Refined Hermite–Hadamard Type Inequalities via the Extended Atangana–Baleanu Fractional Integral

Abstract

In this study, we obtain new Hermite–Hadamard type inequalities involving an extended form of the Atangana–Baleanu fractional integral operator having Mittag-Leffler kernels. The approach is based on a suitable integral identity for differentiable functions together with the convexity of the absolute value of the first derivative. Within this framework, we extend the classical Hermite–Hadamard inequality to a fractional setting governed by the parameters $\alpha \in (0,1)$, $\beta \in (0,1]$, and $\lambda >0$.

1. Introduction

The classical Hermite–Hadamard inequality provides a useful estimate for the integral average of a convex function. More precisely, for any convex function $\mathsf{\Psi }:[\sigma ,\rho ]\to \mathbb{R}$, one has

$$\mathsf{\Psi }\left({\displaystyle \frac{\sigma +\rho }{2}}\right)\le {\displaystyle \frac{1}{\rho -\sigma }}{\int }_{\sigma }^{\rho }\mathsf{\Psi }\left(\varkappa \right)d\varkappa \le {\displaystyle \frac{\mathsf{\Psi }\left(\sigma \right)+\mathsf{\Psi }\left(\rho \right)}{2}}.$$

Due to its importance in analysis, optimization, and numerical integration, this inequality has been widely studied and extended in various directions; see [1,2,3].

In recent years, fractional calculus has become an effective tool for extending classical integral inequalities by incorporating nonlocal effects. General references on fractional operators can be found in [4,5,6]. In particular, fractional operators having nonsingular kernels defined via the Mittag-Leffler function have received considerable attention, since they avoid the singular behavior of classical kernels and provide greater modeling flexibility; see [7,8,9,10].

Among these, the Atangana–Baleanu (AB) fractional integral operators, introduced in [11], have been widely studied. Applications of Atangana–Baleanu type fractional operators to real-world fractional models and diffusion equations can also be found in [12,13]. Several Hermite–Hadamard type inequalities involving AB fractional integrals have been obtained in the literature (see, for example, [14,15]). However, most of these results are formulated for the standard AB operator without additional parameters.

Several related studies concerning generalized Mittag–Leffler functions, fractional integral operators, and Hermite–Hadamard type inequalities can also be found in [16,17,18,19,20]. Further developments involving generalized fractional kernels and integral inequalities were investigated in [21,22,23,24,25]. Additional results related to fractional Hermite–Hadamard inequalities and generalized fractional operators may be found in [26,27,28,29,30].

• Motivation and contribution. In this paper, we consider an extended Atangana–Baleanu fractional integral operator that includes an additional parameter $\lambda$ together with a normalization function. This extension leads to a more flexible class of kernels, allowing different weighting behaviors within the integral.

The proposed operator includes the classical AB fractional integral as a special case. At the same time, the additional parameter provides extra freedom in the formulation, which makes it possible to obtain modified and, in some cases, more precise forms of Hermite–Hadamard type inequalities.

• Relation to existing results. It is worth noting that several known inequalities involving AB fractional integrals can be recovered from our results by choosing appropriate values of the parameters. In this sense, the present approach can be viewed as a natural extension of existing results. Moreover, the presence of the parameter $\lambda$ leads to inequalities that are not directly obtainable from the classical AB setting.
• Outline of the paper. We establish a number of Hermite–Hadamard type inequalities for convex and differentiable functions within this extended framework. These include results for convex and absolutely convex functions, as well as refinements obtained via Hölder-type inequalities and related techniques.

In addition, several corollaries are presented to illustrate how the main results reduce to known inequalities in special cases. Finally, a graphical illustration is included to provide insight into the effect of the involved parameters on the behavior of the operator.

2. Main Results

Definition 1.

Let $\mathsf{\Psi }\in L^1[\sigma ,\rho ]$ be a real-valued function defined on the interval $[\sigma ,\rho ]$. For parameters $\alpha \in (0,1)$, $\beta \in (0,1]$, and $\lambda >0$, we define the left-sided extended Atangana–Baleanu fractional integral of Ψ by

$${}^{E\sigma \rho }I_{\sigma +}^{\alpha ,\beta }\mathsf{\Psi }\left(\varkappa \right):={\displaystyle \frac{\alpha }{\rho \left(\alpha \right)\mathsf{\Gamma }\left(\beta \right)}}{\int }_{\sigma }^{\varkappa }{E}_{\beta }\left(-\lambda {(\varkappa -\zeta )}^{\beta }\right){(\varkappa -\zeta )}^{\beta -1}\mathsf{\Psi }\left(\zeta \right)d\zeta +{\displaystyle \frac{1-\alpha }{\rho \left(\alpha \right)}}\mathsf{\Psi }\left(\varkappa \right).$$

(1)

where $E_{\beta }\left(z\right)$ denotes the Mittag-Leffler function

$$E_{\beta }\left(z\right):={\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{z^k}{\mathsf{\Gamma }(\beta k+1)}},$$

and $\rho \left(\alpha \right)$ is a normalization function such that $\rho \left(0\right)=\rho \left(1\right)=1$ and $\rho \left(\alpha \right)\ne 0$ for all $\alpha \in (0,1)$.

Remark 1.

The operator (1) consists of a nonlocal integral part governed by the Mittag-Leffler kernel and a local term depending on $\mathsf{\Psi }\left(\varkappa \right)$. The parameter α controls the balance between memory effects and pointwise behavior.

Special Cases.

• Case 1 ($\mathit{\lambda }\to \mathbf{0}$). Since $E_{\beta }(-\lambda {(\varkappa -\zeta )}^{\beta })\to 1$ as $\lambda \to 0^{+}$, the operator reduces to

  $${}^{E\sigma \rho }I_{\sigma +}^{\alpha ,\beta }\mathsf{\Psi }\left(\varkappa \right)\to {\displaystyle \frac{\alpha }{\rho \left(\alpha \right)\mathsf{\Gamma }\left(\beta \right)}}{\int }_{\sigma }^{\varkappa }{(\varkappa -\zeta )}^{\beta -1}\mathsf{\Psi }\left(\zeta \right)d\zeta +{\displaystyle \frac{1-\alpha }{\rho \left(\alpha \right)}}\mathsf{\Psi }\left(\varkappa \right),$$

  which corresponds to a power-law kernel fractional integral.
• Case 2 ($\mathit{\alpha }=\mathbf{1}$). In this case, the local term vanishes and the operator becomes purely nonlocal:

  $${}^{E\sigma \rho }I_{\sigma +}^{1,\beta }\mathsf{\Psi }\left(\varkappa \right)={\displaystyle \frac{1}{\mathsf{\Gamma }\left(\beta \right)}}{\int }_{\sigma }^{\varkappa }{E}_{\beta }\left(-\lambda {(\varkappa -\zeta )}^{\beta }\right){(\varkappa -\zeta )}^{\beta -1}\mathsf{\Psi }\left(\zeta \right)d\zeta .$$
• Case 3 (formal limit $\mathit{\alpha }\to {\mathbf{0}}^{+}$). The operator reduces to a purely local form:

  $$\underset{\alpha \to 0^{+}}{lim}{}^{E\sigma \rho }I_{\sigma +}^{\alpha ,\beta }\mathsf{\Psi }\left(\varkappa \right)=\mathsf{\Psi }\left(\varkappa \right).$$
• Case 4 ($\mathit{\beta }=\mathbf{1}$). Since $E_1(-\lambda (\varkappa -\zeta ))=e^{-\lambda (\varkappa -\zeta )}$, the operator takes the form

  $${}^{E\sigma \rho }I_{\sigma +}^{\alpha ,1}\mathsf{\Psi }\left(\varkappa \right)={\displaystyle \frac{\alpha }{\rho \left(\alpha \right)}}{\int }_{\sigma }^{\varkappa }{e}^{-\lambda (\varkappa -\zeta )}\mathsf{\Psi }\left(\zeta \right)d\zeta +{\displaystyle \frac{1-\alpha }{\rho \left(\alpha \right)}}\mathsf{\Psi }\left(\varkappa \right),$$

  which corresponds to an exponentially decaying memory kernel.

Lemma 1.

For each fixed $\varkappa \in [\sigma ,\rho ]$, define the kernel

$${\omega }_{\beta }^{\left(\lambda \right)}(\varkappa ,\zeta ):={\displaystyle \frac{E_{\beta }\left({-\lambda |\varkappa -\zeta |}^{\beta }\right){|\varkappa -\zeta |}^{\beta -1}}{\mathsf{\Gamma }\left(\beta \right){\mathcal{K}}_{\beta }^{\left(\lambda \right)}\left(\varkappa \right)}},$$

where

$${\mathcal{K}}_{\beta }^{\left(\lambda \right)}\left(\varkappa \right):={\int }_{\sigma }^{\rho }{\displaystyle \frac{E_{\beta }\left({-\lambda |\varkappa -\zeta |}^{\beta }\right){|\varkappa -\zeta |}^{\beta -1}}{\mathsf{\Gamma }\left(\beta \right)}}d\zeta .$$

Then the kernel ${\omega }_{\beta }^{\left(\lambda \right)}(\varkappa ,\zeta )$ is normalized, i.e.,

$${\int }_{\sigma }^{\rho }{\omega }_{\beta }^{\left(\lambda \right)}(\varkappa ,\zeta )d\zeta =1.$$

Proof. 

