21世纪以来, 由于分数阶积分理论广泛应用于量子力学、高能物理、水动力学、经济学等众多领域, 彰显了其独特优势和不可替代性, 因此其理论和应用深受国内外众多学者的关注. 2013年, Sarikaya M Z等人将Riemann-Liouville分数阶积分与Hermite-Hadamard不等式结合, 建立一类分数阶积分形式的Hermite-Hadamard型不等式^[4]. 随后, 国内外许多数学家对分数阶积分形式的Hermite-Hadamard型不等式从不同凸性和不同定义的分数阶的角度进行了推广和改进以及应用了大量的工作^{[1, 4-6]}.

另一方面, 不确定性问题出现在许多确定性的数学或者计算模型中, 因此区间分析作为一种新的解决不确定性问题的重要工具被广泛应用于各个领域, 如计算物理学、误差分析、机器人技术等. 特别地, 一些经典的积分不等式被推广到区间值函数的形式中, 例如Hermite-Hadamard不等式^[3]、Ostrowski不等式^[7]等. 2019年, Hüseyin引入了区间值函数Riemann-Liouville分数阶积分的定义, 并且证明了一些区间值函数Riemann-Liouville分数阶积分形式的Hermite-Hadamard不等式, 即

$ \begin{equation*} f\left(\frac{a+b}{2}\right) \supseteq\frac{\Gamma(\alpha+1)}{(b-a)^{\alpha}}\big[\mathcal{J}^{\alpha}_{a^{+}}{f}(b)+ {\mathcal{J}}^{\alpha}_{b^{-}}{f}(a)\big]\supseteq\frac{f(a)+f(b)}{2}. \end{equation*} $

受此启发, 本文引入了区间值函数整合分数阶积分的概念, 讨论了其若干重要性质. 将文献[1] 中的Hermite-Hadamard型不等式推广到区间值函数整合分数阶积分的形式中, 同时也推广了文献[2, 3] 的相关结果.

设$ \; [a] = [\underline{a}, \overline{a}]\; $是$ \; \mathbb{R}\; $上的一个有界闭区间, 这里$ \; \underline{a}, \overline{a}\in\mathbb{R}\; $且$ \; \underline{a}\leq\overline{a}.\; $当$ \; \underline{a} = \overline{a}\; $时, 区间$ \; [a]\; $称为退化的; 当$ \; \underline{a}>0\; $时, 区间$ \; [a]\; $称为正的; 我们用$ \; \mathbb{R}^{+}\; $表示$ \; \mathbb{R}\; $上所有正数构成的集合; 用$ \; \mathbb{R}_{\mathcal{I}}\; $表示$ \; \mathbb{R}\; $上所有非空闭区间构成的集合; 用$ \; \mathbb{R}_{\mathcal{I}}^{+}\; $表示$ \; \mathbb{R}\; $上所有正闭区间构成的集合. 任取$ \mathbb{R}_{\mathcal{I}} $中的元素$ [\underline{a}, \overline{a}] $, $ [\underline{b}, \overline{b}] $, 对任意$ \lambda\in\mathbb{\mathbb{R}} $, $ \mathbb{R}_{\mathcal{I}} $空间中的区间运算规定如下:

$ \begin{eqnarray*} [\underline{a}, \overline{a}]+[\underline{b}, \overline{b}] = [\underline{a}+\underline{b}, \overline{a}+\overline{b}]; \end{eqnarray*} $

$ \begin{eqnarray*} [\underline{a}, \overline{a}]-[\underline{b}, \overline{b}] = [\underline{a}-\overline{b}, \overline{b}-\underline{a}]; \end{eqnarray*} $

$ \begin{eqnarray*} [\underline{a}, \overline{a}]\cdot[\underline{b}, \overline{b}] = [\min\{\underline{ab}, \underline{a}\overline{b}, \overline{a}\underline{b}, \overline{ab}\}, \max\{\underline{ab}, \underline{a}\overline{b}, \overline{a}\underline{b}, \overline{ab}\}];\; \end{eqnarray*} $

$ \begin{eqnarray*} [\underline{a}, \overline{a}]/[\underline{b}, \overline{b}] = [\min\{\underline{a}/\underline{b}, \underline{a}/\overline{b}, \overline{a}/\underline{b}, \overline{a}/\overline{b}\}, \max\{\underline{a}/\underline{b}, \underline{a}/\overline{b}, \overline{a}/\underline{b}, \overline{a}/\overline{b}\}];\; \end{eqnarray*} $

$ \begin{eqnarray*} \lambda[\underline{a}, \overline{a}] = \left\{\begin{array}{ll} \left[\lambda\underline{a}, \lambda\overline{a}\right], &\lambda\geq 0, \\ \left[\lambda\overline{a}, \lambda\underline{a}\right], &\lambda<0. \end{array}\right. \end{eqnarray*} $

同时, $ [\underline{a}, \overline{a}] $, $ [\underline{b}, \overline{b}] $之间的包含关系"$ \subseteq $" 定义为:

$ \begin{eqnarray*} [\underline{a}, \overline{a}]\subseteq[\underline{b}, \overline{b}]\Leftrightarrow\underline{b}\leq\underline{a}, \; \overline{a}\leq\overline{b}. \end{eqnarray*} $

在文献[8] 中, Dinghas给出了区间值函数Riemann可积的定义. 我们用$ \mathcal{IR}_{([a, b])} $表示所有Riemann可积的区间值函数构成的集合, 用$ \mathcal{R}_{([a, b])} $表示所有Riemann可积的实函数构成的集合.

