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Generalized Hermite-Hadamard type inequalities involving fractional integral operators

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  • 2, 3,
  • 4Email author and
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Journal of Inequalities and Applications20172017:169

https://doi.org/10.1186/s13660-017-1444-6

  • Received: 14 February 2017
  • Accepted: 30 June 2017
  • Published:

Abstract

In this article, a new general integral identity involving generalized fractional integral operators is established. With the help of this identity new Hermite-Hadamard type inequalities are obtained for functions whose absolute values of derivatives are convex. As a consequence, the main results of this paper generalize the existing Hermite-Hadamard type inequalities involving the Riemann-Liouville fractional integral.

Keywords

  • Hermite-Hadamard inequality
  • convex function
  • Hölder inequality
  • fractional integral operator

MSC

  • 26A33
  • 26D10
  • 26D15
  • 33B20

1 Introduction and preliminaries

During the last century the theory of convexity has emerged as an interesting and fascinating field of mathematics. It plays a pivotal role in optimization theory, functional analysis, control theory and economics etc.

A function \(f:I\subseteq\mathbb{R}\rightarrow\mathbb{R}\) is said to be convex if the inequality
$$\begin{aligned} f \bigl( tx+ ( 1-t ) y \bigr) \leq tf ( x ) + ( 1-t ) f ( y ) \end{aligned}$$
holds for all \(x,y\in I \) and \(t\in [ 0,1 ] \).
The following inequality is a so-called classical Hermite-Hadamard type inequality for convex functions. Let \(f:I=[a,b]\subseteq\mathbb{R} \rightarrow\mathbb{R}\) be a convex function and \(a,b\in I\) with \(a< b\), then
$$\begin{aligned} f \biggl( \frac{a+b}{2} \biggr) \leq\frac{1}{b-a} \int_{a}^{b}f ( x ) \,dx\leq\frac{f ( a ) +f ( b ) }{2}. \end{aligned}$$
(1.1)
This inequality is one of the most useful inequalities in mathematical analysis. For new proofs, noteworthy extension, generalizations and numerous applications on this inequality, see, e.g., [13] where further references are given.

The relationship between theory of convexity and theory of inequalities has motivated many researchers to study these theories in depth. As a consequence of this fact several inequalities have been obtained via convex functions; see [1].

The history of fractional calculus can be traced back to the letter of L’Hospital to Leibniz in which he inquired him about the notation he was using for the nth derivative of the linear function \(f(x)=x\), \(\frac{\mathrm{D}^{n}x}{\mathrm{D}x^{n}}\). L’Hospital asked the question: what would the result be if \(n=\frac{1}{2}\). Leibniz replied: An apparent paradox, from which one day useful consequences will be drawn. Nowadays fractional calculus has become a powerful tool in many branches of mathematics. Sarikaya et al. [4] used the definitions of Riemannn-Liouville integrals and developed a new generalization of Hermite-Hadamard inequality. This result inspired many researchers to study this area. For more details, and for recent results and recently found properties concerning this operator one can consult [412].

We need some definition and mathematical preliminaries of fractional calculus theory for using in this study as follows.

Definition 1.1

Let \(f\in L[a,b]\). The Riemann-Liouville integrals \(J_{a+}^{\alpha}f\) and \(J_{b-}^{\alpha}f\) of order \(\alpha> 0\) with \(a\geq0\) are defined by
$$ J_{a+}^{\alpha}f(x)=\frac{1}{\Gamma(\alpha)} \int_{a}^{x}(x-t)^{ \alpha-1}f(t)\,dt,\quad x>a, $$
and
$$ J_{b-}^{\alpha}f(x)=\frac{1}{\Gamma(\alpha)} \int_{x}^{b}(t-x)^{ \alpha-1}f(t)\,dt,\quad x< b, $$
respectively. Here \(\Gamma(t)\) is the Gamma function and its definition is \(\Gamma(t)=\int_{0}^{\infty}e^{-x}x^{t-1}\,dx\). It is to be noted that \(J_{a+}^{0}f(x)=J_{b-}^{0}f(x)=f(x)\); in the case of \(\alpha=1\), the fractional integral reduces to the classical integral.

In [13], Zhu et al. established a new identity for differentiable convex mappings via the Riemann-Liouville fractional integral.

Lemma 1.1

[13]

Let \(f:[a,b]\rightarrow\mathbb{R}\) be a differentiable mapping on \((a,b)\) with \(a< b\). If \(f^{\prime}\in L[a,b]\), then the following equality for fractional integrals hold:
$$\begin{aligned}& \frac{\Gamma(\alpha+1)}{2(b-a)^{\alpha}}\bigl[\bigl(J_{\alpha^{-}}^{a}f \bigr) (b)+ \bigl(J_{\alpha^{+}}^{b}f\bigr) (a)\bigr]-f \biggl( \frac{a+b}{2} \biggr) \\& \quad =\frac{b-a}{2} \biggl[ \int_{0}^{1}k(t)f^{\prime}\bigl(ta+(1-t)b \bigr)\,dt- \int _{0}^{1}\bigl[(1-t)^{ \alpha}-t^{\alpha} \bigr]f^{\prime}\bigl(ta+(1-t)b\bigr)\,dt\biggr], \end{aligned}$$
(1.2)
where
$$k(t)= \textstyle\begin{cases} 1, & 0< t\leq\frac{1}{2}, \\ -1, & \frac{1}{2}< t\leq1. \end{cases} $$

Using the above identity, they gave the following result for the Riemann-Liouville fractional integral.

