A bounded operator T on a Banach space X is convex cyclic if there exists a vector x such that the convex hull generated by the orbit Tnxn≥0 is dense in X. In this note we study some questions concerned with convex-cyclic operators. We provide an example of a convex-cyclic operator T such that the power Tn fails to be convex cyclic. Using this result we solve three questions posed by Rezaei (2013).

1. Introduction and Main Results

Throughout this paper we denote by L(X) the algebra of all bounded linear operators on a real or complex infinite dimensional Banach space X. An operator T∈L(X) is said to be cyclic if there exists a vector x∈X (later called cyclic vector for T) such that the linear span of the orbit
(1)linearspan({Tnx:n∈ℕ})
is dense in X. If the orbit Orb(T,x):={Tnx:n∈ℕ} is dense itself, without the help of the linear span, then T is called hypercyclic and x is called hypercyclic for T. In the midway stand several notions studied by different authors. For instance, the operator T is said to be supercyclic if the projective orbit is dense in X. We refer to the books [1, 2] and references therein for further information on hypercyclic operators.

When we sometimes abusively say that a polynomial p(z) is a convex polynomial, what we really mean is that p(z)=t0+t1z+t2z2+⋯+tnzn, ti∈ℝ, i=0,…,n, and t0+t1+⋯+tn=1. We will focus our attention on the notion of convex cyclicity introduced by Rezaei in [3]. An operator T is said to be convex cyclic if there exists a vector x∈X such that the real convex hull of the orbit (denoted by co(Orb(T,x)))
(2)co(Orb(T,x))={p(T)x:pconvexpolynomial}
is dense in X.

In [3] are characterized the convex-cyclic matrices in finite dimension, and the author develops the main properties in the infinite dimensional setting.

A result by Ansari [4] states that if T is a hypercyclic operator then Tn is also hypercyclic; this fact is not true for cyclic operators. In this paper we show that Ansari’s result fails also for convex-cyclic operators, solving a question posed in [3].

Another result proved by Bourdon and Feldman on hypercyclic operators says that if the orbit of a vector is somewhere dense, then it is dense (see [5]). From our previous counterexample we can construct a non-convex-cyclic operator T such that the co(Orb(T,x)) has nonempty interior. That is, Bourdon and Feldman’s result is not true in the convex-cyclic setting. Finally we can construct a convex-cyclic operator T such that T is not weakly hypercyclic; that is, its orbit is not dense in the weak operator topology. The later examples solve Questions 5.5 and 5.6 in [3].

2. Powers of a Convex-Cyclic Operator

The first example of hypercyclic operator on Banach spaces was discovered by Rolewicz (see [6]). Throughout this section ℬ=ℓp1≤p<∞ or c0 of complex valued sequences. Rolewicz’s operator μB with |μ|>1 is defined on ℬ by
(3)μB(x0,x1,…,xn,…)=μ(x1,x2,…,xn,…),
where B denotes the backward shift operator.

Lemma 1.

Set α=e2πi/3 and r0>1. For any z0∈ℂ∖{0} there exist k0≥0 and a sequence of polynomials pk(z) such that

pk(z)=(t1,k+t2,kz+t3,kz2)zk for all k≥k0;

ti,k∈[0,1] and t1,k+t2,k+t3,k=1, i∈{1,2,3} and k≥k0;

pk(r0α)=z0, k≥k0.

Proof.

Let us denote by 𝒯 the triangle with vertices {1,r0α,r02α2}. Since |r0α|>1 and 0∈𝒯, there exists k0 such that zk=z0/(r0α)k∈𝒯 for all k≥k0. Then, there exist barycentric coordinates ti,k∈[0,1]i=1,2,3 satisfying t1,k+t2,kr0α+t3,k(r0α)2=zk and t1,k+t2,k+t3,k=1. Then, the polynomials pk(z)=(t1,k+t2,kz+t3,kz3)zk, for all k≥k0, yield the desired result.

Lemma 2.

Let pk be a sequence of polynomials satisfying Conditions (1)–(3) of Lemma 1. Then, there exists a Gδ dense subset Z⊂ℬ of vectors such that {pk(μB)x0}k≥k0 is dense in ℬ for all x0∈Z.

Proof.

We will use some hypercyclicity criterion version for sequence of operators (see [2, Theorem 3.24]); that is, we will show the existence of two dense subsets X andY and a sequence of mappings Sk such that

limkpk(μB)x=0∀x∈X;

pk(μB)Sky=y∀y∈Y;

limkSky=0∀y∈Y.

Let us consider the subsets
(4)X=span{Ker(μB-λI):|λ|<1},Y={Ker(μB-λI):λ∈ℝ,1<λ<|μ|},
which are dense in ℬ (see [2, Example 3.2, page 70]).

If x∈Ker(μB-λI) with |λ|<1, then
(5)pk(μB)x=(t1,k(μB)k+t2,k(μB)k+1+t3,k(μB)k+2)x=(t1,kI+t2,k(μB)+t3,k(μB)2)(μB)kx≤const|λ|k,
which goes to zero when k→∞; therefore, Condition (i) is fulfilled.

