國立交通大學
應用數學系
碩 士 論 文
在正規 Laurent 級數體探討關於 Kurzweil 定理之延伸結果
Refinements of Kurzweil’s Theorem in the Field of
Formal Laurent Series
研 究 生:林佳穎
指導教授:符麥克 教授
在正規 Laurent 級數體探討關於 Kurzweil 定理之延伸結果
Refinements of Kurzweil’s Theorem in the Field of
Formal Laurent Series
研 究 生:林佳穎
Student: Chia-Ying Lin
指導教授:符麥克
Advisor: Michael Fuchs
國 立 交 通 大 學
應用數學系
碩 士 論 文
A Thesis
Submitted to Department of Applied Mathematics National Chiao Tung University
in Partial Fulfillment of the Requirements for the Degree of
Master in
Applied Mathematics June 2014
Hsinchu, Taiwan, Republic of China
R
EFINEMENTS OF
K
URZWEIL
’
S
T
HEOREM IN
THE
F
IELD OF
F
ORMAL
L
AURENT
S
ERIES
Chia-Ying Lin
Department of Applied Mathematics,
National Chiao Tung University
This thesis was supervised by Dr. Michael Fuchs
June 20, 2014
前言
在過去十年裡,已經有了許多關於正規 Laurent 級數體下之賦距 Diophantine 逼近的研究,而最近這項研究有了一個有趣的新研究方向──關於 Kurzweil 定 理的改良。本論文主要的工作就是總結整理這些不同方向的改良以及提供一些新 的貢獻。 其中一項改良是由 Kim、Nakada 和 Natsui 在[6]中所提出,在本文中,我們 將指出他們提出的證明當中,有部分細節是可以被改進的;更精確來說,我們可 以將其中的單調性條件拿掉,而這可以讓我們重新證明 Kurzweil 定理的其中一 個方向。本文另一個主題是關於 Kurzweil 本身在[8]中所提出的在實數體上的另 一個改良,我們將證明這個定理在正規 Laurent 級數體下的相似結果並與另一個 最近由 Kim、Tan、Wang 與 Xu 在[7]提出的與之相似的改良作比較。 本文的主要架構如下:我們將在第一章介紹一些關於 Diophantine 逼近的背 景知識以及本論文的目標。第一節中,我們將回顧有關 Diophantine 逼近的基本 性質和介紹我們所用的符號。在第二節,我們將介紹 Diophantine 逼近和賦距 Diophantine 逼近,這可以分成 homogeneous 和 inhomogeneous 的情形。我們收集 一些關於這兩個情形的結果,尤其是包含所謂的 double-metric 和 single-metric 兩 種情況的 inhomogeneous 情形。最後,我們將在第三節介紹本論文的主要目的。 在第二章,我們將探討 Kim、Nakada 和 Natsui 所提出的改良。我們將在第 一節陳述一些引理和呈現他們在改良中的證明。第二節中,我們將改進前一節中 的證明,並利用改進後的結果來證明 Kurzweil 定理的其中一個方向。第三節, 我們將利用完全不同於第一節的證明方法來證明一個特殊情形。 我們在第三章證明一個與 Kurzweil 在[8]提出之改良相似的結果。因為這個 結果與另一個最近由 Kim、Tan、Wang 與 Xu 提出的改良有些異同處,我們將在 這個章節的最後讓兩者作些比較。 最後,在第四章,我們將提出一些猜想來對本論文做一個總結。Preface
The last decade has witnessed a lot of research about metric Diophantine ap-proximation in the field of formal Laurent series, where a recent new and interesting research direction was concerned with refinements of Kurzweil’s theorem. The pur-pose of this thesis is to summarize these refinements and give some new contributions. One of these refinements was given by Kim, Nakada and Natsui in [6]. In this thesis, we will show that some details of their proofs can be improved. More pre-cisely, we are able to drop the monotonicity condition and this will allow us to reprove one direction of Kurzweil’s theorem. Another topic of this thesis will be concerned with another refinement of Kurzweil’s theorem which in the real case was obtained by Kurzweil himself in [8]. We will prove an analogue of this theorem in the field of for-mal Laurent series and compare it with another refinement of a similar flavour which was recently proved by Kim, Tan, Wang and Xu in [7].
An outline of this thesis is as follows. In Chapter 1, we will introduce some background knowledge on Diophantine approximation and explain our aim of this the-sis. There are three sections in this chapter. In Section 1.1, we will recall some funda-mental properties for formal Laurent series and give some notations. Then, in Section 1.2, we will introduce Diophantine approximation and metric Diophantine approxima-tion. This introduction will be split into homogeneous and inhomogeneous cases. We will collect some results for the two cases, especially the inhomogeneous case which consists of the so-called double-metric and single-metric cases. Finally, we will state the main goal of this thesis in Section 1.3.
In Chapter 2, we will discuss the refinement of Kim, Nakada and Natsui. In Section 2.1, we will state some lemmas and present the proof of their refinement. In Section 2.2, we will give some improvements of the proofs from the previous sec-tion and use them to prove one direcsec-tion of Kurzweil’s theorem. In Secsec-tion 2.3, we will prove a special case of Kim, Nakada and Natsui’s refinement with a completely different method as the one in Section 2.1.
In Chapter 3, we will show an analogue of the refinement proved by Kurzweil in [8]. Since this result and another refinement which was recently proved by Kim, Tan, Wang and Xu have some similarities and differences, we will compare them at the end of this chapter.
致 謝 辭
猶記得剛踏入交大校園時內心澎湃的自己,一眨眼竟即將跨出校門。就讀研 究所的這兩年,受到了許多人的幫助,最感謝的,莫過於我的指導老師──符麥 克教授了,讓我在學習的過程當中,漸漸摸索出自己的方向,並耐心指導我論文 的寫作。此外,我也要特別感謝在口試當中給予誠摯建議的兩位口試委員── 蕭守仁教授以及楊一帆教授,讓我順利地通過口試。 在研究所的路途上,很開心認識珮柔、妃儀及晏昀等知心的好室友,有你們 在的寢室就充滿歡笑,每當我從研究室疲憊地回到宿舍,都能好好放鬆心情。在 修課的過程中,感謝同學們幫助,尤其是惠閔、伊婕和伯融,每週的討論作業時 間,一起努力合作,互相指教,讓我學習到了許多不同方面的見解。另外,謝謝 跟我一起擔任微積分助教的又菱、嬿蓉及家銘,有你們這些夥伴互相合作,才能 更有效率地完成任務。也感謝男朋友書誼在我不安時給予的支持與鼓勵,讓我更 有動力去面對困難。 在跨出校門的那一刻,等著我的是一個嶄新的路途,希望在未來的道路上, 能夠牢牢記住這兩年努力的自己,勇敢面對將來的挑戰。 最後,謹以此文獻給我摯愛的家人。 佳穎 謹誌Contents
1 Introduction 1 1.1 Fundamental Properties . . . 1 1.2 Diophantine Approximation in the Field of Formal Laurent Series . . 5 1.3 Kurzweil’s Theorem and its Refinements . . . 8 2 The Refinement of Kim, Nakada and Natsui 11 2.1 Proof of Proposition 1.3.1 and Proposition 1.3.2 . . . 11 2.2 Theorem 2.1 without the Monotonicity Condition . . . 18 2.3 A Special Case of Theorem 2.2 . . . 21 3 Proof of Theorem 1.9 24 4 Conclusion 34
Chapter 1
Introduction
1.1
Fundamental Properties
In the beginning, we will introduce some essential knowledge which is necessary for this thesis. Let Fq be a finite field of size q = pk(where p is a prime number and
k ∈ N) and Fq[X] be the set that contains all polynomials over Fq. We denote by
Fq(X) the quotient set of Fq[X]. With the above notations, define
Fq X−1 := ( f = N X i=−∞ aiXi : aN 6= 0, ai ∈ Fq ) ∪ {0}
to be the set of formal Laurent series. To turn the set Fq((X−1)) into a field, we define
addition and multiplication on it as for polynomials. Then, we have the following proposition that was proved in [9].
Proposition 1.1.1. (Fq((X−1)) , +, ·) is a field.
Comparing with the real case, Fq[X], Fq(X) and Fq((X−1)) play the roles of
integers, rational numbers and real numbers, respectively. Next, we define a norm for each f = PN0
i=−∞aiXi ∈ Fq((X−1)) with aN0 6= 0
by |f | = qdeg(f ) = qN0 and |0| = 0. The following property shows that the norm is
non-Archimedean.
Proposition 1.1.2. For any f, g ∈ Fq((X−1)), we have
(1) |f | = 0 ⇔ f = 0. (2) |f g| = |f | |g|.
(3) |f + g| ≤ max {|f | , |g|}. Proof.
