Let $H^s$ denote the Sobolev Hilbert space. Let $\mathbb{T}$ be the circle. Let $H^{s,r}$ be the Hilbert space formed by completing $H^s(\mathbb{T})\otimes H^r(0,1)$. This space can be identified as the space of "dominating mixed smoothness": the subset of functions $u(x,y) \in L^2(\mathbb{T} \times [0,1])$ with norm $$\|u\|_{H^{s,r}}^2 = \sum_{\sigma \leq s} \sum_{\rho \leq r} \|\partial_x^{\sigma} \partial_y^{\rho} u\|_{L^2}^2$$ Let $s,r \geq 2$. I am interested in finding a space, say $A$ such that the laplacian operator is an isomorphism from $H^{s,r}$ to $A$, given some boundary conditions, ie the map $\Delta : H^{s,r}\rightarrow A$ is invertible.
I am not sure how I can define this space, ideally it would be in terms of derivatives. Suppose for a moment we are on $\mathbb{T}^2$ instead of the circular strip. Then we can work in the Fourier spectrum, we have $$\|u\|_{H^{s,r}}^2 \sum_{m,n} (1+m^2)^s(1+n^2)^r|\hat{u}_{m,n}|^2$$ and we can define a norm on $A$: $$\|u\|_{A}^2 = \sum_{m,n} \frac{(1+m^2)^s(1+n^2)^r}{(1+m^2+n^2)^2}|\hat{u}_{m,n}|^2$$ then $I-\Delta: H^{s,r}\rightarrow A$ is clearly an isomorphism. I have tried to mess around with this norm, trying to find some equivalent norm in terms of derivatives hoping to get some insight into the case without Fourier series. But I have no success, has anyone some suggestions?
EDIT:
So looking some more at the $\mathbb{T}^2$ case I am reminded that we have $(1-\Delta)^{-1}: H^{-2}(\mathbb{T}^2) \rightarrow L^2(\mathbb{T}^2)$ is an isomorphism. Therefore we could write $$\|u\|_{H^{s,r}}^2 = \sum_{\sigma \leq s} \sum_{\rho \leq r} \|(I-\Delta)^{-1}\partial_x^{\sigma} \partial_y^{\rho} u\|_{L^2}^2 = \sum_{\sigma \leq s} \sum_{\rho \leq r} \|\partial_x^{\sigma} \partial_y^{\rho} u\|_{H^{-2}}^2$$ or actually dividing out $\frac{(1+m^2)^s(1+n^2)^r}{(1+m^2+n^2)^2}$ we get something like: $$\|u\|_{A}^2 = \|u\|_{L^2(\mathbb{T}^2)}^2 + \cdot \cdot \cdot + \|\partial_x^s \partial_y^r u\|_{H^{-2}(\mathbb{T}^2)}$$
So it seems this space $A$ is a subset of $L^2$ where the dominating mixed derivatives are not quite in $L^2$ but rather in $H^{-2}$. So perhaps on other domains where I can't use the Fourier approach, I can go ahead and define $A$ in such a manner. Specifically, define $A^{s,r,t}(\Omega)$ to be subset of $L^2$ whose partial derivatives up to $\partial_x^s \partial_y^r$ are in $H^{-t}(\Omega)$. But $H^{-s}(\Omega)$ is usually the dual of $H_0^s(\Omega)$ and not $H^s(\Omega)$. Not sure why that is.
Perhaps it would be a good exercise to check $\partial_x$ and $\partial_y$ are bounded maps on $H^{1,1} \rightarrow A^{1,1,1}$.