# Eccentric Orbits PDF Free Download

This opens up the possibility of using GPS based navigation for missions in high altitude orbit, such as Geostationary Operational Environmental Satellites (GOES) in a geostationary orbit, and the Magnetospheric MultiScale (MMS) Mission, in highly eccentric orbits extending to 12 Earth radii and higher. Eccentric Orbits. In Order to Read Online or Download Eccentric Orbits Full eBooks in PDF, EPUB, Tuebl and Mobi you need to create a Free account. Get any books you like and read everywhere you want. Fast Download Speed Commercial & Ad Free. We cannot guarantee that every book is in the library! Download Eccentric Orbits Book For Free in PDF, EPUB. In order to read online Eccentric Orbits textbook, you need to create a FREE account. Read as many books as you like (Personal use) and Join Over 150.000 Happy Readers. We cannot guarantee that every book is in the library. In Eccentric Orbits, John Bloom masterfully traces the conception, development, and launching of Iridium and Colussy’s tireless efforts to stop it from being destroyed, from meetings with his motley investor group, to the Clinton White House, to the Pentagon, to the hunt for customers in special ops, shipping, aviation, mining, search.

**Eccentric Orbits**

An Anthology Of Science Fiction Poetry - Volume 1

#### by**Ken Goudsward**

- Publisher : Unknown Publisher
- Release : 2020-04-04
- Pages : 76
- ISBN : 9781999216061
- Language : En, Es, Fr & De

Eccentric Orbits highlights sci-fi poets from around the globe, celebrating the diversity and unity of the poetic form. Featuring the poetry of: Andrew Burton Deborah L. Kelly Erin J. Bauman Jane Jago Ken Goudsward Kimberly Nugent Mike Van Horn R. C. Larlham Stephanie Barr Thomas Van Horn Wendy Van Camp Lee Garratt

**Eccentric Orbits**

- Publisher : Open Road + Grove/Atlantic
- Release : 2016-06-07
- Pages : 560
- ISBN : 0802192823
- Language : En, Es, Fr & De

“Good corporate drama . . . an enlightening narrative of how new communications infrastructures often come about.” —The Economist, “A Book of the Year 2016” In the early 1990s, Motorola developed a revolutionary satellite system called Iridium that promised to be its crowning achievement. Its constellation of 66 satellites in polar orbit was a mind-boggling technical accomplishment, surely the future of communication. The only problem was that Iridium the company was a commercial disaster. Only months after launching service, it was $11 billion in debt, burning through $100 million a month and crippled by baroque rate plans and agreements that forced calls through Moscow, Beijing, Fucino, Italy, and elsewhere. Bankruptcy was inevitable—the largest to that point in American history. And when no real buyers seemed to materialize, it looked like Iridium would go down as just a “science experiment.” That is, until Dan Colussy got a wild idea. Colussy, a former head of Pan-Am now retired and working on his golf game in Palm Beach, heard about Motorola’s plans to “de-orbit” the system and decided he would buy Iridium and somehow turn around one of the biggest blunders in the history of business. Impeccably researched and wonderfully told, Eccentric Orbits is a rollicking, unforgettable tale of technological achievement, business failure, the military-industrial complex, and one of the greatest deals of all time. “Deep reporting put forward with epic intentions . . . a story that soars and jumps and dives and digresses . . . [A] big, gutsy, exciting book.” —The Wall Street Journal, “A Top 10 Nonfiction Book of 2016” “Spellbinding . . . A tireless researcher, Bloom delivers a superlative history . . . A tour de force.” —Kirkus Reviews (starred review)

**Eccentric Orbits**

Science Fiction Short Stories 1999-2011

#### by**Simon Kewin**

- Publisher : Stormcrow Books
- Release : 2013-11-18
- Pages : 165
- ISBN : 9876543210XXX
- Language : En, Es, Fr & De

An astronaut alone in the void of deep space. An alien starship capable of destroying all creation. A DNA Detective in search of the genetic code of The Beatles. A terrorist explosion trapped inside a bubble of space/time. A new life-form found in the quantum echoes of the void. Eccentric Orbits contains seventeen SF stories originally published between 1999 and 2011 and now collected together for the first time. Stories range from the very short up to novella length. Full Contents Terahertz * wolF emiT * The Armageddon Machine * rho-m10 * 22nd Century Genie * A Loop * Good Vibrations * Ten Million Years * Holy Mountains * Not Better Than One * Remembrance Day * Second War of the Worlds * The Thirteenth Labour * An Explosive Relationship * The Long Walk * Time Dilation * Live From The Continuing Explosion

**The Evolution of Highly Eccentric Orbits**

- Publisher : Unknown Publisher
- Release : 1998
- Pages : 506
- ISBN : 9876543210XXX
- Language : En, Es, Fr & De

**Orbit Determination for Medium Altitude Eccentric Orbits Using Global Positioning System Measurements**

- Publisher : Unknown Publisher
- Release : 1998
- Pages : 236
- ISBN : 9876543210XXX
- Language : En, Es, Fr & De

**Eccentric Orbits**

The Iridium Story - How a Single Man Saved the World's Largest Satellite Constellation From Fiery Destruction

#### by**John Bloom**

- Publisher : Atlantic Books Ltd
- Release : 2016-06-14
- Pages : 123
- ISBN : 1611859565
- Language : En, Es, Fr & De

In the early 1990s, Motorola, the legendary American company, made a huge gamble on a revolutionary satellite telephone system called Iridium. Light-years ahead of anything previously put into space, and built on technology developed for Ronald Reagan's 'Star Wars,' Iridium's constellation of sixty-six satellites in six evenly spaced orbital planes meant that at least one satellite was always overhead. Iridium was a mind-boggling technical accomplishment, surely the future of communication. The only problem was that Iridium was also a commercial disaster. Only months after launching service, it was $11 billion in debt, burning through $100 million a month and bringing in almost no revenue. Bankruptcy was inevitable - the largest to that point in American history. It looked like Iridium would go down as just a 'science experiment.' That is, until Dan Colussy got a wild idea. Colussy, a former CEO of Pan Am, heard about Motorola's plans to 'de-orbit' the system and decided he would buy Iridium and somehow turn around one of the biggest blunders in the history of business. Eccentric Orbits masterfully traces the birth of Iridium and Colussy's tireless efforts to stop it from being destroyed, from meetings with his motley investor group, to the Clinton White House, to the Pentagon, to the hunt for customers in special ops, shipping, aviation, mining, search and rescue. Impeccably researched and wonderfully told, Eccentric Orbits is a rollicking, unforgettable tale of technological achievement, business failure, the military-industrial complex and one of the greatest deals of all time.

**Switch Points for Highly Eccentric Orbits: Modelling the Occurrences of Sign Changes in the Rate of Change of the Eccentricity**

- Publisher : Unknown Publisher
- Release : 2015
- Pages : 329
- ISBN : 9876543210XXX
- Language : En, Es, Fr & De

**The Journal of the Astronautical Sciences**

- Publisher : Unknown Publisher
- Release : 1967
- Pages : 329
- ISBN : 9876543210XXX
- Language : En, Es, Fr & De

**Orbit Determination of Highly Eccentric Orbits Using a RAVEN Telescope**

- Publisher : Unknown Publisher
- Release : 2005
- Pages : 35
- ISBN : 9876543210XXX
- Language : En, Es, Fr & De

For the past eight years, the small automated telescope Raven has been tested in detecting and tracking deep space objects. As the Raven has proven successful in tracking this regular and predictable orbit, its one arc-second accuracy made it a perfect candidate to attempt to accurately track the less predictable Highly Eccentric Orbit (HEO) objects. Ranging data was obtained from the Sirius satellite radio company for the Sirius3 satellite (Satellite Control Center (SCC) # 26626). This satellite was chosen for its long dwell time over the United States and for its favorable Raven tracking conditions. Angles-only data obtained from another Raven telescope located at the AMOS Remote Maui Experiment (RME) facility was used to track the satellite of interest. Then the Analytical Graphics, Inc. Satellite Tool Kit Orbit Determination (STK/OD) program was used to compare the accuracy of the orbit prediction using ranging tracking data from Sirius and angles-only tracking data from Raven. This paper shows the improvement in orbit determination uncertainty obtained by adding Raven observations to the ranging data. The Raven angles data improved the orbit plane uncertainty and eccentricity estimate differences by over 80% when used with the range observations.

**The Use of Eccentric and Circular Orbits in the Design of a Mars Network Constellation**

A Book

#### by**National Aeronautics and Space Administration (NASA)**

- Publisher : Createspace Independent Publishing Platform
- Release : 2018-06-19
- Pages : 30
- ISBN : 9781721288786
- Language : En, Es, Fr & De

This study examined different constellation configurations to determine their suitability for the Mars Network. Some variations on the baseline case of four circular orbits were initially studied. Eccentric orbits were then used to determine their effects on several figures of merit that were selected as representative of the design goals. It was eventually found that the use of eccentric orbits in combination with circular orbits can improve aspects of some navigation and communication figures of merit. The ability to use orbits at different inclinations helps smooth coverage over the middle and upper latitudes for these figures of merit. This configuration has more variability than one consisting of circular orbits, so the occurrence of unfavorable arrangements also results in degradation of some figures of merit. Some cases were used which improved different figures of merit, so a solution could be chosen depending on specified requirements. Anderson, Rodney Jet Propulsion Laboratory

**Reconfiguration and Recovery of Formation Flying Spacecraft in Eccentric Orbits**

- Publisher : Unknown Publisher
- Release : 2009
- Pages : 210
- ISBN : 9780494600115
- Language : En, Es, Fr & De

The problem of reference trajectory reconfiguration and long-term uncontrolled recovery of a formation of spacecraft is considered in an eccentric orbit under the influence of the J2 perturbation. Reference trajectories considered are the Projected Circular Orbit, Along-Track Orbit, and their eccentric modifications. Reconfiguration is accomplished using two, finite-pulse thrusts, modeled as impulses. The state transition matrix (STM) is calculated by four methods: (i) analytically from the Hill-Clohessy-Wiltshire equations, (ii) numerical integration using a fourth-order Runge-Kutta method, (iii) from the fundamental matrix of the linearized equations of motion, and (iv) computing the STM for the relative mean orbital elements, the geometric method. Only the geometric method takes into account J 2, and it is shown to perform the transfers most accurately of all the methods. The methods are also applied to the reconfiguration maneuvers of the University of Toronto's CanX 4/5 formation flying mission.

**The Frequency Spectrum of Gravitational Waves from Eccentric Orbits**

- Publisher : Unknown Publisher
- Release : 2019
- Pages : 39
- ISBN : 9876543210XXX
- Language : En, Es, Fr & De

Gravitational waves emitted from binary black hole systems carry energy away from the orbit, and cause the black holes to spiral towards one another and eventually merge. While all such detected systems are effectively circular at coalescence, these binaries may start out with non-negligible eccentricity. Due to certain astrophysical phenomena, it is expected that some binary systems may merge with some residual eccentricity, and so understanding how to measure the waveforms from these systems are of particular interest for future third generation detectors, such as LISA.

