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TOPOLOGICAL TRANSITIVITY OF BOUNDED LINEAR
OPERATORS

presented by

Luns ENRIQUE SALDIVIA

has been accepted towards fulfillment
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6/01 cJCIRC/DateDuepss-pJS

Topological 'I‘ransitivity Of

Bounded Linear Operators

By

Luis Enrique Saldivia

A DISSERTATION

Submitted to
Michigan State University
in partial fulfillment of the requirements

for the degree of
DOCTOR OF PHILOSOPHY
Department of Mathematics

2003

ABSTRACT

Topological Transitivity Of
Bounded Linear Operators
By

Luis Enrique Saldivia

A bounded linear operator on a separable Banach space is said to satisfy the
three-neighborhood condition if for every pair U, V of non-empty open subsets of
X, and each open neighborhood W of zero in X there exists a positive integer n
such that both T"U O W and T"W D V are non-empty. The operator is called
syndetically hypercyclic if for any strictly increasing syndetic sequence of positive
integers {nk}k, {Tnk}k is a hypercyclic sequence of operators. We prove that these
two conditions are equivalent to the Hypercyclicity Criterion. Then we prove the
existence of topologically transitive multipliers on Banach algebras and study some

necessary conditions for a multiplier to be topologically transitive on Banach algebras.

To my wife Maria Alejandra and my brother Cesar,
silent but close companions throughout this journey.

iii

ACKNOWLEDGEMENTS

I would like to thank my advisor, Professor Joel H. Shapiro for his teaching, guidance,
encouragement and patience through the time it took me to complete this thesis. I
will always by deeply grateful to him for introducing me to the fascinating world
of the dynamics of operators. His dedication to his work in mathematics is truly
invaluable.

I would like to thank my friend, Professor Alfredo Peris, for the work we did
together, which is the content of Chapter 3 of this thesis, and for all the personal
communications and advice. Muchas gracias Chamo!

I would like to thank my friend, Professor Ruben Martinez, for all the small sem-
inars we organize together and for all personal communications and encouragement.
Gracias Chamaco!

I would like to thank Professors Michael Frazier, Shlomo Levental, William Sledd
and Clifford Weil for all the seminars and classes they gave through the years and for

serving on my committee.

iv

Contents

1 Hypercyclic Fundamentals

2 The Three-Neighborhoods Condition
3 Syndetically Hypercyclic Operators
4 Topologically Transitive Operators

5 Topologically Transitive Multipliers

6 Final Remarks and Questions

20

27

39

45

54

Introduction

Let X be a separable Banach space and T a bounded linear operator on X; i.e.,
T : X ——> X is bounded and linear (referred to simply as operator). A sequence of
operators {Tn}n21 is said to be a hypercyclic sequence on X if there exists some at E X
such that its orbit

Orb({Tn},,, 3:) 2: {:r,T1:1:,T2:I:, . . . }

is dense in X. In this case the vector :17 is called hypercyclic for the sequence {Tn}...
An operator T is hypercyclic on X if {Tn}n is a hypercyclic sequence of operators.
Note that if {Tn}n is a hypercyclic sequence of operators on X, then X is necessarily
separable.

A sufficient condition for hypercyclicity, the well known Hypercyclicity Criterion,
independently discovered by Kitai [21] and Gethner and Shapiro [16], has been the
fundamental tool for proving hypercyclicity. Throughout this thesis we will be using

the following version of this Criterion.

Definition. (The Hypercyclicity Criterion) An operator T on a separable Banach
space X is said to satisfy the Hypercyclicity Criterion provided there exists a strictly
increasing sequence of positive integers {nkh for which there are

1. a dense subset X0 C X such that Tnkzr —* 0 for every :1: 6 X0, and

2. a dense subset Y0 C X and a sequence of mappings {S}c : Y0 ——+ X }k

such that
a) Sky ——> 0 for every y 6 Y0,
b) Tn’c Sky ——> y for every y 6 Y0.

In the first three chapters of this thesis we study several new characterizations of
the Hypercyclicity Criterion.

In Chapter 1 we explain the importance of hypercyclic operators and of the Hy-
percyclicity Criterion. We also mention some important known results about such
operators related to this work. At the end of Chapter 1 we give our first character-

ization of the Hypercyclicity Criterion. (For terminology and notation see Chapter
1.)

Theorem. 1.4 ( The Hypercyclicity Criterion 11) Let T be an operator on a separable
Banach space X. Suppose that there exists a strictly increasing sequence of positive
integers {nk}k21 for which there are:

1. a dense subset X0 C X and r1 > 0 such that Tnkrc —> B(0,r1) for every :1: E X0.

2. a well distributed Y0 C X, a sequence of mappings S'k : Y0 ——> X and r2 > 0
such that:

a) Sky ——> 0 for every y 6 Y0

b) TnkSky — y ——> B(O, r2) for every y 6 Y0.
Then T hypercyclic on X.

We also prove that the Hypercyclicity Criterion II is equivalent to the Hyper-
cyclicity Criterion. (Theorem 1.7.)
In Chapter 2 we use the following condition given By Godefroy and Shapiro [16,

Corollary 1.2], to get our second characterization of the Hypercyclicity Criterion.

2

Definition. 2.1 An operator T on a separable Banach space X is said to satisfy the
three-neighborhoods condition if for every pair U, V of non-empty open subsets of X,

and each open neighborhood W of zero in X there exists a positive integer n such

that both T"U O W and T"W (1 V are non-empty.

Theorem. 2.3 Let X be a separable Banach space and T an operator in X. Then T
satisfies the Hypercyclicity Criterion if and only if T satisfies the three-neighborhoods

condition.
As a corollary of the proof of Theorem 2.3 we get the following.

Corollary. 2.4 An Operator T on a Banach space X satisfies the Hypercyclicity Cri-
terion if and only if:

(I) T is hypercyclic, and

(2) For any non-empty open subset U C X and any open neighborhood of zero W
there is annE N such that T‘Uflwaéll) andT"WflU7é(ll.

Being hypercyclic is equivalent to a property called topological transitivity. A
sequence of continuous maps {Tn}n on a topological space X is topologically transitive
if for any pair U, V of non empty open subsets of X there is a positive integer no such
that TnO(U) 0 V 75 (l).

A single map T : X —> X is topologically transitive if the sequence {T"},, is
topologically transitive.

The three—neighborhoods condition seems to be stronger than topological transi-
tivity, but by Theorem 2.3, it turn out that proving that it is actually stronger is
equivalent to giving an negative answer to the still open problem that every hyper-

cyclic operator satisfies the Hypercyclicity Criterion.

3

Then we apply Theorem 2.3 to give a new proof of the following result due to

Leon and Montes:

Corollary. 2.5 Every hypercyclic bilateral weighted shift satisfies the Hypercyclicity

Criterion.

In [2], Ansari showed that if an operator T on a separable Banach space X is
hypercyclic, then T" is also hypercyclic on X, for every positive integer n. Moreover,
Ansari also showed that T and T" share the same set of hypercyclic vectors. Motivated
by this result, Bés posed the following question: suppose that T is hypercyclcic on a
separable space X and {nk},‘:":l is such that supk{nk+1—nk} < oo (syndetic sequence).
Is {Tnk} a hypercyclic sequence of operators on X .9

The topic of Chapter 3 is related to this question. We start with two definitions:

Definition. 3.1 An operator T on X is called syndetically hypercyclic if for any
strictly increasing syndetic sequence of positive integers {nk}k, the sequence {Tnk :

X -—> X }k is hypercyclic.

Definition. 3.2 Let X be a topological space and T : X ———> X be a continuous map.

T is called weakly mixing if T X T : X x X —> X X X is topologically transitive.
Then we prove the main result of Chapter 3.

Theorem. 3.4 Let T : X ——> X be an operator on a separable Banach space X.
Then the following are equivalent:
(i) T satisfies the Hypercyclicity Criterion.

(ii) T is syndetically hypercyclic.

Theorem 3.4 is a consequence of the following result, which is also interesting in

its own right.

Proposition. 3.3 Let T : X ——> X be a continuous map on a topological space X.
Then the following are equivalent:

(i) T is weakly mixing.

(ii) For any pair of non-empty open subsets U, V C; X, and for any strictly increas-

ing sequence {me};c with supk{nk+1—nk} < 00, there exists k0 such that T"’°o U 0V # 0.

We then prove that if the sequence {nk},c is such that supk{nk+1 — nk} = 00, then
there are hypercyclic operators (satisfying the Hypercyclicity Criterion) such that

{TM},c is not a hypercyclic sequence of operators.

Proposition. 3.5 Suppose {nk};c is such that supk{nk+1 — nk} = 00. Then there

exists a bounded sequence {wn}n of positive scalars such that the unilateral weighted

backward shift T : l2 —> l2, given by

Ten 2

wnen—l fOT 71.22
0 for n=1,

is hypercyclic but {Tnk}k is not a hypercyclic sequence of operators in X.

We also show that even in the case that T satisfies the Hypercyclicity Criterion
and {nk},c is syndetic, T and T"’° do not have to share the same set of hypercyclic

vectors.

Theorem. 3.7 Let T be a hypercyclic operator on a locally convex space X and let
x E X be a hypercyclic vector for T. Then there exists a sequence of positive integers
{nk}k with supk{nk+1 — nk} = 2 such that {T"’°x},c is somewhere dense but not

everywhere dense.

This establishes a difference not only with Ansari’s result, but also between the

full orbit and the sub—orbit associated with a sequence {nk}k, for an operator T. If the

5

orbit under T of any vector is somewhere dense, then the orbit is everywhere dense!
This is a result of Bourdon and Feldman [10].

The equivalence between hypercyclicity and topological transitivity (on separa-
ble Banach spaces) is used to extend the definition of hypercyclicity to spaces not
necessarily separable via topological transitivity. This is the subject of the last two
chapters of this thesis.

In Chapter 4 we extend some well known properties for hypercyclic operator (sep-
arable case) to topologically transitive ones (non-separable case). In particular we

prove the following.

