Novel Mean-Type Inequalities via Generalized Riemann-Type Fractional Integral for Composite Convex Functions: Some Special Examples

Citation: Mukhtar, M.; Yaqoob, M.; Samraiz, M.; Shabbir, I.; Etemad, S.; De la Sen, M.; Rezapour, S. Novel Mean-Type Inequalities via Generalized Riemann-Type Fractional Integral for Composite Convex Functions: Some Special Examples. Symmetry 2023,15, 479. https://dx.doi.org/10.3390/ sym15020479 Academic Editors: Nicusor Minculete and Shigeru Furuichi Received: 22 January 2023 Revised: 4 February 2023 Accepted: 7 February 2023 Published: 10 February 2023 Copyright: © 2023 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (https:// creativecommons.org/licenses/by/ 4.0/). symmetry S S Article Novel Mean-Type Inequalities via Generalized Riemann-Type Fractional Integral for Composite Convex Functions: Some Special Examples Muzammil Mukhtar1, Muhammad Yaqoob 1, Muhammad Samraiz 2, Iram Shabbir 1, Sina Etemad 3,∗, Manuel De la Sen 4,∗and Shahram Rezapour 3,5,6,∗ 1Department of Mathematics, The Islamia University of Bahawalpur, Bahawalnagar Campus, Bahawalnagar 63100, Pakistan 2Department of Mathematics, University of Sargodha, Sargodha 40100, Pakistan 3Department of Mathematics, Azarbaijan Shahid Madani University, Tabriz 3751-71379, Iran 4Institute of Research and Development of Processes, Department of Electricity and Electronics, Faculty of Science and Technology, University of the Basque Country (UPV/EHU), 48940 Leioa, Bizkaia, Spain 5Department of Mathematics, Kyung Hee University, 26 Kyungheedae-ro, Dongdaemun-gu, Seoul 02447, Republic of Korea 6Department of Medical Research, China Medical University Hospital, China Medical University, Taichung 40402, Taiwan *Correspondence: [email protected] (S.E.); [email protected] (M.D.l.S.); [email protected] (S.R.) Abstract: This study deals with a novel class of mean-type inequalities by employing fractional calculus and convexity theory. The high correlation between symmetry and convexity increases its significance. In this paper, we first establish an identity that is crucial in investigating fractional mean inequalities. Then, we establish the main results involving the error estimation of the Hermite– Hadamard inequality for composite convex functions via a generalized Riemann-type fractional integral. Such results are verified by choosing certain composite functions. These results give wellknown examples in special cases. The main consequences can generalize many known inequalities that exist in other studies. Keywords: mean inequalities; fractional integral; Hölder’s inequality; Minkowski inequality MSC: 26A33; 35J05 1. Introduction Fractional calculus has wide application in mathematics as well as in many other fields of the modern sciences, such as bio-engineering [ 1 – 3 ], biological membranes [ 4 ], medicine [ 5 – 7 ], geophysics [ 8 ], demography [ 9 ], the economy [ 10 ], physics [ 11 ] and also in signal processing. Over the past few decades, scientists have paid attention to the