# Someδ-Tempered Fractional Hermite–Hadamard Inequalities Involving Harmonically Convex Functions and Applications

Authors: Nousheen Akhtar; Muhammad Uzair Awan
Journal: Journal of Mathematics
 (2021)

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## Abstract

The main objective of this paper is to obtain some newδ-tempered fractional versions of Hermite–Hadamard’s inequality using the class of harmonic convex functions. In order to show the significance of the main results, we also discuss some interesting applications.


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## Body

## 1. Introduction and Preliminaries

A functionf:I⊂ℝ⟶ℝ is said to be convex if(1)f1−τν1+τν2≤1−τfν1+τfν2,∀ν1,ν2∈I,τ∈0,1.In recent years, several new extensions of classical convexity have been proposed in the literature. Iscan [1] introduced the notion of harmonically convex functions as follows.A functionf:I⊂0,∞⟶ℝ is said to be harmonically convex if(2)fν1ν2τν1+1−τν2≤1−τfν1+τfν2,∀ν1,ν2∈I,τ∈0,1.Hermite–Hadamard’s inequality is one of the most studied results pertaining to convexity property of the functions. This result of Hermite and Hadamard reads as follows.Letf:I=ν1,ν2⊂ℝ⟶ℝ be a convex function; then,(3)fν1+ν22≤1ν2−ν1∫ν1ν2fxdx≤fν1+fν22.Iscan [1] extended the classical version of Hermite–Hadamard’s inequality using the harmonic convexity property of the function.Letf:I=ν1,ν2⊂0,∞⟶ℝ be a harmonically convex function; then,(4)f2ν1ν2ν1+ν2≤ν1ν2ν2−ν1∫ν1ν2fxdx≤fν1+fν22.The interrelation between theory of convex functions and theory of integral inequalities has attracted several inequality experts and as result several new versions of classical results have been obtained in the literature. For example, Sarikaya et al. [2] used the concepts of fractional calculus in obtaining the fractional analogue of Hermite–Hadamard’s inequality. This idea attracted several researchers and a result number of new refined fractional analogues of classical inequalities have been obtained in the literature. For example, Gurbuz et al. [3] obtained some new refinements of integral inequalities using fractional integral operators of positive real order. İşcan and Wu [4] obtained fractional analogue of Hermite–Hadamard’s inequality using the concept of harmonically convex functions Awan et al. [5] obtained conformable fractional Hermite–Hadamard’s inequality using the harmonic convexity property of the functions. Iftikhar et al. [6] obtained some local fractional Newton’s type inequalities via generalized harmonic convex functions. Recently, Sanli et al. [7] obtained some more new fractional Hermite–Hadamard type of inequalities using the harmonic convexity property of the functions.In recent years, the classical concepts of fractional calculus have been extended and generalized using novel and innovative ideas. For instance, Meerschaert et al. [8] introduced the concepts where power laws are tempered by an exponential factor and showed that this exponential tempering has both mathematical and practical advantages. This inspired Mohammed et al. [9], and they obtained new generalizations of Hermite–Hadamard’s inequality using tempered fractional integrals. Mubeen [10] and Sarikaya and Karaca [11] introduced the notion of δ-fractional integrals. Using the concepts of δ-fractional calculus, Lei et al. [12] obtained Hadamard δ-fractional inequalities of Fejer type using GA-s-convex functions. Luo et al. [13] obtained bounds related to multiparameterized δ-fractional integrals and discussed their applications as well. Awan et al. [14] obtained new fractional analogues of Hermite–Hadamard’s inequality using δ-Appell’s hypergeometric functions and the harmonic convexity property of the functions.The objective of this paper is also to obtain some newδ-tempered fractional versions of Hermite–Hadamard’s inequality using the class of harmonic convex functions. Before we proceed further, let us recall some previously known concepts.Riemann–Liouville fractional integrals are defined as follows.Definition 1.
(see [15]). Let ℱ∈L1ν1,ν2. Then, Riemann–Liouville integrals Jν1+αℱ and Jν2−αℱ of order α>0 with ν1≥0 are defined by(5)Jν1+αℱx=1Γα∫ν1xx−λα−1ℱλdλ,x>ν1,Jν2−αℱx=1Γα∫xν2λ−xα−1ℱλdλ,x<ν2,where(6)Γα=∫0∞e−xxα−1dx,is the well-known Gamma function.The concept ofδ-Riemann–Liouville fractional integral is defined as follows: let ℱ be piecewise continuous on I∗=0,∞ and integrable on any finite subinterval of I=0,∞. Then, for λ>0, we consider δ-Riemann–Liouville fractional integral of ℱ of order α(7)Jν1αkℱx=1δΓδα∫ν1xx−λα/δ−1ℱλdλ,x>ν1,δ>0.Ifδ⟶1, then δ-Riemann–Liouville fractional integrals reduce to classical Riemann–Liouville fractional integral.In [8], the authors have described a new variation on the fractional calculus as follows.Definition 2.
Letℱ∈Lν1,ν2 with λ≥0 and α>0. Then, right- and left-tempered fractional integrals are defined as(8)Jν1+α,λτℱx=1Γα∫ν1xx−μα−1exp−λx−μℱμdμ,x∈ν1,ν2,Jν2−α,λτℱx=1Γα∫xν2μ−xα−1exp−λμ−xℱμdμ,x∈ν1,ν2.We now introduce theδ-tempered fractional integrals.Definition 3.
Letℱ∈Lν1,ν2 with λ≥0,α>0 and δ≥1. Then, right- and left-tempered δ-fractional integrals are defined as(9)Jν1+,δα,λτℱx=1δΓδα∫ν1xx−μα−1exp−λx−μδ/δℱμdμ,x∈ν1,ν2,Jν2−,δα,λτℱx=1δΓδα∫xν2μ−xα−1exp−λμ−xδ/δℱμdμ,x∈ν1,ν2.Definition 4.
For the real numbersα>0,λ≥1 with δ≥1, we define the λ-incomplete δ-gamma function by(10)δγλα,x=∫0xμα−1e−λμδ/δdμ.
Ifλ=0, it reduces the incomplete δ-gamma function:(11)δγα,x=∫0xμα−1e−μδ/δdμ.Remark 1.
For the real numbersα>0,λ≥1 with δ≥1, we have(1)
δγλν2−ν1α,1=∫01μα−1e−λν2−ν1μδ/δdμ=1/ν2−ν1αδγλα,ν2−ν1(2)
∫01δγλν2−ν1α,xdx=δγλα,ν2−ν1/ν2−ν1α−δγλα+1,ν2−ν1/ν2−ν1α+1Proof.
(1)
The proof is straightforward, by using the change of variable techniquex=ν2−ν1μ.(2)
To prove this, we use definition ofλ-incomplete δ-gamma function:(12)∫01δγλν2−ν1α,xdx=∫01∫0xμα−1e−λν2−ν1μδ/δdμdx.
By changing the order of integration, we obtain(13)∫01δγλν2−ν1α,xdx=∫01∫μ1μα−1e−λν2−ν1μδ/xdμdx=∫011−μμα−1e−λν2−ν1μδ/δdμ=∫01μα−1e−λν2−ν1μδ/δdμ−∫01μαe−λν2−ν1μδ/δdμ.
Using Remark1 (1), we have(14)∫01δγλν2−ν1α,xdx=δγλα,ν2−ν1ν2−ν1α−δγλα+1,ν2−ν1ν2−ν1α+1.
This completes the proof.


