Dense Orbits of the Aluthge Transform

DENSE ORBITS OF THE ALUTHGE TRANSFORM

Kevin Rion

A Dissertation

Submitted to the Graduate College of Bowling Green State University in partial fulﬁllment of the requirements for the degree of

DOCTOR OF PHILOSOPHY

January 2011

Committee:

Juan B´es,Advisor

John Hoag, Graduate Faculty Representative

Kit Chan

Craig Zirbel ii ABSTRACT

Juan B´es,Advisor

In this dissertation we investigated the two topics hypercyclicity and the Aluthge transform. Each of these is related to the Invariant Subspace Problem. On a topological vector space X, a linear operator T : X → X is said to be hypercyclic if there is a vector x for which the orbit {x, T x, T 2x, T 3x, . . .} is dense in X. The Aluthge transform is a mapping introduced to transform continuous linear operators defined on a Hilbert space into nicer operators that have the same spectral properties as the original operators. The first chapter is an introduction to hypercyclicity, chaos, and mixing, and is where the Aluthge transform is defined and some of its spectral properties are listed. In the second chapter, we explored whether the dynamical properties of an operator are preserved by the Aluthge transform. We showed for bilateral weighted shifts, an operator T is mixing, chaotic, or hypercyclic if and only if the Aluthge transform ∆(T ) of the operator has the same dynamical property. We also supplied conditions for when the Aluthge transform ∆(T ) of an arbitrary operator T has the same dynamical properties as T . In chapter three we provided a strong form of a counterexample to a conjecture by Jung, Ko, and Percy [26]. They conjectured that

n the iterates of the Aluthge transform {∆ (T )}n≥1 converge to a normal operator for every continuous linear operator T on a Hilbert space X. We used a probabilistic argument to

n show that if T is any bilateral forward shift, then either the sequence {∆ (T )}n≥1 converges to a normal shift in the strong operator topology, or fails to converge in a dramatic sense in that the set of strong operator topology subsequential limits of the sequence is an “interval” of normal shifts. Then we showed for any interval [a, b] ⊂ (0, ∞) there is a bilateral weighted forward shift T for which the set of strong operator topology subsequential limits of the sequence of iterates of the Aluthge transform of T is the set {t · S : a ≤ t ≤ b} where S is the pure forward shift. These results were extended to address complexly weighted shifts, and bilateral backward shifts. In the last chapter, we address where “most” hypercyclic vectors are located relative to the range of a hypercyclic operator. If x is hypercyclic for T , iii then so is T nx for every natural number n, and T nx is in the range of T . Since moreover, the range of T is dense in X, one might expect that for any hypercyclic operator T , most (if not all) hypercyclic vectors for T are in the range of T . To the contrary, we showed for every non-surjective hypercyclic operator T on a Banach space X, the set of hypercyclic vectors for T that are not in the range of T is large in that it is a set of category II. We also provided a sense by which the range of an arbitrary hypercyclic operator is large in the set of hypercyclic vectors for T . iv

To my parents. v ACKNOWLEDGMENTS

In 2003, I came to BGSU with the intention of earning a master’s degree in statistics. I stayed a little longer than I had planned, and in the process met kind professors, made good friends, and met my wife. I will forever carry with me many fond memories of my time in Bowling Green. There are many people I would like to thank for helping me to grow more knowledgeable and more thoughtful along the way. The help sometimes came from direct advice, but more often it came by way of example. I am grateful to my committee members Dr. Juan Bés, Dr. Kit Chan, Dr. John Hoag, and Dr. Craig Zirbel for all of their help and hard work during this process. I would like to thank my advisor Dr. Juan Bèswho, while on sabbatical in Spain, helped me to finish this year by agreeing to a teleconferenced final defense. I would like to thank Dr. Kit Chan from whom I learned much about teaching by being a student in his classes. Moreover, his thoughtfulness and advice were invaluable. I apologize to Dr. Craig Zirbel for asking him to read yet another dissertation on hypercyclicity. As everyone who meets him soon learns, his kindness and good spirit were reliable constants. I wish to thank Dr. Gordon Wade who stood by me in a time of hardship and without whom I would have left long before earning a Ph.D. I am left with fond memories of being a student in the classes of Dr. Arjun Gupta, Dr. John Chen, and Dr. Truc Nguyen. Each was very caring and understands the value of encouragement. I will miss watching the above mentioned statisticians as well as Dr. Hanfeng Chen argue with each other in seminar. The arguments were often informative, and always entertaining. I would also like to thank Ian Clough who first showed me what math is about, and left me with a strong desire to learn more. Lastly, I would like to thank my wife Irina for all of her love and support. I have both learned from and relied upon the strength of her will and the compassion in her heart. I am grounded and guided by her companionship and conversation. She has more confidence in me than I have in myself and I am grateful for her presence in my life. vi

Table of Contents

CHAPTER 1: Introduction 1 1.1 The Invariant Subspace Problem ...... 1 1.2 An Introduction to the Aluthge Transform ...... 2 1.3 An Introduction to Hypercyclicity ...... 4 1.4 An Introduction to Weighted Shifts ...... 9

CHAPTER 2: Dynamical properties of operators under the action of the Aluthge transform 13 2.1 Hypercyclicity of the Aluthge transform of weighted shifts ...... 13 2.1.1 Bilateral Case ...... 13 2.1.2 Unilateral Case ...... 21 2.2 Chaos of the Aluthge transform of weighted shifts ...... 23 2.3 Dynamical Properties of the Aluthge transform of an arbitrary operator . . . 24 2.4 D-hypercyclicity of the Aluthge transform of N weighted shifts ...... 31 2.5 Universality of the Aluthge transform of a sequence of weighted shifts . . . . 37

CHAPTER 3: The Aluthge transform acting on the set of weighted shifts 41 3.1 Properties of the Aluthge transform on the set of weighted shifts ...... 41 3.2 Locally universal shifts under the iterations of the Aluthge transform . . . . 50

CHAPTER 4: Hypercyclicity and the range of an operator 67 vii BIBLIOGRAPHY 74 1

CHAPTER 1

Introduction

1.1 The Invariant Subspace Problem

The two notions of hypercyclicity and the Aluthge transform, which constitute the object of study in my dissertation, are closely related to the Invariant subspace Problem in operator theory. This well known open problem asks whether there exist continuous linear operators whose only invariant closed subspaces are the trivial ones, {0} and the whole space X. Here, a set A is said to be invariant for T if T [A] ⊆ A. Hypercyclicity is the study of continuous linear operators T : X → X that possess a dense orbit Orb(T, x)= {x, T x, T 2x, . . .}, for some x in X. To emphasize the connection of hypercyclicity to the Invariant Subspace Problem we note that if the orbit of every non-zero vector x in X under the action of T is dense in X, then since an invariant closed linear subset F containing x must also contain Orb(T, x), we have that either F = {0} or F = H. So not only would the only invariant subspaces be the trivial subspaces, the only T invariant subsets would be X and {0}. Using X = `1, Enﬂo [16] provided the ﬁrst example of a continuous linear operator on a Banach space whose only invariant closed subspaces are trivial. This answered the Invariance Subspace Problem for Banach Spaces. Later, Read [32] went further by providing an operator on `1 for which every non-zero vector is hypercyclic, thus showing 2 that on a Banach space there are continuous linear operators that have only the trivial invariant closed subsets. However, the problem remains open for Hilbert spaces. It is on these Hilbert spaces that Aluthge [1] introduced in 1990 an operator transform that would aid in answering the Invariant Subspace Problem. Since normal operators have many non-trivial invariant closed subspaces, the idea behind the Aluthge transform is to convert an arbitrary operator T into another operator ∆(T ) that preserves some of the structure of the invariant subspaces of T however ∆(T ) is closer to being normal. After a brief introduction to hypercyclicity, the Aluthge transform and weighted shifts, we focus in Chapter 2 on analyzing some dynamical properties of weighted shift operators under the action of the Aluthge transform. In Chapter 3 we focus our attention on iterates of the Aluthge transform ∆ and deduce some results asserting the existence of a type of

n local density of the sequence {∆ (T )}n≥1 for an operator T in the set of weighted shifts S. Finally, in Chapter 4 we deduce some interesting results about the range of a hypercyclic operator.

1.2 An Introduction to the Aluthge Transform

Let H be a separable inﬁnite dimensional Hilbert space and let B(H) denote the algebra of all bounded linear operators on H. Given that an arbitrary operator T in B(H) has a

∗ 1 unique polar decomposition T = U |T |, where |T | = (T T ) 2 and U is the appropriate partial

1 1 isometry, we define the Aluthge transform ∆ : B(H) → B(H) by setting ∆(T ) = |T | 2 U |T | 2 . Historically, the Aluthge transform was introduced in [1] and has received much attention because of its intimate connection to the Invariant Subspace Problem. The well known open problem raises the question whether every operator in B(H) has a non-trivial, invariant, closed subspace F ⊂ H. Here, a non-trivial subspace means F 6= {0} and F 6= H. On finite dimensional (complex) spaces, every operator has such a non-trivial invariant subspace because of the existence of eigenvectors. However, for separable infinite dimensional Hilbert 3 spaces the problem is still open. It is known that for the class of normal operators the problem has a positive answer and in fact, the idea behind the Aluthge transform in relation to the Invariant Subspace Problem is in rough terms to convert an arbitrary operator in B(H) into another operator which shares the spectral properties of the first but is closer to being a normal operator. To this end, we note first that in [26] Jung, Ko and Pearcy showed that for any T ∈ B(H) the various spectra of the operator T coincide with their counterparts for ∆(T ).

Theorem 1. [26] Let T ∈ B(H) for a Hilbert space H, and let ∆(T ) denote the Aluthge transform of T . Then, (i) the spectrum σ[T ] = σ[∆(T )],

(ii) the point spectrum σp[T ] = σp[∆(T )],

(iii) the approximate point spectrum σap[T ] = σap[∆(T )],

(iv) the essential spectrum σe[T ] = σe[∆(T )],

(v) the left essential spectrum σle[T ] = σle[∆(T )], and

(vi) the right essential spectrum σre[T ] = σre[∆(T )].

Moreover, in [26, Corollary 1.16] the same authors showed that for any T ∈ B(H) if the operator ∆(T ) has a non-trivial, invariant, closed subspace then so does T , however there are examples in which the lattice of invariant subspaces of T , denoted by Lat(T ), is not isomorphic to Lat(∆(T )). In [26] Jung, Ko and Percy conjectured that the iterates of the Aluthge transform

n {∆ (T )}n≥1 converge in norm to a normal operator for every T ∈ B(H). We note that for each non-negative integer n, the nth Aluthge transform ∆n(T ) of T is deﬁned induc- tively by setting ∆n(T ) = ∆[∆n−1(T )], and ∆0(T ) = T . At ﬁrst, results that seemed to support this conjecture appeared. For example, Yamazaki [39] expressed the spectral radius ρ(T ) of any T ∈ B(H) in terms of the Aluthge sequence, by proving ρ(T ) = lim k∆n(T )k. However, the conjecture was shown to be false by Thomson n→∞ (as referenced in [27]), who found an operator whose Aluthge transform sequence does not 4 converge in norm. Furthermore, Yanagida [40] found a forward shift whose Aluthge transform sequence does not even converge in the weak operator topology (WOT). In this dissertation I investigate this conjecture further for weighted shifts and in Section

3.2 show that in fact, if Tw is any bilateral weighted shift, then the Aluthge transform

2 sequence {Tw, ∆(Tw), ∆ (Tw),...} either converges in the strong operator topology (SOT) to a normal shift, or fails to converge in a dramatic sense. More speciﬁcally, the set of SOT subsequential limits of the Aluthge transform sequence either consists of a single normal shift, or is an entire closed “interval” of normal shifts. Furthermore, in Section 3.1 I investigate the continuity and range of the Aluthge transform as a function restricted to the set of all weighted shifts. The following two results about the continuity and the range of ∆ motivate that investigation.

Theorem 2. [2] The Aluthge transform ∆ : B(H) → B(H) is norm to norm continuous at T ∈ B(H), provided T has a closed range.

Finally, Ji, Pong, and Li [25] described some properties about the range Range(∆) = {∆(T ): T ∈ B(H)} of the Aluthge transform.

Theorem 3. [25] The range of the Aluthge transform ∆ : B(H) → B(H) is not closed with respect to the norm, SOT nor WOT topology. Moreover, it is not norm dense, but it is SOT and WOT dense.

For a brief survey on diﬀerent properties of the Aluthge transform see [3].

1.3 An Introduction to Hypercyclicity

Hypercyclicity is the study of continuous linear operators with dense orbits. Since on finite- dimensional spaces linear transformations do not posses dense orbits, hypercyclicity is ex- clusively an infinite dimensional phenomenon. To be precise, we say a continuous linear operator T on a topological vector space X is hypercyclic, if there exists a vector x in X whose orbit Orb(T, x)= {x, T x, T 2x, . . .} is dense 5 in X. We call such a vector x a hypercyclic vector for T and denote the set of hypercyclic vectors for T by HC(T ). Hypercyclicity is closely related to a notion in the field of topological dynamics, which Birkhoff [10] showed to be equivalent to hypercyclicity on second countable Baire spaces.

Theorem 4 (Birkhoﬀ Transitivity Theorem). Let X be a separable F-space and T : X → X be a continuous linear operator. The following are equivalent:

(i) T is hypercyclic;

(ii) T is topologically transitive: that is, for each pair of non-empty open sets U and V in X there exists an n ≥ 1 so that T n(U) ∩ V 6= ∅.

Birkhoff [10] also provided a first instance of a continuous linear operator with a dense orbit. He showed that for the space of entire functions H(C) endowed with the compact-open topology, there exists an entire function f and a non-zero complex number a for which the induced translation operator defined by Ta(f)z = f(z + a) is hypercyclic. In other words, the orbit Orb(Ta, f) = {f(z), f(z + a), f(z + 2a),...} can approximate any entire function g arbitrarily closely in the topology of H(C). As it turns out, if an operator has such a hypercyclic vector then its set of hypercyclic vectors is large in a topological sense.

Theorem 5. [28] Suppose X is a separable F-space and T : X → X a continuous linear operator. If T is hypercyclic, then its set of hypercyclic vectors HC(T ) is a dense Gδ set.

This further implies that the set of hypercyclic vectors is large also in an algebraic sense.

Theorem 6. [22] If X is a separable F-space and T : X → X is a hypercyclic operator, then every x ∈ X is the sum of two hypercyclic vectors.

Since the set of hypercyclic vectors is not closed under addition, it is natural to ask whether the set HC(T ) has a rich algebraic structure. Independently, Bourdon [11] and 6 Herrero [24] showed that HC(T ) ∪ {0} always contains an inﬁnite dimensional linear subset. Such a subset is called a hypercyclic manifold for T , and consists entirely, except for the zero vector, of hypercyclic vectors.

Theorem 7. [38] Let X be a topological vector space and T : X → X a hypercyclic operator.

If x ∈ HC(T ), then K[T ]z := {p(T )x : p is a polynomial} is a hypercyclic manifold for T . In particular, T admits a dense hypercyclic manifold.

For certain classes of hypercyclic operators, the set HC(T ) contains in fact a closed inﬁnite-dimensional subspace. To further outline the development of hypercyclicity, we introduce the well known hypercyclicity criterion. We say a continuous linear operator T on a topological vector space

X satisﬁes the Hypercyclicity Criterion, if there exists an increasing sequence {nk}, two

dense sets D1 and D2 and a sequence of maps Snk : D2 → X such that:

nk (i) T x → 0 for all x ∈ D1;

(ii) Snk y → 0 for all y ∈ D2;

nk (iii) T Snk y → y for all y ∈ D2.

This criterion was obtained in a restricted form by Kitai [28] and Gethner and Shapiro [20]. The version we use here is due to Bes and Peris [9]. While this criterion is only a suﬃcient condition for hypercyclicity, it has proven extremely useful.

Theorem 8. [9] Let X be a separable F-space and T a continuous linear operator. If T satisﬁes the Hypercyclicity Criterion, then T is hypercyclic.

Leon-Saavedra and Montes-Rodriguez [29] obtained a complete characterization of operators admitting a closed hypercyclic subspace provided they satisfy the Hypercyclicity Criterion. By a hypercyclic subspace for an operator T we mean a closed inﬁnite dimensional subspace Y ⊂ X so that Y \{0} ⊂ HC(T ). 7 Theorem 9. [29] Let H be a separable complex Hilbert space and consider a continuous linear operator T satisfying the Hypercyclicity Criterion. The operator T has a closed hypercyclic subspace if and only if the essential spectrum of T intersects the closed unit disk.

In the remainder of this section we introduce some commonly studied variations on the notion of hypercyclicity as well as generalizations of this concept in a few diﬀerent directions. We start by introducing the notion of supercyclicity which only requires a linear operator

T to have a dense projective orbit K · Orb(T, x) = {λT nx : n ∈ N, λ ∈ C} in X. Similarly, we say a linear operator T is cyclic if there exists a vector x in X so that the span of the orbit K[T ]x = span [Orb(T, x)] is dense in X. Clearly, cyclicity is weaker than the notion of supercyclicity, which in turn is weaker than hypercyclicity. A notion stemming from topological dynamics that does not require the linearity of the operator is that of topological mixing. On a topological vector space X, a continuous operator T is said to be (topologically) mixing, if for all non-empty open sets U, V ⊂ X, there exists an N ≥ 1 so that T n(U) ∩ V 6= ∅ for all n ≥ N. We remark that mixing is a strong form of topological transitivity which is known to be equivalent to hypercyclicity of linear operators on a separable F-space. In fact in this setting a linear operator T is mixing if and only if it is hereditarily hypercyclic with respect to the full sequence nk = k for all k in N, a notion we now deﬁne.

