Citation: Eze R. Nwaeze, Muhammad Adil Khan, Ali Ahmadian, Mohammad Nazir Ahmad, Ahmad Kamil Mahmood. Fractional inequalities of the Hermite–Hadamard type for $ m $-polynomial convex and harmonically convex functions[J]. AIMS Mathematics, 2021, 6(2): 1889-1904. doi: 10.3934/math.2021115
The sets $ {\bf T} $ and $ {\bf S}\subseteq \mathbb{R}\setminus\{0\} $ are called convex and harmonically convex, respectively if
$ \begin{equation*} \begin{cases} \varsigma q+(1- \varsigma)z\in {\bf T}\quad\mbox{for all}\quad q, z\in{\bf T}\quad \mbox{and}\quad \varsigma\in[0, 1];\\[2ex] \frac{qz}{ \varsigma q+(1- \varsigma)z}\in {\bf S}\quad\mbox{for all}\quad q, z\in{\bf S}\quad\mbox{and}\quad \varsigma\in[0, 1]. \end{cases} \end{equation*} $ |
Whenever used, we shall always consider $ {\bf T} $ as a convex set and $ {\bf S} $ as a harmonically convex set. Let $ m\in\mathbb{N} $. Recall that a function $ \varphi:{\bf T}\to\mathbb{R} $ is said to be $ m $-polynomial convex [31] on $ {\bf T} $ if
$ \begin{equation*} \varphi\big( \varsigma q+(1- \varsigma)z\big)\leq\frac{1}{m}\sum\limits_{\vartheta = 1}^m\left[1-(1- \varsigma)^\vartheta\right] \varphi(q)+\frac{1}{m}\sum\limits_{\vartheta = 1}^m\left[1-{ \varsigma}^\vartheta\right] \varphi(z) \end{equation*} $ |
for all $ q, z\in{\bf S} $ and $ \varsigma\in [0, 1] $. For this class of functions, Toplu et al. established the following double inequality of the Hermite–Hadamard type.
Theorem 1 ([31]). Let $ \varphi:{\bf T}\to\mathbb{R} $ be an $ m $-polynomial convex function. If $ \xi, \delta\in {\bf T} $ with $ \xi < \delta $, and $ \varphi $ is Lebesgue integrable on $ [\xi, \delta] $, then the following Hermite–Hadamard type inequality holds:
$ \begin{equation} \frac{2^{-1}m}{m+2^{-m}-1} \varphi\left(\frac{ \xi+ \delta}{2}\right)\leq \frac{1}{ \delta- \xi}\int_{ \xi}^{ \delta} \varphi(r)\, dr\leq \frac{ \varphi( \xi)+ \varphi( \delta)}{m}\sum\limits_{\vartheta = 1}^m\frac{\vartheta}{\vartheta+1}. \end{equation} $ | (1.1) |
The inequality (1.1) boils down to the classical Hermite–Hadamard inequality for convex functions if we take $ m = 1 $. Recently, Awan et al. [2] introduced the notion of $ m $-polynomial harmonically convex functions as follows: a real valued function $ \varphi:{\bf S}\to{\mathbb{R}}^+: = [0, \infty) $ is $ m $-harmonically convex if
$ \begin{equation} \varphi\left(\frac{qz}{ \varsigma q+(1- \varsigma)z}\right)\leq \frac{1}{m}\sum\limits_{\vartheta = 1}^m\left[1-(1- \varsigma)^\vartheta\right] \varphi(q)+\frac{1}{m}\sum\limits_{\vartheta = 1}^m\left[1-{ \varsigma}^\vartheta\right] \varphi(z) \end{equation} $ | (1.2) |
for all $ q, z\in{\bf S} $ and $ \varsigma\in [0, 1] $. In the same paper, the authors established the following Hermite–Hadamard type inequality for this class of functions:
Theorem 2 ([2]). Let $ \varphi:{\bf S}\to{\mathbb{R}}^{+} $ be an $ m $-polynomial harmonically convex function. If $ \xi, \delta\in {\bf S} $ with $ 0 < \xi < \delta $, and $ \varphi $ is Lebesgue integrable on $ [\xi, \delta] $, then the following Hermite–Hadamard type inequality holds:
$ \begin{equation*} \frac{2^{-1}m}{m+2^{-m}-1} \varphi\left(\frac{2 \xi \delta}{ \xi+ \delta}\right)\leq \frac{ \xi \delta}{ \delta- \xi}\int_{ \xi}^{ \delta}\frac{ \varphi(r)}{r^2}\, dr\leq \frac{ \varphi( \xi)+ \varphi( \delta)}{m}\sum\limits_{\vartheta = 1}^m\frac{\vartheta}{\vartheta+1}. \end{equation*} $ |
In the sequel, we will denote the sets of all $ m $-polynomial convex and $ m $-polynomial harmonically convex functions from A into B by $ {\bf XP}_m\left(A, B\right) $ and $ {\bf HXP}_m\left(A, B\right) $, respectively. The classical Hermite–Hadamard inequality has generated load of generalizations and extensions to other class of convexity. There are dozens of articles in this direction. We invite the interested reader to see the following articles [3,4,5,6,8,10,11,12,13,14,15,16,17,18,19,20,22,23,24,25,26,27,28,29,30,32,33,34] and the references cited therein.
Now, recall that the left- and right-sided $ \zeta $-Riemann–Liouville fractional integral operators $ _{ \zeta}{\mathcal{J}}_{ \xi^+}^{\epsilon} $ and $ _{ \zeta}{\mathcal{J}}_{ \delta^-}^{\epsilon} $ of order $ \epsilon > 0 $, for a real valued continuous function $ \varphi(r) $, are defined as ([21]):
$ \begin{equation*} \label{DD1} _{ \zeta}{\mathcal{J}}_{ \xi^+}^{ \epsilon} \varphi(r) = \frac{1}{ \zeta \Gamma_ \zeta( \epsilon)} \int_ \xi^r(r- \varsigma)^{\frac{ \epsilon}{ \zeta}-1} \varphi( \varsigma)\, d \varsigma, \quad r \gt \xi, \end{equation*} $ |
and
$ \begin{equation*} \label{DD3} _{ \zeta}{\mathcal{J}}_{ \delta^-}^{ \epsilon} \varphi(r) = \frac{1}{ \zeta \Gamma_ \zeta( \epsilon)}\int_r^ \delta \left( \varsigma-r\right)^{\frac{ \epsilon}{ \zeta}-1} \varphi( \varsigma)\, d \varsigma, \quad r \lt \delta, \end{equation*} $ |
where $ \zeta > 0 $, and $ \Gamma_ \zeta $ is the $ \zeta $-gamma function given by
$ \Gamma_ \zeta(r): = \int_{0}^{\infty} \varsigma^{r-1}e^{-\frac{ \varsigma^ \zeta}{ \zeta}}\, d \varsigma, \quad Re(r) \gt 0, $ |
with the properties $ \Gamma_ \zeta(r+ \zeta) = r \Gamma_ \zeta(r) $ and $ \Gamma_ \zeta(\zeta) = 1 $. If $ \zeta = 1 $, we simply write
$ _{1}{\mathcal{J}}_{ \xi^+}^{ \epsilon} \varphi = {\mathcal{J}}_{ \xi^+}^{ \epsilon} \varphi\quad\mbox{and}\quad _{1}{\mathcal{J}}_{ \delta^-}^{ \epsilon} \varphi = {\mathcal{J}}_{ \delta^-}^{ \epsilon} \varphi. $ |
The beta function $ \mathcal{B} $ is defined by
$ \begin{equation} \mathcal{B}(u, v) = \int_0^1 \varsigma^{u-1}(1- \varsigma)^{v-1}\, d \varsigma \quad\quad\mbox{for}\quad Re(u) \gt 0, Re(v) \gt 0. \end{equation} $ | (1.3) |
Another fractional integral operators of interest is the Caputo–Fabrizio operators [1]: let $ L^2(\xi, \delta) $ be the space of square integrable functions on the interval $ (\xi, \delta) $ and
$ H^1( \xi, \delta): = \Big\{g\, |\, g\in L^2( \xi, \delta)\quad\mbox{and}\quad g'\in L^2( \xi, \delta) \Big\}. $ |
If $ \varphi\in H^1(\xi, \delta) $, $ \xi < \delta $ and $ \mu\in[0, 1] $, then the left- and right-sided Caputo–Fabrizio fractional integral operators $ ^{cf}{\mathcal{I}}_{ \xi}^{\mu} $ and $ ^{cf}{\mathcal{I}}_{ \delta}^{\mu} $ are defined by
$ \begin{equation} ^{cf}{\mathcal{I}}_{ \xi}^{\mu} \varphi(s) = \frac{1-\mu}{B(\mu)} \varphi(s)+\frac{\mu}{B(\mu)}\int_ \xi^{s} \varphi(r)\, dr \end{equation} $ | (1.4) |
and
$ \begin{equation} ^{cf}{\mathcal{I}}_{ \delta}^{\mu} \varphi(s) = \frac{1-\mu}{B(\mu)} \varphi(s)+\frac{\mu}{B(\mu)}\int^ \delta_{s} \varphi(r)\, dr, \end{equation} $ | (1.5) |
where $ B:[0, 1]\to (0, \infty) $ is a normalization function satisfying $ B(0) = B(1) = 1 $.
