$\lim_{\lambda \to \infty} g(\lambda)=0$. One has $$\label{eq:g representation}
g(\lambda)=\int_{(0,\infty)} \frac{1}{\lambda+t}\, {{\mathrm d}}\sigma(t),$$ and, with $\psi = \sqrt{\cdot}$, the pushforward of $\sigma$ by $\psi$ relates $f$ to $g$ via $$\label{eq:f representation}
(\operatorname{{\mathcal F}}^{-1} f)(x)=\int_{(0,\infty)} \exp(-s|x|) \, {{\mathrm d}}\mu(s), \qquad {{\mathrm d}}\mu=\frac{1}{2(\cdot)}{{\mathrm d}}( \psi_*(\sigma)).$$

The main part of this result is known in the theory of subordinate Brownian motions; see [@MR2860308 Proposition 2.14] and [@SchillingSongVondracek12 Example 13.16]. We include a proof for completeness since our result is slightly different as well as phrased in a different language.

Suppose first that $g$ is a Stieltjes function with $\lim_{\lambda \searrow 0} g(\lambda)<\infty$ and $\lim_{\lambda \to \infty} g(\lambda)=0$. By assumption, $g$ is given by with $\int_{(0,\infty)}\frac{{{\mathrm d}}\sigma(t)}{t}<\infty$ (see the remark after Definition \[def:Stieltjes function\]) and hence we obtain from $f(\xi)=g(\xi^2)$ that $$f(\xi) =\int_{(0,\infty)} \frac{1}{\xi^2+t}\, {{\mathrm d}}\sigma(t).$$ From the transform $$\label{eq:exponential transform} 
\operatorname{{\mathcal F}}^{-1} \left( \frac{s}{\xi^2+s^2} \right)(x) =\frac12 \exp(-s\, |x|),$$ with $s = \sqrt{t}$, and an argument as in the proof of Proposition \[prop:completely monotone Fourier\], we get that $$\label{eq:fhat_Stieltjes}
(\operatorname{{\mathcal F}}^{-1} f)(x)= \int_{(0,\infty)} \frac{\exp(-\sqrt{t}\, |x|)}{2\sqrt{t}} \, {{\mathrm d}}\sigma(t).$$ Here, one makes the changes of variables $y = \sqrt{t} x$ in order to justify the use of Fubini’s theorem. Making the change of variables $t \mapsto 