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 浙江大学学报(理学版)  2018, Vol. 45 Issue (5): 555-561  DOI:10.3785/j.issn.1008-9497.2018.05.007 0

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SUN Wenbing. Ostrowski type inequalities for generalized harmonically s-convex functions via local fractional integrals[J]. Journal of Zhejiang University(Science Edition), 2018, 45(5): 555-561. DOI: 10.3785/j.issn.1008-9497.2018.05.007.
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### Fundation item

Supported by the National Natural Science Foundations of China(61672356); Shaoyang City Science and Technology Plan Project (2017GX09)

### About the author

SUN Wenbing(1978-), ORCID:http://orcid.org/0000-0002-5673-4519,male,master,associate professor, the field of interest is analytic inequality

### Article History

Received Date: January 11, 2018
Ostrowski type inequalities for generalized harmonically s-convex functions via local fractional integrals
SUN Wenbing
School of Science, Shaoyang University, Shaoyang 422000, Hunan Province, China
Received Date: January 11, 2018
Fundation item: Supported by the National Natural Science Foundations of China(61672356); Shaoyang City Science and Technology Plan Project (2017GX09)
Abstract: Based on the theory of local fractional calculus on fractal sets, the author established an identity involving local fractional integrals. Using the identity, some generalized Ostrowski type inequalities for generalized harmonically s-convex functions were obtained.
Key words: Ostrowski type inequality    generalized harmonically s-convex function    fractal set    local fractional integral

0 Introduction

Let f: IR, where $I \subseteq R$ is an interval, be a differentiable mapping in I° (the interior of I) and let a, bI° with a < b. If |f′(x)|≤M, for all x∈[a, b], then the following inequality holds:

 $\begin{array}{l} \left| {f\left( x \right) - \frac{1}{{b - a}}\int_a^b {f\left( t \right){\rm{d}}t} } \right| \le \\ \;\;\;\;\;\;M\left( {b - a} \right)\left[ {\frac{1}{4} + \frac{{{{\left( {x - \frac{{a + b}}{2}} \right)}^2}}}{{{{\left( {b - a} \right)}^2}}}} \right] \end{array}$ (1)

for all x∈[a, b]. This inequality is known in the literature as the Ostrowski inequality [1], which gives an upper bound for the approximation of the integral average $\frac{1}{{b - a}}\int_a^b {{\rm{ }}f\left( t \right){\rm{d}}t}$ by the value f(x) at point x∈[a, b]. For more recent results, one can see [2-4] and the references cited therein.

In [3], İȘCAN introduced the definition of harmonically s-convexity as follows.

Definition 1   Let $I \subset (0, \infty )$ be a real interval. If

 $f\left( {\frac{{xy}}{{tx + \left( {1 - t} \right)y}}} \right) \le \left( \ge \right){t^s}f\left( y \right) + {\left( {1 - t} \right)^s}f\left( x \right)$ (2)

for all x, yI and t∈[0, 1] and for some fixed s∈(0, 1], a function f: IR is said to be harmonically s-convex(concave).

Local fractional calculus is one of the useful tools to handle the fractal and continuously nondifferentiable functions. In [6-7], YANG stated the theory of local fractional calculus on fractal space systematically. As a result, more and more researchers extended their studies to fractal space [8-13].

SUN[13] introduced the definition of generalized harmonically s-convex function on fractal space.

Definition 2 [13]   A function $f:I \subset \left( {0, \infty } \right) \to {R^\alpha }(0 < \alpha \le 1)$ is said to be generalized harmonically s-convex, if

 $f\left( {\frac{{xy}}{{tx + \left( {1 - t} \right)y}}} \right) \le {t^{\alpha s}}f\left( y \right) + {\left( {1 - t} \right)^{\alpha s}}f\left( x \right)$ (3)

for all x, yI, t∈[0, 1] and for some fixed s∈(0, 1]. If the inequality in (3) is reversed, then f is said to be generalized harmonically s-concave.

Based on the theory of local fractional calculus and the definition of generalized harmonically s-convex function, this paper established some generalized Ostrowski type inequalities for generalized harmonically s-convex function on fractal space.

1 Preliminaries

Let Rα(0 < α≤1) be the α-type set of the real line numbers. If aα, bα, cαRα, then the following operations satisfy(see[6-7]):

(1) ${a^\alpha } + {b^\alpha } \in {R^\alpha }, {\rm{ }}{a^\alpha }{b^\alpha } \in {R^\alpha }$,

(2) ${a^\alpha } + {b^\alpha } = {b^\alpha } + {a^\alpha } = {\left( {a + b} \right)^\alpha } = {\left( {b + a} \right)^\alpha }$,

(3) ${a^\alpha }{b^\alpha } = {b^\alpha }{a^\alpha } = {\left( {ab} \right)^\alpha } = {\left( {ba} \right)^\alpha }$,

(4) ${a^\alpha }\left( {{b^\alpha } + {c^\alpha }} \right) = {a^\alpha }{b^\alpha } + {a^\alpha }{c^\alpha }$,

(5) ${a^\alpha } + {0^\alpha } = {0^\alpha } + {a^\alpha } = {a^\alpha }, {\rm{ }}{a^\alpha }{1^\alpha } = {1^\alpha }{a^\alpha } = {a^\alpha }$,

(6) ${\left( {a - b} \right)^\alpha } = {a^\alpha } - {b^\alpha }$.

The theory of the local fractional calculus on Rα can refer to references [6-7, 12]. Here, we only make some explanations as follows:

(1) If f(x) is local fractional continuous on (a, b), we denote $f\left( x \right) \in {C_\alpha }(a, b)$.

(2) The local fractional derivative of f(x) of order α at x=x0 is defined by

 ${f^{\left( \alpha \right)}}\left( {{x_0}} \right) = \frac{{{{\rm{d}}^\alpha }f\left( x \right)}}{{{\rm{d}}{x^\alpha }}}\left| {_{x = {x_0}}} \right. = \mathop {\lim }\limits_{x \to {x_0}} \frac{{{\rm{\Gamma }}\left( {\alpha + 1} \right)\left( {f\left( x \right) - f\left( {{x_0}} \right)} \right)}}{{{{\left( {x - {x_0}} \right)}^\alpha }}}.$

If f(x) is local fractional derivable on [a, b], we denote $f\left( x \right) \in {D_\alpha }(a, b)$.

(3) The local fractional integral of function f(x) of order α is defined by

 $\begin{array}{l} {}_aI_b^{\left( \alpha \right)}f\left( x \right) = \frac{1}{{{\rm{\Gamma }}\left( {\alpha + 1} \right)}}\int_a^b {f\left( t \right){{\left( {{\rm{d}}t} \right)}^\alpha }} = \\ \;\;\;\;\;\;\;\;\;\frac{1}{{{\rm{\Gamma }}\left( {\alpha + 1} \right)}}\mathop {\lim }\limits_{\Delta t \to 0} \sum\limits_{j = 0}^{N - 1} {f\left( {{t_j}} \right){{\left( {\Delta {t_j}} \right)}^\alpha }} , \end{array}$

where $a = {t_0} < {t_1} < \ldots < {t_{N - 1}} < {t_N} = b, \left[ {{t_j}, {t_{j + 1}}} \right]$ is a partition of the interval [a, b], ${\rm{\Delta }}{t_j} = {t_{j + 1}} - {t_j}, {\rm{\Delta }}t = {\rm{max}}\{ {\rm{\Delta }}{t_0}, {\rm{\Delta }}{t_1}, \ldots , {\rm{\Delta }}{t_{N - 1}}\}$. We denote $f\left( x \right) \in I_x^{(\alpha )}[a, b]$ if there exits $aI_x^\alpha f(x)$ for any x∈[a, b].

