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

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.

1. Introduction

Throughout the paper, let ℬ(ℋ) denote the algebra of all bounded linear operators acting on a complex Hilbert space (ℋ,⟨⋅,⋅⟩), ℬ_ℎ(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 ℳ_𝑛(ℂ) of all 𝑛×𝑛 matrices with entries in the complex field. An operator 𝐴∈ℬ_ℎ(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∈ℬ_ℎ(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 numbers 𝛼 and 𝛽 with 0≤𝛼≤1≤𝛽, an operator 𝑇 acting on a Hilbert space ℋ is called (𝛼,𝛽)-normal [2, 3] if 𝛼²𝑇^∗𝑇≤𝑇𝑇^∗≤𝛽²𝑇^∗𝑇. (1.1) An immediate consequence of above definition is 𝛼²𝑇^∗𝑇𝑥,𝑥≤𝑇𝑇^∗𝑥,𝑥≤𝛽²𝑇^∗𝑇𝑥,𝑥, (1.2) from which we obtain 𝛼‖𝑇𝑥‖≤𝑇^∗𝑥≤𝛽‖𝑇𝑥‖, (1.3) 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 fixed 𝛼>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.

The numerical radius 𝑤(𝑇) of an operator 𝑇 on ℋ is defined by 𝑤(𝑇)=sup{|⟨𝑇𝑥,𝑥⟩|∶‖𝑥‖=1}. (1.4) Obviously, by (1.4), for any 𝑥∈ℋ we have |⟨𝑇𝑥,𝑥⟩|≤𝑤(𝑇)‖𝑥‖². (1.5) It is well known that 𝑤(⋅) is a norm on the Banach algebra ℬ(ℋ) of all bounded linear operators. Moreover, we have 𝑤(𝑇)≤‖𝑇‖≤2𝑤(𝑇)(𝑇∈ℬ(ℋ)). (1.6) For other results and historical comments on the numerical radius see [4].

The antieigenvalue of an operator 𝑇∈ℬ(ℋ) defined by 𝜇₁(𝑇)∶=inf_𝑇𝑥≠0Re⟨𝑇𝑥,𝑥⟩‖𝑥‖.𝑇𝑥‖‖ (1.7) 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 𝜇₁(𝑇)=inf_{‖𝑥‖=1𝑇𝑥≠0}Re⟨𝑇𝑥,𝑥⟩‖‖.𝑇𝑥 (2.1) We define 𝜇₂(𝑇)∶=sup_{‖𝑥‖=1𝑇𝑥≠0}Re⟨𝑇𝑥,𝑥⟩‖‖.𝑇𝑥 (2.2)

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 0≤𝛼<1<𝛽 and 𝑧∈ℂ. Then 𝑧+𝑇 is (𝛼,𝛽)-normal, if provided one of the following conditions holds:(i)𝜇₁(𝑧𝑇)≥0,(ii)𝜇₁(𝑧𝑇)<0,|𝑧|²≥−2|𝑧|‖𝑇‖𝜇₁(𝑧𝑇).

