lDepartment of Mathematics, Southern Illinois University, Carbondale, Ill.
20n leave from Department of Mathematics, University of Illinois, Urbana.
3
This research was supported in part by the Office of Naval Research, Contract
no. N00014-67-A-032l-0002.
Reproduction in whole or in part is permitted
for any purpose of the
United States Government
..
..
BERRY-ESSEHl BOUNDS MoJO A THEORH1 OF ERDOS ArlO TURAN
m~
UiHFORi,1 DISTRIBUTION
H. Niederreiter
l
~!10D
1
and Walter Phillipp2,3
Department of Statistics
University of North Carolina at Chapel Hill
Institute of Statistias Mimeo Series No. 853
November, 1972
.
..
8ERRY-ESSEE!-~
BOUiWS Aim A THEOREf1 OF ERDOS Arm TURAil
m~
Ul"J I FORlvi 0I STR I BUT IOf-j f:l0D 1
H. Niederreiter and Walter Philipp
Introuuc tion
1.
(x ' n=1,2, ••• > be a sequence of real numbers contained in [0,1).
n
Let
Denote by
A(N,x) the number of
n
~
N
with
is called uniformly distributed (u.d.) if
all 0
~
1
x
~
1.
x
n
x.
<
N- 1A(N,x)
The sequence
x
x
for
+
(In general, (x > is called u.d. mod 1
n
fractional part {x }
n
is u.d.)
I
N- 1A(N,x) - x
Another equivalent condition is the Wey1 criterion:
-1
I
+
n
(For the proof of the basic theorems see [5].)
m
> is
u.d. iff
0
(xn > is u.d. iff
for all h
N
N+
if the sequence of
It is easy to see that (x
def
D*
N = suPO ~ x ~ 1
as
n
€
Z - {O}.
The following theorem due to
Erdos and Turan [3] can be regarded as a quantitative version of the
sufficient part of the Wey1 criterion.
Theorem A:
For any integer
The best constants
c
1
and c
(Niederreiter, unpublished).
m
2
~
1
have been so far
c
1
= 17.2 and
C
z = 4.3.
Much larger values were given by Yudin [9].
1 The second named author was supported in part by ONR Contract N00014-67A-0321-0002.
2
The purpose of this paper is to give various generalizations of this
theorem and to point out their connections with other parts of analysis.
We shall prove the following:
Theorem 1 Let F(x)
= lt
F(l)
and let
be nondecreasing on [0,1] with
=0
F(O)
and
G(x) satisfy a Lipschitz condition on [0,1].
IG(x) - G(y)1 $ Mlx-yl
Suppose that
G(O) = 0
where
and G(l) = 1.
Then for any positive integer
1
F(h)
If we set
F(x)
the Erdos-Turan
Theorem
= f exp(2~ihx)
o
1
1
A
dF(x)
= N-1A(N,x)
and G(h)
= f exp(2~ihx)
o
dG(x) •
=x
for 0 $ x $ 1
we get
and G(x)
theorem as a special case.
has a celebrated continuous analogue, namely the Berry-
Esseen inguality (see [2, p. 206] or [7, p. 285]).
Let
distribution function of a random variable and let
G(x)
Lipschitz-condition
G(_oo)
F(x)
be a
satisfy a
l
IG(x) - G(y)1 $ Mlx-yl
Suppose that
m
=0
and G(oo)
sup IF(x) - G(x)
-oo$x$oo
I
<
-
= 1.
for all xtY
Then for any
24 M +
~ T
~~
fT IF(t)
0
T
~
>
E
R.
0
G(t)
I dt
3
Here and here only we put
00
J
F(t) =
e
itx
00
dF(x)
and
Je
G(t):
_00
itx
dG(x).
_00
It is not difficult to modify the proof of theorem 1 so as to give a
T
~heorem where the integral
v.
J
o
is taken with respect to an arbitrary measure
But the only applications we could think of were the cases v = Lebesgue
measure yielding the Berry-Esseen inequality and
yielding Theorem 1.
v = counting measure
Consequently, we felt that the presentation of this
generalization did not justify the extra effort.
Theorem 1 is also related to the sampling theorem in information theory
[6J.
Sampling theorem:
Let
1
F(x)
Suppose that the jumps of
be of bounded variation on [-Zh,
zn1 J ,
h > O.
F(x) at the endpoints of this interval are equal.
Then
00
f(t)
f(n!h) sin~(ht-n)
=
where
f
f(t) =
1
t
~(ht-n)
n=-oo
E:
R
exp(2nitx) dF(x) •
1
["2h'ZhJ
Putting
h=l
we conclude that the "characteristic
2
function"
f(t) of
1
a random variable, bounded by Z ' is uniquely determined by its values
on the integers.
f(n)
Shifting the scale half a unit to the left we see that this
fact is also implied by Theorem 1.
Of course, the condition on the bounded-
ness of the random variable cannot be entirely omitted.
