aarjaneiro/planetmath huggingface dataset:
proof of Schauder fixed point theorem

proof of Schauder fixed point theorem

The idea of the proof is to reduce to the finite dimensional case where we can apply the

Brouwer fixed point theorem

.

Given

ϵ

>

0

notice that the family of

open sets

{

B

ϵ

⁢

(

x

)

:

x

∈

K

}

is an open covering of

K

. Being

K

compact

there exists a finite

subcover

, i.e. there exists

n

points

x

1

,

…

,

x

n

of

K

such that the balls

B

ϵ

⁢

(

x

i

)

cover the whole set

K

.

Define the functions

g

1

,

…

,

g

n

by

g

i

⁢

(

x

)

:=

{

ϵ

-

∥

x

-

x

i

∥

,

if

∥

x

-

x

i

∥

≤

ϵ

0

,

if

∥

x

-

x

i

∥

≥

ϵ

It is clear that each

g

i

is

continuous

,

g

i

⁢

(

x

)

≥

0

and

∑

i

=

1

n

g

i

⁢

(

x

)

>

0

for every

x

∈

K

.

Thus we can define a function in

K

by

g

⁢

(

x

)

:=

∑

i

=

1

n

g

i

⁢

(

x

)

⁢

x

i

∑

i

=

1

n

g

i

⁢

(

x

)

The above function

g

is a

continuous function

from

K

to the convex hull

K

0

of

x

1

,

…

,

x

n

. Moreover one can easily prove the following

∥

g

⁢

(

x

)

-

x

∥

≤

ϵ

∀

x

∈

K

Now, define the function

B

:=

g

∘

f

. The restriction

B

~

of

B

to

K

0

provides a continuous function

K

0

⟶

K

0

.

Since

K

0

is compact convex subset of a finite dimensional vector space, we can apply the Brouwer fixed point theorem to assure the existence of

z

∈

K

0

such that

B

⁢

(

z

)

=

B

~

⁢

(

z

)

=

z

Therefore

g

⁢

(

f

⁢

(

z

)

)

=

z

and we have the

inequality

∥

f

⁢

(

z

)

-

z

∥

=

∥

f

⁢

(

z

)

-

g

⁢

(

f

⁢

(

z

)

)

∥

≤

ϵ

Summarizing, for each

ϵ

>

0

there exists

z

=

z

⁢

(

ϵ

)

∈

K

such that

∥

f

⁢

(

z

)

-

z

∥

≤

ϵ

. Then

∀

m

∈

ℕ

∃

z

m

∈

K

∥

f

⁢

(

z

m

)

-

z

m

∥

≤

1

m

As

f

⁢

(

z

m

)

is in the

compact space

K

, there is a subsequence

z

m

k

such that

f

⁢

(

z

m

k

)

⟶

x

0

, for some

x

0

∈

K

.

We then have

∥

z

m

k

-

x

0

∥

=

∥

z

m

k

-

f

⁢

(

z

m

k

)

+

f

⁢

(

z

m

k

)

-

x

0

∥

≤

∥

f

⁢

(

z

m

k

)

-

z

m

k

∥

+

∥

f

⁢

(

z

m

k

)

-

x

0

∥

≤

1

m

k

+

∥

f

⁢

(

z

m

k

)

-

x

0

∥

⟶

0

which means that

z

m

k

⟶

x

0

.

As

f

is continuous we have

f

⁢

(

z

m

k

)

⟶

f

⁢

(

x

0

)

. Both limits of

f

⁢

(

z

m

k

)

must coincide, so we conclude that

f

⁢

(

x

0

)

=

x

0

i.e.

f

has a

fixed point

.

