Metric discrepancy results for geometric progressions with large ratios

Abstract

For geometric progressions with common ratios greater than 4, the speed of convergence to the uniform distribution is determined for almost all initial values.

1 Introduction

A sequence $\{x_k\}$ of real numbers is said to be uniformly distributed mod 1 if

$$\begin{aligned} \frac{1}{N} {}^\# \{ k\le N \mid \langle x_k \rangle \in [a, b)\} \rightarrow b-a, \quad (N\rightarrow \infty ), \end{aligned}$$

for all $0\le a< b< 1$, where $\langle x\rangle$ denotes the fractional part $x - [x]$ of a real number x. Since the convergence is uniform in a and b, the following discrepancy is used to measure its speed.

$$\begin{aligned} D_N\{x_k \}&= \sup _{0\le a< b< 1} \left| \frac{1}{N} {}^\# \{ k\le N \mid \langle x_k \rangle \in [a, b)\} - (b-a) \right| . \end{aligned}$$

For an arithmetic progression $\{n_k\}$, the order of convergence of $D_N\{n_k x\}$ was studied by Khintchin [9] and Kesten [8]. For uniformly distributed i.i.d. $\{U_k\}$, Chung-Smirnov theorem asserts the law of the iterated logarithm

$$\begin{aligned} \varlimsup _{N\rightarrow \infty } \frac{ND_N \{U_k\}}{\sqrt{2N\log \log N}} = \frac{1}{2} ,\quad \hbox {a.s.} \end{aligned}$$

By various studies on lacunary series, it is known that a sequence $\{n_kx \}$ behaves like uniformly distributed i.i.d. when $\{n_k\}$ diverges rapidly. Actually Philipp [10] followed Takahashi’s method [11] and proved the result below by assuming the Hadamard gap condition $n_{k+1}/ n_k \ge q > 1$.

$$\begin{aligned} \frac{1}{4\sqrt{2}} \le \varlimsup _{N\rightarrow \infty } \frac{ND_N \{n_k x\}}{\sqrt{2N\log \log N}} \le \frac{1}{\sqrt{2}} \left( 166 + \frac{664}{q^{1/2} -1}\right) , \quad \hbox {a.e.} \end{aligned}$$

Dhompongsa [3] proved that the limsup equals to $\tfrac{1}{2}$ when $\{n_k\}$ satisfies very strong gap condition $\lim _{k\rightarrow \infty }({\log (n_{k+1}/ n_k)})/{ \log \log k} = \infty$. Beside of these results, any concrete value of limsup for exponentially growing sequence was not determined before the recent results below on divergent geometric progressions $\{\theta ^k x\}$.

Theorem 1

[4–7] For any $\theta \notin [-1,1]$, there exists a constant $\varSigma _\theta \ge 1/2$ such that

$$\begin{aligned} \varlimsup _{N\rightarrow \infty } \frac{ND_N \{\theta ^k x\}}{\sqrt{2N\log \log N}} = \varSigma _\theta ,\quad \mathrm{a.e. } \end{aligned}$$

If $\theta ^j \notin \mathbf{Q}$ for any $j \in \mathbf{N}$, then $\varSigma _\theta = \frac{1}{2}$, and otherwise $\varSigma _\theta > \frac{1}{2}$. When $\theta$ satisfies $\theta ^j \in \mathbf{Q}$ for some $j \in \mathbf{N}$, we take p, q, and r as below.

$$\begin{aligned} \theta ^r= p/q \quad \mathrm{where} \quad r=\min \{j\in \mathbf{N}\mid \theta ^j\in \mathbf{Q}\}, \quad p\in \mathbf{Z},\quad q\in \mathbf{N}, \quad \gcd (p,q)=1. \end{aligned}$$

(1)

If p and q are odd, then

$$\begin{aligned} \varSigma _\theta = \frac{1}{2}\sqrt{\frac{|p|q+1}{|p|q-1}}. \end{aligned}$$

If $q=1$, then

$$\begin{aligned} \varSigma _\theta = \frac{1}{2}\sqrt{\frac{|p|+1}{|p|-1}}, \quad \frac{1}{2}\sqrt{ \frac{(|p|+1)|p|(|p|-2)}{(|p|-1)^3}}, \quad \frac{1}{9}{\sqrt{42}}, \quad \mathrm{or}\quad \frac{1}{49}{\sqrt{910}}, \end{aligned}$$

according as p is odd, $|p|\ge 4$ is even, $p=2$, or $p=-2$. If $p=\pm 5$ and $q=2$, then

$$\begin{aligned} \varSigma _\theta = \frac{1}{9}{\sqrt{22}} . \end{aligned}$$

For related works, see [1, 2]. In this paper, we prove the next result and determine $\varSigma _\theta$ when $\theta$ is large.

Theorem 2

Suppose that $\theta$ satisfies (1). If p is odd, q is even and $|p/q|\ge 9/4$, or if p is even, q is odd and $|p/q|\ge 4$, then

$$\begin{aligned} \varSigma _{\theta }= & {} \left( \frac{(|p|q)^{I}+1}{(|p|q)^{I}-1} v\left( \frac{|p|-q-1}{2(|p|-q)}\right) \right. \nonumber \\&\left. +\, \frac{2(|p|q)^{I}}{(|p|q)^{I}-1} \sum _{m=1}^{I-1} \frac{1}{(|p|q)^{m}} v\left( q^m\frac{|p|-q-1}{2(|p|-q)}\right) \right) ^{1/2}, \end{aligned}$$

(2)

where $v(x)=\langle x \rangle (1-\langle x\rangle )$ and $I=\min \{ n\in \mathbf{N}\mid q^n = \pm 1 \mod |p|-q\}$.