Simply consider the
sequence of independent Bernoulli trials centered at expectations, i.e.
e
P(X
k
111
= -2)
= P(X k = Z) = 2
function"
for k = 1,2, •••.
The
X
k
have'tharacteristic
4
00
Fk(t)
=f
00
= cos
exp(2TIit X ) dP
k
TIt •
.
2n+1·
But this coincides on the integers w1th (cos TIt)
,the "characteristic
function
H
of
S2n+1 = Xl + ••• + XZn+1 ' which is a binomially distributed
random variable, centered at expectation.
Since neither
Xl nor S2n+l
have a distribution function satisfying a
Lipschitz condition one could argue that this is the reason why Theorem 1
fails.
But even if the distribution function
G(x)
is assumed to satisfy
a Lipschitz condition, Theorem 1 cannot be true unless the density
has compact support.
X and define a second one
F(t) by
interpolating linearly between the points with integer abscissae.
By Polya's
F(t) is a "characteristic function" coinciding \,rith
e- 1t / on the integers.
function of
e -Itl
Simply consider the iicharacteristic function\!
of a Cauchy-type random variable
theorem [2, p. 169]
G' (x)
But obviously,
F(x)
cannot be the distribution
X.
We are fully aware that the following comments might make sense only
after the reader has compared the proof of Theorem 1 with one of the standard
proofs of the Berry-Esseen inequality (e.g. see Chung [2], pp. 206-208).
The following idea suggests itself immediately.
Simply sum the formulae
obtained from the integration by parts rather than integrate them as is done
in some of the proof s of the Berry-Esseen inequality.
what we are doing.
causes difficulties.
expectation. ~i
But the term with
That i.s, in essence,
h = 0, if not properly disposed of,
These evaporate if we licenter F(x) - G(x) at
A slightly more refined method will even yield a stronger
result (see Theorem 1').
Unfortunately, we were unable to make these methods work in dimension
5
s > 1.
By combining them with ideas from the original proof of the Erdos-
Turan theorem the generalization to higher dimension can be carried out.
In order to state the theorem we introduce some notation.
S
be a distribution function on
whenever some
x
j
=0
(1 ~ j
R
~
with
s).
Let
F(lt ••. tl) = 1
G(Xlt ••• tX )
s
bounded variation in the sense of Vitali
G(B)
=f
be a function of
s
J
B c U
and F(x1t ••• tx ) = 0
s
U t the closed s-dimensional
on
whenever some x.
For any Borel set
Let F(xlt ••• tx )
s
=0
(1 ~ j ~ s) and
s write
dG
B
We shall assume that for any rectangle
we have
(1.1)
IG (a.J
~ x
~ a.
J
j
+
k.
J
(These conditions are satisfied if
with density bounded by
M.)
t
1 ~ j
~ s)1 ~ M j~ k.
_.s J
G is a distribution function on
We note once and for all that (1.1) implies
a Lipschitz condition of the form
~
For any integral vector
R(h)
h
=
= R(m,h)
US
}1
l Ix
j::;s
(hIt ••• ths) write II hi I
'.y.
j
=
J
I
max I h.1
l~j~s
1) }
J
and
'6
where
m is a given positive integer.
I
Let
hjx .
J
j:S:s
product of h and x
=
F(h)
(x1' .•• ,x ).
s
be the inner
Define
= F(h1 , ••• ,hs ) = f
exp(2ni(h,x»
dF(x)
US
and similarly
Theorem 2
G(h).
Under the above hypotheses, we have for any positive integer
IF(x) - G(x)l«
s
IF(h) - G(h)1
1:1
--+
m+1
where the constant implied by
Again we can specialize
«
s
R(h)
depends on
F(x) and
G(x)
s
only.
to obtain the s-dimensiona1
version of the Erdos-Turan theorem, due to Koksma [4] and SzUsz [8].
Unfortunately, our constants implied by
ones known for this particular case [5].
«
s
are much worse than the best
Similarly, there is a continuous
analogue of Theorem 2, or an s-dimensiona1 version of the Berry-Esseen
inequality, due to v. Bahr [1].
We are indebted to Professor Cambanis for providing us with reference
[6] .
2.
The One-Dimensional Case
In this section we present the proof of theorem 1 and some remarks
concerning the values of the constants.
In fact, we shall establish the
following stronger version of Theorem 1.
Theorem 1
F(l)
= 1,
1
Let
and let
F(x) be nondecreasing on [0,1] with
F(O)
G(x) satisfy the Lipschitz condition
=
° and
m
7
TG(x) - G(y)/
Suppose also that
sup
OSx,yS1
S
G(O) = 0
for all 0
Mix - yl
and G(l) = 1.
S
1 •
S
Then for any positive integer
4M
4 m 1
- (F(y)-G(y»I S --- + - L (m+1
'IT h=l h
I (F(x)-G(x»
x, y
1
A
m
A
- ---)
IF(h)-G(h)I
m+1
If, in particular, {x > is a sequence in [0,1), then Theorem l'
n
implies an inequality for the discrepancy
mod 1.
