If all derivative are uniformly bounded, there is a subsequence that converges uniformly to an infinitely differentiable function

Let $\{f_n\}, n \in\mathbb N$, be a sequence of infinitely differentiable functions (smooth function) on $[a,b]$ such that for all integer $k \ge 0$, there exist a real $M_k$ such that $|f_n^{(k)} (x)| \le M_k$ for all $x \in [a,b]$. Show that there exist a subsequence that converges uniformly with all it's derivatives to a infinitely differentiable function.

Since $\{f_n\}$ is bounded I think I could use the Arzela-Ascoli Theorem to show that $\{f_n\}$ has a subsequence that converges uniformly but I'm not sure how to prove equicontinuity.

• This is a standard diagonal method: Using the bound on $|f_n|$ and $|f_n'|$, show that a subsequence $f_{n_k}$ converges uniformly to some $f$. (by Ascoli Arzela), for this subsequence, using the bound on $f_n''$, there is a further subsequence which converges to $f$ in $C^1$ (i.e. first derivative converges)...... – user99914 Dec 4 '15 at 8:57
• You need to somehow use the fact that your derivatives are bounded. You can try with the mean value theorem... – Miguel Dec 4 '15 at 9:01

The family is uniformly bounded: $|f_n|\le M_0$. It is also equicontinuous, thanks to $|f'_n|\le M_1$: $$|f_n(x)-f_n(y)|\le M_1|x-y|$$ Therefore, there is a subsequence $\{f_{n_k}\}$ converging to a continuous function $f$.
Next step, consider $\{f_{n_k}'\}$ which is also a uniformly bounded equicontinuous family. Extract a further subsequence $f'_{n_{k_l}}$ which converges uniformly. Since $f_{n_{k_l}}\to f$ and $f'_{n_{k_l}}$ converge uniformly, it follows that $f$ is differentiable and $f'_{n_{k_l}}\to f'$.
This process of choosing subsequences continues indefinitely, demonstrating that $f$ is infinitely differentiable. It remains to apply the diagonal argument: let $F_n$ be the $n$th term of the $n$th subsequence. Then $F_n$ converge, with all its derivatives, to the corresponding derivatives of $f$.