aarjaneiro/planetmath huggingface dataset:
logarithmic convolution

logarithmic convolution

Definition

The

scale convolution

of two

functions

s

⁢

(

t

)

and

r

⁢

(

t

)

, also known as their

logarithmic convolution

is defined as the function

s

∗

l

r

⁢

(

t

)

=

r

∗

l

s

⁢

(

t

)

=

∫

0

∞

s

⁢

(

t

a

)

⁢

r

⁢

(

a

)

⁢

d

⁢

a

a

when this quantity exists.

Results

The logarithmic convolution can be related to the ordinary

convolution

by changing the

variable

from

t

to

v

=

log

⁡

t

:

s

∗

l

r

⁢

(

t

)

=

∫

0

∞

s

⁢

(

t

a

)

⁢

r

⁢

(

a

)

⁢

d

⁢

a

a

=

∫

-

∞

∞

s

⁢

(

t

e

u

)

⁢

r

⁢

(

e

u

)

⁢

𝑑

u

=

∫

-

∞

∞

s

⁢

(

e

log

⁡

t

-

u

)

⁢

r

⁢

(

e

u

)

⁢

𝑑

u

Define

f

⁢

(

v

)

=

s

⁢

(

e

v

)

and

g

⁢

(

v

)

=

r

⁢

(

e

v

)

and let

v

=

log

⁡

t

, then

s

∗

l

r

⁢

(

v

)

=

f

∗

g

⁢

(

v

)

=

g

∗

f

⁢

(

v

)

=

r

∗

l

s

⁢

(

v

)

.

Title

logarithmic convolution

Canonical

name

LogarithmicConvolution

Date of creation

2013-03-22 14:28:26

Last modified on

2013-03-22 14:28:26

Owner

swiftset (1337)

Last modified by

swiftset (1337)

Numerical id

4

Author

swiftset (1337)

Entry type

Definition

Classification

msc 44A35

Synonym

scale convolution

Related topic

Convolution

Generated on Fri Feb 9 21:33:23 2018 by

LaTeXML
center of gravity of circular sector

center of gravity of circular sector

Consider a

circular sector

with

central angle

2

⁢

α

(in radians) and radius

R

as shown in the diagram below. If we wish to find the

distance

of the

center of gravity

from the center of the sector, we divide the sector into elements of area

d

⁢

A

as illustrated.

In general, the mass of a lamina element is given by

d

⁢

m

=

δ

2

⁢

d

⁢

A

and the

coordinates

of the centers of mass are (given that the mass is evenly distributed over the area):

x

¯

=

∫

∫

A

x

⁢

𝑑

A

A

,

y

¯

=

∫

∫

A

y

⁢

𝑑

A

A

In this case, we will use

polar coordinates

as it would be much easier to carry out the integration, and the boundaries can be defined easily. In polar coordinates

d

⁢

A

=

r

⁢

d

⁢

r

⁢

d

⁢

θ

. The area of the sector is

A

=

1

2

⁢

R

2

⁢

(

2

⁢

α

)

=

α

⁢

R

2

.
Now

x

¯

=

1

α

⁢

R

2

⁢

∫

0

2

⁢

α

∫

0

R

x

⁢

r

⁢

𝑑

r

⁢

𝑑

θ

=

1

α

⁢

R

2

⁢

∫

0

2

⁢

α

∫

0

R

r

2

⁢

cos

⁡

θ

⁢

d

⁢

r

⁢

d

⁢

θ

=

1

α

⁢

R

2

⁢

∫

0

2

⁢

α

R

3

3

⁢

cos

⁡

θ

⁢

d

⁢

θ

=

R

3

⁢

α

⁢

sin

⁡

2

⁢

α

Now we follow a

similar

procedure for the y-coordinate:

y

¯

=

1

α

⁢

R

2

⁢

∫

0

2

⁢

α

∫

0

R

y

⁢

r

⁢

𝑑

r

⁢

𝑑

θ

=

1

α

⁢

R

2

⁢

∫

0

2

⁢

α

∫

0

R

r

2

⁢

sin

⁡

θ

⁢

d

⁢

r

⁢

d

⁢

θ

=

1

α

⁢

R

2

⁢

∫

0

2

⁢

α

R

3

3

⁢

sin

⁡

θ

⁢

d

⁢

θ

=

R

3

⁢

α

⁢

(

1

-

cos

⁡

2

⁢

α

)

