设$X_{1}, \cdot\cdot\cdot, X_{N}$是取值于$R^{d}$的$N$个独立的 (严格) 稳定过程, $X(t)=\sum^{N}\limits_{j=1}X_{j}(t_{j})$我们则称这$N$个参数$R^{d}$值的过程$X=\{X(t), t\in R^{N}_{+}\}$为一可加稳定过程.并记为$X=X_{1}\oplus\cdots\oplus X_{N}$.

本文我们主要研究$X=X_{1}\oplus\cdots\oplus X_{N}$是$X_{1}, \cdots, X_{N}$具有不同指数$\alpha_{1}$, $\cdots$, $\alpha_{N}\in(0,2]$的可加稳定过程.我们假定$X(0)=0$.

具有平稳独立增量且取值与$R^{d}$的随机过程$Z=\{Z(t); t\geq 0\}$称为Lévy过程.众所周知, 对于$t>s\geq 0, Z(t)-Z(s)$的特征函数有下列Lévy-Khintchine公式:

$ E[\exp(i\langle \xi,Z(t)-Z(s)\rangle)]=\exp[-(t-s)\Psi(\xi)], \\ \Psi(\xi)=i\langle a,\xi\rangle+\frac{1}{2}\xi\sum\xi'+\int_{R^{d}}[1-e^{i\langle x,\xi\rangle}+\frac{i\langle x,\xi\rangle}{1+\parallel x\parallel^{2}}]v(dx), \quad \xi\in R^{d}, $

其中$a$是$R^{d}$中的常数, $\sum$是一个非负正定的$d\times d$对称矩阵, $v$是$R^{d}\setminus\{0\}$上满足下式的Borel测度,

$\int_{R^{d}}\frac{\parallel x\parallel^{2}}{1+\parallel x\parallel^{2}}v(dx)<\infty, $

函数$\Psi$称为$Z$的Lévy指数, $v$成为$Z$的Lévy测度.

取值与$R^{d}$的一类特殊的Lévy过程是所谓的 (严格) 稳定过程.指数$\alpha\in(0,2]$的 (严格) 稳定过程$Z$的Lévy指数有以下形式:

$\Psi(\xi)=\sigma\parallel\xi\parallel^{\alpha}\int_{S_{d}}\omega_{\alpha}(\xi,y)\mu(dy), $

其中$\sigma>0$是一个常数,

$ \omega_{\alpha}(\xi,y)=[1-i{\hbox{sgn}}\langle\xi,y\rangle\tan(\pi\alpha/2)]|\langle\xi/\parallel\xi\parallel,y\rangle\mid^{\alpha}, \alpha\neq1,\\ \omega_{1}(\xi,y)=\mid\langle\xi/\parallel\xi\parallel,y\rangle\mid+\frac{2i}{\pi}\langle\xi,y\rangle\log\mid\langle\xi,y\rangle\mid, $

并且$\mu$是$R^{d}$中单位球面$S_{d}$上的概率测度. $\alpha=1$时, $\mu$是以原点为其质量中心, 也就是说, $\displaystyle\int_{S_{d}}y\mu(dy)=0$ (具体参见文献[1]).

如果$p(1,0)>0$ (这里$p(t,x)$是$Z(t)$的密度函数), 则对应的稳定过程称为$A$型, 反之则称为$B$型. Taylor^[2]证明了如果$\alpha\in(0,1)$, $\mu$支撑在一半球, 那么$Z$是$B$型.所有其它严稳定过程 ($\alpha\neq1$) 是$A$型.

Blumenthal和Getoor ^[8]引进上下指数概念, 用于研究Lévy过程的样本轨道性质. Lévy过程$Z$的上指数$\beta$与下指数$\beta^{''}$的定义如下:

$\beta=\inf\{\gamma\geq0:\parallel\xi\parallel^{-\gamma}{\hbox{Re}}\Psi(\xi)\rightarrow\infty, \hspace{0.5cm} \text {当}\|\xi\|\rightarrow0\}, \\ \beta^{''}=\sup\{\gamma\geq0:\parallel\xi\parallel^{-\gamma}{\hbox{Re}}\Psi(\xi)\rightarrow\infty,\hspace{0.5cm} \text{当}\|\xi\|\rightarrow\infty\}. $

我们知道$0\leq\beta^{'}\leq\beta\leq2$, 并且当$Z$为指数$\alpha$的稳定过程是则$\beta^{''}=\beta=\alpha$

关于局部时, 我们记参数 (时间) 空间为$R^{N}_{+}=[0,\infty]^{N}$, 其元素记为$t=(t_{1}, \cdot\cdot\cdot, t_{N})$, 当$t_{1}=t_{2}=\cdots=t_{N}=c$时记为$\langle c\rangle$.在$R^{N}_{+}$中定义一个序“$\preceq$”:如果对$\forall 1\leq\ell\leq N$有$s_{\ell}\leq t_{\ell}$, 则记$s\preceq t$.当$s\preceq t$时, 记$[s,t]=\prod\limits^{N}_{\ell=1}[s_{\ell}, t _{\ell}]$.令$\mathfrak A$表示所有$N$维区间$I=[s,t]\subset R^{N}_{+}$的集合, $\mathfrak A=\{I=[s,t]\in\mathfrak A: 0<s_{\ell}<t_{\ell}, 1\leq\ell\leq N\}$, $\lambda_{N}$表示$N$维Lebesgus测度.

状态空间$R^{d}$赋予$\ell^{2}$欧氏范数$\parallel\cdot\parallel$, 内积$\langle x, y\rangle=\sum\limits^{d}_{j=1} x_{j}y_{j}\ (x, y\in R^{d})$和$\ell^{\infty}$范数$|x|=\max_{1\leq\ell\leq d}|x_{\ell}|$($x\in R^{d}$).

