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1 Document downloaded from: This paper must be cited as: Peris Manguillot, A.; Bernardes, NC.; Bonilla, A.; Müller, V. (203). Distributional chaos for linear operators. Journal of Functional Analysis. 265(9): doi:0.06/j.jfa The final publication is available at Copyright Elsevier

2 Distributional Chaos for Linear Operators N. C. Bernardes Jr., A. Bonilla, V. Müller and A. Peris Abstract We characterize distributional chaos for linear operators on Fréchet spaces in terms of a computable condition (DCC), and also as the existence of distributionally irregular vectors. A sufficient condition for the existence of dense uniformly distributionally irregular manifolds is presented, which is very general and can be applied to many classes of operators. Distributional chaos is also analyzed in connection with frequent hypercyclicity, and the particular cases of weighted shifts and composition operators are given as an illustration of the previous results. Introduction and preliminaries The study of the dynamics of linear operators on infinite-dimensional spaces has attracted attention of many researchers in recent years (we refer the reader to the books [4] and [] for more information about this subject). Hypercyclicity, that is, the existence of vectors x X whose orbit {x, T x, T 2 x,... } is dense in X for an operator T : X X on a topological vector space X, is certainly the most studied phenomenon in this context. It seems that the first time that the term chaos appeared in the mathematical literature was in Li and Yorke s paper [5] on the study of dynamics of interval maps. The notion of chaos derived from [5] concentrates on local aspects of dynamics of pairs. Schweizer and Smítal introduced the concept of distributional chaos in [23] as a natural extension of the notion of chaos given by Li and Yorke. Let f : X X be a continuous map on a metric space X. For each pair x, y X and each n N, the distributional function F n xy : R + [0, ] is defined by F n xy(τ) := n card{0 i n : d(f i (x), f i (y)) < τ}, where card A denotes the cardinality of the set A. Moreover, define F xy (τ) := lim inf F xy(τ) n and Fxy(τ) := lim sup Fxy(τ). n n The following notions were introduced in [23] and [2], respectively. They were considered for linear operators on Banach or Fréchet spaces in [6, 2, 3, 4, 6, 7, 20, 24]. n The first author was partially supported by CAPES: Bolsista - Proc. n o BEX 402/-9. The second author is partially supported by MEC and FEDER, project no. MTM The third author was supported by grant 20/09/0473 of GA CR and RVO: The fourth author was supported in part by MEC and FEDER, Project MTM

3 Definition. A continuous map f : X X on a metric space X is distributionally chaotic if there exist an uncountable set Γ X and ε > 0 such that for every τ > 0 and each pair of distinct points x, y Γ, we have that F xy (ε) = 0 and F xy(τ) =. In this case, the set Γ is a distributionally ε-scrambled set and the pair (x, y) a distributionally chaotic pair. We say that f is densely distributionally chaotic if the set Γ may be chosen to be dense in X. Given A N, its upper and lower densities are defined by dens(a) := lim sup n card(a [, n]) n and dens(a) := lim inf n card(a [, n]), n respectively. With these concepts in mind, one can equivalently say that f is distributionally chaotic on Γ if there exists ε > 0 such that for any x, y Γ, x y, we have dens{n N : d(f n (x), f n (y)) < ε} = 0 and dens{n N : d(f n (x), f n (y)) < τ} =, for every τ > 0. The purpose of this paper is to study the above defined notions for continuous linear operators on Fréchet spaces. Unless otherwise specified, X will denote an arbitrary Fréchet space, that is, a vector space X endowed with an increasing sequence ( k ) k N of seminorms (called a fundamental sequence of seminorms) that defines a metric d(x, y) := k= 2 k min{, x y k} (x, y X), under which X is complete. Moreover, B(X) will denote the set of all continuous linear operators T : X X. The following concept is a generalization to Fréchet spaces of the one introduced by Beauzamy [5] for Banach spaces (see [7]). Definition 2. Given T B(X) and x X, we say that x is an irregular vector for T if there are m N and increasing sequences (n k ) and (j k ) of positive integers such that lim T n k x = 0 and lim T j k x m =. k k Inspired by the notion of a distributionally irregular vector introduced in [6] in the context of Banach spaces, we consider the following generalization to Fréchet spaces. Definition 3. Given T B(X) and x X, we say that x is a distributionally irregular vector for T if there are m N and A, B N with dens(a) = dens(b) = such that lim T n x = 0 and lim T n x m =. n A n B 2