fractional theory of calculus and investigated and modeled many physical real phenomena using fractional calculus theory; for instance, fractional applications in epidemiology [12], the Atangana-Baleanu version of operators in convex analysis [ 13 ], impulsive Langevin equations in fractional settings [ 14 ], the application of fractional operators in inclusion theory [ 15 – 17 ], quantum calculus [ 18 ], variable order fractional engineering models based on thermostat control [19], etc. Mathematical inequalities provide boundedness and uniqueness of solutions of boundary value problems, so they have became the backbone of mathematical methods. Due to their vast use in the field of mathematics as well as in other modern fields of science, their need and importance have inspired mathematicians to turn to more generalized and advanced inequalities [ 20 – 22 ]. Additionally, this group of inequalities has been applied Symmetry 2023,15, 479. https://doi.org/10.3390/sym15020479 https://www.mdpi.com/journal/symmetry Symmetry 2023,15, 479 2 of 18 in most studies studying fractional models, fractional BVPs and IVPs, etc. At present, the list of inequalities is very long and still growing. Studies by Beckenbach [ 23 ] are a good resource to survey these inequalities. The inequalities with general kernels and measures can be studied in the [ 24 , 25 ]. AlNemer et al. [ 26 ] and Zakarya et al. [ 27 ] established some Hardy and Coposn inequalities, respectively. The HH-inequality [ 28 ] is considered the fundamental inequality in the study of convexity. It helps us understand the geometrical aspects of a convex function. It can be written as: Theorem 1. If Φ:[c,d]→Ris a convex function, then Φ(c+d 2)≤1 d−cZd c Φ(x)dx ≤Φ(c) + Φ(d) 2 holds. For the concave function above, inequality holds in the other direction. Taking advantage of fractional operators, Farid et al. utilized a Riemann–Liouville fractional integral to study the error estimation of one of the most basic and famous Hermite– Hadamard (HH) inequalities by using the concept of convexity for strictly monotone mappings [29]. In this paper, to obtain more advanced results, we used a generalized Riemann– Liouville fractional integral [ 30 ] on HH-inequality. We establish the generalized identities and estimate the error of HH-inequality, which is further used in estimating errors of mid-point and trapezoidal inequalities for strictly monotonic convex functions. The inspiration behind this paper is the recent work conducted by Farid et al. in [29]. We develop a generalized identity for Rieman-type fractional integrals and use it to investigate trapezoid-type inequalities for a class of composite convex functions with respect to a strictly monotone function. The basic purpose of this research is to obtain more advanced and refined results than exist in the literature. The organization of the paper is as follows: the preliminaries are stated in Section 2; the main results and special cases, in the form of several examples and applications, are given in Section 3; and conclusive remarks are provided in Section 4. 