(1) |f | = 0 ⇔ deg(f ) = −∞ ⇔ f = 0.
(2) |f g| = qdeg(f g) = qdeg(f )+deg(g) = qdeg(f )qdeg(g) = |f | |g|.
(3) |f + g| = qdeg(f +g)= qmax{deg(f ),deg(g)} = max{qdeg(f ), qdeg(g)} = max {|f | , |g|} .
Now, we are going to introduce a set, denoted by L, which plays the role of the unit interval [0, 1) in Fq((X−1)). It is the subset of Fq((X−1)) which consists of
elements f with deg(f ) < 0. In other words,
L :=f ∈ Fq((X−1)) : |f | < 1 .
By restricting the above norm to this set, we get that L is a compact topological group. Thus, there exists an unique translation-invariant measure called Haar measure which we denote by µ.
Define for all g ∈ L, B g; 1 qk := f ∈ L : |f − g| < 1 qk with k ≥ 1. The following property is a characterization of this measure.
Proposition 1.1.3. Choose b1, b2, . . . , bk ∈ Fq,g ∈ L and k ≥ 1. Then
µ ( f = ∞ X j=1 cjX−j : ci = bi for1 ≤ i ≤ k )! = 1 qk, and µ B g; 1 qk = 1 qk.
Proof.Assume that h = b1X−1+b2X−2+· · ·+bkX−k. Then, for any f =P ∞ j=1cjX
−j ∈
L, we have f ∈ B h; q−k if and only if ci = bi for 1 ≤ i ≤ k. Hence,
( f = ∞ X j=1 cjX−j : ci = bi for 1 ≤ i ≤ k ) = B h; 1 qk . Next, we consider the set L to be the union
[ b1,...,bk∈Fq ( f = ∞ X j=1 cjX−j : ci = bi for 1 ≤ i ≤ k ) .
Then we obtain 1 = µ (L) = µ [ b1,...,bk∈Fq ( f = ∞ X j=1 cjX−j : ci = bi for 1 ≤ i ≤ k ) = X b1,...,bk∈Fq µ ( f = ∞ X j=1 cjX−j : ci = bi for 1 ≤ i ≤ k )! = qkµ ( f = ∞ X j=1 cjX−j : ci = bi for 1 ≤ i ≤ k )! . Therefore, µ B h; 1 qk = µ ( f = ∞ X j=1 cjX−j : ci = bi for 1 ≤ i ≤ k )! = 1 qk.
Since µ is translation-invariant, we have µ B g; 1 qk = µ B h; 1 qk = 1 qk for all g ∈ L
Next, we are going to prove an important property of µ.
Proposition 1.1.4. Each two balls in L are either disjoint or one is contained in the other.
Proof.Let B1 f ; q−k1 and B2 g; q−k2 be two balls in L. Without loss of generality,
we suppose that k1 > k2. If they are not disjoint, then there must exist an element
h ∈ L such that h ∈ B1∩ B2. We have
|f − g| = |f − h + h − g| ≤ max {|f − h| , |h − g|} ≤ 1 qk2.
This implies that f ∈ B2 g; q−k2.
Now, we want to prove that B1 is contained in B2. If there exists h0 ∈ L such
that h0 ∈ B1 \ B2, then we get
|h0− g| = |h0− f + f − g| ≤ max {|h0 − f | , |f − g|} < 1 qk2,
which means that h0 ∈ B2. We have a contradiction. Therefore,
B1 f ; 1 qk1 ⊂ B2 g; 1 qk2 .
The proof is complete.
Similar to the integer part for real numbers, we define for each f =PN
i=−∞aiXi ∈ Fq((X−1)), [f ] = N X i=0 aiXi and {f } = f − [f ].
We call the former the polynomial part and the latter the fractional part of f .
If we fix an element f ∈ Fq((X−1)), then it can be expressed as the sum of its
polynomial part and fractional part. That is, f = [f ] + {f }. We can rewrite this as f = A0+ 1 g1 , where A0 = [f ] and g1 = 1 {f }.
Again, we represent g1 as the sum of its polynomial part, which we denote by A1,
and the reciprocal of 1/{g1}. We let g2 = 1/{g1}. Continuing this iterative process an
infinite number of times or until gkis a polynomial for some k ∈ N, f can be expressed
as a unique continued fraction expansion with Aj = [gj] for gj = 1/{gj−1}, j ≥ 2.
Note that f is irrational if and only if the process does not terminate. Thus, for each irrational f ∈ L, we have
f = 1 A1 + 1 A2+ 1 . .. ,
where the Ai’s are called partial quotients. Moreover, we put
Pk Qk = 1 A1+ 1 A2+ 1 . .. + 1 Ak , (Pk, Qk) = 1 with P0 = 0 and Q0 = 1,
which are called the principle convergent of f with deg(Qk) := nk. We have the
following property.
Lemma 1.1. Fix an irrational f ∈ L. If Pk/Qkare the principle convergents off with
partial quotientsAk, then fork ∈ N, we have the recurrence relation
Pk+1 = Ak+1Pk+ Pk−1
Qk+1 = Ak+1Qk+ Qk−1
, with initial conditions P0 = 0, Q0 = 1, P1 = 1, Q1 = A1.
Consequently,
nk+1− nk = deg(Ak+1),
Lemma 1.2. For any k ≥ 0, we have Pk+1Qk− PkQk+1 = (−1)k and |{Qkf }| = 1 |Qk+1| .
Next, we are going to introduce Diophantine approximation in the field of for-mal Laurent series.
1.2
Diophantine Approximation in the Field of Formal
Laurent Series
The main problem of Diophantine approximation in the field of formal Lau-rent series is as follows: for f ∈ L, find a function ψ fromqk : k ∈ N ∪ {0} into qk
: k ∈ Z such that the Diophantine inequality f − P Q < ψ(|Q|) |Q| with P, Q ∈ Fq[X], Q 6= 0, (1.1) has infinitely many solutions P and Q?
The following theorem which is an analogue of Dirichlet’s theorem is a repre-sentative result in this area.
Theorem 1.1. Let f ∈ L. Then, f −P Q < 1 |Q|2 with P, Q ∈ Fq[X], Q 6= 0,
has infinitely many solutionsP and Q.
In the above theorem, ψ(|Q|) was chosen as 1/|Q| and this result holds for each f ∈ L.
Now, we will discuss the subarea called metric Diophantine approximation in the field of formal Laurent series which asks for properties that hold for almost all f ∈ L. We will give some results from this area. Consider the inequality (1.1) for ψ(|Q|) = q−n−ln with l
n≥ 1 and n = deg(Q). Then, (1.1) becomes
f − P Q < 1
q2n+ln with P, Q ∈ Fq[X], Q 6= 0, Q monic, n = deg(Q). (1.2)
In [4], Inoue and Nakada investigated the condition that makes (1.2) have infinitely many solutions P and Q for almost all f ∈ L.
Theorem 1.2 (K. Inoue and H. Nakada [4]). The inequality (1.2) has infinitely many solutionsP and Q for almost all f ∈ L if and only if
∞
X
n=0
q−ln = ∞.
Furthermore, Nakada and Natsui found the following result on the asymptotic number of solutions Q ∈ Fq[X] in [11].
Theorem 1.3 (H. Nakada and R. Natsui [11]). Define Ψ(N ) := X
n≤N
1 qln.
Iflnsatisfies
(i) lnis non-decreasing and
P∞
n=0q
−ln = ∞,
(ii) there exists a constantD > 1 such that jk+1 > Djkfork ≥ 1, where
j1 := min{n ≥ 2 : ln− ln−1≥ 1},
jk := min{n > jk−1 : ln− ln−1 ≥ 1}, for k ≥ 2.
Then, for almost allf ∈ L, the number of solutions of (1.2) with 0 ≤ deg(Q) ≤ N is asymptotic toΨ(N ).
In addition, Fuchs obtained an improvement of the above theorem by dropping the conditions for lnand adding an error term to this result.
Theorem 1.4 (M. Fuchs [3]). For almost all f ∈ L, the number of solutions of (1.2) with0 ≤ deg(Q) ≤ N satisfies
Ψ(N ) + O Ψ(N )1/2(log Ψ(N ))2+ , whereΨ(N ) is as above and > 0 is an arbitrary constant.
So far, what we have discussed is the so-called homogeneous case. In this case, the inequlaity (1.1) can be rewritten as
|Qf − P | < ψ(|Q|) with P, Q ∈ Fq[X].
If ψ(|Q|) ≤ 1, then we get that the degree of Qf − P is less than zero so that P is the polynomial part of Qf . This implies that this inequality can be simplified to
Now, we are going to be concerned with the metric inhomogeneous Diophantine approximationproblem. For f, g ∈ L, consider the Diophantine inequality
|{Qf } − g| < ψ(|Q|) with Q ∈ Fq[X], (1.3)
where ψ is a function fromqk: k ∈ N ∪ {0} into qk : k ∈ Z . The main question
in this area is again the existence of infinitely many solutions to (1.3).
