Open Access

Notions of generalized s-convex functions on fractal sets

Journal of Inequalities and Applications20152015:312

https://doi.org/10.1186/s13660-015-0826-x

Received: 24 June 2015

Accepted: 15 September 2015

Published: 6 October 2015

Abstract

The purpose of this article is to present some new inequalities for products of generalized convex and generalized s-convex functions on fractal sets. Furthermore, some applications are given.

Keywords

s-convex functions fractal space local fractional derivative

MSC

26A51 26D07 26D15 53C22

1 Introduction

Let \(f\colon I\subset\mathbb{R}\longrightarrow\mathbb{R}^{\alpha} \). For any \(x_{1},x_{2}\in I \) and \(\gamma\in[0,1] \) if the inequality
$$f\bigl(\gamma x_{1}+(1-\gamma)x_{2}\bigr)\leq \gamma^{\alpha}f(x_{1})+(1-\gamma )^{\alpha}f(x_{2}) $$
holds, then f is called a generalized convex function on I [1]. In \(\alpha=1 \), we have convex function, convexity is defined only in geometrical terms as being the property of a function whose graph bears tangents only under it [2].

The convexity of functions plays a significant role in many fields, for example, in biological system, economy, optimization, and so on [35].

In recent years, the fractal theory has received significantly remarkable attention from scientists and engineers. In the sense of Mandelbrot, a fractal set is one whose Hausdorff dimension strictly exceeds the topological dimension [6, 7]. Many researchers studied the properties of functions on fractal space and constructed many kinds of fractional calculus by using different approaches [812]. Particularly, in [13], Yang gave the analysis of local fractional functions on fractal space systematically, which includes local fractional calculus and the monotonicity of function.

Let \(\mathbb{R}^{\alpha} \) be the real line numbers on fractal space. Then by using Gao-Yang-Kang’s concept one can explain the definitions of the local fractional derivative and local fractional integral as in [1216]. Now if \(r_{1}^{\alpha}\), \(r_{2}^{\alpha} \) and \(r_{3}^{\alpha}\in \mathbb{R}^{\alpha} \) (\(0<\alpha\leq1\)), then
  1. (1)

    \(r_{1}^{\alpha}+ r_{2}^{\alpha}\in\mathbb{R}^{\alpha}\), \(r_{1}^{\alpha}r_{2}^{\alpha}\in\mathbb{R}^{\alpha} \),

     
  2. (2)

    \(r_{1}^{\alpha}+ r_{2}^{\alpha}= r_{2}^{\alpha}+ r_{1}^{\alpha}= ( r_{1}+r_{2} ) ^{\alpha}=(r_{2}+r_{1})^{\alpha } \),

     
  3. (3)

    \(r_{1}^{\alpha}+ ( r_{2}^{\alpha}+r_{3}^{\alpha} ) =(r_{1}^{\alpha}+ r_{2}^{\alpha})+r_{3}^{\alpha} \),

     
  4. (4)

    \(r_{1}^{\alpha} r_{2}^{\alpha}= r_{2}^{\alpha} r_{1}^{\alpha }= ( r_{1} r_{2} ) ^{\alpha}= ( r_{2}r_{1} ) ^{\alpha } \),

     
  5. (5)

    \(r_{1}^{\alpha} ( r_{2}^{\alpha}r_{3}^{\alpha} ) = ( r_{1}^{\alpha} r_{2}^{\alpha} ) r_{3}^{\alpha} \),

     
  6. (6)

    \(r_{1}^{\alpha} ( r_{2}^{\alpha}+r_{3}^{\alpha} ) = ( r_{1}^{\alpha} r_{2}^{\alpha} ) + ( r_{1}^{\alpha }r_{3}^{\alpha} ) \),

     
  7. (7)

    \(r_{1}^{\alpha}+0^{\alpha}=0^{\alpha}+r_{1}^{\alpha }=r_{1}^{\alpha} \) and \(r_{1}^{\alpha}\cdot 1^{\alpha}=1^{\alpha }\cdot r_{1}^{\alpha}=r_{1}^{\alpha}\).

     
Let us start with some definitions as regards the local fractional calculus on \(\mathbb{R}^{\alpha} \).

Definition 1.1

[13]

Let y be a local fractional continuous function on the interval \([a_{1},a_{2}] \). The local fractional integral of the function \(y(m) \) of order α is defined by
$$\begin{aligned} {}_{a_{1}}I^{(\alpha)}_{a_{2}}y(m) =& \bigl( \Gamma(1+\alpha) \bigr) ^{-1}\int_{a_{1}}^{a_{2}}y(\mu) (d \mu)^{\alpha} \\ =& \bigl( \Gamma(1+\alpha) \bigr) ^{-1} \lim_{\Delta \mu\longrightarrow 0} \sum_{i=1}^{n} y(\mu_{i}) ( \Delta \mu_{i} ) ^{\alpha } \end{aligned}$$
with \(\Delta \mu_{i}=\mu_{i+1}-\mu_{i} \) and \(\Delta \mu= \max\{\Delta \mu_{i}\colon i=1,2,\ldots,n-1\} \) where \([\mu_{i},\mu_{i+1}]\), \(i=0,1,\ldots,n-1 \), \(\mu_{0}=a_{1}<\mu _{1}<\cdots<\mu_{n-1}<\mu_{n}=a_{2} \) is a partition of the interval \([a_{1},a_{2}] \) and Γ is the well-known Gamma function \({ \Gamma(m)=\int_{0}^{\infty} t^{m-1}e^{-t}\,dt }\).

In [17], Mo and Sui introduced the definitions of two kinds of generalized s-convex functions on fractal sets as follows.

Definition 1.2

  1. (i)
    A function \(f\colon\mathbb{R}_{+}\longrightarrow\mathbb {R}^{\alpha} \), is called a generalized s-convex (\(0< s<1 \)) in the first sense if
    $$ f(\gamma_{1}a_{1}+\gamma_{2}a_{2}) \leq\gamma_{1}^{s\alpha }f(a_{1})+\gamma_{2}^{s \alpha}f(a_{2}) $$
    (1)
    for all \(a_{1},a_{2}\in\mathbb{R}_{+} \) and all \(\gamma_{1},\gamma _{2}\geq0 \) with \(\gamma_{1}^{s}+\gamma_{2}^{s}=1 \). This class of functions is denoted by \(GK^{1}_{s} \).
     
  2. (ii)

    A function \(f\colon\mathbb{R}_{+}\longrightarrow\mathbb {R}^{\alpha} \), is called a generalized s-convex (\(0< s<1 \)) in the second sense if (1) holds for all \(a_{1},a_{2}\in\mathbb {R}_{+} \) and all \(\gamma_{1},\gamma_{2}\geq0 \) with \(\gamma _{1}+\gamma_{2}=1 \). This class of functions is denoted by \(GK^{2}_{s} \).

     

In the same paper, [17], Mo and Sui proved that all functions from \(GK^{2}_{s} \), for \(s\in(0,1) \), are nonnegative.

In this article, recall that \(gs_{1} \) and \(gs_{2} \) are the generic classes consistent for those functions that are generalized s-convex in the first sense, and in the second sense, respectively. It is well known that there are many important established inequalities for the class of generalized convex functions, however, one of the most famous is known as the generalized Hermit-Hadamard inequality, or the ‘generalized Hadamard inequality’ and stated as follows (see [1]): let f be a generalized convex function on \([a_{1},a_{2}]\subseteq\mathbb{R}\), \(a_{1}< a_{2}\), then
$$f \biggl(\frac{a_{1}+a_{2}}{2} \biggr)\leq\frac{\Gamma(1+\alpha )}{(a_{2}-a_{1})^{\alpha}}\, {}_{a_{1}}I^{(\alpha)}_{a_{2}}f(x) \leq\frac {f(a_{1})+f(a_{2})}{2^{\alpha}}. $$

2 Basic results for generalized s-convex functions

Lemma 2.1

If \(f\in GK^{1}_{s} \) or \(f\in GK^{2}_{s} \), then
$$f(\gamma_{1}x_{1}+\gamma_{2}x_{2}) \leq\gamma_{1}^{\alpha s}f(x_{1})+\gamma_{2}^{\alpha s}f(x_{2}) $$
with \(\gamma_{1},\gamma_{2}\in[0,1] \), exclusively.

So, we might re-write the definitions of generalized s-convexity in the first sense and the second sense as follows.

Definition 2.1

A function \(f\colon I\subset\mathbb{R}_{+}\longrightarrow\mathbb {R}^{\alpha} \) is called a generalized s-convex in the first sense if
$$f \bigl( \gamma x_{1}+ \bigl( 1-\gamma^{s} \bigr) ^{\frac {1}{s}}x_{2} \bigr) \leq\gamma^{\alpha s} f(x_{1})+ \bigl( 1-\gamma ^{s} \bigr) ^{\alpha}f(x_{2}) $$
for all \(x_{1},x_{2}\in I \) and for all \(0\leq\gamma\leq1 \).

Definition 2.2

A function \(f\colon I\subset\mathbb{R}_{+}\longrightarrow\mathbb {R}^{\alpha} \) is called a generalized s-convex in the second sense if
$$f\bigl(\gamma x_{1}+(1-\gamma)x_{2}\bigr)\leq \gamma^{\alpha s} f(x_{1})+(1-\gamma )^{\alpha s}f(x_{2}) $$
for all \(x_{1},x_{2}\in I \) and for all \(0\leq\gamma\leq1 \).

Theorem 2.1

The classes \(GK^{1}_{1} \), \(GK^{2}_{1} \) and the class of generalized convex functions are equivalent when the domain is restricted to \(\mathbb{R}_{+} \).

Proof

Simply an issue of applying the definitions. □

Now, some results as regards generalized s-convex functions are given.

Remark 2.1

  1. (i)

    If \(f\in GK^{1}_{s} \), then \(f ( \frac {x_{1}+x_{2}}{2^{\frac{1}{s}}} ) \leq\frac {f(x_{1})+f(x_{2})}{2^{\alpha}}\).

     
  2. (ii)

    If \(f\in GK^{2}_{s} \), then \(f ( \frac {x_{1}+x_{2}}{2} ) \leq\frac{f(x_{1})+f(x_{2})}{2^{\alpha s}}\).