The center of gravity is

(

x

¯

,

y

¯

)

and the distance

d

of the center of gravity from the center of the sector is given by:

d

=

x

¯

2

+

y

¯

2

We substitute for

x

¯

and

y

¯

:

d

=

(

R

⁢

sin

⁡

2

⁢

α

3

⁢

α

)

2

+

(

R

⁢

(

1

-

cos

⁡

2

⁢

α

)

3

⁢

α

)

2

=

R

3

⁢

α

⁢

sin

2

⁡

2

⁢

α

+

(

1

-

cos

⁡

2

⁢

α

)

2

From

trigonometry

, we know that:

2

⁢

sin

2

⁡

α

=

1

-

cos

⁡

2

⁢

α

sin

2

⁡

α

+

cos

2

⁡

α

=

1

sin

⁡

2

⁢

α

=

2

⁢

sin

⁡

α

⁢

cos

⁡

α

Keeping these in mind, we substitute:

d

=

R

3

⁢

α

⁢

4

⁢

sin

2

⁡

α

⁢

cos

2

⁡

α

+

4

⁢

sin

4

⁡

α

=

2

⁢

R

3

⁢

α

⁢

sin

2

⁡

α

⁢

(

cos

2

⁡

α

+

sin

2

⁡

α

)

=

2

⁢

R

⁢

sin

⁡

α

3

⁢

α

In conclusion, the distance of the center of gravity of a circular sector with radius

R

and angle

2

⁢

α

from the center of the sector is given by:

d

=

2

⁢

R

⁢

sin

⁡

α

3

⁢

α

Title

center of gravity of circular sector

Canonical name

CenterOfGravityOfCircularSector

Date of creation

2013-03-22 18:06:30

Last modified on

2013-03-22 18:06:30

Owner

curious (18562)

Last modified by

curious (18562)

Numerical id

6

Author

curious (18562)

Entry type

Topic

Classification

msc 44A99

Related topic

CentreOfMassOfHalfDisc

Generated on Fri Feb 9 21:33:32 2018 by

LaTeXML
Fredholm equation

Fredholm equation

A

Fredholm equation

of the

first kind

is an

integral equation

of the form

∫

a

b

K

⁢

(

x

,

y

)

⁢

f

⁢

(

y

)

⁢

𝑑

y

=

g

⁢

(

x

)

,

∀

x

∈

[

a

,

b

]

,

and a

Fredholm equation of the

second kind

is an integral equation of the form

f

⁢

(

x

)

-

λ

⁢

∫

a

b

K

⁢

(

x

,

y

)

⁢

f

⁢

(

y

)

⁢

𝑑

y

=

g

⁢

(

x

)

,

∀

x

∈

[

a

,

b

]

,

where

K

⁢

f

⁢

(

x

)

=

∫

a

b

K

⁢

(

x

,

y

)

⁢

f

⁢

(

y

)

⁢

𝑑

y

,

∀

x

∈

[

a

,

b

]

is a

compact operator

in some

function space

.

Title

Fredholm equation

Canonical name

FredholmEquation

Date of creation

2013-03-22 15:49:59

Last modified on

2013-03-22 15:49:59

Owner

ncrom (8997)

Last modified by

ncrom (8997)

Numerical id

5

Author

ncrom (8997)

Entry type

Definition

Classification

msc 45B05

Generated on Fri Feb 9 21:33:48 2018 by

LaTeXML
Lagrange multiplier method, proof of

Lagrange multiplier method, proof of

Let

g

⁢

(

x

,

y

)

=

c

and

f

⁢

(

x

,

y

)

