A note on J-sets of linear operators

THEMATICS MA dem y Cze of Scie ch R n epu ces blic Aca Preprint, Institute of Mathematics, AS CR, Prague. 2010-10-19 INSTITU TE of A NOTE ON J-SETS OF LINEAR OPERATORS M. R. AZIMI AND V. MÜLLER Abstract. We construct a Banach space operator T ∈ B(X) such that the set JT (0) has a nonempty interior but JT (0) 6= X. This gives a negative answer to a problem raised by G. Costakis and A. Manoussos. 1. introduction and preliminaries Let X be an infinite dimensional complex Banach space and let B(X) be the algebra of all bounded linear operators on X. For T ∈ B(X) and x ∈ X let Orb(T, x) = {x, T x, T 2 x, ...} be the orbit of T at x. By a result of Bourdon and Feldman [?], if the closure Orb(T, x) has a non-empty interior, then Orb(T, x) = X, and so x is a hypercyclic vector for T . In [?], a weaker concept to that of the limit set of an orbit was introduced and studied. For T ∈ B(X) and x ∈ X, let JT (x) be the set of all vectors y ∈ X such that there exist a strictly increasing sequence (kn ) ⊆ N and a sequence (xn ) ⊆ X with xn → x and T kn xn → y as n → ∞. It is easy to see that the set JT (x) is always closed. In [?], Problem 1, it was asked whether there is an analogue of the BourdonFeldman theorem in the case of J-sets: if the set JT (x) has a nonempty interior, does it imply that JT (x) = X? The goal of this paper is to give a negative answer to this question. Let X be a Banach space, x ∈ X and r > 0. We denote by B(x, r) = {y ∈ X : ky − xk ≤ r} the closed ball with radius r and center x. We denote by int A the interior of any subset A ⊂ X. 2. Main result Example. There exist a Banach space X and an operator T ∈ B(X) such that int JT (0) 6= ∅ and JT (0) 6= X. Construction. Let (kn )∞ n=1 be a fixed fast increasing sequence of positive integers. It is sufficient to assume that kn+1 ≥ 5kn2 for all n ∈ N. Let X be the ℓ1 space with the standard basis   ui : i = 0, 1, ... ∪ vn,j : n ∈ N, 1 ≤ j ≤ kn . 1991 Mathematics Subject Classification. 47A16. Key words and phrases. hypercyclic vectors, orbits, J-sets. The research was supported by grants 201/09/0473 of GA CR and IAA100190903 of GA AV. 1 2 M. R. AZIMI AND V. MÜLLER More precisely, the elements of X can be expressed as x= ∞ X αi ui + kn ∞ X X βn,j vn,j n=1 j=1 i=0 with complex coefficiens αi , βn,j such that kxk := ∞ X |αi | + kn ∞ X X |βn,j | < ∞. n=1 j=1 i=0 Let {wn : n ∈ N} be a countable dense set in B(0, 41). Without loss of generality W we may assume that each wn belongs to the space u0 , u1 , ..., un , vm,j : 1 ≤ m < n, 1 ≤ j ≤ km . We are going to construct an operator T with JT (0) ⊃ B(u0 , 1/4). To this end it is sufficient to have u0 +W wn ∈ T kn B(0, 1/n) for each n. The purpose of the finite-dimensional subspace W {vn,j : 1 ≤ j ≤ kn } is to achieve this relation. The infinite-dimensional subspace {ui : i = 0, 1, . . . } will ensure that JT (0) 6= X. Let T ∈ B(X) be defined by T ui T vn,j = = T vn,kn = 2ui+1 (i = 0, 1, ...), 2vn,j+1 (n ∈ N, 1 ≤ j ≤ kn − 1), n (u0 + wn ) (n ∈ N). 