text:
\section{Introduction}

We consider the cubic periodic fractional Schr\"odinger equation,

\begin{equation}\label{eqn:sch}
\left\{
\begin{array}{l}
iu_{t}+(-\Delta)^{\alpha} u =\gamma|u|^2u, \,\,\,\,  x \in {[0,2\pi]}, \,\,\,\,  t\in \mathbb{R} ,\\
u(x,0)=u_0(x)\in H^{s}([0,2\pi]), \\
\end{array}
\right.
\end{equation}
where $\alpha \in (1/2,1)$ and $\gamma=\pm 1$. The equation is called focusing for $\gamma= 1$ and defocusing for $\gamma=- 1$

On real line, this equation arises as a model in the theory of fractional quantum mechanics, see \cite{laskin}. In \cite{kay}, Kirkpatrick, Lenzmann and  Staffilani derived it as a continuum limit of a model for the interaction of quantum particles on lattice points. Allowing only the nearest points to  interact gives the usual cubic Schr\"odinger equation whereas allowing long range interactions gives rise to the fractional Schr\"odinger equations with paremeter $\alpha$.

For $\alpha=1$, in \cite{bourgain}, Bourgain proved periodic Strichartz's estimates and showed $L^2$ local and global well-posedness for the cubic Schr\"odinger equation. In \cite{bgt}, Burq, Gerard and Tzvetkov noted that this result is sharp since the solution operator is not uniformly continuous on $H^s$ for $s<0$.

The fractional Schr\"odinger equation on real line was recently studied in \cite{kwon}. For $\alpha\in (1/2,1)$, the equation is less dispersive, so one would not expect to  be able to get local well-posedness on $L^2$ level. Indeed, they proved that there is local well-posedness on $H^s$ for $s\geq \frac{1-\alpha}{2}$. They also showed that the solution operator fails to be uniformly continuous in time for $s<\frac{1-\alpha}{2}$.

 In \cite{ourpaper}, we proved that the periodic fractional equation is locally well-posed in $H^s$, for $s>\frac{1-\alpha}{2}$ using direct $X^{s,b}$ estimates. Further, we proved a Strichartz's estimate of the form,
$$
\|e^{it(-\Delta)^\alpha}f\|_{L^4_{t\in \mathbb{R}}L^4_{x\in \mathbb{T}}}\lesssim \|f\|_{H^s},
$$
 for $s>\frac{1-\alpha}{4}$, which also gives local well-posedness for $s>\frac{1-\alpha}{2}$, using the methods in \cite{cat-wang} and \cite{mypaper}.

  Moreover, we proved in \cite{ourpaper} that the defocusing equation is globally well-posed for $s>\frac{5\alpha+1}{6}$, using Bourgain's high-low frequency decomposition introduced in \cite{highlow}. This method uses the decomposition of the equation into the evolutions of the high and the low frequencies of the initial data. Since the low frequancy part is smooth, its evolution is global due to the conservation of the energy. But the same cannot be said for the high frequancy part. To overcome this problem we showed that the nonlinear evolution of the high frequency part is smoother than the initial data. We should mention that for $\alpha=1$ the smoothing coincides with the smoothing estimate for the NLS that was recently obtained in \cite{talbot}.

  After obtaining these local and global well-posedness results, the natural question that arises is how much we can push the global well-posedness range. For example, the cubic periodic Schr\"odinger equation ($\alpha=1$) in $1$-d is locally well-posed in $L^2$, see  \cite{bourgain}, and with the mass conservation, we know that the equation is globally well-posed. That is, conservation laws on the local well-posedness level may give rise to global well-posedness. But then, one can ask whether we can show that the equation is globally well-posed whenever it is locally well-posed. Although when there is no conservation laws on the local well-posedness level, it is not trivial that this statement is true, we can still make sense of the question in a different way. The idea relies on the intuition that the set of 'bad' initial data, where the solutions of the equation with those initial data, may have arbitrarily large norm, should be negligible. This approach of looking at the problem in an 'almost sure' sense originated from the work of Lebowitz, Rose and Spear, \cite{leb}. They were trying to understand the general behavior of a system containing a large number of particles by looking at the values of the observables by taking averages over certain probability distributions containing only a few parameters instead of looking at the individual initial value problems. With this in mind, they constructed probability measures on Sobolev spaces and proved some basic properties of these measures.

  Later, Bourgain. in \cite{bourgain_measure}, proved that the Schr\"odinger equation with power nonlinearity,
\begin{equation}\label{eqn:sch}
\left\{
\begin{array}{l}
iu_{t}-\Delta u =-|u|^{p-2}u, \,\,\,\,  x \in {[0,2\pi]}, \,\,\,\,  t\in \mathbb{R} ,\\
u(x,0)=u_0(x)\in H^{s}([0,2\pi]), \\
\end{array}
\right.
\end{equation}
where $4<p\leq 6$ is locally well-posed in $H^s$ with $s>0$. But for $0<s<1$ there is no conservation law which would easily allow us to extend the local solutions to global ones. He used the idea of Lebowitz, Rose and Spear to construct a probability measure, also known as Gibbs' measure, on $H^s$ for $s<\frac{1}{2}$ which is invariant under the solution flow. Then he showed that for any $\epsilon>0$, there is global in time $H^s$ norm bounds on the solutions with the initial data in $H^s$ up to a set of measure less than $\epsilon$, i.e. the equation is almost surely globally well-posed in $H^s$ for $0<s<\frac{1}{2}$.

The idea of Gibbs' measures and almost sure global well-posedness later have been used to prove similar results for different equations by \cite{bulut}, \cite{bulut2}, \cite{burq}, \cite{coll}, \cite{oh}, \cite{pav}, \cite{oh2}, \cite{oh3}, \cite{richards} and many others.

Our main result here, is the explicit construction of Gibbs' measure for $1$-d fractional periodic cubic Schr\"odinger equation and the proof of almost sure global well-posedness. More precisely, we define an invariant probability measure $\mu$ on $H^s$, for $s<\alpha-\frac{1}{2}$ such that for any $\epsilon>0$ we can find a set $\Omega\subset H^s$ satisfying $\mu(\Omega^c)<\epsilon$ and the solution to the equation \eqref{eqn:sch} exists globally for all initial data in $\Omega$.

  For that, we are going to truncate the equation \eqref{eqn:sch}, and use the idea of invariant measures on finite dimensional Hamiltonian systems. Namely, if we look at the equation,
\begin{equation}\label{trun}
\left\{
\begin{array}{l}
iu^N_{t}+(-\Delta)^{\alpha} u^N =\gamma P_N|u^N|^2u^N,\\
u^N(x,0)=P_Nu_0(x) \\
\end{array}
\right.
\end{equation}
  where $P_N$ is the projection operator onto the first $N$ frequencies, we see that \eqref{trun} is a finite dimensional Hamiltonian system, with the Hamiltonian,
$$
H_N(u)(t)=\frac{1}{2}\sum_{n\leq N}\big||n|^{\alpha} \widehat{u_n(t)}\big|^2-\frac{\gamma}{4}\int_{\mathbb{T}}|\sum_{n\leq N} e^{inx}\widehat{u_n(t)}|^4.
$$

By Liouville's theorem, we know that the Lebesgue measure $\prod_{|n|\leq N} d\widehat{u_n}$ is invariant under the Hamiltonian flow. Thus, by the conservation of the Hamiltonian and the invariance of the Lebesgue measure under the flow, we see that the finite measure,
$$
d\mu_N=e^{-H_N(u)}\prod_{|n|\leq N} d\widehat{u_n},
$$
is invariant under the solution operator, call it S(t).