Theorem. 4.3 Let X be a Banach space (separable or not) and T E L(X). IfT is

topologically transitive on X, 0(T) H {A E (C : [A] = 1} 75 (l).

Generalizing the idea of Rolewicz, who showed that if A E (C is such that [A] > 1
and B is the backward shift in I2 then AB is hypercyclic, J. H. Shapiro [4, pg. 1452]
showed the existence of topologically transitive operators on any Hilbert space. In
Example 4.4 we present a more natural example of topologically transitive operators
on a non-separable Banach space.

In Chapter 5 we study the topological transitivity of a special class of operators
on Banach algebras.

Let A be a Banach algebra. Let T in L(A). T is called a left multiplier if T (uv) =
T(u)v V u, v E A, and T is called a right multiplier if T(uv) = uT(v) Vu, v E A.

First we show the existence of topologically transitive multipliers.

Theorem. 5.1 There exists a Banach algebra A and a E A (the unitization of A),
such that the left multiplier L5 6 L(A), given by L;,(b) 2 db V a E A is topologically

transitive on A.

The Banach algebra of Theorem 5.1 is non-unital and this is not casual.

Theorem. 5.3 No multiplier on a Banach algebra with unit element is topologically

transitive.

Then we consider some special cases when A is a non-unital Banach algebra. We

start with commutative Banach algebras.

Theorem. 5.5 No multiplier on a commutative Banach algebra with a non-zero,

bounded, multiplicative linear functional is topologically transitive.

Using the following Comparison Principle we show that a more general result

holds with multiplication operators by elements in the algebra (multipliers of the

form La(b) = ab or Ra(b) = be for some a E A and for all b E A.)

Lemma. 5.7(A Comparison Principle) Let A1, A2 be Banach algebras and T,- E
L(A,), i = 1,2. Let W : A1 —+ A2 be continuous with dense range such that T2 0 \II =

\II 0 T1 . If T1 is topologically transitive, then so is T2.

Proposition. 5.8 Let A be a Banach algebra with a non-zero, bounded, multiplicative
linear functional and let a E A. The operator multiplication by a on A (from the left

or from the right) is not topologically transitive.

We also show that this result holds for general multipliers if the Banach algebra

contains a bounded left approximate identity.

Theorem. 5.9 No left (right) multiplier on a Banach algebra with non-zero, bounded,
multiplicative linear functional and with a bounded left (right) approximate identity

is topologically transitive.

In Chapter 6 we give some final remarks and questions.

Chapter 1

Hypercyclic Fundamentals

Let X be a Banach space. Throughout this thesis L(X) will denote the algebra of
bounded linear operators, T : X —+ X. The elements of L(X) will be referred to
simply as operators. I

We will also be using the following standard notation. For each x E X and e > 0,
let

Bee) = {y e X = lly—xll < .}

Even though most of the results remain true for metrizable and complete topo-

logical vector space (F-spaces), we will keep the underlying space X as a Banach

space.

Definition 1.1. A sequence of operators {Tn},, is said to be a hypercyclic sequence

on X if there exists some x E X such that its orbit
Orb({Tn},,,x) := {x,T1x,T2x, ...}

is dense in X. In this case the vector x is called hypercyclic for the sequence {Tn}...

An operator T is hypercyclic on X if {T"}n. is a hypercyclic sequence of operators.

Note that if {Tn},, is a hypercyclic sequence of operators on X, then X is neces-

sarily separable.

The importance of hypercyclic operators derives from several sources:

(1) The invariant subset problem. Note that Or—b(T,—x) is the smallest closed set,
invariant under T containing the vector x. Thus, an operator lacks invariant closed
subsets if and only if each non-zero vector is hypercyclic.

(2) Density of hypercyclic vectors. Suppose x is a hypercyclic vector for T. Then
every element in 0rb(T, x) is hypercyclic for T. Moreover, if we let H C(T) denote
the set of hypercyclic vectors for T, and S is any countable dense subset of X, then
H C (T) can be written as

oo oo
HC(T) = 0 fl U{x e X: Hm — 8]] < i}.
sESk=1 n=o »
For fixed 3 E S, k 2 1 and n _>_ 0, the set in braces is T’"B(s, i), so it is open by the
continuity of T". Then if T is hypercyclic the set of hypercyclic vectors is a dense Ga
subset of X.

(3) The study of chaotic operators. In the sense of Devaney [13], an operator T
on a separable Banach space X is chaotic if:

a) T is hypercyclic on X.

b) X has a dense subset of periodic points for T.

c) T has sensitive dependence on initial conditions; i.e., there exists a positive
number 6 such that for every 6 > O and every x E X, there is a point y E B (x, e)
such that d(T"x, Tny) > 6 for the same positive integer n.

Godefroy and Shapiro showed that hypercyclic operators have a dramatic form of

this property.

Theorem. Suppose that X is a a Banach space and T is a hypercyclic operator on

X. Then for every x E X there is a dense G6 set of points S(x) C X, such that the

set of orbit-diflerences {Tnx — T"y : n 2 0} is dense in X for every y E S(x).

Thus to prove that an operator is chaotic on a Banach space it is only necessary
to check properties (a) and (b) in Devaney’s definition.

(4) There exist many hypercyclic operators. It is surprising that an operator can
actually be hypercyclic. In fact, Rolewicz [27, pg. 17] showed that no linear operator
on a finite-dimensional space is hypercyclic. This result can easily be seen by using
the following result due to Kitai.

Given a Banach space X, the dual space of X, denoted by X *, is the space of
continuous linear functionals on X. For T E L(X) the adjoint T‘ : X * -—> X * is

defined by T‘A = A o T, for A E X".

Theorem. [21, Collorary 2.4] If an operator on a Banach space is hypercyclic, then

its adjoint has no eigenvalues.

Since any operator on a finite-dimensional space has eigenvalues, and since the
adjoint of a linear operator on a finite dimensinal space is again an operator on a
finite-dimensional space, we get Rolewicz’s result.

Moreover, Kitai’s result can also be used to show that no compact operator on a
Banach space can be hypercyclic. This provides another proof of Rolewicz’s result
since any operator on a finite-dimensional space is compact.

Kitai also showed that no normal (more generally hyponormal) operator can be
hypercyclic on a Banach space.

Nonetheless, hypercyclic operators are more common than one might expect. In
fact the only restrictions are that the underlying space be infinite—dimensional and

separable.

10

Theorem. (Ansari [2], Berna] [5] ) Every infinite dimensional separable Banach space

carries a hypercyclic operator.

Moreover, Bés and Chan [6] recently showed that the set of hypercyclic operators
on a separable Banach space X is dense in the strong operator topology (S.O.T) of
L(X). (The strong operator topology on L(X) is the topology defined by the basic

neighborhoods:
N(T;A,e) = {R : R E L(X),|(T— R)x| < 6,11: 6 A}

where A is an arbitrary finite subset of X. Thus in the strong operator topology a
net {T a} converges to T if and only if {Tax} converges to Tx for every x E X.)

(5) Dense invariant hypercyclic vector manifolds. The fact that the adjoint T‘
of a hypercyclic operator has no eigenvalues was used by Bourdon [9] to show that
if T is hypercyclic and P(z) is any nonzero polynomial, then the operator P(T) has
dense range. Then he showed that every hypercyclic operator on a Banach space X
has a dense invariant hypercyclic vector manifold. To see this, suppose that x E X

is hypercyclic for T. The invariant manifold
[M = {P(T)x : P is a polynomial}

consists entirely, except for zero, of hypercyclic vectors. Indeed, if 0 # P(T)x, then
Orb(T, P(T)x) = P(T)(Orb(T, x)). But the image of a dense set under an operator
with dense range is again dense. Thus P(T)x is hypercyclic for T.

Note that since Orb(T,x) C .M, then the hypercyclic vector manifold M is also
dense.

(6) We can derive new hypercyclic operators from old ones. First note that hy-

percyclicity is invariant under similarity. Indeed, suppose T is a hypercyclic operator

11

on X and S, T1 E L(X) are such that S is invertible and ST 2 T13. If x E X is
hypercyclic for T, then Sx is hypercyclic for T1.
Another way of deriving new hypercyclic operators is using the next result due to

J. H. Shapiro [29, pg. 111], which can be very useful in establishing hypercyclicity

Proposition. ( The Hypercyclicity Comparison Principle) Let T be a continuous lin-
ear operator on a topological vector space X. Let Y be a topological space such that
Y is dense in X and the identity map Iy : Y —> X is continuous. If le : Y —> Y
is a well defined continuous and hypercyclic operator, then T is hypercyclic on X. In

particular T has a hypercyclic vector in Y.

(7) Topological transitivity. Being hypercyclic, for a single operator as well as for
a sequence of commuting operators with dense range, is equivalent (see, for instance
[18, Theorem 1 and Proposition 1]) to a property called topological transitivity. A
sequence of continuous maps {Tn}n on a topological space X is topologically transitive
if for any pair U, V of non empty open subsets of X there is a positive integer no such
that Tn,(U) n V 74 (i).

A single map T : X ——> X is topologically transitive if the sequence {Tn}n is
topologically transitive.

Note that transitivity means that the orbit under T of any non-empty open set U
is dense in X.

By elementary set theory T"(U) (1 V 7t (0 is equivalent to U (1 T ””V # (2). Then

we get at once the following result due to Kitai.

Corollary. [21, Corollary 2.2] Suppose T is an invertible operator on a Banach space

X. Then T is hypercyclic if and only if T‘1 is hypercyclic.

l2

In applications it is sometimes useful to use the following sequential version of
topological transitivity given by Godefroy and Shapiro [16, pg. 233]. For every pair
of vectors x,y E X there exists a sequence {xk} of vectors convergent to x, and a
subsequence {nk} of positive integers, such that Tnkx,c —) y.

But a sufficient condition for hypercyclicity, the well known Hypercyclicity Cri-
terion, independently discovered by Kitai [21] and Gethner and Shapiro [16], has
been the fundamental tool for proving hypercyclicity. The following version of the

Hypercyclicity Criterion was given by Bés and Peris (see [7]).

