to verify that it applies one can consider the change of variables $y=x/\sqrt{t}$, recalling that $\mu$ is finite. Hence, $$\label{eq:inverse Fourier Bernstein I}
(\operatorname{{\mathcal F}}^{-1} f)(x) =  \int_{(0,\infty)} \frac{\exp(-\frac{x^2}{4 t })}{\sqrt{4\pi t }} \, {{\mathrm d}}\mu(t).$$ This calculation together with Bochner’s theorem implies that $ \operatorname{{\mathcal F}}f \in L^1({\mathbb R})\cap C^\infty({\mathbb R}\setminus\{0\})$, so that $\operatorname{{\mathcal F}}f$ is actually given by a function and not just a measure (in this and the following proof it does not matter if we consider $\operatorname{{\mathcal F}}f$ or $\operatorname{{\mathcal F}}^{-1}f$, since they differ only by a constant factor). The positivity of $\operatorname{{\mathcal F}}f$ is clear from the above formula, too, and we in addition see that $ \operatorname{{\mathcal F}}f$ is monotone for $x>0$. More precisely, $$(\operatorname{{\mathcal F}}^{-1} f)^\prime(x)=-\frac{1}{\sqrt{8\pi}} \int_{(0,\infty)} \frac{x\exp(-\frac{x^2}{4t})}{t^{3/2}}  \,{{\mathrm d}}\mu(t) <0.$$ Finally, the fact that $(\operatorname{{\mathcal F}}^{-1} f)(\sqrt{\cdot})$ is completely monotone is a consequence of Bernstein’s theorem and the computation $$\label{eq:inverse Fourier Bernstein II}
(\operatorname{{\mathcal F}}^{-1} f)(\sqrt{\lambda}) =  \int_{(0,\infty)} \frac{\exp(-\frac{\lambda}{4 t })}{\sqrt{4\pi t }} \, {{\mathrm d}}\mu(t)=
\int_{(0,\infty)} \exp(-s\lambda)\, {{\mathrm d}}\tilde \mu(s),$$ in which the measure $\tilde \mu$ is given by $$\label{eq:tilde mu definition}
{{\mathrm d}}\tilde \mu=\sqrt{\frac{\cdot}{\pi}} {{\mathrm d}}(\psi_*(\mu)),$$ where $\psi(t)=\frac{1}{4t}$ and $\psi_*(\mu)$ is the pushforward of $\mu$ by $\psi$.

Conversely, suppose that 