     
  3. (iii)

    For a function that is both generalized \(gs_{1} \) and \(gs_{2} \), there is a one-to-one correspondence between the set of all pairs of the sort \((\gamma,\eta) \) with respect to \(gs_{1} \) and the set of all pairs of the sort \((\gamma,\eta) \) with respect to \(gs_{2} \). (So we may write each γ as a \(\gamma_{1}^{s} \) and each η as a \(\eta_{1}^{s} \) and vice versa. This happens in the light of the fact that \(0\leq\gamma,\eta\leq1\), \(0\leq s\leq1 \).)

     

Theorem 2.2

If a function \(f \in GK^{1}_{s}\) and \(f\in GK^{2}_{s} \), then
$$f(\gamma_{1}x_{1}+\eta_{1}x_{2})\leq \gamma_{1}^{\alpha s} f(x_{1})+ \eta_{1}^{\alpha s}f(x_{2}) \leq\gamma_{2}^{\alpha s} f(x_{1})+ \eta _{2}^{\alpha s}f(x_{2}) $$
for some \(\lbrace \gamma_{i},\eta_{i}, i=1,2 \rbrace \subset[0,1] \).

Proof

It follows from the one-to-one correspondence proved before. To each \(\gamma_{2}\), \(\eta_{2} \) such that \(\gamma_{2}+\eta_{2}=1 \), there corresponds \(\gamma_{1}\), \(\eta_{1} \) where \(\gamma_{1}^{s}+\eta _{1}^{s}=1 \) and \(\gamma_{2}\geq\gamma_{1} \), \(\eta_{2}\geq\eta_{1} \) since \(\lbrace \gamma_{i},\eta_{i}, i=1,2 \rbrace\subset [0,1] \). □

Theorem 2.3

If a function \(f \in GK^{1}_{s} \) and \(f\in GK^{2}_{s} \) and its domain concurs with its counter-domain, then \(fof \in GK^{2}_{s} \).

Proof

\(f ( \gamma_{1}x_{1}+ ( 1-\gamma_{1}^{s} ) ^{\frac {1}{s}}x_{2} ) \leq\gamma_{1}^{\alpha s}f(x_{1})+ ( 1-\gamma _{1}^{s} ) ^{\alpha} f(x_{2}) \), then
$$\begin{aligned} f \bigl( \gamma_{1}^{\alpha s}f(x_{1})+ \bigl( 1- \gamma_{1}^{s} \bigr) ^{\alpha} f(x_{2}) \bigr) \leq& \bigl( \gamma_{1}^{\alpha s} \bigr) ^{\alpha s} f \bigl(f(x_{1})\bigr)+ \bigl( \bigl( 1-\gamma_{1}^{s} \bigr) ^{\alpha } \bigr) ^{\alpha s} f \bigl( f(x_{2}) \bigr) \\ =& \gamma _{2}^{\alpha s} fof(x_{1})+ \beta_{2}^{\alpha s} fof (x_{2}). \end{aligned}$$
 □

3 Inequalities for generalized s-convex functions

Theorem 3.1

Let \(g\colon[0,1]\longrightarrow\mathbb{R}^{\alpha} \) be a function such that
$$g(\gamma)=\frac{1}{(a_{2}-a_{1})^{2\alpha}(\Gamma(1+\alpha))^{2}} \int_{a_{1}}^{a_{2}}\int _{a_{1}}^{a_{2}} f\bigl(\gamma x_{1}+(1- \gamma)x_{2}\bigr) (dx_{1})^{\alpha}(dx_{2})^{\alpha}, $$
where \(f\colon[a_{1},a_{2}]\longrightarrow\mathbb{R}^{\alpha} \) and \(f\in GK^{2}_{s} \). Then:
  1. (i)

    \(g\in GK^{2}_{s}\) in \([0,1] \). If \(f\in GK^{1}_{s} \), then \(g\in GK^{1}_{s} \).

     
  2. (ii)

    \(2^{\alpha}g(0)=2^{\alpha}g(1)= \frac{2^{\alpha}}{ (a_{2}-a_{1})^{\alpha}\Gamma(1+\alpha)} {}_{a_{1}}I^{(\alpha )}_{a_{2}} f(x_{1}) \) is an upper bound for \(g(\gamma) \).

     
  3. (iii)

    \(2^{\alpha(1-s)}g(\gamma)\geq g(\frac{1}{2})=\frac {1}{(a_{2}-a_{1})^{2\alpha}(\Gamma(1+\alpha))^{2}}\int_{a_{1}}^{a_{2}}\int_{a_{1}}^{a_{2}} f ( \frac{x_{1}+x_{2}}{2} ) (dx_{1})^{\alpha}(dx_{2})^{\alpha} \), \(\gamma\in[0,1] \).

     

Proof

(i) Take \(\lbrace\gamma_{1},\gamma_{2} \rbrace\subset[0,1]\), \(\gamma_{1}+\gamma_{2}=1 \), \(t_{1},t_{2}\in D \) and \(f\in GK^{2}_{s} \), then we have
$$\begin{aligned} & g(\gamma_{1} t_{1}+\gamma_{2} t_{2}) \\ &\quad= \frac{1}{(a_{2}-a_{1})^{2\alpha}(\Gamma(1+\alpha))^{2}} \int_{a_{1}}^{a_{2}}\int _{a_{1}}^{a_{2}} f\bigl( (\gamma_{1} t_{1}+\gamma _{2} t_{2})x_{1} \\ &\qquad{}+\bigl(1-(\gamma_{1} t_{1}+\gamma_{2} t_{2})\bigr)x_{2}\bigr) (dx_{1})^{\alpha}(dx_{2})^{\alpha} \\ &\quad \leq \frac{1}{(a_{2}-a_{1})^{2\alpha }(\Gamma(1+\alpha))^{2}}\int_{a_{1}}^{a_{2}}\int _{a_{1}}^{a_{2}} \bigl[ \gamma_{1}^{\alpha s}f( t_{1} x_{1}+x_{2}-t_{1}x_{2}) \\ &\qquad{}+ \gamma _{2}^{\alpha s}f(t_{2}x_{1} + x_{2}-t_{2}x_{2})\bigr] (dx_{1})^{\alpha }(dx_{2})^{\alpha} \\ &\quad = \gamma_{1}^{\alpha s} \frac{1}{(a_{2}-a_{1})^{2\alpha}(\Gamma(1+\alpha ))^{2}}\int _{a_{1}}^{a_{2}}\int_{a_{1}}^{a_{2}} f\bigl( t_{1} x_{1}+(1-t_{1})x_{2} \bigr) (dx_{1})^{\alpha}(dx_{2})^{\alpha} \\ &\qquad{} + \gamma_{2}^{\alpha s} \frac{1}{(a_{2}-a_{1})^{2\alpha} (\Gamma(1+\alpha))^{2}}\int _{a_{1}}^{a_{2}}\int_{a_{1}}^{a_{2}} f\bigl(t_{2}x_{1} + (1-t_{2})x_{2} \bigr) (dx_{1})^{\alpha}(dx_{2})^{\alpha} \\ &\quad = \gamma_{1}^{\alpha s}g(t_{1}) + \gamma_{2}^{\alpha s}g(t_{2}), \end{aligned}$$
which implies that \(g\in GK^{2}_{s} \) in \([0,1] \).
(ii) Since \(f(\gamma x_{1}+(1-\gamma)x_{2})\leq\gamma^{\alpha s} f(x_{1})+(1-\gamma)^{\alpha s} f(x_{2}) \) we have
$$\begin{aligned} A =& \frac{1}{(\Gamma(1+\alpha))^{2}} \int_{a_{1}}^{a_{2}}\int _{a_{1}}^{a_{2}} f\bigl(\gamma x_{1}+(1- \gamma)x_{2}\bigr) (dx_{1})^{\alpha }(dx_{2})^{\alpha} \\ \leq& \frac{1}{(\Gamma(1+\alpha))^{2}} \int_{a_{1}}^{a_{2}}\int _{a_{1}}^{a_{2}} \gamma^{\alpha s} f(x_{1}) (dx_{1})^{\alpha}(dx_{2})^{\alpha} \\ &{} +\frac{1}{(\Gamma(1+\alpha))^{2}} \int_{a_{1}}^{a_{2}}\int _{a_{1}}^{a_{2}} (1-\gamma)^{\alpha s} f(x_{2}) (dx_{1})^{\alpha }(dx_{2})^{\alpha} \\ =& \gamma^{\alpha s}\frac{1}{\Gamma (1+\alpha)}(a_{2}-a_{1})^{\alpha} {}_{a_{1}}I^{(\alpha)}_{a_{2}} f(x_{1})+(1- \gamma)^{\alpha s} \frac{1}{\Gamma(1+\alpha )}(a_{2}-a_{1})^{\alpha} {}_{a_{1}}I^{(\alpha)}_{a_{2}} f(x_{2}) \\ =& \frac{ ( \gamma^{\alpha s}+(1-\gamma )^{\alpha s} ) }{\Gamma(1+\alpha)}(a_{2}-a_{1})^{\alpha} {}_{a_{1}}I^{(\alpha)}_{a_{2}} f(x_{1}), \end{aligned}$$
since \(\gamma^{\alpha s}\leq1^{\alpha} \). Because \((1-\gamma )^{\alpha s}\leq1^{\alpha} \) as well, we have
$$A\leq\frac{2^{\alpha}}{\Gamma(1+\alpha)} (a_{2}-a_{1})^{\alpha} {}_{a_{1}}I^{(\alpha)}_{a_{2}} f(x_{1}). $$
(iii) Since f is generalized s-convex in the second sense,
$$\frac{f ( \gamma x_{1}+(1-\gamma)x_{2} ) +f ( (1-\gamma )x_{1}+\gamma x_{2} ) }{2^{\alpha s}}\geq f \biggl( \frac {x_{1}+x_{2}}{2} \biggr) $$
for all \(\gamma\in[0,1] \) and \(x_{1},x_{2}\in[a,b] \).
If we integrate on \([a_{1},a_{2}]\times[a_{1},a_{2}] \), we obtain
$$\begin{aligned} & \frac{1}{2^{\alpha s}(\Gamma(1+\alpha))^{2}}\int_{a_{1}}^{a_{2}}\int _{a_{1}}^{a_{2}} \bigl[ f\bigl(\gamma x_{1}+(1- \gamma)x_{2}\bigr)+f\bigl((1-\gamma )x_{1}+\gamma x_{2}\bigr) \bigr] (dx_{1})^{\alpha}(dx_{2})^{\alpha} \\ &\quad \geq\frac{1}{(\Gamma(1+\alpha))^{2}}\int_{a_{1}}^{a_{2}}\int _{a_{1}}^{a_{2}} f \biggl( \frac{x_{1}+x_{2}}{2} \biggr) (dx_{1})^{\alpha}(dx_{2})^{\alpha} \\ &\quad\Rightarrow \\ &\frac{1}{2^{\alpha(s-1)}(a_{2}-a_{1})^{2\alpha}(\Gamma(1+\alpha ))^{2}}\int_{a_{1}}^{a_{2}}\int _{a_{1}}^{a_{2}} f\bigl(\gamma x_{1}+(1-\gamma )x_{2}\bigr) (dx_{1})^{\alpha}(dx_{2})^{\alpha} \\ &\quad\geq\frac{1}{(a_{2}-a_{1})^{2\alpha}(\Gamma(1+\alpha))^{2}} \int_{a_{1}}^{a_{2}}\int _{a_{1}}^{a_{2}} f \biggl( \frac{x_{1}+x_{2}}{2} \biggr) (dx_{1}) ^{\alpha}(dx_{2})^{\alpha}, \end{aligned}$$
which means that
$$2^{\alpha(1-s)}g(\gamma)\geq g \biggl( \frac{1}{2} \biggr) . $$
The proof is complete. □
Assume the following functions:
  1. (i)
    \(g_{F_{1}}\colon[0,1]\longrightarrow\mathbb{R}^{\alpha} \), defined by
    $$g_{F_{1}}(\gamma)=\frac{1}{(\Gamma(1+\alpha))^{2}}\int_{a_{1}}^{a_{2}} \int_{a_{1}}^{a_{2}} f\bigl(\gamma x_{1}+(1- \gamma )x_{2}\bigr)F(x_{1})F(x_{2}) (dx_{1})^{\alpha}(dx_{2})^{\alpha}, $$
    where \(f\colon[a_{1},a_{2}]\subset\mathbb{R}_{+}\longrightarrow \mathbb{R}^{\alpha} \) and \(f\in GK^{1}_{s} \).
     
