# Compactness of unit ball in Holder space [duplicate]

I'm trying to prove that the closed unit ball in the Holder space, $$B^\alpha=\{ f\in C^\alpha[a,b]:\|f\|_\alpha\leq1\}$$, is compact in $$(C[a,b],\|\cdot\|_\infty)$$. I know I should use Arzela-Ascoli Theorem and show that the ball is bounded, equicontinuous and closed. For the closedness part of the proof, I take a convergent sequence $$f_n$$ in $$B^\alpha$$ and want to show that it converges to a point $$f$$ in the ball. Since the sequence is convergent, for every $$\epsilon>0$$, there exists an $$N$$ s.t. $$\epsilon>\|f_n-f\|_\infty\geq |f_n(x)-f(x)|\geq |f(x)|-|f_n(x)|$$ for every $$x$$ in the domain and $$N>n$$.

I'm stuck here! How do I proceed?

• Do you know the Arzela-Ascoli theorem? – WoolierThanThou Oct 9 at 21:51
• @WoolierThanThou Yes. I need to show it's bounded, equicontinuous and closed. I'm stuck on the closedness part. – Teodorism Oct 9 at 21:54
• For that, assume $(f_n)_{n\in\mathbb{N}}\subseteq B^{\alpha}$ convergent in $||\cdot||_{\infty}$ with limit $f$. Then for any $x, y$, $|f(x)-f(y)|=\lim_{n\to\infty} |f_n(x)-f_n(y)|\leq |x-y|^{\alpha}$. – WoolierThanThou Oct 9 at 21:57
• @WoolierThanThou How does that prove that f is in the unit ball? – Teodorism Oct 9 at 22:19
• Well, the unit ball is closed in the uniform topology, so that should be a non-issue? Assuming that $||f||_{\alpha}=||f||_{\infty}+C_{\alpha}$, where $C_{\alpha}$ is the smallest possible Hölder constant (the usual Hölder norm to my knowledge). – WoolierThanThou Oct 9 at 22:25

To see that it's closed, note that $$f_n\rightarrow f$$ uniformly implies pointwise convergence, and so for $$x,y\in [a,b]$$ with $$x\neq y,$$ $$\frac{|f(x)-f(y)|}{|x-y|^\alpha}=\lim_{n\rightarrow\infty} \frac{|f_n(x)-f_n(y)|}{|x-y|^\alpha}\leq 1.$$ In particular, $$\|f\|_\alpha\leq 1$$ (note- this also uses the uniform convergence, where the unit ball is closed). So, $$f\in B^\alpha,$$ showing that the set is closed.
The uniform boundedness in $$C([a,b])$$ is obvious, as $$f\in B^\alpha\implies \|f\|_\infty\leq\|f\|_{\alpha}\leq 1,$$ and equicontinuity follows from the Holder assumption. So, as suggested, the Arzela-Ascoli theorem applies.
• We want to know if it's closed in $(C([a,b]),\|\cdot\|_\infty)$, so the convergence has to be with respect to that topology (i.e. uniform convergence). – cmk Oct 9 at 22:35
• Thanks. One other question. How do we know $f$ is Holder continuous? Because it can be in the unit ball but not continuous? – Teodorism Oct 9 at 23:10
• We automatically know that it’s continuous as the uniform limit of continuous functions. You can see Holder continuity from the first estimate I wrote, but you should multiply by $|x-y|^\alpha$ (like in the comment below the original post). – cmk Oct 10 at 0:16