text:
\section{Introduction}

This paper address the problem of determining the Hausdorff dimension
of the level sets of the solutions of some stochastic PDEs from fluid
dynamics in two space dimensions. Consider the following stochastic PDE,
\[
  \dot\theta
      + \nu (-\Delta)^\alpha\theta
      + u\cdot\nabla\theta
     = \dot\eta,
        \qquad t\geq0,x\in\Torus,
\]
on $[-\pi,\pi]^2$, with periodic boundary conditions and zero spatial
mean, where $\nu$, $\alpha$ and $M$ are suitable parameters, $\dot\eta$
is Gaussian noise and the transport velocity is given by
$u = \nabla^\perp (-\Delta)^{-M}\theta$.
We prove almost sure upper bounds, as well as lower bounds with positive
probability, on the Hausdorff and packing dimension of the level
sets of the random field $\theta(t)$ at any positive time $t>0$.

Our equation \eqref{e:spde} belongs to the class of \emph{Navier--Stokes
$\alpha$--like models} (see for instance \cite{OlsTit2007} and
the reference therein) and, when $\alpha=1$, the 2D Navier--Stokes
equations in the vorticity formulation correspond to $M=1$, the
surface quasi--geostrophic equation corresponds to $M=\tfrac12$,
while $M=-1$ describes the large scale flows of a rotating shallow
fluid (see \cite{Fal2007a}). In this paper we will assume $\alpha>1$
and $M\geq1$. Unfortunately, our results do not cover the above
mentioned cases, although they are the motivating examples.

Our interest in level sets for equations from fluid dynamics is
inspired by the theory developed for the $0$--level set
at a physical and numerical level \cite{Fal2007a}. Our technique
is not powerful enough to be able to say something about the
``physical case'', for a series of reasons: we work at finite
viscosity and not in the vanishing viscosity regime, we assume
hyper--dissipation, due to the limits of the techniques we employ,
we are not able to capture all of the interesting cases of
the transport velocity, we are dealing with rough noise.
On the other hand, this paper is a preliminary work and
extensions are currently being developed.

The results presented here focus on any level set. On the other
hand the zero spatial mean condition gives a privileged status
to the zero level set. It would be expected to obtain stronger
results on the zero level set than on any other level set.
We conjecture (see Remark \ref{r:failed} that indeed the zero
level set should have a ``deterministic'' (that is, determined
almost surely) dimension.

We give a few details on the techniques used to achieve our results.
As in \cite{DalKhoNua2007,DalKhoNua2009}, the non--linear problem
is reduced to the linear problem (namely, the same equation without
the non--linearity or, in other words, the linearization at $0$)
by means of an absolute continuity result between the laws of the
two processes. In the above references the equivalence is provided
by the Girsanov theorem. In our problem Girsanov's transformation
cannot be applied (see Remark \ref{r:girsanov}), and we apply a weaker
result from \cite{DapDeb2004}, using the polynomial moments from
\cite{EssSta2010}. On the one hand this gives equivalence of the
laws at the level of the single time rather than of the full path,
but on the other hand this is enough for our purpose. The equivalence
result is already known from \cite{MatSui2005,Wat2010} and ours is
an alternative proof.

Once the problem is reduced to a linear equation, it
becomes more amenable and one can use the theory developed for
Gaussian processes \cite{Kah1985,Xia1995,Xia1997,WuXia2006} to
show almost sure upper bounds on the Hausdorff and packing dimension,
as well as lower bounds with positive probability.
\subsection{Notations}

Let $\Torus = [-\pi,\pi]^2$ be the $2$--dimensional torus. For every
$\gamma\in\mathbf{R}$ denote by $H^\gamma_\#(\Torus)$ the Sobolev space of
periodic functions on $\Torus$ with mean zero on $\Torus$, defined in
terms of the complex Fourier coefficients with respect to the Fourier basis
$\{\operatorname{e}^{\mathrm{i} k\cdot x} : k\in\mathbf{Z}^2\}$, as
\[
  H^\gamma_\#(\Torus)
    = \bigl\{(\xi_k)_{k\in\mathbf{Z}^d}\subset\mathbf{C}:
        \xi_0=0,\quad
        \|\xi\|_\gamma^2
    \vcentcolon=\sum_{k\in\mathbf{Z}^d}|k|^{2\gamma}|\xi_k|^2<\infty\bigr\},
\]
with norm $\|\cdot\|_\gamma$. In particular, when $\gamma=0$, we use
the standard notation $L^2_\#(\Torus)$.
Define moreover the spaces of divergence--free vector fields $V_\gamma$ as
\[
  V_\gamma
    = \bigl\{(u_k)_{k\in\mathbf{Z}^2_\star}\subset \mathbf{C}^2:
        k\cdot u_k = 0\text{ for all }k,\
        u^i = (u_k^i)_{k\in\mathbf{Z}^d}\in H^\gamma_\#\text{ for i=1,2}\bigr\},
\]
where $\mathbf{Z}^2_\star = \mathbf{Z}^2\setminus\{(0,0)\}$, with norm
$\|u\|_\gamma^2 \vcentcolon= \|u^1\|_\gamma^2 + \|u^2\|_\gamma^2$,
and $u = (u^1,u^2)$. In particular, set $H = V_0$.

Denote by $A$ the realization of the Laplace operator $-\Delta$ on
$L^2_\#(\Torus)$ with periodic boundary conditions. A real orthonormal
basis of $L^2_\#(\Torus)$ (and hence of each $H_\#^\gamma$) of
eigenvectors of $A$ is given as follows. Set
\[
  \mathbf{Z}^2_+
    = \{k\in\mathbf{Z}^2:k_2>0\}
      \cup \{k\in\mathbf{Z}^2:k_1>0,k_2=0\},
\]
$\mathbf{Z}^2_- = -\mathbf{Z}^2_+$ and $\mathbf{Z}^2_\star = \mathbf{Z}^2_+\cup\mathbf{Z}^2_-$, and
$e_k = \mcref{cc:onb}\sin k\cdot x$ for $k\in\mathbf{Z}^2_+$ and
$e_k = \mcref{cc:onb}\cos k\cdot x$ for $k\in\mathbf{Z}^d_-$,
where $\mcdef{cc:onb} = \sqrt{2}(2\pi)^{-1}$, then \cite{BerMaj2002}
$(e_k)_{k\in\mathbf{Z}^2_\star}$ is an orthonormal basis of $L^2_\#(\Torus)$.
With these positions, given $\gamma\in\mathbf{R}$,
$A^\gamma x = \sum_{k\in\mathbf{Z}^2_\star} |k|^{2\gamma} x_k e_k$
for $x = \sum_k x_k e_k$.
Set $E_k = \tfrac{k^\perp}{|k|}e_k$ for
every $k\in\mathbf{Z}^2_\star$, then $(E_k)_{k\in\mathbf{Z}^2_\star}$ is
an orthonormal basis of $H$. With a slight abuse of notations,
we will also denote by $A$ the realization of the Laplace operator
on $H$.

Given $y\in\mathbf{R}$ and a field $v:\Torus\to\mathbf{R}$, denote by $\mathcal{L}_y(v)$
the $y$--level set of $v$, namely $\mathcal{L}_y(v) = \{x\in\Torus: v(x) = y\}$.

We denote by $\dim_H$ and by $\dim_P$ the Hausdorff and the packing dimension,
respectively. We refer to \cite{Fal2003} for their definition and
properties.
\section{Formulation of the problem and main results}

\subsection{Formulation of the problem}

Fix $\nu>0$, $\alpha\geq1$ and $M\in\mathbf{R}$. Consider on $\Torus$ the following
stochastic PDE,
\begin{equation}\label{e:spde}
  \dot\theta
      + u\cdot\nabla\theta
    = - \nu A^\alpha\theta
      + \dot\eta,
        \qquad t\geq0,x\in\Torus,
\end{equation}
with periodic boundary conditions and with $\iint_{\Torus}\theta\,dx = 0$,
where $u$ is given as $u = \nabla^\perp A^{-M}\theta$.
By its definition it turns out that
$\Div(u)=0$ and $\iint u(x)\,dx = 0$.  The 2D Navier--Stokes equations
in the vorticity formulation correspond to $M=1$ \cite{BerMaj2002},
the surface quasi--geostrophic to $M=\tfrac12$, while $M=-1$ describes
the large scale flows of a rotating shallow fluid (see \cite{Fal2007a}).
Unfortunately, none of these values can be covered by our results.
\subsubsection*{The non--linearity}

Set for a vector function $v$, with $\Div(v) = 0$, and a scalar
function $f$,
\[
  B(v, f)
    = v\cdot\nabla f
    = \Div(vf).
\]
We use the same notation $B(v,v')$ when $v$ is a vector and the
operator $B$ is understood component--wise, namely
$[B(v,v')]_i = B(v_i,v')$. Set moreover
\[
  B_M(x)
    = B(\nabla^\perp A^{-M} x, x)
\]
We stress two important properties of the non--linear term.
Their proofs are standard using integration by parts arguments.

\begin{lemma}
  For every $x$ and $v$, with $\Div v = 0$,

  \begin{equation}\label{e:B1}
    \scal{x, B(v,x)}_{L^2}
      = 0,
        \qquad\qquad
    \scal{A^{-M}x, B_M(x)}_{L^2}
      = 0.
  \end{equation}

\end{lemma}
\subsubsection*{The random forcing term}

The random forcing term $\dot\eta$ is modeled as a coloured in space
and white in time Gaussian noise, namely $\eta$ is a Wiener process with
covariance $\mathcal{C}\in\mathscr{L}(L^2_\#(\Torus))$.
For our purposes, we will assume that $\mathcal{C}$ has a smoothing effect.
\begin{assumption}[on the covariance]\label{a:noise}
  The operator $\mathcal{C}$ is positive linear bounded on $L^2_\#(\Torus)$.
  The driving noise is \emph{homogeneous} in space,
  hence $\mathcal{C}$ has the same eigenvectors of the operator $A$.
  Under these assumptions, there are numbers $(\sigma_k)_{k\in\mathbf{Z}^2_\star}$
  such that for $x = \sum_k x_k e_k\in L^2_\#$,
  \[
    \mathcal{C} x
      = \sum_{k\in\mathbf{Z}^d_\star}\sigma_k^2 x_k e_k.
  \]
  Assume additionally that there exists $\delta\in(1-\alpha,2-\alpha)$
  such that

  \begin{equation}\label{e:mainass}
    \frac{\mcdef{cc:noise1}}{|k|^\delta}
      \leq |\sigma_k|
      \leq \frac{\mcdef{cc:noise2}}{|k|^\delta}.
  \end{equation}

\end{assumption}
\subsubsection*{The abstract formulation}

In conclusion \eqref{e:spde} can be recast
in its abstract form as
\begin{equation}\label{e:spde_abs}
  d\theta
    + \bigl(\nu A^\alpha\theta + B(u,\theta)\bigr)\,dt
    = \mathcal{C}^{\frac12}\,dW.
\end{equation}
\subsection{The linear problem}

Our first result gives upper and lower bounds for the dimension
of the level sets of the linear version of the problem under
examination, namely,
\begin{equation}\label{e:z}
  \begin{cases}
    dz + \nu A^\alpha z = \mathcal{C}^{1/2}\,dW,\\
    z(0) = 0.
  \end{cases}
\end{equation}
\begin{theorem}\label{t:linear}
  Let $\alpha\geq1$ and let Assumption \ref{a:noise} be true.
  For every $t>0$ and $y\in\mathbf{R}$,
  \begin{itemize}
    \item $\mathbb{P}[\dim_P(\mathcal{L}_y(z_t))\leq 3-\alpha-\delta] = 1$,
    \item $\mathbb{P}[\dim_P(\mathcal{L}_y(z_t))=\dim_H(\mathcal{L}_y(z_t))= 3-\alpha-\delta]>0$.
  \end{itemize}
\end{theorem}
The proof of the theorem is based on well--known techniques
\cite{Kah1985,Xia1995,Xia1997} for the dimension of level sets.
\begin{remark}\label{r:failed}
  Theorem \ref{t:linear} above says that $\mathcal{L}_y(z_t)$ has
  dimension equal $3-\alpha-\delta$ with positive probability.
  It should not be expected that in general this may hold with
  probability one. Indeed, if $y=0$, the set $\mathcal{L}_y(z_t)$
  is empty with positive probability, since $z_t$ is continuous
  and defined on a compact set. On the other hand if $y=0$ then
  $\mathcal{L}_0(z_t)\neq\emptyset$ with probability one,
  since $z_t$ is non--zero and with zero average. We conjecture
  that $\dim_P(\mathcal{L}_0(z_t))=\dim_H(\mathcal{L}_0(z_t))
  = 3-\alpha-\delta$ with probability one.

  An effective way to prove that a random set has an almost sure
  dimension is to show that the set contains a \emph{limsup
  random fractal} \cite{KhoPerXia2000}. It is easy to construct
  a limsup random fractal contained into $\mathcal{L}_0(z_t)$
  by using the sets $\{|z_t(x_k^n)|\leq\epsilon\}$ as
  building blocks, where $(x_k^n)_{k}$ is a dyadic grid.
  Unfortunately, the correlation between distant blocks is too
  strong to apply \cite[Corollary 3.3]{KhoPerXia2000}.

