s 0024609304003698 a

Bull. London Math. Soc. 37 (2005) 254–264 C2005 London Mathematical Societydoi:10.1112/S0024609304003698

TRANSITIVE AND HYPERCYCLIC OPERATORSON LOCALLY CONVEX SPACES

J. BONET, L. FRERICK, A. PERIS and J. WENGENROTH

Abstract

Solutions are provided to several questions concerning topologically transitive and hypercycliccontinuous linear operators on Hausdorff locally convex spaces that are not Frechet spaces. Amongothers, the following results are presented. (1) There exist transitive operators on the space ϕ ofall finite sequences endowed with the finest locally convex topology (it was already known thatthere is no hypercyclic operator on ϕ). (2) The space of all test functions for distributions, whichis also a complete direct sum of Frechet spaces, admits hypercyclic operators. (3) Every separableinfinite-dimensional Frechet space contains a dense hyperplane that admits no transitive operator.

1. Introduction

A discrete dynamical system is a continuous function f : X −→ X from a Hausdorfftopological space X into itself. Topological transitivity was introduced in 1920 byG. D. Birkhoff, in the following way (see the survey [16]): the discrete dynamicalsystem f : X −→ X is called transitive if for each pair of non-empty open subsetsU, V of X there is an n ∈ N such that fn(U)∩V = ∅. If the space X has no isolatedpoints and there is a point x ∈ X whose orbit O(f, x) := x, f(x), f2(x), . . . isdense in X, then f is transitive. The converse holds if X is a metrizable separableBaire space. While the existence of an element with a dense orbit implies that Xmust be separable, the transitivity of f : X −→ X does not require that the space Xbe separable [3]. Transitivity plays an important role in many definitions of chaos,in particular in Devaney’s [11].

A continuous and linear map T : E −→ E (called an operator from now on)acting on a Hausdorff locally convex space E is called hypercyclic if there exists avector x ∈ E (which is called the hypercyclic vector) such that its orbit O(T, x) isdense in E. As mentioned above, if E is a separable Frechet space, then an operatorT on E is hypercyclic if and only if it is transitive. Although the first examples ofhypercyclic operators were given in the first half of the last century, much researchhas been done concerning hypercyclic operators during recent years, starting withthe investigations of Godefroy and Shapiro [12]. The article [14] and its sequel [15],which covers recent developments, give a comprehensive picture; see also [7].

Since the Baire category theorem is essential in most of the fundamental resultsconcerning hypercyclicity, the space E on which the operators are defined is usually

Received 18 July 2003; revised 2 February 2004.

2000 Mathematics Subject Classification 47A16 (primary), 46A03, 46A04, 46A13, 37D45(secondary).

This paper was completed during a visit by L. Frerick to the Universitat Politecnica de Valenciain late 2002 and early 2003. The support of the Programa de Incentivo a la Investigacion Cientıficade la Universitat Politecnica de Valencia 2002 is gratefully acknowledged. The research of J. Bonetand A. Peris was partially supported by MCYT and FEDER Proyecto no. BFM2001-2670, andby AVCIT Grup 03/050.

transitive and hypercyclic operators 255

assumed to be a Frechet space. However, several recent important results by Ansari[1], Bourdon [8], Costakis and Peris [10, 21] and Bourdon and Feldman [9] hold forarbitrary locally convex spaces. Bonet and Peris [6] have considered the existenceof hypercyclic operators on locally convex spaces that are not metrizable and notBaire. Moreover, Bonet [5] has shown that concrete convolution operators or linearpartial differential operators have interesting dynamical behavior on non-metrizablespaces of distributions or analytic functions. The existence of transitive operatorson non-separable Banach spaces was considered recently by Bermudez and Kalton[3]. This has provided our motivation for investigating several questions concerningthe existence of transitive and hypercyclic operators.