□

Title

proof of Schauder fixed point theorem

Canonical name

ProofOfSchauderFixedPointTheorem

Date of creation

2013-03-22 13:45:22

Last modified on

2013-03-22 13:45:22

Owner

asteroid (17536)

Last modified by

asteroid (17536)

Numerical id

10

Author

asteroid (17536)

Entry type

Proof

Classification

msc 47H10

Classification

msc 46T99

Classification

msc 46T20

Classification

msc 46B50

Classification

msc 54H25

Generated on Fri Feb 9 21:47:42 2018 by

LaTeXML
convex hull of S is open if S is open

convex hull of

S

is open if

S

is open

Theorem

If

S

is an

open set

in a

topological vector space

, then
the convex hull

co

⁡

(

S

)

is open.

As the next example shows, the corresponding result does not hold for a

closed set

.

Example

(Valentine, p. 14)
If

S

=

{

(

x

,

1

/

|

x

|

)

∈

ℝ

2

∣

x

∈

ℝ

∖

{

0

}

}

,

then

S

is closed,
but

co

⁡

(

S

)

is the open

half-space

{

(

x

,

y

)

∣

x

∈

ℝ

,

y

∈

(

0

,

∞

)

}

,
which is not closed (points on the

x

-axis are

accumulation points

not in the set, or also can be seen by checking the

complement

is not open).

□

Reference

F.A. Valentine,

Convex sets

, McGraw-Hill book company, 1964.

Title

convex hull of

S

is open if

S

is open

Canonical name

ConvexHullOfSIsOpenIfSIsOpen

Date of creation

2013-03-22 13:44:47

Last modified on

2013-03-22 13:44:47

Owner

drini (3)

Last modified by

drini (3)

Numerical id

9

Author

drini (3)

Entry type

Theorem

Classification

msc 47L07

Classification

msc 46A55

Generated on Fri Feb 9 21:47:44 2018 by

LaTeXML
proof that the convex hull of S is open if S is open

proof that the convex hull of

S

is open if

S

is open

Let

S

be an open set in some

topological vector space

V

. For any sequence of positive real numbers

Λ

=

(

λ

1

,

…

,

λ

n

)

with

∑

i

=

1

n

λ

i

=

1

define

S

Λ

=

{

x

∈

V

such that

x

=

∑

i

=

1

n

λ

i

s

i

for

s

i

∈

S

}

.

Then since addition and

scalar multiplication

are both open maps, each

S

Λ

is open. Finally, the

convex hull

is clearly just

⋃

Λ

S

Λ

,

which is therefore open.

Title

proof that the convex hull of

S

is open if

S

is open

Canonical name

ProofThatTheConvexHullOfSIsOpenIfSIsOpen

Date of creation

2013-03-22 14:09:48

Last modified on

2013-03-22 14:09:48

Owner

archibal (4430)

Last modified by

archibal (4430)

Numerical id

6

Author

archibal (4430)

Entry type

Proof

Classification

msc 47L07

Classification

msc 46A55

Generated on Fri Feb 9 21:47:46 2018 by

LaTeXML
derivative operator is unbounded in the sup norm

derivative operator is unbounded in the sup norm

Consider

C

∞

⁢

(

[

-

1

,

1

]

)

the vector space of

functions

with

derivatives

or arbitrary order on the set

[

-

1

,

1

]

.

This space admits a norm called the

supremum norm

given by

|

f

|

=

sup

⁡

{

|

f

⁢

(

x

)

|

:

x

∈

[

-

1

,

1

]

}

This norm makes this vector space into a metric space.

We claim that the derivative

operator

D

:

(

D

⁢

f

)

⁢

(

x

)

=

f

′

⁢

(

x

)

is an

unbounded operator

.

All we need to prove is that there exists a succession of functions

f

n

∈

C

∞

⁢

(

[

-

1

,

1

]

)

such that

|

D

⁢

(

f

n

)

|

|

f

n

|

is

divergent

as

n

→

∞

consider

f

n

⁢

(

x

)

=

exp

⁡

(

-

n

4

⁢

x

2

)

(

D

⁢

f

n

)

⁢

(

x

)

=

-

2

⁢

x

⁢

n

4

⁢

exp

⁡

(

-

n

4

⁢

x

2

)

Clearly

|

f

n

|

=

f

n

⁢

(

0

)

=

1

To find

|

D

⁢

f

n

|

we need to find the

extrema

of the derivative of

f

n

, to do that calculate the

second derivative

and equal it to zero. However for the task at hand a crude estimate will be enough.

|

D

⁢

f

n

|

≥

|

(

D

⁢

f

n

)

⁢

(

1

n

2

)

|

=

2

⁢

n

2

e

So we finally get

|

D

⁢

f

n

|

|

f

n

|

≥

2

⁢

n

2

e

showing that the derivative operator is indeed

unbounded

since

2

⁢

n

2

e

is divergent as

n

→

∞

.

