Chung’s functional law of the iterated logarithm for the Brownian sheet

Abstract

In this paper, we investigate functional limit problem for path of a Brownian sheet, Chung's functional law of the iterated logarithm for a Brownian sheet is obtained. The main tool in the proof is large deviation and small deviation for a Brownian sheet.

Keywords

Brownian sheet / Chung's functional law of the iterated logarithm

1 Introduction and main results

The functional limit theorems for Brownian motion and Brownian sheet have been investigated by some scholars. Gao and Wang [2] studied the rate of convergence in the functional limit theorem for increments of a Brownian motion. Lucas [4] obtained Chung's type modulus of continuity for Brownian motion. Chen [1] studied Strassen's the law of the iterated logarithm for the two-parameter Wiener processes. Xu [8] studied quasi sure functional modulus of continuity for a two-parameter Wiener process in Hölder norm, and partial results of [8, Theorem 3.1] were generalized to case of fractional Brownian sheet [9, Theorem 1.2]. The study for Chung's limit theorems of Brownian sheet is complicated, and the result is a fewer. In the existing results, Talagrand obtained Chung's law of the iterated logarithm [6, Theorem 1.3], and Kuelbs and Li [3, Theorem 5.1] obtained Chung's functional law of the iterated logarithm for Brownian sheet in Hölder norm, but these results are rough. In this paper, Chung's law of the iterated logarithm for Brownian sheet is further studied, the more exact Chung's type law of iterated logarithm is obtained.

In this paper, let \{w(s,t);s\ge 0,t\ge 0\} be Brownian sheet, \mathcal{C}=\{f;f:[0,1{]}^2\to \mathbb{R},f\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{u}\mathrm{o}\mathrm{u}\mathrm{s}\}, endowed with the uniform norm \Vert f\Vert :=\underset{(s,t)\in [0,1{]}^2}{sup}|f(s,t)|. Denote {\mathcal{C}}_0:=\{f\in \mathcal{C};f(0,t)=f(s,0)=0\}, , define a function I:{\mathcal{C}}_0\to [0,\mathrm{\infty }] by

I(z)=\{\begin{array}{l}{\displaystyle \frac{\Vert z{\Vert }_{\mathcal{H}}^2}{2}},\mathrm{i}\mathrm{f}z\in \mathcal{H},\\ \mathrm{\infty },\mathrm{o}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{w}\mathrm{i}\mathrm{s}\mathrm{e}.\end{array}

Denote K=\{f\in \mathcal{H};I(f)\le 1\}. We define L_{2}u=\log \log u,L_{3}u=\log \log \log u.

The following theorems are main results of this paper.

Theorem 1.1　 For any f\in K with , we have

(1.1)\underset{u\to \mathrm{\infty }}{lim inf}(L_{2}u)(L_{3}u{)}^{3/2}\Vert \frac{w(u\cdot ,u\cdot )}{\sqrt{u^2(L_{2}u)(L_{3}u{)}^3}}-f(\cdot ,\cdot )\Vert \ge \frac{1}{\sqrt{8C(1-I(f))}},a.s.,

(1.2)\underset{u\to \mathrm{\infty }}{lim inf}\frac{L_{2}u}{(L_{3}u{)}^{3/2}}\Vert \frac{w(u\cdot ,u\cdot )}{\sqrt{u^2(L_{2}u)}}-f(\cdot ,\cdot )\Vert \ge \frac{1}{\sqrt{8C(1-I(f))}},a.s.

Theorem 1.2　(1) Let a positive constant . For any f\in K with , we have

(1.3)\underset{u\to \mathrm{\infty }}{lim inf}(L_{2}u{)}^b(L_{3}u{)}^{3/2}\Vert \frac{w(u\cdot ,u\cdot )}{\sqrt{u^2(L_{2}u{)}^b(L_{3}u{)}^3}}-f(\cdot ,\cdot )\Vert \le \frac{\sqrt{b^{3}C}}{\sqrt{8(1-I(f))}},a.s.

(2) For any f\in K with , we have

(1.4)\underset{u\to \mathrm{\infty }}{lim inf}\frac{L_{2}u}{(L_{3}u{)}^{3/2}}\Vert \frac{w(u\cdot ,u\cdot )}{\sqrt{u^2(L_{2}u)}}-f(\cdot ,\cdot )\Vert \le \frac{\sqrt{C}}{\sqrt{8(1-I(f))}},a.s.

Where C is a positive constant in Theorem 1.1 of [6].

Remark 1.1　In Theorem 1.1 and Theorem 1.2, if f=0, then

(1.5)\frac{1}{\sqrt{8C}}\le \underset{u\to \mathrm{\infty }}{lim inf}\frac{(L_{2}u{)}^{1/2}}{u(L_{3}u{)}^{3/2}}\Vert w(u\cdot ,u\cdot )\Vert \le \frac{\sqrt{C}}{\sqrt{8}},\mathrm{a}.\mathrm{s}.

In this case, Theorem 1.3 in [6] can be seen as the corollary of (1.5).

2 Proofs of main results

In this section, Theorem 1.1 is proved by small deviations, Theorem 1.2 is proved by large deviations and small deviations.

