孙文兵
邵阳学院理学院, 湖南 邵阳 422000
摘要: 建立一些关于（
h，m）-凸函数乘积的新Hadamard-型不等式，得到的结果是对通常凸性、第2种意义下的
s-凸性、
m-凸性、
h-凸性意义下的Hadamard-型不等式的推广.
关键词: Hadamard不等式凸函数(
h,
m)-凸函数2个函数乘积
In recent years, the concept of convex function has been extended by some scholars. For example, Breckner
[1] introduced the concept of
s-convexity, and Varo?anec
[2] defined
h-convex functions. Some results for Hadamard-type inequalities related to the extended convex functions have been obtained
[3-7].
1 Background knowledgeIn 2011, ?zdemir et al.
[8] presented the (
h,
m)-convex function as follows.
Definition 1.1?Let
h:
J??→? be a non-negative function. We say that
f:[0,
b]→? is an (
h,
m)-convex function with
m∈[0, 1], if
f is non-negative and for all
x,
y∈[0,
b] and
α∈(0, 1), we have
| $\begin{array}{*{20}{c}}{f\left( {\alpha x + m\left( {1 - \alpha } \right)y} \right) \le h\left( \alpha \right)f\left( x \right) + }\\{mh\left( {1 - \alpha } \right)f\left( y \right).}\end{array}$ |
If the above inequality is reversed,
f is said to be (
h,
m)-concave function on [0,
b].
Remark 1.11) if we choose
m=1, we have
h-convex functions;
2) if we choose
m=1 and
h(
α)=
α, we obtain non-negative ordinary convex functions;
3) if we choose
m=1 and
h(
α)=
αs, we have
s-convex functions in the second sense;
4) if we choose
h(
α)=
α, we have
m-convex functions.
One of important applications of the concept of convex function is the Hadamard's inequality as follows.
Let
f:
I??→? be a convex function and
a,
b ∈
I with
a <
b, then the inequality
| $f\left( {\frac{{a + b}}{2}} \right) \le \frac{1}{{b - a}} \le \int_a^b {f\left( x \right){\rm{d}}x} \le \frac{{f\left( a \right) + f\left( b \right)}}{2}$ | (1) |
holds, which is well known as Hadamard's inequality.
In Ref. [
9], Dragomir and Fitzpatrick established the Hadamard's type inequalities for
s-convex function as follows.
Theorem 1.1?Suppose that
f: [0, ∞)→[0, ∞) is an
s-convex function in the second sense, where
s∈(0, 1], and let
a,
b∈[0, ∞),
a <
b. If
f∈
L1([
a,
b]), the inequalities
| ${2^{s - 1}}f\left( {\frac{{a + b}}{2}} \right) \le \frac{1}{{b - a}}\int_a^b {f\left( x \right){\rm{d}}x} \le \frac{{f\left( a \right) + f\left( b \right)}}{{s + 1}}$ | (2) |
hold. The constant
$ k = \frac{1}{{s + 1}} $ is the best possible in the second inequality in (2).
In Ref.[
3], Sarkaya proved the Hadamard's type inequalities for class of
h-convex functions as follows.
Theorem 1.2?Let
f:
I??→? be an
h-convex function,
a,
b∈
I,
a <
b and
f∈
L1[
a,
b].Then
| $\begin{array}{*{20}{c}}{\frac{1}{{2h\left( {\frac{1}{2}} \right)}}f\left( {\frac{{a + b}}{2}} \right) \le \frac{1}{{b - a}}\int_a^b {f\left( x \right){\rm{d}}x} }\\{ \le \left[ {f\left( a \right) + f\left( b \right)} \right]\int_0^1 {h\left( t \right){\rm{d}}t} .}\end{array}$ | (3) |
In Ref.[
5], Dragomir and Toader proved the inequality for
m-convex function as follows.
Theorem 1.3 Let
f:[0, ∞)→? be an
m-convex function with
m∈(0, 1]. If 0≤
a <
b < ∞ and
f∈
L1[
a,
b], one has the inequality
| $\begin{array}{*{20}{c}}{\frac{1}{{b - a}}\int_a^b {f\left( x \right){\rm{d}}x} \le }\\{\min \left\{ {\frac{{f\left( a \right) + mf\left( {\frac{b}{m}} \right)}}{2},\frac{{f\left( b \right) + mf\left( {\frac{a}{m}} \right)}}{2}} \right\}.}\end{array}$ | (4) |
Some inequalities of Hadamard-type related to this new class of (
h,
m)-convex functions are given
[8].
Theorem 1.4?Let
f:[0, ∞)→? be an (
h,
m)-convex function with
m∈(0, 1] and
t∈[0, 1]. If 0≤
a <
b < ∞ and
f∈
L1[
a,
b], the inequality
| $\begin{array}{*{20}{c}}{\frac{1}{{b - a}}\int_a^b {f\left( x \right){\rm{d}}x} \le }\\{\min \left\{ {f\left( a \right)\int_0^1 {h\left( t \right){\rm{d}}t} + mf\left( {\frac{b}{m}} \right)\int_0^1 {h\left( {1 - t} \right){\rm{d}}t} ,} \right.}\\{\left. {f\left( b \right)\int_0^1 {h\left( t \right){\rm{d}}t} + mf\left( {\frac{a}{m}} \right)\int_0^1 {h\left( {1 - t} \right){\rm{d}}t} } \right\}}\end{array}$ | (5) |
holds.
In Ref. [
10], Pachpatte established some Hadamard's type inequalities for products of convex functions as follows.
Theorem 1.5 Let
f,
g : [
a,
b] →[0, ∞) be convex functions on [
a,
b] ∈?,
a <
b, then
| $\frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} \le \frac{1}{3}M\left( {a,b} \right) + \frac{1}{6}N\left( {a,b} \right),$ | (6) |
and
| $\begin{array}{*{20}{c}}{2f\left( {\frac{{a + b}}{2}} \right)g\left( {\frac{{a + b}}{2}} \right) \le }\\{\frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} + \frac{1}{6}M\left( {a,b} \right) + \frac{1}{3}N\left( {a,b} \right),}\end{array}$ | (7) |
where
$ M\left( {a, b} \right) = f\left( a \right)g\left( a \right) + f\left( b \right)g\left( b \right) $,
$ N\left( {a, b} \right) = f\left( a \right)g\left( b \right) + f\left( b \right)g\left( a \right) $.
In Ref.[
11], some Hadamard's type inequalities for products of
s-convex functions in the second sense are constructed by Kirmaci et al. as follows.
Theorem 1.6? Let
f,
g:[
a,
b]→?,
a,
b∈[0, ∞),
a <
b, be functions such that
f and
fg are in
L1([
a,
b]). If
f is convex and nonnegative on [
a,
b] and if
g is
s-convex on [
a,
b] for some fixed
s∈(0, 1), then
| $\begin{array}{*{20}{c}}{\frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} \le }\\{\frac{1}{{s + 2}}M\left( {a,b} \right) + \frac{1}{{\left( {s + 1} \right)\left( {s + 2} \right)}}N\left( {a,b} \right),}\end{array}$ | (8) |
where
$ M\left( {a, b} \right) = f\left( a \right)g\left( a \right) + f\left( b \right)g(b) $,
$ N\left( {a, b} \right) = f\left( a \right)g\left( b \right) + f\left( b \right)g(a) $.
