made equality is non-active at theset of solutions to Li−1 , i.e., is strict at any solution to the latter system. But this means exactlythat eliminating this inequality from the system Li−1 we do not extend the set of solutions, sothat we can take as Li the system which is obtained from Li−1 by deleting the inequality inquestion; the resulting system is solvable and has the same set of solutions as that one of Li−1 ;consequently, the set of solutions to Li is contained in that one of the initial system. And, ofcourse, we have reduced by 1 the number of inequalities in our current system.
In no more than m steps of the indicated construction we come to a system Lm which does
not contain inequalities, i.e., is a system of linear equations; our construction ensures that thesystem is solvable and that any solution to it solves the initial system of inequalities. Now, tofind a solution to a system of linear equations it takes, of course, polynomial time. To pass fromthe initial system L0 to Lm it also takes polynomial time: we should solve m feasibility problemsof the same input length L as that one of the initial system. Thus, Linear Programming admitsa polynomial time solution algorithm and is therefore polynomially solvable.
To conclude, let me say that although the Ellipsoid method, in the worst case sense, is incomparably better than the Simplex method, in practical applications everything looks opposite: theEllipsoid method works more or less according to its theoretical efficiency bound, i.e., requires
something like O(n4 + n3 m) arithmetic operations per accuracy digit, while the Simplex method
in real-world applications never behaves itself too bad, God knows why; the observed # of steps
in the case of a Linear Programming problem with n variables and m inequality constraints normally is proportional, with a very moderate coefficient, to n, which is much-much better thanthe complexity of the Ellipsoid method. Numerous experiments with the Ellipsoid method in the
beginning of eighties definitely confirm practical superiority of the Simplex method. But this is
not the end of the story: in 1984, Narendra Karmarkar proposed a completely new polynomial
time algorithm for Linear Programming. The algorithm of Karmarkar turned out to be very
efficient in practice and, in contrast to the Ellipsoid method, indeed became a serious practicalalternative to the Simplex method.

84

LECTURE 4. POLYNOMIAL SOLVABILITY OF LINEAR PROGRAMMING

4.3

Exercises: Around the Simplex method and other Simplices

4.3.1

Example of Klee and Minty

In early sixties, Klee and Minty demonstrated that the worst-case behaviour of the Simplex
method is very bad. Let us reproduce their famous example of a ”hard” family of LP problems.
Consider an LP program P of the form
maximize xn s.t. x ∈ Rn , Ax ≤ bwith m inequality constraints and a nonempty and bounded feasible set Q. Let us call a simplex
path associated with P a sequence v 1 , ..., v N of vertices of the polytope Q such thatthe objective increases along the sequence:(v 1 )n < (v 2 )n < ... < (v N )n ;the vertices v k , v k+1 neighbouring in the sequence are neighbouring in the polytope Q, i.e.,they are linked by an edge of the polytope.
If v 1 , ..., v N is a simplex path, then the Simplex method as applied to the problem may, atleast for certain pivoting rules, generate a sequence of iterates which includes the simplex path
as a fragment. Klee and Minty prove that, given n, one can point out a problem P of theaforementioned type with m = 2n in such a way that the problem would admit a simplex pathof the length N = 2n , so that the Simplex method, at least with certain ”bad” pivoting rules,
would solve the problem in 2n steps. Later on, the result was strengthen: it was shown that not
only ”bad” pivoting rules, but also all standard rules may result, on some problem instances,in an exponential in the number m of constraints and n of variables amount of pivots. It still
remains unknown whether one can invent pivoting rules free of this drawback. This seems to be
an extremely difficult question, due to the following reasons. If there exists a pivoting rule which
results in a polynomial in n and m number of pivots, then for any polytope Q ⊂ Rn defined by mlinear inequalities and for any pair v, v 0 of vertices of the polytope there exists a path of vertices
of Q which starts at v, ends at v 0 , contains a polynomial in m, n number of vertices and is suchthat any pair of vertices which are neighbours in the path are also neighbours in Q (indeed, it
suffices to choose the objective in the LP problem cT x → max | x ∈ Q in a way which makes v 0
the optimal solution and to run the Simplex method, v being the starting vertex). On the other
hand, there is a well-known almost 40-years old conjecture of M. Hirsh (1957) that any pair ofvertices of a polytope of dimensions (m, n) can be linked by a path of the aforementioned typecomprised of no more than m − n + 1 vertices; in numerous attempts to prove the conjecture,
nobody was lucky to prove even a weaker statement - to obtain a polynomial in m, n upperbound on the length of a path. And to find ”polynomial” pivoting rules for the Simplex method
is, of course, more difficult than to prove the weak version of the Hirsh conjecture.
Let us come back to the example of Klee and Minty.
Exercise 4.3.1 Consider an LP problem Pn :xn → max |

n
X
j=1

aij xj ≤ bi , i = 1, ..., 2n,4.3. EXERCISES: AROUND THE SIMPLEX METHOD AND OTHER SIMPLICES

85

the feasible polytope Qn of the problem being contained in the cube
Cn = {x ∈ Rn | 0 ≤ xi ≤ 1, i = 1, ..., n}.
Let the LP problem Pn+ be defined asxn+1 → max |

n
X

1
1
aij xj ≤ bj , j = 1, ..., 2n, xn ≤ xn+1 ≤ 1 − xn .
33
j=1

Prove that+
1) the set of vertices of the (n + 1)-dimensional polytope Q+
n of feasible solutions to Pn is
as follows: and each vertex v = (x1 , ..., xn ) of Qn ”gives birth” a pair of vertices v+ , v− in Q+n,namely,11v+ = (x1 , ..., xn , 1 − xn ), v− = (x1 , ..., xn , xn );332) the ”children” v+ and v− of a vertex v of Qn are neighbouring in Q+n;3) if v 1 , ..., v N is a simplex path for Pn , then the sequence(v 1 )− , (v 2 )− , ..., (v N )− , (v N )+ , (v N −1 )+ , ..., (v 1 )+is a simplex path for the problem Pn+ .
Exercise 4.3.2 (the Klee and Minty ”hard” LP problem) Prove that the problem
11
1
1xn → max | 0 ≤ x1 ≤ 1, x1 ≤ x2 ≤ 1 − x1 , ..., xn−1 ≤ xn ≤ 1 − xn−13333with n variables and 2n inequality constraints admits a simplex path with 2n vertices. What is
the input length of the problem?

4.3.2

The method of outer simplex

Looking at the Ellipsoid method, one may ask: why should we restrict ourselves to ellipsoids
and not to use localizers of some other ”simple” shape? This is a reasonable question. In fact
all properties of the ”ellipsoidal” shape which were important in the Ellipsoid version of the
cutting plane scheme were as follows:(a) all ellipsoids in Rn are affine equivalent to each other, and, in particular, are affine
equivalent to certain ”standard” ellipsoid, the unit Euclidean ball; therefore in order to lookwhat happens after a cut passing through the center of an ellipsoid, it suffices to study thesimplest case when the ellipsoid is the unit Euclidean ball;(b) the part of the unit Euclidean ball which is cut off the ball by a hyperplane passing
through the center of the ball can be covered by an ellipsoid of volume less than that one of theball.
Now, can we point out another ”simple shape” which satisfies the above requirements? The
natural candidate is, of course, a simplex. All n-dimensional simplexes are affine equivalent toeach other, so that we have no problems with (a); the only bottleneck for simplexes could be(b). The below exercises demonstrate that everything is ok with (b) as well.
Let us start with considering the standard simplex (which now plays the role of the unit
Euclidean ball).

86

LECTURE 4. POLYNOMIAL SOLVABILITY OF LINEAR PROGRAMMING

Exercise 4.3.3 + Let∆ = {x ∈ Rn | x ≥ 0,nX

x1 ≤ 1}i=1

be the standard n-dimensional simplex in Rn , let
c=(

1
1 T
, ...,)
n+1
n+1

be the barycenter (≡ the center of gravity, see exercise 1.5) of the simplex, let
g = (γ1 , ..., γn )T ≥ 0
be a nonzero nonnegative vector such thatg T c = 1,
and let¯ = {x ∈ ∆ | g T x ≤ g T c = 1}.
∆
Prove that if ϑ ∈ [0, 1], then the simplex ∆ϑ with the vertices v0 = 0,vi = (1 − ϑ + ϑγi )−1 ei , i = 1, ..., n,¯(ei are the standard orths of the axes) contains ∆.
Prove that under appropriate choice of ϑ one can ensure that
Voln (∆ϑ ) ≤ χn (n)Voln (∆),

χn (n) = 1 +

1
2
n −1

n−1 

1−

1
n+1



≤ 1 − O(n−2 ).

(4.3.1)

Exercise 4.3.4 Let D be a simplex in Rn , v0 , ..., vn be the vertices of D,w=

1(v0 + ... + vn )n+1

be the barycenter of D and g be a nonzero linear functional.
Prove that the set
D̄ = {x ∈ D | (x − w)T g ≤ 0}
can be covered by a simplex D0 such that
Voln (D0 ) ≤ χn (n)Voln (D),with χ(n) given by (4.3.1). Based on this observation, construct a cutting plane method forconvex problems with functional constraints where all localizers are simplexes.
What should be the associated size?
What is the complexity of the method?
Note that the progress in volumes of the subsequent localizers in the method of outer simplex(i.e., the quantity χn (n) = 1−O(n−2 )) is worse than that one κn (n) = 1−O(n−1 ) in the Ellipsoid
method. It does not, anyhow, mean that the former method is for sure worse than the latter
one: in the Ellipsoid method, the actual progress in volumes always equals to κn (n), while inthe method of outer simplex the progress depends on what are the cutting planes; the quantityχn (n) is nothing but the worst case bound for the progress, and the latter, for a given problem,
may happen to be more significant.

Lecture 5

Linearly converging methods for
games
The cutting plane scheme and the methods based on it, e.g., the Center-of-Gravity and the
Ellipsoid one, can be applied not only to the convex optimization problems, but to other problems
with convex structure, I mean solving convex-concave games and variational inequalities withmonotone operators. This is our now topic; today I shall speak about convex-concave games.

5.1

Convex-concave games

The problem of solving a convex-concave game (it is also called the saddle point problem forconvex-concave functions) is as follows. We are given a pair of solids P ⊂ Rp and Q ⊂ Rq anda continuous function
f (x, y) : G = P × Q → R
which is convex in x ∈ P for every fixed y ∈ Q and concave in y ∈ Q for every fixed x ∈ P . The
problem is to approximate a saddle point (x∗ , y ∗ ) of the function, i.e., a point where f attains
its minimum in x and its maximum in y:f (x, y ∗ ) ≥ f (x∗ , y ∗ ) ≥ f (x∗ , y), (x, y) ∈ P × Q.
Let me recall you the standard interpretation of the saddle point problem. There are two players,
let them be you and me. My possibility is to choose a point x ∈ P , and yours - to choose y ∈ Q;
after we have chosen x and y, I pay to you the sum f (x, y). Of course, my goal is to decrease
my payment, and yours - to increase it. A saddle point (x∗ , y ∗ ) is an equilibrium: if you keep
the choice y ∗ , I cannot decrease my payment by passing from x∗ to any other choice, same as
if I keep the choice x∗ , you cannot enforce me to pay more by passing from y ∗ to any otherchoice. Note that the issue of existence of the equilibrium is far from being trivial; in the casein question (f is continuous and convex-concave, P and Q are solids) the existence is given by
the well-known von Neumann Lemma.
The indicated interpretation explains the following construction: letf¯(x) = max f (x, y),y∈Q87f (y) = min f (x, y).
x∈P

88

LECTURE 5. LINEARLY CONVERGING METHODS FOR GAMESf¯(x) is my loss function: if I choose x, you cannot enforce me to pay more than f¯(x), andcan enforce me to pay this sum (if you know x). Similarly, f (y) is your profit function: if you
choose y, I cannot pay you less than f (y) and, if I know y, can pay exactly this sum. Thus, if
I choose as x any minimizer of f¯, I may be sure that I shall not pay more than minx∈P f¯(x);similarly, if you choose as y any maximizer of f (y), you may be sure that your profit will be atleast maxy∈Q f (y). The existence of a saddle point implies that our worst-case expectations areequal:
min f¯(x) = min max f (x, y) = max min f (x, y) = max f (y);x∈Px∈P y∈Qy∈Q x∈Py∈Q

and it is easily seen that the set S ∗ (f ) of all saddle points of f is nothing but the direct productof the optimal sets of our cost functions:S ∗ (f ) = Argmin f¯ × Argmax f .
PQThus, to solve the game is exactly the same as to solve the pair of convex optimization
problems
minimize f¯(x), x ∈ P ; maximize f (y), y ∈ Q(note that f¯ clearly is convex and continuous on P , while f is concave and continuous on Q),
the optimal values in the problems being equal to each other. The difficulty, anyhow, is thatthe objectives in the problems are defined implicitly, this is why we need specific tools to findsaddle points.
Now I can easily motivate the notion of an ε-saddle point. Let ε ∈ (0, 1) be the required
relative accuracy. We say that a pair (x, y) ∈ P × Q is an ε-saddle point, iff¯(x) − f (y) ≤ ε max f − min f
P ×Q

P ×Q



≡ εV (f );note that the left hand side is nothing but(f¯(x) − min f¯) + (max f − f (y))
PQ(since the optimal values minP f¯ and maxQ f are equal to each other), so that the left handside is the sum of the absolute inaccuracy of x regarded as an approximate minimizer to myloss function and that one of y regarded as an approximate maximizer of your profit function.
This sum has a very natural interpretation in the equilibrium terms: if you keep your choice tobe y and I pass from x to any other choice x0 ∈ P , I cannot save more than f¯(x) − f (y), andsimilarly for you.
In what follows we assume that we have an access to the first-order oracle which, given on
input a pair (x, y) ∈ int (P × Q), returns on output the value f (x, y) of the cost function f , its
subgradient fx0 (x, y) in x and its supergradient fy0 (x, y) in y. It is convenient to assemble the”gradient part” of the answer in a single vector
g(x, y) = (fx0 (x, y), −fy0 (x, y)) ∈ Rp × Rq .
The crucial property of this vector is as follows:

5.2. CUTTING PLANE SCHEME FOR GAMES: UPDATING LOCALIZERS

89

Lemma 5.1.1 One has(z 0 − z)T g(z 0 ) ≥ f (x0 , y) − f (x, y 0 ), z 0 = (x0 , y 0 ) ∈ int G, z = (x, y) ∈ G, G = P × Q.

(5.1.1)

In particular, if z = (x, y) is a saddle point of f , then(z 0 − z)T g(z 0 ) ≥ 0, z 0 ∈ int G.

(5.1.2)The proof is immediate:f (x, y 0 ) − f (x0 , y 0 ) ≥ (x − x0 )T fx0 (x0 , y 0 )andf (x0 , y) − f (x0 , y 0 ) ≤ (y − y 0 )T fy0 (x0 , y 0 ),whencef (x0 , y) − f (x, y 0 ) ≤ (z 0 − z)T g(z 0 );thus, we come to (5.1.1). To prove (5.1.2), it suffices to note that if (x, y) is a saddle point, thenf (x0 , y) ≥ f (x, y) ≥ f (x, y 0 )and therefore the right hand side of (5.1.1) is nonnegative.

5.2

Cutting plane scheme for games: updating localizers

Now assume that, given an access to the first-order oracle, we are asked to find an ε-saddle pointof f . Looking at relation (5.1.2), we immediately get the idea to use the cutting plane scheme:
having chosen a search point z = (x, y) ∈ int (P × Q) and given the answer g(z) of the oracle atthe input z, we can point out a half-space, namelyΠz = {w ∈ Rp+q | (w − z)T g(z) ≤ 0},
which for sure contains the saddle set of f ; the boundary hyperplane of the half-space passes
through the search point, and we may forget about all points of P × Q which belong to thecomplementary half-space, thus reducing the current localizer. To be more exact, we shouldtake into account that Πz is not necessarily a half-space; if g(z) is zero, this is the whole space.
But this is the minor difficulty: if g(z) is zero, that means that f (x0 , y) attains its minimum
in x0 ∈ P at x0 = x, while f (x, y 0 ) attains its maximum in y 0 ∈ Q at y 0 = y (recall that thecomponents of g are sub- and minus supergradient of f with respect to x and y), so that z is asaddle point, and we are done.
Thus, we come to the following cutting plane scheme for saddle point problems:
Initialization. Choose the initial localizer - a solid G0 ⊃ G.
i-th step. Given the previous localizer - a solid Gi−1 1) choose i-th search point zi = (xi , yi ) ∈ int Gi−1 ;2) check whether zi ∈ int G. If it is not the case, call the step i non-productive, choose anonzero vector ei which separates zi and G:(z − zi )T ei ≤ 0 ∀z ∈ G

90

LECTURE 5. LINEARLY CONVERGING METHODS FOR GAMES

and go to 3), otherwise call the oracle, zi being the input. If the returned vector ei ≡ g(zi ) iszero, terminate and claim that zi is a saddle point, otherwise call the step i productive and goto 3).
3) Define i-th intermediate localizer Ḡi as
Ḡi = {z ∈ Gi−1 | (z − zi )T ei ≤ 0}.
Embed Ḡi into a solid Gi , the updated localizer, and loop.
I have presented the scheme in the form which covers both implementations known to us- the Center-of-Gravity version and the Ellipsoid one. We see that the situation is completely
similar to that one arising in convex minimization without functional constraints. In particular,
we know how to decrease at a problem-independent rate an appropriately chosen size of thelocalisers. Namely, in the Center-of-Gravity method we set
G0 = G,

1zi =Vol Gi−1Z

zdz,

Gi = Ḡi ,Gi−1which results ini} AvDiam(G).
p+qIn the Ellipsoid method all Gi are ellipsoids, zi is the center of Gi−1 and Gi is the ellipsoid of
the smallest volume containing the half-ellipsoid Ḡi ; this strategy results in
AvDiam(Gi ) ≤ exp{−O(1)EllOut(Gi ) ≤ exp{−O(1)i} EllOut(G0 ).
(p + q)2

In both the cases O(1) is certain positive absolute constant.

5.3

Cutting plane scheme for games: generating solutions

You might think that we are done, but this is too early to think so – we did not say how to
generate approximate solutions to the problem. In the minimization case we simply take as i-th
approximate solution the best, in terms of the objective, of the search points generated at the
productive steps. Now we also have an objective, our cost function f , but we are not interested
in minimizing it, we are interested in something else, so that there is no reason to look at thevalues of f . In fact the situation is not so simple, as it is seen from the following example:
f (x, y) = xy : [−1, 3] × [−1, 1] → R.
This simple cost function is, of course, convex-concave. The corresponding loss and profit
functions aref¯(x) = max xy = |x|,y∈[−1,1]

f (y) = min xy =x∈[−1,3]

−y,
3y,

y≥0
,
y≤0

the saddle point is unique and this is the origin. Now let us look what happens when we apply
to the problem, say, the Center-of-Gravity method. Note thatg(x, y) = (fx0 (x, y), −fy0 (x, y))T = (y, −x)T ;5.3. CUTTING PLANE SCHEME FOR GAMES: GENERATING SOLUTIONS

91

this vector is orthogonal to (x, y), so that all our cutting lines{z | z T g(zi ) = 0}pass through the origin. Now, the center of gravity of our initial localizer, i.e., of the rectangle[−1, 3] × [−1, 1], is (1, 0), and the first cut results in the rectangular localizer [−1, 3] × [0, 1].
What happens later, it is clear from the picture:

After several steps the localizers Gi become triangles of the type shown on the picture; as i → ∞,these triangles shrink to the segment [−1, 0] of the x-axis, and their centers of gravity, i.e., oursearch points, converge to the point (−2/3, 0). Thus, the method converges, but this is a bad
news, since it does not converge to the saddle point!
It turns out, anyhow, that the situation is not hopeless. As we shall see in a while, it ispossible to pass from the search points to certain convex combinations of these points which doconverge to the saddle set.
The trick we are about to demonstrate is as follows. Assume that we have performed N
steps, and let IN be the set of all indices i ≤ N associated with productive steps; assume that
this set is nonempty. Consider the auxiliary concave functiong N (z) = min(zi − z)T g(zi ) ≡ min gi (z).
i∈In

i∈In

I claim that if the maximum of the function over G is small, then we can form a good approximation to the saddle point set. Indeed, the maximum of g N over G is the same as the maximum
over the set of certain convex combination of the linear components of the function:∃λ∗ = {λ∗i }i∈IN ∈ ∆ = {λ | λi ≥ 0,X

λi = 1} :i∈INXmax g N (z) = max{
z∈G

z∈G

λi gi (z)}.

i∈INThis is the standard duality statement, and the proof is immediate:
max g N (z) ≡ max min gi (z) = max minz∈Gz∈G i∈INz∈G λ∈∆Xi∈IN

λi gi (z) = min max
λ∈∆ z∈GX

λi gi (z)i∈IN(when interchanging minλ and maxz , we have applied the von Neumann Lemma to the bilinearP
in z and λ and therefore concave in z and convex in λ Lagrange function
λi gi (z)). I shallcall the indicated weights λ∗i the optimal Lagrange multipliers associated with N -th step of the
cutting plane scheme.
Now let us prove the following lemma:

92

LECTURE 5. LINEARLY CONVERGING METHODS FOR GAMES

Lemma 5.3.1 Letz N = (xN , y N ) =X

λ∗i zii∈INbe the weighted sum of the search points zi associated with productive steps i ≤ N , the weights
being the optimal Lagrange multipliers. Then z N is feasible, i.e., z N ∈ G, and the absoluteinaccuracy of z N as an approximate saddle point of f admits the following upper bound:f¯(xN ) − f (y N ) ≤ max g N (z).
z∈G

(5.3.1)

Proof. Inclusion z N ∈ G is evident, since zi ∈ G when i ∈ IN and G is convex. Let z = (x, y) ∈G. From Lemma 5.1.1 it follows thatgi (z) ≡ (zi − z)T g(zi ) ≥ f (xi , y) − f (x, yi ), i ∈ IN .
Taking weighted sum of these inequalities, we come toXi∈IN

λ∗i gi (z) ≥X

λ∗i f (xi , y) −X

λ∗i f (x, yi ) ≥ f (xN , y) − f (x, y N )

i∈IN

i∈IN

(the concluding inequality follows from the convexity of f in the first argument and the concavity
in the second). Taking maxima of both sides with respect to z = (x, y) ∈ G and using the originof the optimal Lagrange weights, we come to
max g N (z) ≥ f¯(xN ) − f (y N ),z∈G

as claimed.
Thus, given in advance that the maximum of the function g N is small, we can, looking at
the function, aggregate the associated search points in a good approximation to the saddle setof f . And it turns out that our cutting plane scheme enforces the maximum of g N to go to zero
as N → ∞:Lemma 5.3.2 Let a convex-concave saddle point problem be solved by a cutting plane method,and let Size be a size. Assume that at certain step N we haveεN ≡Size(GN )< 1.
Size(G)

Then the set IN of those indices i ≤ N which are associated with productive steps is nonempty
andεN
max g N (z) ≤V (f ).
z∈G
1 − εN
Proof. To prove that IN is nonempty, note that by construction of the vector ei associatedwith a non-productive step the set
Gi−1 \Gidoes not intersect G (at such a step, we remove from the previous localizer only those points
which do not belong to G). Consequently, if we have performed N steps and all of them were5.3. CUTTING PLANE SCHEME FOR GAMES: GENERATING SOLUTIONS

93

non-productive, then GN must contain G and therefore the size of GN , due to the monotonicityof the size, should be at least the size of G, which in fact is not the case.
Thus, IN is nonempty, so that g N is well-defined. Let z ∗ be a maximizer of the function overG, let α ∈ (εN , 1) and let
Gα = z ∗ + α(G − z ∗ ) ≡ {(1 − α)z ∗ + αz | z ∈ G}.
The size Size(Gα ) is exactly α Size(G), due to the homogeneity of the size with respect tosimilarity transformations. Thus,
Size(Gα ) > Size(GN )and therefore, due to the monotonicity of the size with respect to inclusions, Gα cannot becovered by GN :∃w = (1 − α)z ∗ + αz, z ∈ G : w 6∈ GN .
Since w does not belong to the N -th localizer, it was cut off at one of our steps:(w − zi )T ei > 0
for some i ≤ N . I claim that the step i was productive; indeed, otherwise all points of G, and,in particular, the point w would satisfy the inequality(z − zi )T ei ≤ 0,
which, as we know, is not the case.
Thus, i ∈ IN , so that ei = g(zi ). We come to(w − zi )T g(zi ) > 0,
whencegi (w) = (zi − w)T g(zi ) < 0.
Since w = (1 − α)z ∗ + αz, we come to(1 − α)gi (z ∗ ) ≤ −αgi (z) ≡ −α(zi − w)T g(zi ).
In view of our initial lemma, the quantity (zi − w)T g(zi ) is bounded from below by the differenceof two values of f at certain points of G, i.e., the quantity is ≥ than minus the variation of f
over G:(zi − w)T g(zi ) ≥ −V (f ),and we come to(1 − α)gi (z ∗ ) ≤ αV (f ).
The left hand side in the latter inequality is ≥ g N (z ∗ ) = maxz g N (since g N is the minimum of
all gi , i ∈ IN ), and we come toα
max g N ≤V (f ).
G
1−αSince α ∈ (εi , 1) is arbitrary, we are done.
Combining the results of Lemmas 5.3.1 and 5.3.2, we come to the conclusion as follows.

94

LECTURE 5. LINEARLY CONVERGING METHODS FOR GAMES

Proposition 5.3.1 Assume that we are solving a convex-concave saddle point problem by acutting plane method and let Size be a size. If a step N is such thatεN ≡Size(GN )< 1,
Size(G)

then the set IN of indices i ≤ N of the productive steps is nonempty, and one can point out afeasible approximate saddle point which solves the game within relative inaccuracy εN /(1 − εN ).
To form z N , it suffices to solve the auxiliary piesewise linear convex optimization problem
maximize min (zi − z)T g(zi ) over z ∈ G,i∈INfind the corresponding optimal Lagrange multipliers λ∗i , i ∈ IN and setzN =X

λ∗i zi .

i∈IN

As a corollary, we obtain the following statement:
Proposition 5.3.2 The complexity of solving a convex-concave game within relative accuracyε ∈ (0, 1) by the Center-of-Gravity version of the cutting plane scheme does not exceed
NCG (ε) =cO(1)(p + q) ln(

1+ε
)b;
ε

for the Ellipsoid method, the complexity does not exceed1+εNEll (ε) =cO(1)(p + q) ln
b,
βε
2





β=

EllOut(G).
EllOut(G0 )In both the estimates, O(1) are appropriately chosen absolute constants.
We see that to solve convex-concave saddle point problems is at most as difficult as to minimize
convex functions. At the same time, to solve saddle point problems is at least as difficult as tominimize a convex function over a solid (indeed, the latter problem can be regarded as a saddle
point one with the cost function independent of the choice of one of the players). Thus, in
our now situation we may make the same conclusions on the level of optimality of the methods
as those for the minimization problems. Namely, the Center-of-Gravity method is optimal incomplexity, up to an absolute constant factor, provided that the required accuracy is smallenough; under the latter assumption, the Ellipsoid method is optimal up to a factor of order ofthe dimension of the problem.

5.4

Concluding remarks

As far as actual computations are concerned, the disadvantage of the outlined scheme is thenecessity to solve auxiliary piesewise-linear convex optimization problems. This is not so hardas one can think. One should not solve these problems at each step, since their solutions are
not used in the search process; it suffices to solve the auxiliary problem exactly once, at thetermination step. It also is worthy of note that if P and Q are polytopes, then the auxiliary5.5. EXERCISES: MAXIMAL INSCRIBED ELLIPSOID

95

problems are Linear Programming programs, and one can use the corresponding very efficientpivoting technique.
Let me also note that the outlined scheme is a rather unusual example of an iterative process:the influence of the initial search points does not disappear with time. Indeed, look at our twodimensional example: we cannot allow ourselves to forget about the very first search point evenafter thousands of steps, since after this point is removed the convex hull of the search points
becomes far from the actual saddle point.

5.5

Exercises: Maximal Inscribed Ellipsoid

The below series of exercises deals with the following geometrical problem which is importantfor efficient implementation of the cutting plane scheme:(P): Given a solid G ⊂ Rn , find among all ellipsoids
W = W (B, c) = {x = Bu + c | uT u ≤ 1}

(5.5.1)that one of the maximal volume.
Let us start with the existence issues.
Exercise 5.5.1 # Prove that problem P is solvable.
Our next task is to demonstrate that P is a convex optimization problem. The first step is easy:
Exercise 5.5.2 #∗ Prove that if W is an ellipsoid, then one can find a representation (5.5.1)
of W involving a positive semidefinite symmetric B.
According to Exercise 5.5.2, in what follows we may restrict ourselves with PD-representationsof ellipsoids, i.e., with representations of the form (5.5.1) with symmetric positive definite B.
Thus, from now on all matrices in question, if the opposite is not explicitly stated, are symmetricpositive definite.
Now let us prove that the set W(Q) of pairs (B, c) corresponding to ellipsoids W (B, c)contained in a closed convex domain Q ⊂ Rn , int Q 6= ∅, is convex.
Exercise 5.5.3 #1) Prove that if Q is a half-space
Q = {x ∈ Rn | aT x ≤ b},then W(Q) is open convex set given by the constraints
B is symmetric positive definite;|Ba|2 + aT c ≤ b.
Derive from this observation that W(Q) is convex whenever Q is convex with a nonempty interior.
2) What are the constraints defining W(Q) in the case when Q is a polyhedral set {x ∈ Rn |aTi x ≤ bi , i = 1, ..., n}?

96

LECTURE 5. LINEARLY CONVERGING METHODS FOR GAMESWe have demonstrated that the feasible set of problem P is convex. Now let us look at theobjective. Let
vol W (B, c) = Det B(this is nothing but the ratio of volumes of W (B, c) and the unit Euclidean ball), and let
F (B) ≡ − ln Det B = − ln vol W (B, c).
Note that P can be equivalently rewritten as(P 0 ) : given a solid G ⊂ Rn , find (B, c) ∈ W(G) which minimizes the function F (B). We
already know that the feasible domain of P 0 is convex. Now let us prove that the objective F (·)
also is convex, so that P 0 is a convex optimization problem.
Exercise 5.5.4 #+ Prove that the function F (B) = − ln Det B is convex on the (open) convex
cone Sn+ comprised of n × n positive semidefinite matrices.
Note that the function
EllIn(G) = max{vol1/n W (B, c) | W (B, c) ∈ G} = exp{F ∗ /n},F ∗ being the optimal value in P 0 , clearly is a size. The goal of the remaining exercises is toestablish the following theorem of L. Khachiyan:
Let (B ∗ , c∗ ) be an optimal solution to P 0 and let e be a nonzero vector. Consider the set
Ḡ = {x ∈ G | (x − c∗ )T e ≤ 0}.
ThenωEllIn(Ḡ) ≤ exp{− }EllIn(G),(5.5.2)nω being certain positive absolute constant.
I believe you understand the meaning of this theorem: it says that in the cutting plane scheme
it is possible to decrease certain size, namely, EllIn(·), of current localizers at, basically, thesame rate as that one at which the Center-of-Gravity method reduces the size AvDiam(·). To
decrease the size EllIn at the indicated rate, it suffices to choose the current search point as thecenter of the maximal in volume ellipsoid inscribed into the previous localizer. Since the rateof convergence of a cutting plane method can be expressed in terms of the rate at which themethod decreases (some) size of the localizers, the Inscribed Ellipsoid method which implements
the indicate tactics solves convex problems within relative accuracy ε in O(1)n ln(1/ε) steps,
O(1) being an absolute constant; thus, the method is optimal in complexity, up to an absolute
constant factor, provided than ε is small. Now, what is the advantage of the Inscribed Ellipsoid
Method as compared to the Center-of-Gravity method which also is optimal? The advantage
is that the auxiliary problems arising in the Inscribed Ellipsoid method, i.e., those of finding
the extremal inscribed ellipsoids, are, as we just have seen, convex optimization problems andare therefore computationally tractable, in contrast to computationally untractable problem offinding the center of gravity of a solid.
Of course, one cannot implement the outlined ”inscribed ellipsoid tactics” literally (how tofind the exact extremal inscribed ellipsoid in finite time?), but this is not a great difficulty, since
the statement of Khachiyan’s theorem is stable: the conclusion remains true if W (B ∗ , c∗ ) is an

5.5. EXERCISES: MAXIMAL INSCRIBED ELLIPSOID

97”almost optimal” solution to P 0 , namely, with absolute inaccuracy in terms of the objective F (·)
less than certain once for ever fixed positive quantity (one can take as the inaccuracy something
like 0.01). And to find an almost optimal, in the indicated sense, inscribed ellipsoid, this is aproblem which can be solved in finite time, say, by the known to us Ellipsoid method (which
now would be better to call the ”Circumscribed Ellipsoid” algorithm). It turns out, that if G isa polytope in Rn defined by m linear inequality constraints, then to find an ”almost optimal”
inscribed ellipsoid it takes somewhere O(m8 ) arithmetic operations. Although this is too muchfor practice, theoretically it is an important result: we see that it is possible to solve convex
optimization problems with n variables in the theoretically optimal number O(n ln(1/ε) oracle
calls, ε being the required relative accuracy, and with polynomial in n and ln(1/ε) arithmetic
cost of processing the answers of the oracle. It should be also mentioned that the particularconvex optimization problem P 0 we are interested in can be solved by much more efficienttechnique than that one given by the ”universal” Circumscribed Ellipsoid method; the latest
results state that in order to solve, within relative accuracy ε, a convex optimization problem on
an n-dimensional box, it suffices to perform O(n ln(1/ε)) steps of the slightly modified Inscribed
Ellipsoid method, the arithmetic cost of a step being, up to a logarithmic in n factor, O(n3.5 )operations.
The goal of the remaining exercises is to prove Khachiyan’s theorem. We are not going todeal with the version related to ”almost optimal” inscribed ellipsoids; from the below proof one
can easily see that in fact the reasoning is ”stable” and can be word by word extended onto thecase when W (B ∗ , c∗ ) is ”almost optimal” rather than optimal.
Let us first understand what should be proved. Our goal is to demonstrate that if W ∗ =W (B ∗ , c∗ ) is the extremal inscribed ellipsoid for a solid G, Ḡ is the part of G bounded by a
hyperplane passing through the center c∗ of W ∗ and W is an arbitrary ellipsoid contained in Ḡ,then
vol W ≤ (1 − κ) vol W ∗for certain absolute constant κ > 0. This statement is affine invariant, so that when proving
it, it suffices to consider the case when W ∗ is the unit Euclidean ball centered at the origin.
Now, what we for sure know about W is that the interior of the latter ellipsoid does not containcenter of W ∗ , i.e., the origin (indeed, W is contained in a half-space with the boundary passing
through the center of W ∗ ). Last, we know that W ∗ is the largest ellipsoid contained in theconvex hull G∗ of W ∗ and W .
Thus, we come to the task as follows.
We are given an ellipsoid W with the interior not containing the origin and the unit Euclideanball W ∗ centered at the origin. It is known that W ∗ is the ellipsoid of the maximal volumecontained in the convex hull G∗ of W ∗ and W . We should prove that thenvol W ≤ (1 − κ) vol W ∗ ≡ (1 − κ)for certain absolute constant κ > 0.
Our plan is: let us fix κ ∈ (0, 0.01] and assume that the statement in question is not valid
for this κ, so that the data (i.e., W ) in the above statement satisfy the inequality
vol W > (1 − κ)

(5.5.3)

opposite to that one we should prove. And we shall prove that under this assumption κ cannot
be small, thus coming to the desired result.

98

LECTURE 5. LINEARLY CONVERGING METHODS FOR GAMESIn what follows O(1) denotes various absolute constants, c is the center of W and (1 − λi ),i = 1, ..., n, are the half-axes of the ellipsoid, so that
W = W (B, c),B being a symmetric positive definite matrix with the eigenvalues (1 − λi ), i = 1, ..., n.
Exercise 5.5.5 > Prove that0≤X

λi ≤ O(1)κ

(5.5.4)iandX

λ2i ≤ O(1)κ.

(5.5.5)i

Besides this,√|c|2 ≥ 1 − O(1) κ.

(5.5.6)

Now let V be the (n − 1)-dimensional ellipsoid which is the cross-section of W with the affine
hyperplane Πc = {x | (x − c)T c = 0} passing through c orthogonal to c, and let V ∗ be theintersection of W ∗ with the parallel to Πc hyperplane Π0 = {x | xT c = 0}. Let 1 − µi ,i = 1, ..., n − 1, be the half-axes of the ellipsoid V . It is convenient to choose the coordinate axes
in such a way that the first (n − 1) would be parallel to the principal axes of the ellipsoid V andthe last, n-th axis would be collinear to c. Let us denote by s the collection of the first (n − 1)coordinates of a point with respect to these axes and by t the last coordinate. In the indicatedcoordinates the situation is as follows:
A. The set G∗ contains the ”disk”
V ∗ = {(s, 0) |n−1Xs2i ≤ 1}i=1and the ”ellipsoidal disk”
V = {(s, τ ) | si = (1 + µi )ui ,n−1X

u2i ≤ 1},√
τ = |c|2 ≥ (1 − O(1) κ)

i=1

B. G∗ contains also the unit ballW ∗ = {(s, t) | t2 +n−1X

s2i ≤ 1}i=1and the ballU = {(s, t) | (t − τ )2 +

n−1
X

√
s2i ≤ (1 − O(1) κ)2 }i=1(this latter fact holds true since U , under appropriate choice of the corresponding O(1), is
contained already in W in view of (5.5.5)). Let us add to these two facts the third as follows:

5.5. EXERCISES: MAXIMAL INSCRIBED ELLIPSOID

99

Exercise 5.5.6 ∗ One has
n−1
X
√
√
−O(1) κ ≤µi ≤ O(1) κ(5.5.7)

i=1

and
n−1
X

µ2i ≤ O(1)κ.

(5.5.8)i=1

Now we are prepared to finish our job. To this end let us look at the situation: we know that theunit ball W ∗ is the maximum volume ellipsoid contained in G∗ ; on the other hand, G∗ contains
the convex hull Q of V ∗ and V , which is ”almost” the cylinder C = {(s, t) | |s|2 ≤ 1, 0 ≤ t ≤ τ },and two half-balls, the first of them beingW−∗ = {(s, t) | |s|22 + t2 ≤ 1, t ≤ 0},
and the second being√
W+ = {(s, t) | |s|22 + (t − τ )2 ≤ (1 − O(1) κ)2 }.
Thus, G∗ is ”almost” the set G+ comprised of the cylinder C and two half-balls covering thebases of the cylinder; of course, G+ contains an ellipsoid of the volume ”significantly larger”
than that one of W ∗ (we may make the first (n − 1) half-axes equal to 1 and the last one equal toτ /2 + 1, which would result in almost 50% growth of the volume as compared to that one of theunit ball W ∗ ). The difficulty is that G∗ is not exactly G+ , and we should look carefully whether
or not the difference has serious consequences. The idea, anyhow, is clear: we should construct
an ellipsoid E ∈ G∗ centered at the point c/2 = (0, τ /2) and with the principal axes parallelto the coordinate ones; the n-th half-axis should be ”large”, say, w = 1.1, and the precedinghalf-axis 1 − ωi should be as large as possible under the restriction that E ⊂ G∗ . If we were
lucky to demonstrate that, for small enough κ, it is possible to find such a E withn−1
Y

(1 − ωi ) > (1.1)−1 ,(5.5.9)i=1

we would be done: the latter inequality means that vol E > vol W ∗ , and this is impossible since
E is an ellipsoid contained in G∗ , and W ∗ is the maximal in volume ellipsoid contained in thelatter set. Thus, we would prove that assumption (5.5.3) with small κ leads to a contradiction,
as we desire.
Let us start with the simple part of the task.
Exercise 5.5.7 > Let κ be less than an appropriate absolute constant κ1 , and let the quantities
ωi identifying the above ellipsoid E be less in absolute values than 0.1. Then the parts of theellipsoid E outside the stripe
S = {(s, t) | 0 ≤ t ≤ τ }are covered by the half-balls W−∗ and W+ and therefore are contained in G∗ .
Now the concluding, a slightly more difficult, part of the reasoning:

100

LECTURE 5. LINEARLY CONVERGING METHODS FOR GAMESExercise 5.5.8 > Prove that if κ ≤ κ2 with an appropriately chosen absolute constant κ2 andif ωi are given by
1 − 0.5µi − 1.1rµi√
1 − ωi = min φi (r), φi (r) =,(5.5.10)|r|τ <1
1 − τ 2 r2i.e.,
1 − ωi = φi (ri ), ri =2.2µi,
τ 2 (2 − µi )(5.5.11)then |ωi | ≤ 0.1 and, besides this, the part of the ellipsoid E in the stripe S is contained in G∗ ,so that the whole ellipsoid E is contained in G∗ (see the previous exercise).
Demonstrate that1
(5.5.12)|ωi − µi | ≤ O(1)µ2i .
2Exercise 5.5.9 > Conclude from the previous exercises that the ellipsoid E for small enoughpositive absolute constant κ satisfies (5.5.9) and thus complete the proof of Khachiyan’s theorem.

Lecture 6

Variational inequalities withmonotone operators
We continue considerations related to non-optimization problems with convex structure, andthis lecture is devoted to variational inequalities with monotone operators.

6.1

Variational inequalities with monotone operators

Let F (x) : Rn 7→ Rn be a multi-valued mapping with the domain Dom F ⊂ Rn . These latterwords mean that the image F (x) of x ∈ Dom F is a nonempty set F (x) ⊂ Rn . We shall beinterested in the so called monotone operators, i.e., such that(y − y 0 )T (x − x0 ) ≥ 0∀x, x0 ∈ Dom F∀y ∈ F (x), y 0 ∈ F (x0 ).

In a while we shall look at important examples of these mappings.
Let G be a closed convex subset in Rn with a nonempty interior, and let F be a monotone
mapping with the domain containing the interior of G. The pair (G, F ) defines a variational
inequality
find x∗ ∈ G ∩ Dom F and y∗ ∈ F (x∗ ) s.t. y∗T (x − x∗ ) ≥ 0 ∀x ∈ G.

(6.1.1)

The corresponding x∗ ’s will be called strong solutions to the variational inequality. Note that ifx∗ is a strong solution and x ∈ int G, y ∈ F (x), theny T (x − x∗ ) ≥ y∗T (x − x∗ ) ≥ 0;
thus, if x∗ is a strong solution to the inequality, theny T (x − x∗ ) ≥ 0 ∀x ∈ int G ∀y ∈ F (x).

(6.1.2)The points x∗ ∈ G satisfying this latter relation are called weak solutions to the variational
inequality. As we just have seen, each strong solution is a weak solution as well; the inverse
statement, as we shall see in a while, generally speaking, is not true; it becomes true under very
mild restrictions on the operator. By some reasons (which soon will become clear) it is more
101

102

LECTURE 6. VARIATIONAL INEQUALITIES WITH MONOTONE OPERATORS

reasonable to look for weak solutions, and this is why in what follows I call these weak solutions
simply the solutions to the variational inequality given by G and F .
Now it is time to present examples.
1. Subrgadient of a convex function. Let f (x) be a finite continuous convex function definedon G (f (x) = +∞ outside G), and let
F (x) = ∂f (x)be defined as the set of all subgradients of f at x. Since the subgradients do exist at all interior
points of G, the domain of this multi-valued mapping contains the interior of G, so that thepair (G, F ) defines a variational inequality. Note that the operator in question is monotone: ifx, x0 ∈ G and f 0 (x) ∈ ∂f (x), f 0 (x0 ) ∈ ∂f (x0 ), then(f 0 (x))T (x − x0 ) ≥ f (x) − f (x0 ) ≥ (f 0 (x0 ))T (x − x0 ).
What are the strong solutions to the inequality? Let us demonstrate that these points arenothing but the minimizers of f on G, and that in the case in question weak solutions areexactly the same as strong ones. Indeed, if x∗ is a weak solution, then, for any x ∈ int G one
hasy T (x − x∗ ) ≥ 0, y ∈ ∂f (x);it follows that the convex continuous function fx (t) = f (x∗ + t(x − x∗ )) on the segment 0 ≤ t ≤ 1
possesses nonnegative derivative at any point t > 0 where it is differentiable, and therefore thefunction is monotone, so that fx (1) = f (x) ≥ fx (0) = f (x∗ ), and since x is an arbitrary pointof int G, x∗ is a minimizer of f . Thus, any weak solution to the inequality is a minimizer off . Now, if x∗ is a minimizer of f on G, then 0 is a subgradient of f at x∗ (recall that weare speaking about subgradients on G), and x∗ clearly is a strong solution to the inequality.
Thus, weak solutions are minimizers of f , and minimizers of f are strong solutions; since strong
solutions always are weak ones, we conclude that all three notions - a weak/strong solution to
the variational inequality and a minimizer of f - in the case in question coincide with each other.
Now note that if we ”shrink” the subgradient mapping F , i.e., pass from it to anothermapping x 7→ F̄ (x) ⊂ F (x) such that F̄ (x) is nonempty when x is in the interior of G, thenthe set of weak solutions may only increase - indeed, if y T (x − x∗ ) ≥ 0 for all y ∈ F (x) and allx ∈ int G, then for sure y T (x − x∗ ) ≥ 0 for all y ∈ F̄ (x) ⊂ F (x) for all x ∈ int G; at the sametime, our proof that a weak solution is a minimizer of f still will be valid, so that the set of
weak solutions, which initially - before we have shrunk F - was the set of all minimizers of f ,after passing from F to F̄ will still be the set of all minimizers of f . On the other hand, the setof strong solutions after shrinking the mapping may become smaller, since when passing fromF (x∗ ) to F̄ (x∗ ) we may loose the y which previously demonstrated that x∗ is a strong solution.
In fact the set of strong solutions may even become empty, as is clear from the following example:
G = R, f (x) = |x|.
The mapping F here is {−1},

x<0
F (x) = [−1, 1], x = 0

{1},
x>0

6.1. VARIATIONAL INEQUALITIES WITH MONOTONE OPERATORS

103

and the strong solution is x = 0. If passing from F to F̄ we exclude 0 from the set of values ofthe mapping at x = 0, we loose this unique strong solution, but it still remains a weak one.
This basic example is very instructive. We see that the weak solutions are more stable than
the strong ones; their existence does not depend on whether we took enough care of F (x) to be
not too ”small”. In fact there is a very strong existence theorem for weak solutions:
Proposition 6.1.1 Let G be a solid and F be a monotone operator with the domain containing
the interior of F . Then the set of weak solutions to the variational inequality defined by (G, F )is a nonempty closed convex subset of G.
Proof. The set X∗ of all weak solutions is given by an infinite system of linear inequalities withrespect to w ∈ G:
F = {y T (x − w) ≥ 0 | x ∈ int G, y ∈ F (x)},so that X∗ clearly is closed and convex subset of G. The only thing which should be proved isthat X∗ is nonempty. To this end let us demonstrate that any finite subsystem of the indicatedsystem of inequalities is solvable; this would imply the nonemptiness of X∗ in view of thestandard compactness reasons (”a nested family of compacts has a point in common”).
Thus, consider a finite subsystem
yiT (xi − w) ≥ 0, i = 1, ..., N
of F, so that yi ∈ F (xi ). Assume that the system has no solutions in G; then, due to the von
Neumann Lemma, certain convex combination of the inequalities from the system, the inequality
I(w) ≡

MX

λi yiT (xi − w) ≥ 0

i=1

also has no solutions in G (cf. similar reasoning for games). On the other hand, the pointw=

MX

λ i xii=1

belongs to int G. Let y ∈ F (w), from the monotonicity of F it follows thatMXi=1

λi yiT (xi − w) ≥

M
X
i=1

M
X

λi y T (xi − w) = y T (λi (xi − w)) = y T 0 = 0,i=1

and w turns out to be a solution to the inequality I(w) ≥ 0, which is the desired contradiction.
Let us come back to important examples of variational inequalities with monotone operators.
The inequality associated with a convex function gives an instructive example, but it does not
motivate the importance of the notion of a variational inequality - we know direct ways to deal
with a problem of minimizing a convex function, so that what for should we reduce it to avariational inequality? The below examples give the required motivation.
2. Kuhn-Tucker point of a constrained optimization problem. Consider an optimization
problemminimize f (u) s.t. gi (u) ≤ 0, i = 1, ..., m, u ∈ Rk ,

104

LECTURE 6. VARIATIONAL INEQUALITIES WITH MONOTONE OPERATORS

and assume for the sake of simplicity that the constraints gi , i = 1, ..., m, same as the objective
f are finite and continuously differentiable. A Kuhn-Tucker point of a problem is, by definition,a point x∗ ≡ (u∗ , λ∗ ) ∈ Rn ≡ Rk × Rm such thatf 0 (u∗ ) +mX

λ∗i gi0 (u∗ ) = 0,i=1λ∗i ≥ 0, i = 1, ..., m, gi (u∗ ) ≤ 0, i = 1, ..., m, andλ∗i gi (u∗ ) = 0.
Let us write down a variational inequality which identifies the Kuhn-Tucker points. To this end
set
F (x) ≡ F (u, λ) = (f 0 (u) +mX

λi gi0 (u), −g1 (u), ..., −gm (u)) : Rn → Rn ,i=1

G = {x = (u, λ) ∈ Rn | λ ≥ 0}.
Thus, we have defined a single-valued continuous operator on Rn and a closed convex domain
in the latter space; let us prove that the strong solutions to the associated variational inequality
are exactly the Kuhn-Tucker points of the initial constrained optimization problem.
Indeed, strong solutions to our inequality are exactly the pairs x∗ = (u∗ , λ∗ ) with nonnegative
λ∗ such thatX
X
l(v, h) = v T (f 0 (u∗ ) +λ∗i gi0 (u∗ )) −hi gi (v∗ ) ≥ 0iiwhenever v ∈ Rk and h ∈ Rm are such that hi ≥ 0 for those i with λi∗ = 0.

