SUPERQUADRACITY OF FUNCTIONS AND REARRANGEMENTS OF SETS Superquadracity and Rearrangements SHOSHANA ABRAMOVICH Shoshana Abramovich Department of Mathematics University of Haifa Haifa, 31905, Israel EMail: abramos@math.haifa.ac.il vol. 8, iss. 2, art. 46, 2007 Title Page Contents Received: 26 December, 2006 Accepted: 29 May, 2007 Communicated by: S.S. Dragomir 2000 AMS Sub. Class.: 26D15. Key words: Superquadratic functions, Convex functions, Jensen’s inequality. Abstract: In this paper we establish upper bounds of JJ II J I Page 1 of 11 n X |xi − xi+1 | xi + xi+1 f +f , 2 2 i=1 Go Back Full Screen xn+1 = x1 when the function f is superquadratic and the set (x) = (x1 , . . . , xn ) is given except its arrangement. Close Contents 1 Introduction 3 2 The Main Results 7 Superquadracity and Rearrangements Shoshana Abramovich vol. 8, iss. 2, art. 46, 2007 Title Page Contents JJ II J I Page 2 of 11 Go Back Full Screen Close 1. Introduction We start with the definitions and results of [1] and [5] which we use in this paper. Definition 1.1. The sets (y− ) = y1− , . . . , yn− and (− y) = (− y1 , . . . ,− yn ) are symmetrically decreasing rearrangements of an ordered set (y) = (y1 , . . . , yn ) of n real numbers, if y1− (1.1) ≤ yn− ≤ y2− ≤ ··· ≤ y − Superquadracity and Rearrangements [ n+2 2 ] Shoshana Abramovich vol. 8, iss. 2, art. 46, 2007 and (1.2) − yn ≤ − y1 ≤ − yn−1 ≤ · · · ≤ − y[ n+1 ] . 2 A circular rearrangement of an ordered set (y) = (y1 , . . . , yn ) is a cyclic rearrangement of (y) or a cyclic rearrangement followed by inversion. Definition 1.2. An ordered set (y) = (y1 , . . . , yn ) of n real numbers is arranged in circular symmetric order if one of its circular rearrangements is symmetrically decreasing. Theorem A ([1]). Let F (u, v) be a symmetric function defined for α ≤ u, v ≤ β 2 F (u,v) for which ∂ ∂u∂v ≥ 0. Let the set (y) = (y1 , . . . , yn ), α ≤ yi ≤ β, i = 1, . . . , n be given except its arrangement. Then n X F (yi , yi+1 ) , (yn+1 = y1 ) i=1 is maximal if (y) is arranged in circular symmetrical order. Title Page Contents JJ II J I Page 3 of 11 Go Back Full Screen Close Definition 1.3 ([5]). A function f , defined on an interval I = [0, L] or [0, ∞) is superquadratic, if for each x in I, there exists a real number C (x) such that f (y) − f (x) ≥ C (x) (y − x) + f (|y − x|) for all y ∈ I. A function is subquadratic if −f is superquadratic. Lemma A ([5]). Let f be a superquadratic function with C (x) as in Definition 1.3. Superquadracity and Rearrangements Shoshana Abramovich (i) Then f (0) ≤ 0. vol. 8, iss. 2, art. 46, 2007 (ii) If f (0) = f 0 (0) = 0, then C (x) = f 0 (x) whenever f is differentiable. (iii) If f ≥ 0, then f is convex and f (0) = f 0 (0) = 0. Title Page Contents The following lemma presents a Jensen’s type inequality for superquadratic functions. Lemma B ([6, Lemma Let xr ≥ 0, 1 ≤ P 2.3]). Suppose that f is superquadratic. P r ≤ n and let x = nr=1 λr xr , where λr ≥ 0, and nr=1 λr = 1. Then n X r=1 λr f (xr ) ≥ f (x) + n X λr f (|xr − x|) . r=1 If f (x) is subquadratic, the reverse inequality holds. From Lemma