Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2013, Article ID 124510, 6 pages
http://dx.doi.org/10.1155/2013/124510
Research Article
Some Operator Inequalities on Chaotic Order and
Monotonicity of Related Operator Function
Changsen Yang and Yanmin Liu
College of Mathematics and Information Science, Henan Normal University, Xinxiang 453002, China
Correspondence should be addressed to Changsen Yang; yangchangsen0991@sina.com
Received 24 March 2013; Accepted 24 April 2013
Academic Editor: Yisheng Song
Copyright © 2013 C. Yang and Y. Liu. This is an open access article distributed under the Creative Commons Attribution License,
which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
We will discuss some operator inequalities on chaotic order about several operators, which are generalization of Furuta inequality
and show monotonicity of related Furuta type operator function.
1. Introduction
An operator 𝑇 is said to be positive (denoted by 𝑇 ≥ 0) if
(𝑇𝑥, 𝑥) ≥ 0 for all vectors 𝑥 in a Hilbert space, and 𝑇 is said
to be strictly positive (denoted by 𝑇 > 0) if 𝑇 is positive and
invertible.
Theorem LH (Löwner-Heinz inequality, denoted by (LH)
briefly). If 𝐴 ≥ 𝐵 ≥ 0 holds, then 𝐴𝛼 ≥ 𝐵𝛼 for any 𝛼 ∈ [0, 1].
This was originally proved in [1, 2] and then in [3].
Although (LH) asserts that 𝐴 ≥ 𝐵 ≥ 0 ensures 𝐴𝛼 ≥ 𝐵𝛼 for
any 𝛼 ∈ [0, 1], unfortunately 𝐴𝛼 ≥ 𝐵𝛼 does not always hold
for 𝛼 > 1. The following result has been obtained from this
point of view.
Theorem A. For 𝐴, 𝐵 > 0, the following (i) and (ii) hold:
(i) 𝐴 ≫ 𝐵 holds if and only if 𝐴𝑟 ≥ (𝐴𝑟/2 𝐵𝑝 𝐴𝑟/2 )𝑟/(𝑝+𝑟) for
𝑝, 𝑟 ≥ 0;
(ii) 𝐴 ≫ 𝐵 holds if and only if for any fixed 𝛿 ≥ 0,
𝐹𝐴,𝐵 (𝑝, 𝑟) = 𝐴−𝑟/2 (𝐴𝑟/2 𝐵𝑝 𝐴𝑟/2 )(𝛿+𝑟)/(𝑝+𝑟) 𝐴−𝑟/2 is a decreasing function of 𝑝 ≥ 𝛿 and 𝑟 ≥ 0.
(i) in Theorem A is shown in [9, 10], an excellent proof in
[11], a proof in the case 𝑝 = 𝑟 in [12], (ii) in [9, 10], and so
forth.
Lemma B (see [11]). Let 𝐴 be a positive invertible operator,
and let 𝐵 be an invertible operator. For any real number 𝜆,
𝜆−1
Theorem F (Furuta inequality). If 𝐴 ≥ 𝐵 ≥ 0, then for each
𝑟 ≥ 0,
(i) (𝐵𝑟/2 𝐴𝑝 𝐵𝑟/2 )1/𝑞 ≥ (𝐵𝑟/2 𝐵𝑝 𝐵𝑟/2 )1/𝑞 ,
(ii) (𝐴𝑟/2 𝐴𝑝 𝐴𝑟/2 )1/𝑞 ≥ (𝐴𝑟/2 𝐵𝑝 𝐴𝑟/2 )1/𝑞
hold for 𝑝 ≥ 0 and 𝑞 ≥ 1 with (1 + 𝑟)𝑞 ≥ 𝑝 + 𝑟.
The original proof of Theorem F is shown in [4], an
elementary one-page proof is in [5], and alternative ones are
in [6, 7]. We remark that the domain of the parameters 𝑝, 𝑞,
and 𝑟 in Theorem F is the best possible for the inequalities (i)
and (ii) under the assumption 𝐴 ≥ 𝐵 ≥ 0; see [8].
We write 𝐴 ≫ 𝐵 if log 𝐴 ≥ log 𝐵 for 𝐴, 𝐵 > 0, which is
called the chaotic order.
(𝐵𝐴𝐵∗ ) = 𝐵𝐴1/2 (𝐴1/2 𝐵∗ 𝐵𝐴1/2 )
𝐴1/2 𝐵∗ .
(1)
Definition 1. Let 𝐴 𝑛 , 𝐴 𝑛−1 , . . . , 𝐴 2 , 𝐴 1 , 𝐵 ≥ 0, 𝑟1 , 𝑟2 , . . . , 𝑟𝑛 ≥
0, and 𝑝1 , 𝑝2 , . . . , 𝑝𝑛 ≥ 0 for a natural number 𝑛.
Let 𝐶𝐴 𝑖 ,𝐵 [𝑛] be defined by
𝐶𝐴 𝑖 ,𝐵 [𝑛]
𝑟
/2
𝑟 /2
𝑟 /2
𝑟 /2 𝑝2
𝑟 /2
𝑟 /2
𝑝3
𝑛−1
[⋅ ⋅ ⋅ 𝐴 33 {𝐴 22 (𝐴 11 𝐵𝑝1 𝐴 11 ) 𝐴 22 }
= 𝐴𝑟𝑛𝑛 /2 {𝐴 𝑛−1
𝑟 /2
𝑟
/2
𝑝𝑛
𝑛−1
× 𝐴 33 ⋅ ⋅ ⋅ ] 𝐴 𝑛−1
} 𝐴𝑟𝑛𝑛 /2 .