Theorem. ( The Hypercyclicity Criterion) Let T be an operator on a separable Banach
space X. Suppose that there exists a strictly increasing sequence of positive integers
{nk},c for which there are:
1. a dense subset X0 C X such that Tnkx —> O for every x E X0, and
2. a dense subset Y0 C X and a sequence of mappings {3;c : Y0 ——> X )k such that
a) Sky ——> O for every y E Y0,
b) T"'=Sky —> y for every y E Y0.

Then T is hypercyclic on X.

T is said to satisfy the Hypercyclicity Criterion if it satisfies the hypothesis of last
theorem.

In spite of the complicated statement of the Hypercyclicity Criterion, the result
is often easy to use. To illustrate this, let’s apply the Hypercyclicity criterion to the
first examples of hypercyclic operators on Banach spaces (see Corollary 1.6 in [21]).

For 1 S p < 00, let B be the backward shift on l” defined by B((:r0, x1,:r2, . . .)) =

($1,.T2, . . .)

13

Theorem. (Rolewicz [27, pg. 17]) For every A E C with [A] > 1 the operator AB is

hypercyclic on I? (1 S p < 00).

Proof. Fix 1 S p < co and A E C with |A| > 1. To apply the Hypercyclicity Criterion
we need to find dense subsets X0, Y0, a sequence of positive integers {n.k}:°:1, and a
sequence of mappings Sk : Y0 —+ l” satisfying the hypothesis of this Criterion.

Let X0 be the collection of finite sequences in l”, and {in} = {k} for k = 1, 2,. . ..
Then X0 is dense in l” and for each x E X0 (AB)’°x is eventually zero. Then trivially

(AB)kx —> 0 on X0. Let Y0 be l” itself, and let U denote the forward shift on l”:
U((Io, (131,. . .)) = (0,130, (131, . . ..)

Let 5;, = (A‘IU)’€. Then (AB)’CS;c = I on 1” (where I is the identity operator
on lp). Therefore, again trivially, (AB)"Sky —> y in Y0 = l”. Finally since U is an

isometry and [A] > 1, then Sky —> 0 on Y0 = I”. This completes the proof. 1:]

In fact, every example of hypercyclic operators in the literature so far seems to
satisfy the Hypercyclicity Criterion, but it is still an open question if every hypercyclic
operator satisfies it.

The Hypercyclicity Criterion Problem: Does every hypercyclic operator sat-
isfy the Hypercyclicity Criterion?

This problem has been open for over 15 years. An affirmative answer for this
question would simplify the proof of known results and would answer some still open
problems.

Leén and Miiller [24] recently showed, using a very clever argument for semigroups
of operators, that if T is a hypercyclic operator and A E C with [A] = 1, then AT

is hypercyclic. If T satisfied the Hypercyclicity Criterion the proof of this theorem

14

would be totally trivial; the same dense subsets, sequence of positive integers and
sequence of mappings that satisfy the condition of the Hypercyclicity Criterion for T
would do the job for AT!

In 1992 Herrero [20, Problem 1] proposed the following (still open) problem. Let
T be a hypercyclic operator on a Hilbert space H. Does it follows that the operator
T 63 T on H 69 H is hypercyclic? The answer is clearly (and trivially) positive if T
satisfies the Hypercyclicity Criterion; if X0, Y0, {nk} and 5,, do the job for T, then,
X0 EBXO, Y0 EB Y0, {nk} and 5;, GB 3,. do the job for TEBT on HEB H.

In order to attack the Hypercyclicity Criteron problem, several characterization
of this Criterion have been given.

Related to the Herrerro’s problem, Bés and Peris ( [7], 1999) gave the following

characterization of the Hypercyclicity Criterion.

Theorem. [7, Theorem 2.3] An operator T on a separable Banach space X satisfies

the Hypercyclicity Criterion if and only if T 69 T is hypercyclic on X 63 X.

They also obtained the following result. An operator T on a separable Banach
space X is called hereditarily hypercyclic provided there is a sequence of positive
integers {nk}§‘_’__1 such that for any subsequence {Rafi-:1 of {nk}f__l, {T"":'} is a

hypercyclic sequence of operators on X.

Theorem. [7, Theorem 2.3] An operator on a separable Banach space satisfies the

Hypercyclicity Criterion if and only if it is hereditarily hypercyclic.

Next we will provide an apparently weaker formulation of the Hypercyclicity Cri-
terion. We will also show that this new formulation is equivalent to the usual Hyper-

cyclicity Criterion.

15

Definition 1.2. A subset A g X is called well distributed if there is an r > 0 such

that A H B(x,r) ¢ (0, for every x E X.
We will make use of the following recent result due to Feldman [14].

Theorem. [14, Theorem 2.1]. Let X be a Banach space and T E L(X). Suppose that
there is a vector x E X such that Orb(T,x) := {x,Tx,T2x, ...} is well distributed.

Then T is hypercyclic.

This result is far from obvious since a dense subset is well distributed, but the

converse is false.

Definition 1.3. Let T be an operator on X, A Q X, r > 0 and {72.1,}le be a
sequence of positive integers. We will say that T” converges pointwise to B (0, r) on
A (Tm‘y —> B(O, r) for every y E A) if for each y E A there exists a positive integer
k0 (:2 k0(y)) such that T”’¢(y) E B(0,r) V k 2 k0

Theorem 1.4. ( T he Hypercyclicity Criterion 11) Let T be an operator on a separable
Banach space X. Suppose there exists a strictly increasing sequence of positive integers
{nk}k21 for which there are:
1. a dense subset X0 C X and r1 > 0 such that T"‘°x ——> B(O, r1) for every x E X0.
2. a well distributed Y0 C X, a sequence of mappings Sk : Y0 —-+ X and r2 > 0
such that:
a) Sky —> O for every y E Y0
b) Tn" Sky —- y ——+ B(O, r2) for every y E Y0.

Then T hypercyclic on X.

16

. my

31' iii‘

yam-

Proof. It is enough to prove that under the hypothesis of the theorem there is a
positive number r > 0 and a vector x E X whose orbit under T comes within a
distance r of every point in X. The result will follow using Feldman’s theorem.
By assumption, there is a d > 0 such that Y0 fl B(y, d) # 0, for every y E X. Let P

r0 = r1 + r2 + d and Z = {2,},21 be a countable dense subset of X. Fix 2,- E Z and

I ' 7T 2 hon-l :

let
co.) = Up: 6 X: ||T""x — z.” < r0}.
k=0

Note that C(zi) is open. Moreover,

Claim: G(z,~) is dense in X.
Proof of Claim: let x0 E X and 6 > 0. We want to Show that B (x0, 6) contains points
of G (2,). Since T "kx —> B (0, r1) on the dense subset X0, Sk —> 0 pointwise on the well
distributed set Y0 (which intersect any ball of radius (1), and T” Sk — I yo —+ B (0, r2)
on Y0, for sufficiently large k E N, there exist:

1) x1 E X such that H131- x0||< gandHTnkx1||< r1,

2) yl E X such that ||y1 — 2,-|| < d, [ISkylll < % and ||T"*Sky1— 311” < r2.

For such a k, let x = x1 + Skyl. Then

(5 6
H113 " 5170” S “131— 5130” + ”5ka < 5 + 2 = 5-

Therefore x E B (x0, 6). On the other hand, by linearity of Tn" we have that [|T"*x|| =

||T""x1 + TnkSkylll. Then

”Tn” —’ 22'“ S [[T"“x1|]+ ”kaSkm — 21“ < T1+||Tnksky1- Ill—(31"- will
< T1+||T""Sky1- y1|[+ ”21' — ylll < T1+ 7‘2 + d 2 7'0.
Therefore, x E C(zi). Then G(z,) is dense in X. This concludes the proof of the
claim.

17

But since z,- was an arbitrary element of Z, G(z,-) is dense (and open) for all

i = 1, 2, . . .. By the Baire Category Theorem

We have proved that the set D of vectors in X whose orbit comes within a distance
r0 of a countable dense subset of X is non—empty. Therefore for y E D there is an
r > 0 such that the orbit of y comes within a distance r of any vector of X. Then

Orb(T, x) is well distributed, which implies that T is hypercyclic. El

Remarks 1.5. By the last proof, under the hypothesis of Theorem 1.4 the set D
of vectors whose orbit comes within a distance r of every vector of X is not only
non-empty but dense (by Baire Category Theorem). Feldman [14] also showed if the
orbit of a vector x comes within a distance r of every point of X, then for every 6 > 0,
there exists a vector x6 = fix whose orbit comes within a distance 6 of every point in
X. Therefore, the proof of Theorem 1.4 actually shows that the set of vectors whose

orbit comes within a distance 6 of every point is dense for every 6 > 0.

Definition 1.6. Let T E L(X). We say that T satisfies the Hypercyclicity Criterion
II if T satisfies the conditions of Theorem 1.4 for a dense subset X0, a well dis-
tributed subset Yo, a strictly increasing sequence of positive integers (nk), a sequence

of mappings Sk : Y0 —> X and two positive numbers r1 and r2.

Theorem 1.7. Let T be an operator on a separable Banach space. The following are
equivalent:
1) T satisfies the Hypercyclicity Criterion.

2) T satisfies the Hypercyclicity Criterion II.

18

Proof. 1) implies 2) Follows immediately from the definitions.

On the other hand if T satisfies the Hypercyclicity Criterion II for X0, Y0, the
sequence of positive integers (nk);"_’__1, the sequence of mappings Sc, and r1,r2 > 0,
then TEBT satisfies the Hypercyclicity Criterion II for X0 69 X0, Y0 6}} Y0, the sequence
of positive integers (nkfif’:1 and the sequence of mappings St 619 Sk and r1,r2 > 0.
Therefore, by the Hypercyclicity Criterion II, T {B T is hypercyclic, and then by Bés

and Peris [7, Theorem 2.3], T satisfies the Hypercyclicity Criterion. El

19

 

Chapter 2

The Three-Neighborhoods
Condition

Our second characterization of the Hypercyclicity Criterion is related to the following

sufficient condition for hypercyclicity, given by Godefroy and Shapiro in 1991.