Theorem 1.1

[13]

Let \(f:[a,b]\rightarrow\mathbb{R}\) be a differentiable mapping on \((a,b)\) with \(a< b\). If \(\vert f^{\prime} \vert \) is convex on \([a,b]\), then the following fractional inequality for fractional integrals holds:
$$\begin{aligned}& \biggl\vert \frac{\Gamma(\alpha+1)}{ 2(b-a)^{\alpha}}\bigl[\bigl(J_{\alpha^{-}} ^{a}f\bigr) (b)+ \bigl(J_{\alpha^{+}}^{b}f\bigr) (a) \bigr]-f \biggl( \frac{a+b}{2} \biggr) \biggr\vert \\& \quad \leq\frac{b-a}{4(\alpha+1)} \biggl(\alpha+3-\frac{1}{2^{\alpha-1}} \biggr) \bigl[ \bigl\vert f^{\prime}(a) \bigr\vert + \bigl\vert f^{\prime }(b) \bigr\vert \bigr]. \end{aligned}$$
(1.3)
In [14], Raina introduced a class of functions defined formally by
$$ \mathcal{F}_{\rho,\lambda}^{\sigma}(x) = \mathcal{F}_{\rho,\lambda }^{\sigma(0),\sigma(1),\ldots} (x)= \sum_{k=0}^{\infty} \frac{ \sigma(k)}{\Gamma(\rho k+\lambda)}x^{k} \quad \bigl(\rho,\lambda>0; \vert x \vert < \mathbf{R}\bigr), $$
(1.4)
where the coefficients \(\sigma(k)\) (\(k\in\mathbb{N}=\mathbb{N} \cup\{0\}\)) are a bounded sequence of positive real numbers and R is the set of real numbers. With the help of (1.4), Raina [14] and Agarwal et al. [15] defined the following left-sided and right-sided fractional integral operators, respectively:
$$\begin{aligned}& \bigl( \mathcal{J}_{\rho,\lambda,a+;w}^{\sigma}\varphi \bigr) (x)= \int_{a}^{x}(x-t)^{\lambda-1} \mathcal{F}_{\rho,\lambda}^{\sigma} \bigl[w(x-t)^{\rho}\bigr] \varphi(t)\,dt \quad (x>a>0), \end{aligned}$$
(1.5)
$$\begin{aligned}& \bigl( \mathcal{J}_{\rho,\lambda,b-;w}^{\sigma}\varphi \bigr) (x)= \int_{x}^{b}(t-x)^{\lambda-1} \mathcal{F}_{\rho,\lambda}^{\sigma} \bigl[w(t-x)^{\rho}\bigr] \varphi(t)\,dt \quad (0< x< b), \end{aligned}$$
(1.6)
where \(\lambda,\rho>0\), \(w\in\mathbb{R}\) and \(\varphi(t)\) is such that the integral on the right side exists. Recently some new integral inequalities involving this operator have appeared in the literature (see, e.g., [1522]).
It is easy to verify that \(\mathcal{J}_{\rho,\lambda,a+;w}^{\sigma }\varphi(x)\) and \(\mathcal{J}_{\rho,\lambda,b-;w}^{\sigma}\varphi (x)\) are bounded integral operators on \(L(a,b)\), if
$$ \mathfrak{M}:=\mathcal{F}_{\rho,\lambda+1}^{\sigma} \bigl[w(b-a)^{ \rho}\bigr] < \infty. $$
(1.7)
In fact, for \(\varphi\in L(a,b)\), we have
$$ \bigl\Vert \mathcal{J}_{\rho,\lambda,a+;w}^{\sigma}\varphi(x) \bigr\Vert _{1} \leq \mathfrak{M} (b-a)^{\lambda} \Vert \varphi \Vert _{1} $$
(1.8)
and
$$ \bigl\Vert \mathcal{J}_{\rho,\lambda,b-;w}^{\sigma}\varphi(x) \bigr\Vert _{1} \leq \mathfrak{M} (b-a)^{\lambda} \Vert \varphi \Vert _{1}, $$
(1.9)
where
$$ \Vert \varphi \Vert _{p}:= \biggl( \int_{a}^{b} \bigl\vert \varphi(t) \bigr\vert ^{p} \,dt \biggr) ^{ \frac{1}{p}}. $$

Here, many useful fractional integral operators can be obtained by specializing the coefficient \(\sigma(k)\). For instance the classical Riemann-Liouville fractional integrals \(J_{a+}^{\alpha}\) and \(J_{b-}^{\alpha}\) of order α follow easily by setting \(\lambda=\alpha\), \(\sigma(0)=1\) and \(w=0\) in (1.5) and (1.6).

Motivated by the work in [1315], firstly, we will prove a generalization of the identity given by Zhu et al. using generalized fractional integral operators. Then we will give some new Hermite-Hadamard type inequalities, which are generalizations of the results in [13] to the case \(\lambda=\alpha\), \(\sigma(0)=1\) and \(w=0\). Our results can be viewed as a significant extension and generalization of the previously known results.

2 Results and discussions

In this section, we derive our main results. For the sake of simplicity, we denote
$$ L_{f}(a,b;w;J):=\frac{1}{2(b-a)^{\lambda}} \bigl[ \bigl( \mathcal{J} ^{\sigma}_{\rho,\lambda,b^{-};w}f \bigr) (a)+ \bigl( \mathcal{J}^{ \sigma}_{\rho,\lambda,a^{+};w }f \bigr) (b) \bigr] -\mathcal{F}_{ \rho,\lambda+1}^{\sigma} \bigl[w(b-a)^{\rho}\bigr]f \biggl( \frac{a+b}{2} \biggr) . $$

Lemma 2.1

Let \(f :[a,b]\to\mathbb{R}\) be a differentiable mapping on \((a,b)\) with \(a< b\). If \(f^{\prime} \in L[a,b] \), then the following equality for generalized fractional integral operators holds:
$$\begin{aligned} L_{f}(a,b;w;J) =&\frac{(b-a)}{2} \biggl\{ \int_{0}^{1}k(t)f^{\prime}\bigl(ta+(1-t)b \bigr)\,dt \\ & {} - \int_{0}^{1}(1-t)^{\lambda} \mathcal{F}_{\rho,\lambda+1}^{\sigma} \bigl[w(b-a)^{\rho}(1-t)^{\rho} \bigr] f^{\prime} \bigl( ta+(1-t)b \bigr) \,dt \\ &{}+ \int_{0}^{1}t^{\lambda}\mathcal{F}_{\rho,\lambda+1}^{\sigma} \bigl[w(b-a)^{ \rho}t^{\rho}\bigr] f^{\prime} \bigl( ta+(1-t)b \bigr) \,dt \biggr\} , \end{aligned}$$
where
$$k(t)= \textstyle\begin{cases} \mathcal{F}_{\rho,\lambda+1}^{\sigma} [w(b-a)^{\rho}],& 0< t\leq \frac{1}{2}, \\ -\mathcal{F}_{\rho,\lambda+1}^{\sigma} [w(b-a)^{\rho}],& \frac{1}{2}< t\leq1, \end{cases} $$
\(\rho,\lambda>0\), \(w\in\mathbb{R}\).

