现在,我们准备陈述并证明我们的主要成果。在本节中,任何时间尺度\(\mathbb{T}\)是无界的,我们假设不等式的右边收敛,如果左边收敛。
定理2.1
假设 \(\mathbb{T}\) 是带有的时间刻度 \(0\leqr\in\mathbb{T}\)。此外,假设是这样 (f) 和 λ 是非负的身份证-上的连续函数 \([r,\infty)_{\mathbb{T}}\) 具有 (f) 无衰减。如果 \(第1页) 和 \(\beta\leqsleat 0\),然后针对 \(t \ geq斜面a \ in \ mathbb{t}\) 和 \(伽马\in(0,1]\) 我们有这个
$$开始{aligned}\int_{r}^{infty}\frac{\lambda(t)\Phi^{p-\gamma+1}(t)}{(\lambda^{\rho}(t))^{beta}}\nabla_{a}^{gamma}t\leqslead\int_{r}^{ifty}\frac{\lampda(t{(\lambda^{\rho}(t))^{\beta}}\nabla_{a}^{\gamma}t,\end{aligned}$$
(2.1)
哪里
$$\begin{aligned}\Phi(t)=\int_{t}^{infty}\lambda(s)f$$
证明
作为(f)没有减少,我们有\(x\geq t\geq r)
$$\开始{对齐}F(x,t)=\int_{t}^{x}\lambda(s)F$$
然后
$$\开始{对齐}f(x)f^{p-\gamma}(x,t)\leqsleat\biggl[\int_{t}^{x}\lambda(s)\nabla_{a}^{gamma}s\biggr]^{p-\ gamma}f^{p-\gamma+1}(x)。\结束{对齐}$$
(2.2)
应用链式规则(1.9)和使用\(nabla_{a}^{{gamma,x}}F(x,t)=\lambda(x)F(x)\geq0),其中\(nabla{a}^{{gamma,x}})表示\((伽玛射线,a))-关于的nabla导数x个,我们得到
$$开始{对齐}\nabla_{a}^{\gamma,x}}\bigl(F^{p-\gamma+1}(x,t)\bigr)&=(p-\gama+1)\nabla_{a}^{\gamma,x}}F(x,t)\int_{0}^{1}\bigr[(1-h)F(x、t)+hF\bigl(\rho(x),t\bigr]^{p-\ gamma}\,dh\\&\leqsleat(p-\gamma+1)\lambda(x)F(x)\int_{0}^{1}\bigl[(1-h)F(x,t)+hF(x、t)\bigr]^{p-\gama}\,dh\\&=(p-\gamma+1)\ lambdaF^{p-\gamma}(x,t)。\结束{对齐}$$
(2.3)
组合(2.2)带有(2.3)给予
$$开始{aligned}\nabla_{a}^{\gamma,x}}\bigl(F^{p-\gamma+1}(x,t)\bigr)\leqslate(p-\gama+1)\lambda(x$$
所以(注意\(x\geq t\geq r)因此由于∧是非退化的,\((0\leqsleat(\Lambda^{rho}(x))^{beta}\leqstreat\Lambda~{rho{(t))^}\beta}\))
$$开始{对齐}\frac{\lambda(t)\nabla_{a}^{\gamma,x}}(F^{p-\gamma+1}(x,t))}{(\lambda^{\rho}(t))^{\beta}}和\leqslate\frac}(p-\gama+1)\lambda^(t ^{x}\lambda(s)\nabla_{a}^{gamma}s\biggr]^{p-\gamma}F^{p-\ gamma+1}(x)\\&\leqsleat\frac{(p-\ gama+1)\lambda(t)\lambda(x)}{(\lambda^{\rho}(x))^{\beta}}\biggl[\int_{t}^{x}\lambda。\结束{对齐}$$
在以下方面整合双方x个结束\([t,\infty)_{\mathbb{t}}\)给予
$$开始{对齐}\frac{\lambda(t)\Phi^{p-\gamma+1}(t)}{(\lambda^{\rho}(t))^{\beta}}&=\int_{t}^{\infty}\frac{\lampda(t \nabla_{a}^{gamma}x\\&\leqslate(p-\gamma+1)\nint_{t}^{infty}\frac{\lambda(t)\lambda(x)}{(\lambda^{rho}(x))^{\beta}}\biggl[\int_{t}^{x}\lambda(s)\nabla_{a}^{gamma,x}s\biggr]^{p-\gamma}f^{p-\ gamma+1}(x)\nabla _{a}^{gama}x$$
再次整合双方,但这次涉及吨结束\([r,\infty)_{\mathbb{T}}\),生产
$$\开始{对齐}&\int_{r}^{\infty}\frac{\lambda(t)\Phi^{p-\gamma+1}(t)}{(\lambda^{\rho}(t))^{\beta}}\nabla{a}^{\ gamma}t\\&\quad\leqslate(p-\gama+1)\int_}{r}{\inffy}\biggl[\int _{t}^{infty{\lampda(t)\lambda(x)}{(\lambda^{\rho}(x))^{\beta}}\biggl[\int_{t}^{x}\lambda(s)\nabla_{a}^{gamma}s\biggr]^{p-\gamma}f^{p-\gamma+1}(x)\nabla{a}^{gamma}x\biggr]\nabla{a}^{gama}t.\end{aligned}$$
