Before going on our main result first we prove the following integral inequality.
Lemma 1
Let
\(I\subset \mathbb{R}\)
be an open interval, \(a,b\in I\)
with
\(a< b\)
and
\(f:[a,b]\rightarrow \mathbb{R}\)
be a differentiable function such that
\(f^{\prime }\)
is integrable and
\(0< \alpha \leq 1\)
on
\((a,b)\)
with
\(a< b\). If
\(|f^{\prime }|\)
is convex on
\([a,b]\), then we have the following inequality:
$$\begin{aligned}& \Biggl[ \biggl( \frac{ ( b-a ) ^{\alpha }- ( x-a ) ^{\alpha }}{ ( b-a ) ^{\alpha }}+\frac{ ( b-x ) ^{\alpha }}{ ( b-a ) ^{\alpha }} \biggr) \frac{f ( b ) }{2}+ \biggl( \frac{ ( b-a ) ^{\alpha }- ( b-x ) ^{\alpha }}{ ( b-a ) ^{\alpha }}+\frac{ ( x-a ) ^{\alpha }}{ ( b-a ) ^{\alpha }} \biggr) \frac{f ( a ) }{2} \\& \quad {}-\frac{\varGamma (\alpha +1)}{2(b-a)^{\alpha }} \bigl[ J_{a^{+}}^{ \alpha }f(b)+J_{b^{-}}^{\alpha }f(a) \bigr] =\frac{1}{2} \sum_{k=1}^{4}I_{1k} \Biggr], \end{aligned}$$
where
$$\begin{aligned}& I_{11}=\frac{ ( x-a ) ^{\alpha +1}}{ ( b-a ) ^{\alpha }} \int _{0}^{1} \bigl( t^{\alpha }-1 \bigr) f^{\prime }\bigl(tx+(1-t)a\bigr)\,dt , \\& I_{12}=\frac{ ( b-x ) ^{\alpha +1}}{ ( b-a ) ^{\alpha }} \int _{0}^{1} \bigl( 1-t^{\alpha } \bigr) f^{\prime }\bigl(tx+(1-t)b\bigr)\,dt , \\& I_{13}=\frac{ ( b-x ) ^{\alpha +1}}{ ( b-a ) ^{\alpha }} \int _{0}^{1} \biggl[ \biggl( \frac{a-b}{x-b}-t \biggr) ^{ \alpha }- \biggl( \frac{a-x}{x-b} \biggr) ^{\alpha } \biggr] f^{\prime }\bigl(tx+(1-t)b\bigr)\,dt , \\& I_{14}=\frac{ ( x-a ) ^{\alpha +1}}{ ( b-a ) ^{\alpha }} \int _{0}^{1} \biggl[ \biggl( \frac{b-x}{x-a} \biggr) ^{ \alpha }- \biggl( \frac{b-a}{x-a}-t \biggr) ^{\alpha } \biggr] f^{ \prime }\bigl(tx+(1-t)a\bigr)\,dt . \end{aligned}$$
Proof
Integrating by parts
$$\begin{aligned}& I_{11}=\frac{ ( x-a ) ^{\alpha +1}}{ ( b-a ) ^{\alpha }} \int _{0}^{1} \bigl( t^{\alpha }-1 \bigr) f^{\prime }\bigl(tx+(1-t)a\bigr)\,dt \\& \hphantom{I_{11}}=\frac{ ( x-a ) ^{\alpha +1}}{ ( b-a ) ^{ \alpha }}\biggl\{ \frac{ ( t^{\alpha }-1 ) f(tx+(1-t)a)}{x-a}\bigg|_{0}^{1} \\& \hphantom{I_{11}={}}{}+\frac{\alpha }{x-a} \int _{0}^{1} \bigl( t^{\alpha -1} \bigr) f \bigl(tx+(1-t)a\bigr)\,dt\biggr\} \\& \hphantom{I_{11}}=\frac{ ( x-a ) ^{\alpha +1}}{ ( b-a ) ^{ \alpha }} \biggl\{ \frac{f ( a ) }{x-a}-\frac{\alpha }{x-a} \int _{a}^{x}\frac{ ( u-a ) ^{\alpha -1}}{ ( x-a ) ^{\alpha -1}}\cdot \frac{f(u)\,du}{ ( x-a ) } \biggr\} \\& \hphantom{I_{11}} =\frac{f ( a ) ( x-a ) ^{\alpha }}{ ( b-a ) ^{\alpha }}- \frac{\alpha }{ ( b-a ) ^{\alpha }} \int _{a}^{x} ( u-a ) ^{\alpha -1}f(u)\,du, \\& I_{13}=\frac{ ( b-x ) ^{\alpha +1}}{ ( b-a ) ^{\alpha }} \int _{0}^{1} \biggl[ \biggl( \frac{a-b}{x-b}-t \biggr) ^{ \alpha }- \biggl( \frac{a-x}{x-b} \biggr) ^{\alpha } \biggr] f^{\prime }\bigl(tx+(1-t)b\bigr)\,dt \\& \hphantom{I_{13}}=\frac{ ( b-x ) ^{\alpha +1}}{ ( b-a ) ^{\alpha }}\frac{ [ ( \frac{a-b}{x-b}-t ) ^{\alpha }- ( \frac{a-x}{x-b} ) ^{\alpha } ] f(tx+(1-t)b)\,dt}{x-b}\bigg|_{0} ^{1} \\& \hphantom{I_{13}={}}{} - \int _{0}^{1}-\alpha \biggl( \frac{a-b}{x-b}-t \biggr) ^{\alpha -1} \frac{f(tx+(1-t)b)\,dt}{x-b} \\& \hphantom{I_{13}} =\frac{ ( b-x ) ^{\alpha +1}}{ ( b-a ) ^{\alpha }} \biggl[ \biggl( \frac{a-b}{x-b}-t \biggr) ^{\alpha }- \biggl( \frac{a-x}{b-x} \biggr) ^{\alpha } \biggr] \frac{f ( b ) }{b-x} \\& \hphantom{I_{13}={}}{} +\frac{-\alpha }{b-x} \int _{b}^{x}\frac{ ( u-a ) ^{ \alpha -1}}{ ( b-x ) ^{\alpha -1}}\cdot \frac{f(u)\,du}{ ( b-x ) } \\& \hphantom{I_{13}} =\frac{ ( b-x ) ^{\alpha +1}}{ ( b-a ) ^{\alpha }} \biggl\{ \frac{ ( b-a ) ^{\alpha }- ( x-a ) ^{\alpha }}{ ( b-x ) ^{\alpha +1}}f ( b ) - \frac{ \alpha }{ ( b-x ) ^{\alpha +1}} \int _{x}^{b} ( u-a ) ^{\alpha -1}f(u)\,du \biggr\} . \end{aligned}$$
Analogously
$$\begin{aligned}& I_{12}=\frac{f ( b ) ( b-x ) ^{\alpha }}{ ( b-a ) ^{\alpha }}-\frac{\alpha }{ ( b-a ) ^{\alpha }} \int _{b}^{x} ( b-u ) ^{\alpha -1}f(u)\,du , \\& I_{14}=\frac{ ( x-a ) ^{\alpha +1}}{ ( b-a ) ^{\alpha }} \biggl\{ \frac{- ( b-x ) ^{\alpha }+ ( b-a ) ^{\alpha }}{ ( x-a ) ^{\alpha +1}}f ( a ) - \frac{\alpha }{ ( x-a ) ^{\alpha +1}} \int _{a} ^{x} ( b-u ) ^{\alpha -1}f(u)\,du \biggr\} . \end{aligned}$$
Adding the above equalities, we get
$$\begin{aligned}& I_{11}+I_{13}=\frac{ ( x-a ) ^{\alpha }}{ ( b-a ) ^{\alpha }}f ( a ) + \biggl( 1- \frac{ ( x-a ) ^{\alpha }}{ ( b-a ) ^{\alpha }} \biggr) f ( b ) -\frac{\varGamma (\alpha +1)}{(b-a)^{\alpha }}J_{b^{-}}^{\alpha }f(a) , \\& I_{12}+I_{14}=\frac{ ( b-x ) ^{\alpha }}{ ( b-a ) ^{\alpha }}f ( b ) + \biggl( 1- \frac{ ( b-x ) ^{\alpha }}{ ( b-a ) ^{\alpha }} \biggr) f ( a ) -\frac{\varGamma (\alpha +1)}{(b-a)^{\alpha }}J_{a^{+}}^{\alpha }f(b) . \end{aligned}$$
The proof is completed. □
Theorem 4
Let
\(I\subset \mathbb{R}\)
be an open interval, \(a,b\in I\)
with
\(a< b\)
and
\(f:[a,b]\rightarrow \mathbb{R}\)
be a differentiable function such that
\(f^{\prime }\)
is integrable and
\(0<\alpha \leq 1\)
on
\((a,b)\)
with
\(a< b\). If
\(|f^{\prime }|\)
is convex on
\([a,b]\), then the following inequality for Riemann–Liouville fractional integrals holds:
