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Note
Hypercyclic operators are subspace hypercyclic Nareen Bamerni a,c,∗ , Vladimir Kadets b , Adem Kılıçman a a
Department of Mathematics, Universiti Putra Malaysia, 43400 UPM, Serdang, Selangor, Malaysia Department of Mechanics and Mathematics, Kharkov National University, 4 Svobody Sq., Kharkov, 61077, Ukraine c Department of Mathematics, University of Duhok, Kurdistan Region, Iraq b
a r t i c l e
i n f o
Article history: Received 2 October 2015 Available online xxxx Submitted by Richard M. Aron Keywords: Hypercyclicity Subspace-hypercyclicity
a b s t r a c t In this short note, we prove that for a dense set A ⊂ X (X is a Banach space) there is a non-trivial closed subspace M ⊂ X such that A ∩ M is dense in M. We use this result to answer a question posed in Madore and Martínez-Avendaño (2011) [9]. In particular, we show that every hypercyclic operator is subspace-hypercyclic. © 2015 Elsevier Inc. All rights reserved.
1. Introduction A bounded linear operator T on a separable Banach space X is hypercyclic if there is a vector x ∈ X such that Orb(T, x) = {T n x : n ≥ 0} is dense in X , such a vector x is called hypercyclic for T . The first example of a hypercyclic operator on a Banach space was constructed by Rolewicz in 1969 [12]. He showed that if B is the backward shift on p (N) then λB is hypercyclic if and only if |λ| > 1. The studying of the scaled orbit and disk orbit is motivated by the Rolewicz example [12]. In 1974, Hilden and Wallen [5] defined the concept of supercyclicity. An operator T is called supercyclic if there is a vector x such that the scaled orbit COrb(T, x) is dense in X . Similarly, an operator T is called diskcyclic if there is a vector x ∈ X such that the disk orbit DOrb(T, x) is dense in X , and such a vector x is called diskcyclic for T . For more information about diskcyclic operators, the reader may refer to [4] and [3]. In 2011, Madore and Martínez-Avendaño [9] considered the density of the orbit in a non-trivial subspace instead of the whole space, this phenomenon is called as subspace-hypercyclicity. An operator is called subspace-hypercyclic (or M-hypercyclic, for short) for a subspace M of X if there exists a vector such that the intersection of its orbit and M is dense in M. This concept has been studied in several papers, for example [11] and [6]. * Corresponding author. E-mail addresses:
[email protected] (N. Bamerni),
[email protected] (V. Kadets),
[email protected] (A. Kılıçman). http://dx.doi.org/10.1016/j.jmaa.2015.11.015 0022-247X/© 2015 Elsevier Inc. All rights reserved.
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Doctopic: Functional Analysis
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In the same paper, Madore and Martínez-Avendaño showed that subspace-hypercyclicity does not imply to hypercyclicity (see [9, Example 2.2]). However they asked a question whether hypercyclicity implies to subspace-hypercyclicity for some subspaces. In the same year, Le [7] answered this question but under certain conditions. In 2015, Martínez-Avendaño and Zatarain-Vera [10] proved that hypercyclic coanalytic Toeplitz operators are subspace-hypercyclic under certain conditions. However, the problem of whether every hypercyclic operator is subspace-hypercyclic is still an open problem. Therefore, we answer this problem affirmatively in this paper. In Section 2, we prove that if A is a dense subset of a Banach space X , then there is a non-trivial closed subspace M of X such that A ∩ M is dense in M. As a consequence, we show that if T is hypercyclic, then T is M-hypercyclic which answers the question (iii) that was posed by Madore and Martínez-Avendaño in [9]. As immediate consequences, we get some further properties of subspace-hypercyclic operators. 2. Main result The following theorem is our main result. Theorem 2.1. If A is a dense subset of a Banach space X , then there is a non-trivial closed subspace M such that A ∩ M is dense in M. To prove Theorem 2.1 above, we need the following two lemmas. Lemma 2.2. Let e be a fixed element in X such that e > 1. Then, for every > 0 there is a non-empty open subset U of X \{0}, such that for all x ∈ U x < and dist (e, lin{x}) > 1. Proof. Let us fix f ∈ X ∗ with f = 1 and f (e) = e, and let e − 1 |f (x)| < . U = x ∈ X \{0} : x < and x 1 + e It is clear that U is an open subset of X \{0} and norms of its elements are smaller than . Let us fix x ∈ U and t ∈ C, then we can consider the following two cases. Case 1. If |t| >
1+e x .
