设\begin{document}$f:[a, b]\rightarrow \mathbb{R} $\end{document}在区间\begin{document}$(a, b)$\end{document}上是一个四阶连续可微函数,且\begin{document}$\|f^{(4)}\|_{\infty}=\mathop {\sup }\limits_{x \in (a,b)} |f^{(4)}| < \infty$\end{document},那么有如下不等式成立

(1.1) $\begin{equation} \label{eq:1.1}\left| \frac{1}{3}\left[\frac{f(a)+f(b)}{2}+2f\left(\frac{a+b}{2}\right)\right]- \frac{1}{{b - a}}\int_a^b {f(x)} {\rm d}x\right| \le \frac{1}{2880}\|f^{(4)}\|_{\infty}(b-a)^{4}, \end{equation}$

不等式(1.1)是著名的Simpson不等式.对于该不等式的改进和推广,一般要用到函数凸性的定义,随着对凸性定义的推广研究, Simpson不等式的研究成果也越来越多,读者可以参见文献[1-5]等.

近年来,分形理论作为一门新理论、新科学在科学工程领域得到广泛的应用.分形理论最基本的特点是用分数维度的视角和数学方法描述和研究客观事物,由于分形理论的出现,许多力学等工程应用领域的问题得到了合理的解决和处理.因此,关于分形空间的数学理论的发展也十分迅速,尤其一些学者通过不同的方法构建了分形空间上的微积分理论,受到广泛关注,见文献[6-10].在文献[8, 11-12]中Yang系统阐述了建立在分形空间上的局部分数阶微积分的相关理论.

文献[13]中, Mo等提出了关于分形空间上广义凸函数的定义,并在分形空间上推广了Hermite-Hadamard不等式,结论如下:

定义1.1^[13] 设\begin{document}$f:I\subseteq\mathbb{R} \rightarrow\mathbb{R} ^{\alpha}$\end{document},对任意\begin{document}$x_{1}, x_{2}\in I$\end{document}且\begin{document}$\lambda\in[0, 1]$\end{document},如果以下不等式成立

$f(\lambda x_{1}+(1-\lambda) x_{2})\leq\lambda^{\alpha} f(x_{1})+(1-\lambda)^{\alpha} f(x_{2}), $

则称\begin{document}$f$\end{document}为定义在\begin{document}$I$\end{document}上的广义凸函数.

定理1.1^[13] (广义Hermite-Hadamard不等式) 令\begin{document}$f(x)\in I^{(\alpha)}_{x}[a, b]$\end{document}是区间\begin{document}$[a, b]$\end{document}上的一个广义凸函数, \begin{document}$a \le b$\end{document},则

(1.2) $ \begin{equation} \label{eq:1.2} f(\frac{{a + b}}{2}) \le \frac{\Gamma(1+\alpha)}{{(b - a})^{\alpha}} {}_{a}I_{b}^{(\alpha)}{f(x)} \le \frac{{f(a) + f(b)}}{2^{\alpha}} . \end{equation}$

文献[14]中, Set等建立了一些关于广义拟凸函数的Simpson型积分不等式,并且证明了如下恒等式.

引理1.1设\begin{document}$I\subseteq\mathbb{R} $\end{document}是一个区间, \begin{document}$f:I^{o}\subseteq\mathbb{R} \rightarrow\mathbb{R} ^{\alpha}$\end{document}(\begin{document}$I^{o}$\end{document}是\begin{document}$I$\end{document}的内部)使得\begin{document}$f\in D_{\alpha}(I^{o})$\end{document}且\begin{document}$f^{(\alpha)}\in C_{\alpha}[a, b]$\end{document},其中\begin{document}$a, b\in I^{o}, a < b$\end{document}.则对于所有的\begin{document}$x\in[a, b]$\end{document},有如下等式成立

(1.3) $ \begin{eqnarray} \label{eq:1.3} &&\frac{1}{6^{\alpha}}\left[f(a)+f(b)+4^{\alpha}f\left(\frac{a+b}{2}\right)\right]-\frac{\Gamma(1+\alpha)}{(b-a)^{\alpha}} {}_{a}I_{b}^{(\alpha)}f(x) \nonumber \\&=& \frac{(b-a)^{\alpha}}{2^{\alpha}}\frac{1}{\Gamma(1+\alpha)}\int_{0}^{1}\Bigg[\big(\frac{t}{2}-\frac{1}{3}\big)^{\alpha}f^{(\alpha)}\left(\frac{1+t}{2}b+\frac{1-t}{2}a\right)\nonumber \\&&+\big(\frac{1}{3}-\frac{t}{2}\big)^{\alpha}f^{(\alpha)}\left(\frac{1+t}{2}a+\frac{1-t}{2}b\right)\Bigg]({\rm d}t)^{\alpha}. \end{eqnarray}$

本文在局部分数阶微积分理论的基础上,由分形集上广义凸函数的定义以及广义Hölder不等式等工具对Simpson型不等式进行推广,得到几个分形空间中关于广义凸函数的新Simpson型不等式,文章最后给出了这些不等式在求特殊均值以及求局部分数阶积分的数值积分上的应用.

