Self-adjointness

Question

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.

What is $O$? If you are going to reference another question, please link to it. — Thomas Andrews, Dec 14 '14 at 01:40
@TobiasHurth Then define $O$. Make posts self contained, please. — Pedro, Dec 14 '14 at 01:41
$\mathcal{O}f(x) = \frac{1}{1-x^2}f(x)$. 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 01:42
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 01:43

score 0 · Accepted Answer · edited Apr 13 '17 at 12:21

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I'm going to build on another problem of yours, Tobias: Why is this operator self-adoint .

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.

edited Apr 13 '17 at 12:21

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answered Dec 14 '14 at 02:27

Disintegrating By Parts

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@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 $T$. – Disintegrating By Parts Dec 14 '14 at 03:00
@TobiasHurth : The Schrodinger operators on infinite intervals with periodic potentials are not particularly well-known to me, even though I do know some things. Hopefully someone here will be able to help. I've wanted to learn about this subject myself, for exactly the application that you cite. Good stuff. Hopefully we can learn from someone here. – Disintegrating By Parts Dec 14 '14 at 03:14

Self-adjointness

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