aarjaneiro/planetmath huggingface dataset:
sets where sequence of continuous functions diverge

sets where sequence of continuous functions diverge

Related Exercise from Rudin’s

Real and Complex Analysis

.

Exercise 5.20

(

a

)

Does there exist a

sequence

of

continuous

positive

functions

f

n

on

ℝ

1

such that

{

f

n

⁢

(

x

)

}

is unbounded if and only
if

x

is

rational

?

(

b

)

Replace “rational” by

irrational

in (

a

) and answer the resulting
question.

(

c

)

Replace “

{

f

n

⁢

(

x

)

}

is unbounded”
by “

f

n

⁢

(

x

)

→

∞

as

n

→

∞

”
and answer the resulting analogues of (

a

) and (

b

).

Solution

:
The answer to (

a

) is negative.
This by showing that
the subset of points where such sequence is unbounded must be

G

δ

.
But the rationals cannot be such,
since in

ℝ

dense

G

δ

sets must be of second category.

Rest of the answer not yet ready here

Title

sets where sequence of continuous functions diverge

Canonical name

SetsWhereSequenceOfContinuousFunctionsDiverge

Date of creation

2013-03-22 15:23:34

Last modified on

2013-03-22 15:23:34

Owner

yotam (10129)

Last modified by

yotam (10129)

Numerical id

8

Author

yotam (10129)

Entry type

Derivation

Classification

msc 26A15

Classification

msc 40A30

Generated on Fri Feb 9 19:49:07 2018 by

LaTeXML
sum function of series

sum function of series

Let the terms of a series be real functions

f

n

defined in a certain subset

A

0

of

ℝ

; we can speak of a

function series

.  All points

x

where the series

f

1

+

f

2

+

⋯

(1)

converges

form a subset

A

of

A

0

, and we have the

S

:

x

↦

S

⁢

(

x

)

of (1) defined in

A

.

If the

sequence

S

1

,

S

2

,

…

of the

partial sums

S

n

=

f

1

+

f

2

+

⋯

+

f

n

of the series (1)

converges uniformly

(

http://planetmath.org/LimitFunctionOfSequence

) in the

interval

[

a

,

b

]

⊆

A

to a

function

S

:

x

↦

S

⁢

(

x

)

,  we say that

the series

in this interval.  We may also set the direct

Definition.

The function series (1), which converges in every point of the interval

[

a

,

b

]

having sum function

S

:

x

↦

S

⁢

(

x

)

,
in the interval

[

a

,

b

]

,  if for every positive number

ε

there is an integer

n

ε

such that each value of

x

in the interval

[

a

,

b

]

the

inequality

|

S

n

⁢

(

x

)

-

S

⁢

(

x

)

|

<

ε

when

n

≧

n

ε

.

Note.

One can without trouble be convinced that the term functions of a uniformly converging series converge uniformly to 0 (cf. the necessary condition of

convergence

).

The notion of of series can be extended to the series with complex function terms (the interval is replaced with some subset of

ℂ

).  The significance of the is therein that the sum function of a series with this property and with

continuous

term-functions is continuous and may be integrated termwise.

Title

sum function of series

Canonical name

SumFunctionOfSeries

Date of creation

2013-03-22 14:38:15

Last modified on

2013-03-22 14:38:15

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

18

Author

pahio (2872)

Entry type

Definition

Classification

msc 26A15

Classification

msc 40A30

Related topic

UniformConvergenceOfIntegral

Related topic

SumOfSeries

Related topic

OneSidedContinuityBySeries

Defines

function series

Defines

sum function

Defines

uniform convergence

of series

Generated on Fri Feb 9 19:49:09 2018 by

LaTeXML
termwise differentiation

termwise differentiation

Theorem

.

