# More on(α,β)-Normal Operators in Hilbert Spaces

Authors: Rasoul Eskandari; Farzollah Mirzapour; Ali Morassaei
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

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 ℬ(ℋ) of all bounded linear operators T:ℋ→ℋ, where ℋ is Hilbert space.


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## Body

## 1. Introduction

Throughout the paper, letℬ(ℋ) denote the algebra of all bounded linear operators acting on a complex Hilbert space (ℋ,〈·,·〉), ℬh(ℋ) denote the algebra of all self-adjoint operators in ℬ(ℋ), and I is the identity operator. In case of dimℋ=n, we identify ℬ(ℋ) with the full matrix algebra ℳn(ℂ) of all n×n matrices with entries in the complex field. An operator A∈ℬh(ℋ) is called positive if 〈Ax,x〉≥0 is valid for any x∈ℋ, and then we write A≥0. Moreover, by A>0 we mean 〈Ax,x〉>0 for any x∈ℋ. For A,B∈ℬh(ℋ), we say A≤B if B-A≥0. An operator A is majorized by B, if there exists a constant λ such that ∥Ax∥≤λ∥Bx∥ for all x∈ℋ or equivalently A*A≤λ2B*B [1].For real numbersα and β with 0≤α≤1≤β, an operator T acting on a Hilbert space ℋ is called (α,β)-normal [2, 3] if
(1.1)α2T*T≤TT*≤β2T*T.
An immediate consequence of above definition is
(1.2)α2〈T*Tx,x〉≤〈TT*x,x〉≤β2〈T*Tx,x〉,
from which we obtain
(1.3)α‖Tx‖≤‖T*x‖≤β‖Tx‖,
for all x∈ℋ.Notice that, according to (1.1), if T is (α,β)-normal operator, then T and T* majorize each other.In [3], Moslehian posed two problems about (α,β)-normal operators as follows.For fixedα>0 and β≠1,(i)
give an example of an(α,β)-normal operator which is neither normal nor hyponormal;(ii)
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 [2], 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 ℬ(ℂ2) is an (α,β)-normal with α=(3-5)/2 and β=(3+5)/2.Thenumerical radiusw(T) of an operator T on ℋ is defined by
(1.4)w(T)=sup⁡⁡{|〈Tx,x〉|:‖x‖=1}.
Obviously, by (1.4), for any x∈ℋ we have
(1.5)|〈Tx,x〉|≤w(T)‖x‖2.
It is well known that w(·) is a norm on the Banach algebra ℬ(ℋ) of all bounded linear operators. Moreover, we have
(1.6)w(T)≤‖T‖≤2w(T)(T∈B(H)).
For other results and historical comments on the numerical radius see [4].Theantieigenvalue of an operator T∈ℬ(ℋ) defined by
(1.7)μ1(T)∶=inf⁡Tx≠0Re〈Tx,x〉‖Tx‖‖x‖.
The vector x∈ℋ which takes μ1(T) is called an antieigenvector of T. We refer more study on this matter to [4].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(α,β)-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∈ℂ such that z+T is (α,β)-normal where T is (α,β)-normal.We know by the Cauchy-Schwartz inequality that-1≤μ1(T)≤1. Also we can write
(2.1)μ1(T)=inf⁡‖x‖=1Tx≠0⁡Re〈Tx,x〉‖Tx‖.
We define
(2.2)μ2(T)∶=sup⁡‖x‖=1Tx≠0⁡Re〈Tx,x〉‖Tx‖.We know that ifT is normal operator then z+T is also normal.Theorem 2.1.
LetT be an (α,β)-normal operator on a Hilbert space such that 0≤α<1<β and z∈ℂ. Then z+T is (α,β)-normal, if provided one of the following conditions holds:(i)
μ1(z¯T)≥0,(ii)
μ1(z¯T)<0,|z|2≥-2|z|∥T∥μ1(z¯T).Proof.
In both of above cases, we show that(2.3)|z|2+2Re〈z¯Tx,x〉≥0,∀x∈Hwith‖x‖=1,Tx≠0.
