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Hermite–Hadamard type inequalities for F-convex function involving fractional integrals
Journal of Inequalities and Applications volume 2018, Article number: 359 (2018)
Abstract
In this study, the family F and F-convex function are given with its properties. In view of this, we establish some new inequalities of Hermite–Hadamard type for differentiable function. Moreover, we establish some trapezoid type inequalities for functions whose second derivatives in absolute values are F-convex. We also show that through the notion of F-convex we can find some new Hermite–Hadamard type and trapezoid type inequalities for the Riemann–Liouville fractional integrals and classical integrals.
1 Introduction
A function \(f: I\subseteq \mathbb{R}\to \mathbb{R}\) is said to be convex on the interval I, if for all \(x,y\in I\) and \(t\in (0,1)\) it satisfies the following inequality:
Convex functions play an important role in the field of integral inequalities. For convex functions, many equalities and inequalities have been established, but one of the most important ones is the Hermite–Hadamard’ integral inequality, which is defined as follows [1]:
Let \(f: I\subseteq \mathbb{R}\to \mathbb{R}\) be a convex function with \(a< b\) and \(a, b\in I\). Then the Hermite–Hadamard inequality is given by
In recent years, a number of mathematicians have devoted their efforts to generalizing, refining, counterparting, and extending the Hermite–Hadamard inequality (2) for different classes of convex functions and mappings. The Hermite–Hadamard inequality (2) is established for the classical integral, fractional integrals, conformable fractional integrals and most recently for generalized fractional integrals; see for details and applications [2,3,4,5,6,7,8] and the references therein.
The concepts of classical convex functions have been extended and generalized in several directions, such as quasi-convex [9], pseudo-convex [10], MT-convex [11] strongly convex [12], ϵ-convex [13], s-convex [14], h-convex [15], and \(\lambda _{\varphi }\)-preinvex [16]. Recently, Samet [17] has defined a new concept of convexity that depends on a certain function satisfying some axioms, generalizing different types of convexity, including ϵ-convex functions, α-convex functions, h-convex functions, and so on, as stated in the next section.
2 Review of the family of \(\mathcal{F}\)
We address the family of \(\mathcal{F}\) of mappings \(F:\mathbb{R} \times \mathbb{R}\times \mathbb{R}\times [0,1]\to \mathbb{R}\) satisfying the following axioms:
-
(A1)
If \(u_{i}\in L^{1}(0,1)\), \(i=1,2,3\), then, for every \(\lambda \in [0,1]\), we have
$$\begin{aligned} \int _{0}^{1} F\bigl(u_{1}(t),u_{2}(t),u_{3}(t), \lambda \bigr)\,dt=F \biggl( \int _{0}^{1} u_{1}(t)\,dt, \int _{0}^{1} u_{2}(t)\,dt, \int _{0}^{1} u_{3}(t)\,dt, \lambda \biggr). \end{aligned}$$ -
(A2)
For every \(u\in L^{1}(0,1)\), \(w\in L^{\infty }(0,1)\) and \((z_{1},z_{2})\in \mathbb{R}^{2}\), we have
$$\begin{aligned} \int _{0}^{1} F\bigl(w(t) u(t),w(t)z_{1},w(t)z_{2},t \bigr)\,dt=T_{F,w} \biggl( \int _{0}^{1} w(t) u(t)\,dt,z_{1},z_{2} \biggr), \end{aligned}$$where \(T_{F,w}:\mathbb{R}\times \mathbb{R}\times \mathbb{R}\to \mathbb{R}\) is a function that depends on \((F,w)\), and it is nondecreasing with respect to the first variable.
-
(A3)
For any \((w,u_{1},u_{2},u_{3})\in \mathbb{R}^{4}\), \(u_{4} \in [0,1]\), we have
$$\begin{aligned} wF(u_{1},u_{2},u_{3},u_{4})=F(wu_{1},wu_{2},wu_{3},u_{4})+L_{w}, \end{aligned}$$where \(L_{w}\in \mathbb{R}\) is a constant that depends only on w.
Definition 2.1
Let \(f: [a,b]\to \mathbb{R}\), \((a,b)\in \mathbb{R}^{2}\), \(a< b\), be a given function. We say that f is a convex function with respect to some \(F\in \mathcal{F}\) (or F-convex function) iff
Remark 1
Suppose that \((a,b)\in \mathbb{R}^{2}\) with \(a< b\).
-
(i)
Let \(f:[a,b]\to \mathbb{R}\) be an ε-convex function, that is [18],
$$\begin{aligned} f\bigl(tx+(1-t)y\bigr)\leq tf(x)+(1-t)f(y),\quad (x,y,t)\in [a,b]\times [a,b] \times [0,1]. \end{aligned}$$Define the functions \(F:\mathbb{R}\times \mathbb{R}\times \mathbb{R} \times [0,1]\to \mathbb{R}\) by
$$\begin{aligned} F(u_{1},u_{2},u_{3},u_{4})=u_{1}-u_{4}u_{2}-(1-u_{4})u_{3}- \varepsilon \end{aligned}$$(3)and \(T_{F,w}:\mathbb{R}\times \mathbb{R}\times \mathbb{R}\times [0,1] \to \mathbb{R}\) by
$$\begin{aligned} T_{F,w}(u_{1},u_{2},u_{3})=u_{1}- \biggl( \int _{0}^{1} t w(t)\,dt \biggr)u _{2} - \biggl( \int _{0}^{1} (1-t)w(t)\,dt \biggr)u_{3}- \varepsilon . \end{aligned}$$(4)For
$$\begin{aligned} L_{w}=(1-w)\varepsilon , \end{aligned}$$(5)it will be seen that \(F\in \mathcal{F}\) and
$$\begin{aligned} F\bigl(f\bigl(tx+(1-t)y\bigr),f(x),f(y),t\bigr) &=f\bigl(tx+(1-t)y \bigr)-tf(x)-(1-t)f(y)-\varepsilon \leq 0, \end{aligned}$$that is, f is an F-convex function. Particularly, taking \(\varepsilon =0\) we show that if f is a convex function then f is an F-convex function with respect to F defined above.
