I am having some difficulties proving this fact, since I am not very familiar with this kind of exercises.
Let $f_n : \mathbb{R} \to \mathbb{R}$ a sequence of $C^\infty$ functions. Suppose $f_n$ converge to some $f$ uniformly on compact intervals and that $f_n ''$ are uniformly bounded. Then $f_n$ converge to $f$ in $C^1$ on compact intervals.
I have to prove that $\underset{x\in [-T,T]}{\sup} |f'_n(x) -f'(x)| \to 0$ as $n\to \infty$, but I don't quite know how to proceed. Thanks in advance for any help.
Let $K$ be a compact subset of $\mathbb{R}$ and $\Vert \cdot \Vert_{\infty}^{K}$ defined by :
$$ \Vert f \Vert_{\infty}^{K} = \sup \limits_{x \in K} \vert f(x) \vert $$
Let $n \in \mathbb{N}$. For all $n$, $\varphi := f_{n}-f$ is twice differentiable, with $\varphi$ and $\varphi''$ bounded on $K$. An inequality due to Landau gives you :
$$ \Vert \varphi' \Vert_{\infty}^{K} \leq 2 \big( \Vert \varphi \Vert_{\infty}^{K} \Vert \varphi'' \Vert_{\infty}^{K} \big)^{\frac{1}{2}} $$
Since $(f_{n}'')_{n}$ are uniformly bounded, there exists a constant $C > 0$ such that :
$$ \Vert f_{n}'-f' \Vert_{\infty}^{K} \leq C \big( \Vert f_{n}-f \Vert_{\infty}^{K} \big)^{\frac{1}{2}} $$
Which gives you that $(f_{n}')_{n}$ converges uniformly to $f'$ on every compact. It proves that $(f_{n})$ converges to $f$ in $\mathcal{C}^{1}$ norm on compact intervals since the norm is : $\displaystyle \Vert f \Vert_{\infty ~ , ~ \mathcal{C}^{1}}^{K} = \sup \limits_{x \in K} \big( \vert f(x) \vert + \vert f'(x) \vert \big)$.