Lower bounds for moments of L-functions: symplectic and orthogonal examples

0.16 MB·English

by  Z. Rudnick

LOWER BOUNDS FOR MOMENTS OF L-FUNCTIONS:
SYMPLECTIC AND ORTHOGONAL EXAMPLES
6
0
0
2
Z. Rudnick and K. Soundararajan

n
a
J
1. Introduction
0
2
An important problem in number theory asks for asmptotic formulas for the moments

]
of central values of L-functions varying in a family. This problem has been intensively
T
N studied in recent years, and thanks to the pioneering work of Keating and Snaith [7], and
. the subsequent contributions of Conrey, Farmer, Keating, Rubinstein and Snaith [1], and
h
t Diaconu, Goldfeld and Hoffstein [3] there are now well-established conjectures for these
a
m moments. The conjectured asymptotic formulas take different shapes depending on the
symmetry group attached to the family of L-functions, given in the work of Katz and
[

Sarnak [6], with three classes of formulas depending on whether the group in question is
1
v unitary, orthogonal or symplectic. While there are many known examples of asymptotic
8
formulas dealing with the first few moments of a family of L-functions, in general the
9
4 moment conjectures seem formidable. In [8] we recently gave a simple method to obtain
1 lower bounds of the conjectured order of magnitude in many families of L-functions. In [8]
0
6 we illustrated our method by working out lower bounds for ∗χ (mod q)|L(21,χ)|2k where
0 q is a large prime, and the sum is over the primitive DirichlePt L-functions (mod q). This
/
h was an example of a ‘unitary’ family of L-functions, and in this paper we round out the
t
a picture by providing lower bounds for moments of L-functions arising from orthogonal and
m
symplectic families.
:
v As our first example, we consider k the set of Hecke eigencuspforms of weight k for
H
Xi the full modular group SL(2,Z). We will think of the weight k as being large, and note
that contains about k/12 forms. Given f we write its Fourier expansion as
r k k
a H ∈ H
∞ k−1
f(z) = λf(n)n 2 e(nz),
nX=1
where we have normalized the Fourier coefficients so that the Hecke eigenvalues λ (n)
f
satisfy Deligne’s bound λ (n) τ(n) where τ(n) is the number of divisors of n. Consider
f
| | ≤
the associated L-function
∞
L(s,f) = λ (n)n s = (1 λ (p)p s +p 2s) 1,
f − f − − −
−
nX=1 Yp
The first author is partially supported by a grant from the Israel Science Foundation. The second
author is partially supported by the National Science Foundation.
Typeset by AMS-TEX
1
2 Z. RUDNICK AND K. SOUNDARARAJAN
which converges absolutely in Re (s) > 1 and extends analytically to the entire complex
plane. Recall that L(s,f) satisfies the functional equation
Λ(s,f) := (2π) sΓ(s+ k 1)L(s,f) = ikΛ(1 s,f).
− −2 −
If k 2 (mod 4) then the sign of the functional equation is negative and so L(1,f) = 0.
≡ 2
We will therefore assume that k 0 (mod 4).
≡
While dealing with moments of L-functions in , it is convenient to use the natural
k
H
‘harmonic weights’ that arise from the Petersson norm of f. Define the weight
(4π)k 1 k 1
ω := − f,f = − L(1,Sym2f),
f Γ(k 1)h i 2π2
−
where f,f denotes the Petersson inner product. For a typical f in the harmonic
k
h i H
weight ω is of size about k/12, and so ω 1 is very nearly 1. The weights ω arise
f f k f− f
naturally in connection with the PeterPsson∈Hformula, and the facts mentioned above are
standard and may be found in Iwaniec [4].
For a positive integer r, we are interested in the r-th moment
h 1
L(1,f)r := L(1,f)r.
2 ω 2
X X f
f k f k
∈H ∈H
This family of L-functions is expected to be of ‘orthogonal type’ and the Keating-Snaith
conjectures predict that for any given r N as k with k 0 (mod 4) we have
∈ → ∞ ≡
h
L(1,f)r C(r)(logk)r(r 1)/2,
−
2 ∼
X
f k
∈H
for some positive constant C(r). This conjecture can be verified for r = 1 and r = 2, and
if we permit an additional averaging over the weight k then for r = 3 and 4 also.
