An Asymptotic Distribution Function of the Three-Dimensional Shifted van der Corput Sequence ()
1. Introduction
In this paper we apply the Weyl’s limit relation [1] (p. 1-61)
(1.1)
to the sequence
, where 
is the van der Corput sequence in base 
and 
is the asymptotic distribution function (abbreviated a.d.f.) of 
and
. The van der Corput sequence in base 
is defined as follows: Let 
be the 
-adic expression of a positive integer
. Then
(1.2)
It is well-known that this sequence is uniformly distributed (abbreviated u.d.), see [1] (2.11, p. 2-102), [2] (Theorem 3.5, p. 127), [3] (p. 41).
For 
a motivation for the study of the distribution function (abbreviated d.f.) 
of
, 
is a result of Pillichshammer and Steinerberger in [4] which states that
(1.3)
while in J. Fialová and O. Strauch [5] the relation (1.3) was proved applying (1.1) as
Moreover, in the Unsolved Problems [6] (1.12), the following problem is stated: Find the d.f. 
of the sequence
, 
, in
. Ch. Aistleitner and M. Hofer [7] gave the following theoretical solution:
Theorem 1 Let 
denote the von Neuman-Kakutani transformation described in Figure 1. Define the 
- dimensional curve
, where
. Then the searched a.d.f. is
where 
is the Lebesgue measure of a set
.
The paper consists of the following parts: After definitions (Part 2) we derive the a.d.f. of 
Figure 1. Line segments containing 
The graph of the von Neumann-Kakutani transformation
.
Haoshangban (Part 3), the a.d.f. of 
(Part 4), intervals containing 
in diagonals (Part 5) and an explicit form of a.d.f. 
(Part 6). As an application (Part 7) we compute the limit
(1.4)
for
, 
and
, respectively, see (0.39), (0.41) and (0.46).
2. Definitions and Notations
Let
, 
be a sequence in the unit interval
. Denote
the step distribution function (step d.f.) of the finite sequence 
in
, while
.
A function 
is a distribution function (d.f.) if
(i) 
is nondecreasing;
(ii) 
and
.
A d.f. 
is a d.f. of the sequence
, 
if an increasing sequence of positive integers 
exists such that 
a.e. on
.
A d.f. 
is an asymptotic d.f. (a.d.f.) of the sequence
, 
if 
a.e. on
.
The sequence 
is uniformly distributed (abbreviating u.d.) if its a.d.f. is
.
Similar definitions take place for 
and 
-dimensional sequence
, 
, in
, cf. [1] (1.11, pp. 1-60).
In the sequel the 
-dimensional interval 
we denote by
, where 
are projections on 
axes, respectively.
3. a.d.f. of 
Let 
be an integer.
Lemma 1 Every point
, 
, lies on the diagonals of intervals
(1.5)
(1.6)
Proof. Express an integer 
in the base 
where 
and
. We consider the following two cases:
,
.
Let
Then
,
and by (0.2)
. In this case
Thus such 
lies on the line-segment
(1.7)
Let
Then
and
, where
. Then
. Thus
,
, and we have
and
and
. Thus such 
lies on the segment
(1.8)
Thus, for
, terms of the sequence 
lie on the diagonal of the interval
(1.9)
and for
, after reduction
, terms of the sequence 
lie on the diagonals of the intervals
(1.10)
These intervals are maximal with respect to inclusion.
Adding the maps (1.7) and (1.8) we found the so-called von Neumann-Kakutani transformation
, see Figure 1. Because 
is u.d., the sequence 
has a.d.f. 
of the form1
(1.11)
where 
is the projection of a two dimensional set to the 
-axis.
The sum (1.11) implies
(1.12)
From (1.12) it follows
(1.13)
and for
, the mean equality misses.
4. a.d.f. of 
Let 
be an integer.
Lemma 2 All terms of the sequence
, 
, lie in the diagonals of the following intervals
(1.14)
(1.15)
(1.16)
Proof. Express an integer 
in the base 
(1.17)
where 
and
. We consider three following cases:
,
,
.
Let
Then
,
and
. In this case
and thus such 
lies on the line-segment
(1.18)
Let
. Then
and
, then
. Thus
,
, and we have
.