By (2), we can calculate the concrete values of $\varSigma _\theta$ in the following way.

$$\begin{aligned}&\varSigma _{\pm 7/2}= \frac{1}{5}\sqrt{\frac{1294}{195}}, \quad \varSigma _{\pm 9/2}= \frac{2}{49}\sqrt{\frac{18561}{119}},\\&\quad \varSigma _{\pm 14/3}= \frac{1}{11}\sqrt{\frac{4096559910}{130691231}},\quad \varSigma _{\pm 9/4}= \frac{1}{5}\sqrt{\frac{222}{35}},\dots \end{aligned}$$

When $q=1$ and $|p|\ge 4$ is even, we have $I=1$, and (2) gives the values stated in Theorem 1. It also gives the value $\varSigma _{\pm 5/2}$ in Theorem 1.

We here emphasize that the values of $\varSigma _\theta$ are determined for large $\theta$, say, that satisfying $|\theta |\ge 4$. On the other hand, we can find some smaller p / q for which (2) fails to hold. $\theta =\pm 2$ is one such example.

Before closing this section, we note that the conditions $|p/q|\ge 9/4$ and $|p/q|\ge 4$ are superfluous, and much weaker conditions are sufficient. Set

$$\begin{aligned} A_1(p,q):= & {} 2p^3-(4q+2)p^2-(q^2-2q)p+3q^2,\\ A_2(p,q):= & {} (8 q^2 - 6)p^3 +(- 16 q^3 + 4 q + 24)p^2 +(- q^4 + 10 q^2 - 24 q - 24)p +q^3,\\ A_3(p,q):= & {} p^3-(4q+1)p^2+2pq+2q^2\\ A_4(p,q):= & {} (6 q^2 - 4) p^6 +(- 4 q^3 + 8 q^2 + 2 q + 16)p^5 +(- 25 q^4 - 24 q - 16)p^4\\&+\,(12 q^5 - 24 q^4 \!+\! 25 q^3 \!+\! 8 q^2 \!-\! 2 q)p^3 +(- 4 q^6 + 16 q^5 - 12 q^4 + 32 q^3 - 8 q^2)p^2\\&+\,(4 q^5 - 32 q^4 - 10 q^3)p -4q^4. \end{aligned}$$

When p is odd and q is even, if the conditions $A_1(|p|, q) >0$, $A_2(|p|, q)>0$, and

$$\begin{aligned} |p/q| \ge 2, \end{aligned}$$

(3)

are satisfied, then we have (2). The condition $|p/q|> 9/4$ implies these conditions. Although $A_1(9, 4)< 0$ and $A_2(9, 4)>0$, we can still prove (2) for $p/q=\pm 9/4$. We here note that $|p/q|\ge (2+\sqrt{6})/2 + 1/q$ also implies $A_1(|p|, q) >0$ and $A_2(|p|, q)>0$.

When p is odd and q is even, if (3), $A_3(|p|, q)>0$ and $A_4(|p|, q)>0$ are satisfied, then (2) holds. The condition $|p/q|\ge 4$ implies these conditions.

2 Preliminary

We prepare some results. Proofs can be found in [4, 6, 7]. For a, b, $a'$, $b'\in [0,1)$, put

$$\begin{aligned} V(a, a') \!=\! a\wedge a' - aa' , \quad \widetilde{V}(a, b, a', b') \!=\! V(a, a') \!+\! V(b, b') - V(a, b') - V(b, a') \end{aligned}$$

and

$$\begin{aligned} \sigma _{p/q}^2(a,b)&= \widetilde{V}(a,b,a,b) + 2 \sum _{k=1}^\infty \frac{1}{(p q)^{k} } \widetilde{V}(\langle p^{k} a\rangle , \langle p^{k} b\rangle , \langle q^{k} a\rangle , \langle q^{k} b\rangle ). \end{aligned}$$

(4)

Here we list general properties that we use.

$$\begin{aligned} \widetilde{V}(a, b, a', b')= & {} \widetilde{V}(a', b', a, b) = - \widetilde{V}(b, a, a', b') = - \widetilde{V}(a, b, b', a'), \end{aligned}$$

(5)

$$\begin{aligned} 0\le & {} \widetilde{V}(0, b, 0, b') = V(b, b') \le V(b, b) \le 1/4 , \end{aligned}$$

(6)

$$\begin{aligned} \widetilde{V}(a, b, a, b)= & {} |b-a| - |b-a| ^2 = v(b-a) = v(a-b), \end{aligned}$$

(7)

$$\begin{aligned} \sigma ^2_{p/q}(0,a)= & {} V(a,a) +2\sum _{k=1}^\infty \frac{1}{(p q)^{k} } V(\langle p^k a\rangle , \langle q^k a\rangle ) . \end{aligned}$$

(8)