=

d

. Taking the

derivative

of

f

and

g

with respect to

t

gives:

∂

⁡

f

∂

⁡

t

=

∂

⁡

f

∂

⁡

x

⁢

x

′

⁢

(

t

)

+

∂

⁡

f

∂

⁡

y

⁢

y

′

⁢

(

t

)

=

0

and

∂

⁡

g

∂

⁡

t

=

∂

⁡

g

∂

⁡

x

⁢

x

′

⁢

(

t

)

+

∂

⁡

g

∂

⁡

y

⁢

y

′

⁢

(

t

)

=

0

By letting

r

→

=

x

⁢

(

t

)

⁢

i

^

+

y

⁢

(

t

)

⁢

j

^

,

the

partial derivatives

can be rewritten as follows:

∂

⁡

f

∂

⁡

t

=

grad

⁡

f

⋅

r

′

→

;

∂

⁡

g

∂

⁡

t

=

grad

⁡

g

⋅

r

′

→

This implies that

grad

⁡

f

×

grad

⁡

g

=

0

,

thus

grad

⁡

f

=

λ

⁢

grad

⁡

g

.

Now this equation can be rewritten as

f

x

⁢

i

^

+

f

y

⁢

j

^

=

λ

⁢

(

g

x

⁢

i

^

+

g

y

⁢

j

^

)

.

Since

ℝ

n

↦

ℝ

,

this equation can be separated into two new equations:

f

x

=

λ

⁢

g

x

;

f

y

=

λ

⁢

g

y

Using the above equations, a new

function

,

F

, can be defined:

F

⁢

(

x

,

y

,

λ

)

=

f

⁢

(

x

,

y

)

-

λ

⁢

g

⁢

(

x

,

y

)

which can be generalized as:

F

⁢

(

x

,

y

,

λ

)

=

f

⁢

(

x

,

y

)

-

∑

i

=

1

m

λ

i

⁢

[

g

i

⁢

(

x

,

y

)

]

.

Title

Lagrange multiplier method

, proof of

Canonical name

LagrangeMultiplierMethodProofOf

Date of creation

2013-03-22 15:25:09

Last modified on

2013-03-22 15:25:09

Owner

aplant (12431)

Last modified by

aplant (12431)

Numerical id

6

Author

aplant (12431)

Entry type

Proof

Classification

msc 45C05

Classification

msc 15A42

Classification

msc 15A18

Generated on Fri Feb 9 21:33:57 2018 by

LaTeXML
integral equation

integral equation

An

integral equation

involves an unknown

function

under the .  Most common of them is a

linear integral equation

α

⁢

(

t

)

⁢

y

⁢

(

t

)

+

∫

a

b

k

⁢

(

t

,

x

)

⁢

y

⁢

(

x

)

⁢

𝑑

x

=

f

⁢

(

t

)

,

(1)

where

α

,

k

,

f

are given functions.  The function

t

↦

y

⁢

(

t

)

is to be solved.

Any linear integral equation is

equivalent

(

http://planetmath.org/Equivalent3

) to a

linear differential equation

; e.g. the equation

y

⁢

(

t

)

+

∫

0

t

(

2

⁢

t

-

2

⁢

x

-

3

)

⁢

y

⁢

(

x

)

⁢

𝑑

x

=

1

+

t

-

4

⁢

sin

⁡

t

to the equation

y

′′

⁢

(

t

)

-

3

⁢

y

′

⁢

(

t

)

+

2

⁢

y

⁢

(

t

)

=

4

⁢

sin

⁡

t

with the

initial conditions

y

⁢

(

0

)

=

1

and

y

′

⁢

(

0

)

=

0

.

The equation (1) is of

•

1st kind

if

α

⁢

(

t

)

≡

0

,

•

2nd kind

if

α

⁢

(

t

)

is a nonzero constant,

•

3rd kind

else.

If both limits (

http://planetmath.org/UpperLimit

) of integration in (1) are constant, (1) is a

Fredholm equation

, if one limit is

variable

, one has a

Volterra equation

.  In the case that

f

⁢

(

t

)

≡

0

,  the linear integral equation is .

