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R E S E A R C H

Open Access

Some reverse mean inequalities for

operators and matrices

Chaojun Yang

1

, Yaxin Gao

1

and Fangyan Lu

1*

*Correspondence:fylu@suda.edu.cn 1Department of Mathematics,

Soochow University, Suzhou, P.R. China

Abstract

In this paper, we present some new reverse arithmetic–geometric mean inequalities for operators and matrices due to Lin (Stud. Math. 215:187–194,2013). Among other inequalities, we prove that ifA,BB(H) are accretive and 0 <mI≤ (A),(B)≤MI, then, for every positive unital linear map

Φ

,

Φ

2A+B

2

(

K(h)

)

2

Φ

2

(

(A

B)

)

,

whereK(h) =(h+1)4h2 andh=Mm. Moreover, some reverse harmonic–geometric mean inequalities are also presented.

MSC: Primary 47A63; secondary 15A15

Keywords: Positive linear maps; Arithmetic–geometric–harmonic mean; Sector matrix; Inequality

1 Introduction

Throughout this paper,B(H) stands for all bounded linear operators on a complex Hilbert spaceH. In the finite-dimensional setting,Mndenotes the set of alln×ncomplex matri-ces. ForA,BB(H), we useAB(BA) to mean thatABis positive. The operator norm is denoted by · . An operatorAB(H) is called accretive if in its Cartesian (or Toeptliz) decomposition,A=A+iA, A is positive, whereA=A+2A∗,A=A2Ai∗. A linear mapΦ:B(H)→B(H) is called positive ifΦ(A)≥0 wheneverA≥0. IfΦ(I) =I, whereIdenotes the identity operator, then we say thatΦis unital. We reserveM,mfor scalars. In the finite-dimensional setting, we useInfor the identity.

The numerical range ofA∈Mnis defined by

W(A) =xAx:x∈Cn,xx= 1.

Forα∈[0,π

2),denote the sector regions in the complex plane as follows:

=

z∈C:z≥0,|z| ≤(z)tanα.

Clearly, Ais positive semidefinite if and only ifW(A)⊂S0, and ifW(A),W(B)⊂

for someα∈[0,π2), thenW(A+B)⊂. As 0 /∈, ifW(A)⊂, thenAis

(2)

lar. Moreover,W(A)⊂impliesW(XAX)⊂for any nonzeron×mmatrixX, thus

W(A–1)S

α.

Recent studies on matrices with numerical ranges in a sector can be found in [3–5,8–10,

14] and the references therein.

In [13], Tominaga presented an operator inequality as follows: LetA,Bbe positive op-erators on a Hilbert space with 0 <mIA,BMI, then

A+B

2 ≤S(h)AB, (1)

whereS(h) = h 1 h–1

elogh

1 h–1

is called Specht’s ratio andh=Mm.

Lin [6] found that, for a positive unital linear mapΦbetweenC∗-algebra,

Φ

A+B

2

K(h)Φ(AB) (2)

due to (1) and the following observation [6]:

S(h)≤K(h)≤S2(h) (h≥1),

whereK(h) =(h+1)4h2.

It is well known that, for two general positive operators (or positive definite matrices)

A,B,

AB A2≥B2.

However, Lin [6] showed that (2) can be squared as follows:

Φ2

A+B

2

K2(h)Φ2(AB). (3)

Zhang [15] generalized (3) whenp≥2 as follows:

Φ2p

A+B

2

≤(K(h)(M2+m2))2p

16M2pm2p Φ

2p(AB). (4)

For two accretive operatorsA,BB(H), Drury [3] defined the geometric mean ofAand

Bas follows:

AB=

2

π

0

tA+t–1B–1dt t

–1

. (5)

This new geometric mean defined by (5) possesses some similar properties compared to the geometric mean of two positive operators. For instance, AB=BA, (AB)–1=

A–1B–1. For more information about the geometric mean of two accretive operators, see [3]. Moreover, ifA,B∈MnwithW(A),W(B)⊂, thenW(AB)⊂.

(3)

2 Main results

To reach our goal, we need the following lemmas.

Lemma 2.1([10]) If A,BB(H)are accretive,then

(A)(B)≤ (AB).