2.1 Proof of Theorem 1.1

Our proofs are based on the following lemmas.

Lemma 2.1　 Let f\in \mathcal{H}, f^{\ast }(\cdot ,\cdot )=f(\lambda \cdot ,{\lambda }^{\prime }\cdot ), . Then we have

\Vert f-f^{\ast }\Vert \le \{((1-\lambda )(1-{\lambda }^{\prime }){)}^{1/2}+(1-\lambda {)}^{1/2}+(1-{\lambda }^{\prime }{)}^{1/2}\}\Vert f{\Vert }_{\mathcal{H}}.

Proof　Let f(s,t)={\int }_0^s{\int }_0^{t}g(u,v)\mathrm{d}u\mathrm{d}v,0\le s,t\le 1 and \Vert f{\Vert }_{\mathcal{H}}^2={\int }_0^1{\int }_0^1{g}^2(u,v)\mathrm{d}u\mathrm{d}v. We have

\begin{array}{rl}& \Vert f(\cdot ,\cdot )-f^{\ast }(\cdot ,\cdot )\Vert =\underset{s,t\in [0,1{]}^2}{sup}|f(s,t)-f(\lambda s,{\lambda }^{\prime }t)|\\ & =\underset{s,t\in [0,1{]}^2}{sup}\left|{\int }_0^s{\int }_0^{t}g(u,v)\mathrm{d}u\mathrm{d}v-{\int }_0^{\lambda s}{\int }_0^{{\lambda }^{\prime }t}g(u,v)\mathrm{d}u\mathrm{d}v\right|\\ & =\underset{s,t\in [0,1{]}^2}{sup}\left|{\int }_{\lambda s}^s{\int }_{{\lambda }^{\prime }t}^{t}g(u,v)\mathrm{d}u\mathrm{d}v+{\int }_{\lambda s}^s{\int }_0^{{\lambda }^{\prime }t}g(u,v)\mathrm{d}u\mathrm{d}v+{\int }_0^{\lambda s}{\int }_{{\lambda }^{\prime }t}^{t}g(u,v)\mathrm{d}u\mathrm{d}v\right|\\ & \le \{((1-\lambda )(1-{\lambda }^{\prime }){)}^{1/2}+(1-\lambda {)}^{1/2}+(1-{\lambda }^{\prime }{)}^{1/2}\}\Vert f{\Vert }_{\mathcal{H}}.\end{array}

Lemma 2.1 is proved.

Lemma 2.2　 For f\in K with , there exists a u_n=\exp \left(\frac{n}{(\log n{)}^3}\right), such that

\underset{n\to \mathrm{\infty }}{lim inf}{L}_2{u}_n(L_3{u}_n{)}^{3/2}\Vert \frac{w(u_n\cdot ,u_n\cdot )}{\sqrt{u_n^2(L_2{u}_n)(L_3{u}_n{)}^3}}-f\Vert \ge \frac{1}{\sqrt{8C(1-I(f))}},a.s.

Proof　Let f\in K with , and choose \delta >0, such that {\eta }_0:=I(f)+\frac{1-I(f)}{1-\epsilon }-\delta >1. We have

(2.1)\begin{array}{rl}& P\left((L_2{u}_n)(L_3{u}_n{)}^{3/2}\Vert \frac{w(u_n\cdot ,u_n\cdot )}{\sqrt{u_n^2(L_2{u}_n)(L_3{u}_n{)}^3}}-f\Vert \le \frac{1-\epsilon }{\sqrt{8C(1-I(f))}}\right)\\ & =P\left(\Vert w(\cdot ,\cdot )-(L_2{u}_n{)}^{1/2}(L_3{u}_n{)}^{3/2}f\Vert \le \frac{1-\epsilon }{(L_2{u}_n{)}^{1/2}}\frac{1}{\sqrt{8C(1-I(f))}}\right).\end{array}

By Proposition 4.2 in [5], for \delta >0, n large enough, we have

\begin{array}{rl}& (L_2{u}_n{)}^{-1}(L_3{u}_n{)}^{-3}\log P\left(\Vert w(\cdot ,\cdot )-(L_2{u}_n{)}^{1/2}(L_3{u}_n{)}^{3/2}f\Vert \le \frac{(1-\epsilon )(L_2{u}_n{)}^{-1/2}}{\sqrt{8C(1-I(f))}}\right)\\ & \le (L_2{u}_n{)}^{-1}(L_3{u}_n{)}^{-3}\log P\left(\Vert w(\cdot ,\cdot )\Vert \le \frac{1-\epsilon }{(L_2{u}_n{)}^{1/2}}\frac{1}{\sqrt{8C(1-I(f))}}\right)-I(f)+\delta .\end{array}

