Throughout this section, we write \(\left\| g\right\| _{\infty }= \underset{t\in \left[ a,b\right] }{\sup }\left| g(t)\right|\), for the continuous function \(g:\left[ a,b\right] \mathbb {\rightarrow \mathbb {R} }\).
Lemma 2
If
\(g:\left[ a,b\right] \subseteq \mathbb {R{\setminus } }\left\{ 0\right\} \mathbb {\rightarrow R}\)
is integrable and harmonically symmetric with respect to
\(\frac{2ab}{a+b}\), then
$$J_{\frac{a+b}{2ab}+}^{\alpha }\left( g\circ h\right) (1/a)=J_{\frac{a+b}{2ab} -}^{\alpha }\left( g\circ h\right) (1/b)=\frac{1}{2}\left[ \begin{array}{c} J_{\frac{a+b}{2ab}+}^{\alpha }\left( g\circ h\right) (1/a) \\ +J_{\frac{a+b}{2ab}-}^{\alpha }\left( g\circ h\right) (1/b) \end{array} \right]$$
with
\(\alpha >0\)
and
\(h(x)=1/x\), \(x\in \left[ \frac{1}{b},\frac{1}{a}\right]\).
Proof
Since g is harmonically symmetric with respect to \(\frac{2ab}{a+b}\), using Definition 3 we have \(g\left( \frac{1}{x}\right) =g\left(\frac{1}{(\frac{1 }{a})+(\frac{1}{b})-x}\right)\), for all \(x\in \left[ \frac{1}{b},\frac{1}{a}\right]\). Hence, in the following integral setting \(t=\left(\frac{1}{a})+(\frac{1}{b}\right)-x\) and \(dt=-dx\) gives
$$\begin{aligned} J_{\frac{a+b}{2ab}+}^{\alpha }\left( g\circ h\right) (1/a)&= {} \frac{1}{ \varGamma (\alpha )}\int \nolimits _{\frac{a+b}{2ab}}^{\frac{1}{a}}\left( \frac{1 }{a}-t\right) ^{\alpha -1}g\left( \frac{1}{t}\right) dt \\&= {} \frac{1}{\varGamma (\alpha )}\int \nolimits _{\frac{1}{b}}^{\frac{a+b}{2ab} }\left( x-\frac{1}{b}\right) ^{\alpha -1}g\left( \frac{1}{(1/a)+(1/b)-x} \right) dx \\&= {} \frac{1}{\varGamma (\alpha )}\int \nolimits _{\frac{1}{b}}^{\frac{a+b}{2ab} }\left( x-\frac{1}{b}\right) ^{\alpha -1}g\left( \frac{1}{x}\right) dx=J_{ \frac{a+b}{2ab}-}^{\alpha }\left( g\circ h\right) (1/b). \end{aligned}$$
This completes the proof.
Theorem 4
Let
\(f:I\subseteq \left( 0,\infty \right) \rightarrow \mathbb {R}\)
be a function such that
\(f\in L\left[ a,b\right]\), where
\(a,b\in I\). If
f
is a harmonically convex function on
\(\left[ a,b\right]\), then the following inequalities for fractional integrals holds:
$$\begin{aligned} f\left( \frac{2ab}{a+b}\right) &\le \frac{\varGamma \left( \alpha +1\right) }{ 2^{1-\alpha }}\left( \frac{ab}{b-a}\right) ^{\alpha }\left\{ \begin{array}{c} J_{\frac{a+b}{2ab}+}^{\alpha }\left( f\circ h\right) \left( 1/a\right) \\ +J_{\frac{a+b}{2ab}-}^{\alpha }\left( f\circ h\right) \left( 1/b\right) \end{array} \right\} \\&\le \frac{f\left( a\right) +f\left( b\right) }{2} \end{aligned}$$
(6)
with
\(\alpha >0\)
and
\(h(x)=1/x\), \(x\in \left[ \frac{1}{b},\frac{1}{a}\right]\).
Proof
Since f is a harmonically convex function on \(\left[ a,b\right]\), we have for all \(t\in \left[ 0,1\right]\)
$$\begin{aligned} f\left( \frac{2ab}{a+b}\right)&= {} f\left( \frac{2\left( \frac{ab}{ta+(1-t)b} \right) \left( \frac{ab}{tb+(1-t)a}\right) }{\left( \frac{ab}{ta+(1-t)b} \right) +\left( \frac{ab}{tb+(1-t)a}\right) }\right) \\&\le \frac{f\left( \frac{ab}{ta+(1-t)b}\right) +f\left( \frac{ab}{tb+(1-t)a }\right) }{2}. \end{aligned}$$
(7)
Multiplying both sides of (7) by \(2t^{\alpha -1}\), then integrating the resulting inequality with respect to t over \(\left[ 0, \frac{1}{2}\right]\), we obtain
$$\begin{aligned}&2f\left( \frac{2ab}{a+b}\right) \int _{0}^{\frac{1}{2}}t^{\alpha -1}dt \\&\quad \le \int _{0}^{\frac{1}{2}}t^{\alpha -1}\left[ f\left( \frac{ab}{ta+(1-t)b} \right) +f\left( \frac{ab}{tb+(1-t)a}\right) \right] dt \\&\quad =\int _{0}^{\frac{1}{2}}t^{\alpha -1}f\left( \frac{ab}{ta+(1-t)b}\right) dt+\int _{0}^{\frac{1}{2}}t^{\alpha -1}f\left( \frac{ab}{tb+(1-t)a}\right) dt. \end{aligned}$$
Setting \(x=\frac{tb+(1-t)a}{ab}\) and \(dx=\left( \frac{b-a}{ab}\right) dt\) gives
$$\begin{aligned} \frac{2^{1-\alpha }}{\alpha }f\left( \frac{2ab}{a+b}\right)&\le \left( \frac{ab}{b-a}\right) ^{\alpha }\left\{ \begin{array}{c} \int _{\frac{1}{b}}^{\frac{a+b}{2ab}}\left( x-\frac{1}{b}\right) ^{\alpha -1}f\left( \frac{1}{\frac{1}{a}+\frac{1}{b}-x}\right) dx \\ +\int _{\frac{1}{b}}^{\frac{a+b}{2ab}}\left( x-\frac{1}{b}\right) ^{\alpha -1}f\left( \frac{1}{x}\right) dx \end{array} \right\} \\&= {} \left( \frac{ab}{b-a}\right) ^{\alpha }\left\{ \begin{array}{c} \int _{\frac{a+b}{2ab}}^{\frac{1}{a}}\left( \frac{1}{a}-x\right) ^{\alpha -1}f\left( \frac{1}{x}\right) dx \\ +\int _{\frac{1}{b}}^{\frac{a+b}{2ab}}\left( x-\frac{1}{b}\right) ^{\alpha -1}f\left( \frac{1}{x}\right) dx \end{array} \right\} \\&= {} \left( \frac{ab}{b-a}\right) ^{\alpha }\varGamma (\alpha )\left[ J_{\frac{a+b }{2ab}+}^{\alpha }\left( f\circ h\right) (1/a)+J_{\frac{a+b}{2ab}-}^{\alpha }\left( f\circ h\right) (1/b)\right] \end{aligned}$$
and the first inequality is proved.
For the proof of the second inequality in (6), we first note that, if f is a harmonically convex function, then, for all \(t\in \left[ 0,1\right]\), it yields
$$f\left( \frac{ab}{ta+(1-t)b}\right) +f\left( \frac{ab}{tb+(1-t)a}\right) \le f(a)+f(b).$$
(8)
Then multiplying both sides of (8) by \(t^{\alpha -1}\) and integrating the resulting inequality with respect to t over \(\left[ 0, \frac{1}{2}\right]\), we obtain
$$\begin{aligned}&\int _{0}^{\frac{1}{2}}t^{\alpha -1}f\left( \frac{ab}{ta+(1-t)b}\right) dt+\int _{0}^{\frac{1}{2}}t^{\alpha -1}f\left( \frac{ab}{tb+(1-t)a}\right) dt \\&\quad \le \left[ f(a)+f(b)\right] \int _{0}^{\frac{1}{2}}t^{\alpha -1}dt=\frac{ 2^{1-\alpha }}{\alpha }\frac{f(a)+f(b)}{2} \end{aligned}$$
i.e.
$$\begin{aligned}&\left( \frac{ab}{b-a}\right) ^{\alpha }\varGamma (\alpha )\left[ J_{\frac{a+b }{2ab}+}^{\alpha }\left( f\circ h\right) (1/a)+J_{\frac{a+b}{2ab}-}^{\alpha }\left( f\circ h\right) (1/b)\right] \\&\quad \le \frac{2^{1-\alpha }}{\alpha }\left( \frac{f(a)+f(b)}{2}\right) . \end{aligned}$$
The proof is completed.
Theorem 5
Let
\(f:\left[ a,b\right] \mathbb {\rightarrow R}\)
be a harmonically
convex function with
\(a<b\)
and
\(f\in L\left[ a,b\right]\). If
\(g:\left[ a,b\right] \mathbb {\rightarrow R}\)
is nonnegative, integrable and harmonically symmetric with respect to
\(\frac{2ab}{a+b}\), then the following inequalities for fractional integrals holds:
$$\begin{aligned}&f\left( \frac{2ab}{a+b}\right) \left[ J_{\frac{a+b}{2ab}+}^{\alpha }\left( g\circ h\right) (1/a)+J_{\frac{a+b}{2ab}-}^{\alpha }\left( g\circ h\right) (1/b)\right] \\&\quad \le \left[ J_{\frac{a+b}{2ab}+}^{\alpha }\left( fg\circ h\right) (1/a)+J_{ \frac{a+b}{2ab}-}^{\alpha }\left( fg\circ h\right) (1/b)\right] \\&\quad \le \frac{f(a)+f(b)}{2}\left[ J_{\frac{a+b}{2ab}+}^{\alpha }\left( g\circ h\right) (1/a)+J_{\frac{a+b}{2ab}-}^{\alpha }\left( g\circ h\right) (1/b) \right] \end{aligned}$$
(9)
with
\(\alpha >0\)
and
\(h(x)=1/x\), \(x\in \left[ \frac{1}{b},\frac{1}{a}\right]\).