  A different approach to prove the conjecture could be based
  on existence and regularity of the occupation density of
  $z_t$ at $0$. First, the random field $z_t$ has an
  occupation density $\ell$ due to \cite[Theorem 22.1]{GemHor1980}
  (see also \cite[Theorem 3]{Pit1978}). Moreover, it is not
  difficult to see that $z_t$ satisfies the property of
  \emph{local $2(\alpha+\delta-1)$--nondeterminism}
  (see \cite{MonPit1987} for the definition, and \cite{Xia2008,Xia2009}
  for a recent account) and so, by \cite[Theorem 4]{Pit1978}
  (or \cite[Theorem 26.1]{GemHor1980}), the occupation
  density $\ell$ is H\"older continuous. By \cite[Theorem 1]{MonPit1987},
  in order to prove the ``probability one'' statement, it is
  sufficient to show that $\ell(0)>0$ with probability one.
  For instance this is true if there is a random constant
  $\mcdef{cc:fail} = \mcref{cc:fail}(\omega)>0$ such that
  \[
    \text{Leb}_{\Torus}(\{x\in\Torus:\ |z_t(x)|\leq\epsilon\})
      \geq\mcref{cc:fail}\epsilon.
  \]
  We have been not able to prove that $\mathbb{P}[\ell(0)>0]=1$.
\end{remark}
\subsection{The non--linear problem}

We extend the results on the dimension of level sets by means of
a absolute continuity result. The random perturbation we consider
is not ``strong'' enough (in terms of regularization) to apply
Girsanov's theorem (which is a standard method when dealing with
non--linear terms of order zero, see for instance
\cite{DalKhoNua2007,DalKhoNua2009}).
We use an absolute continuity theorem of \cite{DapDeb2004}
to translate the dimension results on the linear problem to the
non--linear problem.
We remark that another option to prove the absolute continuity
could be given by the idea in \cite{MatSui2005} (see also
\cite{MatSui2008,Wat2010}).
\begin{theorem}\label{t:main}
  Let $\nu>0$, $\alpha>1$, and $M\geq1$, and assume the
  covariance $\mathcal{C}$ satisfies Assumption \ref{a:noise}.
  Let $\theta$ be the solution of problem \eqref{e:spde}
  with $\theta(0)\in L^2_\#$, then for every $y\in\mathbf{R}$ and $t>0$,
  \begin{itemize}
    \item $\mathbb{P}[\dim_P(\mathcal{L}_y(\theta_t))\leq 3-\alpha-\delta] = 1$,
    \item $\mathbb{P}[\dim_P(\mathcal{L}_y(\theta_t))=\dim_H(\mathcal{L}_y(\theta_t))= 3-\alpha-\delta]>0$.
  \end{itemize}
\end{theorem}
As we shall see in the course of the proof of the above result,
the same holds when $\theta$ is the stationary solution.
\section{Linear results}

In this section we prove Theorem \ref{t:linear}. Existence and uniqueness
for the solution $z$ of \eqref{e:z}, as well as of its invariant measure
and strong mixing are a standard matter, see \cite{DapZab1992}. In the
next lemma we summarize a few results concerning point--wise properties
of $z$ that we will need in the rest of the section.
\begin{lemma}\label{l:zprop1}
  Under the assumptions of Theorem \ref{t:linear}, for every $t>0$
  and $x\in\Torus$, $z(t,x)$ is a centred Gaussian random variable
  such that $\mcref{cc:noise1}\sigma_t^2\leq\mathbb{E}|z(t,x)|^2\leq
  \mcref{cc:noise2}\sigma_t^2$, with $\sigma_t>0$. Moreover,
  there is $g_t:\mathbf{R}\to\mathbf{R}$ such that
  \[
    \mathbb{E}[|z(t,x) - z(t,y)|^2]
      = g_t(x-y),
  \]
  and $\mcdef{cc:z1}|x|^{2(\alpha+\delta-1)}\leq g_t(x)\leq
  \mcdef{cc:z2}|x|^{2(\alpha+\delta-1)}$.
\end{lemma}
\begin{proof}
  We can write $z(t,x)$ as
  \[
    z(t,x)
      = \sum_{k\in\mathbf{Z}^2_\star}\Bigl(\sigma_k\int_0^t
          \operatorname{e}^{-\nu|k|^{2\alpha}(t-s)}\,d\beta_k(s)\Bigr)e_k(x),
  \]
  where $(\beta_k)_{k\in\mathbf{Z}^2_\star}$ are independent standard Brownian motions.
  Given $x\in\mathbb{T}_2$, $t>0$, the real valued random variable $z(t,x)$
  is Gaussian and,
  \[
    \operatorname{Var}(z(t,x))
      = \sum_{k\in\mathbf{Z}^2_\star} \frac{\sigma_k^2e_k(x)^2}{2\nu|k|^{2\alpha}}
          \bigl(1 - \operatorname{e}^{-2\nu|k|^{2\alpha}t}\bigr)
      \approx \sigma_t^2
      = \sum_{k\in\mathbf{Z}^2_\star} \frac{1 - \operatorname{e}^{-2\nu|k|^{2\alpha}t}}
          {4\nu|k|^{2(\alpha+\delta)}}.
  \]
  The expectation of the increments yields
  \[
    \mathbb{E}[|z(t,x) - z(t,y)|^2]
     = \sum_{k\in\mathbf{Z}^2_\star}\frac{\sigma_k^2}{2\nu|k|^{2\alpha}}
         \bigl(1 - \operatorname{e}^{-2\nu|k|^{2\alpha}t}\bigr)\sin^2\frac{k}{2}(x-y)
     = g_t(x-y).
  \]
  Using \eqref{e:mainass} and the fact that $(1 - \operatorname{e}^{-2\nu|k|^{2\alpha}t})$
  is bounded from above and below by constants independent of $k$ (but not $t$),
  we see that
  \[
    g_t(x)
      \approx\sum_{k\in\mathbf{Z}^2_\star}\frac1{|k|^{2\alpha+2\delta}}\sin^2\frac{k}{2}(x-y),
  \]
  hence $g_t(x)\approx|x|^{2(\alpha+\delta-1)}$ by Lemma \ref{l:sin}
  below.
\end{proof}
\begin{lemma}\label{l:sin}
  Let $\gamma>0$. Then for all $x\in\mathbf{R}^d$ with $|x|<1$,
  \[
    \sum_{k\in\mathbf{Z}^d_\star} \frac1{|k|^{d+\gamma}}\sin^2(k\cdot x)
      \sim h_\gamma(x),
  \]
  where $h_2(x) = - |x|^2\log|x|$ and $h_\gamma(x) = |x|^{\gamma\wedge2}$
  otherwise.
\end{lemma}
\begin{proof}
  For the upper bound,
  \[

    \sum_{k\in\mathbf{Z}^d_\star} \frac1{|k|^{d+\gamma}}\sin^2\bigl(k\cdot x\bigr)

      \leq \sum_{|k|\cdot|x|\leq1} \frac{|x|^2}{|k|^{d+\gamma-2}}
           + \sum_{|k|\cdot|x|\geq1} \frac1{|k|^{d+\gamma}}
      \leq \memo{s} + \memo{l}.

  \]
  The term \memo{l} can be easily estimated by comparison with an integral,
  yielding $\memo{l}\leq \stepcounter{ConstantsCounter}\ensuremath{c_\text{\theConstantsCounter}}|x|^\gamma$. For \memo{s} we use the elementary
  result,
  \[
    \sum_{|k|\leq A} \frac1{|k|^\beta}
      \sim
        \begin{cases}
          A^{(d-\beta)_+} & \beta\neq d,\\
          \log A  & \beta=d,\\
        \end{cases}
        \qquad A\text{ large,}
  \]
  to obtain immediately that $\memo{s}\sim h_\gamma(x)$.

  For the lower bound, an elementary computation shows that
  there is $\mcdef{cc:gamma} = \mcref{cc:gamma}(\gamma)>0$ such that,
  \[
    \frac1{|k|^{d+\gamma}}
      = \mcref{cc:gamma}\int_0^\infty \operatorname{e}^{-t|k|^2}t^{\frac{d+\gamma-2}2}\,dt,
  \]
  hence
  \[
    \begin{aligned}
    \sum_{k\in\mathbf{Z}^d_\star} \frac1{|k|^{d+\gamma}}\sin^2\bigl(\tfrac12 k\cdot x\bigr)
      &= \mcref{cc:gamma} \int_0^\infty
           \Bigl(\sum_{k\in\mathbf{Z}^d_\star}\operatorname{e}^{-t|k|^2}\sin^2\bigl(\tfrac12 k\cdot x\bigr)\Bigr)
           t^{\frac{d+\gamma-2}2}\,dt\\
      &= \frac12 \mcref{cc:gamma} \int_0^\infty
           \Bigl(\sum_{k\in\mathbf{Z}^d_\star}\operatorname{e}^{-t|k|^2}(1 - \cos(k\cdot x)\Bigr)
           t^{\frac{d+\gamma-2}2}\,dt\\
      &\geq\frac12 \mcref{cc:gamma} \int_0^1
           \bigl(\phi(t,0) - \phi(t,x)\bigr)
           t^{\frac{d+\gamma-2}2}\,dt,
    \end{aligned}
  \]
  where
  \[
    \phi(t,x)
      = \sum_{k\in\mathbf{Z}^d_\star}\operatorname{e}^{-t|k|^2}\cos(k\cdot x).
  \]
  The $\phi$ is the fundamental solution of the heat equation with periodic
  boundary conditions and mean zero. In particular,
  $\phi(t,0) - \phi(t,x)\geq \stepcounter{ConstantsCounter}\ensuremath{c_\text{\theConstantsCounter}} t^{-d/2}\bigl(1\wedge|x|^2/t\bigr)$,
  hence
  \[
    \sum_{k\in\mathbf{Z}^d_\star} \frac1{|k|^{d+\gamma}}\sin^2\bigl(\tfrac12 k\cdot x\bigr)
      \geq \int_0^1 \frac{\stepcounter{ConstantsCounter}\ensuremath{c_\text{\theConstantsCounter}}}{t^{\frac{d}2}}\Bigl(1\wedge\frac{|x|^2}{t}\Bigr)
             t^{\frac{d+\gamma-2}2}\,dt
      \geq \stepcounter{ConstantsCounter}\ensuremath{c_\text{\theConstantsCounter}} h_\gamma(x),
  \]
  by a direct computation.
\end{proof}
\begin{remark}\label{r:distance}
  If we replace the usual Euclidean distance in the statement of Lemma
  \ref{l:zprop1} and Lemma \ref{l:sin} with the ``torus distance'', namely
  $|x-y|_{\Torus} = \inf_{k\in\mathbf{Z}^2} |x-y+2\pi k|$, the conclusions
  of both lemmata still hold true.
\end{remark}
\begin{remark}
  If $d=1$, the above lemma admits a probabilistic proof, using a Fourier
  series expansion of the fractional Brownian motion. Indeed, by \cite{Igl2005}
  it follows, by simple computations that exploit the explicit form of the covariance
  function of the process, that
  \[
    \sum_{k=1}^\infty \frac1{k^{1+2H}}\sin^2(\tfrac12kt)
      \sim t^{2H}.
  \]
  We have not been able to find a similar proof in the multi--dimensional case.
\end{remark}
\subsection{The upper bound}