4 We also recall that, given an operator T B(X), a vector x X is called frequently hypercyclic for T if for every non-empty open subset U of X, the set {n N : T n x U} has positive lower density. The operator T is called frequently hypercyclic if it possesses a frequently hypercyclic vector. T is called topologically mixing if for any pair U, V of non-empty open subsets of X, there exists some n 0 N such that T n (U) V for all n n 0. T is Devaney chaotic if it is hypercyclic and it has a dense set of periodic points, that is, vectors x X such that T n x = x for some n N. 2 Distributional chaos and distributionally irregular vectors Definition 4. Let T B(X) and x X. The orbit of x is said to be distributionally near to 0 if there exists A N with dens(a) = such that lim n A T n x = 0. We say that x has a distributionally unbounded orbit if there exist m N and B N with dens(b) = such that lim n B T n x m =. Whenever we need to emphasize the number m in question, we say that x has a distributionally m-unbounded orbit. With these definitions we have that a vector x X is distributionally irregular if and only if its orbit is both distributionally unbounded and distributionally near to 0. Assume X is a Banach space. If T B(X) and T n, it follows from the Banach- Steinhaus Theorem that there exists a residual set of vectors in X with unbounded orbits. The next example shows that under these conditions we cannot guarantee the existence of a vector with distributionally unbounded orbit. Nevertheless, it was proved in [8] that if < then there exists a vector x X such that T n x (in particular, x T n has distributionally unbounded orbit). Example 5. Assume X = l (N). Given ε (0, ), there exists T B(X) with 5 T n = (n + ) ( ε) (n N) such that no x X has distributionally unbounded orbit. Proof. Let (e k ) k N be the standard basis for X. Define T B(X) by T e = 0 and T e k = ( k k ) ε e k for k >. It is easy to see that T n = T n e n+ = (n + ) ( ε) for all n N. Suppose that there are x X, x =, and N N with card { n N : T n x > 3 ε} N( ε). Then N N T n x 3 ε ( ε) = 3 ε 3. n= 3

5 On the other hand, by writing x = j= α je j we obtain N T n x = n= = = N n= j=n+ N n= j=n+ α j T n e j ( j ) ε α j j n min{n,j } α j j ε (j n) ε j= 2N j= 2N j= n= j α j j ε (j n) ε + n= ε jε α j j ε + 2N, where we have used the following estimations: j j (j n) ε = k ε + n= k= and, for j > 2N, ( j j n ) ε 2 ε 2. So, j=2n+ j α j N n= x ε dx jε ε ( j ) ε j n N N T n x N n= 2N j= α j j ε ε Since ε < 3 ε 3, we have a contradiction. Remark 6. In [7, Thm 2.] it is provided an example of the same kind in a weighted l p -space such that T n and card{ j n : T j x < ε} lim n n for all x X and ε >, which in particular shows that T has no distributionally chaotic pair. Soon we will show that the above example admits no distributionally chaotic pair either. Proposition 7. If T B(X) and m N, then the following assertions are equivalent: (i) there exist ε > 0, a sequence (y k ) in X and an increasing sequence (N k ) in N such that lim k y k = 0 and for all k N; = card { j N k : T j y k m > ε } N k ( k ) (ii) there exists y X with distributionally m-unbounded orbit; (iii) the set of all y X with distributionally m-unbounded orbit is residual in X. 4

6 Proof. (iii) (ii): Trivial. (ii) (i): Let y X be a vector with distributionally m-unbounded orbit. By definition, there exists A N with dens(a) = such that lim n A T n y m =. Set y k := k y. Then y k 0. Choose ε > 0 arbitrary. For each k N, we have dens{j N : T j y k m > ε} = dens{j N : T j y m > εk} dens(a) =. So we can find N k N satisfying (i). (i) (iii): For each k N, let { M k := x X : n N with card { j n : T j x m > k } } n( k ). Clearly M k is open. We show that M k is dense. Let x X, δ > 0 and m N. By (i), there exist u {y, y 2,... } and n N such that u m < C := δε and 4k 2 ( card{ j n : T j u m > ε} n ). 2k Consider the vectors u s := x + δsu 2kC (s = 0,,..., 2k ). Clearly u s x m < δ for all s. We show that there exists s {0,,..., 2k } with u s M k. Let A := { j n : T j u m > ε}. Then card A n( ). For each 2k s = 0,,..., 2k, let B s := { j n : T j u s m k}. If s, t {0,,..., 2k } and s t, then B s B t A =. Indeed, suppose s t and j B s B t A. Then However, T j u s T j u t m = s t δ T j u m 2kC > δε 2kC = 2k. T j u s T j u t m T j u s m + T j u t m 2k, a contradiction. So there exists s 0 {0,,..., 2k } with card(b s0 A) card A. Then 2k card(a \ B s0 ) n( 2k )2 n( k ). For j A \ B s0 we have T j u s0 m > k. Hence u s0 M k and M k is dense. Thus k M k is a residual subset of X. Let x k M k. For each k N, there exists n k N with card A k n k ( k ), where A k := { j n k : T j x m > k}. Let A := k A k. Then dens(a) = and lim j A T j x m =. Proposition 8. If T B(X) then the following assertions are equivalent: (i) there exist ε > 0, a sequence (y k ) in X and an increasing sequence (N k ) in N such that lim k y k = 0 and lim k N k card { j N k : d(t j y k, 0) > ε } = ; (ii) there exists y X with distributionally unbounded orbit; (iii) the set of all y X with distributionally unbounded orbit is residual in X. In the case X is a Banach space, the above assertions are also equivalent to the following: 5

