Fractional Weighted Ostrowski-Type Inequalities and Their Applications

Abstract

An important area in the field of applied and pure mathematics is the integral inequality. As it is known, inequalities aim to develop different mathematical methods. Nowadays, we need to seek accurate inequalities for proving the existence and uniqueness of the mathematical methods. The concept of convexity plays a strong role in the field of inequalities due to the behavior of its definition and its properties. Furthermore, there is a strong correlation between convexity and symmetry concepts. Whichever one we work on, we can apply it to the other one due the strong correlation produced between them, especially in the last few years. In this study, by using a new identity, we establish some new fractional weighted Ostrowski-type inequalities for differentiable quasi-convex functions. Further, further results for functions with a bounded first derivative are given. Finally, in order to illustrate the efficiency of our main results, some applications to special means are obtain. The obtained results generalize and refine certain known results.

1. Introduction

Computational and Fractional Analysis are nowadays more and more in the center of mathematics and of other related sciences either by themselves because of their rapid development, which is based on very old foundations, or because they cover a great variety of applications in the real world. In current years, fractional calculus $(FC)$ applied in many phenomena in applied sciences, fluid mechanics, physics and other biology can be described as very effective using mathematical tools of $FC$. The fractional derivatives have occurred in many applied sciences equations such as reaction and diffusion processes, system identification, velocity signal analysis, relaxation of damping behaviour fabrics and creeping of polymer composites [1,2,3,4].

Definition 1

([5]). Let I be an interval of real numbers. A function $\phi :I\to \mathbb{R}$ is said to be convex, if for all $x,y\in I$ and all $t\in [0,1]$, we have

$$\phi \left(tx+\left(1-t\right)y\right)\le t\phi \left(x\right)+\left(1-t\right)\phi (y).$$

The concept of convex functions has been also generalized in diverse manners. One of them is the quasi-convex function defined as follows:

Definition 2

([6]). A function $\phi :I\to \mathbb{R}$ is said to be quasi-convex, if

$$\phi (tx+(1-t)y)\le max\left\{\phi (x),\phi (y)\right\}$$

holds for all $x,y\in I$ and $t\in [0,1]$.

In 1938 Ostrowski ([7]) proved the following Ostrowski inequality:

Theorem 1

([7]). Let φ be a differentiable function defined on the finite interval $[a,b]$, whose derivative is integrable and bounded over $[a,b],$ $i.e.,$ ${∥{\phi }^{\prime }∥}_{\infty }:={sup}_{x\in (a,b)}\left|{\phi }^{\prime }(x)\right|<\infty ,$ then

$$\left|\phi (x)-\frac{1}{b-a}{\int }_a^b\phi (u)du\right|\le \left[\frac{1}{4}+\frac{{\left(x-\frac{a+b}{2}\right)}^2}{{(b-a)}^2}\right](b-a){∥{\phi }^{\prime }∥}_{\infty }.$$

(1)

Several generalizations, improvement, and variants of such type of inequality have been obtained, we refer readers to [8,9,10,11,12,13,14,15,16,17,18,19,20,21] and references therein.

The following notations will be used in the sequel. We denote, respectively $I^{\circ }$ the interior of I and $L[a,b]$ the set of all integrable functions on $[a,b].$

In [22], Alomari et al. gave the following midpoint type inequalities for differentiable quasi-convex.

Theorem 2

([22]). Let $\phi :I\subset [0,\infty )\to \mathbb{R}$ be a differentiable mapping on $I^{\circ }$ such that ${\phi }^{\prime }\in L[a,b]$, where $a,b\in I$ with $a<b$. If $\left|{\phi }^{\prime }\right|$ is quasi-convex on $[a,b]$, then the following inequality holds:

$$\begin{array}{cc}\hfill & \left|\frac{1}{b-a}\stackrel{b}{\underset{a}{\int }}\phi (u)du-\phi \left(\frac{a+b}{2}\right)\right|\le \frac{b-a}{8}(max\left\{\left|{\phi }^{\prime }\left(a\right)\right|,\left|{\phi }^{\prime }\left(\frac{a+b}{2}\right)\right|\right\}\hfill \\ \hfill & +max\left\{\left|{\phi }^{\prime }\left(\frac{a+b}{2}\right)\right|,\left|{\phi }^{\prime }\left(b\right)\right|\right\}).\hfill \end{array}$$

Theorem 3

([22]). Let $\phi :I\subset [0,\infty )\to \mathbb{R}$ be a differentiable mapping on $I^{\circ }$ such that ${\phi }^{\prime }\in L[a,b]$, where $a,b\in I$ with $a<b$. If ${\left|{\phi }^{\prime }\right|}^q$ is quasi-convex on $[a,b]$, $q>1$ with $\frac{1}{p}+\frac{1}{q}=1$, then the following inequality holds:

$$\begin{array}{cc}\hfill & \left|\frac{1}{b-a}\stackrel{b}{\underset{a}{\int }}\phi (u)du-\phi \left(\frac{a+b}{2}\right)\right|\le \frac{b-a}{4{\left(p+1\right)}^{\frac{1}{p}}}({\left(max\left\{{\left|{\phi }^{\prime }\left(a\right)\right|}^q,{\left|{\phi }^{\prime }\left(\frac{a+b}{2}\right)\right|}^q\right\}\right)}^{\frac{1}{q}}\hfill \\ \hfill & +{\left(max\left\{{\left|{\phi }^{\prime }\left(\frac{a+b}{2}\right)\right|}^q,{\left|{\phi }^{\prime }\left(b\right)\right|}^q\right\}\right)}^{\frac{1}{q}}).\hfill \end{array}$$

Theorem 4

([22]). Let $\phi :I\subset [0,\infty )\to \mathbb{R}$ be a differentiable mapping on $I^{\circ }$ such that ${\phi }^{\prime }\in L[a,b]$, where $a,b\in I$ with $a<b$. If ${\left|{\phi }^{\prime }\right|}^q$ is quasi-convex on $[a,b]$, $q\ge 1$, then the following inequality holds:

$$\begin{array}{cc}\hfill & \left|\frac{1}{b-a}\stackrel{b}{\underset{a}{\int }}\phi (u)du-\phi \left(\frac{a+b}{2}\right)\right|\le \frac{b-a}{8}({\left(max\left\{{\left|{\phi }^{\prime }\left(a\right)\right|}^q,{\left|{\phi }^{\prime }\left(\frac{a+b}{2}\right)\right|}^q\right\}\right)}^{\frac{1}{q}}\hfill \\ \hfill & +{\left(max\left\{{\left|{\phi }^{\prime }\left(\frac{a+b}{2}\right)\right|}^q,{\left|{\phi }^{\prime }\left(b\right)\right|}^q\right\}\right)}^{\frac{1}{q}}).\hfill \end{array}$$

Alomari and Darus in [23] obtained the Ostrowski-type inequalities for differentiable quasi-convex functions:

Theorem 5

([23]). Let $\phi :I\subset [0,\infty )\to \mathbb{R}$ be a differentiable mapping on $I^{\circ }$ such that ${\phi }^{\prime }\in L[a,b]$, where $a,b\in I$ with $a<b$. If $\left|{\phi }^{\prime }\right|$ is quasi-convex on $[a,b]$, then the following inequality holds:

$$\begin{array}{cc}\hfill & \left|\frac{1}{b-a}\stackrel{b}{\underset{a}{\int }}\phi (u)du-\phi \left(x\right)\right|\le \frac{{\left(x-a\right)}^2}{2\left(b-a\right)}max\left\{\left|{\phi }^{\prime }\left(x\right)\right|,\left|{\phi }^{\prime }\left(a\right)\right|\right\}\hfill \\ \hfill & +\frac{{\left(b-x\right)}^2}{2\left(b-a\right)}max\left\{\left|{\phi }^{\prime }\left(x\right)\right|,\left|{\phi }^{\prime }\left(b\right)\right|\right\}.\hfill \end{array}$$

Theorem 6

([23]). Let $\phi :I\subset [0,\infty )\to \mathbb{R}$ be a differentiable mapping on $I^{\circ }$ such that ${\phi }^{\prime }\in L[a,b]$, where $a,b\in I$ with $a<b$. If ${\left|{\phi }^{\prime }\right|}^q$ is quasi-convex on $[a,b]$, $q>1$ with $\frac{1}{p}+\frac{1}{q}=1$, then the following inequality holds:

$$\begin{array}{cc}\hfill & \left|\frac{1}{b-a}\stackrel{b}{\underset{a}{\int }}\phi (u)du-\phi \left(x\right)\right|\le {\left(\frac{{\left(x-a\right)}^{p+1}}{\left(b-a\right)\left(p+1\right)}\right)}^{\frac{1}{p}}{\left(max\left\{{\left|{\phi }^{\prime }\left(x\right)\right|}^q,{\left|{\phi }^{\prime }\left(a\right)\right|}^q\right\}\right)}^{\frac{1}{q}}\hfill \\ \hfill & +{\left(\frac{{\left(b-x\right)}^{p+1}}{\left(b-a\right)\left(p+1\right)}\right)}^{\frac{1}{p}}{\left(max\left\{{\left|{\phi }^{\prime }\left(x\right)\right|}^q,{\left|{\phi }^{\prime }\left(b\right)\right|}^q\right\}\right)}^{\frac{1}{q}}.\hfill \end{array}$$

Theorem 7

([23]). Let $\phi :I\subset [0,\infty )\to \mathbb{R}$ be a differentiable mapping on $I^{\circ }$ such that ${\phi }^{\prime }\in L[a,b]$, where $a,b\in I$ with $a<b$. If ${\left|{\phi }^{\prime }\right|}^q$ is quasi-convex on $[a,b]$, $q\ge 1$, then the following inequality holds:

$$\begin{array}{cc}\hfill & \left|\frac{1}{b-a}\stackrel{b}{\underset{a}{\int }}\phi (u)du-\phi \left(x\right)\right|\le \frac{{\left(x-a\right)}^2}{2\left(b-a\right)}{\left(max\left\{{\left|{\phi }^{\prime }\left(x\right)\right|}^q,{\left|{\phi }^{\prime }\left(a\right)\right|}^q\right\}\right)}^{\frac{1}{q}}\hfill \\ \hfill & +\frac{{\left(b-x\right)}^2}{2\left(b-a\right)}{\left(max\left\{{\left|{\phi }^{\prime }\left(x\right)\right|}^q,{\left|{\phi }^{\prime }\left(b\right)\right|}^q\right\}\right)}^{\frac{1}{q}}.\hfill \end{array}$$

Fractional calculus has been widely studied by many researchers over the past decades. In particular, to generalize classic inequalities. Among the best known and use of these fractional integral operators we recall that of Riemann–Liouville.

Definition 3

([24]). Let $\phi \in L[a,b]$. The Riemann–Liouville integrals $J_{a^{+}}^{\alpha }\phi$ and $J_{b^{-}}^{\alpha }\phi$ of order $\alpha >0$ with $a\ge 0$ are defined by

$$\begin{array}{cc}\hfill J_{a^{+}}^{\alpha }\phi (x)=& \frac{1}{\Gamma \left(\alpha \right)}\stackrel{x}{\underset{a}{\int }}{\left(x-u\right)}^{\alpha -1}\phi (u)du,x>a,\hfill \\ \hfill J_{b^{-}}^{\alpha }\phi (x)=& \frac{1}{\Gamma \left(\alpha \right)}\stackrel{b}{\underset{x}{\int }}{\left(u-x\right)}^{\alpha -1}\phi (u)du,b>x,\hfill \end{array}$$

respectively, where $\Gamma (\alpha )=\stackrel{\infty }{\underset{0}{\int }}$ $e^{-u}{u}^{\alpha -1}du$, is the Gamma function and $J_{a^{+}}^0\phi (x)=J_{b^{-}}^0\phi (x)=\phi (x)$.

