Applied Mathematics Letters 22 (2009) 1345–1350 Contents lists available at ScienceDirect Applied Mathematics Letters journal homepage: www.elsevier.com/locate/aml New inequalities of Ostrowski-like type involving n knots and the Lp -norm of the m-th derivative Vu Nhat Huy a , Quô´c-Anh Ngô a,b,∗ a Department of Mathematics, College of Science, Viêt Nam National University, Hà Nôi, Viêt Nam b Department of Mathematics, National University of Singapore, Science Drive 2, Singapore 117543, Singapore article abstract info Article history: Received 23 April 2008 Received in revised form March 2009 Accepted 30 March 2009 On the basis of recent results due to Nenad Ujević, we obtain some new inequalities of Ostrowski-like type involving n knots and the Lp -norm of the m-th derivative where n, m, p are arbitrary numbers © 2009 Elsevier Ltd All rights reserved Keywords: Inequality Error Integral Taylor Ostrowski Introduction In recent years, a number of authors have considered error inequalities for some known and some new quadrature formulas Sometimes they have considered generalizations of these formulas; see [1,2] and the references therein where the mid-point and trapezoid quadrature rules are considered This work is motivated by some results due to Nenad Ujević Here we recall them Theorem (See [1]) Let I ⊂ R be an open interval such that [a, b] ⊂ I and let f : I → R be a twice-differentiable function such that f is bounded and integrable Then we have b f (x)dx − b−a a +f a+b f a+b √ − 2− ( b − a) √ √ + 2− 7−4 3 (b − a) f ∞ ( b − a) (1) Theorem (See [2]) Let I ⊂ R be an open interval such that [a, b] ⊂ I and let f : I → R be a twice-differentiable function such that f ∈ L2 (a, b) Then we have √ b f (x)dx − a ∗ b−a f a+b − 3− (b − a) Corresponding author at: Department of Mathematics, College of Science, Viêt Nam National University, Hà Nôi, Viêt Nam E-mail addresses: nhat_huy85@yahoo.com (V.N Huy), bookworm_vn@yahoo.com (Q.-A Ngơ) 0893-9659/$ – see front matter © 2009 Elsevier Ltd All rights reserved doi:10.1016/j.aml.2009.03.002 1346 V.N Huy, Q.-A Ngô / Applied Mathematics Letters 22 (2009) 1345–1350 √ a+b +f + 3− 49 (b − a) 80 1√ f − ( b − a) √ In the above mentioned results, constants 7−48 49 80 in (1) and − (2) √ in (2) are sharp in the sense that these cannot be replaced by smaller ones This leads us to strengthen (1) and (2) by enlarging the number of knots (two knots in both (1) and (2)) and replacing the norms · ∞ in (1) and · in (2) Before stating our main result, let us introduce the following notation b f (x)dx I (f ) = a Let m, n < ∞ and p ∞ For each i = 1, n, we assume < xi < such that n x1 + x2 + · · · + xn = , · · · n j j x1 + x2 + · · · + xnj = , j + · · · xm−1 + xm−1 + · · · + xm−1 = n n m Put Q (f , n, m, x1 , , xn ) = n b−a f (a + xi (b − a)) n i=1 Remark With the above notation, (1) reads as follows: I (f ) − Q f , 2, 2, √ , − 2− √ √ + 2− 7−4 3 f ∞ (b − a)3 (3) while (2) reads as follows: √ I (f ) − Q f , 2, 2, − 3− √ , + 3− 49 80 − 1√ f ( b − a) (4) We are in a position to state our main result Theorem Let I ⊂ R be an open interval such that [a, b] ⊂ I and let f : I → R be an m-times-differentiable function such that f (m) ∈ Lp (a, b) Then we have |I (f ) − Q (f , n, m, x1 , , xn )| 1 m! mq + 1 q + (m − 1) q + 1 q f (m) p (b − a)m+ q (5) where p + q =1 and f r := b |f (x)| dx a ess sup |f |, r [a,b] r , when ≤ r < ∞, when r = ∞ Remark It is worth noticing that the right hand side of (5) does not involve xi , i = 1, n, and that m can be chosen arbitrarily This means that our inequality (5) is better in some sense However, the inequality (5) is not sharp due to the restriction of the technique that we use We hope that we will soon see some responses on this problem V.N Huy, Q.