We immediately

conclude thatf (u∗ ) +X

λ∗i gi0 (u∗ ) = 0i(otherwise we could make l(v, 0) negative by an appropriate choice of v),
gi (u∗ ) ≤ 0, i = 1, ..., m,(otherwise we could make l(0, h) negative by an appropriate choice of h > 0) andλi∗ > 0 ⇒ gi (u∗ ) = 0(otherwise we could make l(0, h) negative by an appropriate choice of h, since the componentsof h associated with positive λi∗ have no restrictions on their signs). Thus, any strong solution
is a Kuhn-Tucker point.
Now, if x∗ = (u∗ , λ∗ ) is a Kuhn-Tucker point, then the vector F (x∗ ) is(0, −g1 (u∗ ), ..., −gm (u∗ ));if x = (u, λ) ∈ G, then(F (x∗ ))T (x − x∗ ) = −Xi(λi − λi∗ )gi (u∗ ) = −Xiλi gi (u∗ ) ≥ 0

6.1. VARIATIONAL INEQUALITIES WITH MONOTONE OPERATORS

105(the second equality comes from the complementary slackness λi∗ gi (u∗ ) = 0, and the concluding
inequality - from the fact that u∗ is feasible for the optimization problem and therefore gi (u∗ ) ≤0). Thus, Kuhn-Tucker points are strong solutions to the inequality, and the summary of ourobservations is that in the case in question strong solutions are exactly the same as the KuhnTucker points of the initial problem.
Of course, the operator F should not necessarily be monotone; anyhow, if the problem inquestion is convex (f and all gi are convex), then the operator is monotone on G:(u − v)T (f 0 (u) +X

λi gi0 (u) − f 0 (v) −Xiµi gi0 (v)) −i= (u − v)T (f 0 (u) − f 0 (v)) +X(λi − µi )(gi (u) − gi (v)) =i

Xλi (u − v)T (gi0 (u) − gi0 (v))+

i

+

X



λi − µi )(gi (u) − gi (v) − (u − v)T gi0 (v) ≥ 0.

i

It is easily seen that in the case of an operator coming from a convex problem weak solutions
are strong ones, since the operator is single-valued and continuous; and if F is a single-valued
and continuous operator on a closed convex domain G, then, passing to limit in the inequality
F T (x∗ + t(x − x∗ ))(x − x∗ ) ≥ 0(x∗ is a weak solution to the variational inequality and x ∈ int G) as t → 0, we come toF T (x∗ )(x − x∗ ) ≥ 0, x ∈ int G, and consequently F T (x∗ )(x − x∗ ) ≥ 0, x ∈ G, so that x∗ is astrong solution.
Thus, solving a convex optimization problem can be reduced to finding a weak solution toa variational inequality with monotone operator. This also is not a good motivation, since weknow direct ways to deal with constrained convex problems. My next (and last) example is freeof this shortcoming.
3. Nash equilibrium. Consider a game with k participants; i-th of the participants choosesa point xi ∈ Gi ⊂ Rni , Gi being certain solid; the payment of i-th player is a functionfi (x1 , ..., xk )of his choice and choices of all other players.
Q
A point z∗ = (x1∗ , ..., xk∗ ) ∈ G = ki=1 Gi is called Nash equilibrium, if nobody can improve
his position by his separate actions:kxi∗ ∈ Argmin fi (x1∗ , ..., x∗i−1 , xi , xi+1
∗ , ..., x∗ ), i = 1, ..., k.
xi ∈GiThis is a very natural extension of the notion of a saddle point in a game of two players we haveconsidered last time; the saddle points in the 2-player game with the cost function f (x, y) arenothing but Nash equilibriums in the game withf1 (x, y) = f (x, y), f2 (x, y) = −f (x, y)(this is an equivalent reformulation of the initial game which is symmetric with respect to bothof the players; previously we were saying that I wish to decrease f and you wish to increase

106

LECTURE 6. VARIATIONAL INEQUALITIES WITH MONOTONE OPERATORS

it, but this is the same as to say that I wish to decrease f1 by controlling x and you wish to
decrease f2 by controlling y). Note that the Nash game coming from the usual 2-player game isa game with zero sum:f1 (x, y) + f2 (x, y) ≡ 0.
This is not the case in a general Nash game: the sums(z) =kXfi (z)

i=1

should not be zero. Note also that the notion of Nash equilibrium reflects a very specific understanding of the goals of the game: each player thinks only of himself, there is no cooperation.
For games with nonzero sum (or even for games with zero sum and more than two players) thisformalization of the motivations of the players might be not so reasonable, so that there is a
deep theory of cooperative games where the definition of equilibrium differs from that one given
by Nash, but this is not the issue we are interested in now. My goal is to demonstrate thatthe problem of finding the Nash equilibrium, under reasonable convexity assumptions, can be
reduced to solving a variational inequality with a monotone operator. The related assumptions
are as follows:(i) The cost function fi of i-th player is Lipschitz continuous on G, convex in the variableix ∈ Gi controlled by the player and is concave with respect to the collection of variables
controlled by the remaining players (note: with respect to the collection {xj , j 6= i}, not to xj ’s
separately), i = 1, ..., k;(ii) The sum s(z) of the cost functions fi (z) is convex on G.
Given a game possessing the indicated properties, let us associate with the game a mapping
F (z) : G → Rn1 × ... × Rnk defined as follows:
F (x1 , ..., xk ) = ∂f1 (·, x2 , x3 , ..., xk )(x1 )×∂f2 (x1 , ·, x3 , ..., xk )(x2 ) × ... × ∂fk (x1 , ..., xk−1 , ·)(xk ),where ∂f (·, y)(x) denotes the subgradient set of a convex with respect to the correspondingargument function at the value x of the argument. Note that for the case of 2-player game F (z)is exactly the set of all possible values of the vector g(z) which we used last time.
Let us make the following observation:
Proposition 6.1.2 Under assumptions (i), (ii) the mapping F : G → Rn , n = n1 + ... + nk ,is monotone, and each weak solution to the variational inequality defined by (G, F ) is a Nashequilibrium. Besides this, F is semibounded on G:
V(F ) = sup sup sup (z 0 − z)T y ≤ V ≡z∈int G z 0 ∈G y∈F (z)Xi(max fi − min fi ).
GG(6.1.3)

Proof. First of all, let us prove monotonicity. Let u = (u1 , ..., uk ), v = (v 1 , ..., v k ) ∈ G and let
U = (U1 , ..., Uk ) ∈ F (u), V = (V1 , ..., Vk ) ∈ F (v). Let us set
wi =ui + v i i ui − v i
, δ =
.
2
2

6.1. VARIATIONAL INEQUALITIES WITH MONOTONE OPERATORS107

Due to the origin of F , we havefi (u1 , ..., ui−1 , wi , ui+1 , ..., uk ) ≥ fi (u) − (δ i )T Ui ,fi (v 1 , ..., v i−1 , wi , v i+1 , ..., v k ) ≥ fi (v) + (δ i )T Vi ,whencefi (u1 , ..., ui−1 , wi , ui+1 , ..., uk ) + fi (v 1 , ..., v i−1 , wi , v i+1 , ..., v k ) ≥ fi (u) + fi (v) − (δi )T (Ui − Vi ).
Since fi is concave with respect to the collection of all arguments excluding the i-th of them,
the left hand side in this inequality is ≤ 2fi (w), and we come to
2fi (w) ≥ fi (u) + fi (v) − (δi )T (Ui − Vi ).
Taking sum in i, we come to1
2s(w) ≥ s(u) + s(v) − (u − v)T (U − V ),2or1(u − v)T (U − V ) ≥ s(u) + s(v) − 2s(w) ≥ 0
2(the latter inequality follows from the convexity of s), as required.
Now, if x∗ = (x1∗ , ..., xk∗ ) is a weak solution to the variational inequality defined by (G, F ),iki+1
then for any i, any xi ∈ int Gi and any gi ∈ ∂fi (x1∗ , ..., xi−1
∗ , ·, x∗ , ..., x∗ )(x ) there exists
1
i−1
i
i+1
k
g ∈ F (x̄i ), x̄i = F (x∗ , ..., x∗ , x , x∗ , ..., x∗ ), such that gi is the i-th component of g; from thefact that x∗ is a weak solution and from Lipschitz continuity of fi it can be easily derived thatg T (x̄i − x∗ ) = giT (xi − xi∗ ) ≥ 0,
so that xi∗ is a weak solution to the variational inequality associated with the monotone operatori+1
ki
xi 7→ ∂fi (x1∗ , ..., xi−1
∗ , ·, x∗ , ..., x∗ )(x );i i+1kas we already know, this means that xi∗ is a minimizer of fi (x1∗ , ..., xi−1
∗ , x , x∗ , ..., x∗ ) over Gi ,and therefore x∗ indeed is a Nash equilibrium.
It remains to verify (6.1.3), but this is immediate: if u = (u1 , ..., uk ) ∈ int G, v = (v 1 , ..., v k ) ∈G and U = (u1 , ..., uk ) ∈ F (u), then(v i − ui )T Ui ≤ fi (u1 , ..., ui−1 , v i , ui+1 , ..., uk ) − fi (u) ≤ max fi − min fi ,Gwhence(v − u)T U ≤ V,
as claimed.

G

108

LECTURE 6. VARIATIONAL INEQUALITIES WITH MONOTONE OPERATORS

6.2

Cutting plane scheme for variational inequalities

I think that the presented examples demonstrate that variational inequalities with monotone
operators form an important class of problems; in a sense, this is the widest class of problems
with convex structure. What we are interested in is, of course, how to solve these general
problems. As always, we restrict ourselves with a ”bounded” case, i.e., that one of a variational
inequality
find x∗ ∈ G such that y T (x − x∗ ) ≥ 0 ∀x ∈ int G ∀y ∈ F (x)with G being a solid and F being monotone and semibounded:V(F ) ≡ sup{y T (u − x) | x ∈ int G, y ∈ F (x), u ∈ G} < ∞.
As always, we should start with choosing a good accuracy measure for candidate solutions.
To this end let us note that a weak solution to variational inequality is, by definition, a pointx∗ such that all linear forms y T (x − x∗ ) coming from the operator are nonnegative at x∗ or,which is the same, all linear forms y T (x∗ − x) are nonpositive at x∗ . Thus, the weak solutions
are exactly the points of G where the convex function
ν(u) = sup{y T (u − x) | x ∈ int G, y ∈ F (x)}is nonpositive; note that due to semiboundedness this function if finite on G (and V(F ) is exactly
the upper bound of this function over u ∈ G). Thus, it is natural to regard this function as
the measure of absolute accuracy of candidate solutions and to define a solution of relative
inaccuracy ε ∈ (0, 1) as any point x ∈ G such that
ν(x) ≤ εV(F ).
In fact we wish to minimize the convex function ν(x), but, same as in the case of games, thedifficulty is that the function is not given explicitly; moreover, in contrast to the case of games
even to compute the value of ν at a point is an untractable problem, since the maximization
problem involved into the definition of ν is, generally speaking, non-convex.
Nevertheless, the problem is quite tractable. Assume that we are given ε and have an accessto an oracle which, given on input x ∈ int G, reports on output a vector g(x) ∈ F (x), and ourgoal is to solve the variational inequality within relative accuracy ε. How should we act, it isclear: by definition, the solution set is comprised of points x satisfying, for all z ∈ int G, linear
inequalities (x − z)T g(z) ≤ 0, and we may use these inequalities for cuts. Thus, we come tothe following generic scheme which does not differ from that one for the case of games and is asfollows:
Initialization. Choose the initial localizer - a solid G0 ⊃ G.
i-th step. Given the previous localizer - a solid Gi−1 1) choose i-th search point xi ∈ int Gi−1 ;2) check whether xi ∈ int G. If it is not the case, call the step i non-productive, choose anonzero vector ei which separates xi and G:(x − xi )T ei ≤ 0 ∀x ∈ G

6.2. CUTTING PLANE SCHEME FOR VARIATIONAL INEQUALITIES

109

and go to 3), otherwise call the oracle, xi being the input. If the returned vector ei ≡ g(xi ) iszero, terminate and claim that xi is an exact (and even strong) solution, otherwise call the stepi productive and go to 3).
3) Define i-th intermediate localizer Ḡi as
Ḡi = {x ∈ Gi−1 | (x − xi )T ei ≤ 0}.
Embed Ḡi into a solid Gi , the updated localizer, and loop.
As always, with the Center-of-Gravity or the Ellipsoid implementation we may ensure certain
rate of decreasing appropriately chosen sizes of the sequential localizers. The point, same as in
the case of games, is how to aggregate search points xi into approximate solutions. You should
not be surprised by the fact that this question can be resolved exactly as in the case of games.
Assume that we have performed N steps of a cutting plane method and among these steps
there were productive ones; let IN be the set of indices of these productive steps. Same as in
the case of games, we form the functiong N (x) = min gi (x) ≡ min (xi − x)T g(xi ),
i∈IN

i∈IN

find the optimal Lagrange multipliers λ∗i ≥ 0, i ∈ IN , with the unit sum:max g N (x) = max
x∈G

x∈G

X

λ∗i gi (x)i∈INand define N -th approximate solution as the corresponding convex combination of the search
points:X
λ∗i xi .
x̄N =
i∈IN

In the case of games our reasoning was as follows: first, we demonstrated that this aggregate
solves the game within absolute inaccuracy maxG g N , and, second, proved that the cutting plane
scheme automatically decreases the latter maximum at the same rate at which it decreases
certain size of localizers. Now we shall act completely similarly.
Lemma 6.2.1 One hasν(x̄N ) ≤ max g N .

(6.2.1)

G

Proof. Let x be an arbitrary point of int G, and let y ∈ F (x). From monotonicity of F it
follows thaty T (xi − x) ≤ g T (xi )(xi − x) = gi (x),whencey T (x̄N − x) ≤X

λ∗i gi (x) ≤ max g N
Gi∈IN(the concluding inequality follows from the origin of λ∗i ). Since the resulting inequality is valid
for all x and all y ∈ F (x), we come toν(x̄N ) ≤ max g N ,G

as claimed.
The fact that the cutting plane scheme decreases maxG g N at the same rate at which it
decreases certain size of localizers can be formulated as follows:

110

LECTURE 6. VARIATIONAL INEQUALITIES WITH MONOTONE OPERATORS

Lemma 6.2.2 Let a variational inequality with a semibounded monotone operator be solved by
a cutting plane method, and let Size be a size. Assume that at certain step N we haveεN ≡Size(GN )< 1.
Size(G)

Then the set IN of those indices i ≤ N which are associated with productive steps is nonempty
andεNV(F ).
(6.2.2)
max g N (x) ≤x∈G
1 − εN
There is actually nothing to prove here, since one can word by word repeat the reasoning used
for the saddle point case. Indeed, the latter reasoning was completely independent of what is
the origin of the linear forms gi , only the fact that these are exactly the forms used at the
productive steps of the method and that all these forms are uniformly bounded from below;
minus this lower bound is exactly the constant factor which appears in the right hand side of(6.2.2). In our case the lower bound for gi (x) = (xi − x)T g(xi ) can be taken equal to −V(F ),simply by definition of the latter quantity, and this is the origin of (6.2.2).
Now we can summarize our considerations as follows:
Proposition 6.2.1 Assume that we are solving a variational inequality with a monotone semibounded operator by a cutting plane method and let Size be a size. If a step N is such thatεN ≡Size(GN )< 1,
Size(G)

then the set IN of indices i ≤ N of the productive steps is nonempty, and one can point out afeasible point x̄N which solves the inequality within relative inaccuracy εN /(1 − εN ). To formx̄N , it suffices to solve the auxiliary piecewise linear convex optimization problem
maximize min (xi − x)T g(xi ) over x ∈ G,i∈INfind the corresponding optimal Lagrange multipliers λ∗i , i ∈ IN and setx̄N =λ∗i xi .

X
i∈IN

As a corollary, we obtain the following statement:
Proposition 6.2.2 The complexity of solving an n-dimensional variational inequality within
relative accuracy ε ∈ (0, 1) by the Center-of-Gravity version of the cutting plane scheme does
not exceed
1+ε
NCG (ε) =cO(1)n ln()b;εfor the Ellipsoid method, the complexity does not exceed
NEll (ε) =cO(1)n2 ln



1+ε
b,
βε


β=

EllOut(G).
EllOut(G0 )In both the estimates, O(1) are appropriately chosen absolute constants.

6.3. EXERCISES: AROUND MONOTONE OPERATORS

111

Of course, the methods possess the same optimality properties as their ”optimization” parents.
To conclude, let me explain why we separately studied the case of saddle point problems;
now it is completely clear that this was a particular case of variational inequalities. By the same
reason one might ask what for should we be interested in convex minimization problems, which
also can be reduced to solving variational inequalities with monotone operators. The answer
is as follows: as far as the cutting plane scheme is concerned, there actually is no difference
between the saddle point problems and variational inequalities, and there is almost no difference
between the case of these inequalities and optimization problems without functional constraints(the only difference is that in the optimization case we should not solve auxiliary piecewise
linear problems). Nevertheless, the conclusions we make are not completely identical, since they
relate to different inaccuracy measures. It is easily seen that our now inaccuracy measure, being
restricted to saddle point problems, is dominated by the inaccuracy in terms of the corresponding
cost function. In other words, an approximate saddle point of certain absolute inaccuracy solves,
within the same absolute inaccuracy, the variational inequality associated with the saddle point
problem, but the inverse statement is not true. Thus, last time we dealt with a particular caseof the variational problem and have obtained for this case stronger results - although they lookexactly as our today ones - because these former results dealt with a more ”strict” inaccuracy
measure.
Now we say ”good bye” to the cutting plane scheme and methods with linear instanceindependent convergence for convex problems. This is not because there is nothing to say more
- there are other implementations of the scheme, e.g., the Inscribed Ellipsoid method which has
the same optimal information-based complexity as the Center-of-Gravity method and at thesame time, similarly to the Ellipsoid method, results in reasonable auxiliary problems which canbe solved in polynomial time; those who have looked through the problems distributed last timeknow what the method is. There are also other ”optimal” in complexity implementations of thescheme, but I shall not speak about these more sophisticated constructions. What should be
stressed is that the cutting plane scheme, although extremely simple in its basic ideas, demonstrates ability to solve, with theoretically quite reasonable complexity, very general problems of
convex structure. This is due to the intrinsic simplicity of convex optimization, simplicity which
is not seen so clear in more traditional optimization schemes.

6.3

Exercises: Around monotone operators

The below problems are motivated by the following natural question: when any weak solution
to a variational inequality with a monotone operator is a strong one? This question is not toointeresting, but it gives us the possibility to get acquainted with the cornerstone notion of a
maximal monotone operator.
To motivate this notion, let us look at the example of variational inequality given by thedata
G = R, F∗ (x) = ∂|x|;
here we are able to ”kill” the (unique) strong solution x = 0 by deleting from the set of valuesof F∗ at x = 0 the required value 0. The resulting operator F still is monotone, but it is ”bad”- we can extend the graph of the operator, i.e., the set
G(F ) = {(x, y) | x ∈ Dom F, y ∈ F (x)}112

LECTURE 6. VARIATIONAL INEQUALITIES WITH MONOTONE OPERATORS

to a larger set which still is a graph of a monotone operator.
Note, by the way, that an operator F : Rn → Rn is monotone if and only if its graph is amonotone set in Rn × Rn , i.e.,(x, y), (x0 , y 0 ) ∈ G(F ) ⇒ (y − y 0 )T (x − x0 ) ≥ 0.
In particular, an operator inverse to a monotone one (i.e., with the domain formed by the
projection of G(F ) onto the y-space and naturally defined sets of values) also is monotone,
since both the operators have ”symmetric” graphs (or, better to say, the same graph), and theproperty of the monotonicity of a graph is kept under swapping x and y.
Thus, one of the reasons for a weak solution not to be a strong one is ”non-maximality”
of the corresponding operator. The fundamental notion of a maximal monotone operator isdefined as follows: a monotone operator F : Rn → Rn with a domain Dom F is called maximal
monotone, if its graph is a maximal monotone set in Rn × Rn , i.e., if it is impossible to pointout a pair (x, y) not belonging to G(F ) such that the set
G(F ) ∪ {(x, y)}also would be monotone. Note that a monotone operator is maximal if and only if its inverse ismaximal.
By standard sophistic (Zorn’s Lemma) any monotone operator can be extended (generally
speaking, not uniquely) to a maximal monotone one, i.e., the graph of the initial operator canbe embedded into the graph of a maximal monotone operator.
Exercise 6.3.1 #∗ Let F : Rn → Rn be a maximal monotone operator. Detect which of thefollowing statements are for sure true and which are, generally speaking, false:
1) Dom F is a closed convex set
2) Dom F is closed
3) Dom F is convex
4) G(F ) is a closed convex set5) G(F ) is closed6) G(F ) is convex
7) F (x) is a closed convex set for any x ∈ Dom F
8) F (x) is a closed set for any x ∈ Dom F
9) F (x) is a convex set for any x ∈ Dom F
The domain of a maximal monotone operator not necessarily is convex, but its non-convexity is
nothing but an illusion:
Exercise 6.3.2 #∗ Let F : Rn → Rn be a maximal monotone operator, and let G be the closed
convex hull of Dom F . Prove that int G ⊂ Dom F .
A basic example of a monotone operator is as follows: let f : Rn → R+ ≡ R ∪ {+∞} bea convex function, and let Dom f be the set of those x with finite f (x). Of course, Dom f is aconvex set. Assume that this set possesses a nonempty interior; in this case let us say that f is
a proper convex function of n variables. Let x 7→ f 0 (x) = ∂f (x) be the mapping which maps x

6.3. EXERCISES: AROUND MONOTONE OPERATORS

113

into the set of all subgradients of f at x (this set, by definition, is empty for x 6∈ Dom f ; it canbe also empty if x ∈ Dom f , but it for sure is nonempty when x ∈ int Dom f ); thus,
int Dom f ⊂ Dom f 0 ⊂ Dom f.

Exercise 6.3.3 #∗ Prove that if f is a proper convex function of n variables, then f 0 is amonotone operator. Is this operator maximal monotone?
Exercise 6.3.4 #∗ Let f be a proper convex function of n variables. Consider the following
properties of f :1) f 0 is maximal monotone;
2) f is lower semicontinuous;3) the graph G(f ) = {(x, t) | x ∈ Dom f, t ≥ f (x)} is a closed set;4) for all x ∈ Rn one has
f (x) =sup{f (u) + v T (x − u)}.

(u,v)∈G(f 0 )Prove that 2)⇔3)⇔4)⇒1) (concluding ⇒ is not a misprint).
The latter exercise immediately provides us with a number of interesting examples of maximal
monotone operators, e.g., with the following one: let G be a closed convex domain in Rn(as always, ”domain” means nonemptyness of the interior). Given a finite continuous convex
function f on G, we may extend it outside G by the value +∞, and the resulting function, ofcourse, will be convex and lower semicontinuous (and proper). The subdifferential mapping f 0of the resulting function is, as we now know, a maximal monotone operator.
Exercise 6.3.5 Let us apply the aforementioned construction to the function f ≡ 0 (recall that
this defines f only on G; outside G f = +∞). Prove that the resulting monotone mapping f 0 is
as follows: the domain of it is G; f 0 (x) = {0} for x ∈ int G, and at the boundary of G there isa ”crown”: if x ∈ ∂G, then f 0 (x) is comprised of all functionals e such that eT (z − x) ≤ 0 for
all z ∈ G; nonzero functionals e with this property are called normal to G at x, and the set ofthese normal functionals is nonempty at any x ∈ ∂G.
The maximal monotone operator associated with a closed convex domain G in the way presented
in the latter exercise will be denoted NG .
Given a pair of monotone operators F1 , F2 : Rn → Rn with Dom F1 ∩ Dom F2 6= ∅, one candefine the sum F1 + F2 of the operators as follows:
Dom(F1 + F2 ) = Dom F1 ∩ Dom F2 ;(F1 + F2 )(x) = {y1 + y2 | y1 ∈ F1 (x), y2 ∈ F2 (x)}, x ∈ Dom(F1 + F2 ).

Exercise 6.3.6 # Prove that the sum of monotone operators also is monotone.

114

LECTURE 6. VARIATIONAL INEQUALITIES WITH MONOTONE OPERATORSOne of the basic results in the theory of variational inequalities is given by the following
Theorem of Rockafellar:
let F1 and F2 be maximal monotone, and let int Dom F1 intersect int Dom F2 . Then theoperator F1 + F2 also is maximal monotone.
With the Theorem of Rockafellar in hands, we can resolve the question we started with.
Exercise 6.3.7 #∗ Prove that if G is a closed convex domain in Rn and F is a maximal
monotone operator such that Dom F ⊃ int G, and F is semibounded on int G:
sup{(u − x)T y | u ∈ G, (x, y) ∈ G(F )} < ∞,then any weak solution to the variational inequality associated with (G, F ) is a strong solution
to the inequality.
The latter exercise gives rather mild sufficient conditions for a weak solution to be a strong one.
The concluding exercise deals with stronger conditions; as a compensation for the weaker result,the proof is much simpler and does not use any serious tools.
Exercise 6.3.8 # Let G be a closed convex domain in Rn and F : G → Rn be a single-valued
and continuous monotone operator. Prove that then every weak solution to the variational
inequality given by (G, F ) is a strong solution to the inequality.

Lecture 7

Large-scale optimization problems
7.1

Goals and motivations

Today we start a new topic - complexity and efficient methods of large-scale convex optimization.
Thus, we come back to problems of the type(p)

minimize f (x) s.t. gi (x) ≤ 0, i = 1, ..., m, x ∈ G,

where G is a given solid in Rn and f, g1 , ..., gm are convex continuous on G functions. The familyof all consistent problems of the indicated type was denoted by Pm (G), and we are interested
in finding ε-solution to a problem instance from the family, i.e., a point x ∈ G such thatf (x) − min f ≤ ε(max f − min f ), gi (x) ≤ ε(max gi )+ , i = 1, ..., m.
G

G

G

G

We have shown that the complexity of the family in question satisfies the inequalitiesbO(1)n ln(1/ε)c ≤ Compl(ε) ≤c2.182n ln(1/ε)b,(7.1.1)

where O(1) is an appropriate positive absolute constant; what should be stressed that the uppercomplexity bound holds true for all ε ∈ (0, 1), while the lower one is valid only for not too large
varepsilon, namely, forε < ε(G).
The ”critical value” ε(G) depends, as we remember, on affine properties of G; for the box it is1
2 , and for any n-dimensional solid G one has
ε(G) ≥

1.
2n3Thus, our complexity bounds identify complexity, up to an absolute constant factor, only forsmall enough values of ε; there is an initial interval of values of the relative accuracy∆(G) = [ε(G), 1)
where we to the moment have only an upper bound on the complexity and have no lower bound.
Should we be bothered by this incompleteness of our knowledge? I think we should. Indeed,
115

116

LECTURE 7. LARGE-SCALE OPTIMIZATION PROBLEMS

what is the initial segment, it depends on G; if G is a box, then this segment is once for ever
fixed, so that there, basically, is nothing to worry about - one hardly might be interested insolving optimization problems within relative inaccuracy ≥ 1/2, and for smaller ε we know thecomplexity. But if G is a more general set than a box, then there is something to think about:
all we can say about an arbitrary n-dimensional G is that ε(G) ≥ 1/(2n3 ); this lower boundtends to 0 as the dimension n of the problem increases, so that for large n ∆(G) can in fact
”almost cover” the interval (0, 1) of all possible values of ε. On the other hand, when solving
large-scale problems of real world origin, we often are not interested in too high accuracy, andit may happen that the value of ε we actually are interested in is exactly in ∆(G), where we do
not know what is the complexity and what are the optimal methods. Thus, we have reasons,both of theoretical and practical origin, to be interested in the pre-asymptotic behaviour of thecomplexity.
The difficulty in investigating the behaviour of the complexity in the initial range of valuesof the accuracy is that this behaviour depends on affine properties of the domain G, and thisis something too diffuse for quantitative description. This is why it is reasonable to restrict
ourselves with certain ”standard” domains G. We already know what happens when G is aparallelotope, or, which is the same, a box - in this case there, basically, is no initial segment.
And, of course, the next interesting case is when G is an ellipsoid, or, which is the same, an
Euclidean ball (all our notions are affine invariant, so to speak about Euclidean balls is the sameas to speak about arbitrary ellipsoids). This is the case we shall focus on. In fact we shallassume G to be something like a ball rather than a ball exactly. Namely, let us fix a real α ≥ 1
and assume that the asphericity of G is at most α, i.e., there is a pair of concentric Euclidean
balls Vin and Vout with the ratio of radii not exceeding α and such that the smaller ball is insideG, and the larger one contains G:
Vin ⊂ G ⊂ Vout .

7.2

The main result

My main goal is to establish the following
Theorem 7.2.1 The complexity of the family Pm (G) of convex problems with m functional
constraints on an n-dimensional domain G of asphericity α satisfies the bounds
min{n, b

1
α2
−
0c}
≤
Compl(ε)≤
b4c + 1, 0 < ε < 1;(2αε)2ε2(7.2.1)

here ba − 0c is the largest integer which is smaller than a real a.
Comment. Before proving the theorem, let us think what the theorem says. First, it says thatthe complexity of convex minimization on a domain similar to an Euclidean ball is bounded fromabove uniformly in the dimension by a function O(1)α2 ε−2 ; the asphericity α is responsible forthe level of similarity between the domain and a ball. Second, we see that in the large-scalecase, when the dimension of the domain is large enough for given α and ε, or, which is the same,
when the inaccuracy ε is large enough for a given dimension (and asphericity), namely, when
ε≥

1
√ ,
2α n

(7.2.2)

7.3. UPPER COMPLEXITY BOUND: THE GRADIENT DESCENT

117

n lnn
n

Figure 7.1: Complexity of convex minimization over an n-dimensional Euclidean ball. The
whole range [1, ∞) of values of 1/ε can be partitioned into three segments:√• the initial segment [1, 2 n]; within this segment the complexity, up to an absolute constantfactor, is ε−2 ; at the right endpoint of the segment the complexity is equal to n; in this initial
segment the complexity is independent on the dimension and is in fact defined by the affine
geometry of G;
1• the final segment [ ε(G), ∞) = [2n, ∞); here the complexity, up to an absolute constant factor,
is n ln(1/ε), this is the standard asymptotics known to us; in this final segment the complexity
forgets everything about the geometry of G;√• the intermediate segment [2 n, 2n]; at the left endpoint of this segment the complexity isO(n), at the right endpoint it is O(n ln n); within this segment we know complexity up to afactor of order of its logarithm rather than up to an absolute constant factor.
then the complexity admits a lower bound O(1)α−2 ε−2 which differs from the aforementioned
upper bound by factor O(1)α−4 which depends on asphericity only. Thus, in the large-scale case
(7.2.2) our upper complexity bound coincides with the complexity up to a factor depending on
asphericity only; if G is an Euclidean ball (α = 1), then this factor does not exceed 16.
Now, our new complexity results combined with the initial results related to the case of small
inaccuracies gives us basically complete description of the complexity in the case when G is an
Euclidean ball. The behaviour of complexity in this case is as shown on Figure 7.1.
Now let me prove the theorem.

7.3

Upper complexity bound: the Gradient Descent

The upper complexity bound is associated with one of the most traditional methods for nonsmooth convex minimization - the short-step Subgradient Descent method. The generic schemeof the method is as follows:
given a starting point x1 ∈ int G and the required relative accuracy ε ∈ (0, 1), form thesequencexi+1 = xi − γi Rei ,(7.3.1)
where γi > 0 is certain stepsize, R is the radius of the Euclidean ball Vout covering G and theunit vector ei is defined as follows:

118

LECTURE 7. LARGE-SCALE OPTIMIZATION PROBLEMS(i) if xi 6∈ int G, then ei is an arbitrary unit vector which separates xi and G:(x − xi )T ei ≤ 0, x ∈ G;(ii) if xi ∈ int G, but there is a constraint gj which is ”ε-violated” at xi , i.e., is such that

gj (xi ) > ε max{gj (xi ) + (x − xi )T gj0 (xi )}
x∈G

then
ei =

,

(7.3.2)+

1
g 0 (xi );0|gj (xi )|2 j(iii) if xi ∈ int G and no constraint is ε-violated at xi , i.e., no inequality (7.3.2) is satisfied,then
1ei = 0f 0 (xi ).
|f (xi )|2
Note that the last formula makes sense only if f 0 (xi ) 6= 0; if in the case of (iii) we meet withf 0 (xi ) = 0, then we simply terminate and claim that xi is the result of our activity.
Same as in the cutting plane scheme, let us say that search point xi is productive, if at i-th
step we meet the case (iii), and non-productive otherwise, and let us define i-th approximate
solution x̄i as the best (with the smallest value of the objective) of the productive search points
generated in course of the first i iterations (if no productive search point is generated, x̄i isundefined).
The efficiency of the method is given by the following.
Proposition 7.3.1 Let a problem (p) from the family Pm (G) be solved by the short-step Subgradient Descent method associated with accuracy ε, and let N be a positive integer such that2 + 12PN

2
j=1 γjPN

j=1 γj<ε.
α(7.3.3)

Then either the method terminates in course of N steps with the result being an ε-solution to(p), or x̄N is well-defined and is an ε-solution to (p).
In particular, if
γi ≡ ε/α,then (7.3.3) is satisfied byα2N ≡ N (ε) = b4 2 c + 1,εand with the indicated choice of the stepsizes we can terminate the method after the N -th step; theresulting method solves any problem from the class within relative accuracy ε with the complexity
N (ε), which is exactly the upper complexity bound stated in Theorem 7.2.1.
Proof. Let me make the following crucial observation: let us associate with the method the
localizers
Gi = {x ∈ G | (x − xj )T ej ≤ 0, 1 ≤ j ≤ i}.
(7.3.4)
Then the presented method fits our generic cutting plane scheme for problems with functional
constraints, up to the fact that Gi now should not necessarily be solids (they may possess empty

7.3. UPPER COMPLEXITY BOUND: THE GRADIENT DESCENT

119

interior or even be themselves empty) and xi should not necessarily be an interior point of Gi−1 .
But all these particularities were not used in the proof of the general proposition on the rate ofconvergence of the scheme (Lecture 2, Proposition 2.3.1), and in fact there we have proved thefollowing:
Proposition 7.3.2 Assume that we are generating a sequence of search points xi ∈ Rn andassociate with these points vectors ei and approximate solutions x̄i in accordance to (i)-(iii). Let
the sets Gi be defined by the pairs (xi , ei ) according to (7.3.4), and let Size be a size. Assume
that in course of N steps we either terminate due to vanishing the subgradient of the objectiveat a productive search point, or this is not the case, butSize(GN ) < ε Size(G)(if GN is not a solid, then, by definition, Size(GN ) = 0). In the first case the result formed at
the termination is an ε-solution to the problem; in the second case such a solution is x̄N (which
is for sure well-defined).
Now let us apply this latter proposition to our short-step Subgradient Descent method and to
the size
InnerRad(Q) = max{r | Q contains an Euclidean ball of radius r}.
We know in advance that G contains an Euclidean ball Vin of the radius R/α, so that
InnerRad(G) ≥ R/α.

(7.3.5)

Now let us estimate from above the size of i-th localizer Gi , provided that the localizer is welldefined (i.e., that the method did not terminate in course of the first i steps due to vanishing
the subgradient of the objective at a productive search point). Assume that Gi contains an
Euclidean ball of certain radius r > 0, and let x+ be the center of the ball. Since V is containedin Gi , we have(x − xj )T ej ≤ 0, x ∈ V, 1 ≤ j ≤ i,
whence(x+ − xj )T ej + hT ej ≤ 0, |h|2 ≤ r, 1 ≤ j ≤ i,and since ej is a unit vector, we come to(x+ − xj )T ej ≤ −r, 1 ≤ j ≤ i.

(7.3.6)

Now let us write down the cosine theorem:|xj+1 − x+ |22 = |xj − x+ |22 + 2(xj − x+ )T (xj+1 − xj ) + |xj+1 − xj |22 == |xj − x+ |22 + 2γj R(x+ − xj )T ej + γj2 R2 ≤ |xj − x+ |22 − 2γj Rr + γj2 R2 .
We see that the squared distance from xj to x+ is decreased with j at least by the quantity
2γj R − γj2 R2 at each step; since the squared distance cannot become negative, we come to
R(2r

i
X
j=1

γj −

i
X
j=1

γj2 R) ≤ |x1 − x+ |22 ≤ 4R2

120

LECTURE 7. LARGE-SCALE OPTIMIZATION PROBLEMS(we have used the fact that G is contained in the Euclidean ball Vout of the radius R). Thus,
we come to the estimateP
2 + 12 ij=1 γj2r≤RPi
j=1 γjThis bound acts for the radius r of an arbitrary Euclidean ball contained in Gi , and we come toInnerRad(Gi ) ≤2 + 12Pi2
j=1 γjR.

(7.3.7)2 + 2 j=1 γj2InnerRad(Gi )α,≤
Pi
InnerRad(G)j=1 γj(7.3.8)Pi

j=1 γjCombining this inequality with (7.3.5), we come to
1 Pi

and due to the definition of N , we come toInnerRad(GN )< ε,InnerRad(G)Thus, the conclusion of the Theorem follows from Proposition 7.3.2.

7.4

The lower boundThe lower bound in (7.2.1) is given by a simple reasoning which is in fact already known to us.
Due to similarity reasons, we without loss of generality may assume that G is contained in the
Euclidean ball of the radius R = 21 centered at the origin and contains the ball of the radius1
r = 2αwith the same center. It, of course, suffices to establish the lower bound for the case ofproblems without functional constraints. Besides this, due to monotonicity of the complexity inε, it suffices to prove that if ε ∈ (0, 1) is such thatM =b1
− 0c ≤ n,(2αε)2then the complexity Compl(ε) is at least M . Assume that this is not the case, so that there
exists a method M which solves all problems from the family in question in no more than M − 1step. We may assume that M solves any problem exactly in M steps, and the result always is
the last search point. Let us set
1√ − ε,
δ=
2α M
so that δ > 0 by definition of M . Now consider the family F0 comprised of functions
f (x) = max (ξi xi + di )1≤i≤M

where ξi = ±1 and 0 < di < δ. Note that these functions are well-defined, since M ≤ n andtherefore we have enough coordinates in x.
Now consider the following M -step construction.

7.4. THE LOWER BOUND

121

The first step:
let x1 be the first search point generated by M; this point is instance-independent. Let i1 besign of the largest in absolute value of the coordinates of x1 , ξi∗1 be the index of the coordinate
and let d∗i1 = δ/2. Let F1 be comprised of all functions from F with ξi1 = ξi∗1 , di1 = d∗i1 anddi ≤ δ/4 for all i 6= i1 . It is clear that all the functions of the family F1 possess the same local
behaviour at x1 and are positive at this point.
The second step:
let x2 be the second search point generated by M as applied to a problem from the family F1 ;this point does not depend on the representative of the family, since all these representatives
have the same local behaviour at the first search point x1 . Let i2 be the index of the largest in
absolute value of the coordinates of x2 with indices different from i1 , let ξi∗2 be the sign of thecoordinate, and let d∗i2 = δ/4. Let F2 be comprised of all functions from F1 such that ξi2 = ξi∗2 ,di2 = d∗i2 and di ≤ δ/8 for all i different from i1 and i2 . Note that all functions from the family
coincide with each other in a neighbourhood of the two-point set {x1 , x2 } and are positive at
this set.
Now it is clear how to proceed. After k steps of the construction we have a family Fkcomprised of all functions from F with the parameters ξi and di being set to certain fixed values
for k values i1 , ..., ik of the index i and all di for the remaining i being ≤ 2−(k+1) δ; the family
satisfies the following predicate
Pk : the first k points x1 , ..., xk of the trajectory of M as applied to any function from thefamily do not depend on the function, and all the functions from the family coincide with eachother in certain neighbourhood of the k-point set {x1 , ..., xk } and are positive at this set.
From Pk it follows that the (k + 1)-th search point xk+1 generated by M as applied to a function
from the family Fk is independent of the function. At the step k + 1 wefind the index ik+1 of the largest in absolute value of the coordinates of xk+1 with indicesdifferent from i1 , ..., ik ,define ξi∗k+1 as the sign of the coordinate,
set d∗ik+1 = 2−(k+1) δ,anddefine Fk+1 as the set of those functions from Fk for which ξik+1 = ξi∗k+1 , dik+1 = d∗ik+1 anddi ≤ 2−(k+2) for i different from i1 , ..., ik+1 .
It is immediately seen that the resulting family satisfies the predicate Pk+1 , and we mayproceed in the same manner.
Now let us look what will be found after M step of the construction. We will end with afamily FM which consists of exactly one function
f = max (ξi∗ xi + d∗i )1≤i≤Msuch that f is positive along the sequence x1 , ..., xM of search points generated by M as applied
to the function. On the other hand, G contains the ball of the radius r = 1/(2α) centered at
the origin, and, consequently, contains the point
∗

x =−

M
X

ξi∗
√ ei ,
2α
M
i=1

122

LECTURE 7. LARGE-SCALE OPTIMIZATION PROBLEMS

ei being the basic orths in Rn . We clearly havef ∗ ≡ min f (x) ≤ f (x∗ ) < −
x∈G

1
√

2α M

+ δ ≤ −ε(the concluding inequality follows from the definition of δ). On the other hand, f clearly is
Lipschitz continuous with constant 1 on G, and G is contained in the Euclidean ball of theradius 1/2, so that the variation (maxG f − minG f ) of f over G is ≤ 1. Thus, we havef (xM ) − f ∗ > 0 − (−ε) = ε ≥ ε(max f − min f );GGsince, by construction, xM is the result obtained by M as applied to f , we conclude that M
does not solve the problem f within relative accuracy ε, which is the desired contradiction withthe origin of M .

7.5

Exercises: Around Subgradient Descent

The short-step version of the Subgradient Descent presented in Lecture 7 (I hope you are familiar
with the lecture) is quite appropriate for proving the upper complexity bound; as a computational scheme, it is not too attractive. The most unpleasant property of the scheme is that it
actually is a short-step procedure: one should from the very beginning tune it to the desired
accuracy ε, and the stepsizes γi = ε/α associated with the upper complexity bound stated in
Theorem 7.2.1 should be of order of ε. For the sake of simplicity, let us assume for a momentthat G is an Euclidean ball of radius R, so that α = 1. With the choice of stepsizes γi = ε/α = ε,the method will for sure be very slow, since to pass from the starting point x1 to a reasonableneighbourhood of the optimal solution it normally requires to cover a distance of order of thediameter of the ball G, i.e., of order of R, and it will take at least M = O(1/ε) steps (since asingle step moves the point by γi R = εR) even in the ideal case when all directions −ei lookdirectly to the optimal solution. The indicated observation does not contradict the theoreticaloptimality of the method in the large-scale case, where the worst-case complexity, as we know,
is at least is O(ε−2 ), which is much worse that the above M , namely, something like M 2 . Note,anyhow, that right now we were comparing the best possible worst-case behaviour of a method,
the complexity of the family, and the best possible ideal-case behaviour of the short-step Subgradient Descent. And there is nothing good in the fact that even in the ideal case the method
is slow.
There are, anyhow, more or less evident possibilities to make the method computationallymore reasonable. The idea is not to tune the method to the prescribed accuracy in advance,thus making the stepsizes small from the very beginning, but to start with ”large” stepsizes
and then decrease them at a reasonable rate. To implement the idea, we need an auxiliary tool
(which is important an interesting in its own right), namely, projections.
Let Q be a closed and nonempty convex subset in Rn . The projection πQ (x) of a pointx ∈ Rn onto Q is defined as the closest, with respect to the usual Euclidean norm, to x point of
Q, i.e., as the solution to the following optimization problem(Px ) :

minimize |x − y|22 over y ∈ Q.

7.5. EXERCISES: AROUND SUBGRADIENT DESCENT

123

Exercise 7.5.1 # Prove that πQ (x) does exist and is unique.
Exercise 7.5.2 # Prove that a point y ∈ Q is a solution to (Px ) if and only if the vector x − yis such that(u − y)T (x − y) ≤ 0 ∀u ∈ Q.
(7.5.1)
Derive from this observation the following important property:|πQ (x) − u|22 ≤ |x − u|22 − |x − πQ (x)|22 .

(7.5.2)Thus, when we project a point onto a convex set, the point becomes closer to any point u of theset, namely, the squared distance to u is decreased at least by the squared distance from x to Q.
Derive from (7.5.1) that the mappings x 7→ πQ (x) and x 7→ x−πQ (x) are Lipschitz continuouswith Lipschitz constant 1.
Now consider the following modification of the Subgradient Descent scheme:
given a solid Q which covers G, a starting point x1 ∈ int Q and a pair of sequences {γi > 0}
(stepsizes) and {εi ∈ (0, 1)} (tolerances), form the sequencexi+1 = πQ (xi − γi ρei ),(7.5.3)where 2ρ is the Euclidean diameter of Q and the unit vector ei is defined as follows:(i) if xi 6∈ int G, then ei is an arbitrary unit vector which separates xi and G:(x − xi )T ei ≤ 0, x ∈ G;(ii) if xi ∈ int G, but there is a constraint gj which is ”εi -violated” at xi , i.e., is such that

gj (xi ) > εi max{gj (xi ) + (x − xi )T gj0 (xi )}
x∈G

then
ei =

,

(7.5.4)+1
g 0 (xi );|gj0 (xi )|2 j(iii) if xi ∈ int G and no constraint is εi -violated at xi , i.e., no inequality (7.5.4) is satisfied,then1
f 0 (xi ).
ei = 0|f (xi )|2
If in the case of (iii) f 0 (xi ) = 0, then ei is an arbitrary unit vector.
After ei is chosen, loop.
The modification, as we see, is in the following:1) We add projection onto a covering G solid Q into the rule defining the updating xi 7→ xi+1
and use the half of the diameter of Q as the scale factor in the steplength (in the basic versionof the method, there was no projection, and the scale factor was the radius of the ball Vout );2) We use time-dependent tactics to distinguish between search points which ”almost satisfy”
the functional constraints and those which ”significantly violate” a constraint.
3) If we meet with a productive point xi with vanishing subgradient of the objective, wechoose as ei an arbitrary unit vector and continue the process. Note that in the initial versionof the method in the case in question we terminate and claim that xi is an ε-solution; now wealso could terminate and claim that xi is an εi -solution, which in fact is the case, but εi could
be large, not the accuracy we actually are interested in.

124

LECTURE 7. LARGE-SCALE OPTIMIZATION PROBLEMS

Exercise 7.5.3 # Prove the following modification of Proposition 7.3.1 (Lecture 7):
Let a problem (p) from the family Pm (G) be solved by the aforementioned Subgradient Descentmethod with nonincreasing sequence of tolerances {εi }. Assume that for a pair of positive integers
N > N 0 one has
P2
2 + 12 N
rin
j=N 0 γj0Γ(N : N ) ≡< εN ,(7.5.5)PNρ
j=N 0 γjwhere rin is the maximal of radii of Euclidean balls contained in G. Then among the search
points xN 0 , xN 0 +1 , ..., xN there were productive ones, and the best of them (i.e., that one with thesmallest value of the objective) point x̄N 0 ,N is an εN 0 -solution to (p).
Derive from this result that in the case of problems without functional constraints (m = 0),
where εi do not influence the process at all, the relationε∗ (N ) ≡min{ρΓ(M 0 : M )/rin } < 1

N ≥M ≥M 0 ≥1

(7.5.6)

implies that the best of the productive search points found in course of the first N steps is welldefined and is an ε∗ (N )-solution to (p).
Looking at the statement given by Exercise 7.5.3, we may ask ourselves what could be areasonable way to choose the stepsizes γi and the tolerances εi . Let us start with the case ofproblems without functional constraints, where we can forget about the tolerances - they do not
influence the process. What we are interested in is to minimize over stepsizes the quantitiesε∗ (N ). For a given pair of positive integers M ≥ M 0 the minimum of the quantity0Γ(N : N ) =2 + 21

PM

2
j=M 0 γjPM

j=M 0 γj220√over positive γj is attained when γj = √M −M; thus,0 +1 , M ≤ j ≤ M , and is equal to
M −M 0 +1

to minimize ε∗ (N ) for a given i, one should set γj = √2N , j = 1, ..., N , which would result inε∗ (N ) = 2N −1/2

ρ
.
rin

This is, basically, the choice of stepsizes we used in the short-step version of the Subgradient
Descent; an unpleasant property of this choice is that it is ”tied” to N , and we would like to
avoid necessity to fix in advance the number of steps allowed for the method. A natural idea is
to use the recommendation γj = 2N −1/2 in the ”sliding” way, i.e., to set
γj = 2j −1/2 , j = 1, 2, ...
Let us look what will be the quantities ε∗ (N ) for the stepsizes (7.5.7).
Exercise 7.5.4 # Prove that for the stepsizes (7.5.7) one hasε∗ (N ) ≤ Γ(]N/2[: N )ρρ≤ κN −1/2
rinrinwith certain absolute constant κ. Compute the constant.

(7.5.7)

7.5. EXERCISES: AROUND SUBGRADIENT DESCENT

125

We see that the stepsizes (7.5.7) result in optimal, up to an absolute constant factor, rate ofconvergence of the quantities ε∗ (N ) to 0 as N → ∞. Thus, when solving problems without
functional constraints, it is reasonable to use the aforementioned Subgradient Descent withstepsizes (7.5.7); according to the second statement of Exercise 7.5.3 and Exercise 7.5.4, for all
N such thatρε(N ) ≡ κN −1/2
<1rin
the best of the productive search points found in course of the first N steps is well-defined andsolves the problem within relative accuracy ε(N ).
Now let us look at problems with functional constraints. It is natural to use here the same
rule (7.5.7); the only question now is how to choose the tolerances εi . A reasonable policy would
be something likeρεi = min{0.9999, 1.01κi−1/2 },(7.5.8)rin
Exercise 7.5.5 # Prove that the Subgradient Descent with stepsizes (7.5.7) and tolerances(7.5.8), as applied to a problem (p) from the family Pm (G), possesses the following convergenceproperties: for all N such thatρ
< 0.99κN −1/2
rinamong the search points x]N/2[ , x]N/2[+1 , ..., xN there are productive ones, and the best (with thesmallest value of the objective) of these points solves (p) within relative inaccuracy not exceedingε]N/2[ ≤ χN −1/2

ρ
,
rin

χ being an absolute constant.
Note that if one chooses Q = Vout (i.e., ρ = R, so that ρ/rini = α is the asphericity of G),
then the indicated rate of convergence results in the same (up to an absolute constant factor)
as for the basic short-step Subgradient Descent complexity of solving problems from the family
within relative accuracy ε.

126

LECTURE 7. LARGE-SCALE OPTIMIZATION PROBLEMSLecture 8

Subgradient Descent and Bundle
methods
This lecture, same as the previous one, is devoted to large-scale nonsmooth convex problems.
For the sake of simplicity I restrict myself to problems without functional constraints:(f )minimize f (x) s.t. x ∈ G ⊂ Rn .

From now on we assume that G is a closed and bounded convex subset in Rn , possibly, withempty interior, and that the objective is convex and Lipschitz continuous on G:|f (x) − f (y)| ≤ L(f )|x − y|, x, y ∈ G,
where L(f ) < ∞ and | · | is the usual Euclidean norm in Rn . Note that the subgradient set of f
at any point from G is nonempty and contains subgradients of norms not exceeding L(f ); fromnow on we assume that the oracle in question reports such a subgradient at any input pointx ∈ G.
We would like to solve the problem within absolute inaccuracy ≤ ε, i.e., to find x ∈ G suchthatf (x) − f ∗ ≡ f (x) − min f ≤ ε.
G

8.1

Subgradient Descent method

The simplest way to solve the problem is to apply the standard Subgradient Descent method
which generates the sequence of search points {xi }∞i=1 according to the rulexi+1 = πG (xi − γi g(xi )), g(x) = f 0 (x)/|f 0 (x)|,(8.1.1)

where x1 ∈ G is certain starting point, γi > 0 are positive stepsizes andπG (x) = argmin{|x − y| | y ∈ G}
is the standard projector onto G. Of course, if we meet a point with f 0 (x) = 0, we terminate
with optimal solution at hands; from now on I ignore this trivial case.
127

128

LECTURE 8. SUBGRADIENT DESCENT AND BUNDLE METHODSAs always, i-th approximate solution x̄i found by the method is the best - with the smallest
value of f - of the search points x1 , ..., xi ; note that all these points belong to G.
It is easy to investigate the rage of convergence of the aforementioned routine. To this end
let x∗ be the closest to x1 optimal solution to the problem, and let
di = |xi − x∗ |.
We are going to see how di vary. To this end let us start with the following simple and important
observation:
Lemma 8.1.1 Let x ∈ Rn , and let G be a closed convex subset in Rn . Under projection onto
G, x becomes closer to any point u of G, namely, the squared distance from x to u decreases atleast by the squared distance from x to G:|πG (x) − u|2 ≤ |x − u|2 − |x − πG (x)|2 .

(8.1.2)

Proof is immediate. There is nothing to do if x ∈ G, since then πG (x) = x. Now, if x 6∈ G,
then from the optimality conditions for the problem|x − y| → min, y ∈ G
it follows that (u − πG (x))T (x − πG (x)) ≤ 0 (it is also evident geometrically). Thus, in thetriangle (u πG (x) x) the angle opposite to the side [ux] is acute, and therefore the square of thisside is ≥ the sum of squares of two other sides.
From Lemma 8.1.1 it follows thatd2i+1 ≡ |xi+1 − x∗ |2 = |πG (xi − γi g(xi )) − x∗ |2 ≤ |xi − γg(xi ) − x∗ |2 == |xi − x∗ |2 − 2γi (xi − x∗ )T f 0 (xi )/|f 0 (xi )| + γi2 ≤≤ d2i − 2γi (f (xi ) − f ∗ )/|f 0 (xi )| + γi2(the concluding inequality is due to the convexity of f ). Thus, we come to the recurrenced2i+1 ≤ d2i − 2γi (f (xi ) − f ∗ )/|f 0 (xi )| + γi2 ;(8.1.3)in view of the evident inequality
f (xi ) − f ∗ ≥ f (x̄i ) − f ∗ ≡ εi ,and since |f 0 (x)| ≤ L(f ), the recurrence implies thatd2i+1 ≤ d2i − 2γiεi+ γi2 .
L(f )(8.1.4)The latter inequality allows to make several immediate conclusions.
1) From (8.1.4) it follows that
2

N
X
i=1

γi εi ≤ L(f )

d21 +

N
X
i=1

!

