BMAA3-2-21. 338KB Jun 04 2011 12:03:32 AM

ISSN: 1821-1291, URL: http://www.bmathaa.org Volume 3 Issue 2(2011), Pages 213-228.

ON THE CLASS OF n-POWER QUASI-NORMAL OPERATORS

ON HILBERT SPACE

(COMMUNICATED BY SALAH MECHERI)

OULD AHMED MAHMOUD SID AHMED

Abstract. LetT be a bounded linear operator on a complex Hilbert space H. In this paper we investigate some, properties of the class ofn-power quasi-normal operators , denoted [nQN], satisfyingTn

|T|2_{− |T|}2_Tn

= 0 and some relations betweenn-normal operators andn-quasinormal operators.

1. INTRODUCTION AND TERMINOLOGIES

A bounded linear operator on a complex Hilbert space, is quasi-normal ifT and T∗_{T commute. The class of quasi-normal operators was first introduced and} studied by A.Brown [5] in 1953. From the definition, it is easily seen that this class contains normal operators and isometries. In [9] the author introduce the class of n-power normal operators as a generalization of the class of normal operators and study sum properties of such class for different values of the parametern. In particular for n= 2 and n= 3 (see for instance [9,10]). In this paper, we study the bounded linear transformationsT of complex Hilbert space H that satisfy an identity of the form

Tn_T∗_{T =T}∗_{T T}n_{, (1.1)} for some integer n. Operators T satisfying (1.1) are said to be n-power quasi-normal.

LetL(H) =L(H, H) be the Banach algebra of all bounded linear operators on a complex Hilbert spaceH. For T ∈ L(H), we use symbolsR(T) , N(T) and T∗ the range , the kernel and the adjoint ofT respectively,.

LetW(T) = { hT x|xi :x∈H,kxk= 1} the numerical range ofT. A subspace M ⊂ H is said to be invariant for an operator T ∈ L(H) if T M ⊂ M, and in this situation we denote by T|M the restriction of T to M. Let σ(T), σa(T) and σp(T), respectively denote the spectrum, the approximate point spectrum and point spectrum of the operatorT.

2000Mathematics Subject Classification. 47B20, 47B99.

Key words and phrases. Operator, Quasi-normal,n-normal; reducing subspace, Hilbert space. c

2008 Universiteti i Prishtin¨es, Prishtin¨e, Kosov¨e. Submitted July , 2010. Published June 2011.

For any arbitrary operatorT ∈ L(H),|T|= (T∗_T)1 2 and

[T∗_{, T] =T}∗_{T −T T}∗_=|T|2_{− |T}∗_|2 (the self-commutator of T).

An operator T is normal if T∗_{T =T T}∗_{, positive-normal (posinormal) il there} ex-its a positive operatorP ∈ L(H) such that T T∗ _=T∗_{P T, hyponormal if [T}∗_{, T]} is nonnegative(i.e. |T∗_|2 _{≤ |T|}2_{, equivalently kT}∗_{xk ≤ kT xk, ∀ x ∈ H),} quasi-hyponormal if T∗_[T∗_{, T]T is nonnegative, paranormal if kT xk}2 _{≤ kT}2_{xk for all}

x∈H, n-isometry if T∗nTn− n₁

T∗n−1Tn−1+ n₂

T∗n−2Tn−2...+ (−1)nI= 0, m-hyponormal if there exists a positive numberm, such that

m2(T −λI)∗(T −λI)−(T−λI)(T−λI)∗≤0; for all λ∈C_{, .}

Let [N]; [QN]; [H]; and (m−H) denote the classes constituting of normal, quasi-normal, hypoquasi-normal, and m-hyponormal operators. Then

[N]⊂[QN]⊂[H]⊂[m−H]. For more details see [1, 2, 3, 11, 14 ,15].

Definition 1.1. ([7]) An operatorT ∈ L(H)is called(α, β)-normal (0≤α≤1≤ β)if

α2T∗_{T ≤T T}∗_≤β2_T∗_T.

or equivalently

αkT xk ≤ kT∗_{xk ≤βkT xk for all x∈H.}

Definition 1.2. ([9]) Let T ∈ L(H). T is said n-power normal operator for a positive integern if

TnT∗=T∗Tn.

The class of alln-normal operators is denoted by [nN].

Proposition 1.3. ([9]) LetT ∈ L(H), thenT is of class[nN]if and only ifTn _is

normal for any positif integern.