By Theorem 1.1 in [6],

\begin{array}{rl}& \log P\left(\Vert w(\cdot ,\cdot )\Vert \le \frac{1-\epsilon }{(L_2{u}_n{)}^{1/2}}\frac{1}{\sqrt{8C(1-I(f))}}\right)\\ & \le -\frac{1}{C}\frac{8C(1-I(f))L_2{u}_n}{(1-\epsilon {)}^2}{\left(\log \frac{\sqrt{8C(1-I(f))}}{1-\epsilon }+\frac{1}{2}{L}_3{u}_n\right)}^3,\end{array}

thus

(L_2{u}_n{)}^{-1}(L_3{u}_n{)}^{-3}\log P\left(\Vert w(\cdot ,\cdot )\Vert \le \frac{1-\epsilon }{(L_2{u}_n{)}^{1/2}}\frac{1}{\sqrt{8C(1-I(f))}}\right)\le -\frac{1-I(f)}{1-\epsilon },

which implies

\begin{array}{rl}& (L_2{u}_n{)}^{-1}(L_3{u}_n{)}^{-3}\log P\left(\Vert w(\cdot ,\cdot )-(L_2{u}_n{)}^{1/2}(L_3{u}_n{)}^{3/2}f\Vert \le \frac{(1-\epsilon )(L_2{u}_n{)}^{-1/2}}{\sqrt{8C(1-I(f))}}\right)\\ & \le -\frac{1-I(f)}{1-\epsilon }-I(f)+\delta =-{\eta }_0.\end{array}

We obtain

(2.2)\begin{array}{rl}& P\left(\Vert w(\cdot ,\cdot )-(L_2{u}_n{)}^{1/2}(L_3{u}_n{)}^{3/2}f\Vert \le \frac{1-\epsilon }{(L_2{u}_n{)}^{1/2}}\frac{1}{\sqrt{8C(1-I(f))}}\right)\\ & \le (\log u_n{)}^{-{\eta }_0(L_3{u}_n{)}^3}.\end{array}

By (2.1) and (2.2), we have

By Borel-Cantelli's lemma, Lemma 2.2 is proved.

Proof of (1.1)　Let h(u)=u^2(L_{2}u)(L_{3}u{)}^3, {\psi }_u(x,y)=\frac{w(ux,uy)}{\sqrt{h(u)}},x,y\in [0,1], and let u_n be defined as in Lemma 2.2, such that . Let X(u)=(L_{2}u)(L_{3}u{)}^{3/2}\Vert {\psi }_u(\cdot ,\cdot )-f(\cdot ,\cdot )\Vert, . Then for any \epsilon >0, by definition of infimum, there exists a T_n\in [u_n,u_{n+1}), such that X_n\ge X(T_n)-\epsilon. We have

(2.3)\begin{array}{rl}& \Vert {\psi }_{u_n}(\cdot ,\cdot )-f(\cdot ,\cdot )\Vert =\Vert \frac{\sqrt{h(T_n)}}{\sqrt{h(u_n)}}{\psi }_{T_n}\left(\frac{u_n\cdot }{T_n},\frac{u_n\cdot }{T_n}\right)-f(\cdot ,\cdot )\Vert \\ & =\Vert \frac{\sqrt{h(T_n)}}{\sqrt{h(u_n)}}\{{\psi }_{T_n}\left(\frac{u_n\cdot }{T_n},\frac{u_n\cdot }{T_n}\right)-f\left(\frac{u_n\cdot }{T_n},\frac{u_n\cdot }{T_n}\right)\}+\frac{\sqrt{h(T_n)}}{\sqrt{h(u_n)}}f\left(\frac{u_n\cdot }{T_n},\frac{u_n\cdot }{T_n}\right)-f(\cdot ,\cdot )\Vert \\ & \le \frac{\sqrt{h(T_n)}}{\sqrt{h(u_n)}}\Vert {\psi }_{T_n}(\cdot ,\cdot )-f(\cdot ,\cdot )\Vert +\left|\frac{\sqrt{h(T_n)}}{\sqrt{h(u_n)}}-1\right|\Vert f(\cdot ,\cdot )\Vert +\Vert f\left(\frac{u_n\cdot }{T_n},\frac{u_n\cdot }{T_n}\right)-f\left(\cdot ,\cdot \right)\Vert .\end{array}

By Lemma 2.1

(2.4)\begin{array}{rl}\Vert f\left(\frac{u_n\cdot }{T_n},\frac{u_n\cdot }{T_n}\right)-f\left(\cdot ,\cdot \right)\Vert & \le \left(1-\frac{u_n}{T_n}\right)+2{\left(1-\frac{u_n}{T_n}\right)}^{1/2}\\ & \le \left(1-\frac{u_n}{u_{n+1}}\right)+2{\left(1-\frac{u_n}{u_{n+1}}\right)}^{1/2},\end{array}

moreover,

(2.5)\left|\frac{\sqrt{h(T_n)}}{\sqrt{h(u_n)}}-1\right|\le \left|\frac{\sqrt{u_{n+1}^2(L_2{u}_{n+1})(L_3{u}_{n+1}{)}^3}}{\sqrt{u_n^2(L_2{u}_n)(L_3{u}_n{)}^3}}-1\right|\le |(\frac{u_{n+1}}{u_n}{)}^3-1|.