设\begin{document}$\mathbb{R} ^{\alpha}$\end{document}为分形实线的\begin{document}$\alpha(0 < \alpha\leq1)$\end{document}型集合,称为一个分形集合,它与实数集\begin{document}$\mathbb{R} $\end{document}构成一对一的关系^[8],相关运算律如下:

若\begin{document}$a^{\alpha}, b^{\alpha}, c^{\alpha}\in\mathbb{R} ^{\alpha} (0 < \alpha\leq1)$\end{document},则

(1) \begin{document}$a^{\alpha}+b^{\alpha}\in\mathbb{R} ^{\alpha}$\end{document}, \begin{document}$a^{\alpha}b^{\alpha}\in\mathbb{R} ^{\alpha}$\end{document},

(2) \begin{document}$a^{\alpha}+b^{\alpha}=b^{\alpha}+a^{\alpha}=(a+b)^{\alpha}=(b+a)^{\alpha}$\end{document},

(3)\begin{document}$a^{\alpha}+(b^{\alpha}+c^{\alpha})=(a+b)^{\alpha}+c^{\alpha}$\end{document},

(4) \begin{document}$a^{\alpha}b^{\alpha}=b^{\alpha} a^{\alpha}=(ab)^{\alpha}=(ba)^{\alpha}$\end{document},

(5) \begin{document}$a^{\alpha}(b^{\alpha}c^{\alpha})=(a^{\alpha}b^{\alpha})c^{\alpha}$\end{document},

(6)\begin{document}$a^{\alpha}(b^{\alpha}+c^{\alpha})=a^{\alpha}b^{\alpha}+a^{\alpha}c^{\alpha}$\end{document},

(7) \begin{document}$a^{\alpha}+0^{\alpha}=0^{\alpha}+a^{\alpha}=a^{\alpha}$\end{document}且\begin{document}$a^{\alpha}1^{\alpha}=1^{\alpha}a^{\alpha}=a^{\alpha}$\end{document}.

利用Gao-Yang-Kang的方法给出局部分数阶导数和局部分数阶积分的定义^[8-9].

定义2.1^[8] 设\begin{document}$f:\mathbb{R} \rightarrow\mathbb{R} ^{\alpha}, x\rightarrow f(x)$\end{document}是一个不可微函数,如果对于任意的\begin{document}$\varepsilon>0$\end{document},总存在\begin{document}$\delta>0$\end{document},其中\begin{document}$\epsilon, \delta\in\mathbb{R} $\end{document},使得当\begin{document}$|x-x_{0}| < \delta$\end{document}时有

$|f(x)-f(x_{0})|<\epsilon^{\alpha}$

成立,则称不可微函数\begin{document}$f$\end{document}在\begin{document}$x_{0}$\end{document}处局部分数阶连续.如果\begin{document}$f(x)$\end{document}在区间\begin{document}$(a, b)$\end{document}上局部分数阶连续，记为\begin{document}$f(x)\in C_{\alpha}(a, b)$\end{document}.

定义2.2^[8] 若

$f^{(\alpha)}(x_{0})=\frac{{\rm d}^{\alpha}f(x)}{{\rm d}x^{\alpha}}\bigg|_{x=x_{0}}=\lim\limits_{x\rightarrow x_{0}}\frac{\Gamma(\alpha+1)(f(x)-f(x_{0}))}{(x-x_{0})^{\alpha}}, $

则称之为\begin{document}$f(x)$\end{document}在\begin{document}$x=x_{0}$\end{document}处的\begin{document}$\alpha$\end{document}阶局部分数阶导数,记\begin{document}$D_{x}^{(\alpha)}=f^{(\alpha)}(x)$\end{document}.

当\begin{document}$x\in(a, b)$\end{document},存在\begin{document}$ f^{(\alpha)}(x)=D_{x}^{(\alpha)}$\end{document},记为\begin{document}$f^{(\alpha)}(x)\in D_{x}^{(\alpha)}(a, b)$\end{document}. \begin{document}$D_{\alpha}(a, b)$\end{document}称为\begin{document}$\alpha$\end{document} -阶局部分数阶导数的集合.

定义2.3^[8] 设\begin{document}$f(x)\in C_{\alpha}[a, b], a=t_{0} < t_{1} < \cdot\cdot\cdot < t_{N-1} < t_{N}=b, [t_{j}, t_{j+1}]$\end{document}是区间\begin{document}$[a, b]$\end{document}的一个划分, \begin{document}$\Delta t_{j}=t_{j+1}-t_{j}, \Delta t= \max\{\Delta t_{0}, \Delta t_{1} \cdot\cdot\cdot\Delta t_{N-1}\}$\end{document},则\begin{document}$f(x)$\end{document}的\begin{document}$\alpha$\end{document}阶局部分数阶积分定义为

${}_{a}I_{b}^{(\alpha)}f(x)=\frac{1}{\Gamma(\alpha+1)}\int_{a}^{b}f(t)({\rm d}t)^{\alpha}=\frac{1}{\Gamma(\alpha+1)}\lim\limits_{\Delta t\rightarrow 0}\sum\limits_{j=0}^{N-1}f(t_{j})(\Delta t_{j})^{\alpha}.$

注2.1 \begin{document}${}_{a}I_{a}^{(\alpha)}f(x)=0$\end{document},当\begin{document}$a < b$\end{document}时, \begin{document}${}_{a}I_{b}^{(\alpha)}f(x)=-{}_{b}I_{a}^{(\alpha)}f(x)$\end{document}.若\begin{document}$x\in[a, b]$\end{document}时, \begin{document}${}_{a}I_{x}^{(\alpha)}f(x)$\end{document}存在,则记为\begin{document}$f(x)\in I_{x}^{(\alpha)}[a, b]$\end{document}.

