我们依靠以下引理来展示我们的其他主要发现。基于这一发现,我们发展了各种积分不等式,这些不等式与比例Caputo-hybrid算子的Hermite–Hadamard型不等式的左侧有关。
引理1
让 \(\psi:I\subet\mathbb{R}^{+}\rightarrow\mathbb{R}\) 是上的二次可微函数 \(我^{o}\),间隔I的内部,哪里 \(I^{o}中的\ zeta,\ vartheta) 令人满意的 \(\zeta<\vartheta),然后让 \(L_{1}[\zeta,\vartheta]\中的\psi,\psi^{\prime},\ps2^{\prime}\).然后满足以下身份:
$$开始{对齐}和\varrho^{2}(\vartheta-\zeta)^{\varrho+1}2^{-\varrho2}\biggl[\int_{0}^{\frac{1}{2}}\mathfrak{s}\psi^{\prime}\bigl \frac{1}{2}}^{1}(\mathfrak{s} -1个)\psi^{prime}\bigl(\mathfrak{s}\zeta+(1-\mathfrak{s})\vartheta\bigr)\,d\mathfrak{s}\ biggr]\\&\quad\quad{}+(1-\ varrho)(\vartheta-\zeta)^{2-\varrho}2^{\varrho-3}\\&\quid ^{2-2\varrho}-1\biger)\biggl[\psi^{\prime\prime}\biggl(\frac{1+\mathfrak{s}}{2}\zeta+\frac{1-\mathfrak{s}}{2}\vartheta\biggr)-\psi^{\prime\prime}\biggl(\frac}1-\matchfrak{s}}{2}\zeta+\fracc{1+\mathfrak{s}{2{2\vartheta\figgr)\biggr]\,d\mathfrak{s}\&quad=-\varrho^{2}(\vartheta-\zeta)^{\varrho}2^{-\varrho+1}\psi\biggl(\frac{\zeta+\vartheta}{2}\biggr)-(1-\varrho)(\vartheta-\zeta)^{1-\varrro}2^{\varrho-1}\psi^{\prime}\biggl^{个人电脑}D_{(\frac{\zeta+\vartheta}{2})}^{\varrho}\psi\biggl(\frac{\zeta+\varheta}{2}\biggr)+{}_{\vartheta ^{-}}^{个人电脑}D_{(\frac{\zeta+\vartheta}{2})}^{\varrho}\psi\biggl(\frac{\zeta+\varheta}{2}\biggr)\biggr]。\结束{对齐}$$
(2)
证明
通过部件集成,我们
$$\开始{aligned}&\int_{0}^{1}(\mathfrak{s} -1个)\psi^{prime}\biggl(\frac{1-\mathfrak{s}}{2}\zeta+\fracc{1+\mathfrak{s}{2{\vartheta\biggr)\,d\mathfrak{s}\\&\quad=\frac}2}{\vartheta-\zeta}\psi\biggal(\frac{\zeta+/vartheta}{2neneneep \bigger)-\frac[2]{\varheta-\ziggr ta}\int_{0}^{1}\psi\biggl(\frac{1-\mathfrak{s}}{2}\zeta+\frac}1+\mathfrak{s}{2neneneep \vartheta\biggr)\,d\mathfrac{s}\结束{对齐}$$
和
$$\begon{aligned}&&int _{0}^{1}\bigl(\mathfrak{s}^{2-2\varrho}-1\bigr)\psi^{\prime\prime}\biggl(\frac{1-\mathfrak{s}}{2}\zeta+\frac{1+\mathfrak{s}}{2}\vartheta\biggr)\,d\mathfrak{s}\\&&quad=\frac{2}{\vartheta-\zeta}\psi^{\prime}\biggl(\frac a+\vartheta}{2}\biggr)-\frac{4(1-\varrho)}{\vartheta-\zeta}\int _{0}^{1}\矩阵{s}^{1-2\varrho}\psi^{prime}\biggl(\frac{1-\mathfrak{s}}{2}\zeta+\fracc{1+\mathfrak{s}{2{vartheta\biggr)\,d\mathfrak{s}。\结束{对齐}$$
通过利用变量的变化,将结果乘以\(\varrho^{2}(\vartheta-\zeta)^{\varrho+1}2^{-\varrho-1})和\((1-\varrho)(\vartheta-\zeta)^{2-\varrho}2^{\varrho-3}),并将其并排合并,我们获得以下结果:
$$\begin{aligned}&\varrho^{2}(\vartheta-\zeta)^{\varrho+1}2^{-\varrho-1}\int _{0}^{1}(\mathfrak{s} -1个)\psi^{prime}\biggl(\frac{1-\mathfrak{s}}{2}\zeta+\fracc{1+\mathfrak{s}{2{\vartheta\biggr)\,d\mathfrak{s}\\&\quad\quadro{}+(1-\varrho)(\vartheta-\zeta)^{2-\varrho}2^{\varrho-3}\int_{0}^{1}\bigl(\mathbrak{s}^{2-2\varrho}-1\biger)\psi^{prime\prime}\biggl(\frac{1-\mathfrak{s}}{2}\zeta+\fracc{1+\mathfrak{s}{2{\vartheta\biggr)\,d\mathfrak{s}\\&\quad=\varrho^{2}c{\zeta+\vartheta}{2}\biggr)\\&\quad\quad{}-\frac{2^{1-\varrho}}{(\vartheta-\zeta)^{1-\ varrho{}\int_{\frac{\zeta+\vartheta}{2}}^{\vartheta}\biggl[\varrho^{2}\bigl \psi^{\prime}(\tau)\biggr]\\&\quad\quad{}\times\biggl(\tau-\frac{\zeta+\vartheta}{2}\biggr)^{-\varrho}\,迪托。\结束{对齐}$$
