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?



Continuity of a multiplication operator in fractional Sobolev space

> 
> **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.


