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

c) Every subspace of $L^2(G)$ stable under the mappings $f\mapsto \varepsilon_x*f$ for every $x\in G$ is of the form $E_M$ for some measurable subset M of $\widehat{G}$.

This result will be proved later (cf. V, to appear).

Since $\mathscr{K}(\widehat{G})$ is dense in $L^2(\widehat{G})$ and the Fourier transform of $L^2(G)$ is an isometry onto $L^2(\widehat{G})$ (th. 1 of II, p. 215), the subspace V of $L^2(G)$ formed by the $f\in L^2(G)$ such that $\mathscr{F}_G(f)\in \mathscr{K}(\widehat{G})$ is dense in $L^2(G)$.

Let $g\in L^1(G)$. There exist $g_1, g_2\in L^2(G)$ such that $g=g_1g_2$ (one may for example take $g_1=|g|^{1/2}$, and $g_2(x) = 0$ if $g(x) = 0,g_2(x) =$ $g(x)/g_1(x)$ otherwise). It follows therefore from what precedes that $g$ is the limit of a sequence of functions of the form $h_1h_2$, where $h_1$ and $h_2$ belong to V. Now $\mathscr{F}_G(h_1h_2) =\mathscr{F}_G(h_1)*\mathscr{F}_G(h_2)$ (II, p. 223, prop. 14), and $\mathscr{F}_G(h_1)*$ $\mathscr{F}_G(h_2)$ belongs to $\mathscr{K}(G)$. The lemma follows.

Let $\varepsilon >0$. There exists a function $g\in L^1(G)$ such that $\|f-f*g\|_1<$ $\varepsilon$ (prop. 8 of II, p. 211 (iv)). Let $h\in L^1(G)$ be such that the support of $\overline{\mathscr{F}}_G(h)$ is compact and $\|f\|_1\|g-h\|_1< \varepsilon$ (lemma 1). We have

By the assumption on $f$, the function $\overline{\mathscr{F}}_G(f*h) =\overline{\mathscr{F}}_G(f)\overline{\mathscr{F}}_G(h)$ has compact support disjoint from V(I), which implies that $f*h\in I$ (remark 2 above). Since $\varepsilon$ is arbitrarily small, we have $f\in$ I = I.

Comme the algebra $L^1(G)$ is regular (prop. 2 of II, p. 219) and without radical (cor. of prop. 22 of I, p. 126), and since the set of functions whose Fourier cotransform has compact support is dense in $L^1(G)$ (lemma 1), cor. 1 of I, p. 92 shows that the ideal I is contained in a regular maximal ideal of $L^1(G)$, that is to say in the kernel of a character $\widehat{y}$ of $L^1(G)$ (th. 2 of I, p. 30); one can then take $\widehat{x}=\widehat{y}^{-1}($cf. formula (8) of II, p. 207).

Corollary 1 (Wiener Tauberian Theorem)

Let $f\in L^1(G)$. If the Fourier transform of $f$ does not vanish, the functions $f*\varepsilon_x:g\mapsto f(gx^{-1})$, where $x$ runs through G, form a total set in $L^1(G)$ (EVT, I, p. 12, def. 1).

Let V be the closed vector subspace of $L^1(G)$ generated by the $f*\varepsilon_x$. By INT, VIII, §4, cor. of prop. 20, the space V is a closed ideal of $L^1(G)$. By th. 1 we have $V = L^1(G)$.

a) For every function $h\in L^1(G)$ such that $\int_Gh(x)dx= 1$, the limit of $h*g$ following Φ is equal to $\alpha$;

b) Suppose in addition that $g$ is slowly oscillating following Φ. Then $g$ tends to $\alpha$ following Φ.

Replacing $g$ by $g-\alpha$, we reduce to the case where $\alpha = 0$. Let I be the set of functions $h\in L^1(G)$ such that $h*g$ tends to 0 following Φ. The set I is a vector subspace of $L^1(G)$ invariant under translation. It is a closed space. Indeed, let $h\in \overline{I}$. For every function $h_0\in L^1(G)$ and every $x\in G$, we have

For every $\varepsilon >0$, there exists $h_0\in I$ such that $\|h-h_0\|_1\|g\|_{\infty}< \varepsilon$. Let $M\in$ Φ such that $|(h_0*g)(x)|< \varepsilon$ for every $x\in M$. We then have $|(h*g)(x)|<2\varepsilon$ for every $x\in M$, hence $h*g$ converges to 0 following Φ. This shows that $h\in I$.

The space I is therefore a closed ideal of $L^1(G)$. We have $f\in I$ by assumption, hence $I = L^1(G)$ by th. 1 since the Fourier transform of $f$ does not vanish. This implies a).

Let us place ourselves under the hypotheses of b). Let $\varepsilon >0$. Since $g$ is slowly oscillating according to Φ, there exists $M\in \Phi$ and a compact neighbourhood V of $e$ such that

$x\in M$ and $y\in V =\Rightarrow |g(y^{-1}x)-g(x)|\leqslant \varepsilon$.

Let $\varphi$ be the characteristic function of V and $\mu=\int\varphi (x)dx$. For every $x\in G$, we have

Thus for every $x\in M$, we have

Since, by a), the limit of $\varphi *g$ according to Φ is zero, we have lim sup$_{\Phi}|g|\leqslant \varepsilon$. Since $\varepsilon$ is arbitrary, we conclude that the limit of $g$ according to Φ is zero, which proves b).

b) the function $\mathscr{F}_G(j)$ is equal to 1 in a neighbourhood of the identity element of $\widehat{G}$;

c) for every $x\in K$, we have $\|j-j*\varepsilon_x\|_1\leqslant \eta$.

