Banach Journal of Mathematical Analysis ISSN: (electronic)

Banach J. Math. Anal. 4 (200), no., 59 69 Banach Journal of Mathematical Analysis ISSN: 735-8787 (electronic) www.emis.de/journals/bjma/ IMPROVEMENT OF JENSEN STEFFENSEN S INEQUALITY FOR SUPERQUADRATIC FUNCTIONS SHOSHANA ABRAMOVICH SLAVICA IVELIĆ2 AND JOSIP E. PEČARIĆ3 This paper is dedicated to Professor Lars-Erik Persson on the occasion of his 65th birthday. Communicated by S. S. Dragomir Abstract. In this paper, improvements for superquadratic functions of Jensen Steffensen s and related inequalities are discussed. For superquadratic functions which are not convex we get inequalities analog to Jensen Steffensen s inequality for convex functions. For superquadratic functions which are convex (including many useful functions), we get improvements and extensions of Jensen Steffensen s inequality and related inequalities.. Introduction and preliminaries In this paper we refine results derived from Jensen Steffensen s inequality and its extension for superquadratic functions. These functions were introduced in [4] and [5] and dealt with in numerous papers (see for example [9]). Jensen Steffensen s inequality states that if f : I R is convex, then ( ) f a i x i a i f (x i ) i= i= Date: Received: 26 September 2009; Accepted: 23 March 200. Corresponding author. 2000 Mathematics Subject Classification. Primary 39B82; Secondary 44B20, 46C05. Key words and phrases. Jensen Steffensen inequality, superquadratic function, convex function. 59

60 S. ABRAMOVICH, S. IVELIĆ, J.E. PEČARIĆ holds, where I is an interval in R, x = (x,..., x n ) is any monotonic n-tuple in I n and a = (a,..., a n ) is real n-tuple that satisfies 0 A j, j =,..., n, > 0, j A j = a i, A j = a i, j =,..., n, (.) i= (see [, p. 57]). In [0], J. Pečarić proved a refinement of Slater s inequality established in [2]. J. Pečarić proved that under the same conditions leading to Jensen Steffensen s inequality, together with a i f i= (x i ) 0 and M = a ix i f (x i ) i= a I, (.2) if (x i ) the inequality i= i=j a i f (x i ) f (M) i= holds. Now we quote some definitions and state a list of basic properties of superquadratic functions. Definition.. [4, Definition 2.] A function f : [0, b) R is superquadratic provided that for all 0 x < b, there exists a constant C (x) R such that f (y) f (x) f ( y x ) C (x) (y x) for all y [0, b). We say that f is subquadratic if f is a superquadratic function. Lemma.2. [5, Lemma 2.3] Suppose that f is a superquadratic function on [0, b), x i [0, b), i =,..., n, and a i 0, i =,..., n, are such that = i= a i > 0. Then a i f (x i ) f (x) a i f ( x i x ), (.3) i= where x = i= a ix i. Lemma.3. [4, Lemma 2.] Let f be a superquadratic function with C (x) as in Definition.. (i) Then f(0) 0. (ii) If f(0) = f (0) = 0, then C(x) = f (x) whenever f is differentiable at 0 < x < b. (iii) If f 0, then f is convex and f(0) = f (0) = 0. Lemma.4. [4, Lemma 3.] Suppose that f : [0, b) R is a continuously differentiable function and f(0) 0. If f then f is superquadratic. i= is superadditive or f (x) x We quote now some theorems that we refine in the sequel. is nondecreasing,

IMPROVEMENT OF JENSEN STEFFENSEN S INEQUALITY 6 Theorem.5. [, Theorem ] Let f : [0, b) R be a differentiable superquadratic and nonnegative function. Let x i [0, b), i =,..., n, be such that x x 2... x n or x x 2... x n. Let a be a real n-tuple satisfying (.) and x = i= a ix i. Then k a i f (x i ) f (x) A i f(x i+ x i ) + A k f(x x k ) i= i= where k {,..., n } satisfies + A k+ f(x k+ x) + ( k A i + i= i=k+ i=k+2 A i ) f A i f(x i x i ) ( n i= a i x i x k i= A i + ( i= (n ) f a ) i ( x i x ), (n ) x k x x k+. i=k+ A i Lemma.6. [6, Lemma a] Let f : (a, b) R, where a < b, be a convex function. Let z (a, b) be fixed. Then the function : (a, b) R defined by (y) = f(y) f(z) f (z)(y z) is nonnegative on (a, b), nonincreasing on (a, z] and nondecreasing on [z, b). Theorem.7. [6, Theorem ] Let f : (a, b) R be a convex function and a i R, i =,..., n, be such that (.) holds. Then for any x i (a, b), i =,..., n, such that the inequalities f(c) + f (c)(x c) x x 2... x n or x x 2... x n, a i f (x i ) f(d) + i= hold for all c, d (a, b), where x = i= a ix i. a i f (x i )(x i d) Theorem.8. [6, Theorem 4] Suppose that all the conditions of Theorem.7 are satisfied and additionally assume that f is differentiable on I = (a, b) and M satisfies (.2). If f is nondecreasing and f is concave on I, or if f is nonincreasing and f is convex on I, then a i f (x i ) f(x) + a i f (x i )(x i x) f(m). (.4) i= i= i= )

