Hypercyclic Extensions of Bounded Linear Operators

HYPERCYCLIC EXTENSIONS OF BOUNDED LINEAR OPERATORS

George Romeo Turcu

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

December 2013

Committee

Kit Chan, Advisor

Ron Lancaster, Graduate Faculty Representative

Juan Bes

Craig Zirbel ii ABSTRACT

KIT CHAN, Advisor

If X is a topological vector space and T : X → X is a continuous linear operator, then

T is said to be hypercyclic when there is a vector x in X such that the set {T nx : n ∈ N} is dense in X. If a hypercyclic operator has a dense set of periodic points it is said to be chaotic. This paper is divided into five chapters. In the first chapter we introduce the hypercyclicity phenomenon. In the second chapter we study the range of a hypercyclic operator and we find hypercyclic vectors outside the range. We also study arithmetic means of hypercyclic operators and their convergence. The main result of this chapter is that for a chaotic operator it is possible to approximate its periodic points by a sequence of arithmetic means of the first iterates of the orbit of a hypercyclic vector. More precisely, if z is a periodic point of multiplicity α, that is T αz = z then there exists a hypercyclic vector of T such that 1 A x = (z + T αz + ··· + T (n−1)αz) converges to the periodic point z. In the third chapter n,α n we show that for any given operator T : M → M on a closed subspace M of a Hilbert space H with infinite codimension it has an extension A : H → H that is chaotic. We conclude the chapter by observing that the traditional Rota model for operator theory can be put in the hypercyclicity setting. In the fourth chapter, we show that if T is an operator on a closed subspace M of a Hilbert space H, and P : H → M is the orthogonal projection onto M, then there is an operator A : H → H such that P AP = T , PA∗P = T ∗ and both A, A∗ are hypercyclic. iii

I dedicate this dissertation to Georgia. iv ACKNOWLEDGMENTS

I owe my gratitude to all those people who have made this dissertation possible. My deepest gratitude is to my advisor Dr. Kit Chan, it would have been next to impossible to write this dissertation without his excellent guidance. His patience and support helped me overcome many crisis situations and finish this dissertation. I would like to thank Dr. Craig Zirbel for his constant support and encouragement. He is one of the most helpful teachers I have had in my life. I want to thank him for reading and recommending improvements on my dissertation. Dr. Juan Bés’insightful comments and constructive criticisms helped me focus my ideas. I am grateful to him for carefully reading and commenting on revisions of this manuscript. I am grateful to Dr. Alexander Izzo for his encouragement and for the warm and heartfelt friendliness. Dr. Corneliu Hoffman was the first person to introduce me to the world of mathematics. I would like to thank him for this and for the pleasent conversations we had over the years. My mother Julieta, my father Dumitru, my brother Raul, my sister in law Tatiana and my friends Irina and Catalin Seceleanu, Kevin Rion and Dr. Warren McGovern have helped me stay sane through these difficult years. I greatly value their friendship. None of this have been possible without the love and the emotional support of my wife, Andrea. v

TABLE OF CONTENTS

CHAPTER 1: Introduction 1

CHAPTER 2: Hypercyclic means 8

CHAPTER 3: Extensions to chaotic operators 21

CHAPTER 4: Compressions of hypercyclic operators 38

BIBLIOGRAPHY 49 1

CHAPTER 1

Introduction

The invariant subspace problem is one of the most puzzling unsolved problems in operator theory. The problem is to determine whether every bounded linear operator T : H → H defined on an infinite dimensional Hilbert space H has a non-trivial invariant subspace, that is a closed linear subspace M of H which is different from 0 and H such that TM ⊂ M. The question has been open for over eighty years. There are, though, special cases for which the question has been answered in the case that the underlying space H is a Banach space. In the case where the vector space is finite dimensional over the complex field, every operator has an eigenvector, so it has an invariant subspace. Also the spectral theorem proves that all normal operators (an operator that commutes with its adjoint is said to be normal) admit invariant subspaces. It is easy to see that if the vector space H is not separable then the question can be answered in the affirmative; say, take any nonzero vector x, the closed linear subspace generated by {x, T x, T 2x, . . .} is invariant under T and it is different from X. Recall that a topological vector space is a vector space with a topology with respect to which the two vector space operations, namely vector addition and scalar multiplications are continuous. A Hilbert space is a topological vector space whose topology is induced by an inner product, which makes the space complete. A topological space is separable if it has a countable dense subset. 2 Definition 1. Let T : X → X be an operator on a topological vector space X and let x be a vector in X. Then we define the orbit of x under T to be the set {x, T x, T 2x, T 3x, . . .} and we denote it by orb(x, T ).

Suppose X is nonseparable. The orbit of any non-zero vector x ∈ X is an invariant subset of T . It is clear that span(orb(x, T )), the closed subspace generated by the orbit of x under T, is an invariant subspace of T , in fact the smallest closed subspace containing the vector x. Since the orbit is countable, the closed subspace generated by the orbit is separable and thus a nontrivial invariant subspace of X.

Deﬁnition 2. Let T : X → X be a continuous linear operator on a topological vector space X and x be a vector in X. We call the vector x a cyclic vector for T if the linear span of its orbit under T is dense in X; that is, if

span(orb(x, T )) = X.

The operator T is called a cyclic operator if it has a cyclic vector.

Deﬁnition 3. Let T : X → X be a continuous linear operator on a topological vector space

X over a scalar ﬁeld F and x be a vector in X. We call the vector x a supercyclic vector for T if the set of the scalar multiples of the vectors in the orbit is dense in X; that is, if

[ λ · orb(x, T ) = X. λ∈F

The operator T is called a supercyclic operator if it has a supercyclic vector.

Deﬁnition 4. Let T : X → X be a continuous linear operator on a topological vector space X and x a vector in X. We call the vector x a hypercyclic vector for T if its orbit under T is dense in X; that is, if orb(x, T ) = X.

The operator T is called a hypercyclic operator if it has a hypercyclic vector. 3 Notice that if all vectors of a linear space X are cyclic then X does not have an invariant nontrivial subspace and the invariant subspace problem can be rephrased in terms of cyclic vectors. The answer to the invariant subspace problem is negative if and only if there is an operator for which all the nonzero vectors are cyclic. In the same way, a hypercyclic operator having all vectors hypercyclic if and only if it has no nontrivial invariant closed subset. This suggests that hypercyclicity theory is strongly connected to, if not generated by, the interest in solving the invariant subspace problem.

Deﬁnition 5. Let T be a linear operator on a topological vector space X. The operator T is called a chaotic operator if it is hypercyclic and it possesses a dense set of periodic vectors.

Chaotic operators suggest a connection to ergodic theory and an evolution of hypercyclicity into the theory of linear dynamical systems. Unfortunately we do not have enough space in the dissertation to give more details but we recommend [11] for further reading on the subject of chaotic dynamical systems. Historically the term “hypercyclic” was first used in its modern acception by Beauzamy [4]. There are, though, examples of avant la lettre hypercyclic operators. In 1929 Birkhoff [7] essentially showed the hypercyclicity of the translation operator on the Fréchet space of entire functions H(C) = {f : C → C : f is analytic}. More precisely he showed the existence of an entire function f with the property that the set {f(z), f(z+1), f(z+2), f(z+3),...} is dense in the space of entire functions H(C) endowed with the topology of uniform convergence on compact sets. The translation operator T : H(C) → H(C), defined by

T (f)(z) = f(z + 1) was shown by Birkhoff to have a dense orbit and thus to be hypercyclic. More than twenty years later, in the same area of complex analysis MacLane [21] proved that there exists an entire function f such that the set {f, f 0, f 00,...} is dense in H(C). This proves that the 4 differentiation operator D : H(C) → H(C) defined by

D(f)(z) = f 0(z) is hypercyclic. In his detailed and almost exhaustive survey on universal families and hypercyclic operators [15] Grosse-Erdmann credited Rolewicz to be the initiator of the general theory of hypercyclicity. In 1969 Rolewicz [23] produced a very simple example of a hypercyclic operator, the backward shift on the Banach space `p where p ≥ 1. If B : `p → `p is the backward shift deﬁned by

B(a0, a1, a2,...) = (a1, a2, a3,...), then the operator λB is hypercyclic whenever |λ| > 1. The magnitude of the scalar λ needs to be strictly greater than one because the backward shift is a contraction and so any vector’s orbit under B would be bounded and hence not hypercyclic. One of the most useful results in hypercyclicity is the so called Kitai’s Hypercyclicity Criterion due to Kitai [17] and it was rediscovered a few years later in much greater generality by Gethner and Shapiro [13]. Later B´esand Peris [6] weakened the suﬃcient condition of the criterion and this is the form in which we choose to state it.

Theorem 6. The Hypercyclicity Criterion. Let X be a separable F -space and T be a continuous linear operator on X. Suppose that there are dense subsets X0 and Y0 of X, an increasing sequence (nk) of positive integers and (possibly nonlinear and discontinuous)

mappings Snk : Y0 → X such that

nk 1. for every x ∈ X0, T x → 0

2. for every y ∈ Y0,Snk y → 0

nk 3. for every y ∈ Y0, (T ◦ Snk )y → y.

