考虑经典的Wiener空间$(B, H, \mu)$, 设$D^{r, p}$是Wiener泛函的Sobolev空间, 即

$\;\;D^{r, p}=(1-\mathcal{L})^{-\frac{r}{2}}\mathbf{L}^p, \|F\|_{r, p}=\|(1-\mathcal{L})^{r/2}F\|_p, F\in \mathbf{L}^p, r\geqslant 0, 1\leqslant p＜\infty, $

其中$\mathbf{L}^p$记为$(B, \mu )$上的实值函数的$L^p$空间, $\mathcal L$是$(B, H, \mu)$上的Ornstein-Uhlenbeck算子.对$r\geqslant 0, p＞1$, $(r, p)$-容度定义如下

$\;\;C_{r, p}(O)=\inf\{\|F\|_{r, p}^p; F\in D_{r, p}, F\geqslant1, \mu-{\rm{a.s.}}\mbox{在}O \mbox{上}\}, \quad \mbox{对开集} O\subset B, $

且对任意集合$A\subset B$有, $C_{r, p}(A)=\inf\{C_{r, p}(O);A\subset O\subset B, O\mbox{是开集}\}.$设${\mathcal{C}}^d$为从[0, 1]到$\mathbb{R}^d$的连续函数空间, 赋予上确界范数$\|f\|=\sup\limits_{0\leqslant t \leqslant 1}|f(t)|$.记$\mathcal C^d_0=\{f\in \mathcal C^d; f(0)= 0\}$, $\mathcal H^d=\{f\in\mathcal C^d_0; f(t)=\displaystyle\int _0^t \dot {f}(s)ds, \|f\|_{\mathcal H^d}^2=\displaystyle\int_0^1|\dot{f}(t)|^2dt＜\infty \}$, $\mathcal{H}^d$是一定义如下内积的Hilbert空间, $\langle r_1, r_2\rangle_{\mathcal{H}^d}=\displaystyle\int^1_0(\dot{r}_1(s), \dot{r}_2(s))ds$.设$\mu$是$\mathcal{C}^d_0$上的Wiener测度, $(\mathcal{C}^d_0, \mathcal{H}^d, \mu)$是一经典Wiener空间.下面考虑如下两个Banach空间

$ \;\;\;{\mathcal{C}}^\alpha =\{f\in {\mathcal{C}}_0^d; \|f(\cdot)\|_\alpha =\mathop{\sup}\limits_{\mathop {s,~t\in [0,1]}\limits_{s\neq t}}\frac{|f(t)-f(s)|}{|t-s|^\alpha}＜\infty\},\\ \;\;\;{\mathcal{C}}^{\alpha,~ 0}=\{f\in {\mathcal{C}}^\alpha; \mathop{\lim}\limits_{ \delta\rightarrow 0}\mathop{\sup}\limits_{\mathop{s,~t\in [0,1]}\limits_{0＜|t-s|＜\delta}}\frac{|f(t)-f(s)|}{|t-s|^\alpha} =0\}, $

其中$0＜\alpha＜\frac{1}{2}$.则有$(\mathcal{C}^{\alpha, 0}, \mathcal{H}^d, \mu)$也是一经典Wiener空间(见文[2]定理2.4).设$w\in{\mathcal{C}}^{\alpha, 0}$为一标准Brown运动, 记$K=\{f\in\mathcal{H}^{d}; I(f)\leqslant 1\}$, 其中$I: B\to [0, \infty]$定义为若$z\in \mathcal{H}^d$, $I(z)=\|z\|^2_{\mathcal{H}^d}/2$; 否则$I(z)=\infty$.

