In the article, a functional inequality in abstract spaces is established, which gives a new affirmative answer to an open problem posed by Feng Qi in Several integral inequalities which appeared in J. Inequal. Pure Appl.

Math. 1 (2000), no. 2, Art. 19. Moreover, some integral inequalities and a discrete inequality involving sums are deduced

Mazouzi, S

Qi, Feng

English

Victoria University Eprints Repository

ON AN OPEN PROBLEM BY FENG QI REGARDING ANINTEGRAL INEQUALITYS. MAZOUZI AND FENG QIAbstract. In the article, a functional inequality in abstract spaces is estab-lished, which gives a new affirmative answer to an open problem posed byFeng Qi in Several integral inequalities appeared in J. Inequal. Pure Appl.Math. 1 (2000), no. 2, Art. 19 (Available online at http://jipam.vu.edu.au/v1n2/001_00.html) and RGMIA Res. Rep. Coll. 2 (1999), no. 7, Art. 9,1039–1042 (Available online at http://rgmia.vu.edu.au/v2n7.html). More-over, some integral inequalities and a discrete inequality involving sums arededuced.1. IntroductionUnder what condition does the inequality∫ ba[f(x)]t dx ≥(∫ baf(x) dx)t−1(1)hold for t > 1?This problem was proposed by the second author, F. Qi, in [8] after the followinginequality was proved:∫ ba[f(x)]n+2 dx ≥ (∫ baf(x) dx)n+1, (2)where f(x) has continuous derivative of the n-th order on the interval [a, b], f (i)(a) ≥0 for 0 ≤ i ≤ n− 1, and f (n)(x) ≥ n!.In the joint paper [12], K.-W. Yu and F. Qi obtained an answer to the aboveproblem by using the integral version of Jensen’s inequality and a property of2000 Mathematics Subject Classification. Primary 39B62; Secondary 26D15.Key words and phrases. functional inequality, integral inequality, Jensen’s inequality.The second author was supported in part by NNSF (#10001016) of China, SF for the Promi-nent Youth of Henan Province (#0112000200), SF of Henan Innovation Talents at Universities,NSF of Henan Province (#004051800), Doctor Fund of Jiaozuo Institute of Technology, CHINA.This paper was typeset using AMS-LATEX.1convexity: Inequality (1) is valid for all f ∈ C([a, b]) such that ∫ baf(x) dx ≥ (b −a)t−1 for given t > 1.Let [x] denote the greatest integer less than or equal to x, f (−1)(x) =∫ xaf(s) ds,f (0)(x) = f(x), γ(t) = t(t − 1)(t − 2) · · · [t − (n − 1)] for t ∈ (n, n + 1], andγ(t) = 1 for t < 1, where n is a positive integer. In [11], N. Towghi provided othersufficient conditions for inequality (1) to be valid: If f (i)(a) ≥ 0 for i ≤ [t− 2] andf [t−2](x) ≥ γ(t − 1)(x − a)(t−[t]), then ∫ baf(x) dx ≥ (b − a)t−1 and inequality (1)holds.T. K. Poga´ny in [7], by avoiding the assumptions of differentiability used in [8, 11]and the convexity criteria used in [12], and instead using the classical integral in-equalities due to Ho¨lder, Nehari, Barnes and their generalizations by Godunova andLevin, established some inequalities which generalize, reverse, or weight inequality(1).In this paper, by employing a functional inequality introduced in [5], which is anabstract generalization of the classical Jensen’s inequality [9], we further establishthe following functional inequality (4) from which inequality (1), some integralinequality, and an interesting discrete inequality involving sums can be deduced.Theorem 1. Let L be a linear vector space of real-valued functions, p and q be tworeal numbers such that p ≥ q ≥ 1. Assume that f and g are two positive functionsin L and G is a positive linear form on L such that(1) G(g) > 