Some inequalities for Heinz operator mean
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Dragomir, Sever S (2020) Some inequalities for Heinz operator mean.
Mathematica Moravica, 24 (1). pp. 71-82. ISSN 1450-5932
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doi: 10.5937/MatMor2001071S
Vol. 24, No. 1 (2020), 71–82
Some inequalities for Heinz operator mean
Silvestru Sever Dragomir
Abstract. In this paper we obtain some new inequalities for Heinz operator mean.
1. Introduction
Throughout this paperA, Bare positive invertible operators on a complex Hilbert space (H,h·,·i). We use the following notations for operators and ν ∈[0,1]
A∇νB := (1−ν)A+νB, theweighted operator arithmetic mean,and
A]νB :=A1/2
A−1/2BA−1/2 ν
A1/2,
the weighted operator geometric mean. When ν = 12 we write A∇B and A]B for brevity, respectively.
Define theHeinz operator mean by Hν(A, B) := 1
2(A]νB+A]1−νB). The following interpolatory inequality is obvious
(1) A]B≤Hν(A, B)≤A∇B
for any ν∈[0,1].
The famousYoung inequalityfor scalars says that ifa, b >0andν ∈[0,1], then
(2) a1−νbν ≤(1−ν)a+νb
with equality if and only ifa=b. The inequality (2) is also calledν-weighted arithmetic-geometric mean inequality.
2010Mathematics Subject Classification. Primary: 47A63, 47A30; Secondary: 26D15;
26D10.
Key words and phrases. Young’s Inequality, Convex functions, Arithmetic mean- Geometric mean inequality, Heinz means.
Full paper. Received 1 October 2019, revised 27 February 2020, accepted 2 March 2020, available online 9 March 2020.
c
2020 Mathematica Moravica
71
Some inequalities for Heinz operator mean
We recall that Specht’s ratio is defined by [11]
(3) S(h) :=
h
1 h−1
eln
hh−11
if h∈(0,1)∪(1,∞), 1 ifh= 1.
It is well known thatlimh→1S(h) = 1, S(h) =S h1
>1 for h >0, h6= 1.
The function is decreasing on (0,1)and increasing on(1,∞).
The following inequality provides a refinement and a multiplicative reverse for Young’s inequality:
(4) Sa
b r
a1−νbν ≤(1−ν)a+νb≤Sa b
a1−νbν, wherea, b >0,ν∈[0,1], r = min{1−ν, ν}.
The second inequality in (4) is due to Tominaga [12] while the first one is due to Furuichi [4].
The operator version is as follows [4], [12] : For two positive operatorsA, B and positive real numbersm, m0, M, M0 satisfying either of the following conditions:
(i)0< mI ≤A≤m0I < M0I ≤B ≤M I, (ii)0< mI≤B≤m0I < M0I ≤A≤M I we have
(5) S h0r
A]νB ≤A∇νB ≤S(h)A]νB, whereh:= Mm, h0 := Mm00 and ν ∈[0,1].
We observe that, if we write the inequality (5) for 1−ν and add the obtained inequalities, then we get by division with2that
S h0r
Hν(A, B)≤A∇B ≤S(h)Hν(A, B) that is equivalent to
(6) S−1(h)A∇B≤Hν(A, B)≤S−1 h0r A∇B, whereh:= Mm, h0 := Mm00 and ν ∈[0,1].
We consider theKantorovich’s constant defined by
(7) K(h) := (h+ 1)2
4h , h >0.
The function K is decreasing on (0,1) and increasing on[1,∞), K(h)≥1 for any h >0and K(h) =K 1h
for anyh >0.
The following multiplicative refinement and reverse of Young inequality in terms of Kantorovich’s constant holds:
(8) Kra
b
a1−νbν ≤(1−ν)a+νb≤KRa b
a1−νbν, wherea, b >0,ν∈[0,1], r= min{1−ν, ν} and R= max{1−ν, ν}.
The first inequality in (8) was obtained by Zou et al. in [13] while the second by Liao et al. [10].
The operator version is as follows [13], [10]: For two positive operatorsA, Band positive real numbersm, m0, M, M0 satisfying either of the conditions (i) or (ii) above, we have
(9) Kr h0
A]νB ≤A∇νB ≤KR(h)A]νB,
whereh:= Mm, h0:= Mm00,ν∈[0,1]r = min{1−ν, ν}andR = max{1−ν, ν}. We observe that, if we write the inequality (9) for 1−ν and add the obtained inequalities, then we get by division with2that
Kr h0
Hν(A, B)≤A∇B≤KR(h)Hν(A, B) that is equivalent to
(10) K−R(h)A∇B ≤Hν(A, B)≤K−r h0 A∇B, whereh:= Mm, h0 := Mm00 and ν ∈[0,1].
