1 Introduction
In literature, inequalities are very important for convex functions especially the integral inequalities for convex functions originated form Hermite and Hadamard (see [11, p. 137]). The researchers have worked on Hermite–Hadamard type inequalities since 1893 [4]. The classical Hermite–Hadamard inequality reads as follows: if \(f :I\rightarrow \mathbb{R}\) is convex on the interval I of real numbers and \({a,b}\in I\) with \(a< b\), then We note that the Hermite–Hadamard inequality may be assessed as a treatment of the conception of convexity. The Hermite–Hadamard inequality for convex functions has conferred revived awareness in the latest years and some unusual variations of essential and conclusion have been established (see, for example, [5, 6, 14, 17]). In the last few years, the theory of inequalities has progressed very fast. The evolution of the hypothesis associated with ancient inequalities has developed in a resumption of attentiveness in this field. In many classical inequalities, the Hermite–Hadamard inequality is one of the important inequality of analysis. Such an inequality has been applied for different types of problems of fractional calculus (see [1, 2, 8‐10, 15, 16]). In this paper, as a continuation of the study of the Hermite–Hadamard inequality, we establish some results for k-Riemann–Liouville fractional integral by using the definition of convex functions via fractional calculus.
$$ f \biggl(\frac{a+b}{2} \biggr)\leq \frac{1}{b-a} \int _{a}^{b}f (\mu )\,d\mu \leq \frac{f (a)+f (b)}{2}. $$
Below, let us recall first some basic concepts and some earlier results.
Anzeige
Definition 1
A function \(f :[a,b]\rightarrow \mathbb{R}\) is said to be convex on an interval \([a,b]\subseteq \mathbb{R}\), if holds for \(\xi ,\eta \in [a,b]\) and \(\tau \in [0,1]\).
$$ f \bigl(\tau \xi + (1-\tau ) \eta \bigr) \leq \tau f (\xi )+ (1- \tau ) f (\eta ) $$
(1)
Definition 2
([13])
Let \(f \in L_{1}[a,b]\). The Riemann–Liouville integrals \(I_{a^{+}}^{\lambda }{f }\) and \(I_{b^{-}}^{\lambda }{f }\) of order \({\lambda }>0\) with \(a\geq 0\) are defined by and respectively, where Γ is the classical Gamma function and \(I_{a^{+}}^{0}{f }=I_{b^{-}}^{0}{f }=f (\xi )\).
$$ I_{a^{+}}^{\lambda }{f (\xi )}=\frac{1}{\Gamma (\lambda )} \int _{a}^{\xi }(\xi -\mu )^{\lambda -1}f (\mu )\,d\mu ,\quad \xi >a, $$
$$ I_{b^{-}}^{\lambda }{f (\xi )}=\frac{1}{\Gamma (\lambda )} \int _{\xi }^{b}(\mu -\xi )^{\lambda -1}f (\mu )\,d\mu ,\quad \xi < b, $$
Theorem 1
([14])
Anzeige
Consider \(f :[a,b]\rightarrow \mathbb{R}\) a positive mapping with \(0 \leq a< b\) and \(f \in L_{1}[a,b]\). If f is a convex function on \([a,b]\), then
$$ f \biggl(\frac{a+b}{2} \biggr) \leq \frac{\Gamma (\lambda + 1)}{2(b-a)^{{\lambda }}} \bigl[I^{\lambda }_{a^{+}}f (b)+I^{ \lambda }_{b^{-}}f (a) \bigr] \leq \frac{f (a)+f (b)}{2}. $$
(2)
Lemma 1.1
([14])
Suppose \(f :[a,b]\rightarrow \mathbb{R}\) is a differentiable function on \((a,b)\) with \(a< b\). If \(f ^{\prime }\in \L [a,b]\), then
$$\begin{aligned}& \frac{f (a)+f (b)}{2}- \frac{\Gamma (\lambda +1)}{2(b-a)^{{\lambda }}} \bigl[I^{\lambda }_{a^{+}}f (b)+I^{ \lambda }_{b^{-}}f (a) \bigr] \\& \quad = \frac{b-a}{2} \int _{0}^{1} \bigl[(1-\mu )^{{\lambda }}-\mu ^{{ \lambda }} \bigr]f ^{\prime } \bigl(\mu a+(1-\mu )b \bigr)\,d\mu . \end{aligned}$$
(3)
Theorem 2
([14])
Assume that \(f :[a,b]\rightarrow \mathbb{R}\) is a differentiable function on \((a,b)\) with \(a< b\). If \(|f ^{\prime }|\) is convex on \([a,b]\), then
$$\begin{aligned}& \biggl\vert \frac{f (a)+f (b)}{2}- \frac{{\Gamma (\lambda +1)}}{2(b-a)^{\lambda }} \bigl[I^{\lambda }_{a^{+}}f (b)+I^{ \lambda }_{b^{-}}f (a) \bigr] \biggr\vert \\& \quad \leq \frac{b-a}{2({\lambda }+1)} \biggl(1-\frac{1}{2^{\lambda }} \biggr) \bigl[f ^{\prime }(a)+f ^{\prime }(b) \bigr]. \end{aligned}$$
(4)
Theorem 3
([12])
Let \(f :[a,b]\rightarrow \mathbb{R}\) be a positive mapping with \(0 \leq a< b\) and \(f \in L_{1}[a,b]\). If f is a convex function on \([a,b]\), then
$$ f \biggl(\frac{a+b}{2} \biggr) \leq \frac{2^{{\lambda }-1}\Gamma (\lambda + 1)}{(b-a)^{{\lambda }}} \bigl[I^{ \lambda }_{(\frac{a+b}{2})^{+}}f (b)+I^{\lambda }_{(\frac{a+b}{2})^{-}}f (a) \bigr]\leq \frac{f (a)+f (b)}{2}. $$
(5)
Lemma 1.2
([12])
Suppose \(f :[a,b]\rightarrow \mathbb{R}\) is a differentiable function on \((a,b)\) with \(a< b\). If \(f ^{\prime }\in \L [a,b]\), then
$$\begin{aligned}& \frac{2^{{\lambda }-1}{\Gamma }({\lambda }+1)}{(b-a)^{{\lambda }}} \bigl[I^{ \lambda }_{({\frac{a+b}{2}})^{+}}f (b)+I^{\lambda }_{({\frac{a+b}{2}}) ^{-}}f (a) \bigr]-f \biggl(\frac{a+b}{2} \biggr) \\& \quad = \frac{b-a}{4} \biggl[ \int _{0}^{1}\mu ^{{\lambda }}f ^{ \prime } \biggl(\frac{\mu }{2}a+\frac{2-\mu }{2}b \biggr)\,d\mu - \int _{0}^{1}\mu ^{{\lambda }}f ^{\prime } \biggl(\frac{\mu }{2}b+ \frac{2-\mu }{2}a \biggr)\,d\mu \biggr]. \end{aligned}$$
(6)
Theorem 4
([12])
Consider \(f :[a,b]\rightarrow \mathbb{R,}\) a differentiable function on \((a,b)\) with \(a< b\). If \(|f ^{\prime }|^{h}\) is convex on \([a,b]\) for \(h\geq {1}\), then
$$\begin{aligned}& \biggl\vert \frac{2^{{\lambda }-1}{\Gamma }(\lambda +1)}{(b-a)^{{\lambda }}} \bigl[I^{ \lambda }_{({\frac{a+b}{2}})^{+}}f (b)+I^{\lambda } _{({\frac{a+b}{2}})^{-}}f (a) \bigr]-f \biggl( \frac{a+b}{2} \biggr) \biggr\vert \\& \quad \leq \frac{b-a}{4({\lambda }+1)} \biggl(\frac{1}{2({\lambda }+2)} \biggr)^{\frac{1}{h}} \bigl\{ \bigl(({\lambda }+1) \bigl\vert f ^{\prime }(a) \bigr\vert ^{h}+({ \lambda }+3) \bigl\vert f ^{\prime }(b) \bigr\vert ^{h} \bigr) ^{\frac{1}{h}} \\& \qquad {} + \bigl(({\lambda }+3) \bigl\vert f ^{\prime }(a) \bigr\vert ^{h}+({\lambda }+1) \bigl\vert f ^{ \prime }(b) \bigr\vert ^{h} \bigr)^{\frac{1}{h}} \bigr\} . \end{aligned}$$
(7)
2 Hermite–Hadamard’s inequalities for k-fractional integrals
Definition 3
Let k and \(\mathbb{R}(v)\) be positive. Then the k-gamma function is defined by following integral:
$$ \Gamma _{k}(\xi )= \int _{0}^{\infty }v^{\xi -1}\exp \biggl(- \frac{v^{k}}{k} \biggr)\,dv. $$
Definition 4
([7])
If \(k>0\), Let \(f \in L_{1}(a,b)\), \(a\geq 0\), then k-Riemann–Liouville fractional integrals \(I_{a^{+},k}^{\lambda }{f }\) and \(I_{b^{-},k}^{\lambda }{f }\) of order \({\lambda }>0\) for a real-valued continuous function \(f(\mu )\) are defined by and respectively. Here \(\Gamma _{k}\) is the k-Gamma function.