  2. (ii)
    \(g_{F_{2}}\colon[0,1]\longrightarrow\mathbb{R}^{\alpha} \), defined by
    $$g_{F_{2}}(\gamma)=\frac{1}{(\Gamma(1+\alpha))^{2}}\int_{a_{1}}^{a_{2}} \int_{a_{1}}^{a_{2}} f\bigl(\gamma x_{1}+(1- \gamma )x_{2}\bigr)F(x_{1})F(x_{2}) (dx_{1})^{\alpha}(dx_{2})^{\alpha}, $$
    where \(f\colon[a_{1},a_{2}]\subset\mathbb{R}_{+}\longrightarrow \mathbb{R}^{\alpha} \) and \(f\in GK^{2}_{s} \).
     

Theorem 3.2

The following holds:
  1. (i)

    \(g_{F_{1}} \) and \(g_{F_{2}} \) are both symmetric about \(\gamma=\frac{1}{2} \).

     
  2. (ii)

    \(g_{F_{2}} \in GK^{2}_{s} \) in \([0,1] \).

     
  3. (iii)
    We have the upper bound
    $$2^{\alpha}g_{F_{2}}(1)= \frac{2^{\alpha}}{(\Gamma(1+\alpha))^{2}}\int_{a_{1}}^{a_{2}} \int_{a_{1}}^{a_{2}} f(x_{1})F(x_{1})F(x_{2}) (dx_{1})^{\alpha}(dx_{2})^{\alpha} $$
    for the function \(g_{F_{2}}(\gamma) \).
     

Proof

(i) \(g_{F_{1}} \) and \(g_{F_{2}} \) are both symmetric about \(\gamma=\frac{1}{2} \) because \(g_{F_{1}}(\gamma)= g_{F_{1}}(1-\gamma) \) and \(g_{F_{2}}(\gamma)= g_{F_{2}}(1-\gamma) \).

(ii) We get
$$\begin{aligned} g_{F_{2}}\bigl(\gamma t_{1}+(1-\gamma)t_{2}\bigr) =& \frac{1}{(\Gamma(1+\alpha ))^{2}}\int_{a_{1}}^{a_{2}}\int _{a_{1}}^{a_{2}} f\bigl(\bigl(\gamma t_{1}+(1- \gamma)t_{2}\bigr)x_{1} \\ &{} +\bigl(1-\bigl(\gamma t_{1}+(1-\gamma)t_{2} \bigr)x_{2}\bigr)\bigr)F(x_{1})F(x_{2}) (dx_{1})^{\alpha }(dx_{2})^{\alpha}. \end{aligned}$$
But
$$\begin{aligned} &\bigl(\gamma t_{1}+(1-\gamma)t_{2}\bigr)x_{1}+(1- \bigl(\gamma t_{1}+(1-\gamma )t_{2}\bigr)x_{2} \\ &\quad= \gamma t_{1}x_{1}+(1-\gamma)t_{2}x_{1}+ \gamma x_{2}+(1-\gamma)x_{2} -\gamma t_{1}x_{2}-(1- \gamma)t_{2}x_{2} \\ &\quad = \gamma (t_{1}x_{1}+x_{2}-t_{1}x_{2})+(1- \gamma ) (t_{2}x_{1}+x_{2}-t_{2}x_{2}). \end{aligned}$$
Since \(f\in GK^{2}_{s} \),
$$\begin{aligned} & g_{F_{2}}\bigl(\gamma t_{1}+(1-\gamma)t_{2}\bigr) \\ &\quad= \frac{1}{(\Gamma(1+\alpha ))^{2}}\int_{a_{1}}^{a_{2}}\int _{a_{1}}^{a_{2}} f\bigl(\bigl(\gamma t_{1}+(1- \gamma)t_{2}\bigr)x_{1} \\ &\qquad{} +\bigl(1-\bigl(\gamma t_{1}+(1-\gamma)t_{2} \bigr)x_{2}\bigr)\bigr)F(x_{1})F(x_{2}) (dx_{1})\alpha (dx_{2})^{\alpha} \\ &\quad \leq\gamma^{\alpha s} \frac{1}{(\Gamma (1+\alpha))^{2}}\int_{a_{1}}^{a_{2}} \int_{a_{1}}^{a_{2}}f\bigl(t_{1}x_{1}+(1-t_{1})x_{2} \bigr)F(x_{1})F(x_{2}) (dx_{1})^{\alpha }(dx_{2})^{\alpha} \\ &\qquad{} +(1-\gamma)^{\alpha s}\frac{1}{(\Gamma (1+\alpha))^{2}}\int_{a_{1}}^{a_{2}} \int_{a_{1}}^{a_{2}}f\bigl(t_{2}x_{1}+(1-t_{2})x_{2} \bigr)F(x_{1})F(x_{2}) (dx_{1})^{\alpha }(dx_{2})^{\alpha} \\ &\quad =\gamma^{\alpha s}g_{F_{2}}(t_{1})+(1-\gamma )^{\alpha s}g_{F_{2}}(t_{2}), \end{aligned}$$
which proves that \(g_{F_{2}}\in GK^{2}_{s} \).
(iii) From the definition and the assumptions, we get
$$\begin{aligned} g_{F_{2}}(\gamma) =&\frac{1}{(\Gamma(1+\alpha))^{2}}\int_{a_{1}}^{a_{2}} \int_{a_{1}}^{a_{2}}f\bigl(\gamma x_{1}+(1- \gamma )x_{2}\bigr)F(x_{1})F(x_{2}) (dx_{1})^{\alpha}(dx_{2})^{\alpha} \\ \leq&\gamma^{\alpha s}\frac{1}{(\Gamma(1+\alpha))^{2}}\int_{a_{1}}^{a_{2}} \int_{a_{1}}^{a_{2}} f( x_{1})F(x_{1})F(x_{2}) (dx_{1})^{\alpha}(dx_{2})^{\alpha} \\ &{}+ (1-\gamma)^{\alpha s}\frac{1}{(\Gamma(1+\alpha))^{2}}\int_{a_{1}}^{a_{2}} \int_{a_{1}}^{a_{2}} f(x_{2})F(x_{1})F(x_{2}) (dx_{1})^{\alpha}(dx_{2})^{\alpha} \\ =& \bigl(\gamma^{\alpha s}+(1-\gamma)^{\alpha s}\bigr) \frac{1}{(\Gamma(1+\alpha ))^{2}}\int_{a_{1}}^{a_{2}}\int _{a_{1}}^{a_{2}} f( x_{1})F(x_{1})F(x_{2}) (dx_{1})^{\alpha}(dx_{2})^{\alpha} \\ \leq& \frac{2^{\alpha}}{(\Gamma(1+\alpha))^{2}}\int_{a_{1}}^{a_{2}}\int _{a_{1}}^{a_{2}}f( x_{1})F(x_{1})F(x_{2}) (dx_{1})^{\alpha }(dx_{2})^{\alpha} \\ =& 2^{\alpha}g_{F_{2}}(1). \end{aligned}$$
 □

