AAA Abstract and Applied Analysis 1687-0409 1085-3375 Hindawi Publishing Corporation 204031 10.1155/2012/204031 204031 Research Article More on (α,β)-Normal Operators in Hilbert Spaces Eskandari Rasoul Mirzapour Farzollah Morassaei Ali Teschke Gerd 1 Department of Mathematics Faculty of Sciences University of Zanjan Zanjan 45195-313 Iran znu.ac.ir 2012 16 8 2012 2012 18 11 2011 03 07 2012 20 07 2012 2012 Copyright © 2012 Rasoul Eskandari et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We study some properties of (α,β)-normal operators and we present various inequalities between the operator norm and the numerical radius of (α,β)-normal operators on Banach algebra () of all bounded linear operators T:, where is Hilbert space.

1. Introduction

Throughout the paper, let () denote the algebra of all bounded linear operators acting on a complex Hilbert space (,·,·), h() denote the algebra of all self-adjoint operators in (), and I is the identity operator. In case of dim=n, we identify () with the full matrix algebra n() of all n×n matrices with entries in the complex field. An operator Ah() is called positive if Ax,x0 is valid for any x, and then we write A0. Moreover, by A>0 we mean Ax,x>0 for any x. For A,Bh(), we say AB if B-A0. An operator A is majorized by B, if there exists a constant λ such that AxλBx for all x or equivalently A*Aλ2B*B .

For real numbers α and β with 0α1β, an operator T acting on a Hilbert space is called (α,β)-normal [2, 3] if (1.1)α2T*TTT*β2T*T. An immediate consequence of above definition is (1.2)α2T*Tx,xTT*x,xβ2T*Tx,x, from which we obtain (1.3)αTxT*xβTx, for all x.

Notice that, according to (1.1), if T is (α,β)-normal operator, then T and T* majorize each other.

In , Moslehian posed two problems about (α,β)-normal operators as follows.

For fixed α>0 and β1,

give an example of an (α,β)-normal operator which is neither normal nor hyponormal;

is there any nice relation between norm, numerical radius, and spectral radius of an (α,β)-normal operator?

Dragomir and Moslehian answered these problems in , as more as, they propounded a nice example of (α,β)-normal operator that is neither normal nor hyponormal, as follows.

The matrix (1011) in (2) is an (α,β)-normal with α=(3-5)/2 and β=(3+5)/2.

The numerical radius w(T) of an operator T on is defined by (1.4)w(T)=sup{|Tx,x|:x=1}. Obviously, by (1.4), for any x we have (1.5)|Tx,x|w(T)x2. It is well known that w(·) is a norm on the Banach algebra () of all bounded linear operators. Moreover, we have (1.6)w(T)T2w(T)(TB(H)). For other results and historical comments on the numerical radius see .

The antieigenvalue of an operator T() defined by (1.7)μ1(T)=infTx0ReTx,xTxx. The vector x which takes μ1(T) is called an antieigenvector of T. We refer more study on this matter to .

In this paper, we prove some properties of (α,β)-normal operators and state various inequalities between the operator norm and the numerical radius of (α,β)-normal operators in Hilbert spaces.

2. Some Properties of <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M73"><mml:mo stretchy="false">(</mml:mo><mml:mi>α</mml:mi><mml:mo>,</mml:mo><mml:mi>β</mml:mi><mml:mo stretchy="false">)</mml:mo></mml:math></inline-formula>-Normal Operators

In this section, we establish some properties of (α,β)-normal operators. It is easy to see that if T is an (α,β)-normal (α>0) then T* is (1/β,1/α)-normal. We find numbers z such that z+T is (α,β)-normal where T is (α,β)-normal.

We know by the Cauchy-Schwartz inequality that -1μ1(T)1. Also we can write (2.1)μ1(T)=infx=1Tx0ReTx,xTx. We define (2.2)μ2(T)=supx=1Tx0ReTx,xTx.

