1. Introduction
The classical or the usual convexity is defined as follows:
A function \(f:\emptyset\neq I\subseteq \mathbb{R} \longrightarrow \mathbb{R}\), is said to be convex on \(I\) if inequality
\begin{equation*}
f\left(tx+\left(1-t\right)y\right) \leq tf(x)+\left(1-t\right)f(y)
\end{equation*}
holds for all \(x,y \in I \) and \(t\in \left[0,1\right]\).
A number of papers have been written on inequalities using the classical
convexity and one of the most fascinating inequalities in mathematical
analysis is stated as follows
\begin{equation}
f\left( \frac{a+b}{2}\right) \leq \frac{1}{b-a}\int\limits_{a}^{b}f(x)dx\leq
\frac{f(a)+f(b)}{2}\text{,}
\end{equation}
(1)
where \(f:I\subseteq\mathbb{R}\longrightarrow\mathbb{R}\) be a convex mapping and \(a,b\in I\) with \(a\leq b\) . Both the inequalities
hold in reversed direction if \(f\) is concave. The inequalities stated in (1) are known as Hermite-Hadamard inequalities.
For more results on (1) which provide new proof, significantly
extensions, generalizations, refinements, counterparts, new
Hermite-Hadamard-type inequalities and numerous applications, we refer the
interested reader [
1,
2,
3,
4] and the references therein.
Many researchers extend their studies to Hermite-Hadamard type inequalities
involving fractional integrals not limited to integer integrals.
The usual notion of convex function have been generalized in diverse
manners. Some of them is the so called harmonically \(s\)-convex function and harmonically convex function is stated in the definition below.
Definition 1.1 [5] A function \(f:I\subset \left( 0,\infty \right) \longrightarrow\mathbb{R}\) is said to be harmonically-\(s\)-convex function if
\begin{equation*}
f\left( \frac{xy}{tx+(1-t)y}\right) \leq t^{s}f\left( y\right) +\left(
1-t\right) ^{s}f\left( x\right)
\end{equation*}
holds for all \(x,y\in I\) and \(t\in \left[ 0,1\right]\), and for some fixed
\(s\in \left( 0,1\right]\).
In [6], İ.İşcan gave definition of harmonically convex functions
and established following Hermite-Hadamard type inequality for harmonically
convex functions as follows:
Definition 1.2 [6] A function \(f:I\subseteq \mathbb{R} \backslash \left\{ 0\right\} \longrightarrow \mathbb{R}\) is said to be harmonically-convex function on \(I\) if
\begin{equation*}
f\left( \frac{xy}{tx+(1-ty)}\right) \leq tf\left( y\right) +\left(
1-t\right) f\left( x\right)
\end{equation*}
holds for all \(x,y\in I\) and \(t\in \left[ 0,1\right]\). If the inequality is
reversed, then \(f\) is said to be harmonically concave.
Proposition 1.3 [6] Let \(I \subset \mathbb{R}\backslash \{0\}\) be a real interval and \(f:I\rightarrow\mathbb{R}\) is function, then:
- if \(I\subset (0,\infty )\) and \(f\) is convex and nondecreasing function then
\(f\) is harmonically convex.
- if \(I\subset (0,\infty )\) and \(f\) is harmonically convex and nonincreasing
function then \(f\) is convex.
- if \(I\subset (-\infty ,0)\) and \(f\) is harmonically convex and nondecreasing
function then \(f\) is convex.
- if \(I\subset (-\infty ,0)\) and \(f\) is convex and nonincreasing function then
\(f\) is harmonically convex.
For the properties of harmonically-convex functions and
harmonically-\(s\)-convex function, we refer the reader to [
5,
6,
7,
8,
9,
10,
11,
12,
13,
14] and the
reference there in.
Most recently, a number of findings have been seen on Hermite-Hadamard type
integral inequalities for harmonically-convex and for harmonically-\(s\)-convex
functions.
Definition 1.4 A function \(g:\left[ a,b\right] \subseteq \mathbb{R}\backslash \{0\}\longrightarrow\mathbb{R}\) is said to be harmonically symmetric with respect to \(2ab/a+b\) if
\begin{equation*}
g\left( x\right) =g\left( \frac{1}{\frac{1}{a}+\frac{1}{b}-\frac{1}{x}}\right)
\end{equation*}
holds for all \(x\in \lbrack a,b]\).
Definition 1.5
Let \(f\in L[a,b]\). The right-hand side and left-hand side Hadamard
fractional integrals \(J_{a^{+}}^{\alpha }f\) and \(J_{b^{-}}^{\alpha }f\) of
order \(\alpha >0\) with \(b>a\geq 0\) are defined by
\begin{equation*}
J_{a^{+}}^{\alpha }f(x)=\frac{1}{\Gamma (\alpha )}\overset{x}{\underset{a}{
\int }}\left( x-t\right) ^{\alpha -1}f(t)dt,\ x>a,
\end{equation*}
\begin{equation*}
J_{b^{-}}^{\alpha }f(x)=\frac{1}{\Gamma (\alpha )}\overset{b}{\underset{x}{
\int }}\left( t-x\right) ^{\alpha -1}f(t)dt,\ x< b
\end{equation*}
respectively where \(\Gamma (\alpha )\) is the Gamma function defined by \(\Gamma (\alpha )=\overset{\infty }{\underset{0}{\int }}e^{-t}t^{\alpha -1}\),
\(J_{a^{+}}^{0}f(x)=J_{b^{-}}^{0}f(x)=f(x)\).
Theorem 1.6 [5] Let \(f:I\subset \mathbb{R}\backslash \{0\}\mathbb{\rightarrow R}\) be a harmonically convex function and \(a,b\in I\) with \(a< b \). If \(\ f\in L\left[ a,b\right] \) then the following inequalities hold:
\begin{equation}
f\left( \frac{2ab}{a+b}\right) \leq \frac{ab}{b-a}\int\limits_{a}^{b}\frac{
f(x)}{x^{2}}dx\leq \frac{f(a)+f(b)}{2}.