Lemma 1 [6]

(1) Suppose that $f\left( x \right) = {g^{(\alpha )}}\left( x \right) \in {C_\alpha }[a, b]$, then $_aI_b^{\left( \alpha \right)}f\left( x \right) = g\left( b \right) - g(a)$.

(2) Suppose that $f\left( x \right), g\left( x \right) \in {D_\alpha }[a, b]$, and ${f^{(\alpha )}}\left( x \right), {g^{(\alpha )}}\left( x \right) \in {C_\alpha }[a, b]$, then

 $_aI_b^{\left( \alpha \right)}f\left( x \right){g^{(\alpha )}}\left( x \right) = f\left( x \right)g\left( x \right)|_a^b{ - _a}I_b^{\left( \alpha \right)}{f^{(\alpha )}}\left( x \right)g\left( x \right).$

Lemma 2 [6]

 $\frac{{{{\rm{d}}^\alpha }{x^{k\alpha }}}}{{{\rm{d}}{x^\alpha }}} = \frac{{{\rm{\Gamma }}\left( {1 + k\alpha } \right)}}{{{\rm{\Gamma }}\left( {1 + \left( {k - 1} \right)\alpha } \right)}}{x^{\left( {k - 1} \right)\alpha }},$
 $\begin{array}{l} \frac{1}{{{\rm{\Gamma }}\left( {\alpha + 1} \right)}}\int_a^b {{x^{k\alpha }}{{\left( {{\rm{d}}x} \right)}^\alpha }} = \\ \;\;\;\;\;\;\;\;\frac{{{\rm{\Gamma }}\left( {1 + k\alpha } \right)}}{{{\rm{\Gamma }}\left( {1 + \left( {k + 1} \right)\alpha } \right)}}\left( {{b^{\left( {k + 1} \right)\alpha }} - {a^{\left( {k + 1} \right)\alpha }}} \right),k > 0. \end{array}$

Lemma 3 [6, 14](Generalized Hölder's inequality) Let $f, g \in {C_\alpha }\left[ {a, b} \right], p, q > 1,$, with $\frac{1}{p} + \frac{1}{q} = {\rm{1}}$, then

 $\begin{array}{l} \frac{1}{{{\rm{\Gamma }}\left( {\alpha + 1} \right)}}\int_a^b {\left| {f\left( x \right)g\left( x \right)} \right|{{\left( {{\rm{d}}x} \right)}^\alpha }} \le \\ \;\;\;\;\;\;\;\;\;\;{\left( {\frac{1}{{{\rm{\Gamma }}\left( {\alpha + 1} \right)}}\int_a^b {{{\left| {f\left( x \right)} \right|}^p}{{\left( {{\rm{d}}x} \right)}^\alpha }} } \right)^{1/p}} \times \\ \;\;\;\;\;\;\;\;\;\;{\left( {\frac{1}{{{\rm{\Gamma }}\left( {\alpha + 1} \right)}}\int_a^b {{{\left| {g\left( x \right)} \right|}^q}{{\left( {{\rm{d}}x} \right)}^\alpha }} } \right)^{1/q}}. \end{array}$

Lemma 4 [6]   $_aI_b^{(\alpha )}1 = \frac{{{{\left( {b - a} \right)}^\alpha }}}{{{\rm{\Gamma }}(1 + \alpha )}}$.

2 Main results

In order to simply the representation of results, we will introduce the following special functions on fractal sets.

(1) The Beta function is given by

 $\begin{array}{l} {{\rm{B}}_\alpha }\left( {x,y} \right) = \frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {{t^{\left( {x - 1} \right)\alpha }}{{\left( {1 - t} \right)}^{\left( {y - 1} \right)\alpha }}{{\left( {{\rm{d}}t} \right)}^\alpha }} ,\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x > 0,y > 0. \end{array}$

(2) The hypergeometric function is given by

 $\begin{array}{l} {}_2F_1^\alpha \left( {a,b;c;z} \right) = \frac{1}{{{{\rm{B}}_\alpha }\left( {b,c - b} \right)}}\frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}} \times \\ \;\;\;\;\;\;\;\int_0^1 {{t^{\left( {b - 1} \right)\alpha }}{{\left( {1 - t} \right)}^{\left( {c - b - 1} \right)\alpha }}{{\left( {1 - zt} \right)}^{ - a\alpha }}{{\left( {{\rm{d}}t} \right)}^\alpha }} ,\\ \;\;\;\;\;\;\;c > b > 0,\left| z \right| < 1. \end{array}$

Lemma 5   Let $I \subset ({\rm{0}}, \infty )$ be an interval, f: $I^\circ \subset {\rm{ }}({\rm{0}}, \infty )$Rα(I° is the interior of I) such that $f \in {D_\alpha }\left( {I^\circ } \right)$ and ${f^{(\alpha )}} \in {C_\alpha }(a, b)$ for a, bI° with a < b. Then the following equality involving local fractional integrals holds:

 $\begin{array}{l} \left[ {\frac{{{{\left( {x - a} \right)}^\alpha }}}{{{a^\alpha }{x^\alpha }}} + \frac{{{{\left( {b - x} \right)}^\alpha }}}{{{b^\alpha }{x^\alpha }}}} \right]f\left( x \right) - {\rm{\Gamma }}\left( {1 + \alpha } \right){}_aI_b^{\left( \alpha \right)}\frac{{f\left( u \right)}}{{{u^{2\alpha }}}} = \\ \;\;\;\;\;\;\;{\left( {x - a} \right)^{2\alpha }}\frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {\frac{{{t^\alpha }}}{{{{\left( {ta + \left( {1 - t} \right)x} \right)}^{2\alpha }}}}} \times \\ \;\;\;\;\;\;\;{f^{\left( \alpha \right)}}\left( {\frac{{ax}}{{ta + \left( {1 - t} \right)x}}} \right){\left( {{\rm{d}}t} \right)^\alpha } - {\left( {b - x} \right)^{2\alpha }}\frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}} \times \\ \;\;\;\;\;\;\;\int_0^1 {\frac{{{t^\alpha }}}{{{{\left( {tb + \left( {1 - t} \right)x} \right)}^{2\alpha }}}}{f^{\left( \alpha \right)}}\left( {\frac{{bx}}{{tb + \left( {1 - t} \right)x}}} \right){{\left( {{\rm{d}}t} \right)}^\alpha }} . \end{array}$ (4)

Proof   Let

 $\begin{array}{l} {I_1} = {\left( {x - a} \right)^{2\alpha }}\frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}} \times \\ \int_0^1 {\frac{{{t^\alpha }}}{{{{\left( {ta + \left( {1 - t} \right)x} \right)}^{2\alpha }}}}{f^{\left( \alpha \right)}}\left( {\frac{{ax}}{{ta + \left( {1 - t} \right)x}}} \right){{\left( {{\rm{d}}t} \right)}^\alpha }} . \end{array}$ (5)