Lemma 2.2([10]) If A,BB(H)are accretive,then

A–1+B–1 2

–1

(A)–1+(B)–1 2

–1 .

Lemma 2.3([9]) If A∈Mnhas a positive definite real part,then

A–1≤ (A)–1.

Lemma 2.4([4]) If A∈Mnwith W(A)⊂,then

sec2(α) A–1≥ (A)–1.

It is easy to verify that((A–1+2B–1)–1)(AB)(A+B

2 ) does not persist for two accre-tive operatorsAandB. However, Lin presented the following extension of the arithmetic– geometric mean inequality.

Lemma 2.5([9]) Let A,B∈Mnbe such that W(A),W(B)⊂.Then

(AB)≤sec2(α)

A+B

2

. (6)

Lemma 2.6([2]) Let A,BB(H)be positive.Then

AB ≤ 1

4A+B 2.

Lemma 2.7([1]) Let AB(H)be positive.Then,for every positive unital linear mapΦ,

Φ–1(A)≤Φ A–1.

Lemma 2.8([1]) Let A,BB(H)be positive.Then,for1≤r< +∞,

Ar+Br≤(A+B)r.

An operator Kantorovich inequality obtained by Marshall and Olkin [12] reads as fol-lows.

Let 0 <mIAMI, then, for every positive unital linear mapΦ,

Φ A–1≤K(h)Φ(A)–1, (7)

(4)

Lin [7] showed that (7) can be squared as follows:

Φ2 A–1≤ K(h)2Φ(A)–2, (8)

whereK(h) =(h+1)4h2 andh=M m.

LetA∈Mnhave a positive definite real part, 0 <mIn≤ (A)≤MInandΦbe a unital

positive linear map. By (7) and Lemma2.3, we can obtain the following inequality:

Φ A–1≤K(h)Φ (A)–1, (9)

whereK(h) =(h+1)4h2 andh=Mm.

As an analog of inequality (8), we show that inequality (9) can be squared nicely as fol-lows.

Theorem 2.9 If A∈Mnhas a positive definite real part and0 <mIn≤ (A)≤MIn,then,

for every positive unital linear mapΦ,

Φ2 A–1≤ K(h)2Φ (A)–2, (10)

where K(h) =(h+1)4h2 and h=M m.

Proof Since

mIn≤ (A)≤MIn,

we have

MIn–(A) mIn–(A)

(A)–1≤0,

which is equivalent to

(A) +Mm(A)–1≤(M+m)In. (11)

By Lemma2.3and (11), we get

(A) +Mm A–1

≤ (A) +Mm(A)–1

≤(M+m)In. (12)

Inequality (10) is equivalent to

Φ A–1Φ (A)≤K(h).

By computation, we have

(5)

≤1

4MmΦ A

–1+Φ (A)2 (by Lemma2.6)

≤1

4(M+m)

2 by (12).

That is,

Φ A–1Φ (A)≤K(h).

This completes the proof.

LetA,BB(H) be accretive, 0 <mI≤ (A),(B)≤MIandΦbe a unital positive linear map. By inequality (2) and Lemma2.1, we can obtain the following inequality:

Φ

A+B

2

K(h)Φ (AB), (13)

whereK(h) =(h+1)4h2 andh=Mm.

Following an idea of Lin [6], we give a squaring version of inequality (13) below.

Theorem 2.10 If A,B∈Mn with W(A),W(B)⊂ and0 <mIn≤ (A),(B)≤MIn,

then,for every positive unital linear mapΦ,

Φ2

A+B

2

≤ sec4(α)K(h)2Φ2 (AB), (14)

where K(h) =(h+1)4h2 and h=Mm.

Proof From Theorem2.9we have

1 2(A) +

1

2Mm(A) –11

2(M+m)In (15)

and

1 2(B) +

1

2Mm(B) –11

2(M+m)In. (16)

Summing up inequalities (15) and (16), we get

A+B

2

+Mm

(A)–1+(B)–1 2

≤(M+m)In. (17)

Inequality (14) is equivalent to

Φ

A+B

2

Φ–1 (AB)≤sec4(α)K(h).

