text:
\begin{document}
\author{Guan Huang}
\address{Yau Mathematical Sciences Center, Tsinghua University, Beijing, China}
\email{huangguan@tsinghua.edu.cn}
\author{Sergei Kuksin}
\address{Institut de Math\'emathiques de Jussieu--Paris Rive Gauche, CNRS, Universit\'e Paris Diderot, UMR 7586, Sorbonne Paris Cit\'e, F-75013, Paris, France \& School of Mathematics, Shandong University, Jinan, Shandong,  China \& Saint Petersburg State University, Universitetskaya nab., St. Petersburg, Russia}
\email{Sergei.Kuksin@imj-prg.fr}
\maketitle
\numberwithin{equation}{section}
\begin{abstract}
We consider a damped/driven  nonlinear Schr\"odinger  equation in an $n$-cube $K^{n}\subset\mathbb{R}^n$, $n$ is arbitrary,
under Dirichlet boundary conditions
\[
u_t-\nu\Delta u+i|u|^2u=\sqrt{\nu}\eta(t,x),\quad x\in K^{n},\quad u|_{\partial K^{n}}=0, \quad \nu>0,
\] where $\eta(t,x)$ is a random force that is white in time and smooth in space.  It
is known that the Sobolev norms of solutions satisfy
$
\| u(t)\|_m^2 \le C\nu^{-m},
$
uniformly in $t\ge0$ and  $\nu>0$. In this work we  prove that for small  $\nu>0$ and  any initial data, with large probability  the Sobolev norms $\|u(t,\cdot)\|_m$ of the solutions with $m>2$  become large at least to the order of $\nu^{-\kappa_{n,m}}$ with $\kappa_{n,m}>0$, on  time intervals of order $\mathcal{O}(\frac{1}{\nu})$. \end{abstract}

\section{Introduction}
In this work we study a damped/driven nonlinear Schr\"odinger equation
\begin{equation}\label{m-eq}
u_t-\nu\Delta u+i|u|^2u=\sqrt{\nu}\eta(t,x),\quad x\in K^{n},
\end{equation}
where $n$ is any,
$0<\nu \le1$ is the viscosity constant and the random force $\eta$ is white in time $t$ and regular in $x$. The equation is considered under the
 odd periodic boundary conditions,
\[u(t,\dots,x_j,\dots)=u(t,\dots,x_j+2\pi,\dots)=-u(t,\dots,x_j+\pi,\dots),\quad j=1,\dots,n.
\]
The latter implies that $u$ vanishes on the boundary of the cube of half-periods $K^{n}= [0, \pi]^n$,
$$
u\mid_{\partial K^{n}} =0.
$$
We denote by $\{\varphi_d(\cdot),\; d=(d_1,\dots,d_n)\in\mathbb{N}^n\}$ the  trigonometric basis in the space of odd periodic functions,
$$
\varphi_d(x)=(\tfrac{2}{\pi})^{\frac{n}{2}}\sin(d_1x_1)\cdots\sin(d_nx_n).
$$

The basis  is orthonormal with respect to the scalar product $\llangle \cdot,\cdot \rrangle$ in
$L_2(K^{n}, \pi^{-n} dx)$,
$$
 \llangle u,v\rrangle =  \int_{K^n} \lan u(x),  v(x)\ran \pi^{-n} dx,
$$

where $\lan \cdot, \cdot\ran$ is the real scalar product in $\C$, $\lan u, v \ran =\Re u\bar v$. It is formed by
eigenfunctions of the Laplacian:
$$
(-\Delta)\varphi_d=|d|^2\varphi_d.
$$

 The force $\eta(t,x)$ is a random field of the form
\[\eta(t,x)=\frac{\partial}{\partial t}\xi(t,x),\quad \xi(t,x)=\sum_{d\in\mathbb{N}^n}b_d\beta_d(t)\varphi_d(x).\]
Here $\beta_d(t)=\beta^R_d(t)+i\beta^I_d(t)$, where $\beta_d^R(t)$, $\beta_d^I(t)$ are independent real-valued standard Brownian motions, defined on a complete probability space $(\Omega,\mathcal{F},\mathbb{P})$ with a filtration  $\{\mathcal{F}_t;t\geqslant0\}$.
The set of real numbers $\{b_d,\;d\in\mathbb{N}^n\}$ is assumed to form a non-zero sequence, satisfying
\begin{equation}\label{B_m}
0<B_{m_*}<\infty, \quad m_*= \min\{ m\in\Z: m>n/2\},
\end{equation}
where for $k\in \R$ we set
$$
B_k:=\sum_{d\in\mathbb{N}^n}|d|^{2k}  |b_d|^2\le \infty.
 $$

 For $m\ge 0$ we  denote by $H^m$ the Sobolev space of order $m$, formed by complex odd periodic functions, equipped with the homogeneous norm,
\[\|u\|_m=\|(-\Delta)^{\frac{m}{2}}u\|_0,\]
where $\|\cdot\|_0$ is the $L^2$-norm on  $K^{n}$,
$
\|u\|^2_0 =  \llangle u,u\rrangle.
$
 If we write $u\in H^m$ as Fourier series,
$u(x)=\sum_{d\in\mathbb{N}^n}u_d\varphi_d(x),$ then
$
\|u\|_m^2=\sum_{d\in\mathbb{N}^n}|d|^{2m}| u_d|^2.
$

Equation \eqref{m-eq} with small $\nu>0$ is a natural model for the small-viscosity Navier--Stokes system with a random
force in dimensions 2 and 3, describing  2d and 3d turbulence. Indeed, the former is obtained from the latter by replacing the
Euler equation
$\
u_t + (u\cdot \nabla) u +\nabla p=0, \; \text{div}\; u=0,
$
which is a 2--homogeneous Hamiltonian PDE, with  the 3-homogeneous Hamiltonian system
$
u_t +i |u|^2u=0.
$
See more in \cite{LO, k-gafa1999}. So a progress in the study of eq.~\eqref{m-eq} with small $\nu$ should help to understand turbulence a bit better.

\smallskip

The global solvability of eq.~\eqref{m-eq} for any space dimension $n$
is established in~\cite{kuk1999, Kuk-N2013}. It is proved there  that if
\begin{equation}\label{ic1}
u(0,x)=u_0(x),
\end{equation}
where $u_0\in H^m$, $m>\frac{n}{2}$, and if $B_m<\infty$, then the problem \eqref{m-eq}, \eqref{ic1} has a unique strong solution $u(t,x)$  in $H^m$
which we write as $u(t,x;u_0)$ or $u(t;u_0)$. Its norm satisfies
$$
\EE \|u(t;u_0)\|_m^2 \le C_m \nu^{-m},\quad t\ge0,
$$
where $C_m$ depends on $\|u_0\|_m, |u_0|_\infty$ and $B_m, B_{m_*}$.
  Furthermore, denoting by $C_0(K^{n})$ the space of continuous complex functions on $K^{n}$, vanishing at $\partial K^{n}$, we have that
the solutions $u(t,x)$
  define a Markov process in the space $C_0(K^{n})$.
  Moreover, if the noise $\eta(t,\cdot)$ is sufficiently non-degenerate (depending on $\nu$, see Theorem \ref{mix-measure}), then this  process is mixing.

Our goal is to  study the growth of higher Sobolev norms for solutions of the equation~\eqref{m-eq} as $\nu\to0$ on
 time intervals of order $\mathcal{O}(\frac{1}{\nu})$.
The main result of this work is the following.

 \begin{theorem}\label{m-theorem}
For any real number $m>2$, in 	addition to \eqref{B_m},  assume that $B_m<\infty$. Then
  there exists $\kappa_{n,m}>0$ such that for   every  fixed quadruple  $(\delta,\kappa,\mathscr{K},
 T_0)$, where
$$
 \kappa\in(0,\kappa_{n,m}),\quad \delta\in(0,\tfrac{1}{8}), \quad
  \mathscr{K},   T_0>0,
$$
there exists $\nu_0>0$ with the property that if $0<\nu\le \nu_0$, then for every $u_0\in H^m \cap C_0(K^{n})$, satisfying
\be\label{initial}
 |u_0|_\infty\leqslant \mathscr{K}, \quad \| u_0\|_m \le \nu^{-\kappa m},
 \ee

 the solution $u(t,x;u_0)$ is such that

\begin{enumerate}
\item
\[
\mathbb{P}\big\{ \sup_{t \in [t_0, t_0+ T_0\nu^{-1}]}
\|u_\nu^\omega(t)\|_m> \nu^{-m\kappa }\big\} \ge 1-\delta, \quad \forall\, t_0\ge 0.
\]
\item If  $m$ is an integer, $m\ge3$, then a possible choice of $\kappa_{n,m}$ is $\kappa_{n,m}= \tfrac1{35}$, and
 there exists  $C\ge1$, depending  on $\kappa< \tfrac1{35}$,
  $\mathscr{K},  m,B_{m_*}$ and  $B_m$, such that
\begin{equation}\label{avb}
C^{-1} \nu^{-2m\kappa +1}\leqslant\mathbb{E}\Big(\nu \int_{t_0}^{t_0+\nu^{-1}}\|u_\nu(s)\|_m^2ds\Big)\leqslant C \nu^{-m},
\quad \forall\, t_0\ge 0.
\end{equation}
\end{enumerate}
\end{theorem}

A similar result holds for the classical $C^k$-norms of  solutions:
 \begin{proposition}\label{m-theorem2}
For any integer  $m \ge2$  in 	addition to \eqref{B_m}  assume that $B_m<\infty$. Then
for   every  fixed triplet $K, \cK, T_0>0$  and any $0<\kappa<1/16$ we have
\be\label{conver}
\mathbb{P}\big\{ \sup_{t \in [t_0, t_0+ T_0\nu^{-1}]}
|u_\nu^\omega(t;u_0)|_{C^m} >  K \nu^{-m\kappa }\big\} \to1 \quad\text{as} \quad \nu\to0,

\ee
for each $t_0\ge0$, if $u_0$ satisfies $|u_0|_\infty\le \cK$,   $|u_0|_{C^m} \le \nu^{-\kappa m}$. The rate of convergence depends
only on the triplet  and $\kappa$.
\end{proposition}

The result is proved by adapting the argument from \cite[Section 5]{k-gafa1999} to the current settings
and arguing similar to when proving Theorem~\ref{m-theorem}; see in Appendix \ref{appen-cm}.
Due to \eqref{conver}, for any $m>2+n/2$ we have
$$
\PP \Big\{ \sup_{T_0 \le t \le t_0+{T_0}{\nu^{-1}}}  \| u(t)\|_{m} \ge K \nu^{-  \lfloor m-\frac{n}{2} \rfloor  \kappa}
\Big\} \to 1\quad \text{as} \quad \nu\to0,
 $$
 for every $K>0$ and  $0<\kappa <1/16$, where for  $a\in\R$ we denote
 $
 \lfloor a \rfloor = \max\{ n\in\Z: n<a\}.
 $
 This result improves the first assertion of Theorem~\ref{m-theorem} for large $m$.

We have the following  two corollaries from Theorem~\ref{m-theorem},
 valid if the Markov process defined by the equation \eqref{m-eq} is mixing:

\begin{corollary}\label{c_3}
Assume that $B_m<\infty$ for all $m$ and $b_d\ne0$ for all $d$. Then eq.~\eqref{m-eq} is mixing and for any
$\kappa<1/35$ and $0<\nu\le\nu_0$ its
  unique stationary measure
$\mu_\nu$ satisfies
\be\label{bb1}
C^{-1} \nu^{-2m\kappa+1} \le \int \| u\|_m^2\mu_\nu(du) \le C\nu^{-m}, \quad 3\le m\in\Nn.
\ee
Here $C$ and $\nu_0$ are as in Theorem \ref{m-theorem}.
\end{corollary}

\begin{corollary}\label{c_4}
Under the assumptions of Corollary \ref{c_3}, for any $u_0\in C^\infty$ we have
$$
\tfrac12 C^{-1} \nu^{-2m\kappa+1} \le \EE \| u (s;u_0)\|_m^2 \le 2C\nu^{-m}, \quad 3\le m\in\Nn,
$$
if $s\ge T(\nu, u_0, \kappa, B_m, B_{m_*})$, where $C$ is the same as in  \eqref{bb1}.
\end{corollary}
\begin{remark}
The exponent in the lower bound here can be slightly improved to $-(2m-7.001)\kappa$, see Remark \ref{stationary-lower}.
\end{remark}

Theorem \ref{m-theorem} rigorously establishes the energy cascade to high frequencies for solutions of eq. \eqref{m-eq} with small $\nu$.
Indeed, if $u_0(x)$ and $\eta(t,x)$ are smooth functions of $x$ (or even trigonometric polynomials of $x$), then in view of
\eqref{avb} for $0<\nu\ll1$ and $t \gtrsim \nu^{-1}$ a substantial part of the energy
$
\frac12 \sum |u_d(t)|^2
$
of a solution $u(t,x;u_0)$ is carried by high modes $u_d$, $|d|\gg1$. Relation \eqref{avb} (valid for all integer $m\ge3$) also means that the
averaged in time space-scale $l_x$ of solutions for \eqref{m-eq} satisfies
$
l_x \in [\nu^{1/2}, \nu^{1/35}],
$
and goes to zero with $\nu$ (see \cite{k-gafa1999} and \cite[Section 7.1]{BK}).  We recall that the energy cascade to high frequencies
and formation of short space-scale is the driving force of the Kolmogorov theory of turbulence, see \cite{Fr} and
\cite[Section 6]{BK}.

Two-sided estimates, similar to \eqref{avb}, are known for solutions of the stochastic Burgers equation on $S^1$:
\be\label{B}
 u_t -\nu u_{xx} +uu_x = \eta(t,x), \quad x\in S^1,\;\; \int u\,dx= \int\eta\,dx=0
\ee
(in view of the strong dissipativity in the small-viscosity Burgers equation,
to get solutions of order one the force should be $\sim1$,
rather then $\sim\sqrt\nu$ as in \eqref{m-eq}). Due to some remarkable features of the Burgers equation, the lower and upper estimates
for Sobolev norms of solutions for \eqref{B} are asymptotically sharp in the sense that they contain $\nu$ in the same negative degree.
Moreover, they hold after averaging on time--intervals of order one; see \cite[Section 2]{BK}.
\medskip

Deterministic versions of the result of Theorem \ref{m-theorem} for eq.~\eqref{m-eq} with $\eta=0$, where  $\nu$ is a small
non-zero complex number such that $\Re\nu\ge0$ 	and $\Im\nu\le0$ are known, see \cite{k-gafa1999}. In particular, if $\nu$
is a positive  real number and $u_0$ is a smooth function of order one, then for any integer $m\ge4$
 a solution $u_\nu(t,x;u_0)$ satisfies  estimates \eqref{avb} with the averaging
$
\nu \EE\int_t^{t+\nu^{-1}}\! \!\dots ds
$
replaced by
$
\nu^{1/3} \int_0^{\nu^{-1/3}}\! \!\dots ds,
$
with the same upper bound and with the  lower bound $C_m \nu^{-\kappa_m m}$, where
 $\kappa_m\to1/3$  as $m\to\infty$. Moreover, it was then shown in \cite{Bir09} that the
 lower bounds remain true with $\kappa=1/3$,

  and that the estimates
 $
 \sup_{t\in [0, |\nu|^{-1/3}]} \| u(t)\|_{C^m} \ge C_m |\nu|^{-m/3}, m\ge2,
 $
 hold for smooth solutions of  equation \eqref{m-eq} with $\eta=0$ and any non-zero complex ``viscosity" $\nu$.

