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In this work we obtain integral inequalities of the Hermite-Hadamard type, using weighted integral, via \((h,m)\)-convex modified functions of second type. In the work we show that several results known from the literature can be derived from ours as particular cases.
The concept of convex function is a notion that is among the most well-known functional classes in Mathematical Sciences today. This is due to their properties, geometric characteristics, interpretation and the variety of areas where they are used and applied.
The essence of its definition is the relationship and comparison between two midpoints, that of the interval \([a,b]\) and that of the interval \([f(a),f(b)]\). Below we remember this definition.
Definition 1. A function \(f: I \rightarrow \mathbb{R}\) is said to be convex, if the inequality \[\psi \left( \tau \xi +(1-\tau )\varsigma \right) \leq \tau \psi (\xi)+(1-\tau )\psi \left(\varsigma\right)\] holds for all \(\xi ,\varsigma \in I\) and \(\tau \in [0,1]\).
The interest in this notion has resulted in a large number of extensions and ramifications that make it practically impossible to follow. Interested readers can consult [1] to get a more precise idea of this development.
One of the most important inequalities, for convex functions, is the very well known Hermite–Hadamard inequality, that is the inequality \[\begin{aligned} \label{Hadamard} \psi \left( \frac { a+b }{ 2 } \right) \leq \frac { 1 }{ b-a } \int _{a}^{b}\psi (x)dx\leq \frac { \psi (a)+\psi (b) }{ 2 } \end{aligned}\tag{1}\] holds for any function \(\psi\) convex on the interval \([a,b]\). This inequality was published by Hermite ([2]) in 1883 and, independently, by Hadamard in 1893 ([3]), it gives an estimation of the mean value of a convex function. Various extensions and generalizations can be consulted in [4-19] and references therein.
In [7], the following definitions were presented:
Definition 2. Let \(h:[0,1]\rightarrow \mathbb{R}\) be a nonnegative function, \(h\neq 0\) and \(\psi :I=[0,+\infty )\rightarrow \lbrack 0,+\infty )\). If inequality \[\psi \left( \tau \xi +m(1-\tau )\varsigma \right) \leq h^{s}(\tau )\psi (\xi)+m(1-h^{s}(\tau ))\psi \left(\varsigma\right) \label{e:hmc1}\] holds for all \(\xi ,\varsigma \in I\) and \(\tau \in \lbrack 0,1]\), where \(m\in \lbrack 0,1]\), \(s\in \lbrack -1,1]\). Then a function \(\psi\) is called a \((h,m)\)-convex modified of the first type on \(I\).
Definition 3. Let \(h:[0,1]\rightarrow \mathbb{R}\) nonnegative functions, \(h\neq 0\) and \(\psi :I=[0,+\infty )\rightarrow \lbrack 0,+\infty )\). If inequality \[\psi \left( \tau \xi +m(1-\tau )\varsigma \right) \leq h^{s}(\tau )\psi (\xi)+m(1-h(\tau ))^{s}\psi\left(\varsigma\right) \label{e:hmc11}\] holds for all \(\xi ,\varsigma \in I\) and \(\tau \in \lbrack 0,1]\), where \(m\in \lbrack 0,1]\), \(s\in \lbrack -1,1]\). Then a function \(\psi\) is called a \((h,m)\)-convex modified of the second type on \(I\).
Remark 1. From Definitions 2 and 3, we can define \(N_{h,m}^{s}[a,b]\), where \(a,b\in \lbrack 0,+\infty )\), as the set of functions \((h,m)\)-convex modified, for which \({\psi }(a)\geq 0\), characterized by the triple \((h(\tau ),m,s)\). Note that if:
\((h(\tau ),0,0)\) we have the increasing functions ([20]).
\((\tau ,0,s)\) we have the \(s-\)starshaped functions ([20]).
\((\tau ,0,1)\) we have the starshaped functions ([20]).
\((\tau ,1,1)\) then \(\psi\) is a convex function on \([0,+\infty )\) ([20]).
\((1,1,s)\) then \(\psi\) is a P-convex function on \([0,+\infty )\) ([21]).
\((\tau ,m,1)\) then \(\psi\) is a \(m-\)convex function on \([0,+\infty )\) ([22]).
\((\tau ,1,s)\) \(s\in (0,1]\) then \(\psi\) is a \(s-\)convex function on \([0,+\infty )\) ([23,24]).
\((\tau ,1,s)\) \(s\in [-1,1]\) then \(\psi\) is a \(s-\)convex extended function on \([0,+\infty )\).
\((\tau ,m,s)\) \(s\in (0,1]\) then \(\psi\) is a \((s,m)\)-convex extended function on \([0,+\infty )\) ([25]).
