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