**Two Highly Eccentric Orbits**

- Publisher : Unknown Publisher
- Release : 1928
- Pages : 9
- ISBN : 9876543210XXX
- Language : En, Es, Fr & De

**Attitude Control for Small Satellites on Inclined Eccentric Orbits**

- Publisher : Unknown Publisher
- Release : 1990
- Pages : 18
- ISBN : 9876543210XXX
- Language : En, Es, Fr & De

**Appearance of Emitting Sources on Eccentric Orbits in General Relativity**

A Book

#### by**G. Bao,Erlend Østgaard,P. Hadrava,Universitetet i Trondheim**

- Publisher : Unknown Publisher
- Release : 1993
- Pages : 29
- ISBN : 9876543210XXX
- Language : En, Es, Fr & De

**Spaceflight Mechanics**

Proceedings of the AAS/AIAA Spaceflight Mechanics Meeting

#### by**Anonim**

- Publisher : Unknown Publisher
- Release : 2003
- Pages : 329
- ISBN : 9876543210XXX
- Language : En, Es, Fr & De

**Eccentric Orbits**

- Publisher : Unknown Publisher
- Release : 2017
- Pages : 81
- ISBN : 9781366209979
- Language : En, Es, Fr & De

This thesis deals with what the author terms hypnagogic hallucination; visions and illusions witnessed on the cusp of sleep.

Ubuntu Desktop 14.04 16.10. Windows 10macOS. Windows 7. Red ink pdf free download. macOS 10.13 or higherLinux. Windows 8.

**Newton's apsidal precession theorem and eccentric orbits**

A Book

#### by**Sree Ram Valluri,Curtis Wilson,William Harper**

- Publisher : Unknown Publisher
- Release : 1997
- Pages : 27
- ISBN : 9876543210XXX
- Language : En, Es, Fr & De

**Corrections Due to General Relativity on Radial-velocity Curves for Eccentric Orbits Around Compact Objects**

A Book

#### by**P. Hadrava,G. Bao,Erlend Østgaard,Universitetet i Trondheim**

- Publisher : Unknown Publisher
- Release : 1993
- Pages : 28
- ISBN : 9876543210XXX
- Language : En, Es, Fr & De

## Eccentric Orbitz Pdf Free Download Free

**Eccentric Orbits**

An Anthology Of Science Fiction Poetry, Volume 2

#### by**Ken Goudsward**

- Publisher : Unknown Publisher
- Release : 2021-04-15
- Pages : 90
- ISBN : 9781989940204
- Language : En, Es, Fr & De

Eccentric Orbits is back with more cosmic verse encompassing the globe, celebrating the diversity and unity of the poetic form. Featuring: CATHERINE BROGDON FARUK BUZHALA DALE CHAMPLIN DEBORAH L. KELLY PETER J. KING LEE GARRATT KEN GOUDSWARD ALEX HERNANDEZ AKUA LEZLI HOPE JULEIGH HOWARD-HOBSON JACK MASSA MICHELE MEKEL AMINATH NEENA CHROME OXIDE RK RUGG FABRICE STEPHAN LISA TIMPF WENDY VAN CAMP JEFF YOUNG

## Eccentric Orbit Means

Contents lists available at ScienceDirect

J. Math. Anal. Appl. www.elsevier.com/locate/jmaa

Hypercyclic tuples of operators and somewhere dense orbits Nathan S. Feldman Mathematics Department, Washington & Lee, Lexington, VA 24450, USA

a r t i c l e

i n f o

a b s t r a c t In this paper we prove that there are hypercyclic (n + 1)-tuples of diagonal matrices on Cn and that there are no hypercyclic n-tuples of diagonalizable matrices on Cn . We use the last result to show that there are no hypercyclic subnormal tuples in inﬁnite dimensions. We then show that on real Hilbert spaces there are tuples with somewhere dense orbits that are not dense, but we also give suﬃcient conditions on a tuple to insure that a somewhere dense orbit, on a real or complex space, must be dense. © 2008 Elsevier Inc. All rights reserved.

Article history: Received 21 December 2007 Available online 18 April 2008 Submitted by J.H. Shapiro Keywords: Tuple Hypercyclic Semigroup Orbit Somewhere dense orbit

1. Introduction By an n-tuple of operators we mean a ﬁnite sequence of length n of commuting continuous linear operators on a locally k k k convex space X . If T = ( T 1 , T 2 , . . . , T n ) is an n-tuple of operators, then we will let F = F T = { T 11 T 22 · · · T nn : ki 0002 0} be the semigroup generated by T . If x ∈ X , then the orbit of x under the tuple T or under F is Orb( T , x) = Orb(F , x) = { T x: T ∈ F }. It will also be convenient to write Orb({ T i }i ∈ J , x) for the orbit of x under the semigroup generated by the set of operators { T i }i ∈ J . The tuple T is hypercyclic if there is a vector x ∈ X whose orbit under T is dense in X . When n = 1, then orbits of single operators and hypercyclic operators have been widely studied. The classic example is twice the backward shift on 00022 (N), it was shown to be hypercyclic by Rolewicz [18] in 1969. In fact many natural operators are hypercyclic; they arise within the classes of weighted shifts [19], composition operators [5], co-analytic Toeplitz operators [11], and adjoints of subnormal and hyponormal operators [10]. An important result in hypercyclicity is the so-called “Somewhere Dense Theorem” by Bourdon and Feldman [4]; it states that if T is a continuous linear operator on a locally convex space X , then an orbit for T is either dense in X or nowhere dense in X . In other words, if an orbit is somewhere dense in X , then it must actually be dense in X . Recall that a set E ⊆ X is somewhere dense in X if the closure of E has nonempty interior in X . A set E is nowhere dense in X if the closure of E has empty interior in X . In this paper we give some examples of hypercyclic tuples of operators. Surprisingly hypercyclic n-tuples can arise in ﬁnite dimensions when n > 1, something that does not happen for single operators. For example we prove that there is an (n + 1)-tuple of diagonal matrices that is hypercyclic on Cn and that no n-tuple of commuting diagonalizable matrices on Cn can be hypercyclic. We also prove that there are no hypercyclic tuples of normal operators on an inﬁnite dimensional Hilbert space and no hypercyclic tuples of subnormal operators having commuting normal extensions, on an inﬁnite dimensional space. Surprisingly, examples are given of n-tuples of operators on real Hilbert spaces that have somewhere dense orbits that are not dense! On complex Hilbert spaces the issue is not completely settled, but we present a result that says, with a little extra hypothesis, a somewhere dense orbit must be dense.

E-mail address: [email protected] 0022-247X/$ – see front matter doi:10.1016/j.jmaa.2008.04.027

© 2008 Elsevier Inc.

All rights reserved.

N.S. Feldman / J. Math. Anal. Appl. 346 (2008) 82–98

83

2. Preliminary results and examples Throughout the paper, unless otherwise stated, X will denote a locally convex topological vector space over the complex numbers. One example where we might require X to be a separable Banach space is if the Baire Category Theorem is needed. We begin with some elementary results that are well known for single operators (when applicable to single operators). Some of the standard proofs will not be given. Proposition 2.1. If T = ( T 1 , . . . , T n ) is a commuting n-tuple and if the semigroup F T contains a hypercyclic operator, then T is hypercyclic. Example 2.2. If B is the unilateral backward shift on 00022 (N), then T = (2I , 12 B ) is hypercyclic on 00022 (N). Why? Because 2B is hypercyclic on 00022 (N) and 2B ∈ F T . Proposition 2.3. If F is a set of operators on a separable Banach space X , then F is hypercyclic if for any two open sets U , V there exists T ∈ F such that T (U ) ∩ V 0005= ∅.

0002

Proof. Let { V n } be a countable basis of open sets for X . By assumption, for each n the set T ∈F T −1 ( V n ) is a dense open 0003∞ 0002 set in X . Hence by the Baire Category Theorem, n=1 T ∈F T −1 ( V n ) is a dense G δ set in X consisting of hypercyclic vectors for F . 2 Proposition 2.4. Suppose that T = ( T 1 , . . . , T n ) is a hypercyclic tuple on a separable Banach space X . (1) If T k is invertible for each k, then the tuple ( T 1−1 , . . . , T n−1 ) is also hypercyclic on X . (2) If M is an invariant subspace for T , then the quotient of T is hypercyclic on X /M. (3) Every orbit of T ∗ = ( T 1∗ , . . . , T n∗ ) is unbounded. Proposition 2.5 (Hypercyclicity Criterion). Suppose that ( T 1 , T 2 ) is a pair of operators on a separable Banach space Z . Suppose also that there exist two strictly increasing sequences of positive integers {n j } and {k j }, dense sets X and Y in Z and functions S j : Y → Z such that: nj

kj

nj

kj

(1) For each x ∈ X , T 1 T 2 x → 0 as j → ∞. (2) For each y ∈ Y , S j y → 0 as j → ∞. (3) For each y ∈ Y , T 1 T 2 S j y → y as j → ∞. Then ( T 1 , T 2 ) is a hypercyclic pair. Proof. If U and V are two nonempty open sets in Z , then choose x ∈ X ∩ U and y ∈ V ∩ Y and let z j = x + S j y. Then nj

kj

nj

kj

nj

kj

nj

kj

as j → ∞, z j → x and T 1 T 2 z j = T 1 T 2 x + T 1 T 2 S j y → y. Thus for large j we have z j ∈ U and T 1 T 2 z j ∈ V . Thus Proposition 2.3 implies that the pair ( T 1 , T 2 ) is hypercyclic. 2 Remark. If T 2 is the identity, then the conditions in Proposition 2.5 reduce to the well-known “Hypercyclicity Criterion” for a single operator. Corollary 2.6. If ( T 1 , T 2 ) satisﬁes the hypercyclicity criterion, then

(T 1 ⊕ T 1 , T 2 ⊕ T 2 ) also satisﬁes the hypercyclicity criterion, hence is a hypercyclic pair. Example 2.7. Let T = ( T 1 , T 2 , T 3 ) = (2I 1 , 13 I 1 , e i θ I 1 ), where I 1 is the identity operator on C and θ is an irrational multiple of π . Then T is hypercyclic on C, but T does not satisfy the hypercyclicity criterion. Proof. It follows from Corollary 4.2 and the fact that {e inθ : n ∈ N} is dense in { z: z = 1} that T is hypercyclic on C. However, T does not satisfy the hypercyclicity criterion since ( T 1 ⊕ T 1 , T 2 ⊕ T 2 , T 3 ⊕ T 3 ) = (2I 2 , 13 I 2 , e i θ I 2 ) is not hypercyclic on C2 , where I 2 is the identity operator on C2 .

2

We saw above that the pair (2I , 12 B ) is hypercyclic on 00022 (N), it is also easy to check that this pair satisﬁes the hypercyclicity criterion with respect to any two sequences {n j } and {k j } where n j − k j → ∞.

84

N.S. Feldman / J. Math. Anal. Appl. 346 (2008) 82–98

Example 2.8. Let T 1 = 2I and let T 2 = 12 B where B is the backward shift on 00022 (N). Also let S be the forward shift on 00022 (N). Then T = ( T 1 , T 2 ) satisﬁes the hypercyclicity criterion on 00022 (N). Furthermore, (1) T 1n T 2k = 2n−k B k .

(2) If X is the set of vectors with ﬁnite support, then for each x ∈ X , T 1n T 2k x → 0 as k → ∞.

(3) S n,k := 2k−n S k is a right inverse for T 1n T 2k on all of 00022 (N). (4) If x is a nonzero vector in 00022 (N), then S n,k x → 0 if and only if (n − k) → ∞.