It is shown in [6] and [14] that the space ϕ of all finite scalar sequences admitsno hypercyclic operator. We prove in Theorem 2.2 that it has transitive operators.In contrast to this, we give examples of metrizable locally convex spaces withouttransitive operators: any infinite-dimensional separable Banach space or, moregenerally, any infinite-dimensional separable Frechet space contains a dense hyper-plane that admits no transitive operator defined on it; see Theorem 3.2. It is known(see [2, 4, 6]) that every infinite-dimensional separable Frechet space admits ahypercyclic operator. Using essentially the same proof that works in the caseof Frechet spaces, Bonet and Peris [6] have shown that, in addition, countableinductive limits of Banach spaces in which one step is already dense, also admithypercyclic operators. The proof, which uses a comparison principle and Baire’stheorem, cannot be extended to the case of strict inductive limits of Banach orFrechet spaces, such as the space of test functions for distributions on an openset. This space has a similar topological structure to the space ϕ, but surprisinglyit admits hypercyclic operators, by Theorem 4.2. As an aside, which could be ofindependent interest in connection with recent results of Grivaux [13], we give alinearly independent dense sequence in the countable product ω of copies of thescalar field K that cannot coincide exactly with the orbit of a hypercyclic operatoron ω; see Proposition 3.3. At several points, our results require techniques that havenot previously been used in the context of hypercyclic operators.

We use standard notation for locally convex spaces; see, for example, [18], [20]and [23]. The vector spaces are defined over the field K of real or complex numbers.

2. Transitive operators commuting with representing spectra

We consider as the main example, the space

ϕ := x = (xn)n∈N ∈ KN : xn = 0 for finitely many indexes n =

⊕n∈N

K

equipped with its natural topology: that is, the strong topology with respect tothe dual pair (ϕ, ω), where ω := K

N, which coincides in this case with the finestlocally convex topology. The existence of transitive operators on ϕ is proved byrepresenting ϕ as a non-countable projective limit of weighted 1-spaces and usingthe result of Salas [24, Theorem 3.3] that the operator I + T is hypercyclic on 1

for any weighted backward shift T with positive bounded weights.A projective spectrum X consists of spaces Xα for α in the directed index set I

and continuous linear spectral maps

αβ : Xβ −→ Xα for α β, with α

β βγ = α

γ and αα = idXα

.

256 j. bonet, l. frerick, a. peris and j. wengenroth

The projective limit is

ProjX =

(xα)α∈I ∈∏α∈I

Xα : αβxβ = xα

,

endowed with the restriction of the product topology, and α denotes the restrictionto ProjX of the projection onto the component with index α. Such a X is stronglyreduced if for each α there is a larger β such that α

βXβ is contained in the closureof Im(α) in Xα.

A family (Tα)α∈I of continuous linear mappings on Xα is an endomorphism ofX if it commutes with the spectral maps: Tα α

β = αβ Tβ . The projective limit

of the morphism is defined by T (xα)α∈I = (Tαxα)α∈I . We refer the reader to [28]for more information about projective limits. A proof similar to that of the nextproposition can be seen in [22, 2.1].

Proposition 2.1. Let X be a strongly reduced projective spectrum, and let(Tα)α∈I be an endomorphism with transitive components. Then T is a transitiveoperator on X := ProjX .

Proof. Let x, y ∈ X and U ∈ U0(X) be given. Then there exist α ∈ I and anopen set V ∈ U0(Xα) with U ⊇ (α)−1(V ), and there exists

β α, with αβXβ ⊆ αX.

For each W ∈ U0(Xα), we obtain

αβXβ ⊆ αX + W,

and thus Xβ ⊆ βX +(αβ )−1(W ), which means that the image of β is dense in Xβ

with respect to the vector space topology S having (αβ )−1(W ) : W ∈ U0(Xα) as

a basis of the 0-neighbourhood filter. Moreover, Tβ is continuous on (Xβ ,S) sinceTβ((α

β )−1(W )) ⊆ (αβ )−1(Tα(W )), and Tβ is transitive with respect to S since S is

coarser than the original topology on Xβ .Hence there is a k ∈ N such that βx + (α

β )−1(V ) meets T−kβ (βy + (α

β )−1(V ))and, since the intersection of these sets is open with respect to S, there is a z ∈ Xsuch that βz belongs to this intersection.

Then we have α(z − x) = αβ(βz − βx) ∈ V , and hence z ∈ x + U , and

α(T kz) = αβ(β(T kz)) = α

β(T kβ (βz)) ∈ αy + V,

which gives T kz ∈ y + U . This proves that x + U meets T−k(y + U), and thus T istransitive.

For a sequence w = (wn)n∈N of weights, we denote by Tw the weighted backwardshift (xn)n∈N0 → (wn+1xn+1)n∈N0 .

Theorem 2.2. Let w be a sequence of positive weights. Then I+Tw is transitiveon ϕ.

Proof. We denote by I the set of all increasing sequences α = (αn)n∈N0 ofnatural numbers such that vn+1(α) = wn+1αn/αn+1 defines a bounded sequencev(α) = (vn(α))n∈N. With the natural order inherited from N

N0 , this is a directedset such that ϕ = Proj 1(α), where 1(α) = (xn)n∈N0 : (αnxn)n∈N0 ∈ 1(the spectral maps are the identity embeddings).