引理2.1^[8]

(1)设\begin{document}$f(x)=g^{(\alpha)}(x)\in C_{\alpha}[a, b]$\end{document},则

${}_{a}I_{b}^{(\alpha)}f(x)=g(b)-g(a).$

(2)设\begin{document}$f(x), g(x)\in D_{\alpha}[a, b]$\end{document},且\begin{document}$f^{(\alpha)}(x), g^{(\alpha)}(x)\in C_{\alpha}[a, b]$\end{document}则

${}_{a}I_{b}^{(\alpha)}f(x)g^{(\alpha)}(x)=f(x)g(x)\Big |_{a}^{b}-{}_{a}I_{b}^{(\alpha)}f^{(\alpha)}(x)g(x).$

引理2.2^[8]

$\frac{{\rm d}^{\alpha}x^{k\alpha}}{{\rm d}x^{\alpha}}=\frac{\Gamma(1+k\alpha)}{\Gamma(1+(k-1)\alpha)}x^{(k-1)\alpha};$

$\frac{1}{\Gamma(\alpha+1)}\int_{a}^{b}x^{k\alpha}({\rm d}x)^{\alpha}=\frac{\Gamma(1+k\alpha)}{\Gamma(1+(k+1)\alpha)}(b^{(k+1)\alpha}-a^{(k+1)\alpha}), k>0.$

引理2.3^[8] (广义Hölder不等式) 设\begin{document}$f, g\in C_{\alpha}[a, b], p, q >1$\end{document},且\begin{document}$\frac{1}{p}+\frac{1}{q}=1$\end{document},则

$\begin{eqnarray*} && \frac{1}{\Gamma(\alpha+1)}\int_{a}^{b}|f(x)g(x)|({\rm d}x)^{\alpha}\\&\leq&\left(\frac{1}{\Gamma(\alpha+1)}\int_{a}^{b}|f(x)|^{p}({\rm d}x)^{\alpha}\right)^{1/p}\left(\frac{1}{\Gamma(\alpha+1)}\int_{a}^{b}|g(x)|^{q}({\rm d}x)^{\alpha}\right)^{1/q}. \end{eqnarray*}$

定理3.1 假设\begin{document}$f$\end{document}满足引理1.1的条件,若\begin{document}$\left|f^{(\alpha)}\right|^{q}$\end{document}在\begin{document}$[a, b]$\end{document}上是广义凸函数,则有如下局部分数阶不等式成立

$ \begin{eqnarray} \label{eq: 3.1}&&\left|\frac{1}{6^{\alpha}}\left[f(a)+f(b)+4^{\alpha}f\left(\frac{a+b}{2}\right)\right]-\frac{\Gamma(1+\alpha)}{(b-a)^{\alpha}} {}_{a}I_{b}^{(\alpha)}f(x)\right| \nonumber \\&\leq&\frac{(b-a)^{\alpha}}{2^{\alpha(1+\frac{1}{q})} (\Gamma(1+\alpha))^{\frac{1}{q}}}\left(\frac{2^{\alpha}(2^{(p+1)\alpha}+1)}{6^{(p+1)\alpha}}\frac{\Gamma(1+p\alpha)}{\Gamma(1+(p+1)\alpha)}\right)^{\frac{1}{p}}\nonumber \\&& \times\Bigg[\left( |f^{(\alpha)}(b)|^{q}+ |f^{(\alpha)}(\frac{b+a}{2})|^{q}\right)^{\frac{1}{q}} +\left(|f^{(\alpha)}(a)|^{q}+ |f^{(\alpha)}(\frac{a+b}{2})|^{q}\right)^{\frac{1}{q}}\Bigg], \end{eqnarray}$

其中\begin{document}$p, q>1, $\end{document}并且\begin{document}$\frac{1}{p}+\frac{1}{q}=1.$\end{document}

证 根据引理1.1, (1.3)式两边同时取模,利用模的运算性质和广义Hölder不等式(引理2.3),可得

(3.2) $ \begin{eqnarray} \label{eq: 3.2}&&\left|\frac{1}{6^{\alpha}}\left[f(a)+f(b)+4^{\alpha}f\left(\frac{a+b}{2}\right)\right]-\frac{\Gamma(1+\alpha)}{(b-a)^{\alpha}} {}_{a}I_{b}^{(\alpha)}f(x)\right| \nonumber \\&\leq&\frac{(b-a)^{\alpha}}{2^{\alpha}}\Bigg\{\left(\frac{1}{\Gamma(1+\alpha)}\int_{0}^{1}\left|\big(\frac{t}{2}-\frac{1}{3}\big)^{\alpha}\right|^{p}({\rm d}t)^{\alpha} \right)^{\frac{1}{p}} \\&&\times \left(\frac{1}{\Gamma(1+\alpha)}\int_{0}^{1}\left|f^{(\alpha)}\left(\frac{1+t}{2}b+\frac{1-t}{2}a\right)\right|^{q}({\rm d}t)^{\alpha}\right)^{\frac{1}{q}} \\&&+\left(\frac{1}{\Gamma(1+\alpha)}\int_{0}^{1}\left|\big(\frac{1}{3}-\frac{t}{2}\big)^{\alpha}\right|^{p}({\rm d}t)^{\alpha}\right)^{\frac{1}{p}}\\&&\times \left(\frac{1}{\Gamma(1+\alpha)} \int_{0}^{1}\left|f^{(\alpha)}\left(\frac{1+t}{2}a+\frac{1-t}{2}b\right)\right|^{q}({\rm d}t)^{\alpha}\right)^{\frac{1}{q}}\Bigg\}. \end{eqnarray} $

由引理2.2以及分形集\begin{document}$\mathbb{R} ^{\alpha}$\end{document}上的运算性质得到

(3.3) $\begin{eqnarray} \label{eq: 3.3}\frac{1}{\Gamma(\alpha+1)}\int_{0}^{1}\left|\big(\frac{t}{2}-\frac{1}{3}\big)^{\alpha}\right|^{p}({\rm d}t)^{\alpha} &=&\frac{1}{\Gamma(\alpha+1)}\int_{0}^{1}\left|\big(\frac{1}{3}-\frac{t}{2}\big)^{\alpha}\right|^{p}({\rm d}t)^{\alpha} \nonumber \\&=&\frac{2^{\alpha}(2^{(p+1)\alpha}+1)}{6^{(p+1)\alpha}}\frac{\Gamma(1+p\alpha)}{\Gamma(1+(p+1)\alpha)}.\end{eqnarray} $

