Proposition 3

For every measure $\mu\in \mathscr{M}^1(G)$, the functions $\mathscr{F}_G(\mu)$ and $\overline{\mathscr{F}}_G(\mu)$ are continuous and bounded. The mappings $\mathscr{F}_G:\mu\mapsto$ $\mathscr{F}_G(\mu)$ and $\overline{\mathscr{F}}_G:\mu\mapsto \overline{\mathscr{F}}_G(\mu)$ are continuous morphisms of the involutive algebra $\mathscr{M}^1(G)$ into the involutive algebra of bounded continuous functions on $\widehat{G}$ (Example 1 of I, p. 99).

Let $\mu\in \mathscr{M}^1(G)$. For every $\chi \in \widehat{G}$, one has

therefore the Fourier transform of $\mu$ is bounded. Analogously, one verifies that $\overline{\mathscr{F}}(\mu)$ is bounded.

If $\chi$ tends to $\chi_0$ in $\widehat{G}$, the function $\chi$ on G tends to $\chi_0$ uniformly on every compact while remaining bounded by the constant function 1 which belongs to $L^1(G, \mu)$. By the Lebesgue theorem (INT, IV, §3, n$^o7$, th. 6), it follows that $\mathscr{F}(\mu)(\chi )$ tends to $\mathscr{F}(\mu)(\chi_0)$. Hence $\mathscr{F}(\mu)$ is continuous. The same is true of $\overline{\mathscr{F}}(\mu)$.

For every $\chi \in \widehat{G}$, the mapping $\mu\mapsto \chi (\mu) =\int\langle \chi , x\rangle d\mu(x)$ is a hermitian character of $\mathscr{M}^1(G)$ (Lemma 1 of II, p. 202). This implies that $\mathscr{F}$ and $\overline{\mathscr{F}}$ are morphisms of involutive algebras from $\mathscr{M}^1(G)$ into the involutive algebra of bounded continuous functions on $\widehat{G}$. Inequality (7) proves that these morphisms are continuous.

The Fourier transform of G (resp. the Fourier cotransform of G) is the mapping $\mu\mapsto \mathscr{F}_G(\mu)$ (resp. the mapping $\mu\mapsto$ $\overline{\mathscr{F}}_G(\mu)$) of $\mathscr{M}^1(G)$ into $\mathscr{C}_b(\widehat{G})$.

Let us note some useful formulae for $\mu\in \mathscr{M}^1(G),x\in G$ and $\chi \in \widehat{G}:$

(10)

(in particular $\mathscr{F}(\varepsilon_e) =\overline{\mathscr{F}}(\varepsilon_e) = 1$),

(11)

(12)

Formulae (8), (9), (10) and (11) follow from the definitions. Let us prove the first of formulae (12), the second being analogous. For every $\xi$ in $\widehat{G}$ one has the equalities

Let us note moreover that for every $\chi \in \widehat{G}$ and every measures $\mu$ and $\nu$ in $\mathscr{M}^1(G)$, one has

since the two members of this equality are functions on $\widehat{G}$ whose value at $\xi \in \widehat{G}$ is

Let H be a locally compact commutative group and let $\varphi : G\rightarrow H$ be a continuous morphism. Let $\mu\in \mathscr{M}^1(G)$. The image measure $\varphi (\mu)$ is defined (INT, V, §6, n$^o1$, remark 1), and one gets $\mathscr{F}_H(\varphi (\mu)) =\mathscr{F}_G(\mu)\circ$ $\widehat{\varphi}($cf. INT, V, §6, n$^o4$, prop. 7).

By restriction to the subalgebra $L^1(G)$ of $\mathscr{M}^1(G)$, one obtains the definition of the Fourier transform and of the Fourier cotransform on $L^1(G)$. One thus has for $f\in L^1(G)$ and $\chi \in \widehat{G}:$

In particular, $\mathscr{F}_G(f) =\overline{\overline{\mathscr{F}}_G(\overline{f})}$. One also has

for every $f\in L^1(G)$ and every $\chi \in \widehat{G}$.

Let $\sigma$ be an automorphism of G and Δ the module of $\sigma$ (INT, VII, §1, n$^o4$, def. 4). For $f\in L^1(G)$, one has

(cf. loc. cit., formula (31)).

If one identifies $\widehat{G}$ and $\mathsf{X}(L^1(G))$ (prop. 1 of II, p. 202), the Fourier cotransform is nothing but the Gelfand transform of the Banach algebra $L^1(G)$ (I, p. 7, def. 5).