\end{equation}
(2)
In [
10], İ. İşcan and Wu represented Hermite-Hadamard’s
inequalities for harmonically convex functions in fractional integral form
as follows:
Theorem 1.7 [ 10] Let \(f:I\subseteq \mathbb{R}^{+}\mathbb{\rightarrow R}\) be
differentiable on \(I^{\circ }\), and \(a,b\in I\) with \(a< b \) and \(f\in L\left[
a,b\right]\). If \(f\) is harmonically-convex on \(\left[ a,b\right]\), then
the following inequalities for fractional integrals hold:
\begin{eqnarray}
f\left( \frac{2ab}{a+b}\right) &\leq &\frac{\Gamma (\alpha +1)}{2}\left( \frac{ab}{a+b}\right) ^{\alpha }\left\{
\begin{array}{c}
J_{1/a^{-}}^{\alpha }(f\circ h)(1/b) \\
+J_{1/b^{+}}^{\alpha }(f\circ h)(1/a)
\end{array}
\right\}
&\leq & \frac{f(a)+f(b)}{2}.\label{1.4}
\end{eqnarray}
(3)
with \(\alpha >0\) and \(h(x)=1/x\).
In [
12], Chan and Wu represented Hermite-Hadamard-Fejér inequality
for harmonically convex functions as follows:
Theorem 1.8
Suppose that \(f:I\subseteq\mathbb{R} \backslash \{0\}\longrightarrow\mathbb{R}\) be harmonically-convex function and \(a,b\in I\), with \(a< b\). If \(f\in L
\left[ a,b\right]\) and \(g:[a,b]\subseteq\mathbb{R}\backslash \{0\}\longrightarrow\mathbb{R}\) is nonnegative, integrable and harmonically symmetric with respect to \(2ab/a+b\), then
\begin{equation}
f\left( \frac{2ab}{a+b}\right) \int\limits_{a}^{b}\frac{g(x)}{x^{2}}dx\leq
\int\limits_{a}^{b}\frac{f(x)g(x)}{x^{2}}dx\leq \frac{f(a)+f(b)}{2}
\int\limits_{a}^{b}\frac{g(x)}{x^{2}}dx
\end{equation}
(4)
In [
8], İ. İşcan and Kunt represented Hermite-Hadamard-Fejér
type inequality for harmonically convex functions in fractional integral
forms and established following identity as follow
Theorem 1.9 Let \(f:[a,b]\longrightarrow \mathbb{R}\) be harmonically convex function with \(a< b\) and \(f\in L\left[ a,b\right]\).
If \(g:[a,b]\longrightarrow\mathbb{R}\) is nonnegative, integrable and harmonically symmetric with respect to
\(2ab/a+b\), then the following inequalities for fractional integrals hold:
\begin{equation}
f\left( \frac{2ab}{a+b}\right) \left[ J_{1/a^{-}}^{\alpha }(g\circ
h)(1/b)+J_{1/b^{+}}^{\alpha }(g\circ h)(1/a)\right] \label{1.7}
\end{equation}
\begin{equation*}
\leq \left[ J_{1/a^{-}}^{\alpha }(fg\circ h)(1/b)+J_{1/b^{+}}^{\alpha
}(fg\circ h)(1/a)\right]
\end{equation*}
\begin{equation*}
\leq \frac{f(a)+f(b)}{2}\left[ J_{1/a^{-}}^{\alpha }(g\circ
h)(1/b)+J_{1/b^{+}}^{\alpha }(g\circ h)(1/a)\right]
\end{equation*}
(5)
with \(\alpha >0\) and \(h(x)=1/x\), \(x\in \left[ \frac{1}{b},\frac{1}{a}\right]\).
In [
2] D. Y. Hwang found out a new identity and by using this
identity, established a new inequalities. Then in [
14] İ. İşcan and S. Turhan used this identity for harmonically convex
functions and obtain generalized new inequalities. In this paper, we
established a new inequality similar to inequality in [
14] and
then we obtained some new and general integral inequalities for differentiable harmonically-convex functions using this lemma. The following
sections, let the notion, \(L\left( t\right) =\frac{aH}{tH+(1-t)a}\), \(U\left(
t\right) =\frac{bH}{tH+(1-t)b}\) and \(H=H\left( a,b\right) =\frac{2ab}{a+b}\).
2. Main Results
Throughout this section, let \(\left\Vert g\right\Vert _{\infty }=\sup_{x\in
\left[ a,b\right] }\left\vert g(x)\right\vert \), for the continuous function
\(g:[a,b]\longrightarrow\mathbb{R}\) be differentiable mapping \(I^{o}\), where \(a,b\in I\) with \(a\leq b\), and \(h:
\left[ a,b\right] \longrightarrow \left[ 0,\infty \right)\) be
differentiable mapping.
Lemma 2.1.
Let \(f:I\subseteq \mathbb{R}_{+}=(0,\infty )\longrightarrow \mathbb{R}\) be a
differentiable function on \(I^{o}\), \(a,b\in I^{o}\) with \(a< b\). If \(h:\left[
a,b\right] \longrightarrow \left[ 0,\infty \right)\) is a differentiable
function and \(f^{\prime }\in L\left( [a,b]\right)\), the following
inequality holds:
\begin{equation}
\begin{split}
\left( \frac{f(a)+f(b)}{2}\right) h(a)-f(H)h(b)\\+\frac{b-a}{4ab}\left\{
\begin{array}{c}
\int\limits_{0}^{1}\left[ h^{\prime }(L(t))\left( L(t)\right)^{2}
+h^{\prime}(U(t))\left( U(t)\right) ^{2}\right] \\
\times \left[ f(L(t))+f(U(t))\right] dt
\end{array}
\right\} \
\end{split}
\end{equation}
\begin{equation*}
=\frac{b-a}{4ab}\left\{
\begin{array}{c}
\int\limits_{0}^{1}\left[ h(L(t))-h(U(t))+h(b)\right]\\
\times \left[ -f^{\prime }(L(t))\left( L(t)\right) ^{2}+f^{\prime
}(U(t))\left( U(t)\right) ^{2}\right] dt
\end{array}
\right\}.
\end{equation*}
(6)
Proof.
