Inequalities play a central and fundamental role in the fields of probability and measure theory, classical analysis, optimization theory, mathematical finance and economics. One of the most well-known inequalities for the class of convex functions is the Hermite–Hadamard inequality (see, e.g., [3], p. 137). This inequality states that if f:I=[a,b]→R is a convex function on the interval I of real numbers and a,b∈I with a<b, then

(1.1)fa+b2≤1b-a∫abf(x)dx≤f(a)+f(b)2.

Hermite–Hadamard's inequality for convex functions has received renewed attention in recent years and a remarkable variety of refinements and generalizations have been found; see, for example, [8, 2, 13, 16] and the references cited therein. This inequality for the fractional integral was proved by M. Z. Sarikaya et al. [15]. A general version of the Hermite–Hadamard inequality for pseudo fractional integrals and also for co-ordinated convex functions using fractional integral operator was obtained by [9, 1].

Many elementary inequalities are closely related to entropy mapping and mutual information, which are important in information theory and coding [4], but do not have fractional versions. In this manuscript we present the Hermite-Hadamard inequality for the function f(t)=tp-1,p∈(0,1] using the Riemann–Liouville fractional integral and special functions. Using this we then define fractional cases of the Tsallis relative operator entropy operator, and find bounds for it. This generalized definition facilitates a connection between entropy and fractional calculus.

Some notions and definitions that are useful in this paper are given in Section 2. In Section 3, we state the main results of this paper.

Preliminaries from the fractional calculus and the Tsallis relative operator entropy are outlined here.

Definition 2.1. [3] A function f:[a,b]⟶R is said to be convex if the following inequality holds

f(λx+(1-λ)y)≤λf(x)+(1-λ)f(y)

for all x, y∈[a, b] and λ∈[0, 1].

Using the inequalities (1.1), we have the following corollary.

Corollary 2.2. [11] If f(t)=tp-1,p∈(0,1], then

(2.1)x+12p-1(x-1)≤xp-1p≤xp-1+12(x-1), x≥1.

In the following, we recall that some special functions as well as some of their required properties ([14] p. 2–7). Let a,b>0. The gamma and beta functions were defined by

Γ(a)=∫0∞e-tta-1dt, Γ(x)=Γ(x+1)x, -1<x<0,B(a,b)=∫01ta-1(1-t)b-1dt,

respectively. The incomplete beta function of t∈R was denoted by B(t,a,b) and

B(t,a,b)=∫0tsa-1(1-s)b-1ds.

It should be noted that B(t,a,b)=Γ(a)Γ(b)Γ(a+b)ta+b-1 and therefore,

B(1,a,b)=B(a,b)=B(b,a)=Γ(a)Γ(b)Γ(a+b).

It was also given by the series

B(t,a,b)=tα∑n=0∞(1-b)nn!(a+n)tn,

where (x)n is Pochhammer symbol, in the other words

(x)n=x(x+1)⋯(x+n-1)n times=Γ(x+n)Γ(x).

The Gauss hypergeometric function F12(a,b c z) is defined as the sum of the hypergeometric series ([10], P. 27):

(2.2)F12(a,b c z)=∑n=0∞(an)(b)n(c)nznn!,

where |z|<1 a,b∈C c∈C∖Z0-. For other values of z, the Gauss hypergeometric function is defined as an analytic continuation of the series 2.2. One such analytic continuation is given by the Euler integral representation:

(2.3)F12(a,b c z)=Γ(c)Γ(a)Γ(c-b)∫01tb-1(1-t)c-b-1(1-zt)-adt,

where 0<R(b)<R(c) |arg(1-z)|<π.The Gauss hypergeometric function has the following simple properties,

(2.4) F12[a,b c z]=F12[b,a c z],
(2.5) F12[a,b c 0]=F12[0,b c z]=1,
(2.6) F12[a,b b z]=(1-z)-a,
(2.7) F12[a,b c 1]=Γ(c)Γ(c-a-b)Γ(c-a)Γ(c-b), R(c-a-b)>0,

and the following differentiation relation:

(2.8)dndzn(F12[a,b c z])=(a)n(b)n(c)n2F1[a+n,b+n c+ z], n∈N.

