aa r X i v : . [ m a t h . F A ] J a n TWO PROBLEMS ON WEIGHTED SHIFTS IN LINEAR DYNAMICS
FR´ED´ERIC BAYART
Abstract.
We show that an invertible bilateral weighted shift is strongly structurallystable if and only if it has the shadowing property. We also exhibit a K¨othe sequencespace supporting a frequently hypercyclic weighted shift, but no chaotic weighted shifts. Introduction
Linear dynamics began in the 1980’s with the thesis of Kitai [11] and the paper of Gethnerand Shapiro [8]. Its aim is to study the dynamical properties of a (bounded) operator T acting on some (complete) linear space X , as one studies more usually the dynamics of acontinuous map acting on a (compact) metric space X .At its beginning, linear dynamics was essentially devoted to the study of the density oforbits, thanks to its link with the invariant subspace/subset problem leading to the notionsof hypercyclicity, supercyclicity and their variants. More recently, various notions arisingin classical dynamics have also been investigated in the linear context, for instance rigidity,the specification property, the shadowing property, and so on...One of the interests of linear dynamics is that we have a lot of examples of operators atour disposal to illustrate the definitions and to provide examples and counterexamples.Among these examples, the most studied class is certainly that of weighted shifts: givena sequence ( w n ) n of positive real numbers, the corresponding weighted shifts are formallydefined by B w ( x n ) n ≥ = ( w n +1 x n +1 ) n ≥ (unilateral weighted shifts) B w ( x n ) n ∈ Z = ( w n +1 x n +1 ) n ∈ Z (bilateral weighted shifts) . From the characterization of Salas [15] of hypercyclic weighted shifts on ℓ p or c , theirdynamical properties have been widely studied. It turns out that, very recently, theyappear in two problems where open questions still exist.1.1. Hyperbolicity and strong structural stability.
An operator T on a complexBanach space X is said to be hyperbolic if its spectrum σ ( T ) does not intersect the unitcircle. It is said strongly structurally stable if for every ε >
0, there exists δ > α ∈ Lip( X ), with k α k ∞ ≤ δ and Lip( α ) ≤ δ , there is ahomeomorphism ϕ : X → X with k ϕ k ∞ ≤ ε such that T ◦ ( I + ϕ ) = ( I + ϕ ) ◦ ( T + α ) . Date : January 11, 2021.2020
Mathematics Subject Classification.
Key words and phrases. hypercyclic operators, weighted shifts, strong structural stability, shadowingproperty.The author was partially supported by the grant ANR-17-CE40-0021 of the French National ResearchAgency ANR (project Front).
Namely, a small perturbation T + α of T is conjugated to T via a homeomorphism closeto the identity operator. It is known from the 1960’s that every hyperbolic operator isstrongly structurally stable (see [10, 13, 14] and for a recent advance [4]). That the converseis false was shown only very recently in [5] where the following theorem is obtained. Theorem (Bernardes, Messaoudi).
Let X = ℓ p ( Z ) , p ∈ [1 , + ∞ ) or X = c ( Z ) . Let w = ( w n ) n ∈ Z be a bounded and bounded below sequence of positive integers and let B w bethe associated weighted shift. If lim n → + ∞ sup k ∈ N ( w − k w − k − · · · w − k − n ) /n < and lim n → + ∞ inf k ∈ N ( w k w k +1 · · · w k + n ) /n > then B w is strongly structurally stable and not hyperbolic. This result leaves open the characterization of the strongly structurally stable weightedshifts on ℓ p ( Z ) or c ( Z ). We provide such a characterization. Theorem 1.1.
