# Extension of Jensen's Inequality for Operators without Operator Convexity

Authors: Jadranka Mićić; Zlatko Pavić; Josip Pečarić
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

We give an extension of Jensen's inequality forn-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.


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## Body

## 1. Introduction

We recall some notations and definitions. Letℬ(H) be the C*-algebra of all bounded linear operators on a Hilbert space H and 1H stands for the identity operator. We define bounds of a self-adjoint operator A∈ℬ(H):
(1.1)mA=inf⁡∥x∥=1〈Ax,x〉,MA=sup⁡∥x∥=1〈Ax,x〉
for x∈H. If Sp(A) denotes the spectrum of A, then Sp(A) is real and Sp(A)⊆[mA,MA].Mond and Pečarić in [1] proved the following version of Jensen's operator inequality
(1.2)f(∑i=1nwiΦi(Ai))≤∑i=1nwiΦi(f(Ai)),
for operator convex functions f defined on an interval I, where Φi:ℬ(H)→ℬ(K), i=1,…,n, are unital positive linear mappings, A1,…,An are self-adjoint operators with the spectra in I, and w1,…,wn are nonnegative real numbers with ∑i=1nwi=1.Hansen et al. gave in [2] a generalization of (1.2) for a unital field of positive linear mappings. Recently, Mićić et al. in [3] gave a generalization of this results for a not-unital field of positive linear mappings.Very recently, Mićić et al. gave in [4, Theorem  1] a version of Jensen's operator inequality without operator convexity as follows.Theorem A.
Let(A1,…,An) be an  n-tuple of self-adjoint operators  Ai∈ℬ(H) with bounds  mi and  Mi,  mi≤Mi,  i=1,…,n. Let  (Φ1,…,Φn) be an  n-tuple of positive linear mappings  Φi:ℬ(H)→ℬ(K),  i=1,…,n, such that  ∑i=1nΦi(1H)=1K. If
(1.3)(mC,MC)∩[mi,Mi]=∅fori=1,…,n,
where mC and MC, mC≤MC, are bounds of the self-adjoint operator C=∑i=1nΦi(Ai), then
(1.4)f(∑i=1nΦi(Ai))≤∑i=1nΦi(f(Ai))
holds for every continuous convex function f:I→ℝ provided that the interval I contains all mi,Mi.
Iff:I→ℝ is concave, then the reverse inequality is valid in (1.4).In the same paper [4], we study the quasiarithmetic operator mean:
(1.5)Mφ(A,Φ,n)=φ-1(∑i=1nΦi(φ(Ai))),
where (A1,…,An) is an n-tuple of self-adjoint operators in ℬ(H) with the spectra in I, (Φ1,…,Φn) is an n-tuple of positive linear mappings Φi:ℬ(H)→ℬ(K) such that ∑i=1nΦi(1H)=1K, 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,…,An) and  (Φ1,…,Φn) be as in the definition of the quasiarithmetic mean (1.5). Let  mi and  Mi,  mi≤Mi, be the bounds of  Ai,  i=1,…,n. Let  φ,ψ:I→ℝ be continuous strictly monotone functions on an interval  I which contains all  mi,Mi. Let  mφ and  Mφ,  mφ≤Mφ, be the bounds of the mean  ℳφ(A,Φ,n), such that
(1.6)(mφ,Mφ)∩[mi,Mi]=∅,fori=1,…,n.
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(1.7)Mφ(A,Φ,n)≤Mψ(A,Φ,n).
