TITLE: Arzelà-Ascoli theorem and Hölder spaces
QUESTION [5 upvotes]: Let $B\subset \mathbb{R}^n$ be a open ball. Let $\{f_i\}$ be a sequence of functions bounded in the Hölder norm $C^{k,\alpha}(B)$ for a given integer $k\geq 0$ and $\alpha\in (0,1)$.
Does there exist a subsequence which converges to a function $f$ (necessarily $f\in C^{k,\alpha}(B)$) in the norm $C^{k,\alpha/2}(\bar B')$ for any closed ball $\bar B'\subset B$?
A reference would be very helpful.

REPLY [12 votes]: For completeness, let's mention a simpler and more general statement:  For $\Omega\subset\mathbb{R}^n$ a bounded open set, $k\in\mathbb{N}$ and $0<\beta<\alpha\le1$ there is a compact embedding
$$ C^{k,\alpha}(\Omega) \to  C^{k,\beta}(\Omega) . $$
Some details:
1. For a compact metric space $(E,d)$, and for $0<\beta<\alpha\le1$ we have a compact embedding of the space of the $\alpha$-Hölder functions into the space of $\beta$-Hölder functions:
$$\big( C^\alpha(E),\|\cdot\|_{\alpha,E}\big)\to\big( C^\beta(E),\|\cdot\|_{\beta,E}\big).$$
Here   $\|u\|_{\alpha,E}:= \|u \|_\infty+|u|_{\alpha,E}$  and
$$|u|_{\alpha,E}:=\sup_{x\neq y\in E} \frac{|u(x)-u(y)|}{d(x,y)^\alpha} .  $$
Indeed, let $(u_k)_{k\in\mathbb{N}}\subset C^\alpha(E)$ be a $\|\cdot\|_{\alpha,E}$-bounded sequence, that is, it is uniformly bounded and equicontinuous w.r.to a common modulus of continuity $Ct^\alpha$. By  Ascoli-Arzelà, some subsequence  $(u_{k_j})$ converges uniformly to some $u$ with the same modulus of continuity, so that  $u\in C^\alpha(E)$. We may assume w.l.o.g. that $u$ is the null function (for we just replace  $(u_{k_j} )$ with $(u_{k_j}-u)$). The thesis then follows since for $j\to\infty$ we have  $\|u_{k_j}\|_\infty= o(1)$ and
$$\left|\frac{u_{k_j}(x)-u_{k_j}(y)}{d(x,y)^\beta}\right|=\left| \frac{u_{k_j}(x)-u_{k_j}(y)}{d(x,y)^\alpha}\right|^{\frac{\beta}{\alpha}} 
 \left|u_{k_j}(x)-u_{k_j}(y)\right|^{1-\frac{\beta}{\alpha}} $$whence also
$$ |u_{k_j}|_\beta\le |u_{k_j}|_\alpha^{\frac{\beta}{\alpha}}\left(2\|u_{k_j}\|_\infty \right)^{1-\frac{\beta}{\alpha}}=o(1).$$
2. The same compact embedding holds true if $(E,d)$ is only assumed totally bounded: its completion $(\tilde E,\tilde d)$ is compact, and the map "extension by density of uniformly continuous functions" gives an isometry (whose inverse map is the restriction to $E$)
$$C^\alpha(E)\to C^\alpha(\tilde E).$$
3. For $\Omega\subset\mathbb{R}^n$ a bounded open set, $k\in\mathbb{N}$ and $0<\beta<\alpha\le1$ the analogous compact embedding
$$\big( C^{k,\alpha}(\Omega),\|\cdot\|_{k,\alpha}\big)\to\big( C^{k,\beta}(\Omega),\|\cdot\|_{k,\beta}\big) $$
follows from the case $k=0$, because of the usual closed-range embedding
$$  C^{k,\alpha}(\Omega) \to C^{0,\alpha}(\Omega)^N$$
given by $u\mapsto \big( \partial^\nu u \big)_{\nu\in\mathbb{N^n},|\nu|\le k}$, for $N:=\big({k+n-1\atop k}\big)$.