Positive definite functions, completely monotone functions, the Bernstein-Widder theorem, and Schoenberg’s theorem

1 Linear operators

For a complex Hilbert space $H$ let $ℒ(H)$ be the bounded linear operators $H\to H$. It is a fact that $A\in ℒ(H)$ is self-adjoint if and only if $⟨Ah,h⟩\in ℝ$ for all $h\in H$. For a bounded self-adjoint operator $A$ it is a fact that

$$\parallel A\parallel =\underset{\parallel h\parallel =1}{sup}|⟨Ah,h⟩|.$$

$A\in ℒ(H)$ is called positive if it is self-adjoint and

$$⟨Ah,h⟩\ge 0,h\in H;$$

because we have taken $H$ to be a complex Hilbert space, for $A$ to be positive it suffices that the inequality is satisfied.

For $A,B\in ℒ({ℂ}^n)$, we define their Hadamard product $A*B\in ℒ(H)$ by

$$(A*B)e_i=\sum _{j=1}^n⟨A{e}_i,e_j⟩⟨B{e}_i,e_j⟩e_j.$$

So,

$$⟨(A*B)e_i,e_j⟩=⟨A{e}_i,e_j⟩⟨B{e}_i,e_j⟩.$$

The Schur product theorem states that if $A,B\in ℒ({ℂ}^n)$ are positive then their Hadamard product $A*B$ is positive.

2 Positive definite functions

Let $X$ be a real or complex linear space, let $f:X\to ℂ$ be a function, and for $x_1,\mathrm{\dots },x_n\in X$, define $F_{f;x_1,\mathrm{\dots },x_n}\in ℒ({ℂ}^n)$ by

$$F_{f;x_1,\mathrm{\dots },x_n}{e}_i=\sum _{j=1}^{n}f(x_i-x_j)e_j,$$

where $\{e_1,\mathrm{\dots },e_n\}$ is the standard basis for ${ℂ}^n$. Thus for $u={\sum }_{i=1}^n{u}_i{e}_i\in {ℂ}^n$,

$⟨F_{f;x_1,\mathrm{\dots },x_n}u,u⟩$ $=⟨{\displaystyle \sum _{i=1}^n}{u}_i{\displaystyle \sum _{j=1}^n}f(x_i-x_j)e_j,{\displaystyle \sum _{k=1}^n}{u}_k{e}_k⟩$

$={\displaystyle \sum _{i=1}^n}{u}_i{\displaystyle \sum _{j=1}^n}f(x_i-x_j)⟨e_j,{\displaystyle \sum _{k=1}^n}{u}_k{e}_k⟩$

$={\displaystyle \sum _{i=1}^n}{\displaystyle \sum _{j=1}^n}{u}_i\overline{u_j}f(x_i-x_j).$

We call $f$ positive definite if for all $x_1,\mathrm{\dots },x_n\in X$, $F_{f;x_1,\mathrm{\dots },x_n}$ is a positive operator, i.e. for $u\in {ℂ}^n$,

$$⟨F_{f;x_1,\mathrm{\dots },x_n}u,u⟩\ge 0.$$

We call $f$ strictly positive definite for all distinct $x_1,\mathrm{\dots },x_n\in X$ and nonzero $u\in {ℂ}^n$,

$$⟨F_{f;x_1,\mathrm{\dots },x_n}u,u⟩>0.$$

3 Completely monotone functions

A function $f:[0,\mathrm{\infty })\to ℝ$ is called completely monotone if

1. $f\in C[0,\mathrm{\infty })$

2. $f\in C^{\mathrm{\infty }}(0,\mathrm{\infty })$

3. ${(-1)}^k{f}^{(k)}(x)\ge 0$ for $k\ge 0$ and $x\in (0,\mathrm{\infty })$

Because a completely monotone function is continuous, $f(x)$ tends to $f(0)$ as $x\downarrow 0$. Because a completely monotone function is nonincreasing and convex, $f(x)$ has a limit, which we call $f(\mathrm{\infty })$, as $x\uparrow \mathrm{\infty }$.

