2.1 Hermite–Hadamard type inequalities involving Riemann–Liouville fractional integral
We proved the Riemann–Liouville fractional Hermite–Hadamard inequalities for \(\mathbb{B}\)-convex functions which were given in the following theorems for left-sided integral and right-sided integral, respectively, in [36].
Theorem 3
([36])
Let
\(f: [a,b ]\subset \mathbb{R}_{+}\rightarrow \mathbb{R} _{+}\)
and
\(f\in L_{1} [a,b ]\). If f is a
\(\mathbb{B}\)-convex function on
\([a,b ]\), then the following inequality for fractional integrals holds:
$$ J^{\alpha }_{a+}f (b )\leq \textstyle\begin{cases} \frac{f (a ) (b-a )^{\alpha }}{\varGamma (\alpha +1 )}, & 1\leq \frac{f (a )}{f (b )}, \\ \frac{b^{\alpha } (f (b )-f (a ) ) ^{\alpha +1}+ (\alpha +1 )f (a ) (f (b ) ) ^{\alpha } (b-a )^{\alpha }}{\varGamma (\alpha +2 ) (f (b ) )^{\alpha }}, & 0\leq \frac{f (a )}{f (b )}< 1 \end{cases} $$
(9)
with
\(\alpha >0\).
Theorem 4
([36])
Let
\(f: [a,b ]\subset \mathbb{R}_{+}\rightarrow \mathbb{R} _{+}\)
and
\(f\in L_{1} [a,b ]\). If f is a
\(\mathbb{B}\)-convex function on
\([a,b ]\), then the following inequality for fractional integrals holds:
$$ J^{\alpha }_{b-}f (a )\leq \textstyle\begin{cases} \frac{f (a ) (b-a )^{\alpha }}{\varGamma (\alpha +1 )}, & 1\leq \frac{f (a )}{f (b )}, \\ \frac{ (bf (a )-af (b ) )^{\alpha +1}+ (f (b ) )^{\alpha +1} (b-a )^{ \alpha } (\alpha b+a )}{b\varGamma (\alpha +2 ) (f (b ) )^{\alpha }}, & \frac{a}{b}\leq \frac{f (a )}{f (b )}< 1, \\ \frac{f (b ) (b-a )^{\alpha } (\alpha b+a )}{b \varGamma (\alpha +2 )}, & 0\leq \frac{f (a )}{f (b )}< \frac{a}{b} \end{cases} $$
(10)
with
\(\alpha >0\).
In Theorem 3 and Theorem 4, the inequalities were obtained by getting the multiplier \((1-\lambda )^{\alpha -1}\) and \((\lambda -\frac{a}{b} )^{\alpha -1}\). We derive nearly the same inequalities with changing the multipliers. For example, if we take the multiplier \(\frac{ (\max \{a, \lambda b\} )'}{ [b-\max \{a, \lambda b\} ]^{1-\alpha }}\) for left-sided integral, we deduce inequality (11) and the same inequality given in (10) to take the multiplier \(\frac{ (\max \{a, \lambda b\} )'}{ [\lambda b-a ]^{1-\alpha }}\) for the right-sided integral.
Theorem 5
Let
\(f: [a,b ]\subset \mathbb{R}_{+}\rightarrow \mathbb{R} _{+}\)
and
\(f\in L_{1} [a,b ]\). If f is a
\(\mathbb{B}\)-convex function on
\([a,b ]\), then the following inequality holds:
$$ J^{\alpha }_{a+}f (b )\leq \textstyle\begin{cases} \frac{f (a ) (b-a )^{\alpha }}{\varGamma (\alpha +1 )}, & 1\leq \frac{f (a )}{f (b )}, \\ \frac{b^{\alpha } (f (b )-f (a ) ) ^{\alpha +1}+ (\alpha +1 )f (a ) (f (b ) ) ^{\alpha } (b-a )^{\alpha }}{\varGamma (\alpha +2 ) (f (b ) )^{\alpha }}, & \frac{a}{b}\leq \frac{f (a )}{f (b )}< 1, \\ \frac{f (b ) (b-a )^{\alpha } (\alpha a+b )}{b \varGamma (\alpha +2 )}, & 0\leq \frac{f (a )}{f (b )}< \frac{a}{b} \end{cases} $$
(11)
with
\(\alpha >0\).
Proof
Let f be a \(\mathbb{B}\)-convex function. Thus, inequality (7) holds. For desired inequality, we will multiply both sides of inequality (7) by \(\frac{ (\max \{a, \lambda b\} )'}{ [b- \max \{a, \lambda b\} ]^{1-\alpha }}\) and then integrate with respect to λ over \([0,1 ]\). For the left-hand side of inequality, we get that
$$\begin{aligned} & \int ^{1}_{0} \frac{ (\max \{a, \lambda b \} )'}{ [b-\max \{a, \lambda b \} ]^{1-\alpha }}f \bigl(\max \{a, \lambda b \} \bigr)\,d\lambda \\ &\quad = \int ^{\frac{a}{b}}_{0}\frac{ (a )'}{ [b-a ] ^{1-\alpha }}f (a )\,d \lambda + \int ^{1}_{\frac{a}{b}}\frac{ (\lambda b )'}{ [b-\lambda b ]^{1-\alpha }}f ( \lambda b )\,d\lambda \\ &\quad = \int ^{b}_{a}\frac{f (t )}{ [b-t ]^{1-\alpha }}\,dt = \varGamma ( \alpha )J^{\alpha }_{a^{+}}f (b ) . \end{aligned}$$
For the right-hand side of inequality, we have to examine two cases of \(1\leq \frac{f (a )}{f (b )}\) and \(0\leq \frac{f (a )}{f (b )}<1\). For the first case, we have
$$\begin{aligned} & \int ^{1}_{0} \frac{ (\max \{a, \lambda b \} )'}{ [b-\max \{a, \lambda b \} ]^{1-\alpha }} \max \bigl\{ f (a ), \lambda f (b ) \bigr\} \,d \lambda \\ &\quad = \int ^{1}_{0}\frac{ (\max \{a, \lambda b \} )'}{ [b-\max \{a, \lambda b \} ]^{1-\alpha }}f (a )\,d \lambda \\ &\quad =f (a ) \biggl[ \int ^{\frac{a}{b}}_{0}\frac{ (a )'}{ [b-a ]^{1-\alpha }}\,d\lambda + \int ^{1}_{\frac{a}{b}}\frac{ (\lambda b )'}{ [b-\lambda b ]^{1-\alpha }}\,d \lambda \biggr] \\ &\quad =f (a ) b^{\alpha } \int ^{1}_{\frac{a}{b}} [1- \lambda ]^{\alpha -1}\,d \lambda = \frac{f (a ) (b-a ) ^{\alpha }}{\alpha } . \end{aligned}$$
Thus the inequality is
$$ J^{\alpha }_{a^{+}}f (b )\leq \frac{f (a ) (b-a ) ^{\alpha }}{\varGamma (\alpha +1 )} . $$
(12)
For the second case, it can be
-
(i)
\(\frac{a}{b}\leq \frac{f (a )}{f (b )}<1\) or
-
(ii)
\(0\leq \frac{f (a )}{f (b )}<\frac{a}{b}\).