\((\tau^{a},1,s)\) with \(a\in (0,1]\), then \(\psi\) is a \((a,s)\)-convex function on \([0,+\infty )\) ([26]).
\((\tau^{a},m,1)\) with \(a\in (0,1]\), then \(\psi\) is a \((a,m)\)-convex function on \([0,+\infty )\) ([27]).
\((\tau^{a},m,s)\) with \(a\in (0,1]\), then \(\psi\) is a \(s-(a,m)\)-convex function on \([0,+\infty )\) ([28]).
\((h(\tau ),m,1)\) then we have a variant of the \((h,m)-\)convex function on \([0,+\infty )\) ([29]).
Next we present the weighted integral operators, which will be the basis of our work ([30,31]).
Definition 4. Let \(\phi \in L\left( [a,b]\right)\) and let \(w\) be a continuous and positive function, \(w:[0,1]\rightarrow [0,+\infty )\), with second order derivatives integrable on \(I\). Then the weighted fractional integrals are defined by (right and left respectively): \[\!J_{a+}^{w}\phi (r)=\int_{a}^{r}w^{\prime \prime }\left( \frac{r-\sigma }{ b-a}\right) \phi (\sigma )d\sigma ,\text{ \ \ }r>a\] and \[\!J_{b-}^{w}\phi (r)=\int_{r}^{b}w^{\prime \prime }\left( \frac{\sigma -r}{% b-a}\right) \phi (\sigma )d\sigma ,\text{ \ \ }r<b.\]
Remark 2. To have a clearer idea of the amplitude of the Definition 4, let’s consider some particular cases of the kernel \(w''\):
Putting \(w''(t)\equiv \ 1\), we obtain the classical Riemann integral.
If \(w''(t)=\frac{t^{(\alpha – 1)}}{\Gamma(\alpha) }\), then we obtain the Riemann-Liouville fractional integral right, and left can be obtained similarly.
With convenient kernel choices \(w''\) we can get the \(k\)-Riemann-Liouville fractional integral right and left of ([32]), the right-sided fractional integrals of a function \({\psi }\) with respect to another function \(h\) on \([a,b]\), the right and left integral operator of [33], the right and left sided generalized fractional integral operators of [34] and the integral operators of [35] and [36], can also be obtained from above Definition by imposing similar conditions to \(w^{\prime \prime }\).
Of course there are other known integral operators, fractional or not, that can be obtained as particular cases of the previous one, but we leave it to interested readers (see [37],[38]).
In this work, we present new results related to the Hermite-Hadamard Inequality, for \((h,m)\)-convex functions modified of second type, in the framework of Weighted Integral of Definition 4. We will see that several well-known publications in the literature are particular cases of those obtained by us.
As a basis for proving our main results, we need the following equality.
Lemma 1. Let \(f:I\subset \mathbb{R}\rightarrow \mathbb{R}\) be a differentiable mapping on \(I^{\circ}\) where \(a,b \in I\) with \(a<b\). If \(f'' \in L[a,b]\) and \(w\) has second derivative integrable on \(I\), then the followng equality holds: \[\begin{aligned} \label{e:1}\nonumber &\frac{(b-a)^2}{4}\left[w(0)\left(f^{\prime}(b)-f^{\prime}(a)\right)\right] \\\nonumber &\quad+ \left[w^{\prime}(1)f\left(\frac{a+b}{2}\right)+w^{\prime}(0)\frac{(f(b)+f(a))}{2}\right]\\ \nonumber &\quad-\frac{1}{b-a}\left(J^{w}_{a+}f\left(\frac{a+b}{2}\right)+J^{w}_{b^{-}}f\left(\frac{a+b}{2}\right)\right)\\ \nonumber &=\frac{(b-a)^2}{8}\biggl[ \int_{0}^{1} w(t)f''\left(\frac{a+b}{2}t+(1-t)a\right)dt \\ &\quad+ \int_{0}^{1} w(1-t)f''\left(bt+(1-t)\frac{a+b}{2}\right)dt \biggr].\ \end{aligned}\]
