在整篇论文中,我们假设\(在(0,1]\)中为α,\(\beta\in\mathbb{R}^{+}\)。
引理1
让 \(\mathcal{F}:[\sigma,\delta]\rightarrow\mathbb{R}\) 是上的可微映射 \((\σ,\δ)\) 具有 \(β>0\) 和 \(在(0,1]\)中为α。如果 \(L[\sigma,\delta]\中的\mathcal{F}^{\prime}\),则以下FCIO身份保持不变以下为:
$$\beart{aligned}&&frac{3^{\alpha\beta-1}\alpha^{\beta}}{(\delta-\sigma)^{\alpha\beta}}}\Gamma(\beta+1)\biggl[_{+}^{\beta}\mathcal{日本}_{\sigma}^{\alpha}\mathcal{F}\biggl(\frac{2\sigma+\delta}{3}\biggr)+_{+}^{\beta}\mathcal{日本}_{\frac{2\sigma+\delta}{3}}^{\alpha}\mathcal{F}\biggl(\frac{\sigma+2\delta{3}\bigr)+_{+}^{\ beta}\mathcal{日本}_{\frac{\sigma+2\delta}{3}}^{\alpha}\mathcal{F}(\delta)\biggr]\\&\qquad{}-\frac}{8}\biggl[\mathcal{F}(\sigma)+3\mathcali{F}\bigbl}\biggr)+\mathcal{F}(\delta)\biggr]\\&\quad=\frac{(\delta-\sigma)\alpha^{\beta}}{9}[I{1}+I{2}+I_{3}],\结束{对齐}$$
(5)
哪里 \(\Gamma(\beta)\) 是Euler Gamma函数
$$\begin{aligned}&I{1}=\int_{0}^{1}\biggl(\biggal(\frac{1-(1-\mu)^{\alpha}{\alfa}\bigr)^}\beta}-\frac}5}{8\alpha^{\beta{}\bigcr)\mathcal{F}^{\prime}\bigl(\mu\sigma+(1-\mo)\biggl(\frac{2\sigma+\delta}{3}\biggr)\bigger)\,d\mu,\\&I_{2}=\int_{0}^{1}\bigl(\biggl(\frac{1-(1-\mu)^{alpha}}{\alpha}\biggr}\biggl(\biggl(\frac{1-(1-\mu)^{\alpha}}{\alfa}\biggr)^{β}-\frac}3}{8\alpha^{\beta}}\bigr)\mathcal{F}^{\prime}\biggl(\mu\biggl(\frac{\sigma+2\delta}{3}\bigr)+(1-\mu)\delta\biggr)\,d\mu。\结束{对齐}$$
证明
使用部件集成和更改变量\(x=\mu\sigma+(1-\mu)(\frac{2\sigma+\delta}{3}),我们获得
$$开始{对齐}I_{1}={}&\int_{0}^{1}\biggl(\biggal(\frac{1-(1-\mu)^{\alpha}}{\alfa}\bigr)^β}-\frac}5}{8\alpha^{\beta}}\bigcr)\mathcal{F}^{\prime}\bigl(\mu\sigma+(1-\mu)\biggl(\frac{2\sigma+\delta}{3}\biggr)\,d\mu\\={}&\frac{3}{\delta-\sigma}\bigl(\biggl(\frac}1-(1-\mu)^{\alpha}}{\alpha}\biggr)^{\beta}-\frac{5}{8\alpha^{\beta}}\bigr)\mathcal{F}\bigl \biggl(\frac{1-(1-\mu)^{\alpha}}{\alfa}\biggr)^{β-1}(1-\mo)^{\ alpha-1}\mathcal{F}\biggl(\mu\sigma+(1-\mu)\biggl(\frac{2\sigma+\delta}{3}\biggr)\bigr)\,d\mu\={}&-\biggl[\frac}9}{8\alpha^{beta}(\delta-\sigma)}\mathcal{F}(\ sigma压裂{2\sigma+\delta}{3}\biggr)\biggr]\\&{}+\biggl(压裂{3}{\delta-\sigma}\bigr)^{1+\alpha\beta}\beta\int_{\sigma}^{\frac{2\sigma+\delta}{3}\biggl(\frac}(\frac{\delta-\sigma}{3{)\={}&-\biggl[\frac{9}{8\alpha^{\beta}(\delta-\sigma)}\mathcal{F}(\sigma)+\frac{15}{8\alpha^{\beta}(\delta-\sigma)}\mathcal{F}\biggl(\frac{2\sigma+\delta}{3}\bigr)\biggr]\\&{}+\frac}3^{1+\alpha\beta{\Gamma(\beta+1)}{{日本}_{\sigma}^{\alpha}\mathcal{F}\biggl(\frac{2\sigma+\delta}{3}\bigr)\biggr]。\结束{对齐}$$
与上述过程类似,使用\(x=\mu(\frac{2\sigma+delta}{3})+(1-\mu)和\(x=\mu(\frac{\sigma+2\delta}{3})+(1-\mu)\delta\),我们有