Theorem 1.7 Let
f,
g:[
a,
b]→?,
a,
b∈[0, ∞),
a <
b, be functions such that
f and
fg in
L1([
a,
b]). If
f is
s1-convex and
g is
s2-convex connegative on [
a,
b] for some fixed
s1,
s2∈(0, 1), then
| $\begin{array}{*{20}{c}}{\frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} \le }\\{\frac{1}{{{s_1} + {s_2} + 1}}M\left( {a,b} \right) + B\left( {{s_1} + 1,{s_2} + 1} \right)N\left( {a,b} \right) = }\\{\frac{1}{{{s_1} + {s_2} + 1}}\left[ {M\left( {a,b} \right) + {s_1}{s_2}\frac{{\Gamma \left( {{s_1}} \right)\Gamma \left( {{s_2}} \right)}}{{\Gamma \left( {{s_1} + {s_2} + 1} \right)}}N\left( {a,b} \right)} \right],}\end{array}$ | (9) |
where
$ B\left( {x, y} \right) = \smallint _{_0}^{^1}{t^{x-1}}{\left( {1-t} \right)^{y-1}}{\rm{d}}t = \frac{{\Gamma \left( x \right)\Gamma \left( y \right)}}{{\Gamma (x + y)}} $.
Theorem 1.8? Let
f, g:[
a, b]→?, a, b∈[0, ∞),
a <
b, be functions such that
f and
fg are in
L1 ([
a,
b]). If
f is convex and nonnegative on [
a,
b] and if
g is
s-convex on [
a,
b] for some fixed
s∈(0, 1), then
| $\begin{array}{*{20}{c}}{{2^s}f\left( {\frac{{a + b}}{2}} \right)g\left( {\frac{{a + b}}{2}} \right) - \frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} \le }\\{\frac{1}{{\left( {s + 1} \right)\left( {s + 2} \right)}}M\left( {a,b} \right) + \frac{1}{{s + 2}}N\left( {a,b} \right).}\end{array}$ | (10) |
The main purpose of this work is to establish some new Hadamard-type inequalities, similar to the above inequalities, for products of convex functions and (
h,
m)-convex functions, which are generalizations of the above inequalities.
2 Main results and applications
Theorem 2.1 Let
h:[0, 1]→(0, ∞),
f,
g:[0, ∞)→?, be functions such that
h∈
L1([0, 1]),
f g∈
L1([
a,
b]) with
a,
b∈[0, ∞),
a <
b. If
f is convex and nonnegative on [0, ∞), and if
g is (
h,
m)-convex and nonnegative on [0, ∞) with
m∈(0, 1] and
t∈[0, 1], then the following inequality holds,
| $\begin{array}{*{20}{c}}{\frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} }\\{ \le \min \left\{ {\left[ {f\left( a \right) - f\left( b \right)} \right]P\left( {t,h\left( t \right),h\left( {1 - t} \right)} \right) + } \right.}\\{f\left( b \right)P\left( {1,h\left( t \right),h\left( {1 - t} \right)} \right),}\\{\left[ {f\left( b \right) - f\left( a \right)} \right]Q\left( {t,h\left( t \right),h\left( {1 - t} \right)} \right) + }\\{\left. {f\left( a \right)Q\left( {1,h\left( t \right),h\left( {1 - t} \right)} \right)} \right\},}\end{array}$ | (11) |
where
| $\begin{array}{l}\;\;\;\;\;P\left( {t,h\left( t \right),h\left( {1 - t} \right)} \right)\\ = g\left( a \right)\int_0^1 {th\left( t \right){\rm{d}}t} + mg\left( {\frac{b}{m}} \right)\int_0^1 {th\left( {1 - t} \right){\rm{d}}t} ,\end{array}$ |
| $\begin{array}{l}\;\;\;\;\;P\left( {1,h\left( t \right),h\left( {1 - t} \right)} \right)\\ = g\left( a \right)\int_0^1 {h\left( t \right){\rm{d}}t} + mg\left( {\frac{b}{m}} \right)\int_0^1 {h\left( {1 - t} \right){\rm{d}}t} ,\end{array}$ |
| $\begin{array}{l}\;\;\;\;\;Q\left( {t,h\left( t \right),h\left( {1 - t} \right)} \right)\\ = g\left( b \right)\int_0^1 {th\left( t \right){\rm{d}}t} + mg\left( {\frac{a}{m}} \right)\int_0^1 {th\left( {1 - t} \right){\rm{d}}t} ,\end{array}$ |
| $\begin{array}{l}\;\;\;\;\;Q\left( {t,h\left( t \right),h\left( {1 - t} \right)} \right)\\ = g\left( b \right)\int_0^1 {h\left( t \right){\rm{d}}t} + mg\left( {\frac{a}{m}} \right)\int_0^1 {h\left( {1 - t} \right){\rm{d}}t} .\end{array}$ |
Proof ?Since
f is convex and nonnegative on [0, ∞), we have
| $f\left( {ta + \left( {1 - t} \right)b} \right) \le tf\left( a \right) + \left( {1 - t} \right)f\left( b \right)$ |
and
| $f\left( {tb + \left( {1 - t} \right)a} \right) \le tf\left( b \right) + \left( {1 - t} \right)f\left( a \right).$ |
From
g is (
h,
m)-convex and nonnegative on [0, ∞), that is
| $g\left( {tx + m\left( {1 - t} \right)y} \right) \le h\left( t \right)g\left( x \right) + mh\left( {1 - t} \right)g\left( y \right),$ |
for all
x,
y∈[0, ∞). It follows that, for all
t∈[0, 1],
| $g\left( {ta + \left( {1 - t} \right)b} \right) \le h\left( t \right)g\left( a \right) + mh\left( {1 - t} \right)g\left( {\frac{b}{m}} \right),$ |
and
| $g\left( {tb + \left( {1 - t} \right)a} \right) \le h\left( t \right)g\left( b \right) + mh\left( {1 - t} \right)g\left( {\frac{a}{m}} \right).$ |
By the nonnegativeness of
f and
g, we obtain
| $\begin{array}{l}f\left( {ta + \left( {1 - t} \right)b} \right)g\left( {ta + \left( {1 - t} \right)b} \right)\\ \le \left[ {tf\left( a \right) + \left( {1 - t} \right)f\left( b \right)} \right]\left[ {h\left( t \right)g\left( a \right) + } \right.\\\left. {mh\left( {1 - t} \right)g\left( {\frac{b}{m}} \right)} \right]\\ = \left[ {f\left( a \right) - f\left( b \right)} \right]\left[ {g\left( a \right)th\left( t \right) + } \right.\\\left. {mg\left( {\frac{b}{m}} \right)th\left( {1 - t} \right)} \right] + \\f\left( b \right)\left[ {g\left( a \right)h\left( t \right) + mg\left( {\frac{b}{m}} \right)h\left( {1 - t} \right)} \right],\end{array}$ |
and
| $\begin{array}{l}f\left( {tb + \left( {1 - t} \right)b} \right)g\left( {tb + \left( {1 - t} \right)a} \right)\\ \le \left[ {tf\left( b \right) + \left( {1 - t} \right)f\left( a \right)} \right]\left[ {h\left( t \right)g\left( b \right) + } \right.\\\left. {mh\left( {1 - t} \right)g\left( {\frac{a}{m}} \right)} \right]\\ = \left[ {f\left( b \right) - f\left( a \right)} \right]\left[ {g\left( b \right)th\left( t \right) + } \right.\\\left. {mg\left( {\frac{a}{m}} \right)th\left( {1 - t} \right)} \right] + \\f\left( a \right)\left[ {g\left( b \right)h\left( t \right) + mg\left( {\frac{a}{m}} \right)h\left( {1 - t} \right)} \right].\end{array}$ |
Integrating the above two inequalities on [0, 1], with respect to
t, we obtain
| $\begin{array}{l}\int_0^1 {f\left( {ta + \left( {1 - t} \right)b} \right)g\left( {ta + \left( {1 - t} \right)b} \right){\rm{d}}t} \\ \le \left[ {f\left( a \right) - f\left( b \right)} \right]\left[ {g\left( a \right)\int_0^1 {th\left( t \right){\rm{d}}t} + } \right.\\\left. {mg\left( {\frac{b}{m}} \right)\int_0^1 {th\left( {1 - t} \right){\rm{d}}t} } \right] + \\f\left( b \right)\left[ {g\left( a \right)\int_0^1 {\left( {h\left( t \right){\rm{d}}t + mg\left( {\frac{b}{m}} \right)\int_0^1 {h\left( {1 - t} \right){\rm{d}}t} } \right)} } \right],\end{array}$ |
and
| $\begin{array}{l}\int_0^1 {f\left( {tb + \left( {1 - t} \right)a} \right)g\left( {tb + \left( {1 - t} \right)a} \right){\rm{d}}t} \\ \le \left[ {f\left( b \right) - f\left( a \right)} \right]\left[ {g\left( b \right)\int_0^1 {th\left( t \right){\rm{d}}t} + } \right.\\\left. {mg\left( {\frac{b}{m}} \right)\int_0^1 {th\left( {1 - t} \right){\rm{d}}t} } \right] + \\f\left( a \right)\left[ {g\left( b \right)\int_0^1 {\left( {h\left( t \right){\rm{d}}t + mg\left( {\frac{b}{m}} \right)\int_0^1 {h\left( {1 - t} \right){\rm{d}}t} } \right)} } \right],\end{array}$ |
It is easy to see
| $\begin{array}{l}\int_0^1 {f\left( {ta + \left( {1 - t} \right)b} \right)g\left( {ta + \left( {1 - t} \right)b} \right){\rm{d}}t} \\ = \int_0^1 {f\left( {tb + \left( {1 - t} \right)a} \right)g\left( {tb + \left( {1 - t} \right)a} \right){\rm{d}}t} \\ = \frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} .\end{array}$ |
Using the above inequalities and equality, we obtain the required result.
