Applied Mathematical Sciences, Vol. 3, 2009, no. 8, 399 - 402

A Note on Gr¨ uss Type Inequality Gao-Hui Peng1 College of Mathematics and Information Science North China Institute of Water Conservancy and Hydroelectric Power Henan Province 450011, P.R. China [email protected] Yu Miao College of Mathematics and Information Science Henan Normal University, Henan Province, 453007, P.R. China [email protected]

Abstract. In this short note, we establish a new form of the inequality of Gr¨ uss type for functions whose first and second derivatives are absolutely continuous and third derivative is bounded both above and below almost everywhere. Mathematics Subject Classification: 26D15 Keywords: Gr¨ uss inequality, absolutely continuous 1. Introduction Let f and g be two bounded functions defined on [a, b] with γ1 ≤ f (x) ≤ Γ1 and γ2 ≤ g(x) ≤ Γ2 , where γ1 , γ2 , Γ1 , Γ2 are four constants. Then the classic Gr¨ uss inequality reads as follows:  b  b  b 1 1 1 1 f (x)g(x)dx− f (x)dx g(x)dx ≤ (Γ1 −γ1 )(Γ2 −γ2 ). b−a a b−a a b−a a 4 In the years thereafter, numerous generalizations, extensions and variants of Gr¨ uss inequality have appeared in the literature (see [1, 2, 3, 4, 5, 6, 7, 8, 9]). The purpose of the present note is to establish a new form of the inequality of Gr¨ uss type for functions whose first and second derivatives are absolutely continuous and third derivative is bounded both above and below almost everywhere. 1

This research is supported by the youth scientific research funds of North China Institute of Water Conservancy and Hydroelectric Power (HSQJ2005015).

400

Gao-Hui Peng and Yu Miao

¨ss inequality 2. Gru In this section, we shall obtain the following main result. Theorem 2.1. Let f : [a, b] → (−∞, ∞) be a function such that the derivative  f , f  is absolutely continuous on [a, b]. Assume that there exist constants  γ, Γ ∈ (−∞, ∞) such that γ ≤ f (x) ≤ Γ a.e. on [a, b]. Then we have     2 (a + ba + b2 )(bf ” (a) − af ” (b)) − 3(b2 f (b) − a2 f (a))  b   f (x)dx + 6(bf (b) − af (a)) − a

≤(Γ − γ)

4

b + 3C

4/3

− 4bC

4

.

where (b + a)(b2 + a2 ) C= . 4 Proof. Firstly, it is easy to check 



(a2 + ba + b2 )(bf ” (a) − af ” (b)) − 3(b2 f (b) − a2 f (a))  b f (x)dx + 6(bf (b) − af (a)) − 3



3



2



2

a



= b f (b) − a f (a) − 3(b f (b) − a f (a)) + 6(bf (b) − af (a))  b 2 2 ” ” f (x)dx − (b + a)(b + a )[f (b) − f (a)] − a   b  b 1  3 3 = x dx f (x)dx. x − b−a a a Let

 A=

1 x ∈ [a, b] : x ≥ b−a 3

 c

A =

1 x ∈ [a, b] : x < b−a 3



b

 x dx ;

b

 x dx .

3

a



3

a

Then we have   b  b 1  3 3 x dx f (x)dx x − b−a a a    b  b     1 1 3 3 3 3 x − x − x dx dx + γ x dx dx ≤Γ b−a a b−a a A Ac

A note on Gr¨ uss type inequality

and

Since

401

  b 1  3 x dx f (x)dx x − b−a a a    b  b     1 1 3 3 3 3 x dx dx + Γ x dx dx. ≥γ x − x − b−a a b−a a A Ac  b

  A

3

1 x − b−a 3



b

  x dx dx = − 3

Ac

a



1 x − b−a 3



b

 x dx dx, 3

a

it follows that

 b     b   1  3 3   (2.1) − x dx f (x)dx x   b − a a a     b 1 3 3 ≤ (Γ − γ) x dx dx x − b−a a A     b 1 3 3 x dx dx. x − = (γ − Γ) b−a a Ac Therefore, it is enough to discuss the following integral,     b 1 3 3 x − (2.2) x dx dx. b−a a A From the definition of the set A, it follows that    2 2 3 (b + a)(b + a ) ≤x≤b , A = x ∈ [a, b]; 4 and we can claim that



(b + a)(b2 + a2 ) ≤ b, ∀ a < b. 4 In fact, we can assume b = ka, where k is chosen from R based on a. If a ≥ 0 which implies b > 0, then k > 1 and the inequality (2.3) is equivalent to (2.3)

a≤

3

(k + 1)(k 2 + 1) ≤ k3 4 which is obvious. Similarly if a < 0, b ≤ 0, then 0 ≤ k ≤ 1 and the inequality (2.3) is equivalent to  2 3 (k + 1)(k + 1) (2.4) ≥ k, 1≥ 4 if a < 0, b ≥ 0, then k ≤ 0 and the inequality (2.3) is equivalent also to  2 3 (k + 1)(k + 1) (2.5) ≥ k, 1≥ 4 1≤

402

Gao-Hui Peng and Yu Miao

It is easy to see (2.4) and (2.5) hold correspondingly. Hence the integral (2.2) can be obtained,     b 1 3 3 (2.6) x dx dx x − b−a a A    b  b 1 3 3 =  x dx dx x − 3 (b+a)(b2 +a2 ) b−a a 4

4/3  (b+a)(b2 +a2 ) 4

1/3  b − 4 (b + a)(b2 + a2 ) (a + b)(a2 + b2 ) = − b− 4 4 4

4/3 2 2 2 2 − 4b (b+a)(b4 +a ) b4 + 3 (b+a)(b4 +a ) = . 4 The desired result can be obtained. References [1] X. L. Cheng and J. Sun, A note on the perturbed trapezoid inequality, JIPAM. J. Inequal. Pure Appl. Math. 3(2), Article 29, (2002). [2] S. S. Dragomir, A companion of the Gr¨ uss inequality and applications, Appl. Math. Lett. 17 429-435 (2004). [3] S. S. Dragomir, P. Cerone and A. Sofo, Some remarks on the trapezoid rule in numerical integration, Indian J. Pure Appl. Math. 31(5), 475-494 (2000). [4] S. S. Dragomir, Y. J. Cho and S. S. Kim, Inequalities of Hadamard’s type for Lipschitizian mappings and their applications, J. Math. Anal. Appl. 245, 489-501 (2000). [5] S. S. Dragomir and S. Wang, An inequality of Ostrowski-Gr¨ uss’ type and its applications to the estimation of error bounds for some special means and for home numerical quadrature rules, Computers Math. Applic. 33 (11), 15-20 (1997). [6] N. Elezovi´c, Lj. Maranguni´c and J. Pe˘cari´c, Some improvements of Gr¨ uss type inequality. J. Math. Inequal. 1(3), 425-436 (2007). [7] Z. Liu, A sharp integral inequality of Ostrowski-Gr¨ uss type, Soochow J. Math. 32(2), 223-231 (2006). [8] M. Mati´c, J. Pe˘cari´c and N. Ujevi´c, Improvement and further generalization of inequalities of Ostrowski-Gr¨ uss type, Computers Math. Applic. 39 (3/4), 161-175 (2000). [9] A. M. Mercer, An improvement of the Gr¨ uss inequality. JIPAM. J. Inequal. Pure Appl. Math. 6(4), Article 93 (2005).

Received: July, 2008