By the inequality \exp (-x)\ge 1-x,x>0, we have

\frac{u_n}{u_{n+1}}\ge 1-\frac{n+1}{(\log (n+1){)}^3}+\frac{n}{(\log n{)}^3},

(\frac{u_n}{u_{n+1}}{)}^3\ge 1-\frac{3(n+1)}{(\log (n+1){)}^3}+\frac{3n}{(\log n{)}^3},

which implies

(2.6)1-\frac{u_n}{u_{n+1}}\le \frac{1}{(\log n{)}^3},

(2.7)1-(\frac{u_n}{u_{n+1}}{)}^3\le \frac{3}{(\log n{)}^3}.

By (2.3)−(2.7) and Lemma 2.2, we have

\underset{n\to \mathrm{\infty }}{lim inf}X(T_n)\ge \frac{1}{\sqrt{8C(1-I(f))}},\mathrm{a}.\mathrm{s}.

Since \underset{u\to \mathrm{\infty }}{lim inf}X(u)\ge \underset{n\to \mathrm{\infty }}{lim inf}{X}_n\ge \underset{n\to \mathrm{\infty }}{lim inf}X(T_n)-\epsilon, (1.1) is proved.

The proof of (1.2) is similar to that of (1.1). Hereby omitted.

2.2 Proof of Theorem 1.2

Our proofs are based on the following lemmas.

Lemma 2.3 [7, Lemma 2.1]　 For any closed set F\subset {\mathcal{C}}_0 and for any open set G\subset {\mathcal{C}}_0, we have

\underset{\epsilon \to 0}{lim sup}\epsilon \log P(\sqrt{\epsilon }w\in F)\le -\underset{f\in F}{inf}I(f);\underset{\epsilon \to 0}{lim inf}\epsilon \log P(\sqrt{\epsilon }w\in G)\ge -\underset{f\in G}{inf}I(f).

Lemma 2.4　 Let u_n=n^n. Then we have

\underset{n\to \mathrm{\infty }}{lim sup}\Vert \frac{w(u_n\cdot ,u_n\cdot )}{\sqrt{u_n^2(L_2{u}_n)(L_3{u}_n{)}^3}}-K\Vert =0,a.s.

Proof　LetA=\{g\in \mathcal{C};\Vert g(\cdot )-K\Vert \ge \epsilon \}. Then set A is closed. For any g\in A, there exists a \delta >0, such that {\eta }_0:=\underset{g\in A}{inf}I(g)>1+\delta. Since

\begin{array}{rl}P\left(\Vert \frac{w(u_n\cdot ,u_n\cdot )}{\sqrt{u_n^2(L_2{u}_n)(L_3{u}_n{)}^3}}-K\Vert \ge \epsilon \right)& =P\left(\Vert \frac{w(\cdot ,\cdot )}{\sqrt{(L_2{u}_n)(L_3{u}_n{)}^3}}-K\Vert \ge \epsilon \right)\\ & =P\left(\frac{w(\cdot ,\cdot )}{\sqrt{(L_2{u}_n)(L_3{u}_n{)}^3}}\in A\right),\end{array}

by Lemma 2.3, we have

P\left(\Vert \frac{w(u_n\cdot ,u_n\cdot )}{\sqrt{u_n^2(L_2{u}_n)(L_3{u}_n{)}^3}}-K\Vert \ge \epsilon \right)\le {\left(\frac{1}{\log u_n}\right)}^{{\eta }_0(L_3{u}_n{)}^3}.

By Borel-Cantelli's lemma, Lemma 2.4 is proved.

Proof of (1.3)　For u large enough, there exists an increasing sequence u_n=n^n such that . Then it is sufficient to show that: for any f\in K with ,

(2.8)\underset{n\to \mathrm{\infty }}{lim inf}(L_2{u}_n{)}^b(L_3{u}_n{)}^{3/2}\Vert \frac{w(u_n\cdot ,u_n\cdot )}{\sqrt{u_n^2(L_2{u}_n{)}^b(L_3{u}_n{)}^3}}-f(\cdot ,\cdot )\Vert \le \frac{\sqrt{b^{3}C}}{\sqrt{8(1-I(f))}},\mathrm{a}.\mathrm{s}.

The proof of (2.8) is as follows. We have

\begin{array}{rl}& \Vert \frac{w(u_n\cdot ,u_n\cdot )}{\sqrt{u_n^2(L_2{u}_n{)}^b(L_3{u}_n{)}^3}}-f(\cdot ,\cdot )\Vert \\ & =\underset{0\le s\le 1}{sup}\underset{0\le t\le 1}{sup}\left|\frac{w(u_{n}s,u_{n}t)}{\sqrt{u_n^2(L_2{u}_n{)}^b(L_3{u}_n{)}^3}}-f(s,t)\right|\\ & \le \underset{0\le s\le \frac{u_{n-1}}{u_n}}{sup}\underset{0\le t\le \frac{u_{n-1}}{u_n}}{sup}\left|\frac{w(u_{n}s,u_{n}t)}{\sqrt{u_n^2(L_2{u}_n{)}^b(L_3{u}_n{)}^3}}-f(s,t)\right|\\ & +\underset{\frac{u_{n-1}}{u_n}\le s\le 1}{sup}\underset{\frac{u_{n-1}}{u_n}\le t\le 1}{sup}\left|\frac{w(u_{n}s,u_{n}t)}{\sqrt{u_n^2(L_2{u}_n{)}^b(L_3{u}_n{)}^3}}-f(s,t)\right|\\ & =\Vert \frac{w(u_{n-1}\cdot ,u_{n-1}\cdot )}{\sqrt{u_n^2(L_2{u}_n{)}^b(L_3{u}_n{)}^3}}-f\left(\frac{u_{n-1}}{u_n}\cdot ,\frac{u_{n-1}}{u_n}\cdot \right)\Vert \\ & +\underset{\frac{u_{n-1}}{u_n}\le s\le 1}{sup}\underset{\frac{u_{n-1}}{u_n}\le t\le 1}{sup}\left|\frac{w(u_{n}s,u_{n}t)}{\sqrt{u_n^2(L_2{u}_n{)}^b(L_3{u}_n{)}^3}}-f(s,t)\right|.\end{array}

