Hypercyclic Operators and Their Orbital Limit Points

HYPERCYCLIC OPERATORS AND THEIR ORBITAL LIMIT POINTS

Irina Seceleanu

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

August 2010

Committee:

Kit Chan, Advisor

Arthur Yeh, Graduate Faculty Representative

Juan B´es

Craig Zirbel ii ABSTRACT

Kit Chan, Advisor

Hypercyclicity is the study of linear and continuous operators that possess a dense orbit. Given a separable, inﬁnite dimensional topological vector space X, we say a continuous linear operator T : X → X is hypercyclic if there exists a vector x in X such that its orbit Orb(T, x)= {x, T x, T 2x, . . .} is dense in X. Many interesting phenomena appear when analyzing the behavior of iterates of linear and continuous operators, in particular we emphasize the existence of several zero-one laws.

We first note that, if an operator T has a hypercyclic vector, it has a dense Gδ set of such vectors, and hence the set of hypercyclic vectors for an operator is either empty or very large in a topological sense. Next, by proving that a somewhere dense orbit is everywhere dense, P. S. Bourdon and N. S. Feldman showed a second zero-one law which states that either an orbit Orb(T, x) is nowhere dense or it is dense in the whole space. In my dissertation we uncovered the existence of another such zero-one law for certain classes of operators. We showed that for a weighted backward shift on `p to be hypercyclic it suffices to require the operator to have an orbit Orb(T, x) with a single non-zero limit point, thus relaxing Bourdon and Feldman’s condition of having a dense orbit in some open subset of X. However, our condition does not guarantee that the original orbit Orb(T, x) is dense in X, nonetheless we can demonstrate how to construct a hypercyclic vector for T by using the non-zero limit point of the orbit. Even more interestingly, the condition above can be relaxed to simply requiring that the orbit has infinitely many members in a ball whose closure avoids the zero vector. To summarize this behavior of weighted backward shifts, we emphasize that a shift T is not hypercyclic if and only if every set of the form Orb(T, x)∪{0} is closed in `p. Thus we showed the existence of a zero-one law for the hypercyclicity of these shifts, which states that either no orbit has a non-zero limit point in `p or some orbit has every vector in `p as a limit point. iii Furthermore we showed that this zero-one law for the hypercyclic behavior of shifts is also shared by other classes of operators, in particular the adjoints of the multiplication operators on the Bergman space A2(Ω) for an arbitrary region Ω ⊂ C. To achieve this we cannot borrow techniques used for the shift operators, but instead we have to take a function theoretical approach. However, we also showed that this behavior does not generalize to all classes of operators, namely we provided an example of a linear fractional composition operator on the Hardy space H2(D) that is not hypercyclic, and yet it has an orbit with a non-constant limit point. To summarize the importance of our results, we would like to point out that in our endeavor to study the phenomena of hypercyclicity it is important to understand how an operator fails to be hypercyclic. Having proved that for certain classes of operators, a non- hypercyclic operator can at most have the zero vector as an orbital limit point, we have shown that these operators fail at having a dense orbit in quite a dramatic way. Thus we described the hypercyclic behavior of certain operators as a zero-one law of orbital limit points, and so we have uncovered another facet of hypercyclicity associated with dichotomous behavior. iv

To my mother v ACKNOWLEDGMENTS

I would like to thank my advisor Dr. Kit Chan for his constant support and guidance over the years. His valued suggestions throughout my research have made this dissertation possible. I have greatly benefitted from having him as my advisor in that he carefully planned my development and patiently taught me how to think about mathematics. It was a privilege to have such a skillful researcher and teacher guide me along the way. Lastly, in the pursuit of my PhD I have grown both as a mathematician and as a person, and so I would like to also thank Dr. Chan for being a wonderful mentor during my journey. His advice and our insightful discussions will always stay with me over the years. I owe warm gratitude to my committee member Dr. Craig Zirbel for his valuable suggestions and continued encouragement during my study. His wonderful pedagogy has benefited me both as a student who appreciated his clear instruction, and as a teacher who learned from his methods and approach to teaching. Most importantly, the example he sets has taught me to always strive to be the best that I can. I wish to thank my other committee members Dr. Juan Bes and Dr. Arthur Yeh for all their helpful comments and suggestions. I would also like to express my gratitude to Dr. Alexander Izzo who through his infectious enthusiasm has helped me develop a deep appreciation for mathematics, and has shaped my mathematical thinking. I will always warmly recall his stories and the wonderful memories I take with me from his classes. I would like to thank my friends in Bowling Green for all the good times. In particular, I am very grateful to Marcy and Ngoc for their friendship and all the memories over the years. My sincere thanks also go to Cyndi and Anita for all their help and wonderful conversations. Also my love goes to my dear friends Helen, Bubu and Senghiul. I would also like to thank my mother and father for their love and unconditional support throughout my education. Knowing that they will always be there for me has helped me overcome the difficult times in my life. It is nice to know that somewhere in the world there vi is a wall with a nail in it that is mine. My gratitude also goes to my brother whose support (and secret stash of chocolate) has always come in handy over the years. My thanks also go to my aunt. Lastly, I would like to thank Kevin for his love, encouragement and patience during my study years at Bowling Green. I am touched by his incredible support and sacrifice over the years, and am very thankful for the role he has played in my development into a mathematician. I am grateful for the many wonderful memories and know that with him by my side I have learned to be a better person. vii

Table of Contents

CHAPTER 1: A BRIEF HISTORY OF HYPERCYCLICITY 1 1.1 Hypercyclicity in the context of analysis ...... 1 1.2 How to prove an operator is hypercyclic ...... 4 1.3 How many vectors are hypercyclic? ...... 6 1.4 How many operators are hypercyclic? ...... 7 1.5 How big can a non-dense orbit be? ...... 8

CHAPTER 2: HYPERCYCLICITY OF SHIFTS AS A ZERO-ONE LAW OF ORBITAL LIMIT POINTS 10 2.1 Elementary properties and hypercyclicity of shifts ...... 10 2.2 A Zero-One Law for the hypercyclicity of shifts ...... 16 2.2.1 Introductory remarks ...... 16 2.2.2 The unilateral weighted backward shift ...... 17 2.2.3 The bilateral weighted backward shift ...... 28

CHAPTER 3: ORBITAL LIMIT POINTS AND HYPERCYCLICITY OF OPERATORS ON ANALYTIC FUNCTION SPACES 41 3.1 Introductory remarks ...... 41 3.2 The adjoints of multiplication operators ...... 42 3.2.1 Elementary properties and hypercyclicity of the adjoints of multiplication operators ...... 42 viii 3.2.2 A Zero-One Law for the hypercyclicity of the adjoints of multiplication operators ...... 48 3.3 The linear fractional composition operators ...... 52 3.3.1 Elementary properties and hypercyclicity of linear fractional composition operators ...... 52 3.3.2 Orbital limit points and hypercyclicity of linear fractional composition operators ...... 57 3.4 Further remarks ...... 59

BIBLIOGRAPHY 61

APPENDIX 64 ix

List of Tables

Table 1. Entries in the ﬁrst block for the hypercyclic vector construction 25 Table 2. Entries in the second block for the hypercyclic vector construction 26 Table 3. General term entries for the hypercyclic vector construction 27 Table 4. An example of a bilateral weighted shift 40 Table 5. Summary of the hypercyclic behavior of composition operators 56 1

CHAPTER 1

A BRIEF HISTORY OF HYPERCYCLICITY

1.1 Hypercyclicity in the context of analysis

One of the primary pursuits in the ﬁeld of analysis is that of approximating objects to any desired degree of accuracy. The notion of universality in analysis refers to the existence of a single object which, via a usually countable process, allows us to approximate a maximal class of objects. In the course of time many such objects that exhibit this extraordinary behavior have been discovered. It seems the ﬁrst example is due to Fekete [28] in 1914, who ∞ X j showed the existence of a universal power series ajx on [−1, 1] that not only diverges j=1 at every non-zero x, but it does so in the worst possible way. Namely, for every continuous nk X j function g on [−1, 1] with g(0) = 0 there is a sequence nk % ∞ so that ajx → g(x) j=1 uniformly as k → ∞, i.e. every continuous function vanishing at zero can be uniformly approximated by this universal power series. Another interesting example of a universal object was provided in 1935 by Marcinkiewicz

[26], who showed the existence of a universal primitive. That is, given a sequence {hn} ⊂ R

with hn → 0, there is a continuous function f on [0, 1] so that for every measurable function f(x + h ) − f(x) g, the sequence nk → g(x) almost everywhere on [0, 1], for some sequence hnk 2

{nk} that depends on g. However, one of the most remarkable universalities is due to Menshov [27], who in 1945 proved the existence of a universal trigonometric series whose coeﬃcients converge to zero. Stated in greater generality, Menshov’s result says that given a complete orthonormal system ∞ 2 X {ψn} in L [0, 1], there exists a series ajψj with an ∈ R and an → 0 having the property j=1 that for every measurable function f on [0, 1] there exists a sequence nk % ∞ so that n Xk ajψj(t) → f(t) almost everywhere in [0, 1]. j=1 Although one would expect universality to be a rare phenomenon in analysis, the above examples show that quite the opposite is the case. In fact, any process in analysis that diverges or behaves irregularly is likely to produce a universal element. Furthermore, when a process exhibits universality, then in most cases almost every element (in a Baire category sense) is universal. Thus universality is a generic phenomenon in analysis. One particular type of universality studied in the setting of topological vector spaces is the phenomenon of hypercyclicity. In this particular setting, the countable process by which the universal vector x allows us to approximate all other vectors in the space is the iterated application of an operator T to the vector x.

Deﬁnition 1. Given a topological vector space X and a continuous linear operator T : X → X, we say the vector x ∈ X is a hypercyclic vector for T if the orbit Orb(T, x)= {x, T x, T 2x, . . .} is dense in X. We call T a hypercyclic operator and denote the set of hypercyclic vectors for T by HC(T ).

Thus hypercyclicity is the study of linear and continuous operators that possess a dense orbit. The name of hypercyclicity was suggested in 1986 by Beauzamy because of its connection to the much older concept in operator theory of a cyclic operator. As the name suggests, hypercyclicity is a much stronger notion of cyclicity, which only requires the operator T to have an orbit with a dense linear span. A ﬁrst example of a hypercyclic operator was provided in 1929 by Birkhoﬀ [6], who showed the existence of an entire function f whose successive translates by a non-zero constant a are 3 arbitrarily close to any function in the space of entire functions H(C). That is, there exists an entire function f and a non-zero constant a ∈ C, so that for every g ∈ H(C), the sequence f(z + nka) → g(z) locally uniformly on C, for some sequence {nk} that depends on the function g. Thus, the translation operator Ta has a dense orbit {f(z), f(z +a), f(z +2a),...} in H(C). A second example of a hypercyclic operator was provided in 1952 by MacLane [25], who showed that there exists a universal entire function f whose successive derivatives are dense in H(C). Namely, there is an entire function f having the property that for every g ∈ H(C)

(nk) there exists a sequence nk % ∞ so that f (z) → g(z) locally uniformly in C, and hence the differentiation operator D on H(C) is hypercyclic. As it turns out, hypercyclicity is not just a mere curiosity but a wide-spread phenomenon in analysis as there are many natural operators that are hypercyclic, and in fact we will see that every separable, infinite dimensional Fréchet space supports a hypercyclic operator. Hypercyclicity is also closely related to topological dynamics, which studies among other things continuous maps with dense orbits on compact topological spaces. However, hypercyclicity carries an additional linear structure which allows us to bring into play many tools from the field of functional analysis. As a result, the underlying space in hypercyclicity is never compact. Nevertheless, the diversity of ideas and techniques involved in the study of the linear dynamics of an operator have led to a rich body of interesting results. The importance of the field of hypercyclicity is made evident by its connection to the famous Invariant Subspace Problem in operator theory, which asks whether every continuous linear operator T on X has a non-trivial closed subspace Y ⊂ X which is T -invariant. Although the problem is still open for Hilbert space operators, Enflo [16] offered a negative solution for a specific Banach space, which was later transferred by Read [31] to an operator on `1(N) that has every non-zero vector hypercyclic. Given that a closed T -invariant subspace containing x must also contain the closure of the Orb(T, x), the hypercyclicity of every non-zero vector for the Read operator yields that T has only the trivial closed T -invariant 4 subspaces {0} and `1(N), and hence the Invariant Subspace Problem has a negative answer for the space `1(N). The field of hypercyclicity was born in 1982 with the Ph.D. dissertation of Kitai [24] and later with the paper of Gethner and Shapiro [19] and in more than two decades since, it has become the focus of active research by numerous authors.

1.2 How to prove an operator is hypercyclic

Our ﬁrst characterization of hypercyclicity is due to Birkhoﬀ [6], and it relates the concept of hypercyclicity to the well known notion of topological transitivity.

Theorem 2 (Birkhoﬀ Transitivity Theorem). Let X be a separable F-space and T : X → X 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 n ∈ N such that T n(U) ∩ V 6= ∅.

We remark that the implication (i) ⇒ (ii) holds for an arbitrary topological vector space X, however the converse follows from a direct application of the Baire category theorem, and therefore requires that the space X be a second countable Baire space. In her dissertation, Kitai [24] further remarked that an invertible operator is topologically transitive if and only if its inverse T −1 is topologically transitive. Thus, a continuous linear operator T on an F-space is hypercyclic if and only if T −1 is hypercyclic, however T and T −1 may not share the same hypercyclic vectors. While the above result of Birkhoff characterizes hypercyclicity, the following sufficient condition has proved extremely useful in applications. Although many variants of the criterion have been obtained, the Hypercyclicity Criterion is due, independently, to Kitai [24] and Gethner and Shapiro [19]. We note that the version we adopt here is due to Bésand Peris [5]. 5 Theorem 3 (Hypercyclicity Criterion). Let X be a separable F-space and T : X → X a continuous linear operator. If there exists a 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,

then T is hypercyclic and we say T satisﬁes the Hypercyclicity Criterion.

For a long time, the question whether the above suﬃcient condition is in fact necessary for the hypercyclicity of an operator, evaded all attempts at being resolved. In a relatively recent paper, Read [15] answered the question in the negative by constructing a remarkable operator T that is hypercyclic but does not satisfy the Hypercyclicity Criterion. Nevertheless, one should not underestimate the importance of the above theorem, as nearly all known hypercyclic operators were shown to have this property by using this criterion. Finally, we would like to point out a very useful result of Shapiro [34], which shows that in the search for hypercyclic operators on a space X it can be useful to ﬁnd hypercyclic operators on smaller spaces.

Theorem 4 (Hypercyclicity Comparison Principle). Suppose X and Y are two normed spaces, Y is continuously and densely embedded in X, and T is a linear transformation on X that maps the smaller space Y to itself and is continuous in the topology of each space, then T is hypercyclic on the larger space X whenever T is hypercyclic on Y . In particular,

if x is a hypercyclic vector for T |Y , then x is a hypercyclic vector for T .

T | Y −−−→Y Y     y y X −−−→T X 6 1.3 How many vectors are hypercyclic?

The answer to this question is given by the following remarkable zero-one law for the set of hypercyclic vectors for an operator T , which states that either the set HC(T ) is empty or it is a dense Gδ set. In other words, as soon as an operator is hypercyclic, its set of hypercyclic vectors is very large in a topological sense.

Theorem 5. [24] 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.

The above standard result follows from an application of the Baire category theorem and ∞ ∞ \ [ −n describing the set of hypercyclic vectors by HC(T ) = T (Uj), where {Uj}j∈N forms j=1 n=0 a countable basis for X. Maybe the most remarkable result in the ﬁeld of hypercyclicity is that of Read [15], who showed that this zero-one law for the set of hypercyclic vectors can be taken to the extreme. Read was able to construct an operator for which every non-zero vector is hypercyclic, and thus for this operator the set of hypercyclic vectors is not just large in a topological sense but it is maximal. Furthermore, the preceding result about the topological largeness of HC(T ) also implies that the set of hypercyclic vectors is quite big in an algebraic sense. Since both sets HC(T ) and x − HC(T ) are dense Gδ in X, the Baire category theorem yields that their intersection is non-empty. Thus every vector in X can be written as the sum of two hypercyclic vectors.

Theorem 6. [22] If X is a separable F-space and T : X → X is a hypercyclic operator, then X = HC(T ) + HC(T ).

