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:

    \[ r_{j}\left(t\right):=\sign\left(\sin\left(2\pi2^{j}t\right)\right),\quad0<t<1,\quad j=0,1,2,\dots. \]

여기서 sample space는 \left(0,1\right)이며 probability measure는 \left[0,1\right]에서의 Lebesgue measure이다. 그러면 모든 n\ge1에 대하여

    \[ m\left(\left\{ r_{j_{1}}=\varepsilon_{1},\dots,r_{j_{n}}=\varepsilon_{n}\right\} \right)=2^{-n} \]

를 얻는다. 여기서 \varepsilon_{j}\in\left\{ -1,1\right\}이다. 이 사실은 \left\{ r_{j}\right\}가 independent random variable이라는 것을 보여준다.

Lemma. r_{j}를 Rademacher sequence라 하자. 자연수 N에 대하여 \left\{ a_{j}\right\} _{j=1}^{\infty}\subset\mathbb{C}라 하자. 그러면

    \[ \prob\left(\left\{ \left|\sum_{j=1}^{N}r_{j}a_{j}\right|>\lambda\left(\sum_{j=1}^{N}\left|a_{j}\right|^{2}\right)^{\frac{1}{2}}\right\} \right)\le4e^{-\frac{\lambda^{2}}{2}} \]

이다.

Proof. 우선 a_{i}\in\mathbb{R}이라 가정하자. Rademancher function의 정의는

    \[ r_{j}\left(t\right)=\mathrm{sign}\left(\sin\left(2\pi2^{j}t\right)\right)\quad0<t<1\quad j=0,1,\dots \]

이다. 그러므로 임의의 양수 \rho에 대하여 \left\{ r_{j}\right\}가 independent이므로

    \[ \Exp\left[e^{\rho S_{N}}\right]=\prod_{j=1}^{N}\Exp\left[e^{\rho r_{j}a_{j}}\right] \]

를 얻는다.

그리고 A_{j}=\left\{ 0<t<1:r_{j}\left(t\right)=1\right\}, B_{j}=\left\{ 0<t<1:r_{j}\left(t\right)=-1\right\}라 하면 m\left(A_{j}\right)=\frac{1}{2}, m\left(B_{j}\right)=\frac{1}{2}이므로 임의의 1\le j\le N에 대하여

    \begin{align*} \Exp\left[e^{\rho r_{j}a_{j}}\right] & =\int_{0}^{1}e^{\rho r_{j}\left(t\right)a_{j}}dt\\ & =\frac{1}{2}\left[e^{a_{j}\rho}+e^{-a_{j}\rho}\right]\\ & =\cosh\left(a_{j}\rho\right) \end{align*}

를 얻는다. 그리고 임의의 실수 x에 대하여 \cosh x\le e^{\frac{x^{2}}{2}}이다. 이는

    \[ e^{\frac{x^{2}}{2}}-\cosh x=\sum_{n=0}^{\infty}\left(\frac{1}{n!2^{n}}-\frac{1}{\left(2n\right)!}\right)x^{2n} \]

이로부터 \left(2n\right)!\ge n!2^{n}이 음이 아닌 정수 n에 대하여 성립하기 때문에 부등식을 얻는다. 그러므로

    \begin{align*} \Exp\left[e^{\rho r_{j}a_{j}}\right] & \le\prod_{j=1}^{N}e^{\rho^{2}a_{j}^{2}/2}\\ & =\exp\left(\rho^{2}\sum_{j=1}^{N}\frac{1}{2}a_{j}^{2}\right) \end{align*}

이다. \sigma^{2}=\sum_{j=1}^{n}a_{j}^{2}이라 하자. 그러면 양수 \lambda>0에 대하여

    \begin{align*} &\relphantom{=} \prob\left(\left\{ S_{N}>\lambda\sigma\right\} \right)\\ & =\prob\left(\left\{ \rho S_{N}>\lambda\rho\sigma\right\} \right)\\ & =\prob\left(\left\{ e^{\rho S_{N}}>e^{\lambda\rho\sigma}\right\} \right)\\ & \le\frac{\Exp\left[e^{\rho S_{N}}\right]}{e^{\lambda\rho\sigma}}\le e^{\rho^{2}\sigma^{2}/2}e^{-\lambda\rho\sigma}\le e^{-\frac{\lambda^{2}}{2}} \end{align*}

를 얻는다. 여기서 마지막 부등식은 \rho에 대해서 극댓값을 얻어내는 결과다.