We know that if T is normal operator then z+T is also normal.

Theorem 2.1.

Let T be an (α,β)-normal operator on a Hilbert space such that 0α<1<β and z. Then z+T is (α,β)-normal, if provided one of the following conditions holds:

μ1(z¯T)0,

μ1(z¯T)<0,|z|2-2|z|Tμ1(z¯T).

Proof.

In both of above cases, we show that (2.3)|z|2+2Rez¯Tx,x0,xHwith  x=1,Tx0. By the assumption (i), μ1(z¯T)0, we have Rez¯Tx,x/|z|Tx0 for every x with x=1 and Tx0, consequently we get Rez¯Tx,x0, and therefore (2.3) is valid. On the other hand, if (ii) holds and we set B=μ1(z¯T) then we get BRez¯Tx,x/|z|Tx for every x with x=1 and Tx0, consequently: (2.4)inf{BTx:x=1,Tx0}inf{TxRez¯Tx,x|z|Tx:x=1,Tx0}. Since B<0, we obtain (2.5)-Binf{-Tx:x=1,Tx0}inf{TxRez¯Tx,x|z|Tx:x=1,Tx0}, and so (2.6)Bsup{Tx:x=1,Tx0}inf{TxRez¯Tx,x|z|Tx:x=1,Tx0}. Now, by using the last inequality, we have (2.7)|z|2+2|z|Tμ1(z¯T)=|z|2+2|z|(supx=1Tx0Tx)(infx=1Tx0{Rez¯Tx,x|z|Tx})|z|2+2|z|infx=1{TxRez¯Tx,x|z|Tx}=|z|2+2infx=1{Rez¯Tx,x}. This shows that (2.3) holds for (ii), too. Thus, for any x with x=1 we have (2.8)α2(z+T)*(z+T)x,x=α2[|z|2x,x+z¯Tx,x+zT*x,x]+α2T*Tx,x|z|2x,x+z¯Tx,x+zT*x,x+TT*x,x=(z+T)(z+T)*x,xβ2[|z|2x,x+z¯Tx,x+zT*x,x]+β2T*Tx,x=β2(z+T)*(z+T)x,x and this completes the proof.

Corollary 2.2.

Let T be an (α,β)-normal operator. We have the following.

If μ1(T)0 then z+T is (α,β)-normal operator for any z>0.

If μ2(T)0 then z+T is (α,β)-normal operator for any z<0.

Proof.

(i) By the definition of the first antieigenvalue of T, for all z>0 we have (2.9)μ1(z¯T)=μ1(zT)=μ1(T)0. By using Theorem 2.1(i) we imply that z+T is an (α,β)-normal.

(ii) If z<0, then (2.10)μ1(z¯T)=-μ2(T)0. By using Theorem 2.1(i) we imply that z+T is an (α,β)-normal.

Corollary 2.3.

Let T be an injective and (α,β)-normal operator with α>0. Then

(T) is dense,

T* is injective,

if T is surjective then T-1 is also (α,β)-normal.

Proof.

Since the inequality (1.3) is valid, we obtain 𝒩(T*)=𝒩(T), and therefore (T)=𝒩(T*)=𝒩(T)=0, thus (T) is a dense subspace of and T* is injective. This proves (i) and (ii).

To prove (iii), we note that since T is surjective, we imply that T is invertible. On the other hand we have (T*)-1=(T-1)*. Also we know that if A and B are two positive and invertible operators with 0<AB then B-1A-1. Since T is (α,β)-normal, by taking inverse from all sides of (1.1), we get (2.11)1β2T-1(T*)-1(T*)-1T-11α2T-1(T*)-1. This means that (T-1)* is (1/β,1/α)-normal, thus T-1 is (α,β)-normal.

Example 2.4.

Consider the following matrix T in (2): (2.12)T=(1011).T is an (α,β)-normal operator, with parameters α=(3-5)/2 and β=(3+5)/2. Then T-1=(10-11) is (α,β)-normal.

