Local Strassen’s law of the iterated logarithm for increments of a Brownian motion in Hölder norm

Abstract

With the help of large deviation of Brownian motion in the Hölder norm, local Strassen’s law of the iterated logarithm for increments of a Brownian motion in the Hölder norm is investigated. This paper promotes the corresponding results by Gantert and Wei.

Keywords

Brownian motion / increments / local Strassen’s law of the iterated logarithm / Hölder norm

1 Introduction and main results

The limit theorems for Brownian motion and its increments are a widely studied, and many profound results have been established. For example, Gao and Wang [4] have studied the functional limit convergence rate of Brownian motion increments. Baldi et al. [1, 2, 5] have investigated Strassen’s theorem for Brownian motion under the Hölder norm. Wei [7] extended the results of Baldi et al. [1, Theorem 3.1] and obtained a functional limit theorem for C-R type increments of k-dimensional Brownian motion under the Hölder norm. There has also been research on the local limit theorem of Brownian motion. Wei [6] derived the functional modulus of continuity for Brownian motion under the Hölder norm. Gantert established the local Strassen law of the iterated logarithm for Brownian motion; see [3]. This paper investigates the local Strassen law of the iterated logarithm for the increments of Brownian motion under the Hölder norm. The results obtained by this paper generalize Theorem 3 in [3] and also extend the corresponding results of Wei [6, 7].

Let \left\{w\left(t\right);t\ge 0\right\} be a d-dimensional standard Brownian motion. Define C_0\left[0,1\right]=\{f;f:\left[0,1\right]\to {\mathbb{R}}^d,f\left(0\right)=0, and f is \mathrm{c}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{u}\mathrm{o}\mathrm{u}\mathrm{s}\} equipped with the uniform norm \parallel f\parallel ={\mathrm{s}\mathrm{u}\mathrm{p}}_{0\le t\le 1}\left|f\left(t\right)\right|. The following two Banach spaces are considered:

where Define the space:

Then, H is a Hilbert space with the inner product:

⟨f,g⟩={\int }_0^1\dot{f}\left(x\right)\dot{g}\left(x\right)\mathrm{d}x,f,g\in H.

Define the mapping I:{\mathcal{C}}^{\alpha ,0}\to \left[0,\mathrm{\infty }\right] as follows:

I\left(f\right)=\{\begin{array}{ll}{\displaystyle \frac{1}{2}}{{\displaystyle \int }}_0^1{\left|\dot{f}\left(t\right)\right|}^2\mathrm{d}t,& f\in H,\\ +\mathrm{\infty },& \text{o}\text{therwise}\text{.}\end{array}

Throughout this paper, let a_u,b_u be two non-decreasing continuous functions from (0,1) to (0,1) satisfying

(i)a_u\le b_u, u\in \left(0,1\right), and {\mathrm{l}\mathrm{i}\mathrm{m}}_{u\to 0}{a}_u=0;

(ii)\frac{b_u}{a_u} is non-increasing.

For u\in \left(0,1\right) and 0\le t\le 1, define the trajectory \mathrm{\Delta }\left(t,u\right) as

\mathrm{\Delta }\left(t,u\right)\left(s\right)=w\left(b_{u}t+a_{u}s\right)-w\left(b_{u}t\right),s\in \left[0,1\right].

Define

{\beta }_u={\left(2a_u\mathrm{l}\mathrm{o}\mathrm{g}\frac{b_u\mathrm{l}\mathrm{o}\mathrm{g}{b}_u^{-1}}{a_u}\right)}^{-\frac{1}{2}},u\in \left(0,1\right),

and let

K=\left\{\phi \in H:2I\left(\phi \right)\le 1\right\}.

The main results of this paper are stated as follows.

