On the lower bound of the discrepancy of Halton’s sequence II

Abstract

Let $(H_s(n))_{n \geqslant 1}$ be an s-dimensional generalized Halton’s sequence. Let $D^{*}_N$ be the discrepancy of the sequence $(H_s(n) )_{n = 1}^{N}$. It is known that as $N \rightarrow \infty$. In this paper, we prove that this estimate is exact. Namely, there exists a constant $C(H_s)>0$ such thats

1 Introduction

Let $(\beta _{n})_{n \geqslant 1}$ be a sequence in the unit cube , ,

(1)

where $\mathbf{1}_{B_{\mathbf{y}}}(\mathbf{x}) =1$ if $\mathbf{x}\in B_{\mathbf{y}}$, and $\mathbf{1}_{B_{\mathbf{y}}}(\mathbf{x}) =0$ if $\mathbf{x}\notin B_{\mathbf{y}}$.

We define the star discrepancy of an N-point set $(\beta _{n})_{n=1}^{N}$ as

(2)

In 1954, K. Roth proved that

$$\begin{aligned} \limsup _{N \rightarrow \infty } N (\ln N)^{-s/2} D^{*}\bigl ((\beta _{n})_{n=1}^{N}\bigr )>0 . \end{aligned}$$

According to the well-known conjecture (see, e.g., [1, p. 283]), this estimate can be improved to

$$\begin{aligned} \limsup _{N \rightarrow \infty } N (\ln N)^{-s} D^{*}\bigl ((\beta _{n})_{n=1}^{N}\bigr )>0 . \end{aligned}$$

(3)

In 1972, W. Schmidt proved this conjecture for $s=1$. For $s=2$, Faure and Chaix [4] proved (3) for a class of (t, s)-sequences. See [2] for the most important results on this conjecture.

There exists another conjecture on the lower bound for the discrepancy function: there exists a constant $\dot{c}_3>0$ such that

for all N-point sets $(\beta _{k,N})_{k=0}^{N-1}$ (see [2, p. 147]).

Definition

An s-dimensional sequence $((\beta _{n})_{n \geqslant 1})$ is of low discrepancy (abbreviated l.d.s.) if for .

Let $p\geqslant 2$ be an integer,

van der Corput proved that $(\phi _p(n))_{n\geqslant 0}$ is a 1-dimensional l.d.s. (see [12]). Let

$$\begin{aligned} \widehat{H}_{s}(n)= \bigl (\phi _{\hat{p}_1}(n),\ldots ,\phi _{\hat{p}_s}(n)\bigr ),\qquad n=0,1,2,\dots , \end{aligned}$$

where $\widehat{p}_1,\ldots ,\widehat{p}_s\geqslant 2$ are pairwise coprime integers. Halton proved that $( \widehat{H}_{s}(n))_{n\geqslant 0}$ is an s-dimensional l.d.s. (see [6]). For other examples of l.d.s. see, e.g., [1, 5, 11]. In [9], we proved that Halton’s sequence satisfies (3). In this paper we generalize this result.

Let $Q=(q_1,q_2,\dots )$ and $Q_j=q_1q_2\cdots q_j$, where $q_j\geqslant 2$, $j=1,2,\dots$, is a sequence of integers. Consider Cantor’s expansion of :

The Q-adic representation of x is then unique. We define the odometer transform as

(4)

$n=2,3,\dots$, $T_Q^0(x)=x$, where .

For $Q=(q,q,\dots )$, we obtain von Neumann–Kakutani’s q-adic adding machine (see, e.g., [5]). As is known, the sequence $(T_{Q}^n(x))_{n\ge 1}$ coincides for $x=0$ with the van der Corput sequence (see, e.g., [5, Section 2.5]).