Using these fractional integral operators in (1.4) and (1.5), Gürbüz et al. established the following fractional version of the Hermite–Hadamard inequality:
Theorem 3 ([7]). Let $ \varphi:{\bf T}\to\mathbb{R} $ be a convex function on $ {\bf T} $. If $ \xi, \delta\in {\bf T} $ with $ \xi < \delta $, and $ \varphi $ is Lebesgue integrable on $ [\xi, \delta] $, then the following double inequality holds:
$ \begin{equation*} \varphi\left(\frac{ \xi+ \delta}{2}\right)\leq \frac{B(\mu)}{\mu( \delta- \xi)}\left[^{cf}{\mathcal{I}}_{ \xi}^{\mu} \varphi(s)+^{cf}{\mathcal{I}}_{ \delta}^{\mu} \varphi(s)-\frac{2(1-\mu)}{B(\mu)} \varphi(s)\right]\leq \frac{ \varphi( \xi)+ \varphi( \delta)}{2}, \end{equation*} $ |
where $ \mu\in[0, 1] $, $ s\in [\xi, \delta] $ and $ B(\mu) > 0 $ is a normalization function.
Since the classes of convexity introduced here are new, much work have not been done in this sense. This work is geared towards further development around inequalities for these classes. In view of this, we aim to achieve the following objectives:
1. To establish new Hermite–Hadamard type inequalities for the class of $ m $-polynomial convex functions involving the Caputo–Fabrizio integral operators. Our first result in this direction generalizes and extends Theorem 3.
2. To obtain inequalities of the Hermite–Hadamard type for functions that are $ m $-polynomial harmonically convex functions via the $ \zeta $-Riemann–Liouville fractional integral operators. This, in turn, also complement and generalize some existing results in the literature.
Inequalities of the Hermite–Hadamard type, for $ m $-polynomial convex functions, are hereby presented. The results, presented herein, involve the Caputo–Fabrizio operators.
Theorem 4. Let $ \varphi:{\bf T}\to \mathbb{R} $ be a Lebesgue integrable function on $ [\xi, \delta] $ with $ \xi < \delta $ and $ \xi, \delta\in {\bf T} $. If $ \varphi\in{\bf XP}_m\left({\bf T}, \mathbb{R}\right) $, then
$ \begin{equation*} \begin{aligned} \frac{2^{-1}m}{m+2^{-m}-1}\, \varphi\left(\frac{ \xi+ \delta}{2}\right)&\leq \frac{B(\mu)}{\mu( \delta- \xi)}\left[^{cf}{\mathcal{I}}_{ \xi}^{\mu} \varphi(s)+^{cf}{\mathcal{I}}_{ \delta}^{\mu} \varphi(s)-\frac{2(1-\mu)}{B(\mu)} \varphi(s)\right]\\ &\leq \frac{ \varphi( \xi)+ \varphi( \delta)}{m}\sum\limits_{\vartheta = 1}^m\frac{\vartheta}{\vartheta+1}, \end{aligned} \end{equation*} $ |
where $ \mu\in(0, 1] $, $ s\in [\xi, \delta] $ and $ B(\mu) > 0 $ is a normalization function.
Proof. Given that $ \varphi\in{\bf XP}_m\left({\bf T}, \mathbb{R}\right) $, it follows from (1.1) that
$ \begin{equation} \begin{aligned} \frac{m}{m+2^{-m}-1} \varphi\left(\frac{ \xi+ \delta}{2}\right)&\leq \frac{2}{ \delta- \xi}\int_{ \xi}^{ \delta} \varphi(r)\, dr\\ & = \frac{2}{ \delta- \xi}\left[\int_{ \xi}^{s} \varphi(r)\, dr+ \int_{s}^{ \delta} \varphi(r)\, dr\right]. \end{aligned} \end{equation} $ | (2.1) |
Multiplying both sides of (2.1) by $ \frac{\mu(\delta- \xi)}{2B(\mu)} $ gives:
$ \begin{equation} \begin{aligned} \frac{\mu( \delta- \xi)}{2B(\mu)} \frac{m}{m+2^{-m}-1} \varphi\left(\frac{ \xi+ \delta}{2}\right)&\leq \frac{\mu}{B(\mu)}\left[\int_{ \xi}^{s} \varphi(r)\, dr+ \int_{s}^{ \delta} \varphi(r)\, dr\right]. \end{aligned} \end{equation} $ | (2.2) |
By adding $ \frac{2(1-\mu)}{B(\mu)} \varphi(s) $ to both sides of (2.2), we get:
$ \begin{equation*} \begin{aligned} &\frac{2(1-\mu)}{B(\mu)} \varphi(s)+\frac{\mu( \delta- \xi)}{2B(\mu)} \frac{m}{m+2^{-m}-1} \varphi\left(\frac{ \xi+ \delta}{2}\right)\\ &\leq \frac{2(1-\mu)}{B(\mu)} \varphi(s)+\frac{\mu}{B(\mu)}\left[\int_{ \xi}^{s} \varphi(r)\, dr+ \int_{s}^{ \delta} \varphi(r)\, dr\right]\\ & = \left[ \frac{(1-\mu)}{B(\mu)} \varphi(s)+\frac{\mu}{B(\mu)}\int_{ \xi}^{s} \varphi(r)\, dr\right]\\ &\quad\quad\quad+\left[ \frac{(1-\mu)}{B(\mu)} \varphi(s)+\frac{\mu}{B(\mu)}\int_{s}^{ \delta} \varphi(r)\, dr\right]\\ & = ^{cf}{\mathcal{I}}_{ \xi}^{\mu} \varphi(s)+^{cf}{\mathcal{I}}_{ \delta}^{\mu} \varphi(s). \end{aligned} \end{equation*} $ |
This implies that
$ \begin{equation} \begin{aligned} &\frac{2(1-\mu)}{B(\mu)} \varphi(s)+\frac{\mu( \delta- \xi)}{2B(\mu)} \frac{m}{m+2^{-m}-1} \varphi\left(\frac{ \xi+ \delta}{2}\right)\\[1.5ex] &\leq {^{cf}{\mathcal{I}}}_{ \xi}^{\mu} \varphi(s)+{^{cf}{\mathcal{I}}}_{ \delta}^{\mu} \varphi(s). \end{aligned} \end{equation} $ | (2.3) |
On the other hand, we also get from (1.1) the following inequality:
$ \begin{equation} \begin{aligned} \frac{2}{ \delta- \xi}\int_{ \xi}^{ \delta} \varphi(r)\, dr\leq \frac{ \varphi( \xi)+ \varphi( \delta)}{m}\sum\limits_{\vartheta = 1}^m\frac{2\vartheta}{\vartheta+1}. \end{aligned} \end{equation} $ | (2.4) |
If we multiply (2.4) by $ \frac{\mu(\delta- \xi)}{2B(\mu)} $ and then add $ \frac{2(1-\mu)}{B(\mu)} \varphi(s) $ to the resulting inequality, we obtain:
$ \begin{equation} \begin{aligned} {^{cf}{\mathcal{I}}}_{ \xi}^{\mu} \varphi(s)+{^{cf}{\mathcal{I}}}_{ \delta}^{\mu} \varphi(s)\leq \frac{\mu( \delta- \xi)}{B(\mu)}\frac{ \varphi( \xi)+ \varphi( \delta)}{m}\sum\limits_{\vartheta = 1}^m\frac{\vartheta}{\vartheta+1}+\frac{2(1-\mu)}{B(\mu)} \varphi(s). \end{aligned} \end{equation} $ | (2.5) |
Hence, the desired result is obtained by combining (2.3) and (2.5).
Remark 1. By taking $ m = 1 $, Theorem 4 becomes Theorem 3.
