Journal of Inequalities in Pure and Applied Mathematics AN INEQUALITY IMPROVING THE SECOND HERMITE-HADAMARD INEQUALITY FOR CONVEX FUNCTIONS DEFINED ON LINEAR SPACES AND APPLICATIONS FOR SEMI-INNER PRODUCTS volume 3, issue 3, article 35, 2002. S.S. DRAGOMIR School of Communications and Informatics Victoria University of Technology PO Box 14428 Melbourne City MC 8001 Victoria, Australia EMail: sever@matilda.vu.edu.au URL: http://rgmia.vu.edu.au/SSDragomirWeb.html Received 16 November, 2001; accepted 21 February, 2002. Communicated by: A. Lupaş Abstract Contents JJ J II I Home Page Go Back Close c 2000 Victoria University ISSN (electronic): 1443-5756 080-01 Quit Abstract An inequality for convex functions defined on linear spaces is obtained which contains in a particular case a refinement for the second part of the celebrated Hermite-Hadamard inequality. Applications for semi-inner products on normed linear spaces are also provided. 2000 Mathematics Subject Classification: Primary 26D15, 26D10; Secondary 46B10. Key words: Hermite-Hadamard integral inequality, Convex functions, Semi-Inner products. Contents An Inequality Improving the Second Hermite-Hadamard Inequality for Convex Functions Defined on Linear Spaces and Applications for Semi-Inner Products S.S. Dragomir 1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 2 The Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 3 Applications for Semi-Inner Products . . . . . . . . . . . . . . . . . . . . 14 References Title Page Contents JJ J II I Go Back Close Quit Page 2 of 18 1. Introduction Let X be a real linear space, a, b ∈ X, a 6= b and let [a, b] := {(1 − λ) a + λb, λ ∈ [0, 1]} be the segment generated by a and b. We consider the function f : [a, b] → R and the attached function g (a, b) : [0, 1] → R, g (a, b) (t) := f [(1 − t) a + tb], t ∈ [0, 1]. It is well known that f is convex on [a, b] iff g (a, b) is convex on [0, 1], and the following lateral derivatives exist and satisfy 0 (i) g± (a, b) (s) = (5± f [(1 − s) a + sb]) (b − a), s ∈ (0, 1) 0 (ii) g+ (a, b) (0) = (5+ f (a)) (b − a) 0 (iii) g− (a, b) (1) = (5− f (b)) (b − a) where (5± f (x)) (y) are the Gâteaux lateral derivatives, we recall that f (x + hy) − f (x) , (5+ f (x)) (y) : = lim h→0+ h f (x + ky) − f (x) , x, y ∈ X. (5− f (x)) (y) : = lim k→0− k The following inequality is the well known Hermite-Hadamard integral inequality for convex functions defined on a segment [a, b] ⊂ X : Z 1 f (a) + f (b) a+b (HH) f ≤ f [(1 − t) a + tb] dt ≤ , 2 2 0 An Inequality Improving the Second Hermite-Hadamard Inequality for Convex Functions Defined on Linear Spaces and Applications for Semi-Inner Products S.S. Dragomir Title Page Contents JJ J II I Go Back Close Quit Page 3 of 18 which easily follows by the classical Hermite-Hadamard inequality for the convex function g (a, b) : [0, 1] → R Z 1 1 g (a, b) (0) + g (a, b) (1) g (a, b) ≤ g (a, b) (t) dt ≤ . 