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 $H$.

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ Weak boundedness implies strong boundedness... $\endgroup$ – paul garrett Jun 12 '16 at 23:02
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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 $H$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.