设$(B, H, \mu)$是抽象的Wiener空间.容度是$B$上的集合函数,在$\mu$ -零测集上有可能取正值.因此,一个有意义的问题是:布朗运动在几乎处处意义下成立的一些性质是否在容度意义下也成立?在本文中,我们将研究这一问题.

在近期文章[1]中,高付清等建立了Brown运动增量的局部泛函极限定理.在本文中,我们讨论类似结果,但是概率被容度取代.我们的结果也推广了王文胜^[4]的相应结果.

让我们考虑经典的Wiener空间$(W, H, \mu)$如下:

$ \begin{eqnarray} &&W=\bigg\{w \in C([0, \infty ), \mathbb R^D);w(0)=0, \lim\limits_{t\to\infty}\frac{|w(t)|}{t}=0\bigg\}, \\&& H=\bigg\{h \in W; h(t)=\int_0^t{\dot{h}(s)}{\rm d}s, \dot{h}\in L^2([0, \infty), \mathbb R^D)\bigg\}, \\&&\mbox{$\mu$是$W$上的Wiener测度};\\&&\|w\|_{W}=\sup\limits_{t\ge 0}\frac{|w(t)|}{1+ t}, \quad w \in W; \quad\| h\|_{H}^2=\int_0^\infty\langle\dot{h}(t), \dot{h}(t)\rangle {\rm d}t. \end{eqnarray} $

设$D^{r, p}, r>0, 1\leq p <\infty $是Sobolev空间,即:

$ D^{r, p}=(1-{\cal L})^{-\frac{r}{2}}L^p, \quad\|F\|_{r, p}=\|(1-{\cal L} )^{\frac{r}{2}}F\|_{L^p}, ~F \in L^p, $

其中$L^p$记$(W, \mu)$上实值函数组成的$L^p$ -空间, ${\cal L} $是$(W, H, \mu)$上的Ornstein-Uhlenbeck算子.对$r>0, p> 1$, $(r, p)$ -容度被定义如下：

$ C_{r, p}(O)=\inf\{\|F\|_{r, p}^p; F\geq 1, \mu\mbox{-a.s. on}~O\}, \quad\mbox{对开集} \;\; O \subset W, $

对任意集$A\subset W$,

$ C_{r, p}(A)=\inf\{C_{r, p}(O);A \subset O\subset W, ~ O~\mbox{开集}\}. $

设${\cal C}^D$表示从$[0, 1]$到$R^D$的连续函数空间，赋予上确界范数$ \|f\|:=\sup\limits_{0\leq t \leq 1}|f(t)|$.并且记${\cal C}^D_0:=\{f\in {\mathcal C}^D; f(0)=0\}$, ${\cal H}^D:=\{f\in {\cal C}^D_0; f(t)=\int _0^t\dot{f}(s){\rm d}s, \|f\|_{{\cal H}^D}^2:= \int_0^1|\dot{f}(t)|^2{\rm d}t <\infty \}$, $K:=\{f\in{\mathcal{H}}^D; 2I(f)\leq 1\}$,其中

$ I(f)=\left\{\begin{array}{ll} \frac{1}{2}\int^1_0|\dot{f}(t)|^2{\rm d}t, & f ~ \mbox{是绝对连续的};\\[2mm]+\infty, & \mbox{否则}.\end{array}\right. $

全文中,设$a_u, b_u $是从$(0, 1)$到$(0, e^{-1})$的两个非减的连续函数,满足

(ⅰ) $a_u\leq b_u$,对任何$u\in(0, 1)$且$\lim\limits_{u\to0}a_u=0$;

(ⅱ) $\frac{b_u}{a_u}$是非增的.

