3
$\begingroup$

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 $s$?

Improve this question
$\endgroup$
7
  • $\begingroup$ 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
    May 2 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
    May 2 at 19:52
  • $\begingroup$ Can you give a reference to the theorem you cited? $\endgroup$
    – Iosif Pinelis
    May 2 at 20:10
  • $\begingroup$ math.iit.edu/~fass/603_ch2.pdf $\endgroup$
    – Alexey S
    May 2 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
    May 2 at 20:24

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.