In the following we present the definitions of the Riemann–Liouville fractional integrals.

Definition 2.3. [14, 10] Let f∈L1[a,b]. The Riemann–Liouville fractional integrals Ja+αf and Jb-αf of order α>0 are defined by

(2.9)Ja+αf(x)=1Γ(α)∫ax(x-s)α-1f(s)ds, x>a

and

(2.10)Jb-αf(x)=1Γ(α)∫xb(s-x)α-1f(s)ds, x<b,

respectively. The Γ(α) is the Gamma function and Ja+0f(x)=Jb-0f(x)=f(x).

Proposition 2.4. [14] If f(t)=tp-1,p>0, then J0+αf(x)=Γ(p)Γ(p+α)xα+p-1.

In [15], Sarikaya et al. first proved the following important Hermite–Hadamard type inequalities utilizing the Riemann–Liouville fractional integrals.

Theorem 2.5. f:[a,b]→R be a positive function with 0≤a<b and f∈L1[a,b]. If f is a convex function on [a,b], then

(2.11)fa+b2≤Γ(α+1)2(b-a)α{Ja+αf(b)+Jb-αf(a)}≤f(a)+f(b)2.

Now, we give the following definitions of the relative operator entropy and the Tsallis relative operator entropy.

Definition 2.6. [11] Let A,B be two invertible positive operators in a Hilbert space H and p∈(0,1], the p-power mean of A and B is defined by A#pB=A12(A−12BA−12)pA12. If p=12, then A#12B is denoted by A#B and A#B=A12(A-12BA-12)12A12.

Definition 2.7. [11] Let A,B be two invertible positive operators in a Hilbert space H and the solidarity s for an operator monotone function f is defined by AsB=A12f(A-12BA-12)A. If f(t)=logt, then the relative operator entropy S(A|B) is defined by

S(A|B)=A12log(A-12BA-12)A12.

Very recently, the Tsallis relative operator entropy Tp(A|B) is defined in [11] by

(2.12)Tp(A|B)=A12(A−12BA−12)pA12−Ap, p∈(0,1],

and Tp(A|B) can be rewritten by using the notion of the p-power mean as follows

(2.13)Tp(A|B)=A#pB-Ap, p∈(0,1].

The Tsallis relative operator entropy also can be rewritten as

(2.14)Tp(A|B)=A♮pB-Ap, p∈R,

where A♮pB=A12(A−12BA−12)pA12,p∈ℝ.

The study of the Tsallis relative operator entropy is often strongly connected to the study of the p-weighted geometric operator mean. It is known that [6]:

(2.15)A-AB-1A≤Tp(A|B)≤B-A,

for strictly positive operators A,B and p∈[-1,0)∩(0,1] and limp→0Tp(A|B)=S(A|B).

In addition to introducing the definition of Tsallis relative operator entropy to the fractional case, two main theorems of this article are generalizations of the following theorems to fractional cases. H. R. Moradi et al. obtained the following estimate on the Tsallis relative operator entropy by using the Hermite-Hadamard inequality which is an improvement of the estimate (2.15).

Theorem 2.8. [11] For any invertible positive operator A and B such that A≤B and -1≤p≤1 with p≠0 we have

A12 A−12 BA−12 +I2p−1A−12BA−12−IA12≤Tp(A|B)≤12A#pB−A♮p−1B+B−A,

Theorem 2.9. [7] Let A and B be srtictly positive operators such that uA≤uB with u≥1 and let -1≤p≤1 with p≠0. then, we have

(2.16)Tp(A|B)-Tp-1(A|B)2≤4Tp(A|A+B2)-Tp-1(A|A+B2) ≤Tp(A|B)-T1(A|B)p-1 ≤Tp(A|B)-Tp-1(A|B)2+A#2(B-A)4.

More details for relative operator entropy and Tsallis relative operator entropy, one can refer to [11, 5, 6].

In this section, firstly we obtain some results about the fractional integral and its relevance in the Hermite-Hadamard inequality that are useful for our main purposes in which some special functions play an important role.

