Weighted Ostrowski, trapezoid and midpoint type inequalities for RiemannLiouville fractional integrals

Hüseyin Budak; Ebru Pehlivan; Hüseyin Budak; Ebru Pehlivan

doi:10.3934/math.2020131

AIMS Mathematics

2020, Volume 5, Issue 3: 1960-1984. doi: 10.3934/math.2020131

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Research article

Weighted Ostrowski, trapezoid and midpoint type inequalities for RiemannLiouville fractional integrals

Hüseyin Budak ^,,
Ebru Pehlivan

Department of Mathematics, Faculty of Science and Arts, Düzce University, Düzce, Turkey

Received: 07 October 2019 Accepted: 13 February 2020 Published: 20 February 2020
MSC : 26B25, 26D10, 26D15

Our first aim is to establish two new identities for differentiable function involving Riemann-Liouville fractional integrals. Then, we obtain some new weighted versions of fractional trapezoid and Ostrowski type inequalities. Moreover, we give some weighted fractional midpoint type inequalities as special cases.

Keywords:

Citation: Hüseyin Budak, Ebru Pehlivan. Weighted Ostrowski, trapezoid and midpoint type inequalities for RiemannLiouville fractional integrals[J]. AIMS Mathematics, 2020, 5(3): 1960-1984. doi: 10.3934/math.2020131

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Abstract

1. Introduction

In recent years, the Hermite-Hadamard inequality, which is the first fundamental result for convex mappings with a natural geometrical interpretation and many applications, has drawn attention much interest in elementary mathematics.

The inequalities discovered by C. Hermite and J. Hadamard for convex functions are considerable significant in the literature (see, e.g., [,, p.137]). These inequalities state that if $f:I\rightarrow \mathbb{R}$ is a convex function on the interval $I$ of real numbers and $a, b\in I$ with $a < b$ , then

$\begin{equation} f\left( \frac{a+b}{2}\right) \leq \frac{1}{b-a}\int\limits_{a}^{b}f(x)dx\leq \frac{f\left( a\right) +f\left( b\right) }{2}. \end{equation}$

(1.1)

Both inequalities hold in the reversed direction if $f$ is concave.

Over the last twenty years, the numerous studies have focused on to obtain new bound for left hand side and right and side of the inequality (1.1). For some examples, please refer to (^{[2,4,7,9,10,23,33,34,36,37,44]})

The overall structure of the paper takes the form of four sections including introduction. The remainder of this work is organized as follows: we first give weighted version of (1.1) and definitions of Riemann-Liouville fractional integral operators. We also mention fractional Hermite-Hadamard type inequalities obtained in earlier works. In Section 2, we establish two important weighted equalities for differentiable functions involving fractional integrals. Using an identity given Section 2, we obtain some weighted fractional trapezoid type inequalities in Section 3. In Section 4, we first mention classical and fractional Ostrowski inequalities for bounded variation. Then, utilizing other identity given Section 2, we obtain some fractional Ostrowski type inequalities for convex functions. We also give some fractional midpoint type inequalities.

The weighted version of the inequalities (1.1), so-called Hermite-Hadamard-Fejér inequalities, was given by Fejer in ^[13] as follow:

Theorem 1.1. $f:[a, b]\rightarrow \mathbb{R}$ , be a convex function, then the inequality

$\begin{equation} f\left( \frac{a+b}{2}\right) \int\limits_{a}^{b}g(x)dx\leq \int\limits_{a}^{b}f(x)g(x)dx\leq \frac{f(a)+f(b)}{2}\int \limits_{a}^{b}g(x)dx \end{equation}$

(1.2)

holds, where $g:[a, b]\rightarrow \mathbb{R}$ is non-negative, integrable, and symmetric about $x = \frac{a+b}{2}$ (i.e., $g(x) = g(a+b-x)$ ) $.$

Tseng et al. give the following Lemma and by using this Lemma they obtain several weighed inequalities in ^[51].

Lemma 1. Let $f:\left[a, b\right] \rightarrow \mathbb{R}$ be a differentiable mapping on $(a, b)$ with $a < b$ and let $g:\left[a, b\right] \rightarrow \mathbb{R}$ . If $f^{\prime }, g\in L\left[a, b\right],$ then for all $x\in \left[a, b\right]$ we have the following equality for fractional integrals

$\begin{equation} f(a)\int\limits_{a}^{x}g(t)dt+f(b)\int\limits_{x}^{b}g(t)dt-\int \limits_{a}^{b}f(t)g(t)dt = \int\limits_{a}^{b}\left( \int\limits_{x}^{t}g(s)ds\right) f^{\prime }(t)dt. \end{equation}$

(1.3)

In the following we will give some necessary definitions and mathematical preliminaries of fractional calculus theory which are used further in this paper.

Definition 1. Let $f\in L_{1}[a, b].$ The Riemann-Liouville integrals $J_{a+}^{\alpha }f$ and $J_{b-}^{\alpha }f$ of order $\alpha > 0$ with $a\geq 0$ are defined by

$\begin{equation*} J_{a+}^{\alpha }f(x) = \frac{1}{\Gamma (\alpha )}\int_{a}^{x}\left( x-t\right) ^{\alpha -1}f(t)dt, \ \ x \gt a \end{equation*}$

and

$\begin{equation*} J_{b-}^{\alpha }f(x) = \frac{1}{\Gamma (\alpha )}\int_{x}^{b}\left( t-x\right) ^{\alpha -1}f(t)dt, \ \ x \lt b \end{equation*}$

respectively. Here, $\Gamma (\alpha)$ is the Gamma function and $J_{a+}^{0}f(x) = J_{b-}^{0}f(x) = f(x).$

For more information about fraction calculus please refer to ^{[14,21,27,35]}.

In ^[40], Sarikaya et al. first gave the following interesting integral inequalities of Hermite-Hadamard type involving Riemann-Liouville fractional integrals.

Theorem 2. Let $f:\left[a, b\right] \rightarrow \mathbb{R}$ be a positive function with $0\leq a < b$ and $f\in L_{1}\left[a, b\right].$ If $f$ is a convex function on $[a, b]$ , then the following inequalities for fractional integrals hold:

$\begin{equation} f\left( \frac{a+b}{2}\right) \leq \frac{\Gamma (\alpha +1)}{2\left( b-a\right) ^{\alpha }}\left[ J_{a+}^{\alpha }f(b)+J_{b-}^{\alpha }f(a)\right] \leq \frac{f\left( a\right) +f\left( b\right) }{2} \end{equation}$

(1.4)

with $\alpha > 0.$

On the other hand, Işcan gave following Lemma and using this Lemma he proved the following Fejer type inequalities for Riemann-Liouville fractional integrals in ^[18].

Lemma 2. If $g:\left[a, b\right] \rightarrow \mathbb{R}$ is integrable and symmetric to $\left(a+b\right) /2$ with $\ a < b$ , then

$\begin{equation*} J_{a+}^{\alpha }g(b) = J_{b-}^{\alpha }g(a) = \frac{1}{2}\left[ J_{a+}^{\alpha }g(b)+J_{b-}^{\alpha }g(a)\right] \end{equation*}$

with $\alpha > 0.$

Theorem 3. Let $f:\left[a, b\right] \rightarrow \mathbb{R}$ be convex function with $0\leq a < b$ and $f\in L_{1}\left[a, b\right]$ . If $g:\left[ a, b\right] \rightarrow \mathbb{R}$ is non-negative, integrable and symmetric to $(a+b)/2$ , then the following inequalities for fractional integrals hold

$\begin{eqnarray} f\left( \frac{a+b}{2}\right) \left[ J_{a+}^{\alpha }g(b)+J_{b-}^{\alpha }g(a) \right] &\leq &\left[ J_{a+}^{\alpha }\left( fg\right) (b)+J_{b-}^{\alpha }\left( fg\right) (a)\right] \\ && \\ &\leq &\frac{f\left( a\right) +f\left( b\right) }{2}\left[ J_{a+}^{\alpha }g(b)+J_{b-}^{\alpha }g(a)\right] \end{eqnarray}$

(1.5)

with $\alpha > 0$

Whereupon Sarikaya et al. obtained the Hermite-Hadamard inequality for Riemann-Lioville fractional integrals, many authors have studied to generalize this inequality and establish Hermite-Hadamard inequality other fractional integrals such as $k$ -fractional integral, Hadamard fractional integrals, Katugampola fractional integrals, Conformable fractional integrals, etc. For some of them, please see (^{[3,8,12,17,18,19,20,22,28,32,41,42,43,46,47,52,53,56]}).

2. Some weighted fractional equalities

In this section, we prove two identities which will be used frequently in Section 3 and Section 4 to obtain weighted fractional inequalities.

Lemma 3. Let $f:\left[a, b\right] \rightarrow \mathbb{R}$ be a differentiable mapping on $(a, b)$ with $a < b.$ If $f^{\prime }, g\in L\left[ a, b\right],$ then for all $x\in \left[a, b\right]$ and $\alpha > 0$ we have the following equality for fractional integrals

$\begin{eqnarray} &&f(a)J_{a+}^{\alpha }g\left( x\right) +f(b)J_{b-}^{\alpha }g\left( x\right) -\left[ J_{a+}^{\alpha }\left( fg\right) \left( x\right) +J_{b-}^{\alpha }\left( fg\right) \left( x\right) \right] \\ && \\ & = &\frac{1}{\Gamma \left( \alpha \right) }\left[ \int\limits_{a}^{x}\left( \int\limits_{x}^{t}\left( x-s\right) ^{\alpha -1}g(s)ds\right) f^{\prime }(t)dt+\int\limits_{x}^{b}\left( \int\limits_{x}^{t}\left( s-x\right) ^{\alpha -1}g(s)ds\right) f^{\prime }(t)dt\right] . \end{eqnarray}$

(2.1)

Proof. Integrating by parts, we have

$\begin{eqnarray} &&\int\limits_{a}^{x}\left( \int\limits_{x}^{t}\left( x-s\right) ^{\alpha -1}g(s)ds\right) f^{\prime }(t)dt \\ && \\ & = &\left. \left( \int\limits_{x}^{t}\left( x-s\right) ^{\alpha -1}g(s)ds\right) f(t)\right\vert _{a}^{x}-\int\limits_{a}^{x}\left( x-t\right) ^{\alpha -1}g(t)f(t)dt \\ && \\ & = &\left( \int\limits_{a}^{x}\left( x-s\right) ^{\alpha -1}g(s)ds\right) f(a)-\int\limits_{a}^{x}\left( x-t\right) ^{\alpha -1}g(t)f(t)dt \\ && \\ & = &\Gamma \left( \alpha \right) \left[ f(a)J_{a+}^{\alpha }g\left( x\right) -J_{a+}^{\alpha }\left( fg\right) \left( x\right) \right] . \end{eqnarray}$

(2.2)

Similarly, we get

$\begin{equation} \int\limits_{x}^{b}\left( \int\limits_{x}^{t}\left( s-x\right) ^{\alpha -1}g(s)ds\right) f^{\prime }(t)dt = \Gamma \left( \alpha \right) \left[ f(b)J_{b-}^{\alpha }g\left( x\right) -J_{b-}^{\alpha }\left( fg\right) \left( x\right) \right] . \end{equation}$

(2.3)

By the identities (2.2) and (2.3), we obtain the required result (2.1).

Remark 1. If we choose $\alpha = 1$ in Lemma 3, then the identity (2.1) reduces to identity (1.3) proved by Tseng et al. in ^[51].

Corollary 1. In Lemma 3, let $g$ be symmetric to $\frac{a+b}{2}$ and let $x = \frac{ a+b}{2}$ . Then (2.1) can be written as

$\begin{eqnarray*} &&\left[ J_{a+}^{\alpha }g\left( \frac{a+b}{2}\right) +J_{b-}^{\alpha }g\left( \frac{a+b}{2}\right) \right] \frac{f(a)+f(b)}{2} \\ && \\ &&-\left[ J_{a+}^{\alpha }\left( fg\right) \left( \frac{a+b}{2}\right) +J_{b-}^{\alpha }\left( fg\right) \left( \frac{a+b}{2}\right) \right] \\ && \\ & = &\frac{1}{\Gamma \left( \alpha \right) }\left[ \int\limits_{a}^{\frac{a+b}{ 2}}\left( \int\limits_{\frac{a+b}{2}}^{t}\left( \frac{a+b}{2}-s\right) ^{\alpha -1}g(s)ds\right) f^{\prime }(t)dt+\int\limits_{\frac{a+b}{2} }^{b}\left( \int\limits_{\frac{a+b}{2}}^{t}\left( s-\frac{a+b}{2}\right) ^{\alpha -1}g(s)ds\right) f^{\prime }(t)dt\right] . \end{eqnarray*}$

Proof. If we take $x = \frac{a+b}{2}$ in Lemma 3, since $g$ is symmetric to $\frac{a+b}{2},$ then we have

$\begin{eqnarray*} J_{a+}^{\alpha }g\left( \frac{a+b}{2}\right) & = &\frac{1}{\Gamma \left( \alpha \right) }\int\limits_{a}^{\frac{a+b}{2}}\left( \frac{a+b}{2}-s\right) ^{\alpha -1}g(s)ds \\ && \\ & = &\frac{1}{\Gamma \left( \alpha \right) }\int\limits_{\frac{a+b}{2} }^{b}\left( u-\frac{a+b}{2}\right) ^{\alpha -1}g(a+b-u)ds \\ && \\ & = &\frac{1}{\Gamma \left( \alpha \right) }\int\limits_{\frac{a+b}{2} }^{b}\left( u-\frac{a+b}{2}\right) ^{\alpha -1}g(u)ds \\ && \\ & = &J_{b-}^{\alpha }g\left( \frac{a+b}{2}\right) . \end{eqnarray*}$

This completes the proof.

