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\title[A Study On Hermite-Hadamard Type Inequalities]{A Study On Hermite-Hadamard Type Inequalities For  s-convex Functions  Via Conformable Fractional Integrals}
\author{Erhan Set}
\address{Department of Mathematics, \\ Faculty of Arts and Sciences,\\ Ordu University,\\
52200 Ordu, \\
Turkey}
\email{erhanset@yahoo.com}
%
\author{Abdurrahman G\"{o}zp{\i}nar}
\address{Department of Mathematics, \\ Faculty of Arts and Sciences,\\ Ordu University,\\
52200 Ordu, \\
Turkey}
\email{abdurrahmangozpinar79@gmail.com}
%

\subjclass{26A33,26A51,26D10,26D15}
\keywords{s-convex functions, Hermite-Hadamard inequality, conformable fractional integrals}
\begin{abstract}
In the present note, firstly we established a generalization of Hermite Hadamard's inequality for s-convex functions via conformable fractional integrals which generalized Riemann-Liouville fractional integrals. Secondly, we proved new identity involving conformable fractional integrals via beta and incompleted beta functions.Then, by using this identity, some Hermite Hadamard type integral inequalities for s-convex functions in the second sense  are obtained.
\end{abstract}
\maketitle

\section{Introduction}


%------------------------------her.had tanım------------------------------------------------------------------------------
One of the most famous inequality for convex functions is so called Hermite-Hadamard inequality as follows:
Let $f:I\subseteq \mathbb{R}\to \mathbb{R}$ be a convex function and $a,b\in I$ with  $a<b$, then
\begin{equation}\label{H.H}
f\bigg(\frac{a+b}{2}\bigg) \leq \frac{1}{b-a}\int_{a}^{b} f(x)dx \leq \frac{f(a)+f(b)}{2}
\end{equation}
 This famous inequality  discovered by C. Hermite and J. Hadamard is
  important in the literature. For more studies via Hermite Hadamard type inequalities see   \cite{dragomir-2} in the references.
 \begin{definition}
 Let $\ f:I\subseteq\mathbb{R}\rightarrow\mathbb{R}$ be a  function and $a,b\in I$ with $a<b$, the
function $\ f:I\subseteq\mathbb{R}\rightarrow\mathbb{R}$ is said to be convex if the inequality
\begin{eqnarray*}
f\left( tx+\left( 1-t\right) y\right) \leq tf\left( x\right) +\left(1-t\right) f\left( y\right)
\end{eqnarray*}
holds for all $x,y\in I\ $ and $t\in \left[ 0,1\right] $.
\end{definition}

%-------------------s-convex-2------------------------------------------------------------------------------------------
\begin{definition}\cite{Breckner,Hudzik}
A function $f:\mathbb{R_{+}\rightarrow\mathbb{R}}$ is said to be s-convex in the second sense if
\begin{eqnarray}
f( \alpha x+ \beta y)\leq \alpha^{s}f(x)+ \beta^{s}f(y)\nonumber
\end{eqnarray}
for all $x,y \in \mathbb{\mathbb{R_{+}}}$ and all $\alpha,\beta\geq 0 $ with $\alpha+\beta=1 $.
\end{definition}
We denote this by $K_{s}^{2}$. It is obvious that the s-convexity means just the convexity when $s=1$.

%--------------------- for s convex hermite hadamard---------------burayı bul-------------------------------------------------------------

In \cite{dragomir} Dragomir and Fitzpatrick proved a variant of Hermite-Hadamard inequality which holds for s-convex functions in the second sense.
\begin{theorem}
Suppose that $f:[0,\infty)\rightarrow[0,\infty)$ is an s-convex function in the second sense, where $s\in(0,1]$ and let $a,b\in[0,\infty), a<b.$ If $f\in L^{1}[a,b],$ then the following inequality hold:
\begin{eqnarray}
\label{h.h.s-conv}
2^{s-1}f\left(\frac{a+b}{2}\right) \leq \frac{1}{b-a}\int_{a}^{b}f\left( x\right)
dx\leq \frac{f\left( a\right) +f\left( b\right) }{s+1}  \label{H.H.s-conv}
\end{eqnarray}
\end{theorem}
The constant $k=\frac{1}{s+1}$ is the best possible in the second inequality in (\ref{h.h.s-conv}). For more study related to s-convexity in the second sense,  see, e.g, (for example)(\cite{Alomari-2},\cite{Avci},\cite{dragomir-3}).

%-------------------------------------------------------------------------------------------------------------

