Proposition 10

a) Let M be a measurable subset of $\widehat{G}$. The space $E_M$ is a subrepresentation of the representation $\boldsymbol{\gamma }_G$;

b) Let M and N be measurable subsets of $\widehat{G}$. One has $E_M= E_N$ if and only if M and N are equal up to a locally negligible set;

c) Every subrepresentation of $\boldsymbol{\gamma }_G$ is of the form $E_M$ for a measurable subset M of $\widehat{G}$.

Let $\eta$ be the canonical mapping of G into $\widehat{\widehat{G}} ($cf. II, p. 216, Remark 1). Since $E_M$ is closed, assertion a) follows from the formulae

for $x\in G$ and $f\in L^2(G, \mu)$.

Let M and N be measurable subsets equal up to a locally negligible set. Since $\varphi_M$ is then equal to $\varphi_N$ in $L^{\infty}(\widehat{G},\widehat{\mu})$, this condition implies that $E_M= E_N$.

Suppose conversely that M and N are not equal up to a locally negligible set. Possibly interchanging the roles of M and N, there then exists a compact subset K such that the set $L = K\cap (M$ - $(M\cap N))$ is not negligible. Let $\varphi_L\in L^2(\widehat{G},\widehat{\mu})$ be the class of the characteristic function of L, and put $f=\overline{\mathscr{F}}_{\widehat{G}}(\varphi_L)\in L^2(G, \mu)$; one has $\mathscr{F}_G(f) =\varphi_L$ (Corollary to Theorem 2 of II, p. 220). Then $f$ belongs to $E_N$, since $L\cap N$ is empty, but not to $E_M$, since $\varphi_M\mathscr{F}(f) =\varphi_M\varphi_L=\varphi_L$. Hence $E_M\not = E_N$, which proves b).

Let now E be a subrepresentation of $\boldsymbol{\gamma }_G$. Let $p_E$ be the orthogonal projector of $L^2(G, \mu)$ with image E and let $q_E=\mathscr{F}_G\circ p_E\circ \mathscr{F}_G^{-1}$. The projector $p_E$ belong to Hom$_G(\boldsymbol{\gamma }_G,\boldsymbol{\gamma }_G)$ (prop. 4 of V, p. 383), hence it commutes with $\boldsymbol{\gamma }_G(f)$ for every $f\in L^1(G, \mu) ($cf. V, p. 401). This means that it commutes with the endomorphisms $\varphi \mapsto f*\varphi$ for $f\in L^1(G, \mu)$ (lemma 4 of V, p. 407). Consequently, the endomorphism $q_E$ of $L^2(\widehat{G},\widehat{\mu})$ commutes with the multiplication endomorphism by $g$ for every function $g\in \mathscr{C}_0(\widehat{G})$ belonging to the image of the Fourier transform of $L^1(G, \mu)$ in $\mathscr{C}_0(\widehat{G})$ (prop. 14 of II, p. 223). Since the image of the Fourier transform is dense in $\mathscr{C}_0(\widehat{G})$ (corollary of prop. 5 of II, p. 209), the continuity of the morphism $g\mapsto m_g$ implies that $q_E$ commutes with $m_g$ for every function $g\in \mathscr{C}_0(\widehat{G})$.

By the corollary of proposition 7 of IV, p. 188, there therefore exists a function $\varphi \in \mathscr{L}^{\infty}(\widehat{G},\widehat{\mu})$ such that $q_E=m_{\varphi}$. Since $q_E=q^2_E$, we have $m_{\varphi}=m^2_{\varphi}=m_{\varphi^2}$, whence $\varphi =\varphi^2$ in $L^{\infty}(\widehat{G},\widehat{\mu})$ (prop. 5 of IV, p. 186); this means that $\varphi$ is equal in $L^{\infty}(\widehat{G},\widehat{\mu})$ to the class of the characteristic function of a measurable subset N of $\widehat{G}$. Put $M =\widehat{G}-$ N. Let $f\in L^2(G, \mu)$. We have $f\in E$ if and only if $p_E(f) =f$, if and only if $\varphi \mathscr{F}_G(f) =\mathscr{F}_G(f)$, which is equivalent to $f\in E_M$.