$I \subset R^{N}$ Borel集, 下设$I=[0,1]^{N}$, 记$\mathcal{B}(I)$, $\mathcal{B}(R^{N})$分别为$I$, $R^{N}$上的Borel $\sigma$代数, 若$X(t)$: $I\rightarrow R^{d}$是Borel函数, 则称$X$为$(N,d)$向量场对于$\forall A\in\mathcal{B}(R^{d})$, $\forall B\in\mathcal{B}(I)$, 令

$\mu(A,B)=\lambda_{N}\{X^{-1}(A)\cap B\}=\lambda_{N}\{t:\in B,X(t)\in A\}=\int_{B}1_{A}(X(t))dt, $

其中$\lambda_{N}$是$R^{N}$中的Lebesgus测度.固定$B$时, $\mu(\cdot,B)$是$\mathcal{B}(R^{d})$上的一个测度, 若将$B$看成时间, 则$\mu(A,B)$可以看成在$B$这段时间内, $X(t)$在$A$中停留的时间.我们称$\mu(\bullet,B)$为$X(t) (t\in B)$的逗留时分布, 记$\mu(\bullet)=\mu(\bullet,I)$.

如果$\mu(\bullet)$关于$R^{d}$上的Lebesgue测度$\lambda_{d}$绝对连续, 即$\mu\ll\lambda_{d}$, 则称$X$具有局部时.此时对$\forall B\in\mathcal{B}(I)$, $\mu(\bullet, B)\ll\lambda_{d}$, 我们将其Radon-Nikodym导数$\displaystyle\frac{d\mu(\bullet,B)}{d\lambda_{d}}$称为$X(t)(t\in B)$的局部时, 记为$L(x,B)$由Radon-Nikodym定理, $L(x,B)$是$(R^{d}, \mathcal{B}(R^{d}))$上的可测函数, 且

$\mu(A,B)=\int_{A}L(x,B)dx,\ \ \ \ \forall A\in\mathcal B(R^{d}), $

其中$L(x,B)$可以看成$X(t)\ (t\in B)$在$x$处的逗留时间.若$B=[0,t]$, 则记$L(x,B)=L(x,t).$

由鞅和单调性等经典理论, 我们能推断出局部时存在一个可测的修正满足下面的占有密度公式:对任意的$B\in \mathcal{B}(R^{N})$和任意的可测函数$f$: $R^{d}\rightarrow R$有

$\int_{B}f(X(t))dt=\int_{R^{d}}f(x)L(x,B)dx.$

因此可以在$T=\prod\limits^{N}_{i=1}[a_{i}, a_{i}+h_{i}]\in\mathfrak{A}$找到一个连续的修正使得

$R^{d}\mathop \prod \limits_{i = 1}^N [0,h_{i}] \ni(x,t_{1},\cdots,t_{N})\mapsto L(x,\mathop \prod \limits_{i = 1}^N [a_{i},a_{i}+t_{i}]),$

则称$X$在$T$上的局部时联合连续.

假定$X(t)$(也记为$X_{t}$) 为一可加稳定过程.令

$Z(T)=Z(t_{1},\cdots,t_{r})=(X_{t_{2}}-X_{t_{1}},\cdots,X_{t_{r}}-X_{t_{r-1}}), $

其中$T=(t_{1},\cdots,t_{r})\in R_{+}^{Nr}$, $t_{m}=(t_{m,1},\cdots,t_{m,N})\in R_{+}^{N}\quad (1\leq m\leq r)$, 则称$\{Z(T), T\in R_{+}^{Nr}\}$为$X$的汇合过程.如果$Z$的局部时存在 (记为$L(x,I)$), 则称$L(x,I)$为$X$的自相交局部时,

$I=\prod\limits_{m = 1}^r {{I_m}} (I_{m}=\mathop \prod \limits_{l = 1}^m [a_{m,l},a_{m,l}+h], a_{m,l}+h<a_{m+1,l}, h>0)$

为$R^{Nr}_{+}$中的超立方体, $\mathfrak{R}$表示$R^{Nr}_{+}$中所有超立方体$I$的集合.对于$T\in R^{Nr}_{+}$, 这里$T=(t_{1},\cdots,t_{r})$, $t_{m}=(t_{m,1},\cdots,t_{m,N})$ $1\leq m\leq r$.当$t_{m,l}=c$, $1\leq m\leq r$, $1\leq l\leq N$, 记$T=\langle c\rangle$. $Z(T)$的局部时称为$X(t)$的自相交局部时.

令$\bar{\alpha}=\max\{\alpha_{1},\cdots,\alpha_{N}\}, \underline{\alpha}=\min\{\alpha_{1},\cdots,\alpha_{N}\},\alpha'=\sum^{N}\limits_{l=1}\alpha_{l}/N$, $C_{1},C_{2},\cdots$为不同的常数.

本文主要研究$A$型的不同指数可加稳定过程的自相交局部时.