Proof

It suffices to note that
$$\begin{aligned} I =& \int_{0}^{\frac{1}{2}}\mathcal{F}_{\rho,\lambda+1}^{\sigma} \bigl[w(b-a)^{ \rho}\bigr]f^{\prime}\bigl(ta+(1-t)b\bigr)\,dt \\ & {} - \int_{\frac{1}{2}}^{1}\mathcal{F}_{\rho,\lambda+1}^{\sigma} \bigl[w(b-a)^{ \rho}\bigr]f^{\prime}\bigl(ta+(1-t)b\bigr)\,dt \\ &{}- \int_{0}^{1}(1-t)^{\lambda} \mathcal{F}_{\rho,\lambda+1}^{\sigma } \bigl[w(b-a)^{\rho}(1-t)^{\rho} \bigr] f^{\prime} \bigl( ta+(1-t)b \bigr) \,dt \\ &{}+ \int_{0}^{1}t^{\lambda}\mathcal{F}_{\rho,\lambda+1}^{\sigma} \bigl[w(b-a)^{ \rho}t^{\rho}\bigr] f^{\prime} \bigl( ta+(1-t)b \bigr) \,dt \\ :=&I_{1}+I_{2}+I_{3}+I_{4}. \end{aligned}$$
(2.1)
Changing variables with \(x=ta+(1-t)b\), we get
$$\begin{aligned}& \begin{aligned} I_{1} &= \int_{0}^{\frac{1}{2}}\mathcal{F}_{\rho,\lambda+1}^{\sigma } \bigl[w(b-a)^{\rho}\bigr]f^{\prime}\bigl(ta+(1-t)b\bigr)\,dt \\ &=\frac{\mathcal{F}_{\rho,\lambda+1}^{\sigma} [w(b-a)^{\rho}]}{b-a} \biggl[ f(b)-f \biggl( \frac{a+b}{2} \biggr) \biggr] , \end{aligned} \\& \begin{aligned} I_{2} &=- \int_{\frac{1}{2}}^{1}\mathcal{F}_{\rho,\lambda+1}^{\sigma } \bigl[w(b-a)^{\rho}\bigr]f^{\prime}\bigl(ta+(1-t)b\bigr)\,dt \\ &=\frac{\mathcal{F}_{\rho,\lambda+1}^{\sigma} [w(b-a)^{\rho}]}{b-a} \biggl[ f(a)-f \biggl( \frac{a+b}{2} \biggr) \biggr] . \end{aligned} \end{aligned}$$
Integrating by parts, we have
$$\begin{aligned} I_{3} =&- \int_{0}^{1}(1-t)^{\lambda} \mathcal{F}_{\rho,\lambda+1} ^{\sigma} \bigl[w(b-a)^{\rho}(1-t)^{\rho} \bigr] f^{\prime} \bigl( ta+(1-t)b \bigr) \,dt \\ =&\frac{1}{b-a}(1-t)^{\lambda}\mathcal{F}_{\rho,\lambda+1}^{ \sigma} \bigl[w(b-a)^{\rho}(1-t)^{\rho}\bigr] f \bigl( ta+(1-t)b \bigr) \big|_{0} ^{1} \\ &{}+\frac{1}{b-a} \int_{0}^{1}(1-t)^{\lambda-1} \mathcal{F}_{\rho, \lambda}^{\sigma} \bigl[w(b-a)^{\rho}(1-t)^{\rho} \bigr]f \bigl( ta+(1-t)b \bigr) \,dt \\ =&-\frac{1}{b-a}\mathcal{F}_{\rho,\lambda+1}^{\sigma} \bigl[w(b-a)^{ \rho}\bigr]f(b) \\ &{}+\frac{1}{b-a} \int_{a}^{b} \biggl( \frac{x-a}{b-a} \biggr) ^{\lambda-1} \mathcal{F}_{\rho,\lambda}^{\sigma} \biggl[w(b-a)^{\rho} \biggl( \frac {x-a}{b-a} \biggr) ^{\rho}\biggr]\frac{f(x)}{b-a}\,dx \\ =&-\frac{1}{b-a}\mathcal{F}_{\rho,\lambda+1}^{\sigma} \bigl[w(b-a)^{ \rho}\bigr]f(b) +\frac{1}{(b-a)^{\lambda+1}} \bigl( \mathcal{J}^{\sigma} _{\rho,\lambda,b^{-};w}f \bigr) (a). \end{aligned}$$
(2.2)
Analogously
$$\begin{aligned} I_{4} =& \int_{0}^{1}t^{\lambda}\mathcal{F}_{\rho,\lambda+1}^{ \sigma} \bigl[w(b-a)^{\rho}t^{\rho}\bigr] f^{\prime} \bigl( ta+(1-t)b \bigr) \,dt \\ =&-\frac{1}{b-a}t^{\lambda}\mathcal{F}_{\rho,\lambda+1}^{\sigma} \bigl[w(b-a)^{\rho}t^{\rho}\bigr] f \bigl( ta+(1-t)b \bigr) \big|_{0}^{1} \\ &{}+\frac{1}{b-a} \int_{0}^{1}t^{\lambda-1}\mathcal{F}_{\rho,\lambda }^{\sigma} \bigl[w(b-a)^{\rho}t^{\rho}\bigr]f \bigl( ta+(1-t)b \bigr) \,dt \\ =&-\frac{1}{b-a}\mathcal{F}_{\rho,\lambda+1}^{\sigma} \bigl[w(b-a)^{ \rho}\bigr]f(a) \\ &{}+\frac{1}{b-a} \int_{a}^{b} \biggl( \frac{b-x}{b-a} \biggr) ^{\lambda-1} \mathcal{F}_{\rho,\lambda}^{\sigma} \biggl[w(b-a)^{\rho} \biggl( \frac {b-x}{b-a} \biggr) ^{\rho}\biggr]\frac{f(x)}{b-a}\,dx \\ =&-\frac{1}{b-a}\mathcal{F}_{\rho,\lambda+1}^{\sigma} \bigl[w(b-a)^{ \rho}\bigr]f(a) +\frac{1}{(b-a)^{\lambda+1}} \bigl( \mathcal{J}^{\sigma} _{\rho,\lambda,a^{+};w}f \bigr) (b). \end{aligned}$$
(2.3)
 □
Substituting the resulting equalities into equality (2.1), we have
$$\begin{aligned} I =&\frac{-2}{b-a}\mathcal{F}_{\rho,\lambda+1}^{\sigma} \bigl[w(b-a)^{ \rho}\bigr]f\biggl(\frac{a+b}{2}\biggr) \\ &{} +\frac{1}{(b-a)^{\lambda+1}} \bigl[ \bigl( \mathcal{J}^{\sigma}_{ \rho,\lambda,b^{-};w}f \bigr) (a)+ \bigl( \mathcal{J}^{\sigma}_{ \rho,\lambda,a^{+};w}f \bigr) (b) \bigr] . \end{aligned}$$
(2.4)
Thus, multiplying both sides by \(\frac{(b-a)}{2}\), the result is obtained.