In so-called fractional calculus, there are not only global derivatives (for example, Riemann–Liouville and Caputo), but also local fractional derivatives (Khalil, Almeida, among others), see [25,26]. Regarding some papers dealing with fractional integral inequalities via different types of fractional integral operators, we refer readers to [27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43].

In this paper, we establish a new identity and then apply it to derive new weighted Ostrowski-type inequalities for quasi-convex functions. Further results for functions with a bounded first derivative will be given. In order to illustrate the efficiency of our main results, some applications to special means will be obtain.

2. Main Results

For brevity, we will used in the sequel $J=[a,b].$ In order to find our main results, we need to prove the following lemma.

Lemma 1.

Let$\phi :J\to \mathbb{R}$be differentiable function on$J^{\circ }$, $0\le a<b$, and$\omega :J\to \mathbb{R}$a continuous function. If${\phi }^{\prime },\omega \in$$L(J)$, then

$$\begin{array}{cc}\hfill & J_{x^{-}}^{\alpha }\omega \phi (a)+J_{x^{+}}^{\alpha }\omega \phi (b)-\left[J_{x^{-}}^{\alpha }\omega (a)+J_{x^{+}}^{\alpha }\omega (b)\right]\phi \left(x\right)\hfill \\ \hfill =& \frac{{\left(b-x\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}{p}_1\left(\tau \right){\phi }^{\prime }\left(\tau b+\left(1-\tau \right)x\right)d\tau -\frac{{\left(x-a\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}{p}_2\left(\tau \right){\phi }^{\prime }\left(\tau a+\left(1-\tau \right)x\right)d\tau ,\hfill \end{array}$$

where

$$p_1\left(\tau \right):=\stackrel{1}{\underset{\tau }{\int }}{\left(1-\sigma \right)}^{\alpha -1}\omega \left(\sigma b+\left(1-\sigma \right)x\right)d\sigma$$

(2)

and

$$p_2\left(\tau \right):=\stackrel{1}{\underset{\tau }{\int }}{\left(1-\sigma \right)}^{\alpha -1}\omega \left(\sigma a+\left(1-\sigma \right)x\right)d\sigma .$$

(3)

Proof. 

We integrate by parts and change the variable $\left(u=\sigma b+\left(1-\sigma \right)x\right)$ to obtain

$$\begin{array}{cc}\hfill & \stackrel{1}{\underset{0}{\int }}{p}_1\left(\tau \right){\phi }^{\prime }\left(\tau b+\left(1-\tau \right)x\right)d\tau \hfill \\ \hfill =& \stackrel{1}{\underset{0}{\int }}\left(\stackrel{1}{\underset{\tau }{\int }}{\left(1-\sigma \right)}^{\alpha -1}\omega \left(\sigma b+\left(1-\sigma \right)x\right)d\sigma \right){\phi }^{\prime }\left(\tau b+\left(1-\tau \right)x\right)d\tau \hfill \\ \hfill =& \frac{1}{b-x}{\left.\left(\stackrel{1}{\underset{\tau }{\int }}{\left(1-\sigma \right)}^{\alpha -1}\omega \left(\sigma b+\left(1-\sigma \right)x\right)d\sigma \right)\phi \left(\tau b+\left(1-\tau \right)x\right)\right|}_{\tau =0}^{\tau =1}\hfill \\ \hfill & +\frac{1}{b-x}\stackrel{1}{\underset{0}{\int }}{\left(1-\tau \right)}^{\alpha -1}\omega \left(\tau b+\left(1-\tau \right)x\right)\phi \left(\tau b+\left(1-\tau \right)x\right)d\tau \hfill \\ \hfill =& -\frac{1}{b-x}\left(\stackrel{1}{\underset{0}{\int }}{\left(1-\sigma \right)}^{\alpha -1}\omega \left(\sigma b+\left(1-\sigma \right)x\right)d\sigma \right)\phi \left(x\right)\hfill \\ \hfill & +\frac{1}{b-x}\stackrel{1}{\underset{0}{\int }}{\left(1-\tau \right)}^{\alpha -1}\omega \left(\tau b+\left(1-\tau \right)x\right)\phi \left(\tau b+\left(1-\tau \right)x\right)d\tau \hfill \\ \hfill =& -\frac{1}{{\left(b-x\right)}^{\alpha +1}}\left(\stackrel{b}{\underset{x}{\int }}{\left(b-u\right)}^{\alpha -1}\omega \left(u\right)du\right)\phi \left(x\right)\hfill \\ \hfill & +\frac{1}{{\left(b-x\right)}^{\alpha +1}}\stackrel{b}{\underset{x}{\int }}{\left(b-u\right)}^{\alpha -1}\omega \left(u\right)\phi \left(u\right)du\hfill \\ \hfill =& -\frac{\Gamma \left(\alpha \right)}{{\left(b-x\right)}^{\alpha +1}}\left(J_{x^{+}}^{\alpha }\omega (b)\right)\phi \left(x\right)+\frac{\Gamma \left(\alpha \right)}{{\left(b-x\right)}^{\alpha +1}}\left(J_{x^{+}}^{\alpha }\omega \phi (b)\right).\hfill \end{array}$$

(4)

Now, multiplying (4) by $\frac{{\left(b-x\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}$, we get

$$\frac{{\left(b-x\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}{p}_1\left(\tau \right){\phi }^{\prime }\left(\tau b+\left(1-\tau \right)x\right)d\tau =J_{x^{+}}^{\alpha }\omega \phi (b)-\left(J_{x^{+}}^{\alpha }\omega (b)\right)\phi \left(x\right).$$

(5)

Similar work give

$$\begin{array}{cc}\hfill & \stackrel{1}{\underset{0}{\int }}{p}_2\left(\tau \right){\phi }^{\prime }\left(\tau a+\left(1-\tau \right)x\right)d\tau \hfill \\ \hfill =& \frac{\Gamma \left(\alpha \right)}{{\left(x-a\right)}^{\alpha +1}}\left(J_{x^{-}}^{\alpha }\omega (a)\right)\phi \left(x\right)-\frac{\Gamma \left(\alpha \right)}{{\left(x-a\right)}^{\alpha +1}}{J}_{x^{-}}^{\alpha }\omega \phi (a).\hfill \end{array}$$

(6)

Multiplying (6) by $\frac{{\left(x-a\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}$, we obtain

$$\frac{{\left(x-a\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}{p}_2\left(\tau \right){\phi }^{\prime }\left(\tau a+\left(1-\tau \right)x\right)d\tau =\left(J_{x^{-}}^{\alpha }\omega (a)\right)\phi \left(x\right)-J_{x^{-}}^{\alpha }\omega \phi (a).$$

(7)

By taking the difference between (5) and (7), we get

$$\begin{array}{cc}\hfill & \frac{{\left(b-x\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}{p}_1\left(\tau \right){\phi }^{\prime }\left(\tau b+\left(1-\tau \right)x\right)d\tau -\frac{{\left(x-a\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}{p}_2\left(\tau \right){\phi }^{\prime }\left(\tau a+\left(1-\tau \right)x\right)d\tau \hfill \\ \hfill =& J_{x^{-}}^{\alpha }\omega \phi (a)+J_{x^{+}}^{\alpha }\omega \phi (b)-\left[J_{x^{-}}^{\alpha }\omega (a)+J_{x^{+}}^{\alpha }\omega (b)\right]\phi \left(x\right),\hfill \end{array}$$

which is the desired result. □

Theorem 8.

Consider a differentiable function$\phi :J\to \mathbb{R}$with${\phi }^{\prime }\in L(J)$where$0\le a<b$, and let$\omega :J\to \mathbb{R}$be continuous function. If$\left|{\phi }^{\prime }\right|$is quasi-convex, then

$$\begin{array}{cc}\hfill & \left|J_{x^{-}}^{\alpha }\omega \phi (a)+J_{x^{+}}^{\alpha }\omega \phi (b)-\left[J_{x^{-}}^{\alpha }\omega (a)+J_{x^{+}}^{\alpha }\omega (b)\right]\phi \left(x\right)\right|\hfill \\ \hfill \le & \frac{{\left(b-x\right)}^{\alpha +1}}{\Gamma \left(\alpha +2\right)}max\left\{\left|{\phi }^{\prime }\left(x\right)\right|,\left|{\phi }^{\prime }\left(b\right)\right|\right\}{∥\omega ∥}_{\left[x,b\right],\infty }\hfill \\ \hfill & +\frac{{\left(x-a\right)}^{\alpha +1}}{\Gamma \left(\alpha +2\right)}max\left\{\left|{\phi }^{\prime }\left(x\right)\right|,\left|{\phi }^{\prime }\left(a\right)\right|\right\}{∥\omega ∥}_{\left[a,x\right],\infty }.\hfill \end{array}$$

Proof. 

From Lemma 1, properties of modulus and quasi-convexity of $\left|{\phi }^{\prime }\right|$, we have

$$\begin{array}{cc}\hfill & \left|J_{x^{-}}^{\alpha }\omega \phi (a)+J_{x^{+}}^{\alpha }\omega \phi (b)-\left[J_{x^{-}}^{\alpha }\omega (a)+J_{x^{+}}^{\alpha }\omega (b)\right]\phi \left(x\right)\right|\hfill \\ \hfill \le & \frac{{\left(b-x\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}\left|p_1\left(\tau \right)\right|\left|{\phi }^{\prime }\left(\tau b+\left(1-\tau \right)x\right)\right|d\tau \hfill \\ \hfill & +\frac{{\left(x-a\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}\left|p_2\left(\tau \right)\right|\left|{\phi }^{\prime }\left(\tau a+\left(1-\tau \right)x\right)\right|d\tau \hfill \\ \hfill \le & \frac{{\left(b-x\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}max\left\{\left|{\phi }^{\prime }\left(x\right)\right|,\left|{\phi }^{\prime }\left(b\right)\right|\right\}\stackrel{1}{\underset{0}{\int }}\left|\stackrel{1}{\underset{\tau }{\int }}{\left(1-\sigma \right)}^{\alpha -1}\omega \left(\sigma b+\left(1-\sigma \right)x\right)d\sigma \right|d\tau \hfill \\ \hfill & +\frac{{\left(x-a\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}max\left\{\left|{\phi }^{\prime }\left(x\right)\right|,\left|{\phi }^{\prime }\left(a\right)\right|\right\}\stackrel{1}{\underset{0}{\int }}\left|\stackrel{1}{\underset{\tau }{\int }}{\left(1-\sigma \right)}^{\alpha -1}\omega \left(\sigma a+\left(1-\sigma \right)x\right)d\sigma \right|d\tau \hfill \\ \hfill \le & \left(\frac{{\left(b-x\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}max\left\{\left|{\phi }^{\prime }\left(x\right)\right|,\left|{\phi }^{\prime }\left(b\right)\right|\right\}{∥\omega ∥}_{\left[x,b\right],\infty }\right.\hfill \\ \hfill & +\left.\frac{{\left(x-a\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}max\left\{\left|{\phi }^{\prime }\left(x\right)\right|,\left|{\phi }^{\prime }\left(a\right)\right|\right\}{∥\omega ∥}_{\left[a,x\right],\infty }\right)\stackrel{1}{\underset{0}{\int }}\left|\stackrel{1}{\underset{\tau }{\int }}{\left(1-\sigma \right)}^{\alpha -1}d\sigma \right|d\tau \hfill \\ \hfill =& \frac{{\left(b-x\right)}^{\alpha +1}}{\Gamma \left(\alpha +2\right)}max\left\{\left|{\phi }^{\prime }\left(x\right)\right|,\left|{\phi }^{\prime }\left(b\right)\right|\right\}{∥\omega ∥}_{\left[x,b\right],\infty }\hfill \\ \hfill & +\frac{{\left(x-a\right)}^{\alpha +1}}{\Gamma \left(\alpha +2\right)}max\left\{\left|{\phi }^{\prime }\left(x\right)\right|,\left|{\phi }^{\prime }\left(a\right)\right|\right\}{∥\omega ∥}_{\left[a,x\right],\infty }.\hfill \end{array}$$

The proof is completed. □

Corollary 1.