Remark 2.1 If we choose
f(
x)=1 in (11) for all
x∈[
a,
b], we obtain the inequality (5).
Remark 2.2 If we choose
f(
x)=1 and
h(
t)=
t in (11) for
x∈[
a,
b], we obtain the inequality (4).
Remark 2.3 If we choose
f(
x)=1,
m=1 and
h(
t)=
t in (11), we obtain the right hand side of the Hadamard's inequality (1). If we choose
f(
x)=1,
m=1 and
h(
t)=
ts in (11), we obtain the right hand side of the inequality (2).
Theorem 2.2 Let
h1,
h2:[0, 1]→(0, ∞),
f,
g:[0, ∞)→?, be functions such that
h1h2∈
L1 ([0, 1]),
fg∈
L1([
a,
b]) with
a,
b∈[0, ∞),
a <
b. If
f is (
h1,
m1)-convex and nonnegative on [0, ∞), and if
g is (
h2,
m2)-convex and nonnegative on [0, ∞) with
m1,
m2∈(0, 1] and
t∈[0, 1], then the following inequality holds,
| $\begin{array}{l}\frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} \\ \le \min \left\{ {f\left( a \right)P\left( {{h_1}\left( t \right),{h_2}\left( t \right),{h_2}\left( {1 - t} \right)} \right) + } \right.\\{m_1}f\left( {\frac{b}{{{m_1}}}} \right)P\left( {{h_1}\left( {1 - t} \right),{h_2}\left( t \right),{h_2}\left( {1 - t} \right)} \right),\\f\left( b \right)Q\left( {{h_1}\left( t \right),{h_2}\left( t \right),{h_2}\left( {1 - t} \right)} \right) + \\\left. {{m_1}f\left( {\frac{a}{{{m_1}}}} \right)Q\left( {{h_1}\left( {1 - t} \right),{h_2}\left( t \right),{h_2}\left( {1 - t} \right)} \right)} \right\},\end{array}$ | (12) |
where
| $\begin{array}{*{20}{c}}{P\left( {{h_1}\left( t \right),{h_2}\left( t \right),{h_2}\left( {1 - t} \right)} \right)}\\{ = g\left( a \right)\int_0^1 {{h_1}\left( t \right){h_2}\left( t \right){\rm{d}}t} + }\\{{m_2}g\left( {\frac{b}{{{m_2}}}} \right)\int_0^1 {{h_1}\left( t \right){h_2}\left( {1 - t} \right){\rm{d}}t} ,}\end{array}$ |
| $\begin{array}{*{20}{c}}{P\left( {{h_1}\left( {1 - t} \right),{h_2}\left( t \right),{h_2}\left( {1 - t} \right)} \right)}\\{ = g\left( a \right)\int_0^1 {{h_1}\left( {1 - t} \right){h_2}\left( t \right){\rm{d}}t} + }\\{{m_2}g\left( {\frac{b}{{{m_2}}}} \right)\int_0^1 {{h_1}\left( {1 - t} \right){h_2}\left( {1 - t} \right){\rm{d}}t} ,}\end{array}$ |
| $\begin{array}{*{20}{c}}{Q\left( {{h_1}\left( t \right),{h_2}\left( t \right),{h_2}\left( {1 - t} \right)} \right)}\\{ = g\left( b \right)\int_0^1 {{h_1}\left( t \right){h_2}\left( t \right){\rm{d}}t} + }\\{{m_2}g\left( {\frac{a}{{{m_2}}}} \right)\int_0^1 {{h_1}\left( t \right){h_2}\left( {1 - t} \right){\rm{d}}t} ,}\end{array}$ |
| $\begin{array}{*{20}{c}}{Q\left( {{h_1}\left( {1 - t} \right),{h_2}\left( t \right),{h_2}\left( {1 - t} \right)} \right)}\\{ = g\left( b \right)\int_0^1 {{h_1}\left( {1 - t} \right){h_2}\left( t \right){\rm{d}}t} + }\\{{m_2}g\left( {\frac{a}{{{m_2}}}} \right)\int_0^1 {{h_1}\left( {1 - t} \right){h_2}\left( {1 - t} \right){\rm{d}}t} .}\end{array}$ |
Proof Since
f is (
h1,
m1)-convex and nonnegative on [0, ∞), that is
| $\begin{array}{*{20}{c}}{f\left( {tx + {m_1}\left( {1 - t} \right)y} \right) \le {h_1}\left( t \right)f\left( x \right) + }\\{{m_1}{h_1}\left( {1 - t} \right)f\left( y \right)}\end{array}$ |
for all
x,
y∈[0, ∞), we have
| $f\left( {ta + \left( {1 - t} \right)b} \right) \le {h_1}\left( t \right)f\left( a \right) + {m_1}{h_1}\left( {1 - t} \right)f\left( {\frac{b}{{{m_1}}}} \right),$ |
and
| $f\left( {tb + \left( {1 - t} \right)a} \right) \le {h_1}\left( t \right)f\left( b \right) + {m_1}{h_1}\left( {1 - t} \right)f\left( {\frac{a}{{{m_1}}}} \right).$ |
From
g is (
h2,
m2)-convex and nonnegative on [0, ∞), that is
| $g\left( {tx + {m_2}\left( {1 - t} \right)y} \right) \le {h_2}\left( t \right)g\left( x \right) + {m_2}{h_2}\left( {1 - t} \right)g\left( y \right),$ |
for all
x,
y∈[0, ∞). It follows that, for all
t∈[0, 1],
| $g\left( {ta + \left( {1 - t} \right)b} \right) \le {h_2}\left( t \right)g\left( a \right) + {m_2}{h_2}\left( {1 - t} \right)g\left( {\frac{b}{{{m_2}}}} \right)$ |
and
| $g\left( {tb + \left( {1 - t} \right)a} \right) \le {h_2}\left( t \right)g\left( b \right) + {m_2}{h_2}\left( {1 - t} \right)g\left( {\frac{a}{{{m_2}}}} \right).$ |
By the nonnegativeness of
f and
g, we obtain
| $\begin{array}{*{20}{c}}{f\left( {ta + \left( {1 - t} \right)b} \right)g\left( {ta + \left( {1 - t} \right)b} \right)}\\{ \le \left[ {{h_1}\left( t \right)f\left( a \right) + {m_1}{h_1}\left( {1 - t} \right)f\left( {\frac{b}{{{m_1}}}} \right)} \right] \times }\\{\left[ {{h_2}\left( t \right)g\left( a \right) + {m_2}{h_2}\left( {1 - t} \right)g\left( {\frac{b}{{{m_2}}}} \right)} \right]}\\{ = f\left( a \right)\left[ {g\left( a \right){h_1}\left( t \right){h_2}\left( t \right) + } \right.}\\{\left. {{m_2}g\left( {\frac{b}{{{m_2}}}} \right){h_1}\left( t \right){h_2}\left( {1 - t} \right)} \right] + }\\{{m_1}f\left( {\frac{b}{{{m_1}}}} \right)\left[ {g\left( a \right){h_1}\left( {1 - t} \right){h_2}\left( t \right) + } \right.}\\{\left. {{m_2}g\left( {\frac{b}{{{m_2}}}} \right){h_1}\left( {1 - t} \right){h_2}\left( {1 - t} \right)} \right],}\end{array}$ |
and