Theorem 1. For any givenevennatural number r, and weight k 12 with k 0 (mod 4),
≥ ≡
we have
h
L(1,f)r (logk)r(r 1)/2.
2 ≫r −
X
f k
∈H
In fact, with more effort our method could be adapted to give lower bounds as in
Theorem 1 for all rational numbers r 1, rather than just even integers.
≥
Our other example involves the family of quadratic Dirichlet L-functions. Let d denote
a fundamental discriminant, and let χ denote the corresponding real primitive charac-
d
ter with conductor d . We are interested in the class of quadratic Dirichlet L-functions
| |
L(s,χ ). Recall that, with a = 0 or 1 depending on whether d is positive or negative, these
d
L-functions satisfy the functional equation
q s+a s+a
2
Λ(s,χ ) := Γ L(s,χ ) = Λ(1 s,χ ).
d d d
π 2 −
(cid:16) (cid:17) (cid:16) (cid:17)
LOWER BOUNDS FOR MOMENTS 3
Notice that the sign of the functional equation is always positive, and it is expected that
the central values L(1,χ ) are all positive although this remains unknown. This family
2 d
is expected to be of ‘symplectic’ type and the Keating-Snaith conjectures predict that for
any given k N and as X we have
∈ → ∞
♭
L(1,χ )k D(k)X(logX)k(k+1)/2,
2 d ∼
X
d X
| |≤
for some positive constant D(k), where the ♭ indicates that the sum is over fundamental
discriminants. Jutila [5] established asymptotics for the first two moments of this family,
and the third moment was evaluated in Soundararajan [10].
Theorem 2. For every even natural number k we have
♭
L(1,χ )k X(logX)k(k+1)/2.
2 d ≫k
X
d X
| |≤
As with Theorem 1, our method can be used to obtain lower bounds for these moments
for all rational numbers k, taking care to replace L(1,χ )k by L(1,χ ) k when k is not
2 d | 2 d |
an even integer. In the case of the fourth moment we are able to get a lower bound
(D(4)+o(1))X(logX)10, which matches exactly the asymptotic conjectured by Keating
≥
and Snaith. The details of this calculation will appear elsewhere.
2. Proof of Theorem 1
1
Let x := k2r and consider
λ (n)
f
A(f) := A(f,x) = .
√n
X
n x
≤
We will consider
h h
S := L(1,f)A(f)r 1, and S := A(f)r.
1 2 − 2
X X
f k f k
∈H ∈H
Then H¨older’s inequality gives, keeping in mind that r is even so that A(f) r = A(f)r,
| |
h r h h r 1
L(1,f)A(f)r 1 L(1,f)r A(f)r − ,
−
2 ≤ 2
(cid:16)X (cid:17) (cid:16)X (cid:17)(cid:16)X (cid:17)
f k f k f k
∈H ∈H ∈H
so that
h Sr
L(1,f)r 1 .
2 ≥ Sr 1
fXk 2−
∈H
We will prove Theorem 1 by finding the asymptotic orders of magnitude of S and S .
1 2
4 Z. RUDNICK AND K. SOUNDARARAJAN
We begin with S . To evaluate this, we must expand out A(f)r and group terms using
2
the Hecke relations. To do this conveniently, let us denote by the ring generated over
H
the integers by symbols x(n) (n N) subject to the Hecke relations
∈
mn
x(1) = 1, and x(m)x(n) = x .
d2
X (cid:16) (cid:17)
d(m,n)
|
Thus isapolynomialringonx(p)wherepruns overallprimes. UsingtheHeckerelations
H
we may write
x(n ) x(n ) = b (n ,... ,n )x(t),
1 r t 1 r
···
t| Xrj=1nj
Q
for certain integers b (n ,... ,n ). Note that b (n ,... ,n ) is symmetric in the variables
t 1 r t 1 r
n , ...,n ,andthatb (n ,... ,n )isalwaysnon-negative, andfinallythatb (n ,... ,n )
1 r t 1 r t 1 r
≤
τ(n ) τ(n ) (n n )ǫ. Of special importance for us will be the coefficient of x(1)
1 r 1 r
··· ≪ ···
namely b (n ,... ,n ). It is easy to see that b satisfies a multiplicative property: if
1 1 r 1
r r
( m , n ) = 1 then
j=1 j j=1 j
Q Q
b (m n ,... ,m n ) = b (m ,... ,m )b (n ,... ,n ).
