Adjoint of the boundary trace operator? 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!
 A: If you want to use the $L^2(\partial \Omega)$-inner product in your application (whatever it is), write
$$ \gamma: H^1(\Omega) \to L^2(\partial \Omega). $$
This is a bounded operator by the trace theorem, if we use, e.g., the basic $H^1(\Omega)$- and $L^2(\partial \Omega)$ inner products/norms. Then there should exist unique $\gamma^*$, as stated in the other answer.
With this choice of inner products, the "equation for $\gamma^*$" is
$$
 (\gamma u, v)_{L^2(\partial \Omega)} = (u, \gamma^* v)_{H^1(\Omega)} 
$$
for all $u \in H^1(\Omega), v \in L^2(\partial \Omega)$. Denote $w := \gamma^* v$. Then the above equation in a more explicit form is: Find $w \in H^1(\Omega)$ such that
$$
 (u, v)_{L^2(\partial \Omega)} = (\nabla u, \nabla w)_{L^2(\Omega)} + (u, w)_{L^2(\Omega)},
$$
for all $u \in H^1(\Omega)$. This is the weak form of the Laplacian-like PDE with Neumann b.c:
$$ \Delta w + w = 0 \text{ in } \Omega, $$
$$ \partial_n w = v \text{ on } \partial \Omega.$$
In other words, the adjoint of the trace, $\gamma^* v$, is a solution operator to some Laplacian-like thing, where $v$ becomes the Neumann b.c. And what this means, no idea (at this moment right now).
(If some PDE is involved in your application, you might want to use the corresponding weak form bilinear form/inner product instead instead of the basic $H^1$-inner product. Or you might want to change $L^2(\partial \Omega)$ to $H^{1/2}(\partial \Omega)$. Then $\gamma^*$ changes into... something else.)
A: 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. 