Proof
Since f is a harmonically convex function on \(\left[ a,b\right]\), multiplying both sides of (7) by \(2t^{\alpha -1}g\left( \frac{ab}{tb+(1-t)a}\right)\), then integrating the resulting inequality with respect to t over \(\left[ 0,\frac{1}{2}\right]\), we obtain
$$\begin{aligned}&2f\left( \frac{2ab}{a+b}\right) \int _{0}^{\frac{1}{2}}t^{\alpha -1}g\left( \frac{ab}{tb+(1-t)a}\right) dt \\&\quad \le \int _{0}^{\frac{1}{2}}t^{\alpha -1}\left[ f\left( \frac{ab}{ta+(1-t)b} \right) +f\left( \frac{ab}{tb+(1-t)a}\right) \right] g\left( \frac{ab}{ tb+(1-t)a}\right) dt \\&\quad =\int _{0}^{\frac{1}{2}}t^{\alpha -1}f\left( \frac{ab}{ta+(1-t)b}\right) g\left( \frac{ab}{tb+(1-t)a}\right) dt \\&\quad \quad +\int _{0}^{\frac{1}{2}}t^{\alpha -1}f\left( \frac{ab}{tb+(1-t)a}\right) g\left( \frac{ab}{tb+(1-t)a}\right) dt. \end{aligned}$$
Since g is harmonically symmetric with respect to \(\frac{2ab}{a+b}\), using Definition 3 we have \(g\left( \frac{1}{x}\right) =g\left(\frac{1}{(\frac{1 }{a})+(\frac{1}{b})-x}\right)\), for all \(x\in \left[ \frac{1}{b},\frac{1}{a}\right]\). Setting \(x=\frac{tb+(1-t)a}{ab}\) and \(dx=\left( \frac{b-a}{ab}\right) dt\) gives
$$\begin{aligned}&2\left( \frac{ab}{b-a}\right) ^{\alpha }f\left( \frac{2ab}{a+b}\right) \int _{\frac{1}{b}}^{\frac{a+b}{2ab}}\left( x-\frac{1}{b}\right) ^{\alpha -1}g\left( \frac{1}{x}\right) dx \\&\quad \le \left( \frac{ab}{b-a}\right) ^{\alpha }\left\{ \begin{array}{c} \int _{\frac{1}{b}}^{\frac{a+b}{2ab}}\left( x-\frac{1}{b}\right) ^{\alpha -1}f\left( \frac{1}{\frac{1}{a}+\frac{1}{b}-x}\right) g\left( \frac{1}{x} \right) dx \\ +\int _{\frac{1}{b}}^{\frac{a+b}{2ab}}\left( x-\frac{1}{b}\right) ^{\alpha -1}f\left( \frac{1}{x}\right) g\left( \frac{1}{x}\right) dx \end{array} \right\} \\&\quad =\left( \frac{ab}{b-a}\right) ^{\alpha }\left\{ \begin{array}{c} \int _{\frac{a+b}{2ab}}^{\frac{1}{a}}\left( \frac{1}{a}-x\right) ^{\alpha -1}f\left( \frac{1}{x}\right) g\left( \frac{1}{\frac{1}{a}+\frac{1}{b}-x} \right) dx \\ +\int _{\frac{1}{b}}^{\frac{a+b}{2ab}}\left( x-\frac{1}{b}\right) ^{\alpha -1}f\left( \frac{1}{x}\right) g\left( \frac{1}{x}\right) dx \end{array} \right\} \\&\quad =\left( \frac{ab}{b-a}\right) ^{\alpha }\left\{ \begin{array}{c} \int _{\frac{a+b}{2ab}}^{\frac{1}{a}}\left( \frac{1}{a}-x\right) ^{\alpha -1}f\left( \frac{1}{x}\right) g\left( \frac{1}{x}\right) dx \\ +\int _{\frac{1}{b}}^{\frac{a+b}{2ab}}\left( x-\frac{1}{b}\right) ^{\alpha -1}f\left( \frac{1}{x}\right) g\left( \frac{1}{x}\right) dx \end{array} \right\} . \end{aligned}$$
Therefore, by Lemma 2 we have
$$\begin{aligned}&\left( \frac{ab}{b-a}\right) ^{\alpha }\varGamma (\alpha )f\left( \frac{2ab}{ a+b}\right) \left[ J_{\frac{a+b}{2ab}+}^{\alpha }\left( g\circ h\right) (1/a)+J_{\frac{a+b}{2ab}-}^{\alpha }\left( g\circ h\right) (1/b)\right] \\&\quad \le \left( \frac{ab}{b-a}\right) ^{\alpha }\varGamma (\alpha )\left[ J_{ \frac{a+b}{2ab}+}^{\alpha }\left( fg\circ h\right) (1/a)+J_{\frac{a+b}{2ab} -}^{\alpha }\left( fg\circ h\right) (1/b)\right] \end{aligned}$$
and the first inequality is proved.
For the proof of the second inequality in (9) we first note that if f is a harmonically convex function, then, multiplying both sides of (8) by \(t^{\alpha -1}g\left( \frac{ab}{ tb+(1-t)a}\right)\) and integrating the resulting inequality with respect to t over \(\left[ 0,\frac{1}{2}\right]\), we obtain
$$\begin{aligned}&\int _{0}^{\frac{1}{2}}t^{\alpha -1}f\left( \frac{ab}{ta+(1-t)b}\right) g\left( \frac{ab}{tb+(1-t)a}\right) dt \\&\quad\quad +\int _{0}^{\frac{1}{2}}t^{\alpha -1}f\left( \frac{ab}{tb+(1-t)a}\right) g\left( \frac{ab}{tb+(1-t)a}\right) dt \\&\quad \le \left[ f(a)+f(b)\right] \int _{0}^{\frac{1}{2}}t^{\alpha -1}g\left( \frac{ab}{tb+(1-t)a}\right) dt \end{aligned}$$
i.e.
$$\begin{aligned}&\left( \frac{ab}{b-a}\right) ^{\alpha }\varGamma (\alpha )\left[ J_{\frac{a+b }{2ab}+}^{\alpha }\left( fg\circ h\right) (1/a)+J_{\frac{a+b}{2ab}-}^{\alpha }\left( fg\circ h\right) (1/b)\right] \\&\quad \le \left( \frac{ab}{b-a}\right) ^{\alpha }\varGamma (\alpha )\left( \frac{ f(a)+f(b)}{2}\right) \left[ J_{\frac{a+b}{2ab}+}^{\alpha }\left( g\circ h\right) (1/a)+J_{\frac{a+b}{2ab}-}^{\alpha }\left( g\circ h\right) (1/b) \right] . \end{aligned}$$
The proof is completed.
Remark 1
In Theorem 5,
-
(i)
if we take \(\alpha =1\), then inequality (9) becomes inequality (5) of Theorem 3.
-
(ii)
if we take \(g(x)=1\), then inequality (9) becomes inequality (6) of Theorem 4.
-
(iii)
if we take \(\alpha =1\) and \(g(x)=1\), then inequality (9) becomes inequality (4) of Theorem 2.
Lemma 3
Let
\(f:I\subset \left( 0,\infty \right) \rightarrow \mathbb {R}\)
be a differentiable function on
\(I {{}^\circ }\)
, the interior of
I
, such that
\(f^{\prime }\in L\left[ a,b\right]\)
, where
\(a,b\in I\)
. If
\(g:\left[ a,b\right] \mathbb {\rightarrow R}\)
is integrable and harmonically symmetric with respect to
\(\frac{2ab}{a+b}\)
, then the following equality for fractional integrals holds:
$$\begin{aligned}&f\left( \frac{2ab}{a+b}\right) \left[ J_{\frac{a+b}{2ab}+}^{\alpha }\left( g\circ h\right) (1/a)+J_{\frac{a+b}{2ab}-}^{\alpha }\left( g\circ h\right) (1/b)\right] \\&\quad \quad -\left[ J_{\frac{a+b}{2ab}+}^{\alpha }\left( fg\circ h\right) (1/a)+J_{ \frac{a+b}{2ab}-}^{\alpha }\left( fg\circ h\right) (1/b)\right] \\&\quad =\frac{1}{\varGamma (\alpha )}\left[ \begin{array}{c} \int _{\frac{1}{b}}^{\frac{a+b}{2ab}}\left( \int _{\frac{1}{b}}^{t}\left( s- \frac{1}{b}\right) ^{\alpha -1}\left( g\circ h\right) (s)ds\right) \left( f\circ h\right) ^{\prime }(t)dt \\ -\int _{\frac{a+b}{2ab}}^{\frac{1}{a}}\left( \int _{t}^{\frac{1}{a}}\left( \frac{1}{a}-s\right) ^{\alpha -1}\left( g\circ h\right) (s)ds\right) \left( f\circ h\right) ^{\prime }(t)dt \end{array} \right] \end{aligned}$$
(10)
with
\(\alpha >0\)
and
\(h(x)=1/x\), \(x\in \left[ \frac{1}{b},\frac{1}{a}\right]\).