Theorem. [17, Corollary 1.3] An operator T on a separable Banach space X is hy-
percyclic if for every pair U, V of non-empty open subsets of X, and each open neigh-
borhood W of zero in X there exists a positive integer n such that both T"U n W and

T"W {1 V are non-empty.

Proof. We will show that the hypothesis imply the sequential version of topological
transitivity. Let x and y be vectors in X. The hypothesis of this theorem imply that
there are sequences (x2) converging to x, and (xZ) converging to 0, and a subsequence

{nk} of positive integers such that
T""x[c —> 0 and Tnkxz —> y.
Let xk = x]c + xl’. By linearity of the operators T",

Trikxk:Tfllk$;c+TnL-x’k’__)0+y:y_

20

 

Definition 2.1. An operator on a separable Banach space is said to satisfy the three-

neighborhoods condition if it satisfies the hypothesis of last theorem.

Remark 2.2. The three-neighborhoods condition is equivalent to the apparently
stronger requirement that there are infinitely many positive integers n such that both
T"U n W and T"(W) (1 V are non-empty. Indeed, let U1 and V1 be non—empty open
subsets of X and W1 an open neighborhood of zero in X. If n is any fixed positive
integer, by continuity of T" and the three neighborhoods condition applied to the

sets U = U1, V = T‘"V1 and W = T‘"W1 (1 W1, there is an no such that
T"°U1fl(T""W1 (1 W1) 3A (0 and T"°(T—"W1 (1 W1) (1 T'"V1 75 (b.
In particular,
TmoUl fl T—nWI # 0 and T"°W1 DT—nl/l 7g 0.

By elementary set theory we get
Tn0+nU1 fl W1 7E 0 and T"°+"W1r1 V1 # (I).

It is natural to ask whether any hypercyclic operator on a separable Banach space
satisfies the three-neighborhoods condition. By the next theorem an affirmative an-

swer to this question will give an affirmative answer to the Hypercyclicity Criterion

Problem.

Theorem 2.3. Let X be a separable Banach space and T an operator on X. Then T
satisfies the Hypercyclicity Criterion if and only if T satisfies the three-neighborhoods

condition.

Proof. Suppose that T satisfies the Hypercyclicity Criterion. Let U and V be non-

empty open subsets of X and W an open neighborhood of zero in X. Let X 0 and Y0 be

21

the dense subsets of X, {nk}‘,::1 the sequence of positive integers, and S, : Y0 —> X
the sequence of mappings in the hypothesis of the Hypercyclicity Criterion. Since
T""x —+ O pointwise in X0, Sky ——> 0 pointwise in Y0 and T"'°Sky —> y pointwise in

Y0, we can choose xu E U D X0, yv E V (1 Yo and nk large enough such that
Tnkxu E W, Skyv E W, and Tm‘Skyv E V.
Therefore for such nk,
T"kUnW7é(0 and kaansélb.

Then T satisfies the three-neighborhoods condition.

Conversely suppose that T satisfies the three-neighborhoods condition. Then
by Godefroy and Shapiro’s theorem, T is hypercyclic on X. Moreover, the set of
hypercyclic vectors for T (H C (T)) is a dense 05 set.

Claim: we can choose 2 E X hypercyclic for T and {7%}le a sequence of positive
integers such that T"kz —> 0 and T""(B(0, 715)) F1 B(z, %) ¢ 91, for all k E N.

To see how the claim implies the Hypercyclicity Criterion proceed as follows.
Pick such a hypercyclic vector 2 E X and such sequence of positive integers {nk};:1.
Recall that to show that T satisfies the Hypercyclicity Criterion we need to find X0, Y0
dense subsets of X, a sequence of positive integers {nk}2:1 and a sequence of mappings
Sk : Y0 ——> X satisfying the conditions of the Hypercyclicity Criterion. The sequence of
positive integers has already been chosen. Let X0 = Y0 = {Tnz : n = 0, 1, . . ..} Since
2 is a hypercyclic vector for T, X0 and Y0 are dense subsets of X. Also since z satisfies
the claim, for every k E N, T"‘=(B(0, %))flB(z, %) 7t (0, then, T""*B(z, %)flB(O, 71:) 75 (I).
For each k E N, pick wk E T”"*B(z, %) fl B(O, %) and define Sk : Y0 -—> X by

Sanz 2 ank. Since Orb(T, z) is dense, T"°z 74 Tmz if no 75 n1. (In fact, it can

22

be shown that Orb(T, z) a a linearly independent subset of X.) Therefore S, is well
defined.
By the choice of z and {nk}, using the claim we know that Tnkz —> 0. Therefore

for any fixed n E N, by continuity of T"
T""T"z = T"T""z -—> 0 (as k —> 00).

Then T” —> 0 pointwise on X0. On the other hand, since wk —-> 0 and SkT"z = ank,
again by continuity of T”, Sk —> 0 pointwise on Y0.

Finally, since TnkSkz = Tniw;c E B(z, %) for any k E N,
T""Skz —> 2 (as k -+ 00).
Moreover, for a fixed it E N, by continuity of T", and since
TnkSkTmz = fI’""iT"wic = T"T""w,c ——+ T"z,

we have that T” S. converges pointwise to the identity on Y0. Therefore T satisfies
the Hypercyclicity Criterion.
Proof of Claim: Let

co 00
P = B(ngl T“"(B(O, 715)) n {x e X : T"(B(O, [n n B(x, %) 7i 0})
We are going to show that P is a dense G5 set. Then, by the Baire Category
Theorem, H C (T) (1 P will also be a dense G5 set and the proof is complete.
Fix k E N. Since T is continuous, the set in parenthesis (Gk) is open. we want
to show that it is also dense. Let x E X and e > 0. We can assume, without loss
of generality, that e < 311:. We want to show that B(x, e) O Gk 79 (ll. By the three-

neighborhoods condition applied to the sets U = V = B (x, e) and W = B (0, 3, there

23

exists an n E N such that

ruse, a) n B(0, 1?) 7e (2) and T"(B(0, in n B(x, e) 7e 0.

Let y] E B(x,e) with Tnyl E B(0, 1) and y2 E B(x,e) with yg E T"(B(0, %)).
Subclaim: yl E Gk.

Proof of Subclaim: Note that yg E B (x, 6). Therefore, since 26 < %,

1

1
”(J2 E T"B(0, k) (1 B(yi, k)

(and y1 E T‘"(B(O, i)». Then by definition of Gk, y] E Gk. Therefore
yl E B(x,c) fl Gk

which implies that Gk is dense. This completes the proof of the Claim. C]
From the proof of Theorem 2.3 we get the following corollary.

Corollary 2.4. An operator T on a Banach space X satisfies the Hypercyclicity
Criterion if and only if:

(1) T is hypercyclic, and

( 2) For any non-empty open subset U C X and any open neighborhood of zero W

there is an n E N such that TnUflW 7t (0 and T"WflU 75 lb.

The result of Theorem 2.3 can be used to show that a very important class of
hypercyclic operators satisfy the Hypercyclicity Criterion. Namely, let H = l2(Z).
The operator T is a bilateral (forward) weighted shift with respect to the canonical
basis {en : n E Z} if Ten 2 a,,e,,+1 where the sequence of weights {on : n E Z} is

a bounded subset of C \ {0}. Since hypercyclicity is invariant under similarity, and

24

since any bilateral weighted shift is similar to a bilateral weighted shift with positive
weights, we can assume, without loss of generality, that each an is positive.
In 1995, Salas [28] gave necessary and sufficient conditions (on the sequence of

weights) for a bilateral weighted shift to be hypercyclic on H.

Theorem. [28, Theorem 2.1] Let T be a bilateral weighted shift with positive weight
sequence {an}. Then T is hypercyclic if and only if given 6 > 0 and q E N, there

exists n E N arbitrarily large such that for all | j I S q

n-l n 1

gas“ < e and Ears > e
In the proof of this theorem, Salas showed that hypercyclic bilateral weighted
shifts satisfy the following condition ( the S-condition.) If e > 0, and the vectors

g, h E H are in the span of {ej : [j] S q}, then there exists an arbitrarily large n and

a vector u in the span of {ej : ——q —- n S j S q — n} such that
[lull < e, |[T"u—g|| < e and [lTnhH < 6.

Leon and Montes[23, pg. 251] showed that every bilateral weighted shift satisfies
the Hypercyclicity Criterion. Their proof is based on a direct application of the
Hypercyclicity Criterion. Here we provide a proof using the three-neighborhoods

condition.

Corollary 2.5. Every hypercyclic bilateral weighted shift satisfies the Hypercyclicity

Criterion.

Proof. It is enough to show that the S-condition implies the three-neighborhood con-
dition on H = l2(Z). Let U, V be open subsets of H and W an open neighborhood

of zero in H. since the set of finite sequences is dense in H, we can pick h E U,

25

ql E N and cl > 0 such that h E span{ej : [j] S ql} and B(h, 61) C U. Similarly, pick
9 E V, (12 E N and 52 > 0 such that g E span{ej : [j] S (12} and B(g,cg) C V. Let
q = max{q1, q2} and e = min{e1, 62}. Without loss of generality we may assume that
B(0,e) C W. Therefore, h, g E span{ej : I j | S q}. By the S-condition, there exists

an n E N (arbitrarily large) and u E span{ej : —q — n S j S q — n} such that:

Hill] < c, (then u E W)
[T"u — g|| < c, (then T"u E V) and

”Teri” < c, (then T"h e W).

