Hypercyclic Extensions of an Operator on a Hilbert Subspace with Prescribed Behaviors

HYPERCYCLIC EXTENSIONS OF AN OPERATOR ON A HILBERT SUBSPACE WITH PRESCRIBED BEHAVIORS

Gokul R Kadel

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 2013

Committee:

Kit Chan, Advisor

Rachel Reinhart, Graduate Faculty Representative

Juan B`es

So-Hsiang Chou ii

ABSTRACT

Kit Chan, Advisor

A continuous linear operator T : X → X on an inﬁnite dimensional separable topological vector space X is said to be hypercyclic if there is a vector x in X whose orbit under T , orb(T, x) = {T nx : n ≥ 0} = {x, T x, T 2x, . . .} is dense in X. Such a vector x is said to be a hypercyclic vector for T . While the orbit of a hypercyclic vector goes everywhere in the space, there may be other vectors whose orbits are indeed ﬁnite. Such a vector is called a periodic point. More precisely, we say a vector x in X is a periodic point for T if T nx = x for some positive integer n depending on x. The operator T is said to be chaotic if T is hypercyclic and has a dense set of periodic points.

Let M be a closed subspace of a separable, inﬁnite dimensional Hilbert space H with dim(H/M) = ∞. We say that T : H → H is a chaotic extension of A : M → M if T is chaotic and T |M = A.

In this dissertation, we provide a criterion for the existence of an invertible chaotic extension. Indeed, we show that a bounded linear operator A : M → M has an invertible chaotic extension T : H → H if and only if A is bounded below. Motivated by our result, we further show that A : M → M has a chaotic Fredholm extension T : H → H if and only if A is left semi-Fredholm.

Our further investigation of hypercyclic extension results is on the existence of dual hypercyclic extension. The operator T : H → H is said to be a dual hypercyclic extension of A : M → M if T extends A, and both T : H → H and T ∗ : H → H are hypercyclic. We actually give a complete characterization of the operator having dual hypercyclic extension on a separable, inﬁnite dimensional Hilbert spaces. We show that a bounded linear operator A : M → M has a dual hypercyclic extension T : H → H if and only if its adjoint A∗ : M → M is hypercyclic. iii

To my mother iv

ACKNOWLEDGMENTS

I would like to express my sincere appreciation to my advisor Dr.Kit Chan for his patience, guidance, and continuous support in preparation of this dissertation. He introduced me to the research area of hypercyclicity and taught me everything I needed to know to become a matured mathematician. I would never have been what I am today without his carefully planned advising.

I wish to thank my committee members Dr. Rachael Reinhart, Dr. So-Hsiang Chou and Dr. Juan B`esfor their time and advice. Dr. Chou has always helped me in identifying my potential and grow my conﬁdence level. I would like to thank Dr. Tong Sun for his support as a graduate coordinator.

I would like to thank the faculty and staﬀ of the Mathematics and Statistics Department for all of their help, guidance, and advice. Many thanks go to department secretaries Barbara, Marcia, and Mary who were always there to help me in the administrative and academic needs whenever needed. I would also like to thank Bowling Green State University for the ﬁnancial support, without which I would have never completed my degree.

Special thanks go to my friends Leo Pinheiro and Swarup Ghosh. It can never be over- stated how much I learned in company with these talented young mathematicians. Leo has always stood by my side in my good and bad times and has taught me not only mathematics but also the art of living.

Lastly, I wish to thank my mother, my sister Subidya Mishra, my brothers Pradhyumna and Shreedhar for all the support. Their constant source of inspiration made me who I am today. Most importantly, I wish to thank my wife Usha for her patience, love, support and encouragement. Without Usha, none of this would be possible. TABLE OF CONTENTS

CHAPTER 1: HYPERCYCLICITY AND EXTENSION RESULTS .... 1 1.1 INTRODUCTION ...... 1 1.2 HYPERCYCLIC OPERATORS ...... 2 1.3 HYPERCYCLICITY CRITERION ...... 4 1.4 HYPERCYCLIC VECTORS ...... 5 1.5 HYPERCYCLIC EXTENSION OF OPERATORS ...... 6

CHAPTER 2: RESTRICTIONS OF AN INVERTIBLE CHAOTIC OPER- ATOR TO ITS INVARIANT SUBSPACES ...... 7 2.1 INTRODUCTION ...... 7 2.2 INVERTIBILITY OF A CHAOTIC EXTENSION ...... 9 2.3 FREDHOLM OPERATORS ...... 25 2.4 HYPERCYCLIC SUBSPACES ...... 27

CHAPTER 3: DUAL HYPERCYCLIC EXTENSION FOR AN OPERA- TOR ON A HILBERT SUBSPACE ...... 29 3.1 INTRODUCTION ...... 29 3.2 TECHNICAL RESULTS FOR CONSTRUCTING A HYPERCYCLIC VEC- TOR...... 31 3.3 MAIN THEOREM ON DUAL HYPERCYCLIC EXTENSION ...... 61 vi

BIBLIOGRAPHY ...... 71

APPENDIX ...... 75 1

CHAPTER 1

HYPERCYCLICITY AND EXTENSION RESULTS

1.1 INTRODUCTION

Hypercyclicity is the study of continuous linear operators that possess a dense orbit. Al- though the first examples of hypercyclic operators date back to the first half of the last century, a systematic study of this concept has only been undertaken since the mid-eighties. Seminal papers like the unpublished but widely disseminated thesis of Kitai [21], a highly original and broad investigation by Godefroy and Shapiro [13] and deep operator-theoretic contributions by Herrero [19, 20] were instrumental in creating a flourishing new area of analysis.

Deﬁnition 1. A continuous linear operator T : X → X on a separable Fr´echetspace X is said to be hypercyclic if if there is some x ∈ X whose orbit under T , orb(T, x):={T nx : n ≥ 0} = {x, T x, T 2x, . . .}, is dense in X. In such a case, x is called a hypercyclic vector for T . The set of hypercyclic vectors for T is denoted by HC(T ).

The origin of this terminology is easily explained. For a long time, operator theorists 2 have been studying cyclic vectors in connection with the invariant subspace problem. Here, a vector x ∈ X is said to be cyclic for T if the linear span of its orbit, span {T nx : n ≥ 0}, is dense in X. A vector x ∈ X is called supercyclic for T if its projective orbit, {λT nx : n ≥

0, λ ∈ K}, is dense in X. Operators that possess a cyclic (or supercyclic) vector are called cyclic (or supercyclic, respectively). This suggested the name of hypercyclicity for the case when the orbit itself is dense.

The invariant subspace problem, which is open to this day, asks whether every operator on a separable infinite dimensional Hilbert space possesses an invariant closed subspace other than the trivial ones given by {0} and the whole space. Counterexamples do exist for operators on non-reflexive spaces like `1. Enflo [11] offered a negative solution for a specific Banach space, which was later transferred by Read [27] to an operator on `1 that has every non-zero vector hypercyclic.

1.2 HYPERCYCLIC OPERATORS

The very ﬁrst example of hypercyclic operator is given by Birkhoﬀ [4]. In 1929, he proved the existence of an entire function f whose successive translates by a non-zero constant a are 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).

MacLane [23], in 1952, gave another example of a hypercyclic operator. He 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) there

(nk) 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. 3

Another early example of hypercyclicity was given by Rolewicz [26] in 1969. This was the ﬁrst example of hypercyclicity for an operator on a Banach space. The setting is the sequence space `p for 1 ≤ p < ∞, and the operator is constructed from the backward shift B deﬁned on `p as follows:

B(x1, x2, x3,...) = (x2, x3,...),

p p p where (x1, x2, x3,...) ∈ ` . B itself is a contraction on ` (kBxk ≤ kxk for each x ∈ ` ), so it cannot be hypercyclic. However, Rolewicz proved that the operator T = λB with λ > 1 is hypercyclic.

The ﬁeld of hypercyclicity was actually born in 1982 with the Ph.D. dissertation of Kitai [21] and later with the paper of Gethner and Shapiro [12] and in more than two decades since, it has become the focus of active research by numerous authors.

Hypercyclicity is a wide-spread phenomenon in analysis. Perhaps the reason one might expect otherwise lies in the simple realization that ﬁnite-dimensional spaces do not support hypercyclic operators. The following result of Kitai [21] clariﬁes this point.

Theorem 2 (Kitai, [21]). There are no hypercyclic operators on ﬁnite dimensional spaces.

On the other hand, Ansari [1] showed that space with some nice structure always support a hypercyclic operator.

Theorem 3 (Ansari, [1]). Every separable, inﬁnite dimensional Fr´echetspace carries a hypercyclic operator.

Finally, we mention here a surprising result of Grivaux [14] which conﬁrms a large supply of hypercyclic operators in spaces with nice structures.

Theorem 4 (Grivaux, [14]). (i) Every operator on a separable, inﬁnite dimensional Hilbert space is the sum of two hypercyclic operators. 4

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

1.3 HYPERCYCLICITY CRITERION

In this section, we will discuss techniques of showing an operator to be hypercyclic. First, we mention here a theorem by Birkhoﬀ [4] which connects the concept of hypercyclicity to the well known notion of topological transitivity.

Theorem 5 (Birkhoﬀ Transitivity Theorem). Let X be a separable and T : X → X be a continuous linear operator. The following are equivalent:

(i) T is hypercyclic.

(ii) T is topologically transitive: that is, for each pair of non-empty open sets U and V in

X there exists n ∈ N such that T n(U) ∩ V 6= ∅.

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.

For an invertible operator T , Kitai [21, Corollary 2.2] showed that T is topologically transitive if and only if T −1 is. Thus, a continuous linear operator T on a Fr´echet 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 theorem by Birkhoff (Theorem 5) characterizes hypercyclic operators, in prac- tice, we use the following very useful Hypercyclicity Criterion to prove that an operator is hypercyclic. The Criterion was first given by Kitai [21] and then later independently by Gethner and Shapiro [12] in more generality. Later, the criterion was proved in even more generality by Bèsand Peris [3] which is the following: 5

Theorem 6 (Hypercyclicity Criterion). Let X be a separable Fr´echetspace 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, it was not known whether the above condition was necessary. In 2009, De La Rosa and Read [10] answered the question in the negative by constructing an 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.

1.4 HYPERCYCLIC VECTORS

It is clear that the orbit of any vector under the identity operator consists only of the point itself. This means that identity operator is not hypercyclic and therefore has no hypercyclic vector. However, if an operator T : X → X has a hypercyclic vector x, then there are plenty of supply of hypercyclic vectors for the same operator T . Precisely speaking, each of the vector of the form T nx where n = 0, 1, 2,... is a hypercyclic vector for T . That is the set of hypercyclic vectors of a hypercyclic operator T is a big set in a topological sense, and the following theorem summarizes this.

Theorem 7 (Kitai, [21]). Suppose X is a separable Fr´echetspace 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. 6

Since both HC(T ) and x − HC(T ) are dense Gδ sets, their intersection is nonempty by Baire Category Theorem. This immediately implies the following result which tells that the set of hypercyclic vectors is also algebraically large.

Theorem 8. [18] If X is a separable Fr´echetspace and T : X → X is a hypercyclic operator, then X = HC(T ) + HC(T ).

1.5 HYPERCYCLIC EXTENSION OF OPERATORS

In this dissertation, we investigate the hypercyclic extension of an operator on a closed subspace of an infinite dimensional Hilbert space. For this, we let H be an infinite dimensional, separable Hilbert spaces and M be a closed subspace of H with infinite codimension, that is, dim(H/M) = ∞.

In section 2.2 (Theorem 10), we show the construction of an invertible chaotic extension. Later in Corollary 12, we give a characterization of restrictions of invertible chaotic extensions on their closed invariant subspaces. Our further investigation of the hypercyclic extensions results provides the complete characterization of the restriction of the dual hypercyclic operators on closed invariant subspaces (see Theorem 19 on section 3.3). 7

CHAPTER 2

RESTRICTIONS OF AN INVERTIBLE CHAOTIC OPERATOR TO ITS INVARIANT SUBSPACES

2.1 INTRODUCTION

We recall from Definition 1 that a bounded linear operator T : H → H on a separable, infinite dimensional Hilbert space H is said to be hypercyclic if there is a vector h in H such that its orbit orb(T , h):= {h, T h, T 2h, . . .} is dense in H. Such a vector h is called a hypercyclic vector for T . While the orbit of a hypercyclic vector goes “everywhere” in the Hilbert space, there may be other vectors whose orbits are indeed finite. Such a vector is called a periodic point. More precisely, we say a vector h in H is a periodic point for T if T nh = h for some positive integer n depending on h. The operator T is said to be chaotic if T is hypercyclic and has a dense set of periodic points. 8

A closed subspace M of H is said to be an invariant subspace for T if TM ⊂ M. In the present chapter, we study the prescribed behavior of a chaotic or hypercyclic operator T on its invariant subspace M. To facilitate a deeper discussion, we introduce the following deﬁnition for a bounded linear operator A : M → M on a closed subpsace M of H.A bounded linear operator T : H → H is said to be a hypercyclic extension of A, if T is hypercyclic and T |M = A. If in addition, the set of periodic points of T is dense in H, then T is said to be a chaotic extension of A.

Grivaux [15, Prop 1] and Chan and Turcu [9, Theorem 2] showed that if M has infinite codimension in H, then any operator A : M → M has a chaotic extension T : H → H. They proved the result using different techniques but the extension T they obtained are the same. To be more precise, Grivaux [15] used the analytic function theory techniques to construct an extension T which satisfies the hypothesis of a hypercyclicity result of Godefroy and Shapiro

[13]: If H+(T ) = span {ker(T − λI): |λ| > 1}, and H−(T ) = span {ker(T − λI): |λ| < 1} are dense subspaces of H, then T is hypercyclic.

Chan and Turcu [9] used elementary Hilbert space techniques to show their extension T has an additional property that it satisﬁes the Hypercyclicity Criterion in the strongest form as stated in Theorem 9 below. However, an operator T constructed using the above result involving H+(T ) and H−(T ) is not guaranteed to have that property.

