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Article

Generalized Refinements of Reversed AM-GM Operator Inequalities for Positive Linear Maps

College of Mathematics and Information Science, Henan Normal University, Xinxiang 453007, China
Axioms 2023, 12(10), 977; https://doi.org/10.3390/axioms12100977
Submission received: 9 September 2023 / Revised: 5 October 2023 / Accepted: 9 October 2023 / Published: 17 October 2023
(This article belongs to the Special Issue Current Research on Mathematical Inequalities II)

Abstract

:
We shall present some more generalized and further refinements of reversed AM-GM operator inequalities for positive linear maps due to Xue’s and Ali’s publications.
MSC:
47A64; 47A30; 47B02

1. Introduction

Let m , M be scalars and I be the identity operator. B ( H ) denote all bounded linear operators acting on a Hilbert space ( H , · , · ) . In addition, A 0 means the operator A is positive. We say A B if A B 0 . A linear map Φ : B ( H ) B ( H ) is called positive (strictly positive) if Φ ( A ) 0 ( Φ ( A ) > 0 ) whenever A 0 ( A > 0 ), and Φ is said to be unital if Φ ( I ) = I .
If A , B B ( H ) are two positive operators, then the operator weighted arithmetic mean and geometric mean are defined as
A v B = ( 1 v ) A + v B and A v B = A 1 2 A 1 2 B A 1 2 v A 1 2
for v [ 0 , 1 ] , respectively. Denoted A v B by A B and A v B by A B when v = 1 2 for brevity. The Kantorovich constant and Specht’s ratio are defined by K ( h , 2 ) = ( h + 1 ) 2 4 h and S ( h ) = h 1 h 1 e ln h 1 h 1 when h > 0 . If there is no special explanations, we always default to a , b > 0 and v [ 0 , 1 ] in this paper.
It is well known that the famous Young’s inequality reads
a 1 v b v ( 1 v ) a + v b .
Furuichi [1] improved (1) with Specht’s ratio
S b a r a 1 v b v ( 1 v ) a + v b ,
where r = min { v , 1 v } . Zuo et al. [2] further improved (2) as
S b a r a 1 v b v K r b a a 1 v b v ( 1 v ) a + v b .
Sababheh and Moslehian [3] gave a refinement of (3) in the following form
K h 2 N , 2 β N ( v ) a 1 v b v + S N ( v ; b , a ) ( 1 v ) a + v b ,
where N N + , h = a b , β N ( v ) = min { α N ( v ) , 1 α N ( v ) } with α N ( v ) = 1 + [ 2 N v ] 2 N v and
S N ( v ; b , a ) = j = 1 N s j ( v ) a 2 j 1 k j ( v ) b k j ( v ) 2 j b k j ( v ) + 1 a 2 j 1 k j ( v ) 1 2 j 2
for j = 1 , 2 , , N , k j ( v ) = [ 2 j 1 v ] , r j ( v ) = [ 2 j v ] and s j ( v ) = ( 1 ) r j ( v ) 2 j 1 v + ( 1 ) r j ( v ) + 1 [ r j ( v ) + 1 2 ] .
Taking v = 1 2 in (1), we can get the following AM-GM operator inequality for any two positive operators A and B,
A B A B .
Lin [4] gave a reverse of inequality (6) involving unital positive linear maps Φ :
Φ ( A B ) K ( h , 2 ) Φ ( A B )
for 0 < m I A , B M I and h = M m . In generally, for any two positive operators A , B and P > 1 ,
A B A P B P .
For example, putting P = 2 , A = 2 1 1 1 and B = 1 1 1 1 . However, Lin [4] showed that the inequality (7) can be squared under the same conditions as in it,
Φ 2 ( A B ) K 2 ( h , 2 ) Φ 2 ( A B )
and
Φ 2 ( A B ) K 2 ( h , 2 ) Φ ( A ) Φ ( B ) 2 .
Moreover, the author [4] found that Specht’s ratio and Kantorovich constant have the following relations
S ( h ) K ( h , 2 ) S 2 ( h )
for h > 1 . Also, Lin [4] conjectured K ( h ) can be replaced by S ( h ) in (9) and (10). Xue [5] solved Lin’s conjecture under some conditions: 0 < m I A , B M I , M m 2.314 and h = M m , she got
Φ 2 ( A B ) K ( h , 2 ) Φ 2 ( A B )
and
Φ 2 ( A B ) K ( h , 2 ) ( Φ ( A ) Φ ( B ) ) 2 .
Recently, Ali et al. [6] gave some refinements of inequalities (12) and (13) as follows:
Theorem 1.
Let 0 < m M , M m 2.314 . For every positive unital linear map Φ,
(1) 
if 0 < m I A , B M + m 2 I , then
Φ 2 A B + M + m 2 m A 1 + B 1 2 ( A 1 B 1 ) M + m 2 M m 2 Φ 2 ( A v B ) ;
Φ 2 A B + M + m 2 m A 1 + B 1 2 ( A 1 B 1 ) M + m 2 M m 2 Φ ( A ) v Φ ( B ) 2 ;
(2) 
if 0 < M + m 2 I A , B M I , then
Φ 2 A B + M + m 2 M A 1 + B 1 2 ( A 1 B 1 ) M + m 2 M m 2 Φ 2 ( A v B ) ;
Φ 2 A B + M + m 2 M A 1 + B 1 2 ( A 1 B 1 ) M + m 2 M m 2 Φ ( A ) v Φ ( B ) 2 ;
(3) 
if 0 < m I A < M + m 2 I B M I , then
Φ 2 A B + M + m 2 m A 1 + M B 1 2 ( m A 1 M B 1 ) M + m 2 M m 2 Φ 2 ( A v B ) ;
Φ 2 A B + M + m 2 m A 1 + M B 1 2 ( m A 1 M B 1 ) M + m 2 M m 2 Φ ( A ) v Φ ( B ) 2 ;
(4) 
if 0 < m I B M + m 2 I A M I , then
Φ 2 A B + M + m 2 M A 1 + m B 1 2 ( M A 1 m B 1 ) M + m 2 M m 2 Φ 2 ( A v B ) ;
Φ 2 A B + M + m 2 M A 1 + m B 1 2 ( M A 1 m B 1 ) M + m 2 M m 2 Φ ( A ) v Φ ( B ) 2 ;
For more information about operator inequalities involving positive linear maps, we refer the readers to [7,8,9,10,11] and references therein.
In this paper, we shall give some more generalized and further refinements of reversed AM-GM operator inequalities for positive linear maps due to Ali’s results.

