Extension of Jensen's Inequality for Operators without Operator Convexity

Abstract

We give an extension of Jensen's inequality for 𝑛-tuples of self-adjoint operators, unital 𝑛-tuples of positive linear mappings, and real-valued continuous convex functions with conditions on the operators' bounds. We also study operator quasiarithmetic means under the same conditions.

1. Introduction

We recall some notations and definitions. Let ℬ(𝐻) be the C^∗-algebra of all bounded linear operators on a Hilbert space 𝐻 and 1_𝐻 stands for the identity operator. We define bounds of a self-adjoint operator 𝐴∈ℬ(𝐻):? 𝑚_𝐴=inf_‖𝑥‖=1⟨𝐴𝑥,𝑥⟩,𝑀_𝐴=sup_‖𝑥‖=1⟨𝐴𝑥,𝑥⟩ (1.1) for 𝑥∈𝐻. If 𝖲𝗉(𝐴) denotes the spectrum of 𝐴, then 𝖲𝗉(𝐴) is real and 𝖲𝗉(𝐴)⊆[𝑚_𝐴,𝑀_𝐴].

Mond and Pecaric in [1] proved the following version of Jensen's operator inequality 𝑓^𝑛_𝑖=1𝑤_𝑖Φ_𝑖𝐴_𝑖≤^𝑛_𝑖=1𝑤_𝑖Φ_𝑖𝑓𝐴_𝑖, (1.2) for operator convex functions 𝑓 defined on an interval 𝐼, where Φ_𝑖∶ℬ(𝐻)→ℬ(𝐾), 𝑖=1,…,𝑛, are unital positive linear mappings, 𝐴₁,…,𝐴_𝑛 are self-adjoint operators with the spectra in 𝐼, and 𝑤₁,…,𝑤_𝑛 are nonnegative real numbers with ∑^𝑛_𝑖=1𝑤_𝑖=1.

Hansen et al. gave in [2] a generalization of (1.2) for a unital field of positive linear mappings. Recently, Micic et al. in [3] gave a generalization of this results for a not-unital field of positive linear mappings.

Very recently, Micic et al. gave in [4, Theorem??1] a version of Jensen's operator inequality without operator convexity as follows.

Theorem A. Let (𝐴₁,…,𝐴_𝑛) be an 𝑛-tuple of self-adjoint operators 𝐴_𝑖∈ℬ(𝐻) with bounds 𝑚_𝑖 and 𝑀_𝑖, 𝑚_𝑖≤𝑀_𝑖, 𝑖=1,…,𝑛. Let (Φ₁,…,Φ_𝑛) be an 𝑛-tuple of positive linear mappings Φ_𝑖∶ℬ(𝐻)→ℬ(𝐾), 𝑖=1,…,𝑛, such that ∑^𝑛_𝑖=1Φ_𝑖(1_𝐻)=1_𝐾. If 𝑚_𝐶,𝑀_𝐶∩𝑚_𝑖,𝑀_𝑖=∅for𝑖=1,…,𝑛, (1.3) where 𝑚_𝐶 and 𝑀_𝐶, 𝑚_𝐶≤𝑀_𝐶, are bounds of the self-adjoint operator 𝐶=∑^𝑛_𝑖=1Φ_𝑖(𝐴_𝑖), then 𝑓^𝑛_𝑖=1Φ_𝑖𝐴_𝑖≤^𝑛_𝑖=1Φ_𝑖𝑓𝐴_𝑖 (1.4) holds for every continuous convex function 𝑓∶𝐼→ℝ provided that the interval 𝐼 contains all 𝑚_𝑖,𝑀_𝑖.
If 𝑓∶𝐼→ℝ is concave, then the reverse inequality is valid in (1.4).

In the same paper [4], we study the quasiarithmetic operator mean: ℳ_𝜑(𝐀,𝚽,𝑛)=𝜑^−1𝑛_𝑖=1Φ_𝑖𝜑𝐴_𝑖, (1.5) where (𝐴₁,…,𝐴_𝑛) is an 𝑛-tuple of self-adjoint operators in ℬ(𝐻) with the spectra in 𝐼, (Φ₁,…,Φ_𝑛) is an 𝑛-tuple of positive linear mappings Φ_𝑖∶ℬ(𝐻)→ℬ(𝐾) such that ∑^𝑛_𝑖=1Φ_𝑖(1_𝐻)=1_𝐾, and 𝜑∶𝐼→ℝ is a continuous strictly monotone function.

The following results about the monotonicity of this mean are proven in [4, Theorem??3].

Theorem B. Let (𝐴₁,…,𝐴_𝑛) and (Φ₁,…,Φ_𝑛) be as in the definition of the quasiarithmetic mean (1.5). Let 𝑚_𝑖 and 𝑀_𝑖, 𝑚_𝑖≤𝑀_𝑖, be the bounds of 𝐴_𝑖, 𝑖=1,…,𝑛. Let 𝜑,𝜓∶𝐼→ℝ be continuous strictly monotone functions on an interval 𝐼 which contains all 𝑚_𝑖,𝑀_𝑖. Let 𝑚_𝜑 and 𝑀_𝜑, 𝑚_𝜑≤𝑀_𝜑, be the bounds of the mean ℳ_𝜑(𝐀,Φ,𝑛), such that 𝑚_𝜑,𝑀_𝜑∩𝑚_𝑖,𝑀_𝑖=∅,for𝑖=1,…,𝑛. (1.6) If one of the following conditions(i)𝜓∘𝜑⁻¹ is convex and 𝜓⁻¹ is operator monotone,(i')𝜓∘𝜑⁻¹ is concave and −𝜓⁻¹ is operator monotone,
is satisfied, then ℳ_𝜑(𝐀,𝚽,𝑛)≤ℳ_𝜓(𝐀,𝚽,𝑛). (1.7) If one of the following conditions (ii)𝜓∘𝜑⁻¹ is concave and 𝜓⁻¹ is operator monotone, (ii')𝜓∘𝜑⁻¹ is convex and −𝜓⁻¹ is operator monotone,
is satisfied, then the reverse inequality is valid in (1.7).