비슷한 방법으로

    \[ \prob\left(\left\{ S_{N}<-\lambda\sigma\right\} \right)\le e^{-\frac{\lambda^{2}}{2}} \]

를 얻어내므로

    \[ \prob\left(\left|S_{N}\right|>\lambda\sigma\right)\le2e^{-\frac{\lambda^{2}}{2}} \]

를 얻는다.

이제 a_{j}\in\mathbb{C}라 하면 실수부와 허수부로 나누어서 공략을 하면

    \[ \prob\left(\left|S_{N}\right|>\lambda\sigma\right)\le4e^{-\frac{\lambda^{2}}{2}} \]

를 얻는다.


Theorem (Khinchin’s equality). 1\leq p<\infty이라면 임의의 자연수 N\left\{ a_{j}\right\} _{j=1}^{N}\subset\mathbb{C}
대하여

    \[ \Exp\left[\left|\sum_{j=1}^{N}r_{j}a_{j}\right|^{p}\right]\approx_{p,d}\left(\sum_{j=1}^{N}\left|a_{j}\right|^{2}\right)^{\frac{p}{2}} \]

이다.

Proof. (i) p=2일 때는 직교성에 의해서 쉽게 보인다. 즉

    \begin{align*} \Exp\left[\left|\sum_{j=1}^{N}r_{j}a_{j}\right|^{2}\right] & =\sum_{i,j=1}^{n}\Exp\left[r_{j}\overline{r_{i}}a_{i}\overline{a_{j}}\right]\\ & =\sum_{i=1}^{n}\left|a_{i}\right|^{2}\Exp\left[\left|r_{i}\right|^{2}\right]\\ & =\sum_{i=1}^{N}\left|a_{i}\right|^{2} \end{align*}

를 얻는다.

(ii) 1<p<\infty일 때 upper bound를 구해보도록 하자. \sum_{j=1}^{N}\left|a_{j}\right|^{2}=1이라 하자. S_{N}을 이전과 같이 S_{N}=\sum_{j=1}^{N}a_{j}r_{j}이라 하자. 그러면 앞의 Lemma에 의하여

    \begin{align*} & \relphantom{=}\Exp\left[\left|S_{N}\right|^{p}\right]\\ & =\int_{0}^{\infty}\prob\left(\left\{ \left|S_{N}\right|>\lambda\right\} \right)p\lambda^{p-1}d\lambda\\ & \le\int_{0}^{\infty}4e^{-\frac{\lambda^{2}}{2}}p\lambda^{p-1}d\lambda=:C\left(p\right)<\infty \end{align*}

이다.

(iii) 1<p<\infty일 때 lower bound를 구해보도록 한다. 이는 (i)과 (ii), duality에 의해 얻는다. 그러면 (i)과 Hölder’s inequality에 의하여

    \begin{align*} \sum_{j=1}^{N}\left|a_{j}\right|^{2}= & \Exp\left[\left|\sum_{j=1}^{N}r_{j}a_{j}\right|^{2}\right]\\ & \le\Exp\left(\left|\sum_{j=1}^{N}a_{j}r_{j}\right|^{p}\right)^{\frac{1}{p}}\Exp\left(\left|\sum_{j=1}^{N}a_{j}r_{j}\right|^{p^{\prime}}\right)^{\frac{1}{p^{\prime}}} \end{align*}

을 얻는다.