## 2. Results and Discussion

In this section, we discuss our main results.


### 2.1. A New Version of Hermite–Hadamard’s Inequality

We now derive a newδ-tempered fractional Hermite–Hadamard inequality via harmonically convex function.Theorem 1.
Letf:ν1,ν2⟶ℝ be an harmonically convex function on ν1,ν2 with ν1<ν2; then,(15)f2ν1ν2ν1+ν2≤ν1ν2ν2−ν1αδΓδα2δγλν2−ν1α,1J1/ν1−,δα,λν1ν2τf∘g1ν2+J1/ν2+,δα,λν1ν2τf°g1ν1≤fν1+fν22,where gx=1/x and for all α>0andλ≥1.Proof.
Sincef is a harmonic convex function, then(16)f2xyx+y≤fx+fy2.
This implies(17)2f2ν1ν2ν1+ν2≤fν1ν21−τν1+τν2+fν1ν2τν1+1−τν2.
Multiplying both sides of the above inequality byτα−1e−λν2−ν1τδ/δ and integrating with respect to τ on 0,1, we have(18)2f2ν1ν2ν1+ν2∫01τα−1e−λν2−ν1τδ/δdτ≤∫01τα−1e−λν2−ν1τδ/δfν1ν21−τν1+τν2dτ+∫01τα−1e−λν2−ν1τδ/δfν1ν2τν1+1−τν2dτ.
This implies(19)2δγλν2−ν1α,12ν1ν2ν1+ν2≤ν1ν2αν2−ν1α∫1/ν21/ν1x−1ν2α−1e−λν1ν2x−1/ν2δ/δf1xdx+ν1ν2αν2−ν1α∫1/ν21/ν11ν1−xα−1e−λν1ν21/ν1−xδ/δf1xdx.
Thus, we have(20)f2ν1ν2ν1+ν2≤ν1ν2αδΓδα2ν2−ν1αδγλν2−ν1α,1J1/ν1−,δα,λν1ν2τf°g1ν2+J1/ν2+,δα,λν1ν2τf°g1ν1.
Also,(21)fν1ν21−τν1+τν2≤τfν1+1−τν2,(22)fν1ν2τν1+1−τν2≤1−τfν1+τfν2.
Adding (21) and (22), we have(23)fν1ν21−τν1+τν2+fν1ν2τν1+1−τν2≤fν1+fν2.
Multiplying the above inequality byτα−1e−λν2−ν1τδ/δ and integrating with respect to τ on 0,1, we have(24)∫01τα−1e−λν2−ν1τδ/δfν1ν21−τν1+τν2dτ+∫01τα−1e−λν2−ν1τδ/δfν1ν2τν1+1−τν2dτ≤fν1+fν2∫01τα−1e−λν2−ν1τδ/δdτ.
This implies(25)ν1ν2ν2−ν1α∫1/ν21/ν1x−1ν2α−1e−λν1ν2x−1/ν2δδf1xdx+ν1ν2ν2−ν1α∫1/ν21/ν11ν1−xα−1e−λν1ν21/ν1−xδδf1xdx≤δγλν2−ν1α,1fν1+fν2.
This implies(26)ν1ν2ν2−ν1αδΓδα2δγλν2−ν1α,1J1/ν1−,δα,λν1ν2τf°g1ν2+J1/ν2+,δα,λν1ν2τf°g1ν1≤fν1+fν22.
Combining (20) and (26), we get the required inequality (15)Remark 2.
(1)
Forα=1=k and λ=0 in (15), we obtain the classical Hermite–Hadamard’s type inequality for the notion of harmonic convexity(2)
Fork=1 and λ=0 in 2.1, we obtain the fractional version of Hermite–Hadamard’s type inequality for the notion of harmonic convexity


### 2.2. New Auxiliary Results

We now derive two new auxiliary results. These results will play a significant role in the development of our next results.Lemma 1.
Letf:ν1,ν2⟶ℝ be L1 function; then,(27)fν1+fν22−ν1ν2ν2−ν1αδΓδα2δγλν2−ν1α,1J1/ν1−,δα,λν1ν2τf°g1ν2+J1/ν2+,δα,λν1ν2τf°g1ν1=ν1ν2ν2−ν12δγλν2−ν1α,1∫01δγλν2−ν1α,1−τ−δγλν2−ν1α,τ1−τν1+τν22f′ν1ν21−τν1+τν2dτ,for all α>0,λ≥0.Proof.
Consider(28)I=ν1ν2ν2−ν12δγλν2−ν1α,1∫01δγλν2−ν1α,1−τ+δγλν2−ν1α,τ1−τν1+τν22f′ν1ν21−τν1+τν2dτ=ν1ν2ν2−ν12δγλν2−ν1α,1I1−I2.
Now,(29)I1=∫01δγλν2−ν1α,1−τ1−τν1+τν22f′ν1ν21−τν1+τν2dτ=δγλν2−ν1α,1fν2−∫011−τα−1eλν2−ν11−τδδfν1ν21−τν1+τν2dτ=δγλν2−ν1α,1fν2−ν1ν2ν2−ν1α∫1/ν21/ν11ν1−xα−1eλν2−ν11/ν1−xδδf1xdx=δγλν2−ν1α,1fν2−ν1ν2ν2−ν1αδΓδατJ1/ν2+,δα,λν1ν2f°g1ν1.
Similarly,(30)I2=∫01δγλν2−ν1α,τ1−τν1+τν22f′ν1ν21−τν1+τν2dτ=−δγλν2−ν1α,1fν1+∫01τα−1eλν2−ν1τδ/δfν1ν21−τν1+τν2dτ=−δγλν2−ν1α,1fν2+ν1ν2ν2−ν1α∫1/ν21/ν1x−1ν2α−1eλν2−ν1x−1/ν2δ/δf1xdx=−δγλν2−ν1α,1fν2+ν1ν2ν2−ν1αδΓδατJ1/ν1−,δα,λν1ν2f°g1ν2.
Substituting the values ofI1 and I2 in (28), we get the required result.Lemma 2.
Letf:ν1,ν2⟶ℝ be L1 function; then,(31)2α−1ν1ν2αδΓδαδγλα,ν2−ν1Jν1+ν2/2ν1ν2−,δα,2λν1ν2τf°g1ν2+Jν1+ν2/2ν1ν2+,δα,2λν1ν2τf°g1ν1−f2ν1ν2ν1+ν2=ν2−ν1α2δγλα,ν2−ν1∫012ν1ν2ν2−ν1δγλν2−ν1α,τ2−τν1+τν22f′2ν1ν22−τν1+τν2dτ−∫012ν1ν2ν2−ν1δγλν2−ν1α,ττν1+2−τν22f′2ν1ν2τν1+2−τν2dτ,where α>0,λ≥1.Proof.
Consider(32)I=ν2−ν1α2δγλα,ν2−ν1∫012ν1ν2ν2−ν1δγλν2−ν1α,τ2−τν1+τν22f′2ν1ν22−τν1+τν2dτ−∫012ν1ν2ν2−ν1δγλν2−ν1α,ττν1+2−τν22f′2ν1ν2τν1+2−τν2dτ=ν2−ν1α2δγλα,ν2−ν1I1−I2.
Using Remark1 (1), we have(33)I1=∫012ν1ν2ν2−ν1δγλν2−ν1α,τ2−τν1+τν22f′2ν1ν22−τν1+τν2dτ=−δγλν2−ν1α,1f2ν1ν2ν1+ν2+∫01τα−1e−λν2−ν1τδ/δf2ν1ν22−τν1+τν2dτ=δγλν2−ν1α,1f2ν1ν2ν1+ν2+2ν1ν2αν2−ν1α∫1/ν2ν1+ν2/2ν1ν2x−1ν2α−1e−2ν1ν2λx−1/ν2δ/δf1xdx=−δγλα,1f2ν1ν2ν1+ν2+2ν1ν2αδΓαν2−ν1ατJν1+ν2/2ν1ν2−,δα,2ν1ν2λf°g1ν2.
Similarly,(34)I2=∫012ν1ν2ν2−ν1δγλν2−ν1α,ττν1+2−τν22f′2ν1ν2τν1+2−τν2dτ=δγλν2−ν1α,1f2ν1ν2ν1+ν2−∫01τα−1e−λν2−ν1τδδf2ν1ν2τν1+2−τν2dτ=δγλν2−ν1α,1f2ν1ν2ν1+ν2−2ν1ν2αν2−ν1α∫ν1+ν2/2ν1ν21/ν11ν1−xα−1e−2ν1ν2λ1/ν1−xδ/δf1xdx=δγλα,1f°g2ν1ν2ν1+ν2−2ν1ν2αδΓαν2−ν1ατJν1+ν2/2ν1ν2+,δα,2ν1ν2λf°g1ν1.
Substituting the values ofI1 and I2 in (32) and using Remark 1 (1), we get the required result.