The Bernstein-Widder theorem states that a function $f$ satisfying $f(0)=1$ is completely monotone if and only if it is the Laplace transform of a Borel probability measure on $[0,\mathrm{\infty })$.

Theorem 1 (Bernstein-Widder theorem).

A function $f:[0,\mathrm{\infty })\to ℝ$ satisfies $f(0)=1$ and is completely monotone if and only if there is a Borel probability measure $\mu$ on $[0,\mathrm{\infty })$ such that

$$f(x)={\int }_0^{\mathrm{\infty }}{e}^{-xt}𝑑\mu (t),x\in [0,\mathrm{\infty }).$$

Proof.

If $f$ is the Laplace transform of some probability measure $\mu$ on ${ℬ}_{[0,\mathrm{\infty })}$, then using the dominated convergence theorem yields that $f$ is continuous and by induction that $f\in C^{\mathrm{\infty }}(0,\mathrm{\infty })$. For $k\ge 0$ and for $x\in (0,\mathrm{\infty })$,

$$f^{(k)}(x)={\int }_0^{\mathrm{\infty }}{(-t)}^k{e}^{-xt}𝑑\mu (t),$$

as ${\int }_0^{\mathrm{\infty }}{t}^k{e}^{-xt}𝑑\mu (t)\ge 0$ so ${(-1)}^k{f}^{(k)}(x)\ge 0$. Hence $f$ is completely monotone, and $f(0)={\int }_0^{\mathrm{\infty }}𝑑\mu (t)=1$.

If $f$ satisfies $f(0)=1$ and is completely monotone, then for each $k\ge 0$, the function ${(-1)}^k{f}^{(k)}:(0,\mathrm{\infty })\to ℝ$ is nonnegative and is nonincreasing, so for $k\ge 1$ and $t\in (0,\mathrm{\infty })$, using that ${(-1)}^k{f}^{(k)}$ is nondecreasing and that ${(-1)}^{k-1}{f}^{(k-1)}$ is nonnegative,

${(-1)}^k{f}^{(k)}(t)$ $\le {\displaystyle \frac{2}{t}}{\displaystyle {\int }_{t/2}^t}{(-1)}^k{f}^{(k)}(u)𝑑u$

$={\displaystyle \frac{2}{t}}{(-1)}^k\left(f^{(k-1)}(t)-f^{(k-1)}(t/2)\right)$

$\le {\displaystyle \frac{2}{t}}{(-1)}^{k-1}{f}^{(k-1)}(t/2).$

Doing induction, for any $k\ge 1$,

${(-1)}^k{f}^{(k)}(t)$ $\le {\displaystyle \prod _{j=1}^{k-1}}\left({\displaystyle \frac{2^j}{t}}\right)\cdot f^{\prime }\left({\displaystyle \frac{t}{2^{k-1}}}\right)$

$\le {\displaystyle \prod _{j=1}^{k-1}}\left({\displaystyle \frac{2^j}{t}}\right)\cdot {\displaystyle \frac{2^k}{t}}\left(f\left({\displaystyle \frac{t}{2^{k-1}}}\right)-f\left({\displaystyle \frac{t}{2^k}}\right)\right)$

$=t^{-k}{2}^{k(k-1)/2}\left(f\left({\displaystyle \frac{t}{2^{k-1}}}\right)-f\left({\displaystyle \frac{t}{2^k}}\right)\right).$

Because $f(x)\to f(0)$ as $x\downarrow 0$,

$$f\left(\frac{t}{2^{k-1}}\right)-f\left(\frac{t}{2^k}\right)\to 0,t\downarrow 0,$$

and because $f(x)\to f(\mathrm{\infty })$ as $x\uparrow \mathrm{\infty }$,

$$f\left(\frac{t}{2^{k-1}}\right)-f\left(\frac{t}{2^k}\right)\to 0,t\uparrow \mathrm{\infty }.$$