Denoting by qk(z)=t1,k+t2,kz+t3,kz2, since t1,k, t2,k, t3,k are barycentric coordinates of a triangle, then qk(λ) lies in the degenerate triangle with vertices {1,λ,λ2}, in particular qk(λ)≥1.

Let us take y∈Ker(μB-λI) with λ∈ℝ and 1<λ<|μ|, and let us define the mapping Sk on y as
(6)Sky=1λkqk(λ)y
and we extend linearly Sk on Y. Clearly Sky→0 as k→∞ for all y∈Y and pk(μB)Sky=y for all y∈Y. Thus, by the hypercyclicity criterion there exists a Gδ dense subset of vectors x0∈ℬ such that {pk(μB)x0}k≥k0 is dense in ℬ.

Now, let us prove the main result of this section, which solves Question 5.6 in [3].

Theorem 3.

The operator T=r0αIℂ⊕μB is convex cyclic on ℂ⊕ℬ; however T3 is not.

Proof.

If p is a polynomial, then p(T)=p(r0α)⊕p(μB). Let us observe that the first coordinate of the powers of (T3)n are only real numbers. Take x=∑n=0∞x0e0∈ℂ⊕ℬ. If f⋆ is the projection on the first coordinate,
(7){f⋆(co(Orb(T3,x)))=tx0,∀t≥r0}
which is not dense in ℂ. Therefore, T3 is not convex cyclic.

Now, let us prove that T is a convex-cyclic operator using a direct application of the Baire category theorem (see, for instance, [2, Theorem 1.57]). Thus T is convex cyclic if for any nonempty open subsets U,V⊂ℂ⊕ℬ, there exists a convex polynomial p(z) such that p(T)(U)∩V≠∅.

Indeed, let U=G1×W1 and V=G2×W2 open subsets of ℂ⊕ℬ, where Gi⊂ℂ and Wi⊂ℬ, i=1,2, are nonempty open subsets. Let z1∈G1 and z2∈G2 with z1z2≠0. Set z0=z2/z1 and pk(z) the sequence of polynomials which guarantees Lemma 1. Hence we have pk(r0α)=z2/z1 and therefore pk(r0α)z1=z2 (this fact will imply that pk(T) acting on G1 will intersect G2). Now we apply Lemma 2 and we obtain a Gδ dense subset Z⊂ℬ of hypercyclic vectors for the sequence{pk(μB)}. Thus there exist x0∈W1 and a subsequence {nk} such that pnk(μB)x0∈W2. Therefore pnk(T)(U)∩V≠∅, which yields the desired result.

Remark 4.

If we take α=e2πi/n with n≥4, using similar arguments as in Theorem 3, we can show that T=r0α⊕μB is convex cyclic on ℂ⊕ℬ but Tn is not (if n=4 the operator T is convex cyclic but T2 is not).

Remark 5.

If we consider the Rolewicz operator on real spaces ℓp, 1≤p<∞ or c0, then we can get that the operator T=-r0⊕μB (r0>1) is convex cyclic on ℝ⊕ℓp1≤p<∞ or ℝ⊕c0, but clearly T2 is not. Lemma 1 can be adapted clearly to the real case. Now it is well known that if we consider Rolewicz’s operator on real spaces ℓp, 1≤p<∞ or c0, then its complexification can be identified with the same operator on the corresponding spaces of complex sequences. With some slight modification in the proof of Corollary 2.51 in [2] we can obtain that Lemma 2 continues being true on real spaces. The rest of the proof is straightforward.

Remark 6.

Another difference between hypercyclic operators and convex-cyclic operators is the following: hypercyclic operators are invariant under unimodular multiplications (see [7]); that is, if T is hypercyclic, then λT is also with |λ|=1. However this is not true for convex-cyclic operators; the previous counterexample T=r0αIℂ⊕μB is convex cyclic; however α-T is not.

Now, let α=e2πi/3. Let us consider the operator T=αIℂ⊕μB, that is, the same operator provided by Theorem 3 but without the multiplier factor r0. Then, an easy check shows that the set S={pk(α):pk(z)convexpolynomial} is contained in the unit disk. Moreover, the set S has nonempty interior in ℂ; for instance, S contains the triangle 𝒯 with vertices {1,α,α2}. Using the arguments of Theorem 3 we can find a vector x0∈ℬ such that the convex orbit
(8){p(T)(1⊕x0):pconvexpolynomial}
is dense in 𝒯⊕ℬ. Therefore the convex orbit has nonempty interior. However the operator T is not convex cyclic. This solves Question 5.5 in Rezaei’s paper.

Proposition 7.

Bourdon and Feldman’s result fails for convex-cyclic operators.

The adjoint of the operator T in Theorem 3 has an eigenvalue; therefore, T cannot be weakly hypercyclic. This solves Question 5.4 in [3].

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The authors are indebted to the referees for alerting them to some details that they had to take care of in the proof of the first version. Fernando León-Saavedra and María del Pilar Romero de la Rosa were partially supported by Junta de Andalucía FQM-257.

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