引理 1^[8]  若$ \; f:[a, b]\rightarrow \mathbb{R}_{\mathcal{I}} $且$ \; f = \left[\underline{f}, \overline{f}\right] $, 则$ \; f\in \mathcal{IR}_{([a, b])}\; $当且仅当$ \underline{f}(t), \overline{f}(t)\in \mathcal{R}_{([a, b])}. $进一步, 有

$ \begin{eqnarray*} (\mathcal{IR})\int_{a}^{b}f(t)dt = \left[(\mathcal{R})\int_{a}^{b}\underline{f}(t)dt, (\mathcal{R})\int_{a}^{b}\overline{f}(t)dt\right]. \end{eqnarray*} $

定义 1^[3]  设$ \; f:[a, b]\rightarrow \mathbb{R}_{\mathcal{I}}^{+}\; $且$ \; f(t) = \left[\underline{f}(t), \overline{f}(t)\right], h:[0, 1]\rightarrow \mathbb{R}^{+} $. 若对任意的$ \; x, y\in[a, b], t\in[0, 1], $有

$ \begin{eqnarray} f\left(tx+(1-t)y\right)\supseteq h(t)f(x)+h(1-t)f(y), \end{eqnarray} $

(2.1)

则称$ \; f(t)\; $是$ \; [a, b]\; $上的$ h $-凸函数. 我们用$ SX(h, [a, b], \mathbb{R}_{\mathcal{I}}^{+})\; $表示$ \; [a, b]\; $上所有区间$ h $-凸函数的全体. 若(1) 式的包含符号反向, 则称$ f(t) $是$ \; [a, b]\; $上的$ h $-凹函数. 类似地, 我们用$ SV(h, [a, b], \mathbb{R}_{\mathcal{I}}^{+})\; $表示$ \; [a, b]\; $上所有区间$ h $-凹函数的全体.

定义 2^[9]  设$ f:[a, b]\rightarrow \mathbb{R}_{\mathcal{I}}\; $且$ f(t) = \left[\underline{f}(t), \overline{f}(t)\right] $. 若$ f\in\mathcal{IR}_{([a, b])}, $那么区间值Riemann-Liouville分数阶积分定义为：

$ \begin{eqnarray*} \mathcal{J}_{a^{+}}^{\alpha}f(t) = \frac{1}{\Gamma(\alpha)}\int_{a}^{t}(t-\tau)^{\alpha-1}f(\tau)d\tau, \\ \mathcal{J}_{b^{-}}^{\alpha}f(t) = \frac{1}{\Gamma(\alpha)}\int_{t}^{b}(\tau-t)^{\alpha-1}f(\tau)d\tau. \end{eqnarray*} $

其中$ \Gamma $是Gamma函数.

在文献[1] 中, Ahmad推广了Riemann-Liouville积分, 引入了整合分数阶积分. 为进一步推广, 我们定义了区间整合分数阶积分.

定义 3   设$ f:[a, b]\rightarrow \mathbb{R}_{\mathcal{I}}^{+} $且$ f(t) = \left[\underline{f}(t), \overline{f}(t)\right] $. 若$ \; f\in\mathcal{IR}_{([a, b])} $, 那么整合分数阶积分在$ a $和$ b $的定义为：

$ \begin{eqnarray*} (\mathcal{I}^{a}_{\alpha}f)(t) = \frac{1}{n!}\int_{a}^{t}(t-x)^{n}(x-a)^{\alpha-n-1}f(x)dx, \\ (^{b}\mathcal{I}_{\alpha}f)(t) = \frac{1}{n!}\int_{t}^{b}(x-t)^{n}(b-x)^{\alpha-n-1}f(x)dx. \end{eqnarray*} $

注 1   当$ \; \alpha = n+1 $时, 定义3即为定义2.

推论 1   设$ f:[a, b]\rightarrow \mathbb{R}_{\mathcal{I}}^{+} $且$ f = [\underline{f}, \overline{f}] $. 若$ f\in \mathcal{IR}_{([a, b])} $, 那么

$ \begin{equation*} (\mathcal{I}^{a}_{\alpha}f)(t) = \left[(\mathcal{I}^{a}_{\alpha}\underline{f})(t), (\mathcal{I}^{a}_{\alpha}\overline{f})(t)\right] \end{equation*} $

$ \begin{equation*} ({^{b}\mathcal{I}_{\alpha}}f)(t) = \left[{(^{b}\mathcal{I}_{\alpha}}\underline{f})(t), ({^{b}\mathcal{I}_{\alpha}}\overline{f})(t)\right]. \end{equation*} $

证  由引理1和定义3即可证得.

引理 2   设$ \xi:[a, b]\rightarrow \mathbb{R}_{\mathcal{I}}^{+}\; $且$ \xi = [\underline{\xi}, \overline{\xi}] $. 若$ \; \xi\in\mathcal{IR}_{([a, b])}\; $且关于$ \frac{a+b}{2} $对称, 则

$ \begin{equation*} (\mathcal{I}^{a}_{\alpha}\xi)(b) = ({^{b}\mathcal{I}_{\alpha}}\xi)(a) = \frac{1}{2}\left[(\mathcal{I}^{a}_{\alpha}\xi)(b)+({^{b}\mathcal{I}_{\alpha}}\xi)(a)\right]. \end{equation*} $

证  由于$ \xi $关于$ \frac{a+b}{2} $对称, 即对任意的$ x\in[a, b] $, $ \xi(a+b-x) = \xi(x) $.因此, 令$ t = a+b-x $, 有