In this paper we study an extension of Jensen's inequality given in Theorem A. As an application of this result, we give an extension of Theorem B for a version of the quasiarithmetic mean (1.5) with an 𝑛-tuple of positive linear mappings which is not unital.

2. Extension of Jensens Operator Inequality

In Theorem A we prove that Jensen's operator inequality holds for every continuous convex function and for every 𝑛-tuple of self-adjoint operators (𝐴₁,…,𝐴_𝑛), for every 𝑛-tuple of positive linear mappings (Φ₁,…,Φ_𝑛) in the case when the interval with bounds of the operator 𝐴=∑^𝑛_𝑖=1Φ_𝑖(𝐴_𝑖) has no intersection points with the interval with bounds of the operator 𝐴_𝑖 for each 𝑖=1,…,𝑛, that is, when 𝑚_𝐴,𝑀_𝐴∩𝑚_𝑖,𝑀_𝑖=∅,for𝑖=1,…,𝑛, (2.1) where 𝑚_𝐴 and 𝑀_𝐴, 𝑚_𝐴≤𝑀_𝐴, are the bounds of 𝐴, and 𝑚_𝑖 and 𝑀_𝑖, 𝑚_𝑖≤𝑀_𝑖, are the bounds of 𝐴_𝑖, 𝑖=1,…,𝑛.

It is interesting to consider the case when (𝑚_𝐴,𝑀_𝐴)∩[𝑚_𝑖,𝑀_𝑖]=∅ is valid for several 𝑖∈{1,…,𝑛}, but not for all 𝑖=1,…,𝑛. We study it in the following theorem.

Theorem 2.1. Let (𝐴₁,…,𝐴_𝑛) be an 𝑛-tuple of self-adjoint operators 𝐴_𝑖∈𝐵(𝐻) with the bounds 𝑚_𝑖 and 𝑀_𝑖, 𝑚_𝑖≤𝑀_𝑖, 𝑖=1,…,𝑛. Let (Φ₁,…,Φ_𝑛) be an 𝑛-tuple of positive linear mappings Φ_𝑖∶𝐵(𝐻)→𝐵(𝐾), such that ∑^𝑛_𝑖=1Φ_𝑖(1_𝐻)=1_𝐾. For 1≤𝑛₁<𝑛, one denotes 𝑚=min{𝑚₁,…,𝑚_𝑛1}, 𝑀=max{𝑀₁,…,𝑀_𝑛1}, and ∑^𝑛1_𝑖=1Φ_𝑖(1_𝐻)=𝛼1_𝐾, ∑^𝑛_𝑖=𝑛1+1Φ_𝑖(1_𝐻)=𝛽1_𝐾, where 𝛼,𝛽>0, 𝛼+𝛽=1. If (𝑚,𝑀)∩𝑚_𝑖,𝑀_𝑖=∅,for𝑖=𝑛₁+1,…,𝑛, (2.2) and one of two equalities 1𝛼^𝑛1_𝑖=1Φ_𝑖𝐴_𝑖=1𝛽^𝑛_𝑖=𝑛1+1Φ_𝑖𝐴_𝑖=^𝑛_𝑖=1Φ_𝑖𝐴_𝑖 (2.3) is valid, then 1𝛼^𝑛1_𝑖=1Φ_𝑖𝑓𝐴_𝑖≤^𝑛_𝑖=1Φ_𝑖𝑓𝐴_𝑖≤1𝛽^𝑛_𝑖=𝑛1+1Φ_𝑖𝑓𝐴_𝑖 (2.4) holds for every continuous convex function 𝑓∶𝐼→ℝ provided that the interval 𝐼 contains all 𝑚_𝑖,𝑀_𝑖, 𝑖=1,…,𝑛.

If 𝑓∶𝐼→ℝ is concave, then the reverse inequality is valid in (2.4).

