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 $T$.
I do not think it is self-adjoint in general. If you define the operator $\mathcal{O}$ by $\mathcal{O}f(x) = \dfrac{f(x)}{1-x^2}$, then $G = T\mathcal{O}$. $\mathcal{O}$ is a self-adjoint operator which you can see pretty easily. It is a general result that if $A,B$ are self-adjoint, $AB$ is self-adjoint if and only if $A$ and $B$ commute. Hence $G = T\mathcal{O}$ is self-adjoint if and only if $T$ and $\mathcal{O}$ commute. Without knowing what $T$ is exactly, there's no way to guarantee that $G$ is self-adjoint. If $T = -i\frac{d}{dx}$, then $G$ isn't necessarily self-adjoint but if $Tf(x) = t(x)f(x)$ for some function $t$, then $G$ is self-adjoint.
• could you explain why the operator $O$ is self-adjoint? – user159356 Dec 14 '14 at 0:27
• It is self-adjoint because $$\langle \mathcal{O}f,g\rangle = \int_{-1}^1 \mathcal{O}f(x)\overline{g(x)}\,dx = \int_{-1}^1\frac{f(x)}{1-x^2}\overline{g(x)}\,dx.$$ You can peel the $\dfrac{1}{1-x^2}$ off of $f$ and put it on $g$ (note that complex conjugating $\dfrac{1}{1-x^2}$ doesn't change the function). – Cameron Williams Dec 14 '14 at 0:58