$$开始{对齐}I_{2}={}&\int_{0}^{1}\biggl(\biggal(\frac{1-(1-\mu)^{\alpha}}{\alfa}\bigr)^β}-\frac}1}{2\alpha^{\beta}}\bigcr)\mathcal{F}^{\prime}\bigl(\mu\biggl(\frac{2\sigma+\delta}{3}\bighr)+(1-\mu)\biggl(\frac{\sigma+2\delta}{3}\biggr)\bigr)\,d\mu\\={}&-\frac{3}{(\delta-\sigma)}\比格尔(\biggl(\frac{1-(1-\mu)^{\alpha}}{\alfa}\biggr)^{\taba}-\frac}{1}{2\alpha^{\beta}}\bigr)\mathcal{F}\bigl(\mu\biggl(\frac{2\sigma+\delta}{3}\bigcr)+(1-\m)\bigg gr)\bigg\vert_{0}^{1}\\&{}+\frac{3\beta}{(delta-\sigma)}\int_{0}^{1{\biggl(\frac}1-(1-\mu)^{\alpha}}{\ alpha}\biggr)^{\beta-1}(1-\mu)3}{2\alpha^{\beta}(\delta-\sigma)}\mathcal{F}\biggl(\frac{\sigma+2\delta}{3}\bigr)+\frac}3}{2\alpha^{\beta}(\delta-\sigma)}\mathcal{F}\biggl(\frac{2\sigma+\delta}{3}\bigr)\biggr]\\&{}+\biggl \biggl(\frac{(\frac{\delta-\sigma}{3})^{\alpha}-(x-\frac}2\sigma+\delta}{3{)^}\alpha}}{\alpha}\biggr)^{\beta-1}\frac{\mathcal{F}c{3}{2\alpha^{beta}(\delta-\sigma)}\mathcal{F}\biggl(\frac{2\sigma+\delta}{3}\bigr)\biggr]\\&{}+{\frac{3^{1+\alpha\beta}\Gamma(\beta+1)}{(\delta-\sigma)^{1+\alpha\beta}}}\biggl[_{+}^{\beta}\mathcal{日本}_{\frac{2\sigma+\delta}{3}}^{\alpha}\mathcal{F}\biggl(\frac{\sigma+2\delta{3}\bigr)\biggr]\end{aligned}$$
和
$$\beart{aligned}I_{3}={}&{{\frac{3}{\delta-\sigma}\biggl}{\alpha}\biggr)^{\beta}-\frac{3}{8\alpha^{\beta}}\bigr)\mathcal{F}\bigl \biggl(\frac{1-(1-\mu)^{\alpha}}{\alfa}\biggr)^{β-1}(1-\mo)^{\ alpha-1}\mathcal{F}\biggl(\mu\biggl(\frac{\sigma+2\delta}{3}\biggr)+(1-\mu)\delta \biggr)\,d\mu\\={}&-\biggl(\frac{9}{8\alpha^{\beta}(\delta-\sigma)}\mathcal{F}(\delta-\sigma)+\frac{15}{8\alpha^{\beta}(\delta-\sigma)}\mathcal{F}\biggl(\frac{\sigma+2\delta}{3}\biggr)\biggr]\\&{}+\bigl(\frac{3}{\delta-\sigma}\biggr)^{1+\alpha\beta}\beta\int_{\frac{\sigma+2\delta}{3}}^{\delta{\biggl a+2\delta}{3})^{1-\alpha}}\,dx\\={}&-\biggl[\frac{9}{8\alpha^{\beta}(\delta-\sigma)}\mathcal{F}(delta)+\frac{15}{8\alpha^{beta}(\delta-\sigma)}\matchcal{F}\biggl(\frac}\sigma+2\delta}{3}\bigr)\biggr]\\&{}+\frac{3^{1+\alpha\beta}\Gamma}\马塔尔{日本}_{\frac{\sigma+2\delta}{3}}^{\alpha}\mathcal{F}(\delta)\bigr]。\结束{对齐}$$
最后,如果我们相乘\(I_{1}+I_{2}+I_3}\)通过\(\frac{(\delta-\sigma)\alpha^{\beta}}{9}\),那么我们有(5). 这就完成了引理的证明1. □
定理1
假设引理的所有假设1持有。此外,让 \(\vert\mathcal{F}^{prime}\vert\) 是上的凸函数 \([\sigma,\delta]\)。然后,我们有
$$开始{对齐}和\biggl\vert\frac{3^{\alpha\beta-1}\alpha^{\beta}}{(delta-\sigma)^{\alpha\beta}}\Gamma(\beta+1)\biggl[_{+}^{\beta}\mathcal{日本}_{\sigma}^{\alpha}\mathcal{F}\biggl(\frac{2\sigma+\delta}{3}\bigr)+_{+}^{beta}\matchcal{日本}_{\frac{2\sigma+\delta}{3}}^{\alpha}\mathcal{F}\biggl(\frac{\sigma+2\delta{3}\bigr)+_{+}^{\ beta}\mathcal{日本}_{\frac{\sigma+2\delta}{3}}^{\alpha}\mathcal{F}(\delta)\biggr]\\&\qquad{}-\frac}{8}\biggl[\mathcal{F}(\sigma)+3\mathcali{F}\bigbl}\biggr)+\mathcal{F}(\delta)\biggr]\biggr\vert\\&\quad\leq\frac{(\delta-\sigma)\alpha^{\beta}}{27}\bigl[\bigl(2A{2}(\alpha,\beta)+A{1}(\ alpha、\betaβ)+2A{4}(α,β)-A{3}(β)(\alpha,\beta)-A_{5}(\alfa,\beta)\bigr)\bigl\vert\mathcal{F}^{prime}(\ delta)\bigr\vert\bigr],\end{aligned}$$
