with a remainder term R satisfying
(2.9)
R
d,k,
qd/2 r2T
+
Rp r2p
1 + a r
pqp+d/2 + 18/d r2, T
T
qd/2 r2
,
for any > 0.
Notice, that the estimate (2.9) is uniform in s. The measure (or its support B(R) Zd) represents the main box of size R from which lattice points are taken. The convolution of with µk(·; r) is
a somewhat smoother lattice measure than . Note though that the weights
assigned by µ to the lattice points near the boundary of the box B(R) become
smaller when the points approach the boundary of the box B(R + k r). The
weights assigned to lattice points in B(R  k r) remain unchanged. Later on
we shall choose the size r of the smoothing measure µ(·; r) smaller in compar
ison with R, that is, we shall assume that R ck r with a sufficiently large
constant c. This smoothing near the boundary simplifies the derivation of ap
proximations and helps to avoid extra logarithmic factors in the estimates of
errors. The corresponding measure is the continuous counterpart of µ with the dominating counting measure on Zd replaced by the Lebesgue measure on Rd. Theorem 2.1 allows a generalization. The measure can be replaced by
an arbitrary uniform lattice measure with support in a cube of size R, see
Theorem 2.2 below. Theorem 2.1 is a partial case of Theorem 2.2. We shall
prove Theorem 2.2 in Section 3. In order to formulate that result, we extend
our notation.
We shall denote = (·; 1/2). The measure has the density d =
I x 1/2
with
respect
to
the
Lebesgue
measure
in
Rd,
so
that
dx (dx)
=
I x 1/2 dx. The measure (·; r) has the density (2r)d I x r .
Notice as well that (·; r) = µ(·; r).
Henceforth will denote a probability measure on Rd such that (A) = 1,
for some subset A B(R) Zd and
(2.10)
{x} = 1/ card A,
for all x A.
We do not impose restrictions on the structure of A except that A B(R)Zd and A = . Write = . It is easy to see that has the density
(2.11)
d dx
=
1
I
card A yA
x  y 1/2
.
LATTICE POINTS
989
We define measures µ and as in (2.4), and denote distribution functions of Q[x  a] with respect to µ and as F and F0 respectively. Notice that the measure has the density
(2.12)
D(x) d=ef d = d
d( · ; r)
k ,
dx
dx
dx
where f g denotes the convolution of functions f and g,
f g(x) = f (x  y) g(y) dy.
Using the Fourier transform, we can easily verify that the density D admits continuous bounded partial derivatives D(x) d,k r d1, for  k  2 (see Lemma 7.1).
Theorem 2.2. Theorem 2.1 holds with and defined by (2.10) and (2.11) respectively.
Let us now define the functions Fj, for j 2N. Let = (1, . . . , m)
denote a multiindex with entries 1, . . . , m N. Write
for the sum
: 1=j
which extends over all possible representations of the even number j as a sum
j = 1 + · · · + m of even 1, . . . , m 2, for all possible m 1. For example,
for j = 6, we have 6 = 6, 6 = 4 + 2, 6 = 2 + 4 and 6 = 2 + 2 + 2. Introduce the
functions
(2.13)
Dj(x) =
Dj(x)
: 1=j
with (2.14)
Dj(x) =
(1)m !
···
D(j)(x)u11 . . . umm
m (k+1)(dul),
l=1
where the density D is defined by (2.12), and where we use the notation (1.22) for the Frґechet derivatives. For example, we have
D2(x)
=

1 2
D (x)u2 (k+1)(du),
and D4(x) = D44(x) + D422(x) with
D44(x)
=

1 24
D422(x) =
1 4
D(4)(x)u4 (k+1)(du), D(4)(x)u21u22 (k+1)(du1) (k+1)(du2).
Let j denote the signed measure on Rd with density Dj. We define the function Fj, for j 2N, as the distribution function of Q[x  a] with respect to the signed measure j; that is,
(2.15) Fj(s) = j x Rd : Q[x  a] s = I Q[x  a] s Dj(x) dx.
990
V. BENTKUS AND F. GOЁ TZE
The function Fj : R R is a function of bounded variation, Fj() = Fj() = 0 and
(2.16)
sup Fj(s) s
Rj j,d r2j
1 + a
j qj+d/2,
r
for j < d/2;
see Lemma 7.4. In the elliptic case the choice of is immaterial as long as the support of contains a sufficiently massive box of lattice points. Thus we shall simply choose and as in (2.4). The same choice of is appropriate for the estimation of maximal gaps (cf. Theorem 1.5) in the hyperbolic case. The choice of a general as possible is appropriate for proving refinements of (1.14). We shall restrict ourselves to the following special generated by a starshaped closed bounded set (see (1.13)) whose nonempty interior contains zero. Define
(2.17)
volZ C (R )
(C) =
volZ(R )
and let in accordance with (2.10) the set A be given by A = (R ) Zd. The
measure is again defined by (2.11). In order to guarantee that x Zd :
{x} > 0 B(R), we shall assume throughout that B(1); this is not
a restriction of generality. Hence, for the Minkowski functional of the set we have
(2.18)
x M (x) m x, for all x Rd,
with some m 1. The inequalities (2.18) are equivalent to B(1/m) B(1). The modulus of continuity
(2.19)
() = sup M (x + y)  M (x) y, x=1
of M satisfies lim () = 0. For and in (2.4) we have = B(1), M (x) = 0 x and () = . Let () d=ef + B() be a neighborhood of the boundary of . Then, introducing the weight p0 = 1/ volZ(R), writing for a while = k r/R and assuming that is defined by (2.17), we have
(2.20)
µ(C) = p0 volZ C, 0 µ(C) p0 volZ C (R ) , µ(C) = 0,
for C R \ () , for C Rd, for C Rd \ (R).
LATTICE POINTS
991
Notice that p0 = (2R)d for defined after (2.4). In order to prove (2.20), it suffices to consider the case when the set C is a one point set, and to use elementary properties of convolutions. Similarly, for measurable C Rd, we have
(2.21)
(C) = p0 dx, C
0 (C) p0
dx,
C(R )
(C) = 0,
for C R \ () , for C Rd, for C Rd \ (R),
where now = (k r + 1)/R. Using (2.19), it is easy to see that (2.22) R() x Rd : 1  () M (x/R) 1 + () , for any > 0.
We conclude the section by deriving all results of the introduction as corollaries of Theorem 2.1; the refinement of (1.14), Theorem 2.6, is implied by Theorem 2.2.
The elliptic case. Let us start with the following corollary of Theorem 2.1.
Corollary 2.3. Assume that the operator Q is positive and T 1.
Then we have
(2.23) volZ(Es+a)vol Es
d, (s+1)d/2 qp+d/2
1 sp/2
+
1 sT
+ 18/d
s, T
T , s
for d 9, 2 p < d/2, p N and any > 0. The quantity (s, T ) is defined
in (1.7) (cf. (2.6) and (2.7)).
Proof. The bound (2.23) obviously holds for s 1. Therefore proving (2.23) we shall assume that s > 1. We assume as well that a 1. This assumption does not restrict generality since
(2.24)
volZ(Es + a) = volZ(Es + a  m),
for any m Zd,
and in (2.23) we can replace a by a  m with some m Zd such that a  m 1. Choose the measure as in (2.4), and k = 2p + 2, r = s, R = 2k r. Obviously R k s + 1, for s > 1. Therefore the bound (2.9) of Theorem 2.1 implies (2.23) provided that we verify that our choices yield (2.25) F (s) = (2R)d volZ(Es + a), F0(s) = (2R)d vol Es, Fj(s) = 0, for j = 0.
992
V. BENTKUS AND F. GOЁ TZE
Let us prove the first equality in (2.25). The ellipsoid E1 is contained in the unit ball, that is, E1 x 1 B(1), since the modulus of the minimal eigenvalue q0 is 1. Therefore we have Es B s . Due to our choice of R, r and k 6, we have R k r 6 s. Thus, the inequality a 1, the relations Es + a B 1 + s and F (s) = µ(Es + a) together with (2.20) imply the first equality in (2.25). Let us prove the second equality in (2.25). Using (2.12) and (2.21) we see that the density D is equal to zero outside the set B R + k r + 1 , and D(x) (2R)d, for x B(R  k r  1). The ellipsoid Es + a is a subset of B(R  k r  1), that yields the second equality in (2.25). For the proof of Fj(s) = 0 notice that the derivatives of D vanish in B(R  k r  1), hence in the ellipsoid Es + a as well. Proof of Theorem 1.3. This theorem is implied by Corollary 2.3. Indeed, the estimate (1.10) is obvious for s 1. For s > 1, the estimate (1.10) is implied by (2.23) estimating qp qd/2, choosing p = 2 and T = T (s) as in the condition of Theorem 1.3.
Proof of Corollary 1.4. We have to prove (1.11) and (1.12). The proof of (1.12) reduces to proving that volZ(E+ + a)  volZ(E + a) > 0, for c(d, )q3d/2 0(s) with a sufficiently large constant c(d, ). Using (1.11) it suffices to verify that R 1/2, which is obviously fulfilled. Let us prove (1.11). Consider an interval (, + ] with s 1. We shall apply the bound of Theorem 1.3 which for s 1 yields
(2.26)
volZ(Es + a)  vol Es d, qd s1+d/2 0(s).
