Let $(a,b)\subseteq \mathbb{R}$ bounded, let $K(x,y):(a,b)\times(a,b)\to\mathbb{R}$ be a kernel, consider the operator $T:L^2(a,b)\to L^2(a,b)$, $f\mapsto g(x)=\int_{(a,b)} f(y)K(x,y) \, dy$ if $$K\in L^2(a,b)\times(a,b)\Bigg(i.e.\, \int_a^b\int_a^b |K(x,y)|^2\,dxdy<+\infty \Bigg)$$ then $T$ is well-defined meaning $\mathcal{I}mT\subseteq L^2(a,b)$, linear and continuous $$Tf(x)=\int_a^bf(y)K(x,y)dy\leq\int_a^b|f(y)||K(x,y)|dy\leq(\int_a^b|f(y)|^2dy)^{\frac{1}{2}}(\int_a^b|K(x,y)|^2dy)^{\frac{1}{2}}$$ hence $$|Tf(x)|^2\leq\|f\|^2_{L^2}\int_a^b|K(x,y)|^2dy$$ then we take the integral of both sides: $$\int_a^b|Tf(x)|^2dx\leq\int_a^b(\|f\|^2_{L^2}\int_a^b|K(x,y)|^2dy)dx$$ I do not understand if we can put out the integrals $\|f\|^2_{L^2}$ and why, and then apply Fubini-Tonelli to get on the r.h.s $$\|f\|^2_{L^2}\int_a^b\int_a^bK(x,y)dydx$$ My questions are:
- Can I pull (and why) the norm of $f$ out of the integral and why (I feel like it is already a number and does not vary under the integral but since it is an integral itself by definition I have doubts and feel uncomfortable because of it)
- when I wrote in the beginning "be a kernel" I was not sure, I suppose that it is positive-definite kernel, is it right?
- (EDITED) I am trying to narrow what is a Fredholm operator. I see that $T$ should satisfy $$\|T\|^2\leq\|K\|^2_{L^2}$$ I do not understand what does the kernel here for $K$ mean and what are the constraints for $K,$ now I see only $L^2$ integrability