Considering $x=\frac{a+b}{2}$ in Theorem 8, we obtain

$$\begin{array}{cc}\hfill & \left|J_{{\frac{a+b}{2}}^{-}}^{\alpha }\omega \phi (a)+J_{{\frac{a+b}{2}}^{+}}^{\alpha }\omega \phi (b)-\left[J_{{\frac{a+b}{2}}^{-}}^{\alpha }\omega (a)+J_{{\frac{a+b}{2}}^{+}}^{\alpha }\omega (b)\right]\phi \left(\frac{a+b}{2}\right)\right|\hfill \\ \hfill \le & \frac{{\left(b-a\right)}^{\alpha +1}}{2^{\alpha +1}\Gamma \left(\alpha +2\right)}{∥\omega ∥}_{\left[a,b\right],\infty }\hfill \\ \hfill & \times \left(max\left\{\left|{\phi }^{\prime }\left(\frac{a+b}{2}\right)\right|,\left|{\phi }^{\prime }\left(b\right)\right|\right\}+max\left\{\left|{\phi }^{\prime }\left(\frac{a+b}{2}\right)\right|,\left|{\phi }^{\prime }\left(a\right)\right|\right\}\right).\hfill \end{array}$$

Corollary 2.

Choosing $\omega \left(u\right)=\frac{1}{b-a}$ in Theorem 8, we get

$$\begin{array}{cc}\hfill & \left|J_{x^{-}}^{\alpha }\phi (a)+J_{x^{+}}^{\alpha }\phi (b)-\frac{{\left(b-x\right)}^{\alpha }+{\left(x-a\right)}^{\alpha }}{\Gamma \left(\alpha +1\right)}\phi \left(x\right)\right|\hfill \\ \hfill \le & \frac{{\left(b-x\right)}^{\alpha +1}}{\Gamma \left(\alpha +2\right)}max\left\{\left|{\phi }^{\prime }\left(x\right)\right|,\left|{\phi }^{\prime }\left(b\right)\right|\right\}+\frac{{\left(x-a\right)}^{\alpha +1}}{\Gamma \left(\alpha +2\right)}max\left\{\left|{\phi }^{\prime }\left(x\right)\right|,\left|{\phi }^{\prime }\left(a\right)\right|\right\}.\hfill \end{array}$$

Moreover, if we take $x=\frac{a+b}{2}$, we obtain

$$\begin{array}{cc}\hfill & \left|\frac{2^{\alpha -1}\Gamma \left(\alpha +1\right)}{{\left(b-a\right)}^{\alpha }}\left(J_{{\frac{a+b}{2}}^{-}}^{\alpha }\phi (a)+J_{{\frac{a+b}{2}}^{+}}^{\alpha }\phi (b)\right)-\phi \left(\frac{a+b}{2}\right)\right|\hfill \\ \hfill \le & \frac{b-a}{4\left(\alpha +1\right)}\left(max\left\{\left|{\phi }^{\prime }\left(\frac{a+b}{2}\right)\right|,\left|{\phi }^{\prime }\left(b\right)\right|\right\}+max\left\{\left|{\phi }^{\prime }\left(\frac{a+b}{2}\right)\right|,\left|{\phi }^{\prime }\left(a\right)\right|\right\}\right).\hfill \end{array}$$

Corollary 3.

Let $\alpha =1$ in Theorem 8, then

$$\begin{array}{cc}\hfill & \left|\stackrel{b}{\underset{a}{\int }}\omega \left(u\right)\phi (u)du-\left(\stackrel{b}{\underset{a}{\int }}\omega \left(u\right)du\right)\phi \left(x\right)\right|\hfill \\ \hfill \le & \frac{{\left(b-x\right)}^2}{2}max\left\{\left|{\phi }^{\prime }\left(x\right)\right|,\left|{\phi }^{\prime }\left(b\right)\right|\right\}{∥\omega ∥}_{\left[x,b\right],\infty }\hfill \\ \hfill & +\frac{{\left(x-a\right)}^2}{2}max\left\{\left|{\phi }^{\prime }\left(x\right)\right|,\left|{\phi }^{\prime }\left(a\right)\right|\right\}{∥\omega ∥}_{\left[a,x\right],\infty }.\hfill \end{array}$$

Moreover, for $x=\frac{a+b}{2}$, we get

$$\begin{array}{cc}\hfill & \left|\stackrel{b}{\underset{a}{\int }}\omega \left(u\right)\phi (u)du-\left(\stackrel{b}{\underset{a}{\int }}\omega \left(u\right)du\right)\phi \left(\frac{a+b}{2}\right)\right|\le \frac{{\left(b-a\right)}^2}{8}{∥\omega ∥}_{\left[a,b\right],\infty }\hfill \\ \hfill & \times \left(max\left\{\left|{\phi }^{\prime }\left(\frac{a+b}{2}\right)\right|,\left|{\phi }^{\prime }\left(b\right)\right|\right\}+max\left\{\left|{\phi }^{\prime }\left(\frac{a+b}{2}\right)\right|,\left|{\phi }^{\prime }\left(a\right)\right|\right\}\right).\hfill \end{array}$$

Remark 1.

In Corollary 3, if we choose $\omega \left(u\right)=\frac{1}{b-a}$, we obtain ([23], Theorem 2). Moreover, if we take $x=\frac{a+b}{2}$, we get ([22], Theorem 7).

Theorem 9.

Consider a differentiable function $\phi :J\to \mathbb{R}$ with ${\phi }^{\prime }\in L(J)$ where $0\le a<b$, and let $\omega :J\to \mathbb{R}$ be continuous function. If ${\left|{\phi }^{\prime }\right|}^q$ is quasi-convex, $q>1$ and $\frac{1}{p}+\frac{1}{q}=1$, then

$$\begin{array}{cc}\hfill & \left|J_{x^{-}}^{\alpha }\omega \phi (a)+J_{x^{+}}^{\alpha }\omega \phi (b)-\left[J_{x^{-}}^{\alpha }\omega (a)+J_{x^{+}}^{\alpha }\omega (b)\right]\phi \left(x\right)\right|\hfill \\ \hfill \le & \frac{{\left(b-x\right)}^{\alpha +1}}{{\left(\alpha p+1\right)}^{\frac{1}{p}}\Gamma \left(\alpha +1\right)}{∥\omega ∥}_{\left[x,b\right],\infty }{\left(max\left\{{\left|{\phi }^{\prime }\left(x\right)\right|}^q,{\left|{\phi }^{\prime }\left(b\right)\right|}^q\right\}\right)}^{\frac{1}{q}}\hfill \\ \hfill & +\frac{{\left(x-a\right)}^{\alpha +1}}{{\left(\alpha p+1\right)}^{\frac{1}{p}}\Gamma \left(\alpha +1\right)}{∥\omega ∥}_{\left[a,x\right],\infty }{\left(max\left\{{\left|{\phi }^{\prime }\left(x\right)\right|}^q,{\left|{\phi }^{\prime }\left(a\right)\right|}^q\right\}\right)}^{\frac{1}{q}}.\hfill \end{array}$$

Proof. 

Applying properties of modulus, Lemma 1, Hölder’s inequality, and quasi-convexity of ${\left|{\phi }^{\prime }\right|}^q$, we obtain

$$\begin{array}{cc}\hfill & \left|J_{x^{-}}^{\alpha }\omega \phi (a)+J_{x^{+}}^{\alpha }\omega \phi (b)-\left[J_{x^{-}}^{\alpha }\omega (a)+J_{x^{+}}^{\alpha }\omega (b)\right]\phi \left(x\right)\right|\hfill \\ \hfill \le & \frac{{\left(b-x\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}{\left(\stackrel{1}{\underset{0}{\int }}{\left|p_1\left(\tau \right)\right|}^{p}d\tau \right)}^{\frac{1}{p}}{\left(\stackrel{1}{\underset{0}{\int }}{\left|{\phi }^{\prime }\left(\tau b+\left(1-\tau \right)x\right)\right|}^{q}d\tau \right)}^{\frac{1}{q}}\hfill \\ \hfill & +\frac{{\left(x-a\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}{\left(\stackrel{1}{\underset{0}{\int }}{\left|p_2\left(\tau \right)\right|}^{p}d\tau \right)}^{\frac{1}{p}}{\left(\stackrel{1}{\underset{0}{\int }}{\left|{\phi }^{\prime }\left(\tau a+\left(1-\tau \right)x\right)\right|}^{q}d\tau \right)}^{\frac{1}{q}}\hfill \\ \hfill \le & \frac{{\left(b-x\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}{\left(\stackrel{1}{\underset{0}{\int }}{\left|\stackrel{1}{\underset{\tau }{\int }}{\left(1-\sigma \right)}^{\alpha -1}\omega \left(\sigma b+\left(1-\sigma \right)x\right)d\sigma \right|}^{p}d\tau \right)}^{\frac{1}{p}}{\left(max\left\{{\left|{\phi }^{\prime }\left(x\right)\right|}^q,{\left|{\phi }^{\prime }\left(b\right)\right|}^q\right\}\right)}^{\frac{1}{q}}\hfill \\ \hfill & +\frac{{\left(x-a\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}{\left(\stackrel{1}{\underset{0}{\int }}{\left|\stackrel{1}{\underset{\tau }{\int }}{\left(1-\sigma \right)}^{\alpha -1}\omega \left(\sigma a+\left(1-\sigma \right)x\right)d\sigma \right|}^{p}d\tau \right)}^{\frac{1}{p}}{\left(max\left\{{\left|{\phi }^{\prime }\left(x\right)\right|}^q,{\left|{\phi }^{\prime }\left(a\right)\right|}^q\right\}\right)}^{\frac{1}{q}}\hfill \\ \hfill \le & \frac{{\left(b-x\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}{∥\omega ∥}_{\left[x,b\right],\infty }{\left(\stackrel{1}{\underset{0}{\int }}{\left|\stackrel{1}{\underset{\tau }{\int }}{\left(1-\sigma \right)}^{\alpha -1}d\sigma \right|}^{p}d\tau \right)}^{\frac{1}{p}}{\left(max\left\{{\left|{\phi }^{\prime }\left(x\right)\right|}^q,{\left|{\phi }^{\prime }\left(b\right)\right|}^q\right\}\right)}^{\frac{1}{q}}\hfill \\ \hfill & +\frac{{\left(x-a\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}{∥\omega ∥}_{\left[a,x\right],\infty }{\left(\stackrel{1}{\underset{0}{\int }}{\left|\stackrel{1}{\underset{\tau }{\int }}{\left(1-\sigma \right)}^{\alpha -1}d\sigma \right|}^{p}d\tau \right)}^{\frac{1}{p}}{\left(max\left\{{\left|{\phi }^{\prime }\left(x\right)\right|}^q,{\left|{\phi }^{\prime }\left(a\right)\right|}^q\right\}\right)}^{\frac{1}{q}}\hfill \\ \hfill =& \frac{{\left(b-x\right)}^{\alpha +1}}{{\left(\alpha p+1\right)}^{\frac{1}{p}}\Gamma \left(\alpha +1\right)}{∥\omega ∥}_{\left[x,b\right],\infty }{\left(max\left\{{\left|{\phi }^{\prime }\left(x\right)\right|}^q,{\left|{\phi }^{\prime }\left(b\right)\right|}^q\right\}\right)}^{\frac{1}{q}}\hfill \\ \hfill & +\frac{{\left(x-a\right)}^{\alpha +1}}{{\left(\alpha p+1\right)}^{\frac{1}{p}}\Gamma \left(\alpha +1\right)}{∥\omega ∥}_{\left[a,x\right],\infty }{\left(max\left\{{\left|{\phi }^{\prime }\left(x\right)\right|}^q,{\left|{\phi }^{\prime }\left(a\right)\right|}^q\right\}\right)}^{\frac{1}{q}}.\hfill \end{array}$$

The proof is completed. □

Corollary 4.