Let us examine (i). If \(\frac{a}{b}\leq \frac{f (a )}{f (b )}<1\), then the following equality is obtained:
$$\begin{aligned} & \int ^{1}_{0} \frac{ (\max \{a, \lambda b \} )'}{ [b-\max \{a, \lambda b \} ]^{1-\alpha }} \max \bigl\{ f (a ), \lambda f (b ) \bigr\} \,d \lambda \\ &\quad = \int ^{\frac{a}{b}}_{0}\frac{ ( a )'}{ [b-a ] ^{1-\alpha }}f (a )\,d \lambda + \int ^{\frac{f (a )}{f (b )}}_{\frac{a}{b}}\frac{ (\lambda b )'}{ [b-\lambda b ]^{1-\alpha }}f (a )\,d \lambda + \int ^{1}_{ \frac{f (a )}{f (b )}}\frac{ (\lambda b )'}{ [b-\lambda b ]^{1-\alpha }}\lambda f (b )\,d \lambda \\ &\quad =b^{\alpha }f (a ) \int ^{\frac{f (a )}{f (b )}}_{\frac{a}{b}} [1- \lambda ]^{\alpha -1}\,d \lambda +b^{\alpha }f (b ) \int ^{1}_{\frac{f (a )}{f (b )}}\lambda [1- \lambda ]^{\alpha -1}\,d\lambda \\ &\quad =\frac{b^{\alpha } (f (b )-f (a ) ) ^{\alpha +1}+ (\alpha +1 )f (a ) (f (b ) ) ^{\alpha } (b-a )^{\alpha }}{\alpha (\alpha +1 ) (f (b ) )^{\alpha }} . \end{aligned}$$
Hence, we obtain the inequality
$$ J^{\alpha }_{a^{+}}f (b )\leq \frac{b^{\alpha } (f (b )-f (a ) )^{\alpha +1}+ (\alpha +1 )f (a ) (f (b ) )^{\alpha } (b-a )^{\alpha }}{\varGamma (\alpha +2 ) (f (b ) )^{\alpha }} . $$
(13)
Let us introduce (ii). For the case of \(0\leq \frac{f (a )}{f (b )}<\frac{a}{b}\), we have
$$\begin{aligned} & \int ^{1}_{0} \frac{ (\max \{a, \lambda b \} )'}{ [b-\max \{a, \lambda b \} ]^{1-\alpha }} \max \bigl\{ f (a ), \lambda f (b ) \bigr\} \,d \lambda \\ &\quad = \int ^{\frac{f (a )}{f (b )}}_{0}\frac{ ( a )'}{ [b-a ]^{1-\alpha }}f (a )\,d \lambda + \int ^{\frac{a}{b}}_{\frac{f (a )}{f (b )}}\frac{ ( a )'}{ [b-a ]^{1-\alpha }}\lambda f (b )\,d \lambda + \int ^{1}_{\frac{a}{b}}\frac{ (\lambda b )'}{ [b- \lambda b ]^{1-\alpha }}\lambda f (b )\,d\lambda \\ &\quad =b^{\alpha }f (b ) \int ^{1}_{\frac{a}{b}}\lambda [1- \lambda ]^{\alpha -1}\,d\lambda = \frac{f (b ) (b-a ) ^{\alpha } (\alpha a+b )}{b \alpha (\alpha +1 )} . \end{aligned}$$
Then we deduce that
$$ J^{\alpha }_{a^{+}}f (b )\leq \frac{f (b ) (b-a ) ^{\alpha } (\alpha a+b )}{b \varGamma (\alpha +2 )} . $$
(14)
From (12), (13), and (14), we get the desired inequality. □
For the right-sided integral, the inequality can be proved with the same method of the above theorem by using the multiplier \(\frac{ (\max \{a, \lambda b\} )'}{ [\lambda b-a ]^{1-\alpha }}\).
2.2 Hermite–Hadamard type inequalities involving Hadamard fractional integral
In this section, we introduce Hermite–Hadamard type inequalities via Hadamard fractional integral.