(3)
使用类似方法得出的另一个结果如下:
$$\begin{aligned}&\varrho^{2}(\vartheta-\zeta)^{\varrho+1}2^{-\varrho-1}\int _{0}^{1}(\mathfrak{s} -1个)\psi^{prime}\biggl(\frac{1+\mathfrak{s}}{2}\zeta+\fracc{1-\mathfrak{s}{2{\vartheta\biggr)\,d\mathfrak{s}\\&\quad\quadro{}+(1-\varrho)(\vartheta-\zeta)^{2-\varrho}2^{\varrho-3}\int_{0}^{1}\bigl(\mathbrak{s}^{2-2\varrho}-1\biger)\psi^{prime\prime}\biggl(\frac{1+\mathfrak{s}}{2}\zeta+\frac{1-\mathfrak{s}{2{\vartheta\biggr)\,d\mathfrak{s}\\&\quad=-\varrho^{2}c{\zeta+\vartheta}{2}\biggr)\\&\quad\quad{}+\frac{2^{1-\varrho}}{(\vartheta-\zeta)^{1-\ varrho{}\int_{\zeta}^{\frac{\zeta+\vartheta}{2}}\biggl[\varrho^{2}\bigl(\frac}\zeta+\vartheta}{2}-\tau\biggr)^{\varrho}\psi(\tau)+(1-\varrho)^{2}\biggl(\frac{\zeta+\vartheta}{2}-\tau\biggr)^{1-\varrho}\psi^{\prime}(\tau)\biggr]\\&\quad\quad{}\times\biggl(\frac{\zeta+\vartheta}{2}-\tau\biggr)^{-\varrho}\,d\tau。\结束{对齐}$$
(4)
减法(4)来自(三),我们得到
$$\开始{对齐}和\varrho^{2}(\vartheta-\zeta)^{\varrho+1}2^{-\varrho-1}\int_{0}^{1}(\ mathfrak{s} -1个)\biggl[\psi^{\prime}\biggl d\mathfrak{s}\\&\quad\quad{}+(1-\varrho)(\vartheta-\zeta)^{2-\varrho}2^{\varrho-3}\\&\ quad\quid{}\times\int_{0}^{1}\bigl(\mathfrak{s}^{2-2\varrho}-1\biger+\mathfrak{s}}{2}\vartheta\biggr)\biggr]\,d\mathfrack{s}\\&\quad=-\varrho^{2}(\vartheta-\zeta)^{\varrho}2^{-\varrho+1}\psi\biggl(\frac{\zeta+\vartheta}{2}\biggr)-(1-\varrho)(\vartheta-\zeta){\varrho-1}(\vartheta-\zeta)^{-\varrho+1}}\biggl[{}_{\zeta^{+}}^{个人电脑}D_{(\frac{\zeta+\vartheta}{2})}^{\varrho}\psi\biggl(\frac{\zeta+\varheta}{2}\biggr)+{}_{\vartheta ^{-}}^{个人电脑}D_{(\frac{\zeta+\vartheta}{2})}^{\varrho}\psi期((的(。\结束{对齐}$$
因此,利用平等
$$\开始{aligned}&\int_{0}^{1}(\mathfrak{s} -1个)\biggl[\psi^{\prime}\biggl d\mathfrak{s}\\&\quad=4\biggl[\int_{0}^{\frac{1}{2}}\mathfrak{s}\psi^{\prime}\bigl(\mathfrak{s}\ zeta+(1-\mathbrak{s{)\vartheta\biger)\,d\mathfrak{s}+\int_{\frac{1}{2}}^{1}(\mathfrak{s} -1个)\psi^{\prime}\bigl(\mathfrak{s}\zeta+(1-\mathfrak{s})\vartheta\bigr)\,d\mathfrack{s}\ biggr],\end{aligned}$$
我们得出了证据的结论。□
备注1
通过将限制视为\(\varrho\rightarrow 1\)在引理中1,可以推断出
$$\begin{aligned}&&\frac{1}{\vartheta-\zeta}\int _{\zeta}^{\vartheta}\psi(\varkappa)\,d\varkappa-\psi\biggl(\frac{\zeta+\vartheta}{2}\biggr)\\\quad=(\vartheta-\zeta)\biggl[\int _{0}^{\frac{1}{2}}\mathfrak{s}\psi ^{\prime}\bigl(\mathfrak{s}\zeta+(1-\mathfrak{s}\zeta+frak{s})\vartheta\bigr)\,d\mathfrak{s}+\int _{\frac{1}{2}}^{1}(\mathfrak{s} -1个)\psi^{\prime}\bigl(\mathfrak{s}\zeta+(1-\mathfrak{s})\vartheta\bigr)\,d\mathfrack{s}\ biggr],\end{aligned}$$
如Körmac所示[25].