We get (2.27)
volZ (E+ + a) \ (E + a)  vol E+ \ E
d, qd ( + )d/21 (0( + ) + 0( )).
The estimate (2.27) implies (1.11). Just note that 0( ) 0(s), for s, divide both sides of (2.27) by vol E+ \ E and use
vol E+ \ E = ( + )d/2  d/2 vol E1,
+
( +)d/2  d/2 d
u1+d/2 du d ( +/2)1+d/2 d ( +)1+d/2,
+/2
vol E1 = vol x Rd : Q[x] 1
vol
x
Rd
:
x
1/ q
= cd qd/2.
LATTICE POINTS
993
Proof of Corollary 1.2. It suffices to use (1.9) and (1.12). Proof of Theorem 1.1. Assuming the irrationality of Q, the bound o(s1) is implied by Theorem 1.3 and (1.9) since vol Es d qd/2 sd/2. The bound o(s1) in (1.2) implies d(, Q, 0) 0, as , which is impossible for rational Q. The hyperbolic case. For an interval I = (, ] R we write
(2.28)
F (I) = F ()  F (),
Fj(I) = Fj()  Fj().
Notice that F (I) = F (s) in the case I = (, s]. Theorem 2.2 has the following obvious corollary.
Corollary 2.4. Under the conditions of Theorem 2.2 we have
(2.29)
F (I)
=
F0 (I )
+
j2 N,
Fj (I ) j
+
R
with remainder term R which satisfies (2.9).
Lemma 2.5. Let Q[x] be an indefinite quadratic form of dimension d 9.
Let p, k and satisfy the conditions of Theorem 2.1. Assume that M (x) = x and = B(1). Let c1 = c1(d, ) denote a sufficiently small positive constant. Finally, assume that
(2.30)
Rr 1 , c1
, [c1 R2, c1 R2],
r c1 ,
R
k
q
a
c1
R.
Then
(2.31)
F (I)  1 F0 (I )
q3d/2 p,k,d, r2
Rd rd
+
qp+d R2 
1 r2 T
+
R2p r3p
+ 18/d
r2, T
T r2
.
Proof. The result follows from Corollary 2.4 dividing (2.29) by F0(I) and using the estimates
(2.32)
Fj (I ) j2 N, j
k,d (  ) qd2 r2d Rd2,
(2.33)
F0(I) k,d (  ) qd/2 R2.
Let us prove (2.32). Using (2.15), (2.28), (2.30), (7.1), applying the estimate (8.9) of Lemma 8.2 with
M (x) = x, R = R + k r, and m = = 1,
994
V. BENTKUS AND F. GOЁ TZE
using the bound a0 q a, we obtain
(2.34) Fj (I )
k,d r jd I x R + k r I Q[x  a] I dx
k,d (  ) q(d2)/2 r jd
1 + q a R+kr
d2 (R + k r)d2
k,d (  ) qd2 rjd Rd2.
In the proof of (2.34) the condition R r 1/c1 allowed us to replace R and r by R and r respectively. Summation in j, 2 j < p, of the inequalities (2.34) yields (2.32). Let us prove (2.33). Using (2.12), (2.21), the lower bound (8.10) of Lemma 8.2 and conditions (2.30), we have
F0(I) d Rd I x R  k r I Q[x  a] I dx d (  ) qd/2 R2.
Proof of Theorem 1.5. We shall apply Lemma 2.5 choosing T = T (r2), R = r k/c1, the maximal p and minimal k such that the conditions of Theorem 2.1 are fulfilled. In this particular case we can rewrite (2.31) as
(2.35)
F (I)  1 F0 (I )
d, q3d/2 r2 +
qd+p 
(r2).
In order to estimate the maximal gap, it suffices to show that any interval
I = (, ] contains at least one value of the quadratic form (i.e., F (I) > 0) provided that and satisfy  d, qd+p (r2). The inequality F (I) > 0 holds if the righthand side of (2.35) is bounded from above by a sufficiently
small constant which can depend on d and .
Next we formulate and prove some refinements of (1.14). Recall that the set satisfies B(1/m) B(1) (see (2.18)), and that denotes the modulus of continuity of the Minkowski functional M of the set (see (1.13) and (2.19)). Write
(2.36)
W = x Rd : Q[x  a] I ,
I = (, ],
(2.37) and (2.38)
= volZ W (R ) vol W (R )
1 ,
0 = r/R, 3 = (0/c2),
1
=
q
a/R,
4 = (1 q/c2),
2 = ( + )/R2, 5 = (2 q/c2),
where c2 = c2(d, m) denotes a positive constant.
LATTICE POINTS
995
Theorem 2.6. Assume that the form Q[x] is indefinite and d 9. Let r 1/c2, 0 < < 1  8/d, T 1 and
(2.39)
j c2, for j = 0, 1, 2, 3, j c2 q1/2, for j = 4, 5.
Then we have
(2.40)
d,,m qd1
1+
1 r2 d0
(3 + 4 + 5)
+
q3d/2 
1 20 T
+
1 r3+d/2 d0+2
+ 18/d
r2, T
T 20
provided that the constant c2 is sufficiently small.
Proof of Corollary 1.6. Let us apply Theorem 2.6 choosing c2 = c2(, q, d, m) sufficiently small depending on and q as well. Choose R = C r. Then we have 0 = C1 and the last two inequalities in (1.18) guarantee conditions (2.39). In particular, we have j c2, for j = 3, 4, 5. Hence, we can apply the bound (2.40). Choosing T = 1, = (1  8/d)/2 and estimating 1, we obtain
(2.41)
q,d,m c2 (1 + r2 Cd) + (C2 + r1 Cd+2)/(  ),
for r 1. The estimates (1.17) follow if the righthand side of (2.41) is bounded from above by . But this holds in view of the first two inequalities in (1.18) and our choices of C, C0 and c2.
Proof of Corollary 1.7. During the proof we shall write T = T (r2) and = r2, T . Choose = (1  8/d)/2. Since R = r T 1/4 with T , we have 0 = T 1/4 and the conditions (2.39) are fulfilled for sufficiently large r. Moreover, j 0, for j = 3, 4, 5. Hence, the bound (2.40) yields (1.19) with
g(r) = (1 + r2 T d/4) (3 + 4 + 5), h(r) = T 1/2 + r1 T (d+2)/4 + (18/d)/2 T 14/d. 9 9 If g(r) 0 or h(r) 0, we can choose (if necessary) T = T (r2) growing somewhat slower such that g(r) 0 or h(r) 0 (cf. a similar redefinition of T (s) in the case of (1.9)).
For the proof of Theorem 2.6 we shall need the following Lemma 2.7, which allows us to estimate using Theorem 2.2. Write
(2.42)
= volZ W (R )  vol W (R ) .
By µR = µk(·; r) and R = µk(·; r), where = , we shall denote the measures µ and , emphasizing the dependence on the parameter R which enters in the definition (2.17) of .
996
V. BENTKUS AND F. GOЁ TZE
Lemma 2.7. Let 0 = (k r + 1)/R. Assume that (20) < 1. Write
(2.43) s1 = R/ 1 + (20) , s2 = R/ 1  (20) , p2 = 1/ volZ(s2 ).
Then we have
(2.44) where

p2 1
max s=s1,s2
s
+
v,
v = vol W s1 1  (20) M (x) s2 1 + (20) ,
and
(2.45)
s = I x W µs(dx)  I x W s(dx).
If (20) < 1/4 then
(2.46)
v vol W 1  3 (20) M (x/R) 1 + 3 (20) .
Proof. We omit the elementary proof of (2.46).
Using s1 = R/ 1 + (20) R/2 and the fact that is a nondecreasing function, it is easy to see that
(2.47)
s1 1 + (1) R s2 1  (2) ,
with 1 =
kr +1 , s1
2 =
kr +1 . s2
Let us prove (2.44). Assume first that 0. Applying (2.20)(2.22) to
µs and s and using (2.47), we have
(2.48)
µs2(C) = p2 volZ C, for C R ,
s2(C) p2 vol C s2 (1 + (20)) , for any C.
Using (2.48), we have
(2.49) volZ W (R ) = p2 1 µs2 W (R ) p2 1 I x W p2 1 I x W s2(dx) vol W (R ) + v,
µs2 (dx),
proving the lemma in the case 0. Assume now that < 0. Write p1 = 1/ volZ(s1 ). Relations (2.20) (2.22) together with (2.47) yield
vol W (R ) vol W s1(1  (20)) + v,
vol W (s1(1  (20))) p1 1 volZ W (R ) p1 1 I x W and using p2 p1 we obtain (2.44).