(2)
2
Abstract and Applied Analysis
Theorem 4. If 𝐴 𝑛 ≫ 𝐴 𝑛−1 ≫ ⋅ ⋅ ⋅ ≫ 𝐴 2 ≫ 𝐴 1 ≫ 𝐵 and
𝑟1 , 𝑟2 , . . . , 𝑟𝑛 ≥ 0, 𝑝1 , 𝑝2 , . . . , 𝑝𝑛 ≥ 0 for a natural number 𝑛.
Then the following inequality holds:
For example,
𝑟 /2
𝑟 /2 𝑝2
𝑟 /2
𝑟 /2
𝐶𝐴 𝑖 ,𝐵 [2] = 𝐴 22 (𝐴 11 𝐵𝑝1 𝐴 11 ) 𝐴 22 ,
𝑟 /2
𝑟 /2
𝑟 /2
𝑟 /2 𝑝2
𝑟 /2
𝑟 /2
𝑝3
𝐶𝐴 𝑖 ,𝐵 [4] = 𝐴 44 {𝐴 33 [𝐴 22 (𝐴 11 𝐵𝑝1 𝐴 11 ) 𝐴 22 ]
𝑝4
𝑟 /2
× 𝐴 33
𝐴 𝑛 ≫ 𝐶𝐴 𝑖 ,𝐵 [𝑛]1/𝑞[𝑛] ,
𝑟 /2
} 𝐴 44 .
(3)
Let 𝑞[𝑛] be defined by
𝑞 [𝑛] = {⋅ ⋅ ⋅ [(𝑝1 + 𝑟1 ) 𝑝2 + 𝑟2 ] 𝑝3 + ⋅ ⋅ ⋅ + 𝑟𝑛−1 } 𝑝𝑛 + 𝑟𝑛 . (4)
(9)
where 𝐶𝐴 𝑖 ,𝐵 [𝑛] and 𝑞[𝑛] are defined in (2) and (4).
Proof. We will show (9) by mathematical induction. In the
case 𝑛 = 1.
Since 𝐴 1 ≫ 𝐵 implies
𝑟 /2
𝑟 /2 𝑟1 /(𝑝1 +𝑟1 )
𝐴 1 ≫ (𝐴 11 𝐵𝑝1 𝐴 11 )
For example,
𝑞 [1] = 𝑝1 + 𝑟1 ,
𝑞 [2] = (𝑝1 + 𝑟1 ) 𝑝2 + 𝑟2 ,
𝑞 [4] = {[(𝑝1 + 𝑟1 ) 𝑝2 + 𝑟2 ] 𝑝3 + 𝑟3 } 𝑝4 + 𝑟4 .
(5)
For the sake of convenience, we define
𝐶𝐴 𝑖 ,𝐵 [0] = 𝐵,
𝑞 [0] = 1,
(6)
(10)
holds for any 𝑝1 ≥ 0 and 𝑟1 ≥ 0 by Lemma 3, whence (9) for
𝑛 = 1.
Assume that (9) holds for a natural number 𝑘 (1 ≤ 𝑘 <
𝑛). We will show that (9) holds 𝑟1 , 𝑟2 , . . . , 𝑟𝑘 , 𝑟𝑘+1 ≥ 0 and
𝑝1 , 𝑝2 , . . . , 𝑝𝑘 , 𝑝𝑘+1 ≥ 0 for 𝑘 + 1.
Put 𝐷 = 𝐴 𝑘+1 , 𝐸 = 𝐴 𝑘 , and 𝐹 = C𝐴 𝑖 ,𝐵 [𝑘]1/𝑞[𝑘] , and (9)
holds for 𝑛 = 𝑘 implying
and these definitions in (6) may be reasonable by (2) and (4).
𝐷 ≫ 𝐸 ≫ 𝐹 > 0.
Lemma 2. For 𝐴 𝑛 , 𝐴 𝑛−1 , . . . , 𝐴 2 , 𝐴 1 , 𝐵 ≥ 0 and any natural
number 𝑛, we have
Equation (11) yields the following by Lemma 3, for 𝑟 ≥ 0 and
𝑝 ≥ 0:
(i) 𝐶𝐴 𝑖 ,𝐵 [𝑛] = 𝐴𝑟𝑛𝑛 /2 𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 𝐴𝑟𝑛𝑛 /2 ,
(ii) 𝑞[𝑛] = 𝑞[𝑛 − 1]𝑝𝑛 + 𝑟𝑛 .
Proof. (i) and (ii) can be easily obtained by definitions (2) and
(4).
Lemma 3. If 𝐴 ≫ 𝐵, for 𝑝 ≥ 0 and 𝑟 ≥ 0, then 𝐴 ≫
(𝐴𝑟/2 𝐵𝑝 𝐴𝑟/2 )1/(𝑝+𝑟) .
Proof. Since 𝐴 ≫ 𝐵, we can obtain the following inequality.
