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\title[Strassen's LIL for a Wiener process defined on the half-axis]{\large Strassen's law of iterated logarithm for a Wiener process defined on the half-axis}


\author[E.V. EFREMOV]{{\bf E.V. Efremov}\protect\orcidicon{0009-0004-1484-2118}}
\author[A.V. LOGACHOV]{{\bf A.V. Logachov}\protect\orcidicon{0000-0002-0605-8831}}



\address{Egor Vladimirovich Efremov  % Контактные данные всех авторов и место работы указываются только на английском языке
\newline\hphantom{iii} Sobolev Institute of Mathematics,
\newline\hphantom{iii} 4 Acad. Koptyug avenue,
\newline\hphantom{iii} 630090, Novosibirsk, Russia}%
\email{\textcolor{blue}{efremov.eg.v@yandex.ru}}%

\address{Artem Vasilhevich Logachov  % Контактные данные всех авторов и место работы указываются только на английском языке
\newline\hphantom{iii} Sobolev Institute of Mathematics,
\newline\hphantom{iii} 4 Acad. Koptyug avenue,
\newline\hphantom{iii} 630090, Novosibirsk, Russia}%
\email{\textcolor{blue}{omboldovskaya@mail.ru}}

\thanks{\sc Efremov, E.V., Logachov, A.V.,
Strassen's law of iterated logarithm for a Wiener process defined on the half-axis}
\thanks{\copyright \ 2025 Efremov E.V., Logachov, A.V.}
\thanks{\rm The work of E.V. Efremov is supported by the Mathematical Center in Akademgorodok under the agreement No. 075-15-2025-348 with the Ministry of Science and Higher Education of the Russian Federation. A.V. Logachov thanks the Basic Research Program of Siberian Branch of Russian Academy of Sciences, project FWNF-2022-0010.}
\thanks{\it Received ?, ?, ?, Published ?, ?, ?}%


\semrtop \vspace{1cm}
\maketitle {\small

\vspace{-12pt}
\centerline{{\it Communicated by} {\sc N.S. Arkashov}
%({\it filled by Editorial board})
}

\bigskip
\bigskip

\begin{quote}
\noindent{\bf Abstract:} We prove a version of Strassen’s law of iterated logarithm in the space of continuous functions with weighted $\sup$\,--\,metric for a Wiener process defined on the half-axis. We consider weight functions of the form $1/(1+t^{\alpha})$, where $\alpha > 1/2$. The result is unimprovable in the class of power weight functions.\medskip

\noindent{\bf Keywords:} Wiener process, Strassen's law of iterated logarithm.
 \end{quote}
}

\bigskip

\section{Introduction, formulation of the main result}
Let $w(t)$, $t\geqslant 0$ be a Wiener process defined on a probability space $(\Omega,\mathfrak{F},\mathbf{P})$. We will be interested in the limit behavior of the following sequence of random processes
\[
 w_n(t) = \frac{w(nt)}{\sqrt{n}\varphi(n)}, \quad n\in\mathbb{N},
\]
where $\varphi(n) := \sqrt{2\ln\ln(3\vee n)}$. Note that for any $n\in \mathbb{N}$ the random process $\tilde w(t) = w(nt)/\sqrt{n}$ is also a Wiener process.

We will consider trajectories of random processes $w_n$ in the space $\mathbb{C}$ of functions continuous on the half-axis $[0,\infty)$, with a metric
\[
  \rho(f,g) = \rho_{\alpha}(f,g) := \sup\limits_{t\geqslant 0} \frac{\lvert f(t) -  g(t)\rvert}{1 + t^{\alpha}},
\]
where $1/(1 + t^{\alpha})$ is a power weight function and $\alpha>\frac{1}{2}$ is a fixed constant.

Let us denote as $\mathbb{AC}_0[0,\infty)$ and $\mathbb{AC}_0[0,T]$ sets of functions starting from zero and absolutely continuous on intervals
$[0,\infty)$ and $[0,T]$ respectively.

We define the Strassen's ball on the half-axis
\[
  K:=\left\{f\in \mathbb{AC}_0[0,\infty):\int_0^\infty \dot{f}^2(s)ds\leqslant 1\right\},
\]
where $\dot{f}$ is the derivative in the sense of absolute continuity.\par
We also define the Strassen's ball on $[0,T]$
\[
  K_T:=\left\{f\in \mathbb{AC}_0[0,T]:\int_0^T \dot{f}^2(s)ds\leqslant 1\right\}.
\]
The following Theorem is the main result of the paper.

\begin{theorem} \label{t.1} For any $\alpha>\frac{1}{2}$ the set limit points of the sequence $w_n$ in the metric space  $\mathbb{C}$ coincides with the set $K$ a.s.
\end{theorem}

