Arzelà-Ascoli theorem and Hölder spaces 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.
 A: At first, if partial derivatives of order at most $k$ of $f_{n_i}$ converge to those of $f$, than automatically $f\in C^{k,\alpha}(B)$, since $$|(D^k f)(x)-(D^k f)(y)|\leqslant \limsup_i |(D^k f_{n_i})(x)-(D^k f_{n_i})(y)|\leqslant c\cdot |x-y|^\alpha$$
unformly over $x,y\in B$ (here $D^k$ denotes the vector of all partial derivatives of order at most $k$).
Choose closed balls $B_1\subset B_2\subset B_3\ldots$ such that $B=\cup B_i$. It suffices to solve the problem for each $B_i$ separately, then use diagonalization.
For fixed $B_m$ using Arzela -- Ascoli we may suppose that $f_i$ converge to $f$ in $C^k$. Denote $g_i=D^k (f_i-f)$. Then $g_i$ converge to $0$ in $C(B_m)$, and we need to prove that it holds in $C^{\alpha/2}(B_m)$ too. Assume the contrary, then again passing to subsequence we may suppose that $|g_i(x_i)-g_i(y_i)|\geqslant \kappa\cdot |x_i-y_i|^{\alpha/2}$ for fixed $\kappa$ and certain $x_i,y_i\in B_m$. Without loss of generality $x_i\to x_0$, $y_i\to y_0$. Consider two cases.

*

*$x_0\ne y_0$. But then $|g_i(x_0)|+|g_i(y_0)|\geqslant |g_i(x_i)-g_i(y_i)|$, liminf of the last expression is at least $\kappa\cdot |x_0-y_0|^{\alpha/2}>0$, a contradiction.


*$x_0=y_0$. Then $$\|g_i\|_{C^\alpha}\geqslant \frac{|g_i(x_i)-g_i(y_i)|}{|x_i-y_i|^\alpha}\geqslant \kappa\cdot |x_i-y_i|^{-\alpha/2}\to \infty,$$
a contradiction.
A: 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 $N:=\big({k+n-1\atop k}\big)$.