1<p^{\prime}<\infty이므로 (ii)에 의하여

    \[ \Exp\left(\left|\sum_{j=1}^{N}a_{j}r_{j}\right|^{p^{\prime}}\right)^{\frac{1}{p^{\prime}}}\lesssim_{p}\left(\sum_{j=1}^{N}\left|a_{j}\right|^{2}\right)^{\frac{1}{2}} \]

를 얻는다. 그러므로

    \[ \left(\sum_{j=1}^{N}\left|a_{j}\right|^{2}\right)^{\frac{1}{2}}\lesssim_{p}\Exp\left(\left|\sum_{j=1}^{N}a_{j}r_{j}\right|^{p}\right)^{\frac{1}{p}} \]

를 얻는다.

(iv) p=1일 때 Hölder’s inequality에 의하여

    \begin{align*} & \relphantom{=}\Exp\left[\left|S_{N}\right|^{2}\right]\\ & =\Exp\left[\left|S_{N}\right|^{\frac{2}{3}}\left|S_{N}\right|^{\frac{4}{3}}\right]\\ & \le\Exp\left[\left|S_{N}\right|\right]^{\frac{2}{3}}\left(\Exp\left[S_{N}\right]^{4}\right)^{\frac{1}{3}} \end{align*}

를 얻으며 (ii)에 의하여 \left(\Exp\left[S_{N}\right]^{4}\right)^{\frac{1}{3}}\apprle_{p}\left(\sum_{j=1}^{N}\left|a_{j}\right|^{2}\right)^{2}를 얻는다. 그러므로

    \begin{align*} \Exp\left[\left|S_{N}\right|\right]^{\frac{2}{3}}\left(\Exp\left[S_{N}\right]^{4}\right)^{\frac{1}{3}} & \lesssim_{p}\left(\Exp\left|S_{N}\right|\right)^{\frac{2}{3}}\left(\sum_{j=1}^{N}\left|a_{j}\right|^{2}\right)^{\frac{2}{3}}\\ & \lesssim_{p}\left(\Exp\left|S_{N}\right|\right)^{\frac{2}{3}}\left(\Exp\left|S_{N}\right|^{2}\right)^{\frac{2}{3}} \end{align*}

를 얻는다. 따라서

    \[ \Exp\left[\left|S_{N}\right|^{2}\right]\lesssim_{p}\left(\Exp\left|S_{N}\right|\right)^{2} \]

를 얻어서 증명이 끝난다.


Arzela-Ascoli Theorem

1. Motivation of our main Theorem

In metric space, topological compactness and sequential compactness are equivalent. In {\mathbb{C}^{N}}, 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 {\ell^{2}\left(\mathbb{N}\right)} and {\left\{ e_{n}\right\} _{n=1}^{\infty}} where {e_{n}=\left(0,\dots,0,1,0,\dots\right)} has 1 at {n}th coordinates and others are 0. We call this set as `standard basis’ in {\ell^{2}\left(\mathbb{N}\right)}. Note that {\left\{ e_{n}\right\} _{=1}^{\infty}} is bounded sequence since {\left\Vert e_{n}\right\Vert _{2}=1} for all {n}. Also note that {\left\Vert e_{n}-e_{m}\right\Vert _{2}=\begin{cases} 0 & \text{if }n=m\\ \sqrt{2} & \text{if }n\neq m \end{cases}}. Then {\left\{ e_{n}\right\} } 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 {n,} define {f_{n}\left(x\right)=\sin nx} where {x\in\left[0,2\pi\right]}. Does {\left\{ f_{n}\right\} } have pointwise convergent subsequence? Suppose {\left\{ f_{n}\right\} } has a convergent subsequence {\left\{ f_{n_{k}}\right\} } for every {x\in\left[0,2\pi\right]}. Then we have

\displaystyle \lim_{k\rightarrow\infty}\left(\sin n_{k}x-\sin n_{k+1}x\right)=0\quad\left(0\le x\le2\pi\right).