The above result shows that the set of hypercyclic vectors is not closed under addition, nonetheless it always contains a linear subspace consisting entirely, except the zero vector, of hypercyclic vectors. We call such a subspace a hypercyclic manifold for T .

Theorem 7. [7] Let X be a topological vector space and T a hypercyclic operator. If x ∈ HC(T ), then {p(T )x : p polynomial} is a hypercyclic manifold for T . In particular, T admits a dense hypercyclic manifold. 7 1.4 How many operators are hypercyclic?

Perhaps the reason one does not expect hypercyclicity to be a wide-spread phenomenon lies in the simple realization that ﬁnite-dimensional spaces do not support hypercyclic operators.

Theorem 8. [24] There are no hypercyclic operators on ﬁnite dimensional F-spaces.

Nonetheless, Ansari [1] was able to make use of the aforementioned Hypercyclicity Com- parison Principle to show the remarkable fact that an inﬁnite dimensional space with nice enough structure always carries a hypercyclic operator.

Theorem 9. [1] Every separable, inﬁnite dimensional Fr´echetspace carries a hypercyclic operator.

Suppose now that a space admits hypercyclic operators. Motivated by the fact that the existence of a single hypercyclic vector implies the existence of a dense set of such vectors, one might ask if an analogous statement is true for operators. However, the set of all hypercyclic operators is not dense in the Banach algebra of linear and continuous operators B(X) endowed with the norm topology, as every hypercyclic operator has norm greater than one, and in fact (B(X), k.k) is not even separable. The following theorem shows that in the case of Hilbert spaces an even stronger statement holds.

Theorem 10. [32] Let H be an inﬁnite dimensional Hilbert space. Then the set of all cyclic operators is nowhere dense in B(H) with respect to the norm topology.

However, if we endow B(X) with the strong operator topology, Chan [10] showed that the set of hypercyclic operators on X is dense in B(X). Recall that the strong operator topology (SOT) is the weakest topology on B(X) for which the evaluation maps T 7→ T (x), x ∈ X are continuous.

Theorem 11. [4] Given a separable Fr´echetspace X, if T : X → X is hypercyclic, then there exists an SOT-dense linear subspace of B(X) which consists entirely, except the zero operator, of hypercyclic operators. 8 Finally, continuing the analogy to hypercyclic vectors (see Theorem 6), Grivaux [21] showed the following surprising result.

Theorem 12. (i) Every operator on a separable, inﬁnite dimensional Hilbert space is the sum of two hypercyclic operators. (ii) There exists a separable, inﬁnite dimensional Banach space on which not every operator is the sum of two hypercyclic operators.

1.5 How big can a non-dense orbit be?

To better understand the phenomenon of hypercyclicity, we also need to address the question about how big the orbits of an operator T can be without ever becoming dense. We remark that the results we obtain in Chapters 2 and 3 provide some surprising insight into this question. Furthermore, our work is a continuation of a result of Bourdon and Feldman [8], who showed the following remarkable zero-one law for a topological vector space: either an orbit of an operator T is nowhere dense, or it is everywhere dense, in which case T is hypercyclic.

Theorem 13. [8] Suppose T is a bounded linear operator on a topological vector space X. If T has a somewhere dense orbit Orb(T, x), then this orbit is dense in X, and T is hypercyclic.

In fact we were able to show that for certain classes of operators the statement in Theorem 13 can be further relaxed. We show that if an operator T is not hypercyclic, then no orbit of T can have a non-zero limit point; that is if an orbit of T has a limit point, then it must be the zero vector. Thus we uncover the existence of another zero-one law for the hypercyclicity of certain classes of operators, which states that either no orbit has a non-zero limit point or some orbit has every vector in the space as a limit point. Finally, the question above has also been studied in a diﬀerent direction by Chan and Sanders [11], who showed the existence of a weakly dense orbit that is not norm dense. Here 9 the weak sense of density is with respect to the weak topology on the topological vector space X.

Theorem 14. [11] There exists a bounded linear operator T on a Banach space X that is weakly hypercyclic but not norm hypercyclic. 10

CHAPTER 2

HYPERCYCLICITY OF SHIFTS AS A ZERO-ONE LAW OF ORBITAL LIMIT POINTS

2.1 Elementary properties and hypercyclicity of shifts

The class of weighted shift operators constitutes a favorite testing ground in the literature of operator theory. It is therefore natural to begin our analysis of the phenomenon of hypercyclicity as a zero-one law of orbital limit points by showing that this law does indeed hold for weighted shifts. In this section we deﬁne the class of weighted shifts, establish some of the basic properties of the shift operators, and oﬀer a brief summary of the previously established results about the hypercyclicity of shifts. For a classical survey on weighted shifts see Shields [35].

p We deﬁne the space ` (I) for 1 ≤ p < ∞ to be the space of all sequences {xˆ(j)}j∈I with

X p 1 kxk = ( |xˆ(j)| ) p < ∞ for I = N ∪ {0} or I = Z. For notational simplicity we will use j∈I p Z+ to denote N ∪ {0}. We note that ` (I) is a separable, inﬁnite dimensional Banach space that has the canonical basis {en}n∈I , where each en = {eˆn(j)}n∈I withe ˆn(j) = 0 if j 6= n ande ˆn(j) = 1 if j = n. 11 p p A linear operator T : ` (Z+) → ` (Z+) is said to be a unilateral weighted backward shift

if there is a sequence of complex weights {wn}n≥1 such that T en = wnen−1, if n ≥ 1 and

p p T e0 = 0. Similarly, for the canonical basis {en : n ∈ Z}, a linear operator T : ` (Z) → ` (Z) is said to be a bilateral weighted backward shift if there is a sequence of complex weights {w } such that T e = w e for all n ∈ . n n∈Z n n n−1 Z Weighted forward shifts are also deﬁned on the space `p(I) in the obvious way. In what follows we use B to denote the unweighted backward shift and S to denote the unweighted forward shift. The following proposition [35] relates the forward to the backward shift for the Hilbert space `2(I).

Proposition 15. If T is a bilateral forward weighted shift with weight sequence {w } , n n∈Z ∗ ∗ then its adjoint T is given by T en = wn−1en−1, for all n ∈ Z. If T is a unilateral forward weighted shift with weight sequence {w } , then its adjoint T ∗ is given by T ∗e = w e n n∈Z+ n n−1 n−1 ∗ for all n ≥ 1, and T e0 = 0.

Proof. Note that in the bilateral case, hT ej, eii = hwjej+1, eii = wjhej+1, eii and hej, wi−1ei−1i =

wi−1hej, ei−1i. Both expressions are non-zero if j + 1 = i, in which case they equal wj. For

p X X X x, y ∈ ` (Z) with x = xˆ(j)ej and y = yˆ(j)ej, we have hT x, yi = xˆ(j)yˆ(i)hwjej+1, eii = j∈Z j∈Z i,j∈Z X X X xˆ(j)yˆ(j + 1)wj. On the other hand, xˆ(j)yˆ(i)hej, wi−1ei−1i = xˆ(j)yˆ(j + 1)wj. So j∈Z i,j∈Z j∈Z for T ∗ deﬁned as in the proposition, hT x, yi = hx, T ∗yi, for all x, y ∈ `p(Z). A similar argument can be made in the unilateral case.

We remark that given the weighted forward shift T with weights {wn}n∈I , the adjoint

∗ operator T is not the weighted backward shift with weight sequence {wn}n∈I . Nevertheless,

the forward shift with weight sequence {wn+1}n∈I has as its adjoint the weighted backward shift with the desired weights {wn}n∈I . In the following proposition [35] we establish that a weighted backward shift T is bounded if and only if its sequence of weights {wn}n∈I is bounded, in particular kT k =sup{|wn| : n ∈ I}. 12 n Proposition 16. kT k = sup |wk−n+1 · ... · wk| for all n ≥ 1. k∈I

n Proof. Let n ≥ 1. Since T ek = (wk−n+1 · ... · wk)ek−n, we have that |wk−n+1 · ... · wk| =

n n n kT ekk ≤ kT k kekk = kT k. Thus taking supremum over all k ∈ I, we obtain that the

n sup |wk−n+1 · ... · wk| ≤ kT k. k∈I On the other hand,

kT nk = inf{c > 0 : kT nxk ≤ c kxk , x ∈ `p(I)}

n ≤ inf{c > 0 : kT ekk ≤ c kekk , k ∈ I}

= inf{c > 0 : |wk−n+1 · ... · wk| ≤ c, k ∈ I}

= sup |wk−n+1 · ... · wk| . k∈I

As a consequence of the above result, we will assume that the weight sequence {wn}n∈I for the weighted shift T is always bounded above. In the following we will see that a lower bound on the weights yields another important property for the shift operator T . We ﬁrst observe that a unilateral weighted backward shift is never injective, as its kernel is the linear span of the vector e0. On the other hand, a bilateral weighted shift T is injective if and only if none of its weights is zero. We remark that if one of its weights is zero, say

w0 = 0, we can write T = T1 ⊕ T2, where T1 is a unilateral weighted shift on the closed

subspace spanned by {ek : k ≥ 0} and T2 is a weighted shift on the orthogonal complement. We will assume from now on that none of the weights is zero. p X yˆ(j) Moreover, the vector y ∈ `p(I) is in the image of T if and only if < ∞, in wj+1 j∈I which case the vector x given byx ˆ(j) = yˆ(j) for all j ∈ I is the unique vector T sends to wj y. Thus a weighted shift T is surjective if and only if its weight sequence is bounded away from zero. Therefore, a bilateral weighted backward shift T is invertible if and only if there exists m > 0 so that |wj| ≥ m for all j ∈ I. In fact we will show in the following that in the context of hypercyclicity we can assume without loss of generality that the weight sequence 13

is always positive, i.e. wj > 0 for all j ∈ I, and thus the absolute value in the statement above can be dropped.

We say two bounded linear operators A and B on a Banach space X are isometrically isomorphic, if there is a linear homeomorphism U : X → X with kUk = 1 such that UAU −1 = B.

Proposition 17. [35] If {λn} is a sequence of complex numbers with |λn| = 1, then a weighted backward shift T with weight sequence {wn} is isometrically isomorphic to the back-

ward weighted shift operator with weight sequence {λn−1λnwn}.

Proof. Without loss of generality assume that T is a bilateral weighted backward shift, and

p p let U : ` (Z) → ` (Z) be deﬁned by setting Uen = λnen for all n ∈ Z, and extending by

p X linearity and continuity. Since |λn| = 1 for all n ∈ Z, given g ∈ ` (Z) with g = gˆ(n)en n∈I p Xgˆ(n) we can deﬁne a vector h in ` (Z) by setting h = en, and so Uh = g. Thus U is λn n∈I surjective. 1 ! p p p X ˆ X ˆ Also, for any h ∈ ` (Z) we have kUhk = k λnh(n)enk = h(n) = khk, so U n∈I n∈I is injective and has norm kUk = 1. Thus U is an isometric isomorphism.

Finally, deﬁne T˜ on `p(Z) by setting T˜ = UTU −1. Then T˜ is isometrically isomorphic to T . Furthermore, for all n ∈ , we have U −1e = 1 e = λ e . So T˜ e = UTU −1(e ) = Z n λn n n n n n ˜ UT (λnen) = U(λnwnen−1) = (λn−1λnwn)en−1 for all n ∈ Z. Note that T is linear and ˜ continuous. Thus T is a weighted backward shift with weights {λn−1λnwn}n∈Z.

From the above proposition we can conclude that a weighted backward shift T is isometrically isomorphic to a weighted backward shift operator with positive weights.

Corollary 18. [35] A weighted backward shift operator T with weight sequence {wn}n∈I is isometrically isomorphic to the weighted backward shift operator with weight sequence

{|wn|}n∈I . 14

|w1| |w−1| Proof. Let λ0 = 1. Set λ1 = and λ−1 = so that |w1| = λ0λ1w1 and |w0| = λ−1λ0w0. w1 w−1

1 |w2| 1 |w−2| Next, we set λ2 = and λ−2 = , etc. Clearly |λn| = 1 and λn−1λnwn = |wn| for λ1 w2 λ−1 w−2 all n ∈ I, and hence T is isometrically isomorphic to the weighted backward shift T˜ with

weight sequence {|wn|}.

Finally, the following standard proposition shows that in the context of hypercyclicity we can always assume that the weight sequence for the shift operator T is positive.

Proposition 19. Hypercyclicity is preserved under isomorphism; that is, if a hypercyclic operator T is isomorphic to T˜, then T˜ is also hypercyclic.

Proof. Suppose there exists an isomorphism U : X → X so that U −1TU = T˜, and suppose further that there exists a vector x ∈ X with Orb(T, x) = X. Since U is surjective, letx ˜ ∈ X

so that Ux˜ = x, and let y in X. We want to show that there exists a sequence mk % ∞ so

that T˜mk x˜ → y in X. Note that since Uy ∈ X and T is hypercyclic, there exists a sequence

nk nk nk % ∞ so that T x → Uy in X. Since x = Ux˜, we get that T (Ux˜) → Uy in X. Finally, the continuity of U −1 yields that T˜nk x˜ = U −1T nk Ux˜ → y in X, so Orb(T,˜ x˜) = X.

Given our previous considerations about the weight sequence of a shift, we are now ready to give the deﬁnition of unilateral and bilateral shifts that we will use in the sequel.

p Deﬁnition 20. Given the canonical basis {en : n ∈ Z} for ` (Z+) for 1 ≤ p < ∞, a bounded

p p linear operator T : ` (Z+) → ` (Z+) is said to be a unilateral weighted backward shift if there

is a bounded sequence of positive weights {wn}n≥1 such that T en = wnen−1, if n ≥ 1 and

T e0 = 0.

Similarly, a bounded and linear operator T : `p(Z) → `p(Z) is said to be a bilateral weighted backward shift if there is a bounded sequence of positive weights {wn : n ∈ Z} such that T en = wnen−1 for all n ∈ Z. 15 Having defined the weighted shift operators and established some of their elementary properties, we now turn our attention to the hypercyclic behavior of these shifts. We choose the backward direction for a shift, since the unilateral forward shift is never hypercyclic. We first note that relatively early on, Salas [33] characterized the hypercyclicity of weighted shifts by providing a necessary and sufficient condition in terms of the weights.

Theorem 21. A bilateral weighted backward shift T is hypercyclic if and only if for every

> 0 and every q ∈ N there exists n arbitrarily large such that for every j ∈ Z with |j| ≤ q n−1 n Y 1 Y we have w > and w < . j+s j−s s=0 s=1 As an easy consequence, one deduces the following analogous statement for the unilateral weighted backward shift.

Corollary 22. [33] A unilateral weighted backward shift T is hypercyclic if and only if there n Yk exists a sequence nk % ∞ so that wj → ∞. j=1 We note that the above condition for unilateral shifts can be used to characterize the hypercyclic behavior of a shift with greater ease than its bilateral counterpart. Nonetheless, Feldman [17] showed that an analogous condition can be obtained for a bilateral weighted shift T , provided we assume the operator T is invertible.

Proposition 23. A bilateral weighted backward shift T : `p(Z) → `p(Z) with weight sequence

{wn}n∈Z where wn ≥ m for some m > 0 is hypercyclic if and only if there exists a sequence n n Yk Yk nk % ∞ so that wj → ∞ and w−j → 0. j=1 j=1 We note that in fact Feldman [17] showed that the above proposition still holds if we only assume that the negative indexed weights are bounded below, i.e. there exists m > 0 so that wn ≥ m for all n < 0 (or alternatively all n > 0). Finally, Chan and Sanders [11] showed that the condition in Corollary 22 is also equivalent to another weaker notion of hypercyclicity, which is appropriately called weak hypercyclicity. We say an operator T on a separable, inﬁnite dimensional Banach space X is weakly 16 hypercyclic if there is a vector x in X whose orbit Orb(T, x) is dense in X with respect to the weak topology. Since the norm topology is strictly stronger than the weak topology, it follows that every (norm-)hypercyclic operator is also weakly hypercyclic. In [11] Chan and Sanders showed that for a unilateral weighted backward shift the converse also holds true.

p p Theorem 24. Let T : ` (Z+) → ` (Z+) be a unilateral weighted backward shift. Then T is (norm-)hypercyclic if and only if T is weakly hypercyclic.

However, Chan and Sanders [11] also showed that there exists a bilateral weighted back-

ward shift T on `p(Z) that is weakly hypercyclic but not (norm-)hypercyclic.