By the integration by parts, we have
\begin{eqnarray*}
I_{1}&=&\int\limits_{0}^{1}\left[ h(L(t))-h(U(t))+h(b)\right] d\left( f\left(
L(t)\right) \right) \\
&=&\left. \left[ h(L(t))-h(U(t))+h(b)\right] f\left( L(t)\right) \right\vert
_{0}^{1} \\
&-&\int\limits_{0}^{1}f(L(t))\left[ h^{\prime
}(L(t))L^{\prime }(t)-h^{\prime }(U(t))U^{\prime }(t)\right] dt \\
&=& h(a)f(a)-h(b)f(H)-\int\limits_{0}^{1}f(L(t))\left[ h^{\prime
}(L(t))L^{\prime }(t)-h^{\prime }(U(t))U^{\prime }(t)\right] dt
\end{eqnarray*}
and
\begin{eqnarray*}
I_{2}&=&\int\limits_{0}^{1}\left[ h(L(t))-h(U(t))+h(b)\right] d\left( f\left(
U(t)\right) \right) \\
&=&\left. \left[ h(L(t))-h(U(t))+h(b)\right] f\left( U(t)\right) \right\vert
_{0}^{1} \\
&-&\int\limits_{0}^{1}f(U(t))\left[ h^{\prime
}(L(t))L^{\prime }(t)-h^{\prime }(U(t))U^{\prime }(t)\right] dt \\
&=&h(a)f(b)-h(b)f(H)-\int\limits_{0}^{1}f(U(t))\left[ h^{\prime
}(L(t))L^{\prime }(t)-h^{\prime }(U(t))U^{\prime }(t)\right] dt
\end{eqnarray*}
Therefore
\begin{eqnarray}
\frac{I_{1}+I_{2}}{2}
&=&\left( \frac{f(a)+f(b)}{2}\right) h(a)-h(b)f(H)\nonumber\\
&&-\frac{
b-a}{4ab}\left\{
\begin{array}{c}
\int\limits_{0}^{1}\left[ h^{\prime }(L(t))\left( L(t)\right) ^{2}+h^{\prime
}(U(t))\left( U(t)\right) ^{2}\right] \\
\times \left[ f(L(t))+f(U(t))\right] dt
\end{array}
\right\}
\end{eqnarray}
(7)
This complete the proof.
Theorem 2.2.
Let \(f:I\subseteq\mathbb{R}_{+}=\left( 0,\infty \right) \longrightarrow\mathbb{R}\) be differentiable function on \(I^{o}\) and \(a,b\in I^{o}\) with \(a< b\). If \(h:\left[ a,b\right] \longrightarrow \left[ 0,\infty \right)\) is a
differentiable function and \(\left\vert f^{\prime }\right\vert\) is
harmonically convex on \(\left[ a,b\right]\), the following inequality holds
\begin{eqnarray}
&&\left\vert \left( \frac{f(a)+f(b)}{2}\right) h(a)-h(b)f(\frac{2ab}{a+b})\right. \\
&&+\left. \frac{1}{2}\left[ \int\limits_{a}^{b}f(x)h^{\prime
}(x)dx+\int\limits_{a}^{b}f(x)h^{\prime }(\frac{Hx}{2x-H})\left( \frac{H}{
2x-H}\right) ^{2}dx\right] \right\vert \nonumber \\
&&\leq \frac{b-a}{4ab}\left[ \zeta _{1}(a,b)\left\vert f^{\prime
}(a)\right\vert +\zeta _{2}(a,b)\left\vert f^{\prime }(H)\right\vert +\zeta
_{3}(a,b)\left\vert f^{\prime }(b)\right\vert \right] \nonumber
\end{eqnarray}
(8)
where
\begin{equation}
\zeta _{1}\left( a,b\right) =\int\limits_{0}^{1}t\left( L(t)\right)
^{2}\left\vert h(L(t))-h(U(t))+h(b)\right\vert dt,
\end{equation}
(9)
\begin{equation}
\zeta _{2}\left( a,b\right) =\int\limits_{0}^{1}(1-t)\left\vert
h(L(t))-h(U(t))+h(b)\right\vert \left[ \left( L(t)\right) ^{2}+\left(
U(t)\right) ^{2}\right] dt ,
\end{equation}
(10)
\begin{equation}
\zeta _{3}\left( a,b\right) = \int\limits_{0}^{1}t\left( U(t)\right)
^{2}\left\vert h(L(t))-h(U(t))+h(b)\right\vert dt . \label{2.5}
\end{equation}
(11)
Proof.
We get the following inequality take the absolute value to (6):
\begin{eqnarray}
&&\left\vert \left( \frac{f(a)+f(b)}{2}\right) h(a)-h(b)f(\frac{2ab}{a+b}) \right. \label{2.7} \\
&&\left.+\frac{1}{2}\left[ \int\limits_{a}^{b}f(x)h^{\prime
}(x)dx+\int\limits_{a}^{b}f(x)h^{\prime }(\frac{Hx}{2x-H})\left( \frac{H}{%
2x-H}\right) ^{2}dx\right] \right\vert \nonumber \\
&&\leq \frac{b-a}{4ab}\left\{ \int\limits_{0}^{1}\left\vert
h(L(t))-h(U(t))+h(b)\right\vert \left\vert f^{\prime }\left( L(t)\right)
\left( L(t)\right) ^{2}\right\vert dt\ \right. \nonumber \\
&&\left.
+\int\limits_{0}^{1}\left\vert h(L(t))-h(U(t))+h(b)\right\vert \left\vert
f^{\prime }\left( U(t)\right) \left( U(t)\right) ^{2}\right\vert dt\right\}.\nonumber
\end{eqnarray}
(12)
Since \(\left\vert f^{\prime }\right\vert\) is harmonically-convex on \([a,b]\) in
(12), we have for all \(t\in \lbrack a,b]\) that
\begin{eqnarray}
&&\left\vert \left( \frac{f(a)+f(b)}{2}\right) h(a)-h(b)f(\frac{2ab}{a+b})\right. \label{2.8} \\
&&+\left.\frac{1}{2}\left[ \int\limits_{a}^{b}f(x)h^{\prime
}(x)dx+\int\limits_{a}^{b}f(x)h^{\prime }(\frac{Hx}{2x-H})\left( \frac{H}{%
2x-H}\right) ^{2}dx\right] \right\vert \nonumber \\
&&\leq \frac{b-a}{4ab}\left\{ \int\limits_{0}^{1}\left\vert
h(L(t))-h(U(t))+h(b)\right\vert \left[ t\left\vert f^{\prime }(a)\right\vert
+(1-t)\left\vert f^{\prime }(H)\right\vert \right] \left( L(t)\right)
^{2}dt\ \right. \nonumber \\
&&\left.
