LEC ANA2 7

Remark 167.1(Remark).

Recall that a set is compact iff it is complete and totally bounded. Compactness is a topological property, but completeness and totally boundedness are not; they depend on the metric. Let $d_1$ and $d_2$ be metrics on $\mathbb{N}$, as defined here. We have seen that while $(\mathbb{N},d_1)$ and $(\mathbb{N},d_2)$ are the same topological space (they have the same open sets), the former is complete and the latter is not. Also, $(\mathbb{N},d_1)$ is not totally bounded, but $(\mathbb{N},d_2)$ is!

We now have the tools to prove Theorem 142.3 more succinctly:

Theorem 167.2(Theorem).

Let $X$ be compact, $f:X\to Y$ be continuous. Then, $f$ is uniformly continuous.

Proof.

Let$ϵ>0$. For $x\in X$, let $V_x=f^{-1}(B_{ϵ/2}(f(x)))$. $V_x$ is open since $f$ is continuous. $\{V_x{\}}_{x\in X}$ is an open cover of $X$. By the Lebesgue covering lemma, there exists $\delta >0$ such that for any $y\in X$, there exists $V_x$ such that $B_{\delta }(y)\subseteq V_x$. This $\delta$ works: If $z\in B_{\delta }(y)$, then $y,z\in V_x$, so $d(f(y),f(z))<ϵ$.□

Example 167.3(Example).

Closed balls in $l_1,l_2,\dots ,l_{\infty }$ are not compact (we have shown that any ball centered at the origin in these spaces is not totally bounded; any ball in an NLS can be obtained by scaling and translating the unit ball).

We will now show that balls in $C[0,1]$ are not compact. Consider the sequence $\{t^n{\}}_{n=1}^{\infty }\subseteq B_2(0)$. $\{t^n\}$ clearly does not have a converging subsequence: if it did, the subsequence would have to converge pointwise to the limit, and the pointwise limit of any subsequence is the pointwise limit of $\{t^n\}$, which is not continuous.

In fact, unit balls at not compact in any infinite dimensional NLS. Proving this requires some functional analysis.

bd2833

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The Arzelà–Ascoli Theorem

Recall that for compact $X$, $C(X)$ is a Banach space. The motivating question for this section is: For compact $X$, when is a subset of $C(X)$ compact? If $C(X)$ were finite dimensional, the answer is trivial by the Heine Borel Theorem. The Arzelà–Ascoli Theorem provides a characterization of compact subsets of $C(X)$ when $C(X)$ is infinite dimensional: In addition to being closed and bounded, compact subsets of $C(X)$ are equicontinuous.

Equicontinuity is the natural generalization of uniform continuity to families of functions:

Definition 167.4(Definition).

Let $X$ be a compact metric space. $\mathcal{F}\subseteq C(X,Y)$ is a family of equicontinuous functions if for every $ϵ>0$ there exists $\delta >0$ such that $d_X(x,y)<\delta ⟹d_Y(f(x),f(y))<ϵ$ for all $x,y\in X$ and for all $f\in \mathcal{F}$.

For example, a family of Lipschitz mappings with the same Lipschitz constant is equicontinuous (more generally, their it suffices for their Lipschitz constants to be bounded above).

Example 167.5(Example).

Let $(X,d)$, $(Y,\rho )$ be metric spaces. Let $\{f_n{\}}_{n=1}^{\infty }\subseteq C(X,Y)$. There exists an equivalent metric on $X$ such that each $f_n$ is Lipschitz continuous.

Proof.

First, note that for a single function$f:X\to Y$, the metric $d^{\prime }(x,y)=d(x,y)+\rho (f(x),f(y))$ on $X$ makes $f$ Lipschitz with $c=1$. We can use the same idea when working with a countably family of functions. First, define the equivalent bounded metric ${\rho }^{\prime }(x,y)=\rho (x,y)/(1+\rho (x,y))$ on $Y$.

Next, define

$$d^{\prime }(x,y)=d(x,y)+\sum _{n=1}^{\infty }\frac{1}{2^n}{\rho }^{\prime }(f_n(x),f_n(y)).$$

Then, ${\rho }^{\prime }(f_n(x),f_n(y))\le 2^n{d}^{\prime }(x,y)$.

You’ll have to show $d^{\prime }$ and $d$ are equivalent.□

Note, however, that this does not make $\{f_n\}$ equicontinuous: $2^n$ blows up as $n\to \infty$.

Example 167.6(Example).

Let $X$ be a compact metric space. Let $F:X\times X\to Z$ be continuous. Let $f_y:=F(x,y)$. Show that $\mathcal{A}:=\{f_y:y\in X\}$ is equicontinuous.

Proof.

Let$ϵ>0$. Since $X\times X$ is compact, $F$ is uniformly continuous. If we consider the metric $d((x_1,y_1),(x_2,y_2))=d(x_1,x_2)+d(y_1,y_2)$ on $X\times X$, there exists $\delta$ such that $d(x_1,x_2)+d(y_1,y_2)<\delta$ $⟹$ $d_Z(F(x_1,y_1),F(x_2,y_2))<ϵ$ for all $x_1,x_2,y_1,y_2\in X$. In particular, for all $y\in X$, $d(x_1,x_2)<\delta$ $⟹$ $d_Z(f_y(x_1),f_y(x_2))<ϵ$.□

Example 167.7(Example).