Corollary 2. Under assumptions of Lemma 3, we have the following equality

$\begin{eqnarray*} &&f(a)J_{a+}^{\alpha }g\left( b\right) +f(b)J_{b-}^{\alpha }g\left( a\right) -\left[ J_{a+}^{\alpha }\left( fg\right) \left( b\right) +J_{b-}^{\alpha }\left( fg\right) \left( a\right) \right] \\ && \\ & = &\frac{1}{\Gamma \left( \alpha \right) }\left[ \int\limits_{a}^{b}\left( \int\limits_{a}^{t}\left( s-a\right) ^{\alpha -1}g(s)ds\right) f^{\prime }(t)dt+\int\limits_{a}^{b}\left( \int\limits_{b}^{t}\left( b-s\right) ^{\alpha -1}g(s)ds\right) f^{\prime }(t)dt\right] . \end{eqnarray*}$

Proof. If we write the equality (2.1) for $x = a$ and $x = b,$ then we have the identities

$\begin{equation} f(b)J_{b-}^{\alpha }g\left( a\right) -J_{b-}^{\alpha }\left( fg\right) \left( a\right) = \frac{1}{\Gamma \left( \alpha \right) }\int\limits_{a}^{b} \left( \int\limits_{a}^{t}\left( s-a\right) ^{\alpha -1}g(s)ds\right) f^{\prime }(t)dt \end{equation}$

(2.4)

and

$\begin{equation} f(a)J_{a+}^{\alpha }g\left( b\right) -J_{a+}^{\alpha }\left( fg\right) \left( b\right) = \frac{1}{\Gamma \left( \alpha \right) }\int\limits_{a}^{b} \left( \int\limits_{b}^{t}\left( b-s\right) ^{\alpha -1}g(s)ds\right) f^{\prime }(t)dt, \end{equation}$

(2.5)

respectively. By adding the equalities (2.4) and (2.5), we obtain desired result.

Corollary 3. In Lemma 3, let $g(t) = 1$ for all $t\in \left[a, b\right].$ Then we have the following identity

$\begin{eqnarray*} &&\left( x-a\right) ^{\alpha }f(a)+\left( b-x\right) ^{\alpha }f(b)-\Gamma \left( \alpha +1\right) \left[ J_{a+}^{\alpha }f\left( x\right) +J_{b-}^{\alpha }f\left( x\right) \right] \\ && \\ & = &-\left[ \int\limits_{a}^{x}\left( x-t\right) ^{\alpha }f^{\prime }(t)dt+\int\limits_{x}^{b}\left( t-x\right) ^{\alpha }f^{\prime }(t)dt\right] . \end{eqnarray*}$

Lemma 4. Let $f:\left[a, b\right] \rightarrow \mathbb{R}$ be a differentiable mapping on $(a, b)$ with $a < b.$ If $f^{\prime }, g\in L\left[ a, b\right],$ then for all $x\in \left[a, b\right]$ and $\alpha > 0$ we have the following equality for fractional integrals

$\begin{eqnarray} &&\left[ J_{a+}^{\alpha }g\left( x\right) +J_{b-}^{\alpha }g\left( x\right) \right] f(x)-\left[ J_{a+}^{\alpha }\left( fg\right) \left( x\right) +J_{b-}^{\alpha }\left( fg\right) \left( x\right) \right] \\ && \\ & = &\frac{1}{\Gamma \left( \alpha \right) }\left[ \int\limits_{a}^{x}\left( \int\limits_{a}^{t}\left( x-s\right) ^{\alpha -1}g(s)ds\right) f^{\prime }(t)dt+\int\limits_{x}^{b}\left( \int\limits_{b}^{t}\left( s-x\right) ^{\alpha -1}g(s)ds\right) f^{\prime }(t)dt\right] . \end{eqnarray}$

(2.6)

Proof. Integrating by parts, we have

$\begin{eqnarray} &&\int\limits_{a}^{x}\left( \int\limits_{a}^{t}\left( x-s\right) ^{\alpha -1}g(s)ds\right) f^{\prime }(t)dt \\ && \\ & = &\left. \left( \int\limits_{a}^{t}\left( x-s\right) ^{\alpha -1}g(s)ds\right) f(t)\right\vert _{a}^{x}-\int\limits_{a}^{x}\left( x-t\right) ^{\alpha -1}g(t)f(t)dt \\ && \\ & = &\left( \int\limits_{a}^{x}\left( x-s\right) ^{\alpha -1}g(s)ds\right) f(x)-\int\limits_{a}^{x}\left( x-t\right) ^{\alpha -1}g(t)f(t)dt \\ && \\ & = &\Gamma \left( \alpha \right) \left[ f(x)J_{a+}^{\alpha }g\left( x\right) -J_{a+}^{\alpha }\left( fg\right) \left( x\right) \right] . \end{eqnarray}$

(2.7)

Similarly, we get

$\begin{equation} \int\limits_{x}^{b}\left( \int\limits_{b}^{t}\left( s-x\right) ^{\alpha -1}g(s)ds\right) f^{\prime }(t)dt = \Gamma \left( \alpha \right) f(x)\left[ J_{b-}^{\alpha }g\left( x\right) -J_{b-}^{\alpha }\left( fg\right) \left( x\right) \right] . \end{equation}$

(2.8)

By the identities (2.7) and (2.8), we obtain the required result (2.6).

Remark 2. If we choose $\alpha = 1$ in Lemma 4, then we have

$\begin{equation*} f(x)\int\limits_{a}^{b}g(t)dt-\int\limits_{a}^{b}f(t)g(t)dt = \int \limits_{a}^{x}\left( \int\limits_{a}^{t}g(s)ds\right) f^{\prime }(t)dt+\int\limits_{x}^{b}\left( \int\limits_{b}^{t}g(s)ds\right) f^{\prime }(t)dt. \end{equation*}$

Corollary 4. If we choose $g(t) = 1$ for all $t\in \left[a, b\right]$ in Lemma 4, then we have

$\begin{eqnarray*} &&\left( \frac{\left( x-a\right) ^{\alpha }+\left( b-x\right) ^{\alpha }}{b-a }\right) f(x)-\frac{\Gamma \left( \alpha +1\right) }{b-a}\left[ J_{a+}^{\alpha }f\left( x\right) +J_{b-}^{\alpha }f\left( x\right) \right] \\ && \\ & = &\frac{1}{b-a}\left[ \int\limits_{a}^{x}\left( \left( x-a\right) ^{\alpha }-\left( x-t\right) ^{\alpha }\right) f^{\prime }(t)dt+\int\limits_{x}^{b}\left( \left( b-x\right) ^{\alpha }-\left( t-x\right) ^{\alpha }\right) f^{\prime }(t)dt\right] . \end{eqnarray*}$

Corollary 5. Under assumption of Lemma 4, let $g$ be symmetric to $\frac{a+b}{2}.$ Then we have the following identity

$\begin{eqnarray} &&\frac{f(a)+f(b)}{2}\left[ J_{a+}^{\alpha }g\left( b\right) +J_{b-}^{\alpha }g\left( a\right) \right] -\left[ J_{a+}^{\alpha }\left( fg\right) \left( b\right) +J_{b-}^{\alpha }\left( fg\right) \left( a\right) \right] \\ && \\ & = &\frac{1}{\Gamma \left( \alpha \right) }\left[ \int\limits_{a}^{b}\left( \int\limits_{a}^{t}\left( b-s\right) ^{\alpha -1}g(s)ds\right) f^{\prime }(t)dt+\int\limits_{a}^{b}\left( \int\limits_{b}^{t}\left( s-a\right) ^{\alpha -1}g(s)ds\right) f^{\prime }(t)dt\right] . \end{eqnarray}$

(2.9)

Proof. If we apply the identity (2.6) for $x = a$ and $x = b,$ then we obtain the equalities

$\begin{equation} f(a)J_{b-}^{\alpha }g\left( a\right) -J_{b-}^{\alpha }\left( fg\right) \left( a\right) = \frac{1}{\Gamma \left( \alpha \right) }\int\limits_{a}^{b} \left( \int\limits_{b}^{t}\left( s-a\right) ^{\alpha -1}g(s)ds\right) f^{\prime }(t)dt \end{equation}$

(2.10)

and

$\begin{equation} f(b)J_{a+}^{\alpha }g\left( b\right) -J_{a+}^{\alpha }\left( fg\right) \left( b\right) = \frac{1}{\Gamma \left( \alpha \right) }\int\limits_{a}^{b} \left( \int\limits_{a}^{t}\left( b-s\right) ^{\alpha -1}g(s)ds\right) f^{\prime }(t)dt \end{equation}$

(2.11)

respectively. If we add the equalities (2.10) and (2.11), then by using Lemma 2, we establish the desired result (2.9).

The inequality (2.9) is the same result which is given by İşcan in ^[18].

3. Weighted fractional trapezoid type inequalities

In this section, we obtain some weighted trapezoid inequalities involving Riemann-Liouviille fractional integral operators. We also give some special cases of our results.

Theorem 4. Let $f:\left[a, b\right] \rightarrow \mathbb{R}$ be a differentiable mapping on $(a, b)$ with $a < b$ and $f^{\prime }\in L\left[a, b \right].$ If $g:\left[a, b\right] \rightarrow \mathbb{R}$ is continuous and if $\left\vert f^{\prime }\right\vert$ is convex on $\left[a, b\right],$ then for all $x\in \left[a, b\right]$ and $\alpha > 0$ we have the following inequality for fractional integrals,

$\begin{eqnarray} &&\left\vert f(a)J_{a+}^{\alpha }g\left( x\right) +f(b)J_{b-}^{\alpha }g\left( x\right) -\left[ J_{a+}^{\alpha }\left( fg\right) \left( x\right) +J_{b-}^{\alpha }\left( fg\right) \left( x\right) \right] \right\vert \\ && \\ &\leq &\frac{1}{\left( b-a\right) \Gamma \left( \alpha +3\right) } \\ &&\times \left\{ \left[ \left( x-a\right) ^{\alpha +1}\left[ \left( \alpha +2\right) \left( b-x\right) +\left( \alpha +1\right) \left( x-a\right) \right] \left\Vert g\right\Vert _{\left[ a, x\right] , \infty }+\left( b-x\right) ^{\alpha +2}\left\Vert g\right\Vert _{\left[ x, b\right] , \infty } \right] \left\vert f^{\prime }(a)\right\vert \right. \\ && \\ &&\left. +\left[ \left( x-a\right) ^{\alpha +2}\left\Vert g\right\Vert _{ \left[ a, x\right] , \infty }+\left( b-x\right) ^{\alpha +1}\left[ \left( \alpha +1\right) \left( b-x\right) +\left( \alpha +2\right) \left( x-a\right) \right] \left\Vert g\right\Vert _{\left[ x, b\right] , \infty } \right] \left\vert f^{\prime }(b)\right\vert \right\} \\ && \\ &\leq &\frac{\left\Vert g\right\Vert _{\infty }}{\left( b-a\right) \Gamma \left( \alpha +3\right) }\left\{ \left[ \left( \left( x-a\right) ^{\alpha +1} \left[ \left( \alpha +2\right) \left( b-x\right) +\left( \alpha +1\right) \left( x-a\right) \right] +\left( b-x\right) ^{\alpha +2}\right) \left\vert f^{\prime }(a)\right\vert \right. \right. \\ && \\ &&\left. +\left( \left( x-a\right) ^{\alpha +2}+\left( b-x\right) ^{\alpha +1}\left[ \left( \alpha +1\right) \left( b-x\right) +\left( \alpha +2\right) \left( x-a\right) \right] \left\vert f^{\prime }(b)\right\vert \right) \right\} . \end{eqnarray}$

(3.1)

Proof. By taking modulus in Lemma 3, we have

$\begin{eqnarray} &&\left\vert f(a)J_{a+}^{\alpha }g\left( x\right) +f(b)J_{b-}^{\alpha }g\left( x\right) -\left[ J_{a+}^{\alpha }\left( fg\right) \left( x\right) +J_{b-}^{\alpha }\left( fg\right) \left( x\right) \right] \right\vert \\ && \\ &\leq &\frac{1}{\Gamma \left( \alpha \right) }\left[ \left\vert \int\limits_{a}^{x}\left( \int\limits_{x}^{t}\left( x-s\right) ^{\alpha -1}g(s)ds\right) f^{\prime }(t)dt\right\vert +\left\vert \int\limits_{x}^{b}\left( \int\limits_{x}^{t}\left( s-x\right) ^{\alpha -1}g(s)ds\right) f^{\prime }(t)dt\right\vert \right] \\ && \\ &\leq &\frac{1}{\Gamma \left( \alpha \right) }\left[ \int\limits_{a}^{x} \left\vert \int\limits_{x}^{t}\left( x-s\right) ^{\alpha -1}g(s)ds\right\vert \left\vert f^{\prime }(t)\right\vert dt+\int\limits_{x}^{b}\left\vert \int\limits_{x}^{t}\left( s-x\right) ^{\alpha -1}g(s)ds\right\vert \left\vert f^{\prime }(t)\right\vert dt\right] \\ && \\ &\leq &\frac{1}{\Gamma \left( \alpha \right) }\left[ \left\Vert g\right\Vert _{\left[ a, x\right] , \infty }\int\limits_{a}^{x}\left\vert \int\limits_{x}^{t}\left( x-s\right) ^{\alpha -1}ds\right\vert \left\vert f^{\prime }(t)\right\vert dt+\left\Vert g\right\Vert _{\left[ x, b\right] , \infty }\int\limits_{x}^{b}\left\vert \int\limits_{x}^{t}\left( s-x\right) ^{\alpha -1}ds\right\vert \left\vert f^{\prime }(t)\right\vert dt\right] \\ && \\ & = &\frac{1}{\Gamma \left( \alpha +1\right) }\left[ \left\Vert g\right\Vert _{ \left[ a, x\right] , \infty }\int\limits_{a}^{x}\left( x-t\right) ^{\alpha }\left\vert f^{\prime }(t)\right\vert dt+\left\Vert g\right\Vert _{\left[ x, b \right] , \infty }\int\limits_{x}^{b}\left( t-x\right) ^{\alpha }\left\vert f^{\prime }(t)\right\vert dt\right] . \end{eqnarray}$