We see that equation \eqref{eqn:sch} is an infinite dimensional Hamiltonian system on the Fourier side with the Hamiltonian,
$$
H(u(t))=\frac{1}{2}\sum_{n}\big||n|^{\alpha} \widehat{u_n(t)}\big|^2-\frac{\gamma}{4}\int_{\mathbb{T}}|\sum_{n} e^{inx}\widehat{u_n(t)}|^4=H(u_0),
$$

Then we define the limiting measure $\mu$ on $H^s$ as,
$$
d\mu=e^{-H(u)}\prod_{n} d\widehat{u_n}=e^{-\frac{1}{2}\sum_{n}\big||n|^{\alpha} \widehat{u_n(t)}\big|^2+\frac{\gamma}{4}\int_{\mathbb{T}}|\sum_{n} e^{inx}\widehat{u_n(t)}|^4}\prod_{n} d\widehat{u_n},
$$
and show that the measure $\mu$ is indeed the weak limit of $\mu_N$.

 To construct this measure $\mu$ on appropriate $H^s$ spaces, we use the theory of Gaussian measures on Hilbert spaces following Zhidkov's arguments in \cite{zhid}, and first define,
$$
dw=e^{-\frac{1}{2}\sum_{n}\big||n|^{\alpha} \widehat{u_n(t)}\big|^2}\prod_{n} d\widehat{u_n}.
$$
Then we show that the measure $\mu$ is absolutely continuous with respect to the Gaussian measure $w$ under certain conditions and finish the proof of almost sure global well-posedness by constructing the set $\Omega\subset H^s$ as stated above. For the second part we will mainly use Bourgain's arguments in \cite{bourgain_measure}.

\section{Notations and Preliminaries}

Recall that for $s\geq 0$, $H^s(\T)$ is defined as a subspace of $L^2$ via the norm
$$
\|f\|_{H^s(\T)}:=\sqrt{\sum_{k\in\Z} \la k\ra^{2s} |\widehat{f}(k)|^2},
$$
where $\la k\ra:=(1+k^2)^{1/2}$ and $\widehat{f}(k)=\frac{1}{2\pi}\int_0^{2\pi}f(x)e^{-ikx} dx$ are the Fourier coefficients of $f$.
We  use $(\cdot)^+$ to denote $(\cdot)^\epsilon$ for all $\epsilon>0$ with implicit constants depending on $\epsilon$.

   We denote the linear propogator of the equation as $e^{-it(-\Delta)^\alpha}$, which is defined on the Fourier side as $$\widehat{(e^{-it(-\Delta)^\alpha}f)}(n)=e^{it|n|^{2\alpha}}\widehat{f}(n),$$
 and $|\nabla|^{\alpha}$ is defined as $(\widehat{|\nabla|^{\alpha}f)}(n)=  |n|^{ \alpha} \widehat{f}(n)$.

    When we say the equation \eqref{eqn:sch} is locally well-posed in $H^s$, we mean that there exist a time $T_{LWP}=T_{LWP}(\|u_0\|_{H^s})$ such that the solution exists and is unique in $X^{s,b}_{T_{LWP}}\subset C([0,T_{LWP}),H^s)$ and depends continuously on the initial data. We say that the the equation is globally well-posed when $T_{LWP}$ can be taken arbitrarily large. Here $X^{s,b}$ denote the Bourgain spaces, and are defined via the restriction in time, of the norm,
 \begin{equation}
 \|u\|_{X^{s,b}}\dot{=}\|e^{it(-\Delta)^\alpha}u\|_{H^b_t(\mathbb{R})H^s_x(\mathbb{T})}=\|\langle \tau-|n|^{2\alpha} \rangle^{b} \langle n \rangle^s\widehat{u}(n,\tau)\|_{L_{\tau}^2 l^2_{(m,n)}},
 \end{equation}

 and $\langle x \rangle=(1+|x|^2)^{1/2}$

 By Duhamel Principle, we know that the smooth solutions of \eqref{eqn:sch} satisfy the integral equation
 $$u(t,x)=e^{-it(-\Delta)^\alpha}u_0(x)+i\gamma \int_0^t e^{-i(t-\tau)(-\Delta)^\alpha}|u|^2u(\tau,x)d\tau.$$

We note that along with the Hamiltonian conservation, the equation enjoys mass conservation, namely,
   $$M(u)(t)=\int_{\mathbb{T}}|u(t,x)|^2=M(u)(0).$$

\section{Almost Sure Global Well-posedness}

The main result of this paper is,

\begin{theorem}\label{main}
For $s<\alpha-\frac{1}{2}$ and $\epsilon>0$, there exists an invariant probability measure $\mu$ on $H^s$ such that the equation \eqref{eqn:sch} is globally well-posed for any initial data $u_0 \in \Omega\subset H^s$ such that $\mu(\Omega^c)<\epsilon$ with,
$$
\|u(t)\|_{H^s}\lesssim \Big( \log\big( \frac{1+|t|}{\epsilon}
\big) \Big)^{s+} .
$$
\end{theorem}

  As we mentioned above, in the proof of this theorem, we first define the finite dimensional measures $\mu_N$, which are invariant under the solution operator of the truncated equation \eqref{trun}, and we define $\mu$ as the weak limit of these measures. But then we have to show how the equation \eqref{eqn:sch} and the truncated equation \eqref{trun} are related, namely,

\begin{lemma}\label{convergence}
Let $A\in \mathbb{R}$ and  $u_0\in H^s$ be such that $\|u_0\|_{H^s}<A$, and assume that the solution, $u_N$, of \eqref{trun} satisfies $\|u_N(t)\|_{H^s}<A \text{ \ for\ \ } t\leq T.$
Then the equation \eqref{eqn:sch} is well-posed in $[0,T]$ and moreover, for any $s'<s$, we have,
\begin{equation}\label{approx}
\|u(t)-u_N(t)\|_{H^{s'}}\leq e^{C_1(1+A)^{C_2}T}N^{s'-s},
\end{equation}
where $C_1$ and $C_2$ independent of $s$.
\end{lemma}

In the proof of this lemma, we use,

\begin{lemma}[see \cite{gin}] For $b,b'$ such that $0\leq b+b'<1$, $0\leq b'<1/2$, then we have $$\Big\|\int_0^t e^{-i(-\Delta)^\alpha(t-\tau)}f(\tau)d\tau\Big\|_{X_T^{s,b}}\lesssim T^{1-b-b'}\|f\|_{X_T^{s,-b'}},$$ for $T\in [0,1].$
\end{lemma}

\begin{proof}{(Proof of Lemma \eqref{convergence})}
$$
u(t)-u_N(t)=e^{-it(-\Delta)^\alpha}(u_0-P_N u_0)+i\gamma\int\limits_0^te^{-i(t-\tau)(-\Delta)^\alpha}\big( |u|^2u(\tau)-P_N(|u^N|^2u^N)(\tau) \big)d\tau,
$$
and, taking the $ L^\infty( [0,T];H^{s'})$ norms of both sides for $b>\frac{1}{2}$, since $X^{s',b}\subset L^\infty( [0,T], H^{s'})$ for $b>\frac{1}{2}$, we get,
\begin{eqnarray*}
\|u-u_N\|_{ L^\infty( [0,T], H^{s'})}&\leq&\big\|u_0-P_Nu_{0}\big\|_{H^{s'}}+\big\|\int\limits_0^te^{-i(t-\tau)(-\Delta)^\alpha}\big( |u|^2u(\tau)-P_N(|u^N|^2u^N)(\tau) \big)d\tau \big\|_{X^{s',b}}\\
&\leq&\|u_0-P_Nu_{0}\|_{H^{s'}}+(T_{LWP})^{1-b-b'}\big\||u|^2u-P_N|u^N|^2u^N \big\|_{X^{s',-b'}}\\
&\leq&(T_{LWP})^{1-b-b'}\Big(\big\||u|^2u-P_N(|u|^2u)\big \|_{X^{s',-b'}}+\big\|P_N\big(|u|^2u-|u^N|^2u^N\big) \big\|_{X^{s',-b'}}\Big)\\
&&\qquad \qquad+ \|u_0-u_{0,N}\|_{H^{s'}}\\
&\leq&I+II+III,
\end{eqnarray*}
 for $b'<\frac{1}{2}$ such that $b+b'<1$.