Proof. In both of above cases, we show that |𝑧|²+2Re⟨𝑧𝑇𝑥,𝑥⟩≥0,∀𝑥∈ℋwith‖𝑥‖=1,𝑇𝑥≠0. (2.3) By the assumption (i), 𝜇₁(𝑧𝑇)≥0, we have Re⟨𝑧𝑇𝑥,𝑥⟩/|𝑧|‖𝑇𝑥‖≥0 for every 𝑥∈ℋ with ‖𝑥‖=1 and 𝑇𝑥≠0, consequently we get Re⟨𝑧𝑇𝑥,𝑥⟩≥0, and therefore (2.3) is valid. On the other hand, if (ii) holds and we set 𝐵∶=𝜇₁(𝑧𝑇) then we get 𝐵≤Re⟨𝑧𝑇𝑥,𝑥⟩/|𝑧|‖𝑇𝑥‖ for every 𝑥∈ℋ with ‖𝑥‖=1 and 𝑇𝑥≠0, consequently: {𝐵‖‖∶‖𝑥‖}‖⟨inf𝑇𝑥=1,𝑇𝑥≠0≤inf‖𝑇𝑥Re𝑧𝑇𝑥,𝑥⟩|𝑧‖∶‖𝑥‖.|‖𝑇𝑥=1,𝑇𝑥≠0 (2.4) Since 𝐵<0, we obtain {−‖‖∶‖𝑥‖}‖⟨−𝐵inf𝑇𝑥=1,𝑇𝑥≠0≤inf‖𝑇𝑥Re𝑧𝑇𝑥,𝑥⟩|𝑧‖∶‖𝑥‖,|‖𝑇𝑥=1,𝑇𝑥≠0 (2.5) and so ‖∶‖𝑥‖}‖⟨𝐵sup{‖𝑇𝑥=1,𝑇𝑥≠0≤inf‖𝑇𝑥Re𝑧𝑇𝑥,𝑥⟩|𝑧‖∶‖𝑥‖.|‖𝑇𝑥=1,𝑇𝑥≠0 (2.6) Now, by using the last inequality, we have |𝑧|²+2|𝑧|‖𝑇‖𝜇₁(𝑧𝑇)=|𝑧|²+2|𝑧|sup_{‖𝑥‖=1𝑇𝑥≠0}‖𝑇𝑥‖inf_{‖𝑥‖=1𝑇𝑥≠0}Re⟨𝑧𝑇𝑥,𝑥⟩|𝑧‖≤|𝑧||‖𝑇𝑥₂+2|𝑧|inf_‖𝑥‖=1‖⟨‖𝑇𝑥Re𝑧𝑇𝑥,𝑥⟩|𝑧‖=|𝑧||‖𝑇𝑥₂+2inf_‖𝑥‖=1{Re⟨𝑧𝑇𝑥,𝑥⟩}. (2.7) This shows that (2.3) holds for (ii), too. Thus, for any 𝑥∈ℋ with ‖𝑥‖=1 we have 𝛼²(𝑧+𝑇)^∗(𝑧+𝑇)𝑥,𝑥=𝛼²|𝑧|²𝑥,𝑥+⟨𝑧𝑇𝑥,𝑥⟩+𝑧𝑇^∗𝑥,𝑥+𝛼²𝑇^∗≤𝑧|𝑇𝑥,𝑥|²𝑥,𝑥+⟨𝑧𝑇𝑥,𝑥⟩+𝑧𝑇^∗𝑥,𝑥+𝑇𝑇^∗=)𝑥,𝑥(𝑧+𝑇)(𝑧+𝑇^∗𝑥,𝑥≤𝛽²|𝑧|²𝑥,𝑥+⟨𝑧𝑇𝑥,𝑥⟩+𝑧𝑇^∗𝑥,𝑥+𝛽²𝑇^∗𝑇𝑥,𝑥=𝛽²(𝑧+𝑇)^∗(𝑧+𝑇)𝑥,𝑥 (2.8) and this completes the proof.

Corollary 2.2. Let 𝑇 be an (𝛼,𝛽)-normal operator. We have the following.(i)If 𝜇₁(𝑇)≥0 then 𝑧+𝑇 is (𝛼,𝛽)-normal operator for any 𝑧>0.(ii)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 𝜇₁(𝑧𝑇)=𝜇₁(𝑧𝑇)=𝜇₁(𝑇)≥0. (2.9) By using Theorem 2.1(i) we imply that 𝑧+𝑇 is an (𝛼,𝛽)-normal.
(ii) If 𝑧<0, then 𝜇₁(𝑧𝑇)=−𝜇₂(𝑇)≥0. (2.10) 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(i)ℛ(𝑇) is dense,(ii)𝑇^∗ is injective,(iii)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 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 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<𝐴≤𝐵 then 𝐵⁻¹≤𝐴⁻¹. Since 𝑇 is (𝛼,𝛽)-normal, by taking inverse from all sides of (1.1), we get 1𝛽²𝑇⁻¹𝑇^∗−1≤𝑇^∗−1𝑇⁻¹≤1𝛼²𝑇⁻¹𝑇^∗−1. (2.11) 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 ℬ(ℂ²): 𝑇=.1011 (2.12)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∈ℬ(ℋ) we call 𝑟(𝑇)=sup{|𝜆|∶𝜆∈𝜎(𝑇)} (2.13) the spectral radius of 𝑇, where 𝜎(𝑇) is the spectrum of 𝑇 and it is known that 𝑟(𝑇)=lim_𝑛→∞‖𝑇^𝑛‖^1/𝑛 [5, page 102].

