View metadata, citation and similar papers at core.ac.uk brought to you by CORE provided by Victoria University Eprints Repository A PERTURBED TRAPEZOID INEQUALITY IN TERMS OF THE THIRD DERIVATIVE AND APPLICATIONS N.S. BARNETT AND S.S. DRAGOMIR Abstract. Some error estimates in terms of the p−norms of the third deriva- tive for the remainder in a perturbed trapezoid formula are given. Applications to composite quadrature formulae, for the expectation of a random variable and for Hermite-Hadamard divergence in Information Theory are pointed out. 1. Introduction In [4], by the use of Gr¨uss’ integral inequality, the authors have obtained the following perturbed trapezoid inequality. Theorem 1. Let f :[a, b] → R be a twice differentiable function on (a, b) and asume that γ := inf f 00 (x) > −∞ and Γ := sup f 00 (x) < ∞. x∈(a,b) x∈(a,b) Then we have the inequality 1 Z b b − a (b − a)2 0 0 (1.1) f (x) dx − [f (a) + f (b)] + [f (b) − f (a)] b − a a 2 12 (b − a)3 ≤ (Γ − γ) . 12 Using a finer argument based on a pre-Gr¨uss inequality, Cerone and Dragomir [2, p. 121] improved the above result as follows. Theorem 2. Let f have the properties of Theorem 1. Then 1 Z b b − a (b − a)2 0 0 (1.2) f (x) dx − [f (a) + f (b)] + [f (b) − f (a)] b − a a 2 12 1 ≤ √ (b − a)3 (Γ − γ) . 24 5 The main aim of the present work is to obtain some bounds for the left part of (1.2) in terms of the p−norms of f 000 assuming that the function f is twice differentiable on (a, b) and that the second derivative is absolutely continuous on (a, b). A number of applications are also pointed out. Date: January 11, 2001. 1991 Mathematics Subject Classification. Primary 26D15; Secondary 26D10, 41A55. Key words and phrases. Trapezoid inequality. 1 2 N.S. BARNETT AND S.S. DRAGOMIR 2. A Perturbed Trapezoid Formula The following representation lemma holds. Lemma 1. Let f :[a, b] → R be such that the second derivative is absolutely continuous on [a, b]. Then we have the equality: Z b f (a) + f (b) (b − a)2 f (t) dt − (b − a) + [f 0 (b) − f 0 (a)](2.1) a 2 12 1 Z b Z b Z t t + s = u − f 000 (u) du (t − s) dtds. 4 (b − a) a a s 2 Proof. Integrating by parts, we have Z b Z b Z t t + s I : = u − f 000 (u) du (t − s) dtds a a s 2 Z b Z b f 00 (t) + f 00 (s) Z t = (t − s) − f 00 (u) du (t − s) dtds a a 2 s " # Z b Z b f 00 (t)(t − s)2 + f 00 (s)(t − s)2 = − (f 0 (t) − f 0 (s)) (t − s) dtds a a 2 " # 1 Z b Z b Z b Z b = f 00 (t)(t − s)2 dtds + f 00 (s)(t − s)2 dtds 2 a a a a Z b Z b − (f 0 (t) − f 0 (s)) (t − s) dtds. a a By symmetry, Z b Z b Z b Z b J := f 00 (t)(t − s)2 dtds = f 00 (s)(t − s)2 dtds, a a a a and using Korkine’s identity or direct computation, we have Z b Z b K : = (f 0 (t) − f 0 (s)) (t − s) dtds a a " Z b Z b Z b # = 2 (b − a) f 0 (t) tdt − f 0 (t) dt · tdt . a a a Then, I = J − K. Since Z b Z b ! J = f 00 (t) (t − s)2 ds dt a a " # 1 Z b Z b = f 00 (t)(b − t)3 dt + (t − a)3 f 00 (t) dt 3 a a PERTURBED TRAPEZOID INEQUALITY 3 " 1 b Z b 0 3 2 0 = f (t)(b − t) + 3 (b − t) f (t) dt 3 a a # b Z b 0 3 2 0 + f (t)(t − a) − 3 (t − a) f (t) dt a a " " # 1 b Z b 0 3 2 = −f (a)(b − a) + 3 f (t)(b − t) − 2 (b − t) f (t) dt 3 a a " ## b Z b 0 3 2 +f (b)(b − a) − 3 f (t)(t − a) − 2 (t − a) f (t) dt a a " " # 1 Z b = [f 0 (b) − f 0 (a)] (b − a)3 + 3 −f (a)(b − a)2 + 2 (b − t) f (t) dt 3 a " Z b ## − 3 f (b)(b − a)2 − 2 (t − a) f (t) dt a 1 f (a) + f (b) Z b = [f 0 (b) − f 0 (a)] (b − a)3 − (b − a)2 + 2 (b − a) f (t) dt 3 2 a and " " b Z b # 2 2 # b − a K = 2 (b − a) f (t) t − f (t) dt − [f (b) − f (a)] a a 2 " " # # Z b a + b = 2 (b − a) f (b) b − f (a) a − f (t) dt − (b − a)[f (b) − f (a)] a 2 " # Z b a + b a + b = 2 (b − a) f (b) b − f (a) a − f (t) dt − f (b) + f (a) a 2 2 Z b = (b − a)2 [f (a) + f (b)] − 2 (b − a) f (t) dt, a then 1 Z b I = [f 0 (b) − f 0 (a)] (b − a)3 − 2 [f (a) + f (b)] (b − a)2 + 4 (b − a) f (t) dt 3 a Dividing by 4 (b − a) we deduce the desired equality (2.1). The following perturbed version of the trapezoid inequality holds. 4 N.S. BARNETT AND S.S. DRAGOMIR Theorem 3. Let f :[a, b] → R be such that the second derivative is absolutely continuous on [a, b]. Then we have Z b f (a) + f (b) (b − a)2 0 0 (2.2) f (t) dt − (b − a) + [f (b) − f (a)] a 2 12 1 R b R b |t − s|3 kf 000k dtds if f 000 ∈ L [a, b]; 16(b−a) a a [t,s],∞ ∞ b b 1 1 R R 2+ q 000 000 ≤ 1 |t − s| kf k[t,s],p dtds if f ∈ Lp [a, b] , q a a 8(q+1) (b−a) p > 1, 1 + 1 = 1; p q 1 R b R b 2 000 8(b−a) a a (t − s) kf k[t,s],1 dtds 4 (b−a) 000 000 160 kf k[a,b],∞ if f ∈ L∞ [a, b]; 1 2 3+ q (b−a) q 000 000 ≤ 1 kf k[a,b],p if f ∈ Lp [a, b] , 4(3q+1)(4q+1)(q+1) q 1 1 p > 1, p + q = 1; 3 (b−a) 000 48 kf k[a,b],1 , 1 l where khk := R d |h (u)|l du if l ≥ 1 and khk = ess sup |h (u)| . [c,d],l c [c,d],∞ u∈[c,d] (u∈[d,c]) Proof. Denote 1 Z b Z b Z t t + s R (f; a, b) := u − f 000 (u) du (t − s) dtds. 4 (b − a) a a s 2 As Z t Z t t + s 000 000 t + s u − f (u) du ≤ kf k[t,s],∞ u − du s 2 s 2 (t − s)2 = kf 000k , 4 [t,s],∞ Z t t + s 000 u − f (u) du s 2 1 Z t q q 000 t + s ≤ kf k[t,s],p u − du s 2 1 1+ q |t − s| 000 1 1 = 1 kf k[t,s],q , p > 1, + = 1 2 (q + 1) q p q and Z t t + s 000 t + s 000 u − f (u) du ≤ sup u − kf k[t,s],1 s 2 u∈[t,s] 2 (u∈[t,s]) PERTURBED TRAPEZOID INEQUALITY 5 then we can state that Z t t + s 000 (2.3) u − f (u) du s 2 2 (t−s) kf 000k if f 000 ∈ L [a, b]; 4 [t,s],∞ ∞ 1+ 1 |t−s| q 000 000 ≤ 1 kf k[t,s],q if f ∈ Lp [a, b] , 2(q+1) q 1 1 p > 1, + = 1; p q |t−s| 000 2 kf k[t,s],1 . Taking the modulus of R (f; a, b) we get, by (2.3), |R (f; a, b)| Z b Z b Z t 1 t + s 000 ≤ |t − s| u − f (u) du dtds 4 (b − a) a a s 2 1 R b R b |t − s|3 kf 000k dtds if f 000 ∈ L [a, b]; 4 a a [t,s],∞ ∞ b b 1 1 1 R R 2+ q 000 000 ≤ 1 |t − s| kf k[t,s],p dtds if f ∈ Lp [a, b] , q a a 4 (b − a) 2(q+1) p > 1, 1 + 1 = 1; p q 1 R b R b 2 000 2 a a |t − s| kf k[t,s],1 dtds which proves the first inequality in (2.2). Now, consider the double integral Z b Z b Z b "Z t Z b # m m m Im : = |t − s| dtds = (t − s) ds + (s − t) ds dt a a a a t " # Z b (t − a)m+1 + (b − t)m+1 2 (b − a)m+2 = dt = a m + 1 (m + 1) (m + 2) for all m > 0.