$$\begin{aligned}& \biggl[ \biggl( \frac{ ( b-a ) ^{\alpha }- ( x-a ) ^{\alpha }}{ ( b-a ) ^{\alpha }}+\frac{ ( b-x ) ^{\alpha }}{ ( b-a ) ^{\alpha }} \biggr) \frac{f ( b ) }{2}+ \biggl( \frac{ ( b-a ) ^{\alpha }- ( b-x ) ^{\alpha }}{ ( b-a ) ^{\alpha }}+\frac{ ( x-a ) ^{\alpha }}{ ( b-a ) ^{\alpha }} \biggr) \frac{f ( a ) }{2} \\& \qquad {} -\frac{\varGamma (\alpha +1)}{2(b-a)^{\alpha }} \bigl[ J_{a ^{+}}^{\alpha }f(b)+J_{b^{-}}^{\alpha }f(a) \bigr] \biggr] \\& \quad \leq \frac{ ( x-a ) ^{\alpha +1}}{ ( b-a ) ^{\alpha +1}} \bigl[ A \bigl\vert f^{\prime }(x) \bigr\vert +B \bigl\vert f^{\prime }(a) \bigr\vert \bigr] + \frac{ ( b-x ) ^{\alpha +1}}{ ( b-a ) ^{\alpha +1}} \bigl[ A \bigl\vert f^{\prime }(x) \bigr\vert +B \bigl\vert f^{\prime }(b) \bigr\vert \bigr] \\& \qquad {}+\frac{ ( b-x ) ^{\alpha +1}}{ ( b-a ) ^{ \alpha +1}} \bigl[ C \bigl\vert f^{\prime }(x) \bigr\vert +D \bigl\vert f^{\prime }(b) \bigr\vert \bigr] + \frac{ ( x-a ) ^{\alpha +1}}{ ( b-a ) ^{\alpha +1}} \bigl[ E \bigl\vert f^{\prime }(x) \bigr\vert +F \bigl\vert f^{\prime }(a) \bigr\vert \bigr], \end{aligned}$$
(5)
where
$$\begin{aligned}& A = \int _{0}^{1} \bigl\vert 1-t^{\alpha } \bigr\vert t\,dt=\frac{ \alpha }{2 ( \alpha +2 ) }, \\& B = \int _{0}^{1} \bigl\vert 1-t^{\alpha } \bigr\vert ( 1-t ) \,dt=\frac{\alpha }{\alpha +1}-\frac{\alpha }{2 ( \alpha +2 ) }= \frac{\alpha (\alpha +3)}{2(\alpha +1)(\alpha +2)}, \\& \begin{aligned} C &= \int _{0}^{1}t \biggl\vert \biggl( \frac{a-b}{x-b}-t\biggr)^{\alpha }-\biggl( \frac{a-x}{x-b} \biggr)^{\alpha } \biggr\vert \,dt \\ &=-\frac{1}{ ( \alpha +1 ) ( \alpha +2 ) }\biggl( \frac{a-x}{x-b}\biggr)^{\alpha +2}+ \frac{1}{2}\biggl(\frac{a-x}{x-b}\biggr) \\ &\quad {}+\frac{1}{ ( \alpha +1 ) ( \alpha +2 ) }\biggl( \frac{a-b}{x-b}\biggr)^{\alpha +2}- \frac{1}{\alpha +1}\biggl(\frac{a-x}{x-b}\biggr)^{ \alpha +1}-\biggl( \frac{a-x}{x-b}\biggr)^{\alpha }, \end{aligned} \\& \begin{aligned} D &= \int _{0}^{1} \biggl\vert \biggl( \frac{a-b}{x-b}-t\biggr)^{\alpha }-\biggl( \frac{a-x}{x-b} \biggr)^{\alpha } \biggr\vert ( 1-t ) \,dt \\ &=\frac{1}{ ( \alpha +1 ) }\biggl(\frac{a-b}{x-b}\biggr)^{\alpha +1}+ \frac{1}{ ( \alpha +1 ) ( \alpha +2 ) }\biggl(\frac{a-x}{x-b}\biggr)^{ \alpha +2}- \frac{1}{2}\biggl(\frac{a-x}{x-b}\biggr) \\ &\quad {}-\frac{1}{ ( \alpha +1 ) ( \alpha +2 ) }\biggl( \frac{a-b}{x-b}\biggr)^{\alpha +2}. \end{aligned} \end{aligned}$$
Proof
Here, utilizing the properties of the modulus in Lemma 1 and convexity of \(|f^{\prime }|\), we have
$$\begin{aligned}& |K_{1}| =\frac{(x-a)^{\alpha +1}}{(b-a)^{\alpha }} \int _{0} ^{1} \bigl( t^{\alpha }-1 \bigr) f^{\prime }\bigl(tx+(1-t)a\bigr)\,dt, \\& \begin{aligned} |K_{1}| & \leq \frac{(x-a)^{\alpha +1}}{(b-a)^{\alpha }} \int _{0}^{1} \bigl\vert \bigl( 1-t^{\alpha } \bigr) \bigr\vert \bigl\vert f^{ \prime }\bigl(tx+(1-t)a\bigr) \bigr\vert \,dt \\ & \leq \frac{(x-a)^{\alpha +1}}{(b-a)^{\alpha }} \int _{0}^{1} \bigl\vert \bigl( 1-t^{\alpha } \bigr) \bigr\vert \bigl\{ t \bigl\vert f^{\prime }(x) \bigr\vert +(1-t) \bigl\vert f^{\prime }(a) \bigr\vert \bigr\} \,dt, \end{aligned} \\& |K_{1}| =\frac{(x-a)^{\alpha +1}}{(b-a)^{\alpha }} \bigl\{ A \bigl\vert f^{\prime }(x) \bigr\vert +B \bigl\vert f^{\prime }(a) \bigr\vert \bigr\} , \end{aligned}$$