Proof
It suffices to note that
$$\begin{aligned} I&= {} \int _{\frac{1}{b}}^{\frac{a+b}{2ab}}\left( \int _{\frac{1}{b}}^{t}\left( s-\frac{1}{b}\right) ^{\alpha -1}\left( g\circ h\right) (s)ds\right) \left( f\circ h\right) ^{\prime }(t)dt \\&\quad-\int _{\frac{a+b}{2ab}}^{\frac{1}{a}}\left( \int _{t}^{\frac{1}{a}}\left( \frac{1}{a}-s\right) ^{\alpha -1}\left( g\circ h\right) (s)ds\right) \left( f\circ h\right) ^{\prime }(t)dt \\&= {} I_{1}-I_{2}. \end{aligned}$$
By integration by parts and Lemma 2 we get
$$\begin{aligned} I_{1}&= {} \left. \left( \int _{\frac{1}{b}}^{t}\left( s-\frac{1}{b}\right) ^{\alpha -1}\left( g\circ h\right) (s)ds\right) \left( f\circ h\right) (t)\right| _{\frac{1}{b}}^{\frac{a+b}{2ab}} \\&\quad-\int _{\frac{1}{b}}^{\frac{a+b}{2ab}}\left( t-\frac{1}{b}\right) ^{\alpha -1}\left( g\circ h\right) (t)\left( f\circ h\right) (t)dt \\&= {} \left( \int _{\frac{1}{b}}^{\frac{a+b}{2ab}}\left( s-\frac{1}{b}\right) ^{\alpha -1}\left( g\circ h\right) (s)ds\right) f\left(\frac{2ab}{a+b}\right) \\&\quad-\int _{\frac{1}{b}}^{\frac{a+b}{2ab}}\left( t-\frac{1}{b}\right) ^{\alpha -1}\left( g\circ h\right) (t)\left( f\circ h\right) (t)dt \\&= {} \varGamma (\alpha )\left[ f\left(\frac{2ab}{a+b}\right)J_{\frac{a+b}{2ab}-}^{\alpha }\left( g\circ h\right) (1/b)-J_{\frac{a+b}{2ab}-}^{\alpha }\left( fg\circ h\right) (1/b)\right] \\&= {} \varGamma (\alpha )\left[ \begin{array}{l} \frac{f\left(\frac{2ab}{a+b}\right)}{2}\left[ J_{\frac{a+b}{2ab}+}^{\alpha }\left( g\circ h\right) (1/a)+J_{\frac{a+b}{2ab}-}^{\alpha }\left( g\circ h\right) (1/b)\right] \\ -J_{\frac{a+b}{2ab}-}^{\alpha }\left( fg\circ h\right) (1/b) \end{array} \right] \end{aligned}$$
and similarly
$$\begin{aligned} I_{2}&= {} \left. \left( \int _{t}^{\frac{1}{a}}\left( \frac{1}{a}-s\right) ^{\alpha -1}\left( g\circ h\right) (s)ds\right) \left( f\circ h\right) (t)\right| _{\frac{a+b}{2ab}}^{\frac{1}{a}} \\&\quad+\int _{\frac{a+b}{2ab}}^{\frac{1}{a}}\left( \frac{1}{a}-t\right) ^{\alpha -1}\left( g\circ h\right) (t)\left( f\circ h\right) (t)dt \\&= {} -\left( \int _{\frac{a+b}{2ab}}^{\frac{1}{a}}\left( \frac{1}{a}-s\right) ^{\alpha -1}\left( g\circ h\right) (s)ds\right) f\left( \frac{2ab}{a+b}\right) \\&\quad+\int _{\frac{a+b}{2ab}}^{\frac{1}{a}}\left( \frac{1}{a}-t\right) ^{\alpha -1}\left( g\circ h\right) (t)\left( f\circ h\right) (t)dt \\&= {} \varGamma (\alpha )\left[ -f(\frac{2ab}{a+b})J_{\frac{a+b}{2ab}+}^{\alpha }\left( g\circ h\right) (1/a)+J_{\frac{a+b}{2ab}+}^{\alpha }\left( fg\circ h\right) (1/a)\right] \\&= {} \varGamma (\alpha )\left[ \begin{array}{l} -\frac{f(\frac{2ab}{a+b})}{2}\left[ J_{\frac{a+b}{2ab}+}^{\alpha }\left( g\circ h\right) (1/a)+J_{\frac{a+b}{2ab}-}^{\alpha }\left( g\circ h\right) (1/b)\right] \\ +J_{\frac{a+b}{2ab}+}^{\alpha }\left( fg\circ h\right) (1/a) \end{array} \right] . \end{aligned}$$
Thus, we can write
$$\begin{aligned} I=I_{1}-I_{2}=\varGamma (\alpha )\left\{ \begin{array}{l} f\left( \frac{2ab}{a+b}\right) \left[ J_{\frac{a+b}{2ab}+}^{\alpha }\left( g\circ h\right) (1/a)+J_{\frac{a+b}{2ab}-}^{\alpha }\left( g\circ h\right) (1/b)\right] \\ -\left[ J_{\frac{a+b}{2ab}+}^{\alpha }\left( fg\circ h\right) (1/a)+J_{\frac{ a+b}{2ab}-}^{\alpha }\left( fg\circ h\right) (1/b)\right] \end{array} \right\} . \end{aligned}$$
Multiplying both sides by \(\left( \varGamma (\alpha )\right) ^{-1}\) we obtain (10). This completes the proof.
Theorem 6
Let
\(f:I\subset \left( 0,\infty \right) \rightarrow \mathbb {R}\)
be a differentiable function on
\(I {{}^\circ }\), the interior of
I, such that
\(f^{\prime }\in L\left[ a,b\right]\), where
\(a,b\in I\)
and
\(a<b\). If
\(\left| f^{\prime }\right|\)
is harmonically convex on
\(\left[ a,b\right]\), \(g:\left[ a,b\right] \mathbb { \rightarrow R}\)
is continuous and harmonically symmetric with respect to
\(\frac{2ab}{a+b}\), then the following inequality for fractional integrals holds:
$$\begin{aligned}&\left| \begin{array}{c} f\left( \frac{2ab}{a+b}\right) \left[ J_{\frac{a+b}{2ab}+}^{\alpha }\left( g\circ h\right) (1/a)+J_{\frac{a+b}{2ab}-}^{\alpha }\left( g\circ h\right) (1/b)\right] \\ -\left[ J_{\frac{a+b}{2ab}+}^{\alpha }\left( fg\circ h\right) (1/a)+J_{\frac{ a+b}{2ab}-}^{\alpha }\left( fg\circ h\right) (1/b)\right] \end{array} \right| \\&\quad \le \frac{\left\| g\right\| _{\infty }ab\left( b-a\right) }{\varGamma (\alpha +1)}\left( \frac{b-a}{ab}\right) ^{\alpha }\left[ C_{1}\left( \alpha \right) \left| f^{\prime }\left( a\right) \right| +C_{2}\left( \alpha \right) \left| f^{\prime }\left( b\right) \right| \right] \end{aligned}$$
(11)
where
$$\begin{aligned} C_{1}\left( \alpha \right)&= {} \left[ \begin{array}{c} \frac{b^{-2}}{\left( \alpha +1\right) \left( \alpha +2\right) } \begin{array}{c} _{2}F_{1}\left( 2,\alpha +1;\alpha +3;1-\frac{a}{b}\right) \end{array} \\ -\frac{\left( a+b\right) ^{-2}}{\left( \alpha +1\right) \left( \alpha +2\right) } \begin{array}{c} _{2}F_{1}\left( 2,\alpha +1;\alpha +3;\frac{b-a}{b+a}\right) \end{array} \end{array} \right] ,\\ C_{2}\left( \alpha \right)&= {} \left[ \begin{array}{c} \frac{b^{-2}}{\alpha +2} \begin{array}{c} _{2}F_{1}\left( 2,\alpha +2;\alpha +3;1-\frac{a}{b}\right) \end{array} \\ -\frac{2\left( a+b\right) ^{-2}}{\alpha +1} \begin{array}{c} _{2}F_{1}\left( 2,\alpha +1;\alpha +2;\frac{b-a}{b+a}\right) \end{array} \\ +\frac{\left( a+b\right) ^{-2}}{\left( \alpha +1\right) \left( \alpha +2\right) } \begin{array}{c} _{2}F_{1}\left( 2,\alpha +1;\alpha +3;\frac{b-a}{b+a}\right) \end{array} \end{array} \right] , \end{aligned}$$
with
\(0<\alpha \le 1\)
and
\(h(x)=1/x\), \(x\in \left[ \frac{1}{b},\frac{1}{a} \right]\).