Remark. T isn-power normal if and only ifTn _{is (1,1)-normal.}

The outline of the paper is as follows: Introduction and terminologies are de-scribed in first section. In the second section we introduce the class of n-power quasi-normal operators in Hilbert spaces and we develop some basic properties of this class. In section three we investigate some properties of a class of operators denoted by (Zn_{) contained the class [nQN.]}

2. BASIC PROPERTIES OF THE CLASS [nQN]

In this section, we will study some property which are applied for the n−power quasi-normal operators.

Definition 2.1. For n ∈ N_{, an operator T ∈ L(H) is said to be n-power} quasi-normal operator if

We denote the set ofn-power quasi-normal operators by [nQN]. It is obvious that the class of all n-power quasi-normal operators properly contained classes of n-normal operators and quasi-normal operators, i.e., the following inclusions holds

[nN]⊂[nQN] and [QN]⊂[nQN]. Remark.

(1) A 1-power quasi-normal operator is quasi-normal.

(2) Every quasi-normal operator is n-power quasi-normal for eachn.

(3) It is clear that a n-power normal operator is also n-power quasi-normal. That the converse need not hold can be seen by choosingT to be the

unilat-eral shift, that is, if H =l2_{, the matrix T =}



  

0 0 0 . . . 1 0 0 . . . 0 1 0 . . . . . . .



  

.It is

easily verified thatT2_T∗_−T∗_T2_{6= 0 and(T}2_T∗_−T∗_T2_{)T = 0. So thatT}

is not 2-power normal but is a 2-power quasi-normal.

Remark. An operatorT isn-power quasi-normal if and only if

[Tn, T∗_{T] = [T}n_{, T}∗_{]T = 0.}

Remark. An operatorT isn-power quasi-normal if and only if

Tn_|T|2_=|T|2_Tn_. First we record some elementary properties of [nN Q] Theorem 2.2. If T ∈[nQN],then

(1) T is of class [2nQN].

(2) if T has a dense range in H , T is of class [nN]. In particular, if T is invertible, thenT−1 _{is of class [nQN].}

(3) If T and S are of class [nQN]such that [T, S] = [T, S∗_{] = 0,then T S is}

of class[nQN].

(4) If S and T are of class [nQN] such that ST = T S = T∗_{S =ST}∗ _{= 0,,}

thenS+T is of class[nQN]. Proof.

(1) SinceT is of [nQN],then

Tn_T∗_{T =T}∗_{T T}n_{. (2.1)} Multiplying (2.1) to the left byTn_{, we obtain}

T2nT∗_{T =T}∗_{T T}2n_. ThusT is of class [2nQN].

(2) SinceT is of class [nQN], we have fory∈R(T) :y=T x, x∈H, and k(Tn_T∗_−T∗_Tn_)yk=k(Tn_T∗_−T∗_Tn_{)T xk=k(T}n_T∗_{T −T}∗_Tn+1_{)xk= 0.}

Thus, T isn-power normal onR(T) and henceT is of class [nN].In case T invertible, then it is an invertible operator of class [nN] and so

Tn_T∗_=T∗_Tn_. This in turn shows that

which prove the result. (3)

(T S)n(T S)∗_{T S = T}n_Sn_T∗_S∗_{T S=T}n_T∗_{T S}n_S∗_S = T∗_Tn+1_S∗_Sn+1_{= (T S)}∗_{(T S)}n+1_. Hence,T Sis of class [nQN].

(4)

(T+S)n(T+S)∗(T+S) = (Tn+Sn)(T∗T+S∗S) = Tn_T∗_T+Sn_S∗_S = T∗_Tn+1_+S∗_Sn+1 = (T+S)∗_(T+S)n+1_. Which implies thatT +S is of class [nQN].

Proposition 2.3. If T is of class[nQN]such thatT is a partial isometry, thenT

is of class [(n+ 1)QN].

Proof. SinceT is a partial isometry, therefore

T T∗_{T =T [4], p.153). (2.2)} Multiplying (2.2) to the left byT∗_Tn+1_{and using the fact thatT is of class [nN Q],} we get

T∗_Tn+2 _{= T}∗_Tn+2_T∗_T = Tn_T∗_{T.T T}∗_T = Tn+1T∗T, which implies thatT is of class [(n+ 1)QN].

The following examples show that the two classes [2N Q] and [3N Q] are not the same.

Example 2.4. Let H =C3 _{and letT =}





−1 0 0 0 0 0 1 0 1



∈ L(C3).Then by simple

calculations we see thatT is not of class [3QN] but of class[2QN].