The inequalities (10) have been obtained in [10] where other bounds in terms of the weighted operator harmonic mean
A!νB :=
(1−ν)A−1+νB−1−1
were also given.
Motivated by the above results, we establish in this paper some new in- equalities for the Heinz mean. Related inequalities are also provided.
2. Upper and lower bounds for Heinz mean
We start with the following result that provides a generalization for the inequalities (5) and (9):
Theorem 1. Assume that A, B are positive invertible operators and the constantsM > m >0 are such that
(11) mA≤B≤M A
in the operator order. Letν ∈[0,1], r = min{1−ν, ν}andR= max{1−ν, ν}. Then we have the inequalities
(12) ϕr(m, M)A]νB ≤A∇νB≤Φ (m, M)A]νB, where
Φ (m, M) :=
S(m) if M <1,
max{S(m), S(M)} ifm≤1≤M, S(M) if 1< m,
, (13)
Some inequalities for Heinz operator mean
ϕr(m, M) :=
S(Mr) if M <1, 1 if m≤1≤M, S(mr) if 1< m, and
(14) ψr(m, M)A]νB ≤A∇νB ≤ΨR(m, M)A]νB, where
ΨR(m, M) :=
KR(m) if M <1, max
KR(m), KR(M) if m≤1≤M, KR(M) if 1< m,
, (15)
ψr(m, M) :=
Kr(M) ifM <1, 1 if m≤1≤M, Kr(m) if 1< m.
Proof. From the inequality (4) we have (16)
xν min
x∈[m,M]S(xr)≤S(xr)xν ≤(1−ν) +νx≤S(x)xν ≤xν max
x∈[m,M]S(x) wherex∈[m, M],ν∈[0,1], r = min{1−ν, ν}.
Since, by the properties of Specht’s ratioS,we have
x∈[m,M]max S(x) =
S(m) if M <1,
max{S(m), S(M)} if m≤1≤M, S(M) if 1< m,
= Φ (m, M)
and
x∈[m,Mmin ]S(xr) =
S(Mr) if M <1, 1 if m≤1≤M, S(mr) if 1< m,
=ϕr(m, M),
then by (16) we have
(17) xνϕr(m, M)≤(1−ν) +νx≤xνΦ (m, M) for any x∈[m, M]andν ∈[0,1].
Using the functional calculus for the operatorX withmI ≤X≤M I we have from (17) that
(18) Xνϕr(m, M)≤(1−ν)I+νX ≤XνΦ (m, M) for any ν∈[0,1].
If the condition (11) holds true, then by multiplying in both sides with A−1/2 we get mI ≤A−1/2BA−1/2 ≤M I and by takingX =A−1/2BA−1/2 in (18) we get
A−1/2BA−1/2ν
ϕr(m, M)≤(1−ν)I+νA−1/2BA−1/2 (19)
≤
A−1/2BA−1/2ν
Φ (m, M). Now, if we multiply (19) in both sides with A1/2 we get the desired result (12).
The second part follows in a similar way by utilizing the inequality xν min
x∈[m,M]Kr(x)≤Kr(x)xν ≤(1−ν) +νx
≤KR(x)xν ≤xν max
x∈[m,M]KR(x),
which follows from (8). The details are omitted.
Remark 1. If (i) 0 < mI ≤ A ≤ m0I < M0I ≤ B ≤ M I, h = Mm and h0 = Mm00 then we have
A≤ M0
m0A=h0A≤B ≤hA= M mA, and by (12) we get
(20) S h0r
A]νB ≤A∇νB ≤S(h)A]νB.
If (ii)0< mI ≤B ≤m0I < M0I ≤A≤M I,then we have 1
hA≤B ≤ 1 h0A≤A and by (12) we get
S 1
h0 r
A]νB≤A∇νB ≤S 1
h
A]νB, which is equivalent to (20).
If we use the inequality (14) for the operatorsAandB that satisfy either of the conditions (i) or (ii), then we recapture (9).
Remark 2. From (12) we get forν = 12 that
Some inequalities for Heinz operator mean
(21)
S(Mr)A]B if M <1, A]B if m≤1≤M, S(mr)A]B if 1< m,
≤A∇B
≤
S(m)A]B if M <1,
max{S(m), S(M)}A]B if m≤1≤M, S(M)A]B if 1< m.
