Barnes-Godunova-Levin type inequality of the Sugeno integral for an $(\alpha,m)$-concave function

Abstract

In this paper, a Barnes-Godunova-Levin type inequality for the Sugeno integral based on an $( {\alpha,m} )$-concave function is proved. Some examples are given to illustrate the validity of these inequalities. Finally, several important results, as special cases of an $( {\alpha,m} )$-concave function, are also obtained.

1 Introduction

As a tool for modeling non-deterministic problems, the theory of fuzzy measures and fuzzy integrals was introduced by Sugeno in [1]. Many authors generalized the Sugeno integral by using some other operators to replace the special operator(s) ∨ and/or ∧ and introduced Choquet-like integral [2], Shilkret integral [3], ⊥-integral [4], and pseudo-integral [5]. Suárez and Gil [6] presented two families of fuzzy integrals, the so-called seminormed fuzzy integral and semiconormed fuzzy integral. Wang and Klir [7] provided a general overview on fuzzy measurement and fuzzy integration.

Recently, Flores-Franulič et al. [8–21] generalized several classical integral inequalities of the Sugeno integral. Agahi et al. [22] proved a general Barnes-Godunova-Levin type inequality of the Sugeno integral for a concave function. In [23], Mihesan introduced the concept of $( {\alpha,m} )$-convex function. For recent results and generalizations concerning m-convex and $( {\alpha,m} )$-convex functions, we refer to [24–26]. The purpose of this paper is to prove a Barnes-Godunova-Levin type inequality for the Sugeno integral based on an $( {\alpha,m} )$-concave function. Some examples are given to illustrate the results.

After some preliminaries and summarization of previous known results in Section 2, Section 3 deals with a Barnes-Godunova-Levin type inequality for the Sugeno integral, and some examples are given to illustrate the results. Finally, as special cases, some remarks are obtained.

2 Preliminaries

In this section, we recall some basic definitions or properties of a fuzzy integral and an $( {\alpha,m} )$-concave function. For details, we refer the reader to Refs. [1, 7, 23].

Suppose that ℘ is a σ-algebra of the subsets of X, and let $\mu:\wp \to[0,\infty)$ be a non-negative, extended real-valued set function. We say that μ is a fuzzy measure if it satisfies:

• (1) $\mu ( \emptyset ) = 0$;

• (2) $E,F \in\wp$ and $E \subset F$ imply $\mu ( E ) \le\mu ( F)$;

• (3) $\{ {{E_{n}}} \} \subset\wp$, ${E_{1}} \subset {E_{2}} \subset \cdots$, imply ${\lim_{n \to\infty}}\mu ( {{E_{n}}} ) = \mu ( {\bigcup_{n = 1}^{\infty}{{E_{n}}} } )$;

• (4) $\{ {{E_{n}}} \} \subset\wp$, ${E_{1}} \supset {E_{2}} \supset \cdots$, $\mu ( {{E_{1}}} ) < \infty$, imply ${\lim_{n \to\infty}}\mu ( {{E_{n}}} ) = \mu ( {\bigcap_{n = 1}^{\infty}{{E_{n}}} } )$.

Definition 2.1

(Mihesan [23])

The function $f: [ {0,b} ] \to R$ is said to be $( {\alpha,m} )$-concave, where $( {\alpha,m} ) \in { [ {0,1} ]^{2}}$, if for every $x,y \in [ {0,b} ]$ and $t \in [ {0,1} ]$, it satisfies

$$ f \bigl( {tx + m ( {1 - t} )y} \bigr) \ge{t^{\alpha}}f ( x ) + m \bigl( {1 - {t^{\alpha}}} \bigr)f ( y ). $$

(2.1)

Note that for $( {\alpha,m} ) \in \{ { ( {0,0} ), ( {\alpha,0} ), ( {1,0} ), ( {1,m} ), ( {1,1} ), ( {\alpha,1} )} \}$ one obtains the following classes of functions: decreasing, α-starshaped, starshaped, m-concave, concave and α-concave.

If f is a non-negative real-valued function defined on X, we denote the set $\{ x \in X: f ( x ) \ge\alpha \} = \{ {x \in X:f \ge\alpha} \}$ by ${F_{\alpha }}$ for $\alpha \ge0$. Note that if $\alpha \le\beta$ then ${F_{\beta }} \subset {F_{\alpha }}$.

Let $( {X,\wp,\mu} )$ be a fuzzy measure space, we denote by ${M^{+} }$ the set of all non-negative measurable functions with respect to ℘.

Definition 2.2

(Sugeno [1])

Let $( {X,\wp,\mu} )$ be a fuzzy measure space, $f \in{M^{+} }$ and $A \in\wp$. The Sugeno integral (or the fuzzy integral) of f on A, with respect to the fuzzy measure μ, is defined as

$$ (S)\int_{A} {f\,d\mu} = \bigvee _{\alpha \ge0} \bigl[ { \alpha \wedge\mu ( {A \cap{F_{\alpha }}} )} \bigr], $$

(2.2)

when $A = X$,

$$ (S)\int_{X} {f\,d\mu} = \bigvee _{\alpha \ge0} \bigl[ { \alpha \wedge\mu ( {{F_{\alpha }}} )} \bigr], $$

(2.3)

where ∨ and ∧ denote the operations sup and inf on $[ {0,\infty} )$, respectively.

The properties of the Sugeno integral are well known and can be found in [7].

Proposition 2.3

Let $( {X,\wp,\mu} )$ be a fuzzy measure space, $A,B \in \wp$ and $f,g \in{M^{+} }$ then:

• (1) ${ ( S )\int_{A} {f\,d\mu \le\mu ( A )} }$;

• (2) ${ ( S )\int_{A} {k\,d\mu} = k \wedge\mu ( A )}$, k for a non-negative constant;

• (3) ${ ( S )\int_{A} {f\,d\mu \le} ( S )\int_{A} {g\,d\mu} }$ for $f \le g$;

• (4) ${ ( S )\int_{A \cup B} {f\,d\mu \ge} ( S )\int_{A} {f\,d\mu} \vee ( S )\int_{B} {f\,d\mu} }$;

• (5) ${\mu ( {A \cap \{ {f \ge\alpha} \} } ) \ge\alpha \Rightarrow ( S )\int_{A} {f\,d\mu} \ge \alpha}$;

• (6) ${\mu ( {A \cap \{ {f \ge\alpha} \} } ) \le\alpha \Rightarrow ( S )\int_{A} {f\,d\mu} \le \alpha}$;

• (7) ${ ( S )\int_{A} {f\,d\mu} > \alpha \Leftrightarrow}$ there exists $\gamma > \alpha$ such that $\mu ( {A \cap \{ {f \ge\gamma} \}} ) >\alpha$;

• (8) ${ ( S )\int_{A} {f\,d\mu} < \alpha \Leftrightarrow}$ there exists $\gamma < \alpha$ such that $\mu ( {A \cap \{ {f \ge\gamma} \}} ) < \alpha$.

Remark 2.4

Consider the distribution function F associated to f on A, that is, $F ( \alpha ) = \mu ( {A \cap \{ {f \ge \alpha} \}} )$. Then, due to (4) and (5) of Proposition 2.3, we have $F ( \alpha ) = \alpha \Rightarrow ( S )\int_{A} {f\,d\mu} = \alpha$. Thus, from a numerical point of view, the Sugeno integral can be calculated solving the equation $F ( \alpha ) = \alpha$.

Definition 2.5

Functions $f,g:X \to R$ are said to be co-monotone if for all $x,y \in X$,

$$ \bigl( {f ( x ) - f ( y )} \bigr) \bigl( {g ( x ) - g ( y )} \bigr) \ge0, $$

(2.4)

and f and g are said to be counter-monotone if for all $x,y \in X$,

$$ \bigl( {f ( x ) - f ( y )} \bigr) \bigl( {g ( x ) - g ( y )} \bigr) \leq0. $$

(2.5)

It is clear that if f and g are co-monotone, then for any real numbers s, t either ${F_{s}} \subseteq{G_{t}}$ or ${F_{t}} \subseteq {G_{s}}$.

2.1 Barnes-Godunova-Levin type inequality for the Sugeno integral based on an $( {\alpha,m} )$-concave function

The classical Barnes-Godunova-Levin type inequality provides the inequality

$$ { \biggl( {\int_{a}^{b} {{f^{p}} ( x )\,dx} } \biggr)^{\frac{1}{p}}} { \biggl( {\int_{a}^{b} {{g^{q}} ( x )\,dx} } \biggr)^{\frac{1}{q}}} \le B ( {p,q} )\int _{a}^{b} {f ( x )g ( x )\,dx}, $$

(2.6)

where $p,q > 1$, $B ( {p,q} ) = \frac{{6{{ ( {b - a} )}^{{\frac{1 }{ p}} + {\frac{1 }{ q}} - 1}}}}{{{{ ( {1 + p} )}^{{\frac{1 }{ p}}}}{{ ( {1 + q} )}^{{\frac{1 }{ q}}}}}}$ and f, g are non-negative concave functions on $[ a,b ]$.

Unfortunately, the following example shows that the Barnes-Godunova-Levin type inequality for the Sugeno integral is not valid.

Example

Consider $X = [ {0,100} ]$ and $p = q = 4$. Let m be the Lebesgue measure on X. If we take the functions $f ( x ) = g ( x ) = \sqrt[4]{x}$, then $f ( x )$, $g ( x )$ are two $( {{\frac{1 }{2}},{\frac{1}{3}}} )$-concave functions. In fact,

$$\sqrt[4]{x} = f \biggl( {{\frac{x }{{100}}} \cdot100 + {\frac{1}{3}} \biggl( {1 - {\frac{x }{{100}}}} \biggr) \cdot0} \biggr) \ge \sqrt{{ \frac{x }{ {100}}}} f ( {100} ) + {\frac{1}{3}} \biggl( {1 - \sqrt{{ \frac{x}{{100}}}} } \biggr)f ( 0 ) = \sqrt{\frac{x}{{10}}} . $$

A straightforward calculus shows that

$$\begin{aligned}& ( S )\int_{0}^{100} {{f^{4}} ( x )\,dm} = ( S )\int_{0}^{100} {{g^{4}} ( x )\,dm} = 50,\\& ( S )\int_{0}^{100} {f ( x )g ( x )\,dm} = 9.5125,\qquad B ( {4,4} ) = 0.26833. \end{aligned}$$

However,

$$\begin{aligned} 7.0711 &= { \biggl( { ( S )\int_{0}^{100} {{f^{4}} ( x )\,dm} } \biggr)^{{\frac{1}{4}}}} { \biggl( { ( S )\int _{0}^{100} {{g^{4}} ( x )\,dm} } \biggr)^{{\frac{1}{4}}}} \\ &\ge B ( {4,4} ) ( S )\int_{0}^{100} {f ( x )g ( x )\,dm} = 2.5525. \end{aligned}$$

This proves that the Barnes-Godunova-Levin type inequality for the Sugeno integral is not satisfied.

The aim of this work is to show a Barnes-Godunova-Levin type inequality for the Sugeno integral with respect to an $( {\alpha,m} )$-concave function.