An operator T is said to be hereditarily hypercyclic with respect to the sequence {nk}

mk if for all subsequences {mk} of {nk}, there exists a vector x ∈ X so that {T x : k ∈ N} is dense in X. An operator is hereditarily hypercyclic if it is hereditarily hypercyclic with respect to some sequence {nk}. Another very important concept in topological dynamics is the notion of chaos, for which Devaney [15] proposed the following deﬁnition. A continuous function T on an F-space X is called chaotic if:

(i) it has a dense orbit for some x in X, 8 (ii) it has a dense set of periodic points, and

(iii) it depends sensitively on initial conditions.

We say T depends sensitively on initial conditions if there exists a strictly positive number δ such that for every > 0 and every x ∈ X there is a point y ∈ B(x, ) such that d(T nx, T ny) ≥ δ for some non-negative integer n. In other words, orbits that start close together should not stay close together, otherwise the operator would be too regular for it to be truly unpredictable. However, Banks et al. [5] showed that the third property in the deﬁnition is redundant since it turns out that a dense orbit and a dense set of periodic points is suﬃcient for an operator to have sensitivity on initial conditions. In the linear setting, Godefroy and Shapiro [21] showed that hypercyclicity alone implies the sensitivity to initial conditions, and so on an F-space a continuous linear operator T is chaotic provided it is hypercyclic and has a dense set of periodic points. Finally, a more general notion of hypercyclicity is that of universal operator families, in which the dense set is not necessarily generated by the iterated applications of a single

operator. Instead, a family of continuous functions {Tn}n∈N deﬁned on the same topological vector space X into Y is said to be universal if there exists a vector x in X for which the set

{Tnx : n ∈ N} is dense in Y . Hypercyclicity is of course a particular instance of universality

n in which Tn = T for all n ∈ N, and Y = X. In [22], Grosse-Erdmann provides the following suﬃcient condition for universality in the linear case.

Theorem 10. [22, page 352] Suppose X is a Baire topological vector space, Y is a metrizable topological vector space and for each n ∈ N, Tn : X → Y is a continuous linear operator.

If there are dense subsets X0 of X and Y0 of Y , and mappings S1,S2,... from Y0 to X for which

(i) Tn → 0 pointwise on X0,

(ii) there exists n1 < n2 < . . . so that Snk → 0 pointwise on Y0, and 9

(iii) Tn ◦ Sny → y on Y0,

then the sequence {Tn}n∈N has a dense Gδ set of universal vectors.

Furthermore, Grosse-Erdmann shows that if the Tn are well-behaved on a dense set then, similar to hypercyclicity, the set of universal elements is either empty or residual.

Theorem 11. [22, Proposition 6] If the sequence {Tn} converges pointwise on a dense subset of X, then the set of universal elements is either empty or residual.

1.4 An Introduction to Weighted Shifts

The class of weighted shifts is a favorite testing ground in the ﬁeld of operator theory as many properties of these operators can be described in relatively transparent ways in terms of their weight sequence. Since weighted shifts constitute the object of study in this dissertation, we investigate in this section some of their basic properties as well as provide a description of their dynamical behavior.

2 We start by deﬁning the Hilbert space ` (J) consisting of all {xj}j∈J with square summable

2 X 2 entries, that is x ∈ ` (J) provided |xj| < ∞. Here, J is one of Z+ = N ∪ {0} or Z. j∈J In loose terms, the action of a weighted shift operator on `2(J) can be described by ﬁrst

multiplying each entry xj in the sequence by a corresponding weight wj followed by shifting every entry by one position.

2 For J = Z+, we call the weighted shifts deﬁned on ` (Z+) unilateral, since the sequences

2 (x0, x1, x2,...) in ` (Z+) are inﬁnitely indexed only in one direction. On the other hand, for J = Z we introduce the bilateral weighted shifts deﬁned on `2(Z) consisting of sequences

(. . . , x−2, x−1, x0, x1, x2,...) that are inﬁnitely indexed in two directions. A weighted shift is classiﬁed as either a forward or a backward shift depending on whether we shift the sequence in the increasing or decreasing direction of the indices. More precisely,

2 2 for the canonical base {ej : j ∈ Z}, a linear operator T : ` (Z) → ` (Z) is said to be a 10 bilateral weighted forward shift if there is a sequence of complex weights {w } so that j j∈Z

T ej = wjej+1 for all j ∈ Z. On the other hand, T is a bilateral weighted backward shift if

T ej = wjej−1 for all j ∈ Z. We note that the deﬁnition of a bilateral weighted shift depends on the chosen basis.

The standard is to use the canonical base for which ej is the function taking k in Z to 0 if

0 j 6= k and otherwise taking k to 1. Upon reordering the canonical base, with ej = e−j, a forward shift deﬁned with respect to the canonical base is a backward shift when its action

0 is described with respect to the new base {ej; j ∈ Z}. Because of this, it is often, but not always, immaterial which type of shift one wishes to study. In particular, when a property of bilateral forward shifts is characterized in terms of the weight sequence, the characterization transfers to a characterization for bilateral backward shifts. However, this is not the case for unilateral forward and backward shifts.

2 2 For the canonical base {ej : j ∈ Z+}, we say a linear operator T : ` (Z+) → ` (Z+) is a

unilateral weighted forward shift provided there is a sequence of complex weights {wj}j≥0 so

that T ej = wjej+1 for all j ∈ Z+.A unilateral weighted backward shift with weight sequence

{wj}j≥1 has T ej = wjej−1 for j ≥ 1 and T e0 = 0.

Consequently, since a forward unilateral shift acts on a vector x = (x0, x1, x2,...) in

2 ` (Z+) by creating a zero in the ﬁrst component, it can never have a dense orbit and hence cannot be hypercyclic. However, by a classical result of Rolewicz [34], the unilateral backward

shift T for which T (x0, x1, x2,...) = (w1x1, w2x2,...) can in fact have a dense orbit. For these reasons we will consider unilateral backward shifts when investigating the connection between properties of an operator T and its Aluthge transform ∆(T ).

Later, in Section 3.2 we investigate the dynamical properties of the orbit Orb(∆,Tw) =

2 {Tw, ∆(Tw), ∆ (Tw),...} of the Aluthge transform on a shift. Here we will focus on unilateral forward shifts because as we will show, the Aluthge transform of a unilateral forward shift is another unilateral forward shift, but that of a unilateral backward shift is not another backward shift. The results in this section on the unilateral forward shift are then extended 11 to include bilateral shifts, both forward and backward.

In the remainder of this section we summarize some results about the dynamical properties of weighted backward shifts. Since a weighted backward shift with complex weight

sequence {wj}j∈J is isometrically isomorphic to a shift with positive weights {|wj|}j∈J (see [36]), we will assume if not otherwise indicated that the weight sequence for the shift operator is always strictly positive. Finally, we note that the norm of a weighted shift operator

kT k is sup|wj|, and so the operator T is bounded if and only if its weight sequence {wj}j∈J j∈J is bounded. In describing the linear dynamics of weighted backward shifts we start by noting that Salas [35] characterized the hypercyclicity of weighted shifts by providing a necessary and suﬃcient condition in terms of the weight sequence.

Theorem 12. [35] A bilateral weighted backward shift T : `2(Z) → `2(Z) is hypercyclic if and only if for every > 0 and every q ∈ N there exists n arbitrarily large so that for every j ∈ Z n−1 n Y 1 Y with |j| ≤ q we have w > and w < . A unilateral weighted backward shift T j+s j−s s=0 s=1 n Yk is hypercyclic if and only if there exists an increasing sequence {nk} so that wj → ∞. j=1 In [13], Chan and Seceleanu further showed that a bilateral or unilateral weighted backward shift is hypercyclic if in fact the shift operator has an orbit with a non-zero limit point.

Theorem 13. [13] A bilateral or unilateral weighted backward shift T is hypercyclic if and only if there exists a vector x in `2(J) whose orbit Orb(T, x) has a non-zero limit point.

In the case of invertible bilateral weighted shifts, Feldman [17] showed the following useful equivalence. We note that while a unilateral weighted backward shift is never injective as it has a non-trivial kernel, a bilateral weighted backward shift is invertible if and only if its weight sequence is bounded below by some M > 0, that is |wj| ≥ M for all j ∈ Z. 12 Proposition 14. [17] An invertible bilateral weighted backward shift T with weight sequence n Yk {wj}j∈Z is hypercyclic if and only if there exists an increasing sequence {nk} so that wj → j=1 n Yk ∞ and w−j → 0. j=1 Moreover, Chan and Sanders [12] introduced a weaker notion of hypercyclicity, which they compared to the usual norm-hypercyclicity for weighted shift operators. An operator T on a separable, inﬁnite dimensional Hilbert space H is weakly hypercyclic if there is a vector x in H whose orbit Orb(T, x) is dense in H with respect to the weak topology.

Theorem 15. [12] A unilateral weighted backward shift is norm-hypercyclic if and only if it is weakly hypercyclic. However, there exists a bilateral weighted backward shift that is weakly hypercyclic but not norm-hypercyclic. 13

CHAPTER 2

Dynamical properties of operators under the action of the Aluthge transform

2.1 Hypercyclicity of the Aluthge transform of weighted

shifts

2.1.1 Bilateral Case

2 2 Given the canonical basis {ej : j ∈ Z} for ` (Z), we say a bounded linear operator T : ` (Z) → `2(Z) is a bilateral weighted backward shift if there is a bounded sequence of positive weights

{wj : j ∈ Z} so that T ej = wjej−1 for all j ∈ Z. We note that since T is bounded, so is the weight sequence {w } , as kT k = sup {w : j ∈ }. In the following we characterize the j j∈Z j Z hypercyclicity of the Aluthge transform of a shift T in relation to the hypercyclic behavior of T . For a bilateral weighted shift to be hypercyclic, Salas [35] provided a necessary and suﬃcient condition in terms of the weight sequence: for every > 0 and every q ∈ N there n−1 Y 1 exists n arbitrarily large such that for every j ∈ with |j| ≤ q we have w > and Z s+j s=0 14 n Y wj−s < . A much simpler form of the criteria was derived by Feldman [17] for the case s=1 of an invertible bilateral shift T , which states that if there exists a sequence nk % ∞ such n n Yk Yk that wj → ∞ and w−j → 0, then T is hypercyclic. j=1 j=1 Before we can make use of the above criteria to obtain a characterization of the hypercyclicity of ∆(T ), we ﬁrst need to describe the action of the Aluthge transform on the class of bilateral weighted shifts. Recall for an arbitrary operator T in B(H), the

1 1 Aluthge transform is deﬁned by setting ∆(T ) = |T | 2 U |T | 2 where T has a unique po-

∗ 1 lar decomposition T = U |T | with |T | = (T T ) 2 and U is the appropriate partial isom-

1 1 etry. To ﬁnd |T | 2 we use that |T | = (T ∗T ) 2 . Writing the bilateral weighted backward

shift T = BDw as the composition of a diagonal operator Dw with the weights {wj} on the diagonal followed by the unweighted bilateral backward shift B, we obtain that

∗ ∗ ∗ ∗ ∗ T T = (BDw) BDw = DwB BDw = DwSBDw = D|w|2 , where S = B denotes the unweighted bilateral forward shift. Then the unique positive operator whose square is D|w|2

is D|w|, and thus |T | = D|w|. In particular, for a positive weight sequence {wj}, we obtain

√ that |T | = Dw. Hence the unique positive operator whose square is Dw is D w, and so

1 2 √ |T | = D w. To ﬁnd the form of the partial isometry U we require U to preserve norms on (ker T )⊥ =

2 2 ` (Z), the ran U = cl(ran T ) = ` (Z), and T = U |T |. Since T = BDw, we get that U = B.

√ √ √ √ √ √ √ Thus ∆(T )ej = D wBD wej = BD w( wjej) = D w( wjej−1) = wj−1wjej−1, for all j

in Z. Hence ∆(T ) is a bilateral weighted backward shift with weight sequence {bj} given by √ bj = wj−1wj for all j ∈ Z. We summarize these results in the following proposition.

Proposition 16. Given a bilateral weighted backward shift T : `2(Z) → `2(Z) with positive weight sequence {w } , its Aluthge transform ∆(T ) is a bilateral weighted backward shift j j∈Z √ with weight sequence {b } given by b = w w for all j ∈ . Furthermore, we identify j j∈Z j j−1 j Z 1 1 1 1 2 2 2 2 √ |T | and U in the deﬁnition of the Aluthge transform ∆(T ) = |T | U |T | as |T | = D wj √ the diagonal operator with weights { wj}j∈Z on the diagonal, and U = B the unweighted bilateral backward shift. 15 To simplify our arguments in what follows, we now describe the connection between products of consecutive weights wj of a bilateral weighted backward shift T with the products

of the weights bj of ∆(T ). We ﬁrst observe that for all positive integers n

√ √ √ b1 ····· bn = w0 · w1 · w1 · w2 ····· wn−1 · wn (2.1.1) √ w0 = √ · w1 ····· wn, wn

and

√ √ √ b−1 ····· b−n = w−2 · w−1 · w−3 · w−2 ····· w−n−1 · w−n (2.1.2) √ √ = w−1 · w−2 ····· w−n · w−n−1 √ w−n−1 = √ · w−1 ····· w−n. w−1

More generally, for all integers m < n we have

n √ n Y wm−1 Y b = √ w . (2.1.3) j w j m n m

Similarly, for all integers m < n we can express the weights of T in terms of the weights of ∆(T ) by

n √ n Y wn Y w = √ b . (2.1.4) j w j m m−1 m

We now consider the relation between the hypercyclic behavior of an invertible bilateral weighted backward shift T and its Aluthge transform ∆(T ). Recall that a weighted shift is invertible if and only if there exists a δ > 0 so that δ ≤ wj for all j ∈ Z. 16 Proposition 17. Suppose T : `2(Z) → `2(Z) is an invertible bilateral weighted backward shift with weight sequence {w } . Then the Aluthge transform ∆(T ) of T is hypercyclic if j j∈Z and only if T is hypercyclic.

Proof. Given the invertible bilateral weighted backward shift T we note that there is some

δ > 0 so that 0 < δ ≤ wj ≤ kT k for all j ∈ Z. Therefore the form obtained in formula (2.1.3) gives us that for all integers m < n

√ n n p n δ Y Y kT kY p wj ≤ bj ≤ √ wj. kT k m m δ m

To show that an invertible bilateral shift with weight sequence {aj} is hypercyclic, by

Feldman’s criterion it suﬃces to show that there exists a sequence nk % ∞ such that n n Yk Yk aj → ∞ and a−j → 0. Note that we can use this criterion to show both implications j=1 j=1 in the theorem, as once we assume that 0 < δ ≤ wj, the weights bj are also bounded below by δ and thus the Aluthge transform of T is an invertible shift.

Consequently, given a sequence nk % ∞,

n Yk i. if w1 ····· wnk → ∞ as k → ∞, then b1 ····· bnk → ∞ as k → ∞, since bj ≥ 1 √ n δ Yk p wj; kT k 1

n Yk ii. if w−1 ····· w−nk → 0 as k → ∞, then b−1 ····· b−nk → 0 as k → ∞, since b−j = 1 −1 p −1 p nk Y kT k Y kT kY bj ≤ √ wj = √ w−j; δ δ −nk −nk 1

n Yk iii. if b1 ····· bnk → ∞ as k → ∞, then w1 ····· wnk → ∞ as k → ∞, since wj ≥ 1 √ n δ Yk p bj; kT k 1

n Yk iv. if b−1 ····· b−nk → 0 as k → ∞, then w−1 ····· w−nk → 0 as k → ∞, since w−j = 1 17 −1 p −1 p nk Y kT k Y kT kY wj ≤ √ bj = √ b−j. δ δ −nk −nk 1

Thus by Feldman’s criterion we obtain that an invertible bilateral weighted backward shift T is hypercyclic if and only if ∆(T ) is hypercyclic.

From the above proof one can easily deduce an analogous statement for the case of unilateral shifts, as we will see in Section 2.1.2. Furthermore, the proof also shows how the behavior of the weights of T , diverging as a subsequence to inﬁnity for the positive indices and converging using the same subsequence to zero for the negative indices, carries over to the weights of the Aluthge transform of T . However, we cannot use the same idea for the case of a non-invertible bilateral weighted shift since in the non-invertible case there is √ n n √ n δ Y Y kT k Y no δ so that √ wj ≤ bj ≤ √ wj, and so we cannot apply Salas’ criterion on kT k δ m m m hypercyclicity of shifts with the same ease. Instead we use another equivalent statement for the hypercyclicity of shifts due to Chan and Seceleanu [13]. They showed a bilateral weighted backward shift T is hypercyclic if and only if T has an orbit Orb(T, x) with a non-zero limit point.

Theorem 18. Suppose T is a bilateral weighted backward shift on `2(Z). Then T is hypercyclic if and only if its Aluthge transform ∆(T ) is hypercyclic.

Proof. Given a hypercyclic bilateral weighted backward shift T with positive weights {wj},

there exists a vector x and a non-zero vector y ∈ `2(Z) so that y is a limit point of the orbit

nk Orb(T, x). In other words, there exists a sequence nk % ∞ so that T x → y as k → ∞. √ Recall the transform ∆(T ) is the bilateral shift with weights given by bj = wj−1wj.

√ √ √ √ Let z = D wx and h = D wy, where D w is the diagonal operator with weights { wj}

along its diagonal. We note that the vectors z and h are in `2(Z) by the boundedness of √ { w} and furthermore h is non-zero as all the weights wj are positive. Hence by formula (2.1.1) for any integer m, 18

! nk X ((∆(T )) z)(m) = bj−nk+1 ···· bj · z(j) · ej−nk (m) j∈Z = bm+1 ····· bm+n · z(m + nk) √ k wm √ = √ wm+1 ····· wm+nk · wm+nk · x(m + nk) wm+n √ k = wm · wm+1 ····· wm+nk · x(m + nk) √ nk = wm · (T x)(m).