Title

derivative operator is unbounded in the sup norm

Canonical name

DerivativeOperatorIsUnboundedInTheSupNorm

Date of creation

2013-07-14 20:22:55

Last modified on

2013-07-14 20:22:55

Owner

cvalente (11260)

Last modified by

cvalente (11260)

Numerical id

8

Author

cvalente (11260)

Entry type

Proof

Classification

msc 47L25

Generated on Fri Feb 9 21:47:48 2018 by

LaTeXML
examples of bounded and unbounded operators

examples of bounded and unbounded operators

The aim of this page is to list examples of

bounded

(

http://planetmath.org/BoundedOperator

) and unbounded

linear operators

.

Bounded

•

Identity operator

,

Zero operator

•

Shift operators

on

ℓ

p

•

A linear operator is

continuous

if and only if
it is bounded (see this page (

http://planetmath.org/ContinuousLinearMapping

)).

•

Any isometry is bounded.

•

A

multiplication operator

h

⁢

(

t

)

↦

f

⁢

(

t

)

⁢

h

⁢

(

t

)

, where

f

⁢

(

t

)

is continuous and

h

∈

L

p

⁢

[

0

,

1

]

.

•

An integral operator

h

⁢

(

t

)

↦

∫

0

1

K

⁢

(

t

,

s

)

⁢

h

⁢

(

s

)

⁢

𝑑

s

, where

∫

0

1

∫

0

1

|

K

⁢

(

s

,

t

)

|

2

⁢

𝑑

s

⁢

𝑑

t

<

∞

and

h

∈

L

2

⁢

[

0

,

1

]

.
In fact this is a

Hilbert-Schmidt operator

.

•

The Volterra operator

h

⁢

(

t

)

↦

∫

0

t

h

⁢

(

s

)

⁢

𝑑

s

, where

h

∈

L

p

⁢

[

0

,

1

]

.

Unbounded

•

The

derivative

is an

unbounded operator

on the

vector space

of

smooth functions

equipped with the

sup

-norm.

Title

examples of bounded and unbounded operators

Canonical name

ExamplesOfBoundedAndUnboundedOperators

Date of creation

2013-03-22 15:17:37

Last modified on

2013-03-22 15:17:37

Owner

matte (1858)

Last modified by

matte (1858)

Numerical id

12

Author

matte (1858)

Entry type

Example

Classification

msc 47L25

Generated on Fri Feb 9 21:47:50 2018 by

LaTeXML
operator norm

operator norm

Definition

Let

A

:

𝖵

→

𝖶

be a

linear map

between

normed vector spaces

𝖵

and

𝖶

. To each such map (operator)

A

we can assign a non-negative number

∥

A

∥

op

defined by

∥

A

∥

op

:=

sup

𝐯

∈

𝖵

∥

A

⁢

𝐯

∥

∥

𝐯

∥

,

where the supremum

∥

A

∥

op

could be finite or infinite.
Equivalently, the above definition can be written as

∥

A

∥

op

:=

sup

𝐯

∈

𝖵

∥

A

⁢

𝐯

∥

=

sup

𝐯

∈

𝖵

∥

A

⁢

𝐯

∥

.

By convention, if

𝖵

is the

zero vector space

, any operator from

𝖵

to

𝖶

must be the

zero operator

and is assigned zero norm.

∥

A

∥

op

is called the the

operator norm

(or the

induced norm

)
of

A

, for reasons that will be clear in the next .