Hence for each $k\ge 1$,

$$|f(t)|=o_k(t^{-k}),t\downarrow 0,$$ (1)

and

$$|f(t)|=o_k(t^{-k}),t\uparrow \mathrm{\infty }.$$ (2)

Furthermore, for any $x\in (0,\mathrm{\infty })$, $f^{(k)}(t)\to f^{(k)}(x)$ as $t\to x$, so it is immediate that

$${(t-x)}^k{f}^{(k)}(t)\to 0,t\to x.$$ (3)

For $x\ge 0$ and $k\ge 1$, integrating by parts, using (2) and (1) or (3) respectively as $x=0$ or $x>0$,

$f(x)-f(\mathrm{\infty })$ $=-{\displaystyle {\int }_x^{\mathrm{\infty }}}{f}^{\prime }(t)𝑑t$

$=-{(t-x)f^{\prime }(t)|}_x^{\mathrm{\infty }}+{\displaystyle {\int }_x^{\mathrm{\infty }}}{f}^{\prime \prime }(t)(t-x)𝑑t$

$={\displaystyle {\int }_x^{\mathrm{\infty }}}{f}^{\prime \prime }(t)(t-x)𝑑t$

$={{\displaystyle \frac{{(t-x)}^2}{2}}{f}^{\prime \prime }(t)|}_x^{\mathrm{\infty }}-{\displaystyle {\int }_x^{\mathrm{\infty }}}{f}^{\prime \prime \prime }(t){\displaystyle \frac{{(t-x)}^2}{2}}𝑑t$

$=-{\displaystyle {\int }_x^{\mathrm{\infty }}}{f}^{\prime \prime \prime }(t){\displaystyle \frac{{(t-x)}^2}{2}}𝑑t$

$={(-1)}^k{\displaystyle {\int }_x^{\mathrm{\infty }}}{f}^{(k)}(t){\displaystyle \frac{{(t-x)}^{k-1}}{(k-1)!}}𝑑t.$

Hence for $x\ge 0$ and $n\ge 0$,

$$f(x)-f(\mathrm{\infty })=\frac{{(-1)}^{n+1}}{n!}{\int }_x^{\mathrm{\infty }}{f}^{(n+1)}(t){(t-x)}^n𝑑t.$$

Define

$${\varphi }_n(y)={(1-y/n)}^n{1}_{[0,n]}(y).$$

For $n\ge 1$, by change of variables,

$f(x)-f(\mathrm{\infty })$ $={\displaystyle \frac{{(-1)}^{n+1}}{n!}}{\displaystyle {\int }_{x/n}^{\mathrm{\infty }}}{f}^{(n+1)}(nu){(nu-x)}^{n}n𝑑u$

$={\displaystyle \frac{{(-1)}^{n+1}}{(n-1)!}}{\displaystyle {\int }_{x/n}^{\mathrm{\infty }}}{f}^{(n+1)}(nu){(nu)}^n{\left(1-{\displaystyle \frac{x}{nu}}\right)}^n𝑑u$

$={\displaystyle \frac{{(-1)}^{n+1}}{(n-1)!}}{\displaystyle {\int }_0^{\mathrm{\infty }}}{(nu)}^n{\varphi }_n(x/u)f^{(n+1)}(nu)𝑑u$

$={\displaystyle \frac{{(-1)}^{n+1}}{(n-1)!}}{\displaystyle {\int }_0^{\mathrm{\infty }}}{(n/t)}^n{\varphi }_n(xt)f^{(n+1)}(n/t)t^{-2}𝑑t.$

