A monotonic refinement of Levinson s inequality

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1 Jakšetic et al. Journal of Inequalities and Applications (05) 05:6 DOI 0.86/s RESEARCH Open Access A monotonic refinement of Levinson s inequality Julije Jakšetic, Josip Pec aric and Marjan Praljak3* * Correspondence: mpraljak@pbf.hr Faculty of Food Technology and Biotechnology, University of Zagreb, Zagreb, Croatia Full list of author information is available at the end of the article 3 Abstract In this paper we give a monotonic refinement of the probabilistic version of Levinson s inequality based on the monotonic refinement of Jensen s inequality obtained by Cho et al. (Panam. Math. J. :43-50, 00). MSC: 6D5 Keywords: Levinson s inequality; Jensen s inequality; 3-convexity at a point Introduction Levinson s inequality and its converse are summarized in the following result taken from Bullen [ ]. Theorem. (a) If f : [a, b] R is -convex and pi, xi, yi, i,,..., n, are such that pi >, n i pi, a xi, yi b, max(x,..., xn ) min(y,..., yn ) ( ) x + y x + y xn + yn c ( ) and for some c [a, b], then n i pi f (xi ) f (x) n pi f (yi ) f (y), ( ) i where x ni pi xi and y ni pi yi denote the weighted arithmetic means. (b) If for a continuous function f inequality ( ) holds for all n, all c [a, b], all n distinct points xi, yi [a, b] satisfying ( ) and ( ) and all weights pi > such that ni pi, then f is -convex. Levinson [ ] originally proved the inequality for functions f : (, c) R such that f. Popoviciu [ ] showed that the assumption of nonnegativity of the third derivative can be weakened to -convexity of f. Bullen [ ] gave another proof of the inequality 05 Jakšetic et al.; licensee Springer. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License ( which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

2 Jakšetićetal.Journal of Inequalities and Applications (05) 05:6 Page of 7 (rescaled to a general interval [a, b]) as well as its converse given in part (b) of Theorem.. Pečarić and Raşa [4] extended the inequality by using the method of index set functions; in the process they weakened assumption () and obtained a monotonic refinement of the inequality. The above version of the inequality assumes that the sequences x i s and y i s are symmetrically distributed around the point c.mercer[5] made a significant improvement by replacing this condition of symmetric distribution with the weaker one that the variances of the two sequences are equal. Theorem. If f :[a, b] R satisfies f 0 and p i, x i, y i, i,,...,n, are such that p i >0, n p i,a x i, y i b,() holds and n p i (x i x) n p i (y i y), then (3) holds. Witkowski [6] extended Mercer s result to 3-convex functions and a more general probabilistic setting. Baloch et al. [7] showed that inequality (3) holds for a larger class of functions they introduced and called 3-convex functions at a point. Definition.3 Let I be an interval in R and c I. A function f : I R is said to be 3- convex at point c if there exists a constant A such that the function F(s) f (s) A s is concave on I (, c]andconvexoni [c, ). Baloch et al. [7] also proved the converse of the inequality, i.e., 3-convexfunctionsat a point are the largest class of functions for which Levinson s inequality holds under the equal variances assumption. Probabilistic version of Levinson s inequality and its converse are summarized in the following result taken from Pečarić et al. [8]. Theorem.4 (a) Let f :[a, b] R be 3-convex at point c and X : [a, c] and Y : [c, b] be two random variables such that Var(X) Var(Y ). Then E ( f (X) ) f ( E(X) ) E ( f (Y ) ) f ( E(Y ) ). (4) (b) Let f :[a, b] R be continuous and c (a, b) fixed. Suppose that inequality (4) holds for all discrete random variables X and Y taking two values x, x [a, c] and y, y [c, b], respectively, each with probability and such that Var(X)Var(Y )(i.e. x x y y ). Then f is 3-convex at c. Remark.5 Results in [8]were stated for f defined on an arbitrary interval I.In that case, the finiteness of Var(X)Var(Y ), E[f (X)] and E[f (Y )] needs to be assumed. For simplicity, in this paper we will work with the closed interval [a, b] since in this case the function f and all random variables are bounded and the aforementioned finiteness assumptions are satisfied. If X and Y are discrete random variables taking values x i and y i,respectively,withprobabilities p i, then Theorem.4(a) gives Theorem.. In[8] itwasproventhatafunction