N
For
~
x
n
€
J.
where X(J)
intervals
mod 1.
1
A(N,J) - A(J)I
.:f
is the length of
J
be the number of
Then define
D = sup IN-
N
extended over all intervals
N
J mod 1, let A(N,J)
1 and an interval
n, 1 S n S N, with
D
Taking
J
and the supremum is extended over all
-1
F(x) = N
A(N,x)
and
G(x) = x
in Theorem 1',
we arrive at the following version of the Erdos-Turan theorem.
Corollary
For any sequence (xn > in
(u,v] and any positive integer
m we
have
In order to prove Theorem 1', we need the following simple auxiliary
result concerning the Fejer kernel.
Lem~a 1:
Let
numbers with
(2.1)
m be a positive integer, and let
D
S
~ and (m+1)D ~
dx
fD~ Sin2(~+1)TIX
sin TIX
C.
Then
S
8D (1
1
1
+ TIC )
C and
D be positive
Proof:
Since
sin TIX
2x
~
o
for
~
~
x
'21 '
we have
r
2
2
sin (m+1)TIx
<!
sin (m+l)TIx
. 2
dx - 4 D
2
S1n TIx
x
D
f~
(2.2)
dx.
Using integration by parts, we get
~
2
sin (m+1)TIx
1
sin 2(m+l)TID
dx =-+
f
2
2D
2
D
4 (m+1) TID
x
)2
~
1
-!.+
2D
4 (m+1) TID
2
+
dx
f
< 1 +
3 -2D
2(m+l)TIx
D
1 -
f~
sin 2(m+1)TIx
D 2(Iil+1) TIX 3
1
<-!. (l +
2(m+1)TID2 -2D
dx
~
TI~)
In view of (2.2) we obtain (2.1).
Proof of Theorem 1
F(x)
=
1
:
It is convenient to extend F(x) and G(x) by setting
€
R, where [xl is the
x.
We note that F(x)
[xl + F({x}) and G(x) = [xl + G({x}) for x
integral part and {x} is the fractional part of
is nondecreasing on
with constant
M.
R, and that G(x) satisfies a Lipschitz condition on R
We set
H(x)
= F(x)
- G(x) for
x
€
R.
periodic on R with period 1, and
O~x:U~~l IH(x) - H(y)! =
sup IH(x) - H(y)1
x,y€R
We first consider the case where
f
(2.3)
For an integer
(2.4)
1
o
h~
H(x) dx = 0 •
integration by parts yields
F(h) - G(h) = -2TIih
f
1
o
H(x) e
2TIihx
dx
Then H(x) is
9
Choose a positive integer
m and a real number
a
to be determined later.
m
(m+l - Ihl) e 2 ih(x-a»)dx
Then (2.3) and (2.4) imply
m
I*
(m+l - Ihl)
h=-m
I-a
(2.5)
=
J
(I
H(x+a)
(I
J H(x)
=
o
m
-a
(m+l - Ihl)e
h=-m
27Tihx
)dx
h=-m
where the asterisk indicates that
summation.
1
-27Tiha F(h)-G(h)
e
27Tih
h = 0 is deleted from the range of
Because of the periodicity of the integrand, the last integral
may also be taken over
1
[~,
1
2].
m
I
(2.6)
We note that
2
(m+l - Ihl) e
27Tihx = sin (m+l) 7TX
h=-m
sin
2
7TX
where the right-hand side is interpreted as (m+l)
We infer from (2.5) and (2.6)
I
[
~
-"2
S1.n
dx
I ~ 2;
x
=.1.
(2.7)
m
I*
6 = sup IH(x)l.
H(b-)
= -
Since
!
Suppose that
6 or H(b+) = 6
H(x-)
~
for some b
H(x)
E
R.
=6
is an integer.
Ihl)IF(h) - G(h)1
Ihl
~
A
(m+1 _ h) IF(h) - G(h)
h
6 >
I
~l ' for otherwise the
H(x+) for x
E
R , we have either
We treat only the second alter-
native, the first one being almost identical.
H(t) = 6 + H(t) - H(b+)
(m+l
A
xER
theorem is trivial.
x
h=-m
7T h=l
Let us set
in case
that
2
H(x+a) sin. (~+l) x
2
+ (F(t) - F(b+»
For
t > b, we have
- (G(t) - G(b»
~
6 -M(t-b) •
"10
We set
D
/}.
= 2M
and choose
~
H(x+a)
a
= b+D.
M(D-x)
In particular, we see that
D
~
Then from the above we get
for Ixl < D •
21 '
for otherwise
H(x) would be positive
on an interval of length greater than 1, contradicting the fact that
=0
H(n)
~
for n
€
Z. We obtain
D
Mf (D-x) sin
-D
2
(~+l)nx dx _ 2MD
sin nx
(2.8)
= 2MD't Sin2(~+l)nX
o sin nx
dx
4MDf
~
D
-D
~
2
f sin (m+l)nx dx _ ZMDf sin
-~
sin 2nx
D
2
(~+l)nX dx
sin nx
2
sin (~+l)nx dx .
sin nx
m+l
is - - by (2.6).