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 everyn-tuple of self-adjoint operators (A1,…,An), for every n-tuple of positive linear mappings (Φ1,…,Φn) in the case when the interval with bounds of the operator A=∑i=1nΦi(Ai) has no intersection points with the interval with bounds of the operator Ai for each i=1,…,n, that is,  when
(2.1)(mA,MA)∩[mi,Mi]=∅,fori=1,…,n,
where mA and MA, mA≤MA, are the bounds of A, and mi and Mi, mi≤Mi, are the bounds of Ai, i=1,…,n.It is interesting to consider the case when(mA,MA)∩[mi,Mi]=∅ 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,…,An) be an n-tuple of self-adjoint operators Ai∈B(H) with the bounds mi and Mi, mi≤Mi, i=1,…,n. Let (Φ1,…,Φn) be an n-tuple of positive linear mappings Φi:B(H)→B(K), such that ∑i=1nΦi(1H)=1K. For 1≤n1<n, one denotes m=min{m1,…,mn1}, M=max{M1,…,Mn1}, and ∑i=1n1Φi(1H)=α1K, ∑i=n1+1nΦi(1H)=β1K, where α,β>0, α+β=1. If
(2.2)(m,M)∩[mi,Mi]=∅,fori=n1+1,…,n,
and one of two equalities
(2.3)1α∑i=1n1Φi(Ai)=1β∑i=n1+1nΦi(Ai)=∑i=1nΦi(Ai)
is valid, then
(2.4)1α∑i=1n1Φi(f(Ai))≤∑i=1nΦi(f(Ai))≤1β∑i=n1+1nΦi(f(Ai))
holds for every continuous convex function f:I→ℝ provided that the interval I contains all mi,Mi, i=1,…,n.Iff:I→ℝ is concave, then the reverse inequality is valid in (2.4).Proof.
We prove only the case whenf is a convex function.
Let us denote(2.5)A=1α∑i=1n1Φi(Ai),B=1β∑i=n1+1nΦi(Ai),C=∑i=1nΦi(Ai).
It is easy to verify that A=B or B=C or A=C implies A=B=C.
(a) Letm<M. Since f is convex on [m,M] and [mi,Mi]⊆[m,M] for i=1,…,n1, then
(2.6)f(t)≤M-tM-mf(m)+t-mM-mf(M),t∈[mi,Mi]fori=1,…,n1,
but since f is convex on all [mi,Mi] and (m,M)∩[mi,Mi]=∅ for i=n1+1,…,n, then
(2.7)f(t)≥M-tM-mf(m)+t-mM-mf(M),t∈[mi,Mi]fori=n1+1,…,n.
Sincemi1H≤Ai≤Mi1H, i=1,…,n1, it follows from (2.6) that
(2.8)f(Ai)≤M1H-AiM-mf(m)+Ai-m1HM-mf(M),i=1,…,n1.
Applying a positive linear mapping Φi and summing, we obtain
(2.9)∑i=1n1Φi(f(Ai))≤Mα1K-∑i=1n1Φi(Ai)M-mf(m)+∑i=1n1Φi(Ai)-mα1KM-mf(M),
since ∑i=1n1Φi(1H)=α1K. It follows that
(2.10)1α∑i=1n1Φi(f(Ai))≤M1K-AM-mf(m)+A-m1KM-mf(M).
Similarly to (2.10) in the case mi1H≤Ai≤Mi1H, i=n1+1,…,n, it follows from (2.7)
(2.11)1β∑i=n1+1nΦi(f(Ai))≥M1K-BM-mf(m)+B-m1KM-mf(M).
Combining (2.10) and (2.11) and taking into account that A=B, we obtain
(2.12)1α∑i=1n1Φi(f(Ai))≤1β∑i=n1+1nΦi(f(Ai)).
It follows that
(2.13)1α∑i=1n1Φi(f(Ai))=∑i=1n1Φi(f(Ai))+βα∑i=1n1Φi(f(Ai))(byα+β=1)≤∑i=1n1Φi(f(Ai))+∑i=n1+1nΦi(f(Ai))(by(2.12))=∑i=1nΦi(f(Ai))=∑i=1n1Φi(f(Ai))+∑i=n1+1nΦi(f(Ai))≤αβ∑i=n1+1nΦi(f(Ai))+∑i=n1+1nΦi(f(Ai))(by(2.12))=1β∑i=n1+1nΦi(f(Ai))(byα+β=1),
which gives the desired double inequality (2.4).
