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.



Arzelà-Ascoli theorem and Hölder spaces
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.


1. $x\_0\ne y\_0$. 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.
2. $x\_0=y\_0$. 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.