Proof. Let {n j } and {k j } be any two (strictly increasing) sequences of positive integers such that n j − k j → ∞. Let X be the nj

kj

vectors with ﬁnite support and let Y = 00022 (N). Notice that T 1 T 2 x = 0 for all large j when x ∈ X and since n j − k j → ∞, then S j y := S n j ,k j y → 0 as j → ∞. 2 The next result is proven in Feldman [9] and generalizes the example that (2I , 12 B ) is hypercyclic on 00022 (N); it shows that “most” pairs of coanalytic Toeplitz operators are hypercyclic, if not, the symbols are closely related. Theorem 2.9. Let f , g ∈ H ∞ (D) {0} satisfy f ( z) > 1 for all z ∈ D and g ( z) < 1 for all z ∈ D, and let M f and M g be the corresponding multiplication operators on H 2 (D). Then neither M ∗f nor M ∗g is hypercyclic on H 2 (D), but the following are equivalent: (1) The pair T = ( M ∗f , M ∗g ) is hypercyclic on H 2 (D). (2) The semigroup F T contains a hypercyclic operator. (3) There exist n, k ∈ N so that f n g k is nonconstant and ( f n g k )(D) ∩ ∂D 0005= ∅. iθ

(4) There does not exist p > 0 and θ ∈ R such that g ( z) = f e(z) p for all z ∈ D. The next example shows that the study of hypercyclic tuples of operators includes the study of supercyclic operators. Recall that an operator A on a space X is supercyclic if there is a vector x ∈ X such that {α A n x: α ∈ C and n 0002 0} is dense in X . Example 2.10. If A is an operator and T = (2I , 13 I , e i θ I , A ) where θ is an irrational multiple of π and I is the identity operator, then T is a hypercyclic tuple if and only if A is a supercyclic operator. Furthermore, the hypercyclic vectors for T are the same as the supercyclic vectors for A. n1 in2 θ

n1 in2 θ

Proof. The semigroup generated by T is F T = { 2 3en3 A n4 : n1 , n2 , n3 , n4 0002 0}. Since { 2 3en3 : n1 , n2 , n3 0002 0} is dense in C (see Corollary 4.2) it follows that the hypercyclic vectors for T are precisely the same as the supercyclic vectors for A. 2 Notice that by using Proposition 3.1 instead of Corollary 4.2, one can ﬁnd two complex numbers a and b so that the set of hypercyclic vectors of the 3-tuple T = (aI , bI , A ) is the same as the set of supercyclic vectors for A. Example 2.11. If A is a supercyclic operator so that no multiple of A is hypercyclic, then the tuple T = (2I , 13 I , e i θ I , A ) where θ is an irrational multiple of π has the property that the semigroup F T generated by T is a hypercyclic semigroup, but it does not contain any hypercyclic operators. Example 2.12. If A and B are hypercyclic operators, and we let T 1 = A ⊕ I and T 2 = I ⊕ B, then T = ( T 1 , T 2 ) is a hypercyclic pair, but neither T 1 nor T 2 is cyclic. Proof. Let x and y be hypercyclic vectors for A and B, respectively. Notice that T 1n T 2k (x ⊕ y ) = A n x ⊕ B k y. It follows easily that x ⊕ y is a hypercyclic vector for the pair ( T 1 , T 2 ). Since A and B are both hypercyclic, then they act on inﬁnite dimensional spaces and the identity operator I on an inﬁnite dimensional space is not cyclic, thus T 1 and T 2 are not cyclic either. 2 We see in the previous example that T 1 and T 2 need not be cyclic for T to be hypercyclic. However, it is possible that in the above example T 1 T 2 (= A ⊕ B ) is cyclic. We will see in Example 2.16 that given p > 0, there are hypercyclic pairs ( T 1 , T 2 ) where T 1n T 2k , 0 0003 n, k 0003 p, are not cyclic. The following proposition is a slightly more general result than the previous example. Proposition 2.13. Let A and B be hypercyclic operators and let C be an operator with dense range that commutes with B. If we deﬁne T1 = A ⊕ C

and

T2 = I ⊕ B,

then ( T 1 , T 2 ) is a hypercyclic pair.

N.S. Feldman / J. Math. Anal. Appl. 346 (2008) 82–98

85

Proof. Let x and y be hypercyclic vectors for A and B, respectively. We claim that x ⊕ y is a hypercyclic vector for the pair ( T 1 , T 2 ). To prove this let U × V be a basic (product) open set. Notice that T 1n T 2k = An ⊕ C n B k . Now since x is a hypercyclic vector for A, then we can choose an n such that A n x ∈ U . Next since C has dense range, then so does C n , thus the inverse image (C n )−1 ( V ) is a nonempty open set. Since y is a hypercyclic vector for B, then there is k such that B k y ∈ (C n )−1 ( V ). Thus, C n B k y ∈ V . Thus, T 1n T 2k (x ⊕ y ) ∈ U × V . So, ( T 1 , T 2 ) is a hypercyclic pair with hypercyclic vector x ⊕ y. 2 If T is a hypercyclic operator, then T ∗ cannot have any eigenvalues. However, this is not the case for tuples of operators as the next example shows. Deﬁnition 2.14. A hypercyclic tuple T = ( T 1 , . . . , T n ) is said to be minimal if no proper subset of { T 1 , . . . , T n } forms a hypercyclic tuple. Example 2.15. If A is a hypercyclic operator and B is an operator that satisﬁes the hypercyclicity criterion and {λn }n∞=1 is a bounded set of nonzero complex numbers, then deﬁne T 1 = A ⊕ λ1 I ⊕ λ2 I ⊕ · · · ⊕ λn I ⊕ · · ·

and

T2 = I ⊕ B ⊕ B ⊕ B ⊕ · · ·

where I is the identity operator. Then ( T 1 , T 2 ) is a minimal hypercyclic pair and

σ p ( T 1∗ ) = {λn }n∞=1 .

Proof. Let C = (λ1 I ⊕ λ2 I ⊕ · · · ⊕ λn I ⊕ · · ·) and apply Proposition 2.13 with T 1 = A ⊕ C and T 2 = I ⊕ B (∞) where B (∞) = ( B ⊕ B ⊕ B ⊕ · · ·). Notice that B (∞) is hypercyclic because B satisﬁes the hypercyclicity criterion. The pair ( T 1 , T 2 ) is minimal because neither operator T 1 nor T 2 is hypercyclic (because each of their adjoints have eigenvectors). 2 A natural question is that if T = ( T 1 , . . . , T n ) is a hypercyclic tuple of operators, then must the semigroup F T contain a cyclic operator? The following example shows that the ﬁrst ﬁnitely many operators in the semigroup need not be cyclic. Example 2.16. (a) If A is an operator such that both A and A ∗ are hypercyclic and A has a real matrix representation, and we let T 1 = A ⊕ I and T 2 = I ⊕ A ∗ , then ( T 1 , T 2 ) is a hypercyclic pair, but T 1 T 2 is not a cyclic operator. (b) If T 1 = A ⊕ I ⊕ A2 ⊕ I ⊕ · · · ⊕ A p ⊕ I and T 2 = I ⊕ A ∗ ⊕ I ⊕ A ∗2 ⊕ · · · ⊕ I ⊕ A ∗ p , then ( T 1 , T 2 ) is a hypercyclic pair, but the set { T 1n T 2k : 0 0003 n, k 0003 p } does not contain any cyclic operators. Proof. (a) It is a well-known unpublished result of J.A. Deddens (see [14, Proposition 4.2]) that if A has a real matrix representation with respect to some orthonormal basis, then A ⊕ A ∗ cannot be cyclic. Thus, T 1 T 2 = A ⊕ A ∗ is not cyclic. (b) If 0 0003 n, k 0003 p, then since A k is a summand of T 1 and A ∗n is a summand of T 2 , then A kn will be a summand of T 1n and A ∗nk will be a summand of T 2k , thus A kn ⊕ A ∗nk will be a summand of T 1n T 2k and since A kn has a real matrix representation, then A kn ⊕ A ∗nk cannot be cyclic, hence T 1n T 2k cannot be cyclic either.

2

Question 2.17. If T is a hypercyclic tuple, then is there a cyclic operator A in the semigroup generated by T ? Is there a cyclic operator A that commutes with T ? 3. Hypercyclic normal tuples In this section we show that there are hypercyclic tuples of diagonal matrices on Rn and Cn . This was done independently by Kérchy [15]. We also use this result to show that there are no hypercyclic tuples of normal operators on an inﬁnite dimensional Hilbert space, and even no hypercyclic tuples of subnormal operators having commuting normal extensions. It is not currently known if there are hypercyclic tuples of hyponormal operators in inﬁnite dimensions. Proposition 3.1. There exist two complex numbers a, b such that {an bk : n, k ∈ N} is dense in C. Proof. Let f : [0, 1] → [0, 1] be deﬁned by f (x) = frac(10x) where frac(x) denotes the fractional part of the real number x. Also let F = f × f . Thus F maps [0, 1]2 → [0, 1]2 as follows:

0004

0005

F (x1 , x2 ) = f (x1 ), f (x2 ) .

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N.S. Feldman / J. Math. Anal. Appl. 346 (2008) 82–98

It is easy to see that f is a “mixing” function, meaning for any two open sets U and V in [0, 1], there exists an integer N such that f [n] (U ) ∩ V 0005= ∅ for all n 0002 N, where f [n] denotes the nth iterate of f . In fact given any open set U , there is an integer N such that f [n] (U ) = [0, 1] for all n 0002 N. Since f is mixing, it follows easily that F is topologically transitive. Let x = (x1 , x2 ) ∈ [0, 1]2 be a point that has dense orbit under F . Also choose any positive number α and a positive integer q and θ ∈ (0, 1]. Let a = eqα x1 +2π i (qθ x1 +x2 )

and b = e −α −2π i θ .

(1)

Claim. {an bk : n, k ∈ N} is dense in C. Since x has dense orbit under F , then

00060007

frac(10n x1 )

: n∈N

frac(10n x2 ) Thus,

00060007

10n x1 − k

is dense in [0, 1]2 .

10n x2 + m

: n, k ∈ N, m ∈ Z

is dense in R2 .

Notice that k above is only required to be in N since the integer part of 10n x1 can be arbitrarily large and positive. Now multiplying the above set by the invertible diagonal matrix

0007

A=

qα (1 + 2παi θ ) 1

gives that

00060007

10n qα (1 + 2παi θ )x1 − qkα (1 + 2παi θ ) 10n x2 + m

: n, k ∈ N, m ∈ Z

is dense in L × R

where L = {t (1 + 2παi θ ): t ∈ R}. Thus, a small argument shows that

00060007

10n qα 1 +

2π i θ

α

x1 − qkα 1 +

2π i θ

α

000b

+ 2π i 10n x2 + m : n, k ∈ N, m ∈ Z

is dense in C. Simplifying this gives,

000e

10n qα x1 + 10n q2π i θ x1 − qkα − qk2π i θ + 2π i10n x2 + 2π im: n, k ∈ N, m ∈ Z

000f

is dense in C. By taking the exponential of the previous set we get

000f n e 10 [qα x1 +q2π i θ x1 +2π ix2 ] · eqk[−α −2π i θ] · e 2π im : n, k ∈ N, m ∈ Z is dense in C.

000e

Or simplifying gives

000f 000e0004 qα x +2π i (qθ x +x ) 000510n 0004 −α −2π i θ 0005qk 1 2 e 1 e : n, k ∈ N is dense in C. Thus we have established our claim that {an bk : n, k ∈ N} is dense in C.