(b) Letm=M. Since [mi,Mi]⊆[m,M] for i=1,…,n1, then Ai=m1H and f(Ai)=f(m)1H for i=1,…,n1. It follows that
(2.14)1α∑i=1n1Φi(Ai)=m1K,1α∑i=1n1Φi(f(Ai))=f(m)1K.
On the other hand, since f is convex on I, we have
(2.15)f(t)≥f(m)+l(m)(t-m)foreveryt∈I,
where l is the subdifferential of f. Replacing t by Ai for i=n1+1,…,n, applying Φi and summing, we obtain from (2.15) and (2.14) that
(2.16)1β∑i=n1+1nΦi(f(Ai))≥f(m)1K+l(m)(1β∑i=n1+1nΦi(Ai)-m1K)=f(m)1K=1α∑i=1n1Φi(f(Ai)).
So (2.12) holds again. The remaining part of the proof is the same as in the case (a).As a special case of Theorem2.1 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 whenf is a convex function.
We define operatorsBi∈B(H), i=1,…,n+1, by B1=C=∑i=1nΦi(Ai) and Bi=Ai-1, i=2,…,n+1. Then mB1=mC, and MB1=MC are the bounds of B1 and mBi=mi-1, and MBi=Mi-1 are the ones of Bi, i=2,…,n+1. We have
(2.17)(mB1,MB1)∩[mBi,MBi]=∅fori=2,…,n+1,
since (1.3) holds. Also, we define mappings Ψi:B(H)→B(K) by Ψ1(B)=(1/2)B and Ψi(B)=(1/2)Φi-1(B), i=2,…,n+1. Then we have ∑i=1n+1Ψi(1H)=1K and
(2.18)∑i=1n+1Ψi(Bi)=Ψ1(B1)+∑i=2n+1Ψi(Bi)=12∑i=1nΦi(Ai)+12∑i=1nΦi(Ai)=B1.
It follows that
(2.19)2Ψ1(B1)=∑i=1n+1Ψi(Bi)=2∑i=2n+1Ψi(Bi).
Taking into account (2.17) and (2.19), we can apply Theorem 2.1 for n1=1 and Bi, Ψi as above. We get
(2.20)2Ψ1(f(B1))≤∑i=1n+1Ψi(f(Bi))≤2∑i=2n+1Ψi(f(Bi)),
that is,
(2.21)f(∑i=1nΦi(Ai))≤12f(∑i=1nΦi(Ai))+12∑i=1nΦi(f(Ai))≤∑i=1nΦi(f(Ai)),
which gives the desired inequality (1.4).Remark 2.2.
We obtain the equivalent inequality to the one in Theorem2.1 in the case when ∑i=1nΦi(1H)=γ1K, for some positive scalar γ. If α+β=γ and one of two equalities
(2.22)1α∑i=1n1Φi(Ai)=1β∑i=n1+1nΦi(Ai)=1γ∑i=1nΦi(Ai),
is valid, then
(2.23)1α∑i=1n1Φi(f(Ai))≤1γ∑i=1nΦi(f(Ai))≤1β∑i=n1+1nΦi(f(Ai))
holds for every continuous convex function f.Remark 2.3.
Let the assumptions of Theorem2.1 be valid.
(1) We observe that the following inequality(2.24)f(1β∑i=n1+1nΦi(Ai))≤1β∑i=n1+1nΦi(f(Ai)),
holds for every continuous convex function f:I→ℝ.
Indeed, by the assumptions of Theorem2.1 we have
(2.25)mα1H≤∑i=1n1Φi(f(Ai))≤Mα1H,1α∑i=1n1Φi(Ai)=1β∑i=n1+1nΦi(Ai),
which implies that
(2.26)m1H≤∑i=n1+1n1βΦi(f(Ai))≤M1H.
Also (m,M)∩[mi,Mi]=∅ for i=n1+1,…,n and ∑i=n1+1n(1/β)Φi(1H)=1K hold. So we can apply Theorem A on operators An1+1,…,An and mappings (1/β)Φi and obtain the desired inequality.
