1 Introduction
A function \(f:I\rightarrow \mathbb{R}\), where I is an interval of real numbers, is called convex if the following inequality holds:
for all \(a,b\in I\) and \(t\in [0,1]\). Function f is called concave if −f is convex.
$$ f\bigl(ta+(1-t)b\bigr)\leq tf(a)+(1-t)f(b) $$
(1)
The Hermite–Hadamard inequality [4] for convex functions \(f:I\rightarrow \mathbb{R}\) on an interval of real line is defined as
where \(a,b\in I\) with \(a< b\).
$$ f\biggl(\frac{a+b}{2}\biggr)\leq \frac{1}{b-a} \int^{b}_{a}f(x)\,dx\leq \frac{f(a)+f(b)}{2}, $$
(2)
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Since the Hermite–Hadamard inequality has many applications, many authors generalized this inequality. The Hermite–Hadamard inequality is also established for several kinds of convex functions. For more results and generalizations, see [2, 6, 10‐14]. The Hermite–Hadamard inequality (2) is not only established for the classical integral but also for fractional integrals (e.g., see [1, 7, 18, 22]), for conformable fractional integrals (e.g., see [19, 21]), and recently for generalized fractional integrals (e.g., see [8, 9]).
Definition 1.1
([5])
Let \(s\in (0,1]\). A function \(f:I\subset \mathbb{R}_{+}\rightarrow \mathbb{R}\), where \(\mathbb{R}_{+}=[0,\infty)\), is called s-convex function in the second sense if
for all \(a,b\in I\) and \(t\in [0,1]\).
$$ f\bigl(ta+(1-t)b\bigr)\leq t^{s}f(a)+(1-t)^{s}f(b) $$
(3)
Definition 1.2
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A function \(f:[0,b]\rightarrow \mathbb{R}\), with \(b>0\), is said to be m-convex if the following inequality holds:
for all \(a,c\in [0,b]\) and \(t\in [0,1]\) and for all \(m\in [0,1]\). f is m-concave if −f is m-convex.
$$ f\bigl(ta+m(1-t)c\bigr)\leq tf(a)+m(1-t)f(c) $$
(4)
Definition 1.3
([15])
Let \(\alpha >0\) with \(n-1<\alpha \leq n\), \(n\in \mathbb{N}\), and \(1< x< b\). The left- and right-hand side Riemann–Liouville fractional integrals of order α of function f are given by
and
respectively, where \(\Gamma (\alpha)\) is the gamma function defined by \(\Gamma (\alpha)=\int_{0}^{\infty }e^{-t}t^{\alpha -1}\,dt\).
$$ J^{\alpha }_{a+}f(x)=\frac{1}{\Gamma (\alpha)} \int_{a}^{x}(x-t)^{ \alpha -1}f(t)\,dt, $$
$$ J^{\alpha }_{b-}f(x)=\frac{1}{\Gamma (\alpha)} \int_{x}^{b}(t-x)^{ \alpha -1}f(t)\,dt, $$
Definition 1.4
([16])
Let \(\alpha >0\) with \(n-1<\alpha \leq n\), \(n\in \mathbb{N}\), and \(1< x< b\). The left- and right-hand side Hadamard fractional integrals of order α of function f are given by
and
$$ H^{\alpha }_{a+}f(x)=\frac{1}{\Gamma (\alpha)} \int_{a}^{x} \biggl( \ln \frac{x}{t} \biggr) ^{\alpha -1}\frac{f(t)}{t}\,dt, $$
$$ H^{\alpha }_{b-}f(x)=\frac{1}{\Gamma (\alpha)} \int_{x}^{b} \biggl( \ln \frac{t}{x} \biggr) ^{\alpha -1}\frac{f(t)}{t}\,dt. $$
Definition 1.5
([9])
Let \([a,b]\subset \mathbb{R}\) be a finite interval. Then the left- and right-hand side Katugampola fractional integrals of order \(\alpha (>0)\) of \(f\in X^{p}_{c}(a,b)\) are defined by
and
with \(a< x< b\) and \(\rho >0\), where \(X^{p}_{c}(a,b)\) (\(c\in \mathbb{R}\), \(1\leq p\leq \infty \) ) is the space of those complex-valued Lebesgue measurable functions f on \([a,b]\) for which \(\|f\|_{X^{p}_{c}}<\infty \), where the norm is defined by
for \(1\leq p<\infty \), \(c\in \mathbb{R}\) and for the case \(p=\infty \),
where ess sup stands for essential supremum.
$$ {}^{\rho }I^{\alpha }_{a+}f(x)= \frac{\rho^{1-\alpha }}{\Gamma (\alpha)} \int_{a}^{x}\bigl(x^{\rho }-t^{ \rho } \bigr)^{\alpha -1}t^{\rho -1}f(t)\,dt $$
$$ {}^{\rho }I^{\alpha }_{b-}f(x)= \frac{\rho^{1-\alpha }}{\Gamma (\alpha)} \int_{x}^{b}\bigl(t^{\rho }-x^{ \rho } \bigr)^{\alpha -1}t^{\rho -1}f(t)\,dt, $$
$$ \Vert f \Vert _{X^{p}_{c}}= \biggl( \int_{a}^{b} \bigl\vert t^{c}f(t) \bigr\vert ^{p}\frac{dt}{t} \biggr) ^{1/p}< \infty $$
$$ \Vert f \Vert _{X^{\infty }_{c}}= \mathop{\operatorname{ess\,sup}} _{a\leq t\leq b}\bigl[t^{c} \bigl\vert f(t) \bigr\vert \bigr], $$
Theorem 1.6
([9])
Let
\(\alpha >0\)
and
\(\rho >0\). Then, for
\(x>a\),
1.
\(\lim_{\rho \rightarrow 1} ^{\rho }I^{\alpha }_{a+}f(x)=J^{ \alpha }_{a+}f(x)\),
2.
\(\lim_{\rho \rightarrow 0+} ^{\rho }I^{\alpha }_{a+}f(x)=H^{ \alpha }_{a+}f(x)\).
Lemma 1.7
([20])
For
\(0<\alpha \leq 1\)
and
\(0\leq a< b\), we have
$$ \bigl\vert a^{\alpha }-b^{\alpha } \bigr\vert \leq (b-a)^{\alpha }. $$
We recall the classical beta functions:
We introduce the following generalization of beta function:
Note that as \(\rho \rightarrow 1\) then \(^{\rho }\gamma (a,b)\rightarrow \beta (a,b)\).
$$ \beta (a,b)= \int_{0}^{1}x^{a-1}(1-x)^{b-1} \,dx. $$
$$ ^{\rho }\gamma (a,b)= \int_{0}^{1}\bigl(x^{\rho } \bigr)^{a-1}\bigl(1-x^{\rho }\bigr)^{b-1}x ^{\rho -1}\,dx. $$
In this paper, we give the Hermite–Hadamard type inequalities for s-convex functions and for m-convex functions via generalized fractional integral. Throughout the paper, \(X^{p}_{c}(a,b)\) (\(c \in \mathbb{R}\), \(1\leq p\leq \infty \)) is the space as defined in Definition 1.5 and \(L_{1}[a,b]\) stands for the space of Lebesgue integrable over the closed interval \([a,b]\) where a, b are some real numbers with \(a< b\).
2 Hermite–Hadamard type inequalities for s-convex function
In this section we give Hermite–Hadamard type inequalities for s-convex function.