0,(2) fg and gfp ∈ L.If[G(g)]p−1 ≤ [G(gf)]p−q, (3)then[G(gf)]q ≤ G(gfp). (4)The new inequality (4) has the feature that it is stated for summable functionsdefined on a finite measure space (E,Σ, µ) whose L1-norms are bounded from be-low by some constant involving the measure of the whole space E as well as theexponents p and q.22. Lemma and proof of Theorem 1To prove our main result, Theorem 1, it is necessary to recall a functional in-equality in [5], which can be stated as follows.Lemma 1. Let L be a linear vector space of real valued functions and f, g ∈ L withg ≥ 0. Assume that F is a positive linear form on L and ϕ : R → R is a convexfunction such that(1) F (g) = 1,(2) fg and (ϕ ◦ f)g ∈ L.Thenϕ(F (fg)) ≤ F ((ϕ ◦ f)g). (5)Notice that Lemma 1 is in fact a form of the classical Jensen inequality. Thereis a vast literature on this subject, see, e.g., [1, 2, 3, 4, 6, 10] and references therein.Proof of Theorem 1. Define a positive linear form F (u) = G(u)G(g) , then, we obviouslyhave F (g) = 1. From Lemma 1, if we take as a convex function ϕ(x) = xp forp ≥ 1, then[F (gf)]p ≤ F (gfp), (6)that is, [G(gf)G(g)]p≤ G(gfp)G(g)which gives[G(gf)]p−q[G(g)]p−1[G(gf)]q ≤ G(gfp).Since inequality (3) holds, thus inequality (4) follows. 3. Corollaries and remarksAs a new positve and concrete answer to F. Qi’s problem mentioned at thebeginning of this paper, we have the followingCorollary 1. Let (E,Σ, µ) be a finite measure space and let L be the space of allintegrable functions on E. If p and q are two real numbers such that p ≥ q ≥ 1, andf and g are two positive functions of L such that(1)∫Eg dµ > 0,(2) fg and gfp ∈ L,3then (∫Egf dµ)q≤∫Egfp dµ, (7)provided that(∫Egf dµ)p−q ≥ (∫Eg dµ)p−1.Proof. This follows from Theorem 1 by taking G(u) =∫Eu dµ as a positive linearform. Remark 1. We observe that if p = q and G(g) ≤ 1, then inequality (3) is alwaysfulfilled, and accordingly, we have[G(gf)]p ≤ G(gfp)for all p ≥ 1.Remark 2. If L contains the constant functions, then forf =0, p ≥ q ≥ 1,[G(g)]p−q, p > q ≥ 1,1, p = q, G(g) = 1,(8)equality occurs in (4)Remark 3. In fact, inequality (4) holds even if inequality (3), as merely a sufficientcondition, is not satisfied. Let p > q ≥ 1, m = q−1p−q and c =[q(p−qp−1)q−1]1/(p−q). IfE = [a, b] is a finite interval of R and f(x) = c(x−a)m, then (∫ baf dx)q = ∫ bafp dx.On the other hand, inequality (3) is no longer satisfied if q(p−qp−1)p−1< 1. This isdue to the fact that(∫ baf dx)p−q = q( p−qp−1)p−1(b− a)p−1.Corollary 2. Let f ∈ L1(a, b), the space of integrable functions on the interval(a, b) with respect to the Lebesgue measure, such that |f(x)| ≥ k(x) a.e. for x ∈(a, b), where(b− a)(p−1)/(p−q) ≤∫ bak(x) dx <∞ (9)for some p > q ≥ 1, then(∫ ba|f(x)|dx)q≤∫ ba|f(x)|p dx. (10)Proof. This follows from Lemma 1 easily. We now apply Corollary 2 to deduce F. Qi’s main result, Proposition 1.3 in [8],in detail.4Corollary 3. Suppose that f ∈ Cn([a, b]) satisfies f (i)(a) ≥ 0 and f (n)(x) ≥ n! forx ∈ [a, b], where 0 ≤ i ≤ n− 1 and n ∈ N, the set of all positive integers, then∫ ba[f(x)]n+2 dx ≥ (∫ baf(x) dx)n+1. (11)Proof. Since f (n)(x) ≥ n!, then successive integrations over [a, x] givef (n−k)(x) ≥ n!k!(x− a)k, k = 0, 1, . . . , n− 1,hence(x− a)n−kf (n−k)(x) ≥ n!k!(x− a)n, k = 0, 1, . . . , n− 1.On the other hand, Taylor’s expansion applied to f with Lagrange remainderstates thatf(x) = f(a) +n−1∑k=0f (k)(a)k!(x− a)k + f(n)(ξ)n!(x− a)n≥n∑k=0n!k!