Operators with Diskcyclic Vectors Subspaces

0.13 MB·English

by  Nareen Bamerni

Operators with Diskcyclic Vectors Subspaces
Nareen Bamerni ∗1 and Adem Kılıc¸man †2
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1,2Department of Mathematics, University Putra Malaysia, 43400 UPM, Serdang,
1
Selangor, Malaysia
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2

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a
J

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] Abstract
A
In this paper, we prove that if T is diskcyclic operator then the closed unit disk
F
. multiplied by the union of the numerical range of all iterations of T is dense in .
h H
t Also, if T is diskcyclic operator and λ 1, then T λI has denserange. Moreover,
a | |≤ −
m we prove that if α > 1, then α1T is hypercyclic in a separable Hilbert space H if and
[ only if T ⊕αIC is diskcyclic in H⊕C. We show at least in some cases a diskcyclic
operator has an invariant, dense linear subspace or an infinite dimensional closed

1
linear subspace, whose non-zero elements are diskcyclic vectors. However, we give
v
7 some counterexamples to show that not always a diskcyclic operator has such a
3 subspace.
5
2
0 Keywords: Diskcyclicoperator,diskcyclicvector,DiskcyclicityCriterion,conditionB1,numericalrange.
.
1 AMS Subject Classification: Primary47A16;Secondary 47A12,47A15.
0
5
1 1 Introduction
:
v
i
X An operator T is called hypercyclic if there is a vector x such that Orb(T,x) =
∈ H
r Tnx : n N is dense in , such a vector x is called hypercyclic for T. In 1969,
a
{ ∈ } H
Rolewicz [4] constructed the first example of hypercyclic operator in a Banach space. He
proved that if B is a backward shift on the Banach space ℓp(N) then λB is hypercyclic
for any complex number λ; λ > 1. This led Hilden and Wallen [10] to consider the
| |
scaled orbit of an operator. An operator T is supercyclic if there is a vector x such
∈ H
that COrb(T,x) = λTnx : λ C, n N is dense in , where x is called supercyclic
{ ∈ ∈ } H
vector. For more information on hypercyclicity and supercyclicity concepts, one may
∗nareen [email protected]
†[email protected]
1
refer to [1, 2, 3].
By Rolewicz example, a backward shift λB is not hypercyclic whenever λ 1. In
| | ≤
the last case, we can notice that even the multiplication of the closed unit disk D =
x C : x 1 by the orbit of B will not be dense. Therefore, one may ask “ Can the
{ ∈ | | ≤ }
multiplicationoftheclosed unitdisk bytheorbitofanoperatorbedense?” In2003, Zeana
[11] considered the disk orbit of an operator. An operator T is called diskcyclic if there
is a vector x such that the disk orbit DOrb(T,x) = αTnx : n 0,α C, α 1 is
∈ H { ≥ ∈ | | ≤ }
dense in , such a vector x is called diskcyclic for T. She proved that the diskcyclicity
H
is a mid way between the hypercyclicity and the supercyclicity.
Hypercyclicity Diskcyclicity Supercyclicity.
⇒ ⇒
In this paper, all Hilbert spaces are infinite dimensional (unless stated otherwise) separa-
ble over the field C of complex numbers. The set of all diskcyclic operators in a Hilbert
space is denoted by DC( ) and the set of all diskcyclic vectors for an operator T is
H H
denoted by DC(T).
We recall the following facts from [5].
Theorem 1.1 ( Diskcyclic Criterion). Let T ( ). Assume that there exist an
∈ B H
increasing sequence of integers n , two dense sets X,Y and a sequence of maps
k
{ } ⊂ H
S : Y such that:
nk → H
1. lim Tnkx Snky = 0 for all x X, y Y.
k→∞
k kk k ∈ ∈
2. lim S y 0 for all y Y;
k→∞k nk k → ∈
3. TnkS y y for all y Y.
nk → ∈
Then T has a diskcyclic vector.
Proposition 1.2. Let T, S ( ) such that ST = TS and R(S) is dense in . If
∈ B H H
x DC(T), then Sx DC(T).
