设$ \{w(s, t); s\ge 0, t\ge 0\} $是定义在概率空间$ (\Omega, {\cal F}, P) $上的两参数Brown运动. 对这一过程轨道性质的研究已有许多深刻结果, 参见文献[1-3, 5]. 本文中, 我们也对两参数Brown运动Csörg\H{o}-Révész型增量的泛函极限问题进行了研究, 得到了一种局部泛函重对数律. 在文中, 我们设$ {\cal C} = \{f; f:[0, 1]^2\to {{\Bbb R}}, f\;\rm{连续}\} $, 赋予一致范数$ \|f\|: = \mathop{\sup}\limits_{(x, y)\in[0, 1]^2}| f(x, y)| $. 记$ {\cal C}_0: = \{f \in {\cal C}; f(0, t) = f(s, 0) = 0\} $, $ {\cal H} = \{f \in {\cal C}_0; f\ \rm{绝对连续, }\ \|f\|_{\mathcal H}^2 = \int _0^1\int_0^1 |\frac{\partial^2f}{\partial y\partial x}|^2{\rm d}x{\rm d}y < +\infty\} $, 设函数$ I: {\cal C}_0\to [0, \infty] $, 定义如下

$ I(z) = \left\{ \begin{array}{ll} { }\frac{\|z\|^2_{{\cal H}}}{2}, & {\rm{若}}\; z\in {\cal H}, \\ \infty, & \rm{否则}. \end{array} \right. $

本文中, 设$ a_u: (0, 1)\to (0, 1) $是非减连续函数, 满足

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

(ii) $ \frac{u}{a_u} $非增;

(iii) $ \lim\limits_{u\to 0}\frac{\log(u/a_u)}{\log\log u^{-1}} = \frac{r}{2} $, $ r $为非负常数. \\ 对$ 0\le s\le u-a_u, 0\leq t\leq u-a_u $, $ (x, y)\in [0, 1]^2 $, $ \Delta(s, t, a_ux, a_uy) $记下面轨道

$ \Delta(s, t, a_ux, a_uy) = w(s+a_ux, t+a_uy)-w(s+a_ux, t)-w(s, t+a_uy)+w(s, t), $

设$ \ell_u = \log\frac{u^2\log u^{-1}}{{a_u^2}} $, $ \beta_u = (2{a_u^2}{\ell_u})^{-\frac{1}{2}} $. 定义$ Y_{s, t, u}(x, y) = \beta_u\Delta(s, t, a_ux, a_uy) $. 令

$ K_b = \{f\in{\cal C}_0;I(f)\leq\frac{1}{2}b^2\}, {\quad} b: = b(r) = \left\{ \begin{array}{ll} { }\sqrt{\frac{r}{r+1}}, & {\rm{若}}\; 0\leq r<\infty, \\ 1, & \rm{若}\; r = \infty. \end{array} \right. $

本文主要结果陈述如下:

定理 1.1  若条件(i)–(iii)成立, 那么我们有

(1.1) $ \begin{equation} \liminf\limits_{u\to0}\sup\limits_{s, t\in [0, u-a_u]^2}\inf\limits_{f\in K_b}\|Y_{s, t, u}(\cdot, \cdot)-f(\cdot, \cdot)\| = 0, \; \; a.s., \end{equation} $

且对任意的$ f\in K_b $, 有

(1.2) $ \begin{equation} \lim\limits_{u\to0}\inf\limits_{s, t\in [0, u-a_u]^2}\|Y_{s, t, u}(\cdot, \cdot)-f(\cdot, \cdot)\| = 0, \; \; a.s.. \end{equation} $

为证明我们的结果, 需要下面的引理.