(6)
哪里
$$\begin{aligned}&A{1}(\alpha,\beta)=\int_{0}^{1}\mu\biggl\vert\biggl(\frac{1-(1-\mu)^{\alpha}}{\ alpha}\biggr)^{\ beta}-\frac}5}{8\alpha^{\beta}}\bigr\vert\,d\mu\\&\hphantom{A{1{^{\beta}}\biggl[\frac{5}{8}\bigl(C_{1}^{2}-\frac{1}{2}\biggr)+\frac{1}}{\alpha}\bigl(2\mathcal{B}\bigbl(\beta+1,\frac}{\alpha},\biggl(\frac5}{8}\bigr)^{\frac[1}{\beta}}\bighr)\\&\hphantom{A{1}(\alpha,\beta)=}{}-2\mathcal}{B}\ biggl(\beta+1,\frac{1}{\alpha},\biggl(\frac{5}{8}\biggr){\alpha}\biggr)+\mathfrak{B}\bigl beta}}\biggr\vert\,d\mu\\&\hphantom{A{2}(\alpha,\beta)}=\frac{1}{\alpha^{\beta}{\biggl[\frac}5}{8}(2摄氏度_{1}-1)+\frac{1}{\alpha}\biggl(\mathfrak{B}\bigl(\beta+1,\frac}{\alpha}\biggr(\alpha,\beta)=\int_{0}^{1}\mu\biggl\vert\biggl(\frac{1-(1-\mu)^{\alpha}}{\alfa}\biggr)^{beta}-\frac}1}{2\alpha^{beta}}\biggr\vert\,d\mu\\&\hphantom{A{3}(\alpha,\beta)}=\frac{1}{\alpha^}\beta}{\biggl[\frac}{2}\bigl(C_{2}^{2}-\frac{1}{2}\biggr)+\frac{1}}{\alpha}\bigl(2\mathcal{B}\bigbl(\beta+1,\frac}{\alpha},\biggl(\frac[1}{2{\bigger)^{\frac#1}{\beta}}\bigr)\\&\hphantom{A{3}(\alpha,\beta)=}{}-2\mathcal}{B}\ biggl(\beta+1,\frac{1}{\alpha},\biggl(\frac}1}{2}\biggr){\alpha}\biggr)+\mathfrak{B}\bigl beta}}\biggr\vert\,d\mu\\&\hphantom{A{4}(\alpha,\beta)}=\frac{1}{\alpha^{\beta}{\biggl[\frac}1}{2}(2摄氏度_{2}-1)+\frac{1}{\alpha}\biggl(\mathfrak{B}\bigl(\beta+1,\frac}{\alpha}\biggr(\alpha,\beta)=\int_{0}^{1}\mu\biggl\vert\biggl(\frac{1-(1-\mu)^{\alpha}}{\alfa}\biggr)^{beta}-\frac}3}{8\alpha^{\beta}}\biggr\vert\,d\mu\\&&\hphantom{A_{5}(\alpha,\beta)}=\frac{1}{\alpha^{\beta}}\biggl{3}{8}\biggl(C_{3}^{2}-\frac{1}{2}\biggr)+\frac{1}}{\alpha}\bigl(2\mathcal{B}\bigbl(\beta+1,\frac}{\alpha},\biggl(\frac[3]{8}\bigr)^{\frac[1}{\beta}}\bighr)\\&\hphantom{A_{5}(\alpha,\beta)=}{}-2\mathcal}{B}\ biggl(\beta+1,\frac{1}{\alpha},\biggl(\frac}3}{8}\biggr){\alpha}\biggr)+\mathfrak{B}\bigl(\beta+1,\frac{1}{\alfa}\bigr)\biggr]\end{aligned}$$
和
$$\beart{aligned}A_{6}(\alpha,\beta)&=\int _{0}^{1}\biggl\vert\biggl(\frac{1-(1-\mu)^{\alpha}{\alpha}\biggr)^{\beta}-\frac{3}{8\alpha^{\beta}\biggr\vert\,d\mu \\&=\frac{1}{\alpha^{\beta}}\biggl{3}{{8}(2C_{3}-1)+\frac{1}{\alpha}\biggl(\mathfrak{B}\bigl(\beta+1,\frac}{\alpha}\biggr。\结束{对齐}$$
在这里,\(C_{1}=1-(1-(\frac{5}{8})^{\frac{1}{\beta})^{\frac{1}{\alpha}}),\(C_{2}=1-(1-(压裂{1}{2})^{压裂{1{beta}}),\(C_{3}=1-(1-(压裂{3}{8})^{压裂{1}{beta}}),功能 \(\mathfrak{B}(\cdot,\cdot)\) 和 \(\mathcal{B}(\cdot、\cdot和\cdot)\) Beta函数和不完整Beta函数定义为
