Journal of Inequalities and Applications This Provisional PDF corresponds to the article as it appeared upon acceptance Fully formatted PDF and full text (HTML) versions will be made available soon Inequalities for convex and s-convex functions on Delta=[a,b]x[c,d] Journal of Inequalities and Applications 2012, 2012:20 doi:10.1186/1029-242X-2012-20 Muhamet Emin Ozdemir (emos@atauni.edu.tr) Havva Kavurmaci (hkavurmaci@atauni.edu.tr) Ahmet Ocak Akdemir (ahmetakdemir@agri.edu.tr) Merve Avci (merveavci@ymail.com) ISSN Article type 1029-242X Research Submission date May 2011 Acceptance date February 2012 Publication date February 2012 Article URL http://www.journalofinequalitiesandapplications.com/content/2012/1/ This peer-reviewed article was published immediately upon acceptance It can be downloaded, printed and distributed freely for any purposes (see copyright notice below) For information about publishing your research in Journal of Inequalities and Applications go to http://www.journalofinequalitiesandapplications.com/authors/instructions/ For information about other SpringerOpen publications go to http://www.springeropen.com © 2012 Ozdemir et al ; licensee Springer This is an open access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited INEQUALITIES FOR CONVEX AND s-CONVEX FUNCTIONS ON = [a, b] ì [c, d] ă MUHAMET EMIN OZDEMIR1 , HAVVA KAVURMACI∗1 , AHMET OCAK AKDEMIR2 AND MERVE AVCI3 DEPARTMENT OF MATHEMATICS, K.K EDUCATION FACULTY, ATATURK UNIVERSITY, ERZURUM 25240, TURKEY DEPARTMENT OF MATHEMATICS, FACULTY OF SCIENCE AND ARTS, ˘ ˙ ˘ AGRI IBRAHIM CECEN UNIVERSITY, AGRI 04100, TURKEY ¸ ¸ DEPARTMENT OF MATHEMATICS, FACULTY OF SCIENCE AND ARTS, ADIYAMAN UNIVERSITY, ADIYAMAN, TURKEY ∗ CORRESPONDING AUTHOR: HKAVURMACI@ATAUNI.EDU.TR EMAIL ADDRESSES: ME: EMOS@ATAUNI.EDU.TR AO: AHMETAKDEMIR@AGRI.EDU.TR MA: M.AVCI@POSTA.ADIYAMAN.EDU.TR Abstract In this article, two new lemmas are proved and inequalities are established for co-ordinated convex functions and co-ordinated s-convex functions Mathematics Subject Classification (2000): 26D10; 26D15 Keywords: Hadamard-type inequality; co-ordinates; s-convex functions 1 ă M UHAMET EMIN OZDEMIR , HAVVA KAVURMACI , AHMET OCAK A Introduction Let f : I ⊆ R → R be a convex function defined on the interval I of real numbers and a < b The following double inequality; f a+b ≤ b−a b f (x)dx ≤ a f (a) + f (b) is well known in the literature as Hermite–Hadamard inequality Both inequalities hold in