EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 10, No. 2, 2017, 231-237
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

An Extension of Kantorovich Inequality for Sesquilinear
Maps

Hamid Reza Moradi1,∗, Mohsen Erfanian Omidvar2, Mohammad Kazem Anwary3

1 Young Researchers and Elite Club, Mashhad Branch, Islamic Azad University, Mashhad,
Iran
2 Department of Mathematics, Mashhad Branch, Islamic Azad University, Mashhad, Iran
3 Department of Pure Mathematics, Ferdowsi University of Mashhad, Mashhad, Iran

Abstract. By using sesquilinear map we generalize some operator Kantorovich inequalities. Our
results are more extensive than many previous results due to Mond and Pečarić.

2010 Mathematics Subject Classifications: 47A63, 47A30

Key Words and Phrases: Kantorovich inequality, positive linear map, operator inequality

1. Introduction and Preliminaries

For every unit vector x and MI ≥ A ≥ mI > 0, the Kantorovich inequality [4] states

〈x,Ax〉
〈
x,A−1x

〉
≤ (M +m)2

4Mm
. (1)

In [3, Theorem 1.29], the authors obtained the following reverse of Hölder-McCarthy
inequality by the Kantorovich inequality:

Theorem 1. Let A be a positive operator on H satisfying M1H ≥ A ≥ m1H > 0 for
some scalars m < M . Then

〈
A2x, x

〉
≤ (M +m)2

4Mm
〈Ax, x〉2, (2)

for every unit vector x ∈H .

∗Corresponding author.

Email addresses: hrmoradi@mshdiau.ac.ir (H.R. Moradi), erfanian@mshdiau.ac.ir (M.E. Omidvar),
abdh1248@gmail.com (M.K. Anwary)

http://www.ejpam.com 231 c© 2017 EJPAM All rights reserved.



H.R. Moradi, M.E. Omidvar, M.K. Anwary / Eur. J. Pure Appl. Math, 10 (2) (2017), 231-237 232

Many authors have investigated on extensions of the Kantorovich one, such as Liu
et al. [5], Furuta [2] and Ky Fan [1]. Among others, we pay our attentions to the long
research series of Mond-Pečarić method [3].

As customary, we reserve M,m for scalars and 1H for identity operator. Other capital
letters denote general elements of the C∗-algebra B (H ) (with unit) of all bounded linear
operators acting on a Hilbert space (H , 〈·, ·〉). Also, we identify a scalars with the unit
multiplied by this scalar. We write A ≥ 0 to mean that the operator A is positive and
identify A ≥ B (the same as B ≤ A) with A−B ≥ 0. A positive invertible operator A is
naturally denoted by A > 0. For A,B > 0, the geometric mean A#B is defined by

A#B = A
1
2

(
A− 1

2BA− 1
2

) 1
2
A

1
2 .

It is well known that

A#B ≤ A+B

2
.

We use ϕ for sesquilinear map. A map ϕ : B (H )× B (H )→ B (H ) is a sesquilinear
map, if satisfying the following conditions:

(a) ϕ (αA1 + βA2, B) = αϕ (A1, B) + βϕ (A2, B);

(b) ϕ (A,αB1 + βB2) = αϕ (A,B1) + βϕ (A,B2);

(c) ϕ (A,A) ≥ 0;

(d) ϕ (AX,Y ) = ϕ (X,A∗Y );

for all α, β ∈ C and A1, A2, B1, B2, X, Y ∈ B (H ).
Note that, if A ≥ 0 then ϕ (AC,C) ≥ 0 for all C ∈ B (H ). In fact, if A ≥ 0 then

A = B∗B for some B ∈ B (H ). Therefore,

ϕ (AC,C) = ϕ (B∗BC,C) = ϕ (BC,BC) ≥ 0

It turn implies that, if A ≥ B then, ϕ (AC,C) ≥ ϕ (BC,C). Since A−B ≥ 0.
We remark that if we define ϕ (A,B) = B∗A, then above definition coincides with

the ordinal definition of positive operator. In fact, in this case ϕ (AC,C) = C∗AC and
ϕ (BC,C) = C∗BC, hence A ≥ B if and only if C∗AC ≥ C∗BC for any C ∈ B (H ). We
call U ∈ B (H ) is ϕ-unitary if ϕ (U,U) = 1H .

The main results are given in the next section. In this paper, we will present some
operator inequalities which are generalizations of (1) and (2).

2. Proofs of the inequalities

To prove our main results we need the following lemma.

Lemma 1. [3, Lemma 1.24] Let A ∈ B (H ) be positive and satisfying M1H ≥ A ≥
m1H > 0 for some scalars m < M . Then

(M +m) 1H ≥MmA−1 +A.



H.R. Moradi, M.E. Omidvar, M.K. Anwary / Eur. J. Pure Appl. Math, 10 (2) (2017), 231-237 233

The following result is our first main result. It presents a generalization of the Kan-
torovich inequality.

Theorem 2. Let A,C ∈ B (H ) and A be a positive satisfying M1H ≥ A ≥ m1H > 0
for some scalars m < M . Then

ϕ (AC,C) #ϕ
(
A−1C,C

)
≤ M +m

2
√
Mm

ϕ (C,C) . (3)

Proof. By Lemma 1, we have

(M +m) 1H ≥MmA−1 +A.

Since ϕ is sesquilinear map, we obtain

(M +m)ϕ (C,C) ≥Mmϕ
(
A−1C,C

)
+ ϕ (AC,C)

≥ 2
√
Mmϕ

(
A−1C,C

)
#ϕ (AC,C) .

Which is exactly desired result (3).

Example 1. By taking ϕ (A,B) = B∗A in Theorem 2 we infer that

C∗AC#C∗A−1C ≤ M +m

2
√
Mm

C∗C.

In addition, if C is unitary then

C∗AC#C∗A−1C ≤ M +m

2
√
Mm

.

Theorem 3. Let Ai, Ci ∈ B (H ) and Ai be a positive satisfying M1H ≥ Ai ≥ m1H > 0
for some scalars m < M (i = 1, . . . , n). Then(

n∑
i=1

ϕ (AiCi, Ci)

)
#

(
n∑

i=1

ϕ
(
A−1

i Ci, Ci

))
≤ M +m

2
√
Mm

n∑
i=1

ϕ (Ci, Ci).

Proof. Putting

à =

A1 . . . 0
...

. . .
...

0 · · · An

 , C̃ =

C1
...
Cn


then we have sp

(
Ã
)
⊂ [m,M ]. Next we define

ϕ̃ : ⊕B (H )×⊕B (H )→ ⊕B (H )

ϕ̃


A1

...
An

 ,

A1
...
An


 =

n∑
i=1

ϕ (Ai, Ai)
.



H.R. Moradi, M.E. Omidvar, M.K. Anwary / Eur. J. Pure Appl. Math, 10 (2) (2017), 231-237 234

In particular, we have

ϕ̃
(
ÃC̃, C̃

)
= ϕ̃


A1 . . . 0

...
. . .

...
0 · · · An


C1

...
Cn

 ,

C1
...
Cn




= ϕ̃


A1C1

...
AnCn

 ,

C1
...
Cn




=

n∑
i=1

ϕ (AiCi, Ci).

It can be deduced from Theorem 2 that

ϕ̃
(
ÃC̃, C̃

)
#ϕ̃

(
Ã−1C̃, C̃

)
≤ M +m

2
√
Mm

ϕ̃
(
C̃, C̃

)
.

This completes the proof.

The following corollary follows immediately.

Corollary 1. If in Theorem 3, C̃ =

C1
...
Cn

 is a ϕ̃-unitary, then

(
n∑

i=1

ϕ (AiCi, Ci)

)
#

(
n∑

i=1

ϕ
(
A−1

i Ci, Ci

))
≤ M +m

2
√
Mm

.

Theorem 4. Let A be a positive operator on H satisfying M1H ≥ A ≥ m1H > 0 for
some scalars m < M . Then

ϕ
(
A−1C,C

)
− ϕ(AC,C)−1 ≤

(√
M −

√
m
)2

Mm
ϕ (C,C) ,

for every C ∈ B (H ).

Proof. According to Lemma 1, we have

(M +m) 1H ≥MmA−1 +A

and hence

ϕ
(
A−1C,C

)
≤ M +m

Mm
ϕ (C,C)− 1

Mm
ϕ (AC,C) ,



H.R. Moradi, M.E. Omidvar, M.K. Anwary / Eur. J. Pure Appl. Math, 10 (2) (2017), 231-237 235

for every C ∈ B (H ). Then it follows that

ϕ
(
A−1C,C

)
− ϕ(AC,C)−1

≤
(

1

m
+

1

M

)
ϕ (C,C)− 1

Mm
ϕ (AC,C)− ϕ(AC,C)−1

=

(
1√
m
− 1√

M

)2

ϕ (C,C)−
(

1√
Mm

ϕ(AC,C)
1
2 − ϕ(AC,C)−

1
2

)2

≤
(

1√
m
− 1√

M

)2

ϕ (C,C) .

Based on the discussion above, we conclude that

ϕ
(
A−1C,C

)
− ϕ(AC,C)−1 ≤

(√
M −

√
m
)2

Mm
ϕ (C,C) .

We have completed the proof of Theorem 4.

Proposition 1. Let A be a positive operator on H satisfying M1H ≥ A ≥ m1H > 0 for
some scalars m < M . Then

ϕ
(
A2C,C

)
#ϕ (C,C) ≤ M +m

2
√
Mm

ϕ (AC,C) ,

for every C ∈ B (H ).

Proof. Replacing C with A
1
2C in the (3), we have

ϕ
(
AA

1
2C,A

1
2C
)

#ϕ
(
A−1A

1
2C,A

1
2C
)
≤ M +m

2
√
Mm

ϕ
(
A

1
2C,A

1
2C
)

therefore

ϕ
(
A2C,C

)
#ϕ (C,C) ≤ M +m

2
√
Mm

ϕ (AC,C) .

Which completes the proof.

To prove the Theorem 5, we need the following basic lemma.

Lemma 2. Let A be a self-adjoint operator on H satisfying M1H ≥ A ≥ m1H for some
scalars m < M , then

(M1H −A) (A−m1H ) ≤
(
M −m

2

)2

.



REFERENCES 236

Proof. A simple computation yields

(M1H −A) (A−m1H )

= (M +m)A−Mm1H −A2

=
(M −m)2

4
1H −

(
A− M +m

2
1H

)2

≤
(
M −m

2

)2

1H ,

as desired.

Theorem 5. Let A be a self-adjoint operator on H satisfying M1H ≥ A ≥ m1H for
some scalars m < M and ϕ (C,C) = 1H . Then

ϕ
(
A2C,C

)
− ϕ(AC,C)2 ≤ (M −m)2

4
.

Proof. By Lemma 2 we have

ϕ
(
A2C,C

)
− ϕ(AC,C)2

= (M1H − ϕ (AC,C)) (ϕ (AC,C)−m1H )− ϕ ((M1H −A) (A−m1H )C,C)

≤ (M1H − ϕ (AC,C)) (ϕ (AC,C)−m1H )

≤ (M −m)2

4
1H ,

which is exactly what we needed to prove.

Acknowledgements

The authors would like to express their thanks to the referees for their valuable com-
ments and suggestions, which helped to improve the paper.

References

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[2] T. Furuta. Extensions of Hölder-McCarthy and Kantorovich inequalities and their ap-
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[3] T. Furuta, J. Mićić, J.E. Pečarić, and Y. Seo. Mond-Pečarić method in operator in-
equalities. Element, Zegreb, 2005.

[4] L.V. Kantorovich. Functional analysis and applied mathematics. Uspekhi Mat. Nauk,
3(6), 89-185. 1948.



REFERENCES 237

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