%---------------fractional tanımlar-----------------------------------------------------------------------
Theory of convex functions has great importance in various fields of pure and applied sciences. It is known that theory of convex functions is closely related to theory of  inequalities. Many interesting convex functions inequalities established via Riemann-Liouville fractional integrals. Now, lets us give  some necessary definition and mathematical preliminaries of fractional calculus theory as follows,  which are used lots of study. For more details, one can consult (\cite{chen}-\cite{Dahmani2},\cite{Gorenflo}, \cite{imdat1}-\cite{set-1}, \cite{Zhu}).
\begin{definition}
Let $f\in L_{1}[a,b]$. The Riemann-Liouville integrals $J_{a+}^{\alpha}f$ and $J_{b-}^{\alpha}f$ of order $\alpha> 0$ with $a\geq0$ are defined by
$$
J_{a+}^{\alpha}f(x)=\frac{1}{\Gamma (\alpha)}\int_{a}^{x}(x-t)^{\alpha-1}f(t)dt,\,\,\,\,\,\,\,\,\,\,x>a
$$
and
$$
J_{b-}^{\alpha}f(x)=\frac{1}{\Gamma (\alpha)}\int_{x}^{b}(t-x)^{\alpha-1}f(t)dt,\,\,\,\,\,\,\,\,\,\,x<b
$$
respectively. Here $\Gamma(t)$ is the Gamma function and its definition is $\Gamma(t)=\int_{0}^{\infty}e^{-x}x^{t-1}dx.$ It is to be noted that $J_{a+}^{0}f(x)=J_{b-}^{0}f(x)=f(x)$  and in the case of $\alpha=1$, the fractional integral reduces to the classical integral.
\end{definition}
%----------------------beta function------------------------------------------------------------------------------------------------------------------
The beta function defined as follows:
\begin{equation*}
B \left( a,b\right) =\frac{\Gamma (a)\Gamma (b)}{\Gamma (a+b)}%
=\int_{0}^{1}t^{a-1}\left( 1-t\right) ^{b-1}dt,\text{ \ \ \ \ }a,b>0,
\end{equation*}
where
$\Gamma \left( \alpha \right) $ is Gamma function. The incomplete beta function is defined by
$B_{x}(a,b)=\int_{0}^{x}t^{a-1}(1-t)^{b-1}dt$, \, $0\leq x\leq1.$ For $x=1$, the incomplete beta function coincides with the complete beta function.\\
For easy  understanding the computation in our theorems, let us  give some properties of beta and incompleted beta function;
\begin{eqnarray*}
&&B(a,b)=B_{t}(a,b)+B_{1-t}(b,a),  \,\,  i.e  \,\, B(a,b)=B_{\frac{1}{2}}(a,b)+B_{\frac{1}{2}}(b,a)\\
&&B_{x}(a+1,b)=\frac{aB_{x}(a,b)-(x)^{a}(1-x)^{b}}{a+b}\\
&&B_{x}(a,b+1)=\frac{bB_{x}(a,b)+(x)^{a}(1-x)^{b}}{a+b}\\
&&B(a,b+1)+B(a+1,b)=B(a,b)
\end{eqnarray*}
%************************************fractional hermite hadamard***************************************
In \cite {mz} Sar{\i}kaya et al. gave a remarkable integral inequality of Hermite-Hadamard type involving Riemann-Liouville fractional integrals as follows:
\begin{theorem}
\label{h.h-con-frac}
Let $f:[a,b]\rightarrow\mathbb{R}$ be a positive function with $0\leq a<b$ and $f\in L^{1}[a,b]$. If $f$ is convex function on $[a,b]$, then the following inequality for fractional integrals hold:
\begin{eqnarray}
\label{h.h-con-frac1}
&&f\left(\frac{a+b}{2}\right)
\leq \frac{\Gamma(\alpha +1)}{2(b-a)^{\alpha}}[(J_{a^{+}}^{\alpha}f)(b)+ (J_{b^{-}}^{\alpha}f)(a)]
\leq\frac{f\left( a\right) +f\left( b\right)}{2}\\\nonumber
\end{eqnarray}
\end{theorem}
It is obviously seen that, if we take $\alpha=1$ in Theorem \ref{h.h-con-frac}, then the inequality (\ref{h.h-con-frac1}) reduces to well known Hermite-Hadamard inequality as (\ref{H.H}).

%-------s-convex for fractional ***********************************-------------------------------

Hermite-Hadamard type inequalities for s-convex functions via Riemann-Liouville  fractional integral is given in \cite{set} as follows:
\begin{theorem}
Let $f:[a,b]\rightarrow\mathbb{R}$ be a positive function with $0\leq a<b$ and $f\in L_{1}[a,b]$. If $f$ is s-convex mapping in the second sense  on $[a,b]$, then the following inequality for fractional integral with $\alpha>0$ and $s\in(0,1]$ hold:
\begin{eqnarray}
\label{s-cnvx.fract}
2^{s-1}f\left(\frac{a+b}{2}\right)
&\leq& \frac{\Gamma(\alpha +1)}{2(b-a)^{\alpha}}[(J_{a^{+}}^{\alpha}f)(b)+ (J_{b^{-}}^{\alpha}f)(a)] \\
&\leq&\alpha\big[\frac{1}{\alpha +s}+B(\alpha,s+1)]\frac{f\left( a\right) +f\left( b\right)}{2}\nonumber
\end{eqnarray}
where B(a,b) is Euler beta function.
\end{theorem}

%--------------------KULLANDIGIMIZ LEMMA --------yaldız---sarıkaya---------------------------------------------------------------------------
  Sarikaya et al. established an identity which we will  generalize  for conformable fractional integral in section 3 for differentiable convex mappings via  Riemann-Liouville fractional integral. Then they gave  some results by using this identity.

\begin{lemma}\,\cite{mz}
\label{srky lemma}
Let $f:[a,b]\rightarrow\mathbb{R}$ be a differentiable mapping on $(a,b)$ with $a<b$. If $f^{'}\in L[a,b]$, then the following equality for fractional integrals holds:
\begin{eqnarray}
\label{srky lemma-1}
&&\frac{f(a)+f(b)}{2}-\frac{\Gamma(\alpha+1)}{2(b-a)^{\alpha}}[J_{a^{+}}^{\alpha}f(b)+ J_{b_{-}}^{\alpha}f(a)]\\\nonumber
&&=\frac{b-a}{2}\int_{0}^{1}\big[(1-t)^{\alpha}-t^{\alpha}\big]f^{'}(ta+(1-t)b)dt.\\\nonumber
\end{eqnarray}
\end{lemma}
%---------------------------sarıkaya aldığım teorem -----------------------------------------------------------
\begin{theorem}\cite{mz} Let $f:[a,b]\rightarrow\mathbb{R}$ be a differentiable mapping on $(a,b)$ with $a<b$. If $f^{'}\in L[a,b]$, then the following inequality for fractional integrals holds:
\begin{eqnarray}
\label{srky teo-1}
&&\big|\frac{f(a)+f(b)}{2}-\frac{\Gamma(\alpha+1)}{2(b-a)^{\alpha}}[J_{a^{+}}^{\alpha}f(b)+ I_{b_{-}}^{\alpha}f(a)]\big|\\\nonumber
&&\leq\frac{b-a}{2(\alpha+1)}\big(1-\frac{1}{2^{\alpha}})[f^{'}(a)+f^{'}(b)]\\\nonumber
\end{eqnarray}
\end{theorem}
%--------------------------------------------------------------------------------------------------------------------
%------------------------------------------------------------
%-----------------------------conformable giriş-----------------------------------------------------------------------------------------
Recently, some authors started to study on conformable fractional  integral. In \cite{Khalil}, Khalil et al. defined the fractional integral of order $0<\alpha\leq1$ only.
In \cite{Abdelj}, Abdeljawad gave the definition of left and right conformable fractional integrals of any order $\alpha>0$.