𝐴𝑟 ≥ (𝐴𝑟/2 𝐵𝑝 𝐴𝑟/2 )𝑟/(𝑝+𝑟) holds for 𝑝 ≥ 0 and 𝑟 ≥ 0 by (i)
of Theorem A.
Take the logarithm on both sides of the previous inequality; that is,
𝑟/2 𝑟/(𝑝+𝑟)
log 𝐴 ≥ log (𝐴 𝐵 𝐴 )
(12)
,
1
1
1
=
=
,
𝑝 + 𝑟 𝑞 [𝑘] 𝑝𝑘+1 + 𝑟𝑘+1 𝑞 [𝑘 + 1]
.
(8)
(13)
(14)
and we have the following desired (15) by (12) and (13):
(7)
therefor we have
.
Put 𝑟 = 𝑟𝑘+1 , 𝑝 = 𝑞[𝑘]𝑝𝑘+1 in (13), then by (ii) of Lemma 2,
the exponential power 1/(𝑝 + 𝑟) of the right hand side of (13)
can be written as follows:
𝑟
1/(𝑝+𝑟)
,
𝑝/𝑞[𝑘] 𝑟/2
𝐴 𝑘+1 )
𝐴 𝑘+1 ≫ (𝐴𝑟/2
𝑘+1 𝐶𝐴 𝑖 ,𝐵 [𝑘]
𝑝𝑘+1
/2
𝑘+1
𝐴 𝑘+1 ≫ {𝐴 𝑘+1
(𝐶𝐴 𝑖 ,𝐵 [𝑘])
𝐴 ≫ (𝐴𝑟/2 𝐵𝑝 𝐴𝑟/2 )
1/(𝑝+𝑟)
1/(𝑝+𝑟)
We will give some operator inequalities on chaotic order, and
Theorem 5 is further extension of Theorem 3.1 in [13].
𝑟/2 𝑝
𝐷 ≫ (𝐷𝑟/2 𝐹𝑝 𝐷𝑟/2 )
that is,
2. Basic Results Associated with
𝐶𝐴 𝑖 ,𝐵 [𝑛] and 𝑞[𝑛]
𝑟
(11)
= 𝐶𝐴 𝑖 ,𝐵 [𝑘 + 1]
1/𝑞[𝑘+1]
𝑟
/2 1/𝑞[𝑘+1]
𝑘+1
𝐴 𝑘+1
}
,
so that (15) shows that (9) holds for 𝑘 + 1.
(15)
Abstract and Applied Analysis
3
Theorem 5. If 𝐴 𝑛 ≫ 𝐴 𝑛−1 ≫ ⋅ ⋅ ⋅ ≫ 𝐴 2 ≫ 𝐴 1 ≫ 𝐵 and
𝑟1 , 𝑟2 , . . . , 𝑟𝑛 ≥ 0 for a natural number 𝑛. For any fixed 𝛿 ≥ 0,
let 𝑝1 , 𝑝2 , . . . , 𝑝𝑛 be satisfied by
−𝑟
𝑟
/2
×
𝛿 + 𝑟1 + 𝑟2 + ⋅ ⋅ ⋅ + 𝑟𝑘−1
𝑝𝑘 ≥
,
𝑞 [𝑘 − 1]
(16)
(24)
𝑟𝑘+1 /2 (1+𝑟)/(𝑝+𝑟) −𝑟𝑘+1 /2
𝐴 𝑘+1
)
𝐴 𝑘+1 .
Putting 𝑝 = (𝑞[𝑘]𝑝𝑘+1 )/(𝛿 + 𝑟1 + 𝑟2 + ⋅ ⋅ ⋅ + 𝑟𝑘 ) ≥ 1, since
𝑝𝑘+1 ≥ (𝛿 + 𝑟1 + 𝑟2 + ⋅ ⋅ ⋅ + 𝑟𝑘 )/𝑞[𝑘] in (16), then we have
𝑟 /2
..
.
𝑟 /2
𝐴 𝑘𝑘 𝐼𝑘 (𝑝𝑘 , 𝑟𝑘 ) 𝐴 𝑘𝑘
= 𝐵1 = 𝐶𝐴 𝑖 ,𝐵 [𝑘](𝛿+𝑟1 +𝑟2 +⋅⋅⋅+𝑟𝑘 )/𝑞[𝑘]
𝛿 + 𝑟1 + 𝑟2 + ⋅ ⋅ ⋅ + 𝑟𝑛−1
.
𝑝𝑛 ≥
𝑞 [𝑛 − 1]
−𝑟
The operator function 𝐼𝑘 (𝑝𝑘 , 𝑟𝑘 ) for any natural number 𝑘 such
that 1 ≤ 𝑘 ≤ 𝑛 is defined by
−𝑟 /2
−𝑟 /2
𝐼𝑘 (𝑝𝑘 , 𝑟𝑘 ) = 𝐴 𝑘 𝑘 𝐶𝐴 𝑖 ,𝐵 [𝑘](𝛿+𝑟1 +𝑟2 +⋅⋅⋅+𝑟𝑘 )/𝑞[𝑘] 𝐴 𝑘 𝑘 .
