Volume 9 (2008), Issue 4, Article 108, 7 pp. FURTHER DEVELOPMENT OF AN OPEN PROBLEM YU MIAO AND JUNFEN LI C OLLEGE OF M ATHEMATICS AND I NFORMATION S CIENCE H ENAN N ORMAL U NIVERSITY H ENAN P ROVINCE , 453007, C HINA yumiao728@yahoo.com.cn junfen_li@yahoo.com.cn Received 06 March, 2008; accepted 5 October, 2008 Communicated by N.S. Barnett A BSTRACT. In this paper, we generalize an open problem posed by Ngô et al. in [Notes on an Integral Inequality, JIPAM, 7(4) (2006), Art.120] and give some answers which extend the results of Boukerrioua-Guezane-Lakoud [On an open question regarding an integral inequality, JIPAM, 8(3) (2007), Art. 77.] and Liu-Li-Dong [On an open problem concerning an integral inequality, JIPAM, 8 (3) (2007), Art. 74.]. Key words and phrases: Integral inequality, Cauchy inequality, Open problem. 2000 Mathematics Subject Classification. 26D15; 60E15. 1. I NTRODUCTION In [4], the following inequality is found. Theorem A. Let f (x) ≥ 0 be a continuous function on [0, 1] satisfying Z 1 Z 1 (1.1) f (t)dt ≥ tdt, ∀ x ∈ [0, 1], x x then Z 1 f (1.2) α+1 Z 1 xα f (x)dx, (x)dx ≥ 0 0 and Z 1 f (1.3) α+1 Z (x)dx ≥ 0 1 xf α (x)dx, 0 hold for every positive real number α > 0. The authors next proposed the following open problem: The authors are indebted to the referees and the editor for their many helpful and valuable comments. 069-08 2 Y. M IAO AND J. F. L I Open Problem 1. Let f (x) ≥ 0 be a continuous function on [0, 1] satisfying Z 1 Z 1 (1.4) f (t)dt ≥ tdt, ∀ x ∈ [0, 1]. x x Under what conditions does the inequality Z 1 Z α+β (1.5) f (x) ≥ 0 1 xα f β (x)dx 0 hold for α and β? Several answers and extension results have been given to this open problem [1, 3]. In the present paper, we obtain a generalization of the above Open Problem 1 and provide some resolutions to it. Here, and in what follows, we use X to denote a non-negative random variable (r.v.) on [0, ∞) with probability density function p(x). E(X) denotes the mathematical expectation of X and 1A := 1A (X) denotes the indicator function of the event A. Let At = [t, ∞). We now consider the following generalization of Open Problem 1. Open Problem 2. Let f (x) ≥ 0 be a continuous function on [0, ∞). Under what conditions does the inequality (1.6) E(f α+β (X)) ≥ E(X α f β (X)) hold for α and β? Remark 1. Let X possess a uniform distribution on the support interval [0, 1], i.e., the probability density function of X is equal to 1, x ∈ [0, 1] and zero elsewhere, then the above open problem becomes Open Problem 1. For convenience, we assume that all the necessary functions in the following are integrable. 2. M AIN R ESULTS Theorem 2.1. Let f (x) ≥ 0 be a continuous function on [0, ∞) satisfying (2.1) E(f β (X)1At ) ≥ E(X β 1At ), ∀ t ∈ [0, ∞). Then (2.2) E(f α+β (X)) ≥ E(X α f β (X)) holds for every positive real number α and β. Proof. By Fubini’s Theorem and (2.1), we have Z ∞ α β (2.3) E(X f (X)) = xα f β (x)p(x)dx 0 Z Z x 1 ∞ α−1 = t dt f β (x)p(x)dx α 0 0 Z ∞ Z ∞ 1 α−1 β = t f (x)p(x)dx dt α 0 t Z 1 ∞ α−1 = t E(f β (X)1At )dt α 0 J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 108, 7 pp. http://jipam.vu.edu.au/ F URTHER D EVELOPMENT OF AN O PEN P ROBLEM 3 Z 1 ∞ α−1 ≥ t E(X β 1At )dt α 0 Z ∞ Z 1 ∞ α−1 β = t x p(x)dx dt α 0 t Z Z x 1 ∞ α−1 = t dt xβ p(x)dx α 0 0 Z ∞ = xβ+α p(x)dx = E(X β+α ). 