Theorem 5. Let $ \varphi, \upsilon:{\bf T}\to \mathbb{R} $ be two functions such that $ \varphi \upsilon $ is Lebesgue integrable function on $ [\xi, \delta] $ with $ \xi < \delta $ and $ \xi, \delta\in {\bf T} $. If $ \varphi\in{\bf XP}_{m_1}\left({\bf S}, \mathbb{R}\right) $, $ \upsilon\in{\bf XP}_{m_2}\left({\bf T}, \mathbb{R}\right) $, then
$ \begin{equation*} \begin{aligned} &\frac{B(\mu)}{\mu( \delta- \xi)}\left[^{cf}{\mathcal{I}}_{ \xi}^{\mu} \varphi(s) \upsilon(s)+^{cf}{\mathcal{I}}_{ \delta}^{\mu} \varphi(s) \upsilon(s)-\frac{2(1-\mu)}{B(\mu)} \varphi(s) \upsilon(s)\right]\\ &\leq \int_0^1\Big[\Delta _1( \varsigma) \varphi( \xi) \upsilon( \xi)+\Delta _2( \varsigma) \varphi( \xi) \upsilon( \delta)+ \Delta _3( \varsigma) \varphi( \delta) \upsilon( \xi)+\Delta _4( \varsigma) \varphi( \delta) \upsilon( \delta)\Big]\, d \varsigma, \end{aligned} \end{equation*} $ |
where $ \mu\in(0, 1] $, $ s\in [\xi, \delta] $ and $ B(\mu) > 0 $ is a normalization function, and
$ \begin{equation*} \begin{aligned} &\Delta _1( \varsigma): = \frac{1}{m_1}\frac{1}{m_2}\sum\limits_{\vartheta = 1}^{m_1}\left[1-(1- \varsigma)^\vartheta\right]\sum\limits_{\vartheta = 1}^{m_2}\left[1-(1- \varsigma)^\vartheta\right];\\ &\Delta _2( \varsigma): = \frac{1}{m_1}\frac{1}{m_2}\sum\limits_{\vartheta = 1}^{m_1}\left[1-(1- \varsigma)^\vartheta\right]\sum\limits_{\vartheta = 1}^{m_2}\left[1-{ \varsigma}^\vartheta\right];\\ &\Delta _3( \varsigma): = \frac{1}{m_1}\frac{1}{m_2}\sum\limits_{\vartheta = 1}^{m_1}\left[1- \varsigma^\vartheta\right]\sum\limits_{\vartheta = 1}^{m_2}\left[1-(1- \varsigma)^\vartheta\right];\\ &\Delta _4( \varsigma): = \frac{1}{m_1}\frac{1}{m_2}\sum\limits_{\vartheta = 1}^{m_1}\left[1- \varsigma^\vartheta\right]\sum\limits_{\vartheta = 1}^{m_2}\left[1-{ \varsigma}^\vartheta\right]. \end{aligned} \end{equation*} $ |
Proof. Let $ \varphi\in{\bf XP}_{m_1}\left({\bf T}, \mathbb{R}\right) $ and $ \upsilon\in{\bf XP}_{m_2}\left({\bf T}, \mathbb{R}\right) $. Then for $ \varsigma\in[0, 1] $, we have:
$ \begin{equation} \varphi\big( \varsigma \xi+(1- \varsigma) \delta\big)\leq\frac{1}{m_1}\sum\limits_{\vartheta = 1}^{m_1}\left[1-(1- \varsigma)^\vartheta\right] \varphi( \xi)+\frac{1}{m_1}\sum\limits_{\vartheta = 1}^{m_1}\left[1-{ \varsigma}^\vartheta\right] \varphi( \delta) \end{equation} $ | (2.6) |
and
$ \begin{equation} \upsilon\big( \varsigma \xi+(1- \varsigma) \delta\big)\leq\frac{1}{m_1}\sum\limits_{\vartheta = 1}^{m_1}\left[1-(1- \varsigma)^\vartheta\right] \upsilon( \xi)+\frac{1}{m_1}\sum\limits_{\vartheta = 1}^{m_1}\left[1-{ \varsigma}^\vartheta\right] \upsilon( \delta). \end{equation} $ | (2.7) |
Multiplying (2.6) and (2.7) gives:
$ \begin{equation*} \begin{aligned} & \varphi\big( \varsigma \xi+(1- \varsigma) \delta\big) \, \upsilon\big( \varsigma \xi+(1- \varsigma) \delta\big)\\ &\leq \frac{1}{m_1}\frac{1}{m_2}\sum\limits_{\vartheta = 1}^{m_1}\left[1-(1- \varsigma)^\vartheta\right]\sum\limits_{\vartheta = 1}^{m_2}\left[1-(1- \varsigma)^\vartheta\right] \varphi( \xi) \upsilon( \xi)\\ &\quad+\frac{1}{m_1}\frac{1}{m_2}\sum\limits_{\vartheta = 1}^{m_1}\left[1-(1- \varsigma)^\vartheta\right]\sum\limits_{\vartheta = 1}^{m_2}\left[1-{ \varsigma}^\vartheta\right] \varphi( \xi) \upsilon( \delta)\\ &\quad+ \frac{1}{m_1}\frac{1}{m_2}\sum\limits_{\vartheta = 1}^{m_1}\left[1- \varsigma^\vartheta\right]\sum\limits_{\vartheta = 1}^{m_2}\left[1-(1- \varsigma)^\vartheta\right] \varphi( \delta) \upsilon( \xi)\\ &\quad+\frac{1}{m_1}\frac{1}{m_2}\sum\limits_{\vartheta = 1}^{m_1}\left[1- \varsigma^\vartheta\right]\sum\limits_{\vartheta = 1}^{m_2}\left[1-{ \varsigma}^\vartheta\right] \varphi( \delta)w( \delta)\\ &: = \Delta _1( \varsigma) \varphi( \xi) \upsilon( \xi)+\Delta _2( \varsigma) \varphi( \xi) \upsilon( \delta)+ \Delta _3( \varsigma) \varphi( \delta) \upsilon( \xi)+\Delta _4( \varsigma) \varphi( \delta) \upsilon( \delta). \end{aligned} \end{equation*} $ |
This implies that
$ \begin{equation} \begin{aligned} & \varphi\big( \varsigma \xi+(1- \varsigma) \delta\big) \, \upsilon\big( \varsigma \xi+(1- \varsigma) \delta\big)\\[1ex] &\leq \Delta _1( \varsigma) \varphi( \xi) \upsilon( \xi)+\Delta _2( \varsigma) \varphi( \xi) \upsilon( \delta)+ \Delta _3( \varsigma) \varphi( \delta) \upsilon( \xi)+\Delta _4( \varsigma) \varphi( \delta) \upsilon( \delta). \end{aligned} \end{equation} $ | (2.8) |
Integrating both sides of (2.8) with respect to $ \varsigma $ over $ [0, 1] $ results to:
$ \begin{equation*} \begin{aligned} & \frac{2}{ \delta- \xi}\int_{ \xi}^{ \delta} \varphi(r) \upsilon(r)\, dr\\[1ex] &\leq 2\int_0^1\Big[\Delta _1( \varsigma) \varphi( \xi) \upsilon( \xi)+\Delta _2( \varsigma) \varphi( \xi) \upsilon( \delta)+ \Delta _3( \varsigma) \varphi( \delta) \upsilon( \xi)+\Delta _4( \varsigma) \varphi( \delta) \upsilon( \delta)\Big]\, d \varsigma\\ &: = \mathcal{N}( \xi, \delta). \end{aligned} \end{equation*} $ |
That is,
$ \begin{equation} \begin{aligned} & \frac{2}{ \delta- \xi}\left[\int_{ \xi}^{s} \varphi(r) \upsilon(r)\, dr+\int_{s}^{ \delta} \varphi(r) \upsilon(r)\, dr\right]\\[1ex] &\leq\mathcal{N}( \xi, \delta). \end{aligned} \end{equation} $ | (2.9) |
Now, multiplying (2.9) by $ \frac{\mu(\delta- \xi)}{2B(\mu)} $ and then adding $ \frac{2(1-\mu)}{B(\mu)} \varphi(s) \upsilon(s) $ to the result to obtain:
$ \begin{equation*} \begin{aligned} & \frac{\mu}{B(\mu)}\left[\int_{ \xi}^{s} \varphi(r) \upsilon(r)\, dr+\int_{s}^{ \delta} \varphi(r) \upsilon(r)\, dr\right]+\frac{2(1-\mu)}{B(\mu)} \varphi(s) \upsilon(s)\\[1ex] &\leq\frac{\mu( \delta- \xi)}{2B(\mu)}\mathcal{N}( \xi, \delta)+\frac{2(1-\mu)}{B(\mu)} \varphi(s) \upsilon(s). \end{aligned} \end{equation*} $ |
Hence,
$ \begin{equation*} \begin{aligned} &{^{cf}{\mathcal{I}}}_{ \xi}^{\mu} \varphi(s) \upsilon(s)+{^{cf}{\mathcal{I}}}_{ \delta}^{\mu} \varphi(s) \upsilon(s) \\[1ex] &\leq\frac{\mu( \delta- \xi)}{2B(\mu)}\mathcal{N}( \xi, \delta)+\frac{2(1-\mu)}{B(\mu)} \varphi(s) \upsilon(s), \end{aligned} \end{equation*} $ |
from which we get the intended inequality.
Remark 2. Set $ m_1 = m_2 = 1 $ in Theorem 5. Then we recover [7,Theorem 3] .
In this subsection, we present some new Hermite–Hadamard type results involving the $ \zeta $-Riemann–Liouville fractional integral operators.
