國立臺中教育大學數學教育學系碩士班碩士論文
指導教授：陳中川 博士
餘弦算子函數的拓樸混合性
研究生：張 仕 儒 撰
謝誌
首先誠摯的感謝指導教授陳中川博士，老師悉心的教導使我得以一窥
數學領域的深奧，不時的討論並指點我正確的方向，使我在這些年中獲益
匪淺，老師對學問的嚴謹更是我輩學習的典範。
本論文的完成另外亦得感謝國立中央大學的李明憶教授大力協助，及
國立東華大學林英芬教授的支持。因為有你們的體諒及幫忙，使得本論文
能夠更完整而嚴謹。
兩年裡的日子，研究室裡共同的生活點滴，學術上的討論、言不及意
的閒聊、讓人又愛又怕的宵夜、趕作業的革命情感、因為睡太晚而遮遮掩
掩的閃進研究室...，感謝多位學長姐、同學、學弟妹的共同砥礪(墮
落?)，你/妳們的陪伴讓兩年的研究生活變得絢麗多彩。
感謝蘇偉興、黃立餘、江銘倫學長們不厭其煩的指出我研究中的缺失，
且總能在我迷惘時為我解惑，也感謝許敬平、黃之敬、虞翔、高嘉祥、王
中瓔、許哲毓、楊品柔、駱湘芸同學的幫忙，恭喜我們順利走過這兩年。
研究室的蔡松祐、王志文學弟當然也不能忘記，你的幫忙及搞笑我銘
記在心。
女朋友張詩怡在背後的默默支持更是我前進的動力，没有張詩怡的體
諒、包容，相信這兩年的生活將是很不一樣的光景。
最后，謹以此文獻给我摯愛的雙親。
摘要
對於 Banach 空間中的一個有界線性算子，若存在一個非零向量使得其所形 成的軌跡稠密於該空間，則我們稱此算子為超循環算子，而超循環的概念相近於 拓樸性質中的傳遞性。同時，我們也可以考慮算子數列的拓樸傳遞性或超循環性 質。在論文中，我們推演出一些關於拓樸傳遞和拓樸混合算子數列的結果，其中 混合性是一個比傳遞性更強的拓樸性質且此一算子數列可被視為一個餘弦算子 函數。此外，我與我的指導老師已經將論文中的一些結果發表於期刊（[6]）。 在這篇論文當中，首先我們針對一個在 Lebesgue空間中的雙邊加權位移算 子去探究它的拓樸傳遞性以及超循環性質。接著，我們定義了一個餘弦算子函 數，它是由可逆的加權位移算子與其逆算子所構成的一個有界算子數列，而這樣 的啟發是來自於學者 Bonilla 和 Miana 的研究工作。其中，我們透過權重的描 述來刻劃拓樸混合餘弦算子函數，特別的是此一描述亦是其加權位移算子形成拓 撲混合性的充分必要條件。同時我們若放寬權重的條件，則可以得到造成拓樸傳 遞餘弦算子函數的充分條件，且我們也給出拓樸傳遞餘弦算子函數的必要條件。 最後，我們利用此篇論文主要結果的相似證明方法，重新發現 Salas、Costakis 和 Sambarino 在加權位移算子方面的一些結果；從而統一了拓樸混合單一算子與拓 樸混合算子數列的論證方式。 關鍵字：超循環性、拓樸傳遞性、拓樸混合性、餘弦算子函數、加權位移算子 I1
Introduction
A bounded linear operator T on a Banach space X is hypercyclic if there exists a non-zero vector x ∈ X such that its orbit under the operator
orb(T, x) = {x, T x, T2x, T3x, ...}
is dense in X, in which x is called a hypercyclic vector for T . We note that T is said to be topologically transitive if for any pair U , V of nonempty open subsets of X, there exists some n ∈ N such that Tn(U ) ∩ V 6= ∅. If Tn(U ) ∩ V 6= ∅ from some n onwards, then T is topologically mixing. It has been shown in [11] that topological transitivity and hypercyclicity are equivalent. Transitivity and hypercyclicity for a single operator T have been studied by many authors; we refer to [11, 12] for surveys and recent books on this subject.
Let N0 = N ∪ {0}. In the study of hypercyclicity and transitivity, the shift
operators on `p(N0) or `p(Z) play an important role for researchers to construct
and demonstrate the theory of dynamic of linear operators. Let {ej : j ∈ N0}
be the canonical basis of `p(N0), and let (aj)j∈N0 be a bounded sequence of
positive numbers. Then a unilateral weighted forward shift T : `p(N
0) → `p(N0)
is defined by T ej = ajej+1 for all j ∈ N0. Since hypercyclicity can only occur
in infinite-dimensional separable Banach spaces [2, 3], we assume 1 ≤ p < ∞ throughout. In 1969, Rolewicz showed the following result in [14], who is the first author studying transitivity and hypercyclicity on Banach spaces.
