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Let $\Omega \subset \mathbb{R}^n$ be a nonempty bounded open subset (assumed to be sufficiently smooth).

Let $\Gamma$ denote the boundary of $\Omega$, and let $\Gamma_1 \subset \Gamma$ be a (measurable) subset.

For any function $u$ defined a.e. on $\Gamma_1$, let $\tilde{u}$ denote its extension to $\Gamma$ by zero, that is, $\tilde{u} =u$ on $\Gamma_1$ and $\tilde{u} = 0$ on $\Gamma$.

Now I consider the Lions-Magenes space

$$ H^{1/2}_{00} (\Gamma_1) := \{ u \in H^{1/2}(\Gamma_1) \mid \tilde{u} \in H^{1/2}(\Gamma) \}. $$

Is it true that $H^{1/2}_{00} (\Gamma_1)$ is dense in $H^{1/2} (\Gamma_1)$?

Thank you for any assistance.

user309395
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1 Answers1

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I don't think so. Take $n=2$ and assume that a portion of $\Gamma$ is a segment, say, $[-1,1]\times\{0\}$. Take $\Gamma_1=[-1,0]\times\{0\}$. Then the constant function $u=1$ belongs to $H^{1/2}(\Gamma_1)$ but if you extend it to zero outside $\Gamma_1$ you get a $BV$ function, with a jump. If you compute it directly, you should be able to prove that $\tilde u\notin H^{1/2}([-1,1]\times\{0\})$. This has to do with the embedding in fractional Sobolev spaces. In one dimension I think that $BV((0,1))$ is contained in $W^{s,p}((0,1))$ for $sp<1$ but not for $ps\ge 1$.

Update Tartar's book Sobolev has a chapter on the Lions-Magenes spaces. In particular he shows that if $u\in H^{1/2}_{00}(\mathbb{R}_+)$ then $\frac{u}{\sqrt x}\in L^2(\mathbb{R}_+)$ with a continuous bound. If you had density, then for $u\in H^{1/2}(\mathbb{R}_+)$ you could find a sequence $u_n\in H^{1/2}_{00}(\mathbb{R}_+)$ with $u_n\to u$. By pointwise convergence and Fatou's lemma you would have $\Vert \frac{u}{\sqrt x}\Vert_{L^2(\mathbb{R}_+)}\le\liminf_n \Vert \frac{u_n}{\sqrt x}\Vert_{L^2(\mathbb{R}_+)}\le C$. Take a function $u\in H^{1/2}(\mathbb{R}_+)$ such that $u=1$ in $[0,1]$. Then you have a contradiction.

Update 2 Take a look at this paper paper. The space $H^{1/2}_{00}$ is defined as the space of $u\in H^{1/2}$ such that $\frac{u}{d}\in L^2(\Omega)$ and the norm is $$\Vert u\Vert_{H^{1/2}}+\Vert \frac{u}{d}\Vert_{L^{2}}.$$ With this norm it is complete. So it all depends on which norm you take in $H^{1/2}_{00}$.

Gio67
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  • Thank you for your interest and your answer. I do agree that, in your example, $\tilde{u} \notin H^{1/2}$. Nevertheless, my question is not to prove that $H^{1/2}{00}(\Gamma_1) = H^{1/2}(\Gamma_1)$, but to prove that $H^{1/2}{00}(\Gamma_1)$ is dense in $H^{1/2}(\Gamma_1)$. With your example, my question would be: can we approximate the constant function 1 by a sequence of functions in $H^{1/2}_{00}(\Gamma_1)$? Thank you again. – user309395 Jun 27 '17 at 12:09
  • I think that the space $H^{1/2}{00)$ is closed (I don't remember which norm you take in that space though), so exactly because $H^{1/2}{00)\ne H^{1/2}$ you cannot have density. – Gio67 Jun 27 '17 at 12:49
  • Thank you for your answer. Since $H^{1/2}{00} (\Gamma_1) \subset H^{1/2} (\Gamma_1)$, I consider the induced norm ... With this norm, I do not think that $H^{1/2}{00} (\Gamma_1)$ is closed, is it? – user309395 Jun 27 '17 at 12:51
  • I think it is closed. Need to go to work but I will think about it. – Gio67 Jun 27 '17 at 13:13
  • OK thank you for your time. I will think about it. Just a remark: in the case where we work on $\Omega \subset \mathbb{R}^n$ (and not on $\Gamma_1 \subset \Gamma$), then $\mathcal{D}(\Omega)$ is dense in $H^{1/2}(\Omega)$. As a consequence, $H^{1/2}_{00}(\Omega)$ is dense in $H^{1/2}(\Omega)$. The question is: is this result true with $\Gamma_1 \subset \Gamma$? – user309395 Jun 27 '17 at 13:28
  • I updated my answer. – Gio67 Jun 29 '17 at 12:18
  • Thank you Gio67. I do understand your argument. I gave a look in Tartar book. Indeed, it is proved that if $u \in H^{1/2}{00}(\mathbb{R}+)$ then $\frac{u}{\sqrt{x}} \in L^{2}(\mathbb{R}_+)$. Nevertheless, I do not find the "continuous bound". Is that true? – user309395 Jun 30 '17 at 10:41
  • I am a little bit surprised since in the book of Lions-Magenes, it is written that $\mathcal{D}(\Omega)$ is dense in $H^{1/2}(\Omega)$ ... As a consequence, I thought that if $\Gamma_1$ is an open subset of $\Gamma$, then $\mathcal{D}(\Gamma_1)$ is dense in $H^{1/2}(\Gamma_1)$. As a consequence, $H^{1/2}_{00}(\Gamma_1)$ is dense in $H^{1/2}(\Gamma_1)$. Am I wrong somewhere? – user309395 Jun 30 '17 at 10:44
  • he is using embeddings on interpolation spaces. The embeddings are linear and continuous. That's where you get the control on the norms. – Gio67 Jun 30 '17 at 10:48
  • as for your other question, I don't know. I would have to read that proof. – Gio67 Jun 30 '17 at 10:55
  • Tartar says that $H^{1/2}{00}(\Omega)=(H^1_0(\Omega),L^(\Omega)){1/2,2)$. Interpolation spaces are complete. So $H^{1/2}_{00}(\Omega)$ seems to be closed (the norm should be the same, I think....). He also says that $\mathcal D(\Omega)$ is dense in $H^{1/2}(\Omega)$ so yeah, I am definitely missing something.... – Gio67 Jun 30 '17 at 11:18
  • I saw your update 2. The article you sent me is very interesting. Thank you. In my problem, the norm considered on $H^{1/2}_{00}(\Gamma_1)$ is just the induced norm of $H^{1/2}(\Gamma_1)$.

    I do not want to consider another norm. For example, we know that $H^1(\Omega)$ is complete (and thus closed) with its usual norm. On the other hand, it is dense in $L^2(\Omega)$ considering the induced norm of $L^2(\Omega)$.

    My question is exactly the same, but with $H^{1/2}_{00}(\Gamma_1) \subset H^{1/2}(\Gamma_1)$.

     – user309395 Jun 30 '17 at 15:09