Some Bullen-type Inequalities for Co-ordinated s-convex Function
Abstract
2nd Kocaeli Science Congress (KOSC-2025), 19-21 November 2025, Kocaeli, TÜRKİYE https://fefkongre.kocaeli.edu.tr/en
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Some Bullen-type Inequalities for Co-ordinated s-convex Function Mehmet Eyüp Kiriş1, Tuğba Çınar 1 1Department of Mathematics, Faculty of Science and Arts, Afyon Kocatepe University, Afyonkarahisar 03000, Türkiye Corresponding author: [email protected] ORCID IDs: First Author: 0000-0002-6463-5289 Second Author: 0000-0001-9670-4018 DOI : 10.5281/zenodo.18033394 Abstract In this paper,we establish new Bullen-Simpson type integral inequalities for functions of two variables whose mixed partial derivatives are s-convex in the second sense on the coordinates. Several general results and corollaries are obtained by employing Hölder and Power Mean inequalities. Furthermore, the constants involved in these inequalities are explicitly determined,and the classical one-dimensional results are shown to be special cases of our findings.The obtained results provide a comprehensive generalization of the existing Bullen-Simpson type inequalities to the framework of coordinate s-convex functions,offering new insights and potential applications in numerical integration and mean inequalities. Keywords: Bullen-type inequalities, s-convex functions, Co-ordinated convexity, Simpson-type inequalities. 1 Introduction and Motivation Definition 1.1. Let I be an interval of real numbers. A function B : I→R is said to be convex, if for all µ1,µ2ϵI all hϵ [0,1] ([2]) ,we have B(hµ1+(1−h)µ2)≤hB(µ1) + (1 −h)B(µ2) One of the famous inequalities for the class of convex functions is the so-called Hermite-Hadamard inequality, which can be stated as follows: Theorem 1.1. Let B be a convex function on the interval [µ1,µ2] with µ1< µ2 , then we have ([18]) Bµ1+µ2 2≤1 (µ2−µ1) µ1 Z µ1 B(x)dx ≤B(µ1)+B(µ2) 2(1) Since its discovery,several articles related to inequality (1) have been published ([3]-[18]) M34-1
KOSC-2025 Proceedings The concept of convetixy has been also generalized in diverse manners. One of them is the so-called s-convex function or Breckner convex function defined as follows: A nonnegative function B : I⊂[0,∞)→R is said to be s-convex in the second sense for some fixed sϵ (0,1] ,if B(hµ1+(1−h)µ2)≤hsB(µ1) + (1 −h)sB(µ2) holds for all µ1,µ2ϵI and hϵ [0,1] (see [19]) In [ 20 ],Dragomir and Fitzpatrick,proved the following variant of inequality (1) which holds for s-convex functions in the second sense. Definition 1.2. A function B : ∆ ⊆[0,∞)2→R is called s-convex in the second sense on the co-ordinates on ∆if; B(hµ1+(1−h)µ2, tv1+(1−t)v2) ≤hstsB(µ1, v1)+hs(1 −t)sB(µ1, v2) +ts(1 −h)sB(µ2, v1) + (1 −h)s(1 −t)sB(µ2, v2) holds for all h, tϵ [0,1] and ( µ1, v1 ) , ( µ1, v2 ) , ( µ2, v1 ) , ( µ2, v2 ) ϵ ∆ , for some fixed sϵ (0,1]. (see [ 21 ]) Definition 1.3. In [22], Hwang et al. established the following Bullen-type inequality 1 4B(µ1)+2Bµ1+µ2 2+B(µ2)−1 (µ2−µ1) µ1 Z µ1 B(x)dx (2) ≤µ2−µ1 16 