and analogously
$$\begin{aligned}& |K_{2}| =\frac{(b-x)^{\alpha +1}}{(b-a)^{\alpha }} \int _{0}^{1} \bigl( t^{\alpha }-1 \bigr) f^{\prime }\bigl(tx+(1-t)b\bigr)\,dt, \\& \begin{aligned} \vert K_{2} \vert & \leq \frac{(b-x)^{\alpha +1}}{(b-a)^{ \alpha }} \int _{0}^{1} \bigl\vert 1-t^{\alpha } \bigr\vert \bigl\vert f^{\prime }\bigl(tx+(1-t)b\bigr) \bigr\vert \,dt \\ & \leq \frac{(b-x)^{\alpha +1}}{(b-a)^{\alpha }} \int _{0}^{1} \bigl\vert \bigl( 1-t^{\alpha } \bigr) \bigr\vert \bigl\{ t \bigl\vert f^{\prime }(x) \bigr\vert {+}(1-t) \bigl\vert f^{\prime }(b) \bigr\vert \bigr\} \,dt, \end{aligned} \\& \vert K_{2} \vert =\frac{(b-x)^{\alpha +1}}{(b-a)^{ \alpha }} \bigl\{ A \bigl\vert f^{\prime }(x) \bigr\vert +B \bigl\vert f ^{\prime }(b) \bigr\vert \bigr\} , \end{aligned}$$
using the convexity on \(|f^{\prime }|\) and the fact that, for \(\alpha \in (0,1]\) and \(\forall t\in {}[ 0,1]\),
$$\begin{aligned}& \begin{aligned} |K_{3}| &\leq \frac{(b-x)^{\alpha +1}}{(b-a)^{\alpha }} \int _{0}^{1} \biggl\vert \biggl( \frac{a-b}{x-b}-t\biggr)^{\alpha }-\biggl(\frac{a-x}{x-b} \biggr)^{ \alpha } \biggr\vert \bigl\vert f^{\prime }\bigl(tx+(1-t)b \bigr) \bigr\vert \,dt \\ &\leq \frac{(b-x)^{\alpha +1}}{(b-a)^{\alpha }} \int _{0}^{1} \biggl\vert \biggl( \frac{a-b}{x-b}-t\biggr)^{\alpha }-\biggl(\frac{a-x}{x-b} \biggr)^{\alpha } \biggr\vert \bigl\{ t \bigl\vert f^{\prime }(x) \bigr\vert +(1-t) \bigl\vert f ^{\prime }(b) \bigr\vert \bigr\} \,dt, \end{aligned} \\& \vert K_{3} \vert =\frac{(b-x)^{\alpha +1}}{(b-a)^{ \alpha }} \bigl\{ C \bigl\vert f^{\prime }(x) \bigr\vert +D \bigl\vert f ^{\prime }(b) \bigr\vert \bigr\} , \end{aligned}$$
and analogously
$$\begin{aligned} |K_{4}| \leq &\frac{(x-a)^{\alpha +1}}{(b-a)^{\alpha }} \int _{0}^{1} \biggl\vert \biggl( \frac{b-x}{x-a} \biggr) ^{\alpha }- \biggl( \frac{b-a}{x-a}-t \biggr) ^{\alpha } \biggr\vert \bigl\vert f^{\prime }\bigl(tx+(1-t)a\bigr) \bigr\vert \,dt \\ \leq &\frac{ ( x-a ) ^{\alpha +1}}{ ( b-a ) ^{\alpha }} \int _{0}^{1} \biggl\vert \biggl( \frac{b-x}{x-a} \biggr) ^{\alpha }- \biggl( \frac{b-a}{x-a}-t \biggr) ^{\alpha } \biggr\vert \bigl\{ t \bigl\vert f^{\prime }(x) \bigr\vert +(1-t) \bigl\vert f ^{\prime }(a) \bigr\vert \bigr\} \,dt \\ \leq &\frac{ ( x-{a} ) ^{\alpha +1}}{ ( b-a ) ^{\alpha }} \bigl\{ E \bigl\vert f^{\prime }(x) \bigr\vert +F \bigl\vert f ^{\prime }(a) \bigr\vert \bigr\} . \end{aligned}$$
The proof is completed. □
Remark 1
On letting \(\alpha =1\), \(x=\frac{a+b}{2}\) in Theorem 4, inequality (5) reduces to inequality (1).