For $t\ge 0$, define

$$s_n(t)=\frac{{(-1)}^{n+1}}{(n-1)!}{\int }_{1/t}^{\mathrm{\infty }}{(nu)}^n{f}^{(n+1)}(nu)𝑑u,$$

where $s_n(0)=0$, and for let $s_n(t)=0$. $s_n$ is continuous and because $f$ is completely monotone, $s_n$ is nondecreasing, so there is a unique positive measure ${\sigma }_n$ on ${ℬ}_{ℝ}$ such that

$${\sigma }_n((a,b])=s_n(b)-s_n(a),a\le b.$$

On the other hand, $s_n$ is absolutely continuous, so ${\sigma }_n$ is absolutely continuous with respect to Lebesgue measure ${\lambda }_1$, and for ${\lambda }_1$-almost all $t\in ℝ$,

$$\frac{d{\sigma }_n}{d{\lambda }_1}(t)=s_n^{\prime }(t).$$

Now for $t>0$, by the fundamental theorem of calculus and the chain rule,

$$s_n^{\prime }(t)=\frac{{(-1)}^{n+1}}{(n-1)!}{(n/t)}^n{f}^{(n+1)}(n/t)\cdot t^{-2},$$

and therefore

$f(x)-f(\mathrm{\infty })$ $={\displaystyle {\int }_0^{\mathrm{\infty }}}{\varphi }_n(xt)s_n^{\prime }(t)𝑑{\lambda }_1(t)$

$={\displaystyle {\int }_0^{\mathrm{\infty }}}{\varphi }_n(xt){\displaystyle \frac{d{\sigma }_n}{d{\lambda }_1}}(t)𝑑{\lambda }_1(t)$

$={\displaystyle {\int }_0^{\mathrm{\infty }}}{\varphi }_n(xt)𝑑{\sigma }_n(t).$

The total variation of ${\sigma }_n$ is equal to the total variation of $s_n$, and because $s_n$ is nondecreasing,

$$\parallel {\sigma }_n\parallel ={\int }_0^{\mathrm{\infty }}|s_n^{\prime }(t)|𝑑t={\int }_0^{\mathrm{\infty }}{s}_n^{\prime }(t)𝑑t=s_n(\mathrm{\infty })-s_n(0)=s_n(\mathrm{\infty }),$$

which is

$$\parallel {\sigma }_n\parallel =\frac{{(-1)}^{n+1}}{(n-1)!}{\int }_0^{\mathrm{\infty }}{(nu)}^n{f}^{(n+1)}(nu)𝑑u=f(0)-f(\mathrm{\infty }),$$

showing that $\{{\sigma }_n:n\ge 1\}$ is bounded for the total variation norm. We claim that $\{{\sigma }_n:n\ge 1\}$ is tight: for each $ϵ>0$ there is a compact subset $K_{ϵ}$ of $ℝ$ such that for all $n$. Taking this for granted, Prokhorov’s theorem states that there is a subsequence ${\sigma }_{k_n}$ of ${\sigma }_n$ that converges narrowly to some positive measure $\sigma$ on ${ℬ}_{ℝ}$. Finally, the sequence $t\mapsto {\varphi }_n(xt)$ tends in $C_b([0,\mathrm{\infty }))$ to $t\mapsto e^{-xt}$, and it thus follows that

$${\int }_0^{\mathrm{\infty }}{\varphi }_n(xt)𝑑{\sigma }_n(t)\to {\int }_0^{\mathrm{\infty }}{e}^{-xt}𝑑\sigma (t),$$

so

$$f(x)-f(\mathrm{\infty })={\int }_0^{\mathrm{\infty }}{e}^{-xt}𝑑\sigma (t).$$

Let

$$\mu =\sigma +f(\mathrm{\infty }){\delta }_0,$$

with which

$${\int }_0^{\mathrm{\infty }}{e}^{-xt}𝑑\mu (t)={\int }_0^{\mathrm{\infty }}{e}^{-xt}𝑑\sigma (t)+f(\mathrm{\infty }),$$

hence

$$f(x)={\int }_0^{\mathrm{\infty }}{e}^{-xt}𝑑\mu (t).$$

Because $f(0)=1$, ${\int }_0^{\mathrm{\infty }}𝑑\mu (t)=1$, showing that $\mu$ is a probability measure. ∎

4 Fourier transforms

For a topological space $X$ and a positive Borel measure $\mu$ on $X$, $F\subset X$ is called a support of $\mu$ if (i) $F$ is closed, (ii) $\mu (F^c)=0$, and (iii) if $G$ is open and $G\cap F\ne \mathrm{\varnothing }$ then $\mu (G\cap F)>0$. If $F_1$ and $F_2$ are supports of $\mu$, it is straightforward that $F_1=F_2$. It is a fact that if $X$ is second-countable then $\mu$ has a support, which we denote by $\mathrm{supp}\mu$.