\begin{definition}
Let $\alpha \in(n,n+1]$ and set $\beta=\alpha-n$ then the left conformable fractional integral starting at $a$ if order $\alpha$ is defined by
$$
(I_{\alpha}^{a}f)(t)=\frac{1}{n!}\int_{a}^{t}(t-x)^{n}(x-a)^{\beta -1}f(x)dx
$$
Analogously, the right conformable fractional integral is defined by
$$
(^{b}I_{\alpha}f)(t)=\frac{1}{n!}\int_{t}^{b}(x-t)^{n}(b-x)^{\beta -1}f(x)dx.
$$
\end{definition}
Notice that if $\alpha=n+1$ then $\beta=\alpha-n=n+1-n=1$ where $n=0,1,2,3...$ and hence $(I_{\alpha}^{a}f)(t)=(J_{n+1}^{a}f)(t)$.
%----------------------İlkerden gelen genelleştirme---theorem---------------
In \cite{set-2} Set et.al. gave  Hermite-Hadamard inequality for conformable fractional integral as follows:
\begin{theorem}
\label{heha}
Let $f:[a,b]\rightarrow\mathbb{R}$ be a function with $0\leq a<b$ and $f\in L_{1}[a,b]$. If $f$ is a convex function on $[a,b]$, then the following inequalities for conformable fractional integrals hold:
\begin{equation}
\label{hh0}
f\left( \frac{a+b}{2}\right) \leq \frac{\Gamma(\alpha +1)}{2(b-a)^{\alpha}\Gamma(\alpha -n)}[(I_{\alpha}^{a}f)(b)+ (^{b}I_{\alpha}f)(a)]\leq \frac{f\left( a\right) +f\left( b\right)}{2}
\end{equation}
with $\alpha \in (n,n+1]$, where $\Gamma$ is Euler Gamma function.
\end{theorem}
For some studies on conformable fractional integral, see (\cite{Abdelj}, \cite{Abdelj-2}, \cite{anderson}, \cite{benkhettou}). In papers (\cite{set-6}-\cite{set-8}), Set et.al obtained some  Hermite-Hadamard, Ostrowski, Chebyshev, Fejer  type inequalities by using conformable fractional integrals  for various classes of functions. The aim of this study is to establish new Hermite-Hadamard inequalities related to other fractional integral inequalities for conformable fractional integral.

%----------------------------------------------------------------------------------------------------------------------

\section{Hermite-Hadamard's inequalities for conformable fractional integrals}
%----------s-convex teorem------------------------------------------------------------------------------------------------------------
In this section, using the given properties of conformable fractional integrals, we will establish a generalization of Hermite-Hadamard type inequalities for s-convex functions. We will also noticed the relation with fractional and classical Hermite-Hadamard type integral inequalities.
\begin{theorem}
\label{teo.s-cnx}
Let $f:[a,b]\rightarrow\mathbb{R}$ be a function with $ 0\leq a<b, \,\,s\in(0,1]$ and $f\in L_{1}[a,b]$. If $f$ is an s-convex function on $[a,b]$, then the following inequalities for conformable fractional integrals hold:
\begin{eqnarray}
\label{h.h for s-cnvx}
&&\frac{\Gamma(\alpha-n)}{\Gamma(\alpha +1)}f\left(\frac{a+b}{2}\right)\\
&\leq& \frac{1}{(b-a)^{\alpha}2^{s}}[(I_{\alpha}^{a}f)(b)+ (^{b}I_{\alpha}f)(a)]\nonumber\\
&\leq&\bigg[\frac{B(n+s+1,\alpha-n)+B(n+1,\alpha-n+s)}{n!}\bigg] \frac{f\left( a\right) +f\left( b\right)}{2^{s}}\nonumber
\end{eqnarray}
with $\alpha \in (n,n+1],\,\,n=0,1,2,...$ where $\Gamma$ is Euler Gamma function and $B(a,b)$ is a beta function.
\end{theorem}
\begin{proof}
Let $x,y\in[a,b]$. If $f$ is a s-convex function on [a,b],
\begin{equation}
f(\frac{x+y}{2})\leq(\frac{1}{2})^{s}f(x)+(\frac{1}{2})^{s}f(y)\\\nonumber
\end{equation}
if we change the variables with $x=ta+(1-t)b,\,\, y=(1-t)a+tb$,
\begin{equation}
2^{s}f(\frac{a+b}{2})\leq f(ta+(1-t)b)+f((1-t)a+tb).
\end{equation}
Multiplying both sides of above inequality with $\frac{1}{n!}t^{n}(1-t)^{\alpha-n-1}$ and integrating the resulting inequality with respect to $t$ over $[0,1]$, we get
\begin{eqnarray*}
&&\frac{2^{s}}{n!}f(\frac{a+b}{2})\int_{0}^{1}t^{n}(1-t)^{\alpha -n-1}dt\\
&\leq&\frac{1}{n!}\int_{0}^{1}t^{n}(1-t)^{\alpha -n-1}f(ta+(1-t)b)dt\\
&&+\frac{1}{n!}\int_{0}^{1}t^{n}(1-t)^{\alpha -n-1}f((1-t)a+tb)dt\\
&=&\frac{1}{n!} \int_{a}^{b}(\frac{b-x}{b-a})^{n}(\frac{x-a}{b-a})^{\alpha -n-1}f(y)\frac{dx}{b-a}\\
&&+\frac{1}{n!} \int_{a}^{b}(\frac{y-a}{b-a})^{n}(\frac{b-y}{b-a})^{\alpha -n-1}f(y)\frac{dy}{b-a}\\
& =&\frac{1}{(b-a)^{\alpha}}[I_{\alpha}^{a}f(b)+ ^{b}I_{\alpha}f(a)].
\end{eqnarray*}
Note that
\begin{equation}
\label{left side}
f\left(\frac{a+b}{2}\right) \leq \frac{\Gamma(\alpha +1)}{2^{s}(b-a)^{\alpha}\Gamma(\alpha -n)}[I_{\alpha}^{a}f(b)+ ^{b}I_{\alpha}f(a)]
\end{equation}
where
$$
\int_{0}^{1}t^{n}(1-t)^{\alpha -n-1}dt=B(n+1,\alpha -n)=\frac{\Gamma(n+1)\Gamma(\alpha-n)}{\Gamma(\alpha+1)}
$$
which means that the left side of (\ref{h.h for s-cnvx}) is proved.
Since $f$ is s-convex in the second sense, to prove the right side of (\ref{h.h for s-cnvx}) we have the following inequalities;
\begin{eqnarray*}
f(ta+(1-t)b)&\leq & t^{s} f(a)+(1-t)^{s}f(b) \\
f((1-t)a+tb)&\leq & (1-t)^{s}f(a)+t^{s} f(b).
\end{eqnarray*}
Adding these two inequalities, we get
$$
f(ta+(1-t)b)+f((1-t)a+tb)\leq [t^{s}+(1-t)^{s}][f(a)+f(b)].
$$
Multiplying both sides of the resulting inequality with $\frac{1}{n!}t^{n}(1-t)^{\alpha -n-1}$ and integrating  with respect to $t$ over $[0,1]$, we have
\begin{eqnarray}
\label{right side}
&&\frac{1}{(b-a)^{\alpha}}[I_{\alpha}^{a}f(b)+ ^{b}I_{\alpha}f(a)] \\
&\leq&\frac{1}{n!}\int_{0}^{1}t^{n}(1-t)^{\alpha-n-1}[t^{s}+(1-t)^{s}][f(a)+f(b)]dt\nonumber\\
&=&\frac{1}{n!}\Big[B(n+s+1,\alpha-n)+B(n+1,\alpha-n+s)\Big][f(a)+f(b)].\nonumber
\end{eqnarray}
Combining (\ref{left side}) and (\ref{right side}) completes the proof.