(17)
/2
/2
𝑟
/2
−𝑟
/2
𝑟
/2 (1+𝑟)/(𝑝+𝑟)
𝑘+1
𝑘+1
× (𝐴 𝑘+1
𝐶𝐴 𝑖 ,𝐵 [𝑘]((𝛿+𝑟1 +𝑟2 +⋅⋅⋅+𝑟𝑘 )/𝑞[𝑘])𝑝 𝐴 𝑘+1
)
𝑘+1
× 𝐴 𝑘+1
−𝑟
𝑟
/2
𝑘−1
𝑘−1
𝐴 𝑘−1
𝐼𝑘−1 (𝑝𝑘−1 , 𝑟𝑘−1 ) 𝐴 𝑘−1
≥ 𝐼𝑘 (𝑝𝑘 , 𝑟𝑘 )
(18)
for every natural number 𝑘 such that 1 ≤ 𝑘 ≤ 𝑛, where 𝐶𝐴 𝑖 ,𝐵 [𝑛]
and 𝑞[𝑛] are defined in (2) and (4).
Proof. Since 𝐶𝐴 𝑖 ,𝐵 [0] = 𝐵, 𝑞[0] = 1 in (6), we may define
𝐼0 (𝑝0 , 𝑟0 ) = 𝐵𝛿 for 𝑝0 = 𝑟0 = 0.
Because 𝐴 1 ≫ 𝐵, then for any fixed 𝛿 ≥ 0,
𝑟 /2 (𝛿+𝑟1 )/(𝑝1 +𝑟1 )
𝑟 /2
𝑘+1
≥ 𝐴 𝑘+1
/2
−𝑟
𝑘+1
𝑘+1
= 𝐴 𝑘+1
𝐶𝐴 𝑖 ,𝐵 [𝑘 + 1](𝛿+𝑟1 +𝑟2 +⋅⋅⋅+𝑟𝑘 +𝑟𝑘+1 )/(𝑞[𝑘+1]) 𝐴 𝑘+1
Then the following inequality holds:
−𝑟 /2
/2
𝑘+1
𝑘+1
(𝐴 𝑘+1
𝐶𝐴 𝑖 ,𝐵 [𝑘]((𝛿+𝑟1 +𝑟2 +⋅⋅⋅+𝑟𝑘 )/𝑞[𝑘])𝑝
𝐵1 ≥ 𝐴 𝑘+1
𝑝1 ≥ 𝛿,
𝛿 + 𝑟1
𝑝2 ≥
,
𝑝1 + 𝑟1
..
.
𝑟
Putting 𝑟𝑘+1 = 𝑟(𝛿 + 𝑟1 + 𝑟2 + ⋅ ⋅ ⋅ + 𝑟𝑘 ) in (23), then (23)
can be rewritten by
𝐵𝛿 ≥ 𝐴 1 1 (𝐴 11 𝐵𝑝1 𝐴 11 )
−𝑟 /2
𝐴1 1
(19)
/2
= 𝐼𝑘+1 (𝑝𝑘+1 , 𝑟𝑘+1 ) ,
(25)
and we have (18) for 𝑘 such that 1 ≤ 𝑘 ≤ 𝑛 by (25) and (20)
since (20) means (18) for 𝑘 = 1.
Corollary 6. If 𝐴 𝑛 ≫ 𝐴 𝑛−1 ≫ ⋅ ⋅ ⋅ ≫ 𝐴 2 ≫ 𝐴 1 ≫ 𝐵 and
𝑟1 , 𝑟2 , . . . , 𝑟𝑛 ≥ 0 for a natural number 𝑛. For any fixed 𝛿 ≥ 0,
let 𝑝1 , 𝑝2 , . . . , 𝑝𝑛 be satisfied by (16).
Then the following inequalities hold:
for 𝑝1 ≥ 𝛿, 𝑟1 ≥ 0,
since 𝐹𝐴 1 ,𝐵 (𝛿, 𝑟0 ) ≥ 𝐹𝐴 1 ,𝐵 (𝑝1 , 𝑟1 ) holds by (ii) of Theorem A.
And (19) can be expressed as
𝑟 /2
𝑟 /2
𝐵𝛿 = 𝐴 00 𝐼0 (𝑝0 , 𝑟0 ) 𝐴 00
≥ 𝐼1 (𝑝1 , 𝑟1 ) .
(20)
(21)
Since 𝑋 ≫ 𝑌 implies that 𝑋𝑡 ≫ 𝑌𝑡 holds for any 𝑡 ≥ 0, (21)
ensures
𝛿+𝑟 +𝑟2 +⋅⋅⋅+𝑟𝑘
𝐴 𝑘+11
≫ 𝐶𝐴 𝑖 ,𝐵 [𝑘](𝛿+𝑟1 +𝑟2 +⋅⋅⋅+𝑟𝑘 )/𝑞[𝑘] .
(22)
𝛿+𝑟 +𝑟 +⋅⋅⋅+𝑟
𝑘
Putting 𝐴 = 𝐴 𝑘+11 2
, 𝐵1 = 𝐶𝐴 𝑖 ,𝐵 [𝑘](𝛿+𝑟1 +𝑟2 +⋅⋅⋅+𝑟𝑘 )/𝑞[𝑘] and
applying (19) for 𝛿 = 1 and 𝐴 ≫ 𝐵1 , we have
𝑝
𝐵1 ≥ 𝐴−𝑟/2 (𝐴𝑟/2 𝐵1 𝐴𝑟/2 )
holds for 𝑝 ≥ 1 and 𝑟 ≥ 0.
