Let $ \{X(t); t\geq 0\} $ be a standard $ \alpha $-fractional Brownian motion with $ 0<\alpha<1 $ and $ X(0) = 0 $. The $ \{X(t); t\geq 0\} $ has a covariance function

$ R(s, t) = E(X(s)X(t)) = \frac{1}{2}(s^{2\alpha}+t^{2\alpha}-|s-t|^{2\alpha}) $

for $ s, t\geq 0 $, and representation

$ X(t) = \int_{R^1}\frac{1}{k_\alpha}\left\{|x-t|^{(2\alpha-1)/2}-|x|^{(2\alpha-1)/2}\right\}dB(x), $

where

(i) $ k_{\alpha}^2 = \int_{R^1}\left\{|x-t|^{(2\alpha-1)/2}-|x|^{(2\alpha-1)/2}\right\}^2dx, $

(ii) $ \{B(t); -\infty<t<+\infty\} $ is a Brownian motion,

(iii) $ \frac{1}{k_\alpha}\left\{|x-t|^{(2\alpha-1)/2}-|x|^{(2\alpha-1)/2}\right\} $ is interpreted to be $ I_{(0, t]} $ when $ \alpha = \frac{1}{2}. $

$ \{X(t); t\geq 0\} $ has stationary increments with $ E(X(s+t)-X(s))^2 = t^{2\alpha}, s, t\geq 0 $ and is a standard Brownian motion when $ \alpha = \frac{1}{2}. $

Let $ C_0[0, 1] $ be the space of continuous functions from $ [0, 1] $ to $ R $ with value zero at the origin, endowed with usual norm $ \|f\| = \sup\limits_{0\le t\le 1}|f(t)| $, and

$ H = \bigg\{f\in C_0[0, 1]: f\hbox{ is an absolutely continuous function}, \|f\|_H^2 = \int^1_0(\dot{f}(s))^2ds<\infty\bigg\}. $

Then $ H $ is a Hilbert space with respect to the scalar product

$ \langle f, g\rangle = \int^1_0\dot{f}(x)\dot{g}(x)dx \; \; \; \mbox{for}\; \; \; f, g \in H. $

Define a mapping $ I:C_0[0, 1]\to [0, \infty] $ by

$ \begin{equation} I(f) = \left \{ \begin{array}{ll} \frac{1}{2} \int_0^1|\dot{f}(t)|^2dt, & \quad f\in H, \\ +\infty, & \quad \text{otherwise.} \end{array} \right. \end{equation} $

(1.1)

The limit set associated with functional laws of the iterated logarithm for $ \{X(t); t\geq 0\} $ is $ K_\alpha $, the subset of functions in $ C_0[0, 1] $ with the form

$ f(t) = \int_{R^1}\frac{1}{k_\alpha}\{|x-t|^{(2\alpha-1)/2}-|x|^{(2\alpha-1)/2}\}g(x)dx, \; \; \; 0\le t\le 1, $

here the function $ g(x) $ ranges over the unit ball of $ L^2(R^1) $, and hence $ \int_{R^1}g^2(s)ds\le 1 $. The subset $ K $ of $ C_0[0, 1] $ is defined by

$ K = \{f\in H: f\in K_\alpha, 2I(f)\le 1\}. $

For $ 0<h<1 $, $ 0\le s\le 1, 0\leq t\leq 1 $, let

$ l(h) = (2h^{2\alpha}\log(h^{-1}))^{\frac{1}{2}}, \; \; \Delta(t, h)(s) = X(t+hs)-X(t). $

In [1], Monrad and Rootzén gave a Chung's functional law of the iterated logarithm for fractional Brownian motion, as follows, for any $ f\in K $, $ \langle f, f\rangle<1 $,

$ \liminf\limits_{t\to 0}(\log\log t^{-1})^{\alpha+1/2}\bigg\|\frac{X(t\cdot)}{(2t^{2\alpha}\log\log t^{-1})^{1/2}}-f\bigg\| = \gamma(f)\; \; \; {\rm a.s.}, $

where $ \gamma(f) $ is a constant satifying $ 2^{-1/2}c^{\alpha}(1-\langle f, f\rangle)^{-\alpha}\le \gamma(f)\le 2^{-1/2}C^{\alpha}(1-\langle f, f\rangle)^{-\alpha} $, $ c, C $ denote the positive constants in (2.13) of [1].