Proof. Let’s call \(I=I_1+I_2\), so we have \[\begin{aligned} I_{1}=& \int_{0}^{1} w(t)f''\left(\frac{a+b}{2}t+(1-t)a\right)dt \\ =& \frac{2}{b-a} w(t)f^{\prime}\left(\frac{a+b}{2}t+(1-t)a\right) |^{1}_{0}\\ & – \frac{2}{b-a}\int_{0}^{1}w^{\prime}(t)f^{\prime}\left(\frac{a+b}{2}t+(1-t)a\right)dt \\ =&\frac{2}{b-a}\left[w(1)f^{\prime}\left(\frac{a+b}{2}\right)-w(0)f^{\prime}(a)\right]\\ &- \frac{2}{b-a}\biggl[\frac{2}{b-a}w^{\prime}(t)f\left(\frac{a+b}{2}t+(1-t)a\right)|^{1}_{0}\\ &- \frac{2}{b-a} \int_{0}^{1}w''(t)f\left(\frac{a+b}{2}t+(1-t)a\right) \biggr]\\ =&\frac{2}{b-a}\left[w(1)f^{\prime}\left(\frac{a+b}{2}\right)-w(0)f^{\prime}(a)\right] \\ &-\left(\frac{2}{b-a}\right)^{2} \biggl[w^{\prime}(1)f\left(\frac{a+b}{2}\right)-w^{\prime}(0)f(a)\\ &- \frac{2}{b-a} \int_{a}^{\frac{a+b}{2}} w''\left[\frac{u-a}{\frac{a+b}{2}-a}\right]f(u)du \biggr]. \end{aligned}\] Analogously \[\begin{aligned} I_{2}=& \int_{0}^{1} w(1-t)f''\left(bt+(1-t)\frac{a+b}{2}\right)dt \\ =& \frac{2}{b-a} w(1-t)f^{\prime}\left(bt+(1-t)\frac{a+b}{2}\right) |^{1}_{0} \\ &+ \frac{2}{b-a}\int_{0}^{1}w^{\prime}(1-t)f^{\prime}\left(bt+(1-t)\frac{a+b}{2}\right)dt \end{aligned}\]\[\begin{aligned} =& \frac{2}{b-a}\left[w(0)f^{\prime}(b)-w(1)f^{\prime}\left(\frac{a+b}{2}\right)\right] \\ &+ \frac{2}{b-a} \biggl[ \frac{2}{b-a}w^{\prime}(1-t)f^{\prime}\left(bt+(1-t)\frac{a+b}{2}\right)|^{1}_{0}\\ &+ \frac{2}{b-a}\int_{0}^{1} w''(1-t)f\left(bt+(1-t)\frac{a+b}{2}\right)dt \biggr] \\ =&\frac{2}{b-a}\left[w(0)f^{\prime}(b)-w(1)f^{\prime}\left(\frac{a+b}{2}\right)\right] \\ &+\left(\frac{2}{b-a}\right)^{2} \biggl[ w^{\prime}(0)f(b)-w^{\prime}(1)f\left(\frac{a+b}{2}\right)\\ &+\frac{2}{b-a}\int_{0}^{1} w''(1-t)f\left(bt+(1-t)\frac{a+b}{2}\right)dt \biggr]\\ =& \frac{2}{b-a}\left[w(0)f^{\prime}(b)-w(1)f^{\prime}\left(\frac{a+b}{2}\right)\right] \\ &+\left(\frac{2}{b-a}\right)^{2} \biggl[ w^{\prime}(0)f(b)-w^{\prime}(1)f\left(\frac{a+b}{2}\right)\\ &+\frac{2}{b-a}\int_{\frac{a+b}{2}}^{b}w''\left[\frac{b-u}{b-\frac{a+b}{2}}\right]f(u)du \biggr]. \end{aligned}\] Adding the results obtained for \(I_1\) and \(I_2\) we have \[\begin{aligned} I=&\frac{2}{b-a}\left[w(0)\left(f^{\prime}(b)-f^{\prime}(a)\right)\right] \\ &+ \left(\frac{2}{b-a}\right)^{2} \biggl\{ \biggl[ 2w^{\prime}(1)f\left(\frac{a+b}{2}\right)\\ &+ w^{\prime}(0)\left(f(b)-f(a)\right) \biggr] \\ &+\frac{2}{b-a}\left(J^{w}_{a+}f\left(\frac{a+b}{2}\right)+J^{w}_{b^{-}}f\left(\frac{a+b}{2}\right)\right) \biggr\}. \end{aligned}\] Which is the desired result. ◻
From Lemma 1 we have the following result (see Lemma 1 of [39]).
