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
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