Exercise 7.19

Suppose $S$ is uniformly closed, pointwise bounded, and equicontinuous, and let $\{f_n\}$ be a sequence of elements of $S$. By Theorem 7.25, $\{f_n\}$ has a subsequence converging uniformly to $f\in C(K)$, and since $f$ is uniformly closed, we have $f\in S$. Hence by Exercise 2.26, $S$ is compact.

Conversely, suppose that $S$ is compact. By Theorem 2.34 it is closed with respect to the supremum norm, that is, $S$ is uniformly closed. Let $x\in K$ and define the complex-valued function $F_x$ on $C(K)$ by $F_x(f)=f(x)$. $F_x$ is continuous since if $g\to f$, then

Hence $F_x(S)$ is bounded by Theorem 4.15, that is, $S$ is pointwise bounded. Since $S$ is compact, by Theorem 2.37 every infinite subset of $S$ has a limit point in $S$, that is, it has a subsequence which converges uniformly on $K$. If $S$ were not equicontinuous, then there would exist an $𝜀>0$ such that for all positive integers there exists $x_n,y_n\in K$ and $f_n\in S$ such that

Hence $\{f_n\}$ would have no equicontinuous subsequence, so by Theorem 7.24 it would have no subsequence which converges uniformly on $K$, which we have seen contradicts the compactness of $S$.