The following proposition contains the first part of Theorem \ref{t:linear}.
\begin{proposition}
  Under the assumptions of Theorem \ref{t:linear},
  \[
    \dim_H(\mathcal{L}_y(z_t))
      \leq \dim_P(\mathcal{L}_y(z_t))
      \leq 3 - \alpha - \gamma.
  \]
\end{proposition}
\begin{proof}
  We know by Lemma \ref{l:zprop1} and Gaussianity that $z_t$ is
  $\gamma$--H\"older continuous for every $\gamma<\alpha+\delta-1$
  and that the H\"older coefficient $L_\gamma$ has finite polynomial
  moments \cite[Theorem 1.4.1]{Kun1990}. Let $\gamma<\alpha+\delta-1$
  and consider a ball $B_\epsilon(x)$ in $\Torus$, then
  \[
    \mathbb{P}[B_\epsilon(x)\cap\mathcal{L}_y(z_t)]
      \leq \mathbb{P}[y\in z_t(B_\epsilon(x)),\ |z_t(x)-y|\geq\epsilon^{\gamma}]
        + \mathbb{P}[|z_t(x)-y|\leq\epsilon^\gamma].
  \]
  On the first event there is $x_y\in B_\epsilon(x)$ such that $z_t(x_y)=y$,
  hence for a $\gamma'$ such that $\gamma<\gamma'<\alpha+\delta-1$, we have that
  $\epsilon^\gamma\leq|z_t(x)-z_t(x_y)|\leq L_{\gamma'}\epsilon^{\gamma'}$,
  hence for $n$ large enough,
  \[
    \mathbb{P}[B_\epsilon(x)\cap\mathcal{L}_y(z_t)]
      \leq \mathbb{P}[L_{\gamma'}\geq\epsilon^{\gamma-\gamma'}]
        + \stepcounter{ConstantsCounter}\ensuremath{c_\text{\theConstantsCounter}}\epsilon^\gamma
      \leq \epsilon^{n(\gamma'-\gamma)}\mathbb{E}[L_{\gamma'}^n]
        + \stepcounter{ConstantsCounter}\ensuremath{c_\text{\theConstantsCounter}}\epsilon^\gamma
      \leq \stepcounter{ConstantsCounter}\ensuremath{c_\text{\theConstantsCounter}}\epsilon^\gamma.
  \]
  Consider now a covering of $\Torus$ of $2^{2k}$ balls $B_k$ of radius $2^{-k}$,
  and let $N_k$ be the smallest number of balls of radius $2^{-k}$ covering
  $\mathcal{L}_y(z_t)$. Clearly, $N_k\leq\sum \mathbbm{1}_{\{B_k\cap\mathcal{L}_y\neq\emptyset\}}$,
  hence $\mathbb{E}[N_k]\leq \mcdef{cc:box}2^{2k-\gamma k}$. By the first Borel--Cantelli lemma,
  $N_k\leq \mcref{cc:box}2^{2k-\gamma'' k}$ for $k$ large enough, a.s., where $\gamma''<\gamma$.
  Therefore, $\dim_P\mathcal{L}_y(z_t)\leq 2-\gamma''$, and the upper bound
  follows by taking $\gamma''\uparrow\alpha+\delta-1$.
\end{proof}
\subsection{The lower bound}

We prove the second part of Theorem \ref{t:linear}. To this end we
first give an estimate of the two--points covariance.
\begin{lemma}\label{l:cov}
  Under the same assumptions of Theorem \ref{t:linear}, let
  $t>0$. Then there is $\mcdef{cc:cov}>0$ such that
  for every $x,x'\in\Torus$, with $x\neq x'$,
  $\det(q_{xx'})\geq\mcref{cc:cov}|x-x'|^{2(\alpha+\delta-1)}$,
  where $q_{xx'}$ is the covariance matrix of $(z_t(x),z_t(x'))$.
\end{lemma}
\begin{proof}
  Given $t>0$, define the numbers $a_k(t)$ by
  \[
    \sigma_t(x)^2
      = \sum_{k\in\mathbf{Z}^2_\star} \frac{\sigma_k^2}{2\nu|k|^{2\alpha}}
          \bigl(1 - \operatorname{e}^{-2\nu|k|^{2\alpha}t}\bigr)e_k(x)^2
      = \sum_{k\in\mathbf{Z}^2_\star} a_k(t)^2 e_k(x)^2.
  \]
  If $x,x'\in\Torus$, define $\sigma_t(x,x')$ by
  \[
    \sigma_t(x,x')
      = \mathbb{E}[z_t(x)z_t(x')]
      = \sum_{k\in\mathbf{Z}^2_\star} a_k(t)^2 e_k(x)e_k(x').
  \]
  Then a few elementary computations (using the fact that
  $a_{-k} = a_k$ and the symmetries of $\mathbf{Z}^2_\star$)
  show that
  \[
    \det(q_{xx'})
      = \frac12\sum_{m,n\in\mathbf{Z}^2_\star}a_m(t)^2 a_n(t)^2
          \sin^2\bigl((m+n)\tfrac{x-x'}{2}\bigr)
  \]
  By re--arranging the sum, we finally obtain
  \[
    \det(q_{xx'})
      = \sum_{k\in\mathbf{Z}^2_\star} A_k\sin^2(k\cdot \tfrac{x-x'}{2}),
        \quad\text{where}\quad
    A_k
      = \frac12\sum_{m+n=k;m,n\in\mathbf{Z}^2_\star}a_n(t)^2a_m(t)^2.
  \]
  Since $a_k(t)\sim |k|^{-2(\alpha+\delta)}$ and $\alpha+\delta>1$,
  it is easy to see that $A_k\sim |k|^{-2(\alpha+\delta)}$ and
  \[
    \det(q_{xx'})
      \geq \sum_{k\in\mathbf{Z}^2_\star} \frac{\stepcounter{ConstantsCounter}\ensuremath{c_\text{\theConstantsCounter}}}
        {|k|^{2 + 2(\alpha+\delta-1)}}
        \sin^2(k\cdot \tfrac{x-x'}{2})
      \geq \stepcounter{ConstantsCounter}\ensuremath{c_\text{\theConstantsCounter}}|x-x'|^{2(\alpha+\delta-1)}
  \]
  where the last inequality follows from Lemma \ref{l:sin}.
\end{proof}
\begin{proposition}
  Under the assumptions of Theorem \ref{t:linear},
  \[
    \dim_P(\mathcal{L}_y(z_t))
      \geq\dim_H(\mathcal{L}_y(z_t))
      \geq 3 - \alpha - \delta,
  \]
  with positive probability.
\end{proposition}
\begin{proof}
  We use Frostman's ideas, see \cite{Kah1985}
  To this end, given a non--negative measure $\mu$ on $\Torus$ and a
  number $\gamma>0$, define the $\gamma$--energy of $\mu$ as
  \[
    \|\mu\|_\gamma
      = \int_{\Torus}\int_{\Torus}
        \frac{\mu(dx)\mu(dx')}{|x-x'|^\gamma},
  \]
  where $|\cdot|$ denotes the distance on $\Torus$ (Remark
  \ref{r:distance}). We proceed as in \cite{WuXia2006} (see also
  \cite{Xia1995,Xia1997}) and define the measures
  $\mu_n = \sqrt{2\pi n}\exp({-\frac12n|z_t(x)-y|^2})\,dx$.
  For our purposes, it is sufficient to show that there are
  $\mcdef{cc:frost1},\mcdef{cc:frost2}$ (independent of $n$),
  such that $\mathbb{E}[\mu_n(\Torus)]\geq\mcref{cc:frost1}$,
  $\mathbb{E}[\mu_n(\Torus)^2]\leq\mcref{cc:frost2}$ and
  $\mathbb{E}[\|\mu_n\|_\gamma]<\infty$ for every $\gamma<3-\alpha-\delta$.
  Indeed, by these facts it follows that there is a sub--sequence
  converging to a measure $\mu$. Moreover, $\mu$ is non--zero
  with probability $\mcref{cc:frost1}^2\mcref{cc:frost2}^{-1}$
  (see \cite{Kah1985}). By continuity of $z_t$ (Lemma \ref{l:zprop1},
  it follows that $\mu$ has support in $\mathcal{L}_y(z_t)$ and hence
  Frostman's lemma \cite[Theorem 10.3.2]{Kah1985} yields
  $\mathbb{P}[\dim_H\mathcal{L}_y(z_t)\geq\gamma]\geq
  \mcref{cc:frost1}^2\mcref{cc:frost2}^{-1}$.
  We notice preliminarily that
  \[
    \sqrt{2\pi n}\operatorname{e}^{-\frac12n|z_t(x)-y|^2}
      = \int_\mathbf{R}\operatorname{e}^{-\frac1{2n}u^2+\mathrm{i} u(z_t(x)-y)}\,du.
  \]
  As in \cite{WuXia2006}, simple computations yield,
  \[
  \begin{multlined}
    \mu_n(\Torus)
      =\int_{\Torus}\int_\mathbf{R} \operatorname{e}^{-\frac1{2n}u^2-\mathrm{i} u y}
        \mathbb{E}[\operatorname{e}^{\mathrm{i} uz_t(x)}]\,du\,dx = \\
      =\int_{\Torus}\int_\mathbf{R} \operatorname{e}^{-\mathrm{i} u y}
        \operatorname{e}^{-\frac12(\frac1n+\sigma_t(x)^2)u^2}\,du\,dx
      =\int_{\Torus} \sqrt{\frac{2\pi}{\frac1n+\sigma_t(x)^2}}
        \operatorname{e}^{-\frac{y^2}{2( \frac1n+\sigma_t(x)^2 )}}\,dx\geq\\
      \geq (1+\mcref{cc:z2}^2\sigma_t^2)^{-\frac12}
        \operatorname{e}^{-\frac{y^2}{2\mcref{cc:z1}^2\sigma_t^2}}
      = \mcref{cc:frost1}.
  \end{multlined}
  \]
  With similar computations, involving this time two dimensional Gaussian
  random variables, we see that
  \[
  \begin{multlined}[.9\linewidth]
    \mu_n(dx)\mu_n(dx')
      = \Bigl(\iint_{\mathbf{R}^2}\operatorname{e}^{-\frac12u\cdot g_{xx'}\cdot u^T}
        \operatorname{e}^{-\mathrm{i} y(u_1+u_2)}\,du\Bigr)\,dx\,dx' = \\
      = \frac{2\pi}{\sqrt{\det g_{xx'}}}
        \operatorname{e}^{-\frac12(y,y)\cdot g_{xx'}^{-1}\cdot(y,y)^T}\,dx\,dx'
      \leq \frac {2\pi}{\sqrt{\det g_{xx'}}}\,dx\,dx',
  \end{multlined}
  \]
  where $g_{xx'}=\frac1n I + q_{xx'}$ and $q_{xx'}$ is the covariance
  matrix of $(z_t(x),z_t(x'))$. By Lemma \ref{l:cov},
  $\det(g_{xx'})\geq\det(q_{xx'})\geq\mcref{cc:cov}|x-x'|^{2(\alpha+\beta-1)}$,
  hence $\mu_n(dx)\mu_n(dx')\leq\stepcounter{ConstantsCounter}\ensuremath{c_\text{\theConstantsCounter}}|x-x'|^{-(\alpha+\beta-1)}$
  and it is immediate to deduce that $\mathbb{E}[\mu_n(\Torus)^2]\leq\mcref{cc:frost2}$.
  Likewise, we deduce that $\mathbb{E}[\|\mu_n\|_\gamma]$ is bounded uniformly
  in $n$ if $\gamma<3-\alpha-\delta$.
\end{proof}
\section{Non--linear results}

We turn to the proof of Theorem \ref{t:main}. Our strategy
is based on the idea that if $\theta$ is the solution of
\eqref{e:spde_abs} and $z$ of \eqref{e:z}, and if the laws of
$\theta(t)$ and $z(t)$ are equivalent measures, then
Theorem \ref{t:linear} immediately implies Theorem
\ref{t:main}.
\begin{remark}\label{r:girsanov}
  A standard way to prove absolute continuity of laws of solutions
  of stochastic PDEs is the Girsanov transformation. In our
  case, to apply Girsanov's transformation, the quantity
  \[
    \int_0^t \|\mathcal{C}^{-1/2}B_M(\theta)\|_{L^2}^2\,ds
      <\infty,
        \qquad\mathbb{P}-{a.s.}
  \]
  should be finite, at the very least. This happens when $\alpha>2$
  (see \cite[Theorem 3]{MatSui2005}).
\end{remark}
The following theorem could be proved by means of the same method
in \cite[Theorem 2]{MatSui2005}, which is indeed the case
$\alpha>1$, $M=1$ (see also \cite{Wat2010}). Here we present
an alternative proof based on the method introduced in
\cite{DapDeb2004} and on the polynomial moments proved
in Lemma \ref{l:improved}.
\begin{theorem}\label{t:equiv_gt1}
  Let $\alpha>1$, $M\geq1$ and assume the covariance is as in
  Assumption \ref{a:noise}. Let $x\in L^2_\#(\Torus)$,
  $\theta(\cdot;x)$ be the solution of \eqref{e:spde} with initial
  condition $x$ and $z$ be the solution of \eqref{e:z} with
  initial condition $z(0) = 0$. Then for every $t>0$ the law of
  $\theta(t;x)$ is equivalent to the law of $z(t)$.
\end{theorem}
We first state some preliminary results that will be necessary for the
absolute continuity theorem given above. The first result ensures
existence and uniqueness for the solutions of \eqref{e:spde_abs}.
Its proof is quite standard and follows the lines of the proof of
\cite[Theorem 2.9]{Fla2008}.
\begin{lemma}\label{l:exist_uniq}
  Let $\alpha\geq1$ and $M\geq1$, and let Assumption~\ref{a:noise}
  be in force. Given a probability $\mu$ on $H^{-1}_\#(\Torus)$
  with all polynomial moments finite in $H^{-1}_\#(\Torus)$, there
  exists a unique (path--wise) solution $\theta$ of \eqref{e:spde_abs}
  with initial distribution $\mu$ such that
  $\theta\in C([0,\infty);H^{-1}_\#(\Torus))\cap
  L^2_{\textrm{\tiny loc}}([0,\infty);L^2_\#(\Torus))$. Moreover for every $m\geq1$
  and $T>0$,
  \[
    \begin{gathered}
      \mathbb{E}\Bigl[\sup_{[0,T]}\|\theta(t)\|_{-M}^{2m}\Bigr]
          + \nu\mathbb{E}\Bigl[\int_0^T\|\theta(t)\|_{-M}^{2m-2}\|\theta(t)\|_{\alpha-M}^2\,dt\Bigr]
        <\infty,\\
      \mathbb{E}\Bigl[\log\Bigl(1 + \sup_{[0,T]}\|\theta(t)\|_{-1}^2
          + \nu\int_0^T\|\theta(t)\|_{L^2}^2\,dt\Bigr)\Bigr]
        <\infty.
    \end{gathered}
  \]
  Denote by $\theta(\cdot;x)$ the solution with initial distribution
  concentrated at $x$. Then the process
  $(\theta(\cdot;x))_{x\in H^{-1}_\#(\Torus)}$
  is a Markov process and the associated transition semigroup is Feller
  in $H^{-1}_\#(\Torus)$.