2. Main Results

We give some lemmas to prove our main results.
Lemma 1.
Let 0 < m I A m I < M I B M I , h = M m , h = M m . Then we have
K h 2 N , 2 β N ( v ) A v B + j = 1 N s j ( v ) A α j ( v ) B + A α j ( v ) + 2 1 j B 2 A α j ( v ) + 2 j B A v B ,
where α j ( v ) = k j ( v ) 2 j 1 , k j ( v ) = [ 2 j 1 v ] and β N ( v ) = min 1 + [ 2 N v ] 2 N v , 2 N v [ 2 N v ] , r j ( v ) = [ 2 j v ] and s j ( v ) = ( 1 ) r j ( v ) 2 j 1 v + ( 1 ) r j ( v ) + 1 r j ( v ) + 1 2 .
Proof. 
Putting a = 1 in (4), we obtain
K 1 b 2 N , 2 β N ( v ) b v + j = 1 N s j ( v ) b α j ( v ) + b α j ( v ) + 2 1 j 2 b α j ( v ) + 2 j ( 1 v ) + v b .
Taking X = A 1 2 B A 1 2 , and then S p ( X ) [ h , h ] ( 1 , + ) . By standard functional calculus and the Kantorovich constant K ( 1 t , 2 ) is a decreasing function on 1 t ( 0 , 1 ) , we get
K 1 h 2 N , 2 β N ( v ) X v + j = 1 N s j ( v ) X α j ( v ) + X α j ( v ) + 2 1 j 2 X α j ( v ) + 2 j ( 1 v ) + v X .
Multiplying A 1 2 on both sides of inequality (15), with the fact K ( h , 2 ) = K 1 h , 2 , we can get (14) directly.  □
Lemma 2.
Under the same conditions as in Lemma 1, we have
A v B + M m K h 2 N , 2 β N ( v ) ( A v B ) 1 + M m S N ( v ; B 1 , A 1 ) M + m ,
where S N ( v ; B 1 , A 1 ) = j = 1 N s j ( v ) A 1 α j ( v ) B 1 + A 1 α j ( v ) + 2 1 j B 1 2 A 1 α j ( v ) + 2 j B 1 .
Proof. 
If 0 < m I A , B M I , then
( M A ) ( A m ) A 1 0 and ( M B ) ( B m ) B 1 0 ,
that is
A + M m A 1 M + m and B + M m B 1 M + m .
So we have
A v B + M m ( A 1 v B 1 ) M + m .
Thus, we obtain
A v B + M m S N ( v ; B 1 , A 1 ) + M m K h 2 N , 2 β N ( v ) ( A v B ) 1 = A v B + M m S N ( v ; B 1 , A 1 ) + K h 2 N , 2 β N ( v ) A 1 v B 1 A v B + M m A 1 v B 1 ( by ( 14 ) ) M + m . ( by ( 18 ) )
 □
Lemma 3
([12]). Let A , B 0 . Then the following norm inequality holds
A B 1 4 A + B 2 .
Lemma 4
([13]). Let Φ be a unital positive linear map and A > 0 . Then
Φ ( A ) 1 Φ ( A 1 ) .
Lemma 5
([14]). (i) If 0 P 1 and A B 0 , then
A P B P .
(ii) Let Φ be a unital positive linear map and A , B > 0 . For v [ 0 , 1 ] , we have
Φ ( A v B ) Φ ( A ) v Φ ( B ) .
Lemma 6
([15]). Let A , B 0 . Then for 1 P < + ,
A P + B P ( A + B ) P .
Theorem 2.
Let 0 < m I M I , M m 2.314 and S N ( v ; B 1 , A 1 ) defined as in Lemma 2. Then for every positive unital linear map Φ and P [ 0 , 2 ] ,
(1) 
if 0 < m I A m 1 I < M 1 I B M + m 2 I , then