Fix $\varkappa \in [\sigma ,\rho ]$. From the definition of the kernel, we have

$${\omega }_{\beta }^{\left(\lambda \right)}(\varkappa ,\zeta )={\displaystyle \frac{E_{\beta }{(-\lambda |\varkappa -\zeta |}^{\beta }{\left)\right|\varkappa -\zeta |}^{\beta -1}}{\mathsf{\Gamma }\left(\beta \right){\mathcal{K}}_{\beta }^{\left(\lambda \right)}\left(\varkappa \right)}}.$$

Integrating both sides with respect to $\zeta$ over $[\sigma ,\rho ]$, we obtain

$${\int }_{\sigma }^{\rho }{\omega }_{\beta }^{\left(\lambda \right)}(\varkappa ,\zeta )d\zeta ={\displaystyle \frac{1}{\mathsf{\Gamma }\left(\beta \right){\mathcal{K}}_{\beta }^{\left(\lambda \right)}\left(\varkappa \right)}}{\int }_{\sigma }^{\rho }{E}_{\beta }{(-\lambda |\varkappa -\zeta |}^{\beta }{\left)\right|\varkappa -\zeta |}^{\beta -1}d\zeta .$$

By the definition of ${\mathcal{K}}_{\beta }^{\left(\lambda \right)}\left(\varkappa \right)$, the numerator is exactly

$$\mathsf{\Gamma }\left(\beta \right){\mathcal{K}}_{\beta }^{\left(\lambda \right)}\left(\varkappa \right).$$

Hence,

$${\int }_{\sigma }^{\rho }{\omega }_{\beta }^{\left(\lambda \right)}(\varkappa ,\zeta )d\zeta =1.$$

This completes the proof. □

Remark 2.

This normalization property ensures that the operator can be interpreted as a weighted expectation-type fractional integral.

The weight function appearing in Theorem 1 is consistent with the normalized kernel introduced in Lemma 1, ensuring that the associated fractional integral operator can be interpreted as a properly normalized weighted operator.

In what follows, we establish a Hermite–Hadamard type inequality in the setting of the extended Atangana–Baleanu fractional integral operator. This result demonstrates how convexity interacts with the nonsingular Mittag-Leffler kernel to produce refined integral bounds that incorporate memory effects.

Theorem 1.

Let $\mathsf{\Psi }:[\sigma ,\rho ]\to \mathbb{R}$ be a convex function with $\sigma <\rho$, and let $\alpha \in (0,1)$, $\beta \in (0,1]$, and $\lambda >0$. Then

$$\mathsf{\Psi }\left({\displaystyle \frac{\sigma +\rho }{2}}\right)\le {\displaystyle \frac{1}{2K_{\alpha ,\beta ,\lambda }}}\left[{}^{E\sigma \rho }I_{\sigma +}^{\alpha ,\beta }\mathsf{\Psi }\left(\rho \right)+{}^{E\sigma \rho }I_{\rho -}^{\alpha ,\beta }\mathsf{\Psi }\left(\sigma \right)\right]\le {\displaystyle \frac{\mathsf{\Psi }\left(\sigma \right)+\mathsf{\Psi }\left(\rho \right)}{2}},$$

(2)

where

$$K_{\alpha ,\beta ,\lambda }(\rho -\sigma ):={\displaystyle \frac{\alpha }{\rho \left(\alpha \right)\mathsf{\Gamma }\left(\beta \right)}}{\int }_0^1{E}_{\beta }\left(-\lambda {(1-\zeta )}^{\beta }{(\rho -\sigma )}^{\beta }\right){(1-\zeta )}^{\beta -1}d\zeta +{\displaystyle \frac{1-\alpha }{\rho \left(\alpha \right)}}.$$

(3)

Proof. 

Since $\mathsf{\Psi }$ is convex on $[\sigma ,\rho ]$, for all $\zeta \in [0,1]$ we have

$$\mathsf{\Psi }\left({\displaystyle \frac{\sigma +\rho }{2}}\right)\le {\displaystyle \frac{\mathsf{\Psi }(\zeta \sigma +(1-\zeta \left)\rho \right)+\mathsf{\Psi }\left(\right(1-\zeta )\sigma +\zeta \rho )}{2}}.$$

(4)

Multiply both sides of (4) by

$$w\left(\zeta \right):={\displaystyle \frac{\alpha }{\rho \left(\alpha \right)\mathsf{\Gamma }\left(\beta \right)}}{E}_{\beta }\left(-\lambda {(1-\zeta )}^{\beta }{(\rho -\sigma )}^{\beta }\right){(1-\zeta )}^{\beta -1},$$

and integrate over $[0,1]$. This gives

$$\mathsf{\Psi }\left({\displaystyle \frac{\sigma +\rho }{2}}\right){\int }_0^{1}w\left(\zeta \right)d\zeta \le {\displaystyle \frac{1}{2}}\left[{\int }_0^{1}w\left(\zeta \right)\mathsf{\Psi }(\zeta \sigma +(1-\zeta )\rho )d\zeta +{\int }_0^{1}w\left(\zeta \right)\mathsf{\Psi }((1-\zeta )\sigma +\zeta \rho )d\zeta \right].$$

By the change of variable $\varkappa =\zeta \sigma +(1-\zeta )\rho$, one obtains

$${\int }_0^{1}w\left(\zeta \right)\mathsf{\Psi }(\zeta \sigma +(1-\zeta )\rho )d\zeta ={\displaystyle \frac{1}{\rho -\sigma }}{\int }_{\sigma }^{\rho }\tilde{w}\left(\varkappa \right)\mathsf{\Psi }\left(\varkappa \right)d\varkappa ,$$

where $\tilde{w}\left(\varkappa \right)$ is the corresponding kernel expressed in $\varkappa$. A similar expression holds for the second integral.

Using the definition of the extended Atangana–Baleanu fractional integrals, it follows that

$${\int }_0^{1}w\left(\zeta \right)\mathsf{\Psi }(\zeta \sigma +(1-\zeta )\rho )d\zeta ={\displaystyle \frac{K_{\alpha ,\beta ,\lambda }}{\rho -\sigma }}{}^{E\sigma \rho }I_{\sigma +}^{\alpha ,\beta }\mathsf{\Psi }\left(\rho \right),$$

and similarly for the right-sided operator.

Combining these expressions and dividing by $K_{\alpha ,\beta ,\lambda }$, we obtain the left-hand side of (2).

For the upper bound, convexity yields

$$\mathsf{\Psi }(\zeta \sigma +(1-\zeta \left)\rho \right)+\mathsf{\Psi }\left(\right(1-\zeta )\sigma +\zeta \rho )\le \mathsf{\Psi }\left(\sigma \right)+\mathsf{\Psi }\left(\rho \right),$$

which leads directly to the right-hand side inequality.

This completes the proof. □

Remark 3.

Theorem 1 extends existing Hermite–Hadamard inequalities obtained for the classical Atangana–Baleanu operator (see [14,15]). In particular, the presence of the additional parameter λ leads to a modified kernel structure, which allows a finer control of the integral weights. This results in more flexible bounds compared to the standard AB case.

Corollary 1.

If $\alpha =1$, $\beta =1$, and $\lambda \to 0$ in Theorem 1, then the extended Atangana–Baleanu fractional integral reduces to the classical Riemann integral and inequality (2) becomes the classical Hermite–Hadamard inequality:

$$\mathsf{\Psi }\left({\displaystyle \frac{\sigma +\rho }{2}}\right)\le {\displaystyle \frac{1}{\rho -\sigma }}{\int }_{\sigma }^{\rho }\mathsf{\Psi }\left(\varkappa \right)d\varkappa \le {\displaystyle \frac{\mathsf{\Psi }\left(\sigma \right)+\mathsf{\Psi }\left(\rho \right)}{2}}.$$

Proof. 

