Multiplicative inequalities for weighted arithmetic and harmonic operator means

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DOI: 10.2478/ausm-2019-0005

Multiplicative inequalities for weighted

arithmetic and harmonic operator means

Sever S. Dragomir

Mathematics, College of Engineering & Science Victoria University, Melbourne City, Australia

email:[email protected] URL:http://rgmia.org/dragomir

School of Computer Science & Applied Mathematics, University of the Witwatersrand,

Johannesburg, South Africa

Abstract. In this paper we establish some multiplicative inequalities for weighted arithmetic and harmonic operator means under various as-sumption for the positive invertible operators A, B. Some applications whenA, Bare bounded above and below by positive constants are given as well.

1 Introduction

Throughout this paper A, B are positive invertible operators on a complex

Hilbert space(H,h·,·i).We use the following notations for operators

A∇_νB:= (1−ν)A+νB,

theweighted operator arithmetic mean,

A]νB:=A1/2

A−1/2BA−1/2νA1/2,

2010 Mathematics Subject Classification:47A63, 47A30, 15A60, 26D15, 26D10 Key words and phrases:Young’s inequality, convex functions, arithmetic mean-harmonic mean inequality, operator means, operator inequalities

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theweighted operator geometric mean and

A!νB:=

(1−ν)A−1+νB−1

−1

theweighted operator harmonic mean, whereν∈[0, 1].

When ν= 1₂, we writeA∇B, A]Band A!Bfor brevity, respectively.

The following fundamental inequality between the weighted arithmetic, ge-ometric and harmonic operator means holds

A!_νB≤A]νB≤A∇νB (1)

for any ν∈[0, 1].

For various recent inequalities between these means we recommend the re-cent papers [3]-[6], [8]-[12] and the references therein.

The following additive double inequality has been obtained in the recent paper [7]:

ν(1−ν) (b−a) 2

max{b, a} ≤Aν(a, b) −Hν(a, b)≤ν(1−ν)

(b−a)2

min{b, a}, (2)

for any a, b > 0 and ν ∈[0, 1],where Aν(a, b) and Hν(a, b) are the scalar

weighted arithmetic mean and harmonic mean, respectively, namely

A_ν(a, b) := (1−ν)a+νb and H_ν(a, b) := ab (1−ν)b+νa.

In particular,

1 4

(b−a)2

max{b, a} ≤A(a, b) −H(a, b)≤ 1 4

(b−a)2

min{b, a}, (3)

where

A(a, b) := a+b

2 and H(a, b) := 2ab b+a.

We consider theKantorovich’s constant defined by

K(h) := (h+1) 2

4h , h > 0. (4)

The function Kis decreasing on (0, 1) and increasing on[1,∞), K(h)≥1for any h > 0and K(h) =K _h1

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Observe that for any h > 0

K(h) −1= (h−1) 2

4h =K

1 h

−1.

Observe that

K

b a

−1= (b−a) 2

4ab fora, b > 0.

Since, obviously

ab=min{a, b}max{a, b} fora, b > 0,

then we have the following version of (2):

4ν(1−ν)min{a, b}

K

b a

−1

≤Aν(a, b) −Hν(a, b) (5)

≤4ν(1−ν)max{a, b}

K

b a

−1

.

for any a, b > 0 and ν∈[0, 1].

For positive invertible operators A, B we define

A∇_∞B:= 1

2(A+B) + 1 2A

1/2 A

−1/2_(B₋_A)_A−1/2 A

1/2

and

A∇₋_∞B:= 1

2(A+B) − 1 2A

1/2 A

−1/2_(B₋_A)_A−1/2 A

1/2_.

If we consider the continuous functions f_∞, f₋_∞: [0,∞)→[0,∞) defined by

f_∞(x) =max{x, 1}= 1

2(x+1) + 1 2|x−1|

and

f₋_∞(x) =max{x, 1}= 1

2(x+1) − 1

2|x−1|,

then, obviously, we have

A∇±_∞B=A1/2f±_∞

A−1/2BA−1

A1/2.

IfA andB are commutative, then

A∇±∞B= 1

2(A+B)± 1

2|B−A|=B∇±∞A.

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Theorem 1 Let A, B be positive invertible operators and M > m > 0 such that the condition

mA≤B≤MA (6) holds. Then we have

4ν(1−ν)g(m, M)A∇₋_∞B≤A∇_νB−A!_νB (7)

≤4ν(1−ν)G(m, M)A∇_∞B,

where

g(m, M) :=

  

K(M) −1 if M < 1, 0 if m≤1≤M, K(m) −1 if 1 < m

and

G(m, M) :=

  

K(m) −1 if M < 1,

max{K(m), K(M)}−1 if m≤1≤M, K(M) −1 if 1 < m.

In particular,

g(m, M)A∇−∞B≤A∇B−A!B≤G(m, M)A∇∞B. (8)

Motivated by the above facts, we establish in this paper some multiplicative inequalities for weighted arithmetic and harmonic operator means under

var-ious assumption for the positive invertible operators A, B. Some applications

when A, B are bounded above and below by positive constants are given as

well.