Then the operator T is hypercyclic. 5 We should note that this condition is only suﬃcient. M. De La Rosa and C. Read [10] showed that there exist hypercyclic Banach space operators which do not satisfy the Hyper- cyclicity Criterion. Recently, F. Bayart and E. Matheron [3] proved that such operators can

p be constructed on a large class of Banach spaces, including c0(N) or ` (N). Another amazing result in hypercyclicity due to Read [22] states the existence of a hypercyclic operator on the Banach space `1 for which every nonzero vector is hypercyclic. This means Read’s result solved the invariant subset problem for `1 in the negative. It is very interesting to note that hypercyclicity appears only in the setting of infinite dimensional spaces. Rolewicz [23] showed that no linear operator on a finite dimensional space is hypercyclic. One would think that bizarre objects such as hypercyclic operators must be very rare. It turns out that this is actually false. In fact, Ansari [2] and Luis Bernal-González[5] showed that every separable infinite dimensional Banach space carries a hypercyclic operator . On the other hand chaotic operators are not as common. Bonet, Martinez-Gimenez and Peris showed that there exists a Banach space which admits no chaotic operator [8]. The set of hypercyclic vectors for a hypercyclic operator is residual and every vector is the sum of two hypercyclic vectors [15]. We remind here that a residual set is the complement of a meager set, or equivalently, the intersection of countably many sets with with dense interiors. Also, an F -space is a topological vector space X whose topology is defined by a translation invariant metric d and such that (X, d) is complete.

Proposition 7. Let X be a separable F-space.

1. If T is a hypercyclic operator on X, then every vector in X is the sum of two hypercyclic vectors for T .

2. If Tn(n ∈ N) are hypercyclic operators on X, then there exists a vector in X that

is hypercyclic for each Tn. In fact, the set of common hypercyclic vectors is a dense

Gδsubset of X and thus residual. 6 In operator theory it is always interesting to find bounded linear extensions A : X → X that preserve or improve the properties of the original operator T : M → M, where M is a closed linear subspace of X. The fact that it is always possible to extend a bounded linear functional λ : M → C to f : X → C with kfk = kλk was proven by Hahn in 1927; see [19]. This is the well known Hahn-Banach theorem. A subnormal operator is defined to be an operator with a normal extension. However, a subnormal operator can never be hypercyclic; see [17]. This also means that none of the techniques in the theory of subnormal operators can be used in hypercyclicity. In this dissertation we show that any bounded linear operator T : M → M defined on a separable, infinite dimensional Banach space M can be extended to a chaotic operator A : X → X, where M is a closed subspace of X. As we pointed out every Banach space admits a hypercyclic operator, but not every Banach space admits a chaotic operator; see [2], [8]. Our result shows that, in fact, every operator can be extended to be chaotic. Sophie Grivaux [14] proved a similar result which we state here. Let X be any Banach space and T is a bounded operator on X. An extension (X,˜ T˜) of the pair (X,T ) consists of a Banach space X˜ in which X embeds isometrically through an isometry i and a bounded operator T˜ on X˜ such that T˜ ◦ i = i ◦ T on X. When X is separable, it is additionally required that X˜ be separable. We say that (X,˜ T˜) is a topologically transitive extension of (X,T ) when T˜ is topologically transitive on X˜, i.e. for every pair (U,˜ V˜ ) of non-empty open subsets of X˜ there exists an integer n such that Tñ(U) ∩ V˜ is nonempty. Grivaux shows that any such pair (X,T ) admits a topologically transitive extension (X,˜ T˜), and that when H is a Hilbert space, (H,T ) admits a topologically transitive extension (H,˜ T˜) where H˜ is also a Hilbert space. Grivaux shows that these extensions are indeed chaotic. In the setting of a Hilbert space Feldman [12] showed that there is a continuous linear operator T on a separable Hilbert space such that if f : X → X is a continuous function on a compact metric space X, then there exists an invariant closed set K for T such that T |K is topologically conjugate to f. 7 In this dissertation we extend an operator T defined on a subspace M of infinite codimension in a Hilbert space H to a chaotic operator defined on H that has a hypercyclic subspace, that is an infinite dimensional closed subspace consisting entirely of hypercyclic vectors, except the zero vector. In the same setting of a Hilbert space H with a subspace M having infinite codimension, we show that for every operator T defined on M, there exist an operator A defined on H that is hypercyclic with a hypercyclic adjoint A∗, so that the compression of A to M is T . That is, P AP = T, where P : H → H is the orthogonal projection onto M. In this dissertation, we also make connection to a very famous theorem in ergodic theory which is known as von Neumann’s mean ergodic theorem [18].

Theorem 8. If T is a contraction in a Hilbert space H, and P the projection on F = {g ∈

−1 Pn−1 k H : T g = g}, then Anf = n k=1 T f converges in norm to P f for f ∈ H as n goes to ∞.

In this dissertation we replace the contraction in the above theorem by a hypercyclic operator which has a dense orbit; also instead of ﬁxed points we look at periodic points. We also look at the range of a hypercyclic operator and we construct a hypercyclic operator such that not all of its hypercyclic vectors are in its range. 8

CHAPTER 2

Hypercyclic means

An operator T is called Cesaro hypercyclic provided there exists a vector x ∈ X so that its

1 2 n−1 Cesaro orbit { n (I + T + T + ··· + T )x : n = 1, 2,...} under T is dense in X. F. Le´on- Saavedra showed that T is Cesaro hypercyclic if and only if there exists a vector y ∈ X such

T n that the sequence { n y : n = 1, 2,... } is dense in X, see [20]. In this chapter, however, we obtain results motivated by von Neumann’s mean ergodic theorem by studying arithmetic

1 2 n−1 means n (I + T + T + ··· + T ) of hypercyclic operators and their convergence. We prefer to call these arithmetic means of hypercyclic operators, simply, hypercyclic means. The main result of this chapter establishes an approximation of the periodic points of a chaotic operator with hypercyclic vectors by the way of hypercyclic means.

Definition 9. Let T be a bounded linear operator defined on a separable, infinite dimensional Banach space. If there is a vector x whose orbit {T nx : n ≥ 0} is dense in X, then we call T a hypercyclic operator and x a hypercyclic vector for T .

We can rephrase the deﬁnition in terms of invariant subsets for T ; that is a vector x is a hypercyclic vector if and only if the smallest closed invariant subset containing x is the whole space X. We show two easy ways in which a vector x fails to be hypercyclic.

Remark 10. Let X be a separable inﬁnite dimensional Banach space and T : X → X be a bounded linear operator with kT k > 1. If there is a vector x in X and an increasing sequence 9

nk nk−nk+1 (nk) of positive integers such that kT xk ≤ kT k then the vector x is not hypercyclic.

Proof. Let m ∈ N be arbitrarily ﬁxed. Then there is k such that nk ≤ m < nk+1 and thus

kT mxk = kT m−nk T nk xk

≤ kT km−nk kT nk xk

< kT knk+1−nk kT nk xk

≤ 1.

So the orbit of x under T is bounded and hence x cannot be hypercyclic.

Remark 11. Let X be a separable Banach space, and T : X → X be a bounded linear

operator with kT k ≥ 1. Let x be vector in X and (nk) be an increasing sequence of positive

nk integers such that M := sup{nk+1 − nk : k ∈ N} < ∞ and (T x) converges to the zero vector. Then the vector x is not hypercyclic.

Proof. Without loss of generality we may assume that kT nk xk < 1 for every k. Let m ∈ N

be arbitrarily ﬁxed. Then there is an integer k such that nk ≤ m < nk+1 and we have

kT mxk = kT m−nk T nk xk

≤ kT km−nk kT nk xk

≤ kT kM kT nk xk

< kT kM ,

and thus the orbit of x is bounded.

The range of a hypercyclic operator is a dense subset of the whole Banach space X; in fact the orbit of every hypercyclic vector is dense. When a hypercyclic operator T is not onto it is simple to see that it has a hypercyclic vector x that is not in the range of T . To see that let HC(T ) be the set of all hypercyclic vectors for T . By Proposition 10 7, HC(T ) + HC(T ) = X. Hence if we assume HC(T ) ⊂ ranT then it would follow that X = HC(T ) + HC(T ) ⊂ ranT + ranT = ranT, and thus T is onto. In the following, we provide a construction of such a hypercyclic vector that is not in the range of T , to give an illustration of how such a vector can arise.

Proposition 12. There is a hypercyclic operator T such that not all of its hypercyclic vectors are in its range.

Proof. We construct a hypercyclic unilateral weighted backward shift T deﬁned on `2(N) together with a hypercyclic vector x which is not in the range of T . Let D = {dn : n ∈ N} be a countable dense subset of `2 formed by vectors with ﬁnitely many nonzero entries. For

Pln ˆ ˆ each dn, we write dn = k=0 dn(k)ek, where dn(ln) 6= 0 and {ek : k = 0, 1, 2, ···} is the canonical basis in `2. A linear operator T : `2 → `2 is a unilateral backward weighted shift

with the sequence of bounded weights (wn) if T en = wnen−1 for all n ≥ 1 and T e0 = 0. For

this we deﬁne inductively a sequence (mk) of increasing positive integers and a sequence (ak)

kd1k 1 |dˆ1(l1)| of real numbers. Let m1 be a positive integer such that 2m1 < 2 and let a1 = 2m1 > 0.

Inductively we deﬁne mk+1 for k ≥ 1, such that

kdk+1k 1 m −k−m < k+1 , (2.1) a1 ··· ak2 k+1 k 2

and

mk+1 > 2(mk + lk + k) (2.2)

and deﬁne ˆ |dk+1(lk+1)| ak+1 = m −k > 0. a1 ··· ak2 k+1

Note that each ak satisﬁes ak < 1 by 2.1. Now, using the sequence (ak) we deﬁne a bounded

sequence of weights (wj) by wj = ak if j = mk + lk + 1 and wj = 2 otherwise. Let T be the unilateral backward weighted shift deﬁned by these weights, and let S : `2 → `2 be the

−1 possibly unbounded unilateral forward shift deﬁned by the sequence of weights (wk ); that 11 −1 is, Sek = wk+1ek+1, whenever k ≥ 0.

2 We deﬁne a vector in ` , using the dense subset D and the sequences (ak) and (nk) by

∞ lk ˆ X X dk(j) x = e m −(k−1) mk+j a1 ··· ak−12 k k=1 j=0

Note that mk + lk < mk+1 for each k by 2.2, and hence by 2.1,

∞ 2 ∞ 2 2 X kdkk X 1 kxk ≤ m −(k−1) ≤ k < ∞, a1 ··· ak−12 k 2 k=1 k=1

and so x ∈ `2. By deﬁnition, for each k, we have that

mk T emk+j = (wmk+j ··· wj+2wj+1)ej.

