Continuity of a multiplication operator in fractional Sobolev space

Question

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?

Answer 1

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.