Suggested Equicontinuity Problem

Question

My professor has suggested the following equicontinuity problem:

Let $X$ be a compact metric space and $C$ be a subset of $C(X)$. If $C$ is compact, then $C$ is closed and bounded (in the metric space $C(X)$). Show that $C$ is also equicontinuous.

Thanks for any help in advance.

Answer 1

Note that you only have to prove that $C$ is bounded if it is compact since in every metric space, compact sets are closed. Also, I guess that you have provided $C(X)$ with the uniform norm.

Assume, to get a contradiction, that $C$ is not bounded. Then there's a sequence of functions $(f_n)$ such that $\Vert f_n\Vert_\infty \geq n$. Since the sequence is infinite and $C$ is compact, it has limit point, say $f$. (Wich we may assume to be a limit taking a subsequence if necessary). Note that there exist an $N$ such that $\Vert f-f_n\Vert _\infty <1$ for $n\geq N$ and this implies that $ \Vert f_n\Vert_\infty -1<\Vert f\Vert_\infty$. Then, $f$ i s not bounded and this is a to contradiction to the fact that $f\in C(X)$ since if were continuous, it will be bounded since $X$ is compact.

Now, as you have commented in a comment, since every function in $C$ it is uniformly continuous. Using this, given $\varepsilon>0$, consider the open balls $B_f= \{g\in C: \Vert g-f\Vert_\infty<\varepsilon/4\}$ an the open cover $U=\{B_f:f\in C\}$. Since $C$ is compact there exist a finite set of functions $f_1, \ldots, f_n$ such that $\{B_{f_1}, \ldots, B_{f_n}\}$ is a cover of $C$. Since each of this functions is uniformly continuous on $X$ to each $f_n$ there's associated a $\delta_n$ such that $$ d(f_n(x),f_n(y ))<\varepsilon/4 \quad \mathrm{if}\quad d(x,y)<\delta_n $$ Take $\delta=\min\limits_{1\leq k\leq n}(\delta_k)$. Can you show that this $\delta$ works for the definition of equicontinuity?

Answer 2

What I wrote in my comment, in more detail: If $C$ is not equicontinuous at $x$, there is an $\varepsilon>0$ such that for every $\delta>0$ there is some $y$ with $d(x,y)<\delta$ and a function $f$ with $|f(y)-f(x)|\ge\varepsilon$. Pick a sequence $\delta_n\to0$ with corresponding $y_n$, $f_n$, and check to see if $(f_n)$ can have a convergent subsequence.