aarjaneiro/planetmath huggingface dataset:
basic criterion for self-adjointness

basic criterion for self-adjointness

Let

A

:

D

⁢

(

A

)

⊂

ℋ

→

ℋ

be a

symmetric operator

on a

Hilbert space

. The following are

equivalent

:

1.

A

=

A

*

(i.e

A

is self-adjoint);

2.

Ker

⁡

(

A

*

±

i

)

=

{

0

}

and

A

is closed;

3.

Ran

⁡

(

A

±

i

)

=

ℋ

.

Remark:

A

+

λ

represents the

operator

A

+

λ

⁢

I

:

D

⁢

(

A

)

⊂

ℋ

→

ℋ

, and

Ker

and

Ran

stand for kernel and range, respectively.

A similar version for essential self-adjointness is an easy corollary of the above. The following are equivalent:

1.

A

¯

=

A

*

(i.e.

A

is essentially self-adjoint);

2.

Ker

⁡

(

A

*

±

i

)

=

{

0

}

;

3.

Ran

⁡

(

A

±

i

)

is dense in

ℋ

.

Title

basic criterion for self-adjointness

Canonical name

BasicCriterionForSelfadjointness

Date of creation

2013-03-22 14:53:02

Last modified on

2013-03-22 14:53:02

Owner

Koro (127)

Last modified by

Koro (127)

Numerical id

5

Author

Koro (127)

Entry type

Theorem

Classification

msc 47B25

Generated on Fri Feb 9 21:47:05 2018 by

LaTeXML
proof of basic criterion for self-adjointness

proof of basic criterion for self-adjointness

1.

(

1

⟹

2

)

If

A

is

self-adjoint

and

A

⁢

x

=

i

⁢

x

, then

i

⁢

∥

x

∥

2

=

(

i

⁢

x

,

x

)

=

(

A

⁢

x

,

x

)

=

(

x

,

A

*

⁢

x

)

=

(

x

,

A

⁢

x

)

=

(

x

,

i

⁢

x

)

=

(

i

⁢

x

,

x

)

¯

=

-

i

⁢

∥

x

∥

2

,

so

x

=

0

. Similarly we prove that

A

⁢

x

=

-

i

⁢

x

implies

x

=

0

. That

A

is closed follows from the fact that the

adjoint

of an

operator

is always closed.

2.

(

2

⟹

3

)

If

2

holds, then

{

0

}

=

Ker

(

A

*

±

i

)

*

=

Ker

(

A

∓

i

)

*

=

Ran

(

A

∓

i

)

⟂

, so that

Ran

⁡

A

∓

i

is dense in

ℋ

. Also, since

A

is symmetric, for

x

∈

D

⁢

(

A

)

,

∥

(

A

+

i

)

⁢

x

∥

2

=

∥

A

⁢

x

∥

2

+

∥

x

∥

2

+

(

A

⁢

x

,

i

⁢

x

)

+

(

i

⁢

x

,

A

⁢

x

)

=

∥

A

⁢

x

∥

2

+

∥

x

∥

2

because

(

A

⁢

x

,

i

⁢

x

)

=

(

x

,

i

⁢

A

*

⁢

x

)

=

(

x

,

i

⁢

A

⁢

x

)

=

-

(

i

⁢

x

,

A

⁢

x

)

.
Hence

∥

x

∥

≤

∥

(

A

+

i

)

⁢

x

∥

, so that given a

sequence

x

n

∈

D

⁢

(

A

)

such that

(

A

+

i

)

⁢

x

n

→

y

, we have that

{

(

A

+

i

)

⁢

x

n

}

is a

Cauchy sequence

and thus

{

x

n

}

itself is a Cauchy sequence. Hence

{

x

n

}

converges

to some

x

∈

ℋ

and since

A

is closed it follows that

x

∈

D

⁢

(

A

)

and

(

A

+

i

)

⁢

x

=

y

. This proves that

y

∈

Ran

⁡

(

A

+

i

)

, so that

Ran

⁡

(

A

+

i

)

is closed (and similarly,

Ran

⁡

(

A

-

i

)

is closed. Thus

Ran

⁡

(

A

±

i

)

=

ℋ

.

3.