\end{proof}
%*******************Remark for s-convex*******************************************************************************************************
\begin{remark}
\label{rmrk-s-cnvx}
If we choose $s=1$ in Theorem (\ref{h.h for s-cnvx}), by using relation between  $\Gamma$ and $B$ functions, the inequality   (\ref{h.h for s-cnvx}) reduced to inequality (\ref{hh0}).
\end{remark}
%***********************conformable=fractional*********************************************
\begin{remark}
If we choose $ \alpha=n+1 $ in Theorem \ref{teo.s-cnx}, the inequality (\ref{rmrk-s-cnvx}) reduced to inequality (\ref{s-cnvx.fract}).
And also  if we  choose $ \alpha, \, s = 1$  in the inequality (\ref{rmrk-s-cnvx}), then we get well-known Hermite-Hadamard inequality as (\ref{H.H.s-conv}).
\end{remark}

%******************************************************************************************************************************

%****************************Main Results*************************************************************************************
\section{Some new Hermite Hadamard type inequalities via conformable integration}
In order to achieve our aim, we will give an important  identity for differentiable  functions involving conformable fractional integrals as follows;
%****************L E M M A********************************************************************
\begin{lemma}
\label{lemma}
Let $f:[a,b]\rightarrow\mathbb{R}$ be a differentiable mapping on $(a,b)$ with $a<b$. If $f^{'}\in L[a,b]$, then the following inequality for conformable fractional integrals holds:
\begin{eqnarray}
\label{lemma-0}
&&B(n+1,\alpha-n)\big(\frac{f(a)+f(b)}{2})-\frac{n!}{2(b-a)^{\alpha}}[I_{\alpha}^{a}f(b)+ ^{b}I_{\alpha}f(a)]\\\nonumber
&&=\frac{(b-a)}{2}\Big\{\int_{0}^{1}\big[B_{1-t}(n+1,\alpha-n)-B_{t}(n+1,\alpha-n)\big]f^{'}(ta+(1-t)b)dt\Big\}\\\nonumber
\end{eqnarray}
where $B(a,b)$, $B_{t}(a,b)$ is Euler beta and incompleted beta functions respectively and $\alpha \in (n,n+1], n=0,1,2,\ldots$.
\end{lemma}
%**************LEMMA PROOF ***************************************************************
\begin{proof}Let
\begin{eqnarray}
\label{lemma-1}
&&I=\int_{0}^{1}\big[B_{1-t}(n+1,\alpha-n)-B_{t}(n+1,\alpha-n)\big]f^{'}(ta+(1-t)b)dt.\nonumber
\end{eqnarray}
Then, integrating by parts and changing variables with $x=ta+(1-t)b$, we can write
\begin{eqnarray}
\label{lemma1}
I_{1}&=&\int_0^{1}B_{1-t}(n+1,\alpha-n)f^{'}(ta+(1-t)b)dt\\
&=&\int_0^{1}\bigg(\int_{0}^{1-t}x^{n}(1-x)^{\alpha-n-1}dx\bigg)f^{'}(ta+(1-t)b)dt\nonumber\\
&=&\bigg(\int_{0}^{1-t}x^{n}(1-x)^{\alpha-n-1}dx\bigg)\frac{f(ta+(1-t)b)dt}{a-b}\bigg|_{0}^{1}\nonumber\\
&&+\int_0^{1}(1-t)^{n}t^{\alpha-n-1}f(ta+(1-t)b)\frac{dt}{a-b}\nonumber\\
&=&\bigg(\int_{0}^{1}x^{n}(1-x)^{\alpha-n-1}dx\bigg)\frac{f(b)}{b-a}\nonumber\\
&&+\frac{1}{b-a}\int_{a}^{b}\big(\frac{x-a}{b-a}\big)^{n}\big(\frac{b-x}{b-a}\big)^{\alpha-n-1}f(x)\frac{dx}{a-b}\nonumber\\
&=&B(n+1,\alpha-n)\frac{f(b)}{b-a}-\frac{n!}{(b-a)^{\alpha+1}}(^{b}I_{\alpha}f)(a)\nonumber
\end{eqnarray}
\begin{eqnarray}\label{lemma2}
I_{2}&=&\int_0^{1}B_{t}(n+1,\alpha-n)f^{'}(ta+(1-t)b)dt\\\nonumber
&=&B_{t}(n+1,\alpha-n)\frac{f(ta+(1-t)b)}{a-b}\bigg|_{0}^{1}\\\nonumber
&&-\int_0^{1}t^{n}(1-t)^{\alpha-n-1}f(ta+(1-t)b)\frac{dt}{a-b}\\\nonumber
&=&-B(n+1,\alpha-n)\frac{f(a)}{b-a}
+\frac{1}{b-a}\int_{a}^{b}\big(\frac{b-x}{b-a}\big)^{n}\big(\frac{x-a}{b-a}\big)^{\alpha-n-1}f(x)\frac{dx}{b-a}\nonumber\\
&=&-B(n+1,\alpha-n)\frac{f(a)}{b-a}+\frac{n!}{(b-a)^{\alpha+1}}(I_\alpha^{a}f)(b).\nonumber
\end{eqnarray}
It means that
$I=I_{1}-I_{2}$.
Thus, by multiplying both sides by $\frac{b-a}{2}$ i.e $\frac{b-a}{2}I=\frac{b-a}{2}I_{1}-\frac{b-a}{2}I_{2}$ we have desired result.
\end{proof}
%**********************ALINAN LEMMAYA  ATIF ******************
\begin{remark}
If we choose $\alpha=n+1$ in Lemma \ref{lemma}, the equality (\ref{lemma-0}) becomes the equality (\ref{srky lemma-1}).
\end{remark}
Now, using the obtained identity, we will establish some inequalities connected with the left part of the inequality (\ref{h.h for s-cnvx})
%----------S-CONVEXS 1.THEOREM------------------------------------------------------------------------------------------------------------