$$\textstyle\begin{cases}\mathfrak{B}(x,y)=\int_{0}^{1} u个^{x-1}(1-u)^{y-1}\,du,\\mathcal{B}(x,y,r)=\int_{0}^{r} u个^{x-1}(1-u)^{y-1}\,du\结束{cases}$$
对于 \(x,y>0) 和 \(r \在[0,1]\中)。
证明
按引理1,按部分积分和的凸性\(\vert\mathcal{F}^{prime}\vert\),我们获得
$$开始{对齐}和\biggl\vert\frac{3^{\alpha\beta-1}\alpha^{\beta}}{(delta-\sigma)^{\alpha\beta}}\Gamma(\beta+1)\biggl[_{+}^{\beta}\mathcal{日本}_{\sigma}^{\alpha}\mathcal{F}\biggl(\frac{2\sigma+\delta}{3}\bigr)+_{+}^{beta}\matchcal{日本}_{\frac{2\sigma+\delta}{3}}^{\alpha}\mathcal{F}\biggl(\frac{\sigma+2\delta{3}\bigr)+_{+}^{\ beta}\mathcal{日本}_{\frac{\sigma+2\delta}{3}}^{\alpha}\mathcal{F}(\delta)\biggr]\\&\qquad{}-\frac}{8}\biggl[\mathcal{F}(\sigma)+3\mathcali{F}\bigbl}\biggr)+\mathcal{F}(\delta)\biggr]\biggr\vert\\&\quad\leq\frac{(\delta-\sigma)\alpha^{\beta}}{9}\biggl[\biggl\vert\int_{0}^{1}\biggl gr)\biggr)\,d\mu\biggr\vert\\&\qquad{}+\biggl\vert\int_{0}^{1}\biggl(\biggl(\frac{1-(1-\mu)^{\alpha}}{\alfa}\biggr)^{\beta}-\frac{1}{2\alpha^{\beta}}\bigr)\mathcal{F}^{\prime}\bigl&\qquad{}+\biggl\vert\int_{0}^{1}\biggl-\frac{3}{8\alpha^{\beta}}\biggr)\mathcal{F}^{\prime}\bigl{1}\biggl\vert\biggl(\frac{1-(1-\mu)^{\alpha}}{\alfa}\bigr)^}\beta}-\frac{5}{8\alpha^{\beta}}\biggr\vert\biggl\vert\mathcal{F}^{\prime}\bigl(\frac{2+\mu}{3}\bigr)\sigma+\biggl(\frac{1-\mu}{3}\ biggr)\delta\biggr{1-(1-\mu)^{\alpha}}{\alalpha}\biggr)^{beta}-\frac{1}{2\alpha^{\beta}}\biggr\vert\biggl\vert\mathcal{F}^{prime}\biggl(\biggr(\frac{1+\mu}{3}\bigr)\sigma+\biggal(\frac{2-\mu}{3}\ biggr)\delta\biggro)\biggr\vert\,d\mu\&\qquad{}+\int_{0}^{1}\bigbl\vert_biggl alpha}}{\alpha}\biggr)^{\beta}-\frac{3}{8\alpha^{\beta}}\bigr\vert\biggl\vert\mathcal{F}^{\prime}\biggl(\frac{\mu}{3}\sigma+\biggl(\frac{3-\mu}{3}\biggr)\delta\biggr)\biggr\vert\,d\mu\biggr]\\&\quad\leq\frac}(\delta-\sigma)\alpha^{\beta}}{9}\biggl[\bigl\vert\mathcal{F}\prime}(\sigma)\bigr\vert\int_{0}^{1}\ biggl(\frac{2+\mu}{3}\biggr)\biggl\vert\biggl\biggr)^{\beta}-\frac{5}{8\alpha^{\beta}}\biggr\vert\,d\mu\\&\qquad{}+\bigl\vert\mathcal{F}^{\prime}(\delta)\bigr\vert\int_{0}^{1}\bigl alpha}\biggr)^{\beta}-\frac{5}{8\alpha^{\beta}}\bigr\vert\,d\mu\\&\qquad{}+\bigl\vert\mathcal{F}^{\prime}(\sigma)\bigr\vert\int_{0}^{1}\biggl(\frac{1+\mu}{3}\bigr)\biggl\vert\biggl F}^{\prime}(\delta)\bigr\vert\int_{0}^{1}\biggl(\frac{2-\mu}{3}\bigr)\biggl\vert\biggl(\frac{1-(1-\mu)^{\alpha}}{\alfa}\biggr)^{β}-\frac}{2\alpha^{\beta}}\bigr\vert\,d\mu\\&\qquad{}+\bigl\vert\mathcal{F}^{\prime}gl(\frac{1-(1-\mu)^{\alpha}}{\alalpha}\biggr)^{beta}-\frac}3}{8\alpha^{\beta}}\bighr\vert\,d\mu\\&\qquad{}+\bigl\vert\mathcal{F}^{prime}(\delta)\bigr\vert\int_{0}^{1}\biggl(\frac{3-\mu}{3}\bigr)\biggl\vert\biggl vert\,d\mu\biggr]\\&\quad=\frac{(\delta-\sigma)\alpha^{\beta}}{27}\bigl[\bigl(2A{2}(α,β)+A{1}(α,贝塔)+A_4}(β,α)+A_3}(δ,β)+A_5}(γ,β)β)-A{3}(α,β)+3A{6}(β,α)-A{5}(\alpha,\beta)\bigr)\bigl\vert\mathcal{F}^{prime}(\delta)\bigr\vert\bigr]。\结束{对齐}$$