(3.2)

Since $\left\vert f^{\prime }\right\vert$ is convex on $\left[a, b\right],$ we get

$\begin{equation} \left\vert f^{\prime }(t)\right\vert = \left\vert f^{\prime }\left( \frac{b-t }{b-a}a+\frac{t-a}{b-a}b\right) \right\vert \leq \frac{b-t}{b-a}\left\vert f^{\prime }(a)\right\vert +\frac{t-a}{b-a}\left\vert f^{\prime }(b)\right\vert . \end{equation}$

(3.3)

By using (3.3) in (3.2), we obtain

(3.4)

By simple calculations, one can establish

$\begin{equation} \int\limits_{a}^{x}\left( x-t\right) ^{\alpha }\left( b-t\right) dt = \frac{ \left( x-a\right) ^{\alpha +1}\left[ \left( \alpha +2\right) \left( b-x\right) +\left( \alpha +1\right) \left( x-a\right) \right] }{\left( \alpha +1\right) \left( \alpha +2\right) }, \end{equation}$

(3.5)

$\begin{equation} \int\limits_{a}^{x}\left( x-t\right) ^{\alpha }\left( t-a\right) dt = \frac{ \left( x-a\right) ^{\alpha +2}}{\left( \alpha +1\right) \left( \alpha +2\right) }, \end{equation}$

(3.6)

$\begin{equation} \int\limits_{x}^{b}\left( t-x\right) ^{\alpha }\left( b-t\right) dt = \frac{ \left( b-x\right) ^{\alpha +2}}{\left( \alpha +1\right) \left( \alpha +2\right) } \end{equation}$

(3.7)

and

$\begin{equation} \int\limits_{x}^{b}\left( t-x\right) ^{\alpha }\left( t-a\right) dt = \frac{ \left( b-x\right) ^{\alpha +1}\left[ \left( \alpha +1\right) \left( b-x\right) +\left( \alpha +2\right) \left( x-a\right) \right] }{\left( \alpha +1\right) \left( \alpha +2\right) }. \end{equation}$

(3.8)

By substituting the equalities (3.5)–(3.8) in (3.4), we obtain

$\begin{eqnarray*} &&\left\vert f(a)J_{a+}^{\alpha }g\left( x\right) +f(b)J_{b-}^{\alpha }g\left( x\right) -\left[ J_{a+}^{\alpha }\left( fg\right) \left( x\right) +J_{b-}^{\alpha }\left( fg\right) \left( x\right) \right] \right\vert \\ && \\ &\leq &\frac{\left\Vert g\right\Vert _{\left[ a, x\right] , \infty }}{\left( b-a\right) \Gamma \left( \alpha +3\right) }\left[ \left( x-a\right) ^{\alpha +1}\left[ \left( \alpha +2\right) \left( b-x\right) +\left( \alpha +1\right) \left( x-a\right) \right] \left\vert f^{\prime }(a)\right\vert +\left( x-a\right) ^{\alpha +2}\left\vert f^{\prime }(b)\right\vert \right] \\ && \\ &&+\frac{\left\Vert g\right\Vert _{\left[ x, b\right] , \infty }}{\left( b-a\right) \Gamma \left( \alpha +3\right) }\left[ \left( b-x\right) ^{\alpha +2}\left\vert f^{\prime }(a)\right\vert +\left( b-x\right) ^{\alpha +1}\left[ \left( \alpha +1\right) \left( b-x\right) +\left( \alpha +2\right) \left( x-a\right) \right] \left\vert f^{\prime }(b)\right\vert \right] \\ && \\ & = &\frac{1}{\left( b-a\right) \Gamma \left( \alpha +3\right) } \\ && \\ &&\times \left\{ \left[ \left( x-a\right) ^{\alpha +1}\left[ \left( \alpha +2\right) \left( b-x\right) +\left( \alpha +1\right) \left( x-a\right) \right] \left\Vert g\right\Vert _{\left[ a, x\right] , \infty }+\left( b-x\right) ^{\alpha +2}\left\Vert g\right\Vert _{\left[ x, b\right] , \infty } \right] \left\vert f^{\prime }(a)\right\vert \right. \\ && \\ &&\left. +\left[ \left( x-a\right) ^{\alpha +2}\left\Vert g\right\Vert _{ \left[ a, x\right] , \infty }+\left( b-x\right) ^{\alpha +1}\left[ \left( \alpha +1\right) \left( b-x\right) +\left( \alpha +2\right) \left( x-a\right) \right] \left\Vert g\right\Vert _{\left[ x, b\right] , \infty } \right] \left\vert f^{\prime }(b)\right\vert \right\} \end{eqnarray*}$

which completes the proof of first inequality in (3.1).

From the facts that

$\begin{equation} \left\Vert g\right\Vert _{\left[ a, x\right] , \infty }\leq \left\Vert g\right\Vert _{\left[ a, b\right] , \infty } = \left\Vert g\right\Vert _{\infty } \text{ and }\left\Vert g\right\Vert _{\left[ x, b\right] , \infty }\leq \left\Vert g\right\Vert _{\left[ a, b\right] , \infty } = \left\Vert g\right\Vert _{\infty } \end{equation}$

(3.9)

for all $x\in \left[a, b\right],$ the proof of the second inequality in (3.1) is obvious.

Remark 3. If we choose $\alpha = 1$ in Theorem 4, then we have the weighted inequality

$\begin{eqnarray*} &&\left\vert f(a)\int\limits_{a}^{x}g(t)dt+f(b)\int\limits_{x}^{b}g(t)dt-\int \limits_{a}^{b}f(t)g(t)dt\right\vert \\ && \\ &\leq &\left[ \left( x-a\right) ^{2}\left( \frac{3b-2a-x}{6\left( b-a\right) }\right) \left\Vert g\right\Vert _{\left[ a, x\right] , \infty }+\frac{\left( b-x\right) ^{3}}{6\left( b-a\right) }\left\Vert g\right\Vert _{\left[ x, b \right] , \infty }\right] \left\vert f^{\prime }(a)\right\vert \\ && \\ &&+\left[ \frac{\left( x-a\right) ^{3}}{6\left( b-a\right) }\left\Vert g\right\Vert _{\left[ a, x\right] , \infty }+\left( b-x\right) ^{2}\left( \frac{2b-3a+x}{6\left( b-a\right) }\right) \left\Vert g\right\Vert _{\left[ x, b\right] , \infty }\right] \left\vert f^{\prime }(b)\right\vert \\ && \\ &\leq &\left[ \left( \frac{\left( x-a\right) ^{2}\left( 3\left( b-x\right) +2\left( x-a\right) \right) +\left( b-x\right) ^{3}}{6\left( b-a\right) } \right) \left\vert f^{\prime }(a)\right\vert \right. \\ && \\ &&\left. +\left( \frac{\left( x-a\right) ^{3}+\left( b-x\right) ^{2}\left( 2\left( b-x\right) +3\left( x-a\right) \right) }{6\left( b-a\right) } \left\vert f^{\prime }(b)\right\vert \right) \right] \left\Vert g\right\Vert _{\infty } \end{eqnarray*}$

which is proved by Tseng et. al in ^[51].

Corollary 6. In Theorem 4, let $g$ be symmetric to $\frac{a+b}{2}$ and let $x = \frac{a+b}{2}$ . Then we have the following "weighted fractional trapezoid inequality"

$\begin{eqnarray*} &&\left\vert \left[ J_{a+}^{\alpha }g\left( \frac{a+b}{2}\right) +J_{b-}^{\alpha }g\left( \frac{a+b}{2}\right) \right] \frac{f(a)+f(b)}{2}- \left[ J_{a+}^{\alpha }\left( fg\right) \left( \frac{a+b}{2}\right) +J_{b-}^{\alpha }\left( fg\right) \left( \frac{a+b}{2}\right) \right] \right\vert \\ && \\ &\leq &\frac{\left( b-a\right) ^{\alpha +1}}{2^{\alpha +2}\Gamma \left( \alpha +3\right) } \\ &&\times \left[ \left( \left( 2\alpha +3\right) \left\Vert g\right\Vert _{ \left[ a, x\right] , \infty }+\left\Vert g\right\Vert _{\left[ x, b\right] , \infty }\right) \left\vert f^{\prime }(a)\right\vert +\left( \left\Vert g\right\Vert _{\left[ a, x\right] , \infty }+\left( 2\alpha +3\right) \left\Vert g\right\Vert _{\left[ x, b\right] , \infty }\right) \left\vert f^{\prime }(b)\right\vert \right] \\ && \\ &\leq &\frac{\left( b-a\right) ^{\alpha +1}}{2^{\alpha }\Gamma \left( \alpha +2\right) }\frac{\left\vert f^{\prime }(a)\right\vert +\left\vert f^{\prime }(b)\right\vert }{2}. \end{eqnarray*}$

Corollary 7. Under assumptions of Theorem 4, we have the following inequality

Proof. If we write the inequality (3.1) for $x = a$ and $x = b,$ then we have the inequalities

$\begin{equation} \left\vert f(b)J_{b-}^{\alpha }g\left( a\right) -J_{b-}^{\alpha }\left( fg\right) \left( a\right) \right\vert \leq \frac{\left( b-a\right) ^{\alpha +1}\left\Vert g\right\Vert _{\infty }}{\Gamma \left( \alpha +3\right) }\left[ \left\vert f^{\prime }(a)\right\vert +\left( \alpha +1\right) \left\vert f^{\prime }(b)\right\vert \right] \end{equation}$

(3.10)

and

$\begin{equation} \left\vert f(a)J_{a+}^{\alpha }g\left( b\right) -J_{a+}^{\alpha }\left( fg\right) \left( b\right) \right\vert \leq \frac{\left( b-a\right) ^{\alpha +1}\left\Vert g\right\Vert _{\infty }}{\Gamma \left( \alpha +3\right) }\left[ \left( \alpha +1\right) \left\vert f^{\prime }(a)\right\vert +\left\vert f^{\prime }(b)\right\vert \right] \end{equation}$

(3.11)

respectively. By adding the inequalities (3.10) and (3.11) and using triangle inequality, we obtain

$\begin{eqnarray*} &&\left\vert f(a)J_{a+}^{\alpha }g\left( b\right) +f(b)J_{b-}^{\alpha }g\left( a\right) -\left[ J_{a+}^{\alpha }\left( fg\right) \left( b\right) +J_{b-}^{\alpha }\left( fg\right) \left( a\right) \right] \right\vert \\ && \\ &\leq &\frac{\left( b-a\right) ^{\alpha +1}\left\Vert g\right\Vert _{\infty } }{\Gamma \left( \alpha +3\right) }\left[ \left\vert f^{\prime }(a)\right\vert +\left( \alpha +1\right) \left\vert f^{\prime }(b)\right\vert \right] +\frac{\left( b-a\right) ^{\alpha +1}\left\Vert g\right\Vert _{\infty }}{\Gamma \left( \alpha +3\right) }\left[ \left( \alpha +1\right) \left\vert f^{\prime }(a)\right\vert +\left\vert f^{\prime }(b)\right\vert \right] \\ && \\ & = &\frac{\left\Vert g\right\Vert _{\infty }\left( b-a\right) ^{\alpha +1}}{ 2^{\alpha }\Gamma \left( \alpha +2\right) }\left[ \left\vert f^{\prime }(a)\right\vert +\left\vert f^{\prime }(b)\right\vert \right] \end{eqnarray*}$

which completes the proof.

Remark 4. If we choose $\alpha = 1$ in Corollary 7, then we have the following weighted trapezoid inequality

$\begin{equation*} \left\vert \frac{f(a)+f(b)}{2}\int\limits_{a}^{b}g(t)dt-\int \limits_{a}^{b}f(t)g(t)dt\right\vert \leq \frac{\left\vert f^{\prime }(a)\right\vert +\left\vert f^{\prime }(b)\right\vert }{8}\left\Vert g\right\Vert _{\infty } \end{equation*}$

which is given by Tseng et al. in ^[51].