As an important part of the Hypercyclicity Criterion, Chan and Turcu’s result provides us with a chaotic extension T which is right invertible by its design but it must have a nontrivial kernel (see [9, p. 417]). Thus their extension T is never invertible regardless of what further assumption we make on the operator A. This naturally leads to a question whether we can construct an invertible chaotic extension T : H → H, particularly when A itself is invertible. In that regard, we observe that if T has a left inverse S, then SA = Id on M and hence A is bounded below. In Section 2.2, we show that the converse also holds true. That is, if A is bounded below on M, we construct an invertible chaotic extension T for A; 9 see Corollary 12 below. To show that, we ﬁrst construct a right invertible chaotic extension T for an operator A having a closed range so that ker T = ker A; see Theorem 10 below. Since it is important for A and T to have the same kernel, our techniques are diﬀerent from those in [9] and [15].

Motivated by our above results on invertibility and chaotic extension, we further study in Section 2.3 the invertibility of the operator in the Calkin algebra, on the Fredholm operators. Speciﬁcally, we show that A : M → M has a chaotic Fredholm extension T : H → H if and only if A is left semi-Fredholm. Lastly in Section 2.4, we show that the right invertible chaotic extension T : H → H in Theorem 10 can be constructed to have a hypercyclic subspace.

2.2 INVERTIBILITY OF A CHAOTIC EXTENSION

The Hypercyclicity Criterion plays a key role in the main theorem [Theorem 10] of this chapter. Kitai [21] in 1982 gave the ﬁrst and simplest version of the Criterion. Later in 1987, Gethner and Shapiro [12] rediscovered the Criterion in much greater generality. Since then hypercyclicity has been attracting a lot of attention as a hot research area. We exhibit in Theorem 10 a chaotic extension that satisﬁes the Criterion in the strongest possible sense. We now state the version of the Criterion, which is a particular case of Theorem 6, here for the sake of completeness.

Theorem 9 (Kitai [21]; Gethner and Shapiro [12]). A bounded linear operator T : H → H is hypercyclic if there is a map (not necessarily bounded and linear) S : H → H such that TS = Identity, and if there is a dense subset D of H such that T nx → 0, and Snx → 0 for each vector x in D.

We are now ready to state and prove the main result of the present chapter. We ﬁrst remark that Chan and Turcu [9, Prop 1] showed that no bounded linear operator A : M → M 10 on a nontrivial closed subspace M of H with dim(H/M) < ∞ can have a hypercyclic extension T to H. Hence, we have to assume dim(H/M) = ∞ in order to construct a hypercyclic extension T on H.

Theorem 10. Let H be a separable, inﬁnite dimensional Hilbert space, and M be a closed subpace of H with dim(H/M) = ∞. Every bounded linear operator A : M → M with a closed range has a right invertible chaotic extension T : H → H with ker A = ker T that satisﬁes the hypothesis of Theorem 9.

Proof. First we note that ranA is a closed subspace of M and so M is an orthogonal sum M = ranA ⊕ (ranA)⊥. In the rest of the proof, we use the symbol ⊕ to denote an orthogonal sum of closed subspaces or an orthogonal sum of vectors. Since M has inﬁnite codimension M in H, we can rename M as M0 and write H as an orthogonal sum Mj where for each j∈Z j ≥ 0, Mj is isomorphic to M, and furthermore for each j < 0, Mj is isomorphic to ranA.

0 00 0 Hence for each j ∈ Z, we can write Mj = Mj ⊕ Mj where each Mj is isomorphic to ranA,

00 ⊥ 00 and if j ≥ 0 then Mj is isomorphic to (ranA) , and if j < 0 then Mj = {0}. Thus, we can M 0 00 write H as the sum Mj ⊕ Mj . j∈Z Each vector h in H is written as

h = hj = . . . , h−2, h−1, h0, h1, h2,... , |{z}

where each hj ∈ Mj and the symbol “ ” indicates the zeroth component; that is, the |{z} 0 00 00 vector in M0. Since Mj = Mj ⊕ Mj for each j ∈ Z, and indeed Mj = {0} whenever j < 0, we can re-write h as

0 0 h = hj = . . . , h−2, h−1, h0, h1, h2,... , |{z}

0 0 where hj ∈ Mj = Mj for each j < 0.

⊥ Since A : M → M is bounded linear, the restriction map A|(ker A)⊥ : (ker A) → ranA 11 is invertible, that is, there exists a bounded linear operator B : ranA → (ker A)⊥ such that AB is the identity on ranA.

Let α > max{1, kAk, kBk}. By additively decomposing each vector hj in Mj as the vector

0 00 0 00 0 hj ⊕ hj in Mj ⊕ Mj , and by suppressing the symbols for the isomorphisms between Mi and

0 00 00 Mj for all integers i, j, and also the isomorphisms between Mi and Mj for all non-negative integers i, j, we can deﬁne T : H → H by

0 0 h−2 h−1 0 T h = T h = ..., , , αh1, Ah0 + αh1, αh2, αh3, αh4 ... . j α α | {z }

Equivalently,   1 h0 if j ≤ −2  α j+1   0  αh1 if j = −1 T h j =  Ah + αh if j = 0  0 1    αhj+1 if j ≥ 1.

Hence by our identiﬁcation of the given subspace M with M0, and the above formula for T ,

we see that T |M = A. One can easily verify that T is bounded linear.

We now prove that T is bounded linear. For linearity, we observe that if h, k are in H, and β is a scalar, then

0 0 0 0 βh + k = . . . , βh−2 + k−2, βh−1 + k−1, βh0 + k0, βh1 + k1, βh2 + k2,... | {z }

so that

βT h + T k 0 0 h−1 0 k−1 0 = β ..., , αh1, Ah0 + αh1, αh2, αh3,... + ..., , αk1, Ak0 + αk1, αk2, αk3,... α | {z } α | {z } 0 0 βh−1 + k−1 0 0 = ..., , α(βh1 + k1),A(βh0 + k0) + α(βh1 + k1), α(βh2 + k2), α(βh3 + k3),... α | {z } = T (βh + k). 12

To show T is bounded we observe that

∞ 0 2 ∞ 2 2 0 2 X h−j X 2 kT hk = kAh0 + αh1k + kαh k + + kαhjk 1 α j=1 j=2 ∞ ∞ 2 2 0 2 1 X 0 2 2 X 2 ≤ 2 kAh0k + 2 kαh1k + kαh k + h + α khjk 1 α2 −j j=1 j=2 ∞ ∞ 2 2 2 2 2 0 2 2 X 0 2 2 X 2 ≤ 2 kAk kh0k + 2α kh1k + 2α kh1k + 2α h−j + 2α khjk j=1 j=2

2 2 0 2 ≤ 2α khk + kh1k

≤ 2α2khk2 + khk2

≤ 4α2khk2.

We now show that ker A = ker T . If x0 ∈ M0 with Ax0 = 0, then T x0 = Ax0 = 0 and so ker A ⊂ ker T . On the other hand, if h = hj ∈ H with T h = 0, then

0 h1 = 0

0 hj = 0 for j = −1, −2, −3 ...

hj = 0 for j = 2, 3,...

Ah0 + αh1 = 0.

0 00 Since Ah0 + αh1 is in M0, and h1 = 0, we see that Ah0 + αh1 = 0. Observe that Ah0 ∈

00 ⊥ 00 ranA ⊂ M0, and αh1 ∈ (ranA) ⊂ M0, and so we get Ah0 = 0 and αh1 = 0. Therefore, if

h ∈ ker T , then h = h0 ∈ ker A.

We now turn our attention to showing that T is right invertible and satisﬁes the hypothesis of the Theorem 9. Deﬁne S : H → H using the operator B : ranA → (ker A)⊥ with AB = I on ranA as follows:

0 0 0 0 0 1 0 00 1 1 Sh = Sh j = . . . , αh−4, αh−3, αh−2,B h0 − h−1 , h−1 ⊕ h0 , h1, h2,... . | {z } α α α 13

Equivalently,   α h0 if j ≤ −1  j−1   0 0  B h0 − h−1 if j = 0 Sh j =  1 h0 ⊕ h00 if j = 1  α −1 0   1  α hj−1 if j ≥ 2. It is easy to check that T Sh = h for any h ∈ H. Indeed,

T Sh 0 0 0 0 0 1 0 00 1 1 = T . . . , αh−4, αh−3, αh−2,B h0 − h−1 , α h−1 ⊕ h0 , αh1, αh2,...  | {z }  αh0 αh0 αh0 h0 ⊕ h00 α α α = ..., −3 , −2 , −1 , AB h0 − h0 + α −1 0 , h , h , h ,...  α α α 0 −1 α α 1 α 2 α 3  | {z } = h.

Thus, T is right invertible. To create a dense set D stated in Theorem 9, we ﬁrst let ∆0 be a countable dense subset of ranA, and ∆00 be a countable dense set of (ranA)⊥. Then,

0 00 0 ∆ = ∆ + ∆ is a countable dense set in M = M0. Via the isomorphism between Mj and

0 0 0 0 ran A, the isomorphic copy ∆j in Mj of ∆ is dense in Mj for each j ∈ Z. Similarly, the

00 00 00 00 00 isomorphic copy ∆j in Mj of ∆ is dense in Mj for each j ≥ 0. By letting ∆j = {0} if

0 00 j < 0, and letting ∆j = ∆j + ∆j for each j ∈ Z, we deﬁne

0 0 C = ..., 0, 0, 0, d−k, . . . , d−1, d0, d1, . . . , dk, 0, 0, 0,... : dj ∈ ∆j and k ≥ 1 . |{z}

By the density of ∆j in Mj and the fact that the norm of h = (hj) in H is given by ∞ 2 X 2 khk = khjk , one can easily show that C is dense in H. To prove this, let h = j=−∞ 0 0 . . . , h−2, h−1, h0, h1, h2,... be a vector in H. Let ε > 0 be given. There exists a k ∈ N |{z} such that ∞ 2 X 0 2 2 ε h + khjk < −j 2 j=k+1 14

Since ∆ = ∆0 + ∆00 is dense in M = ranA ⊕ (ranA)⊥, so by the isomorphism, there is a dense

0 00 0 00 subset ∆j = ∆j + ∆j of Mj = Mj ⊕ Mj for each j ∈ Z. So, for −k ≤ j ≤ −1 there is a

0 0 dj ∈ ∆j such that 2 0 0 2 ε h − d < j j 4k

and for 0 ≤ j ≤ k we can ﬁnd a dj ∈ ∆j such that

ε2 kh − d k2 < j j 4 (k + 1)

So, if we write 0 0 w = ..., 0, d−k, . . . , d−1, d0, d1, . . . , dk, 0,... , |{z} then w is in C, and it satisﬁes

−1 k ∞ 2 X 0 0 2 X 2 X 0 2 2 kw − hk = hj − dj + khj − djk + h−j + khjk j=−k j=0 j=k+1 ε2 ε2 ε2 < k + (k + 1) + 4k 4 (k + 1) 2 = ε2,

which proves the density of C in H.

Let k ∈ N. Consider the following vector x in C given by

0 0 x = ..., 0, 0, 0, d−k, . . . , d−1, d0, d1, . . . , dk, 0, 0, 0,... . |{z}

Observe that for the same integer k, we can write

k k k k k k k T x = T x j = ..., T x −2 , T x −1 , T x 0, T x 1 , T x 2 ,... , | {z } 15 where   0 if j ≤ −2k − 1    α−k d0 if −2k ≤ j ≤ −k − 1  j+k    2j+k+2 0 k α dj+k+1 if −k ≤ j ≤ −1 T x j =   k  X i k−i  α A di if j = 0   i=0   0 if j ≥ 1.

Now, for any m > k, by applying the operator T repeatedly on the vector x we see that the

m m−k k m components of the vector T x = T T x = (T x)j is given by

  0 if j ≤ −m − k − 1     α−m d0 if −m − k ≤ j ≤ −m − 1  j+m    2j+m+2 0  α dj+m+1 if −m ≤ j ≤ −m + k − 1 m  (2.1) (T x)j =   0 if −m + k ≤ j ≤ −1    k  X i m−i  α A di if j = 0   i=0   0 if j ≥ 1.

Now, we let P : M → M denote the orthogonal projections onto ran A. Then we use the operator S on the vector x successively k times to get the vector Skx, which is of the form (2.2) k k S x = S x j k k k k = ..., 0, 0, 0, S x 0 , S x 1 , S x 2 ,..., 0, S x 2k , 0, 0, 0,... . | {z } 16

Here we claim that for 1 ≤ j ≤ k, we have

" j−1 # j j−1 0 X i j−1−i 0 (2.3) S x 0 = B (PB) d0 − α (PB) d−i−1 . i=0

1 We prove this claim by induction. Indeed, for j = 1, the above formula provides (S x)0 = 0 0 B d0 − d−1 , which is true. Assume that (2.3) holds for some integer j = q < k, that is, assume that " q−1 # q q−1 0 X i q−1−i 0 (S x)0 = B (PB) d0 − α (PB) d−i−1 i=0 holds. Then for j = q + 1, we verify that

q+1 q (S x)0 = S (S x) 0 q q = B P (S x)0 − (S x)−1

" q−1 # q 0 X i q−i 0 q 0 = B (PB) d0 − α (PB) d−i−1 − α d−q−1 i=0 " q # q 0 X i q−i 0 = B (PB) d0 − α (PB) d−i−1 , i=0

proving the claim.

m m Observe that if m > k, then S x = (S x)j is a vector of the form

m m m m S x = ..., 0, 0, 0, (S x)0 , (S x)1 ,..., (S x)m+k , 0, 0, 0,... . | {z }

j j We claim that for j ≥ k + 1, the zeroth component (S x)0 of the vector S x is given by

" k−1 # j j−1 0 X i j−1−i 0 (2.4) S x 0 = B (PB) d0 − α (PB) d−i−1 . i=0

k To show this, we ﬁrst observe from equation (2.2) that S x j = 0 for j < 0. So, for j = k+1 17 we get by using equation (2.3),

k+1 k S x 0 = SS x 0

h k k i = B P S x 0 − S x −1

k = B P S x 0

" ( k−1 )# k−1 0 X i k−1−i 0 = B PB (PB) d0 − α (PB) d−i−1 i=0 " k−1 # k 0 X i k−i 0 = B (PB) d0 − α (PB) d−i−1 . i=0

Inductively we get equation (2.4) whenever j ≥ k + 1.

