I encounter the following estimate in several situations, but cannot see how to prove it. Could anyone show me how?
If $u$ is a harmonic fuction on a domian $\Omega$ with smooth boundary, then $$\|u\|_{H^1(\Omega)}\leq C\|u\|_{H^{1/2}(\partial\Omega)}. $$ where $C$ is some constant independent of $u$
Edit: Thanks you, Michał Miśkiewicz, for your answer. Although, I am still curious about whether this estimate can be derived without the use of inverse trace theorem.
Let me use two propositions:
1. $H^{1/2}(\partial \Omega)$ is the trace space of $H^1(\Omega)$, meaning that the trace operator $H^1(\Omega) \to H^{1/2}(\partial \Omega)$ is continuous and onto. For each $v \in H^{1/2}(\partial \Omega)$ there is $w \in H^1(\Omega)$ such that $w = v$ on $\partial \Omega$ and \begin{equation} \label{eq} \tag{$\star$} \| w \|_{H^1(\Omega)} \le C \| v \|_{H^{1/2}(\partial \Omega)}. \end{equation}
2. Harmonic functions are minimizers of the Dirichlet energy. That is, if $v \in H^{1/2}(\partial \Omega)$, then the unique solution of $$ \begin{cases} \Delta u = 0 & \text{ in } \Omega, \\ u = v & \text{ on } \partial \Omega \end{cases} $$ is at the same time the unique minimizer of $\int_{\Omega} |\nabla u|^2$ in the class of $H^1(\Omega)$-functions equal to $v$ on $\partial \Omega$.
Taking $v = u|_{\partial \Omega}$ and choosing some $w \in H^1(\Omega)$ as in 1, we can compare $$ \int_{\Omega} |\nabla u|^2 \le \int_{\Omega} |\nabla w|^2 $$ according to 2. Since $u-w \in H_0^1(\Omega)$, Poincare's inequality yields \begin{align*} \| u \|_{L^2(\Omega)} - \| w \|_{L^2(\Omega)} & \le \| u-w \|_{L^2(\Omega)} \\ & \le C \| \nabla u - \nabla w \|_{L^2(\Omega)} \\ & \le 2 C \| \nabla w \|_{L^2(\Omega)}. \end{align*} Thus, the estimate \eqref{eq} is valid also for $u$, possibly with a larger constant.