Corollary 8. In Theorem 4, let $g(t) = 1$ for all $t\in \left[a, b\right].$ Then we have the following identity

$\begin{eqnarray*} &&\left\vert \left( x-a\right) ^{\alpha }f(a)+\left( b-x\right) ^{\alpha }f(b)-\Gamma \left( \alpha +1\right) \left[ J_{a+}^{\alpha }f\left( x\right) +J_{b-}^{\alpha }f\left( x\right) \right] \right\vert \\ && \\ &\leq &\frac{1}{\left( b-a\right) \left( \alpha +1\right) \left( \alpha +2\right) } \\ &&\times \left[ \left( \left( x-a\right) ^{\alpha +1}\left[ \left( \alpha +2\right) \left( b-x\right) +\left( \alpha +1\right) \left( x-a\right) \right] +\left( b-x\right) ^{\alpha +2}\right) \left\vert f^{\prime }(a)\right\vert \right. \\ && \\ &&+\left( \left( x-a\right) ^{\alpha +2}+\left( b-x\right) ^{\alpha +1}\left[ \left( \alpha +1\right) \left( b-x\right) +\left( \alpha +2\right) \left( x-a\right) \right] \left\vert f^{\prime }(b)\right\vert \right) . \end{eqnarray*}$

Theorem 5. Let $f:\left[a, b\right] \rightarrow \mathbb{R}$ be a differentiable mapping on $(a, b)$ with $a < b$ and $f^{\prime }\in L\left[a, b \right].$ If $g:\left[a, b\right] \rightarrow \mathbb{R}$ is continuous and if $\left\vert f^{\prime }\right\vert ^{q},$ $q > 1,$ is convex on $\left[a, b \right],$ then for all $x\in \left[a, b\right]$ and $\alpha > 0$ we have the following inequality for fractional integrals,

$\begin{eqnarray} &&\left\vert f(a)J_{a+}^{\alpha }g\left( x\right) +f(b)J_{b-}^{\alpha }g\left( x\right) -\left[ J_{a+}^{\alpha }\left( fg\right) \left( x\right) +J_{b-}^{\alpha }\left( fg\right) \left( x\right) \right] \right\vert \\ &\leq &\frac{1}{\Gamma \left( \alpha +1\right) }\left( \frac{1}{p\alpha +1} \right) ^{\frac{1}{p}} \\ &&\times \left[ \left\Vert g\right\Vert _{\left[ a, x\right] , \infty }\left( x-a\right) ^{\alpha +\frac{1}{p}}\left( \left( \frac{x-a}{b-a}\right) \left( b-\frac{a+x}{2}\right) \left\vert f^{\prime }(a)\right\vert ^{q}+\frac{ \left( x-a\right) ^{2}}{2\left( b-a\right) }\left\vert f^{\prime }(b)\right\vert ^{q}dt\right) ^{\frac{1}{q}}\right. \\ &&\left. +\left\Vert g\right\Vert _{\left[ x, b\right] , \infty }\left( b-x\right) ^{\alpha +\frac{1}{p}}\left( \frac{\left( b-x\right) ^{2}}{ 2\left( b-a\right) }\left\vert f^{\prime }(a)\right\vert ^{q}+\left( \frac{ b-x}{b-a}\right) \left( \frac{x+b}{2}-a\right) \left\vert f^{\prime }(b)\right\vert ^{q}\right) ^{\frac{1}{q}}\right] \\ && \\ &\leq &\frac{\left\Vert g\right\Vert _{\infty }}{\Gamma \left( \alpha +1\right) }\left( \frac{1}{p\alpha +1}\right) ^{\frac{1}{p}}\left[ \left( x-a\right) ^{\alpha +\frac{1}{p}}\left( \left( \frac{x-a}{b-a}\right) \left( b-\frac{a+x}{2}\right) \left\vert f^{\prime }(a)\right\vert ^{q}+\frac{ \left( x-a\right) ^{2}}{2\left( b-a\right) }\left\vert f^{\prime }(b)\right\vert ^{q}dt\right) ^{\frac{1}{q}}\right. \\ &&\left. +\left( b-x\right) ^{\alpha +\frac{1}{p}}\left( \frac{\left( b-x\right) ^{2}}{2\left( b-a\right) }\left\vert f^{\prime }(a)\right\vert ^{q}+\left( \frac{b-x}{b-a}\right) \left( \frac{x+b}{2}-a\right) \left\vert f^{\prime }(b)\right\vert ^{q}\right) ^{\frac{1}{q}}\right] \end{eqnarray}$

(3.12)

where $\frac{1}{p}+\frac{1}{q} = 1.$

Proof. By using the well-known Hölder inequality in (3.2), we have

$\begin{eqnarray} &&\left\vert f(a)J_{a+}^{\alpha }g\left( x\right) +f(b)J_{b-}^{\alpha }g\left( x\right) -\left[ J_{a+}^{\alpha }\left( fg\right) \left( x\right) +J_{b-}^{\alpha }\left( fg\right) \left( x\right) \right] \right\vert \\ && \\ &\leq &\frac{1}{\Gamma \left( \alpha +1\right) }\left[ \left\Vert g\right\Vert _{\left[ a, x\right] , \infty }\int\limits_{a}^{x}\left( x-t\right) ^{\alpha }\left\vert f^{\prime }(t)\right\vert dt+\left\Vert g\right\Vert _{\left[ x, b\right] , \infty }\int\limits_{x}^{b}\left( t-x\right) ^{\alpha }\left\vert f^{\prime }(t)\right\vert dt\right] \\ && \\ &\leq &\frac{1}{\Gamma \left( \alpha +1\right) }\left[ \left\Vert g\right\Vert _{\left[ a, x\right] , \infty }\left( \int\limits_{a}^{x}\left( x-t\right) ^{p\alpha }dt\right) ^{\frac{1}{p}}\left( \int\limits_{a}^{x}\left\vert f^{\prime }(t)\right\vert ^{q}dt\right) ^{ \frac{1}{q}}\right. \\ && \\ &&\left. +\left\Vert g\right\Vert _{\left[ x, b\right] , \infty }\left( \int\limits_{x}^{b}\left( t-x\right) ^{p\alpha }dt\right) ^{\frac{1}{p} }\left( \int\limits_{x}^{b}\left\vert f^{\prime }(t)\right\vert ^{q}dt\right) ^{\frac{1}{q}}\right] \\ && \\ & = &\frac{1}{\Gamma \left( \alpha +1\right) }\left[ \left\Vert g\right\Vert _{ \left[ a, x\right] , \infty }\frac{\left( x-a\right) ^{\alpha +\frac{1}{p}}}{ \left( p\alpha +1\right) ^{\frac{1}{p}}}\left( \int\limits_{a}^{x}\left\vert f^{\prime }(t)\right\vert ^{q}dt\right) ^{\frac{1}{q}}\right. \\ && \\ &&\left. +\left\Vert g\right\Vert _{\left[ x, b\right] , \infty }\frac{\left( b-x\right) ^{\alpha +\frac{1}{p}}}{\left( p\alpha +1\right) ^{\frac{1}{p}}} \left( \int\limits_{x}^{b}\left\vert f^{\prime }(t)\right\vert ^{q}dt\right) ^{\frac{1}{q}}\right] . \end{eqnarray}$

(3.13)

Since $\left\vert f^{\prime }\right\vert ^{q}$ is convex on $\left[a, b \right],$ we get

$\begin{equation} \left\vert f^{\prime }(t)\right\vert ^{q} = \left\vert f^{\prime }\left( \frac{ b-t}{b-a}a+\frac{t-a}{b-a}b\right) \right\vert ^{q}\leq \frac{b-t}{b-a} \left\vert f^{\prime }(a)\right\vert ^{q}+\frac{t-a}{b-a}\left\vert f^{\prime }(b)\right\vert ^{q}. \end{equation}$

(3.14)

Then it follows that

$\begin{eqnarray*} &&\left\vert f(a)J_{a+}^{\alpha }g\left( x\right) +f(b)J_{b-}^{\alpha }g\left( x\right) -\left[ J_{a+}^{\alpha }\left( fg\right) \left( x\right) +J_{b-}^{\alpha }\left( fg\right) \left( x\right) \right] \right\vert \\ && \\ &\leq &\frac{1}{\Gamma \left( \alpha +1\right) }\left( \frac{1}{p\alpha +1} \right) ^{\frac{1}{p}} \\ &&\times \left[ \left\Vert g\right\Vert _{\left[ a, x\right] , \infty }\left( x-a\right) ^{\alpha +\frac{1}{p}}\left( \int\limits_{a}^{x}\left( \frac{ \left( b-t\right) \left\vert f^{\prime }(a)\right\vert ^{q}+\left( t-a\right) \left\vert f^{\prime }(b)\right\vert ^{q}}{b-a}\right) dt\right) ^{\frac{1}{q}}\right. \\ &&\left. \left\Vert g\right\Vert _{\left[ x, b\right] , \infty }\left( b-x\right) ^{\alpha +\frac{1}{p}}\left( \int\limits_{x}^{b}\left( \frac{ \left( b-t\right) \left\vert f^{\prime }(a)\right\vert ^{q}+\left( t-a\right) \left\vert f^{\prime }(b)\right\vert ^{q}}{b-a}\right) dt\right) ^{\frac{1}{q}}\right] \\ &\leq &\frac{1}{\Gamma \left( \alpha +1\right) }\left( \frac{1}{p\alpha +1} \right) ^{\frac{1}{p}} \\ &&\times \left[ \left\Vert g\right\Vert _{\left[ a, x\right] , \infty }\left( x-a\right) ^{\alpha +\frac{1}{p}}\left( \left( \frac{x-a}{b-a}\right) \left( b-\frac{a+x}{2}\right) \left\vert f^{\prime }(a)\right\vert ^{q}+\frac{ \left( x-a\right) ^{2}}{2\left( b-a\right) }\left\vert f^{\prime }(b)\right\vert ^{q}dt\right) ^{\frac{1}{q}}\right. \\ &&\left. +\left\Vert g\right\Vert _{\left[ x, b\right] , \infty }\left( b-x\right) ^{\alpha +\frac{1}{p}}\left( \frac{\left( b-x\right) ^{2}}{ 2\left( b-a\right) }\left\vert f^{\prime }(a)\right\vert ^{q}+\left( \frac{ b-x}{b-a}\right) \left( \frac{x+b}{2}-a\right) \left\vert f^{\prime }(b)\right\vert ^{q}\right) ^{\frac{1}{q}}\right] . \end{eqnarray*}$

This completes the proof of first inequality in (3.12). The proof of second inequality is obvious from the inequalities (3.9).

Remark 5. If we choose $\alpha = 1$ in Theorem 5, then we obtain the following weighted inequality

$\begin{eqnarray*} &&\left\vert f(a)\int\limits_{a}^{x}g(t)dt+f(b)\int\limits_{x}^{b}g(t)dt-\int \limits_{a}^{b}f(t)g(t)dt\right\vert \\ &\leq &\left( \frac{1}{p+1}\right) ^{\frac{1}{p}}\left[ \left\Vert g\right\Vert _{\left[ a, x\right] , \infty }\left( x-a\right) ^{1+\frac{1}{p} }\left( \left( \frac{x-a}{b-a}\right) \left( b-\frac{a+x}{2}\right) \left\vert f^{\prime }(a)\right\vert ^{q}+\frac{\left( x-a\right) ^{2}}{ 2\left( b-a\right) }\left\vert f^{\prime }(b)\right\vert ^{q}dt\right) ^{ \frac{1}{q}}\right. \\ && \\ &&\left. +\left\Vert g\right\Vert _{\left[ x, b\right] , \infty }\left( b-x\right) ^{1+\frac{1}{p}}\left( \frac{\left( b-x\right) ^{2}}{2\left( b-a\right) }\left\vert f^{\prime }(a)\right\vert ^{q}+\left( \frac{b-x}{b-a} \right) \left( \frac{x+b}{2}-a\right) \left\vert f^{\prime }(b)\right\vert ^{q}\right) ^{\frac{1}{q}}\right] \\ && \\ &\leq &\left( \frac{1}{p1+1}\right) ^{\frac{1}{p}}\left[ \left( x-a\right) ^{1+\frac{1}{p}}\left( \left( \frac{x-a}{b-a}\right) \left( b-\frac{a+x}{2} \right) \left\vert f^{\prime }(a)\right\vert ^{q}+\frac{\left( x-a\right) ^{2}}{2\left( b-a\right) }\left\vert f^{\prime }(b)\right\vert ^{q}dt\right) ^{\frac{1}{q}}\right. \\ &&\left. +\left( b-x\right) ^{1+\frac{1}{p}}\left( \frac{\left( b-x\right) ^{2}}{2\left( b-a\right) }\left\vert f^{\prime }(a)\right\vert ^{q}+\left( \frac{b-x}{b-a}\right) \left( \frac{x+b}{2}-a\right) \left\vert f^{\prime }(b)\right\vert ^{q}\right) ^{\frac{1}{q}}\right] \left\Vert g\right\Vert _{\infty }. \end{eqnarray*}$

Corollary 9. In Theorem 5, let $g$ be symmetric to $\frac{a+b}{2}$ and let $x = \frac{a+b}{2}$ . Then we have the following weighted fractional trapezoid inequality

$\begin{eqnarray} &&\left\vert \left[ J_{a+}^{\alpha }g\left( \frac{a+b}{2}\right) +J_{b-}^{\alpha }g\left( \frac{a+b}{2}\right) \right] \frac{f(a)+f(b)}{2}- \left[ J_{a+}^{\alpha }\left( fg\right) \left( \frac{a+b}{2}\right) +J_{b-}^{\alpha }\left( fg\right) \left( \frac{a+b}{2}\right) \right] \right\vert \\ &\leq &\frac{1}{\Gamma \left( \alpha +1\right) }\left( \frac{1}{p\alpha +1} \right) ^{\frac{1}{p}}\left( \frac{b-a}{2}\right) ^{\alpha +1} \\ &&\times \left[ \left\Vert g\right\Vert _{\left[ a, \frac{a+b}{2}\right] , \infty }\left( \frac{3\left\vert f^{\prime }(a)\right\vert ^{q}+\left\vert f^{\prime }(b)\right\vert ^{q}dt}{4}\right) ^{\frac{1}{q}}+\left\Vert g\right\Vert _{\left[ x, b\right] , \infty }\left( \frac{\left\vert f^{\prime }(a)\right\vert ^{q}+3\left\vert f^{\prime }(b)\right\vert ^{q}}{4}\right) ^{ \frac{1}{q}}\right] \\ && \\ &\leq &\frac{\left\Vert g\right\Vert _{\infty }}{\Gamma \left( \alpha +1\right) }\left( \frac{1}{p\alpha +1}\right) ^{\frac{1}{p}}\left( \frac{b-a }{2}\right) ^{\alpha +1} \\ &&\times \left[ \left( \frac{3\left\vert f^{\prime }(a)\right\vert ^{q}+\left\vert f^{\prime }(b)\right\vert ^{q}dt}{4}\right) ^{\frac{1}{q} }+\left( \frac{\left\vert f^{\prime }(a)\right\vert ^{q}+3\left\vert f^{\prime }(b)\right\vert ^{q}}{4}\right) ^{\frac{1}{q}}\right] \\ && \\ &\leq &\frac{\left\Vert g\right\Vert _{\infty }}{\Gamma \left( \alpha +1\right) }\left( \frac{4}{p\alpha +1}\right) ^{\frac{1}{p}}\left( \frac{b-a }{2}\right) ^{\alpha +1}\left[ \left\vert f^{\prime }(a)\right\vert +\left\vert f^{\prime }(b)\right\vert \right] \end{eqnarray}$

