Kernel of an integral transform

Remarks

Example: the Laplace transform $F(s) = \int_0^\infty f(t) e^{-st}\, dt$ has kernel $K(t, s) = e^{-st}$ on $[0, \infty) \times (0, \infty)$; this is read off by comparing with the formula above. Non-example: the function $f$ itself is not a kernel of $T$; the kernel does not depend on the input $f$, only on the transform. Suppose the transform is defined on all bounded functions with compact support. Then the kernel is determined by the transform up to changes on a set of measure zero in $I \times J$. A kernel is symmetric when $I = J$ and $K(t, u) = K(u, t)$.