By computation, we have

sec4(α)MmΦ

A+B

2

(6)

≤1

4

sec4(α)Φ

A+B

2

+MmΦ–1 (AB) 2

(by Lemma2.6)

≤1

4

sec4(α)Φ

A+B

2

+MmΦ (AB)–1 2

(by Lemma2.7)

≤1

4

sec4(α)Φ

A+B

2

+sec2(α)MmΦ A–1B–1

2

(by Lemma2.4)

≤1

4

sec4(α)Φ

A+B

2

+sec4(α)MmΦ

A–1+B–1 2

2 by (6)

=1 4

sec4(α)Φ

A+B

2

+Mm

A–1+B–1 2 2 ≤1 4

sec4(α)Φ

A+B

2

+Mm

(A)–1+(B)–1 2

2 (by Lemma2.3)

≤1

4sec

8(α)(M+m)2 by (17).

That is, Φ

A+B

2

Φ–1 (AB)≤sec4(α)K(h).

This completes the proof.

Next we give apth (p≥2) powering of inequality (14).

Theorem 2.11 If A,B∈Mnwith W(A),W(B)⊂, 0 <mIn≤ (A),(B)≤MIn, 1 <β

2and p≥2β,then,for every positive unital linear mapΦ,

Φp

A+B

2

≤(sec2β(α)K(h)

β

2(Mβ+mβ)) 2p

β

16Mpmp Φ

p (AB), (18)

where K(h) =(h+1)4h2 and h=Mm.

Proof Since

mInΦ

A+B

2

MIn,

we have

MβmβΦβ

A+B

2

+Φβ

A+B

2

+. (19)

By (14) and the L-H inequality [1], we obtain

Φβ (AB)≤ sec4(α)K(h)βΦβ

A+B

2

(7)

Inequality (18) is equivalent to

Φp2

A+B

2

Φp2 (AB) ≤(sec

2β(α)K(h)β2(Mβ+mβ))βp

4Mp2m p 2

.

By computation, we have

Mp2m p 2Φ p 2

A+B

2

Φp2 (AB)

≤1

4

sec4(α)K(h) p 4Φp2

A+B

2

+

M2m2 sec4(α)K(h)

p 4

Φp2 (AB) 2

≤1

4

sec4(α)K(h) β 2Φβ

A+B

2

+

M2m2 sec4(α)K(h)

β

2

Φβ (AB) p β

≤1

4

sec4(α)K(h) β 2Φβ

A+B

2

+ sec4(α)K(h) β 2Mβ

mβΦβ

A+B

2 p β =1 4

sec4(α)K(h) β 2 Φβ

A+B

2

+MβmβΦβ

A+B

2 p β ≤1 4 sec 2β

(α)K(h)β2 Mβ+mβ p β,

where the first inequality is by Lemma2.6, the second one is by Lemma2.8, the third one is by (20) and the last one is by (19).

That is,

Φp2

A+B

2

Φp2 (AB) ≤(sec

2β(α)K(h)β2(Mβ+mβ))βp

4Mp2m p 2

.

This completes the proof.

We are not satisfied with the factor (sec4(α)K(h))2 in Theorem2.10, the ideal factor should be (K(h))2. We shall prove it in the following theorem.

Theorem 2.12 If A,BB(H)are accretive and0 <mI≤ (A),(B)≤MI,then,for every

positive unital linear mapΦ,

Φ2

A+B

2

K(h)2Φ2 (AB), (21)

where K(h) =(h+1)4h2 and h=Mm.

Proof From Theorem2.10one can get

A+B

2

+Mm

(A)–1+(B)–1 2

(8)

Inequality (21) is equivalent to

Φ

A+B

2

Φ–1 (AB)K(h).

By computation, we have

MmΦ

A+B

2

Φ–1 (AB)

≤1

4

Φ

A+B

2

+MmΦ–1 (AB) 2

(by Lemma2.6)

≤1

4

Φ

A+B

2

+MmΦ (AB)–1 2

(by Lemma2.7)

≤1

4

Φ

A+B

2

+MmΦ (A)(B)–1 2

(by Lemma2.1)

=1 4

Φ

A+B

2

+MmΦ (A)–1(B)–1 2

≤1

4

Φ

A+B

2

+MmΦ

(A)–1+(B)–1 2

2 (by AM-GM inequality)

≤1

4(M+m)

2 by (22).