$ \begin{eqnarray*} \begin{split} &\mathcal{I}^{a}_{\alpha}\xi(b) = \frac{1}{n!}\int_{a}^{b}(b-x)^{n}(x-a)^{\alpha-n-1}\xi(x)dx\\ = &\frac{1}{n!}\int_{a}^{b}(b-x)^{n}(x-a)^{\alpha-n-1}\left[\underline{\xi}(x), \overline{\xi}(x)\right]dx\\ = &\left[\frac{1}{n!}\int_{a}^{b}(b-x)^{n}(x-a)^{\alpha-n-1}\underline{\xi}(x)dx, \frac{1}{n!}\int_{a}^{b}(b-x)^{n}(x-a)^{\alpha-n-1}\overline{\xi}(x)dx\right]\\ = &\left[\frac{1}{n!}\int_{a}^{b}(t-a)^{n}(b-t)^{\alpha-n-1}\underline{\xi}(a+b-t)dt, \frac{1}{n!}\int_{a}^{b}(t-a)^{n}(b-t)^{\alpha-n-1}\overline{\xi}(a+b-t)dt\right]\\ = &\left[\frac{1}{n!}\int_{a}^{b}(t-a)^{n}(b-t)^{\alpha-n-1}\underline{\xi}(t)dt, \frac{1}{n!}\int_{a}^{b}(t-a)^{n}(b-t)^{\alpha-n-1}\overline{\xi}(t)dt\right]\\ = &\left[^{b}\mathcal{I}_{\alpha}\underline{\xi}(a), ^{b}\mathcal{I}_{\alpha}\overline{\xi}(a)\right]\\ = &^{b}\mathcal{I}_{\alpha}\xi(a). \end{split} \end{eqnarray*} $

证毕.

定理 1   设$ f\in SX(h, [a, b], \mathbb{R}_{\mathcal{I}}^{+}) $且$ f = [\underline{f}, \overline{f}], \; h:[0, 1]\rightarrow \mathbb{R}^{+}. $若$ f\in\mathcal{IR}_{([a, b])} $, 那么

$ \begin{eqnarray} \begin{aligned} &\frac{\Gamma(\alpha-n)} {\Gamma(\alpha+1)h(\frac{1}{2})}f\left(\frac{a+b}{2}\right)\\ \supseteq&\frac{1}{(b-a)^{\alpha}}\big[\mathcal{I}_{\alpha}^{a}{f}(b)+{^{b}\mathcal{I}}_{\alpha}{f}(a)\big]\\ \supseteq&\frac{f(a)+f(b)}{n!}\int_{0}^{1}\nu^{n}(1-\nu)^{\alpha-n-1}\big[h(\nu)+h(1-\nu)\big]d\nu. \end{aligned} \end{eqnarray} $

(3.1)

其中$ \; \alpha\in(n, n+1] $.

证  由于$ f\in SX(h, [a, b], \mathbb{R}_{\mathcal{I}}^{+}) $, 那么对任意的$ t\in[0, 1] $, 有

$ \begin{eqnarray} \begin{aligned} f\left(\frac{a+b}{2}\right)& = f\left(\frac{ta+(1-t)b+tb+(1-t)b}{2}\right)\\ &\supseteq h\left(\frac{1}{2}\right)\left[{f\left(ta+(1-t)b\right)+f\left(tb+(1-t)a\right)}\right]. \end{aligned} \end{eqnarray} $

(3.2)

在(3.2) 式两边同乘以$ \frac{1}{n!}\nu^{n}(1-\nu)^{\alpha-n-1} $并在$ [0, 1] $上关于$ t $取积分,

$ \begin{equation*} \begin{split} & \frac{\Gamma(\alpha-n)} {\Gamma(\alpha+1)h(\frac{1}{2})}f\left(\frac{a+b}{2}\right)\\ = &\frac{1} {h(\frac{1}{2})} f \left(\frac{a+b}{2}\right)\frac{1}{n!}\int_{0}^{1}\nu^{n}(1-\nu)^{\alpha-n-1}d\nu\\ \supseteq & \left[\frac{1}{n!}\int_{0}^{1}\nu^{n}(1-\nu)^{\alpha-n-1}f(\nu a+(1-\nu) b)d\nu+\frac{1}{n!}\int_{0}^{1}\nu^{n}(1-\nu)^{\alpha-n-1}f((1-\nu)a+\nu b)d\nu \right]\\ = &\left[\frac{1}{n!}\int_{0}^{1}\nu^{n}(1-\nu)^{\alpha-n-1}(\underline{f}(\nu a+(1-\nu) b)+\underline{f}((1-\nu) a+\nu b))d\nu, \right.\\& \; \left.\frac{1}{n!}\int_{0}^{1}\nu^{n}(1-\nu)^{\alpha-n-1}(\overline{f}(\nu a+(1-\nu) b)+\overline{f}((1-\nu) a+\nu b))d\nu\right]\\ = & \left[\frac{1}{n!}\int_{b}^{a}\left(\tau(\mu)\right)^{n}\left(1-\tau(\mu)\right)^{\alpha-n-1}\underline{f}(\mu)\frac{d\mu}{a-b}+ \frac{1}{n!}\int_{a}^{b}\left(1-\tau(\mu)\right)^{n}\left(\tau(\mu)\right)^{\alpha-n-1}\underline{f}(\mu)\frac{d\mu}{b-a}, \right.\\ & \; \left.\frac{1}{n!}\int_{b}^{a}\left(\tau(\mu)\right)^{n}\left(1-\tau(\mu)\right)^{\alpha-n-1}\overline{f}(\mu)\frac{d\mu}{a-b}+ \frac{1}{n!}\int_{a}^{b}\left(1-\tau(\mu)\right)^{n}\left(\tau(\mu)\right)^{\alpha-n-1}\overline{f}(\mu)\frac{d\mu}{b-a}\right]\\ = &\frac{1}{(b-a)^{\alpha}} \left[\mathcal{I}_{\alpha}^{a}{\underline{f}}(b)+{^{b}\mathcal{I}}_{\alpha}{\underline{f}}(a), \mathcal{I}_{\alpha}^{a}{\overline{f}}(b)+{^{b}\mathcal{I}}_{\alpha}{\overline{f}}(a)\right].\\ = & \frac{1}{(b-a)^{\alpha}}\big[\mathcal{I}_{\alpha}^{a}{f}(b), {^{b}\mathcal{I}}_{\alpha}{f}(a)\big]. \end{split} \end{equation*} $

