research article inequalities of convex functions and self...

Research ArticleInequalities of Convex Functions and Self-Adjoint Operators

Zlatko PaviT

Mechanical Engineering Faculty in Slavonski Brod, University of Osijek, Trg Ivane Brlic Mazuranic 2, 35000 Slavonski Brod, Croatia

Correspondence should be addressed to Zlatko Pavic; [email protected]

Received 28 November 2013; Accepted 4 January 2014; Published 9 February 2014

Academic Editor: Palle E. Jorgensen

Copyright © 2014 Zlatko Pavic. This is an open access article distributed under the Creative Commons Attribution License, whichpermits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

The paper offers generalizations of the Jensen-Mercer inequality for self-adjoint operators and generally convex functions. Theobtained results are applied to define the quasi-arithmetic operator means without using operator convexity. The version of theharmonic-geometric-arithmetic operator mean inequality is derived as an example.

1. Introduction

Throughout the paper we will use a real interval I with thenonempty interior and real segments [𝑎, 𝑏] and (𝑎, 𝑏)with 𝑎 <𝑏.

We briefly summarize a development path of the operatorformof Jensen’s inequality. LetH andK beHilbert spaces, letB(H) and B(K) be associated C∗-algebras of bounded linearoperators, and let 1

𝐻and 1𝐾be their identity operators.

Combining the results from [1, 2], it follows that everyoperator convex function 𝑓 : I → R satisfies the Schwarzinequality

𝑓 (Φ (𝐴)) ≤ Φ (𝑓 (𝐴)) , (1)

where Φ : B(H) → B(K) is a positive linear mapping suchthatΦ(1

𝐻) = 1𝐾and𝐴 ∈ B(H) is a self-adjoint operator with

the spectrum Sp(𝐴) ⊆ I. The above inequality was extendedin [3] to the inequality

𝑓(

𝑛

∑𝑖=1

Φ𝑖(𝐴𝑖)) ≤

𝑛

∑𝑖=1

Φ𝑖(𝑓 (𝐴

𝑖)) , (2)

where Φ𝑖: B(H) → B(K) are positive linear mappings

such that ∑𝑛𝑖=1

Φ𝑖(1𝐻) = 1𝐾and 𝐴

𝑖∈ B(H) are self-adjoint

operators with spectra Sp(𝐴𝑖) ⊆ I. The operator inequality

of (2) was formulated for convex (without operator) con-tinuous functions in [4] assuming the spectral conditions:Sp(𝐴) ⊆ [𝑎, 𝑏] and Sp(𝐴

𝑖) ∩ (𝑎, 𝑏) = 0 for all 𝐴

𝑖, where

𝐴 = ∑𝑛

𝑖=1Φ𝑖(𝐴𝑖).

Including positive operators 𝑃𝑖

∈ B(H) satisfying∑𝑛

𝑖=1Φ𝑖(𝑃𝑖) = 1

𝐾, we have that every convex continuous

function 𝑓 : I → R satisfies the inequality

𝑓(

𝑛

∑𝑖=1

Φ𝑖(𝑃1/2

𝑖𝐴𝑖𝑃1/2

𝑖)) ≤

𝑛

∑𝑖=1

Φ𝑖(𝑃1/2

𝑖𝑓 (𝐴𝑖) 𝑃1/2

𝑖) (3)

if provided the spectral conditions: Sp(𝐴) ⊆ [𝑎, 𝑏] andSp(𝐴𝑖) ∩ (𝑎, 𝑏) = 0 for all self-adjoint operators 𝐴

𝑖, and

the operator sum 𝐴 = ∑𝑛

𝑖=1Φ𝑖(𝑃1/2

𝑖𝐴𝑖𝑃1/2

𝑖). The inequality

in (3) is possible because the operators 𝑃1/2

𝑖𝐴𝑖𝑃1/2

𝑖and

𝑃1/2

𝑖𝑓(𝐴𝑖)𝑃1/2

𝑖are self-adjoint.