Let $x=\frac{a+b}{2}$ in Theorem 9, then

$$\begin{array}{cc}\hfill & \left|J_{{\frac{a+b}{2}}^{-}}^{\alpha }\omega \phi (a)+J_{{\frac{a+b}{2}}^{+}}^{\alpha }\omega \phi (b)-\left[J_{{\frac{a+b}{2}}^{-}}^{\alpha }\omega (a)+J_{{\frac{a+b}{2}}^{+}}^{\alpha }\omega (b)\right]\phi \left(\frac{a+b}{2}\right)\right|\hfill \\ \hfill \le & \frac{{\left(b-a\right)}^{\alpha +1}}{2^{\alpha +1}{\left(\alpha p+1\right)}^{\frac{1}{p}}\Gamma \left(\alpha +1\right)}{∥\omega ∥}_{\left[a,b\right],\infty }\left({\left(max\left\{{\left|{\phi }^{\prime }\left(\frac{a+b}{2}\right)\right|}^q,{\left|{\phi }^{\prime }\left(b\right)\right|}^q\right\}\right)}^{\frac{1}{q}}\right.\hfill \\ \hfill & +\left.{\left(max\left\{{\left|{\phi }^{\prime }\left(\frac{a+b}{2}\right)\right|}^q,{\left|{\phi }^{\prime }\left(a\right)\right|}^q\right\}\right)}^{\frac{1}{q}}\right).\hfill \end{array}$$

Corollary 5.

Choosing $\omega \left(u\right)=\frac{1}{b-a}$ in Theorem 9, we get

$$\begin{array}{cc}\hfill & \left|J_{x^{-}}^{\alpha }\phi (a)+J_{x^{+}}^{\alpha }\phi (b)-\frac{{\left(b-x\right)}^{\alpha }+{\left(x-a\right)}^{\alpha }}{\Gamma \left(\alpha +1\right)}\phi \left(x\right)\right|\hfill \\ \hfill \le & \frac{{\left(b-x\right)}^{\alpha +1}}{{\left(\alpha p+1\right)}^{\frac{1}{p}}\Gamma \left(\alpha +1\right)}{\left(max\left\{{\left|{\phi }^{\prime }\left(x\right)\right|}^q,{\left|{\phi }^{\prime }\left(b\right)\right|}^q\right\}\right)}^{\frac{1}{q}}\hfill \\ \hfill & +\frac{{\left(x-a\right)}^{\alpha +1}}{{\left(\alpha p+1\right)}^{\frac{1}{p}}\Gamma \left(\alpha +1\right)}{\left(max\left\{{\left|{\phi }^{\prime }\left(x\right)\right|}^q,{\left|{\phi }^{\prime }\left(a\right)\right|}^q\right\}\right)}^{\frac{1}{q}}.\hfill \end{array}$$

Moreover, if we take $x=\frac{a+b}{2}$, we obtain

$$\begin{array}{cc}\hfill & \left|\frac{2^{\alpha -1}\Gamma \left(\alpha +1\right)}{{\left(b-a\right)}^{\alpha }}\left(J_{{\frac{a+b}{2}}^{-}}^{\alpha }\phi (a)+J_{{\frac{a+b}{2}}^{+}}^{\alpha }\phi (b)\right)-\phi \left(\frac{a+b}{2}\right)\right|\hfill \\ \hfill \le & \frac{b-a}{4{\left(\alpha p+1\right)}^{\frac{1}{p}}}\left({\left(max\left\{{\left|{\phi }^{\prime }\left(\frac{a+b}{2}\right)\right|}^q,{\left|{\phi }^{\prime }\left(b\right)\right|}^q\right\}\right)}^{\frac{1}{q}}\right.\hfill \\ \hfill & +\left.{\left(max\left\{{\left|{\phi }^{\prime }\left(\frac{a+b}{2}\right)\right|}^q,{\left|{\phi }^{\prime }\left(a\right)\right|}^q\right\}\right)}^{\frac{1}{q}}\right).\hfill \end{array}$$

Corollary 6.

For $\alpha =1$ in Theorem 9, we have

$$\begin{array}{cc}\hfill & \left|\stackrel{b}{\underset{a}{\int }}\omega \left(u\right)\phi (u)du-\left(\stackrel{b}{\underset{a}{\int }}\omega \left(u\right)du\right)\phi \left(x\right)\right|\hfill \\ \hfill \le & \frac{{\left(b-x\right)}^2}{{\left(p+1\right)}^{\frac{1}{p}}}{∥\omega ∥}_{\left[x,b\right],\infty }{\left(max\left\{{\left|{\phi }^{\prime }\left(x\right)\right|}^q,{\left|{\phi }^{\prime }\left(b\right)\right|}^q\right\}\right)}^{\frac{1}{q}}\hfill \\ \hfill & +\frac{{\left(x-a\right)}^2}{{\left(p+1\right)}^{\frac{1}{p}}}{∥\omega ∥}_{\left[a,x\right],\infty }{\left(max\left\{{\left|{\phi }^{\prime }\left(x\right)\right|}^q,{\left|{\phi }^{\prime }\left(a\right)\right|}^q\right\}\right)}^{\frac{1}{q}}.\hfill \end{array}$$

Moreover, considering $x=\frac{a+b}{2}$, we get

$$\begin{array}{cc}\hfill & \left|\stackrel{b}{\underset{a}{\int }}\omega \left(u\right)\phi (u)du-\left(\stackrel{b}{\underset{a}{\int }}\omega \left(u\right)du\right)\phi \left(\frac{a+b}{2}\right)\right|\hfill \\ \hfill \le & \frac{{\left(b-a\right)}^2}{4{\left(p+1\right)}^{\frac{1}{p}}}{∥\omega ∥}_{\left[a,b\right],\infty }\left({\left(max\left\{{\left|{\phi }^{\prime }\left(\frac{a+b}{2}\right)\right|}^q,{\left|{\phi }^{\prime }\left(b\right)\right|}^q\right\}\right)}^{\frac{1}{q}}\right.\hfill \\ \hfill & +\left.{\left(max\left\{{\left|{\phi }^{\prime }\left(\frac{a+b}{2}\right)\right|}^q,{\left|{\phi }^{\prime }\left(a\right)\right|}^q\right\}\right)}^{\frac{1}{q}}\right).\hfill \end{array}$$

Corollary 7.

In Corollary 6, if we choose $\omega \left(u\right)=\frac{1}{b-a}$, we obtain

$$\begin{array}{cc}\hfill & \left|\frac{1}{b-a}\stackrel{b}{\underset{a}{\int }}\phi (u)du-\phi \left(x\right)\right|\hfill \\ \hfill \le & \left(\frac{{\left(b-x\right)}^2}{\left(b-a\right){\left(p+1\right)}^{\frac{1}{p}}}{\left(max\left\{{\left|{\phi }^{\prime }\left(x\right)\right|}^q,{\left|{\phi }^{\prime }\left(b\right)\right|}^q\right\}\right)}^{\frac{1}{q}}\right.\hfill \\ \hfill & +\left.\frac{{\left(x-a\right)}^2}{\left(b-a\right){\left(p+1\right)}^{\frac{1}{p}}}{\left(max\left\{{\left|{\phi }^{\prime }\left(x\right)\right|}^q,{\left|{\phi }^{\prime }\left(a\right)\right|}^q\right\}\right)}^{\frac{1}{q}}\right).\hfill \end{array}$$

For $x=\frac{a+b}{2}$, we get ([22], Theorem 8).

Theorem 10.

Consider a differentiable function $\phi :J\to \mathbb{R}$ with ${\phi }^{\prime }\in L(J)$ where $0\le a<b$, and let $\omega :J\to \mathbb{R}$ be continuous function. If ${\left|{\phi }^{\prime }\right|}^q$ is quasi-convex, where $q\ge 1$, then

$$\begin{array}{cc}\hfill & \left|J_{x^{-}}^{\alpha }\omega \phi (a)+J_{x^{+}}^{\alpha }\omega \phi (b)-\left[J_{x^{-}}^{\alpha }\omega (a)+J_{x^{+}}^{\alpha }\omega (b)\right]\phi \left(x\right)\right|\hfill \\ \hfill \le & \frac{{\left(b-x\right)}^{\alpha +1}}{\Gamma \left(\alpha +2\right)}{∥\omega ∥}_{\left[x,b\right],\infty }{\left(max\left\{{\left|{\phi }^{\prime }\left(x\right)\right|}^q,{\left|{\phi }^{\prime }\left(b\right)\right|}^q\right\}\right)}^{\frac{1}{q}}\hfill \\ \hfill & +\frac{{\left(x-a\right)}^{\alpha +1}}{\Gamma \left(\alpha +2\right)}{∥\omega ∥}_{\left[a,x\right],\infty }{\left(max\left\{{\left|{\phi }^{\prime }\left(x\right)\right|}^q,{\left|{\phi }^{\prime }\left(a\right)\right|}^q\right\}\right)}^{\frac{1}{q}}.\hfill \end{array}$$

Proof. 