Proof. We prove only the case when 𝑓 is a convex function.
Let us denote 1𝐴=𝛼^𝑛1_𝑖=1Φ_𝑖𝐴_𝑖1,𝐵=𝛽^𝑛_𝑖=𝑛1+1Φ_𝑖𝐴_𝑖,𝐶=^𝑛_𝑖=1Φ_𝑖𝐴_𝑖. (2.5) It is easy to verify that 𝐴=𝐵 or 𝐵=𝐶 or 𝐴=𝐶 implies 𝐴=𝐵=𝐶.
(a) Let 𝑚<𝑀. Since 𝑓 is convex on [𝑚,𝑀] and [𝑚_𝑖,𝑀_𝑖]⊆[𝑚,𝑀] for 𝑖=1,…,𝑛₁, then 𝑓(𝑡)≤𝑀−𝑡𝑓(𝑚)+𝑀−𝑚𝑡−𝑚𝑓(𝑀)𝑚𝑀−𝑚,𝑡∈_𝑖,𝑀_𝑖for𝑖=1,…,𝑛₁, (2.6) but since 𝑓 is convex on all [𝑚_𝑖,𝑀_𝑖] and (𝑚,𝑀)∩[𝑚_𝑖,𝑀_𝑖]=∅ for 𝑖=𝑛₁+1,…,𝑛, then 𝑓(𝑡)≥𝑀−𝑡𝑓(𝑚)+𝑀−𝑚𝑡−𝑚𝑓(𝑀)𝑚𝑀−𝑚,𝑡∈_𝑖,𝑀_𝑖for𝑖=𝑛₁+1,…,𝑛. (2.7)
Since 𝑚_𝑖1_𝐻≤𝐴_𝑖≤𝑀_𝑖1_𝐻, 𝑖=1,…,𝑛₁, it follows from (2.6) that 𝑓𝐴_𝑖≤𝑀1_𝐻−𝐴_𝑖𝑓(𝑚)+𝐴𝑀−𝑚^𝑖−𝑚1_𝐻𝑓(𝑀)𝑀−𝑚,𝑖=1,…,𝑛₁. (2.8) Applying a positive linear mapping Φ_𝑖 and summing, we obtain ^𝑛1_𝑖=1Φ_𝑖𝑓𝐴_𝑖≤𝑀𝛼1_𝐾−∑^𝑛1_𝑖=1Φ_𝑖𝐴_𝑖𝑓(𝑚)+∑𝑀−𝑚^𝑛1𝑖=1Φ_𝑖𝐴_𝑖−𝑚𝛼1_𝐾𝑓(𝑀)𝑀−𝑚, (2.9) since ∑^𝑛1_𝑖=1Φ_𝑖(1_𝐻)=𝛼1_𝐾. It follows that 1𝛼^𝑛1_𝑖=1Φ_𝑖𝑓𝐴_𝑖≤𝑀1_𝐾−𝐴𝑓(𝑚)+𝑀−𝑚𝐴−𝑚1_𝐾𝑓(𝑀)𝑀−𝑚. (2.10)
Similarly to (2.10) in the case 𝑚_𝑖1_𝐻≤𝐴_𝑖≤𝑀_𝑖1_𝐻, 𝑖=𝑛₁+1,…,𝑛, it follows from (2.7) 1𝛽^𝑛_𝑖=𝑛1+1Φ_𝑖𝑓𝐴_𝑖≥𝑀1_𝐾−𝐵𝑓(𝑚)+𝑀−𝑚𝐵−𝑚1_𝐾𝑓(𝑀)𝑀−𝑚. (2.11) Combining (2.10) and (2.11) and taking into account that 𝐴=𝐵, we obtain 1𝛼^𝑛1_𝑖=1Φ_𝑖𝑓𝐴_𝑖≤1𝛽^𝑛_𝑖=𝑛1+1Φ_𝑖𝑓𝐴_𝑖. (2.12) It follows that 1𝛼^𝑛1_𝑖=1Φ_𝑖𝑓𝐴_𝑖=^𝑛1_𝑖=1Φ_𝑖𝑓𝐴_𝑖+𝛽𝛼^𝑛1_𝑖=1Φ_𝑖𝑓𝐴_𝑖≤by𝛼+𝛽=1_𝑛1𝑖=1Φ_𝑖𝑓𝐴_𝑖+^𝑛_𝑖=𝑛1+1Φ_𝑖𝑓𝐴_𝑖()=by2.12_𝑛𝑖=1Φ_𝑖𝑓𝐴_𝑖=^𝑛1_𝑖=1Φ_𝑖𝑓𝐴_𝑖+^𝑛_𝑖=𝑛1+1Φ_𝑖𝑓𝐴_𝑖≤𝛼𝛽^𝑛_𝑖=𝑛1+1Φ_𝑖𝑓𝐴_𝑖+^𝑛_𝑖=𝑛1+1Φ_𝑖𝑓𝐴_𝑖()=1by2.12𝛽^𝑛_𝑖=𝑛1+1Φ_𝑖𝑓𝐴_𝑖by𝛼+𝛽=1, (2.13) which gives the desired double inequality (2.4).
(b) Let 𝑚=𝑀. Since [𝑚_𝑖,𝑀_𝑖]⊆[𝑚,𝑀] for 𝑖=1,…,𝑛₁, then 𝐴_𝑖=𝑚1_𝐻 and 𝑓(𝐴_𝑖)=𝑓(𝑚)1_𝐻 for 𝑖=1,…,𝑛₁. It follows that 1𝛼^𝑛1_𝑖=1Φ_𝑖𝐴_𝑖=𝑚1_𝐾,1𝛼^𝑛1_𝑖=1Φ_𝑖𝑓𝐴_𝑖=𝑓(𝑚)1_𝐾. (2.14) On the other hand, since 𝑓 is convex on 𝐼, we have 𝑓(𝑡)≥𝑓(𝑚)+𝑙(𝑚)(𝑡−𝑚)forevery𝑡∈𝐼, (2.15) where 𝑙 is the subdifferential of 𝑓. Replacing 𝑡 by 𝐴_𝑖 for 𝑖=𝑛₁+1,…,𝑛, applying Φ_𝑖 and summing, we obtain from (2.15) and (2.14) that 1𝛽^𝑛_𝑖=𝑛1+1Φ_𝑖𝑓𝐴_𝑖≥𝑓(𝑚)1_𝐾(𝑚)1+𝑙𝛽^𝑛_𝑖=𝑛1+1Φ_𝑖𝐴_𝑖−𝑚1_𝐾=𝑓(𝑚)1_𝐾=1𝛼^𝑛1_𝑖=1Φ_𝑖𝑓𝐴_𝑖. (2.16) So (2.12) holds again. The remaining part of the proof is the same as in the case (a).

As a special case of Theorem 2.1 we can obtain Theorem A. We give this proof as follows.

