设$(B,H,\mu)$是抽象的Wiener空间, 具有Ornstein-Uhlenbeck算子$\mathcal{L}$.用$D^{r,p}=(1-\mathcal{ L})^{-\frac{r}{2}} L^p$记Sobolev空间$D^{r,p}$, 赋予范数

$\begin{eqnarray*} \|F\|_{r,p}=\|(1-\mathcal{L})^{r/2}F\|_p,\quad F\in D^{r,p}, r>0,1\le p<\infty, \end{eqnarray*}$

其中$L^p$记$(B,\mu)$上实值函数$L^p$ -空间.

$(r,p)$ -容度的定义如下:对$B$上开集$O$,

$C_{r,p}(O)=\inf\{\|F\|_{r,p}^p : F\ge 1,\,\mu\hbox{- a.s.} {\rm{on}} O\},$

对任意集$A \subset B$,

$C_{r,p}(A) = \inf \{C_{r,p} (O): A \subset O \subset B,O \mbox{是开集}\}.$

容度是$B$上的集函数具有性质:它可以取正值即使对$\mu$ -零集, 而容度为零的集合总有$\mu$ -测度零.容度$C_{r,p}$有下面性质:

$\bullet$若$A_n\uparrow$, 则$C_{r,p}(\cup_n A_n )=\sup_nC_{r,p}(A_n)$;

$\bullet$ $C_{r.p}(\cup_{n=1}^\infty A_n)\leq\sum\limits_{n=1}^\infty C_{r.p}(A_n)$;

$\bullet$第一Borel-Cantelli引理成立:若$\sum\limits_{n=1}^\infty C_{r.p}(A_n)<\infty$, 则$C_{r,p}(\limsup_n A_n)=0$.容度$C_{r,p}$和$\mu$之间的重要差别是对$C_{r,p}$第二Borel-Cantelli引理不成立, 而对$\mu$成立.

设$\{w(t): t\ge 0\}$是$d$ -维标准Brown运动. $C_{r,p}$ -容度大偏差原理被Yoshida ^[1]建立, Brown运动在Hölder范数下的收敛速率被Baldi和Roynette ^[2]得到.近几年Brown运动在Hölder范数下拟必然泛函极限定理被广泛研究.例如, Chen和Balakrishnan得出了Brown运动在Hölder范数下关于$C_{r,p}$ -容度的Strassen泛函重对数律.本文中, 我们研究了Brown运动在Hölder范数下关于$C_{r,p}$ -容度的Strassen泛函重对数律的收敛速率, 同时加强了文[2]中类似结果.

设$\mathcal C$记从[0, 1]到$\mathbb{R}^d$连续函数空间赋予通常范数$\|f\|:= \sup\limits_{0 \le t \le 1}|f(t)|$.记${\mathcal C}_0:=\{f \in {\mathcal C}: f(0) = 0\}$,

$\begin{eqnarray*} \mathcal{H}^d:=\left\{f \in {\mathcal C}_0\, :\, f(t)=\int_0^t \dot {f}(s)ds, \| f\|_{\mathcal H^d}^2:=\int_0^1|\dot{f}(t)|^2dt < \infty \right\}. \end{eqnarray*}$

显然$\mathcal{H}^d$是Hilbert空间, 具有内积

$\begin{eqnarray*} \langle r_1, r_2\rangle_{\mathcal{H}^d}=\int^1_0(\dot{r}_1(s), \dot{r}_2(s))ds. \end{eqnarray*}$

如$\mu$是Wiener测度, 则$( \mathcal{C}_0,\mathcal{H}^d,\mu )$构成抽象的Wiener空间.对$0<\alpha<\frac{1}{2}$, 定义两个Banach空间如下

$\begin{eqnarray*} {\mathcal{C}}^\alpha &=& \left\{f\in {\mathcal{C}}_0\,:\,\|f\|_\alpha =\mathop{\sup}\limits_{s,t\in [0, 1],s\neq t}\frac{|f(t)-f(s)|}{|t-s|^\alpha} <\infty\right\};\\ {\mathcal{C}}^{\alpha,0} &=& \left\{f\in {\mathcal{C}}^\alpha\,:\,\mathop{\lim}\limits_{ \delta\rightarrow 0} \mathop{\sup}\limits_{s,t\in [0, 1],0<|t-s| <\delta}\frac{|f(t)-f(s)|}{|t-s|^\alpha} =0\right\},\end{eqnarray*}$

则${\mathcal{C}}^{\alpha,0}$是${\mathcal{C}}^{\alpha}$的闭凸子空间.由文[6, 定理2.4], 易证$({\mathcal{C}}^{\alpha,0},\mathcal{H}^d,\mu )$也是抽象的Wiener空间.

