# Solving the Variational Inequality Problem Defined on Intersection of Finite Level Sets

Authors: Songnian He; Caiping Yang
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

Consider the variational inequalityVI(C,F) of finding a point x*∈C satisfying the property 〈Fx*,x-x*〉≥0, for all x∈C, where C is the intersection of finite level sets of convex functions defined on a real Hilbert space H and F:H→H is an L-Lipschitzian and η-strongly monotone operator. Relaxed and self-adaptive iterative algorithms are devised for computing the unique solution of VI(C,F). Since our algorithm avoids calculating the projection PC (calculating PC by computing several sequences of projections onto half-spaces containing the original domain C) directly and has no need to know any information of the constants L and η, the implementation of our algorithm is very easy. To prove strong convergence of our algorithms, a new lemma is established, which can be used as a fundamental tool for solving some nonlinear problems.


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## Body

## 1. Introduction

The variational inequality problem can mathematically be formulated as the problem of finding a pointx*∈C with the property
(1)〈Fx*,x-x*〉≥0,∀x∈C,
where H is a real Hilbert space with inner product 〈·,·〉 and norm ∥·∥, C is a nonempty closed convex subset of H, and F:C→H is a nonlinear operator. Since its inception by Stampacchia [1] in 1964, the variational inequality problem VI(C,F) has received much attention due to its applications in a large variety of problems arising in structural analysis, economics, optimization, operations research and engineering sciences; see [1–23] and the references therein. Using the projection technique, one can easily show that VI(C,F) is equivalent to the fixed-point problem (see, for example, [15]).Lemma 1.
x
*
∈
C is a solution of VI(C,F) if and only if x*∈C satisfies the fixed-point relation:
(2)x*=PC(I-λF)x*,
where λ>0 is an arbitrary constant, PC is the orthogonal projection onto C, and I is the identity operator on H.Recall that an operatorF:C→H is called monotone, if
(3)〈Fx-Fy,x-y〉≥0∀x,y∈C.
Moreover, a monotone operator F is called strictly monotone if the equality “=” holds only when x=y in the last relation. It is easy to see that VI(C,F) (1) has at most one solution if F is strictly monotone.For variational inequality (1), F is generally assumed to be Lipschitzian and strongly monotone on C; that is, for some constants L,η>0,  F satisfies the conditions
(4)∥Fx-Fy∥≤L∥x-y∥,∀x,y∈C,〈Fx-Fy,x-y〉≥η∥x-y∥2,∀x,y∈C.
In this case, F is also called an L-Lipschitzian and η-strongly monotone operator. It is quite easy to show the simple result as follows.Lemma 2.
Assume thatF satisfies conditions (4) and λ and μ are constants such that λ∈(0,1) and μ∈(0,2η/L2), respectively. Let Tμ=PC(I-μF) (or I-μF) and Tλ,μ=PC(I-λμF) (or I-λμF). Then Tμ and Tλ,μ are all contractions with coefficients 1-τ and 1-λτ, respectively, where τ=(1/2)μ(2η-μL2).Using Banach's contraction mapping principle, the following well-known result can be obtained easily from Lemmas1 and 2.Theorem 3.
