Fourier transform distributing over convolution of measure and Lp function

Question

Let $G$ be an infinite compact abelian group with normalized Haar measure $\lambda$ (i.e. $\lambda(G)=1$), let $M(G)$ be the collection of all complex regular Borel measures on $G$, and let $L^p(G)$ be the collection of $f : G \rightarrow \mathbb{C}$ such that

$||f||_p = (\int_G |f|^p d \lambda)^{1/p} < \infty$

I know that for $\mu , \nu \in M(G)$,

$\widehat{\mu * \nu} = \widehat{\mu} \cdot \widehat{\nu}$

and that for $f, g\in L^p(G)$,

$\widehat{f * g} = \widehat{f} \cdot \widehat{g}$

where $*$ is the convolution operator, since $L^p(G) \subseteq L^1(G)$ since $G$ is compact. But is it true that

$\widehat{\mu * f} = \widehat{\mu} \cdot \widehat{f}$ ?

Answer 1

Let $\nu = fd\lambda$. Then $\nu \in M(G)$ and $\widehat{\nu}$=$\widehat{f}$. So we can apply the formula $\widehat{\mu \ast \nu}=\widehat{\mu} \cdot \widehat{\nu}$, which you wrote down.