For T() we call (2.13)r(T)=sup{|λ|:λσ(T)} the spectral radius of T, where σ(T) is the spectrum of T and it is known that r(T)=limnTn1/n [5, page 102].

Theorem 2.5.

Let T be an (α,β)-normal operator such that T2n is (α,β)-normal operator for every n, too. Then, we have (2.14)1βTr(T)T.

Proof.

For any T() we have (2.15)T*T=T2. In particular, if T is a self-adjoint operator then T2=T2. Thus, by the definition of (α,β)-normal operator, we have (2.16)T*2T21β2(T*T)2=1β2T4. By induction on n, we imply that (2.17)T*2nT2n1β2n+1-2T2n+1, from which we obtain (2.18)r(T)2=r(T*)r(T)=limn(T*2nT2n)1/2nlimnT*2nT2n1/2nlimn(1β2n+1-2T2n+1)1/2n=1β2T2limn1β-2/2n=1β2T2. Therefore, we get (1/β)Tr(T)T. This completes the proof.

Below, we give an example of (α,β)-normal operator such that it satisfies in Theorem 2.5.

Example 2.6.

Assume that is a separable Hilbert space and {en:n} is an orthonormal basis for . We define the operator T() as follows: (2.19)Ten={en-1,n0  (mod  3),12en-1,n1  (mod  3),2en-1,n2  (mod  3), so (2.20)T*en={12en+1,n0  (mod  3),2en+1,n1  (mod  3),en+1,n2  (mod  3), and by simple computation we get (2.21)TT*en={14en,n0  (mod  3),4en,n1  (mod  3),en,n2  (mod  3),T*Ten={en,n0  (mod  3),14en,n1  (mod  3),4en,n2  (mod  3). Consequently, T is (1/4,4)-normal operator and also Tn is (1/4,4)-normal operator, for any integer n0. Thus we have T=2 and r(T)=1, hence (2.14) is valid.

3. Inequalities Involving Norms and Numerical Radius

In this section we state some inequalities involving norms and numerical radius.

Theorem 3.1.

Let T() be an (α,β)-normal operator.

For positive real numbers p and q with p2 and (1/p)+(1/q)=1 we have (3.1)T+T*p+T-T*p2(1+αq)p-1Tp.

If 0p1 or p2, then we have (3.2)(T+T*2+T-T*2)pT2pφ(α,p), where φ(α,p)=2p[(1+αp)2+(2p-22)αp].

If 𝒩(T)=0 and for any x with x=1 we have (3.3)TxT*x-T*xTxρ, then, we obtain (3.4)αT2ω(T2)+ρ22βT2.

Proof.

(i) We use the following known inequality: (3.5)a+bp+a-bp2(aq+bq)p-1, which is valid for any a,b where is a Hilbert space.

Now, if we take a=Tx and b=T*x in (3.5), then for any x we get (3.6)Tx+T*xp+Tx-T*xp2(Txq+T*xq)p-12(Txq+αqTxq)p-1=2(1+αq)p-1Txq(p-1)=2(1+αq)p-1Txp. Taking the supremum in (3.6) over x with x=1, we get the desired result (3.1).

(ii) We use the following inequality [6, Theorem 8, page 551]: (3.7)(a+b2+a-b2)p2p((ap+bp)2+(2p-22)apbp), where a and b are two vectors in a Hilbert space and 0p1 or p2.

Now, if we put a=Tx and b=T*x in (3.7), then we obtain (3.8)(Tx+T*x2+Tx-T*x2)p2p((Txp+T*xp)2+(2p-22)TxpT*xp),2p(Tx2p(1+αp)2+(2p-22)αpTx2p)=2pTx2p[(1+αp)2+(2p-22)αp]=Tx2pφ(α,p). Now, taking the supremum over x=1 in (3.8), we get the desired result (3.2).