(2.4)
利用时间尺度上的Fubini定理,不等式(2.4)可以重写为
$$\开始{对齐}&\int_{r}^{\infty}\frac{\lambda(t)\Phi^{p-\gamma+1}(t)}{(\lambda^{\rho}(t))^{\beta}}\nabla_{a}^{\ gamma}t\\&\quad\leqslate(p-\gama+1)\\&\qquad{}\times\int_{r\}^{infty{\lampda(x)\bigl(\lambda^}rho}(x)\bigr)^{-\beta}f^{p-\gamma+1}(x)\biggl[\int_{r}^{x}\lambda(t)\bigl[\int_{t}^{x}\lampda(s)\nabla{a}^{gamma}s\biggr]^{p-\gamma}\nabla{a}^{gama}t\biggr]$$
(2.5)
现在,根据链式法则(1.10),存在\(c在[\rho(t),t]\中)这样(这里\(nabla{a}^{{gamma,t}})表示\((伽玛射线,a))-关于的nabla导数吨)
$$\开始{对齐}\nabla_{a}^{\gamma,t}}\biggl[-\biggl(\int_{t}^{x}\lambda(s)\nabla{a}^{\gamma}\bigr)^{p-\gamma+1}\bighr]&=(p-\gama+1)\lambda(t)\biggl/(int_{c}^{x}\lamda(s)\nabla}^}\gamma}s\biggr]^{p-\gamma}\\&\geqsland(p-\gama+1)\lambda(t)\biggl[\int_{t}^{x}\lambda\s)\nabla_{a}^{\gammaneneneep s\biggr]^{p-\gamma}。\结束{对齐}$$
(2.6)
替换(2.6)到(2.5)导致
$$\开始{对齐}&\int_{r}^{\infty}\frac{\lambda(t)\Phi^{p-\gamma+1}(t)}{(\lambda^{\rho}(t))^{\beta}}\nabla_{a}^{\ gamma}t\\&\quad\leqslead\int_{r}^{\finfty{\lamda(x)\bigl(\lambda^{\rho}(x)\ bigr)^{-\beta{f ^{p-\gamma+1}(x)\biggl[\int_{r}^{x}-\nabla_{a}^{{gamma,t}}\biggl[\biggl(\int_{t}^{x}\lambda(s)\nabla{a}^{gamma}s \biggr)^{p-\gamma+1}\bigr]\nabla-{a}{gammaneneneep t\biggr]\nabra_{a{{gama}x\\quad=\int_}r}^{infty}\frac{\lambda(x)\lambda(x)da^{p-\gamma+1}(x)f^{p-\ gamma+1(x)}{(\lambda^{\rho}(x))^{\beta}\nabla_{a}^{\gamma}x$$
这表明了不等式的有效性(2.1). □
现在,作为我们结果的特例,我们给出了连续、离散和量子α-共形不等式。也就是说,在时间尺度的情况下\(\mathbb{T}=\mathbb{R}\),\(\mathbb{T}=h\mathbb{Z}\),\(\mathbb{T}=\mathbb{Z}\)、和\(\mathbb{T}=q^{mathbb{无}_{0}}\)。
推论2.2
如果 \(\mathbb{T}=\mathbb{R}\) 在定理中2.1,不平等(2.1)减少到
$$开始{对齐}\int_{r}^{\infty}\frac{\lambda(t)\Phi^{p-\gamma+1}(t)}{\lambda^{\beta}(t)}(t-a)^{\gamma-1}\,dt\leqslead\int_{r}^{\finfty{\lampda(t \结束{对齐}$$
哪里
$$\begin{aligned}\Phi(t)=\int_{t}^{infty}\lambda$$
推论2.3
如果 \(\mathbb{T}=h\mathbb{Z}\) 在定理中2.1,不平等(2.1)减少到
$$开始{对齐}和\sum_{t=\frac{r}{h}}^{\infty}\frac}\lambda(ht)\Phi^{p-\gamma+1}(ht)}{\lambda^{\beta}(ht-h)}\bigl(\rho^{\gamma-1}{\infty}\frac{\lambda(ht)\lambda^{p-\gamma+1}(ht)f^{p-\ gamma+1(ht)}{\lambda({\beta}(ht-h)}\bigl(\rho^{\gamma-1}(ght)-a \biger){h}^{(\gamma-1)},\end{aligned}$$
哪里
$$\开始{对齐}&\Phi(t)=h\sum_{s=\frac{t}{h}}^{infty}\lambda(hs)f(hs{h} -1个}\λ(hs)\bigl(\rho^{gamma-1}(hs)-a\bigr){h}^{(\gamma-1)}。\结束{对齐}$$
推论2.4
对于 \(\mathbb{T}=\mathbb{Z}\),我们只需要 \(h=1) 在推论中2.3。在这种情况下,不平等(2.1)减少到
$$开始{对齐}\sum_{t=r}^{\infty}\frac{\lambda(t)\Phi^{p-\gamma+1}(t)}{\lambda^{\beta}(t-1)}\bigl f^{p-\gamma+1}(t)}{\lambda^{\beta}(t-1)}\bigl(\rho^{\gamma-1}(t)-a\bigr)^{(\gamma-1)},\end{aligned}$$