2. Discrete and Operator Inequalitiesfor Convex Functions and TrinomialAffine Combinations

2.1. Discrete Variants. Every number 𝑥 ∈ R can be uniquelypresented as the binomial affine combination

𝑥 =𝑏 − 𝑥

𝑏 − 𝑎𝑎 +

𝑥 − 𝑎

𝑏 − 𝑎𝑏 (4)

which is convex if and only if the number 𝑥 belongs to theinterval [𝑎, 𝑏]. Given the function 𝑓 : R → R, let 𝑓line

{𝑎,𝑏}:

R → R be the function of the chord line passing through thepoints 𝐴(𝑎, 𝑓(𝑎)) and 𝐵(𝑏, 𝑓(𝑏)) of the graph of 𝑓. Applyingthe affinity of 𝑓line

{𝑎,𝑏}to the combination in (4), we get

𝑓line{𝑎,𝑏}

(𝑥) =𝑏 − 𝑥

𝑏 − 𝑎𝑓 (𝑎) +

𝑥 − 𝑎

𝑏 − 𝑎𝑓 (𝑏) . (5)

Hindawi Publishing CorporationJournal of OperatorsVolume 2014, Article ID 382364, 5 pageshttp://dx.doi.org/10.1155/2014/382364

2 Journal of Operators

If the function 𝑓 is convex, then we have the inequality

𝑓 (𝑥) ≤ 𝑓line{𝑎,𝑏}

(𝑥) if 𝑥 ∈ [𝑎, 𝑏] , (6)

and the reverse inequality

𝑓 (𝑥) ≥ 𝑓line{𝑎,𝑏}

(𝑥) if 𝑥 ∉ (𝑎, 𝑏) . (7)

Let 𝛼, 𝛽, 𝛾 ∈ R be coefficients such that 𝛼 + 𝛽 − 𝛾 = 1.Let 𝑎, 𝑏, 𝑐 ∈ R be points where 𝑎 < 𝑏. We consider the affinecombination 𝛼𝑎 + 𝛽𝑏 − 𝛾𝑐. Inserting the affine combination𝑐 = 𝜆𝑎+𝜇𝑏 assuming that 𝜆+𝜇 = 1, we get the binomial form

𝛼𝑎 + 𝛽𝑏 − 𝛾𝑐 = (𝛼 − 𝛾𝜆) 𝑎 + (𝛽 − 𝛾𝜇) 𝑏. (8)

Lemma 1. Let 𝛼, 𝛽, 𝛾 ∈ [0, 1] be coefficients such that 𝛼 + 𝛽 −𝛾 = 1. Let 𝑎, 𝑏, 𝑐 ∈ R be points such that 𝑎 < 𝑏 and 𝑐 ∈ [𝑎, 𝑏].

Then the affine combination

𝛼𝑎 + 𝛽𝑏 − 𝛾𝑐 ∈ [𝑎, 𝑏] , (9)

and every convex function 𝑓 : [𝑎, 𝑏] → R satisfies theinequality

𝑓 (𝛼𝑎 + 𝛽𝑏 − 𝛾𝑐) ≤ 𝛼𝑓 (𝑎) + 𝛽𝑓 (𝑏) − 𝛾𝑓 (𝑐) . (10)

Proof. The condition 𝑐 = 𝜆𝑎 + 𝜇𝑏 ∈ [𝑎, 𝑏] involves 𝜆, 𝜇 ∈

[0, 1]. Then the binomial combination of the right-hand sidein (8) is convex since its coefficients 𝛼−𝛾𝜆 ≥ 𝛼−𝛾 = 1−𝛽 ≥ 0

and also𝛽−𝛾𝜇 ≥ 0. So, the combination 𝛼𝑎+𝛽𝑏−𝛾 belongs to[𝑎, 𝑏]. Applying the inequality in (6) and the affinity of 𝑓line

{𝑎,𝑏},

we get

𝑓 (𝛼𝑎 + 𝛽𝑏 − 𝛾𝑐) ≤ 𝑓line{𝑎,𝑏}

(𝛼𝑎 + 𝛽𝑏 − 𝛾𝑐)

= 𝛼𝑓 (𝑎) + 𝛽𝑓 (𝑏) − 𝛾𝑓line{𝑎,𝑏}

(𝑐)

≤ 𝛼𝑓 (𝑎) + 𝛽𝑓 (𝑏) − 𝛾𝑓 (𝑐)

(11)

because 𝑓line{𝑎,𝑏}

(𝑐) ≥ 𝑓(𝑐).

