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LONDON MATHEMATICAL SOCIETY LECTURE NOTE SERIES Managing Editor: Professor J.W.S. Cassels, Department of Pure Mathematics and Mathematical Statistics, University of Cambridge, 16 Mill Lane, Cambridge CB2 1SB, England The books in the series listed below are available from booksellers, or, in case of difficulty, from Cambridge University Press. 34 36 39
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Representation theory of Lie groups, M.F. ATIYAH et al Homological group theory, C.T.C. WALL (ed) Affine sets and affine groups, D.G. NORTHCOTT Introduction to Hp spaces, P.J. KOOSIS Graphs, codes and designs, P1. CAMERON & J.H. VAN LINT Recursion theory: its generalisations and applications, F.R. DRAKE & S.S. WAINER (eds) p-adic analysis: a short course on recent work, N. KOBLITZ Finite geometries and designs, P. CAMERON, J.W.P. HIRSCHFELD & D.R. HUGHES (eds) Commutator calculus and groups of homotopy classes, H.J. BAUES Synthetic differential geometry, A. KOCK Techniques of geometric topology, R.A. FENN Applicable differential geometry, M. CRAMPIN & F.A.E. PIRANI Economics for mathematicians, J.W.S. CASSELS Several complex variables and complex manifolds II, M.J. FIELD Representation theory, I.M. GELFAND et al Symmetric designs: an algebraic approach, E.S. LANDER Spectral theory of linear differential operators and comparison algebras, H.O. CORDES Isolated singular points on complete intersections, E.J.N. LOOLIENGA A primer on Riemann surfaces, A.F. BEARDON Probability, statistics and analysis, J.F.C. KINGMAN & G.E.H. REUTER (eds) Introduction to the representation theory of compact and locally compact groups, A. ROBERT Skew fields, P.K. DRAXL Surveys in combinatorics, E.K. LLOYD (ed) Homogeneous structures on Riemannian manifolds, F. TRICERRI & L. VANHECKE Topological topics, I.M. JAMES (ed) Surveys in set theory, A.R.D. MATHIAS (ed) FPF ring theory, C. FAITH & S. PAGE An F-space sampler, N.J. KALTON, N.T. PECK & J.W. ROBERTS Polytopes and symmetry, S.A. ROBERTSON Classgroups of group rings, M.J. TAYLOR Representation of rings over skew fields, A.H. SCHOFIELD Aspects of topology, I.M. JAMES & E.H. KRONHEIMER (eds) Representations of general linear groups, G.D. JAMES Low-dimensional topology 1982, R.A. FENN (ed) Diophantine equations over function fields, R.C. MASON Varieties of constructive mathematics, D.S. BRIDGES & F. RICHMAN Localization in Noetherian rings, A.V. JATEGAONKAR Methods of differential geometry in algebraic topology, M. KAROUBI & C. LERUSTE Stopping time techniques for analysts and probabilists, L. EGGHE Groups and geometry, ROGER C. LYNDON Surveys in combinatorics 1985, I. ANDERSON (ed) Elliptic structures on 3-manifolds, C.B. THOMAS A local spectral theory for closed operators, I. ERDELYI & WANG SHENGWANG Syzygies, E.G. EVANS & P. GRIFFITH
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London Mathematical Society Lecture Note Series. 126
Van der Corput's Method of Exponential Sums S. W. Graham Michigan Technological University, USA
and G. Kolesnik California State University, Los Angeles, USA
The right of the University of C
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Published in the United States of America by Cambridge University Press, New York www.cambridge.org Information on this title: www.cambridge.org/9780521339278 © Cambridge University Press 1991
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A catalogue record for this publication is available from the British Library ISBN 978-0-521-33927-8 paperback
TABLE OF CONTENTS
1 Introduction
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2 3
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5
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1.1 Basic Definitions . . . . . . . . . . 1.2 Historical Overview . . . . . . . . 1.3 Two Dimensional Sums . . . . . . . 1.4 The method of Bombieri and Iwaniec . 1.5 Notation . . . . . . . . . . . . .
2 The Simplest Van Der Corput Estimates 2.1 Estimates Using First and Second Derivatives 2.2 Some Simple Inequalities . . . . . . . . . 2.3 The Weyl-van der Corput Inequality . . . . 2.4 Iterating Weyl-Van der Corput . . . . . . 2.5 Upper Bounds for the Riemann Zeta-function 2.6 Notes . . . . . . . . . . . . . . . . .
