By INT, VIII, §3, No$^o3$, Prop. 11, the mapping $\mu\mapsto \chi (\mu)$ is a character of the involutive Banach algebra $\mathscr{M}^1(G)$. Further, we have:

and therefore this character is hermitian.

Thus we have defined a mapping of $\widehat{G}$ into $\mathsf{X}(\mathscr{M}^1(G))$; it will be called canonical.

Let $\chi \in \widehat{G}$. The restriction of $\mu\mapsto \chi (\mu)$ to $L^1(G)$ is non-zero (cf. INT, VIII, §2, No$^o7$, Prop. 10). By restriction to the involutive Banach subalgebra $L^1(G)$, we therefore obtain a mapping of $\widehat{G}$ into $\mathsf{X}(L^1(G))$, called canonical. It associates to $\chi \in \widehat{G}$ the hermitian character

of $L^1(G)$.

Let us denote this canonical mapping by ev and, for $\chi \in \widehat{G}$, let us denote by ev$_{\chi}$ the image of the character $\chi$ under ev, that is to say, the hermitian character $f\mapsto \chi (f)$ of $L^1(G)$. Considered as a mapping of $\widehat{G}$ into the dual of $L^1(G)$, the mapping ev is the composite of the injection of $\widehat{G}$ into $L^{\infty}(G)$, endowed with the weak topology $\sigma (L^{\infty}(G),L^1(G))$, and the injection of $L^{\infty}(G)$ into the dual of $L^1(G)$, endowed with the topology of simple convergence. Since $\widehat{G}$ is a bounded subset of $L^{\infty}(G)$, the first mapping is continuous by the Lebesgue theorem (INT, IV, §3, No$^o7$, Th. 6). The second is likewise continuous, by definition. This proves that the mapping ev is continuous.

If $\chi \in \widehat{G}$ and if $f\in L^1(G)$, we have ev$_{\chi}(\varepsilon_x*f) =\chi (x$)ev$_{\chi}(f)$. Taking $f$ such that ev$_{\chi}(f)\not= 0$, we deduce that the mapping ev is injective.

Let $\zeta \in \mathsf{X}(L^1(G))$ and let $f\in L^1(G)$ be such that $\zeta (f)\not= 0$. Define a mapping $\chi : G\rightarrow \mathbf{C}$ by putting, for $x\in G$:

We have $\chi (e) = 1$. Since the mapping $x\mapsto \varepsilon_x*f=\boldsymbol{\gamma }(x)(f)$ of G into $L^1(G)$ is continuous (INT, VIII, §2, n$^o5$, Proposition 8), the mapping $\chi$ is continuous. It is bounded because, for every $x\in G$, we have

(Theorem 1 of I, p. 29 and INT, VIII, loc. cit.).

Now let $\mathfrak{B}$ be a basis of the filter of neighbourhoods of $e$ consisting of compact neighbourhoods. For every $V\in \mathfrak{B}$, let $g_V$ be a positive continuous function, zero outside V and of integral equal to 1 (Lemma 1 of II, p. 200). For every function $h\in L^1(G)$, one then has

$\varepsilon_x*h=$ lim$_{V,\mathfrak{B}}(\varepsilon_x*h)*g_V=$ lim$_{V,\mathfrak{B}}(\varepsilon_x*g_V*h)$,

in $L^1(G)$, the limit being taken with respect to the filter of sections of $\mathfrak{B}$ (INT, VIII, §4, n$^o7$, Proposition 20). In particular, since $\zeta (\varepsilon_x*g_V*f) =$ $\zeta (\varepsilon_x*g_V)\zeta (f)$, it follows that

$\chi (x) =$ lim$_{V,\mathfrak{B}}\zeta (\varepsilon_x*g_V)$.

For every $h\in L^1(G)$, we obtain

$\zeta (\varepsilon_x*h) =$ lim$_{V,\mathfrak{B}}\zeta (\varepsilon_x*g_V*h) =\zeta (h)$ lim$_{V,\mathfrak{B}}\zeta (\varepsilon_x*g_V) =\chi (x)\zeta (h)$.

Consequently, for $x, y\in G$ we have:

which proves that $\chi$ is a homomorphism of G into $\mathbf{C}^*$. Since $\chi$ is bounded and continuous, it is a unitary character of G. Moreover, if $g\in L^1(G)$, we have

in $L^1(G)$ (INT, VIII, §1, n$^o5$, Proposition 7), whence

$=\zeta (f)\int_G\chi (x)g(x)dx=$ ev$_{\chi}(g)\zeta (f)$ (INT, VI, §1, n$^o1$, Proposition 1), which shows that $\zeta =$ ev$_{\chi}$. Consequently, ev is surjective, hence bijective.