Theorem 2.5. Let 𝑇 be an (𝛼,𝛽)-normal operator such that 𝑇^2𝑛 is (𝛼,𝛽)-normal operator for every 𝑛∈ℕ, too. Then, we have 1𝛽‖𝑇‖≤𝑟(𝑇)≤‖𝑇‖. (2.14)

Proof. For any 𝑇∈ℬ(ℋ) we have 𝑇^∗𝑇=‖𝑇‖². (2.15) In particular, if 𝑇 is a self-adjoint operator then ‖𝑇²‖=‖𝑇‖². Thus, by the definition of (𝛼,𝛽)-normal operator, we have 𝑇^∗2𝑇²≥1𝛽²𝑇^∗𝑇²=1𝛽²‖𝑇‖⁴. (2.16) By induction on 𝑛, we imply that 𝑇^∗2𝑛𝑇^2𝑛≥1𝛽^2𝑛+1−2‖𝑇‖^2𝑛+1, (2.17) from which we obtain 𝑟(𝑇)²=𝑟𝑇^∗𝑟(𝑇)=lim_𝑛→∞𝑇^∗2𝑛𝑇^2𝑛1/2𝑛≥lim_𝑛→∞𝑇^∗2𝑛𝑇^2𝑛1/2𝑛≥lim_𝑛→∞1𝛽^2𝑛+1−2‖𝑇‖^{2𝑛+11/2𝑛}=1𝛽²‖𝑇‖²lim_𝑛→∞1𝛽^−2/2𝑛=1𝛽²‖𝑇‖². (2.18) 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 that ℋ is a separable Hilbert space and {𝑒_𝑛∶𝑛∈ℤ} is an orthonormal basis for ℋ. We define the operator 𝑇∈ℬ(ℋ) as follows: 𝑇𝑒_𝑛=𝑒_𝑛−1(),1,𝑛≡0mod32𝑒_𝑛−1,𝑛≡1(mod3),2𝑒_𝑛−1,𝑛≡2(mod3), (2.19) so 𝑇^∗𝑒_𝑛=12𝑒_𝑛+1,𝑛≡0(mod3),2𝑒_𝑛+1(),𝑒,𝑛≡1mod3_𝑛+1,𝑛≡2(mod3), (2.20) and by simple computation we get 𝑇𝑇^∗𝑒_𝑛=14𝑒_𝑛,𝑛≡0(mod3),4𝑒_𝑛(),𝑒,𝑛≡1mod3_𝑛(),𝑇,𝑛≡2mod3^∗𝑇𝑒_𝑛=𝑒_𝑛(),1,𝑛≡0mod34𝑒_𝑛,𝑛≡1(mod3),4𝑒_𝑛,𝑛≡2(mod3). (2.21) Consequently, 𝑇 is (1/4,4)-normal operator and also 𝑇^𝑛 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. Let 𝑇∈ℬ(ℋ) be an (𝛼,𝛽)-normal operator.(i)For positive real numbers 𝑝 and 𝑞 with 𝑝≥2 and (1/𝑝)+(1/𝑞)=1 we have 𝑇+𝑇^∗𝑝+𝑇−𝑇^∗𝑝≥21+𝛼^𝑞𝑝−1‖𝑇‖^𝑝. (3.1)(ii)If 0≤𝑝≤1 or 𝑝≥2, then we have 𝑇+𝑇^∗2+𝑇−𝑇^∗2𝑝≥‖𝑇‖^2𝑝𝜑𝛼,𝑝, (3.2) where 𝜑(𝛼,𝑝)=2^𝑝[(1+𝛼^𝑝)²+(2^𝑝−2²)𝛼^𝑝]. (iii)If 𝒩(𝑇)=0 and for any 𝑥∈ℋ with ‖𝑥‖=1 we have 𝑇𝑥‖𝑇^∗𝑥‖−𝑇^∗𝑥‖𝑇𝑥‖≤𝜌, (3.3) then, we obtain 𝛼‖𝑇‖²≤𝜔𝑇²+𝜌²2𝛽‖𝑇‖². (3.4)