推论1
当我们考虑 ϱ 接近0在引理中 1,我们发现了
$$\开始{对齐}&\frac{(\vartheta-\zeta)^{2}}{8}\biggl(\int_{0}^{1}\bigl(\mathfrak{s}^{2}-1\bigr)\biggl[\psi^{\prime\prime}\biggl(\frac{1+\mathfrak{s}}{2}\zeta+\frac}1-\mathbrak{s{}{2{\vartheta\biggr)-\psiqu{\prime \prime}\bigl)\biggr]\,d\mathfrak{s}\biggr)\\&\quad=-\frac{(\vartheta-\zeta)}{2}\psi^{\prime}\bigl(\frac}\zeta+\vartheta}{2} \biggr)+\bigl[\psi(\vartheta)-\psi(\ zeta)\bigr]\\&\quad\quad{}+\frac{2}{\vartheta-\zeta}\biggl(\int_{\zeta{^{\frac{\zeta+\varthetta}{2}\psi(\farkappa)\,d\varkappa-\int_}{\frac:zeta+\varheta}{2{}^{\varheta}\psi pa)\,d\varkappa\biggr)。\结束{对齐}$$
此外,通过选择 \(\varrho=\压裂{1}{2}\),平等(2)采取形式
$$开始{aligned}和\frac{1}{\vartheta-\zeta}\biggl\{-\psi\biggl(\frac}\zeta+\vartheta}{2}\bigr)-\psi^{\prime}\bigl)\,d\varkappa+\psi(\vartheta)-\psi(\ zeta)\biggr]\biggr\}\\&\quad=\int_{0}^{\frac{1}{2}}\mathfrak{s}\psi^{\prime}\bigl(\mathfrak{s}\ zeta+(1-\mathfrak{s})\vartheta\bigr)\,d\mathfrake{s}+\int_{\frac{1}{2}}^{1}(\mathfrak{s} -1个)\psi ^{\prime}\bigl(\mathfrak{s}\zeta+(1-\mathfrak{s})\vartheta\bigr)\,d\mathfrak{s}\\&&quad\quad{}+\frac{1}{4}\int _{0}^{1}(\mathfrak{s} -1个)\biggl[\psi^{\prime\prime}\biggl]\,d\mathfrak{s}。\结束{对齐}$$
定理4
让 \(\psi:I\subset\mathbb{R}^{+}\rightarrow\mathbb{R}\) 是上的二次可微函数 \(我^{o}\),区间I的内部,哪里 \(I^{o}中的\ zeta,\ vartheta) 令人满意的 \(\zeta<\vartheta),然后让 \(L_{1}[\zeta,\vartheta]\中的\psi,\psi^{\prime},\ps2^{\prime}\).如果 \(|\psi^{\prime}|^{q}\) 和 \(|\psi^{\prime\prime}|^{q}\) 是凸的 \([\zeta,\vartheta]\) 对于 \(q \ geq 1 \),那么下面的不等式成立:
$$开始{对齐}&\biggl\vert\frac{\Gamma(1-\varrho)}{2^{\varrho-1}(\vartheta-\zeta)^{-\varrho+1}}\biggl[{}_{\zeta^{+}}^{个人电脑}D_{(\frac{\zeta+\vartheta}{2})}^{\varrho}\psi\biggl(\frac{\zeta+\varheta}{2}\biggr)+{}_{\vartheta ^{-}}^{个人电脑}D_{(\frac{\zeta+\vartheta}{2})}^{\varrho}\psi\biggl-\varrho)(\vartheta-\zeta)^{1-\varrho}2^{\varrho-1}\psi^{\prime}\biggl(\frac{\zeta+\vartheta}{2}\biggr)\biggr\vert\\&\quad\leq\varrho^{2}{\prime}(\vartheta)\bigr\vert\bigr)\\&\quad\quad{}+(1-\varrho)(\vartheta-\zeta)^{2-\varrho}2^{\varrho-3}\biggl\{\biggl(\frac{2-2\varrho}{3-2\varrho}\biggr){2}-\frac{1}{3-2\varrho}-\frac}1}{4-2\varrho}\biggr)\\&\quad\quad{}+\fracc{|\psi^{\prime\prime}(\vartheta)|^{q}}{2