B we get an immediate result which we state in the following lemma. JJ II J I Page 4 of 11 Go Back Full Screen Close Lemma C. Let f (x) be superquadratic on [0, L] and let x, y ∈ [0, L], 0 ≤ λ ≤ 1, then λf (x) + (1 − λ) f (y) ≥ f (λx + (1 − λ) y) + λf ((1 − λ) |y − x|) + (1 − λ) f (λ |y − x|) t−1 X k f 2λ (1 − λ) |1 − 2λ| |x − y| ≥ f (λx + (1 − λ) y) + k=0 + λf (1 − λ) |1 − 2λ|t |x − y| + (1 − λ) f λ |1 − 2λ|t |x − y| . Superquadracity and Rearrangements Shoshana Abramovich vol. 8, iss. 2, art. 46, 2007 If f is positive superquadratic we get that: t−1 X λf (x)+(1 − λ) f (y) ≥ f (λx + (1 − λ) y)+ f 2λ (1 − λ) |1 − 2λ|k |x − y| Title Page Contents k=0 More results related to superquadracity were discussed in [2] to [6]. In this paper we refine the results in [7] by showing that for positive superquadratic functions we get better bounds than in [7]. Theorem B ([7, Thm. 1.2]). If f is a convex function and x1 , x2 , . . . , xn lie in its domain, then n X i=1 x1 + · · · + xn n n−1 x1 + x2 xn−1 + xn xn + x1 ≥ f + ··· + f +f . n 2 2 2 f (xi ) − f JJ II J I Page 5 of 11 Go Back Full Screen Close Theorem C ([7, Thm. 1.4]). If f is a convex function and a1 , . . . , an lie in its domain, then (n − 1) [f (b1 ) + · · · + f (bn )] ≤ n [f (a1 ) + · · · + f (an ) − f (a)] , where a = a1 +···+an n and bi = na−ai , n−1 i = 1, . . . , n. Superquadracity and Rearrangements Shoshana Abramovich vol. 8, iss. 2, art. 46, 2007 Title Page Contents JJ II J I Page 6 of 11 Go Back Full Screen Close 2. The Main Results Theorem 2.1. Let f (x) be a superquadratic function on [0, L] . Then for xi ∈ [0, L] , i = 1, . . . , n, where xn+1 = x1 , ! !! n n n X X n−1X xi + xi+1 |xi − xi+1 | (2.1) f +f n i=1 2 2 i=1 i=1 ! ! ! n n n n X X X 1X x xi j ≤ f (xi ) − f − f xi − n n n i=1 i=1 i=1 Superquadracity and Rearrangements Shoshana Abramovich vol. 8, iss. 2, art. 46, 2007 j=1 holds. If f 000 (x) ≥ 0 too, then n (2.2) n−1X f n i=1 n X xi + xi+1 ! n X |xi − xi+1 | !! +f 2 2 i=1 i=1 n n−1X x bi + x bi+1 |b xi − x bi+1 | ≤ f +f n i=1 2 2 ! ! ! n n n n X X X 1X x xi j − f xi − ≤ f (xi ) − f , n n n i=1 j=1 i=1 i=1 where (b x) = (b x1 , . . . , x bn ) is a circular symmetrical rearrangement of (x) = (x1 , . . . , xn ) . Contents JJ II J I Page 7 of 11 Go Back Full Screen Close Example 2.1. The functions f (x) = xn , n ≥ 2, x ≥ 0, and the function f (x) = Title Page x2 log x, x > 0, 0, x=0 are superquadratic with an increasing second derivative and therefore (2.2) holds for these functions. Proof. Let f be a superquadratic function on [0, L] . Then by Lemma B we get for 0 ≤ α ≤ 1, 1 ≤ k ≤ n and xi ∈ [0, L] , xn+1 = x1 , (2.3) n X f (xi ) Superquadracity and Rearrangements i=1 n n kX n−k X = f (xi ) + f (xi ) n i=1 n i=1 Shoshana Abramovich vol. 8, iss. 2, art. 46, 2007 n n n−k X kX = (αf (xi ) + (1 − α) f (xi+1 )) + f (xi ) n i=1 n i=1 n n−k X ≥ f (αxi + (1 − α) xi+1 ) n i=1 n n−k X (αf ((1 − α) |xi+1 − xi |) + (1 − α) f (α |xi+1 − xi |)) + n i=1 ! Pn X Pn n x x 1 i i i=1 +k f + f xi − i=1 . n n n i=1 Contents JJ II J I Page 8 of 11 Go Back Full Screen Close For k = 1 and α = 21 we get that (2.1) holds. 