Theorem 2.1
Let
\(\alpha >0\)
and
\(\rho >0\). Let
\(f:[a^{\rho },b^{\rho }]\subset \mathbb{R}_{+}\rightarrow \mathbb{R}\)
be a positive function with
\(0\leq a< b\)
and
\(f\in X^{p}_{c}(a^{\rho },b^{\rho })\). If
f
is also an
s-convex function on
\([a^{\rho },b^{\rho }]\), then the following inequalities hold:
where the fractional integrals are considered for the function
\(f(x^{\rho })\)
and evaluated at
a
and
b, respectively.
$$\begin{aligned} 2^{s-1}f \biggl( \frac{a^{\rho }+b^{\rho }}{2} \biggr) &\leq \frac{\rho^{\alpha }\Gamma (\alpha +1)}{2(b^{\rho }+a^{\rho })^{ \alpha }} \bigl[ {}^{\rho }I^{\alpha }_{a+}f \bigl(b^{\rho }\bigr)+{}^{\rho }I^{ \alpha }_{b-}f \bigl(a^{\rho }\bigr) \bigr] \\ & \leq \biggl[ \frac{\alpha }{\alpha +s}+\alpha \beta (\alpha,s+1) \biggr] \frac{f(a^{\rho })+f(b^{\rho })}{2}, \end{aligned}$$
(5)
Proof
Let \(t\in [0,1]\). Consider \(x,y\in [a,b]\), \(a\geq 0\), defined by \(x^{\rho }=t^{\rho }a^{\rho }+(1-t^{\rho })b^{\rho }\), \(y^{\rho }=t ^{\rho }b^{\rho }+(1-t^{\rho })a^{\rho }\). Since f is an s-convex function on \([a^{\rho },b^{\rho }]\), we have
Then we have
Multiplying both sides of (6) by \(t^{\alpha \rho -1}\), \(\alpha >0\) and then integrating the resulting inequality with respect to t over \([0,1]\), we obtain
This establishes the first inequality. For the proof of the second inequality in (5), we first observe that for an s-convex function f, we have
and
By adding these inequalities, we get
Multiplying both sides of (8) by \(t^{\alpha \rho -1}\), \(\alpha >0\) and then integrating the resulting inequality with respect to t over \([0,1]\), we obtain
Since
and by choosing the change of variable \(t^{\rho }=z\), we have
Thus (9) becomes
Thus (7) and (10) give (5). □
$$ f \biggl( \frac{x^{\rho }+y^{\rho }}{2} \biggr) \leq \frac{f(x^{\rho })+f(y ^{\rho })}{2^{s}}. $$
$$ 2^{s}f \biggl( \frac{a^{\rho }+b^{\rho }}{2} \biggr) \leq f \bigl(t^{\rho }a^{ \rho }+\bigl(1-t^{\rho } \bigr)b^{\rho }\bigr)+f\bigl(t^{\rho }b^{\rho }+ \bigl(1-t^{\rho }\bigr)a ^{\rho }\bigr). $$
(6)
$$\begin{aligned} \frac{2^{s}}{\alpha \rho }f \biggl( \frac{a^{\rho }+b^{\rho }}{2} \biggr) \leq& \int^{1}_{0}t^{\alpha \rho -1}f\bigl(t^{\rho }a^{\rho }+ \bigl(1-t^{\rho }\bigr)b ^{\rho }\bigr)\,dt + \int^{1}_{0}t^{\alpha \rho -1}f\bigl(t^{\rho }b^{\rho }+ \bigl(1-t ^{\rho }\bigr)a^{\rho }\bigr)\,dt \\ =& \int_{b}^{a} \biggl( \frac{b^{\rho }-x^{\rho }}{b^{\rho }-a^{\rho }} \biggr) ^{\alpha -1}f\bigl(x^{\rho }\bigr)\frac{x^{\rho -1}}{a^{\rho }-b^{\rho }}\,dx \\ &{}+ \int_{a}^{b} \biggl( \frac{y^{\rho }-a^{\rho }}{b^{\rho }-a^{\rho }} \biggr) ^{\alpha -1}f\bigl(y^{\rho }\bigr)\frac{y^{\rho -1}}{b^{\rho }-a^{\rho }}\,dy \\ =&\frac{\rho^{\alpha -1}\Gamma (\alpha)}{(b^{\rho }+a^{\rho })^{ \alpha }} \bigl[ {}^{\rho }I^{\alpha }_{a+}f \bigl(b^{\rho }\bigr)+{}^{\rho }I^{ \alpha }_{b-}f \bigl(a^{\rho }\bigr) \bigr] . \end{aligned}$$
(7)
$$ f\bigl(t^{\rho }a^{\rho }+\bigl(1-t^{\rho } \bigr)b^{\rho }\bigr)\leq \bigl( t^{\rho } \bigr) ^{s}f \bigl(a^{\rho }\bigr)+ \bigl( 1-t^{\rho } \bigr) ^{s}f \bigl(b^{\rho }\bigr) $$
$$ f\bigl(t^{\rho }b^{\rho }+\bigl(1-t^{\rho } \bigr)a^{\rho }\bigr)\leq \bigl( t^{\rho } \bigr) ^{s}f \bigl(b^{\rho }\bigr)+ \bigl( 1-t^{\rho } \bigr) ^{s}f \bigl(a^{\rho }\bigr). $$
$$ f\bigl(t^{\rho }a^{\rho }+\bigl(1-t^{\rho } \bigr)b^{\rho }\bigr)+f\bigl(t^{\rho }b^{\rho }+\bigl(1-t ^{\rho }\bigr)a^{\rho }\bigr)\leq \bigl( \bigl( t^{\rho } \bigr) ^{s}+ \bigl( 1-t ^{\rho } \bigr) ^{s} \bigr) \bigl[ f\bigl(a^{\rho }\bigr)+f\bigl(b^{\rho }\bigr) \bigr] . $$
(8)
$$\begin{aligned}& \frac{\rho^{\alpha -1}\Gamma (\alpha)}{(b^{\rho }+a^{\rho })^{\alpha }} \bigl[ {}^{\rho }I^{\alpha }_{a+}f \bigl(b^{\rho }\bigr)+{}^{\rho }I^{\alpha }_{b-}f\bigl(a ^{\rho }\bigr) \bigr] \\& \quad \leq \int^{1}_{0}t^{\alpha \rho -1} \bigl( \bigl( t^{ \rho } \bigr) ^{s}+ \bigl( 1-t^{\rho } \bigr) ^{s} \bigr) \bigl[ f\bigl(a^{ \rho }\bigr)+f\bigl(b^{\rho } \bigr) \bigr]\,dt. \end{aligned}$$
(9)
$$ \int^{1}_{0}t^{\alpha \rho +s\rho -1}\,dt= \frac{1}{\rho (\alpha +s)}, $$
$$ \int^{1}_{0}t^{\alpha \rho -1} \bigl( 1-t^{\rho } \bigr) ^{s}\,dt= \frac{ \beta (\alpha,s+1)}{\rho }. $$
$$ \frac{\rho^{\alpha -1}\Gamma (\alpha)}{(b^{\rho }+a^{\rho })^{\alpha }} \bigl[ {}^{\rho }I^{\alpha }_{a+}f \bigl(b^{\rho }\bigr)+{}^{\rho }I^{\alpha }_{b-}f\bigl(a ^{\rho }\bigr) \bigr] \leq \frac{1}{\rho } \biggl[ \frac{1}{\alpha +s}+ \beta ( \alpha,s+1) \biggr] \bigl( f\bigl(a^{\rho }\bigr)+f \bigl(b^{\rho }\bigr) \bigr). $$
(10)