### 2.3. Some More New Hermite–Hadamard’s Type of Integral Inequalities

Now using auxiliary results obtained in the previous section, we derive some new Hermite–Hadamard’s type of integral inequalities.Theorem 2.
Letf:ν1,ν2⟶ℝ be L1 function. If f′q,q≥1, is harmonically convex on ν1,ν2 with ν1<ν2, then(35)fν1+fν22−ν1ν2ν2−ν1αδΓδα2δγλν2−ν1α,1J1/ν1−,δα,λν1ν2τf°g1ν2+τJ1/ν2+,δα,λν1ν2τf°g1ν1≤ν1ν2ν2−ν12δγλν2−ν1α,1∫01δγλν2−ν1α,1−τ−δγλν2−ν1α,τ1−τν1+τν22pdτ1/pf′ν1q+f′ν2q21/q.Proof.
Using Lemma1, the harmonic convexity of f′q, and Hölder’s inequality, we have(36)fν1+fν22−ν1ν2ν2−ν1αδΓδα2δγλν2−ν1α,1J1/ν1−,δα,λν1ν2τf°g1ν2+J1/ν2+,δα,λν1ν2τf°g1ν1≤ν1ν2ν2−ν12δγλν2−ν1α,1∫01δγλν2−ν1α,1−τ−δγλν2−ν1α,τ1−τν1+τν22f′ν1ν21−τν1+τν2dτ≤ν1ν2ν2−ν12δγλν2−ν1α,1∫01δγλν2−ν1α,1−τ−δγλν2−ν1α,τ1−τν1+τν22pdτ1/p×∫01f′ν1ν21−τν1+τν2qdτ1/q≤ν1ν2ν2−ν12δγλν2−ν1α,1∫01δγλν2−ν1α,1−τ−δγλν2−ν1α,τ1−τν1+τν22pdτ1/p×∫01τf′ν1q+1−τf′ν2qdτ1/q=ν1ν2ν2−ν12δγλν2−ν1α,1∫01δγλν2−ν1α,1−τ−δγλν2−ν1α,τ1−τν1+τν22pdτ1/pf′ν1q+f′ν2q21/q.
This completes our proof.Theorem 3.
Letf:ν1,ν2⟶ℝ be an harmonically convex function on ν1,ν2 with ν1<ν2; then, we have(37)f2ν1ν2ν1+ν2≤ν1ν2ν2−ν1α2α−1δΓδαδγλα,ν2−ν1Jν1+ν2/2ν1ν2−,δα,2λν1ν2τf°g1ν2+Jν1+ν2/2ν1ν2+,δα,2λν1ν2τf°g1ν1≤fν1+fν22,for all α>0,λ≥1.Proof.
Sincef is an harmonically convex function, then we have(38)f2xyx+y≤fx+fy2.
This implies(39)2f2ν1ν2ν1+ν2≤f2ν1ν22−τν1+τν2+f2ν1ν2τν1+2−τν2.
Multiplying both sides of the above inequality byτα−1e−λν2−ν1τδ/δ and integrating with respect to τ on 0,1, we have(40)2f2ν1ν2ν1+ν2∫01τα−1e−λν2−ν1τδ/δdτ≤∫01τα−1e−λν2−ν1τδ/δdτ,f2ν1ν22−τν1+τν2dτ+∫01τα−1e−λν2−ν1τδ/δf2ν1ν2τν1+2−τν2dτ.
Thus, we have(41)2δγλν2−ν1α,12ν1ν2ν1+ν2≤ν1ν2ν2−ν1α∫1/ν2ν1+ν2/2ν1ν2x−1ν2α−1e−λ2ν1ν2x−1/ν2δδf1xdx+ν1ν2ν2−ν1α∫ν1+ν2/2ν1ν21/ν11ν1−xα−1e−2λν1ν21/ν1−xδδf1xdx.
Using Remark1 (1), we have(42)f2ν1ν2ν1+ν2≤2α−1ν1ν2αδΓδαδγλα,ν2−ν1Jν1+ν2/2ν1ν2−,δα,2λν1ν2τf°g1ν2+Jν1+ν2/2ν1ν2+,δα,2λν1ν2τf°g1ν1.
Also,(43)f2ν1ν22−τν1+τν2+f2ν1ν2τν1+2−τν2≤fν1+fν2.
Multiplying the above inequality byτα−1e−λν2−ν1τδ/δ and integrating with respect to τ on 0,1, we have(44)∫01τα−1e−λν2−ν1τδ/δf2ν1ν22−τν1+τν2dτ+∫01τα−1e−λν2−ν1τδ/δf2ν1ν2τν1+2−τν2dτ≤fν1+fν2∫01τα−1e−λν2−ν1τδ/δdτ.
Using Remark1 (1), we have(45)2ν1ν2ν2−ν1α∫1/ν2ν1+ν2/2ν1ν2x−1ν2α−1e−λ2ν1ν2x−1/ν2δδf1xdx+2ν1ν2ν2−ν1α∫ν1+ν2/2ν1ν21/ν11ν1−xα−1e−2λν1ν21/ν1−xδδf1xdx≤δγλα,ν2−ν1ν2−ν1αfν1+fν2.
Thus, we have(46)ν1ν2ν2−ν1α2α−1δΓδαδγλα,ν2−ν1Jν1+ν2/2ν1ν2−,δα,2λν1ν2τf1ν2+Jν1+ν2/2ν1ν2+,δα,2λν1ν2τf°g1ν1≤fν1+fν22.
Combining (42) and (46), we get the required inequality (37).Theorem 4.