Another Proof of Theorem A. Let the assumptions of Theorem A be valid. We prove only the case when 𝑓 is a convex function.
We define operators 𝐵_𝑖∈𝐵(𝐻), 𝑖=1,…,𝑛+1, by 𝐵₁=𝐶=∑^𝑛_𝑖=1Φ_𝑖(𝐴_𝑖) and 𝐵_𝑖=𝐴_𝑖−1, 𝑖=2,…,𝑛+1. Then 𝑚_𝐵1=𝑚_𝐶, and 𝑀_𝐵1=𝑀_𝐶 are the bounds of 𝐵₁ and 𝑚_𝐵𝑖=𝑚_𝑖−1, and 𝑀_𝐵𝑖=𝑀_𝑖−1 are the ones of 𝐵_𝑖, 𝑖=2,…,𝑛+1. We have 𝑚_𝐵1,𝑀_𝐵1∩𝑚_𝐵𝑖,𝑀_𝐵𝑖=∅for𝑖=2,…,𝑛+1, (2.17) since (1.3) holds. Also, we define mappings Ψ_𝑖∶𝐵(𝐻)→𝐵(𝐾) by Ψ₁(𝐵)=(1/2)𝐵 and Ψ_𝑖(𝐵)=(1/2)Φ_𝑖−1(𝐵), 𝑖=2,…,𝑛+1. Then we have ∑^𝑛+1_𝑖=1Ψ_𝑖(1_𝐻)=1_𝐾 and ^𝑛+1_𝑖=1Ψ_𝑖𝐵_𝑖=Ψ₁𝐵₁+^𝑛+1_𝑖=2Ψ_𝑖𝐵_𝑖=12^𝑛_𝑖=1Φ_𝑖𝐴_𝑖+12^𝑛_𝑖=1Φ_𝑖𝐴_𝑖=𝐵₁. (2.18) It follows that 2Ψ₁𝐵₁=^𝑛+1_𝑖=1Ψ_𝑖𝐵_𝑖=2^𝑛+1_𝑖=2Ψ_𝑖𝐵_𝑖. (2.19) Taking into account (2.17) and (2.19), we can apply Theorem 2.1 for 𝑛₁=1 and 𝐵_𝑖, Ψ_𝑖 as above. We get 2Ψ₁𝑓𝐵₁≤^𝑛+1_𝑖=1Ψ_𝑖𝑓𝐵_𝑖≤2^𝑛+1_𝑖=2Ψ_𝑖𝑓𝐵_𝑖, (2.20) that is, 𝑓^𝑛_𝑖=1Φ_𝑖𝐴_𝑖≤12𝑓^𝑛_𝑖=1Φ_𝑖𝐴_𝑖+12^𝑛_𝑖=1Φ_𝑖𝑓𝐴_𝑖≤^𝑛_𝑖=1Φ_𝑖𝑓𝐴_𝑖, (2.21) which gives the desired inequality (1.4).

Remark 2.2. We obtain the equivalent inequality to the one in Theorem 2.1 in the case when ∑^𝑛_𝑖=1Φ_𝑖(1_𝐻)=𝛾1_𝐾, for some positive scalar 𝛾. If 𝛼+𝛽=𝛾 and one of two equalities 1𝛼^𝑛1_𝑖=1Φ_𝑖𝐴_𝑖=1𝛽^𝑛_𝑖=𝑛1+1Φ_𝑖𝐴_𝑖=1𝛾^𝑛_𝑖=1Φ_𝑖𝐴_𝑖, (2.22) is valid, then 1𝛼^𝑛1_𝑖=1Φ_𝑖𝑓𝐴_𝑖≤1𝛾^𝑛_𝑖=1Φ_𝑖𝑓𝐴_𝑖≤1𝛽^𝑛_𝑖=𝑛1+1Φ_𝑖𝑓𝐴_𝑖 (2.23) holds for every continuous convex function 𝑓.

Remark 2.3. Let the assumptions of Theorem 2.1 be valid.
(1) We observe that the following inequality 𝑓1𝛽^𝑛_𝑖=𝑛1+1Φ_𝑖𝐴_𝑖≤1𝛽^𝑛_𝑖=𝑛1+1Φ_𝑖𝑓𝐴_𝑖, (2.24) holds for every continuous convex function 𝑓∶𝐼→ℝ.
Indeed, by the assumptions of Theorem 2.1 we have 𝑚𝛼1_𝐻≤^𝑛1_𝑖=1Φ_𝑖𝑓𝐴_𝑖≤𝑀𝛼1_𝐻,1𝛼^𝑛1_𝑖=1Φ_𝑖𝐴_𝑖=1𝛽^𝑛_𝑖=𝑛1+1Φ_𝑖𝐴_𝑖, (2.25) which implies that 𝑚1_𝐻≤^𝑛_𝑖=𝑛1+11𝛽Φ_𝑖𝑓𝐴_𝑖≤𝑀1_𝐻. (2.26) Also (𝑚,𝑀)∩[𝑚_𝑖,𝑀_𝑖]=∅ for 𝑖=𝑛₁+1,…,𝑛 and ∑^𝑛_𝑖=𝑛1+1(1/𝛽)Φ_𝑖(1_𝐻)=1_𝐾 hold. So we can apply Theorem A on operators 𝐴_𝑛1+1,…,𝐴_𝑛 and mappings (1/𝛽)Φ_𝑖 and obtain the desired inequality.
(2) We denote by 𝑚_𝐶 and 𝑀_𝐶 the bounds of 𝐶=∑^𝑛_𝑖=1Φ_𝑖(𝐴_𝑖). If (𝑚_𝐶,𝑀_𝐶)∩[𝑚_𝑖,𝑀_𝑖]=∅, 𝑖=1,…,𝑛₁ or 𝑓 is an operator convex function on [𝑚,𝑀], then the double inequality (2.4) can be extended from the left side if we use Jensen's operator inequality (see [3, Theorem 2.1]): 𝑓^𝑛_𝑖=1Φ_𝑖𝐴_𝑖1=𝑓𝛼^𝑛1_𝑖=1Φ_𝑖𝐴_𝑖≤1𝛼^𝑛1_𝑖=1Φ_𝑖𝑓𝐴_𝑖≤^𝑛_𝑖=1Φ_𝑖𝑓𝐴_𝑖≤1𝛽^𝑛_𝑖=𝑛1+1Φ_𝑖𝑓𝐴_𝑖. (2.27)