-
(ii)
Let \(f:[a,b]\to \mathbb{R}\) be \(\lambda _{\varphi }\)-preinvex function according to φ and bifunction η, \(0\leq \varphi \leq \frac{\pi }{2}\), \(\lambda \in (0,\frac{1}{2} ]\), that is [16],
$$\begin{aligned} &f \bigl(u+te^{i\varphi }\eta (v,u) \bigr) \\ &\quad \leq \frac{\sqrt{t}}{2 \sqrt{1-t}}f(v) + \frac{(1-\lambda )\sqrt{1-t}}{2\lambda \sqrt{t}}f(u), \quad (u,v,t) \in [a,b]\times [a,b]\times (0,1). \end{aligned}$$Define the functions \(F:\mathbb{R}\times \mathbb{R}\times \mathbb{R} \times [0,1]\to \mathbb{R}\) by
$$\begin{aligned} F(u_{1},u_{2},u_{3},u_{4})=u_{1}- \frac{\sqrt{u_{4}}}{2\sqrt{1-u _{4}}}u_{3} -\frac{(1-\lambda )\sqrt{1-u_{4}}}{2\lambda \sqrt{u _{4}}}u_{2} \end{aligned}$$(6)and \(T_{F,w}:\mathbb{R}\times \mathbb{R}\times \mathbb{R}\times [0,1] \to \mathbb{R}\) by
$$\begin{aligned} \begin{aligned}[b] T_{F,w}(u_{1},u_{2},u_{3})&=u_{1}- \biggl( \int _{0}^{1} \frac{\sqrt{t}}{2 \sqrt{1-t}}w(t)\,dt \biggr)u_{3} \\ &\quad {}-\frac{1-\lambda }{\lambda } \biggl( \int _{0}^{1} \frac{\sqrt{1-t}}{2\sqrt{t}}w(t)\,dt \biggr)u_{2}. \end{aligned} \end{aligned}$$(7)For \(L_{w}=0\), it will be seen that \(F\in \mathcal{F}\) and
$$\begin{aligned} &F \bigl(f \bigl(u+te^{i\varphi }\eta (v,u) \bigr),f(u),f(v),t \bigr) \\ &\quad =f \bigl(u+te^{i\varphi }\eta (v,u) \bigr)-\frac{\sqrt{t}}{2 \sqrt{1-t}}f(v) - \frac{(1-\lambda )\sqrt{1-t}}{2\lambda \sqrt{t}}f(u)- \varepsilon \leq 0, \end{aligned}$$that is f is an F-convex function.
-
(iii)
Let \(h:I\to \mathbb{R}\) be a given function which is not identical to 0, where I is an interval in \(\mathbb{R}\) such that \((0,1)\subseteq I\). Let \(f:[a,b]\to [0,\infty )\) be an h-convex function, that is,
$$\begin{aligned} f\bigl(tx+(1-t)y\bigr)\leq h(t)f(x)+\frac{1-\lambda }{\lambda }h(1-t)f(y), \quad (x,y,t) \in [a,b]\times [a,b]\times [0,1]. \end{aligned}$$Define the functions \(F:\mathbb{R}\times \mathbb{R}\times \mathbb{R} \times [0,1]\to \mathbb{R}\) by
$$\begin{aligned} F(u_{1},u_{2},u_{3},u_{4})=u_{1}-h(u_{4})u_{3}- \frac{1-\lambda }{ \lambda }h(1-u_{4})u_{2} \end{aligned}$$(8)and \(T_{F,w}:\mathbb{R}\times \mathbb{R}\times \mathbb{R}\times [0,1] \to \mathbb{R}\) by
$$\begin{aligned} T_{F,w}(u_{1},u_{2},u_{3})=u_{1}- \biggl( \int _{0}^{1} h(t)w(t)\,dt \biggr)u _{3} - \frac{1-\lambda }{\lambda } \biggl( \int _{0}^{1} h(1-t)w(t)\,dt \biggr)u _{2}. \end{aligned}$$(9)For \(L_{w}=(1-w)\varepsilon \), it will be seen that \(F\in \mathcal{F}\) and
$$\begin{aligned} &F\bigl(f\bigl(tx+(1-t)y\bigr),f(x),f(y),t\bigr)\\ &\quad =f\bigl(tx+(1-t)y \bigr)-h(t)f(x)-\frac{1-\lambda }{ \lambda }h(1-t)f(y) -\varepsilon \leq 0, \end{aligned}$$that is f is an F-convex function.
Recently Samet [17] established some integral inequalities of Hermite–Hadamard type via F-convex functions.
Theorem 1
([17, Theorem 3.1])
Let \(f: [a,b]\to \mathbb{R}\), \((a,b)\in \mathbb{R}^{2}\), \(a< b\), be an F-convex function, for some \(F\in \mathcal{F}\). Suppose that \(F\in L^{1}(a,b)\). Then
Theorem 2
([17, Theorem 3.4])
Let \(f: I^{o}\subseteq \mathbb{R}\to \mathbb{R}\) be a differentiable mapping on \(I^{o}\), \((a,b)\in I^{o}\times I^{o}\), \(a< b\). Suppose that
-
(i)
\(|f'|\) is F-convex on \([a,b]\), for some \(F\in \mathcal{F;}\)
-
(ii)
the function \(t\in (0,1)\to L_{w(t)}\) belongs to \(L^{1}(a,b)\), where \(w(t)=|1-2t|\).
Then
Theorem 3
([17, Theorem 3.5])
Let \(f: I^{o}\subseteq \mathbb{R}\to \mathbb{R}\) be a differentiable mapping on \(I^{o}\), \((a,b)\in I^{o}\times I^{o}\), \(a< b\) and let \(p>1\). Suppose that \(|f'|^{p/(p-1)}\) is F-convex on \([a,b]\), for some \(F\in \mathcal{F}\) and \(F\in L^{p/(p-1)}(a,b)\). Then
where
As consequences of the above theorems, the author obtained some integral inequalities for ε-convexity, α-convexity, and h-convexity.
Theorem 4
([17, Corollary 4.3])
Let \(f: I^{o}\subseteq \mathbb{R}\to \mathbb{R}\) be a differentiable mapping on \(I^{o}\), \((a,b)\in I^{o}\times I^{o}\), \(a< b\). Suppose that the function \(|f'|\) is ε-convex on \([a,b]\), \(\varepsilon \geq 0\). Then
Theorem 5
([17, Corollary 4.9])
Let \(f: I^{o}\subseteq \mathbb{R}\to \mathbb{R}\) be a differentiable mapping on \(I^{o}\), \((a,b)\in I^{o}\times I^{o}\), \(a< b\). Suppose that the function \(|f'|\) is α-convex on \([a,b]\), \(\alpha \in (0,1]\). Then
Theorem 6
([17, Corollary 4.14])
Let \(f: I^{o}\subseteq \mathbb{R}\to \mathbb{R}\) be a differentiable mapping on \(I^{o}\), \((a,b)\in I^{o}\times I^{o}\), \(a< b\). Suppose that the function \(|f'|\) is h-convex on \([a,b]\). Then
For more recent results on integral inequalities of Hermite–Hadamard type concerning the F-convex functions, we refer the interested reader to [19] and the references therein.
In the sequel, we recall the concepts of the left-sided and right-sided Riemann- Liouville fractional integrals of the order \(\alpha >0\).
