Define a "Neumann" trace of a harmonic function on bounded domain Let us assume $\Omega$ is a smooth bounded domain. It is known that the solution map of Dirichlet problem of elliptic equation $M: H^{1/2}(\partial\Omega) \rightarrow H^1(\Omega)$ is well defined in the following way: $u = M(f)$
$ -\Delta u = 0$ in $\Omega$
$ u = f$ on $\partial \Omega$
further more, $M$ is bounded. The trace theorem gurantees that $M^{-1}: \mathcal{R}(M)\rightarrow H^{1/2}(\partial \Omega)$ is bounded as the range of $M$ is a subset of $ H^1(\Omega)$.
We could also consider the Neumann problem of elliptic equation $N: H^{-1/2}(\partial \Omega) \rightarrow H^{1}(\Omega)/ \mathbb{R}$ which is defined in the following way: $N(g) = u $ mod $\mathbb{R}$ (where we shift $u$ to make it mean zero)
$ -\Delta u = 0$ in $\Omega$
$ \frac{\partial u}{\partial n} = g$ on $\partial \Omega$
This map $N$ is also well defined and bounded. I am wondering if there is a way to define $N^{-1}: \mathcal{R}(N)\subset H^1(\Omega)\rightarrow H^{-1/2}(\partial \Omega)$ and prove it is bounded. 
 A: If $u \in C^2(\Omega),$ then the divergence theorem gives an integral characterization for the Neumann trace $\partial u / \partial n$: for arbitrary $\phi \in C^1(\Omega)$ we have 
$$\int_{\partial \Omega}\frac{\partial u}{\partial n}\phi=\int_{\Omega}\operatorname{div}(\phi \nabla u)=\int_{\Omega}\phi \Delta u+\int_\Omega\nabla \phi\cdot\nabla u.$$
If $u$ is merely $H^1$ but is known to be harmonic, the RHS still makes sense; so we can attempt to use this equation to define a Neumann trace operator valued in $H^{-1/2}(\partial \Omega)$. Since this requires treating test functions $\phi \in H^{1/2}(\partial \Omega),$ we will need an extension operator $H^{1/2}(\partial \Omega) \to H^1(\Omega),$ i.e. a bounded right-inverse of the Dirichlet trace operator. The Dirichlet solution operator $M$ suffices for this purpose.
Let $X = \{ u \in H^1(\Omega) : \Delta u = 0, \int_\Omega u =0 \}$ and define $T_n : X \to H^{-1/2}(\partial \Omega)$ by the formula $$(T_nu)(\phi) = \int_\Omega \nabla (M\phi) \cdot \nabla u.$$
Since $E$ is bounded, we have $$|(T_n u)(\phi)| \le \|M \phi\|_{H^1} \| u\|_{H^1} \le \|M\|\|u\|_{H^1}\|\phi \|_{H^{1/2}};$$ so it's true that $T_n u \in H^{-1/2}(\partial \Omega).$  Furthermore, this inequality tells us $\|T_n u\|_{H^{-1/2}} \le \|M\|\|u\|_{H^1},$ so $T_n$ is bounded with $\|T_n\| \le \|M \|.$ 
Since $X \cap C^2 \subset X$ is a dense subspace and $T_n u = \partial u/\partial n$ for $u \in X \cap C^2$, this is the unique bounded extension of the normal derivative operator to $X.$ (You should be able to show that a very similar construction works with only the assumption $\Delta u \in L^2$, so you can handle the Poisson equation too.)
If we interpret $\partial u/ \partial n$ as $T_n u$ in your definition of $N,$ then we can show that $(X,T_n)$ is your desired $(\mathcal R(N),N^{-1}).$ By the definition of $N$ we immediately see that $\mathcal R(N) \subset X$ and $T_n N = \operatorname{id}$. Since any $u\in X$ is a solution to the Neumann problem with data $T_n u$, the uniqueness of solutions (with mean zero) tells us $N T_n u = u$. Thus we have the counterparts $X \subset \mathcal R(N)$ and $NT_n = \operatorname{id}.$
