0
$\begingroup$

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.

$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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). $\endgroup$
    – frog
    Jul 24, 2014 at 9:52
  • $\begingroup$ 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. $\endgroup$
    – daw
    Jul 24, 2014 at 10:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.