Lemma 1 is trivially true if 𝑎 = 𝑏. It is also valid for𝛾 ∈ [−1, 1] because then the observed affine combinationswith 𝛾 ≤ 0 become convex, and associated inequalities followfrom Jensen’s inequality. The similar combinations including𝛾 ∈ [−1, 1]were observed in [5, Corollary 11 andTheorem 12]additionally using a monotone function 𝑔. If 𝛼 = 𝛽 = 𝛾 =

1, then the inequality in (10) is reduced to simple Mercer’svariant of Jensen’s inequality obtained in [6].

Lemma 2. Let 𝛼, 𝛽, 𝛾 ∈ [1,∞) be coefficients such that 𝛼+𝛽−𝛾 = 1. Let 𝑎, 𝑏, 𝑐 ∈ R be points such that 𝑎 < 𝑏 and 𝑐 ∉ (𝑎, 𝑏).

Then the affine combination

𝛼𝑎 + 𝛽𝑏 − 𝛾𝑐 ∉ (𝑎, 𝑏) , (12)

and every convex function 𝑓 : I → R, where I =

conv{𝑎, 𝑏, 𝑐} satisfies the inequality

𝑓 (𝛼𝑎 + 𝛽𝑏 − 𝛾𝑐) ≥ 𝛼𝑓 (𝑎) + 𝛽𝑓 (𝑏) − 𝛾𝑓 (𝑐) . (13)

Proof. The condition 𝑐 = 𝜆𝑎 + 𝜇𝑏 ∉ (𝑎, 𝑏) entails 𝜆 ≤ 0 or𝜆 ≥ 1, and the coefficients of the binomial form of (8) satisfy𝛼 − 𝛾𝜆 ≥ 𝛼 ≥ 1 if 𝜆 ≤ 0, or 𝛼 − 𝛾𝜆 ≤ 𝛼 − 𝛾 = 1 − 𝛽 ≤ 0 if𝜆 ≥ 1. So, the combination𝛼𝑎+𝛽𝑏−𝛾does not belong to (𝑎, 𝑏).Applying the inequality in (7), we get the series of inequalitiesas in (11) but with the reverse inequality signs.

It is not necessary to require 𝛾 ∈ [1,∞) in Lemma 2,because it follows from the other coefficient conditions.

2.2. Operator Variants. We write 𝐴 ≤ 𝐵 for self-adjointoperators 𝐴, 𝐵 ∈ B(H) if the inner product inequality⟨𝐴𝑥, 𝑥⟩ ≤ ⟨𝐵𝑥, 𝑥⟩ holds for every vector 𝑥 ∈ H. A self-adjoint operator 𝐴 is positive (nonnegative) if it is greaterthan or equal to null operator (𝐴 ≥ 0). If Sp(𝐴) ⊆ I and𝑓, 𝑔 :I → R are continuous functions such that 𝑓(𝑥) ≤ 𝑔(𝑥) forevery 𝑥 ∈ Sp(𝐴), then the operator inequality 𝑓(𝐴) ≤ 𝑔(𝐴)

is valid. The bounds of a self-adjoint operator 𝐴 are definedwith

𝑎𝐴= inf‖𝑥‖=1

⟨𝐴𝑥, 𝑥⟩ , 𝑏𝐴= sup‖𝑥‖=1

⟨𝐴𝑥, 𝑥⟩ , (14)

and its spectrum Sp(𝐴) is contained in [𝑎𝐴, 𝑏𝐴] wherein we

have the operator inequality

𝑎𝐴1𝐻≤ 𝐴 ≤ 𝑏

𝐴1𝐻. (15)

More details on the theory of bounded operators and theirinequalities can be found in [7]. The operator versions ofLemmas 1 and 2 follow.