γi2

;

8.1. SUBGRADIENT DESCENT METHOD

129

since εi clearly do not increase with i, we come to"#2
L(f ) |x1 − x∗ |2 + N
i=1 γiεN ≤ νN ≡.
PN
2
i=1 γiP(8.1.5)

The right hand side in this inequality clearly tends to 0 as N → ∞, provided that∞Xγi = ∞, γi → 0, i → ∞i=1(why?), which gives us certain general statement on convergence of the method as applied to a
Lipschitz continuous convex function; note that we did not use the fact that G is bounded.
Of course, we would like to choose the stepsizes resulting in the best possible estimate (8.1.5).
The simplest way to do it is to minimize the right hand side in (8.1.5) in γi for some given N .
This optimization results in
γi = γ ∗ ≡|x1 − x∗ ||x1 − x∗ |∗√, i = 1, ..., N ; νN ≡ νN
= L(f ) √;NN√
this is the 1/ N -rate of convergence already known to us. Of course, we cannot use the above
stepsizes literally, since they depend on the unknown initial distance |x1 − x∗ | to the optimal
set. What we could use are the stepsizes
γi = DN −1/2 , i = 1, ..., N,(8.1.6)

D being an a priori upper bound on the diameter of G; for these stepsizes we obtain from (8.1.5)
the following upper bound on the absolute accuracy of N -th approximate solution:
εN ≤ L(f )DN −1/2 .

(8.1.7)A disadvantage of stepsizes (8.1.6) is that we should tune them to certain chosen in advancenumber of steps N , and for large N enforce the stepsizes to be small from the very beginning.
A reasonable way to overcome this disadvantage is to use rule (8.1.6) in a ”sliding” mode, i.e.,
to setγi = Di−1/2 , i = 1, 2, ...
(8.1.8)For these stepsizes, (8.1.5) results in an estimate of the typeln N
εi ≤ νN = O(1)L(f )D √ ,N(8.1.9)which is by logarithmic factor worse than (8.1.8). This logarithmic factor comes from the factP
2
2
that the series N
i=1 γi in our case is of order D ln N . It turns out, anyhow, that the logarithmic
factor comes from bad reasoning, not from the fact that the stepsizes (8.1.7) actually are bad.
Indeed, note that our basic recurrence (8.1.3) implies that for any N ≥ M ≥ 1 one has
2εN

N
X
i=M

γi ≤ L(f )

d2M +

N
X
i=M

!

γi2

2

≤ L(f ) D +

N
X
i=m

!

γi2

,

130

LECTURE 8. SUBGRADIENT DESCENT AND BUNDLE METHODS

whence"#2
L(f ) D2 + N
i=M γiεN ≤;PM
2
i=N γiP

with M = bN/2c and γi = Di−1/2 the right hand side in the latter inequality does not exceed
O(1)DN −1/2 , and we come to the optimal, up to an absolute constant factor, estimate
L(f )DεN ≤ O(1) √, N = 1, 2, ...
N(8.1.10)(O(1) is an easily computable absolute constant). I call this rate optimal, since the lower
complexity bound we have obtained at the previous lecture it fact says that if G is an Euclideanball of diameter D in Rn and L is a given constant, then the complexity at which one canminimize over G, within absolute accuracy ε, an arbitrary Lipschitz continuous with constant
L convex function f is at least
(

 )
LD 2
min n; O(1),εso that in the large-scale case, when


n≥

LD
ε

2

,

the lower complexity bound coincides, within absolute constant factor, with the upper boundgiven by (8.1.10).
Thus, we can choose the stepsizes γi according to (8.1.7) and obtain dimension-independent
rate of convergence (8.1.10); this rate of convergence does not admit ”significant” uniform in
the dimension improvement, provided that G is an Euclidean ball.
2) The stepsizes (8.1.7) are theoretically optimal and more or less reasonable form the practical viewpoint, provided that you deal with a domain G of reasonable diameter, i.e., the diameterof the same order of magnitude as the distance from the starting point to the optimal set. If thelatter assumption is not satisfied (as it often is the case), the stepsizes should be chosen more
carefully. A reasonable idea here is as follows. Our rate-of-convergence proof in fact was basedon a very simple relation
d2i+1 ≤ d2i − 2γi (f (xi ) − f ∗ )/|f 0 (xi )| + γi2 ;let us choose as γi the quantity which results in the strongest possible inequality of this type,
namely, that one which minimizes the right hand side:γi =f (xi ) − f ∗.
|f 0 (xi )|(8.1.11)

Of course, this choice is possible only when we know the optimal value f ∗ . Sometimes this is
not a problem, e.g., when we reduce a system of convex inequalities
fi (x) ≤ 0, i = 1, ..., m,to the minimization off (x) = max fi (x);i8.1. SUBGRADIENT DESCENT METHOD

131

here we can take f ∗ = 0. In more complicated cases people use some on-line estimates of f ∗ ; I
would not like to go in details, so that I assume that f ∗ is known in advance. With the stepsizes(8.1.11) (proposed many years ago by B.T. Polyak) our recurrence becomes
d2i+1 ≤ d2i − (f (xi ) − f ∗ )2 |f 0 (xi )|−2 ≤ d2i − ε2i L−2 (f ),whencePN222
i=1 εi ≤ L (f )d1 , and we immediately come to the estimate

εN ≤ L(f )|x1 − x∗ |N −1/2 .

(8.1.12)This estimate seems the best one, since it involves the actual distance |x1 − x∗ | to the optimal
set rather than the diameter of G; in fact G might be even unbounded. I believe that whenever
one can use the Polyak stepsizes, this is the best possible tactics for the Subgradient Descentmethod.
Summary. Let me make certain summary. We see that the Subgradient Descent, which welast lecture were exploiting in order to obtain an optimal method for large scale convex minimization over Euclidean ball, can be applied to minimization of a convex Lipschitz continuousfunction over an arbitrary n-dimensional closed convex domain G; if G is bounded, then, underappropriate choice of stepsizes, one can ensure the inequalities
εN ≡ min f (xi ) − f ∗ ≤ O(1)L(f )D(G)N −1/2 ,1≤i≤N(8.1.13)

where O(1) is a moderate absolute constant, L(f ) is the Lipschitz constant of f and D(G) is thediameter of G. If the optimal value of the problem is known, then one can use stepsizes whichallow to replace D(G) by the distance |x1 − x∗ | from the starting point to the optimal set; inthis latter case, G should not necessarily be bounded. And the rate of convergence is optimal,
I mean, it cannot be improved by more than an absolute constant factor, provided that G is ann-dimensional Euclidean ball and n > N .
Note also that if G is a ”simple” set, say, an Euclidean ball, or a box, or the standard simplex{x ∈ Rn+ |nX

xi = 1},i=1then the method is computationally very cheap - a step costs only O(n) operations in addition
to those spent by the oracle. Theoretically all it looks perfect. It is not a problem to speak
about an upper accuracy bound O(N −1/2 ) and about optimality of this bound in the large scale
case; but in practice such a rate of convergence would result in tens of thousands of steps, whichis too much for the majority of applications. Note that in practice we are interested in ”typical”
behaviour of a method rather than in its worst case behaviour and worst case optimality. And
from this practical viewpoint the Subgradient Descent is far from being optimal: there areother methods with the same worst case theoretical complexity bound, but with significantly
better ”typical” performance; needless to say that these methods are more preferable in actual
computations. What we are about to do is to look at a certain family of methods of this latter
type.

132

8.2

LECTURE 8. SUBGRADIENT DESCENT AND BUNDLE METHODS

Bundle methods

Common sense says to us that the weak point in the Subgradient Descent is that when runningthe method, we almost loose previous information on the objective; the whole ”prehistory” iscompressed to the current iterate (this was not the case with the cutting plane methods, where
the prehistory was memorized in the current localizer). Generally speaking, what we actuallyknow about the objective after we have formed a sequence of search points xj ∈ G, j = 1, ..., i?
All we know is the bundle - the sequence of affine formsf (xj ) + (x − xj )T f 0 (xj )reported by the oracle; we know that every form from the sequence underestimates the objective
and coincides with it at the corresponding search point. All these affine forms can be assembled
into a single piecewise linear convex function - i-th model of the objectivefi (x) = max {f (xj ) + (x − xj )T f 0 (xj )}.
1≤j≤iThis model underestimates the objective:fi (x) ≤ f (x), x ∈ G,

(8.2.1)fi (xj ) = f (xj ), j = 1, ..., i.

(8.2.2)and is exact at the points x1 , ..., xi :And once again - the model accumulates all our knowledge obtained so far; e.g., the information
we possess does not contradict the hypothesis that the model is exact everywhere. Since themodel accumulates the whole prehistory, it is reasonable to formulate the search rules for amethod in terms of the model. The most natural and optimistic idea is to trust in the model
completely and to take, as the next search point, the minimizer of the model:
xi+1 ∈ Argmin fi .
GThis is the Kelley cutting plane method - the very first method proposed for nonsmooth convex
optimization. The idea is very simple - if we are lucky and the model is good everywhere, notonly along the previous search points, we would improve significantly the best found so far valueof the objective; and if the model is bad, then it will be corrected at the ”right” place. Fromcompactness of G one can immediately derive that the method does converge and is even finite if
the objective is piecewise linear. Unfortunately, it turns out that the rate of convergence of themethod is a disaster; one can demonstrate that the worst-case number of steps required by the
Kelley method to solve a problem f within absolute inaccuracy ε (G is the unit n-dimensionalball, L(f ) = 1) is at least (n−1)/2
1
O(1)
.
ε
We see how dangerous is to be too optimistic, and it is clear why: even in the case of smoothobjective the model is close to the objective only in a neighbourhood of the search points; until
the number of these points becomes very-very large, this neighbourhood forms a ”negligible”

8.2. BUNDLE METHODS

133

part of the domain G, so that the global characteristic of the model - its minimizer - is very
unstable and until the termination phase has small in common with the actual optimal set. Letme say that the Kelley method in practice is much better than one could think looking at its
worst-case complexity (a method with practical behaviour like this estimate simply could not
be used even in the dimension 10), but the qualitative conclusions from the estimate are moreor less valid also in practice - the Kelley method sometimes is too slow.
A natural way to improve the Kelley method is as follows. All we may hope on is thatthe model approximates the objective in a neighbourhood of the search points. Therefore it
is reasonable to enforce the next search point to be not too far from the previous ones, moreexactly, from the ”most perspective”, the best of them, since the latter, as the method goes on,
hopefully will become close to the optimal set. To forbid the new iterate to move far away, let
us choose xi+1 as the minimizer of the penalized model:
xi+1 = argmin{fi (x) +Gdi2|x − x+i | },
2

where x+i is what is called the current prox center, and the prox coefficient di > 0 is certain
parameter. When di is large, we enforce xi+1 to be close to the prox center, and when it is small,
we act almost as in the Kelley method. What is displayed, is the generic form of the bundle
methods; to specify a method from this family, one need to indicate the policies of updating theprox centers and the prox coefficients. There is a number of reasonable policies of this type, andamong these policies there are those resulting in methods with very good practical performance.
I would not like to go in details here; let me say only that, first, the best theoretical complexity
estimate for the traditional bundle methods is something like O(ε−3 ); although non-optimal, thisupper bound is incomparably better than the lower complexity bound for the method of Kelley.
Second, there is more or less unique reasonable policy of updating the prox center, in contrast to
the policy for updating the prox coefficient. Practical performance of a bundle algorithm heavily
depends on this latter policy, and sensitivity to the prox coefficient is, in a sense, the weak pointof the bundle methods. Indeed, even without addressing to computational experience we canguess in advance that the scheme should be sensitive to di - since in the limiting cases of zeroand infinite prox coefficient we get, respectively, the Kelley method, which can be slow, and the”method” which simply does not move from the initial point. Thus, both small and large prox
coefficients are forbidden; and it is unclear how to choose the ”golden middle” - our information
has nothing in common with any quadratic terms in the model, these terms are invented by us.

8.2.1

The Level method

What I am going to do is to present to you the Level method, a new method from the bundle
family; this method, in a sense, is free from the aforementioned shortcomings of the traditional
bundle scheme. Namely, the method possesses the optimal complexity bound O(ε−2 ), and the
difficulty with tuning the prox coefficient in the method is resolved in a funny way - this problem
does not occur at all.
To describe the method, let me introduce several simple quantities. Given i-th model fi (·),
we can compute its optimum, same as in the Kelley method; but now we are interested not in
the point where the optimum is attained, but in the optimal value
fi− = min fi
G134

LECTURE 8. SUBGRADIENT DESCENT AND BUNDLE METHODSof the model. Since the model underestimates the objective, the quantity fi− is a lower boundfor the actual optimal value; and since the models clearly increase with i at every point, theirminima also increase, so thatf1− ≤ f2− ≤ ... ≤ f ∗ .
(8.2.3)On the other hand, let fi+ be the best found so far value of the objective:fi+ = min f (xj ) = f (x̄i ),1≤j≤i(8.2.4)

where x̄i is the best (with the smallest value of the objective) of the search point generated sofar. The quantities fi+ clearly decrease with i and overestimate the actual optimal value:f1+ ≥ f2+ ≥ ... ≥ f ∗ .

(8.2.5)

It follows that the gaps
∆i = fi+ − fi−are nonnegative and nonincreasing and bound from above the inaccuracy of the best found so
far approximate solutions:f (x̄i ) − f ∗ ≤ ∆i , ∆1 ≥ ∆2 ≥ ... ≥ 0.

(8.2.6)

Now let me describe the method. Its i-th step is as follows:
1) solve the piecewise linear problem
minimize fi (x) s.t. x ∈ G
to get the minimum fi− of the i-th model;
2) form the levelli = (1 − λ)fi− + λfi+ ≡ fi− + λ∆i ,λ ∈ (0, 1) being the parameter of the method (normally, λ = 1/2), and define the new iterate
xi+1 as the projection of the previous one xi onto the level set
Qi = {x ∈ G | fi (x) ≤ li }of the i-th model, the level set being associated with li :xi+1 = πQi (xi ).

(8.2.7)

Computationally, the method requires solving two auxiliary problems at each iteration. The
first is to minimize the model in order to compute fi− ; this problem arises in the Kelley methodand does not arise in the bundle ones. The second auxiliary problem is to project xi onto Qi ;this is, basically, the same quadratic problem which arises in bundle methods and does not arisein the Kelley one. If G is a polytope, which normally is the case, the first of these auxiliary
problems is a linear program, and the second is a convex linearly constrained quadratic program;to solve them, one can use the traditional efficient simplex-type technique.

8.2. BUNDLE METHODS

135

Let me note that the method actually belongs to the bundle family, and that for this method
the prox center always is the last iterate. To see this, let us look at the solutiondx(d) = argmin{fi (x) + |x − xi |2 }2
Gof the auxiliary problem arising in the bundle scheme as at a function of the prox coefficient d.
It is clear that x(d) is the closest to x point in the set {x ∈ G | fi (x) ≤ fi (x(d))}, so that x(d)is the projection of xi onto the level set{x ∈ G | fi (x) ≤ li (d)}of the i-th model associated with the level li (d) = fi (x(d)) (this latter relation gives us certain
equation which relates d and li (d)). As d varies from 0 to ∞, x(d) moves along certain path
which starts at the closest to xi point in the optimal set of the i-th model and ends at the prox
center xi ; consequently, the level li (d) varies from fi− to f (xi ) ≥ fi+ and therefore, for certain
value di of the prox coefficient, we have li (di ) = li and, consequently, x(di ) = xi+1 . Note thatthe only goal of this reasoning was to demonstrate that the Level method does belong to thebundle scheme and corresponds to certain implicit control of the prox coefficient; this control
exists, but is completely uninteresting for us, since the method does not require knowledge ofdi .
Now let me formulate and prove the main result on the method.
Theorem 8.2.1 Let the Level method be applied to a convex problem (f ) with Lipschitz continuous, with constant L(f ), objective f and with a closed and bounded convex domain G ofdiameter D(G). Then the gaps ∆i converge to 0; namely, for any positive ε one has
L(f )D(G)i > c(λ)ε


wherec(λ) =

2⇒ ∆i ≤ ε,(8.2.8)1.
(1 − λ)2 λ(2 − λ)In particular,L(f )D(G)i > c(λ)ε
2⇒ f (x̄i ) − f ∗ ≤ ε.

Proof. The ”in particular” part is an immediate consequence of (8.2.8) and (8.2.6), so that all
we need is to verify (8.2.8). To this end assume that N is such that ∆N > ε, and let us boundN from above.
10 . Let us partition the set I = {1, 2, ..., N } of iteration indices in groups I1 , ..., Ik as follows.
The first group ends with the index i(1) ≡ N and contains all indices i ≤ i(1) such that∆i ≤ (1 − λ)−1 ∆N ≡ (1 − λ)−1 ∆i(1) ;since, as we know, the gaps never increase, I1 is certain final segment of I. If it differs from thewhole I, we define i(2) as the largest of those i ∈ I which are not in I1 , and define I2 as the setof all indices i ≤ i(2) such that∆i ≤ (1 − λ)−1 ∆i(2) .

136

LECTURE 8. SUBGRADIENT DESCENT AND BUNDLE METHODS

I2 is certain preceding I1 segment in I. If the union of I1 and I2 is less than I, we define i(3) as
the largest of those indices in I which are not in I1 ∪ I2 , and define I3 as the set of those indicesi ≤ i(3) for which∆i ≤ (1 − λ)−1 ∆i(3) ,and so on.
With this process, we partition the set I of iteration indices into sequential segments I1 ,..,Ik(Is follows Is+1 in I). The last index in Is is i(s), and we have∆i(s+1) > (1 − λ)−1 ∆i(s) , s = 1, ..., k − 1(8.2.9)(indeed, if the opposite inequality would hold, then i(s + 1) would be included into the group
Is , which is not the case).
20 . My main (and very simple) observation is as follows:
the level sets Qi of the models corresponding to certain group of iterations Is have a pointin common, namely, the minimizer, us , of the last, i(s)-th, model from the group.
Indeed, since the models increase with i, and the best found so far values of the objective
decrease with i, for all i ∈ Is one has+−− ∆i(s) ≤ fi+ − ∆i(s) ≤ fi+ − (1 − λ)∆i ≡ li
= fi(s)fi (us ) ≤ fi(s) (us ) = fi(s)(the concluding ≤ in the chain follows from the fact that i ∈ Is , so that ∆i ≤ (1 − λ)−1 ∆i(s) ).
30 . The above observation allows to estimate from above the number Ns of iterations in thegroup Is . Indeed, since xi+1 is the projection of xi onto Qi and us ∈ Qi for i ∈ Is , we conclude
from Lemma 8.1.1 that|xi+1 − us |2 ≤ |xi − us |2 − |xi − xi+1 |2 , i ∈ Is ,whence, denoting by j(s) the first element in Is ,X

|xi − xi+1 |2 ≤ |xj(s) − us |2 ≤ D2 (G).

(8.2.10)i∈Is

Now let us estimate from below the steplengths |xi − xi+1 |. At the point xi the i-th model fi
equals to f (xi ) and is therefore ≥ fi+ , and at the point xi+1 the i-th model is, by construction of
xi+1 , less or equal (in fact is equal) to li = fi+ − (1 − λ)∆i . Thus, when passing from xi to xi+1 ,the i-th model varies at least by the quantity (1 − λ)∆i , which is, in turn, at least (1 − λ)∆i(s)(the gaps may decrease only!). On the other hand, fi clearly is Lipschitz continuous with thesame constant L(f ) as the objective (recall that, according to our assumption, the oracle reports
subgradients of f of the norms not exceeding L(f )). Thus, at the segment [xi , xi+1 ] the Lipschitz
continuous with constant L(f ) function fi varies at least by (1 − λ)∆i(s) , whence|xi − xi+1 | ≥ (1 − λ)∆i(s) L−1 (f ).
From this inequality and (8.2.10) we conclude that the number Ns of iterations in the group Is
satisfies the estimate
Ns ≤ (1 − λ)−2 L2 (f )D2 (G)∆−2i(s) .

8.2. BUNDLE METHODS

137

40 . We have ∆i(1) > ε (the origin of N ) and ∆i(s) > (1 − λ)−s+1 ∆i(1) (see (8.2.9)), so thatthe above estimate of Ns results in
Ns ≤ (1 − λ)−2 L2 (f )D2 (G)(1 − λ)2(s−1) ε−2 ,whenceN=X

Ns ≤ c(λ)L2 (f )D2 (G)ε−2 ,s

as claimed.

8.2.2

Concluding remarks

The theorem we have proved says that the level method is optimal in complexity, provided
that G is an Euclidean ball of a large enough dimension. In fact there exists ”computational
evidence”, based on many numerical tests, that the method is also optimal in complexity in a
fixed dimension. Namely, as applied to a problem of minimization of a convex function over ann-dimensional solid G, the method finds an approximate solution of absolute accuracy ε in no
more thanV (f )
c(f ) n lnε

iterations, where V (f ) = maxG f − minG f is the variation of the objective over G and c(f ) is
certain problem-dependent constant which never is greater than 1 and typically is somethingaround 0.2 - 0.5. Let me stress that this is an experimental fact, not a theorem; I have strong
doubts that there exists a theorem of this type, but empirically this ”law” is supported byhundreds of tests.
To illustrate this point, let me present numerical results related to one of the standard
test problems called MAXQUAD. This is a small, although difficult, problem of maximizing
the maximum of 5 convex quadratic forms of 10 variables. In the below table you see theresults obtained on this problem by the Subgradient Descent and by the Level methods. In the
Subgradient Descent the Polyak stepsizes were used (to this end, the method was equipped withthe exact optimal value, so that the experiment was in favour of the Subgradient Descent). The
results are as follows:
Subgradient Descent: 100,000 steps, best found value -0.8413414 (absolute inaccuracy 0.0007),
running time 5400 ;Level: 103 steps, best found value -0.8414077 (absolute inaccuracy ¡0.0000001), running time
200 , ”complexity index” c(f ) = 0.47.
Runs:
Subgradient Descent

Level

138

LECTURE 8. SUBGRADIENT DESCENT AND BUNDLE METHODS
i
1
2

fi+
5337.066429
98.595071

8316.295622
0.198568

41

201
4001
5001
17001
25001
50001
75001
100000ifi+
1 5337.066429
6
1698.586465
7.27811539−0.674044

54
73
81
103

−0.811759−0.841058−0.841232−0.841408−0.221810−0.801369∗ − 0.839771−0.840100−0.841021−0.841144−0.841276−0.841319−0.481341∗ marks the result obtained by the Subgradient Descent after 200 (the total CPU time of theLevel method); to that moment the Subgradient Descent has performed 4,000 iterations, but
has restored the solution within 2 accuracy digits rather then 6 digits given by Level.
The Subgradient Descent with the ”default” stepsizes γi = O(1)i−1/2 in the same 100,000
iterations was unable to achieve value of the objective less than -0.837594, i.e., found the solution
within a single accuracy digit.

8.3

Exercises: Mirror Descent

Looking at the 3-line convergence proof for the standard Subgradient Descent:xi+1 = πG (xi − γi f 0 (xi )/|f 0 (xi )|) ⇒ |xi+1 − x∗ |2 ≤ |xi − x∗ |2 − 2γi (xi − x∗ )T f 0 (xi )/|f 0 (xi )| + γi2⇒ |xi+1 − x∗ |2 ≤ |xi − x∗ |2 − 2γi (f (xi ) − f ∗ )/|f 0 (xi )| + γi22|x1 − x∗ |2 + N
i=1 γi⇒ min f (xi ) − f ≤PNi≤N
2 i=1 γi∗P

one should be surprised. Indeed, all of us know the origin of the gradient descent: if f is smooth,a step in the antigradient direction decreases the first-order expansion of f and therefore, fora reasonably chosen stepsize, increases f itself. Note that this standard reasoning has nothingin common with the above one: we deal with a nonsmooth f , and it should not decreasein the direction of an anti-subgradient independently of how small is the stepsize; there is a
subgradient in the subgradient set which actually possesses the desired property, but this is notnecessarily the subgradient used in the method, and even with the ”good” subgradient you could
say nothing about the amount the objective can be decreased by. The ”correct” reasoning deals

8.3. EXERCISES: MIRROR DESCENT

139

with algebraic structure of the Euclidean norm rather than with local behaviour of the objective,
which is very surprising; it is a kind of miracle. But we are interested in understanding, not in
miracles. Let us try to understand what is behind the phenomenon we have met.
First of all, what is a subgradient? Is it actually a vector? The answer, of course, is ”no”.
Given a convex function f defined on an n-dimensional vector space E and an interior point xof the domain of f , you can define a nonempty set of support functionals - linear forms f 0 (x)[h]of h ∈ E which are support to f at x, i.e., such thatf (y) ≥ f (x) + f 0 (x)[y − x], y ∈ Dom f ;these forms are intrinsically associated with f and x. Now, having chosen somehow an Euclidean
structure (·, ·) on E, you may associate with linear forms f 0 (x)[h] vectors f 0 (x) from E in sucha way thatf 0 (x)[h] = (f 0 (x), h), h ∈ Rn ,thus coming from support functionals to subgradients-vectors. The crucial point is that these
vectors are not defined by f and x only; they also depend on what is the Euclidean structure
on E we use. Of course, normally we think of an n-dimensional space as of the coordinate space
Rn with once for ever fixed Euclidean structure, but this habit sometimes is dangerous; theproblems we are interested in are defined in affine terms, not in the metric ones, so why should
we always look at the problems via certain once for ever fixed Euclidean structure which has
nothing in common with the problem? Developing systematically this evident observation, one
may come to the most advanced and recent convex optimization methods like the polynomial
time interior point ones. Our now goal is much more modest, but we also shall get profit fromthe aforementioned observation. Thus, once more: the ”correct” objects associated with f andx are not vectors from E, but elements of the dual to E space E ∗ of linear forms on E. Of
course, E ∗ is of the same dimension as E and therefore it can be identified with E; but there
are many ways to identify these spaces, and no one of them is ”natural”, more preferable thanothers.
Since the support functionals f 0 (x)[h] ”live” in the dual space, the Gradient Descent cannot
avoid the necessity to identify somehow the initial - primal - and the dual space, and this isdone via the Euclidean structure the method is related to - as it was already explained, thisis what allows to associate with a support functional - something which ”actually exists”, but
belongs to the dual space - a subgradient, a vector belonging to the primal space; in a sense,
this vector is a phantom - it depends on the Euclidean structure on E. Now, is a Euclidean
structure the only way to identify the dual and the primal spaces? Of course, no, there are manyother ways. What we are about to do is to consider certain family of ”identifications” of E andE ∗ which includes, as particular cases, all identifications given by Euclidean structures. This
family is as follows. Let V (φ) be a smooth (say, continuously differentiable) convex function on
E ∗ ; its support functional V 0 (φ)[·] at a point φ is a linear functional on the dual space. Due tothe well-known fact of Linear Algebra, every linear form L[η] on the dual space is defined by avector l from the primal space, in the sense that
L[η] = η[l]for all η ∈ E ∗ ; this vector is uniquely defined, and the mapping L 7→ l is a linear isomorphism
between the space (E ∗ )∗ dual to E ∗ and the primal space E. This isomorphism is ”canonical” -140

LECTURE 8. SUBGRADIENT DESCENT AND BUNDLE METHODS

it does not depend on any additional structures on the spaces in question, like inner products,
coordinates, etc., it is given by intrinsic nature of the spaces. In other words, in the quantity
φ[x] we may think of φ being fixed and x varying over E, which gives us a linear form on E
(this is the origin of the quantity); but we also can think of x as being fixed and φ varying overE ∗ , which gives us a linear form on E ∗ ; and Linear Algebra says to us that every linear form on
E ∗ can be obtained in this manner from certain uniquely defined x ∈ E. Bearing in mind thissymmetry, let us from now on rename the quantity φ[x] as hφ, xi, thus using more ”symmetric”
notation.
Thus, every linear form on E ∗ corresponds to certain x ∈ E; in particular, the linear form0
V (φ)[·] on E ∗ corresponds to certain vector V 0 (φ) ∈ E:
V 0 (φ)[η] ≡ η, V 0 (φ) , η ∈ E ∗ .
We come to certain mapping
φ 7→ V 0 (φ) : E ∗ → E;
this mapping, under some mild assumptions, is a continuous one-to-one mapping with continuous
inverse, i.e., is certain identification (not necessarily linear) of the dual and the primal spaces.
Exercise 8.3.1 Let (·, ·) be certain Euclidean structure on the dual space, and let
1
V (φ) = (φ, φ).
2
The aforementioned construction associates with V the mapping φ 7→ V 0 (φ) : E → E ∗ ; on theother hand, the Euclidean structure in question itself defines certain identification of the dual
and the primal space, the identification I given by the identity
φ[x] = (φ, I −1 x), φ ∈ E ∗ , x ∈ E.
Prove that I = V 0 .
Assume that {ei }ni=1 is a basis in E and {e∗i } is the biorthonormal basis in E ∗ (so that∗ei [ej ] = δij ), and let A be the matrix which represents the inner product in the coordinates ofE ∗ related to the basis {e∗i }, i.e., Aij = (e∗i , e∗j ). What is the matrix of the associated mapping
V 0 taken with respect to the {e∗i }-coordinates in E ∗ and the {ei }-coordinates in E?
We see that all ”standard” identifications of the primal and the dual spaces, i.e., those given
by Euclidean structures, are covered by our mappings φ 7→ V 0 (φ); the corresponding V ’s are,up to the factor 1/2, squared Euclidean norms. A natural question is what are the mappings
associated with other squared norms.
Exercise 8.3.2 Let k · k be a norm on E, letk φ k∗ = max{φ[x] | x ∈ E, k x k≤ 1}
be the conjugate norm on and assume that the function
Vk·k∗ (φ) =1k φ k2∗2

8.3. EXERCISES: MIRROR DESCENT

141

is continuously differentiable. Then the mapping φ 7→ V 0 (φ) is as follows: you take a linear
form φ 6= 0 on E, maximize it on the k · k-unit ball of E; the maximizer is unique and is exactly
V 0 (φ); and, of course, V 0 (0) = 0. In other words, V 0 (φ) is nothing but the ”direction of thefastest growth” of the functional φ, where, of course, the rate of growth in a direction is definedas the ”progress in φ per k · k-unit step in the direction”.
Prove that the mapping φ 7→ φ0 is a continuous mapping form E onto E 0 . Prove that thismapping is a continuous one-to-one correspondence between E and E 0 if and only if the function
Wk·k (x) =1k x k2 : E → R
2is continuously differentiable, and in this case the mapping0x 7→ Wk·k(x)
0 .
is nothing but the inverse to the mapping given by Vk·k∗Now, every Euclidean norm k · k on E induces, as we know, a Subgradient Descent method
for minimization of convex functions over closed convex domains in E. Let us write down thismethod in terms of the corresponding function Vk·k∗ . For the sake of simplicity let us ignore
for the moment the projector πG , thus looking at the method for minimizing over the whole E.
The method, as it is easily seen, would become
φi+1 = φi − γi f 0 (xi )/ k f 0 (xi ) k∗ , xi = V 0 (φi ), V ≡ Vk·k∗ .

(8.3.1)

Now, in the presented form of the Subgradient Descent there is nothing from the fact that k · k isa Euclidean norm; the only property of the norm which we actually need is the differentiability of
the associated function V . Thus, given a norm k · k on E which induces a differentiable outside
0 conjugate norm on the conjugate space, we can write down certain method for minimizing
convex functions over E. How could we analyse the convergence properties of the method? In
the case of the usual Subgradient Descent the proof of convergence was based on the fact thatthe anti-gradient direction f 0 (x) is a descent direction for certain Lyapunov function, namely,|x − x∗ |2 , x∗ being a minimizer of f . In fact our reasoning was as follows: since f is convex, wehave
f 0 (x), x∗ − x ≤ f (x) − f (x∗ ) ≤ 0,
(8.3.2)and the quantity hf 0 (x), x − x∗ i is, up to the constant factor 2, the derivative of the function|x − x∗ |2 in the direction f 0 (x). Could we say something similar in the general case, where,according to (8.3.1), we should deal with the situation x = V 0 (φ)? With this substitution, theleft hand side of (8.3.2) becomes
f 0 (x), x∗ − V 0 (φ) =d|t=0 V + (φ − tf 0 (x)), V + (ψ) = V (ψ) − hψ, x∗ i .
dt

Thus, we can associate with (8.3.1) the function+
V + (φ) ≡ Vk·k(φ) = Vk·k∗ (φ) − hφ, x∗ i ,∗(8.3.3)

142

LECTURE 8. SUBGRADIENT DESCENT AND BUNDLE METHODSx∗ being a minimizer of f , and the derivative of this function in the direction −f 0 (V 0 (φ)) of the
trajectory (8.3.1) is nonpositive:−f 0 (V 0 (φ)), (V + )0 (φ) ≤ f (x∗ ) − f (x) ≤ 0, φ ∈ E ∗ .

(8.3.4)

Now we may try to reproduce the reasoning which leads to the rate-of-convergence estimate for
the Subgradient Descent for our now situation, where we speak about process (8.3.1) associatedwith an arbitrary norm on E (the norm should result, of course, in a continuously differentiable
V ).
For the sake of simplicity, let us restrict ourselves with the simple case when V possesses
a Lipschitz continuous derivative. Thus, from now on let k · k be a norm on E such that the
mapping0
V(φ) ≡ Vk·k(φ) : E ∗ → E∗
is Lipschitz continuous, and let
L ≡ Lk·k = sup{k V(φ0 ) − V(φ00 ) k| φ0 6= φ00 , φ0 , φ00 ∈ E ∗ }.
k φ0 − φ00 k∗

For the sake of brevity, from now on we write V instead of Vk·k∗ .
Exercise 8.3.3 Prove that
V (φ + η) ≤ V (φ) + hη, V(φ)i + LV (η), φ, η ∈ E ∗ .

(8.3.5)

Now let us investigate process (8.3.1).
Exercise 8.3.4 ∗ Let f : E → R be a Lipschitz continuous convex function which attains its
minimum on E at certain point x∗ . Consider process (cf. (8.3.1))
φi+1 = φi − γi f 0 (xi )/|f 0 (xi )|∗ , xi = V 0 (φi ), φ1 = 0,

(8.3.6)and let x̄i be the best (with the smallest value of f ) of the points x1 , ..., xi and let εi = f (x̄i ) −minE f . Prove that then2|x∗ |2 + L N
i=1 γiεN ≤ Lk·k (f ), N = 1, 2, ...
PN
2 i=1 γiP

(8.3.7)

where Lk·k (f ) is the Lipschitz constant of f with respect to the norm k · k. In particular, themethod converges, provided thatXγi = ∞, γi → 0, i → ∞.

iNote that the basic result explains what is the origin of the ”Subgradient Descent miracle”
which motivated our considerations; as we see, this miracle comes not from the very specificalgebraic structure of the Euclidean norm, but from certain ”robust” analytic property of thenorm (the Lipschitz continuity of the derivative of the conjugate norm), and we can fabricate
similar miracles for arbitrary norms which share the indicated property. In fact you could use theoutlined Mirror Descent scheme, with necessary (and more or less straightforward) modifications,in order to extend everything what we know about the usual - ”Euclidean” - Subgradient Descent(I mean, the versions for optimization over a domain rather than over the whole space and for
optimization over solids under functional constraints) onto the general ”non-Euclidean” case,but we skip here these issues.

Lecture 9

Large-scale games and variational
inequalities
In this lecture we extend convex minimization methods with dimension-independent complexity
on saddle point problems and variational inequalities with monotone operators.
In what follows we will be interested in finding a weak solution to a variational inequality
with monotone operator, i.e., in solving a problem as follows:
find z ∗ ∈ G such that ξ T (w − z ∗ ) ≥ 0 for all w ∈ G, ξ ∈ F (w),(9.0.1)where
G ⊂ Rn is closed and bounded convex set;
w 7→ F (w) is a multi-valued monotone operator defined on G. Recall that it means that for
any w ∈ G F (w) is a nonempty subset of Rn and(ξ − ξ 0 )T (w − w0 ) ≥ 0 ∀w, w0 ∈ G ∀ξ ∈ F (w), ξ 0 ∈ F (w0 ).
From now on we assume that F is bounded:L(F ) = sup{|ξ| | ξ ∈ F (w), w ∈ G} < ∞;here and further | · | is the standard Euclidean norm on Rn . Thus, compared to our firstinvestigation of variational inequalities in Lecture 6, now we do not require the domain G of
the inequality to possess a nonempty interior; as a compensation, we assume that the operator
is well-defined on the whole domain, not on the interior of the domain only, and assume theoperator to be bounded rather than semibounded.
Our goal is to solve the inequality within prescribed absolute accuracy ε, i.e., to find z ∈ Gsuch that
ν(z) ≡ sup{ξ T (z − w) | w ∈ G, ξ ∈ F (w)} ≤ ε.
Note that ν(z) is exactly the inaccuracy measure we dealt with in Lecture 6; the only differenceis that there we were looking for a solution of a given relative accuracy ε, i.e., such that
ν(z) ≤ ε sup ν.
G

143

144

LECTURE 9. LARGE-SCALE GAMES AND VARIATIONAL INEQUALITIESRecall that we were especially interested in a particular class of variational inequalities withmonotone operators, i.e., those coming from saddle point problems. Namely, let G = P × Q
be a direct product of two closed and bounded convex subsets, and let f (x, y) be a Lipschitz
continuous function on G which is convex in x ∈ P and concave in y ∈ Q; in Lecture 5 we haveassociated with this function the saddle point problem
find z ∗ = (x∗ , y ∗ ) ∈ G s.t. f (x, y ∗ ) ≥ f (x∗ , y ∗ ) ≥ f (x∗ , y), (x, y) ∈ G;

(9.0.2)

we have seen that the solutions to the saddle point problem are exactly the pairs of optimal
solutions to the following two convex programs:
minimize f¯(x) = max f (x, y) over x ∈ P,y∈Q

maximize f (x) = min f (x, y) over y ∈ Q,x∈P

the optimal values of the problems being equal to each other. In the case of games we used a
specific accuracy measureνg (x, y) = f¯(x) − f (y) = {f¯(x) − min f¯} + {max f − f (y)},PQthe sum of inaccuracies of x and y regarded as approximate solutions to the problems of minimizing f¯ and of maximizing f . Recall that we have associated with the saddle point problem
in question the variational inequality with monotone operator
Ff (z) = {(fx0 , −fy0 ) | fx0 ∈ ∂x f (z), fy0 ∈ ∂y f (z)};here the sub- and supergradient sets of f are comprised of sub- and supergradients of norms
not exceeding the corresponding Lipschitz constants of f , so that the operator Ff is bounded.
From Lecture 6 (Proposition 6.1.2) we know that this operator is monotone and that the weaksolutions to the variational inequality associated with the operator are exactly the saddle pointsof f .
My goal is to demonstrate that the methods of the subgradient and the bundle types, i.e.,
methods for convex optimization with dimension-independent rate of convergence, can be naturally extended onto large-scale variational inequalities. Let me start with the Subgradient
Descent method.

9.1

Subrgadient Descent method for variational inequalities

The Subgradient Descent method, as applied to variational inequality (9.0.1), generates search
points zi according tozi+1 = πG (zi − γi ξi /|ξi |), ξi ∈ F (zi );(9.1.1)
here the starting point z1 is a point of G, γi are positive stepsizes and πG is the projector ontoG:
πG (z) = argmin{|z − w| | w ∈ G}.
As always, (9.1.1) makes sense only if ξi 6= 0; if we meet with the situation ξi = 0, then zi is anexact solution to the inequality and we can terminate. In what follows I ignore this trivial case.

9.1. SUBRGADIENT DESCENT METHOD FOR VARIATIONAL INEQUALITIES145

The indicated scheme is very old; for saddle point problems it was proposed by Arrow,
Hurvits and Uzava somewhere in mid-fifties. It is easily seen that trajectory (9.1.1), generally
speaking, does not converge to the solution set, independently of how one chooses the stepsizes.
Indeed, consider the variational inequality associated with the simplest 2-player gamef (x, y) = xy, P = Q = [−1, 1].
We have considered this particular game in Lecture 5 and know that the saddle point here isunique and is at the origin. On the other hand, here
F (x, y) = (y, −x)Tis a single-valued mapping with the value at a point (x, y) orthogonal to the vector (x, y); it
follows that|(zi − γi ξi /|ξi |)|2 = |zi |2 + γi2 ,so that if there were no projector, the trajectory would go further and further from the saddle
point. The presence of projector prevents the trajectory of going too far, but, as it is easily
seen, does not enforce its convergence to the origin, i.e., to the unique solution to the variational
inequality in question. There were many papers on how to ensure convergence of the method;sometimes people imposed additional restrictions on the operator in question, sometimes modified the algorithm. What we shall see in a while is that the divergence of the algorithm is an
illusion: the method converges, but in the so called ergodic, or, better to say, in the Cezarisense. Namely, it turns out that one can say something quite reasonable about the weighted
search points
N
z̄M
=

N
X
i=M

γi |ξi |−1

!−1 N
X

γi |ξi |−1 zi .

i=M

Proposition 9.1.1 . Let a variational inequality with bounded monotone operator be solved bya Subgradient Descent method with positive stepsizes γi . Then for all integer M, N , M ≤ N ,one hasP
2
D2 (G) + N
i=M γiNν(z̄M ) ≤ L(F ),(9.1.2)PN
2 i=M γi
D(G) being the diameter of G. If the variational inequality comes from a 2-player game withN ).
zero sum, then the same estimate is valid for the quantity νg (z̄M
Proof. Let z be an arbitrary point of G, and let ξ ∈ F (z). Let us look at the quantities|zi − z|2 :|zi+1 − z|2 = |πG (zi − γi |ξi |−1 ξi ) − z|2 ≤ |zi − γi |ξi |−1 ξi − z|2 == |zi − z|2 − 2γi |ξi |−1 ξiT (zi − z) + γi2 ≤≤ |zi − z|2 + γi2 − 2γi |ξi |−1 ξ T (zi − z)
(we have sequentially used the basic property of the projection nd the monotonicity of theoperator). Thus, we come to
ξ

T

N
X
i=M

!

γi |ξi |−1(zi − z)≤ |zM − z|2 − |zN +1 − z|2 +

N
X
i=M

γi2 ≤ D2 (G) +

N
X
i=M

γi2 .

146

LECTURE 9. LARGE-SCALE GAMES AND VARIATIONAL INEQUALITIESThe resulting relation can be rewritten as
ξ

T

P
P
2
2
D2 (G) + N
D2 (G) + N
i=M γii=M γiN(z̄M − z) ≤≤ L(F );PNP
2 i=M γi |ξi |−12 N
i=M γithis inequality is valid for all z ∈ G and all ξ ∈ F (z), and taking maximum over z, ξ, we cometo (9.1.2).
Now consider the case when the variational inequality in question comes from a 2-player zerosum game with the cost function f . Here, as in the general case, we have|zi+1 − z|2 ≤ |zi − z|2 + γi2 − 2γi |ξi |−1 ξiT (zi − z).
As we know from Lemma 5.1.1 (Lecture 5), for the monotone operator associated with a convexconcave saddle point problem one hasξiT (zi − z) ≥ f (xi , y) − f (x, yi );here z = (x, y), zi = (xi , yi ). Now, same as above, we came to the inequality
2

N
X

N
X

γi |ξi |−1 (f (xi , y) − f (x, yi )) ≤ D2 (G) +

γi2 .

i=M

i=M

Due to the Jensen inequality, the left hand side here is greater or equal to
2

N
X

!

γi |ξi |−1Nf (x̄NM , y) − f (x, ȳM ) ,i=M

and we come to2
D2 (G) + N
i=M γi.
PN
2 i=M γi
The concluding inequality holds true for all z = (x, y) ∈ G = P × Q, and taking maximum in(x, y) ∈ G, we come toP2
D2 (G) + N
i=M γiNνg (z̄M) ≤ L(F ),PN
2 i=M γi
as claimed.
We see that the accuracy estimate for the Subgradient Descent as applied to a variational
inequality with a monotone operator or to a saddle point problem, looks exactly similar to theestimate for the case of minimizing a convex Lipschitz continuous function. Therefore we alreadyknow what is the optimal rate of convergence and what are the corresponding stepsizes. The
latter are√
γi = D(G)/ i,Nf (x̄NM , y) − f (x, ȳM ) ≤ L(F )P

and the associated accuracy estimate (related to M = [N/2]) is
−1/2
ν(x̄N
, N = 1, 2, ...,
[N/2] ) ≤ O(1)L(F )D(G)N

O(1) being an absolute constant. Thus, in the large-scale case we have basically the same possibilities to solve variational inequalities with bounded monotone operators (or to find saddle
points of Lipschitz continuous convex-concave functions) as those to minimize Lipschitz continuous convex functions; recall that the same situation was with linearly converging methods in a
fixed dimension.

9.2. LEVEL METHOD FOR VARIATIONAL INEQUALITIES AND GAMES

9.2

147

Level method for variational inequalities and games

As applied to variational inequalities with monotone operators or to games, the Subgradient Descent possesses optimal worst-case behaviour in the large-scale case, but the ”typical” behaviourof the method, same as in the case of convex minimization problems, is far from being appropriate. And same as in the case of convex minimization, to improve the practical behaviour of themethod, it is natural to use the bundle scheme, where we use the whole information collected
so far. Let us start with the saddle point problem.

9.2.1

Level method for games

Consider the saddle point problem (9.0.2) with Lipschitz continuous with constant L(f ) function
f (x, y), and assume that we have already generated search points z1 , ..., zi ∈ G and know thesub- and supergradients fx0 (zj ), fy0 (zj ), j = 1, ..., i; then we can form the modelsf¯i (x) = max {f (zj ) + (fx0 (zj ))T (x − xj )}
1≤j≤i

and
f i (y) = min {f (zj ) + (fy0 (zj ))T (y − yj )}1≤j≤iof the cost functionsf¯(x) = max f (x, y), f (y) = min f (x, y)
y∈Q

x∈P

of the players; from convexity-concavity of f it follows that the function f¯i underestimates f¯,
while f i overestimates f . It follows that the function
νi (x, y) = f¯i (x) − f i (y)underestimates the function
νg (x, y) = f¯(x) − f (y).
Note that the optimal set of the latter function on G is exactly the saddle set of f , and the
optimal value of νg is 0. Since νi underestimates νg , the optimal value−∆i ≡ min νi = min f¯i − max f i
GPQis nonpositive; we shall call ∆i i-th gap. It is clear from our construction that
ν1 (·, ·) ≤ ν2 (·, ·) ≤ ... ≤ νg (·, ·),(9.2.1)∆1 ≥ ∆2 ≥ ... ≥ 0.

(9.2.2)

Note also that f¯i (xi ) ≥ f (zi ), f i (yi ) ≤ f (zi ), and therefore
νi (zi ) ≥ 0.
The Level method for saddle point problems is as follows:(9.2.3)

148

LECTURE 9. LARGE-SCALE GAMES AND VARIATIONAL INEQUALITIES

i-th step: given z1 , ..., zi , form the models f¯i , f i and νi and compute the optimal value −∆iof the model νi . Define the next iterate zi+1 aszi+1 = πGi (zi ), Gi = {z ∈ G | νi (z) ≤ −(1 − λ)∆i }and loop.
The quantity λ in the updating formula is a once for ever fixed quantity from (0, 1).
Note that an iteration of the method, same as in the minimization case, requires to compute−∆i , i.e., to solve a pair of minimization problems with piecewise linear objectives f¯i (thisfunction should be minimized over P ) and −f i (this function should be minimized over Q).
Besides this, one should compute the projection of zi onto the convex set Gi . If P and Q, as
it normally is the case, are polytopes, these problems are, respectively, a Linear Programming
and a linearly constrained Quadratic Programming programs.
Note that in course of minimizing a piecewise linear function f¯i , one can find the corresponding Lagrange multipliers - nonnegative λj , j = 1, ..., i, with unit sum such thatmin f¯i (x) = min
x∈P

x∈P

i
X





λj f (zj ) + (fx0 (zj ))T (x − xj ) ;j=1

similarly, when maximizing f i over Q, one can find the corresponding Lagrange multipliersµj ≥ 0, j = 1, ..., i, with unit sum:max f i (y) = max
y∈Q

y∈Q

i
X





µj f (zj ) + (fy0 (zj ))T (y − yj ) .

j=1

We define i-th approximate solution generated by the method asz̄i = (x̄i =iXµj xj , ȳi =

j=1

i
X

λj yj ).

j=1

Thus, we use the Lagrange multipliers related to f i to aggregate the x-components of thesearch points, while the Lagrange multipliers related to f¯i are the weights in aggregating the
y-components.
The convergence properties of the method are given by the following
Proposition 9.2.1 Let the Level method be applied to the saddle point problem associated witha convex-concave Lipschitz continuous function f (x, y) : P × Q → R. Then for any i one hasνg (z̄i ) ≤ ∆i ,(9.2.4)and the right hand side of the latter inequality is ≤ ε wheneverL(f )D(G)i > c(λ)
ε


2

,

where L(f ) is the Lipschitz constant of f , D(G) is the diameter of G andc(λ) =

2(1 − λ)2 λ(2 − λ).

(9.2.5)

9.2. LEVEL METHOD FOR VARIATIONAL INEQUALITIES AND GAMES

149

Proof. Let us start with proving (9.2.4). To this end let us fix (x, y) ∈ G. We havef (xj , y) ≤ f (zj ) + (fy0 (zj ))T (y − yj ),f (x, yj ) ≥ f (zj ) + (fx0 (zj ))T (x − xj ),and taking weighted sums of these inequalities and subtracting the resulting inequalities fromeach other, we come toiX{λj f (x, yj ) − µj f (xj , y)} ≥

j=1

≥

i
X

λj {f (zj ) + (fx0 (zj ))T (x − xj )} −

j=1

i
X

µj {f (zj ) + (fy0 (zj ))T (y − yj )}.

(9.2.6)j=1

The left hand side in this inequality, due to convexity-concavity of f , is less than or equal tof (x, ȳi ) − f (x̄i , y), and we come to the inequality
f (x, ȳi ) − f (x̄i , y) ≥iX

λj {f (zj ) + (fx0 (zj ))T (x − xj )} −i
Xµj {f (zj ) + (fy0 (zj ))T (y − yj )}.

j=1

j=1

Taking minimum over (x, y) ∈ G and taking into account the origin of the Lagrange multipliers,
we come to−νg (z̄i ) ≥ min f¯i − max f i ≡ −∆i ,PQ

as required in (9.2.4).
It remains to demonstrate that ∆i for sure becomes ≤ ε when i satisfies (9.2.5). Indeed, let
N be an integer such that ∆N > ε; let us prove that then N does not exceed the right handside of (9.2.5). Same as in the proof presented at the previous lecture, we can partition the set
I = {1, ..., N } of iteration numbers ≤ N into groups I1 , ..., Ik as follows: the group I1 ends with
the iteration i1 = N and is comprised of the iterations i such that∆i ≤ (1 − λ)−1 ∆i1 ;since, as we know, the gaps do not increase with i, this is a segment of the set I. If this segment
does not coincide with I, let i2 be the last iteration in I which does not belong to I1 , and let I2
be comprised of all indices i ≤ i2 such that∆i ≤ (1 − λ)−1 ∆i2 .
If I1 ∪ I2 does not coincide with I, let i3 be the last index in I which does not belong neither to
I1 , nor to I2 , and let I3 be comprised of all indices i ≤ i3 such that∆i ≤ (1 − λ)−1 ∆i3 ,and so on. With this construction, I will be partitioned into finitely many segments I1 , ..., Ik(the segments are counted from right to left) in such a way that∆i ≤ (1 − λ)−1 ∆is , i ∈ Is ,(9.2.7)

150

LECTURE 9. LARGE-SCALE GAMES AND VARIATIONAL INEQUALITIES

and∆is+1 > (1 − λ)−1 ∆is .
My claim is that the sets Gi associated with iterations i from a group Is have a point incommon, namely, the minimizer ws of the last, is -th, model νis of the group. Indeed, since themodels grow with i, we haveνi (ws ) ≤ νis (ws ) = −∆is ≤ −(1 − λ)∆i(we have used (9.2.7)), so that ws ∈ Gi , as claimed.
Now we can complete the proof in the same way as it was done for the optimization versionof the method. As we know from (9.2.3), νi (zi ) ≥ 0, so that zi 6∈ Gi ; it follows that νi (zi+1 ) =−(1 − λ)∆i andνi (zi ) − νi (zi+1 ) ≥ (1 − λ)∆i ≥ (1 − λ)∆is .
(9.2.8)Since, as it is immediately seen, νi is Lipschitz continuouswith constant L(f ) with respect to√x and to y, it is Lipschitz continuous with constant 2L(f ) with respect to z, and we conclude
from (9.2.8) that√|zi − zi+1 | ≥ (1 − λ)∆is L−1 (f )/ 2.
(9.2.9)At the same time, since ws ∈ Gi , i ∈ Is , we have|zi+1 − ws |2 = |πGi (zi ) − ws |2 ≤ |zi − ws |2 − |zi − zi+1 |2 ≤ |zi − ws |2 −

1
2L2 (f )(1 − λ)2 ∆2is ,and we conclude from this recurrence that the number Ns of indices in Is satisfies the estimate
Ns ≤ 2D2 (G)L2 (f )(1 − λ)−2 ∆−2is .
From (9.2.7) it follows that∆is > (1 − λ)−(s−1) ∆i1 > (1 − λ)−(s−1) ε,and we come to
Ns ≤ 2D2 (G)L2 (f )(1 − λ)−2 (1 − λ)2(s−1) ε−2 ,so that
N=

X
s

Ns ≤

2
(1 − λ)2 λ(2 − λ)D2 (G)L2 (f )ε−2 ,as claimed.