Inspired by the arguments of Monrad and Rootzén, in the present paper, we obtain a liminf result for Lévy's modulus of continuity of a fractional Brownian motion. The main result is stated as follows.

Theorem 1.1 For each $ f\in K $ with $ \langle f, f\rangle<1 $, then

$ \begin{equation} \liminf\limits_{h\rightarrow 0}(\log h^{-1})^{\alpha+\frac{1}{2}} \inf\limits_{t\in [0, 1]}\bigg\|\frac{\Delta(t, h)}{l(h)}-f\bigg\| = b(f)\quad {\rm a.s.}, \end{equation} $

(1.2)

where $ b(f) $ is a constant satisfying

$ 2^{-1/2}c^{\alpha}(1-\langle f, f\rangle)^{-\alpha}\leq b(f)\leq 2^{-1/2}C^{\alpha}(1-\langle f, f\rangle)^{-\alpha}, $

here $ c $ and $ C $ denote the positive constants in (2.13) of [1].

Remark When $ \alpha = \frac{1}{2} $, $ \{X(t); t\geq 0\} $ is a standard Brownian motion, in this case $ c = C = \frac{\pi^2}{8} $, the result in Theorem 1.1 is the exact approximation rate on the modulus of continuity for Brownian motion.

Our proofs are based on the following lemmas.

In order to prove (3.1) below, we need the following Lemma 2.1.

Lemma 2.1 (see (3.14) of [2]) \label{lem1} Let $ \{X(t);t\ge 0\} $ be fractional Brownian motion as above, $ \sigma^2(u) = E(X(t+u)-X(t))^2 $, we have that for any $ \varepsilon>0 $, there exists a positive constant $ k_0 = k_0(\varepsilon) $ such that

$ \begin{align*} P\left(\sup\limits_{0\le t\le T}\sup\limits_{0\le s\le u}|X(t+s)-X(t)|\ge (1+\varepsilon) x \sigma(u)\right) \le \frac{k_0T}{u}\exp\left(-\frac{x^2}{2}\right) \end{align*} $

for any $ T, 0<u\le T $ and $ x\ge x_0 $ with some $ x_0>0 $.

In order to prove (6) below, we need the following Lemma 2.2 and Lemma 2.3.

Lemma 2.2 (see Lemma 2.3 in [3]) Let $ 0<\alpha<1, 0\le q_0<1 $ and fix $ 0<q_0<q<\alpha $. Let $ d_k = k^{k+(1-r)} $, $ s_k = k^{-k} $ for $ k\ge 1 $ and $ 0<r<1 $. Let

$ \begin{equation} Y_k(s_k, t) = \int_{|x|\notin I_k}\frac{1}{k_\alpha}\left\{|x-s_kt|^{(2\alpha-1)/2}-|x|^{(2\alpha-1)/2}\right\}dB(x), \; \; \; \; 0\le t\le 1, \end{equation} $

(2.1)

where $ I_k = (s_kd_{k-1}, s_kd_r] $. Let $ 0<\beta<r $. Then, for

$ \delta = \min\{2\beta(\alpha-q), r-\beta, (1-r)(2-2\alpha), (2\alpha-2q)r\}, $

there is a constant $ C'>0 $ depending only on $ \alpha $ such that uniformly in $ t, u, k $,