$$ I_{a^{+},k}^{\lambda }{f (\xi )}=\frac{1}{k\Gamma _{k}(\lambda )} \int _{a}^{\xi }(\xi -\mu )^{\frac{\lambda }{k}-1}f (\mu )\,d\mu ,\quad \xi >a, $$
$$ I_{b^{-},k}^{\lambda }{f (\xi )}=\frac{1}{k\Gamma _{k}(\lambda )} \int _{\xi }^{b}(\mu -\xi )^{\frac{\lambda }{k}-1}f (\mu )\,d\mu ,\quad \xi < b, $$
Theorem 5
Let \(k>0\), \(I^{\lambda }_{a^{+},k}f \) and \(I^{\lambda }_{b^{-},k}f \) be the left and right sided k-Riemann–Liouville fractional integral of order \({\lambda }>0\). Let \(f :[a,b]\rightarrow \mathbb{R}\) be positive mapping with \(0\leq a< b\), \(f \in L_{1}[a,b]\). If f is convex on \([a,b]\), then
$$ f \biggl(\frac{a+b}{2} \biggr) \leq \frac{\Gamma _{k}(\lambda +k)}{2(b-a)^{\frac{\lambda }{k}}} \bigl[I^{\lambda }_{a^{+},k}f (b)+I^{\lambda }_{b^{-},k}f (a) \bigr] \leq \frac{f (a)+f (b)}{2} . $$
(8)
Proof
As f is convex mapping on \([a,b]\), we have, for \(\xi ,\eta \in [a,b]\) with \(\tau =\frac{1}{2}\) in (1), Now let \(\xi =\mu a+(1-\mu )b\) and \(\eta =\mu b+(1-\mu )a\), then (9) becomes Multiplying both sides of (10) by \(\mu ^{\frac{\lambda }{k}-1}\), then integrating with respect to μ over \({[0,1]}\), we get where and By taking \(\mu a+(1-\mu )b=\phi \) in \(I_{1}\) and \(\mu b+(1-\mu )a=\omega \) in \(I_{2}\), we get and Substituting the values of \(I_{1}\) and \(I_{2}\) from (12) and (13) in (11), we get which implies that This completes the first inequality in (8). To complete the second inequality, we note that if f is convex, then, for \({\tau }\in [0,1]\), it yields that and By adding the above two inequalities, we get Multiplying by \(\mu ^{\frac{\lambda }{k}-1}\) on both sides of (15), then integrating with regard to μ over \([0,1]\), we get We denote and Putting \(\phi =\mu a+(1-\mu )b\) in \(K_{1}\), and \(\omega =\mu b+(1-\mu )a\) in \(K_{2}\), we obtain and Substituting the values of \(K_{1}\) and \(K_{2}\) from (17) and (18) in (16), we get which implies that By combining (17), and (19), we get (8). □
$$ f \biggl(\frac{\xi +\eta }{2} \biggr) \leq \frac{f (\xi )+f (\eta )}{2}. $$
(9)
$$ 2f \biggl(\frac{a+b}{2} \biggr)\leq f \bigl(\mu a+(1- \mu )b \bigr)+f \bigl(\mu b+(1- \mu )a \bigr). $$
(10)
$$\begin{aligned} \frac{2k}{\lambda }f \biggl(\frac{a+b}{2} \biggr) \leq & \int _{0}^{1} \mu ^{\frac{\lambda }{k}-1}f { \bigl(\mu a+(1-\mu )b \bigr)}\,d\mu + \int _{0}^{1} \mu ^{\frac{\lambda }{k}-1}f { \bigl(\mu b+(1-\mu )a \bigr)}\,d\mu \\ =&I_{1}+I_{2}, \end{aligned}$$
(11)
$$ I_{1} = \int _{0}^{1}\mu ^{\frac{\lambda }{k}-1}f { \bigl(\mu a+(1- \mu )b \bigr)}\,d\mu $$
$$ I_{2} = \int _{0}^{1}\mu ^{\frac{\lambda }{k}-1}f { \bigl(\mu b+(1- \mu )a \bigr)}\,d\mu . $$
$$ I_{1}=\frac{1}{(b-a)^{\frac{\lambda }{k}}} \int _{a}^{b}(b-\phi )^{ \frac{\lambda }{k}-1}f (\phi )\,d\phi $$
(12)
$$ I_{2}=\frac{1}{(b-a)^{\frac{\lambda }{k}}} \int _{a}^{b}(\omega -a)^{ \frac{\lambda }{k}-1}f ( \omega )\,d\omega . $$
(13)
$$ \frac{2k}{\lambda }f \biggl(\frac{a+b}{2} \biggr)\leq \frac{1}{(b-a)^{\frac{\lambda }{k}}} \biggl[ \int _{a}^{b}(b- \phi )^{\frac{\lambda }{k}-1}f (\phi )\,d\phi + \int _{a}^{b}( \omega -a)^{\frac{\lambda }{k}-1}f ( \omega )\,d\omega \biggr], $$
$$ f \biggl(\frac{a+b}{2} \biggr)\leq \frac{\Gamma _{k}(\lambda +k)}{2(b-a)^{\frac{\lambda }{k}}} \bigl[I^{\lambda }_{a^{+}, k}f (b)+I^{\lambda }_{b^{-},k}f (a) \bigr]. $$
(14)
$$ f { \bigl(\mu a+(1-\mu )b \bigr)}\leq {\mu f (a)+(1-\mu )f (b)} $$
$$ f { \bigl(\mu b+(1-\mu )a \bigr)}\leq {\mu f (b)+(1-\mu )f (a)}. $$
$$ f { \bigl(\mu a+(1-\mu )b \bigr)}+f { \bigl(\mu b+(1-\mu )a \bigr)}\leq f (a)+f (b) . $$
(15)
$$ \int _{0}^{1}\mu ^{\frac{\lambda }{k}-1}f { \bigl(\mu a+(1-\mu )b \bigr)}\,d\mu + \int _{0}^{1}\mu ^{\frac{\lambda }{k}-1}f { \bigl(\mu b+(1-\mu )a \bigr)}\, d\mu \leq \frac{k}{\lambda } \bigl[f (a)+f (b) \bigr] . $$