(3.15)

where $\frac{1}{p}+\frac{1}{q} = 1.$

Proof. The proof of the first and second inequalities in (3.15) is obvious. For the proof of third inequality, let $a_{1} = 3\left\vert f^{\prime }\left(a\right) \right\vert ^{q},$ $b_{1} = \left\vert f^{\prime }\left(b\right) \right\vert ^{q},$ $a_{2} = \left\vert f^{\prime }\left(a\right) \right\vert ^{q}$ and $b_{2} = 3\left\vert f^{\prime }\left(b\right) \right\vert ^{q}.$ Using the facts that,

$\begin{equation*} \sum\limits_{k = 1}^{n}\left( a_{k}+b_{k}\right) ^{s}\leq \sum\limits_{k = 1}^{n}a_{k}^{s}+\sum\limits_{k = 1}^{n}b_{k}^{s}, \text{ }0\leq s \lt 1 \end{equation*}$

and $3^{\frac{1}{q}}+1\leq 4,$ the desired result can be obtained straightforwardly.

Corollary 10. Under assumptions of Theorem 4, we have the following equality

Proof. If we write the equality (3.1) for $x = a$ and $x = b,$ then we have the inequalities

$\begin{equation} \left\vert f(b)J_{b-}^{\alpha }g\left( a\right) -J_{b-}^{\alpha }\left( fg\right) \left( a\right) \right\vert \leq \frac{\left\Vert g\right\Vert _{\infty }}{\Gamma \left( \alpha +1\right) }\left( \frac{1}{p\alpha +1} \right) ^{\frac{1}{p}}\left( b-a\right) ^{\alpha +1}\left( \frac{\left\vert f^{\prime }(a)\right\vert ^{q}+\left\vert f^{\prime }(b)\right\vert ^{q}}{2} \right) ^{\frac{1}{q}} \end{equation}$

(3.16)

and

$\begin{equation} \left\vert f(a)J_{a+}^{\alpha }g\left( b\right) -J_{a+}^{\alpha }\left( fg\right) \left( b\right) \right\vert \leq \frac{\left\Vert g\right\Vert _{\infty }}{\Gamma \left( \alpha +1\right) }\left( \frac{1}{p\alpha +1} \right) ^{\frac{1}{p}}\left( b-a\right) ^{\alpha +1}\left( \frac{\left\vert f^{\prime }(a)\right\vert ^{q}+\left\vert f^{\prime }(b)\right\vert ^{q}}{2} \right) ^{\frac{1}{q}}, \end{equation}$

(3.17)

respectively. By adding the inequalities (3.16) and (3.17) and using triangle inequality, we obtain

$\begin{eqnarray*} &&\left\vert J_{a+}^{\alpha }g\left( b\right) f(a)+J_{b-}^{\alpha }g\left( a\right) f(b)-\left[ J_{a+}^{\alpha }\left( fg\right) \left( b\right) +J_{b-}^{\alpha }\left( fg\right) \left( a\right) \right] \right\vert \\ && \\ &\leq &\frac{2\left\Vert g\right\Vert _{\infty }}{\Gamma \left( \alpha +1\right) }\left( \frac{1}{p\alpha +1}\right) ^{\frac{1}{p}}\left( b-a\right) ^{\alpha +1}\left( \frac{\left\vert f^{\prime }(a)\right\vert ^{q}+\left\vert f^{\prime }(b)\right\vert ^{q}}{2}\right) ^{\frac{1}{q}}. \end{eqnarray*}$

This completes the proof.

Remark 6. If we choose $\alpha = 1$ in Corollary 10, then we have the following weighted trapezoid inequality

$\begin{equation*} \left\vert \frac{f(a)+f(b)}{2}\int\limits_{a}^{b}g(t)dt-\int \limits_{a}^{b}f(t)g(t)dt\right\vert \leq \left( \frac{1}{p+1}\right) ^{ \frac{1}{p}}\left( b-a\right) ^{2}\left( \frac{\left\vert f^{\prime }(a)\right\vert ^{q}+\left\vert f^{\prime }(b)\right\vert ^{q}}{2}\right) ^{ \frac{1}{q}}\left\Vert g\right\Vert _{\infty } \end{equation*}$

which is given by Tseng et al. in ^[51].

Corollary 11. In Theorem 5, let $g(t) = 1$ for all $t\in \left[a, b\right].$ Then we have the following inequality

$\begin{eqnarray*} &&\left\vert \left( x-a\right) ^{\alpha }f(a)+\left( b-x\right) ^{\alpha }f(b)-\Gamma \left( \alpha +1\right) \left[ J_{a+}^{\alpha }f\left( x\right) +J_{b-}^{\alpha }f\left( x\right) \right] \right\vert \\ && \\ &\leq &\left( \frac{1}{p\alpha +1}\right) ^{\frac{1}{p}}\left[ \left( x-a\right) ^{\alpha +\frac{1}{p}}\left( \left( \frac{x-a}{b-a}\right) \left( b-\frac{a+x}{2}\right) \left\vert f^{\prime }(a)\right\vert ^{q}+\frac{ \left( x-a\right) ^{2}}{2\left( b-a\right) }\left\vert f^{\prime }(b)\right\vert ^{q}dt\right) ^{\frac{1}{q}}\right. \\ && \\ &&\left. +\left( b-x\right) ^{\alpha +\frac{1}{p}}\left( \frac{\left( b-x\right) ^{2}}{2\left( b-a\right) }\left\vert f^{\prime }(a)\right\vert ^{q}+\left( \frac{b-x}{b-a}\right) \left( \frac{x+b}{2}-a\right) \left\vert f^{\prime }(b)\right\vert ^{q}\right) ^{\frac{1}{q}}\right] . \end{eqnarray*}$

4. Weighted fractional Ostrowski and midpoint type inequalities

In this section, we obtain some weighted fractional Ostrowski type inequalities and we give some weighted fractional midpoint type inequalities as special cases.

First, we give the classical and fractional Ostrowski inequalities:

Ostrowski ^[31] gave the following classical integral inequality associated with the differentiable mappings:

Theorem 6. Let $f:[a, b]\mathbb{\rightarrow R}$ be a differentiable mapping on $(a, b)$ whose derivative $f^{^{\prime }}:(a, b)\mathbb{\rightarrow R}$ is bounded on $(a, b),$ i.e., $\left\Vert f^{\prime }\right\Vert _{\infty } = \underset{t\in (a, b)}{\overset{}{\sup }}\left\vert f^{\prime }(t)\right\vert < \infty.$ Then, the inequality holds:

$\begin{equation} \left\vert f(x)-\frac{1}{b-a}\int\limits_{a}^{b}f(t)dt\right\vert \leq \left[ \frac{1}{4}+\frac{\left( x-\frac{a+b}{2}\right) ^{2}}{(b-a)^{2}}\right] (b-a)\left\Vert f^{\prime }\right\Vert _{\infty } \end{equation}$

(4.1)

for all $\ x\in \lbrack a, b].$ The constant $\frac{1}{4}$ is the best possible.

In ^[45], Set obtained the following Ostrowski inequality for fractional integrals:

Theorem 7. Let $f:\left[a, b\right] \rightarrow \mathbb{R}$ be a differentiable mapping on $(a, b)$ with $a < b$ and $f^{\prime }\in L\left[a, b\right].$ If $\left\vert f^{\prime }\right\vert$ is convex on $\left[a, b\right]$ and $\left\vert f^{\prime }(x)\leq M\right\vert,$ $x\in \left[a, b\right]$ then the following inequality for fractional integrals with $\alpha > 0$ holds:

In recent years, several papers have devoted to Ostrowski type inequalities for several type fractional integrals, for some of them please see ^{[1,5,6,11,15,16,24,26,29,38,39,48,50,25,49,54,55]}.

Theorem 8. Let $f:\left[a, b\right] \rightarrow \mathbb{R}$ be a differentiable mapping on $(a, b)$ with $a < b$ and $f^{\prime }\in L\left[a, b \right].$ If $g:\left[a, b\right] \rightarrow \mathbb{R}$ is continuous and if $\left\vert f^{\prime }\right\vert$ is convex on $\left[a, b\right],$ then for all $x\in \left[a, b\right]$ and $\alpha > 0$ we have the following "weighted fractional Ostrowski inequality"

(4.2)

Proof. From Lemma 4, we have

$\begin{eqnarray} &&\left\vert \left[ J_{a+}^{\alpha }g\left( x\right) +J_{b-}^{\alpha }g\left( x\right) \right] f(x)-\left[ J_{a+}^{\alpha }\left( fg\right) \left( x\right) +J_{b-}^{\alpha }\left( fg\right) \left( x\right) \right] \right\vert \\ && \\ &\leq &\frac{1}{\Gamma \left( \alpha \right) }\left[ \left\vert \int\limits_{a}^{x}\left( \int\limits_{a}^{t}\left( x-s\right) ^{\alpha -1}g(s)ds\right) f^{\prime }(t)dt\right\vert +\left\vert \int\limits_{x}^{b}\left( \int\limits_{b}^{t}\left( s-x\right) ^{\alpha -1}g(s)ds\right) f^{\prime }(t)dt\right\vert \right] \\ && \\ &\leq &\frac{1}{\Gamma \left( \alpha \right) }\left[ \int\limits_{a}^{x} \left\vert \int\limits_{a}^{t}\left( x-s\right) ^{\alpha -1}g(s)ds\right\vert \left\vert f^{\prime }(t)\right\vert dt+\int\limits_{x}^{b}\left\vert \int\limits_{b}^{t}\left( s-x\right) ^{\alpha -1}g(s)ds\right\vert \left\vert f^{\prime }(t)\right\vert dt\right] \\ && \\ &\leq &\frac{1}{\Gamma \left( \alpha +1\right) }\left[ \left\Vert g\right\Vert _{\left[ a, x\right] , \infty }\int\limits_{a}^{x}\left( \left( x-a\right) ^{\alpha }-\left( x-t\right) ^{\alpha }\right) \left\vert f^{\prime }(t)\right\vert dt\right. \\ && \\ &&\left. +\left\Vert g\right\Vert _{\left[ x, b\right] , \infty }\int\limits_{x}^{b}\left( \left( b-x\right) ^{\alpha }-\left( t-x\right) ^{\alpha }\right) \left\vert f^{\prime }(t)\right\vert dt\right] \\ && \\ &\leq &\frac{\left\Vert g\right\Vert _{\infty }}{\Gamma \left( \alpha +1\right) }\left[ \int\limits_{a}^{x}\left( \left( x-a\right) ^{\alpha }-\left( x-t\right) ^{\alpha }\right) \left\vert f^{\prime }(t)\right\vert dt+\int\limits_{x}^{b}\left( \left( b-x\right) ^{\alpha }-\left( t-x\right) ^{\alpha }\right) \left\vert f^{\prime }(t)\right\vert dt\right] . \end{eqnarray}$

(4.3)

Since $\left\vert f^{\prime }\right\vert$ is convex on $\left[a, b\right],$ then we get

$\begin{eqnarray} &&\int\limits_{a}^{x}\left( \left( x-a\right) ^{\alpha }-\left( x-t\right) ^{\alpha }\right) \left\vert f^{\prime }(t)\right\vert dt \\ && \\ &\leq &\frac{1}{b-a}\int\limits_{a}^{x}\left( \left( x-a\right) ^{\alpha }-\left( x-t\right) ^{\alpha }\right) \left[ \left( b-t\right) \left\vert f^{\prime }(a)\right\vert +\left( t-a\right) \left\vert f^{\prime }(b)\right\vert \right] dt \\ && \\ & = &\frac{\left\vert f^{\prime }(a)\right\vert }{b-a}\int\limits_{a}^{x} \left( \left( x-a\right) ^{\alpha }-\left( x-t\right) ^{\alpha }\right) \left( b-t\right) dt \\ && \\ &&+\frac{\left\vert f^{\prime }(b)\right\vert }{b-a}\int\limits_{a}^{x} \left( \left( x-a\right) ^{\alpha }-\left( x-t\right) ^{\alpha }\right) \left( t-a\right) dt \\ && \\ & = &\frac{\left( x-a\right) ^{\alpha +1}}{b-a}\left[ \left( b-\frac{a+x}{2} \right) -\frac{\left( \alpha +2\right) \left( b-x\right) +\left( \alpha +1\right) \left( x-a\right) }{\left( \alpha +1\right) \left( \alpha +2\right) }\right] \left\vert f^{\prime }(a)\right\vert \\ && \\ &&+\frac{\left( x-a\right) ^{\alpha +2}}{b-a}\left[ \frac{1}{2}-\frac{1}{ \left( \alpha +1\right) \left( \alpha +2\right) }\right] \left\vert f^{\prime }(b)\right\vert . \end{eqnarray}$

(4.4)

Similarly we can obtain

$\begin{eqnarray} &&\int\limits_{x}^{b}\left( \left( b-x\right) ^{\alpha }-\left( t-x\right) ^{\alpha }\right) \left\vert f^{\prime }(t)\right\vert dt \\ && \\ &\leq &\frac{1}{b-a}\int\limits_{a}^{x}\left( \left( b-x\right) ^{\alpha }-\left( t-x\right) ^{\alpha }\right) \left[ \left( b-t\right) \left\vert f^{\prime }(a)\right\vert +\left( t-a\right) \left\vert f^{\prime }(b)\right\vert \right] dt \\ && \\ & = &\frac{\left( b-x\right) ^{\alpha +2}}{b-a}\left[ \frac{1}{2}-\frac{1}{ \left( \alpha +1\right) \left( \alpha +2\right) }\right] \left\vert f^{\prime }(a)\right\vert \\ && \\ &&+\frac{\left( b-x\right) ^{\alpha +1}}{b-a}\left[ \left( \frac{x+b}{2} -a\right) -\frac{\left( \alpha +1\right) \left( b-x\right) +\left( \alpha +2\right) \left( x-a\right) }{\left( \alpha +1\right) \left( \alpha +2\right) }\right] \left\vert f^{\prime }(b)\right\vert . \end{eqnarray}$

(4.5)

If we substitute the equalities (4.4) and (4.5) in (4.3), then we obtain the required inequality (4.2).

