This is a "sequel" to that question where I explain why I need the spectrum of an operator given as the sum of a convolution and a function multiplication. Here, I am considering the following theorem, which would solve the issue.
Theorem. Let $L_1$ and $L_2$ be two bounded self-adjoint operators on $L^2(\mathbb{R})$. Assume the spectra of $L_1$ and $L_2$, respectively, are the intervals $[0,1]$ and $[a,b]$. Then we have the following
- $a\leq \frac{\left<v,(L_1+L_2)v\right>}{||v||^2}\leq 1+b$
- $\sigma(L_1+L_2)\subset [a,1+b],$
where $<.,.>$ and $||.||$ denote the $L^2(\mathbb{R})$ inner product and norm.
Idea of proof. Since $L_i$ are self-adjoint, and given their spectra, we have that
- $0\leq \frac{\left<v,L_1v\right>}{||v||^2}\leq 1$
- $a\leq \frac{\left<v,L_2v\right>}{||v||^2}\leq b.$
Then statement 1 in the theorem is implied from 1 and 2 above. Because $L_i$ are self-adjoint, statements 1 and 2 in the theorem are equivalent.
Ideas and questions. I found this paper that does it in general (not necessarily s-a or bounded) for the "Fredholm spectrum" under a condition ($L_1L_2$ compact) I can verify. I checked that the Fredholm is the essential spectrum, which is what I have.
Then I found this more recent paper about bounded s-a operators, where the authors improve a result that says that under the same condition as above, the essential spectrum of the sum is the sum of the essential spectra (minus the zero eigenvalue).
Is what I need in the theorem trivial? Does my idea of proof make sense? Is the result in the second paper what I should be using?
If I understand the second paper correctly, that would mean that the inequalities in my theorem are actually equalities and $\subset$ can be replaced by $=$.
It seems like everything works out well but I wanted someone who feels more confident with those topics to take a look at that.