 The better quality of the lower bounds for solutions of the
  the deterministic equations  is due to an extra difficulty which occurs in the stochastic case: when time grows, simultaneously with increasing of high Sobolev
 norms of a solution,  its $L_2$-norm may decrease, which accordingly
 would weaken the mechanism, responding to the energy transfer to
 high modes. Significant part of the proof of Theorem~\ref{m-theorem} is devoted to demonstration that the  $L_2$-norm of a
 solution  cannot
 go down without sending up the second Sobolev norm.

 If $\eta=0$ and $\nu=i\delta\in i\R$, then \eqref{m-eq} is a Hamiltonian PDE (the defocusing Schr\"odinger equation), and   the
 $L_2$-norm is  its  integral of motion. If this integral is of order one, then the results of  \cite{k-gafa1999}
 (see there Appendix~3)  imply
 that   at some point of   each time-interval of order $\delta^{-1/3}$ the $C^m$-norm of a  corresponding solution
  will become  $\ \gtrsim \delta^{-m\kappa}$ if $m\ge2$, for
  any $\kappa<1/3$.
    Furthermore,    if $n=2$ and $\delta=1$, then
  due to  \cite{Tao} for $m>1$ and any $M>1$ there exists a $T=T(m,M)$ and a smooth $u_0(x)$ such that
  $\|u_0\|_m< M^{-1}$ and $\| u(T;u_0)\|_m>M$.
 \medskip

The paper is organized as follows.  In Section \ref{apriori-estimate}, we recall the results   from \cite{kuk1999, Kuk-N2013}
on  upper estimates for solutions of the equation \eqref{m-eq}.
Next we show in Section~\ref{l2-norm-section} that if the noise $\eta$
  is non-degenerate, the $L^2$-norm of  a solution of eq.~\eqref{m-eq} cannot stay too small on  time
   intervals of order $\mathcal{O}(\frac{1}{\nu})$ with high probability, unless its  $H^2$-norm gets very large

    (see Lemma~\ref{l2low}). Then in  Section~\ref{section-proof} we derive from this fact the assertion~(1)
    of Theorem~\ref{m-theorem}, and prove  assertion~(2) and both corollaries     in Section \ref{proof-section-2}.

    Constants in estimates never depend on $\nu$, unless otherwise stated. For a metric space $M$ we denote by $\B(M)$ the Borel $\sigma$-algebra on $M$, and by
    ${\mathcal P}(M)$ -- the space of probability Borel measures on $M$. By $\cD(\xi)$ we denote the law of a r.v. $\xi$.
    \smallskip

 We submit  to a journal an abridged version of this  paper, where we removed the proof of the lower bounds for the classical  $C^m$-norms of solutions for eq.~\eqref{m-eq},  presented in Appendix \ref{appen-cm}, and made some other small diminutions.

\section{Solutions and  estimates}\label{apriori-estimate}

Strong
 solutions for the equation \eqref{m-eq} are defined in the usual way:

\begin{definition} \label{def-strong-sol}Let $0<T<\infty$ and  $(\Omega, \mathcal{F},\{\mathcal{F}_t\}_{t\geqslant0}, \mathbb{P})$  be the
filtered   probability space as in the introduction.
 Let $u_0$ in \eqref{ic1} be a r.v., measurable in $\mathcal{F}_0$ and independent from the Wiener
process $\xi$. Then a random process $u(t)=u(t,\cdot)\in C_0(K^{n}) $, $t\in[0,T]$,
adapted to the filtration,    is called a strong solution of \eqref{m-eq}, \eqref{ic1}, if
\begin{enumerate}

\item a.s.  its trajectories $u(t)$  belong to the space
\[\mathcal{H}([0,T]):=C([0,T],C_0(K^{n}))\cap L^2([0,T], H^1);\]
\item  we have
\[u(t)=u_0+\int_0^t(\nu\Delta u-i|u|^2u)ds+\sqrt{\nu}\,\xi(t),\quad \forall t\in[0,T], \; a.s.,\]
where both sides are regarded as elements of $H^{-1}$.
\end{enumerate}
If (1)-(2) hold for every $T<\infty$, then $u(t)$ is called a strong solution for $t\in [0,\infty)$. In this case a.s.
 $u\in \mathcal{H}([0,\infty))$, where
$$
 \mathcal{H}([0,\infty)) = \{ u(t), t\ge0: u\mid_{[0,T]} \in  \mathcal{H}([0,T]) \;\; \forall T>0\}.
$$
\end{definition}

Everywhere below when we talk about solutions for the problem  \eqref{m-eq}, \eqref{ic1} we assume that   the  r.v. $u_0$ is as in the definition above.
In particular,  $u_0(x)$ may be  a non-random function.

The global well-posedness of eq.~\eqref{m-eq} was established in \cite{kuk1999, Kuk-N2013}:

\begin{theorem}\label{existence}
For any  $u_0\in C_0(K^{n})$  the problem \eqref{m-eq}, \eqref{ic1} has a unique strong solution $u^\om(t,x;u_0)$,  $t\ge0$.
 The family of   solutions $\{ u^\om(t;u_0)\}$  defines in the space $C_0(K^{n})$ a Fellerian Markov process.
\end{theorem}

In  \cite{kuk1999, Kuk-N2013} the theorem above is proved when \eqref{B_m} is replaced by the weaker assumption $B_*<\infty$, where
$\
B_* =\sum |b_d|
$
(note that $B_*\le C_n B_{m_*}^{1/2}$).

  The transition probability for the obtained Markov process in $C_0(K^{n})$ is

$$
P_t(u,\Gamma)=\mathbb{P}\{u(t;u)\in\Gamma\}, \quad u\in C_0(K^{n}),\; \Gamma\in \mathscr{B}(C_0(K^{n})),
$$
and the corresponding Markov semigroup in the space $\mathscr{P}(C_0(K^{n}))$ of Borel measures on $C_0(K^{n})$
 is formed by the operators $\{ \mathcal{B}_t^*, t\ge0\}$,

\[ \mathcal{B}_t^*\mu(\Gamma)=\int_{C_0(K^{n})}P_t(u,\Gamma)\mu(du),\quad t\in\mathbb{R}.
\]
\medskip
Then  $\mathcal{B}_t^*\mu = \cD u(t;u_0)$ if $u_0$ is a r.v., independent from $\xi$ and such that   $\cD(u_0)=\mu$.

Introducing the slow time  $\tau=\nu t$ and denoting $v(\tau,x)=u(\frac{\tau}{\nu},x)$, we rewrite  eq.~\eqref{m-eq} in the form below, more convenient for some calculations:
\begin{equation}\label{the_eq}
\frac{\partial v}{\partial \tau}-\Delta v+i\nu^{-1}|v|^2v=\tilde{\eta}(\tau,x),
\end{equation}
where \[\tilde{\eta}(\tau,x)=\frac{\partial}{\partial\tau}\tilde{\xi}(\tau,x),\quad \tilde{\xi}(\tau,x)=\sum_{d\in\mathbb{N}^n}b_d\tilde{\beta}_d(\tau)\varphi_d(x),\]
and $\tilde{\beta}_d(\tau):=\nu^{1/2}\beta_d(\tau\nu^{-1})$, $d\in \mathbb{N}^d$, is another set of independent
standard complex Brownian motions.

 Let $\Upsilon\in C^\infty(\mathbb{R})$ be any smooth function such
\[\Upsilon(r)=\begin{cases}0 ,&\text{ for }r\leqslant \frac{1}{4};\\
r, &\text{ for }r\geqslant \frac{1}{2}. \end{cases}\]
Writing $v\in\C$ in the polar form $v=re^{i\Phi}$, where $r=|v|$, and recalling that $\lan\cdot,\cdot\ran$
stands for the real scalar product in $\C$, we apply
 It\^{o}'s  formula to $\Upsilon(|v|)$ and obtain that the process $\Upsilon(\tau):=\Upsilon(|v(\tau)|)$ satisfies
\be\label{ups}
\begin{split} \Upsilon(\tau)&=\Upsilon_0+\int_0^\tau\Big[\Upsilon'(r)(\nabla r-r|\nabla \Phi|^2)\\
&\quad+\frac{1}{2}\sum_{d\in\mathbb{N}^n}b_d^2\Big(\Upsilon''(r)\langle e^{i\Phi},\varphi_d\rangle^2+
\Upsilon'(r)\frac{1}{r}(|\varphi_d|^2-\langle e^{i\Phi},\varphi_d\rangle^2)\Big)\Big]ds+\mathbb{W}(\tau),\end{split}
\ee
where $\Upsilon_0=\Upsilon(|v(0)|)$ and $\mathbb{W}(\tau)$ is the stochastic integral
\[\mathbb{W}(\tau)=\sum_{d\in\mathbb{N}^n}\int_0^\tau\Upsilon'(r)b_d\varphi_d\langle e^{i\Phi}, d\tilde{\beta}_d(s)\rangle.\]

In \cite{Kuk-N2013} eq. \eqref{the_eq} is considered with $\nu=1$ and, following \cite{kuk1999}, the norm $|v(t)|_\infty$ of a solution $v$ is
estimated via $\Upsilon(t)$ (since $|v| \le \Upsilon +1/2$). But the nonlinear term $i\nu^{-1}|v|^2v$ does not contribute to
eq.~\eqref{ups}, which is the same as  the $\Upsilon$-equation
(2.3) in \cite{Kuk-N2013} (and as the corresponding equation in \cite[Section 3.1]{kuk1999}).
So the estimates on $|\Upsilon(t)|_\infty$ and the resulting estimates on $|v(t)|_\infty$, obtained in \cite{Kuk-N2013}, remain true for solutions of
 \eqref{the_eq} with any $\nu$. Thus  we get  the following upper bound  for a quadratic  exponential moments of  the
	$L_\infty$-norms of solutions:\footnote{In \cite{kuk1999} polynomial moments of the random variables $\sup_{\tau\leqslant s\leqslant \tau+T}|v(s)|_\infty^2)$
are estimated, and in  \cite{Kuk-N2013} these results are strengthened to the exponential bounds \eqref{main_esti}.	}

\begin{theorem} \label{sup-norm} For any $T>0$ there are constants $c_*>0$ and $C>0$, depending only on $B_*$ and $T$, such
 that for any  r.v.   $v_0^\omega\in C_0(K^{n})$ as in Definition~\ref{def-strong-sol},
  any $\tau\geqslant0$ and any $c\in(0,c_*]$, a solution $v(\tau; v_0)$ of eq.~\eqref{the_eq} satisfies
\be\label{main_esti}
\mathbb{E}\exp(c \sup_{\tau\leqslant s\leqslant \tau+T}|v(s)|_\infty^2)\leqslant C\, \EE\exp(5c\ |v_0|^2_\infty)\le\infty.
\ee
\end{theorem}

In \cite{Kuk-N2013} the result above is proved for a deterministic initial data $v_0$. The theorem's assertion follows by averaging the result
of  \cite{Kuk-N2013} in $v_0^\omega$.

The estimate \eqref{main_esti} is crucial for  derivation of further properties of solutions, including the  given below
upper bounds  for their Sobolev norms, obtained in  the work  \cite{kuk1999}. Since the scaling of the equation in \cite{kuk1999} differs
from that in  \eqref{the_eq} and
the result there is a bit less general than in the theorem below, a sketch of the proof is given in Appendix~\ref{app_B}.

\begin{theorem}\label{upper-bound}
 Assume that $B_m<\infty$ for some  $ m\in\Nn$, and $v_0=v_0^\nu \in H^m\cap C_0(K^n)$ satisfies
$$
|v_0|_\infty\le M, \quad \|v_0\|_m \le M_m \nu^{-m}, \quad 0<\nu\le1.
$$
Then
\be\label{sob_esti}
\mathbb{E}\|v(\tau;v_0)\|_m^2\leqslant C_{m}\nu^{-m},\quad   \forall \tau\in[0,\infty),
\ee
where $C_{M,m}$ also
depends on $M$, $M_m$ and $B_m$, $B_{m_*}$.
\end{theorem}

Neglecting the dependence on $\nu$, we have that if $B_m<\infty$, $m\in\Nn$,  and a   r.v.
 $v_0^\omega \in H^m\cap C_0(K^n)$ satisfies  $\EE \| v_0\|_m^2<\infty$ and $\EE \exp(c\ |v_0|^2_\infty)<\infty$ for some $c>0$,
 then eq.~\eqref{the_eq} has a solution, equal $v_0$ at $t=0$, such that
\be\label{m_est}
\EE \| v(\tau;v_0)\|_m^2 \le e^{- t}   \EE  \| v_0\|_m^2 +C, \quad \tau\ge0,
\ee
\be\label{m_est1}
\EE \sup_{0\le\tau\le T} \| v(\tau;v_0)\|_m^2 \le  C' ,
\ee
where $ C>0$ depend on  $c, \nu, \EE \exp(c\ |v_0|^2_\infty),  B_{m_*}$ and $B_m$, while $C'$ also depends on $\EE \| v_0\|_m^2<\infty$ and~$T$.
 See Appendix~\eqref{app_B}.

  As it is shown  in \cite{Kuk-N2013}, the estimate \eqref{main_esti} jointly with an abstract theorem from \cite{Kuk_Shi2012}, imply that under a mild
  nondegeneracy   assumption on the random force   the Markov process  in the space $C_0(K^{n})$,  constructed in Theorem \ref{existence},
  is  mixing:

\begin{theorem}\label{mix-measure}
For each $\nu>0$, there is an integer $N=N(B_*,\nu)> 0$ such that if $b_d\neq0$ for $|d|\leqslant N$, then the equation~\eqref{m-eq} is mixing. I.e. it has
a unique stationary measure $\mu_\nu\in\mathscr{P}(C_0(K^{n}))$,  and for any probability measure $\lambda\in\mathscr{P}(C_0(K^{n}))$ we have $\mathcal{B}^*_t\lambda\rightharpoonup\mu_\nu$ as $t\to\infty$.
\end{theorem}

Under the assumption of Theorem  \ref{upper-bound}, for any $u_0\in H^m$ the law $\cD u(t;u_0)$ of a solution $u(t;u_0)$
is a measure in $H^m$. The mixing property in Theorem~\ref{mix-measure}  and \eqref{sob_esti} easily imply

\begin{corollary}\label{sob_mix}
If under  the assumptions of Theorem \ref{mix-measure}
$B_m<\infty$ for some $m>n/2$  and $u_0\in H^m$,  then $\cD(u(t;u_0)) \strela \mu_\nu$ in $\Pe(H^m)$.
\end{corollary}

In view of Theorems  \ref{sup-norm}, \ref{upper-bound} with $v_0=0$  and the established mixing, we have:

\begin{corollary}\label{cor-sup}
Under the assumptions of Theorem \ref{mix-measure}, if  $v^{st}(\tau)$ is  the stationary solution of the equation,  then

\[
\mathbb{E}\exp(c_*\sup_{\tau\leqslant s\leqslant \tau+T}|v^{st}(s)|_\infty^2)\leqslant\mathcal{C},
\]
where the constant $\mathcal{C}>0$ depends only on $T$ and $B_*$. If in addition $B_m<\infty$ for some $m\in \Nn\cup\{0\}$, then

$
\mathbb{E}\|v^{st}(\tau)\|_m^2\leqslant C_{m}\nu^{-m},
$
where $C_m$ depends on $B_*$ and $B_m$.
\end{corollary}

Finally we note that applying  It\^{o}'s  formula to  $\| v^{st}(\tau)\|_0^2$, where $v^{st}$ is a stationary solution
 of \eqref{the_eq}, and taking the expectation we get the balance relation
\be\label{ito_stat}
\EE \| v^{st}(\tau)\|^2_1 = B_0.
\ee
We cannot prove that $\EE \| v^{st}(\tau)\|_0^2 \ge B'>0$ for some $\nu$-independent constant $B'$,
and cannot bound from below
the energy $\tfrac12 \EE \| v(\tau; v_0)\|^2_0$
of a solution $v$ by a positive $\nu$-independent quantity. Instead in next section we get a
weaker conditional  lower bound on the energies  of solutions.