其中$ \tau(\mu) = \frac{b-\mu}{b-a}. $

因此

$ \begin{equation*} \frac{\Gamma(\alpha-n)} {\Gamma(\alpha+1)h(\frac{1}{2})}f\left(\frac{a+b}{2}\right)\supseteq\frac{n!}{(b-a)^{\alpha}}\big[\mathcal{I}_{\alpha}^{a}{f}(b)+{^{b}\mathcal{I}}_{\alpha}{f}(a)\big]. \end{equation*} $

即证得(3.1) 式中的第一个不等式.

类似地, 我们有

$ \begin{eqnarray} f\left(ta+(1-t)b\right)+f\left(tb+(1-t)a\right)\supseteq \left[h(t)+h(1-t)\right]f(a)+f(b). \end{eqnarray} $

(3.3)

在(3.3) 式两边同时乘以$ \frac{1}{n!}\nu^{n}(1-\nu)^{\alpha-n-1} $并在$ [0, 1] $上关于$ t $取积分,

$ \begin{equation*} \begin{split} \frac{1}{(b-a)^{\alpha}}\big[\mathcal{I}_{\alpha}^{a}{f}(b)+ {^{b}\mathcal{I}}_{\alpha}{f}(a)\big] \supseteq \frac{f(a)+f(b)}{n!}\int_{0}^{1}\nu^{n}(1-\nu)^{\alpha-n-1}\big[h(\nu)+h(1-\nu)\big]d\nu. \end{split} \end{equation*} $

证毕.

推论 2   设$ f\in SV(h, [a, b], \mathbb{R}_{\mathcal{I}}^{+}) $且$ f = [\underline{f}, \overline{f}], \; h:[0, 1]\rightarrow \mathbb{R}^{+}. $若$ f\in\mathcal{IR}_{([a, b])} $, 那么

$ \begin{eqnarray*} \begin{aligned} &\frac{\Gamma(\alpha-n)} {\Gamma(\alpha+1)h(\frac{1}{2})}f\left(\frac{a+b}{2}\right)\\ \subseteq&\frac{1}{(b-a)^{\alpha}}\big[\mathcal{I}_{\alpha}^{a}{f}(b)+{^{b}\mathcal{I}}_{\alpha}{f}(a)\big]\\ \subseteq&\frac{f(a)+f(b)}{n!}\int_{0}^{1}\nu^{n}(1-\nu)^{\alpha-n-1}\big[h(\nu)+h(1-\nu)\big]d\nu. \end{aligned} \end{eqnarray*} $

注 2   若定理1中$ \; h(t) = t $, 则

$ \begin{eqnarray*} \begin{aligned} f\left(\frac{a+b}{2}\right) \supseteq\frac{\Gamma(\alpha+1)}{2(b-a)^{\alpha}\Gamma(\alpha-n)}\big[\mathcal{I}_{\alpha}^{a}{f}(b)+{^{b}\mathcal{I}}_{\alpha}{f}(a)\big]\supseteq\frac{f(a)+f(b)}{2}. \end{aligned} \end{eqnarray*} $

注 3   若定理1中$ \; \alpha = n+1, h(t) = t $, 则得到文献[2] 中的定理2.

注 4   若定理1中$ \; n = 1, \; \alpha = 2 $, 则得到文献[3] 中的定理4.1.

注 5   若定理1中$ \; \underline{f} = \overline{f} $且$ h(t) = t $, 则得到文献[1] 中的定理2.1.

定理 2   设$ f\in SX(h, [a, b], \mathbb{R}_{\mathcal{I}}^{+}) $且$ f = [\underline{f}, \overline{f}] $, $ \xi:[a, b]\rightarrow \mathbb{R}_{\mathcal{I}}^{+} $关于$ \frac{a+b}{2} $对称, $ \; h:[0, 1]\rightarrow \mathbb{R}^{+}. $若$ f\xi\in\mathcal{IR}_{([a, b])} $, 那么

$ \begin{eqnarray} \begin{aligned} &\frac{n!}{2h\left(\frac{1}{2}\right)(b-a)^{\alpha}}f\left(\frac{a+b}{2}\right)\left[\mathcal{I}^{a}_{\alpha}\xi(b)+{^{b}\mathcal{I}_{\alpha}}\xi(a)\right]\\ \supseteq&\frac{n!}{(b-a)^{\alpha}}\left[\mathcal{I}^{a}_{\alpha}(f\xi)(b)+{^{b}\mathcal{I}_{\alpha}}(f\xi)(a)\right]\\ \supseteq&\left[f(a)+f(b)\right]\int_{0}^{1}\left[h(t)+h(1-t)\right]t^{n}(1-t)^{\alpha-n-1}\xi(tb+(1-t)a)dt, \end{aligned} \end{eqnarray} $