Remark 2.1

Choosing \(\lambda=\alpha\), \(\sigma(0)=1\) and \(w=0\) in Lemma 2.1, equality (2.1) reduces to equality (1.2).

Theorem 2.1

Let \(f :[a,b]\to\mathbb{R}\) be a differentiable function on \([a,b]\) with \(a< b\). If \(\vert f^{\prime} \vert \) is convex on \((a,b)\), then the following inequality for generalized fractional integral operators holds:
$$\begin{aligned} \bigl\vert L_{f}(a,b;w;J) \bigr\vert \leq& \frac{b-a}{4}\mathcal{F}_{\rho, \lambda+2}^{\sigma_{1}} \bigl[ \vert w \vert (b-a)^{\rho}\bigr] \bigl[ \bigl\vert f^{\prime }(a) \bigr\vert + \bigl\vert f^{\prime}(b) \bigr\vert \bigr] , \end{aligned}$$
(2.5)
where
$$ \sigma_{1}(k)=\sigma(k)\biggl( \lambda+\rho k+3-\frac{1}{2^{ \lambda+\rho k-1}} \biggr) , $$
\(\rho,\lambda>0\), \(w\in\mathbb{R}\), \(s\in(0,1]\).

Proof

Using Lemma 2.1 and convexity of \(\vert f^{\prime} \vert \), we have
$$\begin{aligned}& \bigl\vert L_{f}(a,b;w;J) \bigr\vert \\& \quad \leq \frac{b-a}{2} \biggl\{ \biggl\vert \int_{0}^{\frac{1}{2}}\mathcal{F} _{\rho,\lambda+1}^{\sigma} \bigl[w(b-a)^{\rho}\bigr]f^{\prime}\bigl(ta+(1-t)b\bigr)\,dt \\& \quad\quad{} + \int_{\frac{1}{2}}^{1}\mathcal{F}_{\rho,\lambda+1}^{\sigma } \bigl[w(b-a)^{\rho}\bigr]f^{\prime}\bigl(ta+(1-t)b\bigr)\,dt \biggr\vert \\& \quad\quad{} + \biggl\vert \int_{0}^{1}(1-t)^{\lambda} \mathcal{F}_{\rho,\lambda+1} ^{\sigma} \bigl[w(b-a)^{\rho}(1-t)^{\rho} \bigr] f^{\prime} \bigl( ta+(1-t)b \bigr) \,dt \\& \quad\quad{} - \int_{0}^{1}t^{\lambda}\mathcal{F}_{\rho,\lambda+1}^{\sigma} \bigl[w(b-a)^{ \rho}t^{\rho}\bigr] f^{\prime} \bigl( ta+(1-t)b \bigr) \,dt \biggr\vert \biggr\} \\& \quad \leq \frac{b-a}{2} \Biggl\{ \int_{0}^{1} \bigl\vert \mathcal{F}_{\rho, \lambda+1}^{\sigma} \bigl[w(b-a)^{\rho}\bigr] \bigr\vert \bigl( t \bigl\vert f^{\prime}(a) \bigr\vert +(1-t) \bigl\vert f ^{\prime}(b) \bigr\vert \bigr) \,dt \\& \quad\quad{} +\sum_{k=0}^{\infty}\frac{\sigma(k) \vert w \vert ^{k}(b-a)^{\rho k}}{\Gamma (\rho k+\lambda+1)} \biggl[ \int_{0}^{\frac{1}{2}} \bigl((1-t)^{\lambda +\rho k}-t^{\lambda+\rho k} \bigr) \bigl( t \bigl\vert f^{\prime}(a) \bigr\vert +(1-t) \bigl\vert f^{\prime}(b) \bigr\vert \bigr) \,dt \\& \quad\quad{} + \int_{\frac{1}{2}}^{1} \bigl(t^{\lambda+\rho k}-(1-t)^{\lambda+ \rho k} \bigr) \bigl( t \bigl\vert f^{\prime}(a) \bigr\vert +(1-t) \bigl\vert f^{\prime}(b) \bigr\vert \bigr) \,dt \biggr] \Biggr\} \\& \quad \leq \frac{b-a}{2} \sum_{k=0}^{\infty} \frac{\sigma(k) \vert w \vert ^{k}(b-a)^{ \rho k}}{\Gamma(\rho k+\lambda+1)} \times \biggl\{ \biggl( \frac{ \vert f ^{\prime}(a) \vert + \vert f^{\prime}(b) \vert }{2} \biggr) \\& \quad\quad{} + \bigl\vert f^{\prime}(a) \bigr\vert \int_{0}^{\frac{1}{2}} \bigl(t(1-t)^{\lambda+\rho k}-t ^{\lambda+\rho k+1} \bigr)\,dt \\& \quad\quad{} + \bigl\vert f^{\prime}(b) \bigr\vert \int_{0}^{\frac{1}{2}} \bigl((1-t)^{\lambda+\rho k+1}-t ^{\lambda+\rho k}(1-t) \bigr)\,dt \\& \quad\quad{} + \bigl\vert f^{\prime}(a) \bigr\vert \int_{\frac{1}{2}}^{1} \bigl(t^{\lambda+\rho k+1}-t(1-t)^{ \lambda+\rho k} \bigr)\,dt \\& \quad\quad{} + \bigl\vert f^{\prime}(b) \bigr\vert \int_{\frac{1}{2}}^{1} \bigl((1-t)t^{\lambda+\rho k}-(1-t)^{ \lambda+\rho k+1} \bigr)\,dt \biggr\} \\& \quad = \frac{b-a}{2} \sum_{k=0}^{\infty} \frac{\sigma(k) \vert w \vert ^{k}(b-a)^{ \rho k}}{\Gamma(\rho k+\lambda+1)} \biggl[ \frac{1}{2}+\frac{1}{ \lambda+\rho k+1} \biggl( 1- \frac{1}{2^{\lambda+\rho k}} \biggr) \biggr] \bigl[ \bigl\vert f^{\prime}(a) \bigr\vert + \bigl\vert f^{\prime }(b) \bigr\vert \bigr] \\& \quad = \frac{b-a}{4}\sum_{k=0}^{\infty} \frac{\sigma (k) \vert w \vert ^{k}(b-a)^{ \rho k}}{\Gamma(\rho k+\lambda+2)} \biggl( \lambda+\rho k+3-\frac{1}{2^{ \lambda+\rho k-1}} \biggr) \bigl[ \bigl\vert f^{\prime}(a) \bigr\vert + \bigl\vert f^{\prime}(b) \bigr\vert \bigr] \\& \quad = \frac{b-a}{4}\mathcal{F}_{\rho,\lambda+2}^{\sigma_{1}} \bigl[ \vert w \vert (b-a)^{ \rho}\bigr] \bigl[ \bigl\vert f^{\prime}(a) \bigr\vert + \bigl\vert f^{\prime }(b) \bigr\vert \bigr] , \end{aligned}$$
(2.6)
using the facts that
$$\begin{aligned} \int_{0}^{\frac{1}{2}} \bigl(t(1-t)^{\lambda+\rho k}-t^{\lambda+ \rho k+1} \bigr)\,dt =& \int_{\frac{1}{2}}^{1} \bigl((1-t)t^{\lambda+ \rho k}-(1-t)^{\lambda+\rho k+1} \bigr)\,dt \\ =&\frac{2^{\lambda+\rho k+1}-(\lambda+\rho k+2)}{(\lambda+\rho k+1)( \lambda+\rho k+2)2^{\lambda+\rho k+1}} \end{aligned}$$
and
$$\begin{aligned} \int_{0}^{\frac{1}{2}} \bigl((1-t)^{\lambda+\rho k+1}-t^{\lambda+ \rho k}(1-t) \bigr)\,dt =& \int_{\frac{1}{2}}^{1} \bigl(t^{\lambda+ \rho k+1}-t(1-t)^{\lambda+\rho k} \bigr)\,dt \\ =& \frac{1}{\lambda+\rho k+2}-\frac{1}{(\lambda+\rho k+2)2^{\lambda +\rho k+2}} \\ &{}- \frac{\lambda+\rho k+3}{(\lambda+\rho k+1)(\lambda+\rho k+2) 2^{ \lambda+\rho k+2}}. \end{aligned}$$

