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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

2 Answers2

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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.

Cameron Williams
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  • could you explain why the operator $O$ is self-adjoint? –  Dec 14 '14 at 00:27
  • 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 00:58
  • I think this just shows that the operator is symmetric on its maximal domain, but how does this show that the operator is self-adjoint? –  Dec 14 '14 at 01:07
  • Well a real-valued operator that is symmetric is self-adjoint (on its domain of course). – Cameron Williams Dec 14 '14 at 01:10
  • I have never heard this theorem before and highly doubt it, do you have a reference? Sorry, it is not that I don't believe you per se, but I find this statement so counterintuitive. –  Dec 14 '14 at 01:15
  • I don't see why it is counter-intuitive. Let me complete my argument. $$\langle \mathcal{O}f,g\rangle = \int_{-1}^1 \mathcal{O}f(x)\overline{g(x)},dx = \int_{-1}^1\frac{f(x)}{1-x^2}\overline{g(x)},dx \ = \int_{-1}^1 f(x)\overline{g(x)}{1-x^2},dx = \int_{-1}^1 f(x)\overline{\mathcal{O}g(x)},dx = \langle f,\mathcal{O}g\rangle.$$ So clearly $\mathcal{O}$ is self-adjoint. – Cameron Williams Dec 14 '14 at 01:17
  • I mean, we agree that $O$ is not a bounded operator? Then this just shows that $O$ is symmetric and from this we have $O \subset O^*$. I assume that $\operatorname{dom}(O) = {f \in L^2; \frac{1}{\sqrt{1-x^2}}f \in L^2}$ which is the maximal domain and of course $O$ is also densely-defined as this domain contains all the testfunctions. –  Dec 14 '14 at 01:19
  • You know what? I've never once heard the term symmetric operator to mean what you mean it to mean. I've always understood symmetric to mean the real analogue of self-adjoint. $\mathcal{O}$ may very well not be bounded (and isn't). That said, just consider the problem restricted to the intersection of the domains and work there. The result I quoted about self-adjoint operators also works if they are densely defined. – Cameron Williams Dec 14 '14 at 01:24
  • @CameronWillaims You know what? I think I will open a new thread that discuss self-adjointness and we look what happens. Apparently, we are both completely sure that we are right and therefore it will be interesting to see who is right. I will inform you as soon as I have opened the thread. –  Dec 14 '14 at 01:26
  • You're right here about symmetric operators. That was my bad. There is a pretty good question here about it: http://math.stackexchange.com/questions/788412/the-difference-between-hermitian-symmetric-and-self-adjoint-operators – Cameron Williams Dec 14 '14 at 01:34
  • how sure are you about the "composition is self-adjoint iff the commutator vanishes"? –  Dec 14 '14 at 01:57
  • If $AB$ is self adjoint, $(AB)^* = AB$ but $(AB)^* = B^* A^* = BA$ so $AB = BA$. Likewise if $A$ and $B$ commute, then $AB = BA$ but $BA = B^* A^* = (AB)^$ so $AB = (AB)^$. – Cameron Williams Dec 14 '14 at 02:04
  • and you know that $(AB)^* = B^A^$ even in the unbounded case? - I mean, I don't, but I am new to this field. –  Dec 14 '14 at 02:12
  • Do you want to move this to chat? – Cameron Williams Dec 14 '14 at 02:16
  • Let us continue this discussion in chat. – Cameron Williams Dec 14 '14 at 02:16
  • check this question http://math.stackexchange.com/questions/663828/adjoint-of-unbounded-operators-product-and-sum , apparently, also your other statement was wrong, so I think you should delete your answer, as this might cause confusion. (no offense). –  Dec 14 '14 at 03:54
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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 $T$.

TZakrevskiy
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