By the local fractional integration by parts, we have

 $\begin{array}{l} {I_1} = \frac{{{{\left( {x - a} \right)}^{2\alpha }}}}{{{a^\alpha }{x^\alpha }}}\left[ {f\left( x \right) - } \right.\\ \;\;\;\;\;\;\;\left. {\frac{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {f\left( {\frac{{ax}}{{ta + \left( {1 - t} \right)x}}} \right){{\left( {{\rm{d}}t} \right)}^\alpha }} } \right]. \end{array}$

Setting $u = \frac{{ax}}{{ta + \left( {1 - t} \right)x}}$, we obtain

 ${I_1} = \frac{{{{\left( {x - a} \right)}^{2\alpha }}}}{{{a^\alpha }{x^\alpha }}}f\left( x \right) - {\rm{\Gamma }}{\left( {1 + \alpha } \right)_a}I_x^{\left( \alpha \right)}\frac{{f\left( u \right)}}{{{u^{2\alpha }}}}.$ (6)

Let

 $\begin{array}{l} {I_2} = - {\left( {b - x} \right)^{2\alpha }}\frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}} \times \\ \int_0^1 {\frac{{{t^\alpha }}}{{{{\left( {tb + \left( {1 - t} \right)x} \right)}^{2\alpha }}}}{f^{\left( \alpha \right)}}\left( {\frac{{bx}}{{tb + \left( {1 - t} \right)x}}} \right){{\left( {{\rm{d}}t} \right)}^\alpha }} . \end{array}$ (7)

Similarly, we can get

 ${I_2} = \frac{{{{\left( {b - x} \right)}^\alpha }}}{{{b^\alpha }{x^\alpha }}}f\left( x \right) - {\rm{\Gamma }}{\left( {1 + \alpha } \right)_x}I_b^{\left( \alpha \right)}\frac{{f\left( v \right)}}{{{v^{2\alpha }}}}.$ (8)

Adding (6) and (8), we obtain

 $\begin{array}{l} {I_1} + {I_2} = \left[ {\frac{{{{\left( {x - a} \right)}^\alpha }}}{{{a^\alpha }{x^\alpha }}} + \frac{{{{\left( {b - x} \right)}^\alpha }}}{{{b^\alpha }{x^\alpha }}}} \right]f\left( x \right) - \\ \;\;\;\;\;\;\;\;\;\;\;\;\;{\rm{\Gamma }}{\left( {1 + \alpha } \right)_a}I_b^{\left( \alpha \right)}\frac{{f\left( u \right)}}{{{u^{2\alpha }}}}. \end{array}$

This completes the proof.

Theorem 1    Let $I \subset (0, \infty ){\rm{ }}$ be an interval, f: I°→Rα (I° is the interior of I) such that fDα(I°) and f(α)Cα[a, b] for a, bI° with a < b. If |f(α)|q is generalized harmonically s-convex on [a, b] for q>1, then for all x∈[a, b], the following inequality holds:

 $\begin{array}{l} \left| {\left[ {\frac{{{{\left( {x - a} \right)}^\alpha }}}{{{a^\alpha }{x^\alpha }}} + \frac{{{{\left( {b - x} \right)}^\alpha }}}{{{b^\alpha }{x^\alpha }}}} \right]f\left( x \right) - {\rm{\Gamma }}{{\left( {1 + \alpha } \right)}_a}I_b^{\left( \alpha \right)}\frac{{f\left( u \right)}}{{{u^{2\alpha }}}}} \right| \le \\ {\left( {\frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}} \right)^{1 - \frac{1}{q}}}\left[ {{{\left( {x - a} \right)}^{2\alpha }}\left( {C_1^\alpha \left( {a,x,s,q,q} \right){{\left| {{f^{\left( \alpha \right)}}\left( x \right)} \right|}^q} + } \right.} \right.\\ {\left. {C_2^\alpha \left( {a,x,s,q,q} \right){{\left| {{f^{\left( \alpha \right)}}\left( a \right)} \right|}^q}} \right)^{\frac{1}{q}}} + \\ {\left( {b - x} \right)^{2\alpha }}\left( {C_3^\alpha \left( {b,x,s,q,q} \right){{\left| {{f^{\left( \alpha \right)}}\left( x \right)} \right|}^q} + } \right.\\ \left. {{{\left. {C_4^\alpha \left( {b,x,s,q,q} \right){{\left| {{f^{\left( \alpha \right)}}\left( b \right)} \right|}^q}} \right)}^{\frac{1}{q}}}} \right], \end{array}$ (9)

where

 $\begin{array}{l} C_1^\alpha \left( {a,x,s,\nu ,q} \right) = \\ \;\;\;\;\;\;\;\;\frac{{{B_\alpha }\left( {\rho + s + 1,1} \right)}}{{{x^{2\nu \alpha }}}}{}_2F_1^\alpha \left( {2\nu ,\rho + s + 1;\rho + s + 2;1 - \frac{a}{x}} \right), \end{array}$
 $\begin{array}{l} C_2^\alpha \left( {a,x,s,\nu ,q} \right) = \\ \;\;\;\;\;\;\;\;\frac{{{B_\alpha }\left( {\rho + 1,s + 1} \right)}}{{{x^{2\nu \alpha }}}}{}_2F_1^\alpha \left( {2\nu ,\rho + 1;\rho + s + 2;1 - \frac{a}{x}} \right), \end{array}$
 $\begin{array}{l} C_3^\alpha \left( {a,x,s,\nu ,q} \right) = \\ \;\;\;\;\;\;\;\;\frac{{{B_\alpha }\left( {1,\rho + s + 1} \right)}}{{{b^{2\nu \alpha }}}}{}_2F_1^\alpha \left( {2\nu ,1;\rho + s + 2;1 - \frac{x}{b}} \right), \end{array}$
 $\begin{array}{l} C_4^\alpha \left( {a,x,s,\nu ,q} \right) = \\ \;\;\;\;\;\;\;\;\frac{{{B_\alpha }\left( {s + 1,\rho + 1} \right)}}{{{b^{2\nu \alpha }}}}{}_2F_1^\alpha \left( {2\nu ,s + 1;\rho + s + 2;1 - \frac{x}{b}} \right). \end{array}$

Proof    Taking modulus in equality (4), by the generalized Hölder's inequality (lemma 3) and the generalized harmonically s-convexity of |f(α)|q on [a, b], we have