The following result contains two upper and lower bounds for the Heinz operator mean in terms of the operator arithmetic meanA∇B :
Corollary 1. With the assumptions of Theorem 1 we have the following upper and lower bounds for the Heinz operator mean
(22) Φ−1(m, M)A∇B ≤Hν(A, B)≤ϕ−1r (m, M)A∇B and
(23) Ψ−1R (m, M)A∇B ≤Hν(A, B)≤ψ−1r (m, M)A∇B.
Remark 3. If the operatorsA and B satisfy either of the conditions (i) or (ii) from Remark 1, then we have the inequality
(24) S−1(h)A∇B ≤Hν(A, B)≤S−1 h0r A∇B and
(25) K−R(h)A∇B ≤Hν(A, B)≤K−r h0 A∇B.
The following result provides an upper and lower bound for the Heinz mean in terms of the operator geometric meanA]B :
Theorem 2. With the assumptions of Theorem 1 we have (26) ω(m, M)A]B≤Hν(A, B)≤Ω (m, M)A]B, where
(27) Ω (m, M) :=
S m|2ν−1|
if M <1, max
S m|2ν−1|
, S M|2ν−1| if m≤1≤M, S M|2ν−1|
if 1< m,
and
(28) ω(m, M) :=
S
M|ν−12|
if M <1, 1 if m≤1≤M, S
m|ν−12|
if 1< m, where ν∈[0,1].
Proof. From the inequality (4) we have for ν = 12
(29) S
rc d
√
cd≤ c+d
2 ≤Sc d
√ cd, for any c, d >0.
If we take in (29) c=a1−νbν and d=aνb1−ν then we get
(30) S
a b
12−ν√
ab≤ a1−νbν +aνb1−ν
2 ≤S
a b
1−2ν√ ab, for any a, b >0 for anyν ∈[0,1].
This is an inequality of interest in itself.
If we take in (30) a=x and b= 1,then we get
(31) S
x12−ν
√
x≤ x1−ν +xν
2 ≤S x1−2ν√ x, for any x >0.
Now, ifx∈[m, M]⊂(0,∞) then by (31) we have
(32) √
x min
x∈[m,M]S x12−ν
≤ x1−ν +xν
2 ≤√
x max
x∈[m,M]S x1−2ν , for any x∈[m, M].
Ifν ∈ 0,12 ,then
x∈[m,M]max S x1−2ν
=
S m1−2ν
if M <1, max
S m1−2ν
, S M1−2ν if m≤1≤M, S M1−2ν
if 1< m, and
x∈[m,M]min S x12−ν
=
S
M1−2ν2
ifM <1, 1 if m≤1≤M, S
m1−2ν2
if 1< m.
Some inequalities for Heinz operator mean
Ifν ∈ 12,1 , then
x∈[m,Mmax ]S x1−2ν
= max
x∈[m,M]S x2ν−1
=
S m2ν−1
if M <1, max
S m2ν−1
, S M2ν−1 if m≤1≤M, S M2ν−1
if 1< m, and
x∈[m,Mmin ]S
x12−ν
= min
x∈[m,M]S
xν−12
=
S
M2ν−12
if M <1, 1if m≤1≤M, S
m2ν−12
if 1< m.
Then by (32) we have
(33) ω(m, M)√
x≤ x1−ν +xν
2 ≤Ω (m, M)√ x, for any x∈[m, M].
IfX is an operator with mI≤X≤M I,then by (33) we have (34) ω(m, M)X1/2≤ X1−ν +Xν
2 ≤Ω (m, M)X1/2.
If the condition (11) holds true, then by multiplying in both sides withA−1/2 we getmI ≤A−1/2BA−1/2 ≤M I and by taking X =A−1/2BA−1/2 in (34) we get
ω(m, M)
A−1/2BA−1/21/2
(35)
≤ 1 2
A−1/2BA−1/21−ν
+
A−1/2BA−1/2ν
≤Ω (m, M)
A−1/2BA−1/21/2
.
Now, if we multiply (35) in both sides with A1/2 we get the desired result
(26).
Corollary 2. For two positive operatorsA, B and positive real numbers m, m0, M, M0 satisfying either of the following conditions:
(i) 0< mI≤A≤m0I < M0I ≤B≤M I, (ii) 0< mI≤B≤m0I < M0I ≤A≤M I, we have forh= Mm and h0 = Mm00 that
(36) S
h0|ν−12|
A]B≤Hν(A, B)≤S
h|2ν−1|
A]B,
where ν∈[0,1].
3. Related results We call Heron means, the means defined by
Fα(a, b) := (1−α)
√
ab+αa+b 2 , wherea, b >0 and α∈[0,1].