3 The Method of Exponent Pairs
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3.1 Introduction . . . . . . . . . . 3.2 Lemmas on Exponential Integrals . 3.3 Heuristic Arguments and Definitions 3.4 Proof of the A-Process . . . . . . 3.5 Proof of the B-Process . . . . . . 3.6 Notes . . . . . . . . . . . . .
4 Applications of Exponent Pairs
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22 29 32 35 37
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5 Computing Optimal Exponent Pairs
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6 8 10 13 16 20
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4.1 The Riemann Zeta-function . . . . . . . . . 4.2 Sums Involving 0 . . . . . . . . . . . . . 4.3 The Dirichlet Divisor Problem . . . . . . . . 4.4 The Circle Problem . . . . . . . . . . . . 4.5 Gaps Between Squarefree Numbers . . . . . . 4.6 The Piatetski-Shapiro Prime Number Theorem 4.7 Notes . . . . . . . . . . . . . . . . . .
5.1 Introduction . . . . 5.2 Preliminary Lemmas . 5.3 The Algorithm . . . 5.4 Applications . . . . 5.5 Notes . . . . . . .
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38 39 40 42 44 46 53
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54 55 57 63 69
Table of Contents
6 Two Dimensional Exponential Sums
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6.1 Introduction . . . . . . . . . . . . . 6.2 Generalized Weyl-van der Corput Inequality 6.3 Omega Conditions . . . . . . . . . . . 6.4 The AB Theorem . . . . . . . . . . .
7 New Exponent Pairs
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7.1 Introduction . . . . . . . 7.2 Preliminaries . . . . . . . 7.3 The Airy-Hardy Integral . . 7.4 Gauss Sums . . . . . . . . 7.5 Lemmas on Rational Points . 7.6 Semicubical Powers of Integers 7.7 Proof of the Theorem . . . 7.8 Notes . . . . . . . . . .
Appendix
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Bibliography
Index
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70 75 79 83
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86 87 90 93 96 101 104 110
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111
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117
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120
Acknowledgments
Several people read the preliminary drafts of this book, corrected errors, and made suggestions. It is our pleasure to thank Michael Filaseta, Mary Graham, Roger Heath-Brown, Martin Huxley, Matti Jutila, and Jeff Vaaler for their assistance. We also thank Hugh Montgomery for allowing us to use some unpublished material. We thank our editor, David Tranah, for his assistance and unflagging patience. Finally, the first author would like to thank Julian Gevirtz for introducing him to the world of electronic typesetting.
1. INTRODUCTION
1.1 BASIC DEFINITIONS In this monograph, we will give an account of van der Corput's method in one and two dimensions. The purpose of this method is to obtain bounds for exponential sums, particularly those exponential sums that arise in number-theoretic problems. An exponential sum is a sum of the form
e(f(n)),
(1.1.1)
a 0, (569
31 +e'56+E1
is an exponent pair. We will prove this result in Chapter 7.
Introduction
5
The line of research opened up by Bombieri and Iwaniec has proved promising,
and we expect further progress along the same lines in the future. Iwaniec and Mozzochi (1988) proved that A(x) . For functions
f and g with g taking non-negative real values, f « g means If I < CIgI for some unspecified constant C. If f is also non-negative, f >> g means g « f. We also use
the Landau notation f = 0(g); this is equivalent to f « g. The notation f means that f « g and g A > 0 on I then
E e(f (n)) « A. nEl
Proof. Since
I Ee(f(n))I = I Ee(-f(n))I, nEI
nEI
we may assume that' f' is increasing. Our hypotheses imply that for some integer k,
k + A < f'(n) 6, and < aIIjA+1
other intervals, each of length < 26/A. We apply Theorem 2.1 to the former set of intervals, and the trivial estimate to the latter. We get
E e(f (n)) « (alIJA + 1)(1/6 + 6/A + 1). nEl
We choose 6 = A1/2; this proves the desired result if A < 1/4. If A > 1/4, the result follows from the trivial estimate.
2.2 SOME SIMPLE INEQUALITIES In this section, we will derive a couple of results that we shall be using throughout this book. Both are simple but, nonetheless, useful.