(3.4)

其中$ \alpha\in(n, n+1]. $

证  由于$ f\in SX(h, [a, b], \mathbb{R}_{\mathcal{I}}^{+}) $, 那么在(3.2) 式两边同乘以$ \frac{1}{h\left(\frac{1}{2}\right)}t^{n}(1-t)^{\alpha-n-1}\xi(tb+(1-t)a) $且在$ [0, 1] $上关于$ t $取积分,

$ \begin{equation*} \begin{split} &\frac{1}{h\left(\frac{1}{2}\right)}f\left(\frac{a+b}{2}\right)\int_{0}^{1}t^{n}(1-t)^{\alpha-n-1}\xi(tb+(1-t)a)dt\\ \supseteq&\int_{0}^{1}t^{n}(1-t)^{\alpha-n-1}\left[f\left(ta+(1-t)b\right)+f\left(tb+(1-t)a\right)\right]\xi(tb+(1-t)a)dt\\ = &\int_{0}^{1}t^{n}(1-t)^{\alpha-n-1}f\left(ta+(1-t)b\right)\xi(tb+(1-t)a)dt\\ &\; +\int_{0}^{1}t^{n}(1-t)^{\alpha-n-1}f\left(tb+(1-t)a\right)\xi(tb+(1-t)a)dt. \end{split} \end{equation*} $

令$ x = tb+(1-t)a $, 我们有

$ \begin{equation*} \begin{split} &\frac{1}{{h\left(\frac{1}{2}\right)}(b-a)^{\alpha}}f\left(\frac{a+b}{2}\right)\int_{a}^{b}(x-a)^{n}(b-x)^{\alpha-n-1}\xi(x)dx\\ \supseteq&\frac{1}{(b-a)^{\alpha}}\left\{\int_{a}^{b}(x-a)^{n}(b-x)^{\alpha-n-1}f\left(a+b-x\right)\xi(x)dx+\int_{a}^{b}(x-a)^{n}(b-x)^{\alpha-n-1}f(x)\xi(x)dx\right\}\\ = &\frac{1}{(b-a)^{\alpha}}\left\{\int_{a}^{b}(x-a)^{n}(b-x)^{\alpha-n-1}f(x)\xi\left(a+b-x\right)dx+\int_{a}^{b}(x-a)^{n}(b-x)^{\alpha-n-1}f(x)\xi(x)dx\right\}\\ = &\frac{1}{(b-a)^{\alpha}}\left\{\int_{a}^{b}(x-a)^{n}(b-x)^{\alpha-n-1}f(x)\xi(x)dx+\int_{a}^{b}(x-a)^{n}(b-x)^{\alpha-n-1}f(x)\xi(x)dx\right\}. \end{split} \end{equation*} $

由引理2, 有

$ \begin{equation*} \frac{n!}{2{h\left(\frac{1}{2}\right)}(b-a)^{\alpha}}f\left(\frac{a+b}{2}\right)\left[\mathcal{I}^{a}_{\alpha}\xi(b)+{^{b}\mathcal{I}_{\alpha}}\xi(a)\right]\supseteq\frac{n!}{(b-a)^{\alpha}}\left[\mathcal{I}^{a}_{\alpha}(f\xi)(b)+{^{b}\mathcal{I}_{\alpha}}(f\xi)(a)\right], \end{equation*} $

即证得(3.4) 式中的第一个不等式. 类似地,

$ \begin{eqnarray} f\left(ta+(1-t)b\right)+f\left(tb+(1-t)a\right)\supseteq \left[h(t)+h(1-t)\right]\left[f(a)+f(b)\right]. \end{eqnarray} $

(3.5)

在(3.5) 式两边同乘以$ t^{n}(1-t)^{\alpha-n-1}\xi(tb+(1-t)a) $并在$ [0, 1] $上关于$ t $取积分, 得

$ \begin{equation*} \begin{split} &\int_{0}^{1}t^{n}(1-t)^{\alpha-n-1}f(ta+(1-t)b)\xi(tb+(1-t)a)dt\\ &\; \; \; +\int_{0}^{1}t^{n}(1-t)^{\alpha-n-1}f(tb+(1-t)a)\xi(tb+(1-t)a)dt\\ \supseteq&\left[f(a)+f(b)\right]\int_{0}^{1}\left[h(t)+h(1-t)\right]t^{n}(1-t)^{\alpha-n-1}\xi(tb+(1-t)a)dt. \end{split} \end{equation*} $

因此,

$ \begin{equation*} \begin{split} &\frac{n!}{(b-a)^{\alpha}}\left[\mathcal{I}^{a}_{\alpha}(f\xi)(b)+{^{b}\mathcal{I}_{\alpha}}(f\xi)(a)\right]\\ \supseteq&\left[f(a)+f(b)\right]\int_{0}^{1}\left[h(t)+h(1-t)\right]t^{n}(1-t)^{\alpha-n-1}\xi(tb+(1-t)a)dt. \end{split} \end{equation*} $

证毕.

