Is  $C^{k+1}(X)$ compactly contained in $C^{k}(X)$  for a closed manifold $X$? Let $X$ be a closed manifold, let $k$ be a nonnegative integer and let $C^k(X)$ denote the space of $k$-times continuously differentiable functions equipped with the C$^k$ norm.
Is $C^{k+1}(X)$ compactly contained in $C^k(X)$? Does this follow from Arzela-Ascoli?
Thanks.
 A: Yes.
Choose a sequence $f_n$ bounded in $C^{k+1}(X)$-norm.  First let us assume all  $f_n$ are supported in a compact subset of a coordinate neighborhood $U$, with coordinates $x_1, \dots, x_d$.   Since the $f_n$ are bounded in $C^{k+1}$, there is a uniform bound on their partial derivatives up to  order $k+1$.  It then follows from the mean value theorem that for each $i_1, \dots, i_m$, $m \le k$, the set $\{\partial_{i_1} \dots \partial_{i_m} f_n : n \ge 1\}$ is equicontinuous.  So by Arzelá-Ascoli it has a uniformly convergent subsequence.  Indeed, passing to subsequences finitely many times, we can find a single subsequence $f_{n_j}$ so that $\partial_{i_1} \dots \partial_{i_m} f_{n_j}$ converges uniformly for every $i_1, \dots, i_m$.  That is, the subsequence converges in $C^k(X)$.
To remove the assumption on the support of $f_n$, choose a smooth partition of unity $\psi_1, \dots, \psi_N$ such that each $\psi_i$ is compactly supported inside some coordinate neighborhood.  By passing to subsequences $N$ times, we can find a single subsequence $f_{n'_j}$ such that $f_{n'_j} \psi_i$ converges in $C^k(X)$ for each $i$, which means the sum $f_{n'_j} = f_{n'_j} \psi_1 + \dots + f_{n'_j} \psi_N$ converges in $C^k(X)$ as well.