I xW µs1 (dx),
s1 (dx),
LATTICE POINTS
997
Proof of Theorem 2.6. The result follows from the bound
(2.50)
d,,m,p,k qd1
1+
1 r2
Rd rd
(20) + 4 + 5)
+
qd+p 
R2 r2T
+
R2p+2 r3p
+ 18/d
r2, T
T
R2 r2
choosing the maximal p < d/2, k = 2 p + 2 and using the notation (2.38). Recall, that we assume that the constant c2 in (2.38) is sufficiently small. In particular, this assumption guarantees that (20) is as small as will be required in the auxiliary lemmas used below. Let us prove (2.50). Dividing the bound (2.44) of Lemma 2.7 by v1 d=ef vol W (R ) , we obtain

1 p2 v1
max s=s1,s2
s

+
v. v1
Hence, in order to prove (2.50) it suffices to show that
(2.51) (2.52) (2.53)
p2 1 = volZ(s2 ) d Rd, v1 d,m (  ) qd/2 Rd2, v d,m (  ) (2 0) + 4 + 5 q(d2)/2 Rd2,
and, for s = s1, s2,
(2.54)
s s,1 + s,2
with s,1 and s,2 defined below by (2.57) such that
(2.55)
s,1
d,k,
qd/2 r2T
+
R2p r3p
qp+d/2
+ 18/d
r2, T
T
qd/2 r2
.
(2.56) s,2 d,m (  ) (2 0) + 4 + 5 q(d2)/2 r4 (R/r)d2.
To prove (2.51) it suffices to notice that s2 2R since B(1) and we
assume that (20) is sufficiently small.
ing
The estimate (2.52) = 1 and using the
follows from estimate a0
/(1R.13)aqnda(/8R.10=)
of Lemma 8.2 choos1 1/(2m) which is
guaranteed by the assumption that the constant c2 is sufficiently small.
The bound (2.53) follows from (2.46) and Lemma 8.3 with = 3(20)
1/4 which is fulfilled since c2 is small. Applying Lemma 3.8 we use the esti
mate 1,0 1.
Let us prove (2.54). Let F (I; R) and Fj(I; R) denote the functions F (I)
and Fj(I) defined by (2.28) with the underlying parameter R which enters into
the definitions of the measures µ = µR and = R. Then (2.54) holds with
(2.57)
s,1 = F (I; s)  F0(I; s)  s,2,
s,2
=
j2 N,
Fj (I ; j
s).
998
V. BENTKUS AND F. GOЁ TZE
For the estimation of s,1 we shall apply Theorem 2.2. Our choice of the constant c2 yields (20) 1/2. Therefore s = s1, s2 is asymptotically equivalent to R. The functions s F (I; s) and s Fj(I; s) are differences of the corresponding distribution functions. Furthermore, a R. Hence, Theorem 2.2 yields (2.55). It remains to prove (2.56). Let us estimate Fj(I; s). Using s R, r r and (7.4), we have Fj (I; s) d,p v2 r jd, where v2 = vol W x Rd : M (x/s)  1 (ck r/R) . By Lemma 8.3 and a suitable choice of c2, the volume v2 allows the same upper bound as the volume v in (2.53) since s R. Hence, summing these bounds over j, we get (2.56).
3. Proof of Theorems 2.1 and 2.2
We shall use the following approximate (see, e.g., relation (8.4) in [BG4]) and precise (see, e.g., Chung [Ch]) formulas for the FourierStieltjes inversion. For any T > 0 and any distribution function F of a normalized nonnegative measure with the FourierStieltjes transform
F (t) = e tx dF (x),
t R,
we have (3.1)
F (x) = 1 +
i
T V. P. e
x t
F (t) dt + R
2 2
T
t
with remainder term R such that
R 1
T F (t) dt.
T T
Here V. P. f (t) dt = lim f (t) dt denotes the Principal Value of the inte 0t> gral. Furthermore, for any function F : R R of bounded variation such that F () = 0 and 2F (x) = F (x+) + F (x), for all x R, we have
(3.2)
F (x) = 1 F () + i lim V. P.
e{xt} F (t) dt .
2
2 M
tM
t
The formula is wellknown for distribution functions. To functions of bounded variation it extends by linearity arguments.
LATTICE POINTS
999
Theorem 2.1 is implied by Theorem 2.2 and we have to prove Theorem 2.2 only. The expansion (2.8) yields
(3.3)
R =
F

j2 N0,
Fj j
.
Let us prove Theorem 2.2 assuming that r 1. Using (3.3), Lemma 7.4 to bound Fj, and the obvious estimates F  1 and F0 1, we obtain
(3.4)
R
d 1+ 1j
Rj r2j
1 + a r
j qj+d/2
Rp d r2p
1 + a r
p qp+d/2
since q 1, 1 R/r, 1/r 1 and j < p < d/2. The estimate (3.4) implies
the theorem in the case r 1. Therefore in the remaining part of the proof
we can and shall assume that r 1.
Using the representation (3.3), representing F by the approximate Fourier
Stieltjes inversion (3.1) and Fj by the inversion formula (3.2), splitting the
intervals of integration, and using the triangle inequality and the obvious esti
mate
1
F (t) dt
F (t)
T 1/rtT
1/rtT
dt t
,
we obtain
(3.5) with
R I1 + I2 + I3 + j2 N0, j
I1 = t r1
F (t) 
Fj (t)
j2 N0, j
dt t
,
I2 =
1 T
F (t) dt,
t r1
I3 =
F (t) dt , t
t r1, tT
Ij+4 =
Fj (t)
dt , t
t r1
j 2 N0, j < p.
The estimate (3.5) shows that in order to prove the theorem it suffices to prove
that
(3.6)
I1, Ij+4
Rp d,k r2p
1 + a r
p qp+d/2,
j 2 N0, j < p,
(3.7)
I2
d
qd/2 r2T
,
1000
V. BENTKUS AND F. GOЁ TZE
(3.8)
I3
d, 18/d r2, T
T
, qd/2 r2
for any > 0.
Let us estimate I3. Changing variables, we have
Splitting (3.9)
F (t) = e tQ[x  a] µ(dx). µ = µ3(·; r) with = µ(k3)(·; r),
we obtain
(3.10)
I3 I3,
where a is given by
I3
d=ef
sup aRd
r t1, tT
a(t; r2)
dt t
,
(3.11)
a(t; r2) = e tQ[x  a] µ3(dx; r) ,
(cf. (2.6)). The function a satisfies the following inequalities (see Theorem 5.1 in [BG4])
(3.12)
a(t; r2) a(t + ; r2) d qd/2 Md/2( ; r2),
(3.13)
R a( ; r2) d qd/2 Md/2( ; r2), t,
involving the function M(t; r2) introduced in (1.21). The inequality (3.12) allows us to apply Theorem 5.1 of the present paper. Choosing in this theorem
= cd qd/2, = d/2, s = r2, = 1,
and using ln x x, for x 1 and > 0, we obtain
(3.14)
r2, T
I3 d,
qd/2
18/d (1 + ln T )
qd/2 r2
.
Estimating in (3.14) 1 + ln T T , for T 1, and q 1, and using (3.10), we derive the desired bound (3.8) for I3. Let us estimate I2. Similarly to the estimation of I3 we obtain
(3.15)
I2
I2 , T
I2
d=ef
sup aRd
r t1
a(t; r2) dt.
The bound (3.13) for a and the definition (1.21) of the function M(t; r2) together with the inequality a 1 yield
(3.16)
a(t; r2) d qd/2 min 1; (r2 t)d/2 ,
for r t 1.
Hence, we have
I2
d qd/2 min 1; (r2 t)d/2 R
dt =
qd/2 r2
min R
1; td/2
dt
which together with (3.15) yields the bound (3.7) for I2.
d
, qd/2 r2
LATTICE POINTS
1001
Let us prove (3.6) for I1. Applying the bound (4.10) to R defined by (3.3), bounding by a the function in (4.10) and using (3.16), we obtain
I1
q p+d/2 Rp
d,k
r2p
1 + a r
p I1
with
(3.17)
I1 = r t1
t r2 (p) + t r2 p min 1; (t r2)d/2 dt . t
By change of variables tr2 = we get
(3.18)
I1
( (p) + p) min 1; d/2
d
0
d,p 1,
since 0 < (p) p < d/2, see (4.7). The bounds (3.17) and (3.18) yield (3.6) for I1. Let us prove (3.6) for Ij+4. Using (7.13) we have
Ij+4
Rj d rj+d/2
1 + a
j qj+d/2;
r
an application of the inequalities
r 1, R/r 1, j < p < d/2, q 1
implies (3.6) for Ij+4, thus concluding the proof of Theorem 2.2.
4. An expansion for the FourierStieltjes transform F
By change of variables we can write the FourierStieltjes transforms of the distribution functions F and F0 as
(4.1) F (t) = e tQ[x  a] µ(dx),
F0(t) = e tQ[x  a] (dx).
Similarly, for the FourierStieltjes transforms of the functions of bounded variation Fj (see (2.13)(2.15)) we have
(4.2)
Fj(t) =
Fj(t),
: 1=j
where the sum
is defined as in (2.13), and
: 1=j
(4.3)
Fj(t) =
(1)m !