哪里
$$\begin{aligned}\Phi(t)=\sum_{s=t}^{\infty}\lambda(s)f(s)\bigl(\rho^{\gamma-1}(s)-a\bigr)^{(\gamma-1)}\quad\textit{和}\quade\lambda(t)=\sum_{s=r}^{t-1}\lampda(s。\结束{对齐}$$
推论2.5
如果 \(\mathbb{T}=q^{mathbb{无}_{0}}\) 在定理中2.1,不平等(2.1)减少到
$$开始{对齐}和\sum_{t\in(r,\infty)}\frac{t(\rho^{\gamma-1}(t)-a){\tilde{q}}^{(\gamma-1)}\lambda(t)\Phi^{p-\gamma+1}{\gamma-1}(t)-a){\tilde{q}}^{(\gamma-1)}\lambda(t)\lambda^{p-\gamma+1})},\结束{对齐}$$
哪里
$$\开始{对齐}&\Phi(t)=(\波浪号{q} -1个)\sum_{s\in(t,\infty)}s\lambda(s)f(s)\bigl(\rho^{\gamma-1}(s)-a\bigr)_{\tilde{q}}^{(\gamma-1)}\quad\textit{和}\\&\lambda(t)=(\tilde{q} -1个)\sum{s\in(r,t)}s\lambda(s)\bigl(\rho^{\gamma-1}(s)-a\bigr){\tilde{q}}^{(\gamma-1)}。\结束{对齐}$$
讨论不等式很有趣(2.1)改变积分极限后\(\int_{r}^{t}\lambda(s)\nabla_{a}^{gamma}}s\)来自吨至∞。让我们在下面的定理中这样做。
定理2.6
在定理的相同假设下2.1具有 \(β>1\),那么我们就有了
$$\begin{aligned}\int _{r}^{\infty}\frac{\lambda(t)\Phi^{p-\gamma+1}(t)}{\Omega ^{β-\gamma+1}(t)}\nabla _{a}^{\gamma}t\leqslant\frac{p-\gamma+1}{β-\gamma}\int _{r}^{\infty}\lambda(t)\Omega ^{p-\gamma+1}(t)\nabla _{a}^{\gamma}t,\结束{对齐}$$
(2.7)
哪里
$$\beagin{aligned}\Phi(r)=0,\qquad\Phi(t)=\int _{t}^{\infty}\lambda(s)f(s)\nabla _{a}^{\gamma}s\quad\textit{and}\quad\Omega(t)=\int _{t}^{\infty}\lambda(s)\nabla _{a}^{\gamma}s \end{aligned}$$
证明
自(f)没有减少,我们有\(t \geq x \geq r)
$$\begin{aligned}\Phi(x)=\int_{x}^{infty}\lambda(s)f(s)\nabla_{a}^{gamma}s\geqslatef(x)\Omega(x)。\结束{对齐}$$
所以,
$$\开始{对齐}f(x)\Phi^{p-\gamma}(x)\ geqslide\Omega^{p-\ gamma}(x)f^{p-\gamma+1}(x)。\结束{对齐}$$
(2.8)
利用链式法则(1.9)和使用\(\nabla_{a}^{gamma}(\Phi(x))=-\lambda(x)f(x)\leq0),我们得到
$$\开始{对齐}\nabla_{a}^{\gamma}\bigl(\Phi^{p-\gamma+1}(x)\bigr)&=(p-\gama+1)\nabla _{a}^{\gamma}\bigl(\ Phi(x)\ bigr leqslate-(p-\gamma+1)\lambda(x)f(x)\int_{0}^{1}\bigl[h\Phi(x)+(1-h)\Phi,dh\\&=-(p-\gamma+1)\lambda(x)f(x)\Phi^{p-\garma}(x)。\结束{对齐}$$
(2.9)
发件人(2.8)和(2.9)我们得到了
$$\begin{aligned}\nabla_{a}^{gamma}\bigl(\Phi^{p-\gamma+1}(x)\biger)\leqslate-(p-\gama+1)\lambda(x)\ Omega^{p-\ gamma}(x)f^{p-\gamma+1}(z)。\结束{对齐}$$
因此,
$$开始{对齐}\frac{\lambda(t)\nabla_{a}^{gamma}(\Phi^{p-\gamma+1}(x)(t)}。\结束{对齐}$$
因此,在整合双方x个结束\([r,t]_{\mathbb{t}}\),
$$开始{对齐}\frac{\lambda(t)[\Phi^{p-\gamma+1}(t)}\nabla{a}^{gamma}x\\&\leqslate-(p-\gamma+1)\int_{r}^{t}\frac{\lambda(t)\lambda(x)\Omega^{p-\gama}(x)f^{p-\gamma+1}(x)}{\Omega^{\beta-\gamma+1}(t)}\nabla{a}^{\gamma}x.\end{aligned}$$
自\(\Phi^{p-\gamma+1}(r)=0\),我们有
$$开始{对齐}\frac{\lambda(t)\Phi^{p-\gamma+1}(t)}{\Omega^{\beta-\gamma+1}}\nabla{a}^{gamma}x.end{aligned}$$
然后,通过整合双方吨结束\([r,\infty)_{\mathbb{T}}\),我们得到