2 F (u,v) If f 000 (x) ≥ 0, then ∂ ∂u∂v ≥ 0, where F (u, v) = f (u + v) + f (|u − v|) , Title Page u, v ∈ [0, L] . Therefore according to Theorem A, the sum n X |xi + xi+1 | xi + xi+1 f +f , 2 2 i=1 xn+1 = x1 , is maximal for (b x) = (b x1 , . . . , x bn ) , which is the circular symmetric rearrangement of (x) . Therefore in this case (2.2) holds as well. Remark 1. For a positive superquadratic function f , which according to Lemma A is also a convex function, (2.1) is a refinement of Theorem B. If f 000 (x) ≥ 0, (2.2) is a refinement of Theorem B as well. Remark 2. Theorem B is refined by ! Pn n n X X n−1 x bi + x bi+1 i=1 xi f (xi ) − f ≥ f n n 2 i=1 i=1 n n−1X xi + xi+1 ≥ f , n i=1 2 because a convex function f satisfies the conditions of Theorem A for F (u, v) = f (u + v) . The following inequality is a refinement of Theorem C for a positive superquadratic function f , which is therefore also convex. The inequality results easily from Lemma B and the identity ! n n n X X 1 X f (ai ) = f (aj ) (1 − δij ) n − 1 j=1 i=1 i=1 (where δij = 1 for i = j and δij = 0 for i 6= j), therefore the proof is omitted. Superquadracity and Rearrangements Shoshana Abramovich vol. 8, iss. 2, art. 46, 2007 Title Page Contents JJ II J I Page 9 of 11 Go Back Full Screen Close Theorem 2.2. Let f be a superquadratic function on [0, L] , and let xi ∈ [0, L], i = 1, . . . , n. Then ! ! n n X X n f (yi ) f (xi ) − f (x) − n−1 i=1 i=1 ! n n n XX 1 1 X ≥ f (|yi − xj |) (1 − δij ) + f (|x − xi |) , n − 1 i=1 j=1 n − 1 i=1 where x = Pn xi i=1 n , yi = nx−xi n−1 , i = 1, . . . , n. Superquadracity and Rearrangements Shoshana Abramovich vol. 8, iss. 2, art. 46, 2007 Title Page Contents JJ II J I Page 10 of 11 Go Back Full Screen Close References [1] S. ABRAMOVICH, The increase of sumsand products dependent on (y1 , . . . , yn ) by rearrangement of this set, Israel J. Math., 5(3) (1967). [2] S. ABRAMOVICH, S. BANIĆ AND M. MATIĆ, Superquadratic functions in several variables, J. Math. Anal. Appl., 327 (2007), 1444–1460. [3] S. ABRAMOVICH, S. BANIĆ AND M. KLARICIĆ BACULA, A variant of Jensen-Steffensen’s inequality for convex and superquadratic functions, J. Ineq. Pure & Appl. Math., 7(2) (2006), Art. 70. [ONLINE: http://jipam.vu. edu.au/article.php?sid=687]. [4] S. ABRAMOVICH, S. BANIĆ, M. MATIĆ AND J. PEČARIĆ, JensenSteffensen’s and related inequalities for superquadratic functions, to appear in Math. Ineq. Appl. [5] S. ABRAMOVICH, G. JAMESON AND G. SINNAMON, Refining Jensen’s inequality, Bull. Sci. Math. Roum., 47 (2004), 3–14. Superquadracity and Rearrangements Shoshana Abramovich vol. 8, iss. 2, art. 46, 2007 Title Page Contents JJ II J I [6] S. ABRAMOVICH, G. JAMESON AND G. SINNAMON, Inequalities for averages of convex and superquadratic functions, J. Ineq. Pure & Appl. Math., 5(4) (2004), Art. 91. [ONLINE: http://jipam.vu.edu.au/article.php? sid=444]. Page 11 of 11 [7] L. BOUGOFFA, New inequalities about convex functions, J. Ineq. Pure & Appl. Math., 7(4) (2006), Art. 148. [ONLINE: http://jipam.vu.edu. au/article.php?sid=766]. Close Go Back Full Screen