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.



Factorisation of positive definite functions
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](https://books.google.com/books?id=sXfO4pjF9QYC&pg=PA254&lpg=PA254&dq=positive%20definite%20semigroup&source=bl&ots=20O26jW0c8&sig=WfMJLvQp_VscW_2QZETkwdPykBc&hl=en&sa=X&ved=0ahUKEwj5z6KL5enWAhXnyFQKHUhDDQ8Q6AEISTAH#v=onepage&q=positive%20definite%20semigroup&f=false)). 
(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. 