(16)
$$ K_{1} = \int _{0}^{1}\mu ^{\frac{\lambda }{k}-1}f { \bigl(\mu a+(1- \mu )b \bigr)}\,d\mu $$
$$ K_{2}= \int _{0}^{1}\mu ^{\frac{\lambda }{k}-1}f { \bigl(\mu b+(1- \mu )a \bigr)}\,d\mu . $$
$$ K_{1}= \frac{1}{(b-a)^{\frac{\lambda }{k}}} \int _{a}^{b}(b-\phi )^{ \frac{\lambda }{k}-1}f (\phi )\,d\phi $$
(17)
$$ K_{2}=\frac{1}{(b-a)^{\frac{\lambda }{k}}} \int _{a}^{b}(\omega -a)^{ \frac{\lambda }{k}-1}f ( \omega )\,d\omega . $$
(18)
$$ \frac{1}{(b-a)^{\frac{\lambda }{k}}} \biggl[ \int _{a}^{b}(b- \phi )^{\frac{\lambda }{k}-1}f (\phi )\,d\phi + \int _{a}^{b} ( \omega -a)^{\frac{\lambda }{k}-1}f ( \omega )\,d\omega \biggr]\leq \frac{k}{\lambda } \bigl[f (a)+f (b) \bigr], $$
$$ \frac{\Gamma _{k}(\lambda +k)}{2(b-a)^{\frac{\lambda }{k}}} \bigl[I^{\lambda }_{a^{+},k}f (b)+I^{\lambda }_{b^{-},k}f (a) \bigr]\leq \frac{f (a)+f (b)}{2}. $$
(19)
Lemma 2.1
Let \(k>0\), \(I^{\lambda }_{a^{+},k}f \) and \(I^{\lambda }_{b^{-},k}f \) be defined as Definition 4. Let \(f :[a,b]\rightarrow \mathbb{R}\) be a differentiable mapping on \((a,b)\) with \(a< b\). If \(f ^{\prime }\in \L [a,b]\), then
$$\begin{aligned}& \frac{(b-a)}{2} \int _{0}^{1} \bigl[(1-\mu )^{ \frac{\lambda }{k}}- \mu ^{\frac{\lambda }{k}} \bigr]f ^{\prime } \bigl(\mu a+(1- \mu )b \bigr)\,d\mu \\& \quad = \frac{f (a)+f (b)}{2}- \frac{{\Gamma }_{k}({\lambda }+k)}{2(b-a)^{\frac{\lambda }{k}}} \bigl[I^{ \lambda }_{a^{+}, k}f (b)+I^{\lambda }_{b^{-},k}f (a) \bigr] . \end{aligned}$$
(20)
Proof
Let us consider which, we can write as Integrating \(I_{1}\) by parts, we get Setting \(\xi =\mu a+(1-\mu )b\), then after some calculation, we get Now integrating \(I_{2}\) by parts to get Setting \(\xi =(\mu a+(1-\mu )b)\), after some calculation, we get Applying (22) and (23) in (21), it follows that or Multiplying both sides of (24) by \(\frac{b-a}{2}\) to get the required result. □
$$ I= \int _{0}^{1} \bigl[(1-\mu )^{\frac{\lambda }{k}}- \mu ^{ \frac{\lambda }{k}} \bigr]f ^{\prime } \bigl(\mu a+(1-\mu )b \bigr)\,d\mu , $$
$$\begin{aligned} I =& \biggl[ \int _{0}^{1}(1-\mu )^{\frac{\lambda }{k}}f ^{\prime } \bigl( \mu a+(1-\mu )b \bigr)\,d\mu \biggr]+ \biggl[- \int _{0}^{1}\mu ^{\frac{\lambda }{k}} \bigl(f ^{\prime } \bigl(\mu a+(1-\mu )b \bigr) \,d\mu \bigr) \biggr] \\ =&I_{1}+I_{2}. \end{aligned}$$
(21)
$$ I_{1} = \frac{f (b)}{b-a}-\frac{\lambda }{k(b-a)} \int _{0}^{1}(1- \mu )^{\frac{\lambda }{k}-1}f \bigl( \mu a+(1-\mu )b \bigr)\,d\mu . $$
$$ I_{1}=\frac{f (b)}{b-a}- \frac{{\Gamma }_{k}({\lambda }+k)}{(b-a)^{\frac{\lambda }{k}+1}} \bigl[I^{ \lambda }_{b^{-},k}f (a) \bigr] . $$
(22)
$$ I_{2} = \frac{f (a)}{b-a}-\frac{\lambda }{k(b-a)} \int _{0}^{1} \mu ^{\frac{\lambda }{k}-1}f \bigl(\mu a+(1-\mu )b \bigr)\,d\mu . $$
$$ I_{2}=\frac{f (a)}{b-a}- \frac{{\Gamma }_{k}({\lambda }+k)}{(b-a)^{\frac{\lambda }{k}+1}} \bigl[I^{ \lambda }_{a^{+},k}f (b) \bigr] . $$
(23)
$$ I=\frac{f (b)}{b-a}+\frac{f (a)}{b-a}- \frac{{\Gamma }_{k}({\lambda }+k)}{(b-a)^{\frac{\lambda }{k}+1}} \bigl[I^{ \lambda }_{a^{+},k}f (b)+I^{\lambda }_{b^{-},k}f (a) \bigr], $$
$$\begin{aligned}& \int _{0}^{1} \bigl[(1-\mu )^{\frac{\lambda }{k}}- \mu ^{ \frac{\lambda }{k}} \bigr]f ^{\prime } \bigl(\mu a+(1-\mu )b \bigr)\,d\mu \\& \quad = \frac{f (b)}{b-a}+\frac{f (a)}{b-a}- \frac{{\Gamma }_{k} ({\lambda }+k)}{(b-a)^{\frac{\lambda }{k}+1}} \bigl[I^{ \lambda }_{a^{+},k}f (b)+I^{\lambda }_{b^{-},k}f (a) \bigr] . \end{aligned}$$
(24)
Theorem 6
Let \(k>0\), \(I^{\lambda }_{a^{+},k}f \) and \(I^{\lambda }_{b^{-},k}f \) be defined as Definition 4. Let \(f :[a,b]\rightarrow \mathbb{R}\) be differentiable on \((a,b)\) with \(a< b\). If \(|f ^{\prime }|\) is convex on \([a,b]\), then