Since αnwn+1xn+1 = vn+1(α)αn+1xn+1, the boundedness of v(α) implies thatTw acts continuously on 1(α). If g denotes the isomorphism

1(α) −→ 1, (xn)n∈N0 → (αnxn)n∈N0 ,

then we have, for Tv(α) : 1 −→ 1 and Tw : 1(α) −→ 1(α), the identity g Tw =Tv(α) g. Since v(α) is bounded, Tv(α) defines a continuous operator on 1, andby Salas’s theorem mentioned above, I + Tv(α) is hypercyclic and hence transitive,from which we deduce by [17, Lemma 2.1] that Sα = I + Tw : 1(α) −→ 1(α) istransitive, too. Clearly, (Sα)α∈I is a morphism, and the proposition above impliesthat S = I + Tw : ϕ −→ ϕ is transitive.

One way to obtain hypercyclic or transitive operators is to decrease the space onwhich a natural operator is defined. Generally, it is more difficult to show that afixed operator is hypercyclic or transitive on a small space. The advantage of doingso is that the comparison principle of Shapiro [14, Proposition 9] (or its generalizedversion, given in [17]) then helps us to conclude the hypercyclicity of the operatoron a large space, as soon as it is well defined and continuous on it. Since there areno transitive (or hypercyclic) operators on finite-dimensional spaces, the space ϕis the smallest space on which the natural operator T , defined as the sum of theidentity and a weighted backward shift as considered above, could be transitive. Infact, the comparison principle and Theorem 2.2 yield the following consequence.

Corollary 2.3. Let λ be a locally convex sequence space such that ϕ ⊂ λ ⊂ ωwith continuous inclusions, and such that ϕ is dense in λ. If w is a strictly positiveweight such that the weighted backward shift Tw is continuous on λ, then I + Tw

is a transitive operator on λ.

Continuous, hypercyclic and chaotic weighted backward shifts on Kothe echelonspaces are characterized in [17].

3. Locally convex spaces without transitive operators

Let I be an arbitrary infinite index set. Bermudez and Kalton [3, Section 3,Example] show that there exist transitive operators on the Banach space p(I), p ∈[1,∞). A similar proof shows that every infinite product of copies of a topologicalspace X admits a continuous transitive map that is a transitive operator in the casewhere X is a locally convex space. Indeed, if we write the product of the form XI

as a countable product (XJ)N, where X is an arbitrary topological space, then thebackward shift f(x1, x2, . . .) = (x2, x3, . . .) is a transitive operator.

In [3, Theorem 3.4] it is shown that there is no transitive operator on ∞ or onL(2). The proof depends on the observation (which remains valid for locally convexspaces) [3, 3.3] that if an operator T on a locally convex space is transitive, thenits transpose T t admits no eigenvalues. Wark constructs in [27] a non-separablereflexive Banach space X such that every operator T : X −→ X is of the formT = λI+S, and S has separable range. It is easy to see that the transpose T t of everyoperator T on X has an eigenvalue; consequently, the space X constructed by Warkadmits no transitive operator. We show here that there are a lot of separable locallyconvex spaces – and even separable pre-Hilbert spaces – that admit no transitiveoperators. To show this, we need a lemma, which may also be of independentinterest.


Lemma 3.1. If E is a separable infinite-dimensional Frechet space, there existsa hyperplane Y in E such that every operator T : Y −→ Y has the form T = λI+F ,where λ is a scalar and F is a finite-dimensional operator. In particular, any operatoron Y has an eigenvalue.

Proof. By a result of Valdivia (see [20, 6.3.11]), there exists a dense hyperplaneY in E that contains no infinite-dimensional continuously embedded Frechet space.Since Y is a hyperplane in E, there is an e ∈ E with E = Ke ⊕ Y . If we extendT uniquely to an operator acting on E, there are a λ ∈ K and a w ∈ Y withT (e) = λe + w. This shows that T −λI maps E into Y , and that T −λI : E −→ Yis an operator. But (T − λI)(E) equipped with the induced quotient topology isa Frechet space continuously embedded in Y , and it has to be finite-dimensional.Hence (T − λI)(Y ) is also finite-dimensional, from which the conclusion follows.