The term $III$ is easier to estimate,

\begin{equation*}
III= \big\|\sum_{|n|>N}e^{inx}\widehat{(u_0)}_n\big\|_{H^{s'}}\leq  N^{s'-s}\|u_0\|_{H^s}\leq N^{s'-s}A.
\end{equation*}

For the term $I$, we first observe that $P_N\big( |v|^2v \big)= |v|^2v $ for $v=P_{\frac{N}{3}}u$, from the convolution property of frequency restriction. Then we write,
\begin{eqnarray*}
I&\leq&\big\||u|^2u-P_N(|v|^2v) \big\|_{X^{s',b'}}+\big\|P_N(|v|^2v-|u|^2u)\big \|_{X^{s',-b'}}\\
&=&\big\||u|^2u-|v|^2v \big\|_{X^{s',b'}}+\big\|P_N(|v|^2v-|u|^2u) \big\|_{X^{s',-b'}}\\
&=&I_1+I_2\leq 2 I_1,
\end{eqnarray*}

Estimating term $I_ 1$ using $X^{s,b}$ estimates and local well-posedness theory, see \cite{ourpaper}, we see that,

\begin{eqnarray*}
I_1&\lesssim&  (T_{LWP})^{1-b-b'} \big(\|u\|_{X^{s',b}}+\|v\|_{X^{s',b}} \big)^2\|u-v\|_{X^{s',b}}\\
&\lesssim& (T_{LWP})^{1-b-b'}A^2\|u-P_{\frac{N}{3}}u\|_{X^{s',b}}\\
&\lesssim& (T_{LWP})^{1-b-b'}A^2\|u_0-P_{\frac{N}{3}}u_0\|_{H^{s'}}\\
&\lesssim& (T_{LWP})^{1-b-b'}A^3N^{s'-s}.
\end{eqnarray*}

Thus we get,
\begin{equation*}
I\lesssim  (T_{LWP})^{1-b-b'}A^3N^{s'-s}.
\end{equation*}
Similarly, for the second term we have,
\begin{equation*}
II\lesssim (T_{LWP})^{1-b-b'} \big(\|u\|_{X^{s',b}}+\|u^N\|_{X^{s',b}} \big)^2\|u-u^N\|_{X^{s',b}}\lesssim  (T_{LWP})^{1-b-b'} A^2\|u-u^N\|_{X^{s',b}},
\end{equation*}

and collecting all the terms, we get,
\begin{eqnarray*}
\|u-u_N\|_{X^{s',b}}&\leq& C N^{s'-s}A+C_2 (T_{LWP})^{1-b-b'} A^2\|u-u^N\|_{X^{s',b}}\\
&&\qquad\qquad+ C_1(T_{LWP})^{1-b-b'}A^3N^{s'-s},\\
&\leq& CAN^{s'-s}+\frac{1}{2}\|u-u^N\|_{X^{s',b}}\\
&\leq&2CAN^{s'-s}.
\end{eqnarray*}
for $T_{LWP}$ small enough independent of $N,s$ and $s'$. Repeating this argument, since the implicit constant $C$ can be taken independent of $T_{LWP}$ and $N$, we see that at any $T_{LWP}$ time the norm at most doubles and thus, at time $T$ we get,
$$
\|u-u_N\|_{H^{s'}}\lesssim 2^{\frac{T}{T_{LWP}}}CAN^{s'-s}\approx e^{C'(1+A)^{\delta}T}AN^{s'-s},
$$
which gives the result.
\end{proof}
 Now, we define a probability measure on $H^s$ using the Hamiltonian. For that we will mainly follow Zhidkov's arguments.

   \subsection{Construction of the Measure on $H^s$:}

  First we will fix the notation that we will use for the rest of the paper. Let $F=(-\Delta)^{\alpha-s}$ on $H^s$. We see that $F$ has the orthonormal eigenfunctions $e_n=e^{inx}/\langle n \rangle^{{s}}$ in $H^s$ with the eigenvalues $|n|^{2\alpha-2s}$. We also denote $u_n=(u,e_n)_{H^s}$.

  \begin{defin}

 A set $M \subset H^s$ is called cylindrical if there exists an integer $k\geq 1$ such that,
$$M=\{ u\in H^s :[u_{-k},\ldots, u_{-2},u_{-1}, u_1, u_2,\ldots , u_k ]\in D \},$$
for a Borel set $D\subset \mathbb{R}^{2k}$.

  \end{defin}

 We denote by $\mathcal{A}$, the algebra containing all such cylindrical sets. Then we define the additive normalized measure $w$ on the algebra $\mathcal{A}$ as follows: For $M\subset \mathcal{A}$, cylindrical,
$$
 w(M)=(2\pi)^{-{k}}\prod_{|n|=1}^k |n|^{\alpha-s}\int_D e^{-\frac{1}{2}\sum_{n=1}^{k}|n|^{2\alpha-2s}|u_n|^2}\prod_{|n|=1}^k du_n.
$$

By the definition of the cylindrical sets, we see that the minimal $\sigma-algebra$ $\overline{\mathcal{A}}$ containing $\mathcal{A}$ is the Borel $\sigma-algebra$, see \cite{zhid}. Although the measure is additive by definition, it doesn't necessarily follow that it is countably additive. Indeed,

\begin{theorem}\label{cont-add}
The Gaussian measure $w$ is countably additive on $\mathcal{A}$ if and only if $\sum_n |n|^{2s-2\alpha}<\infty$, i.e. $s<\alpha-\frac{1}{2}$.
\end{theorem}

\begin{proof}{(cf. \cite{zhid})}
Let $\sum_n |n|^{2s-2\alpha}<\infty$. Wel first show that for any $\epsilon>0$, there exists a compact set $K_\epsilon\subset H^s$ with $w(M)<\epsilon$ for any cylindrical set $M$ such that $M\cap K_\epsilon=\emptyset$.

Let $b_n=|n|^{\tilde \epsilon}$ such that $a=\sum_n |n|^{2s-2\alpha+\tilde \epsilon}<\infty$. Then for an arbitrary $R>0$ take the cylindrical sets of the form,
$$
M=\Big\{ u\in H^s: [u_{-k},\ldots,u_{-2},u_{-1}, u_1,\ldots, u_k]\in D \text{,\quad where }\ \sum_{|n|=1}^k |n|^{\tilde \epsilon}u_n^2>R^2\Big\}.
$$

 Then we see that,
\begin{eqnarray*}
w(M)&=&(2\pi)^{-{k}}\prod_{|n|=1}^k |n|^{\alpha-s}\int_{\sum_{n=1}^k |n|^{\tilde \epsilon}u_n^2>R^2} e^{-\frac{1}{2}\sum_{n=1}^{k}|n|^{2\alpha-2s}|u_n|^2} \prod_{|n|=1}^k du_n\\
&\leq&(2\pi)^{-{k}}\prod_{|n|=1}^k |n|^{\alpha-s}\int_{\mathbb{R}^n}\sum_{n=1}^k \big(\frac{|n|^{\tilde \epsilon}}{R^2}u_n^2\big) e^{-\frac{1}{2}\sum_{n=1}^{k}|n|^{2\alpha-2s}|u_n|^2} \prod_{|n|=1}^k du_n\\
&\leq& R^{-2}\sum_n |n|^{2s-2\alpha+\tilde \epsilon}\\
&=&aR^{-2},
\end{eqnarray*}
here, to pass to the third line we used integration by parts with $f=\frac{-u_n}{|n|^{2\alpha-2s}}$ and $dg=-|n|^{2\alpha-2s}u_ne^{-\frac{1}{2}|n|^{2\alpha-2s}u_n^2}du_n$. Then, for $R>\sqrt{\frac{a}{\epsilon}}$, we have $w(M)<\epsilon$.