Theorem 1.1　 If conditions (i) and (ii) are satisfied, then with probability 1, the family \left\{{\beta }_u\mathrm{\Delta }\left(t,u\right);u\in \left(0,1\right)\right\} is relatively compact in C^{\alpha ,0} as u\to 0, and its set of limit points is K. That is,

\underset{u\to 0+}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{s}\mathrm{u}\mathrm{p}}\underset{t\in \left[0,1-\frac{a_u}{b_u}\right]}{\mathrm{s}\mathrm{u}\mathrm{p}}\parallel {\beta }_u\mathrm{\Delta }\left(t,u\right)(\cdot )-K{\parallel }_{\alpha }=0,a.s.

and

(1.1)\underset{u\to 0+}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{f}}\underset{t\in \left[0,1-\frac{a_u}{b_u}\right]}{\mathrm{i}\mathrm{n}\mathrm{f}}\parallel {\beta }_u\mathrm{\Delta }\left(t,u\right)(\cdot )-\phi {\parallel }_{\alpha }=0,a.s.\text{for any}\phi \in K.

If the following additional condition holds:

(iii) {\mathrm{l}\mathrm{i}\mathrm{m}}_{u\to 0}\frac{\mathrm{l}\mathrm{o}\mathrm{g}\frac{b_u}{a_u}}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{l}\mathrm{o}\mathrm{g}{b}_u^{-1}}=+\mathrm{\infty },

there is

\underset{u\to 0+}{\mathrm{l}\mathrm{i}\mathrm{m}}\underset{t\in \left[0,1-\frac{a_u}{b_u}\right]}{\mathrm{i}\mathrm{n}\mathrm{f}}\parallel {\beta }_u\mathrm{\Delta }\left(t,u\right)(\cdot )-\phi {\parallel }_{\alpha }=0,a.s.\text{for any}\phi \in K.

2 Proof of the theorem

The proof of the theorem is completed by the following lemmas.

Lemma 2.1 [1, Theorem 2.1]　 For any Borel set A\subset {\mathcal{C}}^{\alpha ,0},

\mathrm{l}\mathrm{i}\mathrm{m}\underset{\epsilon \to 0}{\mathrm{s}\mathrm{u}\mathrm{p}}{\epsilon }^2\mathrm{l}\mathrm{o}\mathrm{g}P\left(\epsilon w\subset A\right)\le -\Lambda \left(\overline{A}\right), \underset{\epsilon \to 0}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{f}}{\epsilon }^2\mathrm{l}\mathrm{o}\mathrm{g}P\left(\epsilon w\subset A\right)\ge -\Lambda \left(\stackrel{\circ }{A}\right),

where \Lambda \left(A\right)={\mathrm{i}\mathrm{n}\mathrm{f}}_{f\in A}I\left(f\right).

Lemma 2.2 [7, Lemma 3]　 Let F\subset {\mathcal{C}}^{\alpha ,0}. Then

\begin{array}{rl}& \begin{array}{c}\underset{\epsilon \to 0}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{s}\mathrm{u}\mathrm{p}} {\epsilon }^2\end{array}\left\{\mathrm{l}\mathrm{o}\mathrm{g}P\left\{\underset{0\le t\le T-h}{\begin{array}{c}\bigcup _{0\le t\le T-h} \end{array}\left(\frac{\epsilon }{\sqrt{h}}\left(w\left(t+h\cdot \right)-w\left(t\right)\right)\in F\right)}\right\}+\mathrm{l}\mathrm{o}\mathrm{g}\frac{h}{T}\right\}\\ & \le -\Lambda \left(F\right),\end{array}

where \Lambda \left(F\right)={\mathrm{i}\mathrm{n}\mathrm{f}}_{f\in F}I\left(f\right), .

Lemma 2.3　 There exists a non-increasing sequence \left\{u_n\in \left(0,1\right),n\in \mathbb{N}\right\} such that {\mathrm{l}\mathrm{i}\mathrm{m}}_{n\to \mathrm{\infty }}{u}_n\to 0, and for any \epsilon >0,

P\left(\underset{t\in \left[0,b_{u_n}-a_{u_{n+1}}\right]}{\mathrm{s}\mathrm{u}\mathrm{p}}\parallel {\beta }_{u_n}\left(w\left(t+a_{u_n}\cdot \right)-w\left(t\right)\right)-K{\parallel }_{\alpha }\ge \epsilon ,i.o.\right)=0.

