aarjaneiro/planetmath huggingface dataset:
bounded linear extension of an operator

bounded linear extension of an operator

0.1

Bounded Linear Extension

Let

X

and

Y

be

normed vector spaces

and denote by

X

~

and

Y

~

their

completions

.

Theorem

1 -

Every

bounded linear operator

T

:

X

⟶

Y

can be extended to a bounded linear operator

T

~

:

X

~

⟶

Y

~

. Moreover, this

extension

is unique and

∥

T

∥

=

∥

T

~

∥

.

In particular, if

Y

is a

Banach space

and

S

⊆

X

is a (not necessarily closed (

http://planetmath.org/ClosedSet

))

subspace

of

X

, an

operator

T

:

S

⟶

Y

has an extension

T

~

:

S

¯

⟶

Y

to

S

¯

(the

closure

(

http://planetmath.org/Closure

) of

S

), which is unique and such that

∥

T

∥

=

∥

T

~

∥

.

0.2

Functorial Property of the Extension

The extension of bounded linear operators between two normed vector spaces to their completions is functorial. More precisely, let

𝐍𝐕𝐞𝐜

be the

category

of normed vector spaces (whose

morphisms

(

http://planetmath.org/Category

) are the bounded linear operators) and

𝐁𝐚𝐧

the categroy of Banach spaces (whose are also the bounded linear operators). We have that

Theorem 2 -

The completion

~

:

𝐍𝐕𝐞𝐜

⟶

𝐁𝐚𝐧

, which associates each normed vector space

X

with its completion

X

~

and each bounded linear operator

T

with its extension

T

~

, is a

covariant functor

.

This, in particular, implies that

T

1

⁢

T

2

~

=

T

1

~

⁢

T

2

~

.

0.3

Extensions in Spaces with Additional Structure

When the normed vector spaces

X

and

Y

have some additional

structure

(for example, when

X

and

Y

are

normed algebras

) it is interesting to know if the (unique) extension of a morphism

T

:

X

⟶

Y

preserves the additional structure. The following theorem states that this indeed the case for normed algebras or normed *-

algebras

.

Theorem 3 -

If

X

and

Y

be normed vector spaces that are also normed algebras (normed *-algebras) and

T

:

X

⟶

Y

is a

bounded

homomorphism

(bounded *-homomorphism), then the unique bounded linear extension

T

~

of

T

is also an homomorphism (*-homomorphism).

Thus, completion is also a covariant functor from the category of normed algebras (normed *-algebras) to category of

Banach algebras

(Banach *-algebras).

Title

bounded linear extension of an operator

Canonical name

BoundedLinearExtensionOfAnOperator

Date of creation

2013-03-22 17:35:17

Last modified on

2013-03-22 17:35:17

Owner

asteroid (17536)

Last modified by

asteroid (17536)

Numerical id

11

Author

asteroid (17536)

Entry type

Theorem

Classification

msc 46B99

Classification

msc 47A05

Synonym

continuous

extension of an operator

Defines

completion of

normed spaces

is a covariant functor

Defines

continuous extension of a normed algebra homomorphism

Generated on Fri Feb 9 21:37:08 2018 by

LaTeXML
bounded operator

bounded operator

Definition

[

1

]

1.

Suppose

X

and

Y

are

normed vector spaces

with norms

∥

⋅

∥

X

and

∥

⋅

∥

Y

. Further, suppose

T

is a

linear map

T

:

X

→

Y

. If there is a

C

∈

𝐑

such that for all

x

∈

X

we have

∥

T

⁢

x

∥

Y

≤

C

⁢

∥

x

∥

X

,

then

T

is a

bounded operator

.

2.

Let

X

and

Y

be as above, and let

T

:

X

→

Y

be a bounded
operator. Then the

norm

of

T

is defined as the

real number

∥

T

∥

:=

sup

⁢

{

∥

T

⁢

x

∥

Y

∥

x

∥

X

|

x

∈

X

∖

{

0

}

}

.

Thus the

operator norm

is the smallest

constant

C

∈

𝐑

such that

∥

T

⁢

x

∥

Y

≤

C

⁢

∥

x

∥

X

.