+\int\limits_{0}^{1}\left\vert h(L(t))-h(U(t))+h(b)\right\vert \left[
t\left\vert f^{\prime }(b)\right\vert +(1-t)\left\vert f^{\prime
}(H)\right\vert \right] \left( U(t)\right) ^{2}dt\right\} , \nonumber
\end{eqnarray}
(13)
this proof is completed.
Corollary 2.3.
Suppose that \(g:\left[ a,b\right] \longrightarrow \left[ 0,\infty \right)\)
is a continuous positive mapping and geometrically symmetric with respect to
\(\frac{2ab}{a+b}\) (i.e. \(g(x)=g\left(\frac{1}{\frac{1}{a}+\frac{1}{b}-\frac{1}{x}}\right)\) holds for all \(x\in
\lbrack a,b]\) with \(a< b \). Taking $$h(t)=\int\limits_{1/t}^{1/a}\left[ \left( x-\frac{1}{b}\right) ^{\alpha
-1}+\left( \frac{1}{a}-x\right) ^{\alpha -1}\right] \left( g\circ \varphi
\right) (x)dx$$ \(1/t\in \lbrack \frac{1}{b},\frac{1}{a}],\alpha <0 \)
and \(\varphi \left( x\right) =\frac{1}{x}\) in
Theorem 1, we get
\begin{eqnarray}
&&\left\vert \left[ J_{1/b^{+}}^{\alpha }\left( fg\circ \varphi \right) \left(
1/a\right) +J_{1/a^{-}}^{\alpha }\left( fg\circ \varphi \right) \left(
1/b\right) \right] \right. \label{2.9} \\
&&-\left. f\left( \frac{2ab}{a+b}\right) \left[ J_{1/b^{+}}^{%
\alpha }\left( g\circ \varphi \right) (1/a)+J_{1/a^{-}}^{\alpha }\left(
g\circ \varphi \right) (1/b)\right] \right\vert \nonumber \\
&&\leq \frac{(b-a)^{\alpha +1}\left\Vert g\right\Vert _{\infty }}{%
2(ab)^{\alpha +1}\Gamma \left( \alpha +1\right) }\left[ C_{1}\left( \alpha
\right) \left\vert f^{\prime }(a)\right\vert +C_{2}\left( \alpha \right)
\left\vert f^{\prime }(H)\right\vert +C_{3}\left( \alpha \right) \left\vert
f^{\prime }(b)\right\vert \right] \nonumber
\end{eqnarray}
(14)
where
\begin{eqnarray*}
C_{1}\left( \alpha \right) &=&\int\limits_{0}^{1}\left[ 1-(\frac{1+t}{2}
)^{\alpha }+(\frac{1-t}{2})^{\alpha }\right] t\left( L(t)\right) ^{2}dt ,\\
C_{2}\left( \alpha \right) &=&\int\limits_{0}^{1}\left[ 1-(\frac{1+t}{2}
)^{\alpha }+(\frac{1-t}{2})^{\alpha }\right] \left( 1-t\right) \left[ \left(
L(t)\right) ^{2}+\left( U(t)\right) ^{2}\right] dt ,\\
C_{3}\left( \alpha \right) &=&\int\limits_{0}^{1}\left[ 1-(\frac{1+t}{2}
)^{\alpha }+(\frac{1-t}{2})^{\alpha }\right] t\left( U(t)\right) ^{2}dt .
\end{eqnarray*}
Proof.
If we write
\(h(t)=\int\limits_{1/t}^{1/a}\left[ \left( x-\frac{1}{b}\right) ^{\alpha-1}\right.\)
\(+\left. \left( \frac{1}{a}-x\right) ^{\alpha -1}\right] g\circ \varphi (x)dx\)
for all \(1/t\in \lbrack \frac{1}{b},\frac{1}{a}]\) and \(\varphi (x)=1/x\) in (13), we
have
\begin{eqnarray}
&&\left\vert \frac{1}{2}\left[ \int\limits_{a}^{b}\left[ \left( \frac{1}{a}%
-x\right) ^{\alpha -1}+\left( x-\frac{1}{b}\right) ^{\alpha -1}\right] \frac{%
f(x)g(x)}{x^{2}}dx \right.\right. \label{2.10} \\
&&\left.\left. +\int\limits_{a}^{b}\left[ \left( \frac{1}{a}-x\right)
^{\alpha -1}+\left( x-\frac{1}{b}\right) ^{\alpha -1}\right] \frac{f(x)g(%
\frac{Hx}{2x-H})}{x^{2}}dx\right] \right. \nonumber
\end{eqnarray}
(14)
\begin{eqnarray*}
&&\left. -f(\frac{2ab%
}{a+b})\int\limits_{a}^{b}\left[ \left( \frac{1}{a}-x\right) ^{\alpha
-1}+\left( x-\frac{1}{b}\right) ^{\alpha -1}\right] \frac{g(x)}{x^{2}}
dx\right\vert .