Example 2.5. LetH =C3_{and letS =} 



1 1 1

0 0 0

−1 0 −1



∈ L(C3).Then by simple

calculations we see thatS is not of class [2QN] but of class [3QN]. Proposition 2.6. LetT ∈ L(H)such thatT is of class[2QN]T

[3QN],thenT is of class[nQN] for all positive integern≥4.

Proof. We proof the assertion by using the mathematical induction. Forn= 4 it is a consequence of Theorem 2.2. 1 .

We prove this forn= 5. SinceT ∈[2QN] ,

T2T∗_{T =T}∗_T3_{, (2.3)}

multiplying (2.3) to the left byT3 _{we get}

Thus we have

T5T∗_{T = T}3_T∗_T3 = T∗_T4_T2 = T∗_T6_. Now assume that the result is true forn≥5 i.e

Tn_T∗_{T =T}∗_{T T}n_, then

Tn+1T∗T = T T∗Tn+1 = T T∗_T3_Tn−2 = T3T∗T Tn−2 = T∗_T4_T∗(n−2) = T∗_Tn+2_. ThusT is of class [(n+ 1)QN].

Proposition 2.7. If T is of class [nQN] such that N(T∗_{)⊂N(T), then T is of}

class[nN].

Proof. In view of the inclusion N(T∗_{) ⊂ N(T), it is not difficult to verify the} normality ofTn_.

Next couple of results shows that [nQN] is not translation invariant Theorem 2.8. If T andT−I are of class [2QN], thenT is normal.

Proof. First we see that the condition onT−I implies

T2(T∗_T)−T2_T∗_−2T(T∗_{T) + 2T T}∗_{= (T}∗_T)T2_−T∗_T2_−2(T∗_{T)T+ 2T}∗_T. SinceT is of class [2QN], we have

−T2_T∗₋

2T(T∗T) + 2T T∗=−T∗

T2−2(T∗T)T+ 2T∗T, or

−T T∗2−2(T∗T)T∗+ 2T T∗ =−T∗2_T−2T∗

(T∗T) + 2T∗T (2.4) We first show that (2.4) implies

N(T∗_{)⊂N(T) (2.5)}

SupposeT∗_{x= 0.From (2.4), we get}

−3T∗2T x+ 2T∗T x= 0. (2.6) Then

−3T∗3_{T x+ 2T}∗2_{T x= 0.} Therefore, asT is of class [2QN],

−3T∗_{T T}∗2_{x+ 2T}∗2_{T x= 0} and hence

2T∗2_{T x= 0.}

Consequently, (2.6) gives 2T∗_{T x= 0 orT x= 0.This proves (2.5). As observe in} Proposition 2.7 and Proposition 1.3T2 _{is normal. This along with (2.4) gives}

or

T∗_(T∗_{T−T T}∗_{) =T}∗_{T−T T}∗_{. (2.7)} IfN(T∗_{−I) ={0}, then (2.7) impliesT is normal.}

Now assume thatN(T∗_{−I) is non trivial. LetT}∗_{x=x.Then (2.6) gives} T∗2T x−T∗T x=T∗T x−T x.

SinceT∗2_{T =T T}∗2_{,we have}

T∗_{T x=T x.} Therefore

||T x||2=< T∗_{T x|x >=< T x|x >=< x|T}∗_{x >=||x||}2_. Hence

||T x−x||2 _{= ||T x||}2_+||x||2_{−2Re < T x|x >} = ||T x||2− ||x||2

= 0.

OrT x=x.ThusN(T∗_{−I)⊂N(T−I).This along with (2.7), yields} T(T∗_{T−T T}∗_{) =T}∗_{T−T T}∗

and so

T(T∗_{T−T T}∗_{)T = (T}∗_{T−T T}∗_)T or

T T∗_T2_−T2_T∗_{T =T}∗_T2_{−T T}∗_T.

SinceT2_T∗ _=T∗_T2 _andT3_T∗ _=T∗_T3 _{we deduce that T}∗_T2 _{=T T}∗_{T.Thus T is} quasinormal. From (2.5), the normality ofT follows.

In attempt to extend the above result for operators of class [nQN], we prove Theorem 2.9. If T is of class [2QN]∩[3QN] such that T−I is of class [nQN], thenT is normal.