Iff:ν1,ν2⟶ℝ is L1 function, using the harmonic convexity property of f′q,q≥1, on ν1,ν2 with ν1<ν2, we have(47)2α−1ν1ν2δΓδααδγλα,ν2−ν1Jν1+ν2/2ν1ν2,δα,2λν1ν2τf°g1ν2+Jν1+ν2/2ν1ν2+,δα,2λν1ν2τf°g1ν1−f2ν1ν2ν1+ν2≤ν2−ν1α2δγλα,ν2−ν1ν1ν2ν2−ν1lnay+2−yν22−yν1+yb−4ν12ν21−y2−yν1+ybya+2−yν2δγλν2−ν1α,1f′ν1,+ν1ν2ν2−ν1ln2−yν1+ybya+2−yν2−4ν1ν221−y2−yν1+ybya+2−yν2δγλν2−ν1α,1f′ν2.Proof.
Using Lemma2 and the harmonic convexity property of f′, we have(48)2α−1ν1ν2αδΓδαδγλα,ν2−ν1Jν1+ν2/2ν1ν2,δα,2λν1ν2τf°g1ν2+Jν1+ν2/2ν1ν2+,δα,2λν1ν2τf°g1ν1−f2ν1ν2ν1+ν2≤ν2−ν1α2δγλα,ν2−ν1∫012ν1ν2ν2−ν1δγλν2−ν1α,τ2−τν1+τν22f′2ν1ν22−τν1+τν2dτ+∫012ν1ν2ν2−ν1δγλν2−ν1α,ττν1+2−τν22f′2ν1ν2τν1+2−τν2dτ≤ν2−ν1α2δγλα,ν2−ν1∫012ν1ν2ν2−ν1δγλν2−ν1α,τ2−τν1+τν22τ2f′ν1+2−τ2f′ν2dτ+∫012ν1ν2ν2−ν1δγλν2−ν1α,ττν1+2−τν222−τ2f′ν1+τ2f′ν2dτ.
Changing the order of integration, we have(49)=ν2−ν1α2δγλα,ν2−ν1ν1ν2ν2−ν1lnay+2−yν22−yν1+yb−4ν12ν21−y2−yν1+ybya+2−yν2δγλν2−ν1α,1f′ν1,+ν1ν2ν2−ν1ln2−yν1+ybya+2−yν2−4ν1ν221−y2−yν1+ybya+2−yν2δγλν2−ν1α,1f′ν2.
This completes our proof.Theorem 5.
Letf:ν1,ν2⟶ℝ be L1 function. If f′q,q≥1, is harmonically convex on ν1,ν2 with ν1<ν2, then we have(50)2α−1ν1ν2αδΓδαδγλα,ν2−ν1Jν1+ν2/2ν1ν2,δα,2λν1ν2τf°g1ν2+Jν1+ν2/2ν1ν2+,δα,2λν1ν2τf°g1ν1−f2ν1ν2ν1+ν2≤ν2−ν1α2δγλα,ν2−ν1M1α,λ1−1/qM2α,λf′ν1q+M3α,λf′ν2q1/q+M4α,λ1−1/qM5α,λf′ν1q+M6α,λf′ν2q1/q,where(51)M1α,λ=δγλν2−ν1α,12ν1ν22−yν1+yb−2ν1ν2ν1+ν2,M2α,λ=δγλν2−ν1α,12ν12ν2ν2−ν11ν1+ν2−12−yν1+yb+ν1ν2ν2−ν1lnν1+ν22−yν1+yb,M3α,λ=δγλν2−ν1α,12ν1ν22ν2−ν112−yν1+yb−1ν1+ν2−ν1ν2ν2−ν1lnν1+ν22−yν1+yb,M4α,λ=γλν2−ν1α,1δ2ν1ν2ν1+ν2−2ν1ν2ya+2−yν2,M5α,λ=γλν2−ν1α,1δ2ν12ν2ν2−ν11ya+2−yν2−1ν1+ν2−ν1ν2ν2−ν1lnν1+ν2ya+2−yν2,M6α,λ=δγλν2−ν1α,12ν1ν22ν2ν2−ν11ν1+ν2−1ya+2−yν2+ν1ν2ν2−ν1lnν1+ν2ya+2−yaν2.Proof.
Using Lemma2, the harmonic convexity property of f′, and power mean inequality, we have(52)2α−1ν1ν2αδΓδαδγλα,ν2−ν1Jν1+ν2/2ν1ν2,δα,2λν1ν2τf°g1ν2+Jν1+ν2/2ν1ν2+,δα,2λν1ν2τf°g1ν1−f2ν1ν2ν1+ν2≤ν2−ν1α2δγλα,ν2−ν1∫012ν1ν2ν2−ν1δγλν2−ν1α,τ2−τν1+τν22f′2ν1ν22−τν1+τν2dτ+∫012ν1ν2ν2−ν1δγλν2−ν1α,ττν1+2−τν22f′2ν1ν2τν1+2−τν2dτ≤ν2−ν1α2δγλα,ν2−ν1∫012ν1ν2ν2−ν1δγλν2−ν1α,τ2−τν1+τν22dτ1−1/q∫012ν1ν2ν2−ν1δγλν2−ν1α,τ2−τν1+τν22f′2ν1ν22−τν1+τν2qdτ1/q+∫012ν1ν2ν2−ν1δγλν2−ν1α,ττν1+2−τν22dτ1−1/q∫012ν1ν2ν2−ν1δγλν2−ν1α,ττν1+2−τν22f′2ν1ν2τν1+2−τν2qdτ1/q≤ν2−ν1α2δγλα,ν2−ν1∫012ν1ν2ν2−ν1δγλν2−ν1α,τ2−τν1+τν22dτ1−1/q∫012ν1ν2ν2−ν1δγλν2−ν1α,τ2−τν1+τν22τ2f′ν1q+2−τ2f′ν2qdτ1/q+∫012ν1ν2ν2−ν1δγλν2−ν1α,ττν1+2−τν22dτ1−1/q∫012ν1ν2ν2−ν1δγλν2−ν1α,ττν1+2−τν222−τ2f′ν1q+τ2f′ν2qdτ1/q.
Changing the order of integration and after simple calculation, we get the required result.Corollary 1.
All the assumption of Theorem2 are true; for α=1=k and λ=0, we have(53)fν1+fν22−1ν2−ν1∫ν1ν2fxx2dx≤ν1ν2ν2−ν12×ν1−2p2p+1F122p,1,p+2;ν1−ν22ν1+F122p,p+1,p+2;ν1−ν2ν1+ν21/pf′ν1q+f′ν2q21/q.


