# Some Inequalities for Operatorp,h-Convex Function

Authors: Zahra Omrani; Omid Pourbahri Rahpeyma; Hamiderza Rahimi
Journal: Journal of Mathematics
 (2022)

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## Abstract

In this paper, we introduce operatorp,h-convex functions and establish a Hermite–Hadamard inequality for these functions. As application, we obtain several trace and singular value inequalities of operators.


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## Body

## 1. Introduction

In recent years, several extensions and generalizations have been considered for classical convexity and the theory of inequalities has made essential contributions to many areas of mathematics.In 1973, Elliott Lieb published a ground-breaking article on operator inequalities [1]. This and a subsequent article by Lieb and Ruskai [2] have had a profound effect on quantum statistical mechanics and more recently on quantum information theory. Since then, a number of attempts have been made to elucidate and extend these results. Two elegant examples are those of Nielsen and Petz [3] and Ruskai [4], which use the analytic representations for operator convex functions. In addition, Hansen [5] has developed a powerful theory that utilizes geometric means of positive operators. The latter notion was formulated by Pusz and Woronowicz [6] and subsequently investigated by Ando [7] (see the discussion in Section 3) and by Kubo and Ando [8].We recall some concepts of convexity that is well-known in the literature.The following inequality holds for any convex functionf defined on ℝ and a,b∈ℝ, with a<b,(1)fa+b2≤1b−a∫abfxdx≤fa+fb2.It was firstly discovered by Hermite in 1881 in the journal Mathematics (see [9]) and independetly proved in 1893 by Hadamard in [10]. The inequality (1) is known in the literature as the Hermite–Hadamard inequality.The Hermite–Hadamard inequality has several applications in nonlinear analysis and the geometry of Banach space (see [11]).The Hermite–Hadamard inequality has been the subject of intensive research; many applications, generalizations, and improvement of them can be found in the literature (see [12]).In recent years, many scholars have been interested in modifying and extending the Hermite–Hadamard inequality.In [13], Dragomir and Fitzpatrick proved the following version of Hermite–Hadamard inequality for s-convex functions in the second sense: let f:0,∞⟶0,∞ be an s-convex function, where s∈0,1 and a,b∈0,∞, a<b. If f∈L1a,b, then the following inequalities hold:(2)2s−1fa+b2≤1b−a∫abfxdx≤fa+fbs+1.The authors of [13] defined the operator s-convex and proved the following inequality for operator s-convex function. He proved that if f:I⊆0,∞⟶ℝ is an operator s-convex function, then the following inequalities hold:(3)2s−1fA+B2≤1b−a∫abfλA+1−λBdλ≤fA+fBs+1.The following inequalities due to the authors [14] give the Hermite–Hadamard inequalities for operator h-convex function.Letf be an operator h-convex function. Then,(4)12h1/2fA+B2≤∫01ftA+1−tBdt≤fA+fB∫01htdt,for any self-adjoint operators A and B with spectra in K.Motivated by the above results, we investigate in this paper the