Are Matérn class kernels universal kernels or not?

Question

This is a question that I can't find the solution. I don't know it is a open question or it is a well-known result that can be attained from several lemmas.

Here are the definition of Matérn class kernel:

$M_{\sigma^2,\nu,\rho}(x,y)=\sigma^2\frac{2^{1-\nu}}{\Gamma(\nu)}(\sqrt{2\nu}\frac{||x-y||}{\rho})^{\nu}K_{\nu}(\sqrt{2\nu}\frac{||x-y||}{\rho})$

where $\sigma^2$ is a length scale parameter, $\Gamma(.)$ is gamma function, $K_{\nu}$ is the modified Bessel function of the second kind, and $\nu,\rho$ are non-negative parameters of covariance.

I use the definition of universal kernel in Universal Kernels(Charles A. Micchelli):

Given any prescribed compact subset Z of a Hausdorff topological space X, A kernel is universal if for any positive number $\epsilon$ and any function f ∈ C(Z) (continuous function on Z) there is a function g ∈ K(Z) (RKHS space generated by the kernel) such that sup norm $||f-g||_{Z}<\epsilon$

I know that frequently used Gaussian and Laplacian kernels are universal kernels and they are special case of Matérn class so I am wondering what kind of $\nu$ will lead to a universal kernel.

I am interesting in this problem because I am now studying Gaussian Process and Bayesian Optimization [Niranjan Srinivas,Jonathan Scarlett, Jasper Snoek] and I find that not only in theory they consider Matérn/Gaussian Kernel but also in real-world implementation. The default Kernel choice of many Bayesian optimization tools are Matérn 5/2. I don't think it is reasonable if it's not universal but I don't know how to prove it.

Answer 1

Yes, Matérn kernels are universal, for any parameters.

First note that $M_{\sigma^2,\nu,\rho}(x, y) = \Psi(x - y)$, i.e. the kernel is translation-invariant. It is also bounded and continuous.

The first main result we'll need is from:

Sriperumbudur, Fukumizu, and Lanckiret. Universality, Characteristic Kernels and RKHS Embedding of Measures. JMLR 12(Jul):2389−2410, 2011.

They show that the notion they call $cc$-universal is equivalent to the universality of Michhelli et al. (Remark 3). They also show that for bounded, continuous, translation-invariant kernels on $\mathbb R^d$, being characteristic implies being $cc$-universal (Figure 1 part 3).

It was previously shown that bounded continuous translation-invariant kernels on $\mathbb R^d$ are characteristic if and only if the Fourier transform of $\Psi$, which we know to be a nonnegative measure from Bochner's theorem, has support on all of $\mathbb R^d$: Theorem 7 of

Sriperumbudur, Gretton, Fukumizu, Lanckriet, and Schölkopf. Injective Hilbert Space Embeddings of Probability Measures. 21st Annual Conference on Learning Theory, 2008.

But the Fourier transform of $\Psi$ for the Matérn kernel, from equation (4.15) of Rasmussen and Williams, Gaussian Processes in Machine Learning (chapter 4 pdf), is the density $$ S(s) = \sigma^2 \frac{2^d \pi^{d/2} \Gamma(\nu + d/2) (2 \nu)^\nu}{\Gamma(\nu) \rho^{2 \nu}} \left( \frac{2 \nu}{\rho^2} + 4 \pi^2 \lVert s \rVert^2 \right)^{-\nu + d/2} $$ where $d$ is the dimension. This of course has support on all of $\mathbb R^d$, so the Matérn kernel is characterestic, so it's $cc$-universal.