First of all, we prove that

(2.9)\underset{n\to \mathrm{\infty }}{lim sup}(L_2{u}_n{)}^b(L_3{u}_n{)}^{3/2}\Vert \frac{w(u_{n-1}\cdot ,u_{n-1}\cdot )}{\sqrt{u_n^2(L_2{u}_n{)}^b(L_3{u}_n{)}^3}}-f\left(\frac{u_{n-1}}{u_n}\cdot ,\frac{u_{n-1}}{u_n}\cdot \right)\Vert =0,\mathrm{a}.\mathrm{s}.

In fact,

\begin{array}{rl}& \Vert \frac{w(u_{n-1}\cdot ,u_{n-1}\cdot )}{\sqrt{u_n^2(L_2{u}_n{)}^b(L_3{u}_n{)}^3}}-f\left(\frac{u_{n-1}}{u_n}\cdot ,\frac{u_{n-1}}{u_n}\cdot \right)\Vert \\ & \le \frac{u_{n-1}}{u_n}\frac{\sqrt{(L_2{u}_{n-1})(L_3{u}_{n-1}{)}^3}}{\sqrt{(L_2{u}_n{)}^b(L_3{u}_n{)}^3}}\Vert \frac{w(u_{n-1}\cdot ,u_{n-1}\cdot )}{\sqrt{u_{n-1}^2(L_2{u}_{n-1})(L_3{u}_{n-1}{)}^3}}\Vert +\Vert f\left(\frac{u_{n-1}}{u_n}\cdot ,\frac{u_{n-1}}{u_n}\cdot \right)\Vert ,\end{array}

by Lemma 2.4, \Vert \frac{w(u_{n-1}\cdot ,u_{n-1}\cdot )}{\sqrt{u_{n-1}^2(L_2{u}_{n-1})(L_3{u}_{n-1}{)}^3}}\Vert is almost surely bounded, thus

\begin{array}{rl}& (L_2{u}_n{)}^b(L_3{u}_n{)}^{3/2}\frac{u_{n-1}}{u_n}\frac{\sqrt{(L_2{u}_{n-1})(L_3{u}_{n-1}{)}^3}}{\sqrt{(L_2{u}_n{)}^b(L_3{u}_n{)}^3}}\Vert \frac{w(u_{n-1}\cdot ,u_{n-1}\cdot )}{\sqrt{u_{n-1}^2(L_2{u}_{n-1})(L_3{u}_{n-1}{)}^3}}\Vert \\ & =\frac{u_{n-1}}{u_n}(L_2{u}_n{)}^{b/2}\sqrt{(L_2{u}_{n-1})(L_3{u}_{n-1}{)}^3}\Vert \frac{w(u_{n-1}\cdot ,u_{n-1}\cdot )}{\sqrt{u_{n-1}^2(L_2{u}_{n-1})(L_3{u}_{n-1}{)}^3}}\Vert \to 0,(n\to \mathrm{\infty }).\end{array}

Moreover, for f\in K, we have

\Vert f\left(\frac{u_{n-1}}{u_n}\cdot ,\frac{u_{n-1}}{u_n}\cdot \right)\Vert \le \frac{u_{n-1}}{u_n},

which follows that

(L_2{u}_n{)}^b(L_3{u}_n{)}^{3/2}\Vert f\left(\frac{u_{n-1}}{u_n}\cdot ,\frac{u_{n-1}}{u_n}\cdot \right)\Vert \le (L_2{u}_n{)}^b(L_3{u}_n{)}^{3/2}\frac{u_{n-1}}{u_n}\to 0,(n\to \mathrm{\infty }).

