OON SUPER-RECURRENT OPERATORS
MOHAMED AMOUCH AND OTMANE BENCHIHEB
Abstract.
In this paper, we introduce and study the notion of super-recurrence of operators. Weinvestigate some properties of this class of operators and show that it shares some characteristics withsupercyclic and recurrent operators. In particular, we show that if T is super-recurrent, then σ ( T )and σ p ( T ∗ ), the spectrum of T and the point spectrum of T ∗ respectively, have some noteworthyproperties. Introduction and preliminaries
Throughout this paper, X will denote a Banach space over the field C of complex numbers. By anoperator, we mean a linear and continuous map acting on X .The most important and studied notions in the linear dynamical system are those of hypercyclicityand supercyclicity:An operator T acting on X is said to be hypercyclic if there exists a vector x whose orbit under T ; Orb ( T, x ) := { T n x : n ∈ N } , is dense in X. The vector x is called a hypercyclic vector for T . The set ofall hypercyclic vectors for T is denoted by HC ( T ) . One of the first examples of hypercyclic operatorson the Banach space setting was given in 1969 by Rolewicz [20].Birkhoff introduced an equivalent notion of the hypercyclicity called topological transitivity: anoperator T acting on a separable Banach space is hypercyclic if and only if it is topologically transitive,that is, for each pair ( U, V ) of nonempty open subsets of X there exists some positive integer n suchthat T n ( U ) ∩ V (cid:54) = ∅ , see [4].In 1974, Hilden and Wallen in [16] introduced the concept of supercyclicity. An operator T actingon X is said to be supercyclic if there exists some vector x whose scaled orbit under T ; C Orb ( T, x ) := { λT n x : λ ∈ C , n ∈ N } , is dense in X . Such a vector x is called a supercyclic vector for T . The setof all supercyclic vectors for T is denoted by SC ( T ) . As in the case of the hypercyclicity, there existsa characterization of the supercyclicity basing on the open subsets of X. An operator T acting on aseparable Banach space is supercyclic if and only if for each pair ( U, V ) of nonempty open subsets of X there exist λ ∈ C and n ∈ N such that λT n ( U ) ∩ V (cid:54) = ∅ . For more information about hypercyclic and supercyclic operators and their proprieties, see thebook [12] by KG. Grosse-Erdmann and A. Peris , the book [3] by F. Bayart and E. Matheron, and thesurvy article [13] by KG. Grosse-Erdmann.Another notion in the dynamical system that has a long story is that of recurrence which is in-troduced by Poincar´e in [19]. A systematic study of recurrent operators goes back to the work ofGottschalk and Hedlund [14] and also the work of Furstenberg [10]. Recently, recurrent operators havebeen studied in [7].An operator T acting on X is said to be recurrent if for each open subset U of X , there exists somepositive integer n such that T n ( U ) ∩ U (cid:54) = ∅ . A vector x ∈ X is called a recurrent vector for T if thereexists an increasing sequence ( n k ) of positive integers such that T n k x −→ x as k −→ ∞ . The set of all
Mathematics Subject Classification.
Key words and phrases.
Hypercyclicity, supercyclicity, recurrence, super-recurrence. a r X i v : . [ m a t h . F A ] F e b M. AMOUCH AND O. BENCHIHEB recurrent vectors for T is denoted by Rec ( T ), and we have that T is recurrent if and only if Rec ( T ) isdense in X. For more information about this classe of operators, see [1, 5, 8, 11, 21, 17, 15, 6].Motivated by the relationship between hypercyclic and recurrent operators, we introduce in thispaper a new class of operators called super-recurrent operators which is related to the supercyclicityand recurrence.In section 2, we introduce the notion of super-recurrence for operators. We show that every recurrentoperator is super-recurrent but the converse is false. We also prove that every supercyclic operatoris super-recurrent and that there exists an operator which is super-recurrent but not supercyclic.In section 3, we prove some proprieties for super-recurrent operators, we prove that if T ∈ B ( X )admits a super-recurrent vector, then it admits an invariant subspace consisting except for zero, ofsuper-recurrent vectors. Also, we prove that T is super-recurrent if and only if T admits a densesubset of super-recurrent vectors. Moreover, we prove that T is super-recurrent if and only if T p issuper-recurrent, for every nonzero positive integer p .In section 4, we focus on the spectral proprieties of super-recurrent operators. We prove that if T issuper-recurrent, then σ p ( T ∗ ) and σ ( T ) have almost the same proprieties as supercyclic operators. Inparticular, we show that there exists R > T intersect the circle { z ∈ C : | z | = R } . Moreover, we prove that the σ p ( T ∗ ) is completely contained ina circle of center 0. Finally, we show that if λ ∈ σ p ( T ∗ ), then one can find a T -invariant hyperplane X such that λ − T /X is recurrent on X .2. Super-recurrent operators
Definition 2.1.