Using the same isomorphisms between Mj that we have been using, we deﬁne a vector

0 dm ∈ Mm as the isomorphic image of the vector dm ∈ M0 given by

k 1 1 X (2.5) d0 = − Am−k T kx = − αi Am−id , m αm 0 αm i i=0

0 0 and another vector d−m ∈ M−m as the isomorphic image of the vector d−m ∈ M0 given by

" k−1 # 1 1 X (2.6) d0 = P Sm−1x = (PB)m−1 d0 − αi(PB)m−1−i d0 . −m αm−1 0 αm−1 0 −i−1 i=0

0 We remark that dm and d−m are deﬁned in terms of the components of the vector x. We let

xm = ..., 0, 0, 0,..., 0, dm, 0,... , |{z}

in which dm ∈ Mm, and

0 0 x−m = ..., 0, d−m, 0,..., 0, 0, 0,... |{z} 18

0 in which d−m ∈ M−m, and deﬁne a set D by

0 0 0 D = x−m + x + xm : x = ..., 0, d−k, . . . , d−1, d0, d1, . . . , dk, 0,... ∈ C and m > k . |{z}

0 Since α has been chosen so that α > max {1, kAk, kBk}, we see that both kdmk and kd−mk go to zero as m goes to inﬁnity. Hence D is dense in the set C, which in turn is dense in H. Thus D is dense in H also.

For the details, let x ∈ C. Then we can choose an integer m > k large enough, and deﬁne

0 dm ∈ Mm and d−m ∈ M−m as given by equations (2.5) and (2.6) so that

ε2 ε2 kd k2 < and kd0 k2 < . m 2 −m 2

To see that this can be done, we observe the following from equation (2.5):

m 1 m−k k kAk −k k kdmk = A T x ≤ kAk T x , αm 0 α 0 which goes to 0 as m goes to inﬁnity because α is chosen to satisfy α > max(1, kAk, kBk). Similarly, by equation (2.6)

" k−1 # 1 X kd0 k = (PB)m−1 d0 − αi(PB)m−1−i d0 −m αm−1 0 −i−1 i=0 k−1 k(PB)km−k X ≤ (PB)k−1 d0 − αi(PB)k−1−i d0 αm−1 0 −i−1 i=0 m k−1 kBk X ≤ α kBk−k (PB)k−1 d0 − αi(PB)k−1−i d0 , α 0 −i−1 i=0 which goes to 0 as m goes to inﬁnity.

0 0 With these d−m and dm, we can now deﬁne the vectors x−m and xm as before. So, if we 19

0 0 write y = x−m + x + xm, then y is in D and since x−m and xm are orthogonal we get

2 2 2 0 2 2 ε ε 2 kx − yk = d + kdmk < + = ε , −m 2 2

showing that D is dense in C, and hence in H because C is dense in H.

We now turn our attention to showing T , S, D satisfy the hypothesis of Theorem 9 by showing that both T n and Sn go to 0 pointwise on D as n → ∞. For this, let us consider

0 y = x−m + x + xm in D.

m m m 0 For the vector T y, we note that its zeroth component (T y)0 vanishes because T x−m 0 = 0, and by equation (2.1) and (2.5) we have

k m X i m−i m m (T x)0 = α A di = −α dm = − (T xm)0 . i=0

Thus the vector T my is of the form (2.7) T my

m 0 m m = T x−m + T x + T xm m m m m = ..., 0, (T y)−2m , 0,..., 0, (T y)−m−k ,..., (T y)−m+k−1 , 0,..., 0, (T y)−1 , 0, 0,... . |{z}

m m m For the vector S y, we note that its zeroth component (S y)0 vanishes because (S xm)0 = 0 and by the deﬁnition of S and equation (2.4), we see that

" k−1 # m 0 m m−1 0 m−1 0 X i m−1−i 0 S x−m 0 + (S x)0 = B −α d−m + B (PB) d0 − α (PB) d−i−1 i=0 " k−1 # m−1 0 m−1 0 X i m−1−i 0 = B −α d−m + (PB) d0 − α (PB) d−i−1 i=0 = 0, 20 where the last equality follows from equation (2.6). Thus

m m 0 m m S y = S x−m + S x + S xm m m m (2.8) = ..., 0, 0, (S y)1 ,..., (S y)m+k , 0,..., 0, (S y)2m , 0,... . |{z}

Thus if n > m, we use the deﬁnitions for T and S to see that

−m+k−1 ! 2 2 1 m 2 m 2 X m kT nyk = (T y) + (T y) + (T y) , αn−m −2m −1 j j=−m−k and m+k ! 2 2 1 2 X kSnyk = k(Smy) k + (Smy) . αn−m 2m j j=1

Since α > 1, both kT nyk and kSnyk go to 0, as the integer n goes to inﬁnity. This shows that both T n and Sn go to 0 pointwise on D and thus the hypothesis of Theorem 9 is satisﬁed. Hence T is hypercyclic.

0 To find periodic points for T , we continue to use vectors y = (x−m + x + xm) in D. Let ∞ X z = y + T i(2m+1)y + Si(2m+1)y. To explain why this infinite sum defines a vector z in i=1 H, we now show that the sum is absolutely convergent. We first notice that for each i ≥ 1,

i(2m+1) (i−1)(2m+1) m+1 m T y = T T T y ,

m and by equation (2.7), (T y)j = 0 for all nonnegative integer j. We use the formula for T to see that

m i(2m+1) 1 1 m α m T y = T y = kT yk . α(i−1)(2m+1) αm+1 αi(2m+1) 21

Thus,

∞ ∞ m X i(2m+1) X α m (2.9) T y = kT yk < ∞. αi(2m+1) i=1 i=1

Similarly, for each i ≥ 1

i(2m+1) (i−1)(2m+1) m+1 m S y = S S S y ,

m and by equation (2.8), (S y)j = 0 for all integers j ≤ 0. We use the formula for S to see that m i(2m+1) 1 1 m α m S y = S y = kS yk . α(i−1)(2m+1) αm+1 αi(2m+1)

Therefore,

∞ ∞ m X i(2m+1) X α m (2.10) S y = kS yk < ∞. αi(2m+1) i=1 i=1

To show that z is a periodic point, we observe that

∞ ! X T 2m+1z = T 2m+1 y + T i(2m+1)y + Si(2m+1)y i=1 ∞ ∞ X X = T 2m+1y + T i(2m+1)y + S(i−1)(2m+1)y i=2 i=1 ∞ ∞ X X = T i(2m+1)y + y + Si(2m+1)y i=1 i=1 = z,

which shows that z is a periodic point of T with period (2m + 1). ( ∞ ) X Finally, we need to show that the set Z = z = y + T i(2m+1)y + Si(2m+1)y : y ∈ D i=1 ∞ X i(2m+1) is dense in H. This actually follows from equations (2.9) and (2.10) since T y → 0 i=1 22

∞ X i(2m+1) and S y → 0 as m → ∞. i=1 0 Indeed, we let x ∈ C and ε > 0. Deﬁne vectors d−m and dm using equations (2.5) and (2.6). Choose m > k large enough so that

ε2 ε2 kd k2 < and kd0 k2 < . m 4 −m 4

0 0 Now we use the vectors d−m and dm to deﬁne the vectors x−m and xm as before. So, if we

0 write y = x−m + x + xm, then y is in D. ∞ X With this y, we now deﬁne z = y + T i(2m+1)y + Si(2m+1)y. Then z is in Z, and thus i=1 we observe that

∞ 2 0 2 2 X h i(2m+1) 2 i(2m+1) 2i kz − xk = d−m + kdmk + T y + T y i=1 ε2 ε2 ε2 ε2 < 4 + 4 + 4 + 4 = ε2.

This shows the density of Z in C. But, C is dense in H, so Z is dense in H.

For the rest of the chapter, we state and prove two corollaries of Theorem 10, and continue to use H as a separable, inﬁnite dimensional Hilbert space and M as a closed subspace of H with dim(H/M)=∞. We ﬁrst show that the extenstion T : H → H preserves the right invertibility of its restriction T |M on M.

Corollary 11. Every right invertible bounded linear operator A : M → M has a right invertible chaotic extension T : H → H satisfying ker A = ker T and the hypothesis of Theorem 9.

Proof. Since A is right invertible, ran A = M, a closed subspace of H. So by Theorem 10, A has a right invertible chaotic extension T : H → H with kerA = kerT and that satisﬁes the 23 hypothesis of Theorem 9.

Although Corollary 11 is proved using Theorem 10, we remark that Corollary 11 can be proved using a simpler construction than the one in the proof of Theorem 10. To see that we decompose H additively into an orthogonal sum L M where M is identiﬁed with j∈Z j

M0 and each Mj is isomorphic to M. Since A is right invertible, there exists an operator B : M → M such that AB is the identity on M. Let α > max {1, kAk, kBk}. Then for any vector h = . . . , h−2, h−1, h0 , h1, h2,... in H where hj ∈ Mj for each j ∈ Z, we |{z} deﬁne T : H → H by

h−2 h−1 T h = ..., , , αh1, Ah0 + αh1, αh2, αh3, αh4,... . α α | {z }

It is easy to check that T |M = A and ker A = ker T .

Now we deﬁne an operator S : H → H by

h−1 h1 h2 Sh = . . . , αh−4, αh−3, αh−2,B(h0 − h−1), , , ,... , | {z } α α α

where h = . . . , h−2, h−1, h0 , h1, h2,... is in H and hj ∈ Mj for each j ∈ Z. One |{z} can verify that T Sh = h for all h ∈ H, and therefore S is a right inverse of T .

By considering a countable dense set ∆ of M = M0 and its isomorphic copies ∆j in Mj

for each j ∈ Z, we construct a set C given by

C = ..., 0, 0, d−k, . . . , d−1, d0 , d1, . . . , dk, 0, 0,... : dj ∈ ∆j and k ≥ 1 . |{z}

Note that C is dense in H. Let x = ..., 0, 0, d−k, . . . , d−1, d0 , d1, . . . , dk, 0, 0,... |{z} be a vector in C. For m > k we construct vectors dm in Mm and d−m in M−m such that dm 24

0 0 is the isomorphic copy of dm in M0 and d−m is the isomorphic copy of d−m in M0 given by

−1 1 d0 = Am−k T kx and d0 = Bm−k−1 Skx m αm 0 −m αm−1 0

k k k k where T x 0 and S x 0 are the zeroth components of the vectors T x and S x respectively.

Now we deﬁne a set

D = x−m + x + xm : x = ..., 0, d−k, . . . , d−1, d0 , d1, . . . , dk, 0,... ∈ C and m > k , |{z}

where x−m = (..., 0, d−m, 0,..., 0, 0 , 0,...) and xm = (..., 0, 0 , 0,..., 0, dm, 0,...) |{z} |{z} are vectors in H. It can be easily shown that D is dense in C and hence in H. Moreover, both T n and Sn go to zero pointwise on D as n → ∞ using an argument similar to the one in the proof of Theorem 10. Hence T is hypercyclic by Theorem 9.

The idea of showing that T is chaotic by ﬁnding a dense set of periodic points is just the same as in Theorem 10.

In the next corollary we prove the existence of invertible chaotic extension T : H → H for a given invertible bounded linear operator A : M → M on the closed subspace M of inﬁnite codimension.

The following result gives a criterion for the existence of invertible chaotic extension.

Corollary 12. A bounded linear operator A : M → M has an invertible chaotic extension T : H → H if and only if A is bounded below.

Proof. Since A is bounded below, ran A is a closed subspace of H, and A is one to one. So by Theorem 10, A has a right invertible chaotic extension T : H → H satisfying ker T = ker A = {0} and the hypothesis of Theorem 9. Hence T is invertible.

Conversely, suppose that T is invertible and extends A. In particular T is left invertible. So, there exists an operator S : H → H such that ST = identity. Thus for any vector x in 25

M, we have kxk = kST xk ≤ kSkkT xk = kSkkAxk, and hence A is bounded below.

As a special case of Corollary 12, we provide the following statement: Every bounded, invertible, linear operator A : M → M has an invertible chaotic extension T : H → H that satisﬁes the hypothesis of Theorem 9.

2.3 FREDHOLM OPERATORS

To continue our discussion, we provide results on semi-Fredholm operators which are analogous to Corollaries 11 and 12. We continue to use M to denote a closed subspace of a separable inﬁnite dimensional Hilbert space H with dim(H/M) = ∞.

We observe that if A : M → M is left semi-Fredholm, then ran A is closed and dim ker A < ∞. Theorem 10 gives a right invertible chaotic extension T : H → H of A such that ker A = ker T . So dim ker T < ∞. Moreover, since T is right invertible ran T is closed and dim ker T ∗ = 0. Thus T is Fredholm, and

ind T = dim ker T − dim ker T ∗ = dim ker A ≥ dim ker A − dim ker A∗ = ind A.

Thus we have shown half of the the following statement.

Corollary 13. A bounded linear operator A : M → M has a chaotic Fredholm extension T : H → H if and only if A is left semi-Fredholm. Moreover, the indices of T and A satisfy ind T ≥ ind A.

Proof. Suppose that T is a Fredholm operator that extends A. Then we have ker A ⊂ ker T and dim ker T < ∞. Therefore dim ker A ≤ dim ker T < ∞. It remains to show that ran A is closed, and hence A is left semi-Fredholm. 26

⊥ If we let N = ker T , then the operator T |N ⊥ : N → ran A is invertible. In particular,

⊥ T |N ⊥ is bounded below and so there exists a C > 0 such that kT xk ≥ Ckxk for all x ∈ N . Since N = ker T has ﬁnite dimension, the subspace Y := M + N is also closed. Let

B : Y → H be deﬁned by Bx = T x. That is, B = T |Y . Thus if x ∈ M, then Bx = T x = Ax and if x ∈ N, then Bx = T x = 0. Hence ran B = ran A, and it suﬃces to show ran B is closed.

Suppose (xn) ⊂ Y and Bxn → b. We must show b ∈ ran B; that is, there exists an a ∈ Y such that Ba = T a = b.

Let P : H → H be the orthogonal projection onto N ⊥ = (ker T )⊥. For all x ∈ Y we write x = P x + (I − P )x. Then Bx = T x = T (P x + (I − P )x) = T P x. Therefore

T P xn = Bxn → b. So (T P xn) is a Cauchy sequence.

Moreover, since kT P xn − T P xmk = kT (P xn − P xm)k ≥ C kP xn − P xmk for some posi-

tive constant C, we see that (P xn) is a Cauchy sequence. Note that xn = P xn + (I − P )xn.

Thus xn − (I − P )xn = P xn gives a Cauchy sequence.