If in the

open interval

I

,
all the of the series

f

1

⁢

(

x

)

+

f

2

⁢

(

x

)

+

⋯

(1)

have

continuous

derivatives

,
the series

converges

having sum

S

⁢

(

x

)

and
the differentiated series

f

1

′

⁢

(

x

)

+

f

2

′

⁢

(

x

)

+

⋯

converges uniformly

(

http://planetmath.org/SumFunctionOfSeries

) on the interval

I

,
then the series (1) can be differentiated termwise, i.e. in every point of

I

the sum

function

S

⁢

(

x

)

is

differentiable

and

d

⁢

S

⁢

(

x

)

d

⁢

x

=

f

1

′

⁢

(

x

)

+

f

2

′

⁢

(

x

)

+

⋯

The situation implies also that the series (1) converges uniformly on

I

.

Title

termwise differentiation

Canonical name

TermwiseDifferentiation

Date of creation

2013-03-22 14:38:38

Last modified on

2013-03-22 14:38:38

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

9

Author

Mathprof (13753)

Entry type

Theorem

Classification

msc 26A15

Classification

msc 40A30

Synonym

differentiating a series

Related topic

PowerSeries

Related topic

IntegrationOfLaplaceTransformWithRespectToParameter

Related topic

IntegralOfLimitFunction

Generated on Fri Feb 9 19:49:11 2018 by

LaTeXML
testing continuity via filters

testing continuity via filters

Proposition 1

.

Let

X

,

Y

be

topological spaces

. Then a

function

f

:

X

→

Y

is

continuous

iff it sends converging filters to converging filters.

Proof.

Suppose first

f

is continuous. Let

𝔽

be a filter in

X

converging to

x

. We want to show that

f

⁢

(

𝔽

)

:=

{

f

⁢

(

F

)

∣

F

∈

𝔽

}

converges

to

f

⁢

(

x

)

. Let

N

be a

neighborhood

of

f

⁢

(

x

)

. So there is an

open set

U

such that

f

⁢

(

x

)

∈

U

⊆

N

. So

f

-

1

⁢

(

U

)

is open and contains

x

, which means that

f

-

1

⁢

(

U

)

∈

𝔽

by

assumption

. This means that

f

⁢

f

-

1

⁢

(

U

)

∈

f

⁢

(

𝔽

)

. Since

f

⁢

f

-

1

⁢

(

U

)

⊆

U

⊆

N

, we see that

N

∈

f

⁢

(

𝔽

)

as well.

Conversely

, suppose

f

preserves converging filters. Let

V

be an open set in

Y

containing

f

⁢

(

x

)

. We want to find an open set

U

in

X

containing

x

, such that

f

⁢

(

U

)

⊆

V

. Let

𝔽

be the neighborhood filter of

x

. So

𝔽

→

x

. By assumption,

f

⁢

(

𝔽

)

→

f

⁢

(

x

)

. Since

V

is an open neighborhood of

f

⁢

(

x

)

, we have

V

∈

f

⁢

(

𝔽

)

, or

f

⁢

(

F

)

⊆

V

for some

F

∈

𝔽

. Since

F

is a neighborhood of

x

, it contains an open neighborhood

U

of

x

. Furthermore,

f

⁢

(

U

)

⊆

f

⁢

(

F

)

⊆

V

. Since

x

is arbitrary,

f

is continuous.
∎

Title

testing continuity via filters

Canonical name

TestingContinuityViaFilters

Date of creation

2013-03-22 19:09:31

Last modified on

2013-03-22 19:09:31

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

4

Author

CWoo (3771)

Entry type

Result

Classification

msc 26A15

Classification

msc 54C05

Related topic

Filter

Generated on Fri Feb 9 19:49:20 2018 by

LaTeXML
testing for continuity via basic open sets

testing for continuity via basic open sets

Proposition 1

.

Let

X

,

Y

be

topological spaces

, and

f

:

X

→

Y

a

function

. The following are

equivalent

:

1.

f

is

continuous

;

2.

f

-

1

⁢

(

U

)

is open for any

U

in a basis (

U

called a basic

open set

) for the

topology

of

Y

;

3.

f

-

1

⁢

(

U

)

is open for any

U

is a

subbasis

for the topology of

Y

.

Proof.