Example 2.4. If neither assumptions that(𝑚_𝐶,𝑀_𝐶)∩[𝑚_𝑖,𝑀_𝑖]=∅, 𝑖=1,…,𝑛₁ nor 𝑓 is operator convex in Remark 2.3(2). is satisfied and if 1<𝑛₁<𝑛, then (2.4) cannot be extended by Jensen's operator inequality, since it is not valid. Indeed, for 𝑛₁=2 we define mappings Φ₁,Φ₂∶𝑀₃(ℂ)→𝑀₂(ℂ) by Φ₁((𝑎_𝑖𝑗)_{1≤𝑖,𝑗≤3})=(𝛼/2)(𝑎_𝑖𝑗)_{1≤𝑖,𝑗≤2}, Φ₂=Φ₁. Then Φ₁(𝐼₃)+Φ₂(𝐼₃)=𝛼𝐼₂. If 𝐴₁=2101001111,𝐴₂=2100000000, (2.28) then 1𝛼Φ₁(𝐴₁1)+𝛼Φ₂(𝐴₂)⁴=1𝛼⁴≰160100𝛼=804014024𝛼Φ₁𝐴⁴₁+1𝛼Φ₂𝐴⁴₂ (2.29) for every 𝛼∈(0,1). We observe that 𝑓(𝑡)=𝑡⁴ is not operator convex and (𝑚_𝐶,𝑀_𝐶)∩[𝑚_𝑖,𝑀_𝑖]≠∅, since 𝐶=𝐴=(1/𝛼)Φ₁(𝐴₁)+(1/𝛼)Φ₂(𝐴₂)=(1/𝛼)²⁰₀₀, [𝑚_𝐶,𝑀_𝐶]=[0,2/𝛼], [𝑚₁,𝑀₁]⊂[−1.60388,4.49396], and [𝑚₂,𝑀₂]=[0,2].

With respect to Remark 2.2, we obtain the following obvious corollary of Theorem 2.1.

Corollary 2.5. Let (𝐴₁,…,𝐴_𝑛) be an 𝑛-tuple of self-adjoint operators 𝐴_𝑖∈𝐵(𝐻) with the bounds 𝑚_𝑖 and 𝑀_𝑖, 𝑚_𝑖≤𝑀_𝑖, 𝑖=1,…,𝑛. For some 1≤𝑛₁<𝑛, one denotes 𝑚=min{𝑚₁,…,𝑚_𝑛1}, 𝑀=max{𝑀₁,…,𝑀_𝑛1}. Let (𝑝₁,…,𝑝_𝑛) be an 𝑛-tuple of nonnegative numbers, such that 0<∑^𝑛1_𝑖=1𝑝_𝑖=𝐩_𝐧𝟏<𝐩_𝐧=∑^𝑛_𝑖=1𝑝_𝑖. If (𝑚,𝑀)∩𝑚_𝑖,𝑀_𝑖=∅for𝑖=𝑛₁+1,…,𝑛, (2.30) and one of two equalities 1𝐩_𝐧𝟏^𝑛1_𝑖=1𝑝_𝑖𝐴_𝑖=1𝐩_𝐧^𝑛_𝑖=1𝑝_𝑖𝐴_𝑖=1𝐩_𝐧−𝐩_𝐧𝟏^𝑛_𝑖=𝑛1+1𝑝_𝑖𝐴_𝑖 (2.31) is valid, then 1𝐩_𝐧𝟏^𝑛1_𝑖=1𝑝_𝑖𝑓𝐴_𝑖≤1𝐩_𝐧^𝑛_𝑖=1𝑝_𝑖𝑓𝐴_𝑖≤1𝐩_𝐧−𝐩_𝐧𝟏^𝑛_𝑖=𝑛1+1𝑝_𝑖𝑓𝐴_𝑖 (2.32) holds for every continuous convex function 𝑓∶𝐼→ℝ provided that the interval 𝐼 contains all 𝑚_𝑖,𝑀_𝑖, 𝑖=1,…,𝑛.
If 𝑓∶𝐼→ℝ is concave, then the reverse inequality is valid in (2.32).

3. Quasiarithmetic Means

In this section we study an application of Theorem 2.1 to the quasiarithmetic mean with weight. For a subset {𝐴_𝑝1,…,𝐴_𝑝2} of {𝐴₁,…,𝐴_𝑛} one denotes the quasiarithmetic mean by ℳ_𝜑𝛾,𝐀,𝚽,𝑝₁,𝑝₂=𝜑⁻¹1𝛾^𝑝2_𝑖=𝑝1Φ_𝑖𝜑𝐴_𝑖, (3.1) where (𝐴_𝑝1,…,𝐴_𝑝2) are self-adjoint operators in ℬ(𝐻) with the spectra in 𝐼, (Φ_𝑝1,…,Φ_𝑝2) are positive linear mappings Φ_𝑖∶ℬ(𝐻)→ℬ(𝐾) such that ∑^𝑝2_𝑖=𝑝1Φ_𝑖(1_𝐻)=𝛾1_𝐾, and 𝜑∶𝐼→ℝ is a continuous strictly monotone function.

The following theorem is an extension of Theorem B.