Since xn ∈ Y and (I − P )xn ∈ N ⊂ Y , we have xn − (I − P )xn ∈ Y . Since Y is closed,

there exists an a ∈ Y such that xn − (I − P )xn → a. Thus B(xn − (I − P )xn) → Ba. On the other hand,

B(xn − (I − P )xn) = T (xn − (I − P )xn) = T xn = Bxn → b.

Hence, Ba = b.

The above corollary makes us wonder what extension results one can have for a right semi-Fredholm operator A. In fact, it easily follows from Theorem 10 that if A : M → M is right semi-Fredholm, then there exists an extension T : H → H of A which is chaotic and right invertible and satisﬁes ker A = ker T . Moreover, the indices of A and T satisfy ind T ≥ ind A. 27

2.4 HYPERCYCLIC SUBSPACES

Chan and Turcu [9, Corollary 4] proved that the extension T : H → H of an arbitrary operator A : M → M with dim(H/M) = ∞ can have a hypercyclic subspace. By a hypercyclic subspace of an operator T we mean a closed, infinite dimensional subspace of H which consists of entirely, except for the zero vector, of hypercyclic vectors of T . To prove this, they used a result of Montes-Rodr´ıguez[24]: T has a hypercyclic subspace if T satisfies the hypothesis of Theorem 9, and there is an infinite dimensional closed subspace K of H such that every vector f in K satisfies lim T nf = 0. Actually Montes-Rodr´ıguezproved this result in a Banach space setting and his proof was very hard. Later much easier proofs were given by several authors, including Bonet et al. [5], Chan [6], Chan and Taylor [8] and León-Saavedra and Müller[22].

In this section, we show that the right invertible chaotic extension T in the statement of Theorem 10 can be taken to have a hypercyclic subspace by modifying the construction for T . To show that, we remark that dim(H/M) = ∞ and so H can be additively decomposed

as the orthogonal sum H = H1 ⊕ H2 of inﬁnite dimensional closed subspaces H1 and H2 of

H with M being a closed subspace of H1 with dim(H1/M) = ∞.

We use the exact same technique in the proof of Theorem 10 to obtain a right invertible chaotic extension T1 : H1 → H1 of A : M → M with ker T1 = ker A. M Next we decompose H2 as an orthogonal sum H2 = Nj where each Nj is an inﬁnite j∈Z dimensional closed subspace of H2 and that all Nj’s are isomorphic. For any vector x ∈ H2, we write x = . . . , x−2, x−1, x0 , x1, x2,... |{z}

where xj ∈ Nj for each j. We deﬁne an operator T2 : H2 → H2 by

x−2 x−1 x0 T2 . . . , x−2, x−1, x0 , x1, x2,... = ..., , , , 2x1 , 2x2, 2x3,... . |{z} 2 2 2 |{z} 28

Then the formula for the inverse, S2 : H2 → H2, of T2 is given by

x0 x1 x2 S2 . . . , x−2, x−1, x0 , x1, x2,... = ..., 2x−3, 2x−2, 2x−1, , , ,... . |{z} |{z} 2 2 2

One can easily check that T2S2 = S2T2.

Let ∆0 be a countable dense subset of N0. Since Nj’s are isomorphic, there is an iso-

morphic countable dense image ∆j of ∆0 in each Nj. Now if we deﬁne a set D2 of H2 by

D2 = ..., 0, 0, d−k, . . . , d−1, d0 , d1, . . . , dk, 0, 0,... : dj ∈ ∆j and k ≥ 1 , |{z}

n n then D2 is dense in H2. One can easily check that both T2 and S2 go to zero on D2.

Therefore, T2 satisﬁes the hypothesis of Theorem 9.

To ﬁnd a dense set of periodic points for T2, we continue using vectors x ∈ D2 and then P∞ jm P∞ jm consider vectors of the form z = j=1 T2 x + x + j=1 S2 x. Since S2 is the inverse of

T2, one can easily see that

∞ ∞ ! m m X jm X jm T2 z = T2 T2 x + x + S2 x = z, j=1 j=1

showing that z is a periodic point of T2 with period m. It is easy to show that the set nP∞ mj P∞ mj o Z = j=1 T2 x + x + j=1 S2 x : x ∈ D2 is dense in H2. Hence T2 is chaotic on H2.

Finally, we let T = T1 ⊕ T2 with the notion that T1|H2 = 0 and T2|H1 = 0. Then T is a

bounded linear operator on H. Since both T1 and T2 satisfy the hypothesis of Theorem 9, one can easily check that T satisﬁes the Hypercyclicity Criterion and hence T is hypercyclic, and indeed T is chaotic on H.

n Notice that T goes to zero on each of the inﬁnite dimensional closed subspaces Nj of H. Thus T has a hypercyclic subspace. 29

CHAPTER 3

DUAL HYPERCYCLIC EXTENSION FOR AN OPERATOR ON A HILBERT SUBSPACE

3.1 INTRODUCTION

A bounded linear operator T : H → H on a separable, inﬁnite dimensional Hilbert space H is said to be dual hypercyclic if both T and its adjoint T ∗ : H → H are hypercyclic.

Herrero [19] asked whether a dual hypercyclic operator exists. Salas [29], [32] gave con- crete examples of a dual hypercyclic operator. Later Petersson [25] took Salas’ result to a more general Banach space setting: For any inﬁnite-dimensional Banach space X with a shrinking symmetric basis, such as c0 and any `p, (1 < p < ∞), there is an operator T : X → X, where both T and T ∗ : X∗ → X∗ are hypercyclic. Later Salas [30] proved that any Banach space X with a separable dual space X∗ admits a dual hypercyclic operator on X. Then Shkarin [33] studied dual hypercyclic tuples of operators on Banach spaces, and Salas [31] studied dual disjoint hypercyclic operators. More recently, B`es,Martin, Peris, 30

Shkarin [2] studied disjoint dual hypercyclic tuples of operators.

Chan [7] showed that if M is a closed subspace of a separable, inﬁnite dimensional Hilbert space with dim(H/M) = ∞, then every bounded linear operator A : M → M is the compression of a dual hypercyclic operator T : H → H. By A being a compression of T , we mean PTP |M = A, where P : H → H is the orthogonal projection onto M. That result naturally leads to a general question whether an arbitrary operator A : M → M can be

the restriction T |M of a dual hypercyclic operator T : H → H. In the present chapter, we investigate this question. It turns out that the answer is positive if an only if A∗ : M → M is hypercyclic; see Theorem 19 below. To prove that, we need some techniques from [7]. However, the dual hypercyclic operator T that we construct must be very diﬀerent from the one in [7]. We have to use new ideas and make use of the hypothesis on the hypercyclicity of the adjoint A∗. For instance, we have to state a lemma in terms of a hypercyclic vector

∗ x0 of A ; see Lemma 17 below.

To discuss our question further, we introduce the following deﬁnition. If M is a closed subspace of H, and A : M → M is a bounded linear operator, then T : H → H is said to be

∗ a dual hypercyclic extension of A provided T |M = A, and both T and T are hypercyclic. If

T : H → H is hypercyclic or chaotic and T |M = A, then we say T is a hypercyclic extension or respectively a chaotic extension of A. Here, a chaotic operator is a hypercyclic operator with a dense set of periodic points x of H. A vector x in H is said to be a periodic point if T nx = x for some positive integer n depending on x.

The first result on chaotic extension is given by Grivaux [15], who showed that every operator A : M → M on a closed subspace M of H with dim(H/M) = ∞ has a chaotic extension T : H → H that satisfies a simple form of the well-known hypercyclicity criterion. Then Chan and Turcu [9] showed the same chaotic extension actually satisfies the simplest form of the criterion. Indeed they also showed that every closed invariant subspace of a hypercyclic operator is necessarily of infinite codimension. This justifies our assumption of 31 dim(H/M) = ∞ on our subspace M.

In the present chapter, we investigate only dual hypercyclic extension. As remarked in [7], a chaotic operator cannot be dual hypercyclic. Consequently, a dual chaotic extension can never exist, and we have to introduce techniques diﬀerent from those in [9] to tackle the problem here.

3.2 TECHNICAL RESULTS FOR CONSTRUCTING

A HYPERCYCLIC VECTOR

Throughout this section, we let H be a separable infinite dimensional Hilbert space, and M be a closed infinite dimensional subspace of H with infinite codimension, that is, dim(H/M) = ∞. We are to construct a dual hypercyclic extension of a bounded linear operator on M. To do that we now give H a structure that facilitates our construction. First, we rename M as M0 and write H as an orthogonal sum of closed subspaces Mj with −∞ < j < ∞, where

each Mj is separable and has inﬁnite dimension. That is,

∞ M H = Mj, j=−∞

where Mi ⊥ Mj whenever i 6= j, and M0 is identiﬁed with M. Furthermore, for each j ≥ 1,

we decompose Mj as an orthogonal sum of Mi,j starting with index i = 0, where each Mi,j is isomorphic to M. That is, ∞ M Mj = Mi,j, i=0

0 where Mi,j is isomorphic to M for each integer i ≥ 0, and Mi,j ⊥ Mi0,j whenever i 6= i . On

the other hand, for each j ≤ −1, we decompose Mj as an orthogonal sum of Mi,j starting 32

with index i = 1 instead of 0, where Mi,j is isomorphic to M. That is,

∞ M Mj = Mi,j, i=1

0 where Mi,j is isomorphic to M for each integer i ≥ 1, and Mi,j ⊥ Mi0,j whenever i 6= i . The structure of these orthogonal subspaces Mi,j is used in our construction of a dual hypercyclic operator T . We provide a diagram below showing the action of T on these subspaces after we give a formal deﬁnition of T .

A vector h in H is written as

∞ h = (hj)j=−∞ = . . . , h−2, h−1, h0 , h1, h2,... , |{z}

where hj is in Mj for each integer j. We use the symbol “ ” to indicate a vector in the |{z} zeroth component; that is, a vector in M0.

For j ≥ 1, we can write

∞ hj = (hi,j)i=0 = (h0,j, h1,j, h2,j,...) ,

where hi,j is in Mi,j for each integer i ≥ 0. Furthermore, for each j ≤ −1, we write

∞ hj = (hi,j)i=1 = (h1,j, h2,j, h3,j,...) ,

where each hi,j is in Mi,j for each integer i ≥ 1.

Suppose A : M → M is a bounded linear operator whose adjoint A∗ : M → M is hypercyclic. To construct a dual hypercyclic extension T : H → H of A, we let α >

max{1, kAk}. For each integer j, we appropriately choose a weight wj satisfying 0 < wj ≤ α,

n 1 o as explained in Lemma 15 and 17 below. For any such choice of w1, we let w > max w1, w1 1 and deﬁne weights wi,j by taking wi,j = wi+j . 33

Using the isomorphisms between Mi,j and M, we can view H as an orthogonal sum of

inﬁnite copies of M and each vector hi,j in Mi,j as a vector in M.

Convention. In the rest of the proof we suppress the isomorphism symbols for the sake of notational simplicity. To avoid the confusion it may cause we specify the subspace to which a vector belongs.

Hence for vectors hi−1,1 ∈ Mi−1,1 and hi,1 ∈ Mi,1, we still use the symbols hi−1,1 and hi,1 to denote their isomorphic images in Mi,−1. With this convention, for each i ≥ 1, we now

0 deﬁne a vector ui ∈ Mi,−1 given by

0 ui = wi−1,i hi−1,1 + wi,i hi,1 1 1 = h + h w2i−1 i−1,1 w2i i,1

∈ Mi,−1.

We now deﬁne an operator T : H → H so that both T and its adjoint T ∗ are hypercyclic by taking

∞ ! M 0 (3.1) T h = . . . , w−2h−2, w−1h−1, ui, Ah0 + w1h0,1, w2h2, w3h3, w4h4,... . i=1 | {z }

The following ﬁgure is helpful in understanding the decomposition of the Hilbert space H into an orthogonal sum and the action of operator T on H as given by equation (3.1). 34

A M M M M 0 w1 0,1 w2 0,2 w3 0,3 w4

1/w

M1,−3 M1,−2 M1,−1 M1,1 M1,2 M1,3 w−3 w−2 w−1 1/w2 w2 w3 w4

1/w3

M2,−3 M2,−2 M2,−1 M2,1 M2,2 M2,3 w−3 w−2 w−1 1/w4 w2 w3 w4

1/w5

M3,−3 M3,−2 M3,−1 M3,1 M3,2 M3,3 w−3 w−2 w−1 1/w6 w2 w3 w4

...... Fig. Model showing the decomposition of H as an orthogonal sum of subspaces and the

corresponding action of the operator T : H → H.

One can easily verify that this operator T is linear and bounded. For boundedness, we observe that if h ∈ H, then by using the deﬁnition of T and the orthogonality we have

kT hk2 = hT h, T hi 2 ∞ X 2 2 M 0 2 = wj khjk + ui + kAh0 + w1 h0,1k . j∈Z\{0,1} i=1

2 2 2 2 2 First we note that kAh0 + w1 h0,1k ≤ (kAh0k + kw1h0,1k) ≤ 2 kAh0k + w1 kh0,1k . Moreover, by orthogonality, we note that

∞ 2 ∞ ∞ 2 ∞ 2 2 ! M 0 X 0 2 X hi−1,1 hi,1 X khi−1,1k khi,1k u = ku k = + ≤ 2 + . i i w2i−1 w2i w4i−2 w4i i=1 i=1 i=1 i=1 35

So,

kT hk2

∞ 2 2 ! X 2 2 X khi−1,1k khi,1k ≤ w2 kh k2 + 2 kAh k + 2w2 kh k + 2 + j j 0 1 0,1 w4i−2 w4i j∈Z\{0,1} i=1 ∞ X 2 1 2 X 1 1 2 = w2 kh k2 + 2 kAh k + 2 w2 + kh k + 2 + kh k . j j 0 1 w2 0,1 w4i w4i+2 i,1 j∈Z\{0,1} i=1

1 1 Also from the deﬁnition of w, it is clear that w < w1 and thus w < α. Moreover, it is also 1 1 clear that w > 1 and so for each i = 1, 2, 3,..., the terms w4i and w4i+2 are bounded by 1. Therefore,

∞ 2 X 2 2 2 2 2 X 2 kT hk < wj khjk + 2 kAh0k + 4α kh0,1k + 4 khi,1k . j∈Z\{0,1} i=1

Since α > 1, we have kT hk2 < 4α2khk2.

That is, kT k < 2α.