Theorem 6
Let
\(\alpha >0\). If
f
is a
\(\mathbb{B}\)-convex function on
\([a,b ]\), then
$$ \mathbf{J}^{\alpha }_{a^{+}}f (b )\leq \textstyle\begin{cases} \frac{f (a ) (\ln \frac{b}{a} )^{\alpha }}{ \varGamma (\alpha +1 )}, & 1\leq \frac{f (a )}{f (b )}, \\ \frac{f (a ) [ (\ln \frac{b}{a} )^{\alpha }- (\ln \frac{f (b )}{f (a )} )^{ \alpha } ]}{\varGamma (\alpha +1 )}+\frac{f (b )}{ \varGamma (\alpha )}\int ^{1}_{\frac{f (a )}{f (b )}} (\ln \frac{1}{\lambda } )^{ (\alpha -1 )}\,d \lambda , & \frac{a}{b}\leq \frac{f (a )}{f (b )}< 1, \\ \frac{f (b )}{\varGamma (\alpha )}\int ^{1}_{ \frac{a}{b}} (\ln \frac{1}{\lambda } )^{ (\alpha -1 )}\,d \lambda , & 0\leq \frac{f (a )}{f (b )}< \frac{a}{b} . \end{cases} $$
(15)
Proof
Let us multiply inequality (7) that comes from \(\mathbb{B}\)-convexity of f by \(\frac{ (\max \{a, \lambda b\} )'}{ \max \{a, \lambda b\} [\ln b-\ln (\max \{a, \lambda b\} ) ] ^{1-\alpha }}\), then integrate with respect to λ over \([0,1 ]\). Hereby, for the left-hand side of inequality, we have that
$$\begin{aligned} & \int ^{1}_{0} \frac{ (\max \{a, \lambda b\} )'}{\max \{a, \lambda b\} [\ln b-\ln (\max \{a, \lambda b\} ) ] ^{1-\alpha }} f \bigl( \max \{a, \lambda b \} \bigr)\,d \lambda \\ &\quad = \int ^{\frac{a}{b}}_{0}\frac{ (a )'}{a [\ln b- \ln a ]^{1-\alpha }}f (a )\,d \lambda + \int ^{1}_{ \frac{a}{b}}\frac{ (\lambda b )'}{\lambda b [\ln b- \ln \lambda b ]^{1-\alpha }}f ( \lambda b )\,d\lambda \\ &\quad = \int ^{b}_{a}\frac{f (t )}{t [\ln \frac{b}{t} ] ^{1-\alpha }}\,dt = \varGamma ( \alpha )\mathbf{J}^{\alpha } _{a^{+}}f (b ) . \end{aligned}$$
If we introduce the right-hand side of the inequality, we come across two cases of \(\frac{f (a )}{f (b )}\). One of these cases is \(1\leq \frac{f (a )}{f (b )}\), and in this situation it is
$$\begin{aligned} & \int ^{1}_{0} \frac{ (\max \{a, \lambda b\} )'}{\max \{a, \lambda b\} [\ln b-\ln (\max \{a, \lambda b\} ) ] ^{1-\alpha }} \max \bigl\{ f (a ), \lambda f (b ) \bigr\} \,d \lambda \\ &\quad = \int ^{1}_{0}\frac{ (\max \{a, \lambda b\} )'}{\max \{a, \lambda b\} [\ln b-\ln (\max \{a, \lambda b\} ) ] ^{1-\alpha }} f (a ) \,d \lambda \\ &\quad =f (a ) \biggl[ \int ^{\frac{a}{b}}_{0}\frac{ (a )'}{a [\ln b- \ln a ]^{1-\alpha }}\,d\lambda + \int ^{1}_{\frac{a}{b}}\frac{ (\lambda b )'}{ \lambda b [\ln b-\ln \lambda b ]^{1-\alpha }} \,d\lambda \biggr] \\ &\quad =f (a ) \int ^{1}_{\frac{a}{b}}\frac{1}{\lambda } \biggl(\ln \frac{1}{ \lambda } \biggr)^{\alpha -1}\,d\lambda = \frac{f (a ) (\ln \frac{b}{a} )^{\alpha }}{\alpha } . \end{aligned}$$
Thence, with these calculations the first part of requested inequality is
$$ \mathbf{J}^{\alpha }_{a^{+}}f (b )\leq \frac{f (a ) (\ln \frac{b}{a} )^{\alpha }}{\varGamma (\alpha +1 )} . $$
(16)
The other case is \(0\leq \frac{f (a )}{f (b )}<1\) and it must be analyzed in the following separations:
(i) Let \(\frac{a}{b}\leq \frac{f (a )}{f (b )}<1\). We attain the following:
$$\begin{aligned} & \int ^{1}_{0} \frac{ (\max \{a, \lambda b\} )'}{\max \{a, \lambda b\} [\ln b-\ln (\max \{a, \lambda b\} ) ] ^{1-\alpha }} \max \bigl\{ f (a ), \lambda f (b ) \bigr\} \,d\lambda \\ &\quad = \int ^{\frac{a}{b}}_{0} \frac{ (a )'}{a [\ln b- \ln a ]^{1-\alpha }}f (a )\,d \lambda + \int ^{\frac{f (a )}{f (b )}}_{\frac{a}{b}} \frac{ (\lambda b )'}{\lambda b [\ln b-\ln \lambda b ] ^{1-\alpha }} f (a )\,d \lambda \\ &\qquad {}+ \int ^{1}_{\frac{f (a )}{f (b )}} \frac{ (\lambda b )'}{ \lambda b [\ln b-\ln \lambda b ]^{1-\alpha }} \lambda f (b ) \,d\lambda \\ &\quad =f (a ) \int ^{\frac{f (a )}{f (b )}} _{\frac{a}{b}}\frac{1}{\lambda } \biggl(\ln \frac{1}{\lambda } \biggr) ^{\alpha -1} \,d\lambda +f (b ) \int ^{1}_{\frac{f (a )}{f (b )}} \biggl(\ln \frac{1}{\lambda } \biggr)^{\alpha -1}\,d \lambda \\ &\quad =\frac{f (a ) [ (\ln \frac{b}{a} )^{ \alpha }- (\ln \frac{f (b )}{f (a )} ) ^{\alpha } ]}{\alpha }+f (b ) \int ^{1}_{\frac{f (a )}{f (b )}} \biggl(\ln \frac{1}{\lambda } \biggr)^{\alpha -1}\,d \lambda . \end{aligned}$$