We say that an operator T is super-recurrent if, for every nonempty open subset U of X there exists some n ≥ λ ∈ C such that λT n ( U ) ∩ U (cid:54) = ∅ . A vector x ∈ X \ { } is called a super-recurrent vector for T if there exist a strictly increasingsequence of positive integers ( k n ) n ∈ N and a sequence ( λ k n ) n ∈ N of complex numbers such that λ k n T k n x −→ x as n −→ + ∞ . We will denote by SRec ( T ) the set of all super-recurrent vectors for T. Remarks . (1) The supercyclicity implies the super-recurrence. However, the converse doesnot hold in general. Indeed, let n ∈ N and λ , . . . , λ n be nonzero complex numbers such that | λ i | = | λ j | = R for some strictly positive real number R , for 1 ≤ i , j ≤ n . We define anoperator T on C n by T : C n −→ C n ( x , . . . , x n ) (cid:55)−→ ( λ x , . . . , λ n x n ) . Let U be a nonempty open subset of X and x ∈ U . Since | R − λ i | = 1, for all 1 ≤ i ≤ n , itfollows that there exists a strictly increasing sequence of positive integers ( k n ) n ∈ N such that (cid:0) R − λ i (cid:1) k n −→
1, for all 1 ≤ i ≤ n . Let λ k = R − k n , for all k , then λ k T k n x −→ x. as k −→ ∞ . Since x ∈ U and U is an open subset of X , it follows that there exists k suchthat λ k T n k x ∈ U . Hence λ k T n k ( U ) ∩ U (cid:54) = ∅ . N SUPER-RECURRENT OPERATORS 3
This means that T is a super-recurrent operators. However, T cannot be supercyclic whenever n ≥
2, since a Banach space X supports supercyclic operators if and only if dim( X ) = 1 ordim( X ) = ∞ , see [16].(2) A recurrent operator is super-recurrent, but the converse does not hold in general. Indeed, if T is the operator defned in (1), then T is recurrent if and only if | λ i | = 1, for all 1 ≤ i ≤ n ,see [7].We have the following diagram showing the relationships among super-recurrence, recurrence andsupercyclicity. Hypercyclic RecurrentSupercyclic Super-recurrent (cid:54)↑ [3, Example 1.15] (cid:56) [7, section 4] (cid:56) Remarks 2.2 (cid:54)↑
Remarks 2.2 Some properties of super-recurrent operators
In the following, we give some properties satisfies by super-recurrent operators.
Proposition 3.1. If S ∈ B ( X ) is an operator such that T S = ST , then SRec ( T ) is invariant under S. Proof.
Let x ∈ SRec ( T ). Then there exist a strictly increasing sequence of positive integers ( k n ) n ∈ N and a sequence ( λ k n ) n ∈ N of complex numbers such that λ k n T k n x −→ x as n −→ + ∞ . Since S is continuous and T S = ST , it follows that λ k n T k n Sx −→ Sx as n −→ + ∞ . This means that Sx ∈ SRec ( T ). (cid:3) We are now ready to deduce an important result on the algebraic structure of the set of super-recurrent vectors.Recall that if p ( z ) = (cid:80) ni =0 λ i z i and T ∈ B ( X ), then p ( T ) = (cid:80) ni =0 λ i T i . Theorem 3.2. If x is a super-recurrent vector for T, then { p ( T ) x : p is a polynomial } \ { } ⊂ SRec ( T ) . In particular, If T has a super-recurrent vector, then it admits an invariant subspace consisting, exceptfor zero, of super-recurrent vectors.Proof. For a nonzero polynomial p , let S = p ( T ). Then ST = T S.