Using the deﬁnition of the weight sequence (wj), one can check that if 0 ≤ j ≤ lk, then

mk−(k−1) wmk+j ··· wj+2wj+1 = a1 ··· ak−12 .

It follows that

lk ˆ ! lk dk(j) mk X X ˆ T em +j = dk(j)ej = dk. a ··· a 2mk−(k+1) k j=0 1 k−1 j=0

Thus using 2.1 we get,

∞ kd k 2 mk 2 X i kT x − dkk ≤ m −m −(i−1) a1 ··· ai−12 i k i=k+1 ∞ 2 X 1 ≤ 2i i=k+1 (2.3) 1 1 = · 3 4k 1 < 4k 12 To show that the orbit of x under T is dense in `2, let y be a vector in `2 and let > 0. There is a positive integer N such that if k > N then

1 < , (2.4) 4k 2

and there is k > N such that kd − yk < . (2.5) k 2

The triangle inequality together with 2.3, 2.4 and 2.5 prove that the vector x has a dense orbit and thus hypercyclic. We will show that x is not in the range of T . Notice that for each k S(e ) = w−1 e = a−1e mk+lk mk+lk+1 mk+lk+1 k mk+lk+1 and hence ∞ 2 ∞ 2 X −1 |dk(lk)| X kS(x)k ≥ ak m −(k−1) = 1 = ∞. a1 ··· ak−12 k k=1 k=1 Suppose that there were a vector y in `2 such that T y = x it would follow that y = Sx and the norm of y would be inﬁnite. We have constructed a vector x that is hypercyclic for T but it is not in its range.

We deﬁne the arithmetic mean of a hypercyclic operator in the manner suggested by the ergodic means.

Deﬁnition 13. Let X be a Banach space and let T ∈ B(X), we deﬁne the n−th arithmetic 1 mean of T to be the operator A = (I + T + T 2 + ··· + T n−1) and for a hypercyclic vector n n

x we call Anx the hypercyclic mean, when the hypercyclic operator T is understood.

We state the simplest version of the Hypercyclicity Criterion due to Kitai [17] that we are using in the rest of this chapter. This version is deﬁnitely simpler than the one we stated in Theorem 6, but it already provides all we need in this chapter.

Theorem 14. Kitai’s Hypercyclicity Criterion. Let T be a bounded linear operator on 13 a separable inﬁnite dimensional Banach space X. Suppose D is a countable dense subset of X and S is a right inverse for T such that:

1. T n → 0 pointwise on D

2. Sn → 0 pointwise on D

then the operator T is hypercyclic.

The next proposition is inspired by von Neumann’s ergodic theorem.

Theorem 15. If T is a contraction in a Hilbert space H, and P the projection on F = {g ∈

−1 Pn−1 k H : T g = g}, then Anf = n k=1 T f converges in norm to P f for f ∈ H as n goes to ∞.

We replace the contraction by a hypercyclic operator, which can never be a contraction since its orbit is a dense subset. We prove the result for periodic points, but notice that a ﬁxed point is a particular kind of periodic point.

Proposition 16. Let T be a bounded linear operator on a separable inﬁnite dimensional Banach space X that satisﬁes the Hypercyclicity Criterion and let α be a positive integer.

1 α 2α (n−1)α Then there is a hypercyclic vector x in X such that An,αx = n (x+T x+T x+···+T x) converges to the zero vector in X.

Proof. Let {d1, d2, d3, ··· , dn, ···} be an enumeration of the dense subset D. We deﬁne

inductively a sequence (bn) of positive integer multiples of α, bn = αcn, where cn is an

m 1 m 1 integer. Let b1 be such that kS d1k < 2 and kT d1k < 2 for all m ≥ b1. Let b2 be such that m 1 m 1 m 1 b2 > 2b1, kS d2k < 22 and kT d1 < 22 k and kT d2k < 22 for all m ≥ b2.

Suppose we have constructed b1, b2, ··· , bk−1. We deﬁne bk to be large enough such that the following conditions are satisﬁed:

bk > 2bk−1, for all k > 2, (2.6) 14

1 kSmd k < , for all m ≥ b − b and all i ≤ k + 1, (2.7) i 2k k k−1

1 kT md k < , for all m ≥ b − b , and all i ≤ k, (2.8) i 2k k k−1

1 1 bk−1 α bk−1 kS dik + ··· + kdik + kT dik + ··· + kT dik < k , for all i ≤ k + 1. (2.9) ck 2

We deﬁne a vector x by ∞ X bi x = S di. i=1

P∞ bi P∞ 1 Observe that x is well deﬁned since by (2.7), kxk = k i=1 S dik < i=1 2i = 1. To show that the vector x is hypercyclic let y be a vector in X and let > 0. Since D is

k a dense subset of X there is a positive integer k such that ky − dkk < 2 and 2k < 2 . On the other hand

∞ bk bk X bi kT x − dkk = kT S di − dkk i=1 ∞ X bk bi (2.10) = k T S di − dkk i=1

bk−b1 bk−bk−1 bk+1−bk = kT d1 + ··· + T dk−1 + S dk+1 + · · · k, and using the triangle inequality and inequality (2.6) we get

k−1 ∞ bk X bk−bi X bi−bk kT x − dkk ≤ k T dik + k S dik i=1 i=k+1 ∞ 1 X 1 < (k − 1) + (2.11) 2k 2i i=k+1 k = . 2k 15 It is clear now that T bn x approximates the vector y since

bn bn kT x − yk < kT x − dkk + kdk − yk <

and thus x is a hypercyclic vector.

To show that the hypercyclic mean An,αx converges to the zero vector in X, let n be a positive integer. Then there is a positive integer k such that bk ≤ (n − 1)α < bk+1 and

1 A x = (x + T αx + T 2αx + ··· + T (n−1)αx) n,α n ∞ ∞ ∞ 1 X X X = ( Sbi d + T α Sbi d + ··· + T (n−1)α Sbi d ) n i i i i=1 i=1 i=1 ∞ 1 X = (Sbi d + T αSbi d + ··· + T (n−1)αSbi d ) n i i i i=1 k−1 (2.12) 1 X = (Sbi d + Sbi−αd + ··· + d + T αd + ··· + T (n−1)α−bi d ) n i i i i i i=1 1 + (Sbk d + Sbk−αd + ··· + d + T αd + ··· T (n−1)α−bk d ) n k k k k k ∞ 1 X + (Sbi d + Sbi−αd + ··· + Sbi−(n−1)αd ). n i i i i=k+1

We approximate each of the three summands above separately. We use conditions (2.9) 16 and (2.8) and the fact that bk ≥ 2bk−1 to approximate the ﬁrst summand:

k−1 1 X k (Sbi d + Sbi−αd + ··· + d + T αd + ··· + T bi d + T bi+αd + ··· + T (n−1)α−bi )d k n i i i i i i i i=1 k−1 1 X ≤ k (Sbi d + Sbi−αd + ··· + d + T αd + ··· + T bk−1 d k n i i i i i i=1 k−1 1 X + k T bk−1+αd + ··· + T (n−1)α−bi d )k n i i i=1 1 n < + 2k n2k 1 1 = + 2k 2k 2 = 2k 1 = . 2k−1 (2.13)

For the second summand we use the conditions (2.7), (2.8) and (2.9) to get:

1 kSbk d + Sbk−αd + ··· + d + ··· + T (n−1)α−bk d k n k k k k 1 ≤ kSbk d + Sbk−αd + ··· + Sbk−1+αd k n k k k 1 bk−1 bk−1−α α bk−1 + kS dk + S dk + ··· + dk + T dk + ··· + T dkk n (2.14) 1 + kT bk−1+αd + T bk−1+2αd + ··· + T (n−1)α−bk d k n k k k 1 1 1 1 1 < · n · + + · n · n 2k 2k n 2k 3 ≤ . 2k 17 We use condition (2.7) to approximate the third summand:

∞ 1 X k (Sbi d + Sbi−αd + ··· + Sbi−(n−1)αd )k n i i i i=k+2 ∞ 1 X 1 < · n · (2.15) n 2i i=k+1 1 = . 2k−1

We use these three approximations and (2.12) to show that An,αx converges to zero.

1 3 1 kAn,αxk < k−1 + k + k−1 2 2 2 (2.16) 7 = . 2k

We let n go to inﬁnity and thus we let k go to inﬁnity and from (2.16) we get the desired

result An,αx → ∞.

Next we give a counterexample, that is an example of a hypercyclic operator, twice the unilateral backward shift, and a hypercyclic vector for which the hypercyclic mean does not converge.

Example 17. Let T = 2B deﬁned on `2, where B is the unilateral backward shift. There

2 exists a hypercyclic vector x in ` such that its hypercyclic means An,1x do not converge to 0.

Proof. Let D be the set of all sequences in `2 with ﬁnite number of nonzero entries all of which are rational. Since D is countable, we let D = {x1, x2,...} and observe that D is a dense set of vectors in `2. We will denote by δ the length of a vector x in `2; that is, if x = (ˆx(0), xˆ(1),...) and j is the largest integer so thatx ˆ(j) 6= 0, then the length δ of x is

said to be δ = j + 1. Choose a sequence of integers (an) such that:

kxnk 1 ≤ , for all n ∈ N. 2an 2n 18 2 2 1 Let F : ` → ` be the unilateral forward shift and S = 2 F. Deﬁne bn by bn = a1 + δ1 + ··· +

P∞ bn an−1 + δn−1 + an and let x = n=1 S xn. Then

∞ X 1 kxk ≤ < ∞ 2n n=1 and x is a well deﬁned vector in `2. Notice that if T = 2B, then TS is the identity and

bn T xi = 0 for all i ≤ n − 1. Hence

∞ bn X bi−bn kT x − xnk = kxn + S xi − xnk i=n+1 ∞ X kxik ≤ 2ai i=n+1 ∞ X 1 ≤ 2i i=n+1 1 = , 2n and so the vector x is hypercyclic.