\section{Conditional lower bound for the $L^2$-norm of solutions}\label{l2-norm-section}
In this section we prove the following result:

\begin{lemma}\label{l2low} Let $m\geqslant2$ and $B_m<\infty$. Let $u(\tau)\in H^m$ be a solution of \eqref{the_eq}. Take any constants
$
\chi>0, \Gamma\ge1, \tau_0\ge0,
$
and define the stopping time
$$
\tau_\Gamma:=\inf\{\tau \ge \tau _0 : \|u(\tau )\|_2\geqslant\Gamma\}
$$
(as usual, $\tau_\Gamma =\infty$ if the set under the $\inf$-sign is empty).
Then
\be\label{HG}
\mathbb{E}\int_{\tau_0}^{\tau\wedge\tau_\Gamma} \mathbb{I}_{[0,\chi]} (\|u(s)\|_0)ds\leqslant 2(1+\tau)B_0^{-1} \chi \Gamma,

\ee
for any $\tau>\tau_0$.
Moreover,  if $u(\tau)$ is a stationary solution of the equation~\eqref{the_eq} and  $\mathbb{E}\|u(\tau )\|_2\leqslant\Gamma,$ then
\begin{equation}\label{stat2}
\mathbb{P}\Big(\{\|u(\tau)\|_0\leqslant \chi\}\Big)\leqslant2B_0^{-1}\chi\Gamma,
\quad \forall \tau \geqslant0. \end{equation}
\end{lemma}

\begin{proof} We establish  the result by adapting the proof from  \cite{Shir} (also see
\cite[Theorem~5.2.12]{ Kuk_Shi2012}) to non-stationary solutions.
The argument  relies on the concept of local time for semi-martingales (see e.g. \cite[Chapter VI.1]{RY} for  details of the  concept).
By $[\cdot]_b$ we denote the quasinorm
$
[u]_b^2 = \sum_d |u_d|^2 b_d^2.
$

  Without loss of generality we assume $\tau_0=0$, otherwise we just need to replace $u(
  \tau,x)$ by  the process $\tilde u(\tau,x):=u(\tau+\tau_0,x)$, apply the lemma with $\tau_0=0$ and with
   $u_0$ replaced by the  initial data
   $\tilde u_0^\omega=u^\omega(\tau_0; u_0)$, and then   average  the estimate in the random $\tilde u_0^\omega$.    Let  us write the solution  $u(\tau;u_0)$ as
$u(\tau,x)=$ $\sum_{d\in\mathbb{N}^n}u_d(\tau)\varphi_d(x)$.
For any fixed function $g\in C^2(\mathbb{R})$,  consider the process
$$
f(\tau )=g(\|u(\tau \wedge\tau_\Gamma)\|_0^2).
$$
Since
\[
\partial_ug(\|u\|_0^2)=2g'(\|u\|_0^2)\llangle u,\cdot\rrangle, \quad

\partial_{uu}g(\|u\|_0^2)=4g''(\|u\|_0^2)\llangle u,\cdot\rrangle \llangle u,\cdot\rrangle+2g'(\|u\|^2_0)\llangle \cdot,\cdot\rrangle,
\]
then by It\^{o}'s formula we have
\begin{equation}\label{ito-f}f(\tau)=f(0)+  \int_0^{\tau\wedge\tau_\Gamma}A(s)ds+\sum_{d\in\mathbb{N}^n}b_d\int_0^{\tau\wedge\tau_\Gamma}
2g'(\|u(s)\|_0^2)\lan u_d(s),
 d\beta_d(s)\ran,\end{equation}
where
\begin{equation}\label{A-expression}
\begin{split}
 A(s)&=2g'(\|u\|_0^2)\llangle u, \Delta u-\frac{1}{\nu}i|u|^2u\rrangle +2 \sum_d b_d^2 \big( g''(\| u\|_0^2 ) |u_d|^2 +g'(\| u\|_0^2)\big) \\

&=- 2g'(\|u\|^2_0)\| u\|_1^2+ 2g''(\|u\|_0^2)

[u]_b^2
+2g'(\|u\|_0^2) B_0, \quad u=u(s).
\end{split} \end{equation}

\noindent
{\bf Step 1: } We firstly show that for any bounded measurable set  ${G}\subset\mathbb{R}$, denoting by
$\mathbb{I}_G$ its indicator function, we have the following equality
\begin{equation}\label{expectation-key}
\begin{split}
2\mathbb{E} \int_0^{\tau\wedge\tau_\Gamma}&\mathbb{I}_G(f(s))  \,\big(g'(\|u(s)\|_0^2)\big)^2 \! [u(s)]_b^2 ds

=\int_{-\infty}^\infty\mathbb{I}_G(a)\\&\Big[\mathbb{E}(f(\tau)-a)_+-\mathbb{E}(f(0)-a)_+-\mathbb{E}\int_0^{\tau\wedge\tau_\Gamma}\mathbb{I}_{(a+\infty)}(f(s))A(s)ds\Big]da.
\end{split}\end{equation}

Let $L(\tau,a)$, $(\tau,a)\in[0,\infty)\times\mathbb{R}$, be the local time for the semi martingale $f(\tau)$ (see e.g. \cite[Chapter VI.1]{RY}).
Since in view of \eqref{ito-f} the quadratic variation of the process $f(\tau)$ is
$$
d  \langle f,f \rangle_s = \sum_d (2g'(\| u\|_0^2) |u_d| b_d)^2 =4\big( g'(\| u\|_0^2)\big)^2 [u]_b^2,
$$
then for  any bounded measurable set  $G\subset\mathbb{R}$,   we have the following equality (known as the occupation time formula,
see  \cite[Corollary~VI.1.6]{RY}),
\begin{equation}\label{localtime1}\int_0^{\tau\wedge\tau_\Gamma}\mathbb{I}_G(f(s))
4\big(g'(\|u(s)\|_0^2)\big)^2\, [u(s)]_b^2

ds=\int_{-\infty}^\infty \mathbb{I}_{{G}}(a)L(\tau,a)da.\end{equation}
For the local time $L(\tau,a)$, due to Tanaka's formula (see \cite[Theorem VI.1.2]{RY}) we have
\begin{equation}\label{localtime2}\begin{split}(f(\tau)-a)_+=&(f(0)-a)_+\\
&+\sum_{d\in\mathbb{N}^n} b_d\int_0^{\tau\wedge\tau_\Gamma}\mathbb{I}_{(a,+\infty)}\big(f(s)\big)2g'(\|u(s)\|_0^2)  \langle u_d (s), d\beta_d(s)\rangle \\
&+\int_0^{\tau\wedge\tau_\Gamma}\mathbb{I}_{(a,+\infty)}(f(s))A(s)ds+\frac{1}{2}L(\tau,a).
\end{split}\end{equation}
Taking expectation of both sides of \eqref{localtime1} and \eqref{localtime2} we  obtain the required  equality~\eqref{expectation-key}.
\medskip

\noindent
{\bf Step 2: }
Let us choose ${G}=[\rho_0,\rho_1]$ with $\rho_1>\rho_0>0$,  and $g(x) =g_{\rho_0} (x)
\in C^2(\mathbb{R})$ such that $g'(x)\geqslant0$,
$g(x)=\sqrt{x}$ for $x\geqslant \rho_0$ and  $g(x)=0$ for $x\leqslant 0$.

Then due to the factors $\mathbb{I}_G(f)$ and $\mathbb{I}_G(a)$ in \eqref{expectation-key}, we may there
replace $g(x)$ by $\sqrt x$, and accordingly replace
$
g(\|u\|^2_0), g'(\|u\|^2_0)$ and  $ g''(\|u\|^2_0)
$
 by $\| u\|_0$,   $ \tfrac12\| u\|_0^{-1}$ and $ -\tfrac14\| u\|_0^{-3}$.  So the relation \eqref{expectation-key}   takes the form
\[\begin{split}
&\mathbb{E}\int_0^{\tau\wedge\tau_\Gamma} \mathbb{I}_G(f(s))
\|u(s)\|_0^{-2} [u(s)]_b^2

=2\int_{\rho_0}^{\rho_1}\Big[\mathbb{E}(f(\tau)-a)_+-\mathbb{E}(f(0)-a)_+\Big]da\\
&-2\int_{\rho_0}^{\rho_1}\Big\{\mathbb{E}\int_0^{\tau\wedge\tau_\Gamma}\mathbb{I}_{(a,+\infty)}\Big(f(s)\Big)\Big[\frac{2}{2\|u(s)\|_0}(B_0-\|u(s)\|_1^2)
-\frac{2}{4\|u(s)\|_0^3}
[u(s)]_b^2

\Big]ds\Big\}da.
\end{split}\]
Since the l.h.s. of  the above equality is  non-negative, we have
\begin{equation}\label{main_est}
\begin{split}
&\int_{\rho_0}^{\rho_1}\Big[\mathbb{E}\int_0^{\tau\wedge\tau_\Gamma}\mathbb{I}_{(a,+\infty)}\Big(f(s)\Big)\frac{1}{\|u(s)\|_0^3}\Big(B_0\|u(s)\|_0^2-\tfrac{1}{2} [u(s)]_b^2
\Big)ds\Big]da\\
&\leqslant \int_{\rho_0}^{\rho_1} \mathbb{E}  \Big[\big( (f(\tau)-a)_+  - (f(0)-a)_+\big)
+
\int_0^{\tau\wedge \tau_\Gamma}\mathbb{I}_{(a,+\infty)}\big(f(s)\big)\frac{\|u(s)\|_1^2}{\|u(s)\|_0} ds\Big]da.
\end{split}
\end{equation}

Noting that
\[
B_0\|u\|_0^2-\tfrac{1}{2} [u(s)]_b^2

 =\sum_{d\in\mathbb{N}^n} (B_0-\frac{1}{2}b_d^2)|u_d|^2\geqslant \frac{B_0}{2} \|u\|_0^2,
\]
that  by the definition of the stopping time $\tau_\Gamma$
\[
 (f(\tau)-a)_+-(f(0)-a)_+\leqslant \Gamma,
\]
and that by interpolation, \[\int_0^{\tau\wedge\tau_\Gamma}\frac{\| u(s)\|_1^2}{\|u(s)\|_0}ds\leqslant \int_0^{\tau\wedge\tau_\Gamma}\|u(s)\|_2ds\leqslant (\tau\wedge \tau_\Gamma) \Gamma,
\]
we derive from \eqref{main_est} the relation
$$
\frac{B_0}2\! \int_{\rho_0}^{\rho_1} \big( \mathbb{E}\int_0^{\tau\wedge\tau_\Gamma}\mathbb{I}_{(a,+\infty)}\big(f(s)\big) \|u(s)\|_0^{-1}

ds\big)  da \le (\rho_1-\rho_0) \Gamma \big(1 \,+ \,   \tau ) .
$$
 When $\rho_0 \to0$, we have $g(x) \to \sqrt x$ and $f(\tau) \to \| u(\tau \wedge \tau_\Gamma)\|_0$.
So sending $\rho_0$ to 0 and using Fatou's lemma
 we get from the last estimate that
$$
\int_0^{\rho_1}\mathbb{E}\int_0^{\tau\wedge\tau_\Gamma}\mathbb{I}_{(a,\infty)}\Big(\|u(s)\|_0\Big)\|u(s)\|_0^{-1}dsda\leqslant 2\rho_1(1+\tau)B_0^{-1}\Gamma.
$$
As the l.h.s. above is not smaller than
\[\begin{split}

 \frac{1}{\chi}\int_0^{\rho_1}\mathbb{E}\int_0^{\tau\wedge\tau_\Gamma}\mathbb{I}_{(a,\chi]}(\|u(s)\|_0)dsda,
\end{split}\]
then
\be\label{hrrr}
\frac{1}{\rho_1}\int_0^{\rho_1}
\mathbb{E}\int_0^{\tau\wedge\tau_\Gamma}\mathbb{I}_{(a,\chi]}(\|u(s)\|_0)dsda\leqslant2(1+\tau)B_0^{-1}\Gamma\,\chi.
\ee
By the monotone convergence theorem
$$
\lim_{a\to0}  \mathbb{E}\int_0^{\tau\wedge\tau_\Gamma}\mathbb{I}_{(a,\chi]}(\|u(s)\|_0)ds =
 \mathbb{E}\int_0^{\tau\wedge\tau_\Gamma}\mathbb{I}_{(0,\chi]}(\|u(s)\|_0)ds,
$$
so we get from \eqref{hrrr} that
\be\label{hihh}
\mathbb{E}\int_0^{\tau\wedge\tau_\Gamma}\mathbb{I}_{(0,\chi]}(\|u(s)\|_0)ds\leqslant 2(1+\tau)B_0^{-1}\Gamma\chi.
\ee
\medskip

\noindent
{\bf Step 3: }
We continue to verify that
 \begin{equation}\label{probability0}
 \mathbb{E}\int_0^{\tau\wedge\tau_\Gamma}\mathbb{I}_{\{0\}}(\|u(s)\|_0)ds=0.