 $\begin{array}{l} \left| {\left[ {\frac{{{{\left( {x - a} \right)}^\alpha }}}{{{a^\alpha }{x^\alpha }}} + \frac{{{{\left( {b - x} \right)}^\alpha }}}{{{b^\alpha }{x^\alpha }}}} \right]f\left( x \right) - {\rm{\Gamma }}{{\left( {1 + \alpha } \right)}_a}I_b^{\left( \alpha \right)}\frac{{f\left( u \right)}}{{{u^{2\alpha }}}}} \right| \le \\ \;\;\;\;\;\;{\left( {x - a} \right)^{2\alpha }}{\left( {\frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {\left. {1{{\left( {{\rm{d}}t} \right)}^\alpha }} \right)} } \right)^{1 - \frac{1}{q}}} \times \\ \;\;\;\;\;\;\left( {\frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {\frac{{{t^{q\alpha }}}}{{{{\left( {ta + \left( {1 - t} \right)x} \right)}^{2q\alpha }}}}\left[ {{t^{s\alpha }}{{\left| {{f^{\left( \alpha \right)}}\left( x \right)} \right|}^q} + } \right.} } \right.\\ \;\;\;\;\;\;{\left. {{{\left( {1 - t} \right)}^{s\alpha }}{{\left| {{f^{\left( \alpha \right)}}\left( a \right)} \right|}^q}{{\left( {{\rm{d}}t} \right)}^\alpha }} \right)^{\frac{1}{q}}} + \\ \;\;\;\;\;\;{\left( {b - x} \right)^{2\alpha }}{\left( {\frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {1{{\left( {{\rm{d}}t} \right)}^\alpha }} } \right)^{1 - \frac{1}{q}}}\left( {\frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}} \times } \right.\\ \;\;\;\;\;\;\int_0^1 {\frac{{{t^{q\alpha }}}}{{{{\left( {tb + \left( {1 - t} \right)x} \right)}^{2q\alpha }}}}} \left[ {{t^{s\alpha }}{{\left| {{f^{\left( a \right)}}\left( x \right)} \right|}^q} + } \right.\\ \;\;\;\;\;\;{\left. {\left. {{{\left( {1 - t} \right)}^{s\alpha }}{{\left| {{f^{\left( a \right)}}\left( b \right)} \right|}^q}} \right]{{\left( {{\rm{d}}t} \right)}^\alpha }} \right)^{\frac{1}{q}}}. \end{array}$ (10)

By lemma 4, it is easily check that

 $\frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {1{{\left( {{\rm{d}}t} \right)}^\alpha }} = \frac{{{1^\alpha }}}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}.$ (11)

Further calculation, we have

 $\begin{array}{l} \frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {\frac{{{t^{\left( {q + s} \right)\alpha }}}}{{{{\left( {ta + \left( {1 - t} \right)x} \right)}^{2q\alpha }}}}{{\left( {{\rm{d}}t} \right)}^\alpha } = } \\ \;\;\;\;\;\;\frac{{{B_\alpha }\left( {q + s + 1,1} \right)}}{{{x^{2q\alpha }}}}{}_2F_1^\alpha \left( {2q,q + s + 1;q + s + 2;1 - \frac{a}{x}} \right) = \\ \;\;\;\;\;\;C_1^\alpha \left( {a,x,s,q,q} \right), \end{array}$ (12)
 $\begin{array}{l} \frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {\frac{{{t^{q\alpha }}{{\left( {1 - t} \right)}^{s\alpha }}}}{{{{\left( {ta + \left( {1 - t} \right)x} \right)}^{2q\alpha }}}}{{\left( {{\rm{d}}t} \right)}^\alpha }} = \\ \frac{{{B_\alpha }\left( {q + 1,s + 1} \right)}}{{{x^{2q\alpha }}}}{}_2F_1^\alpha \left( {2q,q + 1;q + s + 2;1 - \frac{a}{x}} \right) = \\ C_2^\alpha \left( {a,x,s,q,q} \right). \end{array}$ (13)

Changing the variable with 1-t=r, we have

 $\begin{array}{l} \frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {\frac{{{t^{\left( {q + s} \right)\alpha }}}}{{{{\left( {tb + \left( {1 - t} \right)x} \right)}^{2q\alpha }}}}{{\left( {{\rm{d}}t} \right)}^\alpha }} = \\ \frac{{{B_\alpha }\left( {1,q + s + 1} \right)}}{{{b^{2q\alpha }}}}{}_2F_1^\alpha \left( {2q,1;q + s + 2;1 - \frac{x}{b}} \right) = \\ C_3^\alpha \left( {b,x,s,q,q} \right). \end{array}$ (14)

Similarly,

 $\begin{array}{l} \frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {\frac{{{t^{q\alpha }}{{\left( {1 - t} \right)}^{s\alpha }}}}{{{{\left( {tb + \left( {1 - t} \right)x} \right)}^{2q\alpha }}}}{{\left( {{\rm{d}}t} \right)}^\alpha }} = \\ \frac{{{B_\alpha }\left( {s + 1,q + 1} \right)}}{{{b^{2q\alpha }}}}{}_2F_1^\alpha \left( {2q,s + 1;q + s + 2;1 - \frac{x}{b}} \right) = \\ C_4^\alpha \left( {b,x,s,q,q} \right). \end{array}$ (15)

Thus, using (11)~(15) in (10), we can obtain the inequality of (9).

This completes the proof.

Theorem 2    Let $I \subset (0, \infty ){\rm{ }}$ be an interval, f: I°→Rα (I° is the interior of I) such that fDα(I°) and f(α)Cα[a, b] for a, bI° with a < b. If |f(α)|q is generalized harmonically s-convex on [a, b] for q>1, then for all x∈[a, b], the following inequality holds.

 $\begin{array}{l} \left| {\left[ {\frac{{{{\left( {x - a} \right)}^\alpha }}}{{{a^\alpha }{x^\alpha }}} + \frac{{{{\left( {b - x} \right)}^\alpha }}}{{{b^\alpha }{x^\alpha }}}} \right]f\left( x \right) - {\rm{\Gamma }}{{\left( {1 + \alpha } \right)}_a}I_b^{\left( \alpha \right)}\frac{{f\left( u \right)}}{{{u^{2\alpha }}}}} \right| \le \\ {\left( {x - a} \right)^{2\alpha }}{\left( {C_5^\alpha \left( {a,x} \right)} \right)^{1 - \frac{1}{q}}}\left( {C_1^\alpha \left( {c,x,s,1,1} \right){{\left| {{f^{\left( \alpha \right)}}\left( x \right)} \right|}^q} + } \right.\\ {\left. {C_2^\alpha \left( {c,x,s,1,1} \right){{\left| {{f^{\left( \alpha \right)}}\left( a \right)} \right|}^q}} \right)^{\frac{1}{q}}} + \\ {\left( {b - x} \right)^{2\alpha }}{\left( {C_5^\alpha \left( {b,x} \right)} \right)^{1 - \frac{1}{q}}}\left( {C_3^\alpha \left( {b,x,s,1,1} \right){{\left| {{f^{\left( \alpha \right)}}\left( x \right)} \right|}^q} + } \right.\\ {\left. {C_4^\alpha \left( {b,x,s,1,1} \right){{\left| {{f^{\left( \alpha \right)}}\left( b \right)} \right|}^q}} \right)^{\frac{1}{q}}}, \end{array}$ (16)

where C1α, C2α, C3α, C4α are defined in theorem 1, and

 $C_5^\alpha \left( {a,x} \right) = \frac{1}{{{{\left( {x - a} \right)}^\alpha }}}\left[ {\frac{1}{{{a^\alpha }}} - \frac{{{{\ln }_\alpha }\left( {{x^\alpha }} \right) - {{\ln }_\alpha }\left( {{a^\alpha }} \right)}}{{{{\left( {x - a} \right)}^\alpha }}}} \right],$
 $C_5^\alpha \left( {b,x} \right) = \frac{1}{{{{\left( {b - x} \right)}^\alpha }}}\left[ {\frac{{{{\ln }_\alpha }\left( {{b^\alpha }} \right) - {{\ln }_\alpha }\left( {{x^\alpha }} \right)}}{{{{\left( {b - x} \right)}^\alpha }}} - \frac{1}{{{b^\alpha }}}} \right],$

lnα(xα) denotes the inverse function of Mittag-Leffler function ${E_\alpha }\left( {{x^\alpha }} \right) = \sum\limits_{k = 0}^\infty {\frac{{{x^{k\alpha }}}}{{\Gamma \left( {1 + k\alpha } \right)}}}$ [6] on fractal set.