When $\theta$ satisfies (1), $\varSigma _\theta$ does not depend on r and is given by

$$\begin{aligned}&\varSigma _\theta = \varSigma _{p/q} = \sup _{0\le a'< a\le 1} \sigma _{p/q}(a',a), \quad \hbox {and}\nonumber \\&\quad \varSigma _\theta = \varSigma _{p/q} = \sup _{0\le a\le 1/2} \sigma _{p/q}(0,a) \quad \hbox {if}\quad p>0. \end{aligned}$$

(9)

Put $b_i:= \frac{i}{p-q}$, $c_1:= \frac{q-2}{2q}$, $c_2 := \frac{p-3}{2p}$, $c_3 := \frac{p-1}{2p}$, $c_4 := b_{(p-q-1)/2} = \frac{(p-q-1)/2}{p-q}$, $c_5 := \frac{q-1}{2q}$, $c_6 := \frac{p-2}{2p}$, and $c_7 := b_{(p-q-3)/2} = \frac{(p-q-3)/2}{p-q}$. When p and q are positive and satisfying $p/q\ge 2$, one can verify $c_1<c_2 < c_4 < c_3$ and $c_5 \le c_6 \le c_4$. By $p=q$ mod $p-q$, we have

$$\begin{aligned} \langle p^k b_i\rangle = \langle q^k b_i\rangle \quad \hbox {and}\quad V\left( \langle p^k b_i \rangle , \langle q^k b_i \rangle \right) = v(p^k b_i) = v(q^k b_i). \end{aligned}$$

(10)

Thanks to $v(-x)= v(x)$, we have $v(q^{m+nI} c_4) = v(\pm q^{m} c_4)=v(q^{m} c_4)$. Hence $\sigma _{p/q}(0,c_4)$ equals to the left hand side of (2).

In the next two sections, we prove $\varSigma _{p/q} = \sigma _{p/q}(0,c_4)$ in the case p is positive and p / q is large. By assuming this, we here prove $\varSigma _{-p/q}= \varSigma _{p/q}$. In [6] we proved $\varSigma _{-p/q} \le \varSigma _{p/q}$. If we find $0\le \widehat{b}< \widetilde{b}\le 1$ with $\sigma _{-p/q}(\widehat{b}, \widetilde{b}) = \sigma _{p/q}(0, c_4)=\varSigma _{p/q}$, we have the equality.

We put $\widehat{b} = (1-c_4)p/({p+q})$ and $\widetilde{b} = \widehat{b} + c_4$. It holds that $\left\langle (-p)^{n} \widehat{b} \,\right\rangle =\left\langle q^{n} \widehat{b} \,\right\rangle$ and $\left\langle (-p)^{n} \widetilde{b} \,\right\rangle =\left\langle q^{n} \widetilde{b} \, \right\rangle$ if n is even, and that $\left\langle (-p)^{n} \widehat{b} \,\right\rangle =\left\langle q^{n} \widetilde{b} \,\right\rangle$ and $\left\langle (-p)^{n} \widetilde{b} \,\right\rangle =\left\langle q^{n} \widehat{b} \,\right\rangle$ if n is odd. Actually, we have $(p+q)\widehat{b} =p-p c_4 = -p c_4 = -qc_4$ mod 1, which implies $-p \widehat{b} = q\widetilde{b}$, $-p\widetilde{b} = q\widehat{b}$ mod 1, and $(-p)^2 \widehat{b} = -pq\widetilde{b} = q^2 \widehat{b}$ mod 1, and so on. By noting (5), (7), and $\left\langle \langle q^{n} \widetilde{b} \,\rangle - \langle q^{n} \widehat{b} \,\rangle \right\rangle = \langle q^{n} c_4\rangle$, we have

$$\begin{aligned} \widetilde{V}(\langle (-p)^{n} \widehat{b} \,\rangle , \langle (-p)^{n} \widetilde{b} \,\rangle , \langle q^{n} \widehat{b} \rangle , \langle q^{n} \widetilde{b}\rangle )= & {} - \widetilde{V}(\langle q^{n} \widehat{b} \,\rangle , \langle q^{n} \widetilde{b} \,\rangle , \langle q^{n} \widehat{b} \rangle , \langle q^{n} \widetilde{b}\rangle )\\= & {} -v(q^{n} c_4) \end{aligned}$$

for odd n, and

$$\begin{aligned} \widetilde{V}(\langle (-p)^{n} \widehat{b} \,\rangle , \langle (-p)^{n} \widetilde{b} \,\rangle , \langle q^{n} \widehat{b} \rangle , \langle q^{n} \widetilde{b}\rangle ) =v(q^{n} c_4) \end{aligned}$$

for even n, to have $\sigma _{-p/q}(\widehat{b}, \widetilde{b}) = \sigma _{p/q}(0, c_4)$.

We prepare some inequalities to prove $\varSigma _{p/q} = \sigma _{p/q}(0,c_4)$. We denote $\sigma _{p/q}(0,a)$ simply by $\sigma (a)$.

