Let $f: [0, 1] \to \mathbb R$ be a continuous function, and $1 <p \leq \infty$. Suppose $u_n \in W^{1, p}$ are such that $u_n \to f$ uniformly. Is it true that if $f$ fails to be absolutely continuous, then $\u_n\_{W^{1, p}} \to \infty$ for $p > 1$?
1 Answer
If $\u_n\_{1,p}$ does not diverge to $+\infty$, some subsequence $u_{n_j}$ converges weakly$W^{1,p}(I)$ to some $g\in W^{1,p}(I)$, and still in $L^\infty(I)$ to $f$. So e.g. $u_{n_j}$ converges to both $f$ and $g$ in the weak topology of $L^1(I)$, which is Hausdorff, so $f=g$ by uniqueness of the limit. But then $f$ is AC and differentiable a.e.

1$\begingroup$ For $p=1$ the space $W^{1,1}(I)$ is not reflexive but it is a closed subspace of $BV(I)$. So the same argument applies, with now $g\in BV(I)$, therefore differentiable a.e. $\endgroup$ Oct 1 at 14:43