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I'm reading a hand-waving argument in a proof of Chapter 7 of the Navier-Stokes Equations by Constantin and Foias. I would like to know if I understand it correctly.

Let $\Omega\subset{\mathbb{R}^n}$ be an open set with $\partial \Omega$ being $C^k$, $k\geq 2$. Let $\mathcal{V}$ be the space

$$\displaystyle \mathcal{V}=\{u\in C_0^\infty(\Omega)\mid \nabla\cdot u=0\}.$$

Let $ H=\overline{\mathcal{V}}^{\|\cdot\|_{L^2(\Omega)^n}} $ and $ V=\overline{\mathcal{V}}^{\|\cdot\|_{H_0^1(\Omega)^n}}. $

Is the following statement true?

Suppose $u_m\to u$ weakly in $V$. Then there exist a subsequence $u_{m'}\to u$ strongly in $H$.

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    $\begingroup$ I think this would be true if the inclusion from $V$ into $H$ is a compact linear map $\endgroup$ – Yemon Choi Jun 12 '16 at 22:05
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    $\begingroup$ Weak boundedness implies strong boundedness... $\endgroup$ – paul garrett Jun 12 '16 at 23:02
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    $\begingroup$ It is true if $\Omega$ is bounded, since then $H^1_0$ imbeds compactly into $L^2$. $\endgroup$ – Michael Renardy Jun 13 '16 at 8:11
  • $\begingroup$ @paulgarrett: Thanks for your comment. A quick search on Google for "Weak boundedness implies strong boundedness" returns your lecture notes on Functional Analysis. Are you saying that "$u_m$ converges weakly in $V$" implies it is weakly bounded and thus by the theorem in your note, it is strongly bounded in $H$? With this argument, it seems that one does not need the assumption that $\Omega$ is bounded? $\endgroup$ – Jack Jun 13 '16 at 22:14
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    $\begingroup$ Jack, indeed, weak convergence of a sequence implies that the sequence, as a set, is weakly bounded, hence bounded. But bounded does not imply compact. This weak-to-strong trick is just a frequently helpful little extra fact. The key point is the compactness, as @MichaelRenardy notes. $\endgroup$ – paul garrett Jun 13 '16 at 22:27
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I think it is correct. since $u_m \to u$ weakly in V, so $u_m$ is bounded ,then (by the compact embedding theorem) $H$ can be embedding $V$,thus $u_{{m}^{\prime}} \to u $ strongly in $H$.

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