\end{equation}
To do this  let us fix any index $d\in\mathbb{N}^n$ such that $b_d\neq0$. The  process $u_d(\tau)$ is a semimartingale,
$\
du_d = v_d ds +b_d d\beta_d, \,
$
where  $v_d(s)$ is the $d$-th
 Fourier coefficient of $\, \Delta u+\frac{1}{\nu}i|u|^2u\,$ for the solution  $u(\tau)=\sum_d u_d(\tau) \phi_d$   which we discuss. Consider the stopping time
 $$
 \tau_R = \inf\{ s\le {\tau\wedge\tau_\Gamma}: |u(s)|_\infty \ge R\}.
 $$
 Due to \eqref{main_esti} and \eqref{m_est1}, $ \mathbb{P} (\tau_R= \tau\wedge\tau_\Gamma) \to1$ as $R\to\infty$. Let us denote
 $
 u_d^R(\tau) = u_d(\tau\wedge\tau_R).
 $
 To prove \eqref{probability0} it suffices to verify that
 $$
 \pi(\delta) :=   \mathbb{E}\int_0^{\tau\wedge\tau_\Gamma}\mathbb{I}_{\{ |u_d(s)| <\delta\}} ds

  \to 0\quad\text{as} \quad \delta\to0.
 $$
 If we replace above $u_d$ by $u_d^R$, then the obtained new quantity $ \pi^R(\delta)$ differs from  $ \pi(\delta)$ at most
 by $  \mathbb{P} (\tau_R<  \tau\wedge\tau_\Gamma)$. The process $u_d^R$ is an Ito process with a bounded drift. So by \cite[Theorem 2.2.2, p. 52]{Kry},
  $ \pi^R(\delta)$ goes to zero with $\delta$. Thus,  given any $\eps>0$, we firstly choose $R$ sufficiently big and then  $\delta$
  sufficiently small to achieve $\pi(\delta) <\eps$, for a suitable $\delta(\eps)>0$. So \eqref{probability0} is verified.  Jointly
   with \eqref{hihh} this proves~\eqref{HG}.
   \medskip

{\bf Step 4:} We now consider the stationary case. Let $u(\tau)$ be a stationary solution of \eqref{the_eq}. Then applying It\^o's formula to the process $f(\tau):=g(\|u(\tau)\|_0^2)$, following the same argument as   in Step 1, we obtain for any bounded measure set $G\subset\mathbb{R}$,
\[\begin{split}
&2\mathbb{E}\int_0^{\tau}\mathbb{I}_G(f(s))  \,\big(g'(\|u(s)\|_0^2)\big)^2 \! [u(s)]_b^2\\
&=\int_{-\infty}^\infty\mathbb{I}_G(a)\Big[\mathbb{E}(f(\tau)-a)_+-\mathbb{E}(f(0)-a)_+-\mathbb{E}\int_0^{\tau}\mathbb{I}_{[a+\infty)}(f(s))A(s)ds\Big]da,
\end{split}\]
(which is exactly \eqref{expectation-key} without the stoping time $\tau_\Gamma$).
From  the stationarity of the solution, we have
\[\mathbb{E}(f(\tau)-a)_+-\mathbb{E}(f(0)-a)_+=0,\]
\[\mathbb{E}\int_0^{\tau}\mathbb{I}_G(f(s))  \,\big(g'(\|u(s)\|_0^2)\big)^2 \! [u(s)]_b^2ds=\tau\mathbb{E}\Big(\mathbb{I}_G(f(s))  \,\big(g'(\|u(s)\|_0^2)\big)^2 \! [u(s)]_b^2\Big),\;\forall s\geqslant0,\]
and
\[\mathbb{E}\int_0^{\tau}\mathbb{I}_{[a+\infty)}(f(s))A(s)ds=\tau\mathbb{E}\Big(\mathbb{I}_{[a+\infty)}(f(s))A(s)\Big),\;\forall s\geqslant0.\]
Therefore,  with \eqref{A-expression}, we have
\[\begin{split}&\mathbb{E}\int_G\mathbb{I}_{[a,\infty)}\big(g(\|u\|_0^2)\big)\Big(2g'(\|u\|_0^2)(B_0-\|u\|_1^2)+2g''(\|u\|_0^2)
[u]_b^2
\Big)da\\
&+\sum_{d\in\mathbb{N}^n}\mathbb{E}\Big(\mathbb{I}_{{G}}\big(g(\|u\|_0^2)\big)\big(2g'(\|u\|_0^2)|u_d|\big)^2\Big)=0.\end{split}\]
Then proceeding as in Step 2,  we obtain the inequality \eqref{stat2}. Thus, we finished the proof of the theorem.
\end{proof}

\section{Lower bounds for  Sobolev norms of solutions}\label{section-proof}
In this section we  work with eq.~\eqref{m-eq} in the original time scale $t$ and  provide lower bounds for the $H^m$-norms
of its solutions with  $m>2$. This will prove the assertion (1) of Theorem \ref{m-theorem}. As always, the constants do not depend on $\nu$, unless otherwise
stated.

\begin{theorem}\label{t_sobolev} For any    $m\geqslant3$, if  $B_m<\infty$
  and
  $$
  0<\kappa<\tfrac{1}{35},\quad T_0\ge0, \quad T_1>0,
  $$
  then for any   r.v. $u_0(x)\in H^m\cap C_0(K^n)$,   satisfying
  \be\label{new_cond}
  \EE \| u_0\|_m^2<\infty, \quad \EE \exp(c\ |u_0|^2_\infty) \le C
  <\infty
  \ee
 for some $c, C >0$,   we have
\be\label{1.0_sobolev}
\PP \Big\{ \sup_{T_0 \le t \le T_0+{T_1}{\nu^{-1}}}  \| u(t;u_0)\|_{m} \ge K \nu^{-m\kappa}
\Big\} \to 1\quad \text{as} \quad \nu\to0,

\ee
for every $K>0$.
\end{theorem}

\begin{proof}
 Consider the complement to the event in \eqref{1.0_sobolev}:
\[
 Q^\nu=\{\sup_{T_0\leqslant t \leqslant T_0+\frac{T_1}{\nu}}\|u(t )\|_{m}<K\nu^{-m\kappa}\}.
\]
  We will prove the assertion \eqref{1.0_sobolev} by contradiction. Namely, we
  assume that there exists a $\gamma>0$ and a sequence $\nu_j\to0$ such that
\be\label{contra}
\PP( Q^{\nu_j})\geqslant 5\gamma \quad\text{for}\quad j=1,2,\dots,
\ee
and will derive a contradiction.
Below we write $ Q^{\nu_j}$ as $ Q$ and always suppose that
$$
\nu \in \{\nu_1, \nu_2, \dots\}.
$$
The constants below may depend on $\cK, K, \gamma$, $B_{\,m\vee m_*}$,
 but not on $\nu$.

Without lost of generality   we assume that  $T_1=1$.    For any $T_0>0$, due to \eqref{m_est} and \eqref{main_esti} the r.v.
$
\tilde u_0 := u(T_1)
$
satisfies \eqref{new_cond}  with $c$ replaced by $c/5$. So considering $\tilde u(t,x) = u(t+T_0,x)$ we may assume that $T_0=0$.

 Let us denote $J_1=[0,\frac{1}{\nu}]$. Due to Theorem \ref{sup-norm},
 \[\mathbb{P}(Q_1)\geqslant 1-\gamma, \quad Q_1=\{\sup_{t\in J_1}|u(t)|_\infty\leqslant C_1(\gamma)\},
 \]
 uniformly in $\nu$, for a suitable $C_1(\gamma)$. Then, by the definition of $Q$ and  Sobolev's  interpolation,
 \be\label{hih}
 \|u^\omega(t)\|_l\leqslant C_{l,\gamma}\nu^{-l\kappa}, \quad \omega\in Q\cap Q_1, \; t\in J_1,
 \ee
 for $l\in[0,m]$ (and any $\nu\in \{\nu_1,\nu_2,\dots\}$).

 Denote $J_2=[0,\frac{1}{2\nu}]$ and consider the stopping time
 \[
 \tau_1=\inf\{t\in J_2:\|u(t)\|_2\geqslant C_{2,\gamma}\nu^{-2\kappa}\} \le \tfrac{1}{2\nu}.
 \]
 Then  $\tau_1=\frac{1}{2\nu}$ for $\omega\in Q\cap Q_1$. So due to \eqref{HG} with $\Gamma=C_{2,\gamma}\nu^{-2\kappa}$, for any $\chi>0$, we have
 \[\begin{split}\mathbb{E}\big(\nu\int_{J_2}\mathbb{I}_{[0,\chi]}(\|u(s)\|_0)ds\mathbb{I}_{Q\cap Q_1}(\omega)\big)&=\mathbb{E}\big(\nu\int_0^{\frac{1}{2\nu}\wedge\tau_1}\mathbb{I}_{[0,\chi]}(\|u(s)\|_0)ds\mathbb{I}_{Q\cap Q_1}(\omega)\big)\\
 &\leqslant \mathbb{E}\big(\nu\int_0^{\frac{1}{2\nu}\wedge\tau_1}\mathbb{I}_{[0,\chi]}(\|u(s)\|_0)ds\big)\leqslant C \nu^{-2\kappa}\chi.
 \end{split}\]
 Consider the event
 \[\Lambda=\{\omega\in Q\cap Q_1:\|u(s)\|_0\leqslant \chi, \;\forall s\in J_2\}.
  \]
Due to the above,  we have,
 \[\mathbb{P}(\Lambda)\leqslant   2 \mathbb{E}\big(\nu\int_{J_2}\mathbb{I}_{[0,\chi]}(\|u(s)\|_0)ds\mathbb{I}_{Q\cap Q_1}(\omega)\big)\leqslant 2{C}\nu^{-2\kappa}\chi.\]
 So $\mathbb{P}(\Lambda)\leqslant\gamma$ if we choose
 \begin{equation}\label{chi}
 \chi=c_3(\gamma)\nu^{2\kappa}, \quad c_3(\gamma)=\gamma(2{C})^{-1}.
 \end{equation}
 Let us set
 \begin{equation}\label{q2-set}
 Q_2=(Q\cap Q_1)\setminus \Lambda, \quad \mathbb{P}(Q_2)\geqslant 3\gamma,
 \end{equation}
 and for  $\chi$ as in \eqref{chi}, consider the stopping time

$$
 \tilde\tau_1=\inf\{t\in J_2:\|u(t)\|_0\geqslant \chi\}.
 $$

 Then $\tilde\tau_1\leqslant \frac{1}{2\nu}$ for all $\omega\in Q_2$.
 Consider the function
$$
v(t,x) := u (\tilde\tau_1+t,x), \quad t \in [0,\tfrac1{2\nu}].
$$
It solves eq.~\eqref{m-eq} with modified Wiener processes and with  initial data $v_0(x) = u^\omega(\tilde\tau_1,x)$, satisfying
\be\label{lower_b}
\| v_0^\omega\|_0 \ge \chi=c\nu^{2\kappa} \quad \text{if}\quad \omega\in  Q_2.
\ee
Now we
introduce another stopping time, in terms of $v(t,x)$:
\[\tau_2=\inf\{t\in[0,\tfrac1{2\nu}]: \|v(t)\|_m\geqslant K\nu^{-m\kappa}\}\le \tfrac1{2\nu}.
 \]
 For $\omega\in  Q_2$,  $\tau_2=\tfrac1{2\nu}$ and in view of \eqref{hih}
\begin{equation}\label{a-bound}
\|v^\omega(t)\|_l\leqslant C_3(\gamma)\nu^{-l\kappa}, \quad t\in[0,\tfrac1{2\nu}],\; l\in[0,m], \quad \forall\, \omega \in Q_2.
\end{equation}

\noindent
 \textbf{Step 1}: Let us estimate from above   the increment $\mathscr{E}(t,x)=|v(t\wedge\tau_2,x)|^2-|v_0(x)|^2$.
  Due to  It\^o's formula, we have that
\[
\begin{split}\mathscr{E}(t,x)&=2\nu\int_{0}^{t\wedge\tau_2}\Big(\langle v(s,x),\Delta v(s,x)\rangle +\sum_{d\in\mathbb{N}^n}b_d^2\varphi^2_d(x)\Big)ds + \sqrt{\nu} \,  M(t,x), \\
& M(t,x) = \int_{0}^{t\wedge \tau_2}\sum_{d\in\mathbb{N}^n}b_d\varphi_d(x)\langle v(s,x),d\beta_d(s)\rangle. \,
\end{split}
\]

We treat $ M$ as a martingale $M(t)$
in the space $H^1$. Since in view of  \eqref{a2} for $0\le s<\tau_2$ we have
$$
\| v(s) \phi_d\|_1 \le C\big( |v(s)|_\infty \| \phi_d\|_1 +  |v(s)\|_1 |\phi_d|_\infty \big) \le
 C(\zeta d+\zeta^{(m-1)/m} \nu^{-\kappa}),

$$
  where $\zeta =  \sup_{0\leqslant s\leqslant \frac{1}{\nu}}|u(s)|_\infty $
(the assertion is empty if $\tau_2=0$), then for any $0<T_*\le \tfrac1{2\nu}$
 \be\label{Ito_int}
 \mathbb{E}\| M(T_*)\|_1^2 \le
 \int_0^{T_*}
  \EE\, \sum_d b_d^2 \| \phi_d v(s)\|_1^2 ds \le C T_* \nu^{-2\kappa},
   \ee
 where we  used that $B_1<\infty$.  So
 by  Doob's inequality
\be\label{Doob1}
\mathbb{P}\Big(\sup_{0\leqslant s\leqslant T_*} \| M(s)\|_1^2\geqslant r^2\Big)\leqslant C T_*r^{-2}  \nu^{-2\kappa},
 \quad \forall r>0.
\ee

Let us choose
$$
T_*=\nu^{-b}, \quad b\in (0, 1),
$$
where $b$ will be  specified  later.
Then
$
1\le T_*\le \tfrac1{2\nu}
$
if $\nu$ is sufficiently small, so due to \eqref{Doob1}
\[
\mathbb{P}( Q_3)\geqslant1-\gamma,  \quad Q_3=\{\sup_{0\leqslant\tau\leqslant T_*}\|  M(\tau)\|_1\leqslant C_4(\gamma)\nu^{-\kappa}\sqrt{T_*}\},
\]
for a suitable $C_4(\gamma)$ (and for $\nu\ll1$);  thus   $ \mathbb{P} (Q_2\cap Q_3) \ge 2\gamma$.
Since $ \| \lan v , \Delta v\ran \|_1 \le C|v|_\infty \| v\|_3$ by \eqref{a3} and
$
 \| \sum_d  b_d \phi_d\|_1 \le C,
$
then in view of \eqref{a-bound}  and the definition of $Q_3$,
\begin{equation}\label{l2-upper}
\|\mathscr{E}^\omega(\tau)\|_1\leqslant C(\gamma)(\nu^{1-3\kappa}T_*+\nu^{\frac{1}{2}-\kappa} T_*^{1/2}), \qquad \forall\,
 \tau\in[0,T_*] , \;\; \forall\,\omega\in Q_2\cap Q_3.
\end{equation}
\smallskip