由于\begin{document}$\left|f^{(\alpha)}\right|^{q}$\end{document}在\begin{document}$[a, b]$\end{document}上是广义凸函数,由广义Hermite-Hadamard不等式(即定理1.1),可得

(3.4) $\begin{eqnarray} \label{eq: 3.4}\int_{0}^{1}\left|f^{(\alpha)}\left( \frac{1+t}{2}b+\frac{1-t}{2}a\right)\right|^{q}({\rm d}t)^{\alpha}&=&\frac{2^{\alpha}}{(b-a)^{\alpha}}\int_{\frac{b+a}{2}}^{b}\left|f^{(\alpha)}(x)\right|^{q}({\rm d}x)^{\alpha}\nonumber\\ &\leq&\frac{|f^{(\alpha)}(b)|^{q}+ |f^{(\alpha)}(\frac{b+a}{2})|^{q}}{2^{\alpha}}, \end{eqnarray}$

(3.5) $\begin{eqnarray} \label{eq: 3.5}\int_{0}^{1}\left|f^{(\alpha)}\left( \frac{1+t}{2}a+\frac{1-t}{2}b\right)\right|^{q}({\rm d}t)^{\alpha}&=&\frac{2^{\alpha}}{(a-b)^{\alpha}}\int_{\frac{a+b}{2}}^{a}\left|f^{(\alpha)}(x)\right|^{q}({\rm d}x)^{\alpha}\nonumber\\ &\leq&\frac{|f^{(\alpha)}(a)|^{q}+ |f^{(\alpha)}(\frac{a+b}{2})|^{q}}{2^{\alpha}}.\end{eqnarray}$

将不等式(3.3)-(3.5)代入不等式(3.2),可得

(3.6) $\begin{eqnarray} \label{eq: 3.6}&&\left|\frac{1}{6^{\alpha}}\left[f(a)+f(b)+4^{\alpha}f\left(\frac{a+b}{2}\right)\right]-\frac{\Gamma(1+\alpha)}{(b-a)^{\alpha}} {}_{a}I_{b}^{(\alpha)}f(x)\right| \nonumber \\&\leq&\frac{(b-a)^{\alpha}}{2^{\alpha}(\Gamma(1+\alpha))^{\frac{1}{q}}}\left(\frac{2^{\alpha}(2^{(p+1)\alpha}+1)}{6^{(p+1)\alpha}}\frac{\Gamma(1+p\alpha)}{\Gamma(1+(p+1)\alpha)}\right)^{\frac{1}{p}} \nonumber \\&& \times\Bigg[\left(\frac{|f^{(\alpha)}(b)|^{q}+ |f^{(\alpha)}(\frac{b+a}{2})|^{q}}{2^{\alpha}}\right)^{\frac{1}{q}} +\left( \frac{|f^{(\alpha)}(a)|^{q}+ |f^{(\alpha)}(\frac{a+b}{2})|^{q}}{2^{\alpha}}\right)^{\frac{1}{q}}\Bigg]. \end{eqnarray}$

定理得证.

推论3.1 假设满足引理1.1的条件,若\begin{document}$\left|f^{(\alpha)}\right|^{q}$\end{document}在\begin{document}$[a, b]$\end{document}上是广义凸函数,则有如下局部分数阶不等式成立

(3.7) $ \begin{eqnarray} \label{eq: 3.7} &&\left|\frac{1}{6^{\alpha}}\left[f(a)+f(b)+4^{\alpha}f\left(\frac{a+b}{2}\right)\right]-\frac{\Gamma(1+\alpha)}{(b-a)^{\alpha}} {}_{a}I_{b}^{(\alpha)}f(x)\right| \nonumber \\&\leq&\frac{(b-a)^{\alpha}}{2^{\alpha(1+\frac{2}{q})} (\Gamma(1+\alpha))^{\frac{1}{q}}}\left(\frac{2^{\alpha}(2^{(p+1)\alpha}+1)}{6^{(p+1)\alpha}}\frac{\Gamma(1+p\alpha)}{\Gamma(1+(p+1)\alpha)}\right)^{\frac{1}{p}}\nonumber \\&& \times\Bigg[\left( 3^{\alpha}|f^{(\alpha)}(b)|^{q}+ |f^{(\alpha)}(a)|^{q}\right)^{\frac{1}{q}} +\left(3^{\alpha}|f^{(\alpha)}(a)|^{q}+ |f^{(\alpha)}(b)|^{q}\right)^{\frac{1}{q}}\Bigg], \end{eqnarray} $

其中\begin{document}$p, q>1, $\end{document}并且\begin{document}$\frac{1}{p}+\frac{1}{q}=1.$\end{document}

证 由\begin{document}$\left|f^{(\alpha)}\right|^{q}$\end{document}在\begin{document}$[a, b]$\end{document}上是广义凸函数,可得

$\left |f^{(\alpha)}\left(\frac{b+a}{2}\right)\right|^{q}\leq\frac{|f^{(\alpha)}(b)|^{q}+|f^{(\alpha)}(a)|^{q}}{2^{\alpha}}, $

将其代入定理3.1的(3.1)式可得结论成立.