(

3

⟹

1

)

Suppose

3

. If

y

∈

D

⁢

(

A

*

)

, then there is

x

∈

D

⁢

(

A

)

such that

(

A

+

i

)

⁢

x

=

(

A

*

-

i

)

⁢

y

. Since

A

is symmetric,

(

A

+

i

)

⁢

x

=

(

A

*

+

i

)

⁢

x

=

(

A

-

i

)

*

⁢

x

, so that

(

A

*

-

i

)

⁢

(

x

-

y

)

=

0

. But since

Ker

(

A

*

-

i

)

=

Ran

(

A

+

i

)

⟂

=

{

0

}

, it follows that

x

=

y

, so that

y

∈

D

⁢

(

A

)

. Hence

D

⁢

(

A

)

=

D

⁢

(

A

*

)

, and therefore

A

is self-adjoint.

Title

proof of basic criterion for self-adjointness

Canonical name

ProofOfBasicCriterionForSelfadjointness

Date of creation

2013-03-22 14:53:05

Last modified on

2013-03-22 14:53:05

Owner

Koro (127)

Last modified by

Koro (127)

Numerical id

5

Author

Koro (127)

Entry type

Proof

Classification

msc 47B25

Generated on Fri Feb 9 21:47:07 2018 by

LaTeXML
multiplication operator on L^2

multiplication operator on

L

2

Let

(

X

,

𝒜

,

μ

)

be a

measure space

and

f

:

X

→

𝕂

a

measurable function

. Then

M

f

:

ϕ

↦

f

⁢

ϕ

is the

multiplication operator

with

f

defined on the

subspace

D

⁢

o

⁢

m

⁢

(

M

f

)

=

{

ϕ

∈

L

𝕂

2

⁢

(

X

,

𝒜

,

μ

)

:

f

⁢

ϕ

∈

L

𝕂

2

⁢

(

X

,

𝒜

,

μ

)

}

. It plays an important role in quantum mechanics where the

multiplication

with the coordinates on

ℝ

n

is the position

operator

.

Title

multiplication operator on

L

2

Canonical name

MultiplicationOperatorOnL2

Date of creation

2013-03-22 15:42:28

Last modified on

2013-03-22 15:42:28

Owner

scineram (4030)

Last modified by

scineram (4030)

Numerical id

8

Author

scineram (4030)

Entry type

Definition

Classification

msc 47B38

Related topic

operator

Defines

multiplication operator

Generated on Fri Feb 9 21:47:09 2018 by

LaTeXML
functional calculus for Hermitian matrices

functional calculus for Hermitian matrices

Let

I

⊂

ℝ

be a real interval,

f

a real-valued function on

I

, and let

M

be an

n

×

n

real

symmetric

(and thus

Hermitian

) matrix whose

eigenvalues

are contained in

I

.

By the

spectral theorem

, we can

diagonalize

M

by an

orthogonal matrix

O

, so we can write

M

=

O

⁢

D

⁢

O

-

1

where

D

is the

diagonal matrix

consisting of the eigenvalues

{

λ

1

,

λ

2

,

…

,

λ

n

}

. We then define

f

⁢

(

A

)

=

O

⁢

f

⁢

(

D

)

⁢

O

-

1

,

where

f

⁢

(

D

)

denotes the diagonal matrix whose

diagonal

entries are given by

f

⁢

(

λ

i

)

.

It is easy to verify that

f

⁢

(

A

)

is well-defined, i.e. a

permutation

of the eigenvalues corresponds to the same definition of

f

⁢

(

A

)

.

Title

functional calculus for Hermitian matrices

Canonical name

FunctionalCalculusForHermitianMatrices

Date of creation

2013-03-22 14:40:12

Last modified on

2013-03-22 14:40:12

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

4

Author

mathcam (2727)

Entry type

Definition

Classification

msc 47C05

Related topic

FunctionalCalculus

Generated on Fri Feb 9 21:47:15 2018 by

LaTeXML
Baer ring

Baer ring

Baer Rings

Let

R

be a ring with

multiplicative identity

1

. Then

R

is called a left

Baer ring

if, for any subset

S

of

R

, the

left annihilator

of

S

is left principal, generated by an

idempotent

:

l

.

ann

⁡

(

S

)

:=

{

r

∈

R

∣

r

⁢

S

=

0

}

=

R

⁢

e

.