Theorem 2.6

Let $X= [ {0,1} ]$, $\alpha,m \in ( {0,1} )$ and f, g be $(\alpha,m )$-concave functions for all $x \in X$. If m is a Lebesgue measure on X, then

Case (i). If $f ( 0 ) \le f ( 1 )$ and $g ( 0 ) \le g ( 1 )$, then

$$\begin{aligned} ( S ) \int_{0}^{1} {f ( x )g ( x )\,dx} \ge{t_{1}} {t_{2}} \wedge \biggl( 1 - \biggl( { \frac{{{t_{1}} - mf ( 0 )}}{{f ( 1 ) - mf ( 0 )}}} \biggr)^{\frac{1}{\alpha}} \biggr) \wedge \biggl( {1 - { \biggl( {\frac{{{t_{2}} - mg ( 0 )}}{{g ( 1 ) - mg ( 0 )}}} \biggr)^{\frac{1}{\alpha}}}} \biggr), \end{aligned}$$

(2.7)

where ${t_{1}} = { ( { ( S )\int_{0}^{1} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{0}^{1} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case (ii). If $f ( 0 ) > f ( 1 )$ and $g ( 0 ) > g ( 1 )$, then

Case (a). If $\frac{{f ( 1 )}}{{f ( 0 )}} < \frac{{g ( 1 )}}{{g ( 0 )}}$, then

Case 1. If $m \in ( {0,{\frac{{f ( 1 )} }{{f ( 0 )}}}} )$, then

$$\begin{aligned} ( S ) \int_{0}^{1} {f ( x )g ( x )\,dx} \ge{t_{1}} {t_{2}} \wedge \biggl( {1 - {{ \biggl( { \frac{{{t_{1}} - mf ( 0 )}}{{f ( 1 ) - mf ( 0 )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr) \wedge \biggl( {1 - {{ \biggl( {\frac{{{t_{2}} - mg ( 0 )}}{{g ( 1 ) - mg ( 0 )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr), \end{aligned}$$

(2.8)

where ${t_{1}} = { ( { ( S )\int_{0}^{1} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{0}^{1} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case 2. If $m = {\frac{{f ( 1 )} }{{f ( 0 )}}}$, then

$$ ( S )\int_{0}^{1} {f ( x )g ( x )\,dx} \ge {t_{1}} {t_{2}}\wedge1 \wedge f ( 1 ) \wedge \biggl( {1 - {{ \biggl( {\frac{{{t_{2}} - mg ( 0 )}}{{g ( 1 ) - mg ( 0 )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr), $$

(2.9)

where ${t_{1}} = { ( { ( S )\int_{0}^{1} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{0}^{1} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case 3. If $m \in ( {{\frac{{f ( 1 )} }{{f ( 0 )}}},{\frac{{g ( 1 )} }{{g ( 0 )}}}} )$, then

$$ ( S ) \int_{0}^{1} {f ( x )g ( x )\,dx} \ge{t_{1}} {t_{2}}\wedge{ \biggl( {\frac{{{t_{1}} - mf ( 0 )}}{{f ( 1 ) - mf ( 0 )}}} \biggr)^{\frac{1}{\alpha}}} \wedge \biggl( {1 - {{ \biggl( {\frac{{{t_{2}} - mg ( 0 )}}{{g ( 1 ) - mg ( 0 )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr), $$

(2.10)

where ${t_{1}} = { ( { ( S )\int_{0}^{1} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{0}^{1} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case 4. If $m = {\frac{{g ( 1 )} }{{g ( 0 )}}}$, then

$$ ( S )\int_{0}^{1} {f ( x )g ( x )\,dx} \ge {t_{1}} {t_{2}}\wedge{ \biggl( {\frac{{{t_{1}} - mf ( 0 )}}{{f ( 1 ) - mf ( 0 )}}} \biggr)^{\frac{1}{\alpha}}} \wedge 1\wedge g ( 1 ), $$

(2.11)

where ${t_{1}} = { ( { ( S )\int_{0}^{1} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{0}^{1} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case 5. If $m \in ( {{\frac{{g ( 1 )} }{{g ( 0 )}}},1} )$, then

$$ ( S )\int_{0}^{1} {f ( x )g ( x )\,dx}\ge {t_{1}} {t_{2}}\wedge{ \biggl( {\frac{{{t_{1}} - mf ( 0 )}}{{f ( 1 ) - mf ( 0 )}}} \biggr)^{\frac{1}{\alpha}}} \wedge { \biggl( {\frac{{{t_{2}} - mg ( 0 )}}{{g ( 1 ) - mg ( 0 )}}} \biggr)^{\frac{1}{\alpha}}}, $$

(2.12)

where ${t_{1}} = { ( { ( S )\int_{0}^{1} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{0}^{1} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case (b). If $\frac{{f ( 1 )}}{{f ( 0 )}} = \frac{{g ( 1 )}}{{g ( 0 )}}$, then

Case 1. If $m \in ( {0,{\frac{{f ( 1 )} }{ {f ( 0 )}}}} )$, then

$$\begin{aligned} ( S ) \int_{0}^{1} {f ( x )g ( x )\,dx} \ge{t_{1}} {t_{2}} \wedge \biggl( {1 - {{ \biggl( { \frac{{{t_{1}} - mf ( 0 )}}{{f ( 1 ) - mf ( 0 )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr) \wedge \biggl( {1 - {{ \biggl( {\frac{{{t_{2}} - mg ( 0 )}}{{g ( 1 ) - mg ( 0 )}}} \biggr)}^{\frac {1}{\alpha}}}} \biggr), \end{aligned}$$

(2.13)

where ${t_{1}} = { ( { ( S )\int_{0}^{1} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{0}^{1} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case 2. If $m = {\frac{{f ( 1 )} }{{f ( 0 )}}}$, then

$$ ( S )\int_{0}^{1} {f ( x )g ( x )\,dx} \ge {t_{1}} {t_{2}} \wedge f ( 1 )\wedge g ( 1 )\wedge1, $$

(2.14)

where ${t_{1}} = { ( { ( S )\int_{0}^{1} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{0}^{1} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case 3. If $m \in ( {{\frac{{f ( 1 )} }{{f ( 0 )}}},1} )$, then

$$ ( S )\int_{0}^{1} {f ( x )g ( x )\,dx} \ge {t_{1}} {t_{2}} \wedge{ \biggl( {\frac{{{t_{1}} - mf ( 0 )}}{{f ( 1 ) - mf ( 0 )}}} \biggr)^{\frac{1}{\alpha}}} \wedge { \biggl( {\frac{{{t_{2}} - mg ( 0 )}}{{g ( 1 ) - mg ( 0 )}}} \biggr)^{\frac{1}{\alpha}}}, $$

(2.15)

where ${t_{1}} = { ( { ( S )\int_{0}^{1} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{0}^{1} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case (c). If $\frac{{f ( 1 )}}{{f ( 0 )}} > \frac{{g ( 1 )}}{{g ( 0 )}}$, then

Case 1. If $m \in ( {0,{\frac{{g ( 1 )} }{ {g ( 0 )}}}} )$, then

$$\begin{aligned} ( S ) \int_{0}^{1} {f ( x )g ( x )\,dx} \ge{t_{1}} {t_{2}} \wedge \biggl( {1 - {{ \biggl( { \frac{{{t_{1}} - mf ( 0 )}}{{f ( 1 ) - mf ( 0 )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr) \wedge \biggl( {1 - {{ \biggl( {\frac{{{t_{2}} - mg ( 0 )}}{{g ( 1 ) - mg ( 0 )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr), \end{aligned}$$

(2.16)

where ${t_{1}} = { ( { ( S )\int_{0}^{1} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{0}^{1} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case 2. If $m = {\frac{{g ( 1 )} }{{g ( 0 )}}}$, then

$$ ( S )\int_{0}^{1} {f ( x )g ( x )\,dx} \ge {t_{1}} {t_{2}} \wedge \biggl( {1 - {{ \biggl( { \frac{{{t_{1}} - mf ( 0 )}}{{f ( 1 ) - mf ( 0 )}}} \biggr)}^{\frac {1}{\alpha}}}} \biggr) \wedge g ( 1 )\wedge1, $$

(2.17)

where ${t_{1}} = { ( { ( S )\int_{0}^{1} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{0}^{1} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case 3. If $m \in ( {{\frac{{g ( 1 )} }{{g ( 0 )}}},{\frac{{f ( 1 )} }{{f ( 0 )}}}} )$, then

$$ ( S ) \int_{0}^{1} {f ( x )g ( x )\,dx} \ge{t_{1}} {t_{2}} \wedge \biggl( {1 - {{ \biggl( { \frac{{{t_{1}} - mf ( 0 )}}{{f ( 1 ) - mf ( 0 )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr) \wedge { \biggl( { \frac{{{t_{2}} - mg ( 0 )}}{{g ( 1 ) - mg ( 0 )}}} \biggr)^{\frac{1}{\alpha}}}, $$

(2.18)

where ${t_{1}} = { ( { ( S )\int_{0}^{1} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{0}^{1} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case 4. If $m = {\frac{{f ( 1 )} }{{f ( 0 )}}}$, then

$$ ( S )\int_{0}^{1} {f ( x )g ( x )\,dx} \ge {t_{1}} {t_{2}} \wedge f ( 1 ) \wedge{ \biggl( { \frac{{{t_{2}} - mg ( 0 )}}{{g ( 1 ) - mg ( 0 )}}} \biggr)^{\frac{1}{\alpha}}}, $$

(2.19)

where ${t_{1}} = { ( { ( S )\int_{0}^{1} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{0}^{1} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case 5. If $m \in ( {{\frac{{f ( 1 )} }{{f ( 0 )}}},1} )$, then

$$ ( S )\int_{0}^{1} {f ( x )g ( x )\,dx} \ge {t_{1}} {t_{2}} \wedge{ \biggl( {\frac{{{t_{1}} - mf ( 0 )}}{{f ( 1 ) - mf ( 0 )}}} \biggr)^{\frac{1}{\alpha}}} \wedge { \biggl( {\frac{{{t_{2}} - mg ( 0 )}}{{g ( 1 ) - mg ( 0 )}}} \biggr)^{\frac{1}{\alpha}}}, $$

(2.20)

where ${t_{1}} = { ( { ( S )\int_{0}^{1} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{0}^{1} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Proof

Let $p,q \in ( {0,\infty} )$, ${ ( { ( S )\int_{0}^{1} {{f^{p}} ( x )\,dx} } )^{{\frac{1 }{ p}}}} = {t_{1}}$ and ${ ( { ( S )\int_{0}^{1} {{g^{q}} ( x )\,dx} } )^{{\frac{1 }{ q}}}} = {t_{2}}$. Since $f,g: [ {0,1} ] \to [ {0,\infty} )$ are two $( {\alpha,m} )$-concave functions for $x \in [ {0,1} ]$, we have

$$\begin{aligned}& f ( x ) = f \bigl( {m ( {1 - x} ) \cdot0 + x \cdot 1} \bigr) \ge m \bigl( {1 - {x^{\alpha}}} \bigr)f ( 0 ) + {x^{\alpha}}f ( 1 ) = {h_{1}}( x ), \end{aligned}$$

(2.21)

$$\begin{aligned}& g ( x ) = g \bigl( {m ( {1 - x} ) \cdot0 + x \cdot 1} \bigr) \ge m \bigl( {1 - {x^{\alpha}}} \bigr)g ( 0 ) + {x^{\alpha}}g ( 1 ) = {h_{2}} ( x ). \end{aligned}$$

(2.22)