Similar to the above, by suppressing the notation for the isomorphisms between the

subspaces, for any vectors h0 ∈ M0 and h1,−1 ∈ M1,−1, we continue to write h0 and h1,−1 to

denote their isomorphic images in the subspace M0,1 respectively. Similarly, for each i ≥ 1,

we also write hi,−1 and hi+1,−1 to denote their isomorphic images in the subspace Mi,1 of

the vectors hi,−1 ∈ Mi,−1 and hi+1,−1 ∈ Mi+1,−1 respectively. With this convention, we now deﬁne a vector ui ∈ Mi,1 given by

  w1h0 + w0,1h1,−1 if i = 0, ui =  wi,i hi,−1 + wi,i+1 hi+1,−1 if i ≥ 1. 36

We now claim that T ∗ : H → H is given by

∞ ! ∗ ∗ M (3.2) T h = . . . , w−3h−4, w−2h−3, w−1h−2,A h0, ui, w2h1, w3h2,... . | {z } i=0

To understand the action of the operator T ∗ on H, one may simply reverse all the arrows in

∗ the above diagram for T with A : M0 → M0 and the exact same weight sequences.

To verify that the operator T ∗ deﬁned above in (3.2) is indeed the adjoint of T , we show

∗ that hf, T gi = hT f, gi for each pair of vectors f = (. . . , f−2, f−1, f0 , f1, f2,...) and |{z} g = (. . . , g−2, g−1, g0 , g1, g2,...) in H. We use equations (3.2) and (3.1) respectively |{z} to compute hf, T ∗gi and hT f, gi, and show that they are equal:

hf, T ∗gi

* ∞ + ∗ M = (. . . , f−2, f−1, f0 , f1, f2,...), (. . . , w−2g−3, w−1g−2,A g0, ui, w2g1,...) |{z} |{z} i=0 * ∞ + M = ··· + hw−2f−2, g−3i + hw−1f−1, g−2i + hAf0, g0i + f1, ui + hw2f2, g1i + ··· i=0 where ui is a vector in Mi,1 for each i ≥ 0, and is given by

 g w g + 1,−1 if i = 0,  1 0 w ui = gi,−1 gi+1,−1  + if i ≥ 1, w2i w2i+1 and

hT f, gi

* ∞ M 0 = (. . . , w−2f−2, w−1f−1, ui, Af0 + w1f0,1, w2f2, w3f3, w4f4,...), i=1 | {z } (. . . , g−3, g−2, g−1, g0 , g1, g2, g3,...) |{z} 37

* ∞ + M 0 = ··· + hw−2f−2, g−3i + hw−1f−1, g−2i + ui, g−1 i=1

+ hAf0 + w1f0,1, g0i + hw2f2, g1i + hw3f3, g2i + ···

f f where, for each integer i ≥ 1, the vector u0 = i−1,1 + i,1 is in M . i w2i−1 w2i i,−1 To establish the equality, it suﬃces to observe that

* ∞ + * ∞ + M M 0 f1, ui = hw1f0,1, g0i + ui, g−1 . i=0 i=1

Indeed, if we write f1 = (f0,1, f1,1, f2,1,...) and g−1 = (g1,−1, g2,−1, g3,−1,...), then

* ∞ + M f1 , ui i=0

= hf0,1 , u0i + hf1,1 , u1i + hf2,1 , u2i + ···

D g E D g g E D g g E = f , w g + 1,−1 + f , 1,−1 + 2,−1 + f , 2,−1 + 3,−1 + ··· 0,1 1 0 w 1,1 w2 w3 2,1 w4 w5 f f f f f f = hw f , g i + 0,1 + 1,1 , g + 1,1 + 2,1 , g + 2,1 + 3,1 , g + ··· 1 0,1 0 w w2 1,−1 w3 w4 2,−1 w5 w6 3,−1

* ∞ + M 0 = hw1f0,1, g0i + ui, g−1 . i=1

It immediately follows from the deﬁnition of T that T |M0 = A, and similarly if we let

∗ ∗ P : H → H be the orthogonal projection onto M0, then PT |M0 = A . Indeed, if x ∈ M0

∗ ∗ ∗ ∗ then T x = ..., 0, 0,A x, w1x, 0, 0,... so that PT x = A x. |{z} In the following lemma, we use the deﬁnition of the operator T to show that for any

given vector in M−1 ⊕ M0, there exists a vector in M1 whose image under T approximates the given vector.

Lemma 14. Let ε > 0. 38

(i) Given a vector e0 ∈ M0, there exists a vector x ∈ M1 such that T x = e0 + p, where p

is a vector in M−1 with kpk < ε.

(k) (ii) Given a vector bk ∈ Mk,−1 for some k ∈ N, there exists a vector x ∈ M1 such that

(k) T x = bk + pk, where pk is a vector in M−1 with kpkk < ε.

(iii) Let k0 ∈ N. Given vectors bi ∈ Mi,−1 whenever 1 ≤ i ≤ k0 and a vector e ∈ M0, there Pk exists a vector y ∈ M1 such that T y = e + i=1 bi + f, where f is a vector in M−1 with kfk < ε.

|j|−1 (iv) Let zj ∈ Mj with j ≤ −1. There exists a vector y−1 ∈ M−1 such that T y−1 = zj.

n 1 o Proof. Let ε > 0. To prove (i), we ﬁrst recall w has been chosen so that w > max w1, ≥ w1

e0 1 and note that there exists a large enough integer N ∈ N so that N < ε. For each w w1 integer i ≥ 0, deﬁne a vector xi ∈ Mi,1 by

 i  i w e0 (−1) if 0 ≤ i ≤ N − 1, w1 (3.3) xi =  0 i ≥ N.

L∞ Now if we write x = x0 ⊕ x1 ⊕ x2 ⊕ · · · , then x is a vector in M1 = i=0 Mi,1. So, by using the deﬁnition of T , we get

∞ ! M 0 T x = ..., 0, 0, ui, w1x0, 0, 0,... , i=1 |{z}

0 1−2i −2i where ui = wi−1,i xi−1 + wi,i xi = w xi−1 + w xi is a vector in the subspace Mi,−1.

First we recall that P : H → H is the orthogonal projection onto M0 and note that e0 P T x = w1x0 = w1 = e0. For the index i with 1 ≤ i ≤ N − 1, we use the deﬁnition of w1 0 ui and equation (3.3) to ﬁnd that

i−1 i 0 1−2i −2i 1−2i i−1 w e0 −2i i w e0 ui = w xi−1 + w xi = w (−1) + w (−1) = 0. w1 w1 39

0 Let us write p = uN . Since xN = 0, we have

N−1 N−1 N−1 0 xN−1 xN xN−1 (−1) w e0 (−1) e0 p = uN = 2N−1 + 2N = 2N−1 = 2N−1 = N . w w w w w1 w w1

Thus by our choice of N, we have kpk < ε, and ﬁnally, whenever i ≥ N + 1, we have

x x u0 = i−1 + i = 0. i w2i−1 w2i

Hence,

T x = e0 + p.

bk To prove (ii), we first find a large enough integer N = N(k) ∈ N so that wN < ε. (k) L∞ Define a vector x = x0 ⊕ x1 ⊕ x2 ⊕ · · · in M1 = i=0 Mi,1, with each xi in Mi,1 given by

  0 if 0 ≤ i ≤ k − 1,   i+k i+k (3.4) xi = (−1) w bk if k ≤ i ≤ k + N − 1,    0 if i ≥ k + N.

∞ ! (k) M 0 0 Using the deﬁnition of T , we get T x = ..., 0, 0, ui, 0 , 0, 0,... , where ui = i=1 |{z} 1−2i −2i wi−1,i xi−1 + wi,i xi = w xi−1 + w xi is a vector in the subspace Mi,−1.

First, we note that P T x(k) = 0. Now, for the index i with 1 ≤ i ≤ k − 1, by using the

0 0 1−2i −2i deﬁnition of ui and equation (3.4), we see that ui = w xi−1 + w xi = 0.

When i = k, we have

0 1−2k −2k −2k 2k 2k uk = w xk−1 + w xk = w (−1) w bk = bk. 40

When k + 1 ≤ i ≤ k + N − 1, we have

x x (−1)i+k−1 wi+k−1 b (−1)i+k wi+k b u0 = i−1 + i = k + k = 0. i w2i−1 w2i w2i−1 w2i

0 Moreover, if we write pk = uk+N , then

x x (−1)2k+N−1 w2k+N−1 b (−1)N−1 b p = u0 = k+N−1 + k+N = k = k , k k+N w2k+2N−1 w2k+2N w2k+2N−1 wN

and thus kpkk < ε, by our choice of N. Finally, whenever i ≥ k + N + 1, we have

0 1−2i −2i ui = w xi−1 + w xi = 0.

Hence,

T x = bk + pk.

To prove (iii), we note that by (i), there exists a vector x ∈ M1 such that T x = e+p, with

ε (i) p ∈ M−1 and kpk < . By (ii), for each i with 1 ≤ i ≤ k0, there exists a vector x ∈ M1 k0+1

(i) ε Pk0 (i) such that T x = bi + pi with pi ∈ M1 and kpik < . Thus the vector y = x + x k0+1 i=1

is in M1 such that

k0 k0 k0 X (i) X X T y = T x + T x = e + p + (bi + pi) = e + bi + f, i=1 i=1 i=1

Pk0 where f = p + i=1 pi is in M1 with kfk < ε.

Finally to prove (iv), for a given vector zj ∈ Mj with j ≤ −1, we deﬁne a vector y−1 in M by −1   z−1 if j = −1,  y−1 = zj  if j ≤ −2. w−1w−2 ··· wj+2wj+1  41

|j|−1 Then by deﬁnition of T , it is easy to see that T y−1 = zj.

The previous lemma is not yet suﬃcient for us to prove the main theorem. We need to further provide estimations on the norms of vectors and a way to determine the positive

weights wj to make T dual hypercyclic. Speciﬁcally, we show that for any given vector in

H whose nonzero component are in Mj with |j| ≤ k for some nonnegative integer k, we can

construct a vector having its only nonzero components in ﬁnitely many Mj with j > k such that its image under T approximates the given vector. Moreover, the vector we construct can have arbitrarily small norm.

Lemma 15. Suppose k is a positive integer, and for each integer j with |j| ≤ k we are given

a positive weight wj satisfying 0 < wj ≤ α and also vectors xj and yj in Mj. Let ε > 0.

There exist an integer n ≥ 2k + 2, and weights wj with 0 < wj ≤ α and also vectors xj ∈ Mj for all integers j with 1 + k ≤ |j| ≤ n + k such that if T : H → H is as given by (3.1), then

n−k !

n X (a) T xi < ε, i=−k

n+k ! k

n X X (b) T xi − yi < ε, i=n−k+1 i=−k

n+k X 2 (c) kxik < ε, and i=1+k

(d) xi = 0 whenever −n − k ≤ i ≤ −1 − k.

Proof. Before we determine a desired integer n, let us choose an integer m with m ≥ 1 + k.

In terms of the orthogonal projection P : H → H onto M0, let e be a vector in M0 given by

k ! k ! m X m−k k X (3.5) e = −PT xi = −A PT xi , i=−k i=−k

where the last equality follows from the deﬁnition of T . 42

With this vector e in M0, we apply Lemma 14(i) to ﬁnd a vector a1 in M1 such that

(3.6) T a1 = e + p,

where p ∈ M−1 and kpk < ε. Then for this vector a1 in M1, we deﬁne a vector xm in Mm by

a1 (3.7) xm = ∈ Mm w2 ··· wkw1+k ··· wm

m−1 so that T xm = a1. Then,

m m−1 PT xm = PTT xm = P T a1 = P (e + p) = e ∈ M0,

because p ∈ M−1. Thus by using equation (3.5), we have

k ! k m X m m X PT xm + xi = PT xm + PT xi = e + (−e) = 0. i=−k i=−k

m Pk On the other hand, m has been chosen so that m ≥ 1+k, and so T xm + i=−k xi ⊥ Mj whenever j ≥ 1. Hence,

k ! m X (3.8) T xm + xi ⊥ Mj, whenever j ≥ 0. i=−k

Now we deﬁne wj = α for all indices j with 1 + k ≤ j ≤ m. To estimate the norm of

the vector xm, we ﬁrst note that the operator T |Mi : Mi → H is given by z 7→ wiz ∈ Mi−1 for each integer i 6= 0, 1. Therefore, the vector T xm ∈ Mm−1 and is given by T xm = wmxm. Hence

2 1 2 1 2 kxmk = 2 kT xmk = 2 kT xmk . wm α 43

By using (3.6) and (3.7) we get

" 2 # " 2 2 # 2 1 kT a1k 1 kek + kpk kxmk = = , 2 2(m−k) 2 2 2(m−k) 2 α α (w2 ··· wk) α α (w2 ··· wk) which along with equation (3.5), gives

2  k  PT k P x 2(m−k) 2 2(m−k) 1 j=−k j kAk kpk 1 kx k2 ≤  +  . m 2  2 2  α (w2 ··· wk) α (w2 ··· wk) α

Thus we can ﬁrst choose a large enough integer m, even before we determine the desired integer n in the statement of the lemma, so that we have

ε (3.9) kx k2 < . m 2

Then we determine a large enough integer n with n ≥ m + k + 1 ≥ 2k + 2 so that if we deﬁne

1 w−1−k = ··· = w−n−k = α < 1, then it follows from (3.8) and the deﬁnition of T that

k !

n X T xm + xi < ε. i=−k

Thus if we define x1+k = ··· = xm−1 = 0 and xm+1 = ··· = xn−k = 0, then (a) is clearly satisfied. We note that this holds for any sufficiently large integer n which satisfies n ≥ m + k ≥ 2k + 1.

Next we want to deﬁne k vectors xn−k+1, . . . , xn. For each j with −k ≤ j ≤ −1, the

given vector yj can be expressed as yj = (y1,j, y2,j, y3,j,...), where yi,j ∈ Mi,j for each i ≥ 1.

P∞ 2 ε 2 There exists an integer N > 0 such that i=1+N kyi,jk < 2k for each j. Then we deﬁne 0 a vector yj = (y1,j, y2,j, . . . yN,j, 0, 0 ...) for each j, so that

−1 −1 ∞ X X X ε2 ky − y0 k2 = ky k2 < . j j i,j 4k j=−k j=−k i=1+N 44

" −1 #2 −1 X 0 X 0 2 Since yj − yj ≤ k yj − yj , we have j=−k j=−k

k X 0 ε yj − y < . j 2 j=1

0 Now for each vector yj in Mj whenever −k ≤ j ≤ −2, we apply Lemma 14(iv) to get a (j) vector z−1 ∈ M−1 such that

|j|−1 (j) 0 T z−1 = yj.

