In another thread it was claimed that the operator $O : \operatorname{dom}(O) \subset L^2(-1,1) \rightarrow L^2(-1,1)$ is self-adjoint.

$$Of(x)= \frac{f(x)}{{1-x^2}}$$ It is obvious that $$\langle O f,g \rangle = \int_{-1}^{1} Ofg = \int_{-1}^{1} f \overline{Og} = \langle f,Og\rangle$$

This just shows symmetry and since $\operatorname{dom}(O) = \{f \in L^2: Of \in L^2\}$ is dense (contains all testfunctions) we just know from this that $O \subset O^*$.

So how can I prove the converse.

• I fail to identify the question here. – Pedro Tamaroff Dec 14 '14 at 1:39
• What is $O$? If you are going to reference another question, please link to it. – Thomas Andrews Dec 14 '14 at 1:40
• @TobiasHurth Then define $O$. Make posts self contained, please. – Pedro Tamaroff Dec 14 '14 at 1:41
• $\mathcal{O}f(x) = \frac{1}{1-x^2}f(x)$. I think that since it is a multiplication operator and $\frac{1}{1-x^2}$ is real, $D(\mathcal{O}^*)$ should be $D(\mathcal{O})$. – Cameron Williams Dec 14 '14 at 1:42
• Yeah I keep forgetting that ideas are a little bit different between the bounded and unbounded setting. It's a little annoying keeping the details straight between the two. – Cameron Williams Dec 14 '14 at 1:43

In the above problem, it is shown that, if $A : \mathcal{D}(A)\subseteq H\rightarrow H$ is symmetric with $(A\pm iI)$ surjective, then $A$ is densely-defined and selfadjoint.
As you noted, your operator $O$ is symmetric on its domain. To see that $(O\pm iI)$ are surjective, suppose $f \in L^{2}(-1,1)$ and define $$g_{\pm} = \frac{(1-x^{2})f}{x\pm i}.$$ Clearly $g_{\pm}$ are in $L^{2}(-1,1)$, but also $g_{\pm}\in\mathcal{D}(O)$ with $$(O\pm iI)g_{\pm} = f.$$ Conclusion: $\mathcal{O}$ is densely-defined and selfadjoint.
• @TobiasHurth : Yes, that little result comes in very handy in showing that domains are dense and operators are selfadjoint. I think you can why very it is very generally true that multiplication by real functions on $L^{2}$ spaces are densely-defined and selfadjoint. Concerning the example you reference in the previous problem: $\frac{1}{1-x^{2}}T\frac{1}{1-x^{2}}f$ is symmetric on its domain, whatever that domain may be, but no guarantees of selfadjointness for a general selfadjoint $T$. – DisintegratingByParts Dec 14 '14 at 3:00