First, note that

(

1

)

⇒

(

2

)

⇒

(

3

)

, since every basic open set is open, and every

element

in a subbasis is in the basis it generates. We next prove

(

3

)

⇒

(

2

)

⇒

(

1

)

.

•

(

2

)

⇒

(

1

)

. Suppose

ℬ

is a basis for the topology of

Y

. Let

U

be an open set in

Y

. Then

U

is the union of elements in

ℬ

. In other words,

U

=

⋃

{

U

i

∈

ℬ

∣

i

∈

I

}

,

for some

index set

I

. So

f

-

1

⁢

(

U

)

=

f

-

1

⁢

(

⋃

{

U

i

∈

ℬ

∣

i

∈

I

}

)

=

⋃

{

f

-

1

⁢

(

U

i

)

∣

i

∈

I

}

.

By

assumption

, each

f

-

1

⁢

(

U

i

)

is open, so is their union

f

-

1

⁢

(

U

)

.

•

(

3

)

⇒

(

2

)

. Suppose now that

𝒮

is a subbasis, which generates the basis

ℬ

for the topology of

Y

. If

U

is a basic open set, then

U

=

⋂

i

=

1

n

U

i

,

where each

U

i

∈

𝒮

. Then

f

-

1

⁢

(

U

)

=

f

-

1

⁢

(

⋂

i

=

1

n

U

i

)

=

⋂

i

=

1

n

f

-

1

⁢

(

U

i

)

.

By assumption, each

f

-

1

⁢

(

U

i

)

is open, so is their (finite)

intersection

f

-

1

⁢

(

U

)

.

∎

Title

testing for continuity via basic open sets

Canonical name

TestingForContinuityViaBasicOpenSets

Date of creation

2013-03-22 19:08:55

Last modified on

2013-03-22 19:08:55

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

4

Author

CWoo (3771)

Entry type

Result

Classification

msc 26A15

Classification

msc 54C05

Generated on Fri Feb 9 19:49:30 2018 by

LaTeXML
uniformly continuous

uniformly continuous

Let

f

:

A

→

ℝ

be a

real function

defined on a subset

A

of the real line. We say that

f

is

uniformly continuous

if, given an arbitrary small

positive

ε

, there exists a positive

δ

such that whenever two points in

A

differ by less than

δ

, they are mapped by

f

into points which differ by less than

ε

. In symbols:

∀

ε

>

0

⁢

∃

δ

>

0

⁢

∀

x

,

y

∈

A

⁢

|

x

-

y

|

<

δ

⇒

|

f

⁢

(

x

)

-

f

⁢

(

y

)

|

<

ε

.

Every uniformly continuous function is also

continuous

, while the converse does not always hold. For instance, the

function

f

:

]

0

,

+

∞

[

→

ℝ

defined by

f

⁢

(

x

)

=

1

/

x

is continuous in its domain, but not uniformly.

A more general

definition

of uniform continuity applies to functions between

metric spaces

(there are even more general environments for uniformly continuous functions, i.e.

uniform spaces

).
Given a function

f

:

X

→

Y

, where

X

and

Y

are metric spaces with distances

d

X

and

d

Y

, we say that

f

is uniformly continuous if

∀

ε

>

0

⁢

∃

δ

>

0

⁢

∀

x

,

y

∈

X

⁢

d

X

⁢

(

x

,

y

)

<

δ

⇒

d

Y

⁢

(

f

⁢

(

x

)

,

f

⁢

(

y

)

)

<

ε

.

Uniformly continuous functions have the

property

that they map

Cauchy sequences

to Cauchy sequences and that they

preserve

uniform convergence

of

sequences

of functions.

Any

continuous function

defined on a

compact space

is uniformly continuous (see Heine-Cantor theorem).