Proof
From Lemma 3 we have
$$\begin{aligned}&\left| \begin{array}{c} f\left( \frac{2ab}{a+b}\right) \left[ J_{\frac{a+b}{2ab}+}^{\alpha }\left( g\circ h\right) (1/a)+J_{\frac{a+b}{2ab}-}^{\alpha }\left( g\circ h\right) (1/b)\right] \\ -\left[ J_{\frac{a+b}{2ab}+}^{\alpha }\left( fg\circ h\right) (1/a)+J_{\frac{ a+b}{2ab}-}^{\alpha }\left( fg\circ h\right) (1/b)\right] \end{array} \right| \\&\quad \le \frac{1}{\varGamma (\alpha )}\left[ \begin{array}{c} \int _{\frac{1}{b}}^{\frac{a+b}{2ab}}\left( \int _{\frac{1}{b}}^{t}\left( s- \frac{1}{b}\right) ^{\alpha -1}\left| \left( g\circ h\right) (s)\right| ds\right) \left| \left( f\circ h\right) ^{\prime }(t)\right| dt \\ +\int _{\frac{a+b}{2ab}}^{\frac{1}{a}}\left( \int _{t}^{\frac{1}{a}}\left( \frac{1}{a}-s\right) ^{\alpha -1}\left| \left( g\circ h\right) (s)\right| ds\right) \left| \left( f\circ h\right) ^{\prime }(t)\right| dt \end{array} \right] \\&\quad \le \frac{\left\| g\right\| _{\infty }}{\varGamma (\alpha )}\left[ \begin{array}{c} \int _{\frac{1}{b}}^{\frac{a+b}{2ab}}\left( \int _{\frac{1}{b}}^{t}\left( s- \frac{1}{b}\right) ^{\alpha -1}ds\right) \left| \left( f\circ h\right) ^{\prime }(t)\right| dt \\ +\int _{\frac{a+b}{2ab}}^{\frac{1}{a}}\left( \int _{t}^{\frac{1}{a}}\left( \frac{1}{a}-s\right) ^{\alpha -1}ds\right) \left| \left( f\circ h\right) ^{\prime }(t)\right| dt \end{array} \right] \\&\quad =\frac{\left\| g\right\| _{\infty }}{\varGamma (\alpha )}\left[ \begin{array}{c} \int _{\frac{1}{b}}^{\frac{a+b}{2ab}}\frac{\left( t-\frac{1}{b}\right) ^{\alpha }}{\alpha }\frac{1}{t^{2}}\left| f^{\prime }(\frac{1}{t} )\right| dt \\ +\int _{\frac{a+b}{2ab}}^{\frac{1}{a}}\frac{\left( \frac{1}{a}-t\right) ^{\alpha }}{\alpha }\frac{1}{t^{2}}\left| f^{\prime }(\frac{1}{t} )\right| dt \end{array} \right] . \end{aligned}$$
Setting \(t=\frac{ub+(1-u)a}{ab}\) and \(dt=\left( \frac{b-a}{ab}\right) du\) gives
$$\begin{aligned}&\left| \begin{array}{c} f\left( \frac{2ab}{a+b}\right) \left[ J_{\frac{a+b}{2ab}+}^{\alpha }\left( g\circ h\right) (1/a)+J_{\frac{a+b}{2ab}-}^{\alpha }\left( g\circ h\right) (1/b)\right] \\ -\left[ J_{\frac{a+b}{2ab}+}^{\alpha }\left( fg\circ h\right) (1/a)+J_{\frac{ a+b}{2ab}-}^{\alpha }\left( fg\circ h\right) (1/b)\right] \end{array} \right| \\&\quad \le \frac{\left\| g\right\| _{\infty }ab\left( b-a\right) }{\varGamma (\alpha +1)}\left( \frac{b-a}{ab}\right) ^{\alpha }\left[ \begin{array}{c} \int _{0}^{\frac{1}{2}}\frac{u^{\alpha }}{\left( ub+\left( 1-u\right) a\right) ^{2}}\left| f^{\prime }(\frac{ab}{ub+\left( 1-u\right) a} )\right| du \\ +\int _{\frac{1}{2}}^{1}\frac{\left( 1-u\right) ^{\alpha }}{\left( ub+\left( 1-u\right) a\right) ^{2}}\left| f^{\prime }\left( \frac{ab}{ub+\left( 1-u\right) a}\right) \right| du \end{array} \right] . \end{aligned}$$
(12)
Since \(\left| f^{\prime }\right|\) is harmonically convex on \(\left[ a,b\right]\), we have
$$\begin{aligned} \left| f^{\prime }\left(\frac{ab}{ub+(1-u)at}\right)\right| \le u\left| f^{\prime }\left( a\right) \right| +\left( 1-u\right) \left| f^{\prime }\left( b\right) \right| . \end{aligned}$$
(13)
If we use (13) in (12) , we have
$$\begin{aligned}&\left| \begin{array}{c} f\left( \frac{2ab}{a+b}\right) \left[ J_{\frac{a+b}{2ab}+}^{\alpha }\left( g\circ h\right) (1/a)+J_{\frac{a+b}{2ab}-}^{\alpha }\left( g\circ h\right) (1/b)\right] \\ -\left[ J_{\frac{a+b}{2ab}+}^{\alpha }\left( fg\circ h\right) (1/a)+J_{\frac{ a+b}{2ab}-}^{\alpha }\left( fg\circ h\right) (1/b)\right] \end{array} \right| \\&\quad \le \frac{\left\| g\right\| _{\infty }ab\left( b-a\right) }{\varGamma (\alpha +1)}\left( \frac{b-a}{ab}\right) ^{\alpha } \\&\quad \quad \times \left[ \begin{array}{c} \int _{0}^{\frac{1}{2}}\frac{u^{\alpha }}{\left( ub+\left( 1-u\right) a\right) ^{2}}\left[ u\left| f^{\prime }\left( a\right) \right| +\left( 1-u\right) \left| f^{\prime }\left( b\right) \right| \right] du \\ +\int _{\frac{1}{2}}^{1}\frac{\left( 1-u\right) ^{\alpha }}{\left( ub+\left( 1-u\right) a\right) ^{2}}\left[ u\left| f^{\prime }\left( a\right) \right| +\left( 1-u\right) \left| f^{\prime }\left( b\right) \right| \right] du \end{array} \right] . \end{aligned}$$
(14)
Calculating the following integrals by Lemma 1, we have
$$\begin{aligned}&\int _{0}^{\frac{1}{2}}\frac{u^{\alpha +1}}{\left( ub+(1-u)a\right) ^{2}} du+\int _{\frac{1}{2}}^{1}\frac{\left( 1-u\right) ^{\alpha }u}{\left( ub+(1-u)a\right) ^{2}}du \\&\quad =\int _{0}^{1}\frac{\left( 1-u\right) ^{\alpha }u}{\left( ub+(1-u)a\right) ^{2}}du-\int _{0}^{\frac{1}{2}}\frac{\left( 1-u\right) ^{\alpha }-u^{\alpha } }{\left( ub+(1-u)a\right) ^{2}}udu \\&\quad \le \int _{0}^{1}\frac{\left( 1-u\right) ^{\alpha }u}{\left( ub+(1-u)a\right) ^{2}}du-\int _{0}^{\frac{1}{2}}\frac{\left( 1-2u\right) ^{\alpha }}{\left( ub+(1-u)a\right) ^{2}}udu \\&\quad =\int _{0}^{1}\frac{\left( 1-u\right) ^{\alpha }u}{\left( ub+(1-u)a\right) ^{2}}du-\frac{1}{4}\int _{0}^{1}\frac{\left( 1-u\right) ^{\alpha }}{\left( \frac{u}{2}b+(1-\frac{u}{2})a\right) ^{2}}udu \\&\quad =\int _{0}^{1}\left( 1-u\right) u^{\alpha }b^{-2}\left( 1-u\left( 1-\frac{a }{b}\right) \right) ^{-2}du \\&\quad \quad -\frac{1}{4}\int _{0}^{1}\left( 1-v\right) v^{\alpha }\left( \frac{a+b}{2} \right) ^{-2}\left( 1-v\left( \frac{b-a}{b+a}\right) \right) ^{-2}dv \\&\quad =\left[ \begin{array}{c} \frac{b^{-2}}{\left( \alpha +1\right) \left( \alpha +2\right) } \begin{array}{c} _{2}F_{1}\left( 2,\alpha +1;\alpha +3;1-\frac{a}{b}\right) \end{array} \\ -\frac{\left( a+b\right) ^{-2}}{\left( \alpha +1\right) \left( \alpha +2\right) } \begin{array}{c} _{2}F_{1}\left( 2,\alpha +1;\alpha +3;\frac{b-a}{b+a}\right) \end{array} \end{array} \right] \\&\quad =C_{1}\left( \alpha \right) \end{aligned}$$
(15)
and similarly we get
$$\begin{aligned}&\int _{0}^{\frac{1}{2}}\frac{u^{\alpha }}{\left( ub+(1-u)a\right) ^{2}} \left( 1-u\right) du+\int _{\frac{1}{2}}^{1}\frac{\left( 1-u\right) ^{\alpha } }{\left( ub+(1-u)a\right) ^{2}}\left( 1-u\right) du \\&\quad =\int _{0}^{1}\frac{\left( 1-u\right) ^{\alpha +1}}{\left( ub+(1-u)a\right) ^{2}}du-\int _{0}^{\frac{1}{2}}\frac{\left( 1-u\right) ^{\alpha }-u^{\alpha } }{\left( ub+(1-u)a\right) ^{2}}\left( 1-u\right) du \\&\quad \le \int _{0}^{1}\frac{\left( 1-u\right) ^{\alpha +1}}{\left( ub+(1-u)a\right) ^{2}}du-\int _{0}^{\frac{1}{2}}\frac{\left( 1-2u\right) ^{\alpha }}{\left( ub+(1-u)a\right) ^{2}}\left( 1-u\right) du \\&\quad =\int _{0}^{1}\frac{\left( 1-u\right) ^{\alpha +1}}{\left( ub+(1-u)a\right) ^{2}}du-\int _{0}^{\frac{1}{2}}\frac{\left( 1-2u\right) ^{\alpha }}{\left( ub+(1-u)a\right) ^{2}}du \\&\quad \quad +\int _{0}^{\frac{1}{2}}\frac{u\left( 1-2u\right) ^{\alpha }}{\left( ub+(1-u)a\right) ^{2}}du \\&\quad =\int _{0}^{1}\frac{u^{\alpha +1}}{\left( ua+(1-u)b\right) ^{2}}du-\frac{1}{ 2}\int _{0}^{1}\frac{\left( 1-u\right) ^{\alpha }}{\left( \frac{u}{2}b+(1- \frac{u}{2})a\right) ^{2}}du \\&\quad \quad +\frac{1}{4}\int _{0}^{1}\frac{u\left( 1-u\right) ^{\alpha }}{\left( \frac{u }{2}b+(1-\frac{u}{2})a\right) ^{2}}du \\&\quad =\int _{0}^{1}\frac{u^{\alpha +1}}{\left( ua+(1-u)b\right) ^{2}}du-\frac{1}{ 2}\int _{0}^{1}v^{\alpha }\left( \frac{a+b}{2}\right) ^{-2}\left( 1-v\left( \frac{b-a}{b+a}\right) \right) ^{-2}dv \\&\quad \quad +\frac{1}{4}\int _{0}^{1}\left( 1-v\right) v^{\alpha }\left( \frac{a+b}{2} \right) ^{-2}\left( 1-v\left( \frac{b-a}{b+a}\right) \right) ^{-2}dv \\&\quad =\left[ \begin{array}{c} \frac{b^{-2}}{\alpha +2} \begin{array}{c} _{2}F_{1}\left( 2,\alpha +2;\alpha +3;1-\frac{a}{b}\right) \end{array} \\ -\frac{2\left( a+b\right) ^{-2}}{\alpha +1} \begin{array}{c} _{2}F_{1}\left( 2,\alpha +1;\alpha +2;\frac{b-a}{b+a}\right) \end{array} \\ +\frac{\left( a+b\right) ^{-2}}{\left( \alpha +1\right) \left( \alpha +2\right) } \begin{array}{c} _{2}F_{1}\left( 2,\alpha +1;\alpha +3;\frac{b-a}{b+a}\right) \end{array} \end{array} \right] \\&\quad =C_{2}\left( \alpha \right) . \end{aligned}$$
(16)
If we use (15) and (16) in (14) , we have (11). This completes the proof.