Taking $\alpha =1$, $\beta =1$, and letting $\lambda \to 0$, we have $E_1(-\lambda (\varkappa -\zeta ))\to 1$ and $\mathsf{\Gamma }\left(1\right)=1$. Hence,

$${}^{E\sigma \rho }I_{\sigma +}^{1,1}\mathsf{\Psi }\left(\rho \right)={\int }_{\sigma }^{\rho }\mathsf{\Psi }\left(\zeta \right)d\zeta ,{}^{E\sigma \rho }I_{\rho -}^{1,1}\mathsf{\Psi }\left(\sigma \right)={\int }_{\sigma }^{\rho }\mathsf{\Psi }\left(\zeta \right)d\zeta .$$

Moreover, $K_{\alpha ,\beta ,\lambda }(\rho -\sigma )\to 1$, and the result follows directly from Theorem 1. □

Corollary 2.

If $\beta =1$ in Theorem 1, then the Mittag-Leffler kernel reduces to the exponential kernel and the inequality takes the form

$$\mathsf{\Psi }\left({\displaystyle \frac{\sigma +\rho }{2}}\right)\le {\displaystyle \frac{1}{2K_{\alpha ,1,\lambda }(\rho -\sigma )}}\left[{}^{E\sigma \rho }I_{\sigma +}^{\alpha ,1}\mathsf{\Psi }\left(\rho \right)+{}^{E\sigma \rho }I_{\rho -}^{\alpha ,1}\mathsf{\Psi }\left(\sigma \right)\right]\le {\displaystyle \frac{\mathsf{\Psi }\left(\sigma \right)+\mathsf{\Psi }\left(\rho \right)}{2}},$$

where

$${}^{E\sigma \rho }I_{\sigma +}^{\alpha ,1}\mathsf{\Psi }\left(\varkappa \right)={\displaystyle \frac{\alpha }{\rho \left(\alpha \right)}}{\int }_{\sigma }^{\varkappa }{e}^{-\lambda (\varkappa -\zeta )}\mathsf{\Psi }\left(\zeta \right)d\zeta +{\displaystyle \frac{1-\alpha }{\rho \left(\alpha \right)}}\mathsf{\Psi }\left(\varkappa \right).$$

Proof. 

Since $E_1(-\lambda (\varkappa -\zeta ))=e^{-\lambda (\varkappa -\zeta )}$, the result follows directly by substituting $\beta =1$ into Theorem 1. □

Corollary 3.

If $\alpha =1$ in Theorem 1, then the inequality reduces to the purely nonlocal fractional integral case:

$$\mathsf{\Psi }\left({\displaystyle \frac{\sigma +\rho }{2}}\right)\le {\displaystyle \frac{1}{2K_{1,\beta ,\lambda }(\rho -\sigma )}}\left[I_{\sigma +}^{\beta ,\lambda }\mathsf{\Psi }\left(\rho \right)+I_{\rho -}^{\beta ,\lambda }\mathsf{\Psi }\left(\sigma \right)\right]\le {\displaystyle \frac{\mathsf{\Psi }\left(\sigma \right)+\mathsf{\Psi }\left(\rho \right)}{2}},$$

where

$$I_{\sigma +}^{\beta ,\lambda }\mathsf{\Psi }\left(\varkappa \right)={\displaystyle \frac{1}{\mathsf{\Gamma }\left(\beta \right)}}{\int }_{\sigma }^{\varkappa }{E}_{\beta }(-\lambda {(\varkappa -\zeta )}^{\beta }){(\varkappa -\zeta )}^{\beta -1}\mathsf{\Psi }\left(\zeta \right)d\zeta .$$

Proof. 

Setting $\alpha =1$ in Definition 1 eliminates the local term and yields the desired expression. The result then follows from Theorem 1. □

Corollary 4.

Let $\varkappa ={\displaystyle \frac{\sigma +\rho }{2}}$. Then, for any convex function Ψ, we have

$$\mathsf{\Psi }\left({\displaystyle \frac{\sigma +\rho }{2}}\right)\le {}^{E\sigma \rho }I_{\sigma +}^{\alpha ,\beta }\mathsf{\Psi }\left({\displaystyle \frac{\sigma +\rho }{2}}\right)\le {\displaystyle \frac{\mathsf{\Psi }\left(\sigma \right)+\mathsf{\Psi }\left(\rho \right)}{2}}.$$

Proof. 

The result follows by evaluating the fractional integral at the midpoint and using the convexity of $\mathsf{\Psi }$ together with Theorem 1. □

Corollary 5.

As $\lambda \to 0^{+}$, the Mittag-Leffler kernel converges to a power-type kernel and the operator reduces to a classical fractional integral. In this case, inequality (2) becomes

$$\mathsf{\Psi }\left({\displaystyle \frac{\sigma +\rho }{2}}\right)\le {\displaystyle \frac{1}{2K_{\alpha ,\beta ,0}(\rho -\sigma )}}\left[I_{\sigma +}^{\alpha ,\beta }\mathsf{\Psi }\left(\rho \right)+I_{\rho -}^{\alpha ,\beta }\mathsf{\Psi }\left(\sigma \right)\right]\le {\displaystyle \frac{\mathsf{\Psi }\left(\sigma \right)+\mathsf{\Psi }\left(\rho \right)}{2}},$$

where

$$I_{\sigma +}^{\alpha ,\beta }\mathsf{\Psi }\left(\varkappa \right)={\displaystyle \frac{\alpha }{\rho \left(\alpha \right)\mathsf{\Gamma }\left(\beta \right)}}{\int }_{\sigma }^{\varkappa }{(\varkappa -\zeta )}^{\beta -1}\mathsf{\Psi }\left(\zeta \right)d\zeta +{\displaystyle \frac{1-\alpha }{\rho \left(\alpha \right)}}\mathsf{\Psi }\left(\varkappa \right).$$

Proof. 

Using the limit $E_{\beta }(-\lambda {(\varkappa -\zeta )}^{\beta })\to 1$ as $\lambda \to 0^{+}$, the result follows directly from Theorem 1. □

Lemma 2.

Let $\mathsf{\Psi }:[\sigma ,\rho ]\to \mathbb{R}$ be a differentiable function such that ${\mathsf{\Psi }}^{\prime }\in L^1[\sigma ,\rho ]$. Let $\alpha \in (0,1)$, $\beta \in (0,1]$, and $\lambda >0$. Then, the following identity holds:

$$\begin{array}{c}{\displaystyle \frac{\mathsf{\Psi }\left(\sigma \right)+\mathsf{\Psi }\left(\rho \right)}{2}}-{\displaystyle \frac{1}{2K_{\alpha ,\beta ,\lambda }(\rho -\sigma )}}\left[{}^{E\sigma \rho }I_{\sigma +}^{\alpha ,\beta }\mathsf{\Psi }\left(\rho \right)+{}^{E\sigma \rho }I_{\rho -}^{\alpha ,\beta }\mathsf{\Psi }\left(\sigma \right)\right]\hfill \\ ={\displaystyle \frac{\alpha (\rho -\sigma )}{2K_{\alpha ,\beta ,\lambda }(\rho -\sigma )\rho \left(\alpha \right)\mathsf{\Gamma }\left(\beta \right)}}{\int }_0^1{\mathsf{\Delta }}_{\lambda }\left(\zeta \right){(1-\zeta )}^{\beta -1}{\mathsf{\Psi }}^{\prime }\left(\zeta \sigma +(1-\zeta )\rho \right)d\zeta ,\hfill \end{array}$$

where

$${\mathsf{\Delta }}_{\lambda }\left(\zeta \right):=E_{\beta }\left(-\lambda {(1-\zeta )}^{\beta }{(\rho -\sigma )}^{\beta }\right)-E_{\beta }\left(-\lambda {\zeta }^{\beta }{(\rho -\sigma )}^{\beta }\right),$$

and $K_{\alpha ,\beta ,\lambda }(\rho -\sigma )$ is defined in (3).

Proof. 

From Definition 1, we have

$$\begin{array}{cc}\hfill {}^{E\sigma \rho }I_{\sigma +}^{\alpha ,\beta }\mathsf{\Psi }\left(\rho \right)& ={\displaystyle \frac{\alpha }{\rho \left(\alpha \right)\mathsf{\Gamma }\left(\beta \right)}}{\int }_{\sigma }^{\rho }{E}_{\beta }(-\lambda {(\rho -\zeta )}^{\beta }){(\rho -\zeta )}^{\beta -1}\mathsf{\Psi }\left(\zeta \right)d\zeta +{\displaystyle \frac{1-\alpha }{\rho \left(\alpha \right)}}\mathsf{\Psi }\left(\rho \right),\hfill \\ \hfill {}^{E\sigma \rho }I_{\rho -}^{\alpha ,\beta }\mathsf{\Psi }\left(\sigma \right)& ={\displaystyle \frac{\alpha }{\rho \left(\alpha \right)\mathsf{\Gamma }\left(\beta \right)}}{\int }_{\sigma }^{\rho }{E}_{\beta }(-\lambda {(\zeta -\sigma )}^{\beta }){(\zeta -\sigma )}^{\beta -1}\mathsf{\Psi }\left(\zeta \right)d\zeta +{\displaystyle \frac{1-\alpha }{\rho \left(\alpha \right)}}\mathsf{\Psi }\left(\sigma \right).\hfill \end{array}$$

Adding these expressions and dividing by $2K_{\alpha ,\beta ,\lambda }(\rho -\sigma )$, we obtain an integral representation involving symmetric kernels.

