Symmetrized convexity and Hermite-Hadamard type inequalities

Abstract

In this paper we extend the Hermite-Hadamard inequality to the class of symmetrized convex functions. The corresponding version for h-convex functions is also investigated. Some examples of interest are provided as well. 1. Introduction The following inequality holds for any convex function f de ned on R (1.1) f a+ b 2 1 b a Z b a f(x)dx f(a) + f(b) 2 ; a; b 2 R, a 6= b: It was rstly discovered by Ch. Hermite in 1881 in the journal Mathesis (see [42]). But this result was nowhere mentioned in the mathematical literature and was not widely known as Hermite’s result. E. F. Beckenbach, a leading expert on the history and the theory of convex functions, wrote that this inequality was proven by J. Hadamard in 1893 [5]. In 1974, D. S. Mitrinovic found Hermite’s note in Mathesis [42]. Since (1.1) was known as Hadamard’s inequality, the inequality is now commonly referred as the Hermite-Hadamard inequality. For related results, see [10]-[19], [21]-[24], [31]-[34] and [45]. In this paper we show that the Hermite-Hadamard inequality can be extended to a larger class of functions containing the convex functions. The corresponding version for h-convex functions is also investigated. Some examples of interest are provided as well. 2. Symmetrized Convexity For a function f : [a; b] ! C we consider the symmetrical transform of f on the interval [a; b] ; denoted by f[a;b] or simply f , when the interval [a; b] is implicit, which is de ned by f (t) := 1 2 [f (t) + f (a+ b t)] ; t 2 [a; b] : The anti-symmetrical transform of f on the interval [a; b] is denoted by ~ f[a;b]; or simply ~ f and is de ned by ~ f (t) := 1 2 [f (t) f (a+ b t)] ; t 2 [a; b] : 1991 Mathematics Subject Classi cation. 26D15; 25D10. Key words and phrases. Convex functions, Hermite-Hadamard inequality, Integral inequalities, h-Convex functions. 1 2 S. S. DRAGOMIR It is obvious that for any function f we have f + ~ f = f: If f is convex on [a; b] ; then for any t1; t2 2 [a; b] and ; 0 with + = 1 we have f ( t1 + t2) = 1 2 [f ( t1 + t2) + f (a+ b t1 t2)] = 1 2 [f ( t1 + t2) + f ( (a+ b t1) + (a+ b t2))] 1 2 [ f (t1) + f (t2) + f (a+ b t1) + f (a+ b t2)] = 1 2 [f (t1) + f (a+ b t1)] + 1 2 [f (t2) + f (a+ b t2)] = f (t1) + f (t2) ; which shows that f is convex on [a; b] : Consider the real numbers a 0 and strictly concave on [a; b] if a+b 2 0; then we can conclude that f0 is not convex on [a; b] while f0 is convex on [a; b] : We can introduce the following concept of convexity. De nition 1. We say that the function f : [a; b] ! R is symmetrized convex (concave) on the interval [a; b] if f is convex (concave) on [a; b] : Now, if we denote by Con [a; b] the closed convex cone of convex functions de ned on [a; b] and by SCon [a; b] the class of symmetrized convex functions, then from the above remarks we can conclude that (2.1) Con [a; b] SCon [a; b] : Also, if [c; d] [a; b] and f 2 SCon [a; b] ; then this does not imply in general that f 2 SCon [c; d] : Theorem 1. Assume that f : [a; b] ! R is symmetrized convex on the interval [a; b] : Then we have the Hermite-Hadamard inequalities (2.2) f a+ b 2 1 b a Z b a f (t) dt f (a) + f (b) 2 : Proof. Since f : [a; b] ! R is symmetrized convex on the interval [a; b] ; then by writing the Hermite-Hadamard inequality for the function f we have (2.3) f a+ b 2 1 b a Z b a f (t) dt f (a) + f (b) 2 : However f a+ b 2 = f a+ b 2 ; f (a) + f (b) 2 = f (a) + f (b) 2 ; SYMMETRIZED CONVEXITY 3