Now for any

x

∈

X

∖

{

0

}

, if we let

y

=

x

/

∥

x

∥

, then linearity implies that

∥

T

⁢

y

∥

Y

=

∥

T

⁢

(

x

∥

x

∥

X

)

∥

Y

=

∥

T

⁢

x

∥

Y

∥

x

∥

X

and thus it easily follows that

∥

T

∥

=

sup

⁢

{

∥

T

⁢

x

∥

Y

∥

x

∥

X

|

x

∈

X

∖

{

0

}

}

=

sup

⁢

{

∥

T

⁢

y

∥

Y

|

x

∈

X

∖

{

0

}

,

y

=

x

∥

x

∥

}

=

sup

⁢

{

∥

T

⁢

y

∥

Y

|

y

∈

X

,

∥

y

∥

=

1

}

.

In the special case when

X

=

{

𝟎

}

is the

zero vector space

, any linear
map

T

:

X

→

Y

is the

zero map

since

T

⁢

(

𝟎

)

=

T

⁢

(

𝟎𝟎

)

=

0

⁢

T

⁢

(

𝟎

)

=

0

. In this
case, we define

∥

T

∥

:=

0

.

3.

To avoid cumbersome notational stuff usually one can simplify the symbols
like

||

x

||

X

and

||

T

⁢

x

||

Y

by writing only

||

x

||

,

||

T

⁢

x

||

since there is a little danger in confusing which is space about calculating norms.

0.0.1

TO DO:

1.

The defined norm for

mappings

is a norm

2.

Examples:

identity operator

, zero operator: see

[

1

]

.

3.

Give alternative

expressions

for norm of

T

.

4.

Discuss boundedness and

continuity

Theorem

[

1

,

2

]

Suppose

T

:

X

→

Y

is a
linear map between normed vector spaces

X

and

Y

. If

X

is

finite-dimensional

, then

T

is

bounded

.

Theorem

Suppose

T

:

X

→

Y

is a
linear map between normed vector spaces

X

and

Y

. The following are

equivalent

:

1.

T

is

continuous

in some point

x

0

∈

X

2.

T

is

uniformly continuous

in

X

3.

T

is bounded

Lemma

Any bounded operator with a finite dimensional kernel and cokernel has a closed image.

Proof

By Banach’s isomorphism theorem.

References

1

E. Kreyszig,

Introductory

Functional Analysis

With Applications

,
John Wiley & Sons, 1978.

2

G. Bachman, L. Narici,

Functional analysis

, Academic Press, 1966.

Title

bounded operator

Canonical

name

BoundedOperator

Date of creation

2013-03-22 14:02:17

Last modified on

2013-03-22 14:02:17

Owner

bwebste (988)

Last modified by

bwebste (988)

Numerical id

19

Author

bwebste (988)

Entry type

Definition

Classification

msc 46B99

Related topic

VectorNorm

Generated on Fri Feb 9 21:37:12 2018 by

LaTeXML
closure of a vector subspace is a vector subspace

closure of a vector subspace is a vector subspace

Theorem 1

.

In a

topological vector space

the

closure

(

http://planetmath.org/Closure

) of a

vector subspace

is a vector subspace.

Proof.

Let

X

be the topological vector space over

𝔽

where

𝔽

is either

ℝ

or

ℂ

, let

V

be a vector subspace
in

X

, and let

V

¯

be the closure of

V

.
To prove that

V

¯

is a vector subspace of

X

, it suffices
to prove that

V

¯

is non-empty, and

λ

⁢

x

+

μ

⁢

y

∈

V

¯

whenever

λ

,

μ

∈

𝔽

and

x

,

y

∈

V

¯

.

First, as

V

⊆

V

¯

,

V

¯

contains the

zero vector

,
and

V

¯

is non-empty.
Suppose

λ

,

μ

,

x

,

y

are as above.
Then there are nets

(

x

i

)

i

∈

I

,

(

y

j

)

j

∈

J

in

V

converging to

x

,

y

, respectively.
In a topological vector space,

addition

and

multiplication

are

continuous

operations

. It follows that there is a net

(

λ

⁢

x

k

+

μ

⁢

y

k

)

k

∈

K

that

converges

to

λ

⁢

x

+

μ

⁢

y

.