Operator norm is in fact a norm

Definition -

If

∥

A

∥

op

is finite, we say that

A

is a

. Otherwise, we say that

A

is

.

It turns out that, for

bounded operators

,

∥

⋅

∥

op

satisfies all the properties of a norm
(hence the name

operator norm

). The proof follows immediately from the definition:

Positivity:

Since

∥

A

⁢

𝐯

∥

≥

0

, by definition

∥

A

∥

op

≥

0

. Also,

∥

A

⁢

𝐯

∥

=

0

identically only if

A

=

0

.
Hence

∥

A

∥

op

=

0

only if

A

=

0

.

Absolute homogeneity:

Since

∥

λ

⁢

A

⁢

𝐯

∥

=

|

λ

|

⁢

∥

A

⁢

𝐯

∥

, by definition

∥

λ

⁢

A

∥

op

=

|

λ

|

⁢

∥

A

∥

op

.

Triangle inequality:

Since

∥

(

A

+

B

)

⁢

𝐯

∥

=

∥

A

⁢

𝐯

+

B

⁢

𝐯

∥

≤

∥

A

⁢

𝐯

∥

+

∥

B

⁢

𝐯

∥

, by definition

∥

A

+

B

∥

op

≤

∥

A

∥

op

+

∥

B

∥

op

.

The set

L

⁢

(

𝖵

,

𝖶

)

of bounded linear maps from

𝖵

to

𝖶

forms a

vector space

and

∥

⋅

∥

op

defines a norm in it.

Example

Suppose that

𝖵

=

(

ℝ

n

,

∥

⋅

∥

p

)

and

𝖶

=

(

ℝ

n

,

∥

⋅

∥

p

)

, where

∥

⋅

∥

p

is the

vector p-norm

. Then the operator norm

∥

⋅

∥

op

=

∥

⋅

∥

p

is the

matrix p-norm

.

Title

operator norm

Canonical name

OperatorNorm

Date of creation

2013-03-22 12:43:20

Last modified on

2013-03-22 12:43:20

Owner

asteroid (17536)

Last modified by

asteroid (17536)

Numerical id

15

Author

asteroid (17536)

Entry type

Definition

Classification

msc 47L25

Classification

msc 46A32

Classification

msc 47A30

Synonym

induced norm

Related topic

VectorNorm

Related topic

OperatorTopologies

Related topic

HomomorphismsOfCAlgebrasAreContinuous

Related topic

CAlgebra

Defines

bounded linear map

Defines

unbounded linear map

Defines

bounded operator

Defines

unbounded operator

Generated on Fri Feb 9 21:47:52 2018 by

LaTeXML
Fuglede-Putnam-Rosenblum theorem

Fuglede-Putnam-Rosenblum theorem

Let

A

be a

C

∗

-algebra with unit

e

.

The Fuglede-Putnam-Rosenblum

theorem

makes the assertion that for a

normal element

a

∈

A

the kernel of the commutator

mapping

[

a

,

-

]

:

A

→

A

is a

∗

-

closed set

.

The general formulation of the result is as follows:

Theorem.

Let

A

be a

C

∗

-algebra with unit

e

. Let two normal elements

a

,

b

∈

A

be given and

c

∈

A

with

a

⁢

c

=

c

⁢

b

.
Then it follows that

a

∗

⁢

c

=

c

⁢

b

∗

.

Lemma.

For any

x

∈

A

we have that

exp

⁡

(

x

-

x

∗

)

is a element of

A

.

Proof.

We have for

x

∈

A

that

exp

(

x

-

x

∗

)

∗

exp

(

x

-

x

∗

)

=

exp

(

x

∗

-

x

+

x

-

x

∗

)

=

exp

(

0

)

=

e

.
And similarly

exp

(

x

-

x

∗

)

exp

(

x

-

x

∗

)

∗

=

e

. ∎

With this we can now give a proof the Theorem.

Proof.