(iii) We use the following reverse of Schwarz's inequality: (3.9)(0)ab-|a,b|ab-Rea,b12ρ2ab, which is valid for a,b{0} and ρ>0, with (a/b)-(b/a)ρ (see ). We take a=Tx and b=T*x in (3.9) to get (3.10)TxT*x|Tx,T*x|+12ρ2TxT*x. Thus, we obtain (3.11)αTx2|Tx,T*x|+12ρ2βTx2. Now, taking the supremum over x=1 in recent inequality, we get the desired result (3.4).

Theorem 3.2.

Assume that T is an (α,β)-normal operator. Then, we have (3.12)(1+α2)T212T-T*2+ω(T2).

Proof.

By [2, Theorem 3.1], we have (3.13)2(1+αp)Tp12[T+T*p+T-T*p], and also (3.14)T*T+TT*2p/214[T+T*p+T-T*p]. On the other hand, it is known  that for A,B() we have (3.15)A+B2212[A*A+B*B2+ω(B*A)]. By using this inequality we get (3.16)T+T*2212[T*T+TT*2+ω(T2)]. If we put p=2 in (3.14), we obtain (3.17)T+T*2212[14(T+T*2+T-T*2)+ω(T2)]=12[T+T*22+T-T*22+ω(T2)]. Thus we get (3.18)12T+T*2212T-T*22+ω(T2)2. Now, we take p=2 in (3.13) to obtain (3.19)(1+α2)T2T-T*22+T-T*22+ω(T2)=12T-T*2+ω(T2). This completes the proof.

Theorem 3.3.

Assume that T is an (α,β)-normal operator. Then for any real s with 0s1, we have (3.20)((1-s)1β2+s)((1-s)+s1β2)T4[1-s+sβ2]T2T-T*2+w(T2)2.

Proof.

By [9, Theorem 2.6] (see also [10, Theorem 2.4]), we have (3.21)[(1-s)a2+sb2][(1-s)b2+sa2]-|a,b|2[(1-s)a2+sb2][(1-s)b-ta2+stb-a2], where 0s1, t and a,b. By taking t=1,a=Tx, and b=T*x in (3.21), we get (3.22)[(1-s)Tx2+sT*x2][(1-s)T*x2+sTx2]-|Tx,T*x|2[(1-s)Tx2+sT*x2][(1-s)T*x-Tx2+sT*x-Tx2], thus, we have (3.23)[(1-s)β2T*x2+sT*x2][(1-s)T*x2+sβ2T*x2]-|T2x,x|2[(1-s)Tx2+sT*x2][(1-s)T*x2+sTx2]-|T2x,x|2[(1-s)Tx2+sT*x2][(1-s)T*x-Tx2+sT*x-Tx2][(1-s)Tx2+sβ2Tx2]T*x-Tx2. Finally, we take supremum over x=1 from both sides of (3.24)((1-s)β2+s)((1-s)+sβ2)T*x4[(1-s)Tx2+sβ2Tx2]T*x-Tx2+|T2x,x|2  , and we use triangle inequality for supremums to complete the proof.

Corollary 3.4.

Let T be an (α,β)-normal operator. Then, we have (3.25)1βT2TT-T*+ω(T2).

Proof.

By using the inequality (3.21) we get (3.26)((1-s)+sα2)((1-s)α2+s)T4[1-s+sα2]T2T-T*2+w(T2)2. We take s=0 in inequalities (3.20) and (3.26) to imply (3.27)max{1β2,α2}Tx4Tx2T-T*2+ω(T2)2. Thus, max{1/β,α}Tx2TxTx-T*x+ω(T2). Now, taking supremum overall x with x=1, the desired inequality is obtained.

Acknowledgments

The authors would like to sincerely thank the anonymous referee for several useful comments improving the paper and also Professor Mehdi Hassani for a useful discussion.

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