62 S. ABRAMOVICH, S. IVELIĆ, J.E. PEČARIĆ In this paper we obtain Lemma.2 and related results for superquadratic functions, but this time the coefficients a i (i =,..., n) are not only nonnegative. There, in Lemma.2, the coefficients satisfy a i 0 (i =,..., n) and = a i > 0. Here we deal with coefficients that satisfy (.) and 0 x x 2... x n < b. We generalize also part of the theorems proved in [3]. In particular we extend the following version of [3, Theorem 2.a], as most of the theorems there, in [3], result from this theorem. Theorem.9. [3, Theorem 2.a] Let 0 x k < b, a k > 0, k =,..., n. If f is superquadratic on [0, b), then ( i= a i f (x i ) f a ) ( ( ix i i=m+ a i f (x i ) A m+ f a ) ix ) i i= i=m+ i= A m+ ( m a i f x j= i a ) jx j ( n i=m+ + A m+ f a ix i i= a ) ix i. (.5) i= A m+ This type of inequality, but for convex functions, was dealt with in [7]. In the sequel, all the results dealing with 0 x x 2... x n < b hold also for b > x x 2... x n 0. 2. Main results Analogous to Lemma.6 and Theorem.7 we prove the following Lemma 2. and Theorem 2.2. We use a technique similar to the one used in [6]. Lemma 2.. Let f be a continuously differentiable function on [0, b) and f be superadditive on [0, b). Then the function D : [0, b) R, defined by D (y) = f (y) f (z) f (z) (y z) f ( y z ) + f (0) is nonnegative on [0, b), nonincreasing on [0, z) and nondecreasing on [z, b), for 0 z < b. Proof. Since f is superadditive, then for 0 z y < b we have 0 y and for 0 y z < b, 0 z (f (t) f (z) f (t z)) dt = f(y) f(z) f (z)(y z) f(y z) + f(0) z y (f (z) f (t) f (z t)) dt = f (z)(z y) f(z) + f(y) + f(0) f(z y). Together these show that for any y, z [0, b) the inequality f (y) f (z) f (z) (y z) f ( y z ) + f (0) 0

IMPROVEMENT OF JENSEN STEFFENSEN S INEQUALITY 63 holds. So we conclude that the function D is nonnegative on [0, b). Since D (y) = f (y) f (z) f ( y z ) sgn (y z), and as f is superadditive, then for 0 y z we have and for z y < b, This completes the proof. D (y) = f (y) f (z) + f (z y) f (y) f (z) + f (z) f (y) = 0 D (y) = f (y) f (z) f (y z) f (y) f (z) f (y) + f (z) = 0. Now we can present the main results of this section where we show that inequality (.3) is satisfied not only for nonnegative coefficients but also when (.) is satisfied and x x 2... x n. Theorem 2.2. Let f be a continuously differentiable function on [0, b) and f be superadditive on [0, b). Let a be a real n-tuple satisfying (.). Let x i [0, b), i =,..., n, be such that x x 2... x n and x = n i= a ix i. Then (a) the inequality f (c) f (0) + f (c) (x c) + a i f ( x i c ) a i f (x i ) (2.) i= holds for all c [0, b). In particular, a i f (x i ) f (x) + f (0) i= i= a i f ( x i x ). (2.2) (b) If in addition f (0) 0, then f is superquadratic and a i f (x i ) f (c) + f (c) (x c) + a i f ( x i c ). (2.3) i= In particular, a i f (x i ) f (x) + i= i= i= a i f ( x i x ). (2.4) (c) If in addition f 0 and f(0) = f (0) = 0, then f is superquadratic and convex increasing and a i f (x i ) f (c) f (c) (x c) a i f ( x i c ) 0. (2.5) i= i= i=