Thus the proof is completed. □

Remark 2.2

Choosing \(\lambda=\alpha\), \(\sigma(0)=1\) and \(w=0\) in Theorem 2.1, inequality (2.5) reduces to inequality (1.3).

Theorem 2.2

Let \(f :[a,b]\to\mathbb{R}\) be a differentiable function on \((a,b)\) with \(a< b\). If \(\vert f^{\prime} \vert ^{q}\) is convex and \(q>1\) with \(\frac{1}{p}+\frac{1}{q}=1\), then the following inequality for generalized fractional integral operators holds:
$$\begin{aligned}& \bigl\vert L_{f}(a,b;w;J) \bigr\vert \\& \quad \leq \frac{b-a}{2} \biggl[\mathcal{F}_{\rho,\lambda+1}^{\sigma} \bigl[ \vert w \vert (b-a)^{ \rho}\bigr] \biggl( \frac{ \vert f^{\prime}(a) \vert ^{q}+ \vert f^{\prime}(a) \vert ^{q}}{2} \biggr) ^{ \frac{1}{q}} \\& \quad\quad{} +\mathcal{F}_{\rho,\lambda+1}^{\sigma_{2}} \bigl[ \vert w \vert (b-a)^{\rho}\bigr] \biggl( \biggl[ \frac{1}{8} \bigl\vert f^{\prime}(a) \bigr\vert ^{q}+\frac {3}{8} \bigl\vert f^{\prime}(b) \bigr\vert ^{q} \biggr] ^{\frac{1}{q}} + \biggl[ \frac{3}{8} \bigl\vert f^{\prime}(a) \bigr\vert ^{q}+\frac {1}{8} \bigl\vert f^{\prime}(b) \bigr\vert ^{q} \biggr] ^{\frac{1}{q}} \biggr)\biggr] \\& \quad \leq \mathcal{F}_{\rho,\lambda+1}^{\sigma_{3}} \bigl[ \vert w \vert (b-a)^{\rho}\bigr] \bigl[ \bigl\vert f^{\prime}(a) \bigr\vert + \bigl\vert f^{\prime }(b) \bigr\vert \bigr] , \end{aligned}$$
(2.7)
where
$$\begin{aligned}& \sigma_{2}(k)=\sigma(k) \biggl[ \frac{1}{p(\lambda+\rho k)+1} \biggl( 1- \frac{1}{2^{p( \lambda+\rho k)}} \biggr) \biggr] ^{\frac{1}{p}}, \\& \sigma_{3}(k)=\sigma(k) \biggl( \frac{1}{2} \biggr) ^{\frac{1}{q}} \biggl(1+\biggl[ \frac{4}{p( \lambda+\rho k)+1} \biggl( 1-\frac{1}{2^{p(\lambda+\rho k)}} \biggr) \biggr] ^{\frac{1}{p}}\biggr), \end{aligned}$$
\(\rho,\lambda>0\) and \(w\in\mathbb{R}\).