Example.

Solve the Volterra equation

y

⁢

(

t

)

+

∫

0

t

(

t

-

x

)

⁢

y

⁢

(

x

)

⁢

𝑑

x

=

1

by using

Laplace transform

.

Using the

convolution

(

http://planetmath.org/LaplaceTransformOfConvolution

), the equation may be written

y

⁢

(

t

)

+

t

*

y

⁢

(

t

)

=

1

.  Applying to this the Laplace transform, one obtains

Y

⁢

(

s

)

+

1

s

2

⁢

Y

⁢

(

s

)

=

1

s

,  whence

Y

⁢

(

s

)

=

s

s

2

+

1

.  This corresponds the function

y

⁢

(

t

)

=

cos

⁡

t

,  which is the solution.

http://eqworld.ipmnet.ru/en/solutions/ie.htm

Solutions on some integral equations in EqWorld.

Title

integral equation

Canonical

name

IntegralEquation

Date of creation

2013-03-22 18:03:57

Last modified on

2013-03-22 18:03:57

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

8

Author

pahio (2872)

Entry type

Definition

Classification

msc 45D05

Classification

msc 45A05

Related topic

Equation

Defines

linear integral equation

Defines

Volterra equation

Generated on Fri Feb 9 21:33:59 2018 by

LaTeXML
balanced set

balanced set

Definition

[

1

,

2

,

3

,

4

]

Let

V

be a

vector space

over

ℝ

(or

ℂ

),
and let

S

be a subset of

V

. If

λ

⁢

S

⊂

S

for all scalars

λ

such
that

|

λ

|

≤

1

, then

S

is a

balanced set

in

V

.
The

balanced hull

of

S

,
denoted by

eq

⁡

(

S

)

, is the smallest
balanced set containing

S

.

In the above,

λ

⁢

S

=

{

λ

⁢

s

∣

s

∈

S

}

,
and

|

⋅

|

is the

absolute value

(in

ℝ

),
or the

modulus

of a

complex number

(in

ℂ

).

0.0.1

Examples and properties

1.

Let

V

be a

normed space

with norm

|

|

⋅

|

|

. Then the unit ball

{

v

∈

V

∣

||

v

||

≤

1

}

is a balanced set.

2.

Any

vector subspace

is a balanced set. Thus, in

ℝ

3

, lines and planes passing
through the origin are balanced sets.

0.0.2

Notes

A balanced set is also sometimes called

circled

[

3

]

.
The term

balanced evelope

is also used for the balanced hull

[

2

]

.
Bourbaki uses the term

équilibré

[

2

]

, c.f.

eq

⁡

(

A

)

above. In

[

5

]

, a balanced set is defined as above, but with the condition

|

λ

|

=

1

instead of

|

λ

|

≤

1

.

References

1

W. Rudin,

Functional Analysis

,
McGraw-Hill Book Company, 1973.

2

R.E. Edwards,

Functional Analysis: Theory and Applications

,
Dover Publications, 1995.

3

J. Horváth,

Topological Vector Spaces

and

Distributions

,
Addison-Wsley Publishing Company, 1966.

4

R. Cristescu,

Topological vector spaces

,
Noordhoff International Publishing, 1977.

5

M. Reed, B. Simon,

Methods of Modern Mathematical Physics: Functional Analysis I

,
Revised and enlarged edition, Academic Press, 1980.

Title

balanced set

Canonical name

BalancedSet

Date of creation

2013-03-22 15:33:16

Last modified on

2013-03-22 15:33:16

Owner

matte (1858)

Last modified by

matte (1858)

Numerical id

5

Author

matte (1858)

Entry type

Definition

Classification

msc 46-00

Related topic

AbsorbingSet

Defines

balanced subset

Defines

balanced hull

Defines

balanced evelope

Defines

circled

Defines

équilibré

Generated on Fri Feb 9 21:34:16 2018 by

LaTeXML
bounded function

bounded function

Definition

Suppose

X

is a

nonempty set

. Then a

function

f

:

X

→

ℂ

is a

if there exist a

C

<

∞

such that

|

f

⁢

(

x

)

|

<

C

for all

x

∈

X

. The
set of all

bounded functions

on

X

is usually
denoted by

B

⁢

(

X

)

(

[

1

]

, pp. 61).