operator version of the Hermite–Hadamard inequality for operatorα,h-preinvex function.Letℬℋ stand for the C∗-algebra of all bounded linear operators on a complex separable Hilbert space ℋ,,. An operator A∈ℬℋ is positive if Ax,x≥0 for all x∈ℋ (we write A≥0). A positive invertible operator A is naturally denoted by A>0. Let ℬℋ+ stand for all positive operators ℬℋ. For self-adjoint operators A,B∈ℬℋ, we write B≥A if B−A≥0. A linear map φ:ℬℋ⟶ℬℋ is positive if φA≥0 whenever A≥0 and φ is said to be unital if φI=I.LetA be a self-adjoint operator in ℬℋ. The Gelfand map establishes a ∗-isometrically isomorphism Φ between the set CspA of all continuous functions defined on the spectrum of A denoting spA and C∗-algebra C∗A generated by A and the identity operator 1ℬℋ on ℋ as follows:For anyf,g∈CspA and any α,β∈ℂ, we have(i)
Φαf+βg=αΦf+βΦg(ii)
Φfg=ΦfΦg and Φf¯=Φf∗(iii)
Φf=f:=supt∈spAft(iv)
Φf0=1ℋ and Φf1=A, where f0t=1 and f1t=t for all t∈spAIff is a continuous complex valued function on spA, the element Φf of C∗A is denoted by fA and we call it the continuous functional calculus for a bounded self-adjoint operator A. If A is a bounded self-adjoint operator and f is a real-valued continuous function on spA, then ft≥0 for any t∈spA implies that fA≥0,i.e.,fA is a positive operator on ℋ. Moreover, if both f and g are real-valued functions on spA such that ft≤gt for any t∈spA, then fA≤gA in the operator order in ℬℋ.An important and useful class of functions are called operator convex functions. A real-valued continuous functionf on interval K⊆0,∞ is said to be operator convex (operator concave) if(5)fλA+1−λB≤≥λfA+1−λfB,in the operator order in ℬℋ, for all λ∈0,1 and for every bounded self-adjoint operators A and B in ℬℋ whose spectra are contained in K.K⊆ℝ+ is p-convex set if(6)spλAp+1−λBp1/p⊆K,for every positive operators A,B∈ℬℋ+ with spectra in K, λ∈0,1 and p>0.LetJ⊆ℝ+ such that 0,1⊂J. A function h:J⟶ℝ+ is called super-multiplicative function if hxy≥hxhy for all x,y∈J.In this paper, we assume thatK⊆ℝ+ is a p-convex set. We introduce operator p,h-convex function. We establish some properties of operator p,h-convex function. This paper is organized as follows: In Section 2, we will show some properties of operator p,h-convex functions. In Section 3, a new refinement of the Hermite–Hadamard type inequality is presented for operator p,h-convex functions. If f is an operator p,h-convex function, then(7)12h1/2fAp+Bp21/p≤∫01fλAp+1−λBp1/pdλ≤fA+fB∫01hλdλ.Our results enable us to obtain a new inequality for positive operators onℬℋ. For example, let 0<s≤p and φ be an unital linear operator positive, then(8)12φpA+φpBs/p≤∫01λφpA+1−λφpBs/pdλ≤pp+sφsA+φsB,for any positive operators A and B belonging to Mℋ with spectra in K.In some special cases, we show that our result gives a generalized estimation for operatorp,h-convex functions than the corresponding results obtained in [15].In Section4, we will show that if f is an operator p,h-convex function, then we have(9)12h1/2TrfAp+Bp21/p≤∫01TrftAp+1−tBp1/pdt≤TrfA+TrfB∫01htdt,for any self-adjoint operators A and B. We establish several trace inequalities for positive operator on ℬℋ.