设$w\in W$,对$ u\in (0, 1)$, $0\leq t\leq 1$, $\Delta(t, u)$记下面轨道:

$ \Delta(t, u)(s)=w(b_ut+a_us)-w(b_ut), ~s\in [0, 1]. $

令

$ \beta_u=\left(2a_u\log\frac{b_u\log b_u^{-1}}{a_u}\right)^{-1/2}. $

本文主要结果陈述如下：

定理1.1  设$(r, p)\in [0, \infty)\times [1, \infty)$.若条件(ⅰ)和(ⅱ)成立,则

(1.1) $ \limsup\limits_{u\rightarrow 0}\sup\limits_ {t\in[0, 1-\frac{a_u}{b_u}]}\|\beta_u\Delta(t, u)(\cdot)-K\|=0, ~~~~C_{r, p}\mbox{-q.s.}, $

且

(1.2) $ \liminf\limits_{u\rightarrow 0}\inf\limits_ {t\in[0, 1-\frac{a_u}{b_u}]}\|\beta_u\Delta(t, u)(\cdot)-\varphi\|=0, ~\mbox{对所有}~\varphi \in K, ~~~~ C_{r, p}\mbox{-q.s.}. $

而且,若条件

(ⅲ) $\lim\limits_{u\rightarrow 0}\frac{\log{(b_u/{a_u})}}{\log\log b^{-1}_u}=\infty$也成立,则

(1.3) $ \lim\limits_{u \to 0} \inf\limits_{t\in [0, 1-\frac{a_u}{b_u}]}\|\beta_u \Delta(t, u)(\cdot)-\varphi\|=0, ~\mbox{对任意}~\varphi \in K, ~~~~C_{r, p}\mbox{-q.s.}. $

推论1.1  设$M_{t, h}(x)=\frac{w(t+hx)-w(t)}{\sqrt{2h\log h^{-1}}}, 0\le x\le 1, 0\le t\le 1-h.$对$(r, p)\in [0, \infty)\times [1, \infty)$,我们有

(1.4) $ \label{cor1}\lim\limits_{h\rightarrow 0}\sup\limits_ {t\in[0, e^{-2}-h]}\|M_{t, h}(\cdot)-K\|=0, ~~~~C_{r, p}\mbox{-q.s.}, $

且

(1.5) $ \label{cor2}\lim\limits_{h\rightarrow 0}\inf\limits_{t\in[0, e^{-2}-h]}\|M_{t, h}(\cdot)-\varphi\|=0~\mbox{对任何}~~\varphi \in K. ~~~~ C_{r, p}\mbox{-q.s.}. $

引理2.1^{[4,引理2.4]}   设$(W, H, \mu)$是经典Wiener空间,那么对任何闭子集$F\subset W$,任何$T>0$和$0 <h\le T$,下式成立

$ \lim\limits_{\varepsilon\to 0}\varepsilon\left(\log C_{r, p}\left(\bigcup\limits_{0\le t\le T-h}\left\{\sqrt{\frac{\varepsilon}{h}}(w(t+h\cdot)-w(t))\in F\right\}\right)+\log\frac{h}{T}\right)\le -\inf\limits_{f\in F}I(f). $

引理2.2^{[3,引理2.2]}   设$1\leq k\in Z$, $q_1, q_2, \in(1, \infty)$被给定，使得$\frac{1}{p}=\frac{1}{q_1}+\frac{1}{q_2}$.对任何$f\in K$,令

$ F^{(i)}_\varepsilon(w)=\left\|\varepsilon \left(\frac{w(t_i+\cdot h_i)-w(t_i)}{\sqrt{h_i}}\right)- f\right\|, ~ i=1, 2, \cdot\cdot\cdot, n, \varepsilon>0, $

其中$0\leq t_i <\infty, h_i>0, i=1, 2, \cdot\cdot\cdot, n$.那么,存在一个常数$c=c(k, p, q_1, f, d)>0$,对任何$\delta\in(0, 1], \varepsilon\in(0, 1]$,我们有

$ \begin{eqnarray} &&C_{k, p}\left(\bigcap^n_{i=1}\{z; a_i<F^{(i)}_\varepsilon(z)<b_i\}\right)^{1/p} \leq c \delta^{-2k^2-k}n^k\mu\left(\bigcap^n_{i=1}\{z; a_i-\delta<F^{(i)}_\varepsilon(z)<b_i+ \delta\}\right)^{1/q_2}. \end{eqnarray} $