Now we will do a brief review of results obtained earlier. The case of $\alpha=1$ was considered in the paper \cite[Theorem 1.4.1]{DeuStr} for multidimensional Wiener process; the result then was used to prove the Strassen's law of iterated logarithm for small time in \cite[Theorem 3]{Gantert}. The paper \cite[Theorem 1]{BudMakh} is also dedicated to the case, when $\alpha=1$, but a class of more general normalizing functions than $\varphi(n)= \sqrt{2\ln\ln(3\vee n)}$ was considered. We note that the methods of proof used in this paper differ significantly form those proposed in \cite{BudMakh}. We also mention the paper \cite{Mog}. In that paper the moderate deviations principle for diffusion processes was obtained for the case when $\alpha\geqslant 1$, which is the main tool for proving the Strassen's law of iterated logarithm.
\par

\begin{remark}
  In the case when $\alpha\leqslant\frac{1}{2}$, the Strassen's law of iterated logarithm cannot be obtained, since for such $\alpha$, due to the Khinchin's law of iterated logarithm, the following holds
  \begin{equation}\label{lln_lower_bound}
    \limsup\limits_{t\rightarrow\infty} \frac{\lvert \tilde{w}(t)\rvert}{1 + t^{\alpha}}
    \geqslant\limsup\limits_{t\rightarrow\infty} \frac{\lvert \tilde{w}(t)\rvert}{1 + \sqrt{t}}=\infty \ \ \ \text{a.s.}
  \end{equation}
  Also, due to the Cauchy--Bunyakovsky--Schwarz inequality, for any $f\in K$ the following bound holds
  \begin{equation}\label{f_t_bound}
    |f(t)|\leqslant\int_0^{t}|\dot{f}(s)|ds\leqslant \sqrt{t}\left(\int_{0}^{t}\dot{f}^2(s)ds\right)^{1/2}
    \leqslant \sqrt{t}.
  \end{equation}
  Using formulas (\ref{lln_lower_bound}) and (\ref{f_t_bound}) we can conclude that for any $\alpha\leqslant\frac{1}{2}$, any function $f\in K$ and any $n\in\mathbb{N}$ the following holds
  \begin{align*}
  \rho_{\alpha}(w_n,f)&=\sup\limits_{t\geqslant 0}\frac{\lvert w_n(t) - f(t)\rvert}{1+t^{\alpha}}\geqslant \sup\limits_{t\geqslant 0} \left(\frac{\lvert w_n(t)\rvert}{1+t^{\alpha}} - \frac{\lvert f(t)\rvert}{1+t^{\alpha}}\right)\\
  &\geqslant \sup\limits_{t\geqslant 0} \left(\frac{\lvert w_n(t)\rvert}{1+t^{\alpha}} - \frac{ \sqrt{t}}{1+t^{\alpha}}\right)=\sup\limits_{t\geqslant 0} \frac{1+\sqrt{t}}{1+t^{\alpha}}\left(\frac{\lvert w_n(t)\rvert}{1+\sqrt{t}} - \frac{ \sqrt{t}}{1+\sqrt{t}}\right)\\
  &\geqslant \sup\limits_{t\geqslant 0} \frac{1+\sqrt{t}}{1+t^{\alpha}}\left(\frac{\lvert  w_n(t)\rvert}{1+\sqrt{t}} - 1\right) = \infty \quad \text{a.s.}
  \end{align*}
  Thus, the result of Theorem \ref{t.1} is unimprovable in the class of power weight functions.
\end{remark}

Let us introduce some more new notations.
For fixed $T\in (0,\infty)$ we will denote the space of functions continuous on the interval $[0,T]$ as $\mathbb{C}[0,T]$. On that space we define the following $\sup$\,--\,metric
\[
  \rho_{T}(f,g) := \sup\limits_{t\in[0,T]} \frac{\lvert f(t) -  g(t)\rvert}{1 + t^{\alpha}},
\]
where  $\alpha>\frac{1}{2}$; and the uniform metric
\[
  \rho_{T,U}(f,g):=\sup\limits_{t\in[0,T]}\left|f(t)-g(t)\right|.
\]

The rest of the paper consists of Section 2 and Section 3. In Section 2 we prove the main result. In Section 3 we formulate and prove auxiliary results.

\section{Proof of the main result}
\begin{proof}
We divide the proof into two steps.