2kn −1 It is easy to see that kT k = 2. For each n ∈ N we have T kn (n−1 vn,1 ) = 2kn −1 n−1 T vn,kn = u0 + wn . This implies that B(u0 , 14 ) ⊂ JT (0). Indeed, let z ∈ X with kzk ≤ 41 and let (ni ) be an increasing sequence in N satisfying wni → z as i → ∞. Then n−1 i vni ,1 → 0 and −1 kni limi→∞ T (ni vni ,1 ) = limi→∞ (u0 + wni ) = u0 + z. In particular, int JT (0) 6= ∅. It remains to show that JT (0) 6= X. Suppose on the contrary that JT (0) = X. In particular, it means that v1,1 ∈ JT (0), and so there exist k ∈ N and y ∈ X, kyk ≤ 1 with 1 kT k y − v1,1 k < . (1) 4 Moreover, we may assume that k > k2 + k1 . Write mn = kn + kn−1 + · · · + k1 . Since ki+1 ≥ 5ki2 ≥ 5ki , we have mn ≤ 5k4n , and so kn ≤ mn ≤ 45 kn . W Let n ∈ N satisfy mn−1 ≤ k < mnW. By assumption, n ≥ 3. Write X0 = {ui : i = 0, 1, . . . }. For n ∈ N let Xn W = {vn,i : 1 ≤ i ≤ kn }. Let Pj be the natural projection onto Xj , i.e., ker Pj = i6=j Xi . Clearly kPj k = 1 for each j. P  n−1 Write y = y0 + y1 + x + y2 , where y0 = P0 y, y1 = P y, x = Pn y and i i=1 P  ∞ y2 = i=n+1 Pi y. We have ky0 k + ky1 k + kxk + ky2 k = kyk ≤ 1. Obviously T k y0 ∈ X0 and T k y1 ∈ T k n−1 _ i=1 Finally, P n−1 i=0  n−2  _
Xi ⊂ T k−kn−1 Xi ⊂ · · · ⊂ T k−kn−1 −···−k1 X0 ⊂ X0 . i=0  Pi T k y2 ≤ k 2 (n+1) 2kn+1 −1 ≤ n+1 2kn+1 −mn ≤ n+1 2 kn < 14 . A NOTE ON J-SETS OF LINEAR OPERATORS 3 If mn−1 ≤ k < mn − 2mn−1 = kn − mn−1 , then 2k n 1 1 n 1 + ≤ mn−1 + < . 2kn −1 4 2 4 2 1 1 k k So kT y − v1,1 k ≥ kP1 (T y − v1,1 )k ≥ 1 − 2 = 2 , a contradiction with (1). So we may assume that kn − mn−1 ≤ k ≤ kn + mn−1 = mn . Write for short m = W Sn−1 mn−1 . For j = 1, 2, . . . let Yj = {u(j−1)m , . . . , ujm−1 }. Write also Y0 = i=1 Xi . Let Qj be the natural projection onto Yj (j = 0, 1, . . . ). Note Wthat k − m ≥ 16 2 2 k 2 2 kn − 2m P≥ 5kn−1  − 2m ≥ 5 m − 2m ≥ m , and so T (y0 + y1 ) ∈ {ui : i ≥ m }. m k Thus i=0 Qj T (y0 + y1 ) = 0 and P  P  m m k Q (T x − v ) = Q (T k (y0 + y1 + x) − v1,1 ) j 1,1 j j=0 j=0 P  P  (2) m m 1 1 1 k k ≤ j=0 Qj (T y − v1,1 ) + j=0 Qj T y2 ≤ 4 + 4 = 2 . Pi0 −1 Pkn αi vn,i . Let i0 = kn − k + 1 and x0 = i=1 αi vn,i (if i0 ≤ 1 then Let x = i=1 x0 = 0). For j = 1, ..., m let kP1 T k yk ≤ kP1 T k xk + kP1 T k y2 k ≤ i0 +jm−1 X xj = αi vn,i . i=i0 +(j−1)m We have T k x0 ∈ Xn , and so P m j=0 i0 +jm−1 X T k xj =  Qj T k x0 = 0. For j = 1, ..., m, we have αi T k vn,i = = αi 2kn −i T k−kn +i vn,kn i i=i0 +(j−1)m X X kn −i αi i 2 n k−kn +i−1 T (u0 + wn ) = sj + qj , 2kn −1 where i0 +jm−1 X