Since x ∈ SRec ( T ), it follows byProposition 3.1, that p ( T ) x ∈ SRec ( T ) . (cid:3) Remark . If T is a super-recurrent operator, then it is of dense range.Let X and Y be two Banach spaces. If T and S are operators acting on X and Y respectively, then T and S are called quasi-conjugate or quasi-similar if there exists some operator φ : X −→ Y withdense range such S ◦ φ = φ ◦ T. If φ can be chosen to be a homeomorphism, then T and S are calledconjugate or similar, see [12, Definition 1.5]. Proposition 3.4.
Assume that T ∈ B ( X ) and S ∈ B ( Y ) are quasi-similar. Then, T is super-recurrentin X implies that S is super-recurrent in Y .Proof. Suppose that T is super-recurrent. If U is a nonempty open subset of Y , then φ − ( U ) is anonempty open subset of X . Since T is super-recurrent, it follows that there exist n ∈ N , λ ∈ C and x ∈ X such that x ∈ φ − ( U ) and λT n x ∈ φ − ( U ), this means that φ ( x ) ∈ U and λφ ◦ T n ( x ) ∈ U . Since M. AMOUCH AND O. BENCHIHEB T and S are quasi-similar, it follows that φ ( x ) ∈ U and λS n ◦ φ ( x ) ∈ U . Hence, S is super-recurrentin Y . (cid:3) Remark . Assume that T ∈ B ( X ) and S ∈ B ( Y ) are similar. Then, T is super-recurrent in X ifand only if S is super-recurrent in Y .The following theorem gives necessary and sufficient conditions of super-recurrence of operators. Theorem 3.6.
The following assertions are equivalent : (1) T is super-recurrent;(2) for each x ∈ X, there exist a sequence ( n k ) of positive integers, a sequence ( x n k ) of elementsof X and a sequence ( λ n k ) of nonzero complex numbers such that x n k −→ x and λ n k T n k ( x n k ) −→ x ; (3) for each x ∈ X and for W a neighborhood of zero, there exist z ∈ X , λ ∈ C , and n ∈ N suchthat λT n ( z ) − x ∈ W and z − x ∈ W. Proof. (1) ⇒ (2) Let x ∈ X . For all k ≥
1, let U k = B ( x, k ). Then U k is a nonempty open subset of X . Since T is super-recurrent, there exist n k ∈ N and λ n k such that λ n k T n k ( U k ) ∩ U k (cid:54) = ∅ . For all k ≥
1, let x n k ∈ U k such that λ n k T n k ( x n k ) ∈ U k , then (cid:107) x n k − x (cid:107) < k and (cid:107) λ n k T n k ( x n k ) − x (cid:107) < k which implies that x n k −→ x and λ n k T n k ( x n k ) −→ x. (2) ⇒ (3) : It is clear;(3) ⇒ (1) Let U be a nonempty open subsets of X and x ∈ U . Since for all k ≥ W k = B (0 , k ) isa neighborhood of zero, there exist z k ∈ X , n k ∈ N and λ n k ∈ C such that (cid:107) λ n k T n k ( z k ) − x (cid:107) < k and (cid:107) x − z k (cid:107) < k . This implies that z k −→ x and λ n k T n k ( z k ) −→ x, which implies the result. (cid:3) Proposition 3.7.
Assume that T ⊕ S is super-recurrent in X ⊕ Y . Then T and S are super-recurrenton X and Y respectively.Proof. If U and U are nonempty open set of X and Y respectively, then U ⊕ U is a nonempty openset of X ⊕ Y . Since T ⊕ S is super-recurrent, there exist n ∈ N and λ ∈ C such that ( λT n ⊕ S n )( U ⊕ U ) ∩ ( U ⊕ U ) (cid:54) = ∅ , which means that λT n ( U ) ∩ U (cid:54) = ∅ and λS n ( U ) ∩ U (cid:54) = ∅ . Hence T and S aresuper-recurrent. (cid:3) The next theorem gives the relationship between super-recurrent vectors and super-recurrent oper-ators.