\textbf{Step 1}. Let us prove that the set of limit points of the sequence $w_{n}$ is contained in $K$ a.s. To this end we show that the conditions of Lemma $\ref{l.2}$ are met for $w_{n}$ a.s.\par
From  [1, appendix 8, Theorem 4] and the inequality $\rho_T\left(w_{n_r(c)},K_T\right)\leqslant\rho_{T,U}\left(w_{n_r(c)},K_T\right)$
it follows that for any $c>1$ and $T>0$ the following equality holds
\begin{equation}\label{05.04.18}
  \mathbf{P}\left(\limsup\limits_{r\rightarrow\infty}\rho_{T}\left(w_{n_r(c)},K_T\right)=0\right)=1,
\end{equation}
where $n_r(c) = \lfloor c^r\rfloor$. That is, the condition (1) of Lemma \ref{l.2} is satisfied a.s.\par
Now we show that for any $\varepsilon>0$ there is $T_\varepsilon>0$ such that for any $c>1$ the following equality holds
\begin{equation}\label{05.04.19}
  \mathbf{P}\left(\limsup\limits_{r\rightarrow\infty}\left(\sup\limits_{t\geqslant T_\varepsilon}\frac{\max\limits_{s\in[0,t]}\left|w_{n_r(c)}(s)\right|}{1+t^\alpha}\right)<\varepsilon\right)=1.
\end{equation}
For $T>0$ we define a set of events as follows
\[
  A_r := \left\{\sup\limits_{t\geqslant T} \frac{\max\limits_{s\in[0,t]}\left\lvert w_{n_r(c)}(s)\right\rvert}{1+t^{\alpha}}\geqslant\varepsilon\right\},\quad r\in\mathbb{N}.
\]
Using Lemma \ref{prob_bound_lemma}, we obtain the following bound
\begin{multline*}
  \sum\limits_{r=1}^{\infty}\mathbf{P}(A_r) \leqslant C(T,\varepsilon)\sum\limits_{r=1}^{\infty}2\exp \left\{\frac{-T^{2\alpha - 1}\varphi^2(n_r(c))\varepsilon^2}{4}\right\}\\
  = C(T,\varepsilon)\sum\limits_{r=1}^{\infty}\frac{2}{(\ln(3\vee n_r(c)))^{T^{2\alpha-1}\varepsilon^2/2}},
\end{multline*}
from which it follows that the series $\sum\limits_{r=1}^{\infty}\mathbf{P}(A_r)$ converges if
\[
  T_{\varepsilon} >\left(\frac{2}{\varepsilon^2}\right)^{1/(2\alpha - 1)}.
\]
Therefore, from Borel--Cantelli Lemma it follows that the equality (\ref{05.04.19}) holds, and consequently the condition (2) of Lemma \ref{l.2} is satisfied a.s.\par
Thus, all the conditions of Lemma \ref{l.2} are satisfied a.s., and we can conclude that the set of limit points of the sequence $w_{n}$ is contained in $K$ a.s.

\textbf{Step 2}. It remains to show that for any function $f\in K$ and almost all $\omega\in\Omega$ there is a sequence $n_k(\omega)$ such that $\lim\limits_{k\rightarrow\infty}\rho(w_{n_k(\omega)},f)=0$.\par

We fix $\varepsilon > 0$. Let us choose $T_{1,\varepsilon}$ such that the following equality holds
\begin{equation}\label{06.04.5}
  \mathbf{P}\left(\limsup\limits_{n\rightarrow\infty}\left(\sup\limits_{t\geqslant T_\varepsilon}\frac{\max\limits_{s\in[0,t]}\left|w_{n}(s)\right|}{1+t^\alpha}\right)<\varepsilon\right)=1.
\end{equation}
We recall that such choice is possible due to the equality (\ref{05.04.19}) and the Lemma \ref{l.1}.

Let $f\in K$. For any $T>0$ the following bound holds (see \ref{f_t_bound})
\[
  \lvert f(T)\rvert \leqslant \sqrt{T}.
\]
Therefore we can choose $T_{2,\varepsilon}>0$ such that for any $T\geqslant T_{2,\varepsilon}$ the following holds
\begin{equation}\label{06.04.1}
  \sup\limits_{t\geqslant T}\frac{|f(T)|}{1+t^\alpha}<\varepsilon.
\end{equation}

Since $f$ is in $K$, for any $\varepsilon>0$ there is $T_{3,\varepsilon}>0$ such that
\begin{equation}\label{06.04.2}
  \int_{T_{3,\varepsilon}}^{\infty}\dot{f}^2(s)ds\leqslant \varepsilon^2.
\end{equation}

It is easy to see that for $T_\varepsilon:=\max(T_{1,\varepsilon},T_{2,\varepsilon},T_{3,\varepsilon})$
the inequalities (\ref{06.04.5}), (\ref{06.04.1}) and (\ref{06.04.2}) hold simultaneously.