Theorem 5
Let
\(I\subset \mathbb{R}\)
be an open interval, \(a,b\in I\)
with
\(a< b\)
and
\(f:[a,b]\rightarrow \mathbb{R}\)
be a differentiable function such that
\(f^{\prime }\)
is integrable and
\(0<\alpha \leq 1\)
on
\((a,b)\)
with
\(a< b\). If
\(|f^{\prime }|^{q}\)
is convex on
\([a,b]\), \(q\geq 1\)
then the following inequality holds:
$$\begin{aligned}& \biggl[ \biggl( \frac{ ( b-a ) ^{\alpha }- ( x-a ) ^{\alpha }}{ ( b-a ) ^{\alpha }}+\frac{ ( b-x ) ^{\alpha }}{ ( b-a ) ^{\alpha }} \biggr) \frac{f ( b ) }{2}+ \biggl( \frac{ ( b-a ) ^{\alpha }- ( b-x ) ^{\alpha }}{ ( b-a ) ^{\alpha }}+\frac{ ( x-a ) ^{\alpha }}{ ( b-a ) ^{\alpha }} \biggr) \frac{f ( a ) }{2} \\& \qquad {} + -\frac{\varGamma (\alpha +1)}{2(b-a)^{\alpha }} \bigl[ J_{a ^{+}}^{\alpha }f(b)+J_{b^{-}}^{\alpha }f(a) \bigr] \biggr] \\& \quad \leq \biggl[ \frac{(x-a)^{\alpha +1}}{(b-a)^{\alpha }} ( \gamma _{1} ) ^{1-1/q} \bigl( A \bigl\vert f^{\prime }(x) \bigr\vert ^{q}+B \bigl\vert f^{ \prime }(a) \bigr\vert ^{q} \bigr) ^{1/q} \\& \qquad {}+ \frac{(b-x)^{\alpha +1}}{(b-a)^{ \alpha }} ( \gamma _{2} ) ^{1-1/q} \bigl( A \bigl\vert f^{\prime }(x) \bigr\vert ^{q}+B \bigl\vert f ^{\prime }(b) \bigr\vert ^{q} \bigr) ^{1/q} \\& \qquad {}+ \frac{(b-x)^{\alpha +1}}{(b-a)^{\alpha }} ( \gamma _{3} ) ^{1-1/q} \bigl( C \bigl\vert f^{\prime }(x) \bigr\vert ^{q}+D \bigl\vert f^{\prime }(b) \bigr\vert ^{q} \bigr) ^{1/q} \\& \qquad {}+ \frac{(x-a)^{\alpha +1}}{(b-a)^{\alpha }} ( \gamma _{4} ) ^{1-1/q} \bigl( E \bigl\vert f^{\prime }(x) \bigr\vert ^{q}+F \bigl\vert f ^{\prime }(a) \bigr\vert ^{q} \bigr) ^{1/q} \biggr], \end{aligned}$$
(6)
where
$$\begin{aligned}& \begin{aligned} E &= \int _{0}^{1}t \biggl\vert \biggl( \frac{b-a}{x-a}-t\biggr)^{\alpha }-\biggl( \frac{b-x}{x-a} \biggr)^{\alpha } \biggr\vert \,dt \\ &=\biggl(\frac{b-x}{2 ( x-a ) }\biggr)-\frac{1}{ ( \alpha +1 ) }\biggl(\frac{b-x}{x-a} \biggr)^{\alpha +1}-\frac{b-x}{(x-a)} \\ &\quad {}-\frac{1}{ ( \alpha +1 ) ( \alpha +2 ) }\biggl( \frac{b-x}{x-a}\biggr)^{\alpha +2}+ \frac{1}{ ( \alpha +1 ) ( \alpha +2 ) }\biggl(\frac{b-a}{x-a}\biggr)^{\alpha +2} , \end{aligned} \\& \begin{aligned} F &= \int _{0}^{1} \biggl\vert \biggl( \frac{b-a}{x-a}-t\biggr)^{\alpha }-\biggl( \frac{b-x}{x-a} \biggr)^{\alpha } \biggr\vert ( 1-t ) \,dt \\ &=\frac{1}{ ( \alpha +1 ) }\biggl(\frac{b-a}{x-a}\biggr)^{\alpha +2}- \frac{b-x}{2(x-a)}+\frac{1}{ ( \alpha +1 ) ( \alpha +2 ) }\biggl(\frac{b-x}{x-a} \biggr)^{\alpha +2}-E, \end{aligned} \\& \gamma _{1} = \int _{0}^{1} \bigl\vert t^{\alpha }-1 \bigr\vert \,dt=\frac{ \alpha }{ ( \alpha +1 ) }, \\& \gamma _{2} = \int _{0}^{1} \bigl\vert 1-t^{\alpha } \bigr\vert \,dt=\frac{ \alpha }{ ( \alpha +1 ) }, \\& \begin{aligned} \gamma _{3} &= \int _{0}^{1} \biggl\vert \biggl( \frac{a-b}{x-b}-t\biggr)^{\alpha }-\biggl( \frac{a-x}{x-b} \biggr)^{\alpha } \biggr\vert \,dt \\ &=-\frac{1}{ ( \alpha +1 ) }\biggl(\frac{a-x}{x-b}\biggr)^{\alpha +1}-\biggl( \frac{a-x}{x-b}\biggr)^{\alpha }+\frac{1}{ ( \alpha +1 ) }\biggl( \frac{a-b}{x-b}\biggr)^{\alpha } , \end{aligned} \\& \begin{aligned} \gamma _{4} &= \int _{0}^{1} \biggl\vert \biggl( \frac{b-x}{x-a}-t\biggr)^{\alpha }-\biggl( \frac{b-x}{x-a} \biggr)^{\alpha } \biggr\vert \,dt \\ &=\biggl(\frac{b-x}{x-a}\biggr)^{\alpha }+\frac{1}{ ( \alpha +1 ) }\biggl( \frac{b-x}{x-a}\biggr)^{\alpha +1}-\frac{1}{ ( \alpha +1 ) }\biggl( \frac{b-a}{x-a}\biggr)^{\alpha +1}. \end{aligned} \end{aligned}$$
Proof
By using the properties of the modulus in Lemma 1, we have
$$\begin{aligned}& \biggl\vert \biggl[ \biggl( \frac{ ( b-a ) ^{\alpha }- ( x-a ) ^{\alpha }}{ ( b-a ) ^{\alpha }}+\frac{ ( b-x ) ^{\alpha }}{ ( b-a ) ^{\alpha }} \biggr) \frac{f ( b ) }{2}+ \biggl( \frac{ ( b-a ) ^{ \alpha }- ( b-x ) ^{\alpha }}{ ( b-a ) ^{\alpha }}+\frac{ ( x-a ) ^{\alpha }}{ ( b-a ) ^{\alpha }} \biggr) \frac{f ( a ) }{2} \\& \quad {}+-\frac{\varGamma (\alpha +1)}{2(b-a)^{\alpha }} \bigl[ J_{a^{+}}^{\alpha }f(b)+J_{b^{-}}^{\alpha }f(a) \bigr] \biggr] \biggr\vert \leq \sum_{k=1}^{4}|J_{k}| \end{aligned}$$
and using convexity of \(|f^{\prime }|\), we have
$$\begin{aligned} |J_{1}| & \leq \frac{(x-a)^{\alpha +1}}{(b-a)^{\alpha }} \int _{0}^{1} \bigl( 1-t^{\alpha } \bigr) \bigl\vert f^{\prime }\bigl(tx+(1-t)a\bigr) \bigr\vert \,dt \\ & \leq \frac{(x-a)^{\alpha +1}}{(b-a)^{\alpha }} \biggl( \int _{0}^{1} \bigl( 1-t^{\alpha } \bigr) \,dt \biggr) ^{1-\frac{1}{q}} \biggl( \int _{0}^{1} \bigl( 1-t^{\alpha } \bigr) \bigl\vert f^{\prime }\bigl(tx+(1-t)a\bigr) \bigr\vert ^{q}\,dt \biggr) ^{\frac{1}{q}} \\ & =\frac{(x-a)^{\alpha +1}}{(b-a)^{\alpha }} ( \gamma _{1} ) ^{{1-\frac{1}{q}}} \bigl[ A \bigl\vert f^{\prime }(x) \bigr\vert ^{q}+B \bigl\vert f^{\prime } ( a ) \bigr\vert ^{q} \bigr] ^{\frac{1}{q}} \end{aligned}$$
and analogously
$$\begin{aligned} |J_{3}| \leq& \frac{(b-x)^{\alpha +1}}{(b-a)^{\alpha }} \int _{0}^{1} \biggl\vert \biggl( \frac{a-b}{x-b}-t\biggr)^{\alpha }-\biggl(\frac{a-x}{x-b} \biggr)^{\alpha } \biggr\vert \bigl\vert f ^{\prime }\bigl(tx+(1-t)b \bigr) \bigr\vert \,dt \\ \leq& \frac{(b-x)^{\alpha +1}}{(b-a)^{\alpha }} \biggl( \int _{0}^{1} \biggl\vert \biggl( \frac{a-b}{x-b}-t\biggr)^{\alpha }-\biggl(\frac{a-x}{x-b} \biggr)^{\alpha } \biggr\vert \,dt \biggr) ^{1-\frac{1}{q}} \\ &{} \times \biggl( \int _{0}^{1} \biggl\vert \biggl( \frac{a-b}{x-b}-t\biggr)^{\alpha }-\biggl(\frac{a-x}{x-b} \biggr)^{\alpha } \biggr\vert \bigl\vert f^{\prime }\bigl(tx+(1-t)b \bigr) \bigr\vert ^{q}\,dt \biggr) ^{\frac{1}{q}} \\ =&\frac{(b-x)^{\alpha +1}}{(b-a)^{\alpha }} ( \gamma _{3} ) ^{{1-\frac{1}{q}}} \bigl[ C \bigl\vert f^{\prime }(x) \bigr\vert ^{q}+D \bigl\vert f^{\prime } ( b ) \bigr\vert ^{q} \bigr] ^{\frac{1}{q}}, \end{aligned}$$
using the convexity and the fact that, for \(\alpha \in (0,1]\) and \(\forall t\in {}[ 0,1]\),