··· e t Q[x  a] D(j)(x)u11 . . . umm dx m (k+1)(dul). l=1
Introducing the notation
(4.4)
g(x) = exp h(x) ,
h(x) = itQ[x  a],
1002
V. BENTKUS AND F. GOЁ TZE
integrating by parts and using D(x) dx = (dx), we have
(4.5)
F (t) = g(x) µ(dx),
F0(t) = g(x) (dx)
and (4.6)
Fj(t) =
(1)m !
··· g(j)(x)u11 . . . umm (dx) m (k+1)(dul). l=1
In the proof of (4.6) we used that the function D has compact support and the derivatives D are continuous functions, for  k  2. Write
(4.7) (s) = s/2, for even s, and (s) = (s + 1)/2, for odd s,
and
(4.8)
(t) = sup e tQ[x] + a, x µ[k/(p+1)](dx; r) , aRd
(4.9)
0(t) = sup e tQ[x] + a, x [k/(p+1)](dx; r) . aRd
Lemma 4.1. Let 2 p k  1 and q 1. Then, for R r 1, we have
(4.10) F (t)  F0(t)  j2N, j
and, for j 2N0, j < p,
(t) Rp qp
d,k
r2p
1 + a r
(4.11)
Fj (t)
0(t) Rj qj
d,k
r2j
1 + a j r
p t r2 (p) + t r2 p t r2 (j) + t r2 j .
Proof. Let g : Rd C denote a sufficiently smooth complex valued function. Assume that x, u1, . . . , up Rd. We shall use the following decomposition
(4.12)
g(x) = g(x + u1) + g1 + · · · + gp,
where (4.13) (4.14)
gj =
c() g(j)(x + um+1)u11 . . . umm , for 1 j < p,
: 1=j
1
gp =
cp() (1  )m g(p)(x + um)u11 . . . umm d,
: 1=p
0
and
(4.15)
c() = (1)m , 1! . . . m!
cp() =
(1)m 1! . . . m1! (m
 1)!
.
Here the sum
is taken over all representations of j as a sum j = 1 +· · ·+
N : 1=j m, m j, of integers 1, . . . , m . For example, in the case j = 3 we have
LATTICE POINTS
1003
the following four representations: 3 = 3, 3 = 2 + 1, 3 = 1 + 2, 3 = 1 + 1 + 1. In particular, if p = 1 then
and if p = 2 then
1 g1 =  (1  ) g (x + u1)u1 d, 0
g1 = g (x + u2)u1,
1
1
g2 = (1  ) g (x + u2)u1u2 d  (1  )2 g (x + u1)u21 d.
0
0
In order to prove (4.12) it suffices to iteratively apply Taylor expansions.
In the first step we use the Taylor expansion of the function g:
(4.16)
p1 g(x) = g(x + u1)  1=1
1 1!
g(1)(x)u11

1 (p  1)!
1 (1  )p g(p)(x + u1)up1 d. 0
In the second step we apply expansions of type (4.16) to the functions x g(1)(x)u11, for 1 1 < p, using u2 instead of u1. After p such steps we arrive at (4.12).
For the derivatives of the composite function g(x) (see (4.4)), we shall use
the following formula, which follows from a general formula (see Theorem 2.5 in Averbuh and Smoljanov [AS]). Let v1, . . . , vs Rd. Let A = A1, . . . , A denote a partition of the set v1, . . . , vs = A1 · · · A into disjoint subsets Aj such that 1 card Aj 2, for all j. Notice that (s) s, where (s) is defined by (4.7). For a subset A v1, . . . , vs , say A = vi1, . . . , vil , introduce the derivative Ah(x) d=ef h(l)(x)vi1 . . . vil. Then
(4.17)
g(s)(x)v1 . . . vs = g(x) A1h(x) . . . Ah(x), A
where the sum extends over all partitions A with properties specified above. A One can easily prove (4.17) using h (x) 0 and induction in s.
The following identities
(4.18)
(·; r) = µ(·; r) ,
= µ (k+1),
are obvious (see (2.2), (2.3) and (2.4) for the definitions of measures which appear in (4.18)). For example, for any integrable function u : Rd R, we have
u(x) (dx; r) =
(2r)d u(x+y) (dx) = u(x+y) (dx) µ(dy; r),
yZdB(r)
1004
V. BENTKUS AND F. GOЁ TZE
proving the first identity in (4.18). Integrating both sides of the identity (4.12) with respect to the measures µ(dx), (k+1)(du1), . . . , (k+1)(dup) and using (4.5), (4.6), (4.13), (4.14), (4.18), we obtain
(4.19)
F (t) = F0(t) + f1(t) + · · · + fp1(t) + fp(t),
with F and F0 defined by (4.5),
(4.20) fj(t) =
c() ··· g(j)(x)u11 . . . umm (dx) m (k+1)(dus),
: 1=j
s=1
for 1 j < p, and
(4.21) fp(t) =
1 cp() (1  )m fp(t; ) d,
: 1=p
0
fp(t; ) d=ef ··· g(p)(x + um)u11 . . . umm µ(dx) m (k+1)(dul). l=1
The sums in (4.20) and (4.21) are the same as in (4.13) and (4.14) respectively. However, notice that fj = 0, for odd j < p, since the measure is symmetric. Moreover, fj = Fj, for even j < p, since all terms in the sum (4.20) vanish unless all 1, . . . , m are even, due to the symmetry of . Hence, (4.19) yields
(4.22)
F (t)  F0(t)  j2N, j
Now we can return to the proof of (4.10). The equality (4.22) shows that it
suffices to verify that fp(t) is bounded from above by the righthand side of (4.10). Using (4.21), we have to verify that any of the suprema sup fp(t; ) 0 1 is bounded by the righthand side of (4.10), for all allowable 1, . . . , s. Define v1, . . . , vp repeating u1 1 times, followed by u2 2 times, etc. For example, vj = u1, for all 1 j 1. Using (4.17) and (4.21) with m = p, we obtain
(4.23)
sup fp(t; ) 0 1
m
p
sup 0 1
max A
···
sup IA (k+1)(dul), z1 l=1
where
(4.24) IA = g(x + z) A1h(x + z) . . . Ah(x + z) µ(dx) ,
z d=ef um,
and (p) p. Fix a partition A = A1, . . . , A of type used in (4.17) and (4.23). Let denote the number of 1point sets in this partition. Without loss of generality we can assume that A1, . . . , A are 1point sets, and that A+1, . . . , A are 2point sets. Then
(4.25)
Aj h(x + z) = 2it x  a + z, Qwj , for 1 j ,
LATTICE POINTS
1005
and
(4.26)
Aj h(x + z) = 2it wj, Qwj , for < j ,
with some wj, wj u1, . . . , um . Notice that wj 1 and wj 1. Furthermore, using (4.25), (4.26) and (4.24), substituting
x  a + z, Qwj = x, Qwj  a  z, Qwj ,
multiplying, applying the triangle inequality and reenumerating wj if necessary, we obtain
(4.27)
IA
p
t
max 0
I
where (4.28)
I =
g(x + z) x, Qwj B µ(dx) , j=1
and (4.29)
B d=ef z  a, Q wj
wj, Qwj .
j=+1
j=+1
Using Qw q w d q, we have
(4.30)
B d,p q z + a
p q 1 + a
since z = um 1 (see (4.24)) and wj 1, wj 1. Let us split the measure µ = µk(·; r) as follows (4.31) µ = 0 (p+1), 0 = µ( · ; r)(k(p+1) [k/(p+1)]), = µ[k/(p+1)]( · ; r),
where [u] denotes the integer part of u. Then we have
(4.32)
U (x) µ(dx) = U (x0 + · · · + xp+1) 0(dx0) (dx1) . . . (dxp+1),
for any integrable function U . Applying (4.32) to (4.28), writing x = x0 + · · · + xp+1 and using
p+1 p+1
x, Qwj = · · ·
xj1 , Q w1 . . . xj , Q w ,
j=1
j1=0 j=0
we obtain
(4.33) with (4.34) Ij =
I
p
max 0j1,...,jp+1
Ij ,
p+1 ··· g(x + z) xj1, Qw1 . . . xj, Qw B (dxs) 0(dx0). s=1
1006
V. BENTKUS AND F. GOЁ TZE
Given the variables xj1, . . . , xj, p, we find among x1, . . . , xp+1 at least one variable, say xl, such that l / j1, . . . , j . Without loss of generality we can assume that l = 1. Then (4.34) yields
(4.35) with
p+1
Ij
··· Qxj1, w1 . . . Qxj, w B J 0(dx0) (dxs)
s=2
J = g(x + z) (dx1) ,
x = x0 + · · · + xp+1.
Recall (see (4.4)) that g(x) = e tQ[x  a] . Therefore
(4.36)
J sup e tQ[x] + L, x µ[k/(p+1)](dx; r) = (t) LRd
with defined by (4.8). Hence, the bound (4.35) combined with (4.30) and (4.36) yields
(4.37)
Ij d,p (t) q 1 + a Jj
with
(4.38)
Jj = ···
We have
p+1 Qxj1 , w1 . . . Qxj , w 0(dx0) (dxs). s=2
(4.39)
Q xj1 , w1 . . . Q xj , w
k q R
since we assume that R r, and since the variables xs in the integral (4.38) satisfy xs k r + R R. Combining (4.37)(4.39), we get
(4.40)
Ij k (t) R q 1 + a .