$$开始{对齐}和\int_{r}^{\infty}\frac{\lambda(t)\Phi^{p-\gamma+1}(t)}{\Omega^{\beta-\gamma+1}da(x)\Omega^{p-\gamma}(x)f^{p-\ gamma+1}(x)}{\Omega ^{beta-\gamma+1{(t)}\nabla_{a}^{gamma}x\biggr]\nabla_{a}^{\gamma}t.\end{aligned}$$
(2.10)
借助于Fubini时间尺度定理,不等式(2.10)可以重写为
$$开始{对齐}和\int_{r}^{\infty}\frac{\lambda(t)\Phi^{p-\gamma+1}(t)}{\Omega^{\beta-\gamma+1}}(x)\biggl[\int_{x}^{\infty}-\lambda(t)\Omega^{-(beta-\gamma+1)}(t)\nabla_{a}^{gamma}t\biggr]\nabla{a}^{gamma}x.end{aligned}$$
(2.11)
来自链式规则(1.10),存在\(d\in[\rho(t),t]\)具有
$$开始{对齐}-\nabla_{a}^{\gamma}\bigl。\结束{对齐}$$
(2.12)
组合(2.12)和(2.11)收益率
$$\开始{对齐}&\int_{r}^{\infty}\frac{\lambda(t)\Phi^{p-\gamma+1}(t(x)f^{p-\gamma+1}(x)\biggl[\int_{x}^{infty}-\nabla_{a}^{gamma}\bigl(\Omega^{-\beta+\gamma}(t)\biger)\nabla{a}^{gamma}t\biggr]\nabla{a}^{ga玛}x \\&\quad=\frac{p-\gamma+1}{\beta-\gamma}\int_{r}^{infty}\lambda(x)\Omega^{p-\beta}(x)f^{p-\ gamma+1}(x)\nabra{a}{gamma x,\end{aligned}$$
从哪个不等式(2.7)如下。□
我们准备将我们的结果的几个特殊情况呈现给连续、离散和量子α-共形不等式。也就是说,在时间尺度的情况下\(\mathbb{T}=\mathbb{R}\),\(\mathbb{T}=h\mathbb{Z}\),\(\mathbb{T}=\mathbb{Z}\)、和\(\mathbb{T}=q^{mathbb{无}_{0}}\)。
推论2.7
如果 \(\mathbb{T}=\mathbb{R}\) 在定理中2.6,然后是不等式(2.7)归结为
$$开始{aligned}\int_{r}^{infty}\frac{\lambda(t)\Phi^{p-\gamma+1}(t)}{\Omega^{\beta-\gamma+1}(t)}(t-a)^{\gamma-1}\,dt\leqsleat\frac{p-\gamma+1}{\beta-\gamma}\int_{r}{infty}\lambda^{(t p-\gamma+1}(t)(t-a)^{\gamma-1}\,dt,\end{aligned}$$
哪里
$$\begin{aligned}\Phi(t)=\int_{t}^{infty}\lambda$$
推论2.8
如果 \(\mathbb{T}=h\mathbb{Z}\) 在定理中2.6,然后是不等式(2.7)归结为
$$开始{对齐}&\sum_{t=\frac{r}{h}}^{\infty}\frac}\lambda(ht)\Phi^{p-\gamma+1}(ht)}{\Omega^{\beta-\gamma+1}\gamma}\sum_{t=\frac{r}{h}}^{\infty}\lambda(ht)\Omega^{p-\gamma}(ht)f^{p-\ gamma+1}(tt)\bigl(\rho^{\gamma-1}(ght)-a \biger){h}^{(\gamma-1)},\end{aligned}$$
哪里
$$\开始{对齐}&\Phi(t)=h\sum_{s=\frac{t}{h}}^{infty}\lambda(hs)f(hs \rho^{\gamma-1}(hs)-a\biger){h}^{(\gamma-1)}。\结束{对齐}$$
推论2.9
对于 \(\mathbb{T}=\mathbb{Z}\),我们只需要 \(h=1) 在推论中2.8。在这种情况下,不平等(2.7)归结为
$$开始{对齐}&\sum_{t=r}^{\infty}\frac{\lambda(t)\Phi^{p-\gamma+1}(t)}{\Omega^{\beta-\gamma+1}\infty}\lambda(t)\Omega^{p-\beta}(t)f^{p-\gamma+1}(t)\bigl(\rho^{\gamma-1}(c)-a\bigr)^{(\gamma-1)},\结束{对齐}$$
哪里
$$\begin{aligned}\Phi(t)=\sum_{s=t}^{\infty}\lambda(s)f(s)\bigl(\rho^{\gamma-1}(s)-a\bigr)^{(\gamma-1)}\quad\textit{和}\quade\Omega(t)=\sum_{s=t}^{\infty}\lamda(s)\ bigl第页。\结束{对齐}$$
推论2.10
如果 \(\mathbb{T}=q^{mathbb{无}_{0}}\) 在定理中2.6,然后是不等式(2.7)归结为
$$\开始{对齐}和\sum_{t\in(r,\infty)}\frac{t\lambda(t)\Phi^{p-\gamma+1}(t sum_{t\in(r,\infty)}t\lambda(t)\Omega^{p-\beta}(t)f^{p-\γ+1}(t)\bigl(\rho^{\γ-1}(c)-a \biger){\波浪线{q}}^{(\gamma-1)},\end{aligned}$$
