Skip to main content

New Hadamard-type inequalities for E-convex functions involving generalized fractional integrals

Abstract

In this article, we establish some new Hadamard-type inequalities for E-convex functions involving generalized fractional integrals. These inequalities include a generalized Hadamard-type inequality and the corresponding right Hadamard-type inequalities for E-convex functions. The results presented here are generalizations of some of the results discussed in the recent literature.

Introduction and basic definitions

The theory of convexity is not only important in itself but also it contributes to almost all areas of mathematics. Convexity gives rise to inequalities, the Hadamard inequality is the first consequence of convex functions. The book by Hardy [1] has played a key role in popularizing the subject of convex analysis. Over the years, the idea of convex sets and convex functions has been largely generalized. Today, the study of convex functions has evolved into a broader theory of functions including quasiconvex functions [2, 3], coordinated convex functions [4, 5], preinvex functions [6], GA-convex functions [7], strongly convex functions [8], \((g, \varphi _{h}) \)-convex functions [9], E-convex functions [10] and so on. Youness [10] defined the E-convex set and the corresponding function as follows:

Definition 1

A set \(S\subset \mathbb{R} \) is called E-convex if and only if there is a function \(E:\mathbb{R} \longrightarrow \mathbb{R} \) such that \(t E(\zeta) + (1-t) E(\eta)\in S \) for each \(\zeta,\eta \in S\) and \(t\in [0,1]\).

Definition 2

A function \(f: \mathbb{R}\longrightarrow \mathbb{R} \) is called E-convex on a set \(S \subseteq \mathbb{R}\) if and only if there is a map \(E: \mathbb{R}\longrightarrow \mathbb{R} \) such that S is an E-convex set and

$$\begin{aligned} f \bigl(t E(\zeta) + (1 - t)E(\eta) \bigr)\leq t f \bigl(E(\zeta) \bigr) +(1- t) f \bigl(E(\eta) \bigr) \end{aligned}$$
(1.1)

holds for each \(\zeta,\eta \in S\) and \(t\in [0,1]\). On the other hand, if the inequality sign in the inequality (1.1) is reversed then f is called E-concave on the set S.

Every convex function f on a convex set S is an E-convex function provided that E is an identity function. For a detailed explanation of E-convex functions see [10]. The Hadamard-type inequality for E-convex given in [11] is as follows:

Theorem 1

Let \(E : J\subset \mathbb{R}\longrightarrow \mathbb{R} \) be a continuous increasing function and \(\zeta , \eta \in J\) with \(\zeta <\eta \). Let \(f:I\subseteq \mathbb{R} \longrightarrow \mathbb{R}\) be an E-convex function on \([\zeta , \eta ]\), then we have

$$\begin{aligned}& f \biggl( \frac{E(\zeta )+E(\eta )}{2} \biggr)\leq \frac{1}{E(\eta )-E(\zeta )} \int _{E(\zeta )}^{E(\eta )}f\bigl(E(t)\bigr) \,dE(t)\leq \frac{f(E(\zeta ))+f(E(\eta ))}{2}, \end{aligned}$$
(1.2)

inequality (1.2) is Hadamard’s inequality for E-convex functions.

Convexity is mixed with other mathematical concepts such as; optimization [12], time scale [13, 14], quantum and postquantum calculus [15, 16], and fractional calculus [3, 11, 1719]. Fractional calculus is basically a generalization of integer-order calculus. Strictly speaking, it is a generalization of operators beyond the integral order to real or complex order. Many fractional models have been proposed so far [2027]. The key drivers behind such proposals are identified with the various real data corresponding to different systems under consideration requiring different kernels. Raina [27] and Agarwal [26] defined the following generalized fractional operators:

Definition 3

Let \(f\in L(\zeta , \eta ) \), then for \(\sigma , \rho >0\), \(\omega \in \mathbb{R} \) the right-handed and left-handed generalized fractional integrals of f are, respectively, defined as follows:

$$\mathcal{J}_{\sigma ,\rho ,\zeta +;\omega }^{\alpha }f(s)= \int _{ \zeta }^{s}(s-t)^{\rho -1} \mathcal{F}^{\alpha }_{\sigma ,\rho }\bigl[\omega (s-t)^{ \sigma } \bigr]f(t)\,dt \quad ( s>\zeta ), $$

and

$$\mathcal{J}_{\sigma ,\rho , \eta -;\omega }^{\alpha }f(s)= \int _{s}^{ \eta }(t-s)^{\rho -1} \mathcal{F}^{\alpha }_{\sigma ,\rho }\bigl[\omega (t-s)^{ \sigma } \bigr]f(t)\,dt \quad (s< \eta ), $$

where \(\mathcal{F}_{\sigma , \rho }^{\alpha }(s) \) is defined in [27] as follows:

$$\mathcal{F}_{\sigma , \rho }^{\alpha }(s)=\mathcal{F}_{\sigma , \rho }^{ \alpha (0), \alpha (1), \alpha (2), \ldots }(s)= \sum_{n=0}^{\infty } \frac{\alpha (n)}{\Gamma (\sigma n+\rho )}s^{n} \quad \bigl(\sigma , \rho >0, \vert s \vert < R\bigr), $$

where R is a real positive constant. The coefficients \(\alpha (n)\) (\(n\in N_{0} = N\cup \{0\}\)) are terms of a bounded sequence of positive real numbers and \(\mathbb{R} \) is the set of real numbers. Moreover, the operators \(\mathcal{J}_{\sigma ,\rho ,\zeta +;\omega }^{\alpha }f \) and \(\mathcal{J}_{\sigma ,\rho , \eta +;\omega }^{\alpha }f \) are bounded on \(L(\zeta , \eta ) \), i.e.,

$$\bigl\Vert \mathcal{J}_{\sigma ,\rho ,\zeta +;\omega }^{\alpha }f(s) \bigr\Vert \leq \mathfrak{P}(\eta -\zeta )^{\rho } \Vert f \Vert _{1} $$

and

$$\bigl\Vert \mathcal{J}_{\sigma ,\rho , \eta +;\omega }^{\alpha }f(s) \bigr\Vert \leq \mathfrak{P}(\eta -\zeta )^{\rho } \Vert f \Vert _{1}, $$

where \(\mathfrak{P}:= \mathcal{F}^{\alpha }_{\sigma ,\rho }[\omega (t-s)^{ \sigma }]<\infty \) and \(\| f\| _{1}= \int _{\zeta }^{\eta }|f(t)|\,dt\).

These fractional integrals are really important because of their generality. Many other fractional operators can be obtained by specifying the coefficients \(\alpha (n) \). For instance, if we set \(n=0\), \(\alpha (0)=1 \) and \(\omega =0\), we obtain the well-known Riemann–Liouville fractional operators

$$J_{\zeta +}^{\lambda }f(s)=\frac{1}{\Gamma (\lambda )} \int _{\zeta }^{s}(s-t)^{ \lambda -1}f(t)\,dt \quad ( s>\zeta ) $$

and

$$J_{\eta -}^{\lambda }f(s)=\frac{1}{\Gamma (\lambda )} \int _{s}^{\eta }(t-s)^{ \lambda -1}f(t)\,dt \quad (s< \eta ). $$

Lemma 1

([28, 29])

For \(0<\alpha \leq 1\) and \(0\leq x < y\), we have

$$\begin{aligned} \bigl\vert x^{\alpha }-y^{\alpha } \bigr\vert \leq (y-x)^{\alpha }. \end{aligned}$$

Fractional calculus has useful applications in almost all areas of applied mathematics and other sciences, see [30] and the references therein. In the present work, notions of E-convexity and generalized fractional operators are joined together. These ideas are independently utilized before, however, in combined form we obtain even more generalized results.

Main outcomes

In this section, mainly the Hadamard inequality for E-convex function (1.2) is extended using Definition 3 of generalized fractional integrals. Then, an identity is established for differentiable functions that is used to develop right Hadamard-type inequalities for the said extended Hadamard-type inequality. Likewise, another important identity is developed for twice-differentiable functions that is further used to develop more right Hadamard-type inequalities for the said extended Hadamard-type inequality for E-convex functions.

In the following, we use J to represent the interval of nonnegative real numbers and I to represent the interval of real numbers. Moreover, we use the following notations for brevity;

$$\begin{aligned}& \Delta :=\eta -\zeta ,\qquad E(\Delta ):=E(\eta )-E(\zeta ), \\& \mathcal{M}(J):=\frac{f(\zeta )+f(\eta )}{2}- \frac{\Gamma (\lambda +1)}{2(\Delta )^{\lambda }} \bigl[{J}_{\zeta +}^{ \lambda }f(\eta )+{J}_{\eta -}^{\lambda }f( \zeta ) \bigr], \\& \mathcal{M}(\mathcal{J}) := \frac{f(\zeta ) + f(\eta )}{2} - \frac{1}{2(\Delta )^{\rho }\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (\Delta )^{\sigma }]} \bigl[ \mathcal{J}_{\sigma ,\rho , \zeta +;\omega }^{\alpha }f(\eta )+ \mathcal{J}_{\sigma ,\rho , \eta -;\omega }^{\alpha }f( \zeta ) \bigr], \\& \begin{aligned} \mathcal{M}(\mathcal{J}_{E}):={}& \frac{f(E(\zeta ))+f(E(\eta ))}{2} \\ &{}- \frac{1}{2(E(\Delta ))^{\rho }\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]} \bigl[\mathcal{J}_{\sigma ,\rho , E(\zeta ) +;\omega }^{\alpha }f \bigl(E( \eta )\bigr)+\mathcal{J}_{\sigma ,\rho , E(\eta ) -;\omega }^{\alpha }f\bigl( E( \zeta )\bigr) \bigr]. \end{aligned} \end{aligned}$$

Theorem 2

Let \(E: J \longrightarrow \mathbb{R}\) be a continuous increasing function and \(\zeta , \eta \in J\) with \(\zeta < \eta \). Let \(f : I\longrightarrow \mathbb{R}\) be a function such that \(f\in L[E(\zeta ),E(\eta )] \), where \(E(\zeta ),E(\eta )\in I \). If f is an E-convex function on \([\zeta , \eta ]\), then the following inequality holds for generalized fractional integral operators

$$\begin{aligned}& f \biggl(\frac{E(\zeta )+E(\eta )}{2} \biggr) \\& \quad \leq \frac{1}{2 (E(\Delta ) )^{\rho }\mathcal{F}_{\sigma ,\rho +1}^{\alpha }[\omega (E(\Delta ) )^{\sigma }]} \bigl[\mathcal{J}_{\sigma ,\rho , E(\zeta ) +;\omega }^{\alpha }f \bigl(E( \eta )\bigr)+\mathcal{J}_{\sigma ,\rho , E(\eta ) -;\omega }^{\alpha }f\bigl( E( \zeta )\bigr) \bigr] \\& \quad \leq \frac{f (E(\zeta ) )+f (E(\eta ) )}{2}, \end{aligned}$$
(2.1)

for all \(\sigma , \rho \in \mathbb{R}^{+}\) and \(\omega \in \mathbb{R}\).