Using the Cauchy--Bunyakovsky--Schwarz inequality we get

\begin{multline}\label{06.04.3}
  \sup\limits_{t\geqslant T_\varepsilon}\frac{|f(t)-f(T_\varepsilon)|}{1+t^\alpha}
  \leqslant\sup\limits_{t\geqslant T_\varepsilon}\frac{\int_{T_\varepsilon}^{t}|\dot{f}(s)|ds}{1+t^\alpha}
  \leqslant\sup\limits_{t\geqslant T_\varepsilon}\int_{T_\varepsilon}^{t}\frac{|\dot{f}(s)|}{1+s^\alpha}ds\\
  \leqslant\int_{T_\varepsilon}^{\infty}\frac{|\dot{f}(s)|}{1+s^\alpha}ds
  \leqslant\left(\int_{T_\varepsilon}^{\infty}\dot{f}^2(s)ds\right)^{1/2}
  \cdot\left(\int_{T_\varepsilon}^{\infty}\frac{ds}{(1+s^\alpha)^2}\right)^{1/2}
  \leqslant\varepsilon C_\alpha.
\end{multline}

From inequalities (\ref{06.04.1}) and (\ref{06.04.3}) it follows that
\begin{multline}\label{06.04.7}
  \rho(w_n,f)\leqslant\sup\limits_{t\in[0,T_\varepsilon]}\frac{|w_n(t)-f(t)|}{1+t^\alpha}+ \sup\limits_{t\geqslant T_\varepsilon}\frac{|w_n(t)-f(t)|}{1+t^\alpha} \leqslant\sup\limits_{t\in[0,T_\varepsilon]}\frac{|w_n(t)-f(t)|}{1+t^\alpha}\\
  +\sup\limits_{t\geqslant T_\varepsilon}\frac{|f(t)-f(T_\varepsilon)|}{1+t^\alpha} +\sup\limits_{t\geqslant T_\varepsilon}\frac{|w_n(t)|}{1+t^\alpha}+\sup\limits_{t\geqslant T_\varepsilon}\frac{|f(T_\varepsilon)|}{1+t^\alpha}\\
  \leqslant\sup\limits_{t\in[0,T_\varepsilon]}\frac{|w_n(t)-f(t)|}{1+t^\alpha} +\sup\limits_{t\geqslant T_\varepsilon}\frac{|w_n(t)|}{1+t^\alpha} +(1+C_\alpha)\varepsilon.
\end{multline}

Using the equality (\ref{06.04.5}) we can conclude that for almost all $\omega\in\Omega$ there is $n(\varepsilon,\omega)$ such that for all $n\geqslant n(\varepsilon,\omega)$ the following equality holds
\begin{equation}\label{06.04.8}
  \sup\limits_{t\geqslant T_\varepsilon}\frac{|w_n(t)|}{1+t^\alpha}
  <2\varepsilon.
\end{equation}

From \cite[appendix 8, Theorem 4]{BulShir} (the law of iterated logarithm for the Wiener process on the interval $[0,T_\varepsilon]$) it follows that for any function $f\in K_{T_\varepsilon}$ and almost all $\omega\in\Omega$ exists a sequence $n_k(\omega)$  such that $\lim\limits_{k\rightarrow\infty}\rho_{T_\varepsilon,U}(w_{n_k(\omega)},f)=0$.
Therefore for almost all $\omega\in\Omega$
\begin{equation}\label{06.04.10}
  \lim\limits_{k\rightarrow\infty}\sup\limits_{t\in[0,T_\varepsilon]}\frac{|w_{n_k(\omega)}(t)-f(t)|}{1+t^\alpha}
  \leqslant \lim\limits_{k\rightarrow\infty}\rho_{T_\varepsilon,U}(w_{n_k(\omega)},f)=0.
\end{equation}
From (\ref{06.04.7}), (\ref{06.04.8}) and (\ref{06.04.10}) it follows that for any function $f\in K$, $\varepsilon>0$ and almost all $\omega\in\Omega$ there is a subsequence $n_k(\omega)$ such that
$$
\limsup\limits_{k\rightarrow\infty}\rho(w_{n_k(\omega)},f)<(3+C_\alpha)\varepsilon.
$$
The Theorem \ref{t.1} is fully proven.
\end{proof}