引理2.1 ^[5] 假定$X=X_{1}\oplus\cdots\oplus X_{N}$是$R^{d}$上的可加稳定过程, 其中$X_{1},\cdots,X_{N}$的指数分别为$\alpha_{1},\cdots,\alpha_{N}\in(0,2]$, $Z(T)=(X_{t_{2}}-X_{t_{1}},\cdots,X_{t_{r}}-X_{t_{r-1}})$.如果$Nr\underline{\alpha}>d(r-1)$那么对每个$I\in\mathfrak{R}$, $Z$有联合连续的局部时a.s.且令$\gamma\in(0,1\bigwedge(Nr\underline{\alpha}-d(r-1))/2)$存在有限常数$M_{1}$, $M_{2}$使得任何$I\in\mathfrak{R}$所有$x$, $y\in R^{d(r-1)}$和偶数$k$有

$E[L(Z(t)+x,I)/\lambda_{Nr}(I)^{1-d(r-1)/Nr\underline{\alpha}}]^{k}\leq M_{1}^{k}(k!)^{Nr}$

和

$E[\frac{L(Z(t)+x,I)-L(Z(t)+y,I)}{\lambda_{Nr}(I)^{1-(d+2\gamma) (r-1)/Nr\underline{\alpha}}\|x-y\|^{\gamma}}]^{k}\leq M_{2}^{k}(k!)^{Nr}, $

其中$t=0$或$t=I^{t}$.

引理2.2 假定$X=X_{1}\oplus\cdots\oplus X_{N}$是$R^{d}$上的可加稳定过程, 其中$X_{1},\cdots,X_{N}$的指数分别为$\alpha_{1},\cdots,\alpha_{N}\in(0,2]$, $Z(T)=(X_{t_{2}}-X_{t_{1}},\cdots,X_{t_{r}}-X_{t_{r-1}})$.如果$Nr\underline{\alpha}>d(r-1),$那么$\forall\gamma\in(0,1\bigwedge(Nr\underline{\alpha}-d(r-1))/2)$, 存在有与$I\in\mathfrak{R}$, $x$, $y\in R^{d(r-1)}$无关的正常数$b_{1}$, $b_{2}$, $M_{3}$, $M_{4}$使得任何$u>0$, 有

$P\{L(Z(t)+x,I)\geq\lambda_{Nr}(I)^{1-d(r-1)/Nr\underline{\alpha}}u^{Nr}\}\leq M_{3}e^{-b_{1}u}$

和

$P\{\frac{\|L(Z(t)+x,I)-L(Z(t)+y,I)\|}{\lambda_{Nr}(I)^{1-(d+2\gamma)(r-1)/Nr\underline{\alpha}}\|x-y\|^{\gamma}}\leq u^{Nr}\}\leq M_{4}e^{-b_{2}u}, \quad \forall x,y\in R^{d(r-1)}, x\neq y, $

其中$\tau=0$或$\tau=I^{t}$.

证 记$\Lambda$为$L(Z(t)+x,I)/\lambda_{Nr}(I)^{1-d(r-1)/Nr\underline{\alpha}}]^{k}$或

$|L(Z(t)+x,I)-L(Z(t)+y,I)|/\lambda_{Nr}(I)^{1-(d+2\gamma)(r-1)/Nr\underline{\alpha}}\|x-y\|^{\gamma}$

中任何一项, 那么由引理1.1偶数$k$有

$E\Lambda^{\frac{k}{Nr}}\leq (E\Lambda^{k})^{\frac{1}{Nr}}\leq C_{1}^{\frac{k}{Nr}}k! (k\text{为偶数}).$

由Jensen不等式有

$E\Lambda^{\frac{(k-1)}{Nr}}\leq(E\Lambda^{\frac{k}{Nr}})^{(k-1)/k}\leq [C_{1}^{\frac{k}{Nr}}k!]^{(k-1)/k}\leq C_{1}^{\frac{k-1}{Nr}}k!,$

也就是对任何的整数$m\geq0$有

$E\Lambda^{\frac{m}{Nr}}\leq C_{1}^{\frac{m}{Nr}}(m+1)!\leq C_{2} (m+1)!,$

那么当$M$足够大时, 可得

$E\exp(\Lambda^{\frac{1}{Nr}}/M)\leq E(1+\mathop \sum \limits_{m = 1}^\infty \frac{1}{m!}(\frac{\Lambda^{\frac{1}{Nr}}}{M})^{m})\leq C_{2}\mathop \sum \limits_{m = 0}^\infty (m+1)(\frac{1}{M})^{m}\leq C_{3}.$

由Chebyshev不等式得

$P(\Lambda\geq(Mu)^{\frac{1}{Nr}})=P(\frac{\Lambda^{\frac{1}{Nr}}}{M}\geq u)=P(e^{\frac{\Lambda^{\frac{1}{Nr}}}{M}}\geq e^{u})\leq \frac{1}{e^{u}}E(e^{\frac{\Lambda^{\frac{1}{Nr}}}{M}})\leq C_{3} e^{-u}.$

令$Mu$换成$u$便得要证的结论.

引理2.3 设$X=X_{1}\oplus\cdots\oplus X_{N}$是$R^{d}$上的可加稳定过程, 其中$X_{1},\cdots,X_{N}$的指数分别为$\alpha_{1},\cdots,\alpha_{N}$.那么存在一个正常数$M_{5}$使得对时间$I=[0,a]\in\mathfrak{A}$和$0<\xi<1$,

$P\{\mathop {\sup }\limits_{t \in I} \|X(t)\|\geq\xi\}\leq M_{5}|a|\xi^{-\bar{\alpha}}.$

证 由$\alpha$阶stable过程的一般结论 (Bertoin ^[9], P221) 可以有

$ P\{\mathop {\sup }\limits_{t \in I} \|X(t)\|\geq\xi\} \leq P\{\mathop {\sup }\limits_{t \in I}\mathop \sum \limits_{l = 1}^N |X_{l}(t_{l})|\geq\xi\} \le \mathop \sum \limits_{l = 1}^N P\{\mathop {\sup }\limits_{{t_l} \in [0,{a_l}]} t_{l}^{\frac{1}{\alpha_{l}}}|X_{l}(1)|\geq\xi\}\\ \le \mathop \sum \limits_{l = 1}^N P\{\mathop {\sup }\limits_{{t_l} \in [0,{a_l}]}|X_{l}(1)|\geq\xi t_{l}^{-\frac{1}{\alpha_{l}}}\} \leq \mathop \sum \limits_{l = 1}^N P\{|X_{l}(1)|\geq\xi|a|^{-\frac{1}{\alpha_{l}}}\} \sim \mathop \sum \limits_{l = 1}^N C_{4}[\xi|a|^{-\frac{1}{\alpha_{l}}}]^{-\alpha_{l}}\\ \leq M_{5}|a|\xi^{-\bar{\alpha}}. $