哪里
$$\开始{对齐}&\Phi(t)=(\波浪号{q} -1个)\sum_{s\in(t,\infty)}s\lambda(s)f(s)\bigl(\rho^{\gamma-1}(t)-a\bigr)_{\tilde{q}}^{(\gamma-1)}\quad\textit{和}\\&\Omega(t)=(\tilde{q} -1个)\sum{s\in(t,\infty)}s\lambda(s)\bigl(\rho^{\gamma-1}(t)-a\bigr){\tilde{q}}^{(\gamma-1)}。\结束{对齐}$$
定理2.11
在定理的相同假设下2.1具有 \(0\leqslant\β\leqslant1\),我们有这个
$$开始{aligned}\int_{r}^{infty}\frac{\lambda(t)\Psi^{p-\gamma+1}(t)}{\lambda^{\beta}(t)}\nabla{a}^{\gamma}t\leqsleat\frac{p-\gamma+1}{\beta-1}\int_{r}{\infty{\lampda(t gamma+1}(t)\nabla{a}^{gamma}t,\end{aligned}$$
(2.13)
哪里
$$\开始{aligned}\Psi(t)=\int_{r}^{t}\lambda(s)f$$
证明
作为(f)没有减少,我们有\(x\geqr)
$$\开始{对齐}\Psi(x)=\int_{r}^{x}\lambda(s)f$$
然后
$$\开始{对齐}f(x)\Psi^{p-\gamma}(x)\倾斜f^{p-\ gamma+1}(x)\Lambda^{p-\fgamma}(x)。\结束{对齐}$$
(2.14)
使用链式法则(1.9)事实上\(\nabla_{a}^{gamma}(\Psi(x))=\lambda(x)f(x)\geq0\),我们得到
$$\开始{对齐}\nabla_{a}^{\gamma}\bigl(\Psi^{p-\gamma+1}(x)\bigr)&=(p-\gama+1)\nabla _{a}^{\gamma}\bigl(\ Psi(x)\ bigr leqslate(p-\gamma+1)\lambda(x)f(x)\int_{0}^{1}\bigl[h\Psi(x)+(1-h)\Psi\lambda(x)f(x)\Psi^{p-\gamma}(x)。\结束{对齐}$$
(2.15)
组合(2.14)带有(2.15)给予
$$开始{aligned}\nabla_{a}^{gamma}\bigl$$
因此
$$开始{对齐}\frac{\lambda(t)\nabla_{a}^{\gamma}(\Psi^{p-\gamma+1}(x))}{\lambda^{\beta}(t)}\leqsleat\frac{(p-\gama+1)\lambda\(t)\ lambda。\结束{对齐}$$
因此,
$$开始{对齐}\frac{\lambda(t)\nabla_{a}^{\gamma}\Psi^{p-\gamma+1}(t)}{\lambda^{\beta}(t)}&=\int_{r}^{t}\frac{\lampda(t x\\&\leqslate(p-\gamma+1)\int_{r}^{t}\frac{\lambda(t)\lambda(x)\lambda^{p-\gama}(x)f^{p-\ gamma+1}(x)}{\Lambda^{\beta}(t)}\nabla_{a}^{\gamma}x,\end{aligned}$$
因此
$$开始{对齐}和\int_{r}^{\infty}\frac{\lambda(t)\Psi^{p-\gamma+1}(t)}{\lambda^{\beta}^{p-\gamma}(x)f^{p-\ gamma+1}(x)}{\lambda^{\beta}(t)}\nabla_{a}^{\gamma{x\biggr]\nabla_{a}^{\gamma}t.\结束{对齐}$$
(2.16)
利用时间尺度上的Fubini定理,不等式(2.16)可以重写为
$$开始{对齐}和\int_{r}^{\infty}\frac{\lambda(t)\Psi^{p-\gamma+1}(t)}{\lambda^{\beta}gl[\int_{x}^{infty}\lambda(t)\lambda^{-\beta}(t)\nabla_{a}^{gamma}t\biggr]\nabla_a}^{gamma}x.\结束{对齐}$$
(2.17)
再次使用链式法则(1.10),有\(c在[\rho(t),t]\中)这样的话
$$开始{aligned}\nabla_{a}^{gamma}\Lambda^{-\beta+1}(t)&=(1-\beta)\Lambda ^{-\ beta}。\结束{对齐}$$
(2.18)
替换(2.18)到(2.17)给予
$$开始{aligned}和\int_{r}^{infty}\frac{\lambda(t)\Psi^{p-\gamma+1}(t)}{\lambda^{\beta}(t)}\nabla{a}^{\gamma}t\&\quad\leqsleat\frac{p-\gamma+1}{1-\beta{\int_r}{\infty{\lampda(x)\lambda ^{p-\ gamma}(x)f^{p-\gamma+1}(x)\biggl[\int_{x}^{infty}\nabla_{a}^{gamma}\lambda^{-\beta+1}t\biggr]\nabla_{a}^{gamma}x\\&\quad=\frac{p-\gamma+1}{\beta-1}\int_{r}^{\infty}\lambda(x)\lambda^{p-\gamma-\beta+1}(x)f^{p-\ gamma+1}(x)\nabla{a}^{\gamma}x.\end{aligned}$$