Theorem 3.3

Let \(f_{1},f_{2}\colon[a_{1},a_{2}]\subset\mathbb {R}_{+}\longrightarrow\mathbb{R}^{\alpha}\), \(a_{1}< a_{2} \), be nonnegative, and generalized s-convex functions in the second sense. If \(f_{1}\in GK^{2}_{s_{1}} \) and \(f_{2}\in GK^{2}_{s_{2}} \) on \([a_{1},a_{2}] \) for some \(\gamma\in[0,1] \) and \(s_{1}, s_{2}\in ( 0,1 ] \), then
$$\begin{aligned} &\frac{1}{(\Gamma(1+\alpha))^{3}}\int_{a_{1}}^{a_{2}}\int _{a_{1}}^{a_{2}}\int_{0}^{1}f_{1} \bigl(\gamma x_{1}+(1-\gamma )x_{2}\bigr)f_{2} \bigl(\gamma x_{1}+(1-\gamma)x_{2}\bigr) (d \gamma)^{\alpha }(dx_{2})^{\alpha}(dx_{1})^{\alpha} \\ &\quad\leq\frac{2^{\alpha }\Gamma(1+(\gamma_{1}+\gamma_{2})\alpha)}{\Gamma(1+(\gamma_{1}+\gamma _{2}+1)\alpha)}\frac{1}{(\Gamma(1+\alpha))^{2}}(a_{2}-a_{1})^{\alpha } \int_{a_{1}}^{a_{2}}f_{1}(x_{1})f_{2}(x_{1}) (dx_{1})^{\alpha} \\ &\qquad{}+ \frac{1}{2^{\alpha}(\Gamma(1+\alpha))^{3}}\beta_{\alpha }(a_{2}-a_{1})^{2\alpha} \bigl[ T_{1}(a,b)+T_{2}(a,b) \bigr], \end{aligned}$$
where
$$\begin{aligned}& \beta_{\alpha}= \int_{0}^{1} \gamma^{\alpha s_{1}} (1-\gamma)^{\alpha s_{2}} (d\gamma)^{\alpha}, \\& T_{1}(a_{1},a_{2})= f_{1}(a_{1}) f_{2}(a_{1})+f_{1}(a_{2})f_{2}(a_{2}), \end{aligned}$$
and
$$T_{2}(a_{1},a_{2})= f_{1}(a_{1}) f_{2}(a_{2})+f_{1}(a_{2})f_{2}(a_{1}). $$

Proof

Since \(f_{1}\in GK^{2}_{s_{1}} \) and \(f_{2}\in GK^{2}_{s_{2}} \) on \([a_{1},a_{2}] \),
$$\begin{aligned}& f_{1}\bigl(\gamma x_{1}+(1-\gamma)x_{2}\bigr) \leq\gamma^{\alpha s_{1}}f_{1}(x_{1})+(1- \gamma)^{\alpha s_{1}}f_{1}(x_{2}), \\& f_{2}\bigl(\gamma x_{1}+(1-\gamma)x_{2}\bigr) \leq\gamma^{\alpha s_{2}}f_{2}(x_{1})+(1- \gamma)^{\alpha s_{2}}f_{2}(x_{2}), \end{aligned}$$
for all \(\gamma\in[0,1] \). Since \(f_{1} \) and \(f_{1} \) are nonnegative,
$$\begin{aligned} &f_{1}\bigl(\gamma x_{1}+(1-\gamma)x_{2} \bigr)f_{2}\bigl(\gamma x_{1}+(1-\gamma )x_{2} \bigr) \\ &\quad\leq \gamma^{\alpha(s_{1}+s_{2})} f_{1}(x_{1}) f_{2}(x_{1})+(1-\gamma)^{\alpha(s_{1}+s_{2})} f_{1}(x_{2}) f_{2}(x_{2}) \\ &\qquad{} + \gamma^{\alpha s_{1}}(1-\gamma)^{\alpha s_{2}} f_{1}(x_{1}) f_{2}(x_{2})+\gamma ^{\alpha s_{2}}(1-\gamma)^{\alpha s_{1}} f_{1}(x_{2}) f_{2}(x_{1}). \end{aligned}$$
Integrating both sides of the above inequality over \([0,1] \), we obtain
$$\begin{aligned} & \frac{1}{\Gamma(1+\alpha)}\int_{0}^{1}f_{1} \bigl(\gamma x_{1}+(1-\gamma )x_{2}\bigr)f_{2} \bigl(\gamma x_{1}+(1-\gamma)x_{2}\bigr) (d \gamma)^{\alpha} \\ &\quad\leq \frac{1}{\Gamma(1+\alpha)}f_{1}(x_{1}) f_{2}(x_{1})\int_{0}^{1} \gamma ^{\alpha(s_{1}+s_{2})} (d\gamma)^{\alpha} \\ &\qquad{}+\frac{1}{\Gamma(1+\alpha)} f_{1}(x_{2}) f_{2}(x_{2})\int_{0}^{1}(1- \gamma)^{\alpha (s_{1}+s_{2})}(d\gamma)^{\alpha} \\ &\qquad{}+ \frac {1}{\Gamma(1-\alpha)} f_{1}(x_{1}) f_{2}(x_{2})\int_{0}^{1} \gamma ^{\alpha s_{1}}(1-\gamma)^{\alpha s_{2}}(d\gamma)^{\alpha} \\ &\qquad{}+\frac {1}{\Gamma(1+\alpha)} f_{1}(x_{2}) f_{2}(x_{1})\int_{0}^{1} \gamma ^{\alpha s_{2}}(1-\gamma)^{\alpha s_{1}}(d\gamma)^{\alpha} \\ &\quad = \frac{\Gamma(1+(s_{1}+s_{2})\alpha)}{\Gamma (1+(s_{1}+s_{2}+1)\alpha)} f_{1}(x_{1})f_{2}(x_{1})+ \frac{\Gamma (1+(s_{1}+s_{2})\alpha)}{\Gamma(1+(s_{1}+s_{2}+1)\alpha)} f_{1}(x_{2})f_{2}(x_{2}) \\ &\qquad{}+\frac{1}{\Gamma (1+\alpha)}\beta_{\alpha }\bigl[f_{1}(x_{1})f_{2}(x_{2})+f_{1}(x_{2})f_{2}(x_{1}) \bigr]. \end{aligned}$$
If we integrate both sides of the above inequalities on \([a_{1},a_{2}]\times[a_{1},a_{2}] \), we obtain
$$\begin{aligned} &\frac{1}{(\Gamma(1+\alpha))^{3}}\int_{a_{1}}^{a_{2}}\int _{a_{1}}^{a_{2}}\int_{0}^{1}f_{1} \bigl(\gamma x_{1}+(1-\gamma )x_{2}\bigr)f_{2} \bigl(\gamma x_{1}+(1-\gamma)x_{2}\bigr) (d \gamma)^{\alpha }(dx_{2})^{\alpha}(dx_{1})^{\alpha} \\ &\quad \leq \frac{\Gamma(1+(s_{1}+s_{2})\alpha)}{\Gamma(1+(s_{1}+s_{2}+1)\alpha )}\frac{1}{(\Gamma(1+\alpha))^{2}} \biggl[ \int _{a_{1}}^{a_{2}}\int_{a_{1}}^{a_{2}}f_{1}(x_{1})f_{2}(x_{1}) (dx_{2})^{\alpha }(dx_{1})^{\alpha} \\ &\qquad{} +\int_{a_{1}}^{a_{2}}\int _{a_{1}}^{a_{2}}f_{1}(x_{2})f_{2}(x_{2}) (dx_{2})^{\alpha }(dx_{1})^{\alpha} \biggr] \\ &\qquad{} + \frac {1}{(\Gamma(1+\alpha))^{3}}\beta_{\alpha}\biggl[ \int _{a_{1}}^{a_{2}}\int_{a_{1}}^{a_{2}}f_{1}(x_{1})f_{2}(x_{2}) (dx_{2})^{\alpha }(dx_{1})^{\alpha} \\ &\qquad{}+\int_{a_{1}}^{a_{2}}\int_{a_{1}}^{a_{2}}f_{1}(x_{2})f_{2}(x_{1}) (dx_{2})^{\alpha }(dx_{1})^{\alpha}\biggr] \\ &\quad =\frac{\Gamma (1+(s_{1}+s_{2})\alpha)}{\Gamma(1+(s_{1}+s_{2}+1)\alpha)}\frac {1}{(\Gamma(1+\alpha))^{2}}(a_{2}-a_{1})^{\alpha} \\ &\qquad{}\times\biggl[ \int_{a_{1}}^{a_{2}}f_{1}(x_{1})f_{2}(x_{1}) (dx_{1})^{\alpha}+\int_{a_{1}}^{a_{2}}f_{1}(x_{2})f_{2}(x_{2}) (dx_{2})^{\alpha}\biggr] \\ &\qquad{} +\frac{1}{(\Gamma(1+\alpha))^{3}}\beta _{\alpha}\biggl[ \int _{a_{1}}^{a_{2}}f_{1}(x_{1}) (dx_{1})^{\alpha}\int_{a_{1}}^{a_{2}}f_{2}(x_{2}) (dx_{2})^{\alpha} \\ &\qquad{}+\int_{a_{1}}^{a_{2}}f_{1}(x_{2}) (dx_{2})^{\alpha}\int_{a_{1}}^{a_{2}}f_{2}(x_{1}) (dx_{1})^{\alpha}\biggr]. \end{aligned}$$
By using the right half of the generalized Hadamard inequality on the right side of the above inequality, we have
$$\begin{aligned} &\frac{1}{(\Gamma(1+\alpha))^{3}}\int_{a_{1}}^{a_{2}}\int _{a_{1}}^{a_{2}}\int_{0}^{1}f_{1} \bigl(\gamma x_{1}+(1-\gamma )x_{2}\bigr)f_{2} \bigl(\gamma x_{1}+(1-\gamma)x_{2}\bigr) (d \gamma)^{\alpha }(dx_{2})^{\alpha}(dx_{1})^{\alpha} \\ &\quad \leq\frac{2^{\alpha}\Gamma (1+(s_{1}+s_{2})\alpha)}{\Gamma(1+(s_{1}+s_{2}+1)\alpha)}\frac {1}{(\Gamma(1+\alpha))^{2}}(a_{2}-a_{1})^{\alpha} \int_{a_{1}}^{a_{2}}f_{1}(x_{1})f_{2}(x_{1}) (dx_{1})^{\alpha} \\ &\qquad{} +\frac{1}{2^{\alpha}(\Gamma(1+\alpha))^{3}}\beta_{\alpha }(a_{2}-a_{1})^{2\alpha } \bigl[T_{1}(a_{1},a_{2})+T_{2}(a_{1},a_{2}) \bigr]. \end{aligned}$$
The proof is complete. □