Let σ : N → N be a permutation and (ξn) and (αn) be the sequences given by ξn = xσ(n) and αn = aσ(n). If we deﬁne βn = α1 + δσ(1) + ··· + αn−1 + δσ(n−1) + αn then the vector

P∞ βn ξ = n=1 S ξn is still hypercyclic. Indeed since

∞ 2 ∞ 2 X kxnk X kξnk = , 2an 2αn n=1 n=1 the vector ξ is hypercyclic. The sequence (xn) has a subsequence (nk) in N and a subsequence

(x ) with x = (x1 , 0, 0, ··· ) and 2nk−1 < kx k ≤ 2nk , and a = 2n for all k and nk nk nk nk nk k also n1 6= 1. We construct a new sequence (yn) obtained by permuting the terms of the sequence (xn). Define y1 = x1. For all j∈ / {nk : k ∈ N} define yj = xl and αj = al, where l = min{i ∈ N : xi ∈/ {y1, ··· , yj−1}} and this condition makes our new sequence (yj) a permutation of our original sequence (x ). Define y = x , where k ∈ {n : i ∈ } is i nj nkj j i N such that a ≥ α + δ + ··· + α + δ , and |y1 + y1 + ··· + y1 | ≥ |y1 | for all j ∈ . nkj 1 1 nj −1 nj −1 1 2 nj nj N 19 Also define α = a . We get that for any j there is a k such that nj nkj j

nk ky k ky k kxnk k 2 j nj ≥ nj = j ≥ > 1, for all n ≥ 6. β 2α 2a 8n k nj nj nkj kj

Now the vector y is hypercyclic in `2 and for any j large we have

β | < y + ··· + T nj y, e > | |y | ky k 1 ≥ nj = nj > 1. βnj βnj βnj

β ky + ··· + T nj yk So cannot converge to 0 and we constructed a hypercyclic vector without βnj y such that its hypercyclic means Any do not converge to zero.

Proposition 18. Let x be a vector in a Banach space X and T ∈ B(X). If An,αx converges to a vector a in X, then a is a periodic point of multiplicity α, that is T αa = a.

α α Proof. Suppose An,α(x) converges to a, then T (An,αx) converges to T a. On the other hand we have:

α α T (An,αx) = An,αT x 1 = (T αx + ··· + T nαx) n x 1 = − + (x + T αx + ··· + T nαx) n n x n + 1 = − + A x n n n+1,α → 0 + 1a

= a

Hence, T αa = a.

Now we have all the tools necessary to state and prove the main result of this chapter. Chaotic operators satisfy the Hypercyclicity Criterion, see [6]. Here, we impose an extra condition such that they satisfy Kitai’s version of the criterion stated above. We prove that 20 for any periodic point of a chaotic operator there is hypercyclic vector whose hypercyclic mean converges to this periodic point.

Theorem 19. Approximation of a periodic point by a hypercyclic vector Let X be a Banach space, and T : X → X be a chaotic operator that satisﬁes the Kitai’s Hypercyclicity Criterion. If z is a periodic point of multiplicity α, that is T αz = z then there exists a hypercyclic vector of T such that An,αx converges to the periodic point z.

Proof. First note that z is a ﬁxed point vector for An,α since

1 nz A z = (z + T αz + ··· + T (n−1)αz) = = z. n,α n n

Also note that by Proposition 16 there exists a hypercyclic vector y such that An,αy → 0. We deﬁne the vector x by x = z − y. Then we have that

An,αx = An,α(z − y) = An,αz − An,αy = z − An,αy.

Since An,αy converges to 0, An,αx must converge to z. To show that x is a hypercyclic vector for T we ﬁrst use a result of Ansari [1] that T α is hypercyclic whenever T is. Let f be a vector in X and let > 0. Since z − f ∈ X and since

αnk y is hypercyclic there exists (nk) so that kT y − (z − f)k < . Now we have that

kT αnk x − fk = kT αnk (z − y) − fk = kT αnk y − (z − f)k < and we show that we can obtain a periodic point by using a hypercyclic vector.

We have seen some properties of hypercyclic vectors and periodic points of a hypercyclic operator. These two sets of vectors are seemingly to have totally totally diﬀerent properties, but nonetheless, they can co-exist as interesting properties of an operator; each of these two sets of vectors is dense. Theorem 19 provided an interesting way of approximating vectors in one set by means of vectors in another. 21

CHAPTER 3

Extensions to chaotic operators

In this chapter we prove some results concerning extensions of bounded linear operators to hypercyclic operators. First we show that any bounded linear operator can be extended to a chaotic operator. Grivaux proved a somewhat similar result; that is, any bounded linear operator on a Banach space has a topologically transitive extension, see [14]. We start with a uniformly bounded sequence of operators (Tj)j on a Banach space X. We construct a larger Banach space X˜ which contains X as a complemented subspace, say X˜ = X L Y, ˜ ˜ ˜ where Y is a closed subspace of X and we denote by Tj : X → X, the operator obtained by ˜ extending Tj : X → X to X by deﬁning it to be identically zero on Y . We show that there ˜ ˜ ˜ ˜ exists and an operator T on X such that the operator obtained by adding T to any Tj is a chaotic operator on X.

Theorem 20. Simultaneous hypercyclic extension Let X be an inﬁnite dimensional separable Banach space. Given a uniformly bounded sequence (Tj) of linear operators in B(X), there exists an inﬁnite dimensional separable Banach space X˜ and a bounded linear operator on T˜ : X˜ → X˜ such that the following are true:

1. X is a complemented subspace of X˜

˜ ˜ 2. operators T + Tj have a common hypercyclic vector 22 ˜ ˜ ˜ 3. (T + Tj)|X = Tj|X for each j

˜ ˜ 4. T + Tj is a chaotic operator for each j.

Proof. We deﬁne the vector space X˜ to be the direct sum of inﬁnite many copies of X; that is, we let each Xi = X and let

∞ ∞ ∞ ˜ M Y X X = Xi = {x ∈ Xi : kxk = kxˆ(i)k < ∞}, i=0 i=0 i=0

˜ where we denote x = (ˆx(0), xˆ(1),...) with eachx ˆ(i) ∈ Xi = X. The vector space X is a P∞ ˜ Banach space with the norm given by kxk = i=0 kxˆ(i)k, for all x ∈ X. For simplicity we use the same notation for the two diﬀerent norms since it is obvious from the context to which one we refer. Now we deﬁne a positive weight a by

a = 3(sup(kTjk) + 1) and the operator T˜ : X˜ → X˜ by

T˜(ˆx(0), xˆ(1), xˆ(2), xˆ(3),...)) = (axˆ(1), axˆ(2), axˆ(3),...), for any x = (ˆx(0), xˆ(1),...) ∈ X˜. Clearly T˜ is bounded with kT˜k = a. From now on we identify X/X˜ with the closed subspace of X˜ that consists of vectors x in X˜ of the form ˜ ˜ ˜ x = (0, xˆ(1), xˆ(2),...). We extend Tj : X → X trivially to an operator Tj : X → X given by

 T x, whenever x ∈ X ˜  j Tjx =  0, whenever x ∈ X/X.˜

By taking X0 to be the original given Banach space X, we see condition (3) is satisﬁed and ˜ ˜ since T |X0 = 0 it is clear that the above extension of Tj on X to the whole Banach space X 23 remains a bounded linear operator on X˜. The operator T˜ on X˜ has a norm a because for any x = (ˆx(0), xˆ(1),...) in X˜, we have

kT˜ xk = kT˜(ˆx(0), xˆ(1),...)k

= kaxˆ(1), axˆ(2),...)k

= akxˆ(1), xˆ(2),...)k

≤ kxk,

furthermore for any x = (0, xˆ(1), xˆ(2),...) in X/X˜

kT˜ xk = kT˜(0, xˆ(1), xˆ(2),...)k

= kaxˆ(1), axˆ(2),...)k

= akxˆ(1), xˆ(2),...)k

= kxk.

˜ ˜ Thus T + Tj is also a bounded linear operator for each j. ˜ ˜ We remark here that, we can have a better uniform estimate for the norm of kT + Tjk. ˜ ˜ We claim that kT + Tjk = a for each j. To prove that, we let x = (ˆx(0), xˆ(1), xˆ(2),...) be a vector in X,˜ then

˜ ˜ kT + Tjk = k(Tjxˆ(0) + axˆ(1), axˆ(2), axˆ(3))k ∞ X = kTjxˆ(0) + axˆ(1)k + akxˆ(j)k. j=2 24 Using the triangle inequality we get

∞ ˜ ˜ X kT + Tjk ≤ kTjxˆ(0)k + kaxˆ(1)k + a kxˆ(j)k j=2 ∞ X ≤ kTjkkxˆ(0)k + akxˆ(1)k + kxˆ(j)k j=2 ∞ X ≤ akxˆ(0)k + akhatx(1)k + akxˆ(j)k j=2 ∞ X ≤ a kxk. j=0

˜ ˜ ˜ ˜ Thus kT + Tjk ≤ a and since k(T + Tj)(0, xˆ(1), 0, 0, 0,...)k = akxˆ(1)k we conclude that ˜ ˜ kT + Tjk = a. It is easy to see that conditions (1) and (3) are satisfied by the way we constructed the Banach space X˜ and the operator T˜. A vector x = (ˆx(0), xˆ(1), xˆ(2),...) is said to have finite length if there is a nonnegative integer K such thatx ˆ(j) = 0, whenever j ≥ K. The smallest such integer K is called the length of the vector x. Fixing a countable dense subset ∆ of X that contains the origin, we let D be the countable subset of X˜ consisting of vectors x = (ˆx(0), xˆ(1), xˆ(2),...) of finite length and with each componentx ˆ(j) ∈ ∆. We now show that D is a dense subset of the Banach space X˜. For that D is dense let x ∈ X and let > 0. Then since kxk < ∞ there is a positive integer k

