Apply Arzela-Ascoli theorem to smooth function sequence let $U\subseteq \mathbb{R}^n$ be open, bounded and convex and $(f_i)$ is a sequence in $C^{\infty}(\bar{U})$. Suppose that for each $m\in\mathbb{N}$, there is a constant $C_m>0$ such that for all $i\in\mathbb{N}$, the inequality $$\sup_{\bar{U}}|D^mf_i|\leq C_m$$ holds.
I want to show that $(f_i)$ has uniformly convergent subsequence in $C^{\infty}(\bar{U})$.
My attempt is that by Arzela-Ascoli theorem, $(f_i)$ should have a uniformly convergent subsequence
(call it $(f_{j^\prime})$ and the convengent to $f$). This is what I can do.
But what I can not do yet is that how to prove $f$ is $C^{\infty}$ (although it should be obversly)
and how to prove is subsequence satisfies that $$D^{\alpha}f_{j^\prime}\to D^{\alpha}f$$ uniformly on $\bar{U}$ for every multiindex $\alpha$
 A: we can first pick a subsequence of $(f_n)$ (call it $(f_n^0)$)
such that $f_n^0\rightrightarrows f$ and therefore $\sup_{\bar{U}}|f|\leq C_0$
then sililarly like Problem 2 we can pick a subsequence of $(f_n^0)$ (call it $f_n^1$)\
such that $Df_n^1\rightrightarrows  Df$ and therefore $\sup_{\bar{U}}|Df|\leq C_1$
again we can pick a subsequence of $(f_n^1)$ (call it $(f_n^2)$) 
such that $D^\alpha f_n^2\rightrightarrows D^\alpha f$ where $|\alpha|=2$ and therefore $\sup_{\bar{U}}|D^2f|\leq C_2$ 
and so on
then we get something like a matrix that
$$\begin{matrix}
f_1^0 & f_1^1 & f_1^2 & \ldots\\
f_2^0 & f_2^1 & f_2^2 & \ldots\\
f_3^0 & f_3^1 & f_3^2 &       \\
\vdots & \vdots& &\ddots
\end{matrix}$$
define $g_n=f_n^{n-1}$ which is the diagonal of the matrix\
claim that $(g_n)$ is the subsequence we want\
proof of the claim: 
firstly, since $(g_n)$ is a subsequence of $(f_n^0)$ so
$$g_n\rightrightarrows f$$
for any multiinex $\alpha$ and assume that $|\alpha|=k$
by the construction of $(g_n)$ we have that $$D^\alpha g_n\rightrightarrows D^\alpha f$$
and  $$\sup_{\bar{U}}|D^kg_n|\leq C_k\mbox{ which also leads to }\sup_{\bar{U}}|D^kf|\leq C_k$$
