Homogeneous Fredholm Integral Equation of Second Kind with Boundary Conditions
0
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^Tint_0^TK(0,x)K(x,x_1)f(x_1),mathrm{d}x_1mathrm{d}x \ &=int_0^Tint_0^Tint_0^T K(0,x)K(x,x_1)K(x_1,x_2)f(x_2),mathrm{d}x_2mathrm{d}x_1mathrm{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 ...mat...
