0 Introduction

Let f: I→R, where $I \subseteq R $ is an interval, be a differentiable mapping in I° (the interior of I) and let a, b∈I° 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, y∈I and t∈[0, 1] and for some fixed s∈(0, 1], a function f: I→R 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, y∈I, 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 R^α(0 < α≤1) be the α-type set of the real line numbers. If a^α, b^α, c^α∈R^α, 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 R^α 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 ) $ →R^α(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, b∈I° 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°→R^α (I° is the interior of I) such that f∈D_α(I°) and f^(α)∈C_α[a, b] for a, b∈I° 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°→R^α (I° is the interior of I) such that f∈D_α(I°) and f^(α)∈C_α[a, b] for a, b∈I° 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 C₁^α, C₂^α, C₃^α, C₄^α 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], $

ln_α(x^α) 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°→R^α such that f∈D_α(I°) and f^(α)∈C_α[a, b] for a, b∈I° 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 C₁^α, C₂^α, C₃^α, C₄^α 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°→R^α such that f∈D_α(I°) and f^(α)∈C_α[a, b] for a, b∈I° 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°→R^α such that f∈D_α(I°) and f^(α)∈C_α[a, b] for a, b∈I° 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.