$=d(u)\int_Gf(g$)Tr($u(gh^{-1})$)$dg=d(u)\int_Gf(gh$)Tr($u(g)$)$dg$. (10)

For all $u\in \widehat{G}$ choose an orthonormal basis $Bu$ of $E_u$, and denote by $(u_{ij}(g))$ the matrix of $u(g)$ in this basis. Prop. 1 also means that the family of functions $\surd\overline{d(u)}u_{ij}$, for $u$ in $\widehat{G}$ and $i, j$ in $B_u$, is an orthonormal basis of the space $L^2(G)$.

If $f$ is an integrable function on G such that the family $(f_u)$ is uniformly summable, then the sum of this family is a continuous function which coincides almost everywhere with $f$; in other words, if we assume in addition that $f$ is continuous, then for all $h\in G$, $f(h) =\sum_{u\in\widehat{G}}d(u)\int_Gf(gh$)Tr($u(g)$)$dg$. (11)

Conversely, let $A\in F( \widehat{G})$; if the family $(\mathscr{F}_uA)_{u\in\widehat{G}}$ is uniformly summable, the function

$g \rightarrow \sum_{u\in\widehat{G}}(\mathscr{F}_uA)(g) =\sum_{u\in\widehat{G}}d(u$)Tr(A$_uu(g)^{-1}$)

is a continuous function on G whose Fourier cotransform is A.

Let $f$ be an integrable function on G, and let $s\in G$. Denote by $\gamma (s)f$ and $\delta (s)f$ the functions on G defined by $\gamma (s)f=\varepsilon_s*f,\delta (s)f=f*\varepsilon_{s^{-1}}$, that is,

$(\gamma (s)f)(g) =f(s^{-1}g),(\delta (s)f)(g) =f(gs)$ for $g\in G$,

(Chap. III, §3, no. 4 and Integration, Chap. VII, §1, no. 1). We have

so

and similarly

When G is commutative, $\widehat{G}$ is the underlying set of the dual group of G (Spectral Theories, Chap. II, §1, no. $1$)$,d(u) = 1$ for all $u\in \widehat{G}$, and we recover the definitions of the Fourier transform given in Spectral Theories, Chap. II.

a) The following conditions are equivalent:

(i) There exists an integer $n >0$ such that $\varphi (u)\preccurlyeq (\|\lambda (u)\|+ 1)^n$ (resp. for every integer $n >0$, we have $\varphi (u)\preccurlyeq (\|\lambda (u)\|+ 1)^{-n}$).

(ii) There exists an integer $n >0$ such that $\varphi (u)\preccurlyeq ( \widetilde{\Gamma}(u) + 1)^n$ (resp. for every integer $n >0$, we have $\varphi (u)\preccurlyeq ( \widetilde{\Gamma}(u) + 1)^{-n}$).

b) If G is semi-simple, conditions (i) and (ii) are also equivalent to:

(iii) There exists an integer $n >0$ such that $\varphi (u)\preccurlyeq d(u)^n$ (resp. for every integer $n >0$, we have $\varphi (u)\preccurlyeq d(u)^{-n}$).

Note first of all that condition (i) is clearly independent of the choice of norm. Thus we can use the norm defined by the quadratic form $Q_{\Gamma}$ associated to $\Gamma ($§7, no. 6, Prop. 4). Then

so $\widetilde{\Gamma}(u) + 1\preccurlyeq (\|\lambda (u)\|+ 1)^2\preccurlyeq \widetilde{\Gamma}(u) + 1$, hence a).

Further, if G is semi-simple,

$\|\lambda (u) +\rho \|\preccurlyeq d(u)\preccurlyeq \|\lambda (u) +\rho \|^N$, where $N = 1/$2(dim $G-$ dim T)

(§7, no. 5, Cor. 1 of Th. 3), so $\|\lambda (u)\|+ 1\preccurlyeq d(u)\preccurlyeq (\|\lambda (u)\|+ 1)^N$, hence b).

It follows from Prop. 2 that condition (i) is independent of the choice of maximal torus, chamber, and norm, and that condition (ii) is independent of the choice of Casimir element. A function $\varphi$ satisfying conditions (i) and (ii) is said to be moderately increasing (resp. rapidly decreasing). The product of two moderately increasing functions is moderately increasing; the product of a moderately increasing function and a rapidly decreasing function is rapidly decreasing. If $\varphi$ is rapidly decreasing, the family $(\varphi (u))_{u\in\widehat{G}}$ is summable.