Theorem 3.8.
Let T be an operator acting on X . The following assertion are equivalent :(1) T admits a dense subset of super-recurrent vectors; (2) T is super-recurrent.Proof. (1) ⇒ (2) : Let U be a nonempty open subset of X , then there is a T -super-recurrent vector x such that x ∈ U . There exist a increasing sequence ( n k ) of positive integers and an sequence ( λ n k ) ofcomplex numbers such that λ n k T n k x −→ x as k −→ + ∞ . Since U is open and x ∈ U , it follows thatthere exist λ ∈ C and n ∈ N such that λT n ( U ) ∩ U (cid:54) = ∅ , this means that T is super-recurrent.(2) ⇒ (1) : For a fixed element x ∈ X and a fixed strictly positive numbers ε >
0, let B := B ( x, ε ) . N SUPER-RECURRENT OPERATORS 5
Since T is super-recurrent, there exist some positive integer k and some number λ such that λ T − k ( B ) ∩ B (cid:54) = ∅ . Let x ∈ X such that x ∈ λ T − k ( B ) ∩ B . Since T is continuous, thereexists ε < such that B := B ( x , ε ) ⊂ λ T − k ( B ) ∩ B. Again, since T is super-recurrent, there exist some k ∈ N and some λ ∈ C such that λ T − k ( B ) ∩ B (cid:54) = ∅ . Let x ∈ X such that x ∈ λ T − k ( B ) ∩ B . By continuity of T , there exists ε < suchthat B := B ( x , ε ) ⊂ λ T − k ( B ) ∩ B . Continuing inductively, we construct a sequence ( x n ) n ∈ N of elements of X , a sequence ( λ n ) n ∈ N ofcomplex numbers, a strictly increasing sequence of positive integers ( k n ) n ∈ N and a sequence of positivereal numbers ε n < n , such that B ( x n , ε n ) ⊂ B ( x n − , ε n − ) and λ n T n k ( B ( x n , ε n )) ⊂ B ( x n − , ε n − ) . Since X is a Banach space, then by Cantor’s Theorem, there exists some vector y ∈ X such that(3.1) (cid:92) n ∈ N B ( x n , ε n ) = { y } . Since y ∈ B , we need only to show that y is T -super-recurrent. By (3.1), we have y ∈ B ( x n , ε n ) forall n , which implies that(3.2) (cid:107) x n − y (cid:107) < ε n . On the other hand, λ n T n k y ∈ B ( x n , ε n ). Indeed, we have y ∈ B ( x n +1 , ε n +1 ). This implies that λ n T n k y ∈ λ n T n k ( B ( x n +1 , ε n +1 )) ⊂ λ n T n k ( B ( x n , ε n )) ⊂ B ( x n , ε n ) . Hence,(3.3) (cid:107) λ n T n k y − x n (cid:107) < ε n . Now, by using (3.2) and (3.3) we conclude that (cid:107) λ n T n k y − y (cid:107) ≤ (cid:107) λ n T n k y − x n (cid:107) + (cid:107) x n − y (cid:107) < n − . Hence, λ n T n k y −→ y , that is y is a T -super-recurrent vector. Hence each open ball of X contains a T -super-recurrent vector. Thus the set of all super-recurrent vectors for T is dense in X . (cid:3) Theorem 3.8 shows that any super-recurrent operator on a Banach space admits super-recurrentvectors. However, an operator may has super-recurrent vectors without being super-recurrent as weshow in the following example.
Example 3.9.
Let X be a Banach space and let ( e i ) i ∈ I be a basis of X . Let i ∈ I and λ ∈ C anonzero fixed number. We define an operator T on X by: T e i = λe i and T e i = 0, for all i ∈ I \ { i } . It is clear that e i is a T -super-recurrent vector for T . However, T itself is not super-recurrent since itis not of dense range and super-recurrent operators are of dense range by Remark 3.3. Remark . If T is super-recurrent, then λT is super-recurrent for all λ ∈ C ∗ . Moreover, T and λT have the same super-recurrent vectors.The next theorem gives the relationship between the super-recurrence of an operator and its iterates. M. AMOUCH AND O. BENCHIHEB
Theorem 3.11.
