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# 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:

## 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 bounds
• refinement of an integral inequality
• a monotone form of l'Hospital's rule
• Mitrinović-Pečarić inequality

## 1 Introduction

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

Theorem 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
$\underset{0}{\overset{b}{\int }}{t}^{k}p\left(t\right)dt\ge {\left(\frac{b}{a}\right)}^{k+2}\underset{0}{\overset{a}{\int }}{t}^{k}p\left(t\right)dt$
(1)

holds.

In this note, we shall show that the convexity of the function p(t) may be replaced by the condition that $\frac{p\left(t\right)}{t}$ 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$\frac{p\left(t\right)}{t}$be a monotone function on (0, T ]. Let$A=\underset{x\to {0}^{+}}{\text{lim}}\frac{p\left(x\right)}{x}.$Then for 0 < x ≤ b ≤ T and k ≥ 0 the double inequality
$\alpha \le {\left(\frac{b}{x}\right)}^{k+2}\underset{0}{\overset{x}{\int }}{t}^{k}p\left(t\right)dt\le \beta$
(2)

holds so that

(i) when$\frac{p\left(t\right)}{t}$is increasing, we have$\alpha =\frac{{b}^{k+2}A}{k+2}$, $\beta =\underset{0}{\overset{b}{\int }}{t}^{k}p\left(t\right)dt$;

(ii) when$\frac{p\left(t\right)}{t}$is decreasing, we have$\alpha ={\int }_{0}^{b}{t}^{k}p\left(t\right)dt$, $\beta =\frac{{b}^{k+2}A}{k+2}$.

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

## 2 Two lemmas

Lemma 1. ([, 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 ${g}^{\prime }\ne 0$ on(a, b). If${f}^{\prime }/{g}^{\prime }$is increasing (or decreasing) on (a, b), then the functions $\frac{f\left(x\right)-f\left(b\right)}{g\left(x\right)-g\left(b\right)}$ and $\frac{f\left(x\right)-f\left(a\right)}{g\left(x\right)-g\left(a\right)}$ are also increasing (or decreasing) on (a, b).

Lemma 2. ([, Mitrinović-Pečarić inequality]) If f is increasing function and p satisfies the conditions$0\le {\int }_{a}^{x}p\left(t\right)dt\le {\int }_{a}^{b}p\left(t\right)dt$for x [a, b], and for some$c\in \left[a,\phantom{\rule{2.77695pt}{0ex}}b\right]$, ${\int }_{a}^{c}p\left(t\right)dt>0$, ${\int }_{c}^{b}p\left(t\right)dt>0$, then we have
$\frac{{\int }_{a}^{c}p\left(t\right)f\left(t\right)dt}{{\int }_{a}^{c}p\left(t\right)dt}\le \frac{{\int }_{a}^{b}p\left(t\right)f\left(t\right)dt}{{\int }_{a}^{b}p\left(t\right)dt}\le \frac{{\int }_{c}^{b}p\left(t\right)f\left(t\right)dt}{{\int }_{c}^{b}p\left(t\right)dt}.$
(3)

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

## 3 A concise proof of Theorem 2

Let $H\left(t\right)=\frac{{\int }_{0}^{t}{b}^{k+1}{s}^{k}p\left(bs\right)ds}{{t}^{k+2}}=\frac{{f}_{1}\left(t\right)}{{g}_{1}\left(t\right)},$ where ${f}_{1}\left(t\right)={\int }_{0}^{t}{b}^{k+1}{s}^{k}p\left(bs\right)ds\phantom{\rule{0.3em}{0ex}}$, g1(t) = tk + 2, and 0 < t ≤ 1. Then $\frac{{f}_{1}^{\text{'}}\left(t\right)}{{g}_{1}^{\text{'}}\left(t\right)}=\frac{{b}^{k+1}p\left(bt\right)}{\left(k+2\right)t}$.
1. (a)
When $\frac{p\left(t\right)}{t}$ is increasing, we have $\frac{{f}_{1}^{\text{'}}\left(t\right)}{{g}_{1}^{\text{'}}\left(t\right)}$ is also increasing, and $H\left(t\right)=\frac{{f}_{1}\left(t\right)}{{g}_{1}\left(t\right)}=\frac{{f}_{1}\left(t\right)-{f}_{1}\left(0\right)}{{g}_{1}\left(t\right)-{g}_{1}\left(0\right)}$ is increasing by Lemma 1. At the same time, $\underset{t\to {0}^{+}}{\text{lim}}H\left(t\right)=\underset{t\to {0}^{+}}{\text{lim}}\frac{{b}^{k+1}p\left(bt\right)}{\left(k+2\right)t}=\frac{{b}^{k+2}A}{k+2}$, and $\underset{t\to 1}{\text{lim}}H\left(t\right)={\int }_{0}^{1}{b}^{k+1}{t}^{k}p\left(bt\right)dt={\int }_{0}^{b}{u}^{k}p\left(u\right)du$. So we obtain
$\frac{{b}^{k+2}A}{k+2}\le \frac{{\int }_{0}^{t}{b}^{k+1}{s}^{k}p\left(bs\right)ds}{{t}^{k+2}}\le \underset{0}{\overset{b}{\int }}{u}^{k}p\left(u\right)du,$
(4)

$\frac{{b}^{k+2}A}{k+2}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\mathsf{\text{and}}\phantom{\rule{0.3em}{0ex}}{\int }_{0}^{b}{u}^{k}p\left(u\right)du$ are the best constants in (4).