$$\begin{aligned}& \biggl\vert \frac{f (a)+f (b)}{2}- \frac{\Gamma _{k}(\lambda +k)}{2(b-a)^{\frac{\lambda }{k}}} \bigl[I^{\lambda }_{a^{+},k}f (b)+I^{\lambda }_{b^{-} ,k}f (a) \bigr] \biggr\vert \\& \quad \leq \frac{(b-a)}{2(\frac{\lambda }{k}+1)} \biggl(1- \frac{1}{2^{\frac{\lambda }{k}}} \biggr) \bigl( \bigl\vert f ^{\prime }(a) \bigr\vert + \bigl\vert f ^{ \prime }(b) \bigr\vert \bigr). \end{aligned}$$
(25)
Proof
By using Lemma 2.1 and the definition of a convex function of \(|f ^{\prime }|\), we have where and We calculate \(K_{1}\) to get Similarly we can calculate \(K_{2}\) and get Substituting the values of \(K_{1}\) and \(K_{2}\) in (26) and after some calculations, we get (25). □
$$\begin{aligned}& \biggl\vert \frac{f (a)+f (b)}{2}- \frac{\Gamma _{k}(\lambda +k)}{2(b-a)^{\frac{\lambda }{k}}} \bigl[I^{\lambda }_{a^{+},k}f (b)+I^{\lambda } _{b^{-},k}f (a) \bigr] \biggr\vert \\& \quad \leq \frac{(b-a)}{2} \int _{0}^{1} \bigl\vert (1-\mu )^{ \frac{\lambda }{k}}-\mu ^{\frac{\lambda }{k}} \bigr\vert \bigl\vert f ^{\prime } \bigl( \mu a+(1-\mu )b \bigr) \bigr\vert \,d\mu \\& \quad \leq \frac{(b-a)}{2} \int _{0}^{1} \bigl\vert (1-\mu )^{ \frac{\lambda }{k}}-\mu ^{\frac{\lambda }{k}} \bigr\vert \bigl[\mu \bigl\vert f ^{ \prime }(a) \bigr\vert +(1-\mu ) \bigl\vert f ^{\prime }(b) \bigr\vert \bigr]\,d\mu \\& \quad = \frac{(b-a)}{2} \biggl[ \int _{0}^{1} \bigl\vert (1-\mu )^{\frac{\lambda }{k}}-\mu ^{\frac{\lambda }{k}} \bigr\vert \mu \bigl\vert f ^{\prime }(a) \bigr\vert \,d\mu + \int _{0}^{1} \bigl\vert (1-\mu )^{\frac{\lambda }{k}} -\mu ^{\frac{\lambda }{k}} \bigr\vert (1-\mu ) \bigl\vert f ^{\prime }(b) \bigr\vert \,d\mu \biggr] \\& \quad = \frac{(b-a)}{2} (K_{1}+K_{2} ), \end{aligned}$$
(26)
$$\begin{aligned} K_{1} =& \bigl\vert f ^{\prime }(a) \bigr\vert \biggl[ \int _{0}^{\frac{1}{2}}\mu (1- \mu )^{\frac{\lambda }{k}}\,d\mu - \int _{0}^{\frac{1}{2}}\mu ^{ \frac{\lambda }{k}+1}\,d\mu \biggr] \\ &{}+ \bigl\vert f ^{\prime }(b) \bigr\vert \biggl[ \int _{0}^{\frac{1}{2}}(1-\mu )^{ \frac{\lambda }{k}+1}\,d\mu - \int _{0}^{\frac{1}{2}}(1-\mu )\mu ^{\frac{\lambda }{k}}\,d\mu \biggr] \end{aligned}$$
$$\begin{aligned} K_{2} =& \bigl\vert f ^{\prime }(a) \bigr\vert \biggl[ \int _{\frac{1}{2}}^{1}\mu ^{ \frac{\lambda }{k}+1}\,d\mu - \int _{\frac{1}{2}}^{1}\mu (1-\mu )^{\frac{\lambda }{k}}\,d\mu \biggr] \\ &{}+ \bigl\vert f ^{\prime }(b) \bigr\vert \biggl[ \int _{\frac{1}{2}}^{1}(1-\mu )\mu ^{ \frac{\lambda }{k}}\,d\mu - \int _{\frac{1}{2}}^{1}(1-\mu )^{ \frac{\lambda }{k}+1}\,d\mu \biggr]. \end{aligned}$$
$$ K_{1}= \bigl\vert f ^{\prime }(a) \bigr\vert \biggl[ \frac{1}{({\frac{\lambda }{k}+1})({\frac{\lambda }{k}+2})}- \frac{(\frac{1}{2})^{\frac{\lambda }{k}+1}}{(\frac{\lambda }{k}+1)} \biggr]- \bigl\vert f ^{\prime }(b) \bigr\vert \biggl[ \frac{1}{(\frac{\lambda }{k}+2)}- \frac{(\frac{1}{2})^{\frac{\lambda }{k}+1}}{({\frac{\lambda }{k}+1})} \biggr] . $$
(27)
$$ K_{2}= \bigl\vert f ^{\prime }(a) \bigr\vert \biggl[ \frac{1}{(\frac{\lambda }{k}+2)}- \frac{(\frac{1}{2})^{\frac{\lambda }{k}+1}}{(\frac{\lambda }{k}+1)} \biggr]+ \bigl\vert f ^{\prime }(b) \bigr\vert \biggl[ \frac{1}{({\frac{\lambda }{k}+1})({\frac{\lambda }{k}+2})}- \frac{(\frac{1}{2})^{\frac{\lambda }{k}+1}}{(\frac{\lambda }{k}+1)} \biggr] . $$
(28)
3 Some more fractional inequalities for convex functions
Definition 5
Let \(f \in L_{1}[a,b]\). The k-Riemann–Liouville integrals \(I^{\lambda }_{(\frac{a+b}{2})^{+}, k}f \) and \(I^{\lambda }_{(\frac{a+b}{2})^{-}, k}f \) of order \({\lambda }>0\) and \(k>0\) with \(a\geq 0\) are defined by and respectively. Here \({\Gamma _{k}{(\lambda )}}\) is the k-Gamma function.