(1+𝑟)/(𝑝+𝑟)
𝐴−𝑟/2
(23)
𝑟 /2
−𝑟 /2
−𝑟 /2
𝑟 /2 (𝛿+𝑟1 )/(𝑝1 +𝑟1 )
−𝑟 /2
𝐴1 1
≥ 𝐴1 1 𝐴2 2
𝑟 /2
𝑟 /2
−𝑟 /2
−𝑟 /2
𝑟 /2 𝑝2
𝑟 /2
× [𝐴 22 (𝐴 11 𝐵𝑝1 𝐴 11 ) 𝐴 22 ]
We can apply Theorem 4, and we have the following (21) for
any natural number 𝑘 such that 1 ≤ 𝑘 ≤ 𝑛:
𝐴 𝑘+1 ≫ 𝐴 𝑘 ≫ 𝐶𝐴 𝑖 ,𝐵 [𝑘]1/𝑞[𝑘] .
−𝑟 /2
𝐵𝛿 ≥ 𝐴 1 1 (𝐴 11 𝐵𝑝1 𝐴 11 )
(𝛿+𝑟1 +𝑟2 )/((𝑝1 +𝑟1 )𝑝2 +𝑟2 )
× 𝐴2 2 𝐴1 1
..
.
−𝑟 /2
−𝑟 /2
−𝑟 /3
−𝑟
/2
𝑛−1
𝑛 /2
𝐴−𝑟
≥ 𝐴 1 1 𝐴 2 2 𝐴 3 3 ⋅ ⋅ ⋅ 𝐴 𝑛−1
𝑛
× 𝐶𝐴 𝑖 ,𝐵 [𝑛](𝛿+𝑟1 +𝑟2 +⋅⋅⋅+𝑟𝑛 )/𝑞[𝑛]
−𝑟
/2
−𝑟 /3
−𝑟 /2
−𝑟 /2
𝑛−1
𝑛 /2
× 𝐴−𝑟
𝐴 𝑛−1
⋅ ⋅ ⋅ 𝐴3 3 𝐴2 2 𝐴1 1 ,
𝑛
(26)
where 𝐶𝐴 𝑖 ,𝐵 [𝑛], 𝑞[𝑛], and 𝐼𝑘 (𝑝𝑘 , 𝑟𝑘 ) (1 ≤ 𝑘 ≤ 𝑛) are defined in
(2), (4), and (17).
4
Abstract and Applied Analysis
Proof. Applying (18) of Theorem 5 for 𝑘 such that 1 ≤ 𝑘 ≤ 𝑛,
we have
We can obtain the following inequality from the hypothesis (28) for the case 𝑛 = 𝑘:
𝑟
𝐴 𝑘𝑘 ≥ 𝐶𝐴 𝑖 ,𝐵 [𝑘]𝑟𝑘 /𝑞[𝑘] ,
𝐵𝛿 = 𝐴𝑟0 /2 𝐼0 (𝑝0 , 𝑟0 ) 𝐴𝑟0 /2
≥ 𝐼1 (𝑝1 , 𝑟1 )
−𝑟 /2
𝑟 /2 (𝛿+𝑟1 )/(𝑝1 +𝑟1 )
𝑟 /2
= 𝐴 1 1 (𝐴 11 𝐵𝑝1 𝐴 11 )
≥
−𝑟 /2
𝐴 1 1 𝐼2
=
−𝑟 /2 −𝑟 /2
𝐴1 1 𝐴2 2
hence we have 𝐴 𝑘+1 ≫ 𝐴 𝑘 ≫ 𝐶𝐴 𝑖 ,𝐵 [𝑘]1/𝑞[𝑘] , and (i) of
Theorem A ensures
−𝑟 /2
𝐴1 1
𝑟/(𝑝+𝑟)
𝑝/𝑞[𝑘] 𝑟/2
𝐴𝑟𝑘+1 ≥ (𝐴𝑟/2
𝐴 𝑘+1 )
𝑘+1 𝐶𝐴 𝑖 ,𝐵 [𝑘]
−𝑟 /2
(𝑝2 , 𝑟2 ) 𝐴 1 1
×
−𝑟 /2
𝑟
−𝑟 /2
(27)
−𝑟 /2 −𝑟 /2 −𝑟 /3
𝐴1 1 𝐴2 2 𝐴3 3
−𝑟
/2
−𝑟𝑛−1 /2
⋅ ⋅ ⋅ 𝐴 𝑛−1
𝐼𝑛
−𝑟 /3
−𝑟 /2
−𝑟 /2
(𝑝𝑛 , 𝑟𝑛 )
−𝑟 /2
−𝑟 /3
−𝑟
/2 𝑟𝑘+1 /(𝑞[𝑘]𝑝𝑘+1 +𝑟𝑘+1 )
(32)
= 𝐶𝐴 𝑖 ,𝐵 [𝑘 + 1]𝑟𝑘+1 /𝑞[𝑘+1] ,
𝑛 /2
𝑛 /2
𝐼𝑛 (𝑝𝑛 , 𝑟𝑛 ) = 𝐴−𝑟
𝐶𝐴 𝑖 ,𝐵 [𝑛](𝛿+𝑟1 +𝑟2 +⋅⋅⋅+𝑟𝑛 )/𝑞[𝑛] 𝐴−𝑟
𝑛
𝑛
/2
𝑛−1
𝑛 /2
= 𝐴 1 1 𝐴 2 2 𝐴 3 3 ⋅ ⋅ ⋅ 𝐴 𝑛−1
𝐴−𝑟
𝑛
/2
−𝑟 /3
−𝑟 /2
(33)
is a decreasing function of both 𝑟𝑛 ≥ 0 and 𝑝𝑛 which satisfies
× 𝐶𝐴 𝑖 ,𝐵 [𝑛](𝛿+𝑟1 +𝑟2 +⋅⋅⋅+𝑟𝑛 )/𝑞[𝑛]
−𝑟
𝑟
Theorem 8. If 𝐴 𝑛 ≫ 𝐴 𝑛−1 ≫ ⋅ ⋅ ⋅ ≫ 𝐴 2 ≫ 𝐴 1 ≫ 𝐵 and
𝑟1 , 𝑟2 , . . . , 𝑟𝑛 ≥ 0 for a natural number 𝑛. For any fixed 𝛿 ≥ 0,
let 𝑝1 , 𝑝2 , . . . , 𝑝𝑛 be satisfied by (16).