这是定理的期望结果1. □
定理2
假设引理的所有假设1持有。假设也是这样 \(\vert\mathcal{F}^{\prime}\vert^{q}\) 是上的凸函数 \([\sigma,\delta]\),哪里 \(压裂{1}{p}+压裂{1{q}=1\) 具有 \(p,q>1)。然后,我们有
$$开始{对齐}和\biggl\vert\frac{3^{\alpha\beta-1}\alpha^{\beta}}{(delta-\sigma)^{\alpha\beta}}\Gamma(\beta+1)\biggl[_{+}^{\beta}\mathcal{日本}_{\sigma}^{\alpha}\mathcal{F}\biggl(\frac{2\sigma+\delta}{3}\bigr)+_{+}^{beta}\matchcal{日本}_{\frac{2\sigma+\delta}{3}}^{\alpha}\mathcal{F}\biggl(\frac{\sigma+2\delta{3}\bigr)+_{+}^{\ beta}\mathcal{日本}_{\frac{\sigma+2\delta}{3}}^{\alpha}\mathcal{F}(\delta)\biggr]\\&\qquad{}-\frac}{8}\biggl[\mathcal{F}(\sigma)+3\mathcali{F}\bigbl}\biggr)+\mathcal{F}(\delta)\biggr]\biggr\vert\\&\quad\leq\frac{(\delta-\sigma)\alpha^{\beta}}{9}\biggl[A_{7}^{\frac{1}{p}}(\alpha,\beta,p)\biggl(\frac}5\vert\mathcal{F}^{prime}(\sigma)\vert^{q}+\vert\mathcal{F1}^{prime}1}{p}}(\alpha,\beta,p)\biggl(\frac{\vert\mathcal{F}^{prime}(\sigma)\vert^{q}+\vert\mathcal{F}^{prime}(δ)\vert^{q}}{2}\biggr)^{\frac{1}{q}{\\&\qquad{}+A_{9}^{\frac{1{p}}(alpha,beta,p)\biggl(\frac}\vert\mathcal{F}{prime}(\sigma)\vert_q}+5\vert\mathcal{F1}^{prime{6}\biggr)^{\frac{1}{q}}\biggr],\end{aligned}$$
(7)
哪里
$$开始{aligned}&A_{7}(\alpha,\beta,p)=\int_{0}^{1}\biggl\vert\biggl(\frac{1-(1-\mu)^{\alpha}}{\alfa}\bigr)^}\beta}-\frac}5}{8\alpha^{\beta{}\vert^{p}\,d\mu,\\&A_}8}1}\biggl\vert\biggl(\frac{1-(1-\mu)^{\alpha}}{\alfa}\biggr)^{beta}-\frac}{2\alpha^{\beta}}\biggr\vert^{p}\,d\mu\end{aligned}$$
和
$$开始{对齐}A_{9}(\alpha,\beta,p)&=\int_{0}^{1}\biggl\vert\biggl(\frac{1-(1-\mu)^{\alpha}}{\alba}\bigr)^}\beta}-\frac}3}{8\alpha^{\beta{}}\vert^{p}\,d\mu。\结束{对齐}$$
证明
如果我们考虑引理1,然后我们可以很容易地获得
$$开始{对齐}和\biggl\vert\frac{3^{\alpha\beta-1}\alpha^{\beta}}{(delta-\sigma)^{\alpha\beta}}\Gamma(\beta+1)\biggl[_{+}^{\beta}\mathcal{日本}_{\sigma}^{\alpha}\mathcal{F}\biggl(\frac{2\sigma+\delta}{3}\bigr)+_{+}^{beta}\matchcal{日本}_{\frac{2\sigma+\delta}{3}}^{\alpha}\mathcal{F}\biggl(\frac{\sigma+2\delta{3}\bigr)+_{+}^{\ beta}\mathcal{日本}_{\frac{\sigma+2\delta}{3}}^{\alpha}\mathcal{F}(\delta)\biggr]\\&\qquad{}-\frac}{8}\biggl[\mathcal{F}(\sigma)+3\mathcali{F}\bigbl}\biggr)+\mathcal{F}(\delta)\biggr]\biggr\vert\\&\quad\leq\frac{(\delta-\sigma)\alpha^{\beta}}{9}\biggl[\int_{0}^{1}\biggl\vert\biggl压裂{1-\mu}{3}\biggr)\delta\bigger)\biggr\vert\,d\mu\\&\qquad{}+\int_{0}^{1}\biggl\vert\biggl(\frac{1-(1-\mu)^{\alpha}}{\alalpha}\biggr))\delta\biggr)\biggr\vert\,d\mu\\&\qquad{}+\int_{0}^{1}\biggl\vert\biggl(\frac{1-(1-\mu)^{\alpha}}{\alpha}\biggr)^{\beta}-\frac{3}{8\alpha^{\beta}}\bigr\vert\biggl\vert\mathcal{F}^{\prime}\bigl。\结束{对齐}$$