设函数$I: B\to [0,\infty]$定义为$I(z)= \frac{\|z\|^2_{\mathcal{H}^d}}{2}$, 若$z\in \mathcal{H}^d$, $=\infty$否则.用文[3, 定理2.1]和文[4, 定理2.2], 有下面结果:

定理2.1 设$\{S_\varepsilon\}_{\varepsilon>0}$是$\mathcal{C}^{\alpha,0}$上一簇双射, 连续线性算子, 使得对所有Borel子集$A\subset \mathcal{C}^{\alpha,0}$和$\varepsilon>0$,

$\mu(S^{-1}_{\varepsilon}A)=\mu(\varepsilon^{-1/2}A).$

那么对任何$A \subset \mathcal{C}^{\alpha,0}$和$(r,p)\in [0,\infty)\times(1,\infty),$下面结论成立

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

设$K=\{f\in \mathcal{H}^{d}\,:\,I(f)\le 1\}$.在文[2]中, 作者们证明存在常数$k(\alpha )>0$使得

$\begin{equation}\label{k-alpha} \lim\limits_{\varepsilon\to 0}\varepsilon^{2/(1-2\alpha)}\log \mu\{\| w \|_\alpha \leq \varepsilon\}=-k(\alpha ), \end{equation}$

(2.1)

并且对每个$f\in K$, $\gamma=(1-2\alpha)/2$

$\begin{eqnarray}\label{k-alpha-2} \lim_{\varepsilon\to 0}\varepsilon^{1/\gamma}\log \mu\left(\left\|w- \frac{f} {\varepsilon^{1/(2\gamma)}}\right\|_\alpha \le r \varepsilon\right)=-I(f)-\frac{k(\alpha)}{r^{1/\gamma}}. \end{eqnarray}$

(2.2)

陈述本节的主要结果如下:

定理2.2 设$0 <\alpha <\frac{1}{2},\,\gamma=\frac{1}{2}-\alpha$, $f\in K=\{f\in \mathcal{H}^{d}\,:\,I(f)\le 1\}$, $k(\alpha)>0$如式(2.1) 中定义.那么, 对任何$\tau>0$, $ t\geq 0$, 我们有

$\begin{eqnarray*} &&\lim_{\varepsilon\to 0}\varepsilon^{1/\gamma}\log C_{r,p} \left(\left\|\frac{w(t+h\cdot)-w(t)}{\sqrt{h}}-\frac{f}{\varepsilon^{1/(2\gamma)}} \right\|_\alpha\le \varepsilon\tau\right)\\ &=& \lim_{\varepsilon \to 0}\varepsilon^{1/\gamma}\log \mu \left(\left\|\frac{w(t+h\cdot)-w(t)}{\sqrt{h}}-\frac{f}{\varepsilon^{1/(2\gamma)}} \right\|_\alpha \le \varepsilon\tau\right)\\ &=& -\frac{k(\alpha)}{\tau^{1/\gamma}}-I(f). \end{eqnarray*}$

为证明定理2.2, 下面的引理被用到.

引理2.1 (见文[7, 引理2.1])设$k$是一个自然数, 给定$q_1,q_2\in(1,\infty)$使得$\frac{1}{p}=\frac{1}{q_1}+\frac{1}{q_2}$.那么, 存在常数$c=c(k,p,q_1,q_2)>0$, 使得对任何$\delta\in (0,1),$ $F_i\in D^{k,kq_1}$和$-\infty<a_i<b_i<\infty$,

$\begin{eqnarray*} &&C_{k,p}\left (\bigcap_{i=1}^n\left \{a_i<\tilde{F_i}(z)<b_i\right\}\right )^{1/p}\\ &\le&c\left (\frac{n}{\delta}\right )^k\left (1+\max_{1\le i\le n}\|F_i\|_{k,kq_1}\right )^k\mu\left (\bigcap_{i=1}^n\left\{a_i-\delta < F_i(z)< b_i+\delta\right \}\right)^{1/q_2} \end{eqnarray*}$

成立, 其中$\tilde{F_i}$记$F_i$的一个拟必然修正.

