Are bounded sequences in $C^{1,\alpha}(B_1)$ precompact in $C^{1,\beta}(B_1)$ for $\alpha>\beta$? Are bounded sequences in $C^{1,\alpha}(B_1)$ precompact in $C^{1,\beta}(B_1)$ for $0<\beta<\alpha<1$?
I.e., if $(u_n)_n\subset C^{1,\alpha}(B_1)$ with $||u_n||_{C^{1,\alpha}(B_1)}\leq C$ then there exists $u\in C^{1,\beta}(B_1)$ and a subsequence $(u_{n_k})_k$ such that $u_{n_k}\to u$ in $C^{1,\beta}(B_1)$.
I know this is the case for $\beta=0$, i.e. in $C^1$. I also saw this in wikipedia, but I cant find a proof or a more reliable source. Thanks!
 A: Yes.
First,  it's enough to prove the corresponding result for $\newcommand\L{\text{Lip}}$ $\L_\alpha$ and $\L_\beta$; the result you want then follows by applying this result to $f_n'$.
Say $f_n\in\L_\alpha([0,1])$ and $||f_n||_\alpha\le1$. The sequence $f_n$ is equicontinuous so it has a uniformly convergent subsequence; we make things easier to  type  by saying WLOG $f_n\to f$ uniformly. It's clear that $||f||_\alpha\le1$, so $||g_n||_\alpha\le2$ if we set $g_n=f_n-f$.
Let $\epsilon\in(0,1)$, and choose $N$ so  $$|g_n(t)|\le\epsilon\quad(n>N).$$Suppose $n>N$ and $h>0$. $\newcommand\f[3]{\left|\frac{#1-#2}{#3}\right|}$ If $h<\epsilon$ then $$\f{g_n(t+h)}{g_n(t)}{h^\beta}=h^{\alpha-\beta}\f{g_n(t+h)}{g_n(t)}{h^\alpha}\le||g_n||_\alpha\epsilon^{\alpha-\beta}\le 2\epsilon^{\alpha-\beta}.$$Otoh if $h\ge\epsilon$ then $$\f{g_n(t+h)}{g_n(t)}{h^\beta}\le 2\epsilon^{1-\beta}<2\epsilon^{\alpha-\beta}.$$So $$||g_n||_\beta\le 2\epsilon^{\alpha-\beta}\quad(n>N),$$which says $$||g_n||_\beta\to0,$$since $\alpha-\beta>0$.