In [1], Bhatia obtained the following interesting inequality between the Heinz and Heron means
(37) Hν(a, b)≤F(2ν−1)2(a, b) wherea, b >0 and α∈[0,1].
This inequality can be written as (38) (0≤)Hν(a, b)−√
ab≤(2ν−1)2
a+b 2 −√
ab
, wherea, b >0 and α∈[0,1].
Making use of a similar argument to the one in the proof of Theorem 1 we can state the following result as well:
Theorem 3. Assume that A, B are positive invertible operators and ν ∈ [0,1]. Then
(39) (0≤)Hν(A, B)−A]B≤(2ν−1)2(A∇B−A]B).
Moreover, if there exists the constants M > m >0 such that the condition (11) is true, then we have the simpler upper bound
(40) (0≤)Hν(A, B)−A]B ≤ 1
2(2ν−1)2√
M−√ m2
.
Kittaneh and Manasrah [5], [6] provided a refinement and an additive reverse for Young inequality as follows:
(41) r√ a−√
b2
≤(1−ν)a+νb−a1−νbν ≤R√ a−√
b2
wherea, b >0,ν∈[0,1], r= min{1−ν, ν} and R= max{1−ν, ν}. If we replace in (41)νwith1−ν, add the obtained inequalities and divide by 2,then we get
(42) r
√ a−√
b 2
≤ a+b
2 −Hν(a, b)≤R √
a−√ b
2
, wherea, b >0,ν∈[0,1].
We also have by (42) that, see [7] and [8]:
Theorem 4. Assume that A, B are positive invertible operators and ν ∈ [0,1]. Then
(43) 2r(A∇B−A]B)≤Hν(A, B)−A]B ≤2R(A∇B−A]B).
Some inequalities for Heinz operator mean
Since (2ν−1)2 ≤ 2 max{1−ν, ν} for any ν ∈[0,1], it follows that the inequality (39) is better than the right side of (43).
In [2], by using the equality
(44) a+b
2 + 2ab a+b−2
√ ab=
√ a−√
b 4
2 (a+b) ≥0 for a, b >0,the authors obtained the interesting inequality
(45) 1
2[A(a, b) +H(a, b)]≥G(a, b),
where A(a, b) is the arithmetic mean, H(a, b) is the harmonic mean and G(a, b) is the geometric mean of positive numbersa, b.
Now, if we replace aby a1−νbν and b by aνb1−ν in (45) then we get the following result for Heinz means
(46) 1
2
Hν(a, b) +Hν−1 a−1, b−1
≥G(a, b) for any fora, b >0and ν ∈[0,1].
Since
1
2 max{a, b} ≤ 1
a+b ≤ 1 2 min{a, b}, then by (44) we have
(47) 1 4
√ a−√
b 4
max{a, b} ≤ 1
2[A(a, b) +H(a, b)]−G(a, b)≤ 1 4
√ a−√
b 4
min{a, b} , for any fora, b >0.
Since√ a−√
b2
= 2 [A(a, b)−G(a, b)], √
a−√ b
2
max{a, b} =
√ a−√
b 2
max
n√ a,√
b o2 =
1− min
n√ a,√
b o
max n√
a,√ b
o
2
and
√ a−√
b2
min{a, b} =
maxn√ a,√
bo minn√
a,√ bo −1
2
, then the inequality (47) can be written as
1 2
1−
minn√ a,√
bo maxn√
a,√ bo
2
[A(a, b)−G(a, b)]
(48)
≤ 1
2[A(a, b) +H(a, b)]−G(a, b)
≤ 1 2
maxn√ a,√
bo minn√
a,√ bo −1
2
[A(a, b)−G(a, b)], for any fora, b >0.
Ifa, b∈[m, M]⊂(0,∞),then by (48) we get
1 2
1−
rm M
2
[A(a, b)−G(a, b)]≤ 1
2[A(a, b) +H(a, b)]−G(a, b) (49)
≤ 1 2
rM m −1
!2
[A(a, b)−G(a, b)]. Similar results may be stated for the corresponding operator means, how- ever the details are nor presented here.
References
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[4] S. Furuichi,Refined Young inequalities with Specht’s ratio, Journal of the Egyptian Mathematical Society, 20 (2012), 46–49.
[5] F. Kittaneh and Y. Manasrah, Improved Young and Heinz inequalities for matrix, Journal of Mathematical Analysis and Applications, 361 (2010), 262–269.
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Some inequalities for Heinz operator mean
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Silvestru Sever Dragomir
Mathematics, College of Engineering & Science Victoria University, PO Box 1442
Melbourne City, MC 8001 Australia
E-mail address: [email protected]