Theorem 1.1 ([14]) Let T be a unilateral weighted forward shift on `p(N
0), with
aj ≡ 1 for all j ∈ N0. Then λT is hypercyclic (transitive) if | λ |> 1.
Similarly, let {ej : j ∈ Z} be the canonical basis of `p(Z) for 1 ≤ p < ∞,
and let (aj)j∈Z be a bounded sequence of positive numbers. Then a bilateral
weighted forward shift T : `p(Z) → `p(Z) is defined by T e
j = ajej+1 for all
j ∈ Z. If (aj) is also bounded away from 0, then T is invertible and its inverse
T−1 = S : `p(Z) → `p(Z) is given by Se
j = aj−11 ej−1. It has been shown in [11]
that an invertible operator is transitive if, and only if, its inverse is transitive. In [15], Salas successfully gave the sufficient and necessary conditions for bilateral weighted forward shifts on `p(Z) to be topologically transitive.
Theorem 1.2 ([15]) Let T be a bilateral weighted forward shift on `p(Z) with a
positive weight sequence (aj)j∈Z. Then T is transitive if, and only if, given ε > 0
and q ∈ N, there exists n arbitrarily large such that for all |j| ≤ q,
n−1 Y s=0 aj+s< ε and n Y s=1 aj−s> 1 ε.
Costakis and Sambarino in [9] characterized topologically mixing bilateral weighted forward shifts on `p(Z) as below.
Theorem 1.3 ([9]) Let T be a bilateral weighted forward shift on `p(Z) with a
positive weight sequence (aj)j∈Z. Then T is topologically mixing if, and only if,
given ε > 0 and q ∈ N, there exists some N ∈ N such that for all | j |≤ q and n ≥ N , n−1 Y s=0 aj+s< ε and n Y s=1 aj−s> 1 ε.
We note that the weight conditions in Theorem 1.2 and 1.3 should be satisfied for a subsequence of numbers and a sequence of numbers respectively. After the work of Salas, Costakis and Sambarino, many related results for weighted shifts were obtain affluently. For example, Feldman simplified Salas’ characterization for invertible bilateral weighted shifts in [10]. Le´on-Saavedra gave the descriptions for unilateral and bilateral weighted shifts on `p(N0) and `p(Z) to be Ces´aro
hypercyclic in [13]. Common hypercyclic backward shifts on `p(N
0) were studied
in [1]. Recently, Salas’ results were extended to weighted translation operators on the Lebesgue space of locally compact groups in [7, 8], which provided a class of topologically transitive and mixing operators on Banach spaces.
In fact, the notion of hyperyclicity can be considered for a sequence of bounded linear operators (Tn). That is, (Tn) is hypercyclic if {x, T1x, T2x, T3x, ...} is dense
in a Banach space X for some vector x ∈ X. One can also study transitivity of a sequence of operators. We call (Tn) topologically transitive if for any nonempty
open subsets U and V of X, Tn(U ) ∩ V 6= ∅ for some n. If Tn(U ) ∩ V 6= ∅ from
some n onwards, then (Tn) is topologically mixing . It should be noted that in
this case, (Tn) is transitive if, and only if, (Tn) is hypercyclic and (Tn) has a
dense set of hypercyclic vectors [11]. In the thesis, we will consider a sequence of bounded operators, generated by an invertible weighted shift and its inverse, and give sufficient and necessary conditions for such a sequence of operators to be topologically mixing. We will make use of the equivalence below to obtain the main result.
Theorem 1.4 ([11]) Let X be a Banach space, and (Tn) be a sequence of bounded
linear operators on X. Then the following conditions are equivalent. (i) (Tn) is topologically mixing.
(ii) For every x, y ∈ X, there is a sequence (xn) in X such that
xn→ x and Tnxn → y
as n → ∞.
Also, (Tn) is topologically transitive if, and only if, for every x, y ∈ X, there
are sequences (xk) in X and (nk) in N0 such that xk → x and Tnkxk → y as
k → ∞.
In what follows, let 1 ≤ p < ∞ and let T be an invertible bilateral weighted forward shift on `p(Z), defined above, with its inverse S. Then we define a sequence of bounded operators (Cn) by
Cn=
1 2(T
n
+ Sn)
for all n ∈ Z in which T−n := (T−1)n = Sn. Then (C
n) can be regarded as a
cosine operator function.