## 2. Some Properties of Operatorp,h-Convex Functions

In order to obtain the main result of this section, we need the following known definitions.Letf be a real-valued continuous function defined on an interval K⊆ℝ+. We say that f is of operator P-class function on K if(10)fλA+1−λB≤fA+fB,for all self-adjoint operators A, B with spectra in K and all λ∈0,1.For some properties of this class of operators, see [16].Letf be a continuous real-valued function defined on an interval K⊆ℝ+. We say that f is an operator Q-class function on K if(11)fλA+1−λB≤fAλ+fB1−λ,for all self-adjoint operators A, B with spectra in K and all λ∈0,1.See [17] for some results and inequalities on operator Q-class function.A continuous functionf:K⊆0,∞⟶ℝ+ is said to be operator s-convex on K if(12)fλA+1−λB≤λsfA+1−λsfB,for all λ∈0,1 and for every positive operators A and B in ℬℋ+, whose spectra are contained in K for some fixed s∈0,1.For some properties about operators-convex function, see [18].LetK,J⊆ℝ+, 0,1⊆J, and h:J⟶ℝ be a nonnegative function nonidentical to 0. We say that a continuous function f:K⟶ℝ+ is said be a operator h-convex function on K if(13)fλA+1−λB≤hλfA+h1−λfB,for every A,B∈ℬℋ+ whose spectra are contained in K for all λ∈0,1.For some results on operatorh-convex, see [14].Lemma 1 (see [19]).
LetA,B∈ℬℋ+, then AB+BA≥0 if and only if fA+B≤fA+fB for all nonnegative operator monotone function f on 0,∞.Lemma 2 (see [20]).
Letφ be a unital positive linear map on ℋ and f an operator monotone function on 0,∞, then for every A≥0,(14)φfA≤fφA.Lemma 3 (Davis-Choi-Jensen’s inequality).
Letφ be a unital positive linear map on ℋ and g an operator convex function on 0,∞, then for every A≥0,(15)φgA≥gφA.The following lemma is the special case of Lemmas2 and 3.Lemma 4 (see [20]).
Letφ be a unital positive linear map on ℋ, then(i)
φAr≤φAr for every A≥0 and 0≤r≤1(ii)
φAr≤φAr for every A>0 and 1≤r≤2 or −1≤r≤0The following lemma is a consequence of theorem Hansen-Pedersen-Yensen’s inequality.Lemma 5 (see [21]).
Letf be a continuous function mapping the positive half-line 0,∞ into itself. Then, f is the operator monotone if and only if it is operator concave.Definition 1 (see [22]).
Letp>0 and K,J⊆ℝ+, 0,1⊂J, and h:J⟶ℝ+ be a nonnegative function nonidentical to 0. A continuous function f:K⟶ℝ+ is said to be operator p,h-convex (concave) if(16)fλAp+1−λBp1/p≤≥hλfA+h1−λfB,for any positive operators A,B with spectra in K and λ∈0,1.
This class contains several well-known classes of nonnegative operator convex function, operatorP-class function, operator Q-class function, operator s-convex function, and operator h-convex functions on K. We can see some results operator p,h-convex functions by Dinh and Khue in [22].Lemma 6 (see [22]).
Letφ be a unital positive linear map on ℬℋ, A a positive operator in ℋ, and f an operator p,h-convex function on ℝ+ such that f0=0. Let h be a nonnegative and nonzero super-multiplicative function on J satisfying 2h1/2≤λ−1hλλ∈0,1. Then,(17)fφAp1/p≤2h12φfA.Now, let us prove some properties of operatorp,h-convex functions. In this paper, we suppose 0,1⊂J and J,K⊆ℝ+.Proposition 1.
Letλ∈0,1 and h:J⟶ℝ+ be a nonzero function and f:K⟶ℝ+ be an operator p,h-convex. Therefore, for positive operator A with spectra in K,(i
Ifhλ+h1−λ>1, then fA≥0(ii)
Ifhλ+h1−λ<1, then fA≤0Proof.
(i)
Suppose thatA is a positive operator with spectra in K and A=B in (16). Then, fA≤hλfA+h1−λfA. Therefore, fA≤hλ+h1−λfA. Hence, hλ+h1−λ−1fA≥0 and so fA≥0.(ii)
On account of the mentioned,i.Proposition 2.
Leth:J⟶ℝ+ be a nonzero function; therefore,(i)
Letf:K⟶ℝ+ be an operator p,h-concave. If g:0,∞⟶0,∞ is an operator monotone and hλ+h1−λ=1, then g°f is an operator p,h-concave.(ii)