7 (i ) there exist ε > 0, a sequence (y k ) in X and an increasing sequence (N k ) in N such that lim k y k = 0 and for all k N; card { j N k : T j y k > ε } εn k (ii ) there exist y X and A N with dens(a) > 0 such that lim j A T j y =. Proof. (iii) (ii): Trivial. (ii) (i) and (i) (iii): Follow from the previous proposition. Now, assume X is a Banach space. (ii) (ii ): Obvious. (ii ) (i ): Choose ε (0, dens A) and argue as in the proof of (ii) (i) in the previous proposition. (i ) (i): Let δ > 0. We consider the following property P (δ): For every L > 0, there exist y X, y =, and n N such that card{ j n : T j y > L} δn (equivalently, there are y X, y ε L, and n N with card{j n : T j y > ε} δn). Let δ 0 be the supremum of all δ for which P (δ) is true. By (i ), δ 0 ε > 0. Let k N. Let δ > 0 satisfy δ 0 δ δ 0 + δ k. Since P (δ 0 + δ ) is not true, there exists L > 0 such that card{ j n : T j y > L} < (δ 0 + δ )n for all y X, y =, and all n N. Clearly we may assume that L > max{k, T k }. Since P (δ 0 δ ) is true, there exist u X, u =, and n N such that Set card{ j n : T j u > L 2 } (δ 0 δ )n. A := { j n : T j u L}, A 2 := { j n : L < T j u L 2 }, A 3 := { j n : T j u > L 2 }. Since T < L, we have A. Let A = {r,..., r d } with = r < r 2 < < r d. Set formally r d+ = n +. For each s =,..., d, let B s := A 2 ]r s, r s+ [ and C s := A 3 ]r s, r s+ [. We have d card C s = card A 3 (δ 0 δ )n s= and d card(c s B s ) = card(a 3 A 2 ) (δ 0 + δ )n. s= 6

8 So there exists s 0, s 0 d, such that C s0 and Since r s0 A, we have that card C s0 card(c s0 B s0 ) δ 0 δ δ 0 + δ k. T rs 0 +j u T j T rs 0 u T k L < L 2, for j =,..., k. Hence, {r s0 +,..., r s0 + k} C s0 =, which implies that n := r s0 + r s0 > k. That is, since k N was arbitrary, we have that n is arbitrarily large. Set v := εt rs 0 u k T rs 0 u ( v = ε/k). For each j C s 0, we have T j rs 0 v v = T j u T rs 0 u L. So card { j n : T j v > ε } = card { j n : T j v v > k} which completes the proof. card { r s0 j < r s0 + : T j u > k T rs 0 u } card C s0 n( k ), Proposition 9. Let T B(X) and suppose that there exists a dense subset X 0 of X such that the orbit of each x X 0 is distributionally near to 0. Then the set of all vectors with orbits distributionally near to 0 is residual. Proof. For each k, m N, let M k,m := { x X : n N with card{ j n : T j x m < k } n( k ) }. Clearly each M k,m is open and dense (since M k,m X 0 ). So the set k,m M k,m is residual and consists of vectors with orbits distributionally near to 0. Definition 0. Let T B(X). We say that T satisfies the Distributional Chaos Criterion (DCC) if there exist sequences (x k ), (y k ) in X such that: (a) There exists A N with dens(a) = such that lim n A T n x k = 0 for all k. (b) y k span{x n : n N}, lim k y k = 0 and there exist ε > 0 and an increasing sequence (N k ) in N such that for all k N. card{ j N k : d(t j y k, 0) > ε} N k ( k ) Remarks.. We can assume that (x k ) and (y k ) are the same sequence because there exists a sequence ( x k ) in span{x n : n N} \ {0} that satisfies condition (a) by linearity and condition (b) by density. 7