Hence, √ √ nk 2 X nk 2 k(∆(T )) z − hk = | wm · (T x)(m) − wm · y(m)| m∈Z X 2 ≤ kT k · |(T nk x − y)(m)| m∈Z = kT k kT nk x − yk2 → 0, as k → ∞.

This shows h is a non-zero limit point of Orb(∆(T ), z), and so the result of Chan and Seceleanu gives us that ∆(T ) is hypercyclic.

To show the converse, suppose T is a bilateral shift whose Aluthge transform ∆(T ) is

2 hypercyclic. Then there exists an increasing sequence nk, non-zero vectors u and z in ` (Z) so that [∆(T )]nk u → z as k → ∞.

1 n 1 Now, since T = U |T | we obtain that U |T | 2 [∆(T )] k = T nk U |T | 2 for all k ≥ 1. Thus, from the form we deduced in Proposition 16 for the Aluthge transform of a bilateral shift

1 √ 2 nk √ nk √ and the continuity of BD w we have U |T | [∆(T )] u = BD w [∆(T )] u → BD w(z) as

√ 2 k → ∞. Note that BD w(z) is a non-zero vector in ` (Z), as the bilateral backward shift

√ √ B and the diagonal operator D w with positive weights wj have trivial kernels. Hence,

1 1 nk √ nk 2 2 nk √ T BD wu = T U |T | (u) = U |T | [∆(T )] u → BD w(z) as k → ∞ and therefore the

√ √ bilateral shift T has an orbit Orb(T,BD w(u)) with the non-zero limit point BD w(z). By Chan and Seceleanu’s result it follows that T is hypercyclic. 19 Having observed that the hypercyclic behavior of a shift is preserved under the action of the Aluthge transform, we want to better understand how ∆ acts on the set of weighted shifts. We focus our attention on answering the question: For a given bilateral weighted backward shift Tb with bounded weight sequence {bj}j∈Z, ﬁnd all bounded sequences {wj}j∈Z so that

∆(Tw) = Tb, where Tw is the bilateral weighted backward shift of weights {wj}j∈Z.

We ﬁrst note that the sequence of weights {bj}j∈Z that we start with takes values in

√ i·Arg(wj ) C \{0}, and if ∆(Tw) = Tb, then bj = wj−1wj · e for all j ∈ Z. Thus, for any p sequence {wj}j∈Z satisfying ∆(Tw) = Tb, we have Arg(wj) = Arg(bj) and |bj| = |wj−1wj|, and so any solution w must have the angles given by b. Consequently, we only need to ﬁnd

Z Z the w in (0, ∞) that satisﬁes ∆(Tw) = Tb for b in (0, ∞) . The actual solution set for a

iArg(bj ) complex valued sequence b is {{wje }j∈Z : ∆(Tw) = T|b|}, where |b|(j) = |b(j)| for all j ∈ Z.

Z Z Now, suppose that ∆(Tw) = Tb and b is in (0, ∞) and we want to ﬁnd w ∈ (0, ∞) . Let √ w0 be a non-zero constant. From bj = wj−1wj we obtain the following forward equations 2 2 bj bj wj = and backward equations wj−1 = . Solving these equations gives the following wj−1 wj solution for w: (a) from the backward equations we obtain for k ≤ −1

0 2 2 0 2 Y b2m−1 b2k Y b2m w2k = w0 · and w2k−1 = · ; b2m w0 b2m−1 m=k+1 m=k+1

(b) from the forward equations we obtain for k ≥ 1

k 2 2 k 2 Y b2m b Y b2m−1 w = w · and w = 2k+1 · . 2k 0 b 2k+1 w b m=1 2m−1 0 m=1 2m

The formal solution oﬀered above provides us with a necessary condition on the weight sequence w so that ∆(Tw) can equal a given Tb.

However, for the solution Tw to be a bounded linear operator, we must also require the weight sequence w to be bounded. We now show that in fact there is a bilateral weighted shift Tb that is not in the range of the Aluthge transform on the space of bilateral weighted 20 shifts, that is there exists a sequence b ∈ `∞(Z) for which there is no solution w ∈ `∞(Z) to

2 the equation ∆(Tw) = Tb, hence no Tw ∈ B(` (Z)) can be mapped by ∆ to Tb.

Proposition 19. There exists a bilateral weighted shift Tb that is not in the range of the

Aluthge transform ∆ acting on the set of bilateral weighted shifts on `2(Z).

∞ Proof. Consider the sequence b ∈ ` (Z) with b2k = 2 and b2k+1 = 3 for all k ∈ Z. Suppose √ that the sequence of non-zero weights w satisﬁes bj = wj−1wj for all j ∈ Z. Using the above backward and forward equations we obtain for k ≥ 1,

k 2 2k 2 k 2 2 2k Y b2m 2 b Y b2m−1 3 3 w = w · = w · and w = 2k+1 · = · . 2k 0 b 0 3 2k+1 w b w 2 m=1 2m−1 0 m=1 2m 0 Also for k ≤ −1

0 2 2|k| 2 0 2 2 2|k| Y b2m−1 3 b2k Y b2m 2 2 w2k = w0 · = w0 · and w2k−1 = · = · . b2m 2 w0 b2m−1 w0 3 m=k+1 m=k+1

Thus as k → ∞ we have w2k → 0 and w2k+1 → ∞, and as k → −∞ we have w2k → ∞ and w2k+1 → 0. Hence the weight sequence w is not bounded and so the induced bilateral weighted shift Tw is not a bounded operator.

We observe that the example provided above is that of an invertible bilateral weighted shift that is not hypercyclic. To produce an invertible hypercyclic shift not in the range of the Aluthge transform we simply set the positively indexed weights as in the example above,

1 the negatively indexed weights to be bj = 2 for all j ≤ −1, and observe that once again any induced solution w to the equation ∆(Tw) = Tb is not a bounded sequence.

Question 1: Which shifts Tb are in the range of the Aluthge transform on the space of bilateral weighted shifts?

Now that we have characterized the formal solutions Tw to the equation ∆(Tw) = Tb for a given sequence of non-zero bounded weights b, we can oﬀer a description of the equivalence 21 classes on the space of bilateral weighted shifts induced by the equivalence relation w ∼ v if and only if ∆(Tw) = ∆(Tv).

We note that given two non-zero bounded sequences w and v we have ∆(Tw) = ∆(Tv) if and only if there exists α > 0 with w (i) j = α for all j odd, vj w 1 (ii) j = for all j even. vj α Z One can see this by observing that two sequences w and v in (0, ∞) with ∆(Tw) = ∆(Tv) √ √ wj vj−1 have wj−1wj = vj−1vj for all j ∈ Z. Equivalently, = for all j ∈ Z, so letting vj wj−1 w α = 1 gives the desired description. v1 Using the result in Theorem 18, we note that the Aluthge transform induces an equivalence relation on the space of bilateral weighted shifts that is sensitive to the dynamical properties of these operators, that is only shifts that share the same hypercyclic behavior are in the same equivalence class. In fact, in Section 2.3 we will see that more generally, equivalent shifts share other dynamical properties.

2.1.2 Unilateral Case

2 Let {en : n ≥ 0} be the canonical base for ` (Z+). A bounded and linear operator T :

2 2 ` (Z+) → ` (Z+) is said to be a unilateral weighted backward shift if there is a sequence of

positive weights {wj}j≥1 such that T ej = wjej−1, if j ≥ 1 and T e0 = 0. As before, the

weight sequence {wj}j≥1 is bounded because of the boundedness of T . We note that there is great similarity between the form of a bilateral shift under the Aluthge transform and that of a unilateral shift under the same transform, however there is

one important diﬀerence. Given a unilateral shift T with weights {wj}j≥1 and proceeding as ∗ ∗ ∗ ∗ we did in Section 2.1.1, T T = (BDw) BDw = DwB BDw = DwSBDw, where B denotes the backward shift and S the forward bilateral shift. However, we can no longer conclude

∗ that T T equals D|w|2 . This follows from observing that the composition SB no longer

2 equals the identity operator on ` (Z+), as was the case for the bilateral shift. Instead, since 22 2 the ﬁrst component of the vector SBx for any x ∈ ` (Z+) is always zero, the operator SB

⊥ is now the projection onto the orthogonal complement he0i of he0i := span{e0}. So the

2 polar decomposition is T = U |T | with U = B and |T | = SBD|w|, since |T | = DwSBDw = √ SBD|w|2 . If we use Ta to denote the unilateral shift with weights aj = wj−1wj, then

2 ∆(T ) = STa and hence we can say that ∆(T ) is a unilateral shift on ` (Z+) with weight √ sequence bj = wj−1wj, if j ≥ 2 and b1 = 0. For a unilateral weighted backward shift T to be hypercyclic, Salas [35] provided a nec- n Y essary and suﬃcient condition on the weights, namely that sup wj = ∞. It is easy to n≥1 j=1 see that parts (i) and (ii) in the proof of Proposition 17 give us that if T is a hypercyclic unilateral shift, then its Aluthge transform ∆(T ) has an orbit that is dense in its range

⊥ 2 he0i ⊂ ` (Z+). Hence, even though the hypercyclic behavior of T does not carry over to the operator ∆(T ), the density of orbits is still preserved on a subspace of co-dimension one. We now show that in fact for a unilateral weighted shift to be hypercyclic it suﬃces for

⊥ 2 its Aluthge transform to have a dense orbit in the space he0i ⊂ ` (Z+).

2 Proposition 20. Suppose T is a unilateral weighted backward shift on ` (Z+). If ∆(T ) has

⊥ an orbit that is dense in he0i , then T is hypercyclic.

√ 2 1 √ 1 Proof. We ﬁrst observe that (SBD w) = SBDw = |T |, so |T | 2 = SBD w and U|T | 2 =

√ √ √ BSBD w = BD w, which is the unilateral weighted backward shift with weights { wj},

√ denoted by T w in the following.

√ ⊥ √ ⊥ Since c00 = span{ej : j ≥ 0} ⊆ T w he0i , we obtain T w he0i is a dense subset of

2 √ ⊥ √ ` (Z+) . Thus, since T w is continuous, any dense subset of he0i is sent by T w to a dense

2 subset of ` (Z+).

√ 1 1 Now we use that T w[Orb(∆(T ), z)] = U|T | 2 [Orb(∆(T ), z)] =Orb(T,U|T | 2 z) to infer

2 ⊥ the latter orbit is dense in ` (Z+) whenever the former orbit is dense in he0i . Thus if the

⊥ 1 2 orbit Orb(∆(T ), z) is dense in he0i , then Orb(T,U|T | 2 z) is dense in ` (Z+). 23 2.2 Chaos of the Aluthge transform of weighted shifts

Having looked at the hypercyclic behavior of the Aluthge transform of a shift T , we now turn our attention to another interesting dynamical property of an operator called chaos. Using Devaney’s deﬁnition [15] we say an operator T on a Hilbert space is chaotic if the operator is hypercyclic, exhibits sensitivity to initial conditions and has a dense set of periodic points. Banks et. al. [5] were able to show sensitivity to initial conditions follows from the other two conditions. In fact, for a bilateral weighted backward shift T with weight sequence {wj} Grosse-Erdmann [23, Theorem 9] showed T is chaotic if and only ∞ 2 ∞ X 1 X 2 if < ∞ and (w · w · ... · w ) < ∞. We now show that w · w · ... · w 0 −1 −j+1 j=1 1 2 j j=1 just as in the case of hypercyclicity, the chaotic behavior of a shift transfers to its Aluthge transform.

Proposition 21. Suppose T : `2(Z) → `2(Z) is a bilateral weighted backward shift with weight sequence {w } . If T is chaotic, then its Aluthge transform ∆(T ) is chaotic. j j∈Z

Proof. As we remarked in Section 2.1.1, given a bilateral weighted backward shift T with weight sequence {wj} its Aluthge transform ∆(T ) is also a weighted backward shift T with √ weight sequence bj = wj−1wj for all j ≥ 1. Then since T is chaotic,

∞ 2 ∞ 2 X 1 X 1 = √ √ b · b · ... · b w · w · ... · w · w j=1 1 2 j j=1 0 1 j−1 j ∞ √ 2 1 X wj = · w w · w · ... · w · w 0 j=1 1 2 j−1 j ∞ 2 1 X 1 ≤ · kT k · < ∞. w w · w · ... · w 0 j=1 1 2 j Furthermore, 24

∞ X 2 √ 2 √ √ 2 √ √ 2 (b0 · b−1 · ... · b−j+1) = ( w0w−1) + ( w0w−1 w−2) + ( w0w−1w−2 w−3) + ... j=1 2 2 2 2 ≤ w0w−1 + kT k · [(w−1) + (w−1w−2) + (w−1w−2w−3) + ...]

2 2 2 2 ≤ kT k · [1 + (w−1) + (w−1w−2) + (w−1w−2w−3) + ...] ∞ kT k2 X = · (w · w · ... · w )2 < ∞. w2 0 −1 −j+1 0 j=1

Thus by Theorem 9 in [23] we have that ∆(T ) is chaotic.

We note that in fact, in the next section we show that the converse to the above proposition also holds.

2.3 Dynamical Properties of the Aluthge transform of

an arbitrary operator

As we saw in the previous section, the hypercyclic behavior of a bilateral weighted shift is preserved under the action of the Aluthge transform. However we also observed that the same result does not hold in the case of unilateral weighted backward shifts. In this section we determine for which operators T the Aluthge transform ∆ preserves the dynamical properties of T . We note that the notion of semi-conjugacy plays an essential role in answering this question.

Deﬁnition 22. Let X and Y be topological spaces and T : X → X and S : Y → Y be functions. Suppose A : X → Y is continuous and SA = AT .

X −−−→T X     yA yA Y −−−→S Y 25 (i) If A is surjective then it is called a conjugacy from X to Y and S is called a factor of T . (ii) If A has dense range then it is called a semi-conjugacy from X to Y and S is called a quasi-factor of T .

We recall that if T is a bilateral weighted shift with a strictly positive weight sequence

1 2 √ √ w = {wj}j∈Z, then |T | = D w, where D w is the diagonal operator sending the vector

√ √ en to wnen for all n ∈ Z. The range of D w contains c00, the set of all sequences with

2 1 only ﬁnitely many non-zero entries. Since c00 is dense in ` (Z), we obtain that |T | 2 is a semi-conjugacy whenever w is a strictly positive sequence.

1 On the other hand, in the case of a unilateral weighted shift we have that |T | 2 (x0, x1, x2,...)

√ √ 1 = (0, w1x1, w2x2,...) and so |T | 2 never has a dense range in either the norm or the weak

2 topology on ` (Z+).

1 1 1 1 Now, since ∆(T ) = |T | 2 U|T | 2 and T = U|T |, we obtain ∆(T )|T | 2 = |T | 2 T . So for a bilateral shift T we have that ∆(T ) is a quasi-factor of T , but not if T is a unilateral shift. For completeness, we state the following theorem summarizing previously known facts about the preservation of dynamical properties by a semi-conjugacy.

Theorem 23. Let X and Y be topological vector spaces and T : X → X and S : Y → Y be continuous linear operators. If S is a quasi-factor of T with semi-conjugacy A, then

(i) AT n = SnA for all n ≥ 1.

(ii) For all x ∈ X, we have A[Orb(T, x)] = Orb(S, Ax), and so if z is a limit point of Orb(T, x), then Az is a limit point of Orb(S, Ax).

(iii) If T is hypercyclic, then S is hypercyclic.

(iv) A[HC(T )] ⊆ HC(S).

(v) If T is topologically transitive, then S is topologically transitive. 26 (vi) If T is (weakly) mixing, then S is (weakly) mixing.

(vii) If T is hereditarily hypercyclic with respect to {nk}, then S is hereditarily hypercyclic

with respect to the same sequence {nk}.

(viii) If x is a periodic vector for T , then Ax is a periodic vector for S.

(ix) If T has a dense set of periodic points, then S has a dense set of periodic points.

(x) If T is chaotic, then S is chaotic.

Proof. (i) By hypothesis AT = SA, so suppose n ≥ 1 and AT n = SnA. Now, AT n+1 = AT nT = SnAT = SnSA = Sn+1A, so AT n = SnA must hold for all n ≥ 1.

(ii) A takes x to Ax, and for n ≥ 1, AT nx = SnAx by (i). Furthermore, if T n1 x, T n2 x, . . .

converges to z we can use the continuity of A to obtain Az = lim AT nk x = lim Snk Ax. k→∞ k→∞

(iii) Since A is continuous and has a dense range, it takes dense subsets of X to dense subset of X. To see this, we note that for a non-empty open set U, A−1[U] is non-empty and open in X since the range of A is dense. Let D ⊆ X be dense and x ∈ A−1[U] ∩ D and observe Ax ∈ U ∩ A−1[D]. Thus A[D] is dense in Y if D is dense in X. Thus if Orb(T, x) is dense in X, Orb(S, Ax) is dense in Y .

(iv) For x ∈ HC(T ), A[Orb(T, x)] = Orb(S, Ax) and is dense since Orb(T, x) is dense. So Ax ∈ HC(S) if x ∈ HC(T ).

(v) Let U and V be non-empty open subsets of Y . Since A has dense range, A−1[U] and A−1[V ] are non-empty and open subsets of X.

Using that T is topologically transitive, let n ≥ 1 and x ∈ X satisfy x ∈ T n [A−1[U]] ∩ A−1[V ].

Now apply A and use that AT n = SnT to obtain 27

Ax ∈ A T n A−1[U] ∩ A−1[V ]

⊆ AT n A−1[U] ∩ AA−1[V ]

⊆ SnA A−1[U] ∩ V

⊆ Sn[U] ∩ V.

Thus S is topologically transitive if T is.