That is,

Φ

A+B

2

Φ–1 (AB)K(h).

This completes the proof.

Remark2.13 LettingA,B≥0 in Theorem2.12, inequality (21) coincides with inequality (3).

Next we give apth (p≥2) powering of inequality (21) along the same line as in Theo-rem2.11.

Theorem 2.14 If A,BB(H)are accretive and0 <mI≤ (A),(B)≤MI1 <β≤2and

p≥2β,then,for every positive unital linear mapΦ,

Φp

A+B

2

≤(K(h)

β

2(+)) 2p

β

16Mpmp Φ

p (AB), (23)

where K(h) =(h+1)4h2 and h=M m.

Remark2.15 LettingA,B≥0 andβ= 2 in Theorem2.14, inequality (23) coincides with inequality (4).

(9)

Theorem 2.16 Let A,B∈Mnbe such that W(A),W(B)⊂,then

A–1+B–1

2

–1

≤sec4(α)(AB). (24)

Proof We can get

A–1+B–1 2

–1

≤sec2(α) A–1B–1–1 (25)

along the same line as Liu et al. did in [11] by Lemma2.1and Lemma2.5. Thus we have

A–1+B–1 2

–1

A–1+B–1 2

–1

(by Lemma2.3)

≤sec2(α) A–1B–1–1 by (25)

=sec2(α) A–1–1 B–1–1

≤sec4(α) (A)(B) (by Lemma2.4)

≤sec4(α)(AB) (by Lemma2.1).

This completes the proof.

Remark2.17 Maybe it is just a clerical error in Theorem 1.2 of their work [11]. However, the authors present the following inequalities in their proof:

A–1+B–1 2

–1

≤sec2(α) A–1–1 B–1–1

≤sec2(α) (A)(B).

Obviously, such a deduction in their proof collapses given the property of geometric mean for positive definite matrices. Thus we give Theorem2.16and the proof.

LetA,B∈MnwithW(A),W(B)⊂, 0 <mIn≤ (A–1),(B–1)≤MInandΦbe a unital

positive linear map. As a complement of inequalities (13) and (24), we have the following reverse harmonic–geometric mean inequality:

Φ (AB)≤sec2(α)K(h)Φ

A–1+B–1 2

–1

, (26)

whereK(h) =(h+1)4h2 andh=M m.

Proof Compute

(AB) = A–1B–1–1

A–1B–1–1

K(h)

A–1+B–1 2

(10)

≤sec2(α)K(h)

A–1+B–1

2

–1 ,

in which the first inequality is by Lemma2.3, the second one is by inequality (13) and the last one is by Lemma2.4.

ImposingΦon both sides of the inequalities above, we thus obtain inequality (26).

As an analog of Theorem2.12, we shall present a squaring version of inequality (26).

Theorem 2.18 If A,B∈Mnwith W(A),W(B)⊂Sαand0 <mIn≤ (A–1),(B–1)≤MIn,

then,for every positive unital linear mapΦ,

Φ2 (AB)≤ sec4(α)K(h)2Φ2

A–1+B–1

2

–1

, (27)

where K(h) =(h+1)4h2 and h=Mm.

Proof From Theorem2.10we have

1

2 A

–1+1

2Mm A

–1–11

2(M+m)In (28)

and

1

2 B

–1+1

2Mm B

–1–11

2(M+m)In. (29)

Summing up inequalities (28) and (29), we get

A–1+B–1

2

+Mm

A+B

2

A–1+B–1 2

+Mm

(A–1)–1+(B–1)–1 2

≤(M+m)In.

Inequality (27) is equivalent to

Φ (AB)Φ–1

A–1+B–1 2

–1

≤sec4(α)K(h).

By computation, we have

MmΦ (AB)Φ–1

A–1+B–1 2

–1

≤1

4

MmΦ (AB)+Φ–1

A–1+B–1

2

–1

2 (by Lemma2.6)

≤1

4

MmΦ (AB)+Φ

A–1+B–1 2

–1–1

2 (by Lemma2.7)

≤1

4

MmΦ (AB)+sec2(α)Φ

A–1+B–1 2

(11)

≤1

4

sec2(α)MmΦ

A+B

2

+sec2(α)Φ

A–1+B–1

2

2 by (6)

=1 4

sec2(α)Φ

Mm

A+B

2

+

A–1+B–1 2

2

≤1

4sec

4(α)(M+m)2.