Let X = ℓ p ( Z ) , p ∈ [1 , + ∞ ) or X = c ( Z ) . Let w = ( w n ) n ∈ Z be a boundedand bounded below sequence of positive integers and let B w be the associated weighted shift.Then B w is strongly structurally stable if and only if one the following conditions holds:(A) lim n → + ∞ sup k ∈ Z ( w k · · · w k + n ) /n < ;(B) lim n → + ∞ inf k ∈ Z ( w k · · · w k + n ) /n > ;(C) lim n → + ∞ sup k ∈ N ( w − k · · · w − k − n ) /n < and lim n → + ∞ inf k ∈ N ( w k · · · w k + n ) /n > . This characterization should be compared with the characterization of weighted shiftshaving the shadowing property, a property of dynamical systems that we recall now.A sequence ( x n ) n ∈ Z in X is called a δ -pseudotrajectory of T ∈ L ( X ), with δ >
0, if k T x n − x n +1 k ≤ δ for all n ∈ Z . An invertible operator T is said to have the shadowingproperty if for every ε >
0, there exists δ > δ -pseudotrajectory is ε -shadowed by a real trajectory, namely there exists x ∈ X such that k T n x − x n k < ε for all n ∈ Z . Comparing Theorem 1.1 and [5, Theorem 18], we get the following corollary:
Corollary 1.2.
Let B w be an invertible weighted shift acting on ℓ p ( Z ) , p ∈ [1 , + ∞ ) or c ( Z ) . Then B w is strongly structurally stable if and only if B w has the shadowing property. It is unknown whereas strong structural stability implies the shadowing property or if theconverse holds. Unfortunately, weighted shifts on ℓ p or c cannot provide a counterexam-ple.1.2. Frequently hypercyclic and chaotic weighted shifts.
An operator T on a sep-arable Fr´echet space X is called chaotic if it admits a dense orbit and if it has a denseset of periodic points. It is said frequently hypercyclic if there exists a vector x ∈ X ,called a frequently hypercyclic vector for T , such that, for any nonempty open subset U of X , dens { n ≥ T n x ∈ U } > A ⊂ N , dens( A ) = lim inf N → + ∞ N +1 card { n ≤ N : n ∈ A } denotes the lowerdensity of A . The relationship between chaos and frequent hypercyclicity has been the subject of manyinvestigations. Bayart and Grivaux in [2] have built a frequently hypercyclic operatorwhich is not chaotic; their counterexample is a unilateral backward shift on c . In [12],Menet exhibited operators on any ℓ p or on c that are chaotic and not frequently hyper-cyclic.Let us come back to (unilateral) weighted shifts. A natural context to study them is thatof K¨othe sequence spaces. Let A = ( a m,n ) m ≥ ,n ≥ be a matrix of strictly positive numberssuch that, for all m ≥ n ≥ a m,n ≤ a m +1 ,n . The K¨othe sequence space of order p isdefined as λ p ( A ) = x = ( x n ) n ≥ : ∀ m ≥ , X n ≥ | x n | p a m,n < + ∞ while the K¨othe sequence space of order 0 is given by c ( A ) = (cid:26) x = ( x n ) n ≥ : ∀ m ≥ , lim n → + ∞ | x n | a m,n = 0 (cid:27) . Chaotic and frequently hypercyclic weighted shifts on K¨othe sequence spaces are studiedin depth in [7]. Whereas it is known for a long time that every chaotic weighted shift isfrequently hypercyclic (compare [9] and [1]), it is pointed out in [7] that there exist • K¨othe sequence spaces that do not support any chaotic or frequently hypercyclicweighted shifts ([7, Example 3.8]); • K¨othe sequence spaces where every frequently hypercyclic weighted shift is chaoticand that support such shifts ( ℓ p or H ( D ), see [3] and [7, Proposition 3.12]); • K¨othe sequence spaces that support a chaotic weighted shift and a frequentlyhypercyclic non chaotic weighted shift ( c or H ( C ), see [2] and [7, Proposition3.15]);Thus there remains an intriguing case [7, Question 3.17]: is there a K¨othe sequence spacesupporting frequently hypercyclic weighted shifts but no chaotic weighted shifts? We willanswer this question in Section 3: Theorem 1.3.
There exists a K¨othe sequence space supporting frequently hypercyclicweighted shifts but no chaotic weighted shifts. Strong structural stability of weighted shifts
We begin the proof of Theorem 1.1 by the following technical lemma. We fix X = ℓ p ( Z ), p ∈ [1 , + ∞ ), or X = c ( Z ) and w = ( w n ) n ∈ Z a bounded and bounded below sequence ofpositive integers. Lemma 2.1.