9 2. In the case X is a Banach space, it follows from Proposition 8 that condition (b) in the above definition of (DCC) can be replaced by (b ) y k span{x n : n N}, y k 0 and there exist ε > 0 and an increasing sequence (N k ) in N such that for all k N. card{ j N k : T j y k > ε} εn k Theorem 2. If T B(X) then the following assertions are equivalent: (i) T satisfies (DCC); (ii) T has a distributionally irregular vector; (iii) T is distributionally chaotic; (iv) T admits a distributionally chaotic pair. Proof. (i) (ii): Let X 0 := {x X : lim n A T n x = 0}. Then X 0 is a subspace of X, T (X 0 ) X 0 and T (X 0 ) X 0. Moreover, x k X 0 and y k X 0 for all k N. By Proposition 9, the set of all vectors x X 0 with orbits distributionally near to 0 is residual in X 0. By Proposition 8, the set of all vectors x X 0 with distributionally unbounded orbit is residual in X 0. So the set of all distributionally irregular vectors is residual in X 0. In particular, there exists a distributionally irregular vector. (ii) (iii): Let u X be a distributionally irregular vector. Then {λu : λ K} is an uncountable distributionally ε-scrambled set for a certain ε > 0. (iii) (iv): Trivial. (iv) (i): Let (x, y) X X be a distributionally chaotic pair for T and set u := x y. There exists ε > 0 such that and dens{j N : d(t j u, 0) > ε} = () dens{j N : d(t j u, 0) < δ} = for each δ > 0. So there is an increasing sequence (n k ) in N such that card{ j n k : d(t j u, 0) < k } n k ( k ). Let A k := { j n k : d(t j u, 0) < k } (k N) and A := k= A k. Then dens(a) = and lim n A T n u = 0. For each k N, let x k := T k u. Clearly lim T n x k = 0 n A (k N). Now, choose s k such that T s k u k < k and let y k := T s k u. Then yk 0. By (), we have dens{j N : d(t j y k, 0) > ε} = for all k N. Hence there is an increasing sequence (N k ) in N such that card{ j N k : d(t j y k, 0) > ε} N k ( k ) for all k N. This proves that T satisfies (DCC). 8

10 3 Dense distributional chaos We devote this section to the existence of dense sets (manifolds) of distributionally irregular vectors. Concerning necessary conditions, we observed in [7] that the existence of a dense set of just irregular vectors for T implies that the adjoint operator T admits no eigenvalues λ with λ. This result is sharp, even for dense distributional chaos. Remark 3. There are operators T : l p (v) l p (v) for certain weight sequences v such that every non-zero vector x l p (v) is distributionally irregular, and the set of eigenvalues of T is D := {λ : λ < }. Indeed, let S be the bilateral forward shift on a weighted space l p (w, Z) such that w i+ /w i and every non-zero vector x l p (w, Z) is distributionally irregular, as constructed in [7]. Let v := (w i ) i N and consider the natural inclusion l p (v) l p (w, Z), (x, x 2,... ) (..., 0, 0, x, x 2,... ). The operator T := S l p (v) clearly satisfies the desired properties. In what follows in this section we will consider several sufficient conditions for dense distributional chaos. Definition 4. Given T B(X), a vector subspace Y of X is called a uniformly distributionally irregular manifold for T if there exists m N such that the orbit of every nonzero vector y in Y is simultaneously distributionally m-unbounded and distributionally near to 0. It is easy to see that such a Y is a distributionally 2 m+ -scrambled set for T. Hence the existence of a dense uniformly distributionally irregular manifold implies dense distributional chaos. Theorem 5. Assume X separable. Suppose that T B(X) satisfies T n x 0 for all x X 0, where X 0 is a dense subset of X. Then the following assertions are equivalent: (i) T is distributionally chaotic; (ii) T is densely distributionally chaotic; (iii) T admits a dense uniformly distributionally irregular manifold; (iv) T admits a distributionally unbounded orbit. We recall that Proposition 8 contains a very useful computable characterization of the existence of a distributionally unbounded orbit. It will be used several times in applications of the above theorem. Proof. (iii) (ii) (i): Obvious. (i) (iv): Follows from Theorem 2. (iv) (iii): Without loss of generality we may assume that X 0 is a dense subspace of X and T x k x k+ for all x X and k N. By hypothesis, there is a vector y X with distributionally unbounded orbit. So there exist m N and B N with dens(b) = such that lim T n y m =. n B 9