The set $U_1$ of the elements $\widehat{x}\in \widehat{G}$ such that

for every $x\in$ K is a neighbourhood of $e$ in $\widehat{G}$. Let U $\subset U_1$ be an open, symmetric, and integrable neighbourhood for the Haar measure $m=d\widehat{x}$ of $\widehat{G}$ dual to the measure $dx$. Let $V\subset U$ be a compact symmetric neighbourhood of $e$ such that $m(V)\geqslant \frac{1}{2}m(U)$. Let us denote by $\varphi_U$ (resp. $\varphi_V$) the characteristic function of U (resp. of V). Since $\varphi_U$ belongs to $L^2(G)$, there exists $u\in L^2(G)$ such that $\varphi_U=\mathscr{F}_G(u)$ (th. 1 of II, p. 215). Analogously, there exists a function $v\in L^2(G)$ such that $\varphi_V=\mathscr{F}_G(v)$. We now prove that the function $j=\frac{1}{m(V)}uv$ verifies the required properties. We have $j\in L^1(G)$.

a) By the Plancherel theorem and the condition $m(V)\geqslant \frac{1}{2}m(U)$, we have

b) There exists a neighbourhood W of $e$ in $\widehat{G}$ such that WV $\subset U$ (TG, II, p. 31, prop. 4). For every $\widehat{x}\in W$, we have $\widehat{x}V\subset U$, and prop. 14 of II, p. 223 implies

since V is symmetric.

c) If $x\in K$, we have

since $U\subset U_1$, and analogously $\|v-v*\varepsilon_x\|^2_2\leqslant m(V)(\frac{\eta}{4})^2$. Hence

Let $\chi$ be a character of $L^1(G)$. Distinguish two cases according as $\chi$ is zero or nonzero. If $\chi$ is zero, it must be verified that for every function $f\in L^1(G)$, there exists a sequence $(f_n)_{n\geqslant 1}$ in $L^1(G)$ such that $\mathscr{F}(f_n)$ vanishes outside a compact subset of $\widehat{G}$ and such that $f_n*f$ tends to $f$ in $L^1(G)$. The existence of such a sequence follows from Lemma 1 above and Proposition 8 of II, p. 211.

Suppose now that $\chi$ is nonzero, hence $\chi \in \mathsf{X}(L^1(G)) =\widehat{G}$ (Proposition 1 of II, p. 202). Let $f\in L^1(G)$ be such that $\mathscr{G}_{L^1(G)}(f)(\chi ) =$ $\overline{\mathscr{F}}(f)(\chi ) = 0$. It is a question of proving the existence of a sequence $(f_n)_{n\geqslant 1}$ in $L^1(G)$ such that $f*f_n$ converges to $f$ in $L^1(G)$ and such that $\overline{\mathscr{F}}(f_n)$ vanishes in a neighbourhood of $\chi$. We may suppose that $\|f\|_1= 1$. By translation in $\widehat{G}$, we reduce to the case where $\chi =e$.

Let $K_n$ be a compact subset of G such that

Let $u_n\in L^1(G)$ be a function $\geqslant 0$ such that $\|u_n\|_1= 1$ and

(cf. Proposition 8 of II, p. 211, (iii)). According to Lemma$\surd$ 2, there exists a function $j_n$ in $L^1(G)$ such that $\|j_n\|_1\leqslant \overline{2}$, whose Fourier cotransform is equal to 1 in a neighbourhood of $e$, and moreover such that $\|j_n-j_n*\varepsilon_x\|_1\leqslant n^{-1}$ for every $x\in K_n$. Put

We shall prove that the sequence $(f_n)_{n\geqslant 1}$ has the required properties. First of all, we have

hence the Fourier transform of $f_n$ vanishes in a neighbourhood of $\chi =e$. On the other hand,

1

Now, for almost every $y\in G$, we have

since, by assumption, we have $\mathscr{F}(f)(e) =\int_Gf(x)dx= 0$. Hence

$G-K_n$

$1\int\int$ 5 $\leqslant |f(x)|dx+ 4|f(x)|dx\leqslant$.

$n_{K_n}G-K_{_n}n$

Finally, $\|f*f_n-f\|_1\leqslant 6n^{-1}$ and therefore $f*f_n$ converges in $f$ in $L^1(G)$, as desired.

Applying prop. 5, we obtain the following result:

For an arbitrary closed ideal of $L^1(G)$, the conclusion of th. 2 is in general false (cf. exer. 12 of II, p. 314). More precisely, one can now prove that, if G is noncompact, there exists a closed ideal of $L^1(G)$ which is not self-adjoint (see for example W. Rudin, Fourier analysis on groups, Interscience tracts in pure and applied mathematics, theorem 7.7.1.)

We have seen that $A(W) = V(W^{\circ})$. Since $W\not= 0$, we have $W^{\circ}\not= L^1(G)$, and then $V(W^{\circ})$ is nonempty by th. 1 of II, p. 252.

a) For every neighbourhood U of A(W) in $\widehat{G}$, every function of W is the weak limit of linear combinations of characters belonging to U;

b) If the boundary of A(W) contains no nonempty perfect set, every function of W is the weak limit of linear combinations of characters belonging to W.

To prove a), it is enough by the bipolar theorem to show that if $f$ is a function of $L^1(G)$ orthogonal to the elements of U, then $f$ is orthogonal to W. Now the Fourier cotransform $\overline{\mathscr{F}}(f)$ then vanishes on the neighbourhood U of $A(W) = V(W^{\circ})$, so that prop. 2 of II, p. 252 shows indeed that $f\in W^{\circ}$. Assertion b) is established similarly, using th. 2 of II, p. 257 instead of prop. 2 of II, p. 252.