Theorem 6. Let $ \varphi:{\bf S}\to {\mathbb{R}}^{+} $ be a Lebesgue integrable function on $ [\xi, \delta] $ with $ 0 < \xi < \delta $ and $ \xi, \delta\in {\bf S} $. If $ \varphi\in{\bf HXP}_m\left({\bf S}, {\mathbb{R}}^{+}\right) $ and $ \zeta, \epsilon > 0 $, then
$ \begin{equation*} \begin{aligned} &\frac{1}{m+2^{-m}-1}\, \, \varphi\left(\frac{2 \xi \delta}{ \xi+ \delta}\right)\\ &\leq \frac{ \Gamma_ \zeta( \epsilon+ \zeta)}{m}\left(\frac{ \xi \delta}{ \delta- \xi}\right)^{\frac{ \epsilon}{ \zeta}}\left[_{ \zeta}{\mathcal{J}}_{\frac{1}{ \delta}^+}^{ \epsilon}\big( \varphi\circ \tilde{ \varphi}\big)\left(\frac{1}{ \xi}\right)+_{ \zeta}{\mathcal{J}}_{\frac{1}{ \xi}^-}^{ \epsilon}\big( \varphi\circ \tilde{ \varphi}\big)\left(\frac{1}{ \delta}\right)\right]\\ &\leq \frac{ \varphi( \xi)+ \varphi( \delta)}{m^2}\sum\limits_{\vartheta = 1}^{m}\left[2-\frac{ \epsilon}{ \epsilon+ \zeta \vartheta}-\frac{ \epsilon}{ \zeta}\mathcal{B}\left(\frac{ \epsilon}{ \zeta}, \, \vartheta+1\right)\right], \end{aligned} \end{equation*} $ |
where $ \tilde{ \varphi}(r) = \frac{1}{r} $ and $ \mathcal{B} $ is the beta function defined by (1.3).
Proof. Given that $ \varphi\in{\bf HXP}_m\left({\bf S}, {\mathbb{R}}^{+}\right) $, we get the following relation:
$ \varphi\left(\frac{qz}{\frac{1}{2}q+\frac{1}{2}z}\right)\leq \frac{1}{m}\sum\limits_{\vartheta = 1}^m\left[1-\frac{1}{2^\vartheta}\right] \varphi(q)+\frac{1}{m}\sum\limits_{\vartheta = 1}^m\left[1-\frac{1}{2^\vartheta}\right] \varphi(z). $ |
This implies that for all $ q, z\in {\bf S} $:
$ \begin{equation} \varphi\left(\frac{2qz}{q+z}\right)\leq\frac{1}{m}\sum\limits_{\vartheta = 1}^m\left[1-\frac{1}{2^\vartheta}\right]\Big( \varphi(q)+ \varphi(z)\Big). \end{equation} $ | (2.10) |
Now, let $ q = \frac{ \xi \delta}{ \varsigma \xi+(1- \varsigma) \delta} $ and $ z = \frac{ \xi \delta}{ \varsigma \delta+(1- \varsigma) \xi} $. Then (2.10) becomes:
$ \begin{equation} \varphi\left(\frac{2 \xi \delta}{ \xi+ \delta}\right)\leq\frac{1}{m}\sum\limits_{\vartheta = 1}^m\left(1-\frac{1}{2^\vartheta}\right)\left\{ \varphi\left(\frac{ \xi \delta}{ \varsigma \xi+(1- \varsigma) \delta}\right)+ \varphi\left(\frac{ \xi \delta}{ \varsigma \delta+(1- \varsigma) \xi}\right)\right\}. \end{equation} $ | (2.11) |
Multiplying both sides of (2.11) by $ \varsigma^{\frac{ \epsilon}{ \zeta}-1} $ and integrating with respect to $ \varsigma $ over $ [0, 1] $, we get:
$ \begin{equation*} \label{E2} \begin{aligned} &\int_{0}^{1} \varsigma^{\frac{ \epsilon}{ \zeta}-1} \varphi\left(\frac{2 \xi \delta}{ \xi+ \delta}\right)\, d \varsigma \\ &\leq \frac{1}{m}\sum\limits_{\vartheta = 1}^m\left(1-\frac{1}{2^\vartheta}\right)\int_0^1 \varsigma^{\frac{ \epsilon}{ \zeta}-1}\left\{ \varphi\left(\frac{ \xi \delta}{ \varsigma \xi+(1- \varsigma) \delta}\right)+ \varphi\left(\frac{ \xi \delta}{ \varsigma \delta+(1- \varsigma) \xi}\right)\right\}\, d \varsigma\\ & = \frac{1}{m}\sum\limits_{\vartheta = 1}^m\left(1-\frac{1}{2^\vartheta}\right)\left[\int_0^1 \varsigma^{\frac{ \epsilon}{ \zeta}-1} \varphi\left(\frac{ \xi \delta}{ \varsigma \xi+(1- \varsigma) \delta}\right)\, d \varsigma\right.\\ &\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\left.+\int_0^1 \varsigma^{\frac{ \epsilon}{ \zeta}-1} \varphi\left(\frac{ \xi \delta}{ \varsigma \delta+(1- \varsigma) \xi}\right)\, d \varsigma\right]\\ & = \frac{1}{m}\sum\limits_{\vartheta = 1}^m\left(1-\frac{1}{2^\vartheta}\right)\left[\left(\frac{ \xi \delta}{ \delta- \xi}\right)^{\frac{ \epsilon}{ \zeta}}\int_{\frac{1}{ \delta}}^{\frac{1}{ \xi}}\left(\frac{1}{ \xi}-r\right)^{\frac{ \epsilon}{ \zeta}-1} \varphi\left(\frac{1}{r}\right)\, dr\right.\\ &\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\left.+\left(\frac{ \xi \delta}{ \delta- \xi}\right)^{\frac{ \epsilon}{ \zeta}}\int_{\frac{1}{ \delta}}^{\frac{1}{ \xi}}\left(r-\frac{1}{ \delta}\right)^{\frac{ \epsilon}{ \zeta}-1} \varphi\left(\frac{1}{r}\right)\, dr\right]\\ & = \frac{ \zeta \Gamma_ \zeta( \epsilon)}{m}\sum\limits_{\vartheta = 1}^m\left(1-\frac{1}{2^\vartheta}\right)\left(\frac{ \xi \delta}{ \delta- \xi}\right)^{\frac{ \epsilon}{ \zeta}}\left[\frac{1}{ \zeta \Gamma_ \zeta( \epsilon)}\int_{\frac{1}{ \delta}}^{\frac{1}{ \xi}}\left(\frac{1}{ \xi}-r\right)^{\frac{ \epsilon}{ \zeta}-1} \varphi\left(\frac{1}{r}\right)\, dr\right.\\ &\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\left.+\frac{1}{ \zeta \Gamma_ \zeta( \epsilon)}\int_{\frac{1}{ \delta}}^{\frac{1}{ \xi}}\left(r-\frac{1}{ \delta}\right)^{\frac{ \epsilon}{ \zeta}-1} \varphi\left(\frac{1}{r}\right)\, dr\right]\\ & = \frac{ \zeta \Gamma_ \zeta( \epsilon)}{m}\sum\limits_{\vartheta = 1}^m\left(1-\frac{1}{2^\vartheta}\right)\left(\frac{ \xi \delta}{ \delta- \xi}\right)^{\frac{ \epsilon}{ \zeta}}\left[_{ \zeta}{\mathcal{J}}_{\frac{1}{ \delta}^+}^{ \epsilon}\big( \varphi\circ \tilde{ \varphi}\big)\left(\frac{1}{ \xi}\right)+_{ \zeta}{\mathcal{J}}_{\frac{1}{ \xi}^-}^{ \epsilon}\big( \varphi\circ \tilde{ \varphi}\big)\left(\frac{1}{ \delta}\right)\right], \end{aligned} \end{equation*} $ |
where $ \tilde{ \varphi}(r) = \frac{1}{r} $. This implies that