引理2.4 假定$X=X_{1}\oplus\cdots\oplus X_{N}$是$R^{d}$上的可加稳定过程, 其中$X_{1},\cdots,X_{N}$的指数分别为$\alpha_{1},\cdots,\alpha_{N}\in(0,2]$, $Z(T)=(X_{t_{2}}-X_{t_{1}},\cdots,X_{t_{r}}-X_{t_{r-1}})$.如果$Nr\underline{\alpha}>d(r-1)$那么存在正常数$M_{6}$, 使得对所有$I\in\mathfrak{R}$, $0<\xi\leq 1$, 有

$P(\mathop {\sup }\limits_{s \in I} \|Z(s)-Z(I^{t})\|\geq \xi)\leq M_{6} \xi^{-\bar{\alpha}}h, $

其中$I=\prod\limits^{r}_{m=1}I_{m}$, $I_{m}=\prod\limits^{N}_{l=1}[a_{ml},a_{ml}+h]$.

证 由定义和引理2.3有

$ P\{\mathop {\sup }\limits_{s \in I} \|Z(s)-Z(I^{t})\|\geq\xi\}\\ = P\{\mathop {\sup }\limits_{s \in I} \|(X(s_{2})-X(I^{t}_{2}))-(X(s_{1})-X(I^{t}_{1})),\cdots,\\ (X(s_{r})-X(I^{t}_{r}))-(X(s_{r-1})-X(I^{t}_{r-1}))\|\geq\xi\}\\ \leq P\{\bigcup\limits_{1 \le m \le r - 1}\mathop {\sup }\limits_{s \in I} \|(X(s_{m+1})-X(I^{t}_{m+1}))-(X(s_{m})-X(I^{t}_{m}))\|\geq\frac{\xi}{\sqrt{r-1}}\}\\ \le \mathop \sum \limits_{m = 1}^{r - 1} P\{\mathop {\sup }\limits_{s \in I} \|(X(s_{m+1})-X(I^{t}_{m+1}))-(X(s_{m})-X(I^{t}_{m}))\|\geq\frac{\xi}{\sqrt{r-1}}\}\\ \le \mathop \sum \limits_{m = 1}^{r - 1} [P\{\mathop {\sup }\limits_{{s_{m + 1}} \in {I_{m + 1}}} \|(X(s_{m+1})-X(I^{t}_{m+1}))\|\geq\frac{\xi}{\sqrt{2(r-1)}}\}\\ +P\{\mathop {\sup }\limits_{{s_m} \in {I_m}} \|X(s_{m})-X(I^{t}_{m})\|\geq\frac{\xi}{\sqrt{2(r-1)}}\}]\\ \leq 2(r-1)C_{5}(\frac{\xi}{\sqrt{2(r-1)}})^{-\bar{\alpha}}h\\ = M_{6}\xi^{-\bar{\alpha}}h. $

定理2.1 假定$X=X_{1}\oplus\cdots\oplus X_{N}$是$R^{d}$上的可加稳定过程, 其中$X_{1},\cdots,X_{N}$的指数分别为

$\alpha_{1},\cdots,\alpha_{N}\in(0,2], Z(T)=(X_{t_{2}}-X_{t_{1}},\cdots,X_{t_{r}}-X_{t_{r-1}}).$

记$L^{*}(B)=\sup\limits_{x\in R^{d(r-1)}}L(x,B),B\in \mathfrak{R}$.如果$ Nr\underline{\alpha}>d(r-1)$, 那么对

$\displaystyle\forall\gamma\in (0,1\wedge\frac{Nr\underline{\alpha}-d(r-1)}{ 2}\wedge\frac{Nr\bar{\alpha}\underline{\alpha}-d\bar{\alpha}(r-1)}{2\bar{\alpha}(r-1)-\underline{\alpha}})$

存在正常数$M_{6},M_{7}$, 使得对每个$s\in(0,\infty)^{Nr}$, $s=(s_{1},\cdots, s_{r})$, $s_{m}=(s_{m1},\cdots,s_{mr})$, $(1\leq m\leq r)$, $s_{m+1 j}>s_{mj}$和每个$I\in \mathfrak{R}$有下列两式成立

$\mathop {\lim \sup }\limits_{u \to 0} \frac{L^{*}([s-\langle u\rangle, s+\langle u\rangle])}{u^{Nr-d(r-1)/\underline{\alpha}}(u^{\frac{\underline{\alpha}-2\bar{\alpha}(r-1)}{Nr\bar{\alpha}\underline{\alpha}}\gamma}\log\log u^{-1})^{Nr}}\leq M_{6} \quad {\hbox{a.s.}}$

(2.1)

和

$\lim_{\varepsilon\rightarrow 0}\mathop {\sup }\limits_{B \in \Re ,B \subset I{\lambda _{Nr}}(B)<\varepsilon } \frac{L^{*}(B)}{\lambda_{Nr}(B)^{1-d(r-1)/Nr\underline{\alpha}}(\log\lambda_{Nr}(B)^{-1})^{Nr}}\leq M_{7} \quad {\hbox{a.s.}}.$