Proposition 3.2. (a) Every separable infinite-dimensional Frechet space Econtains a hyperplane Y such that no operator T : Y −→ Y is transitive. Inparticular, there is a separable pre-Hilbert space admitting no transitive operators.

(b) There is a countable dimensional locally convex space X admitting notransitive operator.

Proof. (a) We take a hyperplane Y of E, as constructed in Lemma 3.1. Let T bean operator on Y . Then T = λI +F , where F is finite-dimensional. Its transpose isof the same form, so it has an eigenvalue. This implies that T cannot be transitive.

(b) We apply Lemma 3.1 to choose a hyperplane Y in ω with the propertiesindicated there. Let X be the space ϕ equipped with the topology σ(ϕ, Y ); that is,the weak topology induced by Y . Then the transpose of each operator T on X isan operator on Y , and hence it has an eigenvalue. This implies that T cannot betransitive.

Remark. In contrast to Proposition 3.2, Grivaux [13] has shown that everycountable dimensional normed space Y admits a hypercyclic operator.

We think it worth noting that neither Grivaux’s main result [13, Theorem 3.1],nor the instrumental [13, Lemma 2.1] holds for non-normable Frechet spaces.

Proposition 3.3. (a) Every countable product of copies of an infinite-dimensional separable Banach space X contains two dense linearly independentsequences of vectors such that their linear spans are not isomorphic.

(b) There is a dense linearly independent sequence in the Frechet space ω of allcomplex sequences that cannot be the orbit of a hypercyclic operator on ω.

Proof. (a) We select a countable linearly independent dense sequence A in XN

which is contained in the direct sum of copies of X, and a linearly independentsequence (yk)k in X. We set zk := (zk

n)n with zkn = 0 if n < k, and zk

n = yk if n k.The set A ∪ zk : k ∈ N is dense and linearly independent in XN, and its linearspan F admits no continuous norm. On the other hand, by a result of Moscatelliand Metafune [19], XN contains a dense subspace G with a continuous norm. SinceX is separable, we can apply [20, 2.2.1] to obtain a dense linearly independent


sequence in G. Let G′ be the linear span of this sequence. Clearly, F and G′ cannotbe isomorphic.

(b) Let V = (vk)k be a linearly independent dense sequence in ω that iscontained in ϕ and is such that vk(1) = 0 for each k. We define, for k ∈ N,wk(j) := 0 if j k and wk(j) = 2jk if j > k. The set A := V ∪ wk : k ∈ N isdense and linearly independent in ω. Assume that there are a continuous operatorT : ω −→ ω and a vector x ∈ A such that the orbit O(T, x) of x by T coincideswith A. In this case, there is a strictly increasing sequence (ki)i of natural numberssuch that T (wki

) ∈ V . The continuity of T implies the existence of m and C > 0such that |T (w)(1)| C supjm |w(j)| for each w ∈ ω. Taking w = wki

for i > m,we reach a contradiction.

4. Hypercyclic operators on direct sums

There are many natural hypercyclic and chaotic operators on non-metrizablespaces; for example, convolution operators either on spaces of infinite differentiablefunctions, or on spaces of distributions [12, 5]. In the case of inductive limits ofBanach spaces, there are examples of weighted spaces of holomorphic functions onwhich the differentiation operator is hypercyclic; see [5]. It is shown in [6] that, onnontrivial inductive limits of Banach spaces in which one step is dense, there existhypercyclic operators. This covers the cases of Kothe co-echelon spaces and spacesof germs of holomorphic functions on compact sets. In contrast to this, the space ϕ,which is the direct sum of copies of the scalar field, admits no hypercyclic operator,as has already been mentioned above.

The proofs of all the positive results depend on the existence of a dense subspacethat is a Frechet space for a stronger topology; hence the conclusion followsfrom Baire’s theorem and a comparison principle. This kind of reduction to theFrechet case fails if one considers, for example, the space D of test functions(which is isomorphic to a direct sum of Frechet spaces) because, in this case, thesubspace would be a fortiori contained in a finite sum as a consequence of theGrothendieck factorization theorem [18]. Accordingly, no examples of direct sumsof Frechet spaces admitting a hypercyclic operator are known. It is not difficultto show that on any countable direct sum of an infinite-dimensional Frechet space,there are transitive operators that are not hypercyclic. This was observed by Grosse-Erdmann; see [5, Note added in proof]. However, if one wants to obtain a hypercyclicoperator in such a direct sum, the hypercyclic vector must be explicitly constructed.Here, an operator on a direct sum of 1 and a hypercyclic vector for this operator areconstructed. After a slight modification, this operator is shown to be hypercyclic onthe spaces of test functions defined on open sets. We believe that our constructionmight be of interest for continuous dynamical systems on non-metrizable topologicalspaces.