1 1 1 r r 1 1 r 1 1 r
Thus it suffices to understand b when the n , ..., n are all powers of some prime
1 1 r
p. Here we note that b (pa1,... ,par) is independent of p, always lies between 0 and
1
(1+a ) (1+a ), and that it equals 0 if a +...+a is odd. If we write
1 r 1 r
···
B (n) = b (n ,... ,n ),
r 1 1 r
n1X,...,nr
n1 nr=n
···
then we find that B (n) is a multiplicative function, that B (n) = 0 unless n is a square,
r r
and that B (pa) is independent of p and grows at most polynomially in a. Finally, and
r
crucially, we note that
B (p2) = r(r 1)/2,
r
−
which follows upon noting that b (p2,1,... ,1) = 0 and that b (p,p,1,... ,1) = 1.
1 1
Returning to S note that
2
1
A(f)r = b (n ,... ,n )λ (t),
t 1 r f
√n n
n1,..X.,nr≤x 1··· r t|nX1···nr
h
andsowerequireknowledgeof λ (t). ThisfollowseasilyfromPetersson’sformula.
f k f
∈H
P
Lemma 2.1. If k is large, and t and u are natural numbers with tu k2/104 then
≤
h
λ (t)λ (u) = δ(t,u)+O(e k),
f f −
X
f k
∈H
LOWER BOUNDS FOR MOMENTS 5
where δ(t,u) is 1 if t = u and is 0 otherwise.
Proof. Petersson’s formula (see [4]) gives
h ∞ S(t,u;c) 4π√tu
λ (t)λ (u) = δ(t,u)+2πik J .
f f k 1
fXk Xc=1 c − (cid:16) c (cid:17)
∈H
Note that if z 2k then (z/2)k 1+ℓ/Γ(k 1+ℓ) (z/2)k 1/Γ(k 1) for all ℓ 0. We
− −
≤ − ≤ − ≥
now use the series representation for J (z) which gives, for z 2k,
k 1
− ≤
J (z) = ∞ (−1)ℓ(z/2)ℓ (z/2)ℓ+k−1 (z/2)k−1ez/2.
k 1
| − | (cid:12)(cid:12)Xℓ=0 ℓ! Γ(ℓ+k −1)(cid:12)(cid:12) ≤ Γ(k −1)
(cid:12) (cid:12)
Therefore, for tu k2/104, we deduce that
≤
4π√tu (2πk/(100c))k 1
J − eπk/50.
k 1
(cid:12) − (cid:16) c (cid:17)(cid:12) ≤ (k 2)!
(cid:12) (cid:12) −
(cid:12) (cid:12)
Using the trivial bound S(t,u;c) c we conclude that
| | ≤
∞ S(t,u;c) 4π√tu πk k 1 1
2πik J − eπk/50 e k,
k 1 −
Xc=1 c − (cid:16) c (cid:17) ≪ (cid:16)50(cid:17) (k −2)! ≪
for large k, as desired.
Since n n xr = √k we see by Lemma 2.1 that
1 r
··· ≤
b (n ,... ,n ) τ(n ) τ(n )τ(n n )
S = 1 1 r +O e k 1 ··· r 1··· r .
2 −
√n n √n n
n1,..X.,nr x 1··· r (cid:16) n1,.X..nr x 1··· r (cid:17)
≤ ≤
The error term is easily seen to be e−kxk = k12e−k, a negligible amount. As for the
≪
main term we see easily that
B (n) b (n ,... ,n ) B (n)
r 1 1 r r
.
√n ≤ √n n ≤ √n
nXx n1,..X.,nr x 1··· r nXxr
≤ ≤ ≤
Recall that B (n) is a multiplicative function with B (p) = 0, B (p2) = r(r 1)/2 and
r r r
−
Br(pa) grows only polynomially in a. Thus the generating function ∞n=1Br(n)n−s can
be compared with ζ(2s)r(r 1)/2, the quotient being a Dirichlet series aPbsolutely convergent
−
in Re(s) > 1. A standard argument (see Theorem 2 of [9]) therefore gives that
4
B (n)
r C (logz)r(r 1)/2,
r −
√n ∼
X
n z
≤
for a positive constant C . It follows that
r
S (logx)r(r 1)/2 (logk)r(r 1)/2.
2 − −
≍ ≍
We now turn to S . To evaluate S we need an ‘approximate functional equation’ for
1 1
L(1,f).