2. Preliminaries We give some preliminaries that are necessary to deal with our main results. Definition 1. A real-valued function Φdefined on [c,d]is called convex if it satisfies Φ(ηx+ (1−η)y)≤ηΦ(x) + (1−η)Φ(y), where 0≤η≤1and x,y∈[c,d]. The HH-inequality and its generalizations have been studied by many authors in [ 31 – 33 ]. Due to advancement and enhancement of effectiveness operators, mathematicians are struggling to invent new efficient mechanisms and extend the existing studies. The convexity of a function w.r.t. a strictly monotone mapping given in [ 34 ] is presented as follows: Definition 2. The function Φ is convex w.r.t. a strictly monotone mapping f if the composite function Φ◦f−1is convex. The following theorem gives the description of HH-inequality under a convex function w.r.t. a strictly monotone mapping [35]. Symmetry 2023,15, 479 3 of 18 Theorem 2. Suppose I1 and I2 are sub-intervals of (−∞ , +∞) , f:I2⊃[c , d]→R is a mapping with strict monotonicity property and Φ:[c,d]⊂I1→R is a convex function w.r.t. f. Then Φ f−1f(c) + f(d) 2!≤1 f(d)−f(c)Zf(d) f(c) Φ(f−1(ν))dν≤Φ(c) + Φ(d) 2. The following definition is an extension of the classical Gamma function. For more details, see [36]. Definition 3. The k-Gamma function denoted by Γkis formulated as Γk(z) = limn→∞n!kn(nk)z k−1 (z)n,kwhere k >0and z ∈C\Z−. Another form is Γk(z) = R∞ 0e−νk kνz−1dν,z∈Cand Re(z)>0. One can easily observe that νΓk(ν) = Γk(ν+k). Definition 4 ([ 37 ]) . The left and right sided fractional RL-integrals (Riemann-Liouville) of G with order w are given as Iw c+G(x) = 1 Γ(w)Zx c(x−ν)w−1G(ν)dν,x>c, Iw d−G(x) = 1 Γ(w)Zd x(ν−x)w−1G(ν)dν,x<d. The generalized RL-integrals introduced in [30] are as follows: Definition 5. The left and right generalized RL-integrals of G with order w are given as: kIw c+G(x) = 1 kΓk(w)Zx c(x−ν)w k−1G(ν)dν,x>c, kIw d−G(x) = 1 kΓk(w)Zd x(ν−x)w k−1G(ν)dν,x<d. Note that the obtained results of the current manuscript are connected with the findings of [38–40]. 3. Main Results This section consists of several novel mean-type inequalities involving the generalized Riemann–Liouville fractional integrals. The following lemma gives an integral identity that will be helpful to study the error estimation (lower and upper bounds estimation) of HH-inequality. Lemma 1. Consider a real function Φ and a strictly monotone real function f defined on [a1 , a2] with a2>a1s.t. (Φ◦f−1)is differentiable and (Φ◦f−1)0∈L[a1,a2]. In this case, Φ(a1) + Φ(a2) 2−Γk(u+k) 2f(a2)−f(a1)u k kIu f(a1)+Φ(a2) + kIu f(a2)−Φ(a1)! =f(a2)−f(a1) 2Z1 0(1−ν)u k−νu k(Φ◦f−1)0νf(a1) + (1−ν)f(a2)dν. (1) Symmetry 2023,15, 479 4 of 18 Proof. First we evaluate the integral Z1 0(1−ν)u k(Φ◦f−1)0νf(a1) + (1−ν)f(a2)dν = (1−ν)u k(Φ◦f−1)νf(a1) + (1−ν)f(a2) f(a1)−f(a2) 1 o +u kZ1 0 (1−ν)u k−1(Φ◦f−1)νf(a1) + (1−ν)f(a2)dν f(a1)−f(a2). =Φ(a2) f(a2)−f(a1)− u k f(a2)−f(a1)Z1 0(1−ν)u k−1(Φ◦f−1)νf(a1) + (1−ν)f(a2)dν. =Φ(a2) f(a2)−f(a1)− u k