What exactly is an integral kernel?

Solution 1:

Intuitively, $K(x,y)$ is a "continuous" matrix $K$ acting on a "continuous" row vector vector $f$: $$ \sum_{x}f(x)K(x,y) = \int f(x)K(x,y)dx $$ The associated quadratic form would be $$ \sum_{x}f(x)K(x,y)g(y)^{\star} = \int\int K(x,y)f(x)\overline{g(y)}dxdy $$ This would be selfadjoint if $K(x,y)=\overline{K(y,x)}$, analogous to a Hermitian matrix, and would lead to an observable. A type of condition that allows everything to work really well is the Hilbert-Schmidt condition: $$ M^{2}= \iint |K(x,y)|^{2}dxdy < \infty. $$ That may be a little restrictive for what you're doing. I'm not sure, but it's very nice from a Mathematical point-of-view because it allows one to define a bounded linear operator $K : L^{2}(\mathbb{R})\rightarrow L^{2}(\mathbb{R})$: $$ Kf = \int K(x,y)f(x)dx. $$ This follows from Cauchy-Schwarz: $$ |Kf(y)|^{2} \le \int |K(x,y)|^{2}dx\int |f(x)|^{2}dx \\ \int |Kf(y)|^{2}dy \le \int\int |K(x,y)|^{2}dxdy\int|f(x)|^{2}dx \\ \|Kf\|^{2} \le \int\int|K(x,y)|^{2}dxdy \|f\|^{2} \\ \|Kf\| \le \left(\int\int|K(x,y)|^{2}dxdy\right)^{1/2}\|f\| \\ \|Kf\| \le M\|f\|. $$ Here $M$ is the Hilbert-Schmidt constant defined above. This describes how Hilbert originally started in generalizing matrices. Eventually von Neumann's operator theory replaced these ideas because (1) Matrices cannot distinguish between different linear operators and (2) linear operators allowed more general objects that were better suited for Quantum.