}\bigl(\frac{1}{2}-\frac{1}{3-2\varrho}+\frac}1}{4-2\varrho}\biggr)\biggr]^{\frac[1}{q}}+\biggl[\frac{|\psi^{\prime\prime}(\zeta)|^{q}{2}\bigl(\frac{1}{2}-\frac{1}{3-2\varrho}+\frac}1}{4-2\varrho}\biggr)\\&\quad\quad{}+\fracc{|\psi^{\prime\prime}(\vartheta)|^{q}}{2}\bigl(\frac[3]{2}-\压裂{1}{3-2\varrho}-\压裂{1}{4-2\varrho}\biggr)\biggr]^{\frac{1}{q}}\bigr)\bigr\}。\结束{对齐}$$
(5)
证明
首先,考虑以下情况\(q=1).通过使用凸性\(|\psi^{\prime}|\)和\(|\psi^{\prime\prime}|\),它来自引理1那个
$$开始{对齐}&\biggl\vert\frac{\Gamma(1-\varrho)}{2^{\varrho-1}(\vartheta-\zeta)^{-\varrho+1}}\biggl[{}_{\zeta^{+}}^{个人电脑}D_{(\frac{\zeta+\vartheta}{2})}^{\varrho}\psi\biggl(\frac{\zeta+\varheta}{2}\biggr)+{}_{\vartheta ^{-}}^{个人电脑}D_{(\frac{\zeta+\vartheta}{2})}^{\varrho}\psi\biggl-\varrho)(\vartheta-\zeta)^{1-\varrho}2^{\varrho-1}\psi^{\prime}\biggl(\frac{\zeta+\vartheta}{2}\biggr)\biggr\vert\\&\quad\leq\varrho^{2}(\vartheta-\zeta)^{\varrho+1}2^{-\varrho2}\biggl[\int_{0}^{\frac{1}{2}}\mathfrak{s}\bigl psi^{\prime}(\vartheta)\bigr|\biger)\,d\mathfrak{s}\\&\quad\quad{}+\int_{\frac{1}{2}}^{1}(1-\mathfrac{s})\bigl(\mathfrak{s}\bigl|\psi^{\prime}(\zeta)\bigr|+(1-\mathfrak{s})\bigl|\psi^{\prime}(\vartheta)\bigr|\bigr)\,d\mathfrak{s}\biggr]\\&&quad\quad{}+(1-\varrho)(\vartheta-\zeta)^{2-\varrho}2^{\varrho-3}\int _{0}^{1}\bigl(1-\mathfrak{s ^{2-2\varrho}\bigr)\\&&\quad\quad{}\times\biggl(\frac{1-\mathfrak{s}}{2}\bigl\vert\psi^{\prime\prime}(\zeta)\bigr\vert+\frac{1+\mathfrak{s}}{2}\bigl\ vert\psis^{\prime\prime}(\ vartheta)\bigr\vert+\frac{1+\ mathfrak{s}{2{\bigl\svert\psia^{\rime\primer}(\seta)\bigr\vert+\fracc{1-\mathfrak{s}{2}\bigl\\vert\psi ^{\prime\prime}(\vartheta)\bigr\vert\biggr)\,d\mathfrak{s}。\结束{对齐}$$
(6)
然后,因为
$$\int_{0}^{\frac{1}{2}}\mathfrak{s}^{2}\,d\mathfrac{s}=\int_}\frac}{2{1}(1-\mathfrak{s})(1-\mathfrak{s})\,d\mathfrak{s}=\int_{\frac{1}{2}}^{1}$$
和
$$\int_{0}^{1}\bigl(1-\mathfrak{s}^{2-2\varrho}\bigr)\,d\mathfrak{s}=\frac{2-2\warrho}{3-2\varrho}$$
我们得到了不等式右边的表达式(6)是
$$开始{对齐}和\frac{\varrho^{2}(\vartheta-\zeta)^{\varhro+1}2^{-\varrho+1}}{4}\biggl 2^{\varrho-2}\biggl(\frac{2-2\varrho}{3-2\varrho}\bigr)\biggl(\frac{|\psi^{\prime\prime}(\zeta)|+|\psi ^{\prime\prime}(\vartheta)|}{2}\biggr)。\结束{对齐}$$
