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−lrti+δ(ti)−ti = ∞.
Finally, recall the sets U{rn} :=
f ∈ L : ∃c > 0 such that
f −P Q
> q−rn+c, ∀P, Q ∈ Fq[X] with deg(Q) = n
. and
W{rn} :=f ∈ L : ∀ln ∈ Ω{rn}, (1.6) has infinitely many solutions Q for almost all g ∈ L .
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
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
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
rti+δ(ti)−rnki+cnk
i−ti = −nki−cnki+δ(ti)+nki+cnki−rnki+cnk
i−(ti+δ(ti)−rti+δ(ti))
≤ −nki− dnki. (3.2) From this and (3.1), we get
rti+δ(ti)− ti ≤ rnki+cnk
i − 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−lrti+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
n=1+rti−1+δ(ti−1)−ti−1
[
n=1+rti−1+δ(ti−1)−ti−1
[
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
Proof.Let us fix a sequence {ln} ∈ Ω{rn}. Choose the fuction δ(n) and the sequence
Assume that (3.3) is false. Then, there exists k0 ∈ N such that
µ nonnegative integer c such that
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+ 2δ0(ti) + rti+δ(ti)− 2(ti+ δ(ti)) ≥ ti+ 2δ0(ti).
Hence, we can define a set
Ls+1 :=
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 these balls are disjoint. Then, the number of {Qf } with deg(Q) ≤ nis+1 belonging to B(g, q−d) is Now, we are going to estimate the number of elements in Ls+1. Let
rts+δ0(ts)−ts Since n − rnis non-increasing,
rn+δ(n)− δ(n) ≥ rn+δ0(n)− δ0(n).
Adding δ(n) to both sides, we obtain
This implies that (3.10) is less than qnis+1−m+1. Hence, the number of elements in Ls+1 is at least qnis+1 − qnis+1−m+1.
Next, we claim that [ 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
0nis+1. On the other hand, since ln0 is non-decreasing and deg(Q2) = u < nis+1, we have q−lu0 ≥ q−l
0
nis+1, a contradiction. Consequently, (3.12) holds.
Now, we consider two cases:
1. If q−nis+1 ≤ q−l
Since |{Q1f } − {Q2f }| = |{(Q1− Q2)f }| ≥ q−nis+1 for Q1, Q2 ∈ Ls+1 with
which when iterated yields a contradiction.
2. If q−nis+1 > q−l
Applying (3.5), (3.6), (3.7) and the property that n − rn is non-increasing, we have
Using (3.7), we get As the series P
sqts+1−lrts+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 =
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 3.4. 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) logqik . 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 < ∞.
Hence, {ln} is not in Ωs, which implies Ω{(s+1)n}\Ωs 6= ∅.
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
qln∗ = ∞, where ln∗ = 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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