Therefore,

TnUflW750 and annvaélb

which implies that T satisfies the three-neighborhoods condition, and then, by The-

orem 2.3, T satisfies the Hypercyclicity Criterion. El

26

Chapter 3

Syndetically Hypercyclic Operators

Let X be a separable Banach space, and T E L(X). Recall that a sequence of

operators {Tn},, is hypercyclic on X if there exists some at E X such that its orbit
Orb({Tn}n,x) := {x,Tlx,T2x, ...}
is dense in X

Definition 3.1. A strictly increasing sequence of positive integers {nk},c is said to
be syndetic if supk{nk+1 — nk} < 00. An operator T on X is called syndetically
hypercyclic if for any strictly increasing syndetic sequence of positive integers {nk}k,

{TM},c is a hypercyclic sequence of operators.

We will show that T E L(X) is syndetically hypercyclic if and only if T satisfies
the Hypercyclicity Criterion. This partially settles a question posed by Bés, who
asked if every hypercyclic operator is syndetically hypercyclic. Bés’ problem was
motivated by a result of Ansari [1, Theorem 2.1], which asserts that the sequence
{Tm},, is hypercyclic for each p E N whenever T is hypercyclic (see also [3, Theorem
25]). Again, by our equivalence, an affirmative answer to Bés’ question would give

an affirmative answer to the Hypercyclicity Criterion problem.

27

We will also show that if {nk} is not syndetic; i.e., if supk{nk+1 — nk} = 00, there
are examples of operators T satisfying the Hypercyclicity Criterion such that {T'"=},c
is not a hypercyclic sequence of operators

In the final part of this chapter we will show that, for any hypercyclic operator
T E L(X) on a general locally convex space X, and for any vectorx hypercyclic for T,
there exists a strictly increasing sequence of positive integers such that supk{nk+1 —
nk} = 2 and {T""x};lc is not dense in X. However, the sequence {Tnkx},c turns out
to be somewhere dense. This establishes a difference between sub-orbits and orbits
of vectors under T. Bourdon and Feldman [10] recently proved that if a full orbit is
somewhere dense, then it is everywhere dense.

We start with some results from topological dynamics related to the topic of this

chapter.

Definition 3.2. Let X be a topological space and T : X ——> X be a continuous map.

T is called weakly mixing if T x T : X x X —> X x X is topologically transitive.
Kirstenberg [15, Prop. II.3] showed that if T is weakly mixing, then for any m E N

TXTx---xT:XxXx---xX—>XxXx--~xX,
\__,__/

m—times

is topologically transitive.

Banks [3] gave several interesting conditions equivalent to weak mixing.

Theorem. Let X be a topological space and f : X —+ X be continuous. The following
are equivalent:

(a) f is weakly mixing.

(b) For any n E N and U1,U2, . . . Un,V1,V2...V,, non-empty open subsets of X,

there exists k E N such that fk(U,-) O V,- ¢ (I) for alli = 1,. ..n.

28

(c) Given U, V non-empty open subsets ofX there exists n E N such that f"(U) fl
V750 andf"(V)flV;£0.

(d) Given U, V non-empty open subsets of X there exists n E N such that f"(U) n
U750 andf"(U)flV75(l).

(6) Given U, V,W non-empty open subsets of X there exists n E N such that
f"(U) n V aé (i) and f"(W) n W 7A (0.

(f) Given U,V1,V2 non-empty open subsets of X there exists n E N such that
f”(U) 0 V1 75 (b and f"(U) flVg 75 0.

(g) Given U, V,W non-empty open subsets of X there exists n E N such that
f"(U) H V # Q) and f"(V) n W' 75 (0.

(h) Given U1,U2,V non-empty open subsets of X there exists n E N such that

f"(U1) n V 7A 0 and f"(U2) n V 75 (2).

Banks’ proof is given in this order:
a (b ) is Furstemberg’ 8 Theorem.

)=>
b) =>(c ), (b) :> (g), (b) => (h) follows trivially.

(
(
(c) :> (d)=> (e) [3, Lemma 2 and Lemma 3]
(e )=> (f ):> (a) [3, Lemma 4 and lemma 4]

(g=>) (Biand (h) => (6) [3 pg 85]

If f is an operator on a separable Banach space, then by Bés and Peris [7, The-
orem 2.3], f is weakly mixing if and only if f satisfies the Hypercyclicity Criterion.
Thus each of the conditions (b) through (h) is also equivalent to f satisfies the Hy-
percyclicity Criterion.

Note that condition (g) in Banks’ theorem is a formally stronger version of the

three-neighborhoods condition of Chapter 2. (Recall that the three—neighborhoods

29

condition defined in Chapter 2 required that the open set in the middle (in this case
V) be a neighborhood of zero.)

Banks also showed [3, Lemma 8] that if a continuous function f on a topological
space X is:

(I) flip topologically transitive; i.e., for any pair of non-empty open subsets U, V
of X there is a positive integer n such that f"(U) n V 75 (0 and f"(V) (1 U 75 (I), and

(2) f2 is transitive,
then f is weakly mixing.

In the case that f is an operator on a separable Banach space, if f is flip topologi-
cally transitive (which implies that f topologically transitive), then f " is topologically
transitive for all positive integer n (by Ansari [2]). Thus, if f is a flip topologically
transitive operator, then f is weakly mixing. But then since weakly mixing is equiv-
alent to the Hypercyclicity Criterion we get that an operator on a separable Banach
space is flip topologically transitive if and only if it satisfies the Hypercyclicity Crite-
rion.

In Corollary 2.4 we proved a similar result. The difference is that in the corollary
we only required that the open set V, in the definition of flip topological transitivity,
be a neighborhood of zero. But we also required that the operator be topologically
transitive. (A condition that is part of the definition of flip topological transitivity.)

The first result of this chapter, which is fundamental for the desired equivalence
between syndetically hypercyclic and the Hypercyclicity Criterion, remains valid for
continuous maps on topological spaces. Thus we work initially within this general

context.

Proposition 3.3. Let T : X —> X be a continuous map on a topological space X.

30

Then the following are equivalent:
(i) T is weakly mixing.
(ii) For any pair of non-empty open subsets U, V Q X, and for any strictly in-

creasing syndetic sequence {nk}k, there exists k0 such that T"'°0U O V 75 (ll.

Implication (f) => (a) in Banks’ Theorem is necessary for the proof. Also (i) =>
(ii) in the proposition is due to Furstenberg [15, Prop. II.11]. We include those proofs

for the sake of completeness.

Lemma. [3, Lemma 5] Let X be a topological space and f : X —> X a continuous
map. If for any open non-empty subsets U, V1, V2 C X there is an n E N such that

f"(U) O V,- 75 Q), fori = 1,2, then f is weakly mixing.

Proof. Given U1, U2, W1 and W2 non-empty open sunsets of X, we need to find k E N
such that fk(U,) n W,- 79 (ll for i = 1, 2. Apply the hypothesis to the sets U1, U2 and

W2. Then there exists m 2 1 such that
S: fm(U1) nUg 3E and T: fm(U1)flW2 79 0.

Then
S: U1 flf—ng aé and T: U1 (if—"'W2 75 (I).

Now apply the hypothesis to the sets S, T and W1. Then there exists k E N such that
fk(S) 0T 75 (I) and fk(S) (1 W1# 0.

Since S Q U1, we have fk(U1) (1 W1 # (I). Also Y = S O f‘k(T) is non-empty and
open. For x E Y Q S Q f’m(U2) we have fm(x) E U2. But x E Y Q f‘k(T); so

fk($) E T Q f—m(W2) and hence f'"(fk(x)) E W2 which gives fk(U2) (1 W2 # (I). D

31

Proof of Proposition 3.3: (i) implies (ii) [Furstenberg]: Suppose {nk},c and U,V
satisfy the hypothesis of (ii). Set m :2 supk{nk+1 — nk}. Since T is weakly mixing,

the m-product map

TXTX---XT:XXXX---XX-—->XXXX»~XX,
\—\,_/

m—times

is transitive. Then there is an n E N such that
T"U n T‘iV aé 9

for all i = 1,. .. ,m. This implies that T"+‘U n V 76 9 for all i = 1,...,m. By the
assumption on {nk}k, we have that {nk : k E N} fl{n+ 1, . . . ,n+m} 7t 9. If we select
nko in this intersection, we get TnkoU (1 V # 9.

(ii) implies (i): We will show that, given non-empty open subsets U, V1, V2 C X, there
is an n E N, such that T"U D V,- ¢ 9, for i = 1, 2. This will imply that T is weakly
mixing by Banks’ result.

Fix m E N such that TmVl (1 V2 # 9. (Such m exists because (ii) is satisfied.)
By continuity, we can find V1 C V1 open and non—empty such that TmV1 C V2.
Assumption (ii) implies the existence of some I E N such that THU (1 VI 75 9, for
all i = O, 1 . . . , m. (Otherwise we would find a strictly increasing sequence of positive
integers {nk},c such that nk+1 — nk S m + 1, and T"’°U H V] = 9 for all k E N). In

particular we have
T'+mU n V, 7e 0), and T‘+’"U n rev, 3 Tm(T’U n V.) 75 (2).
If we fix n := l + m, we conclude

rnv n vl s (2), and T"U n v2 ,5 0),

32

which completes the proof. C]
We notice that condition (ii) can be equivalently formulated as follows. For any
pair of non-empty open subsets U, V C X, and for any m E N, there exists it E N
such that Tm+iUflV aé 9, i = O,...,m.
Recall that Bés and Peris showed that an operator on a Banach space satisfies
the Hypercyclicity Criterion if and only if it is weakly mixing. Combining this result

with the previous proposition we obtain the next theorem.

Theorem 3.4. Let T : X ——> X be an operator on a separable Banach space X.
Then the following are equivalent:
( i ) T satisfies the Hypercyclicity Criterion.

(ii) T is syndetically hypercyclic.

Proof. (1) implies (ii): If T satisfies the Hypercyclicity Criterion, then by [7, Theorem
2.3], T is weakly mixing. By Proposition 3.3, if {nk},c is a syndetic sequence, then
{Tnk},c is topologically transitive and, therefore, hypercyclic.