Inhomogeneous Diophantine approximation consists of two cases: double-metric and single-metric.
Double-metric, as the name suggests, is the case for which f and g are both random. Consider the inequality (1.3) for ψ(|Q|) = q−n−ln with l
n≥ 1 such that
|{Qf } − g| < 1
qn+ln with Q ∈ Fq[X], Q monic, n = deg(Q), (1.4)
where f, g ∈ L. Ma and Su studied (1.4) for the double-metric case in [10]. The following theorem is their result.
Theorem 1.5 (C. Ma and W.-Y. Su [10]). The inequality (1.4) has infinitely many solutionsQ ∈ Fq[X] for almost all (f, g) ∈ L2if and only if
∞
X
n=0
1
qln = ∞.
Moreover, the asymptotic number of solutions Q was estimated by Fuchs in [3]. He derived a strong law of large numbers with error terms for the number of solutions of (1.4) with deg(Q) ≤ N . We have the following result.
Theorem 1.6 (M. Fuchs [3]). For almost all (f, g) ∈ L2, the number of solutions of (1.4) with0 ≤ deg(Q) ≤ N satisfies
Ψ(N ) + O(Ψ(N ))12 (log Ψ(N )) 3 2+ , whereΨ(N ) =P n≤Nq
−ln and > 0 is an arbitrary constant.
Next, we are going to introduce the two single-metric cases: (1) fix g and choose a random f ∈ L,
(2) fix f and choose a random g ∈ L.
As for Case 1, Fuchs in fact showed that Theorem 1.4 holds for any fixed g (not only g = 0).
In this research, we will focus on the Case 2. Our main objective of this thesis will be introduced in the next section.
1.3
Kurzweil’s Theorem and its Refinements
In this section, we will introduce the main topic of this thesis. Consider the Diophantine inequality for f fixed and g random
|{Qf } − g| < 1
qn+ln with Q ∈ Fq[X], n = deg(Q), (1.5)
where {ln} is a given sequence of positive integers. In [5], Kim and Nakada studied
the following problem: for any sequence {ln} with
P∞
n=0q
−ln = ∞, which condition
do we need for f ∈ L such that (1.5) has infinitely many solutions in Q for almost all g ∈ L ? They found that the condition for f satisfying the above property is that f is badly approximablewhose definition we give next:
Definition 1.1. f is badly approximable if and only if there exists a constant c > 0 such that
|{Qf }| > 1
qn+c, n = deg(Q),
for allQ ∈ Fq[X], Q 6= 0.
Then, we have the following result which is Kurzweil’s theorem for formal Lau-rent series.
Theorem 1.7 (D. H. Kim and H. Nakada [5]). f is badly approximable if and only if (1.5) has infinitely many solutionsQ for almost all g ∈ L and all sequence {ln} with
P∞
n=0q
−ln = ∞.
In the sequel, we are going to discuss some refinements of Kurzweil’s theorem in the field of formal Laurent series. Kim, Nakada and Natsui [6] investigated the con-dition for f when adding the adcon-ditional requirement to {ln} that lnis non-decreasing.
Note that this set of f contains all elements that are badly approximable. They ob-tained some partial results even though they did not find the exact set for f . In order to state their results, we need some notations. Define
Ω := ( ln≥ 1 : non-decreasing and ∞ X n=0 1 qln = ∞ ) and
WΩ := {f ∈ L : ∀ln∈ Ω, (1.5) has infinitely many solutions Q for almost all g ∈ L} .
Proposition 1.3.1 (D. H. Kim, H. Nakada and R. Natsui [6]). Let Pk/Qk be the
prin-ciple convergents off with deg(Qk) = nk. If
P∞ k=1 1 nk < ∞, then we have f /∈ WΩ. Note thatP∞ k=1 1
nk = ∞ is satisfied for almost every f ∈ L.
The second result proved in [6] is as folows:
Proposition 1.3.2 (D. H. Kim, H. Nakada and R. Natsui [6]). If there exists a positive integerC such that nk ≤ Ck for all k ∈ N, then we have f ∈ WΩ.
Note that Proposition 1.3.2 is only a sufficient but not necessary condition. We will prove Proposition 1.3.1, Proposition 1.3.2 and improve some details of this result in the next chapter.
Kim, Tan, Wang and Xu put forward another refinement from a new viewpoint in [7]. Consider the inequality which is different from (1.5)
|{Qf } − g| < 1
qln, deg(Q) = n, (1.6)
where f, g, Q and lnare as above. For s ≥ 1, we define
Us := f ∈ L : ∃c > 0 such that |{Qf }| > 1 qsn+c, ∀Q ∈ Fq[X] with n = deg(Q) . It is obvious that Usis the set of badly approximable elements when s = 1.
The main goal of [7] was to search for the set of {ln} such that f ∈ Us is
a necessary and sufficient condition for (1.6) having infinitely many solutions Q for almost all g ∈ L whenever {ln} belongs to this set. The set is defined as
Ωs := ( ln ≥ 1 : ∞ X n=0 qn−sln = ∞ ) . Then, we have the following theorem.
Theorem 1.8 (D. H. Kim, B. Tan, B. Wang and J. Xu [7]). Define
Ws := {f ∈ L : ∀ln∈ Ωs, (1.6) has infinitely many solutions Q for almost all g ∈ L} .
Then, we haveWs= Us.
In addition, Kurzweil also gave a refinement of the same flavour for the real case in [8]. One of the main goals of this thesis is to obtain an analogue of this result in the field of formal Laurent series. We first consider a non-negative and non-decreasing sequence {rn} which fulfils the following conditions:
(1) n − rnis non-increasing, (2) rn ≥ 2n, for all n ∈ N. Define U{rn} := f ∈ L : ∃c > 0 such that f −P Q > q−rn+c , ∀P, Q ∈ Fq[X] with deg(Q) = n . Our goal is similar as the one in Theorem 1.8. However, the set of {ln} now becomes
of course different. More precisely, we define a set Ω{rn}as follows: the sequence {ln}
belongs to Ω{rn}if
(1) lnis non-decreasing, and
(2) there exists an increasing sequence of non-negative integers t1 < t2 < t3 < ...,
and a function δ(n) which is non-decreasing with δ(n) → ∞ as n → ∞ such that ti+1> rti+δ(ti)− ti, and X i≥1 qti−lrt i+δ(ti)−ti = ∞.
Then, we have the following result. Theorem 1.9. Define the set
W{rn} :=f ∈ L : ∀ln ∈ Ω{rn}, (1.6) has infinitely many solutions Q for almost all g ∈ L .
Then, we haveW{rn} = U{rn}.
Note that Us = U{rn} when we take rn = (s + 1)n. In Chapter 3, we are going
to prove Theorem 1.9 and compare the sets Ωsand Ω{rn} which have some similarities
Chapter 2
The Refinement of Kim, Nakada and
Natsui
In this chapter, we will prove Proposition 1.3.1 and Proposition 1.3.2 from the introduction and give some improvements.
2.1
Proof of Proposition 1.3.1 and Proposition 1.3.2
From now on, we assume that f is irrational. In order to prove the two proposi-tions, we need the following lemmas.
Lemma 2.1. {{Qf } : Q ∈ Fq[X]} is dense in L.
Lemma 2.2 (0-1 law). Let E be a measurable set contained in L. If E is invariant under the actionh· + {Qf }i for all Q ∈ Fq[X], then we have
µ (E) = 0 or 1.
The proofs of Lemma 2.1 and Lemma 2.2 are in [1]. From the second lemma, we obtain the following result. Lemma 2.3. Define the set
E := {g ∈ L : |{Qf } − g| < ψ(|Q|) with Q ∈ Fq[X] has infinitely many solutions} .
Then, E is invariant under the action h· + {Qf }i for all Q ∈ Fq[X]. Consequently,
Proof.Let g ∈ E. If we fix a polynomial Q0, then we can find infinitely many Q with deg(Q − Q0) > 0 such that
|{(Q − Q0)f } − g| = |{Qf } − (g + {Q0f })| < ψ(|Q − Q0|) = ψ(|Q|). This means g + {Q0f } ∈ E. That is, E + {Q0f } ⊆ E. On the other hand, since
|{(Q + Q0)f } − g| = |{Qf } − (g − {Q0f })| < ψ(|Q|)
has infinitely many solutions, we have g−{Q0f } ∈ E. Then, g = g−{Q0f }+{Q0f } ∈ E + {Q0f }, which implies that E ⊆ E + {Q0f }. Hence, we get E = E + {Q0f }. Consequently, E is invariant under the action h· + {Qf }i for all Q ∈ Fq[X] so that
µ (E) = 0 or 1 by Lemma 2.2.
