I was reading article named Determining Conductivity by Boundary Measurements by Robert Kohn and Vogelius. In need that I stuck at following one point.
We denote by $H^{s}(\Omega), s \geqq 0$, the standard Sobolev space of order $s$ based on $L^{2}(\Omega)$, with norm $\|\cdot\|_{s, \Omega}$. Auther writes following.
It is well known that $$ C^{\prime}\|\phi\|_{W^{\frac{1}{2}, 2}(\partial \Omega)} \leqq \inf _{\substack{\left.v \in H^{1}(\Omega) \\ v\right|_{\partial\Omega}=\phi}}\|v\|_{W^{1, 2}(\Omega)} \leqq C\|\phi\|_{W^{\frac{1}{2}, 2}(\partial \Omega)}, $$ I know that the trace operator is the bounded linear operator between space $W^{1, 2}(\Omega) \rightarrow W^{\frac{1}{2}, 2}(\partial \Omega)$. So we have $$ C^{\prime}\|\phi\|_{W^{\frac{1}{2}, 2}(\partial \Omega)} \leq \inf _{v \in W^{1, 2}(\Omega),\left.v\right|_{\partial \Omega}=\phi}\|v\|_{W^{1, 2}(\Omega)} $$
But I do not understand that $$ \inf _{\substack{\left.v \in H^{1}(\Omega) \\ v\right|_{\partial\Omega}=\phi}}\|v\|_{W^{1, 2}(\Omega)} \leqq C\|\phi\|_{W^{\frac{1}{2}, 2}(\partial \Omega)} $$
Any help or refernce will be highly appreciated.
The trace operator is not only linear continuous from $H^1$ to $W^{1/2,2}$ (aka $H^{1/2}$), it is also onto, and there exists a linear continuous lifting from $H^{1/2}$ to $W^{1/2,2}$. The inequality you mention corresponds to this continuity.