9.2.2

Level method for variational inequalities

As we know, to solve a saddle point problem is the same as to minimize an implicitly defined
convex function νg . Similarly, to find a weak solution to a variational inequality with monotone
operator (9.0.1) is the same as to minimize the implicitly defined convex function
ν(z) = sup{ξ T (z − w) | w ∈ G, ξ ∈ F (w)};(9.2.10)indeed, by definition of a weak solution, this function is nonpositive at any weak solution, andby setting w = z in the formula defining ν we see that the function is nonnegative everywhere;9.2. LEVEL METHOD FOR VARIATIONAL INEQUALITIES AND GAMES

151

thus, the minimum value of the function is 0, and weak solutions to inequality are exactly theminimizers of ν. We see that we should minimize an implicitly defined convex function withzero optimal value; same as in the case of games, given w ∈ G and ξ ∈ F (w), we know an affine
function ξ T (z − w) of z which underestimates ν, and therefore can use the Level scheme, namely,
as follows. At i-th step of the method we already have generated the search points z1 , ..., zi ∈ G
and know values ξj ∈ F (zj ) of the operator at these points; this information allows us to form
the i-th model
νi (z) = max ξjT (z − zj )1≤j≤iof the function ν; in view of (9.2.10) this function underestimates ν, and since the optimal valueof ν is 0, the optimal value of the model is nonpositive:−∆i ≡ min ν(z) ≤ 0.
z∈G

(9.2.11)

It is clear that the models grow with i:ν1 (·) ≤ ν2 (·) ≤ ... ≤ ν(·),(9.2.12)∆1 ≥ ∆2 ≥ ... ≥ 0.

(9.2.13)and, consequently,
The Level method for variational inequalities is as follows:i-th step: given z1 , ..., zi , form the model νi and compute the optimal value −∆i of themodel. Define the next iterate zi+1 aszi+1 = πGi (zi ), Gi = {z ∈ G | νi (z) ≤ −(1 − λ)∆i }and loop.
The quantity λ in the updating formula is a once for ever fixed quantity from (0, 1).
Note that in course of minimizing a piecewise linear function νi , one can find the corresponding Lagrange multipliers - nonnegative λj , j = 1, ..., i, with unit sum such thatmin νi (z) = min
z∈G

z∈G

i
X

λj ξjT (z − zj ).

j=1

We define i-th approximate solution generated by the method as
z̄i =

i
X

λ j zj ;j=1

note that this is exactly the way of aggregating search points we have used in the cutting plane
scheme for variational inequalities.
The convergence properties of the method are given by the following
Proposition 9.2.2 Let the Level method be applied to variational inequality (9.0.1) associatedwith a monotone and bounded operator F . Then for any i one hasν(z̄i ) ≤ ∆i ,(9.2.14)152

LECTURE 9. LARGE-SCALE GAMES AND VARIATIONAL INEQUALITIESand the right hand side of the latter inequality is ≤ ε wheneverL(F )D(G)i > c(λ)
ε


2

,

(9.2.15)

where L(F ) = sup{|ξ| | ξ ∈ F (z), z ∈ G}, D(G) is the diameter of G andc(λ) =

1(1 − λ)2 λ(2 − λ).

Proof. Relation (9.2.14) is, up to notation, the statement of Lemma 6.2.1 (Lecture 6). The
implicationi > c(λ)L2 (F )D2 (G)ε−2 ⇒ ∆i ≤ ε
is given by word-by-word repeating the proof of Proposition 9.2.1.

9.3

Exercises: Around Level

In Lecture 8 we became acquainted with the family of bundle methods and a particular memberof the family - the Level method. The below exercises present useful additional information on
the Level.

9.3.1”Prox-Level”

Consider a convex optimization problem
minimize f (x) over x ∈ G,

(9.3.1)

where f is a Lipschitz continuous with constant L(f ) convex function and G is a bounded andclosed convex subset of Rn . The Level method, in its basic version, solves the problem as follows:
after i − 1 steps of the procedure we already have generated the search points x1 , ..., xi ∈ G andknow the values f (xj ) and subgradients f 0 (xj ), |f 0 (xj )| ≤ L(f ), of the objective at the points
xj , j = 1, ..., i. At i-th step we1) form the i-th model of the objectivefi (x) = max {f (xj ) + (x − xj )T f 0 (xj )},1≤j≤i

and compute its minimal value over G, let it be called fi− ;2) compute the best found so far value of the objectivefi+ = min f (xj );1≤j≤i3) define i-th levelli = fi− + λ∆i , ∆i = fi+ − fi−(λ ∈ (0, 1) is the parameter of the method)
and, finally,9.3. EXERCISES: AROUND LEVEL

153

4) project the current iterate xi onto the level set
Qi = {x ∈ G | fi (x) ≤ li }of the i-th model to get the new iterate:xi+1 = πQi (xi ).
This scheme can be modified in various directions; e.g., intuitively it is strange that in orderto obtain the next iterate xi+1 we project onto the level set of the model the previous iterate xi
independently of whether xi is good or bad, I mean, independently of whether f (xi ) is the best
found so far value of the objective fi+ or f (xi ) is greater (even significantly greater) than fi+ . It
seems to be more reasonable to move from the best of the iterates - in this case our chances toimprove the best found so far value of the objective look better. It turns out that the strategy”to obtain xi+1 , project onto Qi the best of the search points” also is appropriate - it resultsin basically the same complexity bound as that one for the initial version of the method. To
demonstrate this, we need the following preliminary observation:
Exercise 9.3.1 #+ Let P1 ⊃ P2 ⊃ P3 ⊃ ... be a sequence of closed nonempty convex subsets of
Rn , and let y0 ∈ Rn . Let
yi = πPi (y0 )be the projections of y0 onto Pi . Then|y0 − yi |2 ≥iX|yj−1 − yj |2 .

(9.3.2)

j=1

Now we are able to present and justify the modified version of Level. Let∆i = fi+ − fi−be the same gaps as in the basic version. When running the method, we compute these gaps; as
it is immediately seen, these gaps never increase, independently on how we update the search
points (provided that the latter always are chosen in G). Let us define the ”regeneration steps”
j1 , j2 , ... as follows: the first step is a regeneration one:j1 = 1.
Now, let i ≥ 1 and let j be the last regeneration step preceding the step i. It is possible that∆i < (1 − λ)∆j ;if it is the case, we claim that i also is a regeneration step, otherwise i is not a regeneration step.
Now, the modified version of the method differs from the basic one by replacing rules 3) - 4)with the following ones:
3∗ ) Check whether i is a regeneration step. If it is the case, set i-th level li , same as in thebasic version of the method, equal tofi− + λ∆i ;154

LECTURE 9. LARGE-SCALE GAMES AND VARIATIONAL INEQUALITIES

if i is not a regeneration step, then set the level li equal tomin{li−1 , fi− + λ∆i }.
Note that the difference between the new rule for computing the level and the initial one is thatnow we enforce the level li at a non-regeneration step to be not greater than the level used at
the previous step; if this requirement is satisfied by the ”usual” level fi− + λ∆i , we choose as li
this latter quantity, otherwise we set li = li−1 .
4∗ ) Define the new iterate xi+1 as the projection of the best found so far (i.e., with thesmallest vale of the objective) search point x̄i onto the level set Qi = {x ∈ G | fi (x) ≤ li } :xi+1 = πQi (x̄i ).
If there are several candidates on the role of x̄i (i.e., the smallest value of f on the set {x1 , ..., xi }is attained at more than one point), x̄i is defined as the latest of the candidates.
Exercise 9.3.2 #+ Let Ik is the sequence of iterations which starts with k-th regeneration step
jk and ends with the step ik ≡ jk+1 − 1 (so that the sequence contains exactly one regeneration
step, which starts the sequence).
1) Prove that the level sets Qi , i ∈ Ik , are nonempty and decrease:Qjk ⊃ Qjk +1 ⊃ ... ⊃ Qik ;in particular, all level sets are nonempty, so that the modified method is well-defined;
2) Prove that the number Nk of iterations in the group Ik satisfies the estimate
Nk ≤ D2 (G)L2 (f )(1 − λ)−4 ∆−2
jk ,D(G) being the diameter of G;
3) Derive from 2) that for any positive ε and any N such thatN > c(λ) (L(f )D(G)/ε)2 , c(λ) = (1 − λ)−4 (2 − λ)−1 λ−1 ,one hasf (x̄N ) − min f ≤ ∆N ≤ ε,G

so that the efficiency estimate for the modified method is, up to the value of c(λ), the same as
for the initial version of the method.

9.3.2

Level for constrained optimization

To the moment we know how to apply the Level method to convex problems without functional
constraints, to variational inequalities and games, but do not know how to use the method fora convex problem with functional constraints:
minimize f (x) s.t. fi (x) ≤ 0, i = 1, ..., m, x ∈ G.
The goal of the below exercises is to introduce to you a version of the method for this latterproblem.

9.3. EXERCISES: AROUND LEVEL

155

By setting g(x) = maxi fi (x), we may reduce the situation to the case of a single functional
constraint, i.e., to a problem
minimize f (x) s.t. g(x) ≤ 0, x ∈ G;

(9.3.3)from now on we assume that f and g are Lipschitz continuous with constant L and G is a
closed and bounded convex set in Rn of a diameter D. Of course, we assume that the problem
is feasible (and therefore is solvable). Besides this, it is convenient to assume that g is notcompletely redundant, i.e., that g(x) > 0 although somewhere on G.
Assume that we already have generated search points x1 , ..., xi ∈ G and know from theoracle the values f (xj ), g(xj ) and subgradients f 0 (xj ), g 0 (xj ) (of norms not exceeding L) of
the objective and the constraints. This information can be assembled into the piecewise linear
models of the objective and the constraint:fi (x) = max {f (xj ) + (x − xj )T f 0 (xj )},1≤j≤i

gj (x) = max {g(xj ) + (x − xj )T g 0 (xj )}.
1≤j≤i

Besides this, we can form the model problem(Pi )minimize fi (x) s.t. gi (x) ≤ 0, x ∈ G.

Exercise 9.3.3 Prove that (Pi ) is solvable, and that the optimal value fi∗ of the problem is a
non-increasing function of i which is ≤ the actual optimal value f ∗ of problem (9.3.3).
Let Ti be the set of all pairs (f (xj ), g(xj )), j = 1, ..., i, and let
C(i) = Conv Ti + R2+
be the set of all pairs (u, v) on the plane R2 which are minorized by convex combinations of thepairs from Ti .
Exercise 9.3.4 ∗ Let (u, v) ∈ C(i). Prove that one can easily (i.e., without oracle calls orcomplicated computations) find a point x in G where the pair (f (x), g(x)) of values of the objective
and the constraint is ”not worse” than (u, v):
f (x) ≤ u, g(x) ≤ v.
Now we come to the key construction - to support function hi . This is the function of α ∈ [0, 1]
defined ashi (α) = min {α(f (xj ) − fi∗ ) + (1 − α)g(xj )}.
1≤j≤i

Exercise 9.3.5 Prove that hi is a concave piecewise linear function with Lipschitz constant2LD on the segment [0, 1] and thathi (α) = min{α(u − fi∗ ) + (1 − α)v | (u, v) ∈ C(i)}.
Besides this,h1 (·) ≥ h2 (·) ≥ ...

156

LECTURE 9. LARGE-SCALE GAMES AND VARIATIONAL INEQUALITIESFrom now on we setρ(u, v) = max{(u)+ , (v)+ }.

Exercise 9.3.6 ∗ Prove that∆i ≡ max hi (α) =
0≤α≤1

min

(u,v)∈C(i)ρ(u, v).

Note that hi is defined in terms of the information on the problem accumulated in course of
the first i steps, so that after these steps we can easily compute the i-th gap ∆i . According
to exercise 9.3.6, this gap is nonnegative, and we can point out in the set C(i) (this set also isdefined in terms of the information accumulated during the first i steps) a pair (ui , vi ) such thatui − fi∗ ≤ ∆i , vi ≤ ∆i .
In view of exercise 9.3.4, we can easily associate with the latter pair a point x̄i ∈ G such thatf (x̄i ) − fi∗ ≤ ∆i , g(x̄i ) ≤ ∆i .
Since fi∗ is ≤ the actual optimal value of the problem (exercise 9.3.3), we see thatthe information collected at the first i steps allows us to form an approximate solution x̄i toproblem (9.3.3) with absolute residuals f (x̄i ) − f ∗ and (g(x̄i ))+ in the values of the objective
and the constraint not exceeding the i-th gap ∆i .
Thus, all we interested in is to enforce the gaps ∆i to converge to 0 at the highest possiblerate.
The gap is the maximum of the support function over α ∈ [0, 1], so that if we would be able
to enforce the gaps to tend to 0, we for sure will be able to enforce the values of hi at a particularpoint α to tend to something nonpositive (this values do converge as i → ∞, since the sequence{hi } is monotone nonincreasing sequence of uniformly Lipschitz continuous functions (exercise
9.3.5) with nonnegative maximums over [0, 1] (exercise 9.3.6)). On the other hand, to enforce
lim hi (α) ≤ 0i→∞for a given α ∈ [0, 1] seems to be easier than to ensure
lim max hi ≡ lim ∆i = 0,i→∞ [0,1]i→∞

so it makes sense to start with the easier problem. In fact we know how to solve it:
Exercise 9.3.7 ∗ Given α ∈ [0, 1], let us use the basic Level method to minimize the function
f α (x) = αf (x) + (1 − α)g(x)over G. Prove that after N steps of the method we for sure will havehi (α) ≤ O(1)DLN −1/2 ,where the constant O(1) depends on the parameter λ of the method only.

9.3. EXERCISES: AROUND LEVEL

157

Now we are in the situation as follows: at i-th step we are given certain piecewise linear concaveand Lipschitz continuous with the constant 2LD function hi (α) on the segment [0, 1]; thisfunction depends on our behaviour at the previous steps. It is for sure known that as i grows,
the functions do not increase at every point of the segment. Choosing at certain moment an
α ∈ [0, 1], we know how to behave ourselves in future in a way which, after an arbitrary number
N of steps (counted starting from the moment when α was chosen), would make the value ofh· (α) less than ≤ O(1)LDN −1/2 ; and our goal is to enforce convergence to 0 of the quantities∆i ≡ maxα∈[0,1] hi (α). Given this description of our goals and abilities, could you guess how toachieve the goal? Please try to do it before reading the concluding exercises. The expected rateof convergence of ∆i to 0 should be as follows: to ensure ∆i ≤ ε, it should take no more thanLD
N (ε) = O(1)
ε


2

ln(

LD
)
ε

steps.
The way I know is as follows. The function hi possibly is negative somewhere on [0, 1];consider the segment
δi = {α ∈ [0, 1] | hi (α) ≥ 0}.
Since hi decrease with i, these segments are embedded into each other: δi+1 ⊂ δi . Let us fix aµ ∈ (0, 1/2) and call a point α ∈ [0, 1] µ-centered with respect to δi , if the point belongs to δiand the smaller of two parts into which the point splits the segment δi is at least µ times thesegment itself. Now consider the following policy. We start with
α0 = 1/2
and use the given by exercise 9.3.7 tactics to decrease hi (α0 ). After each i-th step we check
whether α0 is µ-centered with respect to δi . Immediately after this condition turns out to beviolated, we replace α0 by the midpoint α1 of the current segment δi and switch to decreasing
h· (α1 ), using the same tactics given by Exercise 9.3.7. This stage is terminated immediately
after we discover that α1 is no more µ-centered with respect to the current segment δi ; when
it happens, we set our ”reference value of α” to the midpoint α2 of the current segment δi andswitch to decreasing h· (α2 ), and so on.
Exercise 9.3.8 ∗ Prove that the indicated policy for any positive ε ensures that ∆i ≤ ε wheneveri ≥ O(1)(LD/ε)2 ln(LD/ε),with O(1) depending on the parameters λ and µ of the method (recall that λ is the parameterused by our ”working horse” - the basic Level method).
Now let us summarize our construction. To solve (9.3.3), we sequentially apply the basic Level
method to auxiliary problems of the typeminimize αf (x) + (1 − α)g(x), x ∈ G,
from time to time varying α. Needless to say that in practical implementation there is nonecessity to solve the auxiliary problems from scratch; you may use the bundles for f and gcollected so far. The complexity of the method, as we have seen, is O(1)(LD/ε)2 ln(LD/ε),

158

LECTURE 9. LARGE-SCALE GAMES AND VARIATIONAL INEQUALITIES

which is by logarithmic factor worse than the optimal complexity O(1)(LD/ε)2 given by the
Subgradient Descent. In spite of this theoretical shortcoming, the practical behaviour of themethod is significantly better than that one of the Subgradient Descent.
Note also that in principle we could reduce the constrained problem to the saddle point
problem for the Lagrange function
L(x, y) = f (x) + yg(x),which could be solved by the saddle point version of Level. This straightforward approach has asevere drawback: we do not know in advance how large is the optimal Lagrange multiplier (andeven whether it exists - the problem should not necessarily satisfy the Slater condition). Since
all our efficiency estimates depend on the diameters of the domains involved into the problem,
this Lagrange approach would result in bad efficiency estimates. What we in fact use is Fritz
John duality: an optimal solution to a convex problem (9.3.3) is, for some α ∈ [0, 1], a solution
to the problem
minimize αf (x) + (1 − α)g(x), x ∈ G(why?) The advantage of this latter duality is that it does not require any qualification of
constraints and, consequently, is insensitive to the Slater condition.

Lecture 10

Smooth convex minimization
problems
To the moment we have more or less complete impression of what is the complexity of solving general nonsmooth convex optimization problems and what are the corresponding optimal
methods. In this lecture we start a new topic: optimal methods for smooth convex minimization. Today I shall deal with the simplest case of unconstrained problems with smooth convex
objective. Thus, we shall be interested in methods for solving the problem(f )minimize f (x) over x ∈ Rn ,where f is a convex function of the smoothness type C1,1 , i.e., a continuously differentiable with
Lipschitz continuous gradient:|f 0 (x) − f 0 (y)| ≤ L(f )|x − y| ∀x, y ∈ Rn ;from now on | · | denotes the standard Euclidean norm. We assume also that the problem issolvable (i.e., Argmin f 6= ∅) and denote by R(f ) the distance from the origin (which will be
the starting point for the below methods) to the optimal set of (f ). Thus, we have definedcertain family Sn of convex minimization problems; the family is comprised of all programs (f )associated with C1,1 -smooth convex functions f on Rn which are below bounded and attaintheir minimum. As usual, we provide the family with the first-order oracle O which, given on
input a point x ∈ Rn , reports the value f (x) and the gradient f 0 (x) of the objective at the point.
The accuracy measure we are interested in is the absolute inaccuracy in terms of the objective:ε(f, x) = f (x) − minf.
nR

It turns out that complexity of solving a problem from the above family heavily depends on the
Lipschitz constant L(f ) of the objective and the distance R(f ) from the starting point to the
optimal set; this is why in order to analyse complexity it makes sense to speak not on the wholefamily Sn , but on its subfamilies
Sn (L, R) = {f ∈ S | L(f ) ≤ L, R(f ) ≤ R},
where L and R are positive parameters identifying, along with n, the subfamily.
159

160

10.1

LECTURE 10. SMOOTH CONVEX MINIMIZATION PROBLEMS

Traditional methods

Unconstrained minimization of C1,1 -smooth convex functions is, in a sense, the most traditional
area in Nonlinear Programming. There are many methods for these problems: the simplest
Gradient Descent method with constant stepsize, Gradient Descent with complete relaxation,
numerous Conjugate Gradient routines, etc. Note that I did not mention the Newton method,since the method, in its basic form, requires twice continuously differentiable objectives andsecond-order information, and we are speaking about the first-order information and objectives
which are not necessarily twice continuously differentiable. One might think that since the field
is so well-studied, among the traditional methods there for sure are optimal ones; surprisingly,
this is not the case, as we shall see in a while.
To get a kind of a reference point, let me start with the simplest method for smooth convex
minimization - the Gradient Descent. This method, as applied to (f ), generates the sequence of
search points
xi+1 = xi − γi f 0 (xi ), x0 = 0.
(10.1.1)
Here γi are positive stepsizes; various versions of the Gradient Descent differ from each other by
the rules for choosing these stepsizes. In the simplest version - Gradient Descent with constantstepsize - one setsγi ≡ γ.
(10.1.2)
With this rule for stepsizes the convergence of the method is ensured if
0<γ<

2
.
L(f )

(10.1.3)

Proposition 10.1.1 Under assumption (10.1.3) process (10.1.1) - (10.1.2) converges at therate O(1/N ): for all N ≥ 1 one hasf (xN ) − min f ≤R2 (f )
N −1 .
γ(2 − γL(f ))(10.1.4)

In particular, with the optimal choice of γ, i.e., γ = L−1 (f ), one hasf (xN ) − min f ≤L(f )R2 (f ).
N

(10.1.5)

Proof. Let x∗ be the minimizer of f of the norm R(f ). We havef (x) + hT f 0 (x) ≤ f (x + h) ≤ f (x) + hT f 0 (x) +L(f ) 2|h|2(10.1.6)(we have used the convexity of f and the Lipschitz continuity of the gradient of f ). From thesecond inequality it follows thatf (xi+1 ) ≤ f (xi ) − γ|f 0 (xi )|2 +L(f ) 2 0
γ |f (xi )|2 =2= f (xi ) − ω|f 0 (xi )|2 , ω = γ(1 − γL(f )/2) > 0,

(10.1.7)

10.2. COMPLEXITY OF CLASSES SN (L, R)

161

whence|f 0 (xi )|2 ≤ ω −1 (f (xi ) − f (xi+1 ));(10.1.8)

we see that the method is monotone and thatX

|f 0 (xi )|2 ≤ ω −1 (f (x1 ) − f (x∗ )) = ω −1 (f (0) − f (x∗ )) ≤ ω −1i

L(f )R2 (f )2

(10.1.9)(the concluding inequality follows from the second inequality in (10.1.6), where one should set
x = x∗ , h = −x∗ ).
We have|xi+1 − x∗ |2 = |xi − x∗ |2 − 2γ(xi − x∗ )T f 0 (xi ) + γ 2 |f 0 (xi )|2 ≤≤ |xi − x∗ |2 − 2γ(f (xi ) − f (x∗ )) + γ 2 |f 0 (xi )|2(we have used convexity of f , i.e., have applied the first inequality in (10.1.6) with x = xi ,h = x∗ − xi ). Taking sum of these inequalities over i = 1, ..., N and taking into account (10.1.9),
we come toNX(f (xi ) − f (x∗ )) ≤ (2γ)−1 R2 (f ) + γω −1 L(f )R2 (f )/4 =i=1

R2 (f ).
γ(2 − γL(f ))Since the absolute inaccuracies εi = f (xi ) − f (x∗ ), as we already know, decrease with i, we
obtain from the latter inequalityεN ≤R2 (f )
N −1 ,
γ(2 − γL(f ))as required.
In fact our upper bound on the accuracy of the Gradient Descent is sharp: it is easily seen
that, given L > 0, R > 0, a positive integer N and a stepsize γ > 0, one can point out a quadratic
form f (x) on the plane such that L(f ) = L, R(f ) = R and for the N -th point generated by the
Gradient Descent as applied to f one hasf (xN ) − min f ≥ O(1)

10.2

LR2
.
N

Complexity of classes Sn (L, R)

The rate of convergence of the Gradient Descent O(N −1 ) given by Proposition 10.1.1 is dimensionindependent and is twice better in order than the known to us optimal in the nonsmooth large
scale case rate O(N −1/2 ). There is no surprise in the progress: we have passed from a very
wide family of nonsmooth convex problems to the much more narrow family of smooth convex
problems. What actually is a surprise is that the rate O(N −1 ) still is not optimal; the optimal dimension-independent rate of convergence in smooth convex minimization turns out to beO(N −2 ).

162

LECTURE 10. SMOOTH CONVEX MINIMIZATION PROBLEMS

Theorem 10.2.1 The complexity of the family Sn (L, R) satisfies the bounds

O(1) min n,


s



s

LR2 
≤ Compl(ε) ≤c
ε 

4LR2
b
ε

(10.2.1)

with positive absolute constant O(1).
Note that in the large scale case, namely, when
n2 ≥

LR2
,
ε

the upper complexity bound only by absolute constant factor differs from the lower one. Whendimension is not so large, namely, when
n2 <<

LR2
,
ε

both lower and upper complexity bounds are not sharp; the lower bound does not say anything
reasonable, and the upper is valid, but significantly overestimates the complexity. In fact inthis ”low scale” case the complexity turns out to be O(1)n ln(LR2 /ε), i.e., is basically the same
as for nonsmooth convex problems. Roughly speaking, in a fixed dimension the advantages ofsmoothness can be exploited only when the required accuracy is not too high.
Let us prove the theorem. As always, we should point out a method associated with theupper complexity bound and to prove somehow the lower bound.

10.2.1

Upper complexity bound: Nesterov’s method

The upper complexity bound is associated with a relatively new construction developed by Yurii
Nesterov in 1983. This construction is a very nice example which demonstrates the meaning of
the complexity approach; the method with the rate of convergence O(N −2 ) was found mainly
because the investigating of complexity enforced to believe that such a method should exist.
The method, as applied to problem (f ), generates the sequence of search points xi , i = 0, 1, ...,and the sequence of approximate solutions yi , i = 1, 2, ..., along with auxiliary sequences ofvectors qi , positive reals Li and reals ti ≥ 1, according to the following rules:
Initialization: Set
x0 = y1 = q0 = 0; t0 = 1;choose as L0 an arbitrary positive initial estimate of L(f ).
i-th step, i ≥ 1: Given xi−1 , yi , qi−1 , Li−1 , ti−1 , act as follows:
1) set1
xi = yi −(yi + qi−1 )ti−1(10.2.2)

and compute f (xi ), f 0 (xi );2) Testing sequentially the values l = 2j Li−1 , j = 0, 1, ..., find the first value of l such that11
f (xi − f 0 (xi )) ≤ f (xi ) − |f 0 (xi )|2 ;l
2l(10.2.3)

10.2. COMPLEXITY OF CLASSES SN (L, R)

163

set Li equal to the resulting value of l;3) Setyi+1 = xi −qi = qi−1 +

1 0
f (xi ),
Li

ti−1 0f (xi ),
Li

define ti as the larger root of the equationt2 − t = t2i−1
and loop.
Note that the method is as simple as the Gradient Descent; the only complication is a kindof line search in rule 2). In fact the same line search (similar to what is called the ArmiljoGoldstein rule) should be used in the Gradient Descent, if the Lipschitz constant of the gradientof the objective is not given in advance (of course, in practice it is never given in advance).
Theorem 10.2.2 Let f ∈ Sn be minimized by the aforementioned method. Then for any N one
hasb )R2 (f )2L(fb ) = max{2L(f ), L0 }.
f (yN +1 ) − min f ≤, L(f(10.2.4)(N + 1)2
Besides this, first N steps of the method require N evaluations of f 0 and no more than 2N +b )/L0 ) evaluations of f .
log2 (L(fProof. 10 . Let us start with the following simple statement:
A. If l ≥ L(f ), then11f (x − f 0 (x)) ≤ f (x) − |f 0 (x)|2 .
l
2lThis is an immediate consequence of the inequalityf (x − tf 0 (x)) ≤ f (x) − t|f 0 (x)|2 +L(f ) 2 0
t |f (x)|2 ,2which in turn follows from (10.1.6).
We see that if L0 ≥ L(f ), then all Li are equal to L0 , since l = L0 for sure passes all tests(10.2.3). Further, if L0 < L(f ), then the quantities Li never become ≥ 2L(f ). Indeed, in thecase in question the starting value L0 is ≤ L(f ). When updating Li−1 into Li , we sequentiallysubject to test (10.2.3) the quantities Li−1 , 2Li−1 , 4Li−1 , ..., and choose as Li the first of thesequantities which passes the test. It follows that if we come to Li ≥ 2L(f ), then the quantity
Li /2 ≥ L(f ) did not pass one of our tests, which is impossible by A. Thus,b ), i = 1, 2, ...
Li ≤ L(fNow, the total number of evaluations of f 0 in course of the first N steps clearly equals to
N , while the total number of evaluations of f is N plus the total number, K, of tests (10.2.3)
performed during these steps. Among K tests there are N successful (when the tested value ofl satisfies (10.2.3)), and each of the remaining tests increases value of L· by the factor 2. Sinceb )/L0 ), so that
L· , as we know, is ≤ 2L(f ), the number of these remaining tests is ≤ log2 (L(f164

LECTURE 10. SMOOTH CONVEX MINIMIZATION PROBLEMS

b )/L0 ), as
the total number of evaluations of f in course of the first N steps is ≤ 2N + log2 (L(fclaimed.
20 . It remains to prove (10.2.4). To this end let us add to A. the following observation
B. For any i ≥ 1 and any y ∈ Rn one has

f (y) ≥ f (yi+1 ) + (y − xi )T f 0 (xi ) +1 0|f (xi )|2 .
2Li(10.2.5)

Indeed, we havef (y) ≥ (y − xi )T f 0 (xi ) + f (xi ) ≥ (y − xi )T f 0 (xi ) + f (yi+1 ) +1 0|f (xi )|22Li(the concluding inequality follows from the definition of Li and yi+1 ).
30 . Let x∗ be the minimizer of f of the norm R(f ), and let
εi = f (yi ) − f (x∗ )be the absolute accuracy of i-th approximate solution generated by the method.
Applying (10.2.5) to y = x∗ , we come to inequality
0 ≥ εi+1 + (x∗ − xi )T f 0 (xi ) +1 0|f (xi )|2 .
2Li(10.2.6)

Now, the relation (10.2.2) can be rewritten as
xi = (ti−1 − 1)(yi − xi ) − qi−1 ,and with this substitution (10.2.6) becomesT
0 ≥ εi+1 + (x∗ )T f 0 (xi ) + (ti−1 − 1)(xi − yi )T f 0 (xi ) + qi−1
f 0 (xi ) +1 0|f (xi )|2 .
2Li(10.2.7)

At the same time, (10.2.5) as applied to y = yi results in
0 ≥ −δi ≡ f (yi+1 ) − f (yi ) + (yi − xi )T f 0 (xi ) +1 0|f (xi )|2 ,2Liwhich implies an upper bound on (yi − xi )T f 0 (xi ), or, which is the same, a lower bound on thequantity (xi − yi )T f 0 (xi ), namely,(xi − yi )T f 0 (xi ) ≥ εi+1 − εi +1 0
|f (xi )|2 + δi .
2Li

Substituting this estimate into (10.2.7), we come toT0 ≥ ti−1 εi+1 − (ti−1 − 1)εi + (x∗ )T f 0 (xi ) + qi−1
f 0 (xi ) + (ti−1 − 1)δi +ti−1 0|f (xi )|2 .
2Li(10.2.8)

Multiplying this inequality by ti−1 /Li , we come tot2 − ti−1
t2
t2 − ti−10 ≥ i−1
δi + i−1 εi+1 − i−1
εi + (x∗ )T
Li
Li
Li



t2
ti−1 0T ti−1 0
f (xi ) + qi−1
f (xi ) + i−12 |f 0 (xi )|2 .
Li
Li
2Li
(10.2.9)10.2. COMPLEXITY OF CLASSES SN (L, R)

165

We have (ti−1 /Li )f 0 (xi ) = qi − qi−1 and t2i−1 − ti−1 = t2i−2 (here t−1 = 0); thus, (10.2.9) can be
rewritten as0≥

t2i−1 − ti−1
t2
t2
1
T
δi + i−1 εi+1 − i−2 εi + (x∗ )T (qi − qi−1 ) + qi−1
(qi − qi−1 ) + |qi − qi−1 |2 =
Li
Li
Li
2
=

t2
t2
t2i−1 − ti−111
δi + i−1 εi+1 − i−2 εi + (x∗ )T (qi − qi−1 ) + |qi |2 − |qi−1 |2 .
Li
Li
Li
2
2

Since the quantities Li do not decrease with i and εi+1 is nonnegative, we shall only strengthen
the resulting inequality by replacing the coefficient t2i−1 /Li at εi+1 by t2i−1 /Li+1 ; thus, we cometo the inequality0≥

t2i−1 − ti−1
t2
t2
1
1
δi + i−1 εi+1 − i−2 εi + (x∗ )T (qi − qi−1 ) + |qi |2 − |qi−1 |2 .
Li
Li+1
Li
2
2

Taking sum of these inequalities over i = 1, ..., N , we come to
N 2
Xti−1 − ti−1
i=1

Li

δi +

t2N −11
1
εN +1 ≤ −(x∗ )T qN − |qN |2 ≤ |x∗ |2 ≡ R2 (f )/2;
LN +1
2
2

(10.2.10)

N 2
ti−1 − ti−1
LN +1 R2 (f ) X−
δi .
εN +1 ≤ 2
2
Li
tN −1
i=1

(10.2.11)

thus,
#

"

b ), and from the recurrence
As we know, LN +1 ≤ L(fq1ti = (1 + 1 + 4t2i−1 ), t0 = 1
2

it immediately follows that ti ≥ (i + 1)/2; therefore (10.2.11) implies (10.2.4).
Theorem 10.2.2 implies the upper complexity bound announced in (10.2.1); to see this, itsuffices to apply to a problem from Sn (L, R) the Nesterov method with L0 = L; in this case, as
we know from A., all Li are equal to L0 , so that we may skip the line search 2); thus, to run
the method it will be sufficient to evaluate f and f 0 at the search points only, and in view of(10.2.4) to solve the problem within absolute accuracy ε it is sufficient to perform the indicatedin (10.2.1) number of oracle calls.
Let me add some words about the Nesterov method. As it is, it looks as an analytical trick; I
would be happy to present to you a geometrical explanation, but I do not know it. Historically,
there were several predecessors of the method with more clear geometry, but these methods had
slightly worse efficiency estimates; the progress in these estimates made the methods less andless geometrical, and the result is as you just have seen. By the way, I have said that I do
not know geometrical explanation of the method, but I can immediately point out geometricalpresentation of the method. It is easily seen that the vectors qi can be eliminated from theequations describing the method, and the resulting description is simply
xi = (1 + αi )yi − αi yi−1 , αi =ti−2 − 1, i ≥ 1, y0 = y1 = 0,
ti−1

166

LECTURE 10. SMOOTH CONVEX MINIMIZATION PROBLEMS
yi+1 = xi −1 0
f (xi )
Li

(Li are given by rule 2)). Thus, i-th search point is a kind of forecast - it belongs to theline passing through the pair of the last approximate solutions yi , yi−1 , and for large i yi isapproximately the middle of the segment [yi−1 , xi ]; the new approximate solution is obtained
from this forecast xi by the standard Gradient Descent step with Armiljo-Goldstein rule for the
steplength.
Let me make one more remark. As we just have seen, the simplest among the traditional
methods for smooth convex minimization - the Gradient Descent - is far from being optimal;
its worst-case order of convergence can be improved by factor 2. What can be said about
optimality of other traditional methods, like the Conjugate Gradients? Surprisingly, the answer
is as follows: no one of the traditional methods is proved to be optimal, i.e., with the rateof convergence O(N −2 ). For some versions of the Conjugate Gradient family it is known that
their worst-case behaviour is not better, and sometimes is even worse, than that one of theGradient Descent, and no one of the remaining traditional methods is known to be better thanthe Gradient Descent.

10.2.2Lower bound

Surprisingly, the ”most difficult” smooth convex minimization problems are the quadratic ones;
the lower complexity bound in (10.2.1) comes exactly from investigating the complexity of large
scale unconstrained convex quadratic minimization with respect to the first order minimization
methods (I stress this ”first order”; of course, the second order methods, like the Newton one,minimize a quadratic function in one step).
Letσ̄ = {σ1 < σ2 < ... < σn }be a set of n points from the half-interval (0, L] andµ̄ = {µ1 , ..., µn }be a sequence of nonnegative reals such thatnXµi = R 2 .

i=1

Let e1 , ..., en be the standard orths in Rn , and let
A = Diag{σ1 , ..., σn },
b=

n
X
√

µi σ i e i .

i=1Consider the quadratic form
1f (x) ≡ fσ̄,µ̄ (x) = xT Ax − xT b;2

10.2. COMPLEXITY OF CLASSES SN (L, R)

167

this function clearly is smooth, with the gradient f 0 (x) = Ax being Lipschitz continuous withconstant L; the minimizer of the function is the vectornX√µi e i ,(10.2.12)i=1

and the norm of this vector is exactly R. Thus, the function f belongs to Sn (L, R). Now, let
U be the family of all rotations of Rn which remain the vector b invariant, i.e., the family of
all orthogonal n × n matrices U such that U b = b. Since Sn (L, R) contains the function f , itcontains also the family F(σ̄, µ̄) comprised by all rotations
fU (x) = f (U x)of the function f associated with U ∈ U. Note that fU is the quadratic form
1 T
x AU x − xT b, AU = U T AU.
2
Thus, the family F(σ̄, µ̄) is contained in Sn (L, R), and therefore the complexity of this family
underestimates the complexity of Sn (L, R). What I am going to do is to bound from below thecomplexity of F(σ̄, µ̄) and then choose the worst ”parameters” σ̄ and µ̄ to get the best possible,
within the bounds of this approach, lower bound for the complexity of Sn (L, R).
Consider a convex quadratic form
1g(x) = xT Qx − xT b;
2
here Q is a positive semidefinite symmetric n × n matrix. Let us associate with this form the
Krylov subspaces
Ei (g) = Rb + RQb + ... + RQi−1 b
and the quantitiesε∗i (g) =minx∈Ei (Q,b)g(x) − minn g(x).
x∈RNote that these quantities clearly remain invariant under rotations(Q, b) 7→ (U T QU, U T b)
associated with n × n orthogonal matrices U . In particular, the quantities ε∗i (fU ) associatedwith fU ∈ F(σ̄, µ̄) in fact do not depend on U :ε∗i (fU ) ≡ ε∗i (f ).

(10.2.13)

The key to our complexity estimate is given by the following:
Proposition 10.2.1 Let M be a method for minimizing quadratic functions from F(σ̄, µ̄) and
M be the complexity of the method on the latter family. Then there is a problem fU in the familysuch that the result x̄ of the method as applied to fU belongs to the space E2M +1 (fU ); in particular,
the inaccuracy of the method on the family in question is at least ε∗2M +1 (fU ) ≡ ε∗2M +1 (f ).

168

LECTURE 10. SMOOTH CONVEX MINIMIZATION PROBLEMS

The proposition will be proved in the concluding part of the lecture.
Let us derive from Proposition 10.2.1 the desired lower complexity bound. Let ε > 0 be
fixed, let M be the complexity Compl(ε) of the family Sn (L, R) associated with this ε. Then
there exists a method which solves in no more than M steps within accuracy ε all problems
from Sn (L, R), and, consequently, all problems from the smaller families F(σ̄, µ̄), for all σ̄ andµ̄. According to Proposition 10.2.1, this latter fact implies thatε∗2M +1 (f ) ≤ ε(10.2.14)

for any f = fσ̄,µ̄ . Now, the Krylov space EN (f ) is, by definition, comprised of linear combinationsof the vectors b, Ab, ..., AN −1 b, or, which is the same, of the vectors of the form q(A)b, where q
runs over the space PN of polynomials of degrees ≤ N . Now, the coordinates of the vector b are√√σi µi , so that the coordinates of q(A)b are q(σi )σi µi . Thereforef (q(A)b) =nX1{ σi3 q 2 (σi )µi − σi q(σi )µi }.
2
i=1

As we know, the minimizer of f over Rn is the vector with the coordinatesvalue of f , as it is immediately seen, is−
We come to2ε∗N (f ) = minnXq∈PN −1√µi , and the minimaln1X
σi µ i .
2 i=1{σi3 q 2 (σi )µi − 2σi2 q(σi )µi } +

i=1

n
X

σi µi =

i=1
n
X

= min

q∈PN −1σi (1 − σi q(σi ))2 µi .

i=1

Substituting N = 2M + 1 and taking in account (10.2.14), we come to2ε ≥ min

q∈P2M

n
X

σi (1 − σi q(σi ))2 µi .

(10.2.15)

i=1This inequality is valid for all sets σ̄ = {σi }ni=1 comprised of n different points from (0, L] andall sequences µ̄ = {µi }ni=1 of nonnegative reals with the sum equal to R2 ; it follows that2ε ≥ sup max min
σ̄

µ̄

p∈P2M

n
X

σi (1 − σi q(σi ))2 µi ;(10.2.16)

i=1

due to the von Neumann Lemma, one can interchange here the minimization in q and the
maximization in µ̄, which immediately results in
2ε ≥ R2 sup min max σ(1 − σq(σ))2 .
σ̄ q∈P2M σ∈σ̄Now, the quantityν 2 (σ̄) = min max σ(1 − σq(σ))2q∈P2M σ∈σ̄(10.2.17)

10.2. COMPLEXITY OF CLASSES SN (L, R)

169

is nothing but the squared best quality of the approximation, in the uniform (Tschebyshev’s)√
norm on the set σ̄, of the function σ by a linear combination of the 2M functions
φi (σ) = σ i+1/2 , i = 0, ..., 2M − 1
From Exercise 1.3.16 we know that√the quality of the best uniform on the segment [0, L] approximation of the function σ
by a linear combination of 2M functions φi (σ) is the same as the quality of the best uniform
approximation of the function by such a linear combination on certain (2M + 1)-point subset of[0, L].
We see that if 2M + 1 ≤ n, then2ε ≥ R2 sup ν 2 (σ̄) = R2 minσ̄

max σ(1 − σq(σ))2 .

q∈P2M σ∈[0,L]

The concluding quantity in this chain can be computed explicitly, see Exercise 12.6.5; it is equalto (4M + 1)−2 LR2 . Thus, we have established the implication
2M + 1 ≤ n ⇒ (4M + 1)−2 LR2 ≤ 2ε,which immediately results ins n − 1 1 LR2
1
M ≥ min
,
−
;
 2
4
2ε
4

this is nothing but the lower bound announced in (10.2.1).

10.2.3

Appendix: proof of Proposition 10.2.1

Let M be a method for solving problems from the family F(σ̄, µ̄), and let M be the complexityof the method at the family. Without loss of generality one may assume that M always performs
exactly N = M + 1 steps, and the result of the method as applied to any problem from thefamily always is the last, N -th point of the trajectory. Note that the statement in questionis evident if ε∗2M +1 (f ) = 0; thus, to prove the statement, it suffices to consider the case whenε∗2M +1 (f ) > 0. In this latter case the optimal solution A−1 b to the problem does not belong to
the Krylov space E2M +1 (f ) of the problem. On the other hand, the Krylov subspaces evidently
form an increasing sequence,
E1 (f ) ⊂ E2 (f ) ⊂ ...,and it is immediately seen that if one of these inclusions is equality, then all subsequent inclusions
also are equalities. Thus, the sequence of Krylov spaces is as follows: the subspaces in thesequence are strictly enclosed each into the next one, up to some moment, when the sequencestabilizes. The subspace at which the sequence stabilizes clearly is invariant for A, and since it
contains b and is invariant for A = AT , it contains A−1 b (why?). We know that in the case inquestion E2M +1 (f ) does not contain A−1 b, and therefore all inclusions
E1 (f ) ⊂ E2 (f ) ⊂ ... ⊂ E2M +1 (f )are strict. Since it is the case with f , it is also the case with every fU (since there is a rotation of
the space which maps Ei (f ) onto Ei (fU ) simultaneously for all i). With this preliminary observation in mind we can immediately derive the statement of the Proposition from the following

170

LECTURE 10. SMOOTH CONVEX MINIMIZATION PROBLEMS

Lemma 10.2.1 For any k ≤ N there is a problem fUk in the family such that the first k pointsof the trajectory of M as applied to the problem belong to the Krylov space E2k−1 (fUk ).
Proof is given by induction on k.
base k = 0 is trivial.
step k 7→ k + 1, k < N : let Uk be given by the inductive hypothesis, and let x1 , ..., xk+1 be
the first k + 1 points of the trajectory of M on fUk . By the inductive hypothesis, we know that
x1 , ..., xk belong to E2k−1 (fUk ). Besides this, the inclusions
E2k−1 (fUk ) ⊂ E2k (fUk ) ⊂ E2k+1 (fUk )are strict (this is given by our preliminary observation; note that k < N ). In particular, there
exists a rotation V of Rn which is identical on E2k (fUk ) and is such thatxk+1 ∈ V T E2k+1 (fUk ).
Let us set
Uk+1 = Uk V
and verify that this is the rotation required by the statement of the Lemma for the new value ofk. First of all, this rotation remains b invariant, since b ∈ E1 (fUk ) ⊂ E2k (fUk ) and V is identicalon the latter subspace and therefore maps b onto itself; since Uk also remains b invariant, sodoes Uk+1 . Thus, fUk+1 ∈ F(σ̄, µ̄), as required.
It remains to verify that the first k + 1 points of the trajectory of M of fUk+1 belong toE2(k+1)−1 (fUk+1 ) = E2k+1 (fUk+1 ). Since Uk+1 = Uk V , it is immediately seen thatEi (fUk+1 ) = V T Ei (fUk ), i = 1, 2, ...;
since V (and therefore V T = V −1 is identical on E2k (fUk ), it follows thatEi (fUk+1 ) = Ei (fUk ), i = 1, ..., 2k.

(10.2.18)

Besides this, from V y = y, y ∈ E2k (fUk ), it follows immediately that the functions fUk andfUk+1 , same as their gradients, coincide with each other on the subspace E2k−1 (fUk ). This pattersubspace contains the first k search points generated by M as applied to fUk , this is our inductive
hypothesis; thus, the functions fUk and fUk+1 are ”informationally indistinguishable” along thefirst k steps of the trajectory of the method as applied to one of the functions, and, consequently,
x1 , ..., xk+1 , which, by construction, are the first k + 1 points of the trajectory of M on fUk ,are as well the first k + 1 search points of the trajectory of M on fUk+1 . By construction, thefirst k of these points belong to E2k−1 (fUk ) = E2k−1 (fUk+1 ) (see (10.2.18)), while xk+1 belongs
to V T E2k+1 (fUk ) = E2k+1 (fUk+1 ), so that x1 , ..., xk+1 do belong to E2k+1 (fUk+1 ).

Lecture 11

Constrained smooth and strongly
convex problems
In this lecture we continue investigation of optimal methods for smooth large scale convex
problems; what we are interested in now are constrained problems.
In the nonsmooth case, as we know, it is easy to pass from methods for problems without
functional constraints to those with constraints. It is not the case with smooth convex programs;
we do not know a completely satisfactory way to extend the Nesterov method onto problems withsmooth functional constraints. This is why it I shall restrict myself with something intermediate,namely, with composite minimization.

11.1

Composite problem

Consider a problem as follows:
minimize f (x) = F (φ(x)) s.t. x ∈ G;

(11.1.1)

Here G is a closed convex subset in Rn which contains a given starting point x̄. Now, the inner
functionφ(x) = (φ1 (x), ..., φk (x)) : Rn → Rk
is a k-dimensional vector-function with convex continuously differentiable components possessing
Lipschitz continuous gradients:|φ0i (x) − φ0i (x0 )| ≤ Li |x − x0 | ∀x, x0 ,i = 1, ..., k. The outer function
F : Rk → R
is assumed to be Lipschitz continuous, convex and monotone (the latter means that F (u) ≥ F (u0 )whenever u ≥ u0 ). Under these assumptions f is a continuous convex function on Rn , so that
(11.1.1) is a convex programming program; we assume that the program is solvable, and denoteby R(f ) the distance from the x̄ to the optimal set of the problem.
When solving (11.1.1), we assume that the outer function F is given in advance, and the inner
function is observable via the standard first-order oracle, i.e., we may ask the oracle to compute,
171

172

LECTURE 11. CONSTRAINED SMOOTH AND STRONGLY CONVEX PROBLEMS

at any desired x ∈ Rn , the value φ(x) and the collection of the gradients φ0i (x), i = 1, ..., k, ofthe function.
Note that many convex programming problems can be written down in the generic form
(11.1.1). Let me list the pair of the most important examples.
Example 1. Minimization without functional constraints. Here k = 1 and F (u) ≡ u,so that (11.1.1) becomes a problem of minimizing a C1,1 -smooth convex objective over a given
convex set. It is a new problem for us, since to the moment we know only how to minimize asmooth convex function over the whole space.
Example 2. Minimax problems. Let
F (u) = max{u1 , ..., uk };with this outer function, (11.1.1) becomes the minimax problem - the problem of minimizing the
maximum of finitely many smooth convex functions over a convex set. The minimax problems
arise in many applications. In particular, in order to solve a system of smooth convex inequalities
φi (x) ≤ 0, i = 1, ..., k, one can minimize the residual function f (x) = maxi φi (x).
Let me stress the following. The objective f in a composite problem (11.1.1) should notnecessarily be smooth (look what happens in the minimax case). Nevertheless, I speak about
the composite problems in the part of the course which deals with smooth minimization, and weshall see that the complexity of composite problems with smooth inner functions is the same as
the complexity of unconstrained minimization of a smooth convex function. What is the reason
for this phenomenon? The answer is as follows: although f may be non-smooth, we know inadvance the source of nonsmoothness, i.e, we know that f is combined from smooth functions
and how it is combined. In other words, when solving a nonsmooth optimization problem of ageneral type, all our knowledge comes from the answers of a local oracle. In contrast to this,when solving a composite problem, we from the very beginning possess certain global knowledge
on the structure of the objective, namely, we know the outer function F , and only part of therequired information, that one on the inner function, comes from a local oracle.