引理2.3^{[2,引理4.4]}  对$u \ge 3$,定义

$ \xi _u(t, w) = \frac{w(\frac{t}{u})}{\sqrt {\frac{2}{u}\log \logu} }, \quad t \in [0, \infty ), w \in W. $

(1)   对任何$(r, p) \in (0, \infty)\times (1, \infty)$,对$C_{r, p}\mbox{-q.s.} w$, $\{\xi _u(\cdot, w), u \ge 3\}$在$W$中相对紧且其极限点集是$C: = \{f \in H; \| f\|_{H}^2 \le 1\}$当$u\to\infty$.

(2)   对每个$f\in C$且$\| f\|_{H}^2 < 1$,对任何$\eta>0$,存在$0 <c_\varepsilon < 1$使得对所有$0 <c < c_\varepsilon $,

$ C_{r, p} \left(\bigcup\limits_{n=1}^\infty\bigcap\limits_{j=n}^\infty\{\|\xi_{c^{-j}} -f\|_W\geq\eta\}\right) = 0. $

引理2.4  存在非增序列$\{u_n\in(0, 1), n\in{\Bbb N}\}$且$\lim\limits_{n\to\infty}u_n\to0$, $\lim\limits_{n\to \infty}\frac{a_{u_n}}{a_{u_{n+1}}}=1$,使得对任何$\varepsilon>0$,有

$ C_{r, p}\left(\sup\limits_{t\in[0, b_{u_n}-a_{u_{n+1}}]}\|\beta_{u_n}(w(t+a_{u_n})-w(t))-K\|\geq\varepsilon ~ {\rm i.o.}\right)=0. $

证  设$ A=\{f\in C_0[0, 1]: \|f-K\|\geq \varepsilon \} $,显然, $A$是闭集.对任何$f\in A$,存在$\delta>0$使得

$ \inf\limits_{f\in A}I(f)>\frac{1}{2}+\delta. $

情形(Ⅰ)   $\limsup\limits_{u\to 0} \frac{\log \frac{b_u}{a_u}} {\log\log b_{u}^{-1}}<\infty$.

若$\limsup\limits_{u\to 0} \frac{\log \frac{b_u}{a_u}} {\log\log b_{u}^{-1}}<\infty$,则存在$0 <M<\infty$,使得

$ \frac{b_u}{a_u}\le (\log b_u^{-1})^M\le (\log a_u^{-1})^M, $

因此$\frac{1}{\log b_u^{-1}}\le \frac{1}{\log a_u^{-1}-M\log\log a_u^{-1}}$.取$u_n$使得$a_{u_n}=\exp\left(-\frac{n}{(\log n)^3}\right)$,那么

(2.1) $ \label{eq-2-1}\frac{1}{\log b^{-1}_{u_n}}\leq\frac{(\log n)^3}{n-M(\log n)^4}. $

由引理2.1,我们有

(2.2) $\begin{eqnarray}\label{22eq-2-2}&&C_{r, p}\Big(\sup\limits_{0\leq t\leqb_{u_n}-a_{u_{n+1}}}\|\beta_{u_{n}}(w(t+a_{u_n}\cdot)-w(t))-K\|\geq\varepsilon \Big)\\&=&C_{r, p}\Big(\sup\limits_{0\leq t\leqb_{u_n}-a_{u_{n+1}}}\left\|\frac{\Big(2\log \frac{b_{u_{n}}\log b_{u_{n}}^{-1}}{a_{u_{n}}}\Big)^{-1/2}}{\sqrt{a_{u_n}}}\Big(w(t+a_{u_n}\cdot)-w(t)\Big)-K\right\|\geq \varepsilon \Big)\\&=&C_{r, p}\Big(\bigcup\limits_{0\leq t\leq (b_{u_n}+a_{u_n}-a_{u_{n+1}})-a_{u_n}}\Big\{\frac{\Big(2\log \frac{b_{u_{n}}\log b_{u_{n}}^{-1}}{a_{u_{n}}}\Big)^{-1/2}}{\sqrt{a_{u_n}}}\Big(w(t+a_{u_n}\cdot)-w(t)\Big)\in A\Big\}\Big)\\&\leq& \frac{b_{u_{n}}+a_{u_n}-a_{u_{n+1}}}{a_{u_{n}}}\left(\frac{a_{u_n}}{b_{u_n}\log b^{-1}_{u_n}}\right)^{1+2\delta}.\end{eqnarray}$