From now on, we let ψ(|Q|) = q−n−ln throughout this chapter. (In the next
chapter, we will discuss the inequality with ψ(|Q|) = q−ln.)
Next, we state the Borel-Cantelli Lemma which will be used below.
Lemma 2.4 (Borel-Cantelli Lemma). Let Fn be a sequence of events in a probability
space. Then, (i) IfP∞
n=1µ(Fn) < ∞, then µ(lim supn→∞Fn) = 0.
(ii) IfP∞ n=1µ(Fn) = ∞, then µ ∞ \ N =1 ∞ [ n=N Fn ! ≥ lim sup N →∞ PN n=1µ(Fn) 2 PN n=1 PN m=1µ(Fn∩ Fm) for all N ∈ N. Moreover, we have the right hand side is positive if there existsK > 0 such that µ(Fi∩ Fj) ≤ Kµ(Fi)µ(Fj), ∀i, j ∈ N.
Fix f ∈ L. Let Pk/Qk be the principle convergents of f with deg(Qk) = nk.
Then, we have the following result that was proved by Kim and Nakada in [5].
Lemma 2.5 (D. H. Kim and H. Nakada [5]). (i) For each Q ∈ Fq[X] with deg(Q) <
nk+1, there exists a unique decompositionQ = B1Q0+ B2Q1+ · · · + Bk+1Qk
withBi ∈ Fq[X] and deg(Bi) < ni− ni−1.
(ii) For each nonzero Q ∈ Fq[X] with deg(Q) < nk+1, we have|{Qf }| ≥ q−nk+1.
Moreover,
|{Qf }| = 1
qs for 0 < s ≤ nk+1,
if and only ifBi = 0 for 1 ≤ i ≤ m with nm < s ≤ nm+1 anddeg(Bm+1) =
Next, we are going to discuss two lemmas from [6]. Since we will revisit the proof in the next section, we will give them as well.
Let {ln} be a non-decreasing sequence. Define for nk ≤ n < nk+1, l∗n :=
max{nk+1− n, ln}. Let BQ = B({Qf }; q−n−l
∗
n) with n = deg(Q). Note that B
Q is
a ball whose radius is not greater than q−nk+1. We denote by F
kthe union of BQwith
nk ≤ deg(Q) < nk+1. That is,
Fk= [ nk≤n<nk+1 [ deg(Q)=n BQ.
Note that the BQ’s are disjoint. Thus,
µ(Fk) = X nk≤n<nk+1 (q − 1)qn. 1 qn+l∗ n = (q − 1) X nk≤n<nk+1 1 ql∗ n. (2.1)
Lemma 2.6 (D. H. Kim, H. Nakada and R. Natsui [6]). Let Q ∈ Fq[X] with deg(Q) =
n and nk ≤ n < nk+1. Then, form ≥ 1,
µ(BQ∩ Fk+m) = µ(BQ)µ(Fk+m), ifn + l∗n< nk+m, q−nk+m µ(F k+m) − (q − 1)P q−l ∗ s nk+m+1+nk+m−n−l∗n≤s<nk+m+1 ! , ifnk+m ≤ n + ln∗ < nk+m+1, 0, ifn + l∗n≥ nk+m+1.
Proof. Let Q0 be a polynomial with deg(Q0) = n0 and nk+m ≤ n0 < nk+m+1 such
that B({Qf }; q−n−l∗n) ∩ B({Q0f }; q−n 0−l∗ n0) 6= ∅. Since n + l∗ n < n 0 + l∗ n0, we have BQ0 ⊂ BQby Proposition 1.1.4.
In the case n + ln∗ ≥ nk+m+1, we obtain |{(Q − Q0)f }| < q−n−l
∗
n ≤ q−nk+m+1.
Conversely, since deg(Q − Q0) < nk+m+1, we get |{(Q − Q0)f }| ≥ q−nk+m+1 by
Lemma 2.5 (ii), a contradiction. Hence, BQdoes not contain BQ0 in this case. Then,
B {Qf }; 1 qn+l∗ n ∩ B {Q0f }; 1 qn0+ln0∗ = ∅. Therefore, µ(BQ∩ Fk+m) = 0.
Now, we consider the case n + ln∗ < nk+m+1. Choose i ∈ N with 1 ≤ i ≤ m
such that nk+i ≤ n + l∗n< nk+i+1. Then,
|{(Q − Q0)f }| < 1
qn+l∗ n
≤ 1 qnk+i.
By Lemma 2.5, we have
Q0 = Q + Bk+i+1Qk+i+ · · · + Bk+m+1Qk+m,
where Bk+j+1 ∈ Fq[X] with deg(Bk+j+1) < nk+j+1 − nk+j for i ≤ j ≤ m and
Bk+m+1 6= 0.
If Bk+i+1 6= 0, then we use the Lemma 2.5 (ii) again to obtain
|{(Q − Q0)f }| = q
deg(Bk+i+1)
qnk+i+1 .
Thus, we have |{(Q − Q0)f }| < q−n−l∗nif and only if deg(B
k+i+1) < nk+i+1− n − ln∗.
If Bk+i+1 = 0, then there exists r ∈ N with i < r ≤ m such that
|{(Q − Q0)f }| = q deg(Bk+r+1) qnk+r+1 < 1 qnk+r ≤ 1 qnk+i+1 < 1 qn+l∗ n. Therefore, we have BQ∩ Fk+m = [
deg(Bk+i+1)<nk+i+1−n−ln∗
deg(Bk+j+1)<nk+j+1−nk+j,i<j<m
0≤deg(Bk+m+1)<nk+m+1−nk+m
BQ+Bk+i+1Qk+i+···+Bk+m+1Qk+m,
where the union is disjoint. By the definition of BQ, we have
µ(BQ+Bk+i+1Qk+i+···+Bk+m+1Qk+m) =
1
qnk+m+t+l∗nk+m+t,
where t = deg(Bk+m+1).
If i < m (i.e., n + l∗n < nk+m), then the number of Q0 = Q + Bk+i+1Qk+i+
· · · + Bk+m+1Qk+m with deg(Q0) = nk+m+ t such that |{(Q − Q0)f }| < q−n−l
∗ n is
qnk+i+1−n−l∗n · qnk+i+2−nk+i+1· · · qnk+m−nk+m−1 · (q − 1)qt= (q − 1)qnk+m−n−l∗n+t.
Hence, µ(BQ∩ Fk+m) = X 0≤t<nk+m+1−nk+m ]{deg(Q0) = nk+m+ t : |{(Q − Q0)f }| < q−n−l ∗ n} qnk+m+t+l∗nk+m+t = X 0≤t<nk+m+1−nk+m (q − 1)qnk+m−n−l∗n+t qnk+m+t+lnk+m+t∗ = q−n−l∗n X 0≤t<nk+m+1−nk+m (q − 1) ql∗nk+m+t = µ(BQ)µ(Fk+m).
If i = m (i.e., nk+m≤ n + l∗n< nk+m+1), then BQ∩ Fk+m = [ 0≤deg(Bk+m+1)<nk+m+1−n−l∗n BQ+Bk+m+1Qk+m, and µ(BQ∩ Fk+m) = X 0≤t<nk+m+1−n−ln∗ ]{deg(Q0) = nk+m+ t : |{(Q − Q0)f }| < q−n−l ∗ n} qnk+m+t+l∗nk+m+t = X 0≤t<nk+m+1−n−ln∗ (q − 1)qt qnk+m+t+l∗nk+m+t = 1 qnk+m X 0≤t<nk+m+1−n−l∗n q − 1 ql ∗ nk+m+t = 1 qnk+m · µ(Fk+m) − X nk+m+1−n−l∗n≤t<nk+m+1−nk+m q − 1 ql ∗ nk+m+t , where the last equality is by (2.1).
Applying the above lemma, we have the following result.
Lemma 2.7 (D. H. Kim, H. Nakada and R. Natsui [6]). For any k ≥ 0 and m ≥ 1, we have
|µ(Fk∩ Fk+m) − µ(Fk)µ(Fk+m)| ≤
1
qm−1µ(Fk+m).
Proof.By Lemma 2.6, we have µ(BQ∩ Fk+m) ≤ µ(BQ)µ(Fk+m), 1 qnk+mµ(Fk+m), 0, if n + ln∗ < nk+m, if nk+m ≤ n + ln∗ < nk+m+1, if n + ln∗ ≥ nk+m+1.
This implies that
µ(BQ∩ Fk+m) − µ(BQ)µ(Fk+m) ≤
1
qnk+mµ(Fk+m).