Theorem 3.1. Let (𝐴₁,…,𝐴_𝑛) be an 𝑛-tuple of self-adjoint operators in ℬ(𝐻) with the spectra in 𝐼, and let (Φ₁,…,Φ_𝑛) be an 𝑛-tuple of positive linear mappings Φ_𝑖∶ℬ(𝐻)→ℬ(𝐾) such that ∑^𝑛_𝑖=1Φ_𝑖(1_𝐻)=1_𝐾. Let 𝑚_𝑖 and 𝑀_𝑖, 𝑚_𝑖≤𝑀_𝑖 be the bounds of 𝐴_𝑖, 𝑖=1,…,𝑛. Let 𝜑,𝜓∶𝐼→ℝ be continuous strictly monotone functions on an interval 𝐼 which contains all 𝑚_𝑖,𝑀_𝑖. For 1≤𝑛₁<𝑛, one denotes 𝑚=min{𝑚₁,…,𝑚_𝑛1}, 𝑀=max{𝑀₁,…,𝑀_𝑛1}, and ∑^𝑛1_𝑖=1Φ_𝑖(1_𝐻)=𝛼1_𝐾, ∑^𝑛_𝑖=𝑛1+1Φ_𝑖(1_𝐻)=𝛽1_𝐾, where 𝛼,𝛽>0, 𝛼+𝛽=1. Let (𝑚,𝑀)∩𝑚_𝑖,𝑀_𝑖=∅,for𝑖=𝑛₁+1,…,𝑛, (3.2) and let one of two equalities ℳ_𝜑𝛼,𝐀,𝚽,1,𝑛₁=ℳ_𝜑(1,𝐀,𝚽,1,𝑛)=ℳ_𝜑𝛽,𝐀,𝚽,𝑛₁+1,𝑛 (3.3) be valid.
If one of the following conditions(i)𝜓∘𝜑⁻¹ is convex, and 𝜓⁻¹ is operator monotone,(i')𝜓∘𝜑⁻¹ is concave and −𝜓⁻¹ is operator monotone,
is satisfied, then ℳ_𝜓𝛼,𝐀,𝚽,1,𝑛₁≤ℳ_𝜓(1,𝐀,𝚽,1,𝑛)≤ℳ_𝜓𝛽,𝐀,𝚽,𝑛₁+1,𝑛. (3.4)
If one of the following conditions(i)𝜓∘𝜑⁻¹ is concave and 𝜓⁻¹ is operator monotone, (ii')𝜓∘𝜑⁻¹ is convex and −𝜓⁻¹ is operator monotone,
is satisfied, then the reverse inequality is valid in (3.4).

Proof. We only prove the case (i). Suppose that 𝜑 is a strictly increasing function. Since 𝑚1_𝐻≤𝐴_𝑖≤𝑀1_𝐻, 𝑖=1,…,𝑛₁, implies 𝜑(𝑚)1_𝐾≤𝜑(𝐴_𝑖)≤𝜑(𝑀)1_𝐾, then (𝑚,𝑀)∩𝑚_𝑖,𝑀_𝑖=∅for𝑖=𝑛₁+1,…,𝑛 (3.5) implies 𝜑(𝑚),𝜑(𝑀)∩𝜑𝑚_𝑖,𝜑𝑀_𝑖=∅,for𝑖=𝑛₁+1,…,𝑛. (3.6) Also, by using (3.3), we have 1𝛼^𝑛1_𝑖=1Φ_𝑖𝜑𝐴_𝑖=^𝑛_𝑖=1Φ_𝑖𝜑𝐴_𝑖=1𝛽^𝑛_𝑖=𝑛1+1Φ_𝑖𝜑𝐴_𝑖. (3.7) Taking into account (3.6) and the above double equality, we obtain by Theorem 2.11𝛼^𝑛1_𝑖=1Φ_𝑖𝑓𝜑𝐴_𝑖≤^𝑛_𝑖=1Φ_𝑖𝑓𝜑𝐴_𝑖≤1𝛽^𝑛_𝑖=𝑛1+1Φ_𝑖𝑓𝜑𝐴_𝑖 (3.8) for every continuous convex function 𝑓∶𝐽→ℝ on an interval 𝐽 which contains all [𝜑(𝑚_𝑖),𝜑(𝑀_𝑖)]=𝜑([𝑚_𝑖,𝑀_𝑖]), 𝑖=1,…,𝑛.
Also, if 𝜑 is strictly decreasing, then we check that (3.8) holds for convex 𝑓∶𝐽→ℝ on 𝐽 which contains all [𝜑(𝑀_𝑖),𝜑(𝑚_𝑖)]=𝜑([𝑚_𝑖,𝑀_𝑖]).
Putting 𝑓=𝜓∘𝜑⁻¹ in (3.8), we obtain 1𝛼^𝑛1_𝑖=1Φ_𝑖𝜓𝐴_𝑖≤^𝑛_𝑖=1Φ_𝑖𝜓𝐴_𝑖≤1𝛽^𝑛_𝑖=𝑛1+1Φ_𝑖𝜓𝐴_𝑖. (3.9) Applying an operator monotone function 𝜓⁻¹ on the above double inequality, we obtain the desired inequality (3.4).

As a special case of Theorem 3.1 we can obtain Theorem B as follows.

Another Proof of Theorem B. We give the short version of the proof, since it is essentially the same as the one of Theorem A in Section 2.
Let the assumptions of Theorem B be valid, 𝜓∘𝜑⁻¹ is convex and 𝜓⁻¹ is operator monotone.
Let 𝐵₁=𝜑⁻¹(∑^𝑛_𝑖=1Φ_𝑖(𝜑(𝐴_𝑖))) and 𝐵_𝑖=𝐴_𝑖−1, 𝑖=2,…,𝑛+1. Then 𝑚_𝐵1=𝑚_𝜑, and 𝑀_𝐵1=𝑀_𝜑 are the bounds of 𝐵₁, and 𝑚_𝐵𝑖=𝑚_𝑖−1, and 𝑀_𝐵𝑖=𝑀_𝑖−1, are the ones of 𝐵_𝑖, 𝑖=2,…,𝑛+1. We have 𝑚_𝐵1,𝑀_𝐵1∩𝑚_𝐵𝑖,𝑀_𝐵𝑖=∅for𝑖=2,…,𝑛+1, (3.10) since (1.6) holds. Also, we define mappings Θ₁(𝐵)=(1/2)𝐵 and Θ_𝑖(𝐵)=(1/2)Φ_𝑖−1(𝐵), 𝑖=2,…,𝑛+1. Then we have ∑^𝑛+1_𝑖=1Θ_𝑖(1_𝐻)=1_𝐾 and ^𝑛+1_𝑖=1Θ_𝑖𝜑𝐵_𝑖=12^𝑛_𝑖=1Φ_𝑖𝜑𝐴_𝑖+12^𝑛_𝑖=1Φ_𝑖𝜑𝐴_𝑖=𝜑𝐵₁. (3.11) It follows that 𝐵₁=ℳ_𝜑12,𝐁,𝚯,1,1=ℳ_𝜑(1,𝐁,𝚯,1,𝑛+1)=ℳ_𝜑12,𝐁,𝚯,2,𝑛+1. (3.12) So the assumptions of Theorem 3.1 are valid and it follows that 𝐵₁=ℳ_𝜓12,𝐁,𝚯,1,1≤ℳ_𝜓(1,𝐁,𝚯,1,𝑛+1)≤ℳ_𝜓12,𝐁,𝚯,2,𝑛+1 (3.13) holds. Therefore, it follows that 𝜑^−1𝑛_𝑖=1Φ_𝑖𝜑𝐴_𝑖=𝐵₁≤𝜓^−1𝑛_𝑖=1Φ_𝑖𝜓𝐴_𝑖, (3.14) which is the desired inequality (1.7).
In the remaining cases the proof is essentially the same as in the above cases.