Letf:K⟶ℝ+ be an operator p,h-convex and operator monotone. Then, for all 1≤p≤2, f is the operator h-convex function.Proof.
(i)
Letf:K⟶ℝ+ be an operator p,h-concave and A and B are positive operators with spectra in K, then(18)fλAp+1−λBp1/p≥hλfA+h1−λfB.So,(19)gfλAp+1−λBp≥ghλfA+h1−λfB≥hλgfA+h1−λgfB.Therefore,g°f is the operator p,h-concave.(ii)
If1≤p≤2, then rx=xp is the operator convex. For positive operators A and B with spectra in K, we have λA+1−λBp≤λAp+1−λBp. Since 1/2≤1/p≤1, then kx=x1/p is the operator monotone. Hence,(20)λA+1−λB≤λAp+1−λBp1/p.We have(21)fλA+1−λB≤fλAp+1−λBp1/p≤hλfA+h1−λfB.Then,f is the operator h-convex function.Proposition 3.
Letφ be a unital positive linear map on ℬℋ, A and B be positive operators in ℋ with spectra in K, and f:K⟶ℝ+ be an operator p,h-convex function such that f0=0. Let h:J⟶0,∞ be a super-multiplicative function satisfying 2h1/2≤λ−1hλ for λ∈0,1. We have(22)fφλAp+1−λBp1/p≤2h12hλφfA+h1−λφfB.Proof.
By Lemma7, fφAp1/p≤2h1/2φfA. If A=λAp+1−λBp1/p, then(23)fφλAp+1−λBp1/p≤2h12φfλAp+1−λBp1/p≤2h12φhλfA+h1−λfB≤2h12hλφfA+h1−λφfB.Corollary 1.
With conditions, Proposition3, we have(24)fφAp+Bp21/p≤2h212φfA+φfB.Proof.
Putλ=1/2 in inequality (22).Remark 1.
ForφA=A, inequality (24) reduces to the inequality(25)fAp+Bp21/p≤2h212fA+fB.Proposition 4.
Letp>0 and h:J⟶ℝ+ be a nonzero function, then the following statements are equivalent:(i)
fx=xs is an operator p,h-convex function(ii)
fx=xs/p is an operator h-convex functionProof.
i⇒ii: suppose that fx=xs is an operator p,h-convex. Thus, λAp+1−λBps/p≤hλAs+h1−λBs for each positive operator A and B with spectra in K and λ∈0,1. For A=A1/p and B=B1/p, we have λA+1−λBs/p≤hλAs/p+h1−λBs/p. Therefore, fx=xs/p is an operator h-convex.ii⇒i: the proof is similarly i⇒ii.Remark 2.
Letfx=xs/p for 0<s≤p be a nonnegative operator monotone function on 0,∞ and ℳℋ=A,B∈ℬℋ+×ℬℋ+|AB+BA≥0. By Lemma 1 for each A,B∈ℳℋ, we have fA+B≤fA+fB. Put A=λA and B=1−λB, λ∈0,1. Hence, fλA+1−λB≤fλA+f1−λB. Therefore,(26)λA+1−λBs/p≤λs/pAs/p+1−λs/pBs/p.
Therefore,fx=xs/p for hλ=λs/p is an operator h-convex for 0<s≤p.
The following example is a characteristic example of operatorp,h-convex function.Example 1.
Let0<s≤p. By Remark 2, fx=xs/p is an operator h-convex function for hλ=λs/p on Mℋ. By Proposition 4, fx=xs is an operator p,h-convex on Mℋ.
In the following preposition, we eliminate conditionA,B∈Mℋ and we get the new example of operator p,h-convex.Proposition 5.
Letp>0 and fx=xs and hλ=λ, then f is the operator p,h-convex if and only if s∈p,2p.Proof.
Iff is an operator p,h-convex, for each positive operator A and B with spectra in K, λAp+1−λBps/p≤hλAs+h1−λBs. Hence, λA+1−λBs/p≤λAs/p+1−λBs/p. The last inequality means that the function gt=ts/p is the operator convex, which is equivalent to the condition s∈p,2p.Corollary 2.
Letfx=xs, hλ=λ, and p>0, then we have(i)
Ifs∈p,2p∩0,1, then the function f is an operator p,h-convex and operator concave(ii)
Ifs∈p,2p∩1,2, then the function f is an operator p,h-convex and operator convexProof.
Becausefx=xs for 0≤s≤1 is operator concave and fx=xs for 1≤s≤2 is operator convex, then proof is trivial.Proposition 6.
Letf:K⟶ℝ+ be an operator p,h-convex and h:J⟶ℝ+ be a nonzero function and A,B are two positive operators with spectra in K, then φx,A,Bt:0,1⟶ℝ+ defined by φx,A,Bt=ftAp+1−tBp1/px,x is a h-convex function on 0,1 for any x∈ℋ with x=1.Proof.
For eachu,v∈0,1, we have(27)φx,A,Btu+1−tv=ftu+1−tv​Ap+1−tu+1−tvBp1/px,x=ftuAp+1−uBp+1−tvAp+1−vBp1/px,x=ftuAp+1−uBp1/pp+1−tvAp+1−vBp1/pp1/px,x≤htfuAp+1−uBp1/px,x+h1−tfvAp+1−vBp1/px,x=htφx,A,Bu+h1−tφx,A,Bv.