On the other hand,
µ(BQ∩ Fk+m) ≥ µ(BQ)µ(Fk+m), 0, if n + l∗n< nk+m, if n + l∗n≥ nk+m.
1. For n + l∗n< nk+m, we have µ(BQ∩ Fk+m) − µ(BQ)µ(Fk+m) ≥ 0 ≥ − µ(Fk+m) qnk+m . 2. For n + l∗n≥ nk+m, we have µ(BQ∩ Fk+m) − µ(BQ)µ(Fk+m) ≥ −µ(BQ)µ(Fk+m) = −µ(Fk+m) qn+l∗ n ≥ −µ(Fk+m) qnk+m .
Therefore, for all m ≥ 1, we have
|µ(BQ∩ Fk+m) − µ(BQ)µ(Fk+m)| ≤ µ(Fk+m) qnk+m . Notice that µ(Fk∩Fk+m) − µ(Fk)µ(Fk+m) = X nk≤deg(Q)<nk+1 (µ(BQ∩ Fk+m) − µ(BQ)µ(Fk+m)) . Thus, |µ(Fk∩ Fk+m) − µ(Fk)µ(Fk+m)| ≤ X nk≤deg(Q)<nk+1 µ(Fk+m) qnk+m = (qnk+1− qnk)µ(Fk+m) qnk+m ≤ µ(Fk+m) qnk+m−nk+1 ≤ µ(Fk+m) qm−1 .
The proof is complete.
By the above lemma and Lemma 2.4, we have the following result. Lemma 2.8 (D. H. Kim, H. Nakada and R. Natsui [6]).
∞ X n=0 1 ql∗ n = ∞ i.e., ∞ X k=0 µ(Fk) = ∞ ! if and only if|{Qf } − g| < q−n−l∗
n has infinitely many solutions Q with deg(Q) = n
for almost allg ∈ L.
This result implies the following two theorems that were proved in [6].
Theorem 2.1 (D. H. Kim, H. Nakada and R. Natsui [6]). Let {ln} be a non-decreasing
sequence. Then, (1.5) has infinitely many solutionsQ for almost all g ∈ L if and only if f ∈ ( f ∈ L : ∞ X n=0 1 ql∗ n = ∞, where l ∗ n= max{nk+1− n, ln} for nk ≤ n < nk+1 ) .
Theorem 2.2 (D. H. Kim, H. Nakada and R. Natsui [6]). Let {ln} be a non-decreasing
sequence. Then, (1.5) has infinitely many solutionsQ for almost all g ∈ L if and only if f ∈ ( f ∈ L : ∞ X k=0 min{lnk, nk+1− nk} qlnk = ∞ ) .
Now, we are going to recall Proposition 1.3.1, Proposition 1.3.2 and provide the proofs of them via Theorem 2.1 and Theorem 2.2.
Recall the sets Ω := ( ln≥ 1 : non-decreasing and ∞ X n=0 1 qln = ∞ ) and
WΩ := {f ∈ L : ∀ln∈ Ω, (1.5) has infinitely many solutions Q for almost all g ∈ L} .
We have the following results.
Proposition 2.1.1 (D. H. Kim, H. Nakada and R. Natsui [6]). Let Pk/Qk be the
prin-ciple convergents off with deg(Qk) = nk. IfP ∞ k=0 1 nk < ∞, then we have f /∈ WΩ. Proof.Choose ln= blogq(n logqn)c, 1, if n ≥ q, if n < q. Then, we have ∞ X n=0 1 qln ≥ ∞ X n=q 1 n logqn = ∞.
Let nk∗ be the smallest positive integer that is larger or equal to q. Then,
∞ X k=0 min{lnk, nk+1− nk} qlnk ≤ ∞ X k=0 lnk qlnk ≤ k∗−1 X k=0 lnk qlnk + X k≥k∗
logqnk+ logq(logqnk)
q−1· n klogqnk ≤ k∗−1 X k=0 lnk qlnk + ∞ X k=k∗ 2q nk < ∞. By Theorem 2.2, the proof is complete.
Proposition 2.1.2 (D. H. Kim, H. Nakada and R. Natsui [6]). If there exist a positive integerC such that nk ≤ Ck for all k ∈ N, then we have f ∈ WΩ.
Proof.Let {ln} ∈ Ω. Then, ∞ X k=0 min{lnk, nk+1− nk} qlnk ≥ ∞ X k=0 1 qlnk ≥ ∞ X k=0 1 qlCk ≥ 1 C ∞ X k=0 1 qlCk + 1 qlCk+1 + · · · + 1 qlC(k+1)−1 ≥ 1 C ∞ X n=0 1 qln = ∞.
Since {ln} ∈ Ω is arbitrary, we have f ∈ WΩby applying Theorem 2.2.
In the next section, we will give some improvements of this refinement and prove them.
2.2
Theorem 2.1 without the Monotonicity Condition
In this section, we will improve Lemma 2.6 and Lemma 2.7 by dropping the monotonicity condition that {ln} is non-decreasing and use these improvements to
prove one direction of Theorem 2.1 without the monotonicity condition. Moreover, we will show that this theorem also implies one direction of Kurzweil’s theorem.
If we remove the condition that {ln} is non-decreasing, then the main difference
in Lemma 2.6 is that: q−n0−l∗n0, the radius of BQ0, might be greater than q−n−l ∗ n. We
will show the following modification of Lemma 2.6.
Lemma 2.9. Let Q ∈ Fq[X] with deg(Q) = n and nk ≤ n < nk+1. Then, form ≥ 1,
µ(BQ∩ Fk+m) = µ(BQ)µ(Fk+m), ifn + l∗n< nk+m, q−nk+m µ(F k+m) − (q − 1)P q−l ∗ s nk+m+1+nk+m−n−l∗n≤s<nk+m+1 ! , ifnk+m ≤ n + ln∗ < nk+m+1, 0 or µ(BQ), ifn + ln∗ ≥ nk+m+1.
Proof.Let Q0 be a polynomial with deg(Q0) = n0and nk+m ≤ n0 < nk+m+1 such that
B({Qf }; q−n−l∗n) ∩ B({Q0f }; q−n0−l∗n0) 6= ∅.
In the case n + ln∗ < nk+m+1, we have n0 + ln∗0 ≥ nk+m+1 > n + l∗naccording
to the definition of ln∗0. Thus, the method in this case is the same as before so that we
only have to consider the case n + ln∗ ≥ nk+m+1.
1. If n + ln∗ < n0+ l∗n0, then |{(Q − Q0)f }| < q−n−l ∗
n ≤ q−nk+m+1. Since the degree
of Q − Q0 is less than nk+m+1, we have |{(Q − Q0)f }| ≥ q−nk+m+1 by Lemma
2.5 (ii). This implies that this case cannot happen, i.e., such a Q0 does not exist. 2. If n + ln∗ ≥ n0+ l∗ n0, then B {Qf }; 1 qn+l∗ n ⊆ B {Q0f }; 1 qn0+l∗n0 , and from the disjointness of the balls in Fk+m, we get
µ(BQ∩ Fk+m) = µ(BQ).
Overall this implies that either µ(BQ ∩ Fk+m) = 0 or µ(BQ) in the case n + l∗n ≥
nk+m+1 as claimed.
By the above lemma, Lemma 2.7 still holds when dropping the monotonicity condition and the proof is almost the same as before.
Lemma 2.10. For any k ≥ 0 and m ≥ 1, we have |µ(Fk∩ Fk+m) − µ(Fk)µ(Fk+m)| ≤
1
qm−1µ(Fk+m).
Proof.As in Lemma 2.7, our first goal is to show that |µ(BQ∩ Fk+m) − µ(BQ)µ(Fk+m)| ≤
µ(Fk+m)
qnk+m (2.2)
Since the case n + l∗n< nk+m+1 can be proved in the same way as before, we focus on
the case n + l∗n≥ nk+m+1.
Here, we first show that
µ(BQ∩ Fk+m) − µ(BQ)µ(Fk+m) ≤
µ(Fk+m)
qnk+m .
There are two cases:
(i) If µ(BQ∩ Fk+m) = 0, then the claim is trivial.
(ii) If µ(BQ∩ Fk+m) = µ(BQ), then BQ ⊆ Fk+m which means that BQ ⊆ BQ0 with
nk+m ≤ n0 = deg(Q0) < nk+m+1. Observe that
µ(BQ) = 1 qn+l∗ n ≤ 1 qn0+l∗n0 ≤ 1 qnk+m+l∗n0 ≤ 1 qnk+m µ(Fk+m) q − 1 ≤ µ(Fk+m) qnk+m . Consequently, µ(BQ∩ Fk+m) − µ(BQ)µ(Fk+m) = µ(BQ) − µ(BQ)µ(Fk+m) ≤ 1 qnk+mµ(Fk+m).
Thus, the claim is established for this case as well.