(ii) implies (i): Condition (ii) means that, for any syndetic sequence {nk}k, we have
that {T"" }k is hypercyclic; then topologically transitive, and thus, by Proposition 3.3,
T is weakly mixing. Theorem 2.3 of [7] concludes that T satisfies the Hypercyclicity

Criterion. E]

The observation after Proposition 3.3 now yields the following equivalence with
the Hypercyclicity Criterion. Let X be a separable Banach space, U the family of all
non-empty open subsets of X, and T E L(X). Then T satisfies the Hypercyclicity

Criterion if and only if

VU,VEU,VmEN, ElnEN: TUflVsé9, i=n,...,n+m.

33

N on-Syndetic Sequences.

We are going to show that if {nk},c is such that supk{nk+1 - nk} = 00, then there
are hypercyclic unilateral weighted backward shift operators T on the Hilbert space
l2 of square-summable sequences such that {Tm‘},c is not a hypercyclic sequence of
operators. Since hypercyclic weighted backward shifts operators are known to satisfy
the Hypercyclicity Criterion (in fact any hypercyclic operator with dense generalized
kernel satisfies the Hypercyclicity Criterion [7]), then even in the case that T satisfies
the Hypercyclicity Criterion, it is not true that for any sequence {nk}, {Tm} is a
hypercyclic sequence of operators.

Let {wn},, be a bounded sequence of nonzero numbers, and {en}n be the standard
basis of l2. We will be using the following two results, the first due to Salas and the

second to Bés and Peris.

Proposition. [28, Theorem 2.8] Let T be a unilateral weighted backward shift on l2

with positive weight sequence {wn : n E N}. Then T is hypercyclic if and only if

n —
sup" Hk=l wk — 00.

Proposition. [7, Proposition 3.1] Let {nk};c C N, and T be a unilateral weighted
backward shift in l2 with positive weight sequence {wn : n E N}. Then {T""}lc is
hypercyclic if and only if for all e > 0 and all M, q E N, there exists m = m(e,q) E

{nk}k, such that m > M and

I .
wi+l'°‘wi+m > — (1 S 2 S Q)-
6
Proposition 3.5. Suppose {nk};c is such that supk{nk+1 — nk} = 00. Then there

exists a bounded sequence {ru,,},, of positive scalars such that the unilateral weighted

34

backward shift T : l2 ——> l2, given by

wnen_1 for n 2 2
Ten =
0 for n =1,

is hypercyclic but {Tnk},c is not a hypercyclic sequence of operators in X.

Proof. Take an increasing subsequence {nkj}j C {nk};c such that nij — nkj > 23' for

all j Z 1. Define the following sequence of weights. For j = 1, 2, . . . , let

M

if nkj+1SnSnkj+j
g if nkj+j+1SnSnkj+2j
1

E
3
ll

otherwise.

Let T be the unilateral weighted backward shift associated with {wn},,.

1. T is hypercyclic:

By the definition of the sequence of weights we have that wl - - - wm = 2i whenever
m = nkj + j for some j E N. Since j is arbitrarily large, we get supm wl - - - wm = 00
which implies the hypercyclicity of T (using Salas’ result).

2. {TM};c is not a hypercyclic sequence of operators in X:

Note that there are no m E {nk},c in between any sequence of 2’s and {3. Then
for anym E {nk};c we have ’IU2 - nwm =1. Thus w2 - ~~wm+1 S 2. Set q := 1, e := 1/2.
The condition for hypercyclicity of {T""}k given in Bés and Peris’ proposition is not

satisfied. Cl

Finally we will show that even in the case when an operator T satisfies the Hyper-
cyclicity Criterion and {nk},c is a syndetic sequence (then {Tm} is hypercyclic) , the
set of hypercyclic vectors for {TM},c can be strictly contained in the set of hypercyclic
vectors for T. This contrasts with Ansari’s result [2] which shows that if for every
n E N T and T" share the same set of hypercyclic vectors. (Of course that this set

can be empty.)

35

More precisely, we will prove that if T is a hypercyclic operator and x E X is any

hypercyclic vector for T, there exists a syndetic sequence {nk}k, such that the orbit
Orb({T"*}k, x) := {x, me, . . .}

is somewhere dense but not everywhere dense. As we mentioned at the beginning
of the chapter, this establishes a difference between the full orbit and the sub—orbit
associated to a sequence {nk}k, for a single operator T, which should be compared
with the result of Bourdon and Feldman [10].

We begin with a result for standard dynamical systems that came out of a con-

versation with L. Frerick.

Lemma 3.6. Let X be a topological space without isolated points, T : X —-> X a
continuous map, and x E X such that Orb(T , x) is dense in X. Then, for any
syndetic sequence {nk},c of positive integers, the associated orbit Orb({Tnk}k,x) is

somewhere dense.

Proof. If {nk},c is syndetic, we set m :2 supk{nk+1 — nk}. Without loss of generality

n1 > m. Since X has no isolated points and Orb(T, x) is dense in X, we have

 

 

X = {Tnx : n 2 n1} = U{T"k“x : k E N}.
i=0

 

We define M,- := {Tnk—‘x : k E N}, i = 0,... ,m. If int(IVIO) 74 9, then we are done.
If not X = U M,, and this would imply

i=1

m.—

m m—l l
X = T(X) = UT(M,-) = U M,- = U M,.
i=1 i=0 i=1

 

 

By iterating this process we arrive at X = Ill]. Thus X = T(M 1) = IMO, which is a

contradiction. El

36

The next result holds for general locally convex spaces X.

Theorem 3.7. Let T be a hypercyclic operator on a locally convex space X and
let x E X be a hypercyclic vector for T. Then there exists a sequence of positive
integers {nk};c with supk{nk+1 — nk} = 2 such that {Tm‘x};c is somewhere dense but

not everywhere dense.

Proof. Let x E X be a hypercyclic vector for T. Then Orb(T, x) is linearly indepen—
dent. Therefore, T2(x) E span{x,Tx,T3x,T4x}. Then there exists an element x“ in
the dual X’ of X such that <x*,T2x>= land <x*,Tix>== 0 for i = 0, 1, 3,4, where

<x"‘,y> denotes x*(y), for any y E X. Let P: X ——> K3 be given by
P(y) = (<x*,y>, <x*,Ty>, <x*,T2y>)
for all y E X. P is linear, continuous and, since by definition P(x) = (0,0,1),

P(Tx) = (0, 1,0) and P(T2x) = (1,0,0), we have that P is surjective. We define

k if [<x“,T’°+1x>|>[<x*,Tk+2x>|
k = _
k +1 otherwise.

(i) If [<x*,Tk+1x>| > I<x*,Tk+2x>[, then
P(Tnkx) = P(Tkx) = (<x*,Tkx>,<x*,T’°+1x>,<x*,T"+2x>).
(ii) If |<x*,Tk+1x>| S |<x*,Tk+2x>|, then
P(Tnkx) = P(T"+1x) = (<x",Tk+’x>, <x*,T"+2x>, <x*,T“‘+3x>).

Consequently, for any k E N, the second coordinate of P(Tnkx) has magnitude greater
than or equal to the first or the third coordinate of P(kax). By continuity these

inequalities pass on to anything in the closure of the set {P(T"kx)}k. In particular

37

 

(1, 0, 1) E {P(T’Wx) : k E N}. The surjectivity of P implies that {Tnkxh can not be
dense.

To complete the proof note that, by Lemma 3.6, this set is somewhere dense. Cl

38

Chapter 4

Topologically Transitive Operators

Let X be a Banach space (separable or not). The equivalence between hypercyclic
operators and topologically transitive ones (Chapter 1) suggests an extension of the
notion of a hypercyclic operator to Banach spaces which are not necessarily separable.

Recall that an operator T is said to be topologically transitive on X if for any pair

U, V of non-empty open subsets of X there exists a positive integer n such that
T"(U) n V ;£ 0).

It is natural to ask what properties of hypercyclic operators (separable case) can
be extended to topologically transitive ones (non-separable case). For instance the
following two properties, well known for hypercyclic operators, are also enjoyed by
topologically transitive ones:

1. Kitai [21, Corollary 2.4] showed that if T is hypercyclic, then T“ has no eigenval-
ues. Bermudez and Kalton [4, Proposition 3.3], extended this result to topologically
transitive operators.

2. Kitai [21, Corollary 2.8] showed that if T is hypercyclic, then

o(T)fl{AEC:[A|=1}7€9.

39

An argument similar to the one used by Kitai can be used to show that the same
result extends to topologically transitive operators. To prove this we start with two

lemmas. Let r(T) denote the spectral radius of T; i.e., r(T) = {IAI : A E o(T)}.

Lemma 4.1. Let X be a Banach space (separable or not) and T E L(X).
(a) If r(T) < 1, then T is not topologically transitive on X.

(b) If (r(T) C {A E C : [A] > 1}, then T can not be topologically transitive.

Proof. (a) Since r(T) = limn_,+oo [[T"||Tlu, if r(T) < 1, then T is power bounded; i.e.,
there exists a positive constant C such that ||T"|| < C for all n E N. But note
that, as in the separable case, if T is power bounded, then T can not be topologically
transitive on X.

(b) If 0(T) C {A E C : [Al > 1}, then T is invertible and r(T’l) < 1. By (a),
T‘1 is not topologically transitive. But, as we mentioned in Chapter 1, if an invert-
ible operator is topologically transitive, then the inverse is topologically transitive.

Therefore T is not topologically transitive. [:1

Lemma 4.2. Suppose that X1, . . . ,Xn are Banach spaces and that T, E L(Xi) for
i = 1,. ..,n. If T1 619 T2 . . . EB Tn is topologically transitive on X1613 X2 . . . 69 X”, then

T; is topologically transitive on X,-, for alli = 1,... , n.

Proof. Fix i E {1, . . . , ii}. Let U, V be non-empty open subsets of X,. Take

~

U=XICDXQ...EBU,EBX,-+1...(DX,,,

and

V=X1EDX2...@I/IEBXH.1H.CDX".