1
The integral of the Feger kernel over [0, 2"]
2
Therefore from (2.8) and Lemma 1
~
2
f H(x+a) sin (m+l)nx
dx
2
sin nx
-~
~ m;l /}. _ ~(l+~)
where we used
(~l)D
> 1.
~
2
~ sin 2 (m+l) nx
sin (m+l)nx
dx
6MDf
dx
2
2
sin nx
D
sin nx
2MDj
0
> m;l /}. - M ,
Combining the above inequality with (2.7),
we arrive at
2M
2
~
1
~
IF(h)
1
~
I
(2.9)
/}. ~ + - l. (m+l
n h=l h
In
this proves the theorem in case (2.3) is satisfied.
particular~
- -)
m+l
We consider now the case where
f
- G(h)
1
o
H(x) dx f O.
We shall first show
~
11
that there exists
c, 0 < c < 1, such that
1
J H(x)
(2.10)
= H(c)
dx
o
•
1
J H(x) dx
Suppose first
H(s) ~
f
o
O.
>
There exists
1
We note also that
H(x) dx.
o
uities of
H (x)
H(l)
s, 0
= O.
value
J H(x)
o
in the interval [s,l).
dx
If
analogously.
With a number
and
G (x)
1
= G(x+c)
c
- G(c)
for
F(x)
x
R.
€
H(x) must attain the
1
o
H(x) dx < 0, one proceeds
F (x)
1
We observe that
and G(x)
1
1, such that
<
satisfying (2.10), we define
share the properties of
f
f
s
Since the only discontin-
are positive jumps, the function
1
<
respectively.
F1 (x)
=J
- F(c)
and G1 (x)
Furthermore, we have
1
(F (X) - G (x»dx
1
o 1
= F(x+c)
(H(x+c) - H(c)dx
=0
•
0
By what we have already shown, the inequality
(2.11)
:~k F1 (x) - G1 (x)
I
I
5;
holds for any positive integer
I ~\2rrihx
1
1
d(F (x) - G (x»
_ 11I"e2rrih(X-C)
£Or all integers
dH(x)
h.
I
=
2H
m+1 +
m.
I
~ (ll"1 - m+1)
1
1
21Tihx
1T2 h~l
~ e
d(F 1 (x)-G 1 (x»
We note that
I-I (e2dhx
Il!Ce2rribx
dH(x)
d(F(x+c) - G(x+c»
I
=
I ~le2rrihx
It follows from (2.11) that
cH(x)
II - !F(h)-G(h)!
I
12
and so
which is the desired inequality.
An alternative (and somewhat shorter) proof of the weaker Theorem 1
rests on centering
H(x)
at expectation t i.e' t replacing the function
H(x) in the above argument by
H* (x) = H(x)
- It where
I
=f
1
o
H(x) dx.
In
exactly the same waYt one proves then
Choosins
2C
estimate for
= O.
II I 't-lhich
one obtains an estimate for
implies the desired
sup I H(x) I .
xe:R
We add some remarks concerning the values of the constants in the
corollary of Theorem I'.
Suppose
c
l
and c
2
are absolute constants such
that
holds for any sequence (x ) in the [Otl) and any positive integer
n
choosing
N
= 2t
Xl
m.
1
= at and x 2 = 2
t we get
1
2
c1
$; m+1 + c 2
1
I ("h h$;m
1
m+1) •
h even
With
m
=I
we obtain
c
l
~
It and with
m
=3
we obtain c
l
+ c2
~
2.
Then
13
Also, taking
m
= 19
and using c
~
l
4, we get c
Proof of Theorem 2 for s
3.
We shall prove theorem 2 in detail for s
necessary changes for
and define
2
=
0.31.
~
2
=2
s
~
3 in the next section.
~
=
sup
(X,y)EU 2
only and indicate the
D by
IH(x,y)1
Lemma 2 If (U,V)EU 2 satisfies H(u,v) =
~
= 100
MD •
and max (u,v)
~
1 - 50D
then
the conclusion of theorem 2 holds.
Proof:
Without loss of generality we assume that max (u,v)
H(u,l)
=~
=
t~
But F(u,l) and G(u,l) satisfy the hypotheses of Theorem 1.
!~ ~ H(u,l) ~
2
sup
O~x~l
IH(x,l) I
Since the last integral is just
Lemma 3 Let 0
min (u,v)
Proof:
~
<
c ~ 1.
Then
<
-
Hence
4M + TI
4 h__
~Ll(1-m+l)hexP(2nihx)dH(X,1)
h
m+l
0
11 II
H(h,O) the lemma is proved.