(j) (j) For each z−1 in M−1, we apply Lemma 14(iii) to vectors 0 ∈ M0 and z−1 ∈ M−1 to get a (j) vector a1 ∈ M1 such that (j) (j) T a1 = z−1 + pj,

ε where pj is a vector in M−1 with kpjk < 2kkT k|j|−1 . Continuing this process, for the vector (j) a1 ∈ M1, we deﬁne a vector xn+j+1 ∈ Mn+j+1 given by

(j) a1 (3.10) xn+j+1 = w2 ··· wkwk+1 ··· wn+j+1

so that

n+j (j) T xn+j+1 = a1 .

Hence, if −k ≤ j ≤ −2, then

n |j| n+j |j| (j) |j|−1 (j) |j|−1 (j) 0 |j|−1 T xn+j+1 = T T xn+j+1 = T a1 = T T a1 = T z−1 + pj = yj + T pj,

|j|−1 |j|−1 ε where T pj is a vector in Mj with T pj < 2k . Thus,

−2 −2 −2 X n X 0 X |j|−1 (3.11) T xn+j+1 = yj + T pj. j=−k j=−k j=−k 45

We move on to handle the given vectors y0, . . . , yk using lemma 14. We apply the Lemma

0 (−1) 14(iii) to vectors y0 ∈ M0 and y−1 ∈ M−1 to ﬁnd a vector a1 ∈ M1 such that

(−1) 0 0 T a1 = y0 + y−1 + p ,

0 0 ε (−1) where p is a vector in M−1 with kp k < 2k . Then for the vector a1 in M1, we deﬁne a

vector xn ∈ Mn, given by

(−1) a1 (3.12) xn = w2 ··· wkwk+1 ··· wn

so that

n−1 (−1) T xn = a1 .

Hence,

n (−1) 0 0 T xn = T a1 = y0 + y−1 + p .

Therefore, using equation (3.11), we get

n ! −1 n X X n T xj = T xn+j+1 j=n−k+1 j=−k −2 −2 X 0 X |j|−1 0 0 (3.13) = yj + T pj + y0 + y−1 + p j=−k j=−k −1 −2 X 0 X |j|−1 0 = yj + y0 + T pj + p , j=−k j=−k

P−2 |j|−1 0 ε with the property that k j=−k T pj + p k < 2 .

Now we deﬁne vectors xn+j in Mn+j whenever 1 ≤ j ≤ k by

yj (3.14) xn+j = ∈ Mn+j wn+j ··· w1+j 46 so that

k k X n X (3.15) T xn+j = yj. j=1 j=1

To estimate the norms of the vectors xn+j given by equation (3.14), ﬁrst we let wj = α > 1 whenever m + 1 ≤ j ≤ n + k and then for suﬃciently large integer n we get

k X 2 ε (3.16) kx k < . n+j 3 j=1

Therefore, from equations (3.13) and (3.15) we get

n+k ! k −1 −2 ! 0

n X X X 0 X |j|−1 0 X T xi − yi = yj + y0 + T pj + p − yi i=n−k+1 i=−k j=−k j=−k i=−k

−1 −2 X 0 X |j|−1 0 ≤ yj − yj + T pj + p j=−k j=−k

< ε.

This proves (b). To estimate the norms of the vectors xj in the above equation, one proceeds similarly as in the paragraph before (3.9) and choose suﬃciently large n so that the vectors

xn−k+1, . . . , xn+k deﬁned in equations (3.10), (3.12) and (3.14) satisfy

n+k X ε kx k2 < . j 2 j=n−k+1

This along with (3.9) and the fact that x1+k = ··· = xm−1 = xm+1 = ··· = xn−k = 0,

gives (c). Part (d) of the lemma is completed by setting x−1−k = ··· = x−n−k = 0.

Since we are to prove that T is dual hypercyclic, we need to establish two lemmas for the adjoint T ∗, analogous to Lemmas 14 and 15. One major diﬀerence between T and T ∗ is 47 that M is not necessarily an invariant subspace of T ∗ even though M is invariant under T . That is also the reason for the complexity of the structure of our operator T .

In the following lemma, we show that for any given vector in M0 ⊕ M1, there exists a

∗ vector in M−1 whose image under T approximates the given vector.

Lemma 16. Let ε > 0.

∗ (i) Given a vector e ∈ M0, there is a vector x ∈ M−1 such that T x = −w1e + p, where

w1e and p are vectors in M1 with kpk < ε.

(k) (ii) Given a vector bk ∈ Mk,1 for some nonnegative integer k, there exists a vector x ∈

∗ (k) M−1 such that T x = bk + pk, where pk is a vector in M1 with kpkk < ε.

(iii) Given a nonnegative integer k0, and vectors e ∈ M0 and bi ∈ Mi,1 whenever 0 ≤ i ≤

∗ ∗ Pk0 k0, there exists a vector y ∈ M−1 such that T (y + e) = A e + i=0 bi + g, where g

∗ ∗ Pk0 is a vector in M1 with kgk < ε and A e ∈ M0. Indeed, T y = −w1e + i=0 bi + g,

where w1e is in M1.

∗(j−1) (iv) Let zj ∈ Mj with j ≥ 1. There exists a vector y1 ∈ M1 such that T y1 = zj.

n 1 o Proof. Let ε > 0. To prove (i), we ﬁrst recall w has been chosen so that w > max w1, ≥ w1

w1 e 1 and note that there exists a positive integer N such that wN < ε. Now we deﬁne a ∞ M vector x = x1 ⊕ x2 ⊕ x3 ⊕ · · · in M−1 = Mi,−1, where xi ∈ Mi,−1 for each i, and i=1

 (−1)i wi (w e) if 1 ≤ i ≤ N  1 (3.17) xi =  0 if i ≥ N + 1.

Then by using the formula of T ∗ in (3.2), we get

∞ ! ∗ ∗ M T (x + e) = ..., 0, 0,A e, ui, 0, 0,... , |{z} i=0 48

where ui is a vector in Mi,1 for each integer i ≥ 0, and is given by

  −1  w1e + w x1 if i = 0, ui =  −2i −2i−1  w xi + w xi+1 if i ≥ 1.

From the expression for T ∗(x + e), it is clear that PT ∗(x + e) = A∗e, where P is the

orthogonal projection onto M0. By using the deﬁnition of ui and the equation (3.17), we

−1 −1 note that u0 = w1e + w x1 = w1e + w [− w (w1e)] = 0. Similarly, for the index i with 1 ≤ i ≤ N − 1, we have

x x wi (w e) wi+1 (w e) u = i + i+1 = (−1)i 1 + (−1)i+1 1 = 0. i w2i w2i+1 w2i w2i+1

If we write p = uN , then

x x wN (w e) w e p = u = N + N+1 = (−1)N 1 = (−1)N 1 , N w2N w2N+1 w2N wN

with kpk < ε. Furthermore, for the index i with i ≥ N + 1, it is clear that ui = 0. Hence

(3.18) T ∗(x + e) = A∗e + p.

∗ ∗ Moreover, by (3.2) we have T e = A e + w1e with w1e ∈ M1, and so

∗ ∗ ∗ T x = T (x + e) − T e = −w1e + p.

bk To prove (ii), we first find an integer N > 0 such that wN < ε. Define a vector 49

∞ (k) M x = x1 ⊕ x2 ⊕ x3 ⊕ · · · in M−1 = Mi,−1, where xi ∈ Mi,−1 for each i, i=1

  0 if 1 ≤ i ≤ k,   i−k−1 i+k (3.19) xi = (−1) w bk if k + 1 ≤ i ≤ k + N,    0 if i ≥ k + N + 1.

Now by (3.2), we see that

∞ ! ∗ (k) M T x = ..., 0, 0, 0 , ui, 0, 0,... , |{z} i=0

where ui is a vector in Mi,1 for each integer i ≥ 0 and is given by

  −1  w x1 if i = 0, ui =  −2i −2i−1  w xi + w xi+1 if i ≥ 1.

An easy calculation involving the use of the equation (3.19) in the deﬁnition of ui shows that

 b if i = k,  k   N−1 u = (−1) bk i N if i = k + N,  w   0 otherwise.

N−1 (−1) bk If we let pk = wN be a vector in the subspace Mk+N,1, then kpkk < ε and hence

∗ (k) T x = bk + pk.

To prove (iii), we repeat the proof for equation (3.18) to show there exists a vector

∗ ∗ ε x ∈ M−1 such that T (x + e) = A e + p, with p ∈ M1 and kpk < . By (ii), for each i k0+1 50

(i) ∗ (i) with 1 ≤ i ≤ k0, there exists a vector x ∈ M−1 such that T x = bi + pi with pi ∈ M1 and

ε Pk0 (i) kpik < . Thus the vector y = x + x is in M−1 such that k0+1 i=1

k0 k0 k0 ∗ ∗ X ∗ (i) ∗ X ∗ X T (y + e) = T (x + e) + T x = A e + p + (bi + pi) = A e + bi + g, i=1 i=1 i=1

Pk0 where g = p + i=1 pi is a vector in M1 with kgk < ε.

∗ ∗ Moreover, since T e = A e + w1e with w1e ∈ M1 by (3.2), we get

k0 ∗ ∗ ∗ X T y = T (y + e) − T e = −w1e + bi + g i=1

Lastly we prove (iv). For any j ≥ 1 and any vector zj ∈ Mj, we deﬁne a vector y1 in M1 by   z1 if j = 1, y1 =  zj  if j ≥ 2. w2w3 ··· wj−1wj

∗j−1 Then by (3.2), it is easy to see that T y1 = zj.

We are now ready to use the previous lemma to prove the analogous version of Lemma 15 for the adjoint T ∗. We recall that we are to show T : H → H is hypercyclic extension

∗ of A : M0 → M0 whose adjoint A : M0 → M0 is hypercyclic. So we can state the following

∗ lemma in terms of a particular hypercyclic vector x0 of A . This vector x0 plays an important role in the construction of a dual hypercyclic vector in the proof of our main theorem.

We now show that for any given vector in H whose component in Mj is zero whenever j > k for some nonnegative integer k, we can construct a vector having its only nonzero components in ﬁnitely many Mj with j > k such that the image of this vector and a hypercyclic vector for A∗ under T ∗ approximates the given vector. Moreover, the vector we construct can have arbitrarily small norm. 51

∗ Lemma 17. Suppose that A : M0 → M0 is hypercyclic and let x0 ∈ M0 be a hypercyclic vector for A∗. Suppose k is a positive integer, and for each integer j with |j| ≤ k we are

given positive weight wj satisfying 0 < wj ≤ α and also vectors xj and yj in Mj. Let ε > 0.

There exists an integer n ≥ 2k + 2, and there exist weights wj and vectors xj ∈ Mj for all integers j with 1 + k ≤ |j| ≤ n + k such that if T ∗ : H → H is as given by (3.2), then

k !

∗n X (i) T xi − x0 < ε, i=−k

−k−1 ! k

∗n X X (ii) T xi + x0 − yi < ε, i=−n−k i=−k

−1−k X 2 (iii) kxik < ε, and i=−n−k

(iv) xi = 0 whenever 1 + k ≤ i ≤ n + k.

Proof. To prove the lemma, we must construct the weights wj and vectors xj for all integers j with 1+k ≤ |j| ≤ n+k. Before we determine the integer n in the statement of the lemma, we choose an integer m ≥ k + 1 and then deﬁne the vectors

(3.20) x−1−k = ··· = x−m = 0.

∗ Let ε > 0. Let x0 ∈ M0 be a hypercyclic vector for A . So for the given vector y0 there

∗n` ε exists an increasing sequence {n`} of positive integers such that kA x0 − y0k < 5 . Thus we can choose a member n of the increasing sequence such that n ≥ m + k + 1 ≥ 2k + 2 and

ε (3.21) kA∗nx − y k < . 0 0 5

1 To develop the proof of the lemma, we deﬁne the weights wj = α whenever 1 + k ≤ j ≤ n + k, and wj = α whenever −n − k ≤ j ≤ −1 − k. To prove (i), ﬁrst we observe that from

∗ ∗ the formula of T in (3.2), if j ∈ Z \{0, −1} then T takes Mj to Mj+1. Note that the vector 52

k X xj −x0 in Statement (i) of the Lemma is in the subspace M−k ⊕· · ·⊕M−1 ⊕M1 ⊕· · ·⊕Mk, j=−k 1 and we have chosen wj = α whenever 1 + k ≤ j ≤ n + k. It now follows easily from (3.2) that there exists a large enough integer N so that whenever n ≥ N, statement (i) of the Lemma holds true.

Next we move on to prove Statements (ii) and (iii).

For each index j with 1 ≤ j ≤ n − m − k, we now deﬁne a vector x−m−j in M−m−j. Let

(j) ∗(m+j−1) (j) e0 = A x0 ∈ M0 and apply Lemma 16(i) to deﬁne a vector a−1 in M−1 so that

∗ (j) (j) 0 (3.22) T a−1 = −w1e0 + pj,

(j) 0 where both w1e0 and pj are vectors in M1 with

0 ε (3.23) kpjk < 5(n − m − k)(w2 ··· wk)

for each j. Then we deﬁne a vector x−m−j in M−m−j given by

(j) a−1 (3.24) x−m−j = w−m−j+1 ··· w−k−1w−k ··· w−1

so that

∗(m+j−1) (j) T x−m−j = a−1 ∈ M−1 for each j, and thus,

∗n ∗(n−m−j) ∗ ∗(m+j−1) ∗(n−m−j) (j) 0 T x−m−j = T T T x−m−j = T −w1e0 + pj .