B′(µ1)+B′(µ2) Definition 1.4. In [23], Hwang et al. established the following Simpson-type inequality 1 4B(µ1)+2Bµ1+µ2 2+B(µ2)−1 (µ2−µ1) µ1 Z µ1 B(x)dx ≤µ2−µ1 16 B′(µ1)+B′(µ2) 2 Background Over the last two decades,error estimation of quadrature rules via different types of convexity has become an attractive and fascinating area of research and has gained popularity. Consequently, several papers treating integral inequalities under the principle of convexity have been widely studied by mathematicians and researchers. Regarding Newton-Cotes type inequalities involving one point ([ 24 ]-[ 29 ]) two point Newton-Cotes type inequalities ([ 30 ]-[ 33 ]), for three point NewtonCotes type inequalities ([ 34 ]-[ 39 ]) and Newton-Cotes type inequalities involving four points ([40]-[44] M34-2 2nd Kocaeli Science Congress, November 19-21, 2025
3 Mathematical Preliminaries In this section, we establish the mathematical foundations necessary for our subsequent development. We begin with theorem and then proceed to more advanced concepts. Theorem 3.1. Let B : [µ1, µ2]x[v1, v2]→[0,∞) be an integrable and coordinate s-convex function in the second sense, where 0 ≤µ1< µ2 and s∈(0,1] .Then the following four-point inequality holds: (1) ≤(2) ≤(3) ≤(4) (1) = 22s−2Bv1+v2 2,µ1+µ2 2(3) (2) = 2s−2 (µ2−µ1)Zµ2 µ1Zv2 v1Bx, v1+v2 2+Bµ1+µ2 2, ydydx (3) = 1 (µ2−µ1)(v2−v1)Zµ2 µ1Zv2 v1 B(x, y)dydx (4) = 1 (s+1)2B(µ1, v1)+B(µ2, v1)+B(µ1, v2)+B(µ2, v2) 4 in this paper, we propose to study the so called Bullen-Simpson Type İnequalities which is a 25 point Newton Cotes Rule(Boole Quadrature Rule) and can be represented as follows ; Zµ2 µ1Zv2 v1 B(x, y)dydx −F(B, µ, v)≤ϵ(B, s, µ, v) 1) 25-Point boole quadrature rule (F) F(B, µ, v) = (µ2−µ1)(v2−v1) 144 5 X i=1 5 X j=1 Wi,jB(xi, yj) The Wcoefficient is given by the product rule of the Boole coefficient (1,4,2,4,1), W= 1 4 2 4 1 x14241= 1 4 2 4 1 4168161 2 8 4 8 2 4168164 1 4 2 4 1 the points (xi, yj)are the 25 nodes dividing the region [µ1, µ2]x[v1, v2]into 4X4 subregions 2)Error Bound ( ϵ ): The precise formula for the term (upper bound ϵ ( B, s, µ, v )is the main result of the paper typically involving a complex structure that depends on the maximum value of the mixed derivative ∂2B ∂x∂y and constants related to the parameters To this end,we first establish a novel integral identity four double integrals of the BullenSimpson type. Utilizing this identity, we absolute value of the mixed partial derivative ∂2B ∂x∂y q , is a coordinate s-convex function.Furthermore, we adress related results for the cases where this mixed partial derivative is bounded and satisfies the Hölder condition on the coordinates. Finally, 2nd Kocaeli Science Congress, November 19-21, 2025 M34-3