A right Baer ring is defined similarly, by replacing the word left with right above. It turns out that a left Baer ring is a right Baer ring, and vice versa, so we may drop the word left or right in the name.

Clearly, a domain is a Baer ring. And, by

Wedderburn-Artin theorem

, a

semisimple ring

is also a Baer ring. Another example, found in operator theory, is the ring of

bounded linear operators

on a

Hilbert space

.

A closely related

concept

is that of a left (right)

Baer *-ring

: it is a

ring with involution

*

such that the left (right)

annihilator

of any subset is left (right) principal, generated by a projection (an idempotent that is in

addition

a

self-adjoint element

).

A left Baer *-ring is a left Baer ring, a right Baer *-ring is a right Baer ring. And, interestingly, the notion of left and right is also redundent for Baer *-rings. For example, let’s show left means right. Since a left Baer * ring

R

is left, and consequently right Baer, the right annihilator of a subset

S

has the form

r

.

ann

⁡

(

S

)

=

e

⁢

R

for some idempotent

e

∈

R

. Since

S

⁢

e

=

0

,

e

*

⁢

S

*

=

0

or that

l

.

ann

⁡

(

S

*

)

=

R

⁢

e

*

. But

R

is left Baer *,

l

.

ann

⁡

(

S

*

)

=

R

⁢

f

for some projection

f

∈

R

. So

f

⁢

R

=

f

*

⁢

R

*

=

(

R

⁢

f

)

*

=

(

R

⁢

e

*

)

*

=

e

*

*

⁢

R

*

=

e

⁢

R

=

r

.

ann

⁡

(

S

)

.

Rickart Rings

A closely related type of rings is called a left

Rickart ring

. A ring is left Rickart if the left annihilator of any

element

a

∈

R

is left principal, generated by an idempotent:

l

.

a

⁢

n

⁢

n

⁢

(

a

)

:=

{

r

∈

R

∣

r

⁢

a

=

0

}

=

R

⁢

e

where

e

=

e

⁢

e

. A right Rickart ring is similarly defined. Clearly, a Baer ring is both right and left Rickart. However, the

converse

is not always true. In fact, right Rickart and left Rickart are not the same. An example of a left Rickart ring that is not right Rickart can be found in the second reference below. It may be shown that if all idempotents are central (lying in the center) in a left Rickart ring

R

, then

R

is also right Rickart (and vice versa).

References

1

I. Kaplansky,

Rings of

Operators

, W. A. Benjamin, Inc., New York, 1968.

2

T. Y. Lam,

Lectures on Modules and Rings

, Springer, New York, 1998.

Title

Baer ring

Canonical name

BaerRing

Date of creation

2013-03-22 15:52:07

Last modified on

2013-03-22 15:52:07

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

6

Author

CWoo (3771)

Entry type

Definition

Classification

msc 47C10

Classification

msc 47C15

Classification

msc 16U99

Defines

Baer *-ring

Defines

Rickart ring

Generated on Fri Feb 9 21:47:17 2018 by

LaTeXML
partial isometry

partial isometry

Partial isometry is a

generalization

of an

isometry

. Before defining what a partial isometry is, let’s recall two familiar

concepts in linear algebra

: an isometry and the

adjoint

of a

linear map

.

1.

An isometry

T

is a linear

automorphism

over an

inner product space

V

which preserves the

inner product

of any two vectors:

⟨

x

,

y

⟩

=

⟨

T

⁢

x

,

T

⁢

y

⟩

.

2.

The adjoint

T

*

of a linear transformation

T

is linear transformation such that

⟨

T

⁢

x

,

y

⟩

=

⟨

x

,

T

*

⁢

y

⟩

, for any pair of vectors

x

,

y

∈

V

.

If

V

is

non-singular

with respect to the inner product

⟨

⋅

,

⋅

⟩

and that the adjoint

T

*

of a linear transformation

T

exists, it is not hard to show that

T

is an isometry if and only if

T

⁢

T

*

=

I

=

T

*

⁢

T

.

In other words,

T

*

is the

inverse

of

T

.