H(u-,v-) ~ -e~
with (U,V)EU
2
implies
100 e D.
In fact,
u < 100eD leads to a contradiction,
-lOOcl~ = -e~ ~
Then
+ F(u,l) - F(u,v) - (G(u,l) - Gu,v»
~ ~ - 50MD =
(3.1)
v.
F(u-,v-) - (G(u,v) - (G(O,v»
>
-lOOeMD.
I
14
Corollary:
We may assume that
Proof:
D > 1/50 an application of Lemma 3 with
If
is no point (u,v)€U
(u,v)€U
2
with
2
with
= 6.
H(u,v)
D
~
1/50.
H(u-,v-)
= -6.
£ •
1 shows that there
Hence there must exist a point
But then, since 1 - SOD <
a~
max(u,v) , Lemma
2 applies.
Define
2
sin 7f(m+1)t
2
for t
~
Z
for
€
Z
sin 7ft
=
f (t)
t
Then by (2.6)
For
a
~
u
1 put
~
r (w)
u
We observe that for 0
=u
with aO(u,a)
a
h
~
a,S,a,b
1
u
= --J
m+l 0
~
1
f(w-t) dt .
we have
and
(u,a)
=
1
1
Using the fact
J dH(x,y) = 0 we obtain for any choice of u,S,a and b
2
U
15'
I{
I
ra(x-a)rs(y-b)dH(X,y)! =
I
0<
U
ah
II h II Sm
1
(a~a)ah
2
)I
(S,b)H(h 1 ,h 2
A
(3.2)
1:.
1T
<
-
1\
L.
0<
II hi ISm
IH(h)1
R(h)
Our goal is to estimate the integral from below by a nontrivial linear
combination of 8, M!(m+l) and the last sum in (3.2).
Integration by parts
gives
f
U
2
1
ra(x-a)rS(y-b)dH(x,y) = -rS(l-b) oIH(X,l)r~(x-a)dX
1
(3.3)
-r (I-a) IH(l,y)r~(y-b)dy
a
Notice that
0
S
ra(l-a), rS(l-b)
0
1
S
I-'
for any choice of a,S, a and b.
Lemma 4 For any choice of a,S,a and b we have
1
II H(x,l)r~(x-a)dxl
~
I IH(h,O)
I
R(h,O)
8M
+ §.
m+l
1T
8M +
m+l
~1T h=l
I IH(O,h)1
R(O,h)
h=l
and
Iof H(l,y)r~(Y-b)dyl:s
I-'
Proof: We prove the first inequality only.
~l I ~lH(X'l)(f(X_a)
S
s
- f(x-a-a»
~l
2 sup IH(x,l)1
osxsl
m
8M
+ §.
m+l
1T
Y (1
h=l
Its left-hand side equals
_ l)
m+l
dxl
Ilf(x)dX = 2 sup IH(x,l)l
0
O~x:sl
1~(h,O) I
h
16
by (3.1)
and the remark following it.
Lemma 5 If (u,v)€U 2 satisfies H(u,v) ~ ~ ~ and max(u,v) ~ 2D then the
conclusion of Theorem 2 holds.
Proof: Choose a
= b = O.
a.
r (x)
= ~l
=u
+ 2(m+l)
a.
o~
For
f0
x
~
1
a. -
---:,,,,,---:"~
2 (m+l)
we have
11 (2m+2)
f(x-t)dt ~ __1__ f
m+l 0
f(t)dt ~ 2/~2
Renee
Next choose a.
l/(m+l)
~
1
D, we have 0
~
and
a., S
~
13
=v
1
+ 2 (m+1) •
Since we may assume
1/20 in view of the corollary of Lemma 3.
Thus
and
~ ~ ~ H(u,v) = F(u,v) - G(u,v)
4
4
~4 { ro.(x)rS(y) dH (x,y) + ~4 { ro.(x)rS(y) dG(x,y) - G(u,v).
~
U
U
The first integral has been estimated in (3.2).
Because of (1.1) the
second integral is bounded by
M{ ro.(x)rS(y)dx dy
= Mo.S ~ ~(2D
+
2(m~1»
.
U
Finally,
IG(u,v)1
=
IG(u,v) - G(u,O)1
~
2MD •
Hence
A
.!
4
A
u
<.!
-
4 ~
l.!!.llill +.!4 ~ 4 (10- 3
L R(h)
3 1\
A
u
+ .J:...1L) + 21
40 m+l
50
A
u
17
or
. 5
~ s
4
~
3 \ H(h)
M
L R(h) + 4 m+1 •
In view of (3.3), Lemma 4 and the observation made after (3.2) we
have to estimate the last integral in (3.3) from below.
In these estimates
we frequently have to deal with integrals of the following type.
Lenma 6 We have
A
=I
f(x)f(y)dx dy ~ (m+l)2 _ m;l
max(lxl, lyl)SD
B
=!