Corollary 3. Let 𝛼, 𝛽, 𝛾 ∈ [0, 1] be coefficients such that 𝛼 +𝛽 − 𝛾 = 1. Let 𝐴 ∈ B(H) be a self-adjoint operator such thatSp(𝐴) ⊆ [𝑎, 𝑏].

Then

Sp (𝛼𝑎1𝐻+ 𝛽𝑏1

𝐻− 𝛾𝐴) ⊆ [𝑎, 𝑏] , (16)

and every convex continuous function 𝑓 : [𝑎, 𝑏] → R satisfiesthe inequality

𝑓 (𝛼𝑎1𝐻+ 𝛽𝑏1

𝐻− 𝛾𝐴) ≤ 𝛼𝑓 (𝑎) 1

𝐻+ 𝛽𝑓 (𝑏) 1

𝐻− 𝛾𝑓 (𝐴) .

(17)

Proof. The spectral inclusion in (16) follows from the inclu-sion in (9). Using the affinity of the function 𝑓

line{𝑎,𝑏}

and theoperator inequalities 𝑓line

{𝑎,𝑏}(⋅) ≥ 𝑓(⋅), we can replace the

discrete inequalities in (11) with the operator inequalities.

Corollary 4. Let 𝛼, 𝛽, 𝛾 ∈ [1,∞) be coefficients such that 𝛼 +𝛽 − 𝛾 = 1. Let 𝐴 ∈ B(H) be a self-adjoint operator such thatSp(𝐴) ∩ (𝑎, 𝑏) = 0.

Then

Sp (𝛼𝑎1𝐻+ 𝛽𝑏1

𝐻− 𝛾𝐴) ∩ (𝑎, 𝑏) = 0, (18)

and every convex continuous function 𝑓 : I → R, where Icontains Sp(𝐴) and [𝑎, 𝑏], satisfies the inequality

𝑓 (𝛼𝑎1𝐻+ 𝛽𝑏1

𝐻− 𝛾𝐴) ≥ 𝛼𝑓 (𝑎) 1

𝐻+ 𝛽𝑓 (𝑏) 1

𝐻− 𝛾𝑓 (𝐴) .

(19)

Journal of Operators 3

3. Main Results

We want to extend and generalize the inequalities in (17)and (19) including positive operators and positive linearmappings. The main results are Theorems 8 and 9.

Lemma 5. Let Φ𝑖: B(H) → B(K) be linear mappings and

let 𝑃𝑖∈ B(H) be positive linear operators so that∑𝑛

𝑖=1Φ𝑖(𝑃𝑖) =

1𝐾. Let 𝐴

𝑖∈ B(H) be self-adjoint operators.

Then every affine function 𝑔(𝑥) = 𝑢𝑥 + V, where 𝑢 and Vare real constants, satisfies the operator equality

𝑔(

𝑛

∑𝑖=1

Φ𝑖(𝑃1/2

𝑖𝐴𝑖𝑃1/2

𝑖)) =

𝑛

∑𝑖=1

Φ𝑖(𝑃1/2

𝑖𝑔 (𝐴𝑖) 𝑃1/2

𝑖) . (20)

Proof. Applying the affinity of the function𝑔 and the assump-tion ∑𝑛

𝑖=1Φ𝑖(𝑃𝑖) = 1𝐾, it follows that

𝑔(

𝑛

∑𝑖=1

Φ𝑖(𝑃1/2

𝑖𝐴𝑖𝑃1/2

𝑖)) = 𝑢

𝑛

∑𝑖=1

Φ𝑖(𝑃1/2

𝑖𝐴𝑖𝑃1/2

𝑖) + V1

𝐾

= 𝑢

𝑛

∑𝑖=1

Φ𝑖(𝑃𝑖𝐴𝑖) + V𝑛

∑𝑖=1

Φ𝑖(𝑃𝑖)

=

𝑛

∑𝑖=1

Φ𝑖(𝑃1/2

𝑖(𝑢𝐴𝑖+ V1𝐻) 𝑃1/2

𝑖)

=

𝑛

∑𝑖=1

Φ𝑖(𝑃1/2

𝑖𝑔 (𝐴𝑖) 𝑃1/2

𝑖)

(21)

achieving the equality in (20).

