Is $T$ a compact mapping from $W_{0}^{1,2}\left(\Omega\right)$ into itself? Let $\Omega$ be an open bounded subset in $\mathbb{R}^{6}$ and $f$ be in $L^{8}\left(\Omega\right)$. For any $w$ in $W_{0}^{1,2}\left(\Omega\right)$, define $T\left(w\right)$ be in $W_{0}^{1,2}\left(\Omega\right)$ such that
$${\displaystyle \int_{\Omega}\nabla\left(Tw\right)\nabla\phi dx}={\displaystyle \int_{\Omega}f\phi w dx}
 \ \ \ \  \forall\phi\in W_{0}^{1,2}\left(\Omega\right)$$
Is $Tw$ well defined? Is $T$ a compact mapping from $W_{0}^{1,2}\left(\Omega\right)$ into itself? 
 A: By Lax-Milgram there exists only one $\theta\in W_0^{1,2}$ such that
\begin{equation}
\int_\Omega\nabla\theta\cdot\nabla\phi dx=\int_\Omega f\phi dx
\end{equation}
for all $\phi\in W_0^{1,2}$. So in fact $T$ is well defined, linear and continuous from $W_0^{1,2}$ to $W_0^{1,2}$ with norm $|\theta|_{W_0^{1,2}}$. Obviously it is compact, since the range of $T$ has only one element $\theta$.

(Sorry that I made a mistake before. I hope this time every thing should be right.)
As Ian said, we can consider the problem
\begin{equation}
\int_\Omega\nabla\theta\cdot\nabla\phi dx=\int_\Omega f\omega\phi dx
\end{equation}
for all $\phi\in W_0^{1,2}$. First we should prove that the RHS is in fact in $H^{-1}$. We obtain
\begin{align}
|\int_\Omega f\omega\phi dx|&\leq\int_\Omega |f\omega\phi|dx\\
&\leq c\int_\Omega |f|^3+|\omega|^3+|\phi|^3dx\\
&=:S_1+S_2+S_3.
\end{align}
Since $f\in L_8$, $S_1$ is finite, and using the Rellich's embedding theorem we know $W_0^{1,2}$ is embedded in $L_3$ (consider the Sobolev numbers 1-6/2=0-6/3), So the functional at RHS is linear and continuous in $W_0^{1,2}$ and therefore $T$ is well defined.
The second part in still in consideration. (What I wrote before was wrong, sorry about that. $H^2$ is not compactly embedded into $H_0^1$.)

Ok, I found a proof that $T$ is compact if we assume the bdry is lipschitzian: Indeed, we can show that $fw$ is in $L_2$ and see that the dimensions really play a role here. We have
\begin{align}
\int (fw)^2&\leq c\int f^6+w^3+w^3<\infty,
\end{align}
the inequality follows from the same argument as before. By regularity theory we know that the solution $\theta=\bar{T}w$ in fact lies in $H^2$, and the identity operator $I:H^2\rightarrow H^1$ is by Rellich's theorem compact, so the decomposition $T=I\circ \bar{T}:H_0^1\rightarrow H^1$ is compact. We also know that all solutions are really in $H_0^1$, so $T$ is compact between $H_0^1$ and $H_0^1$.
