It is known that the natural trace map $W^{1}(M) \ni \varphi \rightarrow \varphi|_{\partial M} \in W^{1/2}(\partial M)$ is continuous and onto. Since the Dirichlet problem $\Delta \varphi=0$, $\varphi|_{\partial M}=f \in W^{1/2}(\partial M)$ has a unique solution, it follows that by limiting $\varphi \in W^{1}(M)$ to be harmonic the trace map becomes 1:1 and by using the closed graph theorem is easy to show that the inverse map is continuous. Therefore, this map is bicontinuous. Now this map should implies that if $\varphi$ is harmonic $$|\varphi|_{1/2}^{2}=||\varphi||_{1}^{2}$$, where $||\varphi||_{1}^{2}=||\nabla \varphi||_{0}^{2}+ ||\varphi||_{0}^{2}$ is the Sobolev norm in $W^{1}(M)$, $|.|_{1/2}$ is the Sobolev norm in $W^{1/2}(\partial M)$, and $||.||_0$ is the norm in $L^{2}(M)$, and $M$ is a compact riemannian manifold. I don't know why this bicontinuous map implies this equality between the Sobolev norms above.
Thanks for your help.