引理2.2 设$k,p,q_1,q_2$如引理2.1定义.对任何$f\in K$, $\varepsilon>0$, 令

$\begin{eqnarray*} F^{(i)}_\varepsilon(w)=\left\|\varepsilon \left(\frac{w(t_i+\cdot h_i)-w(t_i)}{\sqrt{h_i}}\right)- f\right\|_\alpha,\quad 0\le t_i<\infty,h_i>0,i=1,2,\cdots,n. \end{eqnarray*}$

那么存在一个常数$c=c(k,p,q_1,f)>0$, 对任何$\delta\in(0,1],\varepsilon\in(0,1]$, 有

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

证 类似文[7]中引理2.2的证明.

定理2.2的证明 设$k,p,q_1,q_2$如引理2.1中定义.因为$C_{r,p}(\cdot)\ge \mu(\cdot)$, 只需证明

$\begin{eqnarray*} && \lim_{\varepsilon\to 0}\varepsilon^{1/\gamma}\log C_{r,p} \left(\left\|\frac{w(t+h\cdot)-w(t)}{\sqrt{h}}-\frac{f}{\varepsilon^{1/(2\gamma)}} \right\|_\alpha\le \varepsilon\tau\right)\\&\leq&\lim_{\varepsilon \to 0}\varepsilon^{1/\gamma}\log \mu \left(\left\|\frac{w(t+h\cdot)-w(t)}{\sqrt{h}}-\frac{f}{\varepsilon^{1/(2\gamma)}} \right\|_\alpha \le \varepsilon\tau\right). \end{eqnarray*}$

令$k=[r]+1$.由容度的性质$C_{r_1,p}(\cdot)\le C_{r_2,p}(\cdot) (r_1\le r_2)$与引理2.2, 对任何$1>\delta>0$, $c_0>0$, 有

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

由式(2.2), 得到

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

令$\delta\to 0,q_2\to p$, 完成定理 2.2 的证明.

用记号$LL(t):=\log\log t$.本节主要结果如下.

定理3.1 设$0 <\alpha <\frac{1}{2},\,\gamma=\frac{1}{2}-\alpha$, $f\in K$.若$f$满足$I(f)<1$, 那么有

$\begin{eqnarray}\label{eq3-1} \liminf_{t\to \infty}\Big(LL(t)\Big )^{1-\alpha}\Big\|\frac{w(t\cdot)}{\sqrt{t LL(t)}}-f \Big\|_\alpha=\Big (\frac{k(\alpha)}{1-I(f)}\Big)^\gamma,\qquad C_{r,p}\hbox{-q.s.},\end{eqnarray}$

(3.1)

其中$k(\alpha)>0$如式(2.1) 中定义.

证 用定理3.1中的记号, 需要证明下面两个不等式

$\begin{eqnarray*} \liminf_{t\to \infty}\Big(LL(t)\Big )^{1-\alpha}\Big\|\frac{w(t\cdot)}{\sqrt{t LL(t)}}-f \Big\|_\alpha &\geq &\Big (\frac{k(\alpha)}{1-I(f)}\Big )^\gamma,\qquad C_{r,p}\hbox{-q.s.},\label{geq}\\ \liminf_{t\to \infty}\Big(LL(t)\Big )^{1-\alpha}\Big\|\frac{w(t\cdot)}{\sqrt{t LL(t)}}-f \Big\|_\alpha &\leq &\Big (\frac{k(\alpha)}{1-I(f)}\Big)^\gamma,\qquad C_{r,p}\hbox{-q.s.},\label{leq} \end{eqnarray*}$

在引理3.2和引理3.3中估计上述不等式, 即完成了定理的证明.