$$ I^{\lambda }_{(\frac{a+b}{2})^{+}, k}{f (\xi )}= \frac{1}{k\Gamma _{k}(\lambda )} \int _{\frac{a+b}{2}}^{\xi }( \xi -\mu )^{\frac{\lambda }{k}-1} f ( \mu )\,d\mu ,\quad \xi >{ \frac{a+b}{2}}, $$
$$ I^{\lambda }_{(\frac{a+b}{2})^{-}, k}{f (\xi )}= \frac{1}{k\Gamma _{k}(\lambda )} \int _{\xi }^{\frac{a+b}{2}}( \mu -\xi )^{\frac{\lambda }{k}-1} f ( \mu )\,d\mu ,\quad \xi < { \frac{a+b}{2}}, $$
Theorem 7
Let \(k>0\), \(I^{\lambda }_{(\frac{a+b}{2})^{+}, k}f \) and \(I^{\lambda }_{(\frac{a+b}{2})^{-}, k}f \) be defined in Definition 5. Let \(f :[a,b]\rightarrow \mathbb{R}\) be positive mapping with \(0\leq a< b\) and \(f \in L_{1}[a,b]\). If f is a convex function on \([a,b]\), then
$$ f \biggl(\frac{a+b}{2} \biggr) \leq \frac{2^{\frac{\lambda }{k}-1}\Gamma _{k}(\lambda +k)}{(b-a)^{\frac{\lambda }{k}}} \bigl[I^{ \lambda }_{(\frac{a+b}{2})^{+},k} f (b)+I^{\lambda }_{(\frac{a+b}{2})^{-}, k}f (a) \bigr]\leq \frac{f (a)+f (b)}{2}. $$
(29)
Proof
As f is convex function on \([a,b]\), we have, for \(\xi ,\eta \in [a,b]\) with \(\tau =\frac{1}{2}\), Putting \(\xi =\frac{\mu a}{2}+\frac{(2-\mu )b}{2}\) and \(\eta =\frac{ \mu b}{2}+\frac{(2-\mu )a}{2}\), then (30) becomes Multiplying by \(\mu ^{\frac{\lambda }{k}-1}\) on both sides of (31), then integrating with respect to μ over \({[0,1]}\), we get We set Taking \(\phi =\frac{\mu a}{2}+\frac{(2-\mu )b}{2}\), after some calculations we get and we set Putting \(\omega =\frac{\mu b}{2}+\frac{(2-\mu )a}{2}\) to get Substituting the values of \(I_{1}\) and \(I_{2}\) from (33) and (35) in (32), we get The first part of the inequality is proved. To complete the second inequality, we note that if f is convex function, then, for \({\tau }\in [0,1]\), showing and By adding the above two inequalities, we get Multiplying by \(\mu ^{\frac{\lambda }{k}-1}\) on both sides of (37) and integrating inequalities with respect to μ over \([0,1]\), we get We take and choose \(\phi =\frac{\mu a}{2}+\frac{(2-\mu )b}{2}\), we get after some simple calculations Likewise we take and choose \(\omega =\frac{\mu b}{2}+\frac{(2-\mu )a}{2}\), we get Substituting the values of \(L_{1}\) and \(L_{2}\) from (39) and (40) in (38), we get This implies that From (36) and (41), we get the required result. □
$$ f \biggl(\frac{\xi +\eta }{2} \biggr) \leq \frac{f(\xi )+f (\eta )}{2}. $$
(30)
$$ 2f \biggl(\frac{a+b}{2} \biggr)\leq f \biggl( \frac{\mu a}{2}+ \frac{(2-\mu )b}{2} \biggr)+ f \biggl( \frac{\mu b}{2}+ \frac{(2-\mu )a}{2} \biggr). $$
(31)
$$\begin{aligned} \frac{2k}{\lambda }f \biggl(\frac{a+b}{2} \biggr) \leq& \int _{0}^{1} \mu ^{\frac{\lambda }{k}-1}f \biggl( \frac{\mu a}{2}+ \frac{(2-\mu )b}{2} \biggr)\,d\mu \\ &{}+ \int _{0}^{1}\mu ^{ \frac{\lambda }{k}-1}f \biggl( \frac{\mu b}{2}+\frac{(2-\mu )a}{2} \biggr)\,d\mu . \end{aligned}$$
(32)
$$ I_{1}= \int _{0}^{1}\mu ^{\frac{\lambda }{k}-1}f \biggl( \frac{\mu a}{2}+\frac{(2-\mu )b}{2} \biggr)\,d\mu . $$
(33)
$$ I_{1}=\frac{2^{\frac{\lambda }{k}}}{(b-a)^{\frac{\lambda }{k}}} \int _{\frac{a+b}{2}}^{b}(b-\phi )^{\frac{\lambda }{k}-1}f (\phi )\, d\phi , $$
(34)
$$ I_{2}= \int _{0}^{1}\mu ^{\frac{\lambda }{k}-1}f \biggl( \frac{\mu b}{2}+\frac{(2-\mu )a}{2} \biggr)\,d\mu . $$
$$ I_{2}=\frac{2^{\frac{\lambda }{k}}}{(b-a)^{\frac{\lambda }{k}}} \int _{a}^{\frac{a+b}{2}}(\omega -a)^{\frac{\lambda }{k}-1}f ( \omega )\,d\omega . $$
(35)
$$ f \biggl(\frac{a+b}{2} \biggr) \leq \frac{2^{\frac{\lambda }{k}-1}{\Gamma }_{k}(\lambda + k)}{(b-a)^{\frac{\lambda }{k}}} \bigl[I^{\lambda }_{({\frac{a+b}{2}})^{+},k}f (b)+I^{ \lambda }_{({\frac{a+b}{2}})^{-}, k}f (a) \bigr]. $$
(36)
$$ f \biggl(\frac{\mu }{2}a+\frac{2-\mu }{2}b \biggr)\leq \frac{\mu }{2}f (a)+ \frac{2-\mu }{2}f (b) $$
$$ f \biggl(\frac{\mu }{2}b+\frac{2-\mu }{2}a \biggr)\leq \frac{\mu }{2}f (b)+ \frac{2-\mu }{2}f (a). $$
$$ f \biggl(\frac{\mu }{2}a+ \frac{2-\mu }{2}b \biggr)+f \biggl( \frac{\mu }{2}b+ \frac{2-\mu }{2}a \biggr)\leq f (a)+f (b). $$
(37)
$$ \int _{0}^{1}\mu ^{\frac{\lambda }{k}-1}f \biggl( \frac{\mu }{2}a+ \frac{2-\mu }{2}b \biggr)\,d\mu + \int _{0}^{1}\mu ^{ \frac{\lambda }{k}-1} f \biggl( \frac{\mu }{2}b+\frac{2-\mu }{2}a \biggr)\,d\mu \leq \frac{k}{\lambda } \bigl[f (a)+f (b) \bigr] .$$
(38)
$$ L_{1}= \int _{0}^{1}\mu ^{\frac{\lambda }{k}-1}f \biggl( \frac{\mu a}{2}+\frac{(2-\mu )b}{2} \biggr)\,d\mu $$
$$ L_{1}=\frac{2^{\frac{\lambda }{k}}}{(b-a)^{\frac{\lambda }{k}}} \int _{\frac{a+b}{2}}^{b}(b-\phi )^{\frac{\lambda }{k}-1}f (\phi )\, d\phi . $$
(39)
$$ L_{2}= \int _{0}^{1}\mu ^{\frac{\alpha }{k}-1}f \biggl( \frac{\mu b}{2}+\frac{(2-\mu )a}{2} \biggr)\,d\mu , $$