推论3.2 在推论3.1中,如果\begin{document}$f(a)=f(b)=f\left(\frac{a+b}{2}\right)$\end{document},则有如下局部分数阶不等式成立

(3.8) $ \begin{eqnarray}\label{eq: 3.8} &&\left|f\left(\frac{a+b}{2}\right)-\frac{\Gamma(1+\alpha)}{(b-a)^{\alpha}} {}_{a}I_{b}^{(\alpha)}f(x)\right| \nonumber \\&\leq&\frac{(b-a)^{\alpha}}{2^{\alpha(1+\frac{2}{q})} (\Gamma(1+\alpha))^{\frac{1}{q}}}\left(\frac{2^{\alpha}(2^{(p+1)\alpha}+1)}{6^{(p+1)\alpha}}\frac{\Gamma(1+p\alpha)}{\Gamma(1+(p+1)\alpha)}\right)^{\frac{1}{p}}\nonumber \\&& \times\Bigg[\left( 3^{\alpha}|f^{(\alpha)}(b)|^{q}+ |f^{(\alpha)}(a)|^{q}\right)^{\frac{1}{q}} +\left(3^{\alpha}|f^{(\alpha)}(a)|^{q}+ |f^{(\alpha)}(b)|^{q}\right)^{\frac{1}{q}}\Bigg], \end{eqnarray}$

其中\begin{document}$p, q>1, $\end{document}并且\begin{document}$\frac{1}{p}+\frac{1}{q}=1.$\end{document}

定理3.2 假设\begin{document}$f$\end{document}满足引理1.1的条件,若\begin{document}$\left|f^{(\alpha)}\right|^{q}$\end{document}在\begin{document}$[a, b]$\end{document}上是广义凸函数,则有如下局部分数阶不等式成立

(3.9) $ \begin{eqnarray}\label{eq: 3.9} &&\left|\frac{1}{6^{\alpha}}\left[f(a)+f(b)+4^{\alpha}f\left(\frac{a+b}{2} \right)\right]-\frac{\Gamma(1+\alpha)}{(b-a)^{\alpha}} {}_{a}I_{b}^{(\alpha)}f(x)\right| \nonumber \\ &\leq& \frac{(b-a)^{\alpha}}{2^{\alpha(1+\frac{1}{q})}}\left (\frac{5^{\alpha}\Gamma(1+\alpha)}{18^{\alpha}\Gamma(1+2\alpha)} \right)^{\frac{1}{p}} \Bigg\{ \Bigg[\left(\frac{13^{\alpha}}{54^{\alpha}} \frac{\Gamma(1+\alpha)}{\Gamma(1+2\alpha)}+\frac{11^{\alpha}}{54^{\alpha}} \frac{\Gamma(1+2\alpha)}{\Gamma(1+3\alpha)}\right)\left|f^{\alpha}(b)\right|^{q} \\ && +\left(\frac{25^{\alpha}}{54^{\alpha}}\frac{\Gamma(1+2\alpha)} {\Gamma(1+3\alpha)}-\frac{7^{\alpha}}{54^{\alpha}}\frac{\Gamma (1+\alpha)}{\Gamma(1+2\alpha)}\right)\left|f^{\alpha}(a)\right|^{q} \Bigg]^{\frac{1}{q}}\nonumber \\&& +\Bigg[\left(\frac{13^{\alpha}}{54^{\alpha}}\frac{\Gamma(1+\alpha)}{\Gamma(1+2\alpha)}+\frac{11^{\alpha}}{54^{\alpha}}\frac{\Gamma(1+2\alpha)}{\Gamma(1+3\alpha)}\right)\left|f^{\alpha}(a)\right|^{q} \\ &&+\left(\frac{25^{\alpha}}{54^{\alpha}}\frac{\Gamma(1+2\alpha)}{\Gamma(1+3\alpha)}-\frac{7^{\alpha}}{54^{\alpha}}\frac{\Gamma(1+\alpha)}{\Gamma(1+2\alpha)}\right)\left|f^{\alpha}(b)\right|^{q}\Bigg]^{\frac{1}{q}}\Bigg\}, \end{eqnarray}$

其中\begin{document}$p, q>1, $\end{document}并且\begin{document}$\frac{1}{p}+\frac{1}{q}=1.$\end{document}

证 因为\begin{document}$\big|\frac{1}{3}-\frac{t}{2}\big|^{\alpha}= \big|\frac{1}{3}-\frac{t}{2}\big|^{\alpha(\frac{1}{p}+\frac{1}{q})}$\end{document},根据引理1.1, (1.3)式两边同时取模,并由广义Hölder不等式(引理2.3),可得

(3.10) $\begin{eqnarray} \label{eq: 3.10}&&\left|\frac{1}{6^{\alpha}}\left[f(a)+f(b)+4^{\alpha}f\left(\frac{a+b}{2}\right)\right]-\frac{\Gamma(1+\alpha)}{(b-a)^{\alpha}} {}_{a}I_{b}^{\alpha}f(x)\right| \nonumber \\ &\leq&\frac{(b-a)^{\alpha}}{2^{\alpha}}\Bigg\{\left(\frac{1}{\Gamma(1+\alpha)} \int_{0}^{1}\left|\frac{t}{2}-\frac{1}{3}\right|^{\alpha}({\rm d}t)^{\alpha} \right)^{\frac{1}{p}} \\ &&\times \left(\frac{1}{\Gamma(1+\alpha)}\int_{0}^{1} \left|\frac{t}{2}-\frac{1}{3}\right|^{\alpha}\left|f^{(\alpha)}\left( \frac{1+t}{2}b+\frac{1-t}{2}a\right)\right|^{q}({\rm d}t)^{\alpha} \right)^{\frac{1}{q}}\nonumber \\&& +\left(\frac{1}{\Gamma(1+\alpha)}\int_{0}^{1}\left|\frac{1}{3}-\frac{t}{2}\right|^{\alpha}({\rm d}t)^{\alpha}\right)^{\frac{1}{p}} \\ &&\times\left(\frac{1}{\Gamma(1+\alpha)}\int_{0}^{1}\left|\frac{1}{3}-\frac{t}{2}\right|^{\alpha}\left|f^{(\alpha)}\left(\frac{1+t}{2}a+\frac{1-t}{2}b\right)\right|^{q}({\rm d}t)^{\alpha}\right)^{\frac{1}{q}}\Bigg\}. \end{eqnarray}$