  Hence, if we take $K_\epsilon=\{ u\in H^s : \sum_n |n|^{\tilde{\epsilon}} u_n^2\leq R^2\}$, we get the desired compact set.

Now let $A_1\supset A_2\supset \ldots \supset A_m\supset \ldots$ be a sequence of cylindrical sets in $H^s$ such that $\bigcap_{m=1}^\infty A_m=\emptyset$. Then for any $\epsilon>0$ there exists closed cylindrical sets $C_m\subset A_m$ for all $m$ such that $w(A_m/C_m)<\epsilon 2^{-m-2}$. Let $D_m=\bigcap_{k=1}^m C_k$. Then $w(A_m/D_m)\leq w(\bigcup_{k=1}^m (A_k/C_k))<\epsilon/2$. Let $E_m=D_m\cap K_{\epsilon/2}$, then $E_m$'s are compact with $E_m\subset A_m$ and $w(A_m/E_m)<\epsilon$. Since $\bigcap_m A_m=\emptyset$, $\bigcap_m E_m=\emptyset$, and since $(E_m)$ is a nested sequence of compact sets, we see that $E_m=\emptyset$ for all $m>m_0$ for some $m_0\in \mathbb{N}$.

Hence, $w(A_m)<w(E_m)+\epsilon<\epsilon$, for all $m>m_0$. Thus $w(A_m)\rightarrow 0$, i.e. $w$ is countably additive.

For the converse, assume $w$ is countably additive and also $\sum_n |n|^{2s-2\alpha}=\infty$, i.e. $s\geq \alpha-\frac{1}{2}$. Then consider two cases,

\textbf{Case 1: $(s\leq \alpha)$.} In this case we see that $|n|^{2s-2\alpha}\leq 1$ for any $n$. Consider the cylindrical sets of the form,
$$
M_k=\Big\{ u\in H^s : \big| \sum_{|n|=1}^k (u_n^2)-\lambda_k \big|< 2\sqrt {\lambda_k} \Big\},
$$
where $\lambda_k=\sum_{|n|=1}^k |n|^{2s-2\alpha} $.

Then we have,
\begin{eqnarray*}
w(M_k^c)&=&w(\Big\{ u\in H^s : \big| \sum_{|n|=1}^k (u_n^2)-\lambda_k) \big|\geq 2\sqrt {\lambda_k} \Big\})\\
&\leq&\int_{\mathbb{R}^{2n}}\frac{\big(\sum_{|n|=1}^k (u_n^2)-\lambda_k\big)^2}{4\lambda_k} e^{-\frac{1}{2}\sum_{|n|=1}^{k}|n|^{2\alpha-2s}|u_n|^2} \prod_{|n|=1}^k du_n\\
&=&\frac{1}{4\lambda_k}\int_{\mathbb{R}^{2n}}\Big(\big(\sum_{|n|=1}^k u_n^2\big)^2-2\big(\sum_{|n|=1}^k u_n^2\big)\lambda_k+\lambda_k^2\Big) e^{-\frac{1}{2}\sum_{|n|=1}^{k}|n|^{2\alpha-2s}|u_n|^2} \prod_{|n|=1}^k du_n\\
&=&\frac{1}{4\lambda_k}\Big( \big( \lambda_k^2 + 2 \sum_{|n|=1}^k |n|^{4s-4\alpha}\big)-2\lambda_k.\lambda_k+\lambda_k^2 \Big)\\
&\leq&\frac{1}{2}\frac{ \sum_{|n|=1}^k |n|^{4s-4\alpha}}{\lambda_k}\\
&\leq&\frac{1}{2},
\end{eqnarray*}

where, to pass from the third line to the fourth line we used integration by parts again. Since $\lambda_k\rightarrow \infty$ as $k\rightarrow \infty$, there exist balls $B_{\lambda_k-2\sqrt{\lambda_k}}(0)$ of arbitrarily large radii with $w(B_{\lambda_k-2\sqrt{\lambda_k}}(0))\leq w(M_k^c)\leq \frac{1}{2}$, which contradicts with the countably additivity of $w$.

\textbf{Case 2: $(s>\alpha)$.} In this case, for each $n\geq 1$, consider the cylindrical set
$$
M_k=\{ u\in H^s : |u_i|\leq k,\ \  |i|=1,2,...,a_k\},
$$
where $a_k>0$ is an integer. Then by a change of variables, we have,
\begin{eqnarray*}
w(M_k)&=& (2\pi)^{-{a_k}}\prod_{|n|=1}^{a_k} \Big(\int_{-k|n|^{\alpha-s}}^{k|n|^{\alpha-s}}e^{-\frac{1}{2}|u_n|^2} du_n\Big)\\
&\leq&\Big[  (2\pi)^{-{1}}\int_{-k}^{k}e^{-\frac{1}{2}|x|^2} dx \Big]^{a_k},
\end{eqnarray*}
since $s>\alpha$. By choosing $a_k$ large enough, we can take $w(M_k)\leq 2^{-k-1}$ for each $k$ and that $a_k\rightarrow \infty$ as $k\rightarrow \infty$. Then since $\bigcup_{k=1}^\infty M_k=H^s$ and $w(H^s)=1$, since $H^s$ is a cylindrical set with full measure. But then $w(\bigcup_{k=1}^\infty M_k)\leq \sum_{k=1}^\infty w(M_k)\leq \frac{1}{2}$, which is a contradiction. Hence the theorem follows.
\end{proof}

 Now we define the sequence of finite dimensional measures $(w_k)$ as follows: For any fixed $k\geq 1$, we take the $\sigma-algebra$, $\mathcal{A}_n$, of cylindrical sets in $H^s$ of the form $M_k=\{ u\in H^s : [u_{-k},\ldots, u_{-2}, u_{-1},u_1,\ldots, u_k]\in D \}$, for some Borel set $D\subset \mathbb{R}^{2k}$. Then,
$$
w_k(M_k)=(2\pi)^{-{k}}\prod_{|n|=1}^k |n|^{\alpha-s}\int_D e^{-\frac{1}{2}\sum_{|n|=1}^{k}|n|^{2\alpha-2s}|u_n|^2} \prod_{|n|=1}^k du_n.
$$
Hence we get the sequence of finite dimensional, countably additive measures $w_k$ on the $\sigma-algebra$ $\mathcal{A}_k$. We can also extend these measures to the $\sigma-algebra$ $\overline{\mathcal{A}}$ in $H^s$, by setting,
$$
w_k(A)=w_k(A\cap H^s_k),\ \ \text{for } A\in \overline{\mathcal{A}},
$$
where $H^s_k=span(e_{-k}, \ldots, e_{-1}, e_1,\ldots, e_k)$, since $A\cap H^s_k$ is a Borel subset of $H^s_k$ for $A\in \overline{\mathcal{A}}$, see \cite{zhid}.

