Journal of Inequalities in Pure and Applied Mathematics http://jipam.vu.edu.au/ Volume 6, Issue 3, Article 83, 2005 AN INEQUALITY OF OSTROWSKI TYPE VIA POMPEIU’S MEAN VALUE THEOREM S.S. DRAGOMIR S CHOOL OF C OMPUTER S CIENCE AND M ATHEMATICS V ICTORIA U NIVERSITY PO B OX 14428, MCMC 8001 VIC, AUSTRALIA . sever@csm.vu.edu.au URL: http://rgmia.vu.edu.au/dragomir Received 08 February, 2005; accepted 12 June, 2005 Communicated by B.G. Pachpatte A BSTRACT. An inequality providing some bounds for the integral mean via Pompeiu’s mean value theorem and applications for quadrature rules and special means are given. Key words and phrases: Ostrowski’s inequality, Pompeiu mean value theorem, quadrature rules, Special means. 2000 Mathematics Subject Classification. Primary 26D15, 26D10; Secondary 41A55. 1. I NTRODUCTION The following result is known in the literature as Ostrowski’s inequality [1]. Theorem 1.1. Let f : [a, b] → R be a differentiable mapping on (a, b) with the property that |f 0 (t)| ≤ M for all t ∈ (a, b) . Then !2 Z b a+b x− 2 1 f (x) − 1 (b − a) M, (1.1) f (t) dt ≤ + b−a a 4 b−a for all x ∈ [a, b] . The constant smaller constant. 1 4 is best possible in the sense that it cannot be replaced by a In [2], the author has proved the following Ostrowski type inequality. Theorem 1.2. Let f : [a, b] → R be continuous on [a, b] with a > 0 and differentiable on (a, b) . Let p ∈ R\ {0} and assume that Kp (f 0 ) := sup u1−p |f 0 (u)| < ∞. u∈(a,b) ISSN (electronic): 1443-5756 c 2005 Victoria University. All rights reserved. 032-05 2 S.S. D RAGOMIR Then we have the inequality Z b 1 Kp (f 0 ) (1.2) f (x) − f (t) dt ≤ b−a a |p| (b − a) p p 2x (x − A) + (b − x) Lp (b, x) − (x − a) Lpp (x, a) if p ∈ (0, ∞) ; (x − a) Lpp (x, a) − (b − x) Lpp (b, x) − 2xp (x − A) if p ∈ (−∞, −1) ∪ (−1, 0) × (x − a) L−1 (x, a) − (b − x) L−1 (b, x) − 2 (x − A) if p = −1, x for any x ∈ (a, b) , where for a 6= b, A = A (a, b) := bp+1 − ap+1 Lp = Lp (a, b) = (p + 1) (b − a) a+b , is the arithmetic mean, 2 p1 , is the p − logarithmic mean p ∈ R\ {−1, 0} , and b−a is the logarithmic mean. ln b − ln a Another result of this type obtained in the same paper is: L = L (a, b) := Theorem 1.3. Let f : [a, b] → R be continuous on [a, b] (with a > 0) and differentiable on (a, b) . If P (f 0 ) := sup |uf 0 (x)| < ∞, u∈(a,b) then we have the inequality " " # # Z b b−x P (f 0 ) 1 [I (x, b)] f (x) − f (t) dt ≤ ln + 2 (x − A) ln x (1.3) b−a a b−a [I (a, x)]x−a for any x ∈ (a, b) , where for a 6= b 1 I = I (a, b) := e bb aa 1 b−a , is the identric mean. If some local information around the point x ∈ (a, b) is available, then we may state the following result as well [2]. Theorem 1.4. Let f : [a, b] → R be continuous