Case (i). If $f ( 0 ) \le f ( 1 )$ and $g ( 0 ) \le g ( 1 )$, then by (3) of Proposition 2.3 and the co-monotonicity of ${h_{1}} ( x )$ and ${h_{2}} ( x )$, we have

$$\begin{aligned} & ( S )\int_{0}^{1} {f ( x )g ( x )\,dx} \\ &\quad\ge ( S ) \int_{0}^{1} {{h_{1}} ( x ){h_{2}} ( x )\,dx} \\ &\quad= \bigvee _{\beta \ge0} \bigl( {\beta \wedge\mu \bigl( { [ {0,1} ] \cap \bigl\{ {{h_{1}} ( x ){h_{2}} ( x ) \ge\beta} \bigr\} } \bigr)} \bigr) \\ & \quad\ge{t_{1}} {t_{2}} \wedge\mu \bigl( { [ {0,1} ] \cap \bigl\{ {{h_{1}} ( x ){h_{2}} ( x ) \ge{t_{1}} {t_{2}}} \bigr\} } \bigr) \\ &\quad\ge {t_{1}} {t_{2}} \wedge\mu \bigl( { [ {0,1} ] \cap \bigl\{ {{h_{1}} ( x ) \ge{t_{1}}} \bigr\} \cap \bigl\{ {{h_{2}} ( x ) \ge{t_{2}}} \bigr\} } \bigr) \\ &\quad= {t_{1}} {t_{2}} \wedge\mu \bigl( { [ {0,1} ] \cap \bigl\{ {{h_{1}} ( x ) \ge{t_{1}}} \bigr\} } \bigr) \wedge\mu \bigl( { [ {0,1} ] \cap \bigl\{ {{h_{2}} ( x ) \ge{t_{2}}} \bigr\} } \bigr) \\ &\quad= {t_{1}} {t_{2}} \wedge\mu \bigl( { [ {0,1} ] \cap \bigl\{ {{h_{1}} ( x ) \ge{t_{1}}} \bigr\} } \bigr) \wedge\mu \bigl( { [ {0,1} ] \cap \bigl\{ {{h_{2}} ( x ) \ge{t_{2}}} \bigr\} } \bigr) \\ &\quad= {t_{1}} {t_{2}} \wedge\mu \bigl( { [ {0,1} ] \cap \bigl\{ {m \bigl( {1 - {x^{\alpha}}} \bigr)f ( 0 ) + {x^{\alpha}}f ( 1 ) \ge{t_{1}}} \bigr\} } \bigr) \\ &\qquad{} \wedge\mu \bigl( { [ {0,1} ] \cap \bigl\{ {m \bigl( {1 - {x^{\alpha}}} \bigr)g ( 0 ) + {x^{\alpha}}g ( 1 ) \ge{t_{2}}} \bigr\} } \bigr) \\ &\quad= {t_{1}} {t_{2}} \wedge\mu \biggl( { [ {0,1} ] \cap \biggl\{ {x \ge{{ \biggl( {\frac{{{t_{1}} - mf ( 0 )}}{{f ( 1 ) - mf ( 0 )}}} \biggr)}^{{\frac{1 }{\alpha}}}}} \biggr\} } \biggr) \\ &\qquad{} \wedge\mu \biggl( { [ {0,1} ] \cap \biggl\{ {x \ge{{ \biggl( { \frac{{{t_{2}} - mg ( 0 )}}{{g ( 1 ) - mg ( 0 )}}} \biggr)}^{{\frac{1 }{\alpha}}}}} \biggr\} } \biggr) \\ &\quad= {t_{1}} {t_{2}} \wedge \biggl( {1 - {{ \biggl( { \frac{{{t_{1}} - mf ( 0 )}}{{f ( 1 ) - mf ( 0 )}}} \biggr)}^{{\frac{1 }{\alpha}}}}} \biggr) \wedge \biggl( {1 - {{ \biggl( {\frac{{{t_{2}} - mg ( 0 )}}{{g ( 1 ) - mg ( 0 )}}} \biggr)}^{{\frac{1 }{\alpha}}}}} \biggr). \end{aligned}$$

(2.23)

Case (ii). If $f ( 0 ) > f ( 1 )$ and $g ( 0 ) > g ( 1 )$, then by (3) of Proposition 2.3 and the co-monotonicity of ${h_{1}} ( x )$ and ${h_{2}} ( x )$, we have

$$\begin{aligned} & ( S )\int_{0}^{1} {f ( x )g ( x )\,dx} \\ &\quad\ge ( S ) \int_{0}^{1} {{h_{1}} ( x ){h_{2}} ( x )\,dx} \\ &\quad= \bigvee _{\beta \ge0} \bigl( {\beta \wedge\mu \bigl( { [ {0,1} ] \cap \bigl\{ {{h_{1}} ( x ){h_{2}} ( x ) \ge\beta} \bigr\} } \bigr)} \bigr) \\ &\quad\ge {t_{1}} {t_{2}} \wedge\mu \bigl( { [ {0,1} ] \cap \bigl\{ {{h_{1}} ( x ){h_{2}} ( x ) \ge{t_{1}} {t_{2}}} \bigr\} } \bigr) \\ &\quad\ge {t_{1}} {t_{2}} \wedge\mu \bigl( { [ {0,1} ] \cap \bigl\{ {{h_{1}} ( x ) \ge{t_{1}}} \bigr\} \cap \bigl\{ {{h_{2}} ( x ) \ge{t_{2}}} \bigr\} } \bigr) \\ &\quad= {t_{1}} {t_{2}} \wedge\mu \bigl( { [ {0,1} ] \cap \bigl\{ {{h_{1}} ( x ) \ge{t_{1}}} \bigr\} } \bigr) \wedge\mu \bigl( { [ {0,1} ] \cap \bigl\{ {{h_{2}} ( x ) \ge{t_{2}}} \bigr\} } \bigr) \\ &\quad= {t_{1}} {t_{2}} \wedge\mu \bigl( { [ {0,1} ] \cap \bigl\{ {{h_{1}} ( x ) \ge{t_{1}}} \bigr\} } \bigr) \wedge\mu \bigl( { [ {0,1} ] \cap \bigl\{ {{h_{2}} ( x ) \ge{t_{2}}} \bigr\} } \bigr) \\ &\quad= {t_{1}} {t_{2}} \wedge\mu \bigl( { [ {0,1} ] \cap \bigl\{ {m \bigl( {1 - {x^{\alpha}}} \bigr)f ( 0 ) + {x^{\alpha}}f ( 1 ) \ge{t_{1}}} \bigr\} } \bigr) \\ &\qquad{}\wedge\mu \bigl( { [ {0,1} ] \cap \bigl\{ {m \bigl( {1 - {x^{\alpha}}} \bigr)g ( 0 ) + {x^{\alpha}}g ( 1 ) \ge{t_{2}}} \bigr\} } \bigr) \\ &\quad= {t_{1}} {t_{2}} \wedge\mu \bigl( { [ {0,1} ] \cap \bigl\{ {mf ( 0 ) + \bigl( {f ( 1 ) - mf ( 0 )} \bigr){x^{\alpha}} \ge{t_{1}}} \bigr\} } \bigr) \\ &\qquad{}\wedge\mu \bigl( { [ {0,1} ] \cap \bigl\{ {mg ( 0 ) + \bigl( {g ( 1 ) - mg ( 0 )} \bigr){x^{\alpha}} \ge{t_{2}}} \bigr\} } \bigr). \end{aligned}$$

(2.24)

Case (a). If $\frac{{f ( 1 )}}{{f ( 0 )}} < \frac{{g ( 1 )}}{{g ( 0 )}}$, then by (2.24) we obtain

Case 1. If $m \in ( {0,{\frac{{f ( 1 )} }{ {f ( 0 )}}}} )$, then

$$\begin{aligned} ( S ) \int_{0}^{1} {f ( x )g ( x )\,dx} \ge {t_{1}} {t_{2}} \wedge \biggl( {1 - {{ \biggl( { \frac{{{t_{1}} - mf ( 0 )}}{{f ( 1 ) - mf ( 0 )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr) \wedge \biggl( {1 - {{ \biggl( {\frac{{{t_{2}} - mg ( 0 )}}{{g ( 1 ) - mg ( 0 )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr). \end{aligned}$$

(2.25)

Case 2. If $m = {\frac{{f ( 1 )} }{{f ( 0 )}}}$, then

$$ ( S )\int_{0}^{1} {f ( x )g ( x )\,dx} \ge {t_{1}} {t_{2}} \wedge1 \wedge f ( 1 ) \wedge \biggl( {1 - {{ \biggl( {\frac{{{t_{2}} - mg ( 0 )}}{{g ( 1 ) - mg ( 0 )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr). $$

(2.26)

Case 3. If $m \in ( {{\frac{{f ( 1 )} }{{f ( 0 )}}},{\frac{{g ( 1 )} }{{g ( 0 )}}}} )$, then

$$ ( S ) \int_{0}^{1} {f ( x )g ( x )\,dx} \ge {t_{1}} {t_{2}} \wedge { \biggl( {\frac{{{t_{1}} - mf ( 0 )}}{{f ( 1 ) - mf ( 0 )}}} \biggr)^{\frac {1}{\alpha}}} \wedge \biggl( {1 - {{ \biggl( {\frac {{{t_{2}} - mg ( 0 )}}{{g ( 1 ) - mg ( 0 )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr). $$

(2.27)

Case 4. If $m = {\frac{{g ( 1 )} }{{g ( 0 )}}}$, then

$$ ( S )\int_{0}^{1} {f ( x )g ( x )\,dx} \ge {t_{1}} {t_{2}} \wedge{ \biggl( {\frac{{{t_{1}} - mf ( 0 )}}{{f ( 1 ) - mf ( 0 )}}} \biggr)^{\frac{1}{\alpha}}} \wedge g ( 1 )\wedge1. $$

(2.28)

Case 5. If $m \in ( {{\frac{{g ( 1 )} }{{g ( 0 )}}},1} )$, then

$$ ( S )\int_{0}^{1} {f ( x )g ( x )\,dx} \ge {t_{1}} {t_{2}} \wedge{ \biggl( {\frac{{{t_{1}} - mf ( 0 )}}{{f ( 1 ) - mf ( 0 )}}} \biggr)^{\frac{1}{\alpha}}} \wedge { \biggl( {\frac{{{t_{2}} - mg ( 0 )}}{{g ( 1 ) - mg ( 0 )}}} \biggr)^{\frac{1}{\alpha}}}. $$

(2.29)

Case (b). If $\frac{{f ( 1 )}}{{f ( 0 )}} = \frac{{g ( 1 )}}{{g ( 0 )}}$, then by (2.24) we obtain

Case 1. If $m \in ( {0,{\frac{{f ( 1 )} }{ {f ( 0 )}}}} )$, then

$$ ( S ) \int_{0}^{1} {f ( x )g ( x )\,dx} \ge {t_{1}} {t_{2}} \wedge \biggl( {1 - {{ \biggl( { \frac{{{t_{1}} - mf ( 0 )}}{{f ( 1 ) - mf ( 0 )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr) \wedge \biggl( {1 - {{ \biggl( {\frac{{{t_{2}} - mg ( 0 )}}{{g ( 1 ) - mg ( 0 )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr). $$

(2.30)

Case 2. If $m = {\frac{{f ( 1 )} }{{f ( 0 )}}}$, then

$$ ( S )\int_{0}^{1} {f ( x )g ( x )\,dx} \ge {t_{1}} {t_{2}} \wedge f ( 1 ) \wedge g ( 1 )\wedge1. $$

(2.31)