Second, we prove that

(2.10)\begin{array}{rl}& \underset{n\to \mathrm{\infty }}{lim inf}(L_2{u}_n{)}^b(L_3{u}_n{)}^{3/2}\underset{\frac{u_{n-1}}{u_n}\le s\le 1}{sup}\underset{\frac{u_{n-1}}{u_n}\le t\le 1}{sup}\left|\frac{w(u_{n}s,u_{n}t)}{\sqrt{u_n^2(L_2{u}_n{)}^b(L_3{u}_n{)}^3}}-f(s,t)\right|\\ & \le \frac{\sqrt{b^{3}C}}{\sqrt{8(1-I(f))}}\mathrm{a}.\mathrm{s}.\end{array}

In fact,

\begin{array}{rl}& \underset{\frac{u_{n-1}}{u_n}\le s\le 1}{sup}\underset{\frac{u_{n-1}}{u_n}\le t\le 1}{sup}\left|\frac{w(u_{n}s,u_{n}t)}{\sqrt{u_n^2(L_2{u}_n{)}^b(L_3{u}_n{)}^3}}-f(s,t)\right|\\ & =\underset{\frac{u_{n-1}}{u_n}\le s\le 1}{sup}\underset{\frac{u_{n-1}}{u_n}\le t\le 1}{sup}|\frac{(w(u_{n}s,u_{n}t)-w(u_{n}s,u_{n-1})-w(u_{n-1},u_{n}t)+w(u_{n-1},u_{n-1}))}{\sqrt{u_n^2(L_2{u}_n{)}^b(L_3{u}_n{)}^3}}\\ & -f(s,t)+\frac{1}{\sqrt{u_n^2(L_2{u}_n{)}^b(L_3{u}_n{)}^3}}(w(u_{n}s,u_{n-1})+w(u_{n-1},u_{n}t)-w(u_{n-1},u_{n-1}))|\\ & \le \underset{0\le s\le 1}{sup}\underset{0\le t\le 1}{sup}|\frac{1}{\sqrt{u_n^2(L_2{u}_n{)}^b(L_3{u}_n{)}^3}}\{w((u_n-u_{n-1})s+u_{n-1},(u_n-u_{n-1})t+u_{n-1})\\ & -w((u_n-u_{n-1})s+u_{n-1},u_{n-1})-w(u_{n-1},(u_n-u_{n-1})t+u_{n-1})+w(u_{n-1},u_{n-1})\}\\ & -f\left(\frac{(u_n-u_{n-1})s+u_{n-1}}{u_n},\frac{(u_n-u_{n-1})t+u_{n-1}}{u_n}\right)|\\ & +\underset{\frac{u_{n-1}}{u_n}\le s\le 1}{sup}\underset{\frac{u_{n-1}}{u_n}\le t\le 1}{sup}|\frac{w(u_{n}s,u_{n-1})+w(u_{n-1},u_{n}t)-w(u_{n-1},u_{n-1})}{\sqrt{u_n^2(L_2{u}_n{)}^b(L_3{u}_n{)}^3}}|.\end{array}

Moreover

\begin{array}{rl}& \underset{\frac{u_{n-1}}{u_n}\le s\le 1}{sup}\underset{\frac{u_{n-1}}{u_n}\le t\le 1}{sup}\left|\frac{w(u_{n}s,u_{n-1})+w(u_{n-1},u_{n}t)-w(u_{n-1},u_{n-1})}{\sqrt{u_n^2(L_2{u}_n{)}^b(L_3{u}_n{)}^3}}\right|\\ & \le \underset{\frac{u_{n-1}}{u_n}\le s\le 1}{sup}\underset{\frac{u_{n-1}}{u_n}\le t\le 1}{sup}\left|\frac{\{w(u_{n}s,u_{n-1}t)+w(u_{n-1}s,u_{n}t)-w(u_{n-1}s,u_{n-1}t)\}}{\sqrt{u_n^2(L_2{u}_n{)}^b(L_3{u}_n{)}^3}}\right|\\ & \le \underset{\frac{u_{n-1}}{u_n}\le s\le 1}{sup}\underset{\frac{u_{n-1}}{u_n}\le t\le 1}{sup}\left|\frac{w(u_{n}s,u_{n-1}t)}{\sqrt{u_n^2(L_2{u}_n{)}^b(L_3{u}_n{)}^3}}\right|\\ & +\underset{\frac{u_{n-1}}{u_n}\le s\le 1}{sup}\underset{\frac{u_{n-1}}{u_n}\le t\le 1}{sup}\left|\frac{w(u_{n-1}s,u_{n}t)}{\sqrt{u_n^2(L_2{u}_n{)}^b(L_3{u}_n{)}^3}}\right|\end{array}