$ \begin{equation} \sigma_k^2(t, u) = E\{[Y_k(s_k, t+u)-Y_k(s_k, t)]^2\}\le C'u^{2q}s_k^{2\alpha} k^{-\delta}. \end{equation} $

(2.2)

Lemma 2.3 Let $ \{\Gamma(t): t\ge 0\} $ be a centred Gaussian process with stationary increments and $ \Gamma(0) = 0 $. We assume that $ \sigma^2(u) = E(\Gamma(t+u)-\Gamma(t))^2. $ Let $ T>0 $, we have, for $ x $ large enough, any $ \varepsilon>0 $,

$ P\left\{\sup\limits_{0\le s\le T}|\Gamma(s)|>x\sigma(T)\right\}\le C_1\exp\left(-\frac{x^2}{2+\varepsilon}\right), $

where $ C_1>0 $ is a constant.

Proof This conclusion is from page 49 in [4].

We only need to show the following two claims:

$ \begin{eqnarray} &&\liminf\limits_{h\rightarrow 0}\left(\log h^{-1}\right)^{\alpha+\frac{1}{2}} \inf\limits_{t\in [0, 1]}\|\frac{\Delta(t, h)}{l(h)}-f\| \geq 2^{-1/2}c^{\alpha}(1-\langle f, f\rangle)^{-\alpha}\quad {\rm a.s.}, \end{eqnarray} $

(3.1)

$ \begin{eqnarray} &&\liminf\limits_{h\to 0}\left(\log h^{-1}\right)^{\alpha+\frac{1}{2}}\inf\limits_ {t\in [0, 1]}\|\frac{\Delta(t, h)}{l(h)}-f\|\leq 2^{-1/2}C^{\alpha}(1-\langle f, f\rangle)^{-\alpha}\quad{{\rm a.s.}}. \end{eqnarray} $

(3.2)

3.1 The Proof of (3.1)

Let $ h_n = n^{-d} $, $ \rho(h) = h(\log h^{-1})^{-3-\frac{1}{\alpha}} $. We further set $ k_n = [\frac{1}{\rho(h_{n})}] $, $ t_i = i\rho(h_{n}), i = 0, 1, \cdots, k_n $. Then

$ \begin{equation} \begin{aligned} &\min\limits_{0\leq i\leq k_n+1}\|\frac{\Delta(t_i, h_n)}{l(h_n)}-f\|\\ \leq & 2\sup\limits_{t\in [0, 2]}\sup\limits_{s\in [0, \rho(h_n)]}\frac{|X(t+s)-X(t)|}{l(h_n)} +\inf\limits_{t\in[0, 1]}\|\frac{\Delta(t, h_n)}{l(h_n)}-f\|. \end{aligned} \end{equation} $

(3.3)

For any $ 0<\varepsilon <1 $, choose $ \delta>0 $, such that $ \eta = -\delta+ \langle f, f\rangle+\frac{1-\langle f, f\rangle }{(1-\varepsilon)^{1/\alpha}}>1 $. Then we have

$ \begin{aligned} & P\left(\left(\log h_n^{-1}\right)^{\alpha+\frac{1}{2}}\min\limits_{0\leq i\leq k_n+1}\| \frac{\Delta(t_i, h_n)}{l(h_n)}-f\|\leq (1-\varepsilon )2^{-1/2}c^{\alpha}(1-\langle f, f\rangle)^{-\alpha}\right)\\ \leq & \sum _{0\leq i\leq k_n+1}P\left(\left(\log h_n^{-1}\right)^{\alpha+\frac{1}{2}}\| \frac{\Delta(t_i, h_n)}{l(h_n)}-f\|\leq (1-\varepsilon)2^{-1/2}c^{\alpha}(1-\langle f, f\rangle)^{-\alpha}\right)\\ \leq &(2+k_n)P\left(\left\|\frac{X(h_n\cdot)}{h_n^{\alpha}} -(2\log h_n^{-1})^{\frac{1}{2}}f\right\| \leq \sqrt{2}(\log h_n^{-1})^{-\alpha}(1-\varepsilon)2^{-1/2}c^{\alpha}(1-\langle f, f\rangle)^{-\alpha}\right).\\ \end{aligned} $