\left\{u_n\right\} will be defined in different cases in the proof.

Proof　Define =\left\{\phi \in {\mathcal{C}}^{\alpha ,0}:\parallel \phi -K{\parallel }_{\alpha }\ge \epsilon \right\}, A is a closed set, and \Lambda \left(A\right)>\frac{1}{2}. Thus, there exists a sufficiently small \delta >0such that \Lambda \left(A\right)>\frac{1+\delta }{2} By Lemma 2.2, for sufficiently large n, there is

(2.1)\begin{array}{rl}& P\left(\underset{t\in \left[0,b_{u_n}-a_{u_{n+1}}\right]}{\mathrm{s}\mathrm{u}\mathrm{p}}{\Vert {\beta }_{u_n}\left(w\left(t+a_{u_n}\cdot \right)-w\left(t\right)\right)-K\Vert }_{\alpha }\ge \epsilon \right)\\ & =P\left(\bigcup _{0\le t\le (b_{u_n}+a_{u_n}-a_{u_{n+1}})-a_{u_n}}\left({\beta }_{u_n}\left(w\left(t+a_{u_n}\cdot \right)-w\left(t\right)\right)\in A\right)\right)\\ & \le \frac{b_{u_n}+a_{u_n}-a_{u_n+1}}{a_{u_n}}{\left(\frac{a_{u_n}}{b_{u_n}\mathrm{l}\mathrm{o}\mathrm{g}{b}_{u_n}^{-1}}\right)}^{1+\frac{\delta }{2}}.\end{array}

Case (I):

If , then there exists a constant such that

\frac{b_u}{a_u}\le {\left(\mathrm{l}\mathrm{o}\mathrm{g}{b}_u^{-1}\right)}^M\le {\left(\mathrm{l}\mathrm{o}\mathrm{g}{a}_u^{-1}\right)}^M.

Thus, \frac{1}{\mathrm{l}\mathrm{o}\mathrm{g}{b}_u^{-1}}\le \frac{1}{\mathrm{l}\mathrm{o}\mathrm{g}{a}_u^{-1}-M\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{l}\mathrm{o}\mathrm{g}{a}_u^{-1}}. Choose u_n such that a_{u_n}=\mathrm{e}\mathrm{x}\mathrm{p}\left(-\frac{n}{{\left(\mathrm{l}\mathrm{o}\mathrm{g}n\right)}^3}\right). Then

(2.2)\frac{1}{\mathrm{l}\mathrm{o}\mathrm{g}{b}_{u_n}^{-1}}\le \frac{{\left(\mathrm{l}\mathrm{o}\mathrm{g}n\right)}^3}{n-M{\left(\mathrm{l}\mathrm{o}\mathrm{g}n\right)}^4}.

From Equations (2.1) and (2.2) and the Borel-Cantelli Lemma, there is

P\left(\underset{t\in \left[0,b_{u_n}-a_{u_{n+1}}\right]}{\mathrm{s}\mathrm{u}\mathrm{p}}\Vert {\beta }_{u_n}\left(w\left(t+a_{u_n}\cdot \right)-w\left(t\right)\right)-K{\Vert }_{\alpha }\ge \epsilon ,\text{i.o.}\right)=0.

Case (II): \mathrm{l}\mathrm{i}\mathrm{m}{\mathrm{s}\mathrm{u}\mathrm{p}}_{u\to 0}\frac{\mathrm{l}\mathrm{o}\mathrm{g}\frac{b_u}{a_u}}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{l}\mathrm{o}\mathrm{g}{b}_u^{-1}}=\mathrm{\infty }.