Under standard point-wise

addition

and point-wise

multiplication

by
a scalar,

B

⁢

(

X

)

is a

complex vector space

.

If

f

∈

B

⁢

(

X

)

, then the

sup-norm

, or

uniform norm

, of

f

is defined as

||

f

||

∞

=

sup

x

∈

X

⁡

|

f

⁢

(

x

)

|

.

It is straightforward to check that

|

|

⋅

|

|

∞

makes

B

⁢

(

X

)

into
a

normed vector space

, i.e., to check that

|

|

⋅

|

|

∞

satisfies

the

assumptions

for a norm.

0.0.1

Example

Suppose

X

is a

compact

topological space

. Further, let

C

⁢

(

X

)

be the
set of

continuous

complex-valued functions on

X

(with the same

vector
space

structure

as

B

⁢

(

X

)

). Then

C

⁢

(

X

)

is a

vector subspace

of

B

⁢

(

X

)

.

References

1

C.D. Aliprantis, O. Burkinshaw,

Principles of Real Analysis

,
2nd ed., Academic Press, 1990.

Title

bounded function

Canonical name

BoundedFunction

Date of creation

2013-03-22 13:44:06

Last modified on

2013-03-22 13:44:06

Owner

Koro (127)

Last modified by

Koro (127)

Numerical id

7

Author

Koro (127)

Entry type

Definition

Classification

msc 46-00

Defines

supremum norm

Defines

sup norm

Defines

sup-norm

Defines

uniform norm

Defines

bounded function

Defines

unbounded function

Generated on Fri Feb 9 21:34:18 2018 by

LaTeXML
bounded set (in a topological vector space)

bounded set (in a topological vector space)

Definition

Suppose

B

is a subset of a

topological vector space

V

.
Then

B

is a

if for every

neighborhood

U

of
the zero vector in

V

, there
exists a scalar

λ

such that

B

⊂

λ

⁢

U

.

References

1

W. Rudin,

Functional Analysis

,
McGraw-Hill Book Company, 1973.

2

F.A. Valentine,

Convex sets

, McGraw-Hill Book company, 1964.

3

R. Cristescu,

Topological vector spaces

,
Noordhoff International Publishing, 1977.

Title

bounded set

(in a topological vector space)

Canonical

name

BoundedSetinATopologicalVectorSpace

Date of creation

2013-03-22 13:44:16

Last modified on

2013-03-22 13:44:16

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

8

Author

mathcam (2727)

Entry type

Definition

Classification

msc 46-00

Generated on Fri Feb 9 21:34:19 2018 by

LaTeXML
boundedness in a topological vector space generalizes boundedness in a normed space

boundedness in a topological vector space generalizes boundedness in a normed space

Boundedness in a

topological vector space

is a

generalization

of boundedness in a

normed space

.

Suppose

(

V

,

∥

⋅

∥

)

is a

normed vector space

over

ℂ

, and suppose

B

is

bounded

in the sense of the parent entry. Then for the

unit ball

B

1

⁢

(

0

)

=

{

v

∈

V

:

∥

v

∥

<

1

}

there exists some

λ

∈

ℂ

such that

B

⊆

λ

⁢

B

1

⁢

(

0

)

. Using
this result (

http://planetmath.org/ScalingOfTheOpenBallInANormedVectorSpace

), it follows that

B

⊆

B

|

λ

|

⁢

(

0

)

.