Assume thatF satisfies the conditions (4). Then VI(C,F) has a unique solution. Moreover, for any 0<λ<2η/L2, the sequence {xn} with initial guess x0∈C and defined recursively by
(5)xn+1=PC(I-λF)xn,n≥0,
converges strongly to the unique solution of VI(C,F).However, Algorithm (5) has two evident weaknesses. On one hand, Algorithm (5) involves calculating the mapping PC, while the computation of a projection onto a closed convex subset is generally difficult. If C is the intersection of finite closed convex subsets of H, that is, C=⋂i=1m‍Ci(≠∅), where Ci(i=1,…,m) is a closed convex subset of H, then the computation of PC is much more difficult. On the other hand, the determination of the stepsize λ depends on the constants L and η. This means that in order to implement Algorithm (5), one has first to compute (or estimate) the constants L and η, which is sometimes not an easy work in practice.In order to overcome the above weaknesses of the algorithm (5), a new relaxed and self-adaptive algorithm is proposed in this paper to solve VI(C,F), where C is the intersection of finite level sets of convex functions defined on H and F:H→H is an L-Lipschitzian and η-strongly monotone operator. Our method calculates PC by computing finite sequences of projections onto half-spaces containing the original set C and selects the stepsizes through a self-adaptive way. The implementation of our algorithm avoids computing PC directly and has no need to know any information about L and η.The rest of this paper is organized as follows. Some useful lemmas are listed in the next section; in particular, a new lemma is established in order to prove strong convergence theorems of our algorithms, which can also be used as a fundamental tool for solving some nonlinear problems relating to fixed point. In the last section, a relaxed algorithm (for the case whereL and η are known) and a relaxed self-adaptive algorithm (for the case where L and η are not known) are proposed, respectively. The strong convergence theorems of our algorithms are proved.


## 2. Preliminaries

Throughout the rest of this paper, we denote byH a real Hilbert space and by I the identity operator on H. If f:H→ℝ is a differentiable functional, then we denote by ∇f the gradient of f. We will also use the following notations: (i)
→ denotes strong convergence.(ii)
⇀ denotes weak convergence.(iii)
ωw(xn)={x∣∃{xnk}⊂{xn} such that xnk⇀x} denotes the weak ω-limit set of {xn}.Recall a trivial inequality, which is well known and in common use.Lemma 4.
For allx,y∈H, there holds the following relation:
(6)∥x+y∥2≤∥x∥2+2〈y,x+y〉.Recall that a mappingT:H→H is said to be nonexpansive if
(7)∥Tx-Ty∥≤∥x-y∥,x,y∈H.T:H→H is said to be firmly nonexpansive if, for x,y∈H,
(8)∥Tx-Ty∥2≤∥x-y∥2-∥(I-T)x-(I-T)y∥2.The following are characterizations of firmly nonexpansive mappings (see [7] or [24]).Lemma 5.
LetT:H→H be an operator. The following statements are equivalent.(i)
T is firmly nonexpansive.(ii)
I-T is firmly nonexpansive.(iii)
∥Tx-Ty∥2≤〈x-y,Tx-Ty〉,  x,y∈H.We know that the orthogonal projectionPC from H onto a nonempty closed convex subset C⊂H is a typical example of a firmly nonexpansive mapping [7], which is defined by
(9)PCx:=argminy∈C∥x-y∥2,x∈H.
It is well known that PCx is characterized [7] by the inequality (for x∈H)
(10)PCx∈C,〈x-PCx,y-PCx〉≤0,∀y∈C.It is well known that the following lemma [25] is often used when we analyze the strong convergence of some algorithms for solving some nonlinear problems, such as fixed points of nonlinear mappings, variational inequalities, and split feasibility problems. In fact, this lemma has been regarded as a fundamental tool for solving some nonlinear problems relating to fixed point.Lemma 6 (see [25]).
Assume(an) is a sequence of nonnegative real numbers such that
(11)an+1≤(1-γn)an+γnδn,n≥0,
where (γn) is a sequence in (0,1) and (δn) is a sequence in ℝ such that(i)
∑n=0∞‍γn=∞,(ii)
limsupn→∞δn≤0 or ∑n=0∞‍|γnδn|<∞.
Thenlimn→∞an=0.In this paper, inspired and encouraged by an idea in [26], we obtain the following lemma. Its key effect on the proofs of our main results will be illustrated in the next section and this may show that this lemma is likely to become a new fundamental tool for solving some nonlinear problems relating to fixed point.Lemma 7.