LEMMA 2.3. If X1, ... , Xk are positive numbers and a1, ... , ak are nonnegative numbers such that a1 +... + ak = 1, then
min(X1,...,Xk) < Xi' ...X,E' <max(X1i...,Xk). The proof is obvious. We shall often refer to this as the convexity principle. Our second Lemma generalizes the following well known principle. Suppose we have an estimate of the form
U « AHa + BH-b, where A, B, a, and b are positive constants and H is at our disposal. We obviously
want to take H so as to minimize the right hand side. By choosing H to satisfy AHa = BH-6, we get U «(AbBa)i/(°+b) and this is best possible apart from the value of the implied constant. To generalize this principle, we prove
Simplest van der Corput estimates
9
LEMMA 2.4. Suppose that m
n
BiH-b'
L(H) _ m A'HQ` +
where Ai, B;, ai, and bi are positive. Assume that H1 < H2. Then there is some H with H1 < H < H2 and m
L(H) «
m
n
n
E(A;' Bai )1/(ai+b;) + > A;Hl' + E BiH2 6' . ,=1j=1
ii=1
j=1
The implied constants depend only on m and n. Proof. Define
L+(H) =max(A1Ha1,...,A..Ha-) and
L_(H) =
max(BjH-b...... BnH-b°).
Now L < mL+ + nL_, so it suffices to bound L+ and L_. We observe that L+ is a strictly increasing continuous function, L+(0) = 0, and L+(oo) = oo. Similarly L_ is a strictly decreasing continuous function, L_(0) = oo, and L_(oo) = 0. Therefore there is a unique Ho such that L+(Ho) = L_(Ho). We distinguish three cases: (a) H1 < Ho < H2, (b) Ho < H1 f (c) Ho > H2. If H1 < Ho < H2 then there is some i and some j such that L+(Ho) = A=Ho` = L-(Ho) = BjHa b'. Consequently
L+(Ho) = L-(Ho) = If Ho < H1 then M
L-(H1) < L+(H1) < EA;Hi'* i=1
If Ho > H2 then n
L+(H2) < L_(H2) < EB;HZ b'. j=1
Simplest van der Corput estimates
10
2.3 THE WEYL-VAN DER CORPUT INEQUALITY Theorem 2.2 has the disadvantage of being worse than the trivial estimate if A > 1. This limitation can be overcome if we replace f by a function with smaller derivatives; the next lemma allows us to make such a replacement. LEMMA 2.5. (Weyl-van der Corput) Suppose fi(n) is a complex valued function such that t;(n) = 0 if n 0 I. If H is a positive integer then
(1_ijL)>2e(n)e(n_h).
11: (n)I2 2e(n+k) _ >2>1:(n+k). n
k=1 n
n k=1
The inner sum is empty unless a - H < n < b - 1. By Cauchy's inequality, H Ee(n+k)12
2>257.(n+ k)(n+I). k=1 1=1 n
We collect together the terms with I - k = h to get the result. For applications of Lemma 2.5 in this chapter, we will take
e(n) = f e(f (n)) if n E I; 0
otherwise.
Let S be as in (2.1.1) and define
S1(h)= >2 e(.f(n+h)-.f(n)),
(2.3.1)
nEI(h)
where 1(h)={n:nEI andn+hEI}. We see that ISI2
IS,(h)1.
lhl 0 and for some a > 1, A < If(3'(x)I A2 > 0 on [a, b]. Then r6
e(f(x))dx A} and E2 = {x : If'(x)I < A}. Since f" has constant sign, E1 consists of at most two intervals and E2 consists of at most one interval. By Lemma 3.1, JE, e
(f(x))dx
A2 > 0 and that g"(xo) = 0 for some xo E [a, b]. Finally, assume that there are positive constants A3 and A4 such that Ig131(x)I A2S3, and the expression is monotonic since (g'(x)x2)' _ g"(x)x2 + 2xg'(x) > 0. By Lemma 3.1, e(
2d X 2. Then e
(f (n)) _ >
nEI
HlH2-1-v
so
E min(I f '(a) - vI -1, F-1/2N) 0. For these functions we have f(P+1)(x) =
(-1)P(s)pyx -P.
Here, (s), is defined recursively by the relations
(s)o = 1,(s)k+1 = (s + k)(s)k.
Note that when p> 1, (s)p=s(s+1)...(s+p-1). Definition. Let N, y, s, and a be positive real numbers with e < 1/2, and let P be a non-negative integer. Define F(N, P, s, y, e) to be the set of functions f such
that (1) f is defined and has P continuous derivatives on some interval [a, b], with [a, b] 9 [N, 2N],
Method of Exponent Pairs
31
(2) if 0 0, and every f E F(N, P, s, y, e), the estimate
E e(f ((n)) 1 in an exponent pair, for an exponent pair with It > 0 and 1 > 1 would yield an estimate worse than the trivial estimate provided
Method of Exponent Pairs
32
by the exponent pair (0,1). Similarly, there is no need to take k > 1/2, for then (k, 1) would provide a weaker estimate than the pair (1/2,1/2) of Theorem 2.2.