2 2 0 For other related results see the monograph on line [1]. Now, assume that (X, k·k) is a normed linear space. The function f0 (s) = 1 kxk2 , x ∈ X is convex and thus the following limits exist 2 h i 2 2 (iv) hx, yis := (5+ f0 (y)) (x) = lim ky+txk2t−kyk ; t→0+ (v) hx, yii := (5− f0 (y)) (x) = lim s→0− h ky+sxk2 −kyk2 2s i ; for any x, y ∈ X. They are called the lower and upper semi-inner products associated to the norm k·k. For the sake of completeness we list here some of the main properties of these mappings that will be used in the sequel (see for example [2]), assuming that p, q ∈ {s, i} and p 6= q: (a) hx, xip = kxk2 for all x ∈ X; (aa) hαx, βyip = αβ hx, yip if α, β ≥ 0 and x, y ∈ X; (aaa) hx, yip ≤ kxk kyk for all x, y ∈ X; (av) hαx + y, xip = α hx, xip + hy, xip if x, y ∈ X and α ∈ R; An Inequality Improving the Second Hermite-Hadamard Inequality for Convex Functions Defined on Linear Spaces and Applications for Semi-Inner Products S.S. Dragomir Title Page Contents JJ J II I Go Back Close Quit Page 4 of 18 (v) h−x, yip = − hx, yiq for all x, y ∈ X; (va) hx + y, zip ≤ kxk kzk + hy, zip for all x, y, z ∈ X; (vaa) The mapping h·, ·ip is continuous and subadditive (superadditive) in the first variable for p = s (or p = i); (vaaa) The normed linear space (X, k·k) is smooth at the point x0 ∈ X\ {0} if and only if hy, x0 is = hy, x0 ii for all y ∈ X; in general hy, xii ≤ hy, xis for all x, y ∈ X; (ax) If the norm k·k is induced by an inner product h·, ·i , then hy, xii = hy, xi = hy, xis for all x, y ∈ X. Applying inequality (HH) for the convex function f0 (x) = 12 kxk2 , one may deduce the inequality Z 1 x + y 2 kxk2 + kyk2 2 (1.1) ≤ k(1 − t) x + tyk dt ≤ 2 2 0 for any x, y ∈ X. The same (HH) inequality applied for f1 (x) = kxk , will give the following refinement of the triangle inequality: Z 1 x + y kxk + kyk (1.2) k(1 − t) x + tyk dt ≤ , x, y ∈ X. 2 ≤ 2 0 An Inequality Improving the Second Hermite-Hadamard Inequality for Convex Functions Defined on Linear Spaces and Applications for Semi-Inner Products S.S. Dragomir Title Page Contents JJ J II I Go Back Close In this paper we point out an integral inequality for convex functions which is related to the first Hermite-Hadamard inequality in (HH) and investigate its applications for semi-inner products in normed linear spaces. Quit Page 5 of 18 2. The Results We start with the following lemma which is also of interest in itself. Lemma 2.1. Let h : [α, β] ⊂ R → R be a convex function on [α, β]. Then for any γ ∈ [α, β] one has the inequality (2.1) 1 (β − γ)2 h0+ (γ) − (γ − α)2 h0− (γ) 2 β Z ≤ (γ − α) h (α) + (β − γ) h (β) − h (t) dt α 1 ≤ (β − γ)2 h0− (β) − (γ − α)2 h0+ (α) . 