\begin{theorem}\label{s-teo}
Let $f:[a,b]\rightarrow\mathbb{R}$ be a differentiable mapping on $(a,b)$ with $a<b$. If $f^{'}\in L[a,b]$ and $|f^{'}|$ is s-convex in the second sence with $s\in (0,1]$, then the following inequality for conformable fractional integrals holds:
\begin{eqnarray}\label{teo1}
&&\bigg|B(n+1,\alpha-n)\big(\frac{f(a)+f(b)}{2})-\frac{n!}{2(b-a)^{\alpha}}[I_{\alpha}^{a}f(b)+ ^{b}I_{\alpha}f(a)]\bigg|\\
&\leq&\frac{b-a}{2}\big[\frac{|f^{'}(a)|+|f^{'}(b)|}{s+1}\big]\nonumber\\
&&\bigg\{B_{\frac{1}{2}}(\alpha-n+s+1,n+1)-B_{\frac{1}{2}}(n+1,\alpha-n+s+1)\nonumber\\
&&+B_{\frac{1}{2}}(n+s+2,\alpha-n)-B_{\frac{1}{2}}(\alpha-n,n+s+2)+B(n+1,\alpha-n)\bigg\}\nonumber
\end{eqnarray}
where $B(a,b)$, $B_{t}(a,b)$ is Euler beta and incompleted beta functions respectively and $\alpha \in (n,n+1], n=0,1,2,\ldots$.
\end{theorem}

\begin{proof}
Taking modulus on Lemma \ref{lemma} and using s-convexity of $|f^{'}|$ we get;
\begin{eqnarray}
\label{teo1-1}
&&\big|B(n+1,\alpha-n)\big(\frac{f(a)+f(b)}{2})-\frac{n!}{2(b-a)^{\alpha}}[I_{\alpha}^{a}f(b)+ ^{b}I_{\alpha}f(a)]\big|\\
&=&\frac{b-a}{2}\bigg|\int_{0}^{1}\big[B_{1-t}(n+1,\alpha-n)-B_{t}(n+1,\alpha-n)\big]f^{'}(ta+(1-t)b)dt\bigg|\nonumber\\
&\leq&\frac{b-a}{2}\int_{0}^{1}\big|\big[B_{1-t}(n+1,\alpha-n)-B_{t}(n+1,\alpha-n)\big]\big|\big|f^{'}(ta+(1-t)b)\big|dt\nonumber\\
&=&\frac{b-a}{2}\int_{0}^{\frac{1}{2}}\big[B_{1-t}(n+1,\alpha-n)-B_{t}(n+1,\alpha-n)\big]\big|f^{'}(ta+(1-t)b)\big|dt\nonumber\\
&&+\int_{\frac{1}{2}}^{1}\big|\big[B_{t}(n+1,\alpha-n)-B_{1-t}(n+1,\alpha-n)\big]\big|\big|f^{'}(ta+(1-t)b)\big|dt\nonumber\\
&\leq&\frac{b-a}{2}\Bigg\{\int_{0}^{\frac{1}{2}}B_{1-t}(n+1,\alpha-n)\big(t^{s}|f^{'}(a)|+(1-t)^{s}|f^{'}(b)|\big)dt\nonumber\\
&&-\int_{0}^{\frac{1}{2}}
B_{t}(n+1,\alpha-n)
\big(t^{s}|f^{'}(a)|+(1-t)^{s}|f^{'}(b)|\big)dt\nonumber\\
&&+\int_{\frac{1}{2}}^{1}B_{t}(n+1,\alpha-n)\big(t^{s}|f^{'}(a)|+(1-t)^{s}|f^{'}(b)|\big)dt\nonumber\\
&&-\int_{\frac{1}{2}}^{1}B_{1-t}(n+1,\alpha-n)
\big(t^{s}|f^{'}(a)|+(1-t)^{s}|f^{'}(b)|\big)dt\Bigg\}\nonumber
\end{eqnarray}