Theorem 2.3
Let
\(\alpha >0\)
and
\(\rho >0\). Let
\(f:[a^{\rho },b^{\rho }]\subset \mathbb{R}_{+}\rightarrow \mathbb{R}\)
be a differentiable mapping on
\((a^{\rho },b^{\rho })\)
with
\(0\leq a< b\). If
\(|f'|\)
is
s-convex on
\([a^{\rho },b^{\rho }]\), then the following inequality holds:
$$\begin{aligned}& \biggl\vert \frac{f(a^{\rho })+f(b^{\rho })}{2}-\frac{\rho^{\alpha }\Gamma (\alpha +1)}{2(b^{\rho }+a^{\rho })^{\alpha }} \bigl[{} ^{\rho }I^{ \alpha }_{a+}f\bigl(b^{\rho } \bigr)+{}^{\rho }I^{\alpha }_{b-}f\bigl(a^{\rho }\bigr) \bigr] \biggr\vert \\& \quad \leq \frac{b^{\rho }-a^{\rho }}{2} \biggl[ \frac{1}{\alpha +s+1}+ \beta (\alpha +1,s+1) \biggr] \bigl( \bigl\vert f'\bigl(a^{\rho }\bigr) \bigr\vert + \bigl\vert f'\bigl(b^{\rho }\bigr) \bigr\vert \bigr). \end{aligned}$$
(11)
Proof
From (7) one can have
Integrating by parts, we get
By using the triangle inequality and s-convexity of \(|f'|\) and the change of variable \(t^{\rho }=z\), we obtain
□
$$\begin{aligned} &\frac{\rho^{\alpha -1}\Gamma (\alpha)}{(b^{\rho }+a^{\rho })^{ \alpha }} \bigl[ {}^{\rho }I^{\alpha }_{a+}f \bigl(b^{\rho }\bigr)+{}^{\rho }I^{ \alpha }_{b-}f \bigl(a^{\rho }\bigr) \bigr] \\ &\quad = \int^{1}_{0}t^{\alpha \rho -1}f\bigl(t^{\rho }a^{\rho }+ \bigl(1-t^{\rho }\bigr)b ^{\rho }\bigr)\,dt + \int^{1}_{0}t^{\alpha \rho -1}f\bigl(t^{\rho }b^{\rho }+ \bigl(1-t ^{\rho }\bigr)a^{\rho }\bigr)\,dt. \end{aligned}$$
(12)
$$\begin{aligned}& \frac{f(a^{\rho })+f(b^{\rho })}{\alpha \rho }-\frac{\rho^{\alpha -1} \Gamma (\alpha)}{(b^{\rho }+a^{\rho })^{\alpha }} \bigl[ {}^{\rho }I ^{\alpha }_{a+}f\bigl(b^{\rho }\bigr)+{}^{\rho }I^{\alpha }_{b-}f \bigl(a^{\rho }\bigr) \bigr] \\& \quad =\frac{b^{\rho }-a^{\rho }}{\alpha } \int_{0}^{1}t^{\rho (\alpha +1)-1} \bigl[ f'\bigl(t^{\rho }b^{\rho }+\bigl(1-t^{\rho } \bigr)a^{\rho }\bigr)-f'\bigl(t^{\rho }a^{ \rho }+ \bigl(1-t^{\rho }\bigr)b^{\rho }\bigr) \bigr]\,dt. \end{aligned}$$
(13)
$$\begin{aligned} & \biggl\vert \frac{f(a^{\rho })+f(b^{\rho }}{\alpha \rho }-\frac{ \rho^{\alpha -1}\Gamma (\alpha)}{(b^{\rho }+a^{\rho })^{\alpha }} \bigl[{} ^{\rho }I^{\alpha }_{a+}f\bigl(b^{\rho } \bigr)+{}^{\rho }I^{\alpha }_{b-}f\bigl(a ^{\rho }\bigr) \bigr] \biggr\vert \\ &\quad \leq \frac{b^{\rho }-a^{\rho }}{\alpha } \int_{0}^{1}t^{\rho (\alpha +1)-1} \bigl\vert \bigl[ f'\bigl(t^{\rho }b^{\rho }+\bigl(1-t^{\rho } \bigr)a^{\rho }\bigr) -f'\bigl(t ^{\rho }a^{\rho }+ \bigl(1-t^{\rho }\bigr)b^{\rho }\bigr) \bigr] \bigr\vert \,dt \\ &\quad \leq \frac{b^{\rho }-a^{\rho }}{\alpha } \int_{0}^{1}t^{\rho (\alpha +1)-1} \bigl[ \bigl\vert f'\bigl(t^{\rho }b^{\rho }+\bigl(1-t^{\rho } \bigr)a^{\rho }\bigr) \bigr\vert + \bigl\vert f'\bigl(t ^{\rho }a^{\rho }+\bigl(1-t^{\rho }\bigr)b^{\rho } \bigr) \bigr\vert \bigr]\,dt \\ &\quad \leq \frac{b^{\rho }-a^{\rho }}{\alpha } \int_{0}^{1}t^{\rho (\alpha +1)-1} \bigl[ \bigl(t^{\rho }\bigr)^{s} \bigl\vert f' \bigl(b^{\rho }\bigr) \bigr\vert +\bigl(1-t^{\rho } \bigr)^{s} \bigl\vert f'\bigl(a^{\rho }\bigr) \bigr\vert \\ &\qquad {}+\bigl(t^{\rho }\bigr)^{s} \bigl\vert f'\bigl(a^{\rho }\bigr) \bigr\vert +\bigl(1-t^{\rho } \bigr)^{s} \bigl\vert f'\bigl(b^{\rho }\bigr) \bigr\vert \bigr]\,dt \\ &\quad =\frac{b^{\rho }-a^{\rho }}{\alpha } \int_{0}^{1}t^{\rho (\alpha +1)-1} \bigl[ \bigl(t^{\rho }\bigr)^{s}+\bigl(1-t^{\rho } \bigr)^{s} \bigr] \bigl[ \bigl\vert f' \bigl(a^{\rho }\bigr) \bigr\vert + \bigl\vert f'\bigl(b ^{\rho }\bigr) \bigr\vert \bigr]\,dt \\ &\quad =\frac{b^{\rho }-a^{\rho }}{\alpha \rho } \biggl[ \frac{1}{\alpha +s+1}+ \beta (\alpha +1,s+1) \biggr] \bigl[ \bigl\vert f'\bigl(a^{\rho }\bigr) \bigr\vert + \bigl\vert f'\bigl(b^{\rho }\bigr) \bigr\vert \bigr] . \end{aligned}$$
(14)
Corollary 2.4
Under the same assumptions of Theorem 2.3.
1.
If
\(\rho =1\), then
$$\begin{aligned} & \biggl\vert \frac{f(a)+f(b)}{2}- \frac{\Gamma (\alpha +1)}{2(b+a)^{\alpha }} \bigl[ J^{\alpha }_{a+}f(b)+J ^{\alpha }_{b-}f(a) \bigr] \biggr\vert \\ &\quad \leq \frac{b-a}{2} \biggl[ \frac{1}{\alpha +s+1}+\beta (\alpha +1,s+1) \biggr] \bigl( \bigl\vert f'(a) \bigr\vert + \bigl\vert f'(b) \bigr\vert \bigr). \end{aligned}$$
(15)
2.
If
\(\rho =s=1\), then
$$\begin{aligned} & \biggl\vert \frac{f(a)+f(b)}{2}- \frac{\Gamma (\alpha +1)}{2(b+a)^{\alpha }} \bigl[ J^{\alpha }_{a+}f(b)+J ^{\alpha }_{b-}f(a) \bigr] \biggr\vert \\ &\quad \leq \frac{b-a}{2} \biggl[ \frac{1}{\alpha +2}+\beta (\alpha +1,2) \biggr] \bigl( \bigl\vert f'(a) \bigr\vert + \bigl\vert f'(b) \bigr\vert \bigr). \end{aligned}$$
(16)
3.