Case 3. If $m \in ( {{\frac{{f ( 1 )} }{{f ( 0 )}}},1} )$, then

$$ ( S )\int_{0}^{1} {f ( x )g ( x )\,dx} \ge {t_{1}} {t_{2}} \wedge{ \biggl( {\frac{{{t_{1}} - mf ( 0 )}}{{f ( 1 ) - mf ( 0 )}}} \biggr)^{\frac{1}{\alpha}}} \wedge { \biggl( {\frac{{{t_{2}} - mg ( 0 )}}{{g ( 1 ) - mg ( 0 )}}} \biggr)^{\frac{1}{\alpha}}}. $$

(2.32)

Case (c). If $\frac{{f ( 1 )}}{{f ( 0 )}} > \frac{{g ( 1 )}}{{g ( 0 )}}$, then by (2.24) we obtain

Case 1. If $m \in ( {0,{\frac{{g ( 1 )} }{ {g ( 0 )}}}} )$, then

$$\begin{aligned} ( S ) \int_{0}^{1} {f ( x )g ( x )\,dx} \ge{}& {t_{1}} {t_{2}} \wedge \biggl( {1 - {{ \biggl( { \frac{{{t_{1}} - mf ( 0 )}}{{f ( 1 ) - mf ( 0 )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr) \\ &{}\wedge \biggl( {1 - {{ \biggl( {\frac{{{t_{2}} - mg ( 0 )}}{{g ( 1 ) - mg ( 0 )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr). \end{aligned}$$

(2.33)

Case 2. If $m = {\frac{{g ( 1 )} }{{g ( 0 )}}}$, then

$$ ( S )\int_{0}^{1} {f ( x )g ( x )\,dx} \ge {t_{1}} {t_{2}} \wedge \biggl( {1 - {{ \biggl( { \frac{{{t_{1}} - mf ( 0 )}}{{f ( 1 ) - mf ( 0 )}}} \biggr)}^{\frac {1}{\alpha}}}} \biggr) \wedge g ( 1 )\wedge1. $$

(2.34)

Case 3. If $m \in ( {{\frac{{g ( 1 )} }{{g ( 0 )}}},{\frac{{f ( 1 )} }{{f ( 0 )}}}} )$, then

$$ ( S ) \int_{0}^{1} {f ( x )g ( x )\,dx} \ge {t_{1}} {t_{2}} \wedge \biggl( {1 - {{ \biggl( { \frac{{{t_{1}} - mf ( 0 )}}{{f ( 1 ) - mf ( 0 )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr) \wedge { \biggl( { \frac{{{t_{2}} - mg ( 0 )}}{{g ( 1 ) - mg ( 0 )}}} \biggr)^{\frac{1}{\alpha}}}. $$

(2.35)

Case 4. If $m = {\frac{{f ( 1 )} }{{f ( 0 )}}}$, then

$$ ( S )\int_{0}^{1} {f ( x )g ( x )\,dx} \ge {t_{1}} {t_{2}} \wedge1\wedge f ( 1 ) \wedge{ \biggl( { \frac{{{t_{2}} - mg ( 0 )}}{{g ( 1 ) - mg ( 0 )}}} \biggr)^{\frac{1}{\alpha}}}. $$

(2.36)

Case 5. If $m \in ( {{\frac{{f ( 1 )} }{{f ( 0 )}}},1} )$, then

$$ ( S )\int_{0}^{1} {f ( x )g ( x )\,dx} \ge {t_{1}} {t_{2}} \wedge{ \biggl( {\frac{{{t_{1}} - mf ( 0 )}}{{f ( 1 ) - mf ( 0 )}}} \biggr)^{\frac{1}{\alpha}}} \wedge { \biggl( {\frac{{{t_{2}} - mg ( 0 )}}{{g ( 1 ) - mg ( 0 )}}} \biggr)^{\frac{1}{\alpha}}}. $$

(2.37)

This completes the proof. □

Example

Consider $X = [ {0,1} ]$ and $p = 2$, $q = 4$. If we take the functions $f ( x ) =\sqrt[2]{x}$, $g ( x ) = \sqrt[4]{x}$, then $f ( x )$, $g ( x )$ are two $( {{\frac{2 }{3}},{\frac{1 }{3}}} )$ -concave functions. In fact, $\sqrt[t]{x} = f ( {x \cdot1 + {\frac{1 }{3}} ( {1 - x} ) \cdot0} ) \ge{x^{{\frac{2 }{3}}}}f ( 1 ) + {\frac{1 }{3}} ( {1 - {x^{{\frac{2 }{ 3}}}}} )f ( 0 ) = {x^{{\frac{2 }{3}}}}$ for $t \ge{\frac{3 }{2}}$. Let m be the Lebesgue measure on X. A straightforward calculus shows that

$$( S )\int_{0}^{1} {{f^{2}} ( x )\,dm} = ( S )\int_{0}^{1} {{g^{4}} ( x )\,dm} = 0.5,\qquad ( S )\int_{0}^{1} {f ( x )g ( x )\,dm} = 0.5497. $$

By Theorem 2.6, we have

$$\begin{aligned} 0.5497 ={}& ( S )\int_{0}^{1} f ( x )g ( x )\,dx \ge \biggl( ( S )\int_{0}^{1} {{f^{2}} ( x )\,dx} \biggr)^{\frac{1}{2}} \biggl( ( S )\int _{0}^{1} {{g^{4}} ( x )\,dx} \biggr)^{\frac{1}{4}} \\ &{} \wedge \biggl( 1 - \biggl( \biggl( { ( S )\int_{0}^{1} {{f^{2}} ( x )\,dx} } \biggr)^{\frac{1}{2}} \biggr)^{\frac{3}{2}} \biggr) \wedge \biggl( 1 - \biggl( \biggl( { ( S )\int _{0}^{1} {{g^{4}} ( x )\,dx} } \biggr)^{\frac{1}{4}} \biggr)^{\frac{3}{2}} \biggr) \\ ={}& 0.0156 \wedge0.8750 \wedge0.9844=0.0156. \end{aligned}$$

(2.38)

Now, we will prove the general cases of Theorem 2.6.

Theorem 2.7

Let $X= [ {a,b} ]$, $\alpha,m \in ( {0,1} )$ and f, g be $( {\alpha,m} )$-concave functions for all $x \in X$. If μ is a Lebesgue measure on X, then

Case (i). If $f ( a ) \le f ( b )$ and $g ( a ) \le g ( b )$, then

$$\begin{aligned} ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge{}& {t_{1}} {t_{2}} \wedge \biggl( { ( {b - ma} ) \biggl( {1 - {{ \biggl( {\frac{{{t_{1}} - mf ( a )}}{{f ( b ) - mf ( a )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr)} \biggr) \\ &{}\wedge \biggl( { ( {b - ma} ) \biggl( 1 - \biggl( {\frac {{{t_{2}} - mg ( a )}}{{g ( b ) - mg ( a )}}} \biggr)^{\frac{1}{\alpha}} \biggr)} \biggr), \end{aligned}$$

(2.39)

where ${t_{1}} = { ( { ( S )\int_{a}^{b} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{a}^{b} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case (ii). If $f (a ) > f ( b )$ and $g ( a ) > g ( b )$, then

Case (a). If $\frac{{f ( b )}}{{f ( a )}} < \frac{{g ( b )}}{{g ( a )}}$, then

Case 1. If $m \in ( {0,{\frac{{f ( b )} }{ {f ( a )}}}} )$, then

$$\begin{aligned} ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge{}& {t_{1}} {t_{2}} \wedge \biggl( { ( {b - ma} ) \biggl( {1 - {{ \biggl( {\frac{{{t_{1}} - mf ( a )}}{{f ( b ) - mf ( a )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr)} \biggr) \\ &{}\wedge \biggl( { ( {b - ma} ) \biggl( {1 - {{ \biggl( {\frac {{{t_{2}} - mg ( a )}}{{g ( b ) - mg ( a )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr)} \biggr), \end{aligned}$$

(2.40)

where ${t_{1}} = { ( { ( S )\int_{a}^{b} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{a}^{b} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case 2. If $m = {\frac{{f ( b )} }{{f ( a )}}}$, then

$$\begin{aligned} ( S ) \int_{a}^{b} {f ( x )g ( x )\,dx} \ge{}&{t_{1}} {t_{2}} \wedge ( {b - a} ) \wedge f ( b ) \\ &{}\wedge \biggl( { ( {b - ma} ) \biggl( {1 - {{ \biggl( {\frac{{{t_{2}} - mg ( a )}}{{g ( b ) - mg ( a )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr)} \biggr), \end{aligned}$$

(2.41)

where ${t_{1}} = { ( { ( S )\int_{a}^{b} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{a}^{b} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case 3. If $m \in ( {{\frac{{f ( b )} }{{f ( a )}}},{\frac{{g ( b )} }{{g ( a )}}}} )$, then

$$ \begin{aligned}[b] ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge{}& {t_{1}} {t_{2}} \wedge \biggl( { ( {b - ma} ){{ \biggl( { \frac{{{t_{1}} - mf ( a )}}{{f ( b ) - mf ( a )}}} \biggr)}^{\frac{1}{\alpha}}} + ma-a} \biggr)\\ &{}\wedge \biggl( { ( {b - ma} ) \biggl( {1 - {{ \biggl( {\frac {{{t_{2}} - mg ( a )}}{{g ( b ) - mg ( a )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr)} \biggr), \end{aligned} $$

(2.42)

where ${t_{1}} = { ( { ( S )\int_{a}^{b} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{a}^{b} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case 4. If $m = {\frac{{g ( b )} }{{g ( a )}}}$, then

$$\begin{aligned} ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge {t_{1}} {t_{2}} \wedge g ( b ) \wedge ( {b - a} ) \wedge \biggl( { ( {b - ma} ){{ \biggl( {\frac{{{t_{1}} - mf ( a )}}{{f ( b ) - mf ( a )}}} \biggr)}^{\frac {1}{\alpha}}} + ma-a} \biggr), \end{aligned}$$

(2.43)

where ${t_{1}} = { ( { ( S )\int_{a}^{b} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{a}^{b} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case 5. If $m \in ( {{\frac{{g ( b )} }{{g ( a )}}},1} )$, then

$$\begin{aligned} ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge{}& {t_{1}} {t_{2}} \wedge \biggl( { ( {b - ma} ){{ \biggl( { \frac{{{t_{1}} - mf ( a )}}{{f ( b ) - mf ( a )}}} \biggr)}^{\frac{1}{\alpha}}} + ma-a} \biggr) \\ &{}\wedge \biggl( { ( {b - ma} ){{ \biggl( {\frac{{{t_{1}} - mg ( a )}}{{g ( b ) - mg ( a )}}} \biggr)}^{\frac {1}{\alpha}}} + ma-a} \biggr), \end{aligned}$$

(2.44)

where ${t_{1}} = { ( { ( S )\int_{a}^{b} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{a}^{b} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case (b). If $\frac{{f ( b )}}{{f ( a )}} = \frac{{g ( b )}}{{g ( a )}}$, then

Case 1. If $m \in ( {0,{\frac{{f ( b )} }{ {f ( a )}}}} )$, then

$$\begin{aligned} ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge{}& {t_{1}} {t_{2}} \wedge \biggl( { ( {b - ma} ) \biggl( {1 - {{ \biggl( {\frac{{{t_{1}} - mf ( a )}}{{f ( b ) - mf ( a )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr)} \biggr) \\ &{}\wedge \biggl( { ( {b - ma} ) \biggl( {1 - {{ \biggl( {\frac {{{t_{2}} - mg ( a )}}{{g ( b ) - mg ( a )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr)} \biggr), \end{aligned}$$