(2.2)

证 先证明 (2.1) 式, 对于固定的$s$, 令$s^{n}$为一点列, 每个$s^{n}$具有$s$一样的条件, 满足$s^{n}_{m+1 l}>s^{n}_{m l} (\forall n,1\leq m\leq r,1\leq l\leq N)$和$\lim\limits_{n\rightarrow\infty}s^{n}=s$.定义一系列超立方体

$I_{n}=[s^{n},s^{n}+\langle 2^{-n}\rangle]\quad (n\geq 1). $

令

$g(u)=u^{Nr-d(r-1)/\underline{\alpha}}(u^{\frac{\underline{\alpha}-2\bar{\alpha}(r-1)}{Nr\bar{\alpha}\underline{\alpha}}\gamma}\log\log u^{-1})^{Nr}\quad (u\text{充分小}),$

对于$0<\gamma<1\wedge\displaystyle\frac{Nr\underline{\alpha}-d(r-1)}{2}\wedge\frac{Nr\bar{\alpha}\underline{\alpha}-d\bar{\alpha}(r-1)}{2\bar{\alpha}(r-1)-\underline{\alpha}}$, $u$充分小时

$ g'(u)=[u^{Nr-d(r-1)/\underline{\alpha}+\frac{\underline{\alpha} -2\bar{\alpha}(r-1)}{\bar{\alpha}\underline{\alpha}}\gamma}(\log\log\frac{1}{u})^{Nr}]'\\ =\frac{Nr\bar{\alpha}\underline{\alpha}-d(r-1)\bar{\alpha}-(2\bar{\alpha}(r-1) -\underline{\alpha})\gamma}{\bar{\alpha}\underline{\alpha}} u^{\frac{Nr\bar{\alpha}\underline{\alpha}-d(r-1)\bar{\alpha}-(2\bar{\alpha}(r-1) -\underline{\alpha})\gamma}{\bar{\alpha}\underline{\alpha}}-1}(\log\log\frac{1}{u})^{Nr}\\ +u^{\frac{Nr\bar{\alpha}\underline{\alpha}-d(r-1)\bar{\alpha}-(2\bar{\alpha}(r-1) -\underline{\alpha})\gamma}{\bar{\alpha}\underline{\alpha}}}Nr(\log\log\frac{1}{u})^{Nr-1} \frac{1}{\log u^{-1}}\frac{1}{u^{-1}}(-\frac{1}{u^{2}})\\ =(\frac{Nr\bar{\alpha}\underline{\alpha} -d(r-1)\bar{\alpha}-(2\bar{\alpha}(r-1)-\underline{\alpha})\gamma}{\bar{\alpha}\underline{\alpha}} -Nr\frac{1}{\log\log u^{-1}\log u^{-1}})\\ u^{\frac{Nr\bar{\alpha}\underline{\alpha} -d(r-1)\bar{\alpha}-(2\bar{\alpha}(r-1)-\underline{\alpha})\gamma}{\bar{\alpha}\underline{\alpha}} -1}(\log\log\frac{1}{u})^{Nr} >0, \\ g(2u)/g(u)=2^{Nr-d(r-1)/\underline{\alpha}+\frac{\underline{\alpha}-2\bar{\alpha}(r-1)} {\bar{\alpha}\underline{\alpha}}\gamma}(\frac{\log\log(2u)^{-1}}{\log\log u^{-1}})^{Nr}\\ \leq 2^{Nr-d(r-1)/\underline{\alpha}+\frac{\underline{\alpha}-2\bar{\alpha}(r-1)} {\bar{\alpha}\underline{\alpha}}\gamma} \leq C_{6}. $

则$g(r)$为一单调增, 限制增长的函数.因此只需证$\displaystyle\mathop {\lim \sup }\limits_{u \to 0} \frac{L(I_{n})}{g(2^{-n})}\leq M_{6}$.记$Z_{n}=Z(s^{n})$.分几步来证明:

(1) 由引理1.4得当$n$足够大时$2^{-n/\bar{\alpha}}n^{\beta}<1$对任何的$\beta>0$有

$ P(\mathop {\sup }\limits_{s \in {I_n}} \|Z(s)-Z_{n}\|\geq2^{-n/\bar{\alpha}}n^{\beta})\\ =P(\mathop {\sup }\limits_{t \in {{[0,{2^{ - n}}]}^N}} \|Z(s)\|\geq2^{-n/\bar{\alpha}}n^{\beta}) \leq C_{7} 2^{-n}(2^{-n/\bar{\alpha}}n^{\beta})^{-\bar{\alpha}} =C_{7} n^{-\bar{\alpha}\beta}. $

选取$\beta > 1/\bar{\alpha}$那么由Borel--Cantelli引理有多足够大的$n$

$\mathop {\sup }\limits_{s \in {I_n}} \|Z(s)-Z_{n}\|< 2^{-n/\bar{\alpha}}n^{\beta}\quad {\hbox{a.s.}}.$

(2) 令$\theta_{n}=2^{-n/\bar{\alpha}}$, $n\geq1$定义

$G_{n}=\{x\in R^{d(r-1)}:x=\theta_{n}p,\text{对某个}p\in \bar{Z}^{d(r-1)},\|x\|\leq 2^{-n/\bar{\alpha}}n^{\beta}\}.$