Remark 3. Putting \(w(t)=z^2\) in (5), we obtain
Corollary 1. Let \(f:I\subset \mathbb{R}\rightarrow \mathbb{R}\) be a differentiable mapping on \(I^{\circ}\) where \(a,b \in I\) with \(a<b\). If \(f'' \in L[a,b]\), then the following equality holds: \[\begin{aligned} & \frac{1}{b-a}\int_{a}^{b}f(u)du-f\left ( \frac{a+b}{2} \right )\\ &=\int_{0}^{1}t^2f''\left(\frac{a+b}{2}t+(1-t)a\right)dt\\ &\quad+\int_{0}^{1}(1-t)^2f''\left(bt+(1-t)\frac{a+b}{2}\right)dt.\ \end{aligned}\]
Proof. From \(I_1\) we have \[\begin{aligned} &\int_{0}^{1}t^2f''\left(\frac{a+b}{2}t+(1-t)a\right)dt\\ &=\frac{2}{b-a}f'\left ( \frac{a+b}{2} \right )-2\left ( \frac{2}{b-a} \right )^2f\left ( \frac{a+b}{2} \right )\\ &\quad +\frac{16}{(b-a)^3}\int_{a}^{\frac{a+b}{2}}f(u)du, \end{aligned}\] and from \(I_2\): \[\begin{aligned} & \int_{0}^{1}(1-t)^2f''\left(bt+(1-t)\frac{a+b}{2}\right)dt\\ &= -\frac{2}{b-a}f'\left ( \frac{a+b}{2} \right )-2\left ( \frac{2}{b-a} \right )^2f\left ( \frac{a+b}{2} \right )\\ &\quad +\frac{16}{(b-a)^3}\int_{a}^{\frac{a+b}{2}}f(u)du. \end{aligned}\] Adding these two results we obtain the searched result. ◻
A new result for fractional Riemann-Liouville integrals can be obtained from Lemma 1 as follows.
Remark 4. Putting \(w(t)=\frac{z^{\alpha+1}}{\Gamma(\alpha)}\) in (5) we obtain the following result:
Let \(f:I\subset \mathbb{R}\rightarrow \mathbb{R}\) be a differentiable mapping on \(I^{\circ}\) where \(a,b \in I\) with \(a<b\). If \(f'' \in L[a,b]\), then the followng equality holds: \[\begin{aligned} &-\frac{1+\alpha }{\Gamma (\alpha )}f\left ( \frac{a+b}{2} \right )+\frac{\alpha (1+\alpha)2^{\alpha -1} }{(b-a)^\alpha }\biggl( J_{a+}^{\alpha } f\left ( \frac{a+b}{2} \right )\\ &+J_{b-}^{\alpha } f\left ( \frac{a+b}{2} \right )\biggr)\\ &=\frac{(b-a)^2}{8\Gamma(\alpha)} \biggl[ \int_{0}^{1}t^{\alpha+1}f''\left(\frac{a+b}{2}t+(1-t)a\right)dt\\ &\quad+\int_{0}^{1}(1-t)^{\alpha+1}f''\left(bt+(1-t)\frac{a+b}{2}\right)dt \biggr]. \end{aligned}\]
Theorem 1. Let \(f:I\subset [0,d)\rightarrow \mathbb{R}\) b a differentiable mapping on \(I^{\circ}\), such that \(f'' \in L[a,b]\), where \(a,b \in I\) with \(0\leq a<b<\infty\) and \(w\) has second derivative integrable on \(I\). If \(\left| f'' \right|\) is \((h,m)\)-convex on \([a,b]\), for some fixed \(m \in (0,1]\), \(t \in [0,1]\), then the following inequality holds: \[\begin{aligned} \label{e:2}\nonumber \left|L \right| \leq& \frac{(b-a)^{2}}{8} \biggl\{ \int_{0}^{1}\biggl[ w(t)\left|f''\left(\frac{a+b}{2}\right)\right|\\ \nonumber &+ w(1-t)\left|f''\left(b\right)\right|\biggr]h^{s}(t)dt \\ \nonumber &+ m \int_{0}^{1}\left[w(t)\left|f''\left(\frac{a}{m}\right)\right|+w(1-t)\left|f''\left(\frac{a+b}{2m}\right)\right|\right]\\ &\cdot (1-h(t))^{s}dt \biggr\} \end{aligned}\] with \[\begin{aligned} L =& \frac{b-a}{4}\left[w(0)\left(f^{\prime}(b)-f^{\prime}(a)\right)\right]+\biggl[w^{\prime}(1)f\left(\frac{a+b}{2}\right)\\ &+ w^{\prime}(0)\frac{f(a)+f(b)}{2}\biggr]\\ &-\frac{1}{b-a}\left(J^{w}_{a+}f\left(\frac{a+b}{2}\right)+J^{w}_{b^{-}}f\left(\frac{a+b}{2}\right) \right).\ \end{aligned}\]
Proof. From (6) we have \[\label{e:3} \left| L \right| = \frac{(b-a)^{2}}{8}\left| I_{1}+I_{2} \right|\leq \frac{(b-a)^{2}}{8}\left(\left| I_{1} \right|+\left| I_{2} \right| \right),\] with \[\begin{aligned} I_{1} &= \int_{0}^{1} w(t)f''\left(\frac{a+b}{2}t+(1-t)a\right)dt,\\ I_{2} &= \int_{0}^{1} w(1-t)f''\left(tb+(1-t)\frac{a+b}{2}\right)dt.\ \end{aligned}\] So using the \((h,m)\)-convexity of \(\left| f'' \right|\) we have \[\begin{aligned} \label{e:4}\nonumber \left| I_{1} \right|=& \left| \int_{0}^{1} w(t)f''\left(\frac{a+b}{2}t+(1-t)a\right)dt \right| \\\nonumber \leq & \int_{0}^{1} w(t)\left|f''\left(\frac{a+b}{2}t+(1-t)a\right) \right|dt \\ \leq & \int_{0}^{1} w(t) \biggl(\left|f''\left(\frac{a+b}{2}\right)\right| h^{s}(t)\notag\\ &+ m \left|f''\left(\frac{a}{m}\right)\right|\left(1-h(t)\right)^{s}\biggr) dt. \end{aligned}\] Analogously \[\begin{aligned} \label{e:5} \left| I_{2} \right| \leq& \int_{0}^{1} w(1-t) \biggl(\left|f''\left(b\right)\right| h^{s}(t)\notag\\ &+m \left|f''\left(\frac{a+b}{2m}\right)\right|\left(1-h(t)\right)^{s}\biggr) dt. \ \end{aligned}\] Substituting 8 and 9 into 7 we obtain the desired result. ◻
Remark 5. With \(s=1\), and \(w(z)=z^2\) we have the Theorem 2.2 of [40], obviously under the above assumptions the Remark 2.3 of above paper is also valid.