\begin{array}{rl}& +\underset{\frac{u_{n-1}}{u_n}\le s\le 1}{sup}\underset{\frac{u_{n-1}}{u_n}\le t\le 1}{sup}\left|\frac{1}{\sqrt{u_n^2(L_2{u}_n{)}^b(L_3{u}_n{)}^3}}w(u_{n-1}s,u_{n-1}t)\right|\\ & \le \underset{0\le s\le 1}{sup}\underset{0\le t\le 1}{sup}\left|\frac{w(u_{n}s,u_{n-1}t)}{\sqrt{u_n^2(L_2{u}_n{)}^b(L_3{u}_n{)}^3}}\right|+\underset{0\le s\le 1}{sup}\underset{0\le t\le 1}{sup}\left|\frac{w(u_{n-1}s,u_{n}t)}{\sqrt{u_n^2(L_2{u}_n{)}^b(L_3{u}_n{)}^3}}\right|\\ & +\underset{0\le s\le 1}{sup}\underset{0\le t\le 1}{sup}\left|\frac{1}{\sqrt{u_n^2(L_2{u}_n{)}^b(L_3{u}_n{)}^3}}w(u_{n-1}s,u_{n-1}t)\right|\\ & =\sqrt{\frac{u_{n-1}}{u_n}}\frac{\sqrt{L_2{u}_n}}{\sqrt{(L_2{u}_n{)}^b(L_3{u}_n{)}^3}}\Vert \frac{1}{\sqrt{u_n{u}_{n-1}{L}_2{u}_n}}w(u_n\cdot ,u_{n-1}\cdot )\Vert \\ & +\sqrt{\frac{u_{n-1}}{u_n}}\frac{\sqrt{L_2{u}_n}}{\sqrt{(L_2{u}_n{)}^b(L_3{u}_n{)}^3}}\Vert \frac{1}{\sqrt{u_n{u}_{n-1}{L}_2{u}_n}}w(u_{n-1}\cdot ,u_n\cdot )\Vert \\ & +\frac{u_{n-1}}{u_n}\frac{\sqrt{L_2{u}_n}}{\sqrt{(L_2{u}_n{)}^b(L_3{u}_n{)}^3}}\Vert \frac{1}{\sqrt{u_{n-1}^2{L}_2{u}_n}}w(u_{n-1}\cdot ,u_{n-1}\cdot )\Vert .\end{array}

Similar to the proof of Lemma 2.4, we see that \Vert \frac{w(u_n\cdot ,u_{n-1}\cdot )}{\sqrt{u_n{u}_{n-1}{L}_2{u}_n}}\Vert, \Vert \frac{w(u_{n-1}\cdot ,u_n\cdot )}{\sqrt{u_n{u}_{n-1}{L}_2{u}_n}}\Vert, \Vert \frac{w(u_{n-1}\cdot ,u_{n-1}\cdot )}{\sqrt{u_{n-1}^2{L}_2{u}_n}}\Vert are almost surely bounded. We have:

\begin{array}{rl}& (L_2{u}_n{)}^b(L_3{u}_n{)}^{3/2}\underset{\frac{u_{n-1}}{u_n}\le s\le 1}{sup}\underset{\frac{u_{n-1}}{u_n}\le t\le 1}{sup}\left|\frac{w(u_{n}s,u_{n-1})+w(u_{n-1},u_{n}t)-w(u_{n-1},u_{n-1})}{\sqrt{u_n^2(L_2{u}_n{)}^b(L_3{u}_n{)}^3}}\right|\\ & \le (L_2{u}_n{)}^b(L_3{u}_n{)}^{3/2}(\sqrt{\frac{u_{n-1}}{u_n}}\frac{\sqrt{L_2{u}_n}}{\sqrt{(L_2{u}_n{)}^b(L_3{u}_n{)}^3}}\Vert \frac{w(u_n\cdot ,u_{n-1}\cdot )}{\sqrt{u_n{u}_{n-1}{L}_2{u}_n}}\Vert \\ & +\sqrt{\frac{u_{n-1}}{u_n}}\frac{\sqrt{L_2{u}_n}}{\sqrt{(L_2{u}_n{)}^b(L_3{u}_n{)}^3}}\Vert \frac{w(u_{n-1}\cdot ,u_n\cdot )}{\sqrt{u_n{u}_{n-1}{L}_2{u}_n}}\Vert \\ & +\frac{u_{n-1}}{u_n}\frac{\sqrt{L_2{u}_n}}{\sqrt{(L_2{u}_n{)}^b(L_3{u}_n{)}^3}}\Vert \frac{w(u_{n-1}\cdot ,u_{n-1}\cdot )}{\sqrt{u_{n-1}^2{L}_2{u}_n}}\Vert )\to 0,(n\to \mathrm{\infty }).\end{array}

Let

\begin{array}{rl}{w}_n(s,t):=& \underset{0\le s\le 1}{sup}\underset{0\le t\le 1}{sup}|\frac{1}{\sqrt{u_n^2(L_2{u}_n{)}^b(L_3{u}_n{)}^3}}\{w((u_n-u_{n-1})s\\ & +u_{n-1},(u_n-u_{n-1})t+u_{n-1})-w((u_n-u_{n-1})s+u_{n-1},u_{n-1})\\ & -w(u_{n-1},(u_n-u_{n-1})t+u_{n-1})+w(u_{n-1},u_{n-1})\}\\ & -f\left(\frac{(u_n-u_{n-1})s+u_{n-1}}{u_n},\frac{(u_n-u_{n-1})t+u_{n-1}}{u_n}\right)|,\end{array}

\begin{array}{rl}{\tilde{w}}_n(s,t):=& \frac{1}{\sqrt{(L_2{u}_n{)}^b(L_3{u}_n{)}^3}(u_n-u_{n-1})}\{w((u_n-u_{n-1})s\\ & +u_{n-1},(u_n-u_{n-1})t+u_{n-1})-w((u_n-u_{n-1})s+u_{n-1},u_{n-1})\\ & -w(u_{n-1},(u_n-u_{n-1})t+u_{n-1})+w(u_{n-1},u_{n-1})\},\end{array}

g(s,t):=\frac{u_n}{u_n-u_{n-1}}f\left(\frac{(u_n-u_{n-1})s+u_{n-1}}{u_n},\frac{(u_n-u_{n-1})t+u_{n-1}}{u_n}\right).