By Proposition 4.2 in [1], we have for any $ \delta>0 $ and $ n $ large enough

$ \begin{aligned} &\left(\log h_n^{-1}\right)^{-1}\log P\left(\left\|\frac{X(h_n\cdot)}{h_n^{\alpha}} -(2\log h_n^{-1})^{\frac{1}{2}}f\right\| \leq\sqrt{2}(\log h_n^{-1})^{-\alpha}(1-\varepsilon)2^{-1/2}c^{\alpha}(1-\langle f, f\rangle)^{-\alpha}\right)\\ \leq&\left(\log h_n^{-1}\right)^{-1}\log P\left(\left\|\frac{X(h_n\cdot)}{h_n^{\alpha}}\right\| \leq \sqrt{2}(\log h_n^{-1})^{-\alpha}(1-\varepsilon)2^{-1/2}c^{\alpha}(1-\langle f, f\rangle)^{-\alpha}\right)-\langle f, f\rangle+\delta.\\ \end{aligned} $

By Corollary 2.2 in [1],

$ \begin{aligned} &\left(\log h_n^{-1}\right)^{-1}\log P\left(\left\|\frac{X(h_n\cdot)}{h_n^{\alpha}}\right\| \leq \sqrt{2}(\log h_n^{-1})^{-\alpha}(1-\varepsilon)2^{-1/2}c^{\alpha}(1-\langle f, f\rangle)^{-\alpha}\right)-\langle f, f\rangle+\delta\\ \leq&\left(\log h_n^{-1}\right)^{-1}\left(-2^{-1/(2\alpha)}c(1-\varepsilon)^{-1/\alpha}(2^{-1/2}c^{\alpha}(1-\langle f, f\rangle)^{-\alpha})^{-1/\alpha}\log h_n^{-1}\right) -\langle f, f\rangle +\delta = -\eta. \end{aligned} $

Thus

$ \begin{aligned} & P\left(\left(\log h_n^{-1}\right)^{\alpha+\frac{1}{2}}\min\limits_{0\leq i\leq k_n+1}\| \frac{\Delta(t_i, h_n)}{l(h_n)}-f\|\leq (1-\varepsilon )2^{-1/2}c^{\alpha}(1-\langle f, f\rangle)^{-\alpha}\right)\\ \leq &\frac{2+\rho(h_{n})}{\rho(h_{n})}h_n^{\eta}.\\ \end{aligned} $

Choose $ d>(\eta-1)^{-1} $, then

$ \sum\limits_{n = 1}^\infty\left(1+\frac{2}{\rho(h_{n})}\right)h_n^{\eta}<\infty, $

which implies, by the Borel-Cantelli lemma,

$ \begin{equation} \mathop{\liminf}\limits_{n\rightarrow\infty}\left(\log h_n^{-1}\right)^{\alpha+\frac{1}{2}} \mathop{\min}\limits_{0\leq i\leq k_n+1 }\|\frac{X(t_i+h_n\cdot)-X(t_i)}{l(h_n)}-f\|\geq \frac{2^{-1/2}c^{\alpha}}{(1-\langle f, f\rangle)^{\alpha}}\; \; {\rm a.s.}. \end{equation} $

(3.4)

