I am having difficulties defining the adjoint of the ``restriction-to-boundary'' operator for continuous functions, to a more general Sobolev setting and the notion of the trace. I'll illustrate the idea first with continuous functions.
Assume $u\in C^1(\overline{\Omega})$ and define $\text{Tr}~u = u\big|_{\partial\Omega}$. Very formally, if I take $v\in C^0(\partial\Omega)$ and use $L^2$ inner products $\langle,\rangle$:
$ \langle \text{Tr}~u,v\rangle_{L^2(\partial\Omega)} = \int_{\partial\Omega} (\text{Tr}~u)v~dS_x \overset{(1)}{=} \int_\Omega u (\text{Tr}^\dagger~v)~dx = \langle u, \text{Tr}^\dagger~v\rangle_{L^2(\Omega)}, $
where in (1) I am defining $\text{Tr}^\dagger~v$ as
$ \text{Tr}^\dagger~v = \left\{\begin{array}{lr} v(x),& x\in\partial\Omega,\\ 0, & \text{otherwise.}\end{array}\right. $
Of course this is formal as I'm ignoring the fact that $\partial\Omega$ has measure $0$ and assigning pointwise values to a function that only exists in $L^2$, but hopefully the main idea comes across easy enough.
I'd like to know if this can be made precise somehow for mappings between Sobolev spaces? That is, given the definition of the trace operator $\text{Tr}: H^1(\Omega)\mapsto H^{1/2}(\partial\Omega)$, is there an operator $\text{Tr}^\dagger: H^{1/2}(\partial\Omega)\mapsto H^1(\Omega)$ such that $ \text{Tr}^\dagger v = 0\text{ in $H^1(\Omega)$ and $\text{Tr}(\text{Tr}^\dagger v)= v$?}$ Such an operator would also need to satisfy $\langle \text{Tr}~u,v\rangle_{H^{1/2}(\partial\Omega)} = \langle u, \text{Tr}^\dagger v\rangle_{H^1(\Omega)}.$ Any help and/or references you can provide is greatly appreciated!
Here is a more or less direct argument.
Let $\gamma : H^1 \to H^{1/2}$ on a sufficiently nice domain denote the trace operator. It is continuous/bounded. Is there an operator $\gamma^* : H^{1/2} \to H^1$ such that $$ (\gamma u, g)_{1/2} = (u, \gamma^* g)_1 \quad \forall (u, g) \in H^1 \times H^{1/2} \qquad(\star) $$ ?
For any given $g \in H^{1/2}$, the mapping $v \mapsto (\gamma v, g)_{1/2}$ defines a bounded linear functional on $H^1$. Hence there exists a function $\gamma^* g \in H^1$ such that $(\star)$ holds (Riesz representation thm on $H^1$), and $\|\gamma^* g\|_{1} \leq C \| g \|_{1/2}$. The mapping $g \mapsto \gamma^* g$ is linear and bounded.
The requirement $(\star)$ defines $\gamma^*$ uniquely. If $u \in H_0^1$ then the LHS of $(\star)$ vanishes. Therefore, $\gamma^* g$ is in $H^1 / H_0^1$, i.e. the $H^1$-orthogonal complement of the closed subspace $H_0^1$ in $H^1$. This, of course, does not mean that $\gamma^* g = 0$ in $H^1$. Also, I do not see how this implies that $\gamma^* g|_K = 0$ on compact subsets $K$ of the domain: we have the $H^1$ scalar product in $(\star)$, not the $L_2$ scalar product.