(8)
现在,我们考虑右边的积分(8). 使用的凸性\(\vert\mathcal{F}^{\prime}\vert^{q}\)还有众所周知的Hölder不等式
$$\开始{aligned}&\int_{0}^{1}\biggl\vert\biggl(\frac{1-(1-\mu)^{\alpha}}{\alfa}\bigr)^}\beta}-\frac}5}{8\alpha^{\beta{}}\bigcr\vert\bigl\vert\mathcal{F}^{\prime}\bigl(\frac{2+\mu}{3}\bighr)\sigma+\biggr gl(\frac{1-\mu}{3}\biggr)\delta\bigger)\biggr\vert\,d\mu\\&\quad\leq\biggl(\int_{0}^{1}\biggl\vert\biggl(\frac{1-(1-\mu)^{\alpha}}{\alpha}\biggr)^{\beta}-\frac{5}{8\alpha ^{\beta}}\biggr\vert^{p}\,d\mu\biggr)^{\frac{1}{p}}}\&\qquad{}\times\biggl(\int _{0}^{1}\biggl\vert\mathcal{F}^{\prime}\biggl(\frac{2+\mu}{3}\ biggr)\西格玛+\比格尔(\frac{1-\mu}{3}\biggr)\德尔塔\比格尔)\比格尔\ vert^{q}\,d\mu\biggr)^{\frac{1}{q}}\\&\quad\leq A_{7}^{\frac{1{p}}(\alpha,\beta,p)\biggl(\int_{0}^1}\biggl(\frac}2+\mu}{3}\bigl\vert\mathcal{F}^{prime}(\sigma)\bigr\vert^{q}+\frac[1-\mu}}{3{\bigl \vert\mathcal{F}^{\prime}(\delta)\bigr\vert^{q}\biggr)\,d\mu\bigger)^{\frac{1}{q}}\\&\quad=A_{7}^{\frac{1}}{p}}(\alpha,\beta,p)\biggl[\frac{5\vert\mathcal{F}^{prime}(\sigma)\vert^{q}+\vert\mathcal{F1}^{prime}。\结束{对齐}$$
(9)
以类似的方式,我们很容易获得
$$开始{aligned}和\int_{0}^{1}\biggl\vert\biggl(\frac{1-(1-\mu)^{\alpha}}{\alfa}\bigr)^}\beta}-\frac}1}{2\alpha^{\beta{}}\bigbr\vert\ biggl\ vert\mathcal{F}^{\prime}\bigl(\frac{1+\mu}{3}\bigcr)\sigma+\biggr gl(\frac{2-\mu}{3}\biggr)\delta\bigger)\biggr\vert\,d\mu\\&\quad\leq\biggl(\int_{0}^{1}\biggl\vert\biggl(\frac{1-(1-\mu)^{\alpha}}{\alpha}\biggr)^{\beta}-\frac{1}{2\alpha^{\beta}}\biggr\vert^{p}\,d\mu\biggr)^{\frac{1}{p}}}\&\qquad{}\times\biggl(\int _{0}^{1}\biggl\vert\mathcal{F}^{\prime}\biggl(\frac{1+\mu}{3}\ biggr)\西格玛+\比格尔(\frac{2-\mu}{3}\biggr)\德尔塔\比格尔)\比格尔\ vert^{q}\,d\mu\biggr)^{\frac{1}{q}}\\&\quad\leq A_{8}^{\frac{1{p}}(\alpha,\beta,p)\biggl(\int_{0}^1}\biggl(\frac}1+\mu}{3}\bigl\vert\mathcal{F}^{prime}(\sigma)\bigr\vert^{q}+\frac[2-\mu}}{3{\bigl \vert\mathcal{F}^{\prime}(\delta)\bigr\vert^{q}\biggr)\,d\mu\bigger)^{\frac{1}{q}}\\&\quad=A_{8}^{\frac{1}}{p}}(\alpha,\beta,p)\biggl[\frac{\vert\mathcal{F}{\prime}(\sigma)\vert^{q}+\vert\mathcal{F}^{\prime}(\ delta)\vert_q}}{2}\biggr]^{\frac}1}{q}}\end{aligned}$$
(10)
和
$$开始{aligned}&\int_{0}^{1}\biggl\vert\biggl(\frac{1-(1-\mu{3}\biggr)\delta\biggr)\biggr\vert\,d\mu\\&\quad\leq\biggl(\int_{0}^{1}\biggl\vert\biggl \mu}{3}\sigma+\biggl(\frac{3-\mu}{3}\ biggr)\delta\biggr}\\&\quad\leq A_{9}^{\frac{1}{p}}(\alpha,\beta,p)\biggl(\int_{0}^{1}\biggal(\frac}\mu}{3}\bigl\vert\mathcal{F}^{prime}(\ sigma)\bigr\vert^{q}+\frac{3-\mu}{3}\bigl vert\matchcal{F}^{prime}^{q}\biggr)\,d\mu\bigger)^{frac{1}{q}}\\&\quad=A{9}^{frac{1}{p}}(\alpha,\beta,p)\biggl[\frac{\vert\mathcal{F}^{\prime}(\sigma)\vert^{q}+5\vert\mathcal{F}^{prime}(\delta)\vert\q}}{6}\biggr]^{\frac}1}{q}}。\结束{对齐}$$