We have proven that

λ

⁢

x

+

μ

⁢

y

∈

V

¯

, so

V

¯

is a vector subspace.
∎

Title

closure of a vector subspace is a vector subspace

Canonical name

ClosureOfAVectorSubspaceIsAVectorSubspace

Date of creation

2013-03-22 15:00:19

Last modified on

2013-03-22 15:00:19

Owner

loner (106)

Last modified by

loner (106)

Numerical id

8

Author

loner (106)

Entry type

Theorem

Classification

msc 46B99

Classification

msc 15A03

Classification

msc 54A05

Related topic

ClosureOfAVectorSubspaceIsAVectorSubspace

Related topic

ClosureOfSetsClosedUnderAFinitaryOperation

Generated on Fri Feb 9 21:37:14 2018 by

LaTeXML
compact operator

compact operator

Let

X

and

Y

be two

Banach spaces

.
A

compact operator

(completely continuous

operator

) is a

linear operator

T

:

X

→

Y

that maps the

unit ball

in

X

to a set in

Y

with compact closure. It can be shown that a compact operator is necessarily a

bounded operator

.

The set of all compact operators on

X

, commonly denoted by

𝕂

⁢

(

X

)

,
is a closed

two-sided ideal

of the set of all bounded operators on

X

,

𝔹

⁢

(

X

)

.

Any bounded operator which is the norm limit of a

sequence

of finite rank operators is

compact

.
In the case of

Hilbert spaces

, the converse is also true.
That is, any compact operator on a Hilbert space is a norm limit of finite rank operators.

Example 1

(Integral operators)

Let

k

⁢

(

x

,

y

)

, with

x

,

y

∈

[

0

,

1

]

, be a

continuous function

.
The operator defined by

(

T

⁢

ψ

)

⁢

(

x

)

=

∫

0

1

k

⁢

(

x

,

y

)

⁢

ψ

⁢

(

y

)

⁢

d

y

,

ψ

∈

C

⁢

(

[

0

,

1

]

)

is compact.

Title

compact operator

Canonical name

CompactOperator

Date of creation

2013-03-22 14:26:59

Last modified on

2013-03-22 14:26:59

Owner

mhale (572)

Last modified by

mhale (572)

Numerical id

8

Author

mhale (572)

Entry type

Definition

Classification

msc 46B99

Synonym

completely continuous

Generated on Fri Feb 9 21:37:16 2018 by

LaTeXML
continuous linear mapping

continuous linear mapping

If

(

V

1

,

∥

⋅

∥

1

)

and

(

V

2

,

∥

⋅

∥

2

)

are

normed vector spaces

, a linear mapping

T

:

V

1

→

V

2

is

continuous

if it is continuous in the metric induced by the norms.

If there is a nonnegative

constant

c

such that

∥

T

⁢

(

x

)

∥

2

≤

c

⁢

∥

x

∥

1

for each

x

∈

V

1

, we say that

T

is

. This should not be confused with the usual terminology referring to a

bounded function

as one that has

bounded

range. In fact,

bounded linear mappings

usually have

unbounded

ranges.

The

expression

bounded linear mapping

is often used in

functional analysis

to refer to continuous linear mappings as well. This is because the two

definitions

are

equivalent

:

If

T

is bounded, then

∥

T

⁢

(

x

)

-

T

⁢

(

y

)

∥

2

=

∥

T

⁢

(

x

-

y

)

∥

2

≤

c

⁢

∥

x

-

y

∥

1

, so

T

is a

Lipschitz function

. Now suppose

T

is continuous. Then there exists

r

>

0

such that

∥

T

⁢

(

x

)

∥

2

≤

1

when

∥

x

∥

1

≤

r

. For any

x

∈

V

1

, we then have

r

∥

x

∥

1

⁢

∥

T

⁢

(

x

)

∥

2

=

∥

T

⁢

(

r

∥

x

∥

1

⁢

x

)

∥

2

≤

1

,

hence

∥

T

⁢

(

x

)

∥

2

≤

r

⁢

∥

x

∥

1

; so

T

is bounded.

It can be shown that a linear mapping between two

topological vector spaces

is continuous if and only if it is continuous at (

http://planetmath.org/Continuous

)

0

[

1

]

.

References

1

W. Rudin,

Functional Analysis

,
McGraw-Hill Book Company, 1973.