Then I(g)\le I(f), thus g\in K. We need only to prove that

(2.11)\underset{n\to \mathrm{\infty }}{lim inf}(L_2{u}_n{)}^b(L_3{u}_n{)}^{3/2}{w}_n(s,t)\le \frac{\sqrt{b^{3}C}}{\sqrt{8(1-I(f))}},\mathrm{a}.\mathrm{s}.

We have

\begin{array}{rl}& P\left((L_2{u}_n{)}^b(L_3{u}_n{)}^{3/2}{w}_n(s,t)\le \frac{\sqrt{b^{3}C}}{\sqrt{8(1-I(f))}}(1+\epsilon )\right)\\ & =P\left((L_2{u}_n{)}^b(L_3{u}_n{)}^{3/2}\frac{u_n-u_{n-1}}{u_n}\Vert {\tilde{w}}_n(\cdot ,\cdot )-g\Vert \le \frac{\sqrt{b^{3}C}}{\sqrt{8(1-I(f))}}(1+\epsilon )\right)\\ & \ge P\left((L_2{u}_n{)}^b(L_3{u}_n{)}^{3/2}\Vert {\tilde{w}}_n(\cdot ,\cdot )-g\Vert \le \frac{\sqrt{b^{3}C}}{\sqrt{8(1-I(f))}}(1+\epsilon )\right)\\ & =P\left((L_2{u}_n{)}^b(L_3{u}_n{)}^{3/2}\Vert \frac{w(\cdot ,\cdot )}{\sqrt{(L_2{u}_n){)}^b(L_3{u}_n{)}^3}}-g\Vert \le \frac{\sqrt{b^{3}C}}{\sqrt{8(1-I(f))}}(1+\epsilon )\right)\\ & =P\left(\Vert w(\cdot ,\cdot )-(L_2{u}_n{)}^{b/2}(L_3{u}_n{)}^{3/2}g\Vert \le \frac{\sqrt{b^{3}C}}{\sqrt{8(1-I(f))}}\frac{(1+\epsilon )}{(L_2{u}_n{)}^{b/2}}\right).\end{array}

For f\in K with , we have . Choose \delta >0, such that . By Proposition 4.2 in [5] and Theorem 1.1 in [6], for n large enough

\begin{array}{rl}& (L_2{u}_n{)}^{-b}(L_3{u}_n{)}^{-3}\log P\left(\Vert w(\cdot ,\cdot )-(L_2{u}_n{)}^{b/2}(L_3{u}_n{)}^{3/2}g\Vert \le \frac{\sqrt{b^{3}C}}{\sqrt{8(1-I(f))}}\frac{(1+\epsilon )}{(L_2{u}_n{)}^{b/2}}\right)\\ & \ge (L_2{u}_n{)}^{-b}(L_3{u}_n{)}^{-3}\log P\left(\Vert w(\cdot ,\cdot )\Vert \le \frac{\sqrt{b^{3}C}}{\sqrt{8(1-I(f))}}\frac{(1+\epsilon )}{(L_2{u}_n{)}^{b/2}}\right)-I(g)-\delta \\ & \ge (L_2{u}_n{)}^{-b}(L_3{u}_n{)}^{-3}\left(-C\frac{8(1-I(f))}{b^{3}C}\frac{(L_2{u}_n{)}^b}{(1+\epsilon {)}^2}(\log \frac{\sqrt{8(1-I(f))(L_2{u}_n{)}^b}}{\sqrt{b^{3}C(1+\epsilon {)}^2}}{)}^3\right)\\ & -I(g)-\delta \\ & \ge -\frac{1-I(f)}{1+\epsilon }-I(g)-\delta =-{\eta }^{\prime },\end{array}

thus

P\left(\Vert w(\cdot ,\cdot )-(L_2{u}_n{)}^{b/2}(L_3{u}_n{)}^{3/2}g\Vert \le \frac{\sqrt{b^{3}C}}{\sqrt{8(1-I(f))}}\frac{(1+\epsilon )}{(L_2{u}_n{)}^{b/2}}\right)\ge (\log u_n{)}^{-{\eta }^{\prime }\frac{(L_3{u}_n{)}^3}{(\log u_n{)}^{1-b}}},

Which follows that

\sum _{n=1}^{\mathrm{\infty }}P\left((L_2{u}_n{)}^b(L_3{u}_n{)}^{3/2}{w}_n(s,t)\le \frac{\sqrt{b^{3}C}}{\sqrt{8(1-I(f))}}(1+\epsilon )\right)=\mathrm{\infty }.

By Borel-Cantelli's lemma,

\underset{n\to \mathrm{\infty }}{lim inf}(L_2{u}_n{)}^b(L_3{u}_n{)}^{3/2}{w}_n(s,t)\le \frac{\sqrt{b^{3}C}}{\sqrt{8(1-I(f))}},\mathrm{a}.\mathrm{s}.

Hither, (2.8) is proved. It follows that (1.3) is proved.

The proof of (1.4) is similar to that of (1.3). Hereby omitted.

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