Title

boundedness in a topological vector space generalizes boundedness in a normed space

Canonical name

BoundednessInATopologicalVectorSpaceGeneralizesBoundednessInANormedSpace

Date of creation

2013-03-22 15:33:29

Last modified on

2013-03-22 15:33:29

Owner

PrimeFan (13766)

Last modified by

PrimeFan (13766)

Numerical id

7

Author

PrimeFan (13766)

Entry type

Result

Classification

msc 46-00

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

LaTeXML
delta distribution

delta distribution

Let

U

be an open subset of

ℝ

n

such that

0

∈

U

.
Then the

delta distribution

is the mapping

δ

:

𝒟

⁢

(

U

)

→

ℂ

u

↦

u

⁢

(

0

)

.

Claim

The delta distribution is a

distribution

of zeroth order, i.e.,

δ

∈

𝒟

′

⁣

0

⁢

(

U

)

.

Proof.

With obvious notation, we have

δ

⁢

(

u

+

v

)

=

(

u

+

v

)

⁢

(

0

)

=

u

⁢

(

0

)

+

v

⁢

(

0

)

=

δ

⁢

(

u

)

+

δ

⁢

(

v

)

,

δ

⁢

(

α

⁢

u

)

=

(

α

⁢

u

)

⁢

(

0

)

=

α

⁢

u

⁢

(

0

)

=

α

⁢

δ

⁢

(

u

)

,

so

δ

is linear. To see that

δ

is

continuous

, we
use condition (3) on this this page (

http://planetmath.org/Distribution4

).
Indeed, if

K

is a

compact set

in

U

, and

u

∈

𝒟

K

, then

|

δ

⁢

(

u

)

|

=

|

u

⁢

(

0

)

|

≤

||

u

||

∞

,

where

|

|

⋅

|

|

∞

is the

supremum norm

.

□

Title

delta distribution

Canonical name

DeltaDistribution

Date of creation

2013-03-22 13:45:52

Last modified on

2013-03-22 13:45:52

Owner

matte (1858)

Last modified by

matte (1858)

Numerical id

6

Author

matte (1858)

Entry type

Definition

Classification

msc 46-00

Classification

msc 46F05

Related topic

ExampleOfDiracSequence

Generated on Fri Feb 9 21:34:26 2018 by

LaTeXML
every locally integrable function is a distribution

every locally integrable function is a distribution

Suppose

U

is an open set in

ℝ

n

and

f

is a

locally
integrable function

on

U

, i.e.,

f

∈

L

loc

1

⁢

(

U

)

.
Then the

mapping

T

f

:

𝒟

⁢

(

U

)

→

ℂ

u

↦

∫

U

f

⁢

(

x

)

⁢

u

⁢

(

x

)

⁢

𝑑

x

is a zeroth order

distribution

. (See parent entry for notation

𝒟

⁢

(

U

)

.)

(proof) (

http://planetmath.org/T_fIsADistributionOfZerothOrder

)

If

f

and

g

are both locally integrable functions on an open set

U

,
and

T

f

=

T

g

, then it follows (see
this page (

http://planetmath.org/TheoremForLocallyIntegrableFunctions

)),
that

f

=

g

almost everywhere. Thus, the mapping

f

↦

T

f

is a linear

injection

when

L

loc

1

is equipped with
the usual

equivalence relation

for an

L

p

-space. For this reason,
one usually writes

f

for the distribution

T

f

.

Title

every locally integrable function is a distribution

Canonical name

EveryLocallyIntegrableFunctionIsADistribution

Date of creation

2013-03-22 13:44:25

Last modified on

2013-03-22 13:44:25

Owner

matte (1858)

Last modified by

matte (1858)

Numerical id

8

Author

matte (1858)

Entry type

Theorem

Classification

msc 46-00

Classification

msc 46F05

Generated on Fri Feb 9 21:34:31 2018 by

LaTeXML
localization for distributions

localization for distributions

Definition

Suppose

U

is an

open set

in

ℝ

n

and

T

is a

distribution

T

∈

𝒟

′

⁢

(

U

)

. Then we say that

T

vanishes

on
an open set

V

⊂

U

, if the

restriction

of

T

to

V

is
the zero distribution on

V

. In other words,

T

vanishes on

V

, if

T

⁢

(

v

)

=

0

for all

v

∈

C

0

∞

⁢

(

V

)

. (Here

C

0

∞

⁢

(

V

)

is the set
of

smooth function with compact support

in

V

.) Similarly, we say that
two distributions

S

,

T

∈

𝒟

′

⁢

(

U

)

are

equal

, or

coincide

on

V

, if

S

-

T

vanishes on

V

. We then
write:

S

=

T

on

V

.