Lemma 6. Let Φ𝑖: B(H) → B(K) be positive linear

mappings and let 𝑃𝑖∈ B(H) be positive linear operators so

that∑𝑛𝑖=1

Φ𝑖(𝑃𝑖) = 1𝐾. Let𝐴

𝑖∈ B(H) be self-adjoint operators

such that Sp(𝐴𝑖) ⊆ [𝑎, 𝑏].

Then the spectrum of the operator sum 𝐴 =

∑𝑛


𝑖𝐴𝑖𝑃1/2

𝑖) is contained in [𝑎, 𝑏].

Proof. Applying the positive operators 𝑃𝑖and the positive

mappingsΦ𝑖to the assumed spectral inequalities

𝑎1𝐻≤ 𝐴𝑖≤ 𝑏1𝐻, (22)

we get

𝑎Φ𝑖(𝑃𝑖) ≤ Φ

𝑖(𝑃1/2

𝑖𝐴𝑖𝑃1/2

𝑖) ≤ 𝑏Φ

𝑖(𝑃𝑖) . (23)

Summing the above inequalities and using the assumption∑𝑛

𝑖=1Φ𝑖(𝑃𝑖) = 1𝐾, we have

𝑎1𝐾≤

𝑛

∑𝑖=1

Φ𝑖(𝑃1/2

𝑖𝐴𝑖𝑃1/2

𝑖) ≤ 𝑏1

𝐾(24)

which provides that Sp(𝐴) ⊆ [𝑎, 𝑏].

Corollary 7. Let Φ𝑖: B(H) → B(K) be positive linear


that∑𝑛𝑖=1




Then every convex continuous function 𝑓 : [𝑎, 𝑏] → R

satisfies the inequality

max{𝑓(𝑛

∑𝑖=1

Φ𝑖(𝑃1/2

𝑖𝐴𝑖𝑃1/2

𝑖)) ,

𝑛

∑𝑖=1

Φ𝑖(𝑃1/2


𝑖)}

≤ 𝑓line{𝑎,𝑏}

(

𝑛

∑𝑖=1

Φ𝑖(𝑃1/2

𝑖𝐴𝑖𝑃1/2

𝑖))

≤ max {𝑓 (𝑎) 1𝐾, 𝑓 (𝑏) 1

𝐾} .

(25)

Proof. The inequality in (25) is the consequence of Lemmas 5and 6, and the discrete inequality

max{𝑓(𝑛

∑𝑖=1

𝑝𝑖𝑥𝑖) ,

𝑛

∑𝑖=1

𝑝𝑖𝑓 (𝑥𝑖)}

≤ 𝑓line{𝑎,𝑏}

(

𝑛

∑𝑖=1

𝑝𝑖𝑥𝑖)

≤ max {𝑓 (𝑎) , 𝑓 (𝑏)} ,

(26)

where 𝑥𝑖∈ [𝑎, 𝑏] are points and 𝑝

𝑖∈ [0, 1] are coefficients of

the sum equal to 1.

Theorem 8. Let 𝛼, 𝛽, 𝛾 ∈ [0, 1] be coefficients such that 𝛼 +

𝛽 − 𝛾 = 1. Let Φ𝑖: B(H) → B(K) be positive linear


that∑𝑛𝑖=1




Then the spectrum of the operator

𝐴 = 𝛼𝑎1𝐾+ 𝛽𝑏1

𝐾− 𝛾

𝑛

∑𝑖=1

Φ𝑖(𝑃1/2

𝑖𝐴𝑖𝑃1/2

𝑖) (27)

is contained in [𝑎, 𝑏], and every convex continuous function 𝑓 :[𝑎, 𝑏] → R satisfies the inequality

𝑓 (𝐴) ≤ 𝛼𝑓 (𝑎) 1𝐾+ 𝛽𝑓 (𝑏) 1

𝐾− 𝛾

𝑛

∑𝑖=1

Φ𝑖(𝑃1/2


𝑖) .