-A Ngô / Applied Mathematics Letters 22 (2009) 1345–1350 1347 Proofs Before proving our main theorem, we need an essential lemma below It is well-known in the literature as Taylor’s formula or Taylor’s theorem with the integral remainder Lemma (See [3]) Let f : [a, b] → R and let r be a positive integer If f is such that f (r −1) is absolutely continuous on [a, b], x0 ∈ (a, b), then for all x ∈ (a, b) we have f (x) = Tr −1 (f , x0 , x) + Rr −1 (f , x0 , x) where Tr −1 (f , x0 , ·) is Taylor’s polynomial of degree r − 1, that is, r −1 Tr −1 (f , x0 , x) = f (k) (x0 ) (x − x0 )k k! k=0 and the remainder can be given by x Rr −1 (f , x0 , x) = x0 (x − t )r −1 f (r ) (t ) dt (r − 1)! (6) By a simple calculation, the remainder in (6) can be rewritten as x −x Rr −1 (f , x0 , x) = (x − x0 − t )r −1 f (r ) (x0 + t ) dt (r − 1)! which helps us to deduce a similar representation of f as follows: r −1 f ( x + u) = uk (k) f (x) + k! k=0 u (u − t )r −1 (r ) f (x + t ) dt (r − 1)! (7) Proof of Theorem Define x F (x) = f (x)dx a By the Fundamental Theorem of Calculus I (f ) = F (b) − F (a) Applying Lemma to F (x) with x = a and u = b − a, we get m F (b) = F (a) + k=1 (b − a)k (k) F ( a) + k! b−a (b − a − t )m (m+1) F (a + t ) dt m! which yields (b − a)k (k) F (a) + k! m I (f ) = k=1 b−a (b − a − t )m (m+1) F (a + t ) dt m! Equivalently, m−1 I (f ) = k=0 For each i (b − a)k+1 (k) f (a) + (k + 1)! b−a (b − a − t )m (m) f (a + t ) dt m! n, applying Lemma to f (x) with x = a and u = xi (b − a), we get xki (b − a)k (k) f (a) + k! k=0 xi (b−a) m−1 f (a + xi (b − a)) = xki (b − a)k (k) f (a) + k! k=0 m−1 = b−a (xi (b − a) − t )m−1 (m) f (a + t ) dt (m − 1)! m−1 xm i ( b − a − u) f (m) (a + xi u) du (m − 1)! (8) 1348 V.N Huy, Q.-A Ngô / Applied Mathematics Letters 22 (2009) 1345–1350 By applying (8) to i = 1, n and then summing, we deduce that xki (b − a)k (k) f (a) + k! i=1 k=0 i=1 m−1 n n f (a + xi (b − a)) = i=1 n n xki (b − a)k m−1 i=1 = k! k=0 n f (k) (a) + n (b − a)k (k) f (a) + (k + 1)! k=0 i =1 n b−a m−1 xm i (b − a − u) f (m) (a + xi u) du (m − 1)! m−1 xm i ( b − a − u) f (m) (a + xi u) du (m − 1)! b −a i =1 m−1 = b−a m−1 xm i ( b − a − u) f (m) (a + xi u) du (m − 1)! (9) Thus, m−1 Q (f , n, m, x1 , , xn ) = k=0 b−a (b − a)k+1 (k) f ( a) + n (k + 1)! n b−a i =1 m−1 xm i ( b − a − u) f (m) (a + xi u) du (m − 1)! Therefore, |I (f ) − Q (f , n, m, x1 , , xn )| b−a b−a (b − a − t )m (m) f (a + t ) dt − m! n b−a b−a (b − a − t ) (m) f (a + t ) dt + m! n = 0 b = a n b−a (b − x) (m) f (x)dx + m! n n (b − ·) (m) f m! m + n n a b −a i=1 xm i m−1 xm i ( b − a − u) f (m) (a + xi u) du (m − 1)! m−1 xm i (b − a − u) f (m) (a + xi u) du (m − 1)! (b − x)m−1 (m) f ((1 − xi ) a + xi x) dx (m − 1)! (b − ·)m−1 (m) f ((1 − xi ) a + xi ·) (m − 1)! n b−a i=1 b i =1 m m b−a xm i i=1 Using the Hölder inequality, we get (b − ·)m (m) f m! = 1 (b − ·) f m m! We see that f (m) ((1 − xi ) a + xi ·) ∞ f (m) p = f m! (m) (b − ·) p f (m) ∞ while, for b ((1 − xi ) a + xi ·) (m) f (m) m q = m! (b − a)mq+1 mq + p < ∞, we have p ((1 − xi ) a + xi x) dx p a = p xi = b f p p = 1 p f (m) (t ) a b f (m) a f (m) p xi xi ((1 − xi ) a + xi x) d ((1 − xi ) a + xi x) (1−xi )a+xi b xi p a xi (m) f (m) p p (t ) dt p p p dt p q f (m) p (10) V.N Huy, Q.-A Ngô / Applied Mathematics Letters 22 (2009) 1345–1350 1349 This helps us to deduce that n m−1 xm i (b − ·) f (m) ((1 − xi )a + xi ·) (m − 1)! n b−a i=1 = n b−a n n n n i =1 n xm i f (m) (m − 1)! xi −1 xm i i =1 f (m) f (m) n b−a f (m) ((1 − xi ) a + xi ·) (m − 1)! n b−a = xm i i =1 (m − 1)! q m! q q q (b − a)(m−1)q+1 (m − 1) q + p (b − ·)(m−1) p (b − a)(m−1)q+1 (m − 1) q + p (b − a)mq+1 (m − 1) q + p (b − ·)m−1 f (m) ((1 − xi ) a + xi ·) (m − 1)! i =1 b−a = xm i It follows that q (b − a)mq+1 mq + 1 |I (f ) − Q (f , n, m, x1 , , xn )| m! f f (m) (m) p + p m! (b − a)mq+1 (m − 1) q + 1 q which completes our proof Examples In this section, by applying our main theorem, we will obtain some new inequalities which cannot be easily obtained from [1,2] Example When n = 3, m = and √ 0< 21 − 12 √ < x1 < + x2 21 12