By properties of modulus, applying Lemma 1, power mean inequality and quasi-convexity of ${\left|{\phi }^{\prime }\right|}^q$, we obtain

$$\begin{array}{cc}\hfill & \left|J_{x^{-}}^{\alpha }\omega \phi (a)+J_{x^{+}}^{\alpha }\omega \phi (b)-\left[J_{x^{-}}^{\alpha }\omega (a)+J_{x^{+}}^{\alpha }\omega (b)\right]\phi \left(x\right)\right|\hfill \\ \hfill \le & \frac{{\left(b-x\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}{\left(\stackrel{1}{\underset{0}{\int }}\left|p_1\left(\tau \right)\right|d\tau \right)}^{1-\frac{1}{q}}{\left(\stackrel{1}{\underset{0}{\int }}\left|p_1\left(\tau \right)\right|{\left|{\phi }^{\prime }\left(\tau b+\left(1-\tau \right)x\right)\right|}^{q}d\tau \right)}^{\frac{1}{q}}\hfill \\ \hfill & +\frac{{\left(x-a\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}{\left(\stackrel{1}{\underset{0}{\int }}\left|p_2\left(\tau \right)\right|d\tau \right)}^{1-\frac{1}{q}}{\left(\stackrel{1}{\underset{0}{\int }}\left|p_2\left(\tau \right)\right|{\left|{\phi }^{\prime }\left(\tau a+\left(1-\tau \right)x\right)\right|}^{q}d\tau \right)}^{\frac{1}{q}}\hfill \\ \hfill \le & \frac{{\left(b-x\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}{\left(\stackrel{1}{\underset{0}{\int }}\left|p_1\left(\tau \right)\right|d\tau \right)}^{1-\frac{1}{q}}{\left(max\left\{{\left|{\phi }^{\prime }\left(x\right)\right|}^q,{\left|{\phi }^{\prime }\left(b\right)\right|}^q\right\}\stackrel{1}{\underset{0}{\int }}\left|p_1\left(\tau \right)\right|d\tau \right)}^{\frac{1}{q}}\hfill \\ \hfill & +\frac{{\left(x-a\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}{\left(\stackrel{1}{\underset{0}{\int }}\left|p_2\left(\tau \right)\right|d\tau \right)}^{1-\frac{1}{q}}{\left(max\left\{{\left|{\phi }^{\prime }\left(x\right)\right|}^q,{\left|{\phi }^{\prime }\left(a\right)\right|}^q\right\}\stackrel{1}{\underset{0}{\int }}\left|p_2\left(\tau \right)\right|d\tau \right)}^{\frac{1}{q}}\hfill \\ \hfill =& \frac{{\left(b-x\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}\left(\stackrel{1}{\underset{0}{\int }}\left|p_1\left(\tau \right)\right|d\tau \right){\left(max\left\{{\left|{\phi }^{\prime }\left(x\right)\right|}^q,{\left|{\phi }^{\prime }\left(b\right)\right|}^q\right\}\right)}^{\frac{1}{q}}\hfill \\ \hfill & +\frac{{\left(x-a\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}\left(\stackrel{1}{\underset{0}{\int }}\left|p_2\left(\tau \right)\right|d\tau \right){\left(max\left\{{\left|{\phi }^{\prime }\left(x\right)\right|}^q,{\left|{\phi }^{\prime }\left(a\right)\right|}^q\right\}\right)}^{\frac{1}{q}}\hfill \\ \hfill \le & \frac{{\left(b-x\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}{∥\omega ∥}_{\left[x,b\right],\infty }\left(\stackrel{1}{\underset{0}{\int }}\left(\stackrel{1}{\underset{\tau }{\int }}{\left(1-\sigma \right)}^{\alpha -1}d\sigma \right)d\tau \right){\left(max\left\{{\left|{\phi }^{\prime }\left(x\right)\right|}^q,{\left|{\phi }^{\prime }\left(b\right)\right|}^q\right\}\right)}^{\frac{1}{q}}\hfill \\ \hfill & +\frac{{\left(x-a\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}{∥\omega ∥}_{\left[a,x\right],\infty }\left(\stackrel{1}{\underset{0}{\int }}\left(\stackrel{1}{\underset{\tau }{\int }}{\left(1-\sigma \right)}^{\alpha -1}d\sigma \right)d\tau \right){\left(max\left\{{\left|{\phi }^{\prime }\left(x\right)\right|}^q,{\left|{\phi }^{\prime }\left(a\right)\right|}^q\right\}\right)}^{\frac{1}{q}}\hfill \\ \hfill =& \frac{{\left(b-x\right)}^{\alpha +1}}{\Gamma \left(\alpha +2\right)}{∥\omega ∥}_{\left[x,b\right],\infty }{\left(max\left\{{\left|{\phi }^{\prime }\left(x\right)\right|}^q,{\left|{\phi }^{\prime }\left(b\right)\right|}^q\right\}\right)}^{\frac{1}{q}}\hfill \\ \hfill & +\frac{{\left(x-a\right)}^{\alpha +1}}{\Gamma \left(\alpha +2\right)}{∥\omega ∥}_{\left[a,x\right],\infty }{\left(max\left\{{\left|\phi \left(x\right)\right|}^q,{\left|{\phi }^{\prime }\left(a\right)\right|}^q\right\}\right)}^{\frac{1}{q}}.\hfill \end{array}$$

The proof is completed. □

Corollary 8.

For $x=\frac{a+b}{2}$ in Theorem 10, we get

$$\begin{array}{cc}\hfill & \left|J_{{\frac{a+b}{2}}^{-}}^{\alpha }\omega \phi (a)+J_{{\frac{a+b}{2}}^{+}}^{\alpha }\omega \phi (b)-\left[J_{{\frac{a+b}{2}}^{-}}^{\alpha }\omega (a)+J_{{\frac{a+b}{2}}^{+}}^{\alpha }\omega (b)\right]\phi \left(\frac{a+b}{2}\right)\right|\hfill \\ \hfill \le & \frac{{\left(b-a\right)}^{\alpha +1}}{2^{\alpha +1}\Gamma \left(\alpha +2\right)}{∥\omega ∥}_{\left[a,b\right],\infty }\left({\left(max\left\{{\left|{\phi }^{\prime }\left(\frac{a+b}{2}\right)\right|}^q,{\left|{\phi }^{\prime }\left(b\right)\right|}^q\right\}\right)}^{\frac{1}{q}}\right.\hfill \\ \hfill & +\left.{\left(max\left\{{\left|{\phi }^{\prime }\left(\frac{a+b}{2}\right)\right|}^q,{\left|\phi \left(a\right)\right|}^q\right\}\right)}^{\frac{1}{q}}\right).\hfill \end{array}$$

Corollary 9.

Choosing $\omega \left(u\right)=\frac{1}{b-a}$ in Theorem 10, we have

$$\begin{array}{cc}\hfill & \left|J_{x^{-}}^{\alpha }\phi (a)+J_{x^{+}}^{\alpha }\phi (b)-\frac{{\left(b-x\right)}^{\alpha }+{\left(x-a\right)}^{\alpha }}{\Gamma \left(\alpha +1\right)}\phi \left(x\right)\right|\hfill \\ \hfill \le & \frac{{\left(b-x\right)}^{\alpha +1}}{\Gamma \left(\alpha +2\right)}{\left(max\left\{{\left|{\phi }^{\prime }\left(x\right)\right|}^q,{\left|{\phi }^{\prime }\left(b\right)\right|}^q\right\}\right)}^{\frac{1}{q}}\hfill \\ \hfill & +\frac{{\left(x-a\right)}^{\alpha +1}}{\Gamma \left(\alpha +2\right)}{\left(max\left\{{\left|{\phi }^{\prime }\left(x\right)\right|}^q,{\left|{\phi }^{\prime }\left(a\right)\right|}^q\right\}\right)}^{\frac{1}{q}}.\hfill \end{array}$$

Moreover, if we take $x=\frac{a+b}{2}$, we obtain

$$\begin{array}{cc}\hfill & \left|\frac{2^{\alpha -1}\Gamma \left(\alpha +1\right)}{{\left(b-a\right)}^{\alpha }}\left(J_{{\frac{a+b}{2}}^{-}}^{\alpha }\phi (a)+J_{{\frac{a+b}{2}}^{+}}^{\alpha }\phi (b)\right)-\phi \left(\frac{a+b}{2}\right)\right|\le \frac{b-a}{4\left(\alpha +1\right)}\hfill \\ \hfill & \times \left({\left(max\left\{{\left|{\phi }^{\prime }\left(\frac{a+b}{2}\right)\right|}^q,{\left|{\phi }^{\prime }\left(b\right)\right|}^q\right\}\right)}^{\frac{1}{q}}+{\left(max\left\{{\left|{\phi }^{\prime }\left(\frac{a+b}{2}\right)\right|}^q,{\left|{\phi }^{\prime }\left(a\right)\right|}^q\right\}\right)}^{\frac{1}{q}}\right).\hfill \end{array}$$

Corollary 10.

Considering $\alpha =1$ in Theorem 10, we get

$$\begin{array}{cc}\hfill & \left|\stackrel{b}{\underset{a}{\int }}\omega \left(u\right)\phi (u)du-\left(\stackrel{b}{\underset{a}{\int }}\omega \left(u\right)du\right)\phi \left(x\right)\right|\hfill \\ \hfill \le & \frac{{\left(b-x\right)}^2}{2}{∥\omega ∥}_{\left[x,b\right],\infty }{\left(max\left\{{\left|{\phi }^{\prime }\left(x\right)\right|}^q,{\left|{\phi }^{\prime }\left(b\right)\right|}^q\right\}\right)}^{\frac{1}{q}}\hfill \\ \hfill & +\frac{{\left(x-a\right)}^2}{2}{∥\omega ∥}_{\left[a,x\right],\infty }{\left(max\left\{{\left|{\phi }^{\prime }\left(x\right)\right|}^q,{\left|{\phi }^{\prime }\left(a\right)\right|}^q\right\}\right)}^{\frac{1}{q}}.\hfill \end{array}$$

Moreover, for $x=\frac{a+b}{2}$, we have

$$\begin{array}{cc}\hfill & \left|\stackrel{b}{\underset{a}{\int }}\omega \left(u\right)\phi (u)du-\left(\stackrel{b}{\underset{a}{\int }}\omega \left(u\right)du\right)\phi \left(\frac{a+b}{2}\right)\right|\hfill \\ \hfill \le & \frac{{\left(b-a\right)}^2}{8}{∥\omega ∥}_{\left[a,b\right],\infty }\left({\left(max\left\{{\left|{\phi }^{\prime }\left(\frac{a+b}{2}\right)\right|}^q,{\left|{\phi }^{\prime }\left(b\right)\right|}^q\right\}\right)}^{\frac{1}{q}}\right.\hfill \\ \hfill & +\left.{\left(max\left\{{\left|{\phi }^{\prime }\left(\frac{a+b}{2}\right)\right|}^q,{\left|{\phi }^{\prime }\left(a\right)\right|}^q\right\}\right)}^{\frac{1}{q}}\right).\hfill \end{array}$$

Remark 2.

Choosing $\omega \left(u\right)=\frac{1}{b-a}$ in Corollary 10, we obtain ([23], Theorem 4). Moreover, for $x=\frac{a+b}{2}$, we get ([22], Theorem 9).

3. Further Results

In this section, we will prove the following results.

Theorem 11.

Consider a differentiable function $\phi :J\to \mathbb{R}$ with ${\phi }^{\prime }\in L(J)$, $0\le a<b$, and let $\omega :J\to \mathbb{R}$ be continuous function. If there exist constants $m<M$ such that $-\infty <m\le {\phi }^{\prime }\left(x\right)\le M<+\infty$ for all $x\in J$, then

$$\left|{\mathsf{\Lambda }}^{\alpha }\left(a,x,b,\omega ,\phi \right)\right|\le \frac{\left(M-m\right){\left(b-x\right)}^{\alpha +1}}{2\Gamma \left(\alpha +2\right)}{∥\omega ∥}_{\left[x,b\right],\infty }+\frac{\left(M-m\right){\left(x-a\right)}^{\alpha +1}}{2\Gamma \left(\alpha +2\right)}{∥\omega ∥}_{\left[a,x\right],\infty },$$

where

$$\begin{array}{cc}\hfill {\mathsf{\Lambda }}^{\alpha }\left(a,x,b,\omega ,\phi \right):=& J_{x^{-}}^{\alpha }\omega \phi (a)+J_{x^{+}}^{\alpha }\omega \phi (b)-\left[J_{x^{-}}^{\alpha }\omega (a)+J_{x^{+}}^{\alpha }\omega (b)\right]\phi \left(x\right)\hfill \\ \hfill & -\frac{\left(M+m\right)\left[{\left(b-x\right)}^{\alpha +1}-{\left(x-a\right)}^{\alpha +1}\right]}{2\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}\left(p_1(\tau )+p_2(\tau )\right)d\tau ,\hfill \end{array}$$

(8)

and $p_1,p_2$ are defined as in (2) and (3), respectively.

Proof. 

