• $d\in\mathbb N$
  • $\Lambda\subseteq\mathbb R^d$ be open
  • $\mathcal V:=\left\{\phi\in C_c^\infty(\Lambda,\mathbb R^d):\nabla\cdot\phi=0\right\}$ and $$H:=\overline{\mathcal V}^{\left\|\;\cdot\;\right\|_{L^2(\Lambda,\:\mathbb R^d)}}$$
  • $\operatorname P_H$ denote the orthogonal projection from $L^2(\Lambda,\mathbb R^d)$ onto $H$

In Remark 1.6 of Navier-Stokes Equations: Theory and Numerical Analysis by Roger Temam, the author is stating that $\text P_HH_0^1(\Lambda,\mathbb R^d)\subseteq H_0^1(\Lambda,\mathbb R^d)$. I don't think that this is trivial. How can we prove it?

  • $\begingroup$ I'm afraid your questions are not quite up to research level and I suggest you read up on Sobolev spaces (density, embedding, extension, restriction, characterization in the frequency space, etc) from a textbook. $\endgroup$ – Fan Zheng Dec 12 '16 at 23:21
  • $\begingroup$ @FanZheng Do you know the answer? $\endgroup$ – 0xbadf00d Dec 12 '16 at 23:39
  • $\begingroup$ I'm only remarking that this is not the typical question asked on mathoverflow; it would be more suitable on math stackexchange. $\endgroup$ – Fan Zheng Dec 13 '16 at 0:19
  • 1
    $\begingroup$ This does not seem true. Are your sure the source you cite really says that? $\endgroup$ – Michael Renardy Dec 13 '16 at 0:48
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    $\begingroup$ The argument that $P_H$ maps $H^1$ to itself is given in Temam's book, and I shall not repeat it beyond pointing out the correction as I did before. $P_H$ does not map $H^m_0$ to itself. It preserves regularity, but not the boundary conditions. $\endgroup$ – Michael Renardy Dec 13 '16 at 15:38

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