# Refinement of an integral inequality

## 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

## 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$α= b k + 2 A k + 2$, $β= ∫ 0 b t k p ( t ) dt$;

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

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$0≤ ∫ a x p ( t ) dt≤ ∫ a b p ( t ) dt$for x [a, b], and for some$c ∈ [ a , b ]$, $∫ a c p ( t ) dt>0$, $∫ c b p ( t ) dt>0$, 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 $∫ 0 x b b k + 1 s k p ( b s ) ds= ∫ 0 x u k p ( u ) du$, 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 $β= ∫ 0 b u k p ( u ) du$ 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 $G ( x ) = ( b x ) k + 2 ∫ 0 x t k p ( t ) dt$. Since $lim x → 0 + G ( x ) =α$ and G(b) = β, we obtain that is α and β are the best constants in (2).

## References

1. 1.

Fischer P, Slodkowski Z: Mean value inequalities for convex and star-shaped sets. Aequationes Math 2005, 70: 213–224. 10.1007/s00010-005-2797-3

2. 2.

Anderson GD, Vamanamurthy MK, Vuorinen M: Inequalities for quasiconformal mappings in space. Pacific J Math 1993, 160: 1–18.

3. 3.

Vamanamurthy MK, Vuorinen M: Inequalities for means. J Math Anal Appl 1994, 183: 155–166. 10.1006/jmaa.1994.1137

4. 4.

Anderson GD, Vamanamurthy MK, Vuorinen M: Conformal Invariants, Inequalities, and Quasiconformal Maps. Wiley, New York; 1997.

5. 5.

Anderson GD, Qiu S-L, Vamanamurthy MK, Vuorinen M: Generalized elliptic integral and modular equations. Pacific J Math 2000, 192: 1–37. 10.2140/pjm.2000.192.1

6. 6.

Mitrinović DS, Pečarić JE: Monotone funkcije i njihove nejednakosti I. Naucna knjiga, Belgrade; 1990:294.

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Correspondence to Ling Zhu.

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Zhu, L. Refinement of an integral inequality. J Inequal Appl 2012, 103 (2012). https://doi.org/10.1186/1029-242X-2012-103