自从Yoshida^[1]首先得到了关于容度$C_{r, p}$意义的大偏差结论, 近年来关于Brown运动在Hölder范数下的拟必然泛函极限理论开始受到关注.Baldi与Roynette^[2]利用大偏差得到了Brown运动在Hölder范数下的收敛速度, 并证明了存在$k=k(\alpha)\geq0$, 使得

$\;\;\mathop{\lim}\limits_{\varepsilon\rightarrow 0}\varepsilon^{2/(1-2\alpha)}\log P\{\| w\|_\alpha \leq \varepsilon\}=-k.$

(1)

进一步, 对任意$f\in K$与$\gamma=(1-2\alpha)/2$有

$\;\;\mathop{\lim}\limits_{\varepsilon\rightarrow 0}\varepsilon^{1/\gamma}\log P\left(\left\|w-\frac{f} {\varepsilon^{1/(2\gamma)}}\right\|_\alpha \le r\varepsilon\right)=-I(f)-\frac{k}{r^{1/\gamma}}.$

(2)

后来, Chen与Balakrishnan ^[3]得到Brown运动在Hölder范数与容度$C_{r, p}$意义下泛函重对数律的极限定理.本文利用Hölder范数下的大偏差小偏差, 得到了Brown运动增量在Hölder范数下, 关于容度$C_{r, p}$意义下局部泛函极限的收敛速度.主要结果如下.

定理1  设$r_u$定义为从$R^+$到$R^+$的单调不减函数, 满足$0＜r_u\le u$且$u/r_u$单调不减.记$\sigma_u=\log\frac{u\log u}{r_u}, $若$\lim\limits_{u\to\infty}\frac{\log(u/r_u)}{\log u}=\infty$, 则在$C_{r, p}$-q.s.意义下有

$\;\;\label{eq002}\mathop{\lim}\limits_{u\to \infty}\left(\sigma_u\right)^{1-\alpha}\mathop{\inf}\limits_{t\in [0, 1-\frac{r_u}{u}]}\|\left(r_u\sigma_u\right)^{-1/2} (w(ut+r_u\cdot)-w(ut))\|_\alpha=k^{\gamma}, \notag$

其中$\gamma=(1-2\alpha)/2$, $k>0$如式(1) 中所定义.

定理1的证明可由引理5与引理6得到, 在此之前叙述已有的相关结果.

引理1  (见文献[3]定理2.1) 设$\{S_\varepsilon\}_{\varepsilon>0}$是$\mathcal{C}^{\alpha, 0}$上一双射线性算子, 使得对任意$\varepsilon>0$及$A\subset \mathcal{C}^{\alpha, 0}$有$\mu(S^{-1}_{\varepsilon}A)=\mu(\varepsilon^{-1/2}A), $则对$(r, p)\in [0, \infty)\times(1, \infty), $有

$\;\;-\mathop{\inf}\limits_{f\in \mathop{A}\limits^\circ}I(f)\leqslant \mathop{\underline{\lim}}\limits_{\varepsilon\to 0}\varepsilon \log C_{r, p}(S^{-1}_\varepsilon A)\leqslant \mathop{\overline{\lim}}\limits_{\varepsilon\to 0}\varepsilon \log C_{r, p}(S^{-1}_\varepsilon A)\leqslant-\mathop{\inf}\limits_{f\in\bar{A}}I(f).$

引理2  (见文献[6]引理2.1) 设$k\in\mathrm{N}$, $q_1, q_2\in(1, \infty)$满足$1/p=1/q_1+1/q_2$, 则存在常数$c=c(k, p, q_1, q_2)>0$使得对任意$-\infty＜a_i＜b_i＜\infty$, $\delta\in (0, 1), $及$F_i\in D^{k, kq_1}$有下式成立

$\begin{array}{*{20}{l}} {\;\;\;{C_{k,p}}{{\left( {\bigcap\limits_{i = 1}^n {\left\{ {{a_i} < {{\tilde F}_i}\left( z \right) < {b_i}} \right\}} } \right)}^{1/p}}}\\ { \le c{{\left( {\frac{n}{\delta }} \right)}^k}{{\left( {1 + \mathop {{\rm{max}}}\limits_{1 \le i \le n} \parallel {F_i}{\parallel _{k,k{q_1}}}} \right)}^k}\mu {{\left( {\bigcap\limits_{i = 1}^n {\left\{ {{a_i} - \delta < {F_i}\left( z \right) < {b_i} + \delta } \right\}} } \right)}^{1/{q_2}}},} \end{array}$

其中$\tilde{F_i}$为$F_i$的拟连续修正.