A cosine operator function on a Banach space X is a mapping C from the real line into the space of continuous operators on X satisfying C(0) = I, and the d’Alembert functional equation 2C(t)C(s) = C(t + s) + C(t − s) for all s, t ∈ R, which implies C(t) = C(−t) for all t ∈ R. Cosine operator functions are crucial for the investigation of semigroups, and we refer to [5, 16] for some references. The motivation for the study of (Cn) is inspired by the work in [4]. In [4], Bonilla
and Miana obtained a sufficient condition for a cosine operator function C(t), defined by
C(t) = 1
2(T (t) + T (−t)),
to be transitive where T is a strongly continuous translation group on some weighted Lebesgue space Lp(R).
Here we define a cosine operator function C from Z into the space of continuous operators by letting C(n) = Cn. Since Cn = C−n for all n ∈ Z, we will consider
the sequence of operators (Cn)n∈N0, and characterize transitive and mixing cosine
operator functions (Cn)n∈N0.
2
Topologically mixing cosine operator functions
In this chapter, we obtain some new results for mixing cosine operator functions generated by shifts, which has been published in [6]. Before discussing the details, we first have the following simple observation.
Lemma 2.1 Let T be a bilateral weighted forward shift on `p(Z) with its inverse
S. Then Cn= 12(Tn+ Sn) is a cosine operator function.
Proof.
Let B(`p(Z)) be the space of bounded operators on `p(Z). Let C : Z → B(`p(Z))
be defined by C(n) = Cn= 12(Tn+ Sn). Then C(0) = C0 = 1 2(T 0 + S0) = I and 2C(n)C(m) = 2 1 2(T n+ Sn) 1 2(T m+ Sm) = 1 2(T n+m + Sn+m+ TnSm+ SnTm) = 1 2(T n+m + Sn+m+ Tn−m+ Sn−m) = C(n + m) + C(n − m)
which implies Cn is a cosine operator function.
♦ Now we are ready to give the main result. The description below is the necessary and sufficient conditions for the cosine operator function Cn, generated
by the weighted shift T and its inverse S, to be topologically mixing.
Theorem 2.2 Let T be a bilateral weighted forward shift on `p(Z) with its inverse
S, and let Cn= 12(Tn+ Sn). Then the following conditions are equivalent.
(i) (Cn) is topologically mixing.
(ii) Given ε > 0 and q ∈ N, there exists some N ∈ N such that for all | j |≤ q and n ≥ N , n−1 Y s=0 aj+s< ε and n Y s=1 aj−s> 1 ε. Proof.
(i) ⇒ (ii). Let (Cn) be topologically mixing. Let ε > 0 and q ∈ N, and choose
δ ∈ (0, 1) such that 1−δ2δ < ε. Then there exist a vector x ∈ `p(Z) and some N ,
such that x − X |j|≤q ej < δ and k Cnx − 0 k< δ for all n ≥ N . Let x = P
j∈Zαjej. Then this inequality k x −P|j|≤qej k< δ implies αj >
1 − δ, if | j |≤ q, and | αj |< δ otherwise. Let E+ = {j ∈ Z : αj > 0} and
E− = {j ∈ Z : αj < 0}. Then we define x+ = X j∈E+ αjej and x−= − X j∈E− αjej which says x = x+− x−.
Now consider k Cnx k< δ. We have
δ >k Cnx k ≥ k Cnx+ k = 1 2(T nx++ Snx+) ≥ 1 2 k T nx+k which implies 2δ > k Tnx+ k 6
≥ Tn X |j|≤q αjej = X |j|≤q αjTnej = X |j|≤q αj n−1 Y s=0 aj+sej+n = X |j|≤q αj n−1 Y s=0 aj+sej+n . Therefore we have, for all n ≥ N ,
| αj | n−1 Y s=0 aj+s < 2δ (| j |≤ q). Hence n−1 Y s=0 aj+s< 2δ | αj | < 2δ 1 − δ < ε (| j |≤ q). Similarly, 2δ > k Snx+ k ≥ Sn X |j|≤q αjej = X |j|≤q αjSnej = X |j|≤q αj 1 Qn s=1aj−s ej−n = X |j|≤q αj 1 Qn s=1aj−s ej−n . Hence for all n ≥ N , we have
n Y s=1 aj−s> | αj | 2δ > 1 − δ 2δ > 1 ε (| j |≤ q). 7
(ii) ⇒ (i). Let D = span{ej : j ∈ Z} which is dense in `p(Z). Let U and V be
open subsets of `p(Z). Then pick f ∈ U ∩ D, and g ∈ V ∩ D with
supp f ∪ supp g = {| j |≤ q}
for some q ∈ N. Then we may assume
f = X |j|≤q αjej and g = X |j|≤q βjej. Let M = max|j|≤q{| αj |, | βj |}.
