# Convolution on noncommutative group algebras

If $G$ is a non-Abelian locally compact group, and $f$ is in $L^1{(G)}$ and $u$ is in $L^{\infty}(G)$, and $f\ast u=0$ can it be concluded that $u\ast f=0$?

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The answer is no, and one can find a counterexample already when $G$ is finite, in fact when $G = S_3$ (or any finite non-abelian group). This MO question has a relevant discussion, and the following answer borrows from there:
If $G$ is finite, then the group ring $\mathbb C[G]$ (which coincides with both $L^1$ and $L^{\infty}$ for finite $G$) is a product of matrix rings. If $G$ is non-abelian, then at least one of these matrix rings consists of $n\times n$ matrices for $n > 1$. But in such a matrix ring, there are elements $A$ and $B$ such that $AB= 0$ but $BA \neq 0$. Thus $\mathbb C[G]$ will contain elements $f$ and $u$ with the same property, namely such that $f*u = 0$ but $u*f \neq 0.$