# Showing that the unit ball in $(C^\alpha[a, b], \|\cdot\|_\alpha)$ is compact in $(C[a, b], \|\cdot\|_\infty)$

Let $$(C[a, b], \|\cdot\|_\infty)$$ be the usual Banach space of continuous functions on $$[a, b]$$ and for $$\alpha\in(0,1]$$ and $$f\in C[a, b]$$ define $$[f]_\alpha = \sup_{x,y\in[a,b];x\neq y}\frac{|f(x)-f(y)|}{|x - y|^\alpha}$$ Let $$C^\alpha[a, b]$$ be the set of functions $$f$$ in $$C[a, b]$$ for which $$[f]_\alpha < \infty$$, and endow $$C^\alpha[a, b]$$ with the norm $$\|f\|_\alpha = \|f\|_\infty + [f]_\alpha$$ It is known that $$(C^\alpha[a, b], \|\cdot\|_\alpha)$$ is a Banach space.

I've been asked to show that the unit ball $$B^\alpha := \{f\in C^\alpha[a, b]\ :\ \|f\|_\alpha\le 1\}$$ is compact in $$(C[a, b], \|\cdot\|)$$. Not just precompact, but compact. I've already shown that it's precompact using Arzela-Ascoli, so all that's left is to show that $$B^\alpha$$ is closed in $$(C[a, b], \|\cdot\|_\infty)$$.

Suppose $$f_n\in B^\alpha$$ converges wrt $$\|\cdot\|_\infty$$ to $$f\in C[a, b]$$. We know that $$\|f_n\|_\alpha \le 1$$, and hence $$\|f_n\|_\infty \le \|f_n\|_\alpha \le 1$$. We can use this to show that $$\|f\|_\infty \le 1$$ as well. What can we do to show $$\|f\|_\alpha\le 1$$? This would show that $$B^\alpha$$ contains its $$\|\cdot\|_\infty$$-limit points, and hence is closed.

• It is late for me but $𝐵^\alpha$ is not a subspace, is it? – hal4math Oct 9 at 2:13
• @hal4math Ahhhhhh of course. What a silly error on my part. Thanks! – user3002473 Oct 9 at 2:17
• You are welcome :). Though, I still have my doubts that this statement is true. If you find a proof I would love to see a sketch of it. – hal4math Oct 9 at 2:18

Take a sequence $$(f_n)$$ in $$B^\alpha$$ that converges uniformly to some continuous $$f$$. Take $$x,y,z\in [a,b]$$ with $$x\ne y$$. Then $$|f(z)|\le |f(z)-f_n(z)| + |f_n(z)|$$ and $$\frac{|f(x)-f(y)|}{|x-y|^\alpha} \le\frac{|f(x)-f_n(x)|}{|x-y|^\alpha} +\frac{|f_n(x)-f_n(y)|}{|x-y|^\alpha} +\frac{|f_n(y)-f(y)|}{|x-y|^\alpha}.$$ Take $$\epsilon>0$$. Due to uniform convergence, there is $$N=N(x,y,z)$$ such that $$|f(z)-f_n(z)| + \frac{|f(x)-f_n(x)|}{|x-y|^\alpha} + \frac{|f_n(y)-f(y)|}{|x-y|^\alpha}\le \epsilon$$ for all $$n>N$$. This implies for $$n>N$$ $$|f(z)| + \frac{|f(x)-f(y)|}{|x-y|^\alpha} \le |f_n(z)| +\frac{|f_n(x)-f_n(y)|}{|x-y|^\alpha} + \epsilon \le \|f_n\|_\alpha + \epsilon\le 1 + \epsilon.$$ Taking the supremum of the left-hand side over $$z$$, $$x\ne y$$, yields $$\|f\|_\alpha \le 1+\epsilon.$$ Now $$\epsilon>0$$ was arbitrary, hence $$\|f\|_\alpha \le 1$$.

• Thanks for grinding out the details! This was actually far more contrived than I was expecting to be, I was looking for some cheeky functional analysis argument using fancy weaponry like the Bounded Inverse Theorem, but sometimes you've just gotta use the triangle inequality a lot. – user3002473 Oct 10 at 0:37

Daw already gave a good answer, but I want to highlight the lower semicontinuity of the Hölder norm as crucial ingredient, because it can be useful in many different situations.

Let $$(f_n)$$ be a sequence converging uniformly to $$f$$ and $$x,y\in [a,b]$$ with $$a\neq b$$. Then $$\frac{|f(x)-f(y)|}{|x-y|^\alpha}=\lim_{n\to\infty}\frac{f_n(x)-f_n(y)|}{|x-y|^\alpha}\leq \liminf_{n\to\infty}\sup_{x'\neq y'}\frac{|f_n(x')-f_n(y')|}{|x'-y'|^\alpha}=\liminf_{n\to\infty}\,[f_n]_\alpha.$$ Taking the supremum over $$x,y$$, we get $$[f]_\alpha\leq \liminf_{n\to\infty} [f_n]_\alpha$$. Hence the map $$C([a,b])\to[0,\infty],\,f\mapsto [f]_\alpha$$ is lower semicontinuous. Clearly, $$\|\cdot\|_\infty$$ is continuous. Thus $$\|\cdot\|_\alpha=\|\cdot\|_\infty+[\cdot]_\alpha$$ is lower semicontinuous. Therefore the sublevel sets (such as $$B_\alpha$$ are closed).

Note that we only used pointwise convergence in the first part and one can see by a similar argument that $$\|\cdot\|_\infty$$ is also lower semicontinuous with respect to pointwise convergence. Hence $$B_\alpha$$ is even closed with respect to pointwise convergence.

• Interesting detail, thanks for pointing it out, quite counterintuitive that $B_\alpha$ is even closed with respect to pointwise convergence! – user3002473 Oct 10 at 0:35
• In the big inequality, some indices $n$ are missing – daw Oct 10 at 6:05
• @daw Thank you, I fixed them. – MaoWao Oct 10 at 7:01