Lemma 3.10. x≥1, α>0 and p∈(0,1). If f(t)=tp-1, then

• (i)

  J1+αf(x)=xα+p-1Γ(p)Γ(α+p)-B(1x,p,α)Γ(α)=xα+p-1Γ(α)B(p,α)-B(1x,p,α),

• (ii)

  Jx-αf(1)=1Γ(α)B(1-α-p,α)-B(1x,1-α-p,α).

Proof. Both identities can be proved directly by using the definitions of gamma, beta, incomplete beta functions and their intermediate, i.e., Using definition of the incomplete beta function, we have

(3.1)B(1x,p,α)=∫01xtp-1(1-t)α-1dt.

By employing variables substitution t=1xs in (3.1) and using Proposition 2.4 we arrive at the formula

B(1x,p,α)=x1-α-p∫01sp-1(x-s)α-1ds =x1-α-p∫0xsp-1(x-s)α-1ds-∫1xsp-1(x-s)α-1ds =Γ(α)x1-α-pJ0+αf(x)-J1+αf(x) =Γ(α)x1-α-pΓ(p)Γ(p+α)xα+p-1-J1+αf(x) =Γ(α)Γ(p)Γ(p+α)-Γ(α)x1-α-pJ1+αf(x)=B(p,α)-Γ(α)x1-α-pJ1+αf(x).

and

Jx−αf(1)=1Γ(α)∫1x (s−1) α−1 sp−1ds=1Γ(α)∫1x (1− 1 s) α−1 sα+p−2ds=−1Γ(α)∫1x (1−u) α−1 ( 1u )α+p−2(1 u2)du=1Γ(α)∫1x1 (1−u) α−1 u−α−pdu=1Γ(α){∫01 (1−u) α−1 u−α−pdu−∫01x (1−u) α−1 u−α−pdu}=1Γ(α)B(1−α−p,α)−B(1x,1−α−p,α).

This completes the proof.

Corollary 3.11. x≥1, α>0 and p∈(0,1) such that 0≤α+p≤1. If f(t)=tp-1, then

x+12p-1≤Γ(α+1)2(x-1)αΓ(p)Γ(p+α)(xα+p-1-1)-Γ(1-α-p)Γ(1-p)(xp-1) ≤xp-1+12.

Proof. It should be noted that, Lemma 3.10 yields

(3.2)J1+αf(x)+Jx-αf(1)=xα+p-1Γ(α)B(p,α)-B(1x,p,α) +1Γ(α)B(1-α-p,α)-B(1x,1-α-p,α) =xα+p-1Γ(α)Γ(p)Γ(α)Γ(p+α)-Γ(p)Γ(α)Γ(p+α)x1-α-p +1Γ(α)Γ(1-α-p)Γ(α)Γ(1-p)-Γ(1-α-p)Γ(α)Γ(1-p) xp =Γ(p)Γ(p+α)xα+p-1(1-x1-α-p)+Γ(1-α-p)Γ(1-p)(1-xp) =Γ(p)Γ(p+α)(xα+p-1-1)-Γ(1-α-p)Γ(1-p)(xp-1).

The results directly follow from Theorem 2.5 and Lemma 3.10.

Now, we introduce a definition of the fractional Tsallis relative operator entropy which is a generalization of Eqn. (2.12). Indeed, we generalize the definition of the Tsallis relative operator entropy for the fractional case.

Definition 3.12. α>0 and p∈(0,1]. Then for any invertible positive operator A,B, we define the fractional Tsallis relative operator entropy as follows

(3.3) Tp,α(A|B)= α2{B(p,α)(A#α+p-1B-A)-B(1-α-p,α)(A#pB-A)}.

Remark 3.13. If we assume α=1, then Eqn. (3.14) coincides with Eqn. (2.12).

The following result shows a comparison of Tp(A|B) and Tp,α(A|B).

Corollary 2.14. For any invertible positive operator A and B, A≤B and α,p∈(0,1] such that 0<α+p<1, the operators Tp(A|B) and Tp,α(A|B) satisfy the following identitiy,

Tp,α(A|B)+pα2B(1-α-p,α)Tp(A|B)=αB(p,α)2(A#α+p-1B-A) =α(α+p-1)B(p,α)2Tα+p-1(A|B)

Proof. It follows from Definition 3.12 that

Tp,α(A|B)=αB(p,α)2A12 A−12 BA−12 α+p−1A12 −A−pα2B(1−α−p,α)A12 A−12 pBA−12 pA12−Ap=αB(p,α)2A12 A−12 BA−12 α+p−1A12 −A−pα2B(1−α−p,α)Tp(A|B)=α(α+p−1)B(p,α)2Tα+p−1(A|B)−pα2B(1−α−p,α)Tp(A|B)

Therefore the result is concluded.