On the other hand, for any $ \delta>0 $,

$ \begin{align*} &P\bigg(\left(\log h_n^{-1}\right)^{\alpha+\frac{1}{2}}\sup\limits_{0\leq t\leq 2}\sup\limits_{0\leq s \leq \rho(h_{n})}\frac{|X(t+s)-X(t)|}{l(h_n)}\geq \delta\bigg)\\ = &P\bigg(\left(\frac{(\log h_n^{-1})^{\alpha}}{\sqrt{2}h_n^{\alpha}}\right)\sup\limits_{0\leq t \leq 2}\sup\limits_{0\leq s\leq \rho(h_{n})}|X(t+s)-X(t)|\geq \delta\bigg)\\ = &P\bigg(\left(\log h_n^{-1}\right)^{-2\alpha-1}\sup\limits_{0\leq t\leq \frac{{2}}{\rho(h_{n})}}\sup\limits_{0\leq s\leq 1}|X(t+s)-X(t)|\geq\sqrt{2} \delta \bigg) \\\leq& \frac{2+\rho(h_{n})}{\rho(h_{n})} P\bigg(\left(\log h_n^{-1}\right)^{-2\alpha-1}\sup\limits_{0\leq t \leq 1}\sup\limits_{0\leq s\leq 1}|X(t+s)-X(t)|\geq \sqrt{2}\delta\bigg)\\ = &\frac{2+\rho(h_{n})}{\rho(h_{n})}P\bigg(\sup\limits_{0\leq t\leq 1}\sup\limits_{0\leq s\leq 1}|X(t+s)-X(t)|\geq 2\delta\left(\log h_n^{-1}\right)^{2\alpha+1}\bigg). \end{align*} $

By Lemma 2.1, we have that for any $ \varepsilon>0 $, there exists a positive $ k_0 = k_0(\varepsilon) $ such that

$ \begin{align*} &P\bigg(\sup\limits_{0\leq t\leq 1}\sup\limits_{0\leq s\leq 1}|X(t+s)-X(t)|\geq \sqrt{2}\delta\left(\log h_n^{-1}\right)^{4\alpha+2}\bigg)\\ \leq& k_0\exp\left(-\left(\frac{\delta}{(1+\varepsilon)}\right)^2\left(\log h_n^{-1}\right)^{4\alpha+2}\right). \end{align*} $

Taking into account $ \log h_n^{-1}\rightarrow\infty $ as $ n\rightarrow\infty $, we have

$ k_0\sum\limits_{n = 1}^\infty \frac{2+\rho(h_{n})}{\rho(h_{n})}h_n^{\left(\frac{\delta}{(1+\varepsilon)}\right)^2 \left(\log h_n^{-1}\right)^{4\alpha+1}}<\infty. $

By the Borel-Cantelli lemma,

$ \begin{equation} \limsup\limits_{n\rightarrow\infty}\left(\log h_n^{-1}\right)^{\alpha+\frac{1}{2}}\sup\limits_{0\leq t\leq 2}\sup\limits_{0\leq s\leq \rho(h_{n})}\frac{|X(t+s)-X(t)|}{l(h_n)} = 0\; \; {\rm a.s.}. \end{equation} $

(3.5)

By (3.3)–(3.5), we get

$ \begin{equation} \lim\limits_{n\to \infty}\left(\log h_n^{-1}\right)^{\alpha+\frac{1}{2}} \inf\limits_{t\in[0, 1]}\|\frac{X(t+h_n\cdot)-X(t)}{l(h_n)}-f\|\geq \frac{2^{-1/2}c^{\alpha}}{(1-\langle f, f\rangle)^\alpha}\; \; {\rm a.s.}. \end{equation} $

(3.6)

Remark that $ h_n $ is ultimately strictly decreasing to 0, so for any small $ h $, there is a unique $ n $ such that $ h\in(h_{n+1}, h_{n}]. $ Let $ \phi_{t, h}(s) = \frac{X(t+hs)-X(t)}{l(h)}, s\in [0, 1], t\in[0, 1] $. We define

$ \xi(h) = (\log h^{-1})^{\frac{1}{2}+\alpha}\mathop{\inf}\limits_{t\in [0, 1]} \|\phi_{t, h}(\cdot)-f(\cdot)\|, \; \; \xi_n = \mathop{\inf}\limits_{h_{n+1}\leq h\leq h_{n}}\xi(h). $