\noindent
 \textbf{Step 2}: For any $x\in K^n$, denoting
 $
 R(t) = |v(t,x)|^2, \, a(t) = \Delta v(t,x)
 $
 and $\xi(t) = \xi(t,x)$,
 we write the equation for $v(t) := v(t,x)$ as an It\^o process:
 \be\label{v_ito}
 dv(t) = (-i R v +\nu a) dt + \sqrt\nu\, d\xi(t).
 \ee
  Setting $w(t) = e^{i\int_0^t R(s)ds} v(t)$, we observe that $w$ also is an It\^o process, $ w(0) =v_0$  and
 $
 dv = e^{-i\int_0^t R(s)ds} dw -i Rv\, dt.
 $
 From here and \eqref{v_ito},
 $$
 w(t) = v_0 +\nu \int_0^t e^{i\int_0^s R(s')ds'} a(s) ds + \sqrt\nu\, \int_0^t e^{i\int_0^s R(s')ds'}  d \xi(s).
 $$
 So $v(t\wedge \tau_2)= v(t\wedge\tau_2,x)$ can be written as
 \be\label{deco}
v(t\wedge \tau_2,x)=  I_1(t\wedge \tau_2,x) + I_2(t\wedge \tau_2,x)+ I_3(t\wedge \tau_2,x),
\ee
where
\[\begin{split}
I_1(t,x)= e^{-i\int_0^{t}|v(s,x)|^2ds}v_0,\quad
I_2(t,x)=\nu\int_0^{t}e^{-i\int_{s}^{t}|v(s',x)|^2ds'}\Delta v(s,x)ds, \\
I_3(t,x)=
\sqrt{\nu} e^{-i\int_0^{t}|v(s',x)|^2ds'}
\int_0^{t}e^{i\int_0^{s}|v(s',x)|^2ds'}d\xi(s,x) .
\end{split}
\]

Our next goal is to obtain a lower bound for $ \| v(T_*)\|_1$ when $\om \in Q_2\cap Q_3$,
using the above decomposition \eqref{deco}.
\medskip

\noindent
 \textbf{Step 3}: We first deal with the stochastic term    $I_3(t)$.  For $0 \leqslant s\leqslant s_1\leqslant T_*\wedge\tau_2$ we set
 \begin{equation}\label{W-def}
 \begin{split}
 W(s,s_1, x):= \exp( i  \int_{s}^{s_1}|v(s',x)|^2ds'), \quad
 F(s,s_1,x) :=
 \int_{s}^{s_1}|v(s',x)|^2ds';
 \end{split}
 \end{equation}
 then  $ W(s,s_1, x) = \exp \big(i F(s,s_1, x)\big)$. The functions $F$ and $W$ are periodic in $x$, but not odd. Speaking about them
 we understand $\|\cdot\|_m$ as the non-homogeneous Sobolev norm, so
 $
 \| F\|_m^2  =  \| F\|_0^2 +  \|( -\Delta)^{m/2}  F\|_0^2,
 $
 etc.  We  write $I_3$ as
 \be\label{I_3}
 I_3(t) =\sqrt\nu\  \overline W ( 0,t \wedge  \tau_2 ,x) \int_0^{t\wedge \tau_2} W(0,s,x) d\xi(s,x).
 \ee
 \noindent
 In view of \eqref{a1},
  \be\label{esti}
  \| \exp(iF(s,s_1 \cdot))\| _{k} \le C _{k}(1+ |F(s,s_1, \cdot)|_\infty)^{k-1} \| F(s,s_1,\cdot)\| _{k}, \quad k \in \Nn.

  \ee
\noindent

For any $s\in J =[ 0, T_*\wedge \tau_2)$, by \eqref{a2} and
 the definition of $ \tau_2$, we have that $v:=v(s)$ satisfies
\be\label{hoh}
 \| |v|^2\|_{1} \le C |v|_\infty \| v\|_{1}  \le C |v|_\infty \| v\|_0^{1- 1/m} \| v\|_m^{1/m}
 \le  C' |v|_\infty^{2-1/m} \nu^{-\kappa}
\ee
 (this assertion is empty if $ \tau_2=0$ since then $J=\emptyset$).
So for $s,s_1 \in J$,
$$
|F(s,s_1,\cdot)|_\infty\leqslant   |s_1-s| \sup_{s'\in J}|v(s')|^2_\infty,\quad \|F (s,s_1,\cdot)\|_{k}
 \le C  \nu^{ -\kappa k} |s_1-s|\big(\sup_{s' \in J} | v(s')|_\infty\big)^{2-k/m}\;\;
$$
for $k\le m$.
Then, due to \eqref{esti},
   \be\label{w_est}
   \|W(0,s\wedge\tau_2,\cdot)\|_{1} \leqslant  C'  T_* \nu^{-\kappa} (1+ \sup_{s \in J} |v(s)|_\infty^{2} ).
   \ee
     Consider the stochastic integral in \eqref{I_3},
   $$
    N(t,x)=\int_0^{t}W(0,s,x) d\xi(s,x).
   $$
  The process
  $
  t \mapsto W(0,t,x)
  $
  is adapted to the filtration $\{ \cF_t\}$, and
   \[dW(0,t,x)=i|v(t,x)|^2W(0,t,x)dt.\]
   So integrating by parts (see, e.g., \cite[Proposition IV.3.1]{RY}) we re-write $N$ as
   \[N(t,x)=W(0,t,x)\xi(t,x)- i \int_0^t\xi(s,x)  |v(s,x)|^2W(0,s,x)ds,
   \]
   and we see from  \eqref{I_3} that
   \begin{equation}\label{I_3-2}I_3(t)=\sqrt{\nu}\xi(t\wedge\tau_2,x)+ i\sqrt{\nu}\int_0^{t\wedge\tau_2}\xi(s,x)  |v(s,x)|^2W(s,t\wedge\tau_2,x)ds.
   \end{equation}
  Due to \eqref{B_m} and since $B_m<\infty$, the   Wiener process $\xi(t,x)$  satisfies
  \[\mathbb{E}\|\xi(T_*,x)\|_1^2\leqslant CB_1T_*,\]
  and
  \[\mathbb{E}\sup_{0\leqslant t\leqslant T_*}|\xi(t,\cdot)|_\infty\leqslant \sum_{d\in\mathbb{N}^n}b_d(\mathbb{E}\sup_{0\leqslant t\leqslant T_*}|\beta_d(t)\varphi_d|_\infty)\leqslant CB_*\sqrt{T_*},\]
  (we recall that $B_*=\sum_{d\in\mathbb{N}^n}|b_d|<\infty$).
  Therefore,
   \[ \mathbb{P}(Q_4)\geqslant 1-\gamma, \quad Q_4=\{ \sup_{0\leqslant t\leqslant T_*}
   ( \|\xi(t)\|_1\vee |\xi(t)|_\infty)
   \leqslant CT_*^{1/2}\},\]
 with a suitable $C=C(\gamma)$.
Let
$$
\tilde Q=\bigcap_{i=1}^4 Q_i,
$$
then $\PP(\tilde Q)\geqslant\gamma$.  As $\tau_2 = T_*$ for $\omega \in \tilde Q$, then due  to  \eqref{hoh},   \eqref{w_est},
\eqref{I_3-2} and \eqref{a2},
 for $\omega\in \tilde Q$ we have
\begin{equation}\label{i3-upper}
\begin{split}
\sup_{0\leqslant t\leqslant T_*} \| I^\omega_3(t)\|_1 &\le   \sqrt\nu\,  \sup_{0\leqslant t\leqslant T_*}
\Big(\|\xi^\omega(t)\|_1+\int_0^t\|\xi^\omega(s)|v^\omega(s)|^2W^\omega(s,t)\|_1ds\Big)\\
 &\leqslant   {C} T_*^{5/2}\nu^{\frac{1}{2}-\kappa}.
 \end{split}
 \end{equation}

 \noindent
\textbf{Setp 4}: We then consider the term
$
I_{2}= \nu\int_0^{t\wedge \tau_2} \bar W(s, t\wedge \tau_2,x)

\Delta v(s,x)ds
$.
To bound   its $H^1$-norm we need to estimate $\|W\Delta v\|_1$.
Since
\[\| \partial_x^a W\partial_x^bv\|_0\leqslant C \|W\|_3^{1/3}\|v\|_3^{2/3}|v|_\infty^{1/3}\quad \text{if} \;\;
 |a|=1,|b|=2,\]
(see \cite[Proposition  3.6]{Tay}),
we have
\[\|W\Delta v\|_1\leqslant C(\|v\|_3+ \|W\|_3^{1/3}\|v\|_3^{2/3}|v|_\infty^{1/3}).\]
Then in view of \eqref{esti} and \eqref{a-bound},   for $\omega\in\tilde Q$
$$
\|W\Delta v\|_1 \le C\big( \nu^{-3\kappa} +(T_*^3 \nu^{-3\kappa})^{1/3} \nu^{-2\kappa} \big)
 \le C \nu^{-3\kappa} T_*,
$$
and accordingly
\begin{equation}\label{i1-upper}
\sup_{0\leqslant t\leqslant T_*}\| I^\omega_2(t)\|_1\leqslant \nu \sup_{0\leqslant t\leqslant T_*}\!\int_0^{t}\|W^\omega(s,T_*)\Delta v^\omega(s)\|_1ds
\leqslant C \nu^{1-3\kappa}T_*^{2},\;\; \;
\forall\,\omega\in\tilde Q.
\end{equation}
\smallskip

\noindent
{\bf Step 5: } Now we estimate from below  the $H^1$-norm of the term $I^\omega_{1}(T_*,x)$, $\omega\in \tilde  Q$.
Writing it as $
 I^\omega_1(T_*,x) = e^{-i T_*|v_0(x)|^2} e^{ -i \int_0^{T_*}  \mathscr{E} (s,x) ds} v_0(x)
$
wee see that
 \[
\| I^\omega_1(T_*) \|_1\geqslant \|\nabla(\exp(-iT_*|v_0|^2)v_0\|_0-\|\nabla(\exp(-i\int_0^{T_*} \mathscr{E} (s)ds))v_0\|_0-\|v_0\|_1.
\]
This first term on the r.h.s is
\[T_*\|v_0\nabla(|v_0|^2)\|_0=T_*\tfrac{2}{3}\|\nabla|v_0|^3\|_0\geqslant CT_* \||v_0|^3\|_0\geqslant CT_*\|v_0\|_0^3\ge
C T_* \nu^{6\kappa},\quad C>0,
\]
where we have used the fact that $u|_{\partial K^{n}}=0$,   Poincar\'e's inequality and \eqref{lower_b}.

For $\om\in \tilde Q$ and $0\le s\le T_*$, in view of \eqref{l2-upper}, the second term is bounded by
\[
\begin{split}
\|(\int_0^{T_*}\nabla \mathscr{E}(s)ds)v_0\|_0  \leqslant CT_*|v_0|_\infty \sup_{0\leqslant s\leqslant T_*}\|\mathscr{E}(s)\|_1
 \le
{C} T_*( \nu^{1-3\kappa} T_* + \nu^{\frac{1}{2}-\kappa} T_*^{1/2} ).

\end{split}
\]
Therefore, using \eqref{l2-upper}, we get for the term $I^\omega_1(T_*)$ the following lower bound:
 \[
 \|I_{1}^\omega(T_*)\|_1\geqslant C  \Big(\nu^{6\kappa}T_*-  T_*\big(\nu^{1-3\kappa}T_*
+\nu^{\frac{1}{2}-\kappa} T_*^{1/2}\big)

-\nu^{-\kappa} \Big).
\]
Recalling $T_*=\nu^{-b}$ we see that if  we assume that
\begin{equation}\label{restrict1}
\begin{cases}
6\kappa-b<-\kappa,\\
6\kappa -b < 1-3\kappa -2b,

\\
 6\kappa-b <
1/2 -\kappa-\frac{3}{2}b ,
\end{cases}
\end{equation}
then for $\omega\in\tilde Q$,
\begin{equation}\label{i2-below1}\|I_{1}^\omega(T_*)\|_1\geqslant C \nu^{6\kappa}T_*, \quad  C >0,
\end{equation}
provided that $\nu$ is sufficiently small.
\smallskip

\noindent
{\bf Step 6: }Finally, remembering that $\tau_2=T_*$ for $\omega \in \tilde Q$ and
combining the relations  \eqref{i3-upper}, \eqref{i1-upper} and \eqref{i2-below1} to estimate the terms of \eqref{deco}, we see that
for $\omega\in\tilde Q$ we have
\be\label{new_esti}
\|v^\omega(T_*)\|_1\geqslant\|I_{1}^\omega(T_*)\|_1-\|I^\omega_2(T_*)\|_1-\|I^\omega_3(\tau_*)\|_1\geqslant\tfrac{1}{2}C_1
\nu^{6\kappa-b}, \quad  C_1>0,
\ee
if we assume in addition to \eqref{restrict1} that
\begin{equation}\label{restrict2}
6\kappa-b<\frac{1}{2}-\kappa-\frac{5}{2}b,
\end{equation}
and $\nu$ is small.
Note that this relation  implies
the last two  in \eqref{restrict1}.

Combining  \eqref{a-bound} and \eqref{new_esti}
we get that
\begin{equation}\label{restrict3}\nu^{-b+7\kappa}\leqslant C^{-1}_2,
\end{equation}
for all sufficiently small $\nu$.
Thus we  have obtained a contradiction with the existence of the sets $Q^{\nu_j}$ 	as at the beginning of the proof
 if (for a chosen $\kappa$) we can find a $b\in(0,1)$ which
  meets \eqref{restrict1},  \eqref{restrict2} and
$$
-b+7\kappa<0.
$$
Noting that this is nothing but the first relation in  \eqref{restrict1}, we see that  we have obtained a contradiction if
$$
\kappa< \tfrac17 b, \quad  \kappa< \tfrac1{14} -\tfrac3{14} b,
$$
for some $b\in(0,1)$.
We see immediately that such a $b$ exists if  and only if $\kappa < \tfrac1{35} $.