Furthermore
and
.
Thus in this case 
lies on the line-segment
(1.19)
Let
.
Then
and
, then
. Thus
,
, and we have
.
Furthermore
and
.
This gives
(1.20)
Summary, if the 
satisfies
, then 
is contained in the diagonal of
(1.14)
for 
in the diagonal of
(1.15)
and for 
in the diagonal of
(1.16).
Proof. Express an integer 
in the base 
(1.17)
where 
and
. We consider three following cases:
, 
,
.
Let
Then
, 
and
. In this case
and thus such 
lies on the linesegment
(1.18)
Let
. Then 
and
, then
. Thus
,
, and we have
.
Furthermore
and
.
Thus in this case 
lies on the line-segment
(1.19)
Let
. then 
and
, then
. Thus
,
, and we have 
.
Furthermore
and
.
This gives
(1.20)
Summary, if the 
satisfies
, then 
is contained in the diagonal of
(1.14) for 
in the diagonal of
(1.15)
and for 
in the diagonal of
(1.16).
Composition of the maps (0.18), (0.19) and (0.20) of 
forms the second iteration 
of the von Neumann-Kakutani transformation
. The diagonals of (1.14), (1.16) and (1.15) yield the following graph of 
in Figure 2.
Here the interval 
on 
-axis is decomposed in
, 
, and the interval 
is decomposed in
,
. On 
-axis the interval 
is decomposed in
, 
and the interval 
is decomposed in
,
. Note that for
, the interval 
has a zero length and is missing.
Exchange for a moment the axis 
by
. Similarly as in (1.11), we have that the a.d.f. 
of the sequence 
is
(1.21)
Decompose 
as the following figure shows:
Then by Figure 3 we have
Figure 3. Decomposition of the Vunit square to parts with fixed expression of
.
(1.22)
Let
.
In this case we find a.d.f 
from (1.22) omitting 
In Part 7. Applications we need to find 
from 
in (1.22):
For 
(1.23)
For 
(1.24)
For 
(1.25)
Note that for 
the term 
is omitted.
5. a.d.f. of 
Let 
be an integer.
Lemma 3 Every point 
is contained in diagonals of the intervals
(1.26)
(1.27)
(1.28)
where 
if
. These intervals are maximal with respect to inclusion.
Proof. Every maximal 
-dimensional interval 
containing points 
will be written as
, where 
are projections of 
to the
, axes, respectively. Moreover if 
then 
and
. From u.d. of 
follows that the lengths
. Combining intervals (1.5), (1.14), (1.15), (1.16), (1.6) of equal lengths by following Figure 3.
We find (1.26), (1.27), and (1.28).
Now, let 
be the union of diagonals of (1.27), (1.28) and (1.26). Again, as in (1.11), the a.d.f. 
has2 the form
(1.29)
and it can be rewritten as
(1.30)
To calculate minimums in (1.30) we can use the following Figure 4 (here
):
As an example of application of (1.30) and Figure 4, we compute 
for 
without using the knowledge of
,3
(1.31)
Proof.
1. Let
.
Then
, 
, 
, consequently
.
2. Let
. Then
, 
, 
, consequently
.
3. Let
. Then
, 
, consequently
.
4. Let
.
Specify
,
. Then
, 
for
. Thus (1.30) implies
For 
we have
(1.32)
6. Explicit Form of 
Let 
be an integer.
Motivated by the Figure 4 we decompose the unit interval 
on
, 
and 
axes in the Figure 5
Figure 4. Projections of intervals 
on axes
.
Figure 5. Divisions of the unit intervals.
intervals (here
):
In this decomposition, for
, we have 
possibilities. We shall order choices of 
from the left to the right. Detailed proofs are included only in non-trivial cases.
1. Let
. Then
.
Proof. We have
, 
,
. Then, by (1.30),
.
Similarly, in the following cases 2-9.
2. Let
. Then
.
3. Let
. Then
.
4. Let
. Then
.
5. Let
. Then
.
6. Let
. Then
.
7. Let
. Then
.
8. Let
. Then
.
9. Let
. Then
.
Proof. We use
.
10. Let
. Then
.
