\sqrt{t} = \psi(t)$ in the integral, we obtain in the variable $s = \sqrt{t}$. In particular, $\operatorname{{\mathcal F}}f$ is completely monotone by Bernstein’s theorem (${{\mathrm d}}\mu(s)/s$ is finite). The evenness of $\operatorname{{\mathcal F}}f$ follows immediately from the evenness of $g(\xi^2)$, and the fact that $\operatorname{{\mathcal F}}f \in L^1({\mathbb R})$ is a consequence of Proposition \[prop:completely monotone Fourier\].

Conversely, suppose that $\operatorname{{\mathcal F}}f\in L^1({\mathbb R})$ is even and completely monotone. Then $$(\operatorname{{\mathcal F}}^{-1} f)(x)=\int_{[0, \infty)} \exp(-|x|s)\,{{\mathrm d}}\mu(s),$$ for some Borel measure $\mu$ on $[0,\infty)$ and the integrability of $\operatorname{{\mathcal F}}f$ implies that ${{\mathrm d}}\mu(s)/s$ is a finite measure. In particular, $\mu$ has no mass at $0$ so that the left-most equality in holds. Thus, we have with $\sigma$ defined by , and together with Fubini’s theorem yield . Moreover, it is easily seen that ${{\mathrm d}}\sigma(t)/t$ is finite so that $g$ is a Stieltjes function with the desired properties.

Positivity and monotonicity properties: the Whitham kernel {#positivity-and-monotonicity-properties-the-whitham-kernel .unnumbered}
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Note that we can write the Whitham symbol as $m(\xi)=g(\xi^2)$, where $$\label{eq:gm}
g(\lambda)=\sqrt{\frac{\tanh \sqrt{\lambda}}{\sqrt{\lambda}}}, \qquad \lambda\ge 0.$$

\[prop:g is Stieltjes\] $(g(\lambda))^{2\alpha}$ is a Stieltjes function for any $\alpha \in (0, 1]$.

To see this, note that the reciprocal $$\lambda \to  \frac{\sqrt{\lambda}}{\tanh \sqrt{\lambda}}$$ 