\end{proof}

\noindent
{\bf Amplification.} If we replace the condition $m\geqslant3$ with the weaker assumption
\[
\R\ni  m>2,
\]
then the  statement \eqref{1.0_sobolev} remains true for $0<\kappa<\kappa(n,m)$ with a suitable (less explicit) constant
$\kappa(n,m)>0$. In this case  we obtain a contradiction with the assumption \eqref{contra}
 by deriving a lower bound for $\|v(T_*)\|_\alpha$,  where
$\alpha=\min\{1,m-2\}$, using the decomposition \eqref{deco}.
The proof remains almost identical except that  now, firstly, we

 bound $\|I_{2}\|_\alpha$ ($\alpha<1$) from above
using the following estimate from \cite[Theorem 5, p.~206]{RS} (also see there p.~14):
\[\|W\Delta u\|_\alpha\leqslant C\|u\|_{2+\alpha}(|W|_\infty+|W|_\infty^{1-\frac{2\alpha}{n}}\|W\|_2^{\frac{2\alpha}{n}});\]
and, secondly,    estimate $\|I_{1}^\omega(T_*)\|_\alpha$ ($\alpha<1$) from below as
\[
\|I_{1}^\omega(T_*)\|_\alpha\geqslant {\|I_{1}^\omega(T_*)\|_1^{2-\alpha}} \,{\|I^\omega_2(T_*)\|_2^{-1+\alpha}},
\]
which directly follows from Sobolev's  interpolation. The lower bound for $\|I^\omega_2(T_*)\|_1$ in \eqref{i2-below1} stays valid,  so to bound $\|I_2^\omega(T_*)\|_\alpha$ $(\alpha<1)$ from below, we just need to estimate the upper bound of $\|I_2^\omega(T_*)\|_2$.  Since  $I^\omega_2(T_*,x)=\overline W(0,T_*,x)v_0(x)$ (see \eqref{W-def}),
then in view of \eqref{a2}  we have
\[\|I_2^\omega(T_*)\|_2\leqslant C (|\overline W(T_*)|_{\infty}\|v_0\|_2+\|\overline W(T_*)\|_2|v_0|_\infty).\]
So by \eqref{w_est},
$
\|I_2^\omega(T_*)\|_2\leqslant C_{8,\gamma}\nu^{-2\kappa}T_*^2.
$
Therefore,
\begin{equation}\label{i2-below}
\|I_2^\omega(T_*)\|_\alpha\geqslant C_{9,\gamma}\frac{(\nu^{6\kappa}T_*)^{2-\alpha}}{(\nu^{-2\kappa}T_*^2)^{1-\alpha)}}\geqslant C_{9,\gamma}\nu^{(14-8\alpha)\kappa-b\alpha}.\end{equation}
We then can complete the proof by an argument,  similar to that at  Step 6.

\begin{remark}\label{stationary-lower} If  eq. \eqref{m-eq} is mixing and $u(t)$ is a stationary solution, then if $\mathbb{E}\|u(t)\|_2\leqslant \nu^{-2\kappa}$, we have for $m\geqslant3$,
\begin{equation}\label{lower-stat1}\mathbb{P}\Big\{\sup_{T_0\leqslant t\leqslant T_0+T_1\nu^{-7.001\kappa}}\|u(t)\|_m\geqslant K\nu^{-m\kappa}\Big\}\to1\quad\text{as}\quad \nu\to0.\end{equation}
Indeed, due to \eqref{stat2}, in the stationary case, we can choose $Q_2$ as in \eqref{q2-set} such that for the stopping time $\tilde \tau_1$ defined in \eqref{stop-stat}, $\tilde\tau_1(\omega)=0$, if $\omega\in Q_2$. Then the same argument gives the above assertion.
\end{remark}

\medskip

 \section{Lower bounds for time-averaged Sobolev norms}\label{proof-section-2}
In this section we prove the assertion (2)  of Theorem \ref{m-theorem}.  We provide a space $H^r$, $r\ge0$, with the scalar product
 $$
 \llangle u,v\rrangle_r:=\llangle (-\Delta)^{\frac{r}{2}}u,(-\Delta)^{\frac{r}{2}}v\rrangle,
 $$

 corresponding to the norm $\|u\|_r$.   Let
$
u(t) =\sum u_d(t) \phi_d
$
 be a solution of eq.~\eqref{m-eq}.  Applying It\^{o}'s formula to the functional $\|u\|_m^2$,
 we have for any $0\leqslant t<t'<\infty$ the relation
\begin{equation}\label{balance1}\begin{split}\|u(t')\|_m^2=&\|u(t)\|_m^2+2\int_{t}^{t'}\llangle u(s), \nu \Delta u(s) -
i|u(s)|^2u(s)\rrangle_mds\\
&+2\nu B_m(t'-t)+2\sqrt{\nu} M(t,t'),
\end{split}\end{equation}
where $M$ is the stochastic integral
\[M(t,t'):= \int_t^{t'}\sum_{d\in\mathbb{N}^n} b_d|d|^{2m} \langle u_d(s), d\beta_d(s)\rangle .

\]
Let us fix a $\gamma\in(0,\frac{1}{8})$.  Due to Theorem \ref{sup-norm} and  \ref{t_sobolev}, for small enough $\nu$ there exists an event
$\Omega_1\subset\Omega$,
$
\PP(\Omega_1)\geqslant1-\gamma/2,
$
such that  for all $\om \in \Omega_1$ we have:

a) $\sup_{0\leqslant t\leqslant\frac{1}{\nu}}|u^\omega(t)|_\infty\leqslant C(\gamma)$, for a suitable $C(\gamma)>0$;

b) there exist  $t_\omega\in[0,\frac{1}{3\nu}]$ and  $t_\omega'\in[\frac{2}{3\nu},\frac{1}{\nu}]$ satisfying
\be\label{relation}
\|u^\omega(t_\omega)\|_m,\;\|u^\omega(t_\omega')\|_m\geqslant \nu^{-m\kappa}.
\ee

Since for  the martingale $M(0,t)$  we have that
$$
\mathbb{E}|M(0,\tfrac{1}{\nu})|^2\leqslant B_m \mathbb{E}\int_0^\frac{1}{\nu}  \|u(s)\|_m^2ds =: X_m,
$$
then by Doob's inequality
\[
\PP(\Omega_2)\geqslant1-\frac{\gamma}{2}, \qquad
\Omega_2=\Big\{ \sup_{0\leqslant t\leqslant\frac{1}{\nu}}|M(0,t)|\leqslant c(\gamma)
X_m^{1/2}

\Big\}.
 \]
Now let us set
$\ \hat\Omega=\Omega_1\cap\Omega_2.$
Then
$\PP(\hat\Omega)\geqslant1-\gamma$ for small  enough $\nu$, and for
 any  $\omega\in \hat\Omega$ there are two alternatives:

i) there exists a $t_\omega^0\in[0,\frac{1}{3\nu}]$ such that $\|u^\omega(t^0_\omega)\|_m=\frac{1}{3}\nu^{-\kappa m}$.  Then
from \eqref{balance1} and \eqref{relation} in view of \eqref{a4} we get

\[\begin{split}
  \frac{8}{9}\nu^{-2m\kappa}+2\nu \int_{t_\omega^0}^{t'_\omega}\|u^\omega(s)\|_{m+1}^2ds
\leqslant C(m,\gamma)  \int_0^{\frac{1}{\nu}} \|u^\omega(s)\|_m^2ds+2B_m+2\sqrt{\nu}c(\gamma)
X_m^{1/2}.
\end{split}\]

ii) There  exists no $t\in[0,\frac{1}{3\nu}]$ with $\|u^\omega(t)\|_m=\frac{1}{3}\nu^{-\kappa m}$. In this case,
since $\|u^\omega(t)\|_m$ is continuous with respect to $t$, then due to \eqref{relation}
$
\|u^\omega(t)\|_m > \frac{1}{3}\nu^{-m\kappa}
$
for all
$ t\in[0,\frac{1}{3\nu}]$,
which  leads to the relation
\[
\tfrac{1}{27}\nu^{-2m\kappa-1}\leqslant \int_0^{\frac{1}{\nu}}\|u^\omega(s)\|_m^2ds.
\]

In both cases  for $\omega\in\hat\Omega$ we have:
\[\frac{1}{27}\nu^{-2m\kappa}\leqslant C'(m,\gamma)  \int_0^{\frac{1}{\nu}}
\|u(s)\|_m^2ds+2B_m
+{\nu} c(\gamma)^2 + X_m.
\]

This implies that
\[\mathbb{E}\nu\int_0^{\frac{1}{\nu}}\|u(\tau)\|_m^2d\tau\geqslant C\nu^{-2m\kappa+1}
\]
(for small enough $\nu$), and
 gives the lower bound in \eqref{avb}. The upper bound follows directly from Theorem~\ref{upper-bound}.
 \medskip

  \noindent
  {\it Proof of Corollaries \ref{c_3} and \ref{c_4}}:
  Since $B_k<\infty$ for each $k$ and all coefficients $b_d$ are non-zero, then eq.~\eqref{m-eq} is
   mixing in the spaces $H^M$,  see
Theorem~\ref{mix-measure}.  As the stationary solution $v^{st} $ satisfies Corollary~\ref{cor-sup} with any $m$, then for
each $\mu\in \Nn$ and $M>0$, interpolating the norm $\|u\|_\mu$ via $\| u\|_0$ and $\| u\|_m$ with $m$ sufficiently
large

 we get that the stationary measure $\mu_\nu$  satisfies
\be\label{mu_nu}
\int \| u\|_\mu^M \mu_\nu(du) < \infty \quad \forall\, \mu\in \Nn, \;\forall M>0.
\ee
Similar, in view of \eqref{m_est} and Theorem  \ref{sup-norm},
\be\label{mu_int}
\EE \| u(t;u_0)\|_\mu^M \le C_\nu(u_0) \quad \forall t\ge0,
\ee
for each $u_0\in C^\infty$ and every $\mu$ and $M$ as in \eqref{mu_nu}.
 Now let us consider the integral in \eqref{avb} and write it as
$$
J_t := \nu \int_t^{t+\nu^{-1}} \EE \| u(s)\|_m^2ds.
$$
Replacing the integrand in $J_t$ with $\EE( \| u_\nu(s)\|_m \wedge N)^2$, $ N\ge1$, using the convergence
\be\label{weak}
\EE\big(  \|u(s;v_0) \|_m  \wedge N\big)^2 \to \int \big(  \|u \|_m  \wedge N\big)^2 \mu_\nu(du)
\quad \text{as} \quad s\to\infty \quad \forall\,N,
\ee
which follows from Corollary \ref{sob_mix},  and the  estimates  \eqref{mu_nu},  \eqref{mu_int} we get that
\begin{equation}\label{weak2}
J_t \, \rightarrow\, \int \| u\|_m^2 \,\mu_\nu(du)\quad \text{as} \quad  t\to\infty.
\end{equation}
This convergence and \eqref{avb} imply the assertion of  Corollary~\ref{c_3}.
\smallskip

Now convergence  \eqref{weak} jointly with estimates   \eqref{mu_nu},  \eqref{mu_int}  and \eqref{conver} imply  Corollary~\ref{c_4}.

\appendix
\section{Some estimates}\label{section-appendix}

    For any integer $l\in \Nn$ and $F\in H^l$
    we have that
  \be\label{a1}
  \| \exp(iF(x))\|_{l} \le C_{l} (1+ |F|_\infty)^{l-1} \| F\|_{l}.
  \ee
\noindent
Indeed, to verify \eqref{a1} it suffices to check that for any non-zero  multi-indices $\beta_1,\dots,\beta_{l'}$, where $1\le l'\leqslant l$ and
 $
 |\beta_1| +\cdots+ |\beta_{l'}| =l,
 $
we have
   \be\label{a3}
   \| \p_x^{\beta_1} F\cdots \p_x^{\beta_{l'}}  F
 \|_0 \le C  |F|_\infty^{l'-1} \| F\|_l.
 \ee
 But this is the assertion of Lemma 3.10 in \cite{Tay}. Similarly,
 \be\label{a2}
\|F G \|_r
\leqslant C_r ( |F|_\infty\|G\|_r + |G|_\infty\|F\|_r), \quad F,G \in H^r,\;\; r\in \Nn,
\ee
see \cite[Proposition 3.7]{Tay} (this relation is known as Moser's estimate).  Finally, since for $| \beta| \le m$ we have
$
| \p_x^\beta v|_{2m/ \beta|} \le C |v|_\infty^{1- |\beta| /m} \| v\|_m^{|\beta| /m}
$
(see relation (3.17) in \cite{Tay}), then
\be\label{a4}
|\llangle |v|^2 v, v\rrangle_m| \le C_m \|v\|_m^2 |v|_\infty^2, \qquad
|\llangle |v|^2 v, v\rrangle_m| \le C'_m \|v\|_{m+1}^{\frac{2m}{m+1}}  |v|_\infty^{\frac{2m+4}{m+1}} .
\ee

\section{Proof of Theorem \ref{upper-bound}}\label{app_B}
Applying  Ito's formula to a solution $v(\tau)$ of eq. \eqref{the_eq} we get a slow time  version of the relation \eqref{balance1}:
\begin{equation}\label{b10}
\begin{split}
\|v(\tau )\|_m^2=\|v_0\|_m^2+2 \int_{0}^{\tau } \big( -\| v \|^2_{m+1} - \nu^{-1} \llangle
i|v |^2v , v \rrangle_m \big)ds
+2 B_m \tau+2 M(\tau ),
\end{split}\end{equation}
where
$M(\tau )= \int_0^{\tau }\sum_{d} b_d|d|^{2m} \langle v_d(s), d\beta_d(s)\rangle .$ Since in view of \eqref{a4}
$$
 \EE \big| \llangle  |v|^2v, v\rrangle_m \big| \le
  C_m \big( \EE \| v\|_{m+1}^{2}\big)^{\frac{m}{m+1}}  \EE\big( |v|_\infty^{2m+4}\big)^{\frac{1}{m+1}},
$$
then denoting
$
\EE \| v(\tau)\|_r^2 =: g_r(\tau), \ r \in \Nn \cup\{0\},
$
 taking expectation of \eqref{b10},  differentiating the result and using \eqref{main_esti}, we get that
\be\label{b1}
\frac{d}{d\tau} g_m \le -2 g_{m+1} +C_m \nu^{-1} g_{m+1}^{\frac{m}{m+1}} +2B_m \le -2 g_{m+1}\big( 1- C'_m  \nu^{-1} g_{m}^{-\frac{1}{m}} +2B_m\big),
\ee
since
$
g_m \le g_0^{1/(m+1)}  g_{m+1}^{m/(m+1)}  \le C_m g_{m+1}^{m/(m+1)} .
$
We see that if $g_m \ge (2\nu^{-1} C'_m)^m$, then the r.h.s. of \eqref{b1} is
\be\label{b2}
\le -g_{m+1} +2B_m \le -C_m^{-1} g_m^{(m+1)/m} +2B_m  \le - \bar C_m \nu^{-m-1} + 2B_m,
\ee
which is negative if $\nu\ll1$. So if
\be\label{b3}
g_m(\tau) <  (2\nu^{-1} C'_m)^m
\ee
at $\tau=0$, then \eqref{b3} holds for all $\tau\ge0$ and \eqref{sob_esti} follows. If  $g_m(0)$ violates \eqref{b3}, then
 in view of \eqref{b1} and \eqref{b2},
for $\tau\ge0$,  while \eqref{b3} is false,  we have that
$$
\frac{d}{d\tau} g_m \le -C_m  g_{m}^{(m+1)/m} +2B_m,
$$
which again implies \eqref{sob_esti} (see details of this argument  in the proof of Theorem 2.2.1 in \cite{BK}).
Note that in view of \eqref{b1},
$$
\frac{d}{d\tau} g_m \le - g_{m} +C_m (\nu, |v_0|_\infty, B_{m_*},B_m).
$$
This relation immediately implies \eqref{m_est}.

\medskip
Now let us return to eq. \eqref{b10}. Using Doob's inequality and   \eqref{sob_esti} we find that
$$
\EE(\sup_{0\le \tau\le T} |M(\tau)|^2 ) \le C<\infty.
$$
Next, applying \eqref{a4} and Young's inequality we get that
$$
\int_{0}^{\tau } \big( -\| v \|^2_{m+1} - \nu^{-1} \llangle i|v |^2v , v \rrangle_m \big)ds \le
C_m \int_{0}^{\tau } | v(s)|_\infty^{2m+3} ds, \quad \forall \ 0\le\tau\le T.
$$
 Finally,
using in \eqref{b10} the last two displayed formulas jointly with \eqref{main_esti} we obtain \eqref{m_est1}.