引理3  设$k, p, q_1, q_2$如引理2中定义.对任意$\varepsilon>0, t_i\ge 0, h_i>0, i=1, 2, \cdots, n, $及$f\in K$, 设

$\;\;F^{(i)}_\varepsilon(w)=\left\|\varepsilon\left(\frac{w(t_i+ h_i\cdot)-w(t_i)}{\sqrt{h_i}}\right)-f\right\|_\alpha, \quad$

则存在一常数$c=c(k, p, q_1, f)>0$, 对任意$\delta\in(0, 1], \varepsilon\in(0, 1]$, 有

$\begin{array}{l} \;\;\;{C_{k,p}}{\left( {\bigcap\limits_{i = 1}^n {\left\{ {z:{a_i} < F_\varepsilon ^{\left( i \right)}\left( z \right) < {b_i}} \right\}} } \right)^{\frac{1}{p}}}\\ \le c{\delta ^{ - 2{k^2} - k}}{n^k}\mu {\left( {\bigcap\limits_{i = 1}^n {\left\{ {z:{a_i} - \delta < F_\varepsilon ^{\left( i \right)}\left( z \right) < {b_i} + \delta } \right\}} } \right)^{\frac{1}{{{q_2}}}}} \cdot \end{array}$

证  利用引理2, 类似文献[6]中引理2.2易证.

引理4  设$0 ＜\alpha ＜\frac{1}{2}, \, \gamma=\frac{1}{2}-\alpha$, $t\geq 0$, $f\in K$, $k>0$如(1) 中所定义, 则对任意$\tau>0$有

$\;\;\mathop{\lim}\limits_{\varepsilon\to 0}\varepsilon^{1/\gamma}\log C_{r, p}\left(\left\|\frac{w(t+h\cdot)-w(t)}{\sqrt{h}}-\frac{f}{\varepsilon^{1/(2\gamma)}}\right\|_\alpha\le\varepsilon\tau\right)=-\frac{k}{\tau^{1/\gamma}}-I(f).$

证  考虑到容度具有性质$C_{r, p}(\cdot)\ge \mu(\cdot)$, 结合(2) 式只需证明

$\;\;\mathop{\lim}\limits_{\varepsilon\to 0}\varepsilon^{1/\gamma}\log C_{r, p}\left(\left\|\frac{w(t+h\cdot)-w(t)}{\sqrt{h}}-\frac{f}{\varepsilon^{1/(2\gamma)}}\right\|_\alpha\le \varepsilon\tau\right)\leq-\frac{k}{\tau^{1/\gamma}}-I(f).$

对任意$1>\delta>0$, $c_0>0$, 令$k=[r]+1$, 根据引理3, 有

$\;\;\; \;\;\;C_{r, p}\left(\left\|\frac{w(t+h\cdot)-w(t)}{\sqrt{h}}-\frac{f}{\varepsilon^{1/(2\gamma)}}\right\|_\alpha \le \varepsilon\tau\right)^{1/p}\\\;\;\;\le c_0(\varepsilon^{1/(2\gamma)+1}\delta)^{-2k^2-k}\mu\left(\left\|\varepsilon^{1/(2\gamma)}\frac{w(t+h\cdot)-w(t)}{\sqrt{h}}-f\right\|_\alpha\le \varepsilon^{1/(2\gamma)+1}(\tau+\delta)\right)^{1/q_2}, $