Let δ > , m ∈ N , t ∈ N . There exist α ∈ Lip( X ) and u ∈ X satisfyingthe following properties: • k α k ∞ ≤ δ ; • Lip( α ) ≤ δ ; • α k + t (cid:0) ( B w + α ) m − k − u (cid:1) = δ for all k = 0 , . . . , m − . FR´ED´ERIC BAYART
Proof.
To simplify the notations we first prove the case t = 0. For k = 0 , . . . , m −
1, we set a m,k = w m w m − · · · w m − k +1 . We fix κ > j = k in { , . . . , m − } , k κa m,k e m − k − κa m,j e m − j k ≥ . We also consider the function ρ : R → [0 ,
1] defined by • ρ = 0 on ( −∞ , −
1) and on (1 , + ∞ ); • ρ (0) = 1 and ρ is affine on [ − ,
0] and on [0 , ρ ) ≤
1. We then construct by induction on j = 0 , . . . , m − α ( j ) ∈ Lip( X ) and vectors y ( j ) ∈ X so that, for all j = 0 , . . . , m − k ≤ j , (cid:0) B w + α ( j ) (cid:1) k ( κe m ) = κa m,k e m − k + y ( k ) ;(b) α ( j ) l = 0 provided l / ∈ { m − − j, . . . , m − } ;(c) for all k ∈ { , . . . , j } , for all x ∈ X , α ( j ) m − − k ( x ) = δρ (cid:16) k x − κa m,k e m − k − y ( k ) k (cid:17) . (d) y ( j ) ∈ span( e m − j , . . . , e m ).Let us proceed with the construction. For j = 0, we simply set y (0) = 0, α (0) l = 0 for l = m − α (0) m − ( x ) = δρ ( k x − κe m k ) . Assume that the construction has been done until j ≤ m − j + 1. We first define y ( j +1) . We know that (cid:0) B w + α ( j ) (cid:1) j +1 ( κe m ) = (cid:0) B w + α ( j ) (cid:1) ( κa m,j e m − j + y ( j ) (cid:1) = κa m,j +1 e m − ( j +1) + α ( j ) (cid:0) κa m,j e m − j + y ( j ) (cid:1) + B w (cid:0) y ( j ) (cid:1) . This leads us to set(1) y ( j +1) = α ( j ) (cid:0) κa m,j e m − j + y ( j ) (cid:1) + B w (cid:0) y ( j ) (cid:1) . Since y ( j ) ∈ span( e m − j , . . . , e m ) and α ( j ) ( X ) ⊂ span( e m − − j , . . . , e m ), we indeed get that y ( j +1) ∈ span( e m − ( j +1) , . . . , e m ). We then observe that (b) and (c) define uniquely α ( j +1) and that α ( j +1) l = α ( j ) l for l = m − − ( j + 1). An important point of these definitions isthat, for all 0 ≤ k ≤ j ≤ m − α ( j ) (cid:0) κa m,k e m − k + y ( k ) (cid:1) = α ( j +1) (cid:0) κa m,k e m − k + y ( k ) (cid:1) . Indeed, α ( j +1) l = α ( j ) l except for l = ( m − − ( j + 1) and for this value of l , α ( j ) m − − ( j +1) = 0whereas α ( j +1) m − − ( j +1) ( κa m,k e m − k + y ( k ) ) = δρ (cid:16) k κa m,k e m − k + y ( k ) − κa m,j +1 e m − ( j +1) − y ( j +1) k (cid:17) and this is equal to 0 by the definition of ρ , that of κ , and Property (d).Thus the construction is done and we already obtained that (b), (c) and (d) are true.It