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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.

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    $\begingroup$ It is late for me but $𝐵^\alpha$ is not a subspace, is it? $\endgroup$ – hal4math Oct 9 at 2:13
  • $\begingroup$ @hal4math Ahhhhhh of course. What a silly error on my part. Thanks! $\endgroup$ – user3002473 Oct 9 at 2:17
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    $\begingroup$ 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. $\endgroup$ – hal4math Oct 9 at 2:18
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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 $\|f\|_\alpha \le 1$.

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  • $\begingroup$ 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. $\endgroup$ – user3002473 Oct 10 at 0:37
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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.

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

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