Finally, we observe that if (k, 1/2) is an exponent pair, then k = 1/2. Assume that (k, 1/2) is an exponent pair. Let H be a positive integer. Let t be defined by the equation t2 = l.c.m.{1, 2, ... , H}, and let N = t2H-2. Define f (x) = 2tx-2. Then
E e(f (n)) < HkNI'2 + H-1.
(3.3.5)
N log N, (ii) 1 < L < log N. We begin with case (i).
Since (k, 1) is an exponent pair, we know that there exists P > 0 and e(0 < e < 1/2) such that if f E F(N, P, s, y, e) then (yN-*)'N' + y-1 Na.
Ee(f (n)) < nEI
We will show that we may take P1 = P + 1 and e1 = e/3.
Assume that H < min(b - a, 2eN/(s + P)). In the notation of (2.3.4), we have I512
«
+ Fr
I
1 0.
Method of Exponent Pairs
37
Assume that y > 0, s > 0, N > 0, and that L = yN-' > 1. We want to find P1 and el such that if f E F(N, PI, s, y, el) then
S = E e(f (n)) < L"Nat nEI
Since (k, 1) is an exponent pair, we know that there exist P > 0 and e(0 < e < 1/2) such that if f E F(N, P, s, y, e) then (3.3.4) holds. We will show that we may take P1 = P and el = e/C, where C = C(s, P) is the constant occurring in Lemma 3.9 Since f satisfies the hypothesis of Lemma 3.6 with F = LN,we may write
e(-¢(.
S= a 0 and L > 1, the first term dominates, and the result is proved.
3.6 NOTES Here, we have presented the method of exponent pairs along very much the same lines as Phillips (1934). One difference is in the use of Lemma-3.4, which is somewhat stronger than the corresponding result of Phillips. Several variants of this result have appeared in the literature; see, for example Lemma 4.6 and pages 90-91 of Titchmarsh (1951), Lemma 10 of Heath-Brown (1983), and pages 86-91 of Vinogradov (1985). The proof given here is due to H.L. Montgomery (unpublished).
4. APPLICATIONS OF EXPONENT PAIRS
4.1 THE RIEMANN ZETA-FUNCTION For our first application of exponent pairs, we shall improve the results we obtained in Chapter 2 on the Riemann zeta-function.
THEOREM 4.1. Suppose that t > 3, and that (k)) is an exponent pair such that k + 21 > 3/2. Let 6(k,1) = (2k + 21 - 1)/4. Then C(1/2 + it) ,G(f(n)) 1 be a parameter to be chosen later. We use Vaaler's trigonometric
polynomial approximation to to t', which is given in Appendix A as Theorem A.6. For our purposes here, it is convenient to interpert Vaaler's theorem as stating that there exist coefficients 7(j) such that
7(j)e(jw)
,P(w) T v. After multiplying both sides of the equation in Lemma 4.12 by e(jn1) and summing over n, we get
E A(n)e(jn7)
1 <j<J N 0, so inf B = inf BA.
In the next corollary, we collect three special cases of Theorem 5.6. We omit the proofs since they are very similar to the discussion in the previous paragraph. COROLLARY 5.7. Let 8 and r be as in Theorem 5.5. If any of the following hypotheses hold, then inf 8 = inf OA.
(a) inf 8BA = inf BBABA and min(rw + v - u, 6w/7 + v - u) > 0. (b) inf 8BA = inf BBABA2 and min(rw + v - u, 5w/6 + v - u) > 0. (c) inf 8BA = inf BBABA3BA and min(rw + v - u, 73w/88 + v - u) > 0. The main part of our algorithm consists of examining C(8) and using Theorem 5.6 or Corollary 5.7 to determine whether inf 8 = inf BA or inf 8 = inf BBA. We then replace C(8) by the appropriate choice of C(8A) or e(8BA), and we repeat the process. After an appropriate number of iterations, we expect to have inf 8 = inf BBj AQ1 BA92 B ... A9 BA, where j = 0 or 1 and ql, q2, ... , q, are positive integers. However, it is possible to have
inf 8 = inf 8A9
for every q > 0. The next theorem gives us an easy way of recognizing when this happens.
Optimal Exponent Pairs
60
THEOREM 5.8. Let 9, u, v, and w be as in Theorem 5.5. The following conditions are equivalent.