\section{Lower bound for  $C^m$-norms of solutions}\label{appen-cm}
In this appendix we work with eq. \eqref{the_eq}, in the time scale $\tau$. Our goal  is to prove the following result:

\begin{theorem} If  $m\geq 2$ is an integer
 and $\kam<\tfrac1{16}$, then for any $\tau_0\ge0$ and $\tau'>0$, every  solution $u(\tau, x)$
of \eqref{the_eq} with a smooth initial data $u_0(x)$
 satisfies
\be\label{1.0}
\PP \Big\{ \sup_{\tau_0 \le\tau \le\tau_0+\tau'}  | u(\tau)|_{C^m} \ge K \nu^{-m\kam}
\Big\} \to 1\quad \text{as} \quad \nu\to0,
\ee
for each $K>0$.
 \end{theorem}

\begin{proof}
{\bf Step 1}  (preliminaries):
Consider the complement to the event in \eqref{1.0}:
$$
Q=Q^\nu =\big\{ \sup_{\tau_0 \le\tau \le\tau_0+\tau'}  | u(\tau)|_{C^m} < K \nu^{-m \kam}  \big\} .
$$
To prove  \eqref{1.0} we assume that there exists a $\gamma>0$ and a sequence $\nu_j\to0$ such that
\be\label{1.1}
\PP (Q^{\nu_j} ) \ge 5\gamma \quad \text{for} \quad
 \, j =1,2, \dots,
\ee
and will derive a contradiction. Below we write $Q^{\nu_j}$ as $Q$ and always suppose that
$$
\nu \in \{\nu_1, \nu_2, \dots\}.
$$
Without lost of generality we assume that $\tau_0=0$ and $\tau'=1$. The constants below may depend on $K$, on the norms $|u_0|_\infty$ and $\| u_0\|_m$,
 but not on $\nu$.

Let us denote $J_1 = [0, 1]$. Due to Theorem \ref{sup-norm},
$$
\PP (Q_1) \ge 1-\gamma, \qquad Q_1 = \{\sup_{ \tau\in J_1} | u(t)\Co \le C_1(\gamma) \}
$$
uniformly in  $\nu$, for a suitable $C_1(\g)$. Then, due to the Hadamard-Landau-Kolmogorov interpolation
inequality,
\be\label{1.2}
| u(t) \Cm \le  C_\mu(\gamma) \nu^{-\mu \kam}\quad \forall\, \omega \in Q\cap Q_1,\; \; t\in J_1,
\ee
for any integer $\mu \in [0,m]$ (and any $\nu \in \{\nu_1, \nu_2, \dots\}$).

Denote $J_2 =[ 0,\frac{1}{2}]$ and consider the stopping time
\be\label{t1}
\tau_1 = \inf\{ \tau\in J_2: |u(\tau)\Cdv \ge   {K}  \nu^{-2 \kam}\}.
\ee
Then $\tau_1\leqslant \frac{1}{2}$ and $ \tau_1=\frac{1}{2}$ for  $\omega \in Q\cap Q_1$. So due to \eqref{HG} with $\Gamma= {K}  \nu^{-2 \kam}$, for any $\chi>0$ we have
\[\begin{split}
\EE\big( \!\int_{J_2} \mathbb{I}_{[0, \chi]} (\| u(s)\|_0) ds\, \mathbb{I}_{Q\cap Q_1} (\omega) \big)
= \EE\big( \!\int_0^{\min(\frac{1}{2}, \tau_1)}  \mathbb{I}_{[0, \chi]} (\| u(s)\|_0) ds\, \mathbb{I}_{Q\cap Q_1} (\omega) \big)\\
\le  \EE  \!\int_0^{\min(\frac{1}{2}, \tau_1)}  \mathbb{I}_{[0, \chi]} (\| u(s)\|_0) ds
\le  \EE  \!\int_0^{\min(\frac{1}{2}, \tau_\Gamma)}  \mathbb{I}_{[0, \chi]} (\| u(s)\|_0) ds \le  C {K}  \nu^{-2 \kam}\, \chi.
\end{split}\]
Consider the event
$$
\Lambda = \{ \omega \in Q\cap Q_1: \| u(s)\|_0 \le \chi  \;\; \forall \, s\in J_2\}.
$$
Due to the above,
$$
\PP(\Lambda ) \le  2\EE\big( \!\int_{J_2} \mathbb{I}_{[0, \chi]} (\| u(s)\|_0) ds\, \mathbb{I}_{Q\cap Q_1} (\omega) \big) \le 2 C {K}  \nu^{-2 \kam}\, \chi.
$$
So $\PP(\Lambda ) \le \g$ if we choose
\be\label{chi-a}
\chi = c_3(\gamma)  \nu^{2 \kam}, \qquad
 c_3(\gamma) =  \g
 (2C {K})^{-1} .
\ee
Consider the event
$$
Q_2 = (Q \cap Q_1) \setminus \Lambda, \qquad \PP(Q_2) \ge 3\gamma.
$$

For $\chi$ as in \eqref{chi-a} consider the stopping time
\begin{equation}\label{stop-l2}
\tau_2 = \inf\{ \tau \in J_1:\| u(s)\|_0 \ge \chi\}.
\end{equation}
Then $\tau_2 \leqslant\frac{1}{2}$ for all $\omega \in Q_2$. Now we consider the function
$$
v(\tau,x) := u^\omega (\tau_2+\tau,x), \quad \tau \in J_2 =[0,\tfrac{1}{2}].
$$
It satisfies the equation \eqref{the_eq} with modified Wiener processes and with  initial data $v_0(x) = u^\omega(\tau_2,x)$.
\medskip

 {\bf Step 2}  (the radius-function $|v|(\tau, x)$):
 If $\om\in Q_2$, then
$
| v_0(\cdot)|_\infty \ge \chi.
$
We define the stopping time $\tau_1\in J_2$ by relation \eqref{t1} with $u$ replaced by $v$. Then $\tau_1\leqslant\frac{1}{2}$ and $\tau_1 =\frac{1}{2} $ for $\om\in Q_2$.

Since $v_0(0)=0$, we can find a point $x_0\in K^n\subset \R^n$ such that $| v_0(x_0)|=\chi$. Considering the ray $R$ in $\R^n$
through this point,
$R:= \R_+  x_0 \sim \R_+$,  we, firstly, find there the smallest point $x_2$
where  $| v_0(x_2)|=\chi$, and, secondly, find on $[0,x_2]$ the biggest point $x_1$ such that  $| v_0(x_1)|= \tfrac12\chi$. We are interested in the behaviour of $v^\om(t,x)$ for $x$ in  the segment
$$
L=L(v_0^\om) = [x_1^\om,x_2^\om] \subset \{ x: \tfrac12 \chi \le |v_0(x)| \le \chi\}\cap K^n,
 \qquad  L\subset R.
$$
We will study this behaviour for $\tau$ from a time-interval $J_2 =[0, \tau_*]$ such that there  still
\be\label{still}
 \tfrac14 \chi \le |v^\om(\tau,x)\!\mid_{L}|  \le 2\chi \quad \text{for} \quad x\in L.
 \ee

 Applying Ito's formula to $I(\tau, x) =
 | v( \tau\wedge \tau_1,x)|^2$ we have
 \be\label{ito_I}
 \begin{split}
 I(\tau,x) - I(0,x) +\int_0^\tau \mathbb{I}_{\{ s\le\tau_1\}}
 \big[ -2\langle \Delta \vs, \vs\rangle +2 \sum_d b_d^2 \| \phi_d(x)|^2\big] ds\\
 =2 \int_0^\tau \mathbb{I}_{\{ s\le\tau_1\}}  \sum b_d \, \phi_d(x) \langle  \vs , d\tilde\beta_d(s)\rangle =: M(\tau,x).
 \end{split}
 \ee
 Now we extend the segment $L=[x_1,x_2] \subset R\sim \R_+$ to the segment
 $$
 L^+ =[x_1, \max(x_2,x_1+1)] \subset \R_+, \qquad 1\le |L^+| \le \sqrt{n\pi},
 $$
 and
 consider the space
 $\cH^\rho =H^\rho(L^+)$, $1/2< \rho\le1$. We will  regard $I$ as a semimartingale  $I(\tau)\in \cH^\rho$ and $M$ -- as a martingale in $\cH^\rho$.  As
 $$
|     \phi_d(\cdot ) v(s,\cdot)  |_{\cH^\rho} \le C_\rho | \phi_d  |_{C^1} | v(s,\cdot) |_{\cH^\rho} ,

$$
and since for $s\le \tau_1$
$$
 | v(s,\cdot) |_{\cH^\rho} \le  \| v(s,\cdot) \|_{L_2(L^+)}^{1-\rho}  \| v(s,\cdot) \|_{H^1(L^+)}^\rho \le C\,
   | v(s,\cdot) |_{\infty}^{1-\rho}  \| v(s,\cdot) \|_{C^1(K^n)}^\rho  \le C|v(s,\cdot)|_\infty^{1-\frac{\rho}{m}} \nu^{-\rho\kappa},
$$
noting that
\[\sup_{0\leqslant \tau\leqslant\frac{1}{2}}|v(\tau)|_\infty\leqslant \sup_{0\leqslant\tau\leqslant 1}|u(\tau)|_\infty,\]
 then in view of Theorem \ref{sup-norm} we have
\be\label{Doob2}
\EE | M(\tau)|_{\cH^\rho} ^2
\le C_{\rho } \int_0^\tau  \nu^{-2\rho\kappa}  \sum b_d^2\, |d|^2ds
 =  C_{\rho} B_{1} \nu^{-2\rho\kappa} \tau.
\ee
So by Doob's inequality,
$$
\PP\left( \sup_{0\le \tau \le\tau_*}  | M(\tau)|_{\cH^\rho} ^2\ge r^2\right)
\le { C'_{\rho} \nu^{-2 \rho\kappa} \tau_*}\, {r^{-2}} .
$$
Since $\cH \subset C^0(L^+)$, then
\be\label{Q3}
\PP(Q_3) \ge1-\gamma, \qquad Q_3= \{  \sup_{0\le \tau \le\tau_*}  | M(\tau)|_{C^0(L^+) } \le C^3(\g) \nu^{-\rho\kappa} \sqrt{\tau_*} \,\},
 \ee
 for a suitable $C^3(\g)$ (depending on $\rho$).

Now let us choose
 \be\label{tau_st}
\tau_* = c^4(\g) \nu^{2(4+\rho)\kappa}<\frac{1}{2}.
 \ee
 Then from \eqref{ito_I} and \eqref{1.2},
 \be\label{I_est}
 | I(\tau,x) - I(0,x) |  \le C_\g (\tau_* \nu^{-2\kam} + \nu^{-\rho\kam}  \sqrt{\tau_*} ) \le \tfrac1{8} \chi^2, \quad \tau\in J_3:=[0, \tau_*], \; x\in L,
  \ee
  for $\om \in Q_2\cap Q_3$,
if $c^4(\g) \ll 1$.  This implies \eqref{still} for $\om\in Q_2\cap Q_3$ with $\tau_*$ as in \eqref{tau_st}, since  $|v_0| \ge \tfrac12 \chi$ on $L$.
\medskip

 {\bf Step 3}  (the  angles):  Finally
 we examine the behaviour of the angles Arg$\,v^\om(\tau,x)$ for $x$ in $L(v_0)$. To do this we consider the angle-function on the   annulus
$$
Ann = \{z: \tfrac14 \chi \le |z| \le 2 \chi\},
$$
i.e.
$$
\Phi: Ann \mapsto S^1, \quad z \mapsto \text{Arg} \,z,
$$
and define the stopping time
$\
\tau_x = \inf\{ \tau \ge 0: v^\om(\tau,x) \in \overline{\C\setminus  Ann }\}.
$
Then $\tau_x=0$  if $x=0$ and $\tau_x \ge \tau_*$ if $x\in L(v_0)$ and $\om \in Q_2\cap Q_3$. For a fixed ``past" $v_0$ and $x \in L(v_0)$ let us consider the
random process
$
\phi_x^\om(\tau) =$ Arg$\, v^\om(\tau\wedge  \tau_x,x).
$
Applying  to it Ito's formula, we get:

  \begin{equation}\label{ito-phi}
 \begin{split}
 \phi_x&(\tau) -  \phi_x(0) + \nu^{-1}\int_0^{\tau\wedge \tau_x} d\Phi (\vs) (i|v|^2\vs)ds\\
  =&\int_0^{\tau\wedge \tau_x} \Big(d\Phi (\vs) (\Delta \vs )
   +  \sum_d b_d^2 d^2\Phi(\vs) (\phi_d(x), \phi_d(x))\Big)ds\\
   + &\int_0^{\tau\wedge \tau_x} \sum_d b_d
   d\Phi(\vs)  (\phi_d(x) d\tilde\beta_d(s)), \quad \tau\ge0,\;\;  x \in L(v_0).

 \end{split}
  \end{equation}

 Let us denote the stochastic integral in the r.h.s. as $N_x(\tau)$. It is convenient to regard $\phi_x(\tau)$ as a point in the real line rather than in $S^1$. To
 do that, if $\tau_x=0$ for some $x\in L$,
 we take $\phi_x^\om(0)  \in [0,2\pi)$. Otherwise

   we take for  $\phi_{x_1}^\om(0) $ the value of Arg$\, v^\om_0(x) \in [0,2\pi)$, continuously  extend it to a function $\phi_{x}^\om(0) $
   on $L$, and then
  construct  $\phi_x^\om(\tau) $  from \eqref{ito-phi} by continuity.

Since
$$
|d \Phi(z)| \le C\chi^{-1}, \quad |d^2 \Phi(z)| \le C\chi^{-2} \quad \forall\, z\in Ann,
$$
then for $\om\in Q_2\cap Q_3$ and for $\tau \le\tau_*$ the sum of the two deterministic integrals in the r.h.s. of \eqref{ito-phi}  is bounded by
$$
C_\g \tau_*\big[ \chi^{-1} \nu^{-2\kam} + \chi^{-2}\big] \le C_\g \nu^{(4+2\rho)\kappa}.
$$

Now consider the stochastic integral   $N_x(\tau)$. Since
$
\EE | d\Phi(\vs) (\phi_d)|^2 \le C\chi^{-2},
$
then
$
\EE | N_x(\tau)|^2 \le C_\gamma \chi^{-2} \tau.
$
So for any $x\in L$,
$$
\PP( Q_4^x) \ge 1- \g/2, \quad Q_4^x = \{ \sup_{0\le \tau \le\tau_*} |N_x(\tau)| \le C^4_\g \chi^{-1}  \sqrt{\tau_*} \},
$$
for a suitable $C^4_\g$.  Let us define
$
Q_4 =Q_4^{x_1} \cap Q_4^{x_2}
$
and consider
$$
\hat Q = Q_2 \cap Q_3\cap Q_4, \qquad \PP \hat Q \ge \g.
$$
For any $\om\in \hat Q$ and  $j=1,2$
 we have  $\tau_{x_j} \ge \tau_*$,  and
$$
d\Phi (v(s,x_j))( i|v|^2 \vs) =| v(s,x_j)|^2,
$$
for $0\le s \le \tau_*$. Due to \eqref{I_est},
$
| v(s,x_1)|^2 \ge \frac{7}{8}\chi^2 $ and $  | v(s,x_2)|^2 \le \frac{5}{8}\,  \chi.
$
Therefore in view of  the Ito's formula for $\phi_x$,  for any $\om\in\hat Q$,
\[\begin{split}
\phi_{x_1}(\tau_*)& \ge \nu^{-1} \tau_* \tfrac78\chi^2\, - \, C_\g( \nu^{(4+2\rho)\kam} + \chi^{-1}\sqrt{\tau_*} )\\
&\ge  \tfrac78 c^4(\g) C_3(\g)
 \nu^{-1 +\kam (12+2\rho )} - C_\g (\nu^{(4+2\rho)\kam}  + \nu^{\kam(2+\rho) } );
 \\
\phi_{x_2}(\tau_*)& \le
  \tfrac58 c^4(\g) C_3(\g)  \nu^{-1 +\kam (12+2\rho )}
   +C_\g \nu^{\rho\kam}.

