4
$\begingroup$

Let $u_\epsilon \in W^{1,p}(\Omega)$ be a sequence with $\Omega \subset \mathbb{R}^n$ bounded. Let $u_\epsilon \rightharpoonup^* u$ weakly* in $L^\infty (\Omega)$ (for a subsequence) with $u \in W^{1,p}(\Omega)$ and assume $\sup_\epsilon \vert \nabla u_\epsilon\vert_{L^p (\Omega; \mathbb{R}^n)} < C$.

Proof that $\nabla u_\epsilon \rightharpoonup \nabla u$ weakly in $L^p (\Omega; \mathbb{R}^n)$ (for a subsequence).


My idea so far:

$\nabla u_\epsilon$ is bounded, therefore there exists a weak limit (for a subsequence).

$u_\epsilon \rightharpoonup^* u$ weakly* in $L^\infty (\Omega)$, therefore $u_\epsilon \rightharpoonup u$ weakly in $L^p (\Omega)$, which means that $~\int_\Omega \phi \partial_x u_\epsilon \to \int_\Omega \phi \partial_x u ~$ for all $~\phi \in W^{1,p^*}(\Omega)$.

If I am correct, the weak convergence in $L^p (\Omega; \mathbb{R}^n)~$ is $~\int_\Omega \Phi \cdot \nabla u_\epsilon \to \int_\Omega \Phi \nabla u~$ for all $~\Phi \in L^{p^*} (\Omega; \mathbb{R}^n)$, but I can't seem to get that from the lines above.

Any help is appreciated.

$\endgroup$

1 Answer 1

2
$\begingroup$

You are on the right track. $u_\epsilon\rightharpoonup u$ in $L^p(\Omega)$ as you noted. Since the sequence is bounded in $W^{1,p}(\Omega)$, there is a subsequence (denoted by $u_{\epsilon'}$) such that $$ u_{\epsilon'} \rightharpoonup \tilde u \quad \textrm{ in }W^{1,p}(\Omega). $$ Since $W^{1,p}(\Omega)$ is continuously embedded into $L^p(\Omega)$ it follows that both weak limits are the same, $u=\tilde u$. Hence $$ u_{\epsilon'} \rightharpoonup u\quad \textrm{ in }W^{1,p}(\Omega). $$ The gradient mapping $\nabla: W^{1,p}(\Omega)\to L^p(\Omega)$ is linear and continuous, thus weakly continuous, implying $$ \nabla u_{\epsilon'} \rightharpoonup \nabla u\quad \textrm{ in }L^{p}(\Omega). $$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .