Filomat 31:3 (2017), 871–876
DOI 10.2298/FIL1703871L
Published by Faculty of Sciences and Mathematics,
University of Niš, Serbia
Available at: http://www.pmf.ni.ac.rs/filomat
Improved Kantorovich and Wielandt Operator Inequalities
for Positive Linear Maps
Wenshi Liaoa , Junliang Wub
a College
of Mathematics and Physics, Chongqing University of Science and Technology, Chongqing, 401331, P. R.China
b College of Mathematics and Statistics, Chongqing University, Chongqing, 401331, P.R. China
Abstract. This paper improves and generalizes the Kantorovich and Wielandt inequalities for positive
linear maps on Hilbert space operators and presents more general and precise results compared to many
recent results.
1. Introduction
Throughout this paper, we reserve M, m for real numbers and I for the identity operator. Other capital
letters denote general elements of the C∗ -algebra B(H) all bounded linear operators on a complex separable
Hilbert space (h·, ·i, H). k·k denotes the operator norm. An operator A is said to be positive (strictly positive)
if hAx, xi ≥ 0 for all x ∈ H (hAx, xi > 0 for all x ∈ H\{0}) and write A ≥ 0 (A > 0). A ≥ B (A > B) means
1
A − B ≥ 0 (A − B > 0). The absolute value of A is denoted by |A|, that is, |A| = (A∗ A) 2 .
A linear map Φ : B(H) → B(K ) is positive if Φ(A) ≥ 0 whenever A ≥ 0. It is said to be unital if
Φ(I)
= I. We
C∗ -algebras is 2-positive if whenever the 2 × 2 operator matrix
"
# say that a linear
" map Φ between
#
A B
Φ(A) Φ(B)
≥ 0, then so is
≥ 0.
B∗ C
Φ(B∗ ) Φ(C)
In 1948, Kantorovich [7] introduced the well-known Kantorovich inequality. In 1990, an operator
Kantorovich inequality was established by Marshall and Olkin [10]. Recently, Lin [9] proved that the
operator Kantorovich inequality is order preserving under squaring. This result was further generalized
by several authors (see [4, 11]), who obtained
Theorem 1.1. [4, 11] Let 0 < m ≤ A ≤ M. Then for every positive unital linear map Φ,
Φp (A−1 ) ≤
(m + M)2p
Φ(A)−p ,
16mp Mp
Φp (A−1 ) ≤
(m2 + M2 )p
Φ(A)−p ,
16mp Mp
p≥2
(1)
and
p ≥ 4.
(2)
2010 Mathematics Subject Classification. Primary 47A63; Secondary 47A30.
Keywords. Kantorovich inequality, Wielandt inequality, operator inequalities, positive linear maps.
Received: 19 January 2015; Accepted: 27 April 2015
Communicated by Dragana Cvetković Ilić
Research supported by Scientic and Technological Research Program of Chongqing Municipal Education Commission (Grant No.
KJ1713330)
Email addresses: [email protected] (Wenshi Liao), [email protected] (Junliang Wu)
W. S. Liao, J. L. Wu / Filomat 31:3 (2017), 871–876
872
When p ≥ 4, the inequality (2) is tighter than (1). There exsits a vacancy unconsidered for 2 ≤ p ≤ 4.
Motivated by this, we intend to obtain some better results.
In view of positive linear map Φ, Lin [9] proved that
(M + m)2
Φ(A−1 )Φ(A) + Φ(A)Φ(A−1 ) ≤
2Mm
(3)
(M + m)2
.
2Mm
(4)
and
Φ(A−1 )Φ(A) + Φ(A)Φ(A−1 ) ≤
Fu [5] generalized the inequalities (3) and (4) to p-th power. We consider an improvement of Fu’s result.
Finally, we study a conjecture on Wielandt type operator inequalities and obtain some refined results.
2. Kantorovich-Type Inequalities
Firstly, we are devoted to obtain a better bound than (1) and (2). To this end, we need two important
lemmas.
Lemma 2.1. [3, Lemma 2.1] Let A, B ≥ 0. Then the following inequality holds:
kABk ≤
1
kA + Bk2 .
4
(5)
Lemma 2.2. [1, p. 28] Let A, B ≥ 0. Then for 1 ≤ r < +∞,
kAr + Br k ≤ k(A + B)r k.
(6)
We know that kAk ≤ 1 is equivalent to A∗ A ≤ I. Using this fact we have the following result:
Theorem 2.3. Let 0 < m ≤ A ≤ M. Then for every positive unital linear map Φ, 1 ≤ α ≤ 2 and p ≥ 2α,
2p
Φp (A−1 ) ≤
(mα + Mα ) α
Φ(A)−p .
16mp Mp
(7)
Proof. The desired inequality is equivalent to
p
p
α
α α
p
Φ 2 (A−1 )Φ 2 (A) ≤ (m +p M p ) .
4m 2 M 2
Compute
p p p
p p p
2
p
p
m 2 M 2 Φ 2 (A−1 )Φ 2 (A) ≤ 1 m 2 M 2 Φ 2 (A−1 ) + Φ 2 (A)
(by (5))
4
2αp 2
1
≤ mα Mα Φα (A−1 ) + Φα (A) (by (6))
4
p
1 = mα Mα Φα (A−1 ) + Φα (A) α
4
p
1 α
≤ (m + Mα ) α .
4
The last inequality above holds as follows: The condition 0 < m ≤ A ≤ M implies that
Mα mα A−α + Aα ≤ Mα + mα ,
(8)
W. S. Liao, J. L. Wu / Filomat 31:3 (2017), 871–876
873
and hence
Mα mα Φ(A−α ) + Φ(Aα ) ≤ Mα + mα ,
(9)
The well-known inequality (see [1, p. 53]) says
Φα (T) ≤ Φ(Tα )
for every positive unital linear map Φ and T > 0. Then it follows from (9) that
Mα mα Φα (A−1 ) + Φα (A) ≤ Mα + mα .
Therefore
p
p
α
α α
p
Φ 2 (A−1 )Φ 2 (A) ≤ (m +p M p ) .
4m 2 M 2
(10)
So the inequality (7) is obtained.
Remark 2.4. Inequalities (1) and (2) are two special cases of (7) by taking α = 1, 2.
Putting α = 2 and p = 4, the inequality (7) reduces to Lin’s result (see [8, Theorem 4.3]).
We next present the generalizations of (3) and (4). The following lemma is useful in our proof of Theorem
2.6.
Lemma 2.5. For any bounded operator X,
"
#
tI X
|X| ≤ tI ⇔ kXk ≤ t ⇔
≥ 0 (t ≥ 0).
X∗ tI
Theorem 2.6. Let A be a positive operator on a Hilbert space H with 0 < m ≤ A ≤ M and Φ be a positive linear
map on B(H). Then for 1 ≤ α ≤ 2 and p ≥ α,
(Mα + mα ) 2pα
Φp (A−1 )Φp (A) + Φp (A)Φp (A−1 ) ≤
2Mp mp
(11)
and
2p
Φp (A−1 )Φp (A) + Φp (A)Φp (A−1 ) ≤
(Mα + mα ) α
.
2Mp mp
(12)
Proof. By (10) and Lemma 2.5, we deduce


2p
 (Mα +mα ) α I
p
p
−1 

Φ
(A)Φ
(A
)

p
p
4M m
 ≥ 0

2p

 p −1 p
(Mα +mα ) α
Φ (A )Φ (A)
I
4Mp mp
and





2p
α

Φp (A−1 )Φp (A) 
 ≥ 0.
2p

(Mα +mα ) α
Φp (A)Φp (A−1 )
I
4Mp mp
(Mα +mα )
4Mp mp
I
Summing up these two operator matrices, we have

2p
(Mα +mα ) α

I
Φp (A)Φp (A−1 ) + Φp (A−1 )Φp (A)

p
p
2M m

2p
 p −1 p
(Mα +mα ) α
Φ (A )Φ (A) + Φp (A)Φp (A−1 )
2Mp mp I



 ≥ 0.

By Lemma 2.5 again, we obtain (11).
As Φp (A)Φp (A−1 ) + Φp (A−1 )Φp (A) is self-adjoint, (12) follows from the maximal characterization of geometric mean.
W. S. Liao, J. L. Wu / Filomat 31:3 (2017), 871–876
874
Remark 2.7. Taking α = 1 and p = 1, (11) and (12) collapse to (3) and (4), respectively.
Fu showed a special case of Theorem 2.6 for α = 1 in [5, Theorem 4].
When α = 2, Theorem 2.6 implies the following.
Corollary 2.8. Let A be a positive operator on a Hilbert space H with 0 < m ≤ A ≤ M and Φ be a positive linear
map on B(H). Then for p ≥ 2 ,
(M2 + m2 )p
Φp (A−1 )Φp (A) + Φp (A)Φp (A−1 ) ≤
,
2Mp mp
(13)
and
Φp (A−1 )Φp (A) + Φp (A)Φp (A−1 ) ≤
(M2 + m2 )p
.
2Mp mp
(14)
Remark 2.9. When p ≥ 2, the inequalities (13) and (14) is tighter than that of Fu [5, Theorem 4], respectively.
3. Wielandt-Type Inequalities
In 2000, Bhatia and Davis [2] proved an operator Wielandt inequality which states that if 0 < m ≤ A ≤ M
and X, Y are two partial isometries on H whose final spaces are orthogonal to each other, then for every
2-positive linear map Φ on B(H),
M−m
Φ(X AY)Φ(Y AY) Φ(Y AX) ≤
M+m
∗
∗
−1
∗
2
Φ(X∗ AX).
Lin [9, Conjecture 3.4] conjecture that the following assertion could be true:
M − m 2
Φ(X∗ AY)Φ(Y∗ AY)−1 Φ(Y∗ AX)Φ(X∗ AX)−1 ≤
.
M+m
(15)
Recently, Fu and He [4] attempt to solve the conjecture and get a step closer to the conjecture. But
Gumus [6] obtains a better upper bound to approximate the right side of (15) based on
Φ(X∗ AY)Φ(Y∗ AY)−1 Φ(Y∗ AX)Φ(X∗ AX)−1 ≤
(M − m)2
,
√
2(M + m) Mm
which is equivalent to
2
Φ(X∗ AY)Φ(Y∗ AY)−1 Φ(Y∗ AX) ≤
(M − m)4
Φ2 (X∗ AX).
4(M + m)2 Mm
(16)
Soon after, Zhang [11] proved two generalized inequalities
!p
p2
p
1 M−m 2
1
∗
∗
−1
∗
∗
−2 Φ(X AY)Φ(Y AY) Φ(Y AX) Φ(X AX) ≤ 4 M + m M + m ,
p≥2
(17)
and
p
2p
p
M−m p M 2
∗
∗
−1
∗
∗
−2 Φ(X
AY)Φ(Y
AY)
Φ(Y
AX)
Φ(X
AX)
≤
,
M+m
m
Now, Let us give improvements of (17) and (18).
p ≥ 1.
(18)
W. S. Liao, J. L. Wu / Filomat 31:3 (2017), 871–876
875
Theorem 3.1. Let 0 < m ≤ A ≤ M, X and Y be two isometries in H whose final spaces are orthogonal to each other
and Φ be a 2-positive linear map on B(H). Then for 1 ≤ α ≤ 2 and p ≥ 2α,
p
2p
p
(M − m)p (Mα + mα ) α
∗
−2 ∗
∗
−1
∗
Φ(X
AX)
≤
Φ(X
AY)Φ(Y
AY)
Φ(Y
AX)
.
p
3p
3p
p
22+ 2 M 4 m 4 (M + m) 2
(19)
Proof. By using (16), we have
α
Φ(X∗ AY)Φ(Y∗ AY)−1 Φ(Y∗ AX) ≤
(M − m)2α
α
∗
α
α Φ (X AX).
2α (M + m)α M 2 m 2
(20)
Combining (20) with Lemma 2.1 and Lemma 2.2, we get
2p
p p
p
(M − m)p
∗
∗
−1
∗
∗
−
4
4 Φ(X AY)Φ(Y AY)
Φ(Y AX) Φ(X AX) 2 p
p M m
2 2 (M + m) 2
! p2 2
2p
(M − m)2 √
1 ∗
∗
−1
∗
∗
−1
≤ Φ(X AY)Φ(Y AY) Φ(Y AX) +
MmΦ(X AX)
4
2(M + m)
p
α
α
(M − m)2α
α
1 ≤ Φ(X∗ AY)Φ(Y∗ AY)−1 Φ(Y∗ AX) + α
(Mm) 2 Φ(X∗ AX)−α α
4
2 (M + m)
p
α
(M − m)2α
(M − m)2α
α
1
2 Φ(X∗ AX)−α ≤ Φ(X∗ AX)α + α
(Mm)
α
4 2α (Mm) 2 (M + m)α
2 (M + m)α
p
(M − m)2p
Φ(X∗ AX)α + Mα mα Φ(X∗ AX)−α α
=
p
p
22+p M 2 m 2 (M + m)p
p
≤
(M − m)2p (Mα + mα ) α
p
p
22+p M 2 m 2 (M + m)p
.
The last inequality follows from (8) and 0 < m ≤ X∗ AX ≤ M. This proves the inequality (19).
Putting α = 1, 2 in Theorem 3.1, we have
Corollary 3.2. Under the same conditions as in Theorem 3.1, then
p
2p
p
(M − m)p (M + m) 2
∗
−2 ∗
∗
−1
∗
, p≥2
p
3p
3p
Φ(X AY)Φ(Y AY) Φ(Y AX) Φ(X AX) ≤
22+ 2 M 4 m 4
(21)
p
2p
p
(M − m)p (M2 + m2 ) 2
∗
∗
−1
∗
∗
−2 Φ(X
AY)Φ(Y
AY)
Φ(Y
AX)
Φ(X
AX)
≤
, p ≥ 4.
p
3p
3p
p
22+ 2 M 4 m 4 (M + m) 2
(22)
and
Remark 3.3. (22) is better than (21) as p ≥ 4. When
(M − m)2 (M + m)
3
3
8M 2 m 2
M−m
≤
M+m
2
M 1
≤
m
4
M
m
M−m
M+m
≤ 57.7341, a simple computation shows that
2
1
M+
m
!2
,
which points out that the right side of (21) is a better bound than that of (17) for p ≥ 2 and that of (18) for p = 2,
respectively.
Following from (19) and the line of the prove of Theorem 2.6, we have
W. S. Liao, J. L. Wu / Filomat 31:3 (2017), 871–876
876
Theorem 3.4. Under the same conditions as in Theorem 3.1, denote
2p
p
Γ = Φ(X∗ AY)Φ(Y∗ AY)−1 Φ(Y∗ AX) Φ(X∗ AX)− 2 ,
then
p
|Γ + Γ | ≤
∗
(M − m)p (Mα + mα ) α
p
3p
p
3p
21+ 2 M 4 m 4 (M + m) 2
and
p
Γ+Γ ≤
∗
(M − m)p (Mα + mα ) α
p
3p
3p
p
21+ 2 M 4 m 4 (M + m) 2
.
Acknowledgement
The authors thank the handling editor and referees for careful reading of the manuscript and useful
comments.
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