Corollary 1
In Theorem 6:
(1) If we take
\(\alpha =1\)
we have the following Hermite–Hadamard–Fejer inequality for harmonically convex functions which is related to the left-hand side of (
5
):
$$\begin{aligned}&\left| f\left( \frac{2ab}{a+b}\right) \int _{a}^{b}\frac{g\left( x\right) }{x^{2}}dx-\int _{a}^{b}\frac{f\left( x\right) g\left( x\right) }{ x^{2}}dx\right| \\&\quad \le \left\| g\right\| _{\infty }\left( b-a\right) ^{2}\left[ C_{1}\left( 1\right) \left| f^{\prime }\left( a\right) \right| +C_{2}\left( 1\right) \left| f^{\prime }\left( b\right) \right| \right] , \end{aligned}$$
(2) If we take
\(g\left( x\right) =1\)
we have following Hermite–Hadamard type inequality for harmonically convex functions in fractional integral forms which is related to the left-hand side of (
6
):
$$\begin{aligned}&\left| f\left( \frac{2ab}{a+b}\right) -\frac{\varGamma \left( \alpha +1\right) }{2^{1-\alpha }}\left( \frac{ab}{b-a}\right) ^{\alpha }\left\{ \begin{array}{c} J_{\frac{a+b}{2ab}+}^{\alpha }\left( f\circ h\right) \left( 1/a\right) \\ +J_{\frac{a+b}{2ab}-}^{\alpha }\left( f\circ h\right) \left( 1/b\right) \end{array} \right\} \right| \\&\quad \le \frac{ab\left( b-a\right) }{2^{1-\alpha }}\left[ C_{1}\left( \alpha \right) \left| f^{\prime }\left( a\right) \right| +C_{2}\left( \alpha \right) \left| f^{\prime }\left( b\right) \right| \right] , \end{aligned}$$
(3) If we take
\(\alpha =1\)
and
\(g\left( x\right) =1\)
we have the following Hermite–Hadamard type inequality for harmonically convex functions which is related to the left-hand side of (
4
):
$$\left| f\left( \frac{2ab}{a+b}\right) -\frac{ab}{b-a}\int _{a}^{b}\frac{ f\left( x\right) }{x^{2}}dx\right| \le ab\left( b-a\right) \left[ C_{1}\left( 1\right) \left| f^{\prime }\left( a\right) \right| +C_{2}\left( 1\right) \left| f^{\prime }\left( b\right) \right| \right] .$$
Theorem 7
Let
\(f:I\subset \left( 0,\infty \right) \rightarrow \mathbb {R}\)
be a differentiable function on
\(I {{}^\circ }\), the interior of
I, such that
\(f^{\prime }\in L\left[ a,b\right]\), where
\(a,b\in I\). If
\(\left| f^{\prime }\right| ^{q},q\ge 1,\)
is harmonically
convex on
\(\left[ a,b\right]\), \(g:\left[ a,b\right] \mathbb { \rightarrow R}\)
is continuous and harmonically symmetric with respect to
\(\frac{2ab}{a+b}\), then the following inequality for fractional integrals holds:
$$\begin{aligned}&\left| \begin{array}{c} \frac{f(a)+f(b)}{2}\left[ J_{1/b+}^{\alpha }\left( g\circ h\right) (1/a)+J_{1/a-}^{\alpha }\left( g\circ h\right) (1/b)\right] \\ -\left[ J_{1/b+}^{\alpha }\left( fg\circ h\right) (1/a)+J_{1/a-}^{\alpha }\left( fg\circ h\right) (1/b)\right] \end{array} \right| \\&\quad \le \frac{\left\| g\right\| _{\infty }ab\left( b-a\right) }{\varGamma (\alpha +1)}\left( \frac{b-a}{ab}\right) ^{\alpha } \\&\quad \quad \times \left[ \begin{array}{l} C_{3}^{^{1-\frac{1}{q}}}\left( \alpha \right) \left[ \left( \begin{array}{c} C_{4}\left( \alpha \right) \left| f^{\prime }(a)\right| ^{q} \\ +C_{5}\left( \alpha \right) \left| f^{\prime }(b)\right| ^{q} \end{array} \right) \right] ^{\frac{1}{q}} \\ +C_{6}^{1-\frac{1}{q}}\left( \alpha \right) \left[ \left( \begin{array}{c} C_{7}\left( \alpha \right) \left| f^{\prime }(a)\right| ^{q} \\ +C_{8}\left( \alpha \right) \left| f^{\prime }(b)\right| ^{q} \end{array} \right) \right] ^{\frac{1}{q}} \end{array} \right] \end{aligned}$$
(17)
where
$$\begin{aligned} C_{3}\left( \alpha \right)&= {} \frac{\left( a+b\right) ^{-2}}{2^{\alpha -1}\left( \alpha +1\right) } \begin{array}{c} _{2}F_{1}\left( 2,1;\alpha +2;\frac{b-a}{b+a}\right) \end{array} , \\ C_{4}\left( \alpha \right)&= {} \frac{\left( a+b\right) ^{-2}}{2^{\alpha }\left( \alpha +2\right) } \begin{array}{c} _{2}F_{1}\left( 2,1;\alpha +3;\frac{b-a}{b+a}\right) \end{array} , \\ C_{5}\left( \alpha \right)&= {} C_{3}\left( \alpha \right) -C_{4}\left( \alpha \right) , \\ \text { }C_{6}\left( \alpha \right)&= {} \frac{b^{-2}}{2^{\alpha +1}\left( \alpha +1\right) } \begin{array}{c} _{2}F_{1}\left( 2,\alpha +1;\alpha +2;\frac{1}{2}\left( 1-\frac{a}{b}\right) \right) \end{array} , \\ C_{7}\left( \alpha \right)&= {} \left[ \begin{array}{c} \frac{b^{-2}}{2^{\alpha +1}\left( \alpha +1\right) } \begin{array}{c} _{2}F_{1}\left( 2,\alpha +1;\alpha +2;\frac{1}{2}\left( 1-\frac{a}{b}\right) \right) \end{array} \\ -\frac{b^{-2}}{2^{\alpha +2}\left( \alpha +2\right) } \begin{array}{c} _{2}F_{1}\left( 2,\alpha +2;\alpha +3;\frac{1}{2}\left( 1-\frac{a}{b}\right) \right) \end{array} \end{array} \right] , \\ C_{8}\left( \alpha \right)&= {} C_{6}\left( \alpha \right) -C_{7}\left( \alpha \right) , \end{aligned}$$
with
\(\alpha >1\)
and
\(h(x)=1/x\), \(x\in \left[ \frac{1}{b},\frac{1}{a}\right]\).
