0
$\begingroup$

Let $p>1$ and $\Omega$ be an open bounded domain in $\mathbb{R}^N$. Take $(u_n)\subset W_0^{1, p}(\Omega)\cap L^{\infty}(\Omega)$ and $u\in W_0^{1, p}(\Omega)\cap L^{\infty}(\Omega)$.

If $$u_n\longrightarrow u \quad\mbox{ in } W_0^{1, p}(\Omega),$$ it it true that $u_n\longrightarrow u$ in the classical sense (up to subsequences)?

Could anyone please help me to understand?

Thank you in advance!

$\endgroup$
4
  • 2
    $\begingroup$ What do you mean by the "classical" sense? $\endgroup$ Commented Nov 7, 2020 at 13:07
  • $\begingroup$ Pointwise convergence or, at worst, almost everywhere convergence. $\endgroup$
    – C. Bishop
    Commented Nov 7, 2020 at 13:12
  • $\begingroup$ Any thoughts on why you expect this to be true? Can you fix the $n$ notation clash? Have you tried anything? Why did you restrict $p>1$? $\endgroup$ Commented Nov 7, 2020 at 13:37
  • $\begingroup$ @CalvinKhor thanks for the comment, I edited the question fixing the right $n$. So, it is a small part of a bigger problem in which I need $p>1$ and it justifies mi choice. I didn't try anything because I don't know where to start. If you please give me a hint, I will try. $\endgroup$
    – C. Bishop
    Commented Nov 7, 2020 at 13:43

1 Answer 1

1
$\begingroup$

If $u_n\rightarrow u$ in $W^{1,p}_0,$ then (in particular) it converges in $L^p$. If $u_n\rightarrow u$ in $L^p$, then there exists a subsequence $(u_{n_j})$ of $(u_n)$ with the property that $u_{n_j}\rightarrow u$ pointwise almost everywhere. The proof of the latter fact comes in when demonstrating that $L^p$ is complete.

$\endgroup$

You must log in to answer this question.

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