Writing up a report, I want to understand the following solution operator better.
Suppose $\Omega$ is an open bounded domain in $\mathbb{R}^d$ with boundary $\partial\Omega$ of class $\mathcal{C}^1$. $H^1(\Omega)$ is the Sobolev space $W^{1,2}(\Omega)$. Let $\varphi$ be in the range of the Trace Operator $Tr:H^1(\Omega) \to L^2(\Omega)$. Then using Stampacchia, we can show that there exists a unique function, a weak solution $u \in H^1(\Omega)$ such that $$\begin{cases} -\triangle u = 0 \text{ in } \Omega \\ u = \varphi \text{ on } \partial\Omega. \end{cases} $$ If we define the operator $P: Rg(Tr) \to H^1(\Omega)$ by $\varphi \mapsto u$, then $P$ is clearly linear. How can we also prove that it is bounded (i.e. continuous) with respect to the norms $||\varphi||_{L^2(\partial\Omega)}$ and $||\triangledown u||_{L^2(\Omega)}$.
Okay, here's an attempt at an answer.
First I don't think the $L^2(\partial \Omega)$ is the appropriate norm to use here, since the space $H^{1/2}(\partial \Omega)=: \text{Range}(Tr)$ is dense in $L^2(\partial \Omega)$, so continuity in this latter norm would imply a bounded extension to $L^2$, which is not the case. Instead you want to use the norm of $H^{1/2}(\partial \Omega)$ defined as $$ \| f\|_{H^{1/2}}:= \inf\{ \| F \|_{H^1(\Omega)} : f=Tr(F) \}. $$ With this you can write a solution of $$ \begin{cases} -\Delta u = 0 \text{ in } \Omega \\ u = \varphi \text{ on } \partial\Omega. \end{cases} $$ as $u=w+\Phi$ where $\Phi$ is an $H^1$ extension of $\phi$ and $w$ solves $$ \begin{cases} -\Delta w = \Delta \Phi \text{ in } \Omega \\ w = 0 \text{ on } \partial\Omega. \end{cases} $$ With this we can estimate $$ \| \nabla u \|_{L^2(\Omega)} \leq \| \nabla w\|_{L^2(\Omega)} + \| \nabla \Phi\|_{L^2(\Omega)} \leq 2 \| \nabla \Phi \|_{L^2(\Omega)}. $$ Taking now the infimum over $\Phi$ we get the desired result.