The immediate question is whether or not the infinite dimensional Gaussian measure $w$ and the finite measures $w_k$ are related and the answer is,

\begin{prop}
The sequence $w_k$ converge weakly to the measure $w$ on $H^s$ for $s<\alpha-\frac{1}{2}$  as $k\rightarrow \infty$.
\end{prop}

\begin{proof}{(cf. \cite{zhid})}
First, recall that a sequence of measures $\upsilon_m$ is said to converge to a measure $\upsilon$ weakly on $H^s$ if and only if for any continuous bounded functional $\phi$ on $H^s$,
$$\int \phi(u)d\upsilon_m(u)\rightarrow \int \phi(u)d\upsilon(u).$$
Also recall that any $\epsilon>0$, if we take $K_\epsilon\subset H^s$ as in the Theorem \eqref{cont-add}, we see that $w(K_\epsilon)>1-\epsilon$ and moreover, $w_m(K_\epsilon)>1-\epsilon$ for all $n\geq 1$. Now let $\phi$ be an arbitrary continuous bounded functional on $H^s$ with $B=\sup_{u\in H^s}\phi(u)$. Then for any $\epsilon>0$ there exists $\delta=\delta(\epsilon)>0$ such that
\begin{equation}\label{uni-cont}
|\phi(u)-\phi(v)|<\epsilon\ \text{ for any }\ u\in K_\epsilon\text{ and }\ v\in H^s \text{ satisfying }\ \|u-v\|_{H^s}<\delta.
\end{equation}
For any $m$, call $K_m=K_\epsilon\cap H_m^s$. Then by the definition of the measures $w_m$ on $\overline{\mathcal{A}}$, we see that,
\begin{equation}\label{first}
\Big| \int_{H^s} \phi(u)dw_m(u)-\int_{K_m} \phi(u)dw_m(u) \Big|<\epsilon B,
\end{equation}
for any $m\geq1$. Define,
$$
K_{m,\epsilon}=\{ v\in H^s : v=v_1+v_2,\ \  v_1\in H^s_m,\ \  v_2^\perp\in H^s_m,\ \  \|v_2\|_{H^s}<\frac{\delta}{2},\ \   dist(v_1, K_m)<\frac{\delta}{2} \}.
$$
Then, $K_\epsilon\subset K_{m,\epsilon}$ for all sufficiently large $m$'s. Thus, for $m$ large enough,
\begin{equation}\label{second}
\Big| \int_{H^s} \phi(u)dw_m(u)-\int_{K_{m,\epsilon}} \phi(u)dw_m(u) \Big|<\epsilon B.
\end{equation}
We now define the measure $w_m^\perp$ on $(H^s_m)^\perp$ as follows:

For a cylindrical set
$$
M^\perp= \{ u\in (H^s_m)^\perp: [u_{-m-k},\ldots, u_{-m-2}, u_{-m-1}, u_{m+1},u_{m+2}, \ldots, u_{m+k}]\in F \},
$$
where $F\subset \mathbb{R}^{2k}$ is a Borel set, and,
$$
w_m^\perp(M^\perp)=(2\pi)^{-{k}}\prod_{|n|=m+1}^{m+k} |n|^{\alpha-s}\int_F e^{-\frac{1}{2}\sum_{|n|=m+1}^{m+k}|n|^{2\alpha-2s}|u_n|^2}\prod_{|n|=m+1}^{m+k}du_n .
$$
then we see that $w_m^\perp$ is a probability measure on $(H^s_m)^\perp$ and $w=w_m\otimes w_m^\perp$.

Thus we get,
\begin{equation}
\int_{K_{m,\epsilon}}\phi(u)dw(u)=\int_{u_m\in K_{m,\epsilon}}dw_m(u_m)\int_{u_m^\perp \in K_{m,\epsilon}^\perp(u_m)}\phi(u_m+u_m^\perp)dw_m^\perp(u_m^\perp),
\end{equation}
where, $K_{m,\epsilon}^\perp(u_m)= K_{m,\epsilon}\cap \{ u\in H^s : u=u_m+y, y\in (H^s_m)^\perp \}$. Then by \eqref{uni-cont},
\begin{eqnarray*}
\int_{K_{m,\epsilon}}\phi(u)dw(u)&=&\int_{u_m\in K_{m,\epsilon}}dw_m(u_m)\int_{u_m^\perp \in K_{m,\epsilon}^\perp(u_m)}\big(\phi(u_m+u_m^\perp)-\phi(u_m)\big)+\phi(u_m)dw_m^\perp(u_m^\perp)\\
&\leq&C\epsilon+\int_{u_m\in K_{m,\epsilon}}\phi(u_m)dw_m(u_m),
\end{eqnarray*}
for $C$ independent of $m$ and $\epsilon$.

Hence,
\begin{equation}\label{third}
\int_{K_{m,\epsilon}}\phi(u)dw(u)-\int_{u_m\in K_{m,\epsilon}}\phi(u_m)dw_m(u_m)\leq C\epsilon.
\end{equation}

Therefore, combining \eqref{first}, \eqref{second} and \eqref{third}, we get the result.
\end{proof}

  Now, we show that the measure $\mu$ is absolutely continuous with respect to the Gaussian measure $w$. Recall that,
\begin{eqnarray*}
d\mu_N&=&(2\pi)^{-{N}}\prod_{|n|=1}^N |n|^{\alpha-s}e^{-\frac{1}{2}\sum_{|n|\leq N}\big||n|^{\alpha-s} u_n(t)\big|^2-\frac{\gamma}{4}\int_{\mathbb{T}}|\sum_{|n|\leq N} \frac{e^{inx}}{\langle n \rangle^s}u_n(t)|^4}du_0\prod_{1\leq |n|\leq N} du_n\\
&=&e^{-\frac{\gamma}{4}\int_{\mathbb{T}}|\sum_{|n|\leq N} \frac{e^{inx}}{\langle n \rangle^s}u_n(t)|^4}(2\pi)^{-{N}}\prod_{|n|=1}^N |n|^{\alpha-s}e^{-\frac{1}{2}\sum_{0<|n|\leq N}|n|^{2\alpha-2s} |u_n(t)|^2}du_0\prod_{1\leq |n|\leq N} du_n,
\end{eqnarray*}
and thus, $\mu_N $ is a weighted Gaussian measure.

  For the defocusing NLS, since
$$
|u_0|^2\leq \int_{\mathbb{T}}|u|^2 \lesssim \Big( \int_{\mathbb{T}}|u(t)|^4\Big)^{\frac{1}{2}},
$$
 we have,
$$
\int_{u_0\in \mathbb{C}} e^{-\frac{1}{4}\int_{\mathbb{T}}|\sum_{n} \frac{e^{inx}}{\langle n \rangle^s}u_n(t)|^4}du_0\lesssim\int_{u_0\in\mathbb{C}}e^{-\frac{1}{4}|u_0|^4}du_0\lesssim C,
$$
uniformly in $N$. Thus, instead of working with the full measure $\mu_N$ it is enough to work with the measure $w_N$, which is also known as the Wiener measure.

  For the focusing NLS, though, we don't have an a priori control over the weight $e^{\frac{1}{4}\int_{\mathbb{T}}|\sum_{n\leq N} e^{inx}\widehat{u_n(t)}|^4}$. We can overcome this problem by using a lemma of Lebovitz et al., see\cite{leb}, which applies an $L^2$ cut-off to the set of initial data, namely,

  \begin{lemma}
  $e^{\frac{1}{4}\int |\sum_{ 1\leq |n|\leq N} e^{inx}\widehat{u_n(t)}|^4}\chi_{\{\|u\|_{L^2}\leq B\}}\in L^1(dw_N)$ uniformly in $N$, for all $B<\infty$.
  \end{lemma}

  \begin{proof}{(cf. \cite{oh3})}
  \begin{align*}
   \int e^{\frac{1}{4}\int |\sum_{ 1\leq |n|\leq N} e^{inx}\widehat{u_n(t)}|^4}&\chi_{\{\|u\|_{L^2}\leq B\}}dw=\int\limits_{({\int |\sum\limits_{|n|=1 }^N e^{inx}\widehat{u_n(t)}|^4}\leq K)}  e^{\frac{1}{4}\int |\sum_{ 1\leq |n|\leq N} e^{inx}\widehat{u_n(t)}|^4}\chi_{\{\|u\|_{L^2}\leq B\}}dw\\
&+\sum_{i=0}^{\infty} \int\limits_{(\int |\sum\limits_{|n|=1 }^N e^{inx}\widehat{u_n(t)}|^4\in(2^iK,2^{i+1}K])} e^{\frac{1}{4}\int |\sum_{ 1\leq |n|\leq N} e^{inx}\widehat{u_n(t)}|^4}\chi_{\{\|u\|_{L^2}\leq B\}}dw\\
&\leq e^{\frac{1}{4}K^4}+\sum_{i=0}^{\infty}  e^{\frac{1}{4}(2^{i+1}K)^4} w(\{ (\int |\sum\limits_{|n|=1 }^N e^{inx}\widehat{u_n(t)}|^4>2^iK, \|u\|_{L^2}<B \}).
\end{align*}

  Now to estimate the second term on the right hand side, choose $N_0$ dyadic, to be specified later. Now call $N_i=N_0.2^i$ and let $a_i$ be such that $\sum_i a_i=\frac{1}{2}$.

