# Powers of Convex-Cyclic Operators

Authors: Fernando León-Saavedra; María del Pilar Romero-de la Rosa
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

A bounded operatorT on a Banach space X is convex cyclic if there exists a vector x such that the convex hull generated by the orbit T
n
x
n
≥
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 T
n fails to be convex cyclic. Using this result we solve three questions posed by Rezaei (2013).


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## Body

## 1. Introduction and Main Results

Throughout this paper we denote byL
(
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)
linear
span
⁡
(
{
T
n
x
:
n
∈
ℕ
}
)
is dense in X. If the orbit Orb
(
T
,
x
)
:
=
{
T
n
x
:
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 polynomialp
(
z
) is a convex polynomial, what we really mean is that p
(
z
)
=
t
0
+
t
1
z
+
t
2
z
2
+
⋯
+
t
n
z
n, t
i
∈
ℝ, i
=
0
,
…
,
n, and t
0
+
t
1
+
⋯
+
t
n
=
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
:
p
convex
polynomial
}
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 T
n 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
=
ℓ
p
1
≤
p
<
∞ or c
0 of complex valued sequences. Rolewicz’s operator μ
B with |
μ
|
>
1 is defined on ℬ by
(3)
μ
B
(
x
0
,
x
1
,
…
,
x
n
,
…
)
=
μ
(
x
1
,
x
2
,
…
,
x
n
,
…
)
,
where B denotes the backward shift operator.Lemma 1.
Setα
=
e
2
π
i
/
3 and r
0
>
1. For any z
0
∈
ℂ
∖
{
0
} there exist k
0
≥
0 and a sequence of polynomials p
k
(
z
) such that(1)
p
k
(
z
)
=
(
t
1
,
k
+
t
2
,
k
z
+
t
3
,
k
z
2
)
z
k for all k
≥
k
0;
(2)
t
i
,
k
∈
[
0,1
] and t
1
,
k
+
t
2
,
k
+
t
3
,
k
=
1, i
∈
{
1,2
,
3
} and k
≥
k
0;
(3)
p
k
(
r
0
α
)
=
z
0, k
≥
k
0.Proof.
Let us denote by𝒯 the triangle with vertices {
1
,
r
0
α
,
r
0
2
α
2
}. Since |
r
0
α
|
>
1 and 0
∈
𝒯, there exists k
0 such that z
k
=
z
0
/
(
r
0
α
)
k
∈
𝒯 for all k
≥
k
0. Then, there exist barycentric coordinates t
i
,
k
∈
[
0,1
]  i
=
1,2
,
3 satisfying t
1
,
k
+
t
2
,
k
r
0
α
+
t
3
,
k
(
r
0
α
)
2
=
z
k and t
1
,
k
+
t
2
,
k
+
t
3
,
k
=
1. Then, the polynomials p
k
(
z
)
=
(
t
1
,
k
+
t
2
,
k
z
+
t
3
,
k
z
3
)
z
k, for all k
≥
k
0, yield the desired result.Lemma 2.
Letp
k be a sequence of polynomials satisfying Conditions (1)–(3) of Lemma 1. Then, there exists a G
δ dense subset Z
⊂
ℬ of vectors such that {
p
k
(
μ
B
)
x
0
}
k
≥
k
0 is dense in ℬ for all x
0
∈
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 S
k such that (i)
lim
⁡
k
p
k
(
μ
B
)
x
=
0
∀
x
∈
X;
(ii)
p
k
(
μ
B
)
S
k
y
=
y
∀
y
∈
Y;
(iii)
lim
⁡
k
S
k
y
=
0
∀
y
∈
Y.Let us consider the subsets
(4)
X
=
span
⁡
{
Ker
⁡
(
μ
B
-
λ
I
)
:
|
λ
|
<
1
}
,
Y
=
{
Ker
⁡
(
μ
B
-
λ
I
)
:
λ
∈
ℝ
,
1
<
λ
<
|
μ
|
}
,
which are dense in ℬ (see [2, Example 3.2, page 70]).
Ifx
∈
Ker
⁡
(
μ
B
-
λ
I
) with |
λ
|
<
1, then
(5)
p
k
(
μ
B
)
x
=
(
t
1
,
k
(
μ
B
)
k
+
t
2
,
k
(
μ
B
)
k
+
1
+
t
3
,
k
(
μ
B
)
k
+
2
)
x
=
(
t
1
,
k
I
+
t
2
,
k
(
μ
B
)
+
t
3
,
k
(
μ
B
)
2
)
(
μ
B
)
k
x
≤
const
|
λ
|
k
,
which goes to zero when k
→
∞; therefore, Condition (i) is fulfilled.
Denoting byq
k
(
z
)
=
t
1
,
k
+
t
2
,
k
z
+
t
3
,
k
z
2, since t
1
,
k, t
2
,
k, t
3
,
k are barycentric coordinates of a triangle, then q
k
(
λ