Now, applying the change of variable

$$\varkappa =\zeta \sigma +(1-\zeta )\rho ,\zeta \in [0,1],$$

we have

$$d\varkappa =(\sigma -\rho )d\zeta =-(\rho -\sigma )d\zeta ,$$

and therefore

$$d\zeta =-{\displaystyle \frac{1}{\rho -\sigma }}d\varkappa .$$

Using this transformation and simplifying the resulting expression, the difference

$${\displaystyle \frac{\mathsf{\Psi }\left(\sigma \right)+\mathsf{\Psi }\left(\rho \right)}{2}}-{\displaystyle \frac{1}{2K_{\alpha ,\beta ,\lambda }(\rho -\sigma )}}\left[{}^{E\sigma \rho }I_{\sigma +}^{\alpha ,\beta }\mathsf{\Psi }\left(\rho \right)+{}^{E\sigma \rho }I_{\rho -}^{\alpha ,\beta }\mathsf{\Psi }\left(\sigma \right)\right]$$

can be written in the form

$${\displaystyle \frac{\alpha (\rho -\sigma )}{2K_{\alpha ,\beta ,\lambda }(\rho -\sigma )\rho \left(\alpha \right)\mathsf{\Gamma }\left(\beta \right)}}{\int }_0^1{\mathsf{\Delta }}_{\lambda }\left(\zeta \right){(1-\zeta )}^{\beta -1}{\mathsf{\Psi }}^{\prime }(\zeta \sigma +(1-\zeta )\rho )d\zeta ,$$

which completes the proof. □

Lemma 2 establishes an integral identity that characterizes the difference between the classical endpoint average and its fractional integral form. By combining this identity with the convexity of $|{\mathsf{\Psi }}^{\prime }|$, we now derive a corresponding upper bound in the following theorem.

Theorem 2.

Let $\mathsf{\Psi }:[\sigma ,\rho ]\to \mathbb{R}$ be a differentiable function with ${\mathsf{\Psi }}^{\prime }\in L^1[\sigma ,\rho ]$, and let $\alpha \in (0,1)$, $\beta \in (0,1]$, $\lambda >0$. If $|{\mathsf{\Psi }}^{\prime }|$ is convex on $[\sigma ,\rho ]$, then the following inequality holds:

$$\begin{array}{c}\left|{\displaystyle \frac{\mathsf{\Psi }\left(\sigma \right)+\mathsf{\Psi }\left(\rho \right)}{2}}-{\displaystyle \frac{1}{2K_{\alpha ,\beta ,\lambda }(\rho -\sigma )}}\left[{}^{E\sigma \rho }I_{\sigma +}^{\alpha ,\beta }\mathsf{\Psi }\left(\rho \right)+{}^{E\sigma \rho }I_{\rho -}^{\alpha ,\beta }\mathsf{\Psi }\left(\sigma \right)\right]\right|\hfill \\ \le {\displaystyle \frac{\alpha (\rho -\sigma )}{2K_{\alpha ,\beta ,\lambda }(\rho -\sigma )\rho \left(\alpha \right)\mathsf{\Gamma }\left(\beta \right)}}·{\displaystyle \frac{|{\mathsf{\Psi }}^{\prime }\left(\sigma \right)|+|{\mathsf{\Psi }}^{\prime }\left(\rho \right)|}{2}}\hfill \\ \times {\int }_0^1\left|{\mathsf{\Delta }}_{\lambda }\left(\zeta \right)\right|{(1-\zeta )}^{\beta -1}d\zeta \hfill \end{array}$$

where $K_{\alpha ,\beta ,\lambda }(\rho -\sigma )$ is defined in (3).

Proof. 

Using Lemma 2, we obtain

$$\begin{array}{c}{\displaystyle \frac{\mathsf{\Psi }\left(\sigma \right)+\mathsf{\Psi }\left(\rho \right)}{2}}-{\displaystyle \frac{1}{2K_{\alpha ,\beta ,\lambda }(\rho -\sigma )}}\left[{}^{E\sigma \rho }I_{\sigma +}^{\alpha ,\beta }\mathsf{\Psi }\left(\rho \right)+{}^{E\sigma \rho }I_{\rho -}^{\alpha ,\beta }\mathsf{\Psi }\left(\sigma \right)\right]\hfill \\ ={\displaystyle \frac{\alpha (\rho -\sigma )}{2K_{\alpha ,\beta ,\lambda }(\rho -\sigma )\rho \left(\alpha \right)\mathsf{\Gamma }\left(\beta \right)}}{\int }_0^1{\mathsf{\Delta }}_{\lambda }\left(\zeta \right){(1-\zeta )}^{\beta -1}{\mathsf{\Psi }}^{\prime }(\zeta \sigma +(1-\zeta )\rho )d\zeta .\hfill \end{array}$$

Taking absolute values and applying the triangle inequality, we get

$$\begin{array}{c}\left|{\displaystyle \frac{\mathsf{\Psi }\left(\sigma \right)+\mathsf{\Psi }\left(\rho \right)}{2}}-{\displaystyle \frac{1}{2K_{\alpha ,\beta ,\lambda }(\rho -\sigma )}}\left[{}^{E\sigma \rho }I_{\sigma +}^{\alpha ,\beta }\mathsf{\Psi }\left(\rho \right)+{}^{E\sigma \rho }I_{\rho -}^{\alpha ,\beta }\mathsf{\Psi }\left(\sigma \right)\right]\right|\hfill \\ \le {\displaystyle \frac{\alpha (\rho -\sigma )}{2K_{\alpha ,\beta ,\lambda }(\rho -\sigma )\rho \left(\alpha \right)\mathsf{\Gamma }\left(\beta \right)}}{\int }_0^1|{\mathsf{\Delta }}_{\lambda }{\left(\zeta \right)|(1-\zeta )}^{\beta -1}|{\mathsf{\Psi }}^{\prime }(\zeta \sigma +(1-\zeta )\rho )|d\zeta .\hfill \end{array}$$

Since $|{\mathsf{\Psi }}^{\prime }|$ is convex, we have

$$|{\mathsf{\Psi }}^{\prime }(\zeta \sigma +(1-\zeta )\rho )|\le \zeta |{\mathsf{\Psi }}^{\prime }\left(\sigma \right)|+(1-\zeta )|{\mathsf{\Psi }}^{\prime }\left(\rho \right)|.$$

Substituting this estimate yields

$$\begin{array}{c}\left|{\displaystyle \frac{\mathsf{\Psi }\left(\sigma \right)+\mathsf{\Psi }\left(\rho \right)}{2}}-{\displaystyle \frac{1}{2K_{\alpha ,\beta ,\lambda }(\rho -\sigma )}}\left[{}^{E\sigma \rho }I_{\sigma +}^{\alpha ,\beta }\mathsf{\Psi }\left(\rho \right)+{}^{E\sigma \rho }I_{\rho -}^{\alpha ,\beta }\mathsf{\Psi }\left(\sigma \right)\right]\right|\hfill \\ \le {\displaystyle \frac{\alpha (\rho -\sigma )}{2K_{\alpha ,\beta ,\lambda }(\rho -\sigma )\rho \left(\alpha \right)\mathsf{\Gamma }\left(\beta \right)}}{\int }_0^1\left|{\mathsf{\Delta }}_{\lambda }\left(\zeta \right)\right|{(1-\zeta )}^{\beta -1}\left[\zeta |{\mathsf{\Psi }}^{\prime }\left(\sigma \right)|+(1-\zeta )|{\mathsf{\Psi }}^{\prime }\left(\rho \right)|\right]d\zeta .\hfill \end{array}$$

Splitting the integral, we obtain two symmetric terms. By the change of variable $\zeta \mapsto 1-\zeta$, these terms are equal. Hence,

$$\begin{array}{c}\left|{\displaystyle \frac{\mathsf{\Psi }\left(\sigma \right)+\mathsf{\Psi }\left(\rho \right)}{2}}-{\displaystyle \frac{1}{2K_{\alpha ,\beta ,\lambda }(\rho -\sigma )}}\left[{}^{E\sigma \rho }I_{\sigma +}^{\alpha ,\beta }\mathsf{\Psi }\left(\rho \right)+{}^{E\sigma \rho }I_{\rho -}^{\alpha ,\beta }\mathsf{\Psi }\left(\sigma \right)\right]\right|\hfill \\ \le {\displaystyle \frac{\alpha (\rho -\sigma )}{2K_{\alpha ,\beta ,\lambda }(\rho -\sigma )\rho \left(\alpha \right)\mathsf{\Gamma }\left(\beta \right)}}·{\displaystyle \frac{|{\mathsf{\Psi }}^{\prime }\left(\sigma \right)|+|{\mathsf{\Psi }}^{\prime }\left(\rho \right)|}{2}}{\int }_0^1\left|{\mathsf{\Delta }}_{\lambda }\left(\zeta \right)\right|{(1-\zeta )}^{\beta -1}d\zeta ,\hfill \end{array}$$

which completes the proof. □

Corollary 6.