11.2

Gradient mapping

Let me start with certain key construction - the construction of a gradient mapping associatedwith a composite problem. Let A > 0 and x ∈ Rn . Consider the function
fA,x (y) = F (φ(x) + φ0 (x)[y − x]) +A|y − x|2 .
2This is a strictly convex continuous function of y which tends to ∞ as |y| → ∞, and therefore
it attains its minimum on G at a unique point
T (A, x) = argmin fA,x (y).
y∈G

We shall say that A is appropriate for x, if
f (T (A, x)) ≤ fA,x (T (A, x)),(11.2.1)

11.2. GRADIENT MAPPING

173

and if it is the case, we call the vectorp(A, x) = A(x − T (A, x))
the A-gradient of f at x.
Let me present some examples which motivate the introduced notion.
Gradient mapping for a smooth convex function on the whole space. Let k = 1, F (u) ≡ u,
G = Rn , so that our composite problem is simply an unconstrained problem with C1,1 -smoothconvex objective. In this casefA,x (y) = f (x) + (f 0 (x))T (y − x) +A|y − x|2 ,2and, consequently,
T (A, x) = x −1 0
1 0
f (x), fA,x (T (A, x)) = f (x) −|f (x)|2 ;A
2A

as we remember from the previous lecture, this latter quantity is ≤ f (T (A, x)) whenever A ≥L(f ) (item A. of the proof of Theorem 10.2.2), so that all A ≥ L(f ) for sure are appropriate
for x, and for these A the vectorp(A, x) = A(x − T (A, x)) = f 0 (x)is the usual gradient of f .
Gradient mapping for a smooth convex function on a convex set. Let, same as above, k = 1
and F (u) ≡ u, but now let G be a proper convex subset of Rn . We still havefA,x (y) = f (x) + (f 0 (x))T (y − x) +A|y − x|2 ,2

or, which is the same,
1 0
A1|f (x)|2 ] + |y − x̄|2 , x̄ = x − f 0 (x).
2A
2AfA,x (y) = [f (x) −We see that the minimizer of fA,x over G is the projection onto G of the global minimizer x̄ ofthe function:1
T (A, x) = πG (x − f 0 (x)).
A
As we shall see in a while, here again all A ≥ L(f ) are appropriate for any x. Note that for a”simple” G (with an easily computed projection mapping) there is no difficulty in computing
an A-gradient.
Gradient mapping for the minimax composite function. Let
F (u) = max{u1 , ..., uk }.
HerefA,x (y) = max {φi (x) + (φ0i (x))T (y − x)} +1≤i≤kA|y − x|2 ,2

174

LECTURE 11. CONSTRAINED SMOOTH AND STRONGLY CONVEX PROBLEMS

and to compute T (A, x) means to solve the auxiliary optimization problem
T (A, x) = argmin fA,x (y);G

we already met auxiliary problems of this type in connection with the bundle scheme.
The main properties of the gradient mapping are given by the following
Proposition 11.2.1 (i) Let
A(f ) =F (u + tL(f )) − F (u), L(f ) = (L1 , ..., Lk ).
tu∈Rk ,t>0
supThen every A ≥ A(f ) is appropriate for any x ∈ Rn ;(ii) Let x ∈ Rn , let A be appropriate for x and let p(A, x) be the A-gradient of f at x. Thenf (y) ≥ f (T (A, x)) + pT (A, x)(y − x) +1
|p(A, x)|22A

(11.2.2)

for all y ∈ G.
Proof. (i) is immediate: as we remember from the previous lecture,φi (y) ≤ φi (x) + (φ0i (x))T (y − x) +Li|y − x|2 ,2and since F is monotone, we have1
F (φ(y)) ≤ F (φ(x) + φ0 (x)[y − x] + |y − x|2 L(f )) ≤2
A(f )|y − x|2 ≡ fA(f ),x (y)2(we have used the definition of A(f )). We see that if A ≥ A(f ), then
≤ F (φ(x) + φ0 (x)[y − x]) +f (y) ≤ fA,x (y)for all y and, in particular, for y = T (A, x), so that A is appropriate for x.
(ii): since fA,x (·) attains its minimum on G at the point z ≡ T (A, x), there is a subgradientp of the function at z such thatpT (y − z) ≥ 0, y ∈ G.
(11.2.3)
Now, since p is a subgradient of fA,x at z, we havep = (φ0 (x))T ζ + A(z − x) = (φ0 (x))T ζ − p(A, x)
for some ζ ∈ ∂F (u), u = φ(x) + φ0 (x)[z − x]. Since F is convex and monotone, one has
f (y) ≥ F (φ(x) + φ0 (x)[y − x]) ≥ F (u) + ζ T (φ(x) + φ0 (x)[y − x] − u) == F (u) + ζ T φ0 (x)[y − z] = F (u) + (y − z)T (p + p(A, x))(11.2.4)

11.3. NESTEROV’S METHOD FOR COMPOSITE PROBLEMS

175(the concluding inequality follows from (11.2.4)). Now, if y ∈ G, then, according to (11.2.3),(y − z)T p ≥ 0, and we come toy ∈ G ⇒ f (y) ≥ F (u) + (y − z)T p(A, x) = F (φ(x) + φ0 (x)[z − x]) + (y − z)T p(A, x) =A|z − x|2 + (y − x)T p(A, x) + (x − z)T p(A, x) =2
[note that z − x = A1 p(A, x) by definition of p(A, x)]= fA,x (z) −= fA,x (z) + (y − x)T p(A, x) −11|p(A, x)|2 + |p(A, x)|2 =
2A
A

1
1
|p(A, x)|2 ≥ f (T (A, x)) + (y − x)T p(A, x) +|p(A, x)|22A
2A(the concluding inequality follows from the fact that A is appropriate for x), as claimed.
= fA,x (z) + (y − x)T p(A, x) +11.3

Nesterov’s method for composite problems

Now let me describe the Nesterov method for composite problems. It is a straightforward
generalization of the basic version of the method; the only difference is that we replace the usual
gradients by A-gradients.
The method, as applied to problem (11.1.1), generates the sequence of search points xi ,i = 0, 1, ..., and the sequence of approximate solutions yi , i = 1, 2, ..., along with auxiliary
sequences of vectors qi , positive reals Ai and reals ti ≥ 1, according to the following rules:
Initialization: Set
x0 = y1 = x̄; q0 = 0; t0 = 1;choose as A0 an arbitrary positive initial estimate of A(f ).
i-th step, i ≥ 1: given xi−1 , yi , qi−1 , Ai−1 , ti−1 , act as follows:
1) set1
xi = yi −(yi − x̄ + qi−1 )ti−1(11.3.1)

and compute φ(xi ), φ0 (xi );2) Testing sequentially the values A = 2j Ai−1 , j = 0, 1, ..., find the first value of A which isappropriate for x = xi , i.e., for A to be tested compute T (A, xi ) and check whetherf (T (A, xi )) ≤ fA,xi (T (A, xi ));set Ai equal to the resulting value of A;
3) Setyi+1 = xi −qi = qi−1 +1p(Ai , xi ),Ai

ti−1
p(Ai , xi ),Aidefine ti as the larger root of the equationt2 − t = t2i−1(11.3.2)

176

LECTURE 11. CONSTRAINED SMOOTH AND STRONGLY CONVEX PROBLEMS

and loop.
Note that the method requires solving auxiliary problems of computing T (A, x), i.e., problems of the typeminimize F (φ(x) + φ0 (x)(y − x)) +A|y − x|2 s.t. y ∈ G;
2

sometimes these problems are very simple (e.g., when (11.1.1) is the problem of minimizing a
C1,1 -smooth convex function over a simple set), but this is not always the case. If (11.1.1) isa minimax problem, then the indicated auxiliary problems are of the same structure as thosearising in the bundle methods, with the only simplification that now the ”bundle” - the # oflinear components in the piecewise-linear term of the auxiliary objective - is of a once for ever
fixed cardinality k; if k is a small integer and G is a simple set, the auxiliary problems still
are not too difficult. In more general cases they may become too difficult, and this is the main
practical obstacle for implementation of the method.
The convergence properties of the method are given by the following
Theorem 11.3.1 Let problem (11.1.1) be solved by the aforementioned method. Then for any
N one hasb )R2 (f )2A(fb ) = max{2A(f ), A0 }.
f (yN +1 ) − min f ≤, A(f(11.3.3)
G(N + 1)2
Besides this, first N steps of the method require N evaluations of φ0 and no more than M = 2N +b )/A0 ) evaluations of φ (and solving no more than M auxiliary problems of computing
log2 (A(fT (A, x)).
Proof. The proof is given by word-by-word reproducing the proof of Theorem 10.2.2, withthe statements (i) and (ii) of Proposition 11.2.1 playing the role of the statements A, B of theoriginal proof. Of course, we should replace Li by Ai and f 0 (xi ) by p(Ai , xi ). To make thepresentation completely similar to that one given in the previous lecture, assume without lossof generality that the starting point x̄ of the method is the origin.
10 . In view of Proposition 11.2.1.(i), from A0 ≥ A(f ) it follows that all Ai are equal toA0 , since A0 for sure passes all tests (11.3.2). Further, if A0 < A(f ), then the quantities Ai
never become ≥ 2A(f ). Indeed, in the case in question the starting value A0 is ≤ A(f ). When
updating Ai−1 into Ai , we sequentially subject to the test (11.3.2) the quantities Ai−1 , 2Ai−1 ,4Ai−1 , ... and choose as Ai the first of these quantities which passes the test. It follows that if
we come to Ai ≥ 2A(f ), then the quantity Ai /2 ≥ A(f ) did not pass one of our tests, which is
impossible. Thus,b ), i = 1, 2, ...
Ai ≤ A(fNow, the total number of evaluations of φ0 in course of the first N steps clearly equals to
N , while the total number of evaluations of φ is N plus the total number, K, of tests (11.3.2)
performed during these steps. Among K tests there are N successful (when the tested value ofA satisfies (11.3.2)), and each of the remaining tests increases value of A· by the factor 2. Sinceb )/A0 ), so that
A· , as we know, is ≤ 2A(f ), the number of these remaining tests is ≤ log2 (A(fb )/A0 ), as
the total number of evaluations of φ in course of the first N steps is ≤ 2N + log2 (A(fclaimed.

11.3. NESTEROV’S METHOD FOR COMPOSITE PROBLEMS

177

20 . It remains to prove (11.3.3). Let x∗ be the minimizer of f on G of the norm R(f ), andlet
εi = f (yi ) − f (x∗ )be the absolute accuracy of i-th approximate solution generated by the method.
Applying (11.2.2) to y = x∗ , we come to inequality
0 ≥ εi+1 + (x∗ − xi )T p(Ai , xi ) +1|p(Ai , xi )|2 .
2Ai(11.3.4)

Now, relation (11.3.1) can be rewritten as
xi = (ti−1 − 1)(yi − xi ) − qi−1 ,(recall that x̄ = 0), and with this substitution (11.3.4) becomesT
0 ≥ εi+1 +(x∗ )T p(Ai , xi )+(ti−1 −1)(xi −yi )T p(Ai , xi )+qi−1
p(Ai , xi )+1|p(Ai , xi )|2 . (11.3.5)
2Ai

At the same time, (11.2.2) as applied to y = yi results in
0 ≥ f (yi+1 ) − f (yi ) + (yi − xi )T p(Ai , xi ) +1|p(Ai , xi )|2 ,2Aiwhich implies an upper bound on (yi − xi )T p(Ai , xi ), or, which is the same, a lower bound on
the quantity (xi − yi )T p(Ai , xi ), namely,(xi − yi )T p(Ai , xi ) ≥ εi+1 − εi +1|p(Ai , xi )|2 .
2Ai

Substituting this estimate into (11.3.4), we come toT0 ≥ ti−1 εi+1 − (ti−1 − 1)εi + (x∗ )T p(Ai , xi ) + qi−1
p(Ai , xi ) + ti−11|p(Ai , xi )|2 .
2Ai(11.3.6)

Multiplying this inequality by ti−1 /Ai , we come tot2i−1
t2 − ti−1
t2
T ti−1εi+1 − i−1
εi + (x∗ )T (ti−1 /Ai )p(Ai , xi ) + qi−1
p(Ai , xi ) + i−12 |p(Ai , xi )|2 .
Ai
Ai
Ai
2Ai
(11.3.7)
We have (ti−1 /Ai )p(Ai , xi ) = qi − qi−1 and t2i−1 − ti−1 = t2i−2 (here t−1 = 0); thus, (11.3.7) canbe rewritten as
0≥

0≥

t2i−1
t2
1
T
εi+1 − i−2 εi + (x∗ )T (qi − qi−1 ) + qi−1
(qi − qi−1 ) + |qi − qi−1 |2 =
Ai
Ai
2

t2
t2i−1
1
1
εi+1 − i−2 εi + (x∗ )T (qi − q i − 1) + |qi |2 − |qi−1 |2 .
Ai
Ai
2
2
Since the quantities Ai do not decrease with i and εi+1 is nonnegative, we shall only strengthen
the resulting inequality by replacing the coefficient t2i−1 /Ai at εi+1 by t2i−1 /Ai+1 ; thus, we cometo the inequality
=

0≥

t2i−1
t2
1
1
εi+1 − i−2 εi + (x∗ )T (qi − qi−1 ) + |qi |2 − |qi−1 |2 ;
Ai+1
Ai
2
2

178

LECTURE 11. CONSTRAINED SMOOTH AND STRONGLY CONVEX PROBLEMS

taking sum of these inequalities over i = 1, ..., N , we come tot2N −11
1
εN +1 ≤ −(x∗ )T qN − |qN |2 ≤ |x∗ |2 ≡ R2 (f )/2;
AN +1
2
2

(11.3.8)

thus,
εN +1 ≤

AN +1 R2 (f ).
2t2N −1(11.3.9)b ) and ti ≥ (i + 1)/2; therefore (11.3.9) implies (11.3.3).
As we know, AN +1 ≤ 2A(f11.4

Smooth strongly convex problems

In convex optimization, when speaking about a ”simple” nonlinear problem, traditionally it ismeant that the problem is unconstrained and the objective is smooth and strongly convex, sothat the problem is(f )minimize f (x) over x ∈ Rn ,and the objective f is continuously differentiable and such that for two positive constants l < L
and all x, x0 one has
l|x − x0 |2 ≤ (f 0 (x) − f 0 (x0 ))T (x − x0 ) ≤ L|x − x0 |2 .
It is a straightforward exercise in Calculus to demonstrate that this property is equivalent tolLf (x) + (f 0 (x))T h + |h|2 ≤ f (x + h) ≤ f (x) + (f 0 (x))T h + |h|22
2

(11.4.1)

for all x and h. A function satisfying the latter inequality is called (l, L)-strongly convex, andthe quantity
LQ=lis called the condition number of the function. If f is twice differentiable, then the property of(l, L)-strong convexity is equivalent to the relationlI ≤ f 00 (x) ≤ LI, ∀x,where the inequalities should be understood in the operator sense. Note that convexity + C1,1 smoothness of f is equivalent to (11.4.1) with l = 0 and some finite L (L is an upper bound for
the Lipschitz constant of the gradient of f ).
We already know how one can minimize efficiently C1,1 -smooth convex objectives; what
happens if we know in advance that the objective is strongly convex? The answer is given by
the following statement.
Theorem 11.4.1 Let 0 < l ≤ L be a pair of reals with Q = L/l ≥ 2, and let SC n (l, L) be thefamily of all problems (f ) with (l, L)-strongly convex objectives. Let us provide this family by the
relative accuracy measuref (x) − min fν(x, f ) =f (x̄) − min f

11.4. SMOOTH STRONGLY CONVEX PROBLEMS

179(here x̄ is a once for ever fixed starting point) and the standard first-order oracle. The complexityof the resulting class of problems satisfies the inequalities as follows:
O(1) min{n, Q1/2 ln(

1
2)} ≤ Compl(ε) ≤ O(1)Q1/2 ln( ),2εε(11.4.2)

where O(1) are positive absolute constants.
Today I shall not speak about the lower complexity bound - it will be obtained in the mean time
as a byproduct of our considerations related to quadratic problems. What we shall focus on is
the upper bound. This bound is dimension-independent and in the large scale case is sharp - itcoincides with the lower bound within an absolute constant factor. In fact we can say something
reasonable about the lower complexity bound in the case of a fixed dimension as well; namely,
if n ≥ Q1/2 , then, besides the lower bound (11.4.2) (which does not say anything reasonable forsmall ε), the complexity admits also the lower bound√Q
1
Compl(ε) ≥ O(1)
ln( )
ln Q
2ε
which is valid for all ε ∈ (0, 1).
Let me stress that what is important in the above complexity results is not the logarithmic dependence on the accuracy, but the dependence on the condition number Q. A linear
convergence on the family in question is not a great deal; it is possessed by more or less all
traditional methods. E.g., the Gradient Descent with constant step γ = L1 solves problems fromthe family SC n (l, L) within relative accuracy ε in O(1)Q ln(1/ε) steps. It does not mean that
the Gradient Descent as good as a method might be: in actual computations the factor ln(1/ε)
is a quite moderate integer, something less than 20, while you can easily meet with condition
number Q of order of thousands and tens of thousands. In order to compare a pair of methods
with efficiency estimates of the type A ln(1/ε) it is reasonable to look at the value of the factor
denoted by A (usually this value is a function of some characteristics of the problem, like its
dimension, condition number, etc.), since this factor, not log(1/ε), is responsible for the efficiency. And from this viewpoint the Gradient Descent is bad - its efficiency is too sensitive tothe condition number of the problem, it is proportional to Q rather than to Q1/2 , as it shouldbe according to our complexity bounds. Let me add that no traditional method, even among
the conjugate gradient and the quasi-Newton ones, is known to possess the ”proper” efficiency
estimate O(Q1/2 ln(1/ε)).
The upper complexity bound O(1)Q1/2 ln(2/ε) can be easily obtained by applying to the
problem the Nesterov method with properly chosen restarts. Namely, consider the problem(fG )minimize f (x) s.t. x ∈ G,

associated with an (l, L)-strongly convex objective and a closed convex subset G ⊂ Rn whichcontains our starting point x̄; thus, we are interesting in something more general than the simplest unconstrained problem (f ) and less general than the composite problem (11.1.1). Assume
that we are given in advance the parameters (l, L) of strong convexity of f (this assumption,although not too realistic, allows to avoid sophisticated technicalities). Given these parameters,
let us set
s
L
N =4
l

180

LECTURE 11. CONSTRAINED SMOOTH AND STRONGLY CONVEX PROBLEMS

and let us apply to (fG ) N steps of the Nesterov method, starting with the point y 0 ≡ x̄ andsetting A0 = L. Note that with this choice of A0 , as it is immediately seen from the proofof Theorem 11.3.1, we have Ai ≡ A0 and may therefore skip tests (11.3.2). Let y 1 be theapproximate solution to the problem found after N steps of the Nesterov method (in the initial
notation it was called yN +1 ). After y 1 is found, we restart the method, choosing y 1 as the newstarting point. After N steps of the restarted method we restart it again, with the starting pointy 2 being the result found so far, etc. Let us prove that this scheme with restarts ensures thatf (y i ) − min f ≤ 2−i [f (x̄) − min f ], i = 1, 2, ...
G

G(11.4.3)√
Since to pass from y i to y i+1 it requires N = O( Q) oracle calls, (11.4.3) implies the upperbound in (11.4.2).
Thus, we should prove (11.4.3). This is immediate. Indeed, let z ∈ G be a starting point,and z + be the result obtained after N steps of the Nesterov method started at z. From Theorem
11.3.1 it follows that
4LR2 (z)f (z + ) − min f ≤,(11.4.4)
G(N + 1)2where R(z) is the distance between z and the minimizer x∗ of f over G. On the other hand,from (11.4.1) it follows thatlf (z) ≥ f (x∗ ) + (z − x∗ )T f 0 (x∗ ) + R2 (z).
2

(11.4.5)

Since x∗ is the minimizer of f over G and z ∈ G, the quantity (z − x∗ )T f 0 (x∗ ) is nonnegative,
and we come tolR2 (G)f (z) − min f ≥G2This inequality, combined with (11.4.4), implies thatf (z + ) − minG f
8L
1≤≤f (z) − minG fl(N + 1)2
2
(the concluding inequality follows from the definition of N ), and (11.4.3) follows.

Lecture 12

Unconstrained quadratic
optimization
We have studied the complexity and efficient algorithms for nonsmooth and smooth convex
minimization. Of course, the presentation was uncomplete; e.g., we have passed from the generalcase, where the objectives might be simultaneously nonsmooth and degenerate - with ”flatgraphs” - directly to the case of ”very good” objectives, i.e., with Lipschitz continuous gradient,which is the strongest smoothness assumption; last lecture we spoke also about strongly convex
optimization, thus imposing on the objective simultaneously the strongest smoothness and thestrongest nondegeneracy assumptions. The classes of problems studied so far can be regarded
as extreme points in the world of nonlinear convex optimization; there are also ”interior points”
which are obtained by imposing on the objective some intermediate smoothness restrictions, like
Holder continuity of the gradient, and these restrictions can be combined with various hypotheseson the degree of nondegeneracy. For the variety of these ”intermediate” problem classes we also
know the complexity and the optimal methods, but I decided to restrict the course with the basic
results related to the ”extreme classes” of problems. This program is more or less completed;
the remaining topics are rather specific, but I think they are of definite interest.

12.1

Complexity of quadratic problems: motivation

Our today topic is unconstrained quadratic optimization, i.e., methods for minimizing convex
quadratic forms
f (x) = fA,b (x) = xT Ax − 2bT x : Rn → Rover the whole Rn . Everybody knows that above quadratic form is a convex function if and only
if the symmetric matrix A is positive semidefinite; such a convex function attains its minimum
on Rn if and only if it is below bounded, and if and only if the linear equation
Ax = b

(12.1.1)

is solvable; the solutions to this equations are exactly the minimizers of f . Thus, the problemof minimizing a quadratic form is nothing but a problem of solving a linear n × n equation witha symmetric positive semidefinite matrix. It is one of the standard problems of Linear Algebra,
181

182

LECTURE 12. UNCONSTRAINED QUADRATIC OPTIMIZATION

and you may, of course, ask why should we deal with this Linear Algebra problem in our course
devoted to convex minimization and what news in this more than traditional area can result
from our complexity approach. The answer is as follows. When considering convex optimization
problems, we normally spoke about the first-order methods, i.e., about iterative methods which
use the first-order information on the objective – its values and its first-order derivatives. In thecase of a quadratic objective these methods, consequently, need computing the matrix-vector
products Ax and do not require neither a direct access to the matrix A, nor updating this matrix.
In other words, a first-order method for quadratic minimization is an iterative procedure whichsolves a system of linear equations via matrix-vector multiplications, without direct access tothe matrix of the system. For these iterative methods, A is not an explicit part of the input; it
is given implicitly, via an oracle which, given on input a vector x, reports on output the productAx (as far as the right hand side vector b is concerned, it is reasonable to assume that it is an
explicit part of the input - it is given to the method in advance). Thus, what we are about to do
is to investigate the complexity of solving linear systems of equations (with symmetric positive
semidefinite matrices) by iterative methods which use only matrix-vector multiplications. Notethat most of the usual Linear Algebra routines, like Gauss elimination, Choleski decomposition,etc. - are not iterative, I mean, they need a direct access to the matrix and update it whensolving the system.
Now, you can ask: in practice one hardly can meet with a situation when a method indeed has
no access to the matrix of a linear system, so why it is reasonable to investigate the complexityof linear equations with respect to iterative methods of the indicated, rather specific, type?
The answer is as follows: direct methods of Linear Algebra, like Colossi decomposition, spend
something O(n3 ) arithmetic operations in order to solve the problem; this estimate is sharp forthe case of dense matrices, same as sparse, but ”unstructured” matrices; better estimates are
known only for sparse matrices of a good structure. And the situation is as follows: before you
pay O(n3 ) arithmetic operations, you have nothing - no approximate solution, bad or good, atall; after you have paid this cost, you get all - I mean, the exact solution (I am considering theidealized case of exact real arithmetic, just ignoring the rounding errors). In contrast to this,in an iterative method you have certain approximate solution at each step. Let K(ν) be thenumber of steps of the iterative method in question which is sufficient to solve all problems froma given family within a desired accuracy ν; for many interesting families of problems (say, forthose given by bounds on the condition number) this quantity is independent of the size n of
the problem. Whenever we deal with such a family of problems, we know in advance that in
the large-scale case it would be better to use the iterative method: to solve the problem within
the desired accuracy, it suffices to perform K(ν) matrix-vector multiplications, i.e., to perform
M K(ν) arithmetic operations, where M is the cost of a single matrix-vector multiplication. Evenfor dense matrices, M = n2 ; thus, the arithmetic cost of an ν-solution for the iterative method inquestion will be at most O(n2 K(ν)), and for a direct method O(n3 ). Thus, the iterative method
will be more efficient than the direct one whenever n >> K(ν). This advantage of iterative
methods becomes more visible if we deal with large-scale sparse matrices, since here M << n2 ;in the case of ”unstructured” sparsity the arithmetic cost of a direct method still is O(n3 ), andwe get an additional factor M/n2 in favour of the iterative method.
I believe that the above reasoning explains why investigating complexity of linear operator
equations with respect to iterative methods based on matrix-vector multiplications is of someinterest for Linear Algebra; and it is for sure interesting for convex optimization. Roughly12.2. FAMILIES OF SOURCE-REPRESENTABLE QUADRATIC PROBLEMS

183

speaking, it turns out that classes of smooth convex and strongly convex optimization problems
are as difficult as their subclasses formed by quadratic problems, and these subclasses are themain source of lower complexity bounds for related classes of non-quadratic problems.

12.2

Families of source-representable quadratic problems

Thus, we are going to investigate the complexity of linear operator equations with respect
to iterative methods based on matrix-vector multiplications. Let me start with the following
convention. If A is a symmetric positive semidefinite matrix and r is a nonnegative real, thenthe operator Ar is defined in the natural way an is a well-defined operator on the whole Rn . If
r < 0, we define Ar as the operator on the image space of A which is inverse to A−r . In otherwords, if ei , λi are the eigenvectors and the eigenvalues of A, then, for r ≥ 0, the operator Ar
has ei as eigenvectors with the eigenvalues λri (00 = 1). If r < 0, then Ar is defined on the linear
span of those ei which are associated with nonzero λi , and the eigenvalues of Ar associated withthese ei are λri .
Now let me describe the families of equations in question. Let Σ be a closed compact subset of
the nonnegative ray [0, ∞), τ be a real and R be a positive real. Consider the family Un (Σ, τ, R)
comprised of all equations
Ax = b
(12.2.1)
with a symmetric n × n positive semidefinite matrix A, such that(i) the spectrum of A belongs to Σ;(ii) the equation (12.2.1) is solvable, and its normal (i.e., of the minimal Euclidean norm)
solution x∗ (A, b) can be represented asx∗ (A, b) = Aτ u

(12.2.2)

with some u ∈ Rn such that |u| ≤ R.
Relation (12.2.2), by reasons coming from certain physical interpretations, is called the”source-representability” of the solution. The meaning of this assumption is very clear: thematrix A generates the scale of norms |Ar x| on Rn (more exactly, if A is degenerate and r > 0,
these are seminorms on Rn , not norms, and is A is degenerate and r < 0, these are norms not
on Rn , but on the image space of A); assumption (ii) says simply that the norm of the solution
in a fixed norm of this family should be bounded from above by a given quantity.
Let us provide the family with the accuracy measureνω (x; A, b) = |Aω (x − x∗ (A, b))|,if the right hand side is well-defined; otherwise νω (x; A, b) is +∞. Here ω is a real parameter.
Consider typical examples.
(I) The family Un (Σ, 0, R) is comprised of all solvable linear operator equation of a given sizewith the eigenvalues of the (symmetric) matrix of the equation belonging to a given compact
set and the solution of a norm not exceeding R;(Ia) The accuracy ν0 is exactly the distance between the approximate and the exact solutions.
(II) The family Un (Σ, −1, R) is comprised of all solvable linear operator equations of a given
size with the eigenvalues of the (symmetric) matrix of the equation belonging to Σ and the normof the right hand side vector not exceeding R

184

LECTURE 12. UNCONSTRAINED QUADRATIC OPTIMIZATION(indeed, to say that x∗ (A, b) = A−1 u with |u| ≤ R is exactly the same as to say that b = Ax∗ (A, b)
is of the norm ≤ R);(IIa) The accuracy ν1 is exactly the norm |Ax − b| of the residual Ax − b.
(III) The family Un (Σ, − 12 , R) is comprised of all solvable linear operator equations of a given
size with the eigenvalues of the (symmetric) matrix of the equation belonging to Σ and suchthat the initial residual in terms of the associated with the equation quadratic minimization
problem, i.e., the quantity
fA,b (0) − min fA,b (x),xis ≤ R2(indeed, we have fA,b (0) − minx fA,b (x) = (x∗ (A, b))T Ax∗ (A, b) = |A1/2 x∗ (A, b)|2 , so that to saythat x∗ (A, b) = A−1/2 u with |u| ≤ R is the same as to say that |A1/2 x∗ (A, b)| ≤ R, i.e., that(x∗ (A, b))T Ax∗ (A, b) ≤ R2 );
2 (x; A, b) is exactly the(IIIa) The accuracy ν1/2 is the norm of A1/2 (x − x∗ (A, b)), so that ν1/2
residual fA,b (x) − min fA,b in terms of the quadratic objective associated with the equation.
After we have defined the families of problems we are interested in and have equipped thefamilies with accuracy measures, we may pose the question of what is the complexity of a familywith respect to iterative methods and what are the associated optimal methods. We actuallyknow a significant part of the answer.

12.3

Lower complexity bounds

Recall that we already know the following important fact (Proposition 10.2.1, Lecture 10):
Let fA,b be a convex quadratic form on Rn and F be the family of rotations of this form, i.e.,
the family of all quadratic forms of the type fU T AU,U T b (x) ≡ f (U x) associated with orthogonallinear operators U which remain b invariant. Let also M be an iterative first order methodof complexity M for minimizing forms from F. Then there is a problem f = fA0 ,b0 ∈ F suchthat the result x̄ formed by the method M as applied to f belongs to the (2M + 1)-st Krylov
subspace
E2M +1 (A0 , b0 ) = Lin{b0 , A0 b0 , (A0 )2 b0 , ..., (A0 )2M b0 }of the problem fA0 ,b0 ; in other words,x̄ = p(A0 )b0

for certain polynomial p of a degree ≤ 2M .
Now note that a first-order method for minimizing quadratic forms is nothing but an iterative
method, based on matrix-vector multiplications, for the associated linear operator equations.
Consider a family of linear operator equations which is invariant with respect to rotations (sothat (U T AU, U T b) belongs to F whenever U is on orthogonal matrix and (A, b) ∈ F). The above
statement says that if M is an iterative method for solving linear operator equations from Fwith the complexity equal to M , then, for any problem (A, b) from the family, there exists a
rotation (A0 , b0 ) = (U T AU, U T b) of the problem such that the result x̄(A0 , b0 ) of M as appliedto (A0 , b0 ) is of the form p(A0 )b0 for certain polynomial p of a degree ≤ 2M . In particular, ifν(x; A, b) is the accuracy measure in question, then the error ν(M, (A0 , b0 )) of the method on
the problem (A0 , b0 ) is at least the quantityν ∗ (2M, A0 , b0 ) =minp:deg p≤2M

ν(p(A0 )b0 ; A0 , b0 ).

(12.3.1)

12.3. LOWER COMPLEXITY BOUNDS

185

Since F is invariant with respect to rotations, (A0 , b0 ) belongs to F whenever (A, b) does, so thatthe left hand side of (12.3.1) is less than or equal to the inaccuracy of the method on the wholefamily. If, in addition, the accuracy measure also is invariant with respect to rotations:ν(U T x; U T AU, U T b) = ν(x; A, b)
for any (A, b) ∈ F, any x and any rotation U , then the right hand side in (12.3.1) is the sameas if it were A0 = A, b0 = b. Thus, we come to the following
Proposition 12.3.1 Let F be a family of linear operator equations which is invariant withrespect to rotations and ν(x; A, b) be an accuracy measure for the family, which also is invariant
with respect to rotations. Then, for any method M, based on matrix-vector multiplications, ofcomplexity M on the family the worst-case accuracy of M on the family satisfies the inequality
as follows:ν(M) ≥ supmin ν(p(A)b; A, b).
(12.3.2)(A,b)∈F p:deg p≤2MThis proposition can be directly applied to the families Un (Σ, τ, R) and the accuracy measuresνω (x; A, b) which clearly possess the required invariance properties. Moreover, for these families
and accuracy measures, the right hand side in (12.3.2) can be computed more or less explicitly:
Proposition 12.3.2 Consider the case of F = U(Σ; τ, R), ν(x; A, b) = νω (x; A, b). Assume
that 0 is not an isolated point of Σ (i.e., either 0 6∈ Σ, or 0 is a limiting point of Σ); if 0 ∈ Σ,
assume, in addition, that ω + τ > 0.Finally assume that that 2M + 1 < n. Then the right handside in (12.3.2) equals toRν ∗ (2M ; Σ; τ + ω) ≡ R

minmax |t|τ +ω |1 − tp(t)|.

p:deg p≤2M t∈Σ

In particular, the νω -inaccuracy of any method M of the complexity M on the family Un (Σ, τ, R)
satisfies the lower boundν(M) ≥ Rν ∗ (2M, Σ; τ + ω).
(12.3.3)
Proof. Let (A, b) ∈ U(Σ, τ, R), and let x∗ (A, b) = Aτ u, |u| ≤ R, be the corresponding ”source
representation” of the normal solution to the problem. Of course, in this representation u canbe chosen from the image space of A:u=kXui ei ,

i=1

where k ≤ n, ei are certain mutually orthogonal unit eigenvectors of A with mutually disjoint
positive eigenvalues λi . Letµi = u2i ,so thatkXµi ≤ R 2 .

(12.3.4)

i=1

We have∗x (A, b) =

k
X
i=1

ui λτi ei , b = Ax∗ (A, b) =

k
X
i=1

ui λ1+τ
ei ,
i

186

LECTURE 12. UNCONSTRAINED QUADRATIC OPTIMIZATION

whence for a polynomial p one hasp(A)b =kXui p(λi )λ1+τ
ei
i

i=1

and
νω2 (p(A)b; A, b) = |Aω (p(A)b − x∗ (A, b))|2 =

k
X

+2ω
λ2τ
(1 − λi p(λi ))2 µi .
i(12.3.5)i=1

Since λi ∈ Σ, this relation combined with (12.3.4) implies thatνω2 (p(A)b; A, b) ≤ R2 max t2τ +2ω (1 − tp(t))2 ,t∈Σwhenceνω (p(A)b; A, b) ≤ R max tτ +ω |1 − tp(t)|.
t∈Σ(12.3.6)

Choosing as p the polynomial p∗2M of the degree ≤ 2M which solves the saddle point problem
S:

min

max tτ +ω |1 − tp(t)|p:deg p≤2M t∈Σ(the problem clearly is solvable), we make the right hand side of (12.3.6) equal to Rν ∗ (2M, Σ, τ +ω); therefore (12.3.6) implies that the right hand side of (12.3.2) is ≤ Rν ∗ (2M, Σ, τ + ω).
To prove the inverse inequality, let us act as follows. The quantity ν ∗ (2M, Σ, τ + ω), byconstruction, is the quality of the best, in the uniform on Σ norm, approximation of the function
f (t) = tτ +ω
by a linear combination of the 2M + 1 functions
fi (t) = ti+τ +ω , i = 1, ..., 2M + 1;
since τ + ω ≥ 0, all aforementioned functions are continuous on the (compact) set Σ. From
Exercise 1.3.16 we know that there is a finite subset Σ0 of Σ with at most 2M + 2 points suchthat it is as difficult to approximate f by a linear combination of fi on Σ0 as it is on the wholeΣ:ν ∗ (2M, Σ, τ + ω) =min max0 tτ +ω |1 − tp(t)|.
(12.3.7)
p:deg p≤2M t∈ΣLet Σ0 = {λ1 < λ2 < ... < λk }, with k ≤ 2M + 2; note that the cardinality k of the set Σ0 is≤ n in view of our premise. Note also that we may assume that λ1 > 0. Indeed, all λi belong toΣ, so that the inequality in question is evident if 0 6∈ Σ. If 0 ∈ Σ, then, in view of our premise,τ + ω > 0, so that the right hand side of (12.3.7) will remain unchanged if we delete from Σ0 thepoint 0 (if initially it belonged to the set).
Further, (12.3.7) can be rewritten as[ν ∗ (2M, Σ, τ + ω)]2 =min

max t2τ +2ω (1 − tp(t))2 ;p:deg p≤2M t∈Σ0

12.4. COMPLEXITY OF LINEAR OPERATOR EQUATIONS

187

due to the standard duality reasons, there exists a probability distribution of masses π1 , ..., πk
on Σ0 such that the right hand side in the latter equality is
k
X

min

p:deg p≤2M+2ωπi λ2τ
(1 − λi p(λi ))2 ,ii=1

so thatν ∗ (2M, Σ, τ + ω) =

k
X

min

p:deg p≤2M+2ωπi λ2τ
(1 − λi p(λi ))2 .
i(12.3.8)i=1

Now consider the following problem (A, b) ∈ Un (Σ, τ, R): the first k eigenvalues of A are λi ,i = 1, ..., k (the remaining, if any, are arbitrary points of Σ); the right hand side b is given by
b=R

k
X

√
πi ei ,
λ1+τ
i

i=1

where e1 , ..., ek are mutually orthogonal unit eigenvectors of A associated with the eigenvalues
λ1 , ..., λk . It is immediately seen that the problem does belong to the family and that for thisparticular problem the quantityminp:deg p≤2Mνω2 (p(A)b; A, b)

coincides with the right hand side of (12.3.8) (cf. (12.3.5)); thus, (12.3.8) implies that the righthand side of (12.3.2) is ≥ Rν ∗ (2M, Σ, τ + ω). The inverse inequality was already proved.

12.4

Complexity of linear operator equations

Proposition 12.3.2 gives us certain lower bound for the complexity of the classes U(Σ, τ, R);
these bounds, anyhow, are not in our usual form: they are lower bounds for inaccuracy of amethod of a given complexity, not bounds for the complexity of a method of a given inaccuracy.
Of course, there is no problem in reformulating the bounds in our standard form. To this end
it suffices to pass from the functionν ∗ (N, Σ, γ) =min

max tγ |1 − tp(t)|p:deg p≤N t∈Σto the inverse functionN ∗ (ν, Σ, γ) = min{N | ν ∗ (N, Σ, γ) ≤ ν}.
With this definition, Proposition 12.3.2 provides us with the lower bound in the following complexity result:
Theorem 12.4.1 Let us equip the family Un (Σ, τ, R) of linear operator equations with the accuracy measure νω (x; A, b). Assume that 0 is not an isolated point of Σ; in the case of 0 ∈ Σassume also that τ + ω > 0. Under these assumptions, the complexity of the family satisfies theinequalities as follows:min{n − 1 N ∗ (ν/R, Σ, τ + ω);
} ≤ Compl(ν) ≤ min{n; N ∗ (ν/R, Σ, τ + ω)}.
2
2

(12.4.1)

188

LECTURE 12. UNCONSTRAINED QUADRATIC OPTIMIZATION

Proof. The lower bound is an immediate reformulation of Proposition 12.3.2. Indeed, if M
is the complexity of a method M which solves all problems from the family within inaccuracy
ν, then, according to Proposition 12.3.2, either 2M + 1 ≥ n, or ν ∗ (2M, Σ, τ + ω) ≤ ν/R; in thelatter case 2M ≥ N ∗ (ν/R, Σ, τ + ω) by definition of this latter quantity, and we come to thelower bound in (12.4.1).
To prove the upper bound, let us act as follows. The complexity of any family of quadratic
problems is ≤ n, since in n matrix-vector multiplications one can identify the matrix of theequation and then solve the equation without further calls to the oracle. It remains to demonstrate that the complexity is also bounded from above by the quantity N = N ∗ (ν/R, Σ, τ + ω).
Indeed, by definition of this quantity there exists a polynomial p∗N of the degree N such thatmax tτ +ω |1 − tp∗N (t)| ≤ ν/R.
t∈ΣConsider the following Tschebyshev method for finding an approximate solution x̄(A, b) to a
problem (A, b):x̄(A, b) = p∗N (A)b.
To find x̄, it requires exactly N multiplications of vectors by A (look at the Horner scheme).
From inequality (12.3.6) it follows that the inaccuracy of the method on any problem instancefrom the family is ≤ ν, as claimed.
We have obtained tight (coinciding with each other within factor 2) upper and lower bounds
for the complexity of the families of linear operator equations in question, which is fine. As a
byproduct, we have pointed out optimal methods - the Tschebyshev ones. As these methods
appeared in our reasoning, they were tuned to the required accuracy; of course, this is not
so reasonable, and a reasonable version of the method generates the sequential approximate
solutions as
xk (A, b) = p∗k (A)b,where p∗k is the optimal solution to the optimization problem
find the polynomial of the degree k which minimizes the functional max tτ +ω |1 − tp(t)|,t∈Σwhich is the classical problem of the best uniform approximation.
Unfortunately, to use the Tschebyshev methods in practice is not so easy, not only because
one should solve in advance certain nontrivial problems of the best uniform approximation, butmainly because normally you have no data even to pose these problems: to this end, you should
know in advance the localizer Σ of the spectrum of the operator, same as the parameter τ of thefamily in question. Besides this, you should decide in advance what is the accuracy measure you
are interested in, since even for a fixed family of problems different accuracy measures resultin different Tschebyshev algorithms. Thus, Tschebyshev approach is not so attractive fromthe practical viewpoint; we shall see in the next lecture that there is a single method, namely
the Conjugate Gradient one, which behaves itself at any family in question, with respect toany of our accuracy measures, in the optimal way, so that to get an optimal method, there infact is no necessity to use the Tschebyshev approach. The actual meaning of our complexity
result, in contrast to our usual practice, is in indicating tight complexity bounds rather than
optimal methods. To the moment these bounds are represented in a rather sophisticated formof optimal values in certain problems of the best uniform approximation. In more or less all12.4. COMPLEXITY OF LINEAR OPERATOR EQUATIONS

189

interesting for applications cases these optimal values admit good explicit estimates (or even
explicit representation). Let us look at several typical cases.
Example 1: nondegenerate problems. Let Σ = [l, L] with certain positive l < L; theassumption that the spectrum of a symmetric positive semidefinite matrix A belongs to [l, L]means exactly that any quadratic form
fA,b (x) = xT Ax − 2bT x + c
is strongly convex with the parameters 2l, 2L. In order to evaluate the functionν ∗ (N, [l, L], γ) =min

max tγ |1 − tp(t)|p:deg p≤N l≤t≤L

it is reasonable to start with the case of γ = 0, where the function can be computed explicitly;
the optimal polynomial is given by
1 − tp∗N (t) = TN−1+1 (Q+12t − l − L
L)TN +1 (), Q = ,
Q−1
L−l
l

where
Tk (z) = cos(k arccos z)
are the Tschebyshev polynomials (I shall not prove here the statements of this type; this is the
subject of exercises accompanying the lecture). Further,
!N +1
√
Q−1
Q+1
1
√
≥ ν ∗ (N, [l, L], 0) = TN−1+1 (
)≥
Q−1
2
Q+1

!N +1
√
Q−1
√
.
Q+1

(12.4.2)

This is an explicit representation of ν ∗ (·, [l, L], γ) for the case of γ = 0; for the case of an arbitraryγ, one can use simple estimates
Lγ ν ∗ (N, [l, L], 0) ≥ ν ∗ (N, [l, L], γ) ≥ lγ ν ∗ (N, [l, L], 0), γ > 0;

(12.4.3)

for the case of negative γ, you need to invert the inequality signs in this chain. The estimates(12.4.3) identify ν ∗ within the factor Q|γ| .
The bounds on ν ∗ (·, [l, L], γ) immediately imply bounds on the inverse function N ∗ (·, γ)
involved in our complexity result; e.g., from (12.4.2) it follows thatc ln−1!
√Q+1R√
ln( ) ≤ N ∗ (ν/R, [l, L], 0) ≤ C ln−1
ν
Q−1

!
√
Q+1
R
√
ln( ), ν < R/2;
ν
Q−1

(12.4.4)

Consider, for example, the family of quadratic problems Un ([l, L], − 12 , R); this class is comprisedof the [2l, 2L]-strongly convex quadratic forms f (x) on Rn such thatf (0) − min f ≤ R2

(12.4.5)(see (III)). Let ω = 1/2, which, as we know from (IIIa), corresponds to the accuracy measureqf (x) − min f .

190

LECTURE 12. UNCONSTRAINED QUADRATIC OPTIMIZATION

From (12.4.1) and (12.4.4) it follows that the complexity of the family in question (associatedwith ω = 1/2) is, up to an absolute multiplicative constant, equal tomin{n, Q1/2 ln(R/ν)}(I assume that Q ≥ 2 and ν < R/2). On the other hand, consider the family SC n (2l, 2L) of
all, not necessary quadratic, strongly convex with parameters 2l, 2L unconstrained problems
on Rn (we have dealt with this family in the previous lecture). This family clearly contains
Un ([l, L], − 21 , R), and a method which solves all problems from SC n with certain relative accuracy√
ε for sure solves all problems from Un within absolute inaccuracy (measured by ν 1 (·)) R ε;2
consequently, the complexity of the method on Un (and therefore on the larger class SC n ) is at
leastO(1) min{n, Q1/2 ln(1/ε)};since this lower bound is valid for the complexity of an arbitrary method which solves the
problems from SC n (2l, 2L) within relative inaccuracy ε, we come to the lower boundO(1) min{n, Q1/2 ln(1/ε)}on the ε-complexity of the class SC n (2l, 2L) of strongly convex optimization problems; this is
the lower complexity bound announced in the previous lecture.
Example 2: degenerate case. Now let Σ = [0, L]; in other words, we deal with below
bounded convex quadratic forms with Lipschitz continuous, with constant 2L, gradient, but do
not impose on the forms any nondegeneracy assumptions, thus including into our considerations
ill-conditioned linear operator equations. Hereν ∗ (N, [0, L], γ) =min

max tγ |1 − tp(t)|.

p:deg p≤N 0≤t≤L

The exact values of ν ∗ are known for γ = 1/2; the optimal polynomial p is given byqq

1 − tp∗N (t) = (−1)N (2N + 3)−1 L/tT2N +3 ( t/L),
where Tk again are the Tschebyshev polynomials, and1ν ∗ (N, [0, L], ) = L1/2 (2N + 3)−1 .
2(12.4.6)

For positive γ not equal to 1/2 one can use the following estimates:
1 γ
L (1 + (N + 1)/γ)−2γ ≤ ν ∗ (N, [0, L], γ) ≤ Lγ (1 + 2(N + 1)/(2γ + 1))−2γ ,2(12.4.7)

so that the function N ∗ involved into the complexity bounds can be estimated as


cγ

RLγ
ν

1

2γ

RLγ
≤ N (ν, [0, L], γ) ≤ C(γ + 1)
ν
∗



1

2γ

(12.4.8)

with certain positive absolute constants c and C.
Consider a couple of instructive particular cases, where the questions and the answers canbe ”translated to the non-quadratic language”.

12.4. COMPLEXITY OF LINEAR OPERATOR EQUATIONS

191(A): τ = 0, ω = 1/2. The assumption that the convex quadratic form in question belongs
to Un ([0, L], 0, R) means exactly that the form possesses Lipschitz continuous with constant 2L
gradient and that we are interested in the accuracy measure which is the square root of theresidual in terms of the objective. The above relations result in the complexity√
R L
O(1) min{n,};ν
in terms of the residual ε in the objective rather than the square root of this residual ν, we cometo the already known to us ε−1/2 -complexity
LR2
O(1) min{n,
ε

!1/2

},

which leads to the ”proper” lower bound for the complexity of smooth non-quadratic convex
minimization; this bound already was established in Lecture 10 (by the way, via the same proof).
(B): τ = 0, ω = 1. We make the same assumptions on the quadratic form as in (A), but
measure the accuracy by the norm |Ax − b| of the residual in associated equation, i.e. (up to
factor 12 ) by the norm of the gradient of the form at an approximate solution. The complexityis
LR 1/2O(1) min{n,}.
ν
Note that the accuracy measure can be extended on the class of smooth non-quadratic convex
problems, and the corresponding complexity problem is as follows:
given a convex objective f with Lipschitz continuous, with constant L, gradient and suchthat the distance from the origin to the optimal set does not exceed R, what is the complexity,
N , of finding an approximate minimizer x of f such that |f 0 (x)| ≤ ν?
Our complexity bound for the quadratic case says thatN ≥ O(1) min{n,

LR
ν

1/2

};

this lower bound gives us a correct guess of what is N : it turns out thatLR
N ≤ O(1)
ν


1/2

LR
ln
.
ν




We could also consider the case of τ = − 12 (the assumption on the quadratic form f is that it
is convex and possesses Lipschitz continuous with constant 2L gradient and the initial residual
f (0) − min f is ≤ R2 for some given R) and ω = 1 (we measure the inaccuracy by the norm of
the gradient at an approximate solution); the associated complexity is√
R L},
O(1) min{n,ν
and this result, same as in the above cases, admits natural ”non-quadratic” extension.

192

LECTURE 12. UNCONSTRAINED QUADRATIC OPTIMIZATION

Example 3: ”completely continuous operators”.
Let Σ = {σi }∞i=1 ∪ {0} be a
monotonically decreasing converging to 0 sequence completed by the limit 0 of the sequence. It
is easily seen that in this caseγν ∗ (2N, Σ, γ) ≤ σN
(1 + 2(N + 1)/(2γ + 1))−2γ , γ > 0.

In particular, in the case of σi = O(i−α ) we come toN ∗ (ν, Σ, γ) ≤ const ν −1/(γ(2+α)) ;in the case of σi = ci−α this upper bound is sharp in order.

12.5

Ill-posed problems

Consider the case when 0 is a limiting point of Σ, i.e., the case of families containing illconditioned equations. Our complexity results say that there is certain nontrivial rate of convergence if τ + ω > 0, i.e., when the exact solution to the problem is bounded in the norm |A−τ x|which is ”stronger” than the norm |Aω x| in which we measure the inaccuracy. Here ”stronger”
means that the first norm dominates, with a constant factor depending on the norm of A andτ + ω only, the second one:|Aω x| ≤k A kτ +ω |A−τ x|, x ∈ Rn .
The fact that in the case of τ + ω > 0 we have certain nontrivial rate of convergence in the
problem of approximating the solution to an equation from Un (Σ, τ, R) in the norm |Aω x| meansthat the problem is well-posed, in spite of possible ill-conditioness of the matrix of the equation.
What happens if τ + σ ≤ 0? The answer is that here the problem is ill-posed; the complexity of
the family Un (Σ, τ, R) with respect to the accuracy measure νω (·) associated with ω ≤ −τ turns
out to be at least (n − 3)/3 for all ν < R. This means that in the case in question the iterative
methods have no advantages as compared to the direct methods of Linear Algebra: in orderto obtain a progress in accuracy, one should perform no less than O(n) steps, which basicallymeans that he is enforced to identify the matrix of the equation.

12.6

Exercises: Around quadratic forms

The goal of the forthcoming exercises is to prove the statements on the best uniform approximation which were used without proofs in Lecture 12. The problem we are interested in is asfollows:
We are given a nonnegative integer N , a compact subset Σ of the nonnegative ray and a real
γ. It is assumed that 0 is not an isolated point of Σ (it may be a limiting point of the set); inthe case of 0 ∈ Σ it is also assumed that γ > 0. The problem is to find the polynomial p∗ ofdegree ≤ N which minimizes the functionall(p) = max tγ |1 − tp(t)|t∈Σover the space PN of all polynomials of degrees ≤ N .

(P )12.6. EXERCISES: AROUND QUADRATIC FORMS

193

In what follows we always assume (not indicating it each time explicitly) that the cardinalityof Σ is ≥ N + 2, since otherwise the optimal value in (P ) is zero, and the optimal p simply
extrapolates the function 1/t from Σ on the whole axis.
Exercise 12.6.1 Prove that (P ) is solvable.
The following exercise gives the Tschebyshev type characterization of the optimal solution to(P ):Exercise 12.6.2 + Let p be a polynomial of a degree ≤ N .
q(t) = tγ (1 − tp(t)), δ = max |q(t)|.
t∈ΣProve that p is an optimal solution to (P ) if and only if p possesses the ”alternance property”:
there exist N + 2 points σ0 < σ1 < ... < σN +1 in Σ (the points of Tschebyshev’s alternance)such thatq(σi ) = (−1)i δ.
Exercise 12.6.3 Let p∗ be an optimal solution to (P ) (in fact one can write ”the” instead of”an”, since the optimal set in (P ) is a singleton). Prove that the polynomials(t) = 1 − tp(t)has exactly N + 1 positive roots t1 < t2 < ... < tN +1 (which separate the alternance points σi ,i = 0, ..., N + 1) and thats(t) =iY(1 − t/ti ).

i=1

The next exercise justifies the statements announced in Example 1, Lecture 12:
Exercise 12.6.4 ∗ Let Σ = [l, L], 0 < l < L; then the optimal solution to problem (P ) associatedwith γ = 0 is given by the polynomialp∗N (t) = TN−1+1 (L+l
2t − l − L)TN +1 (
),
L−l
L−l

and the optimal value isTN−1+1 (

L+l
),
L−l

Tk being the Tschebyshev polynomial of the degree k. Derive from this the estimates (161) and(162) announced in Lecture 12.
Note that problem (P ) associated with Σ = [l, L] and γ = 0 is in fact the following problem:(P 0 ) : given a segment ∆ on the axis and a point t̄ outside the segment, find among thepolynomials of a given degree N +1 with the absolute value on ∆ not exceeding 1 the polynomialwith the largest possible absolute value at t̄
(in the initial formulation, ∆ = [l, L] and t̄ = 0; instead of looking for the largest at t̄ polynomial
among those with the uniform norm 1 on ∆, we were looking for the smallest in the uniform

194

LECTURE 12. UNCONSTRAINED QUADRATIC OPTIMIZATION

norm on ∆ polynomial among the polynomials with the value 1 at t̄; of course, these are two
equivalent formulations of the same problem).
When solving (P 0 ), we can restrict ourselves with ∆ = [−1, 1]; Exercise 12.6.4 says thatthe solution to this latter version of (P 0 ) is given by the Tschebyshev polynomial of the degree
given in the formulation of the problem (Markov’s Theorem). Surprisingly, the answer does notdepend on the location of the point t̄ ! This is a rare case when the optimal solution to a natural
extremal problem involving parameter is independent of the value of the parameter.
Exercise 12.6.5 ∗ Let Σ = [0, L] with some L > 0, and let γ = 1/2. Prove that the solution to(P ) is given byqq
1 − tp∗N (t) = (−1)N (2N + 3) L/tT2N +3 ( t/L),

Tk (t) being the Tschebyshev polynomial of the degree k.
Exercise 12.6.6 ∗ Prove the upper bound on ν ∗ (·) given in Example 3 of Lecture 12.