Proof

Since f is an E-convex function on \([\zeta , \eta ]\), therefore for \(E(x), E(y) \in I\) we have

$$f \biggl(\frac{E(x)+E(y)}{2} \biggr)\leq \frac{f(E(x))+f(E(y))}{2} $$

and we let \(E(x)=tE(\zeta )+(1-t)E(\eta )\) and \(E(y)=(1-t)E(\zeta )+tE(\eta )\), so that we have

$$ 2 f \biggl(\frac{E(\zeta )+E(\eta )}{2} \biggr)\leq f\bigl(tE(\zeta )+(1-t)E( \eta )\bigr)+f\bigl((1-t)E(\zeta )+tE(\eta )\bigr). $$
(2.2)

On multiplying both sides of inequality (2.2) by \(t^{\rho -1}\mathcal{F}^{\alpha }_{\sigma ,\rho }[\omega (E(\Delta ))^{ \sigma }t^{\sigma }]\) and then integrating the resultant inequality with respect to t over \([0,1]\), we have

$$\begin{aligned}& 2f \biggl(\frac{E(\zeta )+E(\eta )}{2} \biggr) \int _{0}^{1}t^{\rho -1} \mathcal{F}^{\alpha }_{\sigma ,\rho }\bigl[\omega \bigl(E(\Delta ) \bigr)^{\sigma }t^{ \sigma }\bigr]\,dt \\& \quad \leq \int _{0}^{1}t^{\rho -1} \mathcal{F}^{\alpha }_{\sigma ,\rho }\bigl[ \omega \bigl(E(\Delta ) \bigr)^{\sigma }t^{\sigma }\bigr]f \bigl(tE(\zeta )+(1-t)E( \eta ) \bigr)\,dt \\& \qquad {}+ \int _{0}^{1}t^{\rho -1} \mathcal{F}^{\alpha }_{\sigma ,\rho }\bigl[\omega \bigl(E( \Delta ) \bigr)^{\sigma }t^{\sigma }\bigr]f \bigl((1-t)E(\zeta )+tE(\eta ) \bigr)\,dt. \end{aligned}$$

Further suppose that \(u=tE(\zeta )+(1-t)E(\eta )\) and \(v=(1-t)E(\zeta )+tE(\zeta )\) and using the definition of generalized fractional integrals

$$\begin{aligned}& \begin{gathered} 2f \biggl( \frac{E(\zeta )+E(\eta )}{2} \biggr) \mathcal{F}^{ \alpha }_{\sigma ,\rho +1}\bigl[\omega \bigl(E(\Delta ) \bigr)^{\sigma }\bigr] \\ \quad \leq \frac{1}{(E(\Delta ))^{\rho }} \int _{E(\zeta )}^{E(\eta )}\bigl(E( \eta )-u \bigr)^{\rho -1}\mathcal{F}^{\alpha }_{\sigma ,\rho }\bigl[\omega \bigl(E( \Delta )\bigr)^{\sigma }\bigr]f(u)du \\ \qquad {} +\frac{1}{(E(\Delta ))^{\rho }} \int _{E(\zeta )}^{E(\eta )}\bigl(v-E(\zeta ) \bigr)^{ \rho -1}\mathcal{F}^{\alpha }_{\sigma ,\rho }\bigl[\omega \bigl(E(\Delta )\bigr)^{ \sigma }\bigr]f(v)\,dv \\ f \biggl( \frac{E(\zeta ) + E(\eta )}{2} \biggr) \\ \quad \leq \frac{1}{2 (E(\Delta ) )^{\rho } \mathcal{F}_{ \sigma ,\rho +1}^{\alpha }( (E(\Delta ) )^{\sigma })} \bigl[\mathcal{J}_{ \sigma ,\rho , E(\zeta ) +;\omega }^{\alpha }f \bigl(E( \eta )\bigr) + \mathcal{J}_{ \sigma ,\rho , E(\eta ) -;\omega }^{\alpha }f\bigl( E( \zeta )\bigr) \bigr]. \end{gathered} \end{aligned}$$
(2.3)

Considering again the E-convexity of f over the interval \([\zeta , \eta ] \), we have

$$\begin{aligned}& f \bigl(tE(\zeta )+(1-t)E(\eta ) \bigr)\leq t f \bigl(E(\zeta ) \bigr)+(1-t)f \bigl(E(\eta ) \bigr), \end{aligned}$$
(2.4)
$$\begin{aligned}& f \bigl((1-t)E(\zeta )+tE(\eta ) \bigr)\leq (1-t)f \bigl(E( \zeta ) \bigr)+tf \bigl(E(\eta ) \bigr), \end{aligned}$$
(2.5)

and on adding inequality (2.4) and inequality (2.5), we have

$$ f \bigl(tE(\zeta )+(1-t)E(\eta ) \bigr)+f \bigl((1-t)E(\zeta )+tE( \eta ) \bigr)\leq f \bigl(E(\zeta ) \bigr)+f \bigl(E(\eta ) \bigr). $$
(2.6)

On multiplying both sides of inequality (2.6) by \(t^{\rho -1}\mathcal{F}^{\alpha }_{\sigma ,\rho }[\omega (E(\Delta ))^{ \sigma }t^{\sigma }]\), integrating with respect to t over the interval \([0,1] \) and finally using the definition of generalized fractional integrals, we have

$$\begin{aligned}& \int _{0}^{1}t^{\rho -1} \mathcal{F}^{\alpha }_{\sigma ,\rho }\bigl[\omega \bigl(E( \Delta ) \bigr)^{\sigma }t^{\sigma }\bigr]f \bigl(tE(\zeta )+(1-t)E(\eta ) \bigr)\,dt \\& \qquad {} + \int _{0}^{1}t^{\rho -1} \mathcal{F}^{\alpha }_{\sigma ,\rho }\bigl[\omega \bigl(E( \Delta ) \bigr)^{\sigma }t^{\sigma }\bigr]f \bigl((1-t)E(\zeta )+tE(\eta ) \bigr)\,dt \\& \quad \leq \bigl[f \bigl(E(\zeta ) \bigr)+f \bigl(E(\eta ) \bigr) \bigr] \int _{0}^{1}t^{\rho -1} \mathcal{F}^{\alpha }_{\sigma ,\rho }\bigl[\omega \bigl(E( \Delta ) \bigr)^{\sigma }t^{\sigma }\bigr]\,dt, \end{aligned}$$

on letting \(u=tE(\zeta )+(1-t)E(\eta )\) and \(v=(1-t)E(\zeta )+tE(\zeta )\) and then using the definition of generalized fractional integrals, we have

$$\begin{aligned}& \frac{1}{ (E(\Delta ) )^{\rho }} \bigl[\mathcal{J}_{ \sigma ,\rho , E(\zeta ) +;\omega }^{\alpha }f \bigl(E(\eta )\bigr)+ \mathcal{J}_{ \sigma ,\rho , E(\eta ) -;\omega }^{\alpha }f\bigl( E(\zeta )\bigr) \bigr] \\& \quad \leq \mathcal{F}^{\alpha }_{\sigma ,\rho +1} \bigl[ \omega \bigl(E( \Delta )\bigr) \bigr] \bigl[f \bigl(E(\zeta ) \bigr) + f \bigl(E( \eta ) \bigr) \bigr]. \end{aligned}$$
(2.7)

On combining inequality (2.3) and inequality (2.7), we obtain the required result. Hence it is proved. □

Remark 1

If in Theorem 2, the function E is chosen to be an identity function, then the following inequality holds for all \(\sigma , \rho \in \mathbb{R}^{+}\) and \(\omega \in \mathbb{R}\):

$$\begin{aligned}& f \biggl(\frac{\zeta +\eta }{2} \biggr)\leq \frac{1}{2 (\Delta )^{\rho }\mathcal{F}_{\sigma ,\rho +1}^{\alpha }[\omega (\Delta )^{\sigma }]} \bigl[ \mathcal{J}_{\sigma ,\rho , \zeta +;\omega }^{\alpha }f(\eta )+ \mathcal{J}_{\sigma ,\rho , \eta -;\omega }^{\alpha }f( \zeta ) \bigr] \leq \frac{f (\zeta )+f (\eta )}{2}, \end{aligned}$$

which was given in [31].

Remark 2

If in Theorem 2, the function E is chosen to be an identity function, \(\alpha (0)=1\), \(\rho=\lambda\) and \(\omega =0\), then the following inequality holds:

$$\begin{aligned}& f \biggl(\frac{\zeta +\eta }{2} \biggr)\leq \frac{\Gamma (\lambda +1)}{2(\Delta )^{\lambda }} \bigl[{J}_{\zeta +}^{ \lambda }f(\eta )+{J}_{\eta -}^{\lambda }f( \zeta ) \bigr]\leq \frac{f (\zeta )+f (\eta )}{2}, \end{aligned}$$

which was given in [32].

Lemma 2

Let \(E: J \longrightarrow \mathbb{R} \) be a continuous increasing function and \(\zeta , \eta \in J \) with \(\zeta < \eta \). Let \(f : I\longrightarrow \mathbb{R}\) be a differentiable function on \(I^{o} \). If \(f^{\prime }\in L([E(\zeta ),E(\eta )])\) for \(E(\zeta ),E(\eta )\in I \), then the following identity holds for generalized fractional operators:

$$\begin{aligned} \mathcal{M}(\mathcal{J}_{E}) =& \frac{E(\Delta )}{2\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]} \int _{0}^{1} \bigl\{ (1 - t)^{\rho } \mathcal{F}^{ \alpha }_{\sigma ,\rho +1}\bigl[\omega \bigl(E(\Delta ) \bigr)^{\sigma }(1 - t)^{\sigma }\bigr] \\ &{}- t^{\rho }\mathcal{F}^{\alpha }_{\sigma ,\rho +1}\bigl[\omega \bigl(E( \Delta )\bigr)^{\sigma }t^{\sigma }\bigr]\bigr\} f^{\prime }\bigl(tE(\zeta ) + (1 - t)E(\eta )\bigr)\,dt. \end{aligned}$$
(2.8)