Then
𝑛−1
× 𝐴 𝑛−1
⋅ ⋅ ⋅ 𝐴3 3 𝐴2 2 𝐴1 1
−𝑟 /2
/2
so that (32) shows (28) for 𝑘 + 1.
..
.
≥
𝑟
𝑘+1
𝑘+1
𝑘+1
𝐴 𝑘+1
≥ (𝐴 𝑘+1
𝐶𝐴 𝑖 ,𝐵 [𝑘]𝑝𝑘+1 𝐴 𝑘+1
)
(𝛿+𝑟1 +𝑟2 )/((𝑝1 +𝑟1 )𝑝2 +𝑟2 )
× 𝐴2 2 𝐴1 1
for 𝑝, 𝑟 ≥ 0. (31)
Putting 𝑟 = 𝑟𝑘+1 and 𝑝 = 𝑞[𝑘]𝑝𝑘+1 , then we have the following
inequality:
𝑟 /2
𝑟 /2
𝑟 /2 𝑝2
[𝐴 22 (𝐴 11 𝐵𝑝1 𝐴 11 )
𝑟 /2
𝐴 22 ]
(30)
𝑝𝑛 ≥
−𝑟 /2
𝑛−1
𝑛 /2
× 𝐴−𝑟
𝐴 𝑛−1
⋅ ⋅ ⋅ 𝐴3 3 𝐴2 2 𝐴1 1 .
𝑛
𝛿 + 𝑟1 + 𝑟2 + ⋅ ⋅ ⋅ + 𝑟𝑛−1
,
𝑞 [𝑛 − 1]
(34)
where 𝐶𝐴 𝑖 ,𝐵 [𝑛] and 𝑞[𝑛] are defined in (2) and (4).
3. Monotonicity Property on
Operator Functions
We would like to emphasize that the condition of Theorem 7
is stronger than Theorem 5, and moreover when we discuss
monotonicity property on operator functions, we can only
apply Theorem 7.
Theorem 7. If 𝐴 𝑛 ≫ 𝐴 𝑛−1 ≫ ⋅ ⋅ ⋅ ≫ 𝐴 2 ≫ 𝐴 1 ≫ 𝐵 and
𝑟1 , 𝑟2 , . . . , 𝑟𝑛 ≥ 0, 𝑝1 , 𝑝2 , . . . , 𝑝𝑛 ≥ 0 for a natural number 𝑛.
Then the following inequality holds:
𝐴𝑟𝑛𝑛 ≥ 𝐶𝐴 𝑖 ,𝐵 [𝑛]𝑟𝑛 /𝑞[𝑛] ,
(28)
where 𝐶𝐴 𝑖 ,𝐵 [𝑛] and 𝑞[𝑛] are defined in (2) and (4).
Proof. We will show (28) by mathematical induction. In the
case 𝑛 = 1.
Since 𝐴 1 ≫ 𝐵 implies
𝑟 /2
𝑟 /2 𝑟1 /(𝑝1 +𝑟1 )
𝐴 1 ≥ (𝐴 11 𝐵𝑝1 𝐴 11 )
(29)
holds for any, 𝑝1 ≥ 0 and 𝑟1 ≥ 0 by (i) of Theorem A, whence
(28) for 𝑛 = 1.
Assume that (28) holds for a natural number 𝑘 (1 ≤
𝑘 < 𝑛). We will show (28) for 𝑟1 , 𝑟2 , . . . , 𝑟𝑘+1 ≥ 0 and
𝑝1 , 𝑝2 , . . . , 𝑝𝑘 , 𝑝𝑘+1 ≥ 0 for 𝑘 + 1.
Proof. Since the condition (16) with 𝛿 ≥ 0 suffices (28) in
Theorem 7, we have the following inequality by Theorem 7;
see (28).