Remark 6. With \(w(z)=\frac{z^{\alpha+1}}{\Gamma(\alpha)}\) we have the following result for Riemann-Liouville integrals:
Let \(f:I\subset \mathbb{R}\rightarrow \mathbb{R}\) be a differentiable mapping on \(I^{\circ}\) where \(a,b \in I\) with \(a<b\). If \(f'' \in L[a,b]\), then the followng equality holds: \[\begin{aligned} &\bigg|\frac{\alpha (1+\alpha)2^{\alpha -1} }{(b-a)^\alpha }\left ( J_{a+}^{\alpha } f\left ( \frac{a+b}{2} \right )+J_{b-}^{\alpha } f\left ( \frac{a+b}{2} \right )\right )\\ &\quad- \frac{1+\alpha }{\Gamma (\alpha )}f\left ( \frac{a+b}{2} \right ) \bigg| \\ \leq &\frac{(b-a)^2}{8\Gamma(\alpha)} \biggl[ \int_{0}^{1}\left [t^{\alpha+1}\left | f''\left ( \frac{a+b}{2} \right ) \right | +(1-t)^{\alpha+1}\left | f''\left ( b \right ) \right | \right ]\\ &\cdot h^s(t)dt\\ &+ m\int_{0}^{1}\left [t^{\alpha+1}\left | f''\left ( \frac{a}{m} \right ) \right | +(1-t)^{\alpha+1}\left | f''\left ( \frac{a+b}{2} \right ) \right | \right ]\\ &\cdot (1-h(t))^sdt\biggr] .\ \end{aligned}\]
The following result establishes a new refinement to the previous Theorem.
Theorem 2. Let \(f:I\subset [0,d)\rightarrow \mathbb{R}\) b a differentiable mapping on \(I^{\circ}\), such that \(f'' \in L[a,b]\), where \(a\), \(b \in I\) with \(0\leq a<b<\infty\) and \(w\) has second derivative integrable on \(I\). If \(\left| f'' \right|^{q}\) is \((h,m)\)-convex on \([a,b]\), for some fixed \(m \in (0,1]\), \(t \in [0,1]\) and \(q>1\) with \(\frac{1}{p}+\frac{1}{q}=1\), then the following inequality holds: \[\begin{aligned} \label{e:21}\nonumber \left|L \right|\leq& \frac{(b-a)^{2}}{8} \biggl\{ \left(\int_{0}^{1}w^{p}(t)dt\right)^{\frac{1}{p}} \biggl(\left|f''\left(\frac{a+b}{2}\right)\right|^{q} \\ \nonumber &\times \int_{0}^{1}h^{s}(t)dt+ m\left|f''\left(\frac{a}{m}\right)\right|^{q}\int_{0}^{1}(1-h(t))^{s}dt \biggr)^{\frac{1}{q}} \\ \nonumber &+ \left(\int_{0}^{1}w^{p}(1-t)dt\right)^{\frac{1}{p}} \biggl(\left|f''\left(b\right)\right|^{q}\int_{0}^{1}h^{s}(t)dt\\ &+ m\left|f''\left(\frac{a+b}{2m}\right)\right|^{q}\int_{0}^{1}(1-h(t))^{s}dt \biggr)^{\frac{1}{q}} \biggr\} \ \end{aligned}\] with \(L\) as before.