The condition

a

⁢

c

=

c

⁢

b

implies by

induction

that

a

k

⁢

c

=

c

⁢

b

k

holds for each

k

∈

ℕ

.
Expanding in

power series

on both sides yields

exp

⁡

(

a

)

⁢

c

=

c

⁢

exp

⁡

(

b

)

.
This is

equivalent

to

c

=

exp

⁡

(

-

a

)

⁢

c

⁢

exp

⁡

(

b

)

. Set

U

1

:=

exp

⁡

(

a

∗

-

a

)

,

U

2

:=

exp

⁡

(

b

-

b

∗

)

. From the Lemma we obtain that

∥

U

1

∥

A

=

∥

U

2

∥

A

=

1

.
Since

a

commutes with

a

∗

und

b

with

b

∗

we obtain that

exp

⁡

(

a

∗

)

⁢

c

⁢

exp

⁡

(

-

b

∗

)

=

exp

⁡

(

a

∗

)

⁢

exp

⁡

(

-

a

)

⁢

c

⁢

exp

⁡

(

b

)

⁢

exp

⁡

(

b

∗

)

which equals

exp

⁡

(

a

∗

-

a

)

⁢

c

⁢

exp

⁡

(

b

-

b

∗

)

=

U

1

⁢

c

⁢

U

2

.

Hence

∥

exp

⁡

(

a

∗

)

⁢

c

⁢

exp

⁡

(

-

b

∗

)

∥

≤

∥

c

∥

.

Define

f

:

ℂ

→

A

by

f

⁢

(

λ

)

:=

exp

⁡

(

λ

⁢

a

∗

)

⁢

c

⁢

exp

⁡

(

-

λ

⁢

b

∗

)

. If we substitute

a

↦

λ

⁢

a

,

b

↦

λ

⁢

b

in the last estimate we obtain

∥

f

⁢

(

λ

)

∥

≤

∥

c

∥

,

λ

∈

ℂ

.

But

f

is clearly an

entire function

and therefore Liouville’s theorem implies that

f

⁢

(

λ

)

=

f

⁢

(

0

)

=

c

for each

λ

.

This yields the equality

c

⁢

exp

⁡

(

λ

⁢

b

∗

)

=

exp

⁡

(

λ

⁢

a

∗

)

⁢

c

.

Comparing the terms of

first order

for

λ

small finishes the proof. ∎

Title

Fuglede-Putnam-Rosenblum theorem

Canonical name

FugledePutnamRosenblumTheorem

Date of creation

2013-05-08 21:47:27

Last modified on

2013-05-08 21:47:27

Owner

karstenb (16623)

Last modified by

karstenb (16623)

Numerical id

1

Author

karstenb (16623)

Entry type

Theorem

Classification

msc 47L30

Generated on Fri Feb 9 21:47:55 2018 by

LaTeXML
Drazin inverse

Drazin inverse

A Drazin inverse of an operator

A

is an operator,

B

, such that

A

⁢

B

=

B

⁢

A

,

B

⁢

A

⁢

B

=

B

,

A

⁢

B

⁢

A

=

A

-

U

,

where the

spectral radius

r

⁢

(

U

)

=

0

. The Drazin inverse (

B

) is denoted by

A

D

. It exists, if

0

is not an

accumulation point

of

σ

⁢

(

A

)

.

For example, a

projection operator

is its own Drazin inverse,

P

D

=

P

, as

P

⁢

P

⁢

P

=

P

⁢

P

=

P

; for a Shift operator

S

D

=

0

holds.

The following are some other useful properties of the Drazin inverse:

1.

(

A

D

)

*

=

(

A

*

)

D

;

2.

A

D

=

(

A

+

α

⁢

P

(

A

)

)

-

1

⁢

(

I

-

P

(

A

)

)

, where

P

(

A

)

:=

I

-

A

D

⁢

A

is the spectral

projection

of

A

at

0

and

α

≠

0

;

3.

A

†

=

(

A

*

⁢

A

)

D

⁢

A

*

=

A

*

⁢

(

A

⁢

A

*

)

D

, where

A

†

is the

Moore-Penrose pseudoinverse

of

A

;

4.

