Let $\{f_n\}$ be a sequence of real-valued differentiable functions on $[0,1]$ whose derivatives are continuous. Further, suppose $\{f_n\}$ converges pointwise to a function $f$ defined on $[0,1]$ and the sequence of the derivatives $\{f'_n\}$ is uniformly bounded and equicontinuous. Show that $f$ is differntiable on $[0,1]$ and $\{f_n\}$ converges uniformaly to $f$.
I am aware of the similar statement that assumes the uniform convergence of $f'_n$ to some function $g$ and gives the same results here. But this statement here is quite different. In particular, I cannot apply the fundamental theorem of calculus since $f'_n$ is not assumed to be convergent. I also tried applying Arzela-Ascoli on $\{f'_n\}$ but that proved the uniform convergence of a subsequence. Any suggestions?
By Arzela-Ascoli Theorem there is a subsequence $(f_{n_k}')$ converging uniformly to some function. By the arrgument you are familiar with it follows that $f$ is differentiable and $(f_{n_k}') \to f'$ uniformly.
Suppose $f_n$ does not converge uniformly to $f$. There exist $\epsilon >0$, points $x_i \in [0,1]$ and a subsequence $(f_{m_i})$ such that $|f_{m_i}(x_i)-f(x_i)| >\epsilon$ for all $i$. Now repeat the previous argument with $(f_n)$ replaced by $(f_{m_i})$. You will get a subsequence along which $f_{m_i}' $ tends to $f'$ uniformly. But then $f_{m_i}(x_i)-f(x_i)=f_{m_i}(0)-f(0)+\int_0^{x} [f_{m_i}'(t)-f'(t)] dt \to 0$ along that subsequence leading to a contradiction.
[ For differentiability of $f$ here is the argument: $f_{n_k}(x)-f_{n_k}(0)=\int_0^{x} f_{n_k}'(t)dt \to \int_0^{x} h(t)dt $ where $h$ is the uniform limit of $f_{n_k}'$. Since $h$ is continuous and $f(x)-f(0)=\int_0^{x} h(t)dt$ it follows that $f$ is differentiable].