(2.45)

where ${t_{1}} = { ( { ( S )\int_{a}^{b} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{a}^{b} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case 2. If $m = {\frac{{f ( b )} }{{f ( a )}}}$, then

$$ ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge {t_{1}} {t_{2}} \wedge f ( b ) \wedge g ( b ) \wedge ( {b - a} ), $$

(2.46)

where ${t_{1}} = { ( { ( S )\int_{a}^{b} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{a}^{b} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case 3. If $m \in ( {{\frac{{f ( b )} }{ {f ( a )}}},1} )$, then

$$\begin{aligned} ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge{}& {t_{1}} {t_{2}} \wedge \biggl( { ( {b - ma} ){{ \biggl( { \frac{{{t_{1}} - mf ( a )}}{{f ( b ) - mf ( a )}}} \biggr)}^{\frac{1}{\alpha}}} + ma-a} \biggr) \\ &{}\wedge \biggl( { ( {b - ma} ){{ \biggl( {\frac{{{t_{1}} - mg ( a )}}{{g ( b ) - mg ( a )}}} \biggr)}^{\frac {1}{\alpha}}} + ma-a} \biggr), \end{aligned}$$

(2.47)

where ${t_{1}} = { ( { ( S )\int_{a}^{b} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{a}^{b} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case (c). If $\frac{{f ( b )}}{{f ( a )}} > \frac{{g ( b )}}{{g ( a )}}$, then

Case 1. If $m \in ( {0,{\frac{{g ( b )} }{ {g ( a )}}}} )$, then

$$ \begin{aligned}[b] ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge{}& {t_{1}} {t_{2}} \wedge \biggl( { ( {b - ma} ) \biggl( {1 - {{ \biggl( {\frac{{{t_{1}} - mf ( a )}}{{f ( b ) - mf ( a )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr)} \biggr)\\ &{}\wedge \biggl( { ( {b - ma} ) \biggl( {1 - {{ \biggl( {\frac {{{t_{2}} - mg ( a )}}{{g ( b ) - mg ( a )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr)} \biggr), \end{aligned} $$

(2.48)

where ${t_{1}} = { ( { ( S )\int_{a}^{b} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{a}^{b} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case 2. If $m = {\frac{{g ( b )} }{{g ( a )}}}$, then

$$\begin{aligned} ( S ) \int_{a}^{b} {f ( x )g ( x )\,dx} \ge{t_{1}} {t_{2}} \wedge \biggl( { ( {b - ma} ) \biggl( {1 - {{ \biggl( {\frac{{{t_{1}} - mf ( a )}}{{f ( b ) - mf ( a )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr)} \biggr) \wedge g ( b ) \wedge ( {b - a} ), \end{aligned}$$

(2.49)

where ${t_{1}} = { ( { ( S )\int_{a}^{b} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{a}^{b} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case 3. If $m \in ( {{\frac{{g ( b )} }{{g ( a )}}},{\frac{{f ( b )} }{{f ( a )}}}} )$, then

$$\begin{aligned} ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge{}& {t_{1}} {t_{2}} \wedge \biggl( { ( {b - ma} ) \biggl( {1 - {{ \biggl( {\frac{{{t_{1}} - mf ( a )}}{{f ( b ) - mf ( a )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr)} \biggr) \\ &{}\wedge \biggl( { ( {b - ma} ){{ \biggl( {\frac{{{t_{1}} - mg ( a )}}{{g ( b ) - mg ( a )}}} \biggr)}^{\frac {1}{\alpha}}} + ma-a} \biggr), \end{aligned}$$

(2.50)

where ${t_{1}} = { ( { ( S )\int_{a}^{b} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{a}^{b} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case 4. If $m = {\frac{{f ( b )} }{{f ( a )}}}$, then

$$\begin{aligned} ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge {t_{1}} {t_{2}} \wedge ( {b - a} )\wedge f ( b ) \wedge \biggl( { ( {b - ma} ){{ \biggl( {\frac{{{t_{1}} - mg ( a )}}{{g ( b ) - mg ( a )}}} \biggr)}^{\frac {1}{\alpha}}} + ma -a} \biggr), \end{aligned}$$

(2.51)

where ${t_{1}} = { ( { ( S )\int_{a}^{b} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{a}^{b} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case 5. If $m \in ( {{\frac{{f ( b )} }{{f ( a )}}},1} )$, then

$$\begin{aligned} ( S ) \int_{a}^{b} {f ( x )g ( x )\,dx} \ge{}&{t_{1}} {t_{2}} \wedge \biggl( { ( {b - ma} ){{ \biggl( { \frac{{{t_{1}} - mf ( a )}}{{f ( b ) - mf ( a )}}} \biggr)}^{\frac{1}{\alpha}}} + ma-a} \biggr) \\ &{} \wedge \biggl( { ( {b - ma} ){{ \biggl( {\frac{{{t_{1}} - mg ( a )}}{{g ( b ) - mg ( a )}}} \biggr)}^{\frac{1}{\alpha}}} + ma-a} \biggr), \end{aligned}$$

(2.52)

where ${t_{1}} = { ( { ( S )\int_{a}^{b} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{a}^{b} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Proof

Let $p,q \in ( {0,\infty} )$, ${ ( { ( S )\int_{a}^{b} {{f^{p}} ( x )\,dx} } )^{{\frac{1 }{ p}}}} = {t_{1}}$ and ${ ( { ( S )\int_{a}^{b} {{g^{q}} ( x )\,dx} } )^{{\frac{1 }{ q}}}} = {t_{2}}$. Since $f,g: [ {a,b} ] \to [ {0,\infty} )$ are two $( {\alpha,m} )$-concave functions for $x \in [ {a,b} ]$, we have

$$\begin{aligned}& \begin{aligned}[b] f ( x ) &= f \biggl( {m \biggl( {1 - \frac{{x - ma}}{{b - ma}}} \biggr) \cdot a + \frac{{x - ma}}{{b - ma}} \cdot b} \biggr)\\ &\ge m \biggl( 1 - \biggl( {\frac{{x - ma}}{{b - ma}}} \biggr)^{\alpha}\biggr)f ( a ) + { \biggl( {\frac{{x - ma}}{{b - ma}}} \biggr)^{\alpha}}f ( b ) = {h_{1}} ( x ), \end{aligned} \end{aligned}$$

(2.53)

$$\begin{aligned}& \begin{aligned}[b] g ( x ) &= g \biggl( {m \biggl( {1 - \frac{{x - ma}}{{b - ma}}} \biggr) \cdot a + \frac{{x - ma}}{{b - ma}} \cdot b} \biggr)\\ &\ge m \biggl( {1 - { \biggl( {\frac{{x - ma}}{{b - ma}}} \biggr)^{\alpha}}} \biggr)g ( a ) + { \biggl( {\frac{{x - ma}}{{b - ma}}} \biggr)^{\alpha}}g ( b ) = {h_{2}} ( x ). \end{aligned} \end{aligned}$$

(2.54)

Case (i). If $f ( a ) \le f ( b )$ and $g ( a ) \le g ( b )$, then by (3) of Proposition 2.3 and the co-monotonicity of ${h_{1}} ( x )$ and ${h_{2}} ( x )$, we have

$$\begin{aligned} & ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \\ &\quad\ge ( S ) \int_{a}^{b} {{h_{1}} ( x ){h_{2}} ( x )\,dx} \\ &\quad= \bigvee _{\beta \ge0} \bigl( {\beta \wedge\mu \bigl( { [ {a,b} ] \cap \bigl\{ {{h_{1}} ( x ){h_{2}} ( x ) \ge\beta} \bigr\} } \bigr)} \bigr) \\ &\quad\ge {t_{1}} {t_{2}} \wedge\mu \bigl( { [ {a,b} ] \cap \bigl\{ {{h_{1}} ( x ){h_{2}} ( x ) \ge{t_{1}} {t_{2}}} \bigr\} } \bigr) \\ &\quad\ge {t_{1}} {t_{2}} \wedge\mu \bigl( { [ {a,b} ] \cap \bigl\{ {{h_{1}} ( x ) \ge{t_{1}}} \bigr\} \cap \bigl\{ {{h_{2}} ( x ) \ge{t_{2}}} \bigr\} } \bigr) \\ &\quad= {t_{1}} {t_{2}} \wedge\mu \bigl( { [ {a,b} ] \cap \bigl\{ {{h_{1}} ( x ) \ge{t_{1}}} \bigr\} } \bigr) \wedge\mu \bigl( { [ {a,b} ] \cap \bigl\{ {{h_{2}} ( x ) \ge{t_{2}}} \bigr\} } \bigr) \\ &\quad= {t_{1}} {t_{2}} \wedge\mu \bigl( { [ {a,b} ] \cap \bigl\{ {{h_{1}} ( x ) \ge{t_{1}}} \bigr\} } \bigr) \wedge\mu \bigl( { [ {a,b} ] \cap \bigl\{ {{h_{2}} ( x ) \ge{t_{2}}} \bigr\} } \bigr) \\ &\quad= {t_{1}} {t_{2}} \wedge\mu \biggl( { [ {a,b} ] \cap \biggl\{ {m \biggl( {1 - {{ \biggl( {\frac{{x - ma}}{{b - ma}}} \biggr)}^{\alpha}}} \biggr)f ( a ) + {{ \biggl( {\frac{{x - ma}}{{b - ma}}} \biggr)}^{\alpha}}f ( b ) \ge{t_{1}}} \biggr\} } \biggr) \\ &\qquad{} \wedge\mu \biggl( { [ {a,b} ] \cap \biggl\{ {m \biggl( {1 - {{ \biggl( { \frac{{x - ma}}{{b - ma}}} \biggr)}^{\alpha}}} \biggr)g ( a ) + {{ \biggl( { \frac{{x - ma}}{{b - ma}}} \biggr)}^{\alpha}}g ( b ) \ge{t_{2}}} \biggr\} } \biggr) \\ &\quad= {t_{1}} {t_{2}} \wedge\mu \biggl( { [ {a,b} ] \cap \biggl\{ {x \ge{{ \biggl( {\frac{{{t_{1}} - mf ( a )}}{{f ( b ) - mf ( a )}}} \biggr)}^{\frac{1}{\alpha}}} ( {b - ma} ) + ma} \biggr\} } \biggr) \\ &\qquad{} \wedge\mu \biggl( { [ {a,b} ] \cap \biggl\{ {x \ge{{ \biggl( { \frac{{{t_{2}} - mg ( a )}}{{g ( b ) - mg ( a )}}} \biggr)}^{\frac{1}{\alpha}}} ( {b - ma} ) + ma} \biggr\} } \biggr) \\ &\quad= {t_{1}} {t_{2}} \wedge \biggl( { ( {b - ma} ) \biggl( {1 - {{ \biggl( {\frac{{{t_{1}} - mf ( a )}}{{f ( b ) - mf ( a )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr)} \biggr) \\ &\qquad{}\wedge \biggl( { ( {b - ma} ) \biggl( {1 - {{ \biggl( {\frac {{{t_{2}} - mg ( a )}}{{g ( b ) - mg ( a )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr)} \biggr). \end{aligned}$$

(2.55)