(a) inf 9 = inf 9AgB for every q > 0. (b) inf 9 = 0(0, 1).
(c) w+v>uandu 0. Then we can find a sequence {(kq, lq)} of exponent pairs such that (kq, lq) E AqP and inf 9 = lim 9(kq, lq).
qoo
Now we can show by induction that 1
0
0
2Q-2
0
Q
Aq= Q-q-1 1 Q-1
(5.3.1)
where Q = 2q. Therefore, if (kq, lq) = A(k, 1) then
k+l-1
k
(kg'
Q'1
- (2Q - 2)k + Q
It follows that kq < 1/Q and 1 - lq < (q + 1)/2Q; therefore glt00(kq, lq) = (0,1).
Now 9 is uniformly continuous on the convex hull of the triangle with vertices (0, 1/2),
(0,1), and (1/2,1/2), so q
lim 9(kq, lq) = 9(0,1).
Now assume that inf 9 = 9(0,1). For any q > 0, AqP C P, so inf 9 < inf 9Aq. On the other hand, (0,1) = A9 (0,1), so inf 9Aq < 9(0,1) = inf 0.
This completes the proof that (a) and (b) are equivalent. Next, we prove that (b) and (c) are equivalent. By Lemma 5.2, 9(0,1) - 9(k,1) has the same sign as
T=u(1-1-k)-k(w+v-u).
If w + v -u > 0 and u < 0, then T < 0 for all (k, l) E P, so inf 9 = 9(0,1). Conversely, assume that inf 0 = 0(0,1). Then T < 0 for all (k, l) E P. In particular, T < 0 for (k, l) = AgB(0,1). From (5.3.1) we see that AgB(0> 1) _ (4Q1 2,1 - 4Q 12
Optimal Exponent Pairs
where Q = 24. Therefore
61
qu<w+v - u
(5.3.2)
for all non-negative integers q. From this with q = 0, we see that w + v - u > 0. We also see that u < 0, for otherwise (5.3.2) would be false for q sufficiently large. This completes the proof.
The example 0(k, 1) = k shows that it is possible to have inf 0 = 0(0, 1). The example 0(k,1) = 1 shows that it is possible to have inf 0 = 0(2 , ). The following theorem shows that this property holds for no other element of P.2
THEOREM 5.9. Suppose 0 satisfies the conditions of Theorem 5.5, and suppose there is some (k, 1) E P such that inf 0 = 0(k, 1). Then either inf 0 = 0(0, 1) or inf 0 = 0(1/2,1/2). Proof.
Suppose there is some (k, 1) E P such that inf 0 = 0(k, 1). Now (k, 1) must
have the form Bj A9i BA... Aq* B(0,1),
where j = 0 or 1 and the qi's are positive integers. If r = 0 then (k, l) = (1/2,1/2) or (0,1), and we are done. Assume that r > 1, and let 01 = 0B'A4iBA... Av.-1BA9°-' Now inf 0 < inf 01aince
B'A9iBA... A9'-1BA9r-1 P C P. Furthermore, inf 0 = 01 AB(0,1) > inf 01. Therefore, inf 01 = 01(6 , ). s Now ABA3B(0,1) _ (82, s2) and BABA 3B(0,1) _ (41, 4i ). Consequently, 2)<min{01(11,57)
01(1,
63
01( 8,26)}min{01(
627
109
13 4164)'01(4'4)}.
From (5.3.3) through (5.3.5), we see that
12 01(6 , 5 ) =
min{01(
27
109
13
64' 164)'01(4'
4)}.
(5.3.5)
Optimal Exponent Pairs
62
From the proof of Lemma 5.4, we see that this can happen only if 27
0'(
64,
109
12
1
3
164) = 81(6' 3) = 81(4,
4).
This together with (5.3.4) shows that 91(4, 4) = min{91(0,1),01(2, 2)}.
Using Lemma 5.4 again, we get 01(6, 3) = 01(4, 4) = 0(0,1) = 0(2, 2);
therefore inf 91 = 01(0,1). Now 01(0,1) = 9BJA91 B ... Aa.-1 BA9r-1(0,1) = 9Bi Aal B ...
A9r-1 B(0,1).
In other words, inf 9 = 8(k1,11), where (k1,11) = BiAa1 B... AQ'-1 B(0,1).
Repeating this argument, we get inf 0 = inf 0(kr, lr), where (kr, lr) = Bi(0,1) is either (0, 1) or (1/2,1/2). This completes the proof. We can now give a step-by-step description of an algorithm for determining optimal exponent pairs. 1. Check 0 for condition (5.2.2); that is, dk + el + f > 0.