By the assumption (i), μ1(z¯T)≥0, we have Re〈z¯Tx,x〉/|z|∥Tx∥≥0 for every x∈ℋ with ∥x∥=1 and Tx≠0, consequently we get Re〈z¯Tx,x〉≥0, and therefore (2.3) is valid. On the other hand, if (ii) holds and we set B∶=μ1(z¯T) then we get B≤Re〈z¯Tx,x〉/|z|∥Tx∥ for every x∈ℋ with ∥x∥=1 and Tx≠0, consequently:
(2.4)inf⁡⁡{B‖Tx‖:‖x‖=1,Tx≠0}≤inf⁡{‖Tx‖Re〈z¯Tx,x〉|z|‖Tx‖:‖x‖=1,Tx≠0}.
Since B<0, we obtain
(2.5)-Binf⁡{-‖Tx‖:‖x‖=1,Tx≠0}≤inf⁡{‖Tx‖Re〈z¯Tx,x〉|z|‖Tx‖:‖x‖=1,Tx≠0},
and so
(2.6)Bsup⁡{‖Tx‖:‖x‖=1,Tx≠0}≤inf⁡{‖Tx‖Re〈z¯Tx,x〉|z|‖Tx‖:‖x‖=1,Tx≠0}.
Now, by using the last inequality, we have
(2.7)|z|2+2|z|‖T‖μ1(z¯T)=|z|2+2|z|(sup⁡‖x‖=1Tx≠0⁡‖Tx‖)(inf⁡‖x‖=1Tx≠0⁡{Re〈z¯Tx,x〉|z|‖Tx‖})≤|z|2+2|z|inf⁡‖x‖=1{‖Tx‖Re〈z¯Tx,x〉|z|‖Tx‖}=|z|2+2inf⁡‖x‖=1{Re〈z¯Tx,x〉}.
This shows that (2.3) holds for (ii), too. Thus, for any x∈ℋ with ∥x∥=1 we have
(2.8)α2〈(z+T)*(z+T)x,x〉=α2[〈|z|2x,x〉+〈z¯Tx,x〉+〈zT*x,x〉]+α2〈T*Tx,x〉≤〈|z|2x,x〉+〈z¯Tx,x〉+〈zT*x,x〉+〈TT*x,x〉=〈(z+T)(z+T)*x,x〉≤β2[〈|z|2x,x〉+〈z¯Tx,x〉+〈zT*x,x〉]+β2〈T*Tx,x〉=β2〈(z+T)*(z+T)x,x〉
and this completes the proof.Corollary 2.2.
LetT be an (α,β)-normal operator. We have the following.(i)
Ifμ1(T)≥0 then z+T is (α,β)-normal operator for any z>0.(ii)
Ifμ2(T)≤0 then z+T is (α,β)-normal operator for any z<0.Proof.
(i) By the definition of the first antieigenvalue ofT, 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) Ifz<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.
LetT be an injective and (α,β)-normal operator with α>0. Then(i)
ℛ(T) is dense,(ii)
T* is injective,(iii)
ifT 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 ℛ(T)⊥=𝒩(T*)=𝒩(T)=0, thus ℛ(T) is a dense subspace of ℋ and T* is injective. This proves (i) and (ii).
To prove (iii), we note that sinceT 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<A≤B then B-1≤A-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-1≤1α2T-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 matrixT in ℬ(ℂ2):
(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.
ForT∈ℬ(ℋ) we call
(2.13)r(T)=sup⁡{|λ|:λ∈σ(T)}
the spectral radius of T, where σ(T) is the spectrum of T and it is known that r(T)=limn→∞∥Tn∥1/n [5, page 102].Theorem 2.5.
LetT be an (α,β)-normal operator such that T2n is (α,β)-normal operator for every n∈ℕ, too. Then, we have
(2.14)1β‖T‖≤r(T)≤‖T‖.Proof.
For anyT∈ℬ(ℋ) we have
(2.15)‖T*T‖=‖T‖2.
In particular, if T is a self-adjoint operator then ∥T2∥=∥T∥2. Thus, by the definition of (α,β)-normal operator, we have
(2.16)‖T*2T2‖≥1β2‖(T*T)2‖=1β2‖T‖4.
By induction on n, we imply that
(2.17)‖T*2nT2n‖≥1β2n+1-2‖T‖2n+1,
from which we obtain
(2.18)r(T)2=r(T*)r(T)=lim⁡n→∞(‖T*2n‖‖T2n‖)1/2n≥lim⁡n→∞‖T*2nT2n‖1/2n≥lim⁡n→∞(1β2n+1-2‖T‖2n+1)1/2n=1β2‖T‖2lim⁡n→∞1β-2/2n=1β2‖T‖2.