\end{eqnarray*}
From \(g(x)\) is harmonically symmetric with respect to \(x=2ab/a+b\), we get
\begin{eqnarray}
&&\left\vert \int\limits_{a}^{b}\left[ \left( \frac{1}{a}-x\right) ^{\alpha
-1}+\left( x-\frac{1}{b}\right) ^{\alpha -1}\right] \frac{f(x)g(x)}{x^{2}}
dx\right. \label{2.11} \\
&&\left. -f(\frac{2ab}{a+b})\int\limits_{a}^{b}\left[ \left( \frac{1}{a}
-x\right) ^{\alpha -1}+\left( x-\frac{1}{b}\right) ^{\alpha -1}\right] \frac{
g(x)}{x^{2}}dx\right\vert \nonumber \\
&&=\left\vert
\begin{array}{c}
\Gamma (\alpha )\left[ J_{1/b^{+}}^{\alpha }\left( fg\circ \varphi \right)
\left( 1/a\right) +J_{1/a^{-}}^{\alpha }\left( fg\circ \varphi \right)
\left( 1/b\right) \right] \\
-\Gamma (\alpha )\left[ J_{1/b^{+}}^{\alpha }g\circ \varphi
(1/a)+J_{1/a^{-}}^{\alpha }g\circ \varphi (1/b)\right] f\left( \frac{2ab}{a+b%
}\right)
\end{array}
\right\vert . \nonumber
\end{eqnarray}
(15)
On the other hand, right side of inequality (13)
\begin{eqnarray}
\leq \frac{b-a}{4ab}\left\{ \int\limits_{0}^{1}\left\vert
\begin{array}{c}
\overset{1/a}{\int\limits_{1/L(t)}}\left[ \left( \frac{1}{a}-x\right)
^{\alpha -1}+\left( x-\frac{1}{b}\right) ^{\alpha -1}\right] \left( g\circ
\varphi \right) (x)dx \\
-\overset{1/a}{\int\limits_{1/U(t)}}\left[ \left( \frac{1}{a}-x\right)
^{\alpha -1}+\left( x-\frac{1}{b}\right) ^{\alpha -1}\right] \left( g\circ
\varphi \right) (x)dx \\
+\int\limits_{1/b}^{1/a}\left[ \left( \frac{1}{a}-x\right) ^{\alpha
-1}+\left( x-\frac{1}{b}\right) ^{\alpha -1}\right] \left( g\circ \varphi
\right) (x)dx
\end{array}
\right\vert Adt\ \right. \label{2.12}
\end{eqnarray}
(16)
\begin{equation*}
+\left. \int\limits_{0}^{1}\left\vert
\begin{array}{c}
\overset{1/a}{\int\limits_{1/L(t)}}\left[ \left( \frac{1}{a}-x\right)
^{\alpha -1}+\left( x-\frac{1}{b}\right) ^{\alpha -1}\right] \left( g\circ
\varphi \right) (x)dx \\
-\overset{1/a}{\int\limits_{1/U(t)}}\left[ \left( \frac{1}{a}-x\right)
^{\alpha -1}+\left( x-\frac{1}{b}\right) ^{\alpha -1}\right] \left( g\circ
\varphi \right) (x)dx \\
+\int\limits_{1/b}^{1/a}\left[ \left( \frac{1}{a}-x\right) ^{\alpha
-1}+\left( x-\frac{1}{b}\right) ^{\alpha -1}\right] \left( g\circ \varphi
\right) (x)dx
\end{array}
\right\vert B dt \right\}.
\end{equation*}
Where $$A=\left[ tf^{\prime }(a)+(1-t)f^{\prime }(H)\right] \left(
L(t)\right) ^{2}$$ and $$B=\left[ tf^{\prime }(b)+(1-t)f^{\prime }(H)\right] \left(
U(t)\right) ^{2}$$
Since \(g(x)\) is symmetric to \(x=2ab/a+b\), we have
\begin{eqnarray}
&&\left\vert
\begin{array}{c}
\overset{1/a}{\int\limits_{1/L(t)}}\left[ \left( \frac{1}{a}-x\right)
^{\alpha -1}+\left( x-\frac{1}{b}\right) ^{\alpha -1}\right] \left( g\circ
\varphi \right) (x)dx \\
-\overset{1/a}{\int\limits_{1/U(t)}}\left[ \left( \frac{1}{a}-x\right)
^{\alpha -1}+\left( x-\frac{1}{b}\right) ^{\alpha -1}\right] \left( g\circ
\varphi \right) (x)dx \\
+\int\limits_{1/b}^{1/a}\left[ \left( \frac{1}{a}-x\right) ^{\alpha
-1}+\left( x-\frac{1}{b}\right) ^{\alpha -1}\right] \left( g\circ \varphi
\right) (x)dx
\end{array}
\right\vert \nonumber \\
&&=\overset{1/a}{\int\limits_{1/L(t)}}\left[ \left( \frac{1}{a}-x\right)
^{\alpha -1}+\left( x-\frac{1}{b}\right) ^{\alpha -1}\right] \left( g\circ
\varphi \right) (x)dx \nonumber \\
&&+\overset{1/U(t)}{\int\limits_{1/b}}\left[ \left( \frac{
1}{a}-x\right) ^{\alpha -1}+\left( x-\frac{1}{b}\right) ^{\alpha -1}\right]
\left( g\circ \varphi \right) (x)dx \nonumber
\end{eqnarray}
(17)
and
\begin{eqnarray}
&&\overset{1/a}{\int\limits_{1/L(t)}}\left[ \left( \frac{1}{a}-x\right)
^{\alpha -1}+\left( x-\frac{1}{b}\right) ^{\alpha -1}\right] \left( g\circ
\varphi \right) (x)dx \nonumber \\
&&=\overset{1/U(t)}{\int\limits_{1/b}}\left[ \left( \frac{%
1}{a}-x\right) ^{\alpha -1}+\left( x-\frac{1}{b}\right) ^{\alpha -1}\right]
\left( g\circ \varphi \right) (x)dx \label{2.14}
\end{eqnarray}
(18)
for all \(t\in \lbrack 0,1]\) and \(\varphi (x)=1/x\).
By (15)-(18), we have
\begin{eqnarray}
&&\left\vert \left[ J_{a^{+}}^{\alpha }\left( fg\right) \left( b\right)
+J_{b^{-}}^{\alpha }\left( fg\right) \left( a\right) \right] -\left[
J_{a^{+}}^{\alpha }g\left( b\right) +J_{b^{-}}^{\alpha }g\left( a\right)
\right] f(\frac{2ab}{a+b})\right\vert \label{2.15} \\
&&\leq \frac{b-a}{2ab\Gamma \left( \alpha \right) }\left\{
\int\limits_{0}^{1}
\begin{array}{c}
\left\vert \overset{1/a}{\int\limits_{1/L(t)}}
\left[\left( \frac{1}{a}-x\right) ^{\alpha -1}+\left( x-\frac{1}{b}\right)
^{\alpha -1}\right] \left( g\circ \varphi \right) (x)dx\right\vert \\
\left[
tf^{\prime }(a)+(1-t)f^{\prime }(H)\right] \left( L(t)\right) ^{2}dt
\end{array}
\right. \\
&&+\left.