A

D

=

A

m

⁢

(

A

2

⁢

m

+

1

)

†

⁢

A

m

for

m

≥

ind

⁢

(

A

)

, if

ind

⁢

(

A

)

:=

min

⁡

{

k

:

Im

⁡

A

k

=

Im

⁡

A

k

+

1

}

is finite;

5.

If the matrix is represented explicitly by its

Jordan canonical form

, (

Λ

regular

and

N

nilpotent

), then

(

E

⁢

[

Λ

0

0

N

]

⁢

E

-

1

)

D

=

E

⁢

[

Λ

-

1

0

0

0

]

⁢

E

-

1

;

6.

Let

e

λ

A

denote an

eigenvector

of

A

to the

eigenvalue

λ

. Then

e

λ

A

+

t

⁢

(

λ

⁢

I

-

A

)

D

⁢

h

⁢

e

λ

A

+

O

⁢

(

t

2

)

is an eigenvector of

A

+

t

⁢

h

.

Title

Drazin inverse

Canonical name

DrazinInverse

Date of creation

2013-03-22 13:58:05

Last modified on

2013-03-22 13:58:05

Owner

kronos (12218)

Last modified by

kronos (12218)

Numerical id

29

Author

kronos (12218)

Entry type

Definition

Classification

msc 47S99

Related topic

MoorePenroseGeneralizedInverse

Generated on Fri Feb 9 21:47:57 2018 by

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matrix inversion lemma

matrix inversion lemma

These frequently used formulae allow to quickly calculate the

inverse

of a slight

modification

of an

operator

(matrix)

x

, given that

x

-

1

is already known.

The

matrix inversion lemma

states that

(

x

+

s

⁢

σ

⁢

z

*

)

-

1

=

x

-

1

-

x

-

1

⁢

s

⁢

(

σ

-

1

+

z

*

⁢

x

-

1

⁢

s

)

-

1

⁢

z

*

⁢

x

-

1

,

where

x

,

s

,

z

*

and

σ

are operators (matrices) of appropriate size. The formula especially is convenient if the rank of the

regular

σ

is 1, or small in comparison to

x

’s rank.

This identity, involving the inverse of Schur’s complement

d

-

z

*

⁢

x

-

1

⁢

s

(hopefully this may be easily computed) holds as well:

[

x

s

z

*

d

]

-

1

=

[

x

-

1

+

x

-

1

⁢

s

⁢

(

d

-

z

*

⁢

x

-

1

⁢

s

)

-

1

⁢

z

*

⁢

x

-

1

-

x

-

1

⁢

s

⁢

(

d

-

z

*

⁢

x

-

1

⁢

s

)

-

1

-

(

d

-

z

*

⁢

x

-

1

⁢

s

)

-

1

⁢

z

*

⁢

x

-

1

(

d

-

z

*

⁢

x

-

1

⁢

s

)

-

1

]

.

Title

matrix inversion lemma

Canonical name

MatrixInversionLemma

Date of creation

2013-03-22 15:38:44

Last modified on

2013-03-22 15:38:44

Owner

kronos (12218)

Last modified by

kronos (12218)

Numerical id

6

Author

kronos (12218)

Entry type

Result

Classification

msc 47S99

Synonym

Sherman-Morrison formula

Synonym

Woodbury matrix identity

Synonym

Woodbury formula

Synonym

rank-k correction

Related topic

SchurComplement

Generated on Fri Feb 9 21:47:59 2018 by

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essential boundary

essential boundary

Let

E

⊂

𝐑

n

be a

measurable set

. We define the

essential boundary

of

E

as

∂

*

⁡

E

:=

{

x

∈

𝐑

n

:

0

<

|

E

∩

B

ρ

⁢

(

x

)

|

<

|

B

ρ

⁢

(

x

)

|

,

∀

ρ

>

0

}

where

|

⋅

|

is the

Lebesgue measure

.