\section{Auxiliary results}
For function $g\in\mathbb{C}$ we set
\[
  g_n(t):= \frac{g(nt)}{\sqrt{n}\varphi(n)},\quad n \in \mathbb{N}.
\]
\begin{lemma} \label{l.1} Let $g\in\mathbb{C}$
  and let the following inequality hold for some $T>0$
  \begin{equation}\label{05.04.12}
    \limsup\limits_{r\rightarrow\infty}\left(\sup\limits_{t\geqslant T}\frac{\max\limits_{s\in[0,t]}\left|g_{2^r}(s)\right|}{1+t^\alpha}\right)<\varepsilon.
  \end{equation}
  Then for the same $T>0$ the following inequality holds
  $$
  \limsup\limits_{n\rightarrow\infty}\left(\sup\limits_{t\geqslant T}\frac{\max\limits_{s\in[0,t]}\left|g_n(s)\right|}{1+t^\alpha}\right)<2\varepsilon.
  $$
\end{lemma}
\begin{proof}
  Since the inequality in condition (\ref{05.04.12}) is strict, there is $r(\varepsilon)$ such that for all $r\geqslant r(\varepsilon)$ the following holds
  \begin{equation}\label{05.04.3}
    \sup\limits_{t\geqslant T}\frac{\max\limits_{s\in[0,t]}\left|g_{2^r}(s)\right|}{1+t^\alpha}<\varepsilon, \ \ \
    \frac{\varphi(2^{r+1})}{\varphi(2^r)}<\sqrt{2}.
  \end{equation}
  It is easy to see, that for any $n\geqslant 2^{r(\varepsilon)}$ there is $r(n)\geqslant r(\varepsilon)$ such that $2^{r(n)}\leqslant n<2^{r(n)+1}$.
  Therefore, using inequalities (\ref{05.04.3}), for $n\geqslant 2^{r(\varepsilon)}$ we get
  \begin{multline*}
    \sup\limits_{t\geqslant T}\frac{\max\limits_{s\in[0,t]}\left|g_n(s)\right|}{1+t^\alpha} =\sup\limits_{t\geqslant T}\frac{\max\limits_{s\in[0,t]}\left|g(ns)\right|}{(1+t^\alpha)\sqrt{n}\varphi(n)}\leqslant\sup\limits_{t\geqslant T}\frac{\max\limits_{s\in[0,t]}\left|g(2^{r(n)+1}s)\right|}{(1+t^\alpha)\sqrt{2^{r(n)}}\varphi(2^{r(n)})}\\
    =\frac{\sqrt{2^{r(n)+1}}\varphi(2^{r(n)+1})}{\sqrt{2^{r(n)}}\varphi(2^{r(n)})}     \cdot\sup\limits_{t\geqslant T}\frac{\max\limits_{s\in[0,t]}\left|g(2^{r(n)+1}s)\right|}{(1+t^\alpha)\sqrt{2^{r(n)+1}}\varphi(2^{r(n)+1})}\\
    =\frac{\sqrt{2}\varphi(2^{r(n)+1})}{\varphi(2^{r(n)})} \cdot\sup\limits_{t\geqslant T}\frac{\max\limits_{s\in[0,t]}\left|g_{2^{r(n)+1}}(s)\right|}{1+t^\alpha}<2\varepsilon.
  \end{multline*}
\end{proof}

  \begin{lemma}\label{l.2} Let $g\in\mathbb{C}$ and let it satisfy the following conditions:
    \begin{enumerate}
      \item for any $c>1$ and $T>0$
      $$
        \limsup\limits_{r\rightarrow\infty}\rho_T\left(g_{n_r(c)},K_T\right)=0,
      $$
      where $n_r(c)=\lfloor c^r\rfloor$, $r\in \mathbb{N}$,
      $\rho_T\left(g_{n_r(c)},K_T\right):=\inf\limits_{f\in K_T}\rho_T\left(g_{n_r(c)},f\right)$;
      \item for any $\varepsilon>0$ there is $T_\varepsilon>0$ such that for any $c>1$
      \begin{equation}\label{05.04.1}
        \limsup\limits_{r\rightarrow\infty}\left(\sup\limits_{t\geqslant T_\varepsilon}\frac{\max\limits_{s\in[0,t]}\left|g_{n_r(c)}(s)\right|}{1+t^\alpha}\right)<\varepsilon.
      \end{equation}
    \end{enumerate}
    Then
    $$
      \limsup\limits_{n\rightarrow\infty}\rho\left(g_n,K\right)=0.
    $$
  \end{lemma}
  \begin{proof}
    It is suffices to show that for any $\varepsilon>0$
    \begin{equation}\label{05.04.5}
      \limsup\limits_{n\rightarrow\infty}\rho\left(\frac{g(nt)}{\sqrt{n}\varphi(n)},K\right)<3\varepsilon.
    \end{equation}

    Fix $\varepsilon > 0$. Let us denote $C_{T_\varepsilon}:=\{f\in \mathbb{C}:f(t)\equiv f(T_\varepsilon), \text{ for } t\geqslant T_\varepsilon\}$.
    It is obvious, that
    \begin{equation}\label{05.04.7}
      \widetilde{K}_\varepsilon:=K_{T_\varepsilon}\cap C_{T_\varepsilon}\subset K,
    \end{equation}
    and, due to the Cauchy--Bunyakovsky--Schwarz inequality, for any function $f\in K_{T_\varepsilon}$ the following equality holds
    $$
    \sup\limits_{t\in[0,T_\varepsilon]}|f(t)|\leqslant\int_0^{T_\varepsilon}|\dot{f}(s)|ds
    \leqslant\sqrt{T_\varepsilon}\left(\int_0^{T_\varepsilon}\dot{f}^2(s)ds\right)^{1/2}\leqslant \sqrt{T_\varepsilon}.
    $$
    Therefore for any $T_\varepsilon$ large enough the following inequality holds for $f\in\widetilde{K}_\varepsilon$
    \begin{equation}\label{05.04.2}
      \sup\limits_{t\geqslant T_\varepsilon}\frac{\max\limits_{s\in[0,t]}\left|f(s)\right|}{1+t^\alpha} = \sup\limits_{t\geqslant T_\varepsilon}\frac{\max\limits_{s\in[0,T_\varepsilon]}\left|f(s)\right|}{1+t^\alpha} \leqslant\frac{\sqrt{T_\varepsilon}}{1+T_\varepsilon^\alpha}<\varepsilon.
    \end{equation}
    It is easy to see, that if the inequality (\ref{05.04.1}) holds for some $T_\varepsilon>0$, then it also holds for any $T>T_\varepsilon$. For this reason, from now on, we will assume that $T_\varepsilon$ is chosen such that both inequalities (\ref{05.04.1}) and (\ref{05.04.2}) hold.\par