再根据(2) 式得

$\;\;\; \;\;\; \mathop{\lim}\limits_{\varepsilon\to 0}\varepsilon^{1/\gamma}\log C_{r, p}\left(\left\|\frac{w(t+h\cdot)-w(t)}{\sqrt{h}}-\frac{f}{\varepsilon^{1/(2\gamma)}}\right\|_\alpha\le \varepsilon\tau\right)\\\;\;\;\le \frac{p}{q_2}\mathop{\lim}\limits_{\varepsilon\to 0}\varepsilon^{1/\gamma}\log \mu\left(\left\|w(\cdot)-\frac{f}{\varepsilon^{1/(2\gamma)}}\right\|_\alpha\le\varepsilon(\tau+\delta)\right)\\\;\;\; = \frac{p}{q_2}(-k(\tau+\delta)^{-\frac{1}{\gamma}}-I(f)).$

最后令$\delta\to 0, q_2\to p$, 引理4获证.

引理5  设$r_u, \sigma_u$如命题1所定义, 则对 s ∈ [0, 1]在$C_{r, p}$-q.s.意义下有

$\;\;\mathop{\liminf}\limits_{u\to \infty}\left(\sigma_u\right)^{1-\alpha}\mathop{\inf}\limits_{t\in [0, 1-\frac{r_u}{u}]}\|\left(r_u\sigma_u\right)^{-1/2} (w(ut+r_us)-w(ut))\|_\alpha\ge k^\gamma.$

证  设$l(u)=r_u (\sigma_u )^{-\frac{2}{1-2\alpha}}$.对于$1＜\theta＜(1-\varepsilon)^{-2} $, 记$u_n=\theta^n.$选取适当$\delta_1>0$, 使得$\delta_2=1/(\theta^{1/2}(1-\varepsilon))^{1/\gamma}-\delta_1>1$.设$k_n=[\frac{u_{n+1}}{l(u_n)}]$, $t_i=il(u_n), i=0, 1, 2, \cdot\cdot\cdot, k_n$.则有

$\;\;\;\mathop{\min}\limits_{0＜i\le k_n}\|\left(r_{u_{n+1}}\sigma_{u_{n+1}}\right)^{-1/2}(w(t_i+r_{u}\cdot)-w(t_i))\|_\alpha\nonumber\\\;\;\;\le \mathop{\max}\limits_{0\le i\le k_n}\mathop{\sup}\limits_{0\le S\le l(u_n)}\|\left(r_{u_{n+1}}\sigma_{u_{n+1}}\right)^{-1/2}(w(S+(t_i+r_{u}\cdot))-w(t_i+r_{u}\cdot))\|_\alpha\nonumber\\\;\;\;+\mathop{\inf}\limits_{t\in[0, 1-\frac{r_{u_{n+1}}}{u_{n+1}}]}\|\left(r_{u_{n+1}}\sigma_{u_{n+1}}\right)^{-1/2}(w(u_{n+1}t+r_u\cdot)-w(u_{n+1}t))\|_\alpha.$

(3)

由引理4, 对任意$0＜\varepsilon＜1, $当$n$充分大时有

$\;\;\;\;\;\; C_{r, p}\left(\sigma_{u_{n+1}}^{1-\alpha}\mathop{\min}\limits_{0\le i\le k_n}\Bigl\|\left(r_{u_{n+1}}\sigma_{u_{n+1}}\right)^{-1/2}(w(t_i+r_{u}\cdot)-w(t_i))\Bigl\|_\alpha)\le (1-\varepsilon)k^\gamma \right)\\\;\;\;\le \mathop{\sum}\limits_{0\le i\le k_n}C_{r, p}\left(\left\|\frac{w(t_i+r_{u}\cdot)-w(t_i)}{\sqrt{\sigma_{u_{n+1}}}\sqrt{r_{u}}}\right\|_\alpha\le \theta^{1/2}\frac{k^\gamma(1-\varepsilon)}{(\sigma_{u_{n+1}})^{1-\alpha}} \right)\\\;\;\;\le (1+k_n)\exp\left\{-\sigma_{u_{n+1}}\delta_2\right\}\le \frac{u_{n+1}+l(u_{n})}{l(u_{n})}\left(\frac{r_{u_{n+1}}}{u_{n+1}\log u_{n+1}}\right)^{\delta_2}, $

