I was following Wolfgang Bangerth's FEM Lecture: Nonlinear problems, part 2 -- Newton's method for PDEs covering the residual form of the minimum surface equation
\begin{equation} \begin{aligned} R(u) = f + \nabla \cdot \left(\frac{1}{\sqrt{1 + |\nabla u|^2}} \nabla u \right) &= 0\ \text{on}\ \Omega, \\ u &= 0\ \text{on}\ \partial\Omega, \end{aligned} \end{equation}
where Dr. Bangerth defines $R$ as the operator $R: u \in H^1_0 \rightarrow H^{-1}$, and I'm wondering if for example the boundary conditions were instead non-homogeneous (e.g,. $u = g$ on $\partial\Omega$) if it would be correct to write $R: u \in H^1_g \rightarrow H^{-1}$? Note that in the problem in the video, $R$ is used in Newton's method as
$$ \begin{aligned} R(u^k + \delta u) &= R(u^k) + R'(u^k)(\delta u), \\ R'(u^k)(\delta u) &= -R(u^k), \\ u^{k+1} = u^k + \delta u, \end{aligned} $$
where $R'(u^k)(\delta u)$ is the Gateaux derivative and $R'(u^k)$ is the continuous linear operator represented by the Jacobian $J_R(u^k)$ during discretization. An additional thought I had is maybe $R$ cannot map $H^1_g \rightarrow H^{-1}$ because the Gateaux derivative requires that $R$ is defined for topological vector spaces, and while $H^1_0$ is a Sobolev space, and therefore a vector space, I am not so sure about $H^1_g$.
I think the root of my question is two fold:
(1) Since $H^1_g, H^1_0 \subset H^1$, and the dual of $H^s$ is $H^{-s}$, then is the dual of subspaces of $H^1$ also just $H^{-1}$? Maybe this is what is meant by the statement all separable Hilbert spaces are isomorphic to each other?
(2) Does $R$ have to map to the dual space? That is, why is the map defined as say $H^1_0 \rightarrow H^{-1}$ instead of a map between some other arbitrary space (e.g., maybe $R$ just maps to the same space like $H^1_0 \rightarrow H^1_0$... though maybe this doesn't make sense since it implies that the operator $R$ enforces $0$ on the boundary)?