sj = 2k−kn n αi uk−kn +i−1 i=i0 +(j−1)m and qj = i0 +jm−1 X αi 21−i nT k−kn +i−1 wn . i=i0 +(j−1)m Note that ksj k = n2k−kn i0 +jm−1 X |αi | = n2k−kn kxj k i=i0 +(j−1)m and kqj k ≤ i0 +jm−1 X i=i0 +(j−1)m |αi |21−i n2k−kn +i−1 kwn k ≤ 1 ksj k. 4 Note also that T k xj ∈ Yj−1 ∨ Yj ∨ Yj+1 . Write tj = Qj−1 qj , t′j = Qj qj and t′′j = Qj+1 qj . For j = 1, . . . , m − 1, we have j X i=0  Qi (T k x − v1,1 ) = kt1 − v1,1 k + ks1 + t′1 + t2 k + ks2 + t′′1 + t′2 + t3 k + · · · 4 M. R. AZIMI AND V. MÜLLER · · · + ksj−1 + t′′j−2 + t′j−1 + tj k + ksj + t′′j−1 + t′j + tj+1 k ≥ 1 − kt1 k + ks1 k − kt′1 k − kt2 k + ks2 k − kt′′1 k − kt′2 k − kt3 k + · · · · · · + ksj k − kt′′j k − kt′j k − ktj+1 k
≥ 1 + ks1 k − kt1 k − kt′1 k − kt′′1 k + · · ·

· · · + ksj−1 k − ktj−1 k − kt′j−1 k − kt′′j−1 k + ksj k − ktj k − kt′j k − ktj+1 k 3 ksj+1 k ≥ 1 + (ks1 k + ks2 k + · · · + ksj k) − . 4 4 P  j 1 k Since i=0 Qi (T x − v1,1 ) ≤ 2 by (2), we have ksj+1 k ≥ 3(ks1 k + ks2 k + · · · + ksj k) ≥ 3ksj k. So kxj+1 k ≥ 3kxj k. By induction, kxm k ≥ 3kxm−1 k ≥ · · · ≥ 3m−1 kx1 k. Since kxm k ≤ kxk ≤ 1, we have kx1 k ≤ 31−m . Hence kQ0 T k xk = kQ0 T k x1 k = kt1 k ≤ 2k−kn n 2m n 1 kx1 k ≤ 2k−kn −2 n31−m ≤ m ≤ , 4 3 2 which is a contradiction with the fact that kQ0 T k xk ≥ kQ0 v1,1 k − kQ0 (T k x − v1,1 )k ≥ 1 − kT k x − v1,1 k ≥ 3 . 4 Remark. The construction above can be modified easily so that we obtain an operator V ∈ B(Y ) and a non-zero vector y ∈ Y such that int JV (y) 6= ∅ and JV (y) 6= Y. Let X and T ∈ B(X) be as in the previous example. Let Y = X ⊕ ℓ1 and let V = T ⊕ 2S, where S ∈ B(ℓ1 ) is the backward shift. Let y 6= 0 and Sy = 0. Then V (0 ⊕ y) = 0. It is easy to see that JV (0 ⊕ y) = JV (0 ⊕ 0). Clearly JV (0 ⊕ 0) ⊂ JT (0) ⊕ J2S (0). Furthermore, it is easy to see that for all ε > 0, y ′ ∈ ℓ1 and all n sufficiently large there exists yn ∈ ℓ1 with kyn k < ε and (2S)n yn = y ′ . This implies that JV (0 ⊕ 0) = JT (0) ⊕ ℓ1 . Hence int JV (0 ⊕ y) 6= ∅ and JV (0 ⊕ y) 6= Y. Acknowledgment. The first author is extremely grateful to Professor Vladimir Müller for his constant scientific support and also for his hospitality during the first author’s research stay at the Mathematical Institute of the Czech Academy of Sciences, Prague, Czech Republic in 2010. References [BF] P.S. Bourdon, N.S. Feldman, Somewhere dense orbits are everywhere dense, Indiana Univ. Math. J., 52(2003), 811-819. [CM] G. Costakis, A. Manoussos, J-class operators and hypercyclicity, J. Operator Theory, to appear. university of tabriz, faculty of mathematical sciences, 29 Bahnan st., 51666-16471 tabriz, iran E-mail address: mh [email protected] Institute of mathematics, Czech Academy of Sciences, Zitna 25, 115 67 Prague 1, Czech Republic E-mail address: [email protected]