Corollary

The Fourier transform $\mathscr{F}_G$ is injective on $\mathscr{M}^1(G)$.

In fact, if $\alpha \in \mathscr{M}^1(G)$ satisfies $\mathscr{F}_G(\alpha ) = 0$, one deduces from (31) that $\alpha (\mathscr{F}_G(f)) = 0$ for every $f\in L^1(\widehat{G})$; since the image of $L^1(\widehat{G})$ under the Fourier tran$\widehat{s}$form is dense in $\mathscr{C}_0(G)$ (corollary to Proposition 5 of II, p. 209), it follows that $\alpha = 0$.

There exist function spaces on G and $\widehat{G}$, other than $L^2(G)$ and $L^2(\widehat{G})$, on which $\mathscr{F}$ and $\overline{\mathscr{F}}$ are inverse isomorphisms of one another. The following theorem gives an example of this. We denote by B(G) the vector subspace of $L^1(G)$ consisting of the elements $f\in L^1(G)$ such that $\mathscr{F}_G(f)\in L^1(\widehat{G})$. This is a subalgebra of $L^1(G)$. In fact, let $f$ and $g$ be in B(G). Then $f*g\in L^1(G)$ and $\mathscr{F}_G(f*g) =\mathscr{F}_G(f)\mathscr{F}_G(g)\in L^1(\widehat{G})$, since $\mathscr{F}_G(f)\in L^1(\widehat{G})$ and $\mathscr{F}_G(g)\in \mathscr{C}_0(\widehat{G})$.