remains to prove that (a) is equally satisfied. We first observe that (1) and (2) easilyimply, by an induction on j , that, for all 0 ≤ k ≤ m − k ≤ j ≤ m − (3) y ( k +1) = α ( j ) (cid:0) κa m,k e m − k + y ( k ) (cid:1) + B w (cid:0) y ( k ) (cid:1) . We fix j = 0 , . . . , m − k = 0 , . . . , j , since the result istrue for k = 0. We write (cid:0) B w + α ( j ) (cid:1) k +1 ( κe m ) = (cid:0) B w + α ( j ) (cid:1)(cid:0) κa m,k e m − k + y ( k ) (cid:1) (induction hypothesis)= κa m,k +1 e m − ( k +1) + y ( k +1) (by (3)) . We finally set α = α ( m − , u = κe m and we prove that the couple ( α, u ) satisfiesthe conclusions of Lemma 2.1. Observe first that each α k vanishes outside the ball B ( κa m,m − − k e m + k +1 − y ( m − − k ) , ≤ k ≤ m −
1, and that these balls are pairwisedisjoint. Since k α k k ∞ ≤ δ for all k , we immediately get k α k ∞ ≤ δ . For the proof thatLip( α ) ≤ δ , we pick x, y ∈ X . If k x − y k ≥
2, then k α ( x ) − α ( y ) k ≤ δ = δ k x − y k .Suppose now that k x − y k < • either x or y belongs to some ball B ( κa m,m − − k e m + k +1 − y ( m − − k ) , x and y can belong to the same ball B ( κa m,m − − k e m + k +1 − y ( m − − k ) ,
1) butthat they cannot belong to two different balls because of they are disjoint. Itfollows that k α ( x ) − α ( y ) k = | α k ( x ) − α k ( y ) |≤ δ (cid:12)(cid:12) ρ ( k κa m,m − − k e m + k +1 − y ( m − − k ) − x k ) − ρ ( k κa m,m − − k e m + k +1 − y ( m − − k ) − y k ) (cid:12)(cid:12) ≤ δ k x − y k . • neither x nor y belongs to some ball B ( κa m,m − − k e m + k +1 − y ( m − − k ) , α ( x ) = α ( y ) = 0.Finally the last requirement α k (cid:0) ( B w + α ) m − k − u (cid:1) = δ for all k = 0 , . . . , m − j = m −
1. The proof for the other values of t issimilar, but we have to shift everything starting from e m + t instead of e m . (cid:3) We will also need the following results coming from [5, Lemma 19 and Proposition 15].
Lemma 2.2.
Let ( x n ) n ∈ N be a bounded sequence of positive real numbers. Then thefollowing assertions are equivalent:(i) lim n → + ∞ sup k ∈ N ( x k x k +1 · · · x k + n ) /n < ;(ii) sup t ∈ N P + ∞ k =0 x t x t · · · x k + t < + ∞ ;(iii) sup m ∈ N P m − k =0 x m − k x m − k +1 · · · x m < + ∞ . Lemma 2.3.
Let w = ( w n ) n ∈ Z be a bounded and bounded below sequence of positive realnumbers. Assume that lim n → + ∞ sup k ∈ N ( w k · · · w k + n ) /n < and lim n → + ∞ inf k ∈ N ( w − k · · · w − k − n ) /n > . Then B w is not strongly structurally stable. FR´ED´ERIC BAYART
Proof of Theorem 1.1.