(28)

If the function 𝑓 is concave, then the reverse inequality isvalid in (28).

Proof. Taking the operator sum𝐴 = ∑𝑛


𝑖𝐴𝑖𝑃1/2

𝑖), the

spectral inclusion Sp(𝐴) ⊆ [𝑎, 𝑏] follows from Lemma 6 and

4 Journal of Operators

the inclusion in (16). Assuming and applying the convexity of𝑓 and the affinity of 𝑓line

{𝑎,𝑏}according to Lemma 5, we get

𝑓 (𝐴) ≤ 𝑓line{𝑎,𝑏}

(𝛼𝑎1𝐾+ 𝛽𝑏1

𝐾− 𝛾

𝑛

∑𝑖=1

Φ𝑖(𝑃1/2

𝑖𝐴𝑖𝑃1/2

𝑖))

= 𝛼𝑓 (𝑎) 1𝐾+ 𝛽𝑓 (𝑏) 1

𝐾

− 𝛾𝑓line{𝑎,𝑏}

(

𝑛

∑𝑖=1

Φ𝑖(𝑃1/2

𝑖𝐴𝑖𝑃1/2

𝑖))

= 𝛼𝑓 (𝑎) 1𝐾+ 𝛽𝑓 (𝑏) 1

𝐾− 𝛾

𝑛

∑𝑖=1

Φ𝑖(𝑃1/2

𝑖𝑓line{𝑎,𝑏}

(𝐴𝑖) 𝑃1/2

𝑖)

≤ 𝛼𝑓 (𝑎) 1𝐾+ 𝛽𝑓 (𝑏) 1

𝐾− 𝛾

𝑛

∑𝑖=1

Φ𝑖(𝑃1/2


𝑖)

(29)

because 𝑓line{𝑎,𝑏}

(𝐴𝑖) ≥ 𝑓(𝐴

𝑖).

The version of Theorem 8 for 𝛼 = 𝛽 = 𝛾 = 1 and all𝑃𝑖= 1𝐻was obtained in [8] as the main result.

Theorem 9. Let 𝛼, 𝛽, 𝛾 ∈ [1,∞) be coefficients such that𝛼 + 𝛽 − 𝛾 = 1. Let Φ

𝑖: B(H) → B(K) be positive linear


that∑𝑛𝑖=1



such that Sp(𝐴𝑖) ∩ (𝑎, 𝑏) = 0, and let𝐴 = ∑

𝑛


𝑖𝐴𝑖𝑃1/2

𝑖)

be the operator sum such that Sp(𝐴) ∩ (𝑎, 𝑏) = 0.Then the spectrum of the operator


𝐾− 𝛾

𝑛

∑𝑖=1

Φ𝑖(𝑃1/2

𝑖𝐴𝑖𝑃1/2

𝑖) (30)

satisfies the relation Sp(𝐴) ∩ (𝑎, 𝑏) = 0, and every convexcontinuous function 𝑓 : I → R, whereI contains all spectraand [𝑎, 𝑏], satisfies the inequality

𝑓 (𝐴) ≥ 𝛼𝑓 (𝑎) 1𝐾+ 𝛽𝑓 (𝑏) 1

𝐾

− 𝛾

𝑛

∑𝑖=1

Φ𝑖(𝑃1/2


𝑖) .

(31)

If the function 𝑓 is concave, then the reverse inequality is validin (31).

Proof. The relation Sp(𝐴) ∩ (𝑎, 𝑏) = 0 is the consequence ofthe relation in (18). Assuming and using the convexity of 𝑓and the affinity of 𝑓line

{𝑎,𝑏}, as well as the inequalities 𝑓line

{𝑎,𝑏}(𝐴𝑖) ≤

𝑓(𝐴𝑖), we get the series of inequalities as in (29) but with the

reverse inequality signs.

4. Application to Quasi-Arithmetic Means

In applications of convexity to quasi-arithmetic means, weuse strictly monotone continuous functions 𝜑, 𝜓 : I → R

such that the function 𝜓 ∘ 𝜑−1 is convex, in which case we say

that 𝜓 is 𝜑-convex. A similar notation is used for concavity.This terminology is taken from [9, Definition 1.19].