P∞ such that i=0 kxˆ(i)k < 2 . For every i there exists a yi ∈ ∆ such that kyi − xˆ(i)k < 2(k+1) . ˆ ˆ ˆ ˆ ˆ We deﬁne a vector d = (d(0), d(1), d(2),...) in D by d(i) = yi for i ≤ k and d(i) = 0 for i > k. Thus we found a vector in D that approximates x, indeed:

k ∞ X X kx − dk < kyi − xˆik + kxˆ(i)k i=0 i=0 (k + 1) (3.1) < + 2(k + 1) 2 = 25 ˜ and D is a dense subset of X. Let D = {d1, d2, d3,...} be an enumeration of D. ˜ Consider the countable set of all possible ordered pairs {(Ti, dj): i, j ≥ 1} and let ˜ {p1, p2, p3,...} be an enumeration of all such ordered pairs with p1 = (T1, d1). Let lk be the

length of the vector in the second entry of the ordered pair pk. For each pair pk we will

construct a vector xk and a positive integer nk. −1 ˆ −1 ˆ −1 ˆ First let n1 = 1 and x1 = (0, a d1(0), a d1(1), . . . , a d1(l1), 0, 0,...) and observe that ˜ T x1 = d1.

Suppose we have constructed xj and nj for all j ≤ k. Now we construct nk+1 and xk+1 ˜ corresponding to the pair pk+1 = (Tu, dv), where dv has length lk+1.

Let nk+1 be large enough such that:

kdvk kxˆk(0)k kxˆk(1)k kxˆk(nk + lk))k n + n + n −1 + ··· + n −(n −l ) a k+1 3 k+1 3 k+1 3 k+1 k k (3.2) 4 1 < ( a)−nk−1 , 3 2k+1

which is possible because nk+1 appears on the left-hand side and nk appears on the right-hand

side. Now we construct xk+1 such that the following conditions are satisﬁed:

xˆk+1(i) =x ˆk(i), whenever 0 ≤ i < nk+1

1 ˆ nk+1 nk+1−1 xˆk+1(nk+1) = [dv(0) − (Tu xˆk(0) + aTu xˆk(1) + ··· ank+1

nk+lk nk+1−(nk+lk) + a Tu xˆk(nk + lk))] (3.3) 1 ˆ nk+1 nk+1 = [dv(0) − (Tu xˆk+1(0) + aTu xˆk+1(1) + ··· ank+1

nk+lk nk+1−(nk+lk) + a Tu xˆk+1(nk + lk))]

dˆv(i) xˆk+1(nk+1 + i) = ank+1 , whenever 1 ≤ i ≤ lk+1

xˆk+1(i) = 0, whenever i > nk+1 + lk+1 26 Now we deﬁne ∞ X x = x1 + (xk+1 − xk), k=1 and notice that for each m ∈ N,

∞ X x = xm + (xk+1 − xk). k=m

To show that x is well-deﬁned we observe that by (3.2):

1 ˆ nk+1 nk+1−1 kxk+1 − xkk ≤ (kdv(0)k + kTu xˆ(0)k + akTu xˆ(1)k + ··· ank+1

nk+lk nk+1−(nk+lk) + a kTu xˆ(nk + lk)k ˆ ˆ ˆ + kdv(1)k + kdv(2)k + ··· + kdv(lk+1)k)

nk+1 nk+1−1 kdvk kTuk kxˆk(0)k kTuk kxˆk(1)k ≤ n + n + n −1 + ··· (3.4) a k+1 [3(kTuk + 1)] k+1 [3(kTuk + 1)] k+1

nk+1−(nk+lk) kTuk kxˆk(nk + lk)k + n −(n +l ) [3(kTuk + 1)] k+1 k k kd k kxˆ (0)k kxˆ (1)k kxˆ (n + l )k ≤ v + k + k + ··· + k k k ank+1 3nk+1 3nk+1−1 3nk+1−(nk+lk) 1 4 < ( a)−nk−1 2k+1 3 and the sum in the deﬁnition of x converges. Hence x is well-deﬁned. ˜ We now show that for any triple (k, i, j), or in other words a pair pk = (Ti, dj) we have that

˜ ˜ nk (T + Ti) xk = dj

˜ Indeed, using the deﬁnition of the vector x, the deﬁnitions of T and Ti and also the fact thatx ˆk(i) =x ˆk−1(i), whenever 0 ≤ i < nk, andx ˆk(i) = 0 whenever i > nk + lk, a short 27 computation yields:

˜ ˜ nk nk nk−1 nk−1 nk (T + Ti) xk = (Ti xˆk(0) + aTi xˆk(1) + ··· + a Tixˆk(nk − 1) + a xˆk(nk),

nk nk a xˆk(nk + 1), . . . , a xˆk(nk + lk), 0, 0,...)

nk nk−1 nk−1 = (Ti xˆk(0) + aTi xˆk(1) + ··· + a Tixˆk(nk − 1) (3.5) 1 n nk ˆ nk k−1 + a [dj(0) − (Ti xˆk(0) + aTi xˆk(1) + ··· ank n −(n +l ) nk−1+lk−1 k k−1 k−1 ˆ ˆ + a Ti xˆk(nk−l − lk−1))], dj(1), ··· , dj(lk))

= dj.

˜ ˜ To show that x is a common hypercyclic vector we ﬁx Ti0 . Let > 0 and y ∈ X. Then

1 there exists a K such that for any k > K we have: 2k < 2 . Now we consider the inﬁnite ˜ family of pairs pk = (Ti0 , dj0 ) with k > K and choose pk such that: kdj0 − yk < 2 . Since:

∞ ˜ ˜ nk ˜ ˜ nk ˜ ˜ nk X k(Ti0 + T ) xk − dj0 k ≤ k(Ti0 + T ) xk − dj0 k + k(Ti0 + T ) (xs+1 − xs)k s=k ∞ ˜ ˜ nk X ≤ 0 + kTi0 + T k kxs+1 − xsk s=k ∞ a X ≤ (( − 1) + a)nk kx − x k 3 s+1 s s=k (3.6) ∞ 4 X 1 4 ≤ ( a)nk ( a)−ns−1 3 2s+1 3 s=k 1 = 2k < 2

and

˜ ˜ nk ˜ ˜ nk k(Ti0 + T ) xk − yk ≤ |(Ti0 + T ) xk − dj0 k + kdj0 − yk < we get that x is a common hypercyclic vector. ˜ ˜ It remains to be shown that for each Tj there exists a dense subset in X of periodic 28 ˜ vectors. For this we ﬁx Tu and we construct for each vector d in D a sequence of periodic ˆ ˆ ˆ vectors xn that converges to d. Let d = (d(0), d(1),..., d(l), 0, 0 ...) be a vector in D and for

each n > l deﬁne xn = (ˆxn(0), xˆn(1),...) by: ˆ xˆn(0) = d(0)

dˆ(j−kn) xˆn(j) = akn , whenever kn + 1 ≤ j ≤ kn + l and k ≥ 0 dˆ(0) − Pl aiT n−idˆ(i) xˆ (j) = i=0 u , whenever j = kn and k ≥ 1 n akn

xˆn(j) = 0, otherwise.

Notice that for any positive integer m and for any vector y = (ˆy(0), yˆ(1), yˆ(2),...),

˜ ˜ m m k−1 k k (T + Tu) y = (Tu (ˆy(0)) + ··· + a Tuyˆ(k − 1) + a yˆ(k), a yˆ(k + 1),...)

and thus by taking m = n and y = xn, and directly from the deﬁnition of xn we get,

˜ ˜ n (T + Tu) xn

n n−1 n n n = (Tu xˆn(0) + ··· + a Tuxˆn(n − 1) + a xˆn(n), a xˆn(n + 1), a xˆn(n + 2),...) dˆ(0) − Pl aiT n−1dˆ(i) dˆ(n + 1 − n) = (T nxˆ (0) + ··· + an−1T xˆ (n − 1) + an i=0 u , an ,...) u n u n an an+1 dˆ(1) = (dˆ(0), ,...) a

= xn

for each n > l and we constructed a sequence (xn) of periodic vectors. We show now that the sequence (xn) converges to d: 29

∞ l l X dˆ(0) − P aiT n−idˆ(i) X dˆ(i) kx − dk ≤ k ( i=0 u + )k n ajn ajn j=1 i=1 ˆ Pl i n−i ˆ l ˆ kd(0)k + a kTuk kd(i)k X kd(i)k ≤ i=0 + an−1 an−1 i=1 → 0 as n → ∞.

(k) Let D = {dk : k ≥ 1} be an enumeration of the dense subset D and for each k let (xn )n be

(k) a sequence of periodic points converging to dk, then {xn : k ≥ 1, n > lk} is a dense subset ˜ ˜ of periodic points and thus (T + Tu) is a chaotic operator.

Statement (2) in the conclusion of Theorem 20 follows from statement (4). The reason is that the set of hypercyclic vectors HC(A) of any hypercyclic operator A : Y → Y on a ˜ ˜ Banach space Y is always a dense Gδ subset of Y ; see [17], and hence if T +Tj is hypercyclic, T ˜ ˜ then certainly the set of common hypercyclic vectors HC(T + Tj) is nonempty by the Baire Category Theorem. However, the constructive proof that we presented for Theorem 20 tells us exactly how such a vector may look like, which potentially leads to further study the properties of those hypercyclic vectors. Next we observe that when we construct T˜ in ˜ ˜ L∞ the above proof of Theorem 20, its deﬁnition T on X/X = i=1 Xi is a shifting operator which does not depend on the actions of any individual operator Tj on X. The only part of ˜ T that depends on Tj is the shifting constant a, which is 3(sup(kTjk) + 1). Thus, we reach the following conclusion.

