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We have for example here and here that $$\|v_n\|_{H^1}=1,\|f_n\|_{L^2},\|v_n\|_{H^{1/2}(\partial \Omega )},\|\partial _\nu v_n\|_{H^{-1/2}(\partial \Omega )}\to 0.$$

Q1) I don't really know what mean $\|v_n\|_{H^{1/2}(\partial \Omega )}$ and $\|\partial _\nu v_n\|_{H^{-1/2}(\partial \Omega )}$. I know that if $v\in H^1(\Omega )$ then $v|_{\partial \Omega }\in H^{1/2}(\partial \Omega )$. Could someone explain ?

For example, we have the equation $$\int \nabla \varphi\nabla v_n=\int_\Omega \varphi v_n+\int_\Omega \varphi f_n+\int_{\partial \Omega }\partial _\nu v_n \varphi.$$

Q2) Here the fact that $\int_{\partial \Omega }\partial _\nu v_n\varphi\to 0$ when $n\to \infty $ come from the fact that $\|\partial _\nu v_n\|_{H^{-1/2}(\partial \Omega )}\to 0$ ? If yes, why ? The ting I know is $$H^{-1/2}(\partial \Omega )=\{v\text{ linear operator on }H^{1/2}(\partial \Omega ): |\left<v,\varphi\right>|\leq \|\varphi\|_{H^{1/2}(\partial \Omega )}\},$$ but it doesn't really help since $\partial _\nu v_n$ is not an operator. The thing I would do is to set $$T_n(\varphi)=\int_{\partial \Omega }\partial _\nu v_n \varphi $$ that is a linear operator on $H^{1/2}(\partial \Omega )$. Now, do we have that $\|T_n(\varphi)\|\leq \|\partial _\nu v_n\|_{H^{-1/2}(\partial \Omega )} $ ? The thing is $\|T\|_{H^{-1/2}(\partial \Omega )}=\sup_{\|\varphi\|_{H^{1/2}\leq 1}}|T(\varphi)|$, but what is $\|\partial _\nu v_n\|_{H^{-1/2}(\partial \Omega )}$ ?

Q3) To conclude, what is $\|v_n\|_{H^{1/2}(\partial \Omega )}$ ? Would it be $$\int_{\partial \Omega }v_n^2\text{ ?}$$

Michael Hardy
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user349449
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1 Answers1

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Q1) Here $H^{1/2}(\partial \Omega)$ stands for the Sobolev-Slobodeckij-Gagliardo-Aronszajn (usually people choose Sobolev and one of the other three) fractional space $W^{1/2,2}(\partial \Omega)$ with the norm $$ \| v \|^2_{W^{1/2,2}(\partial \Omega)} = \int_{\partial \Omega} v^2 + \int_{\partial \Omega} \int_{\partial \Omega} \frac{|v(x)-v(y)|^2}{|x-y|^n}. $$ I suggest to consult it with Hitchhiker’s guide to the fractional Sobolev spaces, keeping in mind that $\partial \Omega$ is $(n-1)$-dimensional when $\Omega \subseteq \mathbb{R}^n$.

The reason to introduce it here is the following. If $\Omega$ is regular enough, the trace map can be defined continuously not only as $W^{1,p}(\Omega) \to L^p(\partial \Omega)$, but even $W^{1,p}(\Omega) \to W^{1-1/p, p}(\partial \Omega)$. Moreover, this is optimal - for any $u \in W^{1-1/p, p}(\partial \Omega)$ there is an extension $\overline{u} \in W^{1,p}(\Omega)$ such that $\| \overline{u} \|_{W^{1.p}(\Omega)} \le C(\Omega,p) \| u \|_{W^{1-1/p,p}(\Omega)}$.

Q2) Yes, this is the right definition of $H^{-1/2}(\partial \Omega)$. Note that you cannot really define $\partial_\nu v$ as a function on $\partial \Omega$ (not even an "$L^p$-function"), so it's natural to view it as a linear functional on $H^{1/2}(\partial \Omega)$. The following fragment is taken from my answer to your previous question.

For regular $v$, the divergence theorem yields the following identity: $$ \int_{\partial \Omega} \varphi \partial_\nu v = \int_\Omega \operatorname{div}(\varphi \nabla v) = \int_\Omega \nabla \varphi \nabla v + \varphi \Delta v \quad \text{for any } \varphi \in C^1(\overline{\Omega}). $$

If $v \in H^1(\Omega)$ and $\Delta v \in L^2(\Omega)$ (as it is in our case), the RHS can be used to define $\langle \partial_\nu v, \varphi \rangle$ for any $\varphi \in H^{1/2}(\partial \Omega)$, that is $$ \langle \partial_\nu v, \varphi \rangle := \int_\Omega \nabla \overline{\varphi} \nabla v + \overline{\varphi} \Delta v \quad \text{for } \varphi \in H^{1/2}(\partial \Omega). $$ It can be checked that this doesn't depend on the choice of the extension $\overline{\varphi} \in H^{1}(\Omega)$. Look up the discussion here.

By this definition, we have the following bound $$ \left| \langle \partial_\nu v, \varphi \rangle \right| \le \left( \| \nabla v \|_{L^2(\Omega)} + \| \Delta v \|_{L^2(\Omega)} \right) \| \overline{\varphi} \|_{H^1(\Omega)}. $$ As I mentioned earlier, the extension can be chosen such that $\| \overline{\varphi} \|_{H^1(\Omega)} \le C \| \varphi \|_{H^{1/2}(\partial \Omega)}$, so $\partial_\nu v$ is a bounded linear functional on $H^{1,2}(\partial \Omega)$ with the norm estimated by $$ \| \partial_\nu v \| \le C \left( \| \nabla v \|_{L^2(\Omega)} + \| \Delta v \|_{L^2(\Omega)} \right). $$

Michał Miśkiewicz
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