## 2.1. A New Version of Hermite–Hadamard’s Inequality

We now derive a newδ-tempered fractional Hermite–Hadamard inequality via harmonically convex function.Theorem 1.
Letf:ν1,ν2⟶ℝ be an harmonically convex function on ν1,ν2 with ν1<ν2; then,(15)f2ν1ν2ν1+ν2≤ν1ν2ν2−ν1αδΓδα2δγλν2−ν1α,1J1/ν1−,δα,λν1ν2τf∘g1ν2+J1/ν2+,δα,λν1ν2τf°g1ν1≤fν1+fν22,where gx=1/x and for all α>0andλ≥1.Proof.
Sincef is a harmonic convex function, then(16)f2xyx+y≤fx+fy2.
This implies(17)2f2ν1ν2ν1+ν2≤fν1ν21−τν1+τν2+fν1ν2τν1+1−τν2.
Multiplying both sides of the above inequality byτα−1e−λν2−ν1τδ/δ and integrating with respect to τ on 0,1, we have(18)2f2ν1ν2ν1+ν2∫01τα−1e−λν2−ν1τδ/δdτ≤∫01τα−1e−λν2−ν1τδ/δfν1ν21−τν1+τν2dτ+∫01τα−1e−λν2−ν1τδ/δfν1ν2τν1+1−τν2dτ.
This implies(19)2δγλν2−ν1α,12ν1ν2ν1+ν2≤ν1ν2αν2−ν1α∫1/ν21/ν1x−1ν2α−1e−λν1ν2x−1/ν2δ/δf1xdx+ν1ν2αν2−ν1α∫1/ν21/ν11ν1−xα−1e−λν1ν21/ν1−xδ/δf1xdx.
Thus, we have(20)f2ν1ν2ν1+ν2≤ν1ν2αδΓδα2ν2−ν1αδγλν2−ν1α,1J1/ν1−,δα,λν1ν2τf°g1ν2+J1/ν2+,δα,λν1ν2τf°g1ν1.
Also,(21)fν1ν21−τν1+τν2≤τfν1+1−τν2,(22)fν1ν2τν1+1−τν2≤1−τfν1+τfν2.
Adding (21) and (22), we have(23)fν1ν21−τν1+τν2+fν1ν2τν1+1−τν2≤fν1+fν2.
Multiplying the above inequality byτα−1e−λν2−ν1τδ/δ and integrating with respect to τ on 0,1, we have(24)∫01τα−1e−λν2−ν1τδ/δfν1ν21−τν1+τν2dτ+∫01τα−1e−λν2−ν1τδ/δfν1ν2τν1+1−τν2dτ≤fν1+fν2∫01τα−1e−λν2−ν1τδ/δdτ.
This implies(25)ν1ν2ν2−ν1α∫1/ν21/ν1x−1ν2α−1e−λν1ν2x−1/ν2δδf1xdx+ν1ν2ν2−ν1α∫1/ν21/ν11ν1−xα−1e−λν1ν21/ν1−xδδf1xdx≤δγλν2−ν1α,1fν1+fν2.
This implies(26)ν1ν2ν2−ν1αδΓδα2δγλν2−ν1α,1J1/ν1−,δα,λν1ν2τf°g1ν2+J1/ν2+,δα,λν1ν2τf°g1ν1≤fν1+fν22.
Combining (20) and (26), we get the required inequality (15)Remark 2.
(1)
Forα=1=k and λ=0 in (15), we obtain the classical Hermite–Hadamard’s type inequality for the notion of harmonic convexity(2)
Fork=1 and λ=0 in 2.1, we obtain the fractional version of Hermite–Hadamard’s type inequality for the notion of harmonic convexity


## 2.2. New Auxiliary Results

We now derive two new auxiliary results. These results will play a significant role in the development of our next results.Lemma 1.
Letf:ν1,ν2⟶ℝ be L1 function; then,(27)fν1+fν22−ν1ν2ν2−ν1αδΓδα2δγλν2−ν1α,1J1/ν1−,δα,λν1ν2τf°g1ν2+J1/ν2+,δα,λν1ν2τf°g1ν1=ν1ν2ν2−ν12δγλν2−ν1α,1∫01δγλν2−ν1α,1−τ−δγλν2−ν1α,τ1−τν1+τν22f′ν1ν21−τν1+τν2dτ,for all α>0,λ≥0.Proof.
Consider(28)I=ν1ν2ν2−ν12δγλν2−ν1α,1∫01δγλν2−ν1α,1−τ+δγλν2−ν1α,τ1−τν1+τν22f′ν1ν21−τν1+τν2dτ=ν1ν2ν2−ν12δγλν2−ν1α,1I1−I2.
Now,(29)I1=∫01δγλν2−ν1α,1−τ1−τν1+τν22f′ν1ν21−τν1+τν2dτ=δγλν2−ν1α,1fν2−∫011−τα−1eλν2−ν11−τδδfν1ν21−τν1+τν2dτ=δγλν2−ν1α,1fν2−ν1ν2ν2−ν1α∫1/ν21/ν11ν1−xα−1eλν2−ν11/ν1−xδδf1xdx=δγλν2−ν1α,1fν2−ν1ν2ν2−ν1αδΓδατJ1/ν2+,δα,λν1ν2f°g1ν1.
Similarly,(30)I2=∫01δγλν2−ν1α,τ1−τν1+τν22f′ν1ν21−τν1+τν2dτ=−δγλν2−ν1α,1fν1+∫01τα−1eλν2−ν1τδ/δfν1ν21−τν1+τν2dτ=−δγλν2−ν1α,1fν2+ν1ν2ν2−ν1α∫1/ν21/ν1x−1ν2α−1eλν2−ν1x−1/ν2δ/δf1xdx=−δγλν2−ν1α,1fν2+ν1ν2ν2−ν1αδΓδατJ1/ν1−,δα,λν1ν2f°g1ν2.
Substituting the values ofI1 and I2 in (28), we get the required result.Lemma 2.
Letf:ν1,ν2⟶ℝ be L1 function; then,(31)2α−1ν1ν2αδΓδαδγλα,ν2−ν1Jν1+ν2/2ν1ν2−,δα,2λν1ν2τf°g1ν2+Jν1+ν2/2ν1ν2+,δα,2λν1ν2τf°g1ν1−f2ν1ν2ν1+ν2=ν2−ν1α2δγλα,ν2−ν1∫012ν1ν2ν2−ν1δγλν2−ν1α,τ2−τν1+τν22f′2ν1ν22−τν1+τν2dτ−∫012ν1ν2ν2−ν1δγλν2−ν1α,ττν1+2−τν22f′2ν1ν2τν1+2−τν2dτ,where α>0,λ≥1.Proof.
Consider(32)I=ν2−ν1α2δγλα,ν2−ν1∫012ν1ν2ν2−ν1δγλν2−ν1α,τ2−τν1+τν22f′2ν1ν22−τν1+τν2dτ−∫012ν1ν2ν2−ν1δγλν2−ν1α,ττν1+2−τν22f′2ν1ν2τν1+2−τν2dτ=ν2−ν1α2δγλα,ν2−ν1I1−I2.
Using Remark1 (1), we have(33)I1=∫012ν1ν2ν2−ν1δγλν2−ν1α,τ2−τν1+τν22f′2ν1ν22−τν1+τν2dτ=−δγλν2−ν1α,1f2ν1ν2ν1+ν2+∫01τα−1e−λν2−ν1τδ/δf2ν1ν22−τν1+τν2dτ=δγλν2−ν1α,1f2ν1ν2ν1+ν2+2ν1ν2αν2−ν1α∫1/ν2ν1+ν2/2ν1ν2x−1ν2α−1e−2ν1ν2λx−1/ν2δ/δf1xdx=−δγλα,1f2ν1ν2ν1+ν2+2ν1ν2αδΓαν2−ν1ατJν1+ν2/2ν1ν2−,δα,2ν1ν2λf°g1ν2.
Similarly,(34)I2=∫012ν1ν2ν2−ν1δγλν2−ν1α,ττν1+2−τν22f′2ν1ν2τν1+2−τν2dτ=δγλν2−ν1α,1f2ν1ν2ν1+ν2−∫01τα−1e−λν2−ν1τδδf2ν1ν2τν1+2−τν2dτ=δγλν2−ν1α,1f2ν1ν2ν1+ν2−2ν1ν2αν2−ν1α∫ν1+ν2/2ν1ν21/ν11ν1−xα−1e−2ν1ν2λ1/ν1−xδ/δf1xdx=δγλα,1f°g2ν1ν2ν1+ν2−2ν1ν2αδΓαν2−ν1ατJν1+ν2/2ν1ν2+,δα,2ν1ν2λf°g1ν1.
Substituting the values ofI1 and I2 in (32) and using Remark 1 (1), we get the required result.