If
\(\rho =s=\alpha =1\), then
$$ \biggl\vert \frac{f(a)+f(b)}{2}-\frac{1}{b+a} \int_{a}^{b}f(x)\,dx \biggr\vert \leq \frac{b-a}{4} \bigl( \bigl\vert f'(a) \bigr\vert + \bigl\vert f'(b) \bigr\vert \bigr). $$
(17)
In order to prove our further results, we need the following lemma.
Lemma 2.5
Let
\(\alpha >0\)
and
\(\rho >0\). Let
\(f:[a^{\rho },b^{\rho }]\subset \mathbb{R}_{+}\rightarrow \mathbb{R}\)
be a differentiable mapping on
\((a^{\rho },b^{\rho })\)
with
\(0\leq a< b\). Then the following equality holds if the fractional integrals exist:
$$\begin{aligned}& \frac{f(a^{\rho })+f(b^{\rho })}{2}-\frac{\rho^{\alpha }\Gamma ( \alpha +1)}{2(b^{\rho }+a^{\rho })^{\alpha }} \bigl[ {}^{\rho }I^{\alpha }_{a+}f \bigl(b^{\rho }\bigr)+{}^{\rho }I^{\alpha }_{b-}f \bigl(a^{\rho }\bigr) \bigr] \\& \quad =\frac{\rho (b^{\rho }-a^{\rho })}{2} \int_{0}^{1} \bigl[ \bigl(1-t^{\rho } \bigr)^{ \alpha }-\bigl(t^{\rho }\bigr)^{\alpha } \bigr] t^{\rho -1}f'\bigl(t^{\rho }a^{ \rho }+ \bigl(1-t^{\rho }\bigr)b^{\rho }\bigr)\,dt. \end{aligned}$$
(18)
Proof
By using the similar arguments as in the proof of Lemma 2 in [18]. First consider
Similarly, we can show that
Thus from (19) and (20) we get (18). □
$$\begin{aligned} & \int_{0}^{1}\bigl(1-t^{\rho } \bigr)^{\alpha }t^{\rho -1}f'\bigl(t^{\rho }a^{\rho }+ \bigl(1-t ^{\rho }\bigr)b^{\rho }\bigr)\,dt \\ &\quad =\frac{(1-t^{\rho })^{\alpha }f(t^{\rho }a^{\rho }+(1-t^{\rho })b ^{\rho })}{\rho (a^{\rho }-b^{\rho })}\bigg|_{0}^{1} \\ &\qquad {} +\frac{\alpha }{a ^{\rho }-b^{\rho }} \int_{0}^{1}\bigl(1-t^{\rho } \bigr)^{\alpha -1}t^{\rho -1}f\bigl(t ^{\rho }a^{\rho }+ \bigl(1-t^{\rho }\bigr)b^{\rho }\bigr)\,dt \\ &\quad =\frac{f(b^{\rho })}{\rho (b^{\rho }-a^{\rho })}-\frac{\alpha }{b ^{\rho }-a^{\rho }} \int_{b}^{a} \biggl( \frac{x^{\rho }-a^{\rho }}{b ^{\rho }-a^{\rho }} \biggr) ^{\alpha -1}\cdot \frac{x^{\rho -1}}{a^{ \rho }-b^{\rho }}\,dx \\ &\quad =\frac{f(b^{\rho })}{\rho (b^{\rho }-a^{\rho })}-\frac{\rho^{\alpha -1}\Gamma (\alpha +1)}{(b^{\rho }-a^{\rho })^{\alpha +1}}\cdot{}^{ \rho }I_{b-}^{\alpha }f \bigl(x^{\rho }\bigr)\bigg|_{x=a}. \end{aligned}$$
(19)
$$\begin{aligned}& \int_{0}^{1}t^{\rho \alpha }\cdot t^{\rho -1}f'\bigl(t^{\rho }a^{\rho }+\bigl(1-t ^{\rho }\bigr)b^{\rho }\bigr)\,dt \\& \quad = - \frac{f(a^{\rho })}{\rho (b^{\rho }-a^{\rho })}+ \frac{\rho^{\alpha -1} \Gamma (\alpha +1)}{(b^{\rho }-a^{\rho })^{\alpha +1}}\cdot^{\rho }I _{a+}^{\alpha }f \bigl(x^{\rho }\bigr)\bigg|_{x=b}. \end{aligned}$$
(20)
Throughout all other results we denote
$$ I_{f}(\alpha,\rho,a,b)=\frac{f(a^{\rho })+f(b^{\rho })}{2}-\frac{ \rho^{\alpha }\Gamma (\alpha +1)}{2(b^{\rho }+a^{\rho })^{\alpha }} \bigl[ {}^{\rho }I^{\alpha }_{a+}f\bigl(b^{\rho } \bigr)+{}^{\rho }I^{\alpha }_{b-}f\bigl(a ^{\rho }\bigr) \bigr] . $$
Theorem 2.7
Let
\(\alpha >0\)
and
\(\rho >0\). Let
\(f:[a^{\rho },b^{\rho }]\subset \mathbb{R}_{+}\rightarrow \mathbb{R}\)
be a differentiable mapping on
\((a^{\rho },b^{\rho })\)
such that
\(f'\in L_{1}[a,b]\)
with
\(0\leq a< b\). If
\(|f'|^{q}\)
is
s-convex on
\([a^{\rho },b^{\rho }]\)
for some fixed
\(q\geq 1\), then the following inequality holds:
$$\begin{aligned} \bigl\vert I_{f}(\alpha,\rho,a,b) \bigr\vert \leq{}& \frac{\rho (b^{\rho }-a ^{\rho })}{2}\biggl(\frac{1}{\rho (\alpha +1)}\biggr)^{1-1/q} \\ &{}\times \biggl( \biggl( {}^{\rho }\gamma (s+1,\alpha +1)+ \frac{1}{\rho ( \alpha +s+1)} \biggr) \bigl\vert f'\bigl(a^{\rho } \bigr) \bigr\vert ^{q} \\ &{}+ \bigl( {}^{\rho }\gamma (1,\alpha +s+1)+ {}^{\rho }\gamma (\alpha +1,s+1) \bigr) \bigl\vert f'\bigl(b^{\rho }\bigr) \bigr\vert ^{q} \biggr)^{1/q}. \end{aligned}$$
(21)
Proof
Using Lemma 2.5 and the power mean inequality and s-convexity of \(|f'|^{q}\), we obtain
Hence the proof is completed. □
$$\begin{aligned} & \bigl\vert I_{f}(\alpha,\rho,a,b) \bigr\vert \\ &\quad = \biggl\vert \frac{\rho (b^{\rho }-a^{\rho })}{2} \int_{0}^{1} \bigl\{ \bigl(1-t ^{\rho } \bigr)^{\alpha }-\bigl(t^{\rho }\bigr)^{\alpha } \bigr\} t^{\rho -1}f'\bigl(t^{ \rho }a^{\rho }+ \bigl(1-t^{\rho }\bigr)b^{\rho }\bigr)\,dt \biggr\vert \\ &\quad \leq \frac{\rho (b^{\rho }-a^{\rho })}{2} \biggl( \int_{0}^{1} \bigl\vert \bigl(1-t ^{\rho } \bigr)^{\alpha }-\bigl(t^{\rho }\bigr)^{\alpha } \bigr\vert t^{\rho -1}\,dt \biggr) ^{1-1/q} \\ &\qquad {}\times \biggl( \int_{0}^{1} \bigl\vert \bigl(1-t^{\rho } \bigr)^{\alpha }-\bigl(t^{\rho }\bigr)^{ \alpha } \bigr\vert t^{\rho -1} \bigl\vert f'\bigl(t^{\rho }a^{\rho }+ \bigl(1-t^{\rho }\bigr)b^{ \rho }\bigr) \bigr\vert ^{q} \,dt \biggr) ^{1/q} \\ &\quad \leq \frac{\rho (b^{\rho }-a^{\rho })}{2} \biggl( \int_{0}^{1} \bigl\{ \bigl(1-t ^{\rho } \bigr)^{\alpha }+\bigl(t^{\rho }\bigr)^{\alpha } \bigr\} t^{\rho -1}\,dt \biggr) ^{1-1/q} \\ &\qquad {}\times \biggl( \int_{0}^{1} \bigl\{ \bigl(1-t^{\rho } \bigr)^{\alpha }+\bigl(t^{\rho }\bigr)^{ \alpha } \bigr\} t^{\rho -1} \bigl[\bigl(t^{\rho }\bigr)^{s} \bigl\vert f'\bigl(a^{\rho }\bigr) \bigr\vert ^{q}+ \bigl(1-t ^{\rho }\bigr)^{s} \bigl\vert f' \bigl(b^{\rho }\bigr) \bigr\vert ^{q} \bigr]\,dt \biggr) ^{1/q} \\ &\quad =\frac{\rho (b^{\rho }-a^{\rho })}{2}\biggl(\frac{1}{\rho (\alpha +1)} \biggr)^{1-1/q} \\ &\qquad {}\times \biggl( \biggl( {}^{\rho }\gamma (s+1,\alpha +1)+\frac{1}{\rho ( \alpha +s+1)} \biggr) \bigl\vert f'\bigl(a^{\rho }\bigr) \bigr\vert ^{q} \\ &\qquad {}+ \bigl( {}^{\rho }\gamma (1,\alpha +s+1)+ {}^{\rho }\gamma (\alpha +1,s+1) \bigr) \bigl\vert f'\bigl(b^{\rho }\bigr) \bigr\vert ^{q} \biggr)^{1/q}. \end{aligned}$$
(22)
Corollary 2.8
Under the similar conditions of Theorem 2.7.