0 Using Cauchy’s inequality, we have β α f α+β (X) + X α+β ≥ X α f β (X), α+β α+β (2.4) which, by (2.3), yields (2.5) β α E(f α+β (X)) + E(X α+β ) ≥ E(X α f β (X)) ≥ E(X β+α ). α+β α+β Remark 2. If we assume that X possesses a uniform distribution on the support interval [0, 1] (or [0, b]), then the above theorem is Theorem 2.1 (or Theorem 2.4) of Liu-Li-Dong in [3]. Theorem 2.2. Let f (x) ≥ 0 be a continuous function on [0, ∞) satisfying (2.6) E(f (X)1At ) ≥ E(X1At ), ∀ t ∈ [0, ∞). Then (2.7) E(f α+β (X)) ≥ E(X β f α (X)) holds for every positive real number α ≥ 1 and β. Proof. By Fubini’s Theorem and (2.6), we have Z ∞ α+1 (2.8) E(X f (X)) = xα+1 f (x)p(x)dx 0 Z ∞ Z x 1 α = t dt f (x)p(x)dx α+1 0 0 Z ∞ Z ∞ 1 α = t f (x)p(x)dx dt α+1 0 t Z ∞ Z ∞ 1 α ≥ t xp(x)dx dt α+1 0 t Z x Z ∞ 1 α = xp(x) t dt dx = E(X α+2 ). α+1 0 0 Applying Cauchy’s inequality, we get 1 α α−1 α f (X) + X ≥ f (X)X α−1 . α α β Multiplying both sides of (2.9) by X , we have (2.9) (2.10) 1 α−1 E(f α (X)X β ) + E(X α+β ) ≥ E(f (X)X α+β−1 ) ≥ E(X α+β ), α α J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 108, 7 pp. http://jipam.vu.edu.au/ 4 Y. M IAO AND J. F. L I which implies E(f α (X)X β ) ≥ E(X α+β ). (2.11) The remainder of the proof is similar to that of Theorem 2.1. Remark 3. If we assume that X possesses a uniform distribution on the support interval [0, 1] then the above theorem is Theorem 2.3 of Boukerrioua and Guezane-Lakoud in [1]. Next we consider the case “α > 0, β > 2” by using the ideas of Dragomir-Ngô in [2]. Lemma 2.3 ([2]). Let f : [a, b] → [0, ∞) be a continuous function and g : [a, b] → [0, ∞) be non-decreasing, differentiable on (a, b) satisfying Z b Z b f (t)dt ≥ g(t)dt. x x Then Z b β Z f (t)dt ≥ x b g β (t)dt x holds for β > 1. 0 Lemma 2.4. Let f (x) ≥ 0 be a continuous function on [0, ∞) with f (x) ≥ 0 on (0, ∞) and satisfying E(f (X)1At ) ≥ E(X1At ), (2.12) ∀ t ∈ [0, ∞). Then E(f β (X)1At )) ≥ E(X β 1At ) (2.13) holds for every positive real number β ≥ 1. Proof. The proof is a direct extension of Theorem 3 in [2]. 0 Theorem 2.5. Let f (x) ≥ 0 be a continuous function on [0, ∞) with f (x) ≥ 0 on (0, ∞) and satisfying E(f (X)1At ) ≥ E(X1At ), (2.14) ∀ t ∈ [0, ∞). In addition, for every positive real number α > 0, β > 2 satisfying lim f α (x)xβ−1 E[(f (X) − X)1Ax ] = 0, (2.15) x→∞ and lim f α (x)xE[(f β−1 (X) − X β−1 )1Ax ] = 0, (2.16) x→∞ then E(f α+β (X)) ≥ E(X β f α (X)). (2.17) Proof. It is obvious that (f (x) − x)(f β−1 (x) − xβ−1 ) ≥ 0, which implies that (2.18) f β+α (x) ≥ xβ−1 f 1+α (x) + xf β−1+α (x) − xβ f α (x). J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 108, 7 pp. http://jipam.vu.edu.au/ F URTHER D EVELOPMENT OF AN O PEN P ROBLEM 5 Integrating by parts and using (2.14) and (2.15), we have E(f α (X)X β−1 (f (X) − X)) Z ∞ = f α (x)xβ−1 (f (x) − x)p(x)dx 0 Z ∞ Z ∞ α β−1 =− f (x)x d (f (t) − t)p(t)dt dx 0 x Z ∞ ∞ α β−1 = −f (x)x (f (t) − t)p(t)dt 0 x Z ∞ Z ∞ 0 α−1 β−1 β−2 α + αf (x)f (x)x + (β − 1)x f (x) (f (t) − t)p(t)dt dx 0 x Z ∞ Z ∞ 0 α−1 β−1 β−2 α = αf (x)f (x)x + (β − 1)x f (x) (f (t) − t)p(t)dt dx 0 x Z ∞ 0 = αf (x)f α−1 (x)xβ−1 + (β − 1)xβ−2 f α (x) E ((f (X) − X)1Ax ) dx ≥ 0, 0 which yields E(f α+1 (X)X β−1 ) ≥ E(f α (X)X β ). (2.19) Furthermore, by Lemma 2.4 and the condition (2.16), we have E(f α (X)X(f β−1 (X) − X β−1 )) Z ∞ f α (x)x(f β−1 (x) − xβ−1 )p(x)dx = 0 Z ∞ Z ∞ β−1 β−1 α (f (t) − t )p(t)dt dx =− f (x)xd x 0 Z ∞ ∞ β−1 β−1 α (f (t) − t )p(t)dt = − f (x)x 0 x Z ∞ Z ∞ 0 β−1 β−1 α−1 α + (αxf (x)f (x) + f (x)) (f (t) − t )p(t)dt dx x 0 Z ∞ 0 = (αxf (x)f α−1 (x) + f α (x))E (f β−1 (X) − X β−1 )1Ax dx ≥ 0, 0 which yields E(f α+β−1 (X)X) ≥ E(f α (X)X β ). (2.20) From (2.18)-(2.20), inequality (2.17) holds. 