Consequently, we obtain that
$$ \mathbf{J}^{\alpha }_{a^{+}}f (b )\leq \frac{f (a ) [ (\ln \frac{b}{a} )^{\alpha }- (\ln \frac{f (b )}{f (a )} )^{\alpha } ]}{ \varGamma (\alpha +1 )}+\frac{f (b )}{\varGamma (\alpha )} \int ^{1}_{ \frac{f (a )}{f (b )}} \biggl(\ln \frac{1}{ \lambda } \biggr)^{\alpha -1}\,d\lambda . $$
(17)
(ii) For \(0\leq \frac{f (a )}{f (b )}< \frac{a}{b}\), we get that
$$\begin{aligned} & \int ^{1}_{0} \frac{ (\max \{a, \lambda b\} )'}{\max \{a, \lambda b\} [\ln b-\ln (\max \{a, \lambda b\} ) ] ^{1-\alpha }} \max \bigl\{ f (a ), \lambda f (b ) \bigr\} \,d\lambda \\ &\quad = \int ^{\frac{f (a )}{f (b )}}_{0} \frac{ (a )'}{a [\ln b-\ln a ]^{1-\alpha }}f (a )\,d \lambda + \int ^{\frac{a}{b}}_{\frac{f (a )}{f (b )}} \frac{ (a )'}{a [\ln b-\ln a ] ^{1-\alpha }} \lambda f (b )\,d\lambda \\ &\qquad {}+ \int ^{1}_{\frac{a}{b}} \frac{ (\lambda b )'}{\lambda b [\ln b-\ln \lambda b ]^{1-\alpha }} \lambda f (b ) \,d\lambda \\ &\quad =f (b ) \int ^{1}_{\frac{a}{b}} \biggl(\ln \frac{1}{\lambda } \biggr)^{\alpha -1}\,d\lambda . \end{aligned}$$
Finally, the inequality is of the following shape:
$$ \mathbf{J}^{\alpha }_{a^{+}}f (b )\leq \frac{f (b )}{ \varGamma (\alpha )} \int ^{1}_{\frac{a}{b}} \biggl(\ln \frac{1}{ \lambda } \biggr)^{\alpha -1}\,d\lambda . $$
(18)
Hence, inequality (15) can be derived from (16), (17), and (18). □
Theorem 7
Let
\(\alpha >0\). If
f
is a
\(\mathbb{B}\)-convex function on
\([a,b ]\), then
$$ \mathbf{J}^{\alpha }_{b^{-}}f (a )\leq \textstyle\begin{cases} \frac{f (a ) (\ln \frac{b}{a} )^{\alpha }}{ \varGamma (\alpha +1 )}, & 1\leq \frac{f (a )}{f (b )}, \\ \frac{f (a ) (\ln \frac{b f (a )}{a f (b )} )^{\alpha }}{\varGamma (\alpha +1 )}+\frac{f (b )}{\varGamma (\alpha )}\int ^{1}_{\frac{f (a )}{f (b )}} (\ln \frac{\lambda b}{a} ) ^{ (\alpha -1 )}\,d\lambda , & \frac{a}{b}\leq \frac{f (a )}{f (b )}< 1, \\ \frac{f (b )}{\varGamma (\alpha )}\int ^{1}_{ \frac{a}{b}} (\ln \frac{\lambda b}{a} )^{ (\alpha -1 )}\,d \lambda , & 0\leq \frac{f (a )}{f (b )}< \frac{a}{b} . \end{cases} $$
(19)
Proof
Since the function f is \(\mathbb{B}\)-convex, inequality (7) holds. Now, we have to multiply by \(\frac{ (\max \{a, \lambda b \} )'}{\max \{a, \lambda b\} [\ln \lambda b-\ln a ] ^{1-\alpha }}\) and integrate with respect to λ over \([0,1 ]\) to this inequality. Hence, we have
$$\begin{aligned} & \int ^{1}_{0} \frac{ (\max \{a, \lambda b\} )'}{\max \{a, \lambda b\} [\ln \lambda b-\ln a ]^{1-\alpha }} f \bigl( \max \{a, \lambda b \} \bigr)\,d\lambda \\ &\quad = \int ^{\frac{a}{b}}_{0} \frac{ (a )'}{a [\ln \lambda b-\ln a ]^{1-\alpha }}f (a )\,d \lambda + \int ^{1}_{\frac{a}{b}} \frac{ (\lambda b )'}{\lambda b [\ln \lambda b-\ln a ]^{1-\alpha }}f ( \lambda b )\,d\lambda \\ &\quad = \int ^{b}_{a}\frac{f (t )}{t [\ln \frac{t}{a} ] ^{1-\alpha }}\,dt = \varGamma ( \alpha )\mathbf{J}^{\alpha } _{b^{-}}f (a ) . \end{aligned}$$
For the right-hand side of inequality, there are the following two situations that have to be examined. The first is \(1\leq \frac{f (a )}{f (b )}\). Thus,
$$\begin{aligned} & \int ^{1}_{0} \frac{ (\max \{a, \lambda b\} )'}{\max \{a, \lambda b\} [\ln \lambda b-\ln a ]^{1-\alpha }} \max \bigl\{ f (a ), \lambda f (b ) \bigr\} \,d\lambda \\ &\quad = \int ^{1}_{0} \frac{ (\max \{a, \lambda b\} )'}{\max \{a, \lambda b\} [\ln \lambda b-\ln a ]^{1-\alpha }} f (a ) \,d \lambda \\ &\quad =f (a ) \biggl[ \int ^{\frac{a}{b}}_{0} \frac{ (a )'}{a [\ln \lambda b-\ln a ]^{1- \alpha }}\,d\lambda + \int ^{1}_{\frac{a}{b}} \frac{ (\lambda b )'}{ \lambda b [\ln \lambda b-\ln a ]^{1-\alpha }} \,d\lambda \biggr] \\ &\quad =f (a ) \int ^{1}_{\frac{a}{b}}\frac{1}{\lambda } \biggl(\ln \frac{ \lambda b}{a} \biggr)^{\alpha -1}\,d\lambda = \frac{f (a ) (\ln \frac{b}{a} )^{\alpha }}{\alpha } . \end{aligned}$$
So, the inequality is in the following form:
$$ \mathbf{J}^{\alpha }_{b^{-}}f (a )\leq \frac{f (a ) (\ln \frac{b}{a} )^{\alpha }}{\varGamma (\alpha +1 )} . $$
(20)
The second case is \(0\leq \frac{f (a )}{f (b )}<1\), and it should be introduced with the following two cases:
(i) One of the cases is \(\frac{a}{b}\leq \frac{f (a )}{f (b )}<1\). At this stage, we get that
$$\begin{aligned} & \int ^{1}_{0} \frac{ (\max \{a, \lambda b\} )'}{\max \{a, \lambda b\} [\ln \lambda b-\ln a ]^{1-\alpha }} \max \bigl\{ f (a ), \lambda f (b ) \bigr\} \,d\lambda \\ &\quad = \int ^{\frac{a}{b}}_{0} \frac{ (a )'}{a [\ln \lambda b-\ln a ]^{1-\alpha }}f (a )\,d \lambda + \int ^{\frac{f (a )}{f (b )}}_{\frac{a}{b}} \frac{ (\lambda b )'}{\lambda b [\ln \lambda b-\ln a ] ^{1-\alpha }} f (a )\,d \lambda \\ &\qquad {}+ \int ^{1}_{\frac{f (a )}{f (b )}} \frac{ (\lambda b )'}{ \lambda b [\ln \lambda b-\ln a ]^{1-\alpha }} \lambda f (b ) \,d\lambda \\ &\quad =f (a ) \int ^{\frac{f (a )}{f (b )}} _{\frac{a}{b}}\frac{1}{\lambda } \biggl(\ln \frac{\lambda b}{a} \biggr) ^{\alpha -1} \,d\lambda +f (b ) \int ^{1}_{\frac{f (a )}{f (b )}} \biggl(\ln \frac{\lambda b}{a} \biggr)^{\alpha -1}\,d \lambda \\ &\quad =\frac{f (a ) (\ln \frac{b f (a )}{a f (b )} )^{\alpha }}{\alpha }+f (b ) \int ^{1}_{\frac{f (a )}{f (b )}} \biggl(\ln \frac{ \lambda b}{a} \biggr)^{\alpha -1}\,d\lambda . \end{aligned}$$
Therefore, the inequality is obtained
$$ \mathbf{J}^{\alpha }_{b^{-}}f (a )\leq \frac{f (a ) (\ln \frac{b f (a )}{a f (b )} ) ^{\alpha }}{\varGamma (\alpha +1 )}+\frac{f (b )}{ \varGamma (\alpha )} \int ^{1}_{\frac{f (a )}{f (b )}} \biggl(\ln \frac{\lambda b}{a} \biggr)^{\alpha -1}\,d \lambda . $$
(21)
(ii) The other one of the cases is \(0\leq \frac{f (a )}{f (b )}<\frac{a}{b}\). So, we have
$$\begin{aligned} & \int ^{1}_{0} \frac{ (\max \{a, \lambda b\} )'}{\max \{a, \lambda b\} [\ln \lambda b-\ln a ]^{1-\alpha }} \max \bigl\{ f (a ), \lambda f (b ) \bigr\} \,d\lambda \\ &\quad = \int ^{\frac{f (a )}{f (b )}}_{0} \frac{ (a )'}{a [\ln \lambda b-\ln a ]^{1- \alpha }}f (a )\,d \lambda + \int ^{\frac{a}{b}}_{\frac{f (a )}{f (b )}} \frac{ (a )'}{a [\ln \lambda b- \ln a ]^{1-\alpha }} \lambda f (b )\,d\lambda \\ &\qquad {}+ \int ^{1}_{\frac{a}{b}} \frac{ (\lambda b )'}{\lambda b [\ln \lambda b-\ln a ]^{1-\alpha }} \lambda f (b ) \,d\lambda \\ &\quad =f (b ) \int ^{1}_{\frac{a}{b}} \biggl(\ln \frac{\lambda b}{a} \biggr)^{\alpha -1}\,d\lambda . \end{aligned}$$
Thus, we get that
$$ \mathbf{J}^{\alpha }_{b^{-}}f (a )\leq \frac{f (b )}{ \varGamma (\alpha )} \int ^{1}_{\frac{a}{b}} \biggl(\ln \frac{ \lambda b}{a} \biggr)^{\alpha -1}\,d\lambda . $$
(22)
As a result, we have proven inequality (19) by using inequalities (20), (21), and (22). □
2.3 Hermite–Hadamard type inequalities involving fractional integral with respect to the function g
Theorem 8
Let
\(\alpha >0\)
and
\(0\leq a< b<+\infty \), \(g: [a,b ]\rightarrow \mathbb{R}\)
be an increasing and positive monotone function on
\((a,b ]\), having a continuous derivative
\(g' (x )\)
on
\((a,b )\). If
\(f: [a,b ] \rightarrow \mathbb{R}_{+}\)
is a
\(\mathbb{B}\)-convex function and
\(f\in L_{1} [a,b ]\), then
$$ \begin{aligned}[b] &I^{\alpha }_{a^{+};g}f (b )\\ &\quad \leq \textstyle\begin{cases} \frac{f (a ) [g (b )-g (a ) ] ^{\alpha }}{\varGamma (\alpha +1 )}, & 1\leq \frac{f (a )}{f (b )}, \\ \frac{f (a ) [ [g (b )-g (a ) ] ^{\alpha }- [g (b )-g (\frac{bf (a )}{f (b )} ) ]^{\alpha } ]}{\varGamma (\alpha +1 )}+\frac{bf (b )}{\varGamma (\alpha )}\int ^{1}_{\frac{f (a )}{f (b )}}\frac{g' (\lambda b ) \lambda }{ [g (b )-g (\lambda b ) ] ^{1-\alpha }}\,d\lambda , & \frac{a}{b}\leq \frac{f (a )}{f (b )}< 1, \\ \frac{bf (b )}{\varGamma (\alpha )}\int ^{1}_{ \frac{a}{b}}\frac{g' (\lambda b )\lambda }{ [g (b )-g (\lambda b ) ]^{1-\alpha }}\,d\lambda , & 0\leq \frac{f (a )}{f (b )}< \frac{a}{b} . \end{cases}\displaystyle \end{aligned} $$
(23)
Proof
