Let $(S,\circ)$ be a semigroup with identical involution. Of course, we know that products of positive definite functions on $S$ are again positive definite. I'm interested in the other direction, meaning:

Suppose $V\colon S \to\mathbb{R}$ is a positive definite function, which factorises as $V=V_{1}\cdot V_{2}$ (usual pointwise multiplication) with $V_{2}$ being positive definite.

Is it true that $V_{1}$ is positive definite? If not, what are additional assumptions ensuring this statement? I'm grateful for any suggestions.

## 1 Answer

The answer is no, $V_{1}$ is not positive definite in general. E.g., let $(S,\circ)=(\mathbb R,+)$. Let $V(x)=1$ and $V_2(x)=(e^x+e^{-x})/2$ for $x\in\mathbb R$. Then $V$ and $V_2$ are positive definite (with respect to the identical involution).

However, $V_1=V/V_2=1/\cosh=\text{sech}$ is not positive definite. Indeed, if $V_1$ were positive definite, it would be a positive mixture of bounded semicharacters; but in this case the only bounded semicharacter is the constant 1 (see e.g. page 254 in Berg). (More elementarily, $\sum _{j=1}^n \sum _{k=1}^n \text{sech}(x_j+x_k)z_j z_k^*=-8.47\ldots\times10^{-6}<0$ if $n=2$, $x_1= \frac{901}{10},x_2= -\frac{641}{10},z_1= -791+37i,z_2= 546+463 i$.)

This example also suggests that there hardly can be broad natural conditions for such $V_1$ to be positive definite.