Let $q_0 \geqslant 4$, $p_{i,j} \geqslant 2$, $s\geqslant i\geqslant 1$, $j\geqslant 1$, be integers, $\mathrm{g.c.d.}(p_{i,k},p_{j,l})=1$ for $i\ne j$, $\mathscr {P}_i = (p_{i,1},p_{i,2},\dots )$, $\varvec{\mathscr {P}}=(\mathscr {P}_1,\dots ,\mathscr {P}_s)$, $T_{\varvec{\mathscr {P}}}(\mathbf{x}) = (T_{\mathscr {P}_1}(x_1),\dots ,T_{\mathscr {P}_s}(x_s))$,

We note that $H_{\varvec{\mathscr {P}}}(n) =T^n_{\varvec{\mathscr {P}}}(\mathbf{0})$ for $n=0,1,\dots$

Let be a sequence of corresponding permutations $\sigma _{i,j}$ of for $j \geqslant 1$, $\varvec{\mathrm{\Sigma }}=(\mathrm{\Sigma }_1,\dots ,\mathrm{\Sigma }_s)$, $\mathbf{x}=(x_1,\dots ,x_s)$,

We consider the following generalization of Halton’s sequence (see [3, 5, 7]):

$$\begin{aligned} H_{\varvec{\mathscr {P}}}^{\varvec{\mathrm{\Sigma }}}(n,\mathbf{x})= \widetilde{\varvec{\mathrm{\Sigma }}}(T_{\varvec{\mathscr {P}}}^n(\mathbf{x})), \qquad n=0,1,2,\dots \end{aligned}$$

We note that $(H_{\varvec{\mathscr {P}}}^{\varvec{\mathrm{\Sigma }}}(n,\mathbf{x}))_{n \geqslant 0}$ coincides for $\mathbf{x}=\mathbf{0}$ and $s=1$ with the Faure sequence $S_Q^{\mathrm{\Sigma }}$ [3]. Similarly to [11, pp. 29–31], we get that $(H_{\varvec{\mathscr {P}}}^{\varvec{\mathrm{\Sigma }}}(n,\mathbf{x}))_{n \geqslant 0}$ is of low discrepancy.

2 Theorem and its proof

In this section we will prove

Theorem

Let $s \geqslant 2$, $C_1=s q_0^{s+1} \log _2 q_0$, and . Then

(7)

This result supports conjecture (3) (see also [8, 10]).

The proof of Theorem is similar to the proof of [9, Theorem]. The main part of the proof in [9] and in this paper is the construction of the bounded vector $(y_1,\dots ,y_s)$ and the application of the Chinese Remainder Theorem. In the paper [9], we take $y_i = \sum _{j=1}^m p_i^{-\tau _{i,j}}$, $i=1,\dots ,s$, where

(8)

In this paper we take , with some special sequences $(\tau _{i,j})_{1 \leqslant i \leqslant s, j \geqslant 1}$. In order to obtain the ‘periodic’ properties similar to (8), we need a more complicated construction of $(\tau _{i,j})_{s \geqslant i \geqslant 1, j \geqslant 1}$:

• $p_{i,\tau _{i,j}}=p_{i,\tau _{i,1}}$, $j=1,2,\dots$,

• , $j=1,2,\dots$,

• , $j=1,2,\dots$,

in such a way that the sets would receive the greatest length, where , $s \geqslant i \geqslant 1$. We need all these conditions to prove statement (26).

In order to construct $(\tau _{i,j})_{1 \leqslant i \leqslant s, j \geqslant 1}$, we define auxiliary sequences $\mathscr {L}_{i,j}^{(\mathfrak m)}\!, L^{(\mathfrak m)}_i \!$, $l_{i,j}, \mathscr {F}_{i,b}^{(\mathfrak m)}\!,\dots$

2.1 Construction of the sequence $(\tau _{i,j})$

Let , . By (5), we get

Hence

Let $a_{i,j} \equiv \sigma ^{-1}_{i,j}(0) - \sigma ^{-1}_{i,j}(1) \, (\mathrm{mod}\, p_{i,j})$, , ,

It is easy to see that there exist and such that

We enumerate the set $\mathscr {L}_{i,g_{i,\mathfrak m}, \mathfrak a_{i}}^{(\mathfrak m)}$:

For we have

(10)