1 f(x)f(y) dx dy
m+l
< --
D
D<max ( I x I , I y I ) S"'2
and
A + B
=
(m+1)
2
•
Proof: The second integral is bounded by
4
I
J;
J;
~
f(x)dx I f(y)dy s 4(m+l)! dx2 < m+l
D 4x
D
-J;
D
The estimate for
A follows now from the last identity.
Lemma 7 Suppose
H(u-,v-)
.1:.2u
A
S
2
for some (u,v)€U.
Then the conclusion
of Theorem 2 holds.
Proof: Observe that for x
H(x,y)
$
-50~ID
S -
Choose a
=u
min(u~v) ~
S
u
and y
+ (F(x,y) -
~
v
F(u~,v-»
- (G(x,y) - G(u,v»
smm + M(u-x+v-y) •
- D and b
=v
SOD by Lemma 3.
2
- D and observe that (a,b) € U since
Hence
f
:'
18
(3.4)
H(x,y)
-M(48D+(x-a)+(y-b»
$
for (x,y)
€
0 = 0 (a,b)c, U2
Here we introduced the following notation
(3.5)
O(~,n)
= {(x,y)
(m+l)2
f
: max(lx-~I,ly-nl)
D}
$
and 0
c
= U2 \0
Now
U
2
H(x,y)ri(x-a)r~(y-b)dx dy =
~
a
f
U
+
2
f
H(x,y)f(x-a)f(y-b)dx dy +
H(x,y)f(x-a-a)f(y-b-I3) dx dy -
2
U
(3.6)
-f
H(x,y)f(x-a)f(x-b-l3) dx dy -
2
U
-f
H(x,y)f(x-a-a)f(y-b) dx dy
2
u
=I
+ II - III - IV .
Using (3.4) we obtain
I
(3.7)
$
-M
f (48D
o
= -48MDA
We choose a
o$
a,S
$
1
=1
+ (x-a) + (y-b»f(x-a)f(y-b) + 6
o
- u + 2D
~
and 13
=1
- v + 2D
and observe that
50D ,as noted before.
For the estimate of II we may assume that on 0
1
46
,
<
of
and Lemma 6
f
f(x-a)f(y-b)
c
+ 6B.
since min(u,v)
H(x,y)
f
for otherwise we are done by Lemma 5.
=0
(D,D) we have
Hence by the periodicity
19.
II ~
(3.8)
f
D
25MD f(x-I-D)f(y-I-D) + ~ ff(x-I-D)f(y-I-D)
DC
:;; 25MD A +
~B
Next, we observe that on
•
D= D (a,D)
= F(x,y)
H(x,y)
- G(x,y) + G(x,O)
~
-2MD
and hence
(3.9)
III ~ -2MD
f
~
f(x-a)f(y-I-D) -
D
=
-2HD A -
f
f(x-a)f(y-I-D)
DC
~B.
By symmetry we get the same bound for IV.
Putting (3.6) - (3.9) together
and using Lemma 6 we obtain
fH(x,y)r'(x-a)re'(y-b)dx dy :;; -19~ID + 419M/(m+l) :;; 0
2
a
U
since otherwise the Lemma is trivially true.
Hence we conclude from (3.2),
(3.3) and Lemma 4 that
A
191'1D :;; 435M/m +
2.
0<
1T
L l!!.ill..l
II h II:;;m R (h)
which gives the result.
Finally, we can finish the proof of the theorem.
that there exists a point (u,v)
H(ul-,b -)
l
= -~
€
for some (ul,v )
l
2
U with
€
2
U
H(u,v)
= ~,
because if
then Lemma 7 applies.
and for the same reason, we even assume that
and that
We can assume
Actually,
H(x,y) > -~~ for all (x,y)
max (u,v) < 1 - SOD, since otherwise we were done by Lemma 2.
The remainder of the proof is nearly the same as in Lemma 7.
H(x,y)
=~
~
He have
+ F(x,y) - F(u,v) - G(x,y) + G(u,v)
b - M(x-u+y-v)
for all x
~
u
and
y
~
v.
€
2
U
20
Choose
a
=u
(3.10)
+ D and
b
=v
~ M(98D -
H(x,y)
+ D.
Then
(x-a) - (y-b)
for (x,y)
€
0
=0
(a,b)
c U
2
Using the same notation as in (3.6) we obtain
I ~ Mf (98D - (x-a) - (y-b»f(x-a) fey-b) dx dy _!~ f f(x-a)f(y-b) dx dy
o
= 98
2
DC
1
2"~B.
MDA -
By the assumption on lI(x,y)
Next, choose a
1 - 5OD.
Let
=1
II
~
1
- 2"
- u - 2D
>
0 and
D = D (a,
H(x,y)
I-D) and
= R(o,l)
SllI(o,l)
2
~(m+l)
a
s=
=u
1 - v - 2D
+ 2D
<
1.
>
0
as max(u,v) <
Then
+ F(x,y) - F(d,l) - G(x,y) + G(o,l)
I+
for (x,y)
4MD
€
D.