\int\limits_{0}^{1}
\begin{array}{c}
\left\vert \overset{1/a}{\int\limits_{1/L(t)}}
\left[ \left( \frac{1}{a}-x\right)
^{\alpha -1}+\left( x-\frac{1}{b}\right) ^{\alpha -1}\right] \left( g\circ
\varphi \right) (x)dx\right\vert \\
\left[ tf^{\prime }(b)+(1-t)f^{\prime }(H)%
\right] \left( U(t)\right) ^{2}dt
\end{array}
\right\} \\
&&\leq \frac{\left( b-a\right) \left\Vert g\right\Vert _{\infty }}{2ab\Gamma
\left( \alpha \right) }\left\{ \int\limits_{0}^{1}
\begin{array}{c}
\left( \overset{1/a}{
\int\limits_{1/L(t)}}\left[ \left( \frac{1}{a}-x\right) ^{\alpha -1}+\left(
x-\frac{1}{b}\right) ^{\alpha -1}\right] dx\right) \\
\left[ tf^{\prime}(a)+(1-t)f^{\prime }(H)\right] \left( L(t)\right) ^{2}dt\
\end{array}
\right. \\
&&+\left. \int\limits_{0}^{1}
\begin{array}{c}
\left( \overset{
1/a}{\int\limits_{1/L(t)}}\left[ \left( \frac{1}{a}-x\right) ^{\alpha
-1}+\left( x-\frac{1}{b}\right) ^{\alpha -1}\right] dx\right) \\
\left[tf^{\prime }(b)+(1-t)f^{\prime }(H)\right] \left( U(t)\right) ^{2}dt
\end{array}
\right\}. \nonumber
\end{eqnarray}
(19)
In the last inequality, we calculate
\begin{equation}
\overset{1/a}{\int\limits_{1/L(t)}}\left[ \left( \frac{1}{a}-x\right)
^{\alpha -1}+\left( x-\frac{1}{b}\right) ^{\alpha -1}\right] dx=\frac{%
(b-a)^{\alpha }}{\alpha (ab)^{\alpha }}\left[ 1-\left( \frac{1+t}{2}\right)
^{\alpha }+\left( \frac{1-t}{2}\right) ^{\alpha }\right] . \label{2.16}
\end{equation}
(20)
A combination of (19) and (20), we have (14). This
complete is proof.
Corollary 2.4.
In Corollary 2.3,(1)If we take \(\alpha =1\), we obtain following
Hermite-Hadamard-Fej\'{e}r Type inequality for harmonically-convex functions
related to (14):
\begin{eqnarray}
&&\left\vert \overset{b}{\underset{a}{\int }}f(x)\frac{g(x)}{x^{2}}dx-f\left(
\frac{2ab}{a+b}\right) \underset{a}{\overset{b}{\int }}\frac{g(x)}{x^{2}}%
dx\right\vert \label{2.17} \\
&&\leq \frac{\left( b-a\right) ^{2}}{2\left( ab\right) ^{2}}\left\Vert
g\right\Vert _{\infty }\left[ C_{1}(1)\left\vert f^{\prime }\left( a\right)
\right\vert +C_{2}(1)\left\vert f^{\prime }\left( H\right) \right\vert
+C_{3}(1)\left\vert f^{\prime }\left( b\right) \right\vert \right] \nonumber
\end{eqnarray}
(21)
where
\begin{eqnarray*}
C_{1}(1)&=&\overset{1}{\underset{0}{\int }}t\left( 1-t\right) \left(
L(t)\right) ^{2}dt, \\
C_{2}\left( 1\right) &=&\overset{1}{\underset{0}{\int }}(1-t)^{2}\left[ \left(
L(t)\right) ^{2}+\left( U(t)\right) ^{2}\right] dt, \\
C_{3}(1)&=&\overset{1}{\underset{0}{\int }}t\left( 1-t\right) \left(
U(t)\right) ^{2}dt.
\end{eqnarray*}
(2)If we take \(g(x)=1\), we obtain following inequality is related to(14):
\begin{eqnarray}
&&\left\vert \frac{(ab)^{\alpha }\Gamma (\alpha +1)}{2\left( b-a\right)
^{\alpha }}\left[ J_{1/b^{+}}^{\alpha }\left( f\circ \varphi \right) \left(
1/a\right) +J_{1/a^{-}}^{\alpha }\left( f\circ \varphi \right) \left(
1/b\right) \right] -f\left( \frac{2ab}{a+b}\right) \right\vert \nonumber \\
&&\leq \frac{\left( b-a\right) }{4ab}\left[ C_{1}\left( \alpha \right)
\left\vert f^{\prime }(a)\right\vert +C_{2}\left( \alpha \right) \left\vert
f^{\prime }(H)\right\vert +C_{3}\left( \alpha \right) \left\vert f^{\prime
}(b)\right\vert \right] .\label{2.18}
\end{eqnarray}
(22)
(3)If we take \(g(x)=1\) and \(\alpha =1\), we obtain following
inequality is related to (14):
\begin{eqnarray}
&&\left\vert \frac{ab}{b-a}\int\limits_{a}^{b}\frac{f(x)}{x}dx-f\left( \frac{
2ab}{a+b}\right) \right\vert \label{2.19}\\
&&\leq \frac{\left( b-a\right) }{4ab}\left[ C_{1}\left( 1\right) \left\vert
f^{\prime }(a)\right\vert +C_{2}\left( 1\right) \left\vert f^{\prime
}(H)\right\vert +C_{3}\left( 1\right) \left\vert f^{\prime }(b)\right\vert
\right] . \nonumber
\end{eqnarray}
(23)
Theorem 2.5.
