1. ABSTRAC T HARMONIC ANALYSIS: A SURVEY 15

Li(Gv) with the property that $v(a) — 1 on Hv for almost all v. On the restricted

direct product we define a "product" function by setting

V

it should be noted that the product on the right hand side is really finite. Obviously,

$ is a continuous function on any of the open subgroups Gs and hence also on all

of G.

LEMMA 3. With notations and assumptions as above, we have for any finite

set S that contains at least those v for which $V(HV) ^ 1 or JH d\xv ^ 1,

/ 4(a)d/x(a) = TT / §v{av)dnv.

Proof. Since on the group Gs = Ylv @v x ^ 5 ' w e n a v e

dfi = TT dfjiv x dfis and 3(a) = TT $v(av),

it follows that

ves ves

/ $(a)dfi(a) = / $(a)dii,s

JGs

JGS

= I

(Yl^v(av))(Y[dfivxdiJ,s)

J°s ves ves

= TT / ^v(av)dfiv / dns

**SJGV JGS

= TT / ®v{pv)diiv\

this proves the claim.

The following lemma, which allows us to view the integral over G of a product

function as the product of integrals, follows from the above remarks and the simple

fact that the absolute value of a finite product of complex numbers is the product

of the absolute values of each of the factors.

LEMMA 4. Let the functions {&v}ves, $ = TL ^ be as above; suppose that

the product

v

oo.

y J Gv

Then$(a) G Li(G) and

/ $(a)d/Lt = TT / $v(av)dav.

JG

V JG*

The Inversion Formula, and Self-Dual Product Measures

The last result from abstract harmonic analysis that we need states that under

suitable conditions the Fourier transform of a product function on a restricted direct