Put

$$\begin{aligned} X_N(a):= & {} V(a, a) + 2\sum _{n=1}^N \frac{1}{(pq)^n} V\left( \langle p^n a\rangle , \langle q^n a\rangle \right) , \quad Y_N(a) := 2\sum _{n=N+1}^\infty \frac{v(q^n a)}{(pq)^n},\\ T_N:= & {} 2\sum _{n=N+1}^\infty \frac{1}{4(pq)^n}, \end{aligned}$$

$S_t(a):=2v(t^2a)/{(pq)^2}$, and $Z(a) := V(a, a) + 2 v(qa)/{pq}$. Thanks to (6) and (10), we have

$$\begin{aligned}&V\left( \langle p^k a\rangle , \langle q^k a\rangle \right) \le v(q^k a), \quad V\left( \langle p^k a\rangle , \langle q^k a\rangle \right) \le v(p^k a),\\&\quad V\left( \langle p^k a\rangle , \langle q^k a\rangle \right) \le 1/4, \end{aligned}$$

and

$$\begin{aligned} \sigma ^2(a) \!\le \! W_j(a) \quad (1\le j\le 6), \quad \hbox {and}\quad W_1(b_i) \!=\! W_2(b_i) \!=\! \sigma ^2(b_i)\quad (0\le i\le p-q), \end{aligned}$$

(11)

where

$$\begin{aligned} W_1&:= X_1 + Y_1,&W_2&:= X_2 + Y_2,&W_3&:= X_1 + T_1,\\ W_4&:= Z + S_q + T_2,&W_5&:= Z + S_p + T_2,&W_6&:= Z + T_1. \end{aligned}$$

By $\gcd (q, p-q)=1$, $\langle p^k c_4\rangle = \langle q^k c_4\rangle \in \{b_1, \dots , b_{p-q-1}\}$ and $v(b_i)\ge v(b_1)$, we have

$$\begin{aligned} \sigma _{}^2 (c_4) \ge Z(c_4) + 2\sum _{n=2}^\infty \frac{1}{(pq)^n}v(b_1) =:U. \end{aligned}$$

Note that the evaluation of U strongly depends on the parity of p, since $\langle qc_4\rangle = qc_4 - (q-2)/2$ or $\langle qc_4\rangle = qc_4 - (q-1)/2$ according as q is even or odd.

We denote the derivative $\frac{d}{da}f$ of f by Df, the right derivative by $D^+f$, and the left derivative by $D^-f$.

3 Odd p and even q

Assume that q is even, and that p is odd and positive. We divide [0, 1 / 2) into subintervals $[ 0, c_1)$, $[ c_1, c_2)$, $[c_2, c_3)$, and $[c_3, 1/2)$, and prove $\sigma ^2(a) \le \sigma ^2(c_4)$ on each.

3.1 $[c_2, c_3)$ part

We assume (3), $A_1(p,q)>0$, and $a\in [c_2, c_3)$. We have $\langle pa\rangle = pa- (p-3)/2$ and $\langle qa\rangle = qa - (q-2)/2$. Since $\langle pa\rangle <\langle qa\rangle$ holds on $[c_2, c_4)$, and $\langle pa\rangle \ge \langle qa\rangle$ on $[c_4, c_3)$, we can evaluate $X_1$ as

$$\begin{aligned} X_1(a)&= {\left\{ \begin{array}{ll} -3a^2 +(3-3/p)a +3/2p-1/2 &{} a \in [c_2, c_4],\\ -3a^2 +(3-2/q-1/p)a-1/pq+1/q+1/2p-1/2 &{} a \in [ c_4, c_3), \end{array}\right. } \end{aligned}$$

and verify that $D X_1(a)$ decreases on $[c_2, c_3)$. We also have $|D v(q^n a)| \le q^n$ a.e. a, and therefore $|D Y_1| \le 2/{p(p-1)}$ a.e. By combining these, we have

$$\begin{aligned} D W_1(a)> & {} D^- X_1(c_4) - 2/p(p-1)\\= & {} (3pq-2p-q)/{p(p-1)(p-2)} > 0 \quad \hbox { a.e. on }\quad (c_2, c_4], \end{aligned}$$

and

$$\begin{aligned} D W_1(a)< & {} D^+ X_1(c_4) + 2/p(p-1)\\= & {} -A_1(p,q)/p(p-1)(p-2) < 0 \quad \hbox { a.e. on }\quad [ c_4, c_3). \end{aligned}$$

Hence $\sigma _{}^2(a) \le W_1(a) \le W_1(c_4) = \sigma _{}^2(c_4)$ for $a\in [c_2, c_3)$. Here we note that functions appearing here are bounded and absolutely continuous, and the exceptional set of null measure for $D W_1(a) > 0$ or $D W_1(a) < 0$ does not harm the argument to show that $c_4$ is the maximal point.

Here we prove $A_1(p,q)>0$ by assuming $p/q> 9/4$. Since $2p^2 -(4q+2)p + (q^2-2q)$ is increasing in $p\ge (9/4)q$, $A_1(p,q)$ also. Because of $A_1((9/4)q, q) = 3q^2(3q-28)/32>0$ if $q\ge 10$, we see that $A_1(p,q)>0$ if $q\ge 10$ and $p/q> 9/4$. Thanks to $A_1(5, 2)> 0$, $A_1(11, 4)>0$, $A_1(17,6)>0$ and $A_1(19,8)>0$, we see that $A_1(p,q)>0$ if $p/q> 9/4$.