$ \begin{equation} \begin{aligned} &\frac{1}{m+2^{-m}-1}\, \, \varphi\left(\frac{2 \xi \delta}{ \xi+ \delta}\right)\\ &\leq \frac{ \Gamma_ \zeta( \epsilon+ \zeta)}{m}\left(\frac{ \xi \delta}{ \delta- \xi}\right)^{\frac{ \epsilon}{ \zeta}}\left[_{ \zeta}{\mathcal{J}}_{\frac{1}{ \delta}^+}^{ \epsilon}\big( \varphi\circ \tilde{ \varphi}\big)\left(\frac{1}{ \xi}\right)+_{ \zeta}{\mathcal{J}}_{\frac{1}{ \xi}^-}^{ \epsilon}\big( \varphi\circ \tilde{ \varphi}\big)\left(\frac{1}{ \delta}\right)\right]. \end{aligned} \end{equation} $ | (2.12) |
Next, substituting $ q = \xi $ and $ z = \delta $ in (1.2) gives
$ \begin{equation} \varphi\left(\frac{ \xi \delta}{ \varsigma \xi+(1- \varsigma) \delta}\right)\leq\frac{1}{m}\sum\limits_{\vartheta = 1}^m\left[1-(1- \varsigma)^\vartheta\right] \varphi( \xi)+\frac{1}{m}\sum\limits_{\vartheta = 1}^m\left[1-{ \varsigma}^\vartheta\right] \varphi( \delta). \end{equation} $ | (2.13) |
Reversing the role of $ \xi $ and $ \delta $ in (2.13) produces:
$ \begin{equation} \varphi\left(\frac{ \xi \delta}{ \varsigma \delta+(1- \varsigma) \xi}\right)\leq\frac{1}{m}\sum\limits_{\vartheta = 1}^m\left[1-(1- \varsigma)^\vartheta\right] \varphi( \delta)+\frac{1}{m}\sum\limits_{\vartheta = 1}^m\left[1-{ \varsigma}^\vartheta\right] \varphi( \xi). \end{equation} $ | (2.14) |
If we now add (2.13) and (2.15), multiply the resulting inequality by $ \varsigma^{\frac{ \epsilon}{ \zeta}-1} $ and integrate with respect to $ \varsigma\in [0, 1] $, then we obtain:
$ \begin{equation} \begin{aligned} &\int_0^1 \varsigma^{\frac{ \epsilon}{ \zeta}-1}\left\{ \varphi\left(\frac{ \xi \delta}{ \varsigma \xi+(1- \varsigma) \delta}\right)+ \varphi\left(\frac{ \xi \delta}{ \varsigma \delta+(1- \varsigma) \xi}\right)\right\}\, d \varsigma\\[1ex] &\leq \frac{ \varphi( \xi)+ \varphi( \delta)}{m}\sum\limits_{\vartheta = 1}^{m}\int_0^1\left[2 \varsigma^{\frac{ \epsilon}{ \zeta}-1}- \varsigma^{\frac{ \epsilon}{ \zeta}-1}(1- \varsigma)^\vartheta- \varsigma^{\frac{ \epsilon}{ \zeta}+\vartheta-1}\right]\, d \varsigma\\ &\leq \frac{ \varphi( \xi)+ \varphi( \delta)}{m}\sum\limits_{\vartheta = 1}^{m}\left[\frac{2 \zeta}{ \epsilon}-\frac{ \zeta}{ \epsilon+ \zeta \vartheta}-\mathcal{B}\left(\frac{ \epsilon}{ \zeta}, \, \vartheta+1\right)\right]. \end{aligned} \end{equation} $ | (2.15) |
From (2.15), we get:
$ \begin{equation} \begin{aligned} &\frac{ \Gamma_ \zeta( \epsilon+ \zeta)}{m}\left(\frac{ \xi \delta}{ \delta- \xi}\right)^{\frac{ \epsilon}{ \zeta}}\left[_{ \zeta}{\mathcal{J}}_{\frac{1}{ \delta}^+}^{ \epsilon}\big( \varphi\circ \tilde{ \varphi}\big)\left(\frac{1}{ \xi}\right)+_{ \zeta}{\mathcal{J}}_{\frac{1}{ \xi}^-}^{ \epsilon}\big( \varphi\circ \tilde{ \varphi}\big)\left(\frac{1}{ \delta}\right)\right]\\ &\leq \frac{ \varphi( \xi)+ \varphi( \delta)}{m^2}\sum\limits_{\vartheta = 1}^{m}\left[2-\frac{ \epsilon}{ \epsilon+ \zeta \vartheta}-\frac{ \epsilon}{ \zeta}\mathcal{B}\left(\frac{ \epsilon}{ \zeta}, \, \vartheta+1\right)\right]. \end{aligned} \end{equation} $ | (2.16) |
Combining (2.12) and (2.16), we get the desired result.
Remark 3. If we take $ \epsilon = \zeta = 1 $, then Theorem 6 reduces to Theorem 2 . If, on the other hand, we let $ m = 1 $, then we get from Theorem 6 the following corollary:
Corollary 1. Let $ \varphi:{\bf S}\to {\mathbb{R}}^{+} $ be a Lebesgue integrable function on $ [\xi, \delta] $ with $ \xi < \delta $ and $ \xi, \delta\in {\bf S} $. If $ \varphi $ is harmonically convex and $ \zeta, \epsilon > 0 $, then
$ \begin{equation*} \begin{aligned} & \varphi\left(\frac{2 \xi \delta}{ \xi+ \delta}\right)\\ &\leq \frac{ \Gamma_ \zeta( \epsilon+ \zeta)}{2}\left(\frac{ \xi \delta}{ \delta- \xi}\right)^{\frac{ \epsilon}{ \zeta}}\left[_{ \zeta}{\mathcal{J}}_{\frac{1}{ \delta}^+}^{ \epsilon}\big( \varphi\circ \tilde{ \varphi}\big)\left(\frac{1}{ \xi}\right)+_{ \zeta}{\mathcal{J}}_{\frac{1}{ \xi}^-}^{ \epsilon}\big( \varphi\circ \tilde{ \varphi}\big)\left(\frac{1}{ \delta}\right)\right]\\ &\leq \frac{ \varphi( \xi)+ \varphi( \delta)}{2}. \end{aligned} \end{equation*} $ |
Theorem 7. Let $ \varphi, \upsilon:{\bf S}\to {\mathbb{R}}^{+} $ be two functions such that $ \varphi \upsilon $ is Lebesgue integrable function on $ [\xi, \delta] $ with $ 0 < \xi < \delta $ and $ \xi, \delta\in {\bf S} $. If $ \varphi\in{\bf HXP}_{m_1}\left({\bf S}, {\mathbb{R}}^{+}\right) $, $ \upsilon\in{\bf HXP}_{m_2}\left({\bf S}, {\mathbb{R}}^{+}\right) $ and $ \zeta, \epsilon > 0 $, then
$ \begin{equation*} \label{22} \begin{aligned} &\left(\frac{ \xi \delta}{ \delta- \xi}\right)^{\frac{ \epsilon}{ \zeta}}\left[_{ \zeta}{\mathcal{J}}_{\frac{1}{ \delta}^+}^{ \epsilon}\big( \varphi \upsilon\circ \tilde{ \varphi}\big)\left(\frac{1}{ \xi}\right)+_{ \zeta}{\mathcal{J}}_{\frac{1}{ \xi}^-}^{ \epsilon}\big( \varphi \upsilon\circ \tilde{ \varphi}\big)\left(\frac{1}{ \delta}\right)\right]\\ &\leq\frac{\mathcal{D}( \xi, \delta)}{ \zeta \Gamma_ \zeta( \epsilon)}\int_0^1 \varsigma^{\frac{ \epsilon}{ \zeta}-1}\Big[\Delta _1( \varsigma)+\Delta _4( \varsigma)\Big]\, d \varsigma+\frac{\mathcal{F}( \xi, \delta)}{ \zeta \Gamma_ \zeta( \epsilon)}\int_0^1 \varsigma^{\frac{ \epsilon}{ \zeta}-1}\Big[\Delta _2( \varsigma)+\Delta _3( \varsigma)\Big]\, d \varsigma, \end{aligned} \end{equation*} $ |
where $ \mathcal{D}(\xi, \delta): = \varphi(\xi) \upsilon(\xi)+ \varphi(\delta) \upsilon(\delta) $, $ \mathcal{F}(\xi, \delta): = \varphi(\xi) \upsilon(\delta)+ \varphi(\delta) \upsilon(\xi) $, $ \tilde{ \varphi} $ is as defined in Theorem 6, and $ \Delta _j(\varsigma) $, $ j = \overline{1, 4} $, as defined in Theorem 5.