## 3. Hermite–Hadamard Inequality for Operatorp,h-Convex Functions

Now, we are ready to state the main result of this paper.Theorem 1.
Letf be a continuous operator p,h-convex function and h:J⟶ℝ+ be a continuous nonzero function. Then, for any A,B∈ℬℋ+ with spectra in K, we have(28)12h1/2fAp+Bp21/p≤∫01fλAp+1−λBp1/pdλ≤fA+fB∫01hλdλ.Proof.
Letλ=1/2 in Definition 1, then(29)fAp+Bp21/p≤h12fA+fB.
PutA1=λAp+1−λBp1/p and B1=1−λAp+λBp1/p. Then, we have A1p+B1p=Ap+Bp. So,(30)fAp+Bp21/p≤h12fλAp+1−λBp1/p+f1−λAp+λBp1/p.
Therefore,(31)fAp+Bp21/p≤h12∫01fλAp+1−λBp1/p+f1−λAp+λBp1/pdλ=2h12∫01fλAp+1−λBp1/pdλ.
As(32)2h12∫01fλAp+1−λBp1/pdλ≤2h12∫01hλfA+h1−λfBdλ≤2h12fA∫01hλdλ+fB∫01h1−λdλ=2h12fA+fB∫01hλdλ,thus (28) holds.Remark 3.
Ifp=1 in inequality (28), then Theorem 1 produces [14].Corollary 3.3.
Letp>0, s∈p,2p, and φ be an unital linear positive map, then(33)φpA+φpB2s/p≤∫01λφpA+1−λφpBs/pdλ≤φsA+φsB2,for any positive operators A and B with spectra in K.Proof.
By Corollary2, fx=xs is the operator p,h-convex for hλ=λ and s∈p,2p. Put fx=xs in Theorem 1.
In particular, we have the following,Corollary 4.
Let0<s≤p and φ be an unital linear positive map, then(34)12φpA+φpBs/p≤∫01λφpA+1−λφpBs/pdλ≤pp+sφsA+φsB,for any positive operators A and B belonging to Mℋ with spectra in K.Proof.
By Example1, fx=xs is the operator p,h-convex for hλ=λs/p and 0<s≤p. Therefore,(35)121/2s/pAp+Bp2s/p≤∫01λAp+1−λBps/pdλ≤As+Bs∫01λs/pdλ.
So,(36)12Ap+Bps/p≤∫01λAp+1−λBps/pdλ≤As+Bsps+p.
PutA=φA and B=φB in inequality (36).Corollary 5.
Let0<s≤p≤1 and φ be a positive operator, then(37)12φAp+Bps/p≤12φpA+φpBs/p≤ps+pφsA+φsB.for each positive operators A and B in Mℋ.Proof.
If0<s≤p≤1, then φAp+Bp≤φpA+φpB (by Lemma 4), we have(38)12φAp+Bps/p≤12φpA+φpBs/p≤∫01λφpA+1−λφpBs/pdλ≤ps+pφsA+φsB.Definition 2.
Letf,g:K⟶ℝ+ be operator p,h-convex functions and A,B self-adjoint operators on Hilbert space ℋ with spectra in K. We defined real functions MA,B and NA,B on Hilbert space ℋ by(39)MA,Bx=fAx,xgAx,x+fBx,xgBx,x,and(40)NA,Bx=fAx,xgBx,x+fBx,xgAx,x,for any x∈ℋ.Theorem 2.