Φ P A v B + M + m 2 m S N ( v ; B 1 , A 1 ) K P 2 ( h , 2 ) K P β N ( v ) h 1 2 N , 2 Φ P ( A v B ) ;
Φ P A v B + M + m 2 m S N ( v ; B 1 , A 1 ) K P 2 ( h , 2 ) K P β N ( v ) h 1 2 N , 2 Φ ( A ) v Φ ( B ) P ;
where h = M m , h 1 = M 1 m 1 and v [ 0 , 1 ] .
(2) 
if 0 < M + m 2 I A m 2 I < M 2 I B M I , then
Φ P A v B + M + m 2 M S N ( v ; B 1 , A 1 ) K P 2 ( h , 2 ) K P β N ( v ) h 2 2 N , 2 Φ P ( A v B ) ;
Φ P A v B + M + m 2 M S N ( v ; B 1 , A 1 ) K P 2 ( h , 2 ) K P β N ( v ) h 2 2 N , 2 Φ ( A ) v Φ ( B ) P ;
where h = M m , h 2 = M 2 m 2 and v [ 0 , 1 ] .
(3) 
if 0 < m I A m 3 I < M + m 2 I B M I , then
Φ P A v B + M + m 2 S N ( v ; M B 1 , m A 1 ) M m P 2 ( 1 2 v ) K P 2 ( h , 2 ) K P β N ( v ) h 3 2 N , 2 Φ P ( A v B ) ;
Φ P A v B + M + m 2 S N ( v ; M B 1 , m A 1 ) M m P 2 ( 1 2 v ) K P 2 ( h , 2 ) K P β N ( v ) h 3 2 N , 2 Φ ( A ) v Φ ( B ) P ;
where h = M m , h 3 = M + m 2 m 3 and v [ 0 , 1 2 ] .
(4) 
if 0 < m I B M + m 2 I < M 4 I A M I , then
Φ P A v B + M + m 2 S N ( v ; m B 1 , M A 1 ) M m P 2 ( 2 v 1 ) K P 2 ( h , 2 ) K P β N ( v ) h 4 2 N , 2 Φ P ( A v B ) ;
Φ P A v B + M + m 2 S N ( v ; m B 1 , M A 1 ) M m P 2 ( 2 v 1 ) K P 2 ( h , 2 ) K P β N ( v ) h 4 2 N , 2 Φ ( A ) v Φ ( B ) P ;
where h = M m , h 4 = 2 M 4 M + m and v [ 1 2 , 1 ] .
Proof. 
If 0 < m I A m 1 I < M 1 I B M + m 2 I , we obtain
| | Φ A v B + M + m 2 m S N ( v ; B 1 , A 1 ) × M + m 2 m K h 1 2 N , 2 β N ( v ) Φ 1 ( A v B ) | | 1 4 | | Φ A v B + M + m 2 m S N ( v ; B 1 , A 1 ) + M + m 2 m K h 1 2 N , 2 β N ( v ) Φ 1 ( A v B ) | | 2 1 4 | | Φ A v B + M + m 2 m S N ( v ; B 1 , A 1 ) + M + m 2 m K h 1 2 N , 2 β N ( v ) Φ ( A v B ) 1 | | 2 = 1 4 | | Φ A v B + M + m 2 m K h 1 2 N , 2 β N ( v ) ( A v B ) 1 + M + m 2 m S N ( v ; B 1 , A 1 ) | | 2 1 4 M + m 2 + m 2 ,
where the first inequality is by (19), the second one is by (20), and the last inequality comes from (16). That is
| | Φ A v B + M + m 2 m S N ( v ; B 1 , A 1 ) Φ 1 ( A v B ) | | ( M + m 2 + m ) 2 4 M + m 2 m K h 1 2 N , 2 β N ( v ) .
Since 1 M m 2.314 , it follows that M m 1 2 M m 3 2 M m + M m 4 0 , which is equivalent to
M + m 2 + m 2 4 M + m 2 m M + m 2 M m .
So we have
| | Φ A v B + M + m 2 m S N ( v ; B 1 , A 1 ) Φ 1 ( A v B ) | | M + m 2 M m K h 1 2 N , 2 β N ( v ) .
That is
Φ 2 A v B + M + m 2 m S N ( v ; B 1 , A 1 ) K ( h , 2 ) K h 1 2 N , 2 2 β N ( v ) Φ 2 ( A v B ) .
In addition, we can get