Assume(sn) is a sequence of nonnegative real numbers such that
(12)sn+1≤(1-γn)sn+γnδn+βn,n≥0,(13)sn+1≤sn-ηn+αn,n≥0,
where (γn) is a sequence in (0,1), (ηn) is a sequence of nonnegative real numbers and (δn), (αn), and (βn) are three sequences in ℝ such that(i)
∑n=0∞‍γn=∞,(ii)
limn→∞αn=0,(iii)
limk→∞ηnk=0 implies limsupk→∞δnk≤0 for any subsequence (nk)⊂(n),(iv)
limsupn→∞(βn/γn)≤0.
Thenlimn→∞sn=0.Proof.
Following and generalizing an idea in [26], we distinguish two cases to prove sn→0 as n→0.
Case  1. (sn) is eventually decreasing (i.e., there exists k≥0 such that sn>sn+1 holds for all n≥k). In this case, (sn) must be convergent, and from (13) it follows that
(14)ηn≤(sn-sn+1)+αn.
Noting condition (ii), letting n→∞ in (14) yields ηn→0 as n→∞. Using condition (iii), we get that limsupn→∞δn≤0. Noting this together with conditions (i) and (iv), we obtain sn→0 by applying Lemma 6 to (12).
Case  2. (sn) is not eventually decreasing. Hence, we can find an integer n0 such that sn0≤sn0+1. Let us now define
(15)Jn:={n0≤k≤n:sk≤sk+1},n>n0.
Obviously, Jn is nonempty and satisfies Jn⊆Jn+1. Let
(16)τ(n)∶=maxJn,n>n0.
It is clear that τ(n)→∞ as n→∞ (otherwise, (sn) is eventually decreasing). It is also clear that sτ(n)≤sτ(n)+1 for all n>n0. Moreover,
(17)sn≤sτ(n)+1,∀n>n0.
In fact, if τn=n, then inequity (17) is trivial; if τ(n)=n-1, then τ(n)+1=n, and (17) is also trivial. If τ(n)<n-1, then there exists an integer i≥2 such that τ(n)+i=n. Thus we deduce from the definition of τ(n) that
(18)sτ(n)+1>sτ(n)+2>⋯>sτ(n)+i=sn,
and inequity (17) holds again. Since sτ(n)≤sτ(n)+1 for all n>n0, it follows from (14) that
(19)0≤ητ(n)≤ατ(n)⟶0,
so that ητ(n)→0 as n→∞ using condition (ii). Due to the condition (iii), this implies that
(20)limsupn→∞δτ(n)≤0.
Noting sτ(n)≤sτ(n)+1 for all n>n0 again, it follows from (12) that
(21)sτ(n)≤δτ(n)+βτ(n)γτ(n).
Combining (20), (21), and condition (iv) yields
(22)limsupn→∞sτ(n)≤0,
and hence sτ(n)→0 as n→∞. This together with (13) implies that
(23)sτ(n)+1≤sτ(n)-ητ(n)+ατ(n)⟶0,
which together with (17), in turn, implies that sn→0 as n→∞.The following result is just a special case of Lemma7, that is, the case where βn=0 for all n≥0.Lemma 8.
Assume(sn) is a sequence of nonnegative real numbers such that
(24)sn+1≤(1-γn)sn+γnδn,n≥0,sn+1≤sn-ηn+αn,n≥0,
where (γn) is a sequence in (0,1), (ηn) is a sequence of nonnegative real numbers, and (δn) and (αn) are two sequences in ℝ such that(i)
∑n=0∞‍γn=∞,(ii)
limn→∞αn=0,(iii)
limk→∞ηnk=0 implies limsupk→∞δnk≤0 for any subsequence (nk)⊂(n).
Thenlimn→∞sn=0.Recall that a functionf:H→ℝ is called convex if
(25)f(λx+(1-λ)y)≤λf(x)+(1-λ)f(y),∀λ∈(0,1),∀x,y∈H.
A differentiable function f is convex if and only if there holds the following relation:
(26)f(z)≥f(x)+〈∇f(x),z-x〉,∀z∈H.