Proof    Taking modulus in equality (4), by the generalized Hölder's inequality(lemma 3) and the generalized harmonically s-convexity of |f(α)|q on [a, b], we have

 $\begin{array}{l} \left| {\left[ {\frac{{{{\left( {x - a} \right)}^\alpha }}}{{{a^\alpha }{x^\alpha }}} + \frac{{{{\left( {b - x} \right)}^\alpha }}}{{{b^\alpha }{x^\alpha }}}} \right]f\left( x \right) - {\rm{\Gamma }}{{\left( {1 + \alpha } \right)}_a}I_b^{\left( \alpha \right)}\frac{{f\left( u \right)}}{{{u^{2\alpha }}}}} \right| \le \\ {\left( {x - a} \right)^{2\alpha }}{\left( {\frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {\frac{{{t^\alpha }}}{{{{\left( {ta + \left( {1 - t} \right)x} \right)}^{2\alpha }}}}{{\left( {{\rm{d}}t} \right)}^\alpha }} } \right)^{1 - \frac{1}{q}}} \times \\ \left( {\frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {\frac{{{t^\alpha }}}{{{{\left( {ta + \left( {1 - t} \right)x} \right)}^{2\alpha }}}}\left[ {{t^{s\alpha }}{{\left| {{f^{\left( \alpha \right)}}\left( x \right)} \right|}^q} + } \right.} } \right.\\ {\left. {\left. {{{\left( {1 - t} \right)}^{s\alpha }}{{\left| {{f^{\left( \alpha \right)}}\left( a \right)} \right|}^q}} \right]{{\left( {{\rm{d}}t} \right)}^\alpha }} \right)^{\frac{1}{q}}} + \\ {\left( {b - x} \right)^{2\alpha }}{\left( {\frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {\frac{{{t^\alpha }}}{{{{\left( {tb + \left( {1 - t} \right)x} \right)}^{2\alpha }}}}{{\left( {{\rm{d}}t} \right)}^\alpha }} } \right)^{1 - \frac{1}{q}}} \times \\ \left( {\frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {\frac{{{t^\alpha }}}{{{{\left( {tb + \left( {1 - t} \right)x} \right)}^{2\alpha }}}}\left[ {{t^{s\alpha }}{{\left| {{f^{\left( \alpha \right)}}\left( x \right)} \right|}^q} + } \right.} } \right.\\ {\left. {\left. {{{\left( {1 - t} \right)}^{s\alpha }}{{\left| {{f^{\left( \alpha \right)}}\left( b \right)} \right|}^q}} \right]{{\left( {{\rm{d}}t} \right)}^\alpha }} \right)^{\frac{1}{q}}}. \end{array}$ (17)

By calculating, it is easily checked that

 $\begin{array}{l} \frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {\frac{{{t^\alpha }}}{{{{\left( {ta + \left( {1 - t} \right)x} \right)}^{2\alpha }}}}{{\left( {{\rm{d}}t} \right)}^\alpha }} = \\ \;\;\;\;\;\;\;\frac{1}{{{{\left( {x - a} \right)}^\alpha }}}\left[ {\frac{1}{{{a^\alpha }}} - \frac{{{{\ln }_\alpha }\left( {{x^\alpha }} \right) - {{\ln }_\alpha }\left( {{a^\alpha }} \right)}}{{{{\left( {x - a} \right)}^\alpha }}}} \right], \end{array}$ (18)

and

 $\begin{array}{l} \frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {\frac{{{t^\alpha }}}{{{{\left( {tb + \left( {1 - t} \right)x} \right)}^{2\alpha }}}}{{\left( {{\rm{d}}t} \right)}^\alpha }} = \\ \;\;\;\;\;\;\;\frac{1}{{{{\left( {b - x} \right)}^\alpha }}}\left[ {\frac{{{{\ln }_\alpha }\left( {{b^\alpha }} \right) - {{\ln }_\alpha }\left( {{x^\alpha }} \right)}}{{{{\left( {b - x} \right)}^\alpha }}} - \frac{1}{{{b^\alpha }}}} \right]. \end{array}$ (19)

Further calculation, we have

 $\begin{array}{l} \frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {\frac{{{t^{\left( {1 + s} \right)\alpha }}}}{{{{\left( {ta + \left( {1 - t} \right)x} \right)}^{2\alpha }}}}{{\left( {{\rm{d}}t} \right)}^\alpha }} = \\ \;\;\;\;C_1^\alpha \left( {a,x,s,1,1} \right), \end{array}$ (20)
 $\begin{array}{l} \frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {\frac{{{t^\alpha }{{\left( {1 - t} \right)}^{s\alpha }}}}{{{{\left( {ta + \left( {1 - t} \right)x} \right)}^{2\alpha }}}}{{\left( {{\rm{d}}t} \right)}^\alpha }} = \\ \;\;\;\;C_2^\alpha \left( {a,x,s,1,1} \right), \end{array}$ (21)
 $\begin{array}{l} \frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {\frac{{{t^{\left( {1 + s} \right)\alpha }}}}{{{{\left( {tb + \left( {1 - t} \right)x} \right)}^{2\alpha }}}}{{\left( {{\rm{d}}t} \right)}^\alpha }} = \\ \;\;\;\;C_3^\alpha \left( {b,x,s,1,1} \right), \end{array}$ (22)

and

 $\begin{array}{l} \frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {\frac{{{t^\alpha }{{\left( {1 - t} \right)}^{s\alpha }}}}{{{{\left( {tb + \left( {1 - t} \right)x} \right)}^{2\alpha }}}}{{\left( {{\rm{d}}t} \right)}^\alpha }} = \\ \;\;\;\;C_4^\alpha \left( {a,x,s,1,1} \right), \end{array}$ (23)

Thus, using (18)~(23) in (17), we can obtain the inequality of (16).

This completes the proof.