Proof. Given that $ \varphi\in{\bf HXP}_{m_1}\left({\bf S}, {\mathbb{R}}^{+}\right) $ and $ \upsilon\in{\bf HXP}_{m_2}\left({\bf S}, {\mathbb{R}}^{+}\right) $, we get:
$ \begin{equation*} \label{DDD1} \varphi\left(\frac{ \xi \delta}{ \varsigma \xi+(1- \varsigma) \delta}\right)\leq\frac{1}{m_1}\sum\limits_{\vartheta = 1}^{m_1}\left[1-(1- \varsigma)^\vartheta\right] \varphi( \xi)+\frac{1}{m_1}\sum\limits_{\vartheta = 1}^{m_1}\left[1-{ \varsigma}^\vartheta\right] \varphi( \delta) \end{equation*} $ |
and
$ \begin{equation*} \label{DDD2} \upsilon\left(\frac{ \xi \delta}{ \varsigma \xi+(1- \varsigma) \delta}\right)\leq\frac{1}{m_1}\sum\limits_{\vartheta = 1}^{m_1}\left[1-(1- \varsigma)^\vartheta\right] \upsilon( \xi)+\frac{1}{m_1}\sum\limits_{\vartheta = 1}^{m_1}\left[1-{ \varsigma}^\vartheta\right] \upsilon( \delta). \end{equation*} $ |
This implies:
$ \begin{equation*} \begin{aligned} & \varphi\left(\frac{ \xi \delta}{ \varsigma \xi+(1- \varsigma) \delta}\right) \upsilon\left(\frac{ \xi \delta}{ \varsigma \xi+(1- \varsigma) \delta}\right)\\ &\leq \frac{1}{m_1}\frac{1}{m_2}\sum\limits_{\vartheta = 1}^{m_1}\left[1-(1- \varsigma)^\vartheta\right]\sum\limits_{\vartheta = 1}^{m_2}\left[1-(1- \varsigma)^\vartheta\right] \varphi( \xi) \upsilon( \xi)\\ &\quad+\frac{1}{m_1}\frac{1}{m_2}\sum\limits_{\vartheta = 1}^{m_1}\left[1-(1- \varsigma)^\vartheta\right]\sum\limits_{\vartheta = 1}^{m_2}\left[1-{ \varsigma}^\vartheta\right] \varphi( \xi) \upsilon( \delta)\\ &\quad+ \frac{1}{m_1}\frac{1}{m_2}\sum\limits_{\vartheta = 1}^{m_1}\left[1- \varsigma^\vartheta\right]\sum\limits_{\vartheta = 1}^{m_2}\left[1-(1- \varsigma)^\vartheta\right] \varphi( \delta) \upsilon( \xi)\\ &\quad+\frac{1}{m_1}\frac{1}{m_2}\sum\limits_{\vartheta = 1}^{m_1}\left[1- \varsigma^\vartheta\right]\sum\limits_{\vartheta = 1}^{m_2}\left[1-{ \varsigma}^\vartheta\right] \varphi( \delta) \upsilon( \delta)\\ &: = \Delta _1( \varsigma) \varphi( \xi) \upsilon( \xi)+\Delta _2( \varsigma) \varphi( \xi) \upsilon( \delta)+ \Delta _3( \varsigma) \varphi( \delta) \upsilon( \xi)+\Delta _4( \varsigma) \varphi( \delta) \upsilon( \delta). \end{aligned} \end{equation*} $ |
This gives:
$ \begin{equation} \begin{aligned} & \varphi\left(\frac{ \xi \delta}{ \varsigma \xi+(1- \varsigma) \delta}\right) \upsilon\left(\frac{ \xi \delta}{ \varsigma \xi+(1- \varsigma) \delta}\right)\\[1ex] &\leq\Delta _1( \varsigma) \varphi( \xi) \upsilon( \xi)+\Delta _2( \varsigma) \varphi( \xi) \upsilon( \delta)+ \Delta _3( \varsigma) \varphi( \delta) \upsilon( \xi)+\Delta _4( \varsigma) \varphi( \delta) \upsilon( \delta). \end{aligned} \end{equation} $ | (2.17) |
Similarly, we also have
$ \begin{equation} \begin{aligned} & \varphi\left(\frac{ \xi \delta}{ \varsigma \delta+(1- \varsigma) \xi}\right) \upsilon\left(\frac{ \xi \delta}{ \varsigma \delta+(1- \varsigma) \xi}\right)\\[1ex] &\leq\Delta _4( \varsigma) \varphi( \xi) \upsilon( \xi)+\Delta _3( \varsigma) \varphi( \xi) \upsilon( \delta)+ \Delta _2( \varsigma) \varphi( \delta) \upsilon( \xi)+\Delta _1( \varsigma) \varphi( \delta) \upsilon( \delta). \end{aligned} \end{equation} $ | (2.18) |
Adding (2.17) and (2.18), we get
$ \begin{equation} \begin{aligned} & \varphi\left(\frac{ \xi \delta}{ \varsigma \xi+(1- \varsigma) \delta}\right) \upsilon\left(\frac{ \xi \delta}{ \varsigma \xi+(1- \varsigma) \delta}\right)+ \varphi\left(\frac{ \xi \delta}{ \varsigma \delta+(1- \varsigma) \xi}\right) \upsilon\left(\frac{ \xi \delta}{ \varsigma \delta+(1- \varsigma) \xi}\right)\\[1ex] &\leq\Big( \varphi( \xi) \upsilon( \xi)+ \varphi( \delta) \upsilon( \delta)\Big)\Big[\Delta _1( \varsigma)+\Delta _4( \varsigma)\Big]\\[1ex] &\quad\quad\quad+\Big( \varphi( \xi) \upsilon( \delta)+ \varphi( \delta) \upsilon( \xi)\Big)\Big[\Delta _2( \varsigma)+\Delta _3( \varsigma)\Big]. \end{aligned} \end{equation} $ | (2.19) |
Now, multiplying both sides of (2.19) by $ \varsigma^{\frac{ \epsilon}{ \zeta}-1} $ and integrating with respect to $ \varsigma $ over $ [0, 1] $, gives:
$ \begin{equation*} \label{G14} \begin{aligned} & \zeta \Gamma_ \zeta( \epsilon)\left(\frac{ \xi \delta}{ \delta- \xi}\right)^{\frac{ \epsilon}{ \zeta}}\left[_{ \zeta}{\mathcal{J}}_{\frac{1}{ \delta}^+}^{ \epsilon}\big( \varphi \upsilon\circ \tilde{ \varphi}\big)\left(\frac{1}{ \xi}\right)+_{ \zeta}{\mathcal{J}}_{\frac{1}{ \xi}^-}^{ \epsilon}\big( \varphi \upsilon\circ \tilde{ \varphi}\big)\left(\frac{1}{ \delta}\right)\right]\\ & = \int_0^1 \varsigma^{\frac{ \epsilon}{ \zeta}-1} \varphi\left(\frac{ \xi \delta}{ \varsigma \xi+(1- \varsigma) \delta}\right) \upsilon\left(\frac{ \xi \delta}{ \varsigma \xi+(1- \varsigma) \delta}\right)\, d \varsigma\\[1ex] &+\int_0^1 \varsigma^{\frac{ \epsilon}{ \zeta}-1} \varphi\left(\frac{ \xi \delta}{ \varsigma \delta+(1- \varsigma) \xi}\right) \upsilon\left(\frac{ \xi \delta}{ \varsigma \delta+(1- \varsigma) \xi}\right)\, d \varsigma\\[1ex] &\leq\Big( \varphi( \xi) \upsilon( \xi)+ \varphi( \delta) \upsilon( \delta)\Big)\int_0^1 \varsigma^{\frac{ \epsilon}{ \zeta}-1}\Big[\Delta _1( \varsigma)+\Delta _4( \varsigma)\Big]\, d \varsigma\\[1ex] &\quad\quad\quad+\Big( \varphi( \xi) \upsilon( \delta)+ \varphi( \delta) \upsilon( \xi)\Big)\int_0^1 \varsigma^{\frac{ \epsilon}{ \zeta}-1}\Big[\Delta _2( \varsigma)+\Delta _3( \varsigma)\Big]\, d \varsigma\\ &: = \mathcal{D}( \xi, \delta)\int_0^1 \varsigma^{\frac{ \epsilon}{ \zeta}-1}\Big[\Delta _1( \varsigma)+\Delta _4( \varsigma)\Big]\, d \varsigma+\mathcal{F}( \xi, \delta)\int_0^1 \varsigma^{\frac{ \epsilon}{ \zeta}-1}\Big[\Delta _2( \varsigma)+\Delta _3( \varsigma)\Big]. \end{aligned} \end{equation*} $ |
Hence, this completes the proof.
Corollary 2. Let $ \varphi, \upsilon:{\bf S}\to {\mathbb{R}}^{+} $ be two functions such that $ \varphi \upsilon $ is Lebesgue integrable function on $ [\xi, \delta] $ with $ 0 < \xi < \delta $ and $ \xi, \delta\in {\bf S} $. If $ \varphi $ and $ \upsilon $ are harmonically convex and $ \zeta, \epsilon > 0 $, then
$ \left(\frac{ \xi \delta}{ \delta- \xi}\right)^{\frac{ \epsilon}{ \zeta}}\left[_{ \zeta}{\mathcal{J}}_{\frac{1}{ \delta}^+}^{ \epsilon}\big( \varphi \upsilon\circ \tilde{ \varphi}\big)\left(\frac{1}{ \xi}\right)+_{ \zeta}{\mathcal{J}}_{\frac{1}{ \xi}^-}^{ \epsilon}\big( \varphi \upsilon\circ \tilde{ \varphi}\big)\left(\frac{1}{ \delta}\right)\right]\\ \leq\frac{\mathcal{D}( \xi, \delta)}{ \Gamma_ \zeta( \epsilon)}\left[\frac{1}{ \epsilon}+\frac{2}{ \epsilon+2 \zeta}-\frac{2}{ \epsilon+ \zeta}\right]+\frac{\mathcal{F}( \xi, \delta)}{ \Gamma_ \zeta( \epsilon)}\left[\frac{2}{ \epsilon+ \zeta}-\frac{2}{ \epsilon+2 \zeta}\right]. $ |
Proof. Let $ m_1 = m_2 = 1 $. Then, $ \Delta _1(\varsigma) = \varsigma^2 $, $ \Delta _2(\varsigma) = \Delta _3(\varsigma) = \varsigma- \varsigma^2 $ and $ \Delta _4(\varsigma) = 1-2 \varsigma+ \varsigma^2 $. The intended result follows by using Theorem 7.