Since the product of two moderately increasing functions is moderately increasing, it suffices to prove this when $t\in \mathfrak{g}:$ in that case the assertion follows from the Remark in §7, no. 6 and the inequality

$f(h) =\sum_{u\in\widehat{G}}\langle u(h)|u(f)\rangle =\sum_{u\in\widehat{G}}d(u)\int_Gf(g$)Tr($u(gh^{-1})$)$dg$.

b) Let $f$ be an integrable function on G; then $f$ is equal almost everywhere to an infinitely-differentiable function if and only if the function $u \rightarrow \|u(f)\|_{\infty}$ is rapidly decreasing on $\widehat{G}$.

Let $f$ be an infinitely-differentiable function on G, and let $\Gamma$ be a Casimir element for G; by formula (14),

for all $n\geq 0$, and consequently

$\widetilde{\Gamma}(u)^n\|u(f)\|_{\infty}\leq \|(L_{\Gamma})^nf\|_1\leq$ sup$_{g\in G}|((L_{\Gamma})^nf)(g)|$; (15)

thus, the function $u \rightarrow \|u(f)\|_{\infty}$ is indeed rapidly decreasing.

Conversely, let $A = (A_u)_{u\in\widehat{G}}$ be an element of $F( \widehat{G})$ such that the function $u \rightarrow \|A_u\|_{\infty}$ is rapidly decreasing. Put $f_u(g) =\langle u(g)|A_u\rangle$; the function $g \rightarrow f_u(g)$ is analytic, hence infinitely-differentiable. By Chap. III, §3, no. 7, Prop. 27,

for all $x\in \mathfrak{g}$. Let $t\in U(G)$; by the preceding formula,

and consequently

Since $d(u)$ and $\|u(t)\|_{\infty}$ are moderately increasing (Prop. 3) and $\|A_u\|_{\infty}$ is rapidly decreasing, the function $u \rightarrow$ sup$_g|(L_tf_u)(g)|$ is rapidly decreasing; thus, the family $(L_tf_u)_{u\in\widehat{G}}$ is uniformly summable. It follows$^7$ that the sum of the family $(f_u)$ is an infinitely-differentiable function on G, whose Fourier cotransform is $(A_u)$, hence the theorem.

Denote by $\mathscr{S}( \widehat{G})$ the vector subspace of $L^2( \widehat{G})$ consisting of the families $A = (A_u)_{u\in\widehat{G}}$ such that the function $u \rightarrow \|A_u\|_{\infty}$ is rapidly decreasing on $\widehat{G}$. It follows from the theorem that the maps $\mathscr{F}:f \rightarrow (u(f))_{u\in\widehat{G}}$ and $\mathscr{F}: A \rightarrow \sum_{u\in\widehat{G}}\langle u(g)|A_u\rangle$ induce inverse isomorphisms between the complex vector spaces $\mathscr{C}^{\infty}(G;\mathbf{C})$ and $\mathscr{S}( \widehat{G})$. Give the space $\mathscr{C}^{\infty}(G;\mathbf{C})$ the topology of uniform $C^{\infty}$-convergence (§6, no. 4) which can be defined by the family of semi-norms $f \rightarrow$ sup$_{g\in G}|L_tf(g)|$ for $t\in U(G)$, and the space $\mathscr{S}( \widehat{G})$ the topology defined by the sequence of semi-norms $p_n: A \rightarrow$ sup$_{u\in\widehat{G}}( \widetilde{\Gamma}(u) + 1)^n\|A_u\|_{\infty}$. Formula (15) of the preceding proof shows that $\overline{\mathscr{F}}$ is continuous. Let $t\in U(G)$, and let $A = (A_u)_{u\in\widehat{G}}$ be an element of $\mathscr{S}( \widehat{G})$; put $f_n(g) =\langle u(g)|A_u\rangle$. Let $p$ be an integer such that $\sum_{u\in\widehat{G}}\widetilde{\Gamma}(u)^{-p}= M<\infty$. By the preceding proof, there exists a positive integer $m$ such that, for all $g\in G$,

so $|(L_t\mathscr{F}(A))(g)| \leq mMp_m(A)$; this proves that $\mathscr{F}$ is continuous. Consequently:

$^7$ This follows from the fact that the space $\mathscr{C}^{\infty}(G;\mathbf{C})$, with the topology of uniform

$C^{\infty}$-convergence (§6, no. 4), is complete.