That a weighted shift satisfying (A), (B) or (C) is strongly struc-turally stable is already done in [5] (see also [4]): (A) and (B) implies that B w is hyperbolic,whereas (C) implies that B w is generalized hyperbolic and a hyperbolic or generalized hy-perbolic operator is always strongly structurally stable.Thus we start with a strongly structurally stable invertible weighted shift B w and weintend to show that one of the following condition is satisfied:(4) lim n → + ∞ sup k ∈ N ( w k · · · w k + n ) /n < n → + ∞ inf k ∈ N ( w k · · · w k + n ) /n > n → + ∞ sup k ∈ N (cid:18) w k · · · w k + n (cid:19) /n ≥ . By Lemma 2.2 applied with x k = 1 /w k , we know thatsup m ∈ N (cid:18) w · · · w m + 1 w · · · w m + · · · + 1 w m (cid:19) = + ∞ . Let t ∈ N . Snce w is bounded and bounded below, the previous property also impliesthat sup m ∈ N (cid:18) w t · · · w m + t + 1 w t · · · w m + t + · · · + 1 w m + t (cid:19) = + ∞ . Let δ > ε = 1 in the definition of a strongly structurally stable operator.There exists m ∈ N as large as we want such that(6) m − X k =0 w k +1+ t · · · w m + t ≥ δδ . Let ( α, u ) ∈ Lip( X ) × X given by Lemma 2.1 for these values of δ, m, t . There exists ϕ : X → X a homeomorphism with k ϕ k ∞ ≤ I + ϕ ) ◦ ( B w + α ) = B w ◦ ( I + ϕ )which gives ϕ ◦ ( B w + α ) = − α + B w ◦ ϕ which in turn yields for all k ∈ Z ϕ k ◦ ( B w + α ) = − α k + w k +1 ϕ k +1 . By an easy induction we then get that for all n ∈ N , ϕ n + t ( u ) = ϕ t (cid:0) ( B w + α ) n u (cid:1) w t · · · w n + t + n − X k =0 α k + t (cid:0) ( B w + α ) n − − k u (cid:1) w k +1+ t . . . w n + t . We apply this equality for n = m and we use the third property of the map α to get(7) ϕ m + t ( u ) = ϕ t (cid:0) ( B w + α ) m u (cid:1) w t · · · w m + t + δ m − X k =0 w k +1+ t · · · w m + t . Since k ϕ k ∞ ≤ w t · · · w m + t ≥ δ (cid:18) δδ (cid:19) − δ . Multiplying (7) by w t · · · w m + t we get δ m − X k =0 w t · · · w k + t = w t +1 · · · w m + t ϕ m + t ( u ) − ϕ t (cid:0) ( B w + α ) m u (cid:1) so that m − X k =0 w t · · · w k + t ≤ δ . This is true for an infinite number of positive integers m which means thatsup t ∈ N + ∞ X k =0 w t · · · w k + t ≤ δ . Applying again Lemma 2.2, but now to the sequence x k = w k , we get thatlim n → + ∞ sup k ∈ N ( w k · · · w k + n ) /n < , which is (4).We now finish like in the proof of Theorem 18 of [5]. We briefly indicate the method. Let w ′ be defined by w ′ n = w − n +1 for all n ∈ Z . Then ( B w ) − and B w ′ are conjugated by anisometric isomorphism so that B w ′ is strongly structurally stable. Applying what we havedone to B w ′ instead of B w , we get that one of the following condition is satisfied:(8) lim n → + ∞ inf k ∈ N ( w − k · · · w − k − n ) /n > n → + ∞ sup k ∈ N ( w − k · · · w − k − n ) /n < . To conclude we observe that • (A) corresponds to the case where (4) and (9) are both true; • (B) corresponds to the case where (5) and (8) are both true; • (C) corresponds to the case where (5) and (9) are both true; • (4) and (8) cannot be simultaneously true by Lemma 2.3, since B w is assumed tobe strongly structurally stable. (cid:3) Chaotic and frequently hypercyclic weighted shifts
The proof of Theorem 1.3 is based on three tools. Firstly we need to exhibit a sequence ofpairwise disjoint sets with positive lower density which are sufficiently sparse to producefrequently hypercyclic weighted shifts which are not chaotic. Up to our knowledge, thiswas used in all examples of a frequently hypercyclic yet not chaotic weighted shift (see [2]or [6]).
FR´ED´ERIC BAYART
Lemma 3.1.