Thus
III S J(IH(o,l)1 + 4~ID)f(x-a)f(y-1+D) dx dy + ~
o
=
f
f(x-a)f(y-1+D) dx dy
DC
(IR(o,l)1 + 4MD) A + ~B.
Similarly with •
=v
+ 2D
we get
IV S (IH(l,.)1 + 4MD) A + ~B.
We substitute the estimates for I-IV into (3.6) and obtain, using Lamma 6
(3.11)
J R(x,y)rV(x-a)r'(y-b) dx dy
U
~ 40MD - 340M/(m+l) -
2
~
In(o,l) I - IR(l,.) I
4fficID - 348M/(m+l) - 4/n i\'
.lii.iliu
R(h)
zt
by (3.1) and the remark following it, the last term can be assumed to be
positive since otherwise the theorem is trivially true.
Hence by (3.2),
(3.3) and Lemma 4 we obtain
40MD ~ 364H/m + 131T. L\'
.l1!..ili2l
R(h)
which implies the result.
4.
The General Case
In the present section we shall sketch the proof of Theorem 2 for
s
~
3.
We assume throughout that Theorem 2 holds for dimension k < sand
use induction.
Define
D by
6
= sup
IH(X)
I = lOsMD.
We replace Lemma 2 by
XEUS
Lemma 25 Let 10- 8
~ £
6
and max
u.~
J
Lemma 35 Let 0
~
£
~ 1.
1
1 - 210
s
£
D, then the conclusion of Theorem 2 holds.
< £ ~ 1.
implies min u j
Corollary S We may assume that D
We use again the functions
r
a
~ 2 • 10-s.
considered in Section 3.
Then for
(4.1)
A
1
\'
L
1T
0<
Ilhll~m
.l1!..ili2l
R(h)
22
Integrating by parts we obtain
f J-S
.~
IUS
r
a.
J
(x.-a j )dH(x1 ,···,X)1
J
S
~ If
i
H(xl'···'x ).IT< r (x.-a.)dx 1 ···dx
s J-S a . J J
s
J
US
(4.2)
with
. =
H(. 00) IT r~ (x - a )dx •.• dx.,
j
j
j
·· . J "
v
1.~....
Jv
v
U
~v
jk
k
k
1
.1
L.
Ij(V)
Here in
1.
H(. •• ) the coordinates
x
n
with
1 S jl< ••• <j" S
S
v
n:f: jk (lskSv)
are replaced by
Lemma 4 becomes
Lemma 4s
For any choice of a , a (lSjSs) we have
j
j
IIj(V)1··· v I S 2
j
,.
v
L
(A M/(m+1) + B
v
v 0< Ilhllsm
.l.!!J&L)
R(h)
Suppose that there is a point (u ' .•• ,u ) € Us with
s
1
-s+l
H(u 1 , .•• ,us) ~ 10
6 and having at least two of the coordinates u
j
Lemma 5s
bounded by
Proof:
2D.
Then the conclusion of Theorem 2 holds.
We can assume that
to prove.
D S 1/(m+1)
since otherwise there is nothing
1
= 0, a. = u j + 2 (m+1)
J
J
We choose
a.
(1 S j s s) and observe
1
that a. S 1 (1 S j ~ s) • In fact, if u. > 1 2
(m+l)
J
J
s
Lemma 2s applies with E = 10- • Hence, as in Lemma 5
10- s +1 6
S
G2) sf
IT
r
US jSs
(4.3)
~
1
-~
2
aj
(x.-a.)dH(xl, ••• ,x)
J J
s
aj
(x.-a.)dG(xl, ••• ,x) - G(u ,··· ,us) •
1
J J
s
2
+
Cr )8J
US
IT
jS8
r
•
I
23
The first integral is estimated in (4.1).
In view of (1.1) the second
integral does not exceed
M • IT
j~s
1
)2
a j ~ M(2D + 2(m+l)
-s
~ 5M • 10
by corollary
s
1
(ZD + -Z""':(m=-+-l-:-) )
and the assumption on D.
Finally, by (1.1)
Substituting these inequalities into (4.3) we obtain the result
Lemma 65 We have
A=
f
f(xl) ••. f(xs)dxl ••• dx
maxlxjl~D
Proof:
=
A + B
=
fI
(m+l)s
Suppose that
_4- s + l 6
Write
and choose
a.
J
from Lemma 3s
(4.4)
~ (m+l)s - (m+l)s-l
s
')
f()d
1 f (Xl···
X
xl·.· dX < (m+l)s-l ZD
s
s
D <max x 1<
j -2'
B
H(u1-, ••• ,u -) ~
s
then the conclusion of the theorem holds.
Lemma 75
s
= u.
J
- u. + ZD (1
- D and
that a
~
J
for I
a ,
j
H(xl, ••• ,x ) ~ -4
s
-s+l
~
j
~
~
s.