因此根据Borel-Cantelli引理得到在$C_{r, p}$-q.s.意义下有

$\begin{aligned}\;\;\;\liminf_{n\to \infty}\sigma_{u_{n+1}}^{1-\alpha}\min_{0\le i\le k_n}\|\left(r_{u_{n+1}}\sigma_{u_{n+1}}\right)^{-1/2}(w(t_i+r_{u}\cdot)-w(t_i))\|_\alpha\ge k^\gamma.\end{aligned}$

(4)

另一方面, 对任意$\eta>0$有

$ \;\;\;\;\;\;C_{r, p}\left\{\sigma_{u_{n+1}}^{1-\alpha}\mathop{\sup}\limits_{0\le i\le k_n}\mathop{\sup}\limits_{0\le s \le l(u_n) }\left(r_{u_{n+1}}\sigma_{u_{n+1}}\right)^{-1/2}\|w(s+t_i+r_{u}\cdot)- w(t_i+r_{u}\cdot)\|_\alpha>\eta\right\}\\ \;\;\;=C_{r, p}\left\{\frac{\sigma_{u_{n+1}}^{1-\alpha}}{\sqrt{r_{u_{n+1}}\sigma_{u_{n+1}}}} \mathop{\sup}\limits_{0\le i\le k_n}\mathop{\sup}\limits_{0\le z\le 1}\|w(z l(u_n)+t_i+r_{u}\cdot)- w(t_i+r_{u}\cdot)\| _\alpha>\eta\right\}\\ \;\;\; \le\sum\limits_{j=0}^{[\frac{{{r}_{u}}}{l({{u}_{n}})}]} \sum\limits_{i=0}^{k_n}C_{r, p}\left\{\frac{\sigma_{u_{n+1}}^{1-\alpha}} {\sqrt{r_{u_{n+1}}\sigma_{u_{n+1}}}}\frac{r_{u}^\alpha}{(l(u_n))^\alpha} T(j, n)>\eta\right\}\\ \;\;\; \le\sum\limits_{j=0}^{[\frac{{{r}_{u}}}{l({{u}_{n}})}]}{\sum\limits_{i=0}^{{{k}_{n}}}{{{C}_{r, p}}}} \left\{\frac{1}{(\sigma_{u_{n+1}})^{\frac{1}{2}+\alpha}}\theta^{\alpha+\frac{3}{2}} \frac{1}{\sqrt{l(u_n)}} T(j, n)>\eta\right\}, $

其中$T(j, n)=\sup\limits_{0\le z\le1}\|w(zl(u_n)+t_i+jl(u_n)+l(u_n)\cdot)- w(t_i+jl(u_n)+ l(u_n)\cdot)\|_\alpha.$由于

$\;\;\;\;\;\;\mu\left\{\frac{1}{({\sigma_{u_{n+1}}})^{\frac{1}{2}+\alpha}}\theta^{\alpha+\frac{3}{2}}\frac{1}{\sqrt{l(u_n)}} T(j, n)>\eta\right\}\\\;\;\;=\mu\left\{\frac{\sqrt{2}\theta^{\alpha+3/2}}{\left(\sigma_{u_{n+1}}\right)^{\frac{1}{2}+\alpha}}\mathop{\sup}\limits_{0\le z\le 1}\left\|w(\frac{z}{2}+\frac{1}{2}\cdot)-w(\frac{1}{2}\cdot)\right\|_\alpha>\eta\right\}\\\;\;\;=\mu\left\{\frac{\sqrt{2}\theta^{\alpha+3/2}}{\left(\sigma_{u_{n+1}}\right)^{\frac{1}{2}+\alpha}}w\in Q\right\}, $

其中$Q=\{f\in \mathcal{C}^{\alpha, 0}; \sup\limits_{0\le z\le 1}\|f(\frac{1}{2}z+\frac{1}{2}\cdot)-f(\frac{1}{2}\cdot)\|_\alpha\ge \eta\}$, 且$\inf\limits_{f\in A}I(f)\ge \frac{\eta^2}{32}$, 故由引理1知当$n$充分大时有