Theorem

[

1

,

3

]

Suppose

U

is an open set in

ℝ

n

and

{

U

i

}

i

∈

I

is an

open cover

of

U

, i.e.,

U

=

⋃

i

∈

I

U

i

.

Here,

I

is an arbitrary

index set

. If

S

,

T

are distributions on

U

,
such that

S

=

T

on each

U

i

, then

S

=

T

(on U).

Proof.

Suppose

u

∈

𝒟

⁢

(

U

)

. Our aim is to show that

S

⁢

(

u

)

=

T

⁢

(

u

)

.
First, we have

supp

⁡

u

⊂

K

for some

compact

K

⊂

U

.
It follows (

http://planetmath.org/YIsCompactIfAndOnlyIfEveryOpenCoverOfYHasAFiniteSubcover

) that there exist a finite

collection

of

U

i

:s from the open cover,
say

U

1

,

…

,

U

N

, such that

K

⊂

∪

i

=

1

N

U

i

.
By a smooth

partition of unity

, there
are

smooth functions

ϕ

1

,

…

,

ϕ

N

:

U

→

ℝ

such that

1.

supp

⁡

ϕ

i

⊂

U

i

for all

i

.

2.

ϕ

i

⁢

(

x

)

∈

[

0

,

1

]

for all

x

∈

U

and all

i

,

3.

∑

i

=

1

N

ϕ

i

⁢

(

x

)

=

1

for all

x

∈

K

.

From the first

property

, and from a property for the

support

of a

function

(

http://planetmath.org/SupportOfFunction

),
it follows that

supp

⁡

ϕ

i

⁢

u

⊂

supp

⁡

ϕ

i

∩

supp

⁡

u

⊂

U

i

.
Therefore, for each

i

,

S

⁢

(

ϕ

i

⁢

u

)

=

T

⁢

(

ϕ

i

⁢

u

)

since

S

and

T

conicide
on

U

i

.
Then

S

⁢

(

u

)

=

∑

i

=

1

N

S

⁢

(

ϕ

i

⁢

u

)

=

∑

i

=

1

N

T

⁢

(

ϕ

i

⁢

u

)

=

T

⁢

(

u

)

,

and the theorem follows.

□

References

1

G.B. Folland,

Real Analysis

: Modern Techniques and Their Applications

, 2nd ed, John Wiley & Sons, Inc., 1999.

2

W. Rudin,

Functional Analysis

,
McGraw-Hill Book Company, 1973.

3

L. Hörmander,

The

Analysis

of Linear Partial Differential

Operators

I,
(Distribution theory and Fourier Analysis)

, 2nd ed, Springer-Verlag, 1990.

Title

localization for distributions

Canonical

name

LocalizationForDistributions

Date of creation

2013-03-22 13:46:17

Last modified on

2013-03-22 13:46:17

Owner

drini (3)

Last modified by

drini (3)

Numerical id

9

Author

drini (3)

Entry type

Definition

Classification

msc 46-00

Classification

msc 46F05

Generated on Fri Feb 9 21:34:33 2018 by

LaTeXML
modular mappings in vector spaces over the field of complex numbers

modular mappings in vector spaces over the field of complex numbers

Suppose

X

is a

ℂ

-vector space. A mapping

ρ

:

X

→

[

0

,

∞

]

is called

modular

if the following three conditions are satisfied:

1.

ρ

⁢

(

x

)

=

0

if and only if

x

=

0

.

2.

ρ

⁢

(

α

⁢

x

)

=

ρ

⁢

(

x

)

for all

x

∈

X

and for all scalars

α

such that

|

α

|

=

1

.

