Let $A$ and $B$ be two unbounded self-adjoint operators. From this mathoverflow post, for instance, we know that $A + B$ is self-adjoint on $\mathcal{D}(A) \cap \mathcal{D}(B)$ if $A$ and $B$ are commuting and positive operators (Putnam's book is cited as a reference there). My question is: assume $A$ is positive and $A$ and $B$ commute. $B$ is not positive though, but $B$ is a relatively bounded perturbation of $A$. Could we still say $A + B$ is self-adjoint?
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$\begingroup$ I don't know what 'relatively bounded perturbation' means, but maybe the top answer in the linked post would help? That answer suggests you should think of $A$ and $B$ as multiplication by measurable functions on $L^2(X,\mu)$. Is it possible that there would be 'cancellation' that enlarges the domain? $\endgroup$– Alex ZornCommented Apr 15, 2015 at 18:22
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$\begingroup$ Have you looked at Kato's book Perturbation theory for linear operators? It has a whole chapter that deals with questions of this kind. The most used perturbation results are there. $\endgroup$– Liviu NicolaescuCommented Apr 15, 2015 at 18:36
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$\begingroup$ The sum of two commuting s.a. possibly unbounded operators is always self-adjoint, it is just that in the general case one has to use a more subtle definition of the sum. Rather than the algebraic sum as described above, one takes its closure. Both facts become quite transparent if one uses the fact that the operators can be simultaneously diagonalised, i.e., represented as multiplication by measurable functions on an $L^2$-space (spectral theorem). $\endgroup$– reportCommented Apr 15, 2015 at 18:39
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This will work if you take the assumption that $A,B$ commute in a sufficiently strong sense (commuting resolvents would be enough). Then no extra assumption is needed.
There is a version of the spectral theorem that says that there is a projection valued measure that represents both $A$ and $B$: $$ A = \int s\, dE(s,t) , \quad B = \int t\, dE(s,t) $$ See here. Clearly this makes $\int (s+t)\, dE(s,t)$ self-adjoint on a suitable domain. Note, however, that this domain is not necessarily $D(A)\cap D(B)$, as simple examples show (multiplication by $|x|$ and $x$).
answered Apr 15, 2015 at 18:39
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$\begingroup$ Very nice answer! Suppose we take the following example: $A = \Delta$, the Laplacian on the torus, and $B = ik\frac{\partial}{\partial x_1}$, where $k \in \mathbb{R}$. Clearly they commute. But I am not sure whether they strongly commute in the sense you describe. Could you please shed a little light on this point? Thanks! $\endgroup$– amanoCommented Apr 15, 2015 at 18:56
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$\begingroup$ @amano: This definitely works because you can use the Fourier transform to (simultaneously) make multiplication operators out of both of them. (So you in fact don't need any general theory to make $A+B$ self-adjoint on a suitable domain.) $\endgroup$ Commented Apr 15, 2015 at 19:03
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$\begingroup$ @amano $ikX$ is relatively $\Delta$ compact, in particular, it is relatively $\Delta$ bounded with relative bound $0$. Then see my answer below. $\endgroup$– ifwCommented Apr 16, 2015 at 6:48
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$\begingroup$ @ChristianRemling Is there any way to read the spectra of $A$ and $B$ from these two variable projection valued measures? In the single variable case, we know that Spec$(A) = $ supp$ E_\lambda$, where $A = \int \lambda dE_\lambda$. $\endgroup$– SMSCommented Aug 21, 2022 at 20:32
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$\begingroup$ @SMS: I think there's a similar description here: $x\notin\sigma (A)$ if and only if $E(M)=0$ for some set of the form $M=(x-\delta,x+\delta)\times\mathbb R$. $\endgroup$ Commented Aug 22, 2022 at 13:50