2. Compute (O). 3. Use Theorem 5.8 to see if inf 0 = 0(0, 1). If so, then stop. 4. Use Theorem 5.8 on 9B to see if inf 0 = 0(2 , ). If so, then stop. 2 5. Use Theorem 5.5 to see if inf 0 = OA or inf 0 = OBA. If this fails, try Theorem 5.6. If this fails, give up.
6. If inf 0 = inf 9A, replace (0) by ie(9A). If inf 0 = inf 9BA, replace i;(9) by e(OBA). In either case, return to step 5. Theorem 5.9 shows that is unnecessary to repeat steps 2 and 3 after the first loop. It is also unnecessary to repeat step 1 after the first loop. To see why, suppose that 0 satisfies (5.2.2) and that inf 0 = inf 9A, for example. Let b1 cl 9A - at dl el fl J
.
Then
dlk+ell+f =dic+eA+ f, where (k, 1) = A(lc, A). The right hand side is positive since (5.2.2) is satisfied for 0.
Optimal Exponent Pairs
63
5.4 APPLICATIONS The results of the algorithm of the preceding section may be regarded as a sequence of exponent pairs that yield approximations to the desired infimum. The terms in this sequence are A9iB(0,1), AQ'BA92B(0,1), et cetera. We shall call the sequence of exponents (gi,g2,q3.... ) the q-sequence.
Furthermore, we shall use score-by-innings notation and write the q-sequence as gi q2 q3 ...
In this section, we will give the results that. algorithm yields when applied to the problems discussed in Chapter 4. The computations reported here were done with MACSYMA.
Rankin's Constant. The quantity inf k +1
plays an important role in our algorithm. We call this quantity Rankin's Constant, and we denote it by the letter R. The first 50 terms in the q-sequence are 13211
21122
21122
12221
11213
21112
11132
11132
11221
11122.
From these terms, we find that
R = 0.8290213568591335924092397772831120509883432703 ± 8 x 10-43
In some of the other applications discussed in Chapter 4, the optimal results can be expressed in terms of R. For example, let
8k,l= 2k+2 and let a = inf 8. We see from Theorems 4.6 and 4.8 that
0(x) 0. Let 81 (k, 1) = k + 1. As we mentioned in the previous section, inf 81 = inf 81 A. Moreover,
81A(k,1) = 22k+21 = 8(k,l)+ 2.
Optimal Exponent Pairs
64
Therefore a = R - 1/2 = 0.3290213568591335924092397772....
In Theorem 4.13, we showed that if
d(k,l)=
4k+1+3 4k+4
and c < (6(k, l))-1, then
x 1r
(5.4.1)
clogx*
Again, let 81(k,1) = k + 1. Since
9(k,1) = 2(0,A+1) we see that inf 0 = (R + 1)/2. Therefore, (5.4.1) is true for c < (inf 6)-1 = 1.0934809440576886234567678933665528741229168 ....
Heath-Brown's improvement of Piatetski-Shapiro. Lest the reader obtain the impression that all applications of the algorithm reduce to computing R, we mention the following result due to Heath-Brown (Heath-Brown (1983)). If (k, 1) is an exponent pair and 12k+10 O(k' 1) -
12k-21+13'
then 7r, - x/(clogx) for c < (9(k, l))-1. The first 50 terms of the q-sequence are 01222
11111
11121
11211
11121
12121
12123
21211
12211
11111.
It follows that (5.4.1) is true for c < 1.140491364814885561329935222649608554....
Gaps between square free numbers. Let
S(N) = E z/'(xn-2). N 0,
(5.4.3)
optimal Exponent Pairs
68
and if (k, 1) = BA(tc, A) then
7R-3
14R+28
(5.4.5)
Now let 9A(a) = inf V,i ,A and 9B(a) = inf o,,,BA, so that inf o,, = min(gA(a), gB(a)).
From the inequalities (5.4.3) and (5.4.4), we see that gA is an increasing function of a and that gB is a decreasing function of a. From Theorem 5.5, we see that gA(1/3) < gB(1/3) and gB(9/25) < gA(9/25). Since gA and gB are continuous, there is some /3 with 1/3 < /3 < 9/25 and gA(l3) = 9B(fl)Let (kn, In) be a sequence of exponent pairs in AP such that inf t/i,sA = lim (kn, In). n-.oo
Let (kg, l;) be a sequence of exponent pairs in BAP such that inf 0pBA = lim (k;;, 1;). n-too
Since 71n - 10kn - 3 > 0 and 7ln -10k;; - 3 < 0 , we can find a convex combination (kn, ln) of (kn, ln) and (k;;, lrt) such that 71n - l0kn - 3 = 0. We then have YB(kn, 1n) = 14((kn
+ 1n) + 28
Since 0,3(kn, In) < max(zli,s(kn, I' ), 0,B(kri, ln)), statement (5.4.5) follows.
