$f$ is the Fourier transform of an even, integrable function $\operatorname{{\mathcal F}}^{-1} f$ and that $(\operatorname{{\mathcal F}}^{-1} f)(\sqrt{\cdot})$ is completely monotone. Then we can write $(\operatorname{{\mathcal F}}^{-1} f)(\sqrt{\cdot})$ in the form $$(\operatorname{{\mathcal F}}^{-1} f)(\sqrt{\lambda}) =
\int_{[0,\infty)} \exp(-s\lambda)\, {{\mathrm d}}\tilde \mu(s),$$ where $\tilde \mu$ is obtained using Bernstein’s theorem. Consequently, $$(\operatorname{{\mathcal F}}^{-1} f)(x) =
\int_{[0,\infty)} \exp(-sx^2)\, {{\mathrm d}}\tilde \mu(s),$$ and integrating this relation using the change of variables $y=\sqrt{s}x$ yields that ${{\mathrm d}}\tilde \mu(s)/\sqrt{s}$ is a finite measure. In particular, $\tilde \mu$ has no mass at $0$, so that holds, where $\mu$ is the finite measure defined by . Consequently, we have and the calculation is now justified by Fubini’s theorem. Thus, holds and this in turn implies that $g$ is completely monotone with $\lim_{\lambda \searrow 0} g(\lambda)=0$ and $\lim_{\lambda \to \infty} g(\lambda)=0$.

When the function $g$ is Stieltjes one can sharpen these conclusions. It is already clear from Theorem \[thm:characterisation\] and Lemma \[lemma:composition\] that $g(\xi^2)$ is infinitely divisible. In addition, we have:

\[prop:Stieltjes Fourier\] Let $f$ and $g$ be two functions satisfying $f(\xi)=g(\xi^2)$. Then $f$ is the Fourier transform of an even, integrable and completely monotone function if and only if $g$ is Stieltjes with $\lim_{\lambda \searrow 0} g(\lambda)<\infty$ and 