| | Φ A v B + M + m 2 m S N ( v ; B 1 , A 1 ) × M + m 2 m K h 1 2 N , 2 β N ( v ) Φ ( A ) v Φ ( B ) 1 | | 1 4 | | Φ A v B + M + m 2 m S N ( v ; B 1 , A 1 ) + M + m 2 m K h 1 2 N , 2 β N ( v ) Φ ( A ) v Φ ( B ) 1 | | 2 1 4 | | Φ A v B + M + m 2 m S N ( v ; B 1 , A 1 ) + M + m 2 m K h 1 2 N , 2 β N ( v ) Φ 1 ( A v B ) | | 2 1 4 M + m 2 + m 2 ,
where the first inequality is by (19), the second is by (22), and the third is by (32). That is
| | Φ A v B + M + m 2 m S N ( v ; B 1 , A 1 ) Φ ( A ) v Φ ( B ) 1 | | M + m 2 + m 2 4 M + m 2 m K h 1 2 N , 2 β N ( v ) K 1 2 ( h , 2 ) K h 1 2 N , 2 β N ( v ) .
So we have
Φ 2 A v B + M + m 2 m S N ( v ; B 1 , A 1 ) K ( h , 2 ) K h 1 2 N , 2 2 β N ( v ) Φ ( A ) v Φ ( B ) 2 .
We can get (24) and (25) by (34) and (35) with Lemma 5 (i), respectively.
Since M + m 2 + M 2 4 M + m 2 M M + m 2 + m 2 4 M + m 2 m , by 2nd case 0 < M + m 2 I A m 2 I < M 2 I B M I , we can similarly obtain the inequalities (26) and (27) by (16), (17), (19) and (20). So we omit the details.
If 0 < m I A m 3 I < M + m 2 I B M I and v [ 0 , 1 2 ] , then we have
A + M + m 2 m A 1 M + m 2 + m and B + M + m 2 M B 1 M + m 2 + M .
So
A v B + M + m 2 ( m A 1 ) v ( M B 1 ) ( v + 1 2 ) M + ( 3 2 v ) m M + m .
Compute
| | Φ A v B + M + m 2 S N ( v ; M B 1 , m A 1 ) × M + m 2 K h 3 2 N , 2 β N ( v ) m 1 v M v Φ 1 ( A v B ) | | 1 4 | | Φ A v B + M + m 2 S N ( v ; M B 1 , m A 1 ) + M + m 2 K h 3 2 N , 2 β N ( v ) m 1 v M v Φ 1 ( A v B ) | | 2 1 4 | | Φ A v B + M + m 2 S N ( v ; M B 1 , m A 1 ) + M + m 2 K h 3 2 N , 2 β N ( v ) m 1 v M v Φ ( A v B ) 1 | | 2 = 1 4 | | Φ A v B + M + m 2 S N ( v ; M B 1 , m A 1 ) + K h 3 2 N , 2 β N ( v ) ( m A 1 ) v ( M B 1 ) | | 2 1 4 | | Φ A v B + M + m 2 ( m A 1 ) v ( M B 1 ) | | 2 ( M + m ) 2 4 .
where the first inequality is by (19), the second is by (20), the third is by (14), and the last one is by (36). That is
| | Φ A v B + M + m 2 S N ( v ; M B 1 , m A 1 ) Φ 1 ( A v B ) | | ( M + m ) 2 4 M + m 2 m 1 v M v K h 3 2 N , 2 β N ( v ) = M m 1 2 v K 1 2 ( h , 2 ) K h 3 2 N , 2 β N ( v ) .
That is
Φ 2 A v B + M + m 2 S N ( v ; M B 1 , m A 1 ) M m 1 2 v K ( h , 2 ) K h 3 2 N , 2 2 β N ( v ) Φ 2 ( A v B ) .
Moreover,
| | Φ A v B + M + m 2 S N ( v ; M B 1 , m A 1 ) × M + m 2 m 1 v M v K h 3 2 N , 2 β N ( v ) Φ ( A ) v Φ ( B ) 1 | | 1 4 | | Φ A v B + M + m 2 S N ( v ; M B 1 , m A 1 ) + M + m 2 m 1 v M v K h 3 2 N , 2 β N ( v ) Φ ( A ) v Φ ( B ) 1 | | 2 1 4 | | Φ A v B + M + m 2 S N ( v ; M B 1 , m A 1 ) + M + m 2 m 1 v M v K h 3 2 N , 2 β N ( v ) Φ 1 ( A v B ) | | 2 ( M + m ) 2 4 ,
