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 
$$ T: W^{k,p}(\mathbb{R}^2, \omega) \rightarrow W^{k-1,p}(\mathbb{R}^2, \omega)  $$ 
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 $k$ and $p$ 
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 
$W^{k-1,p}(\mathbb{R}^2, \omega)$. 

2) The map $T$ is a contraction mapping of metric spaces, i.e. 
$$ |T(v)| < |v|   \qquad \forall v.$$

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 contraction mapping 
principal. The map $T$ arises as the linearization of some map.