the reversed direction if f is concave In [1], Orlicz defined s-convex function in the second sense as following: Definition A function f : R+ → R, where R+ = [0, ∞), is said to be s-convex in the second sense if f (αx + βy) ≤ αs f (x) + β s f (y) for all x, y ∈ [0, ∞), α, β ≥ with α + β = and for some fixed s ∈ (0, 1] We denote by Ks the class of all s-convex functions Obviously, one can see that if we choose s = 1, both definitions reduced to ordinary concept of convexity For several results related to above definition we refer readers to [2–10] In [11], Dragomir defined convex functions on the co-ordinates as following: Definition Let us consider the bidimensional interval ∆ = [a, b] × [c, d] in R2 with a < b, c < d A function f : ∆ → R will be called convex on the coordinates if the partial mappings fy : [a, b] → R, fy (u) = f (u, y) and fx : [c, d] → R, fx (v) = f (x, v) are convex where defined for all y ∈ [c, d] and x ∈ [a, b] Recall that SOME INTEGRAL INEQUALITIES the mapping f : ∆ → R is convex on ∆ if the following inequality holds, f (λx + (1 − λ)z, λy + (1 − λ)w) ≤ λf (x, y) + (1 − λ)f (z, w) for all (x, y), (z, w) ∈ ∆ and λ ∈ [0, 1] In [11], Dragomir established the following inequalities of Hadamard-type for co-ordinated convex functions on a rectangle from the plane R2 Theorem Suppose that f : ∆ = [a, b] × [c, d] → R is convex on the co-ordinates on ∆ Then one has the inequalities; f (1.1) ≤ ≤ ≤ 1 b−a b f c+d b dx + d−c d f c a+b , y dy d f (x, y)dxdy a c b 1 (b − a) (d − c) x, a (b − a)(d − c) + ≤ a+b c+d , 2 f (x, c)dx + a d f (a, y)dy + c b (b − a) (d − c) f (x, d)dx a d f (b, y)dy c f (a, c) + f (a, d) + f (b, c) + f (b, d) The above inequalities are sharp Similar results can be found in [12–14] In [13], Alomari and Darus defined co-ordinated s-convex functions and proved some inequalities based on this definition Another definition for co-ordinated sconvex functions of second sense can be found in [15] 1 ă M UHAMET EMIN OZDEMIR , HAVVA KAVURMACI , AHMET OCAK A Definition Consider the bidimensional interval ∆ = [a, b]×[c, d] in [0, ∞)2 with a < b and c < d The mapping f : ∆ → R is s-convex on ∆ if f (λx + (1 − λ)z, λy + (1 − λ)w) ≤ λs f (x, y) + (1 − λ)s f (z, w) holds for all (x, y), (z, w) ∈ ∆ with λ ∈ [0, 1] and for some fixed s ∈ (0, 1] In [16], Sarıkaya et al proved some Hadamard-type inequalities for co-ordinated convex functions as followings: Theorem Let f : ∆ ⊂ R2 → R be a partial differentiable mapping