Finally, let us show that the reciprocal mapping ev$^{-1}$ is continuous. Let $\zeta \in \mathsf{X}(L^1(G))$. Let $f\in L^1(G)$ be a function such that $\zeta (f)\not= 0$. The set W of $\xi \in \mathsf{X}(L^1(G))$ such that $\xi (f)\not= 0$ is an open neighbourhood of $\zeta$ in $\mathsf{X}(L^1(G))$. For every $\xi \in W$, what precedes shows that ev$^{-1}(\xi )$ is the character

Let $\mathfrak{F}$ be a filter on $W\subset \mathsf{X}(L^1(G))$ converging to $\zeta$. Since the set $\mathsf{X}(L^1(G))$ is bounded, hence equicontinuous, in $L^{\infty}(G)$, the uniform structure of simple convergence coincides with the uniform structure of compact convergence (TG, X, p. 16, Theorem 1). Let K be a compact subset of G. The set of the $\varepsilon_x*f$ for $x\in K$ is compact in $L^1(G)$ (INT, VIII, §2, No.$^o5$, Proposition 8). We have therefore

lim$_{\xi ,\mathfrak{F}}\xi (\varepsilon_x*f) =\zeta (\varepsilon_x*f)$

uniformly for $x\in K$. It follows that

lim$_{\xi ,\mathfrak{F}}$ ev$^{-1}(\xi ) =$ ev$^{-1}(\zeta )$,

hence ev$^{-1}$ is continuous at $\zeta$. This completes the proof of the proposition.

Henceforth we shall identify a unitary character $\chi$ of G with the character $f\mapsto \int_Gf(x)\chi (x)dx$ of $L^1(G)$.

The first assertion follows from Proposition 1 and Lemma 1 by restriction to $L^1(G)$. The second follows from the first and from the corollary of Proposition 20 of I, p. 124.

In fact, $\mathsf{X}(L^1(G))$ is locally compact (corollary of Theorem 1 of I, p. 29).

We shall identify $\widehat{G}$ with $\mathsf{X}(L^1(G))$ and $\mathsf{X}$(Stell(G)). For $x\in G$ and $\chi \in \widehat{G}$, we shall denote by $\langle \chi , x\rangle$ the complex number $\chi (x)$, which belongs to $\mathbf{U}$.

We say that $x$ and $\chi$ are orthogonal if $\langle \chi , x\rangle = 1$. Let A be a subset of G (resp. of $\widehat{G}$); the set of elements of $\widehat{G}$ (resp. of G) orthogonal to A is a closed subgroup of $\widehat{G}$ (resp. of G), called the orthogonal of A and denoted by $A^{\bot}$. The orthogonal of G is reduced to $e$.

For $x\in G$, let us denote by $\eta (x)$ the mapping of $\widehat{G}$ into $\mathbf{U}$ defined by $\chi \mapsto$ $\langle \chi , x\rangle$. By definition of the multiplication in $\widehat{G}$, the mapping $\eta (x)$ is a group homomorphism. It is continuous since the mapping $(x, \chi )\mapsto \langle \chi , x\rangle$ of $G\times \widehat{G}$ into $\mathbf{U}$ is continuous (TG, X, p. 28, th. 3). Thus we have defined a mapping $\eta$, called canonical, of G into the bidual group $\widehat{\widehat{G}}$; it is a group homomorphism. Moreover, the mapping $\eta$ is continuous (TG, X, p. 28, th. 3). We shall prove later (II, p. 220, th. 2) that $\eta$ is a group isomorphism of G onto $\widehat{\widehat{G}}$.

Let G and H be locally compact commutative groups, and let $\varphi : G\rightarrow H$ be a morphism of topological groups. For every $\chi \in \widehat{H}$, the mapping $\chi \circ \varphi$ is a character of G, denoted by $\widehat{\varphi}(\chi )$. This definition is expressed by the formula

for every $\chi \in \widehat{H}$ and every $x\in G$. It follows that $\widehat{\varphi}$ is a morphism of the topological group $\widehat{H}$ into the topological group $\widehat{G}$; one says that $\widehat{\varphi}$ is the dual of the morphism $\varphi$.

Let K be a locally compact commutative group and let $\psi : H\rightarrow K$ be a morphism of topological groups. The definition shows that $\widehat{\psi}\circ \varphi =\widehat{\varphi}\circ \widehat{\psi}$. If $\varphi$ is the identity mapping of G, then $\widehat{\varphi}$ is the identity mapping of $\widehat{G}$. In particular, if $\varphi$ is a group isomorphism, the same is true of $\widehat{\varphi}$, and $\widehat{\varphi}^{-1}$ is the dual of $\varphi^{-1}$.

By definition, we have $\chi \in$ Ker($\widehat{f}$) if and only if the restriction of $\chi$ to the image of $f$ is trivial.