Let , , $\dot{p}_i =p_0/p_i \leqslant q_0^{s-1}$ and

(11)

We define $F_i, m$ and $b_i=b_i^{(\mathfrak m)}$ as follows:

(12)

We enumerate the set $F_{i,b_i}^{(\mathfrak m)}$:

Bearing in mind that and $C_1=s q_0^{s+1} \log _2 q_0$, we have

(13)

Let $\mathbf{k}=(k_1,\dots ,k_s)$, $\tau _{i, j} =l_{i,f_{i,j}}$, $\varvec{\tau }_{\mathbf{k}} =(\tau _{1, k_1},\dots , \tau _{s, k_s})$, $P_{i,k} = \widetilde{P}_{i,\tau _{i,k}}$,

(14)

Applying (10), we get $\tau _{i,m} = l_{i,f_{i,m}} \leqslant l_{i,F^{(\mathfrak m)}_i} \leqslant l_{i,L^{(\mathfrak m)}_i} \leqslant \mathfrak m$. Let $\mathbf{m}=(m,\dots ,m)$. From (5) and (14), we derive

(15)

We will need the following properties of integers $\mathfrak a_{i}$, $1 \leqslant i \leqslant s$, (see (16), (17)): By (11), we have that $(b_i,\dot{p}_i) =1$ and $(b_j,p_i) =1$ for $i\ne j$, $i,j=1,\dots ,s$. Let $c_i \equiv \prod _{1 \leqslant j \leqslant s,j \ne i} b_j \, (\mathrm{mod} \, p_i)$. According to (10), (11) and (14), we obtain

(16)

Let

. Hence

$$\begin{aligned} \frac{d_i}{\widehat{p}_i} \equiv c_i \frac{\mathfrak a_{i}}{p_i} \, (\mathrm{mod}\,1), \quad ( d_i, \widehat{p}_i)=1, \quad \widehat{p}_i >1, \qquad i=1,\dots ,s.\end{aligned}$$

(17)

2.2 Using the Chinese Remainder Theorem

Let , with , $i=1,\dots ,s$. We define the truncation

If , then the truncation is defined coordinatewise, that is, , where $\mathbf{r}=(r_1,\dots ,r_s)$.

By (6), we have

Applying (14) and the Chinese Remainder Theorem, we get

Now we will find the relation between $T_{\varvec{\mathscr {P}}}^n(\mathbf{x})$ and $H_{\varvec{\mathscr {P}}}(n)$ (see (20). It is easy to verify that if $r_i' \geqslant r_i$, $i=1,\dots ,s$, then . According to (4), we get

From (4), (6) and (18), we obtain

Hence

Let . Therefore

(20)

2.3 Construction of boundary points $y_1,\dots ,y_s$ and $u_1,\dots ,u_s$

Let $\mathbf{y}=(y_1,\dots ,y_s)$ with , and let , $k_i \geqslant 1$, $i=1,\dots ,s$, $\mathbf{k}=(k_1,\dots ,k_s)$,

(21)

We deduce

(22)

Consider the following condition:

$$\begin{aligned} H_{\varvec{\mathscr {P}}}^{\varvec{\mathrm{\Sigma }}}(n,\mathbf{x}) \in B^{(\mathbf{k})}\! . \end{aligned}$$

(23)

In order to express this condition in terms of the sequence $(H_{\varvec{\mathscr {P}}}(n))_{n \geqslant 1}$, we will construct boundary points $u_1,\dots ,u_s$. Next we will construct auxiliary sequences $\mathbf{u}^{(\mathbf{k})}\!,\check{u}^{(\mathbf{k})}\!,A_{\mathbf{k}},\dots$ Applying (18), we will get in (26) the solution of (23).

Let $\mathbf{u}=(u_1,\dots ,u_s)$, $u_i =\sum _{j \geqslant 1}^{\tau _{i,m}} u_{i,j}\widetilde{P}_{i,j}^{-1}$ with $u_{i,j} =\sigma _{i,j}^{-1}(y_{i,j})$, $u_{i,j}^{*} =\sigma _{i,j}^{-1}(0)$,

According to (9)–(14), we have $p_{i, \tau _{i,k_i}}= p_i$, $k_i=1,\dots ,m$, $i=1,\dots ,s$. By (9), we get .

