Journal of Inequalities in Pure and Applied Mathematics AN INEQUALITY IMPROVING THE FIRST HERMITE-HADAMARD INEQUALITY FOR CONVEX FUNCTIONS DEFINED ON LINEAR SPACES AND APPLICATIONS FOR SEMI-INNER PRODUCTS volume 3, issue 2, article 31, 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 15 February, 2002. Communicated by: C.P. Niculescu Abstract Contents JJ J II I Home Page Go Back Close c 2000 Victoria University ISSN (electronic): 1443-5756 082-01 Quit Abstract An integral inequality for convex functions defined on linear spaces is obtained which contains in a particular case a refinement for the first 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 First 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 . . . . . . . . . . . . . . . . . . . . 15 References Title Page Contents JJ J II I Go Back Close Quit Page 2 of 19 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 First 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 19 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 First 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 19 (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 First 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 19 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 Z β ≤ h (t) dt − (β − α) h (γ) α 1 ≤ (β − γ)2 h0− (β) − (γ − α)2 h0+ (α) . 2 An Inequality Improving the First Hermite-Hadamard Inequality for Convex Functions Defined on Linear Spaces and Applications for Semi-Inner Products The constant 12 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 γ Z 0 (t − α) h (t) dt + (2.2) α β 0 Z β (t − β) h (t) dt = h (γ) − γ 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 19 and (2.4) h0 (t) ≥ h0+ (γ) for a.e. t ∈ [γ, β] . If we multiply (2.3) by t − α ≥ 0, t ∈ [α, γ] and integrate on [α, γ] , we get Z γ 1 (2.5) (t − α) h0 (t) dt ≤ (γ − α)2 h0− (γ) 2 α and if we multiply (2.4) by β − t ≥ 0, t ∈ [γ, β], and integrate on [γ, β] , we also have Z β 1 (2.6) (β − t) h0 (t) dt ≥ (β − γ)2 h0+ (γ) . 2 γ An Inequality Improving the First Hermite-Hadamard Inequality for Convex Functions Defined on Linear Spaces and Applications for Semi-Inner Products S.S. Dragomir If we subtract (2.6) from (2.5) and use the representation (2.2), we deduce the first inequality in (2.1). Now, assume that the first inequality (2.1) holds with C > 0 instead of 12 , i.e., Z β 2 0 2 0 (2.7) C (β − γ) h+ (γ) − (γ − α) h− (γ) ≤ h (t) dt − (β − α) h (γ) . α , k > 0, t ∈ [α, β]. Then Consider the convex function h0 (t) := k t − α+β 2 α+β α+β α+β 0 0 h0+ = k, h0− = −k, h0 =0 2 2 2 Title Page Contents JJ J II I Go Back Close Quit Page 7 of 19 and Z β α 1 h0 (t) dt = k (β − α)2 . 4 If in (2.7) we choose h = h0 , γ = α+β , then we get 2 1 1 1 2 2 C (β − α) k + (β − α) k ≤ k (β − α)2 4 4 4 which gives C ≤ 12 and the sharpness of the constant in the first part of (2.1) is proved. 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) h0 (t) ≥ h0+ (α) for a.e. t ∈ [α, γ] (γ may be equal to β) S.S. Dragomir Title Page Contents and (2.9) An Inequality Improving the First Hermite-Hadamard Inequality for Convex Functions Defined on Linear Spaces and Applications for Semi-Inner Products h0 (t) ≤ h0− (β) for a.e. t ∈ [γ, β] (γ may be equal to α) . If we multiply (2.8) by t − α ≥ 0, t ∈ [α, γ] and integrate on [α, γ] , then we deduce Z γ 1 (2.10) (t − α) h0 (t) dt ≥ (γ − α)2 h0+ (α) 2 α JJ J II I Go Back Close Quit Page 8 of 19 and if we multiply (2.9) by β − t ≥ 0, t ∈ [γ, β], and integrate on [γ, β] , then we also have Z β 1 (2.11) (β − t) 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 inequality in (2.1). Now, assume that the second inequality in (2.1) holds with a constant D > 0 instead of 12 , i.e., Z β h (t) dt − (β − α) h (γ) (2.12) α ≤ D (β − γ)2 h0− (β) − (γ − α)2 h0+ (α) . , k > 0, t ∈ [α, β], then If we consider the convex function h0 (t) = k t − α+β 2 we have h00− (β) = k, h00+ (α) = −k and by (2.12) applied for h0 in γ = α+β 2 we get 1 1 1 2 2 2 k (β − α) ≤ D k (β − α) + k (β − α) , 4 4 4 giving D ≥ 12 which proves the sharpness of the constant equality in (2.1). 1 2 in the second in- Corollary 2.2. With the assumptions of Lemma 2.1 and if γ ∈ (α, β) is a point of differentiability for h, then Z β 1 α+β 0 (2.13) − γ h (γ) ≤ h (t) dt − h (γ) . 2 β−α α An Inequality Improving the First 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 19 Now, recall that the following inequality, which is well known in the literature as the Hermite-Hadamard inequality for convex functions, holds Z β α+β 1 h (α) + h (β) ≤ h (t) dt ≤ . (2.14) h 2 β−α α 2 The following corollary provides both a sharper lower bound for the difference, Z β α+β 1 h (t) dt − h , β−α α 2 which we know is nonnegative, and an upper bound. 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 α+β 1 ≤ h (t) dt − h β−α α 2 1 0 ≤ h− (β) − h0+ (α) (β − α) . 8 1 The constant 8 is sharp in both inequalities. Example 2.1. Assume that −∞ < α < 0 < β < ∞ and consider the convex function h : [α, β] → R, h (x) = exp |x| . We have −e−x if x < 0, h0 (x) = x e if x > 0; An Inequality Improving the First 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 19 and h0− (0) = −1, h0+ (0) = 1. Also, Z β Z 0 Z −x h (t) dt = e dx + α Now, if (2.16) α α+β 2 β ex dx = exp (β) + exp (−α) − 2. 0 6= 0, then by (2.15) we deduce the elementary inequality α + β exp (β) + exp (−α) − 2 0 ≤ − exp β−α 2 1 ≤ [exp (β) + exp (−α)] (β − α) . 8 If α+β = 0 and if we denote β = a, a > 0, thus α = −a and by (2.15) we also 2 have An Inequality Improving the First Hermite-Hadamard Inequality for Convex Functions Defined on Linear Spaces and Applications for Semi-Inner Products 1 exp (a) − 1 1 a≤ − 1 ≤ a exp (a) . 2 a 2 S.S. Dragomir The reader may produce other elementary inequalities by choosing in an appropriate way the convex function h. We omit the details. We are now able to state the corresponding result for convex functions defined on linear spaces. Title Page (2.17) 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 (2.18) 1 (1 − s)2 (5+ f [(1 − s) a + sb]) (b − a) 2 −s2 (5− f [(1 − s) a + sb]) (b − a) Contents JJ J II I Go Back Close Quit Page 11 of 19 Z ≤ 1 f [(1 − t) a + tb] dt − f [(1 − s) a + sb] 0 ≤ 1 (1 − s)2 (5− f (b)) (b − a) − s2 (5+ f (a)) (b − a) . 2 The constant 21 is sharp in both inequalities. The second inequality also holds for s = 0 or s = 1. 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 the choices α = 0, β = 1, and γ = s. 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: Z 1 1 − λ (5f (c)) (b − a) ≤ f [(1 − t) a + tb] dt − f (c) . (2.19) 2 0 The following result related to the first Hermite-Hadamard inequality for functions defined on linear spaces also holds. Corollary 2.6. If f is as in Theorem 2.4, then a+b a+b 1 (2.20) 0 ≤ 5+ f (b − a) − 5− f (b − a) 8 2 2 Z 1 a+b ≤ f [(1 − t) a + tb] dt − f 2 0 1 ≤ [(5− f (b)) (b − a) − (5+ f (a)) (b − a)] . 