I am reading "Application number 3" of Banach contraction theorem from this article. I want to replace $L(a,b)$ in this example in the link with $L(0,\infty)$ so that. So we need to apply Banach contraction to:$$f(x)=g(x) + \lambda\int_0^\infty K(x,y)f(y)dy\cdots(1)$$
Suppose $f(x),g(x)\in L(0,\infty)$, and $K(x,y)$ is bounded measurable on $[0,\infty)\times [0,\infty)$ with $\int_0^\infty\int_0^\infty |K(x,y)|^2dydx<\infty$.
Do I need any more conditions on $f,g$ and $K$ such that the proof works in the link?
First I need to show that equation $(1)$ is well-defined, and that it is a self-map on $L(0,\infty)$. It is sufficient to show that $\psi(x)=\int_0^\infty K(x,y)f(y)dy\in L(0,\infty).$
Mimicking the proof in the link, we arrive$$\int_0^\infty |\psi(x)|^2 dx\leq \int_0^\infty\int_0^\infty |K(x,y)|^2dydx \int_0^\infty\int_0^\infty |f(y)|^2 dydx.$$Now, by assumption $\int_0^\infty |f(y)|^2 dy<\infty$ but $\int_0^\infty\int_0^\infty |f(y)|^2 dydx$ is not finite, right?
Please tell me how to proceed to find the bounds.