\begin{eqnarray}
&=&\frac{b-a}{2}\Bigg\{|f^{'}(a)|\int_{0}^{\frac{1}{2}}\big[B_{1-t}(n+1,\alpha-n)-B_{t}(n+1,\alpha-n)\big]
t^{s}dt\nonumber\\
&&+|f^{'}(b)|\int_{0}^{\frac{1}{2}}\big[B_{1-t}(n+1,\alpha-n)-B_{t}(n+1,\alpha-n)\big]
(1-t)^{s}dt\nonumber\\
&&+|f^{'}(a)|\int_{\frac{1}{2}}^{1}\big[B_{t}(n+1,\alpha-n)-B_{1-t}(n+1,\alpha-n)\big]
t^{s}\big)dt\nonumber\\
&&+|f^{'}(b)|\int_{\frac{1}{2}}^{1}\big[B_{t}(n+1,\alpha-n)-B_{1-t}(n+1,\alpha-n)\big]
(1-t)^{s}dt.\nonumber
\end{eqnarray}
On the other hand, using the properties of incompleted beta function we have;

\begin{eqnarray}\label{teo1-2}
&&B_{1-t}(n+1,\alpha-n)-B_{t}(n+1,\alpha-n)\\
&=&\int_{0}^{1-t}x^{n}(1-x)^{\alpha-n-1}dx-\int_{0}^{t}x^{n}(1-x)^{\alpha-n-1}dx\nonumber\\
&=&\int_{t}^{1-t}x^{n}(1-x)^{\alpha-n-1}dx,\,\,\,\,\, where \,\,0 \leq t \leq\frac{1}{2} \nonumber
\end{eqnarray}
and
\begin{eqnarray}\label{teo1-3}
&&B_{t}(n+1,\alpha-n)-B_{1-t}(n+1,\alpha-n)\\
&=&\int_{0}^{t}x^{n}(1-x)^{\alpha-n-1}dx-\int_{0}^{1-t}x^{n}(1-x)^{\alpha-n-1}dx\nonumber\\
&=&\int_{1-t}^{t}x^{n}(1-x)^{\alpha-n-1}dx,\,\,\,\,\, where\,\, \frac{1}{2}\leq t \leq 1\nonumber
\end{eqnarray}

Using \eqref{teo1-2}, \eqref{teo1-3} and Newton Leibnitz formula and integrating by parts we can write the following computation;
\begin{eqnarray}\label{pi1}
\Phi_{1}&=&\int_{0}^{\frac{1}{2}}\bigg(\int_{t}^{1-t}x^{n}(1-x)^{\alpha-n-1}dx\bigg)t^{s}dt\\
&=&\Bigg[\bigg(\int_{t}^{1-t}x^{n}(1-x)^{\alpha-n-1}dx\bigg)\frac{t^{s+1}}{s+1}\Bigg]\bigg|_{0}^{\frac{1}{2}} \nonumber\\
&&-\int_{0}^{\frac{1}{2}}\big(-(1-t)^{n}t^{\alpha-n-1}-t^{n}(1-t)^{\alpha-n-1}\big)\frac{t^{s+1}}{s+1}dt\nonumber\\
&=&\frac{1}{s+1}\Big[\int_{0}^{\frac{1}{2}}t^{\alpha-n+s}(1-t)^{n}dt+\int_{0}^{\frac{1}{2}}t^{n+s+1}(1-t)^{\alpha-n-1}dt\Big]\nonumber\\
&=&\frac{1}{s+1}\Big[B_{\frac{1}{2}}(\alpha-n+s+1,n+1)+B_{\frac{1}{2}}(n+s+2,\alpha-n)\Big],\nonumber
\end{eqnarray}
\begin{eqnarray}\label{pi2}
\Phi_{2}&=&\int_{0}^{\frac{1}{2}}\bigg(\int_{t}^{1-t}x^{n}(1-x)^{\alpha-n-1}dx\bigg)(1-t)^{s}dt\\
&=&\Bigg[\bigg(\int_{t}^{1-t}x^{n}(1-x)^{\alpha-n-1}dx\bigg)\frac{-(1-t)^{s+1}}{s+1}\Bigg]\bigg|_{0}^{\frac{1}{2}} \nonumber\\
&&-\int_{0}^{\frac{1}{2}}\big(-(1-t)^{n}t^{\alpha-n-1}-t^{n}(1-t)^{\alpha-n-1}\big)\frac{-(1-t)^{s+1}}{s+1}dt\nonumber\\
&=&\frac{1}{s+1}\int_{0}^{1}x^{n}(1-x)^{\alpha-n-1}dx  \nonumber\\ &&-\frac{1}{s+1}\Big[\int_{0}^{\frac{1}{2}}t^{\alpha-n-1}(1-t)^{n+s+1}dt+\int_{0}^{\frac{1}{2}}t^{n}(1-t)^{\alpha-n+s}dt\Big]\nonumber\\
&=&\frac{1}{s+1}\Big[B(n+1,\alpha-n)-B_{\frac{1}{2}}(\alpha-n,n+s+2)\nonumber\\
&&-B_{\frac{1}{2}}(n+1,\alpha-n+s+1)\Big],\nonumber
\end{eqnarray}
\begin{eqnarray}\label{pi3}
\Phi_{3}&=&\int_{\frac{1}{2}}^{1}\bigg(\int_{1-t}^{t}x^{n}(1-x)^{\alpha-n-1}dx\bigg)t^{s}dt\\
&=&\Bigg[\bigg(\int_{1-t}^{t}x^{n}(1-x)^{\alpha-n-1}dx\bigg)\frac{t^{s+1}}{s+1}\Bigg]\bigg|_{\frac{1}{2}}^{1} \nonumber\\
&&-\frac{1}{s+1}\int_{\frac{1}{2}}^{1}\big(t^{n}(1-t)^{\alpha-n-1}+t^{\alpha-n-1}(1-t)^{n}\big)t^{s+1}dt\nonumber\\
&=&\frac{1}{s+1}\int_{0}^{1}x^{n}(1-x)^{\alpha-n-1}dx \nonumber \\ &&-\frac{1}{s+1}\Big[\int_{\frac{1}{2}}^{1}t^{n+s+1}(1-t)^{\alpha-n-1}dt+\int_{\frac{1}{2}}^{1}t^{\alpha-n+s}(1-t)^{n}dt\Big]\nonumber\\
&=&\frac{1}{s+1}\Big[B(n+1,\alpha-n)-B_{\frac{1}{2}}(\alpha-n,n+s+2) \nonumber\\
&&-B_{\frac{1}{2}}(n+1,\alpha-n+s+1)\Big]\nonumber
\end{eqnarray}
and
\begin{eqnarray}\label{pi4}
\Phi_{4}&=&\int_{\frac{1}{2}}^{1}\bigg(\int_{1-t}^{t}x^{n}(1-x)^{\alpha-n-1}dx\bigg)(1-t)^{s}dt\\
&=&\Bigg[\bigg(\int_{1-t}^{t}x^{n}(1-x)^{\alpha-n-1}dx\bigg)\frac{-(1-t)^{s+1}}{s+1}\Bigg]\bigg|_{\frac{1}{2}}^{1} \nonumber\\
&&+\int_{\frac{1}{2}}^{1}\big(t^{n}(1-t)^{\alpha-n-1}+t^{\alpha-n-1}(1-t)^{n}\big)\frac{(1-t)^{s+1}}{s+1}dt\nonumber
\end{eqnarray}
\begin{eqnarray}
&=&\Big[\int_{\frac{1}{2}}^{1}t^{n}(1-t)^{\alpha-n+s}dt+\int_{\frac{1}{2}}^{1}t^{\alpha-n-1}(1-t)^{n+s+1}dt\Big]\nonumber\\
&=&\frac{1}{s+1}\Big[B_{\frac{1}{2}}(\alpha-n+s+1,n+1)+B_{\frac{1}{2}}(n+s+2,\alpha-n)\Big],\nonumber
\end{eqnarray}
Using the fact that $B(a,b)=B_{\frac{1}{2}}(a,b)+B_{\frac{1}{2}}(b,a)$ and
combining (\ref{pi1}), (\ref{pi2}), (\ref{pi3}), (\ref{pi4}) with (\ref{teo1-1}) completes the proof.
\end{proof}
%------------------------corollary 1-------------------------------
\begin{corollary}\label{corllry1}
Taking $s=1$ in Theorem \ref{s-teo} i.e $|f^{'}|$ is convex, we get the following result;
\begin{eqnarray}\label{corllry1-1}
&&\bigg|B(n+1,\alpha-n)\big(\frac{f(a)+f(b)}{2})-\frac{n!}{2(b-a)^{\alpha}}[I_{\alpha}^{a}f(b)+ ^{b}I_{\alpha}f(a)]\bigg|\nonumber\\
&\leq&\frac{b-a}{2}\big(\frac{|f^{'}(a)|+|f^{'}(b)|}{2})\\
&&\times\bigg\{B_{\frac{1}{2}}(\alpha-n+2,n+1)-B_{\frac{1}{2}}(n+1,\alpha-n+2)\nonumber\\
&&+B_{\frac{1}{2}}(n+3,\alpha-n)-B_{\frac{1}{2}}(\alpha-n,n+3)+B(n+1,\alpha-n)\bigg\}\nonumber
\end{eqnarray}
\end{corollary}
%------------------------remark  1-------------------------------

