A New Perspective on Milne-type Inequalities involving Riemann-Liouville Fractional Integrals

A New Perspective on Milne-type Inequalities involving Riemann–Liouville Fractional Integrals Hüseyin Budak1, Mehmet Zeki Sarıkaya2, Hasan Kara2 1Department of Mathematics, Faculty of Science and Arts, Kocaeli University, Türkiye 2Department of Mathematics, Faculty of Science and Arts, Duzce University, Türkiye Corresponding author: [email protected] ORCID IDs: First Author: 0000-0001-8843-955X Second Author: 0000-0002-6165-9242 Third Author: 0000-0002-2075-944X DOI : 10.5281/zenodo.18017964 Abstract In this study, we establish the upper and below bounds for fractional Milne-type inequalities by using functions whose second derivatives are bounded. Moreover, we present new inequalities for Riemann integrals as special cases. Keywords: Milne type inequality inequality, integral inequalities, bounded functions. 1 Introduction The Hermite-Hadamard inequality, recognized as one of the foundational results concerning convex functions, possesses a clear geometric interpretation and has found numerous applications across various fields of mathematics. It has attracted significant attention within elementary mathematical analysis, prompting many researchers to explore generalizations, refinements, extensions, and analogous results for broader classes of functions, particularly through the lens of convexity. The inequalities originally formulated by C. Hermite and J. Hadamard for convex functions hold a prominent place in the mathematical literature (see, for example, [ 12 , p.137], [ 5 ]). These inequalities state that if f : I→R is a convex function on the interval I of real numbers and a, b ∈Iwith a<b, then fa+b 2≤1 b−aZb a f(x)dx ≤f(a) + f(b) 2.(1) Both inequalities hold in the reversed direction if fis concave. In [ 7 ] and [ 8 ], Dragomir et al. proved the following results connected with the HermiteHadamard inequality: M22-1 KOSC-2025 Proceedings Theorem 1.1. Let f : [a, b]→R be a twice differentiable mapping such that there exists real constants mand Mso that m≤f′′ ≤M. Then, the following inequalities hold: m(b−a)2 24 ≤1 b−a b Z a f(x)dx −fa+b 2≤M(b−a)2 24 .(2) and m(b−a)2 12 ≤f(a)+f(b) 2−1 b−a b Z a f(x)dx ≤M(b−a)2 12 (3) In the following we will give some necessary definitions and mathematical preliminaries of fractional calculus theory which are used further in this paper. Definition 1.1. Let f∈L1 [ a, b ] . The Riemann-Liouville integrals Jα a+f and Jα b−f of order α > 0 with a≥0are defined by Jα a+f(x) = 1 Γ(α)Zx a (x−t)α−1f(t)dt, x > a and Jα b−f(x) = 1 Γ(α)Zb x (t−x)α−1f(t)dt, x < b respectively. Here, Γ(α)is the Gamma function and J0 a+f(x) = J0 b−f(x) = f(x). For more information about fraction calculus please refer to ([9], [10], [11], [13].) In [ 14 ], Sarikaya et al. first give the following interesting integral inequalities of HermiteHadamard type involving Riemann-Liouville fractional integrals. Theorem 1.2. Let 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 the following inequalities for fractional integrals hold: fa+b 2≤Γ(α+ 1) 2 (b−a)αJα a+f(b)+Jα b−f(a)≤f(a) + f(b) 2(4) with α > 0. Moreover, Dragomir give the following another version