Proof. (i) We use the following known inequality: ‖𝑎+𝑏‖^𝑝+‖𝑎−𝑏‖^𝑝≥2‖𝑎‖^𝑞+‖𝑏‖^𝑞𝑝−1, (3.5) which is valid for any 𝑎,𝑏∈ℋ where ℋ is a Hilbert space.
Now, if we take 𝑎=𝑇𝑥 and 𝑏=𝑇^∗𝑥 in (3.5), then for any 𝑥∈ℋ we get 𝑇𝑥+𝑇^∗𝑥^𝑝+𝑇𝑥−𝑇^∗𝑥^𝑝≥2‖𝑇𝑥‖^𝑞+𝑇^∗𝑥^𝑞𝑝−1≥2‖𝑇𝑥‖^𝑞+𝛼^𝑞‖𝑇𝑥‖^𝑞𝑝−1=21+𝛼^𝑞𝑝−1‖𝑇𝑥‖^{𝑞(𝑝−1)}=21+𝛼^𝑞𝑝−1‖𝑇𝑥‖^𝑝. (3.6) 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]: ‖𝑎+𝑏‖²+‖𝑎−𝑏‖^2𝑝≥2^𝑝‖𝑎‖^𝑝+‖𝑏‖^𝑝2+2^𝑝−2²‖𝑎‖^𝑝‖𝑏‖^𝑝, (3.7) where 𝑎 and 𝑏 are two vectors in a Hilbert space and 0≤𝑝≤1 or 𝑝≥2.
Now, if we put 𝑎=𝑇𝑥 and 𝑏=𝑇^∗𝑥 in (3.7), then we obtain 𝑇𝑥+𝑇^∗𝑥²+𝑇𝑥−𝑇^∗𝑥^2𝑝≥2^𝑝‖𝑇𝑥‖^𝑝+𝑇^∗𝑥^𝑝2+2^𝑝−2²‖𝑇𝑥‖^𝑝𝑇^∗𝑥^𝑝,≥2^𝑝‖𝑇𝑥‖^2𝑝1+𝛼^𝑝2+2^𝑝−2²𝛼^𝑝‖𝑇𝑥‖^2𝑝=2^𝑝‖𝑇𝑥‖^2𝑝1+𝛼^𝑝2+2^𝑝−2²𝛼^𝑝=‖𝑇𝑥‖^2𝑝𝜑𝛼,𝑝. (3.8) 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: (0≤)‖𝑎‖‖𝑏‖−|⟨𝑎,𝑏⟩|≤‖𝑎‖‖𝑏‖−R𝑒⟨⟩≤1𝑎,𝑏2𝜌²‖𝑎‖‖𝑏‖, (3.9) 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 ‖𝑇𝑥‖𝑇^∗𝑥≤𝑇𝑥,𝑇^∗𝑥+12𝜌²‖𝑇𝑥‖𝑇^∗𝑥. (3.10) Thus, we obtain 𝛼‖𝑇𝑥‖²≤𝑇𝑥,𝑇^∗𝑥+12𝜌²𝛽‖𝑇𝑥‖². (3.11) 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 1+𝛼²‖𝑇‖²≤12𝑇−𝑇^∗2+𝜔𝑇². (3.12)