More generally, in a ring with involution

*

, an isometry (or an

unitary element

) is a unit (both a left unit and a right unit)

a

whose

product

with its adjoint

a

*

is 1 (i.e. its inverse is its adjoint). Now, if

a

is not a unit, this product

a

⁢

a

*

will not be 1. The next best thing to hope for is that the product will be an

idempotent

. But because

a

⁢

a

*

is

self-adjoint

, this idempotent is in fact a

projection

. This is how a partial isometry is defined. Formally,

let

R

be a ring with involution

*

, an

element

a

∈

R

is a

partial isometry

if

a

⁢

a

*

and

a

*

⁢

a

are both projections.

Given a partial isometry

a

, the projections

a

*

⁢

a

and

a

⁢

a

*

are respectively called the

initial projection

and

final projection

of

a

.

Examples

. Under this definition,

0

is a partial isometry, and so is any isometry.

This definition can be readily applied to specific (more familiar) situations. For example, if the ring in question is the ring of linear endomorphisms over a

Euclidean space

(real or complex), then a partial isometry is just a map such that its

restriction

to the

complementary subspace

of its kernel is an isometry. If we look at the case when the space is 3 dimensional over the reals, and taking the

standard basis

, the matrix

A

=

(

0

0

0

0

sin

⁡

θ

-

cos

⁡

θ

0

cos

⁡

θ

sin

⁡

θ

)

corresponds to a partial isometry whose kernel is a line

L

. Its restriction to the

complement

of

L

corresponds to the matrix

B

=

(

sin

⁡

θ

-

cos

⁡

θ

cos

⁡

θ

sin

⁡

θ

)

,

which is an isometry (

rotation

).

Remark

. If the ring

R

is a Baer *-ring, an element

a

is a partial isometry iff

a

⁢

a

*

⁢

a

=

a

(so

a

*

⁢

a

⁢

a

*

=

a

*

;

a

and

a

*

are

generalized inverses

of one another).

Title

partial isometry

Canonical name

PartialIsometry

Date of creation

2013-03-22 15:50:50

Last modified on

2013-03-22 15:50:50

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

7

Author

CWoo (3771)

Entry type

Definition

Classification

msc 47C10

Defines

unitary element

Defines

initial projection

Defines

final projection

Generated on Fri Feb 9 21:47:19 2018 by

LaTeXML
partial isometry on Hilbert spaces

partial isometry on Hilbert spaces

Definition 1

.

Let

ℋ

and

𝒦

be

Hilbert spaces

. An

operator

W

∈

L

⁢

(

ℋ

,

𝒦

)

is called a

partial isometry

if

W

is an isometry on

M

=

(

ker

⁡

W

)

⟂

. We then call

M

=

(

ker

⁡

W

)

⟂

the

initial space

and

N

=

W

⁢

M

final space

of

W

.

We need to show that the above definition is compatible with the general definition of partial isometry on rings. Indeed we have the following:

Proposition 1

.

W

∈

L

⁢

(

ℋ

,

𝒦

)

is a partial isometry iff

W

*

⁢

W

is a projection from

H

to

M

.

Proof.

We have:

W

⁢

partial isometry with initial space

⁢

M

⇔

⟨

W

⁢

f

,

W

⁢

g

⟩

=

⟨

f

,

g

⟩

⁢

∀

f

,

g

∈

M

⇔

⟨

W

*

⁢

W

⁢

f

,

g

⟩

=

⟨

f

,

g

⟩

⁢

∀

f

∈

M

,

g

∈

ℋ

⇔

W

*

⁢

W

⁢

f

=

f

,

f

∈

M

and

⁢

W

*

⁢

W

⁢

f

=

0

,

f

∈

M

⟂

=

ker

⁡

W

∎

Remark 1

.

If

W

∈

L

⁢

(

H

,

K

)

is a partial isometry with initial space

M

and final space

N

we have:

W

*

⁢

(

W

⁢

f

)

=

f

⁢

∀

f

∈

M

ker

⁡

W

*

=

(

ran

⁢

W

)

⟂

=

N

⟂

Thus

N

is the initial space and

M

the final space of

W

*

.

Title

partial isometry on Hilbert spaces

Canonical name

PartialIsometryOnHilbertSpaces

Date of creation

2013-03-22 18:35:00

Last modified on

2013-03-22 18:35:00

Owner

karstenb (16623)

Last modified by

karstenb (16623)

Numerical id

7

Author

karstenb (16623)

Entry type

Definition

Classification

msc 47C10

Generated on Fri Feb 9 21:47:22 2018 by

LaTeXML
Dini derivative

Dini derivative

The

upper

Dini derivative

of a

continuous function

,

f

:

𝐑

↦

𝐑

, denoted by

f

+

′

, is defined as

f

+

′

⁢

(

t

)

=

lim

h

→

0

+

⁡

sup

⁡

f

⁢

(

t

+

h

)

-

f

⁢

(

t

)

h

.