11 Hence, for every L > 0 and k N, there exist x X as close to zero as we want and n N as large as we want so that card{ i n : T i x m > L} > n( k ). Clearly we can take x X 0 and assume, without loss of generality, that m =. Thus we can construct inductively a sequence (x k ) of vectors in X 0 with x k k, k N, and an increasing sequence (n k ) of positive integers such that ( card{ i n k : T i x k > k2 k } > n k ), (2) k 2 card{ i n k : T i x s k < k } > n k ( k ), s =,..., k. (3) 2 Given α, β {0, } N, we say that β α if β i α i for all i N. Consider an increasing sequence (r j ) of positive integers such that r j+ + r j + n rj + for all j N. (4) Fix α {0, } N defined by α n = if and only if n = r j for some j N. Given β {0, } N such that β α and β contains an infinite number of s, we define the vector x β := i β i 2 i x i = j β rj 2 r j x r j. Observe that the above series is convergent since x i i for each i N. We will show that the orbit of x β is distributionally -unbounded and distributionally near to 0. Let k N with β rk =. If i n rk, T i x rk > r k 2 r k and T i x s rk < r k for each s < r k, then T i x β T i x 2 r rk β rj k 2 T i x r rj j j k > r k r k 2 x rj +i r j 2 r j j>k r k, j<k where we used the inequalities x rj +i for j > k, which hold because +i +n rk + n rj r j (by (4)). Conditions (2) and (3) above imply that card{ i n rk : T i x β > r k } n rk ( r k ), and so the orbit of x β is distributionally -unbounded. On the other hand, if i n rk + and T i x s rk + < r k + for each s < r k +, then T i x β rk + j k r k + j k < r k +, β rj T i x rj rk + 2 r j 2 + r j j>k 0 + j>k β rj T i x rj rk + 2 r j x rj +rk +i 2 r j

12 where we used the inequalities x rj +rk +i for j > k, which hold because + r k + i + r k + n rk + + r j + n rj + r j (by (4)). Condition (3) above implies that card { i n rk + : T i x β rk + < } nrk +( ), r k + r k + and so the orbit of x β is distributionally near to 0. Since X is separable, we can select a dense sequence (y n ) in X 0 and a countable collection γ n {0, } N (n N) such that each sequence γ n contains an infinite number of s, γ n α for every n N, and the sequences γ n have mutually disjoint supports. We set the sequence of vectors u n := γ n,i i x 2 i i, n N. We already know that the orbit of each u n is distributionally -unbounded and distributionally near to 0. Define now z n := y n + n u n, n N. Since (y n ) is dense in X and the u n s are uniformly bounded in X, we get that the sequence (z n ) is dense in X. We set Y := span{z n : n N}, which is a dense subspace of X. If u Y \ {0}, then we can write u = y 0 + k ρ k 2 k x k, where y 0 X 0 and the sequence of scalars (ρ k ) takes only a finite number of values (each of them infinitely many times). As in the above proof we can show that the orbit of v := k ρ k 2 k x k is distributionally -unbounded and distributionally near to 0. Since y = y 0 + v and lim k T k y 0 = 0, the same is true for the orbit of y. Thus Y is a dense uniformly distributionally irregular manifold for T. Theorem 6. Assume X separable. conditions: Suppose that T B(X) satisfies the following (I) There exists a dense subset X 0 of X with lim n T n x = 0 for all x X 0. (II) One of the following conditions is true: (a) X is a Fréchet space and there exists an eigenvalue λ with λ >. (b) X is a Banach space and T n (c) X is a Hilbert space and T n 2 Lebesgue measure). Then T is densely distributionally chaotic. < (in particular if r(t ) > ). < (in particular if σ p (T ) T has positive Proof. In view of Theorem 5, it is enough to prove the existence of a vector y X with distributionally unbounded orbit. In case (a), it is enough to get an eigenvector y 0 associated to an eigenvalue λ with λ >. In cases (b) and (c), it was proved in [8] that there exists a vector y X such that T n y. In particular, y has distributionally unbounded orbit. By Theorem.(i) in [9] (see also p. 239, Theorem in [9]), < whenever T n 2 σ p (T ) T has positive Lebesgue measure.