Let \(f:I\subseteq\mathbb{R}_{+}=\left( 0,\infty \right) \longrightarrow\mathbb{R}\) be a differentiable function on \(I^{o}\), \(a,b\in I^{o}\) with \(a< b \). If \(h:\left[ a,b\right] \longrightarrow \left[ 0,\infty \right) \) is a differentiable function and \(\left\vert f^{\prime }\right\vert ^{q}\) is
harmonically convex on \(\left[ a,b\right]\) for \(q\geq 1\), the following inequality
holds
\begin{eqnarray}
&&\left\vert \left( \frac{f(a)+f(b)}{2}\right) h(a)-h(b)f\left( \frac{2ab}{a+b}%
\right)\right. \label{2.20} \\
&&+\left. \frac{1}{2}\left[ \int\limits_{a}^{b}f(x)h^{\prime
}(x)dx+\int\limits_{a}^{b}f(x)h^{\prime }(\frac{Hx}{2x-H})\left( \frac{H}{%
2x-H}\right) ^{2}dx\right] \right\vert \nonumber \\
&&\leq \frac{\left( b-a\right) }{4ab}\left\{
\begin{array}{c}
\left( \underset{0}{\int\limits^{1}}\left\vert
h(L(t))-h(U(t))+h(b)\right\vert dt\right) ^{1-\frac{1}{q}}\times \\
\left( \underset{0}{\overset{1}{\int }}
\begin{array}{c}
\left( \left\vert h(L(t))-h(U(t))+h(b)\right\vert \right) \\
\times \left( t\left( L(t)\right) ^{2q}\left\vert f^{\prime }(a)\right\vert
^{q}+(1-t)\left( L(t)\right) ^{2q}\left\vert f^{\prime }(H)\right\vert
^{q}\right)
\end{array}
dt\right) ^{\frac{1}{q}}
\end{array}
\right. \nonumber \\
&&\left.
\begin{array}{c}
+\ \left( \underset{0}{\int\limits^{1}}\left\vert
h(L(t))-h(U(t))+h(b)\right\vert dt\right) ^{1-\frac{1}{q}}\times \\
\left( \underset{0}{\overset{1}{\int }}
\begin{array}{c}
\left( \left\vert h(L(t))-h(U(t))+h(b)\right\vert \right) \\
\times \left( t\left( U(t)\right) ^{2q}\left\vert f^{\prime }(b)\right\vert
^{q}+(1-t)\left( U(t)\right) ^{2q}\left\vert f^{\prime }(H)\right\vert
^{q}\right)
\end{array}
dt\right) ^{\frac{1}{q}}
\end{array}
\right\} \nonumber
\end{eqnarray}
(24)
Proof.
Continuing from (12) in proof of Theorem 2.2, the power mean inequality
and using the fact that \(\left\vert f^{\prime }\right\vert ^{q}\) is
harmonically convex on \(\left[ a,b\right]\), we get the required result. This
completes the proof of the theorem.
Corollary 2.6.
Let \(g:\left[ a,b\right] \longrightarrow \left[ 0,\infty \right)\) be a
positive continuous mapping and harmonically symmetric with respect to \(2ab/a+b\)
(i.e. \(g\left( \frac{1}{\frac{1}{a}+\frac{1}{b}-\frac{1}{x}}\right)=g(x)\) holds for all
\(x\in\lbrack a,b]\) with \(a< b\). If we take
$$h(t)=\int\limits_{1/t}^{1/a}\left[ \left( x-\frac{1}{b}\right) ^{\alpha
-1}+\left( \frac{1}{a}-x\right) ^{\alpha -1}\right] \left( g\circ \varphi
\right) (x)dx$$,
\(t\in \left[ a,b\right]\), \(\varphi \left( x\right) =\frac{1
}{x}\) in Theorem 2.5, we obtain
\begin{eqnarray}
&&\left\vert \left[ J_{1/b^{+}}^{\alpha }\left( fg\circ \varphi \right) \left(
1/a\right) +J_{1/a^{-}}^{\alpha }\left( fg\circ \varphi \right) \left(
1/b\right) \right] \right. \label{2.21} \\
&&\left. -f\left( \frac{2ab}{a+b}\right) \left[ J_{1/b^{+}}^{%
\alpha }\left( g\circ \varphi \right) (1/a)+J_{1/a^{-}}^{\alpha }\left(
g\circ \varphi \right) (1/b)\right] \right\vert \nonumber \\
&&\leq \frac{\left( b-a\right) ^{\alpha +1}\left\Vert g\right\Vert _{\infty }}{%
2(ab)^{\alpha +1}\Gamma \left( \alpha +1\right) }\left[ 2-\frac{4}{\alpha +1}%
+\frac{2^{2-\alpha }}{\alpha +1}\right] ^{1-\frac{1}{q}} \nonumber \\
&&\left[ C_{1}\left(
\alpha ,q\right) \left\vert f^{\prime }(a)\right\vert ^{q}+C_{2}\left(
\alpha ,q\right) \left\vert f^{\prime }(H)\right\vert ^{q}+C_{3}\left(
\alpha ,q\right) \left\vert f^{\prime }(b)\right\vert ^{q}\right] ^{\frac{1}{%
q}} \nonumber
\end{eqnarray}
(25)
where for \(q\geq 1\)
\begin{eqnarray*}
C_{1}\left( \alpha ,q\right) &=& \overset{1}{\underset{0}{\int }}\left[
1-\left( \frac{1+t}{2}\right) ^{\alpha }+\left( \frac{1-t}{2}\right)
^{\alpha }\right] t\left( L(t)\right) ^{2q}dt ,\\
C_{2}\left( \alpha ,q\right) &=&\overset{1}{\underset{0}{\int }}\left[
1-\left( \frac{1+t}{2}\right) ^{\alpha }+\left( \frac{1-t}{2}\right)
^{\alpha }\right] (1-t)\left( \left( L(t)\right) ^{2q}+\left( U(t)\right)
^{2q}\right) dt ,\\
C_{3}\left( \alpha ,q\right) &=&\overset{1}{\underset{0}{\int }}\left[
1-\left( \frac{1+t}{2}\right) ^{\alpha }+\left( \frac{1-t}{2}\right)
^{\alpha }\right] t\left( U(t)\right) ^{2q}dt.
\end{eqnarray*}
Proof .
