# Arzela-Ascoli Theorem

By | November 11, 2015

1. Motivation of our main Theorem

In metric space, topological compactness and sequential compactness are equivalent. In , we have a Heine-Borel property. However, the following example gives that closedness and boundedness do not gurantee the compactness in general.

Example 1 This example is related to sequential compactness. Consider and where has 1 at th coordinates and others are 0. We call this set as `standard basis’ in . Note that is bounded sequence since for all . Also note that . Then has no convergent subsequence since these are all isolated.

Also, if we have a bounded sequence, can we gurantee the existence of subsequence which has pointwise convergent? The following example gives the negative answer to the question.

Example 2 For each define where . Does have pointwise convergent subsequence? Suppose has a convergent subsequence for every . Then we have

So

So by bounded convergence theorem, we have

But the calculation shows

2. Proof of Main Theorem

We want to find an equivalent statements of compactness of function spaces. To find the condition, the following terminologies are useful. Let be a collection of complex functions on a metric space with metric . We say that is equicontinuous if to every , corresponds a such that for every and for all paris of points with .

We say that is pointwise bounded if to every corresponds an such that for every . Arzela-Ascoli Theorem gives a behavior of sequences in a some condition. It is strongly related to the weak type convergence which will be explained after we prove the main theorem. There are many versions on this theorem. The following theorem is more general form which is given in Rudin’s Real and Complex Analysis.

Theorem 1 Suppose that is a pointwise bounded equicontinuous collection of complex functions on a separable metric space . Then every sequence in has then a subsequence that converges uniformly on every compact subset of .

Proof: We divide our proof into two steps. First, we will find a subsequence of . Since is separable, there is a countable dense subset of . Consider . Note that it is bounded sequence in since is pointwise bounded. So there is such that for all . So by Bolzano-Weierstrass property in , it has a convergent subsequence . Since is a subsequence of , it is pointwise bounded. So if we consider , then it has a convergent subsequence . Continue this process. Then we summerize the main result of the above process:

• is a subsequence of at .
• is a subsequence of at for .

If we define , then recall that is a subsequence of . Take any , then , is a subsequence of which converges at . For each , converges for . So exists for all . This ends the first part of proof.

Next, we prove that converges uniformly on every compact subset of . Let be a compact subset of and let be given. Then by equicontinuity, there is a such that implies for all . Cover by . Then by compactness, there is a finite subcover, say which has radius . Since is dense in there are points for . Since , exists. So for , there is an integer such that implies if , . Choose .

Pick . Then for some , and . So

if . So converges uniformly on . So we are done.

Let be a compact metric space. Then is compact if and only if is closed, pointwise bounded and equicontinuous on . Proof: Assume is compact. Then it is closed and bounded. So it suffices to check that is equicontinuous on Let be given. Then by compactness of , there is such that , where .

Since each is continuous on the compact set , so it is uniformly continuous on . So there is such that if in , then . Set . Then implies for all .

To show the equicontinuity, let . Then there is such that . So implies .

Assume is closed, pointwise bounded and equicontinuous on . Let be any sequence and let be given. Then there is such that implies for all

Note that there is a countable dense subset of . So by Arzela-Ascoli theorem, there is a subsequence of such that for each , converges. We claim that converges uniformly in . Since is compact, there exists a finitely many such that

Note that for each , converges. Thus, there exists such that implies for all . Let . Then there is such that . So for all . Thus,

This implies if , then . Thus, is uniformly Cauchy. So this converges uniformly to and because is closed.

3. Weak Convergence

means is continuous and linear. A sequence is norm bounded means there is such that for all and . In other language, is pointwise bounded and equicontinuous.

Theorem 2 Let be separable normed vector space. Then every norm bounded sequence in has convergent subsequence such that for each .

Proof: Let be a norm bounded sequence in . Since is separable and is pointwise bounded and equicontinuous, by Arzelá-Ascoli theorem, has a convergent subsequence which converge uniformly on all compact subset of . Thus, converges pointwise on .

We say a norm bounded sequence in has a weak{*} convergent subsequence if there is a subsequence of such that for each . Now, we will connect these kind of concepts on -spaces.

On and with conjugate exponent with , if we define by for some . Then we get the following result:

Theorem 3 Let with conjugate exponent . Define by

for some . Then is a continuous linear functional on and moreover, .

Proof: (i) . Then . Thus, . To prove the reverse inequality, define . Then since . Hence,

Note that

Thus, by the infimum definition of operator norm.Hence, .

(ii) , then its conjugate exponent is . Then

So .

Now we will show that for any , . Let . Then . We choose any subset of so that . Then put . Then so that . Also,

Thus, .

Theorem 4 (Riesz Representation Theorem) Let . If is a continuous linear funcdtional on , then there is such that and . Hence, for . Then this fact holds for any measure for and true for if the measure is -finite.

Thus, any can be considered as a member of . Hence, we can use Arzela-Ascoli theorem to get a new-type of convergence, weak{*}-convergence.

This site uses Akismet to reduce spam. Learn how your comment data is processed.