(vi) Let U and V be non-empty open subsets of Y . Again, since A has dense range, A−1[U] and A−1[V ] are non-empty open subsets of X.

n −1 −1 Using that T is mixing, let N ≥ 1 and xN , xN+1,... satisfy xn ∈ T [A [U]] ∩ A [V ] for all n ≥ N.

n −1 −1 n Applying A we obtain as above in (v) that Axn ∈ A [T [A [U]] ∩ A [V ]] ⊆ S [U]∩V for all n ≥ N. Thus S is mixing.

The case for a weakly mixing operator T follows similarly upon noting one way to characterize the property of being weakly mixing is that for U and V as above and for

every k in N, there is an N ≥ 1 for which T N+i [U] ∩ [V ] 6= ∅ for all i in {1, ..., k − 1}.

0 (vii) For any subsequence {nk} of {nk},

n0 n0 n0 n0 {S 1 Ax, S 2 Ax, . . .} = {AT 1 x, AT 2 x, . . .}

h n0 n0 i = A {T 1 x, T 2 x, . . .} .

n0 n0 n0 n0 So {S 1 Ax, S 2 Ax, . . .} is dense if {T 1 x, T 2 x, . . .} is dense.

(viii) If T kx = x for some k ≥ 1, then Ax = AT k = SkAx.

(ix) A takes dense sets to dense sets. 28 (x) The property of being chaotic is only deﬁned on metric spaces. So suppose X and Y are metric spaces. If T is topologically transitive and has a dense set of periodic points, then S is topologically transitive by (v) and has a dense set of periodic points by (ix). By a well known result of Devaney [15], this implies S is chaotic when T is.

Remark 24.

0. As noted above, Theorem 23 shows that if a bilateral weighted shift T has one of the dynamical properties listed in the theorem, then its Aluthge transform ∆(T ) will inherit the same dynamical property.

1. Suppose for a given arbitrary operator T we know that its Aluthge transform ∆(T ) has

1 1 a certain dynamical property listed in Theorem 23. We note that since ∆(T ) = |T | 2 U|T | 2

1 1 1 we have that U|T | 2 ∆(T ) = TU|T | 2 . Thus, if U|T | 2 has a dense range, then Theorem 23 gives that the operator T has the same dynamical properties as its Aluthge transform ∆(T ).

1 2. We note for bilateral weighted shifts, the operator U|T | 2 does indeed have a dense range, and so the dynamical properties of the Aluthge transform of a given shift T are also

1 √ inherited by T . Recall that for a bilateral shift U|T | 2 = BD w, which is clearly continuous

and its range contains the dense set of all sequences with ﬁnite support c00.

3. If we further assume in Theorem 23 that A is linear, we can conclude that if T is (super)cyclic, then S is also (super)cyclic. Thus, since for a bilateral weighted shift T the

1 operator |T | 2 is linear and continuous, we have that the (super)cyclicity of a bilateral shift T is also preserved under the action of the Aluthge transform.

To see this suppose that T is supercyclic with say K · Orb(T, x) dense in X. Then, if A is linear 29

n A[K · Orb(T, x)] = A[{λT x : n ≥ 0, λ ∈ K}]

n = {λAT x : n ≥ 0, λ ∈ K}

n = {λS Ax : n ≥ 0, λ ∈ K}], which is dense in Y since A is continuous and has dense range. Thus Ax is a supercyclic vector for S if x is supercyclic for T .

n As for cyclicity, let K[T ](x) := {a0x+a1T x+...+anT x : n ≥ 0, a0, . . . , an ∈ K}. Since n ! n X j X j for any n ≥ 0 and a0, . . . , an ∈ K, A ajT x = ajS Ax, we obtain A[K[T ](x)] = j=0 j=0 K[S](Ax). Thus if x is a cyclic vector for T , then Ax is a cyclic vector for S.

4. Suppose that X and Y are Banach spaces and T : X → X and S : Y → Y bounded linear operators. Suppose further that A : X → Y is continuous with respect to the weak topology of X and Y , respectively, it has a dense range and SA = AT . If T is weakly hypercyclic, then S is weakly hypercyclic. To see this, let x be a weakly hypercyclic vector for T and let U be a non-empty weakly open set in Y . We want to show that the orbit Orb(S, Ax) ∩ U 6= ∅. Now, A has a dense range in X, so A−1(U) is non-empty and open. Let V be a non-empty weakly basic set with V ⊂ A−1(U). Since x is a weakly hypercyclic vector for T , there exists k ≥ 1 so that T kx ∈ V . But AT kx = SkAx, so Sk(Ax) = AT kx ∈ A(V ) ⊆ U. Thus Orb(S, Ax) ∩ U 6= ∅, i.e. the orbit is dense in the weak topology of Y .

This gives us the following result for the weak hypercyclicity of bilateral weighted shifts and the Aluthge transform:

Proposition 25. Suppose T is a bilateral weighted backward shift on `2(Z). Then T is weakly hypercyclic if and only if its Aluthge transform ∆(T ) is weakly hypercyclic. 30 Proof. In view of the above remark, it suﬃces to show that for a bilateral shift T both

1 1 operators |T | 2 and U|T | 2 are continuous in the weak topology of X and Y , respectively, and that they have a dense range.

1 1 Clearly, since we have concluded that for a bilateral shift T both |T | 2 and U|T | 2 have a dense range in the norm topology, it follows that they have a dense range in the weak topology.

1 √ 1 √ 2 Furthermore, since |T | 2 = D w and U|T | 2 = BD w are linear and certainly ` -norm to `2-norm continuous, then it follows that they are also weak to weak continuous (see Theorem 2.5.11, page 214 in Megginson [30]). Thus, the remark above gives us that T is weakly hypercyclic if and only if ∆(T ) is weakly hypercyclic.

5. We note that a simple modiﬁcation of an example of Chan and Sanders in [12] gives us a bilateral weighted shift T that is weakly hypercyclic but ∆(T ) is not norm-hypercyclic.

Consider the bilateral weighted shift Tw induced by the sequence of weights   1, if j ≤ 0 or j = 2k + 1 for k ≥ 1, wj =  4, if j = 2k + 2 for k ≥ 1.

Then, the resulting bilateral shift ∆(Tw) = Tb under the action of the Aluthge transform has the weight sequence   1, if j ≤ 0 bj =  2, if j > 0.

Chan and Sanders [12, Corollary 3.3] showed that Tb is weakly hypercyclic however not norm-hypercyclic. Since we showed in the proposition above that a bilateral weighted shift T is weakly hypercyclic if and only if its Aluthge transform ∆(T ) is weakly hypercyclic,

we can conclude that the bilateral shift above Tw is also weakly hypercyclic. This shows that the suﬃcient condition for weak hypercyclicity provided in [12, Corollary 3.5] that

inf{wj : j ≥ 1} > 1 can be further relaxed. 31 6. We note that for a bounded linear operator T weakly mixing is equivalent to satisfying the Hypercyclicity Criterion [9]. Thus, statement (vi) in Theorem 23 can be rephrased as follows.

Corollary 26. For T and S as in Theorem 23, if T satisﬁes the Hypercyclicity Criterion, then S satisﬁes the Hypercyclicity Criterion.

7. Finally, we investigate how the Aluthge transform preserves the existence of hypercyclic subspaces of an operator. Here a hypercyclic subspace for T is deﬁned to be a closed inﬁnite dimensional linear subset consisting entirely (except the zero vector) of hypercyclic vectors for T . As stated in theorem 1, various spectra of a bounded linear operator T coincide with those of its Aluthge transform ∆(T ) [26]. In particular, the essential spectrum is preserved under the action of the Aluthge transform, i.e. σe(T ) = σe(∆(T )). On the other hand, Proposition 8.2 [6, page 195] says that for an operator satisfying the Hypercyclicity Criterion, having a hypercyclic subspace is equivalent to having the essential spectrum of T intersect the closed unit disk. Thus, since the Aluthge transform ∆(T ) preserves the intersection of the essential spectrum with the closed unit disk, we obtain the following result.

1 Proposition 27. Suppose a bounded linear operator T is weakly mixing and has |T | 2 as a semi-conjugacy. If T has a hypercyclic subspace, then so does its Aluthge transform ∆(T ).

2.4 D-hypercyclicity of the Aluthge transform of N

weighted shifts

Inspired by Furstenberg’s deﬁnition of disjointness in dynamical systems [19], Bernal-Gonzalez [7] and Bes and Peris [9] introduce a notion of disjointness for ﬁnitely many hypercyclic operators acting on a common space. Two operators T1 and T2 are said to be disjoint hypercyclic 32

if the operator T1 × T2 has a hypercyclic vector in the diagonal {(x, x): x ∈ X} of X × X. We note that T and T can never be disjoint hypercyclic as the orbit Orb(T × T, (z, z)) is a subset of the diagonal of X × X and so it can never be dense in X × X. Thus for T1 and T2 to be disjoint hypercyclic there must be a decaying dependence between the operators, that

n n is T1 z and T2 z must be frequently close to each other and frequently far from each other.

n n More clearly, for every (x, y) ∈ X × X, T1 z must be frequently near x while T2 z is near y.

Deﬁnition 28. We say N ≥ 2 hypercyclic operators T1,T2,...,TN acting on a Fr´echet space X are disjoint, or diagonally, hypercyclic (in short d-hypercyclic), provided there is some vector (z, z, . . . , z) in the diagonal of XN = X × X × ... × X such that

2 2 2 {(z, z, . . . , z), (T1z, T2z, . . . , TN z), (T1 x, T2 z, . . . , TN z) ...} is dense in XN . We call the vector z ∈ X a d-hypercyclic vector associated to the operators

T1,T2,...,TN .

In our endeavor to understand how the dynamical properties of an operator T are in- ﬂuenced by the action of the Aluthge transform, it is interesting to study whether this

“near independence” of the dynamical systems (X,T1), (X,T2),..., (X,TN ) is preserved by the Aluthge transform. In the following we will focus on the d-hypercyclicity of powers of bilateral weighted backward shifts. We note that Bes and Peris [9, Theorem 4.7] oﬀered the following characterization of the d-hypercyclicity of shifts by providing a necessary and suﬃcient condition in terms of the weights.

Proposition 29. [9] Suppose T1,T2,...,TN is a sequence of bilateral weighted backward

2 shifts on ` (Z) with weight sequences {wl,j}j∈Z for l = 1,...,N. For any integers 1 ≤ r1 < r2 < . . . < rN the following are equivalent:

r1 r2 rN (i) T1 ,T2 ,...,TN have a dense set of d-hypercyclic vectors.

(ii) For every > 0 and every q ∈ N there exists m ∈ N such that for every j ∈ Z with |j| ≤ q we have: 33 If 1 ≤ l ≤ N,   Qj+m·rl 1  i=j+1 wl,i >  Qj w <  i=j−m·rl+1 l,i

If 1 ≤ s < l ≤ N,

  Qj+m·rl w > 1 Qj+m·rl w  i=j+1 l,i i=j+m·(rl−rs)+1 s,i  Qj+m·rs w < Qj+m·rs w  i=j−m·(rl−rs)+1 l,i i=j+1 s,i

Using the above criterion we now show that the d-hypercyclicity of N invertible bilateral weighted shifts is preserved under the action of the Aluthge transform.

Proposition 30. Suppose T1,T2,...,TN is a sequence of invertible bilateral weighted back-

2 r1 r2 rN ward shifts on ` (Z) and 1 ≤ r1 < r2 < . . . < rN . If T1 ,T2 ,...,TN are d-hypercyclic, then

r1 r2 rN [∆(T1)] , [∆(T2)] ,..., [∆(TN )] are d-hypercyclic.

Proof. Let T1,T2,...,TN be a sequence of invertible bilateral weighted backward shifts having each weight sequence {wl,i}i∈Z for 1 ≤ l ≤ N, and let δ > 0 and M ≥ 1 be such that

r1 r2 rN δ ≤ wl,i ≤ M holds for all i ∈ Z. Suppose T1 ,T2 ,...,TN are d-hypercyclic. We will use the

r1 r2 rN Bes and Peris criterion to show that [∆(T1)] , [∆(T2)] ,..., [∆(TN )] are d-hypercyclic.

Let > 0 and q ∈ N, and assume without loss of generality that N = 2. r δ The Bes and Peris criterion gives us that for 0 = · > 0 and q ∈ as above, there M N exists m ∈ N such that for every j ∈ Z with |j| ≤ q we have: 1 (A1) w · w ····· w > , 1,j+1 1,j+2 1,j+m·r1 0 1 (A2) w · w ····· w > , 2,j+1 2,j+2 2,j+m·r2 0 0 (A3) w1,j−m·r1+1 · w1,j−m·r1+2 ····· w1,j < ,

0 (A4) w2,j−m·r2+1 · w2,j−m·r2+2 ····· w2,j < ,

w2,j+1 · w2,j+2 ····· w2,j+m·r2 1 (B1) > 0 , w1,j+m·(r2−r1)+1 · w1,j+m·(r2−r1)+2 ····· w1,j+m·r2 34 w · w ····· w (B2) 2,j−m·(r2−r1)+1 2,j−m·(r2−r1)+2 2,j+m·r1 < 0. w1,j+1 · w1,j+2 ····· w1,j+m·r1

r1 r2 rN To show that [∆(T1)] , [∆(T2)] ,..., [∆(TN )] are d-hypercyclic, it suﬃces to show that (A1)-(A4) and (B1),(B2) hold for the sequences of weights {bl,i}i∈Z with l = 1, 2 for the ﬁxed and all j ∈ Z with |j| ≤ q. Then, for (A1) we have

√ √ b1,j+1 · b1,j+2 ····· b1,j+m·r1 = w1,j · w1,j+1 · w1,j+2 ····· w1,j+m·r1−1 · w1,j+m·r1

r w1,j = w1,j+1 · w1,j+2 · ... · w1,j+m·r1 w1,j+m·r1 r δ 1 1 > · = . M 0

r δ 1 1 Similarly for (A2) we get that b · b ····· b > · = . 2,j+1 2,j+2 2,j+m·r1 M 0 On the other hand, (A3) gives that

√ √ b1,j−m·r1+1 · b1,j−m·r1+2 ····· b1,j = w1,j−m·r1 · w1,j−m·r1+1 · w1,j−m·r1+2 ····· w1,j−1 · w1,j r w1,j−m·r1 = · w1,j−m·r1+1 · w1,j−m·r1+2 ····· w1,j w1,j r M < · 0 = . δ

r M Similarly, (A4) gives b · b ····· b < · 0 = . 2,j−m·r2+1 2,j−m·r2+2 2,j δ Now, for (B1) we have

b2,j+1 · b2,j+2 ····· b2,j+m·r2 b · b ····· b 1,j+m·(r2−r1)+1√ 1,j+m·(r2−r1)+2 1,j+m·r2 √ w2,j · w2,j+1 · w2,j+2 ····· w2,j+m·r2−1 · w2,j+m·r2 = √ √ w1,j+m·(r −r ) · w1,j+m·(r −r )+1 · w1,j+m·(r −r )+2 ····· w1,j+m·r −1 · w1,j+m·r √ 2 1 √ 2 1 2 1 2 2 w2,j w1,j+m·r2 w2,j+1 · w2,j+2 ····· w2,j+m·r = √ · √ · 2 w2,j+r2·m w1,j+m·(r2−r1) w1,j+m·(r2−r1)+1 · w1,j+m·(r2−r1)+2 ····· w1,j+m·r2 r δ 1 δ > · = · ≥ . M 0 M 35 b · b ····· b M Finally, for (B2) we get that 2,j−m·(r2−r1)+1 2,j−m·(r2−r1)+2 2,j+m·r1 < · 0 = b1,j+1 · b1,j+2 ····· b1,j+m·r δ r 1 M · ≤ , which completes the proof. δ

The question arises whether the above result also holds for non-invertible bilateral weighted shifts. A partial answer to this question is given in the following proposition by using the idea emphasized in Section 2.3 that hypercyclicity is preserved under semi-conjugacy.

Proposition 31. Suppose T1,...,TN : X → X and S1,...,SN : Y → Y are linear and continuous operators, and A is a continuous function from X to Y with a dense range for

which ATi = SiA for all i = 1,...,N. If T1 × T2 ... × TN is d-hypercyclic at z × z × ... × z,

then S1 × S2 ... × SN is d-hypercyclic at Az × Az × ... × Az.

N N Proof. Suppose z = z × z × ... × z in X is a d-hypercyclic vector for T1 × T2 ... × TN ,

N N and let y1 × y2 × ... × yN be in Y . We would like the orbit Orb(S1 × S2 ... × SN , (Az) )

N to be arbitrarily near to y1 × y2 × ... × yN in Y . To see this, let > 0 be given. Using that A has a dense range, we note that each set of

−1 the form A [B(yi, )] is non-empty. Furthermore, the continuity of A yields that each set is open as well.

−1 For each j, let B(xj, δj) ⊂ X be a subset of A [B(yj, )].

N Now using that z is a d-hypercyclic vector for T1 × T2 ... × TN , let k ∈ N satisfy

k k k T1 z × T2 z . . . × TN z ∈ B(x1, δ1) × B(x2, δ2) × B(xN , δN ).

k k k k Noting that ATj z = Sj Az for all j, we get that for each j, dY (Sj Az, yj) = dY (ATj z, yj) <

k −1 since Tj z ∈ B(xj, δj) ⊂ A [B(yj, )].

k k k Thus there is a k ≥ 0 for which S1 Az × S2 Az × ... × SN Az ∈ B(y1, ) × B(y2, ) × ... ×

k k k B(yN , ), and since y1 × y2 × ... × yN is arbitrary, this shows {S1 Az × S2 Az × ... × SN Az : k ≥ 0} is dense in Y × Y × ... × Y . 36 1 1 rj rj rj Setting Tj = T , Sj = [∆(T )] and A = |T | 2 in the proposition and noting [∆(T )] |T | 2 =

1 r |T | 2 T j for all j ≥ 1, we obtain that the Aluthge transform preserves the d-hypercyclicity of

r 1 r r 1 1 the operators {T j : 1 ≤ j ≤ N}. Of course, since U|T | 2 [∆(T )] j = T j U|T | 2 for A = U|T | 2 in the same proposition, we can also conclude the converse.