Remark 3.1

  1. (i)
    If \(\alpha=1 \), in Theorem 3.3, then
    $$\beta=\int_{0}^{1}\gamma^{s_{1}}(1- \gamma)^{s_{2}}(d\gamma)=\frac {\Gamma(s_{1}+1)\Gamma(s_{2}+1)}{\Gamma(s_{1}+s_{2}+2)} $$
    and
    $$\begin{aligned} & \int_{a_{1}}^{a_{2}} \int_{a_{1}}^{a_{2}} \int_{0}^{1}f_{1}\bigl(\gamma x_{1}+(1-\gamma)x_{2}\bigr)f_{2}\bigl(\gamma x_{1}+(1-\gamma)x_{2}\bigr) (d\gamma ) (dx_{1}) (dx_{2}) \\ &\quad \leq\frac {2}{s_{1}+s_{2}+1}(a_{2}-a_{1})\int _{a_{1}}^{a_{2}}f_{1}(x_{1})f_{2}(x_{1})\,dx_{1} \\ &\qquad{} +\frac{\Gamma(s_{1}+1)\Gamma(s_{2}+1)}{2\Gamma (s_{1}+s_{2}+2)}(a_{2}-a_{1})^{2} \bigl[T_{1}(a_{1},a_{2})+T_{2}(a_{1},a_{2}) \bigr]. \end{aligned}$$
     
  2. (ii)
    If \(\alpha=1 \), and \(s_{1}=s_{2}=1 \), then \(\frac{\Gamma (s_{1}+1)\Gamma(s_{2}+1)}{\Gamma(s_{1}+s_{2}+2)}=\frac{1}{6} \). Also,
    $$\beta=\int_{0}^{1}\gamma^{s_{1}}(1- \gamma)^{s_{2}}(d\gamma)=\frac{\Gamma (s_{1}+1)\Gamma(s_{2}+1)}{\Gamma(s_{1}+s_{2}+2)}, $$
    which implies that
    $$\begin{aligned} &\int_{a_{1}}^{a_{2}} \int_{a_{1}}^{a_{2}} \int_{0}^{1}f_{1}\bigl(\gamma x_{1}+(1-\gamma)x_{2}\bigr)f_{2}\bigl(\gamma x_{1}+(1-\gamma)x_{2}\bigr) (d\gamma ) (dx_{2}) (dx_{1}) \\ &\quad\leq\frac{2}{3}(a_{2}-a_{1})\int _{a_{1}}^{a_{2}}f_{1}(x_{1})f_{2}(x_{1})\,dx_{1}\\ &\qquad{} + \frac {1}{12}(a_{2}-a_{1})^{2} \bigl[T_{1}(a_{1},a_{2})+T_{2}(a_{1},a_{2}) \bigr]. \end{aligned}$$
     

Theorem 3.4

Let \(f_{1},f_{2}\colon[a_{1},a_{2}]\subset\mathbb {R}_{+}\longrightarrow\mathbb{R}^{\alpha}\), \(a_{1}< a_{2} \), be nonnegative and generalized s-convex functions in the second sense. If \(f_{1}\in GK^{2}_{s_{1}} \) and \(f_{2}\in GK^{2}_{s_{2}} \) on \([a_{1},a_{2}] \) for some \(\gamma\in[0,1] \) and \(s_{1}, s_{2}\in ( 0,1 ] \), then
$$\begin{aligned} &\frac{1}{(\Gamma(1+\alpha))^{2}}\int_{a_{1}}^{a_{2}}\int _{0}^{1}f_{1} \biggl( \gamma x_{1}+(1-\gamma)\frac{a+b}{2} \biggr) f_{2} \biggl( \gamma x_{1}+(1-\gamma)\frac{a+b}{2} \biggr) (d\gamma )^{\alpha}(dx_{1})^{\alpha} \\ &\quad \leq\frac{\Gamma(1+(s_{1}+s_{2})\alpha )}{\Gamma(1+(s_{1}+s_{2}+1)\alpha)}\frac{1}{\Gamma(1+\alpha)}\int_{a_{1}}^{a_{2}}f_{1}(x_{1})f_{2}(x_{1}) (dx_{1})^{\alpha} \\ &\qquad{} + \frac{1}{2^{\alpha}\Gamma(1+\alpha)} \biggl[ \frac {\Gamma(1+(s_{1}+s_{2})\alpha)}{2^{\alpha}\Gamma(1+(s_{1}+s_{2}+1)\alpha )}+\frac{1}{\Gamma (1+\alpha)}\beta_{\alpha} \biggr] \\ &\qquad{}\times (a_{2}-a_{1})^{\alpha }\bigl[T_{1}(a_{1},a_{2})+T_{2}(a_{1},a_{2}) \bigr], \end{aligned}$$
where \(\beta_{\alpha}\), \(T_{1}(a_{1},a_{2})\), and \(T_{2}(a_{1},a_{2})\) are defined in Theorem  3.3.