Let p be a nonzero positive integer. Then, T is super-recurrent if and only if T p issuper-recurrent. Moreover, T and T p have the same super-recurrent vectors.Proof. We will prove that
SRec ( T ) = SRec ( T p ), for that it is enough to show that SRec ( T ) ⊂ SRec ( T p ). Let x be a T -super-recurrent vector, then there exist a strictly increasing sequence ( k n ) n ∈ N of positive integers and a sequence ( λ n ) n ∈ N of complex numbers such that λ n T k n x −→ x as n −→ + ∞ .Without loss of generality we may suppose that k n > p for all n . Hence, for all n , there exist (cid:96) n ∈ N and v n ∈ { , . . . , p − } such that k n = p(cid:96) n + v n . Since ( v n ) n is bounded, there exists v ∈ { , . . . , p − } and a subsequence of ( v n ) n which convergesto v . Thus, λ k n T p(cid:96) n + v x −→ x for some subsequence of ( (cid:96) n ) ∈ N and a subsequence ( λ k n ) ∈ N which wecall them again ( (cid:96) n ) ∈ N and ( λ k n ) ∈ N . Let U be a nonempty open subset of X such that x ∈ U . Since λ k n T p(cid:96) n + v x −→ x , there exists a positive integer m := (cid:96) n such that λ n T pm + v x ∈ U. We have λ k n λ n T p ( (cid:96) n + m )+2 v x = λ n k λ n T p(cid:96) n + v T pm + v x −→ λ n T pm + v x ∈ U. Thus, we can find a positive integer m := m + (cid:96) n > m such that λ n λ n T pm +2 v x ∈ U. Continuinginductively we can find a positive integer m p = m p − + (cid:96) n p such that λ n . . . λ n p T pm p + pv x ∈ U. Put λ = λ n . . . λ n p , then λ ( T p ) m p + v x ∈ U , which means that x is T p -super-recurrent. Hence, SRec ( T ) = SRec ( T p ). Now it suffices to use Theorem 3.8 to conclude the result. (cid:3) Spectral Proprieties of Super-recurrent Operators
In this section, we show that super-recurrent operators have some noteworthy spectral proprieties.If T is hypercyclic, then Kitai [18] showed that every component of the spectrum of T must intersectsthe unit circle. Later, N. S. Feldman, V. G. Miller, and T. L. Miller gave a similar result for thesupercyclicity case. They proved that if T is supercyclic, then there exists R > { z ∈ C : | z | = R } , called a supercyclicity circle for T , intersects each component of the spectrum of T , see [3, Theorem 1.24] or [9]. Recently, G. Costakis, A. Manoussos, and I. Parissis [7] proved thatthe spectrum of recurrent operators share the same propriety with hypercyclic operators by proventhat if T is recurrent, then every component of the spectrum of T intersects the unit circle. Sincesuper-recurrent operators ”look like” supercyclic operators, it is expected that their spectrums sharethe same propriety. This is the objective of the next theorem. Theorem 4.1.