结合(2.1)和(2.2)式,由Borel-Cantelli引理,我们能得到

$ C_{r, p}\left(\sup\limits_{t\in[0, b_{u_n}-a_{u_{n+1}}]}\|\beta_{u_n}(w(t+a_{u_n})-w(t))-K\|\geq\varepsilon ~{\rm i.o.}\right)=0. $

情形(Ⅱ)    $\limsup\limits_{u\to 0} \frac{\log \frac{b_u}{a_u}} {\log\log b_{u}^{-1}}=\infty$.

若$\limsup\limits_{u\to 0}\frac{\log \frac{b_u}{a_u}}{\log\log b_{u}^{-1}}=\infty$,则我们能选$u_n$使得$\frac{b_{u_n}}{a_{u_n}}=n^p$,其中$p>\frac{1}{\delta}$.和情形(Ⅰ)类似的方法,此引理的证明完成.

引理2.5   若条件(ⅰ)和(ⅱ)成立,则我们有

(2.3) $ \label{22eq123a}\limsup\limits_{u\to 0}\sup\limits_{t\in [0, b_u-a_u]}\|\beta_u(w(t+a_u\cdot)-w(t))-K\|=0, ~~ C_{r, p}\mbox{-q.s.}. $

证   设

$ \psi_{t, u}(s)=\beta_u(w(t+a_us)-w(t)), \quad s\in[0, 1], t\in [0, b_u-a_u]. $

情形(Ⅰ)   $\limsup\limits_{u\to 0} \frac{\log \frac{b_u}{a_u}} {\log\log b_{u}^{-1}}<\infty$.

设$u_n$如引理2.4证明中情形(Ⅰ)定义,则对较小的$u\in (0, 1)$,存在整数$n$使得$u_{n+1} <u\le u_{n}$.注意到$\psi_{t, u}(s)=\frac{\beta_u}{\beta_{u_n}}\psi_{t, u_n}(\frac{a_u}{a_{u_n}}s)$,我们能得到

(2.4) $\begin{eqnarray}\label{2a2}\sup\limits_{t\in [0, b_u-a_u]}\|\psi_{t, u}(\cdot)-K\|&=&\sup\limits_{t\in [0, b_u-a_u]}\inf\limits_{f\in K}\|\psi_{t, u}(\cdot)-f(\cdot)\|\\&\leq& \sup\limits_{t\in [0, b_{u_n}-a_{u_{n+1}}]}\inf\limits_{f\in K}\bigg\|\psi_{t, u_n}(\frac{a_u}{a_{u_n}}\cdot)-f(\frac{a_u}{a_{u_n}}\cdot)\bigg\|\\&&+\left|\frac{\beta_u}{\beta_{u_n}}-1\right|\sup\limits_{t\in [0, b_{u_n}-a_{u_{n+1}}]}\|\psi_{t, u_n}(\cdot)\|+\sup\limits_{f\in K}\bigg\|f(\frac{a_u}{a_{u_n}}\cdot)-f(\cdot)\bigg\|.\end{eqnarray}$

因为$\beta_u, \log\frac{b_u\log b_u^{-1}}{a_u}$非增,故我们有

(2.5) $ \label{eqa}\frac{\beta_u}{\beta_{u_{n+1}}}\geq\frac{\beta_{u_n}}{\beta_{u_{n+1}}}=\left(\frac{a_{u_{n+1}}\log\frac{b_{u_{n+1}}\log b_{u_{n+1}}^{-1}}{a_{u_{n+1}}}}{a_{u_n}\log\frac{b_{u_n}\log b_{u_n}^{-1}}{a_{u_n}}}\right)^{1/2}\geq\left(\frac{a_{u_{n+1}}}{a_{u_n}}\right)^{1/2}. $