这就完成了证明。□
再次,我们将我们的结果的一些特殊情况呈现给连续、离散和量子α-共形不等式。也就是说,在时间尺度的情况下\(\mathbb{T}=\mathbb{R}\),\(\mathbb{T}=h\mathbb{Z}\),\(\mathbb{T}=\mathbb{Z}\)、和\(\mathbb{T}=q^{mathbb{无}_{0}}\)。
推论2.12
如果 \(\mathbb{T}=\mathbb{R}\) 在定理中2.11,不平等(2.13)减少到
$$开始{aligned}\int_{r}^{\infty}\frac{\lambda(t)\Psi^{p-\gamma+1}(t)}{\lambda^{\beta}(t)}(t-a)^{\gamma-1}\,dt\leqsleat\frac{p-\gamma+1}{\beta-1}\int_{r}{\inffy}\lambda^{p-\betaneneneep(t)f^{p-\gamma+1}(t)(t-a)^{α-1},dt,结束{对齐}$$
哪里
$$\begin{aligned}\Psi(t)=\int_{r}^{t}\lambda$$
推论2.13
如果 \(\mathbb{T}=h\mathbb{Z}\) 在定理中2.11,不平等(2.13)减少到
$$开始{对齐}&\sum_{t=\frac{r}{h}}^{\infty}\frac}\lambda(ht)\Psi^{p-\gamma+1}(ht)}{\lambda^{\beta}(int)}\bigl(\rho^{\gamma-1}(ght)-a\bigr)_{h}^{(\gamma-1{t=\frac{r}{h}}^{infty}\lambda(ht)\lambda^{p-\beta}(ht)f^{p-\gamma+1}(ght)\bigl(\rho^{gamma-1}(hr)-a\bigr)_{h}^{(\gamma-1)},\end{对齐}$$
哪里
$$\begin{aligned}&\Psi(t)=h\sum_{s=\frac{r}{h}}^{\frac{t}{h} -1个}\λ(h)f(hs)\bigl(\rho^{\gamma-1}(ht)-a\bigr)_{h}^{(\gamma-1)}\quad\textit{和}\\&\lambda(t)=h\sum_{s=\frac{r}{h}}^{\frac}t}{h} -1个}\lambda(hs)\bigl(\rho^{gamma-1}(ht)-a\bigr){h}^{(\gamma-1)}。\结束{对齐}$$
推论2.14
对于 \(\mathbb{T}=\mathbb{Z}\),我们只需要 \(h=1) 在推论中2.13。在这种情况下,不平等(2.13)减少到
$$\开始{对齐}&\sum_{t=r}^{\infty}\frac{\lambda(t)\Psi^{p-\gamma+1}(t)}{\lambda^{\beta}(t)}\bigl(\rho^{\gamma-1}(d)-a\bigr)^{(\gamma-1)}\&\quad\leqsleat\frac{p-\gamma+1}{\beta-1}\sum_{t=r}^{\infty}\ lambda da(t)\lambda^{p-\β}(t)f^{p-\gamma+1}(t)\bigl(\rho^{\gamma-1}(c)-a\bigr)^{(\gamma-1)},\end{aligned}$$
哪里
$$\begin{aligned}\Psi(t)=\sum_{s=r}^{t-1}\lambda(s)f(s)\bigl(\rho^{\gamma-1}(s)-a\bigr)^{(\gamma-1)}\quad\textit{和}\quadr\lambda(t)=\sum_{s=r}^{t-1}\lampda(s。\结束{对齐}$$
推论2.15
如果 \(\mathbb{T}=q^{mathbb{无}_{0}}\) 在定理中2.11,然后是不等式(2.13)减少到
$$开始{对齐}和\sum_{t\in(r,\infty)}\frac{t\lambda(t)\Psi^{p-\gamma+1}(t)}{\lambda^{\beta}在(r,t)}t \lambda(t)\lambda^{p-\beta}(t)f^{p-\γ+1}(t)\bigl(\rho^{\gamma-1}(c)-a\bigr){\tilde{q}}^{(\gamma-1)}中,\结束{对齐}$$
哪里
$$\开始{对齐}&\Psi(t)=(\波浪号{q} -1个)\sum_{s\in(r,t)}s\lambda(s)f(s)\bigl(\rho^{\gamma-1}(s)-a\bigr)_{\tilde{q}}^{(\gamma-1)}\quad\textit{and}\\&&\lambda(t)=(\tilde{q} -1个)\sum{s\in(r,t)}s\lambda(s)\bigl(\rho^{\gamma-1}(s)-a\bigr){\tilde{q}}^{(\gamma-1)}。\结束{对齐}$$
接下来我们讨论不等式(2.13)对于积分极限\(\int_{r}^{t}\lambda(s)\nabla_{a}^{gamma}s\)更改自吨至∞。
定理2.16
在定理的相同假设下2.1具有 \(贝塔斜面1),那么我们就有了
$$开始{aligned}\int_{r}^{infty}\frac{\lambda(t)\Psi^{p-\gamma+1}(t)}{(\Omega^{\rho}(t))^{\beta}}\nabla{a}^{\gamma}t\leqsleat\frac{p-\gamma+1}{1-\beta{\infty{\lampda(x)\lambda^{p-\ gamma}\(t)\欧米茄^{1-\beta}(t)f^{p-\gamma+1}(t)\nabla_{a}^{gamma}t,\end{aligned}$$