Proof

By using Lemma 2.1, we have
$$\begin{aligned}& \bigl\vert L_{f}(a,b;w;J) \bigr\vert \\ & \quad \leq \frac{b-a}{2} \biggl\{ \biggl\vert \int_{0}^{\frac{1}{2}}\mathcal{F} _{\rho,\lambda+1}^{\sigma} \bigl[w(b-a)^{\rho}\bigr]f^{\prime}\bigl(ta+(1-t)b\bigr)\,dt \\ & \quad\quad{} + \int_{\frac{1}{2}}^{1}\mathcal{F}_{\rho,\lambda+1}^{\sigma} \bigl[w(b-a)^{ \rho}\bigr]f^{\prime}\bigl(ta+(1-t)b\bigr)\,dt \biggr\vert \\ & \quad\quad{} + \biggl\vert \int_{0}^{1}(1-t)^{\lambda} \mathcal{F}_{\rho,\lambda+1} ^{\sigma} \bigl[w(b-a)^{\rho}(1-t)^{\rho} \bigr] f^{\prime} \bigl( ta+(1-t)b \bigr) \,dt \\ & \quad\quad{} - \int_{0}^{1}t^{\lambda}\mathcal{F}_{\rho,\lambda+1}^{\sigma} \bigl[w(b-a)^{ \rho}t^{\rho}\bigr] f^{\prime} \bigl( ta+(1-t)b \bigr) \,dt \biggr\vert \biggr\} \\ & \quad \leq \frac{b-a}{2} \Biggl\{ \sum_{k=0}^{\infty} \frac{\sigma(k) \vert w \vert ^{k}(b-a)^{ \rho k}}{\Gamma(\rho k+\lambda+1)} \int_{0}^{1} \bigl\vert f^{\prime} \bigl(ta+(1-t)b\bigr) \bigr\vert \,dt \\ & \quad\quad{} +\sum_{k=0}^{\infty}\frac{\sigma(k) \vert w \vert ^{k}(b-a)^{\rho k}}{\Gamma (\rho k+\lambda+1)} \biggl[ \int_{0}^{\frac{1}{2}} \bigl((1-t)^{\lambda +\rho k}-t^{\lambda+\rho k} \bigr) \bigl\vert f^{\prime}\bigl(ta+(1-t)b\bigr) \bigr\vert \,dt \\ & \quad\quad{} + \int_{\frac{1}{2}}^{1} \bigl(t^{\lambda+\rho k}-(1-t)^{\lambda+ \rho k} \bigr) \bigl\vert f^{\prime}\bigl(ta+(1-t)b\bigr) \bigr\vert \,dt \biggr] \Biggr\} . \end{aligned}$$
(2.8)
Using the well-known Hölder inequality and convexity of \(\vert f^{\prime} \vert ^{q}\) we get
$$ \int_{0}^{1} \bigl\vert f^{\prime} \bigl(ta+(1-t)b\bigr) \bigr\vert \,dt \leq \biggl( \frac{ \vert f^{\prime}(a) \vert ^{q}+ \vert f ^{\prime}(a) \vert ^{q}}{2} \biggr) ^{\frac{1}{q}}. $$
(2.9)
Thus
$$\begin{aligned}& \int_{0}^{\frac{1}{2}} \bigl((1-t)^{\lambda+\rho k}-t^{\lambda+ \rho k} \bigr) \bigl\vert f^{\prime}\bigl(ta+(1-t)b\bigr) \bigr\vert \,dt \\ & \quad \leq \biggl[ \int_{0}^{\frac{1}{2}} \bigl((1-t)^{\lambda+\rho k}-t ^{\lambda+\rho k} \bigr)^{p}\,dt \biggr] ^{\frac{1}{p}} \biggl[ \int_{0}^{ \frac{1}{2}} \bigl\vert f^{\prime} \bigl(ta+(1-t)b\bigr) \bigr\vert ^{q}\,dt \biggr] ^{\frac{1}{q}} \\ & \quad \leq \biggl[ \int_{0}^{\frac{1}{2}} \bigl((1-t)^{p(\lambda+\rho k)}-t ^{p(\lambda+\rho k)} \bigr)\,dt \biggr] ^{\frac{1}{p}} \biggl[ \int_{0}^{ \frac{1}{2}} \bigl( t \bigl\vert f^{\prime}(a) \bigr\vert ^{q}+(1-t) \bigl\vert f^{\prime}(a) \bigr\vert ^{q} \bigr) \,dt \biggr] ^{\frac{1}{q}} \\ & \quad = \biggl[ \frac{1}{p(\lambda+\rho k)+1} \biggl( 1-\frac{1}{2^{p(\lambda +\rho k)}} \biggr) \biggr] ^{\frac{1}{p}} \biggl[ \frac{1}{8} \bigl\vert f^{\prime}(a) \bigr\vert ^{q}+ \frac{3}{8} \bigl\vert f^{\prime}(b) \bigr\vert ^{q} \biggr] ^{\frac{1}{q}} \end{aligned}$$
(2.10)
and
$$\begin{aligned}& \int_{\frac{1}{2}}^{1} \bigl(t^{\lambda+\rho k}-(1-t)^{\lambda+ \rho k} \bigr) \bigl\vert f^{\prime}\bigl(ta+(1-t)b\bigr) \bigr\vert \,dt \\ & \quad \leq \biggl[ \int_{\frac{1}{2}}^{1} \bigl(t^{\lambda+\rho k}-(1-t)^{ \lambda+\rho k} \bigr)^{p}\,dt \biggr] ^{\frac{1}{p}} \biggl[ \int_{0}^{ \frac{1}{2}} \bigl\vert f^{\prime} \bigl(ta+(1-t)b\bigr) \bigr\vert ^{q}\,dt \biggr] ^{\frac{1}{q}} \\ & \quad \leq \biggl[ \int_{\frac{1}{2}}^{1} \bigl(t^{p(\lambda+\rho k)}-(1-t)^{p( \lambda+\rho k)} \bigr)\,dt \biggr] ^{\frac{1}{p}} \biggl[ \int_{ \frac{1}{2}}^{1} \bigl( t \bigl\vert f^{\prime}(a) \bigr\vert ^{q}+(1-t) \bigl\vert f^{\prime}(a) \bigr\vert ^{q} \bigr) \,dt \biggr] ^{\frac{1}{q}} \\ & \quad = \biggl[ \frac{1}{p(\lambda+\rho k)+1} \biggl( 1-\frac{1}{2^{p(\lambda +\rho k)}} \biggr) \biggr] ^{\frac{1}{p}} \biggl[ \frac{3}{8} \bigl\vert f^{\prime}(a) \bigr\vert ^{q}+ \frac{1}{8} \bigl\vert f^{\prime}(b) \bigr\vert ^{q} \biggr] ^{\frac{1}{q}}, \end{aligned}$$
(2.11)
where we used that \((A-B)^{p}\leq A^{p}-B^{p}\) for any \(A\geq B \geq0\) and \(p \geq1\) in (2.10) and (2.11).
Let \(a_{1}=3 \vert f^{\prime}(a) \vert ^{q}\), \(b_{1}= \vert f^{\prime}(b) \vert ^{q}\), \(a_{2}= \vert f^{\prime }(a) \vert ^{q}\), \(b_{2}=3 \vert f^{\prime}(b) \vert ^{q} \). Here \(0 < \frac{1}{q}< 1\) for \(q>1\). We use the fact that
$$ \sum_{k=1}^{n}(a_{k}+b_{k})^{s} \leq\sum_{k=1}^{n}a_{k}^{s}+ \sum_{k=1} ^{n}b_{k}^{s}. $$
For \(0\leq s<1\), \(a_{1},a_{2},a_{3},\ldots,a_{n}\geq0\), \(b_{1},b_{2},b_{3},\ldots,b _{n}\geq0\). Combining the inequalities (2.10) with (2.11) we obtain
$$\begin{aligned}& \int_{0}^{\frac{1}{2}} \bigl((1-t)^{\lambda+\rho k}-t^{\lambda+ \rho k} \bigr) \bigl\vert f^{\prime}\bigl(ta+(1-t)b\bigr) \bigr\vert \,dt \\& \quad\quad{} + \int_{\frac{1}{2}}^{1} \bigl(t^{\lambda+\rho k}-(1-t)^{\lambda+ \rho k} \bigr) \bigl\vert f^{\prime}\bigl(ta+(1-t)b\bigr) \bigr\vert \,dt \\& \quad \leq \biggl[ \frac{1}{p(\lambda+\rho k)+1} \biggl( 1-\frac{1}{2^{p( \lambda+\rho k)}} \biggr) \biggr] ^{\frac{1}{p}} \biggl(\frac{1}{8}\biggr)^{ \frac{1}{q}} \\& \quad\quad{}\times \bigl( \bigl[ 3 \bigl\vert f^{\prime}(a) \bigr\vert ^{q}+ \bigl\vert f^{\prime}(b) \bigr\vert ^{q} \bigr] ^{ \frac{1}{q}} + \bigl[ \bigl\vert f^{\prime}(a) \bigr\vert ^{q}+3 \bigl\vert f^{\prime }(b) \bigr\vert ^{q} \bigr] ^{ \frac{1}{q}} \bigr) \\& \quad \leq \biggl[ \frac{1}{p(\lambda+\rho k)+1} \biggl( 1-\frac{1}{2^{p( \lambda+\rho k)}} \biggr) \biggr] ^{\frac{1}{p}} \biggl(\frac{1}{8}\biggr)^{ \frac{1}{q}} \bigl(3^{\frac{1}{q}}+1 \bigr) \bigl[ \bigl\vert f^{\prime}(a) \bigr\vert + \bigl\vert f^{\prime }(b) \bigr\vert \bigr] \\& \quad \leq \biggl[ \frac{1}{p(\lambda+\rho k)+1} \biggl( 1-\frac{1}{2^{p( \lambda+\rho k)}} \biggr) \biggr] ^{\frac{1}{p}} \biggl(\frac{1}{8}\biggr)^{ \frac{1}{q}}4 \bigl[ \bigl\vert f^{\prime}(a) \bigr\vert + \bigl\vert f^{\prime }(b) \bigr\vert \bigr] \\& \quad = \biggl[ \frac{4}{p(\lambda+\rho k)+1} \biggl( 1-\frac{1}{2^{p(\lambda +\rho k)}} \biggr) \biggr] ^{\frac{1}{p}} \biggl(\frac{1}{2}\biggr)^{\frac{1}{q}} \bigl[ \bigl\vert f^{\prime}(a) \bigr\vert + \bigl\vert f^{\prime }(b) \bigr\vert \bigr] \end{aligned}$$
(2.12)
and
$$ \biggl( \frac{ \vert f^{\prime}(a) \vert ^{q}+ \vert f^{\prime}(a) \vert ^{q}}{2} \biggr) ^{\frac{1}{q}} \leq\biggl( \frac{1}{2}\biggr)^{\frac{1}{q}} \bigl[ \bigl\vert f^{\prime}(a) \bigr\vert + \bigl\vert f^{\prime }(b) \bigr\vert \bigr] . $$
(2.13)
Thus putting the inequalities (2.9), (2.12) and (2.13) in (2.8), the proof is completed. □