引理 2.1^{[3, 引理2.1]}  对任何闭集$ F\subset {\cal C}_0 $, 有

$ \limsup\limits_{\varepsilon\to 0}\varepsilon\log P(\sqrt{\varepsilon}w\in F)\le -\inf\limits_{f\in F}I(f), $

对任何开集$ G\subset {\cal C}_0 $, 有

$ \liminf\limits_{\varepsilon\to 0}\varepsilon\log P(\sqrt{\varepsilon}w\in G)\ge -\inf\limits_{f\in G}I(f). $

引理 2.2^{[3, 引理2.2]}  对任何$ 0<a, b\le 1 $和闭集$ F\subset {\cal C}_0 $, 有

$ \limsup\limits_{\varepsilon\to 0}\varepsilon\left(\log P\left(\bigcup\limits_{0\le s\le 1-a}\bigcup\limits_{0\le t\le 1-b}\left\{\sqrt{\frac{\varepsilon}{ab}}\Delta(s, t, a\cdot, b\cdot)\in F\right\}\right)+\log(ab)\right)\le-\inf\limits_{f\in F}I(f). $

引理 2.3  设$ f\in {\cal H} $, $ f^*(\cdot, \cdot) = f(\lambda\cdot, \lambda'\cdot) $, $ \lambda, \lambda'<1 $, 那么

$ \|f-f^*\|\le \{((1-\lambda)(1-\lambda'))^{1/2}+(1-\lambda)^{1/2}+(1-\lambda')^{1/2}\}\|f\|_{{\cal H}}. $

证  设$ f(s, t) = \int^s_0\int^t_0g(u, v){\rm d}_u{\rm d}_v, 0\le s, t\le 1 $且$ \|f\|_{{\cal H}}^2 = \int^1_0\int^1_0g^2(u, v){\rm d}_u{\rm d}_v $. 我们有

(2.1) $ \begin{eqnarray} \left\|f(\cdot, \cdot)-f^*(\cdot, \cdot)\right\| & = &\sup\limits_{s, t\in[0, 1]^2}|f(s, t)-f(\lambda s, \lambda't)|{}\\ & = &\sup\limits_{s, t\in[0, 1]^2}\bigg|\int^s_0\int^t_0g(u, v){\rm d}_u{\rm d}_v-\int^{\lambda s}_0\int^{\lambda't}_0g(u, v){\rm d}_u{\rm d}_v\bigg|{}\\ & = &\sup\limits_{s, t\in[0, 1]^2}\bigg|\int^s_{\lambda s}\int^t_{\lambda't}g(u, v){\rm d}_u{\rm d}_v+\int^{s}_{\lambda s}\int^{\lambda't}_0g(u, v){\rm d}_u{\rm d}_v {}\\ && +\int^{\lambda s}_{0}\int^{t}_{\lambda' t}g(u, v){\rm d}_u{\rm d}_v\bigg|{}\\ &\le&\{((1-\lambda)(1-\lambda'))^{1/2}+(1-\lambda)^{1/2}+(1-\lambda')^{1/2}\}\|f\|_{{\cal H}}. \end{eqnarray} $

引理2.3证毕.

引理 2.4^{[4, 引理4]}  设$ \{\xi_n\}_{n\ge 1} $为随机变量序列. 若$ \lim\limits_{n\to \infty}P\{\xi_n\ge \xi_0\} = 0 $, 则存在子列$ \{\xi_{n_k}\} $, 使得$ { }\limsup_{k\to \infty}\xi_{n_k}\le \xi_0, \; a.s. $, 因此$ { }\liminf_{n\to \infty}\xi_n\le \xi_0, \; a.s. $.

(1.1)式的证明分三种情形完成:

(Ⅰ) $ 0<r<\infty $, (Ⅱ) $ r = 0 $, (Ⅲ) $ r = \infty $.

情形(Ⅰ)$ 0<r<\infty $.