Optimal Exponent Pairs
69
5.5 NOTES The material in this chapter is taken from Graham (1985), which was the first published account of computing optimal exponent pairs for general 9. Phillips and Rankin had earlier considered the special case 9(k,1) = k + 1. Phillips (1935) stated that he computed the first 12 terms of the q-sequence. He did not publish them; he stated only that he failed to find any pattern. Rankin (1955) gave the first eleven terms of the q-sequence for inf(k + 1), and this gives the numerical value to 10 decimal
places. He did the calculations during the 1940's on an adding machine. He did not publish the details of his method; the proofs, he said, "are rather long and involve much heavy algebra." He also stated that all the terms in the q-sequence of inf(k + 1) satisfy q < 3. Heath-Brown (private communication) has established a corresponding result for more general 9: if 9 satisfies (5.2.2), then every term in the q-sequence after the first is < 3.
6. TWO DIMENSIONAL EXPONENTIAL SUMS
6.1 INTRODUCTION In this section, we shall give a brief overview of the material in this chapter.
Let D be a subset of [X, 2X] x [Y, 2Y]. In most applications, D is a rectangle, or some set bounded by simple algebraic curves. Let f : D -+ R and define
S = >2
e(f(m,n)).
(m,n)ED
In analogy with our hypotheses for exponent pairs in the one dimensional case, it is appropriate to assume that (Ax-ay-R)
Dx+yi f (x, y) = D,,iyt
. {1 + O(0)}
(6.1.1)
for all i, j less than some appropriate bound, and where ai+i
D. y; =
axiayi
,
A is a non-zero real constant, a < 1, ,0 < 1, cep # 0, and A = A(X,Y) -+ 0 as X -+ oo and Y -+ oo. The primary tools for estimating S are two dimensional analogues of the Poisson summation formula and the Weyl-van der Corput inequality.
First, let us consider the Poisson summation formula. Recall that in Lemma Lemma 3.6, terms of the form I appear, so the usefulness of that lemma is lessened when f" becomes small. In two dimensional sums, the Hessian of f plays a similar role. The Hessian of f is defined by
H f= det
[::=y1
f Dyyf J
Under suitable conditions on f and D, we have
E e(f (m, n)) 2 e(f (E,,q) - µ - vn)I + error terms,
(µ,v)ED""
Two Dimensional Sums
71
(ii) D' is the image of D under µ = Dx f and v = Dy f , (iii) D" is some subset of D , (iv) e = e(µ, v) and n = r1(µ, v) are defined by Dz f (C, rl) = µ and Dy f (e, rl) = v.
The two dimensional Weyl-van der Corput inequality states that if Q < X and R < Y then X2Y2 + XY ISI2
21-klakl qm - SjI
k
i=1 i,j
I(i - (jI = 21-kIP'(Cj)I Jqm r
- cjI.
Now set
E = {(m, n) : X < m < 2X,Y < n < 2Y, and I!-- - Cjml < 2k-16IP'(Cj)I-1m} r
.
(6.2.1)
Then E C By Lemma 6.3, IE; I 0 for all t, the Fourier transform j (u) = 0 if u > b, and f (O) = 2T + b-i . Since
ITI Tt
j
00
TK
b(p)e(p ' t)I2 dtt ... dtK < ... I-TK 157 pEP
t) 12 dt1 dtK,
... PEP
f0000
the result follows.
LEMMA 7.5. Let X and Y be two subsets of RK. Let a(x) and b(y) be arbitrary complex numbers for x E X and y E Y. Let X1, ... , XK, Y1,
, YK
positive numbers. Define the bilinear forms
B(b; X) =
E
57, I b(Y)b(Y') I, y'EY lyk-ykl 0, c > 1, and 1 < N < N1 < 2N. Let 8(h) be the characteristic function of the interval [3,ucN2, 3µcN1 ]. If h is real and non-zero, then
IN
le (µx3 -)
(.-_12µh)1/4
dx = 8(h)
+0 min (µN)-1/2, 13µN2 - c
e(1/8)e
(-Zµ-1/2 (3 h
)3/2) 1-'
+
13µN1
-c
+ O(cl hI
_1)
1
Proof. When 6(h) = 1, this follows by using (7.3.2) together with Lemmas 3.1 and
3.2 to estimate the tails. When 6(h) = 0, this follows from Lemmas 3.1 and 3.2.
