Lower semicontinuity

Let $\Omega\subset\mathbb R^n$ be open and bounded. I consider a sequence $u_k:\Omega\to\mathbb R$ of smooth functions which converges uniformly to a function $u:\Omega\to\mathbb R$. Moreover, the sequence of gradients $\nabla u_k$ is uniformly bounded. Now I would like to show that $$\int_\Omega |\nabla u|~dx\leqslant \liminf_{n\to \infty}\int_{\Omega}|\nabla u_k|~dx.$$ References and counterexamples are very welcome.

EDIT: The integral on the lhs exists since the sequence is uniformly Lipschitz and $u$ has therefore (since it is Lipschitz) a gradient defined almost everywhere.

1 Answer

If the sequence of gradients is uniformly bounded in $L^{1+\epsilon}(\Omega)$, $\epsilon>0$, then $\nabla u_{k'}\rightharpoonup \nabla u$ in $L^{1+\epsilon}(\Omega)$ for a subsequence.

This implies $\nabla u_{k'}\rightharpoonup \nabla u$ in $L^{1}(\Omega)$, and the desired inequality is a consequence of weakly lower semicontinuity of norms.

• Thanks for your answer. In my case $\nabla u_k$ is bounded in $L^\infty$. Is then an analogous reasoning possible with weak* convergence instead of weak convergence? What I mean, do I get a subsequence $\nabla u_{k'}\stackrel{*}{\rightharpoonup}\nabla u$ in $L^\infty$ and can deduce from there the same conclusion? (Even if your answer is already enough of course).
– frog
Jul 24, 2014 at 9:52
• Weak-star convergence in $L^\infty$ implies weak convergence in $L^1$.Then using weak lower semicontinuity of the $L^1$-norm you obtain the same conclusion.
– daw
Jul 24, 2014 at 10:25