∈ ∈
Proposition 1.3. If x is a diskcyclic vector of T, then Tnx is also a diskcyclic vector of
T for all n N.
∈
Corollary 1.4. If T is a diskcyclic operator on a Hilbert space , then the set of all
H
diskcyclic vectors for T is dense in .
H
Proposition 1.5. Let T DC( ). Then T∗ has at most one eigenvalue and that one
∈ H
has modules greater than 1.
Corollary 1.6. A multiple of a unilateral backward shift on ℓ2(N) is hypercylcic if and
only if it is diskcyclic.
2
This paper consists of three sections. In section two, we show that if an operator T is
diskcyclic, then the closed unit disk multiplied by the union of the numerical range of all
iterations of T is dense in C. We show that T λI has dense range for all λ C; λ 1
− ∈ | | ≤
whenever T is diskcyclic. We give a relation between a hypercyclic operator on a Hilbert
space and a diskcyclic operator on the Hilbert space C. In particular, we show
H H ⊕
that if α > 1, then 1T is hypercyclic if and only if T αIC is diskcyclic. Moreover, we
α ⊕
give another diskcyclic criterion with respect to a sequence λ ; λ 1, which is
{ nk} | nk| ≤
equivalent to the main diskcyclic criterion Theorem 1.1.
In section three, we show that if T is a diskcyclic operator and σ (T∗) = φ, then T
p
has an invariant, dense subspace whose non-zero elements are diskcyclic vectors for T.
However, we give the counterexample 3.3 to show that not all diskcyclic operators must
have such a subspaces. Moreover, we show that in some cases a diskcyclic operator
may have an infinite dimensional closed subspace whose non-zero elements are diskcyclic
vectors for T. Particularly, we define the condition and use it to show that whenever
0
B
a diskcyclic operator satisfies the condition and diskcyclic criterion, then there is an
0
B
infinite dimensional closed subspace whose non-zero elements are diskcyclic vectors for
T. In a parallel with supercyclic operators, we show that if an operator satisfies the
diskcyclic criterion and there is a normalized basic sequence u goes to zero as n goes to
n
infinity, then there is an infinite dimensional closed subspace whose non-zero elements are
diskcyclic vectors for T. However, Example 3.8 shows that not every diskcyclic operator
has such a subspace.
2 Diskcyclic operators
To prove our first result we need the following lemma
Lemma 2.1. A vector x DC(T) if and only if x DC(T)
∈ kxk ∈
Proof. The proof is clearly follows from the fact DOrb(T, x ) = 1 DOrb(T,x)
kxk kxk
The numerical range of an operator T is defined as ω(T) = Tx,x : x = 1 .
{h i k k }
Theorem 2.2. Suppose that T DC( ). Then
∈ H
1. D ∞ Tnx,x is dense in C for all vectors x DC(T).
n=0h i ∈
S
2. D ∞ ω(Tn) is dense in C.
n=0
S
Proof. (1): Let x DC(T) and λ C. By lemma 2.1 we can suppose that x = 1.
∈ ∈ k k
Since λx , then there exist an increasing sequence n of non-negative integers and a
k
∈ H
sequence α C; α 1 such that
k k
∈ | | ≤
α Tnkx λx < ǫ
k
k − k
3
Now,
α Tnkx,x λ = α Tnkx,x λ x,x
k k
|h i− | |h i− h i|
= α Tnkx λx,x
k
|h − i|
α Tnkx λx x ǫ
k
≤ k − kk k ≤
It follows that D Tnx,x : n 0 is dense in C.
{ h i ≥ }
(2): Let x DC(T) with x = 1 then by (1), D Tnx ,x : n 0 is dense in C.
0 0 0 0
∈ k k { h i ≥ }
Since D ∞ ω(Tn) = D Tnx,x : x = 1 and n 0 . It follows that D ∞ ω(Tn) is
n=0 {h i k k ≥ } n=0
dense inSC. S
Proposition 2.3. If T DC( ) and λ C; λ 1, then T λI has dense range.
