DSpace at VNU: SOME NEW RESULTS ON THE FEJ´ ER AND HERMITE-HADAMARD INEQUALITIES

ROCKY MOUNTAIN JOURNAL OF MATHEMATICS Volume 43, Number 5, 2013 SOME NEW RESULTS ON ´ AND HERMITE-HADAMARD INEQUALITIES THE FEJER ˆ´C-ANH NGO ˆ VU NHAT HUY AND QUO ABSTRACT The Hermite-Hadamard inequality and its generalization, the Fej´ er inequality, have many applications A simple application is to approximate the definite integral b f (x) dx if the function f is convex In this short note, we a show how to relax the convexity property of the function f , and thus we obtain inequalities that involve a larger class of functions This new study also raises some open questions Introduction The Hermite-Hadamard inequality [5, 6] says that (1) f a+b b−a b a f (t) dt f (a) + f (b) holds for any convex function f : I → R and a, b ∈ I As a generalization of (1), the Fej´er inequality [4] says that (2) b b a+b f (a) + f (b) b p(x) dx f (x)p(x) dx p(x) dx f 2 a a a holds for any convex function f : I → R, where a, b ∈ I and p : [a, b] → R is non-negative integrable and symmetric about x = (a + b)/2 Apparently, inequality (2) goes back to inequality (1) if we put p ≡ 1/(b − a) Inequalities (1) and (2) provide a simple way to evaluate b the integral a f (x) dx These inequalities have many extensions and generalizations, see [1, 2, 9] In this paper we present some new refinements of inequalities (1) and (2) Keywords and phrases Integral inequality, Fej´ er, Hermite-Hadamard This work was partially supported by the Vietnam National Foundation for Science and Technology Development (Project No 101 01 50 09) The second author is the corresponding author Received by the editors on July 28, 2010, and in revised form on February 2, 2011 DOI:10.1216/RMJ-2013-43-5-1625 Copyright c 2013 Rocky Mountain Mathematics Consortium 1625 ˆ´C-ANH NGO ˆ VU NHAT HUY AND QUO 1626 Obviously, inequality (2) can be rewritten as (3) f a+b b f (a) + f (b) − b a p(x) dx a f (x)p(x) dx − f (a) + f (b) b a p(x) dx and b a (4) f (x)p(x) dx − f a+b f (a) + f (b) a+b −f 2 b a p(x) dx b a p(x) dx which says that b a f (x)p(x) dx − f (a) + f (b) b a b a p(x) dx f (x)p(x) dx − f a+b b a p(x) dx We observe that, under certain conditions, we can relax the convexity property of function f This is the aim of the present paper Precisely, both inequalities (1) and (2) require function f to be convex; as a consequence, it is natural to assume that f is twice0 Our first result concerns the differentiable Consequently, f case when f is bounded in [a, b] Note that, we not require f to be non-negative Precisely, we first prove the following result: Theorem Suppose p(x) is symmetric about (a + b)/2 and f : [a, b] → R is a twice-differentiable function such that f is bounded ´ AND HERMITE-HADAMARD INEQUALITIES THE FEJER 1627 in [a, b] Then (a+b)/2 (5) m a+b −x a p(x) dx b a f (x)p(x) dx − f (a+b)/2 M a+b a+b −x a b p(x) dx a p(x) dx and (6) −M (a+b)/2 a (x − a)(b − x)p(x) dx b a f (x)p(x) dx − −m (a+b)/2 a f (a) + f (b) b a p(x) dx (x − a)(b − x)p(x) dx where m = inf f (t), M = sup f (t) t∈[a,b] Remark If f inequality (2) t∈[a,b] 0, then we obtain an improvement of the Fej´er Next we consider the case when f is of class Lp ([a, b]); we also obtain the following estimates: Theorem Let < p < ∞ and p(x) be symmetric about (a + b)/2 and f : [a, b] → R be a twice-differentiable function such that f ∈ Lp ([a, b]) Then b (7) a f (x)p(x) dx − f b a+b q f 2(q + 1) p p(x) dx a (a+b)/2 a (a + b − 2x)(1/q)+1 p(x) dx ˆ´C-ANH NGO ˆ VU NHAT HUY AND QUO 1628 and f (a) + f (b) (8) q f 2(q + 1) b a p(x) dx − (a+b)/2 p a b a f (x)p(x) dx (b−a)(1/q)+1 −(a+b−2x)(1/q)+1 p(x) dx, where q is defined to be p/(p − 1) Finally, it is clear to see that inequality f implies that f is non-decreasing Therefore, in the next result, we assume that (9) f (a + b − x) f (x), for all