2.2 A Zero-One Law for the hypercyclicity of shifts

2.2.1 Introductory remarks

As noted in the previous section, the class of weighted backward shifts has been well studied in the field of hypercyclicity, in fact relatively early on Salas [33] characterized the hypercyclic shifts by offering a necessary and sufficient condition in terms of their weight sequences. In the following we relate the concept of hypercyclicity to the geometry of an orbit, thus obtaining a new equivalent condition for the hypercyclicity of these shifts. We would like to point out that our results are also a continuation of the work of Bourdon and Feldman [8]. The authors proved that if an operator T has a somewhere dense orbit Orb(T, x), then the same orbit will be everywhere dense in X, and thus the operator T is hypercyclic. As it turns out, in the case of weighted backward shifts their remarkable insight can be carried even further. Indeed, in the next sections we show that for a shift to be hypercyclic it suffices to require the operator to have an orbit Orb(T, x) with a single non-zero limit point, thus relaxing Bourdon and Feldman’s condition of having a dense orbit in some open subset of X. However, our condition does not guarantee that the original orbit Orb(T, x) is dense in X, nonetheless we can demonstrate how to construct a hypercyclic 17 vector for T using the non-zero limit point of the orbit. Even more interestingly, the condition above can be relaxed to simply requiring that the orbit has infinitely many members in a ball whose closure avoids the zero vector. To summarize this behavior of weighted backward shifts we emphasize that a shift T is not hypercyclic if and only if every set of the form Orb(T, x)∪{0} is closed in X. Thus we uncover the existence of a zero-one law for the hypercyclicity of these shifts, which states that either no orbit has a non-zero limit point in X or some orbit has every vector in X as a limit point. In Section 2.2.2, we provide several conditions that characterize the hypercyclicity of a unilateral weighted backward shift. We then offer a technique for constructing a hypercyclic vector for a unilateral weighted backward shift T having e0 as a limit point of one of its orbits Orb(T, x). In Section 2.2.3, we give a proof for the bilateral analogue of the above results which calls for different techniques.

2.2.2 The unilateral weighted backward shift

p p Let {en : n ≥ 0} be the canonical base for ` (Z+) for p ≥ 1, denoted by ` in the following. ∞ ∞ p X X p A vector x in ` is denoted by x = (ˆx(0), xˆ(1),...) = xˆ(i)ei, where |xˆ(i)| < ∞.A i=0 i=0 bounded and linear operator T : `p → `p is said to be a unilateral weighted backward shift if there is a sequence of bounded positive weights {wn}n≥1 such that T en = wnen−1, if n ≥ 1 and T e0 = 0. We recall that for a unilateral weighted backward shift T to be hypercyclic, Salas [33] n Y provided a necessary and suﬃcient condition on the weights that sup wj = ∞. Another n≥1 j=1 characterization was obtained by Chan and Sanders [11], who showed that T is hypercyclic if and only if T is weakly hypercyclic, which means that T has an orbit Orb(T, x) that is dense in the weak topology of `p. In the following, we show that the above equivalent conditions can be carried forward to other conditions in terms of the geometry of an orbit. 18 Theorem 25. Let T : `p → `p be a unilateral weighted backward shift, where 1 ≤ p < ∞. The following are equivalent: (i) T is hypercyclic.

(ii) For any vector f in `p, there exists a vector x = x(f) in `p whose orbit under T Orb(T, x)= {x, T x, T 2x, . . .} has f as a limit point.

p (iii) The vector e0 is a limit point of a certain orbit Orb(T, x) with x in ` .

(iv) There exists an x in `p whose orbit Orb(T, x) has a non-zero limit point.

(v) There exists an x in `p whose orbit Orb(T, x) has a non-zero weak limit point.

(vi) There exists a vector x in `p whose orbit Orb(T, x) has inﬁnitely many members in an open ball whose closure avoids the origin; that is, there are a non-zero vector f in `p and a positive r with r < kfk such that Orb(T, x) ∩ B(f, r) is inﬁnite.

Before we provide a proof for the theorem, we have a few remarks to illustrate the result. 1. For statement (iv), the limit point f cannot be chosen to be the zero vector. Take

1 p wj = 2 for all j ≥ 1. Let x = (x0, x1, x2,...) be a vector in ` with inﬁnitely many non-zero n entries xi. Then T x → 0 as n → ∞, so the zero vector is a limit point of Orb(T, x), but T is clearly not hypercyclic. 2. Regarding the equivalence of statements (i) and (iv), we remark that if an orbit Orb(T, x) has a non-zero limit point, the vector x that generates the orbit is not necessarily a hypercyclic vector. Take, for instance, the unilateral weighted backward shift T : `1 → `1

whose weight sequence is given by wj = 2 for all j ≥ 1, and the vector x = (x0, x1, x2,...)

−j k 1 2k given by xj = 2 if j = 2 and xj = 0 otherwise. Clearly x ∈ ` and T x − e0 = ∞ X k j 22 2−2 , which goes to 0 as a limit, as k → ∞. However, Tdnx(0) = 0 if n 6= 2k, and j=k+1 Tdnx(0) = 1 if n = 2k. Hence x is not a hypercyclic vector. In fact, for a similar reason x is not even a supercyclic vector because the scalar multiples of Orb(T, x) cannot approximate

e0 + e1. For a deﬁnition, we say an operator T : X → X is supercyclic if there is a vector

x ∈ X such that the set {λT nx : n ≥ 0, λ ∈ C} is dense in X. 19 3. Statement (vi) cannot be relaxed to consider weakly open sets. That is, a unilateral weighted backward shift T on `p may not be hypercyclic even if T has an orbit Orb(T, x) with infinitely many members inside a weakly open set whose weak closure does not contain zero. ∞ ∞ 2 1 X 1 2 X 1 4 1 Let g = e j+1 . Clearly kgk = = = . Consider the weakly open 2j 3·4 2j 1 − 1 3 j=1 j=1 4 2 1 2 1 set U = f ∈ ` : |hf − g, gi| < 10 . Then if f ∈ U, we have that |hf, gi| − kgk < 10 , and 7 13 2 1 thus 30 < |hf, gi| < 30 . Let V = f ∈ ` : |hf, gi| < 30 . Then if f ∈ V , f∈ / U, and hence U ∩ V = ∅. But 0 ∈ V , so U is a weakly open set whose weak closure avoids the origin. We now proceed to define a bounded positive weight sequence for T as follows. For any positive integer j in an interval of the form 1 + 2 · 4k, 4k+1, where k ≥ 1, we define

1 1 4k w1+2·4k = ... = w3·4k = k·2k−1 , 1 k−1 4k w1+3·4k = ... = w4k+1 = k · 2 .

For those positive integers j outside the intervals of the form 1 + 2 · 4k, 4k+1, for k ≥ 1,

we simply take wj = 1.

n 1 o 4x Note that sup x : x ≥ 1 < ∞, and so {wj}j≥1 is a bounded sequence. Hence the

unilateral weighted backward shift T with the weight sequence {wj}j≥1 is a bounded linear operator. Furthermore, from Salas’ criterion for hypercyclicity for unilateral backward shifts

(see [33]) we immediately see that T is not hypercyclic since w1 · ... · wj ≤ 1 for all integers j ≥ 1. ∞ ∞ 2 ∞ X 1 2 X 1 1X 1 Now, let x = e j+1 . Clearly, kxk = = < ∞. 3j 4 3j 9 j2 j=1 j=1 j=1 Furthermore, for any integer k ≥ 2,

k−1 ∞ 4k 1 2 X 1 T x = e + e k + w j+1 k · ... · w j+1 e j+1 k . 3(k − 1) 0 3 3·4 3j 4 −4 +1 4 4 −4 j=k+1

Since 4j+1 −4k > 3·4j+1 whenever j ≥ k+1, the above summation is obviously orthogonal

D 4k E 2 D 2k−1 E 1 1 1 to g. Hence, T x, g − kgk = e k , g − = − = 0 for all k ≥ 2. So, 3 3·4 3 3 3 T 4k x ∈ U for all k ≥ 2, however T is not hypercyclic. 20 We are now ready to prove Theorem 25.

Proof. It is clear that (i) implies (vi), (ii) implies (iii), (iii) implies (iv), which in turn implies (v).

To show (i) implies (ii), suppose that T is hypercyclic. Then by deﬁnition, there exists x ∈ `p such that Orb(T, x) is dense in `p. Let f ∈ `p. If f∈ / Orb(T, x), then clearly f is a limit point of Orb(T, x). On the other hand, if f ∈ Orb(T, x) and f is not a limit point of the orbit, then there is a neighborhood U of f that contains no point of Orb(T, x) other than f. But then the points of U \{f} are not in the closure of Orb(T, x), which gives a contradiction.

To show (iv) implies (i), we suppose that there exist a vector x and a non-zero vector

p f = (f0, f1, f2, f3,...) in ` such that f is a limit point of the orbit Orb(T, x). Since fj 6= 0

for some j ≥ 0, we assume without loss of generality that f0 6= 0. Hence there exists an

increasing sequence {nk}k≥1 ⊂ N and an N > 0 such that

nk 1 |f0| kT x − fk < 2k < 2 , for all k ≥ N.

p Let x = (x0, x1, x2,...) ∈ ` . Then

nk nk T x = T (x0, x1, x2,...) = (w1 · ... · wnk xnk ,...).

nk Hence kT x − fk ≥ |w1 · ... · wnk xnk − f0|. So there exists a sequence {nk}k≥1 such

|f0| that |w1 · ... · wnk xnk − f0| < 2 , for all k ≥ N.

|f0| |f0| Thus 2 < |w1 · ... · wnk xnk | and so < |xnk | for all k ≥ N. Hence we get that 2(w1·...·wnk ) |f |p 0 < |x |p, for all k ≥ N . p p nk 2 (w1 · ... · wnk ) Now, since x ∈ `p we have

p ∞ ∞ |f0| X 1 X p p ≤ |x | ≤ kxk < ∞. p p nk 2 (w1 · ... · wn ) k=N k k=N 21 1 It follows that p → 0, i.e. there exists an increasing sequence {nk} such that (w1 . . . wnk ) w1 · ... · wnk → ∞ as k → ∞. n Y Thus by Salas’ criterion for hypercyclicity of unilateral backward shifts that sup wj = n≥1 j=1 ∞, we have that T is hypercyclic.

To show (v) implies (iv), we suppose there exists a vector x in `p such that the Orb(T, x)

p has f ∈ ` as a non-zero weak limit point. Since f 6= 0, let k ≥ 0 such that fk 6= 0. Considering the weakly open sets that contain f, we get that for all j ≥ 1 there exists

nj 1 an nj ≥ 1 such that for k as above, |hT x − f, eki| < j . 1 That is wk+1 · ... · wk+nj xk+nj − fk < j , for all j ≥ 1.

Next, we inductively pick a subsequence {njk } of {nj} as follows:

1. Let j1 = 1.

2. Once we have chosen jm we pick jm+1 > jm such that k + njm < njm+1 and ∞ ∞ X p 1 X p p |x | ≤ . This can be done since x ∈ `p, so |x | ≤ kxk < ∞ k+ni p·njm k+ni jm · kT k i=jm+1 i=1 1 and has been ﬁxed in the previous m-th step. p·njm jm · kT k Now, without loss of generality we can assume, by taking a subsequence if necessary, that ∞ X p 1 {n } satisﬁes k + n < n and |x | ≤ . j j j+1 k+ni j · kT kp·nj i=j+1 ∞ X p Let y = xk+ni · ek+ni . Clearly since x is in ` , so is y, as kyk ≤ kxk < ∞. i=1 ∞ nm X nm Then T y = xk+ni · T ek+ni . But k + ni < ni+1 for all i ≥ 1, and so k + ni < nm i=1 for all i < m. Thus since T is a unilateral backward shift we conclude that T nm y = ∞ X nm xk+ni · T ek+ni . i=m nm nm Furthermore, since the vectors T ek+ni and T ek+nj have disjoint support for i 6= j,

nm nm that is T \ek+ni (s) = 0 whenever T \ek+nj (s) 6= 0, we have that 22

nm p Thus T y → fkek in norm as m → ∞, where fkek 6= 0 in ` . So Orb(T , y) has a non-zero limit point.

To show (vi) implies (i), we suppose there exist non-zero vectors x and f in `p, and a positive number r with 0 < r < kfk such that Orb(T, x)∩B(f, r) is infinite. wk For p > 1, we have that `p is reflexive, so the convex unit ball Ball(`p) := Ball(`p) = k·k Ball(`p) is weakly compact. Now, since `p is a Banach space, by the Eberlein-Smulian Theorem, Ball(`p) is weak limit point compact, so every infinite set has a weak limit point. Since Orb(T, x)∩B(f, r) is infinite and included in B(f, r) which is weak limit point compact, we conclude that the Orb(T, x) has a non-zero weak limit point in `p as 0 ∈/ B(f, r). Thus since (v) implies (i) we have that T is hypercyclic. For the remaining case that p = 1, we will show that there exists a vector y ∈ `1 such that Orb(T, y) has a non-zero limit point. Thus, since (iv) implies (i), it follows that T is hypercyclic.

Claim 1: Without loss of generality we can assume that f has at most ﬁnitely many non-zero entries.

Proof of Claim 1. Suppose f has inﬁnitely many non-zero entries. Since the set {h ∈ `1 : h has ﬁnitely many non-zero entries} is dense in `1, we have that

1 kfk−r there exists h ∈ ` with ﬁnitely many non-zero entries such that kf − hk < 2 . 23 1 kfk−r Thus for g ∈ ` with kg − fk < r, we have kg − hk ≤ kg − fk + kf − hk < r + 2 = kfk+r kfk−r 2 < khk (since kfk + 2 < khk). kfk+r 0 kfk+r Therefore B(f, r) ⊂ B(h, 2 ), and hence for r = 2 we have that the intersection Orb(T, x)∩B(h, r0) is inﬁnite with 0 < r0 < khk.

Now, by our Claim 1 assume that there exists an N > 0 such that fk = 0 for all k ≥ N.

1 By assumption there exist a vector x ∈ ` and a strictly increasing sequence {nj} ⊂ N such that T nj x ∈ B(f, r) for all j, where 0 < r < kfk.

Let E = {i ≥ 0 : 0 ≤ i ≤ N, fi 6= 0}.

r·|f | Claim 2: For all j ≥ 0 there exists i ∈ E such that (T n\j x − f)(i) < i . kfk

n \j0 Proof of Claim 2. Suppose that there exists j0 ≥ 0 such that for all i ∈ E, (T x − f)(i) ≥

r·|fi| n X n Xr · |fi| r X . Then kT j0 x − fk ≥ (T \j0 x − f)(i) ≥ = · |f | = r, which gives kfk kfk kfk i i∈E i∈E i∈E n a contradiction with T j0 x ∈ B(f, r).

Now, since E ⊂ {1, 2,...,N} is a ﬁnite set we get by our Claim 2 that there exists an

r·|f | i ∈ E such that for inﬁnitely many j we have (T n\j x − f)(i) < i . kfk

Without loss of generality we can assume, by taking a subsequence of {nj} if necessary,

r·|f | that there exists i ∈ E such that for all j ≥ 0, (T n\j x − f)(i) < i . kfk For notational simplicity assume further that i = 0.

r·|f | kfk−r By the reverse triangle inequality, |f |− T[nj x(0) < 0 , and hence for α := |f |· > 0 kfk 0 kfk r·|f | kfk−r 0 we get that T[nj x(0) > |f | − 0 = |f | · = α > 0. That is, T[nj x(0) > α > 0 for 0 kfk 0 kfk all j ≥ 0. (A)

Next, we pick a subsequence {njk } of {nj} as follows:

1. Set j1 = 1. ∞ X 1 2. Inductively choose jk+1 so that jk+1 > jk and |xni | ≤ nj . Without kT k k · (jk + 1) i=jk+1 ∞ X 1 loss of generality, we assume that |x | ≤ . (B) ni kT knk · (k + 1) i=k+1 24 ∞ X α Let y = xnj enj . Clearly, by (A) we have that kyk ≤ kxk < ∞, and thus nj j=1 T[x(0) y ∈ `1.

|x | · w · w · ... · w · e = T[nm x(0) e . nm 1 2 nm 0 0 ∞ α nm X nm So kT y − αe0k ≤ 0 + xnj T enj . nj j=m+1 T[x(0) ∞ nm X nm Hence (A), (B) and continuity of T give that kT y − αe0k ≤ xnj kT k ≤ j=m+1 nm 1 1 kT k · kT knm ·(m+1) = m+1 for all m ≥ 1.

nm Thus kT y − αe0k → 0 as m → ∞, and hence Orb(T, y) has the non-zero limit point

αe0. So T is hypercyclic.