(2) We denote bymC and MC the bounds of C=∑i=1nΦi(Ai). If (mC,MC)∩[mi,Mi]=∅, 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]):
(2.27)f(∑i=1nΦi(Ai))=f(1α∑i=1n1Φi(Ai))≤1α∑i=1n1Φi(f(Ai))≤∑i=1nΦi(f(Ai))≤1β∑i=n1+1nΦi(f(Ai)).Example 2.4.
If neither assumptions that(mC,MC)∩[mi,Mi]=∅, 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((aij)1≤i,j≤3)=(α/2)(aij)1≤i,j≤2, Φ2=Φ1. Then Φ1(I3)+Φ2(I3)=αI2. If
(2.28)A1=2(101001111),A2=2(100000000),
then
(2.29)(1αΦ1(A1)+1αΦ2(A2))4=1α4(16000)≰1α(80404024)=1αΦ1(A14)+1αΦ2(A24)
for every α∈(0,1). We observe that f(t)=t4 is not operator convex and (mC,MC)∩[mi,Mi]≠∅, since C=A=(1/α)Φ1(A1)+(1/α)Φ2(A2)=(1/α)(2000), [mC,MC]=[0,2/α], [m1,M1]⊂[-1.60388,4.49396], and [m2,M2]=[0,2].With respect to Remark2.2, we obtain the following obvious corollary of Theorem 2.1.Corollary 2.5.
Let(A1,…,An) be an n-tuple of self-adjoint operators Ai∈B(H) with the bounds mi and Mi, mi≤Mi, i=1,…,n. For some 1≤n1<n, one denotes m=min{m1,…,mn1}, M=max{M1,…,Mn1}. Let (p1,…,pn) be an n-tuple of nonnegative numbers, such that 0<∑i=1n1pi=pn1<pn=∑i=1npi. If
(2.30)(m,M)∩[mi,Mi]=∅fori=n1+1,…,n,
and one of two equalities
(2.31)1pn1∑i=1n1piAi=1pn∑i=1npiAi=1pn-pn1∑i=n1+1npiAi
is valid, then
(2.32)1pn1∑i=1n1pif(Ai)≤1pn∑i=1npif(Ai)≤1pn-pn1∑i=n1+1npif(Ai)
holds for every continuous convex function f:I→ℝ provided that the interval I contains all mi,Mi, i=1,…,n.
Iff:I→ℝ is concave, then the reverse inequality is valid in (2.32).


## 3. Quasiarithmetic Means

In this section we study an application of Theorem2.1 to the quasiarithmetic mean with weight. For a subset {Ap1,…,Ap2} of {A1,…,An} one denotes the quasiarithmetic mean by
(3.1)Mφ(γ,A,Φ,p1,p2)=φ-1(1γ∑i=p1p2Φi(φ(Ai))),
where (Ap1,…,Ap2) are self-adjoint operators in ℬ(H) with the spectra in I, (Φp1,…,Φp2) are positive linear mappings Φi:ℬ(H)→ℬ(K) such that ∑i=p1p2Φi(1H)=γ1K, and φ:I→ℝ is a continuous strictly monotone function.The following theorem is an extension of Theorem B.Theorem 3.1.
Let(A1,…,An) be an n-tuple of self-adjoint operators in ℬ(H) with the spectra in I, and let (Φ1,…,Φn) be an n-tuple of positive linear mappings Φi:ℬ(H)→ℬ(K) such that ∑i=1nΦi(1H)=1K. Let mi and Mi, mi≤Mi be the bounds of Ai, i=1,…,n. Let φ,ψ:I→ℝ be continuous strictly monotone functions on an interval I which contains all mi,Mi. For 1≤n1<n, one denotes m=min{m1,…,mn1}, M=max{M1,…,Mn1}, and ∑i=1n1Φi(1H)=α1K, ∑i=n1+1nΦi(1H)=β1K, where α,β>0, α+β=1. Let
(3.2)(m,M)∩[mi,Mi]=∅,fori=n1+1,…,n,
and let one of two equalities
(3.3)Mφ(α,A,Φ,1,n1)=Mφ(1,A,Φ,1,n)=Mφ(β,A,Φ,n1+1,n)
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(3.4)Mψ(α,A,Φ,1,n1)≤Mψ(1,A,Φ,1,n)≤Mψ(β,A,Φ,n1+1,n).