Then,
\begin{equation*}
w(\{ (\|u\|_{L^4}>K, \|u\|_{L^2}<B \})\leq \sum_{i=1}^{\infty}w(\{ (\|P_{\{|n|\approx N_i\}}u\|_{L^4}>a_i K \}),
\end{equation*}
and since by Sobolev embedding we have,
$$
\|P_{\{|n|\approx N_i\}}u\|_{L^4}\lesssim N_i^{\frac{1}{4}}\|P_{\{|n|\approx N_i\}} u\|_{L^2},
$$
  we see that,
\begin{eqnarray*}
w(\{ (\|u\|_{L^4}>K, \|u\|_{L^2}<B \})&\leq& \sum_{i=1}^{\infty}w(\{ \|P_{\{|n|\approx N_i\}}u\|_{L^4}>a_i K \})\\
&\leq&\sum_{i=1}^{\infty}w(\{ \|P_{\{|n|\approx N_i\}}u\|_{L^2}\gtrsim a_i N_i ^{-{\frac{1}{4}}}K \}).
\end{eqnarray*}
Letting $a_i=CN_0^{\epsilon}N_i^{-\epsilon}$ and $N_0$ such that $K \approx N_0^{\frac{1}{4}}B$, i.e.$N_0\approx K^4 B^{-4}$, we get,
\begin{eqnarray*}
w(\{ (\|u\|_{L^4}>K, \|u\|_{L^2}<B \})&\leq& \sum_{i=1}^{\infty}w\Big(\{ \big(\sum_{|n|\approx N_i} |\widehat{u_n}|^2\big)^{\frac{1}{2}}\gtrsim a_i N_i ^{-{\frac{1}{4}}}K \}\Big)\\
&\approx&\sum_{i=1}^{\infty}w\Big(\{ \big(\sum_{|n|\approx N_i} |u_n|^2\big)^{\frac{1}{2}}\gtrsim a_i N_i ^{-{\frac{1}{4}+s}}K \}\Big),
\end{eqnarray*}
and by the estimation of the tail of the Gaussian measure, (cf. \eqref{tail_estimate}), we have,
\begin{eqnarray}
w(\{ (\|u\|_{L^4}>K, \|u\|_{L^2}<B \})&\lesssim&\sum_{i=1}^{\infty} e^{-\frac{1}{4}a_i^2N_i^{(2\alpha-2s)+2s-\frac{1}{2}}K^2}\nonumber\\
&\leq&\sum_{i=1}^{\infty} e^{-\frac{1}{4}N_0^{2\epsilon}N_i^{2\alpha-\frac{1}{2}-2\epsilon}K^2}\nonumber\\
&\leq&\sum_{i=1}^{\infty} e^{-\frac{1}{4}N_0^{2\alpha-\frac{1}{2}}2^{(2\alpha-\frac{1}{2}-2\epsilon)i}K^2}\nonumber\\
&\leq&e^{-\frac{1}{4}}K^2N_0^{2\alpha-\frac{1}{2}}\nonumber\\
&\approx& e^{-\frac{1}{4}K^{2+4(2\alpha-\frac{1}{2})}}B^{4-8s}.
\end{eqnarray}
and collecting terms, we obtain,

 \begin{eqnarray*}
   \int e^{\frac{1}{4}\int |u|^4}\chi_{\{\|u\|_{L^2}\leq B\}}dw&\leq& e^{\frac{1}{4}K^4}+\sum_{i=0}^{\infty}  e^{\frac{1}{4}(2^{i+1}K)^4} w(\{ (\|u\|_{L^4}>2^iK, \|u\|_{L^2}<B \})\\
&\leq& e^{\frac{1}{4}K^4}+\sum_{i=0}^{\infty}  e^{\frac{1}{4}(2^{i+1}K)^4} e^{-\frac{1}{4}(2^iK)^{2+4(2\alpha-\frac{1}{2})}}B^{4-8\alpha}\\
&<&\infty,
\end{eqnarray*}
which proves the lemma.
  \end{proof}
Moreover, observe that for $\|u\|_{L^2}<B$, we get $|u_0|^2\leq \sum_{n}\frac{|u_n|^2}{\langle n\rangle^{2s}}\leq B^2$. Hence, $L^2$ cut off also restricts $u_0$ to the ball $\{u_0\in \mathbb{C}: |u_0|\leq B\}$, uniformly in $N$. Therefore, combining these two results, we get that the measure $\mu_N$ is a weighted Gaussian measure with weight being uniformly in $L^1$ with respect to the Gaussian measure.

By the construction of the Gaussian measure, we see that for any compact set $E\subset H^s$, we have,
$$
w_N(E\cap H^s_N)\rightarrow w(E).
$$
Thus, using the result above we get,
$$
\mu_N(E\cap H^s_N)\rightarrow  \mu(E).
$$

\begin{proof}{(Proof of Theorem \eqref{main})}
 For the proof of the theorem and the invariance of the measure $\mu$, we follow Bourgain's arguments in \cite{bourgain_measure}. First, for any $\epsilon$ we will construct the sets $\Omega_N\subset H^s$ such that $\mu_N(\Omega_N^c)<\epsilon$ and,
\begin{equation}\label{omega}
\|u^N(t)\|_{H^s}\lesssim \Big( \log\big( \frac{1+|t|}{\epsilon}
\big) \Big)^{\frac{1}{2}}.
\end{equation}
For that, we fix a large time $T$ and let $[-T_{LWP},T_{LWP}]$ be the local well-posedness interval for the equation \eqref{eqn:sch}. Then consider the set
$$
\Omega^K=\{ u\in H^s_N: \|u\|_{H^s}\leq K \},
$$
where, again, $H^s_N=span\{e_n: |n|\leq N\}$. We see that,
\begin{eqnarray}
w_N((\Omega^K)^c)&=&(2\pi)^{-\frac{N}{2}}\prod_{|n|=1}^N |n|^{\alpha-s}\int\limits_{\{  u\in H^s_N: \|u\|_{H^s}>K  \}} e^{-\frac{1}{2}\sum_{|n|=1}^{N}|n|^{2\alpha-2s}|u_n|^2} \prod_{|n|=1}^N du_n.\nonumber\\
&=&(2\pi)^{-\frac{N}{2}}\prod_{|n|=1}^N |n|^{\alpha-s}\int\limits_{\{  u\in H^s_N: \sum_{|n|\leq N}|u_n|^2>K^2  \}} e^{-\frac{1}{2}\sum_{|n|=1}^{N}|n|^{2\alpha-2s}|u_n|^2} \prod_{|n|=1}^N du_n.\nonumber\\
&=&(2\pi)^{-\frac{N}{2}}\int\limits_{\{ \sum_{|n|\leq N}\frac{|v_n|^2}{\langle n \rangle^{2\alpha-2s}}>K^2  \}} e^{-\frac{1}{2}\sum_{|n|=1}^{N}|v_n|^2}  \prod_{|n|=1}^N dv_n.\nonumber\\
&\leq&(2\pi)^{-\frac{N}{2}}\int\limits_{\{ \sum_{|n|\leq N}|v_n|^2>K^2  \}} e^{-\frac{1}{2}\sum_{|n|=1}^{N}|v_n|^2}  \prod_{|n|=1}^N dv_n\nonumber\\
&=&(2\pi)^{-\frac{N}{2}} \int\limits_{S_{2N}}\int\limits_K^\infty r^{2N-1}e^{-\frac{1}{2}r^2}drdS_{2N}\nonumber\\
&=&(2\pi)^{-\frac{N}{2}} \int\limits_{S_{2N}}\int\limits_K^\infty r\underbrace{r^{2N-2}e^{-\epsilon(r-\epsilon)-\frac{1}{2}\epsilon^2}}_{\leq C}e^{-\frac{1}{2}(r-\epsilon)^2}drdS_{2N}\nonumber\\
&\lesssim&(2\pi)^{-\frac{N}{2}} \int\limits_{S_{2N}}\int\limits_K^\infty re^{-\frac{1}{2}(r-\epsilon)^2}drdS_{2N}\nonumber\\
&\lesssim&(2\pi)^{-\frac{N}{2}} \int\limits_{S_{2N}}\int\limits_K^\infty (r-\epsilon)e^{-\frac{1}{2}(r-\epsilon)^2}drdS_{2N}\nonumber\\
&\leq& e^{-\frac{1}{2}(K-\epsilon)^2}\lesssim e^{-\frac{1}{4}K^2}\label{tail_estimate},
\end{eqnarray}
for $\epsilon$ small enough. Thus, $\mu_N((\Omega^K)^c)\lesssim e^{-\frac{1}{4}K^2}$.