LEMMA 7.8. Let g(x) be the function whose graph is given below.
New Exponent Pairs
92
.............
N- 1 N
N1
Suppose that either h < 21LcN2 or h > 2µcN1. Then
J g(x)e (µx3
-h cx /
dx 2 then G(a, l; c) = 0. Proof. The first assertion is trivial. For the second assertion, we observe that 1
2'-'
(a(j2 r-1 + k)2 + 1(j2'-1 + k)) 2*
j=0 k=1
E k=1
Since I is odd, the inner sum is 0.
2r
J
' j= =0
21-
New Exponent Pairs
95
LEMMA 7.15. For any positive c, G(1, 0; c) = 2 (1 + i-c)(1 +
cJ
i)cl/2.
Proof. By the Poisson summation formula,
e(')="
x2
e(vx+ c f dx.
With a couple of changes of variable, we see that this is 00
= c
1
Cv2
CO
J e(c(x + vx))dx = c
)
e
v=- o
j
l+v/2 /2
e(cy2)dy
After breaking this last sum into odd and even terms, we see that it is
c(l + i-`) foo e( cy2) dy. By Lemma 3.3, the last integral is e(1/8)(2c)-1/2, and the desired result follows. We can now prove (7.4.2). If p is an odd prime, then IG(a,1; pa)I = pale by Lemmas 7.11(a), 7.12, 7.13, and 7.15. Furthermore, IG(a,1; 2a)I < 2(a+1)/2 by Lemmas 7.11(b), 7.14, and 7.15. This together with Lemma 7.10 proves (7.4.2). Now we are ready to evaluate the incomplete perturbed Gauss sum e (µn3 +
S(µ; a, b; c) _
N 0. The improvement comes form using Holder's inequality with fifth powers rather than fourth powers, and from analogs of Lemmas 7.20 and 7.21 with 10 variables rather than 8.
APPENDIX
SOME AMAZING FUNCTIONS In the late 1930's, A. Beurling observed that the entire function
B(z)
(z - n)-2 +
sine rx
°O
°°
n=0
m=1
(z + m)-2 + 2z-1
satisfies a simple and useful extremal property. Define
sgn(x)=
1
ifx>0;
Ll
ifx sgn (x) for all real x. The function B(z) is entire of exponential type 27r. Beurling showed that if F(z) is any entire function of exponential type 27r satisfying F(x) > sgn (x) for all real x, then r F(x) - sgn (x)dx > 1,
J
ocoo
and equality occurs if and only if F(z) = B(z). In 1974, A. Selberg used the function B to give a simple proof of the large sieve inequality. Selberg noted that if Xx(x) is the characteristic function of the interval I = [a, b] and CI = 2 {B(b - z) + B(z - a)} then
C1(x) -> X1(X)
for all real x. Moreover, the Fourier transform
No = J
C1(x)e(-tx)dx
vanishes outside the interval [-1,1]. This property makes CI particularly nice for applications.
In 1985, J. Vaaler showed how Beurling's function could be used to construct a trigonometric polynomial approximation to O(x). For each positive integer N, Vaaler's construction yields a trigonometric polynomial 0* of degree N which satisfies
I0*(x) - 0(4
0. From the
Proof.
identity
r,
Ir 2 3in27rx
n=-oo
we get
sin 2 ax
1 - H(x) = 2
1
(x - n)2
{2
t
1
2
1
(x } M)2 + 2 M=1 `
To complete the proof, it suffices to show that 1
1
X
2x2
00
1
0, this proves the right hand inequality of (A.2). For the left-hand side, we note that {u} = O(u) + 1/2, so 2
10
{u}du
-
/'du
/ 0
(x + u)3
JO
(x + u)3
+ 2 f°° 0
0(u)du = (x + u)3
1
2x2
+2
00 do, 0
(u)
(x + u)3'
Appendix
where
01 ( u ) = I
Vi
(t) dt =
113
{u}({ 2 } - 1)
Integrating by parts again, we get 2
f
-
{u}du
°°
1
+
°° 601(u)du
1.
10 Proof. Let
HN(z) =
sgn (m)
sine 7rz i
I-I