where the first inequality is by (19), the second is by (22), and the third is by (37). That is
| | Φ A v B + M + m 2 S N ( v ; M B 1 , m A 1 ) Φ ( A ) v Φ ( B ) 1 | | M m 1 2 v K 1 2 ( h , 2 ) K h 3 2 N , 2 β N ( v ) ,
so we have
Φ 2 A v B + M + m 2 S N ( v ; M B 1 , m A 1 ) M m 1 2 v K ( h , 2 ) K h 3 2 N , 2 2 β N ( v ) Φ ( A ) v Φ ( B ) 2 .
We can get (28) and (29) by (38) and (39) with Lemma 5 (i), respectively.
We can similarly obtain the inequalities (30) and (31) under the conditions 0 < m I B M + m 2 I < M 3 I A M I and v [ 1 2 , 1 ] . So we omit the details.
Here we complete the proof.  □
Remark 1.
Putting v = 1 2 , N = 1 and P = 2 in Theorem 2, we can get Theorem 1.
Next, we present the generalizations of Theorem 2 for P 2 .
Theorem 3.
Let 0 < m I M I , M m 2.314 and S N ( v ; B 1 , A 1 ) defined as in Lemma 2. Then for every positive unital linear map Φ and P 2 ,
(i) 
if 0 < m I A m 1 I < M 1 I B M + m 2 I , then
Φ P A v B + M + m 2 m S N ( v ; B 1 , A 1 ) 4 P 2 K P 2 ( h , 2 ) K P β N ( v ) h 1 2 N , 2 Φ P ( A v B ) ;
Φ P A v B + M + m 2 m S N ( v ; B 1 , A 1 ) 4 P 2 K P 2 ( h , 2 ) K P β N ( v ) h 1 2 N , 2 Φ ( A ) v Φ ( B ) P ;
where h = M m , h 1 = M 1 m 1 and v [ 0 , 1 ] .
(ii) 
if 0 < M + m 2 I A m 2 I < M 2 I B M I , then
Φ P A v B + M + m 2 M S N ( v ; B 1 , A 1 ) 4 P 2 K P 2 ( h , 2 ) K P β N ( v ) h 2 2 N , 2 Φ P ( A v B ) ;
Φ P A v B + M + m 2 M S N ( v ; B 1 , A 1 ) 4 P 2 K P 2 ( h , 2 ) K P β N ( v ) h 2 2 N , 2 Φ ( A ) v Φ ( B ) P ;
where h = M m , h 2 = M 2 m 2 and v [ 0 , 1 ] .
(iii) 
if 0 < m I A m 3 I < M + m 2 I B M I , then
Φ P A v B + M + m 2 S N ( v ; M B 1 , m A 1 ) 4 P 2 M m P 2 ( 1 2 v ) K P 2 ( h , 2 ) K P β N ( v ) ( h 3 2 N , 2 ) Φ P ( A v B ) ;
Φ P A v B + M + m 2 S N ( v ; M B 1 , m A 1 ) 4 P 2 M m P 2 ( 1 2 v ) K P 2 ( h , 2 ) K P β N ( v ) h 3 2 N , 2 Φ ( A ) v Φ ( B ) P ;
where h = M m , h 3 = M + m 2 m 3 and v [ 0 , 1 2 ] .
(iv) 
if 0 < m I B M + m 2 I < M 4 I A M I , then
Φ P A v B + M + m 2 S N ( v ; m B 1 , M A 1 ) 4 P 2 M m P 2 ( 2 v 1 ) K P 2 ( h , 2 ) K P β N ( v ) ( h 4 2 N , 2 ) Φ P ( A v B ) ;
Φ P A v B + M + m 2 S N ( v ; m B 1 , M A 1 ) 4 P 2 M m P 2 ( 2 v 1 ) K P 2 ( h , 2 ) K P β N ( v ) h 4 2 N , 2 Φ ( A ) v Φ ( B ) P ;
where h = M m , h 4 = 2 M 4 M + m and v [ 1 2 , 1 ] .