Leth1,h2:J⟶ℝ+ and f,g:K⟶ℝ+ be continuous functions. If f is an operator p,h1-convex and g is an operator p,h2-convex function, then for any positive operators A and B with spectra in K and each x∈ℋ with x=1, the following inequality holds:(41)∫10fλAp+1−λBp1/px,xgλAp+1−λBp1/px,xdλ≤MA,Bx∫01h1λh2λdλ+NA,Bx∫01h1λh21−λdλ.Proof.
On account of the operatorp,h1-convexity of f and p,h2-convexity of g, we give(42)fλAp+1−λBp1/px,x≤h1λfA+h11−λfBx,x,(43)gλAp+1−λBp1/px,x≤h2λgA+h21−λgBx,x,for each λ∈0,1 and x∈ℋ with x=1. We have(44)fλAp+1−λBp1/px,xgλAp+1−λBp1/px,x≤h1λh2λfAx,xgAx,x+h11−λh2λfBx,xgAx,x+h1λh21−λfAx,xgBx,x+h11−λh21−λfBx,xgBx,x.
Therefore,(45)∫01fλAp+1−λBp1/px,xgλAp+1−λBp1/px,xdλ≤fAx,xgAx,x∫01h1λh2λdλ+fBx,xgAx,x∫01h11−λh2λdλ+fAx,xgBx,x∫01h1λh21−λdλ+fBx,xgBx,x∫01h11−λh21−λdλ.
The proof is complete.Corollary 6.
In Theorem2, if h1λ=h2λ=hλ, then(46)∫01fλAp+1−λBp1/px,xgλAp+1−λBp1/px,xdλ≤MA,Bx∫01h2tdt+NA,Bx∫01hth1−tdt.Remark 4.
Ifh1λ=h2λ=λ in Theorem 2, then(47)∫01fλAp+1−λBp1/px,xgλAp+1−λBp1/px,xdλ≤13MA,Bx+16NA,Bx.
Hence, in this case, Theorem2 produces [15], Theorem 3.Theorem 3.
Leth1,h2:J⟶ℝ+ and f,g:K⟶ℝ+ be continuous functions. If f is an operator p,h1-convex and g is an operator p,h2-convex function, then for any positive operators A and B with spectra in K and each x∈ℋ with x=1, the following inequality holds:(48)12h11/2h21/2fAp+Bp21/px,xgAp+Bp21/px,x≤∫01fλAp+1−λBp1/px,xgλAp+1−λBp1/px,xdλ+MA,Bx∫01h1λh21−λdλ+NA,Bx∫01h1λh2λdλ.Proof.
SinceAp+Bp/2=λAp+1−λBp/2+1−λAp+λBp/2, for any x∈ℋ with x=1, we have(49)fAp+Bp21/px,xgAp+Bp21/px,x=fλAp+1−λBp2+1−λAp+λBp2x,xgλAp+1−λBp2+1−λAp+λBp2x,x≤h112fλAp+1−λBp1/px,x+f1−λAp+λBp1/px,xh212gλAp+1−λBp1/px,x+g1−λAp+λBp1/px,x≤h112h212fλAp+1−λBp1/px,xgλAp+1−λBp1/px,x+h112h212f1−λAp+λBp1px,xg1−λAp+λBp1px,x+h112h212h1λfAx,x+h11−λfBx,xh21−λgAx,x+h2λgBx,x+h112h212h11−λfAx,x+h1λfBx,xh2λgAx,x+h21−λgBx,x=h112h212fλAp+1−λBp1/px,xgλAp+1−λBp1/px,x+h112h212f1−λAp+λBp1/px,xg1−λAp+λBp1/px,x+h112h212h1λh21−λ+h11−λh2λMA,Bx+h112h212h1λh2λ+h11−λh21−λNA,Bx.
Integrating both sides of inequality over0,1, we get the required inequality (48).Remark 5.
In Theorem3, if h1λ=h2λ=λ, then(50)fAp+Bp21/px,xgAp+Bp21/px,x≤∫01fλAp+1−λBp1/px,xgλAp+1−λBp1/px,xdλ+112MA,Bx+16NA,Bx.
Therefore, in this case, Theorem3 produces [15], Theorem 4.