The next theorem is a generalization of Theorem 2.8 when we apply Corollary 3.11.

Theorem 3.15. For any invertible positive operator A and B such that A≤B and p∈(0,1] we have

A12A−12 BA−12 +I2p−1 A−12 BA−12 −IαA12≤Tp,α(A|B)≤A12 (A−12 BA−12 )p−1+I2 A−12 BA−12 −IαA12.

Proof. From Corollary 3.11 we get

(3.4)x+12p-1(x-1)α≤Γ(α+1)2Γ(p)Γ(p+α)(xα+p-1-1) -Γ(1-α-p)Γ(1-p)(xp-1) ≤xp-1+12(x-1)α.

If we substitute A-12BA-12 instead of x in the above inequality and so that multiplying both sides in each sentence by A12, then the result follows easily.

Remark 3.16. If α=1, then Theorem 3.15 yields Theorem 2.8.

In the next theorem, we obtain some bounds for Tp,α(A|B) which is a generalization of Theorem 2.8. For this purpose, the following lemma is required.

Lemma 3.17. Suppose that  f:[a,b]⟶R is a twice differentiable function that there exist real constants m and M such that m≤f′′≤M. Then

• (i)

  m(b-a)2Λ1(α)≤Γ(α+1)2(b-a)α{Ja+αf(b)+Jb-αf(a)}-fa+b2≤M(b-a)2Λ1(α),

• (ii)

  m(b-a)2Λ2(α)≤f(a)+f(b)2-Γ(α+1)2(b-a)α{Ja+αf(b)+Jb-αf(a)}≤M(b-a)2Λ2(α),

where

Λ1(α)=α2-α+28(α+1)(α+2),Λ2(α)=α2(α+1)(α+2).

Proof. In fact, both functions g(x)=f(x)-mx22 and h(x)=Mx22-f(x) are convex and so we are able to use the Hermit-Hadamard inequality (2.11) for g and h. Therefore both above inequalities are prove applying the left and right sides of inequality (2.11) with g and h respectively. Therefore all details are given as follows: Applying the left side of inequality (2.11) with g we have

ga+b2≤Γ(α+1)2(b-a)α {Ja+αg(b)+Jb-αg(a)}.

Therefore

fa+b2-m8(a+b)2 ≤Γ(α+1)2(b-a)α1Γ(α)∫ab(b-t)α-1g(t)dt+1Γ(α)∫ab(t-a)α-1g(t)dt ≤Γ(α+1)2(b-a)α{1Γ(α)∫ab(b-t)α-1f(t)dt-m2Γ(α)∫ab(b-t)α-1t2dt +1Γ(α)∫ab(t-a)α-1f(t)dt-m2Γ(α)∫ab(t-a)α-1t2dt} ≤Γ(α+1)2(b-a)α{Ja+αf(b)+Jb-αf(a)} -mα4(b-a)α(b-a)αα(a2+b2)-2(b-a)α+2α(α+1)+4(b-a)α+2α(α+1)(α+2).

Hence

-m8(a+b)2+mα4(b-a)α(b-a)αα(a2+b2)-2(b-a)α+2α(α+1)+4(b-a)α+2α(α+1)(α+2)} ≤Γ(α+1)2(b-a)α{Ja+αf(b)+Jb-αf(a)}-fa+b2,

and therefore, with considering Λ1(α) we obtain

(3.5)m8(α2-α+2)(α+1)(α+2)(b-a)2=mΛ1(α)(b-a)2 ≤Γ(α+1)2(b-a)α{Ja+αf(b)+Jb-αf(a)}-fa+b2.