引理 3.1  若条件$ \rm(i)–(iii) $成立, 则存在$ u_n^{-1} = e^{e^n} $, 使得

(3.1) $ \begin{equation} \lim\limits_{n\to\infty}P\Big(\sup\limits_{(s, t)\in [0, u_n-a_{u_{n+1}}]^2}\inf\limits_{f\in K_b}\|Y_{s, t, u_n}(\cdot, \cdot)-f(\cdot, \cdot)\|\ge \varepsilon\Big) = 0. \end{equation} $

证  设$ A = \{g: \inf\limits_{f\in K_b}\|g-f\|\ge \varepsilon\} $, 则$ A $是闭集, 存在任意小的$ \delta>0 $, 使得$ \inf\limits_{g\in A}I(g)>\frac{1}{2}b^2(r)+\delta $. 为了简单, 我们设$ \frac{a_{u_n}}{u_n-a_{u_{n+1}}+a_{u_n}} = p $, 由引理2.2, 对$ n $足够大, 我们有

(3.2) $ \begin{eqnarray} &&P\left(\sup\limits_{(s, t)\in [0, {u_n}-a_{u_{n+1}}]^2}\inf\limits_{f\in K_b}\|Y_{s, t, u_n}-f(\cdot, \cdot)\|\geq\varepsilon\right){}\\ & = &P\left(\sup\limits_{(s, t)\in [0, {u_n}-a_{u_{n+1}}]^2}\inf\limits_{f\in K_b}\Big\|\frac{1}{\sqrt{2a_{u_n}^2\ell_{u_n}}}\Delta(s, t, a_{u_n}\cdot, a_{u_n}\cdot) -f(\cdot, \cdot)\Big\|\geq\varepsilon\right){}\\ & = &P\left(\sup\limits_{(s, t)\in [0, 1-p]^2}\inf\limits_{f\in K_b}\Big\|\sqrt{\frac{1}{p^2}\frac{1}{2a_{u_n}^2\ell_{u_n}}}\Delta(s, t, p\cdot, p\cdot) -f(\cdot, \cdot)\Big\|\geq\varepsilon\right){}\\ & = &P\left(\bigcup\limits_{s\in [0, 1-p]}\bigcup\limits_{t\in [0, 1-p]} \left\{\sqrt{\frac{1}{p^2}\frac{1}{2a_{u_n}^2\ell_{u_n}}}\Delta(s, t, p\cdot, p\cdot)\in A\right\}\right){}\\ &\leq&\left(\frac{{u_n}-a_{u_{n+1}}+a_{u_n}}{a_{u_n}}\right)^2\left(\frac{a_{u_n}^2}{u_n^2\log u_n^{-1}}\right)^{b^2(r)+\delta}{}\\ &\leq& 4\left(\frac{u_n}{a_{u_n}}\right)^{2(1-b^2(r)-\delta)} \left(\frac{1}{\log u_n^{-1}}\right)^{b^2(r)+\delta}. \end{eqnarray} $

在(3.2)式中, 若$ 1-b^2(r)-\delta\leq 0 $, 则我们有

(3.3) $ \begin{eqnarray} &&P\left(\sup\limits_{(s, t)\in [0, u_n-a_{u_{n+1}}]^2}\inf\limits_{f\in K_b}\|Y_{s, t, u_n}-f(\cdot, \cdot)\|\geq\varepsilon\right){}\\ &\le &4\left(\log u_n^{-1}\right)^{-(b^2(r)+\delta)} = 4e^{-(b^2(r)+\delta)n}\to 0, (n\to\infty). \end{eqnarray} $

否则, 若$ 1-b^2(r)-\delta> 0 $, 由条件(iii), 对任意小的$ \mu>0 $, 我们有

$ \frac{u_n}{a_{u_n}}\leq\left(\log u_n^{-1}\right)^{\frac{r}{2}+\frac{\mu}{2}}, $

