# If $H$ is the closure of the set of solenoidal smooth vecor fields in $L^2$ and $P_H$ denote the orthogonal projection onto $H$, then $P_HH_0^1⊆H_0^1$

Let

• $d\in\mathbb N$
• $\Lambda\subseteq\mathbb R^d$ 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?

• 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. – Fan Zheng Dec 12 '16 at 23:21
• @FanZheng Do you know the answer? – 0xbadf00d Dec 12 '16 at 23:39
• I'm only remarking that this is not the typical question asked on mathoverflow; it would be more suitable on math stackexchange. – Fan Zheng Dec 13 '16 at 0:19
• This does not seem true. Are your sure the source you cite really says that? – Michael Renardy Dec 13 '16 at 0:48
• 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. – Michael Renardy Dec 13 '16 at 15:38