Remark 7. If we choose $\alpha = 1$ in Theorem 8, then we have the following weighted Ostrowski inequality

$\begin{eqnarray*} &&\left\vert f(x)\int\limits_{a}^{b}g(t)dt-\int\limits_{a}^{b}f(t)g(t)dt\right\vert \\ && \\ &\leq &\frac{\left\Vert g\right\Vert _{\infty }}{\left( b-a\right) \Gamma \left( \alpha +1\right) }\left[ \frac{\left( x-a\right) ^{2}}{2}\left( b- \frac{a+2x}{3}\right) +\frac{\left( b-x\right) ^{3}}{3}\right] \left\vert f^{\prime }(a)\right\vert \\ && \\ &&+\left[ \frac{\left( b-x\right) ^{\alpha +1}}{2}\left( \frac{2x+b}{3} -a\right) +\frac{\left( x-a\right) ^{\alpha +2}}{3}\right] \left\vert f^{\prime }(b)\right\vert . \end{eqnarray*}$

Corollary 12. If we choose $x = \frac{a+b}{2}$ in Theorem 8, then we have the following "weighted fractional midpoint inequality"

$\begin{eqnarray*} &&\left\vert \left[ J_{a+}^{\alpha }g\left( \frac{a+b}{2}\right) +J_{b-}^{\alpha }g\left( \frac{a+b}{2}\right) \right] f\left( \frac{a+b}{2} \right) -\left[ J_{a+}^{\alpha }\left( fg\right) \left( \frac{a+b}{2}\right) +J_{b-}^{\alpha }\left( fg\right) \left( \frac{a+b}{2}\right) \right] \right\vert \\ && \\ &\leq &\frac{\left\Vert g\right\Vert _{\infty }}{\Gamma \left( \alpha +1\right) }\left( \frac{b-a}{2}\right) ^{\alpha +1}\left[ \left\vert f^{\prime }(a)\right\vert +\left\vert f^{\prime }(b)\right\vert \right] . \end{eqnarray*}$

Corollary 13. If we choose $g(t) = 1$ for all $t\in \left[a, b\right]$ in Theorem 8, then we have the following new version of fractional Ostrowski type inequality

$\begin{eqnarray*} &&\left\vert \left( \frac{\left( x-a\right) ^{\alpha }+\left( b-x\right) ^{\alpha }}{b-a}\right) f(x)-\frac{\Gamma \left( \alpha +1\right) }{b-a} \left[ J_{a+}^{\alpha }f\left( x\right) +J_{b-}^{\alpha }f\left( x\right) \right] \right\vert \\ && \\ &\leq &\frac{1}{\left( b-a\right) ^{2}\Gamma \left( \alpha +1\right) }\left[ \left( x-a\right) ^{\alpha +1}\left( \left( b-\frac{a+x}{2}\right) -\frac{ \left( \alpha +2\right) \left( b-x\right) +\left( \alpha +1\right) \left( x-a\right) }{\left( \alpha +1\right) \left( \alpha +2\right) }\right) \right. \\ && \\ &&\left. +\left( b-x\right) ^{\alpha +2}\left( \frac{1}{2}-\frac{1}{\left( \alpha +1\right) \left( \alpha +2\right) }\right) \right] \left\vert f^{\prime }(a)\right\vert \\ && \\ &&+\left[ \left( b-x\right) ^{\alpha +1}\left( \left( \frac{x+b}{2}-a\right) -\frac{\left( \alpha +1\right) \left( b-x\right) +\left( \alpha +2\right) \left( x-a\right) }{\left( \alpha +1\right) \left( \alpha +2\right) }\right) \right. \\ && \\ &&\left. +\left( x-a\right) ^{\alpha +2}\left( \frac{1}{2}-\frac{1}{\left( \alpha +1\right) \left( \alpha +2\right) }\right) \right] \left\vert f^{\prime }(b)\right\vert . \end{eqnarray*}$

Corollary 14. Under assumption of Theorem 8, let $g$ be symmetric to $\frac{a+b}{2}.$ Then we have the following inequality

$\begin{eqnarray} &&\left\vert \frac{f(a)+f(b)}{2}\left[ J_{a+}^{\alpha }g\left( b\right) +J_{b-}^{\alpha }g\left( a\right) \right] -\left[ J_{a+}^{\alpha }\left( fg\right) \left( b\right) +J_{b-}^{\alpha }\left( fg\right) \left( a\right) \right] \right\vert \\ && \\ &\leq &\frac{\left( b-a\right) ^{\alpha +1}\left\Vert g\right\Vert _{\infty } }{\Gamma \left( \alpha +1\right) }\left( 1-\frac{1}{\alpha +1}\right) \left[ \left\vert f^{\prime }(a)\right\vert +\left\vert f^{\prime }(b)\right\vert \right] \end{eqnarray}$

(4.6)

Proof. If we apply the inequality (4.2) for $x = a$ and $x = b,$ then we obtain the inequalities

$\begin{eqnarray} &&\left\vert f(a)J_{b-}^{\alpha }g\left( a\right) -J_{b-}^{\alpha }\left( fg\right) \left( a\right) \right\vert \\ && \\ &\leq &\frac{\left( b-a\right) ^{\alpha +1}\left\Vert g\right\Vert _{\infty } }{\Gamma \left( \alpha +1\right) }\left[ \left( \frac{1}{2}-\frac{1}{\left( \alpha +1\right) \left( \alpha +2\right) }\right) \left\vert f^{\prime }(a)\right\vert +\left( \frac{1}{2}-\frac{1}{\alpha +2}\right) \left\vert f^{\prime }(b)\right\vert \right] \end{eqnarray}$

(4.7)

and

$\begin{eqnarray} &&\left\vert f(b)J_{a+}^{\alpha }g\left( b\right) -J_{a+}^{\alpha }\left( fg\right) \left( b\right) \right\vert \\ && \\ &\leq &\frac{\left( b-a\right) ^{\alpha +1}\left\Vert g\right\Vert _{\infty } }{\Gamma \left( \alpha +1\right) }\left[ \left( \frac{1}{2}-\frac{1}{\alpha +2}\right) \left\vert f^{\prime }(a)\right\vert +\left( \frac{1}{2}-\frac{1}{ \left( \alpha +1\right) \left( \alpha +2\right) }\right) \left\vert f^{\prime }(b)\right\vert \right] \end{eqnarray}$

(4.8)

respectively. If we add the inequalities (4.7) and (4.8), then by using triangle inequality and Lemma 2, we have

$\begin{eqnarray*} &&\left\vert \frac{f(a)+f(b)}{2}\left[ J_{a+}^{\alpha }g\left( b\right) +J_{b-}^{\alpha }g\left( a\right) \right] -\left[ J_{a+}^{\alpha }\left( fg\right) \left( b\right) +J_{b-}^{\alpha }\left( fg\right) \left( a\right) \right] \right\vert \\ && \\ &\leq &\frac{\left( b-a\right) ^{\alpha +1}\left\Vert g\right\Vert _{\infty } }{\Gamma \left( \alpha +1\right) }\left[ \left( \frac{1}{2}-\frac{1}{\left( \alpha +1\right) \left( \alpha +2\right) }\right) \left\vert f^{\prime }(a)\right\vert +\left( \frac{1}{2}-\frac{1}{\alpha +2}\right) \left\vert f^{\prime }(b)\right\vert \right] \\ && \\ &&+\frac{\left( b-a\right) ^{\alpha +1}\left\Vert g\right\Vert _{\infty }}{ \Gamma \left( \alpha +1\right) }\left[ \left( \frac{1}{2}-\frac{1}{\alpha +2} \right) \left\vert f^{\prime }(a)\right\vert +\left( \frac{1}{2}-\frac{1}{ \left( \alpha +1\right) \left( \alpha +2\right) }\right) \left\vert f^{\prime }(b)\right\vert \right] \\ && \\ & = &\frac{\left( b-a\right) ^{\alpha +1}\left\Vert g\right\Vert _{\infty }}{ \Gamma \left( \alpha +1\right) }\left( 1-\frac{1}{\alpha +1}\right) \left[ \left\vert f^{\prime }(a)\right\vert +\left\vert f^{\prime }(b)\right\vert \right] \end{eqnarray*}$

which completes the proof.

Theorem 9. Let $f:\left[a, b\right] \rightarrow \mathbb{R}$ be a differentiable mapping on $(a, b)$ with $a < b$ and $f^{\prime }\in L\left[a, b \right].$ If $g:\left[a, b\right] \rightarrow \mathbb{R}$ is continuous and if $\left\vert f^{\prime }\right\vert ^{q},$ $q > 1,$ is convex on $\left[a, b \right],$ then for all $x\in \left[a, b\right]$ and $\alpha > 0$ we have the following inequality for fractional integrals,

(4.9)

where $\frac{1}{p}+\frac{1}{q} = 1.$

Proof. By applying Hölder inequality in (4.3) and by using the fact that

$\begin{equation*} \left( A-B\right) ^{p}\leq A^{p}-B^{p} \end{equation*}$

for any $A > B\geq 0$ and $p\geq 1$ , we obtain

$\begin{eqnarray} &&\left\vert \left[ J_{a+}^{\alpha }g\left( x\right) +J_{b-}^{\alpha }g\left( x\right) \right] f(x)-\left[ J_{a+}^{\alpha }\left( fg\right) \left( x\right) +J_{b-}^{\alpha }\left( fg\right) \left( x\right) \right] \right\vert \\ && \\ &\leq &\frac{\left\Vert g\right\Vert _{\infty }}{\Gamma \left( \alpha +1\right) }\left[ \left( \int\limits_{a}^{x}\left( \left( x-a\right) ^{\alpha }-\left( x-t\right) ^{\alpha }\right) ^{p}dt\right) ^{\frac{1}{p} }\left( \int\limits_{a}^{x}\left\vert f^{\prime }(t)\right\vert ^{q}dt\right) ^{\frac{1}{q}}\right. \\ && \\ &&\left. +\left( \int\limits_{x}^{b}\left( \left( b-x\right) ^{\alpha }-\left( t-x\right) ^{\alpha }\right) ^{p}dt\right) ^{\frac{1}{p}}\left( \int\limits_{x}^{b}\left\vert f^{\prime }(t)\right\vert ^{q}dt\right) ^{ \frac{1}{q}}\right] \\ && \\ &\leq &\frac{\left\Vert g\right\Vert _{\infty }}{\Gamma \left( \alpha +1\right) }\left[ \left( \int\limits_{a}^{x}\left( \left( x-a\right) ^{p\alpha }-\left( x-t\right) ^{p\alpha }\right) dt\right) ^{\frac{1}{p} }\left( \int\limits_{a}^{x}\left\vert f^{\prime }(t)\right\vert ^{q}dt\right) ^{\frac{1}{q}}\right. \\ && \\ &&\left. +\left( \int\limits_{x}^{b}\left( \left( b-x\right) ^{p\alpha }-\left( t-x\right) ^{p\alpha }\right) dt\right) ^{\frac{1}{p}}\left( \int\limits_{x}^{b}\left\vert f^{\prime }(t)\right\vert ^{q}dt\right) ^{ \frac{1}{q}}\right] . \end{eqnarray}$

(4.10)

By simple calculations, we establish

$\begin{equation} \int\limits_{a}^{x}\left( \left( x-a\right) ^{p\alpha }-\left( x-t\right) ^{p\alpha }\right) dt = \left( x-a\right) ^{p\alpha +1}\left( 1-\frac{1}{ p\alpha +1}\right) \end{equation}$

(4.11)

and

$\begin{equation} \int\limits_{x}^{b}\left( \left( b-x\right) ^{p\alpha }-\left( t-x\right) ^{p\alpha }\right) dt = \left( b-x\right) ^{p\alpha +1}\left( 1-\frac{1}{ p\alpha +1}\right) . \end{equation}$

(4.12)

Since $\left\vert f^{\prime }\right\vert ^{q}$ is convex on $\left[a, b \right],$ then we get