Title

uniformly continuous

Canonical

name

UniformlyContinuous

Date of creation

2013-03-22 12:45:38

Last modified on

2013-03-22 12:45:38

Owner

n3o (216)

Last modified by

n3o (216)

Numerical id

14

Author

n3o (216)

Entry type

Definition

Classification

msc 26A15

Related topic

UniformContinuity

Defines

uniformly continuous function

Generated on Fri Feb 9 19:49:43 2018 by

LaTeXML
uniformly continuous on ℝ is roughly linear

uniformly continuous on

ℝ

is roughly linear

Theorem 1

Uniformly continuous functions

defined on

[

a

,

∞

)

for

a

>

0

are roughly linear. More precisely, if

f

:

[

a

,

∞

)

→

R

then there exists

B

such that

|

f

⁢

(

x

)

|

≤

B

⁢

x

for

x

≥

a

.

Proof:

By

continuity

we can choose

δ

>

0

such that

|

x

-

y

|

≤

δ

implies

|

f

⁢

(

x

)

-

f

⁢

(

y

)

|

<

1

.

Let

x

≥

a

, choose

n

to be the smallest positive integer such that

x

≤

(

n

+

1

)

⁢

δ

. Then

f

⁢

(

x

)

-

f

⁢

(

a

)

=

f

⁢

(

x

)

-

f

⁢

(

a

+

n

⁢

δ

)

+

∑

i

=

1

n

f

⁢

(

a

+

i

⁢

δ

)

-

f

⁢

(

a

+

(

i

-

1

)

⁢

δ

)

so that we have

|

f

⁢

(

x

)

|

≤

|

f

⁢

(

x

)

-

f

⁢

(

a

+

n

⁢

δ

)

|

+

∑

i

=

1

n

|

f

⁢

(

a

+

i

⁢

δ

)

-

f

⁢

(

a

+

(

i

-

1

)

⁢

δ

)

|

+

|

f

⁢

(

a

)

|

(1)

≤

n

+

1

+

|

f

⁢

(

a

)

|

.

(2)

Therefore,

|

f

⁢

(

x

)

|

x

≤

|

f

⁢

(

a

)

|

+

n

+

1

n

⁢

δ

(3)

≤

|

f

⁢

(

a

)

|

n

⁢

δ

+

n

+

1

n

⁢

δ

.

(4)

As

n

→

∞

, the rhs

converges

to

1

δ

.
Hence, the sequence defined by

b

n

=

|

f

⁢

(

a

)

|

n

⁢

δ

+

n

+

1

n

⁢

δ

is

bounded

by some
number

B

as desired.

Note we can extend this result to

f

:

[

0

,

∞

)

→

ℝ

if

f

is

differentiable

at 0.

Title

uniformly continuous

on

ℝ

is roughly linear

Canonical name

UniformlyContinuousOnmathbbRIsRoughlyLinear

Date of creation

2013-03-22 15:09:41

Last modified on

2013-03-22 15:09:41

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

22

Author

Mathprof (13753)

Entry type

Theorem

Classification

msc 26A15

Generated on Fri Feb 9 19:49:46 2018 by

LaTeXML
version of the fundamental lemma of calculus of variations

version of the fundamental lemma of calculus of variations

Lemma.

If a real function

f

is

continuous

on the

interval

[

a

,

b

]

and if

∫

a

b

f

⁢

(

x

)

⁢

φ

⁢

(

x

)

⁢

𝑑

x

=

0

for all

functions

φ

continuously differentiable

on the interval and vanishing at its

end points

, then

f

⁢

(

x

)

≡

0

on the whole interval.

Proof.

We make the antithesis that

f

does not vanish identically.  Then there exists a point

x

0

of the open interval

(

a

,

b

)

such that

f

⁢

(

x

0

)

≠

0

;  for example

f

⁢

(

x

0

)

>

0

.  The

continuity

of

f

implies that there are the numbers

α

and

β

such that

a

<

α

<

x

0

<

β

<

b

and

f

⁢

(

x

)

>

0

for all

x

∈

[

α

,

β

]

.  Now the function

φ

0

defined by

φ

0

⁢

(

x

)

:=

{

(

x

-

α

)

2

⁢

(

x

-

β

)

2

for

⁢

α

≦

x

≦

β

,

0

otherwise

fulfils the requirements for the functions

φ

.  Since both

f

and

φ

0

are positive on the open interval

(

α

,

β

)

,  we however have

∫

a

b

f

⁢

(

x

)

⁢

φ

0

⁢

(

x

)

⁢

𝑑

x

=

∫

α

β

f

⁢

(

x

)

⁢

φ

0

⁢

(

x

)

⁢

𝑑

x

>

0

.