We prove the case $p/q=9/4$. Due to the above proof, we see that $\sigma _{}^2(a) \le \sigma _{}^2(c_4)$ for $a\in [c_2, c_4]$. Since we have $T_1= \frac{1}{2520}$ and $X_1$ is decreasing in $a\ge c_4$, we have $\sigma ^2(a) \le W_3(a) \le X_1(\frac{34}{81})+\frac{1}{2520} < \frac{222}{875}= \sigma ^2(c_4)$ on $[\frac{34}{81}, c_3)$. On $[c_4, \frac{33}{81})$, we have $\langle 16 a\rangle \le \langle 81 a\rangle$ and see $D X_2$ is decreasing and $|D Y_2(a)| \le \frac{1}{324}$. Hence we have $D W_2(a) \le D X_2(c_4) + \frac{1}{324}< 0$ and $\sigma ^2(c_4) = W_2(c_4) \ge W_2(a)\ge \sigma ^2(a)$ on $[c_4, \frac{33}{81})$. Note that $Y_2\le \frac{1}{90,720}$. On $[\frac{33}{81}, \frac{27}{65})$, by $\langle 16 a\rangle \ge \langle 81 a\rangle$ we see that $X_2$ equals to a quadratic function having axis at $\frac{881}{2160}$. Hence $\sigma ^2(a) \le X_2(\frac{881}{2160}) + \frac{1}{90,720}< \sigma ^2(c_4)$. On $[\frac{27}{65}, \frac{34}{81})$, by $\langle 16 a\rangle < \langle 81 a\rangle$, $X_2$ is decreasing. Hence $\sigma ^2(a) \le X_2(\frac{27}{65})+ \frac{1}{90,720}< \sigma ^2(c_4)$ for $a\in [\frac{27}{65}, \frac{34}{81})$.

3.2 $[c_3, 1/2)$ part

Let $a\in [c_3, 1/2)$. We have $\langle pa\rangle = pa-(p-1)/2$ and $\langle qa\rangle = qa-(q-2)/2$. We see $\langle qa\rangle - \langle pa\rangle = -(p-q)a + (p-q+1)/2 > 1/2$. Since $W_3(a)$ maximizes at $a_1:= (3p-1)/6p$, we have $W_3(a) \le W_3(a_1)$. By

$$\begin{aligned} U- W_3(a_1) = A_2(p,q)/(12 p^2 q (p-q)^2 (p q - 1))> 0, \end{aligned}$$

we have $\sigma _{}^2(a) < U_{}$ for $a\in [c_3, 1/2)$.

We derive $A_2(p,q)>0$ from $p/q\ge 9/4$. Put $\widehat{A}_2(p,q)=(8 q^2 - 6)p^2 +(- 16q^3+4q)p -q^4$. We see

$$\begin{aligned} A_2(p,q) = p\widehat{A}_2(p,q) + (24p^2-24qp-24p) + (10q^2p+q^3) > \widehat{A}_2(p,q). \end{aligned}$$

We see $\widehat{A}_2(p,q)$ is increasing in p if $p\ge 2q$. By $\widehat{A}_2((9/4)q, q)= q^2(28q^2-171)/8>0$ ($q\ge 4$), and by $\widehat{A}_2(5,2)>0$, the proof is over.

3.3 $[0, c_1)$ part

We consider on $[0, c_1)$. We assume $q\ge 4$, since $c_1=0$ otherwise. Since V(a, a) increases on $[0, c_1)$ and $W_4(a)-V(a,a)$ has period 1 / q, the first equality of

$$\begin{aligned} \sup _{a\in [0, c_1)} W_4(a) = \sup _{a\in [c_1-1/q, c_1)} W_4(a) = \sup _{a\in [c_1-1/q^2, c_1)} W_4(a) = W_4(c_1) < U \end{aligned}$$

(12)

holds. On $[c_1-1/q, c_1)$, we have $\langle q a\rangle = qa-(q-4)/2$, and we see Z equals to a quadratic function having axis at $a_2:=(2q+p-6)/(4q+2p)$. By $a_2-c_1 = (p-q)/q(2q+p)>0$, we see that it is increasing on $[c_1-1/q, c_1)$. Since $S_q(a)$ has period $1/q^2$, we verified the second equality of (12). On $[c_1-1/q^2, c_1)$, we have $\langle q^2 a\rangle = q^2 a-q(q-2)/2+1$. We see that $W_4(a)$ equals to a quadratic function with axis $a_3:=({2 q^2 + (2 p - 4) q + p^2 - 6 p - 2})/({4 q^2 + 4 p q + 2 p^2})$. By

$$\begin{aligned} a_3-c_1\!=\! (p^2-pq-q)/(4 q^2 \!+\! 4 p q \!+\! 2 p^2)> p(p-2q)/(2 q^2 + 2 p q + 1 p^2)\ge 0, \end{aligned}$$

we have the third equality. The rest is proved by

$$\begin{aligned}&4 p^2 q^2 (p - q)^2 (p q - 1) (U_{} - W_4(c_1) )\\&\quad =4 qp^5 +(- 8 q^2 - 4)p^4 +(7 q^3 + 8 q)p^3 +(- 6 q^4 - 7 q^2 + 8 q - 2)p^2\\&\qquad +\,(6 q^3 - 8 q^2 -4 q)p -2q^2, \end{aligned}$$

by noting

$$\begin{aligned} 4qp^5-(8q^2+4)p^4 +2 p^3q^3\ge & {} 4qp^5-9q^2p^4 +2p^3q^3\\= & {} p^3q^3(4p/q-1)(p/q-2)\ge 0,\\ 5q^3p^3-6q^4p^2\ge & {} 0, \quad 8qp^3-7q^2p^2> 0, \quad (8q-2)p^2>0,\\ (5q^3-8q^2-4q)p> & {} 0, \quad q^3p-2q^2> 0. \end{aligned}$$