  If additionally $\mu$ has second moment finite in $L^2_\#(\Torus)$,
  then for every $\gamma<\alpha+\delta-1$,
  \[
    \mathbb{E}\Bigl[\log\Bigl(1 + \sup_{[0,T]}\|\theta(t)\|_{L^2}^2
        + \nu\int_0^T\|\theta(t)\|_\gamma^2\,dt\Bigr)\Bigr]
      <\infty.
  \]
  Moreover, the process $(\theta(\cdot;x))_{x\in L^2_\#(\Torus)}$
  is Feller in $L^2_\#(\Torus)$.
\end{lemma}
The next preliminary ingredient is to prove that there exists an invariant
measure for problem~\eqref{e:spde_abs} which has all polynomial moments
finite in $L^2$ (and better). This is done following
(almost) \cite{EssSta2010}.
\begin{lemma}\label{l:improved}
  Let $\alpha\geq1$, $M\geq1$ and let Assumption~\ref{a:noise} be true.
  Then there exists an invariant measure $\mu$ for the transition
  semigroup associated to problem~\eqref{e:spde_abs}. Moreover,
  for every $\gamma<\delta+\alpha-1$ and $m\geq1$ there is a
  number $\mcdef{cc:improve}>0$ such that

  \begin{equation}\label{e:im_bound}
    \int \|x\|_{L^2}^{2m-2}\|x\|_\gamma^2\,\mu(dx)
      \leq \mcref{cc:improve}.
  \end{equation}

\end{lemma}
\begin{proof}
  Consider the Galerkin approximations of \eqref{e:spde_abs}

  \begin{equation}\label{e:spde_approx}
    d\theta_N + (\nu A^\alpha\theta_N + \pi_NB(u_N,\theta_N))\,dt
      = \pi_N\mathcal{C}^{\frac12}dW,
  \end{equation}

  where $\pi_N$ is the projection onto $\Span[e_k: |k|\leq N]$
  and $u_N = \nabla^\perp A^{-M}\theta_N$.
  It is fairly standard (see \cite{Fla2008}) to prove that for every $N$
  the above system admits an invariant
  measure $\mu_N$. If we are able to prove \eqref{e:im_bound} for
  each $\mu_N$ with a constant $\mcref{cc:improve}$ independent
  of $N$, then the lemma is proved. Indeed, the uniform bound ensures
  tightness of $(\mu_N)_{N\geq1}$ and, consequently, of the laws of
  each of the stationary solution of \eqref{e:spde_approx}
  with initial condition $\mu_N$. The same methods of the previous
  lemma ensure that, up to a sub--sequence, there is a solution of
  \eqref{e:spde_abs} which is limit of stationary laws, hence
  stationary itself. Its marginal $\mu$ at fixed time turns out to
  be an invariant measure for \eqref{e:spde_abs} and a limit point
  of $(\mu_N)_{N\geq1}$. By semi--continuity $\mu$
  verifies \eqref{e:im_bound}.

  It remains to prove \eqref{e:im_bound} for the Galerkin system.
  Given $N\geq1$, let $\theta_N$ be the stationary solution of
  \eqref{e:spde_approx} with marginal law $\mu_N$.

  \emph{Step 1: estimates for the linear part}.
  For every $\lambda>0$ consider the solution $z_{\lambda,N}$
  of the following problem,
  \[
      dz_{\lambda,N}
          + (\nu A^\alpha z_{\lambda,N}
          + \lambda z_{\lambda,N})\,dt
        = \pi_N\mathcal{C}^\frac12 dW,
  \]
  with initial condition $z_{\lambda,N}(0) = 0$, and recall that
  $W(t) = \sum_{k\in\mathbf{Z}^2_\star} \beta_k e_k$, with
  $(\beta_k)_{k\in\mathbf{Z}^2_\star}$ independent standard Brownian motions.
  Set for every $T>0$, $a\in(0,\tfrac12)$,
  $\epsilon\in(0,\alpha+\delta-1)$ and $\beta\in(0,1)$
  such that $\alpha(1-2a(1-\beta))<(\alpha+\delta-1-\epsilon)$,
  \[
    M_{a,\epsilon,\beta}(T)^2
      = \sum_{k\in\mathbf{Z}^2_\star} \frac1{|k|^{4a\alpha(1-\beta)+2\delta-2\epsilon}}
          \Bigl(\sup_{0\leq s< t\leq T}\tfrac{|\beta_k(t) - \beta_k(s)|}{|t - s|^a}\Bigr)^2.
  \]
  From Proposition 2.1 and Corollary 2.2 of \cite{EssSta2010} we have that
  $\mathbb{E}[M_{a,\epsilon,\beta}(T)^{2m}]\leq\stepcounter{ConstantsCounter}\ensuremath{c_\text{\theConstantsCounter}} T^{m(1-2a)}$
  and, $\mathbb{P}-a.s.$,  $\|z_{\lambda,N}(t)\|_\epsilon^2\leq
  \mcdef{cc:zlambda}\lambda^{-2a\beta}M_{a,\epsilon,\beta}(T)^2$.
  For the rest of the proof fix values of $a$ and $\beta$ as required
  above.

  \emph{Step 2: estimates for the non--linear part}.
  Set $\eta_{\lambda,N} = \theta_N - z_{\lambda,N}$, then
  $\eta_{\lambda,N}$ solves
  \[
    \dot\eta_{\lambda,N}
        + \nu A^{\alpha}\eta_{\lambda,N}
        + \pi_N B_M(\theta_N)
      = \lambda z_{\lambda,N},
  \]
  with initial condition $\eta_{\lambda,N}(0) = \theta_N(0)$.
  For every $m\geq1$,
  \[
    \begin{multlined}[.9\linewidth]
      \frac{d}{dt}(1 + \|\eta_{\lambda,N}\|_{L^2}^2)^m
        = 2m(1 + \|\eta_{\lambda,N}\|_{L^2}^2)^{m-1}
          \bigl(
            - \nu\|\eta_{\lambda,N}\|_\alpha^2 + \\
            - \scal{\eta_{\lambda,N}, B_M(\theta_N)}
            + \lambda\scal{z_{\lambda,N},\eta_{\lambda,N}}
          \bigr),
    \end{multlined}
  \]
  hence using \eqref{e:B1}, H\"older's inequality and Sobolev's embeddings,
  \[
    \scal{\eta_{\lambda,N}, B_M(\theta_N)}
      \leq \frac\nu4\|\eta_{\lambda,N}\|_\alpha^2
           + \mcdef{cc:ineq}\|z_{\lambda,N}\|_\epsilon^4
           + \mcref{cc:ineq}\|z_{\lambda,N}\|_\epsilon^2\|\eta_{\lambda,N}\|_{L^2}^2,
  \]
  where $\epsilon\in(0,\alpha+\delta-1)$ can be chosen arbitrarily
  small (and $\mcref{cc:ineq} = \mcref{cc:ineq}(\epsilon)$, although is
  independent from $N$). Young's inequality and the inequalities
  of the previous step yield,
  \[
    \begin{aligned}
      \lefteqn{\frac{d}{dt}(1 + \|\eta_{\lambda,N}\|_{L^2}^2)^m
          + \frac32\nu m(1 + \|\eta_{\lambda,N}\|_{L^2}^2)^{m-1}\|\eta_{\lambda,N}\|_\alpha^2\leq}\qquad\quad\\
      &\leq \mcdef{cc:young}m(1 + \|\eta_{\lambda,N}\|_{L^2}^2)^{m-1}
          (\lambda^4+\|z_{\lambda,N}\|_\epsilon^4
          + \|z_{\lambda,N}\|_\epsilon^2\|\eta_{\lambda,N}\|_{L^2}^2)\\
      &\leq \mcref{cc:young}m(1 + \|\eta_{\lambda,N}\|_{L^2}^2)^{m-1}
          (\lambda^4+\|z_{\lambda,N}\|_\epsilon^4
          + \mcref{cc:zlambda}\lambda^{-2a\beta}M_{a,\epsilon,\beta}(T)^2
          \|\eta_{\lambda,N}\|_{L^2}^2).
  \end{aligned}
  \]
  Consider $\omega\in\{M_{a,\epsilon,\beta}(T)\leq R\}$ and choose
  $\lambda = \lambda_R$ so that
  \[
    \mcref{cc:young}\mcref{cc:zlambda}R^2\lambda_R^{-2a\beta}
      = \frac{\nu}4,
  \]
  then by using the Poincaré inequality and again Young's inequality,
  \[
    \frac{d}{dt}(1 + \|\eta_{\lambda,N}\|_{L^2}^2)^m
        + \nu m(1 + \|\eta_{\lambda,N}\|_{L^2}^2)^{m-1}
          \|\eta_{\lambda,N}\|_\alpha^2
      \leq \mcdef{cc:fine}(1 + \lambda_R^{4m}
        + \|z_{\lambda,N}\|_\epsilon^{4m}).
  \]
  Finally, by integrating in $[0,T]$, on the event
  $\{M_{a,\epsilon,\beta}(T)\leq R\}$,

  \begin{equation}\label{e:pathwise}
    \nu m\int_0^T \|\eta_{\lambda_R,N}\|_{L^2}^{2m-2}\|\eta_{\lambda_R,N}\|_\alpha^2\,dt
      \leq (1 + \|\theta(0)\|_{L^2}^2)^m
        + \mcref{cc:fine}\int_0^T(1 + \lambda_R^{4m} + \|z_{\lambda_R,N}\|_{L^2}^{4m})\,dt.
  \end{equation}

  \emph{Step 3: estimates in $\lambda_R$}.
  Define for every integer $R\geq1$ the events
  $A_R = \{R-1 < M_{a,\epsilon,\beta}(T)\leq R\}$.
  By Fernique's theorem (see for instance \cite{DapZab1992}),
  for every $q\geq2$ and every $\epsilon<\alpha+\delta-1$,
  \[
    \mathbb{E}[\|z_{\lambda_R,N}(t)\|_\epsilon^q]
      \leq \stepcounter{ConstantsCounter}\ensuremath{c_\text{\theConstantsCounter}}\bigl(\mathbb{E}[\|z_{\lambda_R,N}\|_\epsilon^2]\bigr)^{\frac{q}2}
      \leq \stepcounter{ConstantsCounter}\ensuremath{c_\text{\theConstantsCounter}}\Bigl(\sum_{k\in\mathbf{Z}^2_\star}\frac{|k|^{2\epsilon-2\delta}}
        {\lambda_R + \nu|k|^{2\alpha}}\Bigr)^{\frac{q}2}
      \leq \stepcounter{ConstantsCounter}\ensuremath{c_\text{\theConstantsCounter}}\lambda_R^{-\frac{q(\alpha+\delta-1-\epsilon)}{2\alpha}},
  \]
  since $\lambda_R + \nu|k|^2\geq\lambda_R\vee(\nu|k|^2)$.
  By our choice of $\lambda_R$, $a$, and $\beta$, if $q\geq2$,

  \begin{equation}\label{e:stima1}
    \mathbb{E}\Bigl[\sum_{R=1}^\infty \mathbbm{1}_{A_R}\int_0^T
        \|z_{\lambda_R,N}(t)\|_\epsilon^q\,dt\Bigr]
      \leq \mathbb{E}\Bigl[\sum_{R=1}^\infty\int_0^T \|z_{\lambda_R,N}(t)\|_\epsilon^q\,dt\Bigr]
      \leq \stepcounter{ConstantsCounter}\ensuremath{c_\text{\theConstantsCounter}} T.
  \end{equation}

  Likewise,

  \begin{equation}\label{e:stima2}
    \mathbb{E}\Bigl[\sum_{R=1}^\infty \mathbbm{1}_{A_R}\lambda_R^q\Bigr]
      \leq \stepcounter{ConstantsCounter}\ensuremath{c_\text{\theConstantsCounter}}\mathbb{E}\Bigl[(1+M_{a,\beta}(T))^{\frac{q}{a\beta}}\Bigr]
      \leq \stepcounter{ConstantsCounter}\ensuremath{c_\text{\theConstantsCounter}} (1 + T^{q\frac{1-2a}{2a\beta}}).
  \end{equation}