$\;\;C_{r, p}\left\{\frac{\theta^{\alpha+\frac{3}{2}}}{(\sigma_{u_{n+1}})^{\frac{1}{2}+\alpha}}\frac{1}{\sqrt{l(u_n)}} T(j, n)>\eta\right\}\;\;\; \le \exp\left(-\frac{\eta^2}{128\theta^{\alpha+3/2}}\left(\sigma_{u_{n+1}}\right)^{2\alpha}\left(\sigma_{u_{n+1}}\right)\right)\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= \left(\frac{r_{u_{n+1}}}{u_{n+1}\log u_{n+1}}\right)^{\frac{\eta^2}{128\theta^{\alpha+3/2}}\left(\sigma_{u_{n+1}}\right)^{2\alpha}}.$

考虑到当$n\to \infty$, $\sigma_{u_{n+1}}\to \infty$, 因此

$\;\;\sum\limits_{n}\frac{r_{u_{n+1}}}{l(u_n)}\frac{u_{n+1}+l(u_n)}{l(u_n)}\left(\frac{r_{u_{n+1}}}{u_{n+1}\log u_{n+1}}\right)^{\frac{\eta^2}{128\theta^{\alpha+3/2}}\left(\sigma_{u_{n+1}}\right)^{2\alpha}}＜\infty.$

再次利用Borel-Cantelli引理可得在$C_{r, p}$-q.s.意义下有

$\;\;\limsup\limits_{n\to \infty}\sigma_{u_{n+1}}^{1-\alpha}\sup\limits_{0\le i\le k_n}\sup\limits_{0\le s \le l(u_n)}\left(r_{u_{n+1}}\sigma_{u_{n+1}}\right)^{-1/2}\|w(s+t_i+r_{u}\cdot)-w(t_i+r_{u}\cdot)\|_\alpha=0.$

(5)

联合(3), (4), (5) 式得

$\;\;\liminf\limits_{n\to \infty}\sigma_{u_{n+1}}^{1-\alpha}\inf\limits_{t\in[0, 1-\frac{r_{u_{n+1}}}{u_{n+1}}]}\|\left(r_{u_{n+1}}\sigma_{u_{n+1}}\right)^{-1/2}(w(u_{n+1}t+r_u\cdot)-w(u_{n+1}t))\|_\alpha\ge k^\gamma, $

(6)

对于$u\in [u_n, u_{n+1})$, 再设$\phi_{t, u}(s)=\left(r_{u}\sigma_{u}\right)^{-1/2}(w(ut+r_us)-w(ut))$, 从而有

$\;\;\;\;\;\;\;\;\;\sigma_u^{1-\alpha}\inf\limits_{t\in [0, 1-\frac{r_u}{u}]}\|\phi_{t, u}\|_\alpha\nonumber\\\;\;\;\geq \sigma_u^{1-\alpha}\inf\limits_{t\in[0, 1-\frac{r_{u_{n+1}}}{u_{n+1}}]}\frac{\left(r_{u}\sigma_{u}\right)^{-1/2}}{\left(r_{u_{n+1}}\sigma_{u_{n+1}}\right)^{-1/2}}\left\|\phi_{t, u_{n+1}}(\frac{r_u}{r_{u_{n+1}}}\cdot)\right\|_\alpha\nonumber\\\;\;\;\geq \sigma_{u_n}^{1-\alpha}\inf\limits_{t\in [0, 1-\frac{r_{u_{n+1}}}{u_{n+1}}]}\left(r_{u_{n+1}}\sigma_{u_{n+1}}\right)^{-1/2}\|w(u_{n+1}t+r_u\cdot)-w(u_{n+1}t)\|_\alpha\nonumber\\\;\;\;\geq\frac{1}{\theta^{2(1-\alpha)}}\sigma_{u_{n+1}}^{1-\alpha}\inf\limits_{t\in[0, 1-\frac{r_{u_{n+1}}}{u_{n+1}}]}\left(r_{u_{n+1}}\sigma_{u_{n+1}}\right)^{-1/2}\|w(u_{n+1}t+r_u\cdot)-w(u_{n+1}t)\|_\alpha.$