(2.6) $ \label{22eq}\bigg \|f(\cdot)-f(\frac{a_u}{a_{u_n}}\cdot)\bigg\|\leqslant 2\bigg|1-\frac{a_{u_{n+1}}}{a_{u_n}}\bigg|^{\frac{1}{2}}. $

对$a_{u_n}=\exp\big(-\frac{n}{(\log n)^3}\big)$,容易看出$\frac{a_{u_n}}{a_{u_{n+1}}}\to 1$ ($n\to\infty$),由引理2.4, (2.4)-(2.6)式,完成了(2.3)式的证明.

情形(Ⅱ)   $\limsup\limits_{u\to 0} \frac{\log \frac{b_u}{a_u}}{\log\log b_{u}^{-1}}=\infty$.

若$\limsup\limits_{u\to 0}\frac{\log\frac{b_u}{a_u}}{\log\log b_u^{-1}}=\infty$,如引理2.4证明中情形(Ⅱ),我们能选非增序列$\{u_n; n\geq 1\}$使得$\frac{b_{u_n}}{a_{u_n}}=n^p, p>\frac{1}{\delta}$.设$h(n)=\frac{\log \frac{b_{u_n}}{a_{u_n}}}{\log\log b_{u_n}^{-1}}=\frac{p\log n}{\log\log b_{u_n}^{-1}}$,则$b_{u_n}^{-1}=\exp\{n^{\frac{p}{h(n)}}\}$且$h(n)\to\infty, n\to\infty $.而且,对任何小的$\alpha>0, \frac{(n+1)^\alpha}{\log b_{u_{n+1}}^{-1}}\to \infty $,且

(2.7) $ \label{eq-2-4r}1\leq \frac{b_{u_n}}{b_{u_{n+1}}}=\exp\Big((n+1)^{\frac{p}{h(n+1)}}-n^{\frac{p}{h(n)}}\Big)\to 1, \quad\frac{a_{u_n}}{a_{u_{n+1}}}\to 1, \quad n\to\infty. $

由情形(Ⅰ)的证明中类似方法,我们能完成(2.3)式的证明.

引理2.6   若条件(ⅰ)和(ⅱ)成立,则对任何$f\in K$,我们有

(2.8) $ \label{33eqa121}\liminf\limits_{u\to 0}\|\beta_u\Delta(1-\frac{a_u}{b_u}, u)-f\|=0, ~~ C_{r, p}\mbox{-q.s.} $

证  设$\rho=\lim\limits_{u\to 0}\frac{a_u}{b_u}.$若$\rho<1$且$b_u\to b\neq 0, (u\to 0)$,则$\lim\limits_{u\to 0}\frac{\log\frac{b_u}{a_u}}{\log\log b_u^{-1}}=\infty$,在此情形, (2.8)式能从下面引理2.8得到.

这里,我们仅考虑两种情形: (Ⅰ) $\rho<1$且$b_u\to 0, (u\to 0)$; (Ⅱ) $\rho=1$.

情形(Ⅰ)   $\rho<1$且$b_u\to 0, (u\to 0)$.