40

Since U, V are non-empty open subsets of X1 69 X2 . .. EB X", by assumption, there

exists a positive integer no such that
(Tl er,...eTn)"°UnV ¢ (0.
Therefore, TinoU, O V,- # 9. Then T,- is topologically transitive on X,-. D

Theorem 4.3. Let X be a Banach space (separable or not) and T E L(X). If T is

topologically transitive on X, o(T) {1 {A E C : [A] = 1} 75 9.

Proof. If 0(T) {1 {A E C : [A] = 1} = 9, then 0(T) = 01 U 02, where 01 = {A E
C : [A] < 1} and 02 = {A E C : |A| > 1}. We can assume that neither 01 nor 02 is
empty. (If either one is empty, by Lemma 4.1, T is not topologically transitive.) Then
01 and 02 are non-empty, disjoint closed subsets whose union is (r(T). By the Riesz
decomposition theorem [26, pg. 131], there exists subspaces M1 and M2, invariant
under T such that X = All EB M2, and o(T|M,) = o,- for i = 1,2. If we let T,- = TIM,-
for i = 1, 2, then by Lemma 4.1, T,- is not topologically transitive. By Lemma 4.2,

T = T1 GB T2 can not be topologically transitive on X = M1 69 1142. C]

An argument similar to the one used by Bés and Peris ([7]) in their version of the

Hypercyclicity Criterion provides a sufficient condition for topological transitivity.

Proposition. (Topological Transitivity Criterion) Let T be a bounded linear operator
on a complex Banach space X (not necessarily separable). Suppose that there exists
a strictly increasing sequence of positive integers {n1lc}:‘_’__1 for which there are:

(I) A dense subset Xo C X such that T"kx —> 0 for every x E Xo.

(2) A dense subset Yo C X and a sequence of mappings S1,. : Yo —+ X such that:

(a) Sky —-* 0 for every y E Yo.

41

(b) TnkSky ——> y for every y E Yo.

Then T is topologically transitive.

This proposition can be used to show that there is a topologically transitive op-
erator in any Hilbert space H. The idea (provided by J. H. Shapiro, see [4, Example
on pg. 1452]) is as follows. If H is a non-separable Hilbert space, write H = l2(X)
where X is a Hilbert space of the same density character and define T as twice the

backward shift on l2(X); that is,
T(ZL‘1, (172, 1‘3, . . .) 32 2(12, 1133, . . . ).

An argument totally analogous to the one we used in Chapter 1 to show that AB
is hypercyclic in l2(N) whenever |A| > 1, shows that this operator T is topologically
transitive in H.

But there are some differences. As we mentioned in Chapter 1, Ansari and Bernal
showed that any infinite dimensional separable Banach space support hypercyclic
operators. This property does not hold when considering non-separable cases. In [4]
Bermudez and Kalton showed that if X is a non-reflexive quotient of a von Nuemann
algebra (in particular X = loo), then X does not support topologically transitive
operators.

To finish this chapter we present a ”more natural” example of a topologically

transitive operator in a non-separable Banach space.
Example 4.4. Let

A = L3°(R+) = {f E L°°[0,+oo] / V c > 0 3 n E IR : ess supra" |f(x)[ < e}.
A is a subspace of L°°(IR+) and it is easy to check that under the same norm,

llflloo = 888 SUPm W (V f E A),

42

A is closed in L°°(1R+), and therefore it is a Banach subspace of L°°(IR+).
Also note that any function g E L°°[0, 1] can be naturally embedded into A by
setting 5 : g ——-> y where

x 0<x<1
g(x):{g() _ _

0 1<xSoo

Therefore A is a non separable Banach space.

Now, fixa>0anddefineT2Ar——>Aby
f(x) ——+ 2f(x+a) (Vx E [0, +00]).

Then T is obviously in L(A).

Claim: T is topologically transitive in A. .
Proof of Claim:, let f, g E A, c > 0, U = {h E A: [lh — f||00 < e}, and V = {j E
A: “3' -— g||oo < e}. We want to show that 3 m E N such that Tm(U) (1 V 75 9.
Since f E A, there exists no E R such that ess sumen0 | f (x)| < e/ 2. Let m such that

ma > no and 2""‘||g[[00 < 6/2, and

f(fil?) 0 S (E S No
h(x) = 2i,,,g(x — ma) x 2 ma
0 otherwise

Then h E A and, moreover,

llh - flloo = ess supxznolhm - f(x)|

S. 888 suprmalh(z)l + GSS SprZn0[f(.’L‘)|

Therefore h E U.

On the other hand,
Tmh = 2mh(x + ma) 2 g(x), \7’ x 2 0.

43

Then Tmh = g (E V), and therefore T is topologically transitive on A.

Remark 4.5. It is also easy to check that this operator satisfies the Topological
Transitivity Criterion. Indeed, take Xo to be the dense subset of functions in A
with finite support, Yo = A, S : A —> A given by S(f(x)) = g (x — a), nk = k

and 3;, = 3". But given U, V non—empty open subsets of A, the construction in the

previous example gives an explicit vector, h E U and m E N such that T”h E V.

44

Chapter 5

Topologically Transitive
Multipliers

Let A be a Banach algebra and T E L(A). T is called a left multiplier if T(uv) =
T(u)v V u,v E A, and T is called a right multiplier if T(uv) = uT(v) Vu, v E A.

Note that if A is a commutative Banach algebra, a left multiplier is also a right
multiplier, and vice versa.

The most natural examples of multipliers are the multiplication operators. A left
multiplication operator on a Banach algebra A is an operator of the form La(x) = ax
for all x E A. A right multiplication operator on A is an operator of the form Ba (x) =
ax for all x E A.

A Banach space X has the approximation property if for every compact subset
K of X and every 6 > 0 there exists a finite rank operator T E L(X) such that
||Tx — x|| < 6 whenever x E K. Bonet, Peris and Martinez [8, Corollaries 1.3,
1.4] recently showed that if X is a Banach space with the approximation property
such that X’ (X ’ denotes the space of bounded linear functionals on X) is separable
(then X is necessarily separable), T E L(X) satisfies the Hypercyclicity Criterion,

and K is the Banach subalgebra of L(X) of compact operators on X, then the left

45

 

‘r‘"'\‘_fl ““ .— .

multiplication operator LT E L(K) is topologically transitive on K.

Moreover, since any separable Banach space admits compact perturbations of the
identity (operators of the form compact + identity), satisfying the Hypercyclicity
Criterion (see [2] and [5]), and since the unitization K of K is K = K 619 CI, we get

the following.

Theorem 5.1. There exists a Banach algebra A and it E A (the unitization of A),

such that the left multiplication operator Lo E L(A) is topologically transitive on A.

Remarks 5.2. 1) If T in L(X) satisfies the Hypercyclicity Criterion, the same result
holds if we consider the right multiplication operator RT : K —> K.
2) As we mention in the introduction, the Banach algebra on Theorem 5.1 is

necessarily non-unital, as we shall see next.

Suppose now that A is a Banach algebra with unit element e. Any left (right)

multiplier T satisfies
T(v) = T(ev) = T(e)v (T(v) = T(ve) = vT(e)).

So if a = T(e), we get T(v) 2 av (T(v) = va). So any left (right) multiplier T on A

has the form L, (Ra) for some a E A.

Theorem 5.3. No multiplier on a Banach algebra with unit element is topologically

transitive.

Proof. We only show the proof for T a left multiplier. The proof for a right multiplier
is totally analogous.
Let A be a Banach algebra with unit element e. Suppose, in order to get a

contradiction, that T E L(A) is a left multiplier and topologically transitive on A.

46

Let U = {x E A: [Ix — 6” < %} and V = {x E A: ||x|| < i}. By assumption

3 m E N such that Tm(U) (1 V 75 9. Note that if x E U, then x is invertible and

x—1 : 2: (e — x)i.

1.

Then
°° . °° 1 .
llx‘1lls lee-xll‘ s 3,): = 2.
i=0 i=0

Since T is a left multiplier on A (unital), 3 a E A such that T = La. Suppose that
u E U is such that (La)m(u) E V. Then [Iamull < Eli and since u is invertible with

Ila-1H: 2, we get

1
Ilamll = “WW—1H S llamUIIIIU"’|I < 52 =1-

For any positive integer n write n = mk + r for some k E {0, 1,2,...} and some

rE {0,1,...,m—1}. Then
||a"|| = Hamkwll S llamkllllarll S [lam||’°||a"|| < llarll-
Therefore, V n E N,
||a"|| S max{||a'|| :0 S r S m — 1} = M.

But ||T"|| = IILZH S Han” S M, V n E N. Then T is power bounded and therefore

T can not be topologically transitive. Cl

Remark 5.4. Chan [11] showed that if H is a separable Hilbert space and T E
L(H) satisfies the Hypercyclicity Criterion, then LT : L(H) —-> L(H) is topologically
transitive in the strong operator topology. (For the definition of this topology see page
11 of this thesis.) But since L(H) is a unital Banach algebra, LT is not topologically

transitive under the norm topology, for any T E L(H).

47

Next we will consider some special cases when A is a non-unital Banach algebra.
Recall that we have just proved (Theorem 5.3) that no unital Banach algebra supports

topologically transitive multipliers.

Theorem 5.5. No multiplier on a commutative Banach algebra with a non—zero,

bounded, multiplicative linear functional is topologically transitive.

Proof. Let T be a multiplier on a commutative Banach algebra A. Assume that

such a non-zero, bounded, multiplicative linear functional (I) exists. Then there exists

b E A such that <I>(b) = 1. For any a E A
<I>(T(ab)) = (P(aT(b)) = <I>(a)<I>(T(b)).

Similarly, by commutativity of A,

Therefore

 

But <I>(b) = 1; so

Let a = (P(T(b)). Then
<I>(T(a)) = a’¢(a) and T‘<I> = (1(1).

Therefore (I) is an eigenvector for T‘ associated to the eigenvalue 01. Then T is not

t0pologically transitive on A. C]

We previously showed that there exists a Banach algebra A and an element a of

its unitization A such that L5 is topologically transitive (Theorem 5.1). Now we are

48

 

going to show an example of a Banach algebra A such that the set of multipliers
on A’ can also be identified with the unitization A of A, but A does not support

topologically transitive multipliers.