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 thatφ is a strictly increasing function. Since m1H≤Ai≤M1H, i=1,…,n1, implies φ(m)1K≤φ(Ai)≤φ(M)1K, then
(3.5)(m,M)∩[mi,Mi]=∅fori=n1+1,…,n
implies
(3.6)(φ(m),φ(M))∩[φ(mi),φ(Mi)]=∅,fori=n1+1,…,n.
Also, by using (3.3), we have
(3.7)1α∑i=1n1Φi(φ(Ai))=∑i=1nΦi(φ(Ai))=1β∑i=n1+1nΦi(φ(Ai)).
Taking into account (3.6) and the above double equality, we obtain by Theorem 2.1(3.8)1α∑i=1n1Φi(f(φ(Ai)))≤∑i=1nΦi(f(φ(Ai)))≤1β∑i=n1+1nΦi(f(φ(Ai)))
for every continuous convex function f:J→ℝ on an interval J which contains all [φ(mi),φ(Mi)]=φ([mi,Mi]), i=1,…,n.
Also, ifφ is strictly decreasing, then we check that (3.8) holds for convex f:J→ℝ on J which contains all [φ(Mi),φ(mi)]=φ([mi,Mi]).
Puttingf=ψ∘φ-1 in (3.8), we obtain
(3.9)1α∑i=1n1Φi(ψ(Ai))≤∑i=1nΦi(ψ(Ai))≤1β∑i=n1+1nΦi(ψ(Ai)).
Applying an operator monotone function ψ-1 on the above double inequality, we obtain the desired inequality (3.4).As a special case of Theorem3.1 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 Section2.
Let the assumptions of Theorem B be valid,ψ∘φ-1 is convex and ψ-1 is operator monotone.
LetB1=φ-1(∑i=1nΦi(φ(Ai))) and Bi=Ai-1, i=2,…,n+1. Then mB1=mφ, and MB1=Mφ are the bounds of B1, and mBi=mi-1, and MBi=Mi-1, are the ones of Bi, i=2,…,n+1. We have
(3.10)(mB1,MB1)∩[mBi,MBi]=∅fori=2,…,n+1,
since (1.6) holds. Also, we define mappings Θ1(B)=(1/2)B and Θi(B)=(1/2)Φi-1(B), i=2,…,n+1. Then we have ∑i=1n+1Θi(1H)=1K and
(3.11)∑i=1n+1Θi(φ(Bi))=12∑i=1nΦi(φ(Ai))+12∑i=1nΦi(φ(Ai))=φ(B1).
It follows that
(3.12)B1=Mφ(12,B,Θ,1,1)=Mφ(1,B,Θ,1,n+1)=Mφ(12,B,Θ,2,n+1).
So the assumptions of Theorem 3.1 are valid and it follows that
(3.13)B1=Mψ(12,B,Θ,1,1)≤Mψ(1,B,Θ,1,n+1)≤Mψ(12,B,Θ,2,n+1)
holds. Therefore, it follows that
(3.14)φ-1(∑i=1nΦi(φ(Ai)))=B1≤ψ-1(∑i=1nΦi(ψ(Ai))),
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 Theorem3.1 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))
(3.15)Mφ(β,A,Φ,n1+1,n)≤Mψ(β,A,Φ,n1+1,n)
holds.
(2) We denote bymφ and Mφ the bounds of ℳφ(1,A,Φ,1,n). If (mφ,Mφ)∩[mi,Mi]=∅, 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 monotoneis 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:
(3.16)Mφ(1,A,Φ,1,n)=Mφ(1,A,Φ,1,n1)≤Mψ(α,A,Φ,1,n1)≤Mψ(1,A,Φ,1,n)≤Mψ(β,A,Φ,n1+1,n).