Corollary 2.1

Choosing \(\lambda=\alpha\), \(\sigma(0)=1\) and \(w=0\) in Theorem  2.2, inequality (2.7) becomes the following inequality:
$$\begin{aligned}& \biggl\vert \frac{\Gamma(\alpha+1)}{2(b-a)^{\alpha}} \bigl[\bigl(J_{\alpha^{-}} ^{a}f\bigr) (b)+ \bigl(J_{\alpha^{+}}^{b}f\bigr) (a) \bigr]-f \biggl( \frac{a+b}{2} \biggr) \biggr\vert \\& \quad \leq \frac{b-a}{2} \biggl\{ \biggl[ \frac{ \vert f^{\prime}(a) \vert ^{q}+ \vert f^{\prime}(a) \vert ^{q}}{2} \biggr] ^{\frac{1}{q}} \\& \quad\quad{} + \biggl[ \frac{1}{\alpha p+1} \biggl( 1-\frac{1}{2^{\alpha p}} \biggr) \biggr] ^{\frac{1}{p}} \biggl( \biggl[ \frac{1}{8} \bigl\vert f^{\prime}(a) \bigr\vert ^{q}+\frac {3}{8} \bigl\vert f ^{\prime}(b) \bigr\vert ^{q} \biggr] ^{\frac{1}{q}} + \biggl[ \frac{3}{8} \bigl\vert f^{\prime}(a) \bigr\vert ^{q}+ \frac{1}{8} \bigl\vert f^{\prime}(b) \bigr\vert ^{q} \biggr] ^{\frac{1}{q}} \biggr) \biggr\} \\& \quad \leq \frac{b-a}{2} \biggl(1+ \biggl[ \frac{4}{\alpha p+1} \biggl(1- \frac{1}{2^{ \alpha p}} \biggr) \biggr] ^{\frac{1}{p}} \biggr) \biggl( \frac{1}{2} \biggr) ^{\frac{1}{q}} \bigl[ \bigl\vert f^{\prime}(a) \bigr\vert + \bigl\vert f^{\prime}(b) \bigr\vert \bigr] . \end{aligned}$$
(2.14)

3 Conclusion

In this paper, we have obtained a new fractional integral identity. Utilizing this new identity as an auxiliary result, we have obtained some new variants of Hermite-Hadamard type inequalities. The results derived in this paper become natural generalizations of classical results. It is expected that the interested reader may find useful applications of these results and consequently this paper may stimulate further research in this area.