The estimate (4.40) combined with (4.33) and (4.27) yields
(4.41)
IA
(t) t r2 q r2 max R 1 + a . 0
Using the condition R r 1 and + 2(  ) = 2  = p, , we
obtain
(4.42) r2 max R 1 + a 0
r2 max R r 1 + a 
0
r
p rp
R r
1 + a
.
r
The inequalities p, q 1 and (p) p combined with (4.40)(4.42) and (4.21)(4.23) yield (4.10). The proof of (4.11) repeats the proof of (4.10) starting from (4.22) since now we have to estimate Fj = fj (see (4.20)) instead of fp. In this proof
LATTICE POINTS
1007
we have to use (4.20) instead of (4.21) and to replace everywhere p by j.
Furthermore, instead of (4.31) we have to use a similar splitting of of the form = 0 (p+1) with
(4.43)
0 = ( · ; r)(k(p+1) [k/(p+1)]) and = [k/(p+1)]( · ; r).
In particular, the splitting (4.43) yields that the integral corresponding to J in (4.36) now satisfies
J sup e tQ[x] + L, x LRd which leads to the factor 0 in (4.11).
[k/(p+1)](dx; r) = 0(t),
5. The integration procedure for large t
Recall that
(5.1)
M(t; s) = ts 1 I t s1/2 + t I t > s1/2 ,
where s > 0 will be a positive large parameter. For a number T 1 and a family of functions (·) = (·; s) : R R introduce
(5.2)
= s, T d=ef sup (t) : s1/2 t T .
The following Theorem 5.1 sharpens Theorem 6.1 of [BG4] in cases where < 1.
Theorem 5.1. Let (t), t 0 denote a continuous function such that (0) = 1 and 0 1. Assume that, for some > 4 and 1,
(5.3)
(t) (t + ) M (; s), for all t 0 and 0.
Let T 1. Assume that the number defined by (5.2) satisfies
(5.4) > 4 /( 4) s /4, if  1 < 0,
1 + ln 1
/(
4) >4
/(
4) s
/4 (1 + ln s)
/(
4),
Then the integral
T J = (t) t dt s1/2
can be bounded as follows:
if = 1.
(5.5)
J ,
14/ T +1 ,
s
for  1 < 0,
1008
V. BENTKUS AND F. GOЁ TZE
and J
14/ 1 + ln (1 + ln T ) ,
s
for = 1.
If (5.4) is not fulfilled then the following trivial bounds hold
(5.6) J T +1 , for > 1,
J (1 + ln s)(1 + ln T ), for = 1.
Proof. Evaluating the integral J using the method of Lebesgue integration by partitioning the range of in intervals [2l1, 2l], we have to estimate the Lebesgue measure of the corresponding sets Bl = {t : 2l1 (t) 2l} [s1/2, T ].
Using inequality (5.3), we shall show that two points in Bl are either very "close" or far apart. This means that the set Bl consists of `small' clusters of size Ol(s1) separated by "large" gaps of size Ol(1). These constraints on the structure of Bl suffice to bound the measure of Bl well enough in order to estimate the size of J for > 4 as claimed in Theorem 5.1. For trigonometric sums this condition translates to to the assumption that the dimension d satisfies d > 8. Throughout the proof we shall write instead of , . To prove (5.6) it suffices to use (t) and to notice that
T dt s1/2 t
(1 + ln s)(1 + ln T ),
T tdt s1/2
T +1, for > 1.
Let us prove (5.5). The inequality (5.3) implies that (set t = 0, use (0) = 1 and note that 1)
(5.7)
(t) M (t; s) and (t) (t + ) 2 M (; s).
Starting the proof of (5.5) with (5.7) we may assume without loss of generality that = 1, that is, that
(5.8)
(t) M (t; s) and (t) (t + ) M (; s).
Indeed, we may replace in (5.7) (resp. ) by / (resp. by /), and we may integrate over / instead of . Notice that now (0) 1 and the case (0) < 1 is not excluded. Thus assuming (5.8) we have to prove that
(5.9)
T (t) t dt s1/2
14/ F , for s
> 4,
with F = T +1, for 1 < 0, and F1 = (1 + ln T )
1 + ln
1
. While
proving (5.9) we may assume that 1 s. Otherwise (5.9) obviously holds
since 1.
LATTICE POINTS
1009
Let l denote the smallest integer such that 2l . For the integers l l, introduce the sets
Bl = [s1/2, T ] t : 2l1 (t) 2l , Dl = [s1/2, T ] t : (t) 2l1 .
Since the function satisfies 0 (t) , for s1/2 t T , the sets Bl and m Dl are closed and Dm Bl = [s1/2, T ]. Furthermore, (5.8) implies that l=l (t) t , for t s1/2, whence Bl [Ll 1, T ], where Ll = 2(l+1)/ . Recall that (t) 2l, for t Bl, and (t) 2m1, for t Dm. There fore the relation Dm [s1/2, T ] yields
(5.10)
T (t) t dt
m (t) t dt +
(t) t dt
s1/2
Dm
l=l Bl
m
2m G +
2l t dt,
l=l
Bl
where G = ln T + ln s, for = 1, and G = T +1, for 1 < 0. We shall choose m such that
(5.11)
2m G 14/ F s1,
for  1 0.
More precisely, we choose the minimal m such that
(5.12)
m
1 ln 2
ln
s G 14/ F
,
for  1 0.
Using (5.10), (5.11), we see that the estimate (5.9) follows provided that we show that
(5.13)
m Il l=l
14/
F , s
where Il = 2l
t dt.
Bl
Below we shall prove the inequalities
(5.14) Il (l + ln T ) s1 2l+4l/ , Il T +1 s1 2l+4l/ ,
for = 1, for  1 < 0,
for l m. These inequalities imply (5.13). Indeed, in both cases = 1 and > 1 we can apply the bound
2l+4l/ l=l
(2l )14/
14/
1010
V. BENTKUS AND F. GOЁ TZE
since > 4 ensures the convergence of the series, and, according to the definitions of and l, we have 2l1 . In the case = 1 one needs in addition the following estimates. For l 1, we have
l 2l+4l/ l 2l+4l/
1 (2l )14/
14/ ,
l=l
l=0
and, for l c with a sufficiently large constant c , we obtain
l 2l+4l/ l=l
x 2x+4x/ dx l 1
l (2l )14/
14/ 1 + ln 1 .
It remains to prove the inequalities (5.14). For the estimation of Il we need a description of the structure of the sets Bl with l m. Let t, t Bl denote points such that t > t. The inequality (5.8) and the definition of Bl imply
(5.15)
4l1 M (t  t; s).
If t  t s1/2 then by (5.15) and the definition of M(; s) we get
(5.16)
t  t , where = s14(l+1)/ .
If t  t s1/2 then by (5.15) and the definition of M(; s) we have
(5.17)
t  t , where = 4(l+1)/ .
For > 4 and sufficiently large s 1 note that
(5.18)
< , provided l m.
Indeed, using the definitions of and , we see that the inequality (5.18) follows from the inequality 4s /4 < 2m, which is implied by (5.12), the assumption (5.4) and s 1. The estimate (5.18) implies that either t  t or t  t . Therefore it follows from (5.16)(5.18) that
(5.19)
t Bl = Bl (t + , t + ) = ;
that is, that in the interval (t+, t+) the function takes values lying outside of the interval [2l1, 2l].
Let us return to the proof of (5.14). If the set Bl is empty then (5.14) is obviously fulfilled. If Bl is nonempty then define e1 = min t : t Bl . Choosing t = e1 and using (5.19) we see that the interval (e1 + , e1 + ) does not intersect Bl. Similarly, let e2 denote the smallest t e1 + such that t Bl. Then the interval (e2 + , e2 + ) does not intersect Bl. Repeating this procedure we construct a sequence Ll 1 e1 < e2 < · · · < ek T such that
(5.20)
k Bl [ej, ej + ] and ej+1 ej + . j=1
LATTICE POINTS
1011
The sequence e1 < · · · < ek cannot be infinite. Indeed, due to (5.20) we have T ek e1 + (k  1) Ll 1 + (k  1) k , and therefore k T /. Using (5.20) we can finally prove (5.14). We start with the case = 1. Using ln(1 + x) x, for x 0, we have
k ej +
k
Il 2l
dt = 2l
ln
t
j=1 ej
j=1
1+ ej
k
2l
ej
j=1
(l + ln T ) 2l+4l/ s1
since e1 Ll 1 , k T /, and
k 1k
1
1 k 1
j=1 ej
j=1 e1 + (j  1)
j=1 j
ln T + ln 1
(l + ln T ) 4l/ .
Finally, let us prove (5.14) for 1 < 0. We have
k ej +
Il
1 2l
tdt
j=1 ej
k
k
1 2l
ej
2l
j
j=1
j=1
T +1 2l
s1 2l+4 l/ T +1.