(2.19)
哪里
$$\begin{aligned}\Psi(t)=\int_{r}^{t}\lambda(s)f(s)\nabla_{a}^{gamma}s,\qquad\lambda(t)=\int_}r}^{t}\ lambda s.\end{对齐}$$
证明
自(f)没有减少,我们有\(x\geqr)
$$\开始{aligned}\Psi(x)&=\int_{r}^{x}\lambda(s)f$$
然后
$$\开始{对齐}f(x)\Psi^{p-\gamma}(x)\倾斜f^{p-\ gamma+1}(x)\Lambda^{p-\fgamma}(x)。\结束{对齐}$$
(2.20)
使用链式法则(1.9)和使用\(nabla_{a}^{gamma}\Psi(x)=\lambda(x)f(x)\geq0\),我们得到
$$\开始{对齐}\nabla_{a}^{\gamma}\bigl(\Psi^{p-\gamma+1}(x)\bigr)&=(p-\gama+1)\nabla _{a}^{\gamma}\Psi(x).int_{0}^{1}\bigr[h\Psi\bigl(\rho(x)\ bigr伽马+1)\lambda(x)f(x)\int_{0}^{1}\bigl[h\Psi(x)+(1-h)\Psif(x)\Psi^{p-\gamma}(x)。\结束{对齐}$$
(2.21)
组合(2.20)带有(2.21)导致
$$开始{aligned}\nabla_{a}^{gamma}\bigl$$
等等
$$开始{对齐}\frac{\lambda(t)\nabla_{a}^{\gamma}(\Psi^{p-\gamma+1}(x))}{(\Omega^{\rho}(t))^{\beta}}\leqsleat\frac{(p-\gama+1)\lambda(t)\ lambda(t)^{\beta}}。\结束{对齐}$$
因此,
$$开始{对齐}\frac{\lambda(t)\Psi^{p-\gamma+1}(t)}{(\Omega^{\rho}(t))^{\beta}}&=\int_{r}^{t}\frac{\lambeda(t a}^{\gamma}x\\&\leqslate(p-\gamma+1)\int_{r}^{t}\frac{\lambda(t)\lambda(x)\lambda^{p-\gama}(x)f^{p-\ gamma+1}(x)}{(\Omega^{\rho}(t))^{\beta}}\nabla_{a}^{\gamma}x,\end{aligned}$$
因此
$$开始{对齐}和\int_{r}^{\infty}\frac{\lambda(t)\Psi^{p-\gamma+1}(t)}{(\Omega^{\rho}(t))^{\beta}}\nabla{a}^{\ gamma}t\\&\quad\geq(p-\gama+1)\int_}^{{\inffy}\biggl[\int_{r}^{t}\frac{\lamda(t)\ lambda(x)\lambda^{p-\gamma}(x)f^{p-\ gamma+1}(x)}{(\Omega^{\rho}(t))^{\beta}}\nabla_{a}^{\gamma{x\biggr]\nabla_{a}^{\gamma}t.\end{aligned}$$
(2.22)
利用时间尺度上的Fubini定理,不等式(2.22)可以重写为
$$开始{对齐}&\int_{r}^{\infty}\frac{\lambda(t)\Psi^{p-\gamma+1}(t)}{(\Omega^{\rho}(t))^{\beta-\gamma+1}}\nabla{a}^{\ gamma}t\\&\quad\leqsleat(p-\gama+1)\int_}r}{\inffy}\lambda(x)\lambda^{p-\ gamma(x)f ^{p-\gamma+1}(x)\biggl[\int_{x}^{\infty}\lambda(t)\bigl(\Omega^{\rho}(t)\ bigr)^{-\beta}\nabla_{a}^{\gamma}t\biggr]\nabla{a}^{\gama}x.\end{aligned}$$
(2.23)
我们采用链式法则(1.10)再次拥有\(d\in[\rho(t),t]\)这样的话
$$\开始{对齐}\nabla_{a}^{\gamma}\Omega^{-\beta+1}(t)=(1-\beta)\Omega ^{-\ beta}(d)\nabla{a}^{\gamma}\欧米茄(t)\geq(\beta-1)\lambda(t)\ bigl。\结束{对齐}$$
(2.24)
替换(2.24)到(2.23)收益率
$$开始{对齐}和\int_{r}^{infty}\frac{\lambda(t)\Psi^{p-\gamma+1}(t)}{(\Omega^{\rho}(t))^{\beta}}\nabla{a}^{\gamma}t\\&\quad\leqsleat\frac{p-\gamma+1}{\beta-1}\int__{r{{\infty{\lampda(x)\lambda^{p-\ gamma}(x)f^{p-\gamma+1}(x)\biggl[\int_{x}^{infty}\nabla_{a}^{gamma}\bigl(\Omega^{1-\beta}(t)\biger)\nabla_{a}^{gamma}t\biggr]\nabla _{a{gamma x\\&\quad=\frac{p-\gamma+1}{1-\beta}\int_{r}^{infty}\lambda(x)\lambda^{p-\gamma}$$
