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 ] → ℝ be a nonnegative convex function such that p(0) = 0. Then for 0 < a ≤ b ≤ T and k ∈ ℕ⁺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 ] → ℝ 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. ([[2–5], A Monotone form of L'Hospital's rule]) Let f, g : [a, b] → ℝ 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. ([[6], 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$c\in \left[a,b\right]$, ${\int }_a^{c}p\left(t\right)dt>0$, ${\int }_c^{b}p\left(t\right)dt>0$, 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 $f_1\left(t\right)={\int }_0^t{b}^{k+1}{s}^{k}p\left(bs\right)ds$, g₁(t) = t^{k + 2}, and 0 < t ≤ 1. Then $\frac{f_1^{\text{'}}(t)}{g_1^{\text{'}}(t)}=\frac{b^{k+1}p(bt)}{(k+2)t}$.

• (a) When $\frac{p\left(t\right)}{t}$ is increasing, we have $\frac{f_1^{\text{'}}(t)}{g_1^{\text{'}}(t)}$ 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)

$\frac{b^{k+2}A}{k+2}\mathsf{\text{and}}{\int }_0^b{u}^{k}p\left(u\right)du$ 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).

• (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 < t ≤ x ≤ b 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

$$\begin{array}{ll}\hfill \alpha & =\frac{b^{k+2}}{k+2}\underset{x\to 0^{+}}{\text{lim}}\frac{p\left(x\right)}{x}=\underset{x\to 0^{+}}{\text{lim}}{b}^{k+2}F\left(x\right)\le \underset{x\in \left(0,b\right]}{\text{inf}}{b}^{k+2}F\left(x\right)\\ \le {\left(\frac{b}{x}\right)}^{k+2}\underset{0}{\overset{x}{\int }}{t}^{k}p\left(t\right)dt\\ \le \underset{x\in \left(0,b\right]}{\text{sup}}{b}^{k+2}F\left(x\right)=\underset{x\to b}{\text{lim}}{b}^{k+2}F\left(x\right)=\underset{0}{\overset{b}{\int }}{t}^{k}p\left(t\right)dt=\beta \end{array}$$

(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) = t^{k + 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)

• (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.

• (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 $\underset{x\to 0+}{\text{lim}}G\left(x\right)=\alpha$ and G(b) = β, we obtain that is α and β are the best constants in (2).