Theorem 4.1. There is a hypercyclic operator T on⊕

i∈N1. Moreover, we

can choose T in such a way that there is a hypercyclic vector for T whose orbit issequentially dense.

Proof. Let us first explain the construction informally. We put the unit vectorsei,j of the ith 1-summand in the ith row of a matrix for i 2, and we orderthe unit vectors of the first summand in the form of a snake above the matrix.


···

···

···

···

···

···

Figure 1. Construction of a hypercyclic shift.

The operator is a weighted shift, as indicated in Figure 1, and the vector e1,1 ismapped to 0.

We construct a vector that lives on the windings (and hence in the first summandof the direct sum). The left-hand parts of the windings are partially filled withzeros, so that application of the shift pulls the non-zero part of a winding intoan upper-left triangle of the matrix, while at the same time only zeros from allsubsequent windings are pulled into the matrix. To obtain a hypercyclic vector inthis way, we choose a countable dense set of upper-left triangles, and arrange them(appropriately weighted) into the sequence of windings.

More formally, we proceed as follows. We start by defining three sequences, n, mand l, of natural numbers, as follows. Let n(1) := 1 and l(1) := 0. Then, for k 2,we set

m(k) := l(k − 1) + n(k − 1) + 2;l(k) := m(k) + 2(k − 1)2 − k;n(k) := l(k) + n(k − 1) + 1.

We fix λ > 1. On the ith 1-summand, we denote the canonical basis by (e(i,j))j∈N,i ∈ N. We define, for k ∈ N,

T (e(1,1)) := 0;T (e(1,j)) := λe(1,j−1), if n(k) + 1 < j n(k + 1) for some k;

T (e(1,n(k)+1)) := λe(2,2k−1);T (e(2,2k)) := λe(1,n(k+1));T (e(2k,1)) := λe(2k−1,1).


Finally,

T (e(i,j)) :=

λe(i−1,j+1), if i + j is even and i > 2;λe(i+1,j−1), if i + j is odd and i, j > 1.

We extend T by linearity to the span of all unit vectors. Then T is a continuousoperator with respect to the subspace topology coming from the direct sum. Hence,we can extend T uniquely to a continuous operator on the whole direct sum. It hasthe following properties, where k ∈ N:

T l(k−1)(e(1,m(k)−1)) = λl(k−1)e(1,n(k−1)+1);

T l(k)(e(1,m(k))) = λl(k)e(1,1);

T l(k)(e(1,n(k))) = λl(k)e(1,n(k−1)+1).

We have to show that there exists a hypercyclic vector x for T . To construct x,we let vk = (ak

(i,j))(i,j)∈N2 : k 2 be a sequentially dense subset of⊕

i∈N1

satisfying

sup(i,j)∈N2

|ak(i,j)| k,

and

ak(i,j) = 0, if i + j > 2k − 1.

There exist uniquely determined scalars xj , where m(k) j n(k), such that

T l(k)

(n(k)∑

j=m(k)

xje(1,j)

)= vk, k 2.

Since 2l(k) n(k) k, we obtainn(k)∑

j=m(k)

|xj | λ−l(k)n(k)k 4λ−l(k)l(k)2, k 3;

hence

x :=∞∑

k=2

n(k)∑j=m(k)

xje(1,j)

is convergent in 1 ⊕⊕

i20, and hence convergent in⊕

i∈N1. We compute

T l(k)(x) − vk = T l(k)

( ∞∑ν=k+1

n(ν)∑j=m(ν)

xje(1,j)

),

so these vectors are contained in the first summand. Using the inequality l(k+1) 2l(k), we obtain

‖T l(k)(x) − vk‖1 λl(k)∞∑

ν=k+1

4λ−l(ν)l(ν)2

∞∑

ν=k+1

4λ−(1/2)l(ν)l(ν)2 → 0,

when k → ∞. This shows that the orbit of x under T is sequentially dense.


Remark. Clearly, one can replace the space 1 in the theorem by p, p ∈ (1,∞),or c0.