Corollary 32. Suppose T is a bilateral weighted backward shifts on `2(Z). The operators

r1 r2 rN T ,T ,...,T are d-hypercyclic for the integers 1 ≤ r1 < r2 < . . . < rN if and only if [∆(T )]r1 , [∆(T )]r2 ,..., [∆(T )]rN are d-hypercyclic.

We note that we cannot use the same proposition above to answer the general question

r1 r2 rN whether the d-hypercyclicity of the bilateral weighted shifts T1 ,T2 ,...,TN is necessary or

r1 r2 rN suﬃcient for the d-hypercyclicity of the operators [∆(T1)] , [∆(T2)] ,..., [∆(TN )] . This is because we no longer have just one map A, rather for each i there is a distinct semi-

1 conjugacy Ai = U|Ti| 2 , so the argument in the proof of Proposition 31 only gives us that

(A1z × A2z × ... × AN z) is a hypercyclic vector for S1 × S2 × ... × SN , which is d-hypercyclic

only if A1z = A2z = ... = AN z. This however gives rise to another interesting question for these weighted shifts.

Question 3: Given two distinct weight sequences {aj} and {bj} when can we construct a hypercyclic vector for the induced weighted shifts Ta and Tb so that Daz = Dbz? Finally we would like to remark that the following three questions remain open both for the reasons mentioned above and because there does not exist a characterization in terms of the weights for the d-hypercyclicity of weighted shifts T1,T2,...,TN when the operators are not raised to distinct powers.

r1 r2 rN r1 r2 rN Question 4: If T1 ,T2 ,...,TN are d-hypercyclic, are ∆(T1 ), ∆(T2 ),..., ∆(TN ) d- hypercyclic?

r1 r2 rN r1 r2 rN Question 5: If T1 ,T2 ,...,TN are d-hypercyclic, are ∆ (T1), ∆ (T2),..., ∆ (TN ) d-hypercyclic?

Question 6: If T1,T2,...,TN are d-hypercyclic, are ∆(T1), ∆(T2),..., ∆(TN ) d-hypercyclic? 37 2.5 Universality of the Aluthge transform of a sequence

of weighted shifts

Having described how the Aluthge transform preserves the dynamical properties of dynamic systems, we now turn our attention to the notion of universality. For any topological space

X, and index set I, we say a family of continuous functions {Ti : i ∈ I} deﬁned on a common domain X is universal if there is an element x of X for which the set {Ti(x): i ∈ I} is dense in X. Hypercyclicity is the special case in which the family consists of the iterates of a single linear operator T deﬁned on a separable topological vector space. In the following we provide a criteria for determining which sequences of powers of the Aluthge transform of bilateral shifts are universal. Since the Aluthge transform of a bilateral shift is a shift, we

ﬁrst need to understand for which shifts Tk and natural numbers rk is the family of shifts

rk {Tk : k ≥ 1} universal. The following result generalizes Proposition 3.1 of Bes and Peris [8]:

Theorem 33. If {rk} is a strictly increasing sequence of positive integers and {Tk} is a sequence of unilateral weighted backward shifts with weight sequence {wk,j}j∈Z+ for each Tk,

r1 r2 then the sequence {T1 ,T2 ,...} is universal if and only if there is a sequence n1 < n2 < ··· in so that for all j ≥ 0, w · ... · w → ∞ as k → ∞. N nk,j+1 nk,j+rnk

r1 r2 Proof. Let us ﬁrst observe the sequence T1 ,T2 ,... converges pointwise on the span{e0, e1,...}

2 which is a dense subset of ` (Z+). This is because each Tn is a unilateral backward shift

rn rn and rn ≥ n for each n. Thus for each j ≥ 0, Tn ej = 0 if n > j, and so Tn → 0 pointwise

on span{e0, e1,...}. By [22, Proposition 6], the set of universal vectors for this sequence of operators is either empty or dense.

r1 r2 To prove suﬃciency, assume the sequence T1 ,T2 ,... is universal, hence has a dense set of universal vectors.

It is enough to show for every k ∈ N, there is an arbitrarily large nk for which wnk,j+1 · 38 ... · w > k holds for all 0 ≤ j ≤ k, since then one may choose to have n < n < ··· nk,j+rnk 1 2 and clearly from this it would follow that for every j ≥ 0, w · ... · w → ∞, as nk,j+1 nk,j+rnk k → ∞. 1 − δ Towards this end, let k and N be in and let 0 < δ < 1 satisfy k < . Now using N δ Pk that set of universal vectors is dense, let y be a universal vector within δ of i=0 ei, and let r nk Pk nk > N be such that Tnk y is within δ of i=0 ei. This gives us,

(i) rnk > k,

k X (ii) ky − eik < δ, and i=0

k rn X (iii) kTn y − eik < δ. i=0 Pk Suppose 0 ≤ j ≤ k, and let z = i=0 ei. By (i) and (ii), |y(j + rnk )| = |y(j + rnk ) − z(j +

rn kTn y − zk < δ. Hence, for 0 ≤ j ≤ k

rnk 1 − δ < | Tn y (j)| = w · ... · w · |y(j + r )| k nk,j+1 nk,j+rnk nk < w · ... · w · δ. nk,j+1 nk,j+rnk 1 − δ So wn ,j+1 · ... · wn ,j+r > > k holds for all 0 ≤ j ≤ k. k k nk δ

Conversely, assume there is sequence n1 < n2 < . . . for which for all j ≥ 0, wnk,j+1 · ... · w → ∞ as k → ∞. nk,j+rnk

r1 r2 We want to show that the sequence T1 ,T2 ,... is universal. We will use Grosse-Erdmann’s Universality Criterion [22, Theorem 2], with X = Y =

2 ` (Z+), the two dense sets X0 and Y0 equal to the span of {e0, e1,...}, and the maps Sn :

2 1 Y0 → ` (Z+) to be the linear maps deﬁned by Sn(ej) = ej+rn for all j ≥ 0 wn,j+1 · ... · wn,j+rn and all n ≥ 1.

rn As we showed at the start of this proof, Tn → 0 pointwise on span{e0, e1,...}. 39 1 Moreover, for every j ≥ 0, kS (e )k = , and this goes to zero by nk j w · ... · w nk,j+1 nk,j+rnk hypothesis as k → ∞.

rn Finally, using that for any p ≥ 0, Tn ep = wn,p · wn,p−1 · ... · wn,p−rn+1 · ep−rn we obtain

rn for p = j + rn that Tn ej+rn = wn,j+rn · wn,j+rn−1 · ... · wn,j+1 · ej, and so

rn rn 1 Tn Sn(ej) = Tn ej+rn wn,j+1 · ... · wn,j+rn

1 rn = · Tn ej+rn wn,j+1 · ... · wn,j+rn

wn,j+rn · wn,j+rn−1 · ... · wn,j+1 = · ej wn,j+1 · ... · wn,j+rn

= ej.

rn rn Thus (Tn ◦ Sn)y = y for all y in Y0 = span{e0, e1,...}, and so trivially (Tn ◦ Sn)y → y

as n → ∞, for each y in Y0.

r1 r2 Hence the Universality Criterion gives us that {T1 ,T2 ,...} has a dense Gδ-set of universal vectors.

An analogous criterion can be deduced for the bilateral case.

Theorem 34. If {rk} is a strictly increasing sequence of positive integers and {Tk} is a

sequence of bilateral weighted backward shifts with weight sequence {wk,j}j∈Z for each Tk, then

r1 r2 the sequence {T1 ,T2 ,...} is universal if and only if there is a sequence of natural numbers n < n < ··· so that for every j ∈ , w ·····w → ∞ and w ·····w → 1 2 Z nk,j+1 nk,j+rnk nk,j−rnk +1 nk,j 0, as k → ∞.

Consequently, we can deduce the following result about the universality of a sequence of powers of Aluthge transforms of bilateral weighted shifts:

Proposition 35. If {rk} is a strictly increasing sequence of positive integers and {Tk} is

a sequence of bilateral weighted backward shifts with weight sequence {wk,j}j∈Z for each Tk,

r1 r2 then the sequence {[∆(T1)] , [∆(T2)] ,...} is universal if and only if for all j ∈ Z, 40 √ √ w · w ····· w · w → ∞, and nk,j nk,j+1 nk,j+rnk −1 nk,j+rnk √ √ w · w ····· w · w → 0, as k → ∞. nk,j−rnk nk,j−rnk +1 nk,j−1 nk,j

r1 r2 However a natural question to ask is whether the universality of {T1 ,T2 ,...} is necessary

r1 r2 or suﬃcient for the universality of {[∆(T1)] , [∆(T2)] ,...}.

Theorem 36. Suppose {rk} is a strictly increasing sequence of positive integers and {Tk}

is a sequence of invertible bilateral weighted backward shifts with weight sequence {wk,j}j∈Z

for each Tk. If there is a sequence n1 < n2 < ··· in N, a δ > 0, and an M ≥ 1 for which

rn1 rn2 δ ≤ wnk,j ≤ M holds for all k in N and all j in Z, then the sequence {Tn1 ,Tn2 , cldots} is

rn1 rn2 universal if and only if the sequence {[∆(Tn1 )] , [∆(Tn2 )] , ···} is universal.

Proof. Using bnk as the weight sequence for the shift ∆(Tnk ) for each k, recall from the proof

of Proposition 17, for any bounded bilateral shift Tnk whose weights satisfy δ ≤ wnk,j for all j in Z, √ n n p n δ Y Y kTn k Y · w ≤ b ≤ √ k · w p nk,j nk,j nk,j kTnk k j=m j=m δ j=m holds for all m < n in Z.

We are hypothesizing that kTnk k ≤ M for each k, so by proceeding as in the proof of Proposition 17, we obtain

i. b ····· b → ∞ as k → ∞ iﬀ w ····· w → ∞ as k → ∞, and nk,j+1 nk,j+rnk nk,j+1 nk,j+rnk

ii. b ····· b → 0 as k → ∞ iﬀ w ····· w → 0 as k → ∞. nk,j−rnk nk,j nk,j−rnk nk,j

rn k rnk Thus under the given conditions, {Tnk : k ≥ 1} is universal if and only if {[∆(Tnk )] : k ≥ 1} is universal. 41

CHAPTER 3

The Aluthge transform acting on the set of weighted shifts

3.1 Properties of the Aluthge transform on the set of

weighted shifts

Deﬁne the set of shifts S to be the set of all bounded bilateral weighted backward shifts

T : `2(Z) → `2(Z) with non-negative weights. The set S is not a linear manifold in the Banach algebra B(`2(Z)), since multiplication by a negative constant produces a shift that is not in the set S. Nevertheless, S is a cone in B(`2(Z)); that is, S is closed under addition and multiplication by positive scalars and −S ∩ S = ∅. Furthermore, S is not closed under composition, as the composition of two shifts produces a multi-shift.

The Banach algebra B(`2(Z)) is naturally endowed with three topologies called the operator norm topology, the strong operator topology (SOT) and the weak operator topology (WOT). In the following we investigate how the Aluthge transform acts on this set of shifts S with respect to the aforementioned topologies.

We now show that the set S of shifts is closed in the norm topology of B(`2(Z)), so it is a complete metric space. On the other hand, B(`2(Z)) is not metrizable with respect to the 42 the strong or the weak operator topology, nevertheless we show S is both SOT and WOT closed in B(`2(Z)).

Proposition 37. The set S of shifts with non-negative weights is closed in B(`2(Z)) in the norm topology, the strong operator topology, and the weak operator topology.

Proof. We show that S is WOT closed, and hence SOT and norm closed in B(`2(Z)). Let

2 {Ti}i∈I be a net in S and assume Ti → T in the WOT sense in B(` (Z)). For each i ∈ I, let

(i) Ti be the shift with weight sequence w .

By deﬁnition Ti → T in the WOT if and only if h(Ti − T )g, hi → 0 for every g and h in

2 ` (Z). In particular we can choose g and h from the total set {en : n ∈ Z}. (i) Since Tiek+1 = wk+1ek for all i ∈ Z and k ∈ Z we obtain

(i) (i) h(Ti − T )ek+1, eki = wk+1hek, eki − hT ek+1, eki = wk+1 − hT ek+1, eki, (3.1.1)

and for j 6= k

(i) h(Ti − T )ek+1, eji = wk+1hek, eji − hT ek+1, eji = −hT ek+1, eji. (3.1.2)

(i) By (3.1.1), the WOT convergence implies wk+1 → hT ek+1, eki as i → ∞, and by (3.1.2)

it implies hT ek+1, eji = 0 if j 6= k.

So T ek+1 is in the span of ek. Thus for each k ∈ Z, T ek+1 = hT ek+1, ekiek. This shows T (i) is a backward shift and the weights are given by wk := lim w = hT ek, ek−1i for all k ∈ . i→∞ k Z (i) Clearly each wk ≥ 0 since {wk }i∈I is a net in [0, ∞). Moreover, {wk : k ∈ Z} is bounded

2 since |wk| = |hT ek, ek−1i| ≤ kT k by Cauchy-Schwarz and we assumed T ∈ B(` (Z)).

2 Thus the weak limit T of the net {Ti}i∈I is in S, and so S is WOT-closed in B(` (Z)).

Since convergence in the set S of shifts will be at the center of the results in this chapter, it is useful to characterize the three types of convergence on S in terms of the weight sequences of the converging shifts. We begin by showing that convergence in norm is equivalent to the uniform convergence of the weight sequences. 43 2 Proposition 38. If {Ti}i∈I is a net of shifts in B(` (Z)) and for each i ∈ I, Ti has the

(i) (i) weight sequence w , then Ti → Tw in norm if and only if w → w uniformly.

2 Proof. Let {Ti}i∈I be a net of shifts in B(` (Z)) so that for each i ∈ I, Ti has the weight

(i) sequence w , and let Tw be the shift with weight sequence w. We show that kTi − Twk =

(i) kw − wk∞. We ﬁrst note that

kTi − Twk = sup kTix − Twxk kxk=1

≥ supkTiem − Twemk m∈Z (i) = supkwm em−1 − wmem−1k m∈Z (i) = supkwm − wmk m∈Z (i) = kw − wk∞.

Moreover, for any x in `2(Z), X (i) X kTix − Twxk = k wn xnen−1 − wnxnen−1k n∈Z n∈Z X (i) = k (wn − wn)xnen−1k n∈Z (i) X ≤ kw − wk∞ · k xnen−1k n∈Z (i) = kw − wk∞ · kxk, (i) and so kTi − Twk ≤ kw − wk∞.

(i) Thus kTi − Twk = kw − wk∞.

While it is clear from the proof of Proposition 37 that convergence in the WOT (and hence SOT) sense in S implies the pointwise convergence of the weight sequences, we now show that for a bounded net of shifts the converse is also true. This is particularly useful for our analysis of the possible limit points of the orbit Orb(∆,T ) of a shift T , since the orbit is bounded by the norm kT k. 44 2 Proposition 39. A bounded net {Ti}i∈I of shifts in B(` (Z)) converges in the SOT sense

(i) to a shift Tw if and only if the net of weight sequences {w }i∈I converges pointwise to w.

Proof. By Proposition 1.3d of Conway [14, page 256], a bounded net {Ti}i∈I converges in the

SOT-topology to T if and only if for every h in some total set T we have kTih − T hk → 0. Recall that by a total set T , we mean T has the property that only the zero vector is orthogonal to every vector in T .

(i) The set {ek}k∈Z is total, and furthermore kTiek − Twekk = |wk − wk| for all i and k in (i) Z. Thus Ti → Tw in the SOT sense if and only if wk → wk pointwise on Z.

Remark: Since SOT convergence implies WOT convergence, the above proposition now gives that for a bounded net of shifts, the SOT convergence is equivalent to the WOT convergence, which is further equivalent to the pointwise convergence of the weight sequences.

Now that we have described the diﬀerent types of convergence on the set of shifts S, we would like to establish some basic properties of the Aluthge transform acting on S. As noted in Section 2.1.1, the Aluthge transform ∆ : S → S is neither an injective nor a surjective

map. Ji, Pang, Li [25, page 214] have shown that as a map from B(`2(Z)) to B(`2(Z)), the range of the Aluthge transform is not closed with respect to any of the three topologies, and is not norm dense, but it is SOT and WOT dense. We show similar results hold for the Aluthge transform as a map from S to S.

Proposition 40. The range of the Aluthge transform ∆ : S → S is not norm dense in S, however it is SOT or WOT dense in S.

Proof. We ﬁrst show the range of the Aluthge transform on S is not norm dense. Recall, in

Proposition 19 we showed the shift Tb whose weights are 2 at the even entries and 3 at the odd entries is not in the range ∆[S] of the Aluthge transform.

1 In fact, any shift Ta ∈ S with kTa − Tbk < 4 is also in the complement of ∆[S], and so

Tb is an interior point of the complement of ∆[S], in the sense of the norm topology. 45 To see this, we observe for every j in Z,

kTa − Tbk ≥ kTaej − Tbejk

= kajej−1 − bjej−1k

= |aj − bj|.