on [a, b] and differentiable on (a, b) . Let p ∈ (0, ∞) and assume, for a given x ∈ (a, b) , we have that Mp (x) := sup |x − u|1−p |f 0 (u)| < ∞. u∈(a,b) Then we have the inequality Z b 1 1 (1.4) f (x) − f (t) dt ≤ (x − a)p+1 + (b − x)p+1 Mp (x) . b−a a p (p + 1) (b − a) For recent results in connection to Ostrowski’s inequality see the papers [3],[4] and the monograph [5]. The main aim of this paper is to provide some complementary results, where instead of using Cauchy mean value theorem, we use Pompeiu mean Value Theorem to evaluate the integral mean of an absolutely continuous function. Applications for quadrature rules and particular instances of functions are given as well. J. Inequal. Pure and Appl. Math., 6(3) Art. 83, 2005 http://jipam.vu.edu.au/ O STROWSKI T YPE I NEQUALITES 3 2. P OMPEIU ’ S M EAN VALUE T HEOREM In 1946, Pompeiu [6] derived a variant of Lagrange’s mean value theorem, now known as Pompeiu’s mean value theorem (see also [7, p. 83]). Theorem 2.1. For every real valued function f differentiable on an interval [a, b] not containing 0 and for all pairs x1 6= x2 in [a, b] , there exists a point ξ in (x1 , x2 ) such that x1 f (x2 ) − x2 f (x1 ) = f (ξ) − ξf 0 (ξ) . x1 − x2 Proof. Define a real valued function F on the interval 1b , a1 by 1 (2.2) F (t) = tf . t Since f is differentiable on 1b , a1 and 1 1 0 1 0 (2.3) F (t) = f − f , t t t (2.1) then applying the mean value theorem to F on the interval [x, y] ⊂ 1 1 , we get b a F (x) − F (y) = F 0 (η) x−y (2.4) for some η ∈ (x, y) . Let x2 = x1 , x1 = y1 and ξ = η1 . Then, since η ∈ (x, y) , we have x1 < ξ < x 2 . Now, using (2.2) and (2.3) on (2.4), we have 1 xf x − yf y1 x−y 1 1 0 1 =f − f , η η η that is x1 f (x2 ) − x2 f (x1 ) = f (ξ) − ξf 0 (ξ) . x1 − x2 This completes the proof of the theorem. Remark 2.2. Following [7, p. 84 – 85], we will mention here a geometrical interpretation of Pompeiu’s theorem. The equation of the secant line joining the points (x1 , f (x1 )) and (x2 , f (x2 )) is given by y = f (x1 ) + f (x2 ) − f (x1 ) (x − x1 ) . x2 − x1 This line intersects the y−axis at the point (0, y) , where y is f (x2 ) − f (x1 ) (0 − x1 ) x2 − x1 x1 f (x2 ) − x2 f (x1 ) = . x1 − x2 y = f (x1 ) + The equation of the tangent line at the point (ξ, f (ξ)) is y = (x − ξ) f 0 (ξ) + f (ξ) . J. Inequal. Pure and Appl. Math., 6(3) Art. 83, 2005 http://jipam.vu.edu.au/ 4 S.S. D RAGOMIR The tangent line intersects the y−axis at the point (0, y) , where y = −ξf 0 (ξ) + f (ξ) . Hence, the geometric meaning of Pompeiu’s mean value theorem is that the tangent of the point (ξ, f (ξ)) intersects on the y−axis at the same point as the secant line connecting the points (x1 , f (x1 )) and (x2 , f (x2 )) . 