| $\begin{array}{*{20}{c}}{f\left( {tb + \left( {1 - t} \right)a} \right)g\left( {tb + \left( {1 - t} \right)a} \right)}\\{ \le \left[ {{h_1}\left( t \right)f\left( b \right) + {m_1}{h_1}\left( {1 - t} \right)f\left( {\frac{a}{{{m_1}}}} \right)} \right] \times }\\{\left[ {{h_2}\left( t \right)g\left( b \right) + {m_2}{h_2}\left( {1 - t} \right)g\left( {\frac{a}{{{m_2}}}} \right)} \right]}\\{ = f\left( b \right)\left[ {g\left( b \right){h_1}\left( t \right){h_2}\left( t \right) + } \right.}\\{\left. {{m_2}g\left( {\frac{a}{{{m_2}}}} \right){h_1}\left( t \right){h_2}\left( {1 - t} \right)} \right] + }\\{{m_1}f\left( {\frac{a}{{{m_1}}}} \right)\left[ {g\left( b \right){h_1}\left( {1 - t} \right){h_2}\left( t \right) + } \right.}\\{\left. {{m_2}g\left( {\frac{a}{{{m_2}}}} \right){h_1}\left( {1 - t} \right){h_2}\left( {1 - t} \right)} \right],}\end{array}$ |
Integrating the above two inequalities on [0, 1], with respect to
t, we obtain
| $\begin{array}{*{20}{c}}{\int_0^1 {f\left( {ta + \left( {1 - t} \right)b} \right)g\left( {ta + \left( {1 - t} \right)b} \right){\rm{d}}t} }\\{ \le f\left( a \right)\left[ {g\left( a \right)\int_0^1 {{h_1}\left( t \right){h_2}\left( t \right){\rm{d}}t} } \right] + }\\{{m_2}g\left( {\frac{b}{{{m_2}}}} \right)\int_0^1 {{h_1}\left( t \right){h_2}\left( {1 - t} \right){\rm{d}}t} + }\\{{m_1}f\left( {\frac{b}{{{m_1}}}} \right)\left[ {g\left( a \right)\int_0^1 {{h_1}\left( t \right){h_2}\left( {1 - t} \right){\rm{d}}t} } \right] + }\\{\left. {{m_2}g\left( {\frac{b}{{{m_2}}}} \right)\int_0^1 {{h_1}\left( {1 - t} \right){h_2}\left( {1 - t} \right){\rm{d}}t} } \right],}\end{array}$ |
and
| $\begin{array}{*{20}{c}}{\int_0^1 {f\left( {tb + \left( {1 - t} \right)a} \right)g\left( {tb + \left( {1 - t} \right)a} \right){\rm{d}}t} }\\{ \le f\left( b \right)\left[ {g\left( a \right)\int_0^1 {{h_1}\left( t \right){h_2}\left( t \right){\rm{d}}t} } \right] + }\\{{m_2}g\left( {\frac{a}{{{m_2}}}} \right)\int_0^1 {{h_1}\left( t \right){h_2}\left( {1 - t} \right){\rm{d}}t} + }\\{{m_1}f\left( {\frac{a}{{{m_1}}}} \right)\left[ {g\left( b \right)\int_0^1 {{h_1}\left( {1 - t} \right){h_2}\left( t \right){\rm{d}}t} } \right] + }\\{\left. {{m_2}g\left( {\frac{a}{{{m_2}}}} \right)\int_0^1 {{h_1}\left( {1 - t} \right){h_2}\left( {1 - t} \right){\rm{d}}t} } \right],}\end{array}$ |
It is easy to see that
| $\begin{array}{l}\int_0^1 {f\left( {ta + \left( {1 - t} \right)b} \right)g\left( {ta + \left( {1 - t} \right)b} \right){\rm{d}}t} \\ = \int_0^1 {f\left( {tb + \left( {1 - t} \right)a} \right)g\left( {tb + \left( {1 - t} \right)a} \right){\rm{d}}t} \\ = \frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} .\end{array}$ |
Using the above three inequalities and equality, we obtain the required result.
Remark 2.4 If we choose
h1(
t)=
t,
m1=1 in(12), we can obtain the inequality (11).
Corollary 2.1 Let
h:[0, 1] →(0, ∞),
f,
g : [0, ∞)→?, be functions such that
h∈
L1 ([0, 1]),
fg∈
L1([
a,
b]) with
a,
b∈[0, ∞),
a <
b. If
f and
g is (
h,
m)-convex and nonnegative on [0, ∞) with
m∈(0, 1] and
t∈[0, 1], then the following inequality holds,
| $\begin{array}{l}\frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} \\ \le \min \left\{ {f\left( a \right)P\left( {h\left( t \right),h\left( t \right),h\left( {1 - t} \right)} \right) + } \right.\\mf\left( {\frac{b}{m}} \right)P\left( {h\left( {1 - t} \right),h\left( t \right),h\left( {1 - t} \right)} \right),\\f\left( b \right)Q\left( {h\left( t \right),h\left( t \right),h\left( {1 - t} \right)} \right) + \\\left. {mf\left( {\frac{a}{m}} \right)Q\left( {h\left( {1 - t} \right),h\left( t \right),h\left( {1 - t} \right)} \right)} \right\},\end{array}$ | (13) |
where
| $\begin{array}{*{20}{c}}{P\left( {h\left( t \right),h\left( t \right),h\left( {1 - t} \right)} \right)}\\{ = g\left( a \right)\int_0^1 {{h^2}\left( t \right){\rm{d}}t} + mg\left( {\frac{b}{m}} \right)\int_0^1 {h\left( t \right)h\left( {1 - t} \right){\rm{d}}t} ,}\end{array}$ |
| $\begin{array}{*{20}{c}}{P\left( {h\left( {1 - t} \right),h\left( t \right),h\left( {1 - t} \right)} \right)}\\{ = g\left( a \right)\int_0^1 {h\left( {1 - t} \right)h\left( t \right){\rm{d}}t} + mg\left( {\frac{b}{m}} \right)\int_0^1 {h{{\left( {1 - t} \right)}^2}{\rm{d}}t} ,}\end{array}$ |
| $\begin{array}{*{20}{c}}{Q\left( {h\left( t \right),h\left( t \right),h\left( {1 - t} \right)} \right)}\\{ = g\left( b \right)\int_0^1 {{h^2}\left( t \right){\rm{d}}t} + mg\left( {\frac{a}{m}} \right)\int_0^1 {h\left( t \right)h\left( {1 - t} \right){\rm{d}}t} ,}\end{array}$ |
| $\begin{array}{*{20}{c}}{Q\left( {h\left( {1 - t} \right),h\left( t \right),h\left( {1 - t} \right)} \right)}\\{ = g\left( b \right)\int_0^1 {h\left( {1 - t} \right)h\left( t \right){\rm{d}}t} + mg\left( {\frac{a}{m}} \right)\int_0^1 {{h^2}\left( {1 - t} \right){\rm{d}}t} .}\end{array}$ |
Proof From Theorem 2.2 let
h1=
h2=
h and
m1=
m2=
m, so Corollary 2.1 immediately holds.