(j) 0 ∗ Since the vector −w1e0 + pj is in M1, by applying the deﬁnition of T we get

∗n (j) 0 T x−m−j = (w2 ··· wkwk+1 ··· wn−m−j+1) −w1e0 + pj 53

(j) 0 = (w2 ··· wn−m−j+1) −w1e0 + (w2 ··· wkwk+1 ··· wn−m−j+1) pj w ··· w = (w ··· w ) −w e(j) + 2 k p0 , 2 n−m−j+1 1 0 αn−m−j+1−k j

w ··· w where both vectors (w ··· w ) −w e(j) and 2 k p0 are in M 2 n−m−j+1 1 0 αn−m−j+1−k j n−m−j+1 with 1 ≤ j ≤ n − m − k. Hence,

−m−1 ! n−m−k ! ∗n X ∗n X (3.25) T xj = T x−m−j j=−n+k j=1 n−m−k X h (j) w2 ··· wk i = (w ··· w ) −w e + p0 2 n−m−j+1 1 0 αn−m−j+1−k j j=1 n−m n−m X n−j X w2 ··· wk = − (w ··· w ) A∗ x + p0 , 1 j 0 αj−k n−m+1−j j=k+1 j=k+1

n−j w ··· w in which the vectors (w ··· w ) A∗ x and 2 k p0 are in M with k + 1 ≤ 1 j 0 αj−k n−m+1−j j j ≤ n − m.

Now we want to estimate the norms of the vectors x−m−j where 1 ≤ j ≤ n − m − k. For

∗ this, ﬁrst we note that for each integer i with i 6= −1 or 0, the restriction T |Mi : Mi → Mi+1

∗ is the map z 7→ wi+1z. So the vector T x−m−j is in M−m−j+1 and is given by

∗ T x−m−j = w−m−j+1 x−m−j = α x−m−j

for each index j with 1 ≤ j ≤ n − m − k. Hence for each index j with m + 1 ≤ j ≤ n − k, we have

1 (3.26) kx k2 = kT ∗x k2 . −j α2 −j

∗ Now we apply the operator T in the equation (3.24) and then use the deﬁnition of wj for 54 j ≤ −k − 1, and equation (3.22) to see that

2 2 ∗ (j) (j) 0 T a−1 −w1e0 + pj ∗ 2 kT x−m−jk = = . 2(m+j−k−1) 2 2(m+j−k−1) 2 α (w−k ··· w−1) α (w−k ··· w−1)

(j) Hence by using the deﬁnition of e0 , we get

∗m+j−1 2 0 2 2 k−w1 (A x0)k + pj ∗ 2 kT x−m−jk ≤ 2(m+j−k−1) 2 α (w−k ··· w−1)

2(m+j−1) 2 2 kA∗k αkw kx k 2 p0 ≤ 2 1 0 + j . 2(m+j−k−1) 2 α w−k ··· w−1 α (w−k ··· w−1)

Thus,

n−k n−k " ∗ 2(j−1) k 2 2k 0 2 # X 2 X kA k α w1 kx0k α pj−m kT ∗x k = 2 + −j α w ··· w 2 α2(j−1) j=m+1 j=m+1 −k −1 (w−k ··· w−1)

k 2 ∞ ∗ 2(j−1) α w1kx0k X kA k ≤ 2 + w ··· w α −k −1 j=m+1 ∞ 2 α2kε2 X 1 + 25 (w ··· w )2(w ··· w )2 α2(j−1 −k −1 2 k j=m+1 by equation (3.23). Since α > max (1, kA∗k), the above two inﬁnite sums go to zero as m

goes to inﬁnity. So we choose m large enough and then choose n from the sequence {n`} satisfying n ≥ m + k + 1 so that

n−k 2 X 2 α ε kT ∗x k < . −j 3 j=m+1

Hence from equation (3.26) we get

−m−1 X ε (3.27) kx k2 < . j 3 j=−n+k 55

After deﬁning vectors x−n+k ∈ M−n+k, . . . , x−m−1 ∈ M−m−1 in (3.24) and giving an estimate on their norms in (3.27), we now skip a few indices j = −n + k − 1, . . . , j = −n and

proceed to deﬁne vectors x−n−k ∈ M−n−k, . . . , x−n−1 ∈ M−n−1. For each j with 1 ≤ j ≤ k let

y−j (3.28) x−n−j = ∈ M−n−j. w−n−j+1 ··· w−k−1w−k ··· w−j

∗ ∗n Then by using the deﬁnition of T , it follows that T x−n−j = y−j ∈ M−j and hence

−n−1 ! k ! k ∗n X ∗n X X (3.29) T xj = T x−n−j = y−j. j=−n−k j=1 j=1

We also note that for each j with 1 ≤ j ≤ k,

ky−jk ky−jk kx−n−jk = = n+j−k−1 , w−n−j+1 ··· w−k−1w−k ··· w−j α (w−k ··· w−j)

so that kx−n−jk → 0 as n → ∞. In other words, we can choose a large enough integer n

from the sequence {n`} so that

k X 2 ε (3.30) kx k < . −n−j 3 j=1

We now return to deﬁne vectors x−n ∈ M−n, . . . , x−n+k−a ∈ M−n+k−1. For each index j

with 1 ≤ j ≤ k, the given vector yj in Mj can be written as yj = (y0,j, y1,j, y2,j,...) where y ∈ M . There exists an integer n > 0 such that P∞ ky k2 < ε 2 for each j. We i,j i,j 0 i=1+n0 i,j 5k 0 0 deﬁne vectors yj in Mj by yj = (y0,j, y1,j, y2,j, . . . , yn0,j, 0, 0,...) so that

k k ∞ k X X X X ε 2 ε2 ky − y0 k2 = ky k2 < = . j j i,j 5k 25k j=1 j=1 i=1+n0 j=1 56

" k #2 k X 0 X 0 2 Since yj − yj ≤ k yj − yj , we have j=1 j=1

k X 0 ε (3.31) yj − y < . j 5 j=1

0 (j) For each vector yj in Mj, Lemma 16(iv) gives a vector z1 in M1 given by

  0 y1 if j = 1, z(j) = 1 y0  j  if 2 ≤ j ≤ k, w2 ··· wj

∗(j−1) (j) 0 such that T z1 = yj.

∗(n−j) For each j with 1 ≤ j ≤ k, let ej = A x0. We apply Lemma 16(iii) for the vectors ej

(j) (j) and z1 ∈ M1 to ﬁnd a vector x−1 in M−1 such that

∗ (j) (j) (3.32) T x−1 = −w1ej + z1 + pj,

where w1ej and pj are vectors in M1 with

ε  if j = 1 5 (3.33) kpjk <  ε  if j ≥ 2. 5k (w2 ··· wj)

Then we deﬁne a vector x−n+j−1 in M−n+j−1 by

(j) x−1 (3.34) x−n+j−1 = , w−n+j ··· w−1 such that

∗(n−j) (j) T x−n+j−1 = x−1, 57 and thus ∗n ∗(j−1) ∗ ∗(n−j) T x−n+j−1 = T T T x−n+j−1

∗(j−1) (j) = T −w1ej + z1 + pj

0 ∗(j−1) ∗(j−1) = yj + T (−w1ej) + T pj

0 = yj + [w2 ··· wj (−w1ej)] + [w2 ··· wjpj]

is a vector in Mj. Therefore,

−n+k−1 ! k ! ∗n X ∗n X (3.35) T xj = T x−n+j−1 j=−n j=1 k k k X 0 X X = yj − (w1 ··· wj) ej + p1 + (w2 ··· wj) pj j=1 j=1 j=2 k k k X 0 X ∗n−j X = yj − (w1 ··· wj) A x0 + p1 + (w2 ··· wj) pj, j=1 j=1 j=2

∗(n−j) where (w1 ··· wj) A x0 ∈ Mj when 1 ≤ j ≤ k, and also (w2 ··· wj) pj ∈ Mj when

2 ≤ j ≤ k, and p1 ∈ M1.

Next we estimate the norms of the vectors x−n+j−1 where 1 ≤ j ≤ k. For this, we

∗ note that for each integer i with i 6= −1 or 0, the restriction T |Mi : Mi → Mi+1 is the map

∗ z 7→ wi+1z. So the vector T x−n+j−1 is in M−n+j and is given by

∗ T x−n+j−1 = w−n+j x−n+j−1 = α x−n+j−1 ∈ M−n+j,

for each index j with 1 ≤ j ≤ k. Hence for each index j with n − k + 1 ≤ j ≤ n, we have

1 (3.36) kx k2 = kT ∗x k2 . −j α2 −j

∗ Now we apply the operator T on both sides of the equation (3.34), use the deﬁnitions of wj 58

and ej, and use equation (3.32) to see that if 1 ≤ j ≤ k, then

2 ∗(n−j) (j) −w1A x0 + z1 + pj ∗ 2 kT x−n+j−1k = 2(n−k−j) 2 α (w−k ··· w−1) 2 (j) ∗(n−j) 2 2 z + p 2 w1A x0 1 j ≤ + 2(n−k−j) 2 2(n−k−j) 2 α (w−k ··· w−1) α (w−k ··· w−1) 2 2 2j (j) ∗ 2n k 2j α z + pj kA k α w1kx0k α 2 1 ≤ 2 ∗ + 2(n−k) 2 . α w−k ··· w−1 kA k α (w−k ··· w−1)

Thus,

n X ∗ 2 kT x−jk j=n−k+1 k X ∗ 2 = kT x−n+j−1k j=1 2 2 k k 2j (j) ∗ 2n k 2j α z + pj kA k α w1kx0k X α 2 X 1 ≤ 2 + , α w ··· w kA∗k α2(n−k) 2 −k −1 j=1 j=1 (w−k ··· w−1)

( 0 ) (j) 0 yj which goes to 0 as n → ∞ because z1 < max ky1k , , 2≤j≤k w2 ··· wj ε ε ∗ kpjk < max , for each j with 1 ≤ j ≤ k, and α > max {1, kA k}. Hence 2≤j≤k 5 5k(w2 ··· wj) for any suﬃciently large n we have

n 2 X 2 α ε kT ∗x k < . −j 3 j=n−k+1

Hence from (3.36), we get

n X ε (3.37) kx k2 < . −j 3 j=n−k+1

From equations (3.20), (3.27), (3.30) and (3.37), it follows that (iii) holds. 59

To prove (ii), we ﬁrst note that by using the formula of T ∗ in (3.2) repeatedly on the

vector x0 in M0, we get

n ∗n ∗n X ∗(n−j) T x0 = A x0 + (w1 ··· wj)A x0, j=1

∗n ∗(n−j) where A x0 ∈ M0 and (w1 ··· wj)A x0 ∈ Mj. So using this and the equations (3.29), (3.35), (3.25), and (3.20), we get

−k−1 ! ∗n X T xj + x0 j=−n−k −n−1 −n+k−1 −m−1 −k−1 ! ∗n X X X X ∗n = T xj + xj + xj + xj + T x0 j=−n−k j=−n j=−n+k j=−m k k k n−m X X X X w2 ··· wk = y + y0 + p + (w ··· w ) p + p0 + A∗nx −j j 1 2 j j αj−k n−m+1−j 0 j=1 j=1 j=2 j=k+1 n X ∗(n−j) + (w1 ··· wj)A x0 j=n−m+1 k k k X ∗n X 0 X = yj + (A x0 − y0) + yj − yj + p1 + (w2 ··· wj) pj j=−k j=1 j=2 n−m n X w2 ··· wk X + p0 + (w ··· w )A∗(n−j)x . αj−k n−m+1−j 1 j 0 j=k+1 j=n−m+1

Hence,

−k−1 ! k

∗n X X T xj + x0 − yj j=−n−k j=−k

k k ∗n X 0 X ≤ kA x0 − y0k + yj − yj + kp1k + (w2 ··· wj) pj j=1 j=2 n−m n X w2 ··· wk X + p0 + (w ··· w )A∗(n−j)x . αj−k n−m+1−j 1 j 0 j=k+1 j=n−m+1 60

To estimate the last summand above, we replace the index j by n − j and see that

n m−1 X ∗(n−j) X ∗j (w1 ··· wj)A x0 = (w1 ··· wn−j)A x0 j=n−m+1 j=0 m−1 X (w1 ··· wk) = A∗jx αn−j−k 0 j=0 m−1 1 X j ≤ αj w ··· w kA∗k kx k , αn−k 1 k 0 j=0

which goes to 0 as n → ∞. Thus, for a suﬃciently large index n, we have

n X ε (3.38) (w ··· w )A∗(n−j)x < . 1 j 0 5 j=n−m+1

Therefore, using (3.21), (3.23), (3.31), (3.33) and (3.38), we see that

−k−1 ! k

∗n X X T xj + x0 − yj < ε, j=−n−k j=−k which proves (ii).

Taking xj = 0 whenever 1 + k ≤ j ≤ n + k gives (iv), and the proof is complete.

The previous lemma provides a technique to create a weight sequence (wj) and deﬁne

∗ the vectors xj in the subspaces Mj by using the properties of the operator T . Lemma 15 provides a similar technique by using the operator T . We use these lemmas back and forth to construct a hypercyclic vector for both T and T ∗ to prove our main theorem in the next section. 61

3.3 MAIN THEOREM ON DUAL HYPERCYCLIC

EXTENSION

Now we are ready to prove our main result in Theorem 19 which gives a necessary and sufficient condition for the existence of dual hypercyclic extension of an operator on a Hilbert subspace. To give a complete proof of the main theorem, we first prove in Proposition 18 the necessity of the condition and then in Theorem 19 the sufficiency of the condition.

Before we continue our discussion on hypercyclicity, we ﬁrst note the fact that if P :

∗ ∗ H → H is the orthogonal projection onto M, then PT |M = A . This follows from our assumption that T |M = A and so if f, g ∈ M then

hPT ∗f, gi = hT ∗f, P gi = hT ∗f, gi = hf, T gi = hf, Agi = hA∗f, gi.

Using this fact, we now show the following result.

Proposition 18. Let T : H → H be a bounded linear operator. If T |M = A : M → M, and T ∗ is hypercyclic, then A∗ is hypercyclic.

Proof. We show that if f ∈ M and g ∈ M ⊥, and f + g is a hypercyclic vector for T ∗, then f is a hypercyclic vector for A∗. Let P : H → H be the orthogonal projection onto M. Since TM = AM ⊂ M, we have T ∗M ⊥ ⊂ M ⊥, and so T ∗g ∈ M ⊥. Thus

∗ ∗ ∗ ∗ ⊥ T (f + g) = PT f + g1 where g1 = (I − P )T f + T g ∈ M

∗ = A f + g1.

By repeating that argument, we see that

∗n ∗n ⊥ T (f + g) = A f + gn, where gn ∈ M .

It remains to show that {A∗nf : n ≥ 0} is dense in M. For that we let h ∈ M and ε > 0. 62

By the hypercyclicity of f + g for T ∗, there exists an integer n ≥ 0 such that

kT ∗n (f + g) − hk2 < ε.