$\begin{eqnarray} \int\limits_{a}^{x}\left\vert f^{\prime }(t)\right\vert ^{q}dt &\leq &\frac{1 }{b-a}\int\limits_{a}^{x}\left( \left( b-t\right) \left\vert f^{\prime }(a)\right\vert ^{q}+\left( t-a\right) \left\vert f^{\prime }(b)\right\vert ^{q}\right) \\ &\leq &\frac{x-a}{b-a}\left[ \left( b-\frac{a+x}{2}\right) \left\vert f^{\prime }(a)\right\vert ^{q}+\frac{x-a}{2}\left\vert f^{\prime }(b)\right\vert ^{q}\right] \end{eqnarray}$

(4.13)

and similarly

$\begin{equation} \int\limits_{x}^{b}\left\vert f^{\prime }(t)\right\vert ^{q}dt\leq \frac{b-x }{b-a}\left[ \frac{b-x}{2}\left\vert f^{\prime }(a)\right\vert ^{q}+\left( \frac{x+b}{2}-a\right) \left\vert f^{\prime }(b)\right\vert ^{q}\right] . \end{equation}$

(4.14)

If we put (4.11)–(4.11) in (4.10), then we have

$\begin{eqnarray*} &&\left\vert \left[ J_{a+}^{\alpha }g\left( x\right) +J_{b-}^{\alpha }g\left( x\right) \right] f(x)-\left[ J_{a+}^{\alpha }\left( fg\right) \left( x\right) +J_{b-}^{\alpha }\left( fg\right) \left( x\right) \right] \right\vert \\ && \\ &\leq &\frac{\left\Vert g\right\Vert _{\infty }}{\Gamma \left( \alpha +1\right) }\left[ \left( \left( x-a\right) ^{p\alpha +1}\left( 1-\frac{1}{ p\alpha +1}\right) \right) ^{\frac{1}{p}}\left( \frac{x-a}{b-a}\left[ \left( b-\frac{a+x}{2}\right) \left\vert f^{\prime }(a)\right\vert ^{q}+\frac{x-a}{2 }\left\vert f^{\prime }(b)\right\vert ^{q}\right] \right) ^{\frac{1}{q} }\right. \\ && \\ &&\left. +\left( \left( b-x\right) ^{p\alpha +1}\left( 1-\frac{1}{p\alpha +1} \right) \right) ^{\frac{1}{p}}\left( \frac{b-x}{b-a}\left[ \frac{b-x}{2} \left\vert f^{\prime }(a)\right\vert ^{q}+\left( \frac{x+b}{2}-a\right) \left\vert f^{\prime }(b)\right\vert ^{q}\right] \right) ^{\frac{1}{q}} \right] \\ && \\ & = &\frac{\left\Vert g\right\Vert _{\infty }}{(b-a)^{\frac{1}{q}}\Gamma \left( \alpha +1\right) }\left( 1-\frac{1}{p\alpha +1}\right) ^{\frac{1}{p}} \left[ \left( x-a\right) ^{\alpha +1}\left( \left( b-\frac{a+x}{2}\right) \left\vert f^{\prime }(a)\right\vert ^{q}+\frac{x-a}{2}\left\vert f^{\prime }(b)\right\vert ^{q}\right) ^{\frac{1}{q}}\right. \\ && \\ &&\left. +\left( b-x\right) ^{\alpha +1}\left( \frac{b-x}{2}\left\vert f^{\prime }(a)\right\vert ^{q}+\left( \frac{x+b}{2}-a\right) \left\vert f^{\prime }(b)\right\vert ^{q}\right) ^{\frac{1}{q}}\right] \end{eqnarray*}$

which completes the proof.

Remark 8. If we choose $\alpha = 1$ in Theorem 9, then we have the following weighted Ostrowski inequality

$\begin{eqnarray*} &&\left\vert f(x)\int\limits_{a}^{b}g(t)dt-\int\limits_{a}^{b}f(t)g(t)dt\right\vert \\ && \\ &\leq &\frac{\left\Vert g\right\Vert _{\infty }}{(b-a)^{\frac{1}{q}}}\left( 1-\frac{1}{p+1}\right) ^{\frac{1}{p}}\left[ \left( x-a\right) ^{2}\left( \left( b-\frac{a+x}{2}\right) \left\vert f^{\prime }(a)\right\vert ^{q}+ \frac{x-a}{2}\left\vert f^{\prime }(b)\right\vert ^{q}\right) ^{\frac{1}{q} }\right. \\ && \\ &&\left. +\left( b-x\right) ^{2}\left( \frac{b-x}{2}\left\vert f^{\prime }(a)\right\vert ^{q}+\left( \frac{x+b}{2}-a\right) \left\vert f^{\prime }(b)\right\vert ^{q}\right) ^{\frac{1}{q}}\right] . \end{eqnarray*}$

Corollary 15. If we choose $x = \frac{a+b}{2}$ in Theorem 9, then we have the following "weighted fractional midpoint inequality"

$\begin{eqnarray} &&\left\vert \left[ J_{a+}^{\alpha }g\left( \frac{a+b}{2}\right) +J_{b-}^{\alpha }g\left( \frac{a+b}{2}\right) \right] f(\frac{a+b}{2})-\left[ J_{a+}^{\alpha }\left( fg\right) \left( \frac{a+b}{2}\right) +J_{b-}^{\alpha }\left( fg\right) \left( \frac{a+b}{2}\right) \right] \right\vert \\ && \\ &\leq &\frac{\left\Vert g\right\Vert _{\infty }}{\Gamma \left( \alpha +1\right) }\left( 1-\frac{1}{p\alpha +1}\right) ^{\frac{1}{p}}\left( \frac{ b-a}{2}\right) ^{\alpha +1} \\ && \\ &&\times \left[ \left( \frac{3\left\vert f^{\prime }(a)\right\vert ^{q}+\left\vert f^{\prime }(b)\right\vert ^{q}}{4}\right) ^{\frac{1}{q} }+\left( \frac{\left\vert f^{\prime }(a)\right\vert ^{q}+3\left\vert f^{\prime }(b)\right\vert ^{q}}{4}\right) ^{\frac{1}{q}}\right] \\ && \\ &\leq &\frac{\left\Vert g\right\Vert _{\infty }}{\Gamma \left( \alpha +1\right) }\left( 4-\frac{4}{p\alpha +1}\right) ^{\frac{1}{p}}\left( \frac{ b-a}{2}\right) ^{\alpha +1}\left[ \left\vert f^{\prime }(a)\right\vert +\left\vert f^{\prime }(b)\right\vert \right] . \end{eqnarray}$

(4.15)

Proof. The proof of the first inequality in (4.15) is obvious. For the proof of second inequality, let $a_{1} = 3\left\vert f^{\prime }\left(a\right) \right\vert ^{q},$ $b_{1} = \left\vert f^{\prime }\left(b\right) \right\vert ^{q},$ $a_{2} = \left\vert f^{\prime }\left(a\right) \right\vert ^{q}$ and $b_{2} = 3\left\vert f^{\prime }\left(b\right) \right\vert ^{q}.$ Using the facts that

$\begin{equation*} \sum\limits_{k = 1}^{n}\left( a_{k}+b_{k}\right) ^{s}\leq \sum\limits_{k = 1}^{n}a_{k}^{s}+\sum\limits_{k = 1}^{n}b_{k}^{s}, \text{ }0\leq s \lt 1 \end{equation*}$

and $3^{\frac{1}{q}}+1\leq 4,$ the desired result can be obtained straightforwardly.

Corollary 16. If we choose $g(t) = 1$ for all $t\in \left[a, b\right]$ in Theorem 9, then we have

$\begin{eqnarray*} &&\left\vert \left( \frac{\left( x-s\right) ^{\alpha }+\left( b-x\right) ^{\alpha }}{b-a}\right) f(x)-\frac{\Gamma \left( \alpha +1\right) }{b-a} \left[ J\right] _{a+}^{\alpha }f\left( x\right) +J_{b-}^{\alpha }f\left( x\right) \right\vert \\ && \\ &\leq &\frac{\left( 1-\frac{1}{p\alpha +1}\right) ^{\frac{1}{p}}}{(b-a)^{1+ \frac{1}{q}}\Gamma \left( \alpha +1\right) }\left[ \left( x-a\right) ^{\alpha +1}\left( \left( b-\frac{a+x}{2}\right) \left\vert f^{\prime }(a)\right\vert ^{q}+\frac{x-a}{2}\left\vert f^{\prime }(b)\right\vert ^{q}\right) ^{\frac{1}{q}}\right. \\ && \\ &&\left. +\left( b-x\right) ^{\alpha +1}\left( \frac{b-x}{2}\left\vert f^{\prime }(a)\right\vert ^{q}+\left( \frac{x+b}{2}-a\right) \left\vert f^{\prime }(b)\right\vert ^{q}\right) ^{\frac{1}{q}}\right] \end{eqnarray*}$

Corollary 17. Under assumption of Theorem 9, let $g$ be symmetric to $\frac{a+b}{2}.$ Then we have the following inequality

$\begin{eqnarray} &&\left\vert \frac{f(a)+f(b)}{2}\left[ J_{a+}^{\alpha }g\left( b\right) +J_{b-}^{\alpha }g\left( a\right) \right] -\left[ J_{a+}^{\alpha }\left( fg\right) \left( b\right) +J_{b-}^{\alpha }\left( fg\right) \left( a\right) \right] \right\vert \\ && \\ &\leq &\frac{2\left( b-a\right) ^{\alpha +1}}{\Gamma \left( \alpha +1\right) }\left( 1-\frac{1}{p\alpha +1}\right) ^{\frac{1}{p}}\left( \frac{\left\vert f^{\prime }(a)\right\vert ^{q}+\left\vert f^{\prime }(b)\right\vert }{2} ^{q}\right) ^{\frac{1}{q}}\left\Vert g\right\Vert _{\infty }. \end{eqnarray}$

(4.16)

Proof. If we write the inequality (4.9) for $x = a$ and $x = b,$ then we obtain the inequalities

$\begin{equation} \left\vert f(a)J_{b-}^{\alpha }g\left( a\right) -J_{b-}^{\alpha }\left( fg\right) \left( a\right) \right\vert \leq \frac{\left\Vert g\right\Vert _{\infty }}{\Gamma \left( \alpha +1\right) }\left( 1-\frac{1}{p\alpha +1} \right) ^{\frac{1}{p}}\left( b-a\right) ^{\alpha +1}\left( \frac{\left\vert f^{\prime }(a)\right\vert ^{q}+\left\vert f^{\prime }(b)\right\vert }{2} ^{q}\right) ^{\frac{1}{q}} \end{equation}$

(4.17)

and

$\begin{equation} \left\vert f(b)J_{a+}^{\alpha }g\left( a\right) -J_{a+}^{\alpha }\left( fg\right) \left( b\right) \right\vert \leq \frac{\left\Vert g\right\Vert _{\infty }}{\Gamma \left( \alpha +1\right) }\left( 1-\frac{1}{p\alpha +1} \right) ^{\frac{1}{p}}\left( b-a\right) ^{\alpha +1}\left( \frac{\left\vert f^{\prime }(a)\right\vert ^{q}+\left\vert f^{\prime }(b)\right\vert ^{q}}{2} \right) ^{\frac{1}{q}} \end{equation}$

(4.18)

respectively. If we add the inequalities (4.7) and (4.8), then by using triangle inequality and Lemma 2, we have

$\begin{eqnarray*} &&\left\vert \frac{f(a)+f(b)}{2}\left[ J_{a+}^{\alpha }g\left( b\right) +J_{b-}^{\alpha }g\left( a\right) \right] -\left[ J_{a+}^{\alpha }\left( fg\right) \left( b\right) +J_{b-}^{\alpha }\left( fg\right) \left( a\right) \right] \right\vert \\ && \\ &\leq &\frac{\left\Vert g\right\Vert _{\infty }}{\Gamma \left( \alpha +1\right) }\left( 1-\frac{1}{p\alpha +1}\right) ^{\frac{1}{p}}\left( b-a\right) ^{\alpha +1}\left( \frac{\left\vert f^{\prime }(a)\right\vert ^{q}+\left\vert f^{\prime }(b)\right\vert }{2}^{q}\right) ^{\frac{1}{q}} \\ && \\ &&+\frac{\left\Vert g\right\Vert _{\infty }}{\Gamma \left( \alpha +1\right) } \left( 1-\frac{1}{p\alpha +1}\right) ^{\frac{1}{p}}\left( b-a\right) ^{\alpha +1}\left( \frac{\left\vert f^{\prime }(a)\right\vert ^{q}+\left\vert f^{\prime }(b)\right\vert ^{q}}{2}\right) ^{\frac{1}{q}} \\ && \\ & = &\frac{2\left( b-a\right) ^{\alpha +1}}{\Gamma \left( \alpha +1\right) } \left( 1-\frac{1}{p\alpha +1}\right) ^{\frac{1}{p}}\left( \frac{\left\vert f^{\prime }(a)\right\vert ^{q}+\left\vert f^{\prime }(b)\right\vert }{2} ^{q}\right) ^{\frac{1}{q}}\left\Vert g\right\Vert _{\infty }. \end{eqnarray*}$

This completes the proof.

Conflict of interest

The authors declare no conflict of interest.