  Finally, recall \eqref{e:pathwise} and use \eqref{e:stima1}
  and \eqref{e:stima2} to obtain

  \begin{equation}\label{e:stima3}
    \mathbb{E}\Bigl[\sum_{R=1}^\infty\mathbbm{1}_{A_R}\int_0^T
        \|\eta_{\lambda_R,N}\|_{L^2}^{2m-2}\|
        \eta_{\lambda_R,N}\|_\alpha^2\,dt\Bigr]
      \leq \stepcounter{ConstantsCounter}\ensuremath{c_\text{\theConstantsCounter}}(1 + \mathbb{E}_{\mu_N}[\|x\|_{L^2}^{2m}]
        + T + T^{2m\frac{1-2a}{a\beta}}).
  \end{equation}

  \emph{Step 4: conclusion.}
  By assumption $\mathbb{E}_{\mu_N}[\|x\|_{L^2}^{2m}]$ is uniformly bounded
  in $N$ for every $m\geq1$, due to the estimate in $H^{-M}_\#(\Torus)$.
  Fix $\gamma\in(0,\alpha+\delta-1)$, then by \eqref{e:stima1}
  and \eqref{e:stima3},
  \[
    \begin{aligned}
      \lefteqn{T\mathbb{E}_{\mu_N}[\|x\|_{L^2}^{2m-2}\|x\|_\gamma^2]=}\qquad&\\
        &= \mathbb{E}\Bigl[\sum_{R=1}^\infty\mathbbm{1}_{A_R}\int_0^T \|\theta_N(t)\|_{L^2}^{2m-2}\|\theta_N\|_\gamma^2\,dt\Bigr]\\
        &\leq \stepcounter{ConstantsCounter}\ensuremath{c_\text{\theConstantsCounter}}\mathbb{E}\Bigl[\sum_{R=1}^\infty\mathbbm{1}_{A_R}
          \Bigl(\int_0^T \|z_{\lambda_R,N}(t)\|_\gamma^{2m}\,dt
          + \int_0^T\|\eta_{\lambda_R,N}(t)\|_{L^2}^{2m-2}
          \|\eta_{\lambda_R,N}(t)\|_\alpha^2\,dt\Bigr)\Bigr]\\
        &\leq \mcdef{cc:opt}\bigl(1 + T + T^{2m\frac{1-2a}{a\beta}}
          + \mathbb{E}_{\mu_N}[\|x\|_{L^2}^{2m}]\bigr).
    \end{aligned}
  \]
  The above inequality holds for all $T>0$, hence if we take
  $T = 2\mcref{cc:opt}$ and use the Poincaré inequality,
  we obtain \eqref{e:im_bound} for $\mu_N$ (with a constant
  uniform in $N$).
\end{proof}
Next, we show that problem \eqref{e:spde_abs} has a unique invariant
measure which is strongly mixing. Moreover, the strong Feller property
ensures that the law of $\theta(t)$ is equivalent to the law of the
invariant measure, for every $t>0$. This allows us to reduce
absolute continuity of laws at each time to absolute continuity
of the invariant measures. As a marginal remark, we notice that
the next lemma holds also when $\alpha=1$. The assumption $\alpha>1$
simplifies slightly the proof and it is what we need to prove the
theorem.
\begin{lemma}\label{l:uniq}
  Let $\alpha>1$, $M\geq1$ and let Assumption~\ref{a:noise} be
  true. Then the transition semigroup is strong Feller in
  $H^{-1}_\#(\Torus)$ and problem \eqref{e:spde_abs} admits
  a unique invariant measure which is strongly mixing.
\end{lemma}
\begin{proof}
  Uniqueness and strong mixing follow immediately from Doob's theorem
  (see \cite{DapZab1996}) if we prove that the transition
  semigroup is strong Feller and irreducible in $H^{-1}_\#(\Torus)$.
  The proof is very similar to \cite{FlaMas1995}, we give only a sketch
  and the key estimates.

  Preliminarily, we prove the strong Feller property in
  $H^\epsilon_\#(\Torus)$, with $\epsilon>0$ small.
  Let $\chi:[0,\infty)\to\mathbf{R}$ be a smooth non--increasing
  function such that $\chi\equiv1$ on $[0,1]$ and
  $\chi\equiv0$ on $[2,\infty)$ and set $\chi_R(r)=\chi(r/R^2)$
  and $B^R_M(x) = \chi_R(\|x\|_\epsilon^2)B_M(x)$ for
  $\epsilon\in(0,\alpha+\delta-1)$. The cut--off problem is
  \[
    d\theta_R + (\nu A^{\alpha}\theta_R + B^R_M(\theta_R))\,dt
      = \mathcal{C}^{\frac12}dW.
  \]
  It is easy to show, as in Lemma \ref{l:exist_uniq}, that the
  above equation has a unique solution in $C([0,\infty);H^\epsilon_\#)$
  for each initial condition in $H^\epsilon_\#(\Torus)$.
  Strong Feller in $H^\epsilon_\#$ follows from the Bismut--Elworthy--Li
  formula \cite{ElwLix1994}, which yields for every bounded
  measurable $\varphi$ and every $x,h\in H^\epsilon_\#(\Torus)$,
  \[
    \begin{aligned}
      |P_t^R\varphi(x+h) - P_t^R\varphi(x)|
        &\leq \frac{\stepcounter{ConstantsCounter}\ensuremath{c_\text{\theConstantsCounter}}}{\sqrt{t}}\|\varphi\|_\infty
           \mathbb{E}\Bigl[\Bigl(\int_0^t\|\mathcal{C}^{-\frac12}
           D_h\theta_R(s;y)\|_{L^2}^2\,ds\Bigr)^{\frac12}\Bigr]\\
        &\leq\frac{\stepcounter{ConstantsCounter}\ensuremath{c_\text{\theConstantsCounter}}}{\sqrt{t}}\|\varphi\|_\infty
           \mathbb{E}\Bigl[\int_0^t\|D_h\theta_R(s;y)\|_\delta^2\,ds\Bigr]^{\frac12},
    \end{aligned}
  \]
  where $(P_t^R)_{t\geq0}$ is the transition semigroup corresponding to
  $\theta_R$ and $D_h\theta_R(\cdot;y)$ is the Gateaux derivative (with
  respect to the initial condition) of $\theta_R$ in the direction $h$.
  Let $\xi(s;y) = D_h\theta_R(s;y)$, then it is sufficient to compute
  the ``energy estimate'' of $\xi$ in $L^2(\Torus)$, and the ``dissipative''
  term will provide the estimate we need. Clearly, the most troublesome term
  is the non--linearity, which is estimated as follows,
  \[
    \begin{aligned}
      \scal{DB_R(\theta_R)\xi, \xi}_{L^2}
        &\leq\stepcounter{ConstantsCounter}\ensuremath{c_\text{\theConstantsCounter}}\chi_R'(\|\theta_R\|_\epsilon^2)
          \|\theta_R\|_\epsilon^3\|\xi\|_{L^2}\|\xi\|_\alpha
          + \stepcounter{ConstantsCounter}\ensuremath{c_\text{\theConstantsCounter}}\chi_R(\|\theta_R\|_\epsilon^2)
          \|\theta_R\|_\epsilon\|\xi\|_{L^2}\|\xi\|_\alpha\\
        &\leq \nu\|\xi\|_\alpha^2 + \stepcounter{ConstantsCounter}\ensuremath{c_\text{\theConstantsCounter}} R^6\|\xi\|_{L^2}^2.
    \end{aligned}
  \]

  Next, we prove strong Feller in $H^{-1}_\#$, following an idea in
  \cite{Rom10a}. We know that if $\varphi:H^{-1}_\#(\Torus)\to\mathbf{R}$
  is bounded measurable, then $P_t\varphi\in C_b(H^\epsilon_\#)$,
  and we want to prove that $P_t\varphi\in C_b(H^{-1}_\#)$.
  To this end, let $x_n\to x$ in $H^{-1}_\#$, then it is sufficient
  to show that $\theta(t;x_n)\to\theta(t;x)$ {a.s.} in $H^\epsilon_\#$,
  since by the Markov property and the Lebesgue theorem,
  $P_t\varphi(x_n) = \mathbb{E}[P_{t/2}\varphi(\theta(t/2;x_n))]$ converges
  to $\mathbb{E}[P_{t/2}\varphi(\theta(t/2;x))] = P_t\varphi(x)$.

  Set $w_n(t) = \theta(t;x_n) - \theta(t;x) = \theta_n - \theta$,
  and choose $\epsilon$ small enough so that $\epsilon+1-\alpha\leq0$,
  then the energy inequality in $H^\epsilon_\#$ yields
  \[
    \begin{aligned}
    \frac{d}{dt}\|w_n\|_\epsilon^2
         + 2\nu\|w_n\|_{\alpha+\epsilon}^2
      &= - 2\scal{A^\epsilon w_n,B_M(\theta_n) - B_M(\theta)}\\
      &\leq \nu\|w_n\|_{\alpha+\epsilon}^2
         + \mcdef{cc:sf}(\|\theta_n\|_{L^2}^2 + \|\theta\|_{2-2M+\epsilon}^2)\|w_n\|_{L^2}^2,
    \end{aligned}
  \]
  hence by Gronwall's lemma, for every $s\leq t$,
  \[
    \|w_n(t)\|_\epsilon^2
      \leq \|w_n(s)\|_\epsilon^2\operatorname{e}^{\mcref{cc:sf}\int_s^t(\|\theta_n\|_{L^2}^2 + \|\theta\|_{2-2M+\epsilon}^2)}
      \leq \|w_n(s)\|_\epsilon^2\operatorname{e}^{\mcref{cc:sf}\int_0^t(\|\theta_n\|_{L^2}^2 + \|\theta\|_{2-2M+\epsilon}^2)}.
  \]
  Notice that the exponential term is $\mathbb{P}$--{a.~s.} finite by
  the bounds in Lemma \ref{l:exist_uniq}. Integrate the above inequality
  for $s\in[0,t]$ to get
  \[
    \|w_n(t)\|_{L^2}^2
      \leq \frac1{t}\Bigl(\int_0^t\|w_n(s)\|_{L^2}^2\,ds\Bigr)\operatorname{e}^{\int_0^t\|\theta\|_{L^2}^2\,ds}.
  \]
  So it is sufficient to prove that $\int_0^t\|w_n(s)\|_{L^2}^2\,ds\to0$.
  Use \eqref{e:B1}, H\"older's inequality and Sobolev's
  embeddings on the energy inequality in $H^{-1}_\#$ for $w_n$,
  to get
  \[
    \begin{aligned}
      \frac{d}{dt}\|w_n\|_{-1}^2 + 2\nu\|w_n\|_{\alpha-1}^2
        &= -\scal{w_n, B(\nabla^\perp A^{-M}\theta_n, w_n) + B(\nabla^\perp A^{-M}w_n, \theta)}_{-1}\\
        &\leq \nu\|w_n\|_{L^2}^2
          + \mcdef{cc:gron}\|\theta\|_{L^2}^2\|w_n\|_{-1}^2.
    \end{aligned}
  \]
  The Gronwall inequality finally yields
  \[
    \|w_n(t)\|_{-1}^2 + \nu\int_0^t \|w_n(s)\|_{L^2}^2\,ds
      \leq \|x_n - x\|_{-1}\operatorname{e}^{\mcref{cc:gron}\int_0^t\|\theta\|_{L^2}^2\,ds},
  \]
  and the right hand side converges to $0$, $\mathbb{P}$--{a.~s.}, since
  $x_n\to x$ in $H^{-1}_\#$.