Let $X\subset {\mathbb{R}}^n$ be convex and open. Let $\mathcal{A}$ be a family of differentiable functions from $X$ to ${\mathbb{R}}^m$. Assume that there exists $M>0$ such that $\Vert Df(x)\Vert \le M$ for all $f\in A$ and for all $x\in X$. Then, by Theorem 187.2, we have

$$\begin{array}{r}|f(x)-f(y)|\le M|x-y|\end{array}$$

for all $f\in \mathcal{A}$ and $x,y\in X$. Thus, $\mathcal{A}$ is equicontinuous.

Theorem 167.8(Arzelà–Ascoli, Kumaresan (2005) 4.4.8).

Let $X$ be a compact metric space. Then a set $\mathcal{F}\subseteq C(X)$ is compact iff $\mathcal{F}$ is bounded¹, closed, and equicontinuous.

Proof.

$(⟹)$ If $\mathcal{F}$ is compact, it is closed. It is also complete and totally bounded. It only remains to show equicontinuity. Let $ϵ>0$. There exist $f_i\in \mathcal{F}$ for $1\le i\le n$ such that $\mathcal{F}\subset {\bigcup }_i{B}_{ϵ}(f_i)$. Let ${\delta }_i$ be chosen by the uniform continuity of $f_i$ for the given $ϵ$. Let $\delta$ be the minimum of the ${\delta }_i$‘s. Then, for $x,y\in X$ such that $d(x,y)<\delta$, for $f\in \mathcal{F}$ and $f^{\prime }=f_i$ such that $f\in B_{ϵ}(f_i)$, we have

$$\begin{array}{rl}|f(x)-f(y)|& \le |f(x)-f^{\prime }(x)|+|f^{\prime }(x)-f^{\prime }(y)|+|f^{\prime }(y)-f(y)|\\ & <ϵ+ϵ+ϵ.\end{array}$$

$(⟸)$ Suppose $\mathcal{F}$ is closed, bounded, and equicontinuous. Since $\mathcal{F}$ is closed, it is complete. It remains to show that $\mathcal{F}$ is totally bounded. Let $ϵ>0$. Let $\delta$ be chosen by the equicontinuity of $\mathcal{F}$. Since $X$ is compact, there exist $x_1,\dots ,x_n\in X$ such that $X\subseteq {\bigcup }_{i=1}^n{B}_{\delta }(x_i)$. Suppose $\Vert f{\Vert }_{\infty }<M$ for all $f\in \mathcal{F}$. Then, we can restrict the codomain of each $f\in \mathcal{F}$ to be $B=\overline{B(0,M)}\subseteq \mathbb{R}$. $B$ is compact in $\mathbb{R}$, so there exist $y_1,\dots ,y_m\in B$ such that $B\subseteq {\bigcup }_{j=1}^m{B}_{ϵ}(y_j)$. Define

$$\mathcal{A}=\{\alpha :\{x_i{\}}_{i=1}^n\to \{y_j{\}}_{j=1}^m\}.$$

$|\mathcal{A}|=m^n$. For $\alpha \in \mathcal{A}$, define

$$U_{\alpha }=\{f\in C(X):|f(x_i)-\alpha (x_i)|<ϵ\forall i=1,\dots ,n\}.$$

We now claim that

$$\mathcal{F}\subseteq \bigcup _{\alpha \in \mathcal{A}}{U}_{\alpha }.$$E1

Indeed, for $f\in \mathcal{F}$, there exists $\alpha$ such that $\alpha (x_i)=y_j$ if $f(x_i)$ lands in $B_{ϵ}(y_j)$, hence $f\in U_{\alpha }$.

It now remains to show that the diameter of the $U_{\alpha }$‘s are bounded by some constant multiple of $ϵ$. Suppose $f,g\in U_{\alpha }$ and $x\in B_{\delta }(x_i)$ for some $i$. Then,

$$\begin{array}{rl}|f(x)-g(x)|\le & |f(x)-f(x_i)|+|f(x_i)-\alpha (x_i)|+\\ & |\alpha (x_i)-g(x_i)|+|g(x_i)-g(i)|\\ <& 4ϵ,\end{array}$$

by the equicontinuity of $\mathcal{F}$ and the definition of $U_{\alpha }$, so $\Vert f-g\Vert \le 4ϵ$. We’re done: for every $U_{\alpha }$, there exists an open ball $O_{\alpha }$ of diameter $5ϵ$ which contains $U_{\alpha }$, and it follows trivially from (E1) that

$$\mathcal{F}\subseteq \bigcup _{\alpha \in \mathcal{A}}{O}_{\alpha }.$$ □

Corollary 167.9(Corollary).

If $C(X)$ is infinite dimensional, then any compact subset of $C(X)$ must be nowhere dense.

All we need to show is that closed balls are not compact in $C(X)$; this can be done by adding triangle functions with increasing slopes to the function at the center of ball, violating equicontinuity.

This holds in general for any infinite dimensional NLS, since closed balls in infinite dimensional normed linear spaces are not compact.

Footnotes

1. As a subset of $C(X)$! ↩

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References

Kumaresan, S. (2005). Topology of Metric Spaces. Alpha Science International Ltd.