Proof. Using Hölder Inequality from \(I_1\) we have \[\begin{aligned} \label{e:A} \nonumber &\int_{0}^{1} w(t)\left|f''\left(\frac{a+b}{2}t+(1-t)a\right)\right|dt \\ \nonumber &\leq \left(\int_{0}^{1}w^{p}(t)dt\right)^{\frac{1}{p}}\left(\int_{0}^{1}\left|f''\left(\frac{a+b}{2}t+(1-t)a\right)\right|^{q}dt \right)^{\frac{1}{q}}\\ \nonumber &\leq \left(\int_{0}^{1}w^{p}(t)dt\right)^{\frac{1}{p}} \biggl(\left|f''\left(\frac{a+b}{2}\right)\right|^{q}\int_{0}^{1}h^{s}(t)dt\\ &\quad+ m\left|f''\left(\frac{a}{m}\right)\right|^{q}\int_{0}^{1}(1-h(t))^{s}dt \biggr)^{\frac{1}{q}}. \end{aligned}\] Analogously, from \(I_2\) \[\begin{aligned} \label{e:B} \nonumber &\int_{0}^{1} w(1-t)\left|f''\left(bt+(1-t)\frac{a+b}{2}\right)\right|dt \\ \nonumber &\leq \left(\int_{0}^{1}w^{p}(1-t)dt\right)^{\frac{1}{p}} \biggl(\left|f''\left(b\right)\right|^{q}\int_{0}^{1}h^{s}(t)dt\\ &\quad + m\left|f''\left(\frac{a+b}{2m}\right)\right|^{q}\int_{0}^{1}(1-h(t))^{s}dt \biggr)^{\frac{1}{q}}.\ \end{aligned}\] From 11 and 12 in Lemma 1 we arrive at the desired result. ◻
Remark 7. If \(s=1\) and \(w(t)=t^2\) we obtain the Theorem 2.4 of [04], and we say before, the Remark 2.5 of the cited paper still valid.
With \(w(z)=\frac{z^{\alpha+1}}{\Gamma(\alpha)}\) we obtain a new result for Riemann-Liouville integrals:
Corollary 2. Let \(f:I\subset \mathbb{R}\rightarrow \mathbb{R}\) be a differentiable mapping on \(I^{\circ}\) where \(a,b \in I\) with \(a<b\). If \(f'' \in L[a,b]\), then the followng equality holds: \[\begin{aligned} &\bigg|\frac{\alpha (1+\alpha)2^{\alpha -1} }{(b-a)^\alpha }\left ( J_{a+}^{\alpha } f\left ( \frac{a+b}{2} \right )+J_{b-}^{\alpha } f\left ( \frac{a+b}{2} \right )\right )\\ &\quad-\frac{1+\alpha }{\Gamma (\alpha )}f\left ( \frac{a+b}{2} \right ) \bigg|\\ &\leq \frac{(b-a)^2}{8\Gamma(\alpha)} \biggl[ \int_{0}^{1} \biggl[ t^{\alpha+1}\bigg| f''\left( \frac{a+b}{2} \right) \bigg|\\ &\quad+ (1-t)^{\alpha+1} \bigg| f''\left( b \right ) \bigg| \biggr]h^s(t)dt \\ &\quad+ m\int_{0}^{1}\biggl[t^{\alpha+1}\left | f''\left ( \frac{a}{m} \right ) \right| +(1-t)^{\alpha+1}\left| f''\left ( \frac{a+b}{2} \right ) \right | \biggr]\\ &\quad\cdot(1-h(t))^sdt \biggr]. \end{aligned}\]
Theorem 3. Let \(f:I\subset [0,d)\rightarrow \mathbb{R}\) b a differentiable mapping on \(I^{\circ}\), such that \(f'' \in L[a,b]\), where \(a\), \(b \in I\) with \(0\leq a<b<\infty\) and \(w\) has second derivative integrable on \(I\). If \(\left| f'' \right|^{q}\) is \((h,m)\)-convex on \([a,b]\), for some fixed \(m \in (0,1]\), \(t \in [0,1]\) and \(q \geq 1\), then the following inequality holds: \[\begin{aligned} \label{e:31}\nonumber \left|L \right| \leq& \frac{(b-a)^{2}}{8} \biggl\{\left(\int_{0}^{1}w(t)dt\right)^{1-\frac{1}{q}} \biggl(\left|f''\left(\frac{a+b}{2}\right)\right|^{q}\\ \nonumber &\times \int_{0}^{1}w(t)h^{s}(t)dt+ m\left|f''\left(\frac{a}{m}\right)\right|^{q}\int_{0}^{1}w(t)(1-h(t))^{s}dt \biggr)^{\frac{1}{q}} \\ \nonumber &+ \left(\int_{0}^{1}w(1-t)dt\right)^{1-\frac{1}{q}} \biggl(\left|f''\left(b\right)\right|^{q}\int_{0}^{1}w(1-t)h^{s}(t)dt \\ &+ m\left|f''\left(\frac{a+b}{2m}\right)\right|^{q}\int_{0}^{1}w(1-t)(1-h(t))^{s}dt \biggr)^{\frac{1}{q}} \biggr\} \ \end{aligned}\] with \(L\) as before.