Corollary 21. Let X be an inﬁnite dimensional separable Banach space. For any given R > 0, there exists an inﬁnite dimensional separable Banach space X˜ and a bounded linear operator T˜ on X˜ such that:

1. X is a complemented subspace of X˜

2. whenever A : X → X is a bounded linear operator with kAk < R, the operator (T˜ + A˜) 30 is a chaotic operator on X˜

3. (T˜ + A˜)|X = A˜|X, where A˜ : X˜ → X˜ is the trivial extension of A : X → X.

Another review of the proof of Theorem 20 also provides the following statement in addition to the three statements in the above Corollary.

Corollary 22. Each operator T˜+A˜ in Corollary 21 satisﬁes Kitai’s Hypercyclicity Criterion; that is, Theorem 14.

˜ L∞ P∞ Proof. We identify X as i=0 Xi = {(x0, x1,...): x ∈ X and i=0 kxik < ∞} and the ˜ ˜ ˜ ˜ original space X as X0 and so A : X → X is deﬁned by A(x0, x1,...) = (Ax0, 0, 0,...). Let S : X˜ → X˜ be deﬁned by

1 S(h , h , h ,...) = (0, h , h , h ,...). 0 1 2 a 0 1 2

n Since a > 1, limn→∞ S (h0, h1, h2,...) = 0. Note that

˜ ˜ (T + A)S(h0, h1, h2,...) 1 1 1 = (T˜ + A˜)(0, h , h , h ,...) (3.7) a 0 a 1 a 2

= (h0, h1, h2,...).

Thus (T˜ + A˜)S is the identity on X.˜ It suﬃces to ﬁnd a dense subset D of X˜ so that

lim(T˜ + A˜)nd = lim Snd = 0

for each d in D. For that we let ∆ be a countable dense subset of X = X0 = X1 = X2 = ··· . ˜ Note that C = {(d0, d1, d2, ..., dk, 0, 0, 0,...) ∈ X : di ∈ ∆, k ∈ N} is a countable dense subset 31 ˜ of X, also that if n ≥ k and x = (d0, d1, d2, ..., dk, 0, 0, 0,...) ∈ C then

˜ ˜ n (T + A) (d0, d1, d2, ..., dk, 0, 0, 0,...) (3.8) n−k k k−1 k = (A (A d0 + aA d1 + ··· + a dk), 0, 0, 0,... ).

We now observe that if n ≥ k then

Sn(T˜ + A˜)nx

n ˜ ˜ n = S (T + A) (d0, d1, d2, ..., dk, 0, 0, 0,...) (3.9) 1 = (0, 0,..., 0, An−k(Akd + aAk−1d + ··· + akd ), 0, 0, 0,...) an 0 1 k in which the nonzero expansion appears as a member of Xn−1. Hence,

(T˜ + A˜)n(x − Sn(T˜ + A˜)nx)

= (T˜ + A˜)nx − (T˜ + A˜)nSn(T˜ + A˜)nx (3.10) = (T˜ + A˜)nx − (T˜ + A˜)nx

= 0, and thus for all m ≥ n, (T˜ + A˜)m(x − Sn(T˜ + A˜)nx) = 0, and trivially

lim (T˜ + A˜)m(x − Sn(T˜ + qa)nx) = 0. m→∞

Note also that for any x ∈ C

lim kSn(T˜ + A˜)nxk n→∞

1 n−k k k−1 k = lim kA (A d0 + aA d1 + ··· + a dk)k (3.11) n→∞ an = 0.

Since C is dense in X˜, the set D = {x − Sn(T˜ + A˜)nx : x ∈ C} is also dense in X˜. Since 32 we have shown that if y ∈ D, then lim Sny = lim(T˜ + A˜)ny = 0, the hypothesis of Kitai’s Hypercyclicity Criterion is satisﬁed.

Although we have shown in the proof of Theorem 20 that the operator T˜ + A˜ has a dense set of periodic points, it is worth a few lines here to reprove it using the ideas and notation in the proof of Corollary 22. Using the same set D as in the proof of Corollary 22, we let x = (d0, d1, d2, . . . , dk, 0, 0, 0,...) be a vector in D. We are to show that for any > 0, there is a periodic point y with kx − yk < . We have shown in the proof of Corollary 22 that

1 kSn(T˜ + A˜)n(x − Sn(T˜ + A˜)nx)k ≤ kAkn−kkAkd + ··· + akd k (3.12) an 0 k and that (T˜ + A˜)n(x − Sn(T˜ + A˜)nx) = 0. (3.13)

Since x is ﬁxed and a > kAk, there is an integer N1, such that if n > N1 then

kSn(T˜ + A˜)nxk 1 < kAkn−kkAkd + ··· + akdkk (3.14) a2n 0 < . 2

n ˜ ˜ n By (3.12) we can further assume that N1 is chosen large enough so that kS (T + A) xk < 1.

We can now chose an integer N2 > N1 so that

∞ X 1 1 = amn an − 1 m=1 (3.15) < 2(kxk + 1)

To construct a periodic point y in the ball Ball(x, ), we let

∞ X y = x − Sn(T˜ + A˜)nx + Snm(x − Sn(T˜ + A˜)nx), m=1 33 where n is a ﬁxed integer with n > N2.

It folllows that if n > N2, then

ky − xk ∞ X 1 ≤ kSn(T˜ + A˜)nxk + kx − Sn(T˜ + A˜)nxk amn m=1 ∞ X 1 < + (kxk + 1) (3.16) 2 amn m=1 < + (kxk + 1) 2 2(kxk + 1) =

Thus y ∈ Ball(x, ). To show that y is a periodic point, we observe that

(T˜ + A˜)ny ∞ X = (T˜ + A˜)nx − (T˜ + A˜)nSn(T˜ + A˜)nx + Sn(m−1)(x − Sn(T˜ + A˜)nx) m=1 ∞ (3.17) X = Smn(x − Sn(T˜ + A˜)nx) m=0 = y.

The simultaneous chaotic extension is stated for a countable set of prescribed operators on a Banach space X. But if the countable set has precisely one operator, then we can have a very precise statement.

Corollary 23. Chaotic extension for Banach spaces Let X be a separable Banach space, and T be a bounded linear operator on X. Then there is an an operator T¯ on a Banach space X˜ such that the following conditions are satisﬁed:

(i) T¯ : X˜ → X˜ is chaotic

(ii) X is a closed subspace of X˜

(iii) T¯|X = T 34 ˜ L∞ L∞ Proof. We apply the previous theorem to the operator T to ﬁnd T : i=0 Xi → i=0 Xi, ¯ ¯ ˜ where each Xi = X, then we deﬁne the operator T by T = T + T , where T x = x for L∞ x ∈ X = X0 and T x = 0 for x ∈ i=1 Xi. Note that the original X in the statement of the

corollary is identiﬁed with X0. It is now easy to see that the second condition is satisﬁed by the way we constructed the Banach space X˜. First and third conditions follow directly from third and fourth conditions in the theorem:

¯ ¯ ˜ T |X = T |X0 = (T + T )|X0 = T |X0 = T |X = T

and T¯ = T˜ + T is a chaotic operator.

As a consequence of the above result, if T is a hypercyclic operator on a Banach space X that does not satisfy the Hypercyclicity Criterion, then it has an extension that satisfies the Hypercyclicity Criterion. This is because every chaotic operator satisfies the Hypercyclicity Criterion, as proved by Bes and Peris [6]. For Hilbert spaces we can prove a more general result. We extend an operator defined on M, a subspace with infinite codimension of a Hilbert space H.

Corollary 24. Chaotic extension on a Hilbert space Let H be an inﬁnite dimensional separable Hilbert space and let M be a closed subspace with dim H/M = ∞. If T : M → M is a bounded linear operator, then T has a chaotic extension T¯ : H → H.

Proof. Since all separable infinite dimensional Hilbert spaces are isomorphic we may write L∞ H = i=0 Mi, where Mi = M and for each i, M0 is identified with the original M given in the statement of the corollary. Using the previous corollary together with the proof of the chaotic extension theorem with M instead of X and H instead of X˜ we get the desired extension T¯ = T˜ + T . 35 Remark 25. We observe that the condition that the codimension of the subspace M is infinite is required. Indeed, suppose that dim H/M < ∞ and T˜ is a hypercyclic extension. Then we consider the linear map S : H/M → H/M defined by S[x] = [T˜ x]. If h is a hypercyclic vector for T˜, then {h, T˜ h, T˜2h, . . .} is dense in H. If in addition, π : H → H/M is the quotient map defined by πh = [h] then π is a continuous linear map. Thus {πh, π(T˜ h), π(T˜2h),...} is dense in H/M. That is, {[h],S[h],S2[h],S3[h],...} is dense in H/M, which means the linear map S : H/M → H/M is hypercyclic. But it is impossible because dim H/M < ∞.

As a corollary of Remark 25, the same idea shows that no hypercyclic operator can have an invariant subspace of ﬁnite codimension. To see that, we let T : H → H be a hypercyclic operator and M be an invariant subspace of T . If h ∈ H is a hypercyclic vector for T and π : H → H/M is the quotient map, then π{h, T h, T 2h, . . .} = {[h], [T h], [T 2h],...} is dense in M/H. But then it would follow that the linear map A :[x] = [T x] is a hypercyclic map on a ﬁnite dimensional space, which is impossible; see [17]. For our next result we use the fact that an operator satisfying the Hypercyclicity Criterion and an extra condition has a hypercyclic subspace. We state here this result (theorem 3.6 in [9]).