3.4 $[c_1, c_2)$ part

Considering on $[c_1, c_2)$, we have $\langle qa\rangle = qa-(q-2)/2$. Since Z equals to a quadratic function with axis $a_4:=(2q+p-2)/(4q+2p)$, and since $a_4 - c_2 = (6q+p)/2p(2q+p)>0$, it is increasing on $[ c_1, c_2)$.

First, we consider the case when $2q\le p\le \frac{3}{2} q^2 + q$ holds. Since $S_p$ has period $1/p^2$, and since

$$\begin{aligned} c_2 -1/p^2 -c_1 = (2p+q)(p-2q)/2p^2q\ge 0, \end{aligned}$$

we have the first equality of

$$\begin{aligned} \sup _{a\in [c_1, c_2)} W_5(a) = \sup _{a\in [c_2-1/p^2, c_2)} W_5(a) \le W_5(a_5) < U. \end{aligned}$$

(13)

On $[c_2-1/p^2, c_2)$, we have $\langle p^2 a\rangle = p^2 a -p(p-3)/2+1$, and we see that $W_5(a)$ equals to a quadratic function with axis $a_5:=(2q^3 + ( p - 2) q^2 + 2 p^3 - 6 p^2 - 2p)/(4 q^3 + 2 p q^2 + 4 p^3)$, hence we have the third inequality. We have the rest by

$$\begin{aligned}&{4 p^2 q (p - q)^2 (p q - 1)(2q^3+pq^2+2p^3)}(U-W_5(a_5))\\&\quad =-4p^6 + (36 q^3 + 4 q^2 + 8 q +16)p^5 +(-79q^4 +16 q^3 - 40 q^2 \\&\qquad -\, 20 q -16)p^4+(44q^5 - 44 q^4 +79 q^3 -8 q^2 -2q)p^3+(-16q^6 +24q^5\\&\qquad -\,44 q^4+ 52 q^3 - 8 q^2 )p^2 +(16q^5 - 40q^4 - 10q^3)p - 4q^4> 0. \end{aligned}$$

Actually, when $q\ge 4$, we see

$$\begin{aligned}&-4p^6 + (36q^3 + 4q^2 + 8q+16)p^5 -79q^4 p^4 + 16q^5p^3\\&\qquad \ge (36q^3 -2 q^2 )p^5 -79q^4 p^4 + 16q^5p^3\\&\qquad \ge \frac{71}{2} q^3 p^5 -79q^4 p^4 + 16q^5p^3\\&\qquad =\frac{1}{2}\left( \frac{p}{q}-2\right) \left( 71\frac{p}{q}-16\right) \ge 0,\\&\quad (16q^3 -40 q^2 -20q -16)p^4 \ge (16 - 10 - 5/4 -1/4)q^3p^4>0,\\&\quad (12q^5 -44q^4)p^3 \ge (12-11)q^5p^3>0, \quad 16q^5p^3 - 16q^6p^2> 0, \\&\quad (24q^5+52q^3-8q^2)p^2> 0,\\&\quad (79q^3-8q^2-2q)p^3-44q^4p^2 \ge (79-2-1/8)q^3p^3 -44q^4p^2> 0,\\&\quad (16q^5-40q^4-10q^3)p -4q^4 \ge (16-10-5/8)q^5p-4q^4> 0. \end{aligned}$$

When $q=2$, and $p=5$ or 7, one can verify the above inequality.

Next, we consider the case $p> \frac{3}{2} q^2 + q$. We have already seen that Z and $W_6$ are increasing on $[c_1, c_2)$. Note that $p/q > \frac{3}{2} q + 1 \ge 4$ and $(c_2 -1/p) - c_1 = (p/q-5/2)/p> 0$. We can verify

$$\begin{aligned}&(U-W_6(c_2-1/p))(4p^3q(p-q)^2(pq-1))\\&\quad =(8q^2 -2) p^4 + (34q^3 -4 q + 8) p^3 +(-95q^4 -36q^2 -8q -8)p^2\\&\qquad +\,(50 q^5 +95q^3)p -50q^4> 0 \end{aligned}$$

by $8p^3 -(8q+8)p^2 \ge 8p^3 -12qp^2>0$ and

$$\begin{aligned}&(8q^2 -2 )p^4 + (34q^3-4q)p^3 +(-95q^4-36q^2)p^2 + 50q^4p\\&\quad > \frac{15}{2} q^2p^4 + 33q^3p^3 -104q^4p^2 + 50q^5p = q^5p\left( \frac{15}{2}\left( \frac{p}{q}\right) ^3 + 33\left( \frac{p}{q}\right) ^2\right. \\&\qquad \left. -104\left( \frac{p}{q}\right) +50\right) =:q^5p\cdot g\left( \frac{p}{q}\right) > 0 \end{aligned}$$

since $g'(t)>0$ for $t\ge 2$ and $g(2)>0$. Hence we see

$$\begin{aligned} \sigma ^2(a) \le W_6 (a) \le W_6(c_2-1/p) \le U\quad (a\in [ c_1, c_2-1/p]). \end{aligned}$$