13 In [0] it is proved that, if T satisfies the Godefroy-Shapiro Criterion, then T is hypercyclic. As consequence of (a) in Theorem 6 we also obtain that Corollary 7. If T satisfies the Godefroy-Shapiro Criterion then T is densely distributionally chaotic. Since every continuous linear operator on H(C N ) that commutes with any translation operator and is not a scalar multiple of the identity satisfies the Godefroy-Shapiro criterion [0], we get a natural class of densely distributionally chaotic operators. Corollary 8. Every continuous linear operator on H(C N ) that commutes with any translation operator and is not a scalar multiple of the identity is densely distributionally chaotic. A series k= x k in X is said to be unconditionally convergent if for every ε > 0, there exists N such that ( ) d x k, 0 < ε, k F whenever F N is finite and F {, 2,..., N} =. Theorem 9. Assume X separable and let T B(X). Suppose that: (a) There exists a dense subset X 0 of X with lim n T n x = 0 for all x X 0. (b) There exist a subset Y of X, a mapping S : Y Y with T Sy = y on Y, and a vector z Y \ {0} such that n= T n z and n= Sn z converge unconditionally. Then T is densely distributionally chaotic. Proof. If we define w k0 := n= T k0n z + z + n= Sk 0n z, then w k0 0 if k 0 is sufficient large and T k 0 w k0 = w k0. Let y k := n=k Sk 0n z. Then y k 0 and as j. For 0 l < k 0 we have j k T k0j y k = T k0n z + z + n= S k0n z w k0 n= T l w k0 = lim j T l+k 0j y k. Hence {T l w k0 : 0 l < k 0 } are accumulation points of the orbit of y k. Let ε := 2 min{d(t l w k0, 0) : 0 l < k 0 }. Then there exists an increasing sequence (N k ) of positive integers such that lim k N k card{ j N k : d(t j y k, 0) > ε} =. By Proposition 8, T admits a distributionally unbounded orbit. Thus, by Theorem 5, T is densely distributionally chaotic. In [8] it was observed that the Frequent Hypercyclicity Criterion implies that T is frequently hypercyclic, Devaney chaotic and mixing. As a consequence of the above theorem, we obtain that it also implies dense distributional chaos. 2

14 Corollary 20. Assume X separable and let T B(X). Suppose that there are a dense subset X 0 of X and a mapping S : X 0 X 0 such that, for any x X 0, (a) n= T n x converges unconditionally, (b) n= Sn x converges unconditionally, (c) T Sx = x. Then T is frequently hypercyclic, Devaney chaotic, mixing and densely distributionally chaotic. Definition 2. We say that a continuous linear operator T on a Banach space X has a perfectly spanning set of eigenvectors associated to unimodular eigenvalues if there exists a continuous probability measure σ on the unit circle T such that for every σ-measurable subset A of T of σ-measure equal to, the span of λ A ker(t λ) is dense in X. Theorem 22. Let T be a continuous linear operator on a Hilbert space X that has a perfectly spanning set of eigenvectors associated to unimodular eigenvalues such that the measure σ can be chosen to be absolutely continuous with respect to the Lebesgue length measure on the unit circle. Then T is densely distributionally chaotic. Proof. Under the conditions of the theorem, T satisfies the Hypercyclicity Criterion with respect to the sequence (n) (see Theorems 2.4 and 2.5 in []) and thus there exists a dense set X 0 such that lim n T n x = 0 for all x X 0. Moreover, σ p (T ) T has positive Lebesgue measure. Thus, by Theorem 6, T is densely distributionally chaotic. Problem 23. Is the above theorem true for Banach spaces? As consequence of Corollary 20, we have a positive partial answer of the above problem under certain conditions. Indeed, if T has a σ-spanning set of T-eigenvectors, where σ is the length measure on T, then T satisfies the Frequent Hypercyclicity Criterion, and so T is densely distributionally chaotic. The same conclusion holds if the eigenvector fields are C 2 -functions or X does not contain c 0 and the eigenvectors fields are α-holderian for some α > ([3, section 5.8]). 2 Since a composition operator with a parabolic or hyperbolic automorphism in H 2 (D) has a perfectly spanning set of eigenvectors associated to unimodular eigenvalues with respect to the Lebesgue length measure on the unit circle [2], we obtain Corollary 24. Let ϕ be an automorphism of D. Then the composition operator C ϕ : H 2 (D) H 2 (D) is densely distributionally chaotic if and only if ϕ is not an elliptic automorphism. 4 Densely distributionally chaotic weighted shifts The next result characterizes dense distributional chaos for unilateral weighted backward shifts on Fréchet sequence spaces in terms of the existence of a distributionally unbounded orbit. Theorem 25. Let X be a Fréchet sequence space in which (e n ) n N is a basis. Suppose that the unilateral weighted backward shift B ω ( (xn ) n N ) := (wn x n+ ) n N is an operator on X. Then the following assertions are equivalent: 3