Proof

Since \(f_{1}\in GK^{2}_{s_{1}} \) and \(f_{2}\in GK^{2}_{s_{2}} \) on \([a_{1},a_{2}] \),
$$\begin{aligned}& f_{1} \biggl( \gamma x_{1}+(1-\gamma)\frac{a_{1}+a_{2}}{2} \biggr) \leq \gamma^{\alpha s_{1}}f_{1}(x_{1})+(1- \gamma)^{\alpha s_{1}}f_{1} \biggl( \frac{a_{1}+a_{2}}{2} \biggr) , \\& f_{2} \biggl( \gamma x_{1}+(1-\gamma)\frac{a_{1}+a_{2}}{2} \biggr) \leq \gamma^{\alpha s_{2}}f_{2}(x_{1})+(1- \gamma)^{\alpha s_{2}}f_{2} \biggl( \frac{a_{1}+a_{2}}{2} \biggr), \end{aligned}$$
for all \(x_{1}\in[a_{1},a_{2}] \) and all \(\gamma\in[0,1] \). Because \(f_{1} \) and \(f_{1} \) are nonnegative, we have
$$\begin{aligned} & f_{1} \biggl( \gamma x_{1}+(1-\gamma)\frac{a_{1}+a_{2}}{2} \biggr) f_{2} \biggl( \gamma x_{1}+(1-\gamma) \frac{a_{1}+a_{2}}{2} \biggr) \\ &\quad\leq\gamma^{\alpha(s_{1}+s_{2})} f_{1}(x_{1}) f_{2}(x_{1})+ ( 1-\gamma ) ^{\alpha(s_{1}+s_{2})} f_{1} \biggl( \frac {a_{1}+a_{2}}{2} \biggr) f_{2} \biggl( \frac{a_{1}+a_{2}}{2} \biggr) \\ &\qquad{} + \gamma^{\alpha s_{1}}(1-\gamma)^{\alpha s_{2}} f_{1}(x_{1}) f_{2} \biggl( \frac{a_{1}+a_{2}}{2} \biggr) +\gamma ^{\alpha s_{2}}(1- \gamma)^{\alpha s_{1}} f_{1} \biggl( \frac {a_{1}+a_{2}}{2} \biggr) f_{2}(x_{1}). \end{aligned}$$
Integrating both sides of the above inequality over \([0,1] \), we obtain
$$\begin{aligned} & \frac{1}{\Gamma(1+\alpha)}\int_{0}^{1}f_{1} \biggl( \gamma x_{1}+(1-\gamma)\frac{a_{1}+a_{2}}{2} \biggr) f_{2} \biggl( \gamma x_{1}+(1-\gamma)\frac{a_{1}+a_{2}}{2} \biggr) (d\gamma)^{\alpha} \\ &\quad\leq \frac{1}{\Gamma(1+\alpha)}f_{1}(x_{1}) f_{2}(x_{1})\int_{0}^{1} \gamma^{\alpha(s_{1}+s_{2})} (d\gamma)^{\alpha} \\ &\qquad{} +\frac{1}{\Gamma(1+\alpha)} f_{1} \biggl( \frac {a_{1}+a_{2}}{2} \biggr) f_{2} \biggl( \frac{a_{1}+a_{2}}{2} \biggr) \int _{0}^{1}(1-\gamma)^{\alpha(s_{1}+s_{2})}(d \gamma)^{\alpha} \\ &\qquad{} + \frac{1}{\Gamma(1+\alpha)} f_{1}(x_{1}) f_{2} \biggl( \frac{a_{1}+a_{2}}{2} \biggr) \int_{0}^{1} \gamma^{\alpha s_{1}}(1-\gamma )^{\alpha s_{2}}(d\gamma)^{\alpha} \\ &\qquad{} +\frac {1}{\Gamma(1+\alpha)} f_{1} \biggl( \frac{a_{1}+a_{2}}{2} \biggr) f_{2}(x_{1})\int_{0}^{1} \gamma^{\alpha s_{2}}(1-\gamma)^{\alpha s_{1}}(d\gamma)^{\alpha} \\ &\quad = \frac{\Gamma (1+(s_{1}+s_{2})\alpha)}{\Gamma(1+(s_{1}+s_{2}+1)\alpha)} f_{1}(x_{1})f_{2}(x_{1}) \\ &\qquad{} +\frac{\Gamma (1+(s_{1}+s_{2})\alpha)}{\Gamma(1+(s_{1}+s_{2}+1)\alpha)} f_{1} \biggl( \frac{a_{1}+a_{2}}{2} \biggr) f_{2} \biggl( \frac{a_{1}+a_{2}}{2} \biggr) \\ &\qquad{} +\frac{1}{\Gamma(1+\alpha)}\beta_{\alpha } \biggl[ f_{1}(x_{1})f_{2} \biggl( \frac{a_{1}+a_{2}}{2} \biggr) +f_{1} \biggl( \frac{a_{1}+a_{2}}{2} \biggr) f_{2}(x_{1}) \biggr]. \end{aligned}$$
If we integrate both sides of the above inequalities on \([a_{1},a_{2}] \), we obtain
$$\begin{aligned} & \frac{1}{(\Gamma(1+\alpha))^{2}}\int_{a_{1}}^{a_{2}}\int _{0}^{1}f_{1} \biggl( \gamma x_{1}+(1-\gamma)\frac {a_{1}+a_{2}}{2} \biggr) f_{2} \biggl( \gamma x_{1}+(1-\gamma)\frac {a_{1}+a_{2}}{2} \biggr) (d \gamma)^{\alpha}(dx_{1})^{\alpha} \\ &\quad\leq\frac{\Gamma(1+(s_{1}+s_{2})\alpha)}{\Gamma(1+(s_{1}+s_{2}+1)\alpha )}\frac{1}{\Gamma(1+\alpha)} \int_{a_{1}}^{a_{2}}f_{1}(x_{1})f_{2}(x_{1}) (dx_{1})^{\alpha} \\ &\qquad{} +\frac{\Gamma(1+(s_{1}+s_{2})\alpha)}{\Gamma (1+(s_{1}+s_{2}+1)\alpha)}\frac{1}{\Gamma(1+\alpha)} (a_{2}-a_{1})^{\alpha}f_{1} \biggl( \frac{a_{1}+a_{2}}{2} \biggr) f_{2} \biggl( \frac{a_{1}+a_{2}}{2} \biggr) \\ &\qquad{}+ \frac{1}{(\Gamma(1+\alpha))^{2}}\beta_{\alpha} \biggl[ f_{2} \biggl( \frac {a_{1}+a_{2}}{2} \biggr) \int_{a_{1}}^{a_{2}}f_{1}(x_{1}) (dx_{1})^{\alpha }+f_{1} \biggl( \frac{a_{1}+a_{2}}{2} \biggr) \int_{a_{1}}^{a_{2}}f_{2}(x_{1}) (dx_{1})^{\alpha} \biggr] \\ &\quad \leq\frac{\Gamma(1+(s_{1}+s_{2})\alpha)}{\Gamma (1+(s_{1}+s_{2}+1)\alpha)}\frac{1}{\Gamma(1+\alpha)} \int_{a_{1}}^{a_{2}}f_{1}(x_{1})f_{2}(x_{1}) (dx_{1})^{\alpha} \\ &\qquad{} +\frac{\Gamma(1+(s_{1}+s_{2})\alpha)}{\Gamma (1+(s_{1}+s_{2}+1)\alpha)} \frac{1}{\Gamma(1+\alpha )}(a_{2}-a_{1})^{\alpha} \frac{f_{1}(a_{1})+f_{1}(a_{2})}{2^{\alpha }}\frac{f_{2}(a_{1})+f_{2}(a_{2})}{2^{\alpha}} \\ &\qquad{} + \frac{1}{(\Gamma(1+\alpha))^{2}}(a_{2}-a_{1})^{\alpha} \beta _{\alpha} \biggl[ 2^{\alpha}\frac{f_{1}(a_{1})+f_{1}(a_{2})}{2^{\alpha }}\frac{f_{2}(a_{1})+f_{2}(a_{2})}{2^{\alpha}} \biggr] \\ &\quad = \frac{\Gamma(1+(s_{1}+s_{2})\alpha)}{\Gamma (1+(s_{1}+s_{2}+1)\alpha)}\frac{1}{\Gamma(1+\alpha)} \int_{a_{1}}^{a_{2}}f_{1}(x_{1})f_{2}(x_{1}) (dx_{1})^{\alpha} \\ &\qquad{} +\frac{\Gamma(1+(s_{1}+s_{2})\alpha)}{4^{\alpha}\Gamma (1+(s_{1}+s_{2}+1)\alpha)} \frac{1}{\Gamma(1+\alpha )}(a_{2}-a_{1})^{\alpha} \bigl[T_{1}(a_{1},a_{2})+T_{2}(a_{1},a_{2}) \bigr] \\ &\qquad{} + \frac{1}{2^{\alpha}(\Gamma(1+\alpha ))^{2}}(a_{2}-a_{1})^{\alpha} \beta_{\alpha }\bigl[T_{1}(a_{1},a_{2})+T_{2}(a_{1},a_{2}) \bigr]. \end{aligned}$$
Then
$$\begin{aligned} & \frac{1}{(\Gamma(1+\alpha))^{2}}\int_{a_{1}}^{a_{2}}\int _{0}^{1}f_{1} \biggl( \gamma x_{1}+(1-\gamma)\frac {a_{1}+a_{2}}{2} \biggr) f_{2} \biggl( \gamma x_{1}+(1-\gamma)\frac {a_{1}+a_{2}}{2} \biggr) (d \gamma)^{\alpha}(dx_{1})^{\alpha} \\ &\quad \leq\frac{\Gamma(1+(s_{1}+s_{2})\alpha )}{\Gamma(1+(s_{1}+s_{2}+1))\alpha}\frac{1}{\Gamma(1+\alpha)} \int_{a_{1}}^{a_{2}}f_{1}(x_{1})f_{2}(x_{1}) (dx_{1})^{\alpha} \\ &\qquad{} +\frac{1}{2^{\alpha}\Gamma(1+\alpha)} \biggl[ \frac{\Gamma (1+(s_{1}+s_{2})\alpha)}{2^{\alpha}\Gamma(1+(s_{1}+s_{2}+1))\alpha }+\frac{1}{\Gamma(1+\alpha)} \beta_{\alpha} \biggr] \\ &\qquad{}\times(a_{2}-a_{1})^{\alpha } \bigl[T_{1}(a_{1},a_{2})+T_{2}(a_{1},a_{2}) \bigr]. \end{aligned}$$
The proof is complete. □

Remark 3.2

If \(\alpha=1\), then
$$\begin{aligned} & \int_{a_{1}}^{a_{2}}\int_{0}^{1}f_{1} \biggl( \gamma x_{1}+(1-\gamma)\frac{a_{1}+a_{2}}{2} \biggr) f_{2} \biggl( \gamma x_{1}+(1-\gamma)\frac{a_{1}+a_{2}}{2} \biggr) (d\gamma) (dx_{1}) \\ &\quad \leq\frac{1}{s_{1}+s_{2}+1} \int_{a_{1}}^{a_{2}}f_{1}(x_{1})f_{2}(x_{1}) (dx_{1}) \\ &\qquad{} +\frac{1}{2} \biggl[ \frac{1}{2(s_{1}+s_{2}+1)}+\frac{\Gamma (s_{1}+1)\Gamma(s_{2}+1)}{\Gamma(s_{1}+s_{2}+1)} \biggr] (a_{2}-a_{1})\bigl[T_{1}(a_{1},a_{2})+T_{2}(a_{1},a_{2}) \bigr]. \end{aligned}$$

Theorem 3.5

Let \(f_{1},f_{2}\colon[a_{1},a_{2}]\subset\mathbb {R}_{+}\longrightarrow\mathbb{R}^{\alpha}\), \(a_{1}< a_{2}\), be nonnegative and generalized s-convex functions in the second sense. If \(f_{1}\in GK^{2}_{s_{1}} \) and \(f_{2}\in GK^{2}_{s_{2}} \) on \([a_{1},a_{2}] \) for some \(\gamma\in[0,1] \) and \(s_{1}, s_{2}\in ( 0,1 ] \), then
$$\begin{aligned} & \frac{1}{(\Gamma(1+\alpha))^{2}}\int_{a_{1}}^{a_{2}}\int _{0}^{1}f_{1} \biggl( \gamma \frac {a_{1}+a_{2}}{2}+(1-\gamma)x_{2} \biggr) f_{2} \biggl( \gamma\frac {a_{1}+a_{2}}{2}+(1-\gamma)x_{2} \biggr) (d \gamma)^{\alpha }(dx_{2})^{\alpha} \\ &\quad \leq\frac{\Gamma(1+(s_{1}+s_{2})\alpha )}{\Gamma(1+(s_{1}+s_{2}+1)\alpha)}\frac{1}{\Gamma(1+\alpha)}\int_{a_{1}}^{a_{2}}f_{1}(x_{2})f_{2}(x_{2}) (dx_{2})^{\alpha} \\ &\qquad{}+ \frac{1}{2^{\alpha}\Gamma(1+\alpha)} \biggl[ \frac{\Gamma (1+(s_{1}+s_{2})\alpha)}{2^{\alpha}\Gamma(1+(s_{1}+s_{2}+1)\alpha )}+\frac{1}{\Gamma(1+\alpha)} \beta_{\alpha} \biggr]\\ &\qquad{}\times (a_{2}-a_{1})^{\alpha } \bigl[T_{1}(a_{1},a_{2})+T_{2}(a_{1},a_{2}) \bigr], \end{aligned}$$
where \(\beta_{\alpha}\), \(T_{1}(a_{1},a_{2})\) and \(T_{2}(a_{1},a_{2})\) are defined in Theorem  3.3.

Proof

The proof of this theorem can easily be given like in Theorem 3.3. □

Theorem 3.6

Let \(f_{1},f_{2}\colon[a_{1},a_{2}]\subset\mathbb {R}_{+}\longrightarrow\mathbb{R}^{\alpha}\), \(a_{1}< a_{2} \), be nonnegative, and generalized convex functions. If \(f_{1} \) and \(f_{2} \) are generalized convex on \([a_{1},a_{2}] \) for some \(\gamma\in[0,1]\), then
$$\begin{aligned} & \frac{1}{(\Gamma(1+\alpha))^{2}}\int_{a_{1}}^{a_{2}}\int _{0}^{1}f_{1} \biggl( \gamma \frac{a_{1}+a_{2}}{2}+(1-\gamma)x_{2} \biggr) f_{2} \biggl( \gamma\frac{a_{1}+a_{2}}{2}+(1-\gamma)x_{2} \biggr) (d\gamma )^{\alpha}(dx_{2})^{\alpha} \\ &\quad \leq\frac {\Gamma(1+2\alpha)}{\Gamma(1+3\alpha)}\frac{1}{\Gamma(1+\alpha)}\int_{a_{1}}^{a_{2}}f_{1}(x_{2})f_{2}(x_{2}) (dx_{2})^{\alpha} \\ &\qquad{} + \frac{1}{2^{\alpha}\Gamma(1+\alpha)} \biggl[ \frac {\Gamma(1+2\alpha)}{2^{\alpha}\Gamma(1+3\alpha)}+\frac{1}{\Gamma (1+\alpha)} \beta_{\alpha} \biggr] (a_{2}-a_{1})^{\alpha } \bigl[T_{1}(a_{1},a_{2})+T_{2}(a_{1},a_{2}) \bigr], \end{aligned}$$
where \(T_{1}(a_{1},a_{2})\) and \(T_{2}(a_{1},a_{2})\) are defined in Theorem  3.3, but
$$\begin{aligned} \beta_{\alpha}=\int_{0}^{1} \gamma^{\alpha}(1-\gamma)^{\alpha}(d\gamma )^{\alpha} = \frac{\Gamma(1+\alpha)}{\Gamma(1+2\alpha)}-\frac{\Gamma (1+2\alpha)}{\Gamma(1+3\alpha)}. \end{aligned}$$