2

Corollary 3.2. If b ∈ D {0}, then there is a dense set Δb ⊆ { z ∈ C: z > 1} such that for any a ∈ Δb , we have {an bk : n, k ∈ N} is dense in C. Proof. Keeping the notation from the previous proof. Let Δ = {(x1 , x2 ): (x1 , x2 ) has dense orbit under F 2 }. Then Δ is a 1 dense G δ set in [0, 1]2 . Suppose we are given b ∈ D {0}. Then deﬁne α = − ln b and θ = − arg(b) where arg(b) denotes 2π the value of the argument of b that lies in (0, 2π ]. Now let

000e 000f Δb = eqα x1 +2π i (qθ x1 +x2 ) : (x1 , x2 ) ∈ Δ, q ∈ N .

Then by checking Eq. (1) in Proposition 3.1 we see that Δb has the required properties.

2

We also have the following result equivalent to the above result. Corollary 3.3. If a ∈ C and a > 1, then there is a dense set Δa ⊆ D such that for any b ∈ Δa , we have {an bk : n, k ∈ N} is dense in C.

N.S. Feldman / J. Math. Anal. Appl. 346 (2008) 82–98

87

Proof. Given a with a > 1, let b = 1/a and then by the previous corollary we get a set Δb such that if c ∈ Δb , then {cn bk : n, k ∈ N} = {cn (1/a)k : n, k ∈ N} is dense in C. Then it follows that {(1/c )n ak : n, k ∈ N} is dense in C. Thus, let Δa = {1/c: c ∈ Δb }. 2 The next theorem is a natural generalization of Proposition 3.1. Kérchy [15] has independently shown the existence of supercyclic tuples of diagonal matrices on Cn . His techniques can also be used to construct hypercyclic tuples on Cn . The methods used here are different. Theorem 3.4. For each n 0002 1, there exists a hypercyclic (n + 1)-tuple of diagonal matrices on Cn . Proof. Fix an integer p 0002 1 and we will construct a hypercyclic ( p + 1)-tuple of diagonal matrices on C p . As in Proposition 3.1, let f : [0, 1] → [0, 1] be deﬁned by f (x) = frac(10x) and for an integer n let F n = f × f × f × · · · × f (n times). Thus F n maps [0, 1]n → [0, 1]n as follows:

0004

0005

F n (x1 , x2 , . . . , xn ) = f (x1 ), f (x2 ), . . . , f (xn ) . As in Proposition 3.1, since f is mixing, it follows that F n is topologically transitive. 2p p Let x = {x j } j =1 ∈ [0, 1]2p be a point that has dense orbit under F 2p . Also choose ﬁnite sets {α j } j =1 of positive real p

p

numbers, {q j } j =1 of positive integers, and {θ j } j =1 ⊆ (0, 1]. For 1 0003 j 0003 p, let

a j = exp q j α j x2 j −1 + 2π i (q j θ j x2 j −1 + x2 j )

and let b j = exp(−α j − 2π i θ j )

where exp( z) = e z is the exponential function. Claim. For each p 0002 1,

⎧⎡ n k1 ⎤ ⎫ a1 b 1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎢ ⎥ ⎪ ⎪ ⎪ ⎪ ⎨⎢ an2 bk22 ⎥ ⎬ ⎢ ⎥ ⎢ ⎥ : n, k i ∈ N ⎪ ⎢ . ⎥ ⎪ ⎪ ⎪ ⎪⎣ . ⎦ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ n kp ⎭ apbp

is dense in C p . Since x has dense orbit under F 2p , then

⎫ ⎧⎡ ⎤ frac(10n x1 ) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ frac(10n x ) ⎥ ⎪ ⎪ 2 ⎪ ⎢ ⎪ ⎥ ⎪ ⎪ ⎪ ⎪ ⎢ ⎥ ⎪ ⎪ n ⎪ ⎢ frac(10 x3 ) ⎥ ⎪ ⎪ ⎪ ⎪ ⎪ ⎥ ⎬ ⎨⎢ ⎢ frac(10n x ) ⎥ ⎢ ⎥ 4 : n ∈ N is dense in [0, 1]2p . ⎢ ⎥ ⎪ ⎪ ⎪ ⎪ ⎢ ⎥ . ⎪ ⎪ . ⎪ ⎪ ⎢ ⎥ ⎪ ⎪ ⎪ ⎪ ⎢ ⎥ ⎪ ⎪ ⎪ ⎪ ⎢ ⎥ ⎪ ⎪ n ⎪ ⎪ ⎣ frac(10 x2p −1 ) ⎦ ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ n frac(10 x2p ) Thus,

⎧⎡ 10n x − k ⎤ ⎫ 1 1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ n ⎪ ⎢ 10 x2 + m1 ⎥ ⎪ ⎪ ⎪ ⎪ ⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ 10n x − k ⎥ ⎪ ⎪ ⎪ ⎢ ⎪ ⎥ ⎪ 3 2 ⎪ ⎪ ⎨⎢ ⎥ ⎬ n ⎢ 10 x4 + m2 ⎥ ⎢ ⎥ : n, ki ∈ N, mi ∈ Z is dense in R2p . ⎪ ⎪ ⎢ ⎥ ⎪ ⎪ . ⎪ ⎪ ⎢ ⎥ ⎪ ⎪ ⎪ ⎪ ⎢ ⎥ . ⎪ ⎪ ⎪ ⎪ ⎢ ⎥ ⎪ ⎪ ⎪⎣ 10n x ⎪ ⎦ ⎪ ⎪ − k 2p −1 p ⎪ ⎪ ⎩ ⎭ n 10 x2p + m p

(2)

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N.S. Feldman / J. Math. Anal. Appl. 346 (2008) 82–98

Now multiplying the above set by the invertible diagonal matrix

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

⎤

q1 α1 (1 + 2παi θ1 ) 1

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

1 2π i θ2

q2 α2 (1 + α ) 2 1

.

. 2π i θ

q p α p (1 + α p ) p 1

gives that

⎧⎡ 10n q α (1 + 2π i θ1 )x − k q α (1 + 2π i θ1 ) ⎤ ⎫ 1 1 1 1 1 1 α1 α1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ n ⎥ ⎪⎢ ⎪ ⎪ ⎪ 10 x2 + m1 ⎢ ⎥ ⎪ ⎪ ⎪ ⎪ ⎢ ⎥ ⎪ ⎪ 2π i θ2 2π i θ2 n ⎪ ⎪ ⎢ ⎥ ⎪ ⎪ 10 q2 α2 (1 + α )x3 − k2 q2 α2 (1 + α ) ⎪ ⎪ 2 2 ⎥ ⎪ ⎪ ⎨⎢ ⎬ ⎢ ⎥ n 10 x4 + m2 ⎢ ⎥ : n, ki ∈ N, mi ∈ Z ⎢ ⎥ ⎪ ⎪ ⎪ ⎪ ⎢ ⎥ . ⎪ ⎪ ⎪ ⎪ ⎢ ⎥ ⎪ ⎪ . ⎪ ⎪ ⎢ ⎥ ⎪ ⎪ ⎪ ⎪ ⎢ ⎥ 2 π i θ 2 π i θ p p ⎪ ⎪ n ⎪ ⎪ ⎣ 10 q p α p (1 + α )x2p −1 − k p q p α p (1 + α ) ⎦ ⎪ ⎪ p p ⎪ ⎪ ⎩ ⎭ n 10 x2p + m p 2π i θ

is dense in ( L 1 × R) × ( L 2 × R) × · · · × ( L p × R) where L j = {t (1 + α j ): t ∈ R}. j Thus it follows that

⎧⎡ ⎪ ⎪ ⎪ ⎢ ⎪ ⎪ ⎨⎢ ⎢ ⎢ ⎪ ⎢ ⎪ ⎪ ⎣ ⎪ ⎪ ⎩

[10n q1 α1 (1 + 2παi1θ1 )x1 − k1 q1 α1 (1 + 2παi1θ1 )] + 2π i [10n x2 + m1 ] [10n q2 α2 (1 + 2παi2θ2 )x3 − k2 q2 α2 (1 + 2παi2θ2 )] + 2π i [10n x4 + m2 ] . . 2π i θ

2π i θ

[10n q p α p (1 + α p p )x2p −1 − k p q p α p (1 + α p p )] + 2π i [10n x2p + m p ]

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬

⎤

⎥ ⎥ ⎥ ⎥ : n, ki ∈ N, mi ∈ Z ⎪ ⎥ ⎪ ⎪ ⎦ ⎪ ⎪ ⎭

is dense in C p . By taking the exponential of each coordinate in the previous set we get

⎧⎡ ⎪ ⎪ ⎪ ⎪⎢ ⎨ ⎢ ⎢ ⎢ ⎪ ⎪ ⎣ ⎪ ⎪ ⎩

⎤

exp[10n (q1 α1 x1 + q1 2π i θ1 x1 + 2π ix2 )] exp[−q1 k1 (α1 + 2π i θ1 )] exp[2π im1 ] exp[10n (q2 α2 x3 + q2 2π i θ2 x3 + 2π ix4 )] exp[−q2 k2 (α2 + 2π i θ2 )] exp[2π im2 ]

. . exp[10n (q p α p x2p −1 + q p 2π i θ p x2p −1 + 2π ix2p )] exp[−q p k p (α p + 2π i θ p )] exp[2π im p ]

is dense in C p . Or simplifying gives

⎧⎡ ⎪ ⎪ ⎪ ⎪⎢ ⎨ ⎢ ⎢ ⎢ ⎪ ⎪ ⎣ ⎪ ⎪ ⎩

n

(exp[q1 α1 x1 + 2π i (q1 θ1 x1 + x2 )])10 (exp[−(α1 + 2π i θ1 )])q1 k1 n

(exp[q2 α2 x3 + 2π i (q2 θ2 x3 + x4 )])10 (exp[−(α2 + 2π i θ2 )])q2 k2 . . n

(exp[q p α p x2p −1 + 2π i (q p θ p x2p −1 + x2p )])10 (exp[−(α p + 2π i θ p )])q p k p

is dense in C p . Thus using our deﬁnitions of ai , b i we see that

⎧⎡ n k1 ⎤ ⎫ a1 b 1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ n k2 ⎥ ⎪ ⎪ ⎪ ⎨⎢ a2 b2 ⎥ ⎬ ⎢ ⎥ ⎢ ⎥ : n, k i ∈ N . ⎥ ⎪ ⎪ ⎪⎢ ⎪ ⎪ ⎪ ⎣ . ⎦ ⎪ ⎪ ⎪ ⎪ ⎩ n kp ⎭ apbp

is dense in C p .

⎤

⎫ ⎪ ⎪ ⎪ ⎪ ⎬

⎥ ⎥ ⎥ : n, k i ∈ N ⎥ ⎪ ⎪ ⎦ ⎪ ⎪ ⎭

⎫ ⎪ ⎪ ⎪ ⎪ ⎬

⎥ ⎥ ⎥ : n, ki ∈ N, mi ∈ Z ⎥ ⎪ ⎪ ⎦ ⎪ ⎪ ⎭

N.S. Feldman / J. Math. Anal. Appl. 346 (2008) 82–98

89

So if we let

⎡ ⎢ ⎢ A=⎢ ⎢ ⎣

⎤

a1

⎥ ⎥ ⎥, ⎥ ⎦

a2

.

. ap

⎡1 ⎢ ⎢ ⎢ B2 = ⎢ ⎢ ⎢ ⎣

⎡

.

1

.

.

⎡1

⎥ ⎥ ⎥ ⎥, ⎥ ⎥ ⎦

⎥ ⎥ ⎥, ⎥ ⎦

1

⎤

b2

.