2 An Inequality Improving the Second Hermite-Hadamard Inequality for Convex Functions Defined on Linear Spaces and Applications for Semi-Inner Products The constant 21 is sharp in both inequalities. The second inequality also holds for γ = α or γ = β. S.S. Dragomir Proof. It is easy to see that for any locally absolutely continuous function h : (α, β) → R, we have the identity Title Page Z β Z 0 β (t − γ) h (t) dt = (γ − α) h (α) + (β − γ) h (β) − (2.2) α h (t) dt α for any γ ∈ (α, β) , where h0 is the derivative of h which exists a.e. on (α, β) . Since h is convex, then it is locally Lipschitzian and thus (2.2) holds. Moreover, for any γ ∈ (α, β), we have the inequalities (2.3) 0 h (t) ≤ h0− (γ) for a.e. t ∈ [α, γ] Contents JJ J II I Go Back Close Quit Page 6 of 18 and (2.4) h0 (t) ≥ h0+ (γ) for a.e. t ∈ [γ, β] . If we multiply (2.3) by γ − t ≥ 0, t ∈ [α, γ] and integrate on [α, γ] , we get Z γ 1 (2.5) (γ − t) h0 (t) dt ≤ (γ − α)2 h0− (γ) 2 α and if we multiply (2.4) by t − γ ≥ 0, t ∈ [γ, β], and integrate on [γ, β] , we also have Z β 1 (2.6) (t − γ) h0 (t) dt ≥ (β − γ)2 h0+ (γ) . 2 γ Now, if we subtract (2.5) from (2.6) and use the representation (2.2), we deduce the first inequality in (2.1). If we assume that the first inequality (2.1) holds with a constant C > 0 instead of 12 , i.e., (2.7) C (β − γ)2 h0+ (γ) − (γ − α)2 h0− (γ) Z β ≤ (γ − α) h (α) + (β − γ) h (β) − h (t) dt α , k > 0, t ∈ [α, β], then and take the convex function h0 (t) := k t − α+β 2 α+β 0 = k, h0+ 2 α+β 0 = −k, h0− 2 An Inequality Improving the Second Hermite-Hadamard Inequality for Convex Functions Defined on Linear Spaces and Applications for Semi-Inner Products S.S. Dragomir Title Page Contents JJ J II I Go Back Close Quit Page 7 of 18 k (β − α) = h0 (β) , 2 Z β 1 h0 (t) dt = k (β − α)2 , 4 α h0 (α) = and the inequality (2.7) becomes, for γ = α+β , 2 1 1 1 2 2 C (β − α) k + (β − α) k ≤ k (β − α)2 , 4 4 4 giving C ≤ 21 , which proves the sharpness of the constant 12 in the first inequality in (2.1). If either h0+ (α) = −∞ or h0− (β) = −∞, then the second inequality in (2.1) holds true. Assume that h0+ (α) and h0− (β) are finite. Since h is convex on [α, β] , we have (2.8) 0 h (t) ≥ h0+ (α) for a.e. t ∈ [α, γ] (γ may be equal to β) and (2.9) h0 (t) ≤ h0− (β) for a.e. t ∈ [γ, β] (γ may be equal to α) . An Inequality Improving the Second Hermite-Hadamard Inequality for Convex Functions Defined on Linear Spaces and Applications for Semi-Inner Products S.S. Dragomir Title Page Contents JJ J II I Go Back If we multiply (2.8) by γ − t ≥ 0, t ∈ [α, γ] and integrate on [α, γ] , then we deduce Z γ 1 (2.10) (γ − t) h0 (t) dt ≥ (γ − α)2 h0+ (α) 2 α Close Quit Page 8 of 18 and if we multiply (2.9) by t − γ ≥ 0, t ∈ [γ, β], and integrate on [γ, β] , then we also have Z β 1 (2.11) (t − γ) h0 (t) dt ≤ (β − γ)2 h0− (β) . 