Compare the definition of

∂

*

⁡

E

with the definition of the

topological boundary

∂

⁡

E

which can be written as

∂

⁡

E

=

{

x

∈

𝐑

n

:

∅

⊊

E

∩

B

ρ

⁢

(

x

)

⊊

B

ρ

⁢

(

x

)

,

∀

ρ

>

0

}

.

Hence one clearly has

∂

*

⁡

E

⊂

∂

⁡

E

.

Notice that the essential boundary does not depend on the Lebesgue
representative of the set

E

, in the sense that if

|

E

⁢

△

⁢

F

|

=

0

then

∂

*

⁡

E

=

∂

*

⁡

F

. For example if

E

=

𝐐

n

⊂

𝐑

n

is
the set of points with

rational

coordinates, one has

∂

*

⁡

E

=

∅

while

∂

⁡

E

=

𝐑

n

.

Nevertheless one can easily prove that

∂

*

⁡

E

is always a

closed set

(in the usual sense).

Title

essential boundary

Canonical name

EssentialBoundary

Date of creation

2013-03-22 15:01:54

Last modified on

2013-03-22 15:01:54

Owner

paolini (1187)

Last modified by

paolini (1187)

Numerical id

9

Author

paolini (1187)

Entry type

Definition

Classification

msc 49-00

Generated on Fri Feb 9 21:48:17 2018 by

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Lagrange multipliers on Banach spaces

Lagrange multipliers on Banach spaces

Let

U

be open in a real

Banach space

X

,
and

Y

be another real Banach space.
Let

f

:

U

→

ℝ

and

g

:

U

→

Y

be

continuously differentiable functions

.

Suppose that

a

is a

minimum

or

maximum

point of

f

on

M

=

{

x

∈

U

:

g

⁢

(

x

)

=

0

}

,
and the Fréchet

derivative

D

⁡

g

⁢

(

a

)

:

X

→

Y

is surjective. Then there exists a

Lagrange multiplier

vector

λ

∈

Y

*

such
that

D

⁡

f

⁢

(

a

)

=

D

⁡

g

⁢

(

a

)

*

⁢

λ

=

λ

∘

D

⁡

g

⁢

(

a

)

.

(The function

D

⁡

g

⁢

(

a

)

*

:

Y

*

→

X

*

denotes
the pullback or

adjoint

by

D

⁡

g

⁢

(

a

)

on the

continuous duals

,
defined by the second equality.)

If

X

and

Y

are

finite-dimensional

, writing out the above
equation in matrix form shows that

λ

really
is the usual Lagrange multiplier vector. The condition
that

D

⁡

g

⁢

(

a

)

is surjective means that

D

⁡

g

⁢

(

a

)

must have full rank as a matrix.

References

1

Eberhard Zeidler.

Applied

functional analysis

: main principles and their applications

. Springer-Verlag, 1995.

Title

Lagrange multipliers on Banach spaces

Canonical name

LagrangeMultipliersOnBanachSpaces

Date of creation

2013-03-22 15:28:30

Last modified on

2013-03-22 15:28:30

Owner

stevecheng (10074)

Last modified by

stevecheng (10074)

Numerical id

5

Author

stevecheng (10074)

Entry type

Theorem

Classification

msc 49-00

Classification

msc 49K35

Generated on Fri Feb 9 21:48:19 2018 by

LaTeXML
proof of existence and uniqueness of best approximations

proof of existence and uniqueness of best approximations

- Proof of the

theorem

on

existence

and uniqueness of -

( entry (

http://planetmath.org/BestApproximationInInnerProductSpaces

))

Existence :

Without loss of generality we can suppose

x

=

0

(we could simply translate by

-

x

the set

A

).

Let

d

=

d

⁢

(

0

,

A

)

=

inf

⁡

{

∥

a

∥

:

a

∈

A

}

be the

distance

of

A

to the origin. By defintion of

infimum

there exists a

sequence

(

a

n

)

in

A

such that

∥

a

n

∥

⟶

d

Let us see that

(

a

n

)

is a

Cauchy sequence

. By the

parallelogram law

we have

∥

a

n

-

a

m

2

∥

2

+

∥

a

n

+

a

m

2

∥

2

=

1

2

⁢

∥

a

n

∥

2

+

1

2

⁢

∥

a

m

∥

2

i.e.

