Equicontinuity + Compact $\rightarrow$ Uniform eqcontinuity

equicontinuityreal-analysissequence-of-functionuniform-convergence

My professor has defined equicontinuty and uniform equicontinuity to be two things:

$H\subset C(X)$, $X$ metric space is called:

  1. Equicontinuous if: $\forall \epsilon>0\; \;y\in X, \exists \delta_y>0 \ni \; d(x,y)<\delta_y \rightarrow |f(x)-f(y)|<\epsilon \; \forall f\in H$

  2. Uniform equicontinuous if: $\forall \epsilon>0\; \exists \delta>0 \ni \; \forall x\in X\;\forall f\in H $ if $d(x,y)<\delta_x \rightarrow |f(x)-f(y)|<\epsilon \; \forall f\in H$

I am supposed to prove :

if $X$ is compact and $H$ is equicontinuous, then $H$ is infact uniform equicontinuous

I know that $$\exists N \in N \ni X\subseteq \cup_{i=1}^{N} B_{\delta_i}(x_i)$$

Now. for all these $x_i$, using equicontinuity, I will get $k_i>0$
such that if $d(x,x_i)<k_i \rightarrow |f(x)-f(x_i)|<\epsilon \;\forall f\in H$

What $\delta$ should I choose so that

if $x,y\in X$ arbitrary and $d(x,y)<\delta$

then $|f(x)-f(y)|<\epsilon \; \forall f$

Best Answer

I think it's just easier to lift the usual proof by contradiction (of the fact that continuous functions on compact metric spaces are uniformly continuous).

So, suppose some equicontinuous family $H$were not uniformly equicontinuous, given $X$ compact. Then there exists $\varepsilon>0$ and sequences $(f_n)_{n\in \mathbb{N}}\subseteq H,$ $(x_n)_{n\in \mathbb{N}}\subseteq X,$ $(y_n)_{n\in\mathbb{N}}\subseteq X$ such that $d(x_n,y_n)\leq \frac{1}{n}$ and $|f(x_n)-f(y_n)|\geq \varepsilon$. However, by compactness, $(x_n)_{n\in \mathbb{N}}$ has a cluster point $x$. So, thin to a subsequence such that $x_n\to x$ (and hence, $y_n\to x$). Then, we have $$ |f_n(x_n)-f(y_n)|\leq |f_n(x_n)-f_n(x)|+|f_n(x)-f_n(y_n)|, $$ which goes to $0$ by equicontinuity of $H$ at $x$. This is a contradiction.

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