Proof. SinceT−I is of class [nQN], we have n

X

k=1

akTk_T∗_T− n

X

k=1

akTk_T∗_=T∗_T n

X

k=1

akTk_−T∗ n

X

k=1

akTk_{, ak= (−1)}n−k₍n k). Under the condition onT, we have by Proposition 2.6

a1T(T∗T)−(

n

X

k=1

akTk_)T∗_=a

1(T∗T)T−T∗(

n

X

k=1

akTk₎ or

a1(T∗T)T∗−T(

n

X

k=1

akT∗k_{) =a}

1T∗(T∗T)−(

n

X

k=1

akT∗k_{)T. (2.8)} (2.8) implies thatN(T∗_{)⊂N(T). In fact, letT}∗_{x= 0. From (2.8),we have}

a1T∗2T x−(

n

X

k=1

akT∗k)T x= 0. T is of class [2QN] and of class [3QN], we deduce that

and hence

a1T∗3T x−a1T∗2T−a2T∗3T x= 0 Hence

a1T∗2T x.

Consequently (2.9) givesT∗_{T x= 0, which implies thatT x= 0.}

It follows by Proposition 2.7 thatTk _{is normal fork= 2,3, ..., nand hence} T(T∗_{T)−T T}∗_{= (T}∗_T)T−T∗_T

or

T∗(T T∗−T∗T) =T T∗−T∗T. Hence,

(T∗_{−I)(T T}∗_−T∗_{T) = 0.}

A similar argument given in as in the proof of Theorem 2.8 gives the desired result.

Theorem 2.10. IfT andT∗ _{are of class [nQN], thenT}n _{is normal.} First we establish

Lemma 2.11. IfT is of class[nQN],then N(Tn_)⊂N(T∗n_{) forn≥2.} Proof. SupposeTn_{x= 0.Then}

T∗n_(T∗_T)Tn−1_{x= 0.} By hypotheses,

T∗_{T T}∗n_Tn−1_{x= 0,} which implies

T T∗n_Tn−1_{x= 0.} Hence

T∗n_Tn−1_{x= 0.} Under the condition onT, we have

T∗T T∗nTn−2x= 0 Hence

T∗nTn−2x= 0. By repeating this process we can find

T∗n_{x= 0.}

Proof of Theorem 2.10. By hypotheses and Lemma 2.11 N(T∗n_{) =N(T}n_).

SinceT is of [nQN],[Tn_T∗_−T∗_Tn_]Tn _{= 0,i.e. [T}n_T∗_−T∗_Tn_{] = 0 on clR(T).} also the fact thatN(T∗_{) is a subset ofN(T}n_{) gives [T}n_T∗_−T∗_Tn_{] = 0 onN(T}∗_). Hence the result follows.

Theorem 2.12. IfT andT2 _{are of class[2QN], andT is of class[3QN], thenT}2

Proof. The condition thatT2_{is of class [2QN] gives}

T∗4(T∗2T2) = (T∗2T2)T∗4 Implies

T∗5_(T∗_{T)T = (T}∗2_T2_)T∗4 SinceT if of class [3QN],we have

T∗2_(T∗_T)T∗3_{T = (T}∗2_T2_)T∗4 _. And hence

T∗2(T∗T)2T∗2= (T∗2T2)T∗4 [T is of class [2QN]]. Implies

(T∗_T)2_T∗4_{= (T}∗2_T2_)T∗4 _{[Tis of class [2QN]]} or

T4((T∗_T)2_−T∗2_T2_{) = 0.} By Lemma 2.11,

T∗2_T2_((T∗_T)2_−T∗2_T2_{) = 0} or

T2[(T∗T)2−T∗2T2)] = 0. (2.10) Hence

T∗2_[((T∗_T)2_−T∗2_T2_{)] = 0, [N(T}2_{) is a subset of N(T}∗2_)]. Or

[((T∗_T)2_−T∗2_T2_)]T2_{= 0. (2.11)} SinceT is of class [2QN],T2_{commutes with (T}∗_T)2_{. Hence from (2.10) and (2.11),} we get the desired conclusion.

Theorem 2.13. If T and T2 _{are of class [2QN] andN(T)⊂N(T}∗_{), then T}2 _is

quasinormal.