Title

continuous linear mapping

Canonical

name

ContinuousLinearMapping

Date of creation

2013-03-22 13:15:41

Last modified on

2013-03-22 13:15:41

Owner

Koro (127)

Last modified by

Koro (127)

Numerical id

7

Author

Koro (127)

Entry type

Definition

Classification

msc 46B99

Synonym

bounded linear mapping

Related topic

HomomorphismsOfCAlgebrasAreContinuous

Related topic

CAlgebra

Related topic

BoundedLinearFunctionalsOnLpmu

Defines

bounded linear transform

Defines

bounded linear operator

Generated on Fri Feb 9 21:37:18 2018 by

LaTeXML
ell^p

ell^p

Let

𝔽

be either

ℝ

or

ℂ

, and let

p

∈

ℝ

with

p

≥

1

. We define

ℓ

p

to be the set of all sequences

(

a

i

)

i

≥

0

in

𝔽

such that

∑

i

=

0

∞

|

a

i

|

p

converges.

We also define

ℓ

∞

to be the set of all

bounded

(

http://planetmath.org/BoundedInterval

) sequences

(

a

i

)

i

≥

0

with norm given by

∥

(

a

i

)

∥

∞

=

sup

⁡

{

|

a

i

|

:

i

≥

0

}

.

By defining addition and

scalar multiplication

pointwise,

ℓ

p

⁢

(

𝔽

)

and

ℓ

∞

⁢

(

𝔽

)

have a natural

vector space

stucture.
That the sum of two elements on

ℓ

p

⁢

(

𝔽

)

is again an element
in

ℓ

p

⁢

(

𝔽

)

follows from

Minkowski inequality

(see below).
We can make

ℓ

p

into a

normed vector space

, by defining the norm as

∥

(

a

i

)

∥

p

=

(

∑

i

=

0

∞

|

a

i

|

p

)

1

/

p

.

The normed vector spaces

ℓ

∞

and

ℓ

p

for

p

≥

1

are complete under these norms, making them into

Banach spaces

. Moreover,

ℓ

2

is a

Hilbert space

under the

inner product

⟨

(

a

i

)

,

(

b

i

)

⟩

=

∑

i

=

0

∞

a

i

⁢

b

i

¯

where

x

¯

denotes the

complex conjugate

of

x

.

For

p

>

1

the (

continuous

)

dual space

of

ℓ

p

is

ℓ

q

where

1

p

+

1

q

=

1

, and the dual space of

ℓ

1

is

ℓ

∞

.

Properties

1.

If

a

=

(

a

0

,

a

1

,

…

)

∈

ℓ

p

⁢

(

𝔽

)

for

1

≤

p

<

∞

, then

lim

k

→

∞

⁡

a

k

=

0

.
(proof. (

http://planetmath.org/ThenA_kto0IfSum_k1inftyA_kConverges

))

2.

For

1

≤

p

<

∞

,

ℓ

p

⁢

(

𝔽

)

is separable, and

ℓ

∞

⁢

(

𝔽

)

is not separable.

3.

Minkowski inequality. If

a

,

b

∈

ℓ

p

⁢

(

𝔽

)

where

p

≥

1

, then

∥

a

+

b

∥

p

≤

∥

a

∥

p

+

∥

b

∥

p

.

Title

ell^p

Canonical name

Ellp

Date of creation

2013-03-22 12:19:03

Last modified on

2013-03-22 12:19:03

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

25

Author

rspuzio (6075)

Entry type

Definition

Classification

msc 46B99

Classification

msc 54E50

Related topic

EllpXSpace

Defines

ℓ

∞

Defines

ℓ

2

Generated on Fri Feb 9 21:37:21 2018 by

LaTeXML
every finite dimensional normed vector space is a Banach space

every finite dimensional normed vector space is a Banach space

Theorem 1

.

Every finite dimensional

normed vector space

is a

Banach space

.

Proof.

