# 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?

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}