我们选$u_n$使得

$ b_{u_{n+1}}=b_{u_n}-a_{u_n}, u_1=1, n\ge 1. $

设$k=[r]+1$,由引理2.2,有

(2.9) $\begin{eqnarray}\label{qqq22eq} &&C_{r, p}\left(\bigcap\limits_{m=l}^n\Big(\|\beta_{u_m}\Delta(1-\frac{a_{u_m}}{b_{u_m}})-f\|\ge 2\varepsilon\Big)\right)^{1/p}\\&=& C_{r, p}\left(\bigcap\limits_{m=l}^n\Big(\|\frac{w(b_{u_m}-a_{u_m}+a_{u_m}\cdot)-w(b_{u_m}-a_{u_m})}{\sqrt{2a_{u_m}\ell_{u_m}}}-f\|\ge 2\varepsilon\Big)\right)^{1/p}\\&\le & cn^{k}\varepsilon^{-2k^2-k}\mu\left(\bigcap\limits_{m=l}^n\Big(\|\frac{w(b_{u_m}-a_{u_m}+a_{u_m}\cdot)-w(b_{u_m}-a_{u_m})}{\sqrt{2a_{u_m}\ell_{u_m}}}-f\|\ge \varepsilon\Big)\right)^{1/q_2}\\&=& cn^{k}\varepsilon^{-2k^2-k}\prod\limits_{m=l}^n\left\{1-\mu\left(\Big(\|\frac{w(b_{u_m}-a_{u_m}+a_{u_m}\cdot)-w(b_{u_m}-a_{u_m})}{\sqrt{2a_{u_m}\ell_{u_m}}}-f\|< \varepsilon\Big)\right)\right\}^{1/q_2}\\&=& cn^{k}\varepsilon^{-2k^2-k}\prod\limits_{m=l}^n\left\{1-\mu\left(\frac{1}{\sqrt{2\ell_{u_m}}}w(\cdot)\in A\right)\right\}^{1/q_2}, \end{eqnarray}$

其中$A=\{g; \|g-f\|<\varepsilon\}$.因为$2\inf \limits_{g\in A}I(g)<1$,我们能选$\eta>0$,使得$\eta_0=2\inf\limits_{g\in A}I(g)+\eta<1$.而且,由大偏差

(2.10) $ \label{32q22eq} \mu\left(\frac{1}{\sqrt{2\ell_{u_m}}}w(\cdot)\in A\right)\ge \left(\frac{a_{u_m}}{b_{u_m}\log b_{u_m}^{-1}}\right)^{\eta_0}, $

我们有

(2.11) $\begin{eqnarray}\label{eq12345} &&C_{r, p}\left(\bigcap\limits_{m=l}^n\Big(\|\beta_{u_m}\Delta(1-\frac{a_{u_m}}{b_{u_m}})-f\|\ge 2\varepsilon\Big)\right)^{1/p}\\&\le & cn^{k}\varepsilon^{-2k^2-k}\exp\left(-\frac{1}{q_2}\sum\limits_{m=l}^n\left(\frac{a_{u_m}}{b_{u_m}\log b_{u_m}}\right)^{\eta_0}\right).\end{eqnarray}$

注意到,存在$A=A(l)>0$,使得

$ \sum\limits_{m=l}^n\left(\frac{a_{u_m}}{b_{u_m}\log b^{-1}_{u_m}}\right)^{\eta_0}> A(\log b^{-1}_{u_n})^{1-\eta_0}. $

当$n$足够大时,对常数$c'>0$,我们能证明

$ \log b_{u_n}^{-1}> c'n. $

因此,我们有

(2.12) $\begin{eqnarray}\label{aceq12345} &&C_{r, p}\left(\bigcap\limits_{m=l}^n\Big(\|\beta_{u_m}\Delta(1-\frac{a_{u_m}}{b_{u_m}})-f\|\ge 2\varepsilon\Big)\right)^{1/p}\\&\le & c_0n^{2k^2}\exp\left(-\frac{A}{q_2}n^{1-\eta_0}\right)\to 0, ~ n\to \infty, \end{eqnarray}$

其中$c_0$是常数.我们得到

$ C_{r, p}\left(\bigcup\limits_{l=1}^\infty\bigcap\limits_{m=l}^n\Big(\|\beta_{u_m}\Delta(1-\frac{a_{u_m}}{b_{u_m}})-f\|\ge 2\varepsilon\Big)\right)=0. $

因此

(2.13) $ \label{33abeq}\liminf\limits_{u\to 0}\|\beta_u\Delta(1-\frac{a_u}{b_u}, u)-f\|=0, ~~ C_{r, p}\mbox{-q.s.}. $

情形(Ⅱ)   $\rho=1$.