We use the equality (20) of Corollary 2.3 and (24) in Theorem 2.5,
\begin{eqnarray}
&&\qquad\quad \left\vert \left[ J_{1/b^{+}}^{\alpha }\left( fg\circ \varphi \right) \left(
1/a\right) +J_{1/a^{-}}^{\alpha }\left( fg\circ \varphi \right) \left(
1/b\right) \right]\right. \label{2.22} \\
&&\left. -f\left( \frac{2ab}{a+b}\right) \left[ J_{1/b^{+}}^{%
\alpha }\left( g\circ \varphi \right) (1/a)+J_{1/a^{-}}^{\alpha }\left(
g\circ \varphi \right) (1/b)\right] \right\vert
\end{eqnarray}
(26)
\begin{eqnarray}
&&\leq \frac{\left( b-a\right) ^{\alpha +1}\left\Vert g\right\Vert _{\infty }}{%
2(ab)^{\alpha +1}\Gamma \left( \alpha +1\right) }\times\\
&&\left\{
\begin{array}{c}
\left( \underset{0}{\overset{1}{\int }}\left[ 1-\left( \frac{1+t}{2}\right)
^{\alpha }+\left( \frac{1-t}{2}\right) ^{\alpha }\right] dt\right) ^{1-\frac{%
1}{q}}\times \\
\left( \underset{0}{\overset{1}{\int }}
\begin{array}{c}
\left[ 1-\left( \frac{1+t}{2}\right)
^{\alpha }+\left( \frac{1-t}{2}\right) ^{\alpha }\right] \\
\left( t\left(
L(t)\right) ^{2q}\left\vert f^{\prime }(a)\right\vert ^{q}+(1-t)\left(
L(t)\right) ^{2q}\left\vert f^{\prime }\left( H\right) \right\vert
^{q}\right) dt
\end{array}
\right) ^{\frac{1}{q}}
\end{array}
\right. \nonumber \\
&&+\left.
\begin{array}{c}
\left( \underset{0}{\overset{1}{\int }}\left[ 1-\left( \frac{1+t}{2}\right)
^{\alpha }+\left( \frac{1-t}{2}\right) ^{\alpha }\right] dt\right) ^{1-\frac{
1}{q}}\times \\
\left( \underset{0}{\overset{1}{\int }}
\begin{array}{c}
\left[ 1-\left( \frac{1+t}{2}\right)
^{\alpha }+\left( \frac{1-t}{2}\right) ^{\alpha }\right] \\
\left( t\left(
U(t)\right) ^{2q}\left\vert f^{\prime }(b)\right\vert ^{q}+(1-t)\left(
U(t)\right) ^{2q}\left\vert f^{\prime }(H)\right\vert ^{q}\right)
\end{array}
dt\right)^{\frac{1}{q}}
\end{array}
\right\} \nonumber \\
&&\leq \frac{\left( b-a\right) ^{\alpha +1}\left\Vert g\right\Vert _{\infty }
}{2(ab)^{\alpha +1}\Gamma \left( \alpha +1\right) }\left[ 1-\frac{2}{\alpha
+1}+\frac{2^{1-\alpha }}{\alpha +1}\right] ^{1-\frac{1}{q}}\times \nonumber \\
&&\left[
\begin{array}{c}
\left( \underset{0}{\overset{1}{\int }}
\begin{array}{c}
\left[ 1-\left( \frac{1+t}{2}\right) ^{\alpha }+\left( \frac{1-t}{2}\right)
^{\alpha }\right] \times \\
\left[ t\left( L(t)\right) ^{2q}\left\vert f^{\prime }(a)\right\vert
^{q}+(1-t)\left( L(t)\right) ^{2q}\left\vert f^{\prime }(H)\right\vert ^{q}
\right]
\end{array}
dt\right) ^{\frac{1}{q}} \\
+\left( \underset{0}{\overset{1}{\int }}
\begin{array}{c}
\left[ 1-\left( \frac{1+t}{2}\right) ^{\alpha }+\left( \frac{1-t}{2}\right)
^{\alpha }\right] \times \\
\left[ t\left( U(t)\right) ^{2q}\left\vert f^{\prime }(b)\right\vert
^{q}+(1-t)\left( U(t)\right) ^{2q}\left\vert f^{\prime }(H)\right\vert ^{q}
\right]
\end{array}
dt\right) ^{\frac{1}{q}}
\end{array}
\right] \nonumber
\end{eqnarray}
(27)
By the power-mean inequality \(\left( a^{r}+b^{r}0,b>0,\quad r\leq 1\right)\), we have
\begin{eqnarray}
&&\leq \frac{\left( b-a\right) ^{\alpha +1}\left\Vert g\right\Vert _{\infty }}{%
2(ab)^{\alpha +1}\Gamma \left( \alpha +1\right) }\left[ 2-\frac{4}{\alpha +1}%
+\frac{2^{2-\alpha }}{\alpha +1}\right] ^{1-\frac{1}{q}} \label{2.23} \\
&&\times \left[ \overset{1}{\underset{0}{\int }}\left(
\begin{array}{c}
\left[ 1-\left( \frac{1+t}{2}\right) ^{\alpha }+\left( \frac{1-t}{2}\right)
^{\alpha }\right] t\left( L(t)\right) ^{2q}\left\vert f^{\prime
}(a)\right\vert ^{q} \\
+\left[ 1-\left( \frac{1+t}{2}\right) ^{\alpha }+\left( \frac{1-t}{2}\right)
^{\alpha }\right] (1-t)\left(
\begin{array}{c}
\left( L(t)\right) ^{2q} \\
+\left( U(t)\right) ^{2q}%
\end{array}%
\right) \left\vert f^{\prime }(H)\right\vert ^{q} \\
+\left[ 1-\left( \frac{1+t}{2}\right) ^{\alpha }+\left( \frac{1-t}{2}\right)
^{\alpha }\right] t\left( U(t)\right) ^{2q}\left\vert f^{\prime
}(b)\right\vert ^{q}%
\end{array}%
\right) dt\right] ^{\frac{1}{q}}. \nonumber
\end{eqnarray}
(28)
Proof.
When \(\alpha =1\), \(g(x)=1\) is taken in Corollary 2.6, we obtain:
\begin{eqnarray}
&&\left\vert \frac{ab}{b-a}\overset{b}{\underset{a}{\int }}\frac{f(x)}{x^{2}}%
dx-f\left( \frac{2ab}{a+b}\right) \right\vert \label{2.24} \\
&&\leq \frac{\left( b-a\right) }{4ab}\left[ C_{1}\left( 1,q\right) \left\vert
f^{\prime }(a)\right\vert ^{q}+C_{2}\left( 1,q\right) \left\vert f^{\prime
}(H)\right\vert ^{q}+C_{3}\left( 1,q\right) \left\vert f^{\prime
}(b)\right\vert ^{q}\right] ^{\frac{1}{q}}. \nonumber
\end{eqnarray}
(29)
This proof is completed.