Applying Lemma 1, we get

$$\begin{array}{cc}\hfill & J_{x^{-}}^{\alpha }\omega \phi (a)+J_{x^{+}}^{\alpha }\omega \phi (b)-\left[J_{x^{-}}^{\alpha }\omega (a)+J_{x^{+}}^{\alpha }\omega (b)\right]\phi \left(x\right)\hfill \\ \hfill =& \frac{{\left(b-x\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}{p}_1\left(\tau \right){\phi }^{\prime }\left(\tau b+\left(1-\tau \right)x\right)d\tau -\frac{{\left(x-a\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}{p}_2\left(\tau \right){\phi }^{\prime }\left(\tau a+\left(1-\tau \right)x\right)d\tau \hfill \\ \hfill =& \frac{{\left(b-x\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}{p}_1\left(\tau \right)\left[{\phi }^{\prime }\left(\tau b+\left(1-\tau \right)x\right)-\frac{M+m}{2}+\frac{M+m}{2}\right]d\tau \hfill \\ \hfill & -\frac{{\left(x-a\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}{p}_2\left(\tau \right)\left[{\phi }^{\prime }\left(\tau a+\left(1-\tau \right)x\right)-\frac{M+m}{2}+\frac{M+m}{2}\right]d\tau \hfill \\ \hfill =& \frac{{\left(b-x\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}{p}_1\left(\tau \right)\left({\phi }^{\prime }\left(\tau b+\left(1-\tau \right)x\right)-\frac{M+m}{2}\right)d\tau \hfill \\ \hfill & +\frac{\left(M+m\right){\left(b-x\right)}^{\alpha +1}}{2\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}{p}_1\left(\tau \right)d\tau \hfill \\ \hfill & -\frac{{\left(x-a\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}{p}_2\left(\tau \right)\left({\phi }^{\prime }\left(\tau a+\left(1-\tau \right)x\right)-\frac{M+m}{2}\right)d\tau \hfill \\ \hfill & -\frac{\left(M+m\right){\left(x-a\right)}^{\alpha +1}}{2\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}{p}_2\left(\tau \right)d\tau .\hfill \end{array}$$

(9)

From (9), we have

$$\begin{array}{cc}\hfill {\mathsf{\Lambda }}^{\alpha }\left(a,x,b,\omega ,\phi \right)=& \frac{{\left(b-x\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}{p}_1\left(\tau \right)\left({\phi }^{\prime }\left(\tau b+\left(1-\tau \right)x\right)-\frac{M+m}{2}\right)d\tau \hfill \\ \hfill & -\frac{{\left(x-a\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}{p}_2\left(\tau \right)\left({\phi }^{\prime }\left(\tau a+\left(1-\tau \right)x\right)-\frac{M+m}{2}\right)d\tau ,\hfill \end{array}$$

(10)

where ${\mathsf{\Lambda }}^{\alpha }\left(a,x,b,\omega ,\phi \right)$ is defined as in (8). Applying absolute value on both sides of (10) and using the fact that $m\le {\phi }^{\prime }\left(x\right)\le M$ for all $x\in J$, we obtain

$$\begin{array}{cc}\hfill & \left|{\mathsf{\Lambda }}^{\alpha }\left(a,x,b,\omega ,\phi \right)\right|\hfill \\ \hfill \le & \frac{{\left(b-x\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}\left|p_1\left(\tau \right)\right|\left|{\phi }^{\prime }\left(\tau b+\left(1-\tau \right)x\right)-\frac{M+m}{2}\right|d\tau \hfill \\ \hfill & +\frac{{\left(x-a\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}\left|p_2\left(\tau \right)\right|\left|{\phi }^{\prime }\left(\tau a+\left(1-\tau \right)x\right)-\frac{M+m}{2}\right|d\tau \hfill \\ \hfill \le & \frac{\left(M-m\right){\left(b-x\right)}^{\alpha +1}}{2\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}\left|\stackrel{1}{\underset{\tau }{\int }}{\left(1-\sigma \right)}^{\alpha -1}\omega \left(\sigma b+\left(1-\sigma \right)x\right)d\sigma \right|d\tau \hfill \\ \hfill & +\frac{\left(M-m\right){\left(x-a\right)}^{\alpha +1}}{2\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}\left|\stackrel{1}{\underset{\tau }{\int }}{\left(1-\sigma \right)}^{\alpha -1}\omega \left(\sigma a+\left(1-\sigma \right)x\right)d\sigma \right|d\tau \hfill \\ \hfill \le & \frac{\left(M-m\right){\left(b-x\right)}^{\alpha +1}}{2\Gamma \left(\alpha +2\right)}{∥\omega ∥}_{\left[x,b\right],\infty }+\frac{\left(M-m\right){\left(x-a\right)}^{\alpha +1}}{2\Gamma \left(\alpha +2\right)}{∥\omega ∥}_{\left[a,x\right],\infty }.\hfill \end{array}$$

The proof is completed. □

Corollary 11.

For $x=\frac{a+b}{2}$ in Theorem 11, we have

$$\begin{array}{cc}\hfill & \left|J_{{\frac{a+b}{2}}^{-}}^{\alpha }\omega \phi (a)+J_{{\frac{a+b}{2}}^{+}}^{\alpha }\omega \phi (b)-\left[J_{{\frac{a+b}{2}}^{-}}^{\alpha }\omega (a)+J_{{\frac{a+b}{2}}^{+}}^{\alpha }\omega (b)\right]\phi \left(\frac{a+b}{2}\right)\right|\hfill \\ \hfill \le & \frac{\left(M-m\right){\left(b-a\right)}^{\alpha +1}}{2^{\alpha +1}\Gamma \left(\alpha +2\right)}{∥\omega ∥}_{\left[a,b\right],\infty }.\hfill \end{array}$$

Corollary 12.

Putting $\omega \left(u\right)=\frac{1}{b-a}$ in Theorem 11, we get

$$\begin{array}{cc}\hfill & \left|J_{x^{-}}^{\alpha }\phi (a)+J_{x^{+}}^{\alpha }\phi (b)-\frac{{\left(b-x\right)}^{\alpha }+{\left(x-a\right)}^{\alpha }}{\Gamma \left(\alpha +1\right)}\phi \left(x\right)-\frac{\left(M+m\right)\left[{\left(b-x\right)}^{\alpha +1}-{\left(x-a\right)}^{\alpha +1}\right]}{2\Gamma \left(\alpha +2\right)}\right|\hfill \\ \hfill \le & \frac{\left(M-m\right)\left[{\left(b-x\right)}^{\alpha +1}+{\left(x-a\right)}^{\alpha +1}\right]}{2\Gamma \left(\alpha +2\right)}.\hfill \end{array}$$

Moreover, if we take $x=\frac{a+b}{2}$, we obtain

$$\begin{array}{cc}\hfill & \left|\frac{2^{\alpha -1}\Gamma \left(\alpha +1\right)}{{\left(b-a\right)}^{\alpha }}\left(J_{{\frac{a+b}{2}}^{-}}^{\alpha }\phi (a)+J_{{\frac{a+b}{2}}^{+}}^{\alpha }\phi (b)\right)-\phi \left(\frac{a+b}{2}\right)\right|\le \frac{\left(M-m\right)\left(b-a\right)}{4\left(\alpha +1\right)}.\hfill \end{array}$$

Corollary 13.

Choosing $\alpha =1$ in Theorem 11, we have

$$\begin{array}{cc}\hfill & \left|\stackrel{b}{\underset{a}{\int }}\omega \left(u\right)\phi (u)du-\left(\stackrel{b}{\underset{a}{\int }}\omega (u)du\right)\phi \left(x\right)\right.\hfill \\ \hfill & -\left.\frac{\left(M+m\right)\left[{\left(b-x\right)}^2-{\left(x-a\right)}^2\right]}{2}\left(\stackrel{1}{\underset{\tau }{\int }}\left(\omega (\sigma a+(1-\sigma )x)+\omega (\sigma b+(1-\sigma )x)\right)d\sigma \right)d\tau \right|\hfill \\ \hfill \le & \frac{\left(M-m\right)}{4}\left({\left(b-x\right)}^2{∥\omega ∥}_{\left[x,b\right],\infty }+{\left(x-a\right)}^2{∥\omega ∥}_{\left[a,x\right],\infty }\right).\hfill \end{array}$$

Moreover, for $x=\frac{a+b}{2}$, we get

$$\begin{array}{cc}\hfill & \left|\stackrel{b}{\underset{a}{\int }}\omega \left(u\right)\phi (u)du-\left(\stackrel{b}{\underset{a}{\int }}\omega (u)du\right)\phi \left(\frac{a+b}{2}\right)\right|\le \frac{\left(M-m\right){\left(b-a\right)}^2}{8}{∥\omega ∥}_{\left[a,b\right],\infty }.\hfill \end{array}$$

Corollary 14.

Let $\omega \left(u\right)=\frac{1}{b-a}$ in Corollary 13, then

$$\begin{array}{cc}\hfill & \left|\frac{1}{b-a}\stackrel{b}{\underset{a}{\int }}\phi (u)du-\phi \left(x\right)-\frac{\left(M+m\right)\left[{\left(b-x\right)}^2-{\left(x-a\right)}^2\right]}{4\left(b-a\right)}\right|\hfill \\ \hfill \le & \frac{\left[{\left(b-x\right)}^2+{\left(x-a\right)}^2\right]\left(M-m\right)}{4\left(b-a\right)}.\hfill \end{array}$$

For $x=\frac{a+b}{2}$, then

$$\left|\frac{1}{b-a}\stackrel{b}{\underset{a}{\int }}\phi (u)du-\phi \left(\frac{a+b}{2}\right)\right|\le \frac{\left(b-a\right)\left(M-m\right)}{8}.$$

Before giving our next result, we recall that a function $\phi :J\to \mathbb{R}$ is r-H-Hölder (Hölder condition, see [44]), if

$$\left|\phi (x)-\phi (y)\right|\le H{\left|x-y\right|}^r$$

holds for all $x,y\in J$, for some $H>0$ and $r\in \left(0,1\right]$.

Theorem 12.

Consider a differentiable function $\phi :J\to \mathbb{R}$ with ${\phi }^{\prime }\in L(J)$, $0\le a<b$, and let $\omega :J\to \mathbb{R}$ be continuous function. If ${\phi }^{\prime }$ satisfies Hölder condition for some $H>0$ and $r\in \left(0,1\right]$, then

$$\left|{\mathsf{\Xi }}^{\alpha }\left(a,x,b,\omega ,\phi \right)\right|\le H\left(\frac{{\left(b-x\right)}^{\alpha +r+1}}{\left(\alpha +r+1\right)\Gamma \left(\alpha +1\right)}{∥\omega ∥}_{\left[x,b\right],\infty }+\frac{{\left(x-a\right)}^{\alpha +r+1}}{\left(\alpha +r+1\right)\Gamma \left(\alpha +1\right)}{∥\omega ∥}_{\left[a,x\right],\infty }\right),$$

where

$$\begin{array}{cc}\hfill {\mathsf{\Xi }}^{\alpha }\left(a,x,b,\omega ,\phi \right):=& J_{x^{-}}^{\alpha }\omega \phi (a)+J_{x^{+}}^{\alpha }\omega \phi (b)-\left[J_{x^{-}}^{\alpha }\omega (a)+J_{x^{+}}^{\alpha }\omega (b)\right]\phi \left(x\right)\hfill \\ \hfill & -\frac{{\phi }^{\prime }\left(b\right){\left(b-x\right)}^{\alpha +1}-{\phi }^{\prime }\left(a\right){\left(x-a\right)}^{\alpha +1}}{2\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}\left(p_1(\tau )+p_2(\tau )\right)d\tau ,\hfill \end{array}$$

(11)

and $p_1,p_2$ are defined as in (2) and (3), respectively.