Proof. 
The proof of the line (ii) and (iv) are similar to the one presented in (i) and (iii), respectively, thus we omit them. Under the conditions i) 0 < m I A m 1 I < M 1 I B M + m 2 I , we have
| | Φ P 2 A v B + M + m 2 m S N ( v ; B 1 , A 1 ) × M + m 2 P 2 m P 2 K P 2 β N ( v ) h 1 2 N , 2 Φ P 2 ( A v B ) | | 1 4 | | Φ P 2 A v B + M + m 2 m S N ( v ; B 1 , A 1 ) + M + m 2 P 2 m P 2 K P 2 β N ( v ) h 1 2 N , 2 Φ P 2 ( A v B ) | | 2 1 4 | | Φ A v B + M + m 2 m S N ( v ; B 1 , A 1 ) + M + m 2 m K h 1 2 N , 2 β N ( v ) Φ 1 ( A v B ) P 2 | | 2 = 1 4 | | Φ A v B + M + m 2 m S N ( v ; B 1 , A 1 ) + M + m 2 m K h 1 2 N , 2 β N ( v ) Φ 1 ( A v B ) | | P 1 4 M + m 2 + m P .
where the first inequality is by (19), the second is by (23) and the third is by (32). That is
| | Φ P 2 A v B + M + m 2 m S N ( v ; B 1 , A 1 ) Φ P 2 ( A v B ) | | M + m 2 + m P 4 M + m 2 P 2 m P 2 K P 2 β N ( v ) h 1 2 N , 2 4 P 2 1 M + m 2 M m P 2 K P 2 β N ( v ) ( h 1 2 N , 2 ) = 4 P 2 1 K P 4 ( h , 2 ) K P 2 β N ( v ) h 1 2 N , 2 ,
where the second inequality is by (33). So we have
Φ P A v B + M + m 2 m S N ( v ; B 1 , A 1 ) 4 P 2 K P 2 ( h , 2 ) K P β N ( v ) h 1 2 N , 2 Φ P ( A v B ) .
In addition, we can get
| | Φ P 2 A v B + M + m 2 m S N ( v ; B 1 , A 1 ) × M + m 2 P 2 m P 2 K P 2 β N ( v ) h 1 2 N , 2 Φ ( A ) v Φ ( B ) P 2 | | 1 4 | | Φ P 2 A v B + M + m 2 m S N ( v ; B 1 , A 1 ) + M + m 2 P 2 m P 2 K P 2 β N ( v ) h 1 2 N , 2 Φ ( A ) v Φ ( B ) P 2 | | 2 1 4 | | Φ ( A v B + M + m 2 m S N ( v ; B 1 , A 1 ) + M + m 2 m K h 1 2 N , 2 β N ( v ) ( Φ ( A ) v Φ ( B ) ) 1 P 2 | | 2 = 1 4 | | Φ A v B + M + m 2 m S N ( v ; B 1 , A 1 ) + M + m 2 m K h 1 2 N , 2 β N ( v ) Φ ( A ) v Φ ( B ) 1 | | P 1 4 | | Φ ( A v B + M + m 2 m S N ( v ; B 1 , A 1 ) ) + M + m 2 m K h 1 2 N , 2 β N ( v ) Φ 1 ( A v B ) | | P 1 4 M + m 2 + m P .
where the first inequality is by (19), the second is by (23), the third is by (22) and the last is by (32). That is
| | Φ P 2 A v B + M + m 2 m S N ( v ; B 1 , A 1 ) Φ ( A ) v Φ ( B ) P 2 | | 4 P 2 1 K P 4 ( h , 2 ) K P 2 β N ( v ) h 1 2 N , 2 ,
so we have
Φ P A v B + M + m 2 m S N ( v ; B 1 , A 1 ) 4 P 2 K P 2 ( h , 2 ) K P β N ( v ) h 1 2 N , 2 Φ ( A ) v Φ ( B ) P ,
as desired.
If 0 < m I A m 3 I < M + m 2 I B M I and v [ 0 , 1 2 ] , then