So

\displaystyle \lim_{k\rightarrow\infty}\left(\sin n_{k}x-\sin n_{k+1}x\right)^{2}=0\quad\left(0\le x\le2\pi\right).

So by bounded convergence theorem, we have

\displaystyle \lim_{k\rightarrow\infty}\int_{0}^{2\pi}\left(\sin n_{k}x-\sin n_{k+1}x\right)^{2}dx=0.

But the calculation shows

\displaystyle \int_{0}^{2\pi}\left(\sin n_{k}x-\sin n_{k+1}x\right)^{2}dx=2\pi,

a contradiction.

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 {\mathscr{F}} be a collection of complex functions on a metric space {X} with metric {\rho}. We say that {\mathscr{F}} is equicontinuous if to every {\varepsilon>0}, corresponds a {\delta>0} such that {\left|f\left(x\right)-f\left(y\right)\right|<\varepsilon} for every {f\in\mathcal{F}} and for all paris of points {x,y} with {\rho\left(x,y\right)<\delta}.

We say that {\mathscr{F}} is pointwise bounded if to every {x\in X} corresponds an {M_{x}<\infty} such that {\left|f\left(x\right)\right|\le M_{x}} for every {f\in\mathscr{F}}. 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 {\mathscr{F}} is a pointwise bounded equicontinuous collection of complex functions on a separable metric space {X}. Then every sequence {\left\{ f_{n}\right\} } in {\mathscr{F}} has then a subsequence that converges uniformly on every compact subset of {X}.

Proof: We divide our proof into two steps. First, we will find a subsequence of {\left\{ f_{n}\right\} }. Since {X} is separable, there is a countable dense subset {A=\left\{ x_{1},x_{2},\dots\right\} } of {X}. Consider {\left\{ f_{n}\left(x_{1}\right)\right\} _{n\in\mathbb{N}}}. Note that it is bounded sequence in {\mathbb{C}} since {\mathscr{F}} is pointwise bounded. So there is {M_{1}} such that {\left|f_{n}\left(x_{1}\right)\right|\le M_{1}} for all {n}. So by Bolzano-Weierstrass property in {\mathbb{C}}, it has a convergent subsequence {\left\{ f_{1,n}\left(x_{1}\right)\right\} } . Since {\left\{ f_{1,n}\right\} } is a subsequence of {\left\{ f_{n}\right\} }, it is pointwise bounded. So if we consider {\left\{ f_{1,n}\left(x_{2}\right)\right\} _{n\in\mathbb{N}}}, then it has a convergent subsequence {\left\{ f_{2,n}\left(x_{2}\right)\right\} }. Continue this process. Then we summerize the main result of the above process:

  • {\left\{ f_{1,n}\right\} _{n=1}^{\infty}} is a subsequence of {\left\{ f_{n}\right\} _{n=1}^{\infty}} at {x_{1}}.
  • {\left\{ f_{m,n}\right\} _{n=1}^{\infty}} is a subsequence of {\left\{ f_{m-1,n}\right\} _{n=1}^{\infty}} at {x_{1},\dots,x_{m}} for {m\ge2}.

If we define {g_{n}=f_{n,n}}, then recall that {\left\{ g_{n}\right\} _{n=1}^{\infty}} is a subsequence of {\left\{ f_{n}\right\} _{n=1}^{\infty}} . Take any {x_{k}\in A}, then {n\ge k}, {\left\{ f_{n,n}\right\} _{n\ge k}} is a subsequence of {\left\{ f_{n,k}\right\} _{n=1}^{\infty}} which converges at {x_{k}}. For each {n}, {\left\{ g_{n}\left(x_{i}\right)\right\} } converges for {1\le i\le n}. So {\lim_{n\rightarrow\infty}g_{n}\left(x\right)} exists for all {x\in A}. This ends the first part of proof.