Proof
Using (12) , power mean inequality and the harmonically convexity of \(\left| f^{\prime }\right| ^{q}\), it follows that
$$\begin{aligned}&\left| \begin{array}{c} f\left( \frac{2ab}{a+b}\right) \left[ J_{\frac{a+b}{2ab}+}^{\alpha }\left( g\circ h\right) (1/a)+J_{\frac{a+b}{2ab}-}^{\alpha }\left( g\circ h\right) (1/b)\right] \\ -\left[ J_{\frac{a+b}{2ab}+}^{\alpha }\left( fg\circ h\right) (1/a)+J_{\frac{ a+b}{2ab}-}^{\alpha }\left( fg\circ h\right) (1/b)\right] \end{array} \right| \\&\quad \le \frac{\left\| g\right\| _{\infty }ab\left( b-a\right) }{\varGamma (\alpha +1)}\left( \frac{b-a}{ab}\right) ^{\alpha }\left[ \begin{array}{c} \int _{0}^{\frac{1}{2}}\frac{u^{\alpha }}{\left( ub+\left( 1-u\right) a\right) ^{2}}\left| f^{\prime }(\frac{ab}{ub+\left( 1-u\right) a} )\right| du \\ +\int _{\frac{1}{2}}^{1}\frac{\left( 1-u\right) ^{\alpha }}{\left( ub+\left( 1-u\right) a\right) ^{2}}\left| f^{\prime }\left(\frac{ab}{ub+\left( 1-u\right) a}\right)\right| du \end{array} \right] \\&\quad \le \frac{\left\| g\right\| _{\infty }ab\left( b-a\right) }{\varGamma (\alpha +1)}\left( \frac{b-a}{ab}\right) ^{\alpha } \\&\quad \left[ \begin{array}{c} \left( \int _{0}^{\frac{1}{2}}\frac{u^{\alpha }}{\left( ub+\left( 1-u\right) a\right) ^{2}}du\right) ^{1-\frac{1}{q}} \\ \times \left( \int _{0}^{\frac{1}{2}}\frac{u^{\alpha }}{\left( ub+\left( 1-u\right) a\right) ^{2}}\left| f^{\prime }\left(\frac{ab}{ub+(1-u)a} \right)\right| ^{q}du\right) ^{\frac{1}{q}} \\ +\left( \int _{\frac{1}{2}}^{1}\frac{\left( 1-u\right) ^{\alpha }}{\left( ub+\left( 1-u\right) a\right) ^{2}}du\right) ^{1-\frac{1}{q}} \\ \times \left( \int _{\frac{1}{2}}^{1}\frac{\left( 1-u\right) ^{\alpha }}{ \left( ub+\left( 1-u\right) a\right) ^{2}}\left| f^{\prime }\left(\frac{ab}{ ub+(1-u)a}\right)\right| ^{q}du\right) ^{\frac{1}{q}} \end{array} \right] \\&\quad \le \frac{\left\| g\right\| _{\infty }ab\left( b-a\right) }{\varGamma (\alpha +1)}\left( \frac{b-a}{ab}\right) ^{\alpha } \\&\quad \times \left[ \begin{array}{c} \left( \int _{0}^{\frac{1}{2}}\frac{u^{\alpha }}{\left( ub+\left( 1-u\right) a\right) ^{2}}du\right) ^{1-\frac{1}{q}} \\ \times \left( \int _{0}^{\frac{1}{2}}\frac{u^{\alpha }}{\left( ub+\left( 1-u\right) a\right) ^{2}}\left[ u\left| f^{\prime }(a)\right| ^{q}+\left( 1-u\right) \left| f^{\prime }(b)\right| ^{q}\right] du\right) ^{\frac{1}{q}} \\ +\left( \int _{\frac{1}{2}}^{1}\frac{\left( 1-u\right) ^{\alpha }}{\left( ub+\left( 1-u\right) a\right) ^{2}}du\right) ^{1-\frac{1}{q}} \\ \times \left( \int _{\frac{1}{2}}^{1}\frac{\left( 1-u\right) ^{\alpha }}{ \left( ub+\left( 1-u\right) a\right) ^{2}}\left[ u\left| f^{\prime }(a)\right| ^{q}+\left( 1-u\right) \left| f^{\prime }(b)\right| ^{q}\right] du\right) ^{\frac{1}{q}} \end{array} \right] \\&\quad =\frac{\left\| g\right\| _{\infty }ab\left( b-a\right) }{\varGamma (\alpha +1)}\left( \frac{b-a}{ab}\right) ^{\alpha } \\&\quad \times \left[ \begin{array}{c} \left( \int _{0}^{\frac{1}{2}}\frac{u^{\alpha }}{\left( ub+\left( 1-u\right) a\right) ^{2}}du\right) ^{1-\frac{1}{q}} \\ \times \left( \begin{array}{c} \int _{0}^{\frac{1}{2}}\frac{u^{\alpha +1}}{\left( ub+\left( 1-u\right) a\right) ^{2}}du\left| f^{\prime }(a)\right| ^{q} \\ +\int _{0}^{\frac{1}{2}}\frac{u^{\alpha }}{\left( ub+\left( 1-u\right) a\right) ^{2}}\left( 1-u\right) du\left| f^{\prime }(b)\right| ^{q} \end{array} \right) ^{\frac{1}{q}} \\ +\left( \int _{\frac{1}{2}}^{1}\frac{\left( 1-u\right) ^{\alpha }}{\left( ub+\left( 1-u\right) a\right) ^{2}}du\right) ^{1-\frac{1}{q}} \\ \times \left( \begin{array}{c} \int _{\frac{1}{2}}^{1}\frac{\left( 1-u\right) ^{\alpha }}{\left( ub+\left( 1-u\right) a\right) ^{2}}udu\left| f^{\prime }(a)\right| ^{q} \\ +\int _{\frac{1}{2}}^{1}\frac{\left( 1-u\right) ^{\alpha +1}}{\left( ub+\left( 1-u\right) a\right) ^{2}}du\left| f^{\prime }(b)\right| ^{q} \end{array} \right) ^{\frac{1}{q}} \end{array} \right] . \end{aligned}$$
(18)
For the appearing integrals, we have
$$\begin{aligned} \int _{0}^{\frac{1}{2}}\frac{u^{\alpha }}{\left( ub+\left( 1-u\right) a\right) ^{2}}du&= {} \frac{1}{2^{\alpha +1}}\int _{0}^{1}\frac{u^{\alpha }}{ \left( \frac{u}{2}b+(1-\frac{u}{2})a\right) ^{2}}du \\&= {} \frac{1}{2^{\alpha +1}}\int _{0}^{1}\left( 1-v\right) ^{\alpha }\left( \frac{a+b}{2}\right) ^{-2}\left( 1-v\left( \frac{b-a}{b+a}\right) \right) ^{-2}du \\&= {} \frac{\left( a+b\right) ^{-2}}{2^{\alpha -1}\left( \alpha +1\right) } \begin{array}{c} _{2}F_{1}\left( 2,1;\alpha +2;\frac{b-a}{b+a}\right) \end{array} \\&= {} C_{3}\left( \alpha \right) , \end{aligned}$$
(19)
$$\begin{aligned} \int _{0}^{\frac{1}{2}}\frac{u^{\alpha +1}}{\left( ub+\left( 1-u\right) a\right) ^{2}}du&= {} \frac{1}{2^{\alpha +2}}\int _{0}^{1}\frac{u^{\alpha +1}}{ \left( \frac{u}{2}b+(1-\frac{u}{2})a\right) ^{2}}du \\&= {} \frac{1}{2^{\alpha +2}}\int _{0}^{1}\left( 1-v\right) ^{\alpha +1}\left( \frac{a+b}{2}\right) ^{-2}\left( 1-v\left( \frac{b-a}{b+a}\right) \right) ^{-2}du \\&= {} \frac{\left( a+b\right) ^{-2}}{2^{\alpha }\left( \alpha +2\right) } \begin{array}{c} _{2}F_{1}\left( 2,1;\alpha +3;\frac{b-a}{b+a}\right) \end{array} \\&= {} C_{4}\left( \alpha \right) , \end{aligned}$$
(20)
$$\begin{aligned} \int _{0}^{\frac{1}{2}}\frac{u^{\alpha }}{\left( ub+\left( 1-u\right) a\right) ^{2}}\left( 1-u\right) du&= {} C_{3}\left( \alpha \right) -C_{4}\left( \alpha \right) =C_{5}\left( \alpha \right) , \end{aligned}$$
(21)
$$\begin{aligned} \int _{\frac{1}{2}}^{1}\frac{\left( 1-u\right) ^{\alpha }}{\left( ub+\left( 1-u\right) a\right) ^{2}}du&= {} \int _{0}^{\frac{1}{2}}\frac{u^{\alpha }}{ \left( ua+\left( 1-u\right) b\right) ^{2}}du \\&= {} \frac{1}{2^{\alpha +1}}\int _{0}^{1}\frac{u^{\alpha }}{\left( \frac{u}{2} a+\left( 1-\frac{u}{2}\right) b\right) ^{2}}du \\&= {} \frac{1}{2^{\alpha +1}}\int _{0}^{1}u^{\alpha }b^{-2}\left( 1-\frac{u}{2} \left( 1-\frac{a}{b}\right) \right) ^{-2}du \\&= {} \frac{b^{-2}}{2^{\alpha +1}\left( \alpha +1\right) } \begin{array}{c} _{2}F_{1}\left( 2,\alpha +1;\alpha +2;\frac{1}{2}\left( 1-\frac{a}{b}\right) \right) \end{array} \\&= {} C_{6}\left( \alpha \right) , \end{aligned}$$
(22)
$$\begin{aligned} \int _{\frac{1}{2}}^{1}\frac{\left( 1-u\right) ^{\alpha }}{\left( ub+\left( 1-u\right) a\right) ^{2}}udu&= {} \int _{0}^{\frac{1}{2}}\frac{u^{\alpha }\left( 1-u\right) }{\left( ua+\left( 1-u\right) b\right) ^{2}}du \\&= {} \int _{0}^{\frac{1}{2}}\frac{u^{\alpha }}{\left( ua+\left( 1-u\right) b\right) ^{2}}du-\int _{0}^{\frac{1}{2}}\frac{u^{\alpha +1}}{\left( ua+\left( 1-u\right) b\right) ^{2}}du \\&= {} \frac{1}{2^{\alpha +1}}\int _{0}^{1}\frac{u^{\alpha }}{\left( \frac{u}{2} a+\left( 1-\frac{u}{2}\right) b\right) ^{2}}du \\&\quad-\frac{1}{2^{\alpha +2}}\int _{0}^{1}\frac{u^{\alpha +1}}{\left( \frac{u}{2} a+\left( 1-\frac{u}{2}\right) b\right) ^{2}}du \\&= {} \frac{1}{2^{\alpha +1}}\int _{0}^{1}u^{\alpha }b^{-2}\left( 1-\frac{u}{2} \left( 1-\frac{a}{b}\right) \right) ^{-2}du \\&\quad-\frac{1}{2^{\alpha +2}}\int _{0}^{1}u^{\alpha +1}b^{-2}\left( 1-\frac{u}{2} \left( 1-\frac{a}{b}\right) \right) ^{-2}du \\&= {} \left[ \begin{array}{c} \frac{b^{-2}}{2^{\alpha +1}\left( \alpha +1\right) } \begin{array}{c} _{2}F_{1}\left( 2,\alpha +1;\alpha +2;\frac{1}{2}\left( 1-\frac{a}{b}\right) \right) \end{array} \\ -\frac{b^{-2}}{2^{\alpha +2}\left( \alpha +2\right) } \begin{array}{c} _{2}F_{1}\left( 2,\alpha +2;\alpha +3;\frac{1}{2}\left( 1-\frac{a}{b}\right) \right) \end{array} \end{array} \right] \\&= {} C_{7}\left( \alpha \right) , \end{aligned}$$
(23)
$$\begin{aligned} \int _{\frac{1}{2}}^{1}\frac{\left( 1-u\right) ^{\alpha +1}}{\left( ub+\left( 1-u\right) a\right) ^{2}}du&= {} \int _{\frac{1}{2}}^{1}\frac{\left( 1-u\right) ^{\alpha }}{\left( ub+\left( 1-u\right) a\right) ^{2}}du-\int _{\frac{1}{2} }^{1}\frac{\left( 1-u\right) ^{\alpha }}{\left( ub+\left( 1-u\right) a\right) ^{2}}udu \\&= {} C_{6}\left( \alpha \right) -C_{7}\left( \alpha \right) =C_{8}\left( \alpha \right) . \end{aligned}$$
(24)
If we use (19–24) in (18) , we have (17). This completes the proof.