3 Jakšetićetal.Journal of Inequalities and Applications (05) 05:6 Page 3 of 7 defined on an interval is 3-convex if and only if it is 3-convex at every point of the interval. Therefore, the converse stated in Theorem.4(b) strengthens the converse stated in Theorem.(b). Theorem.4 shows that 3-convex functions at a point are characterized by Levinson s inequality in a similar way that convex functions are characterized by Jensen s inequality. Cho et al. [9] constructed two mappings connected to Jensen s inequality and proved their monotonicity and convexity properties. Throughout the rest of the paper denotes a measurable space with a finite measure μ, and we assume all mappings to be measurable. Further, E[ ] andvar( ) denote the expectation and variance operators with respect to the probability measure μ, i.e.,forz : R, μ() E[z] z(s) dμ(s), Var[z] ( ) [ z(s) E[z] dμ(s)e z ] E [z]. μ() The following is a result from [9]. Theorem.6 Let f :[a, b] R be convex, x : [a, b] and H, V :[0,] R the mappings H(t) f ( tx(s)+( t)e[x] ) dμ(s) and V(t) f ( tx(s)+( t)x(u) ) dμ(s) dμ(u). μ() Then: (a) the mappings H and V are convex on [0, ], (b) the mapping H is nondecreasing on [0, ], while the mapping V is nonincreasing on [0, ] and nondecreasing on [,], (c) the following equalities hold: inf H(t)H(0) f ( E[x] ), t [0,] sup H(t)H() E [ f (x) ], t [0,] inf V(t)V t [0,] ( ) f μ() sup V(t)V(0) V() E [ f (x) ], t [0,] (d) the following inequality holds for all t [0, ]: ( x(s)+x(u) ) dμ(s) dμ(u), V(t) max { H(t), H( t) }.

4 Jakšetićetal.Journal of Inequalities and Applications (05) 05:6 Page 4 of 7 Remark.7 Theorem.6 was proven in [9]forthecasewhen is an interval in R and μ is a measure with density, i.e., dμ(s)p(s) ds. But, from the proofs given there, it is obvious that the statements hold under the more general setting given here. If we denote x (t) (s)tx(s)+( t)e[x], then H(t)E[f (x (t) )]. As t ranges from 0 to, the function (i.e., random variable) x (t) ranges from the constant E[x] to the function x itself. In the process the expectation E[f (x (t) )] increases by the monotonicity property from Theorem.6(b). Therefore, for 0 s t, the following monotonic refinement of Jensen s inequality holds: f ( E[x] ) H(0) E [ f (x (s) ) ] E [ f (x (t) ) ] H() E [ f (x) ]. Furthermore, if x and x are two independent identically distributed copies of x on the product space,thenv(t)e[f ( x (t) )], where x (t) tx +( t)x,andtheorem.6(d) can be interpreted as E[f (x (t) )] E[f ( x (t) )]. In this paper we will construct the corresponding two mappings in connection with Levinson s inequality and show their monotonicity and convexity properties. Main results The following is our main result. Theorem. Let f :[a, b] R be 3-convex at point c, x : [a, c] and y : [c, b] such that Var(x) Var(y) and H, V :[0, ] R the mappings H(t) [ ( ) ( )] f ty(s)+( t)e[y] f tx(s)+( t)e[x] dμ(s) and V(t) [ ( ) ( )] f ty(s)+( t)y(u) f tx(s)+( t)x(u) dμ(s) dμ(u). μ() Then: (a) the mappings H and V are convex on [0, ], (b) the mapping H is nondecreasing on [0, ], while the mapping V is nonincreasing on [0, ] and nondecreasing on [,], (c) the following equalities hold: inf H(t)H(0) f ( E[y] ) f ( E[x] ), t [0,] sup H(t)H() E [ f (y) ] E [ f (x) ], t [0,] ( [ inf t [0,] V(t)V ) μ() ( x(s)+x(u) f ( ) y(s)+y(u) f )] dμ(s) dμ(u), sup V(t)V(0) V() E [ f (y) ] E [ f (x) ], t [0,]