Remark 3.2. Let the assumptions of Theorem 3.1 be valid.
(1) We observe that if one of the following conditions (i)𝜓∘𝜑⁻¹ is convex and 𝜓⁻¹ is operator monotone, (i')𝜓∘𝜑⁻¹ is concave and −𝜓⁻¹ is operator monotone,is satisfied, then the following obvious inequality (see Remark 2.3(1)) ℳ_𝜑𝛽,𝐀,𝚽,𝑛₁+1,𝑛≤ℳ_𝜓𝛽,𝐀,𝚽,𝑛₁+1,𝑛 (3.15) holds.
(2) We denote by 𝑚_𝜑 and 𝑀_𝜑 the bounds of ℳ_𝜑(1,𝐀,Φ,1,𝑛). If (𝑚_𝜑,𝑀_𝜑)∩[𝑚_𝑖,𝑀_𝑖]=∅, 𝑖=1,…,𝑛₁, and one of two following conditions (i)𝜓∘𝜑⁻¹ is convex and 𝜓⁻¹ is operator monotone (ii)𝜓∘𝜑⁻¹ is concave and −𝜓⁻¹ is operator monotone is satisfied, or if one of the following conditions (i')𝜓∘𝜑⁻¹ is operator convex and 𝜓⁻¹ is operator monotone, (ii')𝜓∘𝜑⁻¹ is operator concave and −𝜓⁻¹ is operator monotone,is satisfied (see [4, Theorem B]), then the double inequality (3.4) can be extended from the left side as follows: ℳ_𝜑(1,𝐀,𝚽,1,𝑛)=ℳ_𝜑1,𝐀,𝚽,1,𝑛₁≤ℳ_𝜓𝛼,𝐀,𝚽,1,𝑛₁≤ℳ_𝜓(1,𝐀,𝚽,1,𝑛)≤ℳ_𝜓𝛽,𝐀,𝚽,𝑛₁+1,𝑛. (3.16)
(3) If neither assumptions that (𝑚_𝜓,𝑀_𝜓)∩[𝑚_𝑖,𝑀_𝑖]=∅, 𝑖=1,…,𝑛₁ nor 𝜓∘𝜑⁻¹ is operator convex (or operator concave) is satisfied and if 1<𝑛₁<𝑛, then (3.4) cannot be extended from the left side by ℳ_𝜑(1,𝐀,Φ,1,𝑛₁) as above. It is easy to check it with a counterexample similarly to [4, Example??2].

We now give some particular results of interest that can be derived from Theorem 3.1.

Corollary 3.3. Let (𝐴₁,…,𝐴_𝑛) and (Φ₁,…,Φ_𝑛), 𝑚_𝑖, 𝑀_𝑖, 𝑚, 𝑀, 𝛼, and 𝛽 be as in Theorem 3.1. Let 𝐼 be an interval which contains all 𝑚_𝑖,??𝑀_𝑖 and (𝑚,𝑀)∩𝑚_𝑖,𝑀_𝑖=∅,for𝑖=𝑛₁+1,…,𝑛. (3.17) If one of two equalities ℳ_𝜑𝛼,𝐀,𝚽,1,𝑛₁=ℳ_𝜑(1,𝐀,𝚽,1,𝑛)=ℳ_𝜑𝛽,𝐀,𝚽,𝑛₁+1,𝑛 (3.18) is valid, then 1𝛼^𝑛1_𝑖=1Φ_𝑖𝐴_𝑖≤^𝑛_𝑖=1Φ_𝑖𝐴_𝑖≤1𝛽^𝑛_𝑖=𝑛1+1Φ_𝑖𝐴_𝑖 (3.19) holds for every continuous strictly monotone function 𝜑∶𝐼→ℝ such that 𝜑⁻¹ is convex on 𝐼. But, if 𝜑⁻¹ is concave, then the reverse inequality is valid in (3.19).
On the other hand, if one of two equalities 1𝛼^𝑛1_𝑖=1Φ_𝑖𝐴_𝑖=^𝑛_𝑖=1Φ_𝑖𝐴_𝑖=1𝛽^𝑛_𝑖=𝑛1+1Φ_𝑖𝐴_𝑖 (3.20) is valid, then ℳ_𝜑𝛼,𝐀,𝚽,1,𝑛₁≤ℳ_𝜑(1,𝐀,𝚽,1,𝑛)≤ℳ_𝜑𝛽,𝐀,𝚽,𝑛₁+1,𝑛 (3.21) holds for every continuous strictly monotone function 𝜑∶𝐼→ℝ such that one of the following conditions(i)𝜑 is convex and 𝜑⁻¹ is operator monotone, (i')𝜑 is concave and −𝜑⁻¹ is operator monotone,
is satisfied. But, if one of the following conditions (ii)𝜑 is concave and 𝜑⁻¹ is operator monotone, (ii')𝜑 is convex and −𝜑⁻¹ is operator monotone,
is satisfied, then the reverse inequality is valid in (3.21).