Definition 2.2
([20])
Suppose that \(f\in L([a,b])\). The left and right Riemann–Liouville fractional integrals denoted by \(J_{a^{+}}^{\alpha }f\) and \(J_{b^{-}}^{\alpha }f\) of order \(\alpha >0\) are defined by
and
respectively, where \(\varGamma (\alpha )\) is the gamma function defined by \(\varGamma (\alpha )=\int _{0}^{\infty }e^{-t}t^{\alpha -1}\,dt\) and \(J_{b^{-}}^{0}f(x)=J_{b^{-}}^{0}f(x)=f(x)\).
In [21], authors established the following Hermite–Hadamard type inequalities for F-convex functions involving a Riemann–Liouville fractional:
Theorem 7
Let \(I\subseteq \mathbb{R}\) be an interval, \(f: I^{o}\subseteq \mathbb{R}\to \mathbb{R}\) be a differentiable mapping on \(I^{o}\), \(a,b \in I^{o}\), \(a< b\). If f is F-convex on \([a,b]\), for some \(F\in \mathcal{F}\), then we have
where \(w(t)=\alpha t^{\alpha -1}\).
Theorem 8
Let \(I\subseteq \mathbb{R}\) be an interval, \(f: I^{o}\subseteq \mathbb{R}\to \mathbb{R}\) be a differentiable mapping on \(I^{o}\), \(a,b \in I^{o}\), \(a< b\). If f is F-convex on \([a,b]\), for some \(F\in \mathcal{F}\) and the function \(t\in (0,1)\to L_{w(t)}\) belongs to \(L_{1}(a,b)\), where \(w(t)= \vert (1-t)^{\alpha }-t^{\alpha } \vert \). Then we have the inequality
The following definitions will be useful for this study [20].
Definition 2.3
The Euler beta function is defined as follows:
The incomplete beta function is defined by
Note that, for \(x=1\), the incomplete beta function reduces to the Euler beta function.
Also, the following three lemmas are important to obtain our main results.
Lemma 1
([22, Lemma 4])
Let \(f: [a,b]\to \mathbb{R}\) be a once differentiable mappings on \((a,b)\) with \(a< b\), \(\eta (b,a)>0\). If \(f'\in L [a,a+e^{i \varphi }\eta (b,a) ]\), then the following equality for the fractional integral holds:
Lemma 2
([16, Lemma 5])
Let \(f: [a,b]\to \mathbb{R}\) be a once differentiable mappings on \((a,b)\) with \(a< b\), \(\eta (b,a)>0\). If \(f''\in L [a,a+e^{i \varphi }\eta (b,a) ]\), then the following equality for the fractional integral holds:
Lemma 3
([22])
For \(t\in [0,1]\), we have
In this study, using the \(\lambda _{\varphi }\)-preinvexity of the function, we establish new inequalities of Hermite–Hadamard type for differentiable function and some trapezoid type inequalities for function whose second derivatives absolutely values are F-convex.
3 Hermite–Hadamard type inequalities for differentiable functions
In this section, we establish some inequalities of Hermite–Hadamard type for F-convex functions in fractional integral forms.
Theorem 9
Let \(I\subseteq \mathbb{R}\) be an open invex set with respect to bifunction \(\eta : I\times I\to \mathbb{R}\), where \(\eta (b,a)>0\). Let \(f: [0,b]\to \mathbb{R}\) be a differentiable mapping. Suppose that \(|f'|\) is measurable, decreasing, \(\lambda _{\varphi }\)-preinvex function on I, and F-convex on \([a,b]\), for some \(F\in \mathcal{F}\) and the function \(t\in (0,1)\to L_{w(t)}\) belongs to \(L^{1}(0,1)\), where \(w(t)= \vert (1-t)^{\alpha }-t^{\alpha } \vert \). Then
Proof
Since \(|f'|\) is F-convex, we have
Multiplying this inequality by \(w(t)= \vert (1-t)^{\alpha }-t^{\alpha } \vert \) and using axiom (A3), we have
Integrating over \([0,1]\) and using axiom (A2), we get
But from Lemma 1 we have
Because \(T_{F,w}\) is nondecreasing with respect to the first variable so that
This proves (10). □
Remark 2
If we choose \(\eta (b,a)=b-a\) and \(\varphi =0\) in Theorem 9, we get
Corollary 1
Under the assumptions of Theorem 9, if \(|f'|\) is ε-convex, then we have
Proof
Using (5) with \(w(t)= \vert (1-t)^{\alpha }-t^{\alpha } \vert \), we find
From (4) with \(w(t)= \vert (1-t)^{\alpha }-t^{\alpha } \vert \), we have
for \(u_{1}, u_{2}, u_{3}\in \mathbb{R}\). Hence, by Theorem 9, we have
This completes the proof. □
Remark 3
In Corollary 1, if we choose
-
(a)
\(\eta (b,a)=b-a\) and \(\varphi =0\), we get
$$\begin{aligned} & \biggl\vert \frac{f(a)+f(b)}{2}-\frac{\varGamma (\alpha +1))}{2(b-a)^{\alpha }} \bigl[J_{a^{+}}^{\alpha }f(b)+J_{b^{-}}^{\alpha }f(a) \bigr] \biggr\vert \\ &\quad \leq \frac{b-a}{2(\alpha +1)} \biggl(1-\frac{1}{2^{\alpha }} \biggr) \bigl( \bigl\vert f'(a) \bigr\vert + \bigl\vert f'(b) \bigr\vert +2\varepsilon \bigr). \end{aligned}$$ -
(b)
\(\eta (b,a)=b-a\), \(\varphi =0\), and \(\varepsilon =0\), we get
$$\begin{aligned} & \biggl\vert \frac{f(a)+f(b)}{2}-\frac{\varGamma (\alpha +1))}{2(b-a)^{\alpha }} \bigl[J_{a^{+}}^{\alpha }f(b)+J_{b^{-}}^{\alpha }f(a) \bigr] \biggr\vert \\ &\quad \leq \frac{b-a}{2(\alpha +1)} \biggl(1-\frac{1}{2^{\alpha }} \biggr) \bigl( \bigl\vert f'(a) \bigr\vert + \bigl\vert f'(b) \bigr\vert \bigr) \end{aligned}$$which is given by [18].