由引理1.2有

$ P\{\mathop {\sup }\limits_{x \in {G_n}} L(Z_{n}+x,I_{n})\geq g(2^{-n})a_{1}^{Nr}\} \\le \sum\limits_{x \in {G_n}} P \{ L({Z_n} + x,{I_n}) \ge g({2^{ - n}})a_1^{Nr}\} \\ \leq C_{8}(2^{-n/\bar{\alpha}}n^{\beta}/\theta_{n})^{d(r-1)}\exp(-b_{1}a_{1}2^{n\gamma\frac{2\bar{\alpha}(r-1) -\underline{\alpha}}{Nr\bar{\alpha}\underline{\alpha}}}\log\log2^{n}) \leq C_{9} n^{-(b_{1}a_{1}-d\beta(r-1))}. $

取$a_{1}>(1+d\beta(r-1))/b_{1}$再用Borel--Cantelli引理, 就得到对足够大的$n$,

$\mathop {\sup }\limits_{x \in {G_n}} L(Z_{n}+x,I_{n})<a_{1}^{Nr}g(2^{-n})\quad {\hbox{a.s.}}.$

(2.3)

(3) 对任何两个整数$n,h\geq1$和任何$x\in G_{n}$令

$F(n,h,x)=\{y\in R^{d(r-1)}:y=x+\theta_{n}\mathop \sum \limits_{j = 1}^h \varepsilon_{j}2^{-j},\varepsilon_{j}\in\{0,1\}^{d(r-1)},1\leq j\leq k\}.$

选取$\gamma>0$和$\delta>0$满足

$N\delta r<\gamma<1\wedge(Nr\underline{\alpha}-d(r-1))/2\wedge\frac{Nr\bar{\alpha}\underline{\alpha}-d\bar{\alpha}(r-1)}{2\bar{\alpha}(r-1)-\underline{\alpha}}.$

记

$ B_{n}=\bigcup\limits_{x \in {G_n}} {\mathop \cup \limits_{h = 1}^\infty } \{|L(Z_{n}+y_{1},I_{n})-L(Z_{n}+y_{2},I_{n})|\geq\lambda_{Nr}(I_{n})^{1-(d+2\gamma)(r-1)/Nr\underline{\alpha}}\\ \|y_{1}-y_{2}\|^{\gamma}(a_{2}2^{\delta h}\log n)^{Nr},\text{对某个}y_{1},y_{2}\in F(n,h,x)\text{且}y_{1}-y_{2}=\theta_{n}\varepsilon 2^{-h}, \\ \varepsilon\in\{0,1\}^{d(r-1)}\}. $

因为对于足够大的$x$有

$\mathop \sum \limits_{h = 1}^\infty 2^{d(r-1)(h+1)}\exp(-x2^{\delta h})\leq e^{-x/2}, $

则由引理2.2有

$ P(B_{n}) \leq \sum\limits_{x \in {G_n}} \sum \limits_{h = 1}^\infty \sum\limits_{{y_1},{y_2}} P \{ |L({Z_n} + {y_1},{I_n})-L(Z_{n}+y_{2},I_{n})|\geq\lambda_{Nr}(I_{n})^{1-(d+2\gamma)(r-1)/Nr\underline{\alpha}} \\ \|y_{1}-y_{2}\|^{\gamma}(a_{2}2^{\delta h}\log n)^{Nr}\}\\\leq C_{10}(2^{-n/\bar{\alpha}}n^{\beta}/\theta_{n})^{d(r-1)}\mathop \sum \limits_{h = 1}^\infty 2^{d(r-1)(h+1)}\exp(-b_{2}a_{2}2^{\delta h}\log n) \\ \leq C_{10}(2^{-n/\bar{\alpha}}n^{\beta}/\theta_{n})^{d(r-1)}\exp(\frac{-b_{2}a_{2}\log n}{2}) \\ = C_{10} n^{-(b_{2}a_{2}/2-d\beta(r-1))}. $

从而选择足够大的$a_{2}>0$使得$\sum\limits_{n}P(B_{n})<\infty$也就是由Borel--Cantelli引理蕴含$B_{n}\quad$ a.s.发生有限次.

(4) 固定整数$n$和某个$y\in R^{d(r-1)}$满足$\|y\|<2^{-n/\bar{\alpha}}n^{\beta}$, 显然可以把$y$表示成

$y=\lim\limits_{h\rightarrow\infty}y_{h},$

其中

$y_{h}=x+\theta_{n}\sum\limits^{h}_{j=1}\varepsilon_{j}2^{-j},\varepsilon_{j}\in\{0,1\}^{d(r-1)},x\in G_{n}.$

因为局部时是关于空间变量是连续的, 所以我们可以在事件$B_{n}^{c}$上, 应用三角不等式可得

$|L(Z_{n}+y,I_{n})-L(Z_{n}+x,I_{n})| \nonumber\\ \leq \mathop \sum \limits_{h = 1}^\infty |L(Z_{n}+y_{h},I_{n})-L(Z_{n}+y_{h-1},I_{n})| \quad (y_{0}=x)\nonumber\\ \leq \mathop \sum \limits_{h = 1}^\infty \lambda_{Nr}(C_{n})^{1-(d+2\gamma)(r-1)/Nr\underline{\alpha}}\|y_{h}-y_{h-1}\|^{\gamma}(a_{2}2^{\delta h}\log n)^{Nr}\nonumber\\ \leq 2^{-n(Nr-Nr(d+2\gamma)(r-1)/Nr\underline{\alpha})}(a_{2}\log n)^{Nr}\mathop \sum \limits_{h = 1}^\infty [\sqrt{d(r-1)}2^{-n/\bar{\alpha}}]^{\gamma}2^{-h(\gamma-Nr\delta)}\nonumber\\ =2^{-n(Nr-d(r-1)/\underline{\alpha})}2^{n\gamma(\frac{2(r-1)}{\underline{\alpha}}-\frac{1}{\bar{\alpha}})}(\log n)^{Nr}a_{2}^{Nr}\mathop \sum \limits_{h = 1}^\infty \sqrt{d(r-1)}2^{-h(\gamma-Nr\delta)}\nonumber\\ \leq C_{11}\hspace{0.5mm} g(2^{-n}).$

