# Connection between non-constant completely monotone function and strictly positive definite kernels (Schoenberg characterization)

I'm reading this book chapter, where they stated two alternative characterizations of completely monotone functions $$\phi$$ using (1) Laplace transform of a finite, non-negative Borel measure and also using (2) the positive definiteness of the kernel matrix $$K$$ constructed from $$\phi$$, respectively in Theorem 2.5.2 and 2.5.3. The theorems go as follows:

Theorem 2.5.2: (Hausdorff-Bernstein-Widder theorem: Laplace transform characterization of completely monotone functions)

A function $$\phi: [0,\infty) \to \mathbb{R}$$ is completely monotone if and only it is the Laplace transform of a finite non-negative Borel measure $$\mu$$ on $$[0,\infty)$$, i.e. $$\phi$$ is of the form:

$$\phi(r)= \mathcal{L}\mu(r)=\int_{0}^{\infty} e^{-rt}d\mu(t)$$

Theorem 2.5.3: (Theorem relating completely monotone functions and positivity of kernel matrix with radially symmetric kernels)

A function $$\phi$$ is completely monotone on $$[0,\infty)$$ if and only if $$\Phi(x):=\phi(||x||^2)$$ is positive definite and radial on $$\mathbb{R}^d$$, $$\forall d \in \mathbb{N}$$.

My question is: are the following true regarding strict positivity of kernel matrices and the completely monotone function being non-constant?

Guess 1:

A non-constant function $$\phi: [0,\infty) \to \mathbb{R}$$ completely monotone if and only it is the Laplace transform of a finite non-negative Borel measure $$\mu$$ on $$[0,\infty)$$ not of the form $$c\delta_0, c>0$$, i.e. $$\phi$$ is of the form:

$$\phi(r)= \mathcal{L}\mu(r)=\int_{0}^{\infty} e^{-rt}d\mu(t)$$ where $$\mu$$ is not of the form $$c\delta_0, c>0$$

Guess 2:

A non-constant function $$\phi$$ is completely monotone on $$[0,\infty)$$ if and only if $$\Phi(x):=\phi(||x||^2)$$ is strictly positive definite and radial on $$\mathbb{R}^d \forall d \in \mathbb{N}$$?

N.B. It's worthwhile mentioning that in this book and in the relevant literature, positive definite normally means positive semidefinite elsewhere, and strictly positive definite normally means positive definite elsewhere.

Here is an answer to your Question 1: Let us actually prove a bit more: A function $$\phi\colon[0,\infty)\to\mathbb R$$ is non-constant and completely monotone if and only if $$\phi(r)=\int_{0}^{\infty} e^{-rt}d\mu(t)\quad \forall r\ge0\tag{1}$$ with $$\mu\ne c\delta_0$$ for any real $$c$$.

The "only if" part: Suppose $$\phi$$ is non-constant and completely monotone. Then (1) holds with some (unique) finite measure $$\mu$$. If $$\mu=c\delta_0$$ for some real $$c$$, then $$\phi$$ is the constant $$c$$. So, $$\mu\ne c\delta_0$$ for any real $$c$$.

The "if" part: Suppose (1) holds with $$\mu\ne c\delta_0$$ for any real $$c$$. Then $$\phi$$ is completely monotone and $$\mu((0,\infty))>0$$, whence $$\phi'(r)=-\int_{0}^{\infty} te^{-rt}d\mu(t)<0\quad \forall r\ge0,\tag{1.5}$$ so that $$\phi$$ is not constant.

Let us now answer your Question 2, proving again a bit more: A function $$\phi\colon[0,\infty)\to\mathbb R$$ is non-constant and completely monotone if and only if the function $$\Phi$$ defined by the formula $$\Phi(x):=\phi(|x|^2)$$ for all $$x\in\mathbb R^d$$ is strictly positive definite, for each natural $$d$$.

This is based on

Lemma 1: The characteristic function (c.f.) of any probability distribution on $$\mathbb R^d$$ with (say) an uncountable support is strictly positive definite.

This lemma will be proved at the end of the answer.

Using Lemma 1, let us now prove

The "only if" part: Suppose $$\phi$$ is non-constant and completely monotone. Then (1) implies $$\Phi(x)=\int_0^\infty e^{-t|x|^2}\,\mu(dt) \quad\forall x\in\mathbb R^d.\tag{2}$$ For each real $$t>0$$, the function $$x\mapsto e^{-t|x|^2}$$ is the c.f. of a Gaussian distribution on $$\mathbb R^d$$ and hence, by Lemma 1, is strictly positive definite. Also, by the above answer to Question 1, $$\mu((0,\infty))>0$$. So, by (2), $$\Phi$$ is strictly positive definite.

The "if" part: Suppose $$\Phi$$ is strictly positive definite. Then, by your Theorem 2.5.3, $$\phi$$ is completely monotone. So, (1) holds and hence (2) holds. Since $$\Phi$$ is strictly positive definite, we have $$\mu((0,\infty))>0$$ -- because otherwise, by (2), $$\Phi$$ is constant. So, by (1.5), $$\phi$$ is non-constant.

It remains to prove Lemma 1. Let $$f$$ be the c.f. of a random vector $$Z$$ in $$\mathbb R^d$$ with an uncountable support. To obtain a contradiction, suppose that $$f$$ is not strictly positive definite. Then for some natural $$k$$, some complex $$a_1,\dots,a_k$$ not all of which are $$0$$, and some pairwise distinct $$x_1,\dots,x_k$$ in $$\mathbb R^d$$ we have $$E|\sum_1^k e^{ix_j\cdot Z}a_j|^2=\sum_{j,l=1}^k f(x_j-x_l)a_j\bar a_l=0$$, where $$\cdot$$ is the dot product. So, $$h(z):=\sum_1^k e^{ix_j\cdot z}a_j=0$$ for all $$z$$ in the uncountable support of the distribution of $$Z$$. Since the function $$h$$ is entire, it follows that $$h(z)=0$$ for all $$z\in\mathbb R^d$$. So, $$a_1=\dots=a_k=0$$, which is a contradiction.

• Thank you, and upvoted for answerign the Question 1. If you've any notes/links that deal with Question 2, please post here :) I guess for the second question, it suffices to show that the $d \times d$ kernel matrix has no zero eigenvalues, i.e. it's non-singular for any dimension $d$. – Let's talk math Feb 7 at 14:46
• I have now added an answer to Question 2 as well. – Iosif Pinelis Feb 7 at 17:32
• Thank you so much, very much appreciate it! – Let's talk math Feb 7 at 20:48
• @Let'stalkmath : I have added a detail concerning "Since Φ is strictly positive definite, we have μ((0,∞))>0." As for "the characteristic function of a probability distribution is strictly positive definite if and only if the support of the probability distribution is uncountable", this is not true -- e.g., it is enough that the support contain an accumulation point. – Iosif Pinelis Feb 18 at 15:45
• @Let'stalkmath : I don't think that is true either, but can't give a counterexample right away. You can ask any additional questions in separate posts. – Iosif Pinelis Feb 19 at 14:22