Next, to establish the right side of (i), it is enough to set the right side of inequality (2.11) with using h and therefore we have

ha+b2=M8(a+b)2-fa+b2 ≤Γ(α+1)2(b-a)α{Ja+αh(b)+Jb-αh(a)} ≤Mα4(b-a)α∫ab(b-t)α-1t2dt+∫ab(t-a)α-1t2dt -Γ(α+1)2(b-a)α{Ja+αf(b)+Jb-αf(a).

Hence

(3.6)Γ(α+1)2(b-a)αJa+αf(b)+Jb-αf(a)-fa+b2≤M8(α2-α+2)(α+1)(α+2)(b-a)2 =MΛ1(α)(b-a)2.

Then by combining inequalities (3.5) and (3.6), inequality (i) is obtained.

The second part is proved similar the first part when we consider the right side of inequality (2.11) by using g.

Remark 3.18. If α=1 and f(t)=tp-1, then both inequalities (i) and (ii) of Lemma 3.17 convert to both inequalities (i) and (ii) of Theorem 1, [12], respectively.

Now, we provide some bounds for Tp,α(A|B) in the following theorem.

Theorem 3.19. For any invertible positive operators A and B such that A≤B and p∈(0,1], the following inequalities hold

• (i)

  Lp,α(A,B)+Kp,α(A,B)≤Tp,α(A|B)≤Rp,α(A,B)+Kp,α(A,B),

• (ii)

  Jp,α(A,B)-Hp,α(A,B)≤Tp,α(A|B)≤Jp,α(A,B)-Gp,α(A,B),

where

Kp,α(A,B)=A12( A − 1 2 B A − 1 2 +I2)p−1( A − 1 2 B A − 1 2 −I)αA12,Rp,α(A,B)=(p−1)(p−2)Λ1(α) A12( A − 1 2 B A − 1 2 −I)α+2A12,Lp,α(A,B)=(p−1)(p−2)Λ1(α) A12( A − 1 2 B A − 1 2 )p−3( A − 1 2 B A − 1 2 −I)α+2A12,Jp,α(A,B)=A12 (A −12 BA −12 )p−1 +I2 ( A − 1 2 B A − 1 2 −I)αA12,Hp,α(A,B)=(p−1)(p−2)Λ2(α) A12( A − 1 2 B A − 1 2 −I)α+2A12,Gp,α(A,B)=(p−1)(p−2)Λ2(α) A12( A − 1 2 B A − 1 2 )p−3( A − 1 2 B A − 1 2 −I)α+2A12,

and Λ1(α), Λ2(α) are defined in Lemma 3.17.

Proof. Let f(t)=tp-1,p∈(0,1] and 1≤t≤x. Then the left side of Lemma 3.17 (i) gives

mΛ1(α)(x-1)2≤Γ(α+1)2(x-1)α {J1+αf(x)+Jx-αf(1)}-x+12p-1

After shifting some expressions of the above inequality and using Eqn. (3.2) we get

(3.7)mΛ1(α)(x-1)α+2+x+12p-1(x-1)α≤Γ(α+1)2{J1+αf(x)+Jx-αf(1)} =Γ(α+1)2Γ(p)Γ(p+α)(xα+p-1-1)-Γ(1-α-p)Γ(1-p)(xp-1).

Now instead of x substituting A-12BA-12 in inequality (3.11), multiplying both sides in each sentence by A12, and using Definition 3.12 we reach

mΛ1(α)A12(A-12BA-12-I)α+2A12 +A12A-12BA-12+I2p-1(A-12BA-12-I)αA12≤Tp,α(A|B).

Hence,

Λ1(α)(p-1)(p-2)A12(A-12BA-12)p-3(A-12BA-12-I)α+2A12 +A12A-12BA-12+I2p-1(A-12BA-12-I)αA12≤Tp,α(A|B).

Using some notation considered in theorem, the left side of (i) is obtained, i.e

(3.8)Lp,α(A,B)+Kp,α(A,B)≤Tp,α(A|B).

To prove the right side of (i), it is enough to consider the right side of Lemma 3.17 (i) and using f(t)=tp-1,p∈(0,1], 1≤t≤x. In this case, the right side of this inequality is obtained similar to the process from the left side, i.e,

(3.9)Tp,α(A|B)≤Rp,α(A,B)+Kp,α(A,B).