6 + s~1D fM L(a.-x )
j~s J
j
j
~
Now
s).
It follows
2;
Expanding
(4.5)
(m+l)s f H(X ,
l
US
0
••
,X ) r'
s
al
(xl-a ) •. or' (x -a )dx ••• dx
l
l
as s s
s
in the same way as in (3.6) we observe that there are exactly 2s-l terms
of positive sign and 2
s-l
terms of negative sign.
Among those with positive
sign we single out the term
By (4.4) it doeR not exceed
(_4-s+l~ + sND) A + ~B •
(4.6)
Each of the other 2
factors of the form
that
-1 terms with positive sign contains at least two
f(x.- 1 - D)
J
=0
(~l'
•••
f(x j - D)
and
~j
terms we choose
to a factor
s-l
0
= f(x.
'~s)
J
with
= aj
- D).
=D
~j
otherwise.
H(xl,.oo,x ) < 10-s+l~
s
For the estimation of these
for every
corresponding
j
By Lemma 5s we can assume
for (x , .•• ,x )
1
s
E
O(~I' ... '~s).
s-l
This leads to the upper bound for the 2
-1 "positive terms ll
(4.7)
10-s+1 M + l:.B.
Each term with negative sign contains at least one factor
f(x
j
as before and observe that
This leads to the lower bound for the 2s-1
Ii
negative terms
11
- 1 - D)
=
2'5
(4.8)
-2HD A -
~B
.
Putting (4.5) - (4.8) together we obtain
J H(xl,···,x s )r'a
US
l
(xl-al) ••• r' (x -a ) ~
as s s
- 12 4-s~+ s20~1/(m+l) ~
0
The result, follows now in the usual way from (4.1), (4.2) and Lemma 4s.
To finish the proof we can assume that there exists a point
(ul'··.,u s )
s
U
€
and that max u
with
H(ul, ••• ,u )
s
= ~,
that
~ 1 - ~lOsD.
j
and treat
Choose
(m+l)s
as before.
J H(xl, •.• ,x s )r'a
(xl-a ) •.. r' (x -a )dx ••• dx
l
l
as s s
s
l
We obtain
J H(xl,··.,xs)f(xl-al)·.·f(xs-as)dxl···dxs
(4.9)
~ (~-sMD)A - ~B
US
For each of the 2
s-l
-1 remaining "positive terms" we get the lower
bound
(4.1)
For the estimate of the "negative terms" we again observe that each of
them contains at least one factor
we choose
O(~l' ••• '~)
s
a factor and
~.
OJ
+ 2D
and
= u
j
J
= a.J
with
~.
J
J
=1
otherwise.
otherwise.
f(x.- l+D) = f(x. + D)
- D for each
Moreover, put
J
j
0.
J
and accordingly
corresponding to such
=1
if ~j
=1
- D
Then a typical Ilnegative term" is bounded by
(4.11)
~
The result follows now from (4.9) - (4.11), (4.1), (4.2), Lemma 4s and the
induction hypothesis.
26
References
[l]v. Bahr, Bengt, "Hulti-dimensional integral limit theorems," Arkiv for
Mat., Vol. I, 71-88 (1969).
[2]
Chung, Kai Lai, A Course in ProbabiZity Theory, Har.court Brace and World,
1968.
[3]
Erdos. P. and Turan. P., ;IOn a problem in the theory of uniform
distribution I, Indag. Math .• 10, 370-378 (1948).
[4]
Koksma, J.F •• IlSome theorems on diophantine inequalities, Scriptum
no. 5, Math. Centrum, Amsterdam, 1950.
[5]
Kuipers, L. and Niederreiter, H., Uniform Distribution of Sequences,
Interscience Tracts, John Wiley & Sons (in preparation).
[6]
S.P. Lloyd, IiA sampling theorem for stationary (wide sense) stochastic
processes,lI Transactions AMS, 92, 1-12 (1959).
[7]
Loeve, Michel, ProbabiZity Theory, 3rd ed., Van Nostrand, 1963.
[8]
Sziisz, P., "On a problem in the theory of uniform distribution, n
(Hung.), C.R. Premier Congres Hongrois,pp. 461-472, Budapest, 1952.
[9]
Yudin, A •• nNew proof of Erdos-Turan' s theorem. II (Russian), Litovsk.
Mat. Sb., 9, 839-848 (1969).
H. Niederreiter
Department of Mathematics
Southern Illinois University
Carbondale, Ill.
Walter Philipp
Department of Mathematics
University of Illinois
Urbana, Ill.
and
Department of Statistics
University of North Carolina
Chapel Hill, N.C.
Footnote to page 2
This is l~ss restrictive than the usual assumption that IG'(x)1 ~ M.
The proofs given in [2] and [7] require no changes.
Footnote to page 3
In this context the term characteristic function usually is reserved
for the expression
f
-~
in scale.
exp(itx)dF(x).
The factor
2~
affects only a change