If \mathrm{l}\mathrm{i}\mathrm{m}{\mathrm{s}\mathrm{u}\mathrm{p}}_{u\to 0}\frac{\mathrm{l}\mathrm{o}\mathrm{g}\frac{b_u}{a_u}}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{l}\mathrm{o}\mathrm{g}{b}_u^{-1}}=\mathrm{\infty }, then choose u_{\mathrm{n}} such that \frac{b_{u_n}}{a_{u_n}}=n^p, where p>\frac{2}{\delta }. Using an argument similar to the proof of Case (I), the lemma is proved. □

Lemma 2.4　 If conditions (i) and (ii) hold, then there is

(2.3)\underset{u\to 0+}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{s}\mathrm{u}\mathrm{p}}\underset{t\in \left[0,b_u-a_u\right]}{\mathrm{s}\mathrm{u}\mathrm{p}}{\Vert {\beta }_u\left(w\left(t+a_u\cdot \right)-w\left(t\right)\right)-K\Vert }_{\alpha }=0,a.s.

Proof　Define

{\psi }_{t,u}\left(s\right)={\beta }_u\left(w\left(t+a_{u}s\right)-w\left(t\right)\right),s\in \left[0,1\right],t\in \left[0,b_u-a_u\right].

Then,

{\psi }_{t,u}\left(s\right)=\frac{{\beta }_u}{{\beta }_{u_n}}{\psi }_{t,u_n}\left(\frac{a_u}{a_{u_n}}s\right).

For any u\in \left(0,1\right), there exists n such that , and there is

\begin{array}{rl}& \underset{t\in \left[0,b_u-a_u\right]}{\mathrm{s}\mathrm{u}\mathrm{p}}\parallel {\psi }_{t,u}(\cdot )-K{\parallel }_{\alpha }\\ & =\underset{t\in \left[0,b_u-a_u\right]}{\mathrm{s}\mathrm{u}\mathrm{p}}\underset{f\in K}{\mathrm{i}\mathrm{n}\mathrm{f}}{\Vert \frac{{\beta }_u}{{\beta }_{u_n}}{\psi }_{t,u_n}\left(\frac{a_u}{a_{u_n}}\cdot \right)-f(\cdot )\Vert }_{\alpha }\\ & \le \underset{t\in \left[0,b_{u_n}-a_{u_{n+1}}\right]}{\mathrm{s}\mathrm{u}\mathrm{p}}\underset{f\in K}{\mathrm{i}\mathrm{n}\mathrm{f}}{\Vert {\psi }_{t,u_n}\left(\frac{a_u}{a_{u_n}}\cdot \right)-f\left(\frac{a_u}{a_{u_n}}\cdot \right)\Vert }_{\alpha }\\ & +\underset{f\in K}{\mathrm{i}\mathrm{n}\mathrm{f}}{\Vert f\left(\frac{a_u}{a_{u_n}}\cdot \right)-f(\cdot )\Vert }_{\alpha }+\left|\frac{{\beta }_u}{{\beta }_{u_n}}-1\right|\underset{t\in \left[0,b_{u_n}-a_{u_{n+1}}\right]}{\mathrm{s}\mathrm{u}\mathrm{p}}{\Vert {\psi }_{t,u_n}\left(\frac{a_u}{a_{u_n}}\cdot \right)\Vert }_{\alpha }.\end{array}

Since {\beta }_u and \mathrm{l}\mathrm{o}\mathrm{g}\frac{b_u\mathrm{l}\mathrm{o}\mathrm{g}{b}_u^{-1}}{a_u} are non-increasing, there is

\frac{{\beta }_u}{{\beta }_{u_{n+1}}}\ge \frac{{\beta }_{u_n}}{{\beta }_{u_{n+1}}}={\left(\frac{a_{u_{n+1}}\mathrm{l}\mathrm{o}\mathrm{g}\frac{b_{u_{n+1}}\mathrm{l}\mathrm{o}\mathrm{g}{b}_{u_{n+1}}^{-1}}{a_{u_{n+1}}}}{a_{u_n}\mathrm{l}\mathrm{o}\mathrm{g}\frac{b_{u_n}\mathrm{l}\mathrm{o}\mathrm{g}{b}_{u_n}^{-1}}{a_{u_n}}}\right)}^{\frac{1}{2}}\ge {\left(\frac{a_{u_{n+1}}}{a_{u_n}}\right)}^{\frac{1}{2}}.