## 2.3. Some More New Hermite–Hadamard’s Type of Integral Inequalities

Now using auxiliary results obtained in the previous section, we derive some new Hermite–Hadamard’s type of integral inequalities.Theorem 2.
Letf:ν1,ν2⟶ℝ be L1 function. If f′q,q≥1, is harmonically convex on ν1,ν2 with ν1<ν2, then(35)fν1+fν22−ν1ν2ν2−ν1αδΓδα2δγλν2−ν1α,1J1/ν1−,δα,λν1ν2τf°g1ν2+τJ1/ν2+,δα,λν1ν2τf°g1ν1≤ν1ν2ν2−ν12δγλν2−ν1α,1∫01δγλν2−ν1α,1−τ−δγλν2−ν1α,τ1−τν1+τν22pdτ1/pf′ν1q+f′ν2q21/q.Proof.
Using Lemma1, the harmonic convexity of f′q, and Hölder’s inequality, we have(36)fν1+fν22−ν1ν2ν2−ν1αδΓδα2δγλν2−ν1α,1J1/ν1−,δα,λν1ν2τf°g1ν2+J1/ν2+,δα,λν1ν2τf°g1ν1≤ν1ν2ν2−ν12δγλν2−ν1α,1∫01δγλν2−ν1α,1−τ−δγλν2−ν1α,τ1−τν1+τν22f′ν1ν21−τν1+τν2dτ≤ν1ν2ν2−ν12δγλν2−ν1α,1∫01δγλν2−ν1α,1−τ−δγλν2−ν1α,τ1−τν1+τν22pdτ1/p×∫01f′ν1ν21−τν1+τν2qdτ1/q≤ν1ν2ν2−ν12δγλν2−ν1α,1∫01δγλν2−ν1α,1−τ−δγλν2−ν1α,τ1−τν1+τν22pdτ1/p×∫01τf′ν1q+1−τf′ν2qdτ1/q=ν1ν2ν2−ν12δγλν2−ν1α,1∫01δγλν2−ν1α,1−τ−δγλν2−ν1α,τ1−τν1+τν22pdτ1/pf′ν1q+f′ν2q21/q.
This completes our proof.Theorem 3.
Letf:ν1,ν2⟶ℝ be an harmonically convex function on ν1,ν2 with ν1<ν2; then, we have(37)f2ν1ν2ν1+ν2≤ν1ν2ν2−ν1α2α−1δΓδαδγλα,ν2−ν1Jν1+ν2/2ν1ν2−,δα,2λν1ν2τf°g1ν2+Jν1+ν2/2ν1ν2+,δα,2λν1ν2τf°g1ν1≤fν1+fν22,for all α>0,λ≥1.Proof.
Sincef is an harmonically convex function, then we have(38)f2xyx+y≤fx+fy2.
This implies(39)2f2ν1ν2ν1+ν2≤f2ν1ν22−τν1+τν2+f2ν1ν2τν1+2−τν2.
Multiplying both sides of the above inequality byτα−1e−λν2−ν1τδ/δ and integrating with respect to τ on 0,1, we have(40)2f2ν1ν2ν1+ν2∫01τα−1e−λν2−ν1τδ/δdτ≤∫01τα−1e−λν2−ν1τδ/δdτ,f2ν1ν22−τν1+τν2dτ+∫01τα−1e−λν2−ν1τδ/δf2ν1ν2τν1+2−τν2dτ.
Thus, we have(41)2δγλν2−ν1α,12ν1ν2ν1+ν2≤ν1ν2ν2−ν1α∫1/ν2ν1+ν2/2ν1ν2x−1ν2α−1e−λ2ν1ν2x−1/ν2δδf1xdx+ν1ν2ν2−ν1α∫ν1+ν2/2ν1ν21/ν11ν1−xα−1e−2λν1ν21/ν1−xδδf1xdx.
Using Remark1 (1), we have(42)f2ν1ν2ν1+ν2≤2α−1ν1ν2αδΓδαδγλα,ν2−ν1Jν1+ν2/2ν1ν2−,δα,2λν1ν2τf°g1ν2+Jν1+ν2/2ν1ν2+,δα,2λν1ν2τf°g1ν1.
Also,(43)f2ν1ν22−τν1+τν2+f2ν1ν2τν1+2−τν2≤fν1+fν2.
Multiplying the above inequality byτα−1e−λν2−ν1τδ/δ and integrating with respect to τ on 0,1, we have(44)∫01τα−1e−λν2−ν1τδ/δf2ν1ν22−τν1+τν2dτ+∫01τα−1e−λν2−ν1τδ/δf2ν1ν2τν1+2−τν2dτ≤fν1+fν2∫01τα−1e−λν2−ν1τδ/δdτ.
Using Remark1 (1), we have(45)2ν1ν2ν2−ν1α∫1/ν2ν1+ν2/2ν1ν2x−1ν2α−1e−λ2ν1ν2x−1/ν2δδf1xdx+2ν1ν2ν2−ν1α∫ν1+ν2/2ν1ν21/ν11ν1−xα−1e−2λν1ν21/ν1−xδδf1xdx≤δγλα,ν2−ν1ν2−ν1αfν1+fν2.
Thus, we have(46)ν1ν2ν2−ν1α2α−1δΓδαδγλα,ν2−ν1Jν1+ν2/2ν1ν2−,δα,2λν1ν2τf1ν2+Jν1+ν2/2ν1ν2+,δα,2λν1ν2τf°g1ν1≤fν1+fν22.
Combining (42) and (46), we get the required inequality (37).Theorem 4.