Proof. By the condition thatT2 _{is of class [2QN],we have} (T∗2_T2_)T∗4 _{= T}∗4_(T∗2_T2₎

= T∗_T∗4_(T∗_T)T

= T∗_(T∗_T)T∗4_{T [T is of class [2QN]]} = T∗(T∗T)T∗(T∗T)T∗2

Thus we have

{(T∗2_T2_)T∗2_−[T∗_(T∗_T)]2_}T∗2_{= 0} or

T2{T2_(T∗2_T2_)−[(T∗_T)T]2_{}= 0.} Then under the kernel condition

T{T2(T∗2_T2_)−[(T∗_T)T]2_{}= 0} or

{(T∗2_T2_)T∗2_−[T∗_(T∗_T)]2_{}x= 0 for x∈clR(T}∗_). SinceN(T)⊂N(T∗_),

Thus

{(T∗2_T2_)T∗2_−[T∗_(T∗_T)]2_{}= 0} or

T2(T∗2T2) = [(T∗T)T]2 = T∗_T2_T∗_T2 = T∗T2(T∗T)T

= T∗_(T∗_T)T3 _{[Tis of class [2QN]} = (T∗2_T2_)T2_.

This proves the result.

Theorem 2.14. Let T be an operator of class [2QN] with polar decomposition

T =U|T|. If N(T∗_{)⊂N(T), then the operator S with polar decompositionU}2_|T|

is normal.

Proof. It follows by Proposition 2.7 thatT2 _{is normal andN(T}∗_{) =N(T}∗2_{) and} by Lemma 2.11 we have

N(T) =N(T∗). (2.12) As a consequence,U turns out to be normal and it is easy to verify that

|T|U|T|2U∗_|T|=|T|U∗_|T|2_U|T|. Since

N(|T|) =N(U) =N(U∗_), |T|U|T|2_U∗_=|T|U∗_|T|2_U and hence

U|T|2_U∗_=U∗_|T|2_U. Again by the normality ofU , we have

U|T|U∗_=U∗_{|T|U (2.13)}

Also U∗2_U2 _{= U}∗_{U, showing U}2 _{to be normal partial isometry with N(U}2_{) =}

N(|T|). Thus U2_{|T| is the polar decomposition Note that (2.13) the normality} shows thatU2 _{and|T|are commuting. Consequently}

(U2|T|)∗_(U2_{|T|) = |T|U}∗2_U2_|T| = |T|U2_U∗2_|T| = (U2|T|)(U2|T|)∗_. This completes the proof.

Corollary 2.15. If T is of class [2QN]and0∈/W(T),then T is normal

Proof. Since 0∈/ W(T) givesN(T) =N(T∗_{) ={0}and so by our Proposition 2.7} ,T2 _{is normal. Then [T}∗_{T, T T}∗_{] = 0.Now the conclusion follows form [8].}

Theorem 2.16. Let T is of class [2QN] such that [T∗_{T, T T}∗_{] = 0. Then T}2 _is

Proof.

(T∗2_T2_)T2 _{= T}∗_(T∗_T)T3 = T∗_T2_T∗_T2 = (T∗T)(T T∗)T2 = (T T∗_)(T∗_T)T2 = T T∗_T2_T∗_T = T(T∗_{T)(T T}∗_)T = T(T T∗)(T∗T)T = T2(T∗2_T2_). This proves the result.

Theorem 2.17. If T is of class[2QN] and[3QN]with N(T)⊂N(T∗_{), thenT is}

quasinormal.

Proof.

T∗3_(T∗_{T) = T}∗_(T∗_T)T∗2 _{[T is of class [2QN]]} = (T∗_T)T∗3

Hence

[T∗2T−T∗T T∗]T∗2= 0 or

T2[T∗T2−T T∗T] = 0, SinceN(T)⊂N(T∗_{), N(T) =N(T}2_{) and therefore}

T[T∗_T2_{−T T}∗_{T] = 0, or [T}∗2_{T −T}∗_{T T}∗_]T∗_{= 0.} Again byN(T)⊂N(T∗_{),we get the desired conclusion.}

Theorem 2.18. If an operator T of class [2QN] is a 2-isometry , then it is an isometry.

Proof. By the definition of a 2-isometry,

(T∗2T2)(T∗T)−2(T∗T)2+T∗T= 0. SinceT is of class [2QN]

T∗2_(T∗_T)T2_−2(T∗_T)2_+T∗_{T = 0,} that is

T∗3_T3_−2(T∗_T)2_+T∗_{T = 0. (2.14)} Also

T∗_[T∗2_T2_−2T∗_{T+I]T = 0} i.e.