As an easy consequence of the equivalence of statements (i) and (iv) in the above theorem, we have the following result.

Corollary 26. A unilateral weighted backward shift T : `p → `p is not hypercyclic if and only if for every x in `p the set Orb(T, x)∪ {0} is closed.

As we have pointed out in Remark 2, an orbit Orb(T, x) may have a non-zero limit point without having the vector x that generates the orbit be hypercyclic for T . Nonetheless, we can demonstrate how to construct a hypercyclic vector for T using the non-zero limit point

of the orbit, which is assumed to be e0 in the following. Let 1 ≤ p < ∞. Let T : `p → `p be a unilateral weighted backward shift with weight

sequence {wj}j≥1 and let c = kT k < ∞. p Suppose that there exists a vector x in ` such that the orbit Orb(T, x) has e0 as a limit point. We want to construct a hypercyclic vector y for T . 25 p Let D = {(a0, a1, a2,...) ∈ ` : ai ∈ Q and ai = 0 for all but ﬁnitely many i ∈ Z+}.

Clearly D is a dense and countable set, so we can enumerate D = {d1, d2, d3,...}.

Since Orb(T ,x) has e0 as a limit point, there exists a sequence of positive integers nk % ∞

nk 1 1 such that kT x − e0k < 2k < 2 for all k ≥ 1. By our proof in Theorem 25 we get that 1 → 0 as k → ∞. w1·...·wnk

In the next steps we will construct a sequence of positive integers {kj}j≥1 whose terms we will then use to deﬁne the desired hypercyclic vector y for T .

For this purpose we will ﬁrst construct the sequence {kj}j≥1 subject to the restriction that ∞ p X p y ∈ ` , that is |yb(j)| < ∞. Furthermore, we require that there exists a sequence {ml}l≥1 j=0 ml such that for each l ≥ 1 and each > 0 there exists an l0 ≥ 1 having kT 0 y − dlk < . Thus each dl in D can be approximated arbitrarily close by an element in the Orb(T, y). We note that for the second condition it suﬃces to require that there exists a sequence

ml 1 {ml}l≥1 such that kT y − dlk < 2l for all l ≥ 1. For if l ≥ 1 and > 0, there exists l large enough such that 1 < and kd − d k < , by the density of the set D. Then 0 2l0 2 l l0 2 ml0 ml0 kT y − dlk ≤ kT y − dl0 k + kdl − dl0 k < 2 + 2 < .

Step 1. In this step we will choose k1 ∈ N and deﬁne the ﬁrst block of entries of y. For this we require the following two conditions: N X1 (a) Since d1 ∈ D we can write d1 = αj(1)ej for some N1 ≥ 0. j=0

We choose k1 by requiring ﬁrst that nk1 > N1, and thus having the entire ﬁrst block

p defined below fit inside y ∈ ` . The first block of y will have (N1 + 1) entries and will end at position nk1 ∈ {nk}k≥1. We further note that in between the blocks of y defined in each step, we set the vector y to have only zero entries.

Position nk1 − N1 nk1 − (N1 − 1) ... nk1 α (1) Entry α0(1) α1(1) ... N1 w1·w2·...·wn −N w2·w3·...·wn −(N −1) wN +1·wN +2·...wn k1 1 k1 1 1 1 k1

(b) Dealing with the requirement that y ∈ `p, we need 26 |α0(1)| < 1 (1) w1·w2·...·wn −N 2 k1 1 |α1(1)| 1 < 2 (2) w2·w3·...·wn −(N −1) 2 k1 1 . . . .

|αN1 (1)| 1 < N +1 (N1 + 1). wN +1·wN +2·...·wn 2 1 1 1 k1 p P∞ 1 p P∞ 1 Similar conditions for k2, k3,... will give us that kyk = j=1 2j ≤ j=1 2j = 1, so y ∈ `p. We achieve the above ﬁnite number of inequalities by using the fact that the product

1 1 → 0 as k → ∞. Namely if k1 is large enough, can be made as small as w1·...·wn w1·...·wn k k1 needed. Now we note that

|α0(1)|wn −(N −1)·...·wn N1 |α0(1)| = k1 1 k1 ≤ |α0(1)|·c . w1·w2·...·wn −N w1·w2·...·wn −N ·wn −(N −1)·...·wn w1·w2·...·wn k1 1 k1 1 k1 1 k1 k1

1 So condition (1) can be satisﬁed by choosing the term k1 large enough such that w1·w2·...·wn k1 1 < N . Similarly, all conditions (1) trough (N1 + 1) can be satisﬁed by choosing k1 2|α0(1)|·c 1 1 n 1 o large enough such that < P1 , where P1 = min j+1 N : j = 0, ..., N1 . w1·w2·...·wn 2 |αj (1)|·c 1 k1 1 We note that P1 = N +1 N , where M1 = max {|αj(1)| : j = 0, 1,...,N1}. 2 1 M1·c 1

We have now chosen k1 ∈ N.

Step 2. This step is for choosing k2 ∈ N and deﬁning the second block of entries of y. N X2 Write d2 = αj(2)ej for some N2 ≥ 0. We require the following of k2: j=0 (a) To avoid the overlapping of entries in the second block of y ∈ `p with entries in the

ﬁrst block, we need that nk2 − nk1 > N2. Also we would need that nk2 > N2, but this follows

from the previous condition. Clearly we require that k2 > k1. The second block will have

(N2 + 1) entries, ending at position nk2 ∈ {nk}k≥1.

Position nk2 − N2 nk2 − (N2 − 1) ... nk2 α (2) Entry α0(2) α1(2) ... N2 w1·w2·...·wn −N w2·w3·...·wn −(N −1) wN +1·wN +2·...wn k2 2 k2 2 2 2 k2 27 p (b) For the requirement that y ∈ ` we need to choose k2 large enough such that

1 n 1 o < P2 , where P2 = min j+1+(N +1) N : j = 0, 1,...,N2 . w1·w2·...·wn 2 1 |α (2)|·c 2 k2 j 1 We note that P2 = N +N +2 N , where M2 = max{|αj(2)| : j = 0,...N2}. 2 1 2 M2·c 2

m1 1 condition that kT y − d1k < 2 .

Let m1 = nk1 − N1. The next conditions will clearly give us this last requirement.

Shifting the vector y by m1 = nk1 − N1 will produce the following changes to the coeﬃ- cients of the second block:

Position nk2 − N2 − m1 ... nk2 − m1

α0(2)·wn −N −(m −1)·...·wn −N αN (2)·wn −(m −1)·...·wn Entry k2 2 1 k2 2 ... 2 k2 1 k2 w1·w2·...·wn −N wN +1·wN +2·...wn k2 2 2 2 k2

N m But each entry above is in absolute value bounded above by |αj (2)|·c 2 ·c 1 , which can be w1·...·wn k2 made as small as needed.

1 So we now choose k2 large enough such that < Q2 , where Q2 = min w1·w2·...·wn k2 n 1 o 1 2 N +m : j = 0, 1,...,N2 = 2 N +m . 2 (N2+1)|αj (2)|·c 2 1 2 (N2+1)·M2·c 2 1

We have now chosen k2 ∈ N.

Step 3. We will choose the general term kj ∈ N for j ≥ 3 satisfying:

(a) kj > kj−1 and nkj − nkj−1 > Nj;

1 1 (b) < Pj ; w1·w2·...·wn Nj j+ Nt kj Mj ·c ·2 t=1 1 1 (c) < N +m , for all l = 1, 2, . . . , j − 1, where ml = nk − Nl w1·w2·...·wn M ·c j l ·(N +1)·2j−1+l l kj j j

and m1 < m2 < m3 < . . .. Note that the entries of y in the j-th block are: α (j) α0(j) , α1(j) , ..., Nj . w1·w2·...·wn −N w2·w3·...·w wN +1·wN +2·...wn kj j n − N −1 j j kj kj ( j ) p ml Thus in steps 1−3 we have chosen the sequence {kl}l≥1 such that y ∈ ` and kT y − dlk < 1 2l for all l ≥ 1 where dl ∈ D. Thus for all dl in the dense set D, we have that dl ∈ Orb(T, y), so y is a hypercyclic vector for T . 28 2.2.3 The bilateral weighted backward shift

After examining how hypercyclicity relates to having an orbit with a non-zero limit point for the unilateral weighted backward shifts in Section 2.2.2, we turn to the study of bilateral weighted shifts.

p Let {en : n ∈ Z} be the canonical basis for ` (Z) for p ≥ 1. Then, a bounded and linear operator T : `p(Z) → `p(Z) is said to be a bilateral weighted backward shift if there is a sequence of bounded positive weights {wn : n ∈ Z} such that T en = wnen−1 for all n ∈ Z. Analogous to the unilateral weighted shift we have the following result.

Theorem 27. Suppose that T : `p(Z) → `p(Z) for 1 ≤ p < ∞ is a bilateral weighted backward shift. The following are equivalent:

(i) T is hypercyclic.

(ii) There exists an x in `p(Z) whose orbit Orb(T ,x) has a non-zero limit point.

(iii) There exists a vector x in `p(Z) whose orbit Orb(T ,x) has inﬁnitely many members in an open ball B(h, r), where 0 < r < khk.

Before we give a proof of the theorem, we remark that statement (ii) cannot be relaxed to having an orbit with a non-zero weak limit point in `p(Z), as is the case in Theorem 25. Chan and Sanders [11] showed the existence of a bilateral weighted backward shift that is weakly hypercyclic, and thus has a non-zero weak limit point, but the shift is not norm hypercyclic. Recall that for a bilateral weighted shift to be hypercyclic, Salas [33] provided a necessary and suﬃcient condition in terms of the weights: for every > 0 and every q ∈ N there exists n n−1 n Y 1 Y arbitrarily large such that for every j ∈ with |j| ≤ q we have w > and w < . Z s+j j−s s=0 s=1 However, as it turns out, this condition is not as helpful in proving the above theorem as its counterpart for the unilateral case that we have studied in Section 2.2.2. For that reason, we now oﬀer a constructive argument to prove Theorem 27. 29 Proof. It is clear that (i) implies (iii). To show that statement (ii) implies (i), we suppose without loss of generality that x =

p (..., xˆ(−1), xˆ(0), xˆ(1),...) is a vector in ` (Z) such that e0 is the non-zero limit point of the orbit Orb(T ,x). We set p = 2 for notational simplicity.

Hence there is an increasing sequence of positive integers {ni} such that

ni 1 kT x − e0k < 2i for all i ≥ 1.

∞ 2 X 1 < ∞, w · ... · w i=1 1 ni and thus 1 → 0 as i → ∞. (B) w1·...·wni By considering the terms of the vector T nj x with negative indices we have that

j−1 X 2 1 w−(n −n −1) · ... · w0 · ... · wni · xˆ(ni) < . j i 22j i=1 Now, focusing on the i-th term of the above summation and using (A), we see that if

1 1 ≤ i ≤ j − 1, then w−(nj −ni−1) · ... · w0 < 2j−1 . (C)

Claim: Let y = (..., 0, yˆ(0),..., yˆ(k), 0,...) and a = (..., 0, aˆ(−l),..., aˆ(0),...

..., aˆ(l), 0,...) in `2(Z) with k and l ≥ 1. For all positive , there exist an integer m > k + l and a vector z of the form z = (..., 0, zˆ(m − l),..., zˆ(m),..., zˆ(m + l), 0,...) such that:

(1) T mz = a,

(2) kzk < ,

(3) kT szk < , if 1 ≤ s ≤ k,

(4) kT myk < . 30

Proof of Claim. Note that kT k = sup{|wi| : i ∈ Z} > 1.

Choose an integer i such that ni > k + l, where k and l are positive integers given in the

vectors y and a in the statement of the Claim. For this ﬁxed ni and for any j > i we denote

m = nj − ni + k. We ﬁrst observe by(C) that

k m 2 X 2 kT yk = w−(m−r−1) · ... · w−(nj −ni−1) · ... · w0 · ... · wr · yˆ(r) r=0 k 2k 2 X 2 ≤ kT k · kyk · w−(m−r−1) · ... · w−(nj −ni−1) · ... · w0 r=0 k 2k 2 2k X 2 ≤ kT k · kyk · kT k w−(nj −ni−1) · ... · w0 r=0 1 ≤ kT k4k · kyk2 · (k + 1) · . 22(j−1)

m Next we need to ﬁnd z such that T z = a, which is equivalent to having wr+1 ·...·wm+r · zˆ(m + r) =a ˆ(r) whenever −l ≤ r ≤ l. Then, l l 2 l 2 2 X 2 X aˆ(r) 2 X 1 kzk = |zˆ(m + r)| = ≤ kak . wr+1 · ... · wm+r wr+1 · ... · wm+r r=−l r=−l r=−l

Note that if −l ≤ r ≤ l, then m + r ≤ m + l = nj − ni + k + l = nj − (ni − k − l) < nj

since ni > k + l. l 2 2 2 X wm+r+1 · ... · wnj Thus, kzk ≤ kak . wr+1 · ... · wn r=−l j n 1 1 1 o We now let c` =max 1, , ,..., and observe that the numerator w0 w−1w0 w−l+1···...···w0

wm+r+1 · ... · wnj is a product of nj − (m + r) = nj − [nj − ni + k + r] = ni − k − r factors. Hence, kzk2 is bounded above by

0 2 l 2 X 1 X 1 kak2 · kT k2(ni−k+l) · + kak2 · kT k2(ni−k−1) · wr+1 · ... · wn wr+1 · ... · wn r=−l j r=1 j 0 2 l 2 2 2(ni−k+l) 2 X 1 2 2(ni−k−1) X w1 · ... · wr ≤ kak · kT k · c` + kak · kT k · w1 · ... · wn w1 · ... · wn r=−l j r=1 j h i2 n o ≤ 1 kak2 · kT k2(ni−k+l) · c2 · (l + 1) + kak2 · kT k2(ni−k−1) · kT k2l · l . w1·...·wnj ` 31 Note that if 1 ≤ s ≤ k, then kT szk ≤ kT kk kzk. It is now evident that for any given > 0, we can use (B) to choose an integer j > i so that (2), (3) and (4) in the Claim are satisﬁed.

To ﬁnish the proof, it remains to show that our Claim implies that T is hypercyclic.

2 For this, let D = {d1, d2,...} be a dense subset of ` (Z), where each di is of the form ˆ ˆ ˆ di = ..., 0, d(−li),..., d(0),..., d(li), 0,... .

1 First, take y1 = 0 with k1 = 1, a1 = d1 and = 41 as in the statement of the Claim. Then

our Claim gives us that there exists an m1 ∈ N and a vector z1 withz ˆ1(i) = 0 whenever

i ≥ m1 + l1 such that

m1 (1) T z1 = d1,

1 (2) kz1k < 41 ,

1 (3) kT z1k < 41 ,

m1 (4) kT y1k = 0.

Inductively, we take kj = mj−1 + lj−1 and yj = z1 + z2 + ... + zj−1. Clearly, {kj}j≥1 is an increasing sequence, as kj+1 = mj + lj > kj + 2lj for all j ≥ 1.

Thus we note thaty ˆj(i) = 0 whenever i ≥ kj.

1 Let = 4j in the statement of the Claim. Then our Claim implies that there exist mj ∈ N

and a vector zj withz ˆj(i) = 0 whenever i ≥ mj + lj so that

mj (1) T zj = dj,

1 (2) kzjk < 4j ,

s 1 (3) kT zjk < 4j , if 1 ≤ s ≤ kj,

mj 1 (4) kT yjk < 4j . 32 ∞ X Deﬁne the vector b by setting b = zi. Since the sum is absolutely convergent, we have i=1 ∞ ∞ ∞ X X X 1 that kbk = z ≤ kz k < < ∞, so b ∈ `2( ). i i 4i Z i=1 i=1 i=1 ∞ X Also note that b = yj + zj + zi, so i=j+1

∞ ! X mj mj mj mj kT b − djk ≤ kT yjk + kT zj − djk + T zi < i=j+1 ∞ ∞ ∞ X 1 X 1 X 1 < 1 + 0 + kT mj z k < + = −→ 0 as j → ∞. 4j i 4j 4i 4i i=j+1 i=j+1 i=j

mj Thus kT b − djk → 0 as j → ∞, so by density of the set D = {d1, d2,...} we have that b is a hypercyclic vector for T .