2
6 Z. RUDNICK AND K. SOUNDARARAJAN
Lemma 2.2. Define for any positive number ξ the weight
W (ξ) := 1 (2π)−12−sΓ(s+ k2)ξ sds,
k 2πi Z(c) (2π)−12Γ(k2) − s
where the integral is over a vertical line c i to c + i with c > 0. Then, for k 0
− ∞ ∞ ≡
(mod 4),
L(1,f) = 2 ∞ λf(n)W (n) = 2 λf(n)W (n)+O(e k).
2 √n k √n k −
nX=1 nXk
≤
Further the weight W (ξ) satisfies W (ξ) kπ k/ξ for ξ > k, W (ξ) = 1+O(e k) for
k k − k −
| | ≪
ξ < k/100, and W (ξ) 1 for k/100 ξ k.
k
≪ ≤ ≤
Proof. The argument is standard. For 1 c > 1 we consider
≤ 2
I = 1 (2π)−21−sΓ(s+ k2)L(1 +s,f)ds.
2πi Z(c) (2π)−21Γ(k2) 2 s
Expanding out L(1 +s,f) and integrating term by term we see that
2
∞ 1
I = λf(n)n−2Wk(n).
nX=1
On the other hand moving the line of integration to the line Re(s) = c we see that
−
1 Λ(1 +s,f) ds
I = L(1,f)+ 2 ,
2 2πi Z(−c) (2π)−12Γ(k2) s
and using the functional equation Λ(1 +s,f) = Λ(1 s,f), and replacing s by s in the
2 2 − −
integral above, we see that I = L(1,f) I. Thus
2 −
L(1,f) = 2I = 2 ∞ λf(n)W (n).
2 √n k
nX=1
Regarding the weight W (ξ)note that by considering the integral for some large positive
k
integer c we get that
1 Γ(s+ k +1) ds
W (ξ) (2πξ) c| 2 | | |
| k | ≤ 2π Z − Γ(k) s(s+k/2)
(c) 2 | |
Γ(c+1+ k)
(2πξ) c 2 (2πξ) c(k +c)c.
− −
≤ Γ(k) ≤
2
LOWER BOUNDS FOR MOMENTS 7
Taking c = k we obtain that W (ξ) (k/(πξ))k so that if ξ k then W (ξ)
k k
| | ≤ ≥ | | ≤
(k/ξ)π (k 1). This proves the first bound for W (ξ) claimed in the Lemma, and also
− − k
shows that
λ (n) λ (n)
f W (n) kπ k | f | e k.
(cid:12)X √n k (cid:12) ≪ − X n23 ≪ −
(cid:12)n>k (cid:12) n>k
(cid:12) (cid:12)
The other claims on W (ξ) are proved similarly; for the range ξ < k/100 we move the line
k
of integration to c = k +1, for the last range k/100 ξ k just take the integral to be
−2 ≤ ≤
on the line c = 1.
Returning to S note that
1
1
A(f)r 1 = b (n ,... ,n )λ (t).
− t 1 r 1 f
n1,...X,nr−1≤x √n1···nr−1 t|n1X···nr−1 −
Since A(f)r 1 is trivially seen to be xr 1 < √k, we see by Lemma 2.2, that
− −
≪
1 b (n ,... ,n ) h
S1 = 2 Wk(n) t 1 r−1 λf(t)λf(n)+O(√ke−k).
√n √n n
nX≤k n1,...X,nr−1≤xt|n1X···nr−1 1··· r−1 fX∈Hk
Now we appeal to Lemma 2.1. The error term that arises is trivially bounded by ke k
−
≪
which is negligible. In the main term δ(n,t), since t xr 1 < √k we may replace W (n)
− k
≤
by 1+O(e k). It follows that
−
b (n ,... ,n ) 1
S1 = 2 t 1 r−1 +O(ke−k).
√n n √t
n1,...X,nr−1≤xt|n1X···nr−1 1··· r−1
Now observe that b (n ,... ,n ,t) = b (n ,... ,n ) if t divides n n , and oth-
1 1 r 1 t 1 r 1 1 r 1
− − ··· −
erwise b (n ,... ,n ,t) is zero. Therefore, writing n for t, we obtain that
1 1 r 1 r
−
b (n ,... ,n )
S = 2 1 1 r +O(ke k).