Proof. 

Applying Lemma 1, we have

$$\begin{array}{cc}\hfill & J_{x^{-}}^{\alpha }\omega \phi (a)+J_{x^{+}}^{\alpha }\omega \phi (b)-\left[J_{x^{-}}^{\alpha }\omega (a)+J_{x^{+}}^{\alpha }\omega (b)\right]\phi \left(x\right)\hfill \\ \hfill =& \frac{{\left(b-x\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}{p}_1\left(\tau \right){\phi }^{\prime }\left(\tau b+\left(1-\tau \right)x\right)d\tau -\frac{{\left(x-a\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}{p}_2\left(\tau \right){\phi }^{\prime }\left(\tau a+\left(1-\tau \right)x\right)d\tau \hfill \\ \hfill =& \frac{{\left(b-x\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}{p}_1\left(\tau \right)\left[{\phi }^{\prime }\left(\tau b+\left(1-\tau \right)x\right)-{\phi }^{\prime }\left(b\right)+{\phi }^{\prime }\left(b\right)\right]d\tau \hfill \\ \hfill & -\frac{{\left(x-a\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}{p}_2\left(\tau \right)\left[{\phi }^{\prime }\left(\tau a+\left(1-\tau \right)x\right)-{\phi }^{\prime }\left(a\right)+{\phi }^{\prime }\left(a\right)\right]d\tau \hfill \\ \hfill =& \frac{{\left(b-x\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}{p}_1\left(\tau \right)\left({\phi }^{\prime }\left(\tau b+\left(1-\tau \right)x\right)-{\phi }^{\prime }\left(b\right)\right)d\tau \hfill \\ \hfill & +\frac{{\phi }^{\prime }\left(b\right){\left(b-x\right)}^{\alpha +1}}{2\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}{p}_1\left(\tau \right)d\tau \hfill \\ \hfill & -\frac{{\left(x-a\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}{p}_2\left(\tau \right)\left({\phi }^{\prime }\left(\tau a+\left(1-\tau \right)x\right)-{\phi }^{\prime }\left(a\right)\right)d\tau \hfill \\ \hfill & -\frac{{\phi }^{\prime }\left(a\right){\left(x-a\right)}^{\alpha +1}}{2\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}{p}_2\left(\tau \right)d\tau .\hfill \end{array}$$

(12)

From (12), we get

$$\begin{array}{cc}\hfill {\mathsf{\Xi }}^{\alpha }\left(a,x,b,\omega ,\phi \right)=& \frac{{\left(b-x\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}{p}_1\left(\tau \right)\left({\phi }^{\prime }\left(\tau b+\left(1-\tau \right)x\right)-{\phi }^{\prime }\left(b\right)\right)d\tau \hfill \\ \hfill & -\frac{{\left(x-a\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}{p}_2\left(\tau \right)\left({\phi }^{\prime }\left(\tau a+\left(1-\tau \right)x\right)-{\phi }^{\prime }\left(a\right)\right)d\tau ,\hfill \end{array}$$

(13)

where ${\mathsf{\Xi }}^{\alpha }\left(a,x,b,\omega ,\phi \right)$ is defined as in (11). Applying absolute value on both sides of (13) and r-H-Hölder property of ${\phi }^{\prime }$, we obtain

$$\begin{array}{cc}\hfill & \left|{\mathsf{\Xi }}^{\alpha }\left(a,x,b,\omega ,\phi \right)\right|\hfill \\ \hfill \le & \frac{{\left(b-x\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}\left|p_1\left(\tau \right)\right|\left|{\phi }^{\prime }\left(\tau b+\left(1-\tau \right)x\right)-{\phi }^{\prime }\left(b\right)\right|d\tau \hfill \\ \hfill & +\frac{{\left(x-a\right)}^{\alpha +1}}{\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}\left|p_2\left(\tau \right)\right|\left|{\phi }^{\prime }\left(\tau a+\left(1-\tau \right)x\right)-{\phi }^{\prime }\left(a\right)\right|d\tau \hfill \\ \hfill \le & H\left(\frac{{\left(b-x\right)}^{\alpha +r+1}}{\Gamma \left(\alpha +1\right)}{∥\omega ∥}_{\left[x,b\right],\infty }+\frac{{\left(x-a\right)}^{\alpha +r+1}}{\Gamma \left(\alpha +1\right)}{∥\omega ∥}_{\left[a,x\right],\infty }\right)\stackrel{1}{\underset{0}{\int }}{\left(1-\tau \right)}^{\alpha +r}d\tau \hfill \\ \hfill =& H\left(\frac{{\left(b-x\right)}^{\alpha +r+1}}{\left(\alpha +r+1\right)\Gamma \left(\alpha +1\right)}{∥\omega ∥}_{\left[x,b\right],\infty }+\frac{{\left(x-a\right)}^{\alpha +r+1}}{\left(\alpha +r+1\right)\Gamma \left(\alpha +1\right)}{∥\omega ∥}_{\left[a,x\right],\infty }\right).\hfill \end{array}$$

The proof is completed. □

Corollary 15.

Letting $x=\frac{a+b}{2}$ in Theorem 12, we have

$$\begin{array}{cc}\hfill & \left|J_{{\frac{a+b}{2}}^{-}}^{\alpha }\omega \phi (a)+J_{{\frac{a+b}{2}}^{+}}^{\alpha }\omega \phi (b)-\left[J_{{\frac{a+b}{2}}^{-}}^{\alpha }\omega (a)+J_{{\frac{a+b}{2}}^{+}}^{\alpha }\omega (b)\right]\phi \left(\frac{a+b}{2}\right)\right|\hfill \\ \hfill \le & H\frac{{\left(b-a\right)}^{\alpha +r+1}}{2^{\alpha +r}\left(\alpha +r+1\right)\Gamma \left(\alpha +1\right)}{∥\omega ∥}_{\left[a,b\right],\infty }\hfill \\ \hfill & +\frac{{\left(b-a\right)}^{\alpha +1}\left({\phi }^{\prime }\left(b\right)-{\phi }^{\prime }\left(a\right)\right)}{2^{\alpha +2}\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}\left(p_1(\tau )+p_2(\tau )\right)d\tau .\hfill \end{array}$$

Corollary 16.

Choosing $\omega \left(u\right)=\frac{1}{b-a}$ in Theorem 12, we get

$$\begin{array}{cc}\hfill & \left|J_{x^{-}}^{\alpha }\phi (a)+J_{x^{+}}^{\alpha }\phi (b)-\frac{{\left(b-x\right)}^{\alpha }+{\left(x-a\right)}^{\alpha }}{\Gamma \left(\alpha +1\right)}\phi \left(x\right)\right|\hfill \\ \hfill \le & H\frac{{\left(b-x\right)}^{\alpha +r+1}+{\left(x-a\right)}^{\alpha +r+1}}{\left(\alpha +r+1\right)\Gamma \left(\alpha +1\right)}+\frac{{\phi }^{\prime }\left(b\right){\left(b-x\right)}^{\alpha +1}-{\phi }^{\prime }\left(a\right){\left(x-a\right)}^{\alpha +1}}{2\Gamma \left(\alpha +2\right)}.\hfill \end{array}$$

Moreover, if we take $x=\frac{a+b}{2}$, we obtain

$$\begin{array}{cc}\hfill & \left|\frac{2^{\alpha -1}\Gamma \left(\alpha +1\right)}{{\left(b-a\right)}^{\alpha }}\left(J_{{\frac{a+b}{2}}^{-}}^{\alpha }\phi (a)+J_{{\frac{a+b}{2}}^{+}}^{\alpha }\phi (b)\right)-\phi \left(\frac{a+b}{2}\right)\right|\hfill \\ \hfill \le & H\frac{{\left(b-a\right)}^{r+1}}{2^{1+r}\left(\alpha +r+1\right)}+\frac{\left(b-a\right)\left({\phi }^{\prime }\left(b\right)-{\phi }^{\prime }\left(a\right)\right)}{8\left(\alpha +1\right)}.\hfill \end{array}$$

Corollary 17.

For $\alpha =1$ in Theorem 12, we have

$$\begin{array}{cc}\hfill & \left|\stackrel{b}{\underset{a}{\int }}\omega \left(u\right)\phi (u)du-\left(\stackrel{b}{\underset{a}{\int }}\omega \left(u\right)du\right)\phi \left(x\right)\right|\hfill \\ \hfill \le & H\frac{{\left(b-x\right)}^{r+2}{∥\omega ∥}_{\left[x,b\right],\infty }+{\left(x-a\right)}^{r+2}{∥\omega ∥}_{\left[a,x\right],\infty }}{r+2}\hfill \\ \hfill & +\frac{{\phi }^{\prime }\left(b\right){\left(b-x\right)}^2-{\phi }^{\prime }\left(a\right){\left(x-a\right)}^2}{2}\hfill \\ \hfill & \times \stackrel{1}{\underset{0}{\int }}\left(\stackrel{1}{\underset{\tau }{\int }}\left(\omega \left(\sigma a+\left(1-\sigma \right)x\right)+\omega \left(\sigma b+\left(1-\sigma \right)x\right)\right)d\sigma \right)d\tau .\hfill \end{array}$$

Moreover, for $x=\frac{a+b}{2}$, we get

$$\begin{array}{cc}\hfill & \left|\stackrel{b}{\underset{a}{\int }}\omega \left(u\right)\phi (u)du-\left(\stackrel{b}{\underset{a}{\int }}\omega \left(u\right)du\right)\phi \left(\frac{a+b}{2}\right)\right|\hfill \\ \hfill \le & H\frac{{\left(b-a\right)}^{r+2}}{2^{r+1}\left(r+2\right)}{∥\omega ∥}_{\left[a,b\right],\infty }+\frac{{\phi }^{\prime }\left(b\right){\left(b-a\right)}^2-{\phi }^{\prime }\left(a\right){\left(b-a\right)}^2}{8}\hfill \\ \hfill & \times \stackrel{1}{\underset{0}{\int }}\left(\stackrel{1}{\underset{\tau }{\int }}\left(\omega \left(\sigma a+\left(1-\sigma \right)\frac{a+b}{2}\right)+\omega \left(\sigma b+\left(1-\sigma \right)\frac{a+b}{2}\right)\right)d\sigma \right)d\tau .\hfill \end{array}$$

Corollary 18.

In Corollary 17, if we choose $\omega \left(u\right)=\frac{1}{b-a}$, we obtain

$$\begin{array}{cc}\hfill & \left|\frac{1}{b-a}\stackrel{b}{\underset{a}{\int }}\phi (u)du-\phi \left(x\right)\right|\hfill \\ \hfill \le & H\frac{{\left(b-x\right)}^{r+2}+{\left(x-a\right)}^{r+2}}{\left(r+2\right)\left(b-a\right)}+\frac{{\phi }^{\prime }\left(b\right){\left(b-x\right)}^2-{\phi }^{\prime }\left(a\right){\left(x-a\right)}^2}{4\left(b-a\right)}.\hfill \end{array}$$

Moreover, if we take $x=\frac{a+b}{2}$, we have

$$\begin{array}{cc}\hfill & \left|\frac{1}{b-a}\stackrel{b}{\underset{a}{\int }}\phi (u)du-\phi \left(\frac{a+b}{2}\right)\right|\hfill \\ \hfill \le & H\frac{{\left(b-a\right)}^{r+2}}{2^{r+1}\left(r+2\right)}+\frac{\left({\phi }^{\prime }\left(b\right)-{\phi }^{\prime }\left(a\right)\right)\left(b-a\right)}{16}.\hfill \end{array}$$

Corollary 19.