Therefore, we get (1/β)∥T∥≤r(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 thatℋ is a separable Hilbert space and {en:n∈ℤ} is an orthonormal basis for ℋ. We define the operator T∈ℬ(ℋ) as follows:
(2.19)Ten={en-1,n≡0(mod⁡3),12en-1,n≡1(mod⁡3),2en-1,n≡2(mod⁡3),
so
(2.20)T*en={12en+1,n≡0(mod⁡3),2en+1,n≡1(mod⁡3),en+1,n≡2(mod⁡3),
and by simple computation we get
(2.21)TT*en={14en,n≡0(mod⁡3),4en,n≡1(mod⁡3),en,n≡2(mod⁡3),T*Ten={en,n≡0(mod⁡3),14en,n≡1(mod⁡3),4en,n≡2(mod⁡3).
Consequently, T is (1/4,4)-normal operator and also Tn is (1/4,4)-normal operator, for any integer n≥0. 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.
LetT∈ℬ(ℋ) be an (α,β)-normal operator.(i)
For positive real numbersp and q with p≥2 and (1/p)+(1/q)=1 we have
(3.1)‖T+T*‖p+‖T-T*‖p≥2(1+αq)p-1‖T‖p.(ii)
If0≤p≤1 or p≥2, then we have
(3.2)(‖T+T*‖2+‖T-T*‖2)p≥‖T‖2pφ(α,p),
where φ(α,p)=2p[(1+αp)2+(2p-22)αp].(iii)
If𝒩(T)=0 and for any x∈ℋ with ∥x∥=1 we have
(3.3)‖Tx‖T*x‖-T*x‖Tx‖‖≤ρ,
then, we obtain
(3.4)α‖T‖2≤ω(T2)+ρ22β‖T‖2.Proof.
(i) We use the following known inequality:(3.5)‖a+b‖p+‖a-b‖p≥2(‖a‖q+‖b‖q)p-1,
which is valid for any a,b∈ℋ where ℋ is a Hilbert space.
Now, if we takea=Tx and b=T*x in (3.5), then for any x∈ℋ we get
(3.6)‖Tx+T*x‖p+‖Tx-T*x‖p≥2(‖Tx‖q+‖T*x‖q)p-1≥2(‖Tx‖q+αq‖Tx‖q)p-1=2(1+αq)p-1‖Tx‖q(p-1)=2(1+αq)p-1‖Tx‖p.
Taking the supremum in (3.6) over x∈ℋ with ∥x∥=1, we get the desired result (3.1).
(ii) We use the following inequality [6, Theorem 8, page 551]:
(3.7)(‖a+b‖2+‖a-b‖2)p≥2p((‖a‖p+‖b‖p)2+(2p-22)‖a‖p‖b‖p),
where a and b are two vectors in a Hilbert space and 0≤p≤1 or p≥2.
Now, if we puta=Tx and b=T*x in (3.7), then we obtain
(3.8)(‖Tx+T*x‖2+‖Tx-T*x‖2)p≥2p((‖Tx‖p+‖T*x‖p)2+(2p-22)‖Tx‖p‖T*x‖p),≥2p(‖Tx‖2p(1+αp)2+(2p-22)αp‖Tx‖2p)=2p‖Tx‖2p[(1+αp)2+(2p-22)αp]=‖Tx‖2pφ(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≤)‖a‖‖b‖-|〈a,b〉|≤‖a‖‖b‖-Re〈a,b〉≤12ρ2‖a‖‖b‖,
which is valid for a,b∈ℋ∖{0} and ρ>0, with ∥(a/∥b∥)-(b/∥a∥)∥≤ρ (see [7]). We take a=Tx and b=T*x in (3.9) to get
(3.10)‖Tx‖‖T*x‖≤|〈Tx,T*x〉|+12ρ2‖Tx‖‖T*x‖.
Thus, we obtain
(3.11)α‖Tx‖2≤|〈Tx,T*x〉|+12ρ2β‖Tx‖2.
Now, taking the supremum over ∥x∥=1 in recent inequality, we get the desired result (3.4).Theorem 3.2.