From (16), we obtain , $k_i=1,\dots ,m$, $i=1,\dots ,s$. Hence

(25)

with .

Let $\mathbf{w}=(w_1,\dots ,w_s)= H_{\varvec{\mathscr {P}}}^{\varvec{\mathrm{\Sigma }}}(n,\mathbf{x})=\widetilde{\varvec{\mathrm{\Sigma }}}(T_{\varvec{\mathscr {P}}}^n(\mathbf{x}))$. We see from (21) and (24) that

Applying (18), (19), (20), (24) and (25), we have

where $v_m \equiv -W_{\mathbf{m}}(\mathbf{x}) + \check{\mathbf{u}}_{\mathbf{m}} \equiv -W_{\mathbf{m}}(\mathbf{x}) + \check{\mathbf{u}}_{\mathbf{k}} \, (\mathrm{mod} \, P_{\mathbf{k}})$ and .

Hence

(26)

2.4 Completion of the proof of Theorem

Bearing in mind that

we get that it is sufficient to find the lower bound of the main value of discrepancy function to prove Theorem.

Lemma 1

Let

(27)

Then

(28)

Proof

Let $\mathscr {H}_n =H_{\varvec{\mathscr {P}}}^{\varvec{\mathrm{\Sigma }}}(n,\mathbf{x})$. Using (26), we have

(29)

and

with $M_1 \geqslant 0$ and , $M_1,M_2 \in \mathbb Z$. From (1) and (22), we get

(30)

By (27), we obtain

(31)

Bearing in mind (29)–(30), we derive

Using (31), we have

$\square$

Lemma 2

With notations as above,

$$\begin{aligned} |\alpha _{m}| \geqslant \frac{m^s}{4p_0} \qquad \text {for}\quad m\geqslant 2p_0.\end{aligned}$$

Proof

From (16) and (25), we get

Applying (17) and (28), we derive

(32)

$( d_i, \widehat{p}_i)=1$, $\widehat{p}_i >1$, $i=1,\dots ,s$, and is the fractional part of x. We have that if $\widehat{p}_0 =\widehat{p}_1\widehat{p}_2\cdots \widehat{p}_s \not \equiv 0 \, (\mathrm{mod} \, 2)$ then $\alpha \not \equiv 1/2 \, (\mathrm{mod} \, 1)$. Let $\widehat{p}_{\nu } \equiv 0 \, (\mathrm{mod} \, 2)$ for some , and let $\alpha \equiv 1/2 \, (\mathrm{mod} \, 1)$. Then

with $a_1 = \widehat{p}_0(\widehat{p}_{\nu }/2-d_{\nu })/\widehat{p}_{\nu }$ and $a_2 = \sum _{i \ne \nu }\widehat{p}_0d_i/\widehat{p}_i$. Let and $j \ne \nu$. We see that $a_1 \equiv 0 \, (\mathrm{mod} \,\widehat{p}_j)$ and $a_2 \not \equiv 0 \, (\mathrm{mod} \, \widehat{p}_j)$. We get a contradiction. Hence $\alpha \not \equiv 1/2 \, (\mathrm{mod} \, 1)$. We have

Thus with $p_0=p_1\cdots p_s$, $(p_0,\widehat{p}_0)=\widehat{p}_0$.

Bearing in mind that $P_{ \mathbf{k}} \geqslant 2^{k_1+k_2 + \cdots +k_s}$, we obtain from (32) that

(33)

This completes the proof. $\square$

Going back to the proof of Theorem, by (7) and (13), we get

where $C_1=s q_0^{s+1} \log _2 q_0$, and $q_0^s \geqslant p_0$.

Using (15) and (26), we have that $v_m + P_{\varvec{\tau }m} \leqslant 2P_{\mathbf{m}} \leqslant N$. According to (33), (27) and (2), we obtain

Hence Theorem is proved.