| | Φ P 2 A v B + M + m 2 S N ( v ; M B 1 , m A 1 ) × M + m 2 K h 3 2 N , 2 β N ( v ) m 1 v M v P 2 Φ P 2 ( A v B ) | | 1 4 | | Φ P 2 A v B + M + m 2 S N ( v ; M B 1 , m A 1 ) + M + m 2 K h 3 2 N , 2 β N ( v ) m 1 v M v P 2 Φ P 2 ( A v B ) | | 2 1 4 | | Φ ( A v B + M + m 2 S N ( v ; M B 1 , m A 1 ) + M + m 2 K h 3 2 N , 2 β N ( v ) m 1 v M v Φ 1 ( A v B ) ) P 2 | | 2 = 1 4 | | Φ A v B + M + m 2 S N ( v ; M B 1 , m A 1 ) + M + m 2 K h 3 2 N , 2 β N ( v ) m 1 v M v Φ 1 ( A v B ) | | P 1 4 ( M + m ) P .
where the first inequality is by (19), the second is by (23) and the third is by (37). That is
| | Φ P 2 A v B + M + m 2 S N ( v ; M B 1 , m A 1 ) Φ P 2 ( A v B ) | | ( M + m ) P 4 M + m 2 K h 3 2 N , 2 β N ( v ) m 1 v M v P 2 = 2 P 2 M m P 4 ( 1 2 v ) K P 4 ( h , 2 ) K P 2 β N ( v ) h 3 2 N , 2 .
So we have
Φ P A v B + M + m 2 S N ( v ; M B 1 , m A 1 ) 4 P 2 M m P 2 ( 1 2 v ) K P 2 ( h , 2 ) K P β N ( v ) h 3 2 N , 2 Φ P ( A v B ) .
At the same time, we can get
| | Φ P 2 A v B + M + m 2 S N ( v ; M B 1 , m A 1 ) × M + m 2 K h 3 2 N , 2 β N ( v ) m 1 v M v P 2 Φ ( A ) v Φ ( B ) P 2 | | 1 4 | | Φ P 2 A v B + M + m 2 S N ( v ; M B 1 , m A 1 ) + M + m 2 K h 3 2 N , 2 β N ( v ) m 1 v M v P 2 Φ ( A ) v Φ ( B ) P 2 | | 2 1 4 | | Φ A v B + M + m 2 S N ( v ; M B 1 , m A 1 ) + M + m 2 K h 3 2 N , 2 β N ( v ) m 1 v M v Φ ( A ) v Φ ( B ) 1 P 2 | | 2 = 1 4 | | Φ A v B + M + m 2 S N ( v ; M B 1 , m A 1 ) + M + m 2 K h 3 2 N , 2 β N ( v ) m 1 v M v Φ ( A ) v Φ ( B ) 1 | | P 1 4 | | Φ A v B + M + m 2 S N ( v ; M B 1 , m A 1 ) + M + m 2 K h 3 2 N , 2 β N ( v ) m 1 v M v Φ 1 ( A v B ) | | P 1 4 ( M + m ) P .
where the first inequality is by (19), the second is by (23), the third is by (22) and the last is by (37). That is
| | Φ P 2 A v B + M + m 2 S N ( v ; M B 1 , m A 1 ) Φ ( A ) v Φ ( B ) P 2 | | ( M + m ) P 4 M + m 2 m 1 v M v P 2 K P 2 β N ( v ) h 3 2 N , 2 = 2 P 2 M m P 4 ( 1 2 v ) K P 4 ( h , 2 ) K P 2 β N ( v ) h 3 2 N , 2 .
So we have
Φ P ( A v B + M + m 2 ( S N ( v ; M B 1 , m A 1 ) ) ) 4 P 2 ( M m ) P 2 ( 1 2 v ) K P 2 ( h , 2 ) K P β N ( v ) ( h 3 2 N , 2 ) ( Φ ( A ) v Φ ( B ) ) P .
Here we complete the proof.  □
Theorems 2 and 3 implies the following results.
Corollary 1.
Let 0 < m M , M m 2.314 and S N ( v ; B 1 , A 1 ) defined as in Lemma 2. Then for every positive unital linear map Φ and P 0 ,
(i) 
if 0 < m I A m 1 I < M 1 I B M + m 2 I , v [ 0 , 1 ] , then
Φ P A v B + M + m 2 m S N ( v ; B 1 , A 1 ) W 1 Φ P ( A v B ) ;