Proof

Solving the subsequent integral by integration by parts, then using a change of variable and finally the definition of the left generalized fractional integral operator

$$\begin{aligned} I_{1} =& \int _{0}^{1}(1-t)^{\rho } \mathcal{F}^{\alpha }_{\sigma ,\rho +1}\bigl[ \omega \bigl(E(\Delta ) \bigr)^{\sigma }(1-t)^{\sigma }\bigr]f^{\prime }\bigl(tE( \zeta )+(1-t)E( \eta )\bigr)\,dt \\ =&\frac{1}{E(\Delta )}\mathcal{F}^{\alpha }_{\sigma , \rho +1} \bigl[\omega \bigl(E(\Delta )\bigr)^{\sigma }\bigr]f\bigl(E(\eta )\bigr) \\ &{}-\frac{1}{E(\Delta )} \int _{0}^{1}(1-t)^{\rho -1} \mathcal{F}^{ \alpha }_{\sigma ,\rho }\bigl[\omega \bigl(E(\Delta ) \bigr)^{\sigma }(1-t)^{\sigma }\bigr]f\bigl(tE( \zeta )+(1-t)E(\eta )\bigr)\,dt \\ =&\frac{1}{E(\Delta )}\mathcal{F}^{\alpha }_{\sigma , \rho +1} \bigl[\omega \bigl(E(\Delta )\bigr)^{\sigma }\bigr]f\bigl(E(\eta )\bigr) \\ & {} -\frac{1}{(E(\Delta ))^{\rho +1}} \int _{E(\zeta )}^{E(\eta )}\bigl(v-E( \zeta ) \bigr)^{\rho -1}\mathcal{F}^{\alpha }_{\sigma ,\rho }\bigl[\omega \bigl(E( \Delta )\bigr)^{\sigma }\bigl(v-E(\zeta )\bigr)^{\sigma } \bigr]f(v)\,dv \\ =&\frac{1}{E(\Delta )}\mathcal{F}^{\alpha }_{\sigma , \rho +1} \bigl[\omega \bigl(E(\Delta )\bigr)^{\sigma }\bigr]f\bigl(E(\eta )\bigr)- \frac{1}{(E(\Delta ))^{\rho +1}}\mathcal{J}_{\sigma ,\rho , \eta +; \omega }^{\alpha }f\bigl( E(\zeta ) \bigr). \end{aligned}$$
(2.9)

Similarly,

$$\begin{aligned} I_{2} =& \int _{0}^{1}t^{\rho } \mathcal{F}^{\alpha }_{\sigma ,\rho +1}\bigl[ \omega \bigl(E(\Delta ) \bigr)^{\sigma }t^{\sigma }\bigr]f^{\prime }\bigl(tE(\zeta )+(1-t)E(\eta )\bigr)\,dt \\ =& -\frac{1}{E(\Delta )}{F}^{\alpha }_{\sigma ,\rho +1} \bigl[\omega \bigl(E(\Delta )\bigr)^{ \sigma }\bigr]f\bigl(E(\zeta )\bigr)+ \frac{1}{(E(\Delta ))^{\rho +1}}\mathcal{J}_{ \sigma ,\rho , E(\zeta ) +;\omega }^{\alpha }f\bigl(E(\eta ) \bigr) \end{aligned}$$
(2.10)

and on subtracting inequality (2.9) and inequality (2.10), then multiplying by \(\frac{E(\Delta )}{2\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]} \), we obtain

$$\frac{E(\Delta )}{2\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]} [ I_{1}-I_{2} ]=\mathcal{M}( \mathcal{J}_{E}), $$

and on submitting the expressions for \(I_{1} \) and \(I_{2} \), we obtain the required result. □

Theorem 3

Let \(E: J \longrightarrow \mathbb{R} \) be a continuous increasing function and \(\zeta , \eta \in J \) with \(\zeta < \eta \). Let \(f : I\longrightarrow \mathbb{R}\) be a differentiable function on \(I^{o} \) and \(f^{\prime }\in L([E(\zeta ),E(\eta )])\) for \(E(\zeta ),E(\eta )\in I \). If \(|f^{\prime }| \) is an E-convex function on \([\zeta , \eta ]\), then the following inequality holds for generalized fractional integral operators:

$$\bigl\vert \mathcal{M}(\mathcal{J}_{E}) \bigr\vert \leq \frac{E(\Delta )}{2} \frac{\mathcal{F}^{\alpha _{1}}_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]}{\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]} \bigl[ \bigl\vert f^{\prime } \bigl(E(\zeta )\bigr) \bigr\vert + \bigl\vert f^{\prime }\bigl(E(\eta ) \bigr) \bigr\vert \bigr], $$

for all \(\sigma , \rho \in \mathbb{R}^{+}\) and \(\omega \in \mathbb{R}\), where

$$\begin{aligned} \alpha _{1}(n) =& \frac{\alpha (n)}{(\sigma n+\rho +1)} \biggl(1- \frac{1}{2^{\sigma n+\rho }} \biggr)\quad \textit{for }n=0,1,2,\ldots . \end{aligned}$$
(2.11)

Proof

Using Lemma 2, the properties of modulus, and the E-convexity of \(|f^{\prime }|\), respectively, we have

$$\begin{aligned}& \frac{2\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]}{E(\Delta )} \bigl\vert \mathcal{M}(\mathcal{J}_{E}) \bigr\vert \\& \quad = \biggl\vert \int _{0}^{ 1} \bigl\lbrace ( 1 - t)^{\rho } \mathcal{F}^{\alpha }_{\sigma ,\rho +1} \bigl[\omega \bigl(E(\Delta )\bigr)^{\sigma }( 1 - t)^{\sigma }\bigr] \\& \qquad {}- t^{\rho } \mathcal{F}^{\alpha }_{\sigma ,\rho +1} \bigl[\omega \bigl(E(\Delta )\bigr)^{ \sigma }t^{\sigma }\bigr] \bigr\rbrace f^{\prime } \bigl(tE(\zeta ) + ( 1 - t)E(\eta )\bigr)\,dt \biggr\vert \\& \quad = \Biggl\vert \sum_{n=0}^{\infty } \frac{\alpha (n)}{\Gamma (\sigma n+\rho +1)}\bigl[\omega \bigl(E(\Delta )\bigr)^{ \sigma } \bigr]^{n} \\& \qquad {}\times \int _{0}^{1} \bigl[(1 - t)^{\sigma n+\rho } - t^{ \sigma n+\rho } \bigr]f^{\prime } \bigl(tE(\zeta ) + (1 - t)E(\eta ) \bigr)\,dt \Biggr\vert \\& \quad \leq \sum_{n=0}^{\infty } \frac{\alpha (n)}{\Gamma (\sigma n+\rho +1)}\bigl[\omega \bigl(E(\Delta )\bigr)^{ \sigma } \bigr]^{n} \\& \qquad {}\times \int _{0}^{1} \bigl\vert (1 - t)^{\sigma n+\rho } - t^{ \sigma n+\rho } \bigr\vert \bigl\vert f^{\prime } \bigl(tE(\zeta ) + (1 - t)E( \eta ) \bigr) \bigr\vert \,dt \\& \quad \leq \sum_{n=0}^{\infty } \frac{\alpha (n)}{\Gamma (\sigma n+\rho +1)}\bigl[\omega \bigl(E(\Delta )\bigr)^{ \sigma } \bigr]^{n} \\& \qquad {}\times\biggl[ \int _{0}^{ \frac{1}{2}} \bigl[(1 - t)^{\sigma n+\rho } - t^{\sigma n+\rho } \bigr] \bigl[t \bigl\vert f^{\prime } \bigl(E(\zeta ) \bigr) \bigr\vert + (1 - t) \bigl\vert f^{\prime } \bigl(E(\eta ) \bigr) \bigr\vert \bigr] \,dt \\& \qquad {}+ \int _{\frac{1}{2}}^{1} \bigl[t^{ \sigma n+\rho }-(1-t)^{\sigma n+\rho } \bigr] \bigl[t \bigl\vert f^{\prime } \bigl(E(\zeta ) \bigr) \bigr\vert +(1-t) \bigl\vert f^{\prime } \bigl(E(\eta ) \bigr) \bigr\vert \bigr]\,dt\biggr] \\& \quad = \sum_{n=0}^{\infty } \frac{\alpha (n)}{\Gamma (\sigma n+\rho +1)}\bigl[ \omega \bigl(E(\Delta )\bigr)^{\sigma } \bigr]^{n} [I_{3}+I_{4} ]. \end{aligned}$$
(2.12)

Consider the following integral

$$\begin{aligned} I_{3} =& \int _{0}^{\frac{1}{2}} \bigl[(1-t)^{\sigma n+\rho }-t^{ \sigma n+\rho } \bigr] \bigl[t \bigl\vert f^{\prime } \bigl(E(\zeta ) \bigr) \bigr\vert +(1-t) \bigl\vert f^{\prime } \bigl(E(\eta ) \bigr) \bigr\vert \bigr]\,dt \\ =& \int _{0}^{\frac{1}{2}} \bigl[t(1-t)^{\sigma n+\rho }-t^{\sigma n+ \rho +1} \bigr] \bigl\vert f^{\prime } \bigl(E(\zeta ) \bigr) \bigr\vert \,dt \\ & {}+ \int _{0}^{\frac{1}{2}} \bigl[(1-t)^{\sigma n+\rho +1}-t^{\sigma n+ \rho +1}(1-t) \bigr] \bigl\vert f^{\prime } \bigl(E(\eta ) \bigr) \bigr\vert \,dt \\ =& \biggl[\frac{1}{(\sigma n+\rho +1)(\sigma n+\rho +2)}- \frac{1}{2^{\sigma n+\rho +1}(\sigma n+\rho +1)} \biggr] \bigl\vert f^{\prime } \bigl(E(\zeta ) \bigr) \bigr\vert \\ & {}+ \biggl[\frac{1}{(\sigma n+\rho +2)}- \frac{1}{2^{\sigma n+\rho +1}(\sigma n+\rho +1)} \biggr] \bigl\vert f^{\prime } \bigl(E(\eta ) \bigr) \bigr\vert . \end{aligned}$$

Similarly,

$$\begin{aligned} I_{4} =& \int _{\frac{1}{2}}^{1} \bigl[t^{\sigma n+ \rho }-(1-t)^{\sigma n+\rho } \bigr] \bigl[t \bigl\vert f^{\prime } \bigl(E( \zeta ) \bigr) \bigr\vert +(1-t) \bigl\vert f^{\prime } \bigl(E(\eta ) \bigr) \bigr\vert \bigr]\,dt \\ =& \biggl[\frac{1}{(\sigma n+\rho +2)}- \frac{1}{2^{\sigma n+\rho +1}(\sigma n+\rho +1)} \biggr] \bigl\vert f^{\prime } \bigl(E(\zeta ) \bigr) \bigr\vert \\ &{} + \biggl[ \frac{1}{(\sigma n+\rho +1)(\sigma n+\rho +2)}- \frac{1}{2^{\sigma n+\rho +1}(\sigma n+\rho +1)} \biggr] \bigl\vert f^{\prime } \bigl(E(\eta ) \bigr) \bigr\vert \end{aligned}$$

and on submitting values of integrals \(I_{3} \) and \(I_{4}\) into inequality (2.12), we have

$$\begin{aligned}& \frac{2\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]}{E(\Delta )} \bigl\vert \mathcal{M}(\mathcal{J}_{E}) \bigr\vert \\& \quad \leq \sum_{n=0}^{\infty } \frac{\alpha (n)}{\Gamma (\sigma n+\rho +1)}\bigl[\omega \bigl(E(\Delta )\bigr)^{ \sigma } \bigr]^{n} \\& \qquad {}\times\biggl[\frac{1}{\sigma n+\rho +1} \biggl( 1 - \frac{1}{2^{\sigma n+\rho }} \biggr) \bigl\lbrace \bigl\vert f^{\prime } \bigl(E(\zeta ) \bigr) \bigr\vert + \bigl\vert f^{\prime } \bigl(E(\eta ) \bigr) \bigr\vert \bigr\rbrace \biggr] \\& \quad ={\mathcal{F}^{\alpha _{1}}_{\sigma ,\rho +1}\bigl[ \omega \bigl(E( \Delta )\bigr)^{\sigma }\bigr]} \bigl[ \bigl\vert f^{\prime } \bigl(E(\zeta )\bigr) \bigr\vert + \bigl\vert f^{\prime }\bigl(E(\eta ) \bigr) \bigr\vert \bigr], \end{aligned}$$

where \(\alpha _{1}\) is as defined in (2.11). On rearranging we obtain the required result.