Proof. We use
, 
,
. Similarly11. Let
. Then
.
12. Let
. Then
.
Proof. We use
, 
,
.
13. Let
. Then
.
14. Let
. Then
.
15. Let
. Then
.
16. Let
.
Specify
, 
,
. Then
Proof. First observe that
, and
. Thus
and 
Further, for
, we have
Thus, using (1.30), we find
.
17. Let
.
Specify
,
. Then
Proof. We have
, 
, 
, then
18. Let
.
Specify
,
. Then
Proof. We have
19. Let
. Then
.
20. Let
.
Specify
, 
,
. Then we have
Proof. We have
, 
and 
21. Let
.
Specify
,
. Then
Proof.
, 
, 
, 
22. Let
. Then
.
23. Let
.
Specify
,
. Then
Proof.
, 
, 
, 
24. Let
. Then
.
Proof. We have
. Specify
. Then
The first term is 
and the second is
.
25. Let
.
Specify
,
. Then
Proof.
26. Let
.
Specify
, 
and
. Then
Proof. We have
, 
, 
,
. Moreover
which gives
.
The final equation holds if 
and
. It can be seen that it holds also for 
and
. For 
and 
we need to compute this sum separately.
27. Let
. Then
.
Proof. First observe
, 
,
. Thus
.
New specify 
and
. Then we have
and
, 
,
. Then for the sums in (1.30) we have
which gives
.
The above computation of 
holds for
.
Let
.
We have 
and
(1.33)
Thus 
for 
directly follows from 
for 
if we use only such items in 
for which
, 
,
. These are
, i.e.,
.
The non-zero values of 
can also be seen in the following table.
In all other cases
.
7. Applications
The knowledge of the a.d.f. 
of the sequence 
allows us to compute the following limit by the Weyl limit relation (1.1) in dimension
.
(1.34)
where 
is an arbitrary continuous function defined in
. For computing (1.34) we use the following two methods.
7.1. Method I
In the first method in the Riemann-Stieltjes integral (1.34) we apply integration by parts.
Lemma 4 Assume that 
is a continuous in 
and 
is a.d.f. Then
(1.35)
Here
(1.36)
Note that
if the partial derivatives exist.
Exercise 1 Put
. We have
,
The differential 
is non-zero if and only if 
and in this case
.
Proof: For every interval 
and every continuous 
the differential 
is defined as
(1.37)
Putting
, 
, 
we have
Then by (1.35)
(1.38)
For 
and by (1.31) we have
For 
and by (1.32) we have
Therefore for
, by (1.34) and by (1.38) we have
(1.39)
Note that the same result follows from (1.44).
Exercise 2 Put
. Since
, 
if
, 
if
, 
if
, and 
if 
and 
otherwiseapplying (1.34) and (1.35) we have
(1.40)
Here we have 
in (1.13) for
. For 
we use 
in (1.23) if
, (1.24) if 
and (1.25) if 
and for 
we use (1.31) if 
and (1.32) if
. Thus we have:
a) 
for
;
b) 
for
;
c) 
for
;
d) 
for
.
e) 
for
;
f) 
for
.
Putting a)-f) into (1.40) yields
(1.41)
7.2. Method II
In the second method we compute the differential 
directly. It is nonzero only for
. For such 
the a.d.f. 
has the form
(1.42)
Thus 
and moreover 
only on the following straight lines
Considering three possible cases, calculate
(1.43)
Summary
(1.44)
Exercise 3 Put
. By Method I, we have
, similarly
, etc., and by (1.35) we have
(1.45)
Since a computation of (1.45) is complicated we use Method II, by (1.44) we have
Inserting these formulas into (1.44) we have
(1.46)
for
.
8. Conclusion
The problems solved in this paper is significantly more complicated in higher dimensions
. For example, in dimension
, to compute the d.f. 
of the sequence
, it is necessary to investigate 
cases analogous to 
cases for the explicit form 
in the part 0.6. Also Figure 3 would have to be converted to the dimension
. Finally, we would need the third iteration of von Neumann-Kakutani transformation.
Acknowledgements
The paper is sponsored by the project P201/12/2351 of GA Czech Republic.
NOTES
1
is a copula.