⎢ ⎢ B1 = ⎢ ⎢ ⎣

⎤

b1

⎢ ⎢ ⎢ Bp = ⎢ ⎢ ⎢ ⎣

..,

1

⎤ ⎥ ⎥ ⎥ ⎥, ⎥ ⎥ ⎦

1

.

. 1

⎡ ⎤

bp

1

1

⎢1⎥ ⎢ ⎥ ⎥ and v = ⎢ ⎢ . ⎥ ⎣.⎦

(3)

1

then ( A , B 1 , B 2 , . . . , B p ) is a hypercyclic ( p + 1)-tuple of diagonal matrices on C p with v as the hypercyclic vector. In fact, the hypercyclic vectors for the tuple are those vectors where each coordinate is nonzero. 2 p

Corollary 3.5. If b = {b j } j =1 ⊆ D {0}, then there is a dense set Δb ⊆ { z ∈ C p : z j > 1 for all j } such that for any a ∈ Δb , we have the tuple ( A , B 1 , B 2 , . . . , B p ) is hypercyclic on C p where the matrices A and B j are given above in Eq. (3). Proof. Keeping the notation from the previous proof. Let Δ = {(x1 , . . . , x2p ): (x1 , . . . , x2p ) has dense orbit under F 2p }. Then p 1 Δ is a dense G δ set in [0, 1]2p . Suppose we are given b = {b j } j =1 ⊆ D {0}. Then deﬁne α j = − ln b j and θ j = − arg(b j ) 2π where arg(b j ) denotes the value of the argument of b j that lies in (0, 2π ]. Now let ⎧⎡ ⎫ ⎤ exp[q1 α1 x1 + 2π i (q1 θ1 x1 + x2 )] ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎥ ⎪ ⎪ exp[q2 α2 x3 + 2π i (q2 θ2 x3 + x4 )] ⎨⎢ ⎬ ⎢ ⎥ ⎥ : (x1 , . . . , x2p ) ∈ Δ, q j ∈ N . Δb = ⎢ . ⎢ ⎥ ⎪ ⎪ . ⎪ ⎪ ⎣ ⎦ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ exp[q p α p x2p −1 + 2π i (q p θ p x2p −1 + x2p )] Then by checking Eq. (2) in Theorem 3.4 we see that Δb has the required properties. Thus using our deﬁnitions of ai , b i we have that

⎧⎡ n k1 a1 b 1 ⎪ ⎪ ⎪ ⎪ ⎢ ⎪ ⎨⎢ an2 bk22 ⎢ ⎢ . ⎪ ⎪⎢ ⎪ ⎣ . ⎪ ⎪ ⎩

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬

⎤

⎥ ⎥ ⎥ ⎥ : n, k j ∈ N is dense in C p . ⎥ ⎪ ⎪ ⎪ ⎦ ⎪ ⎪ ⎭ kp

anp b p

Hence the diagonal tuple ( A , B 1 , . . . , B p ) is hypercyclic.

2

We now show that the previous result is the best possible in the sense that we cannot reduce the size of the tuple. Theorem 3.6. There does not exist a hypercyclic n-tuple of diagonalizable matrices on Cn . Proof. To keep the notation simple we will prove this for n = 3; this case contains all the main ideas. By way of contradiction, assume that there exists a hypercyclic 3-tuple of diagonalizable matrices on C3 . Then the tuple is simultaneously diagonalizable. Hence we may assume that there is a hypercyclic 3-tuple of diagonal matrices on C3 , call them ( A , B , C ). Suppose that

⎡a A=⎣

⎤

1

⎦,

a2

⎡b B =⎣

a3

Suppose that v =

0007 α β

γ

⎦,

b2

⎡c C =⎣

⎤

1

⎦.

c2

b3

is the corresponding hypercyclic vector. Then

⎧⎡ n k l ⎤ ⎫ a b c α ⎪ ⎪ ⎨ 1 1 1 ⎬ ⎢ n k l ⎥ ⎣ a2 b 2 c 2 β ⎦ : n, k, l 0002 0 ⎪ ⎪ ⎩ n k l ⎭ a3 b 3 c 3 γ

⎤

1

c3

90

N.S. Feldman / J. Math. Anal. Appl. 346 (2008) 82–98

is dense in C3 . Clearly we must have

α 0005= 0, β 0005= 0, γ 0005= 0. Hence by applying the invertible matrix

0007 α −1

β −1

γ −1

to the

above dense set we also have that

⎧⎡ n k l ⎤ ⎫ a b c ⎪ ⎪ ⎨ 1 1 1 ⎬ ⎢ n k l ⎥ n : , k , l 0002 0 a b c ⎣ 2 2 2⎦ ⎪ ⎪ ⎩ n k l ⎭ a3 b 3 c 3

is dense in C3 . Now apply the function log z to each coordinate and we get that

⎧⎡ ⎫ ⎤ ⎨ n log a1 + k log b1 + l log c 1 ⎬ ⎣ n log a2 + k log b2 + l log c 2 ⎦ : n, k, l 0002 0 is dense in R3 . ⎩ ⎭ n log a3 + k log b3 + l log c 3

Let

⎫ ⎧⎡ ⎤ ⎬ ⎨ n S := ⎣ k ⎦ : n, k, l 0002 0 ⎭ ⎩

(4)

⎡ log a log b log c ⎤ 1 1 1 and T := ⎣ log a2 log b2 log c 2 ⎦ log a3 log b3

l

log c 3

then T is a 3 × 3 real matrix and Eq. (4) implies that T ( S ) is a dense set in R3 . Thus T has dense range, hence T is onto, thus T is invertible. However, since T is invertible and T ( S ) is dense in R3 , then S = T −1 ( T ( S )) is also dense in R3 . However clearly S is not dense in R3 , thus we have a contradiction. So, there are no hypercyclic diagonalizable 3-tuples on C3 . And in general there are no hypercyclic diagonalizable n-tuples on Cn . 2 Corollary 3.7. There do not exist positive real numbers ai , b i , c i , 1 0003 i 0003 3, such that

⎧⎡ n k l ⎤ ⎫ a b c ⎪ ⎪ ⎨ 1 1 1 ⎬ 0004 00053 ⎢ n k l ⎥ + n : , k , l ∈ R is dense in R+ . a b c ⎣ 2 2 2⎦ ⎪ ⎪ ⎩ n k l ⎭ a3 b 3 c 3

Proof. Following the proof of Theorem 3.6, take the logarithm of each coordinate, get the matrix T and the set S. In this case, the set S is not a discrete lattice, but S is essentially the ﬁrst octant in R3 . However the important point is that S is not dense in R3 , so T ( S ) cannot be dense either. 2 Corollary 3.8. If H is an inﬁnite dimensional Hilbert space and n 0002 1, then there does not exist a hypercyclic normal n-tuple on H. Proof. By way of contradiction, suppose that A = ( A 1 , A 2 , . . . , A n ) is a hypercyclic normal n-tuple on H. Then the tuple A is cyclic. Thus A is unitarily equivalent to a tuple of multiplication operators. More precisely, there is a ﬁnite positive Borel measure μ on a compact set in Cn so that A is unitarily equivalent to the n-tuple, M = ( M z1 , M z2 , . . . , M zn ), of multiplibe a countable set of functions that are continuous on cation operators on L 2 (μ) by the coordinate functions. Let { f j }∞ j =1

the support of μ and dense in L 2 (μ). Suppose that φ ∈ L 2 (μ) is a hypercyclic vector for the tuple M. Then for each j there is a sequence of multi-indices nk, j such that M nk, j φ → f j in L 2 (μ) as k → ∞ and by passing to a subsequence we may also assume that μ-almost everywhere convergence holds. Let Δ j be a set of full μ measure so that ( M nk, j φ)( z) → f j ( z) as 0003∞ k → ∞ for every z ∈ Δ j . Let Δ = j =1 Δ j , then Δ also has full measure. Since H and hence L 2 (μ) is inﬁnite dimensional, then Δ must be an inﬁnite set. Let w 1 , . . . , w n be n distinct points in Δ. Then

0004

0005

M nk, j φ ( w i ) → f j ( w i ) as k → ∞,

for each j and each i. Since { f j } is dense in L 2 (μ), then the set

⎧⎡ ⎫ ⎤ f j (w1) ⎪ ⎪ ⎪ ⎪ ⎨⎢ ⎬ ⎥ . ⎢ . ⎥: j 0002 1 ⎦ ⎣ . ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ f j (wn )

is dense in Cn and from above we have that for each j 0002 1,

⎡ ⎢ ⎢ ⎣

⎤ ⎤ ⎡ f j (w1) ( M nk, j φ)( w 1 ) ⎥ ⎥ ⎢ . ⎥ → ⎢ . ⎥ ⎦ ⎣ . ⎦ . ( M nk, j φ)( w n ) f j (wn )

as k → ∞.

N.S. Feldman / J. Math. Anal. Appl. 346 (2008) 82–98

91

Now deﬁne n diagonal matrices B 1 , . . . , B n on Cn as:

⎡

⎢ ⎢ Bi = ⎢ ⎢ ⎣

⎤

w 1 (i )

⎥ ⎥ ⎥ ⎥ ⎦

w 2 (i )

.

. w n (i )

where w j (i ) denotes the ith coordinate of the vector w j . Also deﬁne a vector v = (φ( w 1 ), . . . , φ( w n )), then the tuple B = ( B 1 , . . . , B n ) is a hypercyclic n-tuple of diagonal matrices on Cn with hypercyclic vector v. However this contradicts Theorem 3.6. Thus there is no hypercyclic n-tuple of normal operators on an inﬁnite dimensional Hilbert space. 2 The deﬁnition of a “subnormal tuple” is a commuting tuple of subnormal operators that have commuting normal extensions [6]. It is known that there are commuting tuples of subnormal operators that do not have commuting normal extensions (see [16] or [7, p. 79]). Thus one should be careful to distinguish between “subnormal tuple” and “a commuting tuple of subnormal operators.” The next corollary applies to subnormal tuples. Corollary 3.9. On an inﬁnite dimensional Hilbert space, there is no hypercyclic n-tuple of subnormal operators with commuting normal extensions. Proof. The proof is exactly the same as in the normal case except one uses the fact that a commuting cyclic tuple of subnormal operators that has a commuting tuple of normal extensions is unitarily equivalent to a tuple of multiplication operators by the coordinate functions on P 2 (μ), where P 2 (μ) denotes the closure of the analytic polynomials in L 2 (μ) and μ is a compactly support regular Borel measure in Cn [6]. Now proceed as above with { f j } being the set of all polynomials with rational coeﬃcients. 2 Bourdon [3] gave a nice proof that a hyponormal operator on a Hilbert space cannot be supercyclic. The following question if answered aﬃrmatively would give a natural extension of Bourdon’s supercyclicity result (recall Example 2.10). Question 3.10. Is there a hypercyclic tuple of commuting hyponormal operators on an inﬁnite dimensional Hilbert space? 4. Real Hilbert spaces: Somewhere dense orbits that are not dense!