To show that (iii) implies (ii), we suppose that there exists a vector x in `p(Z) whose orbit Orb(T ,x) has inﬁnitely many members in an open ball B(h, r), where 0 < r < khk. For notational simplicity we set p = 2.

2 1−s r 2 khk Let s > 0 such that 1+s > khk . Since h ∈ ` (Z), there exists N ≥ 1 such that (1+s)2 < N X 2 hˆ(i) . Let {n } be an increasing sequence of integers such that T nj x ∈ B(h, r) and j j≥1 i=−N nj set fj = T x.

ˆ We now show that for each fj there exists an integer i with i ≤ N such that 0 < s· h(i) <

fbj(i) whenever |i| ≤ N.

ˆ To do that, we suppose on the contrary that there exists an fj so that fbj(i) ≤ s · h(i) whenever |i| ≤ N. This would imply that, by our choice of N,

2 N khk X 2 (1 − s)2 < (1 − s)2 hˆ(i) (1 + s)2 i=−N N 2 X ˆ ≤ h(i) − fbj(i) i=−N hˆ(i)6=0 2 2 ≤ kh − fjk < r ,

< r2, 33 which would contradict our choice of s. Since there are ﬁnitely many integers i with |i| ≤ N, but inﬁnitely many j ≥ 1, there

ˆ exists an integer i0 with |i0| ≤ N such that 0 < s · h(i0) < fbj(i0) , for inﬁnitely many j.

We now assume i0 = 0 for notational simplicity. In addition, by taking a subsequence of

{nj} if necessary, we further assume that

ˆ 0 < s · h(0) < fbj(0) , whenever j ≥ 1.

nj Recall that fj = T x, and thus

fbj(0) = w1 · ... · wnj xˆ(nj), from which it follows that

∞ 2 X 2 ∞ > kxk ≥ |xˆ(nj)| j=−∞ 2 ∞ X fbj(0) = (w · ... · w )2 j=−∞ 1 nj ∞ 2 2 X 1 ≥ s2 hˆ(0) . w · ... · w j=−∞ 1 nj

Hence 1 → 0 as j → ∞. (D) w1·...·wnj

This means that kT k = sup {wj : j ∈ Z} > 1. We now switch our attention to the weights with negative indices. First, we note that

T nj x ∈ B(h, r) for all j, and hence, there exists M > 0 so that kT nj xk ≤ M for all j. Second we note that if 1 ≤ i < j, then

nj T (ˆx(ni)eni ) = w−(nj −ni−1) · ... · w0 · ... · wni xˆ(ni)eni−nj .

Thus, 34

M 2 ≥ kT nj xk2 j−1 2 X ≥ T nj (ˆx(n )e ) i ni i=1 j−1 X 2 2 = w−(nj −ni−1) · ... · w0 · ... · wni |xˆ(ni)| i=1 j−1 X 2 2 = w · ... · w f (0) , by (B) −(nj −ni−1) 0 bi i=1 j−1 2 X 2 > s2 hˆ(0) w · ... ··· w , by (A). −(nj −ni−1) 0 i=1

In other words, if 1 ≤ i < j, then j−1 2 X 2 M w · ... · w < . (E) −(nj −ni−1) 0 2 2 ˆ i=1 s h(0) Claim: For every > 0 and every integer k ≥ 1, there exist positive integers j and m with m < j so that:

(1) k < m,

(2) 2nm < nj,

(3) w−(n −n −1) · ... · w0 < √ n , j m nkkT k k

1 (4) < 2nm . w1·...·wnj kT k

Proof of Claim. Let > 0 and k ≥ 1 be given. Determine a positive integer t > k + 1 so 2 M 2 h i that 2 < √ n . s2|hˆ(0)| (t−k) nkkT k k 1 For that choice of t, we can determine j > t such that nj > 2nt and < 2nt , by w1·...·wnj kT k (D). Since k + 1 < t < j, we have, by (E),

t 2 X 2 M w · ... · w < . −(nj −ni−1) 0 2 2 ˆ i=k+1 s h(0) 35 The summation on the left-hand side has t − k positive terms, so there must exist an integer m with k + 1 ≤ m ≤ t such that

2 2 M 2 h i w−(n −n −1) · ... · w0 < 2 < √ n , j m s2|hˆ(0)| (t−k) nkkT k k

and hence (3) is satisﬁed. Since k + 1 ≤ m ≤ t < j, we see that (1), (2) and (4) are satisﬁed too.

To ﬁnish the proof, we use our Claim to construct a vector z whose orbit Orb(T, z) has e0 as a limit point.

1 Take k = 1 and = 2 in our Claim. We then have positive integers j and m with j > m so that (1) through (4) are satisﬁed. For notational simplicity, we can assume that m = 2

and j = 3, because we can certainly achieve that by taking a subsequence of {ni} having

1 the ﬁrst three terms nk, nm and nj. Hence, 2n2 < n3, w · ... · w0 < √ n , and −(n3−n2−1) 2 n1kT k 1 1 1 < 2n2 . w1·...·wn3 2kT k − k+1 Inductively, for every odd integer k ≥ 1, we take = 2 2 in the Claim, and by choosing

a subsequence of {ni}, we get three consecutive integers k < m < j such that (1) through

(4) are satisﬁed. In this way we can assume that the original sequence {ni} satisﬁes that for any q ≥ 1:

(1) n2q−1 < n2q,

(2) 2n2q < n2q+1,

1 (3) w · ... · w0 < q√ n2q−1 , −(n2q+1−n2q−1) 2 n2q−1kT k

1 1 (4) < q 2n2q . w1·...·wn2q+1 2 kT k

∞ X 1 Let m = n − n + n for i ≥ 1 and set z = e . Then, i 2i+1 2i 2i−1 w · ... · w mi i=1 1 mi 36

∞ 2 ∞ 2 2 X 1 X wmi+1 · ... · wn2i+1 kzk = = w · ... · w w · ... · w i=1 1 mi i=1 1 n2i+1 ∞ n −n 2 ∞ n 2 X kT k 2i 2i−1 X kT k 2i ≤ ≤ (by (4)) w · ... · w i 2n2i i=1 1 n2i+1 i=1 2 kT k ∞ ∞ X 1 X 1 < < < ∞, i 2n2i 4i i=1 4 kT k i=1

so clearly z is in `2(Z). Now, k−1 ∞ 1 mk X X T z = w−(mk−mi−1) · ... · w0emi−mk + e0 + emi−mk , w1 · ... · wm −m i=1 i=k+1 i k and so k−1 ∞ 2 2 1 mk 2 X X kT z − e0k = w−(mk−mi−1) · ... · w0 + . w1 · ... · wm −m i=1 i=k+1 i k

To estimate the ﬁrst summation, we ﬁrst note that its subindex mk − mi − 1 = n2k+1 − n2k + n2k−1 − mi − 1 = (n2k+1 − n2k − 1) + (n2k−1 − mi).

Since the summation index runs between 1 and k − 1, we have mi ≤ mk−1 = n2k−1 − n2k−2 + n2k−3, and so n2k−1 − mi ≥ 0. Hence, using (3) and the fact that {ni} is increasing and so n2k−1 ≥ 2k − 1 > k − 1, we have that

k−1 X 2 w−(mk−mi−1) · ... · w0 i=1 k−1 X 2(n2k−1−mi) 2 ≤ kT k w−(n2k+1−n2k−1) · ... · w0 i=1 k−1 X 1 1 < < . 4kn 4i i=1 2k−1

To estimate the second summand, we proceed as follows, ∞ 2 ∞ 2 X 1 X wmi−m +1 · ... · wn2i+1 = k . w1 · ... · wm −m w1 · ... · wn i=k+1 i k i=k+1 2i+1 37 Note that the number of weights in the numerator of the previous expression is given by n2i+1 − mi + mk = n2i+1 − (n2i+1 − n2i + n2i−1) + mk < n2i + mk < 2n2i, because i ≥ k + 1 and mk < n2k+1 by its deﬁnition. Hence,

∞ 2 ∞ 2n !2 X 1 X kT k 2i < w1 · ... · wm −m w1 · ... · wn i=k+1 i k i=k+1 2i+1 ∞ 2n !2 X kT k 2i < by (4) i 2n2i i=k+1 2 kT k ∞ X 1 1 = < . 4i 3 · 4k i=k+1

mk 2 1 1 By combining both estimates, we see that kT z − e0k < 4k + 3·4k , which goes to 0 as k → ∞. Hence, e0 is a limit point of the orbit Orb(T, z).

A much simpler argument for showing (ii) implies (i) can be made if we require the

bilateral weighted shift having e0 as a limit point of Orb(T ,x) to be invertible.

nk Since e0 is a limit point, there exists a sequence of integers nk % ∞ such that kT x − e0k <

1 1 1 2k < 2 for all k ≥ 1. Thus we have shown previously that |w1 · w2 · ... · wnk xˆ(nk) − 1| < 2 n Yk for all k ≥ 1, and consequently that wj → ∞. j=1 1 It also follows that w−(nk−n1−1) · ... · w−1 · w0 · w1 · ... · wn1 xˆ(n1) < 2k for all k ≥ 2, and 1 1 1 thus since < 2 we observe that w−(nk−n1−1) · ... · w−1 · w0 < k · 2 = k−1 for all w1·...·wn1 |xˆ(n1)| 2 2 k ≥ 2.

w−nk ·...·w−nk+n1 Furthermore, w−nk · ... · w−1 · w0 < 2k−1 for all k ≥ 2. kT n1+1k Hence 0 < w−nk · ... · w−1 · w0 < 2k−1 for all k ≥ 2. Letting k → ∞ we get that n Yk w−nk · ... · w−1 · w0 → 0, and therefore w−j → 0. j=1 n n Yk Yk Thus there is a sequence nk % ∞ so that wj → ∞ and w−j → 0, so by Feldman’s j=1 j=1 38 criterion for invertible bilateral shifts [17] we have that T is hypercyclic. From this we can now deduce the following statement, using the fact that if T is hypercyclic then so is T −1 (see Kitai [24]).

Corollary 28. If T : `p(Z) → `p(Z) is an invertible bilateral weighted shift having an orbit with a non-zero limit point, then T −1 also has an orbit with a non-zero limit point.

As we have pointed out before, Salas [33] proved that a bilateral weighted shift T is

hypercyclic if and only if for every > 0 and every q ∈ N there exists n arbitrarily large n−1 n Y 1 Y such that for every j ∈ with |j| ≤ q we have w > and w < . We would like Z s+j j−s s=0 s=1 to point out that Theorem 27 oﬀers a new equivalent condition to check whether or not an operator T is hypercyclic. To illustrate the applicability of our condition we oﬀer an example of an operator T for which it is not easy to check hypercyclicity using Salas’ condition, but quite easy using ours.

2 ∞ Let T be a bilateral weighted backward shift on ` (Z) with weight sequence {wj}−∞ deﬁned as follows. Let k1 = 1 and n1 = 1. Recursively we take kj = nj−1 + j and nj =

nj−1 + kj + 1 for all j ≥ 2. Let w0 = 1. We deﬁne the weights with positive indices in

1 consecutive blocks, setting the j-th block to have kj entries of 2’s followed by the entry k . 2 j Similarly, we deﬁne the weights with negative indices in consecutive blocks having the following entries: block 1: one weight with value 1 , followed by l entries of 2’s; 21+l1 1 block 2: 1 , then l entries of 2’s; 1 , then l entries of 2’s; 23+l3 3 22+l2 2 block 3: 1 , then l entries of 2’s; 1 , then l entries of 2’s; 1 , then l entries of 26+l6 6 25+l5 5 24+l4 4 2’s, where

l1 = n2 − n1 − 2,

l2 = n3 − n2 − l1 − 3, l3 = n3 − n1 − (l1 + l2) − 4,

l4 = n4 − n3 − (l1 + l2 + l3) − 5, l5 = n4 − n2 − (l1 + l2 + l3 + l4) − 6, l6 = n4 − n1 − (l1 +

l2 + l3 + l4 + l5) − 7, etc. 39

Finally for all j ≥ 1 we let pj = −(l1 + ... + lj + 1 + ... + j).

We list below the values of the terms of the sequences nj, kj and lj, that we have used

∗ to construct the next two tables, where the nj positions have been marked by :

k1 = 1, n1 = 1,

k2 = 3, n2 = 5, l1 = 2,

k3 = 8, n3 = 14, l2 = 4, l3 = 3,

k4 = 18, n4 = 33, l4 = 5, l5 = 8, l6 = 3. We define the vector x ∈ `2( ) by settingx ˆ(n ) = 1 for i ≥ 1 and 0 otherwise. Z i 2ki Looking at the effect of applying T nj to the coordinates of the vector x, we first note

1 that the product of the positive weights in each block is 1. Thus, since x(nj) = k we have b 2 j that T[nj x(0) = 1, for all j ≥ 1. Furthermore the non-zero coordinate of the vector x in

the (j + 1)-block is shifted by T nj to the value 1 · 2nj = 1 · 2nj = 1 . The other 2kj+1 2nj +j+1 2j+1 non-zero entries with positive indices of T nj x are given by the value 1 , where each kj+s−nj j+1 kj+s − nj > 2 and the sequence {kj+s − nj}s>1 is strictly increasing to inﬁnity as s → ∞.

Considering the entries with negative indices of T nj x, we ﬁnd that shifting by T nj moves

the coordinate xb(nj) in the 0-th position, while xb(ns) for 1 ≤ s < j is moved in the p − 1 position for some p.

Furthermore the non-zero terms with negative indices of the vector T nj x are:

2−[1+2+...+(j+Sj )], 2−[1+2+...+((j−1)+Sj )],..., 2−[1+2+...+(1+Sj )],

j−1 X where Sj = i. Thus as j → ∞ we have that the entries above go to zero. i=1 Finally to avoid any overlapping of the entries with negative indices while shifting by

T nj (that is we want the non-zero entries of T nj to be to the left of the non-zero entries of

nj−1 T ), we verify that kj > nj−1 + 2.

Therefore by the observations listed above, we have that for the sequence {nj}j≥1, the vector x in `2( ) and the bilateral weighted shift T with weights {w } , T nj x → e as Z j j∈Z 0 j → ∞. Hence by our theorem, T is hypercyclic. 40 0 1 0 1 1 1 1 → 2 0 0 0 1 l ← 2 0 0 0 1 3 1 0 0 0 2 p ...... 1 2 2 0 0 → 2 18 1 0 0 0 0 0 l 34 ...... 2 as a limit point ← 0 2 0 0 e 18 1 0 0 0 0 2* 2 2 6 1 0 0 2 p 17 1 2 0 0 0 0 2 → 3 ) with 1 2 0 2 → 13 1 3 0 0 ...... T, x l 2 2 0 0 = 18 4 ← k 2 0 0 2 0 0 0 0 ... ← 3 6 1 0 0 4 2 p 1 2 0 0 0 2 ... 6 1 2 0 2 0 0 0 0 0 2 → 4 8 l 1 0 0 0 0 0 2 ...... 15 ← 2 0 8 1 0 0 0 0 2 2* 4 9 1 0 2 p 7 1 2 0 0 0 0 → 2 10 1 2 3 = 8 2 1 0 0 2 ...... 3 → k 5 l ...... 2 0 0 0 0 ... ← ← 2 0 2 0 0 0 0 ... 5 13 1 0 3 p 2 1 6 0 0 0 0 0 2 15 1 2 3 → 2 1 0 0 0 0 2 2* → 6 l 2 0 = 3 2 1 2 0 0 0 0 2 2 ← k 2 0 ← 2 0 0 0 0 0 6 9 1 0 2 p 1 2 2 0 0 0 0 0 21 1 2 1 2 1 2 0 0 0 0 k 2* 0 1 0 1 1 1 1 ...... ) ) ) ) ) ) ) ) j j j j j j j j ( ( ( ( ( ( ( ( x x x x x x x x ) ) 1 2 3 4 1 2 3 4 j j n n n n n n n n ( ( [ [ [ [ [ [ [ [ ˆ b T T T T T T T T Position Weights Position Weights x x (i) The positive indices Example of a bilateral weighted backward shift having an orbit Orb( (ii) The negative indices 41

CHAPTER 3

ORBITAL LIMIT POINTS AND HYPERCYCLICITY OF OPERATORS ON ANALYTIC FUNCTION SPACES

3.1 Introductory remarks

As we saw in Chapter 2, the result of Bourdon and Feldman [8] that an operator T with a somewhere dense orbit is everywhere dense, can be relaxed to a great extent in the case of weighted backward shifts on the sequence spaces `p, for 1 ≤ p < ∞. Indeed we showed if a unilateral or bilateral weighted backward shift T has an orbit Orb(T, x) with a single non-zero limit point, then T is hypercyclic. In fact, we were able to relax the above condition for the hypercyclicity of shifts even further to simply requiring that the orbit has infinitely many members contained in a ball whose closure avoids the zero vector. A natural question that arises is whether the above results can be generalized to other classes of operators. In the field of hypercyclicity, the adjoints of multiplication operators and the composition operators are two well-studied classes of operators on analytic function spaces. In this chapter we show the above results for the shifts hold true also for the adjoints of multiplication 42 operators on the Bergman spaces. To achieve this we cannot borrow techniques used for the shift operators in Chapter 2, but instead we have to take a function theoretical approach. On the other hand, the situation for the composition operators on the Hardy space is totally different, as we show that a composition operator may have an orbit with a non-zero limit point and yet not be hypercyclic, and in fact not even cyclic. Recall that, an operator T : X → X is said to be cyclic if there is a vector x ∈ X such that the linear span of Orb(T, x) is dense in X. Thus in general, the existence of an orbit with a non-zero limit point does not determine the cyclic behavior of the operator. In Section 3.2.2, we prove that hypercyclicity for the adjoint of a multiplication operator on the Bergman spaces is equivalent to having an orbit with infinitely many members contained in a ball whose closure avoids the zero vector. We then offer some considerations regarding the hypercyclicity of this class of operators on the Dirichlet space. In Section 3.3.2 we construct an example of a linear fractional composition operator on the Hardy space that has an orbit with a non-zero limit point, but the operator is not hypercyclic. Finally, motivated by the work in the previous sections we study in Section 3.4 the set of all vectors x in X whose orbit has a non-zero limit point.