Let T be an operator acting on a complex Banach space X. If T is super-recurrent,then there exists R > such that each connected component of the spectrum of T intersects the circle { z ∈ C : | z | = R } .Proof. Assume that T is super-recurrent. We will produce by contradiction. By [3, Lemma 1.25], thereexist R > C , C two component of σ ( T ) such that C ⊂ D and C ⊂ C \ D . Without loss ofgenerality, we may suppose that R = 1. Indeed, this is since T is super-recurrent if and only R − T is.By [3, Lemma 1.21], there exist σ and σ , two closed and open sets of σ ( T ) such that C ⊂ σ ⊂ D and C ⊂ σ ⊂ C \ D . Set σ = σ ( T ) \ ( σ ∪ σ ). We have then σ ( T ) = σ ∪ σ ∪ σ and the sets σ i are closed and pairwise disjoint. By Reisz decomposition theorem there exist X , X , X and T , T , T such that X = X ⊕ X ⊕ X and T = T ⊕ T ⊕ T , where each X i is a T -invariant subspace, T i = T /X i and σ i = σ ( T i ). Let x ∈ X and y ∈ X . By Theorem 3.6 , there exist ( λ k ) ⊂ C , ( n k ) ⊂ N , N SUPER-RECURRENT OPERATORS 7 ( x k ) ⊂ X and ( y k ) ⊂ X such that x k −→ x , y k −→ y , λ k T n k x k −→ x and λ k T n k y k −→ y. By [3, Lemma 1.20], the last assertion implies that ( | λ k | ) converges into 0 and + ∞ , which is a contra-diction. (cid:3) The adjoint Banach operator of a hypercyclic operator cannot have eigenvalue. This means that σ p ( T ∗ ) = ∅ , see [3, Proposition 1.7]. Unlike the hypercyclicity case, the adjoint of a supercyclicoperator T can have an eigenvalue but not more then one. This means that either we have σ p ( T ∗ ) = ∅ or there exists λ such that σ p ( T ∗ ) = { λ } . For the recurrent operators, it is expected that they havethe same result as hypercyclic operators, but this is not the case, see [7, Example 2.13 and Remark2.15]. So the Banach adjoint operator of a recurrent operator may has eigenvalue. However, no oneof those eigenvalue can be outside of the unit circle. This means that σ p ( T ∗ ) ⊂ T , where T the unitcircle. Since recurrent operators are super-recurrent, it follows that some super-recurrent operatorsmay have eigenvalue. However, all those eigenvalues lie in a circle of form { z ∈ C : | z | = R } , where R >
0. This is the content of the next result.
Theorem 4.2.
The eigenvalues of the adjoint operator of a super-recurrent operator have the sameargument. That is, if T is super-recurrent, then there exists R > such that σ p ( T ∗ ) ⊂ { z ∈ C : | z | = R } . In particular, for all λ ∈ C \ { z ∈ C : | z | = R } the operator T − λI has dense range.Proof. Assume that there exist λ , µ ∈ σ p ( T ∗ ) such that | µ | < | λ | and let m be a nonzero real numbersuch that | µ | < m < | λ | . Since λ , µ ∈ σ p ( T ∗ ), there exist x ∗ , y ∗ ∈ X ∗ such that T ∗ x ∗ = λx ∗ and T ∗ y ∗ = µy ∗ . This implies that x ∗ ( T n z ) = λ n x ∗ ( z ) and y ∗ ( T n z ) = µ n y ∗ ( z ) for all z ∈ X . Since T issuper-recurrent if and only m T is, let z ∈ SRec ( m T ) . By Baire Category Theorem we may supposethat x ∗ ( z ) (cid:54) = 0 and y ∗ ( z ) (cid:54) = 0. Since z is a super-recurrent vector for m T , it follows that there exist( β k ) ⊂ C and ( n k ) ⊂ N such that β k m nk T n k z −→ z as k −→ ∞ . Since x ∗ and y ∗ are continuous,we deduce that β k (cid:18) λm (cid:19) n k x ∗ ( z ) −→ x ∗ ( z ) and β k (cid:16) µm (cid:17) n k y ∗ ( z ) −→ y ∗ ( z ) . Using that x ∗ ( z ) (cid:54) = 0 and y ∗ ( z ) (cid:54) = 0 we conclude that β k (cid:0) λm (cid:1) n k −→ β k (cid:0) µm (cid:1) n k −→ | β k | −→ | β k | −→ ∞ , which is a contradiction. (cid:3) Remark . If T is supercyclic, then T is super-recurrent, but either σ p ( T ∗ ) = ∅ or σ p ( T ∗ ) = { λ } for some nonzero number λ . However, there exist several super-recurrent operators such that Card ( σ p ( T ∗ )) >
1. Indeed, let ( λ n ) n ∈ N be a sequence of nonzero complex numbers of the sameargument. Define in (cid:96) ( N ) an operator T by T ( x , x , . . . ) = ( λx , λ x , . . . ) . Then T is a super-recurrent operator. It’s easy to check that ( λ n ) n ∈ N ⊂ σ p ( T ∗ ) and hence σ p ( T ∗ ) isan infinite set.We already know that if T is supercyclic, then either σ p ( T ∗ ) = ∅ or σ p ( T ∗ ) = { λ } for some nonzeronumber λ . Moreover, in the latter case, one can find a T -invariant hyperplane X ⊂ X such that theoperator T := T /X is hypercyclic on X , see [3, Proposition 1.26]. In the next theorem, we provethat the same relation still true between recurrent and super-recurrent operators. M. AMOUCH AND O. BENCHIHEB
Theorem 4.4.