(2.4)

由 (2.3) 和 (2.4) 式当$n$充分大的时候, 就得

$\mathop {\sup }\limits_{\left\| x \right\| \le {2^{ - n/\bar \alpha }}{n^\beta }} L({Z_n} + x,{I_n}) \le {C_{12}}g({2^{ - n}}), $

也就是

$\mathop {\sup }\limits_{\left\| x \right\| \le {2^{ - n/\bar \alpha }}{n^\beta }}L(x,I_{n})\leq C_{12} g(2^{-n}), $

从而当$n$充分大的时有

$L^{*}(I_{n})=\sup\{L(x,I_{n}); x\in \overline{X(I_{n})}\}\leq M_{6}g(2^{-n}), $

就得到 (2.1) 式.

(2.2) 式的证明与 (2.1) 式的证明大体一致.

首先固定$T=\prod\limits^{r}_{m=1} T_{m}\in\mathfrak{R}$, $T_{m}=\prod\limits^{N}_{l=1}[a_{m,l},b_{m,l}]$, 其中$a_{m,l}<b_{m,l}<a_{m+1,l}$对于每个$n=\prod\limits^{r}_{m=1}(n_{m1},\cdots,n_{mN})$, $n_{mj}(1\leq m\leq r,1\leq j \leq N)$为自然数.定义$2^{\sigma(n)}$个超立方体$(\sigma(n)=\sum\limits^{r}_{m=1}\sum\limits^{N}_{l=1}n_{m,l})$.记

$ {Q_{k,n}} = \prod\limits_{m = 1}^r {\mathop \prod \limits_{l = 1}^N } [a_{m,l}+(b_{m,l}-a_{m,l})(k_{m,l}-1)2^{-n_{m,l}},a_{m,l}+(b_{m,l}-a_{m,l})k_{m,l}2^{-n_{m,l}}], \\ k=(k_{1},\cdots,k_{r}), \\ k_{m}=(k_{m,1},\cdots,k_{m,N})\in \mathop \prod\limits^{N}_{l=1}\{1,2,\cdots,2^{n_{m,l}}\}, \\ k\in \prod\limits_{m = 1}^r {\mathop \prod \limits_{l = 1}^N }\{1,2,\cdots,2^{n_{m,l}}\}\equiv J(n). $

显然$T\in\bigcup_{k,n}Q_{k,n}$.对任何$n$, 定义

$ G_{n} =\{x=(x_{1},\cdots,x_{r-1}):x_{m}=(x_{m,1},\cdots,x_{m,d}),x_{m,l}=p2^{-\sigma(n)}, \text{其中}\\ p \text{是正整数},\text{满足} |p|\leq \prod\limits_{m = 1}^r {\mathop \prod \limits_{l = 1}^N }n_{m,l}2^{m,l},1\leq m\leq r-1,1\leq l\leq d\}. $

注意$G_{n}$是逐渐扩充到$R^{d(r-1)}$的整数点集.对$\forall n,h\geq 1,x\in G_{n}$, 令

$F(n,h,x)=\{y\in R^{d(r-1)}:y=x+2^{-\sigma(n)}\ \mathop \sum \limits_{j = 1}^h \varepsilon_{j}2^{-j},\varepsilon_{j}\in\{0,1\}^{d(r-1)},1\leq j\leq h\}.$

定义事件$A_{n}$, $B_{n}$如下

$ {A_n} = \bigcup\limits_{k \in J(n)} {\bigcup\limits_{x \in {G_n}} {\{ L(} } x,{Q_{k,n}}) \ge {\lambda _{Nr}}({Q_{k,n}})^{1-\frac{d(r-1)}{ Nr\underline{\alpha}}}(a_{1}\log\lambda_{Nr}(Q_{k,n})^{-1})^{Nr}\}, \\ B_{n}={A_n} = \bigcup\limits_{k \in J(n)} {\bigcup\limits_{x \in {G_n}} {\{ |L(} } x,{Q_{k,n}}) \ge {\lambda _{Nr}}({Q_{k,n}})-L(y_{2},Q_{k,n})|\geq\lambda_{Nr}(Q_{k,n})^{1-\frac{d+2\gamma(r-1)}{ Nr\underline{\alpha}}}, \\ \|y_{1}-y_{2}\|^{\gamma}(a_{2}\log 2^{\sigma(n)})^{Nr}2^{\delta hNr},\text{对于}y_{1},y_{2}\in F(n,h,x),\\ y_{1}-y_{2}=2^{-\sigma(n)}\varepsilon 2^{-h},\varepsilon\in\{0,1\}^{d(r-1)}\}. $

注意$J(n)$中$k$的个数为$\prod\limits^{r}_{m=1}\prod\limits^{N}_{l=1}2^{n_{m,l}}$, $G_{n}$中$x$的个数小于$(\prod\limits^{r}_{m=1}\prod\limits^{N}_{l=1} n_{m,l}2^{n_{m,l}+2})^{d(r-1)}$, 因此由引理2.2, 有