Theorem 2.3 Let
h:[0, 1]→(0, ∞),
f,
g:[0, ∞)→?, be functions such that
h∈
L1([0, 1]),
f g∈
L1([
a,
b]) with
a,
b∈[0, ∞),
a <
b. If
f is convex and nonnegative on [0, ∞), and
g is (
h,
m)-convex and nonnegative on [0, ∞) with
m∈(0, 1] and
t∈[0, 1], then the following inequality holds,
| $\begin{array}{l}\frac{2}{{h\left( {\frac{1}{2}} \right)}}f\left( {\frac{{a + b}}{2}} \right)g\left( {\frac{{a + b}}{2}} \right) - \frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} \\ \le \left[ {mg\left( {\frac{b}{m}} \right)f\left( a \right) + mg\left( {\frac{b}{m}} \right)f\left( b \right) + } \right.\\\;\;\;\;\;\left. {f\left( a \right)g\left( a \right)} \right]\int_0^1 {h\left( t \right){\rm{d}}t} + \\\left[ {{m^2}g\left( {\frac{a}{{{m^2}}}} \right)f\left( a \right) + {m^2}g\left( {\frac{a}{{{m^2}}}} \right)f\left( b \right) + } \right.\\\left. {mg\left( {\frac{b}{m}} \right)f\left( a \right)} \right]\int_0^1 {h\left( {1 - t} \right){\rm{d}}t} + \left[ {f\left( b \right) - f\left( a \right)} \right] \times \\\int_0^1 {\left[ {g\left( a \right)th\left( t \right) + mg\left( {\frac{b}{m}} \right)th\left( {1 - t} \right)} \right]{\rm{d}}t} .\end{array}$ | (14) |
Proof We can write
| $\frac{{a + b}}{2} = \frac{{ta + \left( {1 - t} \right)b}}{2} + \frac{{\left( {1 - t} \right)a + tb}}{2}.$ |
Since
f is convex and nonnegative on [0, ∞), and
g is (
h,
m)-convex and nonnegative on [0, ∞), so we have
| $\begin{array}{l}f\left( {\frac{{a + b}}{2}} \right)g\left( {\frac{{a + b}}{2}} \right)\\ = f\left( {\frac{{ta + \left( {1 - t} \right)b}}{2} + \frac{{\left( {1 - t} \right)a + tb}}{2}} \right) \times \\g\left( {\frac{{ta + \left( {1 - t} \right)b}}{2} + \frac{{\left( {1 - t} \right)a + tb}}{2}} \right)\\ \le \frac{{h\left( {\frac{1}{2}} \right)}}{2}\left[ {f\left( {ta + \left( {1 - t} \right)b} \right) + f\left( {\left( {1 - t} \right)a + tb} \right)} \right] \times \\ \left[ {g\left( {ta + \left( {1 - t} \right)b} \right) + mg\left( {\frac{{\left( {1 - t} \right)a}}{m} + \frac{{tb}}{m}} \right)} \right]\\ \le \frac{{h\left( {\frac{1}{2}} \right)}}{2}\{ f\left( {ta + \left( {1 - t} \right)b} \right)g\left( {ta + \left( {1 - t} \right)b} \right) + \\m\left[ {tf\left( a \right) + \left( {1 - t} \right)f\left( b \right)} \right]g\left( {\frac{{\left( {1 - t} \right)a}}{m} + \frac{{tb}}{m}} \right) + \\\left[ {\left( {1 - t} \right)f\left( a \right) + tf\left( b \right)} \right]\left[ {h\left( t \right)g\left( a \right) + mh\left( {1 - t} \right)g\left( {\frac{b}{m}} \right)} \right] + \\\left. {m\left[ {\left( {1 - t} \right)f\left( a \right) + tf\left( b \right)} \right]g\left( {\frac{{\left( {1 - t} \right)a}}{m} + \frac{{tb}}{m}} \right)} \right\}\\ \le \frac{{h\left( {\frac{1}{2}} \right)}}{2}\left\{ {f\left( {ta + \left( {1 - t} \right)b} \right)g\left( {ta + \left( {1 - t} \right)b} \right) + } \right.\\m\left[ {f\left( a \right) + f\left( b \right)} \right] \times \\\left[ {h\left( t \right)g\left( {\frac{b}{m}} \right) + mh\left( {1 - t} \right)g\left( {\frac{a}{{{m^2}}}} \right)} \right] + \\f\left( a \right)\left[ {g\left( a \right)h\left( t \right) + mg\left( {\frac{b}{m}} \right)h\left( {1 - t} \right)} \right] + \\\left. {\left[ {f\left( b \right) - f\left( a \right)} \right]\left[ {g\left( a \right)th\left( t \right) + mg\left( {\frac{b}{m}} \right)th\left( {1 - t} \right)} \right]} \right\}\\ = \frac{{h\left( {\frac{1}{2}} \right)}}{2}\left\{ {f\left( {ta + \left( {1 - t} \right)b} \right)g\left( {ta + \left( {1 - t} \right)b} \right) + } \right.\\\left[ {mg\left( {\frac{b}{m}} \right)f\left( a \right) + mf\left( {\frac{b}{m}} \right)f\left( b \right) + f\left( a \right)g\left( a \right)} \right]h\left( t \right) + \\\left[ {{m^2}g\left( {\frac{a}{{{m^2}}}} \right)f\left( a \right) + {m^2}g\left( {\frac{a}{{{m^2}}}} \right)f\left( b \right)} \right] + \\\left. {mg\left( {\frac{b}{m}} \right)f\left( a \right)} \right]h\left( {1 - t} \right) + \\\left. {\left[ {f\left( b \right) - f\left( a \right)} \right]\left[ {g\left( a \right)th\left( t \right) + mg\left( {\frac{b}{m}} \right)th\left( {1 - t} \right)} \right]} \right\}.\end{array}$ |
Integrating both side of the above inequality on [0, 1], with respect to
t, and by the fact that
| $\begin{array}{*{20}{c}}{\int_0^1 {f\left( {ta + \left( {1 - t} \right)b} \right)g\left( {ta + \left( {1 - t} \right)b} \right){\rm{d}}t} }\\{ = \frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} ,}\end{array}$ |
we obtain
| $\begin{array}{l}f\left( {\frac{{a + b}}{2}} \right)g\left( {\frac{{a + b}}{2}} \right)\\ \le \frac{{h\left( {\frac{1}{2}} \right)}}{2}\left\{ {\frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} + } \right.\\\left[ {mg\left( {\frac{b}{m}} \right)f\left( a \right) + mg\left( {\frac{b}{m}} \right)f\left( b \right) + f\left( a \right)g\left( a \right)} \right]\int_0^1 {h\left( t \right){\rm{d}}t} + \\\left[ {{m^2}g\left( {\frac{a}{{{m^2}}}} \right)f\left( a \right) + {m^2}g\left( {\frac{a}{{{m^2}}}} \right)f\left( b \right)} \right] + \\\left. {mg\left( {\frac{b}{m}} \right)f\left( a \right)} \right]\int_0^1 {h\left( {1 - t} \right){\rm{d}}t} + \\\left[ {f\left( b \right) - f\left( a \right)} \right]\int_0^1 {\left[ {g\left( a \right)th\left( t \right) + mg\left( {\frac{b}{m}} \right)th\left( {1 - t} \right){\rm{d}}t} \right]} ,\end{array}$ |
which completes the proof.