Recall that an element g∈H is said to be a subgradient of f:H→ℝ at x if
(27)f(z)≥f(x)+〈g,z-x〉,∀z∈H.A functionf:H→ℝ is said to be subdifferentiable at x, if it has at least one subgradient at x. The set of subgradients of f at the point x is called the subdifferential of f at x and is denoted by ∂f(x). The last relation above is called the subdifferential inequality of f at x. A function f is called subdifferentiable, if it is subdifferentiable at all x∈H. If a function f is differentiable and convex, then its gradient and subgradient coincide.Recall that a functionf:H→ℝ is said to be weakly lower semicontinuous (w-lsc) at x if xn⇀x implies
(28)f(x)≤liminfn→∞f(xn).


## 3. Iterative Algorithms

In this section, we consider the iterative algorithms for solving a particular kind of variational inequality (1) in which the closed convex subset C is of the particular structure, that is the intersection of finite level sets of convex functions given as follows:
(29)C=⋂i=1m‍{x∈H:ci(x)≤0},
where m is a positive integer and ci:H→ℝ(i=1,…,m) is a convex function. We always assume that ci(i=1,…,m) is subdifferentiable on H and ∂ci(i=1,…,m) is a bounded operator (i.e., bounded on bounded sets). It is worth noting that every convex function defined on a finite-dimensional Hilbert space is subdifferentiable and its subdifferential operator is a bounded operator (see [27, Corollary 7.9]). We also assume that F:H→H is an L-Lipschitzian and η-strongly monotone operator. It is well known that in this case VI(C,F) has a unique solution, henceforth, which is denoted by x*.Without loss of the generality, we will consider only the casem=2; that is, C=C1⋂‍C2, where
(30)C1={x∈H:c1(x)≤0},C2={x∈H:c2(x)≤0}.
All of our results can be extended easily to the general case.The computation of a projection onto a closed convex subset is generally difficult. To overcome this difficulty, Fukushima [21] suggested a way to calculate the projection onto a level set of a convex function by computing a sequence of projections onto half-spaces containing the original level set. This idea is followed by Yang [28] and López et al. [29], respectively, who introduced the relaxed CQ algorithms for solving the split feasibility problem in  finite-dimensional and infinite-dimensional Hilbert spaces, respectively. This idea is also used by Censor et al. [30] in the subgradient extragradient method for solving variational inequalities in a Hilbert space.We are now in a position to introduce a relaxed algorithm for computing the unique solutionx* of VI(C,F), where C=C1⋂‍C2 and Ci(i=1,2) is given as in (30). This scheme applies to the case where L and η are easy to be determined.Algorithm 1.
Choose an arbitrary initial guessx0∈H. The sequence (xn) is constructed via the formula
(31)xn+1=PCn2PCn1(I-λnμF)xn,n⩾0,
where
(32)Cn1={x∈H:c1(xn)≤〈ξn1,xn-x〉},Cn2={x∈H:c2(PCn1xn)≤〈ξn2,PCn1xn-x〉},
where ξn1∈∂c1(xn),  ξn2∈∂c2(PCn1xn), the sequence (λn) is in (0,1), and μ is a constant such that μ∈(0,2η/L2).We now analyze strong convergence of Algorithm1, which also illustrates the application of Lemma 7 (or Lemma 8).Theorem 9.
Assume thatλn→0(n→∞) and ∑n=1+∞‍λn=+∞. Then the sequence (xn) generated by Algorithm 1 converges strongly to the unique solution x* of VI(C,F).Proof.