Lemma 2.

If $\mu$ is a Borel measure on a topological space $X$ and $\mu$ has a support $\mathrm{supp}\mu$, if $f:X\to [0,\mathrm{\infty })$ is continuous and ${\int }_{X}f𝑑\mu =0$ then $f(x)=0$ for all $x\in \mathrm{supp}\mu$.

Proof.

Let $F=\mathrm{supp}\mu$ and let $E=\{x\in X:f(x)\ne 0\}$. $E$ is an open subset of $X$. Suppose by contradiction that there is some $x\in E\cap F$, i.e. that $E\cap F\ne \mathrm{\varnothing }$. Because $f$ is continuous and $f(x)>0$, there is some open neighborhood $G$ of $x$ for which $f(y)>f(x)/2$ for $y\in U$. Then $x\in G\cap F$, so $G\cap F\ne \mathrm{\varnothing }$ and because $F$ is the support of $\mu$, $\mu (G\cap F)>0$ and a fortiori $\mu (G)>0$. Then

$$0={\int }_{X}f𝑑\mu \ge {\int }_{G}f(y)𝑑\mu (y)\ge {\int }_G\frac{f(x)}{2}𝑑\mu (y)=\frac{f(x)}{2}\mu (G)>0,$$

a contradiction. Therefore $E\cap F=\mathrm{\varnothing }$, i.e. for all $x\in F$, $f(x)=0$. ∎

The following lemma asserts that a certain function is nonzero ${\lambda }_d$-almost everywhere, where ${\lambda }_d$ is Lebesgue measure on ${ℝ}^d$.

Lemma 3.

Let $x_1,\mathrm{\dots },x_n$ be distinct points in ${ℝ}^d$, let $u\in {ℂ}^n$ not be the zero vector, and define

$$g(y)=\sum _{j=1}^n{u}_j{e}^{-2\pi i{x}_j\cdot y},y\in {ℝ}^d.$$

For ${\lambda }_d$-almost all $y\in {ℝ}^d$, $g(y)\ne 0$.

The following theorem gives conditions under which the Fourier transform of a Borel measure on ${ℝ}^d$ is strictly positive definite.

Theorem 4.

If $\mu$ is a finite Borel measure on ${ℝ}^d$ and ${\lambda }_d(\mathrm{supp}\mu )>0$, then $\widehat{\mu }:{ℝ}^d\to ℂ$ is strictly positive definite.

Proof.

For distinct $x_1,\mathrm{\dots },x_n\in {ℝ}^d$ and for nonzero $u\in {ℂ}^n$,

${\displaystyle \sum _{j=1}^n}{\displaystyle \sum _{k=1}^n}{u}_j\overline{u_k}\widehat{\mu }(x_j-x_k)$ $={\displaystyle \sum _{j=1}^n}{\displaystyle \sum _{k=1}^n}{u}_j\overline{u_k}{\displaystyle {\int }_{{ℝ}^d}}{e}^{-2\pi i(x_j-x_k)\cdot y}𝑑\mu (y)$

$={\displaystyle {\int }_{{ℝ}^d}}\left({\displaystyle \sum _{j=1}^n}{u}_j{e}^{-2\pi i{x}_j\cdot y}\right)\overline{\left({\displaystyle \sum _{k=1}^n}{u}_k{e}^{-2\pi i{x}_k\cdot y}\right)}𝑑\mu (y)$

$={\displaystyle {\int }_{{ℝ}^d}}{\left|{\displaystyle \sum _{j=1}^n}{u}_j{e}^{-2\pi i{x}_j\cdot y}\right|}^2𝑑\mu (y)$