By the definition of infimum, for any $ \varepsilon>0 $, there exists $ h_n'\in (h_{n+1}, h_{n}] $ such that $ \xi_n\geq \xi(h_n')-\varepsilon. $

For any $ r\in[0, 1] $, let $ x = \frac{rh_{n+1}}{h_n'} $. Then we have $ 0\leq x\leq 1, $

$ \begin{equation} \begin{aligned} &\inf\limits_{t\in [0, 1]} \|\frac{X(t+h_{n+1}\cdot)-X(t)}{l(h_{n+1})}-f\|\\ = & \inf\limits_{t\in [0, 1]}\sup\limits_{0\leq r\leq 1}|\phi_{t, h_{n+1}}(r)-f(r)|\\ \le& \inf\limits_{t\in [0, 1]}\sup\limits_{0\leq x\leq 1}\left|\phi_{t, h_{n+1}}(\frac{h'_n}{h_{n+1}}x)-f(\frac{h'_n}{h_{n+1}}x)\right|\\ \leq& (l(h_{n+1}))^{-1}l(h_n')(\log h_n'^{-1})^{-\alpha-\frac{1}{2}}\xi(h_n')+ \left\|\frac{l(h_{n})}{l(h_{n+1})}-1\right|\|f(\cdot)\|+\left\|f(\cdot)- f(\frac{h'_n}{h_{n+1}}\cdot)\right\|. \end{aligned} \end{equation} $

(3.7)

Noting that

$ \begin{eqnarray} &&\|f(\frac{h_n'}{h_{n+1}}\cdot)-f(\cdot)\|\le \sqrt{1-\frac{h_{n+1}}{h_n}}\le\sqrt{1-(1-\frac{1}{n+1})^d}, \end{eqnarray} $

(3.8)

$ \begin{eqnarray} &&\left|\frac{l(h'_n)}{l(h_{n+1})}-1\right|\le (1+\frac{1}{n})^{d\alpha/2}-1. \end{eqnarray} $

(3.9)

By (3.6)–(3.9), we have

$ \begin{equation*} \begin{aligned} \liminf\limits_{n\to \infty}\xi(h'_n)\ge \frac{2^{-1/2}c^{\alpha}}{(1-\langle f, f\rangle)^\alpha}\; \; {\rm a.s.}. \end{aligned} \end{equation*} $

Since $ \liminf\limits_{h\to 0}\xi(h)\geq\liminf\limits_{n\to \infty}\xi_n\geq\liminf\limits_{n\to \infty}\xi(h'_n)-\varepsilon, $ which ends the proof.

3.2 The Proof of (3.2)

Note that

$ \liminf\limits_{h\to 0}\left(\log h^{-1}\right)^{\alpha+\frac{1}{2}}\inf\limits_{t\in [0, 1]}\| \frac{\Delta(t, h)}{l(h)}-f\|\leq \liminf\limits_{h\to 0}\left(\log h^{-1}\right)^{\alpha+\frac{1}{2}}\|\frac{X(h\cdot)}{l(h)}-f\|\quad{{\rm a.s.}}, $

then it is sufficient to show that

$ \begin{align} \liminf\limits_{n\to \infty}\left(\log h^{-1}_n\right)^{\alpha+\frac{1}{2}}\|\frac{X(h_n\cdot)}{l(h_n)}-f\|\leq 2^{-1/2}C^{\alpha}(1-\langle f, f\rangle)^{-\alpha}\quad{{\rm a.s.}, } \end{align} $

(3.10)

where $ h_n = \frac{1}{n} $.

