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,〈·,·〉), 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 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
(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 fixed α>0 and β≠1,

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 [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
(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].

The antieigenvalue of an operator T∈ℬ(ℋ) defined by
(1.7)μ1(T)∶=infTx≠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 <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∈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≠0Re〈Tx,x〉‖Tx‖.
We define
(2.2)μ2(T)∶=sup‖x‖=1Tx≠0Re〈Tx,x〉‖Tx‖.

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:

μ1(z¯T)≥0,

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.

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

ℛ(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 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
(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 matrix T 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.

For T∈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.

Let T 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 any T∈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)=limn→∞(‖T*2n‖‖T2n‖)1/2n≥limn→∞‖T*2nT2n‖1/2n≥limn→∞(1β2n+1-2‖T‖2n+1)1/2n=1β2‖T‖2limn→∞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 that ℋ 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(mod3),12en-1,n≡1(mod3),2en-1,n≡2(mod3),
so
(2.20)T*en={12en+1,n≡0(mod3),2en+1,n≡1(mod3),en+1,n≡2(mod3),
and by simple computation we get
(2.21)TT*en={14en,n≡0(mod3),4en,n≡1(mod3),en,n≡2(mod3),T*Ten={en,n≡0(mod3),14en,n≡1(mod3),4en,n≡2(mod3).
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.

Let T∈ℬ(ℋ) be an (α,β)-normal operator.

For positive real numbers p 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.

If 0≤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].

If 𝒩(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 take a=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 put a=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 that T 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 that T 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.

Let T 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.

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.

HalmosP. R.DragomirS. S.MoslehianM. S.Some inequalities for (α,β)-normal operators in Hilbert spacesMoslehianM. S.On (α,β)-normal operators in Hilbert spaces, IMAGE 39GustafsonK. E.RaoD. K. M.WeidmannJ.MitrinovićD. S.PečarićJ. E.FinkA. M.DragomirS. S.A potpourri of Schwarz related inequalities in inner product spaces. IDragomirS. S.A survey of some recent inequalities for the norm and numerical radius of operators in Hilbert spacesDragomirS. S.A potpourri of Schwarz related inequalities in inner product spaces. IIDragomirS. S.Some inequalities for normal operators in Hilbert spaces