That is,

Φ (AB)Φ–1

A–1+B–1 2

–1

≤sec4(α)K(h).

This completes the proof.

Obviously, the optimal factor in Theorem2.18should be (sec2(α)K(h))2. We note that it is affirmative under the conditionmIn≤ (A–1)≤ (A)–1≤MInandmIn≤ (B–1)≤ (B)–1MI

nby presenting the following theorem.

Theorem 2.19 If A,B∈Mnwith W(A),W(B)⊂Sαand0 <mIn≤ (A)–1,(B)–1≤MIn,

then,for every positive unital linear mapΦ,

Φ2 (AB)≤ sec2(α)K(h)2Φ2

A–1+B–1 2

–1

, (30)

where K(h) =(h+1)4h2 and h=M m.

Proof From Theorem2.10we obtain

(A)–1+(B)–1

2 +Mm

A+B

2

≤(M+m)In. (31)

Inequality (30) is equivalent to

Φ (AB)Φ–1

A–1+B–1 2

–1

≤sec2(α)K(h).

By computation, we have

sec2(α)MmΦ (AB)Φ–1

A–1+B–1 2 –1 ≤1 4

MmΦ (AB)+sec2(α)Φ–1

A–1+B–1 2

–1

2 (by Lemma2.6)

≤1

4

MmΦ (AB)+sec2(α)Φ

A–1+B–1

2

–1–1

2 (by Lemma2.7)

≤1

4

MmΦ (AB)+sec2(α)Φ

(A)–1+(B)–1 2

2 (by Lemma2.2)

≤1

4

sec2(α)MmΦ

A+B

2

+sec2(α)Φ

(A)–1+(B)–1 2

(12)

=1 4

sec2(α)Φ

Mm

A+B

2

+(A) –1

+(B)–1 2

2

≤1

4sec

4(α)(M+m)2 by (31).

That is,

Φ (AB)Φ–1

A–1+B–1 2

–1

≤sec2(α)K(h).

This completes the proof.

Funding

This research is supported by the National Natural Science Foundation of P.R. China (No. 11571247).

Competing interests

The authors declare that they have no competing interest.

Authors’ contributions

All authors contributed almost the same amount of work to manuscript. All authors read and approved the final manuscript.

Publisher’s Note

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

Received: 7 January 2019 Accepted: 15 April 2019 References

1. Bhatia, R.: Positive Definite Matrices. Princeton University Press, Princeton (2007)

2. Bhatia, R., Kittaneh, F.: Notes on matrix arithmetic–geometric mean inequalities. Linear Algebra Appl.308, 203–211 (2000)

3. Drury, S.: Principal powers of matrices with positive definite real part. Linear Multilinear Algebra63, 296–301 (2015) 4. Drury, S., Lin, M.: Singular value inequalities for matrices with numerical ranges in a sector. Oper. Matrices8,

1143–1148 (2014)

5. Fu, X., Liu, Y., Liu, S.: Extension of determinantal inequalities of positive definite matrices. J. Math. Inequal.11, 355–358 (2017)

6. Lin, M.: Squaring a reverse AM-GM inequality. Stud. Math.215, 187–194 (2013)

7. Lin, M.: On an operator Kantorovich inequality for positive linear maps. J. Math. Anal. Appl.402, 127–132 (2013) 8. Lin, M.: Extension of a result of Haynsworth and Hartfiel. Arch. Math.104, 93–100 (2015)

9. Lin, M.: Some inequalities for sector matrices. Oper. Matrices10, 915–921 (2016)

10. Lin, M., Sun, F.: A property of the geometric mean of accretive operators. Linear Multilinear Algebra65, 433–437 (2017)

11. Liu, J., Wang, Q.: More inequalities for sector matrices. Bull. Iran. Math. Soc.44, 1059–1066 (2018)

12. Marshall, A.W., Olkin, I.: Matrix versions of Cauchy and Kantorovich inequalities. Aequ. Math.40, 89–93 (1990) 13. Tominaga, M.: Specht’s ratio in the Young inequality. Sci. Math. Jpn.55, 583–588 (2002)

References

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