此外,对于\(q>1),通过使用引理1,幂平均不等式,并考虑\(|\psi^{\prime}|^{q}\)和\(|\psi^{\prime\prime}|^{q}\),我们可以得出结论
$$开始{对齐}&\biggl\vert\frac{\Gamma(1-\varrho)}{2^{\varrho-1}(\vartheta-\zeta)^{-\varrho+1}}\biggl[{}_{\zeta^{+}}^{个人电脑}D_{(\frac{\zeta+\vartheta}{2})}^{\varrho}\psi\biggl(\frac{\zeta+\varheta}{2}\biggr)+{}_{\vartheta ^{-}}^{个人电脑}D_{(\frac{\zeta+\vartheta}{2})}^{\varrho}\psi\biggl-\varrho)(\vartheta-\zeta)^{1-\varrho}2^{\varrho-1}\psi^{\prime}\biggl(\frac{\zeta+\vartheta}{2}\biggr)\biggr\vert\\&\quad\leq\varrho^{2}裂缝{1}{2}}\mathfrak{s}\bigl[\mathfrak{s}\sbigl|\psi^{prime}(\zeta)\bigr|^{q}+(1-\mathflak{s{)\bigl|\si^{prime}\bigr|^{q}\bigr]\,d\mathfrak{s}\biggr ^{1}(1-\mathfrak{s})\bigl[\mathfrak{s}\bigl|\psi^{prime}(\zeta)\bigr|^{q}+(1-\mathfrak}s},d\mathfrak{s}\biggr)^{\frac{1}{q}}\bigr\}\\&\quad\quad{}+(1-\varrho)(\vartheta-\zeta)^{2-\varrho}2^{\varrho-3}\bigl\{\biggl \biggr)^{\frac{1}{p}}\\&\quad\quad{}\times\biggl(\int_{0}^{1}\bigl(1-\mathfrak{s}^{2-2\varrho}\bigr)\biggl[\frac{1+\mathfrak{s}}{2}\bigl|\psi^{\prime\prime}(\zeta)\bigr|^{q}+\fracc{1-\mathfrak{s}{2{\bigl |\psi ^{\prime\prime}{}+\biggl(\int_{0}^{1}\bigl(1-\mathfrak{s}^{2-2\varrho}\biger)\,d\mathfrack{s}\biggr)^{\frac{1}{p}}\\&\quad\quad{}\times\biggl(\int_{0}^{1}\bigl(1-\mathfrak{s}^{2-2\varrho}\biger)\biggl[\frac{1-\matchfrak{s}}{2}\bigl|\psi^{\prime\prime}(\zeta)\bigr|^{q}+\fracc{1+\mathfrak{s}{2{\bigl|\psia^{\prime\prime}(\ vartheta)\bigr|^}q}\biggr]\,d\mathfrak{s}\bigcr)^{\frac{1}{q}}\bigr\}。\结束{对齐}$$
因此,自
$$开始{对齐}&\int_{0}^{\frac{1}{2}}\mathfrak{s}\,d\mathfrak{s}=\int_}\frac}{1}}^{1}(1-\mathfrak{s})}\bigr)(1+\mathfrak{s})\,d\mathfrak{s}=\frac{3}{2}-\frac{1}{3-2\varrho}-\frac{1}{4-2\varrho},\end{aligned}$$
和
$$\int_{0}^{1}\bigl(1-\mathfrak{s}^{2-2\varrho}\bigr)(1-\mathfrak}s})\,d\mathfrak{s}=\frac{1}{2}-压裂{1}{3-2\varrho}+\压裂{1{4-2\varrho}$$
不平等(5)我们旨在证明的观点是正确的。□
备注2
通过将极限作为\(\varrho\rightarrow 1\)和设置\(q=1)在定理中4,我们发现
$$\biggl\vert\frac{1}{\vartheta-\ zeta}\int_{\zeta}^{\varheta}\psi(\varkappa)\,d\varkappa-\psi\biggl(\frac{\zeta+\ vartheta}{2}\biggr)\biggr\vert\leq\frac}(\vartheta-\zeta)}{8}\bigl(\bigl|\psi^{\prime}(\ zeta)\bigr|+\bigl| psi^{\prime}(\vartheta)\bigr|\bigr)$$
这一点已被K跏rmac跔证明[25].