3.2 The adjoints of multiplication operators

3.2.1 Elementary properties and hypercyclicity of the adjoints of

multiplication operators

In this section we deﬁne multiplication operators on analytic function spaces, examine some of their basic properties, and brieﬂy characterize the hypercyclic behavior of the adjoints of multiplication operators.

Definition 29. Let Ω be a region in the complex plane C. We define the Bergman space A2(Ω) of Ω to be the space of all analytic functions f :Ω → C that are square integrable with Z respect to the area measure, i.e. kfk2 = |f|2dA < ∞. 43 We note that A2(Ω) is a separable, infinite dimensional Hilbert space with the inner Z product hf, gi = fgdA for all f, g ∈ A2(Ω). Furthermore, it is sometimes useful to express the norm of the Bergman spaces in terms of power series coefficients. The well known ∞ X n proposition below establishes that, given an analytic function f(z) = anz of the unit n=0 ∞ 2 2 X |an| disk , the norm of the Bergman space A2( ) can also be expressed as kfk = π . D D n + 1 n=0 ∞ X n Proposition 30. Let f(z) = anz be an analytic function on the unit disk D. Then n=0 ∞ 2 X |an| f ∈ A2( ) if and only if < ∞. D n + 1 n=0

∞ X n Proof. For f(z) = anz analytic on D we have n=0 Z Z |f(z)|2dA(z) = f(z)f(z)dA(z) D D Z ∞ ! ∞ ! X n X n = anz anz dA(z) D n=0 n=0 Z 2π Z 1 ∞ ! ∞ ! X n inθ X m −imθ = anr e amr e rdrdθ 0 0 n=0 m=0 Z 2π Z 1 ∞ X n+m+1 i(n−m)θ = anamr e drdθ 0 0 n,m=0 Z 1 Z 2π ∞ X n+m+1 i(n−m)θ = anamr e dθdr 0 0 n,m=0 Z 1 ∞ Z 2π X n+m+1 i(n−m)θ = anamr e dθ dr 0 n,m=0 0 Z 1 ∞ X 2n+1 = ananr 2πdr 0 n=0 ∞ Z 1 X 2 2n+1 = 2π |an| r dr n=0 0 ∞ 2 X |an| = π . n + 1 n=0 44 We now proceed in deﬁning the multiplication operators on A2(Ω). In the following we let H denote a Hilbert space of analytic functions on Ω with bounded point evaluations.

Deﬁnition 31. For the Hilbert space H, we say an analytic function ϕ :Ω → C is a multiplier of H, if ϕf ∈ H whenever f ∈ H.

Each multiplier ϕ of the Hilbert space H induces a linear multiplication operator Mϕ on

H by the formula Mϕ(f) = ϕ · f for all f ∈ H. As it turns out, in the case of the Bergman spaces A2(Ω) , if ϕ is a function in H∞(Ω)=

2 {h :Ω → C analytic , khk∞ < ∞}, then for every f ∈ A (Ω) Z Z 2 2 2 |ϕf| dA ≤ kϕk∞ |f| dA < ∞,

and thus ϕf ∈ A2(Ω). Thus every bounded analytic function on Ω is a multiplier of A2(Ω) and it induces a linear multiplication operator Mϕ. Moreover, the reverse inclusion also holds, which was established by Godefroy and Shapiro [20] for the more general setting of the Hilbert space H.

Theorem 32. Every multiplier of the Hilbert space H is a bounded analytic function on Ω.

Proof. Suppose ϕ is a multiplier of H, and let f be an element of H that does not vanish identically on Ω. If z is not a zero of f, then

n n n n |ϕ(z)| · |f(z)| = |Mϕ f(z)| = |hMϕ f, kzi| ≤ kMϕk · kfk · kkzk, where kz is the unique reproducing kernel for z given by the Riesz Representation Theorem

from the boundedness of evaluation maps, i.e. f(z) = hf, kzi for all f ∈ H.

0 Taking n-th roots and letting n → ∞, we obtain |ϕ(z)| ≤ kMϕk for all z in Ω = Ω\{zeros of f}. g Now, since ϕ is a multiplier of H, ϕ · f = g for some g ∈ H, and so ϕ = is analytic on f 0 Ω and bounded by kMϕk.

Consequently, ϕ has an analytic extension on Ω, which is also bounded by kMϕk. 45 The above theorem together with our previous considerations allow us to conclude that in the case of the Bergman spaces A2(Ω), the set of multipliers is in fact equal to H∞(Ω). Furthermore, if ϕ is a multiplier of A2(Ω) then the induced multiplication operator has norm Z Z 2 2 2 2 2 kMϕk = kϕk∞. Clearly, since for every f ∈ A (Ω), kMϕfk = |ϕf| dA ≤ kϕk∞ |f| dA, we have that kMϕk ≤ kϕk∞.

To show the reverse inequality, let kz be the reproducing kernel of z. Then for each f ∈ A2(Ω) and z ∈ Ω we have

|ϕ(z)| · |hf, kzi| = |ϕ(z)| · |f(z)| = |hϕf, kzi| = |hMϕf, kzi| ≤ kMϕk · kfk · kkzk.

Recall that for any g in a Hilbert space, kgk = sup |hg, fi|. Then, by taking supremum kfk=1 2 in the above inequality over all f ∈ A (Ω) with kfk = 1, we obtain |ϕ(z)| ≤ kMϕk for all z ∈ Ω. Thus kϕk∞ ≤ kMϕk, which completes the proof.

While the statement in Theorem 32, as well as the inequality kMϕk ≥ kϕk∞, hold in the general case of the Hilbert space H, we observe that the proof of the converse and the reverse inequality were intrinsically linked to the norm of A2(Ω). Thus the question arises as to whether the above results can be generalized to other analytic function spaces.

We remark that the same conclusion about the norm of Mϕ, that kMϕk = kϕk∞, holds also for the Hardy space H2(D) of the open unit disk D, which consists of all analytic functions Z 1/2 1 iθ 2 f on D with lim f(re ) dθ < ∞. Furthermore, the set of multipliers for the r%1 2π Hardy space H2(D) consists, analogous to the Bergman spaces, of all bounded analytic functions on D. However, this is not the case for all analytic function spaces. We note that in the case

of the Dirichlet space Dir(Ω), consisting of all analytic functions f :Ω → C on the region Z 2 2 1 0 2 ∞ Ω with kfk = |f(0)| + π |f | dA < ∞, not every function in H (Ω) is a multiplier of Dir(Ω). As Theorem 32 clearly states, we certainly have that the set of multipliers is contained in H∞(Ω), but the reverse inclusion does not hold. Furthermore, in section

3.2.2 we show that the multiplication operator induced by ϕ(z) = z on Dir(D) has norm 46 r3 kM k = > 1 = kzk , and so the Dirichlet space fails to share another property with z 2 ∞ the Bergman and Hardy spaces. We now move on to summarizing the hypercyclicity results for the adjoints of the multiplication operators on the Bergman, Hardy and Dirichlet spaces. As is made evident by the next theorem, the relation of the set of multipliers to the space of bounded analytic functions plays a crucial role in the hypercyclicity phenomena on the aforementioned analytic function spaces. However, it is not the multiplication operator

∗ Mϕ that exhibits hypercyclic behavior on these three spaces, but rather its adjoint Mϕ. In fact, no multiplication operator Mϕ can be hypercyclic since with a little work one can show that for any multiplier ϕ having |ϕ(α)|= 6 1 for some α, a function g in the closure of an orbit Orb(Mϕ, f) must vanish at α. We note that convergence in norm implies pointwise convergence for all of the above three spaces of analytic functions. In [20], Godefroy and Shapiro oﬀer the following characterization for the hypercyclicity of the adjoints of multiplication operators.

Theorem 33. Suppose every bounded analytic function ϕ on Ω is a multiplier of the Hilbert

∗ space H and kMϕk = kϕk∞. Then for each such non-constant ϕ, the operator Mϕ is hypercyclic if and only if ϕ(Ω) intersects the unit circle.

We note that the above theorem asserts that in the case of the Bergman and Hardy spaces,

∗ the adjoint of the multiplication operator Mϕ is hypercyclic if and only if ϕ(Ω) ∩ ∂D 6= ∅. On the other hand, in the case of the Dirichlet spaces we cannot once again draw the same conclusion. Whilst [20, Theorem 4.5] states that for a non-constant multiplier ϕ, the adjoint

∗ operator Mϕ is hypercyclic whenever ϕ(Ω) intersects the unit circle, the converse does not

∗ hold. We demonstrate this in the next section by providing an example of an operator Mϕ on Dir(D) that is hypercyclic, but nevertheless has ϕ(D) ∩ ∂D = ∅. Finally, we would like to remark that even though in the case of the Dirichlet spaces,

∗ ϕ(Ω) need not intersect the unit disk if Mϕ is hypercyclic, we can however infer that the 47 closure of ϕ(Ω) must intersect the unit circle. Kitai [24] showed that if T is a hypercyclic operator on a Hilbert space, then the spectrum of T intersects the unit circle. Thus, since

one can show that in the case of the multiplication operator Mϕ on the Dirichlet space

∗ ∗ Dir(Ω) the spectrum σ(Mϕ) = σ(Mϕ) = ϕ(Ω), we can deduce that if Mϕ is hypercyclic then

ϕ(Ω) ∩ ∂D 6= ∅.

2 2 Proposition 34. The spectrum of the multiplication operator Mϕ on A (Ω), H (D) or

Dir(Ω) is σ(Mϕ) = ϕ(Ω).

Proof. In the following we present the proof for the Dirichlet space Dir(Ω), and note that the proof for the other two spaces can be derived in a similar fashion.

Suppose ϕ :Ω → C is a multiplier on Dir(Ω), and consider the induced multiplication operator Mϕ.

Let λ ∈ ϕ(Ω), i.e. there exists z0 ∈ Ω so that ϕ(z0) = λ. Since (Mϕ − λI)(f) =

Mϕf − λf = ϕf − λf = (ϕ − λ)f, we have that (Mϕ − λI)(f)(z0) = (ϕ(z0) − λ) · f(z0) = 0 for all f ∈ Dir(Ω). Since the constant function f = 1 in Dir(Ω) does not vanish at z0, we get that Mϕ − λI is not surjective on Dir(Ω), i.e. λ ∈ σ(Mϕ). Since the spectrum σ(Mϕ) is compact, we can further conclude that ϕ(Ω) ⊂ σ(Mϕ).

1 To show the reverse inclusion, suppose λ∈ / ϕ(Ω). Then < ∞, as otherwise ϕ − λ ∞ there exists a sequence {zn} in Ω so that |ϕ(zn)−λ| → 0, which contradicts λ ∈ ϕ(Ω). Thus, 1 ∈ H∞(Ω). We will now show that M − λI is invertible and hence λ∈ / σ(M ). ϕ − λ ϕ ϕ

We ﬁrst note that if f is in the kernel of the operator Mϕ − λI, then (ϕ − λ) f = 0. But

ϕ(z) − λ 6= 0 for every z ∈ Ω, and thus f = 0 on Ω. So Mϕ − λI is injective. 1 On the other hand, if we are able to show that is a multiplier of Dir(Ω), then ϕ − λ 1 1 for every g ∈ Dir(Ω) the function g ∈ Dir(Ω), and hence (M − λI) g = ϕ − λ ϕ ϕ − λ 1 (ϕ − λ) g = g. Thus M − λI is surjective, and so λ∈ / σ(M ), which gives the reverse ϕ − λ ϕ ϕ inclusion. To complete the proof, we note it suﬃces to show that for g ∈ Dir(Ω), the func- 48 1 0 1 2 1 tion g = − ϕ0g + g0 is square integrable with respect to the ϕ − λ ϕ − λ ϕ − λ area measure. Since ϕ is a multiplier of Dir(Ω) and g ∈ Dir(Ω), we can deduce that (ϕg)0 ∈ L2(dA). Furthermore, Theorem 32 also gives that ϕ ∈ H∞(Ω), and thus we have that ϕg0 ∈ L2(dA). Therefore, ϕ0g = (ϕg)0 − ϕg0 ∈ L2(dA). Consequently, since we have 1 1 shown that ∈ H∞(Ω), it follows that the above derivative of g is also square ϕ − λ ϕ − λ 1 integrable. Thus g ∈ Dir(Ω) for all g ∈ Dir(Ω), which completes the proof. ϕ − λ

3.2.2 A Zero-One Law for the hypercyclicity of the adjoints of

multiplication operators

Let Ω be a region in the complex plane C. Recall that the Bergman space A2(Ω) of Ω is the Hilbert space of all analytic functions f on Ω such that |f|2 is integrable with respect Z to the area measure dA on Ω. The norm of this space is given by kfk2 = |f|2dA. We say an analytic function ϕ on Ω is a multiplier, if ϕ · f is in A2(Ω) whenever f is in A2(Ω). It is

∞ easy to verify that if ϕ is a function in H (Ω)={h :Ω → C analytic | khk∞ < ∞}, then ϕ

2 2 is a multiplier and it induces a bounded linear multiplication operator Mϕ : A (Ω) → A (Ω)

given by Mϕ(f) = ϕ · f. Indeed, one can show that the set of all multipliers for the Bergman space is equal to H∞(Ω). A result of Godefroy and Shapiro states that for a non-constant multiplier ϕ, the adjoint

∗ Mϕ of the multiplication operator is hypercyclic if and only if ϕ(Ω) ∩ ∂D 6= ∅. Using the

∗ above result we will show that if Mϕ has an orbit with a non-zero limit point, then it is hypercyclic. In fact, a more general result holds:

Theorem 35. Let Ω be a region in C and ϕ ∈ H∞(Ω) be a non-constant function. Suppose

∗ 2 2 ∗ Mϕ : A (Ω) → A (Ω) has an orbit Orb(Mϕ,f) which has inﬁnitely many members contained

∗ in an open ball B(h, r) where 0 < r < khk. Then Mϕ is hypercyclic.

Proof. By our hypothesis, there exists an increasing sequence {nk}k≥1 of positive integers 49

∗ nk such that (Mϕ) f − h < r for all k ≥ 1. In light of the above result of Godefroy and

∗ Shapiro, to prove that Mϕ is hypercyclic, it suﬃces to show that ϕ(Ω) ∩ ∂D 6= ∅. By way of contradiction, we suppose ϕ(Ω) ∩ ∂D = ∅. Since ϕ(Ω) is an open connected subset of C, we have only two cases to consider:

Case 1: Suppose |ϕ(z)| < 1 for all z in Ω.