Hence it is proved. □

Remark 3

If in Theorem 3, the function E is chosen to be an identity function, then the following inequality holds for all \(\sigma , \rho \in \mathbb{R}^{+}\) and \(\omega \in \mathbb{R}\):

$$\begin{aligned} \bigl\vert \mathcal{M}(\mathcal{J}) \bigr\vert \leq &\frac{\Delta }{2} \frac{\mathcal{F}^{\alpha _{1}}_{\sigma ,\rho +1}[\omega (\Delta )^{\sigma }]}{\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (\Delta )^{\sigma }]} \bigl[ \bigl\vert f^{\prime }(\zeta ) \bigr\vert + \bigl\vert f^{\prime }(\eta ) \bigr\vert \bigr], \end{aligned}$$

which was given in [31].

Remark 4

If in Theorem 3, the function E is chosen to be an identity function, \(\alpha (0)=1\), \(\rho=\lambda\) and \(\omega =0\), then the following inequality holds:

$$\begin{aligned} \bigl\vert \mathcal{M}(J) \bigr\vert \leq& \frac{\Delta }{2} \biggl[ \frac{1}{\lambda +1} \biggl(1- \frac{1}{2^{\lambda }} \biggr) \biggr] \bigl[ \bigl\vert f^{\prime }(\zeta ) \bigr\vert + \bigl\vert f^{\prime }( \eta ) \bigr\vert \bigr], \end{aligned}$$

which was given in [32].

Theorem 4

Let \(E: J\longrightarrow \mathbb{R}\) be a continuous increasing function and \(\zeta , \eta \in J \) with \(\zeta < \eta \). Let \(f : I\longrightarrow \mathbb{R}\) be a differentiable function on \(I^{o} \) and \(f^{\prime }\in L([E(\zeta ),E(\eta )])\) for \(E(\zeta ),E(\eta )\in I \). If \(\vert f^{\prime } \vert ^{q}, q>1 \), is an E-convex function on \([\zeta , \eta ]\), then the following inequality holds for generalized fractional integral operators:

$$\bigl\vert \mathcal{M}(\mathcal{J}_{E}) \bigr\vert \leq \frac{E(\Delta )}{2} \frac{\mathcal{F}^{\alpha _{2}}_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]}{\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]} \biggl[ \frac{ \vert f^{\prime }(E(\zeta )) \vert ^{q}+ \vert f^{\prime }(E(\eta )) \vert ^{q}}{2} \biggr]^{\frac{1}{q}} $$

for all \(\sigma , \rho \in \mathbb{R}^{+}\) and \(\omega \in \mathbb{R}\), where p, q are conjugate indices and

$$ \alpha _{2}(n)=\alpha (n) \biggl( \frac{2}{p(\sigma n+\rho )+1} \biggl(1- \frac{1}{2^{p(\sigma n+\rho )}} \biggr) \biggr)^{\frac{1}{p}}\quad \textit{for } n=0,1,2,\ldots . $$
(2.13)

Proof

Using Lemma 2, the properties of modulus, and the well-known Hölder’s inequality, respectively,

$$\begin{aligned}& \frac{2\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]}{E(\Delta )} \bigl\vert \mathcal{M}(\mathcal{J}_{E}) \bigr\vert \\& \quad = \biggl\vert \int _{0}^{1} \bigl\lbrace (1 - t)^{\rho } \mathcal{F}^{\alpha }_{\sigma ,\rho +1}\bigl[\omega \bigl(E(\Delta )\bigr)^{\sigma }(1 - t)^{\sigma }\bigr]\\& \qquad {} - t^{\rho }\mathcal{F}^{\alpha }_{\sigma ,\rho +1}\bigl[ \omega \bigl(E(\Delta )\bigr)^{\sigma }t^{\sigma }\bigr] \bigr\rbrace f^{\prime } \bigl(tE( \zeta ) + (1 - t)E(\eta )\bigr)\,dt \biggr\vert \\& \quad = \Biggl\vert \sum_{n=0}^{\infty } \frac{\alpha (n)}{\Gamma (\sigma n+\rho +1)}\bigl[\omega \bigl(E(\Delta )\bigr)^{ \sigma } \bigr]^{n} \\& \qquad {}\times \int _{0}^{1} \bigl[(1-t)^{\sigma n+\rho }-t^{\sigma n+ \rho } \bigr]f^{\prime } \bigl(tE(\zeta )+(1-t)E(\eta ) \bigr)\,dt \Biggr\vert \\& \quad \leq \sum_{n=0}^{\infty } \frac{\alpha (n)}{\Gamma (\sigma n+\rho +1)}\bigl[\omega \bigl(E(\Delta )\bigr)^{ \sigma } \bigr]^{n} \\& \qquad {}\times \int _{0}^{1} \bigl\vert (1-t)^{\sigma n+\rho } -t^{ \sigma n+\rho } \bigr\vert \bigl\vert f^{\prime } \bigl(tE(\zeta )+(1 -t)E( \eta ) \bigr) \bigr\vert \,dt \\& \quad \leq \sum_{n=0}^{\infty } \frac{\alpha (n)}{\Gamma (\sigma n+\rho +1)}\bigl[\omega \bigl(E(\Delta )\bigr)^{ \sigma } \bigr]^{n} \\& \qquad {}\times \biggl( \int _{0}^{1} \bigl\vert (1-t)^{\sigma n+\rho }-t^{\sigma n+\rho } \bigr\vert ^{p}\,dt \biggr)^{\frac{1}{p}} \biggl( \int _{0}^{1} \bigl\vert f^{\prime } \bigl(tE(\zeta )+(1-t)E(\eta ) \bigr) \bigr\vert ^{q}\,dt \biggr)^{ \frac{1}{q}} \\& \quad =\sum_{n=0}^{\infty } \frac{\alpha (n)}{\Gamma (\sigma n+\rho +1)}\bigl[ \omega \bigl(E(\Delta )\bigr)^{\sigma } \bigr]^{n} (I_{5} )^{\frac{1}{p}} (I_{6} )^{\frac{1}{q}}. \end{aligned}$$
(2.14)

Solving the first integral from the right side of inequality (2.14) and using Lemma 1, we have

$$\begin{aligned} I_{5} =& \int _{0}^{1} \bigl\vert (1-t)^{\sigma n+\rho }-t^{\sigma n+\rho } \bigr\vert ^{p}\,dt \\ =& \int _{0}^{\frac{1}{2}} \bigl[(1-t)^{\sigma n+\rho }-t^{\sigma n+\rho } \bigr]^{p}\,dt+ \int _{\frac{1}{2}}^{1} \bigl[t^{\sigma n+\rho }-(1-t)^{ \sigma n+\rho } \bigr]^{p}\,dt \\ =& \int _{0}^{\frac{1}{2}} \bigl[(1-t)^{p(\sigma n+\rho )}-t^{p(\sigma n+ \rho )} \bigr]\,dt+ \int _{\frac{1}{2}}^{1} \bigl[t^{p(\sigma n+\rho )}-(1-t)^{p( \sigma n+\rho }) \bigr]\,dt \\ =&\frac{2}{p(\sigma n+\rho )+1} \biggl(1- \frac{1}{2^{p(\sigma n+\rho )}} \biggr). \end{aligned}$$
(2.15)

Solving the second integral from the right side of inequality (2.14) by using the fact that \(\vert f^{\prime } \vert ^{q}\), for any \(q>1 \), is E-convex, therefore we have

$$\begin{aligned} I_{6} =& \int _{0}^{1} \bigl\vert f^{\prime } \bigl(tE( \zeta )+(1-t)E(\eta ) \bigr) \bigr\vert ^{q}\,dt \\ \leq & \int _{0}^{1} \bigl[t \bigl\vert f^{\prime } \bigl(E(\zeta ) \bigr) \bigr\vert ^{q}+(1-t) \bigl\vert f^{\prime } \bigl(E(\eta ) \bigr) \bigr\vert ^{q} \bigr]\,dt \\ =& \frac{ \vert f^{\prime } (E(\zeta ) ) \vert ^{q}+ \vert f^{\prime } (E(\eta ) ) \vert ^{q}}{2}. \end{aligned}$$

On submitting the values of integrals \(I_{5} \) and \(I_{6} \) on the right side of inequality (2.14), we have

$$\begin{aligned} \bigl\vert \mathcal{M}(\mathcal{J}_{E}) \bigr\vert \leq& \frac{E(\Delta )}{2\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]} \sum_{n=0}^{\infty } \frac{\alpha (n)}{\Gamma (\sigma n+\rho +1)}\bigl[ \omega \bigl(E(\Delta )\bigr)^{\sigma } \bigr]^{n} \\ &{}\times \biggl( \frac{2}{p(\sigma n+\rho )+1} \biggl(1-\frac{1}{2^{p(\sigma n+\rho )}} \biggr) \biggr)^{\frac{1}{p}} \biggl( \frac{ \vert f^{\prime } (E(\zeta ) ) \vert ^{q}+ \vert f^{\prime } (E(\eta ) ) \vert ^{q}}{2} \biggr)^{\frac{1}{q}} \\ =&\frac{E(\Delta )}{2} \frac{\mathcal{F}^{\alpha _{2}}_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]}{\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]} \biggl[ \frac{ \vert f^{\prime }(E(\zeta )) \vert ^{q}+ \vert f^{\prime }(E(\eta )) \vert ^{q}}{2} \biggr]^{\frac{1}{q}}, \end{aligned}$$

where \(\alpha _{2} \) is as defined in (2.13). Hence it is proved. □

Remark 5

If in Theorem 4, the function E is chosen to be an identity function, then the following inequality holds for all \(\sigma , \rho \in \mathbb{R}^{+}\) and \(\omega \in \mathbb{R}\):

$$\begin{aligned} \bigl\vert \mathcal{M}(\mathcal{J}) \bigr\vert \leq& \frac{\Delta }{2} \frac{\mathcal{F}^{\alpha _{2}}_{\sigma ,\rho +1}[\omega (\Delta )^{\sigma }]}{\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (\Delta )^{\sigma }]} \biggl[ \frac{ \vert f^{\prime }(\zeta ) \vert ^{q}+ \vert f^{\prime }(\eta ) \vert ^{q}}{2} \biggr]^{\frac{1}{q}}, \end{aligned}$$

which was given in [33].

