Take a sequence $f^{n}$ in $C^{2}([0,1])$, the space of twice continuously differentiable functions, such that
- $f^{n} \rightarrow f$ in $L^{p}([0,1])$ (the Lebesgue space) for a $f \in L^{p}([0,1])$
- $\sup_{n \in \mathbb N} |f^{n}|_{\infty}+|\partial_{x}f^{n}|_{\infty}+|\partial^{2}_{xx}f^{n}|_{\infty} < \infty$
Then we know that $\{ f^{n} \}_{n}$ and $\{ \partial f^{n} \}_{n}$ are equicontinuous. Hence by Aszela-Ascoli, a subsequence converges in $C^{1}$ and therefore $f \in C^{1}$.
Question
Does the sequence $f^{n}$ also converge in $C^{1}$ to $f$?
Yes, it does. Consider a subsequence $(f^{k_n})$, then Arzela-Ascoli shows that there exists a further subsequence $(f^{k_{l_n}})$ that converges in $C^1$ to some $g$. Since $(f^{k_{l_n}})$ converges to $f$ in $L^p$, a subsequence of it converges to $f$ almost everywhere, therefore $f=g$ almost everywhere, which implies that $(f^{k_{l_n}})$ converges to $f$ in $C^1$.
This argument shows that any subsequence of $(f^n)$ contains a further subsequence that converges to $f$ in $C^1$. Show now that this is enough to deduce that $(f^n)$ converges to $f$ in $C^1$.