Corollary 2.2 Let
f,
g:[0, ∞)→[0, ∞), be convex functions such that
fg∈
L1([
a,
b]) with
a,
b∈[0, ∞),
a <
b, then the following inequality holds,
| $\begin{array}{*{20}{c}}{4f\left( {\frac{{a + b}}{2}} \right)g\left( {\frac{{a + b}}{2}} \right)}\\{ \le \frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} + \frac{2}{3}M\left( {a,b} \right) + \frac{5}{6}N\left( {a,b} \right)}\end{array}$ | (15) |
where
$ M\left( {a, b} \right) = f\left( a \right)g\left( a \right) + f\left( b \right)g(b)$,
$ N\left( {a, b} \right) = f\left( a \right)g\left( b \right) + f\left( b \right)g(a) $.
Proof From Theorem 2.3 let
h(
t)=
t,
m=1, so Corollary 2.2 immediately holds.
Theorem 2.4 Let
h1,
h2:[0, 1]→(0, ∞),
f,
g:[0, ∞)→?, be functions such that
h1h2∈
L1([0, 1]),
$f\;g \in ([\min \{ \frac{a}{{{m_1}}},\frac{a}{{{m_2}}}\} ,\max\{ \frac{b}{{{m_1}}},\frac{b}{{{m_2}}}\} ]) $ with
a,
b∈[0, ∞),
a <
b and
m1,
m2∈(0, 1].If
f is (
h1,
m1)-convex and nonnegative on [0, ∞), and if
g is (
h2,
m2)-convex and nonnegative on [0, ∞) with
t∈(0, 1], then the following inequality holds,
| $\begin{array}{l}\frac{1}{{{h_1}\left( {\frac{1}{2}} \right){h_2}\left( {\frac{1}{2}} \right)}}f\left( {\frac{{a + b}}{2}} \right)g\left( {\frac{{a + b}}{2}} \right)\\ \le \frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} + \\{m_1}{m_2}\int_0^1 {f\left( {\frac{{\left( {1 - t} \right)a}}{{{m_1}}} + \frac{{tb}}{{{m_1}}}} \right)g\left( {\frac{{\left( {1 - t} \right)a}}{{{m_2}}} + \frac{{tb}}{{{m_2}}}} \right){\rm{d}}t} + \\{m_1}{N_1}\left( {a,b} \right)\int_0^1 {{h_1}\left( t \right){h_2}\left( t \right){\rm{d}}t} + \\{m_1}{m_2}{N_2}\left( {a,b} \right)\int_0^1 {{h_1}\left( {1 - t} \right){h_2}\left( {1 - t} \right){\rm{d}}t} + \\{m_1}{M_1}\left( {a,b} \right)\int_0^1 {{h_1}\left( {1 - t} \right){h_2}\left( t \right){\rm{d}}t} + \\{m_2}{M_2}\left( {a,b} \right)\int_0^1 {{h_1}\left( t \right){h_2}\left( {1 - t} \right){\rm{d}}t} ,\end{array}$ | (16) |
when
| ${N_1}\left( {a,b} \right) = {m_1}f\left( {\frac{b}{{{m_1}}}} \right)g\left( a \right) + {m_2}f\left( a \right)g\left( {\frac{b}{{{m_2}}}} \right),$ |
| ${N_2}\left( {a,b} \right) = {m_2}f\left( {\frac{b}{{{m_1}}}} \right)g\left( {\frac{a}{{m_2^2}}} \right) + {m_1}f\left( {\frac{a}{{m_1^2}}} \right)g\left( {\frac{b}{{{m_2}}}} \right),$ |
| ${M_1}\left( {a,b} \right) = {m_1}f\left( {\frac{a}{{m_1^2}}} \right)g\left( a \right) + {m_2}f\left( {\frac{b}{{{m_1}}}} \right)g\left( {\frac{b}{{{m_2}}}} \right),$ |
| ${M_2}\left( {a,b} \right) = {m_2}f\left( a \right)g\left( {\frac{a}{{m_2^2}}} \right) + {m_1}f\left( {\frac{b}{{{m_1}}}} \right)g\left( {\frac{b}{{{m_2}}}} \right).$ |
Proof We can write
| $\frac{{a + b}}{2} = \frac{{ta + \left( {1 - t} \right)b}}{2} + \frac{{\left( {1 - t} \right)a + tb}}{2}.$ |
Since
f is (
h1,
m1)-convex and nonnegative on [0, ∞), and
g is (
h2,
m2)-convex and nonnegative on [0, ∞), so we have
| $\begin{array}{l}f\left( {\frac{{a + b}}{2}} \right)g\left( {\frac{{a + b}}{2}} \right)\\ = f\left( {\frac{{ta + \left( {1 - t} \right)b}}{2} + \frac{{\left( {1 - t} \right)a + tb}}{2}} \right) \times \\g\left( {\frac{{ta + \left( {1 - t} \right)b}}{2} + \frac{{\left( {1 - t} \right)a + tb}}{2}} \right)\\ \le {h_1}\left( {\frac{1}{2}} \right){h_2}\left( {\frac{1}{2}} \right)\left[ {f\left( {ta + \left( {1 - t} \right)b} \right) + } \right.\\\left. {{m_1}f\left( {\frac{{\left( {1 - t} \right)a}}{{{m_1}}} + \frac{{tb}}{{{m_1}}}} \right)} \right] \times \\\left[ {g\left( {ta + \left( {1 - t} \right)b} \right) + {m_2}g\left( {\frac{{\left( {1 - t} \right)a}}{{{m_2}}} + \frac{{tb}}{{{m_2}}}} \right)} \right]\\ \le {h_1}\left( {\frac{1}{2}} \right){h_2}\left( {\frac{1}{2}} \right)\left\{ {f\left( {ta + \left( {1 - t} \right)b} \right)g\left( {ta + \left( {1 - t} \right)b} \right) + } \right.\\{m_1}{m_2}f\left( {\frac{{\left( {1 - t} \right)a}}{{{m_1}}} + \frac{{tb}}{{{m_1}}}} \right)g\left( {\frac{{\left( {1 - t} \right)a}}{{{m_2}}} + \frac{{tb}}{{{m_2}}}} \right) + \\{m_2}\left[ {\left( {{h_1}\left( t \right)f\left( a \right) + {m_1}{h_1}\left( {1 - t} \right)f\left( {\frac{b}{{{m_1}}}} \right)} \right)} \right] \times \\\left[ {g\left( {\frac{{\left( {1 - t} \right)a}}{{{m_2}}} + \frac{{tb}}{{{m_2}}}} \right)} \right] + {m_1}f\left( {\frac{{\left( {1 - t} \right)a}}{{{m_1}}} + \frac{{tb}}{{{m_1}}}} \right) \times \\\left. {\left[ {{h_2}\left( t \right)g\left( a \right) + {m_2}{h_2}\left( {1 - t} \right)g\left( {\frac{b}{{{m_2}}}} \right)} \right]} \right\}\\ \le {h_1}\left( {\frac{1}{2}} \right){h_2}\left( {\frac{1}{2}} \right)\left\{ {f\left( {ta + \left( {1 - t} \right)b} \right)g\left( {ta + \left( {1 - t} \right)b} \right) + } \right.