2 γ Finally, if we subtract (2.10) from (2.11) and use the representation (2.2), we deduce the second part of (2.1). Now, assume that the second inequality in (2.1) holds with a constant D > 0 instead of 12 , i.e., Z (2.12) β (γ − α) f (α) + (β − γ) f (β) − f (t) dt α ≥ D (β − γ)2 f−0 (β) − (γ − α)2 f+0 (α) . If we consider the convex function h0 given above, then we have h0− (β) = k, h0+ (α) = −k and by (2.12) applied for h0 and x = α+β we get 2 1 1 1 2 2 2 k (β − α) ≤ D k (β − α) + k (β − α) , 4 4 4 giving D ≥ 21 , and the sharpness of the constant 1 2 is proved. Corollary 2.2. With the assumptions of Lemma 2.1 and if γ ∈ (α, β) is a point of differentiability for h, then α+β (2.13) − γ h0 (γ) 2 An Inequality Improving the Second Hermite-Hadamard Inequality for Convex Functions Defined on Linear Spaces and Applications for Semi-Inner Products S.S. Dragomir Title Page Contents JJ J II I Go Back Close Quit Page 9 of 18 ≤ γ−α β−α h (α) + β−γ β−α 1 h (β) − β−α Z β h (t) dt. α Now, recall that the following inequality, which is well known in the literature as the Hermite-Hadamard inequality for convex functions, holds Z β 1 h (α) + h (β) α+β ≤ . (2.14) h h (t) dt ≤ 2 β−α α 2 The following corollary provides some bounds for the difference Z β h (α) + h (β) 1 − h (t) dt. 2 β−α α Corollary 2.3. Let h : [α, β] → R be a convex function on [α, β]. Then we have the inequality α+β 1 0 α+β 0 (2.15) 0 ≤ h − h− (β − α) 8 + 2 2 Z β h (α) + h (β) 1 ≤ − h (t) dt 2 β−α α 1 0 ≤ h− (β) − h0+ (α) (β − α) . 8 The constant 1 8 is sharp in both inequalities. We are now able to state the corresponding result for convex functions defined on linear spaces. An Inequality Improving the Second Hermite-Hadamard Inequality for Convex Functions Defined on Linear Spaces and Applications for Semi-Inner Products S.S. Dragomir Title Page Contents JJ J II I Go Back Close Quit Page 10 of 18 Theorem 2.4. Let X be a linear space, a, b ∈ X, a 6= b and f : [a, b] ⊂ X → R be a convex function on the segment [a, b]. Then for any s ∈ (0, 1) one has the inequality 1 (2.16) (1 − s)2 (5+ f [(1 − s) a + sb]) (b − a) 2 −s2 (5− f [(1 − s) a + sb]) (b − a) Z 1 ≤ (1 − s) f (a) + sf (b) − f [(1 − t) a + tb] dt 0 1 ≤ (1 − s)2 (5− f (b)) (b − a) − s2 (5+ f (a)) (b − a) . 2 1 The constant 2 is sharp in both inequalities. The second inequality also holds for s = 0 or s = 1. An Inequality Improving the Second Hermite-Hadamard Inequality for Convex Functions Defined on Linear Spaces and Applications for Semi-Inner Products S.S. Dragomir Proof. Follows by Lemma 2.1 applied for the convex function h (t) = g (a, b) (t) = f [(1 − t) a + tb] , t ∈ [0, 1] , and for the choices α = 0, β = 1, and γ = s. Title Page Contents Corollary 2.5. If f : [a, b] → R is as in Theorem 2.4 and Gâteaux differentiable in c := (1 − λ) a + λb, λ ∈ (0, 1) along the direction (b − a), then we have the inequality: 1 (2.17) − λ (5f (c)) (b − a) 2 Z 1 ≤ (1 − λ) f (a) + λf (b) − f [(1 − t) a + tb] dt. 0 JJ J II I Go Back Close Quit Page 11 of 18 The following result related to the second Hermite-Hadamard inequality for functions defined on linear spaces also holds. Corollary 2.6. If f is as in Theorem 2.4, then 1 a+b a+b (2.18) 5+ f (b − a) − 5− f (b − a) 8 2 2 Z 1 f (a) + f (b) − f [(1 − t) a + tb] dt ≤ 2 0 1 ≤ [(5− f (b)) (b − a) − (5+ f (a)) (b − a)] . 