Case (ii). If $f (a ) > f ( b )$ and $g ( a ) > g ( b )$, then by (3) of Proposition 2.3 and the co-monotonicity of ${h_{1}} ( x )$ and ${h_{2}} ( x )$, we have

$$\begin{aligned} & ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \\ &\quad\ge ( S ) \int_{a}^{b} {{h_{1}} ( x ){h_{2}} ( x )\,dx} \\ &\quad= \bigvee _{\beta \ge0} \bigl( {\beta \wedge\mu \bigl( { [ {a,b} ] \cap \bigl\{ {{h_{1}} ( x ){h_{2}} ( x ) \ge\beta} \bigr\} } \bigr)} \bigr) \\ &\quad\ge {t_{1}} {t_{2}} \wedge\mu \bigl( { [ {a,b} ] \cap \bigl\{ {{h_{1}} ( x ){h_{2}} ( x ) \ge{t_{1}} {t_{2}}} \bigr\} } \bigr) \\ &\quad\ge {t_{1}} {t_{2}} \wedge\mu \bigl( { [ {a,b} ] \cap \bigl\{ {{h_{1}} ( x ) \ge{t_{1}}} \bigr\} \cap \bigl\{ {{h_{2}} ( x ) \ge{t_{2}}} \bigr\} } \bigr) \\ &\quad= {t_{1}} {t_{2}} \wedge\mu \bigl( { [ {a,b} ] \cap \bigl\{ {{h_{1}} ( x ) \ge{t_{1}}} \bigr\} } \bigr) \wedge\mu \bigl( { [ {a,b} ] \cap \bigl\{ {{h_{2}} ( x ) \ge{t_{2}}} \bigr\} } \bigr) \\ &\quad= {t_{1}} {t_{2}} \wedge\mu \bigl( { [ {a,b} ] \cap \bigl\{ {{h_{1}} ( x ) \ge{t_{1}}} \bigr\} } \bigr) \wedge\mu \bigl( { [ {a,b} ] \cap \bigl\{ {{h_{2}} ( x ) \ge{t_{2}}} \bigr\} } \bigr) \\ &\quad= {t_{1}} {t_{2}} \wedge\mu \biggl( { [ {a,b} ] \cap \biggl\{ {m \biggl( {1 - {{ \biggl( {\frac{{x - ma}}{{b - ma}}} \biggr)}^{\alpha}}} \biggr)f ( a ) + {{ \biggl( {\frac{{x - ma}}{{b - ma}}} \biggr)}^{\alpha}}f ( b ) \ge{t_{1}}} \biggr\} } \biggr) \\ &\qquad{} \wedge\mu \biggl( { [ {a,b} ] \cap \biggl\{ {m \biggl( {1 - {{ \biggl( { \frac{{x - ma}}{{b - ma}}} \biggr)}^{\alpha}}} \biggr)g ( a ) + {{ \biggl( { \frac{{x - ma}}{{b - ma}}} \biggr)}^{\alpha}}g ( b ) \ge{t_{2}}} \biggr\} } \biggr) \\ &\quad= {t_{1}} {t_{2}} \wedge\mu \biggl( { [ {a,b} ] \cap \biggl\{ {mf ( a ) + \bigl( {f ( b ) - mf ( a )} \bigr){{ \biggl( {\frac{{x - ma}}{{b - ma}}} \biggr)}^{\alpha}} \ge {t_{1}}} \biggr\} } \biggr) \\ &\qquad{} \wedge\mu \biggl( { [ {a,b} ] \cap \biggl\{ {mg ( a ) + \bigl( {g ( b ) - mg ( a )} \bigr){{ \biggl( {\frac{{x - ma}}{{b - ma}}} \biggr)}^{\alpha}} \ge{t_{2}}} \biggr\} } \biggr). \end{aligned}$$

(2.56)

Case (a). If $\frac{{f ( b )}}{{f ( a )}} < \frac{{g ( b )}}{{g ( a )}}$, then by (2.56) we obtain

Case 1. If $m \in ( {0,{\frac{{f ( b )} }{ {f ( a )}}}} )$, then

$$\begin{aligned} ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge{}& {t_{1}} {t_{2}} \wedge \biggl( { ( {b - ma} ) \biggl( {1 - {{ \biggl( {\frac{{{t_{1}} - mf ( a )}}{{f ( b ) - mf ( a )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr)} \biggr) \\ &{}\wedge \biggl( { ( {b - ma} ) \biggl( {1 - {{ \biggl( {\frac {{{t_{2}} - mg ( a )}}{{g ( b ) - mg ( a )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr)} \biggr). \end{aligned}$$

(2.57)

Case 2. If $m = {\frac{{f ( b )} }{{f ( a )}}}$, then

$$\begin{aligned} ( S ) \int_{a}^{b} {f ( x )g ( x )\,dx} \ge{}&{t_{1}} {t_{2}} \wedge ( {b - a} ) \wedge f ( b ) \\ &{}\wedge \biggl( { ( {b - ma} ) \biggl( {1 - {{ \biggl( {\frac{{{t_{2}} - mg ( a )}}{{g ( b ) - mg ( a )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr)} \biggr). \end{aligned}$$

(2.58)

Case 3. If $m \in ( {{\frac{{f ( b )} }{{f ( a )}}},{\frac{{g ( b )} }{{g ( a )}}}} )$, then

$$\begin{aligned} ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge{}& {t_{1}} {t_{2}} \wedge \biggl( { ( {b - ma} ){{ \biggl( { \frac{{{t_{1}} - mf ( a )}}{{f ( b ) - mf ( a )}}} \biggr)}^{\frac{1}{\alpha}}} + ma-a} \biggr) \\ &{}\wedge \biggl( { ( {b - ma} ) \biggl( {1 - {{ \biggl( {\frac {{{t_{2}} - mg ( a )}}{{g ( b ) - mg ( a )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr)} \biggr). \end{aligned}$$

(2.59)

Case 4. If $m = {\frac{{g ( b )} }{{g ( a )}}}$, then

$$\begin{aligned} ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge{}& {t_{1}} {t_{2}} \wedge g ( b ) \wedge ( {b - a} ) \\ &{}\wedge \biggl( { ( {b - ma} ){{ \biggl( {\frac{{{t_{1}} - mf ( a )}}{{f ( b ) - mf ( a )}}} \biggr)}^{\frac {1}{\alpha}}} + ma-a} \biggr). \end{aligned}$$

(2.60)

Case 5. If $m \in ( {{\frac{{g ( b )} }{{g ( a )}}},1} )$, then

$$\begin{aligned} ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge{}& {t_{1}} {t_{2}} \wedge \biggl( { ( {b - ma} ){{ \biggl( { \frac{{{t_{1}} - mf ( a )}}{{f ( b ) - mf ( a )}}} \biggr)}^{\frac{1}{\alpha}}} + ma-a} \biggr) \\ &{}\wedge \biggl( { ( {b - ma} ){{ \biggl( {\frac{{{t_{1}} - mg ( a )}}{{g ( b ) - mg ( a )}}} \biggr)}^{\frac {1}{\alpha}}} + ma-a} \biggr). \end{aligned}$$

(2.61)

Case (b). If $\frac{{f ( b )}}{{f ( a )}} = \frac{{g ( b )}}{{g ( a )}}$, then by (2.56) we obtain

Case 1. If $m \in ( {0,{\frac{{f ( b )} }{ {f ( a )}}}} )$, then

$$\begin{aligned} ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge{}& {t_{1}} {t_{2}} \wedge \biggl( { ( {b - ma} ) \biggl( {1 - {{ \biggl( {\frac{{{t_{1}} - mf ( a )}}{{f ( b ) - mf ( a )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr)} \biggr) \\ &{}\wedge \biggl( { ( {b - ma} ) \biggl( {1 - {{ \biggl( {\frac {{{t_{2}} - mg ( a )}}{{g ( b ) - mg ( a )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr)} \biggr). \end{aligned}$$

(2.62)

Case 2. If $m = {\frac{{f ( b )} }{{f ( a )}}}$, then

$$ ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge {t_{1}} {t_{2}} \wedge f ( b ) \wedge g ( b ) \wedge ( {b - a} ). $$

(2.63)

Case 3. If $m \in ( {{\frac{{f ( b )} }{{f ( a )}}},1} )$, then

$$\begin{aligned} ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge{}& {t_{1}} {t_{2}} \wedge \biggl( { ( {b - ma} ){{ \biggl( { \frac{{{t_{1}} - mf ( a )}}{{f ( b ) - mf ( a )}}} \biggr)}^{\frac{1}{\alpha}}} + ma-a} \biggr) \\ &{}\wedge \biggl( { ( {b - ma} ){{ \biggl( {\frac{{{t_{1}} - mg ( a )}}{{g ( b ) - mg ( a )}}} \biggr)}^{\frac {1}{\alpha}}} + ma-a} \biggr). \end{aligned}$$

(2.64)

Case (c). If $\frac{{f ( b )}}{{f ( a )}} > \frac{{g ( b )}}{{g ( a )}}$, then by (2.56) we obtain

Case 1. If $m \in ( {0,{\frac{{g ( b )} }{ {g ( a )}}}} )$, then

$$\begin{aligned} ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge{}& {t_{1}} {t_{2}} \wedge \biggl( { ( {b - ma} ) \biggl( {1 - {{ \biggl( {\frac{{{t_{1}} - mf ( a )}}{{f ( b ) - mf ( a )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr)} \biggr) \\ &{}\wedge \biggl( { ( {b - ma} ) \biggl( {1 - {{ \biggl( {\frac {{{t_{2}} - mg ( a )}}{{g ( b ) - mg ( a )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr)} \biggr). \end{aligned}$$

(2.65)

Case 2. If $m = {\frac{{g ( b )} }{{g ( a )}}}$, then

$$\begin{aligned} ( S ) \int_{a}^{b} {f ( x )g ( x )\,dx} \ge{}&{t_{1}} {t_{2}} \wedge \biggl( { ( {b - ma} ) \biggl( {1 - {{ \biggl( {\frac{{{t_{1}} - mf ( a )}}{{f ( b ) - mf ( a )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr)} \biggr) \\ &{} \wedge g ( b ) \wedge ( {b - a} ). \end{aligned}$$

(2.66)

Case 3. If $m \in ( {{\frac{{g ( b )} }{{g ( a )}}},{\frac{{f ( b )} }{{f ( a )}}}} )$, then

$$ \begin{aligned}[b] ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge{}& {t_{1}} {t_{2}} \wedge \biggl( { ( {b - ma} ) \biggl( {1 - {{ \biggl( {\frac{{{t_{1}} - mf ( a )}}{{f ( b ) - mf ( a )}}} \biggr)}^{\frac{1}{\alpha}}}} \biggr)} \biggr)\\ &{}\wedge \biggl( { ( {b - ma} ){{ \biggl( {\frac{{{t_{1}} - mg ( a )}}{{g ( b ) - mg ( a )}}} \biggr)}^{\frac {1}{\alpha}}} + ma-a} \biggr). \end{aligned} $$

(2.67)

Case 4. If $m = {\frac{{f ( b )} }{{f ( a )}}}$, then

$$\begin{aligned} ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge {t_{1}} {t_{2}} \wedge f ( b ) \wedge ( {b - a} ) \wedge \biggl( { ( {b - ma} ){{ \biggl( {\frac{{{t_{1}} - mg ( a )}}{{g ( b ) - mg ( a )}}} \biggr)}^{\frac {1}{\alpha}}} + ma-a} \biggr). \end{aligned}$$