Given ε > 0, there exists some N > 2q such that the weight condition (ii) holds for all n ≥ N . Then
k Tn(ej) kpp = k ajaj+1...aj+(n−1)ej+n kpp = | ajaj+1...aj+(n−1)|pk ej+n kpp = | ajaj+1...aj+(n−1)|p = n−1 Y s=0 aj+s !p < εp, and k Sn(e j) kpp = 1 aj−1aj−2...aj−n ej−n p p = 1 aj−1aj−2...aj−n p k ej−n kpp = 1 aj−1aj−2...aj−n p = 1 (Qn s=1aj−s)p < εp. 8
Let vn= f + Tng + Sng ∈ `p(Z). Then k vn− f kpp = k f + T n g + Sng − f kpp = k Tng + Sng kpp = k Tng kpp + k Sng kpp = Tn X |j|≤q βjej p p + Sn X |j|≤q βjej p p = X |j|≤q βjTn(ej) p p + X |j|≤q βjSn(ej) p p = X |j|≤q | βj |pk Tn(ej) kpp + X |j|≤q | βj |pk Sn(ej) kpp < X |j|≤q Mpεp + X |j|≤q Mpεp = 2 X |j|≤q Mpεp = 2(2q + 1)Mpεp which implies vn → f as n → ∞.
Similarly, one has k Cnvn− g kpp = k Cn(f + Tng + Sng) − g kpp = 1 2T nf + 1 2S nf + 1 2T 2ng + 1 2S nTng +1 2T nSng + 1 2S 2ng − g p p = 1 2p k T nf + Snf + T2ng + g + g + S2ng − 2g kp p = 1 2p k T nf + Snf + T2ng + S2ng kp p = 1 2p n k Tnf kp p + k S nf kp p + k T 2ng kp p + k S 2ng kp p o = 1 2p Tn X |j|≤q αjej p p + Sn X |j|≤q αjej p p + T2n X |j|≤q βjej p p + S2n X |j|≤q βjej p p 9
= 1 2p X |j|≤q αjTnej p p + X |j|≤q αjSnej p p + X |j|≤q βjT2nej p p + X |j|≤q βjS2nej p p = 1 2p X |j|≤q | αj |pk Tnej kpp + X |j|≤q | αj |pk Snej kpp + X |j|≤q | βj |pk T2nej kpp +1 2p X |j|≤q | βj |pk S2nej kpp < 1 2p X |j|≤q Mpεp+ X |j|≤q Mpεp+ X |j|≤q Mpεp+ X |j|≤q Mpεp = 1 2p 4 X |j|≤q Mpεp = 1 2p {4(2q + 1)M pεp} = 1 2p−2(2q + 1)M pεp
which says Cnvn→ g as n → ∞. Since D is dense in `p(Z), we have Cn(U )∩V 6= ∅
for all n ≥ N , and therefore (Cn) is topologically mixing.
♦ Using the similar method in the proof of Theorem 2.2, we obtain the necessary condition for topological transitivity in terms of weaker weight conditions. Corollary 2.3 Let T be a bilateral weighted forward shift on `p(Z) with its in-verse S, and let Cn = 12(Tn + Sn). Let (Cn) be topologically transitive. Then
given ε > 0 and q ∈ N, there exists n arbitrarily large such that for all | j |≤ q,
n−1 Y s=0 aj+s< ε and n Y s=1 aj−s> 1 ε. Proof.
We could use the same way as the proof (i) ⇒ (ii) of Theorem 2.2 to illustrate this issue. Let (Cn) be topologically transitive. Let ε > 0 and q ∈ N, and choose
δ ∈ (0, 1) such that 1−δ2δ < ε. Then there exist a vector x ∈ `p(Z) and some n > 2q, such that x − X |j|≤q ej < δ and k Cnx − 0 k< δ. Let x =P
j∈Zαjej. Then this inequality k x −P|j|≤qej k< δ implies αj > 1 − δ,
if | j |≤ q, and | αj |< δ otherwise. Let E+ = {j ∈ Z : αj > 0} and E− = {j ∈
Z : αj < 0}. Then we define x+ = X j∈E+ αjej and x−= − X j∈E− αjej which says x = x+− x−.
