Recall the Holder norm $(0<\alpha\leq 1) $ $$\|f\|_\alpha = \max\bigg\{ |f(x)| + \frac{|f(x) - f(y)|}{|x-y|^\alpha} : x,y \in [0,1], x\neq y\bigg\}$$
I want to show that the set $\mathcal F :=\{ \|f\|_\alpha \leq 1 \}$ has a compact closure in $C([0,1])$.
My attempt:
Using Arzela-Ascoli theorem, it suffice to show that $\mathcal F$ is equicontinuous and uniformly bounded which implies every sequence in $\mathcal F$ has a Cauchy subsequence.
Uniform boundedness follows from the fact that $\|f\|_\infty \leq \|f\|_\alpha\leq 1$, then $$|f(x)| \leq 1 \text{ on } [0,1] \text{ for each } f\in \mathcal F.$$
For equicontinuity, let $x_0$ and $\epsilon$ be given, we use the fact that $$\frac{|f(x_0) - f(y)|}{|x_0-y|^\alpha} \leq \|f\|_\alpha \leq 1$$ to get $\delta = \epsilon^{1/\alpha}$. Then, for each $f\in \mathcal F$, we get $$|f(x_0) - f(y)| \leq |x_0 - y|^\alpha \leq (\epsilon^{1/\alpha})^\alpha = \epsilon.$$
is this okay? thank you very much!