Due to \(\mathbb{B}\)-convexity of \(f: [a,b ]\rightarrow \mathbb{R}_{+}\), inequality (7) is valid for all \(\lambda \in [0,1 ]\) and \(0< a< b<+\infty \). If we multiply both sides of this inequality by \(\frac{g' (\max \{a, \lambda b \} )}{ [g (b )-g (\max \{a, \lambda b \} ) ] ^{1-\alpha }}\) and integrate the resulting inequality with respect to λ over \([0,1 ]\), we obtain for the left-hand side of inequality the following equation:
$$\begin{aligned} & \int ^{1}_{0} \frac{g' (\max \{a, \lambda b \} )}{ [g (b )-g (\max \{a, \lambda b \} ) ] ^{1-\alpha }}f \bigl(\max \{a, \lambda b \} \bigr)\,d \lambda \\ &\quad = \int ^{\frac{a}{b}}_{0}\frac{g' (a ) (a )'}{ [g (b )-g (a ) ]^{1-\alpha }}f (a )\,d \lambda + \int ^{1}_{\frac{a}{b}}\frac{g' (\lambda b ) (\lambda b )'}{ [g (b )-g (\lambda b ) ]^{1-\alpha }}f ( \lambda b )\,d\lambda \\ &\quad = \int ^{b}_{a}\frac{g' (t )}{ [g (b )-g (t ) ]^{1-\alpha }}f ( t )\,dt \\ &\quad =\varGamma (\alpha )I^{\alpha }_{a^{+};g}f (b ) . \end{aligned}$$
For the right-hand side of inequality, we meet two possibilities. One of these is the possibility of \(1\leq \frac{f (a )}{f (b )}\). In this case, the equality is
$$\begin{aligned} & \int ^{1}_{0} \frac{g' (\max \{a, \lambda b \} )}{ [g (b )-g (\max \{a, \lambda b \} ) ] ^{1-\alpha }}\max \bigl\{ f (a ), \lambda f (b ) \bigr\} \,d \lambda \\ &\quad = \int ^{1}_{0}\frac{g' (\max \{a, \lambda b \} )}{ [g (b )-g (\max \{a, \lambda b \} ) ] ^{1-\alpha }}f (a )\,d \lambda \\ &\quad =f (a ) \biggl[ \int ^{\frac{a}{b}}_{0}\frac{g' (a ) (a )'}{ [g (b )-g (a ) ] ^{1-\alpha }}\,d\lambda + \int ^{1}_{\frac{a}{b}}\frac{g' (\lambda b ) (\lambda b )'}{ [g (b )-g (\lambda b ) ] ^{1-\alpha }}\,d\lambda \biggr] \\ &\quad =f (a ) \int ^{b}_{a}\frac{g' (t )}{ [g (b )-g (t ) ]^{1-\alpha }}\,dt = f (a ) \frac{ [g (b )-g (a ) ]^{\alpha }}{\alpha } . \end{aligned}$$
Then we deduce that
$$\begin{aligned}& \int ^{1}_{0}\frac{g' (\max \{a, \lambda b \} )}{ [g (b )-g (\max \{a, \lambda b \} ) ] ^{1-\alpha }}f \bigl(\max \{a, \lambda b \} \bigr)\,d \lambda \\& \quad \leq \int ^{1}_{0}\frac{g' (\max \{a, \lambda b \} )}{ [g (b )-g (\max \{a, \lambda b \} ) ] ^{1-\alpha }}\max \bigl\{ f (a ), \lambda f (b ) \bigr\} \,d \lambda, \\& \varGamma (\alpha )I^{\alpha }_{a^{+};g}f (b ) \leq f (a ) \frac{ [g (b )-g (a ) ] ^{\alpha }}{\alpha }, \\& I^{\alpha }_{a^{+};g}f (b ) \leq \frac{f (a ) [g (b )-g (a ) ]^{\alpha }}{\varGamma (\alpha +1 )} . \end{aligned}$$
In the possibility of \(0\leq \frac{f (a )}{f (b )}<1\), we have to examine the following two cases:
-
(i)
\(\frac{a}{b}\leq \frac{f (a )}{f (b )}<1\),
-
(ii)
\(0\leq \frac{f (a )}{f (b )}<\frac{a}{b}\).
Let us investigate case (i). If \(\frac{a}{b}\leq \frac{f (a )}{f (b )}<1\), then we have
$$\begin{aligned} & \int ^{1}_{0} \frac{g' (\max \{a, \lambda b \} )}{ [g (b )-g (\max \{a, \lambda b \} ) ] ^{1-\alpha }}\max \bigl\{ f (a ), \lambda f (b ) \bigr\} \,d \lambda \\ &\quad = \int ^{\frac{a}{b}}_{0}\frac{g' (a ) (a )'}{ [g (b )-g (a ) ]^{1-\alpha }}f (a )\,d \lambda + \int ^{\frac{f (a )}{f (b )}}_{ \frac{a}{b}}\frac{g' (\lambda b ) (\lambda b )'}{ [g (b )-g (\lambda b ) ]^{1-\alpha }}f (a )\,d \lambda \\ &\qquad {}+ \int ^{1}_{\frac{f (a )}{f (b )}}\frac{g' (\lambda b ) (\lambda b )'}{ [g (b )-g (\lambda b ) ]^{1-\alpha }}\lambda f (b )\,d \lambda \\ &\quad =\frac{f (a ) [ [g (b )-g (a ) ] ^{\alpha }- [g (b )-g (\frac{bf (a )}{f (b )} ) ]^{\alpha } ]}{\alpha } \\ &\qquad {}+b f (b ) \int ^{1}_{ \frac{f (a )}{f (b )}}\frac{g' (\lambda b ) \lambda }{ [g (b )-g (\lambda b ) ] ^{1-\alpha }}\,d\lambda . \end{aligned}$$
Therefore, we deduce that
$$ I^{\alpha }_{a^{+};g}f (b )\leq \frac{f (a ) [ [g (b )-g (a ) ]^{\alpha }- [g (b )-g (\frac{bf (a )}{f (b )} ) ] ^{\alpha } ]}{\varGamma (\alpha +1 )}+ \frac{bf (b )}{ \varGamma (\alpha )} \int ^{1}_{\frac{f (a )}{f (b )}}\frac{g' (\lambda b )\lambda }{ [g (b )-g (\lambda b ) ]^{1-\alpha }}\,d \lambda . $$
In the case of (ii), \(0\leq \frac{f (a )}{f (b )}<\frac{a}{b}\), we get
