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This is a remark from Brezis' book, ch. 9 (pag. 264).

Let $\Omega \subset \mathbb R^N$ an open set and $(u_n)_n$ a sequence in $W^{1,p}(\Omega)$ such that $u_n \to u$ in $L^p(\Omega)$. If $\nabla u_n$ converges to some limit in $L^p(\Omega)^N$ then $u_n \to u$ in $W^{1,p}$. If $p \in [1,+\infty)$ it suffices to know only that $u_n \to u$ in $L^p(\Omega)$ and that $(\nabla u_n)$ is bounded in $L^p(\Omega)^N$, i.e. there exists $C>0$ s. t. $$ \Vert \nabla u_n \Vert_{L^p} \le C $$ in order to conclude that $u_n \to u$ in $W^{1,p}$.

Why is this true? I have thought to Banach Alaoglu theorem: since $p\ne 1$ I can consider on $L^p$ the weak-$\star$ topology; then, due to compactness of the (unit) ball, from the bounded sequence $(\nabla u_n)$ I can extract a convergent subsequence. But I'm not sure of what I am doing and I need your kind help.

Thanks in advance.

Romeo
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2 Answers2

5

This remark seems to be buggy.

Take any sequence $u_n$ which converges towards $u$ weakly in $W^{1,p}(\Omega)$. By Rellich's theorem, the convergence is strong in $L^p(\Omega)$. By the principle of uniform boundedness, we know that $u_n$ is bounded in $W^{1,p}(\Omega)$, hence, $\nabla u_n$ is bounded in $L^p(\Omega)^N$. Now, the remark implies that $u_n$ converges strongly in $W^{1,p}(\Omega)$. This is certainly not true.

gerw
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4

The first thing to note here is that Brezis does not conclude $u_n\to u$ in $W^{1,p}(\Omega)$. In fact, the only thing that he concludes is $u\in W^{1,p}(\Omega)$. Let me state like him:

Let $1<p\leq\infty$ and suppose that $u_n\to u$ in $L^p(\Omega)$ and $\nabla u_n$ is bounded in $L^p(\Omega)^N$. Then $u\in W^{1,p}(\Omega)$.

To prove this, we can assume without loss of generality that there exist $v\in L^p(\Omega)^N$ such that $\nabla u_n\to v$ weakly in $L^p(\Omega)^N$, but then we can conclude that $v=\nabla u$. I have proved the last assertion here in somewhere, let me try to find it and link here, or if you want, you can take a look on Theorem 1.30 of Heinonen.

Remark: I found the proof.

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Tomás
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