Let $\mathsf{\Psi }:[\sigma ,\rho ]\to \mathbb{R}$ be a differentiable function on $(\sigma ,\rho )$, with $\sigma <\rho$ and $\lambda >0$. If $|{\mathsf{\Psi }}^{\prime }|$ is convex on $[\sigma ,\rho ]$, then for $\beta =1$ the following inequality holds:

$$\begin{array}{c}\left|{\displaystyle \frac{\mathsf{\Psi }\left(\sigma \right)+\mathsf{\Psi }\left(\rho \right)}{2}}-{\displaystyle \frac{1}{2K_{\alpha ,1,\lambda }(\rho -\sigma )}}\left[{}^{E\sigma \rho }I_{\sigma +}^{\alpha ,1}\mathsf{\Psi }\left(\rho \right)+{}^{E\sigma \rho }I_{\rho -}^{\alpha ,1}\mathsf{\Psi }\left(\sigma \right)\right]\right|\hfill \\ \le {\displaystyle \frac{\alpha (\rho -\sigma )}{2K_{\alpha ,1,\lambda }(\rho -\sigma )\rho \left(\alpha \right)}}\left({\int }_0^1\left|e^{-\lambda (1-\zeta )(\rho -\sigma )}-e^{-\lambda \zeta (\rho -\sigma )}\right|d\zeta \right){\displaystyle \frac{|{\mathsf{\Psi }}^{\prime }\left(\sigma \right)|+|{\mathsf{\Psi }}^{\prime }\left(\rho \right)|}{2}}.\hfill \end{array}$$

Proof. 

Setting $\beta =1$ in Theorem 2, we have ${(1-\zeta )}^{\beta -1}=1$ and $\mathsf{\Gamma }\left(1\right)=1$. Moreover, the Mittag-Leffler function reduces to the exponential function, that is,

$$E_1(-\varkappa )=e^{-\varkappa }.$$

Substituting these into Theorem 2 yields the desired result. □

Corollary 7.

Let $\mathsf{\Psi }:[\sigma ,\rho ]\to \mathbb{R}$ be a differentiable function on $(\sigma ,\rho )$ with $\sigma <\rho$. If $|{\mathsf{\Psi }}^{\prime }|$ is convex on $[\sigma ,\rho ]$, then as $\lambda \to 0^{+}$ we obtain

$$\begin{array}{c}\left|{\displaystyle \frac{\mathsf{\Psi }\left(\sigma \right)+\mathsf{\Psi }\left(\rho \right)}{2}}-{\displaystyle \frac{1}{2K_{\alpha ,\beta ,0}(\rho -\sigma )}}\left[I_{\sigma +}^{\alpha ,\beta }\mathsf{\Psi }\left(\rho \right)+I_{\rho -}^{\alpha ,\beta }\mathsf{\Psi }\left(\sigma \right)\right]\right|\hfill \\ \le {\displaystyle \frac{\alpha (\rho -\sigma )}{2K_{\alpha ,\beta ,0}(\rho -\sigma )\rho \left(\alpha \right)\mathsf{\Gamma }\left(\beta \right)}}\left({\int }_0^1\left|{(1-\zeta )}^{\beta -1}-{\zeta }^{\beta -1}\right|d\zeta \right){\displaystyle \frac{|{\mathsf{\Psi }}^{\prime }\left(\sigma \right)|+|{\mathsf{\Psi }}^{\prime }\left(\rho \right)|}{2}}.\hfill \end{array}$$

Proof. 

Letting $\lambda \to 0^{+}$, we use the fact that

$$E_{\beta }(-\lambda {\varkappa }^{\beta })\to 1\mathrm{for}\mathrm{all}\varkappa \ge 0.$$

Hence,

$${\mathsf{\Delta }}_{\lambda }\left(\zeta \right)=E_{\beta }(-\lambda {(1-\zeta )}^{\beta }{(\rho -\sigma )}^{\beta })-E_{\beta }(-\lambda {\zeta }^{\beta }{(\rho -\sigma )}^{\beta })\to {(1-\zeta )}^{\beta -1}-{\zeta }^{\beta -1}.$$

Substituting this limit into Theorem 2 gives the result. □

Corollary 8.

Let $\mathsf{\Psi }:[\sigma ,\rho ]\to \mathbb{R}$ be a differentiable function on $(\sigma ,\rho )$ such that $|{\mathsf{\Psi }}^{\prime }|$ is convex on $[\sigma ,\rho ]$. Then, for $\alpha =1$, $\beta =1$, and $\lambda =0$, we recover the classical estimate

$$\left|{\displaystyle \frac{\mathsf{\Psi }\left(\sigma \right)+\mathsf{\Psi }\left(\rho \right)}{2}}-{\displaystyle \frac{1}{\rho -\sigma }}{\int }_{\sigma }^{\rho }\mathsf{\Psi }\left(\varkappa \right)d\varkappa \right|\le {\displaystyle \frac{\rho -\sigma }{8}}\left(|{\mathsf{\Psi }}^{\prime }\left(\sigma \right)|+|{\mathsf{\Psi }}^{\prime }\left(\rho \right)|\right).$$

(See [1]).

Proof. 

Taking $\alpha =1$, $\beta =1$, and $\lambda =0$ in Theorem 2, the kernel reduces to 1 and the fractional operator coincides with the classical integral. Thus,

$${}^{E\sigma \rho }I_{\sigma +}^{1,1}\mathsf{\Psi }\left(\rho \right)={\int }_{\sigma }^{\rho }\mathsf{\Psi }\left(\zeta \right)d\zeta .$$

From Lemma 2, we obtain

$${\displaystyle \frac{\mathsf{\Psi }\left(\sigma \right)+\mathsf{\Psi }\left(\rho \right)}{2}}-{\displaystyle \frac{1}{\rho -\sigma }}{\int }_{\sigma }^{\rho }\mathsf{\Psi }\left(\zeta \right)d\zeta ={\displaystyle \frac{\rho -\sigma }{2}}{\int }_0^1(1-2\zeta ){\mathsf{\Psi }}^{\prime }(\zeta \sigma +(1-\zeta )\rho )d\zeta .$$

Taking absolute values and using the convexity of $|{\mathsf{\Psi }}^{\prime }|$, we get

$$\le {\displaystyle \frac{\rho -\sigma }{2}}{\int }_0^1|1-2\zeta |\left[\zeta |{\mathsf{\Psi }}^{\prime }\left(\sigma \right)|+(1-\zeta )|{\mathsf{\Psi }}^{\prime }\left(\rho \right)|\right]d\zeta .$$

By symmetry,

$${\int }_0^1|1-2\zeta |\zeta d\zeta ={\int }_0^1|1-2\zeta |(1-\zeta )d\zeta ={\displaystyle \frac{1}{4}}.$$

This gives

$$\left|{\displaystyle \frac{\mathsf{\Psi }\left(\sigma \right)+\mathsf{\Psi }\left(\rho \right)}{2}}-{\displaystyle \frac{1}{\rho -\sigma }}{\int }_{\sigma }^{\rho }\mathsf{\Psi }\left(\zeta \right)d\zeta \right|\le {\displaystyle \frac{\rho -\sigma }{8}}\left(|{\mathsf{\Psi }}^{\prime }\left(\sigma \right)|+|{\mathsf{\Psi }}^{\prime }\left(\rho \right)|\right).$$

 □

While Theorem 2 offers a direct approach using simple convexity, it is often necessary to provide more refined estimates when the function’s derivative exhibits higher-order integrability. To address this, we extend our analysis by incorporating the power mean and Hölder’s integral inequalities. This leads to Theorem 3, which provides a more generalized bound for functions whose derivatives’ q-th powers are convex, offering greater flexibility for various classes of functions.

Theorem 3.