15 (i) B ω is distributionally chaotic; (ii) B ω is densely distributionally chaotic; (iii) B ω admits a dense uniformly distributionally irregular manifold; (iv) B ω admits a distributionally unbounded orbit. Proof. Let X 0 be the set of all finite linear combinations of the basis vectors e n. Clearly X 0 is dense in X and (B ω ) n x 0 for all x X 0. Thus the result is just a special case of Theorem 5. As an application of the previous theorem, we have the following computable sufficient condition for dense distributional chaos for unilateral backward shifts on Fréchet sequence spaces. Theorem 26. Let X be a Fréchet sequence space in which (e n ) n N is a basis. Suppose that the unilateral backward shift B is an operator on X. If there exists a set S N with dens(s) = such that e n converges in X, n S then B is densely distributionally chaotic. Proof. For each k N, let y k := n S,n k Then lim k y k = 0 and B n y k = e + for all n S with n k. Since the functional x x is continuous, there exists ε > 0 such that d(x, 0) ε implies x <. Hence d(b n y k, 0) > ε for all n S with n k. So there is a sequence (N k ) of positive integers increasing to such that lim k e n. N k card { j N k : d(b j y k, 0) > ε } =. By Proposition 8, T admits a distributionally unbounded orbit. Thus, by the previous theorem, B is densely distributionally chaotic. Corollary 27. Let X be a Fréchet sequence space in which (e n ) n N is a basis. Let (ω n ) n N be a sequence of positive weights. Suppose that the unilateral weighted backward shift B ω is an operator on X. If there exists a set S N with dens(s) = such that ( n ) en ω ν converges in X, (5) n S ν= then B ω is densely distributionally chaotic. Proof. Follows from the previous theorem by conjugacy. Corollary 28. Let X be a Fréchet sequence space in which (e n ) n N is an unconditional basis. Suppose that the unilateral weighted backward shift B ω is an operator on X. If B ω has a non-trivial periodic point, then B ω is densely distributionally chaotic. 4

16 Proof. Under these hypothesis (5) holds ([, Theorem 4.8(c)]). Hence it is enough to apply the previous corollary. Now we turn our attention to bilateral weighted forward shifts. Theorem 29. Let X be a Fréchet sequence space over Z in which (e n ) n Z is a basis. Let (ω n ) n Z be positive weights and assume that lim n ( n ω ν )e n = 0. (6) ν= Suppose that the bilateral weighted forward shift F ω ( (xn ) n Z ) := (wn x n ) n Z is an operator on X. Then the following assertions are equivalent: (i) F ω is distributionally chaotic; (ii) F ω is densely distributionally chaotic; (iii) F ω admits a dense uniformly distributionally irregular manifold; (iv) F ω admits a distributionally unbounded orbit. Proof. Let X 0 be the set of all finite linear combinations of the basis vectors e n. Clearly X 0 is dense in X. Moreover, lim n (F ω ) n x = 0 for all x X 0, because lim n (F ω ) n e j = 0 for any j Z (by (6)). Hence it is enough to apply Theorem 5. Theorem 30. Let X be a Fréchet sequence space over Z in which (e n ) n Z is a basis. Suppose that the bilateral forward shift F is an operator on X. If lim n e n = 0 and there exists a set S N with dens(s) = such that e n converges in X, n S then F is densely distributionally chaotic. Proof. For each k N, let y k := n S,n k Then lim k y k = 0 and F n+ y k = + e + for all n S with n k. Since the functional x x is continuous, there exists ε > 0 such that d(x, 0) ε implies x <. Hence d(f n+ y k, 0) > ε for all n S with n k. So there is a sequence (N k ) of positive integers increasing to such that lim k e n. N k card{ j N k : d(f j y k, 0) > ε} =. By Proposition 8, T admits a distributionally unbounded orbit. Thus, by the previous theorem, F is densely distributionally chaotic. 5