若$\rho=1$,则$a_u=b_u$.见引理2.3.

引理2.7  若条件(ⅲ)也成立,则对任何$f\in K$,存在子列$\{u_n, n\ge 1\}$,使得

(2.14) $ \label{33eqacc}\limsup\limits_{n\to \infty}\inf\limits_{t\in [0, b_{u_{n+1}}-a_{u_n}]}\|\beta_{u_n}(w(t+a_{u_n}\cdot)-w(t))-f\|_q=0, ~~ C_{r, p}\mbox{-a.s.}. $

证  因为$\lim\limits_{u\to 0}\frac{\log\frac{b_u}{a_u}}{\log\log b^{-1}_u}=\infty$,故存在子列$\{u_n, n\ge 1\}$使得$\frac{b_{u_n}}{a_{u_n}}=n^{d'}$.设$t_i=ia_{u_n}, $$i=0, 1, 2, \cdot\cdot\cdot$, $k_n=[\frac{b_{u_{n+1}}}{a_{u_n}}]-1$.设$t_i=ia_{u_n}, i=0, 1, 2, \cdot\cdot\cdot$, $k_n=[\frac{b_{u_{n+1}}}{a_{u_n}}]-1$且$h(n)=\frac{\log\frac{b_{u_n}}{a_{u_n}}}{\log\log b^{-1}_{u_{n}}}=\frac{\log n^{d'}}{\log\log b^{-1}_{u_n}}$.我们有$b^{-1}_{u_n}=\exp(n^{\frac{d'}{h(n)}})$且$h(n)\to \infty$ ($n\to \infty$).而且,对任何小$\alpha>0$, $\frac{n^\alpha}{\log b^{-1}_{u_{n}}}\to \infty$, $1\le \frac{b_{u_n}}{b_{u_{n+1}}}=\exp\{(n+1)^{\frac{d'}{h(n+1)}}- n^{\frac{d'}{h(n)}}\}\le\exp(n^{\frac{d'}{h(n)}-1})\to 1, (n\to \infty).$取$k=[r]+1$,对$n$足够大,我们有

(2.15) $\begin{eqnarray}\label{33qeq12345} &&C_{r, p}\left(\inf\limits_{t\in [0, b_{u_{n+1}}-a_{u_n}]}\Big(\|\beta_{u_n}\Big(w(t+a_{u_n}\cdot)-w(t)\Big)-f\|\ge 2\varepsilon\Big)\right)^{1/p}\\&\le & c(1+k_n)^k\varepsilon^{-k^2-k}\mu(\min\limits_{0\le i\le k_n}\|\beta_{u_n}(w(t_i+a_{u_n}\cdot)-w(t_i))-f\|\ge \varepsilon)^{1/q_2}\\&=&c(1+k_n)^k\varepsilon^{-k^2-k}\left\{1-\mu\left(\frac{1}{\sqrt{2\ell_{u_n}}}w(\cdot)\in A\right)\right\}^{\frac{k_n+1}{q_2}}, \end{eqnarray}$

其中$A=\{a; \|g-f\|<\varepsilon\}$.因为$2\inf\limits_{g\in A}I(g)<1$,选$\delta>0$,使得$\mu:=2\inf\limits_{g\in A}I(g)+2\delta<1$.对$n$足够大,由大偏差

$ \mu\left(\frac{1}{\sqrt{2\ell_{u_n}}}w(\cdot)\in A\right)\ge \left(\frac{a_{u_n}}{b_{u_n}\log b^{-1}_{u_n}}\right)^{\mu}. $

因此

(2.16) $\begin{eqnarray}\label{33qq**}& &C_{r, p}\left(\inf\limits_{t\in [0, b_{u_{n+1}}-a_{u_n}]}\Big(\|\beta_{u_n}\Big(w(t+a_{u_n}\cdot)-w(t)\Big)-f\|\ge 2\varepsilon\Big)\right)^{1/p}\\&\le & c(1+k_n)^k\varepsilon^{-k^2-k}\mu(\min\limits_{0\le i\le k_n}\|\beta_{u_n}(w(t_i+a_{u_n}\cdot)-w(t_i))-f\|\ge \varepsilon)^{1/q_2}\\&\le &c'n^{k^2d'}\exp\left\{-\frac{1}{q_2}\left(\frac{a_{u_n}}{b_{u_n}\log b^{-1}_{u_n}}\right)^{\mu}\left[\frac{b_{u_{n+1}}}{a_{u_n}}\right]\right\}, \end{eqnarray}$