8 The constant 1 8 is sharp in both inequalities. An Inequality Improving the Second Hermite-Hadamard Inequality for Convex Functions Defined on Linear Spaces and Applications for Semi-Inner Products Now, let Ω ⊂ Rn be an open convex set in Rn . If F : Ω → R is a differentiable convex function on Ω, then, obviously, for any c̄ ∈ Ω we have S.S. Dragomir n X ∂F (c̄) ∇F (c̄) (ȳ) = · yi , ȳ ∈ Rn , ∂x i i=1 Contents ∂F where ∂x are the partial derivatives of F with respect to the variable xi i (i = 1, . . . , n). Using (2.16), we may state that X n ∂F λā + (1 − λ) b̄ 1 (2.19) −λ · (bi − ai ) 2 ∂xi i=1 Title Page JJ J II I Go Back Close Quit Page 12 of 18 1 ≤ (1 − λ) F (ā) + λF b̄ − F (1 − t) ā + tb̄ dt 0 " # n n X X ∂F b̄ 1 ∂F (ā) (1 − λ)2 · (bi − ai ) − λ2 · (bi − ai ) ≤ 2 ∂x ∂x i i i=1 i=1 Z for any ā, b̄ ∈ Ω and λ ∈ (0, 1). In particular, for λ = 21 , we get (2.20) Z 1 F (ā) + F b̄ 0 ≤ − F (1 − t) ā + tb̄ dt 2 0 ! n X ∂F b̄ ∂F (ā) 1 ≤ − · (bi − ai ) . 8 i=1 ∂xi ∂xi In (2.20) the constant 1 8 An Inequality Improving the Second Hermite-Hadamard Inequality for Convex Functions Defined on Linear Spaces and Applications for Semi-Inner Products S.S. Dragomir is sharp. Title Page Contents JJ J II I Go Back Close Quit Page 13 of 18 3. Applications for Semi-Inner Products Let (X, k·k) be a real normed linear space. We may state the following results for the semi-inner products h·, ·ii and h·, ·is . Proposition 3.1. For any x, y ∈ X and σ ∈ (0, 1) we have the inequalities: (3.1) (1 − σ)2 hy − x, (1 − σ) x + σyis − σ 2 hy − x, (1 − σ) x + σyii Z 1 2 2 ≤ (1 − σ) kxk + σ kyk − k(1 − t) x + tyk2 dt 0 2 2 ≤ (1 − σ) hy − x, yii − σ hy − x, yis . The second inequality in (3.1) also holds for σ = 0 or σ = 1. The proof is obvious by Theorem 2.4 applied for the convex function f (x) = kxk2 , x ∈ X. If the space is smooth, then we may put [x, y] = hx, yii = hx, yis for each x, y ∈ X and the first inequality in (3.1) becomes 1 2 (3.2) Z 1 k(1 − t) x + tyk2 dt. 0 An interesting particular case one can get from (3.1) is the one for σ = 12 , (3.3) 1 [hy − x, y + xis − hy − x, y + xii ] 8 Z 1 kxk2 + kyk2 ≤ − k(1 − t) x + tyk2 dt 2 0 0≤ S.S. Dragomir Title Page Contents (1 − 2σ) [y − x, (1 − σ) x + σy] ≤ (1 − σ) kxk2 + σ kyk2 − An Inequality Improving the Second Hermite-Hadamard Inequality for Convex Functions Defined on Linear Spaces and Applications for Semi-Inner Products JJ J II I Go Back Close Quit Page 14 of 18 ≤ 1 [hy − x, yii − hy − x, xis ] . 