Declarations

Acknowledgements

The authors are thankful to the anonymous referee for his/her valuable comments and suggestions. The authors are pleased to acknowledge the support of Distinguished Scientist Fellowship Program (DSFP), King Saud University, Riyadh, Saudi Arabia. Third author is thankful to HEC, Pakistan for SRGP project 21-985/SRGP/R&D/HEC/2016.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Science and Arts, Ordu University, Ordu, Turkey
(2)
Department of Mathematics, King Saud University, Riyadh, Saudi Arabia
(3)
Department of Mathematics, COMSATS Institute of Information Technology, Park Road, Islamabad, Pakistan
(4)
Department of Mathematics, Government College University, Faisalabad, Pakistan

References

  1. Dragomir, SS, Pearce, CEM: Selected Topics on Hermite-Hadamard Inequalities and Applications. RGMIA Monographs. Victoria University (2000) Google Scholar
  2. Mitrinović, DS, Lacković, IB: Hermite and convexity. Aequ. Math. 28, 229-232 (1985) MathSciNetView ArticleMATHGoogle Scholar
  3. Set, E, Özdemir, ME, Sarıkaya, MZ: Inequalities of Hermite-Hadamard’s type for functions whose derivatives absolute values are m-convex. AIP Conf. Proc. 1309(1), 861-873 (2010) View ArticleGoogle Scholar
  4. Sarıkaya, MZ, Set, E, Yaldız, H, Başak, N: Hermite-Hadamard’s inequalities for fractional integrals and related fractional inequalities. Math. Comput. Model. 57, 2403-2407 (2013) View ArticleMATHGoogle Scholar
  5. Dahmani, Z: New inequalities in fractional integrals. Int. J. Nonlinear Sci. 9(4), 493-497 (2010) MathSciNetGoogle Scholar
  6. Dahmani, Z, Tabharit, L, Taf, S: New generalizations of Grüss inequality using Riemann-Liouville fractional integrals. Bull. Math. Anal. Appl. 2(3), 93-99 (2010) MathSciNetMATHGoogle Scholar
  7. Gorenflo, R, Mainardi, F: Fractional calculus: integral and differential equations of fractional order. In: Fractals and Fractional Calculus in Continuum Mechanics, pp. 223-276. Springer, Wien (1997) View ArticleGoogle Scholar
  8. Noor, MA, Cristescu, G, Awan, MU: Generalized fractional Hermite-Hadamard inequalities for twice differentiable s-convex functions. Filomat 29(4), 807-815 (2015) MathSciNetView ArticleGoogle Scholar
  9. Sarıkaya, MZ, Yıldı rım, H: On Hermite-Hadamard type inequalities for Riemannn-Liouville fractional integrals. Miskolc Math. Notes 17(2), 1049-1059 (2016) MathSciNetView ArticleGoogle Scholar
  10. Set, E, Sarıkaya, MZ, Özdemir, ME, Yıldı rım, H: The Hermite-Hadamard’s inequality for some convex functions via fractional integrals and related results. J. Appl. Math. Stat. Inform. 10(2), 69-83 (2014) MathSciNetMATHGoogle Scholar
  11. Set, E: New inequalities of Ostrowski type for mapping whose derivatives are s-convex in the second sense via fractional integrals. Comput. Math. Appl. 63, 1147-1154 (2012) MathSciNetView ArticleMATHGoogle Scholar
  12. Set, E, İşcan, İ, Sarıkaya, MZ, Özdemir, ME: On new inequalities of Hermite-Hadamard-Fejer type for convex functions via fractional integrals. Appl. Math. Comput. 259, 875-881 (2015) MathSciNetGoogle Scholar
  13. Zhu, C, Feckan, M, Wang, J: Fractional integral inequalities for differentiable convex mappings and applications to special means and a midpoint formula. J. Appl. Math. Stat. Inform. 8(2), 21-28 (2012) MATHGoogle Scholar
  14. Raina, RK: On generalized Wright’s hypergeometric functions and fractional calculus operators. East Asian Math. J. 21(2), 191-203 (2005) MATHGoogle Scholar
  15. Agarwal, RP, Luo, M-J, Raina, RK: On Ostrowski type inequalities. Fasc. Math. 204, 5-27 (2016) MathSciNetMATHGoogle Scholar
  16. Set, E, Gözpınar, A: Some new inequalities involving generalized fractional integral operators for several class of functions. AIP Conf. Proc. 1833, 020038 (2017). doi:10.1063/1.4981686 View ArticleGoogle Scholar
  17. Set, E, Gözpınar, A: Hermite-Hadamard type inequalities for convex functions via generalized fractional integral operators. ResearchGate. https://www.researchgate.net/publication/312378686
  18. Set, E, Akdemir, AO, Çelik, B: On generalization of Fejér type inequalities via fractional integral operator. ResearchGate. https://www.researchgate.net/publication/311452467
  19. Set, E, Çelik, B: On generalization related to the left side of Fejér’s inequalites via fractional integral operator. ResearchGate. https://www.researchgate.net/publication/311651826
  20. Set, E, Choi, J, Çelik, B: A new approach to generalized of Hermite-Hadamard inequality using fractional integral operator. ResearchGate. https://www.researchgate.net/publication/313437121
  21. Usta, F, Budak, H, Sarıkaya, MZ, Set, E: On generalization of trapezoid type inequalities for s-convex functions with generalized fractional integral operators. Filomat (accepted) Google Scholar
  22. Yaldız, H, Sarıkaya, MZ: On the Hermite-Hadamard type inequalities for fractional integral operator. ResearchGate. https://www.researchgate.net/publication/309824275

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© The Author(s) 2017

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