We state the following important inequality (35) for the
forthcoming discussion which is the inequality in (16):
𝑞 [𝑛] = 𝑞 [𝑛 − 1] 𝑝𝑛 + 𝑟𝑛 ≥ 𝛿 + 𝑟1 + 𝑟2 + ⋅ ⋅ ⋅ + 𝑟𝑛−1 + 𝑟𝑛 (35)
because the inequality in (35) follows by (ii) of Lemma 2, and
the inequality follows by
𝑞 [𝑛 − 1] 𝑝𝑛 ≥ 𝛿 + 𝑟1 + 𝑟2 + ⋅ ⋅ ⋅ + 𝑟𝑛−1
(36)
obtained by (34).
(a) Proof of the result that 𝐼𝑛 (𝑝𝑛 , 𝑟𝑛 ) is a decreasing
function of 𝑝𝑛 .
Without loss of generality, we can assume that 𝑝𝑛 > 0.
We can obtain the following inequality by (28) and by (i) of
Lemma 2:
𝑟𝑛 /𝑞[𝑛]
𝐴𝑟𝑛𝑛 ≥ 𝐶𝐴 𝑖 ,𝐵 [𝑛]𝑟𝑛 /𝑞[𝑛] = (𝐴𝑟𝑛𝑛 /2 𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 𝐴𝑟𝑛𝑛 /2 )
= 𝐴𝑟𝑛𝑛 /2 𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 /2
(𝑟𝑛 −𝑞[𝑛])/𝑞[𝑛]
× (𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 /2 𝐴𝑟𝑛𝑛 𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 /2 )
× 𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 /2 𝐴𝑟𝑛𝑛 /2 ,
(37)
Abstract and Applied Analysis
5
and (37) implies
(𝑞[𝑛]−𝑟𝑛 )/𝑞[𝑛]
(𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 /2 𝐴𝑟𝑛𝑛 𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 /2 )
(38)
≥ 𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 .
Put 𝛼 = 𝜔/𝑝𝑛 ∈ [0, 1] for 𝑝𝑛 ≥ 𝜔 ≥ 0, then we raise each side
of (38) to the power 𝛼 = 𝜔/𝑝𝑛 ∈ [0, 1], then
(𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 /2 𝐴𝑟𝑛𝑛 𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 /2 )
by (35) and 𝑞[𝑛] + 𝑞[𝑛 − 1]𝜔 = 𝑞[𝑛 − 1](𝑝𝑛 + 𝜔) + 𝑟𝑛 ≥ 𝑞[𝑛]
by (4), so that 𝐼𝑛 (𝑝𝑛 , 𝑟𝑛 ) is a decreasing function of 𝑝𝑛 .
(b) Proof of the result that 𝐼𝑛 (𝑝𝑛 , 𝑟𝑛 ) is a decreasing
function of 𝑟𝑛 .
Without loss of generality, we can assume that 𝑟𝑛 > 0.
Raise each side of (28) to the power 𝜇/𝑟𝑛 ∈ [0, 1] for 𝑟𝑛 ≥ 𝜇 ≥
0 by LH, then
((𝑞[𝑛]−𝑟𝑛 )𝜔)/(𝑞[𝑛]𝑝𝑛 )
𝜔
(39)
𝐴𝜇𝑛 ≥ (𝐴𝑟𝑛𝑛 /2 𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 𝐴𝑟𝑛𝑛 /2 )
≥ 𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1] .
𝜇/𝑞[𝑛]
.
(41)
Whence we have
We state the following inequality by (ii) of Lemma 3 and (35):
𝐼𝑛 (𝑝𝑛 , 𝑟𝑛 )
𝑛 /2
= 𝐴−𝑟
(𝐴𝑟𝑛𝑛 /2 𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 𝐴𝑟𝑛𝑛 /2 )
𝑛
(𝛿+𝑟1 +𝑟2 +⋅⋅⋅+𝑟𝑛 )/𝑞[𝑛]
𝑛 /2
𝐴−𝑟
𝑛
𝑛 /2
= 𝐴−𝑟
𝑛
×
{ (𝐴𝑟𝑛𝑛 /2 𝐶𝐴 𝑖 ,𝐵 [𝑛
− 1]
𝑞 [𝑛] − (𝛿 + 𝑟1 + 𝑟2 + ⋅ ⋅ ⋅ + 𝑟𝑛 )
= 𝑞 [𝑛 − 1] 𝑝𝑛 + 𝑟𝑛 − (𝛿 + 𝑟1 + 𝑟2 + ⋅ ⋅ ⋅ + 𝑟𝑛 )
𝑝𝑛
(42)
= 𝑞 [𝑛 − 1] 𝑝𝑛 − (𝛿 + 𝑟1 + 𝑟2 + ⋅ ⋅ ⋅ + 𝑟𝑛−1 ) ≥ 0.