\\{m_1}{m_2}f\left( {\frac{{\left( {1 - t} \right)a}}{{{m_1}}} + \frac{{tb}}{{{m_1}}}} \right)g\left( {\frac{{\left( {1 - t} \right)a}}{{{m_2}}} + \frac{{tb}}{{{m_2}}}} \right) + \\{m_2}\left[ {\left( {{h_1}\left( t \right)f\left( a \right) + {m_1}{h_1}\left( {1 - t} \right)f\left( {\frac{b}{{{m_1}}}} \right)} \right)} \right] \times \\\left[ {{h_2}\left( t \right)g\left( {\frac{b}{{{m_2}}}} \right) + {m_2}{h_2}\left( {1 - t} \right)g\left( {\frac{a}{{m_2^2}}} \right)} \right] + \\{m_1}\left[ {{h_1}\left( t \right)f\left( {\frac{b}{{{m_1}}}} \right) + {m_1}{h_1}\left( {1 - t} \right)f\left( {\frac{a}{{m_1^2}}} \right)} \right] \times \\\left. {\left[ {{h_2}\left( t \right)g\left( a \right) + {m_2}{h_2}\left( {1 - t} \right)g\left( {\frac{b}{{{m_2}}}} \right)} \right]} \right\}\\ = {h_1}\left( {\frac{1}{2}} \right){h_2}\left( {\frac{1}{2}} \right)\left\{ {f\left( {ta + \left( {1 - t} \right)b} \right)g\left( {ta + \left( {1 - t} \right)b} \right) + } \right.\\{m_1}{m_2}f\left( {\frac{{\left( {1 - t} \right)a}}{{{m_1}}} + \frac{{tb}}{{{m_1}}}} \right)g\left( {\frac{{\left( {1 - t} \right)a}}{{{m_2}}} + \frac{{tb}}{{{m_2}}}} \right) + \\\left[ {{m_1}f\left( {\frac{b}{{{m_1}}}} \right)g\left( a \right) + {m_2}f\left( a \right)g\left( {\frac{b}{{{m_2}}}} \right)} \right]{h_1}\left( t \right){h_2}\left( t \right) + \\\left[ {m_2^2f\left( a \right)g\left( {\frac{a}{{m_2^2}}} \right) + {m_1}{m_2}f\left( {\frac{b}{{{m_1}}}} \right)g\left( {\frac{b}{{{m_2}}}} \right)} \right]{h_1}\left( t \right){h_2}\left( {1 - t} \right) + \\\left[ {{m_1}{m_2}f\left( {\frac{b}{{{m_1}}}} \right)g\left( {\frac{b}{{{m_2}}}} \right) + m_1^2f\left( {\frac{a}{{m_1^2}}} \right)g\left( a \right)} \right]{h_1}\left( {1 - t} \right){h_2}\left( t \right) + \\\left[ {{m_1}m_2^2f\left( {\frac{b}{{{m_1}}}} \right)g\left( {\frac{a}{{m_2^2}}} \right)} \right] + \\\left. {\left. {m_1^2{m_2}f\left( {\frac{a}{{m_1^2}}} \right)g\left( {\frac{b}{{{m_2}}}} \right)} \right]{h_1}\left( {1 - t} \right){h_2}\left( {1 - t} \right)} \right\}.\end{array}$ |
Integrating both side of the above inequality on [0, 1], with respect to
t, and by the fact that
| $\begin{array}{*{20}{c}}{\int_0^1 {f\left( {ta + \left( {1 - t} \right)b} \right)g\left( {ta + \left( {1 - t} \right)b} \right){\rm{d}}t} }\\{ = \frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} ,}\end{array}$ |
we obtain
| $\begin{array}{l}f\left( {\frac{{a + b}}{2}} \right)g\left( {\frac{{a + b}}{2}} \right)\\ \le {h_1}\left( {\frac{1}{2}} \right){h_2}\left( {\frac{1}{2}} \right)\left\{ {\frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} + } \right.\\{m_1}{m_2}\int_0^1 {f\left( {\frac{{\left( {1 - t} \right)a}}{{{m_1}}} + \frac{{tb}}{{{m_1}}}} \right)g\left( {\frac{{\left( {1 - t} \right)a}}{{{m_2}}} + \frac{{tb}}{{{m_2}}}} \right){\rm{d}}t} + \\\left[ {{m_1}f\left( {\frac{b}{{{m_1}}}} \right)g\left( a \right) + {m_2}f\left( a \right)g\left( {\frac{b}{{{m_2}}}} \right)} \right]\int_0^1 {{h_1}\left( t \right){h_2}\left( t \right){\rm{d}}t} + \\{m_2}\left[ {{m_2}f\left( a \right)g\left( {\frac{a}{{m_2^2}}} \right) + {m_1}f\left( {\frac{b}{{{m_1}}}} \right)g\left( {\frac{b}{{{m_2}}}} \right)} \right] \times \\\int_0^1 {{h_1}\left( t \right){h_2}\left( {1 - t} \right){\rm{d}}t} + {m_1}\left[ {{m_2}f\left( {\frac{b}{{{m_1}}}} \right)g\left( {\frac{b}{{{m_2}}}} \right) + } \right.\\\left. {{m_1}f\left( {\frac{a}{{m_1^2}}} \right)g\left( a \right)} \right]\int_0^1 {{h_1}\left( {1 - t} \right){h_2}\left( t \right){\rm{d}}t} ,\end{array}$ |
which completes the proof.
Remark 2.5 If we choose
m1=
m2=1,
h1(
t)=
t,
h2(
t)=
ts in (16) for some
s∈(0, 1), then we obtain the inequaltiy (10).
Remark 2.6 If we choose
m1=
m2=1 and
h1(
t)=
h2(
t)=
t in (16), we can obtain the inequality (7).
Corollary 2.3 Let
f:[0, ∞)→?, be an (
h,
m)-convex and nonnegative on [0, ∞) with
m∈(0, 1] and
t∈[0, 1]. If
h∈
L1([0, 1]),
f∈
L1([
a,
b]) with
a,
b∈[0, ∞),
a <
b, then the following inequality holds,
| $\begin{array}{*{20}{c}}{\frac{2}{{h\left( {\frac{1}{2}} \right)}}f\left( {\frac{{a + b}}{2}} \right) \le }\\{\frac{1}{{b - a}}\left[ {\int_a^b {f\left( x \right){\rm{d}}x} + m\int_a^b {f\left( {\frac{x}{m}} \right){\rm{d}}x} } \right]\left[ {f\left( a \right) + mf\left( {\frac{b}{m}} \right)} \right] \times }\\{\int_0^1 {h\left( t \right){\rm{d}}t} + m\left[ {f\left( {\frac{b}{m}} \right) + mf\left( {\frac{a}{{{m^2}}}} \right)} \right]\int_0^1 {h\left( {1 - t} \right){\rm{d}}t} .}\end{array}$ | (17) |
Proof We choose
g(
x)=1 for all
x∈[
a,
b], and
h2(
t)=
t,
m2=1 in (16), then we can obtain the following inequality
| $\begin{array}{l}\frac{2}{{{h_1}\left( {\frac{1}{2}} \right)}}f\left( {\frac{{a + b}}{2}} \right)\\ \le \frac{1}{{b - a}}\int_a^b {f\left( x \right){\rm{d}}x} + {m_1}\int_0^1 {f\left( {\frac{{\left( {1 - t} \right)a}}{{{m_1}}} + \frac{{tb}}{{{m_1}}}} \right){\rm{d}}t} + \\\left[ {f\left( a \right) + {m_1}f\left( {\frac{b}{{{m_1}}}} \right)} \right]\int_0^1 {{h_1}\left( {1 - t} \right){\rm{d}}t} + \\{m_1}\left[ {f\left( {\frac{b}{{{m_1}}}} \right) + {m_1}f\left( {\frac{a}{{m_1^2}}} \right)} \right]\int_0^1 {{h_1}\left( {1 - t} \right){\rm{d}}t} .\end{array}$ |
By the fact that
| $\int_0^1 {f\left( {\frac{{\left( {1 - t} \right)a}}{{{m_1}}} + \frac{{tb}}{{{m_1}}}} \right){\rm{d}}t} = \frac{1}{{b - a}}\int_a^b {f\left( {\frac{x}{{{m_1}}}} \right){\rm{d}}x} ,$ |
we can obtain the result.