KOSC-2025 Proceedings we provide several applications to two-dimensional numerical integration rules ans discuss new inequalities involving means based on our coordinate s-convexity results 4 Main theorems and Results In this section, we present our main theoretical contributions. We start with a lemma and use it to derive new theorems and results Lemma 4.1. Let B : [µ1, µ2]x[v1, v2]→R be a function such that the mixed partial derivative ∂2B ∂x∂y exists and belongs to L1(D), then the following equality holds B−I=(µ2−µ1)(v2−v1) 256 4 X i=1 4 X j=1 Z1 0Z1 0 Ki(h)Kj(k)∂2B ∂x∂y (xi(h), yj(k))dkdh (4) 1.The integrals mean (I) I=1 (µ2−µ1)(v2−v1)Zµ2 µ1Zv2 v1 B(x, y)dydx 2.The extended bullen rule approximation (B) The approximation B is the weighted sum of the function B ( x, y )evaluated at the 5x5=25 Bullen quadrature points in Dnormalized by 1 144 ; B=1 144 5 X i=1 5 X j=1 Wi,jB(xi, yj) where the weights are w={1,4,6,4,1} and the quadrature points are; xi∈µ1,3µ1+µ2 4,µ1+µ2 2,µ1+ 3µ2 4, µ2 yi∈v1,3v1+v2 4,v1+v2 2,v1+ 3v2 4, v2 3.The kernel functions (Ki ( h ) , Kj ( k )): Ki ( h )and Kj ( k )represent the single-variable BullenSimpson kernel functions corresponding to the subdivision of the intervals. 4.The parameterization:The terms xi ( h )and yi ( k )linearly parameterize the 4x4 subrectangles of D Proof. The proof of this identity is usually done by taking the integrals Ii,j =Z1 0Z1 0 Ki(h)Kj(k)∂2B ∂x∂y (xi(h), yj(k))dkdh (5) we will use successive integration by parts twice. For example; consider the simpliest term on the right-hand side for i= 1,j= 1; I1,1=Z1 0 K1(h)"Z1 0 K1(k)∂2B ∂x∂y (x1(h), y1(k))dk#dh M34-4 2nd Kocaeli Science Congress, November 19-21, 2025
partial integration of the inner integral with respect to k. Then, K1(h)=h−1/3 x1(h) = (1 −h)µ1+h3µ1+µ2 4 x′ 1(h) = µ2−µ1 4 y1(k) = (1 −k)µ1+k3v1+v2 4 y′ 1(k) = v2−v1 4 u=k−1 3⇒du =dk dv = ∂2B ∂x∂y(x1, y1(k))!dk To find v, we need to integrate Bwith respect to k. Using the reverse chain rule; v=1 y′ 1(k) ∂B ∂x y′ 1(k) = 3v1+v2 4−v1=v2−v1 4 v=4 v2−v1 ∂B ∂x (x1(h), y1(k)) Then, lets do the integration by parts and substitue the integral boundaries. 2 3 4 v2−v1 ∂B ∂x x1(h),3v1+v2 4+1 3 4 v2−v1 ∂B ∂x (x1(h), v1) −4 v2−v1Z1 0 ∂B ∂x (x1(h), y1(k))dk J1,1(h) = 4 3(v2−v1)2∂B ∂x x1(h),3v1+v2 4+∂B ∂x (x1(h), v1) −4 v2−v1Z1 0 ∂B ∂x (x1(h), y1(k))dk Transition to the outer integral, I1,1=Z1 0 K1(h).J1,1(h)dh Apply R1 0K1(h)(...)dh to each term in the expression J1,1(h) I1,1=4 3(v2−v1)2Z1 0 K1(h)∂B ∂x x1(h),3v1+v2 4dh (6) +Z1 0 K1(h)∂B ∂x (x1(h), v1)dh −4 v2−v1Z1 0 K1(h)Z1 0 ∂B ∂x (x1(h), y1(k))dkdh 2nd Kocaeli Science Congress, November 19-21, 2025 M34-5
KOSC-2025 Proceedings Lets take L1=Z1 0 K1(h)∂B ∂x x1(h),3v1+v2 4dh L2=Z1 0 K1(h)∂B ∂x (x1(h), v1)dh L3=Z1 0 K1(h)Z1 0 ∂B ∂x (x1(h), y1(k))dkdh Appyling the second partial integration(L1term) u=h−1 3⇒du =dh dv =∂B ∂x (x1(h), y1(k))dh v=1 x′ 1(h)B(x1(h), Y ) = 4 µ2−µ1 B(x1(h), Y )dh Then, lets do the integration by parts and substitue the integral boundaries. L1=4 3(µ2−µ1)2B3µ1+µ2 4, Y +B(µ1, Y ) −4 µ2−µ1Z1 0 B(x1(h), Y )dh Lets convert the remaining integral in L1to an integral with respect to x; x=x1(h) dh =4 µ2−µ1 dx and boundaries x= 0 ⇒h=µ1 x= 1 ⇒h=3µ1+µ2 4 Z1 0 B(x1(h), Y )dh =Z3µ1+µ2 4 µ1 B(x, Y )4 µ2−µ1 dx Lets rewrite L1 L1=4 3(µ2−µ1)2B3µ1+µ2 4, Y +B(µ1, Y )(7) −4 µ2−µ1Z3µ1+µ2 4 µ1 B(x, Y )dx Then, calculate L2,integrating by parts and substitue the integral boundaries. We get; L2=4 3(µ2−µ1)2B3µ1+µ2 4, v1+B(µ1, v1) −4 µ2−µ1Z1 0 B(x1(h), v1)dh M34-6 2nd Kocaeli Science Congress, November 19-21, 2025