推论2
作为 ϱ 方法0和用于 \(问题1) 在定理中 4,我们获得
$$开始{aligned}&\biggl\vert\psi(\vartheta)-\psi(\zeta)+\frac{2}{\vartheta-\zeta}\biggl(\int_{\zeta{^{\frac}\zeta+\varthetar}{2}}\psi(\ varkappa)\,d\varkappa-\int_}{\frac{\zeta+\varheta}{2{}^{\varheta}\psi \biggr)-\frac{(\vartheta-\zeta)}{2}\psi^{prime}\biggl(\frac}\zeta+\vartheta)}}\biggr)\biggr\vert\\&\quad\leq\frac{(\vartheta-\zeta)^{2}}{8}\biggl(\frac}{3}\bigr)^{\frac\q-1}{q}}\bigl[\biggl[\frac[11}{12}\bigl\vert\psi^{\prime\prime}(\zeta \psi^{\prime\prime}(\vartheta)\bigr\vert^{q}\biggr)^{1/q}\\&\quad\quad{}+\biggl(\frac{5}{12}\bigl\vert\psi^{\prime\prime}(\zeta)\bigr\vert^{q}+\frac{11}{12}\bigl\ vert\psia^{\prime\prime}(\ vartheta)\bigr\vert ^{q{\biggr)^{1/q}\biggr]。\结束{对齐}$$
此外,作为 ϱ 倾向于1和用于 \(问题1),定理中所述的不等式 4简化为
$$开始{aligned}和\biggl\vert\frac{1}{\vartheta-\zeta}\int_{\zeta{^{\varheta}\psi(\varkappa)\,d\varkappa-\psi\biggl(\frac{\zeta+\vartheta}{2}\biggr)\biggr\vert\&\quad\leq\frac}(\varthetta-\zeta)}{8}\bigl(\frac{1+2^{\frac}{q}}}{3^{\frac{1}{q{}}\biggr)\bigl(\bigl\vert\psi^{\prime}(\zeta)\bigr\vert+\bigl\vert\psi^{\prime}(\vartheta)\bigr\ vert\biger)。\结束{对齐}$$
现在,为了说明我们的定理的适用性,我们给出了一个例子。
示例1
如果我们使用函数\(\psi(\varkappa)=2\varkappa^{3})在间隔上定义\([0,2]\),我们可以计算不等式的右边(5)按以下方式:
$$\begin{aligned}&12\varrho^{2}\biggl \biggl(\frac{1}{2}-压裂{1}{3-2\varrho}+\frac{1}{4-2\varrho}\biggr)^{压裂{1{q}}+\biggl(压裂{3}{2}-\压裂{1}{3-2\varrho}-\frac{1}{4-2\varrho}\biggr)^{\frac}1}{q}}\bigbr]。\结束{对齐}$$
此外,我们还表明
$$\begin{aligned}&\biggl\vert\frac{\Gamma(1-\varrho)}{2^{\varrho-1}(\vartheta-\zeta)^{-\varrho+1}}\biggl[{}_{\zeta^{+}}^{个人电脑}D_{(\frac{\zeta+\vartheta}{2})}^{\varrho}\psi\biggl(\frac{\zeta+\varheta}{2}\biggr)+{}_{\vartheta ^{-}}^{个人电脑}D_{(\frac{\zeta+\vartheta}{2})}^{\varrho}\psi\biggl-\varrho)(\vartheta-\zeta)^{1-\varrho}2^{\varrho-1}\psi^{\prime}\biggl(\frac{\zeta+\vartheta}{2}\biggr)\biggr\vert\\&\quad=4\varrho^{2}+\frac{6(1-\varrho)^{2{}{2-\varrhoS}。\结束{对齐}$$
图1清楚地证明了不等式的左边(5)始终位于这个不等式的右侧下方\(\varrho\ in(0,1)\)和\(q \ geq 1 \).
定理5
让 \(\psi:I\subset\mathbb{R}^{+}\rightarrow\mathbb{R}\) 是上的二次可微函数 \(我^{o}\),间隔I的内部,哪里 \(I^{o}中的\ zeta,\ vartheta) 令人满意的 \(\zeta<\vartheta),然后让 \(L_{1}[\zeta,\vartheta]\中的\psi,\psi^{\prime},\ps2^{\prime}\).如果 \(|\psi^{\prime}|^{q}\) 和 \(|\psi^{\prime\prime}|^{q}\) 是凸的 \([\zeta,\vartheta]\) 对于 \(q>1),那么下面的不等式成立:
$$开始{对齐}&\biggl\vert\frac{\Gamma(1-\varrho)}{2^{\varrho-1}(\vartheta-\zeta)^{-\varrho+1}}\biggl[{}_{\zeta^{+}}^{个人电脑}D_{(\frac{\zeta+\vartheta}{2})}^{\varrho}\psi\biggl(\frac{\zeta+\vartheta}{2}\biggr)+{}_{\vartheta ^{-}^{个人电脑}D_{(\frac{\zeta+\vartheta}{2})}^{\varrho}\psi\biggl-\varrho)(\vartheta-\zeta)^{1-\varrho}2^{\varrho-1}\psi^{\prime}\biggl(\frac{\zeta+\vartheta}{2}\biggr)\biggr\vert\\&\quad\leq\frac{\varrho^{2})|^{q}}{8}\biggr{q} }\biggr]\\&\quad\quad{}+(1-\varrho)ta)|^{q}+3|\psi^{prime\prime}(\vartheta)||{q}}{4}\biggr)^{frac{1}{q}}+\biggl(\frac{3|\psi^{\prime\prime}(\zeta)|^{q}+|\ps2^{\prime\prime}(\ vartheta)||^{q}}{4}\biggr)^{\frac}1}{q}}\bigr],\end{aligned}$$
(7)
哪里 \(压裂{1}{p}+压裂{1{q}=1\).