The

lower Dini derivative

,

f

-

′

, is defined as

f

-

′

⁢

(

t

)

=

lim

h

→

0

+

⁡

inf

⁡

f

⁢

(

t

+

h

)

-

f

⁢

(

t

)

h

.

Remark: Sometimes the notation

D

+

⁢

f

⁢

(

t

)

is used instead of

f

+

′

⁢

(

t

)

, and

D

-

⁢

f

⁢

(

t

)

is used instead of

f

-

′

⁢

(

t

)

.

Remark: Like conventional

derivatives

, Dini derivatives do not always exist.

If

f

is defined on a vector space, then the upper Dini derivative at

t

in the direction

d

is denoted

f

+

′

⁢

(

t

,

d

)

=

lim

h

→

0

+

⁡

sup

⁡

f

⁢

(

t

+

h

⁢

d

)

-

f

⁢

(

t

)

h

.

If

f

is locally

Lipschitz

then

D

+

⁢

f

is finite. If

f

is

differentiable

at

t

then the Dini derivative at

t

is the derivative at

t

.

Title

Dini derivative

Canonical name

DiniDerivative

Date of creation

2013-03-22 13:57:00

Last modified on

2013-03-22 13:57:00

Owner

lha (3057)

Last modified by

lha (3057)

Numerical id

11

Author

lha (3057)

Entry type

Definition

Classification

msc 47G30

Generated on Fri Feb 9 21:47:29 2018 by

LaTeXML
any topological space with the fixed point property is connected

any topological space with the fixed point property is connected

Theorem

Any

topological space

with the

fixed-point property

(

http://planetmath.org/FixedPointProperty

) is

connected

.

Proof.

We will prove the contrapositive.
Suppose

X

is a topological space which is not connected.
So there are non-empty disjoint

open sets

A

,

B

⊆

X

such
that

X

=

A

∪

B

. Then there are elements

a

∈

A

and

b

∈

B

, and
we can define a function

f

:

X

→

X

by

f

⁢

(

x

)

=

{

a

,

when

⁢

x

∈

B

,

b

,

when

⁢

x

∈

A

.

Since

A

∩

B

=

∅

and

A

∪

B

=

X

, the function

f

is

well-defined

.
Also,

a

∉

B

and

b

∉

A

, so

f

has no

fixed point

.
Furthermore, if

V

is an open set in

X

, a short calculation shows that

f

-

1

⁢

(

V

)

is

∅

,

A

,

B

or

X

, all of which are open sets.
So

f

is

continuous

, and therefore

X

does not have the fixed-point property.

□

References

1

G.J. Jameson,

Topology

and

Normed Spaces

,
Chapman and Hall, 1974.

2

L.E. Ward,

Topology, An Outline for a First Course

,
Marcel Dekker, Inc., 1972.

Title

any topological space with the fixed point property is connected

Canonical

name

AnyTopologicalSpaceWithTheFixedPointPropertyIsConnected

Date of creation

2013-03-22 13:56:35

Last modified on

2013-03-22 13:56:35

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

12

Author

yark (2760)

Entry type

Theorem

Classification

msc 47H10

Classification

msc 54H25

Classification

msc 55M20

Generated on Fri Feb 9 21:47:31 2018 by

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Brouwer fixed point in one dimension

Brouwer fixed point in one dimension

Theorem 1

[

1

,

adams

]

Suppose

f

is a

continuous function

f

:

[

-

1

,

1

]

→

[

-

1

,

1

]

. Then

f

has a

fixed point

, i.e.,
there is a

x

such that

f

⁢

(

x

)

=

x

.