A continuous function 𝑓 : I → R is said to be operatorincreasing onI if 𝐴 ≤ 𝐵 implies 𝑓(𝐴) ≤ 𝑓(𝐵) for every pairof self-adjoint operators 𝐴, 𝐵 ∈ B(H) with spectra in I. Afunction 𝑓 is said to be operator decreasing if the function−𝑓 is operator increasing.

Take an operator affine combination


𝐾− 𝛾

𝑛

∑𝑖=1

Φ𝑖(𝑃1/2

𝑖𝐴𝑖𝑃1/2

𝑖) (32)

as in Theorem 8. If 𝜑 : [𝑎, 𝑏] → R is a strictly monotonecontinuous function, we define the 𝜑-quasi-arithmetic meanof the combination 𝐴 as the operator

𝑀𝜑(𝐴) = 𝜑

−1

(𝛼𝜑 (𝑎) 1𝐾+ 𝛽𝜑 (𝑏) 1

𝐾

− 𝛾

𝑛

∑𝑖=1

Φ𝑖(𝑃1/2

𝑖𝜑 (𝐴𝑖) 𝑃1/2

𝑖)) .

(33)

The spectrum of the operator 𝑀𝜑(𝐴) is contained in [𝑎, 𝑏]

because the spectrum of the operator

𝐴𝜑= 𝛼𝜑 (𝑎) 1

𝐾+ 𝛽𝜑 (𝑏) 1

𝐾− 𝛾

𝑛

∑𝑖=1

Φ𝑖(𝑃1/2

𝑖𝜑 (𝐴𝑖) 𝑃1/2

𝑖)

(34)

is contained in 𝜑([𝑎, 𝑏]). The quasi-arithmetic means definedin (33) are invariant with respect to the affinity; that is, theequality

𝑀𝑢𝜑+V (𝐴) = 𝑀

𝜑(𝐴) (35)

holds for all pairs of real numbers 𝑢 = 0 and V. Indeed, if𝜓(𝑥) = 𝑢𝜑(𝑥) + V, then

𝐴𝜓= 𝑢𝐴𝜑+ V1𝐾,

𝜓−1

(𝑥) = 𝜑−1

(1

𝑢(𝑥 − V)) ,

(36)

and therefore, it follows that

𝑀𝜓(𝐴) = 𝜓

−1

(𝐴𝜓) = 𝜑−1

(1

𝑢(𝐴𝜓− V1𝐾))

= 𝜑−1

(𝐴𝜑) = 𝑀

𝜑(𝐴) .

(37)

The order of the pair of quasi-arithmetic means 𝑀𝜑

and 𝑀𝜓depends on convexity of the function 𝜓 ∘ 𝜑

−1 andmonotonicity of the function 𝜓. Theorem 8 can be applied tooperator means as follows.

Corollary 10. Let 𝐴 be an affine combination as in (32)satisfying the assumptions ofTheorem 8. Let𝜑, 𝜓 : [𝑎, 𝑏] → R

be strictly monotone continuous functions.

Journal of Operators 5

If 𝜓 is either 𝜑-convex with operator increasing 𝜓−1 or𝜑-concave with operator decreasing 𝜓

−1, then one has theinequality

𝑀𝜑(𝐴) ≤ 𝑀

𝜓(𝐴) . (38)

If 𝜓 is either 𝜑-convex with operator decreasing 𝜓−1 or 𝜑-concave with operator increasing 𝜓−1, then one has the reverseinequality in (38).