6. Trigonometric sums of irrational quadratic forms
In this section we introduce a criterion for the rationality of a quadratic form in terms of the behavior of an associated sequence of trigonometric sums. Recall that a quadratic form Q[x] = Qx, x , x Rd, with a nonzero symmetric matrix Q = (qij), 1 i, j d, is rational if there exists an M R, M = 0, such that the matrix M Q has integer entries; otherwise it is irrational. For a Rd, d 1, consider the polynomial
(6.1)
P (x) = Q[x] + a, x ,
x Rd.
Throughout this section we shall denote by µ the uniform lattice measure in the box B(r) = x r (cf. with the notation µ(·; r) used in other sections). In other words, the measure µ is nonnegative, normalized µ(Rd) = µ Zd B(r) = 1 and assigns equal weights µx d=ef µ({x}) = 2[r] + 1 d to points x Zd B(r).
1012
V. BENTKUS AND F. GOЁ TZE
For k N and t R, introduce the trigonometric sum
f (t) = f (t; r) =
e t P (x) µx(2k+1), µx(2k+1) d=ef µ(2k+1)({x}),
xZd
where µk denotes the kfold convolution of µ. Let µ denote the symmetrization of µ, that is, µ(C) = µ(C + x) µ(dx), for C Bd. Writing µ(2k+1) = µ µk µk and using the symmetrization inequality (see Lemma 6.5 below), we obtain 0 f (t) 2 (t) with
(6.2)
(t) = (t; r) d=ef
e 2t Qx, y µx µyk.
xZd yZd
For a family of functions g = g(·; r) : R C parameterized by r 0, consider the following condition:
(6.3)
for any 0 < 0 < ,
lim sup g(t; r) = 0. r 0t
We shall apply condition (6.3) to the trigonometric sums f and . See (6.4) and (6.6) for equivalent formulations of (6.3). The following characterization result holds without assumptions on the eigenvalues of Q.
Theorem 6.1. Let k 1. The quadratic form Q[x] is irrational if and only if satisfies condition (6.3). If Q is irrational then f satisfies (6.3). If a and Q are rational (that is, there exists M = 0 such that M a and M Q have rational coordinates, resp. entries) then f does not satisfy (6.3).
The proof of Theorem 6.1 will be given later. It is based on an application of the theory of successive minima (see Cassels [Ca], Davenport [Dav]) and techniques in [BG1]. For a given family of functions, the condition (6.3) allows the following equivalent reformulation: there exist functions 0(r) 0 and (r) such that
(6.4)
lim sup g(t; r) = 0. r 0(r)t(r)
Applying a double large sieve type bound (see the estimate (5.22) in [BG4]) we obtain (t) d,k q2dMd(t; r2) with M defined by (1.21). Hence, assuming 0(r) r1, we have (t) d,k,Q 0d(r), for r1 t 0(r), and
(6.5)
lim sup sup (t) = 0 r aRd r1t0(r)
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1013
provided that the eigenvalues of Q are nonzero. Combining (6.4) and (6.5) we see that the irrationality of Q is equivalent to the following condition: there exist (r) such that
(6.6)
lim sup sup (t) = 0 r aRd r1t(r)
provided that the eigenvalues of Q are nonzero. Due to f 2 , relations (6.5) and (6.6) hold for f as well.
Remark 6.2. An inspection of the proof shows that the condition (6.3) for the trigonometric sums f and holds uniformly over compact sets, say Q, of irrational matrices; that is,
lim sup sup (t; r) = 0. r QQ 0t
Let us recall some facts of the theory of successive minima in the ge ometry of numbers (see [Dav]). Let F : Rd [0, ) be a norm, that is, F (x) = F (x), for R, and F (x + y) F (x) + F (y). The successive minima M1 · · · Md of F with respect to the lattice Zd are defined as follows: M1 = inf F (x) : x = 0, x Zd , and Mk is defined as the lower bound of > 0 such that the set x Zd : F (x) < contains k linearly indepen dent vectors. It is easy to see that there exist linearly independent vectors a1, . . . , ad Zd such that F (aj) = Mj.
Lemma linear forms
6.3 ([Dav, Lemma 3]). in Rd such that jk =
Let Lj(x) = kj. Assume
d k=1
jk
xk
,
that P
1 1
j and
d, let
be
denote the distance of the number to the nearest integer. Then the number
of x = (x1, . . . , xd) Zd such that
Lj(x) < P 1, x < P, for all 1 j d,
R is bounded from above by
cd , where M1 · · · Md
M1 · · · Md are the first d of
the 2d successive minima M1 · · · M2d of the norm F : 2d [0, )
defined for y = (x, m) R2d with x, m Rd as
(6.7)
F (y) d=ef max P L1(x)  m1 , . . . , P Ld(x)  md , P 1 x .
Furthermore, M1 P 1. We shall use some wellknown known bounds for trigonometric sums which go back to Weyl [We]. For our purposes we formulate a variation of such bounds as the following Lemma 6.4. Its proof is provided for the sake of completeness.
1014
V. BENTKUS AND F. GOЁ TZE
Lemma 6.4. For z = (z1, . . . , zd) Rd,
g(z) d=ef
e z, y µyk d,k
h(m),
yZd
mZd
where
(6.8)
d h(m) = h0(zj  2mj), j=1
h0(s) = 1 + r2 s2 k,
s Z.
Proof. We have g(z) = vk(z) with
(6.9)
2d
v(z) = e z, y µy =
e z, y µy = u(zj),
yZd
yZd
j=1
where u(x) = D[2r](x)/ 2r 2 and 2r = 2[r] + 1 with the Dirichlet kernel
[r]
D[r](x) =
exp ixn
n=[r]
= sin(x r) , sin(x/2)
for x R.
Using  sin y min 1; y and x2 sin2(x/2), with y = xr and x , we obtain
(6.10)
u(x)
min
1 x2 r2
;
1
1 1 + r2 x2
,
for x .
The function u(x) is even and 2periodic. Hence (6.10) yields
(6.11)
uk(x)
k mZ
I x  2 m k 1 + r2 (x  2 m)2
mZ
1
k,
1 + r2 (x  2 m)2
where I{A} denotes the indicator function of the event A. Combining g(z) = vk(z) and (6.9), (6.11) we conclude the proof.
Proof of Theorem 6.1. Let us show that for rational Q the trigonometric sum does not satisfy (6.3). Assume for simplicity that Q has integer entries. Let µ Ч denote the product measure of measures µ and . Clearly, Qx, y Z, for x, y Zd, and the function
(t) = mZ pm e 2tm ,
Z pm d=ef µ Ч µk (x, y) 2d : Qx, y = m ,
with weights pm such that mZ pm = 1, is a periodic function and () = 1. Similar arguments show that for rational Q and a the trigonometric sum f does not satisfy (6.3). To complete the proof of the theorem, it remains to prove that if f or does not satisfy condition (6.3) then Q is rational. The inequality f 2
LATTICE POINTS
1015
shows that we have to consider the case of only. Since (6.3) is not fulfilled, we have
for some 0 < 0 < ,
lim sup sup ( t; r) > 0. r 0t
Hence, there exist > 0 and sequences tn t0, 0 tn , and rn such that
(6.12)
( tn; rn) > 0, for all n N.
We shall prove that (6.12) implies the rationality of Q.
Henceforth we shall write t and r instead of tn and rn. Without loss
of generality we shall assume throughout the proof that r = rn is sufficiently large, that is, that r c with some sufficiently large constant c = c(d, k, Q, , ).
Furthermore, we shall write instead of d,k,Q,,. The measure µ is concentrated in the cube B(2r) such that
(6.13)
G(x) µx =
G(x) µx ,
xZd
xB(2 r)Zd
µx = 1, xB(2 r)Zd
0 µx
rd,
for any summable function G. Using the representation (6.2) of , Lemma 6.4, the relations (6.12) and (6.13) for the function h(m) defined by (6.8) with z = 2 tQx, we have
(6.14)
rd
h(m).
xB(2 r)Zd mZd
Obviously we may replace h(m) in (6.14) by
(6.15)
h(m) d=ef
d h0(zj  mj),
j=1
h0(s) = 1 + r2 s2 k,
z = z(x) = tQx,
since this yields in the inequality (6.14) an extra factor depending on d and k only. For vectors s = (s1, . . . , sd) Zd such that sj 0 and = (1, . . . , d) Zd such that j = ±1, for all j, introduce the class Hs, of vectors (x, m) Z2d such that
x B(2r), sign zj(x)  mj = j,
sj 4r
zj(x)  mj <
(sj + 1) , 4r
for all 1 j d. Here zj(x) denote the jth coordinate of z(x) = tQx. Moreover, consider the class H of vectors (x, m) Z2d such that
(6.16)
x B(4r),
zj (x)
< 1, 4r
for all 1 j d,
where denotes the distance from the number to the nearest integer. Then
Hs, = (x, m) Zd : x B(2r) . s,
1016
V. BENTKUS AND F. GOЁ TZE
Furthermore, it is easy to see that (x, m), (x, m) Hs, implies that (x  x, m  m) H. Hence, card Hs, card H. Using (6.14) and the definitions introduced, we obtain
(6.17)
rd card Hs, max h(m) : (x, m) Hs, s,
rd card H
d 1 + s2j k
s j=1
rd card H,
for sufficiently large r.