    Using (\ref{05.04.7}) and (\ref{05.04.2}) we get
    \begin{multline}\label{05.04.8}
      \limsup\limits_{n\rightarrow\infty}\rho(g_n,K) \leqslant\limsup\limits_{n\rightarrow\infty}\rho(g_n,\widetilde{K}_\varepsilon)\\
      \leqslant\limsup\limits_{n\rightarrow\infty}\inf_{f\in \widetilde{K}_\varepsilon}\left(\sup\limits_{t\in[0,T_\varepsilon]}\frac{|g_n(t)-f(t)|}{1+t^\alpha} +\sup\limits_{t\geqslant T_\varepsilon}\frac{|g_n(t)-f(t)|}{1+t^\alpha}\right)\\
      \leqslant \limsup\limits_{n\rightarrow\infty}\inf_{f\in \widetilde{K}_\varepsilon}\left(\sup\limits_{t\in[0,T_\varepsilon]}\frac{|g_n(t)-f(t)|}{1+t^\alpha} +\sup\limits_{t\geqslant T_\varepsilon}\frac{\max\limits_{s\in[0,t]}\left|g_n(s)\right|}{1+t^\alpha}+\varepsilon \right)\\
      \leqslant \limsup\limits_{n\rightarrow\infty}\rho_{T_\varepsilon}\left(g_n,K_{T_\varepsilon}\right) +\limsup\limits_{n\rightarrow\infty}\left(\sup\limits_{t\geqslant T_\varepsilon}\frac{\max\limits_{s\in[0,t]}\left|g_n(s)\right|}{1+t^\alpha}\right)+\varepsilon.
    \end{multline}
    It is easy to see that
    $\rho_{T_\varepsilon}\left(g_n,K_{T_\varepsilon}\right)\leqslant\rho_{T_\varepsilon,U}\left(g_n,K_{T_\varepsilon}\right)\leqslant (1+T_{\varepsilon}^{\alpha})\rho_{T_{\varepsilon}}(g_n, K_{T_{\varepsilon}})$.
    We also note that the set $K_{T_\varepsilon}$ is compact in the metric space $(\mathbb{C}[0,T_\varepsilon],\rho_{T_\varepsilon,U})$ (see, for example, \cite[appendix 8, Lemma 2]{BulShir}). Therefore from condition (1) and \cite[appendix 8, Lemma 6]{BulShir} it follows that
    \begin{equation}\label{05.04.10}     \limsup\limits_{n\rightarrow\infty}\rho_{T_\varepsilon}\left(g_n,K_{T_\varepsilon}\right)=0.
    \end{equation}
    Applying condition (2) and Lemma \ref{l.1}, we get
    \begin{equation}\label{05.04.11}
      \limsup\limits_{n\rightarrow\infty}\left(\sup\limits_{t\geqslant T_\varepsilon}\frac{\max\limits_{s\in[0,t]}\left|g_n(s)\right|}{1+t^\alpha}
      \right)< 2\varepsilon.
    \end{equation}
    Inequality (\ref{05.04.5}) follows from formulas (\ref{05.04.8})--(\ref{05.04.11}).
  \end{proof}
  \begin{lemma}\label{lim_sup_lemma}
    Let the function $f(t)$ be continuous on $[0,\infty)$, and let $\varepsilon>0$. Then
    \begin{equation}\label{eq_1}
      \limsup\limits_{t\to\infty}\frac{\max\limits_{s\in[0,t]}\lvert f(s)\rvert}{1+t^{\alpha}}\geqslant \varepsilon
    \end{equation}
    if and only if
    \begin{equation}\label{eq_2}
      \limsup\limits_{t\to\infty}\frac{\lvert f(t)\rvert}{1+t^{\alpha}}\geqslant \varepsilon.
    \end{equation}
  \end{lemma}
  \begin{proof}
    It is easy to see that (\ref{eq_1}) follows from (\ref{eq_2}).\par
    Now we show that from the inequality
    \begin{equation}\label{eq_3}
      \limsup\limits_{t\to\infty}\frac{\lvert f(t)\rvert}{1+t^{\alpha}}<\varepsilon,
    \end{equation}
    it follows that
    \begin{equation}\label{eq_4}
      \limsup\limits_{t\to\infty}\frac{\max\limits_{s\in[0,t]}\lvert f(s)\rvert}{1+t^{\alpha}}< \varepsilon.
    \end{equation}
    Suppose that (\ref{eq_3}) holds, but (\ref{eq_4}) does not. Then there are sequences $s_m$ and $t_m$ such that $0\leqslant s_m \leqslant t_m$ for all $m\in\mathbb{N}$, $\lim\limits_{m\to\infty}t_m=\infty$ and
    \begin{equation}\label{eq_5}
      \lim\limits_{m\to\infty}\frac{\lvert f(s_m)\rvert}{1+t^{\alpha}_m}\geqslant \varepsilon.
    \end{equation}
    If $\liminf\limits_{m\to\infty}s_m<\infty$, then due to the continuity of the function $f$ the inequality (\ref{eq_5}) cannot hold. Therefore $\lim\limits_{m\to\infty}s_m = \infty$. Then, due to the fact, that $s_m \leqslant t_m$, we have
    \[
      \varepsilon \leqslant \lim\limits_{m\to\infty}\frac{\lvert f(s_m)\rvert}{1+t^{\alpha}_m}\leqslant \lim\limits_{m\to\infty}\frac{\lvert f(s_m)\rvert}{1+s^{\alpha}_m}\leqslant \limsup\limits_{t\to\infty}\frac{\lvert f(t)\rvert}{1+t^{\alpha}} <\varepsilon.
    \]
    The resulting contradiction completes the proof.
  \end{proof}