引理3.1 对$f\in K$且$I(f)<1$, 有

$\begin{equation}\label{eqabc} \liminf_{n\to \infty}(LL (t_n))^{1-\alpha}\left \|\frac{w(t_n\cdot)} {\sqrt{t_n LL (t_n)}}-f\right\|_\alpha \ge \left(\frac{k(\alpha)}{1-I(f)}\right)^{\gamma},\quad C_{r,p}\hbox{-q.s.},\end{equation}$

(3.2)

其中

$t_n=\exp\left (\frac{n}{(\log n)^a}\right ),a>0.$

证 对任何$\varepsilon\in(0,1),f\in K$, 因为$I(f)<1$, 选$\delta_1>0$, 使得

$\eta_0=I(f)+\frac{1-I(f)}{(1-\varepsilon)^{1/\gamma}}-\delta_1>1.$

那么由定理2.2, 存在$\varepsilon_0>0$, 使得对任何$0<\tilde{\varepsilon}<\varepsilon_0$,

$\begin{eqnarray*} C_{r,p}\left(\left\|\frac{w(h\cdot)}{\sqrt{h}}-\frac{f}{\tilde{\varepsilon}^{1/(2\gamma)}} \right\|_\alpha \le \tilde{\varepsilon}\tau\right)\le \exp\left\{\tilde{\varepsilon}^{-(1/\gamma)} \left(-\frac{k(\alpha)}{\tau^{1/\gamma}}-I(f)+ \delta_1\right)\right\}. \end{eqnarray*}$

现在取$\tilde{\varepsilon}^{1/\gamma}=(LL (t_n))^{-1}$和$\tau=(1-\varepsilon)\left(\frac{k(\alpha)}{1-I(f)}\right)^\gamma$.那么对足够大的$n$, 有

$\begin{eqnarray*} && C_{r,p}\left((LL(t_n))^{1-\alpha} \left\|\frac{w(t_n\cdot)} {\sqrt{t_nLL (t_n )}}-f\right\|_\alpha\le (1-\varepsilon) \left(\frac{k(\alpha)}{1-I(f)}\right)^{\gamma}\right) \\ &=&C_{r,p}\left( \left\|\frac{w(t_n\cdot)} {\sqrt{t_n }}-(LL (t_n))^{1/2}f\right\|_\alpha\le (LL (t_n))^{-\frac{1}{2}+ \alpha}(1-\varepsilon) \left(\frac{k(\alpha)}{1-I(f)}\right)^{\gamma}\right)\\ &\le&\exp\left\{LL (t_n)\left(-\frac{1-I(f)}{(1-\varepsilon)^{1/\gamma}}-I(f) +\delta_1\right ) \right\} = \left(\frac{1}{\log t_n}\right)^{\eta_0}. \end{eqnarray*}$

由Borel-Cantelli's引理

$\begin{equation}\label{eq2*} \liminf\limits_{n\to \infty}(LL (t_n))^{1-\alpha} \left\|\frac{w(t_n\cdot)}{\sqrt{t_nLL (t_n)}}-f\right\|_\alpha \ge \left(\frac{k(\alpha)}{1-I(f)}\right)^{\gamma},\quad C_{r,p}\hbox{-q.s.}. \end{equation}$

(3.3)

引理3.2 对任何$f\in K$且$I(f)<1$, 有

$\begin{equation}\label{eq0} \liminf\limits_{t\to \infty}(LL(t))^{1-\alpha}\left\|\frac{w(t\cdot)}{\sqrt{tLL( t)}}-f \right\|_\alpha\ge \left(\frac{k(\alpha)}{1-I(f)} \right)^\gamma ,\quad C_{r,p}\hbox{-q.s.}.\end{equation}$

(3.4)

证 设

$\psi_{t}(s)=\frac{w(ts)}{\sqrt{tLL(t)}},$

$s\in[0, 1]$, $t_n$如引理3.1中定义.对$t_{n}< t\le t_{n+1}$, 令

$X(t)=(LL (t))^{1-\alpha}\|\psi_{t}(\cdot)-f\|_\alpha, X_n=\inf_{t_{n}<t\le t_{n+1}}X(t).$

对任何$\varepsilon>0$, 由下确界定义, 存在$T_n\in (t_{n},t_{n+1}]$使得$X_n\ge X(T_n)-\varepsilon.$对任何$u,v\in[0, 1]$, 设

