Weak convergence in $L^p [0,1]$ Let $p>1$ and let $f_n$ be a sequence of measurable functions so that $||f_n||_p \leq B$ for every $n$, where the integral is over $[0,1]$. Let $F_n(x) = \int_0^x f_n(t)dt$. I managed to show using a compactness argument that a subsequence $F_{n_j}$converges to a bounded variation function $F$ almost everywhere on $[0,1]$, and that $F$ is absolutely continuous. I need to show $F'$ is in $L^p$. How can I show this?
 A: Let $q$ the conjugate exponent to $p$. Then for $0 \leqslant x < y \leqslant 1$ we have
$$\lvert F_n(y) - F_n(x)\rvert \leqslant \int_x^y \lvert f_n(t)\rvert\,dt = \int_0^1 \lvert f_n(t)\cdot \chi_{[x,y]}(t)\rvert\,dt \leqslant \lVert f_n\rVert_p\lVert\chi_{[x,y]}\rVert_q \leqslant B\cdot (y-x)^{1/q}$$
by Hölder's inequality. So $\{ F_n : n \in \mathbb{N}\}$ is a uniformly equicontinuous family. Since $F_n(0) = 0$ for all $n$, the family is also uniformly bounded. By the Ascoli-Arzelà theorem, there is a uniformly convergent subsequence $(F_{n_k})_{k\in \mathbb{N}}$. Let $F$ be its limit.
For $p = \infty$, we have the Lipschitz continuity of $F$ from the pointwise convergence and the uniform Lipschitz bound on the $F_n$,
$$\lvert F(y) - F(x)\rvert = \lim_{k\to\infty} \lvert F_{n_k}(y) - F_{n_k}(x)\rvert \leqslant B\cdot \lvert y-x\rvert,$$
and hence $\lvert F'(t)\rvert \leqslant B$ at all points where $F'$ exists, that is, $F' \in L^{\infty}([0,1])$.
For $1 < p < \infty$, by the reflexivity of $L^p([0,1])$ there is a weakly convergent subsequence of $(f_{n_k})_{k\in \mathbb{N}}$. By extracting a further subsequence we may assume that $(f_{n_k})$ converges weakly to $f\in L^p([0,1])$. Since $\chi_{[x,y]} \in L^q([0,1])$ for $0 \leqslant x < y \leqslant 1$, it follows that
$$F(y) - F(x) = \lim_{k\to \infty} F_{n_k}(y) - F_{n_k}(x) = \lim_{k\to \infty} \int_0^1 f_n(t)\chi_{[x,y]}(t)\,dt = \int_0^1 f(t)\chi_{[x,y]}(t)\,dt,$$
i.e.
$$F(y) - F(x) = \int_x^y f(t)\,dt,$$
so we have $F'(t) = f(t)$ almost everywhere, whence $F' \in L^p([0,1])$.