$={\displaystyle {\int }_{{ℝ}^d}}{|g(y)|}^2𝑑\mu (y).$

It is apparent that this is nonnegative. If it is equal to $0$ then because $g$ is continuous we obtain from Lemma 2 that ${|g(y)|}^2=0$ for all $y\in \mathrm{supp}\mu$, i.e. $g(y)=0$ for all $y\in \mathrm{supp}\mu$. In other words,

$$\mathrm{supp}\mu \subset \{y\in {ℝ}^d:g(y)=0\}.$$

But by Lemma 3, ${\lambda }_d(\{y\in {ℝ}^d:g(y)=0\})=0$, so ${\lambda }_d(\mathrm{supp}\mu )=0$, contradicting the hypothesis ${\lambda }_d(\mathrm{supp}\mu )>0$. Therefore

$${\int }_{{ℝ}^d}{|g(y)|}^2𝑑\mu (y)>0,$$

which shows that $\widehat{\mu }$ is strictly positive definite. ∎

5 Schoenberg’s theorem

Let $(X,⟨\cdot ,\cdot ⟩)$ be a real inner product space. We call a function $F:X\to ℝ$ radial when $\parallel x\parallel =\parallel y\parallel$ implies that $F(x)=F(y)$.

An identity that is worth memorizing is that for $y\in ℝ$,

$${\int }_{ℝ}{e}^{-\pi x^2}{e}^{-2\pi ixy}𝑑x=e^{-\pi y^2}.$$

Using this and Fubini’s theorem yields, $y\in {ℝ}^d$,

$${\int }_{{ℝ}^d}{e}^{-\pi {|x|}^2}{e}^{-2\pi ⟨x,y⟩}=e^{-\pi {|y|}^2}.$$

Lemma 5.

For $\alpha >0$ and $y\in {ℝ}^d$,

$${\int }_{{ℝ}^d}{\left(\frac{\pi }{\alpha }\right)}^{d/2}\exp \left(-\frac{{\pi }^2}{\alpha }{|x|}^2\right)e^{-2\pi i⟨x,y⟩}𝑑x=e^{-\alpha {|y|}^2}.$$

Proof.

Define $T:{ℝ}^d\to {ℝ}^d$ by

$$T(x)=\sqrt{\frac{\pi }{\alpha }}x,x\in {ℝ}^d.$$

$T^{\prime }(x)=\sqrt{\frac{\pi }{\alpha }}I\in ℒ({ℝ}^d)$ and $J_T(x)=det{T}^{\prime }(x)={\left(\frac{\pi }{\alpha }\right)}^{d/2}$. Let $u\in {ℝ}^d$ and define $f(x)=e^{-\pi {|x|}^2}{e}^{-2\pi i⟨x,u⟩}$. By the change of variables formula,

$${\int }_{{ℝ}^d}(f\circ T)\cdot |J_T|𝑑{\lambda }_d={\int }_{T({ℝ}^d)}f𝑑{\lambda }_d,$$

and because $T$ is self-adjoint this is

$${\int }_{{ℝ}^d}{e}^{-\pi {|T(x)|}^2}{e}^{-2\pi i⟨x,Tu⟩}{\left(\frac{\pi }{\alpha }\right)}^{d/2}𝑑x={\int }_{{ℝ}^d}{e}^{-\pi {|x|}^2}{e}^{-2\pi i⟨x,u⟩}𝑑x,$$

and therefore

$${\int }_{{ℝ}^d}{\left(\frac{\pi }{\alpha }\right)}^{d/2}\exp \left(-\frac{{\pi }^2}{\alpha }{|x|}^2\right)e^{-2\pi i⟨x,Tu⟩}𝑑x=e^{-\pi {|u|}^2}.$$

For $u=T^{-1}(y)=\sqrt{\frac{\alpha }{\pi }}y$ this is

$${\int }_{{ℝ}^d}{\left(\frac{\pi }{\alpha }\right)}^{d/2}\exp \left(-\frac{{\pi }^2}{\alpha }{|x|}^2\right)e^{-2\pi i⟨x,y⟩}𝑑x=e^{-\alpha {|y|}^2},$$

proving the claim. ∎

We now prove that on a real inner product space, $x\mapsto e^{-\alpha {\parallel x\parallel }^2}$ is strictly positive definite whenever $\alpha >0$.