f(a2)−f(a1)u k+1Z(fa2) f(a1)(z−f(a))u k−1(Φ◦f−1)(z)dz. =Φ(a2) f(a2)−f(a1)−Γk(u+k) f(a2)−f(a1)u k+1 kIu f(a2)−Φ(a1)!. (2) Similarly, integrating by parts, we obtain Z1 0νu k(Φ◦f−1)0νf(a1) + (1−ν)f(a2)dν =−Φ(a2) f(a2)−f(a1)+Γk(u+k) f(a2)−f(a1)u k+1 kIu f(a1)+Φ(a2)!. (3) By substituting (2) and (3) in the (1), we can obtain the desired result. We derive the following error estimate of Theorem 2with the help of Lemma 1. Theorem 3. Consider a real function Φ and a strictly monotone function f defined on [a1 , a2] with a2>a1s.t. Φ◦f−1is differentiable and (Φ◦f−1)0∈L[a1,a2]. Then  Φ(a1) + Φ(a2) 2−Γk(u+k) 2[f(a2)−f(a1)]u k (kIu f(a1)+Φ(a2) +kIu f(a2)−Φ(a1)) ≤|f(a2)−f(a1)| 2(u k+1)1−1 2u k  (Φ◦f−1)0(f(a1)) + (Φ◦f−1)0(f(a2))!, (4) holds whenever |(Φ◦f−1)0|is convex. Symmetry 2023,15, 479 5 of 18 Proof. From Lemma 1with the properties of the absolute value function, the above inequality can be estimated by  Φ(a1) + Φ(a2) 2−Γk(u+k) 2f(a2)−f(a1)u k kIu f(a1)+Φ(a2) +kIu f(a2)−Φ(a1)! ≤|f(a2)−f(a1)| 2Z1 0 (1−ν)u k−νu k (Φ◦f−1)0(νf(a1) + (1−ν)f(a2)) dν. (5) Since |(Φ◦f−1)0| is convex, therefore using this on the right-hand side of (5) will imply the following:  Φ(a1) + Φ(a2) 2−Γk(u+k) 2f(a2)−f(a1)u k kIu f(a1)+Φ(a2) +kIu f(a2)−Φ(a1)! ≤|f(a2)−f(a1)| 2Z1 0 (1−ν)u k−νu kν (Φ◦f−1)0(f(a1)) + (1−ν) (Φ◦f−1)0(f(a2))dν ≤|f(a2)−f(a1)| 2 Z1 2 0 (1−ν)u k−νu kν (Φ◦f−1)0(f(a1)) + (1−ν) (Φ◦f−1)0(f(a2))dν +Z1 1 2 (1−ν)u k−νu kν (Φ◦f−1)0(f(a1)) + (1−ν) (Φ◦f−1)0(f(a2))dν! =|f(a2)−f(a1)| 2  (Φ◦f−1)0(f(a1))Z1 2 0(ν(1−ν)u k−νu k+1)dν + (Φ◦f−1)0(f(a2))Z1 2 0((1−ν)u k+1−νu k(1−ν))dν + (Φ◦f−1)0(f(a1))Z1 1 2 (νu k+1−ν(1−ν)u k)dν + (Φ◦f−1)0(f(a2))Z1 1 2 (νu k(1−ν)−(1−ν)u k+1)dν!. Next, some calculations will imply our desired result. Now, we present some special cases in the context of several examples, all of which have been proved in previous studies. Symmetry 2023,15, 479 6 of 18 Example 1. By setting f(x) = 1/x in (4), we obtain  Φ(a1) + Φ(a2) 2−Γk(u+k) 2a1a2 a2−a1u kkIu (1 a1)−Φ◦g(1 a2 ) + kIu (1 a2)+Φ◦g(1 a1 ) ≤|a1−a2| 2|a1a2|(u k+1)1−1 2u ka2 1 Φ0(a1) +a2 2 Φ0(a2), where g(ν) = 1 ν. Example 2. By setting f(x) = 1 xand u k=1in (4), we obtain  Φ(a1)+Φ(a2) 2−ka1a2 a2−a1Z1 a1 1 a2 (Φ◦g)(ν)dν ≤|a1−a2| 8|a1a2|a2 1 Φ0(a1) +a2 2 Φ0(a2), where g(ν) = 1 ν. Example 3. By setting f(x) = xrwhere r 6=0in (4), we obtain  Φ(a1)+Φ(a2) 2−ru kΓk(u+k) 2(ar 2−ar 1)µ kr kIu a1+Φ(ν)+r kIu a2−Φ(ν) ≤|ar 2−ar 1| 2|r|(u k+1)1−1 2u ka1−r 1 Φ0(a1) +a1−r 2 Φ0(a2). Example 4. By setting f(x) = xrwhere r 6=0and u k=1in (4), we obtain  Φ(a1)+Φ(a2) 2−kr 2(ar 2−ar 1)Za2 a1 νr−1f(ν)dν ≤|ar 2−ar 1| 8|r|a1−r 1 Φ0(a1) +a1−r 2 Φ0(a2). Example 5. By setting f(x) = logex in (4), we obtain  Φ(a1)+Φ(a2) 2−Γk(u+k) 2(ln(a2)−ln(a1))u kkIu ln(a1)+Φ(a2)+kIu ln(a2)−Φ(a1) ≤|ln(a2)−ln(a1)| 2(u k+1)1−1 2u ka1 Φ0(a1) +a2 Φ0(a2). Example 6. By setting f(x) = logex with u k=1in (4), we obtain  Φ(a1)+Φ(a2) 2−k ln(a2)−ln(a1)Za2 a1 Φ(u) udu ≤ln(a2)−ln(a1) 8a1 Φ0(a1) +a2 Φ0(a2). Next we present the following theorem. Symmetry 2023,15, 479 7 of 18 Theorem 4. Consider