$x=\frac{ut_{n}}{T_n},y=\frac{vt_{n}}{T_n},$

那么$0\le x,y\le \frac{t_{n}}{T_n}\le 1.$因此有

$\begin{eqnarray}\label{1234ac} \|\psi_{t_{n}}(\cdot)-f\|_\alpha &=&\sup_{0\le u<v\le 1}\frac{|[\psi_{t_{n}}(u)-f(u)] -[\psi_{t_{n}}(v)-f(v)]|}{|u-v|^\alpha}\nonumber\\ &\le& \sup_{0\le x<y\le \frac{t_{n}}{T_n}}\frac{\left|[\psi_{t_{n}} (\frac{T_n}{t_{n}}x)-f(\frac{T_n}{t_{n}}x)] -[\psi_{t_{n}}(\frac{T_n}{t_{n}}y)-f(\frac{T_n}{t_{n}}y)]\right|}{|x-y|^\alpha} \nonumber\\ &\le &\gamma_n\|\psi_{T_n}(\cdot)-f(\cdot)\|_\alpha + \left|\gamma_n-1\right|\cdot \|f(\cdot)\|_\alpha +\left\|f(\cdot)-f(\frac{T_n}{t_{n}}\cdot)\right\|_\alpha,\end{eqnarray}$

(3.5)

这里用记号

$\gamma_n=\frac{\sqrt{T_nLL (T_n)}}{\sqrt{t_{n}LL (t_{n}})}.$

类似文[2]中式(5.3) 的证明, 有

$\begin{equation}\label{eq5ab} \left\|f(\frac{T_n}{t_{n}}\cdot)-f(\cdot)\right\|_\alpha \le \sqrt{2}\sqrt{\frac{T_{n}}{t_{n}}}\left(\frac{T_{n}}{t_{n}}-1\right)^{\frac{1}{2}-\alpha}\le 2\sqrt{\frac{t_{n+1}}{t_{n}}}\left(\frac{t_{n+1}}{t_{n}}-1\right)^{\frac{1}{2}-\alpha}. \end{equation}$

(3.6)

用不等式$\exp(-x)\ge 1-x$, 有

$\begin{equation*}\label{eqliu} \frac{t_n}{t_{n+1}}\ge 1-\frac{n+1}{(\log(n+1))^a}+\frac{n}{(\log n)^a}. \end{equation*}$

这推出

$\begin{equation}\label{cccccc} 1-\frac{t_n}{t_{n+1}}<(\log n)^{-a}. \end{equation}$

(3.7)

注意到

$\begin{eqnarray}\label{eq6a} \left| \frac{(T_{n}LL( T_{n}))^{1/2}}{(t_{n}LL( t_{n}))^{1/2}}-1\right| \le \left|\frac{t_{n+1}}{t_{n}}-1\right|. \end{eqnarray}$

(3.8)

选适当的$a$, 由式(3.5)-(3.8) 和引理3.1, 有

$\liminf_{n\to\infty}X(T_n)\ge \left(\frac{k(\alpha)}{1-I(f)}\right)^{\gamma},\quad C_{r,p}\hbox{-q.s.}.$

因为$\liminf\limits_{t\to \infty}X(t)\ge \liminf\limits_{n\to\infty}X_n\ge \liminf\limits_{n\to\infty}X(T_n) -\varepsilon,$得到了式(3.4).

引理3.3 对$f\in K$且$I(f)<1$, 有

$\begin{eqnarray*} \liminf_{t\to\infty}(LL(t))^{1-\alpha}\left\|\frac{w(t\cdot)}{\sqrt{tLL(t)}} -f \right\|_\alpha\le \left(\frac{k(\alpha)}{1-I(f)}\right)^{\gamma},\quad C_{r,p}\hbox{-q.s.}. \end{eqnarray*}$

证 只需证明对$f\in K$且$I(f)<1$, $t_n=n^n$, 有

$\begin{eqnarray*} \liminf_{n\to\infty}(LL(t_n))^{1-\alpha}\left\|\frac{w(t_n\cdot)}{\sqrt{t_nLL(t_n)}} -f \right\|_\alpha\le\left (\frac{k(\alpha)}{1-I(f)}\right )^{\gamma},\quad C_{r,p}\hbox{-q.s.}. \end{eqnarray*}$