We consider on $[ c_2-1/p, c_2)$. We have $\langle pa \rangle = pa-(p-5)/2$, $\langle qa \rangle = qa-(q-2)/2$, and can see that $\langle pa \rangle < \langle qa \rangle$ on $[ c_2-1/p, c_7)$ and $\langle pa \rangle \ge \langle qa \rangle$ on $[ c_7, c_2)$. By recalling the bound $|D Y_1| \le 2/p(p-1)$, we can verify that $D X_1$ is decreasing on $[ c_2-1/p, c_2)$,

$$\begin{aligned} D W_1(a)\ge & {} D^- X_1(c_7)-2/{p(p-1)} \\= & {} {(4p^2 +(5q-6)p -3q)}/{p(p-q)(p-1)}> 0, \end{aligned}$$

on $[ c_2-1/p, c_7)$, and

$$\begin{aligned} D W_1 \le D^+ X_1(c_7) +2/{p(p-1)}= & {} (-2p^3 + (8q +2)p^2\\&+\,(3q^2-6q)p -5q^2)/{p(p-q)(p-1)}< 0 \end{aligned}$$

on $[ c_7, c_2)$. Actually, the first inequality is clear and the second is proved by

$$\begin{aligned}&-2p^3 + (8q +2)p^2 +3q^2p \le -2p^3 + 9q p^2 +3q^2p\\&\quad = pq^2\left( -2\left( \frac{p}{q} \right) ^2 + 9 \left( \frac{p}{q}\right) +3\right) < 0 \end{aligned}$$

by $\frac{p}{q} \ge 7$ if $q\ge 4$. In case $q=2$, we have

$$\begin{aligned} -2p^3 + (8q +2)p^2 +(3q^2-6q)p -5q^2= -2(p+1)(p^2 -10p +10)< 0 \end{aligned}$$

for $p\ge 9$. Hence we see the second inequality of

$$\begin{aligned} \sigma ^2(a) \le W_1(a) \le W_1(c_7)\!=\! \sigma ^2(c_7) \le W_4(c_7) < \widetilde{U} < \sigma ^2(c_4) \quad (a\in [ c_2-1/p, c_2)). \end{aligned}$$

Put $\widetilde{U}:= Z(c_4) + S_q(c_4) < \sigma ^2(c_4)$. Because of $0< q^2/2(p-q)< 3q^2/2(p-q)< 1$,

$$\begin{aligned}&q^2 c_4 = (q^2/2-1)+ (1-q^2/2(p-q)), \quad \hbox {and}\\&\quad q^2 c_7 = (q^2/2-1)+ (1-3q^2/2(p-q)), \end{aligned}$$

we see $\langle q^2 c_4\rangle = 1-q^2/2(p-q)$ and $\langle q^2 c_7\rangle = 1-3q^2/2(p-q)$. Hence we can calculate $S_q(c_4)$ and $S_q(c_7)$, and can verify the rest by

$$\begin{aligned}&(\widetilde{U} - W_4(c_7)) (2p^2q^2 (p-q)^2 (pq-1))\\&\quad = (12q^4-4q^3-1) p^2 + (8q^5 + 4q^4 -12 q^3 +4q^2 + 2q)p -8q^4 -4q^3 -q^2\\&\quad \ge 9q^4p^2 + 7q^5 p -11 q^4 > 0 . \end{aligned}$$

4 Even p and odd q

Assume that q is odd and p is even and positive. We may assume $q\ge 3$.

4.1 $[ c_6, 1/2)$ part

We consider on $[ c_6, 1/2)$. We have $\langle pa\rangle = pa-(p-2)/2$, and by $c_5\le c_6$, $\langle qa\rangle = qa - (q-1)/2$. Since $\langle pa\rangle < \langle qa\rangle$ holds on $[ c_6, c_4)$, and $\langle pa\rangle \ge \langle qa\rangle$ on $[ c_4, 1/2)$, we see that $D X_1$ is decreasing on $[ c_6, 1/2)$. Recalling $|D Y_1(a) | \le 2/p(p-1)$, we have

$$\begin{aligned} D W_1(a) \ge D^- X_1(c_4) - 2/{p(p-1)} = ({2q^2 - 2q +p^2 - p})/{(p-1)q(p-q)}\!>\! 0 \end{aligned}$$

on $[c_6, c_4)$, and

$$\begin{aligned} D W_1(a) \le D^+ X_1(c_4) + 2/{p(p-1)} = {-A_3(p,q)}/{(p-1)pq(p-q)}<0 \end{aligned}$$

on $[c_4, 1/2)$. Hence $W_1(a) \le W_1(c_4)= \sigma ^2(c_4)$ for $a\in [c_6, 1/2)$.

We derive $A_3(p,q)>0$ from $p/q\ge 4$. Because of $p\ge 4q$ and p and q are relatively prime, we see $p> 4q$ or $p\ge 4q+2$. Hence we have $A_3(p,q)= p^2(p-4q-1)+2pq+2q^2> 0$.