  Similar computations also yield irreducibility as in \cite{FlaMas1995}
  (see also \cite{Fer1997,Fer1999}).
\end{proof}
We have all elements to prove Theorem \ref{t:equiv_gt1}.
Since both problems \eqref{e:z} and \eqref{e:spde_abs} satisfy
the strong Feller property and have irreducible transition
probabilities, the law of each process at some time $t>0$ and
the corresponding invariant measure are equivalent measures.
It is then sufficient to show equivalence of the invariant
measures. To this end choose $\epsilon\in(0,1)$ such that
$\epsilon>2-\alpha-\delta$ and $\epsilon<\alpha-\delta$
(which is possible since $\alpha>1$),
and let $\theta_\epsilon=A^{-\epsilon/2}\theta$ and
$z_\epsilon=A^{-\epsilon/2}z$. The new process
$\theta_\epsilon$ solves
\begin{equation}\label{e:spdepsilon}
  d\theta_\epsilon
      + \nu A^\alpha\theta_\epsilon
      + B_{M,\epsilon}(\theta_\epsilon)
    = \mathcal{C}_\epsilon^{\frac12}\,dW,
\end{equation}
where $B_{M,\epsilon}(x)=A^{-\epsilon/2}B_M(A^{\epsilon/2}x)$
and $\mathcal{C}_\epsilon=A^{-\epsilon}\mathcal{C}$, and similarly for
$z_\epsilon$. We will prove equivalence of the invariant measures
of $\theta_\epsilon$ and $z_\epsilon$ using
\cite[Theorem 3.4]{DapDeb2004} on $\theta_\epsilon$ and
$z_\epsilon$.
\begin{proof}[Proof of Theorem \ref{t:equiv_gt1}]
  Before checking that the assumptions of the aforementioned
  theorem are satisfied, we show how to deduce the equivalence
  of measures. Indeed, the theorem in \cite{DapDeb2004} shows
  the following formula
  \[
    (\lambda - L_\epsilon)^{-1}
      = (\lambda - N_\epsilon)^{-1}
        - (\lambda - N_\epsilon)^{-1}
        \scal{B_{M,\epsilon}, D(\lambda - L_\epsilon)^{-1}},
      \qquad\lambda>0,
  \]
  where $L_\epsilon$ and $N_\epsilon$ are the generators of
  the Markov semigroups associated to $z_\epsilon$ and
  $\theta_\epsilon$, respectively. To prove equivalence
  it is sufficient to show that for every bounded measurable
  function $f$ on $L^2_\#$, $(\lambda - L_\epsilon)^{-1}f=0$
  if and only if $(\lambda - N_\epsilon)^{-1}f=0$.
  If $(\lambda - L_\epsilon)^{-1}f=0$, the conclusion is
  immediate (it is already in \cite{DapDeb2004}). Assume
  that $(\lambda - N_\epsilon)^{-1}f = 0$ and set
  $\phi_\lambda = (\lambda - L_\epsilon)^{-1}f=0$. By
  the formula above,
  \[
    \phi_\lambda
        + (\lambda - N_\epsilon)^{-1}
        \scal{B_{M,\epsilon}, D\phi_\lambda}
      =0.
  \]
  Clearly, $\phi_\lambda\in D(L_\epsilon)$ and, by
  Proposition 2.4 of \cite{DapDeb2004}, $\phi_\lambda\in C^1_b$.
  Therefore $\phi_\lambda\in D(N_\epsilon)$ and
  $(\lambda - N_\epsilon)\phi_\lambda = (\lambda -
  L_\epsilon)\phi_\lambda - \scal{B_{M,\epsilon},D\phi_\lambda}$.
  By the formula above it follows that
  $(\lambda-L_\epsilon)\phi_\lambda=0$, that is $\phi_\lambda=0$.
  This prove the equivalence.

  We turn to the verification of the assumptions of Theorem 3.4
  of \cite{DapDeb2004}. The assumptions for the
  linear problem \cite[Hypothesis 2.1]{DapDeb2004} are standard
  and can be verified as in Section 4 of the mentioned paper. We
  only give a few details on the fourth hypothesis, namely that
  $\|\Lambda_t\|$ admits Laplace transform defined on $(-1,\infty)$,
  where $\Lambda_t = \mathcal{C}_{\epsilon,t}^{-1/2}\operatorname{e}^{-tA^\alpha}$ and
  \[
    \mathcal{C}_{\epsilon,t}
      = \int_0^t \operatorname{e}^{-sA^\alpha}\mathcal{C}_\epsilon\operatorname{e}^{-s(A^\alpha)^*}\,ds.
  \]

  By our assumptions each $e_k$ (the sine--cosine orthonormal basis)
  is an eigenvector of $\Lambda_t$. Denote by $\lambda_{t,k}$ the
  corresponding eigenvalue. Elementary computations yield
  \[
    \lambda_{t,k}
      = \Bigl(|\sigma_k|^2\frac{1-\operatorname{e}^{-2t|k|^{2\alpha}}}
        {2|k|^{2(\alpha+\epsilon)}}\Bigr)^{-\frac12}\operatorname{e}^{-t|k|^{2\alpha}}
      \sim\frac{\stepcounter{ConstantsCounter}\ensuremath{c_\text{\theConstantsCounter}}}{t^{\frac{\alpha+\delta+\epsilon}{2\alpha}}}
  \]
  for $t$ small. The existence of the Laplace transform follows, since
  $\|\Lambda_t\|=\sup_k\lambda_{t,k}$ and
  $\delta+\epsilon<2-\alpha\leq\alpha$.

  As it regards the assumptions for the non--linear problem,
  Lemma \ref{l:exist_uniq} and Lemma \ref{l:uniq} ensure that
  problem \eqref{e:spdepsilon} has a unique solution and generates
  a Feller semigroup in $L^2_\#$. Moreover the semi--group has
  a unique invariant measure $\mu_\epsilon$ which is strongly mixing.
  Let
  \[
    B_{M,\epsilon,N}(x)
      = \frac{N}{N + \|x\|_{L^2}}\pi_N B_M(\pi_N x),
  \]
  where $\pi_N$ is the projection onto $\Span[e_k: |k|\leq N]$,
  then $B_{M,\epsilon,N}$ is Lipschitz--continuous in
  $L^2_\#$. Moreover, if $\theta_{\epsilon,N}$ is the solution of
  \eqref{e:spdepsilon} when $B_{M,\epsilon}$ is replaced
  by $B_{M,\epsilon,N}$, then $\theta_{\epsilon,N}(t)
  \to\theta_\epsilon(t)$ for all $t>0$ and all
  $x\in L^2_\#$, whenever $\theta_{\epsilon,N}(0)=\theta_\epsilon(0)$.
  It remains to prove the main assumption of \cite[Theorem 3.4]{DapDeb2004},
  namely that $B_{M,\epsilon,N}(x)\to B_{M,\epsilon}(x)$, $\mu$--{a.~s.}
  in $L^2_\#$, and that there is a $g\in L^2(L^2_\#,\mu)$ such that
  $\|B_{M,\epsilon,N}(x)\|_{L^2}\leq g(x)$, $\mu$--{a.~s.}.
  Let $\epsilon_M=0$ if $M>1$, and an arbitrary value in
  $\bigl(0,\epsilon-(2-\delta-\alpha)\bigr)$ if $M=1$, and
  let $g_\epsilon=\mcdef{cc:epsm}\|\cdot\|_{\epsilon+\epsilon_M}
  \|\cdot\|_1$. Lemma \ref{l:improved} and interpolation
  immediately imply that $g_\epsilon\in L^2(L^2_\#,\mu_\epsilon)$
  since,
  \[
    \mathbb{E}^{\mu_\epsilon}[g_\epsilon^2(x)]
      \leq \mcref{cc:epsm}\mathbb{E}^\mu
        [\|x\|_{L^2}^2\|x\|_{1-\epsilon+\epsilon_M}^2].
  \]
  Moreover, by choosing $\mcref{cc:epsm}$ large enough, we have
  that $\|B_{M,\epsilon,N}(x)\|_{L^2}\leq g_\epsilon(x)$.
  Indeed, by using the embedding of $L^\infty$ into $H^{1+\gamma}$,
  for $\gamma\leq\epsilon_M\vee 2(M-1)$, we have that
  \[
    \|B_M(x)\|_{L^2}
      \leq \stepcounter{ConstantsCounter}\ensuremath{c_\text{\theConstantsCounter}}\|x\|_{\epsilon_M}\|x\|_1,
        \qquad
    \|B_M(x)\|_{-1}
      \leq \stepcounter{ConstantsCounter}\ensuremath{c_\text{\theConstantsCounter}}\|x\|_{\epsilon_M}\|x\|_0,
  \]
  and hence $\|B_M(x)\|_{-\epsilon}\leq \stepcounter{ConstantsCounter}\ensuremath{c_\text{\theConstantsCounter}}\|x\|_{\epsilon_M}
  \|x\|_{1-\epsilon}$.
\end{proof}

redpajama_set_name: RedPajamaArXiv


text:
Tips On How To Flirt In Croatian
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Which A single Is Best
Toko Bunga Lombok Toko Bunga Lombok 0821

redpajama_set_name: RedPajamaCommonCrawl


Play free game Keeper Of The Grove on bestgames.pk.
The Peacekeeper is a shooting game from www.bestgames.pk. You can choose your weapons in the shop. You have 1000 golden coins at the beginning. Equip your weapons and buy more items in the shop when you earn enough. Choose the better one for attacking. Hold the mouse to fire. Release mouse to crouch. Use up and down arrow keys to move. This game is totally free. Good luck!


As the Prime Minister returned to the House to update us on her meeting with the European Council I asked her about the jurisdiction of the European Court of Justice. The Prime Minister confirmed what we already knew, that transition would mean everything staying the same, only without us having any say at all including full jurisdiction of the European Court of Justice... So much for taking back control!
Will the Prime Minister please confirm, therefore, that the jurisdiction of the European Court of Justice will apply in precisely the same way now as it will during the transition period?


text:
Q: Chrome redirecting me to Google China When I am accessing Google sites, Chrome is automatically redirecting me to Google China.  This issue seems to be intermittent - occasionally, Chrome will direct me to the regular Google website.

*

*I am in California, USA.

*This is not happening in any other browsers.

*This is not happening for any other websites/search engines.

*I have checked my hosts file, and that seems to be fine.

Any suggestions on what this problem could be and how I can fix it?
Thanks

A: Try going to www.google.com/ncr. This should solve the problem.

A: After searching from Google China, try clicking on the top right corner with a "cog shaped" icon. Select the first option from the drop down menu "搜尋設定" This is the settings. On the left column for the next page select "言語". This is the language tab. From the following drop down menu, select "英文". This is English. The blue button at the bottom is to save. If it works, it should revert back to English.
EDIT: To be fair, you might still be on the Chinese one, though after I tried this, I was redirected to the regular http://www.google.com. If it doesn't redirect, you will at least get English even though the address is .hk
(Pictures are listed below, try opening them in a new page since they are hard to see)

A: Try erasing your history and cookies in Chrome, if the burtek's answer doesn't work for you. Then try going to www.google.com/ncr again and see if it works.

redpajama_set_name: RedPajamaStackExchange


text:
Q: Emacs ipython shell error message foreground color setting In my python shell with white background it looks like this:

As you can see, when there is an error, it's white text on white background. In order to see them, I have to highlight the text. I haven't got any luck finding out what face controls those colors. Anyone has any idea?
Thanks

A: in ~\.ipython\ipythonrc there is this section (About a 1/4 away from the top)
# colors - Coloring option for prompts and traceback printouts.

# Currently available schemes: NoColor, Linux, LightBG.

# This option allows coloring the prompts and traceback printouts. This
# requires a terminal which can properly handle color escape sequences. If you
# are having problems with this, use the NoColor scheme (uses no color escapes
# at all).

# The Linux option works well in linux console type environments: dark
# background with light fonts.

# LightBG is similar to Linux but swaps dark/light colors to be more readable
# in light background terminals.

# keep uncommented only the one you want:
colors Linux
#colors LightBG
#colors NoColor


text:
The 2017 Challenge is offering a new prize category to encourage middle and high school students to make games with real-world impact!
The Games for Change Prize is open to any playable game submission that tackles a social issue, whether through game play or subject matter. This prize is supported by Games for Change (G4C), the leading organization promoting games for social good.
Each winner of the Games for Change Prize will receive a cash prize of $1,000, as well as exposure on the STEM Challenge and G4C websites, mentoring from game industry professionals, and access to an online community of peers.
If you want your playable game to be considered for the Games for Change Prize, you will need to answer the question — What is the social impact of your game? — as part of the STEM Challenge online submission process. View the Challenge Rules to learn more
Pittsburgh, NYC and Dallas Students: Submit to the G4C Student Challenge too!
G4C also runs a youth game design competition — the G4C Student Challenge — in New York City, Dallas and Pittsburgh for students in grades 6-12. If you live in these three cities, you are eligible to submit games to both the National STEM Video Game Challenge and G4C Student Challenge through their respective websites!
The deadline for the G4C Challenge is April 1, 2017. Games must be about one of the three themes — Climate Change, Local Stories & Immigrant Voices, and Future Communities — and playable on a web browser. For more information on how to participate, visit: www.gamesforchange.org/studentchallenge/submit.


text:
FROM kbase/kbase:sdkbase.latest
MAINTAINER KBase Developer
# -----------------------------------------

# Insert apt-get instructions here to install
# any required dependencies for your module.

# RUN apt-get update
RUN cpanm -i Config::IniFiles

# Build transform
RUN cd /kb/dev_container/modules && \
    rm -rf transform && \
    rm -rf kb_model_seed && \
    rm -rf communities_api && \
    rm -rf KBaseFBAModeling && \
    rm -rf ModelSEED && \
    git clone https://github.com/danielolson5/transform && \
    cd transform && \
    git checkout develop && \
    . /kb/dev_container/user-env.sh && \
    cd /kb/dev_container/modules/transform && make && make TARGET=/kb/deployment deploy && cd ../../..