3. E VALUATING THE I NTEGRAL M EAN The following result holds. Theorem 3.1. Let f : [a, b] → R be continuous on [a, b] and differentiable on (a, b) with [a, b] not containing 0. Then for any x ∈ [a, b] , we have the inequality Z b a + b f (x) 1 (3.1) · − f (t) dt 2 x b−a a !2 a+b x− 2 b − a 1 kf − `f 0 k , ≤ + ∞ |x| 4 b−a where ` (t) = t, t ∈ [a, b] . The constant 14 is sharp in the sense that it cannot be replaced by a smaller constant. Proof. Applying Pompeiu’s mean value theorem, for any x, t ∈ [a, b] , there is a ξ between x and t such that tf (x) − xf (t) = [f (ξ) − ξf 0 (ξ)] (t − x) giving (3.2) |tf (x) − xf (t)| ≤ sup |f (ξ) − ξf 0 (ξ)| |x − t| ξ∈[a,b] = kf − `f 0 k∞ |x − t| for any t, x ∈ [a, b] . Integrating over t ∈ [a, b] , we get Z b Z b Z b 0 f (x) (3.3) tdt − x f (t) dt ≤ kf − `f k∞ |x − t| dt a a a " # (x − a)2 + (b − x)2 0 = kf − `f k∞ 2 " 2 # 1 a+b 2 0 = kf − `f k∞ (b − a) + x − 4 2 Rb 2 2 and since a tdt = b −a , we deduce from (3.3) the desired result (3.1). 2 Now, assume that (3.2) holds with a constant k > 0, i.e., Z b a + b f (x) 1 (3.4) f (t) dt 2 · x − b−a a !2 a+b x− 2 b−a kf − `f 0 k , ≤ k+ ∞ |x| b−a for any x ∈ [a, b] . J. Inequal. Pure and Appl. Math., 6(3) Art. 83, 2005 http://jipam.vu.edu.au/ O STROWSKI T YPE I NEQUALITES 5 Consider f : [a, b] → R, f (t) = αt + β; α, β 6= 0. Then kf − `f 0 k∞ = |β| , Z b 1 a+b f (t) dt = · α + β, b−a a 2 and by (3.4) we deduce a + b b−a a+b β k + α+ − α + β ≤ 2 x 2 |x| a+b 2 x− b−a !2 |β| giving (3.5) a + b ≤ (b − a) k + − x 2 x − a+b 2 b−a !2 for any x ∈ [a, b] . If in (3.5) we let x = a or x = b, we deduce k ≥ 41 , and the sharpness of the constant is proved. The following interesting particular case holds. Corollary 3.2. With the assumptions in Theorem 3.1, we have (3.6) Z b ≤ (b − a) kf − `f 0 k . f a + b − 1 f (t) dt ∞ 2 |a + b| 2 b−a a 4. T HE C ASE OF W EIGHTED I NTEGRALS We will consider now the weighted integral case. Theorem 4.1. Let f : [a, b] → R be continuous on [a, b] and differentiable on (a, b) with [a, b] not containing 0. If w : [a, b] → R is nonnegative integrable on [a, b] , then for each x ∈ [a, b] , we have the inequality: (4.1) Z b Z b f (x) f (t) w (t) dt − tw (t) dt x a a Z ≤ kf − `f 0 k∞ sgn (x) a J. Inequal. Pure and Appl. Math., 6(3) Art. 83, 2005 x Z b w (t) dt w (t) dt − x Z b Z x 1 + tw (t) dt − tw (t) dt . |x| x a http://jipam.vu.edu.au/ 6 S.S. D RAGOMIR Proof. Using the inequality (3.2), we have Z b Z b (4.2) f (x) tw (t) dt − x f (t) w (t) dt a a ≤ kf − `f 0 k∞ Z b w (t) |x − t| dt a 0 Z = kf − `f k∞ x Z b w (t) (x − t) dt + a w (t) (t − x) dt x Z x Z x Z b Z b 0 = kf − `f k∞ x w (t) dt − tw (t) dt + tw (t) dt − x w (t) dt a a x x Z x Z b Z b Z x 0 = kf − `f k∞ x w (t) dt − w (t) dt + tw (t) dt − tw (t) dt a x x a from where we get the desired inequality (4.1). Now, if we assume that 0 < a < b, then Rb a ≤ Rab (4.3) a tw (t) dt ≤ b, w (t) dt Rb provided a w (t) dt > 0. With this extra hypothesis, we may state the following corollary. Corollary 4.2. With the above assumptions, we have ! Rb Z b tw (t) dt 1 a − Rb f (t) w (t) dt (4.4) f R b w (t) dt w (t) dt a a a # "R x Rb Rx Rb w (t) tdt − tw (t) dt w (t) dt − w (t) dt x a ≤ kf − `f 0 k∞ a + x . Rb Rb w (t) dt tw (t) dt a a 5. A Q UADRATURE F ORMULA We assume in the following that 0 < a < b. Consider the division of the interval [a, b] given by In : a = x0 < x1 < · · · < xn−1 < xn = b, and ξi ∈ [xi , xi+1 ], i = 0, . . . , n − 1 a sequence of intermediate points. Define the quadrature (5.1) n−1 X f (ξi ) x2i+1 − x2i Sn (f, In , ξ) := · ξi 2 i=0 n−1 X f (ξi ) xi + xi+1 = · · hi , ξi 2 i=0 where hi := xi+1 − xi , i = 0, . . . , n − 1. R bThe following result concerning the estimate of the remainder in approximating the integral f (t) dt by the use of Sn (f, In , ξ) holds. a J. Inequal. Pure and Appl. Math., 6(3) Art. 83, 2005 http://jipam.vu.edu.au/ O STROWSKI T YPE I NEQUALITES 7 Theorem 5.1. Assume that f : [a, b] → R is continuous on [a, b] and differentiable on (a, b) . Then we have the representation b Z f (t) dt = Sn (f, In , ξ) + Rn (f, In , ξ) , (5.2) a where Sn (f, In , ξ) is as defined in (5.1), and the remainder Rn (f, In , ξ) satisfies the estimate n−1 2 xi +xi+1 2 X hi 1 ξi − 2 |Rn (f, In , ξ)| ≤ kf − `f 0 k∞ (5.3) + ξi 4 hi i=0 n−1 2 n−1 X kf − `f 0 k∞ X 2 hi 1 0 ≤ kf − `f k∞ ≤ hi . 2 ξ 2a i=0 i i=0 Proof. Apply Theorem 3.1 on the interval [xi , xi+1 ] for the intermediate points ξi to obtain Z xi+1 f (ξ ) x + x i i i+1 (5.4) f (t) dt − · · h i ξi 2 xi !2 xi +xi+1 ξi − 2 1 1 kf − `f 0 k ≤ h2i + ∞ ξi 4 hi ≤ 1 2 1 hi kf − `f 0 k∞ ≤ h2i kf − `f 0 k∞ 2ξi 2a for each i ∈ {0, . . . , n − 1} . Summing over i from 1 to n − 1 and using the generalised triangle inequality, we deduce the desired estimate (5.3). Now, if we consider the mid-point rule (i.e., we choose ξi = xi +xi+1 2 above, i ∈ {0, . . . , n − 1}) n−1 X xi + xi+1 Mn (f, In ) := f hi , 2 i=0 then, by Corollary 3.2, we may state the following result as well. Corollary 5.2. With the assumptions of Theorem 5.1, we have Z b f (t) dt = Mn (f, In ) + Rn (f, In ) , (5.5) a where the remainder satisfies the estimate: (5.6) n−1 kf − `f 0 k∞ X h2i |Rn (f, In )| ≤ 2 x + xi+1 i=0 i n−1 kf − `f 0 k∞ X 2 ≤ hi . 