故

(3.4) $ \begin{eqnarray} &&P\left(\sup\limits_{s, t\in [0, {u_n}-a_{u_{n+1}}]^2}\inf\limits_{f\in K_b}\|Y_{s, t, u_n}-f(\cdot, \cdot)\|\geq\varepsilon\right){}\\ &\le &4\left(\frac{1}{\log u_n^{-1}}\right)^{-(r+\mu)(1-b^2(r)-\delta)+b^2(r)+\delta} = 4\left(\frac{1}{\log u_n^{-1}}\right)^{(\delta-\frac{1}{r+1})\mu+(r+1)\delta}, \end{eqnarray} $

取$ \delta<\frac{1}{r+1} $, $ \mu\le\frac{1}{2}\frac{(r+1)^2\delta}{1-\delta(r+1)} $, 使得$ (\delta-\frac{1}{r+1})\mu+(r+1)\delta>0 $, 我们有

(3.5) $ \begin{eqnarray} &&P\left(\sup\limits_{s, t\in [0, {u_n}-a_{u_{n+1}}]^2}\inf\limits_{f\in K_b}\|Y_{s, t, u_n}-f(\cdot, \cdot)\|\geq\varepsilon\right){}\\ &\le &4\left(\frac{1}{\log u_n^{-1}}\right)^{(\delta-\frac{1}{r+1})\mu+(r+1)\delta}\to 0, (n\to\infty). \end{eqnarray} $

从而(3.1)式获证. 由引理3.1和引理2.4, 得到(1.1)式.

情形(Ⅱ) $ r = 0 $.

类似情形(Ⅰ)的证明, 结论显然成立.

情形(Ⅲ) $ r = \infty $.

当$ r = \infty $时, $ b = 1 $, 此时$ K_b = \{f\in{\cal C}_0;2I(f)\leq 1\} $. 为证(1.1)式, 只需证明下面的引理.

引理3.2  若条件(i)–(iii)成立, 设$ u_n = \frac{1}{\theta^n}, \theta>1 $, 我们有

(3.6) $ \begin{equation} \limsup\limits_{n\rightarrow \infty}\sup\limits_ {(s, t)\in [0, u_n-a_{u_{n+1}}]^2}\left\|\frac{\Delta(s, t, a_{u_n}\cdot, a_{u_n}\cdot)}{\sqrt{2a_{u_n}^2\ell_{u_n}}}-K_b\right\| = 0, \; \; \; \; a.s.. \end{equation} $

证  设$ A_1 = \{g: \|g-K_b\|\ge \varepsilon\} $, 则$ A_1 $是闭集, 存在$ \delta'>0 $, 使得$ \inf\limits_{g\in A_1}I(g)> \frac{1}{2}+\delta' $.为了简单, 我们设$ \frac{a_{u_n}}{u_n-a_{u_{n+1}}+a_{u_n}} = p $,

(3.7) $ \begin{eqnarray} &&P\left(\sup\limits_ {(s, t)\in [0, u_n-a_{u_{n+1}}]^2}\left\|\frac{1}{(2a_{u_n}^2\ell_{u_n})^{1/2}}\Delta(s, t, a_{u_n}\cdot, a_{u_n}\cdot)-K_b\right\|\ge \varepsilon\right){}\\ & = &P\left(\sup\limits_{(s, t)\in [0, 1-\frac{a_{u_n}}{u_n-a_{u_{n+1}}+a_{u_n}}]^2}\left\|\sqrt{\frac{1}{2a_{u_n}^2\ell_{u_n}p^2}}\Delta\left(s, t, pa_{u_n}\cdot, pa_{u_n}\cdot\right)-K_b\right\|\ge \varepsilon\right){}\\ & = &P\left(\mathop{\bigcup}\limits_{\mathop{0\le s\le 1-p}}\mathop{\bigcup}\limits_{\mathop{0\le t\le 1-p}} \left\{\sqrt{\frac{1}{p}\frac{1}{p}\frac{1}{2\log\frac{u_n^2\log u_n^{-1}}{a_{u_n}^2}}}\Delta\left(s, t, p\cdot, p\cdot\right)\in A_1\right\}\right){}\\ &\le&4\left(\frac{u_n}{a_{u_n}}\right)^2\left(\frac{a_{u_n}^2}{u_n^2\log u_n^{-1}}\right)^{1+\delta'}\le 4\left(\frac{1}{\log u_n^{-1}}\right)^{1+\delta'}, \end{eqnarray} $

由Borel-Cantelli引理, 结束引理3.2的证明.