Next, we prove that {\left\{ g_{n}\right\} _{n=1}^{\infty}} converges uniformly on every compact subset of {X}. Let {K} be a compact subset of {X} and let {\varepsilon>0} be given. Then by equicontinuity, there is a {\delta>0} such that {\rho\left(x,y\right)<\delta} implies {\left|g_{n}\left(x\right)-g_{n}\left(y\right)\right|<\frac{\varepsilon}{3}} for all {n}. Cover {K} by {\left\{ B\left(x,\delta/2\right)\right\} _{x\in K}}. Then by compactness, there is a finite subcover, say {B_{1},\dots,B_{M}} which has radius {\delta/2}. Since {A} is dense in {X,} there are points {x_{i}\in B_{i}\cap A} for {1\le i\le M}. Since {x_{i}\in A}, {\lim_{n\rightarrow\infty}g_{n}\left(x_{i}\right)} exists. So for {i=1,\dots,M}, there is an integer {N_{i}} such that {n\ge N_{i}} implies {\left|g_{m}\left(x_{i}\right)-g_{n}\left(x_{i}\right)\right|<\frac{\varepsilon}{3}} if {m>N_{i}}, {n>N_{i}} . Choose {N=\max_{1\le i\le M}\left\{ N_{i}\right\} }.

Pick {x\in K}. Then {x\in B_{i}} for some {i}, and {\rho\left(x,x_{i}\right)<\delta}. So

    \begin{align*} \left|g_{m}\left(x\right)-g_{n}\left(x\right)\right| & \le\left|g_{m}\left(x\right)-g_{m}\left(x_{i}\right)\right|+\left|g_{m}\left(x_{i}\right)-g_{n}\left(x_{i}\right)\right|+\left|g_{n}\left(x_{i}\right)-g_{n}\left(x\right)\right|\\ &\frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}=\varepsilon  \end{align*}

if {m,n>N}. So {\left\{ g_{n}\right\} } converges uniformly on {K}. So we are done. \Box

Let {K} be a compact metric space. Then {\mathscr{F}\subset\mathscr{C}\left(K\right)} is compact if and only if {\mathscr{F}\subset\mathscr{C}\left(K\right)} is closed, pointwise bounded and equicontinuous on {K}. Proof: Assume {\mathscr{F}} is compact. Then it is closed and bounded. So it suffices to check that {\mathscr{F}} is equicontinuous on {K.} Let {\varepsilon>0} be given. Then by compactness of {\mathscr{F}}, there is {\left\{ f_{1},\dots,f_{N}\right\} } such that {\mathscr{F}\subset\bigcup_{i=1}^{N}B\left(f_{i},\varepsilon\right)}, where {B\left(f_{i},\varepsilon\right)=\left\{ f\in\mathscr{C}\left(K\right):\left\Vert f-f_{i}\right\Vert _{u}<\varepsilon\right\} }.

Since each {f_{i}} is continuous on the compact set {K}, so it is uniformly continuous on {K}. So there is {\delta_{i}>0} such that if {d\left(x,y\right)<\delta} in {K}, then {\left|f_{i}\left(x\right)-f_{i}\left(y\right)\right|<\varepsilon}. Set {\delta=\min_{1\le i\le N}\delta_{i}}. Then {d\left(x,y\right)<\delta} implies {\left|f_{i}\left(x\right)-f_{i}\left(y\right)\right|<\varepsilon} for all {1\le i\le N}.

To show the equicontinuity, let {f\in\mathscr{F}}. Then there is {1\le i\le N} such that {\left\Vert f_{i}-f\right\Vert _{u}<\varepsilon}. So {d\left(x,y\right)<\delta} implies {\left|f\left(x\right)-f\left(y\right)\right|\le\left|f\left(x\right)-f_{i}\left(x\right)\right|+\left|f_{i}\left(x\right)-f_{i}\left(y\right)\right|+\left|f_{i}\left(y\right)-f\left(y\right)\right|<3\varepsilon}.