Let $m$ be the product mapping of $\widehat{G}^n$ into $\widehat{G}$, and for every $j$ such that $1\leqslant j\leqslant n$, let $\pi_j$ be the projection of G onto $G_j$. Let $\mu$ be the morphism of topological groups $m\circ (\widehat{\pi}_j)_j$ from $\prod\widehat{G}_j$ into $\widehat{G}$, such that

The mapping $\mu$ is continuous, and one verifies that $\mu$ and $(\widehat{\lambda}_j)$ are reciprocal bijections of one another. The proposition follows.

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).

The Fourier cotransform is a morphism of involutive algebras of $L^1(G)$ into the algebra of bounded continuous functions on $\widehat{G}$ (prop. 3). Since it identifies with the Gelfand transform, its image is contained in $\mathscr{C}_0(\widehat{G})$ (I, p. 37, prop. 5), and its kernel is the radical of $L^1(G)$ (prop. 8 of I, p. 38), which is zero (cor. of prop. 22 of I, p. 126).

We shall see later (corollary to proposition 13 of II, p. 221) that the cotransform of Fourier on $\mathscr{M}^1(G)$ is also injective.

The cotransform of Fourier extends by continuity to a morphism of stellar algebras from Stell(G) into $\mathscr{C}_0(\widehat{G})$. If one identifies $\widehat{G}$ with $\mathsf{X}$(Stell(G)) (cor. 1 of II, p. 204 and prop. 1 of II, p. 202), this extension is the Gelfand transform of Stell(G). By th. 1 of I, p. 108, it is an isomorphism. The assertion concerning the Fourier transform follows from this.

We shall always denote by $\overline{\mathscr{F}}$ and $\mathscr{F}$ the isomorphisms of prop. 5.

Since $L^1(G)$ is dense in Stell(G), this follows from proposition 5.

Let $\chi \in \widehat{G}$. Since $\varepsilon_y*dx=dx$ for every $y\in G$, one has

by formula (11). If $\chi \not= 1$, there exists $y\in G$ such that $\langle \chi , y\rangle \not= 1$, hence $\mathscr{F}(dx)(\chi ) = 0$. If $\chi = 1$, then $\mathscr{F}(dx)(\chi ) =\int_Gdx= 1$ since the measure $dx$ is normalized.

Let $f\in L^1(G)$. For every $g\in L^2(G)$, one has $f*g\in L^1(G)$ (INT, VIII, §4, n$^o5$, prop. 12). Consequently, the space $L^1(G)\cap L^2(G)$ is an ideal of $L^1(G)$, and the same is true of the space A(G). The ideal A(G) is self-adjoint.

Let $f$ and $g$ be in $L^2(G)$. The convolution product $f*g$ is then the class of the continuous function given by

(INT, VIII, §4, n$^o5$, prop. 15). The second assertion follows.

Since $\chi (f*g) = (\chi f)*(\chi g)$ for $\chi \in \widehat{G},f\in L^1(G)$ and $g\in L^1(G)$ (INT, VIII, §3, No.$^o1$, prop. 6), one has $\chi h\in A(G)$ for every $h\in A(G)$ and $\chi \in \widehat{G}$. Since $\varepsilon_x*f=\boldsymbol{\gamma }(x)f$ and the linear representation $\boldsymbol{\gamma }$ is isometric on $L^p(G)$ for every $p$ (INT, VIII, §2, No.$^o5$, prop. 8), one has $\varepsilon_x*f\in A(G)$ for all $x\in G$ and $f\in A(G)$.

We denote by $\widehat{A}(G)$ the image of A(G) under the Fourier transform. It is a subspace of $\mathscr{C}_0(\widehat{G})$.

(i) For every element $\varphi$ of a set of $\mathfrak{B}$, one has $\|\varphi \|_1= 1$ and $\|\mathscr{F}(\varphi )\|_{\infty}\leqslant 1$ ;

(ii) One has

lim$_{\varphi ,\mathfrak{B}}\varphi \cdot dx=\varepsilon_e$

in the space $\mathscr{C}'(G)$ of measures with compact support on G endowed with the topology of uniform convergence on compact parts of $\mathscr{C}(G)$ ;

(iii) One has

lim$_{\varphi ,\mathfrak{B}}\mathscr{F}(\varphi ) = 1$

for the topology of compact convergence on $\widehat{G}$;

(iv) For $p= 1$ or $p= 2$, and for every $f\in L^p(G)$, one has $\varphi *f\in$ A(G) for every $\varphi$ belonging to a set of $\mathfrak{B}$ and

lim$_{\varphi ,\mathfrak{B}}\varphi *f=f$

in $L^p(G)$.

