another proof of Dini’s theorem

another proof of Dini’s theorem

This is the version of the

Dini’s

theorem

I will prove:
Let

K

be a

compact

metric space

and

(

f

n

)

n

∈

N

⊂

C

⁢

(

K

)

which

converges

pointwise

to

f

∈

C

⁢

(

K

)

.

Besides,

f

n

⁢

(

x

)

≥

f

n

+

1

⁢

(

x

)

∀

x

∈

K

,

∀

n

.

Then

(

f

n

)

n

∈

N

converges uniformly in

K

.

Proof

Suppose that the

sequence

does not converge uniformly. Then, by

definition

,

∃

ε

>

0

⁢

such that

⁢

∀

m

∈

N

⁢

∃

n

m

>

m

,

x

m

∈

K

⁢

such that

|

f

n

m

⁢

(

x

m

)

-

f

⁢

(

x

m

)

|

≥

ε

.

So,

For

⁢

m

=

1

⁢

∃

n

1

>

1

,

x

1

∈

K

⁢

such that

⁢

|

f

n

1

⁢

(

x

1

)

-

f

⁢

(

x

1

)

|

≥

ε

∃

n

2

>

n

1

,

x

2

∈

K

⁢

such that

⁢

|

f

n

2

⁢

(

x

2

)

-

f

⁢

(

x

2

)

|

≥

ε

⋮

∃

n

m

>

n

m

-

1

,

x

m

∈

K

⁢

such that

⁢

|

f

n

m

⁢

(

x

m

)

-

f

⁢

(

x

m

)

|

≥

ε

Then we have a sequence

(

x

m

)

m

⊂

K

and

(

f

n

m

)

m

⊂

(

f

n

)

n

is a

subsequence

of the original sequence of

functions

.

K

is compact, so there is a subsequence of

(

x

m

)

m

which converges in

K

, that is,

(

x

m

j

)

j

such that

x

m

j

⟶

x

∈

K

I will prove that

f

is not

continuous

in

x

(A

contradiction

with one of the

hypothesis

).

To do this, I will show that

f

⁢

(

x

m

j

)

j

does not converge to

f

⁢

(

x

)

, using above’s

ε

.

Let

j

0

⁢

such that

⁢

j

≥

j

0

⇒

|

f

n

m

j

⁢

(

x

)

-

f

⁢

(

x

)

|

<

ε

⁢

/

⁢

4

,
which exists due to the punctual convergence of the sequence. Then,
particularly,

|

f

n

m

j

o

⁢

(

x

)

-

f

⁢

(

x

)

|

<

ε

⁢

/

⁢

4

.

Note that

|

f

n

m

j

⁢

(

x

m

j

)

-

f

⁢

(

x

m

j

)

|

=

f

n

m

j

⁢

(

x

m

j

)

-

f

⁢

(

x

m

j

)

because (using the hypothesis

f

n

⁢

(

y

)

≥

f

n

+

1

⁢

(

y

)

∀

y

∈

K

,

∀

n

) it’s easy to see that

f

n

⁢

(

y

)

≥

f

⁢

(

y

)

∀

y

∈

K

,

∀

n

Then,

f

n

m

j

⁢

(

x

m

j

)

-

f

⁢

(

x

m

j

)

≥

ε

⁢

∀

j

. And also the hypothesis implies

f

n

m

j

⁢

(

y

)

≥

f

n

m

j

+

1

⁢

(

y

)

⁢

∀

y

∈

K

,

∀

j

So,

j

≥

j

0

⇒

f

n

m

j

0

⁢

(

x

m

j

)

≥

f

n

m

j

⁢

(

x

m

j

)

,

which implies

|

f

n

m

j

0

⁢

(

x

m

j

)

-

f

⁢

(

x

m

j

)

|

≥

ε

Now,

|

f

n

m

j

0

⁢

(

x

m

j

)

-

f

⁢

(

x

)

|

+

|

f

⁢

(

x

m

j

)

-

f

⁢

(

x

)

|

≥

|

f

n

m

j

0

⁢

(

x

m

j

)

-

f

⁢

(

x

m

j

)

|

≥

ε

⁢

∀

j

≥

j

0

and so

|

f

⁢

(

x

m

j

)

-

f

⁢

(

x

)

|

≥

ε

-

|

f

n

m

j

0

⁢

(

x

m

j

)

-

f

⁢

(

x

)

|

⁢

∀

j

≥

j

0

.

On the other hand,

|

f

n

m

j

0

⁢

(

x

m

j

)

-

f

⁢

(

x

)

|

≤

|

f

n

m

j

0

⁢

(

x

m

j

)

-

f

n

m

j

0

⁢

(

x

)

|

+

|

f

n

m

j

0

⁢

(

x

)

-

f

⁢

(

x

)

|

And as

f

n

m

j

0

is continuous, there is a

j

1

such that

j

≥

j

1

⇒

|

f

n

m

j

0

⁢

(

x

m

j

)

-

f

n

m

j

0

⁢

(

x

)

|

<

ε

⁢

/

⁢

4

Then,

j

≥

j

1

⇒

|

f

n

m

j

0

⁢

(

x

m

j

)

-

f

⁢

(

x

)

|

≤

|

f

n

m

j

0

⁢

(

x

m

j

)

-

f

n

m

j

0

⁢

(

x

)

|

+

|

f

n

m

j

0

⁢

(

x

)

-

f

⁢

(

x

)

|

<

ε

⁢

/

⁢

2

,

which implies

|

f

⁢

(

x

m

j

)

-

f

⁢

(

x

)

|

≥

ε

-

|

f

n

m

j

0

⁢

(

x

m

j

)

-

f

⁢

(

x

)

|

≥

ε

⁢

/

⁢

2

⁢

∀

j

≥

max

⁡

(

j

0

,

j

1

)

.

Then, particularly,

f

⁢

(

x

m

j

)

j

does not converge to

f

⁢

(

x

)

. QED.

Title

another proof of Dini’s theorem

Canonical

name

AnotherProofOfDinisTheorem

Date of creation

2013-03-22 14:04:37

Last modified on

2013-03-22 14:04:37

Owner

gumau (3545)

Last modified by

gumau (3545)

Numerical id

11

Author

gumau (3545)

Entry type

Proof

Classification

msc 54A20

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