由引理3.2和引理2.4, 得到(1.1)式.

(1.2)式的证明分三种情形完成：

(Ⅰ') $ 0<r<\infty $, (Ⅱ') $ r = 0 $, (Ⅲ$ ' $) $ r = \infty $.

情形(Ⅰ')$ 0<r<\infty $

(1.2)式的证明由下面引理完成.

引理 3.3  如果条件(i)–(iii)成立, 那么对任何$ f\in K_b $, 存在子列$ \{a_{u_n} = \theta^{-n}; \theta>1, $$ n\ge 1\} $, 使得

(3.8) $ \begin{equation} \limsup\limits_{n\to \infty}\inf\limits_{s, t\in [0, u_{n+1}-a_{u_n}]^2}\left\|Y_{s, t, u_n}(\cdot, \cdot)-f\right\| = 0, \; \; a.s.. \end{equation} $

证  设$ s_i = ia_{u_{n}}, t_j = ja_{u_{n}}, i, j = 0, 1, 2, \cdot\cdot\cdot, k_n = [\frac{b_{u_{n+1}}}{a_{u_{n}}}]-1 $. 我们有

(3.9) $ \begin{eqnarray} & & P\left(\inf\limits_{(s, t)\in [0, u_{n+1}-a_{u_n}]^2}\Big(\Big\|Y_{s, t, u_n}(\cdot, \cdot)-f\Big\|\ge \varepsilon\Big)\right){}\\ &\le & P\left(\min\limits_{0\le i, j\le k_n}\Big\|\frac{1}{\sqrt{2a_{u_n}^2\ell_{u_n}}}\Delta(s_i, t_j, a_{u_n}\cdot, a_{u_n}\cdot)-f\Big\|\ge \varepsilon\right){}\\ & = &\left\{1-P\left(\frac{1}{\sqrt{2\ell_{u_n}}}w(\cdot, \cdot)\in B\right)\right\}^{(k_n+1)^2}, \end{eqnarray} $

其中$ B = \{g; \|g-f\|<\varepsilon\} $. 由于$ 2\inf\limits_{g\in B}I(g)<b^2 $, 故存在任意小的$ \delta_0>0 $, 使得$ 2\inf\limits_{g\in B}I(g)<b^2-2\delta_0 $. 对$ n $足够大, 由大偏差(引理), 我们有

$ P\left(\frac{1}{\sqrt{2\ell_{u_n}}}w(\cdot, \cdot)\in B\right)\ge \left(\frac{a_{u_n}^2}{u_n^2\log u_n^{-1}}\right)^{b^2(r)-\delta_0}. $

因此, 有

(3.10) $ \begin{eqnarray} & &P\left(\inf\limits_{(s, t)\in [0, u_{n+1}-a_{u_n}]^2}\Big(\Big\|Y_{s, t, u_n}(\cdot, \cdot)-f\Big\|\ge \varepsilon\Big)\right){}\\ &\le &\exp\left\{-\left(\Big[\frac{u_{n+1}-a_{u_n}}{a_{u_n}}\Big]+1\right)^2\left(\frac{a_{u_n}^2}{u_n^2\log u_n^{-1}}\right)^{b^2(r)-\delta_0}\right\}{}\\ &\le &\exp\left\{-\frac{1}{4\theta^2}\left(\frac{u_n}{a_{u_n}}\right)^{2(1-b^2(r)+\delta_0)}\left(\frac{1}{\log u_n^{-1}}\right)^{b^2(r)-\delta_0}\right\}, \end{eqnarray} $

