Powers of positive semidefinite functions

Question

A function $f : \mathbb R \to \mathbb C$ is positive semidefinite if for any $n \geq 1$, any $z_1, \ldots, z_n \in \mathbb C$, and any $x_1, \ldots, x_n$, the inequality holds: $$ \sum_{k,l = 1}^n z_k \overline z_l f(x_k - x_l) \geq 0. $$ In other words, the matrix whose entries are $f(x_k - x_l)$ is positive semidefinite for any $n$ and any choice of $x_1, \ldots, x_n$. We'll say $f$ is Hermitian if $f(-x) = \overline{f(x)}$ (i.e., the matrix is Hermitian).

My question: Suppose $f$ is positive semidefinite, $f(0) =1$, and the range of $f$ is a simply connected region in $\mathbb C$ that does not contain $0$. Given $r>0$, is the exponentiated function $f^r$ also positive semidefinite? What if $f$ is Hermitian? (Here $f^r$ is a continuous branch if $r$ isn't an integer.)

By the spectral theorem for Hermitian operators, the answer is yes for positive semidefinite Hermitian matrices (if $A$ is Hermitian positive semidefinite, a unique positive semidefinite $n$th root exists, so by continuity, real exponentiation can be canonically defined). But I'm not sure how to show this for functions, as the diagonalization of the matrix $\left(f(x_k - x_l)\right)_{1 \leq k,l \leq n}$ need not be related obviously to the diagonalization of $\left(f(x_k - x_l)^r\right)_{1 \leq k,l \leq n}$.

Any thoughts?

(I included the tag "characteristic functions" because this problem came up for me in the context of trying to use Bochner's theorem to generate a continuous family of probability measures on $\mathbb R$.)

Answer 1

Your comment about tags suggests using Bochner's theorem to answer your question. Since the product of characteristic functions is also a characteristic function, $f^r$ is psd if $r$ is a positive integer. But if $f^r$ is a characteristic function for all $r>0$ it must be the case for all $r$ of form $r=1/n$, which makes $f$ an infinitely divisible characteristic function.

A glance at the Wikipedia article tells you that the uniform distribution is not infinitely divisible, so $f(t)=\sin t / t$ is a counterexample to your inequality in the case $r=1/2$.

Answer 2

Under some additional assumptions the answer is yes. If $f$ is down to be the characteristic function of a finite positive measure $\mu$, that is if $f=\widehat{\mu}$, then then $f^n= \widehat{\mu^{*n}}$ where $\mu^{*n}=\mu*\mu^{*(n-1)}$ (convolution of $mu$ with itself $n$-times. A celebrated result by Bochner states

Theorem: A bounded continuous function $f:\mathbb{R}\rightarrow\mathbb{C}$ is positive definite iff $f$ is the characteristic function of a finite Borel measure on $\mathbb{R}$.

The conclusion of the OP for nonnegative intgegrer powers under continuity follows from that theorem.

For fractional powers, one may have to consider infinite divisible measures: an infinite divisible probability measure $\mu$ on $\mathbb{R}$ is a measure such that for any $n\in\mathbb{N}$, there is another probability measure $\mu_n$ such that $\mu^{*n}_n=\mu$. The extension to real (nonnegative) power then is accomplished via the Lévi-Bochner continuity theorem.