Now consider k Cnx k< δ. We have
δ >k Cnx k ≥ k Cnx+ k = 1 2(T nx++ Snx+) ≥ 1 2 k T nx+k which implies 2δ > k Tnx+ k ≥ Tn X |j|≤q αjej = X |j|≤q αjTnej = X |j|≤q αj n−1 Y s=0 aj+sej+n = X |j|≤q αj n−1 Y s=0 aj+sej+n . Therefore we have | αj | n−1 Y s=0 aj+s < 2δ (| j |≤ q). 11
Hence for some n, we have n−1 Y s=0 aj+s< 2δ | αj | < 2δ 1 − δ < ε (| j |≤ q). Similarly, 2δ > k Snx+ k ≥ Sn X |j|≤q αjej = X |j|≤q αjSnej = X |j|≤q αj 1 Qn s=1aj−s ej−n = X |j|≤q αj 1 Qn s=1aj−s ej−n . Hence for all | j |≤ q, we have
n Y s=1 aj−s> | αj | 2δ > 1 − δ 2δ > 1 ε for some n. ♦ Because topological transitivity is a weaker topological property than mixing, we give the sufficient condition for topological transitivity by loosing the weight condition in Theorem 2.2.
Corollary 2.4 Let T be a bilateral forward weighted shift on `p(Z) with its
in-verse S, and let Cn = 12(Tn + Sn). If given ε > 0 and q ∈ N, there exists n
arbitrarily large such that for all | j |≤ q,
n−1 Y s=0 aj+s< ε and n Y s=1 aj−s> 1 ε, 12
2n−1 Y s=0 aj+s< ε and 2n Y s=1 aj−s> 1 ε, then (Cn) is topologically transitive.
Proof.
We could use the similar way as the proof (ii)⇒(i) of Theorem 2.2 to illustrate this issue. Let D = span{ej : j ∈ Z} which is dense in `p(Z). Let U and V be
open subsets of `p(Z). Then pick f ∈ U ∩ D, and g ∈ V ∩ D with
supp f ∪ supp g = {| j |≤ q}
for some q ∈ N. Then we may assume
f = X |j|≤q αjej and g = X |j|≤q βjej. Let M = max|j|≤q{| αj |, | βj |}.
Given ε > 0, there exists some n > 2q such that the four weight conditions in Corollary 2.4 above are satisfied. Then
k Tn(e j) kpp = k ajaj+1...aj+(n−1)ej+n kpp = | ajaj+1...aj+(n−1)|pk ej+n kpp = | ajaj+1...aj+(n−1)|p = n−1 Y s=0 aj+s !p < εp, and k T2n(e j) kpp = k ajaj+1...aj+(2n−1)ej+2nkpp = | ajaj+1...aj+(2n−1) |pk ej+2nkpp 13
= | ajaj+1...aj+(2n−1) |p = 2n−1 Y s=0 aj+s !p < εp. Also, we have k Sn(ej) kpp = 1 aj−1aj−2...aj−n ej−n p p = 1 aj−1aj−2...aj−n p k ej−n kpp = 1 aj−1aj−2...aj−n p = 1 (Qn s=1aj−s)p < εp, and k S2n(e j) kpp = 1 aj−1aj−2...aj−2n ej−2n p p = 1 aj−1aj−2...aj−2n p k ej−2nkpp = 1 aj−1aj−2...aj−2n p = 1 (Q2n s=1aj−s)p < εp. Let vn= f + Tng + Sng ∈ `p(Z). Then k vn− f kpp = k f + T ng + Sng − f kp p = k Tng + Sng kpp ≤ k Tng kp p + k S ng kp p = Tn X |j|≤q βjej p p + Sn X |j|≤q βjej p p = X |j|≤q βjTn(ej) p p + X |j|≤q βjSn(ej) p p 14
= X |j|≤q | βj |pk Tn(ej) kpp + X |j|≤q | βj |pk Sn(ej) kpp < X |j|≤q Mpεp+ X |j|≤q Mpεp = 2 X |j|≤q Mpεp = 2(2q + 1)Mpεp which implies vnk → f as k → ∞.