Moreover, there exists a constant c>0 such that

{\Vert f\left(\frac{a_u}{a_{u_n}}\cdot \right)-f(\cdot )\Vert }_{\alpha }\le c{\left(\frac{a_{u_n}}{a_{u_{n+1}}}-1\right)}^{-\alpha +\frac{1}{2}}.

Case (I):

If , then b_u\to 0 \left(u\to 0\right). For a_{u_n}=\mathrm{e}\mathrm{x}\mathrm{p}\left(-\frac{n}{{\left(\mathrm{l}\mathrm{o}\mathrm{g}n\right)}^3}\right), it is verified that \frac{a_{u_n}}{a_{u_{n+1}}}\to 1(n\to \mathrm{\infty }). Thus, by Lemma 2.3 and Equation (2.3), the result follows.

Case (II): \mathrm{l}\mathrm{i}\mathrm{m}{\mathrm{s}\mathrm{u}\mathrm{p}}_{u\to 0+}\frac{\mathrm{l}\mathrm{o}\mathrm{g}\frac{b_u}{{\alpha }_u}}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{l}\mathrm{o}\mathrm{g}{b}_u^{-1}}=\mathrm{\infty }.

If \mathrm{l}\mathrm{i}\mathrm{m}{\mathrm{s}\mathrm{u}\mathrm{p}}_{u\to 0+}\frac{\mathrm{l}\mathrm{o}\mathrm{g}\frac{b_{ut}}{{\alpha }_u}}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{l}\mathrm{o}\mathrm{g}{b}_u^{-1}} = \mathrm{\infty }, choose a non-increasing sequence \left\{u_n;n\ge 1\right\} such that \frac{b_{u_n}}{a_{u_n}}=n^p,p>\frac{2}{\delta }. Define h\left(n\right)=\frac{\mathrm{l}\mathrm{o}\mathrm{g}\frac{b{u}_n}{{\alpha }_{u_n}}}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{l}\mathrm{o}\mathrm{g}{b}_{un}^{-1}}=\frac{p\mathrm{l}\mathrm{o}\mathrm{g}n}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{l}\mathrm{o}\mathrm{g}{b}_{u_n}^{-1}}, then b_{u_n}^{-1}=\mathrm{e}\mathrm{x}\mathrm{p}\left\{n^{\frac{p}{h\left(n\right)}}\right\}, h\left(n\right)\to \mathrm{\infty }\left(n\to \mathrm{\infty }\right). Thus, there is

1\le \frac{b_{u_n}}{b_{u_{n+1}}}=\mathrm{e}\mathrm{x}\mathrm{p}\left({\left(n+1\right)}^{\frac{p}{h\left(n+1\right)}}-n^{\frac{p}{h\left(n\right)}}\right)\to 1, \frac{a_{u_n}}{a_{u_{n+1}}}\to 1, n\to \mathrm{\infty }.

Therefore, Equation (2.3) is proved. □

Lemma 2.5　 If conditions (i) and (ii) hold, then for any f\in K, there is

\underset{u\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{f}}{\Vert {\beta }_u\mathrm{\Delta }\left(1-\frac{a_u}{b_u},u\right)-f\Vert }_{\alpha }=0,a.s.

Proof　Define u_1={\mathrm{e}}^{-1} and let \rho ={\mathrm{l}\mathrm{i}\mathrm{m}}_{u\to 0}\frac{a_u}{b_u}.

If and b_u\to b\ne 0 (u\to 0), then {\mathrm{l}\mathrm{i}\mathrm{m}}_{u\to 0}\frac{\mathrm{l}\mathrm{o}\mathrm{g}\frac{b_u}{a_u}}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{l}\mathrm{o}\mathrm{g}{b}_u^{-1}}=+\mathrm{\infty }. In this case, Lemma 2.7 implies that Equation (1.1) holds. Thus, the following two cases are considered: (I) and b_u\to 0 (u\to 0); (II) \rho =1.

Case (I): and b_u\to 0 (u→ 0).

Choose u_k such that

b_{u_{k+1}}=b_{u_k}-a_{u_k},k\ge 1.