Change variables in the remaning integral and if we substitue the transformation we found into L2; L2=4 3(µ2−µ1)2B3µ1+µ2 4, v1+B(µ1, v1)(8) −4 µ2−µ1Z3µ1+µ2 4 µ1 B(x, v1)dx (The structure of this term is the same as that of four term, expect that the y− coordinate is to v, instead of 3µ1+µ2 4) Finally, lets calculate for L3,transformation of the inner integral with respesct to k. Then, L3=Z1 0 K1(h) Z3v1+v2 4 v1 ∂B ∂x (x1(h), y)dy!4 v2−v1 dh Integration by parts of the outer integral with respect to h. As in the previous steps to remove the derivative x , we substitute the derivative x1 ( h )with respect to (x1(h)′) we divide. Integration by parts, write boundaries and make necessary arrangements, L3=16 3(µ2−µ1)(v2−v1)"2Z3v1+v2 4 v1 B3µ1+µ2 4, ydy +Z3v1+v2 4 v1 B(µ1, y)dy#−Z1 0 4 µ2−µ1Z3v1+v2 4 v1 B(x1(h), y)dydh Transformation of the remaning double integral (final term) −16 (µ2−µ1)(v2−v1)Z1 0 Z3v1+v2 4 v1 B(x1(h), y)dy!dh where, x=x1(h) dh =4 µ2−µ1 dx and h= 0 ⇒x=µ1 h= 1 ⇒x=3µ1+µ2 4 −16 (µ2−µ1)(v2−v1)Z3µ1+µ2 4 µ1 Z3v1+v2 4 v1 B(x, y)dxdy!4 µ2−µ1 L3final statement, L3=16 3(µ2−µ1)(v2−v1)Z3v1+v2 4 v12B3µ1+µ2 4, y+B(µ1, y)dy 2nd Kocaeli Science Congress, November 19-21, 2025 M34-7
KOSC-2025 Proceedings −16 (µ2−µ1)2(v2−v1)Z Z 3µ1+µ2 4 µ1 B(x, y)dxdy (9) This term, when combined with previous terms L1 and L2, will give I1,1. When all sixteen terms are added, these complex single integrals cancel out,leaving only the Bullen terms B ( ... )and the term RRB(x, y)dxdy I1,1=4 3(v2−v1)[2L1+L2]−4 v2−v1 L3 We collect the L1and L2terms as 2L1+L2 L1=4 3(µ2−µ1)2B3µ1+µ2 4,3v1+v2 4+Bµ1,3v1+v2 4 −4 (µ2−µ1)2Z3µ1+µ2 4 µ1 Bx, 3v1+v2 4dx L2=4 3(µ2−µ1)2B3µ1+µ2 4, v1+B(µ1, v1) −4 (µ2−µ1)2Z3µ1+µ2 4 µ1 B(x, v1)dx 2L1+L2=4 3(µ2−µ1)4B3µ1+µ2 4,3v1+v2 4+ 2Bµ1,3v1+v2 4 +2B3µ1+µ2 4, v1+B(µ1, v1) 2L1+L2by 4 3(v2−v1)multiplying this sum by the first two rows of I1,1,we get 16 9(µ2−µ1)(v2−v1)4B3µ1+µ2 4,3v1+v2 4+ 2Bµ1,3v1+v2 4 +2B3µ1+µ2 4, v1+B(µ1, v1) If we multiply the final statement of L3by −4 (v2−v1) L3=64 3(µ2−µ1)(v2−v1)2Z3v1+v2 4 v12B3µ1+µ2 4, y+B(µ1, y)dy 64 (µ2−µ1)2(v2−v1)2Z Z 3µ1+µ2 4 µ1 B(x, y)dxdy I1,1=16 9(µ2−µ1)(v2−v1)4B3µ1+µ2 4,3v1+v2 4(10) +2Bµ1,3v1+v2 4+ 2B3µ1+µ2 4, v1+B(µ1, v1) −64 3(µ2−µ1)(v2−v1)2Z3v1+v2 4 v12B3µ1+µ2 4, y+B(µ1, y)dy M34-8 2nd Kocaeli Science Congress, November 19-21, 2025
−64 3(µ2−µ1)2(v2−v1)Z3µ1+µ2 4 µ12Bx, 3v1+v2 4+B(x, v1)dx +256 (µ2−µ1)2(v2−v1)2Z Z 3µ1+µ2 4 µ1 B(x, y)dxdy 1)Bullen Type Terms(Four Corners) 1 9 4 µ2−µ1 4 v2−v1 =16 9(µ2−µ1)(v2−v1)(11) 2)Single Integrals (x-Direction) −1 3 16 (µ2−µ1)2 4 v2−v1 =−64 3(µ2−µ1)2(v2−v1)(12) 3)Single Integrals (y-Direction) −1 3 16 (v2−v1)2 4 µ2−µ1 =−64 3(µ2−µ1)(v2−v1)2(13) 4)Double Integral 16 (µ2−µ1)2 16 (v2−v1)2=256 (µ2−µ1)2(v2−v1)2(14) Due to symmetry,the expression of the remaining terms follows the same structures as the term I1,1 , which we proved in detail. Only the points, boundaries and coefficients of the Peano kernel ( K1 ( h ) = h− 1 / 3 , K1 ( h ) = h− 2 / 3) used change, preserving the coefficients x′ 1(h) = µ2−µ1 4, y′ 1(k) = v2−v1 4.So let’s move directly to the sum Ii,j 4 X i=1 4 X j=1 Ii,j = sum of 16 bullen type terms !