1 So |aj − bj| < 4 holds for every j in Z.

But, for Ta to be in ∆[S], there must be a solution Tw ∈ S to the equation ∆Tw = Ta, however the odd positive weights of such a Tw are not bounded since for each k ≥ 1,

2 k 2 a Y a2m−1 w = 2k+1 · 2k+1 w a 0 m=1 2m 1 1 1 must hold. The inequality |a − b | < holds if and only if b − < a < b + , and so for j j 4 j 4 j j 4 every k ≥ 1, and every m in {1, ...k}, we obtain

2 i. a2k+1 ≥ 7, since b2k+1 = 3, and

1 1 a2m−1 b2m−1 − 4 3 − 4 11 ii. > 1 = 1 = . a2m b2m + 4 2 + 4 9

k 2 7 Y a2m−1 Thus w ≥ · → ∞ as k → ∞, and so w is unbounded. 2k+1 w a 0 m=1 2m

To see that the range of the Aluthge transform on S is SOT (and hence WOT) dense in

(n) S, let Tb ∈ S be an arbitrary shift. We deﬁne the sequence of shifts {Ta } ⊂ S by setting the weight sequence for each n ≥ 1

  (n)  bj, |j| ≤ n aj =  1, |j| > n.

(n) (n) Clearly, aj → bj pointwise on Z as n → ∞. Furthermore, kTa k ≤ max{1, kT k} and (n) so Proposition 39 gives that Ta → Tb in the SOT sense. 46 (n) Finally, to see that {Ta } ⊂ ∆[S] we use the backward and forward equations (a) and

(b) in Section 2.1.1 and observe that a shift with ﬁnitely many weights aj 6= 1 has a shift with bounded weights in the pre-image of the Aluthge transform.

We now show that the range of the Aluthge transform on the set S of shifts is not closed

in any of the three topologies of B(`2(Z)).

Proposition 41. The range of the Aluthge transform ∆ on the set S of shifts is not norm closed (and hence neither SOT nor WOT closed) in S.

1 Proof. Consider the bilateral weighted backward shift T with weights b = 1 + for 2m 10m

m ≥ 1 and bj = 1 otherwise. k 2 Y b2m Using the forward equations we obtain that for every k ≥ 1, w = w · = 2k 0 b m=1 2m−1 k 2 Y 1 w · 1 + must hold of any solution T to the equation ∆(T ) = T . Clearly, the 0 10m w w m=1 weight sequence {wj} is unbounded and hence T is not in the range of the Aluthge transform ∆[S].

On the other hand, we deﬁne the sequence of bilateral weighted shifts {Tn} by setting 1 the weights for each n ≥ 1 to be b(n) = 1 + for 1 ≤ m ≤ n and b(n) = 1 otherwise. 2m 10m j 1 Clearly, sup |b(n) − b | = → 0 as n → ∞, and so kT − T k → 0. j j 10n+1 n Finally, using the backward and forward equations one can see that for each n ≥ 1 only

(n) ﬁnitely many of the ratios in the products deﬁning the sequence of weights {wj } for each (n) Tn are not equal to one, so each sequence of weights {wj } is bounded. Thus each Tn is in the range of the Aluthge transform, and so the range of ∆ on S is not norm closed (and hence not SOT or WOT closed) in S.

Another important question regarding the action of the Aluthge transform on the set S of shifts is what are the types of continuity ∆ exhibits on this set. On the Banach algebra 47 B(H) for an arbitrary Hilbert space H we have the following result of Jung, Ko and Percy [26].

Theorem 42. [26] The Aluthge transform ∆ is norm-norm continuous at T ∈ B(H), provided T has a closed range.

Recall that any bilateral weighted shift T with positive weights contains the dense set c00 = span{ej : j ∈ Z} in its range. If T is not invertible, then T is not surjective and hence its range is a set of category I (see proof of Proposition 58). If the range of a non-invertible shift T were to also be closed, then Range(T ) = `2(Z) which is not a set of of category I. Hence the Range(T ) is not closed if and only if T is not invertible. In the theorem below we show the condition in Theorem 42 is in fact not necessary when we restrict ourselves to S. Since we proved a non-surjective shift never has a closed range, it follows that on the set S of shifts, Theorem 42 states that the Aluthge transform is norm-norm continuous at the invertible shifts. We now show that ∆ is in fact norm-norm continuous at every shift T ∈ S.

Theorem 43. The Aluthge transform ∆ : S → S is norm to norm continuous.

Proof. Suppose T is a bilateral weighted shift with positive weights {wj} so that the sequence

(n) {Tn} ⊆ S converges Tn → T in norm as n → ∞. That is, supj∈Z |wj − wj| → 0 as n → ∞.

We want to show that ∆(Tn) → ∆(T ) in norm as n → ∞, i.e. the induced weights have

(n) (n) n supj∈Z |bj − bj| → 0 as n → ∞, where Tb = ∆ (T ).

Let > 0 and M := sup{kTnk : n ≥ 1} which is ﬁnite. Using that the square root function is uniformly continuous on the interval [0, 2kT k], let δ = δ() > 0 be such that for √ √ every x, y ∈ [0, 2kT k], |x − y| < δ implies | x − y| < √ . 3 M

(n) q (n) √ Hence, for this δ we have that |w − wj| < δ implies sup w − wj < √ for j j∈Z j 2 M (n) all wj , wj ∈ [0, 2kT k]. (n) Now, given δ as above (take δ < kT k if necessary), since supj∈Z |wj −wj| → 0 as n → ∞ (n) we conclude that there exists N > 0 so that supj∈Z |wj − wj| < δ for all n ≥ N. Therefore, 48

q (n) √ for every n ≥ N we have sup w − wj < √ . j∈Z j 2 M So, for each n ≥ 1 and j ∈ Z,

q q q q (n) (n) √ (n) (n) (n) (n) √ w w − wj−1wj = w w − w wj + w wj − wj−1wj j−1 j j−1 j j−1 j−1 q q q (n) (n) √ √ (n) √ = w w − wj + wj w − wj−1 j−1 j j−1 q q q (n) (n) √ √ (n) √ ≤ w w − wj + wj w − wj−1 j−1 j j−1 q q p (n) √ p (n) √ ≤ kTnk w − wj + kT k w − wj−1 j j−1 √ q √ q (n) √ (n) √ ≤ M w − wj + M w − wj−1 . j j−1

Thus,

q √ q √ q (n) (n) √ (n) √ (n) √ sup wj−1wj − wj−1wj ≤ M sup wj − wj + M sup wj−1 − wj−1 j∈Z j∈Z j∈Z √ q (n) √ = 2 M sup wj − wj j∈Z √ < 2 M · √ = , 2 M for every n ≥ N.

q (n) (n) √ Thus, sup w w − wj−1wj → 0, i.e. ∆(Tn) → ∆(T ) in norm, as n → ∞. j∈Z j−1 j Another important topic in the area of research involving the Aluthge transform is (hypo)normality. In the introduction, we explained the motivation behind the deﬁnition of the Aluthge transform is that of producing a new operator with the same spectral properties as the original operator, but which in a sense is closer to being normal. This is illustrated in Proposition 44 below. Shields [36] proved a bilateral weighted shift T is hyponormal if and only if its weight sequence is decreasing, and normal if and only if its weight sequence is constant. We now show that if a shift T ∈ S is (hypo)normal, then its Aluthge transform ∆(T ) is also (hypo)normal. On the other hand, we construct a shift T that is not (hypo)normal and yet its Aluthge transform is. 49 Proposition 44. The set of (hypo)normal shifts N is invariant under the Aluthge transform ∆, however ∆−1[N ] is not contained in N .

Proof. Let T ∈ S be a hyponormal shift. Then its weight sequence has the property that wn ≥ wn+1 for all n ∈ Z. Thus its Aluthge transform ∆(T ) is a weighted shift with weight √ √ √ sequence bn = wn−1wn, where bn = wn−1wn ≥ wnwn+1 = bn+1 for all n ∈ Z. Thus ∆(T ) is hyponormal.

Similarly, if T is normal, then for all n ∈ Z and some constant c we have wn = c for

some constant c ∈ [0, ∞). Thus bn = c for all n ∈ Z, and so ∆(T ) = T , which is normal by hypothesis. We now provide an example of a non-hyponormal shift T ∈ S for which ∆(T ) is hyponormal. Consider the weighted backward shift T with the following weights:

 1  w = 1 − , n ≥ 1  2n 2n  1 w = 2 − , n ≥ 0 2n+1 2n + 1    wn = 2, n ≤ 0.

Clearly, the weight sequence w is not decreasing and so T is not hyponormal. On the other hand, we now show that its Aluthge transform has a decreasing weight sequence. First note

that bn = 2, if n ≤ 0. Furthermore, since for n ≥ 1 the even as well as the odd indexed weights √ √ form a decreasing sequence, we can conclude that bn = wn−1wn ≥ wnwn+1 = bn+1, and so ∆(T ) is hyponormal yet T is not hyponormal.

Finally, consider the weighted backward shift T with weight sequence wn = 2 if n ∈ Z

is even and wn = 1 otherwise. Then its Aluthge transform ∆(T ) is a shift with constant √ √ weights bn = wn−1wn = 2 for all n ∈ Z, and so ∆(T ) is normal yet T is not normal. 50 3.2 Locally universal shifts under the iterations of the

Aluthge transform

Historically, the Aluthge transform was introduced in [1] and has received much attention because of its intimate connection to the Invariant Subspace Problem. Since for the class of normal operators the problem has a positive answer, the idea behind the Aluthge transform in relation to the Invariant Subspace Problem is in rough terms to convert an arbitrary operator T in B(H) into another operator which shares some spectral properties of the ﬁrst but is closer to being a normal operator. In fact, Jung, Ko and Percy [26] conjectured that the iterates of the Aluthge transform

n {∆ (T )}n≥1 converge to a normal operator for every T ∈ B(H). This conjecture was shown to be false by Thomson (as referenced in [27]), who found an operator whose Aluthge transform sequence does not converge in norm. The conjecture was modiﬁed in [27] to assert that the sequence of iterates of the Aluthge transform converges in the SOT sense to a normal operator. That too was shown to be false by Yanahida [40], who found an operator whose orbit Orb(∆,T ) does not converge even with respect to the weak operator topology. In this section we deﬁne an operator T whose orbit Orb(∆,T ) under the iterations of the Aluthge transform is in fact dense in an ”interval” of normal operators. Thus we show that

n the sequence {∆ (T )}n≥1 fails to converge in the WOT sense to a single normal operator in quite a dramatic sense as the orbit picks up a large supply of limit points. However, our local density result in the set of shifts S in some sense reinforces the original idea behind

n the Aluthge transform as all the subsequential limit points of the sequence {∆ (T )}n≥1 are normal. In fact, in this section we show only a normal operator can be an SOT (hence

n WOT or norm) subsequential limit of of the sequence {∆ (T )}n≥1 when T is a bilateral shift. On the other hand, unilateral shifts can in fact never be normal (see [36]), nevertheless

n all subsequential limit points of the sequence {∆ (T )}n≥1 are unilateral shifts with constant weight sequences. 51 We now state the principal theorem in this section.

Theorem 45 (Unilateral Forward Theorem). For any 0 < a ≤ b ∈ R, there is a unilateral

weighted forward shift Tw with positive weight sequence w for which the set of subsequential

2 SOT (equivalently WOT) limit points of the sequence {Tw, ∆(Tw), ∆ (Tw),...} is [a, b] · S :=

{t · S : a ≤ t ≤ b}. Moreover, a = inf{wj : j ∈ Z+} and b = kwk∞.

This result is generalized in Theorem 54 to include bilateral forward shifts, in Theorem 55 to include complex weight sequences, and in Theorem 56 to include bilateral backward shifts. The reason for focusing on the forward shifts in this section is the Aluthge transform of a unilateral forward shift is a unilateral forward shift, which is not the case of backward shifts. We also address the unilateral case with positive weights ﬁrst to simplify the proof. As we will see, the bilateral and complex cases follow easily from that of the unilateral.

We start by deducing a useful formula in terms of the weights for the nth iteration of a shift under the Aluthge transform. The Aluthge transform of a unilateral (respectively bilateral) forward shift is also a unilateral (respectively bilateral) forward shift, speciﬁcally

for a forward shift Tw0 with weight sequence w0 : J → C \{0}, ∆(Tw0 ) = Tw1 with w1,j = p |w0,j+1 · w0,j| · exp[i · Arg(w0,j)] for all j ∈ J. Here J is either Z+ or Z depending on

whether Tw0 is the unilateral or bilateral shift.

Iteratively applying ∆ and using wn = {wn,j : j ∈ J} to denote the weight sequence of

n the shift ∆ (Tw0 ), we obtain

1 " n # 2n Y ( n ) wn,j = |w0,j+m| m · exp[i · Arg(w0,j)], (3.2.1) m=0 for all n ≥ 0 and all j ∈ J.

Proof. For n = 0 the equation asserts w0,j = |w0,j|·exp[i·Arg(w0,j)], so suppose the equation p is correct for n ≥ 0. Applying ∆ to Twn yields wn+1,j = |wn,j · wn,j+1| · exp[i · Arg(wn,j)].

But then Arg(wn+1,j) = Arg(wn,j), which further equals Arg(w0,j) by hypothesis, and 52

1 1 " n # 2n " n # 2n Y ( n ) Y (n) |wn,j · wn,j+1| = |w0,j+m| m · |w0,j+1+k| k m=0 k=0 1 " n ! n ! # 2n (n) Y ( n ) Y ( n ) (n) = |w0,j| 0 |w0,j+m| m · |w0,j+m| m−1 · |w0,j+1+n| n m=1 m=1 1 " n ! # 2n (n) Y ( n )+( n ) (n) = |w0,j| 0 |w0,j+m| m m−1 · |w0,j+1+n| n m=1 1 " n ! # 2n (n) Y (n+1) (n) = |w0,j| 0 |w0,j+m| m · |w0,j+1+n| n m=1 1 " n+1 # 2n Y (n+1) = |w0,j+m| m m=0

n n+1 n+1 n since 0 = 0 = n+1 = n . So upon taking the square root and multiplying by

exp[i · Arg(w0,j)], formula (3.2.1) is obtained.

In like manner we obtain a formula for the weights of the Aluthge iterates of a bilateral p backward shift from the equations w1,j = |w0,j · w0,j−1| · exp[i · Arg(w0,j)] for all j ∈ Z,

n namely for ∆ (Tw0 ) = Twn

1 " n # 2n Y ( n ) wn,j = |w0,j−m| m · exp[i · Arg(w0,j)], (3.2.2) m=0 for all n ≥ 0 and j ∈ Z. In particular, for positive weight sequences these equations become for unilateral and bilateral forward shifts

1 " n # 2n Y ( n ) wn,j = (w0,j+m) m , for all j ∈ J, (3.2.3) m=0 53 and for bilateral backward shifts

1 " n # 2n Y ( n ) wn,j = (w0,j−m) m , for all j ∈ Z. (3.2.4) m=0

Since a bounded sequence of shifts Tw1 ,Tw2 ,... converges to another shift Tv in the SOT sense (equivalently, the WOT sense) if and only if for all j ∈ J, wn,j → vj as n → ∞, our main theorem in this section asserts that given 0 < a ≤ b in R there is a weight sequence w0 for which for every t ∈ [a, b] there is some subsequence n1 < n2 < . . . depending on t having 1 " nk # 2nk Y (nk) the property that for every j ∈ Z+, (w0,j+m) m → t as k → ∞. m=0 To make this goal both conceptually clearer and easier to attain, let α := log w, or equivalently exp[α] = w. That is, for each j ∈ J and n ∈ N let αj := log wj. Upon replacing n (m) n (w0,j±m) with exp m αj±m , we obtain for unilateral and bilateral forward shifts

1 n ! 2n " n # Y n X n w = exp · α = exp 2−n · α , (3.2.5) n,j m j+m m j+m m=0 m=0 for all j ∈ J, and for bilateral backward shifts

1 n ! 2n " n # Y n X n w = exp · α = exp 2−n · α , (3.2.6) n,j m j−m m j−m m=0 m=0 for all j ∈ Z. Clearly, from this and the continuity and injectivity of the exponential function on R nk X nk we see for α = log w that w → t if and only if 2−nk · α → log t. Thus our nk,j m j+m m=0 principal theorem can be proven by ﬁnding an α : Z+ → {log a, log b} so that for every element log t of [log a, log b] there is a sequence n1 < n2 < . . . depending on log t so that for nk X nk every j ∈ , 2−nk · α → log t as k → ∞. This is achieved in Propositions 51 Z+ m j+m m=0 and 52. 54 n X n The reader familiar with probability theory should recognize the sum 2−n α m j+m m=0 n X as the expected value E[α(j + Sn)] := P[Sn = m]αj+m of α(j + Sn) := αj+Sn , where m=0 1 n −n Sn is the Binomial(n, p) random variable with p = 2 . Here P[Sn = m] = m 2 for all non-negative integers n and all integers m in {0, 1, 2, ..., n}. Since it is this probabilistic view that will be exploited in producing such an α, from here forward the notation E[α(j + Sn)] n X n is frequently used to denote 2−n α . When we want to restrict the sum to only m j+m m=0 those terms whose indices are in a set, say A, we write E[α(j + Sn); (Sn ∈ A)]. We will also X write P[Sn ∈ A] to denote P[Sn = m]. m∈A We now introduce two key inequalities in Proposition 46 and 47. The ﬁrst is used to

obtain an α for which the set of subsequential limits of the sequence {E[α(Sn)] : n ≥ 0} is dense in [log a, log b], and the second enables us to infer that whenever E[α(j + Sn)] → log t as k → ∞ for some t > 0, n1 < n2 < . . ., and j ∈ Z+, it actually converges to log t for all j ∈ Z+.

Let J be Z+ or Z, and deﬁne kαk∞,[j,j+n+1] := sup{|α(j + k)| : 0 ≤ k ≤ n + 1} for all j and n.