Assume thatT is an (α,β)-normal operator. Then, we have
(3.12)(1+α2)‖T‖2≤12‖T-T*‖2+ω(T2).Proof.
By [2, Theorem 3.1], we have
(3.13)2(1+αp)‖T‖p≤12[‖T+T*‖p+‖T-T*‖p],
and also
(3.14)‖T*T+TT*2‖p/2≤14[‖T+T*‖p+‖T-T*‖p].
On the other hand, it is known [8] that for A,B∈ℬ(ℋ) we have
(3.15)‖A+B2‖2≤12[‖A*A+B*B2‖+ω(B*A)].
By using this inequality we get
(3.16)‖T+T*2‖2≤12[‖T*T+TT*2‖+ω(T2)].
If we put p=2 in (3.14), we obtain
(3.17)‖T+T*2‖2≤12[14(‖T+T*‖2+‖T-T*‖2)+ω(T2)]=12[‖T+T*2‖2+‖T-T*2‖2+ω(T2)].
Thus we get
(3.18)12‖T+T*2‖2≤12‖T-T*2‖2+ω(T2)2.
Now, we take p=2 in (3.13) to obtain
(3.19)(1+α2)‖T‖2≤‖T-T*2‖2+‖T-T*2‖2+ω(T2)=12‖T-T*‖2+ω(T2).
This completes the proof.Theorem 3.3.
Assume thatT is an (α,β)-normal operator. Then for any real s with 0≤s≤1, we have
(3.20)((1-s)1β2+s)((1-s)+s1β2)‖T‖4≤[1-s+sβ2]‖T‖2‖T-T*‖2+w(T2)2.Proof.
By [9, Theorem 2.6] (see also [10, Theorem 2.4]), we have
(3.21)[(1-s)‖a‖2+s‖b‖2][(1-s)‖b‖2+s‖a‖2]-|〈a,b〉|2≤[(1-s)‖a‖2+s‖b‖2][(1-s)‖b-ta‖2+s‖tb-a‖2],
where 0≤s≤1, t∈ℝ and a,b∈ℋ. By taking t=1,a=Tx, and b=T*x in (3.21), we get
(3.22)[(1-s)‖Tx‖2+s‖T*x‖2][‖(1-s)T*x‖2+s‖Tx‖2]-|〈Tx,T*x〉|2≤[(1-s)‖Tx‖2+s‖T*x‖2][(1-s)‖T*x-Tx‖2+s‖T*x-Tx‖2],
thus, we have
(3.23)[(1-s)β2‖T*x‖2+s‖T*x‖2][(1-s)‖T*x‖2+sβ2‖T*x‖2]-|〈T2x,x〉|2≤[(1-s)‖Tx‖2+s‖T*x‖2][(1-s)‖T*x‖2+s‖Tx‖2]-|〈T2x,x〉|2≤[(1-s)‖Tx‖2+s‖T*x‖2][(1-s)‖T*x-Tx‖2+s‖T*x-Tx‖2]≤[(1-s)‖Tx‖2+sβ2‖Tx‖2]‖T*x-Tx‖2.
Finally, we take supremum over ∥x∥=1 from both sides of
(3.24)((1-s)β2+s)((1-s)+sβ2)‖T*x‖4≤[(1-s)‖Tx‖2+sβ2‖Tx‖2]‖T*x-Tx‖2+|〈T2x,x〉|2,
and we use triangle inequality for supremums to complete the proof.Corollary 3.4.
LetT be an (α,β)-normal operator. Then, we have
(3.25)1β‖T‖2≤‖T‖‖T-T*‖+ω(T2).Proof.
By using the inequality (3.21) we get
(3.26)((1-s)+sα2)((1-s)α2+s)‖T‖4≤[1-s+sα2]‖T‖2‖T-T*‖2+w(T2)2.
We take s=0 in inequalities (3.20) and (3.26) to imply
(3.27)max⁡{1β2,α2}‖Tx‖4≤‖Tx‖2‖T-T*‖2+ω(T2)2.
Thus, max{1/β,α}∥Tx∥2≤∥Tx∥∥Tx-T*x∥+ω(T2). Now, taking supremum overall x with ∥x∥=1, the desired inequality is obtained.



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