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Let $f(x)$ be a positive definite function on $x \in R^d$. Assume $f(x)$ is radial , so $f(x)$ is a function of $|x|$, let's say $g(|x|):=f(x)$. How can I show that $g(|x|^\gamma)$ is positive definite for $0<\gamma<1$?

Any hints will be greatly appreciated. I have found the following Theorem: A continuous function $\varphi:[0, \infty) \rightarrow \mathbb{R}$ is positive definite and radial on $\mathbb{R}^s$ for all $s$ if and only if it is of the form $$ \varphi(r)=\int_0^{\infty} e^{-r^2 t^2} d \mu(t), $$ where $\mu$ is a finite non-negative Borel measure on $[0, \infty)$. $ e^{-r^{2\gamma} t^2}$ is positive definite but I do not know how that helps me. UPDATE: the problem with this theorem, the function has to be positive definite/radial for ALL $s$. At the moment I can only show that my function is positive definite for $s=1$. I suspect it is positive definite for $s=2,3$. Are there other results for specific $s$?

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  • $\begingroup$ 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 $\endgroup$
    – Tanya Vladi
    Commented May 2, 2023 at 19:31
  • $\begingroup$ Since $ e^{-r^{2\gamma} t^2}$ can be represented as a Gaussian mixture, $\phi(r^{\gamma})$ is radial $\endgroup$
    – Alexey S
    Commented May 2, 2023 at 19:52
  • $\begingroup$ Can you give a reference to the theorem you cited? $\endgroup$
    – Iosif Pinelis
    Commented May 2, 2023 at 20:10
  • $\begingroup$ math.iit.edu/~fass/603_ch2.pdf $\endgroup$
    – Alexey S
    Commented May 2, 2023 at 20:18
  • $\begingroup$ @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. $\endgroup$
    – Alexey S
    Commented May 2, 2023 at 20:24

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