Thus the antithesis causes a

contradiction

.  Consequently, we must have

f

⁢

(

x

)

≡

0

.

Title

version of the

fundamental lemma of calculus of variations

Canonical name

VersionOfTheFundamentalLemmaOfCalculusOfVariations

Date of creation

2013-03-22 19:12:04

Last modified on

2013-03-22 19:12:04

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

4

Author

pahio (2872)

Entry type

Theorem

Classification

msc 26A15

Classification

msc 26A42

Synonym

fundamental lemma of calculus of variations

Related topic

EulerLagrangeDifferentialEquation

Generated on Fri Feb 9 19:49:48 2018 by

LaTeXML
Weierstrass’ criterion of uniform convergence

Weierstrass’ criterion of uniform convergence

Theorem

.

Let the real functions

f

1

⁢

(

x

)

,

f

2

⁢

(

x

)

, … be defined in the

interval

[

a

,

b

]

.  If they all the condition

|

f

n

⁢

(

x

)

|

≦

M

n

∀

x

∈

[

a

,

b

]

,

with

∑

n

=

1

∞

M

n

a

convergent series

of , then the function series

f

1

⁢

(

x

)

+

f

2

⁢

(

x

)

+

⋯

converges uniformly

(

http://planetmath.org/SumFunctionOfSeries

) on the interval

[

a

,

b

]

.

The theorem is valid also for the series with complex function terms, when one replaces the interval with a subset of

ℂ

.

Title

Weierstrass’ criterion of

uniform convergence

Canonical name

WeierstrassCriterionOfUniformConvergence

Date of creation

2013-03-22 14:38:21

Last modified on

2013-03-22 14:38:21

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

9

Author

pahio (2872)

Entry type

Theorem

Classification

msc 26A15

Classification

msc 40A30

Synonym

Weierstrass’ M-test

Generated on Fri Feb 9 19:49:50 2018 by

LaTeXML
example of Lipschitz condition

example of Lipschitz condition

Statement 1

.

Let

f

⁢

(

x

)

=

x

2

. Then

f

satisfies

the

Lipschitz condition

on

[

a

,

b

]

⊂

ℝ

for finite

real numbers

a

<

b

.

Proof.

We want to show that for some real

constant

L

, and for all

x

,

y

∈

[

a

,

b

]

,

|

x

2

-

y

2

|

≤

L

⁢

|

x

-

y

|

.

Let

x

,

y

∈

[

a

,

b

]

. Clearly if

x

=

y

, the above

inequality

holds, so assume

x

≠

y

. Since

x

and

y

are interchangable in the above equation, it can be assumed without loss of generality that

x

<

y

.

Since

f

is

differentiable

on

(

a

,

b

)

, by the

mean-value theorem

, there is a

z

∈

(

x

,

y

)

such that

f

⁢

(

x

)

-

f

⁢

(

y

)

x

-

y

=

f

′

⁢

(

z

)

,

that is,

x

2

-

y

2

x

-

y

=

2

⁢

z

.

Taking the

modulus

of both sides gives

|

x

2

-

y

2

|

|

x

-

y

|

=

2

⁢

|

z

|

.

Finally, to find

L

it is necessary to consider all possible values of

z

:

|

x

2

-

y

2

|

|

x

-

y

|

=

2

⁢

|

z

|

≤

2

⁢

sup

⁡

{

|

z

|

:

z

∈

(

a

,

b

)

}

=

2

⁢

max

⁡

{

|

a

|

,

|

b

|

}

.

Thus, for all

x

,

y

∈

[

a

,

b

]

,

|

f

⁢

(

x

)

-

f

⁢

(

y

)

|

≤

2

⁢

max

⁡

{

|

a

|

,

|

b

|

}

⁢

|

x

-

y

|

as required.
∎

Statement 2

.

