I'm reading [this book chapter](http://www.math.iit.edu/~fass/603_ch2.pdf), 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.



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