For $ r = 1, 2, 3, \cdot\cdot\cdot, $ we define

$ \begin{eqnarray} &&Z_r(t) = \int_{|x|\in(d_{r-1}, d_r]}\frac{1}{k_\alpha}\left\{|x-t|^{(2\alpha-1)/2}-|x|^{(2\alpha-1)/2}\right\}dB(x), \end{eqnarray} $

(3.11)

$ \begin{eqnarray} &&\tilde{X}_r(t) = X(t)-Z_r(t) \end{eqnarray} $

(3.12)

for $ 0\le t\le 1, d_r = r^{r+(1-\gamma)}, s_r = r^{-r}, 0<\gamma<1 $. Then $ \{Z_r(\cdot)\}, r = 1, 2, \cdot\cdot\cdot $ are independent and

$ \begin{equation} \{s_r^\alpha\tilde{X}_r(\cdot)\} \mathop{ = }\limits^{\mathcal{D}}\{Y_r(s_r, \cdot)\}, \end{equation} $

(3.13)

where $ {Y_r(s_r, \cdot)} $ is as in Lemma 2.2.

In order to prove (14), we need to prove that for any $ \varepsilon>0 $,

$ \begin{equation} \sum\limits_{n = 1}^\infty P\left\{\left(\log h^{-1}_n\right)^{\alpha+\frac{1}{2}}\|\frac{Z_n(h_n\cdot)}{l(h_n)}-f\|\leq (1+\varepsilon)2^{-1/2}C^{\alpha}(1-\langle f, f\rangle)^{-\alpha}\right\} = \infty \end{equation} $

(3.14)

and

$ \begin{equation} \sum\limits_{n = 1}^\infty P\left\{\left(\log h^{-1}_n\right)^{\alpha+\frac{1}{2}}\|\frac{\tilde{X}_n(h_n\cdot)}{l(h_n)}\|\geq \varepsilon\right\}<\infty. \end{equation} $

(3.15)

First of all, we prove (3.15).

Now using the argument as in Lemma 2.2, we have

$ \sigma^2_n(t, u) = E(Y_n(s_n, t+u)-Y_n(s_n, t))^2\le C'u^{2q}s_n^{2\alpha}n^{-\delta}, $

where $ q, \delta $ are as in Lemma 2.2. For any $ \varepsilon>0 $, we have, by Lemma 2.3,

$ \begin{equation} \begin{aligned} &P\left\{\left(\log h^{-1}_n\right)^{\alpha+\frac{1}{2}}\|\frac{\tilde{X}_n(h_n\cdot)}{l(h_n)}\|\geq \varepsilon\right\}\\ = & P\left\{\|\tilde{X}_n(\cdot)\|\geq\sqrt{2}(\log h_n^{-1})^{-\alpha}\varepsilon\right\}\\ = & P\left\{\sup\limits_{0\le t\le 1}|\tilde{X}_n(t)|\geq\sqrt{2}(\log h_n^{-1})^{-\alpha}\varepsilon\right\}\\ = & P\left\{\sup\limits_{0\le t\le 1}|Y_n(s_n, t)|\geq\sqrt{2}(C'^{1/2} s_n^{\alpha}n^{-\delta/2})C'^{-1/2}(\log h_n^{-1})^{-\alpha}n^{\delta/2}\varepsilon\right\}\\ \le& C_1\exp\left(-\frac{C''n^{\delta}\varepsilon^2}{2+\varepsilon}\left (\log h_n^{-1}\right)^{-2\alpha}\right). \end{aligned} \end{equation} $

(3.16)

Taking $ n $ sufficiently large such that $ \frac{C''n^{\delta/2}\varepsilon^2}{2+\varepsilon}>1 $, we get (19) by the definition of the sequence $ \{h_n:n\ge 1\} $

Second, we prove (3.14).

