For bounded linear operators $A$ and $B$ on a Banach space $X$, I'm looking for results which imply that $r(A) < r(A+B)$ (note the strict inequality), where $r(A)$ denotes the spectral radius of $A$. Are there any assumptions on the operators $A$, $B$ which will guarantee this?

I should not have left this so general; specifically, I have two integral operators $A$, $B$ on $L^1(0,1)$. I know the spectral radius for $A$, but all I know about $B$ is that it's compact with a non-negative kernel. Ideally, I would like to show that $r(A)<r(A + B)$.

Here's some more information regarding the operators: $$(A\psi)(y) := \int_0^1 s(x)g(y,x) \psi(x) \, dx,$$ and $$(B \psi)(y) := \int_0^1 d(y)f(x) \psi(x) \, dx$$ where all component functions are continuous and non-negative (in fact, take $s(x)$, $d(y)$, and $f(x)$ to be strictly positive on $[0,1]$). The function $g(y,x)$ has some properties that allowed me to compute $r(A)$ exactly.

These operators came up in a mathematical ecology model, and none of them have self-adjoint kernels.

  • $\begingroup$ There are easy examples with $\dim(X)=2$ even if you exclude the trivial case $A=0$. I suspect you really want to add in some hypotheses to get answers more relevant to the problem you are working on $\endgroup$ – Yemon Choi May 15 at 22:10
  • $\begingroup$ In fact I wonder if there is a typo in your question, because in its current form there are easy examples with $\dim(X)=1$ :) $\endgroup$ – Yemon Choi May 15 at 22:11
  • $\begingroup$ Regarding the updated question: do you have any positivity properties or self-adjointness for the kernel that defines $A$? (We have to rule out possibilities like $A=Q-B$ where Q is uasinilpotent) $\endgroup$ – Yemon Choi May 16 at 0:00
  • 2
    $\begingroup$ "can give some more context about the problem I'm working on if it helps." The best thing to do is to describe the problem as you have it. So if you just post your $A$, that would be the best. $\endgroup$ – fedja May 16 at 0:01
  • $\begingroup$ Is $g$ positive as well? If not, for all we know, you still may have $d=1$, $f=s$, $g=-1$. $\endgroup$ – fedja May 16 at 6:04

Then you are in good shape. The general "theorem" is the following:

Let $K_A(x,y)\ge 0$, $K_B(x,y)>0$ be any continuous kernels on $[0,1]^2$ and let $A,B$ be the corresponding linear integral operators in $L^1[0,1]$. Then $r(A+B)>r(A)$.

Proof: Since $K_A$ is bounded and $K_B$ is separated from $0$, you have $K_A+K_B\ge cK_A$ pointwise with some $c>1$, so for any non-negative $\psi$, we have $(A+B)^n\psi\ge c^nA^n\psi$. Since the norm of any power of an operator with non-negative kernel can be checked on non-negative functions, we immediately get $r(A+B)\ge cr(A)$. This covers all cases except $r(A)=0$. However, in that case $r(A+B)\ge r(B)>0$.

It all gets much more interesting if the kernels are just such that $A$ and $B$ are bounded in $L^1$, but it seems like you just don't need anything more general than the trivial observation above.

  • $\begingroup$ Thank you! A result like this is exactly what I was grasping for. $\endgroup$ – Matt R. May 17 at 2:29

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