Theorem 3    Let $I \subset (0, \infty )$ be an interval, f: I°→Rα such that fDα(I°) and f(α)Cα[a, b] for a, bI° with a < b. If |f(α)|q is generalized harmonically s-convex on [a, b] for q>1, $\frac{1}{p} + \frac{1}{q} = 1$, then for all x∈[a, b], the following inequality holds:

 $\begin{array}{l} \left| {\left[ {\frac{{{{\left( {x - a} \right)}^\alpha }}}{{{a^\alpha }{x^\alpha }}} + \frac{{{{\left( {b - x} \right)}^\alpha }}}{{{b^\alpha }{x^\alpha }}}} \right]f\left( x \right) - {\rm{\Gamma }}{{\left( {1 + \alpha } \right)}_a}I_b^{\left( \alpha \right)}\frac{{f\left( u \right)}}{{{u^{2\alpha }}}}} \right| \le \\ {\left( {\frac{{{\rm{\Gamma }}\left( {1 + p\alpha } \right)}}{{{\rm{\Gamma }}\left( {1 + \left( {p + 1} \right)\alpha } \right)}}} \right)^{\frac{1}{p}}}\left[ {{{\left( {x - a} \right)}^{2\alpha }}\left( {C_1^\alpha \left( {a,x,s,q,0} \right) \times } \right.} \right.\\ {\left. {{{\left| {{f^{\left( \alpha \right)}}\left( x \right)} \right|}^q} + C_2^\alpha \left( {a,x,s,q,0} \right){{\left| {{f^{\left( \alpha \right)}}\left( a \right)} \right|}^q}} \right)^{\frac{1}{q}}} + \\ {\left( {b - x} \right)^{2\alpha }}\left( {C_3^\alpha \left( {b,x,s,q,0} \right){{\left| {{f^{\left( \alpha \right)}}\left( x \right)} \right|}^q} + } \right.\\ \left. {{{\left. {C_4^\alpha \left( {b,x,s,q,0} \right){{\left| {{f^{\left( \alpha \right)}}\left( b \right)} \right|}^q}} \right)}^{\frac{1}{q}}}} \right], \end{array}$ (24)

where C1α, C2α, C3α, C4α are defined in theorem 1.

Proof   Taking modulus in equality (4), by the generalized Hölder's inequality(lemma 3) and the generalized harmonically s-convexity of |f(α)|q on [a, b], we have

 $\begin{array}{l} \left| {\left[ {\frac{{{{\left( {x - a} \right)}^\alpha }}}{{{a^\alpha }{x^\alpha }}} + \frac{{{{\left( {b - x} \right)}^\alpha }}}{{{b^\alpha }{x^\alpha }}}} \right]f\left( x \right) - {\rm{\Gamma }}{{\left( {1 + \alpha } \right)}_a}I_b^{\left( \alpha \right)}\frac{{f\left( u \right)}}{{{u^{2\alpha }}}}} \right| \le \\ {\left( {x - a} \right)^{2\alpha }}{\left( {\frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {{t^{p\alpha }}{{\left( {{\rm{d}}t} \right)}^\alpha }} } \right)^{\frac{1}{p}}} \times \\ \left( {\frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {\frac{{{1^\alpha }}}{{{{\left( {ta + \left( {1 - t} \right)x} \right)}^{2q\alpha }}}}\left[ {{t^{s\alpha }}{{\left| {{f^{\left( \alpha \right)}}\left( x \right)} \right|}^q} + } \right.} } \right.\\ {\left. {\left. {{{\left( {1 - t} \right)}^{s\alpha }}{{\left| {{f^{\left( \alpha \right)}}\left( a \right)} \right|}^q}} \right]{{\left( {{\rm{d}}t} \right)}^\alpha }} \right)^{\frac{1}{q}}} + \\ {\left( {b - x} \right)^{2\alpha }}{\left( {\frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {{t^p}{{\left( {{\rm{d}}t} \right)}^\alpha }} } \right)^{\frac{1}{p}}} \times \\ \left( {\frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {\frac{{{1^\alpha }}}{{{{\left( {tb + \left( {1 - t} \right)x} \right)}^{2q\alpha }}}}\left[ {{t^{s\alpha }}{{\left| {{f^{\left( \alpha \right)}}\left( x \right)} \right|}^q} + } \right.} } \right.\\ {\left. {\left. {{{\left( {1 - t} \right)}^{s\alpha }}{{\left| {{f^{\left( \alpha \right)}}\left( b \right)} \right|}^q}} \right]{{\left( {{\rm{d}}t} \right)}^\alpha }} \right)^{\frac{1}{q}}}. \end{array}$ (25)

By calculating, it is easy to check that

 $\frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {{t^{p\alpha }}{{\left( {{\rm{d}}t} \right)}^\alpha }} = \frac{{{\rm{\Gamma }}\left( {1 + p\alpha } \right)}}{{{\rm{\Gamma }}\left( {1 + \left( {p + 1} \right)\alpha } \right)}}.$ (26)

Further calculation, we have

 $\begin{array}{l} \frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {\frac{{{t^{s\alpha }}}}{{{{\left( {ta + \left( {1 - t} \right)x} \right)}^{2q\alpha }}}}{{\left( {{\rm{d}}t} \right)}^\alpha }} = \\ \;\;\;\;\;\;\;\;C_1^\alpha \left( {a,x,s,q,0} \right), \end{array}$ (27)
 $\begin{array}{l} \frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {\frac{{{{\left( {1 - t} \right)}^{s\alpha }}}}{{{{\left( {ta + \left( {1 - t} \right)x} \right)}^{2q\alpha }}}}{{\left( {{\rm{d}}t} \right)}^\alpha }} = \\ \;\;\;\;\;\;\;\;C_2^\alpha \left( {a,x,s,q,0} \right), \end{array}$ (28)
 $\begin{array}{l} \frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {\frac{{{t^{s\alpha }}}}{{{{\left( {tb + \left( {1 - t} \right)x} \right)}^{2q\alpha }}}}{{\left( {{\rm{d}}t} \right)}^\alpha }} = \\ \;\;\;\;\;\;\;\;C_3^\alpha \left( {b,x,s,q,0} \right), \end{array}$ (29)

and,

 $\begin{array}{l} \frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {\frac{{{{\left( {1 - t} \right)}^{s\alpha }}}}{{{{\left( {tb + \left( {1 - t} \right)x} \right)}^{2q\alpha }}}}{{\left( {{\rm{d}}t} \right)}^\alpha }} = \\ \;\;\;\;\;\;\;\;C_4^\alpha \left( {b,x,s,q,0} \right), \end{array}$ (30)

Thus, using (26)~(30) in (25), we can obtain the inequality of (24).

This completes the proof.

Theorem 4    Let $I \subset (0, \infty )$ be an interval, f: I°→Rα such that fDα(I°) and f(α)Cα[a, b] for a, bI° with a < b. If |f(α)|q is generalized harmonically s-convex on [a, b] for q>1, $\frac{1}{p} + \frac{1}{q} = 1$, then for all x∈[a, b], the following inequality holds:

 $\begin{array}{l} \left| {\left[ {\frac{{{{\left( {x - a} \right)}^\alpha }}}{{{a^\alpha }{x^\alpha }}} + \frac{{{{\left( {b - x} \right)}^\alpha }}}{{{b^\alpha }{x^\alpha }}}} \right]f\left( x \right) - {\rm{\Gamma }}{{\left( {1 + \alpha } \right)}_a}I_b^{\left( \alpha \right)}\frac{{f\left( u \right)}}{{{u^{2\alpha }}}}} \right| \le \\ {\left( {\frac{{{\rm{\Gamma }}\left( {1 + s\alpha } \right)}}{{{\rm{\Gamma }}\left( {1 + \left( {s + 1} \right)\alpha } \right)}}} \right)^{\frac{1}{q}}}{\left( {\frac{{{\rm{\Gamma }}\left( {1 + p\alpha } \right)}}{{{\rm{\Gamma }}\left( {1 + \left( {p + 1} \right)\alpha } \right)}}} \right)^{\frac{1}{p}}} \times \\ \left[ {{{\left( {x - a} \right)}^{2\alpha }}\left( {\frac{1}{{{x^{2p\alpha }}}}{}_2F_1^\alpha \left( {2p,p + 1;p + 2;1 - \frac{\alpha }{x}} \right)} \right)\frac{1}{p} \times } \right.\\ {\left( {{{\left| {{f^{\left( \alpha \right)}}\left( x \right)} \right|}^q} + {{\left| {{f^{\left( \alpha \right)}}\left( a \right)} \right|}^q}} \right)^{\frac{1}{q}}} + \\ {\left( {b - x} \right)^{2\alpha }}\left( {\frac{1}{{{b^{2p\alpha }}}}{}_2F_1^\alpha \left( {2p,p + 1;p + 2;1 - \frac{x}{b}} \right)} \right)\frac{1}{p} \times \\ \left. {{{\left( {{{\left| {{f^{\left( \alpha \right)}}\left( x \right)} \right|}^q} + {{\left| {{f^{\left( \alpha \right)}}\left( b \right)} \right|}^q}} \right)}^{\frac{1}{q}}}} \right]. \end{array}$ (31)

Proof    Taking modulus in equality (4), by the generalized Hölder's inequality(lemma 3) and the generalized harmonically s-convexity of |f(α)|q on [a, b], we have

 $\begin{array}{l} \left| {\left[ {\frac{{{{\left( {x - a} \right)}^\alpha }}}{{{a^\alpha }{x^\alpha }}} + \frac{{{{\left( {b - x} \right)}^\alpha }}}{{{b^\alpha }{x^\alpha }}}} \right]f\left( x \right) - {\rm{\Gamma }}{{\left( {1 + \alpha } \right)}_a}I_b^{\left( \alpha \right)}\frac{{f\left( u \right)}}{{{u^{2\alpha }}}}} \right| \le \\ {\left( {x - a} \right)^{2\alpha }}{\left( {\frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {\frac{{{t^{p\alpha }}}}{{{{\left( {ta + \left( {1 - t} \right)x} \right)}^{2p\alpha }}}}{{\left( {{\rm{d}}t} \right)}^\alpha }} } \right)^{\frac{1}{p}}} \times \\ {\left( {\frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {{t^{s\alpha }}{{\left| {{f^{\left( \alpha \right)}}\left( x \right)} \right|}^q} + {{\left( {1 - t} \right)}^{s\alpha }}{{\left| {{f^{\left( \alpha \right)}}\left( a \right)} \right|}^q}{{\left( {{\rm{d}}t} \right)}^\alpha }} } \right)^{\frac{1}{q}}} + \\ {\left( {b - x} \right)^{2\alpha }}{\left( {\frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {\frac{{{t^{p\alpha }}}}{{{{\left( {tb + \left( {1 - t} \right)x} \right)}^{2p\alpha }}}}{{\left( {{\rm{d}}t} \right)}^\alpha }} } \right)^{\frac{1}{p}}} \times \\ {\left( {\frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {{t^{s\alpha }}{{\left| {{f^{\left( \alpha \right)}}\left( x \right)} \right|}^q} + {{\left( {1 - t} \right)}^{s\alpha }}{{\left| {{f^{\left( \alpha \right)}}\left( b \right)} \right|}^q}{{\left( {{\rm{d}}t} \right)}^\alpha }} } \right)^{\frac{1}{q}}}. \end{array}$ (32)

By calculating, it is easily checked that

 $\begin{array}{l} \frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {{t^{s\alpha }}{{\left( {{\rm{d}}t} \right)}^\alpha }} = \\ \;\;\;\;\;\;\;\;\;\;\;\frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {{{\left( {1 - t} \right)}^{s\alpha }}{{\left( {{\rm{d}}t} \right)}^\alpha }} = \\ \;\;\;\;\;\;\;\;\;\;\;\frac{{{\rm{\Gamma }}\left( {1 + s\alpha } \right)}}{{{\rm{\Gamma }}\left( {1 + \left( {s + 1} \right)\alpha } \right)}}. \end{array}$ (33)

Further calculation, we have

 $\begin{array}{l} \frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {\frac{{{t^{p\alpha }}}}{{{{\left( {ta + \left( {1 - t} \right)x} \right)}^{2p\alpha }}}}{{\left( {{\rm{d}}t} \right)}^\alpha }} = \\ \;\;\;\;\;\;\;\frac{{\Gamma \left( {1 + p\alpha } \right)}}{{{x^{2p\alpha }}\Gamma \left( {1 + \left( {p + 1} \right)\alpha } \right)}}{}_2F_1^\alpha \left( {2p,p + 1;p + 2;1 - \frac{\alpha }{x}} \right), \end{array}$ (34)

and

 $\begin{array}{l} \frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {\frac{{{t^{p\alpha }}}}{{{{\left( {tb + \left( {1 - t} \right)x} \right)}^{2p\alpha }}}}{{\left( {{\rm{d}}t} \right)}^\alpha }} = \\ \;\;\;\;\;\;\;\frac{{\Gamma \left( {1 + p\alpha } \right)}}{{{b^{2p\alpha }}\Gamma \left( {1 + \left( {p + 1} \right)\alpha } \right)}} \times \\ \;\;\;\;\;\;\;{}_2F_1^\alpha \left( {2p,p + 1;p + 2;1 - \frac{x}{b}} \right). \end{array}$ (35)

Thus, using (33)~(35) in (32), we can obtain the inequality of (31).

This completes the proof.

Theorem 5    Let $I \subset (0, \infty )$ be an interval, f: I°→Rα such that fDα(I°) and f(α)Cα[a, b] for a, bI° with a < b. If |f(α)|q is generalized harmonically s-convex on [a, b] for q>1, $\frac{1}{p} + \frac{1}{q} = 1$, then for all x∈[a, b], the following inequality holds:

