Equicontinuity

Definition (Equicontinuous)

Let $\mathcal{F}$ be a collection of functions on $E\to \mathbb{C}$. This collection is equicontinuous if for any $\epsilon >0,\exists \delta >0$ such that if $|x-y|<\delta$, then $|f(x)-f(y)|<\epsilon$ for all $x,y\in E$ and $f\in \mathcal{F}$.

$$(\forall \epsilon >0)(\exists \delta >0)(\forall x,y\in E)(|x-y|<\delta ⟹|f(x)-f(y)|<\epsilon )$$

Intuitively, for any inputs that are “close” to one another, the outputs are also “close” to one another.

In particular, this guarantees that any $f\in \mathcal{F}$ is uniformly continuous.

Arzela - Ascoli Theorem

We first start by defining an $\epsilon -$web or $\epsilon -$net.

Proposition (Compact Enforces Equicontinuous)

Let $f_n\in \mathcal{C}(K,\mathbb{C})$ with $K$ compact. If $f_n\rightrightarrows f$, then $\{f_n\}$ is equicontinuous.

Proof: By theorem, $f$ is continuous. For $\epsilon >0,\exists N$ such that $|f_n(x)-f(x)|<\epsilon /3$ for any $n>N,\forall x\in K$. Now, since $f_1,f_2,\dots ,f_N,f$ are uniformly continuous, then we need to find some $\delta$ that satisfies all of the functions. So we can examine a few functions.

We know the functions are all uniformly continuous, because they are all continuous functions on a compact metric space. Boundedness is a consequence of Heine-Borel Theorem.

Examining finitely many functions, we can select the minimum of all ${\delta }_i$ for each function $f_i$. So, let $\delta$ be this minimum. Then $\exists \delta >0$ for $f$ such that if $d(x,y)<\delta$ then $|f(x)-f(y)|<\epsilon /3$. So, for $n>N$,

$$|f_n(x)-f_n(y)|<|f_n(x)-f(x)|+|f(x)-f(y)|+|f(y)-f_n(y)|<\epsilon /3+\epsilon /3+\epsilon /3=\epsilon$$

and we are done.

For this proposition, the motivation comes from how we proved that converging sequences are bounded. So, if $a_n\to A$, then $\exists N$ such that if $n>N$/ then $|a_n-A|<\epsilon$. Then $\{a_n\}$ is bounded by $\max |a_1|,|a_2|,\dots ,|a_n|,|A|+\epsilon$.

Proposition (Compact Enforces Total Boundedness)

If $K$ is compact, then $K$ is totally bounded. This means that $\forall \epsilon >0,\exists$ finite $x_1,\dots ,x_n\in K$ such that $\forall y\in K,\exists 1\le i\le n$ such that $d(x_i,y)<\epsilon$. This also means that $y\in {\bigcup }_{i=1}^n{N}_{\epsilon }(x_i)$.

Proof: Let $\epsilon >0$ be arbitrary. We know

$$K\subseteq \bigcup _{x_i\in K}{N}_{\epsilon }(x_i)$$

has a finite subcover (it is a finite union of neighborhoods). Thus $\{x_1,\dots ,x_n\}$ is an $\epsilon -$web.

My professor defined an $\epsilon -$web as just a collection of neighborhoods of $\epsilon$ radius where its union is a compact metric space.

Proposition (Subsets of Compacts)

If $K$ is compact, then $\exists L\subseteq K$ such that

1. $L$ is dense in $K$.
2. $L$ is countable.

Proof: For $\epsilon =1,1/2,1/3,\dots$, pick $\epsilon -$web and union them all together. Since we iterate over $\mathbb{N}$ for $\epsilon$, it is countable. By definition of an $\epsilon -$web, it is dense.

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Let $f_n$ be a sequence of continuous functions on compact set $K$. Then if $\{f_n\}$ is pointwise bounded and equicontinuous, then

• $\{f_n\}$ is uniformly bounded on $K$
• $\exists$ a uniformly convergent subsequence

Proof of Theorem (Arzela - Ascoli Theorem): Part 1: Fix $\epsilon >0$. Since $\{f_n\}$ is equicontinuous, then $\exists \delta >0$ such that if $d(x,y)<\delta$ then

$$|f_n(x)-f_n(y)|<\epsilon /3$$

for all $n\in \mathbb{N}$. Since $K$ is compact, there $\exists$ finite $x_1,\dots ,x_n\in K$ such that $\forall y\in K,\exists 1\le i\le n$ where $d(y,x_i)<\delta$. So, $\forall y\in K$,

$$\begin{array}{rl}|f_n(y)|& \le |f_n(x_i)|+|f_n(x_i)-f_n(y)|\\ & \le \sup |f_n(x_i)|+\epsilon /3\\ & \le M+\epsilon /3\end{array}$$

and so $\{f_n\}$ is uniformly bounded.

Part 2: We know that since $K$ is compact, it has a countable dense subset $L$.

This should be linked to the right problem(s)/theorem(s) in Rudin.

By Theorem (Countable is Pointwise Convergence), there $\exists$ a subsequence that converges on $L$. Denote $f_{n_1}=g_1$. Since $L$ is dense, with the same $\delta$ from Part 1, pick

$$z_1,z_2,\dots ,z_r\in L$$

such that $\forall x\in K,\exists i\in [1,r],d(x,z_i)<\delta$. This gives us a $\delta -$web. The subsequence $\{g_i\}$ converges pointwise on all points in $L$ by Theorem (Countable is Pointwise Convergence). In particular, it converges pointwise on $z_1,\dots ,z_r$ such that

$$(\forall \epsilon >0)(\exists N_i)(\forall n,m>N_i)(|g_n(z_i)-g_m(z_i)|<\epsilon /3)$$

for $1\le i\le r$. Since there are finite points $z_i$, let $N={\max }_{1\le i\le r}{N}_i$. We can repeat the same symbolic argument but with $N$ instead of $N_i$.

Using our $\delta -$web, $\forall y\in K,\exists 1\le i\le r$ such that $d(y,z_i)<\delta$ where

$$\begin{array}{rl}|g_n(y)-g_m(y)|& \le \underset{\text{by equicontinuity}}{\underbrace{|g_n(y)-g_n(z_i)|}}+\underset{\text{by Cauchy}}{\underbrace{|g_n(z_i)-g_m(z_i)|}}+\underset{\text{by equicontinuity}}{\underbrace{|g_m(z_i)-g_m(y)|}}\\ & \le \epsilon /3+\epsilon /3+\epsilon \\ & =\epsilon \end{array}$$

And so by the Cauchy Criterion, $g_n$ uniformly converges.

Uniform boundedness + equicontinuous = uniformly convergent subsequence This is like a compactness condition. This theorem is similar to Theorem (Continuity + Compact = Uniform Continuity).

Theorem (Polynomial Uniform Convergence)

Let $f:[a,b]\to \mathbb{C}$ be continuous. Then $\exists$ polynomials $\{p_n\}$ such that $\{p_n\}\rightrightarrows f$ on $[a,b]$. Furthermore, if $f$ is real-valued, then each $p_n$ has real-valued coefficients.