Corollary 2
Under the assumptions of Theorem 9, if \(|f'|\) is \(\lambda _{\varphi }\)-preinvex, then we have
Proof
Using (7) with \(w(t)= \vert (1-t)^{\alpha }-t^{\alpha } \vert \), we have
for \(u_{1}, u_{2}, u_{3}\in \mathbb{R}\). Hence, by Theorem 9, we have
This leads to
Thus, the proof is done. □
Remark 4
In Corollary 2, if we choose
-
(a)
\(\eta (b,a)=b-a\) and \(\varphi =0\), we get
$$\begin{aligned} & \biggl\vert \frac{f(a)+f(b)}{2}-\frac{\varGamma (\alpha +1))}{2(b-a)^{\alpha }} \bigl[J_{a^{+}}^{\alpha }f(b)+J_{b^{-}}^{\alpha }f(a) \bigr] \biggr\vert \\ &\quad \leq \frac{b-a}{4} \biggl[B_{\frac{1}{2}} \biggl(\frac{1}{2}, \alpha + \frac{1}{2} \biggr) -B_{\frac{1}{2}} \biggl(\alpha + \frac{1}{2}, \frac{1}{2} \biggr) \biggr] \biggl( \bigl\vert f'(a) \bigr\vert +\frac{1- \lambda }{\lambda } \bigl\vert f'(b) \bigr\vert \biggr). \end{aligned}$$ -
(b)
\(\eta (b,a)=b-a\), \(\varphi =0\), and \(\lambda =\frac{1}{2}\), we get
$$\begin{aligned} & \biggl\vert \frac{f(a)+f(b)}{2}-\frac{\varGamma (\alpha +1))}{2(b-a)^{\alpha }} \bigl[J_{a^{+}}^{\alpha }f(b)+J_{b^{-}}^{\alpha }f(a) \bigr] \biggr\vert \\ &\quad \leq \frac{b-a}{4} \biggl[B_{\frac{1}{2}} \biggl( \frac{1}{2},\alpha + \frac{1}{2} \biggr) -B_{\frac{1}{2}} \biggl( \alpha +\frac{1}{2}, \frac{1}{2} \biggr) \biggr] \bigl( \bigl\vert f'(a) \bigr\vert + \bigl\vert f'(b) \bigr\vert \bigr). \end{aligned}$$
Corollary 3
Under the assumptions of Theorem 9, if \(|f'|\) is h-convex, then we have
Proof
Using (9) with \(w(t)= \vert (1-t)^{\alpha }-t^{\alpha } \vert \), we have
for \(u_{1}, u_{2}, u_{3}\in \mathbb{R}\). So, by Theorem 9, we have
This leads to
Thus, the proof is done. □
Theorem 10
Let \(I\subseteq \mathbb{R}\) be an open invex set with respect to bifunction \(\eta : I\times I\to \mathbb{R}\), where \(\eta (b,a)>0\). Let \(f: [0,b]\to \mathbb{R}\) be a differentiable mapping. Suppose that \(|f'|^{{\frac{p}{p-1}}}\) is measurable, decreasing, \(\lambda _{\varphi }\)-preinvex function on I, and F-convex on \([a,b]\), for some \(F\in \mathcal{F}\) and \(|f'|\in L^{{\frac{p}{p-1}}}(a,b)\). Then
where
Proof
Since \(|f'|^{\frac{p}{p-1}}\) is F-convex, we have
With \(w(t)=1\) in (A2), we have
Using Lemma 1 and the Hölder inequality, we get
or, equivalently,
Because \(T_{F,1}\) is nondecreasing with respect to the first variable, we get
Thus, the proof is completed. □
Remark 5
If we choose \(\eta (b,a)=b-a\) and \(\varphi =0\) in Theorem 10, we get
Corollary 4
Under the assumptions of Theorem 10, if \(|f'|^{\frac{p}{p-1}}\) is ε-convex, we have
Proof
Using (5) with \(w(t)=1\), we have
From (4) with \(w(t)=1\), we have
for \(u_{1}, u_{2}, u_{3}\in \mathbb{R}\). Hence, by Theorem 10, we have
This leads to
or, equivalently,
This completes the proof. □
Remark 6
In Corollary 4, if we choose
-
(a)
\(\eta (b,a)=b-a\) and \(\varphi =0\), we get
$$\begin{aligned} & \biggl\vert \frac{f(a)+f(b)}{2}-\frac{\varGamma (\alpha +1))}{2(b-a)^{\alpha }} \bigl[J_{a^{+}}^{\alpha }f(b)+J_{b^{-}}^{\alpha }f(a) \bigr] \biggr\vert \\ &\quad \leq \frac{b-a}{2} \biggl(\frac{2-2^{1-\alpha p}}{\alpha p+1} \biggr) ^{\frac{1}{p}} \biggl(\frac{ \vert f'(a) \vert ^{\frac{p}{p-1}}+ \vert f'(b) \vert ^{\frac{p}{p-1}}}{2}+\varepsilon \biggr)^{\frac{p-1}{p}}. \end{aligned}$$ -
(b)
\(\eta (b,a)=b-a\), \(\varphi =0\), and \(\varepsilon =0\), we get
$$\begin{aligned} & \biggl\vert \frac{f(a)+f(b)}{2}-\frac{\varGamma (\alpha +1))}{2(b-a)^{\alpha }} \bigl[J_{a^{+}}^{\alpha }f(b)+J_{b^{-}}^{\alpha }f(a) \bigr] \biggr\vert \\ &\quad \leq \frac{b-a}{2} \biggl(\frac{2-2^{1-\alpha p}}{\alpha p+1} \biggr) ^{\frac{1}{p}} \biggl(\frac{ \vert f'(a) \vert ^{\frac{p}{p-1}}+ \vert f'(b) \vert ^{\frac{p}{p-1}}}{2} \biggr)^{\frac{p-1}{p}}. \end{aligned}$$
Corollary 5
Under the assumptions of Theorem 10. If \(|f'|^{\frac{p-1}{p}}\) is \(\lambda _{\varphi }\)-preinvex, we have
Proof
Using (7) with \(w(t)=1\), we have
for \(u_{1}, u_{2}, u_{3}\in \mathbb{R}\). So, by Theorem 10, we have
This leads to
Thus, the proof is done. □
Remark 7
In Corollary 5, if we choose
-
(a)
\(\eta (b,a)=b-a\) and \(\varphi =0\), we get
$$\begin{aligned} & \biggl\vert \frac{f(a)+f(b)}{2}-\frac{\varGamma (\alpha +1))}{2(b-a)^{\alpha }} \bigl[J_{a^{+}}^{\alpha }f(b)+J_{b^{-}}^{\alpha }f(a) \bigr] \biggr\vert \\&\quad\leq \frac{b-a}{2} \biggl(\frac{2-2^{1-\alpha p}}{\alpha p+1} \biggr) ^{\frac{1}{p}} \biggl[\frac{\pi }{4} \biggl( \bigl\vert f'(a) \bigr\vert ^{ \frac{p-1}{p}} +\frac{1-\lambda }{\lambda } \bigl\vert f'(b) \bigr\vert ^{ \frac{p-1}{p}} \biggr) \biggr]^{\frac{p-1}{p}}. \end{aligned}$$ -
