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I'm trying to solve an exercise in Brezis' Functional Analysis, i.e.,

Let $(K, d)$ be a compact metric space and $E:=C(K, \mathbb R)$ endowed with the supremum norm $\| \cdot\|_\infty$. Let $F$ be a closed subspace of $E$. Assume that every $u\in F$ is Hölder continuous, i.e., for each $u \in F$ there is $\alpha \in (0, 1]$ and $L>0$ such that $$ |u(x)-u(y)| \le L d(x, y)^{\alpha} \quad \forall x,y \in K. $$

The purpose of this exercise is to show that $F$ is finite-dimensional.

  1. Prove that there exist $\gamma \in (0, 1]$ and $C>0$ such that $$ |u(x)-u(y)| \le C\|u\|_\infty d(x, y)^{\gamma} \quad \forall x,y \in K. $$
  2. Prove that the closed unit ball $B_F$ of $F$ is compact and conclude.

There are possibly subtle mistakes that I could not recognize in below attempt. Could you please have a check on it?

We have $E$ is complete and $F$ closed in $E$, so $F$ is complete.

1.

For $n \ge 1$, let $$ F_n := \{ u\in F : |u(x)-u(y)| \le n d(x, y)^{1/n} \quad \forall x,y \in K\}. $$

Then $F_n$ is closed for $n \ge 1$ and $\bigcup_{n \ge 1} F_n = F$. By Baire category theorem, there is some $m \in \mathbb N^*$ such that $\operatorname{int}_F F_{m} \neq \emptyset$. Here $\operatorname{int}_F$ is the interior w.r.t. the subspace topology of $F$. Then there is $v \in \operatorname{int}_F F_{m}$ and $\varepsilon>0$ such that $$ B_F (v, \varepsilon) :=\{u \in F : \|u-v\|_\infty < \varepsilon\} \subset F_{m}. $$

For any $u \in F$, we have $\frac{\varepsilon}{2 \|u\|_\infty} u + v \in B_F (v, \varepsilon)$. Then for $u \in F$, $$ \left | \frac{\varepsilon}{2 \|u\|_\infty} (u(x)-u(y)) + (v(x)-v(y)) \right | \le m d(x, y)^{1/m} \quad \forall x,y \in K. $$

By triangle inequality, we have for $u \in F$: $$ \frac{\varepsilon}{2 \|u\|_\infty} |u(x)-u(y)| \le m d(x, y)^{1/m} +|v(x)-v(y)| \quad \forall x,y \in K. $$

Because $v\in F_m$, we have for $u \in F$: $$ \frac{\varepsilon}{2 \|u\|_\infty} |u(x)-u(y)| \le 2m d(x, y)^{1/m} \quad \forall x,y \in K. $$

Then for $u\in F$, $$ |u(x)-u(y)| \le \frac{4m}{\varepsilon} \|u\|_\infty d(x, y)^{1/m} \quad \forall x,y \in K. $$

The claim follows by picking $C:=\frac{4m}{\varepsilon}$ and $\gamma := 1/m$.

2.

It follows from (1) that $B_F$ is uniformly equi-continuous. Then $B_F$ is compact by Arzelà–Ascoli theorem and thus $\dim F<\infty$.

Akira
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  • Your reasoning for the first question seems perfectly sound. The only issue I'm seeing lies in the second question I think. It doesn't suffice to have $B_F$ be (uniformly) equicontinuous, the sets $B_F(x) := {u(x) \mid u \in B_F}$ need to also be relatively compact (since the functions $u$ have their values in $\mathbb{R}$, it suffices to require boundedness of course). In your case that is easily fixed by using the first question again though, so you weren't too far off at all. – Bruno B Jun 25 '23 at 22:12
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    @BrunoB From the linked version of Arzelà–Ascoli theorem, I think it works because $B_F$ is closed, uniformly bounded and uniformly equi-continuous. – Akira Jun 25 '23 at 22:14

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