(7)

令$\theta\to 1$, 由式(6)、(7) 证得在$C_{r, p}$-q.s.意义下有

$\;\;\liminf\limits_{u\to \infty}\sigma_u^{1-\alpha}\inf\limits_{t\in[0, 1-\frac{r_u}{u}]}\|\phi_{t, u}\|_\alpha\ge k^\gamma.$

引理6  设$r_u, \sigma_u$如引理5中所定义, 若$\lim\limits_{u\to \infty} \frac{\log ur_u^{-1}}{\log\log u}=\infty$, 则在$C_{r, p}$-q.s.意义下有

$\;\;\limsup\limits_{u\to \infty}\sigma_u^{1-\alpha}\inf\limits_{t\in [0, u-r_{u}]}\|\left(r_u\sigma_u\right)^{-1/2}(w(t+r_u\cdot)-w(t))\|_\alpha\le k^{\gamma}.$

证  由于$\lim\limits_{u\to\infty}\frac{\log\frac{u}{r_u}}{\log\log u}=\infty$, 故存在子列$\{u_n\}$, 满足$\frac{u_n}{r_{u_n}}=n^{p_0}, {p_0}>1$.显然$\{u_n\}$单调递增, 且当$n\to \infty$时$u_n \to \infty$.设$t_i=ir_{u_{n+1}}, \; i=0, 1, 2, \cdots, \; k_n=\left[\frac{u_{n}}{r_{u_{n+1}}}\right]-1, $设$g(n)=\frac{\log\frac{{u_n}}{r_{u_n}}}{\log\log u_n} =\frac{\log n^{p_0}}{\log\log {u_n}}$, 则有$u_n=\exp{(n^{\frac{{p_0}}{g(n)}})}$, 且当$n\to \infty$, $g(n)\to \infty$.进一步, 对任意$\theta>0$, 当$n\to \infty$, $\frac{n^{\theta}}{\log u_n}\to \infty$, $1\leq\frac{u_{n+1}}{{u_n}}=\exp{\{(n+1)^{ \frac{{p_0}}{g(n+1)}}-n^{\frac{{p_0}}{g(n)}}\}}\leq \exp{\{n^{ \frac{{p_0}}{g(n)}-1}\}}\to1$.选取$\delta'>0$使得$\delta''=\frac{1}{(1+\varepsilon)^{1/\gamma}}+\delta'＜1$.令$k=[r]+1$, 根据引理3得

$ \;\;\;\;\;\;C_{r, p}\left(\sigma_{u_{n+1}}^{1-\alpha}\inf\limits_{t\in [0, u_n-r_{u_{n+1}}]}\|\left(r_{u_{n+1}}\sigma_{u_{n+1}}\right)^{-1/2}(w(t+r_{u}\cdot)-w(t))\|_\alpha\ge k^\gamma(1+2\varepsilon) \right)^{1/p}\\ \;\;\;\le C_{r, p}\left( \min\limits_{0\leq i\leq k_n} \|\left(r_{u_{n+1}}\sigma_{u_{n+1}}\right)^{-1/2}(w(t_i+r_{u}\cdot)-w(t_i))\|_\alpha\geq \frac{k^\gamma(1+2\varepsilon)}{\left(\sigma_{u_{n+1}}\right) ^{1-\alpha}} \right)^{1/p}\\ \;\;\;= (1+k_n)^k\left(\frac{\sigma_{u_{n+1}} ^{1-\alpha}}{k^\gamma \varepsilon}\right)^{2k^2+k} \mu\left(\pi_i \geq \frac{k^\gamma(1+\varepsilon)} {\left(\sigma_{u_{n+1}}\right)^{1-\alpha}} \right)^{\frac{(1+k_n)}{q_2}}, $