4.2 $[0, c_5)$ part

We consider on $[0, c_5)$. The condition (3) implies $p\ge 2q+2$. Since V(a, a) is increasing on $[0, c_5)$ and $W_4(a) - V(a,a)$ has period 1 / q, we have the first equality of

$$\begin{aligned} \sup _{a\in [0, c_5)} W_4(a) =\sup _{a\in [c_5-1/q, c_5)} W_4(a) =\sup _{a\in [c_5-1/q^2, c_5)} W_4(a) = W_4(c_5)< U.\quad \end{aligned}$$

(14)

On $[c_5-1/q, c_5)$, we have $\langle qa\rangle = qa -(q-3)/2$ and we see that Z equals to a quadratic function having axis at $a_6:=({p+ 2q-4})/({2p+4q})$. Because of

$$\begin{aligned} a_6 - c_5 = ({p-2q})/({2q(2q+p)})> 0, \end{aligned}$$

it is increasing on $[c_5-1/q, c_5)$. Since $S_q$ has period $1/q^2$, we have the second equality of (14). On $[c_5-1/q^2, c_5)$, we have $\langle q^2a\rangle = q^2a -q(q-1)/2+1$ and we see that $W_4(a)$ equals to a quadratic function having axis at $a_7 := ({2 q^2 + (2 p - 2) q + p^2 - 4 p - 2})/({4 q^2 + 4 p q + 2 p^2})$. Because of

$$\begin{aligned} a_7- c_5 = ((p+1)(p-2q-1)+ 1)/({2 q (2 q^2 + 2 p q + p^2 ))} > 0, \end{aligned}$$

$W_4(a)$ is increasing on $[c_5-1/q^2, c_5)$ and the third equality of (14) is proved. The rest is by

$$\begin{aligned}&(U- W_4(c_5))({4p^2q^2(p-q)^2(pq-1)}) \\&\quad =p(p-2q)\left( p(pq-1)(p+2q) + 4q+2\right) + 4(q-2)p^2 +2(2p^2 -q^2) >0. \end{aligned}$$

4.3 $[c_5, c_6)$ part

On $[c_5, c_6)$, we have $\langle qa \rangle = qa-(q-1)/2$ and we see that Z equals to a quadratic function having axis at 1 / 2. Hence it is increasing on $[c_5, c_6)$. Since $S_p$ has period $1/p^2$, we have the first equality of

$$\begin{aligned} \sup _{a\in [c_5, c_6)} W_5(a) =\sup _{a\in [(c_6-1/p^2)\vee c_5, c_6)} W_5(a) \le W_5(a_8)< U. \end{aligned}$$

(15)

On $[(c_6-1/p^2)\vee c_5, c_6)$, by $\langle p^2a \rangle = p^2 a - p(p-2)/2+1$ we see that $W_5(a)$ is a quadratic function having axis at $a_8 := ({2 q^3 + p q^2 + 2 p^3 - 4 p^2 - 2 p}) /(4 q^3 + 2 p q^3+ 4 p^3)$. Hence we have the middle of (15). Rest is by

$$\begin{aligned} U - W_5(a_8) =A_4(p,q)/4 p^2 q (q - p)^2 (p q - 1) (2 q^3 + p q^2 + 2 p^3 )> 0. \end{aligned}$$

Finally, we prove $A_4(p,q)>0$ by assuming

$$\begin{aligned} \frac{p}{q}\ge a_9:= & {} \left( \frac{6^{-3/2} \sqrt{1019}}{9} + \frac{511}{1458}\right) ^{1/3} + \frac{65}{162}\left( \frac{6^{-3/2} \sqrt{1019}}{9} + \frac{511}{1458}\right) ^{-1/3} \\&+\,\frac{8}{9} =2.206\dots . \end{aligned}$$

We decompose as $3A_4(p,q)=h_1 + \cdots + h_7$. Here

$$\begin{aligned} h_1= & {} (3q^2-2)(p\!+\!2q)p^2q^3\left( 6\left( \frac{p}{q}\right) ^3\!-\!16\left( \frac{p}{q}\right) ^2 \!+\! 7\left( \frac{p}{q}\right) \!-\!2\right) \!\ge \! 0 \quad \hbox { if }\quad \frac{p}{q}\ge a_9,\\ {h_2}= & {} {(24q^2-2q)p^5 -50q^2p^4-72q^4p^3+48q^5p^2}\\\ge & {} {q^5p^2}\left( \frac{70}{3} \left( \frac{p}{q}\right) ^3-\frac{50}{3} \left( \frac{p}{q}\right) ^2 -72\left( \frac{p}{q}\right) +48\right) >0 \end{aligned}$$

since the last cubic function in p / q is increasing for $p/q\ge 2$ and equals to 24 at $p/q=2$,

$$\begin{aligned} h_3= & {} 48p^5-(72q+48)p^4 \ge 48p^5-88qp^4> 0,\\ h_4= & {} 99q^3p^3-44q^4p^2>0,\\ h_5= & {} (24q^2-6q)p^3> 0,\\ h_6= & {} (96q^3-24q^2)p^2 -(96q^4+30q^3)p > 88q^3p^2-106q^4p> 0,\\ h_7= & {} 12q^4(pq-1)> 0. \end{aligned}$$