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.



When does the spectral radius strictly increase?
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.