From which and inequality (3.11) the result is concluded.The proof of the second part of this theorem is easily obtained by using the second part of Lemma 3.17, when f(t)=tp-1,p∈(0,1] and 1≤t≤x.

Remark 3.20. If α=1, then Theorem 3.19 yields [11, Theorem 2].

Lemma 3.21. [15] Let f:[a,b]→R be a differentiable mapping on (a,b) with a<b. If |f′| is convex on [a,b], then the following inequality for fractional integrals holds:

(3.10) |f(a)+f(b)2-Γ(α+1)2(b-a)α[Ja+αf(b)+Jb-αf(a)]| ≤b-a2(α+1)(1-12α)[f′(a)+f′(b)].

The following estimate can be obtained by using the Lemma 3.21.

Theorem 3.22. For any invertible positive operator A and B, A≤B and α,p∈(0,1] such that 0<α+p<1, then there holds the following estimate:

−ΛA,B≤Tp,α(A|B)−12A12[ A−12 BA−12 p−1+I][A−12BA−12−I]αA12≤ΛA,B,

where

ΛA,B=(p−1)2(α+1)(1−12α)A12[(A−12 BA−12 )p−1+I][A−12 BA−12 −I]α+1A12.

Proof. If f(t)=tp-1 then according to 3.10 and by using the 3.2 there holds the following inequality:

xp-1+12(x-1)α-Γ(α+1)2[Γ(p)Γ(p+α)(xα+p-1-1) -Γ(1-α-p)Γ(1-p)(xp-1)(x-1)α]≤ (p-1)(x-1)α+12(α+1)(1-12α)[1+xp-1].

Another representation of the above inequality is the following form which can can be applied to obtain an estimate for Tp,α(A|B). Hence 3.11 yields,

(11)-Λ(x)≤Γ(α+1)2[Γ(p)Γ(p+α)(xα+p-1-1) -Γ(1-α-p)Γ(1-p)(xp-1)(x-1)α]-xp-1+12(x-1)α≤Λ(x)

where

(3.12)Λ(x)=(p-1)(x-1)α+12(α+1)(1-12α)(1+xp-1)

Now instead of x substituting A-12BA-12 in inequality (3.11), multiplying both sides in each sentence by A12 and using Definition 3.12 the result is obtained.

In the next theorem, we prove an inequality that involves some special functions that mentioned them all in the section 2. But first of all, there is a matter for the right side of this inequality that requires the following lemma. This matter leads to consider a conditions between α and p of the theorem. More explanation is observed to proof of the next lemma in the appendix.

Lemma 3.23. Assuming α,p∈R such that 0≤α+p<1, then,

∫0t(x-1)α(xp-2+1)dx=Γ(α+1)Γ(1-α-p)Γ(2-p)-B(1t ,1-α-p ,1+α)+(t-1)α+1α+1

Proof. Set F(t)=∫1t(x-1)α(xp-2+1)dx, then an integrating from a term of F(t), we have

(3.13) F(t)=(t-1)α+1α+1+∫1txp-2(x-1)αdx, t>1 =(t-1)α+1α+1+∫1txp-2+α(1-1x)αdx=(t-1)α+1α+1+∫1t1x-p-α(1-x)αdx.

If 0≤α+p<1, then it follows that both integral,

∫01x-p-α(1-x)αdx  and  ∫01tx-p-α(1-x)αdx.

are convergent. By using the beta function and incomplete beta function F(t) can be rewritten as follows whenever 0≤α+p≤1:

F(t)=(t-1)α+1α+1+∫01x-p-α(1-x)αdx-∫01tx-p-α(1-x)αdx =(t-1)α+1α+1+B(α+1,1-α-p)+B[1t,α+1,1-α-p]

Theorem 3.24. A and B be srtictly positive operators such that uA≤uB with u≥1 and let 0≤α+p<1. Then ,we have

A12(A−12BA−12−I)α+1A12 1+α2F1[1+α,2−p,2+α,12(I−A−12 BA−12 )]A 12 ≤α2B(p−1,α){Tα+p−1(A|B)−T1(A|B)}−α2B(2−α−p,α){Tp(A|B)−T1(A|B)}≤12A 12 B(1−α−p,1+α)−B( ( A − 1 2 B A − 1 2 ) −1 ,1−α−p ,1+α)A 12 +A♮ α+1(B−A)2(α+1)

