Let $(X,d)$ be a compact metric space. Let $V$ be a closed subspace of $C_{\mathbb{R}}(X)$ such that every $f\in V $ is Lipschitz. Show that $V$ is finite dimensional.
Hint: Show that $A_n=\{f\in V: |f(x)-f(y)| \leq n d(x,y) \}$ has interior for some $n$, and hence show that the closed unit ball of $V$ is equicontinuous.
Since every $f\in V$ is Lipschitz, there exists some $M$ such that $|f(x)-f(y)|\leq M d(x,y)$ for $x,y\in X$.
Since the hint mentions equicontinuity, I'm trying to use Arzela-Ascoli which says $V$ is a compact subspace of $C(X)$ iff $V$ is closed, uniformly bounded and equicontinuous. But I still don't understand why we have to show $A_n=\{f\in V: |f(x)-f(y)| \leq n d(x,y) \}$ has interior for some $n$. And how would Arzela-Asocli give us that $V$ is finite dimensional?
By Baire Category Theorem $A_n$ has interior for some $n$. Suppose $B(f_0,r) \subseteq A_n$. The closed ball of radius $r$ around $f_0$ in $V$ is therefore equi-continuous. It is also bounded in sup norm. Hence,by Arzela-Ascoli Theorem this ball is compact. This implies that the closed unit ball of $V$ is also compact and hence $V$ is finite dimensional.