Some inequalities for weighted and integral means of convex functions on linear spaces

Let f be a convex function on a convex subset C of a linear space and x; y 2 C; with x 6= y: If p : [0; 1] ! R is a Lebesgue integrable and symmetric function, namely p (1 t) = p (t) for all t 2 [0; 1] and such that the condition 0 Z 0 p (s) ds Z 1 0 p (s) ds for all 2 [0; 1] holds, then we have 1 R 1 0 p ( ) d Z 1 0 p ( ) f ((1 )x+ y) d Z 1 0 f ((1 )x+ y) d 1 R 1 0 p ( ) d Z 1 0 Z 0 p (s) ds (1 ) d [r fy (y x) r+fx (y x)] 1 2 [r fy (y x) r+fx (y x)] : Some applications for norms and semi-inner products are also provided. 1. Introduction LetX be a real linear space, x; y 2 X, x 6= y and let [x; y] := f(1 )x+ y; 2 [0; 1]g be the segment generated by x and y. We consider the function f : [x; y]! R and the attached function '(x;y) : [0; 1]! R, '(x;y) (t) := f [(1 t)x+ ty], t 2 [0; 1]. It is well known that f is convex on [x; y] i¤ ' (x; y) is convex on [0; 1], and the following lateral derivatives exist and satisfy (i) '0 (x;y) (s) = r f(1 s)x+sy (y x), s 2 [0; 1); (ii) '+(x;y) (0) = r+fx (y x) ; (iii) '0 (x;y) (1) = r fy (y x) ; where r fx (y) are the Gâteaux lateral derivatives, we recall that r+fx (y) : = lim h!0+ f (x+ hy) f (x) h ; r fx (y) : = lim k!0 f (x+ ky) f (x) k ; x; y 2 X: The following inequality is the well-known Hermite-Hadamard integral inequality for convex functions de…ned on a segment [x; y] X : (HH) f x+ y 2 Z 1 0 f [(1 t)x+ ty] dt f (x) + f (y) 2 ; 1991 Mathematics Subject Classi…cation. 26D15; 46B05. Key words and phrases. Convex functions, LInear spaces, Integral inequalities, HermiteHadamard inequality, Féjer’s inequalities, Norms and semi-inner products. 1 2 S. S. DRAGOMIR which easily follows by the classical Hermite-Hadamard inequality for the convex function ' (x; y) : [0; 1]! R '(x;y) 1 2 Z 1 0 '(x;y) (t) dt '(x;y) (0) + '(x;y) (1) 2 : For other related results see the monograph on line [8]. For some recent results in linear spaces see [1], [2] and [9]-[12]. In the recent paper we established the following re…nements and reverses of Féjer’s inequality for functions de…ned on linear spaces: Theorem 1. Let f be an convex function on C and x; y 2 C with x 6= y: If p : [0; 1] ! [0;1) is Lebesgue integrable and symmetric, namely p (1 t) = p (t) for all t 2 [0; 1] ; then 0 1 2 h r+f x+y 2 (y x) r f x+y 2 (y x) i Z 1 0 t 12 p (t) dt (1.1) Z 1 0 f ((1 t)x+ ty) p (t) dt f x+ y 2 Z 1 0 p (t) dt 1 2 [r fy (y x) r+fx (y x)] Z 1 0 t 1 2 p (t) dt and 0 1 2 h r+f x+y 2 (y x) r f x+y 2 (y x) i Z 1 0 1 2 t 1 2 p (t) dt (1.2) f (x) + f (y) 2 Z 1 0 p (t) dt Z 1 0 f ((1 t)x+ ty) p (t) dt 1 2 [r fy (y x) r+fx (y x)] Z 1 0 1 2 t 12 p (t) dt: If we take p 1 in (1.1), then we get 0 1 8 h r+f x+y 2 (y x) r f x+y 2 (y x) i (1.3) Z 1 0 f [(1 t)x+ ty] dt f x+ y 2 1 8 [r fy (y x) r+fx (y x)] that was …rstly obtained in [4], while from (1.2) we recapture the result obtained in [5] 0 1 8 h r+f x+y 2 (y x) r f x+y 2 (y x) i (1.4) f (x) + f (y) 2 Z 1 0 f [(1 t)x+ ty] dt 1 8 [r fy (y x) r+fx (y x)] : Motivated by the above results, we establish in this paper some upper and lower bounds for the di¤erence Z 1 0 p ( ) f ((1 )x+ y) d Z 1