# Perturbation of unbounded self-adjoint operators

In the paper "A CRITERION FOR THE NORMALITY OF UNBOUNDED OPERATORS AND APPLICATIONS TO SELF-ADJOINTNESS" by M. H MORTAD (http://arxiv.org/pdf/1301.0241.pdf), the author states the following theorem

## Theorem 1

Let $A$ and $B$ be two operators with domains $D(A)$ and $D(B)$ respectively, such that $A+B$ is densely defined. Then $A+B$ is

1. self-adjoint on $D(A)$ if $A$ and $B$ are selfadjoint and if B is bounded.

2. self-adjoint on $D(A) \cap D(B)$ whenever $A$ and $B$ are commuting selfadjoint and positive operators.

3. self-adjoint on $D(A) \cap D(B)$ whenever $A$ and $B$ are anti-commuting selfadjoint operators.

4. self-adjoint on $D(A)$ if $B$ is symmteric and $A$-bounded with relative bounded $a<1$, and $A$ is selfadjoint (Kato-Rellich).

My question concerns point 2. In this point the author refers the reader to C. Putnam, Commutation Properties of Hilbert Space Operators, Springer, 1967. I didn't find this theorem in that book, so if somebody knew that it is there for sure I would be really grateful for giving me more precise information about the location of that theorem. I'm curious why the positivity assumption is important there, in 3. for anti-commuting operators no positivity assumption was made.

• If you do not assume positivity, you could, for instance have B=-A, and then A+B=0, which has a larger domain than $D(A)\cap D(B)$. Commented Aug 12, 2013 at 17:00

To understand why he requires $A$ and $B$ to be positive, think of them as multiplication operators on some measure space. If they are commuting self-adjoint operators this is legitimate. So if $A$ is multiplication by $f$ on $L^2(X,\mu)$ and $B$ is multiplication by $g$ on $L^2(X,\mu)$, $A + B$ should be multiplication by $f + g$ --- but what is the domain of this operator? If $f$ and $g$ are positive then it is just the common domain of $A$ and $B$, the $L^2$ functions which when multiplied by either $f$ or $g$ remain $L^2$. But if we don't assume $f$ and $g$ are positive, there can be cancellation which could enlarge the domain. The obvious example is $g = -f$.
If $A$ and $B$ anticommute the picture is totally different. The simplest example of this is $A = \left[\begin{matrix}1&0\cr 0&-1\end{matrix}\right]$ and $B = \left[\begin{matrix}0&1\cr 1&0\end{matrix}\right]$. I guess the intuition could be that $A$ and $B$ are $45^\circ$ from each other. But formally the point is that when you're checking convergence, anticommutation yields $$\|(A + B)v\|^2 = \|Av\|^2 + \|Bv\|^2,$$ so if $(v_n)$ converges and $(A + B)v_n$ converges, then both $(Av_n)$ and $(Bv_n)$ converge. So $A+B$ is already closed on $D(A) \cap D(B)$.