(b)
\(\eta (b,a)=b-a\), \(\varphi =0\), and \(\lambda =\frac{1}{2}\), we get
$$\begin{aligned} & \biggl\vert \frac{f(a)+f(b)}{2}-\frac{\varGamma (\alpha +1))}{2(b-a)^{\alpha }} \bigl[J_{a^{+}}^{\alpha }f(b)+J_{b^{-}}^{\alpha }f(a) \bigr] \biggr\vert \\&\quad\leq \frac{b-a}{2} \biggl(\frac{2-2^{1-\alpha p}}{\alpha p+1} \biggr) ^{\frac{1}{p}} \biggl[\frac{\pi }{4} \bigl( \bigl\vert f'(a) \bigr\vert ^{ \frac{p-1}{p}} + \bigl\vert f'(b) \bigr\vert ^{\frac{p-1}{p}} \bigr) \biggr] ^{\frac{p-1}{p}}. \end{aligned}$$
Corollary 6
Under the assumptions of Theorem 10. If \(|f'|^{\frac{p}{p-1}}\) is h-convex, we have
Proof
From (9) with \(w(t)=1\), we have
for \(u_{1}, u_{2}, u_{3}\in \mathbb{R}\). So, by Theorem 9, we have
that is,
This completes the proof. □
Theorem 11
Let \(I\subseteq \mathbb{R}\) be an open invex set with respect to bifunction \(\eta : I\times I\to \mathbb{R}\), where \(\eta (b,a)>0\). Let \(f: [0,b]\to \mathbb{R}\) be a differentiable mapping. Suppose that \(|f'|^{{\frac{p}{p-1}}}\) is measurable, decreasing, \(\lambda _{\varphi }\)-preinvex function on I, and F-convex on \([a,b]\), for some \(F\in \mathcal{F}\) and \(|f'|\in L^{{\frac{p}{p-1}}}(a,b)\). Then
where
for \(w(t)= \vert (1-t)^{\alpha }-t^{\alpha } \vert \).
Proof
Since \(|f'|^{\frac{p}{p-1}}\) is F-convex, we have
Using (A3) with \(w(t)= \vert (1-t)^{\alpha }-t^{\alpha } \vert \), we obtain
Integrating over \([0,1]\) and using axiom (A2), we obtain
Using Lemma 1 and the power mean inequality, we get
or, equivalently,
Because \(T_{F,w}\) is nondecreasing with respect to the first variable, we find
This completes the proof. □
Remark 8
If we choose \(\eta (b,a)=b-a\) and \(\varphi =0\) in Theorem 11, we get
Corollary 7
Under the assumptions of Theorem 11, if \(|f'|^{\frac{p}{p-1}}\) is ε-convex, we have
Proof
Using (5) with \(w(t)= \vert (1-t)^{\alpha }-t^{\alpha } \vert \), we get
From (4) with \(w(t)= \vert (1-t)^{\alpha }-t^{\alpha } \vert \), we get
for \(u_{1}, u_{2}, u_{3}\in \mathbb{R}\). Hence, by Theorem 10, we have
This implies that
This completes the proof. □
Remark 9
In Corollary 7, if we choose
-
(a)
\(\eta (b,a)=b-a\) and \(\varphi =0\), we get
$$\begin{aligned} & \biggl\vert \frac{f(a)+f(b)}{2}-\frac{\varGamma (\alpha +1))}{2(b-a)^{\alpha }} \bigl[J_{a^{+}}^{\alpha }f(b)+J_{b^{-}}^{\alpha }f(a) \bigr] \biggr\vert \\ &\quad \leq \frac{b-a}{2} \biggl(\frac{2-2^{1-\alpha }}{\alpha +1} \biggr) ^{\frac{1}{p}} \biggl[\frac{2^{\alpha }-1}{2^{\alpha }(\alpha +1)} \bigl( \bigl\vert f'(a) \bigr\vert ^{\frac{p}{p-1}} + \bigl\vert f'(b) \bigr\vert ^{ \frac{p}{p-1}} \bigr)+2\varepsilon \biggr]^{\frac{p-1}{p}}. \end{aligned}$$ -
(b)
\(\eta (b,a)=b-a\), \(\varphi =0\), and \(\varepsilon =0\), we get
$$\begin{aligned} & \biggl\vert \frac{f(a)+f(b)}{2}-\frac{\varGamma (\alpha +1))}{2(b-a)^{\alpha }} \bigl[J_{a^{+}}^{\alpha }f(b)+J_{b^{-}}^{\alpha }f(a) \bigr] \biggr\vert \\ &\quad \leq \frac{b-a}{2} \biggl(\frac{2-2^{1-\alpha }}{\alpha +1} \biggr) ^{\frac{1}{p}} \biggl[\frac{2^{\alpha }-1}{2^{\alpha }(\alpha +1)} \bigl( \bigl\vert f'(a) \bigr\vert ^{\frac{p}{p-1}} + \bigl\vert f'(b) \bigr\vert ^{ \frac{p}{p-1}} \bigr) \biggr]^{\frac{p-1}{p}}. \end{aligned}$$
Corollary 8
Under the assumptions of Theorem 11. If \(|f'|^{\frac{p-1}{p}}\) is \(\lambda _{\varphi }\)-preinvex, we have
Proof
Using (7) with \(w(t)= \vert (1-t)^{\alpha }-t^{\alpha } \vert \), we have
for \(u_{1}, u_{2}, u_{3}\in \mathbb{R}\). Now, by Theorem 11, we have
This leads to
Thus, the proof is done. □
Remark 10
In Corollary 8, if we choose
-
(a)
\(\eta (b,a)=b-a\) and \(\varphi =0\), we get
$$\begin{aligned} & \biggl\vert \frac{f(a)+f(b)}{2}-\frac{\varGamma (\alpha +1))}{2(b-a)^{\alpha }} \bigl[J_{a^{+}}^{\alpha }f(b)+J_{b^{-}}^{\alpha }f(a) \bigr] \biggr\vert \\ &\quad \leq \frac{b-a}{2} \biggl(\frac{2-2^{1-\alpha }}{\alpha +1} \biggr) ^{\frac{1}{p}}\\&\qquad{}\times \biggl[\frac{B_{\frac{1}{2}} (\frac{1}{2},\alpha +\frac{1}{2} ) -B_{\frac{1}{2}} (\alpha +\frac{1}{2}, \frac{1}{2} )}{2} \biggl( \bigl\vert f'(a) \bigr\vert ^{\frac{p}{p-1}}+\frac{1- \lambda }{\lambda } \bigl\vert f'(b) \bigr\vert ^{\frac{p}{p-1}} \biggr) \biggr] ^{\frac{p-1}{p}}. \end{aligned}$$ -