\end{split}\]
Since $\kam<1/16$, then choosing $\rho$ close to $1/2$  we achieve that
\be\label{lastest}
\phi_{x_1}(\tau_*) - \phi_{x_2}(\tau_*) \ge C(\g)  \nu^{-1 +\kam (12+2\rho)} ,
\ee
if $\nu\ll1$. From the other hand, since for $\om\in \hat Q$,
$
|v(\tau_*,x)\Cod  \le K \nu^{-\kam}
$
and
$| v(\tau_*,\cdot)|$
 is
$\ge \tfrac14 \chi$ on the segment $L$, then
$$
| \nabla \text{Arg}\, v(\tau_*,\cdot) \!\mid_{L} | \le 4K \nu^{-\kam} \chi^{-1} = C(\g) \nu^{-3\kam}.
$$
So for any $\om\in \hat Q$ we must have
$
|\phi_{x_1} (\tau_*) - \phi_{x_2} (\tau_*) | \le C(\gamma) \nu^{-3\kam}.
$
Combining here and \eqref{lastest}, we obtain
$$
\nu^{-1 +\kam(15+2\rho)} \le C(\g).
$$
As for $\rho$ we can take any number $>1/2$ and the last  relation  holds  for arbitrarily small $\nu$, then
$
\kam \ge \frac1{16}.
$
This conclusion has been obtained for any $\om$ from the event
$\hat Q$, where  $\PP\hat Q>\g$. The obtained contradiction with the assumption of the theorem proves the assertion \eqref{1.0}.
\end{proof}

\section*{Acknowledgment}
GH is supported by NSFC (Significant project No.11790273) in China and SK  thanks the {\it Russian Science Foundation\/} for support through the grant  18-11-00032.

\bibliography{reference}{}
\bibliographystyle{plain}

\end{document}


TITLE: standing waves on a cylindrical jet
QUESTION [4 upvotes]: as we know, there are some perturbations on a falling jet which are always present and according to their wave number and the radius of the jet, they can grow and decay over time. so, imagine a jet which has those sort of long wavelength waves traveling downward on it, we put an obstacle on the path of the jet and then we would see this :

so, what exactly happens to those waves by the presence of an obstacle?
why have those waves been too larger in amplitude and too short in wavelength in presence of an obstacle?
This photo has taked by sayyedali.m.(  Qom)in Shahid qodosi high school

REPLY [3 votes]: You're seeing the effects of a capillary instability of the jet, known as the Rayleigh instability (sometimes also as the Rayleigh-Plateau instability) which is largely driven by surface tension effects. The technical details are nontrivial, however. A rough overview of the phenomenon is on Wikipedia, a detailed account is given in this book chapter on Springer Link (it looks like the chapter is freely available for download).
Notice that the instability is characterized by a broad spectrum of unstable wavenumbers, see this figure:

Thus the specific wavenumber you observe can be significantly affected by initial and boundary conditions. The waves in the image you show are unusually short for this instability (maximum amplification for the inviscid case is for a wavelength of roughly 4.5 times the diameter of the jet; this increases with viscosity). My guess is that in this case surface waves at the point of impact of the jet excite shorter-wavelength disturbances.
P.S.: An extensive discussion of the physics of liquid jets is here, including downloadable PDF article.


TITLE: Must embedding of $X$ in $Y$ be open ($X$, $Y$ $n$-manifolds)?
QUESTION [1 upvotes]: Is it true, that having an open ball $B(0,1)$ in $\mathbb{R}^n$ (say $n=2$, but the more general answer the better) and a set $C$ such that $0 \in \overline{C}$, there is no open neighbourhood of $0$ in $B(0,1)\setminus C$ homeomorphic with $B(0,1)$.
This comes from something with less technical wording: is it true, that image of every embedding of an $n$-manifold into another $n$-manifold is open?

REPLY [2 votes]: Your question (in the title) is the content of the invariance of domain theorem of Brouwer. For a proof (using the Brouwer fixed point theorem) you can see Terry Tao's blog.


TITLE: Molecular explanation of pressure in water at depth
QUESTION [4 upvotes]: At 10 m under water, pressure is 2 atmospheres.  As temperature is not different, and density is minimally different, how does pressure increase at the molecular level (collisions per unit area)?

REPLY [0 votes]: See this question: How does water pressure manifest itself on the molecular level?
In short, the water density increases greatly as ocean depth increases. It's just that for a liquid, a huge difference in pressure is caused by a small difference in the number of molecules per cubic meter.


TITLE: Calculation of approval rates
QUESTION [0 upvotes]: Total requests=100
Requests Approved=25
Requests Pending=65
Requests Rejected=5
So how to calculate the percentage of approval?? Can someone help please..

REPLY [1 votes]: Correct me if I am wrong, but isn't this just the percentage of requests approved out of the total number of requests:
$$\frac{25}{100} = 0.25 = 25\text{%} $$


TITLE: Given $\begin{vmatrix} p & q & r \\ s & t & u \\ v & w & x \end{vmatrix} = -3$, find $\begin{vmatrix}p&2q&5r+4p\\s&2t&5u+4s\\v&2w&5x+4v\end{vmatrix}$.
QUESTION [2 upvotes]: For the real numbers $p$, $q$, $r$, $s$, $t$, $u$, $v$, $w$, $x$,
$$\begin{vmatrix} p & q & r \\ s & t & u \\ v & w & x \end{vmatrix} = -3.$$
Find
$$\begin{vmatrix} p & 2q & 5r + 4p \\ s & 2t & 5u + 4s \\ v & 2w & 5x + 4v \end{vmatrix}.$$

My first thought was to find a particular matrix with determinant $-3$, but that would take too long with no guaranteed results.  But I don't know what else to do.

REPLY [1 votes]: $\begin{vmatrix} p & 2q & 5r + 4p \\ s & 2t & 5u + 4s \\ v & 2w & 5x + 4v \end{vmatrix} = \begin{vmatrix} p & q & r \\ s & t & u \\ v & w & x \end{vmatrix} \begin{vmatrix}1& 0 &4\\0&2&0\\0&0&5\end{vmatrix}$
And $\det(AB) = \det(A)\det(B)$

REPLY [0 votes]: The important thing to understand is that the determinant is linear in each column (or row) and since a determinant with two equal columns (or rows) is zero, you can add and subtract multiples of a row from another row. Algebraically, if $c_i$ are the columns, and $a,b$ are real numbers we have
$$ \begin{vmatrix} (ac_1 + bc'_1) & c_2 & c_3 \end{vmatrix} = a\begin{vmatrix} c_1 & c_2 & c_3 \end{vmatrix}+b\begin{vmatrix} c_1 & c_2 & c_3 \end{vmatrix} $$
and
$$ \begin{vmatrix} c_1 & c_2+ac_1 & c_3 \end{vmatrix} = \begin{vmatrix} c_1 & c_2 & c_3 \end{vmatrix} + a\begin{vmatrix} c_1 & c_1 & c_3 \end{vmatrix} = \begin{vmatrix} c_1 & c_2 & c_3 \end{vmatrix}. $$
In this case, subtract $4$ of the first column from the third column, and then pull the multiple of $2$ out of the second column and the multiple of $5$ out of the new third column.


TITLE: Sequential version of all-pay auction
QUESTION [1 upvotes]: Sequential version of the all-pay auction. Two bidders alternate in bidding. A prize of \$5 is auctioned. At each move of the game, the bidding player decides whether to raise the current bid by \$1 or to quit. The auction ends when one of the players quits. The winner gets the prize and both players pay the amounts of their most recent bids. Player 1 begins the auction with the bid \$1 (unless Player 1 decides to quit, in which case the prize goes to Player 2 and both players pay nothing). If both players raise their bids forever, they both receive $-\infty$ as their payoffs. There is no discount factor in this case.
(a) Find all Nash equilibria in pure strategies.
(b) Find all subgame-perfect equilibria in pure strategies.
To answer this question, I understand that for finding the Nash Equilibria in pure strategies, these can include actions on information sets that might not occur.
In this sense, the strategy profile (Bidder 1 (B1) always raise, Bidder 2 (B2) always quit) is a pure strategy NE: If B1 raise in the first period, B2 quit in the second period, then the payoffs are P1=4 P2=0. Since B1 action is raise in the third period and B2 is quit in the 4th period (with payoffs of P1=2,-2), then neither of them has incentive to deviate. Similarly, the strategy profile (B1 always quit, B2 always raise) is also a NE for similar reasons.
Is my reasoning correct? Are these the only NE in pure strategies?
On the other hand, I was trying to use some sort of backward induction for answering the second question, but I'm really struggling with the logic. Any hint?

REPLY [0 votes]: There are more NE equilibria in pure strategies. But they are all outcome equivalent to the ones you found. For example, Player 1 quitting in each period and Player 2 raising until $n$-th period with $n>5$ and quitting afterwards are also Nash equilibria. Any strategy profile, in which one of the players quits at her first move and the other player raises until the price is $5$ and quits at a later point, constitutes a NE.
However, only the two equilibria that you describe are subgame perfect. Backward induction does not work here as the game is infinite. Thus, use the definition of subgame perfection. That is, a pair of strategy profiles are subgame perfect if they constitute a NE equilibrium in any subgame. In the game at hand, subgames start in each period. The prices in each period are sunk costs. Thus, in each subgame the game going forward is the same. Therefore, one of Players always quitting and the other player always raising are the only subgame-perfect equilibria.
Side note: if you assume that players discount future payoffs at a discount factor $\delta<1$, the game is continuous at infinity and you can use the single-deviation principle to prove the result.


TITLE: How does one reduce the apparent effect of gravity without falling?
QUESTION [0 upvotes]: There are things like the Vomit Comet to simulate zero-gravity for small periods, but that involves going into a steep dive in an airplane.  How could one get the same effect on a surface of the Earth?
("You can't" is an acceptable answer.)

REPLY [1 votes]: Going under water would have a similar effect, thats how astronauts train for performing operations in 0 gravity like space walks ect i believe.


TITLE: Show that limit of integral equals supremum of its set
QUESTION [0 upvotes]: Let $f$ be continuous on $[a,b]$, let $f(x)$ $\geq$ $0$ for $x$ $\in$ $[a,b]$ , and let $M_n := (\int_a^b f^n)^{1/n}$.
Show that $$\lim(M_n) = \sup \{ f(x) : x \in [a,b] \}$$

REPLY [1 votes]: Set $M = \sup\{f(x): x\in[a,b]\}.$ Let $\varepsilon>0$ and define $S_\varepsilon =\{x\in[a,b]:f(x)\geqslant M-\varepsilon\}$. Then
\begin{align}
M_n&=\left(\int_{[a,b]} f(x)^n\ \mathsf dx\right)^{\frac1n}\\&\geqslant \left(\int_{S_\varepsilon} (M-\varepsilon)^n\ \mathsf dx\right)^{\frac1n}\\
&\geqslant (M-\varepsilon)\mu(S_\varepsilon)^{\frac1n}
\end{align}
where $\mu$ is Lebesgue measure, so that
$$\liminf_{n\to\infty} M_n\geqslant M. $$ For the reverse inequality, if $n>m$ then
\begin{align}
\int_{[a,b]} f(x)^n\ \mathsf dx &= \int_{[a,b]} f(x)^{n-m}f(x)^m\ \mathsf dx\\
&\leqslant\int_{[a,b]}M^{n-m} f(x)^m\ \mathsf dx\\
&=M^{n-m}\int_{[a,b]} f(x)^m\ \mathsf dx
\end{align}
and hence
$$M_n \leqslant\left(M^{n-m}\int_{[a,b]} f(x)^m\ \mathsf dx\right)^{\frac1n}\leqslant M^{\frac{n-m}m}\left(\int_{[a,b]} f(x)^m\ \mathsf dx\right)^{\frac1n},  $$
so that
$$\limsup_{n\to\infty} M_n\leqslant M. $$


TITLE: Proving consistency for an estimator. Limits and Convergence in Probability.
QUESTION [2 upvotes]: I need to show that $U$, as defined below, is a consistent estimator for $\mu^{2}$.
$U=\bar{Y}^{2}-\frac{1}{n}$
By the continuous mapping theorem, which states that,
$X_{n} \stackrel{\mathrm{P}}{\rightarrow} X \Rightarrow g\left(X_{n}\right) \stackrel{\mathrm{P}}{\rightarrow} g(X)$
Then,
$\bar{Y} \stackrel{P}{\longrightarrow} \mu $ gives me $\bar{Y}^{2} \stackrel{P}{\longrightarrow} \mu^{2} .$
And since $\frac{1}{n} \rightarrow 0$ as $n \rightarrow \infty$ the result for conistency seems intuitively obvious.
But I have a confusion with how to show this formally, whether using only the mapping theorem, or if I need something else.  Showing how the $\frac{1}{n} \rightarrow 0$ part leads to consistency is the part that I'm missing, since this is a standard limit and not a convergence in probability.
Any help in completing this is greatly appreciated.

REPLY [3 votes]: $X_n=1/n$ can be thought of as a random variable with a Dirac distribution with mass at $1/n$. $P(X_n\leq x)\rightarrow 1$, for all $x\geq 0$, and $0$ for all $x< 0$, therefore $X_n$ converges in distribution to the random variable with Dirac distribution with mass at zero, which is the same to say that it converges to zero.
By Slutsky's Theorem, $\bar{Y}^2-\frac{1}{n}\rightarrow_d \mu^2-0$, and since convergence in distribution to a constant implies convergence in probability, you have your result.
It is also possible to show this with the Continuous Mapping Theorem, since $(\bar{Y},X_n)$ converges jointly in probability to $(\mu,0)$. Then use function $g(y,x)=y^2-x$.


TITLE: Is a theorem of a consistent and complete extension of L a tautology?
QUESTION [1 upvotes]: If in a consistent and complete system we can deduce $A$, that is $⊢_J A$, what can be said about the valuation $V(A)$ or the truth value of $A$?

REPLY [1 votes]: If a formula $A$ is derivable from a set $J$ of axioms (i.e. it is a theorem of the theory having the said axioms), it is a logical consequence of the axioms, i.e. it is true in every interpretation satisfying the axioms.
It is not necessarily valid (i.e. true in every interpretation): the formula $1+1=2$ is provable from the axioms of arithmetic, but it is not valid.
And it is not necessarily a tautology: the formula $\forall x (x=x)$ is valid but it is not a tautology.