(𝑞[𝑛]+𝑞[𝑛−1]𝜔)/𝑞[𝑛] (𝛿+𝑟1 +𝑟2 +⋅⋅⋅+𝑟𝑛 )/(𝑞[𝑛]+𝑞[𝑛−1]𝜔)
× 𝐴𝑟𝑛𝑛 /2 )
}
Then we have
𝑛 /2
× 𝐴−𝑟
𝑛
𝑛 /2
= 𝐴−𝑟
{𝐴𝑟𝑛𝑛 /2 𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 /2
𝑛
× (𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]
𝑝𝑛 /2
𝐼𝑛 (𝑝𝑛 , 𝑟𝑛 )
𝐴𝑟𝑛𝑛
𝑛 /2
𝑛 /2
𝐶𝐴 𝑖 ,𝐵 [𝑛](𝛿+𝑟1 +𝑟2 +⋅⋅⋅+𝑟𝑛 )/𝑞[𝑛] 𝐴−𝑟
= 𝐴−𝑟
𝑛
𝑛
(𝑞[𝑛−1]𝜔)/𝑞[𝑛]
× 𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 /2 )
× 𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 /2 𝐴𝑟𝑛𝑛 /2 }
𝑛 /2
= 𝐴−𝑟
(𝐴𝑟𝑛𝑛 /2 𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 𝐴𝑟𝑛 /2 )
𝑛
(𝛿+𝑟1 +𝑟2 +⋅⋅⋅+𝑟𝑛 )/(𝑞[𝑛]+𝑞[𝑛−1]𝜔)
𝑛 /2
× 𝐴−𝑟
by Lemma B
𝑛
(𝛿+𝑟1 +𝑟2 +⋅⋅⋅+𝑟𝑛 −𝑞[𝑛])/𝑞[𝑛]
((𝑞[𝑛]−𝑟𝑛 )𝜔)/(𝑞[𝑛]𝑝𝑛 )
× { (𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 /2 𝐴𝑟𝑛𝑛
× 𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 /2
(𝑞[𝑛]+𝜇)/𝑞[𝑛] (𝛿+𝑟1 +𝑟2 +⋅⋅⋅+𝑟𝑛 −𝑞[𝑛])/(𝑞[𝑛]+𝜇)
× 𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 /2 )
(𝛿+𝑟1 +𝑟2 +⋅⋅⋅+𝑟𝑛 )/(𝑞[𝑛]+𝑞[𝑛−1]𝜔) −𝑟 /2
𝐴𝑟𝑛𝑛 /2 }
𝐴𝑛 𝑛
}
× 𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 /2
𝑛 /2
≥ 𝐴−𝑟
(𝐴𝑟𝑛𝑛 /2 𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 /2 𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝜔
𝑛
×
𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 /2
= 𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 /2
× 𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 /2 )
× 𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]
= 𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 /2
× 𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 /2 )
× (𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 /2 𝐴𝑟𝑛𝑛
𝑝𝑛 /2
𝑛 /2
𝐴−𝑟
𝑛
× (𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 /2 𝐴𝑟𝑛𝑛
𝑛 /2
{𝐴𝑟𝑛𝑛 /2 𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 /2
= 𝐴−𝑟
𝑛
×
(𝛿+𝑟1 +𝑟2 +⋅⋅⋅+𝑟𝑛 )/𝑞[𝑛]
(𝛿+𝑟1 +𝑟2 +⋅⋅⋅+𝑟𝑛 )/(𝑞[𝑛−1](𝑝𝑛 +𝜔)+𝑟𝑛 )
𝐴𝑟𝑛𝑛 /2)
𝑛 /2
𝐴−𝑟
𝑛
= 𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 /2
× {𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 /2 𝐴𝑟𝑛𝑛 /2
𝜇/𝑞[𝑛]
× (𝐴𝑟𝑛𝑛 /2 𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 𝐴𝑟𝑛𝑛 /2 )
= 𝐼𝑛 (𝑝𝑛 + 𝜔, 𝑟𝑛 ) ,
(40)
and the last inequality holds by LH because (39) and (𝛿 + 𝑟1 +
𝑟2 + ⋅ ⋅ ⋅ + 𝑟𝑛 )/(𝑞[𝑛 − 1](𝑝𝑛 + 𝜔) + 𝑟𝑛 ) ∈ [0, 1] which is ensured
×𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 /2 }
× 𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 /2
𝐴𝑟𝑛𝑛 /2
(𝛿+𝑟1 +𝑟2 +⋅⋅⋅+𝑟𝑛 −𝑞[𝑛])/(𝑞[𝑛]+𝜇)
6
Abstract and Applied Analysis
≥ 𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 /2
[13] T. Furuta, “Operator functions on chaotic order involving
order preserving operator inequalities,” Journal of Mathematical
Inequalities, vol. 6, no. 1, pp. 15–31, 2012.
× {𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 /2 𝐴𝑟𝑛𝑛 +𝜇
× 𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 /2 }
(𝛿+𝑟1 +𝑟2 +⋅⋅⋅+𝑟𝑛 −𝑞[𝑛])/(𝑞[𝑛]+𝜇)
× 𝐶𝐴 𝑖 ,𝐵 [𝑛 − 1]𝑝𝑛 /2
= 𝐼𝑛 (𝑝𝑛 , 𝑟𝑛 + 𝜇) ,
(43)
and the last inequality holds by LH because (41) and
𝛿 + 𝑟1 + 𝑟2 + ⋅ ⋅ ⋅ + 𝑟𝑛 − 𝑞 [𝑛]
𝑞 [𝑛] + 𝜇
𝑞 [𝑛] − (𝛿 + 𝑟1 + 𝑟2 + ⋅ ⋅ ⋅ + 𝑟𝑛 )
=−
∈ [−1, 0] ,
𝑞 [𝑛] + 𝜇
(44)
so that 𝐼𝑘 (𝑝𝑘 , 𝑟𝑘 ) is a decreasing function of 𝑟𝑛 .
Acknowledgments
This work was supported by the National Natural Science
Foundation of China (1127112; 11201127), Technology and
Pioneering project in Henan Province (122300410110).
References
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