on ∆ := [a, b] × [c, d] in R2 with a < b and c < d If ∂2f ∂t∂s is a convex function on the co-ordinates on ∆, then one has the inequalities: (1.2) |J| ≤ × (b − a)(d − c) 16 ∂2f ∂t∂s (a, c) + ∂2f ∂t∂s (a, d) + ∂2f ∂t∂s ∂2f ∂t∂s (b, c) + (b, d) where J= f (a, c) + f (a, d) + f (b, c) + f (b, d) + (b − a)(d − c) b d f (x, y)dxdy − A a c and A= 1 (b − a) b [f (x, c) + f (x, d)] dx + a (d − c) d [f (a, y)dy + f (b, y)] dy c Theorem Let f : ∆ ⊂ R2 → R be a partial differentiable mapping on ∆ := [a, b] × [c, d] in R2 with a < b and c < d If ∂2f ∂t∂s q , q > 1, is a convex function on SOME INTEGRAL INEQUALITIES the co-ordinates on ∆, then one has the inequalities: (1.3) |J| ≤  × (b − a)(d − c) (p + 1) p ∂2f ∂t∂s q ∂2f ∂t∂s (a, c) + q (a, d) + ∂2f ∂t∂s q ∂2f ∂t∂s (b, c) + q (b, d) where A, J are as in Theorem and p + q q  = Theorem Let f : ∆ ⊂ R2 → R be a partial differentiable mapping on ∆ := ∂2f ∂t∂s [a, b] × [c, d] in R2 with a < b and c < d If q , q ≥ 1, is a convex function on the co-ordinates on ∆, then one has the inequalities: (1.4) |J| ≤  × (b − a)(d − c) 16 ∂2f ∂t∂s q (a, c) + ∂2f ∂t∂s q (a, d) + ∂2f ∂t∂s q ∂2f ∂t∂s (b, c) + q (b, d) q  where A, J are as in Theorem In [17], Barnett and Dragomir proved an Ostrowski-type inequality for double integrals as following: Theorem Let f : [a, b] × [c, d] → R be continuous on [a, b] × [c, d], fx,y = ∂2f ∂x∂y exists on (a, b) × (c, d) and is bounded, that is fx,y ∞ = sup (x,y)∈(a,b)×(c,d) ∂ f (x, y) < ∞, ∂x∂y then we have the inequality; b d d f (s, t)dtds − (b − a) a ≤ c a+b (b − a)2 + x− (1.5) f (x, t)dt − (d − c) c b f (s, y)ds − (b − a)(d − c)f (x, y) a c+d (d − c)2 + y− 2 fx,y 1 ă M UHAMET EMIN OZDEMIR , HAVVA KAVURMACI , AHMET OCAK A for all (x, y) ∈ [a, b] × [c, d] In [18], Sarıkaya proved an Ostrowski-type inequality for double integrals and gave a corollary as following: Theorem Let f : [a, b] × [c, d] → R be an absolutely continuous functions such that the partial derivative of order exist and is bounded, i.e., ∂ f (t, s) ∂t∂s = ∞ sup (x,y)∈(a,b)×(c,d) ∂ f (t, s) 1, is a convex function on the co-ordinates on , ds ds ds ds ă MUHAMET EMIN OZDEMIR1 , HAVVA KAVURMACI∗1 , AHMET OCAK A 12 then the following inequality holds; (2.2) |C| ≤ (b − a) (d − c) (p + 1) p  ∂2f ∂t∂s × q ∂2f ∂t∂s (a, c) + q (b, c) + ∂2f ∂t∂s q (a, d) + ∂2f ∂t∂s (b, d) q q  where C is in the proof of Theorem Proof From Lemma 1, we have |C| ≤ (b − a) (d − c) b d × ∂2f ∂t∂s |p (x, t) q (y, s)| a c b−t t−a d−s