  \begin{lemma}\label{prob_bound_lemma}
    For any $n\in\mathbb{N}$, $T>0$ and $\varepsilon>0$ the following bound holds
    \begin{equation*}
      \mathbf{P} \left(\sup\limits_{t\geqslant T} \frac{\max\limits_{s\in[0,t]}\lvert w_n(s)\rvert}{1+t^{\alpha}}\geqslant \varepsilon\right) \leqslant 2\exp\left\{-\frac{T^{2\alpha - 1}\varphi^2(n)\varepsilon^2}{4}\right\} C(T,\varepsilon),
    \end{equation*}
    where the constant $0<C(T,\varepsilon)<\infty$ depends only on $T$ and $\varepsilon$.
  \end{lemma}
  \begin{proof}
    Let $0<\lambda\in\mathbb{R}$. Applying Doob's martingale inequality (see., for example, \cite[chapter 4, Corollary 5]{BulShir})
    to the martingale  $\exp\left\{ \lambda \tilde w(t)-\frac{\lambda^2}{2}t\right\}$, and using the fact, that the random process $-\tilde w(t)$ is a Wiener process, for any $x>0$ and $y>0$ we get
    \begin{multline*}
      \mathbf{P} \left(\max\limits_{t\in[0,y]} \lvert \tilde w(t)\rvert \geqslant x\right) \leqslant \mathbf{P} \left(\max\limits_{t\in[0,y]} \tilde w(t) \geqslant  x\right) + \mathbf{P} \left(\max\limits_{t\in[0,y]} (-\tilde w(t)) \geqslant x\right)\\
      \leqslant 2\mathbf{P} \left(\max\limits_{t\in[0,y]} \exp\left\{ \lambda \tilde w(t)-\frac{\lambda^2}{2}t\right\} \geqslant \exp\left\{\lambda x-\frac{\lambda^2}{2}y\right\}\right)\\
      \leqslant\frac{2\mathbf{E}\exp\left\{ \lambda \tilde w(y)-\frac{\lambda^2}{2}y\right\}}{\exp\left\{\lambda x-\frac{\lambda^2}{2}y\right\}} =2\exp\left\{-\lambda x+\frac{\lambda^2}{2}y\right\}.
    \end{multline*}
    By choosing $\lambda=\frac{x}{y}$, we will have
    \begin{equation}\label{25.06.1}
      \mathbf{P} \left(\max\limits_{t\in[0,y]} \lvert \tilde w(t)\rvert \geqslant x\right) \leqslant 2\exp\left\{-\frac{x^2}{2y}\right\}.
    \end{equation}
    Using inequality (\ref{25.06.1}), Lemma \ref{lim_sup_lemma} and the Khinchin's law of iterated logarithm, we get
    \begin{align*}
      &\mathbf{P} \left(\sup\limits_{t\geqslant T} \frac{\max\limits_{s\in[0,t]}\lvert w_n(s)\rvert}{1+t^{\alpha}}\geqslant \varepsilon\right)=\mathbf{P} \left(\sup\limits_{t\geqslant T} \frac{\max\limits_{s\in[0,t]}\lvert \tilde w(s)\rvert}{(1+t^{\alpha})\varphi(n)} \geqslant \varepsilon\right)\\
      &\leqslant \mathbf{P} \left(\bigcup\limits_{r=1}^{\infty}\left\{\sup\limits_{t\in[Tr,T(r+1)]} \frac{\max\limits_{s\in[0,t]} \lvert \tilde w(s)\rvert}{(1+t^{\alpha})} \geqslant \varphi(n)\varepsilon\right\}\right)+ \mathbf{P} \left(\limsup\limits_{t\to\infty}\frac{\max\limits_{{s\in[0,t]}}\lvert \tilde w(s)\rvert}{(1+t^{\alpha})\varphi(n)}\geqslant\varepsilon\right)\\