Since $\mu_N$ is invariant under the solution operator, $S_N$ of the truncated equation, if we define the set,
$$
\Omega_N'=\Omega^K\cap S_N^{-1}(\Omega^K)\cap S_N^{-2}(\Omega^K)\cap \ldots \cap S_N^{-\frac{T}{T_{LWP}}}(\Omega^K),
$$
$\Omega_N'$ satisfies the property, $\mu_N((\Omega_N')^c)\leq \frac{T}{T_{LWP}}\mu_N((\Omega^K)^c)<TK^\theta e^{-\frac{1}{4}K^2}$, since the local well-posedness interval $[-T_{LWP},T_{LWP}]$ depends uniformly on the $H^s$ norm of the initial data. Thus if we pick $K= \Big((4+2\theta) \log\big( \frac{T}{\epsilon}
\big) \Big)^{\frac{1}{2}} $, for $\epsilon$ small we get,
$$
\mu_N((\Omega_N')^c)<\epsilon,
$$
and by the construction of the set $\Omega_N'$ we have,
$$
\|u^N(t)\|_{H^s}\lesssim \Big( \log\big( \frac{T}{\epsilon}
\big) \Big)^{\frac{1}{2}},
$$
for all $|t|<T$. Moreover, if we take $T_j=2^j$ and $\epsilon_j=\frac{\epsilon}{2^{j+1}}$, and construct $\Omega_{N,j}$'s, we see that $\Omega_N= \bigcap_{j=1}^\infty \Omega_{N,j}$,  satisfies \eqref{omega}.

Also by the approximation lemma \eqref{approx}, we see that for any $s'<s$ we have,
$$
\|u(t)\|_{H^{s'}}<2A\leq C_{s'} \Big( \log\big( \frac{T}{\epsilon}
\big) \Big)^{\frac{1}{2}} .
$$
Again by taking an increasing sequence of times, we get,
\begin{equation}\label{norm-bound}
\|u(t)\|_{H^{s'}}\leq C_{s'} \Big( \log\big( \frac{1+|t|}{\epsilon}
\big) \Big)^{\frac{1}{2}} .
\end{equation}
Hence, if we intersect this result with an increasing sequence of $s<\alpha-\frac{1}{2}$, and taking $\Omega=\bigcap\limits_N \Omega_N$ where $(\Omega_N)$s are defined as above with $\mu_N(\Omega_N^c)<\frac{\epsilon}{2^N}$, we get that $\mu(\Omega)<\epsilon$ and that the solutions to the equation \eqref{eqn:sch} has the norm growth bound,
$$\|u(t)\|_{H^{s}}\leq C_{s} \Big( \log\big( \frac{1+|t|}{\epsilon}\big) \Big)^{\frac{1}{2}},$$
for initial data $u_0\in \Omega$. Moreover, interpolating this bound with,
$$
\|u(t)\|_{L^2}=\|u_0\|_{L^2},
$$
we have,
$$\|u(t)\|_{H^{s}}\leq C \Big( \log\big( \frac{1+|t|}{\epsilon}\big) \Big)^{s+},$$
 which proves theorem \eqref{main}.
\end{proof}

\subsection{Invariance of $\mu$ Under the Solution Flow}

Let $K$ be a compact set and $B_\epsilon$ denote the $\epsilon$ ball in $H^s$. Let $S$ be the flow map for the equation \eqref{eqn:sch} and $S_N$ be the flow map for the equation \eqref{trun}. Then by the weak convergence of the measure,
$$
\mu(S(K)+B_\epsilon)=\lim_{N\rightarrow \infty}\mu_N\big( (S(K)+B_\epsilon)\cap H^s_N \big).
$$
Also by the uniform convergence of the solutions of \eqref{trun} to \eqref{eqn:sch} in $H^{s_1}$ for any $s_1<s$, we get,
$$
S_N(P_N K)\subset S(K)+B_{\epsilon/2},
$$
for $N\geq N_0$ sufficiently large.
  Then for $\epsilon_1$ small enough,
$$
S_N\big( (K+B_{\epsilon_1})\cap H^s_N \big)\subset S_N(P_N K)+B_{\epsilon/2}\subset S(K)+B_\epsilon.
$$
Hence,
$$
\mu_N\big(S_N\big( (K+B_{\epsilon_1})\cap H^s_N \big)\big)\leq \mu_N\big(S(K)+B_\epsilon\big),
$$
and by the invariance of $\mu_N$, we get,
$$
\mu_N\big( (K+B_{\epsilon_1})\cap H^s_N \big)\leq \mu_N\big(S(K)+B_\epsilon\big),
$$
and letting $N\rightarrow \infty$, by the convergence of the measures $\mu_N$ to $\mu$,
$$
\mu(K)\leq \mu(K+B_{\epsilon_1})\leq \mu(S(K)+B_\epsilon),
$$
which say, by the arbitrariness of $\epsilon$, that $\mu(K)\leq \mu(S(K))$, and by the time reversibility, we also have the inverse inequality and, thus,
$$
\mu(K)=\mu(S(K)),
$$
which gives the invariance of $\mu$ under the solution operator.

\textbf{Acknowledgement:} The author wants to thank Burak Erdo\u{g}an and Nikolaos Tzirakis for their valuable comments and support.