(2.68)

Case 5. If $m \in ( {{\frac{{f ( b )} }{{f ( a )}}},1} )$, then

$$\begin{aligned} ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge{}& {t_{1}} {t_{2}} \wedge \biggl( { ( {b - ma} ){{ \biggl( { \frac{{{t_{1}} - mf ( a )}}{{f ( b ) - mf ( a )}}} \biggr)}^{\frac{1}{\alpha}}} + ma-a} \biggr) \\ &{}\wedge \biggl( { ( {b - ma} ){{ \biggl( {\frac{{{t_{1}} - mg ( a )}}{{g ( b ) - mg ( a )}}} \biggr)}^{\frac {1}{\alpha}}} + ma-a} \biggr). \end{aligned}$$

(2.69)

This completes the proof. □

Example

Consider $X = [ {2,5} ]$ and $p = 2$, $q = 4$. Let m be the Lebesgue measure on X. If we take the function $f ( x ) = \sqrt[2]{{6 - x}}$, $g ( x ) = \sqrt[4]{{6 - x}}$, then $f ( x )$, $g ( x )$ are two $( {{\frac{1 }{2}},{\frac{{\sqrt{5} } }{2}}} )$ -concave functions. In fact,

$$\begin{aligned} \sqrt{6 - x} &= f \biggl( {{\frac{{\sqrt{5} } }{4}} \cdot \biggl( {1 - { \frac{{x - {\frac{{\sqrt{5} } }{4}} \times2} }{ {5 - {\frac{{\sqrt{5} } }{4}} \times2}}}} \biggr)2 + \biggl( {{\frac{{x - {\frac{{\sqrt{5} } }{4}} \times2} }{{5 - {\frac{{\sqrt{5} } }{4}} \times2}}}} \biggr) \cdot5} \biggr) \\ &\ge \biggl( \frac{{x - {\frac{{\sqrt{5} } }{4}} \times 2} }{{5 - {\frac{{\sqrt{5} } }{4}} \times2}} \biggr)^{{\frac{1 }{2}}}f ( 5 ) + { \frac{{\sqrt{5} } }{4}} \biggl( 1 - \biggl( {{\frac{{x - {\frac{{\sqrt{5} } }{4}} \times2} }{{5 - {\frac{{\sqrt{5} } }{4}} \times 2}}}} \biggr)^{{\frac{1 }{2}}} \biggr)f ( 2 ) \\ &= \frac{{ ( {2 - \sqrt{5} } )x + 4\sqrt{5} }}{{10 - \sqrt{5} }} \end{aligned}$$

(2.70)

and

$$\begin{aligned} \sqrt[4]{{6 - x}} &= g \biggl( {{\frac{{\sqrt{5} } }{4}} \cdot \biggl( {1 - { \frac{{x - {\frac{{\sqrt{5} } }{4}} \times2} }{{5 - {\frac{{\sqrt{5} } }{4}} \times2}}}} \biggr) \times 2 + \biggl( {{\frac{{x - {\frac{{\sqrt{5} } }{4}} \times2} }{{5 - {\frac{{\sqrt{5} } }{4}} \times2}}}} \biggr) \times 5} \biggr) \\ &\ge { \biggl( {{\frac{{x - {\frac{{\sqrt{5} } }{4}} \times 2} }{{5 - {\frac{{\sqrt{5} } }{4}} \times2}}}} \biggr)^{{\frac{1 }{2}}}}g ( 5 ) + { \frac{{\sqrt{5} } }{4}} \biggl( 1 - \biggl( {{\frac{{x - {\frac{{\sqrt{5} } }{4}} \times2} }{{5 - {\frac{{\sqrt{5} } }{4}} \times 2}}}} \biggr)^{{\frac{1 }{2}}} \biggr)g ( 2 ) \\ &= \frac{{ ( {4 - \sqrt{10} } )x + 5\sqrt{10} - 5}}{{20 - 2\sqrt{5} }}. \end{aligned}$$

(2.71)

A straightforward calculus shows that

$$( S )\int_{2}^{5} {{f^{2}} ( x )\,dm} = ( S )\int_{2}^{5} {{g^{4}} ( x )\,dm} = 2,\qquad ( S )\int_{2}^{5} {f ( x )g ( x )\,dm} = 1.8040. $$

By Theorem 2.7, we have

$$\begin{aligned} 1.8040 ={}& ( S )\int_{2}^{5} {f ( x )g ( x )\,dx} \ge \biggl( { ( S )\int_{2}^{5} {{f^{2}} ( x )\,dx} } \biggr)^{\frac{1}{2}} \biggl( { ( S )\int_{2}^{5} {{g^{4}} ( x )\,dx} } \biggr)^{\frac{1}{4}} \\ &{}\wedge \biggl( { \biggl( {5 - {\frac{{\sqrt{5} } }{4}} \times2} \biggr) \biggl( { \frac{{{{ ( { ( S )\int_{2}^{5} {{f^{2}} ( x )\,dx} } )}^{\frac{1}{2}}} - {\frac{{\sqrt{5} } }{ 4}}f ( 2 )}}{{f ( 5 ) - {\frac{{\sqrt{5} } }{4}}f ( 2 )}}} \biggr)^{2} + {\frac{{\sqrt{5} } }{4}} \times2 - 2} \biggr) \\ &{}\wedge \biggl( \biggl( {5 - {\frac{{\sqrt{5} } }{4}} \times2} \biggr) \biggl( { \frac{{{{ ( { ( S )\int_{2}^{5} {{g^{4}} ( x )\,dx} } )}^{\frac{1}{4}}} - {\frac{{\sqrt{5} } }{ 4}}g ( 2 )}}{{g ( 5 ) - {\frac{{\sqrt{5} } }{4}}g ( 2 )}}} \biggr)^{2} \biggr) \\ ={}& 1.6818 \wedge23.5607 \wedge34.4227 = 1.6818. \end{aligned}$$

(2.72)

As some special cases of $(\alpha,m )$-concave functions in Theorem 2.7, we have the following results.

Remark 2.8

Let $X= [ {a,b} ]$, $\alpha= m= 0$ and f, g be two decreasing functions for all $x \in X$. If μ is a Lebesgue measure on X, then

$$\begin{aligned} ( S ) \int_{a}^{b} {f ( x )g ( x )\,dx} \ge { \biggl( { ( S ) \int_{a}^{b} {{f^{p}} ( x )\,dx} } \biggr)^{\frac{1}{p}}} { \biggl( { ( S ) \int_{a}^{b} {{g^{q}} ( x )\,dx} } \biggr)^{\frac{1}{q}}} \wedge f ( b ) \wedge g ( b ) \wedge ( {b - a} ). \end{aligned}$$

(2.73)

Remark 2.9

Let $X= [ {a,b} ]$, $\alpha=1$, $m= 0$ and f, g be two starshaped functions for all $x \in X$. If μ is a Lebesgue measure on X, then

$$ ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge {t_{1}} {t_{2}} \wedge b \biggl( {1 - {\frac{{{t_{1}}} }{{f ( b )}}}} \biggr) \wedge b \biggl( {1 - {\frac{{{t_{2}}} }{ {g ( b )}}}} \biggr), $$

(2.74)

where ${t_{1}} = { ( { ( S )\int_{a}^{b} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{a}^{b} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Remark 2.10

Let $X= [ {a,b} ]$, $\alpha=1$, $m\in ({0,1} )$ and f, g be two m-concave functions for all $x \in X$. If μ is a Lebesgue measure on X, then

Case (i). If $f ( a ) \le f ( b )$ and $g ( a ) \le g ( b )$, then

$$\begin{aligned} ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge{}& {t_{1}} {t_{2}} \wedge \biggl( { ( {b - ma} ) \biggl( {1 - {{ \biggl( {\frac{{{t_{1}} - mf ( a )}}{{f ( b ) - mf ( a )}}} \biggr)}}} \biggr)} \biggr) \\ &{}\wedge \biggl( { ( {b - ma} ) \biggl( {1 - {{ \biggl( {\frac {{{t_{2}} - mg ( a )}}{{g ( b ) - mg ( a )}}} \biggr)}}} \biggr)} \biggr), \end{aligned}$$

(2.75)

where ${t_{1}} = { ( { ( S )\int_{a}^{b} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{a}^{b} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case (ii). If $f (a ) > f ( b )$ and $g ( a ) > g ( b )$, then

Case (a). If $\frac{{f ( b )}}{{f ( a )}} < \frac{{g ( b )}}{{g ( a )}}$, then

Case 1. If $m \in ( {0,{\frac{{f ( b )} }{ {f ( a )}}}} )$, then

$$\begin{aligned} ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge{}& {t_{1}} {t_{2}} \wedge \biggl( { ( {b - ma} ) \biggl( {1 - {{ \biggl( {\frac{{{t_{1}} - mf ( a )}}{{f ( b ) - mf ( a )}}} \biggr)}}} \biggr)} \biggr) \\ &{}\wedge \biggl( { ( {b - ma} ) \biggl( {1 - {{ \biggl( {\frac {{{t_{2}} - mg ( a )}}{{g ( b ) - mg ( a )}}} \biggr)}}} \biggr)} \biggr), \end{aligned}$$

(2.76)

where ${t_{1}} = { ( { ( S )\int_{a}^{b} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{a}^{b} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case 2. If $m = {\frac{{f ( b )} }{{f ( a )}}}$, then

$$ ( S ) \int_{a}^{b} {f ( x )g ( x )\,dx} \ge{t_{1}} {t_{2}} \wedge ( {b - a} ) \wedge f ( b ) \wedge \biggl( { ( {b - ma} ) \biggl( {1 - {{ \biggl( {\frac{{{t_{2}} - mg ( a )}}{{g ( b ) - mg ( a )}}} \biggr)}}} \biggr)} \biggr), $$

(2.77)

where ${t_{1}} = { ( { ( S )\int_{a}^{b} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{a}^{b} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case 3. If $m \in ( {{\frac{{f ( b )} }{{f ( a )}}},{\frac{{g ( b )} }{{g ( a )}}}} )$, then

$$\begin{aligned} ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge{}& {t_{1}} {t_{2}} \wedge \biggl( { ( {b - ma} ){{ \biggl( { \frac{{{t_{1}} - mf ( a )}}{{f ( b ) - mf ( a )}}} \biggr)}} + ma-a} \biggr) \\ &{}\wedge \biggl( { ( {b - ma} ) \biggl( {1 - {{ \biggl( {\frac {{{t_{2}} - mg ( a )}}{{g ( b ) - mg ( a )}}} \biggr)}}} \biggr)} \biggr), \end{aligned}$$

(2.78)

where ${t_{1}} = { ( { ( S )\int_{a}^{b} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{a}^{b} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case 4. If $m = {\frac{{g ( b )} }{{g ( a )}}}$, then

$$ ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge {t_{1}} {t_{2}} \wedge g ( b ) \wedge ( {b - a} ) \wedge \biggl( { ( {b - ma} ){{ \biggl( {\frac{{{t_{1}} - mf ( a )}}{{f ( b ) - mf ( a )}}} \biggr)}} + ma-a} \biggr), $$