1 −
√n n
n1,...X,nr−1≤xnrX≤√k 1··· r
Using b 0 we see that S 2S +O(ke k), and moreover, arguing as in the case of S
1 1 2 − 2
≥ ≥
we may see that
S (logk)r(r 1)/2.
1 −
≍
Theorem 1 follows.
3. Proof of Theorem 2
For simplicity, we will restrict ourselves to fundamental discriminants of the form 8d where
d is a positive, odd square-free number with X/16 < d X/8. Let k be a given even
≤
1
number, and set x = X10k. Define
χ (n)
8d
A(8d) := ,
√n
X
n x
≤
8 Z. RUDNICK AND K. SOUNDARARAJAN
and let
S := µ2(2d)L(1,χ )A(8d)k 1, and S := µ2(2d)A(8d)k.
1 2 8d − 2
X X
X/16<d X/8 X/16<d X/8
≤ ≤
An application of H¨older’s inequality gives that
♭ Sk
L(1,χ )k µ2(2d)L(1,χ )k 1 ,
2 8d ≥ 2 8d ≥ Sk 1
dXX X/16X<d X/8 2−
| |≤ ≤
so that to prove Theorem 2 we need only give satisfactory estimates for S and S .
1 2
We start with S . Expanding our A(8d)k we see that
2
1 8d
(3.1) S = µ2(2d) .
2
√n n (cid:18)n n (cid:19)
n1,..X.,nk x 1··· k X/16X<d X/8 1··· k
≤ ≤
Lemma 3.1. Let n be an odd integer, and let z 3 be a real number. If n is not a perfect
≥
square then
8d
µ2(2d) z12n41 log(2n),
(cid:18)n(cid:19) ≪
X
d z
≤
while if n is a perfect square then
8d z p
µ2(2d) = +O(z12+ǫnǫ).
(cid:18)n(cid:19) ζ(2) (cid:18)p+1(cid:19)
X Y
d z p2n
≤ |
Proof. Note that µ(α) = 1 if d is square-free and 0 otherwise. Therefore,
α2 d
|
P
8d 8α2 d
µ2(2d) = .
(cid:18)n(cid:19) (cid:18) n (cid:19) (cid:18)n(cid:19)
nXz αX√z dXz/α2
≤ α≤odd d≤odd
If n is not a square then the inner sum over d is a character sum to a non-principal
characterofmodulus2n(wetake2ntoaccount fordbeingodd), andtheP´olya-Vinogradov
inequality (see [2]) gives that the sum over d is √nlog(2n). Further, the sum over d is
≪
trivially z/α2. Thus, if n is not a square, we get that
≪
8d z
µ2(2d) min √nlog(2n), z12n41 log(2n),
(cid:18)n(cid:19) ≪ α2 ≪
X X (cid:16) (cid:17)
n z α √z
≤ ≤
1 1
upon using the first bound for α z2n−4 and the second bound for larger α.
≤
If n is a perfect square, then 8d = 1 if d is coprime to n, and is 0 otherwise. Thus
n
(cid:0) (cid:1)
8d z p
µ(2d)2 = µ2(d) = +O(z21+ǫnǫ),
(cid:18)n(cid:19) ζ(2) (cid:18)p+1(cid:19)
X X Y
d z d z p2n
≤ (d,2≤n)=1 |
LOWER BOUNDS FOR MOMENTS 9
by a standard argument.
Using Lemma 3.1 in (3.1) we obtain that
X 1 p
S2 = +O X12+ǫxk(43+ǫ) .
16ζ(2) √n n (cid:18)p+1(cid:19)
n1 nn1k,=..X.o,ndkd≤sxquare 1··· k p|2nY1···nk (cid:16) (cid:17)
···
Since x = X101k the error term above is X53. Writing n1 nk = m2 we see that
≪ ···
d (m2) p 1 p
k
m (cid:18)p+1(cid:19) ≤ √n n (cid:18)p+1(cid:19)
mmX2o≤dxd pY|2m n1 nn1k,=..X.o,ndkd≤sxquare 1··· k p|2nY1···nk
···
d (m2) p
k
.
≤ m (cid:18)p+1(cid:19)
mX2 xk pY2m
m≤odd |
A standard argument (see Theorem 2 of [9]) shows that
d (m2) p
k C(k)(logz)k(k+1)/2,
m (cid:18)p+1(cid:19) ∼
X Y
m z p2m
m o≤dd |
for a positive constant C(k). We conclude that
(3.2) S X(logX)k(k+1)/2.