Iff:ν1,ν2⟶ℝ is L1 function, using the harmonic convexity property of f′q,q≥1, on ν1,ν2 with ν1<ν2, we have(47)2α−1ν1ν2δΓδααδγλα,ν2−ν1Jν1+ν2/2ν1ν2,δα,2λν1ν2τf°g1ν2+Jν1+ν2/2ν1ν2+,δα,2λν1ν2τf°g1ν1−f2ν1ν2ν1+ν2≤ν2−ν1α2δγλα,ν2−ν1ν1ν2ν2−ν1lnay+2−yν22−yν1+yb−4ν12ν21−y2−yν1+ybya+2−yν2δγλν2−ν1α,1f′ν1,+ν1ν2ν2−ν1ln2−yν1+ybya+2−yν2−4ν1ν221−y2−yν1+ybya+2−yν2δγλν2−ν1α,1f′ν2.Proof.
Using Lemma2 and the harmonic convexity property of f′, we have(48)2α−1ν1ν2αδΓδαδγλα,ν2−ν1Jν1+ν2/2ν1ν2,δα,2λν1ν2τf°g1ν2+Jν1+ν2/2ν1ν2+,δα,2λν1ν2τf°g1ν1−f2ν1ν2ν1+ν2≤ν2−ν1α2δγλα,ν2−ν1∫012ν1ν2ν2−ν1δγλν2−ν1α,τ2−τν1+τν22f′2ν1ν22−τν1+τν2dτ+∫012ν1ν2ν2−ν1δγλν2−ν1α,ττν1+2−τν22f′2ν1ν2τν1+2−τν2dτ≤ν2−ν1α2δγλα,ν2−ν1∫012ν1ν2ν2−ν1δγλν2−ν1α,τ2−τν1+τν22τ2f′ν1+2−τ2f′ν2dτ+∫012ν1ν2ν2−ν1δγλν2−ν1α,ττν1+2−τν222−τ2f′ν1+τ2f′ν2dτ.
Changing the order of integration, we have(49)=ν2−ν1α2δγλα,ν2−ν1ν1ν2ν2−ν1lnay+2−yν22−yν1+yb−4ν12ν21−y2−yν1+ybya+2−yν2δγλν2−ν1α,1f′ν1,+ν1ν2ν2−ν1ln2−yν1+ybya+2−yν2−4ν1ν221−y2−yν1+ybya+2−yν2δγλν2−ν1α,1f′ν2.
This completes our proof.Theorem 5.
Letf:ν1,ν2⟶ℝ be L1 function. If f′q,q≥1, is harmonically convex on ν1,ν2 with ν1<ν2, then we have(50)2α−1ν1ν2αδΓδαδγλα,ν2−ν1Jν1+ν2/2ν1ν2,δα,2λν1ν2τf°g1ν2+Jν1+ν2/2ν1ν2+,δα,2λν1ν2τf°g1ν1−f2ν1ν2ν1+ν2≤ν2−ν1α2δγλα,ν2−ν1M1α,λ1−1/qM2α,λf′ν1q+M3α,λf′ν2q1/q+M4α,λ1−1/qM5α,λf′ν1q+M6α,λf′ν2q1/q,where(51)M1α,λ=δγλν2−ν1α,12ν1ν22−yν1+yb−2ν1ν2ν1+ν2,M2α,λ=δγλν2−ν1α,12ν12ν2ν2−ν11ν1+ν2−12−yν1+yb+ν1ν2ν2−ν1lnν1+ν22−yν1+yb,M3α,λ=δγλν2−ν1α,12ν1ν22ν2−ν112−yν1+yb−1ν1+ν2−ν1ν2ν2−ν1lnν1+ν22−yν1+yb,M4α,λ=γλν2−ν1α,1δ2ν1ν2ν1+ν2−2ν1ν2ya+2−yν2,M5α,λ=γλν2−ν1α,1δ2ν12ν2ν2−ν11ya+2−yν2−1ν1+ν2−ν1ν2ν2−ν1lnν1+ν2ya+2−yν2,M6α,λ=δγλν2−ν1α,12ν1ν22ν2ν2−ν11ν1+ν2−1ya+2−yν2+ν1ν2ν2−ν1lnν1+ν2ya+2−yaν2.Proof.
Using Lemma2, the harmonic convexity property of f′, and power mean inequality, we have(52)2α−1ν1ν2αδΓδαδγλα,ν2−ν1Jν1+ν2/2ν1ν2,δα,2λν1ν2τf°g1ν2+Jν1+ν2/2ν1ν2+,δα,2λν1ν2τf°g1ν1−f2ν1ν2ν1+ν2≤ν2−ν1α2δγλα,ν2−ν1∫012ν1ν2ν2−ν1δγλν2−ν1α,τ2−τν1+τν22f′2ν1ν22−τν1+τν2dτ+∫012ν1ν2ν2−ν1δγλν2−ν1α,ττν1+2−τν22f′2ν1ν2τν1+2−τν2dτ≤ν2−ν1α2δγλα,ν2−ν1∫012ν1ν2ν2−ν1δγλν2−ν1α,τ2−τν1+τν22dτ1−1/q∫012ν1ν2ν2−ν1δγλν2−ν1α,τ2−τν1+τν22f′2ν1ν22−τν1+τν2qdτ1/q+∫012ν1ν2ν2−ν1δγλν2−ν1α,ττν1+2−τν22dτ1−1/q∫012ν1ν2ν2−ν1δγλν2−ν1α,ττν1+2−τν22f′2ν1ν2τν1+2−τν2qdτ1/q≤ν2−ν1α2δγλα,ν2−ν1∫012ν1ν2ν2−ν1δγλν2−ν1α,τ2−τν1+τν22dτ1−1/q∫012ν1ν2ν2−ν1δγλν2−ν1α,τ2−τν1+τν22τ2f′ν1q+2−τ2f′ν2qdτ1/q+∫012ν1ν2ν2−ν1δγλν2−ν1α,ττν1+2−τν22dτ1−1/q∫012ν1ν2ν2−ν1δγλν2−ν1α,ττν1+2−τν222−τ2f′ν1q+τ2f′ν2qdτ1/q.
Changing the order of integration and after simple calculation, we get the required result.Corollary 1.
All the assumption of Theorem2 are true; for α=1=k and λ=0, we have(53)fν1+fν22−1ν2−ν1∫ν1ν2fxx2dx≤ν1ν2ν2−ν12×ν1−2p2p+1F122p,1,p+2;ν1−ν22ν1+F122p,p+1,p+2;ν1−ν2ν1+ν21/pf′ν1q+f′ν2q21/q.


## 3. Applications

In this section, we investigate some applications to special means and special functions.