∈ H ∈ | | ≤ −
Proof. Suppose that the range of T λI is not dense in , then there exists x DC(T)
0
− H ∈
such that x / (T λI) ; otherwise (T λI) would be dense by Corollary 1.4. By
0
∈ − H − H
the Hahn Banach Theorem there exists a continuous linear functional f on such that
H
f(x ) = 0 and f (T λI) = 0 . Then for all x ,f(Tx) = λf(x) and so
0
6 − H { } ∈ H
(cid:16) (cid:17)
f(Tnx) = λnf(x) for all n N. In particulat, f(Tnx ) = λnf(x ). Since x DC(T),
0 0 0
∈ ∈
then there exist n and α C; α 1 for all k N such that α Tnkx 2x ;
k k k k 0 0
→ ∞ ∈ | | ≤ ∈ →
therefore α f(Tnkx ) 2f(x ) and hence α λnkf(x ) 2f(x ). However, since λ 1
k 0 0 k 0 0
→ → | | ≤
andf(x ) = 0, thenα shouldbegreaterthan1forsomek Nwhichiscontradiction.
0 k
6 ∈
By [9, p.38], σ (T∗) = Γ(T) where Γ(T) is the compression spectrum of T i.e the set of all
p
complex numbers λ such that the range of T λI is not dense. Now, if T DC( ) and
− ∈ H
λ σ (T∗), then by the last proposition λ > 1, which gives another proof of Proposition
p
∈ | |
1.5.
Theorem 2.4. If T ( ) and α is a real number such that α > 1, then the operator
∈ B H
S = T αIC ( C) is diskcyclic if and only if 1T is hypercyclic.
⊕ ∈ B H⊕ α
Proof. Let z be a hypercyclic vector for 1T, we will show that z 1 is diskcyclic vector
α ⊕
for S. Let w λ be an arbitrary vector in C with λ = 0. Since 1T is hypercyclic,
⊕ H⊕ 6 α
then there exist an increasing positive sequence n such that
k
{ }
1 nk 1
T z w 0 as k
(cid:13)(cid:18)α (cid:19) − λ (cid:13) → → ∞
(cid:13) (cid:13)
(cid:13) (cid:13)
Therefore, (cid:13) (cid:13)
1 nk
λ Snk (z 1) w λ 0 as k
(cid:13) (cid:18)α(cid:19) ⊕ − ⊕ (cid:13) → → ∞
(cid:13) (cid:13)
and since α > 1,(cid:13)(cid:13)then λ 1 nk < 1 as k . (cid:13)(cid:13)
α → ∞
If λ = 0, then we can fin(cid:0)d a(cid:1) sequence n such that
k
{ }
1 nk
T z kw 0 as k
(cid:13)(cid:18)α (cid:19) − (cid:13) → → ∞
(cid:13) (cid:13)
(cid:13) (cid:13)
(cid:13) (cid:13)
4
and then
1 1 nk
Snk (z 1) w 0 0 as k
(cid:13)k (cid:18)α(cid:19) ⊕ − ⊕ (cid:13) → → ∞
(cid:13) (cid:13)
and it is clear th(cid:13)(cid:13)at 1 1 nk < 1. Therefore, z (cid:13)(cid:13)1 is diskcyclic vector for S.
k α ⊕
For the other side, sin(cid:0)ce(cid:1)S is diskcyclic then it is supercyclic and the proof follows directly
from [6, Theorem 5.2].
Corollary 2.5. If α is a real number; α > 1 and c C, then z c DC(T αIC) if
∈ ⊕ ∈ ⊕
and only if z HC(T).
∈
Theorem 2.6. Let T ( ), suppose that there exist an increasing sequence of positive
∈ B H
integers n , a sequence λ C 0 such that λ 1 for all k N, two dense
{ k} { nk} ∈ \{ } | nk| ≤ ∈
sets X,Y and a sequence of maps S : Y such that:
⊂ H nk → H
1. λ Tnkx 0 for all x X;
k nk k → ∈
2. 1 S y 0 for all y Y;
(cid:13)λnk nk (cid:13) → ∈
(cid:13) (cid:13)
3. (cid:13)TnkS y (cid:13) y for all y Y.
nk → ∈
Then there is a vector x such that λ Tnkx is dense in . In particular, x is diskcyclic
{ nk } H
vector for T.