Assume {\mathscr{F}\subset\mathscr{C}\left(K\right)} is closed, pointwise bounded and equicontinuous on {K}. Let {\left\{ f_{n}\right\} \subset\mathscr{F}} be any sequence and let {\varepsilon>0} be given. Then there is {\delta>0} such that {d\left(x,y\right)<\delta} implies {\left|f\left(x\right)-f\left(y\right)\right|<\varepsilon} for all {f\in\mathscr{F}.}

Note that there is a countable dense subset {E} of {K}. So by Arzela-Ascoli theorem, there is a subsequence {\left\{ f_{n_{k}}\right\} _{k=1}^{\infty}} of {\left\{ f_{n}\right\} } such that for each {x\in E}, {\left\{ f_{n_{k}}\left(x\right)\right\} _{k=1}^{\infty}} converges. We claim that {\left\{ f_{n_{k}}\right\} } converges uniformly in {K}. Since {K} is compact, there exists a finitely many {\left\{ x_{m}\right\} _{m=1}^{q}\in E} such that

\displaystyle K\subset B\left(x_{1},\delta\right)\cup\cdots\cup B\left(x_{q},\delta\right).

Note that for each {m=1,\dots,q}, {\left\{ f_{n_{k}}\left(x_{m}\right)\right\} _{k=1}^{\infty}} converges. Thus, there exists {N>0} such that {j,k\ge N} implies {\left|f_{n_{j}}\left(x_{m}\right)-f_{n_{k}}\left(x_{m}\right)\right|<\varepsilon} for all {m=1,\dots,q}. Let {x\in K}. Then there is {x_{m}} such that {x\in B\left(x_{m},\delta\right)}. So {\left|f_{n_{k}}\left(x\right)-f_{n_{k}}\left(x_{m}\right)\right|<\varepsilon} for all {k}. Thus,

\displaystyle \left|f_{n_{j}}\left(x\right)-f_{n_{k}}\left(x\right)\right|\le\left|f_{n_{j}}\left(x\right)-f_{n_{j}}\left(x_{m}\right)\right|+\left|f_{n_{j}}\left(x_{m}\right)-f_{n_{k}}\left(x_{m}\right)\right|+\left|f_{n_{k}}\left(x_{m}\right)-f_{n_{k}}\left(x\right)\right|<3\varepsilon.

This implies if {j,k\ge N}, then {\left\Vert f_{n_{j}}-f_{n_{k}}\right\Vert _{u}<3\varepsilon}. Thus, {\left\{ f_{n_{k}}\right\} } is uniformly Cauchy. So this converges uniformly to {f^{*}} and {f^{*}\in\mathscr{F}} because {\mathscr{F}} is closed. \Box

3. Weak Convergence

{T\in X^{*}} means {T:X\rightarrow\mathbb{C}} is continuous and linear. A sequence {\left\{ T_{n}\right\} } is norm bounded means there is {M>0} such that {\left|T_{n}\left(x\right)-T_{n}\left(y\right)\right|\le M\left|x-y\right|} for all {x,y\in X} and {n\in\mathbb{N}}. In other language, {\left\{ T_{n}\right\} } is pointwise bounded and equicontinuous.

Theorem 2 Let {X} be separable normed vector space. Then every norm bounded sequence {\left\{ T_{n}\right\} } in {X^{*}} has convergent subsequence {\left\{ T_{n_{k}}\right\} } such that {T_{n_{k}}\left(x\right)\rightarrow T\left(x\right)} for each {x\in X}.

Proof: Let {\left\{ T_{n}\right\} } be a norm bounded sequence in {X^{*}}. Since {X} is separable and {\left\{ T_{n}\right\} } is pointwise bounded and equicontinuous, by Arzelá-Ascoli theorem, {\left\{ T_{n}\right\} } has a convergent subsequence {\left\{ T_{n_{k}}\right\} } which converge uniformly on all compact subset of {X}. Thus, {\left\{ T_{n_{k}}\right\} } converges pointwise on {X}. \Box

We say a norm bounded sequence {\left\{ T_{n}\right\} } in {X^{*}} has a weak{*} convergent subsequence if there is a subsequence {\left\{ T_{n_{k}}\right\} } of {\left\{ T_{n}\right\} } such that {T_{n_{k}}\left(x\right)\rightarrow T\left(x\right)} for each {x\in X}. Now, we will connect these kind of concepts on {L^{p}}-spaces.