由Hölder范数的定义和文[6]中第179页第11行结果, 有

$\begin{eqnarray*} \left\|\frac{w(t_{n+1}\cdot)}{\sqrt{t_{n+1}LL (t_{n+1})}}-f\right\|_\alpha &=& \sup_{0\leq t<s\leq 1} \frac{\left|\frac{w(t_{n+1}s)}{\sqrt{t_{n+1}LL (t_{n+1})}}-f(s)-[\frac{w(t_{n+1}t)} {\sqrt{t_{n+1}LL (t_{n+1})}}-f(t)]\right|}{|s-t|^\alpha} \\ &\le&\left (\frac{t_{n+1}}{t_{n}}\right)^\alpha\left\|\frac{w(t_n\cdot)}{\sqrt{t_{n+1} LL (t_{n+1})}}- f(\frac{t_n}{t_{n+1}}\cdot)\right\|_\alpha\\ &&+\sup_{\frac{t_n}{t_{n+1}}\leq t<s \leq 1}\frac{\left|\frac{w(t_{n+1}s)}{\sqrt{t_{n+1}LL (t_{n+1})}}-f(s)-[\frac{w(t_{n+1}t)} {\sqrt{t_{n+1}LL (t_{n+1})}}-f(t)]\right|}{|s-t|^\alpha}. \end{eqnarray*}$

易证明

$\begin{equation}\label{eq38} \limsup\limits_{n\rightarrow \infty}(LL (t_{n+1}))^{1-\alpha} \left (\frac{t_{n+1}}{t_{n}}\right )^\alpha \left\|\frac{w(t_n\cdot)}{\sqrt{t_{n+1}LL( t_{n+1})}}- f(\frac{t_n}{t_{n+1}}\cdot)\right\|_\alpha=0,\quad C_{r,p}\hbox{-q.s.}. \end{equation}$

(3.9)

事实上, 注意

$\begin{eqnarray*} \left\|\frac{w(t_n\cdot)} {\sqrt{t_{n+1}LL( t_{n+1})}}- f(\frac{t_n}{t_{n+1}}\cdot)\right\|_\alpha \leq \frac{\sqrt{t_nLL(t_n)}}{\sqrt{t_{n+1}LL(t_{n+1})}} \left\|\frac{w(t_n\cdot)}{\sqrt{t_nLL( t_n)}}\right\|_\alpha + \left\|f(\frac{t_n}{t_{n+1}}\cdot)\right\|_\alpha. \end{eqnarray*}$

由Strassen's律(见文[3]中定理3.2), $\left\|\frac{w(t_n\cdot)}{\sqrt{t_nLL(t_n)}}\right\|_\alpha $是$C_{r,p}\hbox{-q.s.}$有界的, 得到

$\begin{eqnarray*} \left (LL( t_{n+1})\right )^{1-\alpha}\left (\frac{t_{n+1}}{t_n}\right )^\alpha \frac{\sqrt{t_nLL(t_n)}}{\sqrt{t_{n+1}LL(t_{n+1})}} \left\|\frac{w(t_n\cdot)}{\sqrt{t_nLL(t_n})}\right\|_\alpha\to 0,n\to\infty. \end{eqnarray*}$

因为$f\in K$, 对$\gamma\le 1$, 有

$\|f(\gamma\cdot)\|_\alpha\le \|f(\gamma\cdot)\|_{\mathcal{H}^d}\le \gamma^{1/2}\|f\|_{\mathcal{H}^d}$

(见文[2]中定理5.1的证明).因此有

$\begin{eqnarray*} &&(LL(t_{n+1}))^{1-\alpha}\left (\frac{t_{n+1}}{t_n}\right )^\alpha \left\|f( \frac{t_n}{t_{n+1}}\cdot)\right\|_\alpha\\ &<&2(LL(t_{n+1}))^{1-\alpha} \left (\frac{t_{n+1}}{t_n}\right )^\alpha \sqrt{\frac{t_n}{t_{n+1}}}\\ &=&2(\log (n+1)+\log\log (n+1))^{1-\alpha} \left(\frac{n^n}{(n+1)^{(n+1)}}\right)^{1/2-\alpha}\rightarrow 0,n \rightarrow\infty. \end{eqnarray*}$

于是得到式(3.9).