For any \epsilon >0, by the scaling property, there is

where . Since K is a compact set, it suffices to show that for \phi \in K, if , the conclusion holds. If , then choose \eta >0 such that . By Lemma 2.1 (Large Deviation Principle), for sufficiently large k, there is

There exist constants k_0\ge 1 and c>0 such that

\mathrm{l}\mathrm{o}\mathrm{g}\frac{b_{u_k}^{-1}}{b_{u_{k-1}}^{-1}}=-\mathrm{l}\mathrm{o}\mathrm{g}\left(1-\frac{a_{u_{k-1}}}{b_{u_{k-1}}}\right)\le c\frac{a_{u_{k-1}}}{b_{u_{k-1}}}.

Thus,

\begin{array}{rl}& \sum _{k=k_0+1}^{N+1}{\left(\frac{a_{u_{k-1}}}{b_{u_{k-1}}\mathrm{l}\mathrm{o}\mathrm{g}{b}_{u_{k-1}}^{-1}}\right)}^{{\sigma }^{\prime }}\\ & \ge {\left(\frac{1}{\mathrm{l}\mathrm{o}\mathrm{g}{b}_{N+1}^{-1}}\right)}^{{\sigma }^{\prime }}\sum _{k=k_0+1}^{N+1}\frac{a_{u_{k-1}}}{b_{u_{k-1}}}\ge \frac{1}{c}{\left(\frac{1}{\mathrm{l}\mathrm{o}\mathrm{g}{b}_{N+1}^{-1}}\right)}^{{\sigma }^{\prime }}\sum _{k=k_0+1}^{N+1}\mathrm{l}\mathrm{o}\mathrm{g}\frac{b_{u_k}^{-1}}{b_{u_{k-1}}^{-1}}\\ & =\frac{1}{c}{\left(\frac{1}{\mathrm{l}\mathrm{o}\mathrm{g}{b}_{N+1}^{-1}}\right)}^{{\sigma }^{\prime }}\left(\mathrm{l}\mathrm{o}\mathrm{g}{b}_{N+1}^{-1}-\mathrm{l}\mathrm{o}\mathrm{g}{b}_{u_{k_0}}^{-1}\right)\to \mathrm{\infty }\left(N\to \mathrm{\infty }\right).\end{array}

Since the events \left\{\parallel {\beta }_{u_n}\mathrm{\Delta }\left(1-\frac{{\alpha }_{u_n}}{b_{u_n}},u_n\right)-f{\parallel }_{\alpha }\le \epsilon \right\} are independent for n\ge 1, by Borel-Cantelli’s Lemma, there is

\underset{n\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{f}}{\Vert {\beta }_{u_n}\mathrm{\Delta }\left(1-\frac{a_{u_n}}{b_{u_n}},u_n\right)-f\Vert }_{\alpha }=0,\text{a.s.}

Case (II): \rho =1.

If \rho =1, then a_u=b_u. In this case, refer to [6, Theorem 1.2] for further details. □

Lemma 2.6　 If conditions (i)‒(iii) hold, then for any f\in K, there is

\underset{n\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\underset{t\in \left[0,b_{u_{n+1}}-a_{u_n}\right]}{\mathrm{i}\mathrm{n}\mathrm{f}}{\Vert {\beta }_{u_n}\left(w\left(t+a_{u_n}\cdot \right)-w\left(t\right)\right)-f\Vert }_{\alpha }=0,a.s.