Proof. The proof follows by Hypercyclic Criterion [1, Definition 1.5].
If the assumptions of the above theorem hold, we say that T satisfies the Diskcyclic
Criterion for the sequence λ .
{ nk}
Proposition 2.7. If T satisfies the Diskcyclic Criterion for the sequence λ , then T
{ nk}
also satisfies the Diskcyclic Criterion for the sequence α where αnk 0.
{ nk} (cid:12)λnk(cid:12) →
(cid:12) (cid:12)
Proof. Let X,Y be two dense sets and S be the right inverse to T(cid:12), th(cid:12)en there exist a
small positive number ǫ and a large positive number J such that λ Tnkx ǫ and
k nk k ≤
1 S y ǫ for all k > J. Setting α = √ǫλ , it is clear that α 1 and the proof
(cid:13)λnk nk (cid:13) ≤ nk nk | nk| ≤
f(cid:13)ollows. (cid:13)
(cid:13) (cid:13)
Proposition 2.8. Both diskcyclic criteria are equivalent.
Proof. If T satisfies the diskcyclic criterion with respect to the sequence λ , then
{ nk}
it is clear that the conditions (1) and (3) of Proposition 1.1 are satisfied. Now since
1 S y 0 for all y Y and 1 1 , then the condition (2) of Proposition 1.1
(cid:13)λnk nk (cid:13) → ∈ ≤ (cid:12)λnk(cid:12)
(cid:13)holds. (cid:13) (cid:12) (cid:12)
(cid:13) (cid:13) (cid:12) (cid:12)
Conversely, suppose that T satisfies the diskcyclic criterion. Fix an x X and y Y
∈ ∈
then there exist a small positive number ǫ such that
Tnkx Snky < ǫ2,
k kk k
5
and
Snky < ǫ.
k k
Define
1
λ = Snky,
nk ǫ
It follows that λ 1 and 1 Snky < ǫ for a small positive number ǫ . Further-
more, | nk| ≤ (cid:12)λnk(cid:12)k k 1 1
(cid:12) (cid:12)
Tnk(cid:12)x (cid:12)Snky = Tnkx λ ǫ < ǫ2,
k kk k k k| nk|
Thus
λ Tnkx < ǫ.
| nk|k k
which completes the proof.
3 Subspaces of diskcyclic vectors
Definition 3.1. Let T DC( ) and let be a linear subspace of whose non-zero
∈ H A H
elements are diskcyclic vectors for T, then is called diskcyclic subspace for T.
A
Theorem 3.2. Let T DC( ) and σ (T∗) = φ, then T has an invariant, dense
p
∈ H
diskcyclic subspace.
Proof. Let x DC(T) and = p(T)x : p is polynomial . It is clear that is a linear
∈ A { } A
subspace of , invariant under T, and dense in since it contains DOrb(T,x). If the
H H
polynomial p is non-constant then, p(T) = a(T µ )(T µ )..., where a = 0 and
1 k
− − 6
µ ,...,µ C. Since σ (T∗) = φ, each operator T µ has dense range; hence p(T) also
1 k p i
∈ −
has dense range. Morever, it is clear that p(T) commutes with T, then by Proposition
1.2, p(T)x DC(T) that is every element in is a diskcyclic vector for T.
⊂ A
If T satisfies the diskcyclic criterion, then T satisfies the supercyclic criterion. Therefore
σ (T∗) = φ by [7, Proposition 4.3.]. It follows that if T satisfies the diskcyclic criterion,
p
then by the last theorem; T has an invariant, dense diskcyclic subspace .
The next example shows that not all diskcyclic operators have diskcyclic subspaces.