$$\begin{aligned} & \int ^{1}_{0} \frac{g' (\max \{a, \lambda b \} )}{ [g (b )-g (\max \{a, \lambda b \} ) ] ^{1-\alpha }}\max \bigl\{ f (a ), \lambda f (b ) \bigr\} \,d \lambda \\ &\quad = \int ^{\frac{f (a )}{f (b )}}_{0}\frac{g' (a ) (a )'}{ [g (b )-g (a ) ] ^{1-\alpha }}f (a )\,d \lambda + \int ^{\frac{a}{b}}_{\frac{f (a )}{f (b )}}\frac{g' (a ) (a )'}{ [g (b )-g (a ) ]^{1-\alpha }}\lambda f (b )\,d\lambda \\ &\qquad {}+ \int ^{1}_{\frac{a}{b}}\frac{g' (\lambda b ) (\lambda b )'}{ [g (b )-g (\lambda b ) ]^{1-\alpha }}\lambda f (b )\,d\lambda \\ &\quad =b f (b ) \int ^{1}_{\frac{a}{b}}\frac{g' (\lambda b ) \lambda }{ [g (b )-g (\lambda b ) ] ^{1-\alpha }}\,d\lambda . \end{aligned}$$
Consequently, for this case, we obtain the following inequality:
$$ I^{\alpha }_{a^{+};g}f (b )\leq \frac{bf (b )}{ \varGamma (\alpha )} \int ^{1}_{\frac{a}{b}}\frac{g' (\lambda b ) \lambda }{ [g (b )-g (\lambda b ) ] ^{1-\alpha }}\,d\lambda . $$
□
Theorem 9
Let
\(\alpha >0\)
and
\(0< a< b<+\infty \), \(g: [a,b ]\rightarrow \mathbb{R}\)
be an increasing and positive monotone function on
\((a,b ]\), having a continuous derivative
\(g' (x )\)
on
\((a,b )\). If
\(f: [a,b ] \rightarrow \mathbb{R}_{+}\)
is a
\(\mathbb{B}\)-convex function and
\(f\in L_{1} [a,b ]\), then the following inequality for fractional integrals holds:
$$ I^{\alpha }_{b^{-};g}f (a )\leq \textstyle\begin{cases} \frac{f (a ) [g (b )-g (a ) ] ^{\alpha }}{\varGamma (\alpha +1 )}, & 1\leq \frac{f (a )}{f (b )}, \\ \frac{f (a ) [g (\frac{bf (a )}{f (b )} )-g (a ) ]^{\alpha }}{\varGamma (\alpha +1 )}+ \frac{bf (b )}{\varGamma (\alpha )}\int ^{1}_{\frac{f (a )}{f (b )}}\frac{g' (\lambda b ) \lambda }{ [g (\lambda b )-g (a ) ] ^{1-\alpha }}\,d\lambda , & \frac{a}{b}\leq \frac{f (a )}{f (b )}< 1, \\ \frac{bf (b )}{\varGamma (\alpha )}\int ^{1}_{ \frac{a}{b}}\frac{g' (\lambda b )\lambda }{ [g (\lambda b )-g (a ) ]^{1-\alpha }}\,d\lambda , & 0\leq \frac{f (a )}{f (b )}< \frac{a}{b} . \end{cases} $$
(24)
Proof
For the \(\mathbb{B}\)-convex function \(f: [a,b ]\rightarrow \mathbb{R}_{+}\), we have the following inequality:
$$ f \bigl(\max \{a, \lambda b \} \bigr)\leq \max \bigl\{ f (a ), \lambda f (b ) \bigr\} $$
(25)
for all \(\lambda \in [0,1 ]\). To obtain inequality (24), we should multiply both sides of (25) by \(\frac{g' (\max \{a, \lambda b \} )}{ [g (\lambda b )-g (a ) ]^{1-\alpha }}\) and integrate the resulting inequality with respect to λ over \([0,1 ]\). Thus, for the left-hand side of inequality, we get
$$\begin{aligned} & \int ^{1}_{0} \frac{g' (\max \{a, \lambda b \} )}{ [g (\lambda b )-g (a ) ]^{1-\alpha }}f \bigl(\max \{a, \lambda b \} \bigr)\,d\lambda \\ &\quad = \int ^{1}_{\frac{a}{b}}\frac{g' (\lambda b ) (\lambda b )'}{ [g (\lambda b )-g (a ) ]^{1-\alpha }}f (\lambda b )\,d\lambda \\ &\quad = \int ^{b}_{a}\frac{g' (t )}{ [g (t )-g (a ) ]^{1-\alpha }}f (t )\,dt = \varGamma (\alpha ) I^{\alpha }_{b^{-};g}f (a ). \end{aligned}$$
For the right-hand side of inequality, we have to consider two cases as follows.
Firstly, it can be \(1\leq \frac{f (a )}{f (b )}\). In this case, for the right-hand side, we obtain that
$$\begin{aligned} \int ^{1}_{0}\frac{g' (\max \{a, \lambda b \} )}{ [g (\lambda b )-g (a ) ]^{1-\alpha }}\max \bigl\{ f (a ), \lambda f (b ) \bigr\} \,d \lambda =& \int ^{1}_{\frac{a}{b}}\frac{g' (\lambda b ) (\lambda b )'}{ [g (\lambda b )-g (a ) ] ^{1-\alpha }}f (a )\,d \lambda \\ =&\frac{f (a ) [g (b )-g (a ) ] ^{\alpha }}{\alpha } . \end{aligned}$$
Hence,
$$ \begin{aligned}[b] &\varGamma (\alpha ) I^{\alpha }_{b^{-};g}f (a ) \leq \frac{f (a ) [g (b )-g (a ) ] ^{\alpha }}{\alpha }, \\ &I^{\alpha }_{b^{-};g}f (a )\leq \frac{f (a ) [g (b )-g (a ) ]^{\alpha }}{\varGamma (\alpha +1 )} . \end{aligned} $$
(26)
When we handle the second case \(0\leq \frac{f (a )}{f (b )}<1\), we meet the following two possibilities:
-
(i)
\(\frac{a}{b}\leq \frac{f (a )}{f (b )}<1\),
-
(ii)
\(0\leq \frac{f (a )}{f (b )}< \frac{a}{b}\).