Proof. Let us prove the case in which 𝜓 is 𝜑-convex withoperator increasing 𝜓−1. Put [𝑐, 𝑑] = 𝜑([𝑎, 𝑏]). Applying theinequality in (28) of Theorem 8 to the affine combination𝐴𝜑of (34) with Sp(𝐴

𝜑) ⊆ [𝑐, 𝑑] and the convex function

𝑓 = 𝜓 ∘ 𝜑−1

: [𝑐, 𝑑] → R, we get

𝜓 ∘ 𝜑−1

(𝐴𝜑) ≤ 𝐴

𝜓. (39)

Assigning the increasing function𝜓−1 to the above inequality,we attain

𝑀𝜑(𝐴) = 𝜑

−1

(𝐴𝜑) ≤ 𝜓

−1

(𝐴𝜓) = 𝑀

𝜓(𝐴) (40)

which finishes the proof.

Using Corollary 10 we get the following version of theharmonic-geometric-arithmetic mean inequality for opera-tors.

Corollary 11. If 𝐴 is an affine operator combination as in(32) satisfying the assumptions of Theorem 8 with the additionthat [𝑎, 𝑏] ⊂ (0,∞), then one has the harmonic-geometric-arithmetic operator inequality

((𝛼

𝑎+𝛽

𝑏) 1𝐾− 𝛾

𝑛

∑𝑖=1

Φ𝑖(𝑃1/2

𝑖𝐴−1

𝑖𝑃1/2

𝑖))

−1

≤ ln 𝑎𝛼𝑏𝛽1𝐾− 𝛾 exp(

𝑛

∑𝑖=1

Φ𝑖(𝑃1/2

𝑖(ln𝐴𝑖) 𝑃1/2

𝑖))

≤ (𝛼𝑎 + 𝛽𝑏) 1𝐾− 𝛾

𝑛

∑𝑖=1

Φ𝑖(𝑃1/2

𝑖𝐴𝑖𝑃1/2

𝑖) .

(41)

Proof. To prove the left-hand side of the inequality in (41)we use the functions 𝜑(𝑥) = ln𝑥 and 𝜓(𝑥) = 𝑥

−1. Then𝜓∘𝜑−1

(𝑥) = exp(−𝑥) and𝜓−1(𝑥) = 𝑥−1, so 𝜓 is 𝜑-convex and

𝜓−1

(𝑥) = 𝑥−1 is operator decreasing. Applying Corollary 10

to this case, we have𝑀ln 𝑥 (𝐴) ≥ 𝑀

𝑥−1 (𝐴) . (42)

To prove the right-hand side we use the functions 𝜑(𝑥) =ln𝑥 and 𝜓(𝑥) = 𝑥. Then 𝜓 ∘ 𝜑

−1

(𝑥) = exp 𝑥 and 𝜓−1(𝑥) = 𝑥,so 𝜓 is 𝜑-convex and 𝜓

−1

(𝑥) = 𝑥−1 is operator increasing.

Applying the inequality in (38), we get

𝑀ln 𝑥 (𝐴) ≤ 𝑀𝑥(𝐴) . (43)

The double inequality in (41) follows by connecting theinequalities in (42) and (43).

Quasi-arithmetic operator means without applying oper-ator convexity were also investigated in [4, 10].

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper.

References

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[3] F. Hansen, J. Pecaric, and I. Peric, “Jensen’s operator inequalityand its converses,”Mathematica Scandinavica, vol. 100, no. 1, pp.61–73, 2007.

[4] J. Micic, Z. Pavic, and J. Pecaric, “Jensen’s inequality foroperators without operator convexity,” Linear Algebra and itsApplications, vol. 434, no. 5, pp. 1228–1237, 2011.

[5] Z. Pavic, “The applications of functional variants of Jensen’sinequality,” Journal of Function Spaces and Applications, vol.2013, Article ID 194830, 5 pages, 2013.

[6] A. M. Mercer, “A variant of Jensen’s inequality,” Journal ofInequalities in Pure and Applied Mathematics, vol. 4, article 73,2 pages, 2003.

[7] T. Furuta, J. Micic Hot, J. Pecaric, and Y. Seo, Mond-PecaricMethod in Operator Inequalities, vol. 1, Element, Zagreb, Croa-tia, 2005.

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[9] J. E. Pecaric, F. Proschan, and Y. L. Tong,Convex Functions, Par-tial Orderings, and Statistical Applications, vol. 187, AcademicPress, Boston, Mass, USA, 1992.

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