这是我们想要的不等式(2.19). □
现在,作为我们结果的特例,我们将给出连续、离散和量子α-共形不等式。也就是说,在时间尺度的情况下\(\mathbb{T}=\mathbb{R}\),\(\mathbb{T}=h\mathbb{Z}\),\(\mathbb{T}=\mathbb{Z}\)、和\(\mathbb{T}=q^{mathbb{无}_{0}}\)。
推论2.17
如果 \(\mathbb{T}=\mathbb{R}\) 在定理中2.16,然后是不等式(2.19)归结为
$$开始{对齐}和\int_{r}^{\infty}\frac{\lambda(t)\Psi^{p-\gamma+1}(t)}{\Omega^{\beta}(t)}^{gamma-\beta}(t)f^{p-\gamma+1}(t)(t-a)^{gama-1},dt,end{aligned}$$
哪里
$$\开始{对齐}&\Psi(t)=\int_{r}^{t}\lambda(s)f(s)(s-a)^{\gamma-1}\,ds,\qquad\lambda(t)=\ int_{r}^{t}\lampda(s \,ds.\结束{对齐}$$
推论2.18
如果 \(\mathbb{T}=h\mathbb{Z}\) 在定理中2.16,然后是不等式(2.19)归结为
$$\开始{对齐}&\sum_{t=\frac{r}{h}}^{\infty}\frac}\lambda(ht)\Psi^{p-\gamma+1}(ht)}{\Omega^{\beta}(ht-h)}\bigl(\rho(ht)^{\gamma-1}-a\biger)^{(1-\gamma)}_{h}\\&\quad\leqslian\frac{p-\gamma+1}{1-\beta{\sum_{t=\frac{r}{h}}^{\infty}\lambda(ht)\lambda^{p-\gamma}(ht)\欧米茄^{\gamma-\beta}(ght)f^{p-\ gamma+1}(tt)\bigl(\rho(ht)^{\gamma-1}-a\biger)^{(1-\gamma)}_{h},\end{aligned}$$
哪里
$$\begin{aligned}&\Psi(t)=h\sum_{s=\frac{r}{h}}^{\frac{t}{h} -1个}\λ(hs)f(hs{h} -1个}\λ(hs)\bigl(\rho(hs)^{\gamma-1}-a\bigr)^{(1-\gamma)}_{h}\fquad\textit{and}\\&&\Omega(t)=h\sum_{s=\frac{t}{h}}^{\infty}\lambda(hs)\bigl(\rho(hs)^{\gamma-1}-a\bigr)^{(1-\gamma)}_{h}。\结束{对齐}$$
推论2.19
对于 \(\mathbb{T}=\mathbb{Z}\),我们只需要 \(h=1) 在推论中2.18。在这种情况下,不平等(2.19)归结为
$$\begin{aligned}&&sum_{t=r}^{\infty}\frac{\lambda(t)\Psi ^{p-\gamma+1}(t)}{\Omega ^{\beta}(t-1)}\bigl(\rho(t)^{\gamma-1}-a\bigr)^{(1-\gamma)}\\&&quad\leqslant\frac{p-\gamma+1}{1-\beta}\sum _{t=r}^{\infty}\lambda(t)\lambda ^{p-\gamma}(t)\Omega^{\gamma-\beta}(t)f^{p-\gamma+1}(t)\bigl(\rho(t)^{\gamma-1}-a \biger)^{(1-\gamma)},\end{对齐}$$
哪里
$$开始{对齐}&\Psi(t)=\sum_{s=r}^{t-1}\lambda(s)f(s)\bigl(s)^{\gamma-1}-a\bigr Omega(t)=\sum_{s=t}^{\infty}\lambda(s)\bigl(\rho(hs)^{\gamma-1}-a\biger)^{(1-\gamma)}。\结束{对齐}$$
推论2.20
如果 \(\mathbb{T}=q^{mathbb{无}_{0}}\) 在定理中2.16,不平等(2.19)归结为
$$开始{对齐}和\sum_{t\in(r,\infty)}\frac{t\lambda(t)\Psi^{p-\gamma+1}(t)}{\Omega^{\beta}(\rho(t))}\bigl \sum_{t\in(r,\infty)}t\lambda(t)\lambda^{p-\gamma}(t)\欧米茄^{\gamma-\beta}(t)f^{p-\ gamma+1}(t-)\bigl(\rho(t)^{\gamma-1}-a\biger)^{(1-\gamma)}{\tilde{q}},\end{aligned}$$
哪里
$$\开始{对齐}&\Psi(t)=(\波浪号{q} -1个)\sum_{s\in(r,t)}s\lambda(s)f(s)\bigl(\rho(s)^{\gamma-1}-a\bigr)^{(1-\gamma)}_{\tilde{q}},\\&\lambda(t)=(\tilde{q} -1个)\sum_{s\in[r,t]}s\lambda(s)\bigl(\rho(s)^{\gamma-1}-a\bigr)^{(1-\gamma)}_{\tilde{q}}\quad\textit{和}\\&\Omega(t)=(\tilde{q} -1个)\sum_{s\in(r,\infty)}s\lambda(s)\bigl(\rho(s)^{\gamma-1}-a\bigr)^{(1-\gamma)}{\tilde{q}}。\结束{对齐}$$