Remark 6

If in Theorem 4, the function E is chosen to be an identity function, \(\alpha (0)=1\), \(\rho=\lambda\) and \(\omega =0\), then the following inequality holds:

$$\begin{aligned} \bigl\vert \mathcal{M}(J) \bigr\vert \leq &\frac{\Delta }{2} \biggl( \frac{2}{p\lambda +1} \biggl(1-\frac{1}{2^{p\lambda }} \biggr) \biggr)^{ \frac{1}{p}} \biggl[ \frac{ \vert f^{\prime }(\zeta ) \vert ^{q}+ \vert f^{\prime }(\eta ) \vert ^{q}}{2} \biggr]^{\frac{1}{q}}, \end{aligned}$$

which was given in [33].

Theorem 5

Let \(E: J\longrightarrow \mathbb{R}\) be a continuous increasing function and \(\zeta , \eta \in J \) with \(\zeta < \eta \). Let \(f : I\longrightarrow \mathbb{R}\) be a differentiable function on \(I^{o} \) and \(f^{\prime }\in L([E(\zeta ),E(\eta )])\) for \(E(\zeta ),E(\eta )\in I \). If \(\vert f^{\prime } \vert ^{q} , q\geq 1 \), is an E-convex function on \([\zeta , \eta ]\), then the following inequality holds for generalized fractional integral operators:

$$\begin{aligned} \bigl\vert \mathcal{M}(\mathcal{J}_{E}) \bigr\vert \leq& \frac{E(\Delta )}{2} \frac{\mathcal{F}^{\alpha _{3}}_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]}{\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]} \biggl[ \frac{ \vert f^{\prime }(E(\zeta )) \vert ^{q}+ \vert f^{\prime }(E(\eta )) \vert ^{q}}{2} \biggr]^{\frac{1}{q}} \end{aligned}$$

for all \(\sigma , \rho \in \mathbb{R}^{+}\) and \(\omega \in \mathbb{R}\), where

$$\begin{aligned} \alpha _{3}(n) =&\alpha (n)\frac{2}{\sigma n +\rho +1} \biggl(1- \frac{1}{2^{\sigma n+\rho }} \biggr)\quad \textit{for }n=0,1,2,\ldots . \end{aligned}$$
(2.16)

Proof

Using Lemma 2, the well-known power mean inequality, and the E-convexity of \(\vert f^{\prime } \vert ^{q}\), respectively, we have

$$\begin{aligned}& \frac{2\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]}{E(\Delta )} \bigl\vert \mathcal{M}(\mathcal{J}_{E}) \bigr\vert \\& \quad = \biggl\vert \int _{0}^{1} \bigl\lbrace (1 - t)^{\rho }\mathcal{F}^{\alpha }_{\sigma ,\rho +1}\bigl[\omega \bigl(E( \Delta )\bigr)^{\sigma }(1 - t)^{\sigma }\bigr] \\& \qquad {} - t^{\rho }\mathcal{F}^{\alpha }_{ \sigma ,\rho +1}\bigl[\omega \bigl(E(\Delta )\bigr)^{\sigma }t^{\sigma }\bigr] \bigr\rbrace f^{\prime } \bigl(tE(\zeta ) + (1 - t)E(\eta )\bigr)\,dt \biggr\vert \\& \quad = \Biggl\vert \sum_{n=0}^{\infty } \frac{\alpha (n)}{\Gamma (\sigma n+\rho +1)}\bigl[\omega \bigl(E(\Delta )\bigr)^{ \sigma } \bigr]^{n} \\& \qquad {}\times \int _{0}^{1} \bigl[(1-t)^{\sigma n+\rho }-t^{\sigma n+ \rho } \bigr]f^{\prime } \bigl(tE(\zeta )+(1-t)E(\eta ) \bigr)\,dt \Biggr\vert \\& \quad \leq \sum_{n=0}^{\infty } \frac{\alpha (n)}{\Gamma (\sigma n+\rho +1)}\bigl[\omega \bigl(E(\Delta )\bigr)^{ \sigma } \bigr]^{n} \\& \qquad {}\times\int _{0}^{1} \bigl\vert (1-t)^{\sigma n+\rho }-t^{\sigma n+ \rho } \bigr\vert \bigl\vert f^{\prime } \bigl(tE(\zeta )+(1-t)E(\eta ) \bigr) \bigr\vert \,dt \\& \quad \leq \sum_{n=0}^{\infty } \frac{\alpha (n)}{\Gamma (\sigma n+\rho +1)}\bigl[\omega \bigl(E(\Delta )\bigr)^{ \sigma } \bigr]^{n} \biggl( \int _{0}^{1} \bigl\vert (1-t)^{\sigma n+\rho }-t^{ \sigma n+\rho } \bigr\vert \,dt \biggr)^{1-\frac{1}{q}} \\& \qquad {}\times \biggl( \int _{0}^{1} \bigl\vert (1-t)^{\sigma n+\rho }-t^{\sigma n+\rho } \bigr\vert \bigl\vert f^{\prime } \bigl(tE(\zeta )+(1-t)E(\eta ) \bigr) \bigr\vert ^{q}\,dt \biggr)^{\frac{1}{q}} \\& \quad \leq \sum_{n=0}^{\infty } \frac{\alpha (n)}{\Gamma (\sigma n+\rho +1)}\bigl[\omega \bigl(E(\Delta )\bigr)^{ \sigma } \bigr]^{n} \biggl[ \int _{0}^{1} \bigl\vert (1-t)^{\sigma n+\rho }-t^{ \sigma n+\rho } \bigr\vert \,dt \biggr]^{1-\frac{1}{q}} \\& \qquad {}\times \biggl( \int _{0}^{1} \bigl\vert (1-t)^{\sigma n+\rho }-t^{\sigma n+\rho } \bigr\vert \bigl[ t \bigl\vert f^{\prime } \bigl(E(\zeta ) \bigr) \bigr\vert ^{q}+(1-t) \bigl\vert f^{\prime } \bigl(E(\eta ) \bigr) \bigr\vert ^{q} \bigr] \biggr)^{ \frac{1}{q}} \\& \quad \leq \sum_{n=0}^{\infty } \frac{\alpha (n)}{\Gamma (\sigma n+\rho +1)}\bigl[\omega \bigl(E(\Delta )\bigr)^{ \sigma } \bigr]^{n} \biggl[\frac{2}{\sigma n +\rho +1} \biggl(1- \frac{1}{2^{\sigma n+\rho }} \biggr) \biggr]^{1-\frac{1}{q}} \\& \qquad {}\times \biggl[\frac{1}{\sigma n +\rho +1} \biggl(1- \frac{1}{2^{\sigma n+\rho }} \biggr) \bigl\lbrace \bigl\vert f^{\prime }\bigl(E( \zeta )\bigr) \bigr\vert ^{q}+ \bigl\vert f^{\prime }\bigl(E(\eta )\bigr) \bigr\vert ^{q} \bigr\rbrace \biggr]^{\frac{1}{q}} \\& \quad =\sum_{n=0}^{\infty } \frac{\alpha (n)}{\Gamma (\sigma n+\rho +1)}\bigl[\omega \bigl(E(\Delta )\bigr)^{ \sigma } \bigr]^{n} \\& \qquad {}\times\frac{2}{\sigma n +\rho +1} \biggl(1- \frac{1}{2^{\sigma n+\rho }} \biggr) \biggl[ \frac{ \vert f^{\prime }(\zeta ) \vert ^{q}+ \vert f^{\prime }(\eta ) \vert ^{q}}{2} \biggr]^{\frac{1}{q}} \end{aligned}$$
(2.17)

and on rearranging we obtain

$$\bigl\vert \mathcal{M}(\mathcal{J}_{E}) \bigr\vert = \frac{E(\Delta )}{2} \frac{\mathcal{F}^{\alpha _{3}}_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]}{\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]} \biggl[ \frac{ \vert f^{\prime }(\zeta ) \vert ^{q}+ \vert f^{\prime }(\eta ) \vert ^{q}}{2} \biggr]^{\frac{1}{q}}, $$

where \(\alpha _{3} \) is as defined in (2.16).

Hence it is proved. □

Remark 7

If in Theorem 5, the function E is chosen to be an identity function, then the following inequality holds for all \(\sigma , \rho \in \mathbb{R}^{+}\) and \(\omega \in \mathbb{R}\):

$$\begin{aligned} \bigl\vert \mathcal{M}(\mathcal{J}) \bigr\vert \leq& \frac{\Delta }{2} \frac{\mathcal{F}^{\alpha _{3}}_{\sigma ,\rho +1}[\omega (\Delta )^{\sigma }]}{\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (\Delta )^{\sigma }]} \biggl[ \frac{ \vert f^{\prime }(\zeta ) \vert ^{q}+ \vert f^{\prime }(\eta ) \vert ^{q}}{2} \biggr]^{\frac{1}{q}}, \end{aligned}$$

which was given in [33].

Remark 8

If in Theorem 5, the function E is chosen to be an identity function, \(\alpha (0)=1\), \(\rho=\lambda\) and \(\omega =0\), then the following inequality holds:

$$\begin{aligned} \bigl\vert \mathcal{M}(J) \bigr\vert \leq& \frac{\Delta }{\lambda +1} \biggl(1- \frac{1}{2^{\lambda }} \biggr) \biggl[ \frac{ \vert f^{\prime }(\zeta ) \vert ^{q}+ \vert f^{\prime }(\eta ) \vert ^{q}}{2} \biggr]^{\frac{1}{q}}, \end{aligned}$$

which was given in [33].