T∗3_T3_−2T∗2_T2_+T∗_{T = 0. (2.15)} From (2-14) and (2-15)T∗2_T2_{= (T}∗_T)2_{and hence}

(T∗T)2−2(T∗T) +I=T∗2T2−2T∗T+I= (T∗T−I)2= 0 or

Theorem 2.19. If An operatorT is of class[2QN]∩[3QN]is ann-isometry, then

T is an isometry.

Proof. By the definition ofn-isometry, T∗nTnT∗_T− n

1

T∗n−1Tn−1T∗_T+...+(−1)n−2 n n−2

T∗2_T2_T∗_T+(−1)n−1₍n

n−1)T∗T T∗T+(−1)nT∗T = 0. SinceT is of class [2QN]∩[3QN,] we have by Proposition 2.6

T∗n+1_Tn+1₋ n

1

T∗n_Tn_+...+(−1)n−2 n n−2

T∗3_T3₊₍₋₁₎n n n−1

(T∗_T)2₊₍₋₁₎n_T∗_{T = 0.} (2.16) Also

T∗_[T∗n_Tn₋ n

1

T∗n−1_Tn−1_{+...+ (−1)}n−1 n n−1

T∗_{T+ (−1)}n_{I]T = 0} i.e.

T∗n+1_Tn+1₋₍n

1)T∗nTn+....+ (−1)n−1

n n−1)T

∗2_T2_{+ (−1)}n_T∗_{T = 0 (2.17)} From (2.16) and (2.17)T2∗_T2_{= (T}∗_T)2_{.Consequently (T}∗₎k_Tk _{= (T}∗_T)k_{, ∀ k∈}

N_{,and hence}

(T∗_T)n₋ n

1

(T∗_T)n−1_{+....+ (−1)}n−1 n n−1

(T∗_{T) + (−1)}n_{I= 0 = (I−T}∗_T)n_. This completes the proof.

Definition 2.20. An operatorA∈ L(H)is said to be quasi-invertible ifAhas zero kernel and dense range.

Definition 2.21. ([18]) Two operatorsS andT inL(H)are quasi-similar if there are quasi-invertible operatorsA andB in L(H) which satisfy the equations

AS=T A and BT =SB.

IfM is a closed subspace ofH,H =M⊕M⊥_{. IfT is inL(H), thenT can be} written as a 2×2 matrix with operators entries,

T =

W X

Y Z

where W ∈ L(M), X ∈ L(M⊥_{, M), Y ∈ L(M, M}⊥_{), and Z ∈ L(M}⊥_{) (cf.} Conway [6]).

Proposition 2.22. IfS andT are quasi-similarn-power quasi-normal operators in

L(H)such thatN(S) =N(T),N(T)andN(S)are reducing respectively forT and

S, then S1 =S|N(S)⊥ andT1 =T|N(T)⊥ are quasi-similar n-power quasi-normal

operators.

Proof. Since S and T are quasi-similar, there exists quasi-invertible operators A and B such that AS = T A and SB =BT. the N(S) is invariant under both A and B . Thus the matrices of S, T, A and B with respect to decomposition H =N(S)⊕N(S)⊥ _are

S1 O

O O

,

T1 O

O O

,

A1 O

A2 A3

,

B1 O

B2 B3

respectively. It is easy to verify that the ranges ofA1 andB1 are dense inN(S)⊥. We now show thatN(A1) =N(B1) ={0}.

{0}. ThereforeA1 and B1 are quasi-invertible operators onN(S)⊥ and equations

AS =T A and SB = T B imply that A1S1 =T1A1 and S1B1 =B1T1. Hence S1 and T1 are quasi-similar. By a similar way as in [10, Proposition 2.1.(iv)] we can see that the operatorsS1andT1 aren-power quasi-normal.

3. THE (Zn_{)-CLASS OPERATORS}

In this section we consider the class (Zn

α) of operatorsT satisfies |Tn_T∗

T−T∗T Tn|α_≤c2

α(T−λI)∗n(T −λI)n, for all λ∈C

and for a positiveα. The motivation is due to S. Mecheri [13] who considered the class of operatorsT satisfying

|T T∗₋

T∗T|α_≤c2

α(T−λI) ∗

(T−λI)

and A. Uchiyama and T. Yoshino [19] who discussed the class of operators T sat-isfying

|T T∗_−T∗_T|α_≤c2

α(T−λI)(T−λI) ∗_.

