Folland Problem 4.64 relating to Hölder continuity and compactness.

Question

This is problem 4.64 in Folland. Given the section it is placed in I suppose it requires Arzela-Ascoli but I am unable to apply it accurately.

Let $(X,\rho)$ be a metric space. A fucntion $f: X \to \mathbb{R}$ is called Hölder continuous of exponent $\alpha \gt 0$ if

$$N_{\alpha}(f) := \ sup_{x \neq y} \frac{|f(x) - f(y)|}{\rho(x,y)^{\alpha}} \lt \infty.$$

Observe that a Hölder continuous function is uniformly continuous.

Prove that if $X$ is compact, then the set

$$\{f \in C(X) : \|f\|_{\infty} \le 1, N_{\alpha}(f) \le 100 \}$$

is compact in $C(X)$.

Answer 1

Denote by $$ K := \{f \in C(X) : \|f\|_\infty \le 1, N_\alpha(f) \le 100 \} $$ the set in question. Obviously $K$ is bounded in $C(X)$ (namely by $1$), we will show that it is closed and equi-continuous, hence then it is compact by Arzela-Ascoli.

• Closedness: Let $(f_n) \in K^{\mathbf N}$, and $f_n \to f \in C(X)$. Then we have $f_n \to f$ pointwise, and hence $|f(x)|\le 1$ for each $x\in X$. So $\|f\|_\infty \le 1$. For each $x,y \in X$ with $x \ne y$, we have $$ \frac{|f(x) - f(y)|}{\rho(x,y)^\alpha} = \lim_n \frac{|f_n(x) - f_n(y)}{\rho(x,y)^\alpha} \le \limsup_n N_\alpha(f_n) \le 100$$ Taking the supremum gives $$ N_\alpha(f) = \sup_{x\ne y} \frac{|f(x) - f(y)|}{\rho(x,y)^\alpha} \le 100 $$ Hence, $f \in K$ and $K$ is closed.
• Equicontinuity: Let $\epsilon > 0$, choose $\delta = 100^{-1/\alpha}\epsilon^{1/\alpha}$, then for every $x,y\in X$ with $\rho(x,y)< \delta$ and every $f \in K$, we have \begin{align*} |f(x) - f(y)| &\le 100 \rho(x,y)^\alpha\\ &\le 100 \delta^\alpha\\ &= \epsilon \end{align*} Hence $K$ is equicontinuous.

Therefore, $K$ is compact.