Lemma 6.3 shows that the cardinality of the class H defined by (6.16) is
bounded from above by
cd M1 · · · Md
,
where
Mj
are
the
successive
minima
of
the
norm F defined by (6.7) with P = 4r and
(6.18)
d Lj(x) = zj(x) = (tQx)j = tqji xi. i=1
Combining (6.17) with card H (M1 · · · Md)1, we obtain rd M1 · · · Md 1. Hence, using P = 4r and P 1 M1 · · · Md, we have Mj r1, for all 1 j d. Let (as, bs) Z2d denote linearly independent vectors such that F (as, bs) = Ms, for all 1 s 2d. The equalities F (as, bs) = Ms together with (6.7) and relations Ms r1 P 1 imply that
(6.19)
as 1, bs 1, 1 s d.
Combining the relations F (as, bs) = Ms with P = 4r and (6.7), (6.18), we get (recall that t = tn and r = rn; therefore as = a(sn) and bs = b(sn) also depend on n)
(6.20)
d tn qjk as(nk)  b(snj ) k=1
1 rn2
,
1 j d, 1 s d.
TinhZe dinreeqpueaaltitsieins fi(6n.i1te9l)ygoufatreann.teCehtohoastintgheasneqaupepnrcoeproifastyesstuembssequae1(nn)c,e.,
. . , ad(n) we may
assume that a(sn) = as are independent of n. Similarly we may assume that b(sn) = bs are independent of n. Passing to the limit as n in (6.20)
along the subsequence and using tn t0, t0 = 0, we see that the numbers
t0 qj1, . . . , t0 qjd satisfy
(6.21)
d asi t0 qji = bsj , i=1
1 s d,
for all 1 j d. Below we shall prove that the vectors a1, . . . , ad are linearly independent. Therefore the system (6.21) has the unique solution
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1017
t0 qj1, . . . , t0 qjd, which obviously has to be rational by Cramer's rule, for all
1 j d.
To conclude the proof we have to show that the vectors a1, . . . , ad Zd are linearly independent in Zd (or, equivalently, in Rd). If a1, . . . , ad are linearly dependent then there exist vs R not all equal to zero such that vs < 1
d
and vs as = 0. Let us prove that there exist integers u1, . . . , ud not all
s=1
equal to zero such that us
1 and x d=ef
d us as = 0. By the multivariate
Dirichlet
approximation
(see,
for
example,
s=1 [Ca,
§V.10]),
for
any
N
N
there
exist uj Z and an integer 0 < q N such that
(6.22)
vs 
us q
<
1 q N 1/d
,
for all 1 s d.
The inequalities vs < 1 and (6.22) yield us 2N . Since the vectors as have d integer coordinates and as 1, the equation vs as = 0 together with s=1 d (6.22) implies x = us as = 0, for sufficiently large N 1. Hence us 1. s=1 For the vector (x, m) with m d=ef d us bs we have s=1
(6.23)
d F (x, m) usF (as, bs) s=1
1 r
since us 1 and F (as, bs) 1/r. Using (6.7), we see that
(6.24)
d F (x, m) 4r us bsj , s=1
for all 1 j d,
where bs = (bs1, . . . , bsd). Combining (6.23) and (6.24), and using that r is
d
d
sufficiently large and that us bsj are integers, we conclude that us bsj = 0,
s=1
s=1
d
for all 1 j d. In other words, m = us bs = 0, which together with the
s=1
assumption x = d us as = 0 means that the vectors (as, bs) Z2d, 1 s d,
s=1
are linearly dependent, a contradiction.
A symmetrization inequality. The following symmetrization inequality is a generalization of a wellknown classical inequality due to Weyl [We]. For a proof, see [BG4, Lemma 7.1]. Recall that the symmetrization µ of a measure µ is defined by µ(C) = µ(C + x) µ(dx), for C Bd.
1018
V. BENTKUS AND F. GOЁ TZE
Lemma 6.5. Let Q : Rd Rd be a linear symmetric operator, L Rd and C R. Let µ1, µ2, µ3, denote arbitrary probability measures on Rd. Define a real valued polynomial of second order by P (x) = Qx, x + L, x + C, for x Rd.
Then the integral J=
2 e tP (x) µ1 µ2 µ3 (dx)
satisfies 2J J1 + J2, where J1 = e 2t Qx, y J2 = e 2t Qx, z In particular, if µ2 = µ3 then J J1.
µ1(dx) µ2(dy), µ1(dx) µ3(dz).
7. Properties of the distribution functions Fj and the signed measures j
We start by establishing some properties of Fj and j (see (2.13)(2.15) for definitions) using the Fourier transforms of the measures j. Recall that we denote W = x Rd : Q[x  a] I , I = (, ], Fj(I) = Fj()  Fj(). Let var Fj denote the variation of Fj. Write
V = x Rd : M (x/R)  1 (0) , 0 = (k r + 1)/R, r = [r] + 1/2.
Lemma 7.1. The density D : Rd R defined by (2.12) has continuous bounded partial derivatives D, for  k  2. Assume that j k  2. Then we have
(7.1)
Dj(x) k,d r jd I x R + k r ,
and
(7.2)
Fj(s) var Fj k,d r j (R/r)d, for R r.
If the measure is defined by (2.17) then we have in addition
(7.3)
Dj(x) k,d r jd I x V ,
and
(7.4)
Fj(I) k,d r jd vol(W V ).
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1019
Proof. The Fourier transforms of complex valued functions f : Rd C are denoted by f (y) = e y, x f (x) dx.
Let u(x) = I x 1/2 , x Rd, denote density of the measure . The
measure (·; r) has density ur(x) d=ef (2r) d u
x/(2 r)
.
Write U =
d dx
for
density of the measure defined by (2.11). Then has density D = U urk
(see (2.12)), where denotes the convolution of functions. Therefore D = U ukr
and
(7.5)
D(x) d y D(y) dy y ur(y) k dy,
since U  1. Hence, using
d u(y) =
sin(yj /2) ,
j=1 yj /2
ur(y) = u(2r y),
we obtain
(7.6)
D(x) k,d r 1d, for  k  2,
which proves the lemma's assertion about D. Let us prove (7.1). Using the definition (2.13)(2.14) of Dj, wellknown properties of the Fourier transforms, and the equality j = 1 + · · · + m, estimating in (2.14) us d 1 and applying (7.5), (7.6), we obtain
Dj (x)
j max : 1=j j,d · · ·
m
· · · D(j)(x)u11 . . . umm
(k+1)(dul)
l=1 m D(y) y, u1 1 · · · y, um m dy (k+1)(dul)
l=1
k,d yj D(y) dy j,d r jd.
Now (7.1) follows since D(x) = 0, for x > R + k r. Let us prove (7.2). Using (2.15) and (7.1), we obtain Fj(s) Dj(x) dx k,d r jd (R + k r)d k,d r j (R/r)d. Let us prove (7.3). By (2.21) and (2.22) the density D is constant outside the set V . Thus, outside V the derivatives of D vanish and Dj(x) = 0, for x Rd \ V . Thus (7.1) yields (7.3). Similarly, (7.4) follows from (7.1), (7.3) and the inequality F (I) I x W V Dj(x) dx.
1020
V. BENTKUS AND F. GOЁ TZE
For further investigation of Fj we shall use the following double large sieve type bound (Lemma 7.2 below) which is a useful corollary of the large sieve of Bombieri and Iwaniec [BI]. It follows from Corollary 5.3 in [BG4] replacing in that corollary q2 by q. Lemma 7.2. Assume that functions g, h : Rd C satisfy g(x) 1 and h(x) 1. Let µ and denote arbitrary probability measures on Rd such that µ x Rd : x T = 1 and x Rd : x S = 1,
for some T > 0 and S > 0. Write
J=
g(x) h(y) e t Qx, y
2 µ(dx) (dy) ,
t R.
Then there exists a positive constant cd, depending only on the dimension d, such that
J
d qd
1 + S T t
d sup µ xRd
x+
cd B t S
sup xRd
x+
cd B t T
,
where B = x Rd : x 1 .
Corollary 7.3. For an integer m 2, the function
satisfies (7.7)
0(t) = sup e tQ[x] + a, x m(dx; r) , aRd 0(t) d qd/2 min 1; (r2 t)d/2 .
Proof. We shall derive the result using Lemma 7.2. If 1 r2 t then (7.7) is obviously fulfilled since 0(t) 1. Thus we can assume that r2 t > 1. Using the obvious identity of the type
f (x) µ1 µ2 µ3(dx) = f (x + y + z) µ1(dx) µ2(dy) µ3(dz),
the fact that the form Q[x] is quadratic and HoЁlder's inequality, we obtain
(7.8) with
02(t)
sup aRd
J (m2)(dz; r)
2
J=
g(x) h(y) e 2t Qx, y (dx; r) (dy; r) ,
where the functions g 1 and h 1 may depend on z and a. Choose S = T = r. The measure (·; r) has nonnegative density (with respect to the
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1021
Lebesgue measure on Rd) bounded from above by (2r) d. Therefore (A; r) (2r) d vol A, for any measurable set A Rd. Thus Lemma 7.2 and the assumption r2 t > 1 yield
J
d qd 1 + r2 t d r2d t r 2d
d
qd r2d td
,
whence, using (7.8), we derive (7.7).