Suppose

(

V

,

∥

⋅

∥

)

is the normed vector space,
and

(

e

i

)

i

=

1

N

is a basis for

V

.
For

x

=

∑

j

=

1

N

λ

j

⁢

e

j

, we can then define

∥

x

∥

′

=

∑

j

=

1

N

|

λ

j

|

2

whence

∥

⋅

∥

′

:

V

→

ℝ

is a norm for

V

.
Since
all norms on a finite dimensional vector space are

equivalent

(

http://planetmath.org/ProofThatAllNormsOnFiniteVectorSpaceAreEquivalent

),
there is a

constant

C

>

0

such that

1

C

⁢

∥

x

∥

′

≤

∥

x

∥

≤

C

⁢

∥

x

∥

′

,

x

∈

V

.

To prove that

V

is a Banach space, let

x

1

,

x

2

,

…

be a

Cauchy sequence

in

(

V

,

∥

⋅

∥

)

. That is,
for all

ε

>

0

there is an

M

≥

1

such that

∥

x

j

-

x

k

∥

<

ε

,

for all

⁢

j

,

k

≥

M

.

Let us write each

x

k

in this

sequence

in the basis

(

e

j

)

as

x

k

=

∑

j

=

1

N

λ

k

,

j

⁢

e

j

for some constants

λ

k

,

j

∈

ℂ

.
For

k

,

l

≥

1

we then have

∥

x

k

-

x

l

∥

≥

1

C

⁢

∥

x

k

-

x

l

∥

′

≥

1

C

⁢

∑

j

=

1

N

|

λ

k

,

j

-

λ

l

,

j

|

2

≥

1

C

⁢

|

λ

k

,

j

-

λ

l

,

j

|

for all

j

=

1

,

…

,

N

.
It follows that

(

λ

k

,

1

)

k

=

1

∞

,

…

,

(

λ

k

,

N

)

k

=

1

∞

are Cauchy sequences in

ℂ

. As

ℂ

is

complete

, these

converge

to
some

complex numbers

λ

1

,

…

,

λ

N

.
Let

x

=

∑

j

=

1

N

λ

j

⁢

e

j

.

For each

k

=

1

,

2

,

…

, we then have

∥

x

-

x

k

∥

≤

C

⁢

∥

x

-

x

k

∥

′

≤

C

⁢

∑

j

=

1

N

|

λ

j

-

λ

k

,

j

|

2

.

By taking

k

→

∞

it follows that

(

x

j

)

converges to

x

∈

V

.

□

Title

every finite dimensional normed vector space is a Banach space

Canonical name

EveryFiniteDimensionalNormedVectorSpaceIsABanachSpace

Date of creation

2013-03-22 14:56:31

Last modified on

2013-03-22 14:56:31

Owner

matte (1858)

Last modified by

matte (1858)

Numerical id

10

Author

matte (1858)

Entry type

Theorem

Classification

msc 46B99

Generated on Fri Feb 9 21:37:26 2018 by

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every finite dimensional subspace of a normed space is closed

every finite dimensional subspace of a normed space is closed

Theorem 1

Any

finite dimensional

subspace

of a

normed vector space

is closed.

Proof.

Let

(

V

,

∥

⋅

∥

)

be such a normed vector space, and

S

⊂

V

a finite dimensional vector subspace.

Let

x

∈

V

, and let

(

s

n

)

n

be a

sequence

in

S

which converges to

x

. We want to prove that

x

∈

S

. Because

S

has finite

dimension

, we have a basis

{

x

1

,

…

,

x

k

}

of

S

. Also,

x

∈

span

⁡

(

x

1

,

…

,

x

k

,

x

)

. But, as proved in the case when

V

is finite dimensional (see this parent (

http://planetmath.org/EverySubspaceOfANormedSpaceOfFiniteDimensionIsClosed

)), we have that

S

is closed in

span

⁡

(

x

1

,

…

,

x

k

,

x

)

(taken with the norm

induced

by

(

V

,

∥

⋅

∥

)

) with

s

n

→

x

, and then

x

∈

S

. QED.

0.0.1

Notes

The definition of a normed vector space requires the ground field to be the real or

complex numbers

. Indeed, consider the following counterexample if that condition doesn’t hold:

V

=

ℝ

is a

ℚ

-

vector space

, and

S

=

ℚ

is a vector subspace of

V

. It is easy to see that

dim

⁡

(

S

)

=

1

(while

dim

⁡

(

V

)

is infinite), but

S

is not closed on

V

.