17 Corollary 3. Let X be a Fréchet sequence space over Z in which (e n ) n Z is a basis. Let (ω n ) n Z be positive weights. Suppose that the bilateral weighted forward shift F ω is an operator on X. If lim n ( n ν= ω ν)e n = 0 and there exists a set S N with dens(s) = such that ( 0 ) e n ω ν converges in X, n S ν= n+ then F ω is densely distributionally chaotic. Corollary 32. Let X be a Fréchet sequence space over Z in which (e n ) n Z is an unconditional basis. Suppose that the bilateral weighted forward shift F ω is an operator on X. If F ω has a non-trivial periodic point, then F ω is densely distributionally chaotic. 5 Densely distributionally chaotic composition operators We begin this section by recalling the following result, where ϕ n denotes the n-th iterate of ϕ. Theorem 33 (The Denjoy-Wolff Iteration Theorem). Suppose that ϕ is an analytic selfmap of D that is not an elliptic automorphism.. If ϕ has a fixed point p D, then (ϕ n ) converges uniformly on compact sets to p. 2. If ϕ has no fixed point in D, then there is a fixed point p D such that (ϕ n ) converges uniformly on compact sets to p. Now we characterize dense distributional chaos for composition operators on the Fréchet space H(D). Theorem 34. Suppose that ϕ is an analytic self-map of D. Then the composition operator C ϕ : H(D) H(D) is densely distributionally chaotic if and only if it has no fixed point in D. Proof. If ϕ is an elliptic automorphism, then ϕ is conjugated to a rotation, and so C ϕ is not distributionally chaotic. If ϕ is a non-elliptic automorphism with a fixed point p D, then (ϕ n ) converges uniformly on compact sets to p. Thus (f ϕ n ) converges uniformly on compact sets to f(p) and C ϕ is not distributionally chaotic. If ϕ has no fixed point in D, let p be the Denjoy-Wolff point that belongs to D such that (ϕ n ) converges uniformly on compact sets to p. Let X 0 denote the set of all functions that are continuous on D, analytic on D and vanish at p. Then X 0 is dense in H(D) and lim n C n ϕf = 0 for all f X 0. For each k N, let g k (z) := k(p z) Then (g k ) converges to zero and there is a sequence (N k ) of positive integers increasing to such that lim card { j N k : d(c k N ϕg j k, 0) > } =. k 2 By Proposition 8, T admits a distributionally unbounded orbit. Thus, by Theorem 5, C ϕ is densely distributionally chaotic. 6

18 Corollary 35. Let Ω be a simply connected domain in the complex plane and ϕ be an automorphism of Ω. For the composition operator C ϕ : H(Ω) H(Ω), the following assertions are equivalent: (i) C ϕ is chaotic; (ii) C ϕ is mixing; (iii) C ϕ is hypercyclic; (iv) (ϕ n ) is a run-away sequence; (v) ϕ has no fixed point in Ω; (vi) C ϕ is densely distributionally chaotic. Proof. The equivalences (i) (ii) (iii) (iv) (v) are known. If Ω C then the equivalence (v) (vi) follows from the above theorem by conjugation. If Ω = C and ϕ has no fixed point, then C ϕ is a translation operator, and so it is densely distributionally chaotic. If ϕ has a fixed point, then C ϕ is conjugated to C ψ where ψ(z) = az and thus it is not densely distributionally chaotic. 6 Miscellanea It is well-known that if T is invertible and hypercyclic (mixing), then T is hypercyclic (mixing). In contrast, there exist operators T which are invertible and densely distributionally chaotic such that T is not distributionally chaotic. Indeed, let T = F ω : l (Z) l (Z) be a bilateral weighted forward shift such that ω i := 2, i 0, and T l (N) is the unilateral weighted backward shift of Example 5 (Alternatively, one can consider T the unweighted bilateral forward shift on l (v, Z) for v i := 2 i, i 0, and T l (v,n) is the unilateral backward shift of [7, Theorem 2.]). The conditions of Corollary 3 are satisfied, so T is densely distributionally chaotic. Moreover, similar considerations to the ones given in Example 5 show that T admits no distributionally unbounded orbit and, therefore, it is not distributionally chaotic. Also, there are densely distributionally chaotic operators on Banach spaces of the form T = I + K, where K is a compact operator (see [6]). Such a T is neither frequently hypercyclic nor Devaney chaotic. Problem 36. Are there frequently hypercyclic operators which are not distributionally chaotic? For example, in [3] the authors constructed a frequently hypercyclic operator on c 0 that is not Devaney chaotic (and not mixing). Is this operator distributionally chaotic? Problem 37. Are there Devaney chaotic operators which are not distributionally chaotic? Acknowledgement The present work was done while the first author was visiting the Departament de Matemàtica Aplicada at Universitat Politècnica de València (Spain). The first author is very grateful for the hospitality. 7

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