1.
If
\(\rho =1\), then
$$\begin{aligned} & \biggl\vert \frac{f(a)+f(b)}{2}- \frac{\Gamma (\alpha +1)}{2(b+a)^{\alpha }} \bigl[ J^{\alpha }_{a+}f(b)+J ^{\alpha }_{b-}f(a) \bigr] \biggr\vert \\ &\quad \leq \frac{ (b-a)}{2}\biggl(\frac{1}{(\alpha +1)}\biggr)^{1-1/q} \times \biggl( \biggl( \beta (s+1,\alpha +1)+\frac{1}{(\alpha +s+1)} \biggr) \bigl\vert f'(a) \bigr\vert ^{q} \\ &\qquad {}+ \bigl(\beta (1,\alpha +s+1)+ \beta (\alpha +1,s+1) \bigr) \bigl\vert f'(b) \bigr\vert ^{q} \biggr)^{1/q}. \end{aligned}$$
2.
If
\(\rho =s=1\), then
$$\begin{aligned} & \biggl\vert \frac{f(a)+f(b)}{2}- \frac{\Gamma (\alpha +1)}{2(b+a)^{\alpha }} \bigl[ J^{\alpha }_{a+}f(b)+J ^{\alpha }_{b-}f(a) \bigr] \biggr\vert \\ &\quad \leq \frac{ (b-a)}{2}\biggl(\frac{1}{(\alpha +1)}\biggr)^{1-1/q} \times \biggl( \biggl( \beta (2,\alpha +1)+\frac{1}{(\alpha +2)} \biggr) \bigl\vert f'(a) \bigr\vert ^{q} \\ &\qquad {}+ \bigl(\beta (1,\alpha +2)+ \beta (\alpha +1,2) \bigr) \bigl\vert f'(b) \bigr\vert ^{q} \biggr)^{1/q}. \end{aligned}$$
3.
If
\(\rho =s=\alpha =1\), then
$$ \biggl\vert \frac{f(a)+f(b)}{2}-\frac{1}{b+a} \int_{a}^{b} f(x)\,dx \biggr\vert \leq \frac{ (b-a)}{2^{2-1/q}} \times \biggl( \frac{ \vert f'(a) \vert ^{q} + \vert f'(b) \vert ^{q}}{2} \biggr) ^{1/q}. $$
Theorem 2.9
Let
\(\alpha >0\)
and
\(\rho >0\). Let
\(f:[a^{\rho },b^{\rho }]\subset \mathbb{R}_{+}\rightarrow \mathbb{R}\)
be a differentiable mapping on
\((a^{\rho },b^{\rho })\)
such that
\(f'\in L_{1}[a,b]\)
with
\(0\leq a< b\). If
\(|f'|^{q}\)
is
s-convex on
\([a^{\rho },b^{\rho }]\)
for some fixed
\(q\geq 1\), then the following inequality holds:
$$\begin{aligned} & \bigl\vert I_{f}(\alpha, \rho,a,b) \bigr\vert \\ &\quad \leq \frac{\rho^{\frac{1}{q}}(b^{\rho }-a^{\rho })}{2} \biggl( \biggl[ \beta (s+1, \alpha +1)+ \frac{1}{\alpha +s+1} \biggr] \bigl[ \bigl\vert f' \bigl(a^{\rho }\bigr) \bigr\vert ^{q}+ \bigl\vert f'\bigl(b^{ \rho }\bigr) \bigr\vert ^{q}\bigr] \biggr) ^{1/q}. \end{aligned}$$
(23)
Proof
Using Lemma 2.5, the property of modulus, the power mean inequality, and the fact that \(|f'|^{q}\) is an s-convex function, we have
By using the change of variable \(t^{\rho }=z\), we get
and
Thus substituting the values of A and B in (24) and applying the fact that \(\beta (a,b)=\beta (b,a)\), we get the desired result. □
$$\begin{aligned} & \bigl\vert I_{f}(\alpha, \rho,a,b) \bigr\vert \\ &\quad \leq \biggl\vert \frac{\rho (b^{\rho }-a^{\rho })}{2} \int_{0}^{1} \bigl\{ \bigl(1-t ^{\rho } \bigr)^{\alpha }-\bigl(t^{\rho }\bigr)^{\alpha } \bigr\} t^{\rho -1} \bigl\vert f'\bigl(t ^{\rho }a^{\rho }+ \bigl(1-t^{\rho }\bigr)b^{\rho }\bigr) \bigr\vert \,dt \biggr\vert \\ &\quad \leq \frac{\rho (b^{\rho }-a^{\rho })}{2} \biggl( \int_{0}^{1}t^{ \rho -1}\,dt \biggr) ^{1-1/q} \\ &\qquad {}\times \biggl( \int_{0}^{1} \bigl\{ \bigl(1-t^{\rho } \bigr)^{ \alpha }-\bigl(t^{\rho }\bigr)^{\alpha } \bigr\} \bigl\vert f'\bigl(t^{\rho }a^{\rho }+\bigl(1-t ^{\rho }\bigr)b^{\rho }\bigr) \bigr\vert ^{q}\,dt \biggr) ^{1/q} \\ &\quad \leq \frac{\rho (b^{\rho }-a^{\rho })}{2}\frac{1}{\rho^{1-1/q}} \\ &\qquad {}\times \biggl( \int_{0}^{1} \bigl\{ \bigl(1-t^{\rho } \bigr)^{\alpha }+\bigl(t^{\rho }\bigr)^{\alpha } \bigr\} \bigl[ \bigl(t^{\rho }\bigr)^{s} \bigl\vert f' \bigl(a^{\rho }\bigr) \bigr\vert ^{q}+\bigl(1-t^{\rho } \bigr)^{s} \bigl\vert f'\bigl(b ^{\rho }\bigr) \bigr\vert ^{q} \bigr]\,dt \biggr) ^{1/q} \\ &\quad =\frac{\rho^{\frac{1}{q}}(b^{\rho }-a^{\rho })}{2} \biggl( \bigl\vert f' \bigl(a^{ \rho }\bigr) \bigr\vert ^{q} \int_{0}^{1} \bigl\{ \bigl(1-t^{\rho } \bigr)^{\alpha }\bigl(t^{\rho }\bigr)^{s} + \bigl(t^{\rho }\bigr)^{\alpha }\bigl(t^{\rho } \bigr)^{s} \bigr\} \,dt \\ &\qquad {}+ \bigl\vert f'\bigl(b^{\rho }\bigr) \bigr\vert ^{q} \int_{0}^{1} \bigl\{ \bigl(1-t^{\rho } \bigr)^{\alpha }\bigl(1-t ^{\rho }\bigr)^{s} + \bigl(t^{\rho }\bigr)^{\alpha }\bigl(1-t^{\rho } \bigr)^{s} \bigr\} \,dt \biggr)^{1/q} \\ &\quad =\frac{\rho^{\frac{1}{q}}(b^{\rho }-a^{\rho })}{2} \bigl( A \bigl\vert f' \bigl(a^{ \rho }\bigr) \bigr\vert ^{q}+B \bigl\vert f'\bigl(b^{\rho }\bigr) \bigr\vert ^{q} \bigr) ^{1/q}. \end{aligned}$$
(24)
