# Monthly Archives: November 2015

## Rademacher function and Khinchin’s equality

Fair coin에 대하여 이 상황을 설명할 수 있는 또 다른 방법이 있는데, Rademacher function을 이용하는 방법이 있다. 함수는 다음과 같이 정의된다:

여기서 sample space는 이며 probability measure는 에서의 Lebesgue measure이다. 그러면 모든 에 대하여

를 얻는다. 여기서 이다. 이 사실은 가 independent random variable이라는 것을 보여준다.

Lemma. 를 Rademacher sequence라 하자. 자연수 에 대하여 라 하자. 그러면

이다.

Proof. 우선 이라 가정하자. Rademancher function의 정의는

이다. 그러므로 임의의 양수 에 대하여 가 independent이므로

를 얻는다.

그리고 , 라 하면 , 이므로 임의의 에 대하여

를 얻는다. 그리고 임의의 실수 에 대하여 이다. 이는

이로부터 이 음이 아닌 정수 에 대하여 성립하기 때문에 부등식을 얻는다. 그러므로

이다. 이라 하자. 그러면 양수 에 대하여

를 얻는다. 여기서 마지막 부등식은 에 대해서 극댓값을 얻어내는 결과다.

비슷한 방법으로

를 얻어내므로

를 얻는다.

이제 라 하면 실수부와 허수부로 나누어서 공략을 하면

를 얻는다.

Theorem (Khinchin’s equality). 이라면 임의의 자연수
대하여

이다.

Proof. (i) 일 때는 직교성에 의해서 쉽게 보인다. 즉

를 얻는다.

(ii) 일 때 upper bound를 구해보도록 하자. 이라 하자. 을 이전과 같이 이라 하자. 그러면 앞의 Lemma에 의하여

이다.

(iii) 일 때 lower bound를 구해보도록 한다. 이는 (i)과 (ii), duality에 의해 얻는다. 그러면 (i)과 Hölder’s inequality에 의하여

을 얻는다.

이므로 (ii)에 의하여

를 얻는다. 그러므로

를 얻는다.

(iv) 일 때 Hölder’s inequality에 의하여

를 얻으며 (ii)에 의하여 를 얻는다. 그러므로

를 얻는다. 따라서

를 얻어서 증명이 끝난다.

## Arzela-Ascoli Theorem

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.

## upper bound estimate of generalized binomial coefficients

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Proposition. If with , then there is such that

Proof. Note that

Note that for any , we have

So

Since and for all , we get

Since , we see that

where . Hence, we got