∥

a

n

-

a

m

2

∥

2

=

1

2

⁢

∥

a

n

∥

2

+

1

2

⁢

∥

a

m

∥

2

-

∥

a

n

+

a

m

2

∥

2

As

A

is convex,

a

n

+

a

m

2

∈

A

, and therefore

∥

a

n

+

a

m

2

∥

≥

d

So we see that

∥

a

n

-

a

m

2

∥

2

≤

1

2

⁢

∥

a

n

∥

2

+

1

2

⁢

∥

a

m

∥

2

-

d

2

⟶

0

when

⁢

m

,

n

→

∞

which means that

∥

a

n

-

a

m

∥

⟶

0

when

m

,

n

→

∞

, i.e.

(

a

n

)

is a Cauchy sequence.

Since

A

is

complete

(

http://planetmath.org/Complete

),

a

n

⟶

a

0

for some

a

0

∈

A

.

As

a

0

∈

A

its norm must be

∥

a

0

∥

≥

d

. But also

∥

a

0

∥

≤

∥

a

0

-

a

n

∥

+

∥

a

n

∥

⟶

d

which shows that

∥

a

0

∥

=

d

. We have thus proven the existence of best

approximations

(

http://planetmath.org/BestApproximationInInnerProductSpaces

).

Uniqueness :

Suppose there were

a

0

,

b

0

∈

A

such that

∥

a

0

∥

=

∥

b

0

∥

=

d

. Then, by the parallelogram law

∥

a

0

-

b

0

2

∥

2

+

∥

a

n

+

b

0

2

∥

2

=

1

2

⁢

∥

a

0

∥

2

+

1

2

⁢

∥

b

0

∥

2

=

d

2

If

a

0

-

b

0

≠

0

then we would have

∥

a

0

+

b

0

2

∥

2

<

d

2

, which is

contradiction

since

a

0

+

b

0

2

∈

A

(

A

is convex).

Therefore

a

0

=

b

0

, which proves the uniqueness of the .

□

Title

proof of existence and uniqueness of best approximations

Canonical name

ProofOfExistenceAndUniquenessOfBestApproximations

Date of creation

2013-03-22 17:32:22

Last modified on

2013-03-22 17:32:22

Owner

asteroid (17536)

Last modified by

asteroid (17536)

Numerical id

6

Author

asteroid (17536)

Entry type

Proof

Classification

msc 49J27

Classification

msc 46N10

Classification

msc 46C05

Classification

msc 41A65

Classification

msc 41A52

Classification

msc 41A50

Generated on Fri Feb 9 21:48:27 2018 by

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Ekeland’s variational principle

Ekeland’s variational principle

Let

(

M

,

d

)

be a

complete metric space

and let

ψ

:

M

→

(

-

∞

,

+

∞

]

,

ψ

≢

+

∞

, be a

lower semicontinuous

function which is bounded from below. Then the following hold: For every

ε

>

0

and for any

z

0

∈

M

there exists

z

∈

M

such that

(i)

ψ

⁢

(

z

)

≤

ψ

⁢

(

z

0

)

-

ε

⁢

d

⁢

(

z

,

z

0

)

;

(ii)

ψ

⁢

(

x

)

≥

ψ

⁢

(

z

)

-

ε

⁢

d

⁢

(

x

,

z

)

, for any

x

∈

M

.

Title

Ekeland’s variational principle

Canonical name

EkelandsVariationalPrinciple

Date of creation

2013-03-22 15:19:16

Last modified on

2013-03-22 15:19:16

Owner

ncrom (8997)

Last modified by

ncrom (8997)

Numerical id

8

Author

ncrom (8997)

Entry type

Theorem

Classification

msc 49J40

Generated on Fri Feb 9 21:48:28 2018 by

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