For a matrix $A \in \mathbb{R}^{n \times n}$, we have the well-known Perron-Frobenius-Theorem which among other things establishes the following properties:
If $A$ is positive (i.e. $A_{ij}>0 \ \forall i,j$) or $A$ is non-negative and irreducible then we have
- The spectral radius $r(A)>0$
- The eigenvalue $r(A)$ is simple
- There is an eigenvector $v$ for the eigenvalue $r(A)$ such that $v_i >0 \forall i$
- There are no other eigenvectors $v'$ to any eigenvalue such that $v'_i >0 \forall i.$
The generalization of this theorem is the Krein-Rutman-theorem, which works for operators on Banach-spaces.
I am interested in a particularly easy class of such operators, namely those that are given by kernels, i.e. let $V$ be some space of functions on some measure space $(\Omega,\mu)$ and $k : \Omega \times \Omega \to \mathbb{R}$ a kernel, such that the operator $T$ is given by
$$Tf(u) = \int k(u,o) f(o) d\mu.$$
(For the sake of this question, we can choose this space as easy as we want to, e.g. compact intervals on the reals).
Now I was wondering if there is an easy intermediate generalization of the Perron-Frobenius-Theorem to this setting, that is, if we can simply transfer the statements of made above and treat the kernel $k(u,o)$ as a matrix, will everything still hold?
In particular, if $\Omega$ is a topological space, can we just say irreducibility is given if for all $o,o' \in \Omega$ there is a directed path $o\to o'$ such that the kernel stays positive along the path? I.e. something like a map $$p:[0,1] \to \Omega : \forall t \in [0,1] \ \exists \ \epsilon>0 \ s.t. \ k(p(t),p(t+\tau))>0 , \ \forall \tau < \epsilon$$ where $p(0) = o, p(1) = o'$.
(This is just my intuitive generalization of the fact that irreducibility is given if the directed graph obtained from the matrix $A$ is strongly connected.)