x ∈ a, a+b Clearly, if f is non-decreasing, then inequality (9) holds However, it is obvious to see that the reverse statement is not true Theorem Suppose that p(x) is symmetric about (a + b)/2 and f : [a, b] → R is a differentiable function satisfying f (a+b−x) f (x), for all x ∈ [a, (a + b)/2] Then (10) b b a+b f (a) + f (b) b p(x) dx f (x)p(x) dx p(x) dx f 2 a a a holds Remark It is worth noticing that the assumption f is a differentiable function which has been used in the literature; for example, in [3] the authors assumed f is convex on [a, b] They then obtained some refinements of the Hermite-Hadamard inequality (1) By using Theorems 3, it turns out that the question of deriving a sharp version becomes open We hope that we will soon see some responses on this problem ´ AND HERMITE-HADAMARD INEQUALITIES THE FEJER 1629 Proofs Proof of Theorem We firstly prove (5) Since p(x) about (a + b)/2, we have b a b f (x)p(x) dx = a b = a is symmetric f (a + b − x)p(a + b − x) dx f (a + b − x)p(x) dx So b (11) a f (x)p(x) dx = b f (x) + f (a + b − x) p(x) dx, a which gives b a f (x)p(x) dx − f = b a a+b b a p(x) dx f (x) + f (a + b − x) − 2f Since f (x) + f (a + b − x) − 2f a+b a+b p(x) dx p(x) is symmetric about (a + b)/2, one has b a f (x) + f (a + b − x) − 2f (a+b)/2 =2 a a+b p(x) dx f (x) + f (a + b − x) − 2f a+b p(x) dx, a+b p(x) dx which implies b (12) a f (x)p(x) dx − f (a+b)/2 = a a+b b a p(x) dx f (x) + f (a + b − x) − 2f ˆ´C-ANH NGO ˆ VU NHAT HUY AND QUO 1630 Since f (a + b − x) − f a+b and f a+b − f (x) = a+b−x = (a+b)/2 f (t) dt (a+b)/2 f (t) dt, x then f (x) + f (a + b − x) − 2f a+b a+b−x = (a+b)/2 f (t) dt − (a+b)/2 (a+b)/2 f (t) dt x f (a + b − t) dt − = x (a+b)/2 x f (t) dt Therefore, (13) f (x) + f (a + b − x) − 2f a+b (a+b)/2 = x f (a + b − t) − f (t) dt Since f (a + b − t) − f (t) = (14) a+b−t f (y) dy t then for t ∈ [a, (a + b)/2], one has m(a + b − 2t) f (a + b − t) − f (t) M (a + b − 2t) Thus, (a+b)/2 x m(a + b − 2t) dt f (x) + f (a + b − x) − 2f (a+b)/2 x M (a + b − 2t) dt a+b ´ AND HERMITE-HADAMARD INEQUALITIES THE FEJER 1631 A simple calculation shows us that m a+b −x 2 a+b f (x) + f (a + b − x) − 2f a+b −x M Then (a+b)/2 m a+b −x a p(x) dx b a f (x)p(x) dx − f (a+b)/2 M a+b 2 a+b −x a b a p(x) dx p(x) dx This completes the proof of (5) We now prove (6) By using (11), one has b a f (x)p(x) dx − = f (a) + f (b) b a b a p(x) dx f (x) + f (a + b − x) − f (a) + f (b) p(x) dx Since the following function f (x) + f (a + b − x) − f (a) + f (b) p(x) is symmetric about (a + b)/2, one gets b (15) a f (x)p(x) dx − (a+b)/2 = a f (a) + f (b) b a p(x) dx f (x) + f (a + b − x) − f (a) + f (b) Since f (b) − f (a + b − x) = b a+b−x f (t) dt p(x) dx ˆ´C-ANH NGO ˆ VU NHAT HUY AND QUO 1632 and x f (x) − f (a) = f (t) dt, a then we have f (x) + f (a + b − x) − f (a) + f (b) x = a x = a f (t) dt − f (t) dt − b a+b−x x a f (t) dt f (a + b − t) dt Therefore, (16) f (x) + f (a + b − x) − f (a) + f (b) x =− a f (a + b − t) − f (t) dt We also have f (a + b − t) − f (t) = (17) a+b−t t f (y) dy which implies, for t ∈ [a, (a + b)/2], that m(a + b − 2t) f (a + b − t) − f (t) M (a + b − 2t) Hence, − x a M (a + b − 2t) dt f (x) + f (a + b − x) − f (a) + f (b) − x a m(a + b − 2t) dt Thus, −M (x − a)(b − x) f (x) + f (a + b − x) − f (a) + f (b) −m(x − a)(b − x) ´ AND HERMITE-HADAMARD INEQUALITIES THE FEJER 1633 It follows that −M (a+b)/2 a (x − a)(b − x)p(x) dx b a f (x)p(x) dx − −m (a+b)/2 a f (a) + f (b) b a p(x) dx (x − a)(b − x)p(x) dx The proof is complete Proof of Theorem We firstly prove (7) From (12) (13), one has b a a+b f (x)p(x) dx − f = b a (a+b)/2 = a b a p(x) dx f (x) + f (a + b − x) − 2f a+b f (x) + f (a + b − x) − 2f a+b p(x) dx p(x) dx and a+b f (x) + f (a + b − x) − 2f (a+b)/2 = x f (a + b − t) − f (t) dt Note that, by (14), f (a + b − t) − f (t) = where a t a+b−t f (y) dy t (a + b)/2, which implies |f (a + b − t) − f (t)| 1/q a+b−t t t 1/q a+b−t t a+b−t dy dy = (a + b − 2t)1/q f f p p | f (y)|p dy 1/p ˆ´C-ANH NGO ˆ VU NHAT HUY AND QUO 1634 Thus, b a f (x)p(x)dx − f b a+b p(x) dx a (a+b)/2 q f 2(q + 1) p a (a + b − 2x)(1/q)+1 p(x) dx The proof of (7) is complete We now prove (8) From (15) (16), one has b f (a) + f (b) a p(x) dx − b b f (x)p(x) dx a f (x) + f (a + b − x) − f (a) + f (b) p(x) dx a (a+b)/2 a f (x) + f (a + b − x) − f (a) + f (b) p(x) dx and x f (x) + f (a + b − x) − f (a) + f (b) f (a + b − t) − f (t) dt a Note that, by (17), f (a + b − t) − f (t) = where a t a+b−t f (y) dy t (a + b)/2, which implies 1/q a+b−t |f (a + b − t) − f (t)| a+b−t dy t t 1/p 1/q a+b−t dy t |f (y)|p dy f = (a + b − 2t)1/q f p p Therefore, f (a) + f (b) q f 2(q + 1) b a p(x) dx − (a+b)/2 p a b a f (x)p(x) dx (1/q)+1 (b − a) (1/q)+1 −(a + b − 2x) p(x) dx ´ AND HERMITE-HADAMARD INEQUALITIES THE FEJER 1635 Proof of Theorem The proof of Theorem comes from the proofs of Theorems and From (12) and (13), one has b a f (x)p(x) dx − f (a+b)/2 b a+b a (a+b)/2 = a x p(x) dx f (a + b − t) − f (t) dt p(x) dx Similarly, from (15) and (16), one gets f (a) + f (b) b a p(x) dx − b a f (x)p(x) dx (a+b)/2 x = a a f (a + b − t) − f (t) dt p(x) dx Thus, the proof follows from the assumption Acknowledgments The authors wish to express their gratitude to the anonymous referee for a number of valuable comments and suggestions which helped to improve the presentation of the paper REFERENCES J.L Brenner and H Alzer, Integral inequalities for concave functions with applications to special functions, Proc Roy Soc Edinburgh 18 (1991), 173 192 S.S Dragomir, On Hadamards inequalities for convex functions, Math Balkan (1992), 215 222 S.S Dragomir and R.P Agarwal, Two inequalities for differentiable mappings and applications to special means of real numbers and to trapezoidal formula, Appl Math Lett 11 (1998), 91 95 L Fej´ er, Uberdie fourierreihen, II, Math Natur Ungar Akad Wiss 24 (1906), 369 390 ´ J Hadamard, Etude sur les proprietes des fonctions entieres et en particulier d’une fonction consideree par Riemann, J Math Pures Appl 58 (1893), 171 215 J.E Pecaric, F Proschan and Y.L Tong, Convex functions, partial orderings, and statistical applications, Academic Press, New York, 1992 K.L Tseng and C.S Wang, Some refinements of the Fej´ ers inequality for convex functions, Tamsui Oxford J Math Sci 21 (2005), 94 104 G.S Yang, D.Y Hwang and K.L Tseng, Some inequalities for differentiable convex and concave mappings, Comput Math Appl 47 (2004), 207 216 1636 ˆ´C-ANH NGO ˆ VU NHAT HUY AND QUO G.S Yang and K.L Tseng, On certain integral inequalities related to HermiteHadamard inequalities, J Math Anal Appl 239 (1999), 180 187 Department of Mathematics, College of Science, Viˆ et Nam National ` No ˆ i, Viˆ University, Ha et Nam Email address: nhat huy85@yahoo.com Department of Mathematics, National University of Singapore, Block S17 (SOC1), 10 Lower Kent Ridge Road, Singapore 119076 and Laboratoire de Math´ ematiques et de Physique Th´ eorique UFR Sciences et Technologie, Universit´ e Franc ¸ ois Rabelais Parc de Grandmont, 37200 Tours, France Email address: bookworm vn@yahoo.com ... their gratitude to the anonymous referee for a number of valuable comments and suggestions which helped to improve the presentation of the paper REFERENCES J.L Brenner and H Alzer, Integral inequalities. .. Proschan and Y.L Tong, Convex functions, partial orderings, and statistical applications, Academic Press, New York, 1992 K.L Tseng and C.S Wang, Some refinements of the Fej´ ers inequality for convex... [3] the authors assumed f is convex on [a, b] They then obtained some refinements of the Hermite-Hadamard inequality (1) By using Theorems 3, it turns out that the question of deriving a sharp version