### 3.1. Applications to Special Means

We now recall some special means for two nonnegative numbers.(1)
The arithmetic mean:Aν1,ν2:=ν1+ν2/2(2)
The geometric mean:Gν1,ν2:=ν1ν2(3)
The harmonic mean:Hν1,ν2:=2ν1ν2/ν1+ν2(4)
The logarithmic mean:Lν1,ν2:=ν2−ν1/lnν2−lnν1(5)
Thep-logarithmic mean: Lpν1,ν2:=ν2p+1−ν1p+1/p+1ν2−ν11/p,p∈ℝ−−1,0Proposition 1.
Let0<ν1<ν2; then, we have(54)Aν1,ν2−L−1ν1,ν2≤G2ν1,ν2ν2−ν12×ν1−2p2p+12F12p,1,p+2;ν1−ν22ν1+2F12p,p+1,p+2;ν1−ν2ν1+ν21/p.Proof.
The proof is simply obtained by usingfx=x in Corollary 1.Proposition 2.
Let0<ν1<ν2; then, we have(55)Aν1n+2,ν2n+2−G2ν1,ν2Lppν1,ν2≤G2ν1,ν2n+2ν2−ν12×ν1−2p2p+1F122p,1,p+2;ν1−ν22ν1+2F12p,p+1,p+2;ν1−ν2ν1+ν21/pA1/qν1n+1q,ν2n+1q.Proof.
The proof is straightforward, by usingfx=xn+2 in Corollary 1.


### 3.2. Bessel Functions

Consider the functionJv:ℝ⟶1,∞, defined by Jvz=2pΓp+1z−vIvz,z∈ℝ, and we recall modified Bessel functions of first kind Iv and its first derivative formula, respectively:(56)Jv=∑n≥0z/2v+2nn!Γv+n+1,Jv′=z2v+1Jv+1z.Proposition 3.
Forv>−1,ν1,ν2∈ℝ with 0<ν1<ν2, then we have(57)Jvν2−Jvν1ν2−ν1≤ν13Jv+1ν1+ν23Jv+1ν24v+1.For a particular case, we have(58)coshν2−coshν1ν2−ν1≤ν12sinhν1+ν22sinhν22.Proof.
The proof is obtained by takingx⟶x2Jp′x and using it in Corollary 1. For the second inequality, we use relation J−1/2z=coshz and J1/2z=sinz/z.


### 3.3.q-Digamma Functions

Let0<q<1. Then, q-digamma function ψq is defined as(59)ψqx=−ln1−q+lnq∑u=0∞qu−q1−qu+z=−ln1−q+lnq∑u=1∞quz1−quz.Forq>1 and z>1, the q-digamma function ψq is defined as(60)ψqx=−ln1−q+lnqz−12−∑u=0∞q−u+q1−q−u+z=−ln1−q+lnqz−12−∑u=1∞q−uq1−q−uz.Proposition 4.
Forν1,ν2∈ℝ, such that ν1<ν2,(61)H2ν1,ν2ψq′2ν1ν2ν1+ν2≤ψqν2−ψqν1ν2−ν1≤ν12ψq′ν1+ν22ψq′ν12.Proof.
The proof is straightforward, by consideringfx=x2ψq′x and using Remark 2.Proposition 5.
Forν1,ν2∈ℝ, such that ν1<ν2,(62)ν12ψq′ν1+ν22ψq′ν12−ψqν2−ψqν1ν2−ν1≤ν1ν2ν2−ν12×ν1−2p2p+1F122p,1,p+2;ν1−ν22ν1+F122p,p+1,p+2;ν1−ν2ν1+ν21/p2ν1+ψq′ν1+ν12ψq″ν1q+2ν2+ψq′ν2+ν22ψq″ν2q1/q.Proof.
The proof is straightforward, by consideringfx=x2ψq′x and using Corollary 1.


## 3.1. Applications to Special Means

We now recall some special means for two nonnegative numbers.(1)
The arithmetic mean:Aν1,ν2:=ν1+ν2/2(2)
The geometric mean:Gν1,ν2:=ν1ν2(3)
The harmonic mean:Hν1,ν2:=2ν1ν2/ν1+ν2(4)
The logarithmic mean:Lν1,ν2:=ν2−ν1/lnν2−lnν1(5)
Thep-logarithmic mean: Lpν1,ν2:=ν2p+1−ν1p+1/p+1ν2−ν11/p,p∈ℝ−−1,0Proposition 1.
Let0<ν1<ν2; then, we have(54)Aν1,ν2−L−1ν1,ν2≤G2ν1,ν2ν2−ν12×ν1−2p2p+12F12p,1,p+2;ν1−ν22ν1+2F12p,p+1,p+2;ν1−ν2ν1+ν21/p.Proof.
The proof is simply obtained by usingfx=x in Corollary 1.Proposition 2.
Let0<ν1<ν2; then, we have(55)Aν1n+2,ν2n+2−G2ν1,ν2Lppν1,ν2≤G2ν1,ν2n+2ν2−ν12×ν1−2p2p+1F122p,1,p+2;ν1−ν22ν1+2F12p,p+1,p+2;ν1−ν2ν1+ν21/pA1/qν1n+1q,ν2n+1q.Proof.
The proof is straightforward, by usingfx=xn+2 in Corollary 1.


## 3.2. Bessel Functions

Consider the functionJv:ℝ⟶1,∞, defined by Jvz=2pΓp+1z−vIvz,z∈ℝ, and we recall modified Bessel functions of first kind Iv and its first derivative formula, respectively:(56)Jv=∑n≥0z/2v+2nn!Γv+n+1,Jv′=z2v+1Jv+1z.Proposition 3.
Forv>−1,ν1,ν2∈ℝ with 0<ν1<ν2, then we have(57)Jvν2−Jvν1ν2−ν1≤ν13Jv+1ν1+ν23Jv+1ν24v+1.For a particular case, we have(58)coshν2−coshν1ν2−ν1≤ν12sinhν1+ν22sinhν22.Proof.
The proof is obtained by takingx⟶x2Jp′x and using it in Corollary 1. For the second inequality, we use relation J−1/2z=coshz and J1/2z=sinz/z.


## 3.3.q-Digamma Functions

Let0<q<1. Then, q-digamma function ψq is defined as(59)ψqx=−ln1−q+lnq∑u=0∞qu−q1−qu+z=−ln1−q+lnq∑u=1∞quz1−quz.Forq>1 and z>1, the q-digamma function ψq is defined as(60)ψqx=−ln1−q+lnqz−12−∑u=0∞q−u+q1−q−u+z=−ln1−q+lnqz−12−∑u=1∞q−uq1−q−uz.Proposition 4.
Forν1,ν2∈ℝ, such that ν1<ν2,(61)H2ν1,ν2ψq′2ν1ν2ν1+ν2≤ψqν2−ψqν1ν2−ν1≤ν12ψq′ν1+ν22ψq′ν12.Proof.
The proof is straightforward, by consideringfx=x2ψq′x and using Remark 2.Proposition 5.
Forν1,ν2∈ℝ, such that ν1<ν2,(62)ν12ψq′ν1+ν22ψq′ν12−ψqν2−ψqν1ν2−ν1≤ν1ν2ν2−ν12×ν1−2p2p+1F122p,1,p+2;ν1−ν22ν1+F122p,p+1,p+2;ν1−ν2ν1+ν21/p2ν1+ψq′ν1+ν12ψq″ν1q+2ν2+ψq′ν2+ν22ψq″ν2q1/q.Proof.
The proof is straightforward, by consideringfx=x2ψq′x and using Corollary 1.



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