Lemma 7.4. The distribution function s F0(s) has a bounded continuous derivative for d 3. The functions s Fj(s) are functions of bounded variation for 2 j < d/2, and
sup Fj(s) s
Rj j,d r2j
1 + a
j qj+d/2,
r
Moreover, each of the functions s Fj(s), j 2N0, has [(d1)/2]j bounded
continuous derivatives.
Proof. Changing the variables t = r2, we have
(7.9)
tr2 min 1; tr2  dt R
,
1 r2
,
for  1 < <  1,
(7.10)
tr2 min 1; tr2  R
dt t
, 1, for 0 < < .
Using (j) j, the bound (4.11) for Fj(t), the estimate of Corollary 7.3 for 0(t) and (7.9), (7.10), choosing = d/2 and = (j), = j respectively, we obtain
(7.11)
Fj(t) dt R
Rj j,d r2j+2
1 + a
j qj+d/2,
r
for j 2N0 such that j < 1 + d/2. Similarly to (7.11), we obtain
(7.12)
R
Fj (t)
dt t
Rj j,d r2j
1 + a r
for j 2N such that j < d/2. Finally we have
j qj+d/2,
(7.13)
r t1
Fj (t)
dt t
Rj j,d rj+d/2
1 + a r
for r 1 and j 2N0 such that j < d/2, and
j qj+d/2,
(7.14)
t Fj(t) dt < , R
for 0 < j  1 + d/2.
To conclude the proof it suffices to apply the Fourier and FourierStieltjes inversion formulas and to use the bounds (7.11)(7.14).
1022
V. BENTKUS AND F. GOЁ TZE
8. On the Lebesgue volumes of certain bodies related to indefinite quadratic forms
We shall describe the asymptotic behavior of the volume of the set (8.1) A = x Rd : M (x) R I0, Q[x  a] I , R , where M is the Minkowski functional of the set defined by (1.13) and I0 and I = [, ] are finite intervals. Throughout the section we assume that Q[x] is an indefinite quadratic form in Rd, d 3. Reenumerating the eigenvalues of Q, choosing an appropriate orthonormal basis of Rd and denoting the coordinates of x Rd in this basis as x1, . . . , xd, we may assume that Q[x] = q1 x21 + · · · + qd x2d with q1, . . . , qn > 0, qn+1, . . . , qd < 0 and some 1 n d/2. Indeed, if n > d/2, we may replace in (8.1) the matrix Q and the interval I by Q and I respectively. Let S = Sn1 Ч Sdn1 denote a direct product of the unit (n  1)and (d  n  1)dimensional spheres r2 d=ef x21 + . . . x2n = 1 and 2 d=ef x2n+1 + . . . x2d = 1 with area elements d1 and d2 respectively, and d d=ef d1 d2. Write M0(x) = M x1/ q1, . . . , xd/ qd .
Lemma 8.1. The volume of the set A defined by (8.1) satisfies
(8.2)
lim Rd+2 vol A R
=  det Q1/2  ud3 2
I M0(u1 + u2) I0 d du.
0
S
Proof. We shall use the following representation
(8.3)
vol A = 21 Rd2  det Q 1/2 J,
where (8.4)
u2
J = R2
(r, ) I R2 v rn1 dn2 dv du.
0 
(8.5)
(r, ) = I M0(x + a0/R) I0 d, S
x = r 1 + 2,
LATTICE POINTS
1023
and a0 = q1 a1, . . . , qd ad , (8.6) r = r(u, v) = u, = (u, v) = u2  v,
0 u < ,  < v u2.
For the proof of (8.3) it suffices to write
vol A = I M (x + a) R I0 I Q[x] dx,
to change the variables xj = R yj/ qj, 1 j d, to use the polar coordinates r and and to perform the change of variables (8.6). Let us prove (8.2). Assuming that R a0, it is easy to see that
I M0(x + a0/R) I0 = 0 unless 0 r c, 0 c,
with some sufficiently large c = c(M0, I0). Consequently, we have (r, ) = 0 unless 0 u c, v c2. Therefore, for sufficiently large R, we can write
(8.7) with
c J = f (u) du 0
/R2
f (u) = rn1 R2
(r, ) I  v u2 dn2 dv.
/R2
The function (r, ) is a continuous function of the variables u, v and a0/R. Therefore, for any u > 0, we obtain
lim f (u) = ud3 (  ) I
R
S
M0(u1 + u2) I0
d.
The estimate f (u) d cd2  +  allows us to apply the dominated convergence theorem, and (8.7) yields
lim J = (  ) ud3 I
R
0
S
M0(u1 + u2) I0
d du,
which combined with (8.3) concludes the proof of (8.2).
The relation (2.18) yields
(8.8)
(dq)1/2 x M0(x) m x.
Lemma 8.2. Let I0 = [0, ] and = + a0/R, = /m  a0/R. Then the volume of the set A defined by (8.1) satisfies
(8.9)
vol A d (  ) q(d2)/2 d2 Rd2.
If > 0 and  +  2 R2/5 then
(8.10)
vol A d (  ) qd/2 d2 Rd2.
1024
V. BENTKUS AND F. GOЁ TZE
Proof. We shall use the representation (8.3)(8.6). Estimating 1  det Q qd, we see that it suffices to prove that
(8.11) (8.12)
J d (  ) q(d2)/2 d2, J d (  ) d2.
Let us prove (8.11). Using (8.8), we have
M0(x + a0/R) (dq)1/2 x  a0/R
and (r, ) I x2 dq 2 d d I u2 + 2 dq 2 S since r = u and x2 = r2 + 2. Hence, we get
/R2
J d R2 q(d3)/2 d3
I u2 d q 2 dv du d (  ) q(d2)/2 d2,
0 /R2
proving (8.11). Let us prove (8.12). Using (8.8), we have M0(x+a0/R) m x+a0/R and (r, ) I x2 2 d d I 2 u2 + v 2 S
since r = u and 2 = u2  v u2 + v. Hence, using the condition  +  2 R2/5, we obtain
J
d R2 I 0
2 u2 2
4
3
u2
·
I v 2 I v R2 rn1 dn2 dv du
5

d d3R2
I 2 u2 2
4
3
· I v R2 dv du d (  ) d2,
proving (8.12).
The definition (2.19) of the modulus of continuity of M yields
(8.13)
M0(x + y)  M0(x) x d q y/x .
Write
(8.14)
1,0 =
a0 , R
2 =
 R2
+
 R2
,
4 = (2d 1,0 q),
5 = (30d 2 q).
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1025
Lemma 8.3. Let I0 = [1  , 1 + ], 0 1/4. Assume that
(8.15)
1,0 c,
2 c,
4 c q1/2,
5 c q1/2
with some sufficiently small positive constant c = c(d, m) depending on d and m only. Then the volume of the set A defined by (8.1) satisfies
(8.16)
vol A d,m (  ) ( + 4 + 5) Rd2 q(d2)/2.
Proof. We shall write instead of d,m. Using the representation (8.3) (8.6) and the inequality  det Q 1/2 1, we reduce the proof of (8.16) to the verification of
(8.17)
J (  ) ( + 4 + 5) q(d2)/2.
Using x2 = r2 + 2, applying the inequalities (8.8) and the triangle inequality, we see that (r, ) = 0 unless 1   1,0 x q (1  ) + 1,0
since now I0 = [1  , 1 + ] with 0 1/4. Hence the assumptions (8.15)
and q 1 imply that
(8.18)
(r, ) = 0 unless 1/2 x q.
The variable v in (8.4) satisfies v 2 c 1/8. Thus, (8.18) shows that we
can assume r, x
thaqt,
r= and
u in (8.4) we have
satisfies
u 1/4.
Using
(8.18)
we
can
estimate
(8.19)
/R2
J q(d3)/2R2
(r, ) dv du.
1/4 /R2
Using (8.13) and (8.18), we obtain
(8.20)
M0
x+
a0 R
 M0(x)
q d q a0 R x
q 4.
Let us write x = r 1 + r 2 + (  r)2. The condition x 1/2 yields r + 1/2. Therefore, using u = r 1/4 and repeating the arguments used
for the proof of (8.20), we obtain
(8.21)
  r r
=
v r (r + )
8v 82,
M0(x)  uM0(1 + 2)
q 5.
Using (8.20) and (8.21), we can replace the indicator function in (8.5) by
(8.22) I
1
c0
q
4c0
q
5
u
M0(1+2)
1++c0 q
4+c0
q
5
,
where the constant c0 = c0(d, m) depends on d and m only. Now using (8.19), integrating the indicator function (8.22) in the variable u, estimating M0(1 + 2) 1 and integrating in v, we obtain (8.17).
1026
V. BENTKUS AND F. GOЁ TZE
UniversitaЁt Bielefeld, Bielefeld, Germany Email addresses: [email protected] [email protected]
References
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LATTICE POINTS
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(Received December 29, 1997) (Revised December 12, 1998)
V Bentkus, F Gotze