0002n An operator T on a space X is said to be multi-hypercyclic if there is a ﬁnite set of vectors {x1 , . . . , xk } ⊆ X such that k=1 Orb( T , xi ) is dense in X . Herrero [13] conjectured that a multi-hypercyclic operator would be hypercyclic. His conjecture was established independently by Costakis [8] and Peris [17]. Later Bourdon and Feldman [4] proved that if a bounded linear operator on a (real or complex) locally convex space has a somewhere dense orbit, then that orbit must actually be dense. It was well known that this fact can be used to give a simple proof of Herrero’s conjecture. The same deﬁnition of multi-hypercyclic applies to a tuple T . Namely that a tuple T is multi-hypercyclic if there are a ﬁnite number of orbits under T whose union is dense. We will say that T is n-hypercyclic if there are n orbits for T whose union is dense in X . In this section we will show that on Hilbert spaces over the ﬁeld of real numbers, there are tuples of operators that have a somewhere dense orbit that is not dense! Furthermore these tuples of operators are multi-hypercyclic but not hypercyclic! Furthermore, examples are given of tuples T that are hypercyclic, but T n is not hypercyclic!

Theorem 4.1. If a, b > 1 and

ln(a) ln(b)

n

is irrational, then { ak : n, k ∈ N} is dense in R+ . b

Proof. The one-dimensional version of Kronecker’s Theorem states: If θ is a positive irrational number, then {nθ − k: n, k ∈ N} is dense in R, see [12, Theorem 438, p. 375]. Applying Kronecker’s Theorem with θ = ln(a)/ ln(b) gives that

0006

n

ln(a) ln(b)

− k: n, k ∈ N

is dense in R. Multiplying by ln(b) gives that

000e

n ln(a) − k ln(b): n, k ∈ N

000f

is dense in R. Simplifying gives that

0006

ln

an bk

: n, k ∈ N

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is dense in R. Thus by taking the exponential of the above set we see that

0006

an bk

: n, k ∈ N

is dense in R+ .

2 n

Corollary 4.2. If a, b > 1 are relatively prime integers, then { ak : n, k ∈ N} is dense in R+ . b

Proof. We simply need to show that ln(a)/ ln(b) is irrational. By way of contradiction, assume that ln(a)/ ln(b) is rational. Then there are integers p , q ∈ N such that ln(a)/ ln(b) = p /q. So, q ln(a) = p ln(b) or equivalently, aq = b p . If a and b are relatively prime, then the Fundamental Theorem of arithmetic implies that a and b cannot be powers of each other. Thus we have a contradiction. So ln(a)/ ln(b) is irrational. Thus Theorem 4.1 applies. 2 Example 4.3 (Multi-Hypercyclic Tuples of Matrices on Rn ). (1) Let I be the identity operator on the real Hilbert space R and let v 1 = 1 and v 2 = −1. If we let T = ( T 1 , T 2 ) where T 1 = 2I and T 2 = 13 I , then by Corollary 4.2 cl Orb( T , 1) = [0, ∞) and cl Orb( T , −1) = (−∞, 0]. Hence T has somewhere dense orbits that are not dense. Furthermore Orb( T , 1) ∪ Orb( T , −1) is dense in R, but T is not hypercyclic. Thus T is multi-hypercyclic, but not hypercyclic. (2) Let

0007

T1 =

0007

2 1

,

T2 =

T = ( T 1 , T 2 , T 3 , T 4 ), and v i , j =

1

(−1)i (−1) j

0007

1/3

,

T3 =

0007

1 2

,

T4 =

1 1/3

,

where i , j ∈ {0, 1}. Then the vectors v 0,0 , v 1,0 , v 1,1 , v 0,1 lie in the ﬁrst, second,

third, and fourth quadrants of R , respectively. Furthermore, Orb( T , v i , j ) is dense in the quadrant that contains v i , j . Thus we see that T has somewhere dense orbits that are not dense in R2 and that T is multi-hypercyclic but not hypercyclic. In fact, T is 4-hypercyclic, but not 3-hypercyclic (since each of the 4 quadrants are invariant sets for T with disjoint interiors). (3) Generalizing the previous example, it is easy to construct a 2n-tuple on Rn that has somewhere dense orbits that are not dense and is 2n -hypercyclic. To see this simply follow the example above: For each 1 0003 i 0003 n, let A i be the n × n diagonal matrix with ones on the main diagonal except in the (i , i ) position which is 2. Also let B i be the n × n diagonal matrix with ones on the main diagonal except in the (i , i ) position which is 1/3. Then T = ( A 1 , . . . , A n , B 1 , . . . , B n ) has the required properties. 2

Theorem 4.4. There is an (n + 1)-tuple of diagonal matrices on Rn that has an orbit dense in (R+ )n . However, there is no n-tuple of diagonalizable matrices on Rn or Cn that has a somewhere dense orbit. Proof. By taking the absolute values of each entry in the diagonal matrices constructed in Theorem 3.4 we see that there is an (n + 1)-tuple of diagonal matrices on Rn that has an orbit that is dense in (R+ )n . To see that there is no n-tuple with a somewhere dense orbit on Rn or Cn , follow the reasoning in the proof of Theorem 3.6. 2 The example and theorem above are related to work of Bermudez, Bonilla, and Peris [2], where they show that if an operator T on a Banach space X is R-supercyclic, then it must be R+ -supercyclic. Also see Corollary 5.9 where another proof of this fact is given. The above examples can be modiﬁed to give tuples of matrices that are R-supercyclic but not R+ -supercyclic. In fact, the pair in (1) from Example 4.3 is such an example. Example 4.5 (A Multi-Hypercylic Triple in Inﬁnite Dimensions). Let 00022R (N) denote the real Hilbert space of all real sequences that are square summable. Also let B denote the backward shift on 00022R (N). Let H = R⊕00022R (N) and deﬁne operators T 1 = 2I R ⊕ I 00022 , T2 =

1 I 3 R

⊕ I 00022 , T 3 = I R ⊕ 2B. Then T = ( T 1 , T 2 , T 3 ) has a somewhere dense orbit that is not dense and is 2-hypercyclic, but

not hypercyclic. More precisely, if x is a hypercyclic vector for 2B on 00022R (N) and we let v 1 = (1, x) and v 2 = (−1, x), then Orb( T , v 1 ) ∪ Orb( T , v 2 ) is dense in H, yet neither orbit is dense in H. Example 4.6 (A Multi-Hypercylic Pair in Inﬁnite Dimensions). On H = R ⊕ 00022R (N) deﬁne operators T 1 = 2I R ⊕ I 00022 , T 2 = Then T = ( T 1 , T 2 ) is 2-hypercyclic, but not hypercyclic.

1 I ⊕ 2B. 3 R

Proof. We simply use the fact that 2B is a mixing operator and show that the pair ( T 1 , T 2 ) is transitive on each of the two closed invariant subsets: R+ ⊕ 00022R (N) and R− ⊕ 00022R (N), where R+ = [0, ∞) and R− = (−∞, 0]. So, let U 1 , V 1 be

N.S. Feldman / J. Math. Anal. Appl. 346 (2008) 82–98

93

two nonempty open sets in R+ and let U 2 , V 2 be two nonempty open sets in 00022R (N). We must ﬁnd integers n, k 0002 0 so n that T 1n T 2k (U 1 ⊕ U 2 ) ∩ ( V 1 ⊕ V 2 ) 0005= ∅. Notice that T 1n T 2k = 2k I R ⊕ (2B )k . Since 2B is a mixing operator, there exists K 0002 0 3

such that (2B )k (U 2 ) ∩ V 2 0005= ∅ for all k 0002 K . Since U 1 and U 2 are both open sets of nonnegative real numbers and since n {2n /3k : n, k 0002 0} is dense in R+ , then there exist k 0002 K and n 0002 0 such that 2k U 1 ∩ V 1 0005= ∅. With this choice of n and k we have that T 1n T 2k (U 1 ⊕ U 2 ) ∩ ( V 1 ⊕ V 2 ) 0005= ∅.

3

2

Ansari [1] gave a truly original proof to show that if T is a hypercyclic operator, then T n is also hypercyclic for every positive integer n. This also follows from the results of Costakis [8], Peris [17] and Bourdon and Feldman [4], since T n is multi-hypercyclic whenever T is hypercyclic. It is therefore natural to ask if a similar result holds for tuples of operators. If T = ( T 1 , T 2 ) is a commuting pair of operators and n = (n1 , n2 ) is a pair of nonnegative integers (a multi-index), then n n we deﬁne T n to be the pair ( T 1 1 , T 2 2 ). In view of Ansari’s Theorem it is natural to ask if T is a hypercyclic pair and n is a n multi-index, then must T be hypercyclic? The following example shows that it need not be true. However, see Corollary 5.8 for a positive result along these lines. Example 4.7 (Powers of a Hypercyclic Pair). Slightly different than the previous example, deﬁne T 1 = −2I R ⊕ I 00022 , T 2 =

on H = R ⊕ 0002R (N). Then T = ( T 1 , T 2 ) is hypercyclic on H. However, if n = (2, 1), then T = 2

n

( T 12 , T 2 )

1 I ⊕ 2B 3 R

is not hypercyclic on H.

Proof. This follows since the ﬁrst coordinates of any term in a given orbit must all have the same sign, and hence cannot be dense in R. 2 5. When somewhere dense orbits are everywhere dense In this section we will give some suﬃcient conditions for a somewhere dense orbit to be dense. These apply to both real and complex spaces. In view of the results in the previous section, some extra conditions are necessary. As above, T = ( T 1 , T 2 , . . . , T n ) will denote a (commuting) n-tuple of operators and F = F T will be the semigroup they generate. Also, in this section, X will denote a real or complex locally convex space unless otherwise stated. Deﬁne W as the following set of polynomials in n-variables,

000e

k

k

k

000f

W = p: p ( z1 , z2 , . . . , zn ) = λ z11 z22 · · · znn + μ where λ, μ ∈ C, ki 0002 0 . Notice that { p ( T ): p ∈ W } = {λ A + μ I: A ∈ F , λ, μ ∈ C}. Lemma 5.1. Let T = ( T 1 , T 2 , . . . , T n ) be an n-tuple of commuting operators on X . If x ∈ X and U is the interior of the closure of Orb( T , x) and T i does not have dense range for some i ∈ {1, . . . , n}, then U is also equal to the interior of the closure of Orb({ T j } j 0005=i , x). Proof. Suppose that T i does not have dense range for some i ∈ {1, . . . , n}. Let R ( T i ) denote the closure of the range of T i . Since T i does not have dense range, then R ( T i ) is a proper closed subspace of X , hence it is nowhere dense in X . If k k k k j 0002 0 for j 0005= i and ki 0002 1, then T 11 T 22 · · · T nn x ∈ R ( T i ). Thus we have that Orb( T , x) ⊆ Orb({ T j } j 0005=i , x) ∪ R ( T i ). So, U ⊆ cl Orb( T , x) ⊆ cl Orb({ T j } j 0005=i , x) ∪ R ( T i ). Thus, U R ( T i ) ⊆ cl Orb({ T j } j 0005=i , x). Thus, U ⊆ cl(U R ( T i )) ⊆ cl Orb({ T j } j 0005=i , x). So we have that U ⊆ int[cl Orb({ T j } j 0005=i , x)] ⊆ int[cl Orb( T , x)] = U , so U = int[cl Orb({ T j } j 0005=i , x)]. 2 Lemma 5.2. If T = ( T 1 , T 2 , . . . , T n ) is an n-tuple of commuting operators on X and if there exists x ∈ X such that Orb( T , x) is somewhere dense in X and x ∈ int[cl Orb( T , x)], then { p ( T )x: p ∈ W } is dense in X . Proof. Let U be the interior of the closure of Orb( T , x). Then U − {x} := { y − x: y ∈ U } ⊆ cl{( A − I )x: A ∈ F T } = k k cl{ p ( T )x: p = ( z11 · · · znn − 1), ki 0002 0}. Since U − {x} is an open set containing zero, then the set of all multiples of U − {x} is dense in X . Now since { p ( T )x: p ∈ W } is invariant under scalar multiplication and U − {x} ⊆ { p ( T )x: p ∈ W }, then { p ( T )x: p ∈ W } is dense in X . 2 Lemma 5.3. Let T = ( T 1 , T 2 , . . . , T n ) be an n-tuple of commuting operators on X such that T k has dense range for each 1 0003 k 0003 n. Suppose that x ∈ X and U is the interior of the closure of Orb( T , x). Suppose also that for each i and for each k 0002 0, Orb({ T j } j 0005=i , T ik x) p

p

p

k

k

k

is nowhere dense. If A = T 1 1 T 2 2 · · · T n n ∈ F T , then U is equal to the interior of the closure of Orb( T , Ax) = { T 11 T 22 · · · T nn x: ki 0002 p i }. Proof. If Orb( T , x) is dense in X , then U = X and we must show that Orb( T , Ax) is also dense in X . However this follows easily since Orb( T , Ax) = A (Orb( T , x)) and A has dense range. So, we may assume that Orb( T , x) is not dense in X and that U is nonempty. Notice that