2 ∗ nk We note that for any g ∈ A (Ω) with kgk ≤ 1 we have that (Mϕ) f − h, g ≤

∗ nk (Mϕ) f − h < r for all k ≥ 1. (i) Z ∗ nk nk nk nk nk Furthermore, (Mϕ) f, g = hf, Mϕ gi = hf, ϕ gi = f ·ϕ ·g dA, where |f · ϕ · g| ≤ |f · g| · |ϕ|nk < |f · g|, which is a function in L1(dA) by the Cauchy-Schwarz inequality. Z ∗ nk nk Hence by the Dominated Convergence Theorem, we have lim (Mϕ) f, g = lim f · ϕ · k→∞ k→∞ g dA = 0 for all g ∈ A2(Ω) with kgk ≤ 1. This implies along with (i) that |hh, gi | ≤ r for D E all g ∈ A2(Ω) with kgk ≤ 1. But if g = h , then |hh, gi | = h, h = khk ≤ r, which is khk khk a contradiction.

Case 2: Suppose |ϕ(z)| > 1 for all z in Ω.

2 ∗ nk As in Case 1 given any g ∈ A (Ω) with kgk ≤ 1 we have (Mϕ) f − h, g < r for all k ≥ Z 2 1 1. Now, since f ∈ A2(Ω) and ϕ ∈ H∞(Ω) with |ϕ(z)| > 1 on Ω we have that f dA ϕnk Z Z 2 !1/2 2 f 2 f f ≤ |f| dA. Thus n ∈ A (Ω), and consequently lim = lim dA = ϕ k k→∞ ϕnk k→∞ ϕnk f/ϕnk 0. We now want to show that f = 0. If this is not the case, we let gk = kf/ϕnk k be a unit vector in A2(Ω), and consequently

−1 Z Z f/ϕnk f Z ∗ nk nk nk 2 (Mϕ) f, gk = fϕ gkdA = fϕ dA = |f| dA. kf/ϕnk k ϕnk

∗ nk ∗ nk Hence we have that (Mϕ) f, gk → ∞ as k → ∞. But we observe that (Mϕ) f, gk ≤

∗ nk ∗ nk (Mϕ) f, gk − hh, gki + |hh, gki|, and since we assumed (Mϕ) f − h, gk < r for all k ≥ 1 we have that lim |hh, gki| = ∞, which contradicts that kgkk = 1. Thus f = 0 and so k→∞

∗ nk since (Mϕ) f = 0 for all k ≥ 1, we get that 0 ∈ B(h, r), which is a contradiction.

∗ Therefore by Cases 1 and 2, we get that ϕ(Ω) ∩ ∂D 6= ∅ and thus Mϕ is hypercyclic. 50 From the last theorem, one can deduce the following corollary.

Corollary 36. Let Ω be a region in C and ϕ ∈ H∞(Ω) be a non-constant function. If

∗ 2 2 ∗ Mϕ : A (Ω) → A (Ω) has an orbit with a non-zero limit point, then Mϕ is hypercyclic.

∗ While the converse of Corollary 36 is absolutely trivial, if we only require Mϕ to be cyclic

∗ and not hypercyclic we cannot conclude that Mϕ has an orbit with a non-zero limit point.

∗ 2 Example 1. A cyclic Mϕ on A (D) having no non-zero orbital limit point.

∗ 2 Consider the adjoint of the multiplication operator Mϕ on the Bergman space A (D) with

∗ ϕ(z) = z. Since ϕ is a non-constant multiplier we have that Mϕ is supercyclic (see Godefroy

and Shapiro [20, page 248]) and thus cyclic. But |ϕ(z)| < 1 on D, so by Theorem 35 we can

∗ infer that Mϕ does not have an orbit with a non-zero limit point.

We now turn our attention to the Hardy space H2(D) for the open unit disk D, which Z 1/2 1 iθ 2 consists of all analytic functions f on D with lim f(re ) dθ < ∞. This limit r%1 2π is deﬁned to be the norm of f. Clearly, every function ϕ ∈ H∞(D) is a multiplier for H2(D);

∞ 2 2 2 2 that is if ϕ ∈ H (D) then ϕ · H (D) ⊂ H (D), and hence Mϕ : H (D) → H (D) given by

Mϕ(f) = ϕ · f is a bounded linear operator. As in the case of the Bergman spaces given ϕ to be a non-constant multiplier, the same result of Godefroy and Shapiro [20, Theorem

∗ 4.9] gives us that the adjoint Mϕ is hypercyclic if and only if ϕ(Ω) ∩ ∂D 6= ∅. However, the arguments in Cases 1 and 2 of the proof of Theorem 35 do not apply directly for the Hardy space.

Another interesting space is the Dirichlet space Dir(Ω) consisting of all analytic functions Z 2 2 1 0 2 f :Ω → C on the region Ω with kfk = |f(0)| + π |f | dA < ∞. As diﬀerent from the Bergman and the Hardy spaces, not every function in H∞(Ω) is a multiplier of Dir(Ω).

Suppose ϕ ∈ H∞(Ω) and ϕ is a non-constant multiplier of Dir(Ω). If ϕ(Ω) ∩ ∂D 6= ∅,

∗ then a result of Godefroy and Shapiro [20, Theorem 4.5] gives us that the adjoint Mϕ of the multiplication operator Mϕ : Dir(Ω) → Dir(Ω) is hypercyclic, however the converse does not hold and as a result, the argument in the proof of Theorem 35 does not apply. We oﬀer 51 ∗ below an example of an operator Mϕ on Dir(Ω) that is hypercyclic but ϕ(Ω) ∩ ∂D = ∅.

∗ Example 2. A hypercyclic Mϕ on Dir(D) having ϕ(D) ∩ ∂D = ∅. We will show that given the identity function ϕ(z) = z on the open unit disk D we have

∗ that Mϕ : Dir(D) → Dir(D) is hypercyclic, but obviously |ϕ(z)| < 1 for all z in D. Consider

n 2 fn : D → C, with fn(z) = z for n ≥ 1. Then each fn is analytic and we have kfnk = Z 2 Z 1Z 2π n 2 1 n−1 2 n n−1 i(n−1)θ 2 n √ kz k = π n · z dA = r · e r dθ dr = n. Thus kz k = n for π 0 0 n n z √z n ≥ 1. Deﬁne further en(z) = kznk = n if n ≥ 1 and e0(z) = 1 on D. Clearly kejk = 1 for Z 1 √ √ n−1 m−1 all j ≥ 0. Also, if n 6= m we have that hen, emi = π n · m · z · z dA = 0. Thus

{en}n≥1 is an orthonormal basis for Dir(D). zn+1 Let Mz : Dir(D) → Dir(D) be deﬁned by Mzf(z) = z · f(z). Then Mzen = kznk = n+1 zn+1 kz k q n+1 q n+1 kzn+1k · kznk = n en+1 for n ≥ 1. Letting wn = n if n ≥ 2 and w1 = 1, we have

Mzen = wn+1 · en+1 for all n ≥ 0.

2 We can now identify ej in Dir(D) with ej = (0, 0,..., 0, 1, 0,...) in ` , and consequently ∞ X the analytic function f = anen in Dir(D) is identiﬁed with the vector (a0, a1, a2,...) in n=0 2 ` . Finally we can then identify the operator Mz : Dir(D) → Dir(D) deﬁned above with

2 2 Mz : ` → ` , where Mz(a0, a1,...) = (0, w1a0, w2a1,...) is the unilateral forward shift with

weight sequence {wn}n≥1.

2 Note that for (a0, a1, a2,...) and (b0, b1, b2,...) in ` , we have

∗ hMz (b0, b1,...), (a0, a1,...)i = h(b0, b1,...),Mz(a0, a1,...)i ∞ X = h(b0, b1,...), (0, w1a0, w2a1,...)i = wi+1 · ai · bi+1 i=0 = h(w1b1, w2b2,...), (a0, a1,...)i.

2 ∗ So since the above identity holds for any (a0, a1, a2,...) in ` , we conclude that Mz (b0, b1, b2,...) =

(w1b1, w2b2, w3b3,...).

∗ Thus Mz : Dir(D) → Dir(D) is isometrically isomorphic to the unilateral backward

2 q n+1 ∗ weighted shift on ` with weights wn = n if n ≥ 2 and w1 = 1. Note that kMz k = r3 rn + 1 sup{wn} = . So since lim (w1 · ... · wn) = lim = ∞, Salas’ criterion for n≥1 2 n→∞ n→∞ 2 ∗ hypercyclicity of unilateral backward weighted shifts gives us that Mz is hypercyclic. 52 ∗ We would like to point out that the above example of the operator Mz on Dir(D) also shows that the well known Hypercyclicity Comparison Principle (see [34]) cannot be applied to conclude the hypercyclicity of adjoint operators. The Hypercyclicity Comparison Principle states that if X and Y are two normed spaces, Y is continuously and densely embedded in X, and T is a linear transformation on X that maps the smaller space Y to itself and is continuous in the topology of each space, then T is hypercyclic on the larger space X whenever T is hypercyclic on Y .

2 2 Clearly in this setting Dir(D) ,→ H (D) ,→ A (D) continuously and densely, and Mz is a continuous linear operator on each of the these spaces, but it would be erroneous to

∗ ∗ conclude that Mz is hypercyclic on the Hardy or Bergman space, given that Mz is indeed hypercyclic on the Dirichlet space. In fact using Godefroy and Shapiro’s result [20] since the

∗ analytic function ϕ(z) = z satisﬁes ϕ(D) ∩ ∂D = ∅, we have that Mz is not hypercyclic on H2(D), respectively A2(D). Indeed since the operation of taking the adjoint of an operator on a given space is intrinsically linked to the inner product of that space, diﬀerent norms

∗ 2 2 will induce diﬀerent adjoints. Thus restricting the adjoint Mz on H (D) or A (D) to the

∗ Dirichlet space will not produce the desired hypercyclic operator Mz : Dir(D) → Dir(D) and hence one cannot apply the Hypercyclicity Comparison Principle.

3.3 The linear fractional composition operators

3.3.1 Elementary properties and hypercyclicity of linear fractional

composition operators

Having examined the relationship between hypercyclicity and possessing an orbit with a non-zero limit point for the adjoints of multiplication operators on the Bergman spaces, we now turn our attention to the class of composition operators on the Hardy space. In the theory of composition operators it is useful to express the norm of the Hardy 53 space both in terms of the integral of its functional values and in terms of its power series coefficients. Having defined the Hardy space from the first viewpoint in section 3.2, we now

focus on the latter perspective. We say that an analytic function f on the unit disk D having ∞ X n the power series representation fˆ(n)z belongs to the Hardy space H2(D) if the sequence n=0 ∞ 2 2 X ˆ of power series coefficients is square summable, namely kfk = f(n) < ∞. n=0 We define a linear fractional transformation on D to be an analytic map ϕ : D → D az + b with ϕ(z) = subject to the condition ad − bc 6= 0, which is necessary and sufficient cz + d for ϕ to be non-constant. Clearly, a linear fractional transformation ϕ taking D into D is a restriction to the unit disk of an automorphism of the Riemann sphere Cb = C ∪ {∞}. We denote by LF T (D) the set of all linear fractional transformations of D and by Aut(D) the set of automorphisms of D. We note that the class of linear fractional transformations LF T (D) is diverse enough to reflect the character of the more general situation of arbitrary analytic self-maps of the unit disk, and hence in the following we will focus our attention on composition operators induced by these rather well-behaved maps. Given a linear fractional transformation ϕ, we define the linear fractional composition

2 2 operator Cϕ on the Hardy space H (D) by setting Cϕ(f) = f ◦ ϕ for any f in H (D). As a consequence of the Littlewood Subordination Principle (see Shapiro [34, page 13]) we have that whenever f belongs to the Hardy space so does f ◦ ϕ. Thus the linear fractional

2 composition operator Cϕ on H (D) takes the Hardy space to itself. The principle also

2 supplies a uniform estimate from which the boundedness of the linear operator Cϕ on H (D) can be determined.

Theorem 37 (Littlewood’s Theorem). Suppose ϕ is an analytic self-map of the unit disk

2 D. Then Cϕ is a bounded linear operator on the Hardy space H (D), and s 1 + |ϕ(0)| kC k ≤ ϕ 1 − |ϕ(0)|

In [34] Shapiro showed that the hypercyclic behavior of these composition operators is strongly inﬂuenced by the dynamical properties of the inducing linear fractional map ϕ. In 54 particular, we note that the location of the ﬁxed point(s) of the map ϕ plays an essential role

for the hypercyclicity of the operator Cϕ, so it is important to classify the linear fractional transformations with respect to their ﬁxed point location.

We note that if ϕ is a linear fractional transformation on Cb, then ϕ has at least one and at most two fixed points in Cb. A map ϕ is called parabolic if it has a single fixed point in Cb. Suppose ϕ is parabolic and has its fixed point ξ ∈ C. If ψ is a linear fractional transformation that takes ξ to ∞, then ρ = ψ ◦ ϕ ◦ ψ−1 is also a linear fractional transformation on Cb that fixes only the point ∞. Therefore ψ(z) = z + τ for some non-zero complex number τ. Thus, every parabolic linear fractional map is conjugate to a translation.

If the map ϕ is not parabolic, then it has two ﬁxed points ξ1 and ξ2 in Cb. Let ψ be a

−1 linear fractional map that takes ξ1 to 0 and ξ2 to ∞. Then the map ρ = ψ ◦ ϕ ◦ ψ is also

a linear fractional transformation on Cb and it ﬁxes both 0 and ∞, so it must have the form ρ(z) = λz for some complex number λ. Thus, ϕ(z) = ψ−1(λψ(z)) for all z ∈ C, where λ is called the multiplier for ϕ. We note that multipliers give the following classiﬁcation for

non-parabolic maps. If |λ| 6= 1, we say the linear fractional transformation ϕ ∈ LF T (D) is elliptic if |λ| = 1, hyperbolic if λ > 0, and loxodromic if ϕ is neither elliptic nor hyperbolic.

Thus, we identify four categories of linear fractional transformations on the unit disk D: (1) parabolic: ϕ is conjugate to a complex translation and has one attractive ﬁxed point on

∂D; (2) hyperbolic: ϕ is conjugate to a positive dilation and has one attractive fixed point in cl(D) and one repellent fixed point on ∂D or outside cl(D); (3) elliptic: ϕ is conjugate to a rotation and has one fixed point in D and the other fixed point outside cl(D); (4) loxodromic: ϕ is conjugate to a complex dilation and has one attractive fixed point in D and one repellent fixed point outside cl(D). As noted above, the dynamics of the linear fractional transformation ϕ on D determine

the hypercyclic behavior of the induced composition operator Cϕ. The next result of Shapiro [34] establishes that composition operators induced by loxodromic and elliptic maps, as well

as hyperbolic non-automorphisms with a ﬁxed point in D, are never hypercyclic. 55

Proposition 38. If ϕ is an analytic self-map of D that ﬁxes a point ξ in D, then Cϕ is not hypercyclic on H2(D). Moreover, if ϕ is not an elliptic automorphism, then for each

2 f ∈ H (D), only the constant function g = f(ξ) on D is in the closure of the orbit Orb(Cϕ, f).

Proof. Suppose the map ϕ ﬁxes the point ξ in D and f ∈ H2(D) is a hypercyclic vector

2 for Cϕ. Assume now that the function g ∈ H (D) is in the closure of the orbit Orb(Cϕ, f),

nk 2 that is there exists a sequence nk % ∞ with f ◦ ϕ → g in H (D). Since convergence in H2(D) implies pointwise convergence, we have f ◦ ϕnk → g pointwise on D, and in particular

f ◦ ϕnk (ξ) → g(ξ). But ξ is the ﬁxed point of ϕ, and so f ◦ ϕnk (ξ) = f(ξ), which implies

g(ξ) = f(ξ). Therefore, any function g in the closure of Orb(Cϕ, f) must agree with f at

2 the ﬁxed point ξ, and thus Orb(Cϕ, f) 6= H (D), a contradiction. Moreover, if ϕ is not an elliptic automorphism, then ξ ∈ D is an attractive ﬁxed point for ϕ, and so ϕn(z) → ξ for all z ∈ D. Thus by continuity of f, we get that f ◦ ϕn(z) → f(ξ) for

all z ∈ D. Thus, any g in the closure of the orbit Orb(Cϕ, f) has g(z) = f(ξ) for all z ∈ D.

Furthermore, an application of the Hypercyclicity Criterion gives that parabolic and

hyperbolic automorphisms of D always induce a hypercyclic composition operator on H2(D).