其中$\pi_i=\|\left(r_{u_{n+1}}\sigma_{u_{n+1}}\right)^{-1/2}(w(t_i+r_{u}\cdot)-w(t_i))\|_\alpha$.由(2) 式, 当$n$充分大时,

$ \;\;\;\;\;\;C_{r, p}\left(\sigma_{u_{n+1}}^{1-\alpha}\inf\limits_{t\in [0, u_n-r_{u_{n+1}}]}\|\left(r_{u_{n+1}}\sigma_{u_{n+1}}\right)^{-1/2}(w(t+r_{u}\cdot)-w(t))\|_\alpha\ge k^\gamma(1+2\varepsilon) \right)^{1/p}\\ \;\;\;\leq (1+k_n)^k\left(\frac{\left(\sigma_{u_{n+1}}\right) ^{1-\alpha}}{k^\gamma \varepsilon}\right)^{2k^2+k} \left(1-\left( \frac{r_{u_{n+1}}}{u_{n+1}\log u_{n+1}}\right)^{\delta''}\right)^{\frac{1+k_n}{q_2}} \\ \;\;\;\leq c n^{k{p_0}}\left(\log (n+1)\right)^{{(2k^2+k)(1-\alpha)}} \exp\left\{-\frac{1}{q_2}\left(\frac{r_{u_{n+1}}}{u_{n+1}\log u_{n+1}}\right)^{\delta''}\left[ \frac{u_{n}}{r_{u_{n+1}}}\right]\right\}, $

其中常数$c>0$.选取适当的$p_0$可得到

$\;\;\sum\limits_{n=1}^\infty c n^{pk{p_0}}\left(\log (n+1)\right)^{{p(2k^2+k)(1-\alpha)}} \exp\left\{-\frac{p}{q_2}\left( \frac{r_{u_{n+1}}}{u_{n+1}\log u_{n+1}}\right)^{\delta''} \left[ \frac{u_{n}}{r_{u_{n+1}}}\right]\right\}＜\infty, $

由Borel-Cantelli引理, 从而证得存在一递增序列${u_n}, u_n\to \infty$, 使得在$C_{r, p}$-q.s.意义下有

$ \;\;\limsup\limits_{n\to \infty} \sigma_{u_{n+1}}^{1-\alpha} \inf\limits_{t\in [0, u_n-r_{u_{n+1}}]}\|\left(r_{u_{n+1}}\sigma_{u_{n+1}}\right)^{-1/2}(w(t+r_{u}\cdot)-w(t))\|_\alpha\le k^{\gamma}, $

(8)

再设$\psi_{t, u}(s)=\left(r_u\sigma_u\right)^{-1/2}(w(t+r_us)-w(t))$, 经推导有$\psi_{t, u}(s)=\frac{\left(r_u\sigma_u\right)^{-1/2}}{\beta_{u_{n+1}}}\psi_{t, u_{n+1}} \left(\frac{r_u}{r_{u_{n+1}}}s\right), $其中$u\in (u_n, u_{n+1})$, 从而有

$ \;\;\;\;\;\;\inf\limits_{t\in [0, u-r_{u}]}\|\left(r_{u}\sigma_{u}\right)^{-1/2}(w(t+r_us)-w(t))\|_\alpha\\ \;\;\;\le \inf\limits_{t\in [0, u_n-r_{u_{n+1}}]}\frac{\left(r_{u_{n}}\sigma_{u_{n}}\right)^{-1/2}}{\left(r_{u_{n+1}}\sigma_{u_{n+1}}\right)^{-1/2}}\left\|\psi_{t, u_{n+1}}\left(\frac{r_u}{r_{u_{n+1}}}\cdot\right)\right\|_\alpha, $

由(8) 式, 考虑到当$n\to \infty$时$\frac{r_{u_{n}}\sigma_{u_{n}}}{r_{u_{n+1}}\sigma_{u_{n+1}}}\to 1$, 引理6得证