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000e

k

k

k

000f

000e

k

k

k

Orb( T , x) = T 11 T 22 · · · T nn x: ki 0002 p i ∪ T 11 T 22 · · · T nn x: ki 0002 0 and ∃i s.t. ki < p i

000f

000e 000f = Orb( T , Ax) ∪ T 1k1 T 2k2 · · · T nkn x: ki 0002 0 and ∃i s.t. ki < p i .

k

k

k

Claim. E := { T 11 T 22 · · · T nn x: ki 0002 0 and ∃i s.t. ki < p i } is nowhere dense.

0002n

0002 p −1

i (Orb({ T j } j 0005=i , T ik (x))). By assumption, Orb({ T j } j 0005=i , T ik (x)) is nowhere To verify the claim, notice that E = i =1 k= 0 dense for each i and for each k 0002 0, thus E is a ﬁnite union of nowhere dense sets, hence E itself is also nowhere dense in X . It now follows easily that U ⊆ cl Orb( T , Ax). Thus, U ⊆ int[cl Orb( T , Ax)] ⊆ int[cl Orb( T , x)] = U . So, U is equal to the interior of the closure of Orb( T , Ax). 2

The next lemma is similar to one from Bourdon and Feldman [4]. If T = ( T 1 , T 2 , . . . , T n ) is a tuple of operators on X and E ⊆ X , then we say that E is invariant for T if T i ( E ) ⊆ E for each 1 0003 i 0003 n. Lemma 5.4. Let T = ( T 1 , T 2 , . . . , T n ) be an n-tuple of commuting operators on X such that T k has dense range for each 1 0003 k 0003 n. If x ∈ X and U is the interior of the closure of Orb( T , x) and if Orb({ T j } j 0005=i , T ik (x)) is nowhere dense for each i ∈ {1, . . . , n} and for each k 0002 0, then X U is invariant under T . Proof. Let F denote the closure of Orb( T , x) and U the interior of F . Observe that F is invariant under T . If U is empty or U = X , then there is nothing to prove. Suppose that U is nonempty and U 0005= X . Choose an A ∈ F T such that Ax belongs to U and set x0 = Ax. Thus x0 ∈ U . By Lemma 5.3, U is contained in the closure of Orb( T , x0 ) = Orb( T , Ax). Since x0 ∈ U , it follows that x0 is a limit point of Orb( T , x0 ) and that U is equal to the interior of the closure of Orb( T , x0 ). Since x0 ∈ U , Lemma 5.2, implies that { p ( T )x0 : p ∈ W } is dense in X . Suppose, in order to obtain a contradiction, that there exists i ∈ {1, . . . , n} such that T i maps some point y ∈ X U into U . Without loss of generality, we may assume y ∈ X F . Because, X F is open and { p ( T )x0 : p ∈ W } is dense in X , we may ﬁnd a polynomial p ∈ W so that p ( T )x0 is close enough to y to ensure (1) p ( T )x0 ∈ X F and (2) T i ( p ( T )x0 ) ∈ U . Because U is contained in the closed T -invariant set F , it follows that the closure of the orbit under T of T i ( p ( T )x0 ) belongs to F ; that is Orb( T , T i p ( T )x0 ) ⊆ F . However, Orb( T , T i p ( T )x0 ) = p ( T ) Orb( T , T i x0 ). Now by Lemma 5.3, int cl Orb( T , T i x0 ) = int cl Orb( T , x0 ) and x0 is a limit point of int cl Orb( T , x0 ), hence also a limit point of int cl Orb( T , T i x0 ). This together with the continuity of p ( T ) yields p ( T )x0 ∈ F . Thus p ( T )x0 ∈ F and p ( T )x0 ∈ X F , a contradiction. Thus X U is invariant under T . 2 Observe that the preceding lemma shows that if Orb( T , x) is somewhere dense, and if Orb({ T j } j 0005=i , T ik (x)) is nowhere dense for each 1 0003 i 0003 n and for each k 0002 0, then every element of Orb( T , x), including x itself, belongs to the interior of the closure of Orb( T , x). Theorem 5.5 (Somewhere Dense Theorem). Suppose that T = ( T 1 , T 2 , . . . , T n ) is an n-tuple of commuting operators on a locally convex space X over F (= R or C). If x ∈ X and Orb( T , x) is somewhere dense in X , then Orb( T , x) is dense in X provided one of the following holds: If F = C: (a) The set K := C

0002

{σ p ( A ∗ ): A ∈ F T } is nonempty and has a connected component that is unbounded;

or (b) There is a cyclic operator B that commutes with T and satisﬁes σ p ( B ∗ ) has no interior in C and C σ p ( B ∗ ) is connected. If F = R: (a )

σ p ( A ∗ ) ⊆ [−1, 1] for each A ∈ FT ;

or (b ) There is a cyclic operator B that commutes with T and satisﬁes p ( B ) has dense range for every nonzero real polynomial p. Proof. We shall assume that F = C and that either (a) or (b) holds. The real versions are very similar. We shall proceed by induction on n. For n = 1 the result follows from the Bourdon–Feldman Somewhere Dense Theorem (see [4]). So, now assume that n 0002 2. Then our induction hypothesis is that for any (n − 1)-tuple T of commuting operators that satisﬁes either (a) or (b) above (or (a ) or (b ) in the case where F = R) that Orb( T , x) is either dense in X or nowhere dense in X , for every x ∈ X .

N.S. Feldman / J. Math. Anal. Appl. 346 (2008) 82–98

95

Now let T = ( T 1 , T 2 , . . . , T n ) be a commuting n-tuple of operators on X . Let x ∈ X and let F denote the closure of Orb( T , x) and let U denote the interior of F . Suppose that U is nonempty and, by way of contradiction, that F is a proper subset of X . Claim 1. For every i ∈ {1, . . . , n}, T i has dense range. To prove the claim, suppose there exists i such that T i does not have dense range. Then by Lemma 5.1, we have that U is the interior of the closure of Orb({ T j } j 0005=i , x). But this contradicts our inductive assumption because { T j } j 0005=i is a set of (n − 1) commuting operators that satisﬁes either (a) or (b) and has an orbit that is somewhere dense in X , but not dense in X . So the claim holds. Claim 2. ( X U ) is invariant under T ; that is T i ( X U ) ⊆ ( X U ) for all i ∈ {1, . . . , n}. Simply notice that for each i ∈ {1, . . . , n} and for each k 0002 0, Orb({ T j } j 0005=i , T ik (x)) ⊆ F , so Orb({ T j } j 0005=i , T ik (x)) is not dense

in X . Thus by our inductive hypothesis, Orb({ T j } j 0005=i , T ik (x)) is nowhere dense in X . So by Claim 1 and Lemma 5.4 we get that X U is invariant under T . Claim 3. Ax ∈ U , for every A ∈ F T . Suppose not, then Ax ∈ ( X U ). However, by Claim 2, X U is invariant under T , thus by Lemma 5.3 we get U ⊆ cl Orb( T , Ax) ⊆ ( X U ), clearly a contradiction (since U 0005= ∅). So, we must have that Ax ∈ U .

/ ∂U . Claim 4. If B is an operator on X that has dense range and commutes with T i for each i, and A ∈ F T , then B ( Ax) ∈ First notice that by Lemma 5.3, U is the interior of the closure of Orb( T , Ax). Since, by Claim 3, F = cl U , we also have that F = cl Orb( T , Ax). Now, by way of contradiction, suppose that B ( Ax) ∈ ∂ U . Because U is the interior of the closed set F , X U is the closure of X F . Thus by Claim 3, Lemma 5.2 and the fact that A has dense range (Claim 1), we may choose a collection Q ⊆ W of polynomials such that {q( T ) Ax: q ∈ Q} is a dense subset of X U . Since B has dense range, B must map the dense set D := U ∪ {q( T ) Ax: q ∈ Q} to a dense set; however, we will show that B ( D ) ⊆ X U . We will ﬁrst show that B (U ) ⊆ ( X U ). Since B ( Ax) ∈ ∂ U , then B ( Ax) ∈ X U and by Claim 2, X U is invariant under T , thus Orb( T , B Ax) ⊆ ( X U ), so cl Orb( T , B Ax) ⊆ ( X U ). Since U ⊆ cl Orb( T , Ax), we have B (U ) ⊆ B (cl Orb( T , Ax)) ⊆ cl B (Orb( T , Ax)) = cl Orb( T , B Ax) ⊆ ( X U ). Thus we have B (U ) ⊂ ( X U ). Now let q ∈ Q. We claim that Bq( T ) Ax ∈ ( X U ). To see this notice that since q( T ) Ax ∈ ( X U ) and since X U is invariant under T (Claim 2), then Orb( T , q( T ) Ax) ⊆ ( X U ). Using this we get that Bq( T ) Ax = q( T ) B Ax ∈ q( T )(∂ U ) ⊆ q( T ) F = q( T ) cl Orb( T , Ax) ⊆ cl q( T ) Orb( T , Ax) = cl Orb( T , q( T ) Ax) ⊆ ( X U ). Thus we have that B (U ) ⊆ ( X U ) and B (q( T ) Ax) ∈ ( X U ) for every q ∈ Q, thus B ( D ) ⊆ X U . However, this contradicts / ∂ U and Claim 4 follows. the fact that B has dense range. Thus we must have that B Ax ∈ Claim 5. Suppose that C is a set of operators on X such that each operator in C has dense range and commutes with T and such that C v is a connected set in X for each v ∈ X . If there exist C 0 ∈ C and A ∈ F T such that C 0 ( Ax) ∈ U , then C ( Ax) ∈ U for all C ∈ C . Notice that C Ax := {C ( Ax): C ∈ C } is a connected set of vectors that intersects U , and by Claim 4, C Ax does not intersect ∂ U . Thus by the connectivity of C Ax, we must have that C Ax ⊆ U . Claim 6. If F = C, then U is invariant under multiplication by nonzero scalars. If F = R, then U is invariant under multiplication by positive scalars. Let L denote the component of F {0} that contains 1. We will prove that U is invariant under multiplication by scalars in L. To see this simply note that if C = {λ I: λ ∈ L}, then each operator in C has dense range, commutes with T , and for each v ∈ X , C v is a connected set in X . Also since I ∈ C , for every A ∈ F T , I ( Ax) ∈ U (by Claim 3), thus by Claim 5, C ( Ax) ∈ U for all C ∈ C . Since { Ax: A ∈ F T } is dense in U , we must have that C (U ) ⊆ F for all C ∈ C . However, since C is invertible, C (U ) is an open set contained in F and since U is the interior of F , it follows that C (U ) ⊆ U . Also since C is closed under the operation of taking inverses, then in fact we have that C (U ) = U for all C ∈ C . Or, in other words, λU = U for all λ ∈ L. 0002 Now suppose that condition (a) holds, as stated in the theorem. Namely, that the set K := C {σ p ( A ∗ ): A ∈ F T } is nonempty and has an unbounded connected component, call this unbounded component E . With this, we must reach a contradiction. Fix an operator A ∈ F T and let us apply Claim 5 to the following collection C A = {( A − λ I ): λ ∈ E }. By the deﬁnition of K ,