4a i=0 J. Inequal. Pure and Appl. Math., 6(3) Art. 83, 2005 http://jipam.vu.edu.au/ 8 S.S. D RAGOMIR 6. A PPLICATIONS FOR S PECIAL M EANS For 0 < a < b, let us consider the means a+b , √2 G = G (a, b) := a · b, 2 H = H (a, b) := 1 1 , +b a b−a L = L (a, b) := (logarithmic mean), ln b − ln a 1 1 bb b−a I = I (a, b) := (identric mean) e aa A = A (a, b) := and the p−logarithmic mean bp+1 − ap+1 Lp = Lp (a, b) = (p + 1) (b − a) p1 , p ∈ R\ {−1, 0} . It is well known that H ≤ G ≤ L ≤ I ≤ A, and, denoting L0 := I and L−1 = L, the function R 3 p 7→ Lp ∈ R is monotonic increasing. In the following we will use the following inequality obtained in Corollary 3.2, (6.1) Z b (b − a) f a + b − 1 f (t) dt ≤ kf − `f 0 k∞ , 2 b−a a 2 (a + b) provided 0 < a < b. (1) Consider the function f : [a, b] ⊂ (0, ∞) → R, f (t) = tp , p ∈ R\ {−1, 0} . Then f 1 b−a Z a+b 2 = [A (a, b)]p , b f (t) dt = Lpp (a, b) , a (1 − p) ap if p ∈ (−∞, 0) \ {−1} , 0 kf − `f k[a,b],∞ = |1 − p| bp if p ∈ (0, 1) ∪ (1, ∞) . (6.2) Consequently, by (6.1) we deduce (1 − p) ap (b − a) if p ∈ (−∞, 0) \ {−1} , A (a, b) p 1 A (a, b) − Lpp (a, b) ≤ × 4 |1 − p| bp (b − a) if p ∈ (0, 1) ∪ (1, ∞) . A (a, b) J. Inequal. Pure and Appl. Math., 6(3) Art. 83, 2005 http://jipam.vu.edu.au/ O STROWSKI T YPE I NEQUALITES 9 (2) Consider the function f : [a, b] ⊂ (0, ∞) → R, f (t) = 1t . Then a+b 1 f = , 2 A (a, b) Z b 1 1 f (t) dt = , b−a a L (a, b) 2 kf − `f 0 k[a,b],∞ = . a Consequently, by (6.1) we deduce b−a (6.3) 0 ≤ A (a, b) − Lp (a, b) ≤ L (a, b) . 2a (3) Consider the function f : [a, b] ⊂ (0, ∞) → R, f (t) = ln t. Then a+b = ln [A (a, b)] , f 2 Z b 1 f (t) dt = ln [I (a, b)] , b−a a a b 0 kf − `f k[a,b],∞ = max ln . , ln e e Consequently, by (6.1) we deduce A (a, b) b−a a b (6.4) 1≤ ≤ exp max ln . , ln I (a, b) 4A (a, b) e e R EFERENCES [1] A. OSTROWSKI, Über die Absolutabweichung einer differentienbaren Funktionen von ihren Integralmittelwert, Comment. Math. Hel, 10 (1938), 226–227. [2] S.S. DRAGOMIR, Some new inequalities of Ostrowski type, RGMIA Res. Rep. Coll., 5(2002), Supplement, Article 11, [ON LINE:http://rgmia.vu.edu.au/v5(E).html]. Gazette, Austral. Math. Soc., 30(1) (2003), 25–30. [3] G.A. ANASTASSIOU, Multidimensional Ostrowski inequalities, revisited. Acta Math. Hungar., 97(4) (2002), 339–353. [4] G.A. ANASTASSIOU, Univariate Ostrowski inequalities, revisited. Monatsh. Math., 135(3) (2002), 175–189. [5] S.S. DRAGOMIR AND Th.M. RASSIAS (Eds), Ostrowski Type Inequalities and Applications in Numerical Integration, Kluwer Academic Publishers, Dordrecht/Boston/London, 2002. [6] D. POMPEIU, Sur une proposition analogue au théorème des accroissements finis, Mathematica (Cluj, Romania), 22 (1946), 143–146. [7] P.K. SAHOO AND T. RIEDEL, Mean Value Theorems and Functional Equations, World Scientific, Singapore, New Jersey, London, Hong Kong, 2000. J. Inequal. Pure and Appl. Math., 6(3) Art. 83, 2005 http://jipam.vu.edu.au/