Proposition 1

The canonical mapping of $\widehat{G}$ into $\mathsf{X}(L^1(G))$ is a homeomorphism.

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 $L^1(G)$.