Open Access

Refinement of an integral inequality

Journal of Inequalities and Applications20122012:103

https://doi.org/10.1186/1029-242X-2012-103

Received: 19 January 2012

Accepted: 4 May 2012

Published: 4 May 2012

Abstract

In this study, we generalize and sharpen an integral inequality raised in theory for convex and star-shaped sets and relax the conditions on the integrand.

Mathematics Subject Classification (2000): 26D15

Keywords

lower and upper boundsrefinement of an integral inequalitya monotone form of l'Hospital's ruleMitrinović-Pečarić inequality

1 Introduction

In the study [1], which investigated convex and star-shaped sets, the following interesting result was obtained.

Theorem 1. ([[1], Lemma 2.1]) Let p : [0, T ] be a nonnegative convex function such that p(0) = 0. Then for 0 < a ≤ b ≤ T and k +the inequality
0 b t k p ( t ) d t b a k + 2 0 a t k p ( t ) d t
(1)

holds.

In this note, we shall show that the convexity of the function p(t) may be replaced by the condition that p ( t ) t is increasing, sharpen inequality (1), and obtain the following a general result using a monotone form of l'Hospital's rule, a elementary method, and Mitrinović-Pečarić inequality, respectively.

Theorem 2. Let p : [0, T ] be a nonnegative continuous function such that p(0) = 0 and p ( t ) t be a monotone function on (0, T ]. Let A = lim x 0 + p ( x ) x . Then for 0 < x ≤ b ≤ T and k ≥ 0 the double inequality
α b x k + 2 0 x t k p ( t ) d t β
(2)

holds so that

(i) when p ( t ) t is increasing, we have α = b k + 2 A k + 2 , β = 0 b t k p ( t ) d t ;

(ii) when p ( t ) t is decreasing, we have α = 0 b t k p ( t ) d t , β = b k + 2 A k + 2 .

Furthermore, these paired numbers α and β defined in (i) and (ii) are the best constants in (2).

2 Two lemmas

Lemma 1. ([[25], A Monotone form of L'Hospital's rule]) Let f, g : [a, b] be two continuous functions which are differentiable on (a, b). Further, let g 0 on(a, b). If f / g is increasing (or decreasing) on (a, b), then the functions f ( x ) - f ( b ) g ( x ) - g ( b ) and f ( x ) - f ( a ) g ( x ) - g ( a ) are also increasing (or decreasing) on (a, b).

Lemma 2. ([[6], Mitrinović-Pečarić inequality]) If f is increasing function and p satisfies the conditions 0 a x p ( t ) d t a b p ( t ) d t for x [a, b], and for some c [ a , b ] , a c p ( t ) d t > 0 , c b p ( t ) d t > 0 , then we have
a c p ( t ) f ( t ) d t a c p ( t ) d t a b p ( t ) f ( t ) d t a b p ( t ) d t c b p ( t ) f ( t ) d t c b p ( t ) d t .
(3)

If f is decreasing the inequalities (3) are reversed.

3 A concise proof of Theorem 2

Let H ( t ) = 0 t b k + 1 s k p ( b s ) d s t k + 2 = f 1 ( t ) g 1 ( t ) , where f 1 ( t ) = 0 t b k + 1 s k p ( b s ) d s , g1(t) = tk + 2, and 0 < t ≤ 1. Then f 1 ' ( t ) g 1 ' ( t ) = b k + 1 p ( b t ) ( k + 2 ) t .
  1. (a)
    When p ( t ) t is increasing, we have f 1 ' ( t ) g 1 ' ( t ) is also increasing, and H ( t ) = f 1 ( t ) g 1 ( t ) = f 1 ( t ) - f 1 ( 0 ) g 1 ( t ) - g 1 ( 0 ) is increasing by Lemma 1. At the same time, lim t 0 + H ( t ) = lim t 0 + b k + 1 p ( b t ) ( k + 2 ) t = b k + 2 A k + 2 , and lim t 1 H ( t ) = 0 1 b k + 1 t k p ( b t ) d t = 0 b u k p ( u ) d u . So we obtain
    b k + 2 A k + 2 0 t b k + 1 s k p ( b s ) d s t k + 2 0 b u k p ( u ) d u ,
    (4)
     

b k + 2 A k + 2 and 0 b u k p ( u ) d u are the best constants in (4).

Replacing t with x/b in (4), we have 0 t b k + 1 s k p ( b s ) d s = 0 x b b k + 1 s k p ( b s ) d s . Then let bs = u, we obtain 0 x b b k + 1 s k p ( b s ) d s = 0 x u k p ( u ) d u , and
b k + 2 A k + 2 b x k + 2 0 x t k p ( t ) d t 0 b u k p ( u ) d u
(5)
holds. Furthermore α = b k + 2 A k + 2 and β = 0 b u k p ( u ) d u are the best constants in (5).
  1. (b)

    When p ( t ) t is decreasing, we obtain corresponding result by the same way.

     

4 New elementary proof of Theorem 2

Let F ( x ) = 0 x t k p ( t ) d t x k + 2 for x (0, b]. Assume that p ( t ) t is increasing. By a simple calculation and the inequality p ( t ) t p ( x ) x for 0 < txb we have that
x F ( x ) = p ( x ) x - ( k + 2 ) 0 x t k p ( t ) d t x k + 2 p ( x ) x - ( k + 2 ) 0 x t k t p ( x ) x d t x k + 2 = 0 .
So F (x) is increasing and the chain inequality
α = b k + 2 k + 2 lim x 0 + p ( x ) x = lim x 0 + b k + 2 F ( x ) inf x ( 0 , b ] b k + 2 F ( x ) b x k + 2 0 x t k p ( t ) d t sup x ( 0 , b ] b k + 2 F ( x ) = lim x b b k + 2 F ( x ) = 0 b t k p ( t ) d t = β
(6)

holds. Then the double inequality (5) holds, α and β are the best constants in (6) or (2).

The decreasing case can be proved similarly.

5 Other proof of Theorem 2

In what follows, we also assume that p ( t ) t is increasing.

Let p(t) = tk + 1, f ( t ) = p ( t ) t ,c = x, and a = 0 in Lemma 2, we can obtain
0 x t k p ( t ) d t 0 x t k + 1 d t 0 b t k p ( t ) d t 0 b t k + 1 d t x b t k p ( t ) d t x b t k + 1 d t .
(7)
  1. (i)
    The left-side inequality of (7) deduces
    b x k + 2 0 x t k p ( t ) d t 0 b t k p ( t ) d t ,
     
then the right-side inequality of (2) holds.
  1. (ii)
    Let b → 0+ in the right-side inequality of (7), we can obtain
    lim b 0 + 0 b t k p ( t ) d t 0 b t k + 1 d t = lim b 0 + p ( b ) b x 0 t k p ( t ) d t x 0 t k + 1 d t = 0 x t k p ( t ) d t 0 x t k + 1 d t ,
     

then the left-side inequality of (2) holds.

Let G ( x ) = ( b x ) k + 2 0 x t k p ( t ) d t . Since lim x 0 + G ( x ) = α and G(b) = β, we obtain that is α and β are the best constants in (2).

Declarations

Authors’ Affiliations

(1)
Department of Mathematics, Zhejiang Gongshang University

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Copyright

© Zhu; licensee Springer. 2012

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