Let X be a Banach space with dim ( X ) > . Let T be a super-recurrent operator actingon X . Then for all λ ∈ σ p ( T ∗ ) , there exists a ( closed ) T -invariant hyperplane X ⊂ X such that T := λ − T /X is recurrent on X . Proof.
First note that λ (cid:54) = 0 for every λ ∈ σ p ( T ∗ ) since a super-recurrent operator has dense range.Since T is super-recurrent if and only if aT is super-recurrent for every a (cid:54) = 0, we may assume,without loss of generality, that λ = 1 . Choose x ∗ ∈ X ∗ \{ } such that T ∗ x ∗ = x ∗ and let X = Ker ( x ∗ ).Since x ∗ is an eigenvector of T ∗ , it follows that X is a T -invariant hyperplane of X . We can considerthen T := T /X . In the following, we will prove that T is a recurrent operator on X .With a slight abuse of notation, we may write X = C ⊕ X and since T ∗ x ∗ = x ∗ , let T (1 ⊕
0) = 1 ⊕ y for some y ∈ X . It follows then that T (1 ⊕ z ) = 1 ⊕ ( y + T ( z ) for all z ∈ X . By straightforwardinduction, we have T n (1 ⊕ z ) = 1 ⊕ ( y + T ( y ) + · · · + T n − ( y ) + T n ( z ))for all z ∈ X . Note that T − I has dense range. Indeed, assume that ( T − I )( X ) (cid:54) = X and without loss ofgenerality we may suppose that y / ∈ ( T − I )( X ) . By the Hahn-Banach theorem, there exists k ∗ ∈ X ∗ such that k ∗ ( y ) (cid:54) = 0 and k ∗ ( T n z ) = k ∗ ( z ) for every z ∈ X . Choose a super-recurrent vector for T ofthe form 1 ⊕ x . Hence there exist ( µ k ) ⊂ C and a strictly increasing sequence ( n k ) ⊂ N such that µ k T n k (1 ⊕ x ) −→ ⊕ x as k −→ ∞ . Thus µ k (1 ⊕ ( y + T ( y ) + · · · + T n − ( y ) + T n ( x ))) −→ ⊕ x . This implies that µ k −→ y + T ( y ) + · · · + T n k − ( y ) + T n k ( x )) −→ x . Since k ∗ is continuousand k ∗ ( y ) (cid:54) = 0, it follows that n k − −→
0, which is a contradiction.Since T is super-recurrent, there exist a subset A of C and a subset B of X such that. SRec ( T ) = A ⊕ B such that A = C and B = X .Finally, let x be an element of B . By the same method applied to x , we have y + T ( y ) + · · · + T n − ( y ) + T n ( x )) −→ x. Applying ( T − I ), we get T n k ( y + ( T − I ) x ) −→ ( y + ( T − I ) x. This implies that ( y + ( T − I ) x ∈ Rec ( T ). Since ( T − I ) has dense range, we conclude that T isrecurrent on X . (cid:3) The Purpose of the following proposition is to show that a large supply of eigenvectors correspondingto eigenvalues with same argument implies that the operator is super-recurrent.
Proposition 4.5.