由条件(iii), 对任意小的$ \mu'>0 $, 我们有

$ \frac{u_n}{a_{u_n}}\ge (\log u_n^{-1})^{\frac{1}{2}(r-\mu')}, $

故有

(3.11) $ \begin{eqnarray} & &P\left(\inf\limits_{(s, t)\in [0, u_{n+1}-a_{u_n}]^2}\Big(\Big\|Y_{s, t, u_n}(\cdot, \cdot)-f\Big\|\ge \varepsilon\Big)\right){}\\ &\le& \exp\left\{-\frac{1}{4\theta^2}\Big(\log u_n^{-1}\Big)^{(r-\mu')(1-b^2(r)+\delta_0)-b^2(r)+\delta_0}\right\}{}\\ & = &\exp\left\{-\frac{1}{4\theta^2}\Big(\log u_n^{-1}\Big)^{(r+1)\delta_0-(\delta_0+\frac{1}{r+1})\mu'}\right\}, \end{eqnarray} $

在(3.11)式中, 可以取适当的$ \mu' $, 使$ (r+1)\delta_0-(\delta_0+\frac{1}{r+1})\mu'>0 $.

由条件(iii), 对任意小的$ \eta'>0 $, 我们有

$ \frac{u_n}{a_{u_n}}\le (\log u_n^{-1})^{\frac{1}{2}(r+\eta')}\le(\log a_{u_n}^{-1})^{\frac{1}{2}(r+\eta')}, $

因此

$ \log u_n^{-1}\ge \log a_{u_n}^{-1}-\frac{1}{2}(r+\eta')\log\log a_{u_n}^{-1}, $

于是得到

(3.12) $ \begin{eqnarray} &&P\left(\inf\limits_{(s, t)\in [0, u_{n+1}-a_{u_n}]^2}\Big(\Big\|Y_{s, t, u_n}(\cdot, \cdot)-f\Big\|\ge \varepsilon\Big)\right){}\\ & \le& \exp\left\{-\frac{1}{4\theta^2}\Big(\log u_n^{-1}\Big)^{(r+1)\delta_0-(\delta_0+\frac{1}{r+1})\mu'}\right\}{}\\ &\le&\exp\left\{-\frac{1}{4\theta^2}\Big(\log a_{u_n}^{-1}-\frac{1}{2}(r+\eta')\log\log a_{u_n}^{-1}\Big)^{(r+1)\delta_0-(\delta_0+\frac{1}{r+1})\mu'}\right\}, \end{eqnarray} $

故

(3.13) $ \begin{equation} \sum\limits_{n = 1}^{\infty}P\left(\inf\limits_{(s, t)\in [0, u_{n+1}-a_{u_n}]^2}\Big(\Big\|Y_{s, t, u_n}(\cdot, \cdot)-f\Big\|\ge \varepsilon\Big)\right)<\infty. \end{equation} $

从而引理3.3得证.

引理 3.4  若条件(i)–(iii)成立, 则对任何$ f\in K_b $, 我们有

(3.14) $ \begin{equation} \limsup\limits_{u\to 0}\inf\limits_{(s, t)\in[0, u-a_u]^2}\Big\|\frac{\Delta(s, t, a_u\cdot, a_u\cdot)}{\sqrt{2a_u^2\ell_u}}-f\Big\| = 0, \; \; a.s.. \end{equation} $

证  设$ u_n $如引理3.3中定义. 对$ u\in(u_{n+1}, u_n] $, 注意到

$ Y_{s, t, u}(x, y) = \frac{\beta_{u}}{\beta_{u_n}}Y_{s, t, u_n}\Big(\frac{a_u}{a_{u_n}}x, \frac{a_u}{a_{u_n}}y\Big), $