Next, since µ(BQ∩ Fk+m) ≥ 0 for n + l∗n≥ nk+m, we have
µ(BQ∩ Fk+m) − µ(BQ)µ(Fk+m) ≥ −
µ(Fk+m)
qnk+m .
This concludes the proof of (2.2). The remaining proof follows along the same lines as in Lemma 2.7.
This lemma together with Lemma 2.4 shows that Lemma 2.8 holds without the monotonicity condition.
Now, we are going to prove that one direction of Theorem 2.1 also holds without the monotonicity condition.
Theorem 2.3. Let {ln} be a sequence. Define
U := ( f ∈ L : ∞ X n=0 1 ql∗ n = ∞, where l ∗ n = max{nk+1− n, ln} for nk≤ n < nk+1 ) . Iff ∈ U , then (1.5) has infinitely many solutions Q for almost all g ∈ L.
Proof.Let f ∈ U , we haveP∞
n=0q −l∗ n = ∞. Consequently, |{Qf } − g| < 1 qn+l∗ n
has infinitely many solutions Q for almost all g ∈ L by Lemma 2.8.
Next, by the definition of ln∗, we have q−n−l∗n ≤ q−n−ln. Thus, (1.5) has also
infinitely many solutions Q for almost all g ∈ L.
We will show that this theorem implies one direction of Kurzweil’s theorem, namely, the direction that if f is badly approximable and {ln} is a sequence with
P∞
n=0q
−ln = ∞, then (1.5) has infinitely many solutions Q for almost all g ∈ L.
Proof.Let f be badly approximable. Then, there exists a c such that nk+1− nk ≤ c
for all k. Fix a sequence {ln} withP ∞ n=0q
−ln = ∞. Then, we consider the following
cases:
(i) ln≥ c for all large n. Then ln∗ = max{nk+1−n, ln} = lnfor all large n. Thus,we
have ∞ P n=0 q−l∗n = ∞ P n=0
q−ln = ∞. The claim now follows from Theorem 2.3.
(ii) ln< c for infinitely many n. Set eln = max{ln, c}. Then ∞
P
n=1
q− eln = ∞. Applying
part (i) gives that |{Qf } − g| < q−n− eln has infinitely many solutions Q for
This concludes this proof.
In the next section, we will prove a special case of Theorem 2.2. The purpose for doing so is two-fold: first we will prove our result with a completely different method as the one used in Section 2.1 and second this is a kind of warm-up for Chapter 3 where the same method will be applied.
2.3
A Special Case of Theorem 2.2
The main purpose in this section is to discuss the existence of infinitely many solutions to (1.5) when the seriesP∞
k=0q
−lnk diverges. In order to prove this, we need
the following lemma.
Lemma 2.11. Let g ∈ L. Then, the number of {Qf } with deg(Q) < nk+1belonging
toB(g, q−d) is at most max{qnk+1−d, 1}.
Proof.Let Q, Q0 be two different polynomials with deg(Q), deg(Q0) < nk+1. Then,
by Lemma 2.5, we have
|{Qf } − {Q0f }| ≥ 1
qnk+1.
This means that the distance between two points {Qf }, {Q0f } with deg(Q), deg(Q0) < nk+1 is at least q−nk+1. Now, consider two cases:
1. If q−nk+1 ≥ q−d, then there is at most one point in B(g, q−d).
2. If q−nk+1 < q−d, then the number of points in B(g, q−d) is at most qnk+1−d.
Hence, the number of {Qf } with deg(Q) < nk+1 belonging to B(g, q−d) is at most
max{qnk+1−d, 1}.
Applying the above lemma, we can prove the following result. Lemma 2.12. Let {ln} be a non-decreasing sequence with
P∞
k=0q
−lnk = ∞. Then,
for all N≥ 0, we have µ ∞ [ k=N Fk ! > 1 qc, for all c ≥ 2. (2.3)
Proof.Assume that (2.3) is false. Then, there exists N0 ∈ N such that
µ K [ k=N0 Fk ! ≤ 1 qc , for all K ≥ N0. (2.4)
Define a set LnK+1−1 := ( deg(Q) = nK+1− 1 : {Qf } ∈ K [ k=N0 Fk\ K−1 [ k=N0 Fk ) . First, we estimate the number of elements in LnK+1−1. Let
K−1 [ k=N0 Fk = K−1 [ k=N0 [ nk≤n<nk+1 [ deg(Q)=n B {Qf }; 1 qn+l∗ n =[ i B {Qif }; 1 qdi ,
where B({Qif }; q−di) are disjoint for all i. By (2.4), we obtain
1 qc ≥ µ K−1 [ k=N0 Fk ! = µ [ i B {Qif }; 1 qdi ! =X i µ B {Qif }; 1 qdi =X i 1 qdi.
Using Lemma 2.11, the number of {Qf } with deg(Q) < nK+1 belonging to
S i B({Qif }; q−di) is at most P i max{qnK+1−di, 1} = max{qnK+1P i q−di, qnK} ≤
qnK+1−1. Thus, the number of elements in L
nK+1−1is at least
qnK+1−1(q − 1) − qnK+1−1 = qnK+1−1(q − 2).
Next, we claim that [ Q∈LnK+1−1 B {Qf }; 1 qnK+1−1+l∗nK+1−1 ! ⊂ K [ k=N0 Fk\ K−1 [ k=N0 Fk. (2.5)
In order to show this, fix Q1 ∈ LnK+1−1. Suppose there exists a polynomial Q2 with
deg(Q2) = u < nK and B({Q1f }; q
−nK+1+1−l∗nK+1−1
) ∩ B({Q2f }; q−u−l
∗
u) 6= ∅. We
know that {Q1f } does not belong to BQ2. Hence, we have
B {Q2f }; 1 qu+l∗ u ⊂ B {Q1f }; 1 qnK+1−1+l∗nK+1−1 ! . Then, we have |{Q1f } − {Q2f }| < 1 qnK+1−1+lnK+1−1∗ .
By Lemma 2.11, the number of {Qf } with deg(Q) < nK+1 belonging to BQ1 is at
most max{qnK+1−nK+1+1−l∗nK+1−1
, 1} = 1. Thus, we get {Q1f } = {Q2f }, a
Note that any two balls appearing on the left side of (2.5) are disjoint. By (2.5), we obtain µ K [ k=N0 Fk ! ≥ µ K−1 [ k=N0 Fk ! + µ [ Q∈LnK+1−1 B {Qf }; 1 qnK+1−1+l∗nK+1−1 ! ≥ µ K−1 [ k=N0 Fk ! + qnK+1−1(q − 2) 1 qnK+1−1+l∗nK+1−1 = µ K−1 [ k=N0 Fk ! + (q − 2) 1 ql ∗ nK+1−1 ≥ µ K−1 [ k=N0 Fk ! + (q − 2) 1 qlnK+1 ≥ µ K−2 [ k=N0 Fk ! +(q − 2) qlnK + (q − 2) qlnK+1 ≥ · · · ≥ (q − 2) K+1 X s=N0+2 1 qlns.
As the seriesP q−lnk diverges, we have a contradiction for K large enough.
Note that the proof of the above lemma only works for q > 2. However, the case q = 2 can be proved in the same way as in [1].
Now, we are going to prove the main result of this section.
Theorem 2.4. Let {ln} be a non-decreasing sequence. Then, (1.5) has infinitely many
solutionsQ for almost all g ∈ L if f ∈ ( f ∈ L : ∞ X n=0 1 qlnk = ∞ ) . Proof.Let {ln} be a non-decreasing sequence and
P∞ n=0q −lnk = ∞. Our goal is to prove that µ ∞ \ N =0 ∞ [ k=N Fk ! = 1. By Lemma 2.12, we have µ ∞ [ k=N Fk ! > 1 qc > 0, for all N. This implies µ ∞ \ N =0 ∞ [ k=N Fk ! > 0. By Lemma 2.3, the proof is complete.
Chapter 3
Proof of Theorem 1.9
In Chapter 1, we have mentioned that Kurzweil also gave a refinement of his Theorem 1.7 for the real case in [8]. In this chapter, we will prove Theorem 1.9 which is an analogue of this refinement and compare the set of {ln} with the refinement in
Theorem 1.8 which was proved by Kim, Tan, Wang and Xu.
For the sake of convenience, we recall some notation of Theorem 1.9. First recall that {rn} is a sequence which is assumed to be non-negative and non-decreasing.
Moreover, {rn} satisfies that n−rnis non-increasing and rn≥ 2n, for all n ∈ N. Next,
the definition of Ω{rn} was as follows: the sequence {ln} belongs to Ω{rn}if
(1) lnis non-decreasing, and
(2) there exists an increasing sequence of non-negative integers t1 < t2 < t3 < ...
and a function δ(n) which is non-decreasing with δ(n) → ∞ as n → ∞ such that ti+1> rti+δ(ti)− ti, and X i≥1 qti−lrt i+δ(ti)−ti = ∞.
