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An Inequality of Ostrowski Type Via Pompeiu's Mean

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An Inequality of Ostrowski Type Via Pompeiu's Mean AN INEQUALITY OF OSTROWSKI TYPE VIA POMPEIU’S MEAN VALUE THEOREM S.S. DRAGOMIR Abstract. An inequality providing some bounds for the integral mean via Pompeiu’s mean value theorem and applications for quadrature rules and spe- cial means are given. 1. Introduction The following result is known in the literature as Ostrowski’s inequality [1]. Theorem 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 1 Z b 1 x − a+b 2 (1.1) f (x) − f (t) dt ≤ + (b − a) M, b − a a 4 b − a 1 for all x ∈ [a, b] . The constant 4 is best possible in the sense that it cannot be replaced by a smaller constant. In [2], the author has proved the following Ostrowski type inequality. Theorem 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 0 1−p 0 Kp (f ) := sup u |f (u)| < ∞. u∈(a,b) Then we have the inequality 1 Z b K (f 0) p (1.2) f (x) − f (t) dt ≤ b − a a |p| (b − a) 2xp (x − A) + (b − x) Lp (b, x) − (x − a) Lp (x, a) if p ∈ (0, ∞); p p p p p × (x − a) Lp (x, a) − (b − x) Lp (b, x) − 2x (x − A) if p ∈ (−∞, −1) ∪ (−1, 0) −1 −1 2 (x − a) L (x, a) − (b − x) L (b, x) − x (x − A) if p = −1, for any x ∈ (a, b) , where for a 6= b, a + b A = A (a, b) := , is the arithmetic mean, 2 Date: February 27, 2003. 1991 Mathematics Subject Classification. Primary 26D15, 26D10; Secondary 41A55. Key words and phrases. Ostrowski’s inequality, Pompeiu mean value theorem, quadrature rules, Special means. 1 2 S.S. DRAGOMIR 1 bp+1 − ap+1 p L = L (a, b) = , is the p − logarithmic mean p ∈ \ {−1, 0} , p p (p + 1) (b − a) R and b − a L = L (a, b) := is the logarithmic mean. ln b − ln a Another result of this type obtained in the same paper is: Theorem 3. Let f :[a, b] → R be continuous on [a, b] (with a > 0) and differen- tiable on (a, b) . If P (f 0) := sup |uf 0 (x)| < ∞, u∈(a,b) then we have the inequality " " # # 1 Z b P (f 0) [I (x, b)]b−x (1.3) f (x) − f (t) dt ≤ ln + 2 (x − A) ln x x−a b − a a b − a [I (a, x)] for any x ∈ (a, b) , where for a 6= b 1 1 bb b−a I = I (a, b) := , is the identric mean. e aa If some local information around the point x ∈ (a, b) is available, then we may state the following result as well [2]. Theorem 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 n 1−p 0 o Mp (x) := sup |x − u| |f (u)| < ∞. u∈(a,b) Then we have the inequality 1 Z b (1.4) f (x) − f (t) dt b − a a 1 h i ≤ (x − a)p+1 + (b − x)p+1 M (x) . p (p + 1) (b − a) p 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. 2. Pompeiu’s Mean Value Theorem 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 5. 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 x f (x ) − x f (x ) (2.1) 1 2 2 1 = f (ξ) − ξf 0 (ξ) . x1 − x2 OSTROWSKI TYPE INEQUALITES 3 1 1 Proof. Define a real valued function F on the interval b , a by 1 (2.2) F (t) = tf . t 1 1 Since f is differentiable on b , a and 1 1 1 (2.3) F 0 (t) = f − f 0 , t t t 1 1 then applying the mean value theorem to F on the interval [x, y] ⊂ b , a we get F (x) − F (y) (2.4) = F 0 (η) x − y for some η ∈ (x, y) . 1 1 1 Let x2 = x , x1 = y and ξ = η . Then, since η ∈ (x, y) , we have x1 < ξ < x2. Now, using (2.2) and (2.3) on (2.4), we have 1 1 xf x − yf y 1 1 1 = f − f 0 , x − y η η η that is x f (x ) − x f (x ) 1 2 2 1 = f (ξ) − ξf 0 (ξ) . x1 − x2 This completes the proof of the theorem. Remark 1. Following [7, p. 84 – 85], we will mention here a geometrical inter- pretation of Pompeiu’s theorem. The equation of the secant line joining the points (x1, f (x1)) and (x2, f (x2)) is given by f (x2) − f (x1) y = f (x1) + (x − x1) . x2 − x1 This line intersects the y−axis at the point (0, y) , where y is f (x2) − f (x1) y = f (x1) + (0 − x1) x2 − x1 x f (x ) − x f (x ) = 1 2 2 1 . x1 − x2 The equation of the tangent line at the point (ξ, f (ξ)) is y = (x − ξ) f 0 (ξ) + f (ξ) . 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)) . 4 S.S. DRAGOMIR 3. Evaluating the Integral Mean The following result holds. Theorem 6. 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 a + b f (x) 1 Z b (3.1) · − f (t) dt 2 x b − a a !2 b − a 1 x − a+b ≤ + 2 kf − `f 0k , |x| 4 b − a ∞ where ` (t) = t, t ∈ [a, b] . 1 The constant 4 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] 0 = kf − `f k∞ |x − t| for any t, x ∈ [a, b] . Integrating over t ∈ [a, b] , we get Z b Z b Z b 0 (3.3) f (x) tdt − x f (t) dt ≤ kf − `f k∞ |x − t| dt a a a " # (x − a)2 + (b − x)2 = kf − `f 0k ∞ 2 " # 1 a + b2 = kf − `f 0k (b − a)2 + x − ∞ 4 2 R b b2−a2 and since a tdt = 2 , we deduce from (3.3) the desired result (3.1). Now, assume that (3.2) holds with a constant k > 0, i.e., a + b f (x) 1 Z b (3.4) · − f (t) dt 2 x b − a a !2 b − a x − a+b ≤ k + 2 kf − `f 0k , |x| b − a ∞ for any x ∈ [a, b] . OSTROWSKI TYPE INEQUALITES 5 Consider f :[a, b] → R, f (t) = αt + β; α, β 6= 0. Then 0 kf − `f k∞ = |β| , 1 Z b a + b f (t) dt = · α + β, b − a a 2 and by (3.4) we deduce 2 a+b ! a + b β a + b b − a x − 2 α + − α + β ≤ k + |β| 2 x 2 |x| b − a giving 2 a+b ! a + b x − 2 (3.5) − x ≤ (b − a) k + 2 b − a for any x ∈ [a, b] . 1 If in (3.5) we let x = a or x = b, we deduce k ≥ 4 , and the sharpness of the constant is proved. The following interesting particular case holds. Corollary 1. With the assumptions in Theorem 6, we have a + b 1 Z b (b − a) 0 (3.6) f − f (t) dt ≤ kf − `f k∞ . 2 b − a a 2 |a + b| 4. The Case of Weighted Integrals We will consider now the weighted integral case. Theorem 7. 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: Z b f (x) Z b (4.1) f (t) w (t) dt − tw (t) dt a x a " Z x Z b ! 0 ≤ kf − `f k∞ sgn (x) w (t) dt − w (t) dt a x !# 1 Z b Z x + tw (t) dt − tw (t) dt |x| x a 6 S.S.
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