Proof

(Following

[

1

]

)
We can assume that

f

⁢

(

-

1

)

>

-

1

and

f

⁢

(

+

1

)

<

1

, since otherwise
there is nothing to prove. Then, consider the function

g

:

[

-

1

,

1

]

→

ℝ

defined by

g

⁢

(

x

)

=

f

⁢

(

x

)

-

x

. It satisfies

g

⁢

(

+

1

)

<

0

,

g

⁢

(

-

1

)

>

0

,

so by the

intermediate value theorem

, there is a point

x

such that

g

⁢

(

x

)

=

0

, i.e.,

f

⁢

(

x

)

=

x

.

□

Assuming slightly more of the function

f

yields the

Banach fixed point theorem

. In one

dimension

it states the following:

Theorem 2

Suppose

f

:

[

-

1

,

1

]

→

[

-

1

,

1

]

is a function that satisfies the
following condition:

for some constant

C

∈

[

0

,

1

)

, we have for each

a

,

b

∈

[

-

1

,

1

]

,

|

f

⁢

(

b

)

-

f

⁢

(

a

)

|

≤

C

⁢

|

b

-

a

|

.

Then

f

has a unique fixed point in

[

-

1

,

1

]

. In other words, there exists
one and only one point

x

∈

[

-

1

,

1

]

such that

f

⁢

(

x

)

=

x

.

Remarks

The fixed point in Theorem 2 can be found by

iteration

from any

s

∈

[

-

1

,

1

]

as follows:
first choose some

s

∈

[

-

1

,

1

]

.
Then form

s

1

=

f

⁢

(

s

)

, then

s

2

=

f

⁢

(

s

1

)

, and generally

s

n

=

f

⁢

(

s

n

-

1

)

.
As

n

→

∞

,

s

n

approaches the fixed point for

f

. More details
are given on the entry for the Banach fixed point theorem.
A function that satisfies the
condition in Theorem 2 is called a contraction mapping. Such mappings also satisfy the

Lipschitz condition

(

http://planetmath.org/LipschitzCondition

).

References

1

A. Mukherjea, K. Pothoven,

Real and

Functional analysis

,
Plenum press, 1978.

Title

Brouwer fixed point in one dimension

Canonical name

BrouwerFixedPointInOneDimension

Date of creation

2013-03-22 13:46:25

Last modified on

2013-03-22 13:46:25

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

10

Author

mathcam (2727)

Entry type

Proof

Classification

msc 47H10

Classification

msc 54H25

Classification

msc 55M20

Related topic

LipschitzCondition

Generated on Fri Feb 9 21:47:34 2018 by

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proof of Brouwer fixed point theorem

proof of Brouwer fixed point theorem

Proof of the

Brouwer fixed point theorem

:

Assume that there does exist a map from

f

:

B

n

→

B

n

with no

fixed point

. Then let

g

⁢

(

x

)

be the following map: Start at

f

⁢

(

x

)

, draw the ray going through

x

and then let

g

⁢

(

x

)

be
the first

intersection

of that line with the sphere. This map is

continuous

and well defined only
because

f

fixes no point. Also, it is not hard to see that it must be the

identity

on the

boundary

sphere. Thus we have a map

g

:

B

n

→

S

n

-

1

, which is the identity on

S

n

-

1

=

∂

⁡

B

n

, that is, a

retraction

. Now, if

i

:

S

n

-

1

→

B

n

is the

inclusion
map

,

g

∘

i

=

id

S

n

-

1

. Applying the reduced homology

functor

, we find that

g

*

∘

i

*

=

id

H

~

n

-

1

⁢

(

S

n

-

1

)

, where

*

indicates the

induced map

on

homology

.

But, it is a well-known fact that

H

~

n

-

1

⁢

(

B

n

)

=

0

(since

B

n

is

contractible

), and that

H

~

n

-

1

⁢

(

S

n

-

1

)

=

ℤ

. Thus we have an

isomorphism

of a non-zero group onto itself
factoring through a trivial group, which is clearly impossible. Thus we have a

contradiction

,
and no such map

f

exists.