# Get perl MMagic library for identifying compressed types
RUN cpanm File::MMagic

# Get LWP::UserAgent and HTTP::Request needed for handling shock-service redirect
RUN cpanm HTTP::Request
RUN cpanm LWP::UserAgent

# Get NCBI SRATools (for fastq-dump)
RUN cd /kb/dev_container/modules && \
    mkdir NCBI_SRA_tools && cd NCBI_SRA_tools && \
    curl 'https://ftp-trace.ncbi.nlm.nih.gov/sra/sdk/2.7.0/sratoolkit.2.7.0-ubuntu64.tar.gz' -O && \
    tar zxf sratoolkit.2.7.0-ubuntu64.tar.gz && \
    cp sratoolkit.2.7.0-ubuntu64/bin/fastq-dump.2.7.0  /kb/deployment/bin/fastq-dump

# -----------------------------------------

COPY ./ /kb/module
RUN mkdir -p /kb/module/work
RUN chmod 777 /kb/module

WORKDIR /kb/module

RUN make

ENTRYPOINT [ "./scripts/entrypoint.sh" ]

CMD [ ]

redpajama_set_name: RedPajamaGithub


text:
BlogMicrosoftMicrosoft AdvertisingMicrosoft Advertising removing Average Position
Microsoft Advertising removing Average Position
So, we were right; it wasn't going to last. In July 2019, Microsoft announced that they would be keeping Average Position metrics whilst introducing new competitive metrics. Today, they have backtracked and have announced that they have decided to sunset the Average Position metric after all!
What's happened?
After initially announcing that they would keep Average Position, they have now said that from April 2020 onwards they will start to remove this metric from reporting. As they announced last year, Microsoft launched a bunch of new competitive metrics which would give us a more accurate picture of where are ads are placed on the search engine results page:
Top impression share.
Top impression share lost to rank.
Top impression share lost to budget.
Absolute top impression share.
Absolute top impression share lost to rank.
Absolute top impression share lost to budget.
You should now be familiar with these metrics as they have been available for a while in Google Ads. Microsoft has said that this change should make it easier for advertisers now that they have aligned these metrics.
Credit: Microsoft Advertising Blog, 2020
When Microsoft initially said that they were going to keep Average Position, we knew that this wasn't going to be long term
Credit: PPC hubbub, 2019
When Google announced that it was going to start depreciating the Average Position metric, this was surrounded by quite a lot of anger in the industry. The announcement from Microsoft that they were going to keep this metric gave them leverage against Google at the time. Now that the anger has calmed down, and now that everyone probably prefers the new competitive metrics, Microsoft has decided that it's the right time to follow suit with Google.
For now, you don't need to do much. But you should prepare your accounts
If you are still using Average Position in Microsoft, you should really move to the new metrics as soon as you can. This will help ensure that you have proper parity across Google & Bing too.
Founder & Contributor at Search Engine Hubbub
Dan has worked at some of the largest global network agencies. He was awarded Rising Star at The Drum Search Awards and Google voted him as one of the Top 20 Search Specialists in the UK. He founded Search Engine Hubbub (formerly PPC hubbub) after becoming frustrated with the lack of actionable insight from industry blogs.
LinkedIn - https://uk.linkedin.com/in/siondanielroberts
Latest posts by Dan Roberts (see all)
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Google Shopping Ads now available on Gmail
Microsoft Advertising has a new interface
Microsoft Advertising has launched a preview of a new interface design which is aimed at helping advertisers achieve more with […]
Microsoft Advertising keeps Average Position and introduces new metrics
As we all know, Google Ads announced in February that it will soon be sunsetting Average Position metrics in favour […]


The army of the confederation is on the move again. For the Undying High Lord Milo Moray is ready to take the next step in his master plan to reunite all the tribes which centuries ago formed a single, powerful nation known as the United States of America.
Before the Confederation forces lie the Armehnee Mountains, the home of the savage tribes that constantly raid the lowlands, bringing with them destruction and death. But Milo's forces are about to face an even more dangerous enemy than the Armehnee. For the Witchmen—twentieth-century scientists who have achieved a kind of immortality by stealing the living bodies of men while destroying their souls—have long been at work in the mountains. And unbeknownst to Milo, his troops are marching into much more trouble than they bargained for—trouble that could spell the end of the Confederation!
On the lip of the smoking fissure known as the Sacred Hoofprint the Witchmen's trap was at last sprung. A small charge exploded, hurling a barrel-sized charge into the fissure. Far it fell, deeper and deeper into the very bowels of the uneasy mountain. And then, unheard by living ear, the charge exploded.
Many leagues away, Bili Morguhn felt a sudden, terrible uneasiness. And when a living carpet of small, scuttling beasts broke out of the woods to rocket downslope over the edge of the plateau, Bili let his instinct guide him. "Mount!" he roared, and had barely followed his own orders when the very earth began to shudder.


There is a gap between natural theology and revealed theology. Arguing for a prime mover is not the same thing as arguing for any faith tradition.
This is a tough one to tackle in a whole blog entry because I totally agree. Thomas Aquinas and others' "prime mover" argument for God really only gets to the first point of theism. However, if just this initial part of the argument stands, at the very least atheism is false.
So there you have it; there is a gap between natural/general and revealed/special revelation, but it is not a huge gap and easy to cross. Show me another religion that can claim anything near as powerful as the arguments for Christianity and I'll at least give it some thought. Though I've done quite a bit of comparative religious studies and I've found other views wanting.
Denominational differences are another question altogether and doesn't belong in this particular discussion, so I'll leave that for another day.
The colors are arbitrary, just the standard colors from Microsoft Word.
Edit: At long last I have finally found the original article, see here.
Now, the original diagram had parts of the extremes cut off. The original author said he cut off the extremes because there are very few people that hold the extreme positions. The ideas are NOT from current word usage and more from the traditional concepts behind the actual words. The word gnostic in the diagram is in no way related to the term given to the heretical view called gnosticism (from ancient Greek) except to use the same Greek term, "γνῶσις gnōsis, knowledge" (from Wikipedia). Also, this table is not an attempt to make new terms. There is an important distinction to be made though, philosophically. The discussion about the existence or non-existence of God often asks, "On whom is the burden of proof?" Well, the burden of proof is on the person that is making the claim.
If a gnostic-atheist stands up and says, "There is no god, and I know it." That person bears the burden of proving that claim. Also, the strict (original meaning of the term) agnostic, if he/she stands up and claims, "We cannot know whether or not god does or does not exist." That too is a claim and must be defended. Likewise, if the gnostic-theist says, "I know there is a God." Then he/she must prove it. Now, this is all philosophically speaking, and thinking like a debate wherein there are rules by which one must abide. If, you're simply seeking a worldview that makes sense, there aren't really "rules," it becomes more about what evidences and arguments to which one is willing to listen.
To absolutely affirm a negative, is to claim infinite knowledge. To affirm the negative that there is no God, is to claim infinite knowledge that there is no being that has infinite knowledge.
I assume you see the contradiction there, though some don't see how affirming a negative is claiming infinite knowledge. Here's one way to think about it, I affirm the negative that there are no such thing as unicorns (an affirmation of the negative). That means I know for certain that there are no unicorns anywhere or at any time. Ironically, unicorns are often used as examples of things that everyone knows don't exist. There's an interesting conundrum with that claim. Take the initial premise that there are an infinite number of universes, a multiverse. *This is a common claim which runs counter to the claim that the universe is uniquely fine-tuned for the existence of our galaxy and life itself. So, here we are with an infinite number of universes, and within at least one of those universes a unicorn must exist. If you don't think so, then you are misunderstanding the concept of infinite!
Back to the diagram. There are so many positions on this continuum. There's rarely anyone on the far extremes, but there are lots of people that sit somewhere in the middle. People that don't think there's a god, but allow that there might be one. People that think there might be a god, but aren't sure about his/her/its existence. There are people that don't care and sit right in the middle. Let's not redefine terms, let's stick to these ideas. Language is vague and that makes things difficult. People might say, "I'm an atheist," and they're really saying they're an agnostic-atheist. Same with theists, some claim that God exists, and if they're being honest they mean they're pretty sure that God exists, but they're not sure. I often hear internet-atheists claim that atheists aren't making claims at all, which is pure ignorance. If you say you don't believe in god because there's no evidence you're not being honest with yourself or those you're dealing with. There is plenty of evidence, the problem isn't the lack of evidence, it's what you will accept as evidence. Will you accept a simple straightforward argument? If not, if you try to explain it away with some kind of untestable theory (like the aforementioned multiverse) then it's not about a lack of evidence, it's about outright dismissal of evidence.
Edit: After a short conversation with a coworker there may be an edit in order for affirming a negative. Here's a possible exception to the idea that affirming a negative requires infinite knowledge. If I make the claim that there is no such thing as a square circle, I do not think it requires infinite knowledge to affirm a philosophically contradictory claim. So, to say, there is no such thing as a square circle or a married bachelor, is affirming an already established contradiction and does not require infinite knowledge. Here's an interesting point that this route would take an atheist should he/she go down the route of philosophical arguments. Many atheists, including John Loftus (though to be fair, he only speaks out against the philosophy of religion and not philosophy itself), are against philosophy, and for good reason too. Looking at this site, the logical arguments against theism are just weak. I'm only an elementary philosopher and I can see through these arguments like glass. Also, in all my discussions with atheists both face-to-face and online, they have always appealed to science. In fact, all I've ever heard from atheists, including in their books, is the mantra "science, science, science." Philosophy is on God's side, in a manner of speaking.
True Reason is a great offering of reasoned responses to the "New Atheists." With the popularity of these anti-theist writers, it's good to see them called out on their irrational positions. As a Christian myself it's an insult to read these anti-theists' books and be called irrational, delusional, etc. When, if one really takes a good hard look at the arguments, the converse is actually true. The Christian Apologist has much more reasonable answers than these New Atheists. This book does a great job collating and responding to the many fallacious arguments throughout the New Atheists' writings. I highly recommend this book for the apologist looking for reasonable responses to brash New Atheists' claims. It's also great for any ordinary lay person that has heard these New Atheists spouting off with bold claims of truth or anti-truth etc. and wants to hear more about these extraordinary claims. The entire text is filled with great arguments and powerful blows to poor arguments offered up in New Atheists' writings but to be the best chapter is the eighth chapter, The Explanatory Emptiness of Naturalism. Most of the other chapters respond to specific issues in specific arguments, but this chapter combines several components of theistic arguments and the huge holes throughout naturalism. My favorite simple argument is in chapter eight.
1. If science explains things, then naturalism is false.
Secular humanists/atheists/naturalists will try to claim 1. If science explains things, then naturalism is true. However, and the chapter explains this quite thoroughly, Naturalism itself if full of holes for which it will never find answers.
I do have more to say about the text but I haven't the time! This review is posted on Goodreads as well as Amazon.
Just Who is Making Extraordinary Claims?
I just read this great blog entry by +Rob Lundberg and I wanted to put in my thoughts on this interesting challenge.
The idea, as Rob so clearly presents it, comes from the late Carl Sagan and the preface to Peter Boghossian's book, A Manual for Creating Atheists. I can't say that I've actually had this challenge levied at me before, but I've seen it in many an online discussion. Honestly, I don't have nearly as much to say on this matter but I did want to ask a question. Who is making the truly extraordinary claim?
Take this analogy. You're watching a magic show and you see the stage magician pull a live rabbit out of an empty hat, you'd think (and you'd be right) that the magician is making an extraordinary claim. That is, the ability to make something, the rabbit, out of nothing, the air in the hat (which is technically not 'nothing,' but for the sake of this argument let's leave it there). Now, that is not unlike the claim of theists. I know, I know, before all the apologists that happen to read this send me nasty grams, let me explain. Keep the image of the magic show... Only this time, there is no magician. There is no hat. There really is no stage either, but let's stop it there. And, with no intervention by anyone at any time. A rabbit appears on the stage. THAT is the atheist's claim. Now, without any formal philosophical training or anything, just the regular guy doing regular life. Which one seems more extraordinary? Keep in mind, that technically the analogy falls short, it's not a rabbit that is pulled from the hat. It's the entire universe created ex nihilo (from nothing).
So, I know there are going to be some objections. Let me approach some now. Some claim that the universe has always existed. Let's apply the same analogy. This time, there is no magician still, and no hat. Only a rabbit. That never gets old or when it does get old it suddenly implodes and becomes a baby again, and then continues this cycle of getting older and then popping back into youth and never being born and never dying. Is that more reasonable than the previous claim? Is that verifiable? Okay let's look at another. There are an infinite number of universes out there. Now, our stage and the extraordinary claim hasn't gotten any easier or more rational. If anything we've now multiplied the extraordinary claim by an infinite. And I don't know what you learned in math, but I'm pretty sure anything times an infinite is an infinite. So now we have either an infinite number of sequential rabbits popping onto an infinite number of stages one at at time popping into existence living an unknown amount of time and then dying in an unknown way for no reason. Or, we have no magician and no hat and an infinite number of rabbits all popping into existence at the same time.
So again I ask, Who is really making extraordinary claims and is really required to provide extraordinary evidence?
There's millions (maybe I didn't count them) of these locks with wishes/messages of love by Seoul Tower.
So, I've read several reviews of this book and I've seen several reactions to this work, but I wanted to get it straight from the horse's mouth in a manner of speaking. So I'm reading The God Delusion, and I'll give you my notes.
Even in the preface Dawkins has already shown his prejudice. Obviously the title says a lot, "Delusion." He even mentions that psychologists asked him to change it because the word delusion is a scientific term and to apply it to the millions of people that believe in god wouldn't do justice to the word. Of course that doesn't seem to faze Dawkins and his writing about that almost seems boastful, like he's proud of the fact that he's insulted the majority of the world. How about this quote? "... I believe there are plenty of open-minded people out there: people whose childhood indoctrination was not too insidious, or for other reasons didn't 'take,' or whose native intelligence is strong enough to overcome it" (pg6). In case you didn't catch that, if you're open-minded or intelligent you'll give up anything you were taught about God in your childhood and be an atheist, like all the other smart people in the world.
How about this "gem"? Quote from a Roman Catholic bishop who wrote to Einstein, "...He is all wrong. Some men think that because they have achieved a high degree of learning in some field, they are qualified to express opinions in all." Of course, Dawkins' retort is a false analogy based on his presupposition that God is just as fanciful as fairies. Dawkins claims that theology isn't a "proper field" at all. I find this particularly interesting because I've heard a very similar comment levied against Dawkins. He's a world-renowned ethologist and evolutionary biologist, not a philosopher or theologian yet he's written extensively and authoritatively on subjects he has no (formal) education in. No offence to his brilliance in his fields, but what expertise does a animal behaviorist and evolutionary biologist have in the fields of philosophy of religion, theology, philosophy in general? Being brilliant in one particular field of knowledge doesn't give one authority to speak to all other fields of knowledge.
The biggest problem I have so far is that there haven't been any arguments (yet, he keeps hinting that they'll be proven in later chapters). Chapter one, the first section is simply a childish foot-stomping raving that when atheists say "god" they don't mean the word "god." Something akin to a child throwing a fit saying, "I did not say what you think I said!" Alright! When Dawkins and other atheists say "god" they really mean, uh, well... nothing really, they apparently mean force or nature or the universe, or whatever they want it to mean at that particular moment, but they most certainly do not mean a personal intervening god that is worthy of worship. Dawkins talks about how Einstein was NOT a theist as some theists claim, that he was actually a deist or pantheist. The only acceptable belief in god, to Dawkins, is the deists' god that creates the universe then leaves it alone and never interacts with it again; an invisible, intangible, inactive, uncaring, uninvolved, person-less, unintelligent being with no human characteristics at all. The deists' god is the same as the pantheists' god, an impersonal force that has nothing to do with humanity, that's the only acceptable god, one that has nothing to do with the universe.
Then in chapter one, section two, Dawkins offers a half-hearted attempt at an (advance) apology, which comes too late since the only thing he's said so far is that only intelligent people believe in either an impersonal god or that the universe is god, and by extension only idiots believe in a personal god. It's almost comical, Dawkins makes this claim that theists are always trotting out scientists, especially Einstein, that believe/believed in god and then dismisses it as a bad argument. Then, he turns the same fallacious argument around and uses it, saying that this or that brilliant scientist is atheist and the vast majority of Nobel prize winners in science have been atheists, on and on. I'm sorry, but if an argument is fallacious for one side of an argument, it's equally as fallacious for the opposing side. Think about it; if I say, "Thousands of people believe in pink unicorns, therefore I'm a pink unicorn." You cannot say, "Thousands of people don't believe in pink unicorns, therefore there's no such thing as pink unicorns." An appeal to the people or authority is a fallacy for either side.
In chapter two the insults just keep coming. Not only does Dawkins attack those that believe in a personal god, but he begins attacking anyone who believes that we cannot know whether or not there is a god, agnostics. As part of his disdain for agnosticism he attacks the concept of NOMA (Non-Overlapping MAgisteria). He, who has been continuously accusing God of evil acts and intentions, seems to be calling on god to prove himself. It's like Dawkins is calling upon god to submit to scientific inquiry or proof. It's like Dawkins is saying he expects this wicked, sneaky, conniving, evil, all-powerful entity to answer this puny ant's call. Reminds me of a quote from the 2012 Avengers movie, "The ant has no quarrel with the boot." (Don't misunderstand me, I'm not saying God is the boot looking to crush ants like us, merely that Dawkins is making god out to be this crushing, killing, evil boot character then complains when that entity doesn't kowtow to his demands for proof.) Dawkins, who is much less than an ant, is shaking his puny fist at a god he doesn't believe exists saying, "How dare you not prove yourself to me! How dare you presume to break the laws you set by performing miracles, but when I ask you're silent!"
I just finished reading Jenna Miscavige Hill's book about getting out of Scientology, and she had a quote that Dawkins would do well to listen to: "My parents were aware it was my choice to remain in [Scientology], but they also knew I was brainwashed. The last thing you want to tell a person who is brainwashed is that they are brainwashed." If Dawkins really wants to get through to us poor brainwashed theists, telling us we're brainwashed is definitely not the way to go.
Then on pg. 83 Dawkins quotes Norman Malcom in relation to Anselm's ontological argument, which seems to have a loophole in it about existence being more perfect than non-existence. "The doctrine that existence is a perfection is remarkably queer. It makes sense and is true to say that my future house will be a better one if it is insulated than if it is not insulated; but what could it mean to say that it will be a better house if it exists than if it does not?" The answer to this riddle seem painfully obvious. If you need/want shelter, it's much better to have a house that exists than one that doesn't.
In summation: the first three chapters and preface do nothing in the way of arguing against God; hopefully the actual arguments will begin in chapter four. The first three chapters have been nothing but blustering and casually brushing aside arguments for god. Even Aquinas' eloquently framed ways to God were poorly treated and no real arguments or counter-arguments have been made. These first three chapters can be summed up as such: Anyone who believes there is a God and anyone who believes that we cannot know if there is or isn't a god, is an idiot.