(11)
如果我们插入(9)–(11)到(8),那么我们有
$$开始{对齐}和\biggl\vert\frac{3^{\alpha\beta-1}\alpha^{\beta}}{(delta-\sigma)^{\alpha\beta}}\Gamma(\beta+1)\biggl[_{+}^{\beta}\mathcal{日本}_{\sigma}^{\alpha}\mathcal{F}\biggl(\frac{2\sigma+\delta}{3}\bigr)+_{+}^{beta}\matchcal{日本}_{\frac{2\sigma+\delta}{3}}^{\alpha}\mathcal{F}\biggl(\frac{\sigma+2\delta{3}\bigr)+_{+}^{\ beta}\mathcal{日本}_{\frac{\sigma+2\delta}{3}}^{\alpha}\mathcal{F}(\delta)\biggr]\\&\qquad{}-\frac}{8}\biggl[\mathcal{F}(\sigma)+3\mathcali{F}\bigbl}\biggr)+\mathcal{F}(\delta)\biggr]\biggr\vert\\&\quad\leq\frac{(\delta-\sigma)\alpha^{\beta}}{9}\biggl[A_{7}^{\frac{1}{p}}(\alpha,\beta,p)\biggl(\frac}5\vert\mathcal{F}^{prime}(\sigma)\vert^{q}+\vert\mathcal{F1}^{prime}1}{p}}(\alpha,\beta,p)\biggl(\frac{\vert\mathcal{F}^{prime}(\sigma)\vert^{q}+\vert\mathcal{F}^{prime}(δ)\vert^{q}}{2}\biggr c{1}{q}}\biggr]。\结束{对齐}$$
这就完成了定理的证明2. □
定理3
让我们考虑一下引理的所有假设1持有。如果 \(\vert\mathcal{F}^{\prime}\vert^{q}\) 在上是凸的 \([\sigma,\delta]\),哪里 \(问题1),那么我们有以下牛顿-类型不等式
$$开始{对齐}和\biggl\vert\frac{3^{\alpha\beta-1}\alpha^{\beta}}{(delta-\sigma)^{\alpha\beta}}\Gamma(\beta+1)\biggl[_{+}^{\beta}\mathcal{日本}_{\sigma}^{\alpha}\mathcal{F}\biggl(\frac{2\sigma+\delta}{3}\bigr)+_{+}^{beta}\matchcal{日本}_{\frac{2\sigma+\delta}{3}}^{\alpha}\mathcal{F}\biggl(\frac{\sigma+2\delta{3}\bigr)+_{+}^{\ beta}\mathcal{日本}_{\frac{\sigma+2\delta}{3}}^{\alpha}\mathcal{F}(\delta)\biggr]\\&\qquad{}-\frac}{8}\biggl[\mathcal{F}(\sigma)+3\mathcali{F}\bigbl}\biggr)+\mathcal{F}(\delta)\biggr]\biggr\vert\\&\quad\leq\frac{(\delta-\sigma)\alpha^{\beta}}{9}\biggl[A_{2}^{1-\ frac{1}}A 3}\biggr)\bigl\vert\mathcal{F}^{{)\biggr \qquad{}+\biggl(\frac{2A_{4}(\alpha,\beta)-A_{3}(\ alpha、\ beta)}{3}\biggr)\bigl\vert\mathcal{F}^{\prime}(\delta)\bigr\vert^{q}\biggr)^{\frac{1}{q}}\\&\qquad{}+A{6}^{1-\frac{1}{q}}(\alpha,\beta)\biggl(\frac{A_{5}(\alpha,\beta)}{3}\bigl\vert\mathcal{F}^{\prime}(\sigma)\bigr\vert^{q}+\biggl(\frac{3A_{6}(\alpha,\beta)-A_{5}(\alpha,β)}{3}\biggr)\bigl\vert\mathcal{F}^{\prime}(\delta)\bigr\vert^{q}\biggr)^{\frac{1}{q}}\biggr],\end{aligned}$$
(12)
哪里 \(A_{1}(α,β)),\(A_{2}(α,β)),\(A_{3}(α,β)),\(A_{4}(α,β)),\(A_{5}(α,β)),和 \(A_{6}(α,β)) 在定理中定义1。
证明
如果我们考虑\(\vert\mathcal{F}^{\prime}\vert^{q}\)和幂-曼不等式,然后我们得到
