ON M-HYPONORMAL WEIGHTED SHIFTS. Jung Sook Ham, Sang Hoon Lee and Woo Young Lee. 1. Introduction

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1 ON M-HYPONORMAL WEIGHTED SHIFTS Jung Soo Ham, Sang Hoon Lee and Woo Young Lee Abstract Let α {α n} n0 be a weight sequence and let Wα denote the associated unilateral weighted shift on l Z + In this paper we prove that if α is eventually increasing, then W α is M-hyponormal and that if α has exactly two subsequential limits such that the larger one is different from the spectral radius of W α then W α is not M-hyponormal 1 Introduction Let LH be the algebra of bounded linear operators on a separable complex Hilbert space H An operator T LH is called normal if T T T T and hyponormal if T T T T An operator T LH is called M-hyponormal if there exists M > 0 such that T x M T x for all C and for all x H If M 1 then M-hyponormality implies hyponormality The notion of an M-hyponormal operator is due to J Stampfli unpublished see [11] The class of M-hyponormal operators has been studied by many authors cf[1 3],[5 1] However examples of M-hyponormal non-hyponormal operators seem to be scarce from the literature The aim of the present article is to give abundant examples of M-hyponormal non-hyponormal operators Our strategy involves the unilateral weighted shifts Results Recall that given a bounded sequence of positive numbers α : α 0, α 1, called weights, the unilateral weighted shift W α associated with α is the operator on l Z + defined by W α e n : α n e n+1 for all n 0, where {e n } n0 is the canonical orthonormal basis for l It is straightforward to chec that W α can never be normal, and that W α is hyponormal if and only if α n α n+1 for all n 0, ie, α is monononically increasing B Wadha [11] gave an example of an M-hyponormal non-hyponormal weighted shift W α of the form: 0 W α On the other hand, M Radjabalipour [8] showed that the only quasinilpotent M-hyponormal operator is 0 Thus if W α is a weighted shift with weight sequence {α n } converging to 0 then W α is not M-hyponormal In this paper we consider the question: Which weighted shifts are M-hyponormal? Our main theorem now follows: 1991 Mathematics Subject Classification 47B0, 47B37 Key words and phrases M-hyponormal operators, unilateral weighted shifts This wor was partially supported by the KOSEF research project No R Typeset by AMS-TEX

2 Theorem 1 Let T W α be a weighted shift with weight sequence α {α n } n0 If α is eventually increasing then T is M-hyponormal Proof For any x n0 x ne n l and for any C, and T x x 0 e 0 + T x Thus a straightforward calculation shows that α i x i x i+1 e i+1 α i x i+1 x i e i T x T x x 0 + α i x i x i+1 α i x i+1 x i α 0 x 0 + αi αi 1 x i Suppose that α {α n } n0 is monotonically increasing for n We now claim that if c : min {α 0, α 1,, α }, then indeed we have that Thus we have that i1 T x c x for any C : T x T x x α i x i 1 x c x i 1 x c x 11 T x c x for < c Write J : {j 1 : α j < α j 1 } If J then T must be hyponormal Evidently, J {1,, } Write m : J We argue that 1 T x Towards 1 observe that and so it suffices to show that [ j 1 j 1 l0 ] 1 αi l + j j+1 x j j 1 il T x x 0 + α i x i x i+1, j 1 13 x 0 + α i x i x i+1 K 1 j j+1 x j j 1,