References

[1]	R. P. Agarwal, M. J. Luo, R. K. Raina, On Ostrowski type inequalities, Fasciculi Math., 56 (2016), 5-27. doi: 10.1515/fascmath-2016-0001
[2]	M. Alomari, M. Darus, U. S. Kirmaci, Refinements of Hadamard-type inequalities for quasi-convex functions with applications to trapezoidal formula and to special means, Comput. Math. Appl., 59 (2010), 225-232. doi: 10.1016/j.camwa.2009.08.002
[3]	G. A. Anastassiou, General fractional Hermite-Hadamard inequalities using m-convexity and (s, m)-convexity, Front. Time Scales Inequal., 237 (2016), 255.
[4]	A. G. Azpeitia, Convex functions and the Hadamard inequality, Rev. Colombiana Math., 28 (1994), 7-12.
[5]	Y. Başcı, D. Baleanu, Ostrowski type Inequalities Involving ψ-Hilfer fractional Integrals, Mathematics, 7 (2019), 770.
[6]	H. Budak, M. Z. Sarikaya, E. Set, Generalized Ostrowski type inequalities for functions whose local fractional derivatives are generalized s-convex in the second sense, J. Appl. Math. Comput. Mech., 15 (2016), 11-22.
[7]	J. de la Cal, J. Carcamob, L. Escauriaza, A general multidimensional Hermite-Hadamard type inequality, J. Math. Anal. Appl., 356 (2009), 659-663. doi: 10.1016/j.jmaa.2009.03.044
[8]	H. Chen, U. N. Katugampola, Hermite-Hadamard and Hermite-Hadamard-Fejér type inequalities for generalized fractional integrals, J. Math. Anal. Appl., 446 (2017), 1274-1291. doi: 10.1016/j.jmaa.2016.09.018
[9]	S. S. Dragomir, C. E. M. Pearce, Selected Topics on Hermite-Hadamard Inequalities and Applications, RGMIA Monographs, Victoria University, 2000, 2004, Available from: http://rgmia.org/papers/monographs/Master2.pdf.
[10]	S. S. Dragomir, R. P. Agarwal, Two inequalities for differentiable mappings and applications to special means of real numbers and to trapezoidal formula, Appl. Math. lett., 11 (1998), 91-95.
[11]	S. S. Dragomir, Ostrowski type inequalities for Riemann-Liouville fractional integrals of bounded variation, Hölder and Lipschitzian functions, RGMIA Res. Rep. Coll., 20 (2017), 48.
[12]	G. Farid, A. ur Rehman, M. Zahra, On Hadamard type inequalities for k-fractional integrals, Konurap J. Math., 4 (2016), 79-86.
[13]	L. Fejer, Über die Fourierreihen, II. Math. Naturwiss. Anz Ungar. Akad. Wiss., 24 (1906), 369-390.
[14]	R. Gorenflo, F. Mainardi, Fractional calculus: Integral and differential equations of fractional order, Springer Verlag, Wien, 1997, 223-276.
[15]	A. Guezane-Lakoud, F. Aissaoui, New fractional inequalities of Ostrowski type, Transylv. J. Math. Mech., 5 (2013), 103-106
[16]	M. Gúrbúz, Y. Taşdan, E. Set, Ostrowski type inequalities via the Katugampola fractional integrals, AIMS Mathematics, 5 (2020), 42-53. doi: 10.3934/math.2020004
[17]	M. Iqbal, S. Qaisar, M. Muddassar, A short note on integral inequality of type Hermite-Hadamard through convexity, J. Comput. Anal. Appl., 21 (2016), 946-953.
[18]	I. Iscan, Hermite-Hadamard-Fejer type inequalities for convex functions via fractional integrals, Stud. U. Babeş Bol. Math., 60 (2015), 355-366
[19]	İ. İşcan, On generalization of different type integral inequalities for s-convex functions via fractional integrals, Math. Sci. Appl., 2 (2014), 55-67.
[20]	M. Jleli, B. Samet, On Hermite-Hadamard type inequalities via fractional integrals of a function with respect to another function, J. Nonlinear Sci. Appl., 9 (2016), 1252-1260. doi: 10.22436/jnsa.009.03.50
[21]	A. A. Kilbas, H. M. Srivastava, J. J. Trujillo, Theory and Applications of Fractional Differential Equations, Elsevier Science Limited, Amsterdam, 2006.
[22]	M. Kirane, B. T. Torebek, Hermite-Hadamard, Hermite-Hadamard-Fejer, Dragomir-Agarwal and Pachpatte type inequalities for convex functions via fractional integrals, arXiv:1701.00092.
[23]	U. S. Kirmaci, Inequalities for differentiable mappings and applications to special means of real numbers to midpoint formula, Appl. Math. Comput., 147 (2004), 137-146.
[24]	W. Liu, A. Tuna, Diamond-α weighted Ostrowski type and Grüss type inequalities on time scales, Appl. Math. Comput., 270 (2015), 251-260.
[25]	W. Liu, X. Gao, Y. Wen, Approximating the finite Hilbert transform via some companions of Ostrowski's inequalities, B. Malays. Math. Sci. So., 39 (2015), 1499-1513.
[26]	W. Liu, A. Tuna, Y. Jiang, On weighted Ostrowski type, Trapezoid type, Grüss type and Ostrowski- Grüss like inequalities on time scales, Appl. Anal., 93 (2013), 551-571.
[27]	S. Miller, B. Ross, An Introduction to the Fractional Calculus and Fractional Differential Equations, John Wiley & Sons, USA, 1993.
[28]	M. A. Noor, M. U. Awan, Some integral inequalities for two kinds of convexities via fractional integrals, TJMM, 5 (2013), 129-136.
[29]	M. A. Noor, K. I. Noor, M. U. Awan, Fractional Ostrowski inequalities for s-Godunova-Levin functions, Int. J. Anal. Appl., 5 (2014), 167-173.
[30]	J. E. Pečarić, F. Proschan, Y. L. Tong, Convex Functions, Partial Orderings and Statistical Applications, Academic Press, Boston, 1992.
[31]	A. M. Ostrowski, Über die absolutabweichung einer differentiebaren funktion von ihrem integralmitelwert, Comment. Math. Helv. 10 (1938), 226-227.
[32]	M. E. Özdemir, M. Avcı-Ardıç, H. Kavurmacı-Önalan, Hermite-Hadamard type inequalities for s-convex and s-concave functions via fractional integrals, Turk. J. Sci., 1 (2016), 28-40.
[33]	M. E. Ödemir, M. Avci, E. Set, On some inequalities of Hermite-Hadamard-type via m-convexity, Appl. Math. Lett., 23 (2010), 1065-1070. doi: 10.1016/j.aml.2010.04.037
[34]	M. E. Ödemir, M. Avci, H. Kavurmaci, Hermite-Hadamard-type inequalities via (α, m)-convexity, Comput. Math. Appl., 61 (2011), 2614-2620. doi: 10.1016/j.camwa.2011.02.053
[35]	I. Podlubni, Fractional Differential Equations, Academic Press, San Diego, 1999.
[36]	S. Qaisar, S. Hussain, On Hermite-Hadamard type inequalities for functions whose first derivative absolute values are convex and concave, Fasciculi Math., 58 (2017), 155-166. doi: 10.1515/fascmath-2017-0011
[37]	A. Saglam, M. Z. Sarikaya, H. Yildirim, Some new inequalities of Hermite-Hadamard's type, Kyungpook Math. J., 50 (2010), 399-410. doi: 10.5666/KMJ.2010.50.3.399
[38]	M. Z. Sarikaya, H. Budak, Generalized Ostrowski type inequalities for local fractional integrals, P. Am. Math. Soc., 145 (2017), 1527-1538.
[39]	M. Z. Sarikaya, H. Filiz, Note on the Ostrowski type inequalities for fractional integrals, Vietnam J. Math., 42 (2014), 187-190. doi: 10.1007/s10013-014-0056-4
[40]	M. Z. Sarikaya, E. Set, H. Yaldiz, et al. Hermite-Hadamard's inequalities for fractional integrals and related fractional inequalities, Math. Comput. Model., 57 (2013), 2403-2407. doi: 10.1016/j.mcm.2011.12.048
[41]	M. Z. Sarikaya, H. Yildirim, On Hermite-Hadamard type inequalities for Riemann-Liouville fractional integrals, Miskolc Math. Notes, 7 (2016), 1049-1059.
[42]	M. Z. Sarikaya, H. Budak, Generalized Hermite-Hadamard type integral inequalities for fractional integrals, Filomat, 30 (2016), 1315-1326. doi: 10.2298/FIL1605315S
[43]	M. Z. Sarikaya, A. Akkurt, H. Budak, et al. Hermite-hadamard's inequalities for conformable fractional integrals, RGMIA Res. Rep. Coll., 19 (2016), 58140.
[44]	E. Set, M. E. Ozdemir, M. Z. Sarikaya, New inequalities of Ostrowski's type for s-convex functions in the second sense with applications, Facta U. Ser. Math. Inform., 27 (2012), 67-82.
[45]	E. Set, New inequalities of Ostrowski type for mappings whose derivatives are s-convex in the second sense via fractional integrals, Comput. Math. Appl., 63 (2012), 1147-1154. doi: 10.1016/j.camwa.2011.12.023
[46]	E. Set, M. Z. Sarikaya, M. E. Ozdemir, et al. The Hermite-Hadamard's inequality for some convex functions via fractional integrals and related results, J. Appl. Math. Stat. Inform., 10 (2014), 69-83. doi: 10.2478/jamsi-2014-0014
[47]	E. Set,İ. İşcan, M. Z. Sarikaya, et al. On new inequalities of Hermite-Hadamard-Fejért ype for convex functions via fractional integrals, Appl. Math. Comput., 259 (2015), 875-881.
[48]	E. Set, A. O. Akdemir, I. Mumcu, Ostrowski type inequalities for functions whoose derivatives are convex via conformable fractional integrals, J. Adv. Math. Stud., 10 (2017), 386-395.
[49]	E. Set, A. O. Akdemir, A. Gözpınar, et al. Ostrowski type inequalities via new fractional conformable integrals, AIMS Mathematics, 4 (2019), 1684-1697. doi: 10.3934/math.2019.6.1684
[50]	S. F. Tahir, M. Mushtaq, M. Muddassar, A Note on integral inequalities on time scales associated with Ostrowski's type, J. Funct. Space., 2019 (2019), 4748373.
[51]	K. L. Tseng, G. S. Yang, K. C. Hsu, Some inequalities for differentiable mappings and applications to Fejer inequality and weighted trapozidal formula, Taiwanese J. Math., 15 (2011), 1737-1747. doi: 10.11650/twjm/1500406376
[52]	J. Wang, X. Li, M. Fečkan, et al. Hermite-Hadamard-type inequalities for Riemann-Liouville fractional integrals via two kinds of convexity, Appl. Anal., 92 (2012), 2241-2253.
[53]	J. Wang, X. Li, C. Zhu, Refinements of Hermite-Hadamard type inequalities involving fractional integrals, Bull. Belg. Math. Soc. Simon Stevin, 20 (2013), 655-666. doi: 10.36045/bbms/1382448186
[54]	H. Yildirim, Z. Kirtay, Ostrowski inequality for generalized fractional integral and related inequalities, Malaya J. Mat., 2 (2014), 322-329.
[55]	C. Yildiz, M. E. Özdemir, M. Z. Sarikaya, New generalizations of Ostrowski-like type inequalities for fractional integrals, Kyungpook Math. J., 56 (2016), 161-172. doi: 10.5666/KMJ.2016.56.1.161
[56]	Y. Zhang, J. Wang, On some new Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals, J. Inequal. Appl., 2013 (2013), 220.

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5.	Humaira Kalsoom, Muhammad Amer Latif, Some weighted Hadamard and Ostrowski-type fractional inequalities for quasi-geometrically convex functions, 2023, 37, 0354-5180, 5921, 10.2298/FIL2318921K
6.	Fatih Hezenci, Martin Bohner, Hüseyin Budak, Fractional midpoint-type inequalities for twice-differentiable functions, 2023, 37, 0354-5180, 8131, 10.2298/FIL2324131H
7.	Ghulam Farid, Hafsa Tariq, Ferdous M. O. Tawfiq, Jong-Suk Ro, Saira Zainab, A Generalized Convexity and Inequalities Involving the Unified Mittag–Leffler Function, 2023, 12, 2075-1680, 795, 10.3390/axioms12080795
8.	Muhammad Tariq, Sotiris K. Ntouyas, Bashir Ahmad, Ostrowski-Type Fractional Integral Inequalities: A Survey, 2023, 3, 2673-9321, 660, 10.3390/foundations3040040
9.	YUNXIU ZHOU, JIAGEN LIAO, TINGSONG DU, THE WEIGHTED PARAMETERIZED INEQUALITIES IN RELATION TO TWICE DIFFERENTIABLE MAPPINGS IN THE FRACTAL DOMAINS ALONG WITH SOME APPLICATIONS, 2023, 31, 0218-348X, 10.1142/S0218348X23500925
10.	Fatih Hezenci, Inequalities with parameters for twice-differentiable functions involving Riemann-Liouville fractional integrals, 2024, 38, 0354-5180, 3275, 10.2298/FIL2409275H

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AIMS Mathematics

Weighted Ostrowski, trapezoid and midpoint type inequalities for RiemannLiouville fractional integrals

Related Papers:

Abstract

1. Introduction

2. Some weighted fractional equalities

3. Weighted fractional trapezoid type inequalities

4. Weighted fractional Ostrowski and midpoint type inequalities

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AIMS Mathematics

Weighted Ostrowski, trapezoid and midpoint type inequalities for RiemannLiouville fractional integrals

Related Papers:

Abstract

1. Introduction

2. Some weighted fractional equalities

3. Weighted fractional trapezoid type inequalities

4. Weighted fractional Ostrowski and midpoint type inequalities

Conflict of interest

References

This article has been cited by:

Reader Comments

通讯作者: 陈斌, bchen63@163.com

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