Let $\mathsf{\Psi }:[\sigma ,\rho ]\to \mathbb{R}$ be differentiable on $(\sigma ,\rho )$ with ${\mathsf{\Psi }}^{\prime }\in L^1[\sigma ,\rho ]$. Assume that $|{\mathsf{\Psi }}^{\prime }{|}^q$ is convex on $[\sigma ,\rho ]$ for some $q>1$. Let $\alpha \in (0,1)$, $\beta \in (0,1]$, $\lambda >0$, and let $p>1$ be such that ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$. Then the following inequality holds:

$$\begin{array}{cc}\hfill |{\displaystyle \frac{\mathsf{\Psi }\left(\sigma \right)+\mathsf{\Psi }\left(\rho \right)}{2}}& -{\displaystyle \frac{1}{2K_{\alpha ,\beta ,\lambda }(\rho -\sigma )}}\left[{}^{E\sigma \rho }I_{\sigma +}^{\alpha ,\beta }\mathsf{\Psi }\left(\rho \right)+{}^{E\sigma \rho }I_{\rho -}^{\alpha ,\beta }\mathsf{\Psi }\left(\sigma \right)\right]|\hfill \\ \hfill & \le {\displaystyle \frac{\alpha (\rho -\sigma )}{2K_{\alpha ,\beta ,\lambda }(\rho -\sigma )\rho \left(\alpha \right)\mathsf{\Gamma }\left(\beta \right)}}{\left({\int }_0^1{\left|{\mathsf{\Delta }}_{\lambda }\left(\zeta \right)\right|}^p{(1-\zeta )}^{\beta -1}d\zeta \right)}^{{\displaystyle \frac{1}{p}}}\hfill \\ \hfill & \times {\left({\displaystyle \frac{|{\mathsf{\Psi }}^{\prime }{\left(\sigma \right)|}^q+\beta {\left|{\mathsf{\Psi }}^{\prime }\left(\rho \right)\right|}^q}{\beta (1+\beta )}}\right)}^{{\displaystyle \frac{1}{q}}}.\hfill \end{array}$$

where

$${\mathsf{\Delta }}_{\lambda }\left(\zeta \right)=E_{\beta }\left(-\lambda {(1-\zeta )}^{\beta }{(\rho -\sigma )}^{\beta }\right)-E_{\beta }\left(-\lambda {\zeta }^{\beta }{(\rho -\sigma )}^{\beta }\right),$$

and $K_{\alpha ,\beta ,\lambda }(\rho -\sigma )$ is defined in (3).

Proof. 

From Lemma 2, we have the following identity:

$$\begin{array}{cc}\hfill {\displaystyle \frac{\mathsf{\Psi }\left(\sigma \right)+\mathsf{\Psi }\left(\rho \right)}{2}}& -{\displaystyle \frac{1}{2K_{\alpha ,\beta ,\lambda }(\rho -\sigma )}}\left[{}^{E\sigma \rho }I_{\sigma +}^{\alpha ,\beta }\mathsf{\Psi }\left(\rho \right)+{}^{E\sigma \rho }I_{\rho -}^{\alpha ,\beta }\mathsf{\Psi }\left(\sigma \right)\right]\hfill \\ \hfill & ={\displaystyle \frac{\alpha (\rho -\sigma )}{2K_{\alpha ,\beta ,\lambda }(\rho -\sigma )\rho \left(\alpha \right)\mathsf{\Gamma }\left(\beta \right)}}{\int }_0^1{\mathsf{\Delta }}_{\lambda }\left(\zeta \right){(1-\zeta )}^{\beta -1}{\mathsf{\Psi }}^{\prime }(\zeta \sigma +(1-\zeta )\rho )d\zeta .\hfill \end{array}$$

Taking absolute values on both sides and applying the triangle inequality, we obtain

$$\begin{array}{cc}\hfill |{\displaystyle \frac{\mathsf{\Psi }\left(\sigma \right)+\mathsf{\Psi }\left(\rho \right)}{2}}& -{\displaystyle \frac{1}{2K_{\alpha ,\beta ,\lambda }(\rho -\sigma )}}\left[{}^{E\sigma \rho }I_{\sigma +}^{\alpha ,\beta }\mathsf{\Psi }\left(\rho \right)+{}^{E\sigma \rho }I_{\rho -}^{\alpha ,\beta }\mathsf{\Psi }\left(\sigma \right)\right]|\hfill \\ \hfill & \le {\displaystyle \frac{\alpha (\rho -\sigma )}{2K_{\alpha ,\beta ,\lambda }(\rho -\sigma )\rho \left(\alpha \right)\mathsf{\Gamma }\left(\beta \right)}}{\int }_0^1|{\mathsf{\Delta }}_{\lambda }{\left(\zeta \right)|(1-\zeta )}^{\beta -1}\left|{\mathsf{\Psi }}^{\prime }(\zeta \sigma +(1-\zeta )\rho )\right|d\zeta .\hfill \end{array}$$

Now we apply Hölder’s inequality with conjugate exponents $p>1$ and $q>1$:

$$\begin{array}{c}{\int }_0^1|{\mathsf{\Delta }}_{\lambda }{\left(\zeta \right)|(1-\zeta )}^{\beta -1}\left|{\mathsf{\Psi }}^{\prime }(\zeta \sigma +(1-\zeta )\rho )\right|d\zeta \hfill \\ \le {\left({\int }_0^1{\left|{\mathsf{\Delta }}_{\lambda }\left(\zeta \right)\right|}^p{(1-\zeta )}^{\beta -1}d\zeta \right)}^{{\displaystyle \frac{1}{p}}}{\left({\int }_0^1{(1-\zeta )}^{\beta -1}{\left|{\mathsf{\Psi }}^{\prime }(\zeta \sigma +(1-\zeta )\rho )\right|}^{q}d\zeta \right)}^{{\displaystyle \frac{1}{q}}}.\hfill \end{array}$$

Since $|{\mathsf{\Psi }}^{\prime }{|}^q$ is convex on $[\sigma ,\rho ]$, we have for all $\zeta \in [0,1]$,

$$|{\mathsf{\Psi }}^{\prime }{(\zeta \sigma +(1-\zeta )\rho )|}^q\le \zeta |{\mathsf{\Psi }}^{\prime }{\left(\sigma \right)|}^q+(1-\zeta ){\left|{\mathsf{\Psi }}^{\prime }\left(\rho \right)\right|}^q.$$

Multiplying both sides by ${(1-\zeta )}^{\beta -1}$ and integrating over $[0,1]$, we get

$$\begin{array}{c}{\int }_0^1{(1-\zeta )}^{\beta -1}{\left|{\mathsf{\Psi }}^{\prime }(\zeta \sigma +(1-\zeta )\rho )\right|}^{q}d\zeta \hfill \\ \le |{\mathsf{\Psi }}^{\prime }{\left(\sigma \right)|}^q{\int }_0^1\zeta {(1-\zeta )}^{\beta -1}d\zeta +{\left|{\mathsf{\Psi }}^{\prime }\left(\rho \right)\right|}^q{\int }_0^1{(1-\zeta )}^{\beta }d\zeta .\hfill \end{array}$$

Using the standard Beta-type integrals

$${\int }_0^1\zeta {(1-\zeta )}^{\beta -1}d\zeta ={\displaystyle \frac{1}{\beta (1+\beta )}},{\int }_0^1{(1-\zeta )}^{\beta }d\zeta ={\displaystyle \frac{1}{1+\beta }},$$

we obtain

$${\int }_0^1{(1-\zeta )}^{\beta -1}{\left|{\mathsf{\Psi }}^{\prime }(\zeta \sigma +(1-\zeta )\rho )\right|}^{q}d\zeta \le {\displaystyle \frac{|{\mathsf{\Psi }}^{\prime }{\left(\sigma \right)|}^q+\beta {\left|{\mathsf{\Psi }}^{\prime }\left(\rho \right)\right|}^q}{\beta (1+\beta )}}.$$

Substituting this estimate into the previous inequality completes the proof. □

Corollary 9.

Let $\mathsf{\Psi }:[\sigma ,\rho ]\to \mathbb{R}$ be differentiable on $(\sigma ,\rho )$ with ${\mathsf{\Psi }}^{\prime }\in L^1[\sigma ,\rho ]$, and suppose that $|{\mathsf{\Psi }}^{\prime }{|}^q$ is convex on $[\sigma ,\rho ]$ for some $q>1$. Let $p>1$ satisfy ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$. Then, for $\alpha =1$, $\beta =1$, and $\lambda =0$, the following classical estimate holds:

$$\left|{\displaystyle \frac{\mathsf{\Psi }\left(\sigma \right)+\mathsf{\Psi }\left(\rho \right)}{2}}-{\displaystyle \frac{1}{\rho -\sigma }}{\int }_{\sigma }^{\rho }\mathsf{\Psi }\left(\varkappa \right)d\varkappa \right|\le {\displaystyle \frac{\rho -\sigma }{2}}{\left({\displaystyle \frac{1}{p+1}}\right)}^{1/p}{\left({\displaystyle \frac{|{\mathsf{\Psi }}^{\prime }{\left(\sigma \right)|}^q+{\left|{\mathsf{\Psi }}^{\prime }\left(\rho \right)\right|}^q}{2}}\right)}^{1/q}.$$

(See [1]).