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Breitling Overhauls the Navitimer in Honor of Its 70th Birthday
Breitling first launched its legendary chronograph watch in 1952, right at the height of the postwar aviation boom. The useful timepiece was made for pilots and came equipped with a circular slide rule. The Aircraft Owners and Pilots Association named the design as its official timepiece and Breitling stamped the organization's winged logo on the watch's dial and began calling it the Navitimer (navigation timer) by 1959.
Now, 70 years later, the Swiss luxury watch brand is unveiling a redesigned Navitimer for a new era. The new collection consists of a wide array of variants, with three different cases sizes across the range: 46mm, 43mm, and 41mm. The cases come in stainless steel or solid 18-karat red gold options, with either alligator leather straps or seven-link stainless steel bracelets. The new Navitimers are also water resistant up to 3 bars.
Still containing the iconic watch's circular slide rule, baton indexes, trio of chronograph counters, and generously notched bezel, the Navitimer 2022 has a sleeker and more modernized look, with a slew of contemporary dial options in various shades of blue, green, and copper in addition to more traditional white and black panda variants. The date window has been moved to six o'clock, and is color-matched to the sub-register in which it sits. The slide rule has been flattened, which, along with the domed crystal, allows the timepiece to look much more compact. The rotor has also been minimized, allowing for a better view of the nicely decorated COSC-certified Breitling Manufacture Caliber 01 automatic movement through the sapphire exhibition case back.
But the new Navitimer isn't just an exercise in modernity. It also has one nostalgic eye on the past, with the AOPA wings logo returning to the dial at the 12 o'clock position. Available now via Breitling's online store, the Navitimer 2022 collection starts at $9,000 for the 41mm version in steel on leather and goes up from there.
Photo: Breitling
Nike Expresses Its Love for Sterling Silver with Its Tiffany & Co. Collab


Through its API, Twitter allows developers to write scripts and clients that provide a wealth of extra features and functionality to enrich the micro-blogging experience.
Most of these applications are completely legitimate and include things like Twitter for iPhone, Flipboard, bit.ly, Twittercounter and Posterous.
Sometimes we accidentally authorise unwelcom scripts to access our Twitter profiles, which can lead to unwelcome automated tweets and direct messages, and even password exploits and the spread of malware.
If your Twitter profile is behaving strangely, there's a good chance that you've provided access to one or more of these unsavoury guests. Fortunately, it's easy to be rid of them once and for all. You just have to be a little responsible.
First, let's see which applications you've authorised to access your Twitter account.
Scroll down the page. Some of these apps will be read-only, which means they cannot publish anything on your Twitter profile. Others will have read and write access, which means they most definitely can, and these are the ones that need your full attention.
If there's anything in your Applications list that you don't recognise, like or need, click on the Revoke Access button to remove it, and Twitter will immediately take action. Repeat as necessary.
Unless you see something that looks really, really dubious, don't worry. Chances are your account hasn't been exploited in any serious way. If it makes you feel better then go ahead and change your password – doing this a few times a year is good practice anyway – but 99.99 per cent of the crap that people give access to in their Twitter profiles are harmless (if irritating) applications that don't do much more than auto-tweet their spammy marketing messages. Still, best to be rid of them. Otherwise, it makes you look like a spammer.
Also, don't be too concerned if you've got a lot of apps tied to your account – just be sure that they're trustworthy. For example, at the time of writing I have 35 applications that I have authorised to have access to my Twitter profile. That seems like quite a lot, but I can assure you that they are all 100 per cent legitimate.
It's a good idea to review your Twitter applications on regular basis – I recommend a minimum of once each month, and even weekly if you're very active on the network. Lead by example, and encourage your friends to be proactive, too, particularly if they're having problems with their accounts.


If you are a fan of the Gpf of War series then this Travel mug is for you! Inspired by the superhit Series, God of War, this Tumbler with lid features the battleground scene with the soldiers and God of War logo on the other side. This is a fully functional eye catching Travel Mug/ Cup and If you are a fan, then this Tumbler is for you. Don't leave home without your favorite beverage and with our travel mugs, you won't have to. This double wall travel cups/ mugs come with a flat screw-on plastic lid with reusable straw. These mugs are perfect for hydrating on the go and the double wall keeps drinks cold! Hip and functional product for fans of all ages these Cups/ Mugs can hold up to 16oz of your favorite drinks are leakproof, durable, inexpensive, reusable and a 100% BPA free. These Mugs are perfect for serving refreshing drinks such as Juices, Iced Coffee, Iced Tea, Bubble Boba Tea, Lemonade, Smoothies, Parfait, Soda, Beer, Frozen Drinks, Mixed Drinks, Cocktails, Frappuccino's, Lattes and More! Whether going to work or sending your kids off to school or a stroll in the park or grocery store these are perfect take-out cups that are leak-proof and functional for men women, adult and kids alike.


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The establishment of the Kalmar Union in 1397 united the kingdoms of Norway, Denmark, Iceland thus followed Norways integration to that Union and came under Danish rule after Swedens secession from that union in 1523. In the wake of the French revolution and the Napoleonic wars, Icelands struggle for independence took form and culminated in independence in 1918, until the 20th century, Iceland relied largely on subsistence fishing and agriculture, and was among the poorest in Europe. Industrialisation of the fisheries and Marshall Plan aid following World War II brought prosperity, in 1994, it became a part of the European Economic Area, which further diversified the economy into sectors such as finance, biotechnology, and manufacturing.
ВAbbiamo l8217;atomicaв. Dichiarazione sconvolgente da parte dei vertici del Roma Fringe Festival. Le agenzie di tutto il mondo battono la notizia in tempo reale.


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Vijay Shekhar Sharma
One97 Communications
Paytm's parent okays Berkshire's $300 million investment
As a part of the deal, Berkshire Hathaway has acquired 1,702,713 fully paid up equity shares in Paytm. This becomes Berkshire Hathaway's first investment in an Indian company.ET Bureau | Updated: September 26, 2018, 09:05 IST
The management board of One97 Communications, which owns Paytm, has approved the investment of $300 million from the US-based investment firm Berkshire Hathaway according to documents accessed from the Registrar of Companies. In lieu of the investment Berkshire will subscribe to 1.7 million equity shares of Paytm.
ET had written in its August 27 edition that Warren Buffet-owned investment firm Berkshire Hathaway was about to invest close to Rs 2,500 crore in Paytm parent entity One97 Communications. Calculations show that the deal could value Paytm at anywhere around $10 billion, making it one of the most valued internet company in India.
The board has also resolved that Paytm founder Vijay Shekhar Sharma, chief finance officer Madhur Deora and senior vice-president of finance Vikas Garg as those responsible for negotiating this deal on behalf of the company.
As a part of the deal, Berkshire Hathaway has acquired 1,702,713 fully paid up equity shares in Paytm. This becomes Berkshire Hathaway's first investment in an Indian company.
Responding to media reports, Berkshire had also confirmed the investment and released a note to the press.
"A number of people have inquired about whether Berkshire had made an investment in One97 Communications, the parent company of Paytm. This is to confirm that the investment was made and that it was not a transaction in which Mr. Buffett was involved." said Debbie Bosanek, assistant to Warren Buffett, CEO of Berkshire Hathaway.
Berkshire Hathaway joins Paytm's existing backers which include the likes of Alibaba as the largest stakeholder followed by SoftBank and SAIF Partners.
Paytm, which offers digital payments and has now diversified into travel, entertainment and even financial services, was valued at $7 billion in the last fund raise from SoftBank last year. Further it was valued at $10 billion at the time of secondary share sales by the employees of the company.
Read: Paytm confirms Berkshire investment, Todd Combs to join the board
Tags : Internet, paytm, digital payments, berkshire hathaway, Vijay Shekhar Sharma, Softbank, One97 Communications