Then e − tx ≥ ||t| x − e| > 1.
Case 2. If |t| ≤
1+e x .
Then, since |f (e − tx)| ≤ f e − tx ≤ e − tx, we have e − tx ≥ |f (e) − tf (x)| ≥ e − |tf (x)| ≥ e −
e + 1 |f (x)| x
> e − (e + 1) > 1. Thus, we get dist (e, lin{x}) > 1. 2
e − 1 e + 1
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Lemma 2.3. Let A be a dense subset of X and e be a fixed element with e > 1. Then, for every finitedimensional subspace Y ⊂ X with dist(e, Y ) > 1, for every > 0 and y ∈ Y , there is an a ∈ A such that y − a < and dist(e, lin{Y, a}) > 1. Proof. Let q : X → X /Y be the quotient map. By applying Lemma 2.2 to X /Y , q(e) and , we obtain a corresponding non-empty open subset U of X /Y ; that is, for every u ∈ U , u < and dist (q(e), lin{u}) > 1. Then, q −1 (U ) ∩ {x ∈ X : x < } is a not empty open subset of X , thus it intersects the dense set A − y. Consequently, there is an x ∈ (A − y) ∩ q −1 (U ) ∩ {x ∈ X : x < } with x < . Setting a = y + x, then a ∈ (A − y) + y = A and y − a = x < . Moreover, dist (e, lin{Y, a}) ≥ dist (q(e), lin{q(Y ), q(a)}) = dist q(e), lin{q(x)} > 1.
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Proof of Theorem 2.1. Let e ∈ X be a fixed element with e > 1. By Lemma 2.2 one can find a onedimensional subspace Y1 ⊂ X with dist(e, Y1 ) > 1. Decompose the set of all positive integers in disjoint union of infinite subsets Nk , k ∈ Z+ with the property j that k=1 Nk ⊃ {1, . . . , j} for every j. Let D1 = {yi }i∈N1 be a countable dense subset of Y1 . Applying inductively Lemma 2.3, one can construct for each k = 1, 2, . . . an element ak ∈ A with ak − yk < k1 , a subspace Yk+1 = lin{Yk , ak } with dist(e, Yk+1 ) > 1 and a countable dense subset Dk+1 = {yi }i∈Nk+1 of Yk+1 . We claim that M=
∞
Yk
k=1
is the subspace we need. Indeed, since dist(e, M) ≥ 1, then M = X . Moreover, by construction we get that ∞ the set {yi }i∈N = Dk is dense in M and {ak }k∈N ⊂ M ∩ A, so the condition ak − yk < k1 , k = 1, 2, . . . k=1
implies that the set M ∩ A is also dense in M.
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Corollary 2.4. If T is a hypercyclic operator on X , then there is a non-trivial closed subspace M such that T is M-hypercyclic. As a consequence of the above corollary, some open problems are answered partially, and some open problems are also raised. Since an operator is hypercyclic if and only if its inverse is, then there exists a subspace-hypercyclic operator such that its inverse is also subspace-hypercyclic, which answers [9, Problem 1] partially. However, it is natural to ask the following question: Question 1. If T is M1 -hypercyclic and T −1 is M2 -hypercyclic, is there any relation between M1 and M2 ? If T is hypercyclic then λT is hypercyclic for all λ ∈ T (where T is the unit circle) (see [8, Corollary 3]). It follows that there exists a subspace-hypercyclic operator T such that λT is also subspace-hypercyclic, which answers [9, Problem 2] partially. Now, the following question can be asked: Question 2. If T is M1 -hypercyclic and λT is M2 -hypercyclic, is there any relation between M1 and M2 ? If T is hypercyclic then T n is hypercyclic for all n ≥ 1 (see [1, Theorem 1]). It follows that there exists a subspace-hypercyclic operator such that T n is also subspace-hypercyclic for all n ≥ 1, which answers [11, Problem 2] partially. This leads us to the following question:
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Question 3. If T n is Mn -hypercyclic for all n ≥ 1, is there any relation among all Mn ? Finally, since every infinite dimensional separable Banach space admits a hypercyclic operator (see [2, Corollary 1]), then every infinite dimensional separable Banach space admits a subspace-hypercyclic operator as well. Acknowledgment The authors are very grateful to the referee for the valuable suggestions and comments to improve our paper. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12]
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