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.
Best Answer
Your argument is fine. The numbers you use are what is known as the numerical range of an operator. It is a basic fact that the spectrum of an operator is contained in the closure of the numerical range. So you show that the numerical range is contained in $[a,1+b]$, which then implies that $\sigma(L_1+L_2)\subset[a,1+b]$.
The inclusion $\sigma(A+B)\subset\sigma(A)+\sigma(B)$ for commuting $A,B$ holds, but equality usually fails. I can only see the abstract of the second paper you mention; what I can say is that the equality in the abstract is not true in general (note in any case that the paper uses the union of the spectra and not their sum). Consider $$ A_0=\begin{bmatrix} 1&0\\0&0\end{bmatrix},\qquad\qquad B_0=\begin{bmatrix} 3&0\\0&0\end{bmatrix}. $$ Then $\sigma(A_0)=\{0,1\}$, $\sigma(B_0)=\{0,3\}$, and $\sigma(A_0+B_0)=\{0,4\}$. This is neither $\sigma(A_0)+\sigma(B_0)=\{0,1,3,4\}$ nor $\sigma(A_0)\cup\sigma(B_0)=\{0,1,3\}$. You do have $\sigma(A_0+B_0)\subset\sigma(A_0)+\sigma(B_0)$.
You could also consider $$ A_1=\begin{bmatrix} 1&0\\0&0\end{bmatrix},\qquad\qquad B_1=\begin{bmatrix} 0&0\\0&3\end{bmatrix}. $$ Then $\sigma(A_1)=\{0,1\}$, $\sigma(B_1)=\{0,3\}$, and $\sigma(A_1+B_1)=\{1,3\}$. This is not $\sigma(A_0)+\sigma(B_0)=\{0,1,3,4\}$; but you do have $$\sigma(A_1+B_1)\setminus\{0\}=\{1,3\}=(\sigma(A_1)\cup\sigma(B_1))\setminus\{0\}.$$
These examples can be easily turned into examples about the essential spectrum by considering $A=\bigoplus_{n\in\mathbb N}A_0$ and $B=\bigoplus_{n\in\mathbb N}B_0$ and similarly for $A_1$ and $B_1$.
What the examples show is the fact that the spectrum will never tell you the "position" of the operator. A selfadjoint operator is described by its spectrum together with its spectral projections (or spectral measure in infinite-dimension). Two selfadjoint operators that commute will have the same spectral projections, but the location of the eigenvalues can be different, as the two examples show. This leads to different possibilities for the spectrum of the sum.