2
≍
It remains to evaluate S . As before, we need an ‘approximate functional equation’ for
1
L(1,χ ).
2 8d
Lemma 3.2. For a positive number ξ define the weight
1 Γ(s + 1) ds
W(ξ) = 2 4 ξ s ,
−
2πi Z Γ(1) s
(c) 4
where the integral is over a vertical line c i to c+i with c > 0. Then, for any odd,
− ∞ ∞
positive, square-free number d we have
L(1,χ ) = 2 ∞ χ8d(n)W n√π .
2 8d nX=1 √n (cid:16)√8d(cid:17)
The weight W(ξ) is smooth and satisfies W(ξ) = 1+O(ξ21−ǫ) for ξ small, and for large ξ
satisfies W(ξ) e−ξ. Moreover the derivative W′(ξ) satisfies W′(ξ) ξ21−ǫe−ξ.
≪ ≪
Proof. This is given in Lemmas 2.1 and 2.2 of [10], but for completeness we give a sketch.
For some 1 c > 1, we consider
≥ 2
1 (8d/π)2s+41Γ(s + 1) ds
2 4 L(1 +s,χ ) ,
2πi Zc) (8d/π)41Γ(1) 2 8d s
4
10 Z. RUDNICK AND K. SOUNDARARAJAN
and argue exactly as in Lemma 2.2. This gives the desired formula for L(1,χ ). The
2 8d
results on the weight W(ξ) follow upon moving the line of integration to Re(s) = 1 +ǫ
−2
when ξ is small, and taking c to be an appropriately large positive number if ξ is large.
By Lemma 3.2 we see that
(3.3)
∞ 1 1 8d n√π
S = 2 µ2(2d) W .
1
nX=1 √n n1,...X,nk−1 x √n1···nk−1 X/16X<d X/8 (cid:18)nn1···nk−1(cid:19) (cid:18)√8d(cid:19)
≤ ≤
If nn n is not a square, then using Lemma 3.1 and partial summation we may see
1 k 1
··· −
that
8d n√π
µ2(2d) W X12(nn1 nk 1)41+ǫe−n/√X.
X/16X<d X/8 (cid:18)nn1···nk−1(cid:19) (cid:18)√8d(cid:19) ≪ ··· −
≤
If nn n is an odd square then Lemma 3.1 and partial summation gives that the
1 k 1
··· −
sum over d in (3.3) is
X p 2 n√2π
W dt+O(X12+ǫe−n/√X).
16ζ(2) (cid:18)p+1(cid:19)Z (cid:18)√Xt(cid:19)
Y 1
p2nn1 nk−1
| ···
We use these two observations in (3.3). Note that the error terms contribute to (3.3)
an amount X21+ǫx(k−1)(34+ǫ)X83+ǫ X4309. It remains to estimate the main term
≪ ≪
contribution to (3.3). To analyze these terms let us write n n as rs2 where r and
1 k 1
s are odd and r is square-free. Then n must be of the form rℓ·2··whe−re ℓ is odd. With this
notation the main term contribution to (3.3) is
X 1 1 2 p rℓ2√2π
W dt.
8ζ(2) rs2=nX1···nk−1 rs ℓXodd ℓ Z1 p|Y2rsℓ(cid:18)p+1(cid:19) (cid:16) √Xt (cid:17)
n1,...,nk−1 x
≤
Note that r xk−1 < X110, and an easy calculation gives that the sum over ℓ above is
≤
p 1 1 √X
= 1 log +O(1).
(cid:18)p+1(cid:19) − p(p+1) 4 r
Y Y (cid:16) (cid:17)
p2rs p∤2rs
|
It follows that the main term contribution to (3.3) is
1 p
XlogX
≫ rs (cid:18)p+1(cid:19)
rs2=nX1···nk−1 pY|2rs
n1,...,nk−1 x
≤
d (rs2) p
XlogX k−1 X(logX)(logx)k−1+k(k−1)/2,
≫ rs (cid:18)p+1(cid:19) ≫
X Y
r odd and square-free p2rs
s odd |
rs2 x
≤
where the last bound follows by invoking Theorem 2 of [9]. We conclude that
S X(logX)k(k+1)/2,
1
≫
which when combined with (3.2) proves Theorem 2.