(2.79)

where ${t_{1}} = { ( { ( S )\int_{a}^{b} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{a}^{b} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case 5. If $m \in ( {{\frac{{g ( b )} }{{g ( a )}}},1} )$, then

$$\begin{aligned} ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge{}& {t_{1}} {t_{2}} \wedge \biggl( { ( {b - ma} ){{ \biggl( { \frac{{{t_{1}} - mf ( a )}}{{f ( b ) - mf ( a )}}} \biggr)}} + ma-a} \biggr) \\ &{}\wedge \biggl( { ( {b - ma} ){{ \biggl( {\frac{{{t_{1}} - mg ( a )}}{{g ( b ) - mg ( a )}}} \biggr)}} + ma-a} \biggr), \end{aligned}$$

(2.80)

where ${t_{1}} = { ( { ( S )\int_{a}^{b} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{a}^{b} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case (b). If $\frac{{f ( b )}}{{f ( a )}} = \frac{{g ( b )}}{{g ( a )}}$, then

Case 1. If $m \in ( {0,{\frac{{f ( b )} }{ {f ( a )}}}} )$, then

$$\begin{aligned} ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge{}& {t_{1}} {t_{2}} \wedge \biggl( { ( {b - ma} ) \biggl( {1 - {{ \biggl( {\frac{{{t_{1}} - mf ( a )}}{{f ( b ) - mf ( a )}}} \biggr)}}} \biggr)} \biggr) \\ &{}\wedge \biggl( { ( {b - ma} ) \biggl( {1 - {{ \biggl( {\frac {{{t_{2}} - mg ( a )}}{{g ( b ) - mg ( a )}}} \biggr)}}} \biggr)} \biggr), \end{aligned}$$

(2.81)

where ${t_{1}} = { ( { ( S )\int_{a}^{b} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{a}^{b} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case 2. If $m = {\frac{{f ( b )} }{{f ( a )}}}$, then

$$ ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge {t_{1}} {t_{2}} \wedge f ( b ) \wedge g ( b ) \wedge ( {b - a} ), $$

(2.82)

where ${t_{1}} = { ( { ( S )\int_{a}^{b} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{a}^{b} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case 3. If $m \in ( {{\frac{{f ( b )} }{ {f ( a )}}},1} )$, then

$$\begin{aligned} ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge{}& {t_{1}} {t_{2}} \wedge \biggl( { ( {b - ma} ){{ \biggl( { \frac{{{t_{1}} - mf ( a )}}{{f ( b ) - mf ( a )}}} \biggr)}} + ma-a} \biggr) \\ &{}\wedge \biggl( { ( {b - ma} ){{ \biggl( {\frac{{{t_{1}} - mg ( a )}}{{g ( b ) - mg ( a )}}} \biggr)}} + ma-a} \biggr), \end{aligned}$$

(2.83)

where ${t_{1}} = { ( { ( S )\int_{a}^{b} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{a}^{b} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case (c). If $\frac{{f ( b )}}{{f ( a )}} > \frac{{g ( b )}}{{g ( a )}}$, then

Case 1. If $m \in ( {0,{\frac{{g ( b )} }{ {g ( a )}}}} )$, then

$$\begin{aligned} ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge{}& {t_{1}} {t_{2}} \wedge \biggl( { ( {b - ma} ) \biggl( {1 - {{ \biggl( {\frac{{{t_{1}} - mf ( a )}}{{f ( b ) - mf ( a )}}} \biggr)}}} \biggr)} \biggr) \\ &{}\wedge \biggl( { ( {b - ma} ) \biggl( {1 - {{ \biggl( {\frac {{{t_{2}} - mg ( a )}}{{g ( b ) - mg ( a )}}} \biggr)}}} \biggr)} \biggr), \end{aligned}$$

(2.84)

where ${t_{1}} = { ( { ( S )\int_{a}^{b} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{a}^{b} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case 2. If $m = {\frac{{g ( b )} }{{g ( a )}}}$, then

$$ ( S ) \int_{a}^{b} {f ( x )g ( x )\,dx} \ge {t_{1}} {t_{2}} \wedge \biggl( { ( {b - ma} ) \biggl( {1 - {{ \biggl( {\frac{{{t_{1}} - mf ( a )}}{{f ( b ) - mf ( a )}}} \biggr)}}} \biggr)} \biggr) \wedge g ( b ) \wedge ( {b - a} ), $$

(2.85)

where ${t_{1}} = { ( { ( S )\int_{a}^{b} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{a}^{b} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case 3. If $m \in ( {{\frac{{g ( b )} }{{g ( a )}}},{\frac{{f ( b )} }{{f ( a )}}}} )$, then

$$\begin{aligned} ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge{}& {t_{1}} {t_{2}} \wedge \biggl( { ( {b - ma} ) \biggl( {1 - {{ \biggl( {\frac{{{t_{1}} - mf ( a )}}{{f ( b ) - mf ( a )}}} \biggr)}}} \biggr)} \biggr) \\ &{}\wedge \biggl( { ( {b - ma} ){{ \biggl( {\frac{{{t_{1}} - mg ( a )}}{{g ( b ) - mg ( a )}}} \biggr)}} + ma-a} \biggr), \end{aligned}$$

(2.86)

where ${t_{1}} = { ( { ( S )\int_{a}^{b} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{a}^{b} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case 4. If $m = {\frac{{f ( b )} }{{f ( a )}}}$, then

$$\begin{aligned} ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge{}& {t_{1}} {t_{2}} \wedge ( {b - a} )\wedge f ( b ) \\ &{}\wedge \biggl( { ( {b - ma} ){{ \biggl( {\frac{{{t_{1}} - mg ( a )}}{{g ( b ) - mg ( a )}}} \biggr)}} + ma-a} \biggr), \end{aligned}$$

(2.87)

where ${t_{1}} = { ( { ( S )\int_{a}^{b} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{a}^{b} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case 5. If $m \in ( {{\frac{{f ( b )} }{{f ( a )}}},1} )$, then

$$\begin{aligned} ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge{}& {t_{1}} {t_{2}} \wedge \biggl( { ( {b - ma} ){{ \biggl( { \frac{{{t_{1}} - mf ( a )}}{{f ( b ) - mf ( a )}}} \biggr)}} + ma-a} \biggr) \\ &{}\wedge \biggl( { ( {b - ma} ){{ \biggl( {\frac{{{t_{1}} - mg ( a )}}{{g ( b ) - mg ( a )}}} \biggr)}} + ma-a} \biggr), \end{aligned}$$

(2.88)

where ${t_{1}} = { ( { ( S )\int_{a}^{b} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{a}^{b} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Remark 2.11

[22]

Let $X= [ {a,b} ]$, $\alpha=1$, $m=1$ and f, g be two concave functions for all $x \in X$. If μ is a Lebesgue measure on X, then

Case (i). If $f (a ) < f ( b )$ and $g ( a ) < g ( b )$, then

$$ \begin{aligned}[b] ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge{}& {t_{1}} {t_{2}} \wedge \biggl( { ( {b - a} ) \biggl( {1 - \biggl( {\frac{{{t_{1}} - f ( a )}}{{f ( b ) - f ( a )}}} \biggr)} \biggr)} \biggr)\\ &{}\wedge \biggl( { ( {b - a} ) \biggl( {1 - \biggl( {\frac{{{t_{2}} - g ( a )}}{{g ( b ) - g ( a )}}} \biggr)} \biggr)} \biggr), \end{aligned} $$

(2.89)

where ${t_{1}} = { ( { ( S )\int_{a}^{b} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{a}^{b} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case (ii). If $f (a ) = f ( b )$ and $g ( a ) = g ( b )$, then

$$ ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge f ( a )g ( a ) \wedge ( {b - a} ). $$

(2.90)

Case (iii). If $f (a ) > f ( b )$ and $g ( a ) > g ( b )$, then

$$ ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge {t_{1}} {t_{2}} \wedge \biggl( { ( {b - a} ) \biggl( { \frac{{{t_{1}} - f ( a )}}{{f ( b ) - f ( a )}}} \biggr)} \biggr) \wedge \biggl( { ( {b - a} ) \biggl( {\frac{{{t_{1}} - g ( a )}}{{g ( b ) - g ( a )}}} \biggr)} \biggr), $$

(2.91)

where ${t_{1}} = { ( { ( S )\int_{a}^{b} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{a}^{b} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Remark 2.12

Let $X= [ {a,b} ]$, $\alpha\in ({0,1} )$, $m=0$ and f, g be two α-starshaped functions for all $x \in X$. If μ is a Lebesgue measure on X, then

$$ ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge {t_{1}} {t_{2}} \wedge b \biggl( {1 - {{ \biggl( { \frac{{{t_{1}}}}{{f ( b )}}} \biggr)}^{{\frac{1 }{\alpha}}}}} \biggr) \wedge b \biggl( {1 - {{ \biggl( {\frac{{{t_{2}}}}{{g ( b )}}} \biggr)}^{{\frac{1 }{\alpha}}}}} \biggr), $$

(2.92)

where ${t_{1}} = { ( { ( S )\int_{a}^{b} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{a}^{b} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Remark 2.13

Let $X= [ {a,b} ]$, $\alpha\in ({0,1} )$, $m=1$ and f, g be two α-concave functions for all $x \in X$. If μ is a Lebesgue measure on X, then

Case (i). If $f (a ) < f ( b )$ and $g ( a ) < g ( b )$, then

$$ \begin{aligned}[b] ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge{}& {t_{1}} {t_{2}} \wedge \biggl( { ( {b - a} ) \biggl( {1 - {{ \biggl( {\frac{{{t_{1}} - f ( a )}}{{f ( b ) - f ( a )}}} \biggr)}^{{\frac{{{1}} }{\alpha}}}}} \biggr)} \biggr)\\ &{}\wedge \biggl( { ( {b - a} ) \biggl( {1 - {{ \biggl( {\frac {{{t_{2}} - g ( a )}}{{g ( b ) - g ( a )}}} \biggr)}^{{\frac{{{1}} }{\alpha}}}}} \biggr)} \biggr), \end{aligned} $$

(2.93)

where ${t_{1}} = { ( { ( S )\int_{a}^{b} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{a}^{b} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

Case (ii). If $f (a ) = f ( b )$ and $g ( a ) = g ( b )$, then

$$ ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge f ( a )g ( a ) \wedge ( {b - a} ). $$

(2.94)

Case (iii). If $f (a ) > f ( b )$ and $g ( a ) > g ( b )$, then

$$\begin{aligned} ( S )\int_{a}^{b} {f ( x )g ( x )\,dx} \ge {t_{1}} {t_{2}} \wedge \biggl( { ( {b - a} ){{ \biggl( { \frac{{{t_{1}} - f ( a )}}{{f ( b ) - f ( a )}}} \biggr)}^{{\frac{{{1}} }{\alpha}}}}} \biggr) \wedge \biggl( { ( {b - a} ){{ \biggl( {\frac{{{t_{1}} - g ( a )}}{{g ( b ) - g ( a )}}} \biggr)}^{{\frac{{{1}} }{\alpha}}}}} \biggr), \end{aligned}$$

(2.95)

where ${t_{1}} = { ( { ( S )\int_{a}^{b} {{f^{p}} ( x )\,dx} } )^{\frac{1}{p}}}$, ${t_{2}} = { ( { ( S )\int_{a}^{b} {{g^{q}} ( x )\,dx} } )^{\frac{1}{q}}}$.

3 Conclusion

In this paper, we have investigated the Barnes-Godunova-Levin type inequality of the Sugeno integral with respect to an $( {\alpha ,m} )$-concave function. For further investigations, we will continue to explore other integral inequalities for the Sugeno integral related to $( {\alpha,m} )$-concavity.