$$ A= \int_{0}^{1} \bigl\{ \bigl(1-t^{\rho } \bigr)^{\alpha }\bigl(t^{\rho }\bigr)^{s} + \bigl(t^{ \rho }\bigr)^{\alpha }\bigl(t^{\rho } \bigr)^{s} \bigr\} \,dt=\beta (s+1,\alpha +1)+\frac{1}{ \alpha +s+1} $$
$$ B= \int_{0}^{1} \bigl\{ \bigl(1-t^{\rho } \bigr)^{\alpha }\bigl(1-t^{\rho }\bigr)^{s} + \bigl(t^{ \rho }\bigr)^{\alpha }\bigl(1-t^{\rho } \bigr)^{s} \bigr\} \,dt=\beta (\alpha +1,s+1)+\frac{1}{ \alpha +s+1}. $$
Corollary 2.10
Under the similar conditions of Theorem 2.7.
1.
If
\(\rho =1\), then
$$\begin{aligned} & \biggl\vert \frac{f(a)+f(b)}{2}- \frac{\Gamma (\alpha +1)}{2(b+a)^{\alpha }} \bigl[ J^{\alpha }_{a+}f(b)+J ^{\alpha }_{b-}f(a) \bigr] \biggr\vert \\ &\quad \leq \frac{(b-a)}{2} \biggl( \biggl[ \beta (s+1,\alpha +1)+ \frac{1}{ \alpha +s+1} \biggr] \bigl[ \bigl\vert f'(a) \bigr\vert ^{q}+ \bigl\vert f'(b) \bigr\vert ^{q} \bigr] \biggr) ^{1/q}. \end{aligned}$$
2.
If
\(\rho =s=1\), then
$$\begin{aligned} & \biggl\vert \frac{f(a)+f(b)}{2}- \frac{\Gamma (\alpha +1)}{2(b+a)^{\alpha }} \bigl[ J^{\alpha }_{a+}f(b)+J ^{\alpha }_{b-}f(a) \bigr] \biggr\vert \\ &\quad \leq \frac{(b-a)}{2} \biggl( \biggl[ \beta (2,\alpha +1)+ \frac{1}{ \alpha +2} \biggr] \bigl[ \bigl\vert f'(a) \bigr\vert ^{q}+ \bigl\vert f'(b) \bigr\vert ^{q} \bigr] \biggr) ^{1/q}. \end{aligned}$$
3.
If
\(\rho =s=\alpha =1\), then
$$ \biggl\vert \frac{f(a)+f(b)}{2}-\frac{1}{b+a} \int_{a}^{b} f(x)\,dx \biggr\vert \leq \frac{(b-a)}{2} \biggl( \frac{ \vert f'(a) \vert ^{q}+ \vert f'(b) \vert ^{q}}{2} \biggr) ^{1/q}. $$
3 Hermite–Hadamard type inequalities for m-convex function
In this section we give Hermite–Hadamard type inequalities for m-convex function.
Theorem 3.1
Let
\(\alpha >0\)
and
\(\rho >0\). Let
\(f:[a^{\rho },b^{\rho }]\subset \mathbb{R}_{+}\rightarrow \mathbb{R}\)
be a positive function with
\(0\leq a< b\)
and
\(f\in X^{p}_{c}(a^{\rho },b^{\rho })\). If
f
is also an
m-convex function on
\([a^{\rho },b^{\rho }]\), then the following inequalities hold:
$$\begin{aligned} f \biggl( \frac{m^{\rho }(a^{\rho }+b^{\rho })}{2} \biggr) & \leq \frac{\rho^{\alpha }\Gamma (\alpha +1)}{2((mb)^{\rho }-(ma)^{ \rho })^{\alpha }} {}^{\rho }I_{ma+}^{\alpha }f \bigl( (mb)^{\rho } \bigr) +\frac{m ^{\rho }\rho^{\alpha }\Gamma (\alpha +1)}{2(b^{\rho }-a^{\rho })^{ \alpha }} {}^{\rho }I_{b-}^{\alpha }f \bigl( a^{\rho } \bigr) \\ & \leq \frac{m^{\rho }}{2} \bigl( f\bigl(a^{\rho }\bigr)+f \bigl(b^{\rho }\bigr) \bigr). \end{aligned}$$
(25)
Proof
Since f is m-convex, we have
Let \(x^{\rho }=m^{\rho }t^{\rho }a^{\rho }+m^{\rho }(1-t^{\rho })b ^{\rho }\), \(y^{\rho }=t^{\rho }b^{\rho }+(1-t^{\rho })a^{\rho }\) with \(t\in [0,1]\). Then we obtain
Multiplying both sides of (26) by \(t^{\alpha \rho -1}\), \(\alpha >0\) and then integrating the resulting inequality with respect to t over \([0,1]\), we obtain
Now by multiplying both sides of (27) by \(\frac{\alpha \rho }{2}\), we get the first inequality of (25). For the second inequality, using m-convexity of f, we have
Multiplying both sides of (28) by \(t^{\alpha \rho -1}\), \(\alpha >0\) and then integrating the resulting inequality with respect to t over \([0,1]\), we obtain
Now, by multiplying both sides of (29) by \(\frac{\alpha \rho }{2}\), we get the second inequality of (25). □
$$ f \biggl( \frac{x^{\rho }+m^{\rho }y^{\rho }}{2} \biggr) \leq \frac{f(x ^{\rho })+m^{\rho }f(y^{\rho })}{2}. $$
$$ f \biggl( \frac{m^{\rho }(a^{\rho }+b^{\rho })}{2} \biggr) \leq \frac{f(m ^{\rho }t^{\rho }a^{\rho }+m^{\rho }(1-t^{\rho })b^{\rho }) +m^{ \rho }f(t^{\rho }b^{\rho }+(1-t^{\rho })a^{\rho })}{2}. $$
(26)
$$\begin{aligned} &\frac{2}{\rho \alpha }f \biggl( \frac{m^{\rho }(a^{\rho }+b^{\rho })}{2} \biggr) \\ &\quad \leq \int_{0}^{1}t^{\alpha \rho -1}f\bigl(m^{\rho }t^{\rho }a^{\rho }+m ^{\rho }\bigl(1-t^{\rho }\bigr)b^{\rho }\bigr)\,dt+ m^{\rho } \int_{0}^{1}t^{\alpha \rho -1}f\bigl(t^{\rho }b^{\rho }+ \bigl(1-t^{\rho }\bigr)a^{\rho }\bigr)\,dt \\ &\quad = \int_{mb}^{ma} \biggl( \frac{x^{\rho }-(mb)^{\rho }}{(ma)^{\rho }-(mb)^{ \rho }} \biggr) ^{\alpha -1} x^{\rho -1}\frac{dx}{(ma)^{\rho }-(mb)^{ \rho }} \\ &\qquad {}+m^{\rho } \int_{a}^{b} \biggl( \frac{y^{\rho }-a^{\rho }}{b^{\rho }-a ^{\rho }} \biggr) ^{\alpha -1} y^{\rho -1} \frac{dy}{b^{\rho }-a^{\rho }} \\ &\quad =\frac{\rho^{\alpha -1}\Gamma (\alpha)}{((mb)^{\rho }-(ma)^{\rho })^{ \alpha }} {}^{\rho }I_{ma+}^{\alpha }f \bigl( (mb)^{\rho } \bigr) +\frac{m ^{\rho }\rho^{\alpha -1}\Gamma (\alpha)}{(b^{\rho }-a^{\rho })^{ \alpha }} {}^{\rho }I_{b-}^{\alpha }f \bigl( a^{\rho } \bigr). \end{aligned}$$