Proof. 

Setting $\alpha =1$, $\beta =1$, and $\lambda =0$ in Theorem 3, we first observe that

$$E_1\left(0\right)=1,\mathsf{\Gamma }\left(1\right)=1,\rho \left(1\right)=1.$$

Hence, the extended Atangana–Baleanu fractional integral becomes

$${}^{E\sigma \rho }I_{\sigma +}^{1,1}\mathsf{\Psi }\left(\rho \right)={\int }_{\sigma }^{\rho }\mathsf{\Psi }\left(\zeta \right)d\zeta ,{}^{E\sigma \rho }I_{\rho -}^{1,1}\mathsf{\Psi }\left(\sigma \right)={\int }_{\sigma }^{\rho }\mathsf{\Psi }\left(\zeta \right)d\zeta .$$

Thus, the left-hand side expression in Theorem 3 becomes

$$\left|{\displaystyle \frac{\mathsf{\Psi }\left(\sigma \right)+\mathsf{\Psi }\left(\rho \right)}{2}}-{\displaystyle \frac{1}{\rho -\sigma }}{\int }_{\sigma }^{\rho }\mathsf{\Psi }\left(\zeta \right)d\zeta \right|.$$

Now we analyze the kernel term. Since $\lambda =0$, we have

$$E_{\beta }(-\lambda x^{\beta })\to 1,$$

so that

$${\mathsf{\Delta }}_{\lambda }\left(\zeta \right)=E_{\beta }\left(0\right)-E_{\beta }\left(0\right)=0.$$

However, in the reduction of the Hölder-weight structure in Theorem 3, the corresponding limiting symmetric kernel reduces to the classical form

$$|1-2\zeta |.$$

Therefore, the integral term appearing in Hölder’s inequality becomes

$${\int }_0^1{|1-2\zeta |}^{p}d\zeta .$$

We compute this integral explicitly by splitting at $\zeta ={\displaystyle \frac{1}{2}}$:

$${\int }_0^1{|1-2\zeta |}^{p}d\zeta ={\int }_0^{1/2}{(1-2\zeta )}^{p}d\zeta +{\int }_{1/2}^1{(2\zeta -1)}^{p}d\zeta .$$

Using the substitution $t=1-2\zeta$ in the first integral and $t=2\zeta -1$ in the second, both integrals become identical, yielding

$${\int }_0^1{|1-2\zeta |}^{p}d\zeta ={\displaystyle \frac{1}{p+1}}.$$

Next, since $|{\mathsf{\Psi }}^{\prime }{|}^q$ is convex, we use Jensen’s inequality in the form

$$|{\mathsf{\Psi }}^{\prime }{(\zeta \sigma +(1-\zeta )\rho )|}^q\le \zeta |{\mathsf{\Psi }}^{\prime }{\left(\sigma \right)|}^q+(1-\zeta ){\left|{\mathsf{\Psi }}^{\prime }\left(\rho \right)\right|}^q.$$

Combining all these reductions in Theorem 3, and noting that all kernel constants collapse to 1, we obtain

$$\left|{\displaystyle \frac{\mathsf{\Psi }\left(\sigma \right)+\mathsf{\Psi }\left(\rho \right)}{2}}-{\displaystyle \frac{1}{\rho -\sigma }}{\int }_{\sigma }^{\rho }\mathsf{\Psi }\left(\varkappa \right)d\varkappa \right|\le {\displaystyle \frac{\rho -\sigma }{2}}{\left({\displaystyle \frac{1}{p+1}}\right)}^{1/p}{\left({\displaystyle \frac{|{\mathsf{\Psi }}^{\prime }{\left(\sigma \right)|}^q+{\left|{\mathsf{\Psi }}^{\prime }\left(\rho \right)\right|}^q}{2}}\right)}^{1/q}.$$

 □

Corollary 10.

Under the assumptions of Theorem 3, let $\alpha \in (0,1)$, $\beta \in (0,1]$, and $\lambda \to 0^{+}$. Then the inequality reduces to

$$\begin{array}{c}\left|{\displaystyle \frac{\mathsf{\Psi }\left(\sigma \right)+\mathsf{\Psi }\left(\rho \right)}{2}}-{\displaystyle \frac{1}{\rho -\sigma }}{\int }_{\sigma }^{\rho }\mathsf{\Psi }\left(\varkappa \right)d\varkappa \right|\hfill \\ \le {\displaystyle \frac{\alpha (\rho -\sigma )}{2\rho \left(\alpha \right)\mathsf{\Gamma }\left(\beta \right)}}{\left({\int }_0^1{|1-2\zeta |}^p{(1-\zeta )}^{\beta -1}d\zeta \right)}^{1/p}{\left({\displaystyle \frac{|{\mathsf{\Psi }}^{\prime }{\left(\sigma \right)|}^q+{\left|{\mathsf{\Psi }}^{\prime }\left(\rho \right)\right|}^q}{2}}\right)}^{1/q}.\hfill \end{array}$$

Proof. 

Let $\lambda \to 0^{+}$. Using continuity of the Mittag-Leffler function, we have

$$E_{\beta }(-\lambda x^{\beta })\to 1.$$

Therefore,

$${\mathsf{\Delta }}_{\lambda }\left(\zeta \right)=E_{\beta }(-\lambda {(1-\zeta )}^{\beta }{(\rho -\sigma )}^{\beta })-E_{\beta }(-\lambda {\zeta }^{\beta }{(\rho -\sigma )}^{\beta })\to 1-1=0.$$

In the structure of Theorem 3, this limit implies that the fractional kernel loses its exponential/ML memory effect and the remaining dominant geometric structure becomes the classical symmetric weight

$$|1-2\zeta |.$$

Hence, the integral term in Hölder’s inequality becomes

$${\int }_0^1{|1-2\zeta |}^p{(1-\zeta )}^{\beta -1}d\zeta .$$

At the same time, the fractional operators converge to their classical Riemann-type counterparts:

$${}^{E\sigma \rho }I_{\sigma +}^{\alpha ,\beta }\mathsf{\Psi }\left(\rho \right)\to {\displaystyle \frac{\alpha }{\rho \left(\alpha \right)\mathsf{\Gamma }\left(\beta \right)}}{\int }_{\sigma }^{\rho }{(\rho -\zeta )}^{\beta -1}\mathsf{\Psi }\left(\zeta \right)d\zeta ,$$

and similarly for the right-sided operator.

Substituting all these limiting forms into Theorem 3, we obtain

$$\begin{array}{cc}\hfill \left|{\displaystyle \frac{\mathsf{\Psi }\left(\sigma \right)+\mathsf{\Psi }\left(\rho \right)}{2}}-{\displaystyle \frac{1}{\rho -\sigma }}{\int }_{\sigma }^{\rho }\mathsf{\Psi }\left(\varkappa \right)d\varkappa \right|& \le {\displaystyle \frac{\alpha (\rho -\sigma )}{2\rho \left(\alpha \right)\mathsf{\Gamma }\left(\beta \right)}}\hfill \\ \hfill & \times {\left({\int }_0^1{|1-2\zeta |}^p{(1-\zeta )}^{\beta -1}d\zeta \right)}^{1/p}\hfill \\ \hfill & \times {\left({\displaystyle \frac{|{\mathsf{\Psi }}^{\prime }{\left(\sigma \right)|}^q+{\left|{\mathsf{\Psi }}^{\prime }\left(\rho \right)\right|}^q}{2}}\right)}^{1/q}.\hfill \end{array}$$

 □

Corollary 11.

Under the same assumptions, letting $\beta \to 1^{-}$ while keeping $\alpha \in (0,1)$ and $\lambda >0$, we obtain

$$\begin{array}{c}\left|{\displaystyle \frac{\mathsf{\Psi }\left(\sigma \right)+\mathsf{\Psi }\left(\rho \right)}{2}}-{\displaystyle \frac{1}{2K_{\alpha ,1,\lambda }(\rho -\sigma )}}\left[{}^{E\sigma \rho }I_{\sigma +}^{\alpha ,1}\mathsf{\Psi }\left(\rho \right)+{}^{E\sigma \rho }I_{\rho -}^{\alpha ,1}\mathsf{\Psi }\left(\sigma \right)\right]\right|\hfill \\ \le {\displaystyle \frac{\alpha (\rho -\sigma )}{2K_{\alpha ,1,\lambda }(\rho -\sigma )\rho \left(\alpha \right)}}{\left({\int }_0^1{\left|{\mathsf{\Delta }}_{\lambda }\left(\zeta \right)\right|}^{p}d\zeta \right)}^{1/p}{\left({\displaystyle \frac{|{\mathsf{\Psi }}^{\prime }{\left(\sigma \right)|}^q+{\left|{\mathsf{\Psi }}^{\prime }\left(\rho \right)\right|}^q}{2}}\right)}^{1/q}.\hfill \end{array}$$

Proof. 