Proof

Since \(f_{1} \) and \(f_{2} \) are generalized convex on \([a_{1},a_{2}] \),
$$\begin{aligned}& f_{1} \biggl( \gamma\frac{a_{1}+a_{2}}{2}+(1-\gamma)x_{2} \biggr) \leq \gamma^{\alpha}f_{1} \biggl( \frac{a_{1}+a_{2}}{2} \biggr) +(1-\gamma )^{\alpha}f_{1}(x_{2}), \\& f_{2} \biggl( \gamma\frac{a_{1}+a_{2}}{2}+(1-\gamma)x_{2} \biggr) \leq \gamma^{\alpha}f_{2} \biggl( \frac{a_{1}+a_{2}}{2} \biggr) +(1-\gamma )^{\alpha}f_{2}(x_{2}), \end{aligned}$$
for all \(x_{2}\in[a_{1},a_{2}] \), \(\gamma\in[0,1] \). Since \(f_{1} \) and \(f_{1} \) are nonnegative,
$$\begin{aligned} & f_{1} \biggl( \gamma\frac{a_{1}+a_{2}}{2}+(1-\gamma)x_{2} \biggr) f_{2} \biggl( \gamma\frac{a_{1}+a_{2}}{2}+(1-\gamma)x_{2} \biggr) \\ &\quad \leq \gamma^{2\alpha} f_{1} \biggl( \frac {a_{1}+a_{2}}{2} \biggr) f_{2} \biggl( \frac{a_{1}+a_{2}}{2} \biggr) +(1- \gamma)^{2\alpha} f_{1}(x_{2}) f_{2}(x_{2}) \\ &\qquad{} + \gamma^{\alpha}(1-\gamma)^{\alpha} f_{1} \biggl( \frac {a_{1}+a_{2}}{2} \biggr) f_{2}(x_{2})+ \gamma^{\alpha}(1-\gamma)^{\alpha } f_{1}(x_{2}) f_{2} \biggl( \frac{a_{1}+a_{2}}{2} \biggr) . \end{aligned}$$
Integrating both sides of the above inequality over \([0,1] \), we obtain
$$\begin{aligned} & \frac{1}{\Gamma(1+\alpha)}\int_{0}^{1}f_{1} \biggl( \gamma\frac{a_{1}+a_{2}}{2}+(1-\gamma)x_{2} \biggr) f_{2} \biggl( \gamma \frac{a_{1}+a_{2}}{2}+(1-\gamma)x_{2} \biggr) (d\gamma)^{\alpha} \\ &\quad\leq\frac{1}{\Gamma(1+\alpha )}f_{1} \biggl( \frac{a_{1}+a_{2}}{2} \biggr) f_{2} \biggl( \frac {a_{1}+a_{2}}{2} \biggr) \int_{0}^{1} \gamma^{2\alpha} (d\gamma)^{\alpha } \\ &\qquad{} +\frac{1}{\Gamma(1+\alpha)} f_{1}(x_{2}) f_{2}(x_{2})\int_{0}^{1}(1- \gamma)^{2\alpha}(d\gamma)^{\alpha} \\ &\qquad{} + \frac{1}{\Gamma(1+\alpha)} f_{1} \biggl( \frac {a_{1}+a_{2}}{2} \biggr) f_{2}(x_{2})\int_{0}^{1} \gamma^{\alpha }(1-\gamma)^{\alpha}(d\gamma)^{\alpha} \\ &\qquad{}+\frac{1}{\Gamma(1+\alpha)} f_{1}(x_{2}) f_{2} \biggl( \frac {a_{1}+a_{2}}{2} \biggr) \int_{0}^{1} \gamma^{\alpha}(1-\gamma)^{\alpha }(d\gamma)^{\alpha} \\ &\quad = \frac{\Gamma (1+2\alpha)}{\Gamma(1+3\alpha)} f_{1} \biggl( \frac{a_{1}+a_{2}}{2} \biggr) f_{2} \biggl( \frac{a_{1}+a_{2}}{2} \biggr) +\frac{\Gamma(1+2\alpha )}{\Gamma(1+3\alpha)} f_{1}(x_{2})f_{2}(x_{2}) \\ &\qquad{} +\frac{1}{\Gamma(1+\alpha)}\beta_{\alpha} \biggl[ f_{1} \biggl( \frac{a_{1}+a_{2}}{2} \biggr) f_{2}(x_{2})+f_{1}(x_{2})f_{2} \biggl( \frac {a_{1}+a_{2}}{2} \biggr) \biggr]. \end{aligned}$$
If we integrate both sides of the above inequalities on \([a_{1},a_{2}] \), we obtain
$$\begin{aligned} & \frac{1}{(\Gamma(1+\alpha))^{2}}\int_{a_{1}}^{a_{2}}\int _{0}^{1}f_{1} \biggl( \gamma \frac {a_{1}+a_{2}}{2}+(1-\gamma)x_{2} \biggr) f_{2} \biggl( \gamma\frac {a_{1}+a_{2}}{2}+(1-\gamma)x_{2} \biggr) (d \gamma)^{\alpha }(dx_{2})^{\alpha} \\ &\quad\leq\frac{\Gamma(1+2\alpha)}{\Gamma(1+3\alpha)}\frac{1}{\Gamma(1+\alpha )}(a_{2}-a_{1})^{\alpha}f_{1} \biggl( \frac{a_{1}+a_{2}}{2} \biggr) f_{2} \biggl( \frac{a_{1}+a_{2}}{2} \biggr) \\ &\qquad{}+\frac{\Gamma(1+2\alpha)}{\Gamma(1+3\alpha)}\frac{1}{\Gamma(1+\alpha )}\int_{a_{1}}^{a_{2}}f_{1}(x_{2})f_{2}(x_{2}) (dx_{2})^{\alpha} \\ &\qquad{} +\frac{1}{(\Gamma(1+\alpha))^{2}}\beta _{\alpha} \biggl[ f_{1} \biggl( \frac{a_{1}+a_{2}}{2} \biggr) \int_{a_{1}}^{a_{2}}f_{2}(x_{2}) (dx_{2})^{\alpha}+f_{2} \biggl( \frac {a_{1}+a_{2}}{2} \biggr) \int_{a_{1}}^{a_{2}}f_{1}(x_{2}) (dx_{2})^{\alpha } \biggr] \\ &\quad \leq\frac{\Gamma(1+2\alpha )}{\Gamma(1+2\alpha)}\frac{1}{\Gamma(1+\alpha)}(a_{2}-a_{1})^{\alpha } \frac{f_{1}(a_{1}) +f_{1}(a_{2})}{2^{\alpha}}\frac{f_{2}(a_{1})+f_{2}(a_{2})}{2^{\alpha }} \\ &\qquad{} +\frac{\Gamma(1+2\alpha)}{\Gamma (1+3\alpha)}\frac{1}{\Gamma(1+\alpha)}\int_{a_{1}}^{a_{2}}f_{1}(x_{2})f_{2}(x_{2}) (dx_{2})^{\alpha} \\ &\qquad{} +\frac{1}{(\Gamma(1+\alpha))^{2}}\beta_{\alpha} \biggl[ 2^{\alpha} \frac{f_{1}(a_{1})+f_{1}(a_{2})}{2^{\alpha}}\frac {f_{2}(a_{1})+f_{2}(a_{2})}{2^{\alpha}} \biggr] (a_{2}-a_{1})^{\alpha } \\ &\quad = \frac{\Gamma(1+2\alpha)}{\Gamma (1+3\alpha)}\frac{1}{\Gamma(1+\alpha)}\int_{a_{1}}^{a_{2}}f_{1}(x_{2})f_{2}(x_{2}) (dx_{2})^{\alpha}+\frac {1}{2^{\alpha}\Gamma(1+\alpha)} \biggl[ \frac{\Gamma(1+2\alpha)}{2^{\alpha }\Gamma(1+3\alpha)} \\ &\qquad{} + \frac {1}{\Gamma(1+\alpha)}\beta_{\alpha} \biggr] (a_{2}-a_{1})^{\alpha }\bigl[T_{1}(a_{1},a_{2})+T_{2}(a_{1},a_{2}) \bigr]. \end{aligned}$$
The proof is complete. □

Remark 3.3

If \(\alpha=1\), then
$$\begin{aligned} &\int_{a_{1}}^{a_{2}}\int_{0}^{1}f_{1} \biggl( \gamma\frac {a_{1}+a_{2}}{2}+(1-\gamma)x_{2} \biggr) f_{2} \biggl( \gamma\frac {a_{1}+a_{2}}{2}+(1-\gamma)x_{2} \biggr) (d\gamma) (dx_{2}) \\ &\quad \leq\frac{1}{3}\int_{a_{1}}^{a_{2}}f_{1}(x_{2})f_{2}(x_{2}) (dx_{2})+\frac {1}{6}(a_{2}-a_{1}) \bigl[T_{1}(a_{1},a_{2})+T_{2}(a_{1},a_{2}) \bigr]. \end{aligned}$$