Additionally,

L

=

2

⁢

max

⁡

{

|

a

|

,

|

b

|

}

is the Lipschitz constant of

f

.

Proof.

Assume

|

b

|

≥

|

a

|

, since if

|

b

|

<

|

a

|

, it is possible to consider

-

f

instead of

f

. This also implies that

b

>

0

. Let

ε

>

0

be sufficiently small that

a

<

b

-

ε

and that higher powers of

ε

can be ignored. Now,

|

f

⁢

(

b

)

-

f

⁢

(

b

-

ε

)

|

|

b

-

(

b

-

ε

)

|

=

b

2

-

(

b

-

ε

)

2

b

-

(

b

-

ε

)

=

b

2

-

b

2

+

2

⁢

b

⁢

ε

-

ε

2

b

-

b

+

ε

=

2

⁢

b

⁢

ε

ε

=

2

⁢

b

.

By the

assumption

above,

b

=

max

⁡

{

|

a

|

,

|

b

|

}

. Thus, since

b

,

b

-

ε

∈

[

a

,

b

]

and by the definition of the Lipschitz condition,

L

≥

|

f

⁢

(

b

)

-

f

⁢

(

b

-

ε

)

|

|

b

-

(

b

-

ε

)

|

=

2

⁢

max

⁡

{

|

a

|

,

|

b

|

}

.

However, the result from the previous proof gives

|

f

⁢

(

b

)

-

f

⁢

(

b

-

ε

)

|

|

b

-

(

b

-

ϵ

)

|

≤

L

≤

2

⁢

max

⁡

{

|

a

|

,

|

b

|

}

.

Combining these inequalities provides

2

⁢

max

⁡

{

|

a

|

,

|

b

|

}

≤

L

≤

2

⁢

max

⁡

{

|

a

|

,

|

b

|

}

,

and the result follows by

trichotomy

.
∎

Title

example of Lipschitz condition

Canonical name

ExampleOfLipschitzCondition

Date of creation

2013-03-22 17:14:16

Last modified on

2013-03-22 17:14:16

Owner

me_and (17092)

Last modified by

me_and (17092)

Numerical id

10

Author

me_and (17092)

Entry type

Example

Classification

msc 26A16

Generated on Fri Feb 9 19:49:53 2018 by

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Lipschitz condition

Lipschitz condition

A mapping

f

:

X

→

Y

between

metric spaces

is said to satisfy the

Lipschitz condition

, or to be

Lipschitz continuous

or

L

-Lipschitz

if there exists a real constant

L

such
that

d

Y

⁢

(

f

⁢

(

p

)

,

f

⁢

(

q

)

)

≤

L

⁢

d

X

⁢

(

p

,

q

)

,

for all

⁢

p

,

q

∈

X

.

The least constant

L

for which the previous

inequality

holds, is called the

Lipschitz constant

of

f

.
The space of Lipschitz continuous functions is often denoted by

Lip

⁢

(

X

,

Y

)

.

Clearly, every Lipschitz continuous function is

continuous

.

Notes.

More generally, one says that a mapping satisfies
a Lipschitz condition of order

α

>

0

if there exists a real constant

C

such that

d

Y

⁢

(

f

⁢

(

p

)

,

f

⁢

(

q

)

)

≤

C

⁢

d

X

⁢

(

p

,

q

)

α

,

for all

⁢

p

,

q

∈

X

.

Functions which satisfy this condition are also called

Hölder continuous

or

α

-Hölder

. The vector space of such functions is denoted by

C

0

,

α

⁢

(

X

,

Y

)

and hence

Lip

=

C

0

,

1

.