Finally, recall the sets U{rn} := f ∈ L : ∃c > 0 such that f −P Q > q−rn+c, ∀P, Q ∈ F q[X] with deg(Q) = n . and
The goal is to show that U{rn} = W{rn}.
In order to show this, we prove the following lemma.
Lemma 3.1. Fix a non-negative integer n and non-negative integers t, k. If we choose a fractionR/S with |f − R/S| < q−t, then we have
µ [ deg(Q)≤n B {Qf }; 1 qk
≤ max{qs−k, qs+n−t} with s = deg(S). Proof.From the inequality |f − R/S| < q−t, we have
Qf − R 0 S < qn−t,
where R0 = QR and deg(Q) ≤ n. Consider the following cases:
1. If n ≤ t, then |Qf − R0/S| < 1. Let R0 = U S + V with deg(V ) < deg(S), we obtain Qf − R 0 S = Qf − U − V S = {Qf } − V S < qn−t. Hence, for g ∈ B {Qf }; q−k, g − V S = g − {Qf } + {Qf } − V S < max{q−k, qn−t}.
This implies that
B {Qf }; q−k ⊆ B V S; max{q −k , qn−t} , and consequently [ deg(Q)≤n B {Qf }; q−k ⊆ [ deg(V )<deg(S) B V S; max{q −k , qn−t} . Thus, µ [ deg(Q)≤n B {Qf }; 1 qk ≤ max{q −k , qn−t} · qs = max{qs−k, qs+n−t}.
2. If n > t, then the conclusion still holds since qs+n−t> 1. The following proposition is one direction of Theorem 1.9.
Proposition 3.1. If f /∈ U{rn}, then there exists a sequence {ln} ∈ Ω{rn} such that
(1.6) has only finitely many solutionsQ with deg(Q) = n.
Proof.If f /∈ U{rn}, then there exist a non-decreasing sequence {cn} tending to infinity
and a sequence (Pk, Qk) such that
f − Pk Qk ≤ 1 qrnk+cnk, with deg(Qk) = nk.
Let us choose a sequence {dn} satisfies the following conditions:
(i) cn≥ dn ≥ 0 and n ≥ dnfor all n ≥ 0,
(ii) dnis non-decreasing and dn → ∞ as n → ∞,
(iii) cn− dnis non-decreasing and cn− dn → ∞ as n → ∞,
(iv) n − dn→ ∞ as n → ∞. Define δ1(n) = cn−dn− dn−dn, and δ(n) = inf n≤k<∞δ1(n).
From the definition of δ1(n), we have δ(n) is non-decreasing and δ(n) → ∞ as n →
∞. Moreover,
δ(n + dn) ≤ δ1(n + dn) = cn+dn−dn+dn − dn+dn−dn+dn ≤ cn− dn.
Now, select a subsequence (Pki, Qki) of (Pk, Qk) with deg(Qki) = nkisuch that
∞ X i=1 1 qdnki < ∞ and nki+1 > rnki+cnk i − nki. (3.1)
Define ti = nki + dnki, i ∈ N. Since n − rn is non-increasing and cn ≥
dn+ δ(n + dn), we obtain
≤ −nki− dnki. (3.2)
From this and (3.1), we get
rti+δ(ti)− ti ≤ rnki+cnki − nki − dnki < nki+1 < ti+1.
Next, we define
lj = nk1 + dnk1, for 0 ≤ j ≤ rt1+δ(t1)− t1,
lj = nki+ dnki, for rti−1+δ(ti−1)− ti−1< j ≤ rti+δ(ti)− ti.
By the above definition of {lj}, we have
X
i≥1
qti−lrt
i+δ(ti)−ti = ∞.
This implies that {lj} ∈ Ω{rn}.
By Lemma 3.1 and (3.2), we can estimate the measure of the union of the fol-lowing balls µ rti+δ(ti)−ti [ n=1+rti−1+δ(ti−1)−ti−1 [ deg(Q)=n B {Qf }; 1 qln ≤ maxnqnki−nki−dnki, qnki+rti+δ(ti)−ti−rnki+cnk
i o ≤ 1 qdnki. Hence, ∞ X i=2 µ rti+δ(ti)−ti [ n=1+rti−1+δ(ti−1)−ti−1 [ deg(Q)=n B {Qf }; 1 qln ≤ ∞ X i=2 1 qdnki < ∞.
The proof is complete.
The converse inclusion U{rn} ⊆ W{rn} is a consequence of the following result.
Proposition 3.2. If f ∈ U{rn} and{ln} ∈ Ω{rn}, then for allk ≥ 0, we have
µ ∞ [ n=k [ deg(Q)=n B {Qf }, 1 qln > 1 qm (3.3)
Proof.Let us fix a sequence {ln} ∈ Ω{rn}. Choose the fuction δ(n) and the sequence
tiaccording to the definition of Ω{rn}.
Put δ0(n) = bδ(n)/2c and
l00 = l01 = · · · = lr0
t1+δ0(t1)−t1 = lrt1+δ(t1)−t1,
l0n= lrts+δ(ts)−ts for rts−1+δ0(ts−1)− ts−1 < n ≤ rts+δ0(ts)− ts, s ≥ 2.
Assume that (3.3) is false. Then, there exists k0 ∈ N such that
µ N [ n=k0 [ deg(Q)=n B {Qf }; 1 ql0 n ≤ 1 qm, for all N ≥ k0. (3.4)
Let Pi/Qi be the principle convergent of f . Since f ∈ U{rn}, there exists a
nonnegative integer c such that f − Pi Qi = 1 |Qi||Qi+1| > q−rni+c, with n i = deg(Qi).
This implies that
ni+1< rni+c − ni. (3.5)
Let us fix an integer s0 fulfilling the conditions
δ0(ts0) > max{c, m} and ts0 > n0. (3.6)
Then, we choose Pis/Qis as the subsequence of Pi/Qi with deg(Qis) = nis whose
indices forms a sequence {is} defined by
nis−1 < ts ≤ nis, s ≥ s0. (3.7) Obviously, is> 1 for s ≥ s0. Since rn≥ 2n, we have rti+δ0(ti)− ti = ti+ 2δ 0 (ti) + rti+δ(ti)− 2(ti+ δ(ti)) ≥ ti+ 2δ 0 (ti).
Hence, we can define a set Ls+1 := ( k0 ≤ deg(Q) < nis+1 : {Qf } ∈ nis+1−1 [ n=k0 [ deg(Q)=n B {Qf }; 1 ql0 n \ rts+δ0(ts)−ts [ n=k0 [ deg(Q)=n B {Qf }; 1 ql0 n )
for s ≥ s1, where the integer s1 fulfills the conditions s1 ≥ s0 and ts1 > k0.
Next, we want to estimate the number of elements in Ls+1. In order to do this,
we need to find the number of elements {Qf } with deg(Q) ≤ nis+1 contained in a
fixed ball with radius q−d. Since f = Pis+1 Qis+1 + R, with |R| < 1 |Qis+1| 2, we have {Qf } = P 0 Qis+1 + R0, (3.8) where |P0| < |Qis+1| and |R 0| < 1/|Q
is+1| for all Q ∈ Fq[X] with deg(Q) ≤ nis+1. By
(3.8), we know that {Qf } is contained in a ball of the form B(QP0
is+1, q
−nis+1
) and all these balls are disjoint. Then, the number of {Qf } with deg(Q) ≤ nis+1 belonging to
B(g, q−d) is max 1, q −d q−nis+1 = max1, qnis+1−d . (3.9)
Now, we are going to estimate the number of elements in Ls+1. Let rts+δ0(ts)−ts [ n=k0 [ deg(Q)=n B {Qf }; 1 ql0 n =[ j B {Qjf }; 1 qdj , where B({Qjf }; q−dj) are disjoint, for all j. By (3.4), we get
1 qm ≥ µ rts+δ0(ts)−ts [ n=k0 [ deg(Q)=n B {Qf }; 1 ql0 n = µ [ j B {Qjf }; 1 qdj ! =X j µ B {Qjf }; 1 qdj =X j 1 qdj.
Using (3.9), the number of {Qf } with deg(Q) ≤ nis+1belonging to
S jB {Qjf }, q −dj is at most X j
max{qnis+1−dj, 1} = max
( qnis+1 X j q−dj,X j 1 ) ≤ maxqnis+1−m , qrts+δ0(ts)−ts+1 (3.10) Since n − rnis non-increasing, rn+δ(n)− δ(n) ≥ rn+δ0(n)− δ0(n).