5 Jakšetićetal.Journal of Inequalities and Applications (05) 05:6 Page 5 of 7 (d) the following inequality holds for all t [0, ]: V(t) max { H(t), H( t) }. Proof Let the constant A be as in Definition.3, i.e., such that the function F(s)f (s) A s is concave on [a, c]andconvexon[c, b]. Since the function y takes values in [c, b], so does the function y (t) ty +( t)e[y] for every t [0, ]. Furthermore, since the function F is convex on [c, b], by Theorem.6 the mapping H (t) F ( ty(s)+( t)e[y] ) dμ(s) is convex and nondecreasing on [0, ]. We have H (t) f ( ty(s)+( t)e[y] ) dμ(s) A ( ) ty(s)+( t)e[y] dμ(s) μ() E [ f (y (t) ) ] A t E [ y ] At( t)e[y]e[y] A ( t) E [y] E [ f (y (t) ) ] A t( E [ y ] E [y] ) A E [y] E [ f (y (t) ) ] A t Var(y) A E [y]. Similarly, the function x (t) tx +( t)e[x] takes values in [a, c]foreveryt [0, ] and F is convex on [a, c], so by Theorem.6 the mapping H (t) F ( tx(s)+( t)e[x] ) dμ(s) is convex and nondecreasing on [0, ], and we have H (t) μ() + A μ() f ( tx(s)+( t)e[x] ) dμ(s) ( tx(s)+( t)e[x] ) dμ(s) E [ f (x (t) ) ] + A t Var(x)+ A E [x]. Let us also denote the (constant) mapping H 3 (t) A (E [y] E [x]). All three of the mappings H i, i,, 3, are convex and nondecreasing and, therefore, so is their sum. Since Var(x)Var(y), we have H H + H + H 3, and this proves the convexity and monotonicity properties of H from parts (a) and (b), while the first two equalities in (c) follow by simple calculation. As for the mapping V, first of all, it is easy to see that V(t)V( t) for all t [0, ], that is, V is symmetric with respect to t.next,sincey takes values in [c, b]andf is convex

6 Jakšetićetal.Journal of Inequalities and Applications (05) 05:6 Page 6 of 7 on that interval, by Theorem.6 the mapping V (t) F ( ty(s)+( t)y(u) ) dμ(s) dμ(u) μ() is convex on [0, ] and nondecreasing on [,].Wehave V (t) μ() A μ() μ() f ( ty(s)+( t)y(u) ) dμ(s) dμ(u) ( ty(s)+( t)y(u) ) dμ(s) dμ(u) dμ(u) f ( ty(s)+( t)y(u) ) dμ(s) dμ(u) A t E [ y ] At( t)e[y]e[y] A ( t) E [ y ] f ( ty(s)+( t)y(u) ) dμ(s) dμ(u) μ() + At( t) ( E [ y ] E [y] ) A E[ y ] f ( ty(s)+( t)y(u) ) dμ(s) dμ(u) μ() + At( t) Var(y) A E[ y ]. Similarly, since x takes values in [a, c] and F is convex on that interval, by Theorem.6 the mapping V (t) F ( tx(s)+( t)x(u) ) dμ(s) dμ(u) μ() is convex on [0, ] and nondecreasing on [,]andwehave V (t) f ( tx(s)+( t)x(u) ) dμ(s) dμ(u) μ() At( t) Var(x)+ A E[ x ]. Let us also denote the (constant) mapping V 3 (t) A (E[y ] E[x ]). All three of the mappings V i, i,, 3, are convex and nondecreasing on [, ] and, therefore, so is their sum. Since Var(x)Var(y), we have V V +V +V 3.Furthermore,sinceV is symmetric around t, it follows that it is nonincreasing on [0, ], its minimum is attained at t and its maximumis attained at t 0andt. This proves the convexity and monotonicity properties of V. Finally, as for part (d), since V is symmetric around t and H is nondecreasing, it is enough to prove that V(t) H(t) fort [, ]. This inequality holds since V (t) H (t) and V (t) H (t) bytheorem.6(d) and V 3 (t)h 3 (t) sincevar(x)var(y) andthisfinishes the proof.