推论 3   设$ f\in SV(h, [a, b], \mathbb{R}_{\mathcal{I}}^{+}) $且$ f = [\underline{f}, \overline{f}] $, $ \varphi:[a, b]\rightarrow \mathbb{R}_{\mathcal{I}}^{+} $关于$ \frac{a+b}{2} $对称, $ \; h:[0, 1]\rightarrow \mathbb{R}^{+}. $若$ f\varphi\in\mathcal{IR}_{([a, b])} $, 那么

$ \begin{equation*} \begin{split} &\frac{n!}{2h\left(\frac{1}{2}\right)(b-a)^{\alpha}}f\left(\frac{a+b}{2}\right)\left[\mathcal{I}^{a}_{\alpha}g(b)+{^{b}\mathcal{I}_{\alpha}}g(a)\right]\\ \subseteq&\frac{n!}{(b-a)^{\alpha}}\left[\mathcal{I}^{a}_{\alpha}(fg)(b)+{^{b}\mathcal{I}_{\alpha}}(fg)(a)\right]\\ \subseteq&\left[f(a)+f(b)\right]\int_{0}^{1}\left[h(t)+h(1-t)\right]t^{n}(1-t)^{\alpha-n-1}g(tb+(1-t)a)dt. \end{split} \end{equation*} $

注 6   若定理2中$ h(t) = t $, 则

$ \begin{equation*} \begin{split} f\left(\frac{a+b}{2}\right)\left[\mathcal{I}^{a}_{\alpha}g(b)+{^{b}\mathcal{I}_{\alpha}}g(a)\right]&\supseteq\left[\mathcal{I}^{a}_{\alpha}(fg)(b)+{^{b}\mathcal{I}_{\alpha}}(fg)(a)\right]\\ &\supseteq\frac{\left[f(a)+f(b)\right]}{2}\left[\mathcal{I}^{\alpha}_{a}\varphi(b)+{\mathcal{I}^{\alpha}_{b}}\varphi(a)\right]. \end{split} \end{equation*} $

注 7   若定理2中$ \underline{f} = \overline{f} $且$ h(t) = t $, 则得到文献[5] 中的定理2.1.

注 8   若定理1中$ \; \alpha = n+1, h(t) = t $, 则得到文献[10] 中的定理3.3.

定理 3   设$ f\in SX(h_{1}, [a, b], \mathbb{R}_{\mathcal{I}}^{+}) $且$ f = [\underline{f}, \overline{f}] $, $ g\in SX(h_{2}, [a, b], \mathbb{R}_{\mathcal{I}}^{+}) $且$ g = [\underline{g}, \overline{g}], \; h:[0, 1]\rightarrow \mathbb{R}^{+} $. 若$ fg\in\mathcal{IR}_{([a, b])} $, 则

$ \begin{eqnarray} \begin{aligned} &\frac{n!}{(b-a)^{\alpha}}\big[\mathcal{J}_{\alpha}^{a}{f}(b)g(b)+ {^b\mathcal{J}}_{\alpha}{f}(a)g(a)\big]\\ \supseteq &\mathcal{M}(a, b)\int_{0}^{1}\nu^{n}(1-\nu)^{\alpha-n-1}\big[h_1(\nu)h_2(\nu)+h_1(1-\nu)h_2(1-\nu)\big]d\nu\\ &+\mathcal{N}(a, b)\int_{0}^{1}\nu^{n}(1-\nu)^{\alpha-n-1}\big[h_1(\nu)h_2(1-\nu)+h_1(1-\nu)h_2(\nu)\big]d\nu, \end{aligned} \end{eqnarray} $

(3.6)

且

$ \begin{eqnarray} \begin{aligned} &\frac{n!\Gamma(\alpha-n)}{\Gamma(\alpha+1) h_1(\frac{1}{2})h_2(\frac{1}{2})}f\Big(\frac{a+b}{2}\Big)g\Big(\frac{a+b}{2}\Big)\\ \supseteq&\frac{n!}{(b-a)^{\alpha}}\big[\mathcal{J}_{\alpha}^{a}{f}(b)g(b)+ {^{b}\mathcal{J}}_{\alpha}{f}(a)g(a)\big]\\ &+\mathcal{M}(a, b)\int_{0}^{1}\nu^{n}(1-\nu)^{\alpha-n-1}\left[h_1(\nu)h_2(1-\nu)+h_1(1-\nu)h_2(\nu)\right]d\nu\\ &+\mathcal{N}(a, b)\int_{0}^{1}\nu^{n}(1-\nu)^{\alpha-n-1}\left[h_1(\nu)h_2(\nu)+h_1(1-\nu)h_2(1-\nu)\right]d\nu. \end{aligned} \end{eqnarray} $

(3.7)

其中,

$ \begin{equation*} \mathcal{M}(a, b) = f(a)g(a)+f(b)g(b), \; \mathcal{N}(a, b) = f(a)g(b)+f(b)g(a). \end{equation*} $

证  由于$ f\in SX(h_{1}, [a, b], \mathbb{R}_{\mathcal{I}}^{+}) $, $ g\in SX(h_{2}, [a, b], \mathbb{R}_{\mathcal{I}}^{+}) $, 则

$ \begin{equation*} \begin{split} f\left(ta+(1-t)b\right)g\left(ta+(1-t)b\right)\supseteq &h_{1}(t)h_{2}(t)f(a)g(a)+h_{1}(1-t)h_{2}(1-t)f(b)g(b)\\ &+h_{1}(t)h_{2}(1-t)f(a)g(b)+h_{1}(1-t)h_{2}(t)f(b)g(a). \end{split} \end{equation*} $

$ \begin{equation*} \begin{split} f\left((1-t)a+tb\right)g\left((1-t)a+tb\right)\supseteq &h_{1}(1-t)h_{2}(1-t)f(a)g(a)+h_{1}(t)h_{2}(t)f(b)g(b)\\ &+h_{1}(1-t)h_{2}(t)f(a)g(b)+h_{1}(t)h_{2}(1-t)f(b)g(a). \end{split} \end{equation*} $