Let T be an operator acting on X . If there exists R > such that the spacegenerated by X := { x ∈ X : T x = λx for some λ ∈ {| λ | = R }} is dense in X , then T is super-recurrent.Proof. Let (cid:80) ni =1 a i x i ∈ span { X } , where T x i = λ i x i , for certain a i , λ i ∈ C with | λ i | = R for i = 1 , . . . , n. Since each R − λ i is in the unite circle, it follows that there exists a strictly increasingsequence ( n k ) such that (cid:0) R − λ i (cid:1) n k −→ k −→ ∞ . Hence R − n k T n k (cid:32) n (cid:88) i =1 a i x i (cid:33) = n (cid:88) i =1 a i R − n k λ i x i −→ n (cid:88) i =1 a i x i N SUPER-RECURRENT OPERATORS 9 as k −→ ∞ . This means that span { X } ⊂ SRec ( T ). Since span { X } is dense in X , it follows that T is super-recurrent. (cid:3) References [1] E.Akin. Recurrence in topological dynamics. The University Series in Mathematics. Plenum Press, New York, 1997.Furstenberg families and Ellis actions.[2] Ansari SI. Hypercyclic and cyclic vectors. J. Funct. Anal. 1995;128:374-383.[3] Bayart F. Matheron E. Dynamics of linear operators. 2009; New York, NY, USA, Cambridge University Press, 2009.[4] Birkhoff GD. Surface transformations and their dynamical applications. Acta Math. 1922;43:1-119.[5] Bonilla A. Grosse-Erdmann K-G. L´opez-Mart´ınez A. Peris A. Frequently recurrent operators. arXiv:2006.11428v1[math.FA] 19 Jun 2020.[6] R. Cardeccia and S. Muro, Arithmetic progressions and chaos in linear dynamics, arXiv:2003.07161 (2020).[7] Costakis G, Manoussos A, Parissis I. Recurrent linear operators. Complex. Anal. Oper. Th. 2014;8:1601-1643.[8] Costakis G, Parissis I. Szemer´edi’s theorem, frequent hypercyclicity and multiple recurrence. Math. Scand.2012;110:251-272.[9] N. S. Feldman, V. G. Miller and T. L. Miller. Hypercyclic and supercyclic cohyponormal operators. Acta Sci. Math.(Szeged), 68: 303-328, 2002. Corrected reprint: Acta Sci. Math. (Szeged), 68: 965-990, 2002.[10] Furstenberg H. Recurrence in ergodic theory and combinatorial number theory. Princeton: Princeton UniversityPress, M. B. Porter Lectures 1981.[11] V. J. Gal´an, F. Martl´ınez-Gimenez, P. Oprocha and A. Peris, Product recurrence for weighted backward shifts,Appl. Math. Inf. Sci. 9 (2015), 2361-2365.[12] Grosse-Erdmann K. G, Peris A. Linear Chaos. (Universitext). Springer, London 2011.[13] Grosse-Erdmann, K. G. (1999). Universal families and hypercyclic operators. Bulletin of the American MathematicalSociety, 36(3), 345-381.[14] W. H. Gottschalk and G. H. Hedlund, Topological dynamics, American Mathematical Society, Providence, R. I.1955.[15] S. Grivaux, ´E. Matheron and Q. Menet, Linear dynamical systems on Hilbert spaces: Typical properties and explicitexamples, arXiv:1703.01854v1 [math.FA] 6 Mar 2017.[16] Hilden HM, Wallen LJ. Some cyclic and non-cyclic vectors of certain operators. Indiana Univ. Math. J. 1994;23:557-565.[17] S. He, Y. Huang and Z. Yin, J F -class weighted backward shifts, Internat. J. Bifur. Chaos Appl. Sci. Engrg. 28(2018), 1850076, 11 pp.[18] C. Kitai. Invariant closed sets for linear operators. Ph.D. thesis, University of Toronto, Toronto, 1982.[19] H. Poincar´e. Sur le probl`eme des trois corps et les ´equations de la dynamique. Acta mathematica, 13(1), 3-270(1890).[20] Rolewicz, S. (1969). On orbits of elements. Studia Mathematica, 32(1), 17-22.[21] Z. Yin and Y. Wei, Recurrence and topological entropy of translation operators, J. Math. Anal. Appl. 460 (2018),203-215.(Mohamed Amouch and Otmane Benchiheb) University Chouaib Doukkali. Department of Mathematics,Faculty of science Eljadida, Morocco
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