For any $ \; \varepsilon>0 $, choose $ \delta>0 $, such that $ \eta' = \frac{1-\langle f, f\rangle}{(1+\varepsilon)^{1/\alpha}}+\langle f, f\rangle+\delta<1 $. Let $ \beta = 2^{-1/2}C^{\alpha}(1-\langle f, f\rangle)^{-\alpha} $, then

$ \begin{align*} &P\left\{\left(\log h^{-1}_n\right)^{\alpha+\frac{1}{2}}\|\frac{Z_n(h_n\cdot)}{l(h_n)}-f\|\leq (1+2\varepsilon)\beta\right\}\\ \geq& P\left\{\left(\log h^{-1}_n\right)^{\alpha+\frac{1}{2}}\|(2\log h^{-1}_n)^{-1/2}X(\cdot)-f\|\leq (1+\varepsilon)\beta\right\}\\ &-P\left\{\left(\log h^{-1}_n\right)^{\alpha+\frac{1}{2}}\|(2\log h^{-1}_n)^{-1/2}\tilde{X}_n(\cdot)\|\geq\varepsilon \beta\right\}\\ = :&I_1-I_2. \end{align*} $

By Proposition 4.2 in [1], we have for any $ \delta>0 $ and $ n $ large enough,

$ \begin{aligned} &\left(\log h^{-1}_n\right)^{-1}\log P\left(\left\|X(\cdot) -(2\log h^{-1}_n)^{\frac{1}{2}}f\right\| \leq\sqrt{2}(\log h^{-1}_n)^{-\alpha}(1+\varepsilon)\beta\right)\\ \geq&\left(\log h^{-1}_n\right)^{-1}\log P\left(\left\|X(\cdot)\right\| \leq\sqrt{2}(\log h^{-1}_n)^{-\alpha}(1+\varepsilon)\beta\right)-\langle f, f\rangle-\delta.\\ \end{aligned} $

By Corollary 2.2 in [1],

$ \begin{aligned} &\left(\log h^{-1}_n\right)^{-1}\log P\left(\left\|X(\cdot)\right\| \leq\sqrt{2}(\log h^{-1}_n)^{-\alpha}(1+\varepsilon)\beta\right)-\langle f, f\rangle-\delta\\ \geq&\left(\log h^{-1}_n\right)^{-1}\left(-2^{-\frac{1}{2\alpha}}C(1+\varepsilon)^{-1/\alpha}\beta^{-1/\alpha}\log h^{-1}_n\right) -\langle f, f\rangle -\delta. \end{aligned} $

Thus

$ \begin{aligned} I_1: = P\left\{\left(\log h^{-1}_n\right)^{\alpha+\frac{1}{2}}\|(2\log h^{-1}_n)^{-1/2}X(\cdot)-f\|\leq (1+\varepsilon)\beta\right\} \geq \bigg(h_n\bigg)^{\eta'}.\\ \end{aligned} $

Similar to the proof of (3.15), we have the following estimate for $ I_2 $,

$ \begin{align*} I_2&: = P\left\{\left(\log h^{-1}_n\right)^{\alpha+\frac{1}{2}}\|(2(\log h^{-1}_n)^{-1/2}\tilde{X}_n(\cdot)\|\geq\varepsilon\beta\right\}\\ & = P\left\{\|\tilde{X}_n(\cdot)\|\geq\sqrt{2}(\log h^{-1}_n)^{-\alpha}\beta\varepsilon\right\}\nonumber\\ & = P\left\{\sup\limits_{0\le t\le 1 }|Y_n(s_n, t)|\geq\sqrt{2}(C'^{1/2}s_n^{\alpha}n^{-\delta/2})C'^{-1/2}(\log h^{-1}_n)^{-\alpha}n^{\delta/2}\beta\varepsilon\right\}\\ &\le C_1\exp\left(-\frac{C''n^{\delta}(2^{-1/2}C^{\alpha}(1-\langle f, f\rangle)^{-\alpha})^2\varepsilon^2}{2+\varepsilon} \left(\log h^{-1}_n\right)^{-2\alpha}\right). \end{align*} $

Thus $ \sum\limits_{n = 1}^\infty I_2<\infty. $ The proof of (3.14) is completed.