Corollary 2
In Theorem 7:
(1) If we take
\(\alpha =1\) we have the following Hermite–Hadamard–Fejer inequality for harmonically convex functions which is related to the left-hand side of (
5
):
$$\begin{aligned}&\left| f\left( \frac{2ab}{a+b}\right) \int _{a}^{b}\frac{g\left( x\right) }{x^{2}}dx-\int _{a}^{b}\frac{f\left( x\right) g\left( x\right) }{ x^{2}}dx\right| \\&\quad \le \left\| g\right\| _{\infty }\left( b-a\right) ^{2}\left[ \begin{array}{c} C_{3}^{^{1-\frac{1}{q}}}\left( 1\right) \left[ \left( \begin{array}{c} C_{4}\left( 1\right) \left| f^{\prime }(a)\right| ^{q} \\ +C_{5}\left( 1\right) \left| f^{\prime }(b)\right| ^{q} \end{array} \right) \right] ^{\frac{1}{q}} \\ +C_{6}^{1-\frac{1}{q}}\left( 1\right) \left[ \left( \begin{array}{c} C_{7}\left( 1\right) \left| f^{\prime }(a)\right| ^{q} \\ +C_{8}\left( 1\right) \left| f^{\prime }(b)\right| ^{q} \end{array} \right) \right] ^{\frac{1}{q}} \end{array} \right] , \end{aligned}$$
(2) If we take
\(g\left( x\right) =1\)
we have the following Hermite–Hadamard type inequality for harmonically convex functions in fractional integral forms which is related to the left-hand side of (
6
):
$$\begin{aligned}&\left| f\left( \frac{2ab}{a+b}\right) -\frac{\varGamma \left( \alpha +1\right) }{2^{1-\alpha }}\left( \frac{ab}{b-a}\right) ^{\alpha }\left\{ \begin{array}{c} J_{\frac{a+b}{2ab}+}^{\alpha }\left( f\circ h\right) \left( 1/a\right) \\ +J_{\frac{a+b}{2ab}-}^{\alpha }\left( f\circ h\right) \left( 1/b\right) \end{array} \right\} \right| \\&\quad \le \frac{ab\left( b-a\right) }{2^{1-\alpha }}\left[ \begin{array}{l} C_{3}^{^{1-\frac{1}{q}}}\left( \alpha \right) \left[ \left( \begin{array}{c} C_{4}\left( \alpha \right) \left| f^{\prime }(a)\right| ^{q} \\ +C_{5}\left( \alpha \right) \left| f^{\prime }(b)\right| ^{q} \end{array} \right) \right] ^{\frac{1}{q}} \\ +C_{6}^{1-\frac{1}{q}}\left( \alpha \right) \left[ \left( \begin{array}{c} C_{7}\left( \alpha \right) \left| f^{\prime }(a)\right| ^{q} \\ +C_{8}\left( \alpha \right) \left| f^{\prime }(b)\right| ^{q} \end{array} \right) \right] ^{\frac{1}{q}} \end{array} \right] , \end{aligned}$$
(3) If we take
\(\alpha =1\)
and
\(g\left( x\right) =1\)
we have the following Hermite–Hadamard type inequality for harmonically convex functions which is related to the left-hand side of (
4
):
$$\begin{aligned}&\left| f\left( \frac{2ab}{a+b}\right) -\frac{ab}{b-a}\int _{a}^{b}\frac{ f\left( x\right) }{x^{2}}dx\right| \\&\quad \le ab\left( b-a\right) \left[ \begin{array}{l} C_{3}^{^{1-\frac{1}{q}}}\left( 1\right) \left[ \left( \begin{array}{c} C_{4}\left( 1\right) \left| f^{\prime }(a)\right| ^{q} \\ +C_{5}\left( 1\right) \left| f^{\prime }(b)\right| ^{q} \end{array} \right) \right] ^{\frac{1}{q}} \\ +C_{6}^{1-\frac{1}{q}}\left( 1\right) \left[ \left( \begin{array}{c} C_{7}\left( 1\right) \left| f^{\prime }(a)\right| ^{q} \\ +C_{8}\left( 1\right) \left| f^{\prime }(b)\right| ^{q} \end{array} \right) \right] ^{\frac{1}{q}} \end{array} \right] . \end{aligned}$$
We can state another inequality for \(q>1\) as follows:
Theorem 8
Let
\(f:I\subset \left( 0,\infty \right) \rightarrow \mathbb {R}\)
be a differentiable function on
\(I {{}^\circ }\), the interior of
I, such that
\(f^{\prime }\in L\left[ a,b\right]\), where
\(a,b\in I\). If
\(\left| f^{\prime }\right| ^{q},q>1,\)
is harmonically convex on \(\left[ a,b\right]\), \(g:\left[ a,b\right] \mathbb { \rightarrow R}\)
is continuous and harmonically symmetric with respect to
\(\frac{2ab}{a+b}\), then the following inequality for fractional integrals holds:
$$\begin{aligned}&\left| \begin{array}{c} \frac{f(a)+f(b)}{2}\left[ J_{1/b+}^{\alpha }\left( g\circ h\right) (1/a)+J_{1/a-}^{\alpha }\left( g\circ h\right) (1/b)\right] \\ -\left[ J_{1/b+}^{\alpha }\left( fg\circ h\right) (1/a)+J_{1/a-}^{\alpha }\left( fg\circ h\right) (1/b)\right] \end{array} \right| \\&\quad \le \frac{\left\| g\right\| _{\infty }ab\left( b-a\right) }{\varGamma (\alpha +1)}\left( \frac{b-a}{ab}\right) ^{\alpha } \\&\quad\quad \times \left[ \begin{array}{c} C_{9}^{\frac{1}{p}}\left( \alpha \right) \left[ \frac{\left| f^{\prime }(a)\right| ^{q}+3\left| f^{\prime }(b)\right| ^{q}}{8}\right] ^{ \frac{1}{q}} \\ +C_{10}^{\frac{1}{p}}\left( \alpha \right) \left[ \frac{3\left| f^{\prime }(a)\right| ^{q}+\left| f^{\prime }(b)\right| ^{q}}{8} \right] ^{\frac{1}{q}} \end{array} \right] \end{aligned}$$
(25)
where
$$\begin{aligned} C_{9}\left( \alpha \right)&= {} \frac{\left( a+b\right) ^{-2p}}{2^{\alpha p-2p+1}\left( \alpha p+1\right) } \begin{array}{c} _{2}F_{1}\left( 2p,1;\alpha p+2;\frac{b-a}{b+a}\right) \end{array} , \\ C_{10}\left( \alpha \right)&= {} \frac{b^{-2p}}{2^{\alpha p+1}\left( \alpha p+1\right) } \begin{array}{c} _{2}F_{1}\left( 2,\alpha p+1;\alpha p+2;\frac{1}{2}\left( 1-\frac{a}{b} \right) \right) \end{array} , \end{aligned}$$
with
\(\alpha >1\), \(h(x)=1/x\), \(x\in \left[ \frac{1}{b},\frac{1}{a}\right]\)
and
\(1/p+1/q=1\).