4 The inequality (3.3) provides a refinement and a counterpart for the second inequality in (1.1). If we consider now two linearly independent vectors x, y ∈ X and apply Theorem 2.4 for f (x) = kxk, x ∈ X, then we get Proposition 3.2. For any linearly independent vectors x, y ∈ X and σ ∈ (0, 1) , one has the inequalities: 1 2 hy − x, (1 − σ) x + σyis 2 hy − x, (1 − σ) x + σyii (3.4) (1 − σ) −σ 2 k(1 − σ) x + σyk k(1 − σ) x + σyk Z 1 ≤ (1 − σ) kxk + σ kyk − k(1 − t) x + tyk dt 0 1 2 hy − x, yii 2 hy − x, xis (1 − σ) −σ . ≤ 2 kyk kxk Title Page JJ J We note that if the space is smooth, then we have (3.5) S.S. Dragomir Contents The second inequality also holds for σ = 0 or σ = 1. An Inequality Improving the Second Hermite-Hadamard Inequality for Convex Functions Defined on Linear Spaces and Applications for Semi-Inner Products 1 [y − x, (1 − σ) x + σy] −σ · 2 k(1 − σ) x + σyk II I Go Back Z ≤ (1 − σ) kxk + σ kyk − Close 1 k(1 − t) x + tyk dt Quit 0 Page 15 of 18 and for σ = 12 , (3.4) will give the simple inequality +# "* + * x+y x+y 1 2 2 (3.6) − y − x, 0 ≤ y − x, x+y x+y 8 2 2 i s Z 1 kxk + kyk ≤ − k(1 − t) x + tyk dt 2 0 1 y x ≤ y − x, − y − x, . 8 kyk i kxk s The inequality (3.6) provides a refinement and a couterpart of the second inequality in (1.2). Moreover, if we assume that (H, h·, ·i) is an inner product space, then by (3.6) we get for any x, y ∈ H with kxk = kyk = 1 that Z 1 1 (3.7) 0≤1− k(1 − t) x + tyk dt ≤ ky − xk2 . 8 0 The constant 81 is sharp. Indeed, if we choose H = R, ha, bi = a · b, x = −1, y = 1, then we get equality in (3.7). We give now some examples. 1. Let `2 (K) , K = C, R; be the Hilbert space of sequences x = (xi )i∈N with An Inequality Improving the Second Hermite-Hadamard Inequality for Convex Functions Defined on Linear Spaces and Applications for Semi-Inner Products S.S. Dragomir Title Page Contents JJ J II I Go Back Close Quit Page 16 of 18 P∞ i=0 |xi |2 < ∞. Then, by (3.7), we have the inequalities Z (3.8) 1 0 ≤ 1− 0 ∞ X ! 12 |(1 − t) xi + tyi |2 dt i=0 ∞ 1 X · |yi − xi |2 , 8 i=0 ≤ for any x, y ∈ `2 (K) provided P∞ i=0 |xi |2 = P∞ i=0 |yi |2 = 1. 2. Let µ be a positive measure, L2 (Ω) the Hilbert space of µ−measurable functions on R Ω with2 complex values that are 2−integrable on Ω, i.e., f ∈ L2 (Ω) iff Ω |f (t)| dµ (t) < ∞. Then, by (3.7), we have the inequalities An Inequality Improving the Second Hermite-Hadamard Inequality for Convex Functions Defined on Linear Spaces and Applications for Semi-Inner Products S.S. Dragomir Z 1 Z 12 2 |(1 − λ) f (t) + λg (t)| dµ (t) dλ 0 ≤ 1− Ω Z0 1 · |f (t) − g (t)|2 dµ (t) ≤ 8 Ω R R for any f, g ∈ L2 (Ω) provided Ω |f (t)|2 dµ (t) = Ω |g (t)|2 dµ (t) = 1. (3.9) Title Page Contents JJ J II I Go Back Close Quit Page 17 of 18 References [1] I. CIORANESCU, Geometry of Banach Spaces, Duality Mappings and Nonlinear Problems, Kluwer Academic Publishers, Dordrecht, 1990. [2] S.S. DRAGOMIR and C.E.M. PEARCE, Selected Topics on Hermite-Hadamard Inequalities and Applications, RGMIA Monographs, Victoria University, 2000. (ONLINE: http://rgmia.vu.edu.au/monographs) An Inequality Improving the Second Hermite-Hadamard Inequality for Convex Functions Defined on Linear Spaces and Applications for Semi-Inner Products S.S. Dragomir Title Page Contents JJ J II I Go Back Close Quit Page 18 of 18