Theorem 6.

Let $(X,⟨\cdot ,\cdot ⟩)$ be a real inner product space. If $\alpha >0$, then

$$x\mapsto e^{-\alpha {\parallel x\parallel }^2},x\in X,$$

is radial and strictly positive definite.

Proof.

Let $x_1,\mathrm{\dots },x_n$ be distinct points in $X$. There is an $n$-dimensional linear subspace $V$ of $X$ that contains $x_1,\mathrm{\dots },x_n$. By the Gram-Schmidt process, $V$ has an orthonormal basis $\{v_1,\mathrm{\dots },v_n\}$. Define $T:V\to {ℝ}^n$ by $T{v}_j=e_j$, where $\{e_1,\mathrm{\dots },e_n\}$ is the standard basis for ${ℝ}^n$, which is an orthogonal transformation, and define

$$f(u)=e^{-\alpha {|u|}^2},u\in {ℝ}^d.$$

For $u\in {ℂ}^n$, $u\ne 0$,

${\displaystyle \sum _{j=1}^n}{\displaystyle \sum _{k=1}^n}{u}_j\overline{u_k}{e}^{-\alpha {\parallel x_j-x_k\parallel }^2}$ $={\displaystyle \sum _{j=1}^n}{\displaystyle \sum _{k=1}^n}{u}_j\overline{u_k}\exp \left(-\alpha {|T(x_j-x_k)|}^2\right)$

$={\displaystyle \sum _{j=1}^n}{\displaystyle \sum _{k=1}^n}{u}_j\overline{u_k}f(T{x}_j-T{x}_k).$

Now, let $\mu$ be the Borel measure on ${ℝ}^d$ whose density with respect to ${\lambda }_d$ is

$$y\mapsto {\left(\frac{\pi }{\alpha }\right)}^{d/2}\exp \left(-\frac{{\pi }^2}{\alpha }{|y|}^2\right).$$

Because $\mu$ is absolutely continuous with respect to ${\lambda }_d$, ${\lambda }_d(\mathrm{supp}\mu )>0$, so Theorem 4 states that the Fourier transform $\widehat{\mu }:{ℝ}^d\to ℂ$ is strictly positive definite. Applying Lemma 5, the Fourier transform of $\mu$ is

$$\widehat{\mu }(u)={\int }_{{ℝ}^d}{\left(\frac{\pi }{\alpha }\right)}^{d/2}\exp \left(-\frac{{\pi }^2}{\alpha }{|y|}^2\right)e^{-2\pi i⟨y,u⟩}𝑑y=e^{-\alpha {|u|}^2}=f(u),$$

so $f$ is strictly positive definite. Because $T$ is an orthogonal transformation it is in particular one-to-one, so $T{x}_1,\mathrm{\dots },T{x}_n$ are distinct points in ${ℝ}^d$. Thus the fact that $f$ is strictly positive definite means that

$$\sum _{j=1}^n\sum _{k=1}^n{u}_j\overline{u_k}{e}^{-\alpha {\parallel x_j-x_k\parallel }^2}=\sum _{j=1}^n\sum _{k=1}^n{u}_j\overline{u_k}f(T{x}_j-T{x}_k)>0,$$

which establishes that $x\mapsto e^{-\alpha {\parallel x\parallel }^2}$ is strictly positive definite. ∎

The following is Schoenberg’s theorem.

Theorem 7 (Schoenberg’s theorem).