On {\left(E,\mathfrak{M},m\right)} and {1\le p<\infty} with conjugate exponent {q} with {\frac{1}{p}+\frac{1}{q}=1}, if we define {\Phi:L^{p}\left(E\right)\rightarrow\mathbb{R}} by {\Phi\left(f\right)=\int_{E}fgdm} for some {g\in L^{q}\left(E\right)}. Then we get the following result:

Theorem 3 Let {1\le p<\infty} with conjugate exponent {q}. Define {\Phi:L^{p}\left(E\right)\rightarrow\mathbb{C}} by

\displaystyle \Phi\left(f\right)=\int_{E}fgdm

for some {g\in L^{q}\left(E\right)} . Then {\Phi} is a continuous linear functional on {L^{p}\left(E\right)} and moreover, {\left\Vert \Phi\right\Vert =\left\Vert g\right\Vert _{q}}.

Proof: (i) {1<p<\infty}. Then {\left|\Phi\left(f\right)\right|=\left|\int_{E}fg\right|\le\left\Vert f\right\Vert _{p}\left\Vert g\right\Vert _{q}}. Thus, {\left\Vert \Phi\right\Vert \le\left\Vert g\right\Vert _{q}}. To prove the reverse inequality, define {f=\left|g\right|^{\frac{q}{p}}\mathrm{sign} g}. Then {\left|f\right|^{p}=\left|g\right|^{q}=fg} since {fg=\left|g\right|^{\frac{q}{p}}\left|g\right|=\left|g\right|^{q}}. Hence,

    \begin{align*} \left|\Phi\left(f\right)\right| & =\left|\int fgdm\right|=\left|\int\left|g\right|^{q}dm\right|\\ & =\left\Vert g\right\Vert _{q}^{q}\\ & =\left\Vert g\right\Vert _{q}\left\Vert g\right\Vert _{q}^{q-1}.  \end{align*}

Note that

\displaystyle \left\Vert g\right\Vert _{q}^{q-1}=\left(\int_{E}\left|g\right|^{q}dm\right)^{\frac{q-1}{q}}=\left(\int_{E}\left|f\right|^{p}dm\right)^{\frac{q-1}{q}}=\left\Vert f\right\Vert _{p}.

Thus, {\left\Vert g\right\Vert _{q}\le\left\Vert \Phi\right\Vert } by the infimum definition of operator norm.Hence, {\left\Vert \Phi\right\Vert =\left\Vert g\right\Vert _{q}}.

(ii) {p=1}, then its conjugate exponent is {q=\infty}. Then

    \begin{align*} \left|\Phi\left(f\right)\right| & =\left|\int_{E}fgdm\right|\\  & \le\int_{E}\left|fg\right|dm \\ &  \le\left\Vert f\right\Vert _{1}\left\Vert g\right\Vert _{\infty}.\\ \end{align*}

So {\left\Vert \Phi\right\Vert \le\left\Vert g\right\Vert _{\infty}}.

Now we will show that for any {\varepsilon>0}, {\left\Vert \Phi\right\Vert \ge\left\Vert g\right\Vert _{\infty}-\varepsilon}. Let {A=\left\{ x\in E:\left|g\left(x\right)\right|>\left\Vert g\right\Vert _{\infty}-\varepsilon\right\} }. Then {m\left(A\right)>0}. We choose any subset {B} of {A} so that {0<m\left(B\right)<\infty}. Then put {f=\chi_{A}\mathrm{sgn}\left(g\right)}. Then {\left\Vert f\right\Vert _{1}=m\left(B\right)<\infty} so that {f\in L^{1}\left(E\right)}. Also,

\displaystyle \Phi\left(f\right)=\int_{B}\left|g\right|dm\ge\left(\left\Vert g\right\Vert _{\infty}-\varepsilon\right)m\left(B\right)=\left(\left\Vert g\right\Vert _{\infty}-\varepsilon\right)\left\Vert f\right\Vert _{1}.