(27)
$$ f\bigl(m^{\rho }t^{\rho }a^{\rho }+m^{\rho } \bigl(1-t^{\rho }\bigr)b^{\rho }\bigr) +m^{ \rho }f\bigl( \bigl(1-t^{\rho }\bigr)a^{\rho }+t^{\rho }b^{\rho } \bigr) \leq m^{\rho } \bigl[ f\bigl(a^{\rho }\bigr)+f \bigl(b^{\rho }\bigr) \bigr] . $$
(28)
$$\begin{aligned} &\frac{\rho^{\alpha -1}\Gamma (\alpha)}{((mb)^{\rho }-(ma)^{\rho })^{ \alpha }} {}^{\rho }I_{ma+}^{\alpha }f \bigl( (mb)^{\rho } \bigr) +\frac{m ^{\rho }\rho^{\alpha -1}\Gamma (\alpha)}{(b^{\rho }-a^{\rho })^{ \alpha }} {}^{\rho }I_{b-}^{\alpha }f \bigl( a^{\rho } \bigr) \\ &\quad \leq \frac{m^{\rho }}{\rho \alpha } \bigl( f\bigl(a^{\rho }\bigr)+f \bigl(b^{\rho }\bigr) \bigr). \end{aligned}$$
(29)
Corollary 3.2
Under the assumptions of Theorem 3.1, we have
1.
For
\(\rho =1\), then
$$\begin{aligned} &f \biggl( \frac{m(a+b)}{2} \biggr) \\ &\quad \leq \frac{\Gamma (\alpha +1)}{2(mb-ma)^{\alpha }} J_{ma+}^{\alpha }f(mb)+ \frac{m\Gamma (\alpha +1)}{2(b-a)^{\alpha }} J_{b-}^{\alpha }f ( a ) \\ &\quad \leq \frac{m}{2} \bigl( f(a)+f(b) \bigr). \end{aligned}$$
(30)
2.
For
\(\rho =\alpha =1\), then
$$\begin{aligned} f \biggl( \frac{m(a+b)}{2} \biggr) &\leq \frac{1}{2(mb-ma)} \int_{ma}^{mb}f(x)\,dx +\frac{m}{2(b-a)} \int_{a}^{b}f(x)\,dx \\ & \leq \frac{m}{2} \bigl( f(a)+f(b) \bigr). \end{aligned}$$
(31)
Theorem 3.4
Let
\(\alpha >0\)
and
\(\rho >0\). Let
\(f:[a^{\rho },b^{\rho }]\subset \mathbb{R}_{+}\rightarrow \mathbb{R}\)
be a positive function with
\(0\leq a< b\)
and
\(f\in X^{p}_{c}(a^{\rho },b^{\rho })\). If
f
is also an
m-convex function on
\([a^{\rho },b^{\rho }]\). Let
\(F(x^{\rho },y ^{\rho })_{t^{\rho }}:[0,1]\rightarrow \mathbb{R}\)
be defined as
Then we have
$$ F\bigl(x^{\rho },y^{\rho }\bigr)_{t^{\rho }}=\frac{1}{2} \bigl[ f\bigl(t^{\rho }x^{ \rho }+m^{\rho } \bigl(1-t^{\rho }\bigr)y^{\rho }\bigr) +f\bigl(\bigl(1-t^{\rho } \bigr)x^{\rho }+m ^{\rho }t^{\rho }y^{\rho }\bigr) \bigr] . $$
$$\begin{aligned} &\frac{1}{(b^{\rho }-a^{\rho })^{\alpha }} \int_{a}^{b}\bigl(b^{\rho }-u ^{\rho } \bigr)^{\alpha -1}u^{\rho -1} F \biggl( u^{\rho },\frac{a^{\rho }+b ^{\rho }}{2} \biggr) _{ ( \frac{b^{\rho }-u^{\rho }}{b^{\rho }-a^{ \rho }} ) }\,du \\ &\quad \leq \frac{\rho^{\alpha -1}\Gamma (\alpha)}{2(b^{\rho }-a^{\rho })^{ \alpha }} {}^{\rho }I_{a+}^{\alpha }f \bigl(b^{\rho }\bigr)+\frac{m}{2\rho \alpha }f \biggl( \frac{a^{\rho }+b^{\rho }}{2} \biggr). \end{aligned}$$
(32)
Proof
Since f is an m-convex function, we have
and also
Take \(x^{\rho }=t^{\rho }a^{\rho }+(1-t^{\rho })b^{\rho }\), we have
Multiplying both sides of (33) by \(t^{\alpha \rho -1}\), \(\alpha >0\) and then integrating the resulting inequality with respect to t over \([0,1]\), we obtain
Then, by the change of variable \(u^{\rho }=t^{\rho }a^{\rho }+(1-t ^{\rho })b^{\rho }\), we get the desired inequality (32). □
$$\begin{aligned} F\bigl(x^{\rho },y^{\rho } \bigr)_{t^{\rho }} &\leq \frac{1}{2} \bigl[ t^{\rho }f\bigl(x ^{\rho }\bigr)+m^{\rho }\bigl(1-t^{\rho }\bigr)f \bigl(y^{\rho }\bigr) +\bigl(1-t^{\rho }\bigr)f\bigl(x^{ \rho } \bigr)+m^{\rho }t^{\rho }f\bigl(y^{\rho }\bigr) \bigr] \\ & =\frac{1}{2} \bigl[ f\bigl(x^{\rho }\bigr)+m^{\rho }f \bigl(y^{\rho }\bigr) \bigr] , \end{aligned}$$
$$ F \biggl( x^{\rho },\frac{a^{\rho }+b^{\rho }}{2} \biggr) _{t^{\rho }} \leq \frac{1}{2} \biggl[ f\bigl(x^{\rho }\bigr)+m^{\rho }f \biggl( \frac{a^{\rho }+b ^{\rho }}{2} \biggr) \biggr] . $$
$$ F \biggl( t^{\rho }a^{\rho }+\bigl(1-t^{\rho } \bigr)b^{\rho },\frac{a^{\rho }+b ^{\rho }}{2} \biggr) _{t^{\rho }}\leq \frac{1}{2} \biggl[ f\bigl(t^{\rho }a^{ \rho }+ \bigl(1-t^{\rho }\bigr)b^{\rho }\bigr)+m^{\rho }f \biggl( \frac{a^{\rho }+b^{ \rho }}{2} \biggr) \biggr] . $$
(33)
$$\begin{aligned} & \int_{0}^{1}t^{\alpha \rho -1}F \biggl( t^{\rho }a^{\rho }+\bigl(1-t^{\rho }\bigr)b ^{\rho }, \frac{a^{\rho }+b^{\rho }}{2} \biggr) _{t^{\rho }}\,dt \\ &\quad \leq \frac{1}{2} \int_{0}^{1}t^{\alpha \rho -1} \biggl[ f \bigl(t^{\rho }a ^{\rho }+\bigl(1-t^{\rho } \bigr)b^{\rho }\bigr) +m^{\rho }f \biggl( \frac{a^{\rho }+b ^{\rho }}{2} \biggr) \biggr]\,dt. \end{aligned}$$