Let $\beta \to 1^{-}$. We use the continuity properties:

$$E_{\beta }\left(z\right)\to E_1\left(z\right)=e^z,\mathsf{\Gamma }\left(\beta \right)\to 1,{(1-\zeta )}^{\beta -1}\to 1.$$

Therefore, the kernel term becomes

$$E_{\beta }(-\lambda x^{\beta })\to e^{-\lambda x}.$$

Hence,

$${\mathsf{\Delta }}_{\lambda }\left(\zeta \right)\to e^{-\lambda (1-\zeta )(\rho -\sigma )}-e^{-\lambda \zeta (\rho -\sigma )}.$$

Substituting these limits into Theorem 3, we obtain that the fractional operator reduces to the exponential-kernel Atangana–Baleanu form:

$${}^{E\sigma \rho }I_{\sigma +}^{\alpha ,1}\mathsf{\Psi }\left(\rho \right)={\displaystyle \frac{\alpha }{\rho \left(\alpha \right)}}{\int }_{\sigma }^{\rho }{e}^{-\lambda (\rho -\zeta )}\mathsf{\Psi }\left(\zeta \right)d\zeta +{\displaystyle \frac{1-\alpha }{\rho \left(\alpha \right)}}\mathsf{\Psi }\left(\rho \right).$$

Similarly for the right-sided operator.

Finally, inserting all these limiting expressions into Theorem 3 yields

$$\begin{array}{cc}\hfill \left|{\displaystyle \frac{\mathsf{\Psi }\left(\sigma \right)+\mathsf{\Psi }\left(\rho \right)}{2}}-{\displaystyle \frac{1}{2K_{\alpha ,1,\lambda }(\rho -\sigma )}}\left[{}^{E\sigma \rho }I_{\sigma +}^{\alpha ,1}\mathsf{\Psi }\left(\rho \right)+{}^{E\sigma \rho }I_{\rho -}^{\alpha ,1}\mathsf{\Psi }\left(\sigma \right)\right]\right|& \le {\displaystyle \frac{\alpha (\rho -\sigma )}{2K_{\alpha ,1,\lambda }(\rho -\sigma )\rho \left(\alpha \right)}}\hfill \\ \hfill & \times {\left({\int }_0^1{\left|{\mathsf{\Delta }}_{\lambda }\left(\zeta \right)\right|}^{p}d\zeta \right)}^{1/p}\hfill \\ \hfill & \times {\left({\displaystyle \frac{|{\mathsf{\Psi }}^{\prime }{\left(\sigma \right)|}^q+{\left|{\mathsf{\Psi }}^{\prime }\left(\rho \right)\right|}^q}{2}}\right)}^{1/q}.\hfill \end{array}$$

 □

3. Graphical Interpretation of the Extended Atangana–Baleanu Fractional Integral

In this section, we provide a graphical interpretation of the extended Atangana–Baleanu (EAB) fractional integral in order to gain further insight into the influence of the fractional parameters. While the analytical results derived in the previous sections establish rigorous bounds, numerical visualization helps to better understand the effect of the parameters on the behavior of the operator.

To simplify the presentation, we consider the unit interval $[0,1]$, that is, $\sigma =0$ and $\rho =1$. We also fix the parameter $\lambda =1$. For visualization purposes, we evaluate the fractional integral for fixed values of $\varkappa \in [0,1]$ and observe its dependence on the parameters $(\alpha ,\beta )$.

We consider three representative convex functions, namely

$$\mathsf{\Psi }\left(\varkappa \right)={\varkappa }^2,\mathsf{\Psi }\left(\varkappa \right)=\varkappa ,\mathrm{and}\mathsf{\Psi }\left(\varkappa \right)=e^{\varkappa },$$

on the interval $[0,1]$. For each function, we compute the left-sided extended Atangana–Baleanu fractional integral

$${}^{E }I_{0+}^{\alpha ,\beta }\mathsf{\Psi }\left(\varkappa \right)={\displaystyle \frac{\alpha }{\rho \left(\alpha \right)\mathsf{\Gamma }\left(\beta \right)}}{\int }_0^{\varkappa }{E}_{\beta }\left(-\lambda {(\varkappa -\zeta )}^{\beta }\right){(\varkappa -\zeta )}^{\beta -1}\mathsf{\Psi }\left(\zeta \right)d\zeta +{\displaystyle \frac{1-\alpha }{\rho \left(\alpha \right)}}\mathsf{\Psi }\left(\varkappa \right),$$

where $\alpha \in (0,1)$ and $\beta \in (0,1]$.

For visualization purposes, the computed values are represented as surfaces over the $(\alpha ,\beta )$-plane. These surfaces describe how the operator transitions between local and nonlocal behavior.

3.1. Function 1: $\mathsf{\Psi }\left(\varkappa \right)={\varkappa }^2$

For the quadratic function, the resulting surface exhibits a smooth dependence on the parameters. As $\alpha$ increases, the contribution of the integral (nonlocal) part becomes more dominant, leading to a smoother averaged behavior.

On the other hand, smaller values of $\beta$ strengthen the memory effect induced by the Mittag-Leffler kernel. This causes a more noticeable deviation from the classical integral behavior, especially near the endpoints of the interval.

3.2. Function 2: $\mathsf{\Psi }\left(\varkappa \right)=\varkappa$

In the linear case, the EAB operator preserves a relatively stable structure. Since linear functions are less sensitive to averaging effects, the resulting surface remains close to linearity.

However, variations in $\beta$ still affect the behavior of the operator. In particular, for smaller values of $\beta$, the kernel influence becomes more pronounced, leading to visible deviations from the classical integral.

3.3. Function 3: $\mathsf{\Psi }\left(\varkappa \right)=e^{\varkappa }$

The exponential function provides a useful example to observe sensitivity to growth. Due to its rapidly increasing nature, the interaction with the nonlocal kernel becomes more significant.

As $\alpha$ increases, the contribution of the integral term becomes stronger, reflecting the accumulation of past values. Similarly, decreasing $\beta$ enhances the memory effect, which further amplifies the influence of earlier function values. This results in a steeper surface compared to the previous cases.

Overall, these graphical representations illustrate how the EAB fractional integral interpolates between local and nonlocal regimes. The parameter $\alpha$ controls the balance between the local term and the memory-dependent integral, while $\beta$ determines the decay and shape of the Mittag-Leffler kernel.

Such behavior shows that the EAB operator is capable of modeling processes with memory and hereditary properties. This makes it relevant in applications arising in mathematical physics, engineering, and related areas.

The behavior of the extended Atangana–Baleanu fractional integral for the considered test functions is illustrated in Figure 1. As can be observed, the parameters $\alpha$ and $\beta$ significantly influence the transition between local and nonlocal behavior, as well as the overall shape of the resulting surfaces.

The analytical results derived in the previous sections, together with the graphical illustrations presented here, provide a consistent interpretation of the extended Atangana–Baleanu fractional integral. In particular, they show that the operator preserves key properties of convex functions while extending classical integral structures to a framework involving memory effects. This overall behavior is illustrated in Figure 1.

More specifically, Figure 1a corresponds to the quadratic case, Figure 1b to the linear case, and Figure 1c to the exponential case. These examples highlight how the growth rate of the function affects the sensitivity of the operator to the fractional parameters.

4. Conclusions

In this work, we have derived a new class of Hermite–Hadamard type inequalities involving the extended Atangana–Baleanu fractional integral operator having a Mittag-Leffler kernel. The results are obtained by using a suitable integral identity for differentiable functions together with standard convexity assumptions on the absolute value of the first derivative.

The obtained inequalities extend the classical Hermite–Hadamard inequality to a fractional integral setting depending on the parameters $\alpha \in (0,1)$, $\beta \in (0,1]$, and $\lambda >0$. It is also shown that the classical case is recovered when $\alpha =\beta =1$ and $\lambda \to 0^{+}$. In addition, the results include the exponential kernel case as a particular example within the Mittag-Leffler framework.

To obtain sharper estimates, we also use Hölder’s inequality together with the convexity properties of $|{\mathsf{\Psi }}^{\prime }{|}^q$. This leads to bounds involving the kernel difference ${\mathsf{\Delta }}_{\lambda }\left(\zeta \right)$. The corresponding corollaries show that the classical Hermite–Hadamard inequality and several known variants are recovered as special cases under appropriate parameter choices, which confirms the consistency of the proposed approach.

Overall, the results provide a unified framework for Hermite–Hadamard type inequalities in the setting of generalized fractional integrals. Possible future work may include extensions to higher-dimensional settings, weaker convexity assumptions, and applications to fractional differential equations and numerical analysis.