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Observe that given a non negative function $\omega: \mathbb{R^2} \rightarrow [0, \infty)$, we can define the weighted $L^{p}(\mathbb{R}^2, \omega) $ spaces. They are measurable functions $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ such that their weighted $L^p$ norm is finite i.e., $$ \int |f|^p \omega d \mu < \infty. $$ Similarly, we can define the weighted Sobolev spaces $W^{k,p}(\mathbb{R}^2, \omega)$. They are the completion with respect to the weighted $L^p$ norm of the spaces of $k$ times differentiable functions that have finite weighted Sobolev $p$-norm. Now consider the linear operator $$ T: W^{k,p}(\mathbb{R}^2, \omega) \rightarrow W^{k-1,q}(\mathbb{R}^2, \omega) $$ given by $$ T(v(x,y)) := \frac{\partial v(x,y)}{ \partial x}.$$ My question is as follows: does there exist some weight $\omega(x,y)$ and integers $k$, $p$ and $q$ such that the following are true:

1) The functions $(x,y) \rightarrow x$, $(x,y) \rightarrow e^x$, $(x,y) \rightarrow y$ and $(x,y) \rightarrow e^y$ all belong to $W^{k,p}(\mathbb{R}^2, \omega)$ and $T$ maps into $W^{k-1,q}(\mathbb{R}^2, \omega)$.

2) The map $T$ is a bounded linear operator.

3) The map $T$ is surjective.

My guess is that $$ \omega(x,y) := e^{-(x^2+y^2)} $$ should do the trick, but I am not certain. The motivation for this question is to show that a certain pde has a solution, via the Implicit Function Theorem. The map $T$ arises as the linearization of some non linear operator.

$\textbf{Remark:}$ If $\omega(x,y) \equiv 1$, then I believe condition 2) will be satisfied for appropriate choices of $p$ and $q$. I believe this follows from the Sobolev embedding theorem. However the exponential function and projection maps are not part of the Sobolev spaces. Hence I am considering weighted Sobolev spaces.

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I would think that if $\omega$ is symmetric in $ x$ and $y$ (as your example) that it cannot satisfy 2) and 3). No claim of a proof or even any intuition. –  Craig Aug 20 '13 at 1:12
    
If you forget about condition 1), then the weight $w(x,y) =1$ does satisfy condition 2). That is symmetric. I agree surjectivity may be a problem. –  Ritwik Aug 20 '13 at 6:12

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