of Hermite-Hadamard inequality for Riemann-Lioville fractional integrals: Theorem 1.3. [ 6 ] Let f : [a, b]→R be a positive function with a<b and f∈L1[a, b]. If f is a convex function on [a, b],then the following inequalities for fractional integrals hold: fa+b 2≤2α−1Γ(α+ 1) (b−a)αJα afa+b 2+Jα bfa+b 2 (5) ≤f(a)+f(b) 2 Theorem 1.4. [ 1 ] Let f : [ a, b ] →R be a twice differentiable function with a<b and f∈L1 [ a, b ] . If f′′ is bounded, i.e. m≤f′′(t)≤M, t ∈(a, b), m, M ∈R,then we have the inequalities m(b−a)2 8(α+ 2) M22-2 2nd Kocaeli Science Congress, November 19-21, 2025 ≤2α−1Γ (α+ 1) (b−a)αJα a+fa+b 2+Jα b−fa+b 2−fa+b 2 ≤M(b−a)2 8(α+ 2) , and m(b−a)2 4(α+ 2) ≤f(a)+f(b) 2−2α−1Γ (α+ 1) (b−a)αJα a+fa+b 2+Jα b−fa+b 2 ≤M(b−a)2 4(α+ 2) , Now we present some Milne-type inequalities: Theorem 1.5. [ 2 ]Let f : [a, b]→R be a differentiable mapping (a, b) such that f∈L1 ( [a, b] ). If the function |f′|is convex on [a, b],then we have the following Milne-type inequality  1 32f(a)−fa+b 2+ 2f(b)−1 b−a b Z a f(t)dt ≤5 (b−a) 24 f′(a)+f′(b). Theorem 1.6 (Milne-type inequality in the first sense).[ 4 ] Let f : [a, b]→R be a differentiable mapping (a, b) such that f∈L1 ( [a, b] ). If the function |f′| is convex on [a, b], then we have the following Milne-type inequality for Riemann-Liouville fractional integrals  1 32f(a)−fa+b 2+ 2f(b)−2α−1Γ (α+ 1) (b−a)αJα a+b 2+f(b)+Jα a+b 2−f(a) ≤4α+ 1 12 (α+ 1)(b−a)f′(a)+f′(b), for α > 0. Theorem 1.7 (Milne-type inequality in the second sense).[ 2 ] Assume that the assumptions of Theorem 1.5. Then we have  1 32f(a)−fa+b 2+ 2f(b)−2α−1Γ (α+ 1) (b−a)αJα a+fa+b 2+Jα b−fa+b 2 ≤α+ 4 12 (α+ 1)(b−a)f′(a)+f′(b), for α > 0. Theorem 1.8 (Milne-type inequality in the third sense).[ 3 ] Assume that the assumptions of Theorem 1.5. Then we have  1 32f(a)−fa+b 2+ 2f(b)−Γ (α+ 1) 2(b−a)αJα a+f(b) + Jα b−f(a) ≤b−a 2[Υ1(α)+Υ2(α)] f′(a)+f′(b), 2nd Kocaeli Science Congress, November 19-21, 2025 M22-3 KOSC-2025 Proceedings for α > 0,where Υ1(α) = 1 2 Z 0 tα−2 3 dt =           2α α+1 2 31 α+1 +1 2α+1(α+1) −1 3,0≤α≤ln 2 3 ln 1 2 1 3−1 2α+1(α+1) α > ln 2 3 ln 1 2 and Υ2(α) = 1 Z 1 2  tα−1 3 dt =           1 α+1 −1 2α+1(α+1) −1 6,0≤α≤ln 1 3 ln 1 2 2α α+1 1 31 α+1 +1 2α+1(α+1) +1 α+1 −1 2α > ln 1 3 ln 1 2 . 2 Main Results In this section, we will prove an upper and below bound for the difference given in Theorem 1.7 by using thr functions whose second derivatives are bounded. Theorem 2.1. Let f : [a, b]→R be atwice differentiable functions and f∈L1[a, b]. If f′′ is bounded, i.e. m≤f′′(t)≤M,t∈(a, b), m, M ∈R,then we have the inequalities α2+ 6α+ 1 4 (α+ 1) (α+ 2) (b−a)2m(6) ≤1 32f(a)−fa+b 2+ 2f(b) −2α−1Γ(α+ 1) (b−a)αJα a+b 2+f(b)+Jα a+b 2−f(a) ≤α2+ 6α+ 1 4 (α+ 1) (α+ 2) (b−a)2M. Proof. From the definition of Riemann-Liouville fractional integrals, we get 2α−1Γ (α+ 1) (b−a)αJα a+b 2+f(b)+Jα a+b 2−f(a)(7) =2α−1α (b−a)α    b Z a+b 2 (b−x)α−1f(x)dx + a+b 2 Z a (x−a)α−1f(x)dx    =2α−1α (b−a)α a+b 2 Z a [f(x)+f(a+b−x)] (x−a)α−1dx. Using the identity (7), we get 1 32f(a)−fa+b 2+ 2f(b)(8) −2α−1Γ (α+ 1) (b−a)αJα a+b 2+f(b)+Jα a+b 2−f(a) M22-4 2nd Kocaeli Science Congress, November 19-21, 2025 =1 32f(a)−fa+b 2+ 2f(b) −2α−1α (b−a)α a+b 2 Z a [f(x)+f(a+b−x)] (x−a)α−1dx =2α−1α 3 (b−a)α a+b 2 Z a  4f(a)−2fa+b 2+ 4f(b) −3f(x)−3f(a+b−x) (x−a)α−1dx. Using the facts that f(x)−f(a) = x Z a f′(t)dt and f(b)−f(a+b−x) = b Z a+b−x f′(t)dt, we get f(a)+f(b)−f(x)−f(a+b−x)(9) = b Z a+b−x f′(t)dt − x Z a f′(t)dt = x Z a f′(a+b−u)du − x Z a f′(t)dt = x Z af′(a+b−t)−f′(t)dt. On the other hand, we have f(a)−fa+b 2=− a+b 2 Z a f′(t)dt (10) and f(b)−fa+b 2= b Z a+b 2 f′(t)dt. (11) By (10) and (11), we get f(a)+f(b)−2fa+b 2(12) 2nd Kocaeli Science Congress, November 19-21, 2025 M22-5 KOSC-2025 Proceedings = b Z a+b 2 f′(t)dt − a+b 2 Z a f′(t)dt = a+b 2 Z a f′(a+b−t)dt − a+b 2 Z a f′(t)dt = a+b 2 Z af′(a+b−t)−f′(t)dt. We also have f′(a+b−t)−f′(t) = a+b−t Zt f′′(u)du. (13) By using the equality (13) and m<f′′ (u)< M, u ∈(a, b),we obtain, m a+b−t Zt du ≤ a+b−t Zt f′′(u)du ≤M a+b−t Zt du i.e. m(a+b−2t)≤f′(a+b−t)−f′(t)≤M(a+b−2t).(14) By integrating the inequality (14) with respect to ton [a, x],from (9), we get m"b−a 22 −a+b 2−x2#(15) ≤f(a)+f(b)−f(x)−f(a+b−x) ≤M"b−a 22 −a+b 2−x2#. That is, m(x−a) (b−x)(16) ≤f(a)+f(b)−f(x)−f(a+b−x) ≤M(x−a) (b−x). Similarly, integrating the inequality (14) with respect to ton ha, a+b 2i,from (12), we obtain mb−a 22 ≤f(a) + f(b)−2fa+b 2≤Mb−a 22 .(17) M22-6 2nd Kocaeli Science Congress, November 19-21, 2025 By using the inequalities (16) and (17), we get 3m(x−a) (b−x) + mb−a 22 (18) ≤4f(a)−2fa+b 2+ 4f(b)−3f(x)−3f(a+b−x) ≤3M(x−a) (b−x) + Mb−a 22 . Multiplying the inequality (18) by 2α−1α (b−a)α(x−a)α−1 and integrating the resultant inequality with respect to xon ha, a+b 2i, we establish m2α−1α (b−a)α a+b 2 Z a"3 (x−a) (b−x) + b−a 22#(x−a)α−1dx ≤2α−1α 3 (b−a)α × a+b 2 Z a  4f(a)−2fa+b 2+ 4f(b) −3f(x)−3f(a+b−x) (x−a)α−1dx ≤M2α−1α (b−a)α a+b 2 Z a"3 (x−a) (b−x) + b−a 22#(x−a)α−1dx. By the equality (8) and equality 2α−1α (b−a)α a+b 2 Z a"3 (x−a) (b−x) + b−a 22#(x−a)α−1dx =α2+ 6α+ 1 4 (α+ 1) (α+ 2) (b−a)2, we get the desired inequality (6). This completes the proof. Remark 2.1. If we choose α= 1 in Theorem 2.1, then we have the following inequality 3 8(b−a)2m ≤1 32f(a)−fa+b 2+ 2f(b)−1 b−a a+b 2 Z a f(t)dt ≤3 8(b−a)2M. 2nd Kocaeli Science Congress, November 19-21, 2025 M22-7 KOSC-2025 Proceedings 3 Acknowledgments The authors would like to express their sincere gratitude to the anonymous reviewers and the handling editor for their careful reading, valuable comments, and constructive suggestions, which greatly improved the quality of this manuscript. The authors also thank all colleagues who contributed to the development of this work. References [1] Budak, H., Kara, H., Sarikaya, M.Z., and Kiriş, M.E., New extensions of the HermiteHadamard inequalities involving Riemann-Liouville fractional integrals, Miskolc Mathematical Notes, 21(2), 2020, 665–678. [2] H. Budak, P. Kosem, and H. Kara, On new Milne-type inequalities for fractional integrals, Journal of Inequalities and Applications, 2023, Article No. 10. [3] H. Budak and A.A. 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Başak, Hermite–Hadamard’s inequalities for fractional integrals and related fractional inequalities, Mathematical and Computer Modelling, 57, 2013, 2403–2407. DOI: 10.1016/j.mcm.2011.12.048 2nd Kocaeli Science Congress, November 19-21, 2025 M22-9