a real function Φ and a strictly monotone function f defined on [a1 , a2] with a2>a1s.t. Φ◦f−1is differentiable and (Φ◦f−1)0∈L[a1,a2]. Then  Φ(a1) + Φ(a2) 2−Γk(u+k) 2f(a2)−f(a1)u kkIu f(a1)+Φ(a2) +kIu f(a2)−Φ(a1) ≤|f(a2)−f(a1)| 21 q(u k+1)1−1 2u k (Φ◦f−1)0(f(a1)) q + (Φ◦f−1)0(f(a2)) q1 q , (6) whenever |(Φ◦f−1)0|q,q≥1is convex. Proof. In two cases, the proof will be completed: Case(i). For q=1. Via the convexity of |(Φ◦f−1)0| and the properties of the absolute value function in Lemma 1, the above inequality can be obtained. Case (ii): For q>1. We use the power mean inequality and the properties of the absolute value function to R.H.S of Lemma 1. We have  Φ(a1) + Φ(a2) 2−Γk(u+k) 2f(a2)−f(a1)u kkIu f(a1)+Φ(a2) +kIu f(a2)−Φ(a1) ≤|f(a2)−f(a1)| 2Z1 0 (1−ν)u k−νu k1−1 q (7) ×Z1 0 (1−ν)u k−νu k (Φ◦f−1)0νf(a1) + (1−ν)f(a2) q dν1 q . This can be written as Z1 0 (1−ν)u k−νu k dν=Z1 2 0(1−ν)u k−νu kdν+Z1 1 2νu k−(1−ν)u kdν =2 (u k+1)1−1 2u k. (8) Since |(Φ◦f−1)0|qis convex, therefore Z1 0 (1−ν)u k−νu k (Φ◦f−1)0νf(a1) + (1−ν)f(a2) q dν ≤Z1 2 0(1−ν)u k−νu kν (Φ◦f−1)0(f(a1)) q + (1−ν) (Φ◦f−1)0(f(a2)) qdν +Z1 1 2νu k−(1−ν)u kν (Φ◦f−1)0(f(a1)) q + (1−ν) (Φ◦f−1)0(f(a2)) qdν.(9) = (Φ◦f−1)0(f(a1)) q Z1 2 oν(1−ν)u k−νu k+1dν+Z1 1 2 ννu k−(1−ν)u kdv! Symmetry 2023,15, 479 8 of 18 +|(Φ◦f−1)0(f(a2))|q Z1 2 o(1−ν)u k+1−νu k(1−ν)dν+Z1 1 2νu k(1−ν)−(1−v)u k+1dν! = (Φ◦f−1)0(f(a1)) q1 (u k+1)1−1 2u k+ (Φ◦f−1)0(f(a2)) q1 (u k+1)1−1 2u k.(10) Next, some calculations with the use of (10), (9) and (8) in (7) will imply our desired result. Now, the following examples show the application of the conclusion of the above theorem. Example 7. By setting f(x) = x in (6), we obtain  Φ(a1) + Φ(a2) 2−Γk(u+k) 2(a2−a1)u kkIu (a1)+Φ(a2) + kIu (a2)−Φ(a1) ≤|a2−a1| 21 q(u k+1)1−1 2u k Φ0(a1) q + Φ0(a2) q1 q . Example 8. By setting Φ(x) = 1 xin (6), we obtain  Φ(a1)+Φ(a2) 2−Γk(u+k) 2a1a2 a2−a1u kkI u k 1 a1 −Φ◦g(1 a2 ) + kI u k 1 a2 +Φ◦g(1 a1 ) ≤|a1−a2| 21 q|a1a2|(u k+1)1−1 2u ka2q 1 Φ0f(a1) q +a2q 2 Φ0f(a2) q1 q . Example 9. By setting f(x) = ln(x)in (6), we obtain  Φ(a1)+Φ(a2) 2−Γk(u+k) 2(lna2−lna1)u kkIu lna1+Φ(a2) + kIu lnb−Φ(a1) ≤|ln(a2)−ln(a1)| 21 q(u k+1)1−1 2u kaq 1 Φ0f(a1) q +aq 2 Φ0f(a2) q1 q . Example 10. By setting f(x) = xrwhere r 6=0in (6), we obtain  Φ(a1) + Φ(a2) 2−ru kΓk(u+k) 2(ar 2−ar 1)u kk Iu a+ 1 Φ(a2) + kIu a− 2 Φ(a1) ≤|ar 2−ar 1| 21 q|r|(u k+1)1−1 2u ka(1−r)q 1 Φ0f(a1) q +a(1−r)q 2 Φ0f(a2) q1 q . For the next theorem, the following Lemma will be helpful. Lemma 2 ([41]).For y >x≥0and α∈(0, 1), we have  xα−yα ≤y−xα . Symmetry 2023,15, 479 9 of 18 Theorem 5. Consider a real function Φ and a strictly monotone real function f defined on [a1 , a2] with a2>a1 such that Φ◦f−1 is differentiable and (Φ◦f−1)0∈L[a1 , a2] If |(Φ◦f−1)0|q , q≥1is convex, then  Φ(a1) + Φ(a2) 2−Γk(u+k) 2f(a2)−f(a1)u kkIu f(a1)+Φ(a2) +kIu f(a2)−Φ(a1) ≤|f(a2)−f(a1)| 21+1 qup k+11 p (Φ◦f−1)0(f(a1)) q + (Φ◦f−1)0(f(a2)) q1 q , (11) s.t. 1 q+1 p=1. Proof. The absolute value function along with the Holder’s inequality on R.H.S of Lemma 