Proof. By mean of power mean inequality, from \(I_1\) we obtain \[\begin{aligned} \label{A} \nonumber &\int_{0}^{1} w(t)\left|f''\left(\frac{a+b}{2}t+(1-t)a\right)\right|dt \leq \left(\int_{0}^{1}w(t)dt\right)^{1-\frac{1}{q}}\\ \nonumber &\quad\times\left(\int_{0}^{1}w(t)\left|f''\left(\frac{a+b}{2}t+(1-t)a\right)\right|^{q}dt \right)^{\frac{1}{q}}\\ \nonumber &\leq \left(\int_{0}^{1}w(t)dt\right)^{1-\frac{1}{q}} \biggl(\left|f''\left(\frac{a+b}{2}\right)\right|^{q}\int_{0}^{1}w(t)h^{s}(t)dt\\ &\quad+ m\left|f''\left(\frac{a}{m}\right)\right|^{q}\int_{0}^{1}w(t)(1-h(t))^{s}dt \biggr)^{\frac{1}{q}}. \ \end{aligned}\] In the case of \(I_2\) the use of this inequality gives us \[\begin{aligned} \label{B} \nonumber &\int_{0}^{1} w(1-t)\left|f''\left(bt+(1-t)\frac{a+b}{2}\right)\right|dt \\ \nonumber &\leq \left(\int_{0}^{1}w(1-t)dt\right)^{1-\frac{1}{q}}\biggl(\left|f''\left(b\right)\right|^{q}\int_{0}^{1}w(1-t)h^{s}(t)dt\\ &\quad+ m\left|f''\left(\frac{a+b}{2m}\right)\right|^{q}\int_{0}^{1}w(1-t)(1-h(t))^{s}dt \biggr)^{\frac{1}{q}}. \ \end{aligned}\] From (14) and (15) we can arrive to the desired result. ◻
Remark 8. If \(s=1\) and \(w(t)=t^2\) we obtain an extension of Theorem 2.6 of [40] and the Remark 2.7 of above paper is still valid.
Remark 9. For Riemann-Liouville integrals we have the following result, i.e., \(w(t)=\frac{z^{\alpha+1}}{\Gamma(\alpha)}\): \[\begin{aligned} &\bigg|\frac{\alpha (1+\alpha)2^{\alpha -1} }{(b-a)^\alpha }\left ( J_{a+}^{\alpha } f\left ( \frac{a+b}{2} \right )+J_{b-}^{\alpha } f\left ( \frac{a+b}{2} \right )\right )\\ &\quad- \frac{1+\alpha }{\Gamma (\alpha )}f\left ( \frac{a+b}{2} \right ) \bigg| \\ &\leq \frac{(b-a)^{2}}{8} \frac{1}{(2+\alpha)^{1-\frac{1}{q}}} \biggl[ \biggl(\left|f''\left(\frac{a+b}{2}\right)\right|^{q}\\ &\quad\times \int_{0}^{1}t^{1+\alpha}h^{s}(t)dt+ m\left|f''\left(\frac{a}{m}\right)\right|^{q}\\ &\quad\times \int_{0}^{1}t^{1+\alpha}(1-h(t))^{s}dt\biggr)^{\frac{1}{q}}\\ &\quad+ \biggl(\left|f''\left(b\right)\right|^{q}\int_{0}^{1}t^{1+\alpha}h^{s}(t)dt\\ &\quad+ m\left|f''\left(\frac{a+b}{2m}\right)\right|^{q}\int_{0}^{1}t^{1+\alpha}(1-h(t))^{s}dt\biggr)^{\frac{1}{q}} \biggr].\ \end{aligned}\]
Theorem 4. Let \(f:I\subset [0,d)\rightarrow \mathbb{R}\) b a differentiable mapping on \(I^{\circ}\) such that \(f'' \in L[a,b]\), where \(a\), \(b \in I\) with \(0\leq a<b<\infty\). If \(\left| f'' \right|^{q}\) is \((h,m)\)-convex on \([a,b]\), for some fixed \(m \in (0,1]\) and \(p\), \(q>1\) with \(\frac{1}{p}+\frac{1}{q}=1\), then the following inequality holds: \[\begin{aligned} \label{e:41}\nonumber \left|L \right| \leq& \frac{(b-a)^{2}}{8} \biggl( \frac{\int_{0}^{1}w^p(t)dt}{p}+\frac{1}{q} \biggl[ \left|f''\left(\frac{a+b}{2}\right)\right|^{q}\\ \nonumber &\times \int_{0}^{1}h^s(t)dt+m\left|f''\left(\frac{a}{m}\right)\right|^{q}\int_{0}^{1}(1-h(t))^{s}dt\biggr]\\ \nonumber &+\ \frac{\int_{0}^{1}w^p(1-t)dt}{p} +\frac{1}{q}\biggl[ \left|f''\left(b\right)\right|^{q} \int_{0}^{1}h^s(t)dt\\ &+ m\left|f''\left(\frac{a+b}{2m}\right)\right|^{q}\int_{0}^{1}(1-h(t))^{s}dt\biggr]\biggr).\ \end{aligned}\]