By Prop. 1 (no. 1), to say that $f$ is central means that $u(\gamma (s)f) =$ $u(\delta (s^{-1})f)$ for all $s\in G$ and all $u\in \widehat{G}$; but this can also be written as $u(s)u(f) =u(f)u(s)$ for all $s\in G$ and all $u\in \widehat{G}$ (formulas (12) and (13)), hence the first assertion of Prop. 4 (Schur’s lemma). If $u(f)$ is a homothety, then $u(f) =\lambda_u\varepsilon_u$ with

$\lambda_u=\frac{1}{d(u)}$ Tr($u(f)$) $=\frac{1}{d(u)}\int_Gf(g$)Tr($u(g)$)$dg=\frac{1}{d(u)}\int_Gf(g)\chi_u(g)dg$.

Consequently, for $f\in$ ZL$^2(G)$ we have

so

with

$uu$

Conversely, if $\varphi$ is a square-integrable complex function on $\widehat{G}$, the element $(\frac{\varphi(u)}{d(u)}\varepsilon_u)_{u\in\widehat{G}}$ of $F( \widehat{G})$ belongs to $L^2( \widehat{G})$, and we have (formula (9)) $(\mathscr{F}_u(\frac{\varphi(u)}{d(u)}\varepsilon_u))(g) =d(u$)Tr $(\frac{\varphi(u)}{d(u)}\varepsilon_uu(g)^{-1})=\varphi (u)\overline{\chi}_u(g)$,

so

$u\in \widehat{G}$

Note, in particular, that formulas (20) and (21) give, for $u, v$ in $\widehat{G},^8$

$u(\overline{\chi}_v) = 0$ if $u\not=v$, (23)

$u(\overline{\chi}_u) =\frac{\varepsilon_u}{d(u)}\in$ End(E$_u$), (24)

$\overline{\mathscr{F}}(\chi_u) =\frac{\varepsilon_u}{d(u)}\in$ End(E$_u$)$\subset F( \widehat{G})$. (25)

By Th. $1b)$, the function $\overline{f}$ is infinitely-differentiable if and only if the function $u \rightarrow \|u(\overline{f})\|_{\infty}$ is rapidly decreasing; but, by (20),

hence the first assertion, since the functions $d(u)$ and $\frac{1}{d(u)}$ are moderately increasing.

Assume that $f$ is infinitely-differentiable; by Th. $1a),f(g) =\sum_{u\in\widehat{G}}f_u(g)$ for all $g\in G$, so

$f_u(g) =\langle u(g)|u(f)\rangle =d(u$)Tr($u(g)^{-1}.u(f)$) $=d(u$)Tr$(u(g)^{-1}\langle \overline{\chi}_u|f\rangle \frac{\varepsilon_u}{d(u)})$

$=\langle \overline{\chi}_u|f\rangle$Tr($u(g)^{-1}$) $=\langle \overline{\chi}_u|f\rangle \overline{\chi}_u(g)$.

Hence, $f(g) =\sum_{u\in\widehat{G}}\langle \overline{\chi}_u|f\rangle \overline{\chi}_u(g)$; but, for all $u\in \widehat{G}$, the contragredient representation $u'$ of $u$ satisfies $\overline{\chi}_u=\chi_{u'}$ and the map $u \rightarrow u'$ is a permutation of $\widehat{G}$; so we also have $f(g) =\sum_{u\in\widehat{G}}\langle \chi_u|f\rangle \chi_u(g)$, hence the proposition.

$^8$ We embed End(E$_u$) in the product $F( \widehat{G}) =\prod_{v\in\widehat{G}}$ End(E$_v$) by associating to any

$A\in$ End(E$_u$) the family $(A_v)_{v\in\widehat{G}}$ such that $A_u= A$ and $A_v= 0$ for $v\not=u$.

Indeed, by Cor. 4 of §7, no. 4,

$\langle \chi_u|f\rangle =\int_G\overline{\lambda(u)(t)}\varphi (t)dt$, where $\varphi (t) =\prod_{\alpha >0}(1-\alpha (t)^{-1})f(t)$.

If $f|T$ is infinitely-differentiable, so is $\varphi$; by Prop. 5, applied to the group T, the function $\mu \rightarrow \int_T\overline{\mu(t)}\varphi (t)dt$ on $\widehat{T} = X(T)$ is then rapidly decreasing, and so is the function $u \rightarrow \langle \chi_u|f\rangle$; hence the function $f$ is infinitely-differentiable (Prop. 5). The converse is clear.