$$\开始{aligned}&\int_{0}^{1}\biggl\vert\biggl(\frac{1-(1-\mu)^{\alpha}}{\alfa}\bigr)^}\beta}-\frac}5}{8\alpha^{\beta{}}\bigcr\vert\bigl\vert\mathcal{F}^{\prime}\bigl(\frac{2+\mu}{3}\bighr)\sigma+\biggr gl(\frac{1-\mu}{3}\biggr)\delta\bigger)\biggr\vert\,d\mu\\&\quad\leq\biggl(\int_{0}^{1}\biggl\vert\biggl(\frac{1-(1-\mu)^{\alpha}}{\alalpha}\biggr)^{β}-\frac}5}{8\alpha^{\beta}}\bigr\vert\,d\mu\bigger)^{1-\frac[1}{q}}\\&\qquad{}\times\biggl mu)^{\alpha}}{\alfa}\biggr)^{\β}-\frac{5}{8\alpha^{\beta}}\bigr\vert\biggl\vert\mathcal{F}^{\prime}\biggl(\frac{2+\mu}{3}\biggr)\sigma+\biggal{1}\biggl\vert\biggl\时间\biggl\vert\mathcal{F}^{prime}\biggl}(\alpha,\beta)\biggl(\int_{0}^{1}\biggl\vert\biggl{8\alpha^{\beta}}\biggr\vert\\&\qquad{}\times\biggl[\frac{2+\mu}{3}\bigl\vert\mathcal{F}^{prime}(\sigma)\bigr\vert^{q}+\frac}1-\mu}{3}\ bigl\verst\mathcal{F}{prime{(\delta)\biggr c{1}{q}}\\&\quad=A{2}^{1-\frac{1{q}{(alpha,\beta)\biggl(\frac}2A{2}(\alpha+A{1}(\alpha,\beta)}{3}\biggr)\bigl\vert\mathcal{F}^{prime}(\sigma)\bigr\vert^{q}\\&\qquad{}+\biggl(\frac{A{2}(\ alpha、\beta \biggr)^{\frac{1}{q}}。\结束{对齐}$$
(13)
以类似的方式,我们
$$开始{aligned}和\int_{0}^{1}\biggl\vert\biggl(\frac{1-(1-\mu)^{\alpha}}{\alfa}\bigr)^}\beta}-\frac}1}{2\alpha^{\beta{}}\bigbr\vert\ biggl\ vert\mathcal{F}^{\prime}\bigl(\frac{1+\mu}{3}\bigcr)\sigma+\biggr gl(\frac{2-\mu}{3}\biggr)\delta\bigger)\biggr\vert\,d\mu\\&\quad\leq\biggl(\int_{0}^{1}\biggl\vert\biggl(\frac{1-(1-\mu)^{\alpha}}{\alfa}\biggr)mu)^{\alpha}}{\alfa}\biggr)^{\β}-\frac{1}{2\alpha^{\beta}}\bigr\vert\biggl\vert\mathcal{F}^{\prime}\biggl(\frac{1+\mu}{3}\biggr)\sigma+\biggal{1}\biggl\vert\biggl(\frac{1-(1-\mu)^{alpha}}{\alpha}\bigr)}\biggr\vert\biggl\vert\mathcal{F}^{prime}\biggl(\biggal(\frac{1+\mu}{3}\bigr)\sigma+\biggl(\frac{2-\mu}{3}\ biggr)\delta\biggr}{q}}(\alpha,\beta)\biggl(\int_{0}^{1}\biggl\vert\biggl-\frac{1}{2\alpha^{beta}}\biggr\vert\\&\qquad{}\times\biggl[\frac{1+\mu}{3}\bigl\vert\mathcal{F}^{prime}(\sigma)\bigr\vert^{q}+\frac}2-\mu}{3}\ bigl\verst\mathcal{F}(\ delta)\biggr\vert^q}\bighr]\,d\mu\biggr)^{\frac{1}{q}}\\&\quad=A_{4}^{1-\frac}{q{}}(\alpha,\beta)\biggl(\biggl(\frac[A_{4](α,β)+A{3}(α,贝塔)}{3}\biggr \vert^{q}\biggr)^{\frac{1}{q}}\end{aligned}$$
(14)
和
$$开始{aligned}&\int_{0}^{1}\biggl\vert\biggl(\frac{1-(1-\mu{3}\biggr)\delta\biggr)\biggr\vert\,d\mu\\&\quad\leq\biggl(\int_{0}^{1}\biggl\vert\biggl mu)^{\alpha}}{\alfa}\biggr)^{\β}-\frac{3}{8\alpha^{\beta}}\bigr\vert\biggl\vert\mathcal{F}^{\prime}\biggl(\frac{\mu}{3}\sigma+\biggl(\frac{3-\mu}{3}\ biggr)\delta\biggr gl(\frac{1-(1-\mu)^{\alpha}}{\alalpha}\biggr)^{β}-\frac}3}{8\alpha^{\beta}}\bigr\vert\biggl\vert\mathcal{F}^{\prime}\biggl(\frac{\mu}{3}\sigma+\biggl(\frac{3-\mu}{3}\ biggr)\delta\biggr \biggl(\frac{1-(1-\mu)^{\alpha}}{\alalpha}\biggr)\bigl\vert\mathcal{F}^{prime}(\sigma)\bigr\vert^{q}+\frac{3-\mu}{3}\bigl\ vert\mathcal{F{{prime{(\delta)\bigr\vert^}\biggr]\,d\mu\biggr biggl(\frac{A_{5}(\alpha,\beta)}{3}\bigl\vert\mathcal{F}^{prime}(\sigma)\bigr\vert^{q}+\biggl(\frac{3A{6}(\alpha,\beta)-A{5}。\结束{对齐}$$
(15)
如果我们插入(13)–(15)到(8),然后是定理的证明三已完成。□