$ P(A_{n})\leq(\prod\limits^{r}_{m=1}\prod\limits^{N}_{l=1}2^{n_{m,l}})(\prod\limits^{r}_{m=1}\prod\limits^{N}_{l=1} n_{m,l}2^{n_{m,l}+2})^{d(r-1)}M_{4}e^{-b_{1}a_{1}\log\lambda_{Nr}(Q_{k,n})^{-1}}\\ =C_{13} \prod\limits_{m = 1}^r {\prod\limits_{l = 1}^N {n_{m,l}^{d(r - 1)}} } {2^{{n_{m,l}}(d(r - 1) + 1)}}\lambda_{Nr}(Q_{k,n})^{b_{1}a_{1}}\\ =C_{13}\prod\limits_{m = 1}^r {\prod\limits_{l = 1}^N {n_{m,l}^{d(r - 1)}} } 2^{n_{m,l}(d(r-1)+1)}\prod\limits_{m = 1}^r {\prod\limits_{l = 1}^N {{2^{ - {n_{m,l}}{b_1}{a_1}}}} } {({b_{m,l}} - {a_{m,l}})^{{b_1}{a_1}}}\\ =C_{13}\prod\limits_{m = 1}^r {\prod\limits_{l = 1}^N {n_{m,l}^{d(r - 1)}} } 2^{-n_{m,l}(b_{1}a_{1}-d(r-1)-1)}(b_{m,l}-a_{m,l})^{b_{1}a_{1}} $

和

$ P(B_{n})\leq(\prod\limits^{r}_{m=1}\prod\limits^{N}_{l=1}2^{n_{m,l}})(\prod\limits^{r}_{m=1}\prod\limits^{N}_{l=1} n_{m,l} 2^{n_{m,l}+2})^{d(r-1)}M_{4}\mathop \sum \limits_{h = 1}^\infty 2^{(d(r-1)+1)h}e^{-b_{2}a_{2}2^{\delta h}\log 2^{\sigma(n)}}\\ \leq C_{14} \prod\limits_{m = 1}^r {\prod\limits_{l = 1}^N {n_{m,l}^{d(r - 1)}} } 2^{n_{m,l}(d(r-1)+1)}e^{-b_{2}a_{2}/2\log 2^{\sigma(n)}}\\ =C_{14}\prod\limits_{m = 1}^r {\prod\limits_{l = 1}^N {n_{m,l}^{d(r - 1)}} } {2^{{n_{m,l}}(d(r - 1) + 1)}}{2^{ - \sigma (n)\frac{{{b_2}{a_2}}}{2}}}\\ =C_{14} \prod\limits_{m = 1}^r {\prod\limits_{l = 1}^N {n_{m,l}^{d(r - 1)}} } 2^{-n_{m,l}(\frac{b_{2}a_{2}}{2}-d(r-1)-1)}, $

则对于充分大的$a_{1}$, $a_{2}$, 使得$\sum\limits_{n}P(A_{n})$和$\sum\limits_{n}P(B_{n})$收敛.由Borel--Cantelli引理就有

$A_{n}\text{和}B_{n}\text{以概率1仅发生有限多次.}$

对于固定的$n$和满足$|y_{m,l}|\leq\prod\limits^{r}_{m=1}\prod\limits^{d}_{l=1}n_{m,l}2^{n_{m,l}}$的$y=(y_{1},\cdots,y_{r-1})$, $y_{m}=(y_{m,1},\cdots,y_{m,d})$, $1\leq m\leq r-1$, $1\leq l\leq d$, 存在$x\in G_{n}$使得$y-x\in[0,2^{-\sigma(n)}]^{d(r-1)}$.选择

$y_{h}=x+2^{-\sigma(n)}\sum\limits^{h}_{j=1}\varepsilon_{j}2^{-j},$

$\varepsilon_{j}\in\{0,1\}^{d(r-1)}$, 满足$y_{h}\rightarrow y (h\rightarrow \infty)$.由自相交局部时关于空间变量的连续性, 在$B_{n}^{c}$上可得

$ |L(y,Q_{k,n})-L(x,Q_{k,n})|\\ \leq\mathop \sum \limits_{h = 1}^\infty |L(y_{h},Q_{k,n})-L(y_{h-1},Q_{k,n})|\\ \leq \lambda_{Nr}(Q_{k,n})^{1-\frac{(d+2\gamma)(r-1)}{\underline{\alpha} Nr}}\mathop \sum \limits_{h = 1}^\infty \|y_{h}-y_{h-1}\|^{\gamma}a_{2}^{Nr}(\log 2^{\sigma(n)})^{Nr}2^{\delta hNr}\\ \leq \lambda_{Nr}(Q_{k,n})^{1-\frac{(d+2\gamma)(r-1)}{\underline{\alpha} Nr}}\mathop \sum \limits_{h = 1}^\infty (2^{-\sigma(n)}\sqrt{d(r-1)}2^{-h})^{\gamma}(a_{2}\log 2^{\sigma(n)})^{Nr}2^{\delta hNr}\\ \leq \lambda_{Nr}(Q_{k,n})^{1-\frac{(d+2\gamma)(r-1)}{\underline{\alpha} Nr}}(2^{-\sigma(n)}\sqrt{d(r-1)})^{\gamma}(a_{2}\log 2^{\sigma(n)})^{Nr}\mathop \sum \limits_{h = 1}^\infty 2^{-h(\gamma-\delta Nr)}\\ \leq C_{15}\sigma(n)^{Nr}2^{-\sigma(n)\gamma}\rightarrow 0(\sigma(n)\rightarrow\infty), $

则对满足$\displaystyle x_{m,l}\leq \prod\limits_{m = 1}^r {\mathop \prod \limits_{l = 1}^N } ,1\leq m\leq r-1,1\leq l\leq d$的$x$有

$\mathop {\sup }\limits_{k \in J(n)} L(x,Q_{n,k})\leq C_{16}\lambda_{Nr}(Q_{n,k})^{1-d(r-1)/Nr\underline{\alpha}}(\log\lambda_{Nr}(Q_{n,k})^{-1})^{Nr}.$

当$n$充分大时, 便可得到 (2.2) 式.