Similarly, one has k Cnvn− g kpp = k Cn(f + Tng + Sng) − g kpp = 1 2T nf + 1 2S nf + 1 2T 2ng +1 2S nTng + 1 2T nSng +1 2S 2ng − g p p = 1 2p k T nf + Snf + T2ng + g + g + S2ng − 2g kp p = 1 2p k T nf + Snf + T2ng + S2ng kp p ≤ 1 2p n k Tnf kp p + k S nf kp p + k T 2ng kp p + k S 2ng kp p o = 1 2p Tn X |j|≤q αjej p p + Sn X |j|≤q αjej p p + T2n X |j|≤q βjej p p + S2n X |j|≤q βjej p p = 1 2p X |j|≤q αjTnej p p + X |j|≤q αjSnej p p + X |j|≤q βjT2nej p p + X |j|≤q βjS2nej p p = 1 2p X |j|≤q | αj |pk Tnej kpp + X |j|≤q | αj |pk Snej kpp + X |j|≤q | βj |pk T2nej kpp +1 2p X |j|≤q | βj |pk S2nej kpp < 1 2p X |j|≤q Mpεp+ X |j|≤q Mpεp + X |j|≤q Mpεp+ X |j|≤q Mpεp = 1 2p 4 X |j|≤q Mpεp 15
= 1 2p {4(2q + 1)M pεp} = 1 2p−2(2q + 1)M pεp
which says Cnkvnk → g as k → ∞. Since D is dense in `
p(Z), we have C
n(U )∩V 6=
∅ for some n, and therefore (Cn) is topologically transitive.
♦
3
Topologically mixing weighted shifts
In this chapter, we recover the results of Salas [15], Costakis and Sambarino [9] by using the similar argument in the proof of Theorem 2.2.
Theorem 3.1 Let T be an invertible bilateral weighted forward shift with a pos-itive weight sequence (aj)j∈Z. Then T is topologically transitive (hypercyclic) if,
and only if, given ε > 0 and q ∈ N, there exists n arbitrarily large such that for all | j |≤ q, n−1 Y s=0 aj+s< ε and n Y s=1 aj−s> 1 ε. Proof.
(i). We first show the necessary condition for transitive T . Let T be topologically transitive. Let ε > 0 and q ∈ N, and choose δ ∈ (0, 1) such that 1−δδ < ε. Then there exist a vector x ∈ `p(Z) and some n > 2q, such that
x − X |j|≤q ej < δ and Tnx + X |j|≤q ej < δ. Let x =P
j∈Zαjej. Then this inequality k x −P|j|≤qej k< δ implies αj > 1 − δ,
if | j |≤ q, and | αj |< δ otherwise. Let E+ = {j ∈ Z : αj > 0} and E− = {j ∈
Z : αj < 0}. Then we define x+ = X j∈E+ αjej and x−= − X j∈E− αjej which says x = x+− x−. Now consider k Tnx+ k= (Tnx + X |j|≤q ej − X |j|≤q ej)+ ≤ (Tnx + X |j|≤q ej)+ + (− X |j|≤q ej)+ ≤ Tnx + X |j|≤q ej + 0 < δ 17
which implies δ > k Tnx+k≥ Tn X |j|≤q αjej = X |j|≤q αjTnej = X |j|≤q αj n−1 Y s=0 aj+sej+n = X |j|≤q αj n−1 Y s=0 aj+sej+n . Therefore we have | αj | n−1 Y s=0 aj+s < δ (| j |≤ q). Hence n−1 Y s=0 aj+s < δ | αj | < δ 1 − δ < ε (| j |≤ q) for some n.
On the other hand, the inequality
Tnx + X |j|≤q ej < δ implies δ > Tnx + X |j|≤q ej = TnX j∈Z αjej+ X |j|≤q ej = X j∈Z αjTnej+ X |j|≤q ej = X j∈Z αj n−1 Y s=0 aj+sej+n+ X |j|≤q ej = X j∈Z αj−n n−1 Y s=0 aj+s−nej + X |j|≤q ej . Therefore we have αj−n n−1 Y s=0 aj+s−n+ 1 < δ (| j |≤ q) 18
for some n, which implies αj−n n−1 Y s=0 aj+s−n < δ − 1 < 0 (| j |≤ q). Hence αj−n n−1 Y s=0 aj+s−n > 1 − δ (| j |≤ q). Consider k x− k= (x − X |j|≤q ej + X |j|≤q ej)− ≤ (x − X |j|≤q ej)− + (X |j|≤q ej)− ≤ x − X |j|≤q ej + 0 < δ which implies δ >k x− k=k SnTnx−k = k Sn(Tnx)− k ≥ Sn X |j|≤q αj−n n−1 Y s=0 aj+s−n ej = X |j|≤q αj−n n−1 Y s=0 aj+s−n k Sne j k = X |j|≤q αj−n n−1 Y s=0 aj+s−n 1 Qn s=1aj−s ej−n = X |j|≤q αj−n n−1 Y s=0 aj+s−n 1 Qn s=1aj−s k ej−n k . Therefore we have αj−n n−1 Y s=0 aj+s−n 1 Qn s=1aj−s < δ (| j |≤ q) for some n. Hence
1 Qn s=1aj−s < δ αj−n Qn−1 s=0 aj+s−n < δ 1 − δ < ε (| j |≤ q) 19
which implies n Y s=1 aj−s> 1 ε (| j |≤ q) for some n.