2 Multiplicative inequalities

The following result is of interest in itself:

Lemma 1 For any a, b > 0 and ν∈[0, 1]we have

ν(1−ν)

1− min{a, b}

max{a, b}

2

≤ Aν(a, b)

H_ν(a, b) −1≤ν(1−ν)

max{a, b}

min{a, b} −1

2

.

(9) In particular,

1 4

1− min{a, b}

max{a, b}

2

≤ A(a, b)

H(a, b) −1≤ 1 4

max{a, b}

min{a, b} −1

2

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Proof.We have for any a, b > 0 and ν∈[0, 1]that

A_ν(a, b) H_ν(a, b) =

[(1−ν)a+νb] [(1−ν)b+νa] ab

= (1−ν)

2_ab₊_ν₍₁₋_ν)_b2₊_ν₍₁₋_ν)_a2₊_ν2_ab ab

= ν(1−ν) b 2₊_a2

+ 1−2ν+2ν2ab

ab ,

which is equivalent with

A_ν(a, b)

H_ν(a, b) −1=ν(1−ν)

(b−a)2

ab (11)

for any a, b > 0 and ν∈[0, 1].

Since min2{a, b}≤ab≤max2{a, b} hence

ν(1−ν)(b−a) 2

ab ≤ ν(1−ν)

(b−a)2

min2{a, b}

= ν(1−ν)

max{a, b}

min{a, b} −1

2

and

ν(1−ν)(b−a) 2

ab ≥ ν(1−ν)

(b−a)2

max2{_{a, b}}

= ν(1−ν)

1− min{a, b}

max{a, b}

2

and by (11) we get the desired result (9).

We observe that the inequality (9) can be written in an equivalent form as

"

ν(1−ν)

1− min{a, b}

max{a, b}

2

+1

#

H_ν(a, b) (12)

≤A_ν(a, b)

≤

"

ν(1−ν)

max{a, b}

min{a, b} −1

2

+1

#

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for any a, b > 0 and ν∈[0, 1],while (10) as

"

1 4

1− min{a, b}

max{a, b}

2

+1

#

H(a, b) (13)

≤A(a, b)

≤

"

1 4

max{a, b}

min{a, b} −1

2

+1

#

H(a, b)

for any a, b > 0.

Corollary 1 For any a, b∈[k, K]⊂(0,∞) and ν∈[0, 1]we have

A_ν(a, b)

H_ν(a, b) −1≤ν(1−ν)

K k −1

2

. (14)

In particular,

A(a, b) H(a, b) −1≤

1 4

K k −1

2

. (15)

We have the following multiplicative inequality between the weighted arith-metic and harmonic operator means:

Theorem 2 Let A, B be positive invertible operators and M > m > 0 such that the condition (6) holds. Then we have

"

ν(1−ν)

1− min{M, 1}

max{m, 1}

2

+1

#

A!_νB (16)

≤A∇_νB

≤

"

ν(1−ν)

max{M, 1}

min{m, 1} −1

2

+1

#

A!νB

In particular,

"

1 4

1− min{M, 1}

max{m, 1}

2

+1

#

A!B (17)

≤A∇B

≤

"

1 4

max{M, 1}

min{m, 1} −1

2

+1

#

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Proof. If we write the inequality (12) fora= 1 and b=x∈(0,∞) then we have

"

ν(1−ν)

1− min{1, x}

max{1, x}

2

+1

#

1−ν+νx−1

−1

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≤1−ν+νx

≤

"

ν(1−ν)

max{1, x}

min{1, x} −1

2

+1

#

1−ν+νx−1−1.

If x ∈ [m, M] ⊂ (0,∞), then max{m, 1} ≤ max{x, 1} ≤ max{M, 1} and min{m, 1}≤min{x, 1}≤min{M, 1}.

We have

max{1, x}

min{1, x} −1

2

≤

max{M, 1}

min{m, 1} −1

2

and

1− min{M, 1}

max{m, 1}

2

≤

1− min{1, x}

max{1, x}

2

for any x∈[m, M]⊂(0,∞).

Therefore, by (18) we have

"

ν(1−ν)

1− min{M, 1}

max{m, 1}

2

+1

#

1−ν+νx−1−1 (19)

≤1−ν+νx

≤

"

ν(1−ν)

max{M, 1}

min{m, 1} −1

2

+1

#

1−ν+νx−1

−1

,

for any x∈[m, M]and for anyν∈[0, 1].

If we use the continuous functional calculus for the positive invertible

oper-ator XwithmI≤X≤MI, then we have from (19) that

"

ν(1−ν)

1− min{M, 1}

max{m, 1}

2

+1

#

(1−ν)I+νX−1

−1

(20)

≤(1−ν)I+νX

≤

"

ν(1−ν)

max{M, 1}

min{m, 1} −1

2

+1

#

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If we multiply (6) both sides byA−1/2 we getMI≥A−1/2BA−1/2 ≥mI.