3.

ρ

⁢

(

α

⁢

x

+

β

⁢

y

)

≤

ρ

⁢

(

x

)

+

ρ

⁢

(

y

)

for all

x

,

y

∈

X

and for all scalars

α

and

β

such that

α

+

β

=

1

and

α

,

β

≥

0

.

Title

modular mappings in vector spaces over the field of

complex numbers

Canonical name

ModularMappingsInVectorSpacesOverTheFieldOfComplexNumbers

Date of creation

2013-03-22 16:08:12

Last modified on

2013-03-22 16:08:12

Owner

gilbert_51126 (14238)

Last modified by

gilbert_51126 (14238)

Numerical id

11

Author

gilbert_51126 (14238)

Entry type

Definition

Classification

msc 46-00

Defines

modular

Generated on Fri Feb 9 21:34:35 2018 by

LaTeXML
modular space

modular space

Let

ρ

be a modular on a real or

complex vector space

X

. Then the

subspace

X

ρ

:=

{

x

∈

X

:

lim

λ

→

0

⁡

ρ

⁢

(

λ

⁢

x

)

=

0

}

of

X

is called the

modular space

corresponding to the modular

ρ

.

Title

modular space

Canonical name

ModularSpace

Date of creation

2013-03-22 16:15:53

Last modified on

2013-03-22 16:15:53

Owner

gilbert_51126 (14238)

Last modified by

gilbert_51126 (14238)

Numerical id

6

Author

gilbert_51126 (14238)

Entry type

Definition

Classification

msc 46-00

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

LaTeXML
proof of convergence theorem

proof of convergence theorem

Let us show the equivalence of (2) and (3).
First, the proof that (3) implies (2) is a direct calculation.
Next, let us show that (2) implies (3):
Suppose

T

⁢

u

i

→

0

in

ℂ

, and if

K

is a

compact set

in

U

, and

{

u

i

}

i

=

1

∞

is a

sequence

in

𝒟

K

such that
for any multi-index

α

, we have

D

α

⁢

u

i

→

0

in the

supremum norm

∥

⋅

∥

∞

as

i

→

∞

.
For a

contradiction

, suppose there is a compact set

K

in

U

such that for all

constants

C

>

0

and

k

∈

{

0

,

1

,

2

,

…

}

there exists
a

function

u

∈

𝒟

K

such that

|

T

⁢

(

u

)

|

>

C

⁢

∑

|

α

|

≤

k

||

D

α

⁢

u

||

∞

.

Then, for

C

=

k

=

1

,

2

,

…

we obtain functions

u

1

,

u

2

,

…

in

𝒟

⁢

(

K

)

such that

|

T

⁢

(

u

i

)

|

>

i

⁢

∑

|

α

|

≤

i

||

D

α

⁢

u

i

||

∞

.

Thus

|

T

⁢

(

u

i

)

|

>

0

for all

i

, so for

v

i

=

u

i

/

|

T

⁢

(

u

i

)

|

, we have

1

>

i

⁢

∑

|

α

|

≤

i

||

D

α

⁢

v

i

||

∞

.

It follows that

||

D

α

⁢

u

i

||

∞

<

1

/

i

for any multi-index

α

with

|

α

|

≤

i

.
Thus

{

v

i

}

i

=

1

∞

satisfies

our

assumption

, whence

T

⁢

(

v

i

)

should tend to

0

.
However, for all

i

, we have

T

⁢

(

v

i

)

=

1

. This contradiction

completes

the proof.

Title

proof of convergence theorem

Canonical name

ProofOfConvergenceTheorem

Date of creation

2013-03-22 13:44:36

Last modified on

2013-03-22 13:44:36

Owner

matte (1858)

Last modified by

matte (1858)

Numerical id

10

Author

matte (1858)

Entry type

Proof

Classification

msc 46-00

Classification

msc 46F05

Generated on Fri Feb 9 21:34:39 2018 by

LaTeXML