$$ L_{2}=\frac{2^{\frac{\lambda }{k}}}{(b-a)^{\frac{\lambda }{k}}} \int _{a}^{\frac{a+b}{2}}(\omega -a)^{\frac{\lambda }{k}-1}f ( \omega )\,d\omega . $$
(40)
$$ \frac{2^{\frac{\lambda }{k}}}{(b-a)^{\frac{\lambda }{k}}} \biggl[ \int _{\frac{a+b}{2}}^{b}(b-\phi )^{\frac{\lambda }{k}-1}f (\phi )\, d\phi + \int _{a}^{\frac{a+b}{2}}( \omega -a)^{ \frac{\lambda }{k}-1}f ( \omega )\,d\omega \biggr] \leq \frac{k}{\lambda } \bigl[f (a)+f (b) \bigr]. $$
$$ \frac{2^{\frac{\lambda }{k}-1}{\lambda }{\Gamma }_{k}(\lambda )}{(b-a)^{\frac{\lambda }{k}}} \bigl[I^{\lambda }_{(\frac{a+b}{2})^{+},k}f (b)+I^{\lambda } _{({ \frac{a+b}{2}})^{-},k}f (a) \bigr]\leq \frac{f (a)+f (b)}{2} . $$
(41)
Lemma 3.1
Let \(k>0\), \(I^{\lambda }_{(\frac{a+b}{2})^{+}, k}f \) and \(I^{\lambda }_{(\frac{a+b}{2})^{-}, k}f \) be defined as Definition 5. Let \(f :[a,b]\rightarrow \mathbb{R}\) be differentiable on \((a,b)\) with \(a< b\). If \(f ^{\prime }\in \L [a,b]\), then
$$\begin{aligned}& \frac{2^{\frac{\lambda }{k}-1}{\Gamma }_{k}(\lambda +k)}{(b-a)^{\frac{\lambda }{k}}} \bigl[I^{ \lambda }_{({\frac{a+b}{2}})^{+},k}f (b)+I^{\lambda }_{({\frac{a+b}{2}})^{-},k}f (a) \bigr]-f \biggl(\frac{a+b}{2} \biggr) \\& \quad = \frac{b-a}{4} \biggl[ \int _{0}^{1}\mu ^{\frac{\lambda }{k}}f ^{ \prime } \biggl(\frac{\mu }{2}a+\frac{2-\mu }{2}b \biggr)\,d\mu - \int _{0}^{1}\mu ^{\frac{\lambda }{k}}f ^{\prime } \biggl( \frac{\mu }{2}b+\frac{2-\mu }{2}a \biggr)\,d\mu \biggr]. \end{aligned}$$
(42)
Proof
Let Note that Substituting \(\xi =\frac{\mu a}{2}+\frac{(2-\mu )b}{2}\), we get after some computations Similarly we can write for \(I_{2}\) Taking \(\xi =\frac{\mu b}{2}+\frac{(2-\mu )a}{2}\), we get By using (43) and (44), it follows that Thus, multiplying \(\frac{b-a}{4}\) on both sides of the above, we get (42). □
$$\begin{aligned} I =& \biggl[ \int _{0}^{1}\mu ^{\frac{\lambda }{k}}f ^{\prime } \biggl(\frac{\mu }{2}a+\frac{2-\mu }{2}b \biggr)\,d\mu - \int _{0}^{1} \mu ^{\frac{\lambda }{k}}f ^{\prime } \biggl(\frac{\mu }{2}b+\frac{2-\mu }{2}a \biggr)\,d\mu \biggr] \\ =&I_{1}-I_{2}. \end{aligned}$$
$$\begin{aligned} I_{1} =& \int _{0}^{1}\mu ^{\frac{\lambda }{k}}f ^{\prime } \biggl(\frac{\mu }{2}a+\frac{2-\mu }{2}b \biggr)\,d\mu \\ =&-f \biggl(\frac{a+b}{2} \biggr) \biggl(\frac{2}{b-a} \biggr)+ \frac{\lambda }{k} \biggl(\frac{2}{b-a} \biggr) \int _{0}^{1} \mu ^{\frac{\lambda }{k}-1} f \biggl( \frac{\mu }{2}a+\frac{2-\mu }{2}b \biggr)\,d\mu . \end{aligned}$$
$$ I_{1}=-f \biggl(\frac{a+b}{2} \biggr) \biggl( \frac{2}{b-a} \biggr)+ \frac{2^{\frac{\lambda }{k}+1}{\Gamma }_{k} ({{\lambda }+{k}})}{(b-a)^{\frac{\lambda }{k}+1}} \bigl[I^{\lambda }_{({\frac{a+b}{2}})^{+},k}f (b) \bigr]. $$
(43)
$$\begin{aligned} I_{2} =& \int _{0}^{1}\mu ^{\frac{\lambda }{k}}f ^{\prime } \biggl(\frac{\mu }{2}b+\frac{2-\mu }{2}a \biggr)\,d\mu \\ =&f \biggl(\frac{a+b}{2} \biggr) \biggl(\frac{2}{b-a} \biggr)- \frac{\lambda }{k} \int _{0}^{1}\mu ^{\frac{\lambda }{k}-1}f \biggl( \frac{\mu }{2}b +\frac{2-\mu }{2}a \biggr) \biggl( \frac{2}{b-a} \biggr)\,d\mu . \end{aligned}$$
$$ I_{2}=f \biggl(\frac{a+b}{2} \biggr) \biggl( \frac{2}{b-a} \biggr)- \frac{2^{\frac{\lambda }{k}+1}{\Gamma }_{k}({{\lambda }+{k}})}{(b-a)^{\frac{\lambda }{k}+1}} \bigl[I^{\lambda }_{({\frac{a+b}{2}})^{-},k}f (a) \bigr] . $$
(44)
$$ I_{1}-I_{2}= \frac{2^{\frac{\lambda }{k}+1}{\Gamma }_{k} ({{\lambda }+{k}})}{(b-a)^{\frac{\lambda }{k}+1}} \bigl[I^{\lambda }_{({\frac{a+b}{2}})^{+}, k}f (b)+I^{\lambda }_{({ \frac{a+b}{2}})^{-},k}f (a) \bigr]-f \biggl(\frac{a+b}{2} \biggr) \biggl(\frac{4}{b-a} \biggr). $$
Theorem 8
Let \(k>0\), \(I^{\lambda }_{(\frac{a+b}{2})^{+}, k}f \) and \(I^{\lambda }_{(\frac{a+b}{2})^{-}, k}f \) be defined as Definition 5. Let \(f :[a,b]\rightarrow \mathbb{R}\) be a differentiable function on \((a,b)\) with \(a< b\). If \(|f ^{\prime }|^{h}\) is convex on \([a,b]\) for \(h\geq {1}\), then
$$\begin{aligned}& \biggl\vert \frac{2^{\frac{\lambda }{k}-1}{\Gamma }_{k}(\lambda +k)}{(b-a)^{\frac{\lambda }{k}}} \bigl[I^{ \lambda }_{({\frac{a+b}{2}})^{+},k}f (b)+I^{\lambda } _{({\frac{a+b}{2}})^{-},k}f (a) \bigr]-f \biggl( \frac{a+b}{2} \biggr) \biggr\vert \\& \quad \leq \biggl(\frac{b-a}{4}\biggr) \biggl(\frac{1}{\frac{\lambda}{k}+1}\biggr) \biggl(\frac{1}{2(\frac{\lambda }{k}+2)} \biggr)^{ \frac{1}{h}} \biggl[ \biggl( \biggl(\frac{{\lambda }}{k}+1 \biggr) \bigl\vert f ^{\prime }(a) \bigr\vert ^{h}+ \biggl( \frac{{\lambda }}{k}+3 \biggr) \bigl\vert f ^{\prime }(b) \bigr\vert ^{h} \biggr)^{\frac{1}{h}} \\& \qquad {} + \biggl( \biggl(\frac{{\lambda }}{k}+3 \biggr) \bigl\vert f ^{\prime }(a) \bigr\vert ^{h}+ \biggl( \frac{{\lambda }}{k}+1 \biggr) \bigl\vert f ^{\prime }(b) \bigr\vert ^{h} \biggr)^{\frac{1}{h}} \biggr] . \end{aligned}$$