64 S. ABRAMOVICH, S. IVELIĆ, J.E. PEČARIĆ Proof. It was proved in [] and in [2] that x x x n. (a) Let D (x i ) = f (x i ) f (c) f (c) (x i c) f ( x i c )+f (0), i =,..., n. From Lemma 2. we know that D(x i ) 0 for all i =,..., n. Comparing c with x,..., x n we must consider three cases. Case. x n < c < b : In this case x i [0, c) for all i =,..., n. Hence, according to Lemma 2. we have Denoting A 0 = 0 it follows and therefore a i D (x i ) = i= D (x ) D (x 2 )... D (x n ) 0. a i = A i A i, i =,..., n, (A i A i ) D (x i ) i= = A D (x ) + (A 2 A ) D (x 2 ) +... + ( ) D (x n ) n = A i (D (x i ) D (x i+ )) + D (x n ) i= 0. Case 2. 0 c < x : In this case x i (c, b) for all i =,..., n. Hence, according to Lemma 2. we have 0 D (x ) D (x 2 )... D (x n ). Denoting + = 0 it follows A k = a i = A k, k =,..., n, and therefore i=k a i = A i A i+, i =,..., n, a i D (x i ) = i= ( ) Ai A i+ D (xi ) i= = A D (x ) + 0. A i (D (x i ) D (x i )) Case 3. x c x n : In this case there exists k {,..., n } such that x k c x k+. By Lemma 2. we have i=2 D (x ) D (x 2 )... D (x k ) 0 and 0 D (x k+ ) D (x k+2 )... D (x n ).

Then IMPROVEMENT OF JENSEN STEFFENSEN S INEQUALITY 65 a i D (x i ) = i= k a i D (x i ) + i= i=k+ a i D (x i ) k = A i (D (x i ) D (x i+ )) + A k D (x k ) i= + A k+ D (x k+ ) + 0. i=k+2 A i (D (x i ) D (x i )) In all three cases we get a i D (x i ) = a i [f (x i ) f (c) f (c) (x i c) f ( x i c ) + f (0)] i= i= 0, and therefore, the inequality (2.) holds. Inserting c = x in (2.) we get (2.2). (b) From Lemma.4 we get that as f(0) 0, f is superquadratic. The inequality (2.3) follows from (2.). Inserting c = x in (2.3) we get (2.4). (c) First, from Lemma.4 we get that as f(0) = 0, f is superquadratic. Then from Lemma.3 it follows that f is convex increasing. We only need to show that under our conditions the inequality a i f ( x i c ) 0 (2.6) i= holds. We will show that (2.6) holds in the case that x k c x k+, k =,..., n. The other cases we can prove similarly. We use the identity k a i f ( x i c ) = A i (f (c x i ) f (c x i+ )) i= i= + A k f(c x k ) + A k+ f(x k+ c) + A i (f (x i c) f (x i c)). i=k+2 Since the function f is nonnegative and convex, and f(0) = 0, then for 0 x i x i+ c, i =,..., k, we have f (c x i ) f (c x i+ ) f(x i+ x i ) f(0) = f(x i+ x i ) 0

66 S. ABRAMOVICH, S. IVELIĆ, J.E. PEČARIĆ and for c x i x i < b, i = k + 2,..., n, f (x i c) f (x i c) f(x i x i ) f(0) = f(x i x i ) 0. Therefore, as (.) holds, (2.6) is satisfied and together with (2.) it gives (2.5). This completes the proof. So, we see that the inequality (2.4) in Part (b) of Theorem 2.2 extends Lemma.2 to coefficients satisfying (.) and x x 2... x n. In (2.3) we get refinement of Lemma.2. Also, the inequality (2.5) for c = x improves Theorem.5. Remark 2.3. Suppose that f is continuously differentiable on [0, b), f (0) = 0 and f is subadditive on [0, b). Then f is subquadratic and the reverse of (2.3) holds. Theorem 2.4. Let f be continuously differentiable and convex increasing on [0, b), f(0) = f (0) = 0 and f be concave on [0, b). Let a be a real n-tuple satisfying (.) and x i [0, b), i =,..., n, be such that x x 2... x n. Then a i f (x i ) i= min { f(x) + } a i f (x i )(x i x), f(x) + a i f ( x i x ) i= holds, where x = i= a ix i. i= (2.7) Proof. Since f (0) = 0 and f is concave, it follows that f is subadditive and therefore, f is subquadratic. Then by Remark 2.3 the inequality a i f (x i ) f (c) + f (c) (x c) + a i f ( x i c ) i= holds. Choosing x = c we get a i f (x i ) f (x) + i= By combining (2.8) with (.4) it follows (2.7). i= a i f ( x i x ). (2.8) In the following example we show that there are cases where the inequality (2.7), which is derived in Theorem.8 from the convexity of f, gives a better bound than the bound derived from (2.8) for subquadratic f and vice-versa. Example 2.5. Consider the function f(x) = x +α for 0 α and x 0. It s clear that f is convex and f is concave on [0, b). Since f(0) = 0, then f is i=