Firstly, we verify that(xn) is bounded. Indeed, it is easy to see from the subdifferential inequality and the definitions of Cn1 and Cn2 that Cn1⊃C1 and Cn2⊃C2 hold for all n≥0, and hence it follows that Cn1⋂‍Cn2⊃C1⋂‍C2=C. Since the projection operators PCn1 and PCn2 are nonexpansive, we obtain from (31), Lemmas 2 and 4 that
(33)∥xn+1-x*∥2=∥PCn2PCn1(I-λnμF)xn-PCn2PCn1x*∥2≤∥(I-λnμF)xn-(I-λnμF)x*-λnμFx*∥2≤(1-τλn)∥xn-x*∥2-2λnμ〈Fx*,xn-x*-λnμFxn〉≤(1-τλn)∥xn-x*∥2-2λnμ〈Fx*,xn-x*〉+2λn2μ2∥Fx*∥∥Fxn∥,(34)∥xn+1-x*∥2=∥PCn2PCn1(I-λnμF)xn-PCn2PCn1(I-λnμF)x*+PCn2PCn1(I-λnμF)x*-PCn2PCn1x*∥2≤(1-τλn)∥xn-x*∥2+2〈PCn2PCn1(I-λnμF)x*-PCn2PCn1x*,xn+1-x*〉≤(1-τλn)∥xn-x*∥2+2λnμ∥Fx*∥∥xn+1-x*∥≤(1-τλn)∥xn-x*∥2+14τλn∥xn+1-x*∥2+4λnμ2τ∥Fx*∥2,
where τ=(1/2)μ(2η-μL2).
Consequently(35)∥xn+1-x*∥2≤1-τλn1-(1/4)τλn∥xn-x*∥2+(3/4)τλn1-(1/4)τλn16μ23τ2∥Fx*∥2.
It turns out that
(36)∥xn+1-x*∥≤max{∥xn-x*∥,4μ3τ∥Fx*∥},
inductively
(37)∥xn-x*∥≤max{∥x0-x*∥,4μ3τ∥Fx*∥},
and this means that (xn) is bounded. Obviously, (Fxn) is also bounded.
Secondly, since a projection is firmly nonexpansive, we obtain(38)∥PCn2PCn1xn-PCn2PCn1x*∥2≤∥PCn1xn-PCn1x*∥2-∥PCn1xn-PCn2PCn1xn∥2≤∥xn-x*∥2-∥xn-PCn1xn∥2-∥PCn1xn-PCn2PCn1xn∥2;
thus we also have
(39)∥xn+1-x*∥2=∥PCn2PCn1(I-λnμF)xn-PCn2PCn1x*∥2=∥PCn2PCn1(I-λnμF)xn-PCn2PCn1xn+PCn2PCn1xn-PCn2PCn1x*∥2≤∥PCn2PCn1xn-PCn2PCn1x*∥2+2λnμ∥Fxn∥·∥xn-x*∥+λn2μ2∥Fxn∥2≤∥PCn2PCn1xn-PCn2PCn1x*∥2+λnM,
where M is a positive constant such that M≥supn{2μ∥Fxn∥·∥xn-x*∥+λnμ2∥Fxn∥2}. The combination of (38) and (39) leads to
(40)∥xn+1-x*∥2≤∥xn-x*∥2-∥xn-PCn1xn∥2-∥PCn1xn-PCn2PCn1xn∥2+λnM.
Setting
(41)sn=∥xn-x*∥2,γn=τλn,αn=Mλn,δn=-2μτ〈Fx*,xn-x*〉+2λnμ2τ∥Fx*∥∥Fxn∥,ηn=∥xn-PCn1xn∥2+∥PCn1xn-PCn2PCn1xn∥2,
then (33) and (40) can be rewritten as the following forms, respectively:
(42)sn+1≤(1-γn)sn+γnδn,sn+1≤sn-ηn+αn.