Proof. It is sufficient to prove the following inequality for t≥1 and 0≤α+p<1 with p≠0,

(3.14)Kp,α(t)≤Cp,α(t)≤Lp,α(t)+(t-1)α+12(α+1),

where

Lp,α(t):=12B(1-α-p,α+1)-B(1t ,1-α-p ,1+α),
Cp,α(t):=α2B(p-1,α)tα+p-1-1α+p-1-(t-1) -α2B(2-α-p,α)tp-1p-(t-1),

and

Kp,α(t):=(t-1)α+11+α2F1[1+α,2-p,2+α,1-t2].

With considering a convex function f:[1,x]→R defined by f(y)=yp-2,p∈(0,1], then from Theorem 2.5 is concluded the following inequality

(3.15)(x-1)α(x+12)p-2≤Γ(α+1)2 {J1+αf(x)+Jx-αf(1)} ≤(x-1)α(xp-2+12).

It should be noted that same process of Lemma 3.10 and Eqn. 3.2 for function f(y)=yp-2 is repeated here and we have,

(3.16) Γ(α+1)2 {J1+αf(x)+Jx-αf(1)}= α2{B(2-α-p,α)(1-xp-1)+B(p-1,α)(xα+p-2-1)}.

By combining 3.15 and 3.16 and integrating both sides on [1,t], then there holds the following inequality:

(t-1)α+11+α2F1[1+α,2-p,2+α,1-t2] ≤Γ(α+1)Γ(p-1)2Γ(α+p-1)tα+p-1-1α+p-1-t+1 +Γ(α+1)Γ(2-α-p)2Γ(2-p)t-1-tp-1p ≤12Γ(α+1)Γ(1-α-p)Γ(2-p)-B(1t ,1-α-p ,1+α)+(t-1)α+12(α+1).

The above inequality can be rewritten as follows:

(3.17) (t-1)α+11+α2F1[1+α,2-p,2+α,1-t2]
(3.18) ≤α2B(p-1,α)tα+p-1-1α+p-1-(t-1) -α2B(2-α-p,α)tp-1p-(t-1) ≤12{B(1-α-p,1+α)-B(t-1,1-α-p,1+α)}+(t-1)α+12(α+1).

Thus, the inequality (3.17) is equivalent to inequality (3.14) sentence by sentence. With substituting t=A-12BA-12 in 3.17 and multiplying A12 both sides in each terms, then we obtain 3.14. This completes of Theorem 3.24.

Corollary 3.25. A and B be srtictly positive operators such that uA≤uB with u≥1 and let 0≤α+p<0.Then, Theorem 3.24 coincides with Theorem 2.9.

Proof. By using the inequality 3.17 and by employing the special functions introduced in the introduction, the conversion of 3.14 to 2.16 can be done easily. But due to the role of the integral representation of the hypergeometric function among special functions, we show that the left side of 3.14 and the left side of 2.16 are equivalent when α=1. Therefore, if α=1, we obtain an expression from left side of 3.14 by which left side of 2.16 is obtained with substituting t=A-12BA-12.Hence, if α=1 then according to the left side of 3.17 and employing 2.3, we have

1α+1(t-1)α+12F1[1+α,2-p,2+α,1-t2] =(t-1)222F1[2,2-p,3,1-t2] =(t-1)22Γ(3)Γ(2)Γ(1)∫01s(1-12(1-t)s)p-2ds =22-p(t-1)p-1(t+1)p-1-4p(p-1)(t+12)p-1) =4p{(t+12)p-1}-4p-1{(t+12)p-1-1}.

The last sentence of the above expression gives the left side of 2.16 quickly whenever A-12BA-12 is replaced instead of t. By the same procedure the other sentences of 2.16 are obtained from 3.17 whenever α=1 and therefore we refrain more details. It should be noted that due to for representation of F(t) to the beta function and incomplete beta function the condition 0≤α+p<1 is considred in Theorem 3.24. Therefore, if we drop this representation, then condition 0≤α+p<1 can be changed to -1≤p≤1 in the Theorem 3.24. This explanation is also valid for Corollary 3.25.