Let Ω be an open subset of RN . For a compact set K in R

N , we denote byD(K) the Frechet space of all infinite differentiable functions supported in K; thespace D(Ω) of all test functions is defined as the (strict) inductive limit of thesystem D(K), K ⊂ Ω compact. We want to show that D(Ω) admits a hypercyclicoperator T with a hypercyclic vector whose orbit is sequentially dense. Since D(Ω)is isomorphic to

⊕i∈N

s, where

s :=

x = (xn)n∈N ∈ CN : ‖x‖t :=

∞∑n=1

|xn|nt < ∞ for all t ∈ N

is the Frechet space of all rapidly decreasing sequences [25, 26], it is enough to showthat the assertion holds for

⊕i∈N

s. We consider⊕

i∈Ns as a subspace of

⊕i∈N

1.Unfortunately, the operator T appearing in the proof of the previous theoremdoes not map

⊕i∈N

s into itself. The reason is that the sequence n constructedin this proof is not polynomially bounded. To overcome this problem, we modifythe sequences n, m and l in such a way that n(k) 3k2, k ∈ N. Let a strictlyincreasing sequence (kν)ν∈N of natural numbers be given. We again set n(1) := 1and l(1) := 0. If k = kν for all ν, then we set n(k) := n(k − 1) + 2. For ν 2, weset inductively

m(ν) := n(kν − 1) + l(ν − 1) + 2;l(ν) := m(ν) + 2(kν − 1)2 − kν ;

n(kν) := l(ν) + n(kν − 1) + 1.

We choose now a strictly increasing sequence (kν)ν∈N satisfying the following (non-independent!) inequalities:

n(kν − 1) 3kν ;2kν m(ν) 4kν ;

k2ν l(ν) 2k2

ν ;n(kν) 3k2

ν kν+1.

From this, we find that n(k) 3k2, k ∈ N. If the operator T is now defined on thespan of the unit vectors as in the previous proof (according to the new sequencen), we see that it has a unique continuous extension to an operator T acting on⊕

i∈Ns. As in the proof of Theorem 4.1, we have, where ν ∈ N,

T l(ν−1)(e(1,m(ν)−1)) = λl(ν−1)e(1,n(kν −1)+1);

T l(ν)(e(1,m(ν))) = λl(ν)e(1,1);

T l(ν)(e(1,n(kν ))) = λl(ν)e(1,n(kν −1)+1).

We can now follow the proof of the previous theorem. Let vν : ν 2 besequentially dense in

⊕i∈N

s, with the same properties as in the proof of Theorem4.1. There exist uniquely determined scalars xj , m(ν) j n(kν), such that

T l(ν)

(n(kν )∑

j=m(ν)

xje(1,j)

)= vν , ν 2.


As before,

x :=∞∑

ν=2

n(kν )∑j=m(ν)

xje(1,j)

converges in s⊕⊕

i20. To see this, one can use the same arguments, which willbe used to show that T l(ν)(x)− vν → 0 in s⊕

⊕i20 when ν → ∞. To this end,

let ν 2. From |xj | λ−l(s)s, m(s) j n(ks), we obtain

‖T l(ν)(x) − vν‖t λl(ν)∞∑

s=ν+1

n(ks )∑j=m(s)

λ−l(s)sjt

λ2k2ν

∞∑s=ν+1

3k2s∑

j=1

λ−2ks sjt

∞∑

s=ν+1

λ−ks s(3k2s)t+1

∞∑

s=ν+1

λ−s(3s2)t+2.

The last expression tends to zero when ν → ∞. This shows that the orbit of x issequentially dense in

⊕i∈N

s. In fact, we have shown that the following theoremholds.

Theorem 4.2. Let Ω be an open subset of RN . The space D(Ω) of all test

functions admits a hypercyclic operator, and there exists a hypercyclic vector inD(Ω) for this operator whose orbit is sequentially dense.

The operator constructed for the proof of Theorem 4.2 is not a natural operator.In fact, most natural operators on D(Ω) are not transitive. Just to mention anexample, no linear partial differential operator P (D) with constant coefficients ishypercyclic on D(Ω), since the orbit of any vector has a fixed support, and cannotbe dense. Moreover, the transpose of P (D) has eigenvalues on D′(Ω).

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J. Bonet and A. PerisE.T.S. ArquitecturaDepartament de Matematica

AplicadaUniversitat Politecnica de ValenciaE-46022 ValenciaSpain

[email protected]@mat.upv.es

L. FrerickFB MathematikBergische Universitat WuppertalGauß-Straße 20D-42097 WuppertalGermany

[email protected]

J. WengenrothFB IV MathematikUniversitat TrierD-54286 TrierGermany

[email protected]

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