Problem: Let $K,f,g$ be in $calD(bR^d)$, and $K$ is radial (for definition, see Problem 2). Show that $$ex int (K*f)(x)g(x) d x=int f(x)(K*g)(x) d x. eex$$
Proof: $$eex ea int (K*f)(x)g(x) d x &=int sez{int K(x-y)f(y) d y} g(x) d x\ &=int sez{int K(y-x)g(x) d x} f(y) d y\ &=int (K*g)(y)f(y) d y. eea eeex$$