Proof. By [2, Theorem 3.1], we have 21+𝛼^𝑝‖𝑇‖^𝑝≤12𝑇+𝑇^∗𝑝+𝑇−𝑇^∗𝑝, (3.13) and also 𝑇^∗𝑇+𝑇𝑇^∗2^𝑝/2≤14𝑇+𝑇^∗𝑝+𝑇−𝑇^∗𝑝. (3.14) On the other hand, it is known [8] that for 𝐴,𝐵∈ℬ(ℋ) we have 𝐴+𝐵2²≤12𝐴^∗𝐴+𝐵^∗𝐵2+𝜔𝐵^∗𝐴. (3.15) By using this inequality we get 𝑇+𝑇^∗2²≤12𝑇^∗𝑇+𝑇𝑇^∗2+𝜔𝑇². (3.16) If we put 𝑝=2 in (3.14), we obtain 𝑇+𝑇^∗2²≤1214𝑇+𝑇^∗2+𝑇−𝑇^∗2+𝜔𝑇²=12𝑇+𝑇^∗2²+𝑇−𝑇^∗2²+𝜔𝑇². (3.17) Thus we get 12𝑇+𝑇^∗2²≤12𝑇−𝑇^∗2²+𝜔𝑇²2. (3.18) Now, we take 𝑝=2 in (3.13) to obtain 1+𝛼²‖𝑇‖²≤𝑇−𝑇^∗2²+𝑇−𝑇^∗2²+𝜔𝑇²=12𝑇−𝑇^∗2+𝜔𝑇². (3.19) This completes the proof.

Theorem 3.3. Assume that 𝑇 is an (𝛼,𝛽)-normal operator. Then for any real 𝑠 with 0≤𝑠≤1, we have ()11−𝑠𝛽²()1+𝑠1−𝑠+𝑠𝛽²‖𝑇‖⁴≤1−𝑠+𝑠𝛽²‖𝑇‖²𝑇−𝑇^∗2+𝑤𝑇²². (3.20)

Proof. By [9, Theorem 2.6] (see also [10, Theorem 2.4]), we have (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, (3.21) 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 (1−𝑠)‖𝑇𝑥‖²+𝑠𝑇^∗𝑥²(1−𝑠)𝑇^∗𝑥²+𝑠‖𝑇𝑥‖²−𝑇𝑥,𝑇^∗𝑥²≤(1−𝑠)‖𝑇𝑥‖²+𝑠𝑇^∗𝑥²(1−𝑠)𝑇^∗𝑥−𝑇𝑥²+𝑠𝑇^∗𝑥−𝑇𝑥², (3.22) thus, we have (1−𝑠)𝛽²𝑇^∗𝑥²+𝑠𝑇^∗𝑥²(1−𝑠)𝑇^∗𝑥²+𝑠𝛽²𝑇^∗𝑥²−𝑇²𝑥,𝑥²≤(1−𝑠)‖𝑇𝑥‖²+𝑠𝑇^∗𝑥²(1−𝑠)𝑇^∗𝑥²+𝑠‖𝑇𝑥‖²−𝑇²𝑥,𝑥²≤(1−𝑠)‖𝑇𝑥‖²+𝑠𝑇^∗𝑥²(1−𝑠)𝑇^∗𝑥−𝑇𝑥²+𝑠𝑇^∗𝑥−𝑇𝑥²≤(1−𝑠)‖𝑇𝑥‖²+𝑠𝛽²‖𝑇𝑥‖²𝑇^∗𝑥−𝑇𝑥². (3.23) Finally, we take supremum over ‖𝑥‖=1 from both sides of (1−𝑠)𝛽²()+𝑠+𝑠1−𝑠𝛽²𝑇^∗𝑥⁴≤(1−𝑠)‖𝑇𝑥‖²+𝑠𝛽²‖𝑇𝑥‖²𝑇^∗𝑥−𝑇𝑥²+𝑇²𝑥,𝑥², (3.24) 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 1𝛽‖𝑇‖²≤‖𝑇‖𝑇−𝑇^∗+𝜔𝑇². (3.25)

Proof. By using the inequality (3.21) we get (1−𝑠)+𝑠𝛼²(1−𝑠)𝛼²+𝑠‖𝑇‖⁴≤1−𝑠+𝑠𝛼²‖𝑇‖²𝑇−𝑇^∗2+𝑤𝑇²². (3.26) We take 𝑠=0 in inequalities (3.20) and (3.26) to imply 1max𝛽²,𝛼²‖𝑇𝑥‖⁴≤‖𝑇𝑥‖²𝑇−𝑇^∗2+𝜔𝑇²². (3.27) 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.

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.