Consider the Fredholm integral equation of the second kind given by $$ f(x)=g(x)+\int_a^bk(x,y)f(y)\ \mathrm dy. $$ In any source I could find online, including some more advanced ones, existence of solutions (and also, uniqueness, and continuous dependence on initial conditions) is proven under conditions like $\|k\|_{\infty}< 1$ or $\sup_{x\in[a,b]}\int_a^b|k(x,y)|\ \mathrm dy<1$. In the textbook Volterrra Integral and Functional equations by Gripenberg, Londen and Steffans, when working on $L^p[a,b]$, this is proven (pages 227 and 235) under the condition that $$ \|k\|_{p}:=\sup_{\substack{\phi\in L^q[a,b]\\\psi\in L^p[a,b]}}\int_a^b\int_a^b|\phi(x)k(x,y)\psi(y)|\ \mathrm dx\ \mathrm dy<1, $$ where $p $ and $q$ are conjugate, i.e. $\frac1p+\frac1q=1$. I intend to work on $L^1[a,b]$, with a positive kernel $k$, in which case the condition simplifies to $$ \sup_{\psi\in L^1[a,b]}\int_a^b\int_a^b|k(x,y)\psi(y)|\ \mathrm dx\ \mathrm dy<1, $$ i.e. the operator norm $\|T\|_{op}$ of $$T:L^1[a,b]\to L^1[a,b]:f(\cdot)\mapsto \int_a^bk(\cdot,y)f(y)\ \mathrm dy$$ should be bounded by $1$.
My question: to me, it makes sense that essentially we want the operator $T$ to be a contraction. To me, this seems to be the case whenever $\rho(T)<1$, i.e. the spectral radius is bounded by $1$. By Gelfand's formula, this condition is weaker than $\|T\|_{op}<1$. Are there any results on existence (and uniqueness, continuous dependence on initial conditions) under this condition on the spectral radius?
Im not 100% sure if this is what you are looking for, but it seems like a standard application of the Neumann series to me:
Let $X$ be a Banach space and $T:X \to X$ continuous. So in your special case $X=L^1[a,b]$ and $T$ the integral operator in your question.
Given $g \in X$ we are trying to find $f\in X$ that solves: $$ g = (I- T ) f .$$
Now if $ \rho(T) <1$, then the following series converges absolutely (in operator norm) by the root test (geometric series): $$ (I-T)^{-1} = \sum_{n=0}^\infty T^n. $$
Therefore $f = (I-T)^{-1}g $ is the unique solution to the problem and the solution depends continuously on the input $g$, since $(I-T)^{-1}$ is continuous.