I have come across a claim that I am quite certain is true, but would be benefited from seeing a proof sketch.
Suppose we know the following equation: $$ f(z)=\int_0^T K(z,x)f(x)\,\mathrm{d}x, $$ which is a standard homogeneous Fredholm equation of the second kind. Suppose that $K>0$ on $[0,T]\times [0,T]$ except $K(T,x)=0$ for all $x \in [0,T].$ Therefore, any solution satisfies $f(T)=0.$
Assume $f(z)$ is a solution. Intuitively, it seems like if $f(0)=0,$ then $f(z)\equiv 0.$ I say this because we have \begin{align*} 0&=\int_0^T K(0,x)f(x)\,\mathrm{d}x \\ &=\int_0^T\int_0^TK(0,x)K(x,x_1)f(x_1)\,\mathrm{d}x_1\mathrm{d}x \\ &=\int_0^T\int_0^T\int_0^T K(0,x)K(x,x_1)K(x_1,x_2)f(x_2)\,\mathrm{d}x_2\mathrm{d}x_1\mathrm{d}x \\ &\vdots \\ &=\int_0^T...\int_0^T K(0,x)...K(x_{n-1},x_n)f(x_n)\, \mathrm{d}x_n ...\mathrm{d}x. \end{align*}
In all of the integrals, $K>0$ a.e. It seems like by repeating this process, we must have $f\equiv 0.$ Note that $K(z,x)\neq K(x,z)$ and $K$ is not separable. Thank you.
$f\equiv0$ is certainly a solution. The question is wether there are other nontrivial solurtions. Consider the case $K(z,x)=g(z)\,h(x)$ with $g(T)=0$, $g\not\equiv0$. Then $g$ is a solution if $\int_0^Th(x)\,g(x)\,dx=1$.