Title

proof of Brouwer fixed point theorem

Canonical name

ProofOfBrouwerFixedPointTheorem

Date of creation

2013-03-22 13:11:24

Last modified on

2013-03-22 13:11:24

Owner

bwebste (988)

Last modified by

bwebste (988)

Numerical id

6

Author

bwebste (988)

Entry type

Proof

Classification

msc 47H10

Classification

msc 54H25

Classification

msc 55M20

Generated on Fri Feb 9 21:47:36 2018 by

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proof of Brouwer fixed point theorem

proof of Brouwer fixed point theorem

The

n

-dimensional

simplex

𝒮

n

is the following
subset of

ℝ

n

+

1

{

(

α

1

,

α

2

,

…

,

α

n

+

1

)

|

∑

i

=

1

n

+

1

α

i

=

1

,

α

i

≥

0

∀

i

=

1

,

…

,

n

+

1

}

Given an

element

x

=

∑

i

α

i

⁢

e

i

∈

𝒮

n

we denote

[

x

]

i

=

α

i

(i.e., the

i

-th

barycentric coordinate

). We
also denote

F

⁢

(

x

)

=

{

i

|

[

x

]

i

≠

0

}

. An

I

-face of

𝒮

n

is the subset

{

x

|

F

⁢

(

x

)

⊆

I

}

.

As was noted in the statement of the

theorem

, the ’shape’ is
unimportant. Therefore, we will prove the
following variant of the theorem using the

KKM lemma

.

Theorem 1

(Brouwer’s Fixed Point Theorem).

Let

f

:

S

n

→

S

n

be a

continuous function

.
Then,

f

has a

fixed point

, namely, there is an

L

∈

S

n

such that

L

=

f

⁢

(

L

)

.

Proof.

Clearly,

∑

i

=

1

n

[

y

]

i

=

1

for any

y

∈

𝒮

n

and

L

=

f

⁢

(

L

)

if and only if

[

L

]

i

=

[

f

⁢

(

L

)

]

i

for all

i

=

1

,

2

,

…

,

n

+

1

. For each

i

=

1

,

2

,

…

,

n

+

1

we define the
following subset

C

i

of

𝒮

n

:

C

i

=

{

x

∈

𝒮

n

|

[

x

]

i

≥

[

f

⁢

(

x

)

]

i

}

We claim that if

x

is in some

I

-face of

𝒮

n

(

I

⊆

{

1

,

2

,

…

,

n

+

1

}

) then there is an index

i

∈

I

such that

x

∈

C

i

. Indeed, if

x

is in some

I

-face then

F

⁢

(

v

)

⊆

I

. Thus, if

[

x

]

i

≠

0

then

i

∈

I

. This
shows that

∑

i

∈

I

[

x

]

i

=

1

Assuming by

contradiction

that

x

∉

C

i

for all

i

∈

I

implies that

[

x

]

i

<

[

f

⁢

(

x

)

]

i

for all

i

∈

I

. But this leads
to a contradiction as the following

inequality

shows:

1

=

∑

i

∈

I

[

x

]

i

<

∑

i

∈

I

[

f

⁢

(

x

)

]

i

≤

∑

i

=

1

n

[

f

⁢

(

x

)

]

i

=

1

This dicussion establishes that each

I

-face is

contained

in the
union

∪

i

∈

I

C

i

. In

addition

, the subsets

C

i

are all
closed. Therefore, we have shown that the

hypothesis

of the KKM
Lemma holds.

By the KKM lemma there is a point

L

that is in every

C

i

for

i

=

1

,

2

,

…

,

n

+

1

. We claim that

L

is a fixed point of

f

.
Indeed,

[

L

]

i

≥

[

f

⁢

(

L

)

]

i

≥

0

for all

i

=

1

,

2

,

…

,

n

+

1

and
thus:

1

=

[

L

]

1

+

[

L

]

2

+

⋯

+

[

L

]

n

+

1

≥

[

f

⁢

(

L

)

]

1

+

[

f

⁢

(

L

)

]

2

+

⋯

+

[

f

⁢

(

L

)

]

n

+

1

=

1

Therefore,

[

L

]

i

=

[

f

⁢

(

L

)

]

i

for all

i

=

1

,

2

,

…

,

n

+

1

which
implies that

L

=

f

⁢

(

L

)

.
∎

Title

proof of Brouwer fixed point theorem

Canonical name

ProofOfBrouwerFixedPointTheorem1

Date of creation

2013-03-22 18:13:24

Last modified on

2013-03-22 18:13:24

Owner

uriw (288)

Last modified by

uriw (288)

Numerical id

4

Author

uriw (288)

Entry type

Proof

Classification

msc 47H10

Classification

msc 54H25

Classification

msc 55M20

Generated on Fri Feb 9 21:47:39 2018 by

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