Remark 2.7 If in Corollary 2.3 we choose
h(
t)=
ts,
m=1, we can obtain the following inequality for
s-convex functions,
| ${2^s}f\left( {\frac{{a + b}}{2}} \right) \le \frac{1}{{b - a}}\int_a^b {f\left( x \right){\rm{d}}x} + \frac{{f\left( a \right) + f\left( b \right)}}{{s + 1}},$ |
which is the Remark 4 in Ref. [
11].
Theorem 2.5 Let
h1,
h2:[0, 1]→(0, ∞),
f,
g:[0, ∞)→?, be functions such that
h1h2∈
L1([0, 1]),
fg∈
L1([
a,
b]) with
a,
b∈[0, ∞),
a <
b. If
f is (
h1,
m1)-convex and nonnegative on [0, ∞), and if
g is (
h2,
m2)-convex and nonnegative on [0, ∞) with
m1,
m2∈(0, 1] and
t∈[0, 1], then the following inequality holds,
| $\begin{array}{*{20}{c}}{\frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} }\\{ \le f\left( a \right)P\left( {{h_1}\left( t \right),{h_2}\left( t \right),{h_2}\left( {1 - t} \right)} \right) + }\\{{m_1}f\left( {\frac{b}{{{m_1}}}} \right)P\left( {{h_1}\left( {1 - t} \right),{h_2}\left( t \right),{h_2}\left( {1 - t} \right)} \right),}\end{array}$ | (18) |
where
| $\begin{array}{l}P\left( {{h_1}\left( t \right),{h_2}\left( t \right),{h_2}\left( {1 - t} \right)} \right)\\ = g\left( a \right)\int_0^1 {{h_1}\left( t \right){h_2}\left( t \right){\rm{d}}t} + {m_2}g\left( {\frac{b}{{{m_2}}}} \right)\int_0^1 {{h_1}\left( t \right){h_2}\left( {1 - t} \right){\rm{d}}t} ,\end{array}$ |
| $\begin{array}{l}P\left( {{h_1}\left( {1 - t} \right),{h_2}\left( t \right),{h_2}\left( {1 - t} \right)} \right)\\ = g\left( a \right)\int_0^1 {{h_1}\left( {1 - t} \right){h_2}\left( t \right){\rm{d}}t} + \\\;\;\;{m_2}g\left( {\frac{b}{{{m_2}}}} \right)\int_0^1 {{h_1}\left( {1 - t} \right){h_2}\left( {1 - t} \right){\rm{d}}t} .\end{array}$ |
Proof From the proof of Theorem 2.2, we are easy to obtain the inequality (18).
Remark 2.8 If we choose
m1=
m2=1,
h1(
t)=
t and
h2(
t)=
ts in (18), we can obtain the inequality (8).
Remark 2.9 If we choose
m1=
m2=1,
h1(
t)=
ts1 and
h2(
t)=
ts2 in (18), we can obtain the inequality (9).
Remark 2.10 If we choose
m1=
m2=1,
h1(
t)=
t and
h2(
t)=
t in (18), we can obtain the inequality (6).
Remark 2.11 From Remark 1.1, if we choose proper values of
mi and
hi(
t) (
i=1, 2) in inequalities (12), (16) and (18), we can botain the corresponding inequalities under the condition of different convexity. For example,
1) If we choose
h1(
t)=
h2(
t)=
t in (16) and (18), we can obtain the following Hadamard-type inequalities for products of
m-convex function,
| $\begin{array}{l}4f\left( {\frac{{a + b}}{2}} \right)g\left( {\frac{{a + b}}{2}} \right)\\ \le \frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} + \\{m_1}{m_2}\int_0^1 {f\left( {\frac{{\left( {1 - t} \right)a}}{{{m_1}}} + \frac{{tb}}{{{m_1}}}} \right)g\left( {\frac{{\left( {1 - t} \right)a}}{{{m_2}}} + \frac{{tb}}{{{m_2}}}} \right){\rm{d}}t} + \\\frac{1}{3}\left[ {{N_1}\left( {a,b} \right) + {m_1}{m_2}{N_2}\left( {a,b} \right)} \right] + \\\frac{1}{6}\left[ {{m_1}{M_1}\left( {a,b} \right) + {m_2}{M_2}\left( {a,b} \right)} \right],\end{array}$ | (19) |
and
| $\begin{array}{*{20}{c}}{\frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} }\\{ \le \frac{1}{3}\left[ {f\left( a \right)g\left( a \right) + {m_1}{m_2}f\left( {\frac{b}{{{m_1}}}} \right)g\left( {\frac{b}{{{m_2}}}} \right)} \right] + \frac{1}{6}{N_1}\left( {a,b} \right),}\end{array}$ | (20) |
where
N1(
a,
b),
N2(
a,
b),
M1(
a,
b),
M2(
a,
b) are as in Theorem 2.4.
2) If we choose
m1=
m,
h1(
t)=
t;
m2=1,
h2(
t)=
ts in (16) and (18), we can obtain the Hadamard-type inequalities for products of
m-convex function and
s-convex function as follows,
| $\begin{array}{*{20}{c}}{{2^{s + 1}}f\left( {\frac{{a + b}}{2}} \right)g\left( {\frac{{a + b}}{2}} \right)}\\{ \le \frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} + }\\{m\int_0^1 {f\left( {\frac{{\left( {1 - t} \right)a}}{m} + \frac{{tb}}{m}} \right)g\left( {\left( {1 - t} \right)a + tb} \right){\rm{d}}t} + }\\{\frac{1}{{s + 2}}\left[ {{{N'}_1}\left( {a,b} \right) + m{{N'}_2}\left( {a,b} \right)} \right] + }\\{m{{M'}_1}\left( {a,b} \right)B\left( {s + 1,2} \right) + {{M'}_2}\left( {a,b} \right)B\left( {2,s + 1} \right),}\end{array}$ | (21) |
and
| $\begin{array}{*{20}{c}}{\frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} }\\{ \le \frac{1}{{s + 2}}{{N'}_1}\left( {a,b} \right) + f\left( a \right)g\left( b \right)B\left( {2,s + 1} \right) + }\\{mf\left( {\frac{b}{m}} \right)g\left( a \right)B\left( {s + 1,2} \right),}\end{array}$ | (22) |
where
$ B\left( {x, y} \right) = \smallint _{_0}^{^1}{t^{x-1}}{\left( {1-t} \right)^{y-1}}{\rm{d}}t $, and
| ${{N'}_1}\left( {a,b} \right) = mf\left( {\frac{b}{m}} \right)g\left( a \right) + f\left( a \right)g\left( b \right),$ |
| ${{N'}_2}\left( {a,b} \right) = f\left( {\frac{b}{m}} \right)g\left( a \right) + mf\left( {\frac{a}{{{m^2}}}} \right)g\left( b \right),$ |
| ${{M'}_1}\left( {a,b} \right) = mf\left( {\frac{b}{{{m^2}}}} \right)g\left( a \right) + f\left( {\frac{b}{m}} \right)g\left( b \right),$ |
| ${{M'}_2}\left( {a,b} \right) = f\left( a \right)g\left( a \right) + mf\left( {\frac{b}{m}} \right)g\left( b \right).$ |
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