Φ P A v B + M + m 2 m S N ( v ; B 1 , A 1 ) W 1 Φ ( A ) v Φ ( B ) P ;
where h = M m , h 1 = M 1 m 1 , W 1 = max K P 2 ( h , 2 ) K P β N ( v ) h 1 2 N , 2 , 4 P 2 K P 2 ( h , 2 ) K P β N ( v ) h 1 2 N , 2 .
(ii) 
if 0 < M + m 2 I A m 2 I < M 2 I B M I , v [ 0 , 1 ] , then
Φ P A v B + M + m 2 M S N ( v ; B 1 , A 1 ) W 2 Φ P ( A v B ) ;
Φ P A v B + M + m 2 M S N ( v ; B 1 , A 1 ) W 2 Φ ( A ) v Φ ( B ) P ;
where h = M m , h 2 = M 2 m 2 , W 2 = max K P 2 ( h , 2 ) K P β N ( v ) h 2 2 N , 2 , 4 P 2 K P 2 ( h , 2 ) K P β N ( v ) h 1 2 N , 2 .
(iii) 
if 0 < m I A m 3 I < M + m 2 I B M I , v [ 0 , 1 2 ] , then
Φ P A v B + M + m 2 S N ( v ; M B 1 , m A 1 ) W 3 Φ P ( A v B ) ;
Φ P A v B + M + m 2 S N ( v ; M B 1 , m A 1 ) W 3 Φ ( A ) v Φ ( B ) P ;
where h = M m , h 3 = M + m 2 m 3 , W 3 = max M m P 2 ( 1 2 v ) K P 2 ( h , 2 ) K P β N ( v ) h 3 2 N , 2 , 4 P 2 M m P 2 ( 1 2 v ) K P 2 ( h , 2 ) K P β N ( v ) h 3 2 N , 2 .
(iv) 
if 0 < m I B M + m 2 I < M 4 I A M I , v [ 1 2 , 1 ] , then
Φ P A v B + M + m 2 ( S N ( v ; m B 1 , M A 1 ) ) W 4 Φ P ( A v B ) ;
Φ P A v B + M + m 2 ( S N ( v ; m B 1 , M A 1 ) ) W 4 Φ ( A ) v Φ ( B ) P ;
where h = M m , h 4 = 2 M 4 M + m , W 4 = max M m P 2 ( 2 v 1 ) K P 2 ( h , 2 ) K P β N ( v ) h 4 2 N , 2 , 4 P 2 M m P 2 ( 2 v 1 ) K P 2 ( h , 2 ) K P β N ( v ) h 4 2 N , 2 .

Funding

This research received no external funding.

Data Availability Statement

Not applicable.

Conflicts of Interest

The author declares no conflict of interest.

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Ren, Y. Generalized Refinements of Reversed AM-GM Operator Inequalities for Positive Linear Maps. Axioms 2023, 12, 977. https://doi.org/10.3390/axioms12100977

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Ren Y. Generalized Refinements of Reversed AM-GM Operator Inequalities for Positive Linear Maps. Axioms. 2023; 12(10):977. https://doi.org/10.3390/axioms12100977

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Ren, Yonghui. 2023. "Generalized Refinements of Reversed AM-GM Operator Inequalities for Positive Linear Maps" Axioms 12, no. 10: 977. https://doi.org/10.3390/axioms12100977

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Ren, Y. (2023). Generalized Refinements of Reversed AM-GM Operator Inequalities for Positive Linear Maps. Axioms, 12(10), 977. https://doi.org/10.3390/axioms12100977

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