4 Applications

Example 4.1

Let \(a_{1},a_{2}\in\mathbb{R}_{+}\), \(a_{1}< a_{2} \), and \(a_{2}-a_{1}\leq1 \), then
$$\begin{aligned} & \biggl\lbrace 2^{\alpha} \biggl[ \frac{2^{\alpha}}{\Gamma(1+\alpha)} \biggl( \frac{\Gamma(1+2\alpha)}{\Gamma(1+3\alpha)} \biggr) ^{2}\\ &\qquad{}+2^{\alpha } \biggl( \frac{\Gamma(1+\alpha)}{\Gamma(1+2\alpha)}-\frac{\Gamma (1+2\alpha)}{\Gamma(1+3\alpha)} \biggr) \biggl( \frac{\Gamma(1+\alpha )}{\Gamma(1+2\alpha)} \biggr) ^{2} \biggr] \biggr\rbrace K^{2}(a_{1},a_{2}) \\ &\qquad{}+ \biggl\lbrace \frac{2^{\alpha}}{\Gamma(1+\alpha)} \biggl( \frac{\Gamma (1+2\alpha)}{\Gamma(1+3\alpha)} \biggr)^{2}\\ &\qquad{}+4^{\alpha} \biggl(\frac{\Gamma (1+\alpha)}{\Gamma(1+2\alpha)}- \frac{\Gamma(1+2\alpha)}{\Gamma(1+3\alpha )} \biggr) \biggl(\frac{\Gamma(1+\alpha)}{\Gamma(1+2\alpha)} \biggr)^{2} \biggr\rbrace G^{2} (a_{1},a_{2}) \\ &\quad\leq2^{\alpha} \biggl( \frac{\Gamma(1+2\alpha)}{\Gamma(1+\alpha)\Gamma (1+3\alpha)} \biggr)^{2} \frac{(a_{2}^{3\alpha}-a_{1}^{3\alpha })}{(a_{2}-a_{1})^{\alpha}}+\frac{2^{\alpha}}{(\Gamma(1+\alpha ))^{3}}\beta_{\alpha}A^{2}(a_{1},a_{2}), \end{aligned}$$
where
$$\begin{aligned}& \beta_{\alpha}=\int_{0}^{1} \gamma^{\alpha}(1-\gamma )^{\alpha}(d\gamma)^{\alpha}, \\& A(a_{1},a_{2})= \frac{a_{1}^{\alpha}+a_{2}^{\alpha}}{2^{\alpha}},\quad a_{1},a_{2}\geq0, \\& G(a_{1},a_{2})= \bigl(a_{1}^{\alpha}a_{2}^{\alpha } \bigr)^{\frac{1}{2}}, \quad a_{1},a_{2}\geq0, \end{aligned}$$
and
$$K(a_{1},a_{2})= \biggl(\frac{a_{1}^{2\alpha}+a_{2}^{2\alpha}}{2^{\alpha}} \biggr)^{\frac{1}{2}}, \quad a_{1},a_{2}\geq0. $$

Proof

If \(f_{1}\in GK^{2}_{s_{1}} \) and \(f_{2}\in GK^{2}_{s_{2}} \) on \([a_{1},a_{2}] \) for some \(\gamma\in[0,1] \) and \(s_{1},s_{2}\in ( 0,1 ] \), then, by Theorem 3.3, if \(f_{1},f_{2}\colon [0,1] \longrightarrow[0^{\alpha},1^{\alpha}] \), \(f_{1}(x)=x^{\alpha s_{1}} \), \(f_{2}(x)=x^{\alpha s_{2}} \), where \(x\in[a_{1},a_{2}] \), \(s_{1}=s_{2}=1 \) and \(a_{2}-a_{1}\leq1 \), then
$$\begin{aligned} &\frac{1}{ ( \Gamma(1+\alpha) )^{3} }\int_{a_{1}}^{a_{2}}\int _{a_{1}}^{a_{2}}\int_{0}^{1} \bigl(\gamma x_{1}+(1-\gamma )x_{2}\bigr)^{2\alpha}(d \gamma)^{\alpha}(dx_{2})^{\alpha}(dx_{1})^{\alpha} \\ &\quad\leq2^{\alpha}\frac{\Gamma(1+2\alpha)}{\Gamma(1+3\alpha)}\frac {1}{ (\Gamma(1+\alpha) )^{2} }(a_{2}-a_{1})^{\alpha} \int_{a_{1}}^{a_{2}}x_{1}^{2\alpha}(dx_{1})^{\alpha} \\ &\qquad{}+\frac{1}{2^{\alpha} (\Gamma(1+\alpha) )^{3} }\beta _{\alpha} \bigl[ a_{1}^{2\alpha}+a_{2}^{2\alpha}+2^{\alpha}a_{1}^{\alpha }a_{2}^{\alpha} \bigr] (a_{2}-a_{1})^{2\alpha} \\ &\quad\Rightarrow \\ & (a_{2}-a_{1})^{2\alpha} \biggl\lbrace \biggl[\frac{2^{\alpha}}{\Gamma(1+\alpha)} \biggl(\frac{\Gamma (1+2\alpha)}{\Gamma(1+3\alpha)} \biggr)^{2}\\ &\qquad{}+2^{\alpha} \biggl[ \frac {\Gamma(1+\alpha)}{\Gamma(1+2\alpha)}-\frac{\Gamma(1+2\alpha)}{\Gamma (1+3\alpha)} \biggr] \biggl(\frac{\Gamma(1+\alpha)}{\Gamma(1+2\alpha)} \biggr)^{2} \biggr]\bigl(a_{1}^{2\alpha}+a_{2}^{2\alpha} \bigr) \\ &\qquad{} + \biggl[\frac{2^{\alpha}}{\Gamma(1+\alpha)} \biggl(\frac{\Gamma (1+2\alpha)}{\Gamma(1+3\alpha)} \biggr)^{2}+4^{\alpha} \biggl[ \frac {\Gamma(1+\alpha)}{\Gamma(1+2\alpha)}- \frac{\Gamma(1+2\alpha)}{\Gamma (1+3\alpha)} \biggr] \biggl(\frac{\Gamma(1+\alpha)}{\Gamma(1+2\alpha)} \biggr)^{2} \biggr]a_{1}^{\alpha}a_{2}^{\alpha} \biggr\rbrace \\ &\quad\leq2^{\alpha} \biggl( \frac{\Gamma(1+2\alpha)}{\Gamma(1+\alpha)\Gamma (1+3\alpha)} \biggr)^{2}(a_{2}-a_{1})^{\alpha} \bigl(a_{2}^{3\alpha }-a_{1}^{3\alpha}\bigr) \\ &\qquad{}+ \frac{1}{2^{\alpha}(\Gamma(1+\alpha))^{3}}\beta _{\alpha}(a_{2}-a_{1})^{2\alpha} \bigl(a_{2}^{\alpha}+a_{1}^{\alpha} \bigr)^{2} \\ &\quad\Rightarrow \\ &(a_{2}-a_{1})^{2\alpha} \biggl\lbrace 2^{\alpha} \biggl[ \frac{2^{\alpha }}{\Gamma(1+\alpha)} \biggl( \frac{\Gamma(1+2\alpha)}{\Gamma(1+3\alpha )} \biggr)^{2}\\ &\qquad{}+2^{\alpha} \biggl[ \frac{\Gamma(1+\alpha)}{\Gamma(1+2\alpha )}-\frac{\Gamma(1+2\alpha)}{\Gamma(1+3\alpha)} \biggr] \biggl(\frac{\Gamma (1+\alpha)}{\Gamma(1+2\alpha)} \biggr)^{2} \biggr]K^{2}(a_{1},a_{2}) \\ &\qquad{}+ \biggl[\frac{2^{\alpha}}{\Gamma(1+\alpha)} \biggl(\frac{\Gamma (1+2\alpha)}{\Gamma(1+3\alpha)} \biggr)^{2}\\ &\qquad{}+4^{\alpha} \biggl[ \frac {\Gamma(1+\alpha)}{\Gamma(1+2\alpha)}- \frac{\Gamma(1+2\alpha)}{\Gamma (1+3\alpha)} \biggr] \biggl( \frac{\Gamma(1+\alpha)}{\Gamma(1+2\alpha )} \biggr)^{2} \biggr]G^{2}(a_{1},a_{2}) \biggr\rbrace \\ &\quad\leq2^{\alpha} \biggl( \frac{\Gamma(1+2\alpha)}{\Gamma(1+\alpha)\Gamma (1+3\alpha)} \biggr)^{2}(a_{2}-a_{1})^{\alpha} \bigl(a_{2}^{3\alpha }-a_{1}^{3\alpha}\bigr)\\ &\qquad{}+ \frac{2^{\alpha}}{(\Gamma(1+\alpha))^{3}} \beta _{\alpha}(a_{2}-a_{1})^{2\alpha}A^{2}(a_{1},a_{2}). \end{aligned}$$
The proof is complete. □

Remark 4.1

If \(\alpha=1 \), then
$$\begin{aligned} &\frac{22}{36}K^{2}(a_{1},a_{2})+ \frac{14}{36}G^{2}(a_{1},a_{2})\leq \frac {2}{9} \biggl( \frac{a_{2}^{3}-a_{1}^{3}}{a_{2}-a_{1}} \biggr) +\frac {1}{3}A^{2}(a_{1},a_{2}) \\ &\quad\Rightarrow \\ &22K^{2}(a_{1},a_{2})+14G^{2}(a_{1},a_{2}) \leq 8\bigl(a_{2}^{2}+a_{1}a_{2}+a_{1}^{2} \bigr)+12A^{2}(a_{1},a_{2}), \end{aligned}$$
where
$$\begin{aligned}& A(a_{1},a_{2})=\frac{a_{1}+a_{2}}{2},\quad a_{1},a_{2}\geq0, \mbox{is the arithmetic mean}, \\& G(a_{1},a_{2})=(a_{1}a_{2})^{\frac{1}{2}}, \quad a_{1},a_{2}\geq0, \mbox{is the geometric mean, and} \\& K(a_{1},a_{2})= \biggl(\frac{a_{1}^{2}+a_{2}^{2}}{2} \biggr)^{\frac {1}{2}},\quad a_{1},a_{2}\geq0, \mbox{is the quadratic mean}. \end{aligned}$$

Declarations

Acknowledgements

The authors would like to thank the referees for valuable suggestions and comments, which helped the authors to improve this article substantially.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
Department of Mathematics and Institute for Mathematical Research, University Putra Malaysia
(2)
Department of Mathematics, University Putra Malaysia

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© Kılıçman and Saleh 2015