3 3 where j 1 K j : l0 j 1 αi l + j for each j 1,, We use an induction on j First, observe that il x 0 + α 0 x 0 x 1 x 0 + α0 x 0 x 1 α 0 x 0 x 1 α 0 x 0 + x 1 + α0 x 0 α 0 x 1 α 0 + α0 + α0 x 1 + x α0 x 1 4 K 1 x 1, which shows that 13 holds for j 1 We now suppose that 13 holds for j n Then a straightforward calculation shows that x 0 + n α i x i x i+1 Kn 1 n+1 x n + α n x n x n+1 Kn 1 n+1 + αn x n α n x n xn+1 α n x n x n+1 + x n+1 Kn 1 n+1 + αn α n x n+1 x n α n Kn 1 n+1 + αn Kn 1 x n+1 + αn n+1 + x n+1 K 1 n n+1 Kn 1 n+1 + αn n+ x n+1 n l0 n il α i l + n+1 x n+1 K 1 n+1 n+ x n+1, which shows that 13 holds for j n + 1 This proves 13 On the other hand, for c and for each j J we can find a constant γ j > 0 for which 14 [ j 1 j 1 l0 il ] 1 αi l + j j+1 γj, where γ j is independent of with c It thus follows from 1 and 14 that if c then Thus if c then 15 T x γ m T x γ j x j for each j J x j, where γ min γ j

4 4 Write d : max { α j 1 α j} and put Then we claim that M : max M 1 T x d { } md γ, 4d c + 1 x j for all C : indeed if c then by 15, M 1 T x md γ T x md γ γ m x j d x j and if instead < c then by 11, Therefore we have that M 1 T x 4d c T x d x d x j M T x T x M 1 T x + T x T x d x j + α 0 x 0 + α i αi 1 xi i1 α 0 x 0 + α i αi 1 xi 0 i N\J This completes the proof We were unable to decide whether or not the converse of Theorem 1 is true However we conjecture that it is: Conjecture Let W α be a weighted shift with weight sequence α {α n } n0 M-hyponormal if and only if α is eventually increasing We now provide evidence for the validity of the conjecture Then W α is Theorem 3 Let T W α be a weighted shift with weight sequence α {α n } n0 If α has exactly two subsequential limits such that the larger one is different from the spectral radius rt of T, then T is not M-hyponormal Proof Suppose that there are infinite sets B and C such that N B C, where i B and C are disjoint; ii β n : α n if n B and γ n : α n if n C; iii β n β and γ n γ; and iv β < γ Assume to the contrary that T is M-hyponormal Then there exists M 1 such that T x M T x for all C and for all x l Suppose β 0 Since γ > 0 we can choose δ such that γ n δ > 0 for all n C Since β n 0, we can find an N B such that β N < δ M Since C is infinite there exists N 0 > N such that N 0 B and N 0 1 C Thus if we tae x e N0, then M T x T x M α N 0 α N 0 1 < M δ M δ 0,

5 which shows that T is not M-hyponormal We now suppose β > 0 Note that span{e : N} is an invariant subspace for T and the restriction of T to such a subspace still yields a weighted shift But since the restriction of an M-hyponormal operator to an invariant subspace is also M-hyponormal we may assume, without loss of generality, that for sufficiently small ɛ > 0, α 0 β 0, β m < γ, β m β < ɛ and γ γ < ɛ for each m B and C If {γ n } occurs infinitely often, in arbitrary long blocs, then γ must be in the approximate point spectrum of T : indeed if {α n } has the consecutive terms such as γ m+1, γ m+,, γ m+ and if f is a unit vector such as f 1 j1 e m+j then T γf 1 1 γ +γ m+1 γ + +γ m+ 1 γ +γm+ ɛ+ γ 0 as 5 But since cf [4, Solution 91] rt lim sup n 1 α n+i 1 γ, it follows that rt γ, which contradicts to our assumption Thus {γ n } occurs infinitely often, in at most finite length of blocs Suppose h is the largest number of such the lengths Let > β be a positive number and choose a sequence {x n } n0 defined by x 0 1 and x n 1 n 1 n α j n 1,, Consider c n z n, where c 0 : 1 and c n : n 1 j0 α j n 1,, If ρ is the radius of convergence of this power series, then n0 x n will converge whenever 1 < ρ, or > 1 ρ : R Thus x n0 x ne n l for > R Observe that if x n0 x ne n l then n0 j0 M T x T x M x 0 + α n x n x n+1 α n x n+1 x n n0 n0 M 1 α n x n x n+1 + α0 x 0 + M 1 x 0 + α n αn 1 xn A straightforward calculation shows that M T x T x M 1 + β n1 βm β m 1x m 1 γn γ n 1x n + γ p βp 1x p + βq γq 1x, where 1 x m + 1 x n + x p + x n1 x n and 31 x γv γ v +e 1 x p and x βw β w +f p+1 1 x e f