(ii). Now, we prove the sufficient condition for T to be transitive. Let D = span{ej : j ∈ Z} which is dense in `p(Z). Let U and V be open subsets of `p(Z).
Then pick f ∈ U ∩ D, and g ∈ V ∩ D with
supp f ∪ supp g = {| j |≤ q}
for some q ∈ N. Then we may assume
f = X |j|≤q αjej and g = X |j|≤q βjej. Let M = max|j|≤q{| αj |, | βj |}.
Given ε > 0, there exists some n > 2q such that the two weight conditions are satisfied. Then k Tn(e j) kpp = k ajaj+1...aj+(n−1)ej+n kpp = | ajaj+1...aj+(n−1)|pk ej+n kpp = | ajaj+1...aj+(n−1)|p = n−1 Y s=0 aj+s !p < εp, and k Sn(e j) kpp = 1 aj−1aj−2...aj−n ej−n p p 20
= 1 aj−1aj−2...aj−n p k ej−n kpp = 1 aj−1aj−2...aj−n p = 1 (Qn s=1aj−s)p < εp. Let vn= f + Sng ∈ `p(Z). Then k vn− f kpp = k f + S ng − f kp p=k S ng kp p= Sn X |j|≤q βjej p p = X |j|≤q βjSn(ej) p p = X |j|≤q | βj |pk Sn(ej) kpp< X |j|≤q Mpεp = (2q + 1)Mpεp which implies vnk → f as k → ∞.
Similarly, one has k Tnv n− g kpp = k Tn(f + Sng) − g kpp=k Tnf + TnSng − g kpp = k Tnf + g − g kpp=k Tnf kpp = Tn X |j|≤q αjej p p = X |j|≤q αjTnej p p = X |j|≤q | αj |pk Tnej kpp < X |j|≤q Mpεp = (2q + 1)Mpεp which says Tnkv nk → g as k → ∞. Since D is dense in ` p(Z), we have Tn(U )∩V 6=
∅ for some n, and therefore T is topologically transitive.
♦ By strengthening the weight condition in Theorem 3.1, we can also describe the necessary and sufficient conditions for T to be topologically mixing.
Theorem 3.2 Let T be an invertible bilateral weighted forward shift with a posi-tive weight sequence (an)n∈Z. Then T is topologically mixing if, and only if, given
ε > 0 and q ∈ N, there exists some N ∈ N such that for all | j |≤ q and n ≥ N , n−1 Y s=0 aj+s< ε and n Y s=1 aj−s> 1 ε. Proof.
(i). We first show the necessary condition for mixing T . Let T be topologically mixing. Let ε > 0 and q ∈ N, and choose δ ∈ (0, 1) such that 1−δδ < ε. Then there exist a vector x ∈ `p(Z) and some N ∈ N, such that
x − X |j|≤q ej < δ and Tnx + X |j|≤q ej < δ
for all n ≥ N . Let x = P
j∈Zαjej. Then this inequality k x −P|j|≤qej k< δ
implies αj > 1−δ, if | j |≤ q, and | αj |< δ otherwise. Let E+= {j ∈ Z : αj > 0}
and E− = {j ∈ Z : αj < 0}. Then we define
x+ = X j∈E+ αjej and x−= − X j∈E− αjej which says x = x+− x−. Now consider k Tnx+ k= (Tnx + X |j|≤q ej − X |j|≤q ej)+ ≤ (Tnx + X |j|≤q ej)+ + (− X |j|≤q ej)+ ≤ Tnx + X |j|≤q ej + 0 < δ which implies δ > k Tnx+k≥ Tn X |j|≤q αjej = X |j|≤q αjTnej = X |j|≤q αj n−1 Y s=0 aj+sej+n = X |j|≤q αj n−1 Y s=0 aj+sej+n . 22
Therefore we have | αj | n−1 Y s=0 aj+s < δ (| j |≤ q). Hence n−1 Y s=0 aj+s < δ | αj | < δ 1 − δ < ε (| j |≤ q) for all n ≥ N .