 $\begin{array}{l} \left| {\left[ {\frac{{{{\left( {x - a} \right)}^\alpha }}}{{{a^\alpha }{x^\alpha }}} + \frac{{{{\left( {b - x} \right)}^\alpha }}}{{{b^\alpha }{x^\alpha }}}} \right]f\left( x \right) - {\rm{\Gamma }}\left( {1 + \alpha } \right){}_aI_b^{\left( \alpha \right)}\frac{{f\left( u \right)}}{{{u^{2\alpha }}}}} \right| \le \\ \;\;\;\;\;\;{\left( {x - a} \right)^{2\alpha }}{\left( {\frac{{{{\left( {{x^{1 - 2p}} - {a^{1 - 2p}}} \right)}^\alpha }}}{{{{\left( {x - a} \right)}^\alpha }{{\left( {1 - 2p} \right)}^\alpha }{\rm{\Gamma }}\left( {1 + \alpha } \right)}}} \right)^{\frac{1}{p}}} \times \\ \;\;\;\;\;\;\left( {\frac{{{\rm{\Gamma }}\left( {1 + \left( {q + s} \right)\alpha } \right)}}{{{\rm{\Gamma }}\left( {1 + \left( {q + s + 1} \right)\alpha } \right)}}{{\left| {{f^{\left( \alpha \right)}}\left( x \right)} \right|}^q} + } \right.\\ \;\;\;\;\;\;{\left. {{B_\alpha }\left( {q + 1,s + 1} \right){{\left| {{f^{\left( \alpha \right)}}\left( a \right)} \right|}^q}} \right)^{\frac{1}{q}}} + \\ \;\;\;\;\;\;{\left( {b - x} \right)^{2\alpha }}{\left( {\frac{{{{\left( {{b^{1 - 2p}} - {x^{1 - 2p}}} \right)}^\alpha }}}{{{{\left( {b - x} \right)}^\alpha }{{\left( {1 - 2p} \right)}^\alpha }{\rm{\Gamma }}\left( {1 + \alpha } \right)}}} \right)^{\frac{1}{p}}} \times \\ \;\;\;\;\;\;\left( {\frac{{{\rm{\Gamma }}\left( {1 + \left( {q + s} \right)\alpha } \right)}}{{{\rm{\Gamma }}\left( {1 + \left( {q + s + 1} \right)\alpha } \right)}}{{\left| {{f^{\left( \alpha \right)}}\left( x \right)} \right|}^q} + } \right.\\ \;\;\;\;\;\;{\left. {{B_\alpha }\left( {q + 1,s + 1} \right){{\left| {{f^{\left( \alpha \right)}}\left( b \right)} \right|}^q}} \right)^{\frac{1}{q}}}. \end{array}$ (36)

Proof   Taking modulus in equality (4), by the generalized Hölder's inequality(lemma 3) and the generalized harmonically s-convexity of |f(α)|q on [a, b], we have

 $\begin{array}{l} \left| {\left[ {\frac{{{{\left( {x - a} \right)}^\alpha }}}{{{a^\alpha }{x^\alpha }}} + \frac{{{{\left( {b - x} \right)}^\alpha }}}{{{b^\alpha }{x^\alpha }}}} \right]f\left( x \right) - {\rm{\Gamma }}\left( {1 + \alpha } \right){}_aI_b^{\left( \alpha \right)}\frac{{f\left( u \right)}}{{{u^{2\alpha }}}}} \right| \le \\ {\left( {x - a} \right)^{2\alpha }}{\left( {\frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {\frac{{{1^\alpha }}}{{{{\left( {ta + \left( {1 - t} \right)x} \right)}^{2p\alpha }}}}{{\left( {{\rm{d}}t} \right)}^\alpha }} } \right)^{\frac{1}{p}}} \times \\ \left( {\frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {{t^{\left( {q + s} \right)\alpha }}{{\left| {{f^{\left( \alpha \right)}}\left( x \right)} \right|}^q}} + } \right.\\ {\left. {{t^{q\alpha }}{{\left( {1 - t} \right)}^{s\alpha }}{{\left| {{f^{\left( \alpha \right)}}\left( a \right)} \right|}^q}{{\left( {{\rm{d}}t} \right)}^\alpha }} \right)^{\frac{1}{q}}} + \\ {\left( {b - x} \right)^{2\alpha }}{\left( {\frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {\frac{{{1^\alpha }}}{{{{\left( {tb + \left( {1 - t} \right)x} \right)}^{2p\alpha }}}}{{\left( {{\rm{d}}t} \right)}^\alpha }} } \right)^{\frac{1}{p}}} \times \\ \left( {\frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {{t^{\left( {q + s} \right)\alpha }}{{\left| {{f^{\left( \alpha \right)}}\left( x \right)} \right|}^q}} + } \right.\\ {\left. {{t^{q\alpha }}{{\left( {1 - t} \right)}^{s\alpha }}{{\left| {{f^{\left( \alpha \right)}}\left( b \right)} \right|}^q}{{\left( {{\rm{d}}t} \right)}^\alpha }} \right)^{\frac{1}{q}}}. \end{array}$ (37)

By calculating, it is easily checked that

 $\frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {{t^{\left( {q + s} \right)\alpha }}{{\left( {{\rm{d}}t} \right)}^\alpha }} = \frac{{{\rm{\Gamma }}\left( {1 + \left( {q + s} \right)\alpha } \right)}}{{{\rm{\Gamma }}\left( {1 + \left( {q + s + 1} \right)\alpha } \right)}},$ (38)

and

 $\frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {{t^{q\alpha }}{{\left( {1 - t} \right)}^{s\alpha }}{{\left( {{\rm{d}}t} \right)}^\alpha }} = {{\rm{B}}_\alpha }\left( {q + 1,s + 1} \right).$ (39)

Applying the change of the variable ta+(1-t)x=v, we have

 $\begin{array}{l} \frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {\frac{{{1^\alpha }}}{{{{\left( {ta + \left( {1 - t} \right)x} \right)}^{2p\alpha }}}}{{\left( {{\rm{d}}t} \right)}^\alpha }} = \\ \frac{1}{{{{\left( {x - a} \right)}^\alpha }}}\frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_a^x {\frac{{{1^\alpha }}}{{{v^{2p\alpha }}}}{{\left( {{\rm{d}}v} \right)}^\alpha }} . \end{array}$ (40)

Letting $\frac{{{1^\alpha }}}{{{v^{2p - 1}}}} = \xi$ and from $\frac{{{1^\alpha }}}{{{v^{2p{\rm{ }}\alpha }}}}{\left( {{\rm{d}}v} \right)^\alpha } = \frac{{{{\left( {{\rm{d}}\xi } \right)}^\alpha }}}{{{{\left( {1 - 2p} \right)}^\alpha }}}{\rm{ }}$, we have

 $\begin{array}{l} \frac{1}{{{{\left( {x - a} \right)}^\alpha }}}\frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_a^x {\frac{{{1^\alpha }}}{{{v^{2p\alpha }}}}{{\left( {{\rm{d}}v} \right)}^\alpha }} = \\ \;\;\;\frac{{{{\left( {{x^{1 - 2p}} - {a^{1 - 2p}}} \right)}^\alpha }}}{{{{\left( {x - a} \right)}^\alpha }{{\left( {1 - 2p} \right)}^\alpha }{\rm{\Gamma }}\left( {1 + \alpha } \right)}}. \end{array}$ (41)

Similarly, using the change of the variable t b+(1-t)x=w, we have

 $\begin{array}{l} \frac{1}{{{\rm{\Gamma }}\left( {1 + \alpha } \right)}}\int_0^1 {\frac{{{1^\alpha }}}{{{{\left( {ta + \left( {1 - t} \right)x} \right)}^{2p\alpha }}}}{{\left( {{\rm{d}}t} \right)}^\alpha }} = \\ \;\;\;\;\frac{{{{\left( {{b^{1 - 2p}} - {x^{1 - 2p}}} \right)}^\alpha }}}{{{{\left( {b - x} \right)}^\alpha }{{\left( {1 - 2p} \right)}^\alpha }{\rm{\Gamma }}\left( {1 + \alpha } \right)}}. \end{array}$ (42)

Thus, using (38)~(42) in (37), we can obtain the inequality of (36).

This completes the proof.

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