Finally, observing that the conditions λn→0 and ∑n=1∞‍λn=∞ imply αn→0 and ∑n=1∞‍γn=∞, respectively, in order to complete the proof using Lemma 7 (or Lemma 8), it suffices to verify that
(43)limk→∞ηnk=0
implies
(44)limsupk→∞δnk≤0
for any subsequence (nk)⊂(n). In fact, if ηnk→0 as k→∞, then ∥xnk-PCnk1xnk∥→0 and ∥PCnk1xnk-PCnk2PCnk1xnk∥→0 hold. Since ∂c1 and ∂c2 are bounded on bounded sets, we have two positive constants κ1 and κ2 such that ∥ξnk1∥≤κ1 and ∥ξnk2∥≤κ2 for all k≥0 (noting that (PCnk1xnk) is also bounded due to the fact that ∥PCnk1xnk-x*∥=∥PCnk1xnk-PCnk1x*∥≤∥xnk-x*∥). From (32) and the trivial fact that PCnk1xnk∈Cnk1 and PCnk2PCnk1xnk∈Cnk2, it follows that
(45)c1(xnk)≤〈ξnk1,xnk-PCnk1xnk〉≤κ1∥xnk-PCnk1xnk∥,(46)c2(PCnk1xnk)≤〈ξnk2,PCnk1xnk-PCnk2PCnk1xnk〉≤κ2∥PCnk1xnk-PCnk2PCnk1xnk∥.
Now if x′∈ωw(xnk), and (xnk) such that xnk⇀x′ without loss of the generality, then the w-lsc and (45) imply that
(47)c1(x′)≤liminfk→∞c1(xnk)≤0.
This means that x′∈C1 holds. On the other hand, noting ∥xnk-PCnk1xnk∥→0, we can assert that PCnk1xnk⇀x′ and have from the w-lsc and (46) that
(48)c2(x′)≤liminfk→∞c2(PCnk1xnk)≤0.
This, in turn, implies that x′∈C2. Moreover, we obtain that x′∈C1⋂‍C2 and hence ωw(xnk)⊂C1⋂‍C2=C.
Notingx* is the unique solution of VI(C,F), it turns out that
(49)limsupk→∞{-2μτ〈Fx*,xnk-x*〉}=-2μτliminfk→∞〈Fx*,xnk-x*〉=-2μτinfw∈ωw(xnk)〈Fx*,w-x*〉≤0.
Since λn→0 and (Fxn) is bounded, it is easy to see that limsupk→∞δnk≤0.Observing that in Algorithm1 the determination of the stepsize μ still depends on the constants L and η; this means that in order to implement Algorithm 1, one has first to estimate the constants L and η, which is sometimes not an easy work in practice.To overcome this difficulty, we furthermore introduce a so-called relaxed and self-adaptive algorithm, that is, a modification of Algorithm1, in which the stepsize is selected through a self-adaptive way that has no connection with the constants L and η.Algorithm 2.
Choose an arbitrary initial guessx0∈H and an arbitrary element x1∈H such that x1≠x0. Assume that the nth iterate xn  (n≥1) has been constructed. Continue and calculate the (n+1)th iterate xn+1 via the following formula:
(50)xn+1=PCn2PCn1(I-λnμnF)xn,n≥1,
where Cn1 and Cn2 are given as in (32), the sequence (λn) is in (0,1), and the sequence (μn) is determined via the following relation:(51)
μ
n
=
{
〈
F
x
n
-
F
x
n
-
1
,
x
n
-
x
n
-
1
〉
∥
F
x
n
-
F
x
n
-
1
∥
2
,
if
x
n
≠
x
n
-
1
,
μ
n
-
1
,
if
x
n
=
x
n
-
1
,
n
≥
1
.Firstly, we show that the sequence(xn) is well defined. Noting strong monotonicity of F, x1≠x0 implies that Fx1≠Fx0 and μ1 is well defined via the first formula of (51). Consequently, μn(n≥2) is well defined inductively according to (51) and thus the sequence (xn) is also well defined.Next, we estimate(μn) roughly. If xn≠xn-1 (that is, Fxn≠Fxn-1), set
(52)ηn=〈Fxn-Fxn-1,xn-xn-1〉∥xn-xn-1∥2,Ln=∥Fxn-Fxn-1∥∥xn-xn-1∥,n≥1.