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each member of C A has dense range and clearly commutes with T . Also, since E is connected, then C A v is connected in X for each v ∈ X . We claim that there exists λ0 ∈ E such that ( A − λ0 I )x ∈ U . Since x ∈ U and U is open, then ( λ1 Ax − x) ∈ U

for all suﬃciently large λ. Since E is unbounded, there exists a nonzero λ0 ∈ E such that ( λ1 Ax − x) ∈ U . Now by Claim 6, 0

U is invariant under multiplication by λ0 , thus ( A − λ0 I )x = λ0 ( λ1 Ax − x) ∈ U . Thus, C A satisﬁes the hypothesis of Claim 5, 0 thus C A x ⊆ U . Thus, since A ∈ F T was arbitrary, we have ( A − λ I )x ∈ U for all λ ∈ E and for all A ∈ F T . Now ﬁx a number λ1 ∈ E . Then ( A − λ1 I )x ∈ U ⊆ F for all A ∈ F T . Since { Ax: A ∈ F T } is dense in F , by taking limits of ( A − λ1 I )x = ( Ax − λ1 x) we see that F − λ1 x = { y − λ1 x: y ∈ F } ⊆ F . However, λ1 x ∈ U by Claim 6, thus 0 ∈ U − λ1 x ⊆ ( F − λ1 x) ⊆ F . Thus, 0 is in the interior of F , since 0 ∈ U − λ1 x ⊆ F and U − λ1 x is an open set. Since F is invariant under multiplication by positive scalars (Claim 6) it follows that F = X . But this contradicts our initial assumption that F 0005= X . Thus we have the desired contradiction, and we can now conclude that F = X . By induction, the proof is now complete in this case where (a) holds. For the case where F = R and (a ) holds we simply let E = (1, ∞) and proceed as above. We must have E being an interval of positive numbers, since Claim 6 only gives that U is invariant under multiplication by positive scalars. Now suppose that (b) holds. Namely that there is a cyclic operator B that commutes with T and satisﬁes σ p ( B ∗ ) has no interior in C and C σ p ( B ∗ ) is connected. Then, as before we will apply Claim 5 to a certain collection of operators. Let P be the set of all analytic polynomials p such that p ( z) is nonzero on σ p ( B ∗ ). Also let C = { p ( B ): p ∈ P } and 0002 Cn = { p ( B ): p ∈ P and the degree of p is n}. So C = n Cn . Clearly T commutes with C and by deﬁnition of P each operator in C has dense range. Also, for every v ∈ X , the set Cn v is connected in X , because it is the image of the connected set (C {0}) × (C σ p ( B ∗ ))n under the continuous map (c , λ1 , . . . , λn ) 000f→ p ( B ) v = c ( B − λ1 I ) · · · ( B − λn I ) v. Notice that since σ p ( B ∗ ) is a bounded set, then p 000e ,n ( z) = 000e zn + 1 ∈ Cn for all small 000e . Also, if A ∈ F T , then since U is an open set and Ax ∈ U , then p 000e ,n ( B ) Ax = (000e B n + I ) Ax = 000e B n Ax + Ax ∈ U for all small 000e > 0. Thus according to Claim 5, p ( B ) Ax ∈ U for all p ∈ Cn . Hence p ( B ) Ax ∈ U for all p ∈ C . By taking limits, we see that p ( B ) F ⊆ F for every p ∈ P . However, since σ p ( B ∗ ) has empty interior, it follows that P is dense in the set of all analytic polynomials (in the topology of uniform convergence on compact sets), thus p ( B ) F ⊆ F for all polynomials p. Now since σ p ( B ∗ ) has empty interior, it follows that B has a dense set of cyclic vectors (to see this simply note that if v is cyclic for B, then { p ( B ) v: p ∈ P } is a dense set that also consists of cyclic vectors for B). So, let y ∈ U be a cyclic vector for B, then X = cl{ p ( B ) y: p is a polynomial} ⊆ F . So, F = X . But this contradicts our assumption that F 0005= X . With this contradiction, we have that F = X . By induction, the proof is now complete in this case where (b) holds. For the case where F = R and (b ) holds we simply let P be the set of all analytic polynomials with real coeﬃcients and let Pn± be the set of all polynomials of degree n with real coeﬃcients and having a positive (respectively negative) leading coeﬃcient. Now let Cn± = { p ( B ): p ∈ Pn± }, then one can check that Cn+ v and Cn− v are both convex sets (hence connected) and the operators in Cn± all have dense range (by assumption), thus we may proceed as above using p ±000e ,n ( z) = ±000e zn + 1 ∈ Cn± for 000e > 0. This completes the proof. 2 Corollary 5.6. Suppose that T = ( T 1 , . . . , T n ) is a tuple of operators on a real or complex locally convex space X and F is the semigroup they generate. If A ∗ has no eigenvalues for every A ∈ F , then any orbit of T that is somewhere dense in X will be dense in X . Corollary 5.7. Suppose that T = ( T 1 , . . . , T n ) is a tuple of operators on a complex locally convex space X and F is the semigroup they generate. If A ∗ has a countable number of eigenvalues for every A ∈ F , then any orbit of T that is somewhere dense in X will be dense in X . In particular, if T is an n-tuple of matrices on Ck , then every somewhere dense orbit of T will be everywhere dense in Ck . The fact that multi-hypercyclic operators are in fact hypercyclic was proven independently by Costakis [8] and Peris [17]. Ansari [1] gave a truly original proof of the fact that if T is a hypercyclic operator, then T n is also hypercyclic for every n 0002 1. Corollary 5.8. Suppose that T = ( T 1 , T 2 , . . . , T n ) is an n-tuple of commuting operators on a locally convex space X and T satisﬁes the hypothesis from Theorem 5.5, then the following hold: (1) If T is multi-hypercyclic, then T is actually hypercyclic. k k k (2) If T is hypercyclic and k = (k1 , k2 , . . . , kn ) is a multi-index with ki 0002 1 for all i, then T k = ( T 11 , T 22 , . . . , T nn ) is also hypercyclic with the same set of hypercyclic vectors as T . Proof. (1) If a ﬁnite set has dense orbit under T , then some element of that set will have a somewhere dense orbit, hence by Theorem 5.5 its orbit must be dense. Hence T is hypercyclic.

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(2) Suppose that k = (k1 , k2 , . . . , kn ) is given and ki 0002 1 for all i. Let p = ( p 1 , p 2 , . . . , pn ) be any multi-index, then by applying the division algorithm we see that there exist integers q i 0002 0 and 0 0003 r i < ki such that p i = ki qi + r i . Thus, p p k r k r T p x = T 1 1 · · · T n n x = ( T 11 )q1 · · · ( T nn )qn T 11 · · · T nn x. Thus, we see that Orb( T , x) =

001d

0004

r

r

0005

Orb T k , T 11 · · · T nn x .

00002r i <>

Now suppose that x is a hypercyclic vector for T . Then the left-hand side above is dense, hence the right-hand side r r above is also dense. Thus for some choice of r i we must have that Orb( T k , T 11 · · · T nn x) is somewhere dense in X . It then r r n 1 k k follows from Theorem 5.5 that Orb( T , T 1 · · · T n x) is dense in X , hence T is hypercyclic. r r Continuing, let us also show that x is a hypercyclic vector for T k . Since we have that Orb( T k , T 11 · · · T nn x) is dense in X , r r r r r r and since Orb( T k , T 11 · · · T nn x) = T 11 · · · T nn Orb( T k , x), it follows that T 11 · · · T nn has dense range, which implies that T i has s s dense range whenever r i 0005= 0. Now deﬁne si = ki − r i when r i 0005= 0 and deﬁne si = 0 when r i = 0. Then T 11 · · · T nn will s1 r sn r have dense range and so T 1 · · · T n Orb( T k , T 11 · · · T nn x) is dense in X . But this last set is a subset of Orb( T k , x). Thus, x is hypercyclic for T k as well. So T and T k have the same set of hypercyclic vectors. 2 An operator T on a space X is said to be F -supercyclic, where F ⊆ C, if there is a vector x ∈ X such that F · Orb( T , x) = {α T n x: α ∈ F , n 0002 0} is dense in X . In [2], Bermdez, Bonilla, and Peris proved that if an operator T is R-supercyclic, then in fact T is R+ -supercyclic. We give another proof of this fact, because it follows easily from the results above and because it gives a simple example of how tuples of operators can be used to solve problems for a single operator. Corollary 5.9. If T is an operator on a complex locally convex space and T is R-supercyclic, then T is actually R+ -supercyclic. Proof. Consider the tuple of operators S = ( S 1 , S 2 , T ) where S 1 = −2I and S 2 =

1 I. 3

Since S 1 and S 2 are multiples of the

identity operators, then S is a commuting tuple. By Corollary 4.2, {2 /3 } is dense in R+ , from which one can show that {(−2)n /3k } is dense in R. This together with the fact that T is R-supercyclic imply that S is a hypercyclic tuple. Also since T is R-supercyclic, then T is cyclic and T ∗ has at most one eigenvalue. Thus with B = T , we see that S satisﬁes condition (b) of Theorem 5.5. Hence by Corollary 5.8, we have that ( S 12 , S 2 , T ) is also hypercyclic, which implies that T is R+ -supercyclic. 2 n

n

n

k

n

Similarly one can show that if F is the closure of {α1 1 α2 2 · · · αk k : ni 0002 0}, where

αi ∈ C, and T is an operator that is F p n1 p 2 n2 α2 · · · αkpk nk : ni 0002 0}.

supercyclic, and p = ( p 1 , . . . , pk ) ∈ N , then T is also F -supercyclic where F p is the closure of {α1 1 k

p

6. Questions (1) Is there an n-tuple on a complex Hilbert space that has a somewhere dense orbit that is not dense? (2) If T is a hypercyclic tuple, then must the semigroup generated by T contain a cyclic operator? (3) If T is a hypercyclic tuple, then is there a cyclic operator B that commutes with T ? (4) Is there a hypercyclic n-tuple of hyponormal operators on an inﬁnite dimensional Hilbert space? (5) Is there a hypercyclic (nondiagonalizable) n-tuple on Cn ? (6) Are there nondiagonalizable n-tuples on Rk that have somewhere dense orbits? (7) If an orbit of a tuple T is somewhere dense, but not dense in a real locally convex space X , then is the closure of the orbit invariant under multiplication by positive scalars? Acknowledgment The author would like to thank the referee for their careful reading and helpful suggestions.

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