Replacing t with x/b in (4), we have ${\int }_{0}^{t}{b}^{k+1}{s}^{k}p\left(bs\right)ds={\int }_{0}^{\frac{x}{b}}{b}^{k+1}{s}^{k}p\left(bs\right)ds$. Then let bs = u, we obtain ${\int }_{0}^{\frac{x}{b}}{b}^{k+1}{s}^{k}p\left(bs\right)ds={\int }_{0}^{x}{u}^{k}p\left(u\right)du$, and
$\frac{{b}^{k+2}A}{k+2}\le {\left(\frac{b}{x}\right)}^{k+2}\underset{0}{\overset{x}{\int }}{t}^{k}p\left(t\right)dt\le \underset{0}{\overset{b}{\int }}{u}^{k}p\left(u\right)du$
(5)
holds. Furthermore $\alpha =\frac{{b}^{k+2}A}{k+2}$ and $\beta ={\int }_{0}^{b}{u}^{k}p\left(u\right)du$ are the best constants in (5).
1. (b)

When $\frac{p\left(t\right)}{t}$ is decreasing, we obtain corresponding result by the same way.

## 4 New elementary proof of Theorem 2

Let $F\left(x\right)=\frac{{\int }_{0}^{x}{t}^{k}p\left(t\right)dt}{{x}^{k+2}}$ for x (0, b]. Assume that $\frac{p\left(t\right)}{t}$ is increasing. By a simple calculation and the inequality $p\left(t\right)\le t\frac{p\left(x\right)}{x}$ for 0 < txb we have that
$x{F}^{\prime }\left(x\right)=\frac{p\left(x\right)}{x}-\left(k+2\right)\frac{{\int }_{0}^{x}{t}^{k}p\left(t\right)dt}{{x}^{k+2}}\ge \frac{p\left(x\right)}{x}-\left(k+2\right)\frac{{\int }_{0}^{x}{t}^{k}t\frac{p\left(x\right)}{x}dt}{{x}^{k+2}}=0.$
So F (x) is increasing and the chain inequality
$\begin{array}{ll}\hfill \alpha & =\frac{{b}^{k+2}}{k+2}\underset{x\to {0}^{+}}{\text{lim}}\frac{p\left(x\right)}{x}=\underset{x\to {0}^{+}}{\text{lim}}{b}^{k+2}F\left(x\right)\le \underset{x\in \left(0,b\right]}{\text{inf}}{b}^{k+2}F\left(x\right)\phantom{\rule{2em}{0ex}}\\ \le {\left(\frac{b}{x}\right)}^{k+2}\underset{0}{\overset{x}{\int }}{t}^{k}p\left(t\right)dt\phantom{\rule{2em}{0ex}}\\ \le \underset{x\in \left(0,b\right]}{\text{sup}}{b}^{k+2}F\left(x\right)=\underset{x\to b}{\text{lim}}{b}^{k+2}F\left(x\right)=\underset{0}{\overset{b}{\int }}{t}^{k}p\left(t\right)dt=\beta \phantom{\rule{2em}{0ex}}\end{array}$
(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 $\frac{p\left(t\right)}{t}$ is increasing.

Let p(t) = tk + 1,$f\left(t\right)=\frac{p\left(t\right)}{t}$,c = x, and a = 0 in Lemma 2, we can obtain
$\frac{{\int }_{0}^{x}{t}^{k}p\left(t\right)dt}{{\int }_{0}^{x}{t}^{k+1}dt}\le \frac{{\int }_{0}^{b}{t}^{k}p\left(t\right)dt}{{\int }_{0}^{b}{t}^{k+1}dt}\le \frac{{\int }_{x}^{b}{t}^{k}p\left(t\right)dt}{{\int }_{x}^{b}{t}^{k+1}dt}.$
(7)
1. (i)
The left-side inequality of (7) deduces
${\left(\frac{b}{x}\right)}^{k+2}\underset{0}{\overset{x}{\int }}{t}^{k}p\left(t\right)dt\le \underset{0}{\overset{b}{\int }}{t}^{k}p\left(t\right)dt,$

then the right-side inequality of (2) holds.
1. (ii)
Let b → 0+ in the right-side inequality of (7), we can obtain
$\underset{b\to {0}^{+}}{\text{lim}}\frac{{\int }_{0}^{b}{t}^{k}p\left(t\right)dt}{{\int }_{0}^{b}{t}^{k+1}dt}=\underset{b\to {0}^{+}}{\text{lim}}\frac{p\left(b\right)}{b}\le \frac{{\int }_{x}^{0}{t}^{k}p\left(t\right)dt}{{\int }_{x}^{0}{t}^{k+1}dt}=\frac{{\int }_{0}^{x}{t}^{k}p\left(t\right)dt}{{\int }_{0}^{x}{t}^{k+1}dt},$

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

Let $G\left(x\right)={\left(\frac{b}{x}\right)}^{k+2}{\int }_{0}^{x}{t}^{k}p\left(t\right)dt$. Since $\phantom{\rule{0.3em}{0ex}}\underset{x\to 0+}{\text{lim}}G\left(x\right)=\alpha$ and G(b) = β, we obtain that is α and β are the best constants in (2).

## Authors’ Affiliations

(1)
Department of Mathematics, Zhejiang Gongshang University, Hangzhou, Zhejiang, 310018, P. R. China

## References 