2 Theorem 39. If ϕ is a non-elliptic automorphism of D, then Cϕ is hypercyclic on H (D).

The following result of Shapiro [34] completes the characterization of the hypercyclicity

of linear fractional composition operators on H2(D).

Theorem 40. Suppose ϕ is linear fractional non-automorphism of the unit disk D with no ﬁxed point in D.

2 (a) If ϕ is hyperbolic, then Cϕ is hypercyclic on H (D).

2 (b) If ϕ is parabolic, then Cϕ fails to be hypercyclic on H (D) in a very strong sense: Only

constat functions can be limit points of Cϕ orbits. 56 As a final remark about the hypercyclic behavior of this class of operators, we note that a common mistake in the literature is to classify all composition operators induced by hyperbolic linear fractional transformations on the Hardy space as hypercyclic. Bourdon and Shapiro [9] showed that the location of the fixed points of the inducing map determines whether or not these composition operators are hypercyclic. For example, the functions 1 z z ϕ(z) = 2 and γ(z) = are hyperbolic non-automorphisms with fixed points 0, 3 2 − z 2 − z 2 and 0, 1, respectively, but they induce composition operators on the Hardy space that are not z + 2 hypercyclic. On the other hand the function φ(z) = is a hyperbolic non-automorphism 4 − z with fixed points 1, 2 that induces a hypercyclic composition operator on H2(D). To check that the above linear fractional transformations are indeed hyperbolic, one easily verifies that the multiplier λ of each map is positive. We recall that the multiplier λ of a linear fractional z − ξ transformation ρ is a complex number such that Ψ ◦ ρ ◦ Ψ−1(z) = λz, where Ψ(z) = 1 z − ξ2 for ξ1, ξ2 the fixed points of ρ. We summarize the hypercyclicity behavior of composition operators on the Hardy space as follows:

Type of ϕ Fixed point(s) location Hypercyclicity of Cϕ

parabolic automorphism ξ ∈ ∂D Yes

parabolic non-automorphism ξ ∈ ∂D No, in a strong sense*

hyperbolic automorphism ξ1, ξ2 ∈ ∂D Yes

ξ1 ∈ D, ξ2 ∈ ∂D No, in a strong sense*

hyperbolic non-automorphism ξ1 ∈ D, ξ2 ∈/ clD No, in a strong sense*

ξ1 ∈ ∂D, ξ2 ∈/ clD Yes

loxodromic ξ1 ∈ D, ξ2 ∈/ clD No, in a strong sense*

elliptic ξ1 ∈ D, ξ2 ∈/ clD No

*Only constant functions can be limit points of an orbit of Cϕ. 57 3.3.2 Orbital limit points and hypercyclicity of linear fractional

composition operators

In the following we investigate which linear fractional transformations ϕ induce composition

2 operators Cϕ on the Hardy space H (D) that have the property patterned in Corollary 36. As noted in the preceding section, there exists a complete characterization of linear fractional composition operators with respect to their hypercyclic behavior based on the dynamics of the inducing linear fractional transformation ϕ. Bourdon and Shapiro [9] emphasized the existence of a zero-one law for the hypercyclic behavior of these composition operators, wherein a linear fractional composition operator is either hypercyclic or it fails at it in a strong sense. This property makes it easy for us to determine the validity of the implication whether hypercyclicity follows from having an orbit with a non-zero limit point. Firstly, since all parabolic and hyperbolic automorphisms, as well as all hyperbolic non- automorphisms with no interior fixed point, induce hypercyclic composition operators, our implication holds trivially for these classes. Secondly, the composition operators induced by parabolic non-automorphisms, loxodromic linear fractional transformations and hyperbolic non-automorphisms with an interior fixed point fail to be hypercyclic in the strong sense that only constant maps can be limit points of any orbit of Cϕ. This shows that our implication fails for these classes of composition operators, but it does so in a not so aggravating way. Finally, the remaining class of composition operators, those induced by elliptic linear fractional transformations, provides us with a non-trivial example of a composition operator that has an orbit with a non-zero limit point but is not hypercyclic. Since no elliptic composition operator is hypercyclic, it suffices to construct an example of an elliptic linear fractional transformation that induces an operator with an orbit having a non-constant limit point. 58

Example 3. A non-hypercyclic Cϕ with a non-constant orbital limit point. We recall from the study of ergodic theory that irrational rotations are ergodic and measure- preserving for the Lebesque measure (see Petersen [29, page 49]). Thus, for a given irrational

α, the Orb(Tα, x) is dense for almost all x in [0, 1), where Tα : [0, 1) → [0, 1) is deﬁned by setting Tα(x) = x + α (mod 1).

∗ ∗ Let α > 0 be an irrational number and let x be a point in (0, 1) such that Orb(Tα, x )

nk ∗ is dense in [0, 1). Then there exists a sequence nk % ∞ such that Tα x → 0 as k → ∞, or

∗ equivalently e2πi(x +nkα) → e0 = 1.

Now, let ϕ : D → D be deﬁned by ϕ(z) = e2πiα ·z. Clearly, ϕ is an elliptic linear fractional

∗ transformation of the unit disk. Further we let f : D → D be deﬁned by f(z) = e2πix · z, where f is in H2(D) with kfk = 1. We claim that the identity function ψ : D → D, ψ(z) = z in H2(D) is a non-zero 2 nk limit point of Orb(Cϕ, f). To show this let nk % ∞ as above. Then Cϕ f − ψ = ∗ 2 ∗ 2 e2πi(x +nkα)z − z = e2πi(x +nkα) − 1 → 0, as k → ∞.

Hence the identity function is a non-zero limit point of Cϕ, but the operator is not hypercyclic.

We summarize this behavior of composition operators in the following proposition.

2 Proposition 41. There exists a linear fractional composition operator Cϕ on H (D) that has an orbit with a non-constant limit point but Cϕ is not hypercyclic.

In fact, hypercyclicity may fail in a more dramatic way, as the operator need not even be cyclic.

Example 4. A non-cyclic Cϕ with a non-zero orbital limit point.

2 2 Consider the linear fractional composition operator Cϕ : H (D) → H (D) induced by the z hyperbolic non-automorphism ϕ(z) = 2−z with ﬁxed points 0 and 1. A result of Bourdon

and Shapiro [9, Theorem 2.2, see also Table I] gives us that Cϕ is not cyclic. To show that

2 some orbit of Cϕ has a non-zero limit point, consider the function f(z) = 1 + z in H (D). 59 Then, since 0 is the attractive ﬁxed point for ϕ in D, we have that f ◦ ϕn → f(0) pointwise on D. Furthermore, as f is an analytic function deﬁned in a neighborhood of the closed disk cl(D) we have that f ◦ ϕn → f(0) pointwise on ∂D \{1}. Thus, since the sequence

n {f ◦ ϕ }n≥1 is uniformly bounded on ∂D we can conclude by the Dominated Convergence Theorem that kf ◦ ϕnk2 = 1 R f ◦ ϕn(eiθ) 2 dθ → |f(0)|2 = 1. Thus f ◦ ϕn → 1 in the 2π ∂D norm of H2(D); see for instance [18, page 187], and hence the function g = 1 on D is a non-zero limit point of Orb(Cϕ, f) but Cϕ is not cyclic.

3.4 Further remarks

Related to what we have studied in the previous sections, another interesting object to explore for a given operator T is the set LP (T ) = {x ∈ X : Orb(T ,x) has a non-zero limit point}, which is a generalization of the set of hypercyclic vectors. We note that different variations of this set have been studied by Costakis and Manoussos [14] and Pr˘ajitura[30]. Clearly the cardinality of the set LP (T ) is either zero or infinite, as once the vector x belongs to the set LP (T ) we have that T x, T 2x, . . . ∈ LP (T ). In fact, if the set LP (T ) contains the non-zero vector x it necessarily contains the complex lines C {T nx} with the exception of the zero vector. To illustrate the set LP (T ) with different examples, we first note that the unilateral

2 backward shift B with constant weights wj = 1 on ` is an operator having an empty LP (T ) set, as only the zero vector can be a limit point of an orbit. On the other hand, if the operator T is hypercyclic the set LP (T ) is certainly dense in X as it contains the dense set of hypercyclic vectors. Furthermore, the work of Beauzamy [3], Enﬂo [16] and Read [31] provided us with an example of an operator having all non-zero vectors as hypercyclic vectors, so the same containment as above gives us that in this case the set LP (T ) is maximal, meaning that LP (T ) = X \{0}. However, the set LP (T ) can be dense but not maximal as is demonstrated by the unilateral backward weighted shift 2B on `2. 60 Finally, we would like to point out that an operator T may have a dense LP (T ) set even if T is not hypercyclic.

Example 5. A non-hypercyclic Cϕ having a dense LP (Cϕ) set.

2 2 Consider the operator Cϕ : H (D) → H (D) induced by the loxodromic linear fractional −z transformation ϕ(z) = 2z+4 . Since 0 is the attractive ﬁxed point for ϕ in D, we have that for 2 n −z any f in H (D) f ◦ ϕ → f(0) pointwise on D. We now show that for the given ϕ(z) = 2z+4

2 nk and any f in H (D) there exists a sequence nk % ∞ so that f ◦ ϕ → f(0) also in the norm

2 of H (D). For this we ﬁrst note that since kϕk∞ < 1, the induced composition operator

2 Cϕ is compact on H (D); see Shapiro [34, page 23]. Next, since ϕ(0) = 0, the Littlewood

n−1 Subordination Principle gives us that kCϕfk ≤ kfk and thus the sequence {f ◦ ϕ }n≥1

2 |z| |z| n is bounded in H (D). Now, |ϕ(z)| ≤ 4−2|z| < 2 on D and thus ϕ → 0 uniformly on compact subsets of D, so f ◦ ϕn → f(0) uniformly on compact subsets of D. Now, for

n−1 notational simplicity let gn = f ◦ ϕ for all n ≥ 1. By the above we clearly have the

2 sequence {gn}n≥1 bounded in H (D) and so the compactness of the operator Cϕ yields that the sequence {gn ◦ ϕ}n≥1 is relatively compact. Thus there exists an increasing sequence {n } so that g ◦ ϕ → g in H2( ) for some function g in H2( ). But convergence k k≥1 nk e D e D in H2(D) implies uniform convergence on compact subsets of D, so we can conclude that

g ◦ ϕ → g uniformly on compact subsets of . However, g ◦ ϕ = f ◦ ϕnk which as a nk e D nk n subsequence of {f ◦ ϕ }n≥1 converges on compact subsets of D to the constant function f(0).

n 2 Thus ge = f(0) on D and hence f ◦ ϕ k → f(0) in the norm of H (D). Thus we can conclude 2 that the set LP (Cϕ) consists of all non-constant functions f in H (D) with f(0) 6= 0 and

2 consequently is dense in H (D).

In closing we note that for the classes of weighted backward shifts and adjoints of multiplication operators on the Bergman spaces, having an orbit with a non-zero limit point implies hypercyclicity. However, the class of linear fractional composition operators on the Hardy space does not enjoy the same property. Thus the question remains for which operators does the geometry of an orbit determine the hypercyclic behavior of the operators. 61

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APPENDIX

Deﬁnition 42. A Hilbert space H is a complete inner product space.

Deﬁnition 43. A Banach space X is a complete normed space.

Deﬁnition 44. A topological vector space X is a vector space X with a topology τ so that addition of vectors is a continuous operation from X × X into X, and multiplication by scalars is a continuous operation from F × X into X.

Deﬁnition 45. An F-space is a topological vector space X whose topology is induced by a translation invariant metric d so that (X, d) is complete.

Deﬁnition 46. A Fr´echetspace is a locally convex F-space, that is it has a basis consisting of convex sets.

Clearly, a Hilbert space is a Banach space, which is a Fr´echet space, which in turn is an F-space, and they are all topological vector spaces.

2 p The space ` (Z+) deﬁned in Section 2.1 is a Hilbert space, whereas the spaces ` (Z+) for 1 ≤ p < ∞ and p 6= 2 are Banach spaces that are not Hilbert spaces. On the other hand,

p the spaces ` (Z+) for 0 < p < 1 are F-spaces that are not Fréchet spaces as they are not locally convex. Finally, C(−1, 1) given by the metric d with ∞ X 1 kf − gkn d(f, g) = , and kf − gk = sup{|f(x) − g(x)| : x ∈ [−1 + 1 , 1 + 1 ]}, 2n 1 + kf − gk n n n n=1 n is a Fréchet space that is not a Banach space. 65 Definition 47. If H and K are Hilbert spaces and T is in the Banach algebra of bounded linear operators B(H,K), then the unique operator S ∈ B(K,H) satisfying hT h, ki = hh, Ski is called the adjoint of T and is denoted by S = T ∗.

Deﬁnition 48. Given T ∈ B(X), the spectrum σ(T ) = {α ∈ F : T −αI is not invertible}.

Deﬁnition 49. If X is a Banach space, then the strong operator topology (SOT) on

B(X) is the topology deﬁned by the family of seminorms {pf : f ∈ X}, where pf (A) = kAfk.

We note that a basic open set for the SOT topology is given by {A : X → X : k(A −

A0)fik < ε, i = 1, . . . , k}. Also the norm topology on B(X) is ﬁner than the SOT topology, so norm convergence implies SOT convergence. Finally, we note that An → A strongly if and only if Anf → Af for every f ∈ X.

Deﬁnition 50. Given a Banach space X, the weak topology on X is the topology induces

∗ by the seminorms {pα∗ }α∗∈X∗ , where each pα∗ is given by pα∗ (x) = |α (x)|.

∗ We note that a basic weakly open set is given by {x ∈ X : |αi (x−x0)| < ε, i = 1, . . . , k}. Furthermore, we remark that the weak topology is weaker than the norm topology.

Fact: Dir(D) ,→ H2(D) ,→ A2(D) ∞ 2 2 X n 2 To see that Dir(D) ,→ H (D) ,→ A (D) we note that if f(z) = anz then kfkν = n=0 ∞ X 2 2ν 1 |an| (n + 1) , where ν = 2 for the Dirichlet space Dir(D), ν = 0 for the Hardy space n=0 2 1 2 H (D) and ν = − 2 for the Bergman space A (D).

2 2 Fact: Mϕ can never be hypercyclic on Dir(Ω), H (D) or A (Ω)

2 To see that Mϕ can never be hypercyclic on either of the three spaces Dir(Ω), H (D) or A2(Ω), we let ϕ be a non-constant multiplier and g be a function in the closure of an orbit

Orb(Mϕ, f). Then there exists a z0 in the space Ω or D so that |ϕ(z0)| < 1 or |ϕ(z0)| > 1. 66 Since convergence in norm implies pointwise convergence for all of the above three spaces

nk of analytic functions, we have that |ϕ(z0)| |f(z0)| → |g(z0)| for some sequence {nk}. Thus

if |ϕ(z0)| < 1 we get that |g(z0)| = 0, and if |ϕ(z0)| > 1 then |f(z0)| = 0 which in turn

gives that |g(z0)| = 0. Therefore only maps that vanish at z0 are in the closure of an orbit

Orb(Mϕ, f), and hence Mϕ can never be hypercyclic.

Fact: (B(X), k.k) is not separable Let H be a separable inﬁnite dimensional Hilbert space. Then H ' L2[0, 1]. Since we

∞ 2 ∞ 2 2 can identify L [0, 1] as a subset of B(L [0, 1]) (if ϕ ∈ L [0, 1] deﬁne Mϕ : L [0, 1] → L [0, 1]

∞ by Mϕ(f) = ϕf and kMϕk = kϕk∞), it suﬃces to show that L [0, 1] is not separable in the

k.k∞ norm.

Let a ∈ [0, 1) and deﬁne fa = χ[0,a). Suppose by means of contradiction that D is a

∞ ∞ countable dense subset of L [0, 1]. Then for fa ∈ L [0, 1] there exists ga ∈ D so that

1 3 5 1 1 kfa − gak∞ < 4 . Thus, 4 < ga(z) < 4 for all z ∈ [0, a] and − 4 < ga(z) < 4 for all z ∈ (a, 1].

Hence, if the function ga is used to approximate fa, it cannot be used to approximate

1 a+b 1 fb for b 6= a (if kfa − gak∞ < 4 , then for b > a we have |(fa − ga) 2 | > 4 and so 1 kfa − gak∞ ≥ 4 ).

Since there are uncountably many functions fa, there are uncountably many functions ga in D, so D cannot be countable. Therefore (B(H), k.k) is not separable.