Let ${\phi }^{\prime }$ be a L–Lipschitzian function ($H=L$ and $r=1$) in Theorem 12, then

$$\left|{\mathsf{\Xi }}^{\alpha }\left(a,x,b,\omega ,\phi \right)\right|\le L\left(\frac{{\left(b-x\right)}^{\alpha +2}}{\left(\alpha +2\right)\Gamma \left(\alpha +1\right)}{∥\omega ∥}_{\left[x,b\right],\infty }+\frac{{\left(x-a\right)}^{\alpha +2}}{\left(\alpha +2\right)\Gamma \left(\alpha +1\right)}{∥\omega ∥}_{\left[a,x\right],\infty }\right),$$

where ${\mathsf{\Xi }}^{\alpha }$ is defined by (11).

Corollary 20.

For $x=\frac{a+b}{2}$ in Corollary 19, we get

$$\begin{array}{cc}\hfill & \left|J_{{\frac{a+b}{2}}^{-}}^{\alpha }\omega \phi (a)+J_{{\frac{a+b}{2}}^{+}}^{\alpha }\omega \phi (b)-\left[J_{{\frac{a+b}{2}}^{-}}^{\alpha }\omega (a)+J_{{\frac{a+b}{2}}^{+}}^{\alpha }\omega (b)\right]\phi \left(\frac{a+b}{2}\right)\right|\hfill \\ \hfill \le & L\frac{{\left(b-a\right)}^{\alpha +2}}{2^{\alpha +1}\left(\alpha +2\right)\Gamma \left(\alpha +1\right)}{∥\omega ∥}_{\left[a,b\right],\infty }\hfill \\ \hfill & +\frac{{\left(b-a\right)}^{\alpha +1}\left({\phi }^{\prime }\left(b\right)-{\phi }^{\prime }\left(a\right)\right)}{2^{\alpha +2}\Gamma \left(\alpha \right)}\stackrel{1}{\underset{0}{\int }}\left(p_1(\tau )+p_2(\tau )\right)d\tau .\hfill \end{array}$$

Corollary 21.

In Corollary 19, if we choose $\omega \left(u\right)=\frac{1}{b-a}$, we obtain

$$\begin{array}{cc}\hfill & \left|J_{x^{-}}^{\alpha }\phi (a)+J_{x^{+}}^{\alpha }\phi (b)-\frac{{\left(b-x\right)}^{\alpha }+{\left(x-a\right)}^{\alpha }}{\Gamma \left(\alpha +1\right)}\phi \left(x\right)\right|\hfill \\ \hfill \le & L\frac{{\left(b-x\right)}^{\alpha +2}+{\left(x-a\right)}^{\alpha +2}}{\left(\alpha +2\right)\Gamma \left(\alpha +1\right)}+\frac{{\phi }^{\prime }\left(b\right){\left(b-x\right)}^{\alpha +1}-{\phi }^{\prime }\left(a\right){\left(x-a\right)}^{\alpha +1}}{2\Gamma \left(\alpha +2\right)}.\hfill \end{array}$$

Moreover, if we take $x=\frac{a+b}{2}$, we have

$$\begin{array}{cc}\hfill & \left|\frac{2^{\alpha -1}\Gamma \left(\alpha +1\right)}{{\left(b-a\right)}^{\alpha }}\left(J_{{\frac{a+b}{2}}^{-}}^{\alpha }\phi (a)+J_{{\frac{a+b}{2}}^{+}}^{\alpha }\phi (b)\right)-\phi \left(\frac{a+b}{2}\right)\right|\hfill \\ \hfill \le & L\frac{{\left(b-a\right)}^2}{4\left(\alpha +2\right)}+\frac{\left(b-a\right)\left({\phi }^{\prime }\left(b\right)-{\phi }^{\prime }\left(a\right)\right)}{8\left(\alpha +1\right)}.\hfill \end{array}$$

Corollary 22.

Choosing $\alpha =1$ in Corollary 19, we get

$$\begin{array}{cc}\hfill & \left|\stackrel{b}{\underset{a}{\int }}\omega \left(u\right)\phi (u)du-\left(\stackrel{b}{\underset{a}{\int }}\omega \left(u\right)du\right)\phi \left(x\right)\right|\hfill \\ \hfill \le & L\frac{{\left(b-x\right)}^3{∥\omega ∥}_{\left[x,b\right],\infty }+{\left(x-a\right)}^3{∥\omega ∥}_{\left[a,x\right],\infty }}{3}\hfill \\ \hfill & +\frac{{\phi }^{\prime }\left(b\right){\left(b-x\right)}^2-{\phi }^{\prime }\left(a\right){\left(x-a\right)}^2}{2}\hfill \\ \hfill & \times \stackrel{1}{\underset{0}{\int }}\left(\stackrel{1}{\underset{\tau }{\int }}\left(\omega \left(\sigma a+\left(1-\sigma \right)x\right)+\omega \left(\sigma b+\left(1-\sigma \right)x\right)\right)d\sigma \right)d\tau .\hfill \end{array}$$

Moreover, for $x=\frac{a+b}{2}$, we obtain

$$\begin{array}{cc}\hfill & \left|\stackrel{b}{\underset{a}{\int }}\omega \left(u\right)\phi (u)du-\left(\stackrel{b}{\underset{a}{\int }}\omega \left(u\right)du\right)\phi \left(\frac{a+b}{2}\right)\right|\hfill \\ \hfill \le & L\frac{{\left(b-a\right)}^3}{12}{∥\omega ∥}_{\left[a,b\right],\infty }+\frac{{\phi }^{\prime }\left(b\right){\left(b-a\right)}^2-{\phi }^{\prime }\left(a\right){\left(b-a\right)}^2}{8}\hfill \\ \hfill & \times \stackrel{1}{\underset{0}{\int }}\left(\stackrel{1}{\underset{\tau }{\int }}\left(\omega \left(\sigma a+\left(1-\sigma \right)\frac{a+b}{2}\right)+\omega \left(\sigma b+\left(1-\sigma \right)\frac{a+b}{2}\right)\right)d\sigma \right)d\tau .\hfill \end{array}$$

Corollary 23.

For $\omega \left(u\right)=\frac{1}{b-a}$ in Corollary 22, we have

$$\begin{array}{cc}\hfill & \left|\frac{1}{b-a}\stackrel{b}{\underset{a}{\int }}\phi (u)du-\phi \left(x\right)\right|\hfill \\ \hfill \le & L\frac{{\left(b-x\right)}^3+{\left(x-a\right)}^3}{3\left(b-a\right)}+\frac{{\phi }^{\prime }\left(b\right){\left(b-x\right)}^2-{\phi }^{\prime }\left(a\right){\left(x-a\right)}^2}{4\left(b-a\right)}.\hfill \end{array}$$

Moreover, for $x=\frac{a+b}{2}$, we get

$$\begin{array}{cc}\hfill & \left|\frac{1}{b-a}\stackrel{b}{\underset{a}{\int }}\phi (u)du-\phi \left(\frac{a+b}{2}\right)\right|\hfill \\ \hfill \le & L\frac{{\left(b-a\right)}^3}{12}+\frac{\left({\phi }^{\prime }\left(b\right)-{\phi }^{\prime }\left(a\right)\right)\left(b-a\right)}{16}.\hfill \end{array}$$

4. Applications

We shall consider the following special means for different positive real numbers a and $b,$ where $a<b$:

• Arithmetic mean: $\mathcal{A}\left(a,b\right)=\frac{a+b}{2}$.
• p–Logarithmic mean: ${\mathcal{L}}_p\left(a,b\right)={\left(\frac{b^{p+1}-a^{p+1}}{\left(p+1\right)\left(b-a\right)}\right)}^{\frac{1}{p}}$,$p\in \mathbb{Z}\backslash \{0,-1\}$.

Let recall from [23] the following quasi-convex functions, $\phi \left(u\right)=u^n,$ $n\in \mathbb{N},$ $n\ge 2$ and $\phi \left(u\right)=\frac{1}{u}$ for all $u>0,$ respectively.

Using Section 2, we are in position to prove the following results regarding above special means.

Proposition 1.

Let $a,b\in \mathbb{R}$, $0<a<b$, $n\in \mathbb{N}$ and $n\ge 2,$ then

$$\left|{\mathcal{L}}_{n+1}^{n+1}\left(a,b\right)-{\mathcal{A}}^{n+1}\left(a,b\right)\right|\le \frac{nb\left(b-a\right)}{4}\mathcal{A}\left({\mathcal{A}}^{n-1}\left(a,b\right),b^{n-1}\right).$$

Proof. 

Taking $\phi \left(u\right)=u^n,u>0$ for $x=\frac{a+b}{2}$ and $\omega (u)=u$ in Corollary 3, we get the desired result. □

Proposition 2.

Let $a,b\in \mathbb{R}$, $0<a<b$, $n\in \mathbb{N}$ and $n\ge 2,$ then

$$\left|{\mathcal{L}}_{n-1}^{n-1}\left(a,b\right)-\frac{{\mathcal{L}}_n^n\left(a,b\right)}{\mathcal{A}\left(a,b\right)}\right|\le \frac{b^n\left(b-a\right)}{4}\mathcal{A}\left(\frac{1}{a^2},\frac{1}{{\mathcal{A}}^2\left(a,b\right)}\right).$$

Proof. 

Choosing $\phi \left(u\right)=\frac{1}{u},u>0$ for $x=\frac{a+b}{2}$ and $\omega (u)=u^n$ in Corollary 3, we obtain the desired result. □

Proposition 3.

Let $a,b\in \mathbb{R}$, $0<a<b$, $p>1,n\in \mathbb{N}$ and $n\ge 2,$ then

$$\left|{\mathcal{L}}_{n+1}^{n+1}\left(a,b\right)-{\mathcal{A}}^{n+1}\left(a,b\right)\right|\le \frac{nb\left(b-a\right)}{2{(p+1)}^{\frac{1}{p}}}\mathcal{A}\left({\mathcal{A}}^{n-1}\left(a,b\right),b^{n-1}\right).$$

Proof. 

Taking $\phi \left(u\right)=u^n,u>0$ for $x=\frac{a+b}{2}$ and $\omega (u)=u$ in Corollary 6, we get the desired result. □

Proposition 4.

Let $a,b\in \mathbb{R}$, $0<a<b$, $p>1,n\in \mathbb{N}$ and $n\ge 2,$ then

$$\left|{\mathcal{L}}_{n-1}^{n-1}\left(a,b\right)-\frac{{\mathcal{L}}_n^n\left(a,b\right)}{\mathcal{A}\left(a,b\right)}\right|\le \frac{b^n\left(b-a\right)}{2{(p+1)}^{\frac{1}{p}}}\mathcal{A}\left(\frac{1}{a^2},\frac{1}{{\mathcal{A}}^2\left(a,b\right)}\right).$$

Proof. 

Choosing $\phi \left(u\right)=\frac{1}{u},u>0$ for $x=\frac{a+b}{2}$ and $\omega (u)=u^n$ in Corollary 6, we obtain the desired result. □

5. Conclusions

The main results and future research of the article can be summarized as follows:

• A new identity regarding fractional weighted Ostrowski-type is established.
• New fractional weighted Ostrowski-type inequalities for quasi-convex functions using the above identity are deduced.
• Several further results for function with a bounded first derivative are given.
• Some applications to special means are obtained.
• The efficiency of our results is shown.
• As future research, from our results, interested reader can find several new interesting inequalities from many areas of pure and applied sciences. Moreover, they can derive (using our technique) applications to special means for different quasi-convex functions.