证明
使用引理中的绝对值1并应用著名的Hölder不等式和\(|\psi^{\prime}|^{q}\),\(|\psi^{\prime\prime}|^{q}\),我们得到
$$开始{对齐}&\biggl\vert\frac{\Gamma(1-\varrho)}{2^{\varrho-1}(\vartheta-\zeta)^{-\varrho+1}}\biggl[{}_{\zeta^{+}}^{个人电脑}D_{(\frac{\zeta+\vartheta}{2})}^{\varrho}\psi\biggl(\frac{\zeta+\varheta}{2}\biggr)+{}_{\vartheta ^{-}}^{个}D_{(\frac{\zeta+\vartheta}{2})}^{\varrho}\psi\biggl(\frac{\zeta+\varheta}{2\biggr)\biggr]\end{aligned}$$
(8)
$$开始{对齐}和\quad\quad{}-\varrho^{2}θ}{2}\biggr)\biggr\vert\\&\quad\leq\varrho^{2}(\vartheta-\zeta)^{\varrho+1}2^{-\varrho2}\biggl[\biggl(\int_{0}^{\frac{1}{2}}\mathfrak{s}^{p}\,d\mathfrac{s}\biggr)-\mathfrak{s})\bigl\vert\psi^{prime}(\vartheta)\bigr\vert^{q}\bigr]\,d\mathfrak{s}\biggr)^{\frac{1}{q}}\\&\quad\quad{}+\biggl(\int_{\frac{1}{2}}^{1}(1-\mathfrak{s})^{p}\,d\mathfrac{s}\biggr)-\mathfrak{s})\bigl\vert\psi^{prime}(\vartheta)\bigr\vert^{q}\bigr]\,d\mathfrak{s}\biggr)^{\frac{1}{q}}\bighr]\\&\quad\quad{}+(1-\varrho)(\vartheta-\zeta)^{2-\varrho}2^{\varrho-3}\\&\quad\quad{}\times\biggl[\biggl(\int_{0}^{1}\bigl[1-\mathfrak{s}^{2-2\varrhoS}\bigr]^{p}\,d\mathfrak{s}\biggr)^{\frac{1}{p}\bigbl(\int _{0}^{1}\biggl[\frac{1+\mathfrak{s}}{2}\bigl\vert\psi^{prime\prime}(\zeta)\bigr\vert^{q}+\frac}1-\mathfrak{s}{2{\bigl\vert\psi^{\prime\prime}(\vartheta)\bigr\vert^{q}\biggr]\,d\mathfrak{s}\bigr)^{frac{1}{q}}\\&\quad\quad{}+\biggl压裂{1}{p}}\biggl(\int_{0}^{1}\bigl[\frac{1-\mathfrak{s}}{2}\bigle\vert\psi^{\prime\prime}(\zeta)\bigr\vert^{q}+\frac{1+\mathfrak{s}}{2}\bigl\vert\psi^{prime\prime}(\vartheta)\bigr\fort^{q{\biggr]\,d\mathfrak{s}\biggr)^{\frac}1}{q}\bigr]。\结束{对齐}$$
(9)
上述不等式中的积分可以通过
$$开始{aligned}&\int_{0}^{\frac{1}{2}}\mathfrak{s}^{p}\,d\mathfrak{s}=\int_}\frac}{2{1}}^{1}(1-\mathfrak{s})c{1}{2}}\bigl[\mathfrak{s}\bigle\vert\psi^{prime}(\zeta)\bigr\vert^{q}+(1-\mathfrak{s})\bigl\vert\psi^{prime},d\mathfrak{s}=\frac{\vert\psi^{\prime}(\zeta)\vert^{q}+3\vert\psi^}\prime{(\vartheta)\virt^{q}}{8},\\&\int_{\frac}1}{2}}^{1}\bigl[\mathfrak{s}\bigr\vert\psi{\prime}(\ zeta)\ bigr\verst^q}+(1-\mathfrak{s})\bigl\vert\psi^{prime}(\vartheta)\bigr\vert^{q}\bigr]\,d\mathfrak{s}=\frac{3\vert\psi^{prime}(泽塔)\vert^{q}+\vert\psi^{prime}(\vartheta)\bigr\vert^{q}\biggr]\,d\mathfrak{s}=\frac{|\psi^{\prime\prime}(\zeta)|^{q{+3|\psi^{\prime\prime}(\vartheta)|^{q}}{4},\\&\int_{0}^{1}\biggl[\frac{1+\mathfrak{s}}{2}\bigl\vert\psi^}\prime\trime}vert^{q}\biggr]\,d\mathfrak{s}=\frac{3|\psi^{\prime\prime}(\zeta)|^{q{+|\psi ^{\prime\prime}(\ vartheta)|^{q} {4}。\结束{对齐}$$
此外,使用\((A-B)^{p}\leq A^{p} -B类^{p} \)对于\(A>B \geq 0)和\(第1页),我们有
$$\int _{0}^{1}\bigl[1-\mathfrak{s}^{2-2\varrho的的一个中,一一之前在后$$
因此,通过将计算的积分结果替换为不等式(8),可以达到预期的结果。□
备注3
在特殊情况下,其中ϱ定理中的方法15,我们获得
$$\开始{aligned}&\biggl\vert\frac{1}{\vartheta-\zeta}\int_{\zeta{^{\varheta}\psi(\varkappa)\,d\varkappa-\psi\biggl(\frac{\zeta+\vartheta}{2}\biggr)\biggr\vert\&\quad\leq\frac}(\varthetan-\zeta)}{16}\bigl(\frac{4}{p+1}\ biggr)^{\frac{1}{p}}\bigl\{\bigl[\bigl|\psi^{\prime}(\zeta)\bigr|^{q}+3\bigl |\psi ^{\prime}(\vartheta)\bigr|^{q}\bigr]^{\frac{1}{q}}+\bigl[3\bigl|\psi^{\prime}(\zeta)\bigr|^{q}+\bigl|\psi^{\prime}(\vartheta)\bigr|^{q}\bigr]^{\frac{1}{q}}\bigr},\end{aligned}$$
这一点在年被Körmac证明了[25].