Title

Lipschitz condition

Canonical name

LipschitzCondition

Date of creation

2013-03-22 11:57:48

Last modified on

2013-03-22 11:57:48

Owner

paolini (1187)

Last modified by

paolini (1187)

Numerical id

27

Author

paolini (1187)

Entry type

Definition

Classification

msc 26A16

Synonym

Lipschitz

Synonym

Lipschitz continuous

Related topic

RademachersTheorem

Related topic

NewtonsMethod

Related topic

KantorovitchsTheorem

Defines

Holder

Defines

Holder continuous

Defines

Lipschitz constant

Generated on Fri Feb 9 19:49:55 2018 by

LaTeXML
Lipschitz condition and differentiability

Lipschitz condition and differentiability

If

X

and

Y

are

Banach spaces

, e.g.

ℝ

n

, one can inquire about the

relation

between differentiability and the

Lipschitz condition

. If

f

is Lipschitz, the ratio

∥

f

⁢

(

q

)

-

f

⁢

(

p

)

∥

∥

q

-

p

∥

,

p

,

q

∈

X

is

bounded

but is not assumed to

converge

to a limit.

Proposition 1

Let

f

:

X

→

Y

be a

continuously differentiable

mapping

(

http://planetmath.org/DifferentiableMapping

) between
Banach spaces. If

K

⊂

X

is a

compact
subset

, then the

restriction

f

:

K

→

Y

satisfies

the Lipschitz
condition.

Proof.

Let

lin

⁡

(

X

,

Y

)

denote the Banach space of

bounded linear maps

from

X

to

Y

. Recall that the norm

∥

T

∥

of a linear mapping

T

∈

lin

⁡

(

X

,

Y

)

is defined by

∥

T

∥

=

sup

⁡

{

∥

T

⁢

u

∥

∥

u

∥

:

u

≠

0

}

.

Let

D

⁡

f

:

X

→

lin

⁡

(

X

,

Y

)

denote the

derivative

of

f

. By definition

D

⁡

f

is

continuous

, which really means that

∥

D

⁡

f

∥

:

X

→

ℝ

is a

continuous function

. Since

K

⊂

X

is

compact

, there exists a finite

upper bound

B

1

>

0

for

∥

D

⁡

f

∥

restricted to

K

. In particular, this means that

∥

D

⁡

f

⁢

(

p

)

⁢

u

∥

≤

∥

D

⁡

f

⁢

(

p

)

∥

⁢

∥

u

∥

≤

B

1

⁢

∥

u

∥

,

for all

p

∈

K

,

u

∈

X

.

Next, consider the secant mapping

s

:

X

×

X

→

ℝ

defined by

s

⁢

(

p

,

q

)

=

{

∥

f

⁢

(

q

)

-

f

⁢

(

p

)

-

D

⁡

f

⁢

(

p

)

⁢

(

q

-

p

)

∥

∥

q

-

p

∥

q

≠

p

0

p

=

q

This mapping is continuous, because

f

is assumed to be continuously
differentiable. Hence, there is a finite
upper bound

B

2

>

0

for

s

restricted to the compact set

K

×

K

. It
follows that for all

p

,

q

∈

K

we have

∥

f

⁢

(

q

)

-

f

⁢

(

p

)

∥

≤

∥

f

⁢

(

q

)

-

f

⁢

(

p

)

-

D

⁡

f

⁢

(

p

)

⁢

(

q

-

p

)

∥

+

∥

D

⁡

f

⁢

(

p

)

⁢

(

q

-

p

)

∥

≤

B

2

⁢

∥

q

-

p

∥

+

B

1

⁢

∥

q

-

p

∥

=

(

B

1

+

B

2

)

⁢

∥

q

-

p

∥

Therefore

B

1

+

B

2

is the desired Lipschitz constant. QED

Neither condition is

stronger

. For example, the

function

f

:

ℝ

→

ℝ

given by

f

⁢

(

x

)

=

x

2

is

differentiable

but not Lipschitz.

Title

Lipschitz condition and differentiability

Canonical name

LipschitzConditionAndDifferentiability

Date of creation

2013-03-22 11:57:50

Last modified on

2013-03-22 11:57:50

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

34

Author

Mathprof (13753)

Entry type

Theorem

Classification

msc 26A16

Synonym

mean value inequality

Related topic

Derivative2

Generated on Fri Feb 9 19:49:58 2018 by

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