Proof. The proof of (3.19) follows from Theorem 3.1 by replacing 𝜓 with the identity function, while the proof of (3.21) follows from the same theorem by replacing 𝜑 with the identity function and 𝜓 with 𝜑.

As a special case of the quasiarithmetic mean (3.1) we can study the weighted power mean as follows. For a subset {𝐴_𝑝1,…,𝐴_𝑝2} of {𝐴₁,…,𝐴_𝑛} one denotes this mean by 𝑀^[𝑟]𝛾,𝐀,𝚽,𝑝₁,𝑝₂=1𝛾^𝑝2_𝑖=𝑝1Φ_𝑖𝐴^𝑟_𝑖^1/𝑟{0},1,𝑟∈ℝ⧵exp𝛾^𝑝2_𝑖=𝑝1Φ_𝑖ln𝐴_𝑖,𝑟=0, (3.22) where (𝐴_𝑝1,…,𝐴_𝑝2) are strictly positive operators, and (Φ_𝑝1,…,Φ_𝑝2) are positive linear mappings Φ_𝑖∶ℬ(𝐻)→ℬ(𝐾) such that ∑^𝑝2_𝑖=𝑝1Φ_𝑖(1_𝐻)=𝛾1_𝐾.

We obtain the following corollary by applying Theorem 3.1 to the above mean.

Corollary 3.4. Let (𝐴₁,…,𝐴_𝑛) be an 𝑛-tuple of strictly positive operators in ℬ(𝐻) and let (Φ₁,…,Φ_𝑛) be an 𝑛-tuple of positive linear mappings Φ_𝑖∶ℬ(𝐻)→ℬ(𝐾) such that ∑^𝑛_𝑖=1Φ_𝑖(1_𝐻)=1_𝐾. Let 𝑚_𝑖 and 𝑀_𝑖, 0<𝑚_𝑖≤𝑀_𝑖, be the bounds of 𝐴_𝑖, 𝑖=1,…,𝑛. For 1≤𝑛₁<𝑛, one denotes 𝑚=min{𝑚₁,…,𝑚_𝑛1}, 𝑀=max{𝑀₁,…,𝑀_𝑛1}, and ∑^𝑛1_𝑖=1Φ_𝑖(1_𝐻)=𝛼1_𝐾, ∑^𝑛_𝑖=𝑛1+1Φ_𝑖(1_𝐻)=𝛽1_𝐾, where 𝛼,𝛽>0, 𝛼+𝛽=1.(i)If either 𝑟≤𝑠, 𝑠≥1 or 𝑟≤𝑠≤−1 and also one of two equalities ℳ^[𝑟]𝛼,𝐀,𝚽,1,𝑛₁=ℳ^[𝑟](1,𝐀,𝚽,1,𝑛)=ℳ^[𝑟]𝛽,𝐀,𝚽,𝑛₁+1,𝑛 (3.23) is valid, then ℳ^[𝑠]𝛼,𝐀,𝚽,1,𝑛₁≤ℳ^[𝑠](1,𝐀,𝚽,1,𝑛)≤ℳ^[𝑠]𝛽,𝐀,𝚽,𝑛₁+1,𝑛 (3.24) holds.(ii)If either 𝑟≤𝑠, 𝑟≤−1 or 1≤𝑟≤𝑠 and also one of two equalities ℳ^[𝑠]𝛼,𝐀,𝚽,1,𝑛₁=ℳ^[𝑠](1,𝐀,𝚽,1,𝑛)=ℳ^[𝑠]𝛽,𝐀,𝚽,𝑛₁+1,𝑛 (3.25) is valid, then ℳ^[𝑟]𝛼,𝐀,𝚽,1,𝑛₁≥ℳ^[𝑟](1,𝐀,𝚽,1,𝑛)≥ℳ^[𝑟]𝛽,𝐀,𝚽,𝑛₁+1,𝑛 (3.26) holds.

Proof. (i) We prove only the case (i). We take 𝜑(𝑡)=𝑡^𝑟 and 𝜓(𝑡)=𝑡^𝑠 for 𝑡>0. Then 𝜓∘𝜑⁻¹(𝑡)=𝑡^𝑠/𝑟 is concave for 𝑟≤𝑠, 𝑠≤0, and 𝑟≠0. Since −𝜓⁻¹(𝑡)=−𝑡^1/𝑠 is operator monotone for 𝑠≤−1 and (3.23) is satisfied, then by applying Theorem 3.1(i') we obtain (3.24) for 𝑟≤𝑠≤−1.
But, 𝜓∘𝜑⁻¹(𝑡)=𝑡^𝑠/𝑟 is convex for 𝑟≤𝑠, 𝑠≥0, and 𝑟≠0. Since 𝜓⁻¹(𝑡)=𝑡^1/𝑠 is operator monotone for 𝑠≥1, then by applying Theorem 3.1(i) we obtain (3.24) for 𝑟≤𝑠, 𝑠≥1, and 𝑟≠0.
If 𝑟=0 and 𝑠≥1, we put 𝜑(𝑡)=ln𝑡 and 𝜓(𝑡)=𝑡^𝑠, 𝑡>0. Since 𝜓∘𝜑⁻¹(𝑡)=exp(𝑠𝑡) is convex, then similarly as above we obtain the desired inequality.
In the case (ii) we put 𝜑(𝑡)=𝑡^𝑠 and 𝜓(𝑡)=𝑡^𝑟 for 𝑡>0 and we use the same technique as in the case (i).