Proof. Taking into account the Young Inequality we have \[\begin{aligned} & \int_{0}^{1}w(t)\left|f''\left(\frac{a+b}{2}t+(1-t)a\right)\right|dt\\ &+\int_{0}^{1}w(1-t)\left|f''\left(bt+(1-t)\frac{a+b}{2}\right)\right|dt \\ &\leq \int_{0}^{1}\left[\frac{w^{p}(t)}{p}+\frac{\left|f''\left(\frac{a+b}{2}t+(1-t)a\right)\right|^{q}}{q}\right]dt\\ &+ \int_{0}^{1}\left[\frac{w^{p}(1-t)}{p}+\frac{\left|f''\left(tb+(1-t)\frac{a+b}{2}\right)\right|^{q}}{q}\right]dt \end{aligned}\]\[\begin{aligned} &= \int_{0}^{1}\frac{w^{p}(t)}{p}dt+\frac{1}{q}\int_{0}^{1}\left|f''\left(\frac{a+b}{2}t+(1-t)a\right)\right|^{q}dt\\ &+ \int_{0}^{1}\frac{w^{p}(1-t)}{p}dt+\frac{1}{q}\int_{0}^{1}\left|f''\left(tb+(1-t)\frac{a+b}{2}\right)\right|^{q}dt \\ &\leq \int_{0}^{1}\frac{w^{p}(t)}{p}dt+\frac{1}{q}\biggl[\left|f''\left(\frac{a+b}{2}\right)\right|^{q}\int_{0}^{1}h^{s}(t)dt\\ &+ m\left|f''\left(\frac{a}{m}\right)\right|^{q}\int_{0}^{1}(1-h(t))^{s}dt\biggr] \\ &+ \int_{0}^{1}\frac{w^{p}(1-t)}{p}dt+\frac{1}{q}\biggl[\left|f''\left(b\right)\right|^{q}\int_{0}^{1}h^{s}(t)dt\\ &+ m\left|f''\left(\frac{a+b}{2m}\right)\right|^{q}\int_{0}^{1}(1-h(t))^{s}dt\biggr]. \ \end{aligned}\] Result we wanted to achieve. ◻
Remark 10. With \(s=1\) we obtain an extension of Theorem 2.8 of [40] and, as before, the Remark 2.9 of above paper is still valid.
Remark 11. If \(w(t)=t^{1+\alpha}\), we obtain the following result valid for Riemann-Liouville fractional integrals: \[\begin{aligned} &\bigg|\frac{\alpha (1+\alpha)2^{\alpha -1} }{(b-a)^\alpha }\left ( J_{a+}^{\alpha } f\left ( \frac{a+b}{2} \right )+J_{b-}^{\alpha } f\left ( \frac{a+b}{2} \right )\right )\\ &- \frac{1+\alpha }{\Gamma (\alpha )}f\left ( \frac{a+b}{2} \right ) \bigg| \leq \frac{(b-a)^{2}}{8}+ \biggl\{ \frac{1}{(\alpha (p+1)+1)} \\ &\times \biggl[ 2+\frac{1}{q} \biggl( \left( \left|f''\left(\frac{a+b}{2}\right)\right|^{q} +\left|f''\left(b\right)\right|^{q}\right ) \int_{0}^{1}h^s(t)dt\\ &+ m\left ( \left|f''\left(\frac{a}{m}\right)\right|^{q}+ \left|f''\left(\frac{a+b}{2m}\right)\right|^{q}\right )\int_{0}^{1}(1-h(t))^{s}dt\biggr) \biggr] \biggr\}. \end{aligned}\]
In this paper, we have obtained new versions of the well-known Hermite-Hadamard Inequality, via weighted integral. One of the features of the generality of our results is given use of a general weight in the integral operator used, which does not offer us a generalization, but rather “families” of inequalities, since the weight can lead us to classical Riemann integrals or to fractional integrals of the Riemann-Liouville type as we have pointed out above throughout the work. If we add to this the notion of convexity, which encompasses many known definitions, we have an idea of the breadth and scope of the results obtained.
