Arzelà–Ascoli theorem for the space $C_b^k(\overline{\Omega})$

Question

Let $S \subseteq \mathbb{R}^m$ and $C_b^k(S)$, for $k \in \mathbb{N}$, the set of continuous functions from $S$ to $\mathbb{R}$ with bounded and continuous partial derivatives of any order $\leq k$. I need to show that if $\Omega \subseteq \mathbb{R}^m$ is a bounded open set, $\{ u_n\}_n \subseteq C_b^k(\overline{\Omega}) $ is bounded in $C_b^k(\overline{\Omega})$, then there is a subsequence of $\{ u_n\}_n$ which converges in $C_b^k(\overline{\Omega})$ and hence in $H^k(\Omega)$. The norm in $C_b^k(\overline{\Omega})$ is $$||u||_{k, \infty}:= \max_{|\alpha| \leq k}||D^{\alpha}u||_{\infty} $$ There is a hint: use Arzelà–Ascoli theorem. But I don't know how can I use that theorem for that sequence and why convergence in $C_b^k(\overline{\Omega})$ implies convergence in $H^k(\Omega)$. I know that in this case Arzelà–Ascoli theorem implies convergence of a subsequence in $C_b(\overline{\Omega})$, but how can I guarantee convergence in $C_b^k(\overline{\Omega})$?.

Can you help me, please?

Answer 1

The statement is also wrong for very general reasons: $C_b^k(\overline{\Omega})$ is a normed space and if every bounded sequence had a convergent subsequence the unit ball would be compact. But this implies that your space is finite dimensional. This is a rather elementary fact about normed spaces which is, e.g., in Rudin's book.

Answer 2

The statement is false. Take $m=1,\Omega=(0,1),k=1$ and $f_n(x)=\frac {x^{n+1}} {n+1}$. Then $\{f_n\}$ and $\{f_n'\}$ are uniformly bounded on [0,1] but there is no subsequence along which $\{f_n'\}$ converges uniformly on [0,1].