Lemma 3

Let \(E: J\subset \mathbb{R}^{+}\cup \{0\} \longrightarrow \mathbb{R} \) be a continuous increasing function and \(\zeta , \eta \in J \) with \(\zeta < \eta \). Let \(f : I\longrightarrow \mathbb{R}\) be a twice differentiable function on \(I^{o} \). If \(f^{\prime \prime }\in L([E(\zeta ),E(\eta )])\) for \(E(\zeta ),E(\eta )\in I \), then the following identity holds for generalized fractional operators:

$$\begin{aligned} \mathcal{M}(\mathcal{J}_{E}) =& \frac{(E(\Delta ))^{2}}{2\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]} \\ &{}\times\int _{0}^{1} \bigl\lbrace \mathcal{F}^{\alpha }_{ \sigma ,\rho +2}\bigl[\omega \bigl(E(\Delta ) \bigr)^{\sigma }\bigr] - (1 - t)^{\rho +1} \mathcal{F}^{\alpha }_{\sigma ,\rho +2} \bigl[\omega \bigl(E(\Delta )\bigr)^{\sigma }(1 - t)^{\sigma }\bigr] \\ &{} -t^{\rho +1} \mathcal{F}^{\alpha }_{\sigma ,\rho +2}\bigl[ \omega \bigl(E(\Delta )\bigr)^{\sigma }t^{ \sigma }\bigr] \bigr\rbrace f^{\prime \prime }\bigl(tE(\zeta )+(1-t)E(\eta )\bigr)\,dt \end{aligned}$$

for all \(\sigma , \rho > 0 \) and \(\omega \geq 0\).

Proof

Solving the following integral by simple integration

$$\begin{aligned} I_{7} =& \int _{0}^{1} \mathcal{F}^{\alpha }_{ \sigma ,\rho +2} \bigl[\omega \bigl(E(\Delta )\bigr)^{\sigma }\bigr]f^{\prime \prime } \bigl(tE(\zeta )+(1-t)E( \eta )\bigr)\,dt \\ =& \mathcal{F}^{\alpha }_{\sigma ,\rho +2}\bigl[\omega \bigl(E(\Delta ) \bigr)^{ \sigma }\bigr] \int _{0}^{1}f^{\prime \prime }\bigl(tE(\zeta )+(1-t)E(\eta )\bigr)\,dt \\ =&\frac{1}{E(\Delta )}\mathcal{F}^{\alpha }_{\sigma ,\rho +2}\bigl[\omega \bigl(E( \Delta )\bigr)^{\sigma }\bigr] \bigl[f^{\prime }\bigl(E( \eta )\bigr)-f^{\prime }\bigl(E(\zeta )\bigr)\bigr]. \end{aligned}$$

Solving the next integral by applying integration by parts twice, we have

$$\begin{aligned}& I_{8}= \int _{0}^{1}(1-t)^{\rho +1} \mathcal{F}^{ \alpha }_{\sigma ,\rho +2}\bigl[\omega \bigl(E(\Delta ) \bigr)^{\sigma }(1-t)^{\sigma }\bigr] f^{\prime \prime }\bigl(tE( \zeta )+(1-t)E(\eta )\bigr)\,dt, \\& \begin{aligned} I_{8}={}& \frac{\mathcal{F}^{\alpha }_{\sigma ,\rho +2}[\omega (E(\Delta ))^{\sigma }]}{E(\Delta )}f^{\prime } \bigl(E( \eta )\bigr)- \frac{\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]}{(E(\Delta ))^{2}}f\bigl(E( \eta )\bigr) \\ &{} +\frac{1}{(E(\Delta ))^{2}} \int _{0}^{1}(1-t)^{ \rho -1} \mathcal{F}^{\alpha }_{\sigma ,\rho }\bigl[\omega \bigl(E(\Delta ) \bigr)^{ \sigma }(1-t)^{\sigma }\bigr] f\bigl(tE(\zeta )+(1-t)E(\eta ) \bigr)\,dt \\ ={}& \frac{\mathcal{F}^{\alpha }_{\sigma ,\rho +2}[\omega (E(\Delta ))^{\sigma }]}{E(\Delta )}f^{\prime }\bigl(E( \eta )\bigr)- \frac{\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]}{(E(\Delta ))^{2}}f\bigl(E( \eta )\bigr) \\ &{} +\frac{1}{(E(\Delta ))^{\rho +2}} \int _{E( \zeta )}^{E(\eta )}\bigl(x-E(\zeta ) \bigr)^{\rho -1}\mathcal{F}^{\alpha }_{ \sigma ,\rho }\bigl[\omega (E \bigl(x-E(\zeta )\bigr)^{\rho -1}\bigr] f(x)dx \\ ={}& \frac{\mathcal{F}^{\alpha }_{\sigma ,\rho +2}[\omega (E(\Delta ))^{\sigma }]}{E(\Delta )}f^{\prime }\bigl(E( \eta )\bigr)- \frac{\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]}{(E(\Delta ))^{2}}f\bigl(E( \eta )\bigr) \\ &{} +\frac{1}{(E(\Delta ))^{\rho +2}}\mathcal{J}_{ \sigma ,\rho , \eta +;\omega }^{\alpha }f\bigl(E( \zeta )\bigr), \end{aligned} \end{aligned}$$

and similarly

$$\begin{aligned} I_{9}={}& \int _{0}^{1}t^{\rho +1} \mathcal{F}^{\alpha }_{ \sigma ,\rho +2}\bigl[\omega \bigl(E(\Delta ) \bigr)^{\sigma }t^{\sigma }\bigr]f^{\prime \prime }\bigl(tE( \zeta )+(1-t)E(\eta )\bigr)\,dt \\ ={}&{-} \frac{\mathcal{F}^{\alpha }_{\sigma ,\rho +2}[\omega (E(\Delta ))^{\sigma }]}{E(\Delta )}f^{\prime }\bigl(E( \zeta )\bigr)- \frac{\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]}{(E(\Delta ))^{2}}f\bigl(E( \zeta )\bigr) \\ &{} +\frac{1}{(E(\Delta ))^{\rho +2}}\mathcal{J}_{ \sigma ,\rho , E(\zeta ) +;\omega }^{\alpha }f\bigl(E( \eta )\bigr), \end{aligned}$$

on subtracting \(I_{8} \) and \(I_{9} \) from \(I_{7} \), then multiplying by \(\frac{(E(\Delta ))^{2}}{2\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]} \), we obtain

$$\frac{(E(\Delta ))^{2}}{2\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]}[ I_{7}-I_{8}- I_{9}]=\mathcal{M}(\mathcal{J}_{E}) $$

and on submitting the values of \(I_{7} \), \(I_{8} \) and \(I_{9} \) we obtain the required result. □

Theorem 6

Let \(E: J\longrightarrow \mathbb{R}\) be a continuous increasing function and \(\zeta , \eta \in J \) with \(\zeta < \eta \). Let \(f : I\longrightarrow \mathbb{R}\) be a differentiable function on \(I^{o} \) and \(f^{\prime \prime }\in L([E(\zeta ),E(\eta )])\) for \(E(\zeta ),E(\eta )\in I \). If \(\vert f^{\prime \prime } \vert \) is an E-convex function on \([\zeta , \eta ]\), then the following inequality holds for generalized fractional integral operators:

$$\begin{aligned} \bigl\vert \mathcal{M}(\mathcal{J}_{E}) \bigr\vert \leq & \frac{ (E(\Delta ) )^{2}}{2} \frac{\mathcal{F}^{\alpha _{4}}_{\sigma ,\rho +3}[\omega (E(\Delta ))^{\sigma }]}{\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]} \biggl[ \frac{ \vert f^{\prime \prime }(E(\zeta )) \vert + \vert f^{\prime \prime }(E(\eta )) \vert }{2} \biggr], \end{aligned}$$

for all \(\sigma , \rho \in \mathbb{R}^{+}\) and \(\omega \in \mathbb{R}\), where

$$\begin{aligned} \alpha _{4}(n) =&\alpha (n) (\sigma n+\rho )\quad \textit{for } n=0,1,2,\ldots . \end{aligned}$$
(2.18)

Proof

Using Lemma 3, the properties of modulus, and the E-convexity of \(\vert f^{\prime \prime } \vert \), respectively,

$$\begin{aligned}& \bigl\vert \mathcal{M}(\mathcal{J}_{E}) \bigr\vert \\& \quad \leq \frac{(E(\Delta ))^{2}}{2\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]} \sum_{n=0}^{\infty } \frac{\alpha (n)}{\Gamma (\sigma n+\rho +2)}\bigl[ \omega \bigl(E(\Delta )\bigr)^{\sigma } \bigr]^{n} \\& \qquad {}\times \int _{0}^{1} \bigl[1-(1-t)^{\sigma n+\rho +1}-t^{\sigma n+\rho +1} \bigr] \bigl\vert f^{\prime \prime }\bigl(tE(\zeta )+(1-t)E(\eta )\bigr) \bigr\vert \,dt \\& \quad \leq \frac{(E(\Delta ))^{2}}{2\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]} \sum_{n=0}^{\infty } \frac{\alpha (n)}{\Gamma (\sigma n+\rho +2)}\bigl[ \omega \bigl(E(\Delta )\bigr)^{\sigma } \bigr]^{n} \\& \qquad {}\times \int _{0}^{1} \bigl[1-(1-t)^{\sigma n+\rho +1}-t^{\sigma n+\rho +1} \bigr] \bigl[ t \bigl\vert f^{\prime \prime }\bigl(E(\zeta )\bigr) \bigr\vert +(1-t) \bigl\vert f^{\prime \prime }\bigl(tE( \eta )\bigr) \bigr\vert \bigr]\,dt \\& \quad \leq \frac{(E(\Delta ))^{2}}{2\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]} \sum_{n=0}^{\infty } \frac{\alpha (n)}{\Gamma (\sigma n+\rho +2)}\bigl[ \omega \bigl(E(\Delta )\bigr)^{\sigma } \bigr]^{n} \\& \qquad {}\times \biggl[\frac{\sigma n+\rho }{2(\sigma n+\rho +2)} \bigl\lbrace \bigl\vert f^{\prime \prime }\bigl(E(\zeta )\bigr) \bigr\vert + \bigl\vert f^{\prime \prime }\bigl(E(\eta )\bigr) \bigr\vert \bigr\rbrace \biggr] \\& \quad \leq \frac{ (E(\Delta ) )^{2}}{2} \frac{\mathcal{F}^{\alpha _{4}}_{\sigma ,\rho +3}[\omega (E(\Delta ))^{\sigma }]}{\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]} \biggl[ \frac{ \vert f^{\prime \prime }(E(\zeta )) \vert + \vert f^{\prime \prime }(E(\eta )) \vert }{2} \biggr], \end{aligned}$$

where \(\alpha _{4}(n)\) is as defined in (2.18).

Hence it is proved. □

Remark 9

If in Theorem 6, the function E is chosen to be an identity function, then the following inequality holds for all \(\sigma , \rho \in \mathbb{R}^{+}\) and \(\omega \in \mathbb{R}\):

$$\begin{aligned} \bigl\vert \mathcal{M}(\mathcal{J}) \bigr\vert \leq& \frac{ (\Delta )^{2}}{2} \frac{\mathcal{F}^{\alpha _{4}}_{\sigma ,\rho +3}[\omega (\Delta )^{\sigma }]}{\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (\Delta )^{\sigma }]} \biggl[ \frac{ \vert f^{\prime \prime }(E(\zeta )) \vert + \vert f^{\prime \prime }(E(\eta )) \vert }{2} \biggr], \end{aligned}$$

which was given in [33].