) lies in the degenerate triangle with vertices {
1
,
λ
,
λ
2
}, in particular q
k
(
λ
)
≥
1.
Let us takey
∈
Ker
⁡
(
μ
B
-
λ
I
) with λ
∈
ℝ and 1
<
λ
<
|
μ
|, and let us define the mapping S
k on y as
(6)
S
k
y
=
1
λ
k
q
k
(
λ
)
y
and we extend linearly S
k on Y. Clearly S
k
y
→
0 as k
→
∞ for all y
∈
Y and p
k
(
μ
B
)
S
k
y
=
y for all y
∈
Y. Thus, by the hypercyclicity criterion there exists a G
δ dense subset of vectors x
0
∈
ℬ such that {
p
k
(
μ
B
)
x
0
}
k
≥
k
0 is dense in ℬ.Now, let us prove the main result of this section, which solves Question 5.6 in [3].Theorem 3.
The operatorT
=
r
0
α
I
ℂ
⊕
μ
B is convex cyclic on ℂ
⊕
ℬ; however T
3 is not.Proof.
Ifp is a polynomial, then p
(
T
)
=
p
(
r
0
α
)
⊕
p
(
μ
B
). Let us observe that the first coordinate of the powers of (
T
3
)
n are only real numbers. Take x
=
∑
n
=
0
∞
‍
x
0
e
0
∈
ℂ
⊕
ℬ. If f
⋆ is the projection on the first coordinate,
(7)
{
f
⋆
(
co
⁡
(
Orb
(
T
3
,
x
)
)
)
=
t
x
0
,
∀
t
≥
r
0
}
which is not dense in ℂ. Therefore, T
3 is not convex cyclic.
Now, let us prove thatT 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, letU
=
G
1
×
W
1 and V
=
G
2
×
W
2 open subsets of ℂ
⊕
ℬ, where G
i
⊂
ℂ and W
i
⊂
ℬ, i
=
1,2, are nonempty open subsets. Let z
1
∈
G
1 and z
2
∈
G
2 with z
1
z
2
≠
0. Set z
0
=
z
2
/
z
1 and p
k
(
z
) the sequence of polynomials which guarantees Lemma 1. Hence we have p
k
(
r
0
α
)
=
z
2
/
z
1 and therefore p
k
(
r
0
α
)
z
1
=
z
2 (this fact will imply that p
k
(
T
) acting on G
1 will intersect G
2). Now we apply Lemma 2 and we obtain a G
δ dense subset Z
⊂
ℬ of hypercyclic vectors for the sequence{
p
k
(
μ
B
)
}. Thus there exist x
0
∈
W
1 and a subsequence {
n
k
} such that p
n
k
(
μ
B
)
x
0
∈
W
2. Therefore p
n
k
(
T
)
(
U
)
∩
V
≠
∅, which yields the desired result.Remark 4.
If we takeα
=
e
2
π
i
/
n with n
≥
4, using similar arguments as in Theorem 3, we can show that T
=
r
0
α
⊕
μ
B is convex cyclic on ℂ
⊕
ℬ but T
n is not (if n
=
4 the operator T is convex cyclic but T
2 is not).Remark 5.
If we consider the Rolewicz operator on real spacesℓ
p, 1
≤
p
<
∞ or c
0, then we can get that the operator T
=
-
r
0
⊕
μ
B (r
0
>
1) is convex cyclic on ℝ
⊕
ℓ
p
1
≤
p
<
∞ or ℝ
⊕
c
0, but clearly T
2 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 ℓ
p, 1
≤
p
<
∞ or c
0, 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 λ
T is also with |
λ
|
=
1. However this is not true for convex-cyclic operators; the previous counterexample T
=
r
0
α
I
ℂ
⊕
μ
B is convex cyclic; however α
-
T is not.Now, letα
=
e
2
π
i
/
3. Let us consider the operator T
=
α
I
ℂ
⊕
μ
B, that is, the same operator provided by Theorem 3 but without the multiplier factor r
0. Then, an easy check shows that the set S
=
{
p
k
(
α
)
:
p
k
(
z
)
convex
polynomial
} 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
,
α
2
}. Using the arguments of Theorem 3 we can find a vector x
0
∈
ℬ such that the convex orbit
(8)
{
p
(
T
)
(
1
⊕
x
0
)
:
p
convex
polynomial
}
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 operatorT in Theorem 3 has an eigenvalue; therefore, T cannot be weakly hypercyclic. This solves Question 5.4 in [3].



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*Source: 631894-2014-05-08.xml*