7 Jakšetićetal.Journal of Inequalities and Applications (05) 05:6 Page 7 of 7 A monotonic refinement of Levinson s inequality (4) based on Theorem. is the following: if x (t) and y (t) for t [0, ] are as in the proof of Theorem.,thenH(t)E[f (y (t) )] E[f (x (t) )] and for 0 s t itholds f ( E[y] ) f ( E[x] ) H(0) E [ f (y (s) ) ] E [ f (x (s) ) ] E [ f (y (t) ) ] E [ f (x (t) ) ] H() E [ f (y) ] E [ f (x) ]. Remark. The convexity and monotonicity property of the mapping H in the case when x and y are two discrete random variables taking values x i and y i,respectively,withprobabilities p i, i,...,n,wasprovenin[7]. Remark.3 The assumption of equal variances in Theorem. can be weakened. If we denote B A(Var(y) Var(x)), then the assumption Var(x) Var(y) can be relaxedto B 0. Indeed, what we have shown in the proof of Theorem. is that H 4 H i and V 4 V i, where H 4 (t) Bt and V 4 (t)bt(t ).ForB 0 the mapping H 4 is convex and nondecreasing, while the mapping V 4 is convex, symmetric around t and nondecreasing on [, ]. Therefore, the convexity and monotonicity properties of H and V are preserved. Furthermore, V 3 (t) H 3 (t) B,sofromV (t) H (t), V (t) H (t)andv 3 (t)+v 4 (t) H 3 (t) H 4 (t) B( t) 0 it follows that V(t) H(t), i.e., part (d) also holds. Competing interests The authors declare that they have no competing interests. Authors contributions All authors jointly worked on the results, and they read and approved the final manuscript. Author details Faculty of Mechanical Engineering and Naval Architecture, University of Zagreb, Zagreb, Croatia. Faculty of Textile Technology, University of Zagreb, Zagreb, Croatia. 3 Faculty of Food Technology and Biotechnology, University of Zagreb, Zagreb, Croatia. Acknowledgements This work has been fully supported by Croatian Science Foundation under the project Received: November 04 Accepted: 30 April 05 References. Bullen, P: An inequality of N. Levinson. Publ. Elektroteh. Fak. Univ. Beogr., Ser. Mat. Fiz. 4(460), 09- (973). Levinson, N: Generalization of an inequality of Ky Fan. J. Math. Anal. Appl. 8, (964) 3. Popoviciu, T: Sur une inégalité de N. Levinson. Mathematica 6, (964) 4. Pečarić, J, Raşa, I: On an index set function. Southeast Asian Bull. Math. 4, (000) 5. Mercer, A: Short proof of Jensen s and Levinson s inequalities. Math. Gaz. 94, (00) 6. Witkowski, A: On Levinson s inequality. Ann. Univ. Paedagog. Crac. Stud. Math., (03) 7. Baloch, I, Pečarić, J, Praljak, M: Generalization of Levinson s inequality. J. Math. Inequal. 9(), (05) 8. Pečarić, J, Praljak, M, Witkowski, A: Generalized Levinson s inequality and exponential convexity. Opusc. Math. 35(3), (05) 9. Cho, Y, Matić, M, Pečarić, J: Two mappings in connection to Jensen s inequality. Panam. Math. J., (00)