Disclaimer: This is a question in functional analysis, on which I don't have much background. It arose from me trying to prove on my own a folklore result in probability theory.
Consider an integral transform
$$(T f)(x)=\int_{a}^b K(x,y) f(y) dy$$
where $K$ is defined on $[a,b]^2$, and is continuous and positive, meaning that $K(x,y)>0$ for all $x,y\in[a,b]^2$. The domain of $T$ is $C([a,b])$, i.e., the set of continuous functions in $[a,b]$.
Krein-Rutman theorem (or the more basic Jentzsch theorem) tells us that $T$ has a real and positive eigenvalue $\lambda$, which has maximum modulus over all the eigenvalues of $T$. Furthermore, it says that the corresponding eigenvector $u$ is positive, and that there is a spectral gap, meaning that all the other eigenvalues are inside a disc with radius strictly smaller than $\lambda$.
Here is my question: Fix a positive $f\in C([a,b])$. Is it true that for large $n$, the function $T^n f$ is essentially aligned with the eigenvector $u$?
In the finite dimensional case (where Krein-Rutman is just Perron-Frobenius), we immediately get this result by applying Jordan decomposition.
Formally, the result I am looking for is that there exists a real positive sequence $(a_n)_{n\in \mathbb{N}}$ such that $\frac{T^n f}{a_n}$ converges to $u$ in some norm (ideally the sup-norm).