最后, 证明

$\liminf\limits_{n\to \infty}(LL(t_{n+1}))^{1-\alpha} w_n\le (\frac{k(\alpha)}{1-I(f)})^{\gamma},C_{r,p}\hbox{-q.s.},$

其中

$w_n= \sup_{\frac{t_n}{t_{n+1}}\leq t<s \leq 1}\frac{\left|\frac{w(t_{n+1}s)-w(t_n)}{\sqrt{t_{n+1}LL(t_{n+1})}}-f(s)- [\frac{w(t_{n+1}t)-w(t_n)} {\sqrt{t_{n+1}LL(t_{n+1})}}-f(t)]\right|}{|s-t|^\alpha}. $

为完成证明, 令

$\begin{eqnarray*} \widetilde{w}_n(s)&:=&\frac{w[(t_{n+1}-t_n)s+t_n]-w(t_n)}{\sqrt{t_{n+1}-t_n}},\quad s\ge 0,\\ g(s_1)&:=&\left(\frac{t_{n+1}}{t_{n+1}-t_n}\right)^{1/2}\left[f\left( \frac{t_n+s_1(t_{n+1}-t_n)}{t_{n+1}}\right)-f\left(\frac{t_n}{t_{n+1}}\right)\right],\quad s_1\in [0, 1]. \end{eqnarray*}$

那么$\widetilde{w}_n=\{ \widetilde{w}_n(s) : s\ge 0\}$也是标准Brown运动, 且$g\in K$, $I(g)\le I(f)$.显然$\frac{t_{n+1}}{t_{n+1}-t_n}\to 1,$当$n\to \infty$.对任何$\varepsilon>0$, 选$\delta>0$使得

$\sigma:=\frac{1-I( g)}{(1+\varepsilon)^{(\frac{1}{2}+\alpha)/\gamma}}+I(g)+\delta<1.$

由引理2.2,

$\begin{eqnarray*} &&C_{r,p}\left\{\bigcap_{n=n_0}^N\left((LL( t_{n+1}))^{1-\alpha} w_n\ge \left(\frac{k(\alpha)}{1-I(f)}\right)^{\gamma}(1+2\varepsilon)\right)\right\}^ {\frac{1}{p}}\\ &=&C_{r,p}\left\{\bigcap_{n=n_0}^N\left( \left(\frac{t_{n+1}}{t_{n+1}-t_n}\right)^{\alpha-\frac{1}{2}} \|W(\cdot)\|_\alpha \ge\left(\frac{k(\alpha)}{1-I(f)}\right)^{\gamma} \frac{1+2\varepsilon}{(LL(t_{n+1}))^{\frac{1}{2}-\alpha}}\right) \right\}^{\frac{1}{p}}\\ &\le &c N^k\left((LL( t_{N+1}))^{\frac{1}{2}-\alpha}\right)^{2k^2+k}\mu\left\{ \bigcap_{n=n_0}^N\left(\|W(\cdot)\|_\alpha\geq \left(\frac{k(\alpha)}{1-I(g)}\right)^\gamma \frac{(1+\varepsilon)^{\frac{1}{2}+\alpha}}{(LL (t_{n+1}))^{\frac{1}{2}-\alpha}}\right)\right\}^{1/q_2},\end{eqnarray*}$

其中$W(\cdot)=\widetilde{w}_n(\cdot)-\sqrt{LL(t_{n+1})}\,g(\cdot).$对足够大的$n$, 由式(2.2),

$\begin{eqnarray*} &&C_{r,p}\left\{\bigcap_{n=n_0}^N\left((LL(t_{n+1}))^{1-\alpha} w_n\ge (\frac{k(\alpha)}{1-I(f)})^{\gamma}(1+2\varepsilon)\right)\right\}^{1/p}\\ &\le &c N^k((LL (t_{N+1}))^{\frac{1}{2}-\alpha})^{2k^2+k}\prod^N_{n=n_0} (1-\exp(-\sigma LL( t_{n+1})))^{1/q_2} \\ &\le&c_0 N^k((\log N)^{\frac{1}{2}-\alpha})^{2k^2+k} \exp\left(-\frac{1}{q_2}\sum_{n=n_0+1}^{N+1}(n\log n)^{-\sigma}\right) \to 0,N\to \infty . \end{eqnarray*}$

因此

$\begin{eqnarray*} C_{r,p}\left\{\bigcup^\infty_{l=1}\bigcap_{n=l}^\infty\left((LL (t_{n+1}))^{1-\alpha} w_n\ge \left(\frac{k(\alpha)}{1-I(f)}\right)^{\gamma}(1+2\varepsilon)\right)\right\}=0,\end{eqnarray*}$

这得到

$\liminf\limits_{n\to \infty}(LL(t_{n+1}))^{1-\alpha} w_n\le (\frac{k(\alpha)}{1-I(f)})^{\gamma},C_{r,p}\hbox{-q.s.}.$

完成了证明.