Theorem 8. Let $ \varphi, \upsilon:{\bf S}\to {\mathbb{R}}^{+} $ be two functions such that $ \varphi \upsilon $ is Lebesgue integrable function on $ [\xi, \delta] $ with $ 0 < \xi < \delta $ and $ \xi, \delta\in {\bf S} $. If $ \varphi\in{\bf HXP}_{m_1}\left({\bf S}, {\mathbb{R}}^{+}\right) $, $ \upsilon\in{\bf HXP}_{m_2}\left({\bf S}, {\mathbb{R}}^{+}\right) $ and $ \zeta, \epsilon > 0 $, then
$ \begin{equation*} \begin{aligned} &\frac{m_1m_2}{(m_1+2^{-m_1}-1)(m_2+2^{-m_2}-1)}\, \varphi\left(\frac{2 \xi \delta}{ \xi+ \delta}\right) \upsilon\left(\frac{2 \xi \delta}{ \xi+ \delta}\right)\\[1ex] &\leq \Gamma_ \zeta( \epsilon+ \zeta)\left(\frac{ \xi \delta}{ \delta- \xi}\right)^{\frac{ \epsilon}{ \zeta}}\left[_{ \zeta}{\mathcal{J}}_{\frac{1}{ \delta}^+}^{ \epsilon}\big( \varphi \upsilon\circ \tilde{ \varphi}\big)\left(\frac{1}{ \xi}\right)+_{ \zeta}{\mathcal{J}}_{\frac{1}{ \xi}^-}^{ \epsilon}\big( \varphi \upsilon\circ \tilde{ \varphi}\big)\left(\frac{1}{ \delta}\right)\right]\\ &\quad\quad\quad+\frac{ \epsilon}{ \zeta}\int_0^1 \varsigma^{\frac{ \epsilon}{ \zeta}-1}\left\{\left[\Lambda_{m_1}( \varsigma)\tilde{\Lambda}_{m_2}( \varsigma)+\tilde{\Lambda}_{m_1}( \varsigma)\Lambda_{m_2}( \varsigma)\right] \mathcal{D}( \xi, \delta)\right.\\ &\quad\quad\quad\quad\quad\left.+\left[\Lambda_{m_1}( \varsigma)\Lambda_{m_2}( \varsigma)+\tilde{\Lambda}_{m_1}( \varsigma)\tilde{\Lambda}_{m_2}( \varsigma)\right] \mathcal{F}( \xi, \delta)\right\}\, d \varsigma, \end{aligned} \end{equation*} $ |
where $ \tilde{ \varphi} $ is defined in Theorem 6, $ \Lambda_{m}(\varsigma) = \frac{1}{m}\sum_{\vartheta = 1}^m\left[1-(1- \varsigma)^\vartheta\right] $ and $ \tilde{\Lambda}_{m}(\varsigma) = \frac{1}{m}\sum_{\vartheta = 1}^m\left[1- \varsigma^\vartheta\right] $.
Proof. We start by noticing that:
$ \tilde{\Lambda}_{m}\left(\frac{1}{2}\right) = \Lambda_{m}\left(\frac{1}{2}\right): = E_m: = \frac{m+2^{-m}-1}{m}. $ |
Now, let $ \varsigma\in[0, 1] $. Hence, from (2.11), one gets:
$ \begin{equation*} \varphi\left(\frac{2 \xi \delta}{ \xi+ \delta}\right)\leq E_{m_1}\left\{ \varphi\left(\frac{ \xi \delta}{ \varsigma \xi+(1- \varsigma) \delta}\right)+ \varphi\left(\frac{ \xi \delta}{ \varsigma \delta+(1- \varsigma) \xi}\right)\right\} \end{equation*} $ |
and
$ \begin{equation*} \upsilon\left(\frac{2 \xi \delta}{ \xi+ \delta}\right)\leq E_{m_1}\left\{ \upsilon\left(\frac{ \xi \delta}{ \varsigma \xi+(1- \varsigma) \delta}\right)+ \upsilon\left(\frac{ \xi \delta}{ \varsigma \delta+(1- \varsigma) \xi}\right)\right\}. \end{equation*} $ |
Now,
$ \begin{equation*} \begin{aligned} & \varphi\left(\frac{2 \xi \delta}{ \xi+ \delta}\right) \upsilon\left(\frac{2 \xi \delta}{ \xi+ \delta}\right)\\ &\leq E_{m_1}E_{m_2}\left[ \varphi\left(\frac{ \xi \delta}{ \varsigma \xi+(1- \varsigma) \delta}\right) \upsilon\left(\frac{ \xi \delta}{ \varsigma \xi+(1- \varsigma) \delta}\right)\right.\\ &\quad+\left. \varphi\left(\frac{ \xi \delta}{ \varsigma \delta+(1- \varsigma) \xi}\right) \upsilon\left(\frac{ \xi \delta}{ \varsigma \delta+(1- \varsigma) \xi}\right)\right]\\ &\quad+E_{m_1}E_{m_2}\left[ \varphi\left(\frac{ \xi \delta}{ \varsigma \xi+(1- \varsigma) \delta}\right) \upsilon\left(\frac{ \xi \delta}{ \varsigma \delta+(1- \varsigma) \xi}\right)\right.\\ &\quad+\left. \varphi\left(\frac{ \xi \delta}{ \varsigma \delta+(1- \varsigma) \xi}\right) \upsilon\left(\frac{ \xi \delta}{ \varsigma \xi+(1- \varsigma) \delta}\right)\right]\\ &\leq E_{m_1}E_{m_2}\left[ \varphi\left(\frac{ \xi \delta}{ \varsigma \xi+(1- \varsigma) \delta}\right) \upsilon\left(\frac{ \xi \delta}{ \varsigma \xi+(1- \varsigma) \delta}\right)\right.\\ &\quad+\left. \varphi\left(\frac{ \xi \delta}{ \varsigma \delta+(1- \varsigma) \xi}\right) \upsilon\left(\frac{ \xi \delta}{ \varsigma \delta+(1- \varsigma) \xi}\right)\right]\\ &\quad+E_{m_1}E_{m_2}\left\{\left[\Lambda_{m_1}( \varsigma) \varphi( \xi)+\tilde{\Lambda}_{m_1}( \varsigma) \varphi( \delta)\right]\left[\Lambda_{m_2}( \varsigma) \upsilon( \delta)+\tilde{\Lambda}_{m_2}( \varsigma) \upsilon( \xi)\right]\right.\\ &\quad\quad\quad\quad\left.+\left[\Lambda_{m_1}( \varsigma) \varphi( \delta)+\tilde{\Lambda}_{m_1}( \varsigma) \varphi( \xi)\right]\left[\Lambda_{m_2}( \varsigma) \upsilon( \xi)+\tilde{\Lambda}_{m_2}( \varsigma) \upsilon( \delta)\right]\right\}\\ & = E_{m_1}E_{m_2}\left[ \varphi\left(\frac{ \xi \delta}{ \varsigma \xi+(1- \varsigma) \delta}\right) \upsilon\left(\frac{ \xi \delta}{ \varsigma \xi+(1- \varsigma) \delta}\right)\right.\\ &\quad+\left. \varphi\left(\frac{ \xi \delta}{ \varsigma \delta+(1- \varsigma) \xi}\right) \upsilon\left(\frac{ \xi \delta}{ \varsigma \delta+(1- \varsigma) \xi}\right)\right]\\ &\quad+E_{m_1}E_{m_2}\left\{\left[\Lambda_{m_1}( \varsigma)\tilde{\Lambda}_{m_2}( \varsigma)+\tilde{\Lambda}_{m_1}( \varsigma)\Lambda_{m_2}( \varsigma)\right]\left[ \varphi( \xi) \upsilon( \xi)+ \varphi( \delta) \upsilon( \delta)\right]\right.\\ &\quad\quad\quad\quad\left.+\left[\Lambda_{m_1}( \varsigma)\Lambda_{m_2}( \varsigma)+\tilde{\Lambda}_{m_1}( \varsigma)\tilde{\Lambda}_{m_2}( \varsigma)\right]\left[ \varphi( \xi) \upsilon( \delta)+ \varphi( \delta) \upsilon( \xi)\right]\right\}\\ & = E_{m_1}E_{m_2}\left[ \varphi\left(\frac{ \xi \delta}{ \varsigma \xi+(1- \varsigma) \delta}\right) \upsilon\left(\frac{ \xi \delta}{ \varsigma \xi+(1- \varsigma) \delta}\right)\right.\\ &\quad+\left. \varphi\left(\frac{ \xi \delta}{ \varsigma \delta+(1- \varsigma) \xi}\right) \upsilon\left(\frac{ \xi \delta}{ \varsigma \delta+(1- \varsigma) \xi}\right)\right]\\ &\quad+E_{m_1}E_{m_2}\left\{\left[\Lambda_{m_1}( \varsigma)\tilde{\Lambda}_{m_2}( \varsigma)+\tilde{\Lambda}_{m_1}( \varsigma)\Lambda_{m_2}( \varsigma)\right] \mathcal{D}( \xi, \delta)\right.\\ &\quad\quad\quad\quad\left.+\left[\Lambda_{m_1}( \varsigma)\Lambda_{m_2}( \varsigma)+\tilde{\Lambda}_{m_1}( \varsigma)\tilde{\Lambda}_{m_2}( \varsigma)\right] \mathcal{F}( \xi, \delta)\right\}. \end{aligned} \end{equation*} $ |