Lecture 13

Optimality of the Conjugate
Gradient method
In the previous lecture we have established tight complexity bounds for the families of linear
operator equations Un (Σ, τ, R). Recall that a family Un (Σ, τ, R) is comprised of all linear systems
Ax = bwith a given number n of variables and equations, such thatthe spectrum of the symmetric positive semidefinite matrix A belongs to a given compact
set Σ ⊂ R+ ;equation Ax = b is solvable,
andthe normal (i.e., of the minimal Euclidean norm) solution x∗ (A, b) to the equation admits”source representation”x∗ (A, b) = Aτ u
with some u such that |u| ≤ R.
We have equipped the aforementioned family of problems with the accuracy measureνω (x; A, b) = |Aω (x − x∗ (A, b))|and have proved that the corresponding complexity of the family (taken with respect to themethods based on matrix-vector multiplications) satisfies the boundsn − 1 N ∗ (ν/R, Σ, ω + τ )
min
,
2
2




≤ Compl(ν) ≤ min {n, N ∗ (ν/R, Σ, ω + τ )} ,(13.0.1)

where function N ∗ (ε, Σ, γ) is defined as the smallest integer N such that there exists a polynomial
p(t) of degree N with the property
max tγ |1 − tp(t)| ≤ ε.
t∈ΣHere and in what follows we assume that 0 is not an isolated point of Σ, and that τ + ω > 0 in
the case when 0 ∈ Σ.
195

196

LECTURE 13. OPTIMALITY OF THE CONJUGATE GRADIENT METHOD

We have seen that upper complexity bound in (13.0.1) is associated with the Tschebyshev
methodxk = p∗k (A)bwhere p∗k are the optimal solutions to the following best unform approximation problems
Pk : find polynomial p of degree ≤ k which minimizes the deviation maxt∈Σ tω+τ |1 − tp(t)|.
As it was explained in the previous lecture, Tschebyshev’s methods, in spite of their optimality, are not too attractive from the computational viewpoint, since to specify the corresponding
polynomials p∗k , one needs a lot of a priori information on the problem: he should know the
localizer Σ of the spectrum of A, the parameters τ , R of source representation of the solution,
etc.
The goal of this lecture is to demonstrate that there is a single and well-known method - theConjugate Gradient - which behaves itself in the optimal way on every of the classes of linear
operator equations with source-representable solutions with respect to every inaccuracy measure
νω .

13.1

The Conjugate Gradient method

Let me recall to you what is the Conjugate Gradient method (CGM). As applied to linear
operator equation
Ax = bwith symmetric positive semidefinite matrix A, the method generates the sequence of iterates
xk = xk (A, b) ≡ pk (A)b, k = 1, 2, ...
in such a way that xk minimizes the quadratic potential
f (x) ≡ fA,b (x) = xT Ax − 2bT xof the equation over the (k − 1)st Krylov subspace of the pair (A, b):xk ∈ Argmin fA,b (x).

(13.1.1)x∈Ek−1 (A,b)

Recall that the Krylov subspaces of the pair (A, b) are given by
Ei (A, b) = Lin{b, Ab, ..., Ai b} = {p(A)b | deg p ≤ i}.
The aforementioned description of the method says what the method does, but does not say
how CGM does it. The standard description of CGM as an iterative routine is given by therecurrence
pk = rk −rkT qk−1
pk−1
pTk−1 qk−1

(13.1.2)

qk = Ark −rkT qk−1
qk−1
pTk−1 qk−1

(13.1.3)

13.2. MAIN RESULT

197
xk+1 = xk −rkT pkpk
pTk qk

(13.1.4)

rk+1 = rk −rkT pkqk
pTk qk

(13.1.5)(13.1.6)

initialized asx0 = p0 = q0 = 0; r0 = −b.
In (13.1.2), fractions with zero denominators by definition are zeros.
Note that there are many equivalent descriptions of the CGM; I choose that one whichstresses that at a step of the method we perform exactly one matrix-vector multiplication andadd to this multiplication O(n) arithmetic operations to update our iterates xi , ri = Axi − b, pi
and qi = Api .
Of course, from the presented iterative description of CGM it is not seen why the iterate xk
is a minimizer of the quadratic potential fA,b (·) over the Krylov subspace Ek−1 (A, b); this is atheorem which can be found in any textbook where the CGM is presented. In what follows wenever use the presentation of the CGM as an iterative routine; I mentioned it only to demonstratethat CGM indeed is a method based on matrix-vector multiplications, and that the arithmetic
cost of a step in the method, modulo the matrix-vector multiplication, is O(n).
Let me add that the Conjugate Gradient Method is in fact finite: it finds a solution to any
solvable linear operator equation with n variables in no more than n steps. This property of themethod also is proved in any Linear Algebra textbook; in fact it is a more or less immediate
consequence of the ”non-iterative” description of the method, see exercises accompanying thislecture.

13.2

Main result

My goal is to establish the following
Theorem 13.2.1 Consider the family of problems Un (Σ, τ, R) equipped with the accuracy measure νω (x; A, b). Assume, as usual, that 0 is not an isolated point of Σ, and, besides this, that0<ω+τ(13.2.1)

ω ≤ 1/2.

(13.2.2)

andUnder these assumptions, the number of steps NCG (ν) in which the Conjugate Gradient Method
solves any problem instance from the family within accuracy ν satisfies the estimateNCG (ν) ≤ min n;

1 + 2τ
[N ∗ (ν/R, Σ, ω + τ ) + 1] .
2(ω + τ )


(13.2.3)

Let me make some comments.
1) In view of complexity bounds (13.0.1) the theorem demonstrates that the complexity of
1+2τ
the CGM, up to the factor 2c 2(ω+τ) b, coincides with the complexity of the class of problems in

198

LECTURE 13. OPTIMALITY OF THE CONJUGATE GRADIENT METHOD

question. This is a very remarkable property, since to run the method, you need no informationon the particular class the problem belongs to. We see that the CGM possesses very strong”adaptation abilities”.
2) The theorem states that the CGM demonstrates the aforementioned ”adaptive abilities”
in certain restricted range of values of τ and ω. The restrictions defining the ”admissible” valuesof these parameters are formed by the inequality
ω+τ >0and the inequalityω ≤ 1/2.
The first of these inequalities is crucial, provided that 0 ∈ Σ; in this latter case this is our standard ”well-posedness” assumption commented in the previous lecture. In the ”non-degenerate”
case 0 6∈ Σ the restriction ω + τ > 0 can in fact be eliminated at the cost of slight modificationof the conclusion.
As far as the restriction ω ≤ 1/2 is concerned, it relates to the particular version of the Conjugate
Gradient we deal with. When passing from the CGM to a slightly more general method CGMm
which, as applied to a problem instance (A, b), generates the sequence of approximate solutions
xk according to the rulexk ∈ Argmin {xT Am x − 2bT Am−1 x}x∈Ek−1 (A,b)(m > 0 is integer parameter of the method), one has a complexity bound completely similarto (13.2.3), but now in the larger range of values of ω and τ , namely, that one given by the
relationsτ + ω > 0, ω ≤ m/2.

13.3

Proof of the main result

13.3.1

CGM and orthogonal polynomials

From the initial - ”non-iterative” - description of CGM it follows that the method as applied toa problem instance (A, b), generates the sequence of iterates
xk = pk (A)b,
pk being certain polynomial of degree not exceeding k − 1. Let us realize what are the polynomials. Assume that (A, b) ∈ Un (Σ, τ, R), and letx∗ = Aτ u, |u| ≤ R,

(13.3.1)

be the source representation of the normal solution to the equation. Since A is symmetric, its
image space is orthogonal to the kernel of A, and when we project a point x onto the image
space of A, the vector Ax remains unchanged, since the projection varies only the component ofx along the kernel of A. Thus, the projection of a solution to the equation onto the image spaceof A also is a solution; in particular, the projection of the normal solution x∗ onto the image

13.3. PROOF OF THE MAIN RESULT

199

space of A is a solution; since the projection decreases the norm of vector whenever it varies
it and x∗ is the solution of the minimal norm, we conclude that the normal solution belongs
to the image space of A. Now, looking at the source representation (13.3.1) of x∗ , we see that
we always can replace in this representation u by its projection u0 onto the image space of A
(indeed, if τ = 0, then u = x∗ coincides with its projection onto the image space of A, and ifτ 6= 0, then, by definition of the operator Aτ , Aτ u = Aτ u0 , and of course |u0 | ≤ |u| ≤ R). Thus,
we may assume that u belongs to the image space of A, so thatu=mXui ei ,

(13.3.2)

i=1

where ei , i = 1, ..., m, are mutually orthogonal eigenvectors of A with positive eigenvalues λi .
Consequently,x ∗ = Aτ u =mX

λτi ui ei ,(13.3.3)

λ1+τ
ui ei
i

(13.3.4)p(λi )ui ei ,λ1+τi

(13.3.5)λ1+τ
λi p(λi )ui ei .
i

(13.3.6)

b = Ax∗ =

i=1
m
X
i=1

and for any polynomial pp(A)b =
Ap(A)b =

m
X
i=1
m
X
i=1

Now, from the initial description of CGM it follows that kth approximate solution found by
the method is the minimizer of fA,b (x) over x = p(A)b, where p runs over the space Pk−1 ofpolynomials of degree ≤ k − 1. Let us look what is fA,b (p(A)b). We have
fA,b (p(A)b) = [p(A)b]T [Ap(A)b] − 2bT [p(A)b] =[see (13.3.4), (13.3.5) and (13.3.6) and take into account that {ei } form an orthonormal system]
=

m h
X

i

λi3+2τ p2 (λi ) − 2λ2+2τp(λi ) u2i =
i

i=1

=

m
X

λ1+2τ
(1 − λi p(λi ))2 u2i −
i

i=1

m
X

λ1+2τ
u2i .
i

i=1

The second sum here is independent of p; as it is immediately seen, this sum, together with thesign -, is nothing but the optimal value of fA,b :f ∗ ≡ min fA,b (x) = −
x

m
X

λ1+2τ
u2i .
i

i=1

Thus, we come to the relation
dA,b (p(A)b) ≡ fA,b (p(A)b) − min fA,b (x) =
x

m
X
i=1

λ1+2τ
(1 − λi p(λi ))2 u2i .
i(13.3.7)

200

LECTURE 13. OPTIMALITY OF THE CONJUGATE GRADIENT METHOD

Note that dA,b also is a quadratic potential of our equation given by the additional requirementthat the minimal value of the potential is 0.
To simplify notation, let us denote by µ the discrete measure on Σ which is concentrated on2
the set {λi }mi=1 and relates to the point λi the mass ui ; then (13.3.7) can be rewritten as
Z

dA,b (p(A)b) =t1+2τ (1 − tp(t))2 dµ(t).

(13.3.8)

ΣNote that the measure µ(Σ) of the whole Σ is nothing but i u2i = |u|2 ≤ R2 .
Now, as we remember, the iterates xk = pk (A)b of the CGM correspond to the polynomials
pk which minimize the potential fA,b , or, which is the same, the potential dA,b over all x’s of theform p(A)b, p being a polynomial of degree < k. Combining this description of the method and(13.3.8), we come to the conclusion as follows:P

Proposition 13.3.1 Assume that a problem (A, b) ∈ Un (Σ, τ, R) is solved by the Conjugate
Gradient Method, and let
xk = pk (A)bbe the iterates generated by the method. Then there exists a measure µ concentrated on a finite
subset of the set Σ\{0} such thatZdµ(t) ≤ R2

(13.3.9)Σand pk , for every k, is an optimal solution to the following quadratic problemZminimizeΣ

t1+2τ (1 − tp(t))2 dµ(t) s.t. p ∈ Pk−1 ,(13.3.10)

Pk−1 being the space of all polynomials of degree < k.
As an immediate consequence of this observation, we come to
Corollary 13.3.1 The polynomials
qk (t) = 1 − tpk (t)for every k are optimal solutions to the quadratic optimization problem
Qk : minimizeZΣ

t1+2τ q 2 (t)dµ(t) s.t. q ∈ Pk0 ,(13.3.11)

Pk0 being the space of all polynomials of degree ≤ k equal to 1 at the point t = 0.
The polynomials qk satisfy the orthogonality conditionsZΣ

t2+2τ p(t)qk (t)dµ(t) = 0, deg p < k.

(13.3.12)

The proof is immediate: to say that p(t) ∈ Pk−1 is the same as to say that q(t) = 1 − tp(t)is in Pk0 ; (13.3.12) is exactly the optimality condition for the quadratic optimization problem(13.3.10).
Corollary claims that {qk } is the system of orthogonal polynomials with respect to themeasure dζ(t) = t2+2τ dµ(t) on Σ; these polynomials are normalized by the conditions
deg qk ≤ k; qk (0) = 1.

13.3. PROOF OF THE MAIN RESULT

13.3.2

201

Expression for inaccuracy

After we have found certain ”functional-analytic” representation of the trajectory of the CGM,
let us find similar representation for the inaccuracyνω (k) ≡ νω (pk (A)b; A, b)of k-th approximate solution found by the method. We have
x∗ =

k
X

λτi ui ei ,

i=1
k
X

pk (A)b =λ1+τ
pk (λi )ui ei ,ii=1

whenceAω (xk − x∗ ) = Aω [pk (A)b − x∗ ] =mX

λωi λτi (λi pk (λi ) − 1)ui ei

i=1

and
νω2 (k) ≡ |Aω (xk − x∗ )|2 =m
X(1 − λi pk (λi ))2 u2i ,λ2ω+2τ
ii=1

or, which is the same,νω2 (k) =13.3.3ZΣt2(ω+τ ) qk2 (t)dµ(t).

(13.3.13)

Momentum inequality

We find ourselves in the situation as follows. We know something about the polynomials qk ,namely, we know that these polynomials are optimal solutions to the quadratic optimization
problems Qk ; and our goal is to derive from this observation something reasonable about thequantities νω (k) given by (13.3.13). To this end we intend to use the following general statement
Proposition 13.3.2 [The Momentum Inequality] Let ξ be a measure with the compact support
set belonging to the positive ray. Then the function
Z

Momξ (s) =ts dξ(t)is logarithmically convex in s: for any triple s ≤ s0 ≤ s00 with s00 > s one hass00 −s0s0 −sMomξ (s0 ) ≤ [Momξ (s)] s00 −s [Momξ (s00 )] s00 −s .

(13.3.14)

This proposition is nothing but a reformulation of the Holder inequality (see the exercises accompanying this lecture).
Let us fix k and take as ξ the measuredξ(t) = qk2 (t)dµ(t);202

LECTURE 13. OPTIMALITY OF THE CONJUGATE GRADIENT METHOD

since µ, in view of its origin, is a discrete measure concentrated on a finite subset of the positive
ray, ξ fits the requirements of the Momentum Inequality. Now let us set
s = 0, s0 = 2(ω + τ ), s00 = 1 + 2τ ;under the premise of the theorem we are proving, we have0 < 2(ω + τ ) ≤ 1 + 2τ(13.3.15)(see (13.2.1) and (13.2.2)), so that these quantities satisfy the requirements s ≤ s0 ≤ s00 , s00 > s.
With this choice of s, s0 , s00 we haveZ

Momξ (s) ≡ Momξ (0) =Momξ (s0 ) ≡ Momξ (2(ω + τ )) =ZΣΣ

qk2 (t)dµ(t);t2(ω+τ ) qk2 (t)dµ(t) ≡ νω2 (k)(13.3.16)(13.3.17)(see (13.3.13)) andMomξ (s00 ) ≡ Momξ (1 + 2τ ) =ZΣt1+2τ qk2 (t)dµ(t).

(13.3.18)

Applying the Momentum Inequality, we come to certain upper bound on the quantity Momξ (2(ω+τ )) we are interested in:νω2 (k) = Momξ (2(τ + ω)) ≤ [Momξ (0)]α [Momξ (1 + 2τ )]β == [Momξ (0)]α

Z
Σ

β

t1+2τ qk2 (t)dµ(t)2(ω + τ )1 − 2ω
, β =1−α=.
1 + 2τ
1 + 2τ
Our further actions will be as follows. Frits, we shall prove thatα=Momξ (0) ≤

Z

dµ(t)

(≤ R2 ),,
(13.3.19)(13.3.20)

Σ

which is the central point of the whole reasoning. Second, we shall establish certain good upperbound on Momξ (1 + 2τ ), systematically using the fact that this quantity, by construction of theCGM, is the optimal value in problem Qk (see Corollary 13.3.1).

13.3.4

Proof of (13.3.20)

It is possible that qk = 0 at the support set of µ; in this case the left hand side of (13.3.20), inview of (13.3.16), is 0, and (13.3.20) is, of course, valid. Now assume that qk is not identicallyzero at the support set of the measure µ. Then this set (which, by construction, is a finite subsetof the positive ray) is comprised of at least k + 1 points with positive µ-masses (if the numberof these points would be less than k + 1, we could find a polynomial q ∈ Pk0 vanishing at thesepoints and the optimal value in Qk would be zero, and we just have assumed that it is not the13.3. PROOF OF THE MAIN RESULT

203case). Corollary 13.3.1 says that q1 , ..., qk form an orthogonal system of polynomials of degrees
not exceeding 1, 2, ..., k, respectively, in the space L2 (ζ),dζ(t) = t2+2τ dµ(t).
In the case in question the support set of the measure ζ (which coincides with the support setof the measure µ) is comprised of at least k + 1 points, and from the standard properties oforthogonal polynomials (see exercises accompanying this lecture) it follows that qk is a polynomial of the degree exactly k with all roots t1 , ..., tk being positive and distinct. Without loss ofgenerality we may assume that t1 < ... < tk . Since qk (0) = 1, we haveqk (t) =kY(1 − t/ti ).

i=1

Now letp(t) =kY(1 − t/ti ) = q(t)(1 − t/t1 )−1 .

i=2

In view of the orthogonality relation (13.3.12) we haveZΣt2+2τ p(t)qk (t)dµ(t) = 0,

which, in view of
qk (t) = p(t)(1 − t/t1 ),can be rewritten as
Z t1

Z ∞

t2+2τ p2 (t)(1 − t/t1 )dµ(t) =0

t2+2τ p2 (t)(t/t1 − 1)dµ(t).

(13.3.21)t1

We haveZ ∞Momξ (0) =0

qk2 (t)dµ(t) =Z ∞p2 (t)(1 − t/t1 )2 dµ(t) =

0

Z ∞

Z t1

p2 (t)(1 − t/t1 )2 dµ(t) +p2 (t)(1 − t/t1 )2 dµ(t) ≤0
t1
t2+2τ [t/t1 − 1] whenever t ≥ t1 in view of 1 + 2τ > 0, see (13.3.15)][since (t/t1 − 1)2 ≤ t−2−2τ
1
Z t1
Z ∞
≤
p2 (t)(1 − t/t1 )2 dµ(t) + t1−2−2τ
t2+2τ p2 (t)(t/t1 − 1)dµ(t) =0t1=[see (13.3.21)]
Z t1

=
0

h

i

p2 (t)(1 − t/t1 ) (1 − t/t1 ) + t2+2τ t−2−2τdµ(t) ≤
1[since 0 ≤ 1 − t/t1 ≤ 1 and 0 ≤ 1 − t/t1 + t2+2τ t−2−2τ≤ 1, 0 ≤ t ≤ t1 ; recall that 1 + 2τ > 0]
1
≤

Z t1

p2 (t)dµ(t) ≤0[since 0 ≤ p(t) ≤ 1 as 0 ≤ t ≤ t1 by construction of p(t)]≤Z t1dµ(t),
0

as required in (13.3.20).

204

13.3.5LECTURE 13. OPTIMALITY OF THE CONJUGATE GRADIENT METHOD

Concluding the proof of Theorem 13.2.1

Now we are enough equipped to proof Theorem 13.2.1. Let us fix ν > 0. We should prove that
the CGM with
1 + 2τ
k = min{n; κ[N ∗ (ν/R, Σ, τ + ω) + 1]}, κ =cb,
(13.3.22)2(ω + τ )steps solves every problem (A, b) from the family Un (Σ, τ, R) within inaccuracy ν. Without lossof generality we may assume that the minimum in the right hand side of (13.3.22) corresponds
to the second quantity in the brackets, since otherwise the statement in question is evident: theConjugate Gradient Method with n steps finds, as we know, exact solution to any solvable linear
equation. Thus, in what follows we may assume that k given by (13.3.22) is nothing but
k = κ(N + 1), N = N ∗ (ν/R, Σ, ω + τ ).

(13.3.23)

Let (A, b) be a problem from Un (Σ, τ, R) and {xs } be the trajectory of the CGM as applied tothis problem and µ(·) be the corresponding measure given by Proposition 13.3.1. In view of(13.3.19) and (13.3.20) we haveνω2 (xs ; A, b) ≤ R2α

β

Z
Σ

t1+2τ qk2 (t)dµ(t), β =1−α=2(ω + τ ).
1 + 2τ

Thus, to prove the statement in question it suffices to demonstrate that for k given by (13.3.23)
one has
R

2α

Z
Σ

(

β

t1+2τ qk2 (t)dµ(t)

≡R

2α

Z

min

q∈Pk0 Σ

)β
1+τ 2

t

q (t)dµ(t)≤ ν2(13.3.24)(the equivalence in this relation follows from Corollary 13.3.1).
The proof of (13.3.24) is as follows.
By definition of N there exists a polynomial q∗ ∈ PN0 +1 such thatν/R ≥ ε ≡ min max tτ +ω |q(t)| = max tτ +ω |q∗ (t)|.
0
t∈Σ
q∈PN
+1

t∈Σ

(13.3.25)

It is possible, first, that q∗ = 0 µ-almost everywhere, or, which is the same, that ε = 0. Sinceq∗ ∈ PN0 +1 ⊂ Pk0 , the left hand side in (13.3.24) in the case in question is 0, and (13.3.24) is, ofcourse, true.
Now assume that ε > 0. From the standard Tschebyshev-type results (see Exercise 12.6.2)
it follows that there exist N + 2 pointsρ ≡ r0 < r1 < ... < rN +1
in Σ - the points of Tschebyshev alternance - such thatriτ +ω q∗ (ri ) = (−1)i ε.

(13.3.26)

From this ”oscillation” property it follows, in turn, that q∗ has N + 1 distinct positive roots
t1 < ... < tN +1 ,

13.3. PROOF OF THE MAIN RESULT

205

which separate the points r0 , ..., rN +1 ; in particular,ρ < t1 .

(13.3.27)

Since q∗ is a polynomial of degree at most N + 1 with the roots t1 , ..., tN +1 and q∗ (0) = 1, wehaveq∗ (t) =NY+1(1 − t/ti ).

(13.3.28)

i=1

Now let us set
q(t) = q∗κ (t).
We have q(0) = 1 and deg q = κ(N + 1) = k, so that q ∈ Pk0 . Thus, q belongs to the set overwhich the minimum in (13.3.24) is taken; therefore to prove (13.3.24) it suffices to demonstrate
that
1+2τ
(13.3.29)
max t1+2τ q 2 (t) ≤ ε (ω+τ ) ≡ ε2/β .
t∈ΣIndeed, this inequality combined with (13.3.9) would result inZ

t1+2τ q 2 (t)dµ(t)}β ≤ R2α {ε2/β R2 }β = R2 ε2 ≤ ν 2

R2α {

Σ

(the concluding inequality follows from (13.3.25)), as required in (13.3.24).
To prove (13.3.29), let us start with the observation thatρ ≥ ε1/(ω+τ ) .

(13.3.30)

Indeed, (13.3.26) says that ρ = (ε/q∗ (ρ))1/(ω+τ ) , while (13.3.28) and (13.3.27) imply that0 ≤ t ≤ ρ ⇒ 0 < q∗ (t) < 1.

(13.3.31)

Now let t ∈ Σ. Assume, first, that t ≥ ρ. From (13.3.25) it follows that|q∗ (t)| ≤ εt−ω−τ ,whencet1+2τ q 2 (t) = t1+2τ q∗2κ (t) ≤ t1+2τ −2κ(ω+τ ) ε2κ ≤[since 1 + 2τ − 2κ(ω + τ ) ≤ 0 by definition of κ and t ≥ ρ by assumption on t]≤ ρ1+2τ −2κ(ω+τ ) ε2κ ≤[again since 1 + 2τ − 2κ(ω + τ ) ≤ 0 and in view of (13.3.30)]≤ ε(1+2τ −2κ(ω+τ ))/(ω+τ ) ε2κ = ε(1+2τ )/(ω+τ ) .
Thus,{t ∈ Σ, t ≥ ρ} ⇒ t1+2τ q 2 (t) ≤ ε(1+2τ )/(ω+τ ) .
Now let t ∈ Σ and t ≤ ρ. From (13.3.31) and (13.3.25) it follows that
0 < q∗ (t) ≤ min{1; εt−ω−τ },(13.3.32)

206

LECTURE 13. OPTIMALITY OF THE CONJUGATE GRADIENT METHOD

whencet1+2τ q 2 (t) ≤ t1+2τ min{1; ε2κ t−2κ(ω+τ } == min{t1+2τ ; ε2κ t1+2τ −2κ(ω+τ ) } ≤ max min{s1+2τ ; ε2κ s1+2τ −2κ(ω+τ ) } =s>0[recall that 1 + 2τ > 0, 1 + 2τ − 2(ω + τ ) ≤ 0]
1+2τ

= ε ω+τ ,and we come to the relation
1+2τ

{t ∈ Σ, t ≤ ρ} ⇒ t1+2τ q 2 (t) ≤ ε ω+τ ,(13.3.33)

which combined with (13.3.32) proves (13.3.29).

13.4

Exercises: Around Conjugate Gradient Method

Let us start with a series of simple exercises which state some useful facts about the Conjugate
Gradient Method as applied to a linear operator equation
Ax = bwith symmetric n × n positive semidefinite matrix A. In what follows the equation is assumed
to be solvable; we exclude the trivial case b = 0.
Exercise 13.4.1 #+ Let e1 , ..., em be the minimal, with respect to the number of elements,
collection of eigenvectors of A such that b can be represented as a linear combination of thevectors from the collection. Prove that the corresponding eigenvalues λi , i = 1, ..., m, are distinct
and positive and that m is exactly the number of steps in which the CGM finds the exact solution
to the equation. In particular, the number of steps before the exact solution is found never exceeds
the maximal number of different positive eigenvalues of A (and, consequently, never exceeds n).
Exercise 13.4.2 # Assume that Σ is a compact subset of the nonnegative half-ray, 0 is not an
isolated point of Σ and ω and τ are such that ω + τ > 0 and ω ≤ 1/2, as required by the premiseof the Theorem on optimality of the CGM (Lecture 13). Assume that problem (A, b) belongs to
the family Un (Σ∗ , τ, R), where Σ∗ is the union of Σ and a set Σ0 comprised of at most k points
which are to the right of Σ (note this ”to the right”!). Prove that for any ν > 0 the CGM solves(A, b) within νω -inaccuracy ≤ ν in no more thank+c

1 + 2τ
b[N ∗ (µ/R, Σ, ω + τ ) + 1]
2(ω + τ )steps (compare with the statement of Theorem 13.2.1). Thus, everything looks as if the CGM
spends k first steps ”to kill” the eigenvalues of A outside Σ and then switches to solving the
problem with the ”truncated” spectrum.

13.4. EXERCISES: AROUND CONJUGATE GRADIENT METHOD

207

Observation given by Exercise 13.4.1 underlies an important trick in solving linear systems:
the prescaled conjugate gradient scheme. Assume we are interested in solving the ”perturbed”
equation
Ax = b, A = A0 + A1 ,where the ”perturbation” A1 is symmetric and the ”unperturbed” matrix A0 is symmetricpositive definite and is ”simpler” than A (in the sense that we can invert A0 significantly cheaper
than A).
Let us act as follows. Imagine that we have passed from the initial system to the equivalent
−1/2
−1/2
one by substitution x = A0 u and termwise multiplication of the system by A0 . The
resulting system with respect to u would be−1/2(I + B1 )u = b1 , B1 = A0

−1/2

A1 A0

−1/2

, b1 = A0b.

Let uk be the trajectory of the usual CGM as applied to the transformed system and xk =−1/2
A0 uk be the image of this trajectory in the x-variables. This latter sequence is, as it is
1/2
easily seen, governed by a simple recurrence which does not involve B1 , b1 or A0 ; each stepof this recurrence requires nothing but a single multiplication of a vector by A and a single
multiplication of another vector by A−1
0 , plus O(n) arithmetic operations. This recurrence iscalled the Conjugate Gradient Method prescaled by the matrix A0 . Note that the potentials
fA,b (x)of the initial system and
fA0 ,b0 (u)−1/2[A0 = A0

−1/2

AA0

−1/2

= I + B 1 , b 0 = A0b]of the transformed one are related to each other:−1/2

fA,b (A0u) = fA0 ,b0 (u),so that the rates of convergence of both the sequences {xk } and {uk } to the solutions of,respectively, the initial and the transformed problems measured in terms of the residuals in
the corresponding potentials are identical to each other.
The recurrent form of the CGM prescaled by A0 is
pk = A0−1 rk −T A−1 r
qk−1
k
0
pk−1
T
pk−1 qk−1(13.4.1)

qk = AA0−1 rk −T A−1 r
qk−1
k
0
qk−1
pTk−1 qk−1

(13.4.2)

xk−1 = xk −pTk−1 rkpk
pTk qk

(13.4.3)

rk+1 = rk −pTk−1 rkqk
pTk qk

(13.4.4)(a fraction with zero denominator by definition is zero) initialized asx0 = p0 = q0 = 0, r0 = −b.

208

LECTURE 13. OPTIMALITY OF THE CONJUGATE GRADIENT METHOD

Exercise 13.4.3 # Prove that the CGM prescaled by a positive definite symmetric matrix A0indeed is given by the above recurrence. What is the CGM prescaled by the unit matrix?
Now, are there any advantages of the prescaled CGM as compared to the standard one? Of
course, we may expect something reasonable only when the perturbation A1 is ”small”. The
first - and the standard - interpretation of the notion of a ”small” perturbation is that one whenr ≡ Rank A1 is relatively small (r << n). A good example here is an ”almost” block-diagonal
matrix A, i.e., the matrix of the form of the form


q1
p
 2

 p3

 ...
pk

pT2
q2
...

pT3
q3
...

...

pTk

...






... 



qk

(blank space corresponds to zero blocks) which is nothing but a block-diagonal matrix A0 withthe diagonal blocks q1 , ..., qk ”spoiled” by several first dense rows and columns; if l is the numberof these dense rows (the row size of q1 ), then one can decompose A as A0 +A1 with Rank A1 ≤ 2l.
In the case in question the matrix I + B1 of the transformed system differs from the unit matrix
−1/2
−1/2
by the matrix B1 = A0 A1 A0of the rank r and therefore I + B1 has at most r eigenvalues
which differ from 1. It follows that I +B1 has at most r+1 different eigenvalues and therefore thestandard CGM as applied to the transformed system solves it exactly in no more than r+1 << nsteps; since the trajectory of the prescaled CGM reproduces the same behaviour in x-variables,
the prescaled CGM finds the exact solution to the initial system in the same r + 1 << n steps.
The discussed notion of a ”small” perturbation A1 is, in a sense, structural: the entries of A1
as compared to those of A0 should not be small, what is small is the rank of the perturbation ascompared to that one of A0 . There is another way to define a ”relatively small” perturbation,
namely, to say that(ρ−1 − 1)A0 ≥ A1 ≥ (ρ − 1)A0
for some ρ ∈ (0, 1) (≥ is understood in the operator sense, i.e., as positive semidefiniteness of thecorresponding differences). Here, of course, the rank of A1 should not be small (it may coincidewith n), but the quadratic form xT Ax associated with the perturbed matrix A is ”compatible”
with that one associated with the unperturbed matrix A0 :xT Ax
≤ ρ−1 ∀x 6= 0.
x T A0 xρ≤(13.4.5)

Exercise 13.4.4 ∗ Prove that in the case of (13.4.5) the CGM prescaled by A0 solves the system
Ax = b within the relative inaccuracy in terms of the potential
fA,b (x) − min fA,b
fA,b (0) − min fA,b
not exceeding ν ∈ (0, 1/2) in no more thanN = O(1)ρ−1 ln(1/ν)
steps, O(1) being an absolute constant.

13.4. EXERCISES: AROUND CONJUGATE GRADIENT METHOD

209

To comment the result of the latter exercise from the computational viewpoint, let me note that,given a unique ”unstructured” linear system, you hardly can decompose the matrix A of thesystem in an ”unperturbed” part A0 with known or easily computable inverse and a perturbation
A1 which satisfies (13.4.5). In some applications, anyhow, e.g., in the Newton-type methods forsmooth minimization, one is required to solve a series of linear systems with matrices ”slightly”
varying from system to system. In this case one can try the strategy as follows. The first system
A(1) x = b(1) is solved via computing the matrix [A(1) ]−1 by any standard direct method ofLinear Algebra. When solving the second system, A(2) x = b(2) , we try first the CGM prescaledby the matrix A(1) . If the method is able to solve the system at a ”reasonable” arithmetic
cost (here ”reasonable” means, say, 10% of the CPU time spent to invert A(1) ), we pass to thenext system; otherwise we stop the CGM and solve the second system by direct computationof [A(2) ]−1 . When solving the third system, we again start with the CGM prescaled by the last
matrix for which we know the exact inverse (i.e., prescaled by A(1) , if the CGM was successfulwhen solving the second system, or by A(2) , if it failed); if the method was successful, we go
to the next system, if not, solve the third system via direct computation of [A(3) ]−1 and use
A(3) as our new prescaling matrix for the fourth system, an so on. It is clear that even in theworst case (no one of our attempts to save time by running the CGM is successful) this strategy
increases the total CPU time, as compared to that one without trying the CGM, by at most
10%; at the same time if the matrices of the subsequent systems indeed are close to each other
and, consequently, the runs of the CGM will ”normally” be successful (say, will solve 90% of
our systems in 10% of the time required by the direct Linear Algebra), we will reduce the total
CPU time by factor of 5.
The remaining exercises deal with some auxiliary facts used in Lecture 13 without proof.
We start with the Momentum Inequality:
Exercise 13.4.5 #∗ Prove Proposition 13.3.2 (Lecture 13).
Now let us prove the (quite standard) property of the roots of orthogonal polynomials. Let µ bea (finite) measure with the support set contained in some segment [a, b] of the real axis; for thesake of simplicity assume that −∞ < a < b < ∞, and let the number of points in the support
set of µ be at least N + 1. Those not too familiar with the Measure Theory may assume, in
addition, that the measure is discrete, i.e., there are M > N points xi of the segment [a, b] towhich the measure assignees positive weights µi , and for any A ⊂ R the measure µ(A) of theset is simply
the total weight of the points which are in A. Recall that in the latter case theRPintegral f (t)dµ(t) is nothing but M
i=1 f (xi )µi , and L2 (µ) is the Euclidean space comprised of
the real-valued functions on the finite set {x1 , ..., xM } with the inner productZ(f, g) =f (t)g(t)dµ(t) ≡MXf (xi )g(xi )µi .

i=1

Exercise 13.4.6 #∗1) Prove that there exists a (N +1)-term µ-orthogonal system of polynomials, i.e., a sequenceof polynomials q0 ,...,qN of the degrees 0, 1, ..., N , respectively, such thata) qj is a nonzero element L2 (µ);

210

LECTURE 13. OPTIMALITY OF THE CONJUGATE GRADIENT METHODb) the polynomials are µ-orthogonal to each other, i.e.,(qi , qj ) = 0, 0 ≤ i < j ≤ N.
2) Prove that the sequence {qi } is uniquely defined by the aforementioned properties, up to
multiplication of its terms by nonzero reals.
3) Prove that any polynomial q of degree k ≤ N can be uniquely represented as a linear
combination of q0 , ..., qk . Derive from this that qk is µ-orthogonal to any polynomial p of degree< k, i.e., for such a p one has (qk , p) = 0.
4) Prove that if p is a nonzero polynomial of degree ≤ N , then (p, p) > 0. Derive from thisobservation that the natural mapping of the space PN of polynomials of degree ≤ N into L2 (µ)(a polynomial p is mapped into itself regarded as an element of L2 (µ)) is an embedding: different
elements of PN are mapped into different elements of L2 (µ). Is it possible to replace here N
with N + 1?
Let us fix an (N + 1)-term µ-orthogonal sequence of polynomials q0 , ..., qN . Let Ek , k ≤ N ,be the linear span of the polynomials q0 , ..., qk−1 regarded as elements of L2 (µ), so that Ek is
a subspace of L2 (µ). Let also T be the operator in L2 (µ) which maps an element of L2 (µ)
represented by a function f (t) into the element represented by the function tf (t). We denoteby Tk the restriction of T onto Ek , i.e., the operator on Ek defined as follows: for f ∈ Ek Tk f
is the orthogonal projection of the vector T f onto Ek .
Exercise 13.4.7 # Let 1 ≤ k ≤ N . Prove that
1) the operator T and the operators Tk are symmetric:Z(T f, g) = (f, T g) =tf (t)g(t)dµ(t), f, g ∈ L2 (µ);(Tk f, g) = (f, Tk g) = (T f, g), f, g ∈ Ek .
2)+ the spectrum of the operator Tk coincides with the set of real roots of the polynomial qk :if λ is an eigenvalue of Tk , then qk (λ) = 0, and vice versa.
3)+ prove that all eigenvalues of Tk belong to the interval (a, b) and are of multiplicity 1.
Derive from this observation and 2) that all k roots of qk are real, distinct and belong to theopen interval (a, b). Prove that qk−1 and qk have no common roots.
Exercise 13.4.8 #∗ Prove the Separation Property: if ti,1 < ti,2 < ... < ti,i are the roots of qi(we know from the previous exercise that qi has i distinct roots and that they belong to (a, b)),then the roots of qi+1 separate those of qi :a < ti+1,1 < ti,1 < ti+1,2 < ti,2 < ... < ti+1,i < ti,i < ti+1,i+1 < b.

Lecture 14

Convex Stochastic Programming
In this final lecture we shall speak about a problem which is quite different from those considered
so far – about stochastic optimization. A general single-stage Stochastic Programming programis as follows:Z

minimize f0 (x) =F0 (x, ω)dP (ω) s.t. x ∈ G, fj (x) =Z

Fj (x, ω)dP (ω) ≤ 0, j = 1, ..., m.

ΩΩ(14.0.1)
Here• Fj (x, ω) are functions of the design vector x ∈ Rn and parameter w ∈ Ω• G is a subset in Rn• P is a probability distribution on the set Ω.
Thus, a stochastic program is a Mathematical Programming program where the objective andthe constraints are expectations of certain functions depending both of the design vector x andrandom parameter ω.
The main source of programs of the indicated type is optimization in stochastic systems,
like Queuing Networks, where the processes depend not only on the design parameters, like
performances and numbers of serving devices of different types), but also on random factors. As
a result, the characteristics of such a system (e.g., time of serving a client, cost of service, etc.)
are random variables depending, as on parameters, on the design parameters of the system. It isreasonable to measure the quality of the system by the expected values of the indicated randomvariables (in the dynamic systems, we should speak about the steady-state expectations). These
expected values have the form of the objective and the constraints in (14.0.1), so that to optimizea system in question over its design parameters is a program of the type (14.0.1).
Looking at stochastic program (14.0.1), you can immediately ask what are, if any, the specificfeatures of these problems and why these problems need specific treatment. (14.0.1) is, finally,
nothing but the usual Mathematical Programming problem; we know how to solve tractable(convex) Mathematical Programming programs, and to the moment we never took into accountwhat is the origin of the objective and the constraints, are they given by explicit simple formulae,
or are they integrals, as in (14.0.1), or solutions to differential equations, or whatever else. All
what was important for us were the properties of the objective and the constraints – convexity,
smoothness, etc., but not their origin.
211

212

LECTURE 14. CONVEX STOCHASTIC PROGRAMMING

This “indifference to the origin of the problem” indeed was the feature of our approach,but it was its weak point, not the strong one. In actual computations, the performance of analgorithm heavily depends not only on the quality of the method we apply to solve the problem,
but also on the computational effort needed to provide the algorithm by the information on the
problem instance – the job which in our model of optimization was the task of the oracle. We
did not think how to help the oracle to solve his task and took care only of total number oforacle calls – we did our best to reduce this number. In the most general case, when we have noidea of what is the internal structure of the problem, this is the only possible approach. But themore we know about the structure of the problem, the more should we think of how to simplify
the task of the oracle in order to reduce the overall computational expenses.
Stochastic Programming is an example when we know something about the structure of the
problem in question. Namely, let us look at a typical stochastic system, like Queuing Network.
Normally the functions Fj (x, ω) associated with the system are “algorithmically simple” – given
x and ω, we can more or less easily compute the quantities Fj (x, ω) and even their derivatives
with respect to x; to this end it suffices to create a simulation model of the system and run
it at the given by ω realization of the random parameters (arrival and service times, etc.);
even for a rather sophisticated system, a single simulation of this type is relatively fast. Thus,
typically we have no difficulties with simulating realizations of the random quantities Fj (x, ·).
On the other hand, even for relatively simple systems it is, as a rule, impossible to computethe expected values of the indicated quantities in a “closed analytic form”, and the only way
to evaluate these expected values if to use a kind of the Monte-Carlo method: to run not asingle, but many simulations, for a fixed value of the design parameters, and to take the empiric
average of the observed random quantities as an estimate of their expected values. According to
the well-known results on the rate of convergence of the Monte-Carlo method, to estimate theexpected values within inaccuracy  it requires O(1/2 ) simulations, and this is just to get theestimate of the objective and the constraints of problem (14.0.1) at a single point! Now imagine
that we are going to treat (14.0.1) as a usual “black-box represented” optimization problem
and intend to imitate the usual first-order oracle for it via the aforementioned Monte-Carloestimator. In order to get an -solution to the problem, we, even in good cases, need to estimate
within accuracy O() the objective and the constraints along the search points. It means thatthe method will require much more simulations than the aforementioned O(1/2 ): this quantity
should be multiplied by the information-based complexity of the optimization method we aregoing to use. As a result, the indicated approach in most of the cases results in inappropriately
long computations.
An extremely surprising thing is that there exists another way to solve the problem. Thisway, under reasonable convexity assumptions, results in overall number of O(1/2 ) computations
only – as if there were no optimization at all and the only goal were to estimate the objective andthe constraints at a given point. The subject of our today lecture is this other way – Stochastic
Approximation.
To get a convenient framework for presenting Stochastic Approximation, it is worthy tomodify a little the way we are looking at our problem. Assume that when solving it, we areallowed to generate a random sample ω1 , ω2 ,... of “random factors” involved into the problem;the elements of the sample are assumed to be mutually independent and distributed according
to P . Assume also that given x and ω, we are able to compute the values Fj (x, ω) and the213
gradients ∇x Fj (x, ω) of the integrants in (14.0.1). Note that under mild regularity assumptions
the differentiation with respect to x and taking expectation are interchangeable:Z

Fj (x, ω)dP (ω), ∇fj (x) =fj (x) =ΩZ∇x Fj (x, ω)dP (ω).

(14.0.2)Ω

It means that the situation is covered by the following model of an optimization method solving(14.0.1):
At a step i, we (the method) form i-th search point xi and forward it to the oracle which wehave in our disposal. The oracle returns the quantities
Fj (xi , ωi ), ∇x Fj (xi , ωi ), j = 0, ..., m,(in our previous interpretation it means that a single simulation of the stochastic system inquestion is performed), and this answer is the portion of information on the problem we geton the step in question. Using the information accumulated so far, we generate the new search
point xi+1 , again forward it to the oracle, enrich our accumulated information by its answer,
and so on.
The presented scheme is a very natural definition of a method based on stochastic first orderoracle capable to provide the method random unbiased (see (14.0.2)) estimates of the values
and the gradients of the objective and the constraints of (14.0.1). Note that the estimates are
not only unbiased, but also form a kind of Markov chain: the distribution of the answers of theoracle at a point depends only on the point, not on the previous answers (recall that {ωi } areassumed to be independent).
Now, for our further considerations it is completely unimportant that the random estimatesreported by the oracle come from certain integrants in such a way that the observations of fj (x)come from the values, and observations of ∇fj (x) come from the gradients of the integrants. It
suffices to postulate the following:• The goal is to solve the Convex Programming program1
f0 (x) → min | fj (x) ≤ 0, j = 1, ..., m; x ∈ G ⊂ Rn

(14.0.3)[G is closed and convex, fi , i = 1, ..., m, are convex and Lipschitz continuous on G]• The information obtained by an optimization method at i-th step, i = 1, 2, ..., comes
form a stochastic oracle, i.e., is a collection
φj (xi , ωi ) ∈ R, ψj (xi , ωi ) ∈ Rn , j = 0, ..., m,of reals and vectors, where– {ωi } is sequence of independent identically distributed, according to certain probability distribution P , random parameters taking values in certain space Ω;1

Of course, the model we are about to present makes sense not only for convex programs; but the methods
we are interested in will, as always, work well only in convex case, so that we loose nothing when imposing theconvexity assumption from the very beginning

214

LECTURE 14. CONVEX STOCHASTIC PROGRAMMING– xi is i-th search point generated by the method; this point may be an arbitrary
deterministic function of the information obtained by the method so far (i.e., at thesteps 1, ..., i − 1)
It is assumed (and it is crucial) that the information obtained by the method is unbiased:
Eφj (x, ω) = fj (x), fj0 (x) ≡ Eψj (x, ω) ∈ ∂fj (x), j = 0, ..., m, x ∈ G;

(14.0.4)

here E is the expectation with respect to the distribution of the random parameters inquestion.
We have described the generic scheme of the optimization methods we are interested in now.
Let us look what are the abilities of these methods.

14.1

Stochastic Approximation: simple caseLet us start with the simplest (and perhaps the most important for applications) case when
there are no functional constraints in the problem, so that it is simply the problem
f (x) → min | x ∈ G ⊂ Rn

(14.1.1)(we write f instead of f0 and, similarly, drop the index in our “observation functions” φ(x, ω)(the estimate of f (x)) and ψ(x, ω) (the estimate of f 0 (x)).

14.1.1

Assumptions

To get reasonable complexity results, we need to bound somehow the magnitude of the random
noise in the process in the stochastic oracle (as it is always done in all statistical considerations).
Mathematically, the most convenient way to do it is as follows: letn

L = sup E|ψ(x, f )|2

o1/2

.

(14.1.2)

x∈G

From now on we assume that the oracle is such that L < ∞. The quantity L will be called
the intensity of the oracle at the problem in question; in what follows it plays the same roleas the Lipschitz constant of the objective in large-scale minimization of Lipschitz continuous
convex functions (Lecture 8).

14.1.2The Stochastic Approximation method

The method we are interested in is completely similar to the Subgradient Descent method from
Section 8.1. Namely, the method generates search points according to the recurrence (cf. (8.1.1))
xi+1 = πG (xi − γi ψ(xi , ωi )), i = 1, 2, ...,
where
• x1 ∈ G is an arbitrary starting point;(14.1.3)

14.1. STOCHASTIC APPROXIMATION: SIMPLE CASE

215

• γi are deterministic positive stepsizes;• πG (x) = argminy∈G |x − y| is the standard projector on G.
The only difference with (8.1.1) is that now we replace the direction g(xi ) = f 0 (xi )/|f 0 (xi )| of
the subgrade of f at xi by the random estimate ψ(xi , ωi ) of the subgradient.
Note that the search trajectory of the method is governed by the random variables ωi andis therefore random:
xi = xi (ω i−1 ), ω s = (ω1 , ..., ωs ).
Recurrence (14.1.3) defines the sequence of the search points, not that one of the approximate
solutions. In the deterministic case (Section 8.1), we extracted from the search points theapproximate solutions by choosing the best (with the smallest value of the objective) of thesearch points generated so far. The same could work in the stochastic case as well, but here
we meet with the following obstacle: we “do not see” the values of the objective, and therefore
cannot say which of the search point is better. To resolve the difficulty, we use the same trickas in the case of games and variational inequalities (Section 9.1) where we dealt with implicitly
defined objectives, namely, define i-th approximate solution as the sliding average of the search
points:−1xi ≡ xi (ω i−1 ) = 

X

γi 

i/2≤t≤iX

γt xt .

(14.1.4)i/2≤t≤i

The efficiency of the resulting method is given by the following
Theorem 14.1.1 For the aforementioned method one has, for all positive integer N ,D2 + L2 N/2≤i≤N γi2PεN ≡ E[f (x ) − min f (x)] ≤,x∈G
2 N/2≤i≤N γi
P

N(14.1.5)

G being the diameter of G.
In particular, for γi chosen according toγi =one hasD√
L i(14.1.6)LD
εN ≤ O(1) √N(14.1.7)

with appropriately chosen absolute constant O(1).
Proof is completely similar to that one used in Section 8.1 to get the estimate (8.1.5). Namely,
from (14.1.3) and Lemma 8.1 it follows that if x∗ is an optimal solution to (14.1.1), then2
δi+1
≡ |xi+1 (ω i ) − x∗ |2 ≤ δi2 − 2γi (xi (ω i−1 ) − x∗ )T ψ(xi (ω i−1 ), ωi ) + γi2 |ψ(xi (ω i−1 ), ωi )|2 .

Taking expectations of both sides, we get
h

i

h

i

2
d2i+1 ≡ Eδi+1≤ d2i − 2γi E (xi (ω i−1 ) − x∗ )T ψ(xi (ω i−1 ), ωi ) + E |ψ(xi (ω i−1 ), ωi )|2 . (14.1.8)

216

LECTURE 14. CONVEX STOCHASTIC PROGRAMMINGThe expectation of the second term in the right hand side can be computed as follows: first
we can take it with respect to ωi , and then take the expectation of the result with respect toω i−1 . In the product to which we should apply the inner expectation (the one over ωi ) the first
factor is independent of this variable, so that all we should do to get the inner expectation isto replace ψ(xi (ω i−1 ), ωi ) with Eωi ψ(xi (ω i−1 ), ωi ); the latter quantity, due to (14.0.4), is exactly
f 0 (xi (ω i−1 )) ∈ ∂f (xi (ω i−1 )), so that
h

i

h

i

E (xi − x∗ )T φ(xi , ωi ) = E (xi − x∗ )T f 0 (xi ) ≥ E[f (xi ) − f (x∗ )](the concluding inequality follows from the convexity of f ).
Similar reasoning applied to the third term in the right hand side of (14.1.8) (with (14.1.2)
playing the role of (14.0.4), demonstrates that this term does not exceed γi2 L2 . Thus, we getfrom (14.1.8)
d2i+1 ≤ d2i − 2γi E [f (xi ) − f (x∗ )] + γi2 L2 ,whence, due to di ≤ D,




E

X

2γi (f (xi ) − f (x∗ )) = 2

γi E [f (xi ) − f (x∗ )] ≤ D2 + L2

X
N/2≤i≤N

N/2≤i≤NX

γi2 .

N/2≤i≤N

The left hand side in this relation, due to Jensen’s inequality, is not less than
2[

X

h

i

γi ]E f (xN ) − f (x∗ ) ,N/2≤i≤N

and we come to (14.1.5). (14.1.7) is an immediate consequence of (14.1.5) - (14.1.6).

