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May 3, 2023 at 2:24 comment added Alexey S @David Roberts I posted the solution BEFORE Iosif posted his
May 2, 2023 at 21:03 comment added Iosif Pinelis @AlexeyS : How do show this for $s=1$?
May 2, 2023 at 20:24 comment added Alexey S @TanyaVladi I see it now, A function $\varphi$ is completely monotone on $[0, \infty)$ if and only if $\Phi=$ $\varphi\left(\|\cdot\|^2\right)$ is positive definite and radial on $\mathbb{R}^s$ for all $s$, since the composition of completely monotone and Bernstein is completely monotone we are back to positive definite.
May 2, 2023 at 20:18 comment added Alexey S math.iit.edu/~fass/603_ch2.pdf
May 2, 2023 at 20:16 review Close votes
May 17, 2023 at 3:09
May 2, 2023 at 20:15 history edited Alexey S CC BY-SA 4.0
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May 2, 2023 at 20:10 comment added Iosif Pinelis Can you give a reference to the theorem you cited?
May 2, 2023 at 19:52 comment converted from answer Alexey S Since $ e^{-r^{2\gamma} t^2}$ can be represented as a Gaussian mixture, $\phi(r^{\gamma})$ is radial
May 2, 2023 at 19:45 history edited Alexey S CC BY-SA 4.0
added 224 characters in body
May 2, 2023 at 19:40 history edited Alexey S CC BY-SA 4.0
added 224 characters in body
May 2, 2023 at 19:31 comment added Tanya Vladi as a start, a composition $m(b(x))$ of completely monotone function $m(x)$ and Bernstein function $b(x)$ is completely monotone function; $x^\gamma$, $0<\gamma<1$ is Bernstein function
May 2, 2023 at 18:59 history edited Alexey S CC BY-SA 4.0
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S May 2, 2023 at 18:58 review First questions
May 2, 2023 at 20:07
S May 2, 2023 at 18:58 history asked Alexey S CC BY-SA 4.0