通过计算,可得

(3.11) $\begin{eqnarray}\label{eq: 3.11}\frac{1}{\Gamma(1+\alpha)}\int_{0}^{1}\left|\frac{t}{2}-\frac{1}{3}\right|^{\alpha}({\rm d}t)^{\alpha}=\frac{1}{\Gamma(1+\alpha)}\int_{0}^{1}\left|\frac{1}{3}-\frac{t}{2}\right|^{\alpha}({\rm d}t)^{\alpha}=\frac{5^{\alpha}\Gamma(1+\alpha)}{18^{\alpha}\Gamma(1+2\alpha)}.\end{eqnarray}$

因为\begin{document}$\left|f^{(\alpha)}\right|^{q}$\end{document}在\begin{document}$[a, b]$\end{document}上是广义凸函数,可得

(3.12) $ \begin{eqnarray}\label{eq: 3.12} && \frac{1}{\Gamma(\alpha+1)}\int_{0}^{1}\left|\frac{t}{2}-\frac{1}{3}\right|^{\alpha}\left|f^{(\alpha)}\left( \frac{1+t}{2}b+\frac{1-t}{2}a\right)\right|^{q}({\rm d}t)^{\alpha}\nonumber\\&\leq&\frac{1}{2^{\alpha}}\bigg[\frac{1}{\Gamma(\alpha+1)}\int_{0}^{1}\left|\frac{t}{2}-\frac{1}{3}\right|^{\alpha} (1+t)^{\alpha}({\rm d}t)^{\alpha}\left|f^{(\alpha)}(b)\right|^{q} \\ &&+\frac{1}{\Gamma(\alpha+1)}\int_{0}^{1}\left|\frac{t}{2}-\frac{1}{3}\right|^{\alpha}(1-t)^{\alpha}({\rm d}t)^{\alpha}\left|f^{(\alpha)}(a)\right|^{q}\bigg]\end{eqnarray}$

和

(3.13) $\begin{eqnarray}\label{eq: 3.13} &&\frac{1}{\Gamma(\alpha+1)}\int_{0}^{1}\left|\frac{1}{3}-\frac{t}{2}\right|^{\alpha}\left|f^{(\alpha)}\left( \frac{1+t}{2}a+\frac{1-t}{2}b\right)\right|^{q}({\rm d}t)^{\alpha}\nonumber\\&\leq&\frac{1}{2^{\alpha}}\bigg[\frac{1}{\Gamma(\alpha+1)}\int_{0}^{1}\left|\frac{1}{3}-\frac{t}{2}\right|^{\alpha} (1+t)^{\alpha}({\rm d}t)^{\alpha}\left|f^{(\alpha)}(a)\right|^{q} \\ &&+\frac{1}{\Gamma(\alpha+1)}\int_{0}^{1}\left|\frac{1}{3}-\frac{t}{2}\right|^{\alpha}(1-t)^{\alpha}({\rm d}t)^{\alpha}\left|f^{(\alpha)}(b)\right|^{q}\bigg].\end{eqnarray}$

根据引理2.2以及(3.11)式,计算可得

(3.14) $\begin{eqnarray}\label{eq: 3.14}\frac{1}{\Gamma(\alpha+1)}\int_{0}^{1}\left|\frac{1}{3}-\frac{t}{2}\right|^{\alpha} (1+t)^{\alpha}({\rm d}t)^{\alpha}&=&\frac{1}{\Gamma(\alpha+1)}\int_{0}^{1}\left|\frac{t}{2}-\frac{1}{3}\right|^{\alpha} (1+t)^{\alpha}({\rm d}t)^{\alpha}\nonumber\\&=&\frac{13^{\alpha}}{54^{\alpha}}\frac{\Gamma(1+\alpha)}{\Gamma(1+2\alpha)}+\frac{11^{\alpha}}{54^{\alpha}}\frac{\Gamma(1+2\alpha)}{\Gamma(1+3\alpha)}.\end{eqnarray}$

令\begin{document}$1-t=s$\end{document},根据引理2.2,计算可得

(3.15) $\begin{eqnarray}\label{eq:3.15}\frac{1}{\Gamma(\alpha+1)}\int_{0}^{1}\left|\frac{1}{3}-\frac{t}{2}\right|^{\alpha}(1-t)^{\alpha}({\rm d}t)^{\alpha}&=&\frac{1}{\Gamma(\alpha+1)}\int_{0}^{1}\left|\frac{t}{2}-\frac{1}{3}\right|^{\alpha}(1-t)^{\alpha}({\rm d}t)^{\alpha}\nonumber\\&=&\frac{25^{\alpha}}{54^{\alpha}}\frac{\Gamma(1+2\alpha)}{\Gamma(1+3\alpha)}-\frac{7^{\alpha}}{54^{\alpha}}\frac{\Gamma(1+\alpha)}{\Gamma(1+2\alpha)}.\end{eqnarray}$

将(3.11)-(3.15)式代入(3.10)式可得不等式(3.9),定理得证.