14.1.3

Comments

The statement and the proof of Theorem 14.1.1 are completely similar to the related “deterministic” considerations of Section 8.1. The only difference is that now we are estimating from above
the expected inaccuracy of N -th approximate solution; this is quite natural, since the stochastic
nature of the process makes it impossible to say something reasonable about the quality of every
realization of the random vector xN = xN (ω N −1 )It turns out that the rate of convergence established in (14.1.7) is, in certain sense, unimprovable. Namely, it is not difficult to prove the following statement.
Proposition 14.1.1 For every L > 0, any D > 0, any positive integer N and any stochasticoracle-based N -step method M of minimizing univariate convex functions over the segment
G = [0, D] on the axis there exists a linear function f and a stochastic oracle with intensity L
on the function such thatLD
E[f (xN ) − min f ] ≥ O(1) √ ,GN
xN being the result formed by the method as applied to the problem f . Here O(1) is properlychosen positive absolute constant.

14.1. STOCHASTIC APPROXIMATION: SIMPLE CASE

217√
Note that in the deterministic case the rate of convergence O(1/ N ) was unimprovable only inthe large-scale case; in contrast to this, in the stochastic case this rate becomes optimal alreadywhen we are minimizing univariate linear functions.
√
Convergence rate O(1/ N ) can be improved only if the objective is strongly convex. The
simplest and the most important result here is as follows (the proof is completely similar to that
one of Theorem 14.1.1):
Proposition 14.1.2 Assume that convex function f on Rn attains its minimum on G at aunique point x∗ and is such thatθΘf (x∗ ) + |x − x∗ |2 ≤ f (x) ≤ f (x∗ ) + |x − x∗ |2 , x ∈ G,
2
2

(14.1.9)

with certain positive θ and Θ. Consider process (14.1.3) with the stepsizes
γi =

γ
,
i

(14.1.10)

γ being a positive scale factor satisfying the relationγθ > 1.

(14.1.11)

Then

D2 + γ 2 L2,
(14.1.12)
GN
where D is the diameter of G, L is the intensity of the oracle at the problem in question andc(·) is certain problem-independent function on (1, ∞).
εN ≡ E(f (xN ) − min f ) ≤ c(γθ)Θ

The algorithm (14.1.3) with the stepsizes (14.1.10) and the approximate solutions identical to
the search points is called the classical Stochastic Approximation; it originates from Kiefer andWolfovitz.
A good news about the algorithm is its rate of convergence: O(1/N ) instead of√O(1/ N ). A bad news is that this better rate is ensured only in the case of problems satisfying(14.1.9) and that the rate of convergence is very sensitive to the choice of the scale factor γ in thestepsize formula (14.1.10): if this scale factor does not satisfy (14.1.11), the rate of convergence
may become worse in order. To see this, consider the following simple example: the problem is
1
f (x) = x2 , [θ = Θ = 1];
2

G = [−1, 1];

x1 = 1,the observations are given byψ(x, ω) = x + ω[= f 0 (x) + ω],and ωi are the standard Gaussian random variables (Eωi = 0, Eωi2 = 1); we do not specify
φ(x, ω), since it is not used in the algorithm. In this example, the best choice of γ is γ = 1 (inthis case one can make (14.1.11) an equality rather than strict inequality due to the extreme
simplicity of the objective). For this choice of γ one has
Ef (xN +1 ) ≡ E[f (xN +1 ) − min f (x)] ≤x

1, N ≥ 1.
2N

218

LECTURE 14. CONVEX STOCHASTIC PROGRAMMING

In particular, it takes no more than 50 steps to reach expected inaccuracy not exceeding 0.01.
Now assume that when solving the problem we overestimate the quantity θ and choose
stepsizes according to (14.1.10) with γ = 0.1. How many steps do we need in this case toreach the same expected inaccuracy 0.01 – 500, 5000, or what? The answer is astonishing:approximately 1,602,000 steps. And with γ = 0.05 (20 times less than the optimal value of theparameter) the same accuracy costs more than 5.2 × 1014 steps!
We see how dangerous the “classical” rule (14.1.10) for the stepsizes is: underestimatingof γ (≡ overestimating of θ) may kill the procedure completely. And where from, in more orless complicated cases, could we take a reasonable estimate of θ? It should be said that there
exist stable versions of the classical Stochastic Approximation (they, same as our version of
the routine, use “large”, as compared to O(1/i), stepsizes and take, as approximate solutions,certain averages of the search points). These stable version of the method are capable to reach(under assumptions similar to (14.1.9)) the O(1/N )-rate of convergence, even with the optimalcoefficient at 1/N . Note, anyhow, that the nondegeneracy assumption (14.1.9) is crucial for theO(1/N )-rate of convergence;
if it is removed, the best possible rate, as we know from Proposition√
14.1.1, becomes O(1/ N ), and this is the rate given by our “robust” Stochastic Approximation
with “large” steps and averaging.

14.2

MinMax Stochastic Programming problems

Now let us consider problems with constraints. In the case of Stochastic Programming (similarly
to what happens in the case of smooth convex optimization) straightforward extension of theresults and methods from the case without functional constraints to that one when they arepresent causes slight difficulties; this is why we restrict ourselves with the MinMax Stochastic
programminimize f¯(x) = max fj (x) s.t. x ∈ G ∈ Rn .
(14.2.1)
j=1,...,m

Here, as always, G is a closed and bounded convex set and the functions fj are convex and
Lipschitz continuous on G. We assume that the functions fj are observed via stochastic oracle
with observation functions
φj (x, ω), ψj (x, ω)[Eφj (x, ω) = fj (x), Eψj (x, ω) ∈ ∂fj (x), ∀x ∈ G] .

Note that although (14.2.1) is, formally, a convex optimization problem without functional
constraints, our previous approach cannot be used here, since we have unbiased observations for
fj only, not for f¯, and the transformation (f1 , ..., fm ) 7→ f¯ is nonlinear and therefore does notpreserve unbiasedness.
The right way to solve the problem is to convert it into the saddle point programfind saddle point of L(x, y) =mX

yj fj (x) on G × Y, Y = {y ≥ 0 |

i=1

X
j

D being the diameter of the set G.
Since the loss function
L̄(x) ≡ max L(x, y)y∈Yyj = D},(14.2.2)

14.2. MINMAX STOCHASTIC PROGRAMMING PROBLEMS

219

of the first player in the game with the cost function L (see Lecture 5) is exactly Df¯, the xcomponents of saddle points of L are exactly the minimizers of f¯ over G, so that in order toapproximate a minimizer of f¯ it suffices to approximate a saddle point of L.
The advantage of the reduction (14.2.1) ⇒ (14.2.2) is that the answers of the oracle allow
to form unbiased observations of partial derivatives (more exactly, sub- and superdifferentials)of L with respect to x and to y; these observations are, respectively,ψx (x, y, ω) =mXyi ψj (x, ω), ψy (x, y, ω) = (φ1 (x, ω), ..., φm (x, ω))T .

j=1

Now we can approximate the saddle set of L by the stochastic version of the Subgradient Descentmethod for games (Section 9.1) as follows:


xi+1
yi+1





= πG×Yxi − γi ψx (xi , yi , ωi )yi + γi ψy (xi , yi , ωi )



, i = 1, 2, ...,

(14.2.3)

where the starting point (x1 , y1 ) is an arbitrary deterministic point of G × Y , and γi are positive
deterministic stepsizes.
Same as in Section 9.1, the approximate solution to (14.2.1) generated after N steps of themethod is the sliding average−1

xN = 

X

X

γi 

γi xi

N/2≤i≤N

N/2≤i≤Nof the search points xi .
The results on the convergence of the proposed algorithm are given by the following statement(we omit its proof):
Theorem 14.2.1 Let problem (14.2.1) be solved by the method (14.2.3), and assume that theintensity


L=D

−1

sup 
x∈G

m
X

1/2
2

E|φj (x, ω)|+ sup 
x∈G

j=1

m
X

1/2
2

E|ψj (x, ω)|j=1of the stochastic oracle is finite. Then, for every positive integer N , the expected inaccuracy of
the N -th approximate solution to the problem can be estimated from above asεN ≡ E f¯(xN +1 ) − min f¯ ≤ 4

D2 + L2GP

P

2N/2≤i≤N γi

N/2≤i≤N γi.

(14.2.4)

In particular, with the stepsizes given by
γi =

D
√
L i

(14.2.5)

one hasLDεN ≤ O(1) √ , N = 1, 2, ...,
N
with appropriately chosen absolute constant O(1).

(14.2.6)

220

LECTURE 14. CONVEX STOCHASTIC PROGRAMMING

Hints to Exercises
Hints to Section 1.3
Exercise 1.3.4+ : to prove that x∗ (Q) ∈ int Q, note that if x is not in the interior of Q, then,due to the Separation Theorem for convex sets, there exists an affine functional f such thatf (x) > f (y) for all y ∈ int Q, and use the result of exercise 1.3.3.
Exercise 1.3.5+ : note that there exists an invertible affine mapping of the space which performs
a given permutation of the vertices of the simplex, and use the result of exercise 1.3.4.
Exercise 1.3.8:1): this is an immediate consequence of the characterization of the center of gravity given
by exercise 1.3.3
2): Given a positive α, consider the cone
Qα = {(x, t) | |x| ≤ α−1 (t + α)φ(0), −α ≤ t ≤ β(α)},
where β(α) > 0 is defined by the requirement that the volume of the part of Qα to the right of
the hyperplane {t = 0} is equal to the volume of the similar part of Q (of course, such a β does
exist). Prove that for an appropriately chosen α > 0 the volume of the part of Qα to the left ofthe hyperplane {t = 0} is equal to the volume of the similar part of Q.
3)+ : make rigorous the following ”mechanical arguments”: since φ is concave, the t-coordinate
of any point from G = Q∗+ \Q+ is ≥ than the t-coordinate of any point from H = Q+ \Q∗+ . Further, G and H are of the same volumes (by construction), so that to obtain Q∗+ from Q+ weshould move certain masses to the hyperplane {t = 0}, which decreases the momentum of thefunctional t. Apply similar reasoning to Q∗− , Q− .
6): the center of gravity of the cone Q∗ can be computed directly:x∗ (Q∗ ) = (0, −α + n(β + α)/(n + 1)) ≡ (0, τ );it is immediately seen that the part Q0 of Q∗ to the left of the hyperplane {t = τ } is of thevolume (n/(n + 1))n Voln (Q∗ ) = (n/(n + 1))n Voln (Q) (the latter equality follows from 2)). Since5) says that τ ≤ 0, the volume of Q0 does not exceed that one of Q∗− , Q.E.D.
Exercise 1.3.10: let x∗ be the center of gravity of Q, and let δ = [a, b] be a chord passing
through x∗ , [a, x∗ ] being the larger, and [x∗ , b] being the smaller part of the chord. Choose thesupport to Q hyperplane Π passing through b, and consider the hyperplane Π0 passing throughx∗ parallel to Π. It is convenient to think of Π as being horizontal with a being above thehyperplane.

221

222

HINTS TO EXERCISES

Now, imagine that we place a lamp at a and make Π0 not transparent, excluding the transparent hole formed by the intersection of Q and Π0 . Let B be the lightened part of Π; consider
the conic set Q+ with the vertex at a and the base B (i.e., the convex hull of a and B). Provethat the center of gravity of this set cannot be above Π0 , and conclude from this that the ratioof [a, x∗ ] to [x∗ , b] is at most n.
Exercise 1.3.11+ : consider the family of all closed half-spaces of µ-measure > n/(n + 1).
Prove that the family satisfies the premise of Helley’s theorem and verify that any point which
belongs to the intersection of the sets from the family satisfies the conclusion of the second
theorem of Grunbaum.
Exercise 1.3.13+ : use the result stated in exercise 1.3.12
Exercise 1.3.16+ : the statement in question is exactly the following one:let xn+1 denote the ordered collection of n + 1 points x1 , ..., xn+1 of X, and letr(xn+1 ) = minmaxp∈Φ i=1,...,n+1

k f (xi ) − p(xi ) k .

Thenδ 0 ≡ max r(xn+1 ) = δ.
xn+1

To prove the statement, verify that the maximum in the latter statement is achieved (so thatδ 0 is well-defined) and is ≤ δ. To prove the inverse inequality, apply the Helley theorem to thefamily of sets
{Hx = {p ∈ Φ | k f (x) − p(x) k≤ δ 0 }}x∈X .
Exercise 1.3.17+ : note that the sum H1 + H2 of one-dimensional closed and nonempty setscontains the translations H10 , H20 of H1 , H2 , respectively, with the intersection comprised of asingle point.
Exercise 1.3.18+ : we should prove that
Z
T+

k−1
t
rk−1
(H+
)dt ≡ (Volk (H1 + H2 ))1/k ≥≥ (Volk (H1 ))1/k + (Volk (H2 ))1/k ≡

Z
T1

k−1
rk−1
(H1t )dt +

Z
T2

k−1
rk−1
(H2t )dt,

or, which is the same in view of (1.3.7), that
Z ∞

s
0

k−1

1/k

µ+ (s)ds

Z ∞

≥

s
0

k−1

1/k

µ1 (s)ds

Z ∞

+

s

k−1

1/k

µ2 (s)ds

,

0

where
µi (s) = Vol1 ({t ∈ Ti | rk−1 (Hit ) ≥ s}), i = 1, 2,((x, t) are the coordinates in Rk , x being (k − 1)-dimensional and t being scalar, Ti are theimages of Hi under the projection onto the t-axis and Hit are the inverse images of the point(0, ..., 0, t) in Hi under this projection),andtµ+ (s) = Vol1 ({t ∈ T+ | rk−1 (H+) ≥ s}),HINTS TO EXERCISES

223

t being similar objects for H = H + H .
T+ , H++1
2
Prove that Ti is a nonempty compact set and the function rk−1 (Hit ) is upper semicontinuous
on Ti , so that it attains its maximum ρi over Ti , i = 1, 2.
Reduce the situation to the case of positive ρi and prove and use the fact that if

αi = ρi /(ρ1 + ρ2 ), i = 1, 2,
then for all nonnegative s one hasµ+ (s)ds ≥ µ1 (α1 s) + µ2 (α2 s).

224

HINTS TO EXERCISES

Hints to Section 2.4
Exercise 2.4.5+ : note that Vα is contained in the set of solutions to the system of the following
pair of quadratic inequalities:xT x ≤ 1;(14.2.7)(eT x − α)(eT x − 1) ≤ 0;

(14.2.8)

Consequently, Vα satisfies any convex quadratic inequality which can be represented as a weighted
sum of (14.2.7) and (14.2.8), the weights being positive. Every inequality of this type defines
an ellipsoid; compute the volume of the ellipsoid as the function of the weights and optimize itwith respect to the weights.
Exercise 2.4.6+ : use the same construction as that one for exercise 2.4.5
Exercise 2.4.7+ : use the results given by exercises 2.4.5 and 2.4.6
Exercise 2.4.8+ : use item 3. of the list of consequences of the Fritz John Theorem
Exercise 2.4.9+ : setnk x k= |x|1 ≡

X

|xi |

i=1

and consider the sequence of random vectors defined as follows: to generate ξi , choose randomly,
according to the uniform distribution on {1, ..., n}, an integer ν between 1 and n and set all entriesof ξi with the indices different from ν to zero; the ν-th entry is set to ±1, with a randomly chosen
sign (this choice is independent of that one for ν and both the signs have equal probabilities tobe chosen).
Exercise 2.4.12+ : combine (2.4.9) and (2.4.17)

HINTS TO EXERCISES

225

Hints to Section 3.3
Exercise 3.3.2: prove that the volume of every simplex in the partitioning is at most
1 √ n(2 n) .
n!
Exercise 3.3.3: note that if Γ0 is not a singleton, thenπ(x) =X
1
π(x0 ),|I(x)| x0 ∈I(x)here I(x) is the set of all vertices in Γ0 \{x} from which the moving point can reach x in a single
step and |I(x)| is the # of elements in I(x). Think of where the function π(x), x ∈ Γ0 , mightachieve its maximum (it is a discrete version of the Maximum Modulus principle for harmonic
functions, is not it?).
Exercise 3.3.5+ :given the trajectory {xt }Tt=0 with xT ≥ λT x0 , consider the trajectory{x̄t = λ−t }Tt=0 and verify that it satisfies the inequalities
Āx̄t+1 ≤ B x̄t , t = 0, ..., T − 1(here Ā = λA) and the relationx̄T ≥ x̄0 .
Look what Ā and B do with the nonnegative and nonzero vector
x̄ =

−1
1 TX
x̄t .
T t=0

Exercise 3.3.7: follow the proof of the similar statement for the case of convex problems
without functional constraints (Lecture 2).

226

HINTS TO EXERCISES

Hints to Section 4.3
Exercise 4.3.4: without loss of generality we may assume that the linear form g T x attains its
minimum over D at the vertex v0 and that g T (w − v0 ) = 1. Choosing v0 as our new origin andv1 − v0 , ..., vn − v0 as the orths of our new coordinate axes, we come to the situation studied in
exercise 4.3.3.

HINTS TO EXERCISES

227

Hints to Section 5.5
Exercise 5.5.2: note that a nonsingular n × n matrix B can be represented as
B = B 0 U,
where B 0 is positive semidefinite symmetric and U is orthogonal; this is called the polar decomposition of B. Note that by evident reasons
W (B, c) = W (B 0 , c).
If the existence of the polar decomposition is a news for you, here is the proof: the matrix
B 0 = (BB T )1/2 is well-defined symmetric positive definite; if U = (B 0 )−1 B, thenU U T = (B 0 )−1 (BB T )(B 0 )−1 = (B 0 )−1 (B 0 )2 (B 0 )−1 = I,
so that U is orthogonal. By construction, B = B 0 U .
Exercise 5.5.6: use (5.5.4), (5.5.5) and the following basic fact:
0
let ν1 ≥ ν2 ≥ ... ≥ νn be half-axes of an n-dimensional ellipsoid E and let ν10 ≥ ν20 ≥ ... ≥ νn−1
be the half-axes of the (n − 1)-dimensional ellipsoid obtained by intersecting E by a hyperplanepassing through the center of E. Then νi0 separate νi :0
ν1 ≥ ν10 ≥ ν2 ≥ ν20 ≥ ... ≥ νn−1 ≥ νn−1
≥ νn .

For those non-acquainted with this important statement I would add that this is an immediate
geometrical consequence of the following fundamental characterization of the eigenvalues ν1 ≥
ν2 ≥ ... ≥ νn of a symmetric n × n matrix A (Fisher-Courant):
νi = minmax

F ∈Fi x∈F,|x|2 =1

xT Ax,

where Fi is the family of all linear subspaces in Rn of codimension n − i + 1.
If these facts are new for you, it would be a good idea to find the proofs yourself (it is notdifficult).

228

HINTS TO EXERCISES

Hints to Section 6.3
Exercise 6.3.1: the true statements are 5), 7), 8) and 9)
Exercise 6.3.2+ : the statement is trivial if int G = ∅; thus, we may assume that int G 6= ∅.
It suffices to lead to a contradiction the assumption that a point x ∈ int G does not belong
to Dom F ; without loss of generality, we may assume that the point in question is 0. What
we should do is demonstrate that one can find y in such a way that xT (ξ − y) ≥ 0 wheneverx ∈ Dom F and ξ ∈ F (x) (this would mean that one can add to the graph of F the pair(0, y) without violating monotonicity, thus yielding the desired contradiction with the maximal
monotonicity of F ). Thus, we should prove that certain infinite system of linear inequalities
is solvable. To this end prove, first, that each finite subsystem of the system is solvable and,second, that there exists a finite subsystem of the system with a bounded solution set.
Exercise 6.3.3: f 0 is not necessarily maximal monotone; counter-example: f (x) = 0, −1 <x < 1, f (−1) = f (1) = 1, f (x) = +∞ when |x| > 1.
Exercise 6.3.7+ : use the following observation: a weak solution to the variational inequality
defined by G and a maximal monotone operator H with int G ⊂ Dom H is a strong solution to
the inequality, provided that H is semibounded on int G. Apply this observation to the operator
H = F + NG

HINTS TO EXERCISES

229

Hints to Section 7.5
Exercise 7.5.3: follow the line of argument of the original proof of Proposition 7.3.1. Namely,
apply the proof to the ”shifted” process which starts at xN 0 and uses at its i-th iteration, i ≥ 1,
the stepsize γi+N 0 −1 and the tolerance εi+N 0 −1 . This process differs from that one considered in
the lecture in two issues:(1) presence of time-varying tolerance in detecting productivity and an ”arbitrary” step, instead of termination, when a productive search point with vanishing subgradient of the objectiveis met;(2) exploiting the projection onto Q ⊃ G when updating the search points.
To handle (1), prove the following version of Proposition 2.3.1 (Lecture 2):
Assume that we are generating a sequence of search points xi ∈ Rn and associate with thesepoints vectors ei and approximate solutions x̄i in accordance to (i)-(iii). Let
Gi = {x ∈ G | (x − xj )T ej ≤ 0, 1 ≤ j ≤ i},and let Size be a size. Assume that for some M
Size(GM ) < εM Size(G)(if GM is not a solid, then, by definition, Size(GM ) = 0). Then among the search points
x1 , ..., xM there were productive ones, and the best (with the smallest value of the objective) of
these productive points is a ε1 -solution to the problem.
To handle (2), note that when estimating InnerRad(GN ), we used the equalities|xj+1 − x+ |22 = |xj − x+ |22 + ...
and would be quite satisfied if = in these inequalities would be replaced with ≤; in view of
Exercise 7.5.2, this replacement is exactly what the projection does.

230

HINTS TO EXERCISES

Hints to Section 8.3
Exercise 8.3.4∗ : use the result of exercise 8.3.3 and (8.3.4) to demonstrate thatV + (φi+1 ) ≤ V + (φi ) − 2γi (f (xi ) − f ∗ )/|f 0 (xi )|∗ +L 2
γ , V + (φ) = V (φ) − hφ, x∗ i ,2 i

and then act exactly as in the case of the usual Subgradient Descent.

HINTS TO EXERCISES

231

Hints to Section 9.3
Exercise 9.3.4: since (u, v) ∈ C(i), we have(u, v) ≥iX

λj (f (xj ), g(xj ))j=1

for some nonnegative λj with unit sum; setx=X

λj xj .

j

Exercise 9.3.6: note that if max{u, v} ≥ 0, thenρ(u, v) = max {αu + (1 − α)v}0≤α≤1and take into account that if (u, v) ∈ C(i), then max{u−fi∗ , v} ≥ minG max{fi (x)−fi∗ , gi (x)} ≥0 (why?). With these observations, apply the von Neumann Lemma to the convex-concave
functionF (α, (u, v)) = α(u − fi∗ ) + (1 − α)v
regarded as a function on the direct product [0, 1] × C(i).
Exercise 9.3.7: prove that hi (α), for every i, is majorated by the corresponding gap in theLevel process, and use the efficiency estimate for the usual Level.
Exercise 9.3.8: formulate exactly and use the following simple facts:
1) if h is a concave nonnegative function on a segment δ, α is µ-centered with respect to δand h(α) is small, then maxδ h also is small; thus, if in our method we were lucky not to vary
the current value of α for a long time, we make the gap small;
2) if α is the midpoint of a segment δ and δ 0 ⊂ δ is such that α is not µ-centered with respectto δ 0 , then δ 0 is ”significantly” (at least by a factor 1 − O(1)µ) less than δ; thus, each step where
we are enforced to vary the current value of α is accompanied by a significant reduction in thelength of the current segment δi as compared to the length of the similar segment at the step
where the ”old” value of α was set.
3) if the length of δi becomes small, then ∆i also becomes small.

232

HINTS TO EXERCISES

Hints to Section 12.6
Exercise 12.6.4: apply the criterion given by exercise 12.6.2. To compute the optimal value,
use the representation
Tk (t) = ch(k arcch(t)), t ≥ 1of the Tschebyshev polynomial in the region t ≥ 1.
Exercise 12.6.5: apply the criterion given by exercise 12.6.2. To prove (166), look at thevalue of the objective in (P ) at polynomials p given by(1 − tp(t)) = (1 − tp∗m (t))kwith (m + 1)k ≤ N + 1, p∗m (t) being the optimal solution of problem (P ) for the case of N = m
and γ = 1/2.
Exercise 12.6.6: to get a polynomial s(t) = 1 − tp(t) of degree 2N + 1 such that tγ |1 − tp(t)|is small along the sequence {σi }, take it in the form
1 − tp(t) = (1 − tpN (t))

NY

(1 − t/σi ),i=1where pN (t) is the associated with (166) polynomial of the degree N corresponding to the case
L = σN +1 .

HINTS TO EXERCISES

233

Hints to Section 13.4
Exercise 13.4.4: show that the spectrum of the matrix I + B1 of the transformed systembelongs to the segment Σ = [ρ, 1/ρ], so that the condition number of this matrix does not exceed
ρ−2 . Apply subsequently the upper complexity bound for the family of problems Un (Σ, − 21 , R),
R2 = fA,b (0) − min fA,b , given in Lecture 12 and the result on optimality of the usual CGM
given in Lecture 13.
Exercise 13.4.5: use the Holder inequality. Those more experienced in Complex Analysis
might prefer to obtain the statement in question as an immediate consequence of the Hadamard
Theorem:
if a non identically zero function f (z) is analytic and bounded in the stripe a < <z < b, thenthe function
M (x) = ln sup |f (x + iy)|y∈R

is convex in x ∈ (a, b).
Exercise 13.4.6: Prove that the functions 1, t, t2 , ..., tN are linearly independent in L2 (µ);
apply to this sequence the Gramm-Schmidt orthogonalization process.
Exercise 13.4.8: combine the result stated in exercise 13.4.7.3) with the following fundamental
fact:
let A be a symmetric operator in n-dimensional Euclidean space E, and let λ1 ≤ λ2 ≤ ... ≤ λn
be the eigenvalues of A taken with their multiplicities. Now, let E 0 be a subspace in E ofcodimension l, let A0 be the restriction of A onto E 0 (i.e., the operator on E 0 such that A0 x is
the orthogonal projection onto E 0 of the vector Ax) and let µ1 ≤ µ2 ≤ ...µn−l be the eigenvaluesof A0 taken with their multiplicities. Thenλi ≤ µi ≤ λi+l , i = 1, ..., n − l.
Those for whom this general Separation Theorem for eigenvalues is new are welcomed to obtainit from the Courant-Fisher characterization of the eigenvalues of a symmetric operator:
let A be a symmetric operator on n-dimensional Euclidean space E and let λ1 ≤ λ2 ≤ ... ≤ λn
be the eigenvalues of A taken with their multiplicities. Thenλi = maxmin

L∈Ei x∈L,xT x=1

xT Ax,

where Ei denotes the set of all subspaces of E of codimension i − 1.
Those not familiar with the latter extremely important statement, are strongly recommended
to prove it (it is not difficult).

234

HINTS TO EXERCISES

Solutions to exercises
Solutions to Section 1.3
Exercise 1.3.2: in view of exercise 1.3.1, I can give you almost nothing if you choose x outside
the interior of Q. Now assume that you have chosen x ∈ int Q, so that
x=

n+1
X

λi ui

i=1

for certain positive λi with the unit sum. Look what happens if I leave you the part of the pie
given, in the barycentric coordinates2 , as
Q0 = {y ∈ Q | λi∗ (y) ≥ λi∗ (x)}where i∗ is the index of the largest of λi . In other words, look what happens if my cutting
hyperplane is parallel to a properly chosen (n − 1)-dimensional facet of the simplex (and I takethe part neighbouring to the facet).
Exercise 1.3.4: to prove that the center of gravity is the interior point of the domain, assume,
on contrary, that x∗ ≡ x∗ (Q) 6∈ int Q. Then, due to the Separation Theorem for convex sets,
there exists an affine functional f such that f (x∗ ) = 0 and 0 ≥ f (y) for all y ∈ Q, the inequality
being strict for y ∈ int Q. In view of exercise 1.3.3, we have∗

0 = f (x ) = Voln (Q)−1Z

f (y)dy;Q

the right hand side in this equality clearly is strictly negative, which gives the desired contradiction.
The proof of the affine invariance is as follows:Z

Z

A(x)| Det A|dx

ydy =
Q+
2

Q

given n + 1 affinely independent points ui in Rn , one can represent any y ∈ Rn as the linear combinationy = λ1 (y)u1 + ... + λn+1 (y)un+1 ,with the unit sum of coefficients (why?); these coefficients (which, under this latter requirement, are uniquely
defined) are called the barycentric coordinates of y with respect to {ui }. The barycentric coordinates are affine
functions of y (why?)

235

236

SOLUTIONS TO EXERCISES(substitution of variable in an n-dimensional integral). Further,Z
Q

A(x)| Det A|dx =Z

Z(Ax + b)| Det A|dx = | Det A|Voln (Q)b + | Det A|A(Qxdx) =
Q= | det A|Voln (Q)b + | Det A|A(Voln (Q)x∗ (Q)) == (| Det A|Voln (Q))A(x∗ (Q)) = Voln (Q+ )A(x∗ (Q)).
Thus,
Z
Q+

ydy = Voln (Q+ )A(x∗ (Q)),

whence
∗

x (Q

+

+
) ≡ Vol−1n (Q )

Zydy = A(x∗ (Q)).

Q+

Exercise 1.3.5: let v1 , ..., vn+1 be the vertices of the simplex, and let i 7→ σ(i) be a permutationof the indices 1, ..., n + 1. There exists a unique affine transformation Tσ of the space which, for
all i, maps vi to vσ(i) ; of course, this transformation maps the simplex onto itself. If
x = λ1 v1 + ... + λn+1 vn+1
is the representation of a point in the simplex as a convex combination of the vertices, thenTσ (x) =n+1
Xλi vσ(i) =n+1
Xλσ−1 (i) vi .

i=1

i=1

Now, if x is the center of gravity of the simplex, then, in view of the statement given by exercise
1.3.4, we have Tσ x = x (recall that Tσ maps the simplex onto itself). Since the barycentric
coordinates of a point are uniquely defined by the point, we come to
λσ−1 (i) = λifor all i and all permutations σ, whence λ1 = ... = λn+1 = 1/(n + 1), Q.E.D.
Exercise 1.3.6: due to affine invariance of the statement in question and affine invarianceon the center of gravity, it suffices to consider the following special situation: b is at the origin,
the base B of Q belongs to the hyperplane {xn = 0}, a = (0, ..., 0, 1). It is clear that the last
coordinate of a point in Q varies from 0 to 1, so that we have
Z 1

Z

Voln (Q) =dx =QVoln−1 (Qt )dt,
0

where Qt = {x0 ∈ Rn−1 | (x0 , t) ∈ Q}. It is clear that
Qt = (1 − t)B ≡ {(1 − t)x | x ∈ B}
(indeed, (x0 , t) ∈ Q if and only if (x0 , t) = α(0, ..., 0, 1) + (1 − α)(x1 , ..., xn−1 , 0) with someα ∈ [0, 1] and with x00 ≡ (x1 , ..., xn−1 ) ∈ B. It follows that α = t, so that (x0 , t) ∈ Q if and only
if x0 = (1 − t)x00 for some x00 ∈ B).

SOLUTIONS TO EXERCISES

237

Since Qt = (1 − t)B, we have Voln−1 (Qt ) = (1 − t)n−1 Voln−1 (B). Thus,
Z 1

Vol(Q) =n(1 − t)n−1 Vol(B)dt = n−1 Vol(B).
n−1

0

n−1

Similarly,
Voln (Q)(x∗ (Q))n =Z 1

tVoln−1 (Qt )dt =xn dx =t(1 − t)n−1 Voln−1 (B)dt =

0

0

Q

= Voln−1 (B)

Z 1

Z 1

Z

(1 − τ )τ n−1 dτ = Voln−1 (B)(1/n − 1/(n + 1)) = Voln−1 (B)n−1 (n + 1)−1 ,0whence, due to Voln (Q) = n−1 Voln−1 (B),(x∗ (Q))n = 1/(n + 1).
Now, if i < n, then
∗

Z 1 Z

Z

Voln (Q)(x (Q))i =(x0 )i dx0 }dt.

{

xi dx =0Q

Qt

As we know, Qt = (1 − t)B; the substitution x0 = (1 − t)y in the inner integral results inZ(x0 )i dx0 = (1 − t)nQtZyi dy = 0B(the latter equality follows from the assumption that the (n − 1)-dimensional center of gravityof B is 0), and we come toVoln (Q)(x∗ (Q))i = 0, i = 1, ..., n − 1.
Thus,x∗ (Q) = (0, ..., 0, 1/(n + 1)) = (0, ...0, 1)/(n + 1) + n0/(n + 1) = a/(n + 1) + nb/(n + 1),
as claimed.
Exercise 1.3.8 3): from the concavity of φ it follows that α ≥ a (otherwise Q∗− would be aproper part of Q− and therefore would be of less volume than the latter set). On the other hand,the slope ξ 0 of ξ is ≥ φ0 (−0) (since otherwise Q− would be a proper part of Q∗− and therefore
would be of less volume that this latter set). Since φ is concave, it follows that φ(t) ≥ ξ(t) forsmall in absolute values negative t and φ(t) ≤ ξ(t) for all positive t. From this latter observation
it follows that β ≤ b (otherwise Q+ would be a proper part of Q∗+ and these sets could not beof equal volumes).
Now consider the function
g+ (t) = n−1
ξ(t) − φn−1 (t), 0 ≤ t ≤ β−φn−1 (t), β < t ≤ b.

As we just have seen, this function is nonnegative for 0 ≤ t ≤ β and is nonnegative whenβ < t ≤ b. We clearly have
0 = Voln (Q∗+ ) − Voln (Q+ ) = cn

Z bg+ (t)dt0

238

SOLUTIONS TO EXERCISES(cn is the (n − 1)-dimensional volume of the unit Euclidean ball in Rn−1 ), which combined withthe fact that g+ is nonnegative to the left of β and is nonpositive to the right of this point
implies that0 ≥ cnZ b0

tg+ (t)dt = It (Q∗+ ) − It (Q+ );this is the first inequality required in 3).
Now, since α ≥ a, φ is a nonnegative concave function on [−a, 0], ξ is a nonnegative linear
function on [−α, 0] and ξ(0) = φ(0), ξ 0 ≥ φ0 (−0), the functiong− (t) = n−1
ξ(t) − φn−1 (t), −a ≤ t ≤ 0

ξ n−1 (t), −α ≤ t < −ais as follows: there is certain τ ∈ [−α, 0] such that g− (t) is nonnegative when −α ≤ t < τ andis nonpositive when τ < t ≤ 0. Besides this, from Voln (Q∗− ) = Voln (Q− ) it follows thatZ 0−α

It follows thatZ 0−αg− (t)dt = 0.

(−t)g− (t)dt ≥ 0,or, which is the same, thatI−t (Q∗− ) ≥ I−t (Q− ),which completes the proof of 3).
When transforming Q into Q∗ , we move masses from Q\Q∗ to the left, thusincreasing the momentum I−t of the ”left” part of the body and decreasing the
momentum It of the ”right” part of it. As a result, the momentum I−t of the leftpart of the updated body becomes larger than the momentum It of the right part ofit, since before updating the parts were balanced.
Exercise 1.3.9: the left hand side ratio clearly is equal toR1(1 − s2 )(n−1)/2 ds
2 0 (1 − s2 )(n−1)/2 ds
Rα1(to get the n-dimensional volumes, integrate the (n−1)-dimensional volumes of the cross-sectionsof the body by the hyperplanes {xn = s} over s).
If n > 1, then the numerator can be bounded from above as follows:
Z 1(1 − s2 )(n−1)/2 ds ≤

α

Z 1

exp{−
α

<

(n − 1)s2
}ds ≤
2

Z 1

exp{−
α

(n − 1)αs}ds <2(n − 1)α2
2
exp{−
}.
(n − 1)α2

If n > 3, then the denominator can be bounded from below as follows:
Z
Z √
1

2
0

2/(n−1)(1 − s2 )(n−1)/2 ds > 20(1 − s2 )(n−1)/2 ds >

SOLUTIONS TO EXERCISES

239

Z √2/(n−1)

2

(1 −

0

q
2 (n−1)/2)
ds > ω 2/(n − 1)
n−1

with certain absolute constant ω.
Thus, in the case of n > 3 the ratio in question does not exceed1nα2
κ0 √ exp{−
},
α n
2
√
κ0 being an absolute constant. We conclude that if α n ≥ 1 thenVoln (Vα )/Voln (V ) ≤ κ0 exp{−

nα2
};
2

in the opposite case we clearly have
Voln (Vα )/Voln (V ) ≤ 1 ≤ κ00 exp{−

nα2
},
2

where κ00 = exp{1/2}. It remains to set κ = max{κ0 , κ00 }.
Exercise 1.3.10: let [a, b] be a chord of a closed convex domain Q passing through the centerof gravity of Q; without loss of generality, let the latter be at the origin. Assume that [a, 0] is
the larger, and [0, b] is the smaller part of the chord. Now, from Separation Theorem for convex
sets it follows that there exists a nonconstant linear functional which attains its minimum over
Q at b; under appropriately chosen coordinates we may assume that this functional is simply
the last coordinate, xn ; let −β be the value of the functional at b and α be its value at a, sothat α > β > 0, and what should be proved is thatα ≤ nβ.
Let Q0 be the cross-section of Q by the hyperplane Π0 = {xn = 0}, and K be the cone with thevertex at a which has Q0 as its intersection with Π0 :K = {x = a + λ(y − a) | y ∈ Q0 , λ ≥ 0}.
Last, let Q∗ be the part of K which is above the hyperplane Π = {xn = −β}:Q∗ = {x ∈ K | xn ≥ −β}.
Let Q∗− , Q∗+ be the parts of Q∗ in the half-spaces {xn ≤ 0}, {xn ≥ 0}, respectively, and let Q− ,Q+ be similarly defined parts of Q. We claim that Q− is contained in Q∗− , while Q+ contains
Q∗+ .
The claim is ”geometrically evident”, and the rigorous proof is as follows. The first statement:
if x ∈ Q is a point with nonpositive last coordinate, then the segment [a, x] intersects Π0 ; theintersection point x0 (which belongs to Q since a and x belongs to Q and Q is convex) gives
us the ray [a, x0 ) which is contained in the cone K and contains x, so that x belongs to K.
Since x ∈ Q and the whole Q is above the hyperplane Π (i.e., xn ≥ −β), x belongs to Q∗ andconsequently to Q∗− (recall that xn ≤ 0 due to x ∈ Q− ).

240

SOLUTIONS TO EXERCISES

Now let us verify that Q∗+ is contained in Q+ . Indeed, if x = a + λ(x0 − a) ∈ Q∗+ , whereλ ≥ 0 and x0 ∈ Q0 , then0 ≤ xn = an + λ(x0n − an ) = an − λan = (1 − λ)α,so that λ ≤ 1. Therefore x = (1 − λ)a + λx0 is a convex combination of two points from Q and
therefore belongs to Q, and since xn ≥ 0, x belongs to Q+ .
Now, since the center of gravity of Q is at the origin, the momentum of Q+ with respect
to the functional xn is equal to the momentum of Q− with respect to the functional −xn . The
momentum of Q∗+ with respect to xn does not exceed that one of Q+ (since Q∗+ is contained in
Q+ ), and the momentum of Q∗− with respect to the functional −xn is less than that one of Q− .
We conclude that
Ixn (Q∗+ ) ≤ I−xn (Q∗− ),and therefore the center of gravity of Q∗ is below the hyperplane {xn = 0} or on the hyperplane.
But the set Q∗ is conic with the vertex at the hyperplane {xn = α}E and the base in thehyperplane {xn = −β}, and therefore the n-th coordinate of the center of gravity is α/(n +1) − nb/(n + 1) (exercise 1.3.6), and since the coordinate is nonpositive, we come to α ≤ nβ, as
required.
Exercise 1.3.11: let F be the family comprised of all closed half-spaces with µ-measure> n/(n + 1). Let us prove that the family satisfies the premise of the Helley theorem. Let us
verify, first, that every n + 1 sets of the family have a point in common. Indeed, otherwise there
would be n + 1 half-spaces Π1 , ..., Πn+1 from the family with empty intersection, so that theunit of the n + 1 open half-spaces Rn \Πi would be the whole space. Since the µ-measure ofeach of these open half-spaces is < 1/(n + 1) (the definition of the family), we conclude that themeasure of the whole space is < 1, which is a contradiction.
It remains to verify that there exist finitely many sets from F with bounded intersection.
Indeed, for a given unit vector e and large enough α the half-space {x | eT x ≤ α} belongs to
the family; choosing e = ±ei , e1 , ..., en be the basis orths in Rn , we find that the intersection of
the corresponding 2n half-spaces from the family is a bounded parallelotope, as required.
Thus, F satisfies the premise of Helley’s theorem, and due to this theorem all the sets fromthe family have a point in common, let it be x. Let us prove that this is the point required by
the Grunbaum theorem, i.e., that any closed half-space Π containing x is of µ-measure at least
1/(n + 1). Assume that it is not the case and there exists a closed half-space Π containing x ofthe mass < 1/(n + 1). Translating the boundary hyperplane outside Π we obtain a half-space
Π0 which contains x in its interior; a small enough translation results in the mass of Π0 being
still less than 1/(n + 1). It follows that the closed half-space Rn \(int Π0 ) is of mass > n/(n + 1)
and therefore belongs to F; at the same time, this half-space does not contain x, which is acontradiction.
Exercise 1.3.12:
without loss of generality we may assume that the functions fi active
at x0 are f1 , ..., fk . Let Vi = ∂fi (x0 ), i = 1, ..., k; then Vi are closed and bounded convex
sets. Therefore the convex hull of the union U of V1 , ..., Vk is nothing but the set of all convex
combinations of k vectors, i-th of the vectors belonging to Ui :V = {v = λ1 v1 + ... + λk vk | λi ≥ 0,Xiλi = 1, vi ∈ Ui }.

SOLUTIONS TO EXERCISES

241

We should prove that V is ∂f (x0 ). It is clear that V ⊂ ∂f (x0 ); indeed, ifv=

kX

λ i vii=1

is a convex combination of subgradients of f1 , ..., fk at x0 , thenfi (x) ≥ fi (x0 ) + (x − x0 )T vi = f (x0 ) + (x − x0 )T vi(we have used the fact that fi are active at x0 ), whence
f (x) ≥ max fi (x) ≥i=1,...,kk
Xλi fi (x) ≥

i=1

k
X

λi (f (x0 ) + (x − x0 )T vi ) = f (x0 ) + (x − x0 )T v,

i=1

so that v is a subgradient of f at x0 .
It remains to verify that every subgradient g of f at x0 belongs to V . To this end let us
first prove that V is closed. Indeed, V is the image of a closed and bounded (and thereforeP
compact) set V1 × V2 × ... × Vk × ∆ (∆ is the simplex {(λ1 , ..., λk ) | λi ≥ 0, i λi = 1}) under
the continuous mapping(v1 , ..., vk , λ1 , ..., λk ) 7→

k
X

λi vi

i=1

and therefore is a compact and, consequently, a closed set.
Assume that a subgradient g of f at x0 does not belong to the closed and convex set V and
let us lead this assumption to a contradiction. Using the Separation Theorem for convex sets,
we may find a nonzero h such thatg T h > max g T v,
v∈V

whence, due to the origin of V ,g T h > max

max v T h = max lim t−1 (fi (x0 + th) − fi (x0 )) =i=1,...,k v∈∂fi (x0 )

i=1,...,k t→+0

= max lim t−1 (fi (x0 + th) − f (x0 ))
i=1,...,k t→+0

(14.2.9)(we have taken into account that fi (x0 ) = f (x0 ), i = 1, ..., k).
Now note that in a neighbourhood of x0 f coincides with maxi=1,...,k fi (recall that all fi ,i = 1, ..., m, are convex in a neighbourhood of x0 and therefore continuous at x0 and that
f1 , ..., fk are exactly those of fi which are active at x0 ), so that g, being a subgradient of f at
x0 , is also a subgradient of the function f¯(x) = maxi=1,...,k fi (x). It follows thatg T h ≤ lim t−1 (f¯(x0 + th) − f¯(x0 )) = lim t−1 ( max fi (x0 + th) − f (x0 )) =
t→+0

t→+0

i=1,...,klim max t−1 (fi (x0 + th) − f (x0 )) = max lim t−1 (fi (x0 ) − f (x0 )),t→+0 i=1,...,ki=1,...,k t→+0

which is a contradiction to (14.2.9).
Now assume that x0 is a minimizer of f . Then 0 is a subgradient of f at x0 ; according
to already proved statement, this subgradient is a convex combination of certain subgradients

242

SOLUTIONS TO EXERCISES

v1 , ..., vk of f1 , ..., fk at x0 ; in view of the Caratheodory theorem, we may represent 0 as a convex
combination of l ≤ n + 1 of these subgradients, let them be vi1 , ..., vil . Again using our mainstatement, we see that 0 is a subgradient at x0 of the maximum of properly chosen l ≤ n + 1functions of the family f1 , ..., fk , as required by the second statement of the exercise.
Exercise 1.3.13: r(x) clearly is a convex function of x; since all Qi are bounded, r(x) → ∞ as|x| → ∞, so that a minimizer x∗ of r does exist. According to the second statement of exercise
1.3.12, we may select l ≤ n + 1 of the functions dist(x, Qi ) in such a way that their maximumr̄(x) attains its minimum at x∗ , the value of this maximum at x∗ being exactly that one of r.
But the minimal value of r̄ clearly is 0 (since the l ≤ n + 1 sets which participate in forming r̄
have a point in common); thus, 0 = r̄(x∗ ) = r(x∗ ), so that x∗ is a common point of all Qi , asclaimed.
Exercise 1.3.14: given a finite family of closed convex sets satisfying the premise of theHelley theorem, we clearly can find a large enough ball such that the family comprised of theintersections of the initial sets with the ball also satisfies the premise (it suffices to choose apoint in each of intersections of n + 1 members of the family and take the ball which covers theresulting finite set). Due to the statement given in exercise 1.3.13, the ”truncated” sets (andtherefore the initial ones) have a point in common.
Exercise 1.3.15: in the case of a finite family the conclusion is already proved (exercise
1.3.14). It remains to consider the case when the family is infinite; in this case, by assumption
given in the premise of the theorem, there exists a finite subfamily F of our family such that theintersection V of the sets from the subfamily is bounded. Let α denote a finite subfamily of theinitial family containing F, and let Fα be the intersection of all sets from the subfamily α. ThenFα is a nonempty (due to already proved version of the theorem for finite families) and closed
subset of V and is therefore a nonempty compact set. The sets Fα corresponding to all possible
α clearly form a nested family of nonempty compacts (”nested” means that the intersection offinitely many sets of the family is again a set from the family); from the well-known theorem oncompacts the members of such a family have a nonempty intersection, Q.E.D.
Exercise 1.3.16: without loss of generality we may assume that φi are linearly independent(no nontrivial linear combination of φi is identically zero), since otherwise we could replace thefamily {φi } by its maximal linearly independent subfamily without violating the linear span of
the functions. Thus, we may assume that Φ is a n-dimensional linear space. Let k = n + 1,
X k = X × X × ... × X (k direct factors), and let xk = (x1 , ..., xk ) denote a point from X k . Setr(xk ) = min k f − p k{x1 ,...,xk },∞p∈Φbe the minimal deviation of f from Φ on the set comprised of the points from the collection xk .
r(xk ) clearly is the minimum of a family of continuous functions of xk and therefore is uppersemicontinuous function on the compact set X k ; therefore it attains its maximum, let it be δ 0 ,at certain point xk∗ .
The statement we should prove means exactly that δ 0 = δ; the inequality δ ≥ δ 0 is evident (iff can be approximated by an element of Φ, within the uniform error δ, on the whole X, it, ofcourse, can be approximated within the same error at any k-point subset of X). What should
be proved is the inverse inequality
δ 0 ≥ δ.
To prove the inverse inequality, let us act as follows.

SOLUTIONS TO EXERCISES

243

For any x ∈ X let Hx = {p ∈ Φ | k f (x) − p(x) k≤ δ 0 } be the set of all elements of Φ whichapproximate f within accuracy δ 0 at x. It is clear that Hx is a closed and convex subset of the(k − 1)-dimensional linear space Φ. To prove that δ ≥ δ 0 it suffices to prove that all sets ofthe family {Hx }x∈X have a point in common, and to prove this latter statement we, of course,
should use the Helley theorem. Let us verify the premise of the theorem. From the origin of δ 0
it follows that any k of the sets of the family {Hx }x∈X have a point in common. Thus, the only
thing we need is to verify that if X is infinite, then there is a finitely many sets from the familywith a bounded intersection. To this end let us note that any finite subset Y of X such thatthe restrictions of φi onto Y form a linearly independent system of functions gives us a finite
subfamily {Hx }x∈Y of the initial family with a bounded intersection. Indeed, to say that the
restrictions of φi onto Y are linearly independent is the same as to say that the mapping whichmaps an element from Φ into its restriction onto Y is a linear embedding; under this embedding
the intersection of the sets Hx , x ∈ Y , is mapped into a bounded subset of a finite-dimensional
space (comprised of all functions on Y taking values in E), and therefore the intersection itself
also should be bounded.
Thus, we come to the necessity to verify the following fact (which is important in its own
right):
given k linearly independent functions φi : X → E taking values in a linear space E, one
can find a finite subset Y in X such that the restrictions of φi onto Y are linearly independent;
in fact this set can be chosen to have at most k points.
The proof is immediate (induction in k): for k = 1 the statement is evident (a linearly
independent system comprised of a single function, i.e., a function which is not identically zero,has a linearly independent restriction on a properly chosen single point set, namely, on any point
where the function is nonzero). Now the inductive step: given k + 1 linearly independent on X
functions φ1 , ..., φk+1 , we, by the inductive hypothesis, can find a finite subset T (comprised ofat most k points) such that the restrictions of φ1 , ..., φk onto T are linearly independent. It is
possible that the restrictions of φ1 , ..., φk+1 onto T also are linearly independent, and then weare done. Now assume that these k + 1 restrictions are linearly dependent; since the first k ofthem are linearly independent, we haveφk+1 (x) = λ1 φ1 (x) + ... + λk φk (x), x ∈ T.
This latter equality is violated at certain x0 ∈ X (since otherwise φ1 , ..., φk+1 would be linearly
dependent); let us add x0 to T . The extended set T 0 is comprised of at most k + 1 points and therestrictions of φi , i = 1, ..., k + 1 onto T 0 are linearly independent. Indeed, any nontrivial linear
combination of φi which could be identically zero on T 0 could be rewritten as a representationof φk+1 as a linear combination of φi , i ≤ k (since T 0 ⊃ T and the restrictions of φi , i ≤ k,onto T are linearly independent); this linear combination should be exactly the aforementioned
one (again due to linear independence of φi , i ≤ k, on T ); but this latter equality is violated at
x0 ∈ T 0 , which gives the desired contradiction.
Exercise 1.3.17: without loss of generality we may assume that the smallest points in H1
and H2 coincide with zero (since translating any of the sets we simply translate the sum of thesets), and let bi denote the largest point of Hi , i = 1, 2.
Since H1 contains 0, H1 + H2 contains H2 , and since H2 contains b2 , H1 + H2 contains
b2 + H1 . Now, the set H2 is to the left of b2 , and b2 + H1 is to the right of b2 , so that we have244

SOLUTIONS TO EXERCISES

found two subsets in H1 + H2 which are congruent to H1 , H2 , respectively, and have a single
point in common, so that
Vol1 (H1 + H2 ) ≥ Vol1 (H1 ) + Vol1 (H2 ),as required in (SA1 ).
Exercise 1.3.18: we should prove that
(Volk (H1 + H2 ))1/k ≥ (Volk (H1 ))1/k + (Volk (H2 ))1/k ,(14.2.10)

or, which is the same in view of (1.3.7), that
Z ∞

s
0

k−2

1/k

µ+ (s)ds

Z ∞

≥

s

k−2

1/k

µ1 (s)ds

Z ∞

+

0

s

k−2

1/k

µ2 (s)ds,(14.2.11)0

whereµi (s) = Vol1 Ti (s), Ti (s) = {t ∈ Ti | rk−1 (Hit ) ≥ s}), i = 1, 2,((x, t) are the coordinates in Rk , x being (k − 1)-dimensional and t being scalar, Ti are theimages of Hi under the projection onto the t-axis and Hit are the inverse images of the point(0, ..., 0, t) in Hi under this projection),andtµ+ (s) = Vol({t ∈ T+ | rk−1 (H+) ≥ s}),
1t being similar objects for H = H + H .
T+ , H++1
2(14.2.10) is evident in the case when Volk (H1 ) or Volk (H2 ) is zero, since H+ contains translations of both H1 and H2 and therefore the volume of H+ is not less than those of H1 and H2 .
Thus, it suffices to consider the case when both H1 and H2 are of positive volumes.
Note that from compactness and nonemptiness of Hi it follows immediately that Ti is anonempty compact set and rk−1 (Hit ) is upper semicontinuous on Ti , i = 1, 2. Let si be themaximum of rk−1 (Hit ) over t ∈ Ti , i = 1, 2.
Let us first note that whenever t0 ∈ T1 , t00 ∈ T2 , one has t0 + t00 ∈ T+ and
0

00

0

00

t +t
H+
⊃ H1t + H2t ,(14.2.12)

whence, from the inductive hypothesis (SAk−1 ),
0

00

0

00

t +t
rk−1 (H1t ) + rk−1 (H2t ) ≥ rk−1 (H+).