s−c a+ b, c+ d b−a b−a d−c d−c dsdt By applying the well-known Hălder inequality for double integrals, then one has o (2.3)   |C| ≤ (b − a) (d − c)  b d × a Since ∂2f ∂t∂s q c ∂2f ∂t∂s b p d p |p (x, t) q (y, s)| dtds a c b−t t−a d−s s−c a+ b, c+ d b−a b−a d−c d−c is co-ordinated convex function on ∆, we can write ∂2f ∂t∂s (2.4) b−t t−a d−s s−c a+ b, c+ d b−a b−a d−c d−c b−t b−a ≤ d−s d−c ∂2f (a, c) ∂t∂s q b−t b−a s−c d−c ∂2f (a, d) ∂t∂s q + t−a b−a d−s d−c ∂2f (b, c) ∂t∂s q + + t−a b−a s−c d−c ∂2f (b, d) ∂t∂s q q q q dsdt SOME INTEGRAL INEQUALITIES 13 Using the inequality (2.4) in (2.3), we get |C| ≤ (b − a) (d − c) (p + 1) p  × ∂2f ∂t∂s q (a, c) + ∂2f ∂t∂s q ∂2f ∂t∂s (b, c) + q ∂2f ∂t∂s (a, d) + (b, d) q q  where we have used the fact that b p d p |p (x, t) q (y, s)| dtds a = c 1+ p [(b − a) (d − c)] (p + 1) p This completes the proof Remark Suppose that all the assumptions of Theorem are satisfied If we choose ∂2f ∂t∂s is bounded, i.e., ∂ f (t, s) ∂t∂s = ∞ ∂ f (t, s) < ∞, ∂t∂s sup (t,s)∈(a,b)×(c,d) we get (2.5) |C| ≤ (b − a) (d − c) ∂ f (t, s) ∂t∂s (p + 1) p which is the inequality in (1.3) with ∂ f (t,s) ∂t∂s ∞ ∞ Theorem Let f : ∆ = [a, b] × [c, d] → R be a partial differentiable mapping on ∆ = [a, b] × [c, d] If ∂2f ∂t∂s q , q ≥ 1, is a convex function on the co-ordinates on ∆, then the following inequality holds; (2.6) |C| ≤  × (b − a) (d − c) 16 ∂2f ∂t∂s q (a, c) + ∂2f ∂t∂s q (b, c) + where C is in the proof of Theorem ∂2f ∂t∂s q (a, d) + ∂2f ∂t∂s (b, d) q q ă MUHAMET EMIN OZDEMIR1 , HAVVA KAVURMACI∗1 , AHMET OCAK A 14 Proof From Lemma and applying the well-known Power mean inequality for double integrals, then one has |C| ≤ (b − a) (d − c) b d (2.7) × |p (x, t) q (y, s)| a ≤ c  b ∂2f ∂t∂s |p (x, t) q (y, s)| dsdt a c t−a d−s s−c b−t a+ b, c+ d b−a b−a d−c d−c ∂ f ∂t∂s c q dsdt 1− q |p (x, t) q (y, s)| a b−t t−a d−s s−c a+ b, c+ d b−a b−a d−c d−c d d × Since b (b − a) (d − c) ∂2f ∂t∂s q q dsdt is co-ordinated convex function on ∆, we can write ∂2f ∂t∂s (2.8) b−t t−a d−s s−c a+ b, c+ d b−a b−a d−c d−c b−t b−a ≤ d−s d−c ∂2f (a, c) ∂t∂s q q b−t b−a s−c d−c ∂2f (a, d) ∂t∂s q + t−a b−a d−s d−c ∂2f (b, c) ∂t∂s q + + t−a b−a s−c d−c ∂2f (b, d) ∂t∂s q If we use (2.8) in (2.7), we get   |C| ≤ (b − a) (d − c)  b |p (x, t) q (y, s)| dsdt a c |p (x, t) q (y, s)| a c t−a b−a d−s d−c 1− q d d × + b d−s d−c b−t b−a q ∂2f (b, c) + ∂t∂s t−a b−a q ∂2f (a, c) + ∂t∂s s−c d−c b−t b−a ∂2f (b, d) ∂t∂s s−c d−c q q ∂2f (a, d) ∂t∂s q SOME INTEGRAL