functions satisfying $f(\xi)=g(\xi^2)$. Then $f$ is the Fourier transform of an even, integrable function such that $(\operatorname{{\mathcal F}}^{-1} f)(\sqrt{\cdot})$ is completely monotone if and only if $g$ is completely monotone with $\lim_{\lambda \searrow 0} g(\lambda)<\infty$ and $\lim_{\lambda \to \infty} g(\lambda)=0$. In this case, $\operatorname{{\mathcal F}}^{-1} f$ is smooth and monotone outside of the origin.

For a related result by Bochner (on subordinate Brownian motions), see [@MR0072370 Theorem 4.3.3] and [@SchillingSongVondracek12 Example 13.16].

When $g$ is completely monotone and continuous at zero, Bochner’s theorem guarantees us that $f(\xi)=g(\xi^2)$ is the Fourier transform of a measure. In fact, by Bernstein’s theorem (cf. Theorem \[thm:HBW\]), we have that $$f(\xi)=\mu(\{0\})+\int_{(0,\infty)} \exp(-\xi^2 t)\, {{\mathrm d}}\mu(t)$$ for some finite Borel measure $\mu$ on $[0, \infty)$. Since $\lim_{\xi\to \infty} f(\xi)=0$, we infer that $\mu(\{0\})=0$, whence $$\label{eq:Bochner_formula}
f(\xi)=\int_{(0,\infty)} \exp(-\xi^2 t)\, {{\mathrm d}}\mu(t).$$ Noting that $$\exp(-t \xi^2)=\operatorname{{\mathcal F}}\left(\frac{\exp(-\frac{(\cdot)^2}{4 t })}{\sqrt{4\pi  t }}\right)(\xi)=
 \int_{{\mathbb R}} \frac{\exp(-\frac{x^2}{4t})}{\sqrt{4\pi t}}\exp(-i x\xi) \, {{\mathrm d}}x,$$ it follows that $$\label{eq:f Fubini}
\begin{aligned}
f(\xi)&= \int_{(0,\infty)} \left( \int_{{\mathbb R}} \frac{\exp(-\frac{x^2}{4t})}{\sqrt{4\pi t }} \exp(-i x\xi)\, {{\mathrm d}}x \right) \, {{\mathrm d}}\mu(t)\\
&= \int_{{\mathbb R}} \left(\int_{(0,\infty)}  \frac{\exp(-\frac{x^2}{4 t })}{\sqrt{4\pi t }} \, {{\mathrm d}}\mu( t )\right) \exp(-i x\xi)\, {{\mathrm d}}x,
\end{aligned}$$ where we have used Fubini’s theorem; in order 