Theorem 26. Suppose T : H → H is a bounded linear operator satisfying the Hypercyclicity Criterion. If H has a closed inﬁnite dimensional subspace K such that for every vector f in K lim kT nfk = 0, n→∞

then H has an inﬁnite dimensional subspace consisting entirely, except for zero, of hypercyclic vectors of T .

Using Theorem 20 and our previous results, we obtain the following corollary.

Corollary 27. Let H be an inﬁnite dimensional separable Hilbert space and let M be a closed subspace of H with dim H/M = ∞. If T : M → M is a bounded linear operator, then T has a chaotic extension with a hypercyclic 36 subspace; i.e. a closed inﬁnite dimensional subspace consisting entirely of hypercyclic vectors, except the zero vector.

Proof. Let N be an inﬁnite dimensional subspace in M ⊥ such that dim H/(N L M) = ∞. L We deﬁne the bounded linear operator T1 on (N M) by

  0, whenever h ∈ N T1h =  T h, whenever h ∈ M

˜ The previous corollary provides T1 with a chaotic extension T : H → H which satisﬁes ˜ the Hypercyclicity Criterion [6]. Since T1 = 0 on N the extension T = 0 on an inﬁnite dimensional subspace N. By the previous theorem T˜ has a hypercyclic subspace.

As a consequence of the previous result, we have the following corollary. Here we let

B(H) be the algebra of all bounded linear operators T : H → H and Bch(H) be the subset of chaotic operators having a hypercyclic subspace.

Corollary 28. SOT-denseness of Bch(H)

The set Bch(H) is dense in B(H) in the strong operator topology.

Proof. Let > 0 and let A ∈ B(H). Let h1, h2, . . . , hk ∈ H and let U = {T ∈ B(H): k(T −

A)hik < , for all i = 1, . . . , k} be an SOT-basic open set. Let M = span{h1, h2, . . . , hk}

be an ﬁnite dimensional subspace spanned by h1, h2, . . . , hk. So, M is closed. By Corollary 27, there exists a chaotic operator T ∈ B(H) such that T = A on M and hence T ∈ U.

The above result was first obtained by Chan [9]. Our chaotic extension result of Corollary 27 is a generalization of his result. One may wonder what will happen if the Hilbert space H in Corollary 22 is a Banach space X. In that case, we cannot expect to have a chaotic extension because not every Banach space can support a chaotic operator. This result was proved by Bonet, Martinez-Gimenez and Peris [8]. One may then ask whether we can have a hypercyclic extension to X instead, because Ansari [2] and Bernal-González[5] proved that 37 every Banach space supports a hypercyclic operator. However our techniques in Corollary 24 cannot be generalized to a Banach space setting. Clearly Theorem 20 shows us how to extend a bounded linear operator T on a larger Banach space containing X. If we are given a Banach space X and a bounded linear operator T defined on a closed subspace M of X, then it will be much harder to extend T to be a hypercyclic operator on X, analogous to the Hilbert space result in Corollary 24. There is a historical reason for that. The first hypercyclic operator on a Hilbert space was obtained by Rolewicz [23], who also asked whether every Banach space supports a hypercyclic operator. This problem was open for about thirty years, until Ansari [2] and Bernal-González[5] gave a positive answer. Since the Banach space extension analogous to Corollary 24 is asking for a lot more, it potentially is a much harder question. 38

CHAPTER 4

Compressions of hypercyclic operators

We now turn our attention to the compression of a hypercyclic operator. We show in Theorem 30 below that a compression of a hypercyclic operator onto a closed subspace of infinite codimension can coincide with any given operator on the subspace. Let H be an infinite dimensional separable Hilbert space, and M be a closed subspace of H with infinite L∞ codimension. For all i ∈ Z, let Mi = M and we write H = −∞ Mi, where Mi ⊥ Mj

whenever i 6= j. We identify the original closed subspace M with M0 in H. Let T : M → M be a bounded linear operator and let a > 0. Let {ej1, ej2, ej3,...} be an orthonormal basis

of Mj. Hence eji ⊥ emn whenever (j, i) 6= (m, n). Let S : H → H be the unitary operator

given by Sej,i = ej+1,i whenever i, j ∈ Z, then S|Mj is an isomorphism from Mj to Mj+1. We now ﬁrst need a lemma.

Lemma 29. Let H be an inﬁnite dimensional Hilbert space, and M be a closed subspace of

H with inﬁnite codimension. Let k be a positive integer and for each i with |i| ≤ k let wi

be a positive weight such that 0 < wi ≤ a and let xi and yi be vectors in Mi. Then for any > 0 there exists an integer n ≥ 2k + 2 and for all integers i with k + 1 ≤ |i| ≤ n + k

there exist weights wi with 0 < wi ≤ a and vectors xi ∈ Mi (in fact xi = 0 whenever 39 −n − k ≤ i ≤ −k − 1) such that if D is deﬁned by

  wjej+1,i whenever i ∈ Z and |j| ≤ n + k Dej,i = (4.1)  0 otherwise

then n X 1. (D + T ) xi <

−k≤i≤n−k−1

X 2 2. kxik < k+1≤i≤n+k

n+k k n X X 3. (D + T ) xi = yi i=n−k i=−k Proof. Suppose D is given in the form of (4.1) and n is the integer given in its deﬁnition. We ﬁrst make the following observations. for any integer m with k + 1 ≤ m and m + k + 1 ≤ n and xm ∈ Mm, we have m X (D + T ) xi + xm ⊥ Mj −k≤i≤k whenever j ≥ 1. In fact, k X (D + T ) xi ⊥ Mj |i|≤k

whenever j ≥ 1.

Hence if we deﬁne wi = a for all integers i with k +1 ≤ i ≤ m and if we take P0 : H → H to be the orthogonal projection onto M0, and if we let

−1 m m X xm = S P0(D + T ) xi w1 ··· wkwk+1 ··· wm |i|≤k (4.2) −1 m m−k k X = S P0T (D + T ) xi. w1 ··· wkwk+1 ··· wm |i|≤k 40 Then k X (D + T ) xi ⊥ Mj whenever j ≥ 0. (4.3) |i|≤k

Since wk+1 = ··· wm = a > kT k, we can first choose m large enough, before we determine the integer n in the definition of D satisfying n ≥ m + k + 1 so that the definition of xm gives kxmk < √ (4.4) 2

−1 Then for large enough integer n ≥ m + k + 1, if we deﬁne w−k−1 = ··· = w−n−k = a < 1, then it will follow from (4.3) that

n X (D + T ) xi + xm < .

|i|≤k

If we define xk+1 = ··· xm−1 = 0 and xm+1 = ··· = xn−k−1 = 0, then (1) is clearly satisfied. It remains to show that if n is chosen large enough, then (2) and (3) can be satisfied too. For that, we first check how (3) can be satisfied whenever n > 2k by choosing appropriate xn−k, ··· , xn+k. Let wm+1 = ··· = wn+k = a. Let

1 n xn−k = S y−k. w−k+1 ··· wn−k

Furthermore for every integer i with −k + 1 ≤ i ≤ 0 let

1 n xn+i = (S yi−1). wi+1 ··· wn+i

Finally for every integer i with 1 ≤ i ≤ k let

1 n+i xn+i = S yi wi+1 ··· wn+i

To verify that the above deﬁnitions of xn−k, ··· , xn+k satisfy (3),we need to show an inter- 41 mediate step that whenever −k ≤ i ≤ 0,

n+i 1 −i (D + T ) xn+i = S yi ∈ M0, wi+1 ··· w0

which follows easily from the fact that (D + T )|Mi = D|Mi if i 6= 0. Now one can verify (3) inductively for the increasing order of the integer i. Since

wm+1 = ··· = wn+k = a > 1

, and S is an isometry, we can assume that n is chosen large enough so that the deﬁnitions

for xn−k, . . . , xn+k give X kx k2 < . 1 2 n−k≤i≤n+k

This inequality, along with (4.4) and the fact that xk+1 = ··· = xm = xm+1 = ··· = x−n−k =

0, yields (2). The whole proof of the lemma is ﬁnished by letting x−k−1 = ··· = x−n−k = 0.

Using this technical lemma we are now ready to prove our main result on compressions.

Theorem 30. Let H be an inﬁnite dimensional Hilbert space, and M be a closed subspace of H with inﬁnite codimension. Let P : H → H be the orthogonal projection onto M. If T : M → M is a bounded linear operator, then there is an operator A : H → H such that

1. both A and A∗ are hypercyclic

2. P AP |M = T

3. PA∗P |M = T ∗

Proof. Since M has inﬁnite codimension in H, we can use an orthonormal basis argument

∞ to write H as an orthogonal sum H = ⊕j=−∞Mj = {(. . . , m−2, m−1, m0, m1, m2,...): mi ∈

Mi}, where M0 = M, and each Mj is isomorphic to M0. 42 In the rest of the proof, we assume dim M = ∞ and the same argument works if dim M is finite. For that, we let {e 1 , e 2 , e 3 , ···} be an orthonormal basis of M . Since T takes M j1 j1 j1 j to M, we can view T as an operator from H to H with T |M ⊥ = 0. This view allows us to define an operator A : H → H by A = B + T , where B : H → H is a linear map defined by Bej,i = wjej−1,i for all integers i, j and {wj : j ∈ Z} is a bounded two-sided sequence of positive numbers. Thus B takes each Mj to Mj−1 and B|Mj is simply wj times a Hilbert space isomorphism. In fact, if S : H → H is the unitary operator given by Sej,i = ej+1,i

−1 whenever i, j ∈ Z, then S|Mj is an isomorphism from Mj to Mj+1, and B|Mj = wjS |Mj.

One can easily verify that B is a bounded linear operator because {wj} is bounded. In fact, B is a bilateral weighted backward shift with inﬁnite multiplicity, with kBk = sup |wj|.

