Characterization of $H^{1/2}(\partial \Omega)$ norm? I can't understand why $\|g\|_{H_{\partial}^{1/2}} = \|u\|_{H^1} $ where $u$ solves the problem
\begin{equation*}
- \Delta u + u = 0, \quad \Omega\\
u|_{\partial \Omega} = g
\end{equation*}
The weak form should be to find $w \in H^1_0$:
\begin{equation*}
(w,v)_{H^1} = -(G, v)_{H^1} , \quad \forall v \in H^1_0
\end{equation*}
where $ u = w+G$ and $G$ is the lifting of $g$.
$(.,.)_{H^1}$ is the $H^1$ inner-product.
Now I can only get this estimate from the Lax-Milgram lemma:
$$
\|u\| -\|G\| \leq \|w\| \leq \|G\|
$$
And so:
$$
\|g\| \leq \|u\| \leq 2\|g\|
$$
As the $H^{1/2}$ norm is the $\inf$
But how to get equality?
 A: Recall that $H^{1/2}(\Gamma):=\gamma_0(H^1(\Omega))$, where $\gamma_0(u):=u|_{\partial \Omega } $. Then we have that $\ker \gamma = H_0^1(\Omega)$. By the orthogonal descomposition $H^1(\Omega) = H^1_0(\Omega) \oplus H_0^1(\Omega)^{\perp}$.  The restricted trace operator $\tilde {\gamma}_0:H_0^1(\Omega)^{\perp} \to H^{1/2}(\Gamma)$ is a linear isomorphism.
Now for any $g \in H^{1/2}(\Gamma)$ there is a $u = u_0+u^{\perp} \in H^1_0(\Omega) \oplus H_0^1(\Omega)^{\perp}$  with $g = \gamma_0(u)$. In particular $g=\gamma_0(u^{\perp})$, immediately we obtain $\tilde{\gamma}_0^{-1}(g)=u^{\perp}$, then
$$\|g\|_{1/2} = \inf_{u \in H^1(\Omega):\gamma_0(u)=g} \|u\|_{H^1}= \inf_{u \in H^1(\Omega):\gamma_0(u)=g} \|u_0+\tilde{\gamma}^{-1}(g)\|_{H^1}.$$
This infimum reachs when $u_0=0$, then
$$\|g\|_{1/2} = \|\tilde{\gamma}^{-1}(g)\|_{H^1}=\|u^{\perp}\|_{H^1}$$
Finally, is not too complicate to show that $u^{\perp}$ is the weak solution, because
\begin{align}
H_0^1(\Omega)^{\perp} & = \{u^{\perp}:\int_{\Omega}(u^{\perp}w+\nabla u^{\perp}\cdot \nabla w ) = 0 \quad \forall w \in H^1_0(\Omega) \} \\
&= \{u^{\perp} \in \mathcal{D}'(\Omega):-\Delta u^{\perp}+u=0\}.
\end{align}