$$ |J_{2}|\leq \frac{(b-x)^{\alpha +1}}{(b-a)^{\alpha }} ( \gamma _{2} ) ^{{1-\frac{1}{q}}} \bigl[ A \bigl\vert f^{\prime }(x) \bigr\vert ^{q}+B \bigl\vert f^{\prime } ( b ) \bigr\vert ^{q} \bigr] ^{\frac{1}{q}} $$
and similarly
$$\begin{aligned} |J_{4}| \leq &\frac{(b-x)^{\alpha +1}}{(b-a)^{\alpha }} \int _{0}^{1} \biggl\vert \biggl( \frac{b-a}{x-a}-t\biggr)^{\alpha }-\biggl(\frac{b-x}{x-a} \biggr)^{\alpha } \biggr\vert \bigl|f ^{\prime }\bigl(tx+(1-t)a\bigr)\bigr|\,dt \\ \leq& \frac{(b-x)^{\alpha +1}}{(b-a)^{\alpha }} \biggl( \int _{0}^{1} \biggl\vert \biggl( \frac{b-a}{x-a}-t\biggr)^{\alpha }-\biggl(\frac{b-x}{x-a} \biggr)^{\alpha } \biggr\vert \,dt \biggr) ^{1-\frac{1}{q}} \\ &{} \times \biggl( \int _{0}^{1} \biggl\vert \biggl( \frac{b-a}{x-a}-t\biggr)^{\alpha }-\biggl(\frac{b-x}{x-a} \biggr)^{\alpha } \biggr\vert \bigl\vert f^{\prime }\bigl(tx+(1-t)a \bigr) \bigr\vert ^{q}\,dt \biggr) ^{\frac{1}{q}} \\ =&\frac{(b-x)^{\alpha +1}}{(b-a)^{\alpha }} ( \gamma _{4} ) ^{{1-\frac{1}{q}}} \bigl[ E \bigl\vert f^{\prime }(x) \bigr\vert ^{q}+F \bigl\vert f^{\prime } ( a ) \bigr\vert ^{q} \bigr] ^{\frac{1}{q}}. \end{aligned}$$
The proof is completed. □
Corollary 2
On letting
\(\alpha =1\), \(x=\frac{a+b}{2}\)
and
\(\vert f ^{\prime }(a) \vert = \vert f^{\prime } ( b ) \vert \leq M\)
in Theorem 5, inequality (6) reduces to the inequality
$$ \biggl\vert \frac{f(a)+f(b)}{2}-\frac{1}{b-a} \int _{a}^{b}f(x)\,dx \biggr\vert \leq \frac{M}{4} ( b-a ). $$
(7)
Remark 2
The obtained inequality (7) is an improvement of the inequality as in (2).
In the following, we obtain an estimate of the Hermite–Hadamard inequality for concave functions.
Theorem 6
Let
\(f:[a,b]\rightarrow \mathbb{R}\)
be a differentiable function on
\((a,b)\)
such that
\(f^{\prime }\in L_{1}[a,b]\). If
\(|f^{\prime }|^{q}\)
is concave on
\([a,b]\), for some fixed
\(p>1\)
with
\(q=\frac{p}{p-1}\), the following inequality for fractional integrals holds:
$$\begin{aligned}& \biggl\vert \biggl[ \biggl( \frac{ ( b-a ) ^{\alpha }- ( x-a ) ^{\alpha }}{ ( b-a ) ^{\alpha }}+\frac{ ( b-x ) ^{\alpha }}{ ( b-a ) ^{\alpha }} \biggr) \frac{f ( b ) }{2}+ \biggl( \frac{ ( b-a ) ^{ \alpha }- ( b-x ) ^{\alpha }}{ ( b-a ) ^{\alpha }}+\frac{ ( x-a ) ^{\alpha }}{ ( b-a ) ^{\alpha }} \biggr) \frac{f ( a ) }{2} \\& \qquad {}+ -\frac{\varGamma (\alpha +1)}{2(b-a)^{\alpha }} \bigl[ J_{a^{+}}^{\alpha }f(b)+J_{b^{-}}^{\alpha }f(a) \bigr]\biggr] \biggr\vert \\& \quad \leq \biggl[ \biggl\{ \gamma _{1} \biggl\vert f^{\prime } \biggl( ( \alpha +1) \biggl\{ \frac{Ax+Bb}{\alpha } \biggr\} \biggr) \biggr\vert + \gamma _{2} \biggl\vert f^{\prime } \biggl( (\alpha +1) \biggl\{ \frac{Ax+Bb}{ \alpha } \biggr\} \biggr) \biggr\vert \biggr\} \\& \qquad {} +\gamma _{3} \biggl\vert f^{\prime } \biggl( (\alpha +1) \biggl\{ \frac{Cx+Db}{\alpha } \biggr\} \biggr) \biggr\vert +\gamma _{3} \biggl\vert f ^{\prime }(\alpha +1) \biggl\{ \frac{Ex+Fa}{\alpha } \biggr\} \biggr\vert \biggr] . \end{aligned}$$
