1
$\begingroup$

Let $\Gamma$ be a regular boundary of a $C^{k,1}$ domain $\Omega$ and $H^s(\Gamma)$, $s\in(0,1)$, denote the fractional Sobolev space on $\Gamma$. Suppose I define a multiplication operator $M_\phi:H^s(\Gamma)\to H^s(\Gamma)$ where $M_\phi v=\phi v$. What should be the minimal regularity of $\phi$ for the map to be continuous?

I am particularly interested on the case when the map $H^\frac12(\Gamma) \to H^\frac12(\Gamma)$ is continuous. More precisely, I want to know whether the following statement is true.

Let $\Omega \subset \mathbb{R}^2$ be a bounded Lipschitz domain and $\Gamma$ be a non-empty subset of $\Omega$. Then, the map $v \mapsto \phi v$ is continuous in $H^\frac12(\Gamma)$ for any $v \in H^\frac12(\Gamma)$ and $\phi \in C^{0,1}(\Gamma)$.

I tried to argue the validity of the above statement as follows.

By McShane-Whitney extension theorem, we know that we can find a $\tilde{\phi} \in C^{0,1}(\bar{\Omega})$ such that $\tilde{\phi}|_{\Gamma} = \phi$. Then, using [Grisvard, Elliptic Problems in Nonsmooth Domains, Theorem 1.4.1.1, p. 21] and [McLean, Strongly Elliptic Systems and Boundary Integral Equations, Theorem 3.37, p. 102], we have that $v \mapsto \phi v$ a continuous linear map in $H^\frac12(\Gamma)$.

Can someone confirm if my argument is correct?

$\endgroup$
  • $\begingroup$ In your first paragraph, you ask for the "minimal regularity of $v$". Should it be the minimal regularity of $\phi$? $\endgroup$ – Matt Rosenzweig May 7 '18 at 16:17
  • $\begingroup$ Can't you just prove what you want by complex interpolation between the endpoint estimates $s=0$ and $s=1$, which hold for $\phi\in C^{0,1}(\Gamma)$? $\endgroup$ – Matt Rosenzweig May 7 '18 at 16:18
  • $\begingroup$ @MHR Yes, it should be the `minimal regularity of $\phi$'. $\endgroup$ – Julienne Franz May 8 '18 at 2:35
  • $\begingroup$ @Piotr Hajlasz Yes it proves the validity of the quoted statement, but can you also confirm whether my approach on proving it is correct? or if not, can you point out what might be the problem. Thanks! $\endgroup$ – Julienne Franz May 8 '18 at 2:38
  • $\begingroup$ @MHR I am not yet familiar with such approach, but are you telling me that the quoted statement follows from $[H^1(\Gamma), H^0(\Gamma)]_{\frac12} = H^{\frac12}(\Gamma)$ (by Theorem 7.7 of Lions and Magenes, p. 36)? Thank you. $\endgroup$ – Julienne Franz May 8 '18 at 7:14
1
$\begingroup$

Theorem. Multiplication by a Lipschitz function defines a bounded operator in $H^{1/2}(\partial\Omega)$.

First proof. More generally, if $\Omega\subset\mathbb{R}^n$ is a bounded Lipschitz domain, then the fractional Sobolev space $W^{1-1/p,p}(\partial\Omega)$ (the trace space for $W^{1,p}(\Omega)$) is equipped with the norm $$ \Vert u\Vert_{1-1/p,p}=\Vert u\Vert_{L^p(\partial\Omega)}+A(u), \quad A(u)=\left(\int_{\partial\Omega} \int_{\partial\Omega}\frac{|u(x)-u(y)|^p}{|x-y|^{n+p-2}}\, dx\, dy\right)^{1/p}. $$ If $p=2$ we have $W^{1-1/2,2}(\partial\Omega)=H^{1/2}(\partial\Omega)$. Now if $\phi$ is Lipschitz with the Lipschitz constant $L,$ then $$ \Vert\phi u\Vert_{L^p(\partial\Omega)}\leq \Vert\phi\Vert_\infty\Vert u\Vert_{L^p(\partial\Omega)}. $$ Since $$ |\phi(x)u(x)-\phi(y)u(y)|\leq |\phi(x)||u(x)-u(y)|+|u(y)||\phi(x)-\phi(y)|\\ \leq \Vert\phi\Vert_\infty |u(x)-u(y)| + L|u(y)||x-y|, $$ we estimate $A(\phi u)$ as follows (by $C$ we will denote a generic constant; it may have a different value each time it is used): $$ A(\phi u)=\left(\int_{\partial\Omega} \int_{\partial\Omega}\frac{|\phi(x)u(x)-\phi(y)u(y)|^p}{|x-y|^{n+p-2}}\, dx\, dy\right)^{1/p}\\ \leq C\Vert\phi\Vert_\infty A(u)+ CL\left(\int_{\partial\Omega} \int_{\partial\Omega}\frac{|u(y)|^p}{|x-y|^{n-2}}\, dx\, dy\right)^{1/p}\\ \leq C\Vert\phi\Vert_\infty A(u)+CL\Vert u\Vert_{L^p(\partial\Omega)}, $$ because the integral $$ \int_{\partial\Omega}\frac{dx}{|x-y|^{n-2}}\leq C $$ is bounded by a constant $C$ independent of $y$. $\Box$

Second proof. There are bounded linear extension operators $E_1:\operatorname{Lip}(\partial\Omega)\to\operatorname{Lip}(\Omega)$ and a $E_2:H^{1/2}(\partial\Omega)\to H^1(\Omega)$. Also the trace operator $T:H^1(\Omega)\to H^{1/2}(\partial\Omega)$ is bounded.

It is easy to check that if $\tilde{\phi}\in\operatorname{Lip}(\partial\Omega)$, then $u\mapsto\tilde{\phi}u$ is a bounded operator on $H^1(\Omega)$. Let $\phi\in \operatorname{Lip}(\partial\Omega)$. Then the operator $Mv=\phi v$ is bounded $M:H^{1/2}(\partial\Omega)\to H^{1/2}(\partial\Omega)$ because $$ Mv=T((E_1\phi)(E_2 v)) $$ is a composition of bounded operators. $\Box$

Remark. In my opinion the first proof is better as being more straightforward.

$\endgroup$

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.