6 6 for some v, w, e, f Note that if < γ + ɛ then x m βt β t +c 1 c x β + ɛ c x and x n γu γ u +d 1 d x p γ + ɛ h for some t, u, c, d Thus if β + ɛ < < γ + ɛ, then h x p since d h 3 1 x m c 0 j1 β + ɛ j x 0 j1 β + ɛ j x β + ɛ β + ɛ x, 0 where x q0 x 0 1 and 33 Also we have that 1 x n γ + ɛ h h h x p 34 M T x T x M 1 + β0 + ɛ x m + 1 x n 1 + β + ɛ γ ɛ x, n1 + γ + ɛ β ɛ M 1 + β1 + ɛ x n + γ β x p x x p If R 1 and R are the radii of convergence of x p and x, respectively then by 31, 3 and 33, β R 1 R R γ If R 1 < R, tae R Then x p converges and x Since ɛ was arbitrary it follows from 34 that M T x T x < 0 for > R sufficiently close to R, a contradiction If R 1 R then there are two cases to consider Case 1 R < γ: In this case, tae so that R < < γ v1, and hence γ v 1 > 1 Then we have γ β x p x γ β x p x γv1 p γ β 1 γ v 1 x p If we tae R then x p Since ɛ was arbitrary it follows that M T x T x < 0 for > R sufficiently close to R, a contradiction

7 7 Case R γ: In this case, tae so that > R and hence βs 1 < 1 Then we have γ β x p x γ β x p 1 + γ β x p x γ β x p 1 + γ β x βs1 x q γ β x p 1 + γ β βs1 1 x If we tae R then x Since ɛ was arbitrary it follows that M T x T x < 0 for > R sufficiently close to R, a contradiction This completes the proof Example 4 Let 0 β 0 γ 0 W α : β 0 : l l γ 0 Then W α is M-hyponormal if and only if β γ Proof Since rw α βγ, this follows at once from Theorem 3 Acnowledgements The authors would lie to than the referee for the helpful suggestions References 1 SC Arora and R Kumar, M-hyponormal operators, Yoohama Math J , B P Duggal, On dominant operators, Arch Math Basel , C K Fong, On M-hyponormal operators, Studia Math , PR Halmos, A Hilbert Space Problem Boo, Springer, New Yor, I H Jeon, E Ko, and H Y Lee, Weyl s theorem for ft when T is a dominant operator, Glasg Math J , S K Li and X M Chen, M-hyponormal operators, J Fudan Univ Natur Sci , RL Moore, DD Rogers and TT Trent, A note on intertwining M-hyponormal operators, Proc Amer Math Soc , M Radjabalipour, On majorization and normality of operators, Proc Amer Math Soc , JG Stampfli and BL Wadhwa, An asymmetric Putnam-Fuglede theorem for dominant operators, Indiana Univ Math J , A Uchiyama and T Yoshino, Weyl s theorem for p-hyponormal or M-hyponormal operators, Glasg Math J , BL Wadhwa, M-hyponormal operators, Due Math J , Y Yang, Some results on dominant operators, Internat J Math Math Sci , 17 0 Department of Mathematics, Sungyunwan University, Suwon , Korea address: jsham@mathsuacr Department of Mathematics, Sungyunwan University, Suwon , Korea address: shlee@mathsuacr Department of Mathematics, Seoul National University, Seoul , Korea address: wylee@mathsnuacr