Thus, {\left\Vert \Phi\right\Vert \ge\left\Vert g\right\Vert _{\infty}-\varepsilon}. \Box

Theorem 4 (Riesz Representation Theorem) Let {1\le p<\infty}. If {\Phi} is a continuous linear funcdtional on {L^{p}\left(E\right)}, then there is {g\in L^{q}\left(E\right)} such that {\Phi\left(f\right)=\int_{E}fgdm} and {\left\Vert \Phi\right\Vert =\left\Vert g\right\Vert _{q}}. Hence, {\left(L^{p}\left(E\right)\right)^{*}=L^{q}\left(E\right)} for {1\le p<\infty}. Then this fact holds for any measure for {1<p<\infty} and true for {1\le p<\infty} if the measure is {\sigma}-finite.

Thus, any {f\in L^{p}\left(E\right)} {1<p\le\infty} can be considered as a member of {\left(L^{q}\right)^{*}}. 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 \alpha\in\mathbb{C} with \mathrm{Re}\left(\alpha\right)>-1, then there is M>0 such that

    \[ \left|\binom{\alpha}{k}\right|\le\frac{M}{k^{1+\mathrm{Re}\left(\alpha\right)}}. \]

Proof. Note that

    \begin{align*} \left|\binom{\alpha}{k}\right|^{2} & =\prod_{j=1}^{k}\left|1-\frac{1+\alpha}{j}\right|^{2}\\  & \le\left(\frac{1}{k}\sum_{j=1}^{k}\left|1-\frac{1+\alpha}{j}\right|^{2}\right)^{k}. \end{align*}

Note that for any \zeta\in\mathbb{C}, we have

    \[ \left|1-\zeta\right|^{2}=1-2\mathrm{Re}\left(\zeta\right)+\left|\zeta\right|^{2}. \]

So

    \begin{align*}  & \frac{1}{k}\sum_{j=1}^{k}\left|1-\frac{1+\alpha}{j}\right|^{2}\\  & =\frac{1}{k}\sum_{j=1}^{k}\left(1-2\mathrm{Re}\left(\frac{1+\alpha}{j}\right)+\left|\frac{1+\alpha}{j}\right|^{2}\right)\\  & =1+\frac{1}{k}\left[-2\mathrm{Re}\left(1+\alpha\right)\sum_{j=1}^{k}\frac{1}{j}+\left|1+\alpha\right|^{2}\sum_{j=1}^{k}\frac{1}{j^{2}}\right]. \end{align*}

Since \sum_{j=1}^{k}\frac{1}{j}\ge\log\left(1+k\right)>\log k and \sum_{j=1}^{k}\frac{1}{j^{2}}\le2 for all k\in\mathbb{N}, we get

    \[ \sum_{j=1}^{k}\left|1-\frac{1+\alpha}{j}\right|^{2}\le1-2\mathrm{Re}\left(1+\alpha\right)\log k+2\left|1+\alpha\right|^{2}. \]

Since \left(1+\frac{r}{k}\right)^{k}\le e^{r}, we see that

    \begin{align*} \left|\binom{\alpha}{k}\right|^{2} & \le\exp\left(1-2\mathrm{Re}\left(1+\alpha\right)\log k+2\left|1+\alpha\right|^{2}\right)\\  & =\frac{M^{2}}{k^{2\left(1+\mathrm{Re}\alpha\right)}}, \end{align*}

where M^{2}=\exp\left(1+2\left|1+\alpha\right|^{2}\right). Hence, we got

    \[ \left|\binom{\alpha}{k}\right|\le\frac{M}{k^{1+\mathrm{Re}\alpha}}. \]