## 4. Some Trace Functional Inequalities for Operator

LetKℋ be two sided ideal of compact operator ℬℋ. For A∈ℬℋ, let A= sup Ax:x=1 denote the usual operator norm of A and A=A∗A1/2 be the absolute value of A. The direct sum A⊕B denotes the block diagonal matrix A00B defined on ℋ⊕ℋ (see [20]). It is clear that A⊕B= max A,B.For any operatorA, the operator A∗A is always positive and its unique positive square root is denoted by A.We remind some basic properties of trace for operators. Leteii∈I be an orthonormal basis of ℋ; we say that A∈ℬℋ is the trace class if(51)A1=∑i∈IAei,ei<∞.The definition ofA1 does not depend on the choice of the orthonormal basis eii∈I. We denote the set of trace class operators in ℬℋ by ℬ1ℋ. We define the trace of a trace class operator A∈ℬ1ℋ to be(52)TrA≔∑i∈IAei,ei,where eii∈I is an orthonormal basis of ℋ. Tr⋅ is a bounded linear functional on ℬ1ℋ with Tr=1.Lemma 7.
Letf:K⟶ℝ+ be an operator p,h-convex function and h:J⟶ℝ+ be a nonzero function, then gt=TrftAp+1−tBp1/p is the h-convex on 0,1 for self-adjoint operators A and B with spectra in K.Proof.
Because trace functional is convexity and monotonicity, then for eachu,v∈0,1 and 0<α<1, we obtain(53)gαu+1−αv=Trfαu+1−αvAp+αu+1−αvBp1/p=TrfαuAp+1−uBp+1−αvAp+1−vBp1/p=TrfαuAp+1−uBp1/pp+1−αvAp+1−vBp1/pp1/p≤TrhαfuAp+1−uBp1/p+h1−αfvAp+1−vBp1/p=hαTrfuAp+1−uBp1/p+h1−αTrfvAp+1−vBp1/p=hαgu+h1−αgv.
Therefore,g is h-convex.Theorem 4.
Letf:K⟶ℝ+ be an operator p,h-convex function and h:J⟶ℝ+ be a nonzero function, then we have(54)12h1/2TrfAp+Bp21/p≤∫01TrftAp+1−tBp1/pdt≤TrfA+TrfB∫01htdt,for each self-adjoint operators A and B with spectra in K.Proof.
gt=TrftAp+1−tBp1/p is h-convex on 0,1 by Lemma 4. Hence, by inequality (28),(55)12h1/2g0+12≤∫01gtdt≤g0+g12∫01htdt.
Therefore, we have the desired result.Remark 6.
We think that with the new definition operatorp,h-convex function for the Jensen’s inequality and operator convex inequalities in [23], generalization or new inequalities can be obtained.



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*Source: 3836837-2022-02-07.xml*