Perturbation of unbounded self-adjoint operators

Question

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.

Answer 1

I'm not familiar with Putnam's book, but part (2) of this theorem should be available in any of the standard references, e.g., Conway's Course in Functional Analysis or Reed and Simon, Functional Analysis vol. 1.

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 $D(A) \cap D(B)$.

Answer 2

Your point 2. is a special case of Lemma 4.16.1 from Putnam's book.