I'm working through Dietmar's Principles of Harmonic Analysis. He defines a Dirac function $\phi\in L^1(G)$ as a nonnegative $\phi: G\to \mathbb{R}$ such that
- $\displaystyle\int_G\phi\space dx = 1$
- $\phi(x)=\phi(x^{-1})$
He makes the claim (in Lemma 1.6.5) that the convolution of two Dirac functions is Dirac. That the convolution is nonnegative is obvious, and that it has integral $1$ is a consequence of Fubini's Theorem. When attempting to prove symmetry, I proved the following relation: $$ \left(\phi\ast\psi\right)(x) = \int_G \phi(y)\psi(y^{-1}x)dy = \int_G \phi(xy)\psi(y^{-1})dy = \\ = \int_G \phi(y^{-1}x^{-1})\psi(y)dy = \left(\psi\ast\phi\right)(x^{-1})$$ Note: The second equality on the first line follows the invariance of the Haar integral under left shifts.
We see that Deitmar's claim is correct if and only if $\phi\ast\psi=\psi\ast\phi$. This is the case whenever $G$ is an abelian group, as the convolution operator on $L^1(G)$ commutes if and only if $G$ is abelian. However, I can't seem to make the claim work on the non-abelian case. The non-commutativity of the convolution product on $L^1(G)$ does not suffice to disprove the claim either, as the class of Dirac functions is rather small. In particular, it does not separate points, so the logic used in the proof of non-commutativity does not carry over.
Any insights would be welcome.
You are correct. To construct a counterexample, you might as well take $G$ to be a finite non-Abelian group: they are certainly locally compact. One can then identify functions on $G$ with elements of the group algebra. Take $\phi$ and $\psi$ to correspond to group algebra elements $\frac12(g+g^{-1})$ and $\frac12(h+h^{-1})$. In general the product of these won't be invariant under the "antipode" map of the group algebra (that which sends group elements to their inverses).