This implies that
$ \begin{equation} \begin{aligned} & \varphi\left(\frac{2 \xi \delta}{ \xi+ \delta}\right) \upsilon\left(\frac{2 \xi \delta}{ \xi+ \delta}\right)\\ &\leq E_{m_1}E_{m_2}\left[ \varphi\left(\frac{ \xi \delta}{ \varsigma \xi+(1- \varsigma) \delta}\right) \upsilon\left(\frac{ \xi \delta}{ \varsigma \xi+(1- \varsigma) \delta}\right)\right.\\ &\quad+\left. \varphi\left(\frac{ \xi \delta}{ \varsigma \delta+(1- \varsigma) \xi}\right) \upsilon\left(\frac{ \xi \delta}{ \varsigma \delta+(1- \varsigma) \xi}\right)\right]\\ &\quad+E_{m_1}E_{m_2}\left\{\left[\Lambda_{m_1}( \varsigma)\tilde{\Lambda}_{m_2}( \varsigma)+\tilde{\Lambda}_{m_1}( \varsigma)\Lambda_{m_2}( \varsigma)\right] \mathcal{D}( \xi, \delta)\right.\\ &\quad\quad\quad\quad\left.+\left[\Lambda_{m_1}( \varsigma)\Lambda_{m_2}( \varsigma)+\tilde{\Lambda}_{m_1}( \varsigma)\tilde{\Lambda}_{m_2}( \varsigma)\right] \mathcal{F}( \xi, \delta)\right\}. \end{aligned} \end{equation} $ | (2.20) |
Multiplying both sides of (2.20) by $ \varsigma^{\frac{ \epsilon}{ \zeta}-1} $ and integrating with respect to $ \varsigma $ over $ [0, 1] $ to get:
$ \begin{equation*} \label{CC2} \begin{aligned} &\frac{ \zeta}{ \epsilon} \varphi\left(\frac{2 \xi \delta}{ \xi+ \delta}\right) \upsilon\left(\frac{2 \xi \delta}{ \xi+ \delta}\right)\\[1ex] & = \int_0^1 \varsigma^{\frac{ \epsilon}{ \zeta}-1} \varphi\left(\frac{2 \xi \delta}{ \xi+ \delta}\right) \upsilon\left(\frac{2 \xi \delta}{ \xi+ \delta}\right)\, d \varsigma\\ &\leq E_{m_1}E_{m_2}\int_0^1 \varsigma^{\frac{ \epsilon}{ \zeta}-1}\left[ \varphi\left(\frac{ \xi \delta}{ \varsigma \xi+(1- \varsigma) \delta}\right) \upsilon\left(\frac{ \xi \delta}{ \varsigma \xi+(1- \varsigma) \delta}\right)\right.\\ &\quad+\left. \varphi\left(\frac{ \xi \delta}{ \varsigma \delta+(1- \varsigma) \xi}\right) \upsilon\left(\frac{ \xi \delta}{ \varsigma \delta+(1- \varsigma) \xi}\right)\right]\, d \varsigma\\ &\quad+E_{m_1}E_{m_2}\int_0^1 \varsigma^{\frac{ \epsilon}{ \zeta}-1}\left\{\left[\Lambda_{m_1}( \varsigma)\tilde{\Lambda}_{m_2}( \varsigma)+\tilde{\Lambda}_{m_1}( \varsigma)\Lambda_{m_2}( \varsigma)\right] \mathcal{D}( \xi, \delta)\right.\\ &\quad\quad\quad\quad\left.+\left[\Lambda_{m_1}( \varsigma)\Lambda_{m_2}( \varsigma)+\tilde{\Lambda}_{m_1}( \varsigma)\tilde{\Lambda}_{m_2}( \varsigma)\right] \mathcal{F}( \xi, \delta)\right\}\, d \varsigma\\ & = E_{m_1}E_{m_2}\left\{ \zeta \Gamma_ \zeta( \epsilon)\left(\frac{ \xi \delta}{ \delta- \xi}\right)^{\frac{ \epsilon}{ \zeta}}\left[_{ \zeta}{\mathcal{J}}_{\frac{1}{ \delta}^+}^{ \epsilon}\big( \varphi \upsilon\circ \tilde{ \varphi}\big)\left(\frac{1}{ \xi}\right)+_{ \zeta}{\mathcal{J}}_{\frac{1}{ \xi}^-}^{ \epsilon}\big( \varphi \upsilon\circ \tilde{ \varphi}\big)\left(\frac{1}{ \delta}\right)\right]\right\}\\ &\quad\quad\quad+E_{m_1}E_{m_2}\int_0^1 \varsigma^{\frac{ \epsilon}{ \zeta}-1}\left\{\left[\Lambda_{m_1}( \varsigma)\tilde{\Lambda}_{m_2}( \varsigma)+\tilde{\Lambda}_{m_1}( \varsigma)\Lambda_{m_2}( \varsigma)\right] \mathcal{D}( \xi, \delta)\right.\\ &\quad\quad\quad\quad\quad\left.+\left[\Lambda_{m_1}( \varsigma)\Lambda_{m_2}( \varsigma)+\tilde{\Lambda}_{m_1}( \varsigma)\tilde{\Lambda}_{m_2}( \varsigma)\right] \mathcal{F}( \xi, \delta)\right\}\, d \varsigma. \end{aligned} \end{equation*} $ |
The required result follows.
Corollary 3. Let $ \varphi, \upsilon:{\bf S}\to {\mathbb{R}}^{+} $ be two functions such that $ \varphi \upsilon $ is Lebesgue integrable function on $ [\xi, \delta] $ with $ 0 < \xi < \delta $ and $ \xi, \delta\in {\bf S} $. If $ \varphi $ and $ \upsilon $ are harmonically convex and $ \zeta, \epsilon > 0 $, then
$ \begin{equation*} \label{23} \begin{aligned} &\, \varphi\left(\frac{2 \xi \delta}{ \xi+ \delta}\right) \upsilon\left(\frac{2 \xi \delta}{ \xi+ \delta}\right)\\[1ex] &\leq \frac{ \Gamma_ \zeta( \epsilon+ \zeta)}{4}\left(\frac{ \xi \delta}{ \delta- \xi}\right)^{\frac{ \epsilon}{ \zeta}}\left[_{ \zeta}{\mathcal{J}}_{\frac{1}{ \delta}^+}^{ \epsilon}\big( \varphi \upsilon\circ \tilde{ \varphi}\big)\left(\frac{1}{ \xi}\right)+_{ \zeta}{\mathcal{J}}_{\frac{1}{ \xi}^-}^{ \epsilon}\big( \varphi \upsilon\circ \tilde{ \varphi}\big)\left(\frac{1}{ \delta}\right)\right]\\ &\quad\quad\quad+\frac{1}{2}\left[\frac{ \epsilon}{ \epsilon+ \zeta}-\frac{ \epsilon}{ \epsilon+2 \zeta}\right]\mathcal{D}( \xi, \delta)+ \frac{1}{4}\left[1+\frac{2 \epsilon}{ \epsilon+2 \zeta}-\frac{2 \epsilon}{ \epsilon+ \zeta}\right]\mathcal{F}( \xi, \delta). \end{aligned} \end{equation*} $ |
Proof. Let $ m_1 = m_2 = 1 $. Then, $ \Lambda_{m_1}(\varsigma) = \Lambda_{m_2}(\varsigma) = \varsigma $ and $ \tilde{\Lambda}_{m_1}(\varsigma) = \tilde{\Lambda}_{m_2}(\varsigma) = 1- \varsigma $. The intended result follows by using Theorem 8.
Utilizing the Caputo–Fabrizio and generalized Riemann–Liouville fractional integral operators, we proved some inequalities of the Hermite–Hadamard kinds for $ m $-polynomial convex and harmonically convex functions. Our results generalize, extend and complement results in [7,9,31].
The authors declare that there are no conflicts of interest regarding the publication of this paper.
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