Proposition 46. For every j in J, n in N, and α : J → C √ 2 · kαk | [α(j + S )] − [α(j + S )]| < √ ∞√,[j,j+n+1] . E n E n+1 π · n

Proposition 47. For every j in J, n in N, and α : J → C √ 2 2 · kαk | [α(j + S )] − [α(j + 1 + S )]| < √ ∞√,[j,j+n+1] . E n E n π · n

We remark that in particular for bounded√ α we obtain 2 · kαk∞ (i) |E[α(j + Sn)] − E[α(j + Sn+1)]| < √ √ , and π√· n 2 2 · kαk (ii) | [α(j + S )] − [α(j + 1 + S )]| < √ √ ∞ . E n E n π · n The following two lemmas facilitate the proofs of Proposition 46 and 47. 55 n X n n n n Lemma 48. For every n ≥ 1, − = 2 · − 2 where Mn = if n is k k − 1 Mn 2 k=1 (n − 1) even and M = if n is odd. n 2 n Proof. For n = 1, each side of the equation is zero. For n ≥ 2, the value of increases k with increasing k in {0, 1,...,Mn}, and decreases with increasing k in {Mn + 1, . . . , n}. So,

n Mn n X n n X n n X n n − = − − − k k − 1 k k − 1 k k − 1 k=1 k=1 k=Mn+1 n n n n = − − − Mn 0 n Mn n = 2 · − 2 Mn

Furthermore, we produce a bound on the size of 2−n · n . Mn √ n 2 Lemma 49. For each n ≥ 1, 2−n · < √ √ . Mn π · n

Proof. In [18] , Feller derives the following inequality that holds for all n ≥ 1: √ √ 1 1 1 1 n+ 2 −n n+ 2 −n 2π · n · e · exp 12n+1 < n! < 2π · n · e · exp 12n .

This inequality was originally proven by Robbins in [33], and so in what follows it will be called Robbins’ inequality. n √ 2 Let us ﬁrst show Mn < √ √ for even n. 2n π n

Let n = 2m. In this case, Mn = m. Using Robbins’ inequality we obtain

√ 2m+ 1 −2m 1 (2m)! < 2π · (2m) 2 · e · exp 12(2m) √ 2m+ 1 2m+ 1 −2m 1 = 2π · 2 2 · m 2 · e · exp , 24m

and 56

√ 2 m+ 1 −m 1 m! · m! > 2π · m 2 · e · exp 12m + 1 2 = 2π · m2m+1 · e−2m · exp . 12m + 1

So, √ 2m+ 1 2m+ 1 −2m 1 n 2m (2m)! 2π · 2 2 · m 2 · e · exp = = < 24m M m m! · m! 2m+1 −2m 2 n 2π · m · e · exp 12m+1 2m+ 1 2 2 1 2 = √ √ · exp − 2π m 24m 12m + 1 √ 2 36m − 1 = 22m · √ √ · exp − π 2m 24m(12m + 1) √ √ 22m · 2 2n · 2 < √ √ = √ √ , π 2m π n

1 2 −36m + 1 36m − 1 since − = = − < 0. 2m 12m + 1 24m(12m + 1) 24m(12m + 1) n √ 2 Thus for even n ≥ 2, Mn < √ √ . 2n π · n n − 1 2m − 2 To establish the same for odd n = 2m − 1, note that M = = = m − 1. n 2 2 We call upon two basic combinatorial facts: n n − 1 n − 1 (i) (Pascal’s Identity) = + , and k k − 1 k n n (ii) = . k n − k 2m 2m−1 2m−1 2m−1 The ﬁrst fact gives us that m = m−1 + m , and the second tells us m−1 = 2m−1 2m 2m−1 m . Combing these we obtain m = 2 m−1 . n 2m−1 22m−1 2m Thus for n = 2m − 1 we see Mn = m−1 = m−1 = m . 2n √22m−1 22m 22m 2m 2 But we have just shown m < √ √ . 22m π · 2m √ √ n Lastly, since n = 2m − 1 we have 2m = n + 1, and so for odd n we have Mn < √ √ 2n 2 2 √ √ < √ √ as claimed. π · n + 1 π · n 57

Now that the lemmas are proven let’s prove the propositions.

Let α be a function from J to C, and for notational convenience, let αn,j = E[α(j + Sn)] for all j ∈ J and n ≥ 0.

Proposition 47. For every n ≥ 1, all α : J → C, and every j in J, √ 2 · kαk |α (j) − α (j)| < √ ∞√,[j,j+n+1] . n n+1 π · n

Proof. For each j and n ≥ 1,

n n+1 X n X n + 1 |α (j) − α (j)| = 2−n α(k + j) − 2−(n+1) · α(k + j) n n+1 k k k=0 k=0 n n X n X n + 1 n + 1 = 2−(n+1) 2α(k + j) − α(k + j) − α(n + 1 + j) k k n + 1 k=0 k=0 n X n n + 1 = 2−(n+1) 2 − · α(k + j) − α(n + 1 + j) k k k=0

n + 1 n n n n + 1 However, for k ≥ 1, = + and so we have 2 − = k k − 1 k k k n n = − . k k − 1 Thus, n X n n |α (j) − α (j)| = 2−(n+1) − · α(k + j) + α(j) − α(n + 1 + j) n n+1 k k − 1 k=1 " n # −(n+1) X n n ≤ 2 − · |α(k + j)| + |α(j)| + |α(n + 1 + j)| k k − 1 k=1 " n # −(n+1) X n n ≤ 2 − + 2 · kαk∞,[j,j+n+1]. k k − 1 k=1 n X n n n −n n By Lemma 48, − = 2 −2, and by Lemma 49 we have, 2 k k − 1 Mn Mn √ k=1 2 < √ √ . π n So, for all j ∈ J and n ≥ 1 58

−(n+1) n |αn(j) − αn+1(j)| ≤ 2 · 2 · kαk∞,[j,j+n+1] Mn n = 2−n · kαk M ∞,[j,j+n+1] √ n 2 · kαk < √ ∞√,[j,j+n+1] . π · n

We now prove our second proposition.

Proposition 48. For all j ∈ J, n ≥ 1, and α : J → C, √ 2 2 · kαk |α (j) − α (j + 1)| < √ ∞√,[j,j+n+1] . n n π · n Proof. Let j ∈ J and n ≥ 1 be given. Then

n n X n X n |α (j) − α (j + 1)| = 2−n α(k + j) − 2−n α(i + j + 1) n n k i k=0 i=0 n n−1 X n X n n n = 2−n α(k + j) − α(i + j + 1) + α(j) − α(n + j + 1) . k i 0 n k=1 i=0

Now using i = k − 1 in the second sum, we obtain

n n X n X n |α (j) − α (j + 1)| = 2−n α(k + j) − α(k + j) + α(j) − α(n + j + 1) n n k k − 1 k=1 k=1 n X n n = 2−n − · α(k + j) + α(j) − α(n + j + 1) k k − 1 k=1 " n # −n X n n ≤ 2 − · |α(k + j)| + |α(j)| + |α(n + j + 1)| k k − 1 k=1 " n # −n X n n ≤ 2 − + 2 · kαk∞,[j,j+n+1]. k k − 1 k=1

n X n n n Lemma 48 tells us − = 2 − 2, and from this we obtain that k k − 1 Mn k=1 −n n |αn(j) − αn(j + 1)| ≤ 2 · 2 · kαk∞,[j,j+n+1]. Mn 59 √ n 2 Finally, Lemma 49 tells us 2−n < √ √ , so we conclude Mn π n √ 2 2 · kαk |α (j) − α (j + 1)| < √ ∞√,[j,j+n+1] , n n π · n for all j ∈ J and n ≥ 1.

Proposition 50. Let J be Z+ or Z, and for each n ≥ 0 let Sn ∼ Binomial(n, p) where the 1 parameter p = 2 . If α : J → R is a bounded function, then the set D of subsequential limit points of the sequence {E[α(Sn)] : n ≥ 0} is a non-empty closed interval. Here we count a singleton as a closed interval.

Proof. Since α is bounded, the sequence {E[α(Sn)] : n ≥ 0} is bounded, and so the set D is not empty. If there is only one subsequential limit, the sequence converges and the interval is a singleton. So suppose a < b in R are subsequential limit points of {E[α(Sn)] : n ≥ 0}, and c is in (a, b). Let > 0 be given, and m1 < m1 < . . . and n1 < n1 < . . . satisfy

(i) E[α(Smk )] → a, and (ii) [α(S )] → b, as k → ∞. E nk √ 2 · kαk Using the inequality from Proposition 46, | [α(j + S )] − [α(j + S )]| < √ √ ∞ , E n E n+1 π n for all n ≥ 1. √ 2 · kαk∞ Let N be so large that √ √ < . By (i), (ii), and the fact that c ∈ (a, b) there are π N mk ≥ N and nl > mk for which E[α(Smk )] < c < E[α(Snl )].

So there must be an n in {mk, mk+1, . . . , nl} for which |c − E[α(Sn)]| < .

Thus the sequence {E[α(Sn)] : n ≥ 0} visits every neighborhood of c, and we conclude c is one of its subsequential limit points. Since it is always the case that the set of subsequential limit points of a sequence is a closed set, it follows that D is a closed interval.

1 Proposition 51. Let p = 2 and for each n ≥ 0, let Sn ∼ Binomial(n, p). For all c ≤ d ∈ R there is an α : J → {c, d} for which the set of subsequential limit points of the sequence 60

{E[α(Sn)] : n ≥ 0} is [c, d].

Proof. Clearly, {E[α(Sn)] : n ≥ 0} ⊆ [c, d] if the range of α is contained in [c, d], and so it is enough to deﬁne an α for which every element of [c, d] is a subsequential limit point of the sequence {E[Sn]: n ≥ 0}. We begin by deﬁning an α for which each of c and d is a subsequential limit of the sequence {E[Sn]: n ≥ 0} and then appeal to Proposition 50.

First observe, for any countable partition A0,A1,A2,... of [0, ∞) and sequence n0 < n1 <

C n2 < . . . in Z+ for which P[Snk ∈ Ak ] → 0 as k → ∞, we can deﬁne an α via the rule

  c, j ∈ A2k for some k ≥ 0 α(j) =  d, j ∈ A2k+1 for some k ≥ 0

C For such an α, E[α(Snk )] = E[α(Snk ); (Snk ∈ Ak)] + E[α(Snk ); (Snk ∈ Ak )] holds for all k ≥ 0.

C C C But E[α(Snk ); (Snk ∈ Ak )] ≤ E[|α(Snk )|;(Snk ∈ Ak )] ≤ kαk∞ · P[Snk ∈ Ak ] → 0 as k → ∞, and so E[α(Sn2k )] → c and E[α(Sn2k+1 )] → d as k → ∞.

To produce such a partition A0,A1,A2,... and a sequence n1 < n2 < . . . we will use

adjacent intervals A0,A1,A2,... and choose the nk’s so that Snk is in Ak with high probability.

Let n0 = 0 and for given k ≥ 0 assume n0 < n1 < . . . < nk have been deﬁned. p As n → ∞, so does np − (k + 1) · np(1 − p). So there is an nk+1 for which nkp + p p k nkp(1 − p) < nk+1p − (k + 1) nk+1p(1 − p).

Thus there is a sequence 0 = n0 < n1 < n2 < . . . for which the preceding inequality holds for all k ≥ 0. p Using µn = E[Sn] and σn = V ar(Sn) for each n, the preceding inequality becomes

µnk + kσnk < µnk+1 − (k + 1)σnk+1 (3.2.7) for all k ≥ 0. 61

As for the partition let A0 = [0, µn1 − σn1 ) and for k ≥ 1 let Ak = [µnk − kσnk , µnk+1 −

(k + 1)σnk+1 ).

By (3.2.7), [µnk − kσnk , µnk + kσnk ] ⊆ Ak for all k ≥ 0.

C Thus P[Snk ∈ Ak ] ≤ P[|Snk − µnk | > kσnk ] which by Chebyschev’s inequality is bounded

1 C by k2 , and so for this partition P[Snk ∈ Ak ] → 0 as k → ∞.

Proposition 52. Let J be Z+ or Z, α : J → C be bounded, n1 < n2 < . . . be a sequence in

N, and c be a complex number. If the sequence {E[α(j + Snk )] : k ≥ 1} converge to c for some j ∈ J, then it does so for all j ∈ J.

Proof. The inequality in Proposition 47 tells us for every j ∈ J and n ∈ N, the terms √ 2 2·kαk √ √ ∞ E[α(j + Sn)] and E[α(j + 1 + Sn)] are within π· n of each other. So for any j1, j2 in J, √ 2 2·kαk √ √ ∞ the terms E[α(j1 + Sn)] and E[α(j2 + Sn)] are within |j1 − j2| · π· n of each other. √ 2 2·|j −j | √ √1 2 That is, |E[α(j1 + Sn)] − E[α(j2 + Sn)]| ≤ π· n · kαk∞ for all n in N and j1, j2 in J.

So, if E[α(j1 + Snk )] → c as k → ∞, and j2 is any element of J, then

|E[α(j2 + Snk )] − c| ≤ |E[α(j1 + Snk )] − c| + |E[α(j2 + Snk )] − E[α(j1 + Snk )]| √ 2 2 · |j2 − j1| < |E[α(j1 + Snk )] − c| + √ √ · kαk∞ → 0 π · nk as k → ∞.

We now introduce a corollary that summarizes the relationships between α := log w, w, and SOT (equivalently WOT) convergence of the Aluthge iterates of a shift.

Let J be Z+ or Z, and let each of α, w : J → R be bounded sequences for which exp[αj] = wj for all j ∈ J. Let Tw be the (unilateral or bilateral) forward shift with weight

n sequence w. Recall for Twn := ∆ (Tw) we showed wn,j = exp [E[α(j + Sn)]] for all j and n.

For notational convenience let αn,j := E[α(j + Sn)] for all j and n.

Corollary 53. For any t > 0 and sequence n1 < n2 < . . . in N, the following are equivalent 62

(i) αnk,0 → log t as k → ∞,

(ii) αnk,j → log t for some j ∈ J as k → ∞,

(iii) αnk,j → log t for all j ∈ J as k → ∞,

(iv) wnk,0 → t as k → ∞,

(v) wnk,j → t for some j ∈ J as k → ∞,

(vi) wnk,j → t for all j ∈ J as k → ∞,

nk (vii) ∆ (Tw) → t · S in the SOT (equivalently, the WOT) sense.

Proof. That parts (i) through (iii) are equivalent is the content of Proposition 52. Their equivalence to parts (iv) through (vi) follows from the continuity and injectivity of the exponential function on R via the equations wn,j = exp [E[α(j + Sn)]].

Part (vi) is equivalent to part (vii) since a sequence Tw1 ,Tw2 ,... of unilateral shifts

converges to a shift Tv = SDv if and only if wn,j → vj as n → ∞ for all j ∈ J. Here

vj = c for all j in J, so SDv = cS.

Remarks: 1. Corollary 53 makes it clear that only a shift whose weight sequence is constant can be a subsequential limit (in the norm, strong operator, or weak operator topology) of the sequence

2 {Tw, ∆(Tw), ∆ (Tw),...} when Tw is a forward shift with positive weights. This is because

nk ∆ (Tw) → Tv in the WOT sense if and only if wnk,j → vj as k → ∞ for all j in J. But then wnk,0 → v0 as k → ∞, and so by the equivalence of (iv) and (vii), Tv = v0 · S.

∞ 2. For any bounded forward shift Tw, the sequence {wn,0}n=1 is bounded too. In fact the sequence is bounded by kwk∞, see

" n # " n ! # X n X n w = exp 2−n · log w ≤ exp 2−n · log kwk n,0 m m m ∞ m=0 m=0 n X n = exp [log kwk ] = kwk, since 2−n · = 1. ∞ m m=0 63 ∞ 3. By Proposition 50, the set of subsequential limits of {αn,0}n=1 either contains a single point or is a closed interval, so the equivalence of (i) and (vii) tells us the set of

2 subsequential limits of {Tw, ∆(Tw), ∆ (Tw),...} is either a single shift with constant weights, or an ”interval” of such shifts [a, b]·S whenever Tw is a bounded forward shift. In other words,

2 the sequence {Tw, ∆(Tw), ∆ (Tw),...} either converges to a shift with constant weights or diverges to an interval of such shifts. Now we are ready to prove the main theorem introduced at the start of this section.

Proof of Theorem 45. We want to show that for every 0 < a ≤ b ∈ R there is a unilateral forward shift Tw with positive weights for which the set of all SOT subsequential limits of

2 the sequence {Tw, ∆(Tw), ∆ (Tw),...} is {t · S : a ≤ t ≤ b}.

Apply Proposition 51 with c = log a and d = log b to obtain an α : Z+ → {log a, log b} for which the set of subsequential limit points of {E[α(Sn)] : n ≥ 0} is [log a, log b].

Let wj = exp(αj) for all j in Z+ and by appealing to parts (i) and (vii) of Corollary 53 we conclude the set of all SOT subsequential limits of Orb(∆,Tw) is {t · S : a ≤ t ≤ b}.

Since all of the propositions and corollaries in this section held for J equal to Z+ or Z, the principal theorem has a bilateral version.

Theorem 54 (Bilateral Forward Theorem). For every 0 < a ≤ b in R, there is an invertible

continuous bilateral weighted forward shift Tw with positive weight sequence w for which the

2 set of all SOT (equivalently WOT) subsequential limits of the sequence {Tw, ∆(Tw), ∆ (Tw),...} is {t · S : a ≤ t ≤ b}.

Proof. Using Theorem 45, let Tv be a unilateral forward shift for which the set of all SOT

2 subsequential limits of {Tv, ∆(Tv), ∆ (Tv),...} is {t·S : a ≤ t ≤ b}, where S is the unilateral

forward shift. Let wj = vj for all j ≥ 0 and choose wj = 1 if j < 0. Since Corollary 53 held

for J = Z, 64

nk ∆ (Tv) → t · S if and only if vnk,0 → 0,

if and only if wnk,0 → 0,

nk ˜ if and only if ∆ (Tw) → t · S, as k → ∞. Here S is the unilateral forward shift and S˜ is the bilateral forward shift.