Obviously, it turns out that
(53)η≤ηn=〈Fxn-Fxn-1,xn-xn-1〉∥xn-xn-1∥2≤∥Fxn-Fxn-1∥∥xn-xn-1∥=Ln≤L.
Consequently
(54)ηL2≤μn=ηnLn2≤1ηn≤1η.
By the definition of (μn), we can assert that (54) holds for all n≥1.Lemma7 (or Lemma 8) is also important for the proof of the strong convergence of Algorithm 2.Theorem 10.
Assume thatλn→0(n→∞) and ∑n=1+∞‍λn=+∞. Then the sequence (xn) generated by Algorithm 2 converges strongly to the unique solution x* of VI(C,F).Proof.
Settingγn=λnμn and βn=(1/2)(2η-γnL2), it concludes observing λn→0 and (54) that there exists some positive integer n0 such that
(55)0<γn<ηL2,n≥n0,
and consequently
(56)βn≥12η,n≥n0.
Using Lemma 2, we have from (55) that PCn(I-γnF) (so is I-γnF) is a contraction with coefficient 1-γnβn. This concludes that, for all n≥n0,
(57)∥xn+1-x*∥2=∥PCn2PCn1(I-γnF)xn-PCn2PCn1x*∥2≤∥(I-γnF)xn-(I-γnF)x*-γnFx*∥2≤(1-γnβn)∥xn-x*∥2-2γn〈Fx*,xn-x*-γnFxn〉,≤(1-γnβn)∥xn-x*∥2-2γn〈Fx*,xn-x*〉+2γn2∥Fx*∥∥Fxn∥,(58)∥xn+1-x*∥2=∥PCn2PCn1(I-γnF)xn-PCn2PCn1(I-γnF)x*+PCn2PCn1(I-γnF)x*-PCn2PCn1x*∥2≤(1-γnβn)∥xn-x*∥2+2〈PCn2PCn1(I-γnF)x*-PCn2PCn1x*,xn+1-x*〉≤(1-γnβn)∥xn-x*∥2+2γn∥Fx*∥∥xn+1-x*∥≤(1-γnβn)∥xn-x*∥2+γnβn4∥xn+1-x*∥2+4γnβn∥Fx*∥2,
and hence
(59)∥xn+1-x*∥2≤1-γnβn1-(1/4)γnβn∥xn-x*∥2+(3/4)γnβn1-(1/4)γnβn163βn2∥Fx*∥2.
Using (56), it turns out that
(60)∥xn+1-x*∥≤max{∥xn-x*∥,83η∥Fx*∥},n≥n0,
inductively
(61)∥xn-x*∥≤max{∥xn0-x*∥,83η∥Fx*∥},n≥n0,
and this means that (xn) is bounded, so is (Fxn).
By an argument similar to getting (38)–(40), we have
(62)∥xn+1-x*∥2≤∥xn-x*∥2-∥xn-PCn1xn∥2-∥PCn1xn-PCn2PCn1xn∥2+γnM,
where M is a positive constant. Setting
(63)sn=∥xn-x*∥2,δn=-2βn〈Fx*,xn-x*〉+2γnβn∥Fx*∥∥Fxn∥,αn=Mγn,σn=∥xn-PCn1xn∥2+∥PCn1xn-PCn2PCn1xn∥2,
then (57) and (62) can be rewritten as the following forms, respectively:
(64)sn+1≤(1-γnβn)sn+γnβnδn,sn+1≤sn-σn+αn.
Clearly, λn→0 and ∑n=1∞‍λn=∞, together with (54) and (56), imply that αn→0 and ∑n=1∞‍γnβn=∞.
By an argument very similar to the proof of Theorem9, it is not difficult to verify that
(65)limk→∞σnk=0
implies
(66)limsupk→∞δnk≤0
for any subsequence (nk)⊂(n). Thus we can complete the proof by using Lemma 7 (or Lemma 8).



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*Source: 942315-2013-05-08.xml*