我们有

(3.15) $ \begin{eqnarray} & &\inf\limits_{(s, t)\in[0, u-a_u]^2}\|Y_{s, t, u}(\cdot, \cdot)-f(\cdot, \cdot)\|{}\\ & = &\inf\limits_{(s, t)\in[0, u-a_u]^2}\Big\|\frac{\beta_u}{\beta_{u_n}}Y_{s, t, u_n}\Big(\frac{a_u}{a_{u_n}}\cdot, \frac{a_u}{a_{u_n}}\cdot\Big)-f(\cdot, \cdot)\Big\|{}\\ &\le & \inf\limits_{(s, t)\in [0, u_{n+1}-a_{u_n}]^2}\frac{\beta_u}{\beta_{u_n}}\Big\|Y_{s, t, u_n} \Big(\frac{a_u}{a_{u_n}}\cdot, \frac{a_u}{a_{u_n}}\cdot\Big)-f \Big(\frac{a_u}{a_{u_n}}\cdot, \frac{a_u}{a_{u_n}}\cdot\Big)\Big\|{}\\ &&+\left|\frac{\beta_u}{\beta_{u_n}}-1\right|\left\|f\Big(\frac{a_u}{a_{u_n}}\cdot, \frac{a_u}{a_{u_n}}\cdot\Big)\right\| +\Big\|f\Big(\frac{a_u}{a_{u_n}}\cdot, \frac{a_u}{a_{u_n}}\cdot\Big)-f(\cdot, \cdot)\Big\|. \end{eqnarray} $

注意到

(3.16) $ \begin{eqnarray} \frac{\beta_u}{\beta_{u_n}}&\le &\frac{\beta_{u_{n+1}}}{\beta_{u_n}} = \left(\frac{a_{u_n}^2\log\frac{u_n^2\log u_n^{-1}}{a_{u_n}^2}}{a^2_{u_{n+1}}\log\frac{u_{n+1}^2\log u_{n+1}^{-1}}{a_{u_{n+1}}^2}}\right)^{1/2} \le \left(\frac{a^2_{{u_n}}}{a^2_{{u_{n+1}}}}\right)^{1/2} = \theta, \end{eqnarray} $

由引理2.3, 我们有

(3.17) $ \begin{eqnarray} \left\|f\Big(\frac{a_u}{a_{u_n}}\cdot, \frac{a_u}{a_{u_n}}\cdot\Big)-f(\cdot, \cdot)\right\|&\le &\left|1-\frac{a_{u_{n+1}}}{a_{u_n}}\right|+2\left|1-\frac{a_{u_{n+1}}}{a_{u_n}}\right|^{1/2}{}\\ &\le &\left|1-\frac{1}{\theta}\right|+2\left|1-\frac{1}{\theta}\right|^{1/2}. \end{eqnarray} $

令$ \theta\to 1 $, 由(3.15)–(3.17)式和引理3.3, 我们得到(3.14)式. 从而(1.2)式获证.

情形(Ⅱ') $ r = 0 $

类似情形(Ⅰ')的证明, 结论显然成立.

情形(Ⅲ')$ r = \infty $

当$ r = \infty $时, $ b = 1 $, 此时$ K_b = \{f\in{\cal C}_0;2I(f)\leq 1\} $. 为证(2)式, 只需证明下面的引理.

引理 3.5  如果条件$ \rm(i)–(iii) $成立, 那么对任何$ f\in K_b $, 存在子列$ \{u_n, n\ge 1\} $, 使得

(3.18) $ \begin{equation} \limsup\limits_{n\to \infty}\inf\limits_{(s, t)\in[0, u_{n+1}-a_{u_n}]^2}\left\|\frac{\Delta(s, t, a_{u_n}\cdot, a_{u_n}\cdot)}{\sqrt{2a_{u_n}^2\ell_{u_n}}}-f\right\| = 0, \; \; a.s.. \end{equation} $