      &= \mathbf{P} \left(\bigcup\limits_{r=1}^{\infty}\left\{\sup\limits_{t\in[Tr,T(r+1)]} \frac{\max\limits_{s\in[0,t]} \lvert \tilde w(s)\rvert}{(1+t^{\alpha})} \geqslant \varphi(n)\varepsilon\right\}\right)+ \mathbf{P} \left(\limsup\limits_{t\to\infty}\frac{\lvert \tilde w(t)\rvert}{(1+t^{\alpha})\varphi(n)}\geqslant\varepsilon\right)\\
      &=\mathbf{P} \left(\bigcup\limits_{r=1}^{\infty}\left\{\sup\limits_{t\in[Tr,T(r+1)]} \frac{\max\limits_{s\in[0,t]} \lvert \tilde w(s)\rvert}{(1+t^{\alpha})} \geqslant \varphi(n)\varepsilon\right\}\right)\nonumber\\
      &\leqslant \sum\limits_{r=1}^{\infty}\mathbf{P} \left(\sup\limits_{t\in[Tr,T(r+1)]} \frac{\max\limits_{s\in[0,t]} \lvert \tilde w(s)\rvert}{(1+t^{\alpha})} \geqslant \varphi(n)\varepsilon\right)  \leqslant \sum\limits_{r=1}^{\infty}\mathbf{P} \left(\frac{\max\limits_{s\in[0,T(r+1)]} \lvert \tilde w(s)\rvert}{(1+(Tr)^{\alpha})} \geqslant \varphi(n)\varepsilon\right)\nonumber\\
      &\leqslant 2\sum\limits_{r=1}^{\infty}\exp \left\{- \frac{(1+(Tr)^{\alpha})^2\varphi^2(n)\varepsilon^2}{2T(r+1)} \right\} = 2\sum\limits_{r=1}^{\infty}\exp \left\{- \frac{(1+2(Tr)^{\alpha}+(Tr)^{2\alpha})\varphi^2(n)\varepsilon^2}{2T(r+1)} \right\}\nonumber\\
      &= 2\sum\limits_{r=1}^{\infty}\exp \left\{- \frac{(1+2(Tr)^{\alpha}+(Tr)^{2\alpha} + T^{2\alpha}r - T^{2\alpha}r)\varphi^2(n)\varepsilon^2}{2T(r+1)} \right\}\nonumber\\
      &= 2\sum\limits_{r=1}^{\infty}\exp \left\{- \frac{(1+2(Tr)^{\alpha}+(Tr)^{2\alpha}- T^{2\alpha}r)\varphi^2(n)\varepsilon^2}{2T(r+1)} \right\}\exp\left\{\frac{-T^{2\alpha}r\varphi^2(n)\varepsilon^2}{2T(r+1)}\right\}\nonumber\\
      &\leqslant 2\sum\limits_{r=1}^{\infty}\exp \left\{- \frac{(1+2(Tr)^{\alpha}+(Tr)^{2\alpha}- T^{2\alpha}r)\varepsilon^2 2\ln\ln 3}{2T(r+1)} \right\}\exp\left\{\frac{-T^{2\alpha - 1}\varphi^{2}(n)\varepsilon^2}{4}\right\}\nonumber\\
      &=2\exp\left\{\frac{-T^{2\alpha - 1}\varphi^{2}(n)\varepsilon^2}{4}\right\}\sum\limits_{r=1}^{\infty}\exp \left\{- \frac{(1+2(Tr)^{\alpha}+(Tr)^{2\alpha}- T^{2\alpha}r)\varepsilon^22\ln\ln 3}{2T(r+1)} \right\}\nonumber\\
      &=2\exp\left\{\frac{-T^{2\alpha - 1}\varphi^2(n)\varepsilon^2}{4}\right\} C(T,\varepsilon).
    \end{align*}
    Lemma \ref{prob_bound_lemma} is proven.
  \end{proof}

\bigskip
\begin{thebibliography}{1} % Ссылки на литературу приводятся только на английском языке (даже если у источника нет англоязычной версии).

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\bibitem{Mog} F.C. Klebaner, A.V. Logachov, A.A. Mogulskii,
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\end{thebibliography}
\end{document}