Letρi = max rn−1 (Hit ), i = 1, 2,t∈Ti

and letαi = ρi /(ρ1 + ρ2 ),so that αi are positive and α1 + α2 = 1.

(14.2.13)

SOLUTIONS TO EXERCISES

245

Our next observation is that whenever 0 ≤ s ≤ ρ1 + ρ2 , one hasµ+ (s) ≥ µ1 (α1 s) + µ2 (α2 s)(14.2.14)

Indeed, by definitiontµ+ (σ) = Vol1 T+ (σ), T+ (σ) = {t ∈ T+ | rn−1 (H+) ≥ σ},µi (σ) = Vol1 Ti (σ), Ti (σ) = {t ∈ Ti | rn−1 (Hit ) ≥ σ}, i = 1, 2.
Note that the sets Ti (·) are compact, since rk−1 (Hit ) are, as it was already mentioned, uppersemicontinuous functions on the compact sets Ti , i = 1, 2.
To prove (14.2.14), let us fix s ∈ [0, ρ1 +ρ2 ] and note that then si = αi s ≤ ρi , i = 1, 2, so thatthe sets Ti (si ) are nonempty. In view of (14.2.12)-(14.2.13) the sum of these sets is contained in
T+ (s1 + s2 ) = T+ (s); in view of already proved (SA1 ) (exercise 1.3.17), we conclude thatµ+ (s) ≡ Vol1 (T+ (s)) ≥ Vol1 (T1 (α1 s) + T2 (α2 s)) ≥≥ Vol1 (T1 (α1 s)) + Vol1 (T2 (α2 s)) = µ1 (α1 s) + µ2 (α2 s),
as required by (14.2.14).
From (1.3.7) we haverkk (Hi ) = (k − 1)Z ∞sk−2 µi (s)ds ≡ (k − 1)

0

Z ρi

sk−2 µi (s)ds, i = 1, 2,0(note that µi (s) clearly vanishes when s > ρi ), whence
rkk (Hi ) = (k − 1)αik−1

Z ρi /αi0sk−2 µi (αi s)ds = (k − 1)αik−1

Z ρ1 +ρ2

sk−2 µi (αi s)ds.

0

Thus,
Z ρ1 +ρ20

sk−2 µi (αi s)ds = (k − 1)−1 rkk (Hi )αi1−k .

(14.2.15)

Besides this, from (1.3.7) it follows thatZ ρ1 +ρ20

sk−2 µ+ (s)ds ≤ (k − 1)−1 rkk (H+ ).

(14.2.16)

Taking sum of equalities (14.2.15) over i = 1, 2 and taking into account (14.2.14) and (14.2.16),
we come torkk (H1 + H2 ) = (k − 1)

Z ∞0

sk−2 µ+ (s)ds ≥2Xrkk (Hi )αi1−k ,i=1and to prove the required relationrk (H1 + H2 ) ≥ rk (H1 ) + rk (H2 )it suffices to use the elementary inequality
rk /αk−1 + sk /β k−1 ≥ (r + s)kwhenever positive reals r, s, α, β are such that α + β = 1 and k ≥ 2.

246

SOLUTIONS TO EXERCISES

Remark 14.2.1 The elementary inequality mentioned in the last proof is a particular case of
the general inequalitymX
i=1|ri |k αi−k+1 ≥ |

m
X
i=1

ri |k (mXαi )−k+1i=1(αi are positive, ri are real and k ≥ 2).
You are welcome to regard the proof of this general inequality as an additional exercise.

SOLUTIONS TO EXERCISES

247

Solutions to Section 2.4
Exercise 2.4.5: when taking a weighted sum of the inequalities (14.2.7) and (14.2.8), we
may take the first weight being equal to 1 (this is equivalent to certain normalization of theresulting inequality, and the normalization, of course, does not influence the volume of theellipsoid defined by the inequality). Let 2λ ≥ 0 be the weight for the second inequality; the”combined” inequality associated with this weight clearly isxT (I + 2λeeT )x − 2λ(1 + α)eT x + 2λα ≤ 1;
the center c of the ellipsoid is given by(I + 2λeeT )c = λ(1 + α)e,
whenceλ(1 + α)e.
1 + 2λThe inequality rewritten with respect to the center isc=(x − c)T (I + 2λeeT )(x − c) ≤ 1 − 2λα + cT (I + 2λeeT )c ==(1 − 2λα)(1 + 2λ) + λ2 (1 + α)2≡ b(λ) ≡1 + 2λs2 (λ),
1 + 2λThe corresponding ellipsoid is given by≡s(λ) = 1 + (1 − α)λ.

Wλ = {x | (x − c)T Aλ (x − c) ≤ 1},

Aλ = b−1 (λ)(I + 2λeeT );recall that this ellipsoid contains Vα , since the inequality defining the ellipsoid is a weighted sumof the inequalities (14.2.7) and (14.2.8), the weights being 1 and 2λ.
Due to exercise 2.4.2, the volume of the ellipsoid Wλ is given by
vol(Wλ ) = Det
(

=



−1/2
Aλ

s2n (λ)(1 + 2λ)n+1



=

 n

b (λ) 1/2

1 + 2λ=)1/2≡ v 1/2 (λ).

Let us minimize the function v(λ); we should start with finding its critical points given by theequationv 0 (λ) = 0,i.e., by
2ns2n−1 (λ)s0 (λ)(1 + 2λ)n+1 − 2(n + 1)(1 + 2λ)n s2n (λ) = 0;
after immediate simplifications we come to a linear equation which gives
λ=1 + nα.
(n − 1)(1 − α)248

SOLUTIONS TO EXERCISES

In the range (−1/n, 1) of values of α the right hand side turns out to be the minimizer of v(λ)
over λ ≥ 0.
For the optimal value of λ the center of the ellipsoid becomesc=

1 + nα,1+nand the volume of it isvol(Wλ ) = v

1/2

sn (λ)(λ) ==(1 + 2λ)(n+1)/2(

n2
n2 − 1)n/2 sn−1(1 − α2 )(n−1)/2 (1 − α).
n+1

It is easily seen that the right hand side in the latter equality is a decreasing function of α ∈[−1/n, 1] and that the function is equal to 1 at the point α = −1/n. Thus, the ellipsoid given by
the above reasoning is, in the case of 1 > α > −1/n, of smaller volume than the unit Euclideanball, as claimed.
The fact that the ellipsoid Wλ can be represented as the image of the unit Euclidean ballunder the affine mapping indicated in exercise 2.4.5 is quite straightforward.
Exercise 2.4.6: All points of V α clearly satisfy the following pair of quadratic inequalities:xT x ≤ 1;xT eeT x ≤ α2 .
Taking weighted sum of the inequalities with the weights 1 and λ ≥ 0, we observe that V α is,for every λ ≥ 0, covered by the ellipsoidW λ = {x | xT A(λ)x ≤ 1},

A(λ) = (1 + λα2 )−1 (I + λeeT ).

Let us find the best covering of this type, i.e., let us minimize in λ ≥ 0 the quantity1/2

vol(W λ ) = Det −1/2 A(λ) = (1 + λα2 )n /(1 + λ).

Am immediate calculation results in
λ=

1 − nα2(n − 1)α2√(note that in our range |α| < 1/ n of values of α λ is positive). The indicated choice of λ resultsin()(n−1)/2√n(1 − α2 )λvol(W ) = α n;n−1√
the derivative of the right hand side in α is positive when 0 ≤ α < 1/ n and the right hand√√side is equal to 1 when α = 1/ n, so that the right hand side is < 1 for 0 ≤ α < 1/ n.
Exercise 2.4.7:(i): Let W be (an) extremal outer ellipsoid associated with Q. As we know, there exists
an invertible affine transformation of the space which maps this ellipsoid onto the centered at 0
unit Euclidean ball V , and since the extremal ellipsoids ”accompany” the corresponding bodies

SOLUTIONS TO EXERCISES

249

when the latter are subject to invertible affine transformations, V is (an) extremal outer ellipsoidassociated with the image Q0 of Q under this transformation. What we should prove is that theconcentric to V Euclidean ball V 0 of the radius 1/n is contained in Q0 . Assume that it is not
the case; then Q0 does not contain the interior of V 0 (indeed, Q0 is closed, and if containing theinterior of V 0 , it would contain V 0 as well). Thus, there exists a point u such that |u|2 < 1/nand u 6∈ Q0 . Due to the Separation Theorem for convex sets, there exists a linear functionaleT x,
e being a unit vector, such thateT u ≤ eT z, z ∈ Q0 .
Besides this, we know that Q0 is contained in the unit Euclidean ball V . Thus, Q0 is containedin the spherical hat
Vα = {x ∈ V | eT x ≥ α ≡ eT u};
since e is a unit vectors and |u|2 < 1/n, we have α > −1/n. Due to this latter observation,Vα can be covered by an ellipsoid of volume strictly less than that on of V (exercise 2.4.5), andthis ellipsoid, of course, covers Q0 ⊂ Vα as well. This is the desired contradiction, since V is anextremal outer ellipsoid for Q0 .
(ii) Consider all ellipsoids centered at c and containing Q; it is immediately seen that thisfamily of ellipsoids is nonempty and contains at least one ellipsoid W of the smallest volume. (Ofcourse, if we knew that the extremal outer ellipsoid of Q is unique, we could use the symmetry
reasons to prove that W is nothing but the extremal outer ellipsoid associated with Q.) Sameas above, let us perform an invertible affine transformation which maps W onto the centered at
0 unit Euclidean ball V , so that V is (an) ellipsoid of minimal volume among those centered at
the origin ellipsoids which contain the image, Q0 , of Q under the above transformation. Let us√
prove that the centered at the origin Euclidean ball V 00 of the radius 1/ n is contained in Q0 ;this is exactly what we need.
Assume, on contrary, that V 00 is not contained in Q0 . Same as above, it means that there√
exists a point u, |u|2 < 1/ n, in the space which does not belong to Q0 , and, consequently, there
is a linear functionaleT x,
e being a unit vector, such thateT u ≤ eT z, z ∈ Q0 .
Since Q0 is central symmetric with respect to the origin (recall that Q was central symmetricwith respect to the center of W , by definition of W ), the latter inequality implies that−eT u ≥ eT z, z ∈ Q0 ;since also Q0 ⊂ V , we conclude that Q0 is contained in the symmetric spherical stripe
V α = {x ∈ V | |eT x| ≤ α ≡ |eT u|}.
√√Since |u|2 < 1/ n and |e|2 = 1, we have α < 1/ n, and the statement given by exercise 2.4.6
says to us that V α (and therefore Q0 ⊂ V α ) can be covered by a centered at the origin ellipsoidof volume less than that one of V , which is the desired contradiction.

250

SOLUTIONS TO EXERCISES

Exercise 2.4.8: according to item 3. of the list of consequences of the Fritz John Theorem,
there exists a Euclidean norm | · |A such thatn−1/2 |x|A ≤k x k≤ |x|A , x ∈ Rn .

(14.2.17)

From the left inequality it follows thatE{|ξi |2A } ≤ ns2i ,whence, due to the Tschebyshev inequality for a Euclidean norm,
E{|ΞN |2A } ≤ n

N
X

s2i ;

i=1
2 , and we are
the left hand side in this inequality, due to the right inequality in (14.2.17), is ≥ SN
done.
Exercise 2.4.9: let us look what is going on in the situation described in the hint to theexercise. Here, by construction, we have si = 1 (the | · |1 - norm of all ξi is 1). Now, let us
look at k-th entry ΞkN in the sum ΞN . This is a sum of iid random reals with zero mean andthe dispersion σ 2 = n−1 ; due to the Central Limit Theorem, the distribution of the normalizedentryqζN = ΞkN n/N

converges, as N → ∞, to the standard Gaussian distribution; it is easily seen that2
lim E{|ζN |} √
N →∞
2π

Z ∞

2
t exp{−t2 /2}dt = √ .
2π0Consequently,
E{|ΞkN |} =q2
N/n √ (1 + o(1)), N → ∞,2π

whence, due to the choice of the norm,
E{|ΞN |1 } =√2N n √ (1 + o(1)), N → ∞;2π

it remains to note that, due to the Cauchy inequality,
E{|ΞN |21 } ≥ (E{|ΞN |1 })2 .
Exercise 2.4.10: it is clearly seen that|x|∞ = lim |x|p ;p→∞therefore it suffices to demonstrate (2.4.8) for the case of finite p ≥ p0 > 1. Due to the homo0
geneity reasons, we may restrict ourselves with the case of |x|p0 = 1. Let αi = |xi |p , so thatPi αi = 1 and αi ≥ 0. The quantity |x|p is nothing but
!1/p
n
X
p/p0

αi

i=1

.

SOLUTIONS TO EXERCISES

251

The function
f (α) =

n
X
p/p0

αi

i=1

is convex on the simplex {α ≥ 0, i αi = 1}, and due to its symmetry it attains its minimum
over the simplex at its barycenter (i.e., when αi = 1/n, i = 1, ..., n), and attains its maximum
at a vertex of the simplex. Thus,P

0

1 ≥ f (α) ≥ n(1/n)p/p
and consequently
1 ≥ |x|p ≥ n1/p−1/p0whenever |x|p0 = 1, as claimed.
Exercise 2.4.11: it is convenient to denote, for a given p ∈ (2, ∞),π(x) = |x|p , x ∈ Rn , f (x) = π 2 (x), q =pp−1(so that 1/p + 1/q = 1).
The function f (x) clearly is continuously differentiable, with the derivative at a point x 6= 0
given byf 0 (x) = 2π(x)ζ(x) ≡ 2|x|p ζ(x),(14.2.18)
where the vector ζ(x) is defined as follows:ζj (x) = sign(x̄j )|x̄j |p−1 ,

x̄ =

x
.
|x|p

10 . Let us first prove that if u and v are vectors of the unit | · |p -norm, then|ζ(u) − ζ(v)|q ≤ (p − 1)2(p−2)/(p−1) |u − v|p .

(14.2.19)

Indeed, letηj = max{|u|j , |v|j };then one clearly has|η|p ≤ 21/p

(14.2.20)(recall that u and v are of the unit lp -norm). On the other hand, since u and v are of the unit
lp -norm, we haveζj (u) = |uj |p−1 sign uj , ζj (v) = |vj |p−1 sign vj ,and since the absolute value of the derivative of the function |t|p−1 sign t on the segment −ηj ≤t ≤ ηj (which covers both uj and vj ) does not exceed (p − 1)ηjp−2 (recall that p ≥ 2), we cometo|ζj (u) − ζj (v)| ≤ (p − 1)ηjp−2 |uj − vj |.
It follows that if
β = p/q ≡ p − 1 ≥ 1, α = β/(β − 1) =

p
p−1
=
,
p−2
(p − 2)q252

SOLUTIONS TO EXERCISES

then(p − 1)−qnXnX(p−2)q|ζj (u) − ζj (v)|q ≤

ηj

j=1



≤

j=1

1/α 

nX(p−2)qα 
η
j

n
X



j=1

|uj − vj |q ≤

1/β

= |η|pp(p−2)/(p−1) |u − v|p/(p−1)
p|uj − vj |qβ 

j=1(we have used the Holder inequality j aj bj ≤ ( j aαj )1/α ( j bβj )1/β which holds true for all
nonnegative aj , bj and positive α, β such that 1/α + 1/β = 1). We conclude form the above
chain of inequalities thatPPP

|ζ(u) − ζ(v)|q ≤ (p − 1)|η|p−2
p |u − v|p ,which combined with (14.2.20) results in (14.2.19).
20 . Now let x and y be a pair of nonzero vectors from Rn and let x̄, ȳ be their normalizationsof the unit | · |p -norm:xyx̄ =, ȳ =.
π(x)π(y)
We have|x̄ − ȳ|p = |π(y) − π(x)1|π(y) − π(x)|1|x − y|px+(x − y)|p ≤π(x) +|x − y|p ≤ 2,π(x)π(y)π(y)π(x)π(y)π(y)|y|pwhence, in view of (14.2.19),|ζ(x) − ζ(y)|q ≤ (p − 1)2(2p−3)/(p−1) |x − y|p /|y|p(we have taken into account that, by definition of ζ(·), ζ(x) = ζ(x̄), ζ(y) = ζ(ȳ)). The resulting
inequality combined with (14.2.18) results in|f 0 (x) − f 0 (y)|q = 2||x|p ζ(x) − |y|p ζ(y)|q ≤ 2 (|y|p |ζ(x) − ζ(y)|q + ||x|p − |y|p ||ζ(x)|q ) ≤≤ 2 (p − 1)2(2p−3)/(p−1) |x − y|p + |x − y|p |ζ(x)|q = 2((p − 1)2(2p−3)/(p−1) + 1)|x − y|p(we have taken into account that |ζ(x)|q = 1, which is an immediate consequence of the definitionof ζ(·)). Thus,|f 0 (x) − f 0 (y)|q ≤ 2ωf |x − y|p ,ωf = (p − 1)2(2p−3)/(p−1) + 1.

(14.2.21)

When deriving this inequality, x and y were assumed to be nonzero, but actually the inequality
is valid for all x and y, due to the continuity of both the sides in x, y.
30 . It remains to note thatT0|f (x + h) − f (x) − h f (x)| = |

Z 1
0

0

0

T

(f (x + th) − f (x)) hdt| ≤Z 1|f (x + th) − f 0 (x)|q |h|p dt

0(we have used the Holder inequality), whence. in view of (14.2.21),|f (x + h) − f (x) − hT f 0 (x)| ≤ ωf |h|2p = ωf f (h).

SOLUTIONS TO EXERCISES
as claimed.
Exercise 2.4.12:
to253

combining (2.4.9) and (2.4.17) (where one should set p = 2 ln n), we comeE{|ΞN |2∞ } ≤ E{|ΞN |2p } ≤ (8 ln n − 3)nX

E{|ξi |2p } ≤i=1(8 ln n − 3) exp{1}

N
X
i=1

Q.E.D.

E{|ξi |2∞ },254

SOLUTIONS TO EXERCISES

Solutions to Section 3.3
Exercise 3.3.4: the statement in question is evident in the case of h0 (x) = 0, since then, due
to the regularity assumption, x is a minimizer of h over G and L∗h (x) = ∅. Now let h0 (x) 6= 0. It
suffices to prove that Lh (x) ⊂ cl Πh (x) (since L∗h (x) is an open in G subset of the set Lh (x) andthe boundary hyperplane of Πh passes through an interior point of G). The set Lh (x) is convex
and contains x; consequently, if there would be a point y in Lh (x) which is not contained incl Πh (x), then the whole segment [x, y] would belong to Lh (x), and h would be ≤ h(x) on thesegment; this is impossible, since the derivative of h at x in the direction y − x is positive.

SOLUTIONS TO EXERCISES

255

Solutions to Section 4.3
Exercise 4.3.3: The point wi where the hyperplaneΠ = {x ∈ Rn | g T x = 1}
intersects i-th coordinate axis is nothing but γi−1 ei (if γi = 0, then Π does not intersect the i-thaxis). Let I be the set of indices of those points wi which are well-defined and are inside ∆, andlet J be the set of remaining indices from {1, ..., n}; thus,
I = {i : 1 ≤ i ≤ n, γi > 1}.
¯ is the same as to verify that the simplex contains
Now, to prove that the simplex ∆ϑ contains ∆¯ are those vertices ofall vertices of the latter polytope. It is easily seen that the vertices of ∆¯
∆ which are inside ∆ and also the points where Π intersects the edges of ∆ (why?) Thus, the¯ are as follows:
vertices of ∆first, the origin is a vertex; it clearly belongs to ∆ϑ ;second, the vertices ej of the initial simplex ∆ with indices j ∈ J; since for j ∈ J one has
γj ≤ 1 and, consequently, 1/(1 − ϑ + ϑγj ) ≥ 1, all these vertices belong to ∆ϑ ;¯ namely, the point wij where
third, any pair of indices i ∈ I, j ∈ J, produces a vertex in ∆,the hyperplane Π intersects the edge Γij = {xi + xj = 1, x ≥ 0} of the simplex ∆. The i-th andthe j-th coordinates of wij are given byξi =1 − γj,γi − γjξj =γi − 1,γi − γj(14.2.22)

and the remaining coordinates of wij are zeros. In order to prove that wij belongs to ∆ϑ it
suffices to verify that
(1 − ϑ + ϑγi )ξi + (1 − ϑ + ϑγj )ξj ≤ 1(14.2.23)(indeed, the intersection of ∆ϑ with the coordinate plane spanned by the axes i and j is nothing
but the triangle
xi ≥ 0; xj ≥ 0; (1 − ϑ + ϑγi )xi + (1 − ϑ + ϑγj )xj ≤ 1
in the plane). Inequality (14.2.23) is an immediate consequence of (14.2.22). Thus, ∆ϑ does¯
contain ∆.
It remails to evaluate the ratio of volumes of ∆ and ∆ϑ . Since both the simplexes differ onlyin lengths of the edges outgoing from their common vertex 0, not in the directions of the edges,
we haven
YVol(∆)=(1 − ϑ + ϑγi ) ≡ exp{φ(ϑ, g)},ψ(ϑ) ≡
Vol(∆ϑ ) i=1
where
φ(ϑ, t) =nXln(1 − ϑ + ϑti ),i=1

t = (t1 , ..., tn ) ≥ 0. Since the function φ(ϑ, t) is concave in t = (t1 , ..., tn ) ∈ Γ = {t ≥ 0 | i ti =n + 1}, it attains its minimum over Γ at an extreme point of this set, i.e., when one of ti is n + 1
P

256

SOLUTIONS TO EXERCISES

and the remaining are zeros; the minimum is equal to (n − 1) ln(1 − ϑ) + ln(1 + nϑ). Since g ∈ Γ(due to g ≥ 0, g T c = 1), we conclude that
ψ(ϑ) ≥ (1 − ϑ)n−1 (1 + nϑ).
Maximizing the right hand side of the latter relation in ϑ ∈ (0, 1), we come to ϑ = n−2 . With
this value of ϑ one hasψ(ϑ) ≥ 1 −1n2

n−1 

1+1n



≡ χ−n (n) > 1 + O(

1
).
n2

SOLUTIONS TO EXERCISES

257

Solutions to Section 5.5
Exercise 5.5.4: Let B 0 and B 00 be two positive definite symmetric n × n matrices and let
α ∈ (0, 1). We should prove that− ln Det (αB 0 + (1 − α)B 00 ) ≤ α[− ln Det B 0 ] + (1 − α)[− ln Det B 00 ].

(14.2.24)

The linear transformation X 7→ X 0 = (B 0 )−1/2 X(B 0 )−1/2 maps a symmetric positive definite
X into the matrix X 0 of the same type and varies F (X) by an additive constant (equal to−F (B 0 )); this mapping transforms B 0 into the unit matrix I, B 00 into certain matrix C and
B 000 ≡ αB 0 + (1 − α)B 00 into C 000 = αI + (1 − α)C. When passing from (B 0 , B 00 , B 000 ) to (I, C, C 000 ),both sides of (14.2.24) varies by the same constant, so that to establish (14.2.24) is the same as
to check that− ln Det (αI + (1 − α)C) ≤ −α ln Det I − (1 − α) ln Det C.
(14.2.25)
Let λi , i = 1, ..., n, be the eigenvalues of C taken with their multiplicities; the left hand side ofPPP(14.2.25) is − n1 ln(α + (1 − α)λi ), the right hand side is −α n1 ln(1) − (1 − α) n1 ln((1 − α)λi ,and (14.2.25) is an immediate consequence of the fact that the function − ln s is convex on thepositive ray.

258

SOLUTIONS TO EXERCISES

Solutions to Section 6.3
Exercise 6.3.2: we should prove that the system of linear inequalities
xT (ξ − y) ≥ 0, (x, ξ) ∈ G(F )(14.2.26)

with unknown y has a solution, provided that 0 ∈ int Dom F . To this end let us first prove thatany finite subsystem
xTi (ξi − y) ≥ 0, i = 1, ..., N,of the system in question has a solution. On contrary, let the indicated finite subsystem beunsolvable. Then there exist Lagrange multipliers λi , nonnegative with the unit sum, such thatthe linear formX
λi xTi (ξi − y)iis negative everywhere, which means thatX

λi xi = 0,X

λi xTi ξi < 0.

ii

On the other hand,(xi − xj )T (ξi − ξj ) ≥ 0, i, j = 1, ..., N,
and taking average of these inequalities with the weights λi λj , we come to2X

λi xTi ξi ≥ 0

iP(we have used the fact that i λi xi = 0), which is the desired contradiction.
It remains to prove that one of the finite subsystems of (14.2.26) has a bounded solution set(after it is proved, we may refer to the general fact: a family of closed sets in Rn with any finite
subfamily possessing a nonempty intersection also possesses a nonempty intersection, provided
that there is a finite subfamily with compact intersection; to prove the indicated fact, you may
immediately reduce the situation to that one with a family comprised of compact sets; and anested family of compact sets does possess a nonempty intersection). Since 0 ∈ int Dom F , wecan find a finite set {xi ∈ Dom F }i=1,...,N such that 0 belongs to the interior of the convex hull ofxi , i = 1, ..., N ; choosing ξi ∈ F (xi ), we come to a finite subsystem of (14.2.26) which says thatthe inner products of y and all xi are bounded from above by given constants; consequently,
y T x ≤ a for some finite a and all x from a convex hull of xi , i.e., for all x from a smallneighbourhood of 0; of course, the solution set of the subsystem is bounded.
Exercise 6.3.4:
2) ⇔ 3): this is the standard fact of Analysis which has nothing in common with convexity: afunction f : X 7→ R+ which is not identically +∞ (X is a metric space) is lower semicontinuous
if and only if its graph {(t, x) ∈ R × X | x ∈ Dom f, t ≥ f (x)} is closed, and the proof is given
by a quite straightforward verification of definitions.
3) ⇒ 4): the right hand side f¯(x) of the relation in question clearly is a convex lower
semicontinuous function (the latter fact follows from already proved equivalence between 2) and3): the graph of f clearly is closed, since it is an intersection of closed half-spaces); due to thebasic theorem on existence of a subgradient of a convex function everywhere at the interior of its

SOLUTIONS TO EXERCISES

259

domain, we have f (x) = f¯(x), x ∈ int Dom f , and of course f (x) ≥ f¯(x) everywhere (due to thedefinition of a subgradient). Now assume that x is a boundary point of Dom f and let x0 be aninterior point of the domain. Then all points of the half-segment [x0 , x) are the interior pointsof Dom f (since this set is convex), and both f and f¯ are convex and lower semicontinuous on[x0 , x] and coincide with each other and are finite on [x0 , x). A convex function on a segment(taking values in R+ and finite on the interior of the segment) clearly is upper semicontinuous
on the segment (why?), so that f and f¯ are both lower- and upper-semicontinuous on [x0 , x] andtherefore are continuous on the segment; since they coincide on its interior, they coincide at thewhole segment and, in particular, at x.
Thus, f and f¯ coincide on the closure of Dom f . To verify that the functions coincide
everywhere, it remains to prove that if x 6∈ cl Dom f , then f¯(x) = ∞. Indeed, let it not be
the case, and let x̄ be an interior point of Dom f , and let x+ be the point where the segment[x̄, x] intersects the boundary of Dom f . The function f¯ is, as we know, lower semicontinuous
and finite on [x̄, x] and therefore, as we just have seen, is continuous on the segment, and, inparticular, is bounded on it. Now let V be a ball centered at x̄ and contained in int Dom f
and W be the conic hull of V and x. Let also xi ∈ [x̄, x+ ) be a sequence converging to x+ andgi (u) = f (xi ) + ξiT (x − xi ), ξi ∈ ∂f (xi ). The affine functions gi are bounded from above at x
(since they are ≤ f¯) and are uniformly bounded from above on V (since f is bounded fromabove on this ball). It follows that the functions are uniformly bounded from above on W , andsince the sequence gi (xi ) is bounded (we already know that f is finite and continuous on [x̄, x+ ])and xi converge to an interior point of W , we immediately conclude that gi (·) are uniformly
bounded on W (why?). This observation implies that the sequence ∇gi is bounded, and passing
to subsequences we may assume that ∇gi → e as i → ∞. Besides this we already know thatgi (xi ) = f (xi ) → f (x+ ), i → ∞; since f (u) ≥ giT (xi )(u − xi ) + gi (xi ), u ∈ Rn , we come tof (u) ≥ eT (u − x+ ) + f (x+ ).
Thus, x+ ∈ Dom f and ∂f (x+ ) 6= ∅. Now, from the Separation Theorem for convex sets it
follows that there exists a nonzero functional h such thathT (u − x+ ) ≤ 0, u ∈ Dom f ;one clearly has
e + th ∈ ∂f (x+ ), t ≥ 0.
Since x is an interior point of Dom f and h is nonzero, we havehT (x − x+ ) < 0
and consequently hT (x − x+ ) > 0; we come to+∞ > f¯(x) ≥ f (x+ ) + (e + th)T (x − x+ )for all t > 0 (the concluding inequality follows from definition of f¯), which clearly is a contradiction to hT (x − x+ ) > 0. Thus, the implication 3) ⇒ 4) is proved.
4) ⇒ 2): the right hand side in 4) clearly is lower semicontinuous on Rn .
Thus, we have proved the equivalence of 2), 3), 4).

260

SOLUTIONS TO EXERCISES

2) ⇒ 1): assume that f is lower semicontinuous, but f 0 is not maximal monotone, so that
there exists (x, y) 6∈ G(f 0 ): (x − x0 )T (y − y 0 ) ≥ 0 for all (x0 , y 0 ) ∈ G(f 0 ). Passing from f (u) tof (u) − y T u, we can reduce the situation to the case when y = 0; by translation we may also
assume that 0 ∈ int Dom f . Let us look what happens with f at the segment [0, x]. Certain
part ∆ of this segment (of the type [0, x+ ) or [0, x+ ]) is contained in Dom f , and the remaining
part (if any) is outside Dom f . If x0 ∈ [0, x+ ), then (x − x0 )T y 0 ≤ 0 for all y 0 ∈ f 0 (x0 ), so thatf does not increase on [0, x+ ); consequently, x+ ∈ Dom f and f is finite at x+ due to lower
semicontinuity of f on Rn ; since f is convex, finite and lower semicontinuous on [0, x+ ], it is acontinuous finite function on [0, x+ ]. Let us prove that x+ = x. Indeed, let xi ∈ [0, x+ ) converge
to x+ , and let gi (u) = f (xi ) + ξi (u − xi ), ξi ∈ f 0 (xi ). The affine forms gi decrease in the direction
x (we already know that f does not increase on [0, x+ )) and are bounded from above at 0, andtherefore they are bounded from above at x. Besides this, these forms are uniformly boundedfrom above in a ball W centered at 0 and gi (xi ) possesses finite limit as i → ∞. From theseobservations, exactly as when proving the implication 3)⇒ 4), we may conclude that ∇gi form abounded sequence and, consequently, that f 0 (x+ ) is nonempty. Since x+ is a boundary point of
Dom f , f 0 (x+ ) contains functionals with arbitrary large derivative in the direction x, or, whichis the same, in the direction x − x+ (see the reasoning used to prove the implication 3) ⇒ 4)),
which is impossible due to (x − x0 )T y 0 ≤ 0 for all x0 ∈ Dom f 0 and all y 0 ∈ f 0 (x0 ) (look what
happens when x0 = x+ ).
Thus, we come to x ∈ Dom f , and, as we remember, (x − x0 )T y 0 ≤ 0 for all x0 ∈ Dom f 0
and all y 0 ∈ f 0 (x0 ). We conclude that if x0 ∈ int Dom f , then the convex function f is finite on
the segment [x0 , x] and its derivative along the direction x − x0 of the segment is nonpositiveat any point from [x0 , x); due to lower semicontinuity of f we conclude that it is continuous on
the segment and f (x) ≤ f (x0 ). Thus, f (x) ≤ f (x0 ) for any x0 ∈ int Dom f . Let us prove that
actually f (x) ≤ f (x0 ) for any x0 . The inequality is evident if x0 6∈ Dom f , since then f (x0 ) = +∞.
Now, if x0 ∈ Dom f , then, as we have seen, f is continuous and finite on the segment [0, x0 ]; since
we already know that f (x) ≤ f (u) for u ∈ [0, x0 ), we conclude that f (x) ≤ f (x0 ), as claimed.
Thus, x ∈ Dom f is a minimizer of f ; but then f 0 (x) clearly is nonempty and contains 0. We
see that (x, 0) belongs to the graph of f 0 , and at the very beginning we assumed the opposite.
Thus, the implication 2)⇒ 1) is proved.
Exercise 6.3.7: let x̄ be a weak solution to the variational inequality defined by G and F ,and let H = F + NG . In view of the description of NG given in exercise 6.3.5, every point ofG, and, in particular, x̄, is a weak solution to the variational inequality defined by G and NG ;consequently, x̄ is a weak solution to the variational inequality defined by (G, H). Besides this,V(NG ) = 0 (why?), so that H is semibounded on int G. Note also that H is maximal monotone
in view of the Rockafellar Theorem.
Now let us prove the following:
let x̄ be a weak solution to a variational inequality (G, R) involving a monotone semibounded
on int G operator R, int G ⊂ Dom R ⊂ G. Then adding (x̄, 0) to the graph of R we preserve
monotonicity of the operator. In particular, if R is maximal monotone, then x̄ ∈ Dom R and
0 ∈ R(x̄).
Let us prove the latter statement. Without loss of generality we may assume that x̄ = 0;
what we should prove is that y T x ≥ 0 for all x ∈ Dom R and all y ∈ R(x). By definition of a
weak solution, the desired statement does hold true for x ∈ int G; now let x ∈ ∂G and y ∈ R(x).
Assume that y T x < 0, and let us lead the assumption to a contradiction. To this end let z be

SOLUTIONS TO EXERCISES

261

an interior point of G, and let xi = (1 − 1/i)( 21 x) + (1/i)z. Then xi ∈ int G, so that one canfind yi ∈ R(xi ). We know that1(xi − x)T yi ≥ (xi − x)T y →i→∞ − xT y > 0,2
so that(xi − x)T yi ≥ α > 0
for all large enough values of i. On the other hand,
2xi − x =and we come to2(z − xi ),i−1

2 T
y (z − xi ) = yiT (2xi − x) ≥ yiT (xi − x) ≥ α > 0i−1 i(recall that xi ∈ int G and yi ∈ R(xi ), so that yiT xi = yiT (xi − x̄) ≥ 0); the resulting inequality
contradicts the assumption that R is semibounded on int G, and we are done.
Applying the statement we just have proved to the operator H, we see that x̄ ∈ Dom H ⊂Dom F and 0 ∈ H(x̄), i.e., there exists y ∈ F (x̄) such that −y ∈ NG (x̄). By construction of NG
we have y T (x − x̄) ≥ 0, x ∈ G, so that x̄ is a strong solution to the variational inequality givenby (G, F ).

262

SOLUTIONS TO EXERCISES

Solutions to Section 8.3
Exercise 8.3.4: From exercise 8.3.3 we know that
V (φ + η) ≤ V (φ) + η, V 0 (φ) + LV (η);
since V + differs from V by a linear term, we have
V + (φ + η) ≤ V + (φ) + φ, (V + )0 (φ) + LV (η);
substituting φ = φi and η = −γi f 0 (xi )/|f 0 (xi )|∗ , we come to the recurrence
Di+1 ≡ V + (φi+1 ) ≤ Di − γi f 0 (xi ), V 0 (φi ) − x∗ /|f 0 (xi )|∗ +

L 2
γ ,
2 i

and since V 0 (φi ) = xi , we havef 0 (xi ), V 0 (φi ) − x∗ = f 0 (xi ), xi − x∗ ≥ f (xi ) − f ∗ ;
thus,
L 2
γ .
2 i
Since also |f 0 (xi )|∗ ≤ L(f ) ≡ Lk·k (f ) (why?), we immediately come toDi+1 ≤ Di − γi (f (xi ) − f ∗ )/|f 0 (xi )|∗ +εN ≤ L(f )D1 − DN +1 + L2PN

i=1 γi(14.2.27)PN

2
i=1 γi.

It remains to note that, since φ1 = 0, we have
δ = D1 − DN +1 = V (φ1 ) − V (φN +1 ) − hφ1 − φN +1 , x∗ i =1
1= −V (φN +1 ) − hφN +1 , x∗ i ≤ − |φN +1 |2∗ + |φn+1 |∗ |x∗ | ≤ |x∗ |2 .
2
2

(14.2.28)

SOLUTIONS TO EXERCISES

263

Solutions to Section 9.3
Exercise 9.3.1: induction on i. For i = 1 the statement is evident. Now, since yi is the
projection of y0 onto Pi and yi+1 ∈ Pi+1 ⊂ Pi , we have(y0 − yi )T (yi+1 − yi ) ≤ 0,
whence|y0 − yi+1 |2 = |y0 − yi |2 + 2(y0 − yi )T (yi − yi+1 ) + |yi − yi+1 |2 ≥ |y0 − yi |2 + |yi − yi+1 |2 ,which clearly justifies the inductive step.
Exercise 9.3.2:
1): let us verify that the sets Qi , i ∈ Ik , are nonempty, or, which is the same, thatfi− ≤ li , i ∈ Ik .
One clearly has
li = fj− + λ∆j(14.2.29)

for certain j which is ≥ jk and is ≤ i. Since fl− do not decrease with l and fl+ do not increase
with l, we have∆i ≤ ∆j − (fi− − fj− ),whencefi− ≤ ∆j − ∆i + fj− .

(14.2.30)

Since both i and j belong to Ik , we have∆i ≥ (1 − λ)∆jk ≥ (1 − λ)∆j(14.2.31)(we have taken into account that the gaps never increase). Combining (14.2.30) and (14.2.31),
we come tofi− ≤ fj− + λ∆j ,which, in view of (14.2.29), means that fi− ≤ li , as claimed.
The fact that Qi ⊃ Qi+1 , i, i + 1 ∈ Ik , is evident: li ≥ li+1 by construction, while fi ≤ fi+1
by the basic properties of the models.
2): according to rule 4∗ ), the iterations from Ik can be partitioned into sequential fragments
J1 , ..., Js as follows: during the steps i ∈ Jl the best found so far search point x̄i remains constant,
let it be denoted by xl ; at the final step from the group Jl , l < s, we form the new best search
point xl+1 . According to the description of the method, we havexi+1 = πQi (xl ), i ∈ Jl .

(14.2.32)

The points xl are defined for l = 1, 2, ..., s; let us define xs+1 as the search point generated at
the last iteration of the last group Js . Then for all l ≤ s we havexl+1 = πQql (xl ),where ql denotes the last iteration from the group Jl .

(14.2.33)

264

SOLUTIONS TO EXERCISES

Now consider group Jl = {pl , pl + 1, ..., ql }. From 1) we already know that Qi ∈ Qi+1 ,i, i + 1 ∈ Ik , so that from exercise 9.3.1 and (14.2.32), (14.2.33) it follows that|xl − xl+1 |2 ≥X|yi−1 − yi |2 ,(14.2.34)i∈Jlwhere yi = xi , ql ≥ i ≥ pl , and ypl −1 = xl . On the other hand, in view of the standard propertiesof projection and the inclusion xs+1 ∈ Qqs ⊂ Qql from (14.2.33) it follows that|x1 − xs+1 |2 ≥sX|xl − xl+1 |2 ;l=1

combining this observation with (14.2.34), we come to|x1 − xs+1 |2 ≥X

|xi − xi+1 |2(14.2.35)

i∈Ik(note that ypl −1 = xql−1 +1 ).
Now we can use the same reasoning as for the basic version of the method. Namely, fi (xi ) =f (xi ) ≥ fi+ , while li ≤ fi− + λ∆i , so that fi (xi ) ≥ li . It follows that fi (xi+1 ) = li , so that when
passing from xi to xi+1 , we decrease fi at least by (1 − λ)∆i ≥ (1 − λ)2 ∆jk . Since fi clearly is
Lipschitz continuous with constant L(f ), we come to |xi − xi+1 | ≥ (1 − λ)2 ∆jk L−1 (f ). Thus,
the right hand side in (14.2.34) is at least Nk L−2 (f )(1 − λ)4 ∆2jk , while the right hand side is≤ D2 (G), and 2) follows.
3) The inequality ∆i ≥ f (x̄i ) − minG f immediately follows from fi ≤ f . Now assume that
N is such that ∆N > ε. Let k be the first integer such that I1 ∪ ... ∪ Ik ⊃ {1, ..., N }; then thefirst index in Ik is ≤ N , and consequently ∆jk ≥ ∆N > ε. We have, by construction of thegroups Il ,∆jl+1 < (1 − λ)∆jl ,so that∆jl ≥ (1 − λ)−(k−l) ∆jl ≥ (1 − λ)−(k−l) ε, 1 ≤ l ≤ k.
From 2) it now follows
N≤

k
X
l=1

Nl ≤

k
X

(1 − λ)−4 D2 (G)L2 (f )∆−2
l ≤

l=1

≤ (1 − λ)−4 D2 (G)L2 (f )

∞
X
p=0

as required.

ε−2 (1 − λ)2p ≡ D2 (G)L2 (f )ε−2 c(λ),

SOLUTIONS TO EXERCISES

265

Solutions to Section 12.6
Exercise 12.6.2:
1) Let p be an optimal solution to (P ); let us prove that then p possesses the ”alternanceproperty”. Since the cardinality of Σ is > N + 1, δ > 0 (otherwise the polynomial 1 − tp(t)would possess more than N + 1 zeros and would be zero identically, which is impossible due tothe structure of the polynomial). Let Σ0 be the subset of Σ formed by the points t ∈ Σ where|q(t)| = δ. Note that Σ0 is a closed subset of Σ which does not contain 0 (the latter is, of course,
true if 0 6∈ Σ; if 0 ∈ Σ, then, by assumption, γ > 0, so that q is a continuous on [0, ∞) function
with q(0) = 0, and since q(0) = 0 and δ > 0, we have 0 6∈ Σ0 ). We may partition the set Σ0
into sequential parts Σ0 , Σ1 ,... (with Σi being a closed set and Σi+1 being to the right of Σi ) insuch a way that q(t) = κ(−1)i δ for t ∈ Σi , where κ (the sign of q on Σ0 ) is either +1, or −1.
The number K of sets Σi in the partitioning clearly is finite (why?). The ”alternance property”
means exactly that the number of these sets is at least N + 2 and that κ = +1, and this is whatshould be proved.
Assume, first, that K < N + 2. One can find K ≤ N + 1 points (0 <)t0 < ... < tK−2 suchthat Σi ⊂ (ti , ti+1 ), where tK−1 = ∞. Let
π(t) =K−1
Y(t − ti );i=1

since K − 1 ≤ N , π is a polynomial of degree ≤ N . There clearly exists a neighbourhood Σ00 of
the set Σ0 such thatsup tγ |1 − t[p(t) − rκπ(t)]| = sup tγ |s(t) − rκπ(t)| < δ, s(t) = 1 − tp(t),
t∈Σ00

t∈Σ00

for all small enough values of r > 0 (why?), and consequently
max tγ |s(t) − rκπ(t)| < δt∈Σ

for all small enough positive r (why?), which contradicts the optimality of p.
Thus, K ≥ N + 2; it remains to verify that κ = 1. Since the polynomial s(t) is positive at
t = 0 and its sign on Σi is (−1)i κ, i = 0, ..., K, in the case of κ < 0 s(t) would have at leastK ≥ N + 2 changes of the sign on the axis; since the degree of the polynomial is ≤ N + 1, it
would be zero identically, which is not the case.
2) Now assume that p possesses the ”alternance property”; let us prove that then p is optimal.
Indeed, otherwise there would exist a polynomial π of degree ≤ N such that tγ |1 − tπ(t)| <tγ |1 − tp(t)| when t = σi , (σi , i = 0, ..., N + 1, are the points of alternance). Consequently, thepolynomial∆(t) = (1 − tp(t)) − (1 − tπ(t)) = t(π(t) − p(t))
at the points σi , i = 0, ..., N + 1, would be of the same sign as q, i.e., of the signs (−1)i . It would
follow that ∆(t) would have at least N + 1 positive zeros, and since this polynomial is also 0 at
t = 0, it would be zero identically, which is the desired contradiction.

266

SOLUTIONS TO EXERCISES

Solutions to Section 13.4
Exercise 13.4.1 : The normal solution x∗ to a solvable operator equation with symmetricmatrix A belongs, as we know, to the image space E of A; this subspace is invariant for A,
and the restriction of A onto E is a symmetric positive definite operator on E. The right handside vector b also belongs to E (since b = Ax∗ , x∗ ∈ E and E is invariant for A); it follows
that all Krylov subspaces of (A, b) belong to E and, consequently, the trajectory of the CGM
on (A, b) coincides with the trajectory of the method as applied to the natural ”restriction” ofthe equation onto E. Thus, replacing, if necessary, Rn with E, we may assume without loss ofgenerality that A is nonsingular.
Now, equation Ax = b with symmetric positive definite matrix A clearly belongs to the family
Un (Σ, 0, R), where Σ is the finite set comprised of eigenvalues of A (all of them are positive) and
R is the norm of the solution to the equation. Bearing in mind this observation and taking into
account Proposition 13.3.1 (Lecture 13), we come to the following expression for the”inaccuracy in terms of the objective”
dA,b (xk ) ≡ fA,b (xk ) − min fA,bxof k-th approximate solution found by the CGM is given byZ

dA,b (xk ) = minq∈Pk0

tq 2 (t)dµ(t),(14.2.36)

where µ is the measure on Σ given by the Proposition; from the construction given in Lecture
13 it follows that the cardinality of the support set Σ0 of this measure is exactly m.
Now, Σ0 is an m-point subset of the positive ray, and there exists a polynomial qm (t) of
the degree m, qm (0) = 1, which vanishes on Σ0 ; it follows that the right hand side of (14.2.36)
corresponding to k = m is zero, so that dA,b (xm ) = 0 and xm is the exact solution to theequation.
Vice versa, if xk is an exact solution to the equation, then dA,b (xk ) = 0, and from (14.2.36)
it follows that there exists a polynomial q ∈ Pk0 which vanishes on Σ0 and therefore possessing
m = card Σ0 distinct roots; it is possible only if k ≥ deg q ≥ m, so that k ≥ m.
Exercise 13.4.7:2): if λ is an eigenvalue of Tk , then λ is real (since Tk is symmetric, see 1)) and there exist
a nonzero p ∈ Ek such that (Tk − λI)p = 0. By definition of Tk it means that the nonzero
polynomial p of degree ≤ k − 1 is such that the polynomial q(t) = (t − λ)p(t) is µ-orthogonal
to Ek . Since q is of the degree at most k and is nonzero, we conclude from the uniquenessstatement of Exercise 13.4.6 that q is proportional, with a nonzero coefficient, to qk , so that λis a root of qk . Vice versa, if λ is a real root of qk , then the polynomial p(t) = qk (t)/(t − λ) is
well defined and is a nonzero element of Ek ; we have (t − λI)p = qk , and since qk is orthogonal
to Ek , (Tk − λI)p = 0 by definition of Tk , so that λ is an eigenvalue of Tk .
3): as we know from 1), the operator Tk is symmetric; besides this, one clearly hasa(p, p) ≤ (T p, p) = (Tk p, p) ≤ b(p, p), p ∈ Ek ,so thata) the spectrum of Tk belongs to the segment [a, b].

SOLUTIONS TO EXERCISES

267

Let us prove thatb) neither a nor b belong to the spectrum of Tk ;c) every eigenvalue of Tk is of multiplicity 1.
To prove b), assume that a belongs to the spectrum of Tk ; then there exists a nonzero polynomial
p of degree ≤ k − 1 such thata(p, p) = (Tk p, p).
The latter quantity is equal to (T p, p) by construction of Tk , so that ((T − aI)p, p) = 0, or,which is the same,Z(t − a)p2 (t)dµ(t) = 0.

Since the support set of µ belongs to [a, b], the latter relation is possible only if this support set
is covered by the set of roots of the polynomial (t − a)p2 (t); since p is nonzero, the latter set
is comprised of at most k ≤ N points (recall that deg p ≤ k − 1), and from the very beginning
we have assumed that the cardinality of the support set of µ is at least N + 1. Thus, a cannot
belong to the spectrum of Tk . A completely similar reasoning demonstrates that b also does notbelong to this spectrum, as claimed in b).
To prove c), assume, on contrary, that certain eigenvalue λ of Tk is of multiplicity at least
2. Then there exist two linearly independent polynomials p0 , p00 of degrees ≤ k − 1 such that thenonzero polynomials (t − λ)p0 (t) and (t − λ)p00 (t) of degrees at most k are µ-orthogonal to Ek ;same as in the proof of 2), we conclude that these polynomials are proportional, with nonzero
coefficients, to qk . Consequently, there exist a nonzero ω such that (t − λ)p0 (t) ≡ ω(t − λ)p00 (t),or, which is the same, (t−λ)(p0 (t)−ωp00 (t)) ≡ 0. The latter relation implies that p0 = ωp00 , which
is a contradiction (p0 and p00 were assumed to be linearly independent). Thus, c) is proved.
Since the dimension of Ek is exactly k (exercise 13.4.6.4)) and, as we know, Tk is symmetric,
Tk has k real eigenvalues (counted with multiplicities); from c) it follows that all these k eigenvalues are distinct; and from a) and b) we know that these eigenvalues belong to (a, b). Thus,
the spectrum of Tk is comprised of k distinct points from (a, b). As we know from 2), qk vanishes
at the spectrum of Tk , so that qk has k distinct roots in (a, b).
To complete the proof of 3), it suffices to demonstrate that qk and qk−1 cannot have a rootin common. This is evident when k = 1 (since q0 is a nonzero constant). Now assume that1 < k ≤ N and that qk and qk−1 have a common root λ. It means that qk (t) = (t − λ)p(t)and qk−1 (t) = (1 − λ)r(t) with certain polynomials p and r of degrees exactly k − 1 and k − 2,
respectively. Let
p=

k−1
X

ci qi

i=0

be the representation of p in the basis q0 , ..., qk−1 ; then ck−1 6= 0, since p is of the degree k − 1.
We come toqk (t) = (t − λ)p(t) =k−1
Xci (t − λ)qi (t) = (t − λ)2 r(t) +

i=0

k−2
X

ci (t − λ)qi (t).

i=1

Let us take here the inner product with the polynomial r. We come toZ

0 = (qk , r) =Z

r(t)qk (t)dµ(t) = ck−1

2 2

(t − λ) r (t)dµ(t) +

k−2
XZ
i=0

(t − λ)r(t)qi (t)dµ(t) =268

SOLUTIONS TO EXERCISES
Z= ck−1(t − λ)2 r2 (t)dµ(t) +

k−2
X

Z

(qk−1 , qi ) = ck−1

(t − λ)2 r2 (t)dµ(t).

i=0Since ck−1 6= 0, the resulting relation says that (t − λ)2 r2 (t)dµ(t) = 0. We conclude that the
support set of µ is covered by the set comprised of λ and the roots of r; this latter set is ofcardinality at most k − 1 < N , so that the support set of µ is of cardinality < N , which is acontradiction.
R