(b)
\(\eta (b,a)=b-a\), \(\varphi =0\), and \(\lambda =\frac{1}{2}\), we get
$$\begin{aligned} & \biggl\vert \frac{f(a)+f(b)}{2}-\frac{\varGamma (\alpha +1))}{2(b-a)^{\alpha }} \bigl[J_{a^{+}}^{\alpha }f(b)+J_{b^{-}}^{\alpha }f(a) \bigr] \biggr\vert \\ &\quad \leq \frac{b-a}{2} \biggl(\frac{2-2^{1-\alpha }}{\alpha +1} \biggr) ^{\frac{1}{p}}\\&\qquad{}\times \biggl[\frac{B_{\frac{1}{2}} (\frac{1}{2},\alpha +\frac{1}{2} ) -B_{\frac{1}{2}} (\alpha +\frac{1}{2}, \frac{1}{2} )}{2} \bigl( \bigl\vert f'(a) \bigr\vert ^{\frac{p}{p-1}}+ \bigl\vert f'(b) \bigr\vert ^{\frac{p}{p-1}} \bigr) \biggr]^{\frac{p-1}{p}}. \end{aligned}$$
Corollary 9
Under the assumptions of Theorem 11. If \(|f'|^{\frac{p}{p-1}}\) is h-convex, we have
Proof
From (9) with \(w(t)= \vert (1-t)^{\alpha }-t^{\alpha } \vert \), we have
for \(u_{1}, u_{2}, u_{3}\in \mathbb{R}\). So, by Theorem 11, we have
that is,
This completes the proof. □
4 Trapezoid type inequalities for twice differentiable functions
In this section, we establish some trapezoid type inequalities for functions whose second derivatives absolutely values are
Theorem 12
Let \(f: [0,b]\to \mathbb{R}\) be a differentiable mapping and \(|f''|\) is measurable, decreasing, \(\lambda _{\varphi }\)-preinvex function on \([0,b]\) for \(0\leq a< b\), \(\eta (b,a)>0\) and \(\alpha >0\). Suppose that F-convex on \([0,b]\), for some \(F\in \mathcal{F}\) and the function \(t\in (0,1)\to L_{w(t)}\) belongs to \(L^{1}(0,1)\), where \(w(t)=1-(1-t)^{\alpha +1}-t^{\alpha +1}\). Then
Proof
Since \(|f''|\) is F-convex, we can see that
Multiplying this inequality by \(w(t)=1-(1-t)^{\alpha +1}-t^{\alpha +1}\) and using axiom (A3), we have
Integrating over \([0,1]\) and using axiom (A2), we get
Using Lemma 2, we have
Because \(T_{F,w}\) is nondecreasing with respect to the first variable so that
This completes the proof. □
Remark 11
By taking \(\eta (b,a)=b-a\) and \(\varphi =0\) in Theorem 12, we obtain
Corollary 10
Under the assumptions of Theorem 12, if \(|f''|\) is ε-convex, then
Proof
Using (5) with \(w(t)=1-(1-t)^{\alpha +1}-t^{\alpha +1}\), we find
With \(w(t)=1-(1-t)^{\alpha +1}-t^{\alpha +1}\), Eq. (4) gives
for \(u_{1}, u_{2}, u_{3}\in \mathbb{R}\). Hence, by Theorem 12, we have
This completes the proof. □
Remark 12
In Corollary 10, if we take
-
(a)
\(\eta (b,a)=b-a\) and \(\varphi =0\), we get
$$\begin{aligned} & \biggl\vert \frac{f(a)+f(b)}{2}-\frac{\varGamma (\alpha +1))}{2(b-a)^{\alpha }} \bigl[J_{a^{+}}^{\alpha }f(b)+J_{b^{-}}^{\alpha }f(a) \bigr] \biggr\vert \\ &\quad \leq \frac{\alpha (b-a)^{2}}{4(\alpha +1)(\alpha +2)} \bigl( \bigl\vert f''(a) \bigr\vert + \bigl\vert f''(b) \bigr\vert +2\varepsilon \bigr). \end{aligned}$$ -
(b)
\(\eta (b,a)=b-a\), \(\varphi =0\), and \(\varepsilon =0\), we get
$$\begin{aligned} & \biggl\vert \frac{f(a)+f(b)}{2}-\frac{\varGamma (\alpha +1))}{2(b-a)^{\alpha }} \bigl[J_{a^{+}}^{\alpha }f(b)+J_{b^{-}}^{\alpha }f(a) \bigr] \biggr\vert \leq \frac{\alpha (b-a)^{2}}{4(\alpha +1)(\alpha +2)} \bigl( \bigl\vert f''(a) \bigr\vert + \bigl\vert f''(b) \bigr\vert \bigr). \end{aligned}$$
Corollary 11
Under the assumptions of Theorem 12, if \(|f''|\) is \(\lambda _{\varphi }\)-preinvex, then
Proof
Using (7) with \(w(t)=1-(1-t)^{\alpha +1}-t^{\alpha +1}\), we have
for \(u_{1}, u_{2}, u_{3}\in \mathbb{R}\). Hence, by Theorem 12, we get
This leads to
Thus, the proof is completed. □
Remark 13
In Corollary 11, if we choose
-
(a)
\(\eta (b,a)=b-a\) and \(\varphi =0\), we get
$$\begin{aligned} & \biggl\vert \frac{f(a)+f(b)}{2}-\frac{\varGamma (\alpha +1))}{2(b-a)^{\alpha }} \bigl[J_{a^{+}}^{\alpha }f(b)+J_{b^{-}}^{\alpha }f(a) \bigr] \biggr\vert \\ &\leq \frac{(b-a)^{2}}{2(\alpha +1)} \biggl[\frac{\pi }{2}-\beta \biggl( \frac{1}{2},\alpha +\frac{5}{2} \biggr) -\beta \biggl( \frac{3}{2}, \alpha +\frac{3}{2} \biggr) \biggr] \biggl( \bigl\vert f''(a) \bigr\vert +\frac{1- \lambda }{\lambda } \bigl\vert f''(b) \bigr\vert \biggr). \end{aligned}$$ -
(b)
\(\eta (b,a)=b-a\), \(\varphi =0\), and \(\lambda =\frac{1}{2}\), we get