By writing the inequality (20) for X=A−1/2BA−1/2 we obtain

"

ν(1−ν)

1− min{M, 1}

max{m, 1}

2

+1

#

(21)

×

(1−ν)I+νA−1/2BA−1/2−1

−1

≤(1−ν)I+νA−1/2BA−1/2

≤

"

ν(1−ν)

max{M, 1}

min{m, 1} −1

2

+1

#

×

(1−ν)I+νA−1/2BA−1/2−1

−1

,

If we multiply the inequality (21) both sides withA1/2, then we get

"

ν(1−ν)

1− min{M, 1}

max{m, 1}

2

+1

#

(22)

×A1/2

(1−ν)I+ν

A−1/2BA−1/2

−1−1

A1/2

≤(1−ν)A+νB

≤

"

ν(1−ν)

max{M, 1}

min{m, 1} −1

2

+1

#

×A1/2

(1−ν)I+ν

A−1/2BA−1/2

−1−1

A1/2,

Since

A1/2

(1−ν)I+νA−1/2BA−1/2−1

−1

A1/2

=A1/2

(1−ν)I+νA1/2B−1A1/2

−1

A1/2

=A1/2

A1/2

(1−ν)A−1+νB−1

A1/2

−1

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=A1/2

A−1/2

(1−ν)A−1+νB−1

−1

A−1/2

A1/2

=

(1−ν)A−1+νB−1

−1

=A!νB

hence by (22) we get the desired result (16).

We also have:

Theorem 3 Let A, B be positive invertible operators and M > m > 0 such that the condition (6) holds. Then we have

dν(m, M)A!νB≤A∇νB≤Dν(m, M)A!νB (23)

for any ν∈[0, 1],where

dν(m, M) :=4





ν−1 2

2

+ν(1−ν)×   

K(M) if M < 1, 1 if m≤1≤M, K(m) if 1 < m





and

D_ν(m, M)

:=4





ν− 1 2

2

+ν(1−ν)×   

K(m) if M < 1,

max{K(m), K(M)} if m≤1≤M,

K(M) if 1 < m.





In particular, we have

d(m, M)A!B≤A∇B≤D(m, M)A!B (24)

where

d(m, M) :=

  

K(M) if M < 1, 1 if m≤1≤M, K(m) if 1 < m

and

D(m, M) :=

  

K(m) if M < 1,

max{K(m), K(M)} if m≤1≤M,

K(M) if 1 < m.

Proof.From (11) we have for any x∈(0,∞)and for any ν∈[0, 1]that

A_ν(1, x)

H_ν(1, x) −1=ν(1−ν)

(x−1)2

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Since K(x) −1= (x−_4x1)2, x > 0, then by (25) we have

A_ν(1, x) Hν(1, x)

= 1+4ν(1−ν) [K(x) −1]

= 4ν(1−ν)K(x) +4

ν− 1 2

2

= 4

"

ν(1−ν)K(x) +

ν−1 2

2#

or, equivalently,

Aν(1, x) =4

"

ν(1−ν)K(x) +

ν− 1 2

2#

Hν(1, x) (26)

for any x∈(0,∞) and for anyν∈[0, 1].

From (26) we then have for any x∈[m, M]⊂(0,∞) that

4

"

ν(1−ν) min

x∈[m,M]K(x) +

ν− 1 2

2#

Hν(1, x) (27)

≤Aν(1, x)≤4

"

ν(1−ν) max

x∈[m,M]

K(x) +

ν−1 2

2#

Hν(1, x).

Since

min

x∈[m,M]K(x) =   

K(M) ifM < 1, 1ifm≤1≤M, K(m) if1 < m

and

max

x∈[m,M]

K(x) =

  

K(m) ifM < 1,

max{K(m), K(M)} ifm≤1≤M,

K(M) if1 < m,

then by (27) we have

dν(m, M)

1−ν+νx−1

−1

≤ 1−ν+νx (28)

≤ Dν(m, M)

1−ν+νx−1

−1

for any x∈[m, M]and for anyν∈[0, 1].

By a similar argument to the one from Theorem 2 we deduce the desired

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3 Some particular cases

Let A, B positive invertible operators and positive real numbers m, m0, M,

M0 such that the condition0 < mI≤A≤m0I < M0I≤B≤MIholds. Puth:= M_m and h0 := M_m00, then we have

A < h0A= M 0

m0A≤B≤ M

mA=hA.

By (16) we get

"

ν(1−ν)

h0−1 h0

2

+1

#

A!_νB≤A∇_νB (29)

≤hν(1−ν) (h−1)2+1

i

A!νB

By (23) we get

4

"

ν− 1 2

2

+ν(1−ν)K h0

#

A!νB (30)

≤A∇νB≤4

"

ν− 1 2

2

+ν(1−ν)K(h)

#

A!νB

If 0 < mI≤B≤m0I < M0I ≤A≤MI, then for h := M_m and h0 := M_m00 we also have

1

hA≤B≤ 1

h0A < A.

Finally, by (16) we get (29) while from (23) we get (30) as well.

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