(n− k)! (x− a)n= 2n(x− a)n,where ξ ∈ (a, x). But since x is arbitrary and 2n ≥ n+ 1 for all n ∈ N, thenf(x) ≥ (n+ 1)(x− a)n ≥ 0for all x ∈ (a, b). Therefore ∫ baf(x) dx ≥ (b− a)n+1,and inequality (11) follows by virtue of Corollary 2. Remark 4. The following functionf : [a, b]→ R+, x 7→ f(x) = (x− a)n+1(n+ 2)nfor a fixed n ∈ N satisfies f ∈ Cn([a, b]) and f (i)(a) ≥ 0, for 0 ≤ i ≤ n − 1, butf (n)(x) = (n+1)!(n+2)n (x− a) for x ∈ [a, b]. This means that the condition f (n) ≥ n! on[a, b] is no longer fulfilled. However, we have(∫ baf dx)n+2=∫ bafn+3 dx =(b− a)(n+2)2(n+ 2)(n+1)(n+2).Finally, let us apply Corollary 1 to derive a discrete inequality.5Corollary 4. Let E = {a1, . . . , aN}, f : E → R+ defined by f(ai) = bi fori = 1, . . . , N , and let µ be a discrete positive measure given by µ({ai}) = αi > 0 fori = 1, . . . , N . If, for p ≥ q ≥ 1(N∑i=1αi)p−1≤(N∑i=1αibi)p−q, (12)then we have (N∑i=1αibi)q≤N∑i=1αibpi . (13)If, in particular, α1 = · · · = αN = c > 0 satisfiescq−1 ≤ 1Np−1(N∑i=1bi)p−q, (14)then (N∑i=1bi)q≤ 1cq−1N∑i=1bpi . (15)Proof. We observe that(∫Ef dµ)p−q=(N∑i=1f(ai)µ({ai}))p−q=(N∑i=1αibi)p−q≥(N∑i=1αi)p−1≡ [µ(E)]p−1.and thus, the sufficient condition is satisfied. We conclude by Corollary 1 that(∫Ef dµ)q=(N∑i=1αibi)q≤∫Efp dµ =N∑i=1αibpi .The proof of inequality (15) is a particular case of the above argument, and thuswe leave it to the reader. References[1] P. R. Beesack and J. E. Pecaric, On Jensen’s inequality for convex functions, J. Math. Anal.Appl. 110 (1985), 536–552.[2] P. R. Beesack and J. E. Pecaric, On Jensen’s inequality for convex functions, II, J. Math.Anal. Appl. 118 (1986), 125–144.6[3] P. R. Beesack and J. E. Pecaric, On Jensen’s inequality for convex functions, III, J. Math.Anal. Appl. 156 (1991), 231–239.[4] S. S. Dragomir, C. E. M. Pearce, and J. Pecaric, On Jensen’s and related inequalities forisotonic sublinear functionals, Acta Sci. Math (Szeged) 61(1995), 373–382.[5] S. Mazouzi, A functional inequality, Magh. Math. Rev. 3 (1994), no. 1, 83–87.[6] J. E. Pecˇaric´ and I. Rasa, On Jensen’s inequality, Acta Sci. Math. (Szeged) 56 (1992),305–309.[7] T. K. Poga´ny, On an open problem of F. Qi, J. Inequal. Pure Appl. Math. 3 (2002), no. 1,Art. 4. Available online at http://jipam.vu.edu.au/v3n4/016_01.html.[8] F. Qi, Several integral inequalities, J. Inequal. Pure Appl. Math. 1 (2000), no. 2, Art. 19.Available online at http://jipam.vu.edu.au/v1n2/001_00.html. RGMIA Res. Rep. Coll. 2(1999), no. 7, Art. 9, 1039–1042. Available online at http://rgmia.vu.edu.au/v2n7.html.[9] W. Rudin, Real and Complex Analysis, McGraw-Hill, Inc., New York, 1974.[10] J. Sa´ndor, On the Jensen-Hadamard inequality, Studia Univ. Babes-Bolyai Math. 36 (1991),no. 1, 9–15.[11] N. Towghi, Notes on integral inequalities, RGMIA Res. Rep. Coll. 4 (2001), no. 2, Art. 12,277–278. Available online at http://rgmia.vu.edu.au/v4n2.html.[12] K.-W. Yu and F. Qi, A short note on an integral inequality, RGMIA Res. Rep. Coll. 4 (2001),no. 1, Art. 4, 23–25. Available online at http://rgmia.vu.edu.au/v4n1.html.(S. Mazouzi) Department of Mathematics, Faculty of Sciences, University of Annaba,P. O. Box 12, Annaba 23000, ALGERIAE-mail address: mazouzi.s@voila.fr(F. Qi) Department of Applied Mathematics and Informatics, Jiaozuo Institute ofTechnology, Jiaozuo City, Henan 454000, CHINAE-mail address: qifeng@jzit.edu.cn7

On an Open Problem by Feng Qi Regarding an Integral Inequality

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