$$\begin{aligned} & \biggl\vert \frac{f(a)+f(b)}{2}-\frac{\varGamma (\alpha +1))}{2(b-a)^{\alpha }} \bigl[J_{a^{+}}^{\alpha }f(b)+J_{b^{-}}^{\alpha }f(a) \bigr] \biggr\vert \\ &\quad \leq \frac{(b-a)^{2}}{2(\alpha +1)} \biggl[\frac{\pi }{2}-\beta \biggl(\frac{1}{2}, \alpha +\frac{5}{2} \biggr) -\beta \biggl( \frac{3}{2},\alpha + \frac{3}{2} \biggr) \biggr] \bigl( \bigl\vert f''(a) \bigr\vert + \bigl\vert f''(b) \bigr\vert \bigr). \end{aligned}$$
Corollary 12
Under the assumptions of Theorem 12, if \(|f''|\) is h-convex, then we have
Proof
Using (9) with \(w(t)=1-(1-t)^{\alpha +1}-t^{\alpha +1}\), we obtain
for \(u_{1}, u_{2}, u_{3}\in \mathbb{R}\), so Theorem 12 implies that
which can be written as
Thus, the proof is done. □
Theorem 13
Let \(f: [0,b]\to \mathbb{R}\) be a differentiable mapping and \(|f''|^{{\frac{p}{p-1}}}\) is measurable, decreasing, \(\lambda _{\varphi }\)-preinvex function on \([0,b]\) for \(\eta (b,a)>0\) and \(0\leq a< b\). Suppose that \(|f''|^{{\frac{p}{p-1}}}\) is F-convex on \([a,b]\), for some \(F\in \mathcal{F}\) and \(|f''|\in L^{{\frac{p}{p-1}}}(a,b)\), \(p>1\). Then we have
where
Proof
Since \(|f''|^{\frac{p}{p-1}}\) is F-convex, we have
Using (A2) with \(w(t)=1\), we have
Using Lemma 2, Lemma 3 and the Hölder inequality, we get
or, equivalently,
Because \(T_{F,1}\) is nondecreasing with respect to the first variable, we get
Thus, the proof is completed. □
Remark 14
If we choose \(\eta (b,a)=b-a\) and \(\varphi =0\) in Theorem 13, we get
Corollary 13
Under the assumptions of Theorem 13, if \(|f''|^{ \frac{p}{p-1}}\) is ε-convex, then
Proof
Using (12), (13), by Theorem 13, we have
This leads to
This completes the proof. □
Remark 15
In Corollary 13, if we choose
-
(a)
\(\eta (b,a)=b-a\) and \(\varphi =0\), we get
$$\begin{aligned} & \biggl\vert \frac{f(a)+f(b)}{2}-\frac{\varGamma (\alpha +1))}{2(b-a)^{\alpha }} \bigl[J_{a^{+}}^{\alpha }f(b)+J_{b^{-}}^{\alpha }f(a) \bigr] \biggr\vert \\ &\quad \leq \biggl(\frac{ (2^{\alpha }-1 )(b-a)^{2}}{2^{\alpha +1}( \alpha +1)} \biggr) \biggl( \frac{ \vert f''(a) \vert ^{\frac{p}{p-1}} + \vert f''(b) \vert ^{\frac{p}{p-1}}}{2}+\varepsilon \biggr)^{ \frac{p-1}{p}}. \end{aligned}$$ -
(b)
\(\eta (b,a)=b-a\), \(\varphi =0\), and \(\varepsilon =0\), we get
$$\begin{aligned} & \biggl\vert \frac{f(a)+f(b)}{2}-\frac{\varGamma (\alpha +1))}{2(b-a)^{\alpha }} \bigl[J_{a^{+}}^{\alpha }f(b)+J_{b^{-}}^{\alpha }f(a) \bigr] \biggr\vert \\ &\quad \leq \biggl(\frac{ (2^{\alpha }-1 )(b-a)^{2}}{2^{\alpha +1}( \alpha +1)} \biggr) \biggl( \frac{ \vert f''(a) \vert ^{\frac{p}{p-1}}+ \vert f''(b) \vert ^{\frac{p}{p-1}}}{2} \biggr)^{\frac{p-1}{p}}. \end{aligned}$$
Corollary 14
Under the assumptions of Theorem 13. If \(|f''|^{ \frac{p-1}{p}}\) is \(\lambda _{\varphi }\)-preinvex, we have
Proof
Using (7), (14), by Theorem 13, we have
that is,
This proves Corollary 14. □
Remark 16
In Corollary 14, if we choose
-
(a)
\(\eta (b,a)=b-a\) and \(\varphi =0\), we get
$$\begin{aligned} & \biggl\vert \frac{f(a)+f(b)}{2}-\frac{\varGamma (\alpha +1))}{2(b-a)^{\alpha }} \bigl[J_{a^{+}}^{\alpha }f(b)+J_{b^{-}}^{\alpha }f(a) \bigr] \biggr\vert \\ &\quad \leq \biggl(\frac{ (2^{\alpha }-1 )(b-a)^{2}}{2^{\alpha +1}( \alpha +1)} \biggr) \biggl[ \frac{\pi }{4} \biggl( \bigl\vert f''(a) \bigr\vert ^{\frac{p-1}{p}} +\frac{1-\lambda }{\lambda } \bigl\vert f''(b) \bigr\vert ^{\frac{p-1}{p}} \biggr) \biggr]^{\frac{p-1}{p}}. \end{aligned}$$ -
(b)
\(\eta (b,a)=b-a\), \(\varphi =0\), and \(\lambda =\frac{1}{2}\), we get
$$\begin{aligned} & \biggl\vert \frac{f(a)+f(b)}{2}-\frac{\varGamma (\alpha +1))}{2(b-a)^{\alpha }} \bigl[J_{a^{+}}^{\alpha }f(b)+J_{b^{-}}^{\alpha }f(a) \bigr] \biggr\vert \\ &\quad \leq \biggl(\frac{ (2^{\alpha }-1 )(b-a)^{2}}{2^{\alpha +1}( \alpha +1)} \biggr) \biggl[ \frac{\pi \vert f''(a) \vert ^{ \frac{p-1}{p}} +\pi \vert f''(b) \vert ^{\frac{p-1}{p}}}{4} \biggr] ^{\frac{p-1}{p}}. \end{aligned}$$
Corollary 15
Under the assumptions of Theorem 13. If \(|f''|^{ \frac{p}{p-1}}\) is h-convex, then
Proof
Using (9) and by Theorem 13, it can be proved easily. It is omitted. □
Theorem 14
Let \(f: [0,b]\to \mathbb{R}\) be a differentiable mapping and \(|f''|^{{\frac{p}{p-1}}}\) is measurable, decreasing, \(\lambda _{\varphi }\)-preinvex function on \([0,b]\) for \(\eta (b,a)>0\) and \(0\leq a< b\). Suppose that \(|f''|^{{\frac{p}{p-1}}}\) is F-convex on \([a,b]\), for some \(F\in \mathcal{F}\) and \(|f''|\in L^{{\frac{p}{p-1}}}(a,b)\), \(p>1\). Then we have
where
for \(w(t)=1-(1-t)^{\alpha +1}-t^{\alpha +1}\).