Title

every finite dimensional subspace of a normed space is closed

Canonical name

EveryFiniteDimensionalSubspaceOfANormedSpaceIsClosed

Date of creation

2013-03-22 14:58:56

Last modified on

2013-03-22 14:58:56

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

14

Author

Mathprof (13753)

Entry type

Corollary

Classification

msc 46B99

Classification

msc 15A03

Classification

msc 54E52

Generated on Fri Feb 9 21:37:28 2018 by

LaTeXML
finite rank approximation on separable Hilbert spaces

finite rank approximation on separable Hilbert spaces

Theorem

Let

ℋ

be a separable Hilbert space and let

T

∈

L

⁢

(

ℋ

)

. Then

T

is a

compact operator

iff there is a

sequence

{

F

n

}

of finite rank

operators

with

∥

T

-

F

n

∥

→

0

.

Proof.

(

⇒

)

: Assume

T

is

compact

on

ℋ

and

{

e

n

}

is an

orthonormal basis

of

ℋ

. Define:

P

n

⁢

f

=

∑

k

=

0

n

⟨

f

,

e

k

⟩

⁢

e

k

It is clear that the

P

n

have finite rank and that we have

∥

P

n

⁢

f

∥

≤

∥

f

∥

for all

n

∈

ℕ

,

f

∈

ℋ

.

Let

ℬ

be the

unit ball

in

ℋ

. We have that

P

n

→

I

pointwise

. Since the

P

n

are contractive they are

equicontinuous

, hence

P

n

converges uniformly to

I

on compact sets, and in particular on

T

⁢

(

ℬ

)

¯

, which is compact by

assumption

.
Therefore

P

n

⁢

T

→

T

uniformly on

ℬ

, hence

∥

P

n

⁢

T

-

T

∥

→

0

.
Since

P

n

⁢

T

is

bounded

and of finite rank the first direction follows.

(

⇐

)

: Now let

{

F

n

}

be a sequence of

bounded operators

of finite rank with

∥

T

-

F

n

∥

→

0

.
We have to show that

T

⁢

(

ℬ

)

is relatively compact in

ℋ

. This is

equivalent

to

T

⁢

(

ℬ

)

being

totally bounded

in

ℋ

.
So we are left to show that for all

ϵ

>

0

there is an

ϵ

-net

x

1

,

⋯

,

x

n

∈

ℋ

so that:

T

⁢

(

ℬ

)

⊆

⋃

k

=

1

n

B

ϵ

⁢

(

x

k

)

So choose

ϵ

>

0

and

n

∈

ℕ

fixed so that:

∥

F

n

-

T

∥

<

ϵ

2

Choose

x

1

,

⋯

,

x

m

∈

ℋ

with:

F

n

⁢

(

ℬ

)

⊆

⋃

k

=

1

m

B

ϵ

2

⁢

(

x

k

)

Hence (by the

triangle inequality

):

T

⁢

(

ℬ

)

⊆

⋃

k

=

1

m

B

ϵ

⁢

(

x

k

)

and we are done.
∎

Title

finite rank approximation on separable Hilbert spaces

Canonical name

FiniteRankApproximationOnSeparableHilbertSpaces

Date of creation

2013-03-22 18:23:18

Last modified on

2013-03-22 18:23:18

Owner

karstenb (16623)

Last modified by

karstenb (16623)

Numerical id

10

Author

karstenb (16623)

Entry type

Theorem

Classification

msc 46B99

Generated on Fri Feb 9 21:37:34 2018 by

LaTeXML
isometric isomorphism

isometric isomorphism

Let

(

X

,

∥

∥

X

)

and

(

Y

,

∥

∥

Y

)

be

normed vector spaces

. A surjective linear map

T

:

X

→

Y

is called an

isometric isomorphism

between

X

and

Y

if

∥

T

⁢

x

∥

Y

=

∥

x

∥

X

,

for all

⁢

x

∈

X

.

In this case,

X

and

Y

are said to be isometrically isomorphic.

Two isometrically isomorphic normed vector spaces share the same , so they are usually identified with each other.