Proof　Since {\mathrm{l}\mathrm{i}\mathrm{m}}_{u\to 0+}\frac{\mathrm{l}\mathrm{o}\mathrm{g}\frac{b_u}{{\alpha }_u}}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{l}\mathrm{o}\mathrm{g}{b}_u^{-1}}=\mathrm{\infty }, there exists a decreasing subsequence \{u_n;n\ge 1\} such that \frac{b_{u_n}}{a_{u_n}}=n^p. Define t_i=i{a}_{u_n} for i=0,1,2,\dots, and let k_n=\left[\frac{b_{u_{n+1}}}{a_{u_n}}\right]-1. Also, define h\left(n\right)=\frac{\mathrm{l}\mathrm{o}\mathrm{g}\frac{b{u}_n}{a{u}_n}}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{l}\mathrm{o}\mathrm{g}{b}_{u_n}^{-1}}=\frac{\mathrm{l}\mathrm{o}\mathrm{g}{n}^p}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{l}\mathrm{o}\mathrm{g}{b}_{u_n}^{-1}}. Then b_{u_n}^{-1}=\mathrm{e}\mathrm{x}\mathrm{p}\left(n^{\frac{p}{h\left(n\right)}}\right) and h\left(n\right)\to \mathrm{\infty } \left(n\to \mathrm{\infty }\right). Moreover, for any small \alpha >0,\frac{(n+1{)}^{\alpha }}{\mathrm{l}\mathrm{o}\mathrm{g}{b}_{u_{n+1}}^{-1}}\to \mathrm{\infty }, 1\le \frac{b_{u_n}}{b_{u_{n+1}}}=\mathrm{e}\mathrm{x}\mathrm{p}\left\{\right(n+1{)}^{\frac{p}{h(n+1)}}-n^{\frac{p}{h\left(n\right)}}\} ≤ \exp \left\{n^{\frac{p}{h\left(n\right)}}\left[(1+\frac{1}{n}{)}^{\frac{p}{h\left(n\right)}}-1\right]\right\}\le \exp \{n^{\frac{p}{h\left(n\right)}}(1+\frac{1}{n}-1)\}=\exp (n^{\frac{p}{h\left(n\right)}-1})\to 1(n\to \mathrm{\infty }), there is

\begin{array}{rl}& P\left(\underset{t\in [0,b_{u_{n+1}}-a_{u_n}]}{\mathrm{i}\mathrm{n}\mathrm{f}}{\Vert {\beta }_{u_n}\left(w(t+a_{u_n}\cdot )-w\left(t\right)\right)-f\Vert }_{\alpha }\ge \epsilon \right)\\ & \le P\left(\underset{0\le i\le k_n}{\mathrm{m}\mathrm{i}\mathrm{n}} {\Vert {\beta }_{u_n}\left(w(t_i+a_{u_n}\cdot )-w\left(t_i\right)\right)-f\Vert }_{\alpha }\ge \epsilon \right)\\ & =\prod _{i=0}^{k_n} P\left(\parallel {\beta }_{u_n}\left(w(t_i+a_{u_n}\cdot )-w\left(t_i\right)\right)-f{\parallel }_{\alpha }\ge \epsilon \right)\\ & ={\left\{1-P\left({\left(2\mathrm{l}\mathrm{o}\mathrm{g}\frac{b_{u_n}\mathrm{l}\mathrm{o}\mathrm{g}{b}_{u_n}^{-1}}{a_{u_n}}\right)}^{-\frac{1}{2}}w\in A\right)\right\}}^{1+k_n},\end{array}

where . If then choose \delta >0 such that . By Lemma 2.1 (Large Deviation Principle), for sufficiently large n, there is

\begin{array}{rl}& P\left(\underset{t\in [0,b_{u_{n+1}}-a_{u_n}]}{inf} {\Vert {\beta }_{u_n}\left(w(t+a_{u_n}\cdot )-w\left(t\right)\right)-f\Vert }_{\alpha }\ge \epsilon \right)\\ & \le \mathrm{e}\mathrm{x}\mathrm{p}\left\{-{\left(\frac{a_{u_n}}{b_{u_n}\mathrm{l}\mathrm{o}\mathrm{g}{b}_{u_n}^{-1}}\right)}^{\mu }\left[\frac{b_{u_{n+1}}}{a_{u_n}}\right]\right\}.\end{array}

Choosing an appropriate p, it is ensured that . By Borel-Cantelli Lemma, it follws that

\underset{n\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{s}\mathrm{u}\mathrm{p}}\underset{t\in \left[0,b_{u_{n+1}}-a_{u_n}\right]}{\mathrm{i}\mathrm{n}\mathrm{f}} {\Vert {\beta }_{u_n}\left(w\left(t+a_{u_n}\cdot \right)-w\left(t\right)\right)-f\Vert }_{\alpha }=0,\mathrm{a}.\mathrm{s}.