Let $K_0$ be a fixed compact neighbourhood of $e$ in G. Let $\mathfrak{B}_0$ be a basis of the filter of neighbourhoods of $e$ in G consisting of symmetric compact neighbourhoods contained in $K_0$ (cf. TG, III, p. 4). For K $\in \mathfrak{B}_0$, let $X'_K$ be the set of functions $\psi \in \mathscr{K}_+(G)$ such that Supp$(\psi )\subset K$ and $\int\psi (x)dx= 1$ ; it is nonempty (lemma 1 of II, p. 200). Let $X_K$ be the set of functions $\psi *\psi$ for $\psi \in X'_K$. It is nonempty and contained in $A(G)\cap \mathscr{K}_+(G)$. The set $\mathfrak{B}$ whose elements are the sets $X_K$ for K varying in $\mathfrak{B}_0$ is a filter basis on $A(G)\cap \mathscr{K}_+(G)$. Let us prove that $\mathfrak{B}$ satisfies the required properties.

If $X\in \mathfrak{B}$ and $\varphi \in X$, one has $\|\varphi \|_1=\int_G\varphi (x)dx= 1$, whence $\|\mathscr{F}(\varphi )\|_{\infty}\leqslant 1$, which establishes property (i).

Property (ii) follows from INT, VIII, § 2, No.$^o7$, Corollary 1 to Lemma 4. A compact subset of $\widehat{G}$ is a compact subset of $\mathscr{C}(G)$, hence (ii) implies lim$_{\varphi ,\mathfrak{B}}\mathscr{F}(\varphi ) = 1$ for the topology of compact convergence on $\widehat{G}$, that is, (iii).

Finally, let $p= 1$ or $p= 2$. Let $f\in L^p(G)$. We have $\varphi *f\rightarrow f$ in $L^p(G)$ with respect to the filter $\mathfrak{B}$ (INT, VIII, §4, n$^o7$, prop. 20). Moreover, for every K in $\mathfrak{B}_0$ and $\varphi \in X_K$, there exists $\psi \in X'_K$ such that $\varphi =\psi *\psi$, whence $\varphi *f=\psi *(\psi *f)$. We have $\psi \in L^1(G)\cap L^2(G)$ and $\psi *f\in L^1(G)\cap L^2(G)$, hence $\varphi *f\in A(G)$.

Assertion (iv) of the proposition provides the first assertion. Since $L^1(G)$ is dense in Stell(G), the second follows from it, and the last then follows from Prop. 5 of II, p. 209.

This follows from the preceding corollary since, for every $\chi \in \widehat{G}$, the mapping $f\mapsto \mathscr{F}(f)(\chi )$ is a nonzero character of $L^1(G)$.

Recall that the left regular representation $\boldsymbol{\gamma }$ of Stell(G) on $L^2(G)$ (cf. I, p. 125, n$^o13$) is denoted by $\boldsymbol{\gamma }(\varphi )f$ = $\varphi *f$ for $\varphi \in$ Stell(G) and $f\in L^2(G)$.

Moreover, for every $f$ and $g$ in A(G), we have the equality

between the measure with density $\mathscr{F}(f)$ with respect to $\mu_g$ and the measure with density $\mathscr{F}(g)$ with respect to $\mu_f$.

Let $f,g$ be elements of $L^1(G)\cap L^2(G)$. For every $\varphi \in$ Stell(G), one has $\varphi *f\in L^2(G)$ and $\|\varphi *f\|_2\leqslant \|\varphi \|_*\|f\|_2($I, p. 126, formula (8)). Moreover, one has $\varphi *(f*g) = (\varphi *f)*g($loc. cit., formula (9)). This latter function belongs to the closure $\mathscr{C}_0(G)$ of $\mathscr{K}(G)$ in $\mathscr{C}(G)$ (INT, VIII, §4, n$^o5$, prop. 15). Moreover, one has

Since the functions $f*g$ for $f$ and $g$ in $L^1(G)\cap L^2(G)$ generate A(G), it follows that $\varphi *f\in \mathscr{C}_0(G)$ for all $f\in A(G)$ and $\varphi \in$ Stell(G), and that the mapping $\varphi \mapsto (\varphi *f)(e)$ is a continuous linear form on Stell(G). Since $\mathscr{F}$ is an isomorphism of Stell(G) onto $\mathscr{C}_0(\widehat{G})$ (prop. 5 of II, p. 209), the first assertion follows.

Now let $f$ and $g$ be in A(G). For $\varphi \in L^1(G)$, one has

Since $(\varphi *f)*g= (\varphi *g)*f$ and since the image of $L^1(G)$ under the Fourier transform is dense in $\mathscr{C}_0(\widehat{G})$ (cor. of II, p. 210), one deduces from formula (19) that formula (18) is satisfied for all $f$ and $g$ in A(G).