注3.1 类似于推论3.2,在定理3.1和定理3.2中,若\begin{document}$f(a)=f(b)=f\left(\frac{a+b}{2}\right)$\end{document},分别可得新的不等式.

取如下广义均值

$A(a, b)=\frac{a^{\alpha}+b^{\alpha}}{2^{\alpha}};$

$L_{n}(a, b)=\left[\frac{\Gamma(1+n\alpha) }{\Gamma(1+(n+1)\alpha)}\frac{b^{(n+1)\alpha}-a^{(n+1)\alpha}}{(b-a)^{\alpha}}\right]^{1/n}, n\in {\Bbb Z}\backslash\{-1, 0\}, a, b\in\mathbb{R}, a\neq b.$

命题4.1 设\begin{document}$a, b\in\mathbb{R}, a < b, 0 \notin [a, b]$\end{document}且\begin{document}$n\in {\Bbb Z}, |n|\geq2$\end{document}.则对所有的\begin{document}$p, q>1, $\end{document}且\begin{document}$\frac{1}{p}+\frac{1}{q}=1$\end{document}有

(4.1) $\begin{eqnarray} \label{eq: 4.1}&&\left|\frac{1}{6^{\alpha}}\left[2^{\alpha}A(a^{n}, b^{n})+4^{\alpha}[A(a, b)]^{n}\right]-\Gamma(1+\alpha)[L_{n}(a, b)]^{n}\right|\nonumber \\&\leq& \frac{(b-a)^{\alpha}}{2^{\alpha(1+\frac{2}{q})} (\Gamma(1+\alpha))^{\frac{1}{q}}}\left(\frac{2^{\alpha}(2^{(p+1)\alpha}+1)}{6^{(p+1)\alpha}}\frac{\Gamma(1+p\alpha)}{\Gamma(1+(p+1)\alpha)}\right)^{\frac{1}{p}} \frac{\Gamma(1+n\alpha)}{\Gamma(1+(n-1)\alpha)} \nonumber\\&&\times \Bigg[\Big(3^{\alpha}b^{(n-1)\alpha q}+ a^{(n-1)\alpha q}\Big)^{\frac{1}{q}}+\Big(3^{\alpha}a^{(n-1)\alpha q}+ b^{(n-1)\alpha q}\Big)^{\frac{1}{q}} \Bigg]. \end{eqnarray}$

证 在推论3.1中,取\begin{document}$f(x)=x^{n\alpha}, x\in\mathbb{R}, n\in{\Bbb Z}, n\geq2$\end{document},由文献[13]可知\begin{document}$f(x)$\end{document}是一个广义凸函数,则

$\begin{eqnarray*} & &\left|\frac{1}{6^{\alpha}}\left[f(a)+f(b)+4^{\alpha}f\left(\frac{a+b}{2}\right)\right]-\frac{\Gamma(1+\alpha)}{(b-a)^{\alpha}} {}_{a}I_{b}^{(\alpha)}f(x)\right|\nonumber \\ &=&\left|\frac{1}{6^{\alpha}}\left[2^{\alpha}A(a^{n}, b^{n})+4^{\alpha}[A(a, b)]^{n}\right]-\Gamma(1+\alpha)[L_{n}(a, b)]^{n}\right| \nonumber \\ &\leq&\frac{(b-a)^{\alpha}}{2^{\alpha(1+\frac{2}{q})} (\Gamma(1+\alpha))^{\frac{1}{q}}}\left(\frac{2^{\alpha}(2^{(p+1)\alpha}+1)} {6^{(p+1)\alpha}}\frac{\Gamma(1+p\alpha)}{\Gamma(1+(p+1)\alpha)}\right) ^{\frac{1}{p}}\nonumber \\ && \times\Bigg[\left( 3^{\alpha}|f^{(\alpha)}(b)|^{q}+ |f^{(\alpha)}(a)| ^{q}\right)^{\frac{1}{q}}+\left(3^{\alpha}|f^{(\alpha)}(a)|^{q}+ |f^{(\alpha)}(b)|^{q}\right)^{\frac{1}{q}}\Bigg]\nonumber \\&=& \frac{(b-a)^{\alpha}}{2^{\alpha(1+\frac{2}{q})} (\Gamma(1+\alpha))^{\frac{1}{q}}}\left(\frac{2^{\alpha}(2^{(p+1)\alpha}+1)}{6^{(p+1)\alpha}}\frac{\Gamma(1+p\alpha)}{\Gamma(1+(p+1)\alpha)}\right)^{\frac{1}{p}} \frac{\Gamma(1+n\alpha)}{\Gamma(1+(n-1)\alpha)} \nonumber\\&&\times \Bigg[\Big(3^{\alpha}b^{(n-1)\alpha q}+ a^{(n-1)\alpha q}\Big)^{\frac{1}{q}}+\Big(3^{\alpha}a^{(n-1)\alpha q}+ b^{(n-1)\alpha q}\Big)^{\frac{1}{q}} \Bigg]. \end{eqnarray*}$

结论得证.

本文中提出的不等式在局部分数积分的数值积分中有重要的应用.令\begin{document}$I_{n}$\end{document}为区间\begin{document}$[a, b], 0 < a < b$\end{document},上的一个分划,且\begin{document}$I_{n}:a=x_{0} < x_{1} < \cdot\cdot\cdot < x_{n}=b$\end{document},考虑下面的局部分数阶求积公式

(5.1) $\begin{eqnarray} \label{5.1}{}_{a}I_{b}^{(\alpha)}f(x)=T(f, I_{n})+E(f, I_{n}), \end{eqnarray}$

其中, \begin{document}$T(f, I_{n})$\end{document}为逼近局部分数阶积分的推广的Simpson型近似计算公式

$T(f, I_{n})=\frac{1}{6^{\alpha}\Gamma(1+\alpha)}\sum\limits_{i = 0}^{n-1} (x_{i+1}-x_{i})^{\alpha}\left[f(x_{i})+f(x_{i+1})+4^{\alpha}f(\frac{x_{i}+x_{i+1}}{2})\right], $

因此, \begin{document}$|E(f, I_{n})|=\left| T(f, I_{n})-{}_{a}I_{b}^{(\alpha)}f(x)\right|$\end{document}即为积分\begin{document}${}_{a}I_{b}^{(\alpha)}f(x)$\end{document}近似计算公式的误差.