(34)
4 Applications to special means
In this section, we consider some applications to our results. Here we consider the following means:
(1)
The arithmetic mean:
$$ A(a,b)=\frac{a+b}{2}; \quad a,b\in \mathbb{R}. $$
(2)
The logarithmic mean:
$$ L(a,b)=\frac{\ln \vert b \vert -\ln \vert a \vert }{b-a}; \quad a,b\in \mathbb{R}, \vert a \vert \neq \vert b \vert , a,b\neq 0. $$
(3)
The generalized log mean:
$$ L_{n}(a,b)= \biggl[ \frac{b^{n+1}-a^{n+1}}{(n+1)(b-a)} \biggr] ^{1/n} ;\quad a,b\in \mathbb{R}, n\in \mathbb{Z}\setminus \{-1,0\}, a,b\neq 0. $$
Proposition 4.1
Let
\(a,b\in \mathbb{R}\), \(a< b\), \(0\notin [a,b]\), and
\(n\in \mathbb{Z}\), \(|n|\geq 2\), then
$$ \biggl\vert A\bigl(a^{n},b^{n}\bigr)- \frac{b-a}{b+a}L_{n}^{n}(a,b) \biggr\vert \leq \frac{ \vert n \vert (b-a)}{2}A \bigl( \vert a \vert ^{n-1}, \vert b \vert ^{n-1} \bigr). $$
(35)
Proof
By taking \(f(x)=x^{n}\) in Corollary 2.4(3), we get the required result. □
Proposition 4.2
Let
\(a,b\in \mathbb{R}\), \(a< b\), \(0\notin [a,b]\), and
\(n\in \mathbb{Z}\), \(|n|\geq 2\). Then, for
\(q\geq 1\), we have
$$ \biggl\vert A\bigl(a^{n},b^{n}\bigr)- \frac{b-a}{b+a}L_{n}^{n}(a,b) \biggr\vert \leq \frac{ \vert n \vert (b-a)}{2^{2-1/q}}A^{1/q} \bigl( \vert a \vert ^{q(n-1)}, \vert b \vert ^{q(n-1)} \bigr). $$
(36)
Proof
By taking \(f(x)=x^{n}\) in Corollary 2.8(3), we get the required result. □
Proposition 4.3
Let
\(a,b\in \mathbb{R}\), \(a< b\), \(0\notin [a,b]\), and
\(n\in \mathbb{Z}\), \(|n|\geq 2\). Then, for
\(q\geq 1\), we have
$$ \biggl\vert A\bigl(a^{n},b^{n}\bigr)- \frac{b-a}{b+a}L_{n}^{n}(a,b) \biggr\vert \leq \frac{ \vert n \vert (b-a)}{2}A^{1/q} \bigl( \vert a \vert ^{q(n-1)}, \vert b \vert ^{q(n-1)} \bigr). $$
(37)
Proof
By taking \(f(x)=x^{n}\) in Corollary 2.10(3), we get the required result. □
Proposition 4.4
Let
\(a,b\in \mathbb{R}\), \(a< b\), \(0\notin [a,b]\), and
\(n\in \mathbb{Z}\), \(m\in [0,1]\), then we have
$$ f\bigl(mA(a,b)\bigr)\leq \frac{1}{2}L_{n}^{n}(ma,mb)+ \frac{m}{2}L_{n}^{n}(a,b) \leq mA \bigl(a^{n},b^{n}\bigr). $$
(38)
Proof
By taking \(f(x)=x^{n}\) in Corollary 3.2(2), we get the required result. □
Proposition 4.5
Let
\(a,b\in \mathbb{R}\), \(a< b\), \(0\notin [a,b]\), then
$$ \biggl\vert A\bigl(a^{-1},b^{-1}\bigr)- \frac{b-a}{b+a}L(a,b) \biggr\vert \leq \frac{b-a}{2}A \bigl( \vert a \vert ^{-2}, \vert b \vert ^{-2} \bigr). $$
(39)
Proof
By taking \(f(x)=\frac{1}{x}\) in Corollary 2.4(3), we get the required result. □
Proposition 4.6
Let
\(a,b\in \mathbb{R}\), \(a< b\), \(0\notin [a,b]\). Then, for
\(q\geq 1\), we have
$$ \biggl\vert A\bigl(a^{-1},b^{-1}\bigr)- \frac{b-a}{b+a}L(a,b) \biggr\vert \leq \frac{b-a}{2^{2-1/q}}A^{1/q} \bigl( \vert a \vert ^{-2q}, \vert b \vert ^{-2q} \bigr). $$
(40)
Proof
By taking \(f(x)=\frac{1}{x}\) in Corollary 2.8(3), we get the required result. □
Proposition 4.7
Let
\(a,b\in \mathbb{R}\), \(a< b\), \(0\notin [a,b]\). Then, for
\(q\geq 1\), we have
$$ \biggl\vert A\bigl(a^{-1},b^{-1}\bigr)- \frac{b-a}{b+a}L(a,b) \biggr\vert \leq \frac{b-a}{2}A^{1/q} \bigl( \vert a \vert ^{-2q}, \vert b \vert ^{-2q} \bigr). $$
(41)
Proof
By taking \(f(x)=\frac{1}{x}\) in Corollary 2.10(3), we get the required result. □
Proposition 4.8
Let
\(a,b\in \mathbb{R}\), \(a< b\), \(0\notin [a,b]\), and
\(m\in [0,1]\), then we have
$$ f\bigl(mA\bigl(a^{-1},b^{-1}\bigr)\bigr)\leq \frac{1}{2}L(ma,mb)+\frac{m}{2}L(a,b)\leq mA\bigl(a ^{-1},b^{-1}\bigr). $$
(42)
Proof
By taking \(f(x)=\frac{1}{x}\) in Corollary 3.2(2), we get the required result. □
5 Conclusion
In Sect. 2, some Hermite–Hadamard type inequalities for s-convex functions in a generalized fractional form were obtained. In Corollaries 2.4, 2.8, and 2.10, we obtained some new results related to s-convex functions, convex functions via Riemann–Liouville fractional integrals and via classical integrals. In Sect. 3, we established a Hermite–Hadamard type inequality for m-convex functions in generalized fractional integrals. In Corollary 3.2, a new Hermite–Hadamard type inequality for m-convex functions via Riemann–Liouville fractional integrals and via classical integrals was proved.
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