Definition 3.1. For T ∈ L(H) we say thatT belongs to the class (Zn

α) for some α≥1if there is a positive numbercα such that

|Tn_T∗_T−T∗_Tn+1_|α_≤c2

α(T−λI)∗n(T−λI)n for all λ∈C,

or equivalently, if there is a positive numbercα such that

k|TnT∗_T−T∗_{T T}n_|α

2xk ≤cαk(T−λI)nxk,

for allxinH andλ∈C_{. Also, let}

Zn₌ [

α≥1

Zn

α.

Remark. An operatorT of class [nQN], it is of class (Zn_).

In the following examples we give an example of an operator not in the classes

Zn _{,and an operator of these classes, which are not of class [nQN].}

Example 3.2. If f is a sequence of complex numbers,f =hf(0), f(1), f(2), ... iT_.

Thep-Ces´aro operatorsCp acting on the Hilbert space l2 _{of square-summable}

com-plex sequencesf is defined by

(Cpf)(k) = 1 (k+ 1)p

k

X

i=1

f(i) for fixed real p >1 and k= 0,1,2, ....

These operators was studied extensively in [16] where it was shown, that these

op-erators are bounded and (C∗

pf)(k) = ∞

X

i=k 1

(i+ 1)pf(i).

In matrix form, we have

Cp =     

1 0 0 . . . (1

2)p ( 1

2)p 0 . . . (1

3)p ( 1 3)p (

1 3)p . . .

We consider the sequencef defined as follows f(0) = 1 and f(k) =

k

Y

j=1

jp

(1 +j)p₋₁ for k≥1.

In [16] it is verified thatf ∈l2 _{, is eigenvector forCpassociated with eigenvalue 1,} sof ∈N(Cp−I), butf /∈N(C∗

p −I). It follows thatk(Cp−I)nf)k= 0. On the other hand, we have

(CpnC ∗ pCp−C

∗

pCpCpn)f = (Cpn−I)C ∗ pf 6= 0. Hence,Cp is a bounded operator but not of classesZn_.

Example 3.3. Let T be a weighted shift operator onl2_{with weightsα}

1= 2, αk = 1

for allk≥2.That is

Tα(x1, x2, x3, ...) = (0, α1x1, α2x2, ...) and T∗(x1, x2, ...) = (α1x2, α2x3, ...).

A simple computation shows that

(TnT∗T −T∗T Tn)(x) = (0,0, ...,0,6x1,0,0, ...)

with6x1 at the(n+ 1)th place.

Morover

(T∗n_T∗_T−T∗_{T T}∗n_{)(x) = (−6xn}

+1,0,0, ...)and |TnT∗T−T∗T Tn|2x= (−36x1,0,0, ...).

Therefore T is not of class [nQN] and howeverT is of class Zn

4 ⊆Zn.

Lemma 3.4. For each α, βsuch as 1≤α≤β, we haveZn

α⊆Znβ. Proof.

|Tn_T∗

T−T∗Tn+1|β _{= |T}n_T∗

T−T∗Tn+1|α

2|TnT∗T−T∗Tn+1|β−α|TnT∗T−T∗Tn+1|

α

2

≤ kTn_T∗

T −T∗Tn+1kβ−α_|Tn_T∗

T−T∗Tn+1|α ≤ (2kTkn+2₎β−α_c2

α(T −λI)∗n(T−λI)n = c2

β(T−λI)∗n(T−λI)n where

Cβ2= (2kTkn+2)β−αc2α.

There exists an Hilbert space H◦_{: H ⊂ H}◦_{, and an isometric *-homomorphism} preserving order , i.e, for allT, S∈ L(H) andλ, µ∈C_{,we have}

Proposition 3.5. ([6],[13] Berberian technique) LetH be a complex Hilbert space. Then there exists a Hilbert spaceH◦_{⊃H and a map}

Φ :L(H)→ L(H◦_{) :T 7−→T}◦

satisfying: Φis an *-isometric isomorphism preserving the order such that

1. Φ(T∗_{) = Φ(T)}∗_.

2. Φ(λT +µS) =λΦ(T) +µΦ(S).

3. Φ(IH) =IH◦.

4. Φ(T S) = Φ(T)Φ(S).

5. kΦ(T)k=kTk.

6. Φ(T)≤Φ(S) if T ≤S.

7. σ(Φ(T)) =σ(T), σa(T) =σa(Φ(T)) =σp(Φ(T)).

Lemma 3.6. If an operatorT is of class [nQN], then Φ(T)is of class [nQN].