Example 5.6. Let bv be the space of all complex sequences U = (un),,eN, for which

CO
”11”,, ;= sup lunl + 2121,,1— unl < oo.
nEN

n=1

This is the space of sequences of bounded variation. With componentwise operations
bu is a commutative unital Banach algebra.

Note that the elements in bv are convergent sequences; so

@(U) := lim un, V U = (un)n€N E bv

n—+oo .
is a bounded, multiplicative, non-zero linear functional on bv. Let A = ker (I). Then
A is a commutative, non-unital Banach subalgebra of codimension one in bv.

The evaluation functionals (Pk, given by <I>k(U) = uk V U E A, are non-zero,
bounded, multiplicative, linear functionals on A. Therefore no multiplier on A is
topologically transitive. However, the set of multipliers on A can be identified with

bv and bv z A EB C, the unitization of A (See [22], pg. 306).

Suppose now A is a non-unital, non-commutative Banach algebra. In Theorem 5.5
we showed that if a Banach algebra is commutative and there are non-zero, bounded,
multiplicative linear functionals, then no multiplier can be topologically transitive.
The same is true in non—commutative Banach algebras when we consider left or right
multiplications by elements in the algebra. (Note that any algebra can be identified
with a subset of the set of multipliers in the algebra via: a ——+ La V a E A.)

For the proof we need the following version of the Comparison Principle (see

Chapter 1).

49

Lemma 5.7. (A Comparison Principle for Topologically Transitive Operators): Let
A1, A2 be Banach algebras and T,- E L(Ai), i = 1, 2. Let W : A1 —> A2 be continuous
with dense range such that T2 0 \II = \II 0 T1 . If T1 is topologically transitive, then so
is To.

Proof. Let U, V be non-empty open subsets of A2. By the density of the range of \I',
\II‘1(U) and \Il‘1(V) are non-empty open subsets of A1. By topologically transitivity

of T1, 3 no E N such that
nr<r1<U>>n WV) 7e 0.
Therefore
“WOW—1WD 0 ‘1’_I(V)) = WIMP—1W)“ 22—10075 9.
But note that since \IITI = T2\II, we get that \IJTI" = T34! V n = 1, 2, . . .. Therefore
Tg’oqlW’l(U) fl \II\IJ"(V) sé 9.

But \II\II‘1(U) Q U and \II\I"1(V) Q V. Therefore T2"°(U) (1 V # 9. Then To is

topologically transitive on A2. El

Proposition 5.8. Let A be a Banach algebra with a non-zero, bounded, multiplicative
linear functional and a E A. The operator multiplication by a on A (from the left or

from the right) is not topologically transitive.

Proof. Fix a E A. We will give the proof for multiplication by a from the left. The
proof for right multiplication by a is totally analogous.

Let ‘11 be a non-zero, bounded, multiplicative linear functional. We can assume

without loss of generality that \Il(a) 75 0. (If \Il(a) = 0 then \Il(La(X)) = 0, by

50

multiplicativity of ‘11. But Since ‘11 7t 0 then \I' is surjective, and since the image of a
dense set under a functional with dense range is dense in C, we get that La(X) can
not dense in X and then L0 is not topologically transitive on A.)

Let g : C —> C, given by g(a) = \Il(a)a. It is easy to see that g is linear and

continuous. Moreover,
g(‘I/(x)) = \Il(a)\Il(x) = \Il(ax) = \IJ(Lax).

Therefore, g 0 \II = ‘IJLa. So, by Lemma 5.7, if L, were topologically transitive, g
would be hypercyclic. But by Rolewicz (see Chapter 1) there are no hypercyclic

operators on finite dimensional Banach spaces. Cl

Since a Banach algebra is not only always contained (by the identification men-
tioned before) but usually properly contained in the set of multipliers on that algebra,
Proposition 5.8 does not consider every multiplier on the Banach algebra. But we
can extend Proposition 5.8 to general multipliers if the Banach algebra contains a

bounded left approximate identity.

Theorem 5.9. Let A be a Banach algebra with a non-zero, bounded, multiplicative
linear functional and with bounded left (right) approximate identity. Then no left

(right) multiplier on A is topologically transitive.

Proof. Assume that T is a left multiplier on a Banach algebra A. The proof for
right multipliers is totally analogous. Suppose {ea}aeA C A is such that ”80” S M

VaEAandeaa—>a,asoz—+oo,V aEA. Then

 

By continuity of \IJ
‘II(T(eaa)) —> ‘I!(T(a)), as a —> 00.

Therefore

\I'(T(ea))\Il(a) —-> \II(T(a)), as a —> 00.

But

I‘P(T(ea))| S ”‘1’” IITIIM-

So there exists a sub—net {e01} of {ea}ae,\ such that \IJ(T(eaJ.)) is convergent (inde-
pendently of a E A). Say, \II(T(eaj)) ——+ A E C, as j —r 00. Then \I'(T(a)) = A‘Il(a)
V a E A. Therefore, \IJ is an eigenvector of T .* associated with the eigenvalue A, which

implies that T can not be topologically transitive on A. [:1

Example 5.10. Let G be a locally compact abelian group, with group operation +
and Haar measure m. Let L1 (G) consist of all (equivalent classes of ) Borel measurable
complex valued functions on G which are integrable with respect to m. Endowed with
the usual Ll—norm, and with pointwise vector space operations, L1(G) becomes a

Banach space. For f, g E L1(G), the convolution product f at g E L1(G) is defined by

(f * me) := [G f(s — age) dm<t>.

for all s E G. With convolution as a multiplication , L1(G) becomes a commutative
Banach algebra, called the group algebra of G. If the group G is discrete, then L1(G)
has identity in which case no multiplier can be topologically transitive in L1(G). If
the group G is not discrete, L1(G) has no identity, but still, since there are non-zero
multiplicative linear functionals on L1(G), we can see in two ways that no multiplier

can be topologically transitive on L1(G).

52

(1) With convolution as multiplication L1(G) is a commutative Banach algebra.
Then by Theorem 5.5, no multiplier can be topologically transitive on L1(G).

(2) L1(G) always has a bounded approximate identity. Then by Theorem 5.9, no
multiplier can be topologically transitive on L1(G).

A celebrated result by Wendel ([30]) and Helson ([19]), asserts that, the set of
multipliers on L1(G) can be identified with the Banach algebra M (G) of all regular
complex Borel measures on G. In this identification, L1(G) is identified with those

measures in M (G) which are absolutely continuous with respect to Haar measure on

G.

Remark 5.11. If H is a Hilbert space and K is the ideal of compact operators in
L(H), then K has a bounded left approximate identity. Since, by Theorem 5.1, K
admits topologically transitive left multipliers, Theorem 5.9 provides another way of

seen that K does not admit non zero multiplicative linear functionals.

53

Chapter 6

Final Remarks and Questions

Feldman [14] showed that if X is a separable Banach space and y E X is such that
that Orb(T, y) is well distributed, then T is hypercyclic. His proof seems to work
only for the full orbit.

Question 1: Suppose that Orb(T"*,y) is well distributed for some sequence (n,,)1c
and some y E X. Is {TM},c a hypercyclic sequence of operators on X .9

The equivalences to the Hypercyclicity Criterion showed in Chapters 2 and 3 can
be used to reformulate the Hypercyclicity Criterion problem.

Question 2: If T is hypercyclic, does T satisfy the three-neighborhood condition?

Question 3: If T is hypercyclic, is T syndetically hypercyclic?

Condition 2 in Corollary 2.4 seems to be very strong. In fact it might happen
that this condition itself is sufficient for hypercyclicity in which case condition 1 of
this corollary would be redundant.

Question 4: IfT satisfies condition (2) of Corollary 2.4, is T hypercyclic?

Using topological arguments and the fact that for every hypercyclic operator T
there is a manifold consisting entirely, except for zero, of hypercyclic vectors for T,

Ansari [1] showed that if T is hypercyclic, then T" is hypercyclic for any n E N. This

54

gives rise to the following question for the non-separable case.

Question 5: If T is topologically transitive and n E N, is T" topologically transi-
tive?

Since an operator is topologically transitive if and only if any non-empty open
subset has a dense orbit, an affirmative answer to this question can be obtained as a
corollary to the following:

Question 6: Suppose that a non-empty open subset U of a Banach space X is
such that Orb(T, U) is dense in X. Is Orb(T", U) dense in X for any n E N .9

Ansari also showed that Orb(T, x) is dense if and only if Orb(T", x) is dense for
every n E N. So a positive answer to Question 6 would generalize, to some extent,
Ansari’s result.

Recently Chan and Sanders [12] began the study of weakly hypercyclic operators.
An operator T on a Banach space X is called weakly hypercyclic if there is a vector
x E X such that Orb(T, x) is dense in X with respect to the weak topology. Obviously
every hypercyclic operator is weakly hypercyclic, but Chan and Sanders showed that
the converse does not hold.

Chan and Sanders also showed that an invertible weakly hypercyclic operator
need not have weakly hypercyclic inverse [12, Corollary 5.5]. Thus the equivalence
between hypercyclic operators and topologically transitive ones is no longer true when
considering the weak topology. This might not seem surprising given that the main
tool for proving such equivalence for the norm topology is the Baire Category Theorem
which is not available in the weak topology. However the proof of Chan and Sanders
is surprisingly deep.

Nonetheless we can define weak topological transitivity as follows. An operator

55

T on a Banach space X is called weakly topologically transitive if for any non-empty
open subset U C X, Orb(T, U) is dense in X with respect to the weak topology. It

would be interesting to study weakly topologically transitive operators.

56

Bibliography

[1] S. I. Ansari, Hypercyclic and cyclic vectors, J. Funct. Anal. 128 (1995), 374-383.

[2] S. I. Ansari, Existence of hypercyclic operators on topological vector spaces, J.
Func. Anal. 148(2) (1997), 221-232.

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