TITLE: A Schur-like product theorem on groups
QUESTION [5 upvotes]: Let $G$ be a finite group, and consider the composition $X * Y$ on $\mathbb{C}G$ defined by $$(\sum_g \alpha_g u_g) * (\sum_g \beta_g u_g) = \sum_g \alpha_g \beta_g u_g.$$ This composition can be reformulated as a convolution using the Fourier transform.
Question: Let $X,Y \in \mathbb{C}G$ be positive elements. Is $X * Y$ also positive?  
Remark: It is an application of Theorem 4.1 in http://dx.doi.org/10.1090/tran/6582 to finite groups theory, but we are here interested in a purely group-theoretic reference or proof.
It is like to Schur product Theorem because $X * Y$ is an entrywise product like the Schur product.
Bonus question: Can we extend to any locally compact group?

REPLY [4 votes]: The answer is positive. For ease of notation, let me write $a=\sum_g \alpha_g u_g$ and $b = \sum_g \beta_g u_g$.
Let's equip $\mathbb{C}G$ with the positive inner product induced by the standard basis. In terms of the usual normalized trace on a group algebra, this inner product is given by $(x,y)\mapsto\mathrm{tr}(x^*y)$. So when you decompose $\mathbb{C}G$ into a direct sum of matrix algebras, this is just the usual Hilbert-Schmidt inner product. In particular, the cone of positive elements is nicely self-dual: we have $y\geq 0$ if and only if $\mathrm{tr}(xy)\geq 0$ for all $x\geq 0$. Hence it is enough to show that $\mathrm{tr}\left(x(a \ast b)\right)\geq 0$ for all $x\geq 0$.
Now your composition operation $*:\mathbb{C}G\otimes\mathbb{C}G\to\mathbb{C}G$ is the adjoint of the usual comultiplication $\Delta:\mathbb{C}G\to\mathbb{C}G\otimes\mathbb{C}G$, meaning that
$$
\mathrm{tr}\left(x(a \ast b)\right) = \mathrm{tr}(\Delta(x)(a\otimes b)).
$$
Finally, we have $\Delta(x)\geq 0$ since $\Delta$ is a $*$-homomorphism, and $a\otimes b\geq 0$ since $a\geq 0$ and $b\geq 0$ by assumption.
I don't know about the generalization to the locally compact case.