Preserve self-adjoint properties I was thinking about this problem recently:
Let $T$ be a self-adjoint operator on $L^2((-1,1),d x)$. Now you define an operator $G$ by $G(f) := T(\frac{f}{(1-x^2)})$ with $\operatorname{dom}(G):=\{f  \in L^2(-1,1); \frac{f}{(1-x^2)} \in \operatorname{dom}(T)\}$. Is this operator also self-adjoint then?
Of course this question could be easily generalized, but I wanted to ask this example first

Essentially, you need to check that $\forall f,g\in dom(G)\cap dom(T)$ you have 
$$(f,T[g]/\phi) = (f,T[g/\phi]),$$
where $(\cdot,\cdot)$ is a scalar product in $L^2$ and $\phi$ is the function $\frac{1}{1-x^2}$.
I don't quite see how it could be possible for generic $T$.