其中$c, c'$是常数.若选适当的$d'$,则

$ \sum\limits_n c'n^{k^2d'p}\exp\left\{-\frac{p}{q_2}\left(\frac{a_{u_n}}{b_{u_n}\log b^{-1}_{u_n}}\right)^{\mu}\left[\frac{b_{u_{n+1}}}{a_{u_n}}\right]\right\}<\infty, $

由Borel-Cantelli引理

$ \limsup\limits_{n\to\infty}\inf\limits_{t\in[0, b_{u_{n+1}}-a_{u_n}]}\left\|\beta_{u_n}\Big(w(t+a_{u_n}\cdot)-w(t)\Big)-f\right\|=0, ~~ C_{r, p}\mbox{-q.s.}. $

引理2.7证毕.

引理2.8   若条件(ⅲ)也成立,则对任何$f\in K$,我们有

(2.17) $ \label{33eqa**}\limsup\limits_{u\to 0}\inf\limits_{t\in [0, b_u-a_u]}\|\beta_u(w(t+a_u\cdot)-w(t))-f\|=0, ~~ C_{r, p}\mbox{-q.s.}. $

证  设$\phi_{t, u}(s)=\beta_u(w(t+a_us)-w(t)), s\in[0, 1], t\in [0, b_u-a_u], $$u_n$如引理2.7的证明中定义.因为$\phi_{t, u}(s)=\frac{\beta_u}{\beta_{u_n}}\phi_{t, u_n}(\frac{a_u}{a_{u_n}}s)$,我们有

(2.18) $\begin{eqnarray}\label{33qqbcb}& &\inf\limits_{t\in [0, b_u-a_u]}\|\phi_{t, u}(\cdot)-f(\cdot)\|=\inf\limits_{t\in [0, b_u-a_u]}\|\frac{\beta_u}{\beta_{u_n}}\phi_{t, u_n}(\frac{a_u}{a_{u_n}}\cdot)-f(\cdot)\|\\&\le & \inf\limits_{t\in [0, b_{u_{n+1}}-a_{u_n}]}\|\phi(\frac{a_u}{a_{u_n}}\cdot)-f(\frac{a_u}{a_{u_n}}\cdot)\|\\&&+ \sup\limits_{t\in [0, b_{u_{n+1}}-a_{u_n}}\left|\frac{\beta_u}{\beta_{u_n}}-1\right|\left\|\phi_{t, u_n}\Big(\frac{a_u}{a_{u_n}}\cdot\Big)\right\|+\|f(\frac{a_u}{a_{u_n}})-f(\cdot)\|.\end{eqnarray}$

而且

(2.19) $ \label{133qqbcb} \frac{\beta_u}{\beta_{u_n}}\le \frac{\beta_{u_{n+1}}}{\beta_{u_n}}= \left(\frac{a_{u_n}\log\frac{b_{u_n}\log b^{-1}_{u_n}}{a_{u_n}}}{a_{u_{n+1}}\log\frac{b_{u_{n+1}} \log b^{-1}_{u_{n+1}}}{a_{u_{n+1}}}}\right)^{1/2}\le\left(\frac{a_{u_n}}{a_{u_{n+1}}}\right)^{1/2}, $

(2.20) $ \label{233qqbcb} \left\|f(\frac{a_u}{a_{u_n}}\cdot)-f(\cdot)\right\|\le 2\left|1-\frac{a_{u_{n+1}}}{a_{u_n}}\right|^{1/2}. $

我们从(2.18)-(2.20)式和引理2.7能得出(2.17)式.