Example 3.3. Let 1T be hypercyclic operator on a Hilbert space with a hypercyclic
2 H
vector x. Then by theorem 2.4, T 2I is a diskcyclic operator on C with a diskcyclic
⊕ H⊕
vector x 1. By corollary 2.5, we can see that x 2 DC(T 2I). Suppose that is a
⊕ ⊕ ∈ ⊕ A
diskcyclic subspace for T 2I. Since (T 2I)(x 1) DC(T 2I) by Proposition 1.3
⊕ ⊕ ⊕ ∈ ⊕
and since is a subspace, then
A
x 2 (T 2I)(x 1) = (x Tx) 0
⊕ − ⊕ ⊕ − ⊕ ∈ A
however, it is clear that (x Tx) 0 / DC(T 2I). Therefore, there is no subspace
− ⊕ ∈ ⊕
whose non-zero elements are diskcyclic vectors for T 2I.
⊕
6
Not only a diskcyclic subspace can be dense and invariant, sometimes it can be infinite
dimensional closed. In that cases, we say that T DC ( ).
∞
∈ H
Montes-Rodr´ıguez and Salas [7] defined the condition to find a suufficient condition
0
B
for an operator to have an infinite dimensional closed subspace of supercyclic vectors.
In parallel with supercyclicity, we define the condition and use it to find a sufficient
1
B
condition for an operator to be in DC ( ).
∞
H
Definition 3.4. Let T ( ). Suppose that T satisfies the diskcyclicity Criterion with
∈ B H
respect to a sequence λ . If there is an infinite dimensional closed subspace
{ nk} B1 ∈ H
such that λ Tnkz 0 for every z , then we say T satisfies Condition for the
k nk k → ∈ B1 B1
sequence λ .
nk
Theorem 3.5. Let T ( ). Suppose that T satisfies the Diskcyclicity Criterion with
∈ B H
respect to a sequence λ . If one of the conditions below satisfies, then T DC ( ).
{ nk} ∈ ∞ H
1. T satifies condition ;
1
B
2. There is an infinite dimensional closed subspace such that λ Tnkz is
A ∈ H k nk k
bounded for all z .
∈ A
Proof. The proof of (1) follows directly from Theorem 2.6 and [8, Theorem 2.2]. For (2),
Suppose that T satisfies the Diskcyclicity Criterion with respect to a sequence λ and
{ nk}
there is a positive real number M such that λ Tnkz < M for all z and k N. By
k nk k ∈ A ∈
Proposition 2.7, we have T satisfies the diskcyclic criterion with respect to the sequence
α where αnk 0. Therefore, we have
{ nk} (cid:12)λnk(cid:12) →
(cid:12) (cid:12)
(cid:12) (cid:12)
α α
α Tnkz = nk λ Tnkz nk M 0
k nk k (cid:12)λ (cid:12)| nk|k k ≤ (cid:12)λ (cid:12) →
(cid:12) nk(cid:12) (cid:12) nk(cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
Thus, we can say that T satisfies condition and hence the proof is finished.
1
B
Proposition 3.6. Let T ( ). Suppose that T satisfies the diskcyclicity Criterion and
∈ B H
there is a normalized basic sequence u such that lim Tu = 0, then T DC ( ).
m m→∞ m ∞
{ } ∈ H
Proof. The proof is similar to that given in [7, Corollary 3.3.]. Since there is no restriction
on the sequence λ of scalars, we may suppose that λ 1 for all k N.
{ nk} | nk| ≤ ∈
The proof of the following corollary follows directly from Theorem 2.4
Corollary 3.7. Suppose that α R and α > 1, then the operator S = T αIC ( C)
∈ ⊕ ∈ B H⊕
has infinite dimensional closed subspaces of diskcyclic vectors if and only if 1T has an
α
infinite dimensional closed subspace of hypercyclic vectors.
The following example shows that not every diskcyclic operators belong to DC ( )
∞
H
7
Example 3.8. Let λ be a complex number of modulus greater than 1 and B be the unilat-
eral backward shift operator. Since λB is hypercyclic if and only if it is diskcyclic Corollary
1.6, then
1. there exists an invariant, dense linear subspace of diskcyclic vectors for λB [12, p.8]
2. all closed subspaces of diskcyclic vectors for λB are finite dimensional [8, Theorem
3.4].
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