(3) If neither assumptions that(mψ,Mψ)∩[mi,Mi]=∅, 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 ℳφ(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 Theorem3.1.Corollary 3.3.
Let(A1,…,An) and (Φ1,…,Φn), mi, Mi, m, M, α, and β be as in Theorem 3.1. Let I be an interval which contains all mi,  Mi and
(3.17)(m,M)∩[mi,Mi]=∅,fori=n1+1,…,n.
If one of two equalities
(3.18)Mφ(α,A,Φ,1,n1)=Mφ(1,A,Φ,1,n)=Mφ(β,A,Φ,n1+1,n)
is valid, then
(3.19)1α∑i=1n1Φi(Ai)≤∑i=1nΦi(Ai)≤1β∑i=n1+1nΦi(Ai)
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(3.20)1α∑i=1n1Φi(Ai)=∑i=1nΦi(Ai)=1β∑i=n1+1nΦi(Ai)
is valid, then
(3.21)Mφ(α,A,Φ,1,n1)≤Mφ(1,A,Φ,1,n)≤Mφ(β,A,Φ,n1+1,n)
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 {Ap1,…,Ap2} of {A1,…,An} one denotes this mean by
(3.22)M[r](γ,A,Φ,p1,p2)={(1γ∑i=p1p2Φi(Air))1/r,r∈R∖{0},exp⁡(1γ∑i=p1p2Φi(ln⁡(Ai))),r=0,
where (Ap1,…,Ap2) are strictly positive operators, and (Φp1,…,Φp2) are positive linear mappings Φi:ℬ(H)→ℬ(K) such that ∑i=p1p2Φi(1H)=γ1K.We obtain the following corollary by applying Theorem3.1 to the above mean.Corollary 3.4.
Let(A1,…,An) be an n-tuple of strictly positive operators in ℬ(H) and let (Φ1,…,Φn) be an n-tuple of positive linear mappings Φi:ℬ(H)→ℬ(K) such that ∑i=1nΦi(1H)=1K. Let mi and Mi, 0<mi≤Mi, be the bounds of Ai, i=1,…,n. For 1≤n1<n, one denotes m=min{m1,…,mn1}, M=max{M1,…,Mn1}, and ∑i=1n1Φi(1H)=α1K, ∑i=n1+1nΦi(1H)=β1K, where α,β>0, α+β=1.(i)
If eitherr≤s, s≥1 or r≤s≤-1 and also one of two equalities
(3.23)M[r](α,A,Φ,1,n1)=M[r](1,A,Φ,1,n)=M[r](β,A,Φ,n1+1,n)
is valid, then
(3.24)M[s](α,A,Φ,1,n1)≤M[s](1,A,Φ,1,n)≤M[s](β,A,Φ,n1+1,n)
holds.(ii)
If eitherr≤s, r≤-1 or 1≤r≤s and also one of two equalities
(3.25)M[s](α,A,Φ,1,n1)=M[s](1,A,Φ,1,n)=M[s](β,A,Φ,n1+1,n)
is valid, then
(3.26)M[r](α,A,Φ,1,n1)≥M[r](1,A,Φ,1,n)≥M[r](β,A,Φ,n1+1,n)
holds.Proof.
(i) We prove only the case (i). We takeφ(t)=tr and ψ(t)=ts for t>0. Then ψ∘φ-1(t)=ts/r is concave for r≤s, s≤0, and r≠0. Since -ψ-1(t)=-t1/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)=ts/r is convex for r≤s, s≥0, and r≠0. Since ψ-1(t)=t1/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.
Ifr=0 and s≥1, we put φ(t)=lnt and ψ(t)=ts, 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 putφ(t)=ts and ψ(t)=tr for t>0 and we use the same technique as in the case (i).



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*Source: 358981-2011-06-20.xml*