INEQUALITIES 15 Computing the above integrals and using the fact that b 1− q d |p (x, t) q (y, s)| dtds (b − a) (d − c) 16 = c a 1− q This completes the proof Inequalities for co-ordinated s-convex functions To prove our main results we need the following lemma: Lemma Let f : ∆ ⊂ R2 → R be an absolutely continuous function on ∆ where a < b, c < d and t, λ ∈ [0, 1], if ∂2f ∂t∂λ ∈ L (∆), then the following equality holds: f (a, c) + r2 f (a, d) + r1 f (b, c) + r1 r2 f (b, d) (r1 + 1) (r2 + 1) + − − = (b − a)(d − c) b d f (x, y)dxdy a r2 r2 + 1 d−c r2 r2 + 1 b−a c d f (b, y)dy − c b f (x, d)dx − a r1 + 1 d−c r2 + 1 b−a d f (a, y)dy c b f (x, c)dx a (b − a)(d − c) (r1 + 1) (r2 + 1) 1 × ((r1 + 1) t − 1) ((r2 + 1) λ − 1) 0 ∂2f (tb + (1 − t) a, λd + (1 − λ) c) dtdλ ∂t∂λ where D= + − − f (a, c) + r2 f (a, d) + r1 f (b, c) + r1 r2 f (b, d) (r1 + 1) (r2 + 1) (b − a)(d − c) b d f (x, y)dxdy a r2 r2 + 1 d−c r2 r2 + 1 b−a c d f (b, y)dy − c b f (x, d)dx − a r1 + 1 d−c r2 + 1 b−a d f (a, y)dy c b f (x, c)dx a ă MUHAMET EMIN OZDEMIR1 , HAVVA KAVURMACI∗1 , AHMET OCAK A 16 and 1 E= ((r1 + 1) t − 1) ((r2 + 1) λ − 1) 0 ∂2f (tb + (1 − t) a, λd + (1 − λ) c) dtdλ ∂t∂λ for some fixed r1 , r2 ∈ [0, 1] Proof Integration by parts, we get E = ((r2 + 1) λ − 1) × ((r1 + 1) t − 1) = ((r2 + 1) λ − 1) − r1 + b−a ((r2 + 1) λ − 1) − = r1 + b−a ((r1 + 1) t − 1) ∂f (tb + (1 − t) a, λd + (1 − λ) c) (b − a) ∂λ − ∂f (tb + (1 − t) a, λd + (1 − λ) c) dt dλ ∂λ − ((r2 + 1) λ − 1) f (a, λd + (1 − λ) c) b−a d−c r1 + b−a 1 ((r2 + 1) λ − 1) 0 r1 ∂f ∂f (b, λd + (1 − λ) c) + (a, λd + (1 − λ) c) b − a ∂λ b − a ∂λ r1 ((r2 + 1) λ − 1) f (b, λd + (1 − λ) c) b−a d−c + ∂f (tb + (1 − t) a, λd + (1 − λ) c) dt dλ ∂λ = ∂2f (tb + (1 − t) a, λd + (1 − λ) c) dt dλ ∂t∂λ r1 (r2 + 1) (b − a)(d − c) − f (b, λd + (1 − λ) c) dλ (r2 + 1) (b − a)(d − c) f (a, λd + (1 − λ) c) dλ ∂f (tb + (1 − t) a, λd + (1 − λ) c) dλ dt ∂λ Computing these integrals, we obtain E= [f (a, c) + r2 f (a, d) + r1 f (b, c) + r1 r2 f (b, d) (b − a)(d − c) − r1 (r2 + 1) f (b, λd + (1 − λ) c)dλ − (r2 + 1) f (a, λd + (1 − λ) c)dλ 0 − r2 (r1 + 1) f (tb + (1 − t) a, d)dt − (r2 + 1) f (tb + (1 − t) a, c)dt 1 (r1 + 1) (r2 + 1) f (tb + (1 − t) a, λd + (1 − λ) c) dtdλ 0 SOME INTEGRAL INEQUALITIES 17 Using the change of the variable x = tb + (1 − t) a and y = λd + (1 − λ) c for t, λ ∈ [0, 1] and multiplying the both sides by (b−a)(d−c) (r1 +1)(r2 +1) , we get the required result Theorem 10 Let f : ∆ = [a, b] × [c, d] ⊂ [0, ∞)2 → [0, ∞) be