TITLE: Pointwise Convergence in $L^1$ norm
QUESTION [6 upvotes]: Suppose we have a sequence of functions $\left\{f_n\right\}_{n=1}^\infty\subseteq C^1([0,1])$ and $f_n\to f\in C([0,1])$ in the $L^1$ norm and $f_n'\to g\in C([0,1])$ in $L^1$. Does it follow that $f_n(x)\to f(x)$ for all $x\in[0,1]$?
I'm hoping that it is true, and my thoughts have been directed toward trying to use a proof by contradiction. The slightly shady part here might be some counterexample like $f(x)=x^{1/3}$ which is continuous but has an unbounded derivative.
Counterexamples abound for sequences of functions not satisfying the second condition, but I've neither been able to prove nor find a counterexample for this question. In addition, if it turns out this is true, then I will also have proven $f\in C^1([0,1])$ and $f'=g$.
As a side note, the question originally began as homework, at which point I e-mailed the professor... It turns out I misread the question, so this question is just seeing through to the end my original thought.

REPLY [2 votes]: The answer is YES. In fact $f_n\to f$ uniformly in $[0,1]$.
The functions $f_n'$ are continuous, and $f_n'\to g$, in $C[0,1]$, means that the convergence is uniform,
which implies that for every $x\in [0,1]$
$$
\varphi_n(x)=f_n(x)-f_n(0)=\int_0^xf_n'(t)\,dt\to\int_0^x g(t)\,dt=G(x),
$$
also uniformly in $x$. This imlplies that $\varphi_n\to G$, in the  $L^1$-norm, as the uniform norm is stronger.
But $f_n(0)=f-\varphi_n$, and hence the sequence of constant functions converges in the $L^1$-norm as well, which means that $f_n(0)\to a\in\mathbb R$. Hence
$$
f_n(x)=f_n(0)+\int_0^xf_n'(t)\,dt\to G(x)+a=h(x),
$$
and the convergence is uniform, as it is a sum of a constant sequence and a uniformly convergent sequence of functions. Lhus
$$
\|f_n-h\|_{L^1}\to 0 \quad \text{and}\quad \|f_n-f\|_{L^1}\to 0,
$$
implies that $f=h$. Thus $f_n$ converges to $f$ uniformly in $[0,1]$.