推论3
在特殊情况下 ϱ 倾向于0在定理中 5,我们获得
$$\begl{aligned}&\biggl\vert\psi(\vartheta)-\psi(\zeta)+\frac{2}{\vartheta-\zeta}\biggl(\int _{\zeta}^{\frac{\zeta+\vartheta}{2}})\psi(\varkappa)\,d\varkappa-\int _{\frac{\zeta+\vartheta}{2}}^{\vartheta}\psi(\varkappa)\,d\varkappa\biggr)-\frac{(\vartheta-\zeta){2}\psi^{\prime}\biggl(\frac{\zeta+\vartheta}{2}\biggr)\biggr\vert\\&\quad\leq\frac{(\vartheta-\zeta)^{2}}{8}\biggl(\frac}{2p+1}\bigr)^{frac{1}{p}}\bigbl[\biggl[\frac(\frac{|\psi^{\prime\prime}(\zeta}\biggr)^{\frac{1}{q}}+\biggl(\frac{3|\psi^{\prime\prime}(\zeta)|^{q}+|\psi ^{\prime\prime}(\ vartheta)||^{q}}{4}\biggr)^{\frac{1}{q}}\biggr]。\结束{对齐}$$
此外,通过选择 \(\varrho=\压裂{1}{2}\),我们得到
$$开始{aligned}和\biggl\vert\frac{1}{\vartheta-\zeta}\biggl[\int_{\zetaneneneep ^{\varheta}\psi(\varkappa)\,d\varkappa+\psi+\vartheta}{2}\biggr)\biggr]\biggr\vert\\&\quad\leq\frac{(\vartheta-\zeta)}{2} \biggl(\frac{1}{2(p+1)}\biggr)^{\frac}{p}}\bigbl[\biggl[(\frac{|\psi^{prime}(\zeta)|^{q}+3|\psi ^{prime}|^{q}+|\psi^{\prime}(\vartheta)|^{q}}{8}\biggr)^{\frac{1}{q}}\bigr]\\&\quad\quad{}+\frac}\vartheta-\zeta}{4}\bigl(\frac[p}{p+1}\biggr)^{frac{1}{p}\bigbl[\biggl(\frac{|\psi^{prime\prime}(\zeta)|^{q}+3|\ psi^}\prime\prime}psi^{prime\prime}(\vartheta)|^{q}}{4}\biggr)^{frac{1}{q}{biggr]。\结束{对齐}$$
现在,我们给出一个例子来证明定理中建立的不等式的有效性5为了说明这一点。
示例2
考虑到功能ψ示例中定义1不等式右侧的表达式(7)可按如下方式进行评估:
$$24\biggl(压裂{1+3^{压裂{p-1}{p}}{4^{裂缝{p}{}}\biggr)+1}\biggr)^{\frac{1}{p}}\bigbr]$$
另一方面,我们知道
$$开始{对齐}&\biggl\vert\frac{\Gamma(1-\varrho)}{2^{\varrho-1}(\vartheta-\zeta)^{-\varrho+1}}\biggl[{}_{\zeta^{+}}^{个人电脑}D_{(\frac{\zeta+\vartheta}{2})}^{\varrho}\psi\biggl(\frac{\zeta+\varheta}{2}\biggr)+{}_{\vartheta ^{-}}^{个人电脑}D_{(\frac{\zeta+\vartheta}{2})}^{\varrho}\psi\biggl(\frac{\zeta+\vartheta}{2}\biggr)\biggr]\\&&quad\quad{}-\varrho^{2}(\vartheta-\zeta)^{\varrho}2^{-\varrho+1}\psi\biggl(\frac{\zeta+\vartheta}{2}\biggr)-(1-\varrho)(\vartheta-\zeta)^{1-\varrho rho}2^{\varrho-1}\psi^{\prime}\biggl(\frac{\zeta+\vartheta}{2}\biggr)\biggr\vert\\&\quad=4\varrho^{2}+\frac{6(1-\varrho)^{2{}{2-\varrhoS}。\结束{对齐}$$
因此,从图中可以看出2不等式的左边(7)对于以下所有值,始终低于右侧\(\varrho\ in(0,1)\)和\(p>1\).