Lemma 3.7. If an operatorT is of class Zn_{,then Φ(T)is of class Z}n_. Proof. SinceT is of classZn_{, there existsα≥1 andc}

α>0 such that |Tn_T∗_T−T∗_Tn+1_|α_≤c2

α(T−λ)∗n(T−λI)n for all λ∈C. It follows from the properties of the map Φ that

Φ(|Tn_T∗_T−T∗_Tn+1_|α_)≤Φ(c2

α(T −λ)∗n(T−λI)n) for all λ∈C. By the condition 8. above we have

Φ(|Tn_T∗

T−T∗Tn+11|α_{) =|Φ(|T}n_T∗

T−T∗Tn+1|)|α_{, for all α >0.} Therefore

|Φ(T)n_Φ(T∗_{)Φ(T)−Φ(T}∗_)Φ(T)n+1_{| ≤Φ(c}2

α(T−λ)∗n(T−λI)n) for all λ∈C. Hence Φ(T) is of classZn_.

Proposition 3.8. Let T be a class Zn _{operator and assume that there exists a} subspaceM_{that reduces T, thenT|}M_{is of class} Zn _operator.

Proof. SinceT is of classZn_{, there exists an integerp≥1 andcp>0 such that} k|Tn_T∗_T−T∗_Tn+1_|2p−1

xk ≤c2pk(T−λI)nxk, for all x ∈H, for all λ∈C. M_{reducesT,T can be written respect to the compositionH =}M_⊕M⊥_{as follows:}

T =

A O O B

, By a simple calculation we get

TnT∗T−T∗Tn+1=

An_A∗_A−A∗_An+1 _O

O Bn_B∗_B−B∗_Bn+1

By the uniqueness of the square root, we obtain

|Tn_T∗_T−T∗_Tn+1_|=

|An_A∗_A−A∗_An+1_{| O}

O |Bn_B∗_B−B∗_Bn+1_|

. Now by iteration to the order 2p_{, it results that}

|Tn_T∗_T−T∗_Tn+1_|2p−1

= |A

n_A∗_A−A∗_An+1_|2p−1

0

0 |Bn_B∗_B−B∗_Bn+1_|2p−1

!

. Therefore for allx∈M_{, we have}

k|Tn_T∗_T−T∗_Tn+1_|2p−1

xk=k|An_A∗_A−A∗_An+1_|2p−1

xk ≤c2pk(T−λI)xk=k(A−λI)nxk.

HenceA is of classZn₂p⊂Z

n_. Theorem 3.9. Let T of class Z1_.

(1) If λ∈σp(T), λ6= 0, thenλ∈σp(T∗_{), furthermore if λ6=µ, then Eλ (the}

proper subspace associated withλ) is orthogonal to Eµ.

(2) If λ∈σa(T),then λ∈σa(T∗_). (3) T T∗_{T −T}∗_T2 _{is not invertible.}

(1) IfT ∈Z1_{, thenT ∈Z}1

αfor someα≥1 and there exists a positive constant cα such that

|T T∗_T−T∗_T2_|α_{≤cα(T−λI)}∗_{(T−λI) for all λ∈C.} As T x =λx implies|T T∗_{T −T}∗_T2_|α

2x= 0 and (T T∗−T∗T)x= 0 and hence

k(T−λ)∗_{xk=k(T−λ)xk}

λhx|yi=hλx|yi=hT x|yi=hx|T∗_yi

=hx|µyi=µhx|yi. Hence

hx|yi= 0.

(2) Let λ∈σa(T) from the condition 7. above, we have σa(T) =σa(Φ(T)) =σp(φ(T)).

Thereforeλ∈σp(φ(T)).By applying Lemma 3.7 and the above condition 1., we get

λ∈σp(Φ(T)∗_{) =σp(Φ(T}∗_)).

(3) Let T ∈Z1_{. then there exists an integerp≥1 andcp>0 such that}

k|T T∗_{T −T}∗_T2_|2p−1

xk ≤c2

pk(T−λI)xk2 for all x∈H and for all λ∈C. It is know that σa(T) 6= ∅. If λ ∈ σa(T), then there exists a normed sequence (xm) inH such that k(T−λI)xmk −→0 as m−→ ∞.Then

(T T∗_T−T∗_T2_{)xm−→0 as m−→ ∞} and so, (T T∗_T−T∗_T2_{) is not invertible.}

Acknowledgments. The author would like to express his cordial gratitude to the referee for valuable advice and suggestions.

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Mathematics Department , College of Science. Aljouf University Aljouf 2014. Saudi Arabia