\begin{remark}
Taking $\alpha=n+1$ in Corollary \ref{corllry1}, the inequality (\ref{corllry1-1}) reduces to (\ref{srky teo-1}).
\end{remark}
%-----------------------------------------remark-----------------------------
\begin{remark}
Taking $\alpha=n+1$ in Theorem  \ref{s-teo}, the inequality (\ref{teo1}) reduces to Theorem 4 in \cite{set}.
\end{remark}
%-----------------------------HÖLDER İÇİN TEO-----------------
\begin{theorem}\label{teo2}Let $f:[a,b]\rightarrow\mathbb{R}$ be a differentiable mapping on $(a,b)$, $a<b$ and $p>1$ with $ \frac{1}{p}+\frac{1}{q}=1 $. If $f^{'}\in L[a,b]$ and $|f^{'}|^{q}$ is s-convex in the second sense, then the following inequality for conformable fractional integrals holds:
\begin{eqnarray}\label{teo-2}
&&\bigg|B(n+1,\alpha-n)\big(\frac{f(a)+f(b)}{2})-\frac{n!}{2(b-a)^{\alpha}}[I_{\alpha}^{a}f(b)+ ^{b}I_{\alpha}f(a)]\bigg|\nonumber\\
&&\leq\frac{b-a}{2}\Psi^{\frac{1}{p}}\big[\frac{|f^{'}(a)|^{q}+|f^{'}(b)|^{q}}{s+1}\big]^{\frac{1}{q}}.
\end{eqnarray}
where $B(a,b)$ is Euler beta function, $\alpha \in (n,n+1], n=0,1,2,\ldots$ and $$\Psi=2\displaystyle\int_{0}^{\frac{1}{2}}\bigg(\int_{t}^{1-t}x^{n}(1-x)^{\alpha-n-1}dx\bigg)^{p}$$.
\end{theorem}
\begin{proof}
Taking modulus and using H\"{o}lder inequality with a function of $|f^{'}|^{q}$ convexity we get  inequalities as follow;
\begin{eqnarray}
\label{teo2-1}
&&\big|B(n+1,\alpha-n)\big(\frac{f(a)+f(b)}{2})-\frac{n!}{2(b-a)^{\alpha}}[I_{\alpha}^{a}f(b)+ ^{b}I_{\alpha}f(a)]\big|\\
&=&\frac{b-a}{2}\bigg|\int_{0}^{1}\big[B_{1-t}(n+1,\alpha-n)-B_{t}(n+1,\alpha-n)\big]f^{'}(ta+(1-t)b)dt\bigg|\nonumber\\
&\leq&\frac{b-a}{2}\int_{0}^{1}\big|B_{1-t}(n+1,\alpha-n)-B_{t}(n+1,\alpha-n)\big|\big|f^{'}(ta+(1-t)b)\big|dt\nonumber\\
&\leq&\frac{b-a}{2}\Big[\int_{0}^{1}\big|B_{1-t}(n+1,\alpha-n)-B_{t}(n+1,\alpha-n)\big|^{p}dt\Big]^{\frac{1}{p}}\nonumber\\
&&\times\Big[\int_{0}^{1}\big|f^{'}(ta+(1-t)b)\big|^{q}dt\Big]^{\frac{1}{q}}.\nonumber
\end{eqnarray}
It follows that:
\begin{eqnarray}
\label{teo2-2}
\Psi&=&\int_{0}^{1}\big|B_{1-t}(n+1,\alpha-n)-B_{t}(n+1,\alpha-n)\big|^{p}dt\\
&=&\int_{0}^{\frac{1}{2}}\bigg(B_{1-t}(n+1,\alpha-n)-B_{t}(n+1,\alpha-n)\bigg)^{p}dt \nonumber\\
&&+\int_{\frac{1}{2}}^{1}\bigg(B_{t}(n+1,\alpha-n)-B_{1-t}(n+1,\alpha-n)\bigg)^{p}dt\nonumber\\
&=&\int_{0}^{\frac{1}{2}}\bigg(\int_{t}^{1-t}x^{n}(1-x)^{\alpha-n-1}dx\bigg)^{p}dt\nonumber\\
&&+\int_{\frac{1}{2}}^{1}\bigg(\int_{1-t}^{t}x^{n}(1-x)^{\alpha-n-1}dx\bigg)^{p}dt\nonumber\\
&=&2\int_{0}^{\frac{1}{2}}\bigg(\int_{t}^{1-t}x^{n}(1-x)^{\alpha-n-1}dx\bigg)^{p}dt\nonumber
\end{eqnarray}
and
\begin{eqnarray}
\label{teo2-3}
&&\int_{0}^{1}\big|f^{'}(ta+(1-t)b)\big|^{q}dt\leq|f^{'}(a)|^{q}\int_{0}^{1}t^{s}dt+|f^{'}(b)|^{q}\int_{0}^{1}(1-t)^{s}dt\nonumber\\
&=&\frac{1}{s+1}\big(|f^{'}(a)|^{q}+|f^{'}(b)|^{q}\big)
\end{eqnarray}
which completes the proof.
\end{proof}
\begin{corollary}\label{corllry2}
If we take $s=1$ in Theorem \ref{teo2}, the inequality (\ref{teo-2}) reduces to following inequality;
\begin{eqnarray}\label{corllry2-1}
&&\bigg|B(n+1,\alpha-n)\big(\frac{f(a)+f(b)}{2})-\frac{n!}{2(b-a)^{\alpha}}[I_{\alpha}^{a}f(b)+ ^{b}I_{\alpha}f(a)]\bigg|\nonumber\\
&\leq&\frac{b-a}{2}\Psi^{\frac{1}{p}}\big[\frac{|f^{'}(a)|^{q}+|f^{'}(b)|^{q}}{2}\big]^{\frac{1}{q}}
\end{eqnarray}
where $B(a,b)$ is Euler beta function  and \\
$$\Psi=2\displaystyle\int_{0}^{\frac{1}{2}}\bigg(\int_{t}^{1-t}x^{n}(1-x)^{\alpha-n-1}dx\bigg)^{p}$$.
\end{corollary}