命题5.1 设\begin{document}$f:[a, b]\subseteq\mathbb{R} \rightarrow\mathbb{R} ^{\alpha}$\end{document}使得\begin{document}$f\in D_{\alpha}(a, b)$\end{document}且\begin{document}$f\in C_{\alpha}[a, b]$\end{document},对于区间\begin{document}$[a, b]$\end{document}上的任意一个分划\begin{document}$I_{n}$\end{document},由公式(5.1)确定的数值积分,余项\begin{document}$E(f, I_{n})$\end{document}满足如下不等式

(5.2) $ \begin{eqnarray}|E(f, I_{n})| &\leq&\frac{1}{2^{\alpha(1+\frac{2}{q})} (\Gamma(1+\alpha))^{1+\frac{1}{q}}}\left(\frac{2^{\alpha}(2^{(p+1)\alpha}+1)}{6^{(p+1)\alpha}}\frac{\Gamma(1+p\alpha)}{\Gamma(1+(p+1)\alpha)}\right)^{\frac{1}{p}}\nonumber\\ &&\times \sum\limits_{i = 0}^{n-1}(x_{i+1}-x_{i})^{2\alpha} \bigg[\Big(3^{\alpha}|f^{(\alpha)}(x_{i+1})|^{q}+ |f^{(\alpha)}(x_{i})|^{q}\Big)^{\frac{1}{q}} \\ &&+\Big(3^{\alpha}|f^{(\alpha)}(x_{i})|^{q} + |f^{(\alpha)}(x_{i+1})|^{q}\Big)^{\frac{1}{q}} \bigg] . \end{eqnarray}$

证 根据推论3.1,对于分划\begin{document}$I_{n}$\end{document}的每一个子区间\begin{document}$[x_{i}, x_{i+1}]$\end{document}\begin{document}$(i=0, \cdot\cdot\cdot, n-1)$\end{document}上,有

$ \begin{eqnarray*} & &\left| \frac{(x_{i+1}-x_{i})^{\alpha}}{6^{\alpha}\Gamma(1+\alpha)}\left[f(x_{i})+f(x_{i+1})+4^{\alpha}f(\frac{x_{i}+x_{i+1}}{2})\right]- {}_{x_{i}}I_{x_{i+1}}^{\alpha}f(x)\right | \nonumber \\&\leq&\frac{(x_{i+1}-x_{i})^{2\alpha}}{2^{\alpha(1+\frac{2}{q})} (\Gamma(1+\alpha))^{1+\frac{1}{q}}}\left(\frac{2^{\alpha}(2^{(p+1)\alpha}+1)}{6^{(p+1)\alpha}}\frac{\Gamma(1+p\alpha)}{\Gamma(1+(p+1)\alpha)}\right)^{\frac{1}{p}}\nonumber\\ &&\times \Bigg[\Big(3^{\alpha}|f^{(\alpha)}(x_{i+1})|^{q}+ |f^{(\alpha)}(x_{i})|^{q}\Big)^{\frac{1}{q}}+\Big(3^{\alpha}|f^{(\alpha)}(x_{i})|^{q} + |f^{(\alpha)}(x_{i+1})|^{q}\Big)^{\frac{1}{q}} \Bigg] . \end{eqnarray*}$

当\begin{document}$i$\end{document}从\begin{document}$0$\end{document}到\begin{document}$n-1$\end{document}对上式两边求和,并由三角不等式,得

$\begin{eqnarray*}&&\left| T(f, I_{n})-{}_{a}I_{b}^{(\alpha)}f(x)\right|\nonumber \\&=&\left|\frac{1}{6^{\alpha}\Gamma(1+\alpha)}\sum\limits_{i = 0}^{n-1} (x_{i+1}-x_{i})^{\alpha}\left[f(x_{i})+f(x_{i+1})+4^{\alpha}f(\frac{x_{i}+x_{i+1}}{2})\right]- \sum\limits_{i = 0}^{n-1} {}_{x_{i}}I_{x_{i+1}}^{(\alpha)}f(x)\right | \nonumber \\&\leq&\frac{1}{2^{\alpha(1+\frac{2}{q})} (\Gamma(1+\alpha))^{1+\frac{1}{q}}}\left(\frac{2^{\alpha}(2^{(p+1)\alpha}+1)}{6^{(p+1)\alpha}}\frac{\Gamma(1+p\alpha)}{\Gamma(1+(p+1)\alpha)}\right)^{\frac{1}{p}}\nonumber\\ &&\times \sum\limits_{i = 0}^{n-1}(x_{i+1}-x_{i})^{2\alpha} \Bigg[\Big(3^{\alpha}|f^{(\alpha)}(x_{i+1})|^{q}+ |f^{(\alpha)}(x_{i})|^{q}\Big)^{\frac{1}{q}}+\Big(3^{\alpha}|f^{(\alpha)}(x_{i})|^{q} + |f^{(\alpha)}(x_{i+1})|^{q}\Big)^{\frac{1}{q}} \Bigg] . \end{eqnarray*}$

结论得证.