an absolutely ∂2f ∂t∂λ continuous function on ∆ If is s-convex function on the co-ordinates on ∆, then one has the inequality: |D| ≤ (3.1) (b − a)(d − c) 2 (r1 + 1) (r2 + 1) (s + 1) (s + 2) × MN + KM ∂2f ∂2f (a, c) + LN (a, d) ∂t∂λ ∂t∂λ ∂2f ∂2f (b, c) + KL (b, d) ∂t∂λ ∂t∂λ where = s + + (r1 + 1) r1 r1 + s+2 M = s + + (r2 + 1) r2 r2 + s+2 N = r2 (s + 1) + r2 + s+1 L = r1 (s + 1) + r1 + s+1 K − r1 − r2 −1 −1 Proof From Lemma and by using co-ordinated s-convexity of f, we have; |D| ≤ (b − a)(d − c) (r1 + 1) (r2 + 1) 1 × |((r1 + 1) t − 1) ((r2 + 1) λ − 1)| 0 ∂2f (tb + (1 − t) a, λd + (1 − λ) c) dtd t ă MUHAMET EMIN OZDEMIR1 , HAVVA KAVURMACI1 , AHMET OCAK A 18 (b − a)(d − c) (r1 + 1) (r2 + 1) ≤ 1 |((r1 + 1) t − 1) ((r2 + 1) λ − 1)| × 0 ts ∂2f s ∂ f (b, λd + (1 − λ) c) + (1 − t) (a, λd + (1 − λ) c) ∂t∂λ ∂t∂λ dt dλ By calculating the above integrals, we have (3.2) |((r1 + 1) t − 1)| ts s + (1 − t) r1 +1 = ∂2f (b, λd + (1 − λ) c) ∂t∂λ ∂2f (a, λd + (1 − λ) c) ∂t∂λ (1 − (r1 + 1) t) ts s + (1 − t) + r1 +1 s = ∂2f (b, λd + (1 − λ) c) ∂t∂λ ∂2f (a, λd + (1 − λ) c) ∂t∂λ ((r1 + 1) t − 1) ts + (1 − t) r1 (s + 1) + + s + + (r1 + 1) dt ∂2f (b, λd + (1 − λ) c) ∂t∂λ ∂2f (a, λd + (1 − λ) c) ∂t∂λ (s + 1) (s + 2) dt r1 r1 + dt r1 + s+2 − r1 s+1 −1 ∂2f (b, λd + (1 − λ) c) ∂t∂λ ∂2f (a, λd + (1 − λ) c) ∂t∂λ By a similar argument for other integrals, by using co-ordinated s-convexity of ∂2f ∂t∂λ , we get |((r2 + 1) λ − 1)| r2 +1 ≤ ∂2f ∂2f (b, λd + (1 − λ) c) + (a, λd + (1 − λ) c) ∂t∂λ ∂t∂λ (1 − (r2 + 1) λ) λs + r2 +1 ((r2 + 1) λ − 1) λs ∂2f s ∂ f (b, d) + (1 − λ) (b, c) ∂t∂λ ∂t∂λ ∂2f s ∂ f (a, d) + (1 − λ) (a, c) ∂t∂λ ∂t∂λ dλ dλ dλ SOME INTEGRAL INEQUALITIES r2 (s + 1) + (s + 1) (s + 2) = r2 + 1 r2 + 19 s+1 ∂2f (b, d) ∂t∂λ −1 s+1 + r2 (s + 1) + (s + 1) (s + 2) s + + (r2 + 1) (s + 1) (s + 2) r2 r2 + s+2 + s + + (r2 + 1) (s + 1) (s + 2) r2 r2 + s+2 + ∂2f (a, d) ∂t∂λ −1 − r2 ∂2f (b, c) ∂t∂λ − r2 ∂2f (a, c) ∂t∂λ By using these in (3.2), we obtain the inequality (3.1) (1) If we choose r1 = r2 = in (3.1), we have (3.3) 1 f (a, c) + f (a, d) + f (b, c) + f (b, d) − d−c − ≤ 1 b−a b [f (x, d) + f (x, c)] dx + a (b − a)(d − c) (s + 1) (s + 2) × s+ 2s d [f (b, y) + f (a, y)] dy c (b − a)(d − c) b d f (x, y)dxdy a c ∂2f ∂2f ∂2f ∂2f (a, c) + (a, d) + (b, c) + (b, d) ∂t∂λ ∂t∂λ ∂t∂λ ∂t∂λ (2) If we choose r1 = r2 = in (3.1), we have f (a, c) − + ≤ d−c (b − a)(d − c) d f (a, y)dy − c b b−a b f (x, c)dx a d f (x, y)dxdy a c (b − a)(d − c) 2 (s + 1) (s + 2) × (s + 1) ∂2f ∂2f ∂2f ∂2f (a, c) + (s + 1) (a, d) + (s + 1) (b, c) + (b, d) ∂t∂λ ∂t∂λ ∂t∂λ ∂t∂λ Remark If we choose s = in (3.3), we get an improvement for the inequality (1.2) ¨ MUHAMET EMIN OZDEMIR1 , HAVVA KAVURMACI∗1 , AHMET OCAK A 20 Theorem 11 Let f : ∆ = [a, b] × [c, d] ⊂ [0, ∞)2 → [0, ∞) be an absolutely continuous function on ∆ If ∂2f ∂t∂λ p p−1 is s-convex function on the co-ordinates on ∆, for some fixed s ∈ (0, 1] and p > 1, then one has the inequality: p+1 p+1 + r1 p (3.4) |D| ≤  × + r2 p (b − a)(d − c) 1 (r1 + 1) (r2 + 1) (r1 + 1) p (r2 + 1) p (p + 1) p ∂2f ∂t∂λ q (a, c) + ∂2f ∂t∂λ q (a, d) + ∂2f ∂t∂λ q q ∂2f ∂t∂λ (b, c) + (b, d) q  (s + 1) for some fixed r1 , r2 ∈ [0, 1] , where q = p p−1 Proof Let p > From Lemma and using the Hălder inequality for double inteo grals, we can write |D| (b − a)(d − c) (r1 + 1) (r2 + 1) ≤ 1 × 0 1 p p |((r1 + 1) t − 1) ((r2 + 1) λ − 1)| dtdλ 0 q ∂2f (tb + (1 − t) a, λd + (1 − λ) c) dtdλ ∂t∂λ In above inequality using the s-convexity on the co-ordinates of q ∂2f ∂t∂λ q on ∆ and calculating the integrals, then we get the desired result (1) Under the assumptions of Theorem 11, if we choose r1 = r2 = in (3.4), we have (3.5) f (a, c) + f (a, d) + f (b, c) + f (b, d) − + 1 d−c d [f (b, y) + f (a, y)] dy + c (b − a)(d − c) b d f (x, y)dxdy a c b−a b [f (x, d) + f (x, c)] dx a SOME INTEGRAL INEQUALITIES ≤ 21 (b − a)(d − c) (p + 1) p  × ∂2f ∂t∂λ q ∂2f ∂t∂λ (a, c) + q q ∂2f ∂t∂λ (a, d) + (b, c) + q ∂2f ∂t∂λ (b, d) (s + 1) q  (2) Under the assumptions of Theorem 11, if we choose r1 = r2 = in (3.4), we have f (a, c) − + = d−c (b − a)(d − c) d f (a, y)dy − c b b−a b f (x, c)dx a d f (x, y)dxdy a c (b − a)(d − c) (p + 1) p  × ∂2f ∂t∂λ q (a, c) + ∂2f ∂t∂λ q ∂2f ∂t∂λ (a, d) + q (b, c) + q ∂2f ∂t∂λ (s + 1) (b, d) q  Remark If we choose s = in (3.5), we obtain an improvement for the inequality (1.3) Theorem 12 Let f : ∆ = [a, b] × [c, d] ⊂ [0, ∞)2 → [0, ∞) be an absolutely continuous function on ∆ If ∂2f ∂t∂λ q is s-convex function on the co-ordinates on ∆, for some fixed s ∈ (0, 1] and q ≥ 1, then one has the inequality: |D| ≤ 2 + r1 + r2 (r1 + 1) (r2 + 1) (b − a)(d − c) (r1 + 1) (r2 + 1)  × MN ∂2f ∂t∂λ q (a, c) + LN for some fixed r1 , r2 ∈ [0, 1] ∂2f ∂t∂λ q 1− q (a, d) + KM ∂2f ∂t∂λ (s + 1) (s + 2) q (b, c) + KL ∂2f ∂t∂λ q (b, d) q  ¨ MUHAMET EMIN OZDEMIR1 , HAVVA KAVURMACI∗1 , AHMET OCAK A 22 Proof From Lemma and using the well-known Power-mean inequality, we can write |D| ≤ 1 × Since (b − a)(d − c) (r1 + 1) (r2 + 1) ∂2f ∂t∂λ q 1− q |((r1 + 1) t − 1) ((r2 + 1) λ − 1)| dtdλ 0 q ∂2f (tb + (1 − t) a, λd + (1 − λ) c) dtdλ |((r1 + 1) t − 1) ((r2 + 1) λ − 1)| ∂t∂λ is s-convex function on the co-ordinates on ∆, we have ∂2f (tb + (1 − t) a, λd + (1 − λ) c) ∂t∂λ q q ≤ ts ∂2f s ∂ f (b, λd + (1 − λ) c) + (1 − t) (a, λd + (1 − λ) c) ∂t∂λ ∂t∂λ and ∂2f (tb + (1 − t) a, λd + (1 − λ) c) ∂t∂λ ≤ ts λs ∂2f ∂t∂λ +λs (1 − t) q s (b, d) + ts (1 − λ) s ∂2f ∂t∂λ q ∂2f ∂t∂λ q (b, c) q s (a, d) + (1 − λ) (1 − t) s ∂2f ∂t∂λ q (a, c) hence, it follows that (3.6) (b − a)(d − c) |D| ≤ (r1 + 1) (r2 + 1) 1 1− q |((r1 + 1) t − 1) ((r2 + 1) λ − 1)| ts λs × 2 + r1 + r2 (r1 + 1) (r2 + 1) +ts (1 − λ) ∂2f ∂t∂λ s s s + (1 − λ) (1 − t) q s (b, c) + λs (1 − t) ∂2f ∂t∂λ ∂2f ∂t∂λ q q (a, c) dtdλ ∂2f ∂t∂λ q (a, d) q (b, d) q q SOME INTEGRAL INEQUALITIES 23 By a simple computation, one can see that 1 0 +ts (1 − λ) ∂2f ∂t∂λ s q s s =  MN q ∂2f ∂t∂λ ∂2f ∂t∂λ s (b, c) + λs (1 − t) q (b, d) q (a, d) q q ∂2f ∂t∂λ + (1 − λ) (1 − t)  ∂2f ∂t∂λ |((r1 + 1) t − 1) ((r2 + 1) λ − 1)| ts λs (a, c) dtdλ (a, c) + LN ∂2f ∂t∂λ q (a, d) + KM (s + 1) (s + 2) ∂2f ∂t∂λ q (b, c) + KL ∂2f ∂t∂λ q (b, d) q  where K, L, M , and N as in Theorem 10 By substituting these in (3.6) and simplifying we obtain the required result (1) Under the assumptions of Theorem 12, if we choose r1 = r2 = 1, we have f (a, c) + f (a, d) + f (b, c) + f (b, d) − + ≤ 1 d−c d [f (b, y) + f (a, y)] dy + c b (b − a)(d − c)  × MN ∂2f ∂t∂λ q b [f (x, d) + f (x, c)] dx a d f (x, y)dxdy a (b − a)(d − c) b−a c 1− q (a, c) + LN ∂2f ∂t∂λ q (a, d) + KM (s + 1) (s + 2) ∂2f ∂t∂λ q (b, c) + KL ∂2f ∂t∂λ q (2) Under the assumptions of Theorem 12, if we choose r1 = r2 = 0, we have f (a, c) + − d−c (b − a)(d − c) d f (a, y)dy − c b d f (x, y)dxdy a b−a c b f (x, c)dx a (b, d) q ă MUHAMET EMIN OZDEMIR1 , HAVVA KAVURMACI1 , AHMET OCAK A 24 ≤ (b − a)(d − c)  × MN ∂2f ∂t∂λ q 1− q (a, c) + LN ∂2f ∂t∂λ q (a, d) + KM (s + 1) (s + 2) ∂2f ∂t∂λ q (b, c) + KL ∂2f ∂t∂λ q (b, d) Remark Under the assumptions of Theorem 12, if we choose r1 = r2 = and s = 1, we have an improvement for the inequality (1.4) Competing interests The authors declare that they have no competing interests Authors’ contributions HK, AOA and MA carried out the design of the study and performed the analysis MEO (adviser) participated in its design and coordination All authors read and approved the final manuscript References [1] Orlicz, 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