IMPROVEMENT OF JENSEN STEFFENSEN S INEQUALITY 67 also subquadratic. If we choose a = a 2 =, x 2 = 0, x 2 =, from (2.7) it follows { 2 min 2 + + α, +α 4 2 + }. +α 2 +α Then for α = we have { } 3 2 min 4, = 4 4 and for α = 0 { } 3 2 min 4, = 3 4. We see that in the first case we get better bound derived from the subquadracity of f and in the second case the better bound is derived from the convexity of f and the concavity of f. Now we prove an analog to Theorem.9 for coefficients that satisfy (.). Theorem 2.6. Let f be a continuously differentiable function on [0, b), f(0) = 0 and f be superadditive on [0, b). Let a be a real n-tuple satisfying (.) and A j > 0 for all j =,..., n. Then for any x i [0, b), i =,..., n, such that x x 2... x n, the inequality (.5) holds. Proof. It is given that 0 x x 2... x n < b and A j > 0 for all j =,..., n. Therefore, according to Theorem 399 in [8] we get a i (x i x m ) 0 which is equivalent to Denote i=m+ A m+ i=m+ a i x i x m. y i = x i, i =,..., m, y m+ = A m+ b i = a i, i =,..., m, b m+ = A m+, B i = A i, i =,..., m, B m+ = From this notation it follows that and B m+ = m+ i= i=m+ a i x i, 0 B i B m+, i =,..., m, B m+ > 0, m+ i= b i =, 0 y y 2... y m+. b i. (2.9)

68 S. ABRAMOVICH, S. IVELIĆ, J.E. PEČARIĆ Applying (2.4) to y i, b i, B i, i =,..., m +, we get where B m+ m+ i= b i f (y i ) f (y) + m+ B m+ i= y = m+ b i y i. B m+ i= Multiplying inequality (2.0) with B m+ we get m b i f (y i ) + b m+ f (y m+ ) i= B m+ f (y) + From (2.9) it follows that b i f ( y i y ), (2.0) m b i f ( y i y ) + b m+ f ( y m+ y ). (2.) i= ( y = m a i x i + i= i=m+ a i x i ) = a i x i (2.2) and therefore, using (2.9) and (2.2), the inequality (2.) becomes ( ) m a i f (x i ) + A m+ f a i x i i= A m+ i=m+ ( ) ( ) m f a i x i + a i f x i a i x i A i= i= n i= ( ) + A m+ f a i x i a i x i, which is equivalent to (.5). A m+ i=m+ References [] S. Abramovich, S. Banić, M. Matić and J.E. Pečarić, Jensen Steffensen s and related inequalities for superquadratic functions, Math. Inequal. Appl. (2007), no., 23 4. [2] S. Abramovich, M. Klaričić Bakula, M. Matić and J.E. Pečarić, A variant of Jensen Steffensen s inequality and quasi-arithmetic means, J. Math. Anal. Appl. 307 (2005), 370 386. [3] S. Abramovich, B. Ivanković and J.E. Pečarić, Improvement of Jensen Steffensen s inequality, Australian J. Math. Anal. Appl., to appear. [4] S. Abramovich, G. Jameson and G. Sinnamon, Refining Jensen s inequality, Bull. Math. Sc. Math. Roum. 47 (2004), 3 4. [5] S. Abramovich, G. Jameson and G. Sinnamon, Inequalities for averages of convex and superquadratic functions, J. Inequal. Pure and Appl. Math. 5 (2004), no. 4, article 9. [6] M. Klaričić Bakula, M. Matić and J.E. Pečarić, Generalization of the Jensen Steffensen and related inequalities Cent. Eur. J. Math. 7 (2009), no. 4, 787 803. [7] S.S. Dragomir, ew refinement of Jensen s inequality in linear spaces with applications, RGMIA, Vol. 2 2009, online: www.staff.vu.edu.au/rgmia/v2(e).asp. i= i=

IMPROVEMENT OF JENSEN STEFFENSEN S INEQUALITY 69 [8] G.H. Hardy, J.E. Littlewood and G. Pólya, Inequalities, Cambridge University Press, New York, 952. [9] J.A. Oguntuase, L.E. Perssons, E.K. Essel and B.A. Popoola, Refined multidimensional Hardy-type inequalities via superquadracity, Banach J. Math. Anal. 2 (2008), no. 2, 29 39. [0] J.E. Pečarić, A companion inequality to Jensen Steffensen s inequality, J. Approx.Theory 44 (985), no. 3, 289 29. [] J.E. Pečarić, F. Proschan and Y.L. Tong, Convex Functions, Partial Orderings, and Statistical Applications, Academic Press, New York, 992. [2] M.L. Slater, A companion inequality to Jensen s inequality, J. Approx. Theory 32 (98), 06 66. Department of Mathematics, University of Haifa, Haifa, Israel. E-mail address: abramos@math.haifa.ac.il 2 Faculty of Civil Technology and Architecture, University of Split, Matice hrvatske 5, 2000 Split, Croatia. E-mail address: sivelic@gradst.hr 3 Faculty of Textile Technology, University of Zagreb, Prilaz Baruna Filipovic 30, 0000 Zagreb, Croatia. E-mail address: pecaric@element.hr