Let $(X,⟨\cdot ,\cdot ⟩)$ be a real inner product space. If $f:[0,\mathrm{\infty })\to ℝ$ is completely monotone, $f(0)=1$, and $f$ is not constant, then

$$x\mapsto f({\parallel x\parallel }^2),X\to [0,\mathrm{\infty }),$$

is radial and strictly positive definite.

Proof.

Because $f$ is completely monotone, the Bernstein-Widder theorem (Theorem 1) tells us that there is a Borel probability measure $\mu$ on $[0,\mathrm{\infty })$ such that

$$f(t)={\int }_0^{\mathrm{\infty }}{e}^{-st}𝑑\mu (s),t\in [0,\mathrm{\infty }),$$

that is, $f$ is the Laplace transform of $\mu$. Now, the Laplace transform of ${\delta }_0$ is $t\mapsto 1$, and because $f$ is not constant, the Laplace transform of $\mu$ is not equal to the Laplace transform of ${\delta }_0$, which implies that $\mu \ne {\delta }_0$. Therefore $\mu ((0,\mathrm{\infty }))>0$.

Let $x_1,\mathrm{\dots },x_n$ be distinct points in $X$ and let $u\in {ℂ}^n$, $u\ne 0$. Then, because ${\sum }_{j=1}^n{\sum }_{k=1}^n{u}_j\overline{u_k}\ge 0$,

${\displaystyle \sum _{j=1}^n}{\displaystyle \sum _{k=1}^n}{u}_j\overline{u_k}f({\parallel x_j-x_k\parallel }^2)$ $={\displaystyle \sum _{j=1}^n}{\displaystyle \sum _{k=1}^n}{u}_j\overline{u_k}{\displaystyle {\int }_0^{\mathrm{\infty }}}\exp \left(-s{\parallel x_j-x_k\parallel }^2\right)𝑑\mu (s)$

$={\displaystyle {\int }_0^{\mathrm{\infty }}}{\displaystyle \sum _{j=1}^n}{\displaystyle \sum _{k=1}^n}{u}_j\overline{u_k}\exp \left(-s{\parallel x_j-x_k\parallel }^2\right)d\mu (s)$

$={\displaystyle \sum _{j=1}^n}{\displaystyle \sum _{k=1}^n}{u}_j\overline{u_k}\mu (\{0\})$

$+{\displaystyle {\int }_0^{\mathrm{\infty }}}{1}_{(0,\mathrm{\infty })}(s){\displaystyle \sum _{j=1}^n}{\displaystyle \sum _{k=1}^n}{u}_j\overline{u_k}\exp \left(-s{\parallel x_j-x_k\parallel }^2\right)d\mu (s)$

$\ge {\displaystyle {\int }_0^{\mathrm{\infty }}}{1}_{(0,\mathrm{\infty })}(s){\displaystyle \sum _{j=1}^n}{\displaystyle \sum _{k=1}^n}{u}_j\overline{u_k}\exp \left(-s{\parallel x_j-x_k\parallel }^2\right)d\mu (s)$

$={\displaystyle {\int }_0^{\mathrm{\infty }}}g(s)𝑑\mu (s).$

Assume by contradiction that ${\int }_0^{\mathrm{\infty }}g(s)𝑑\mu (s)=0$. Because $g\ge 0$, this implies that $\mu (\{s\in [0,\mathrm{\infty }):g(s)>0\})=0$. By Theorem 6, for each $s>0$,

$$\sum _{j=1}^n\sum _{k=1}^n{u}_j\overline{u_k}\exp \left(-s{\parallel x_j-x_k\parallel }^2\right)>0,$$

so $g(s)>0$ when $s>0$. Thus $\mu ((0,\mathrm{\infty }))=0$, a contradiction. Therefore,

$$\sum _{j=1}^n\sum _{k=1}^n{u}_j\overline{u_k}f({\parallel x_j-x_k\parallel }^2)={\int }_0^{\mathrm{\infty }}g(s)𝑑\mu (s)>0,$$

which shows that $x\mapsto f({\parallel x\parallel }^2)$ is strictly positive definite. ∎