3. F URTHER D ISCUSSION R∞ Let g(x) ≥ 0, 0 < 0 g(x)dx < ∞. If p(x) := R ∞g(x) , then it is easy to check that p(x) is 0 g(x)dx a probability density function on the interval [0, ∞). Thus we have the following: Theorem 3.1. Let f (x) ≥ 0 be a continuous function on [0, ∞) satisfying Z ∞ Z ∞ β (3.1) f (t)g(t)dt ≥ tβ g(t)dt ∀ t ∈ [0, ∞). t J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 108, 7 pp. t http://jipam.vu.edu.au/ 6 Y. M IAO AND J. F. L I Then Z ∞ f (3.2) α+β Z ∞ xβ f α (x)g(x)dx (x)g(x)dx ≥ 0 0 holds for every positive real number α and β. Theorem 3.2. Let f (x) ≥ 0 be a continuous function on [0, ∞) satisfying Z ∞ Z ∞ (3.3) f (t)g(t)dt ≥ tg(t)dt, ∀ t ∈ [0, ∞). t t Then Z ∞ f (3.4) α+β Z (x)g(x)dx ≥ 0 ∞ xβ f α (x)g(x)dx 0 holds for every pair of positive real numbers “α ≥ 1 and β > 0”. Furthermore, for every positive real number “α > 0, β > 2” satisfying (2.15) and (2.16), the inequality (3.4) holds. Two more general results follow. Theorem 3.3. Let f (x) ≥ 0, g(x) ≥ 0 be two continuous functions on [0, ∞) satisfying Z ∞ Z ∞ β (3.5) f (t)dt ≥ g β (t)dt ∀ t ∈ [0, ∞). t t 0 Furthermore, for any positive real numbers α and β, let g(x) be differentiable with [g α (x)] ≥ 0 and g(0) = 0, then Z ∞ Z ∞ α+β (3.6) f (x)dx ≥ g β (x)f α (x)dx. 0 0 α Proof. Denoting the derivative of g (x) by G(x), we obtain, Z ∞ Z ∞ Z x α β (3.7) g (x)f (x)dx = G(t)dt f β (x)dx 0 0 0 Z ∞ Z ∞ β = G(t) f (x)dx dt 0 t Z ∞ Z ∞ β ≥ G(t) g (x)dx dt t 0 Z x Z ∞ β = g (x) G(t)dt dx 0 0 Z ∞ = g β+α (x)dx. 0 Using Cauchy’s inequality, we have β α (3.8) f α+β (x) + g α+β (x) ≥ g α (x)f β (x), α+β α+β which, by (3.7), yields Z ∞ Z ∞ Z ∞ β α α+β α+β (3.9) f (x)dx + g (x)dx ≥ g α (x)f β (x)dx α+β 0 α+β 0 Z0 ∞ ≥ g β+α (x)dx. 0 The desired result then follows. J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 108, 7 pp. http://jipam.vu.edu.au/ F URTHER D EVELOPMENT OF AN O PEN P ROBLEM 7 A similar proof yields the following: Theorem 3.4. Let f (x) ≥ 0, g(x) ≥ 0 be two continuous functions on [0, ∞) satisfying Z ∞ Z ∞ (3.10) f (t)dt ≥ g(t)dt, ∀ t ∈ [0, ∞). t t Furthermore, for every pair of positive real numbers satisfying “α ≥ 1 and β > 0”, let g(x) be 0 differentiable with [g α (x)] ≥ 0 and g(0) = 0, then Z ∞ Z ∞ α+β (3.11) f (x)dx ≥ g β (x)f α (x)dx. 0 0 Additionally, for every positive real number “α > 0, β > 2” satisfying (2.15) and (2.16), inequality (3.11) holds. R EFERENCES [1] K. BOUKERRIOUA AND A. GUEZANE-LAKOUD, On an open question regarding an integral inequality, J. Inequal. Pure Appl. Math., 8(3) (2007), Art. 77. [ONLINE: http://jipam.vu. edu.au/article.php?sid=885]. [2] S.S. DRAGOMIR AND Q.A. NGÔ, On an integral inequality, RGMIA Research Report Collection, 11(1) (2008), Art. 13. [ONLINE: http://www.staff.vu.edu.au/RGMIA/v11n1.asp]. [3] W.J. LIU, C.C. LI AND J.W. DONG, On an open problem concerning an integral inequality, J. Inequal. Pure Appl. Math., 8(3) (2007), Art. 74. [ONLINE: http://jipam.vu.edu.au/ article.php?sid=882]. [4] Q.A. NGÔ, D.D. THANG, T.T. DAT AND D.A. TUAN, Note on an integral inequality, J. Inequal. Pure Appl. Math., 7(4) (2006), Art. 120. [ONLINE: http://jipam.vu.edu.au/article. php?sid=737]. J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 108, 7 pp. http://jipam.vu.edu.au/