2 Thus, the set of all SOT subsequential limit points of the sequence {Tw, ∆(Tw), ∆ (Tw),...} is {t · S˜ : a ≤ t ≤ b}.

Since bilateral weighted shifts with positive weights are normal if and only if their weight sequence is constant, we note that Theorem 54 states that either the sequence

2 {Tw, ∆(Tw), ∆ (Tw),...} converges to a normal shift or it diverges to an interval of normal shifts. There is also a complex generalization for each of the Unilateral and Bilateral Forward Theorems.

Theorem 55 (Complex Generalization). Let J be Z+ or Z, and 0 < a ≤ b in R. For any

2 2 sequence of angles θ : J → [0, 2π) there is a continuous forward shift Tw : ` (J) → ` (J) with non-zero complex weights for which

(i) Arg(wj) = θj for all j in J, and (ii) the set of all SOT (equivalently WOT) subsequential limits of the sequence

2 {Tw, ∆(Tw), ∆ (Tw),...} is {Tv : there is t ∈ [a, b], for every j ∈ J, vj = t · exp[i · θj]}.

Proof. Apply Theorem 45 if J = Z+, or Theorem 54 if J = Z to ﬁnd a forward shift

2 Tu for which the set of all SOT subsequential limit points of {Tu, ∆(Tu), ∆ (Tu),...} is {t · S : a ≤ t ≤ b}.

Now deﬁne wj := uj · exp[i · θj] for all j ∈ J.

Clearly, Arg(wj) = θj for all j in J. 65 1 " n # 2n Y n n (m) Recall for Twn := ∆ (Tw), wn,j = (wj+m) · exp[i · Arg(wj)] holds for all n and m=0 j.

Thus Arg(wn,j) = Arg(wj) = θj for all n and j, and so for any t in [a, b] and sequence

n1 < n2 < . . ., unk,j → t for all j as k → ∞, if and only if wnk,j → t · exp[i · θj] for all j as k → ∞.

Consequently, this Tw has property (ii) of this theorem.

As we mentioned in our comments after stating our main theorem of this section, similar results hold for the bilateral backward shift. In particular Corollary 53 holds for backward shifts with the understanding that now

−n Pn n wn,j = exp 2 · m=0 m αj−m and part (vii) now uses the pure backward shift B, rather

nk than the pure forward shift S, that is ∆ (Tw) → t · B as k → ∞. The speciﬁc results for bilateral backwards shifts are as follows.

Theorem 56 (Bilateral Backward Theorem). Fore every 0 < a ≤ b in R there is an invertible

2 2 continuous bilateral backward shift Tw : ` (Z) → ` (Z) with positive weight sequence w : Z → (0, ∞) for which the set of SOT (equivalently WOT) subsequential limit points of the sequence

2 {Tw, ∆(Tw), ∆ (Tw),...} is {t · B : a ≤ t ≤ b}.

Proof. Let Tu be a bilateral forward shift for which the set of subsequential limit points of

2 the sequence {Tu, ∆(Tu), ∆ (Tu) ...} is {t · S : a ≤ t ≤ b}. The bilateral backward shift

Tw = BDw deﬁned by letting wj = u−j for all j in Z has the required property. This is evident from the fact that the following equations holds for all j in Z and n in N

" n # X n w = exp 2−n · log w n,j m j−m m=0 " n # X n = exp 2−n · log u m −j+m m=0 = un,−j. 66

Thus wnk,j → t as k → ∞ for all j in Z if and only if unk,−j → t as k → ∞ for all j in Z.

Using the same basic proof as in the case of bilateral forward shifts, there is a complex version for bilateral backwards shifts as well.

Theorem 57 (Complex Bilateral Backward Theorem). For all 0 < a ≤ b in R, and any sequence of angles θ : Z → [0, 2π), there is an invertible continuous bilateral backward shift

2 2 Tw : ` (Z) → ` (Z) with non-zero complex weights for which

(i) Argwj = θj for all j in Z, and (ii) the set of all SOT (equivalently WOT) subsequential limits of the sequence

2 {Tw, ∆(Tw), ∆ (Tw),...} is {Tv : there is a t ∈ [a, b], for every j in J, vj = t · exp[i · θj]}. 67

CHAPTER 4

Hypercyclicity and the range of an operator

It is well known that the set HC(T ) of hypercyclic vectors for any operator T is either empty or is a dense Gδ, and hence a set of category II. In this chapter we address where these hypercyclic vectors for an operator T lie in relation to the range of the operator T . The results of this section extend the work of G. Turcu [37] who showed that while for every n, T n(x) is a hypercyclic vector whenever x is hypercyclic, there is a hypercyclic operator T whose range does not contain all of the vectors that are hypercyclic for T . To prove this, he provided an operator T and constructed a vector x that is hypercyclic for T , but is not in the range of T . Using a diﬀerent approach, we are able to say more. Recall, a subset of a topological space is said to be a set of category I (also called a meager set) if it is a countable union of nowhere dense sets. Those sets that are not of category I are said to be of category II. Also, a subset of a topological space is called a residual set if it is the complement of a set of category I, and hence is the intersection of countably many dense sets. Every residual subset of a Banach space is a set of category II. Since for any n ≥ 1, the vector T nx is hypercyclic whenever x is hypercyclic for T it might be expected that the intersection of HC(T ) with the range of a hypercyclic operator 68 T is always large. We show to the contrary that HC(T ) \ Range(T ) is a set of category II whenever T is a non-surjective hypercyclic operator on a Banach space. In fact for such an operator T , HC(T ) ∩ Range(T ) is a set of category I, and so for “most” hypercyclic vectors x there is no hypercyclic vector y 6= x for which x is in the orbit {T ny : n ≥ 0}. We also show there is a sense by which the range of a hypercyclic operator is large in HC(T ) in that for any hypercyclic operator T on an F-space X, the set HC(T )∩

∞ n [∩n=1Range (T )] is dense in X.

Proposition 58. For every non-surjective hypercyclic operator T on a Banach space X the set HC(T ) \ Range(T ) is of second category.

Proof. The set HC(T ) is residual, so it is a set of category II. Since the union of two category I sets is a category I set, HC(T ) \ Range(T ) must be a set of category II whenever Range(T ) is a set of category I. Thus, upon demonstrating that the range of every non-surjective hypercyclic operator on a Banach space X is a category I set, the theorem will be proved. We show the converse, namely that if T is a bounded linear operator on a Banach space X and the range of T is a category II set, then T is surjective.

First, we show the range of T contains an open ball centered at zero. Expressing X as the union of open balls B(0, n) for n ≥ 1, and applying T we obtain " ∞ # ∞ [ [ Range(T ) = T B(0, n) = T [B(0, n)]. n=1 n=1 Since the range of T is a set of second category, at least one set forming this union must be somewhere dense, say T [B(0, n0)]. Using that T [B(0, n0)] is somewhere dense, let z be in X and let δ > 0 be so small that B(z, δ) ⊆ T [B(0, n0)], where the bar denotes the closure.

Moroever, being that z is in this closure, let T x be in B(z, δ) for some x in B(0, n0), and let ρ > 0 be small enough that B(T x, ρ) ⊆ B(z, δ). Shifting the ball to the origin, we obtain 69

B(0, ρ) = B(T x, ρ) − T x

⊆ T [B(0, n0)] − T x

= T [B(0, n0)] − T x

= T [B(−x, n0)].

In the argument that follows, it is convenient to have T applied to a neighborhood of

zero (rather than to B(−x, n0)) to make use of the fact that for a normed linear space, for any r > 0, B(0, r · ρ) ⊆ T [B(0, r · )] whenever B(0, ρ) ⊆ T [B(0, )]. So we observe

B(−x, n0) ⊆ B(0, 2n0), since x is in B(0, n0) and acquire

B(0, ρ) ⊆ T [B(0, 2n0)].

We now show B(0, ρ) is contained in the range of T . Let y be in B(0, ρ) and using that

ρ y is in the closure of T [B(0, 2n0)], let x1 be in B(0, 2n0) and y − T x1 be in B(0, 2 ). 1 ρ Using r = 2 , we see B(0, 2 ) is a subset of T [B(0, n0)], so let x2 be in B(0, n0) and ρ (y − T x1) − T x2 be in B(0, 22 ).

2n0 Continuing in like manner, we obtain a sequence x1, x2,... in X with xn ∈ B 0, 2n−1 = n ! X n0 B 0, n0 and y − T x ∈ B(0, ρ ) for each n ≥ 1. Thus kx k < and ky − 2n−2 i 2n n 2n−2 1 n ! X ρ T x k < for all n ≥ 1. i 2n 1 n ! n X X ∞ Clearly, T xi converges to y. Moreover, the sequence { xi}n=1 converges since 1 1 ∞ ∞ n−1 X X 1 X is a Banach space and kx k < 2n = 2n < ∞. i 0 2 0 n=1 n=1 ∞ ! X Thus by continuity of T , y = T xi and so y is in T [B(0, 2n0)]. n=1

∞ X Remark: In trying to pass to more general spaces, we do obtain xi converges when n=1 70 ∞ X d(xi, 0) < ∞ when the space is complete and the metric d is translation invariant. The n=1 problem is we may not get B(0, r1) ⊆ T [B(0, r2)] when B(0, 1) ⊆ T [B(0, 2)]. This fact

was employed in acquiring the summable sequence of xn’s and in having xn+1 be close to n ! X y − T xi . 1 In particular, a bilateral weighted shift Tw is invertible if and only if it is surjective, which is further equivalent to having its weight sequence {wj} bounded away from zero (see [36]).

Thus any hypercyclic bilateral weighted shift Tw whose weight sequence has zero as a limit point provides us with an example of a hypercyclic operator T for which HC(T ) \ Range(T ) is of second category.

T∞ n We now show that the common range n=1 Range (T ) of the iterates of T is large in the set of hypercyclic vectors for T .

Proposition 59. For every hypercyclic operator T , the common range of the iterates of T contains a dense set of vectors hypercyclic for T .

More speciﬁcally, the proof will show for every hypercyclic operator T on an F-space X, and every non-empty open subset U of X, there is some vector x in U that is hypercyclic

T∞ n for T and is in n=1 Range (T ). The following fact sometimes called the Abstract Version of Mittag-Leﬄer’s Theorem is used (see [4, Theorem 2.4], [31]) .

∞ Abstract Version of Mittag-Leﬄer’s Theorem: Let {(Xn, dn)}n=1 be a sequence of complete metric spaces, and for each n, let fn : Xn+1 → Xn be a continuous function with a dense range. For any non-empty open subset U of X1, there is a sequence x1 ∈ U ⊆ X1, x2 ∈

X2, x3 ∈ X3, ... for which fn (xn+1) = xn for all n.

Proof. Let (X, d) be an F-space and T be hypercyclic on X. Recall, an F-space is a topological vector space whose topology is induced by a complete metric. Moreover, a topological vector space must have its topology induced by an invariant 71 metric. A metric d on X is invariant if d (x + z, y + z) = d (x, y) for all x, y, and z in X.

∞ Let Y be the set of all those vectors x in X for which there is a sequence {xn}n=1 ∈ ∞ Q∞ X = n=1 X with x = x1 and for n ≥ 1, T (xn+1) = xn. By letting Xn be X and fn be T for each n, and applying the preceding Abstract Mittag-Leﬄer theorem to every non-empty open subset of X, it is clear Y is dense in X. Clearly, Y is also a linear manifold.

∞ ∞ Next, deﬁne the subset A of X to be the set of all sequences {xn}n=1 for which

T (xn+1) = xn for all n ≥ 1. This set A is a closed linear manifold.

To show A is closed, let x1, x2, ... be a sequence in its complement. There is a ﬁrst n, say

n0, at which T (xn+1) 6= xn. Let be one-half the distance between T (xn0+1) and xn0 , and using the continuity of T , let δ > 0 be so small that for all w and z in X, d (T (w) ,T (z)) <

∞ if d (w, z) < δ. Let U be the open subset of X consisting of all sequences y1, y2, ... in X for

which d (yn0+1, xn0+1) < δ and d (yn0 , xn0 ) < . ∞ The intersection of U with any open neighborhood V of {xn}n=1 is an open neighborhood ∞ of {xn}n=1 that is disjoint with A since for all sequences y1, y2, ... in this intersection,

d (T (yn0+1) , yn0 ) ≥ d (T (xn0+1) , xn0 ) − d (T (yn0+1) ,T (xn0+1)) − d (yn0 , xn0 )

> 2 − − = 0

Thus, since the space X∞ endowed with the product topology is a Fréchet space (i.e. a lo- ∞ X 1 d (ξ (k) , η (k)) cally convex F-space) with for example the metric ρ given by ρ (ξ, η) = , 2k 1 + d (ξ (k) , η (k)) k=1 A is a Fréchet space as well. There is a natural hypercyclic operator on A induced by T whose hypercyclicity can be exploited to find a vector that is both hypercyclic for T and is in the common range of the iterates of T . Toward this end, let T˜ be defined on A via ˜ T ((x1, x2, ...)) = (T (x1) , x1, x2, ...) = (T (x1) ,T (x2) ,T (x3) , ...). The orbit under T˜ of a vector x in A looks like 72

T˜ (x1, x2, x3, ...) 7→ (T (x1) , x1, x2, x3, ...)

T˜ 2 7→ T (x1) ,T (x1) , x1, x2, ...

T˜ 3 2 7→ T (x1) ,T (x1) ,T (x1) , x1, x2, ... . . ˜ So T simply shifts a vector forward and tacks on a new element of the orbit of x1. Clearly, T˜ maps A to A and is linear. T˜ is even invertible, with the unweighted backward ˜ shift serving as its inverse. Moreover, if x is hypercyclic for T , then x1, the ﬁrst component of x, is hypercyclic for T . In fact, every component of a T˜ hypercyclic vector x in A is hypercyclic for T . Now to prove T˜ is hypercyclic, the equivalent property of being topologically transitive will be established. Let U and V be non-empty open subsets of A. We need to show there is some n for which T˜n [U] T V= 6 ∅. Without loss of generality, T ∞ U = A (U1 × · · · × UN × X × X × · · · ) = {{xk}k=1 ∈ A; xk ∈ Uk, for 1 ≤ k ≤ N}, and T ∞ V = A (V1 × · · · × VN × X × X × · · · ) = {{xk}k=1 ∈ A; xk ∈ Vk, for 1 ≤ k ≤ N}, for some natural number N and non-empty open subsets U1,...,UN ,V1,...,VN of X.

∞ ∞ 0 Let x = {xk}k=1 be in U and y = {yk}k=1 be in V. Let U1 = U1. By the continuity

0 0 0 0 of T there are non-empty sets U2 ⊆ U2,U3 ⊆ U3,...,UN ⊆ UN for which xj ∈ Uj and 0 0 T Uj+1 ⊆ Uj for all j = 1, 2,...,N − 1.

0 0 2 0 0 N−1 0 0 But then T UN ⊆ UN−1,T UN ⊆ UN−2,...,T UN ⊆ U1, and so by proceeding ∞ 0 0 in like manner for y = {yk}k=1 in V, we obtain non-empty open subsets UN and VN of X 0 0 j 0 for which xN ∈ UN ⊆ UN , yN ∈ VN ⊆ VN , and for j = 1, 2,...,N, T UN ⊆ UN−j, and

j 0 0 T VN ⊆ VN−j. Here we use T = T .

Now use that T is topologically transitive on X to obtain an arbitrarily large n0 for which

0 0 0 0 n0 T T −n0 T UN VN 6= ∅. For convenience assume n0 ≥ N. Clearly, UN T VN 6= ∅. So let

00 0 00 0 n0 UN be a non-empty open subset of UN for which T UN ⊆ VN . 73 00 T By the density of Y in X, there is a vector zN in UN Y . Moreover, by deﬁnition of Y , ∞ 00 there is a sequence {zk}k=N in X for which zN ∈ UN , and T (zj+1) = zj for all j ≥ N. ∞ Now, deﬁne z1, z2, . . . , zN−1 so as to obtain an element {zk}k=1 of A by letting zN−1 =

2 N−1 T (zN ) , zN−2 = T (zN ) ,..., and z1 = T (zN ).

∞ 00 0 0 0 0 The vector z = {zk}k=1 just deﬁned is in U, since zN is in UN ⊆ UN , and UN ,UN−1,...,U1

j 0 were chosen so that zN−j = T (zN ) = T (zN−j+1) is in UN−j ⊆ UN−j for j = 1, 2,...,N − 1.

n0 To see for the same n0 as above, T takes U into V, observe for each i = 0, 1, 2,...,N −

1 < n0,

n0−i n0−i N−1 T (z1) = T T (zN )

N−(i+1) n0 = T T (zN )

h h 00 ii N−(i+1) n0 ∈ T T UN

N−(i+1) h 0 i ⊆ T VN

⊆ Vi+1. Thus,

˜n0 ∞ n0 n0−1 n0−(N−1) T ({zk}k=1) = T (z1) ,T (z1) ,...,T (z1) ,...,T (z1) , z1, z2,...

∈ V1 × V2 × ... × VN × X × X × ..., and so T˜n0 [U] T V is not empty and T˜ is hypercyclic on A. Finally, let U be a non-empty open subset of X. Since Y is dense in X, the open subset A T U × X × X... of A is not empty. Now using that the set of vectors hypercyclic for ˜ ∞ T T is actually dense in A, let y = {yk}k=1 be in A U × X × X... and be hypercyclic for ˜ T∞ n T . Clearly y1 is in U. It is also in n=1 Range (T ) since as y is in A, it follows that

n y1 = T (yn+1) for all n ≥ 1. Moreover, y1 is hypercyclic for T since by the hypercyclicity of y in A, for any non-empty and open subset U of X, there is some number n for which Tñ (y) is in A T U × X × X.... But this requires the first coordinate of Tñ (y) to be in U,

n and this ﬁrst coordinate is T (y1). 74

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