Remark 10

If in Theorem 6, the function E is chosen to be an identity function, \(\alpha (0)=1\), \(\rho=\lambda\) and \(\omega =0\), then the following inequality holds:

$$\begin{aligned} \bigl\vert \mathcal{M}(J) \bigr\vert \leq& \frac{ (\Delta )^{2}}{2} \frac{\lambda }{(\lambda +1)(\lambda +2)} \biggl[ \frac{ \vert f^{\prime \prime }(E(\zeta )) \vert + \vert f^{\prime \prime }(E(\eta )) \vert }{2} \biggr], \end{aligned}$$

which was given in [33].

Theorem 7

Let \(E: J\longrightarrow \mathbb{R}\) be a continuous increasing function and \(\zeta , \eta \in J \) with \(\zeta < \eta \). Let \(f : I\longrightarrow \mathbb{R}\) be a differentiable function on \(I^{o} \) and \(f^{\prime \prime }\in L([E(\zeta ),E(\eta )])\) for \(E(\zeta ),E(\eta )\in I \). If \(|f^{\prime \prime }|^{q}, q>1 \), is an E-convex function on \([\zeta , \eta ]\), then the following inequality holds for generalized fractional integral operators:

$$\bigl\vert \mathcal{M}(\mathcal{J}_{E}) \bigr\vert \leq \frac{ (E(\Delta ) )^{2}}{2} \frac{\mathcal{F}^{\alpha _{5}}_{\sigma ,\rho +2}[\omega (E(\Delta ))^{\sigma }]}{\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]} \biggl[ \frac{ \vert f^{\prime \prime }(E(\zeta )) \vert ^{q}+ \vert f^{\prime \prime }(E(\eta )) \vert ^{q}}{2} \biggr]^{\frac{1}{q}}, $$

for all \(\sigma , \rho \in \mathbb{R}^{+}\) and \(\omega \in \mathbb{R}\), where

$$ \alpha _{5}(n)=\alpha (n) \biggl(1- \frac{2}{p(\sigma n+\rho +1)+1} \biggr)^{\frac{1}{p}} \quad \textit{for }n=0,1,2,3, \ldots . $$
(2.19)

Proof

Using Lemma 3 and applying Hölder’s inequality, respectively,

$$\begin{aligned}& \bigl\vert \mathcal{M}(\mathcal{J}_{E}) \bigr\vert \\& \quad \leq \frac{(E(\Delta ))^{2}}{2\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]} \sum_{n=0}^{\infty } \frac{\alpha (n)}{\Gamma (\sigma n+\rho +2)}\bigl[ \omega \bigl(E(\Delta )\bigr)^{\sigma } \bigr]^{n} \\& \qquad {} \times \biggl[ \int _{0}^{1} \bigl\vert 1-(1-t)^{\sigma n+\rho +1}-t^{ \sigma n+\rho +1} \bigr\vert ^{p} \,dt \biggr]^{\frac{1}{p}} \biggl[ \int _{0}^{1} \bigl\vert f^{\prime \prime } \bigl(tE(\zeta )+(1-t)E(\eta )\bigr) \bigr\vert ^{q}\,dt \biggr]^{ \frac{1}{q}} \\& \quad \leq \frac{(E(\Delta ))^{2}}{2\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]} \sum_{n=0}^{\infty } \frac{\alpha (n)}{\Gamma (\sigma n+\rho +2)}\bigl[ \omega \bigl(E(\Delta )\bigr)^{\sigma } \bigr]^{n} \\& \qquad {} \times \biggl[ \int _{0}^{1} \bigl[1 - (1 - t)^{p( \sigma n+\rho +1)} -t^{p(\sigma n+\rho +1)} \bigr] \,dt \biggr]^{ \frac{1}{p}}\\& \qquad {} \times \biggl[ \int _{0}^{1} \bigl[t \bigl\vert f^{\prime \prime }\bigl(E(\zeta )\bigr) \bigr\vert ^{q} + (1 - t) \bigl\vert f^{\prime \prime }\bigl(E(\zeta )\bigr) \bigr\vert ^{q} \bigr] \,dt \biggr]^{ \frac{1}{q}} \\& \quad \leq \frac{(E(\Delta ))^{2}}{2\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]} \sum_{n=0}^{\infty } \frac{\alpha (n)}{\Gamma (\sigma n+\rho +2)}\bigl[ \omega \bigl(E(\Delta )\bigr)^{\sigma } \bigr]^{n} \\& \qquad {} \times \biggl[1-\frac{2}{p(\sigma n+\rho +1)+1} \biggr]^{ \frac{1}{p}} \biggl[ \frac{ \vert f^{\prime \prime }(E(\zeta )) \vert ^{q}+ \vert f^{\prime \prime }(E(\eta )) \vert ^{q}}{2} \biggr]^{\frac{1}{q}} \\& \quad = \frac{ (E(\Delta ) )^{2}}{2} \frac{\mathcal{F}^{\alpha _{5}}_{\sigma ,\rho +2}[\omega (E(\Delta ))^{\sigma }]}{\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]} \biggl[ \frac{ \vert f^{\prime \prime }(E(\zeta )) \vert ^{q}+ \vert f^{\prime \prime }(E(\eta )) \vert ^{q}}{2} \biggr]^{\frac{1}{q}}, \end{aligned}$$

where \(\alpha _{5} \) is as defined in (2.19). Hence it is proved. □

Remark 11

If in Theorem 7, the function E is chosen to be an identity function, then the following inequality holds for all \(\sigma , \rho \in \mathbb{R}^{+}\) and \(\omega \in \mathbb{R}\):

$$\begin{aligned} \bigl\vert \mathcal{M}(\mathcal{J}) \bigr\vert \leq& \frac{ (\Delta )^{2}}{2} \frac{\mathcal{F}^{\alpha _{5}}_{\sigma ,\rho +2}[\omega (\Delta )^{\sigma }]}{\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (\Delta )^{\sigma }]} \biggl[ \frac{ \vert f^{\prime \prime }(\zeta ) \vert ^{q}+ \vert f^{\prime \prime }(\eta ) \vert ^{q}}{2} \biggr]^{\frac{1}{q}}, \end{aligned}$$

which was given in [33].

Remark 12

If in Theorem 7, the function E is chosen to be an identity function, \(\alpha (0)=1\), \(\rho=\lambda\) and \(\omega =0\), then the following inequality holds:

$$\begin{aligned} \bigl\vert \mathcal{M}(J) \bigr\vert \leq& \frac{ (\Delta )^{2}}{2} \frac{1}{\lambda +1} \biggl(1- \frac{2}{p(\lambda +1)+1} \biggr)^{\frac{1}{p}} \biggl[ \frac{ \vert f^{\prime \prime }(\zeta ) \vert ^{q}+ \vert f^{\prime \prime }(\eta ) \vert ^{q}}{2} \biggr]^{\frac{1}{q}}, \end{aligned}$$

which was given in [33].

Theorem 8

Let \(E: J\longrightarrow \mathbb{R}\) be a continuous increasing function and \(\zeta , \eta \in J \) with \(\zeta < \eta \). Let \(f : I\longrightarrow \mathbb{R}\) be a differentiable function on \(I^{o}\). If \(|f^{\prime \prime }|^{q}, q \geq 1 \), is an E-convex function on \([\zeta , \eta ]\), then the following inequality holds for generalized fractional integral operators:

$$\bigl\vert \mathcal{M}(\mathcal{J}_{E}) \bigr\vert \leq \frac{ (E(\Delta ) )^{2}}{2} \frac{\mathcal{F}^{\alpha _{6}}_{\sigma ,\rho +2}[\omega (E(\Delta ))^{\sigma }]}{\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]} \biggl[ \frac{ \vert f^{\prime \prime }(E(\zeta )) \vert ^{q}+ \vert f^{\prime \prime }(E(\eta )) \vert ^{q}}{2} \biggr]^{\frac{1}{q}} $$

for all \(\sigma , \rho \in \mathbb{R}^{+}\) and \(\omega \in \mathbb{R}\), where p, q are conjugate indices and

$$ \alpha _{6}(n)=\alpha (n) \biggl( \frac{\sigma n+\rho }{\sigma n+\rho +2} \biggr) \quad \textit{for }n=0,1,2,\ldots . $$
(2.20)

Proof

Using Lemma 3 and applying the well-known power mean inequality

$$\begin{aligned}& \bigl\vert \mathcal{M}(\mathcal{J}_{E}) \bigr\vert \\& \quad \leq \frac{ (E(\Delta ) )^{2}}{2\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]} \sum_{n=0}^{\infty } \frac{\alpha (n)}{\Gamma (\sigma n + \rho + 2)}\bigl[\omega \bigl(E( \Delta )\bigr)^{\sigma } \bigr]^{n} \\& \qquad {} \times\biggl( \int _{0}^{1} \bigl[1 - (1 - t)^{\sigma n+\rho +1} -t^{\sigma n+\rho +1} \bigr] \,dt \biggr)^{ 1-\frac{1}{q}} \\& \qquad {} \times \biggl( \int _{0}^{1} \bigl[1-(1-t)^{ \sigma n+\rho +1}-t^{\sigma n+\rho +1} \bigr] \bigl\vert f^{\prime \prime } \bigl(tE( \zeta )+(1-t)E(\eta ) \bigr) \bigr\vert ^{q}\,dt \biggr)^{\frac{1}{q}} \\& \quad \leq \frac{ (E(\Delta ) )^{2}}{2\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]} \sum_{n=0}^{\infty } \frac{\alpha (n)}{\Gamma (\sigma n+\rho +2)}\bigl[ \omega \bigl(E(\Delta )\bigr)^{\sigma } \bigr]^{n} \biggl[1- \frac{2}{\sigma n +\rho +2} \biggr]^{1-\frac{1}{q}} \\& \qquad {} \times \biggl[ \frac{\sigma n+\rho }{2(\sigma n+\rho +2)} \bigl\lbrace \bigl\vert f^{\prime \prime }\bigl(E( \zeta )\bigr) \bigr\vert ^{q}+ \bigl\vert f^{\prime \prime }\bigl(E(\eta )\bigr) \bigr\vert ^{q} \bigr\rbrace \biggr]^{\frac{1}{q}} \\& \quad \leq \frac{ (E(\Delta ) )^{2}}{2\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]} \sum_{n=0}^{\infty } \frac{\alpha (n)}{\Gamma (\sigma n+\rho +2)}\bigl[ \omega \bigl(E(\Delta )\bigr)^{\sigma } \bigr]^{n} \\& \qquad {} \times \biggl( \frac{\sigma n+\rho }{\sigma n+\rho +2} \biggr) \biggl[ \frac{ \vert f^{\prime \prime }(E(\zeta )) \vert ^{q}+ \vert f^{\prime \prime }(E(\eta )) \vert ^{q}}{2} \biggr]^{\frac{1}{q}} \\& \quad = \frac{ (E(\Delta ) )^{2}}{2} \frac{\mathcal{F}^{\alpha _{6}}_{\sigma ,\rho +2}[\omega (E(\Delta ))^{\sigma }]}{\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (E(\Delta ))^{\sigma }]} \biggl[ \frac{ \vert f^{\prime \prime }(E(\zeta )) \vert ^{q}+ \vert f^{\prime \prime }(E(\eta )) \vert ^{q}}{2} \biggr]^{\frac{1}{q}}, \end{aligned}$$

where \(\alpha _{6} \) is as defined in (2.20). Hence it is proved. □

Corollary 1

If in Theorem 8, the function E is chosen to be an identity function, then the following inequality holds for all \(\sigma , \rho \in \mathbb{R}^{+}\) and \(\omega \in \mathbb{R}\):

$$\bigl\vert \mathcal{M}(\mathcal{J}) \bigr\vert \leq \frac{ (\Delta )^{2}}{2} \frac{\mathcal{F}^{\alpha _{6}}_{\sigma ,\rho +2}[\omega (\Delta )^{\sigma }]}{\mathcal{F}^{\alpha }_{\sigma ,\rho +1}[\omega (\Delta )^{\sigma }]} \biggl[ \frac{ \vert f^{\prime \prime }(E(\zeta )) \vert ^{q}+ \vert f^{\prime \prime }(E(\eta )) \vert ^{q}}{2} \biggr]^{\frac{1}{q}}. $$

