Arzela-Ascoli Theorem

By | November 11, 2015

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).


\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.

Leave a Reply

Your email address will not be published. Required fields are marked *

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