Proof
Using (12), Hölder’s inequality and the harmonically convexity of \(\left| f^{\prime }\right| ^{q}\), it follows that
$$\begin{aligned}&\left| \begin{array}{c} f\left( \frac{2ab}{a+b}\right) \left[ J_{\frac{a+b}{2ab}+}^{\alpha }\left( g\circ h\right) (1/a)+J_{\frac{a+b}{2ab}-}^{\alpha }\left( g\circ h\right) (1/b)\right] \\ -\left[ J_{\frac{a+b}{2ab}+}^{\alpha }\left( fg\circ h\right) (1/a)+J_{\frac{ a+b}{2ab}-}^{\alpha }\left( fg\circ h\right) (1/b)\right] \end{array} \right| \\&\quad \le \frac{\left\| g\right\| _{\infty }ab\left( b-a\right) }{\varGamma (\alpha +1)}\left( \frac{b-a}{ab}\right) ^{\alpha }\left[ \begin{array}{c} \int _{0}^{\frac{1}{2}}\frac{u^{\alpha }}{\left( ub+\left( 1-u\right) a\right) ^{2}}\left| f^{\prime }(\frac{ab}{ub+\left( 1-u\right) a} )\right| du \\ +\int _{\frac{1}{2}}^{1}\frac{\left( 1-u\right) ^{\alpha }}{\left( ub+\left( 1-u\right) a\right) ^{2}}\left| f^{\prime }(\frac{ab}{ub+\left( 1-u\right) a})\right| du \end{array} \right] \\&\quad \le \frac{\left\| g\right\| _{\infty }ab\left( b-a\right) }{\varGamma (\alpha +1)}\left( \frac{b-a}{ab}\right) ^{\alpha }\left[ \begin{array}{c} \left( \int _{0}^{\frac{1}{2}}\frac{u^{\alpha p}}{\left( ub+\left( 1-u\right) a\right) ^{2p}}du\right) ^{\frac{1}{p}} \\ \times \left( \int _{0}^{\frac{1}{2}}\left| f^{\prime }(\frac{ab}{ ub+(1-u)a})\right| ^{q}du\right) ^{\frac{1}{q}} \end{array} \right. \\&\quad \left. \begin{array}{c} +\left( \int _{\frac{1}{2}}^{1}\frac{\left( 1-u\right) ^{\alpha p}}{\left( ub+\left( 1-u\right) a\right) ^{2p}}du\right) ^{\frac{1}{p}} \\ \quad\times \left( \int _{\frac{1}{2}}^{1}\left| f^{\prime }(\frac{ab}{ ub+(1-u)a})\right| ^{q}du\right) ^{\frac{1}{q}} \end{array} \right] \\&\quad \le \frac{\left\| g\right\| _{\infty }ab\left( b-a\right) }{\varGamma (\alpha +1)}\left( \frac{b-a}{ab}\right) ^{\alpha }\left[ \begin{array}{c} \left( \int _{0}^{\frac{1}{2}}\frac{u^{\alpha p}}{\left( ub+\left( 1-u\right) a\right) ^{2p}}du\right) ^{\frac{1}{p}} \\ \times \left( \int _{0}^{\frac{1}{2}}u\left| f^{\prime }(a)\right| ^{q}+\left( 1-u\right) \left| f^{\prime }(b)\right| ^{q}du\right) ^{ \frac{1}{q}} \end{array} \right. \\&\quad \left. \begin{array}{c} +\left( \int _{\frac{1}{2}}^{1}\frac{\left( 1-u\right) ^{\alpha p}}{\left( ub+\left( 1-u\right) a\right) ^{2p}}du\right) ^{\frac{1}{p}} \\ \times \left( \int _{\frac{1}{2}}^{1}u\left| f^{\prime }(a)\right| ^{q}+\left( 1-u\right) \left| f^{\prime }(b)\right| ^{q}du\right) ^{ \frac{1}{q}} \end{array} \right] \\&\quad =\frac{\left\| g\right\| _{\infty }ab\left( b-a\right) }{\varGamma (\alpha +1)}\left( \frac{b-a}{ab}\right) ^{\alpha } \\&\quad \quad \times \left[ \left( \int _{0}^{\frac{1}{2}}\frac{u^{\alpha p}}{\left( ub+\left( 1-u\right) a\right) ^{2p}}du\right) ^{\frac{1}{p}}\left[ \frac{ \left| f^{\prime }(a)\right| ^{q}+3\left| f^{\prime }(b)\right| ^{q}}{8}\right] ^{\frac{1}{q}}\right. \\&\quad \quad \left. +\left( \int _{\frac{1}{2}}^{1}\frac{\left( 1-u\right) ^{\alpha p}}{ \left( ub+\left( 1-u\right) a\right) ^{2p}}du\right) ^{\frac{1}{p}}\left[ \frac{3\left| f^{\prime }(a)\right| ^{q}+\left| f^{\prime }(b)\right| ^{q}}{8}\right] ^{\frac{1}{q}}\right] . \end{aligned}$$
(26)
For the appearing integrals, we have
$$\begin{aligned} \int _{0}^{\frac{1}{2}}\frac{u^{\alpha p}}{\left( ub+\left( 1-u\right) a\right) ^{2p}}du&= {} \frac{1}{2^{\alpha p+1}}\int _{0}^{1}\frac{u^{\alpha p}}{ \left( \frac{u}{2}b+(1-\frac{u}{2})a\right) ^{2p}}du \\&= {} \frac{1}{2^{\alpha p+1}}\int _{0}^{1}\left( 1-v\right) ^{\alpha p}\left( \frac{a+b}{2}\right) ^{-2p}\left[ 1-v\left( \frac{b-a}{b+a}\right) \right] ^{-2p}dv \\&= {} \frac{\left( a+b\right) ^{-2p}}{2^{\alpha p-2p+1}\left( \alpha p+1\right) } \begin{array}{c} _{2}F_{1}\left( 2p,1;\alpha p+2;\frac{b-a}{b+a}\right) \end{array} \\&= {} C_{9}\left( \alpha \right) . \end{aligned}$$
(27)
Similarly, we have
$$\begin{aligned} \int _{\frac{1}{2}}^{1}\frac{\left( 1-u\right) ^{\alpha p}}{\left( ub+\left( 1-u\right) a\right) ^{2p}}du&= {} \int _{0}^{\frac{1}{2}}\frac{u^{\alpha p}}{ \left( ua+\left( 1-u\right) b\right) ^{2p}}du \\&= {} \frac{1}{2^{\alpha p+1}}\int _{0}^{1}\frac{u^{\alpha p}}{\left( \frac{u}{2} a+\left( 1-\frac{u}{2}\right) b\right) ^{2p}}du \\&= {} \frac{1}{2^{\alpha p+1}}\int _{0}^{1}u^{\alpha }b^{-2p}\left( 1-\frac{u}{2} \left( 1-\frac{a}{b}\right) \right) ^{-2p}du \\&= {} \frac{b^{-2p}}{2^{\alpha p+1}\left( \alpha p+1\right) } \begin{array}{c} _{2}F_{1}\left( 2,\alpha p+1;\alpha p+2;\frac{1}{2}\left( 1-\frac{a}{b} \right) \right) \end{array} \\&= {} C_{10}\left( \alpha \right) . \end{aligned}$$
(28)
If we use (27) and (28) in (26), we have (25). This completes the proof.
Corollary 3
In Theorem 8:
(1) If we take
\(\alpha =1\)
we have the following Hermite–Hadamard–Fejer inequality for harmonically convex functions which is related to the left-hand side of (5):
$$\begin{aligned}&\left| f\left( \frac{2ab}{a+b}\right) \int _{a}^{b}\frac{g\left( x\right) }{x^{2}}dx-\int _{a}^{b}\frac{f\left( x\right) g\left( x\right) }{ x^{2}}dx\right| \\&\quad \le \left\| g\right\| _{\infty }\left( b-a\right) ^{2}\left[ \begin{array}{c} C_{9}^{\frac{1}{p}}\left( 1\right) \left[ \frac{\left| f^{\prime }(a)\right| ^{q}+3\left| f^{\prime }(b)\right| ^{q}}{8}\right] ^{ \frac{1}{q}} \\ +C_{10}^{\frac{1}{p}}\left( 1\right) \left[ \frac{3\left| f^{\prime }(a)\right| ^{q}+\left| f^{\prime }(b)\right| ^{q}}{8}\right] ^{ \frac{1}{q}} \end{array} \right] , \end{aligned}$$
(2) If we take
\(g\left( x\right) =1\)
we have following Hermite–Hadamard type inequality for harmonically convex functions in fractional integral forms which is related to the left-hand side of (6):
$$\begin{aligned}&\left| f\left( \frac{2ab}{a+b}\right) -\frac{\varGamma \left( \alpha +1\right) }{2^{1-\alpha }}\left( \frac{ab}{b-a}\right) ^{\alpha }\left\{ \begin{array}{c} J_{\frac{a+b}{2ab}+}^{\alpha }\left( f\circ h\right) \left( 1/a\right) \\ +J_{\frac{a+b}{2ab}-}^{\alpha }\left( f\circ h\right) \left( 1/b\right) \end{array} \right\} \right| \\&\quad \le \frac{ab\left( b-a\right) }{2^{1-\alpha }}\left[ \begin{array}{c} C_{9}^{\frac{1}{p}}\left( \alpha \right) \left[ \frac{\left| f^{\prime }(a)\right| ^{q}+3\left| f^{\prime }(b)\right| ^{q}}{8}\right] ^{ \frac{1}{q}} \\ +C_{10}^{\frac{1}{p}}\left( \alpha \right) \left[ \frac{3\left| f^{\prime }(a)\right| ^{q}+\left| f^{\prime }(b)\right| ^{q}}{8} \right] ^{\frac{1}{q}} \end{array} \right] , \end{aligned}$$
(3) If we take
\(\alpha =1\)
and
\(g\left( x\right) =1\)
we have the following Hermite–Hadamard type inequality for harmonically convex functions which is related to the left-hand side of (4):
$$\begin{aligned}&\left| f\left( \frac{2ab}{a+b}\right) -\frac{ab}{b-a}\int _{a}^{b}\frac{ f\left( x\right) }{x^{2}}dx\right| \\&\quad \le ab\left( b-a\right) \left[ \begin{array}{l} C_{9}^{\frac{1}{p}}\left( 1\right) \left[ \frac{\left| f^{\prime }(a)\right| ^{q}+3\left| f^{\prime }(b)\right| ^{q}}{8}\right] ^{ \frac{1}{q}} \\ +C_{10}^{\frac{1}{p}}\left( 1\right) \left[ \frac{3\left| f^{\prime }(a)\right| ^{q}+\left| f^{\prime }(b)\right| ^{q}}{8}\right] ^{ \frac{1}{q}} \end{array} \right] . \end{aligned}$$