For (i), let us calculate the right-hand side of inequality and then show the last case.
$$\begin{aligned} & \int ^{1}_{0} \frac{g' (\max \{a, \lambda b \} )}{ [g (\lambda b )-g (a ) ]^{1-\alpha }} \max \bigl\{ f (a ), \lambda f (b ) \bigr\} \,d \lambda \\ &\quad = \int ^{\frac{f (a )}{f (b )}}_{\frac{a}{b}}\frac{g' ( \lambda b ) (\lambda b )'}{ [g (\lambda b )-g (a ) ]^{1-\alpha }}f (a )\,d \lambda + \int ^{1}_{\frac{f (a )}{f (b )}}\frac{g' ( \lambda b ) (\lambda b )'}{ [g ( \lambda b )-g (a ) ]^{1-\alpha }}\lambda f (b )\,d\lambda \\ &\quad =\frac{f (a ) [g (\frac{bf (a )}{f (b )} )-g (a ) ]^{\alpha }}{\alpha } + b f (b ) \int ^{1}_{ \frac{f (a )}{f (b )}}\frac{g' (\lambda b ) \lambda }{ [g (\lambda b )-g (a ) ] ^{1-\alpha }}\,d\lambda . \end{aligned}$$
Thus, we have that
$$ \begin{aligned}[b] &\varGamma (\alpha ) I^{\alpha }_{b^{-};g}f (a ) \leq \frac{f (a ) [g (\frac{bf (a )}{f (b )} )-g (a ) ]^{\alpha }}{\alpha } + b f (b ) \int ^{1}_{ \frac{f (a )}{f (b )}}\frac{g' (\lambda b ) \lambda }{ [g (\lambda b )-g (a ) ] ^{1-\alpha }}\,d\lambda , \\ &I^{\alpha }_{b^{-};g}f (a )\leq \frac{f (a ) [g (\frac{bf (a )}{f (b )} )-g (a ) ]^{\alpha }}{\varGamma (\alpha +1 )}+ \frac{bf (b )}{\varGamma (\alpha )} \int ^{1}_{\frac{f (a )}{f (b )}}\frac{g' (\lambda b ) \lambda }{ [g (\lambda b )-g (a ) ] ^{1-\alpha }}\,d\lambda . \end{aligned} $$
(27)
For case (ii), we have that
$$\begin{aligned} \int ^{1}_{0}\frac{g' (\max \{a, \lambda b \} )}{ [g (\lambda b )-g (a ) ]^{1-\alpha }}\max \bigl\{ f (a ), \lambda f (b ) \bigr\} \,d \lambda =& \int ^{1}_{\frac{a}{b}}\frac{g' ( \lambda b ) (\lambda b )'}{ [g (\lambda b )-g (a ) ] ^{1-\alpha }}\lambda f (b )\,d\lambda \\ =& b f (b ) \int ^{1}_{\frac{a}{b}}\frac{g' (\lambda b ) \lambda }{ [g (\lambda b )-g (a ) ] ^{1-\alpha }}\,d\lambda . \end{aligned}$$
Thus, we deduce
$$ \begin{aligned}[b] &\varGamma (\alpha ) I^{\alpha }_{b^{-};g}f (a ) \leq b f (b ) \int ^{1}_{\frac{a}{b}}\frac{g' (\lambda b )\lambda }{ [g (\lambda b )-g (a ) ] ^{1-\alpha }}\,d\lambda, \\ &I^{\alpha }_{b^{-};g}f (a )\leq \frac{bf (b )}{ \varGamma (\alpha )} \int ^{1}_{\frac{a}{b}}\frac{g' (\lambda b ) \lambda }{ [g (\lambda b )-g (a ) ] ^{1-\alpha }}\,d\lambda . \end{aligned} $$
(28)
Consequently, from (26), (27), and (28), we have inequality (24). □
The given results in Theorem 8 and Theorem 9 are the most general Hermite–Hadamard inequalities for \(\mathbb{B}\)-convex functions.
Corollary 1
Inequalities (23) and (24) are generalized forms of inequalities (11) and (10), respectively.
Indeed, observe that for \(g (x ) = x\), the fractional integral (5) reduces to the left-sided Riemann–Liouville fractional integral (1), and the fractional integral (6) reduces to the right-sided Riemann–Liouville fractional integral (2).
Additionally, these hypotheses are valid in our results. Namely, if we get \(g (x ) = x\) in (23), the inequality returns to (11). Similarly, getting \(g (x ) = x\) in (24) gives inequality (10).
Corollary 2
Hermite–Hadamard inequality for a
\(\mathbb{B}\)-convex function involving Hadamard fractional integral is obtained from inequalities (23) and (24).
This can be seen by using the same method in Corollary 1.
We can give the following examples for inequalities (23) and (24) to be more precise.
Example 1
Let us assume that \(\alpha =2\), \(f: [1,2 ]\rightarrow \mathbb{R}_{+}\), \(f (x )=x^{2}\), and \(g: [1,2 ] \rightarrow \mathbb{R}_{+}\), \(g (x )=x^{3}\). Indeed the function f satisfies inequality (7), namely it is a \(\mathbb{B}\)-convex function. Thus, if we examine inequality (23) under all these assumptions, then we deduce
$$ I^{2}_{1^{+};x^{3}}f (2 )= \frac{1}{\varGamma (2 )} \int ^{2}_{1}\frac{3t^{2}\cdot t^{2}}{ [2^{3}-t^{3} ]^{1-2}}\,dt= \frac{2127}{40} < \frac{2\cdot 2^{2}}{ \varGamma (2 )} \int ^{1}_{\frac{1}{2}}\frac{3 (2\lambda )^{2}\lambda }{ [2^{3}- (2\lambda )^{3} ] ^{1-2}}\,d\lambda = \frac{498}{7} $$
because of \(0\leq \frac{f (a )}{f (b )}= \frac{1}{4}<\frac{a}{b}=\frac{1}{2}\). Additionally, if inequality (24) is analyzed in the present circumstances, then one has
$$ I^{2}_{2^{-};x^{3}}f (1 )= \frac{1}{\varGamma (2 )} \int ^{2}_{1}\frac{3t^{2}\cdot t^{2}}{ [t^{3}-1^{3} ]^{1-2}}\,dt= \frac{3081}{40} < \frac{2\cdot 2^{2}}{ \varGamma (2 )} \int ^{1}_{\frac{1}{2}}\frac{3 (2\lambda )^{2}\lambda }{ [ (2\lambda )^{3}-1^{3} ] ^{1-2}}\,d\lambda = \frac{1209}{14} . $$
Furthermore, it can be seen that all inequalities (10), (11), (15), and (19) are valid for the same function.
For more examples of \(\mathbb{B}\)-convex functions, see [9].