Thus ∗n 2 ∗n 2 2 kA f − hk ≤ kA f − hk + kgnk

∗n 2 = kA f − h + gnk = kT ∗n (f + g) − hk2 < ε.

Hence A∗n f : n ≥ 0 is dense in M.

Remark. It is well known that no linear operator on a ﬁnite dimensional normed space can be hypercyclic; see Kitai [21]. Thus the conclusion of the theorem that A∗ be hypercyclic on M would mean that dim M = ∞.

One may wonder whether the converse of the above proposition holds true. It is indeed the main Theorem of our discussion. To prove that, we have to employ the technical lemmas in the previous section. In particular, we use the lemmas 15 and 17 repeatedly back and forth to construct the weight sequence (wj) and a vector which is hypercyclic for both T and T ∗.

Theorem 19. Let H be a separable, inﬁnite dimensional Hilbert space, and M be a nonzero closed subspace of H with dim(H/M) = ∞. A bounded linear operator A : M → M has a dual hypercyclic extension T : H → H if and only if A∗ is hypercyclic.

Proof. It suﬃces to show that if A∗ is hypercyclic then A has a dual hypercyclic extension, as Proposition 18 provides the converse. To begin, we note that dim M = ∞ by the Remark above our Theorem.

Let ∆ be a countable dense subset of M0, which is identiﬁed with M as in the previous section. We continue to use our notation in the previous section to construct the desired 63

dual hypercyclic extension T : H → H. Via the isomorphism between M0 and Mi,j, the isomorphic copy ∆i,j in Mi,j of ∆ is a countable dense subset of Mi,j. Let v1, v2, v3,... be an enumeration of all vectors in H each of which is a ﬁnite orthogonal sum a1 ⊕ · · · ⊕ a` of which each component am is a vector in some ∆i,j. The set of all such vi is dense in H.

Let k ≥ 1 be an integer such that v ∈ L M . After choosing k , let w = α 1 1 |j|≤k1 j 1 j and x = 0 whenever 0 < |j| ≤ k . Write v = P y with each y ∈ M . Let j 1 1 |j|≤k1 j j j −1 2k1 ε1 = 2α > 0.

Lemma 15 provides an integer n1 and weights wj with 0 < wj ≤ α and vectors xj ∈ Mj whenever 1 + k1 ≤ |j| ≤ n1 + k1 such that Statements (a), (b), (c), (d) of the lemma are satisﬁed.

∗ With the above setting, we choose a hypercyclic vector x0 of A and apply Lemma 17 to the same vector v = P y and continue to deﬁne w and x by setting k0 = n + k and 1 |j|≤k1 j j j 1 1 1 0 −1 0 2k1 0 ε1 = 2α > 0. The lemma provides an integer n1 and weights wj with 0 < wj ≤ α, and

0 0 0 vectors xj ∈ Mj, for all indices j in the range 1 + k1 ≤ |j| ≤ n1 + k1 such that Statements (i), (ii), (iii), (iv) of the lemma are satisﬁed.

−1 0 0 2 2k2 In the second step, we let k2 = n1 +k1, and ε2 = 2 α > 0 and assume that, without loss of generality, v ∈ L M . In the case that v is not in the subspace L M , we 2 |j|≤k2 j 2 |j|≤k2 j can choose vi with the least integer i such that vi is in that subspace. We now use lemma

15 and then lemma 17 as we did in the previous case for v1.

−1 0 0 m 2km Inductively, in the m−th step we take km = nm−1 + km−1 and εm = 2 α > 0 and we can assume, without loss of generality, that v ∈ L M . Together with those w m |j|≤km j j 0 0 and xj already deﬁned for indices j with |j| ≤ nm−1 + km−1 = km in the (m − 1)-th step, we obtain from the lemma 15 an integer nm and weights wj with 0 < wj ≤ α, and vectors xj in

Mj, where |j| ≤ nm + km such that

(3.39) xj = 0 for those indices j in the range − nm − km ≤ j ≤ −1 − km, 64

nm+km X 2 1 (3.40) kxjk < , 2mα2km j=1+km

n −k ! m m nm X 1 (3.41) T xj < , 2mα2km j=−km and

n +k ! m m nm X 1 (3.42) T xj − vm < . 2mα2km j=nm−km+1

Next we apply lemma 17 for the same vector v ∈ L M and continue to deﬁne w m |j|≤Mj j j 0 −1 0 0 m 2km and xj by setting km = km + nm and εm = 2 α > 0. For the exact same hypercyclic

0 vector x0 we have chosen, Lemma 17 provides an integer nm and weights wj with 0 < wj ≤ α

0 0 0 and vectors xj, for all indices j in the range 1 + km ≤ |j| ≤ nm + km such that

0 0 0 (3.43) xj = 0 for those indices j in the range 1 + km ≤ j ≤ nm + km,

0 −1−km X 2 1 (3.44) kxjk < 0 , m 2km 0 0 2 α j=−nm−km

 0  km 0 1 ∗nm X (3.45) T  xj − x0 < 0 , m 2km 0 2 α j=−km 65 and

 0  −1−km 0 1 ∗nm X (3.46) T  xj + x0 − vm < 0 . m 2km 0 0 2 α j=−nm−km

With all vectors x given by the inductive process, we let x = P x , which represents j j∈Z j a vector in H because the terms xj are mutually orthogonal and square summable. To verify

0 0 0 that, we note that km = nm + km, and km+1 = nm + km by their deﬁnitions, and hence

 0 0  ∞ ∞ nm+km nm+km X 2 X X 2 X 2 kxjk =  kxjk + kxjk  0 j=k1+1 m=1 j=1+km j=1+km

∞ nm+km ! X X 2 = kxjk by (3.43)

m=1 j=1+km ∞ X 1 < by (3.40) 2mα2km m=1 < ∞.

Similarly, using (3.39) and (3.44), we obtain

∞ ∞ X 2 X 1 kxik < 0 < ∞, 2mα2km i=−k1−1 m=1

and so the sum P x deﬁnes a vector in H. j∈Z j

Now, we show that the vector x is a hypercyclic vector for T . For this, we ﬁrst use equation (3.42) and then consider

nm 2 (3.47) kT x − vmk

( "n −k # " ∞ # " n +k # )2 mXm X mXm nm nm nm ≤ T xi + T xi + T xi − vm i=−∞ i=nm+km+1 i=nm−km+1 66

 n −k ! 2 ∞ ! 2  mXm X nm nm 2 ≤ 3  T xi + T xi + εm . i=−∞ i=nm+km+1

By orthogonality, the ﬁrst summand on the right hand side of (3.47) becomes

n −k ! 2 n −k ! 2 −n −k −1 ! 2 mXm mXm mXm nm nm nm (3.48) T xi = T xi + T xi . i=−∞ i=−nm−km i=−∞

To estimate the ﬁrst summand on the right side of equation (3.48), we use (3.39) and (3.41) and get n −k ! 2 n −k ! 2 m m m m 2 nm X nm X 1 T xi = T xi < . 2mα2km i=−nm−km i=−km To estimate the second summand of the equation (3.48), we ﬁrst note that the operator

T |Mj : Mj → Mj−1 for each integer j < 0 is given by z 7→ wjz so that T |Mj < α. Then by

0 0 0 using the deﬁnitions km = nm + km and km+1 = nm + km, and then using equations (3.39) and (3.44), we get

2   0  2 −n −k −1 ! ∞ −k −1 −ki+1−1 m m i nm X nm X X X T xj = T   xj + xj

j=−∞ i=m 0 0 j=−ki−ni j=−ki+1−ni+1 2 −k0−1 ∞ i nm X X = T xj

i=m 0 0 j=−ki−ni 0 ∞ −ki−1 2nm X X 2 < α kxjk i=m 0 0 j=−ki−ni ∞ 2nm X 1 = α 2k0 2iα i i=m ∞ X 1 < , (because k0 ≥ k0 = n + k if i ≥ m) 2i i m m m i=m = 21−m.

h −2 i Hence, the ﬁrst summand in (3.47) is bounded above by 3 2mα2km + 21−m , which 67 goes to 0 as m goes to ∞.

To estimate the second summand of the equation (3.47), we note that the operator

T |Mj : Mj → Mj−1 for each integer j > 1 is given by z 7→ wjz so that T |Mj < α. So, by

0 0 0 using the deﬁnitions km = nm + km and km+1 = nm + km, and then using equations (3.43) and (3.40), we get

  0 0  2 ∞ ! 2 ∞ n +k ni+1+ki+1 i i nm X nm X X X T xj = T   xj + xj

i=m+1 0 j=nm+km+1 j=1+ki j=1+ki+1 2 ∞ ni+1+ki+1 nm X X = T xj

i=m+1 j=1+ki+1   ∞ ni+1+ki+1 2nm X X 2 < α  kxjk  i=m+1 j=1+ki+1 ∞ X 1 < α2nm . 2i+1α2ki+1 i=m+1

0 0 0 We note that ki+1 > km+1 = nm + km > km = nm + km > nm whenever i ≥ m + 1. So we

have α2nm < α2ki+1 for all i ≥ m + 1 and hence the the second summand of the equation (3.47) becomes ∞ ! 2 ∞ X X 1 T nm x < = 2−m−1, j 2i+1 j=nm+km+1 i=m+1

nm 2 which goes to 0 as m goes to ∞. Hence, it follows from (3.47) that kT x − vmk → 0 as

2 m → ∞. Since the set {v1, v2, v3,...} is dense in H, so the set {x, T x, T x, . . .} is dense in H. In other words, T is hypercyclic.

Finally, we show that the vector x is a hypercyclic vector for T ∗. For that, we use equation (3.46) and obtain

0 2 ∗nm (3.49) T x − vm 68

  0 0    −nm−km−1 ∞  0 0 ∗nm X ∗nm X ≤ T  xj + T  xj − x0 + 0  j=−∞ j=−km

 0  2 −1−km 0  ∗nm X + T  xj + x0 − vm 0 0 j=−nm−km 

  0 0  2   2  −nm−km−1 ∞ 2 0 0 1 ∗nm X ∗nm X < 3  T  xj + T  xj − x0 + 0  . m 2km 0 2 α j=−∞ j=−km

By orthogonality, the second summand on the right hand side of (3.49) becomes

  2 ∞ 0 ∗nm X (3.50) T  xj − x0 0 j=−km

 0 0  2   2 nm+km ∞ 0 0 ∗nm X ∗nm X = T  xj − x0 + T  xj . 0 0 0 j=−km j=nm+km+1

To estimate the ﬁrst summand on the right side of equation (3.50), we use (3.43) and (3.45) and get

 0 0  2  0 0 0  2 nm+km km nm+km ∗n0 X ∗n0 X X T m  xj − x0 = T m  xj + xj − x0 0 0 0 j=−km j=−km j=1+km

 0  2 km ∗n0 X = T m  xj − x0 0 j=−km 1 2 < 0 . 2mα2km

To estimate the second summand of the equation (3.50), we ﬁrst note that the operator

∗ T |Mj : Mj → Mj+1 for each integer j > 0 is given by z 7→ wj+1z so that T |Mj < α. Then

0 0 0 by using the deﬁnitions km = nm + km and km+1 = nm + km, and then using equations (3.43) 69 and (3.40), we get

2 2    n0 +k0  ∞ ∞ i+1 i+1 ni+1+ki+1 n0 0 ∗ m X ∗nm X X X T  xj = T   xj + xj

0 0 i=m 0 j=nm+km+1 j=1+ki+1 j=1+ki+1   2 ∞ ni+1+ki+1 0 ∗nm X X = T  xj

i=m j=1+ki+1

∞ ni+1+ki+1 0 2nm X X 2 < α kxjk

i=m j=1+ki+1 ∞ 0 X 1 < α2nm 2i+1α2ki+1 i=m ∞ X 1 < (because k ≥ k > n0 if i ≥ m) 2i+1 i+1 m+1 m i=m = 2−m.

m 2k0 −2 −m Hence, the second summand in (3.49) is bounded above by 3 2 α m + 3 (2 ), which goes to 0 as m goes to ∞.

To estimate the ﬁrst summand on the right hand side of the equation (3.49), we note

∗ that the operator T |Mj : Mj → Mj+1 for each integer j < −1 is given by z 7→ wj+1z so that

0 0 0 T |Mj < α. So, by using the deﬁnitions km = nm + km and km+1 = nm + km, and then using equations (3.39) and (3.44), we get

 0 0  2 −nm−km−1 0 ∗nm X T  xj

j=−∞ 2   −1−k0  ∞ i −1−ki 0 ∗nm X X X = T   xj + xj

i=m+1 0 0 j=−ni−ki j=−ni−ki 2  −1−k0  ∞ i 0 ∗nm X X = T  xj

i=m+1 0 0 j=−ni−ki 0 ∞ −1−ki 0 2nm X X 2 < α kxjk i=m+1 0 0 j=−ni−ki 70

∞ 0 1 2nm X < α 2k0 2iα i i=m+1 ∞ X 1 < because k0 ≥ k0 > k > n0 if i ≥ m + 1 2i i m+1 m+1 m i=m+1 = 2−m,

∗n0 2 which goes to 0 as m goes to ∞. Hence, it follows from (3.49) that T m x − vm → 0 as

∗ ∗2 m → ∞. Since the set {v1, v2, v3,...} is dense in H, so the set {x, T x, T x, . . .} is dense in H. In other words, T ∗ is hypercyclic.

In the proof of our main theorem, we construct a vector x that is hypercyclic for both

∗ T and T . Since the set of all hypercyclic vectors of any hypercyclic operator is a dense Gδ

∗ set, the set of all vectors that are hypercyclic for both T and T is again a dense Gδ set, by the Baire Category Theorem.

As a corollary of the main theorem, we note that a closed invariant subspace M of a dual hypercyclic operator T : H → H is necessarily of inﬁnite dimension. This is simply because A∗ : M → M is necessarily hypercyclic [Prop. 18] and hence dim M = ∞ [?, Therorem 1.2].

Furthermore we can determine a criterion for an operator A on a closed subspace M of H with dim(H/M) = ∞, to be hypercyclic. That is, A : M → M is hypercyclic if and only if A∗ has a dual hypercyclic extension. This follows easily by replacing the operator A in the statement of the main theorem by its adjoint A∗. 71

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APPENDIX

Deﬁnition 20. A Hilbert space H is a complete inner product space.

Deﬁnition 21. A Banach space X is a complete normed space.

Deﬁnition 22. 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 23. 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 24. 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.