(45)
Proof
First, we consider the case of \(h=1\). By using Lemma 3.1, and the definition of convex function of \(|f ^{\prime }|\), we obtain Now we consider the case of \(h>1\). By using Lemma 3.1, the Holder inequality and the definition of convex function of \(|f ^{\prime }|^{h}\), we get This completes the proof. □
$$\begin{aligned}& \biggl\vert \frac{2^{\frac{\lambda }{k}-1}{\Gamma }_{k}(\lambda + k)}{(b-a)^{\frac{\lambda }{k}}} \bigl[I^{ \lambda }_{({\frac{a+b}{2}})^{+}, k}f (b)+I^{\lambda }_{({\frac{a+b}{2}})^{-},k}f (a) \bigr]-f \biggl(\frac{a+b}{2} \biggr) \biggr\vert \\& \quad \leq \frac{b-a}{4} \int _{0}^{1}\mu ^{\frac{\lambda }{k}} \biggl[ \biggl\vert f ^{\prime } \biggl(\frac{\mu }{2}a+ \frac{2-\mu }{2}b \biggr) \biggr\vert + \biggl\vert f ^{\prime } \biggl(\frac{\mu }{2}b+ \frac{2-\mu }{2}a \biggr) \biggr\vert \,d\mu \biggr] \\& \quad \leq \frac{b-a}{4} \biggl( \int _{0}^{1}\mu ^{ \frac{\lambda }{k}} \biggl[ \frac{\mu }{2} \bigl\vert f ^{\prime }(a) \bigr\vert + \frac{2-\mu }{2} \bigl\vert f ^{\prime }(b) \bigr\vert + \frac{\mu }{2} \bigl\vert f ^{\prime }(b) \bigr\vert + \frac{2-\mu }{2} \bigl\vert f ^{\prime }(a) \bigr\vert \biggr]\,d\mu \biggr) \\& \quad = \frac{b-a}{4} \biggl( \int _{0}^{1} \biggl[ \frac{\mu ^{\frac{\lambda }{k}+1}}{2} \bigl\vert f ^{\prime }(a) \bigr\vert + \frac{2{\mu ^{\frac{\lambda }{k}}}- \mu ^{\frac{\lambda }{k}+1}}{2} \bigl\vert f ^{ \prime }(b) \bigr\vert \\& \qquad {} +\frac{\mu ^{\frac{\lambda }{k}+1}}{2} \bigl\vert f ^{\prime }(b) \bigr\vert + \frac{2{\mu ^{\frac{\lambda }{k}}}-\mu ^{\frac{\lambda }{k}+1}}{2} \bigl\vert f ^{ \prime }(a) \bigr\vert \biggr]\,d\mu \biggr) \\& \quad = \frac{b-a}{4} \biggl( \biggl[ \frac{ \vert f ^{\prime }(a) \vert }{2(\frac{\lambda }{k}+2)}+ \frac{2{(\frac{\lambda }{k}+2)}-({\frac{\lambda }{k}+1})}{({\frac{\lambda }{k}+1}) ({\frac{\lambda }{k}+2})2} \bigl\vert f ^{\prime }(b) \bigr\vert \\& \qquad {} + \frac{2{(\frac{\lambda }{k}+2)}- ({\frac{\lambda }{k}+1})}{({\frac{\lambda }{k}+1})({\frac{\lambda }{k}+2})2} \bigl\vert f ^{\prime }(b) \bigr\vert + \frac{f ^{\prime }(b)}{2(\frac{\lambda }{k}+2)} \biggr] \biggr) \\& \quad = \frac{b-a}{4} \biggl( \biggl[ \frac{ \vert f ^{\prime }(a) \vert }{(\frac{\lambda }{k}+2)}+ \frac{(\frac{\lambda }{k}+3)}{2({\frac{\lambda }{k}+1}) ({\frac{\lambda }{k}+2})} \bigl\vert f ^{\prime }(b) \bigr\vert \\& \qquad {} + \frac{(\frac{\lambda }{k}+3)}{2({\frac{\lambda }{k}+1})({\frac{\lambda }{k}+2})} \bigl\vert f ^{\prime }(a) \bigr\vert + \frac{ \vert f ^{\prime }(b) \vert }{(\frac{\lambda }{k}+2)} \biggr] \biggr) \\& \quad = \frac{b-a}{4} \biggl[ \biggl(\frac{1}{2{(\frac{\lambda }{k}+2)}}+ \frac{({\frac{\lambda }{k}+3})}{2({\frac{\lambda }{k}+1}) ({\frac{\lambda }{k}+2})} \biggr) \bigl\vert f ^{\prime }(a) \bigr\vert \\& \qquad {} + \biggl( \frac{({\frac{\lambda }{k}+3})}{2({\frac{\lambda }{k}+1}) ({\frac{\lambda }{k}+2})}+ \frac{1}{2{(\frac{\lambda }{k}+2)}} \biggr) \bigl\vert f ^{\prime }(b) \bigr\vert \biggr] \\& \quad = \frac{b-a}{4} \biggl( \frac{2({\frac{\lambda }{k}+2})}{2({\frac{\lambda }{k}+1})({\frac{\lambda }{k}+2})} \bigl[ \bigl\vert f ^{\prime }(a) \bigr\vert + \bigl\vert f ^{\prime }(b) \bigr\vert \bigr] \biggr) \\& \quad = \frac{b-a}{4({\frac{\lambda }{k}+1})} \bigl[ \bigl\vert f ^{\prime }(a) \bigr\vert + \bigl\vert f ^{\prime }(b) \bigr\vert \bigr]. \end{aligned}$$
$$\begin{aligned}& \biggl\vert \frac{2^{\frac{\lambda }{k}-1}{\Gamma }_{k}(\lambda +k)}{(b-a)^{\frac{\lambda }{k}}} \bigl[I^{ \lambda }_{({\frac{a+b}{2}})^{+}, k}f (b)+I^{\lambda }_{({\frac{a+b}{2}})^{-},k}f (a) \bigr]-f \biggl(\frac{a+b}{2} \biggr) \biggr\vert \\& \quad \leq \frac{b-a}{4} \biggl[ \int _{0}^{1}\mu ^{ \frac{\lambda }{k}}f ^{\prime } \biggl(\frac{\mu }{2}a+\frac{2-\mu }{2}b \biggr)\,d\mu \\& \qquad {}+ \int _{0}^{1}\mu ^{\frac{\lambda }{k}}f ^{\prime } \biggl(\frac{\mu }{2}b+\frac{2-\mu }{2}a \biggr)\,d\mu \biggr] \\& \quad \leq \frac{b-a}{4} \biggl[ \biggl( \int _{0}^{1}\mu ^{ \frac{\lambda }{k}} \biggl\vert f ^{\prime } \biggl(\frac{\mu }{2}a+ \frac{2-\mu }{2}b \biggr) \biggr\vert ^{h}\,d\mu \biggr)^{\frac{1}{h}} \\& \qquad {} + \biggl( \int _{0}^{1}\mu ^{\frac{\lambda }{k}} \biggl\vert f ^{\prime } \biggl(\frac{\mu }{2}b+\frac{2-\mu }{2}a \biggr) \biggr\vert ^{h}\,d\mu \biggr)^{\frac{1}{h}} \biggr]\biggl(\int_{0}^{1} \mu^{\frac{\lambda}{k}}\,d\mu \biggr)^{1-\frac{1}{h}} \\& \quad \leq \frac{b-a}{4} \biggl[ \biggl( \int _{0}^{1}\mu ^{ \frac{\lambda }{k}} \biggl( \frac{\mu }{2} \bigl\vert f ^{\prime }(a) \bigr\vert ^{h}+ \frac{2-\mu }{2} \bigl\vert f ^{\prime }(b) \bigr\vert ^{h} \biggr)\,d\mu \biggr)^{ \frac{1}{h}} \\& \qquad {} + \biggl( \int _{0}^{1}\mu ^{\frac{\lambda }{k}} \biggl( \frac{\mu }{2} \bigl\vert f ^{\prime }(b) \bigr\vert ^{h}+ \frac{2-\mu }{2} \bigl\vert f ^{\prime }(a) \bigr\vert ^{h} \biggr)\,d\mu \biggr)^{\frac{1}{h}} \biggr]\biggl(\int_{0}^{1} \mu^{\frac{\lambda}{k}}\,d\mu \biggr)^{1-\frac{1}{h}} \\& \quad = \frac{b-a}{4} \biggl[ \biggl( \int _{0}^{1} \biggl( \frac{\mu ^{\frac{\lambda }{k}+1}}{2} \bigl\vert f ^{\prime }(a) \bigr\vert ^{h}+ \frac{2{\mu ^{\frac{\lambda }{k}}}-{\mu ^{\frac{\lambda }{k}+1}}}{2} \bigl\vert f ^{ \prime }(b) \bigr\vert ^{h} \biggr)\,d\mu \biggr)^{\frac{1}{h}} \\& \qquad {} + \biggl( \int _{0}^{1} \biggl( \frac{\mu ^{\frac{\lambda }{k}+1}}{2} \bigl\vert f ^{\prime }(b) \bigr\vert ^{h}+ \frac{2{\mu ^{\frac{\lambda }{k}}}-{\mu ^{\frac{\lambda }{k}+1}}}{2} \bigl\vert f ^{ \prime }(a) \bigr\vert ^{h} \biggr)\,d\mu \biggr)^{\frac{1}{h}} \biggr]\biggl(\int_{0}^{1} \mu^{\frac{\lambda}{k}}\,d\mu \biggr)^{1-\frac{1}{h}} \\& \quad = \frac{b-a}{4} \biggl[ \biggl( \frac{ \vert f ^{\prime }(a) \vert ^{h}}{2({\frac{\lambda }{k}+2})}+ \frac{({\frac{\lambda }{k}+3})}{2({\frac{\lambda }{k}+1})({\frac{\lambda }{k}+2})} \bigl\vert f ^{\prime }(b) \bigr\vert ^{h} \biggr)^{\frac{1}{h}} \\& \qquad {} + \biggl( \frac{ \vert f ^{\prime }(b) \vert ^{h}}{2({\frac{\lambda }{k}+2})} + \frac{({\frac{\lambda }{k}+3})}{2({\frac{\lambda }{k}+1})({\frac{\lambda }{k}+2})} \bigl\vert f ^{\prime }(a) \bigr\vert ^{h} \biggr)^{\frac{1}{h}} \biggr]\biggl(\int_{0}^{1} \mu^{\frac{\lambda}{k}}\,d\mu \biggr)^{1-\frac{1}{h}} \\& \quad \leq \frac{b-a}{4} \biggl( \frac{1}{2 (\frac{\lambda }{k}+1 )({\frac{\lambda }{k}+2})} \biggr)^{\frac{1}{h}} \biggl[ \biggl( \biggl({\frac{\lambda }{k}+1} \biggr) \bigl\vert f ^{ \prime }(a) \bigr\vert ^{h}+ { \biggl( \frac{\lambda }{k}+3 \biggr)} \bigl\vert f ^{\prime }(b) \bigr\vert ^{h} \biggr)^{ \frac{1}{h}} \\& \qquad {} + \biggl( \biggl({\frac{\lambda }{k}+1} \biggr) \bigl\vert f ^{\prime }(b) \bigr\vert ^{h}+{ \biggl( \frac{\lambda }{k}+3 \biggr)} \bigl\vert f ^{\prime }(a) \bigr\vert ^{h} \biggr)^{\frac{1}{h}} \biggr]\biggl(\frac{1}{\frac{\lambda}{k}+1}\biggr)^{1-\frac{1}{h}}. \end{aligned}$$
4 Applications to quadrature formulas
In this section we apply the obtained results in to the error estimations of quadrature formulas. It is shown that our main results contain as special cases results such as mid-point inequality and trapezoid inequality. Also, the Hermite–Hadamard inequality can be deduced directly from our main results.
Proposition 1
(Hermite–Hadamard inequality)
By using the assumptions of Theorem 5with \(\lambda =1\) and \(k=1\), we get the following
$$ f \biggl(\frac{a+b}{2} \biggr)\leq \frac{1}{b-a} \int _{a}^{b}f (\mu )\,d\mu \leq \frac{f (a)+f (b)}{2}. $$
Proposition 2
(Mid-point inequality)
By using the assumptions of Theorem 8with \(\lambda =1\), \(h=1\) and \(k=1\), we get the following mid-point type inequality:
$$ \biggl\vert \frac{1}{(b-a)} \int _{a}^{b}f (\mu )\,d\mu -f \biggl( \frac{a+b}{2} \biggr) \biggr\vert \leq \frac{b-a}{8} \bigl[ \bigl\vert f ^{\prime }(a) \bigr\vert + \bigl\vert f ^{ \prime }(b) \bigr\vert \bigr]. $$
Proposition 3
(Trapezoid inequality)
By using the assumptions of Theorem 6with \(\lambda =1\) and \(k=1\), we get the following trapezoid inequality:
$$ \biggl\vert \frac{f(a)+(b)}{2}-\frac{1}{b-a} \int _{a}^{b}f (\mu )\,d\mu \biggr\vert \leq \frac{b-a}{8} \bigl[ \bigl\vert f ^{\prime }(a) \bigr\vert + \bigl\vert f ^{\prime }(b) \bigr\vert \bigr]. $$
5 Conclusion
We have discussed some Hermite–Hadamard type inequalities for k-Riemann–Liouville fractional integral using the convexity of differentiable functions. We stated our main results by Theorems 5, 6, 7 and 8, and showed that our results contain some existing results as special cases. As applications we have established two inequalities involving the error estimates of quadrature formulas.
Acknowledgements
All authors are thankful to the careful referee and editor for their suggestions which help us to improve the final version of paper.
Competing interests
The authors declare that they have no competing interests.
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