Corollary 2

If in Theorem 8, the function E is chosen to be an identity function, \(\alpha (0)=1\), \(\rho=\lambda\) and \(\omega =0\), then the following inequality holds:

$$\bigl\vert \mathcal{M}(J) \bigr\vert \leq \frac{ (\Delta )^{2}}{2} \biggl( \frac{\lambda }{\lambda +2} \biggr) \biggl[ \frac{ \vert f^{\prime \prime }(E(\zeta )) \vert ^{q}+ \vert f^{\prime \prime }(E(\eta )) \vert ^{q}}{2} \biggr]^{\frac{1}{q}}. $$

Availability of data and materials

This work does not involve any supplementary data or material, however, all related calculations can be supplied on demand.

References

  1. Hardy, G.H., Litllewood, J.E., Polya, G.: Inequalities. Cambridge University Press, Cambridge (1934)

    Google Scholar 

  2. Dragomir, S.S., Pearce, C.E.M.: Quasi-convex functions and Hadamards inequality. Bull. Aust. Math. Soc. 57, 377–385 (1998)

    MathSciNet  Article  Google Scholar 

  3. Hussain, R., Ali, A., Latif, A., Gulshan, G.: Some k-fractional associates of Hadamard’s inequality for quasi-convex functions and applications to special means. Fract. Differ. Calc. 7, 301–309 (2017)

    MathSciNet  Article  Google Scholar 

  4. Dragomir, S.S.: On the Hadamard’s inequlality for convex functions on the co-ordinates in a rectangle from the plane. Taiwan. J. Math. 5, 775–788 (2001)

    Article  Google Scholar 

  5. Hussain, R., Ali, A., Latif, A., Gulshan, G.: Co-ordinated convex function of three variables and some analogues inequalities with applications. J. Comput. Anal. Appl. 29, 505–551 (2021)

    Google Scholar 

  6. Wang, S.H., Qi, F.: Hermite-Hadamard-type inequalities for n-times differentiable and preinvex functions. J. Inequal. Appl. 2014, 49 (2014). https://doi.org/10.1186/1029-242X-2014-49

    MathSciNet  Article  MATH  Google Scholar 

  7. Zhang, X.M., Chu, Y.M., Zhang, X.H.: The Hermite-Hadamard type inequality of GA-convex functions and its application. J. Inequal. Appl. 2010, 507560 (2010). https://doi.org/10.1155/2010/507560

    MathSciNet  Article  MATH  Google Scholar 

  8. Polyak, B.T.: Existence theorems and convergence of minimizing sequences in extremum problems with restrictions. Sov. Math. Dokl. 7, 72–75 (1966)

    Google Scholar 

  9. z̈demir, M.E., Gürbüz, M., Kavurmacı, H.: Hermite-Hadamard-type inequalities for \((g, \varphi _{h}) \)-convex dominated functions. J. Inequal. Appl. 2013, 184 (2013). https://doi.org/10.1186/1029-242X-2013-184

    MathSciNet  Article  MATH  Google Scholar 

  10. Youness, E.A.: E-convex sets, E-convex functions and E-convex programming. J. Optim. Theory Appl. 102, 439–450 (1999)

    MathSciNet  Article  Google Scholar 

  11. Sarikaya, M.Z., Yaldiz, H.: On Hermite-Hadamard type inequalities for φ-convex functions via fractional integrals. Malaysian J. Math. Sci. 9, 243–258 (2015)

    MathSciNet  Google Scholar 

  12. Bubeck, S.: Convex optimization: algorithms and complexity. Found. Trends Mach. Learn. 8, 231–357 (2015)

    Article  Google Scholar 

  13. Dinu, C.: Hermite-Hadamard inequality on time scales. J. Inequal. Appl. 2008, 287947 (2008) https://doi.org/10.1155/2008/287947

    MathSciNet  Article  MATH  Google Scholar 

  14. Fagbemigun, B.O., Mogbademu, A.A.: Some classes of convex functions on time scales. RGMIA Research Report Collections 22, 1–12 (2019)

    MATH  Google Scholar 

  15. Alp, N., Sarıkaya, M.Z., Kunt, M., Iscan, I.: q-Hermite Hadamard inequalities and quantum estimates for midpoint type inequalities via convex and quasi-convex functions. J. King Saud Univ., Sci. 30, 193–203 (2018)

    Article  Google Scholar 

  16. Sudsutad, W., Ntouyas, S.K., Tariboon, J.: Quantum integral inequalities for convex functions. J. Math. Inequal. 9, 781–793 (2015)

    MathSciNet  Article  Google Scholar 

  17. Farid, G., Mahreen, K., Chu, Y.M.: Study of inequalities for unified integral operators of generalized convex functions. Open J. Math. Sci. 5(1), 80–93 (2021)

    Article  Google Scholar 

  18. Farid, G., Rehman, A.U., Bibi, S., Chu, Y.M.: Refinements of two fractional versions of Hadamard inequalities for Caputo fractional derivatives and related results. Open J. Math. Sci. 5(1), 1–10 (2021)

    Article  Google Scholar 

  19. Set, E., Dragomir, S.S., Gözpinar, A.G.: Some generalized Hermite-Hadamard-type inequalities involving fractional integral operator for functions whose second derivatives in absolute value are s-convex. Acta Math. Univ. Comen. 88, 87–100 (2019)

    MathSciNet  MATH  Google Scholar 

  20. Atangana, A., Baleanu, D.: New fractional derivatives with nonlocal and non-singular kernel: theory and application to heat transfer model. Therm. Sci. 20, 763–769 (2016)

    Article  Google Scholar 

  21. Caputo, M.: Linear models of dissipation whose Q is almost frequency independent–II. Geophys. J. Int. 13, 529–539 (1967)

    Article  Google Scholar 

  22. Farid, G.: Existence of an integral operator and its consequences in fractional and conformable integrals. Open J. Math. Sci. 3(3), 210–216 (2019)

    Article  Google Scholar 

  23. Farid, G.: A unified integral operator and further its consequences. Open J. Math. Anal. 4(1), 1–7 (2020)

    Article  Google Scholar 

  24. Miller, K.S., Ross, B.: An Introduction to the Fractional Calculus and Fractional Differential Equations. Wiley, New York (1993)

    MATH  Google Scholar 

  25. Prabhakar, T.R.: A singular integral equation with a generalized Mittag Leffler function in the kernel. Yokohama Math. J. 19, 7–15 (1971)

    MathSciNet  MATH  Google Scholar 

  26. Agarwal, R.P., Luo, M.J., Raina, R.K.: On Ostrowski type inequalities. Fasc. Math. 56, 5–27 (2016)

    MathSciNet  MATH  Google Scholar 

  27. Raina, R.K.: On generalized Wright’s hypergeometric functions and fractional calculus operators. East Asian Math. J. 21, 191–203 (2005)

    MATH  Google Scholar 

  28. Prudnikov, A.P., Brychkov, Y.A., Marichev, O.I.: Integral and series. In: Elementary Functions, vol. 1. Nauka, Moscow (1981)

    MATH  Google Scholar 

  29. Wang, J., Zhu, C., Zhou, Y.: New generalized Hermite-Hadamard-type inequalities and applications to special means. J. Inequal. Appl. 2013, 325 (2013)

    MathSciNet  Article  Google Scholar 

  30. Ross, B. (ed.): Fractional Calculus and Its Applications: Proceedings of the International Conference Held at the University of New Haven, June 1974, vol. 457. Springer, Berlin (2006)

    Google Scholar 

  31. Yaldiz, H., Sarikaya, M.Z.: On the midpoint type inequalities via generalized fractional integral operators. In: Xth International Statistics Days Conference, pp. 181–189 (2016)

    Google Scholar 

  32. Sarikaya, M.Z., Set, E., Yaldiz, H., Basak, N.: Hermite-Hadamard’s inequalities for fractional integrals and related fractional inequalities. Math. Comput. Model. 57, 2403–2407 (2013)

    Article  Google Scholar 

  33. Usta, F., Budak, H., Sarikaya, M.Z., Set, E.: On generalization of trapezoid type inequalities for s-convex functions with generalized fractional integral operators. Filomat 32, 2153–2171 (2018)

    MathSciNet  Article  Google Scholar 

Download references

Acknowledgements

The authors are grateful to the Mirpur University of Science & Technology for allowing study leave and utilization of educational resources to undertake this work.

Funding

There is no special funding from any source for this work.

Author information

Affiliations

Authors

Contributions

AL: conceptualization, methodology, investigation, writing the original draft, writing, reviewing and editing. RH: problem statement, investigation, supervision, provision of study resources, reviewing and editing. The authors read and approved the final manuscript.

Corresponding author

Correspondence to Asia Latif.

Ethics declarations

Competing interests

The publication of this work is approved by all authors, and the authors declare that they have no competing interests.

Rights and permissions

Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/.

Reprints and Permissions

About this article

Verify currency and authenticity via CrossMark

Cite this article

Latif, A., Hussain, R. New Hadamard-type inequalities for E-convex functions involving generalized fractional integrals. J Inequal Appl 2022, 35 (2022). https://doi.org/10.1186/s13660-022-02771-7

Download citation

  • Received:

  • Accepted:

  • Published:

  • DOI: https://doi.org/10.1186/s13660-022-02771-7

MSC

  • 26A51
  • 39B62
  • 26D15
  • 26D10

Keywords

  • E-Convex functions
  • Generalized fractional integrals
  • Hölder’s integral inequality
  • Power mean inequality