因此,

$ \begin{eqnarray} \begin{aligned} &f\left(ta+(1-t)b\right)g\left(ta+(1-t)b\right)+f\left((1-t)a+tb\right)g\left((1-t)a+tb\right)\\ \supseteq&\left[h_{1}(t)h_{2}(t)+h_{1}(1-t)h_{2}(1-t)\right]\mathcal{M}(a, b)\\ &+\left[h_{1}(t)h_{2}(1-t)+h_{1}(1-t)h_{2}(t)\right]\mathcal{N}(a, b). \end{aligned} \end{eqnarray} $

(3.8)

在(3.8) 式两边同时乘以$ \frac{1}{n!}\nu^{n}(1-\nu)^{\alpha-n-1} $并且在$ [0, 1] $上关于$ t $取积分,

$ \begin{eqnarray} \begin{split} &\frac{1}{n!}\int_{0}^{1}\nu^{n}(1-\nu)^{\alpha-n-1}f\big(\nu a+(1-\nu) b\big)g\big(\nu a+(1-\nu) b\big)d\nu\\ &+\frac{1}{n!}\int_{0}^{1}\nu^{n}(1-\nu)^{\alpha-n-1}f\big((1-\nu)a+\nu b\big)g\big((1-\nu)a+\nu b\big)d\nu\\ \supseteq &\frac{\mathcal{M}(a, b)}{n!}\int_{0}^{1}\nu^{n}(1-\nu)^{\alpha-n-1}\big[h_1(\nu)h_2(\nu)+h_1(1-\nu)h_2(1-\nu)\big]d\nu\\ &+\frac{\mathcal{N}(a, b)}{n!}\int_{0}^{1}\nu^{n}(1-\nu)^{\alpha-n-1}\big[h_1(1-\nu)h_2(\nu)+h_1(\nu)h_2(1-\nu)\big]d\nu. \end{split} \end{eqnarray} $

(3.9)

由定义3, 我们有

$ \begin{eqnarray} \frac{1}{n!}\int_{0}^{1}\nu^{n}(1-\nu)^{\alpha-n-1}f\big(\nu a+(1-\nu) b\big)g\big(\nu a+(1-\nu) b\big)d\nu = \frac{1}{(b-a)^{\alpha}}\mathcal{J}^{a}_{\alpha}f(b)g(b), \end{eqnarray} $

(3.10)

$ \begin{eqnarray} \frac{1}{n!}\int_{0}^{1}\nu^{n}(1-\nu)^{\alpha-n-1}f\big((1-\nu)a+\nu b\big)g\big((1-\nu)a+\nu b\big)d\nu = \frac{1}{(b-a)^{\alpha}}{^{b}\mathcal{J}_{\alpha}}f(a)g(a). \end{eqnarray} $

(3.11)

将不等式(3.10)–(3.11) 式代入到(3.9) 式, 即证得(3.6) 式.

类似地,

$ \begin{eqnarray} \begin{aligned} &f\Big(\frac{a+b}{2}\Big)g\Big(\frac{a+b}{2}\Big)\\ = &f\Big(\frac{\nu a+(1-\nu)b}{2}+\frac{(1-\nu)a+\nu b}{2}\Big)g\Big(\frac{\nu a+(1-\nu)b}{2}+\frac{(1-\nu)a+\nu b}{2}\Big)\\ \supseteq & h_1(\frac{1}{2})h_2(\frac{1}{2})\left[f\big(\nu a+(1-\nu)b\big)g\big(\nu a+(1-\nu)b\big)+f\big((1-\nu)a+\nu b\big)g\big((1-\nu)a+\nu b\big)\right]\\ &+h_1(\frac{1}{2})h_2(\frac{1}{2})\left\{\left[h_1(\nu)h_2(1-\nu)+h_1(1-\nu)h_2(\nu)\right]\mathcal{M}(a, b)\right.\\ &\left.+\left[h_1(\nu)h_2(\nu)+h_1(1-\nu)h_2(1-\nu)\right]\mathcal{N}(a, b)\right\}. \end{aligned} \end{eqnarray} $

(3.12)

在(3.12) 式两边同时乘以$ \frac{1}{n!}\nu^{n}(1-\nu)^{\alpha-n-1} $并且在$ [0, 1] $上关于$ t $取积分,

$ \begin{equation*} \begin{split} &\frac{\Gamma(\alpha-n)}{\Gamma(\alpha+1)}f\Big(\frac{a+b}{2}\Big)g\Big(\frac{a+b}{2}\Big)\\ \supseteq & h_1(\frac{1}{2})h_2(\frac{1}{2})\frac{1}{n!}\int_{0}^{1}\nu^{n}(1-\nu)^{\alpha-n-1}\left[f\big(\nu a+(1-\nu)b\big)g\big(\nu a+(1-\nu)b\big)\right.\\ &\left.+f\big((1-\nu)a+\nu b\big)g\big((1-\nu)a+\nu b\big)\right]d\nu\\ &+h_1(\frac{1}{2})h_2(\frac{1}{2})\frac{1}{n!}\int_{0}^{1}\nu^{n}(1-\nu)^{\alpha-n-1}\left\{\left[h_1(\nu)h_2(1-\nu)+h_1(1-\nu)h_2(\nu)\right]\mathcal{M}(a, b)\right.\\ &\left.+\left[h_1(\nu)h_2(\nu)+h_1(1-\nu)h_2(1-\nu)\right]\mathcal{N}(a, b)\right\}d\nu\\ \end{split} \end{equation*} $

