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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
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
[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).
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
𝑖=1Φ𝑖(𝑃1/2
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
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 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
𝑖𝑓 (𝐴𝑖) 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
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
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
𝑖𝑓 (𝐴𝑖) 𝑃1/2
𝑖) .
(28)
If the function 𝑓 is concave, then the reverse inequality isvalid in (28).
Proof. Taking the operator sum𝐴 = ∑𝑛
𝑖=1Φ𝑖(𝑃1/2
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
𝑖𝑓 (𝐴𝑖) 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
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) = 0, and let𝐴 = ∑
𝑛
𝑖=1Φ𝑖(𝑃1/2
𝑖𝐴𝑖𝑃1/2
𝑖)
be the operator sum such that Sp(𝐴) ∩ (𝑎, 𝑏) = 0.Then the spectrum of the operator
𝐴 = 𝛼𝑎1𝐾+ 𝛽𝑏1
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
𝑖𝑓 (𝐴𝑖) 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
𝐴 = 𝛼𝑎1𝐾+ 𝛽𝑏1
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.
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