+ sum of 16 X-Int terms ! + sum of 16 Y-Int terms !+ sum of 16 double integrals ! then we have, Ii,j =16 9(µ2−µ1)(v2−v1)X bullen B(...)(15) +64 3(µ2−µ1)2(v2−v1)X X−ınt Z(...)dx +64 3(µ2−µ1)(v2−v1)2X Y−ınt Z(...)dy +256 (µ2−µ1)2(v2−v1)2Z Z B(...)dxdy 1)This is the most important part of the proof.The x -directional integral term from Ii,j cancels out because the x -directional integral terms on adjocent intervals (e.g I1,1 ) are of exactly the 2nd Kocaeli Science Congress, November 19-21, 2025 M34-9
KOSC-2025 Proceedings we focus on the second factor R1 0R1 0∂2B ∂x∂y |(µt, vh)|qdtdh . Since ∂2B ∂x∂y q is coordinate (s1, s2) convex, we apply the convexity property on each of the 5x5=25 subrectangles derived from the integral representation. For each term, the integral over the subrectangle (in the (µ, v) ) plane is bounded by: Z1 0Z1 0 ∂2B ∂x∂y |(µt, vh)|qdtdh!≤ 5 X i=1 5 X j=1 Ki(s1) 5.3s1...Kj(s2) 5.3s2... ∂2B ∂µ∂v (µi, vj)q Substituing the results,we obtain the required inequality 1 144 5 X i=1 5 X j=1 Wi,jB(Mi, Nj)−1 (µ2−µ1)(v2−v1)Zµ2 µ1Zv2 v1 B(x, y)dydx ≤(µ2−µ1)(v2−v1) 256 5 1821−1 q5 X i=1 5 X j=1 Ci,j (s1, s2) ∂2B ∂x∂y (µi, vj) q!1/q The proof is completed. References [1] B. Meftah, S. Samoudi, Some Bullen-Simpson type inequalities for differentiable s-convex functions,Math. Morav., 28(1) (2024), 63–85 [2] J. E. Pečarić, F. Proschan and Y. L. Tong, Convex functions, partial orderings,and statistical applications, Mathematics in Science and Engineering, 187, Academic Press, Inc., Boston, MA, 1992 [3] M. A. Ali, C. S. Goodrich and H. Budak, Some new parameterized Newton-type inequalities for differentiable functions via fractional integrals, Journal of Inequalities and Applications, 2023 (2023), Article ID: 49, 1–17 [4] M. U. Awan, M. A. Noor, T. Du and K. I. Noor, New refinements of fractional HermiteHadamard inequality, Revista de la Real Academia de Ciencias Exactas,Físicas y Naturales. Serie A. Matemàticas (RACSAM), 113 (1) (2019), 21–29. [5] M. U. Awan, N. Akhtar, A. Kashuri, M. A. Noor and Y.-M. Chu, 2D approximately reciprocal ρ -convex functions and associated integral inequalities, AIMS Mathematics, 5 (5) (2020), 4662–4680. [6] T. Du, C. Luo and Z. Cao, On the Bullen-type inequalities via generalized fractional integrals and their applications, Fractals, 29 (07) (2021), Article ID: 2150188 [7] F. Hezenci and H. Budak, Novel results on trapezoid-type inequalities for conformable fractional integrals, Turkish Journal of Mathematics, 47 (2) (2023), 425–438 [8] H. Kara, H. Budak and A. O. Akdemir, Some new parameterized inequalities based on Riemann-Liouville fractional integrals, Filomat 37 (23) (2023), 7867–7880 M34-16 2nd Kocaeli Science Congress, November 19-21, 2025
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