$ \begin{equation*} \begin{split} = &\frac{h_1(\frac{1}{2})h_2(\frac{1}{2})}{(b-a)^{\alpha}}\big[\mathcal{J}_{\alpha}^{a}{f}(b)g(b)+ {^{b}\mathcal{J}}_{\alpha}{f}(a)g(a)\big]\\ &+\frac{\mathcal{M}(a, b)h_1(\frac{1}{2})h_2(\frac{1}{2})}{n!}\int_{0}^{1}\nu^{n}(1-\nu)^{\alpha-n-1}\left[h_1(\nu)h_2(1-\nu)+h_1(1-\nu)h_2(\nu)\right]d\nu\\ &+\frac{\mathcal{N}(a, b)h_1(\frac{1}{2})h_2(\frac{1}{2})}{n!}\int_{0}^{1}\nu^{n}(1-\nu)^{\alpha-n-1}\left[h_1(\nu)h_2(\nu)+h_1(1-\nu)h_2(1-\nu)\right]d\nu.\\ \end{split} \end{equation*} $

上式两边同时乘以$ \frac{n!}{ h_1(\frac{1}{2})h_2(\frac{1}{2})} $, 即证得(3.7) 式.证毕.

推论 4   设$ f\in SV(h_{1}, [a, b], \mathbb{R}_{\mathcal{I}}^{+}) $且$ f = [\underline{f}, \overline{f}] $, $ g\in SV(h_{2}, [a, b], \mathbb{R}_{\mathcal{I}}^{+}) $且$ g = [\underline{g}, \overline{g}], \; h:[0, 1]\rightarrow \mathbb{R}^{+} $. 若$ fg\in\mathcal{IR}_{([a, b])} $. 那么

$ \begin{eqnarray*} \begin{aligned} &\frac{n!}{(b-a)^{\alpha}}\big[\mathcal{J}_{\alpha}^{a}{f}(b)g(b)+ {^b\mathcal{J}}_{\alpha}{f}(a)g(a)\big]\\ \subseteq &\mathcal{M}(a, b)\int_{0}^{1}\nu^{n}(1-\nu)^{\alpha-n-1}\big[h_1(\nu)h_2(\nu)+h_1(1-\nu)h_2(1-\nu)\big]d\nu\\ &+\mathcal{N}(a, b)\int_{0}^{1}\nu^{n}(1-\nu)^{\alpha-n-1}\big[h_1(\nu)h_2(1-\nu)+h_1(1-\nu)h_2(\nu)\big]d\nu, \end{aligned} \end{eqnarray*} $

且

$ \begin{eqnarray*} \begin{aligned} &\frac{n!\Gamma(\alpha-n)}{\Gamma(\alpha+1) h_1(\frac{1}{2})h_2(\frac{1}{2})}f\Big(\frac{a+b}{2}\Big)g\Big(\frac{a+b}{2}\Big)\\ \subseteq&\frac{n!}{(b-a)^{\alpha}}\big[\mathcal{J}_{\alpha}^{a}{f}(b)g(b)+ {^{b}\mathcal{J}}_{\alpha}{f}(a)g(a)\big]\\ &+\mathcal{M}(a, b)\int_{0}^{1}\nu^{n}(1-\nu)^{\alpha-n-1}\left[h_1(\nu)h_2(1-\nu)+h_1(1-\nu)h_2(\nu)\right]d\nu\\ &+\mathcal{N}(a, b)\int_{0}^{1}\nu^{n}(1-\nu)^{\alpha-n-1}\left[h_1(\nu)h_2(\nu)+h_1(1-\nu)h_2(1-\nu)\right]d\nu. \end{aligned} \end{eqnarray*} $

注 9   若定理1中$ \alpha = n+1 $, 则得到文献[2] 中的定理3和定理4.

注 10   若定理1中$ n = 1, \alpha = 2 $, 则得到文献[3] 中的定理4.5和定理4.6.

注 11   若定理3中$ h(t) = t $, 则

$ \begin{eqnarray*} \begin{aligned} &\frac{1}{(b-a)^{\alpha}}\big[\mathcal{J}_{\alpha}^{a}{f}(b)g(b)+ {^b\mathcal{J}}_{\alpha}{f}(a)g(a)\big]\\ \supseteq &\frac{\Gamma(\alpha-n)}{\Gamma(\alpha+3)}\left[\mathcal{M}(a, b)\left[(n+2)(n+1)+(\alpha-n+1)(\alpha-n)\right]\right.\\ &\left.+2\mathcal{N}(a, b)(n+1)(\alpha-n)\right]\\ \end{aligned} \end{eqnarray*} $

且

$ \begin{eqnarray*} \begin{aligned} &\frac{\Gamma(\alpha-n)}{\Gamma(\alpha+1) h_1(\frac{1}{2})h_2(\frac{1}{2})}f\Big(\frac{a+b}{2}\Big)g\Big(\frac{a+b}{2}\Big)\\ \supseteq&\frac{1}{(b-a)^{\alpha}}\big[\mathcal{J}_{\alpha}^{a}{f}(b)g(b)+ {^{b}\mathcal{J}}_{\alpha}{f}(a)g(a)\big]+\frac{\Gamma(\alpha-n)}{\Gamma(\alpha+3)}\left[\mathcal{M}(a, b)\left[(n+2)(n+1)\right.\right.\\ &\left.\left.+(\alpha-n+1)(\alpha-n)\right]+2\mathcal{N}(a, b)(n+1)(\alpha-n)\right]\\ \end{aligned} \end{eqnarray*} $