Folland Problem 4.64 relating to Hölder continuity and compactness. 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 $C(X)$.
 A: 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.