8 An Inequality Improving the First 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 12 of 19 The constant 1 8 is sharp in both inequalities. Now, let Ω ⊂ Rn be an open and convex set in Rn . If F : Ω → R is a differentiable convex function on Ω, then, obviously, for any c̄ ∈ Ω we have n X ∂F (c̄) ∇F (c̄) (ȳ) = · yi , ∂xi i=1 ȳ ∈ Rn , ∂F where ∂x are the partial derivatives of F with respect to the variable xi i (i = 1, . . . , n) . Using (2.18), we may state that X n ∂F λā + (1 − λ) b̄ 1 (2.21) −λ · (bi − ai ) 2 ∂x i i=1 Z 1 ≤ F (1 − t) ā + tb̄ dt − F (1 − λ) ā + λb̄ 0 n n X X ∂F b̄ ∂F (ā) 2 ≤ (1 − λ) · (bi − ai ) − λ2 · (bi − ai ) ∂x ∂x i i i=1 i=1 for any ā, b̄ ∈ Ω and λ ∈ (0, 1) . In particular, for λ = 21 , we get Z 1 ā + b̄ (2.22) 0 ≤ F (1 − t) ā + tb̄ dt − F 2 0 ! n ∂F (ā) 1 X ∂F b̄ − · (bi − ai ) . ≤ 8 i=1 ∂xi ∂xi An Inequality Improving the First 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 13 of 19 In (2.22) the constant 1 8 is sharp. An Inequality Improving the First 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 14 of 19 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 ≤ k(1 − t) x + tyk2 dt − k(1 − σ) x + σyk2 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) (1 − 2σ) [y − x, (1 − σ) x + σy] Z 1 ≤ k(1 − t) x + tyk2 dt − k(1 − σ) x + σyk2 . 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 x + y 2 2 ≤ k(1 − t) x + tyk dt − 2 0 0≤ An Inequality Improving the First 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 15 of 19 ≤ 1 [hy − x, yii − hy − x, xis ] . 4 The inequality (3.3) provides a refinement and a counterpart for the first inequality (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 + σyiσ 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 ≤ k(1 − t) x + tyk dt − k(1 − σ) x + σyk 0 1 2 hy − x, yii 2 hy − x, xis (1 − σ) −σ . ≤ 2 kyk kxk The second inequality also holds for σ = 0 or σ = 1. We note that if the space is smooth, then we have (3.5) 1 [y − x, (1 − σ) x + σy] −σ · 2 k(1 − σ) x + σyk Z 1 ≤ k(1 − t) x + tyk dt − k(1 − σ) x + σyk , An Inequality Improving the First 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 0 Page 16 of 19 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 x + y ≤ k(1 − t) x + tyk dt − 2 0 y x 1 ≤ y − x, − y − x, . 8 kyk i kxk s The inequality (3.6) provides a refinement and a counterpart for the first 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 x + y 1 ≤ ky − xk2 . (3.7) 0≤ k(1 − t) x + tyk dt − 8 2 0 1 8 The constant is sharp. Indeed, if H = R, ha, bi = a · b, then taking x = −1, y = 1, we obtain 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 First 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 17 of 19 P∞ i=0 |xi |2 < ∞. Then, by (3.7), we have the inequalities Z (3.8) ∞ X 1 0≤ 0 ! 21 |(1 − t) xi + tyi |2 ∞ X xi + y i 2 2 dt − i=0 ! 12 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 First Hermite-Hadamard Inequality for Convex Functions Defined on Linear Spaces and Applications for Semi-Inner Products S.S. Dragomir 1 Z (3.9) Z 0≤ 0 Ω ! 12 Z f (t) + g (t) 2 dµ (t) 2 Ω − ≤ 1 · 8 21 2 |(1 − λ) f (t) + λg (t)| dµ (t) dλ Z Contents JJ J |f (t) − g (t)|2 dµ (t) Ω for any f, g ∈ L2 (Ω) provided Title Page II I Go Back 2 R Ω |f (t)| dµ (t) = 2 R Ω |g (t)| dµ (t) = 1. Close Quit Page 18 of 19 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 First 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 19 of 19