Pointwise Convergence in $L^1$ norm 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.
 A: Yes, if $f_n \xrightarrow{L^1} f$ and $f_n' \xrightarrow{L^1} g$, then the $f_n$ also converge pointwise to $f$.
Suppose there were a point $x_0\in [0,1]$ such that $f_n(x_0) \not\to f(x_0)$. Then we can extract a subsequence with $\lvert f_{n_k}(x_0) - f(x_0)\rvert \geqslant \varepsilon$ for all $k$ and some $\varepsilon > 0$. Without loss of generality, let that subsequence be the full sequence.
Every convergent sequence in $L^p$ has a subsequence that converges pointwise almost everywhere to the limit function. Without loss of generality, let that again be the full sequence. Then we know there is an $x_1\in [0,1]$ with $f_n(x_1) \to f(x_1)$. But now, for every $x\in [0,1]$ and $n\in\mathbb{N}$ we have
$$f_n(x) - f_n(x_1) = \int_{x_1}^x f_n'(t)\,dt \to \int_{x_1}^x g(t)\,dt,$$
and hence
$$f_n(x) \to f(x_1) + \int_{x_1}^x g(t)\,dt =: F(x).$$
So $f_n$ converges pointwise to the continuous function $F$, and pointwise almost everywhere to the continuous function $f$. Two continuous functions equal almost everywhere are equal everywhere, thus $f \equiv F$.
A: Since $g$ is integrable and $\lVert f'_n-g\rVert_1\to 0$, the family $\{f'_n,n\geqslant 1\}$ is uniformly integrable. Using Arzelà-Ascoli's theorem, we obtain relative compactness of $\{f_n,n\geqslant 1\}$ in $C[0,1]$ endowed with the uniform norm.
Notice that each subsequence has to converge to $f$. Indeed, let $(f_{n_k})_{k\geqslant 1}$ be a subsequence of $(f_n)_{n\geqslant 1}$ with converges uniformly on $[0,1]$ to some $h$. We extract a further subsequence converging almost everywhere to $f$. We thus have $h=f$ almost everywhere, hence everywhere by continuity.
We thus obtained $\sup_{x\in [0,1]}|f_n(x)-f(x)|\to 0$ as $n\to+\infty$.
A: 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 $[0,1]$.