(8)
Proof
Using the concavity of \(|f^{\prime }|^{q}\) and the power-mean inequality, we obtain
$$\begin{aligned} \bigl\vert f^{\prime }\bigl(tx+(1-t)y\bigr) \bigr\vert ^{q} & >t \bigl\vert f^{\prime } ( x ) \bigr\vert ^{q}+(1-t) \bigl\vert f ^{\prime } ( y ) \bigr\vert ^{q} \\ & \geq t \bigl\vert f^{\prime } ( x ) \bigr\vert ^{q}+(1-t) \bigl\vert f^{\prime } ( y ) \bigr\vert ^{q}. \end{aligned}$$
Hence
$$ \bigl\vert f^{\prime }\bigl(tx+(1-t)y\bigr) \bigr\vert \geq t \bigl\vert f^{\prime }(x) \bigr\vert +(1-t) \bigl\vert f^{\prime }(y) \bigr\vert , $$
so \(|f^{\prime }|\) is also concave. By the Jensen integral inequality, we have
$$\begin{aligned} |I_{1}| \leq &\frac{ ( x-a ) ^{\alpha +1}}{ ( b-a ) ^{\alpha +1}} \biggl( \int _{0}^{1} \bigl\vert 1-t^{\alpha } \bigr\vert \,dt \biggr) \biggl\vert f^{\prime } \biggl( \frac{\int _{0}^{1} \vert ( 1-t^{\alpha } ) \vert [ f^{\prime }(tx+(1-t)a) ] \,dt}{\int _{0}^{1} \vert 1-t^{ \alpha } \vert \,dt} \biggr) \biggr\vert \\ =&\frac{ ( x-a ) ^{\alpha +1}}{ ( b-a ) ^{ \alpha +1}} ( \gamma _{1} ) \biggl\vert f^{\prime } \biggl( \frac{Ax+Ba}{ \gamma _{1}} \biggr) \biggr\vert \end{aligned}$$
and similarly
$$\begin{aligned}& |I_{2}|\leq \frac{ ( b-x ) ^{\alpha +1}}{ ( b-a ) ^{\alpha +1}} ( \gamma _{2} ) \biggl\vert f^{\prime } \biggl( \frac{Ax+Bb}{\gamma _{2}} \biggr) \biggr\vert , \\& \begin{aligned} |I_{3}| &\leq \frac{(b-x)^{{\alpha +1}}}{(b-a)^{\alpha }} \biggl( \int _{0}^{1} \biggl\vert \biggl( \frac{a-b}{x-b}-t \biggr) ^{\alpha }- \biggl( \frac{a-x}{x-b} \biggr) ^{\alpha } \biggr\vert \biggr) \\ &\quad {}\times \biggl\vert f^{\prime } \biggl( \frac{ ( \int _{0}^{1} \vert ( \frac{a-b}{x-b}-t ) ^{\alpha }- ( \frac{a- {i}x}{x-b} ) ^{\alpha } \vert ) [ f^{\prime }(tx+(1-t)b) ] \,dt}{\int _{0}^{1} \vert ( \frac{a-b}{x-b}-t ) ^{\alpha }- ( \frac{a-x}{x-b} ) ^{\alpha } \vert \,dt} \biggr) \biggr\vert \\ &\leq (\gamma _{3}) \biggl\vert f^{\prime } \biggl( \frac{Cx+Db}{\gamma _{3}} \biggr) \biggr\vert , \end{aligned} \end{aligned}$$
and
$$\begin{aligned}& \begin{aligned} |I_{4}| &\leq \frac{ ( x-a ) ^{\alpha +1}}{ ( b-a ) ^{\alpha +1}} \biggl( \int _{0}^{1} \biggl\vert \biggl( \frac{b-x}{x-a} \biggr) ^{\alpha }- \biggl( \frac{b-a}{x-a}-t \biggr) ^{\alpha } \biggr\vert \biggr) \\ &\quad {}\times \biggl\vert f^{\prime } \biggl( \frac{\int _{0}^{1} \vert ( \frac{b-x}{x-a} ) ^{ \alpha }- ( \frac{b-a}{x-a}-t ) ^{\alpha } \vert [ f^{\prime }(tx+(1-t)a) ] \,dt}{ ( \int _{0}^{1} \vert ( \frac{b-x}{x-a} ) ^{\alpha }- ( \frac{b-a}{x-a}-t ) ^{\alpha } \vert ) \,dt} \biggr) \biggr\vert , \end{aligned} \\& |I_{4}| \leq (\gamma _{4}) \biggl\vert f^{\prime } \biggl( \frac{Ex+Fb}{ \gamma _{4}} \biggr) \biggr\vert . \end{aligned}$$
The proof is completed. □
Remark 3
On letting \(\alpha =1\), \(x=\frac{a+b}{2}\) in Theorem 6, inequality (8) reduces to inequality (4).