1, give  Φ(a1) + Φ(a2) 2−Γk(u+k) 2f(a2)−f(a1)u kkIu f(a1)+Φ(a2) +kIu f(a2)−Φ(a1) ≤|f(a2)−f(a1)| 2Z1 0 (1−ν)u k−νu k p1 p ×Z1 0 (Φ◦f−1)0νf(a1) + (1−ν)f(a2) q dν1 q , We apply Lemma 2, to obtain the following: Z1 0 (1−ν)u k−νu k p dν≤Z1 0 1−2ν up k dν =Z1 2 01−2νup k dν+Z1 1 22ν−1up k dν =1 up k+1. Now convexity of |(Φ◦f−1)0|qimplies that Z1 0 (Φ◦f−1)0νf(a1) + (1−ν)f(a2) q dν ≤Z1 0 ν (Φ◦f−1) 0 (f(a1)) q + (1−ν) (Φ◦f−1)0(f(a2) q!dν = (Φ◦f−1)0(f(a1)) q + (Φ◦f−1)0(f(a2) q 2. Hence by using the computations above, we can obtain the desired result (11). Symmetry 2023,15, 479 16 of 18 Example 20. By setting f(x) = 1 xin (17), we obtain  2u k−1(Γk(u+k))(a1a2)u k (a2−a1)u kk Iu (a1+a2 2a1a2)−Φ◦g(1 a2 )+kIu (a1+a2 2a1a2)+Φ◦g(1 a1 )−Φ2a1a2 a1+a2 ≤|a1−a2| 41−1 p(up k+1)1 p|a1a2|a2 1 Φ0(a1) +a2 2 Φ0(a2), where g(ν) = 1 ν. Example 21. By setting f(x) = xrin (17), we obtain  2u k−1(Γk(u+k))(r)u k (ar 2−ar 1)u kkIu (ar 1+ar 2 2)1 r+ Φ(a2)+kIu (ar 1+ar 2 2)1 r− Φ(a1)−Φar 1+ar 2 21 r ≤|ar 2−ar 1| 41−1 p|r|(up k+1)1 pa1−r 1 Φ0(a1) +a1−r 2 Φ0(a2). Example 22. By setting f(x) = logex in (17), we obtain  2u k−1Γk(u+k) ln(a2)−ln(a1)kIu (ln(a1)+ln(a2) 2)+Φ(a2) +kIu (ln(a1)+ln(a2) 2)−Φ(a1)−Φexpln(a1) + ln(a2) 2 ≤|ln(a2)−ln(a1)| 41−1 p(up k+1)1 pa1 Φ0(a1) +a2 Φ0(a2). 4. Conclusions The fractional calculus theory and integral operators have been used to yield more generalized inequalities. In this article, we utilized the generalized form of the Riemanntype fractional integral to obtain mean-type inequalities. The main results were based on identity. The consequences were verified to correspond to different choices for certain functions. The findings of this research reduced to the findings of [ 29 ] just by replacing k= 1. Similarly, some other results that exist in the literature were recreated. The proven results in this research are hopefully helpful in the field of modified scientific. In the future, we are committed to obtaining more generalized and refined inequalities for fractional operators. Author Contributions: Conceptualization, M.M., M.Y. and I.S.; formal analysis, M.M., M.Y., M.S., I.S. and M.D.l.S.; Funding acquisition, M.D.l.S.; methodology, M.S., I.S., S.E. and S.R.; software, S.E. and S.R.; All authors have read and agreed to the published version of the manuscript. Funding: The sixth author is grateful to the Basque Government for its support through Grants IT1555-22 and KK-2022/00090 and to MCIN/AEI 269.10.13039/501100011033 for Grant PID20211235430B-C21/C22. Institutional Review Board Statement: Not applicable. Informed Consent Statement: Not applicable. Data Availability Statement: Data sharing is not applicable to this article as no datasets were generated nor analyzed during the current study. Acknowledgments: The fifth and seventh authors would like to thank Azarbaijan Shahid Madani University. Symmetry 2023,15, 479 17 of 18 Conflicts of Interest: The authors declare no conflict of interest. References 1. Magin, R.L. 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