On the other hand, the inequality
Tnx + X |j|≤q ej < δ implies δ > Tnx + X |j|≤q ej = TnX j∈Z αjej+ X |j|≤q ej = X j∈Z αjTnej+ X |j|≤q ej = X j∈Z αj n−1 Y s=0 aj+sej+n+ X |j|≤q ej = X j∈Z αj−n n−1 Y s=0 aj+s−nej + X |j|≤q ej . Therefore we have αj−n n−1 Y s=0 aj+s−n+ 1 < δ (| j |≤ q) for all n ≥ N , which implies
αj−n n−1 Y s=0 aj+s−n < δ − 1 < 0 (| j |≤ q). 23
Hence αj−n n−1 Y s=0 aj+s−n > 1 − δ (| j |≤ q). Consider k x− k= (x − X |j|≤q ej + X |j|≤q ej)− ≤ (x − X |j|≤q ej)− + (X |j|≤q ej)− ≤ x − X |j|≤q ej + 0 < δ which implies δ >k x− k=k SnTnx−k = k Sn(Tnx)− k ≥ Sn X |j|≤q αj−n n−1 Y s=0 aj+s−n ej = X |j|≤q αj−n n−1 Y s=0 aj+s−n k Sne j k = X |j|≤q αj−n n−1 Y s=0 aj+s−n 1 Qn s=1aj−s ej−n = X |j|≤q αj−n n−1 Y s=0 aj+s−n 1 Qn s=1aj−s k ej−n k . Therefore we have αj−n n−1 Y s=0 aj+s−n 1 Qn s=1aj−s < δ (| j |≤ q) for all n ≥ N . Hence
1 Qn s=1aj−s < δ αj−n Qn−1 s=0 aj+s−n < δ 1 − δ < ε (| j |≤ q) which implies n Y s=1 aj−s> 1 ε (| j |≤ q) 24
for all n ≥ N .
(ii). Now, we prove the sufficient condition for T to be mixing. Let D = span{ej :
j ∈ Z} which is dense in `p(Z). Let U and V be open subsets of `p(Z). Then pick f ∈ U ∩ D, and g ∈ V ∩ D with
supp f ∪ supp g = {| j |≤ q}
for some q ∈ N. Then we may assume
f = X |j|≤q αjej and g = X |j|≤q βjej. Let M = max|j|≤q{| αj |, | βj |}.
Given ε > 0, there exists some N > 2q such that the two weight conditions are satisfied. Then for all n ≥ N ,
k Tn(ej) kpp = k ajaj+1...aj+(n−1)ej+n kpp = | ajaj+1...aj+(n−1)|pk ej+n kpp = | ajaj+1...aj+(n−1)|p = n−1 Y s=0 aj+s !p < εp, and k Sn(e j) kpp = 1 aj−1aj−2...aj−n ej−n p p = 1 aj−1aj−2...aj−n p k ej−n kpp = 1 aj−1aj−2...aj−n p = 1 (Qn s=1aj−s)p < εp. 25
Let vn= f + Sng ∈ `p(Z). Then k vn− f kpp = k f + S ng − f kp p=k S ng kp p= Sn X |j|≤q βjej p p = X |j|≤q βjSn(ej) p p = X |j|≤q | βj |pk Sn(ej) kpp< X |j|≤q Mpεp = (2q + 1)Mpεp which implies vn → f as n → ∞.
Similarly, one has k Tnv n− g kpp = k T n(f + Sng) − g kp p=k T nf + TnSng − g kp p = k Tnf + g − g kpp=k Tnf kpp = Tn X |j|≤q αjej p p = X |j|≤q αjTnej p p = X |j|≤q | αj |pk Tnej kpp < X |j|≤q Mpεp = (2q + 1)Mpεp which says Tnv
n→ g as n → ∞. Since D is dense in `p(Z), we have Tn(U )∩V 6= ∅
for all n ≥ N , and therefore T is topologically mixing.
♦ From the results of (Cn) and T , we can observe some relations for the
topo-logical properties of (Cn) and T .
Corollary 3.3 Let T be a bilateral weighted forward shift on `p(Z) with its in-verse S. Let Cn = 12(Tn+ Sn). Then we have the following results.
(i) If (Cn) is topologically transitive, then T is topologically transitive.
(ii) (Cn) is topologically mixing if, and only if, T is topologically mixing.
Proof.
We are concerned about (i), first. Let (Cn) be topologically transitive. Then, by
Corollary 2.3, given ε > 0 and q ∈ N, there exists n arbitrarily large such that for all | j |≤ q, n−1 Y s=0 aj+s< ε and n Y s=1 aj−s> 1 ε,
which is the weight conditions of Theorem 3.1 to make T to be topologically transitive.
Now consider (ii). By Theorem 2.2, (Cn) is mixing if, and only if, given ε > 0
and q ∈ N, there exists some N ∈ N such that for all | j |≤ q and n ≥ N ,
n−1 Y s=0 aj+s< ε and n Y s=1 aj−s> 1 ε,
which is also the necessary and sufficient condition for T to be topologically mixing in Theorem 3.2.
♦
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