\begin{corollary}\label{corllry3}
If we take $\alpha=n+1$ in corollary \ref{corllry2}, the inequality (\ref{corllry2-1}) reduces to following inequality;
\begin{eqnarray}\label{corllry3-1}
&&\bigg|B(\alpha,1)\big(\frac{f(a)+f(b)}{2})-\frac{\Gamma(\alpha)}{2(b-a)^{\alpha}}[J_{a^{+}}^{\alpha}f(b)+ J_{b_{-}}^{\alpha}f(a)]\bigg|\\
&\leq&\frac{b-a}{2}\Psi_{1}^{\frac{1}{p}}\big[\frac{|f^{'}(a)|^{q}+|f^{'}(b)|^{q}}{2}\big]^{\frac{1}{q}},\nonumber
\end{eqnarray}
where  $\Psi_{1}=2\displaystyle \int_{0}^{\frac{1}{2}}\big(\frac{(1-t)^{\alpha}-t^{\alpha}}{\alpha}\big)^{p}dt$.
\end{corollary}

\begin{remark}\label{remark}
If we take $\alpha=1$ in Corollary \ref{corllry3}, the inequality (\ref{corllry3-1}) reduces to following inequality;
\begin{eqnarray}
&&\bigg|\frac{f(a)+f(b)}{2}-\frac{1}{(b-a)}\int_{a}^{b}f(x)dx\bigg|\\
&\leq&\frac{b-a}{2}\big(\frac{1}{p+1}\big)^{\frac{1}{p}}\big[\frac{|f^{'}(a)|^{q}+|f^{'}(b)|^{q}}{2}\big]^{\frac{1}{q}},\nonumber
\end{eqnarray}
which is the same as Theorem 2.3 in  \cite{dragomir}.
\end{remark}
\begin{remark}
If we take $\alpha\in(0,1]$ in Corollary \ref{corllry3}, then the inequality (\ref{corllry3-1}) reduces to Corollary 1 in \cite{Özdemir},  which is the same as
\begin{eqnarray}
&&\bigg|\big(\frac{f(a)+f(b)}{2})-\frac{\Gamma(\alpha)}{2(b-a)^{\alpha}}[J_{a^{+}}^{\alpha}f(b)+ J_{b_{-}}^{\alpha}f(a)]\bigg|\\
&&\leq\frac{b-a}{2}(\frac{1}{\alpha p+1})^{\frac{1}{p}}\big[\frac{|f^{'}(a)|^{q}+|f^{'}(b)|^{q}}{2}\big]^{\frac{1}{q}}.\nonumber
\end{eqnarray}

\end{remark}

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\end{document}