Adding δ(n) to both sides, we obtain rn+δ(n) ≥ rn+δ0(n)− δ0(n) + δ(n) ≥ rn+δ0(n)− δ(n) 2 + δ(n) ≥ rn+δ0(n)+ δ 0 (n). (3.11) By (3.6), (3.7), (3.11) and the definition of ti, we have
rts+δ0(ts)− ts≤ rts+δ(ts)− ts− δ 0 (ts) ≤ rts+δ(ts)− ts− m ≤ ts+1− m ≤ nis+1 − m.
This implies that (3.10) is less than qnis+1−m+1. Hence, the number of elements in L
s+1
is at least qnis+1 − qnis+1−m+1.
Next, we claim that [ Q∈Ls+1 B {Qf }; 1 ql 0 nis+1 ! ⊂ nis+1 [ n=k0 [ deg(Q)=n B {Qf }; 1 ql0 n \ rts+δ0(ts)−ts [ n=k0 [ deg(Q)=n B {Qf }; 1 ql0 n . (3.12) In order to show this, fix Q1 ∈ Ls+1. Suppose there exists a polynomial Q2 with
deg(Q) = u and k0 ≤ u ≤ rts+δ0(ts)− tssuch that
B({Q1f }; q −l0 nis+1) ∩ B({Q 2f }; q−l 0 u) 6= ∅.
We know that {Q1f } does not belong to B({Q2f }; q−l
0 u). Hence, B {Q2f }; 1 ql0 u ⊂ B {Q1f }; 1 ql 0 nis+1 ! . Then, we get q−lu0 < q−l 0
nis+1. On the other hand, since l0
n is non-decreasing and deg(Q2) = u < nis+1, we have q −l0 u ≥ q−l 0 nis+1, a contradiction. Consequently, (3.12) holds.
Now, we consider two cases: 1. If q−nis+1 ≤ q−l 0 nis+1 with s ≥ s 1, then we have [ Q∈Ls+1 B {Qf }; 1 qnis+1 ⊂ [ Q∈Ls+1 B {Qf }; 1 ql 0 nis+1 ! . (3.13)
Since |{Q1f } − {Q2f }| = |{(Q1− Q2)f }| ≥ q−nis+1 for Q1, Q2 ∈ Ls+1 with
Q1 6= Q2, we get B {Q1f }; q−nis+1 ∩ B {Q2f }; q−nis+1 = ∅. Thus, any two
balls inS
Q∈Ls+1B {Qf }, q
−nis+1 are disjoint. By (3.12) and (3.13), we obtain
µ nis+1−1 [ n=k0 [ deg(Q)=n B {Qf }; 1 ql0 n \ rts+δ0(ts)−ts [ n=k0 [ deg(Q)=n B {Qf }; 1 ql0 n ≥ µ [ Q∈Ls+1 B {Qf }; 1 qnis+1 = X Q∈Ls+1 1 qnis+1 ≥ (qnis+1 − qnis+1−m+1 ) 1 qnis+1 = (1 − q−m+1),
which when iterated yields a contradiction. 2. If q−nis+1 > q−l
0
nis+1 with s ≥ s
1, then any two balls in
S Q∈Ls+1B {Qf }; q−l 0 nis+1 are disjoint. Thus, we obtain µ nis+1−1 [ n=k0 [ deg(Q)=n B {Qf }; 1 ql0 n \ rts+δ0(ts)−ts [ n=k0 [ deg(Q)=n B {Qf }; 1 ql0 n ≥ X Q∈Ls+1 µ B {Qf }; 1 ql 0 nis+1 !! = X Q∈Ls+1 1 ql 0 nis+1 ≥ (qnis+1 − qnis+1−m+1 ) 1 ql 0 nis+1.
Applying (3.5), (3.6), (3.7) and the property that n − rn is non-increasing, we
have
nis < rnis−1+c− nis−1 ≤ rts+c− ts≤ rts+δ0(ts)− ts− δ
0
(ts) + c ≤ rts+δ0(ts)− ts.
(3.14) By (3.14) and the definition of ln0,
1 ql 0 nis+1 ≥ 1 qlrts+1+δ(ts+1)−ts+1 .
Using (3.7), we get
qnis+1 − qnis+1−m+1 ≥ qnis+1(1 − q−m+1) ≥ qts+1(1 − q−m+1).
Therefore, µ nis+1−1 [ n=k0 [ deg(Q)=n B {Qf }; 1 ql0 n \ rts+δ0(ts)−ts [ n=k0 [ deg(Q)=n B {Qf }; 1 ql0 n ≥ (1 − q−m+1) · q ts+1 qlrts+1+δ(ts+1)−ts+1 . As the series P sq ts+1−lrt
s+1+δ(ts+1)−ts+1 diverges, we have a contradiction again
by iteration.
Hence, the proof is complete
Finally, Proposition 3.1 and Proposition 3.2 imply Theorem 1.9.
Now, we are going to compare Theorem 1.9 with Theorem 1.8. As we have mentioned at the end of Section 1.3, the sets U{rn} and Us are the same when rn =
(s + 1)n. Thus, we should discuss the relationship between Ω{rn} and Ωs.
Recall the set
Ωs := ( ln ≥ 1 : ∞ X n=0 qn−sln = ∞ ) . When s = 1 (i.e., rn= 2n), it is obvious that Ω{2n} ⊆ Ω1.
On the other hand, the following two results show that the two sets Ω{rn} and
Ωsare not contained in each other when s > 1.
Proposition 3.3. Let s > 1. Choose ln =
jn+log qn+logq(logqn) s k . Then, we have{ln} ∈ Ωs\Ω{(s+1)n}.
Proof.Clearly, {ln} ∈ Ωs. Assume that {ln} ∈ Ω{(s+1)n}. Then, there exists {tn} and
δ(n) from the definition of Ω{(s+1)n}. Since ti+1 > sti + (s + 1)δ(ti) > sti for all
i ≥ 1, we have tn> sn−1for n ≥ 2. Thus, ∞ =X n≥2 qtn−lstn+(s+1)δ(tn) ≤X n≥2 qtn−stn+logq stns +1 ≤ qX n≥2 q−n logq ss = q X n≥2 s−ns < ∞, a contradiction.
Proposition 3.4. For s > 1, we have Ω{(s+1)n}\Ωs 6= ∅.
Proof.Choose any {ti} as in the definition of Ω{(s+1)n} where δ(ti) =
j
s−1
2(s+1) logqi
k . For sti−1+ (s + 1)δ(ti−1) < n ≤ sti+ (s + 1)δ(ti), define ln = ti+ blogqic.
Then, X i≥1 qti−lsti+(s+1)δ(ti) =X i≥1 q−blogqic ≥X i≥1 q− logqi =X i≥1 1 i = ∞. Thus, {ln} ∈ Ω{(s+1)n}. However, X n≥0 qn−sln ≤X i≥1 X sti−1+(s+1)δ(ti−1)<n≤sti+(s+1)δ(ti) qn−sln =X i≥1 q−s(ti+blogqic)· X sti−1+(s+1)δ(ti−1)<n≤sti+(s+1)δ(ti) qn ≤ q q − 1 X i≥1 q−sblogqic+(s+1)δ(ti) ≤ q q − 1 X i≥1
q−sblogqic+s−12 blogqic
≤ q q − 1 X i≥1 1 is+12 < ∞.
Chapter 4
Conclusion
We conclude the thesis with some conjectures.
In Section 2.2, we gave some improvements and proved that one direction of Theorem 2.1 still holds even when dropping the monotonicity condition on {ln}. In
fact, we conjecture that the converse direction also holds if we remove the monotonic-ity condition. Thus, we have the following conjecture.
Conjecture 4.1. Let {ln} be a sequence. Define
U := ( f ∈ L : ∞ X n=0 1 ql∗ n = ∞, where l ∗ n = max{nk+1− n, ln} for nk≤ n < nk+1 ) . Then,f ∈ U if and only if (1.5) has infinitely many solutions Q for almost all g ∈ L
If the above conjecture is true, then this would allow us to prove Kurzweil’s theorem in a particular easy manner.
In Chapter 3, we have proved Theorem 1.9 and compared the sets Ω{rn} with
Ωs when rn = (s + 1)n. Note that the approximation functions of Theorem 1.8 and
Theorem 1.9 are of the same form q−ln (in contrast to the other theorems, the fuction
does not tend to 0 as n tends to infinity). The sequence {ln} in Ω{rn} is assumed to be
non-decreasing. An interesting question is whether or not one can improve Theorem 1.9 by dropping the monotonicity condition on {ln}? If yes, then what can be said
about the relation between the sets Ω{rn} and Ωswhen rn= (s + 1)n?
Overall, there are still interesting questions left concerning inhomogeneous Dio-phantine approximation in the field of formal Laurent series.
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