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Let $\Omega \subset \mathbb{R}$ be a bounded open set with $C%1$ boundary and $H^1(\Omega) = W^{1,2}(\Omega)$ be the Hilbert-Sobolev space. Let ${u_n}$ be a sequence of functions which are uniformly bounded in $H^1(\Omega)$. If I know that $\|u - u_n\|_{L^2(\Omega)} \rightarrow 0$ how do I show that $u \in H^1(\Omega)$ and that \begin{equation} \|u\|_{H^1} \leq \liminf_{n \rightarrow \infty} \|u_n\|_{H^1} \end{equation} I don't know much about Sobolev spaces apart from some readings on the Embedding Theorems so I don't really know how to begin.

So is the main point of this problem is to show that $\|D_{x_i}u\|_{L^2(\Omega)} < \infty$ for all coordinate variables $x_i$ (so that $\|u\|_{H^1} < \infty$ hence $u \in H^1(\Omega)$)? I don't see why the inequality I'm required to show should be true either.

Please guide me.

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The following outlines the necessary steps, but some details are left out:

First choose a subsequence $(u_{n_k})_k$ with $\Vert u_{n_k} \Vert_{H^1} \to \liminf_n \Vert u_n \Vert_{H^1}$ and rename the sequence to $(u_n)_n$ again.

You know that every bounded sequence has a weakly convergent subsequence (because $H^1$ is a Hilbert space).

Hence $u_{n_k} \to v$ for a suitable subsequence and some $v \in H^1$, where the convergence is weak convergence in $H^1$. By weak lower continuity of the $H^1$-norm, we get $\Vert v \Vert_{H^1} \leq \liminf_k \Vert u_{n_k} \Vert_{H^1} = \liminf_n \Vert u_n \Vert_{H^1}$.

Finally, using the continuous embedding $H^1 \hookrightarrow L^2$, we also get $u_{n_k} \to v$ weakly in $L^2$. But also $u_{n_k} \to u$ strongly in $L^2$, hence $v=u$, which easily completes the proof.

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  • $\begingroup$ Thank you for your answer. But I'm still confused, don't we always have $H^1 \hookrightarrow L^2$ because I thought the norm of $L^2$ is bounded above by the norm of $H^1$? So why do I need to know that the boundary of $\Omega$ is $C^1$? (sorry I just realize that there's a typo in the question, the boundary is $C^1$ not $C$). $\endgroup$ – user113988 Oct 7 '14 at 13:07
  • $\begingroup$ You do not need to know that for the part of the argument I gave above. But it is probably used in the proof of the fact that $\Vert u_n - u \Vert_{L^2} \to 0$. But I interpreted your question as stating that this is already given. $\endgroup$ – PhoemueX Oct 7 '14 at 15:22
  • $\begingroup$ Could you explain to me why $u_{n_k} \rightarrow v$ weakly in $H^1$ is the same as $u_{n_k} \rightarrow v$ weakly in $L^2$? Because even though we have the embedding the inner product on $H^1$ is still different from the inner product on $L^2$ and I'm having difficulty showing that weakly convergent in the former space implies weakly convergent in the latter. $\endgroup$ – user113988 Oct 8 '14 at 6:59
  • $\begingroup$ It is not the same, but weak convergence in $H^1$ yields weak convergence in $L^2$. The exact way to show this depends on your definition of weak convergence. One way is as follows: Take $g \in L^2$. Then $u \mapsto <u, g>_{L^2}$ is continuous on $H^1$ (check this, this uses $H^1 \hookrightarrow L^2$). By Riesz representation theorem, there is some $h \in H^1$ with $<u,g>_{L^2} = <u,h>_{H^1}$ for all $u \in H^1$. Hence, $<u_{n_k}, g>_{L^2} = <u_{n_k}, h>_{H^1} \to <u, h>_{H^1} = <u,g>_{L^2}$, which shows $u_{n_k} \to u$ also weakly in $L^2$, since $g\in L^2$ was arbitrary. $\endgroup$ – PhoemueX Oct 8 '14 at 13:27

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