There exist an increasing sequence ( N k ) k ≥ of nonnegative integers with N = 0 and a sequence ( A r ) r ≥ of pairwise disjoint subsets of N with positive lower densityand satisfying the following properties:(a) For r ≥ , if n ∈ A r , then for any k ≥ , N k ≤ n < N k +1 = ⇒ N k + k ≤ n < N k +1 − r. (b) For r, s ≥ , if n ∈ A r , m ∈ A s , n > m , then for any k ≥ , N k ≤ n − m < N k +1 = ⇒ N k + max( r, s ) ≤ n − m < N k +1 − max( r, s ) . Our second tool is an abstract result to prove the frequent hypercyclicity of a shift on aFr´echet sequence space in which ( e n ) is an unconditional basis. Lemma 3.2. ( [6, Theorem 6.2] ) Let X be a Fr´echet sequence space in which ( e n ) is anunconditional basis. Then a weighted shift on X is frequently hypercyclic if and only ifthere exist a sequence ( ε r ) r ≥ of positive numbers tending to zero and a sequence ( A r ) r ≥ of pairwise disjoint subsets of N with positive lower density such that(i) for any r ≥ , X n ∈ A r v n + r e n + r converges in X ; (ii) for any r, s ≥ , any m ∈ A s and any j = 0 , . . . , r , (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) X n ∈ A r , n>m v n − m + j e n − m + j (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) s ≤ min( ε r , ε s ) , where, for n ≥ , v n = 1 w · · · w n . The last preliminary result we need is a criterion for the existence of a chaotic weightedshift. Our counterexample will be a power series space of order 0 and infinite type, namelya space C , ∞ ( α ) = n x = ( x n ) n ≥ : ∀ p ≥ , p α n | x n | n → + ∞ −−−−−→ o where ( α n ) is a nondecreasing sequence of positive real numbers tending to infinity, en-dowed with the family of seminorms k ( x n ) k p = sup n | x n | p α n . For such a space, we have the following characterization of the existence of a chaoticweighted shift.
Lemma 3.3. ( [7, Proposition 4.2] ) The space C , ∞ ( α ) supports a chaotic weighted shift ifand only if lim N → + ∞ P N − n =0 α n α N = + ∞ . Proof of Theorem 1.3.
Let ( N k ) k ≥ and ( A r ) r ≥ be the two sequences given by Lemma3.1. We define a nondecreasing sequence ( α n ) n ≥ by α = 1 and (cid:26) α N k = α + · · · + α N k − α n +1 = α n provided n = N k − k ∈ N . We set X = C , ∞ ( α ). Since P N k − n =0 α n /α N k = 1, it follows from Lemma 3.3 that X does not support a chaotic weighted shift. Let us now prove that it admits a frequentlyhypercyclic weighted shift. We set, for n ≥ ( w n = 2 α n provided n = N k for some k ∈ N w n = αNk αNk − ··· + αNk − provided n = N k . In order to prove that B w is continuous, we just need to show that ∀ p ≥ , ∃ q ≥ , ∃ C ≥ , ∀ n ≥ , w n +1 p α n ≤ q α n +1 . Now, for all values of n , we have w n +1 p α n ≤ (2 p ) α n ≤ (2 p ) α n +1 and we just need to choose q = 2 p .Let us now show that B w is frequently hypercyclic. Observe that the definition of w implies that v n = (cid:18) (cid:19) α Nk + ··· + α n where k is the integer such that N k ≤ n < N k +1 . We shall apply Lemma 3.2 with ε r = 2 − r r. (i) Let r ≥
1. We have to show that, for all p ≥ p α n + r v n + r tends to 0 as n tends to+ ∞ , n being in A r . Let k be such that N k ≤ n < N k +1 . Then N k + k ≤ n < N k +1 − r .Therefore p α n + r v n + r = p α Nk (cid:18) (cid:19) α Nk + ··· + α n ≤ p (cid:18) (cid:19) k ! α Nk and this goes to zero as n (hence k ) tends to + ∞ .(ii) Let r, s ≥ m ∈ A s , j = 0 , . . . , r . We have to show that, for all n > m , n ∈ A r ,then v n − m + j s α n − m + j ≤ min( ε r , ε s ) . Let k be such that N k ≤ n − m < N k +1 . Then N k + max( r, s ) ≤ n − m + j < N k +1 so that v n − m + j s α n − m + j ≤ (cid:18) (cid:19) max( r,s ) s ! α Nk ≤ min( ε r , ε s ) . Hence, B w is a frequently hypercyclic weighted shift of the K¨othe sequence space X supporting no chaotic weighted shifts. (cid:3) Acknowledgement: We thank Q. Menet and K-G. Grosse-Erdmann for useful discussions.
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