Pointwise convergence of Sobolev functions implies uniform convergence

Question

Let $u_n$ be a sequence in $W^{1,1}((0,1))$, which converges pointwise ae to some $u$, not necessarily Sobolev, and assume that there's $h$ in $L^1$ such $|u_n'(t)|\leqslant h(t)$, for ae $t$ in $(0,1).$ Then $u_n \to u$ uniformely.

Now if I were able to prove that $u'_n$ converges pointwise ae to some function, I'd be able to conclude, by dominated convergence, that $u'_n \to u'$ in $L^1$, and that $u$ is Sobolev, with weak derivative $u'$.

But then Sup$|u(x)-u_n(x)|=|(\int_0^x u'-u_n')+c-c_n|\leqslant \int_0^1 |u'-u'_n |+|c-c_n| $ and both terms converge to $0$.

Unfortunately I'm not able to prove that $u'_n$ converges pointwise.

Answer 1

You can use the weak version of the fundamental theorem of calculus:

Let $I$ be an interval (possibly unbounded) and $u\in W^{1,p}(I)$ with $1 \leq p \leq \infty$. Then there exists a function $\tilde{u}\in C(\overline{I})$ such that $u = \tilde{u}$ a.e. on $I$ and $$\int_y^x u'(t)\,dt=\tilde{u}(x)- \tilde{u}(y),\qquad \forall\ x,y\in \overline{I}.$$

As $(u_n)_n$ is in $W^{1, 1}$, it is defined up to a set of measure zero, we can say that $(u_n)_n \subset C([0,1])$. We already know that $u_n(x) \to u(x)$ for all $x \in (0,1)$. Moreover, we have $$|u_n| = \left|\int u'_n\right| \le \int h = M \in \mathbb R$$ so that the sequence $(u_n)_n$ is uniformly bounded in $C([0,1])$ and by Arzelà-Ascoli theorem, there exists $\tilde u \in C([0,1])$ such that $u_n \to \tilde u$ uniformly (actually it's a subsequence of $(u_n)_n$ but we still denote it $(u_n)_n$). And as we already have the pointwise convergence, $\tilde u = u$.