Proof
Since \(|f''|^{\frac{p}{p-1}}\) is F-convex, we have
Using (A3) with \(w(t)=1-(1-t)^{\alpha +1}-t^{\alpha +1}\), we obtain
Integrating over \([0,1]\) and using axiom (A2), we obtain
Using Lemma 2 and the power mean inequality, we get
or, equivalently,
Since \(T_{F,w}\) is nondecreasing with respect to the first variable, we have
This completes the proof. □
Remark 17
Taking \(\eta (b,a)=b-a\) and \(\varphi =0\) in Theorem 14, we get
Corollary 16
Under the assumptions of Theorem 14, if \(|f''|^{ \frac{p}{p-1}}\) is ε-convex, we have
Proof
Using (17), (18) and by Theorem 13, we obtain
This completes the proof. □
Remark 18
In Corollary 16, if we choose
-
(a)
\(\eta (b,a)=b-a\) and \(\varphi =0\), we get
$$\begin{aligned} & \biggl\vert \frac{f(a)+f(b)}{2}-\frac{\varGamma (\alpha +1))}{2(b-a)^{\alpha }} \bigl[J_{a^{+}}^{\alpha }f(b)+J_{b^{-}}^{\alpha }f(a) \bigr] \biggr\vert \\ &\quad \leq \biggl(\frac{(b-a)^{2}}{2(\alpha +1)} \biggr) \biggl(\frac{2^{ \alpha }-1}{2^{\alpha }} \biggr)^{\frac{1}{p}} \biggl[\frac{\alpha }{2( \alpha +2)} \bigl( \bigl\vert f''(a) \bigr\vert ^{\frac{p}{p-1}} + \bigl\vert f''(b) \bigr\vert ^{\frac{p}{p-1}} \bigr)+2\varepsilon \biggr]^{\frac{p-1}{p}}. \end{aligned}$$ -
(b)
\(\eta (b,a)=b-a\), \(\varphi =0\), and \(\varepsilon =0\), we get
$$\begin{aligned} & \biggl\vert \frac{f(a)+f(b)}{2}-\frac{\varGamma (\alpha +1))}{2(b-a)^{\alpha }} \bigl[J_{a^{+}}^{\alpha }f(b)+J_{b^{-}}^{\alpha }f(a) \bigr] \biggr\vert \\ &\quad \leq \biggl(\frac{(b-a)^{2}}{2(\alpha +1)} \biggr) \biggl( \frac{2^{ \alpha }-1}{2^{\alpha }} \biggr)^{\frac{1}{p}} \biggl[\frac{\alpha }{2( \alpha +2)} \bigl( \bigl\vert f''(a) \bigr\vert ^{\frac{p}{p-1}} + \bigl\vert f''(b) \bigr\vert ^{\frac{p}{p-1}} \bigr) \biggr]^{\frac{p-1}{p}}. \end{aligned}$$
Corollary 17
Under the assumptions of Theorem 14. If \(|f''|^{ \frac{p-1}{p}}\) is \(\lambda _{\varphi }\)-preinvex, then
Proof
Using (19), by Theorem 14, we have
This leads to
This ends the proof. □
Remark 19
In Corollary 17, if we choose
-
(a)
\(\eta (b,a)=b-a\) and \(\varphi =0\), we get
$$\begin{aligned} & \biggl\vert \frac{f(a)+f(b)}{2}-\frac{\varGamma (\alpha +1))}{2(b-a)^{\alpha }} \bigl[J_{a^{+}}^{\alpha }f(b)+J_{b^{-}}^{\alpha }f(a) \bigr] \biggr\vert \\ &\quad \leq \biggl(\frac{(b-a)^{2}}{2(\alpha +1)} \biggr) \biggl(\frac{2^{ \alpha }-1}{2^{\alpha }} \biggr)^{\frac{1}{p}} \biggl( \biggl[\frac{ \pi }{2}-\beta \biggl( \frac{1}{2},\alpha +\frac{5}{2} \biggr) -\beta \biggl( \frac{3}{2},\alpha +\frac{3}{2} \biggr) \biggr]\\ &\qquad {}\times \biggl( \bigl\vert f''(a) \bigr\vert ^{\frac{p}{p-1}}+\frac{1-\lambda }{\lambda } \bigl\vert f''(b) \bigr\vert ^{\frac{p}{p-1}} \biggr) \biggr)^{\frac{p-1}{p}}. \end{aligned}$$ -
(b)
\(\eta (b,a)=b-a\), \(\varphi =0\), and \(\lambda =\frac{1}{2}\), we get
$$\begin{aligned} & \biggl\vert \frac{f(a)+f(b)}{2}-\frac{\varGamma (\alpha +1))}{2(b-a)^{\alpha }} \bigl[J_{a^{+}}^{\alpha }f(b)+J_{b^{-}}^{\alpha }f(a) \bigr] \biggr\vert \\ &\quad \leq \biggl(\frac{(b-a)^{2}}{2(\alpha +1)} \biggr) \biggl(\frac{2^{ \alpha }-1}{2^{\alpha }} \biggr)^{\frac{1}{p}} \biggl( \biggl[\frac{ \pi }{2}-\beta \biggl( \frac{1}{2},\alpha +\frac{5}{2} \biggr) -\beta \biggl( \frac{3}{2},\alpha +\frac{3}{2} \biggr) \biggr]\\ &\qquad {}\times \bigl( \bigl\vert f''(a) \bigr\vert ^{\frac{p}{p-1}}+ \bigl\vert f''(b) \bigr\vert ^{\frac{p}{p-1}} \bigr) \biggr) ^{\frac{p-1}{p}}. \end{aligned}$$
Corollary 18
Under the assumptions of Theorem 14. If \(|f''|^{ \frac{p}{p-1}}\) is h-convex, we have
Proof
Using (9), by Theorem 14, it can be proved easily. It is omitted. □
5 Conclusion
In the present paper, using the notion of F and F-convex function (see [17]), we construct some new inequalities of Hermite–Hadamard type for differentiable function via Riemann–Liouville fractional integral. We also established some trapezoid type inequalities for a function of whose second derivatives absolutely values are F-convex. Moreover, we obtained some new inequalities of Hermite–Hadamard type for Riemann–Liouville fractional integrals and via classical integrals. The results presented in this paper would provide generalizations and extension of those given in earlier work.
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Mohammed, P.O., Sarikaya, M.Z. Hermite–Hadamard type inequalities for F-convex function involving fractional integrals. J Inequal Appl 2018, 359 (2018). https://doi.org/10.1186/s13660-018-1950-1
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DOI: https://doi.org/10.1186/s13660-018-1950-1