证  因为$ \lim\limits_{u\to 0}\frac{\log\frac{u}{a_u}}{\log\log u^{-1}} = \infty $, 故存在递减子列$ \{u_n, n\ge 1\} $使得$ \frac{u_n}{a_{u_n}} = n^{d} $. 设$ s_i = ia_{u_{n}}, $$ t_j = ja_{u_{n}}, i, j = 0, 1, 2, \cdot\cdot\cdot $, $ k_n = [\frac{u_{n+1}}{a_{u_{n}}}]-1 $, $ h(n) = \frac{\log\frac{u_n}{a_{u_n}}}{\log\log u_n^{-1}} = \frac{\log n^{d}}{\log\log u_n^{-1}} $. 我们有$ u_n^{-1} = \exp(n^{\frac{d}{h(n)}}) $且$ h(n)\to \infty $, $ n\to \infty $. 而且, 对任意小的$ \alpha>0 $, $ \frac{n^\alpha}{\log u_n^{-1}}\to \infty $, $ 1\le \frac{u_n}{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) $. 对$ n $足够大, 我们有

(3.19) $ \begin{eqnarray} & &P\left(\inf\limits_{(s, t)\in[0, u_{n+1}-a_{u_n}]^2}\Big(\Big\|\frac{1}{\sqrt{2a_{u_n}^2\ell_{u_n}}}\Delta(s, t, a_{u_n}\cdot, a_{u_n}\cdot)-f\Big\|\ge \varepsilon\Big)\right){}\\ &\le & P\left(\min\limits_{0\le i, j\le k_n}\Big\|\frac{1}{\sqrt{2a_{u_n}^2\ell_{u_n}}}\Delta(s_i, t_j, a_{u_n}\cdot, a_{u_n}\cdot)-f\Big\|\ge \varepsilon\right){}\\ & = &\left\{1-P\left(\frac{1}{\sqrt{2\ell_{u_n}}}w(\cdot, \cdot)\in B_1\right)\right\}^{(k_n+1)^2}, \end{eqnarray} $

其中$ B_1 = \{g; \|g-f\|<\varepsilon\} $. 因为$ 2\inf\limits_{g\in B_1}I(g)<1 $, 选$ \delta''>0 $, 使得$ \mu_0: = 2\inf\limits_{g\in B_1}I(g)+2\delta''<1 $. 对$ n $足够大, 由大偏差

$ P\left(\frac{1}{\sqrt{2\ell_{u_n}}}w(\cdot, \cdot)\in B_1\right)\ge \left(\frac{a_{u_n}^2}{u_n^2\log u_n^{-1}}\right)^{\mu_0}. $

因此

(3.20) $ \begin{eqnarray} & & P\left(\inf\limits_{(s, t)\in[0, u_{n+1}-a_{u_n}]^2}\Big(\Big\|\frac{1}{\sqrt{2a_{u_n}^2\ell_{u_n}}}\Delta(s, t, a_{u_n}\cdot, a_{u_n}\cdot)-f\Big\|\ge \varepsilon\Big)\right){}\\ &\le &\exp\left\{-\left(\frac{a_{u_n}^2}{u_n^2\log u_n^{-1}}\right)^{\mu_0}\left(\left[\frac{u_{n+1}}{a_{u_n}}\right]\right)^2\right\}, \end{eqnarray} $

若选适当的$ d $, 那么

$ \sum\limits_n \exp\left\{-\left(\frac{a_{u_n}^2}{u_n^2\log u_n^{-1}}\right)^{\mu_0}\left(\left[\frac{u_{n+1}}{a_{u_n}}\right]\right)^2\right\}<\infty, $

由Borel-Cantelli引理

$ \limsup\limits_{n\to \infty}\inf\limits_{(s, t)\in[0, u_{n+1}-a_{u_n}]^2}\Big\|\frac{\Delta(s, t, a_{u_n}\cdot, a_{u_n}\cdot)}{\sqrt{2a_{u_n}^2\ell_{u_n}}}-f\Big\| = 0, \; \; a.s.. $

至此, 引理3.5获证.

由引理3.5, 类似于引理3.4的证明, 可以得到(1.2)式.