∗ ∗ ∗ ∗ ∗ ∗ In addition, A = B + T , where B is given by B ej,i = wj+1ej+1,i. Thus B takes Mj

∗ to Mj+1 and B |Mj is simply wj+1 times a Hilbert space isomorphism. Clearly, the above

∗ definition of A gives P AP |M = T and we need to choose wj so that A and A are hypercyclic, in order to finish the proof. Let a be a positive number strictly greater than max{1, kT k}, say a = 2 + kT k and let v1, v2, v3,... be an enumeration of all vectors v with finite number of nonzero coefficients hv, ej,ii all of which are rational. The set of vi is dense in H. Let k1 ≥ 1 such that v1 ∈ M M . With that integer k , we let w = a and x = 0 whenever |i| ≤ k and v = P y , i 1 i i 1 1 |i|≤k1 i |i|≤k1 2k1 −1 with each yi ∈ Mi With 1 = (2a ) the lemma provides an integer n1 and weights wi with 0 < wi ≤ a and vectors xi ∈ Mi, whenever k1 + 1 ≤ |i| ≤ n1 + k1 and in fact xi = 0 for those i in the range −n1 − k1 ≤ i ≤ −k1 − 1 such that if D1 is given by

  wjej−1,i whenever i ∈ Z and |j| ≤ n1 + k1 D1ej,i =  0 otherwise then n1+k1 X 2 1 kxik < k 2a1 i=k1+1 43 and n1−k1−1 1 n1 X n1 X (D1 + T ) xi − v1 = (D1 + T ) xi < 2a2k1 |i|≤n1+k1 i=−n1−k1 Although the operator D provided by the lemma is shifting backward, the techniques in the lemma provide an analogous result for an operator that shifts forward. Hence, we can

0 0 1 continue our deﬁnitions of wi and xi by settling k1 = n1 + k1 and 1 = 2 0 . The lemma 2a k1 0 0 0 0 provides an integer n1 and weights wi with 0 < wi ≤ a whenever k1 + 1 ≤ |i| ≤ n1 + k1, and

0 0 0 vectors xi ∈ Mi whenever k1 + 1 ≤ |i| ≤ n1 + k1 and in fact xi = 0 for those i in the range

0 0 0 k1 + 1 ≤ i ≤ n1 + k1 such that

X 2 1 kxik < 0 . 2k 0 0 0 2a 1 −n1−k1≤i≤−k1−1

0 In addition, the lemma provides an operator D1 shifting forward by

  0 0 0 0 0 wj+1ej+1,i if i ∈ Z and −n1 − k1 ≤ j ≤ n1 + k1 − 1 D1ej,i =  0 otherwise

such that

0 k 1 0 0 0 0 1 ∗ n1 X ∗ n1 X (D + T ) xi − v1 = (D + T ) xi < 0 1 1 2k 1 0 0 0 0 2a |i|≤n1+k1 i=−n1+k1+1

0 0 In the second step, we let k2 = n1 + k1 and assume that, without loss of generality, v2 ∈ L M . In the case that v is not in the space, we can choose v with the least integer |i|≤k2 i 2 j

0 j such that vj is in that space. We can deﬁne operator D2 and then D2 as in the previous

0 0 1 case for v1. Inductively, in the m−th step, we take km = nm−1 + km−1 and m = 2ma2km and we can assume, without loss of generality, that v ∈ L M . By the lemma, we have m |i|≤km i

an integer nm and weights wi with 0 < wi ≤ a, where |i| ≤ nm + km and vectors xi in Mi,

where |i| ≤ nm + km and in fact xi = 0 for these i in the range −nm − km ≤ i ≤ −km − 1 44

such that if Dm is given by

  wjej−1,i whenever i ∈ Z and |j| ≤ nm + km Dmej,i =  0 otherwise

then

X 2 1 kxik ≤ 2ma2km km+1≤i≤nm+km and nm−km−1 nm X nm X 1 (Dm + T ) xi − vm = (Dm + T ) xi < 2ma2km |i|≤nm+km −nm−km

0 0 1 0 Next, we let km = nm + km and m = 0 . by the lemma, we have an integer nm 2ma2km 0 0 and weights wi with 0 < wi ≤ a where |i| ≤ nm + km and vectors xi in Mi whenever

0 0 0 0 0 0 km + 1 ≤ |i| ≤ nm + km and in fact xi = 0 for those i in the range km + 1 ≤ i ≤ nm + km, such that

X 2 1 kxik < 0 (†) m 2km 0 0 0 2 a −nm−km≤i≤−km−1

0 In addition, the lemma gives an operator Dm given by

  0 0 0 0 0 wj+1ej+1,i if i ∈ Z and −nm − km ≤ j ≤ nm + km − 1 Dmej,i =  0 otherwise such that

0 0 ∗ nm X (Dm + T ) xi − vm 0 0 |i|≤nm+km 0 km 0 0 ∗ nm X = (Dm + T ) xi 0 0 i=−nm+km+1 1 < 0 2ma2km 45

with all weights wi with 0 < wi ≤ a and all vectors xi given by the inductive process above, we deﬁne an operator B : H → H by B|M = w S−1|M and the vector x by x = P x . j j j i∈Z i Hence,

nm 2 k(B + T ) x − vmk

nm−km−1 2 ∞ 2 (††) nm X nm X = (B + T ) xi + (B + T ) xi .

i=−∞ i=nm+km+1

We continue our computations with the two summands separately. for the ﬁrst summand,

0 we note that km = nm + km and so it is bounded above by

nm−km−1 2 −nm−km−1 2 nm X 2nm X (B + T ) xi + kB k xi

i=−nm−km i=−∞

∞ −nj −kj −1 1 2nm X X 2 < + a kxik , 2ma2km j=m i=−kj+1

0 0 because xi = 0 whenever −n j − kj = −kj ≤ i ≤ −kj+1 − 1 for some j ∈ Z. Note that

0 ∞ −nj −kj −1 ∞ −kj −1 2nm X X 2 2nm X X 2 a kxik = a kxik 0 0 j=m i=−kj+1 j+m i=−nj −kj ∞ 2nm X 1 < a 0 m 2km j=m 2 a ∞ X 1 < 2m j=m 1 = . 2m−1

Hence the ﬁrst summand is bounded above by

1 1 2makm + 2m−1 → 0, as m → ∞. For the second summand of (††) we note that xi = 0 for all

0 0 index i in the range nj + kj + 1 ≤ i ≤ nj + kj = kj+1 and so the second summand is bounded 46 above by

∞ nj+1+kj+1 2 X X Bnm xi

j=m i=kj+1+1

∞ nj+1+kj+1 2nm X X 2 ≤kB k kxik

j=m i=kj+1+1 ∞ 2 X 1 ≤a nm ( ) 2ja2kj j=m ∞ X 1 ≤ ( ) 2j j=m 1 = → 0 as m → ∞. 2m−1

nm 2 Hence, k(B + T ) x − vmk → 0 as m → ∞.

0 ∗ ∗ n 2 The exact argument shows that k(B + T ) m x − vmk → 0 as m → ∞. Hence both B + T and B∗ + T ∗ are hypercyclic.

The existence of a hypercyclic operator with a hypercyclic adjoint was a surprising fact. Herrero [16] asked whether such an operator can ever exist. Salas [25] provided the first example of such an operator, and also a second example using a general result about bilateral weighted shifts in [26]. Theorem 30 shows that it is not uncommon to have a hypercyclic operator whose adjoint is also hypercyclic, and indeed we can choose one whose compression to a closed subspace with infinite codimension can coincide with a prescribed operator on the subspace. We remark that Corollary 23 and Theorem 30 call for two different constructions of an operator, but one may wonder if the two constructions may produce the same operator. The answer is never, because if both T and T ∗ are hypercyclic then T cannot be chaotic. To show that by the way of contradiction, suppose T has a periodic point x, say T nx = x, then (T ∗)n = (T n)∗ is not hypercyclic, which contradicts Ansari’s result [1] that the operator Sn 47 is hypercyclic whenever S is. An operator is said to be a universal model if every operator in B(H) is similar to a multiple part of it. Rota’s Theorem says that canonical backward unilateral shifts are universal models.

Theorem 31. Rota Let T be an operator on a Hilbert space H. If r(T ) < 1, then T is

2 similar to a part of the canonical backward unilateral shift on `+(H).

L∞ Let K = i=0 Hi where each Hi is isomorphic to H, and H0 = H. Suppose B : K → K

is the canonical backward unilateral shift; that is B takes H0 to 0 and for i = 1, 2, 3,...,

B|Hi is an isomorphism from Hi onto Hi−1.

Theorem 32. Let a > 1 and let T : H → H be a bounded linear operator. If its spectral radius r(T ) ≤ 1, then T is similar to a part of a hypercyclic operator aB : K → K.

Proof. The proof is exactly the same as the argument for Rota’s Theorem [24] on the existence of universal models. For all vectors x in H we denote

1 1 W x = (x, T x, ,...). a a2

n 1 n Since r(T ) = limn→∞ kT k n ≤ 1 we have that kT k ≤ 1 for all n. Thus

∞ X 1 kT nxk2 < ∞. a2n n=0

Note that 1 ∞ ! 2 X 1 kxk ≤ kW xk ≤ kxk, a2n n=0 and hence W : H → H is a bounded linear operator that is bounded below. Thus K := ranW is closed in K, and W : H → K is one-to-one and onto. Observe that

1 1 W T x = (T x, T x, T 2x, . . .) a a2 48 and

1 1 2 (aB) W x = aB(x, T x, 2 T x, . . .) a a (4.5) = W T x

Hence ranW is an invariant subspace of aB and

T = W −1 (aB|H) W.

Although the techniques we used here are pretty much standard for the Rota model, it is interesting for us to point out that the operator aB is indeed hypercyclic, which has never been observed before. 49

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