Lemma 2.7　 For any f\in K, define {\varphi }_{t,u}\left(s\right)={\beta }_u\left(w\right(t+a_{u}s)-w(t\left)\right), s\in \left[0,1\right], t\in [0,b_u-a_u]. If

\begin{array}{c}\\ \underset{n\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{s}\mathrm{u}\mathrm{p}}\underset{t\in \left[0,b_{u_{n+1}}-a_{u_n}\right]}{\mathrm{i}\mathrm{n}\mathrm{f}} \parallel {\varphi }_{t,u_n}(\cdot )-f{\parallel }_{\alpha }=0,a.s.,\end{array}

then

\underset{u\to 0+}{\mathrm{l}\mathrm{i}\mathrm{m}}\underset{t\in \left[0,b_u-a_u\right]}{\mathrm{i}\mathrm{n}\mathrm{f}}\parallel {\varphi }_{t,u}(\cdot )-f{\parallel }_{\alpha }=0,a.s.,

where a_{u_n} and b_{u_n} are as defined in Lemma 2.6.

Proof　Since {\phi }_{t,u}\left(s\right)=\frac{{\beta }_u}{{\beta }_{u_n}}{\phi }_{t,u_n}\left(\frac{a_u}{a_{u_n}}s\right), there is

\begin{array}{rl}& \underset{t\in [0,b_u-a_u]}{\mathrm{i}\mathrm{n}\mathrm{f}}\parallel {\varphi }_{t,u}(\cdot )-f(\cdot ){\parallel }_{\alpha }\\ & =\underset{t\in [0,b_u-a_u]}{\mathrm{i}\mathrm{n}\mathrm{f}}{\Vert \frac{{\beta }_u}{{\beta }_{u_n}}{\varphi }_{t,u_n}\left(\frac{a_u}{a_{u_n}}\cdot \right)-f(\cdot )\Vert }_{\alpha }\\ & \le \underset{t\in [0,b_{u_{n+1}}-a_{u_n}]}{inf} {\Vert {\varphi }_{t,u_n}\left(\frac{a_u}{a_{u_n}}\cdot \right)-f\left(\frac{a_u}{a_{u_n}}\cdot \right)\Vert }_{\alpha }\\ & +\underset{t\in [0,b_{u_{n+1}}-a_{u_n}]}{sup} \left|\frac{{\beta }_u}{{\beta }_{u_n}}-1\right|{\Vert {\varphi }_{t,u_n}\left(\frac{a_u}{a_{u_n}}\cdot \right)\Vert }_{\alpha }+{\Vert f\left(\frac{a_u}{a_{u_n}}\cdot \right)-f(\cdot )\Vert }_{\alpha }.\end{array}

Observing that

\frac{{\beta }_u}{{\beta }_{u_n}}\le \frac{{\beta }_{u_{n+1}}}{{\beta }_{u_n}}={\left(\frac{a_{u_n}\mathrm{l}\mathrm{o}\mathrm{g}\frac{b_{u_n}\mathrm{l}\mathrm{o}\mathrm{g}{b}_{u_n}^{-1}}{a_{u_n}}}{a_{u_{n+1}}\mathrm{l}\mathrm{o}\mathrm{g}\frac{b_{u_{n+1}}\mathrm{l}\mathrm{o}\mathrm{g}{b}_{u_{n+1}}^{-1}}{a_{u_{n+1}}}}\right)}^{\frac{1}{2}}\le {\left(\frac{a_{u_n}}{a_{u_{n+1}}}\right)}^{\frac{1}{2}}

and

{\Vert f\left(\frac{a_u}{a_{u_n}}\cdot \right)-f(\cdot )\Vert }_{\alpha }\le 4{\left(\frac{a_{u_n}}{a_{u_{n+1}}}-1\right)}^{-\alpha +\frac{1}{2}}\sqrt{{\int }_0^1{\left|\dot{f}\left(x\right)\right|}^2\mathrm{d}x},

the proof of Lemma 2.7 is complete. □

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