Title

isometric isomorphism

Canonical name

IsometricIsomorphism

Date of creation

2013-03-22 17:34:17

Last modified on

2013-03-22 17:34:17

Owner

Gorkem (3644)

Last modified by

Gorkem (3644)

Numerical id

8

Author

Gorkem (3644)

Entry type

Definition

Classification

msc 46B99

Related topic

Isometry

Defines

isometrically isomorphic

Generated on Fri Feb 9 21:37:36 2018 by

LaTeXML
necessary and sufficient conditions for a normed vector space to be a Banach space

necessary and sufficient conditions for a normed vector space to be a Banach space

Theorem 1 -

Let

(

X

,

∥

⋅

∥

)

be a

normed vector space

.

X

is a

Banach space

if and only if every absolutely
convergent series in

X

is convergent, i.e., whenever

∑

n

∥

x

n

∥

<

∞

,

∑

n

x

n

converges

in

X

.

Theorem 2 -

Let

X

,

Y

be normed vector spaces,

X

≠

0

. Let

B

⁢

(

X

,

Y

)

be the space of

bounded operators

X

⟶

Y

. Then

Y

is a Banach space if and only if

B

⁢

(

X

,

Y

)

is a Banach space.

Title

necessary and sufficient conditions for a normed vector space to be a Banach space

Canonical name

NecessaryAndSufficientConditionsForANormedVectorSpaceToBeABanachSpace

Date of creation

2013-03-22 17:23:04

Last modified on

2013-03-22 17:23:04

Owner

asteroid (17536)

Last modified by

asteroid (17536)

Numerical id

6

Author

asteroid (17536)

Entry type

Theorem

Classification

msc 46B99

Generated on Fri Feb 9 21:37:37 2018 by

LaTeXML
normed vector space

normed vector space

Let

𝔽

be a field which is either

ℝ

or

ℂ

. A

over

𝔽

is a pair

(

V

,

∥

⋅

∥

)

where

V

is a

vector space

over

𝔽

and

∥

⋅

∥

:

V

→

ℝ

is a function such that

1.

∥

v

∥

≥

0

for all

v

∈

V

and

∥

v

∥

=

0

if and only if

v

=

0

in

V

(

positive definiteness

)

2.

∥

λ

⁢

v

∥

=

|

λ

|

⁢

∥

v

∥

for all

v

∈

V

and all

λ

∈

𝔽

3.

∥

v

+

w

∥

≤

∥

v

∥

+

∥

w

∥

for all

v

,

w

∈

V

(the

triangle inequality

)

The function

∥

⋅

∥

is called a

norm

on

V

.

Some properties of norms:

1.

If

W

is a

subspace

of

V

then

W

can be made into a

normed space

by simply restricting the norm on

V

to

W

. This is called the

induced norm

on

W

.

2.

Any

normed vector space

(

V

,

∥

⋅

∥

)

is a

metric space

under the metric

d

:

V

×

V

→

ℝ

given by

d

⁢

(

u

,

v

)

=

∥

u

-

v

∥

. This is called the

metric induced by the norm

∥

⋅

∥

.

3.

It follows that any normed space is a

locally convex topological vector space

, in the

topology

induced by the metric defined above.

4.

In this metric, the norm defines a

continuous map

from

V

to

ℝ

- this is an easy consequence of the triangle inequality.

5.

If

(

V

,

⟨

,

⟩

)

is an

inner product space

, then there is a natural induced norm given by

∥

v

∥

=

⟨

v

,

v

⟩

for all

v

∈

V

.

6.

The norm is a

convex function

of its

argument

.

Title

normed vector space

Canonical name

NormedVectorSpace

Date of creation

2013-03-22 12:13:45

Last modified on

2013-03-22 12:13:45

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

14

Author

rspuzio (6075)

Entry type

Definition

Classification

msc 46B99

Synonym

normed space

Synonym

normed linear space

Related topic

CauchySchwarzInequality

Related topic

VectorNorm

Related topic

PseudometricSpace

Related topic

MetricSpace

Related topic

UnitVector

Related topic

ProofOfGramSchmidtOrthogonalizationProcedure

Related topic

EveryNormedSpaceWithSchauderBasisIsSeparable

Related topic

EveryNormedSpaceWithSchauderBasisIsSeparable2

Related topic

FrobeniusProduct

Defines

norm

Defines

metric induced by a norm

Defines

metric induced by the norm

Defines

induced norm

Generated on Fri Feb 9 21:37:40 2018 by

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