192 PH. MICHEL, ANALYTIC NUMBER THEORY AND FAMILIES OF £ -FUNCTIONS
The bound Hd(l/2) is probably far from the truth: assuming that the Functori-
ality Conjecture for Pairs holds, applying (1.6) to ?r^181 k gives the bounds Hd(l / 2k) for
7r at the unramified places (it holds for the ramified one too), and letting k ____, +oo,
one obtains
Ramanujan/Petersson Conjecture (RPC). Hd(O) is true.
Unconditionally, Proposition 1.1 can be strengthened to the following (non-
trivial) bound:
Theorem 1.1. For any d ~ 1, Hd(e) is true for
1 1
8=8d=----.
2 d^2 +1
We present below a proof due to Serre [Ser3]^1 that works for the non-archi-
medean places and uses the most basic analytic properties of L(7r ® 1T, s ); there
are also two alternative methods developped by Duke/ Iwaniec and Luo/ Rudnick/
Sarnak [Dul, Dul2, LRS, LRS2], which interrestingly both build on families of
twists L(x. 7r ® 1T, s) by appropriate characters. At the end of this lecture, we will
present the method of Luo/ Rudnick/Sarnak in the case of the archimedean place
(but the method works for every place and can be extended to number fields as
well).
Proof. The theorem is a consequence of the following refinment (due to Landau)
of Landau's Lemma [La] (see also [BR]):
Theorem 1.2. Let L (s) = L n .X(n)n- s be a Dirichlet L-series with non-negative
coefficients .X(n), and convergent for ~es sufficiently large. Assume that L(s) has
meromorphic continuation to C with at most poles of finite order at s = 0, 1; assume
also that L ( s ) is of bounded order in the half-plane ~es ~ -1, and that it satisfies a
functional equation of the form
q^8 L 00 (s)L(s) = wql-s L 00 (l - s) L (l - s)
for some constants w, q > 0, where
d
L 00 (s) = rrr(ais +{Ji),
i= l
for some d ~ 1 and ai ~ 0, f3i E C for i E {1 , ... , d}. Setting 'f/ = .Z::::f= 1 ai, one has
as x ____, +oo,
for all c > 0, where Pis some polynomial of degree ords=iL (s) and depends only on
L.
Applying this to L(7r ® 1T, s), one obtains
(1.10)
for some c7r > O; this yields
.A7r07r(N) = 07r,c(N^1 - J+^1 +c)
(^1) which might even go back to Rankin