Bochner's Theorem and Universal Kernels

Question

Bochner's theorem asserts that a shift-invariant and properly scaled continuous kernel $K(x,y) = k(x-y)$ is positive definite (and hence a reproducing kernel of some RKHS) if and only if its Fourier transform $p(w)$ is a probability distribution: $$k(x-y) = \int_{\mathbb{R}^d} p(w) e^{i w^T(x-y)} dw$$ I am now wondering what this expression tells us about the universality of the corresponding RKHS. For which probability distributions $p(w)$ is the kernel $k(x-y)$ universal? Are there necessary and sufficient conditions?

Answer 1

We can write this continuous positive definite kernel as $$k(x-y) = \int_{\mathbb{R}^d} p(w) e^{i w^T(x-y)} dw = \int_{\mathbb{R}^d} e^{i w^T(x-y)} d\mu(w),$$ in which $$\mu(A)=\int_{A} p(w) dw,$$ to each measurable set $A\subset \mathbb{R}^d$, and $\rho:\mathbb{R}^d\to [0,\infty)$ is a measurable function.

If $\rho\in L^1(\mathbb{R}^d)$ and $supp(\mu)$ has positive Lebesgue measure then Theorem 15 of https://jmlr.org/papers/volume7/micchelli06a/micchelli06a.pdf tell us that $K$ is a universal kernel.

Please see also Theorem 3.7 in https://doi.org/10.1007/s11117-016-0414-4 to density in $C_0(\mathbb{R}^d)$.

You can find some related result searching for "\( K(x,y)=f(x-y)\) universal kernel" on SearchOnMath, for instance.