redpajama_set_name: RedPajamaArXiv


Many of you know that I've been working on a fiction project for the last few years.
This last week, friends and family made it possible for me to sneak away to a secluded beach in Mexico to work on my project. The timing couldn't have been worse – I had so many things going on at Square Peg and the last thing I wanted to do was to leave my students and horses not to mention my sweetie on a self serving trip – but I did.
Our group was treated like royalty by the staff at Villa Violeta with amazing hospitality, first class meals and handmade margaritas….. Each morning, I'd get up early, grab a cup of coffee and hide in a spot overlooking the Sea of Cortez. I was visited by lizards and birds and even a squirrel on the sunny deck where I worked. The frigate birds soared overhead and the fishermen waved from their boats as they passed by. It was as idyllic a place as I can ever imagine. I was so far removed from my regular life.
But I was there to work. I issued my challenge to our supporters to help me get this project wrapped up and I'm proud to say that I'm ahead of schedule.
I promised a preview of the work and I'm here to deliver.
Below is a short chapter from Gallop Girl All Rights Reserved by the author, Joell Dunlap May 2012.
Enrique closed the lid on the medicated mud, wiped his hands on the towel hanging from his back pocket and used the wall to steady himself as he stood up in the stall. Mercy Street's legs were slathered in cooling mud and wrapped in clean cotton pillows. He'd have to come back in and paint the bandages with a sauce of habanero and cayenne pepper to keep her from chewing them off overnight. He patted the petite mare and she danced sideways, snorting and blowing. She wasn't the friendly type. He laughed quietly and scratched her withers. She submitted to his touch and even leaned into him so that he could dig into her perpetually itchy shoulders. Both satisfied, he ducked outside the stall door and made his way to the equipment room to fetch the pepper paste. His eye was drawn to the empty stall where the dead filly Chayah Khofesh should have been. The stall was clean and bedded, the straw fluffed and the hay net full. Enrique pictured her powerful frame and curious face as it was this morning before her workout. She'd had some promise and she was easy to be around. She had a habit of peeking around her hay net, curious to see what was happening in the stall next to her. He'd liked her for that curiosity and he missed her now.
In an industry where sentimentality is a luxury, Enrique was a wealthy man who spent lavishly. He knew the quirks of each horse and he indulged them whenever possible. Each animal was a treasure of personality. Some made him laugh, some made him wonder. All of them made him a man satisfied with his work. He knew that horses would come and go, some naturally, some tragically and he cherished the time he had with each one. When it came to people however, Enrique was thrifty with his emotions. He was an island of a man who preferred his own quiet company.
He fished a peppermint out of his pocket, fed one to the barn goat and one for Vaya Con Dios who quietly lifted the candy from Enrique's callused palm and crunched it between giant molars. "Che old man." He rubbed the silky nose and headed back to work.
Things would change at the barn and things would stay the same. He had a good crew, with a couple of exceptions. Most of the guys showed up on time and did their jobs well. A couple of the hot walkers were on notice for laziness. Only one groom worried him, his head groom had travelled home to Mexico for a holiday and had sent his cousin to cover his job in his absence. Enrique had an instant dislike the the cousin. Something about his way with the horses unsettled Enrique. The man had skills and wasn't afraid to work, but he had a shiftiness in him that made both Enrique and the horse wary. He'd keep an eye on him and ask around for some good guys needing work.
He'd worried about the Boss talking to Roxy Ayers about galloping. Enrique remembered a time when all of them were much younger and romance was on the menu for Jude and Roxy. But now he was a big shot trainer and Roxy had aged in a hard edged sort of way. The notion of them being together now was laughable. Both were acting professionally and Enrique hoped that things would work smoothly. The barn was poised for a good year with some decent running stock and some up and coming hopefuls and if, just if they could keep The Big Horse, Leo III from destroying himself or some unfortunate human, great things could happen. Enrique had his doubts that Leo III would do well without Ann. He had worked around female riders for most of his career and he knew that some of them were good, most of them were out to prove something and none of them had the talent that Ann possessed.
Normally,Enrique felt that women interrupted the flow of a barn. Most women created chaos somewhere or drama. They needed too much attention, or demanded respect they weren't willing to earn. Most women you wanted too much to protect. Enrique had to admit that he did have protective feelings about Ann but he never let on. He knew it would just confuse Ann and they had their jobs to do. Enrique thought about visiting Ann in the hospital, but he had no idea what he would say and he didn't like the idea of seeing her laying in any bed. He liked thinking of her as strong and capable and laughing in her special way. Yes, he did feel protective of her. If only he would have waited until later to get the filly to the track – she wouldn't have been at the wrong place at the wrong time. If only he would have put that little cholo rider on her back, then Ann wouldn't be laying broken in a hospital right now. None of it mattered of course and nothing would change the way things were.
Enrique passed by Leo III's stall and noticed that he'd emptied his water bucket completely. Whether he'd drunk it all or whether he'd kicked it out of the bucket was anyone's guess. But no racehorse, especially one as talented as Leo III ever went without fresh water and so he slipped inside the stall, fended off the vicious teeth of the chestnut monster and ducked out quickly with the empty water bucket in hand. He washed the bucket clean of any straw or dirt and filled it being careful not to let the hose end droop into the water. Doctor Connor had been adamant that not only did the horses never share a water bucket again, but that immersing the hose into a water bucket was a sure source of disease spread. He'd primed all of the grooms on the practice and he'd had to keep a close eye on one of the lazier grooms lest the barn be beset with another raging respiratory infection. He fished another peppermint from his pocket and let Leo III see it. His sharp ears heard the cellophane noise and perked forward. Enrique was not fooled, he tossed the peppermint expertly into the horse's grain bucket to distract the animal while he re-affixed the filled water bucket to the wall. Leo III dove for the peppermint and savagely rooted around in his grain bucket to find it. He attacked the grain with and open mouth and crunched hard, feed dripping from his lips. Enrique chuckled appreciatively. Aggression in a racing colt the size and breeding of Leo III was not the horse's fault. It was the product of hundreds of years of racing breeding and a wholly unnatural environment. In the wild, Leo III's rogue behavior would be so over the top that the herd would banish him to perpetual bachelorhood. Instead, if he continued to race as well as he had, he would retire to the breeding shed of a beautiful farm where his every whim was attended to and mares would come in from the world over. But if they couldn't get his feet to harden up, Leo III would be just another failed and unrideable ex racehorse with a lovely pedigree and nowhere to go. Enrique looked at the pale saucer shaped feet and sighed. Supplements, special glue on shoes, training only on the dirt track and racing only on choice grass courses or wet and cold muddy tracks, acupuncture – it seemed they had tried everything and now he would have to manage this beast without Ann.
Next PostNext Front Page of this week's Half Moon Bay Review!


People who have their own computers might rely on these devices for a lot of different things in their lives. Because of this, then, they might want to find a way to make their experience of it more comfortable, which can be done through buying the right computer desk. You will be glad to know that there is a way through which you can find the best desk that is right for you, and you can do so when you find the best guides and reviews for computer desks. Finding the best source of reviews and guides for computer desks, then, is definitely something that you should do because when you do so, you can be sure that the benefits you will gain will certainly be worthwhile as you look for the perfect desk.
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text:
describe RuboCop::Cop::Rails::ApplicationJob do
  let(:msgs) { [described_class::MSG] }

  context 'rails 4', :rails4, :config do
    subject(:cop) { described_class.new(config) }

    it 'allows ApplicationJob to be defined' do
      source = "class ApplicationJob < ActiveJob::Base\nend"
      inspect_source(cop, source)
      expect(cop.offenses).to be_empty
    end

    it 'allows jobs that subclass ActiveJob::Base' do
      source = "class MyJob < ActiveJob::Base\nend"
      inspect_source(cop, source)
      expect(cop.offenses).to be_empty
    end

    it 'allows a single-line class definitions' do
      source = 'class MyJob < ActiveJob::Base; end'
      inspect_source(cop, source)
      expect(cop.offenses).to be_empty
    end

    it 'allows namespaced jobs that subclass ActiveJob::Base' do
      source = "module Nested\n  class MyJob < ActiveJob::Base\n  end\nend"
      inspect_source(cop, source)
      expect(cop.offenses).to be_empty
    end

    it 'allows jobs defined using nested constants' do
      source = "class Nested::MyJob < ActiveJob::Base\nend"
      inspect_source(cop, source)
      expect(cop.offenses).to be_empty
    end

    it 'allows jobs defined using Class.new' do
      source = 'MyJob = Class.new(ActiveJob::Base)'
      inspect_source(cop, source)
      expect(cop.offenses).to be_empty
    end

    it 'allows nested jobs defined using Class.new' do
      source = 'Nested::MyJob = Class.new(ActiveJob::Base)'
      inspect_source(cop, source)
      expect(cop.offenses).to be_empty
    end

    it 'allows anonymous jobs' do
      source = 'Class.new(ActiveJob::Base) {}'
      inspect_source(cop, source)
      expect(cop.offenses).to be_empty
    end

    it 'allows ApplicationJob defined using Class.new' do
      source = 'ApplicationJob = Class.new(ActiveJob::Base)'
      inspect_source(cop, source)
      expect(cop.offenses).to be_empty
    end
  end

  context 'rails 5', :rails5 do
    subject(:cop) { described_class.new }

    it 'allows ApplicationJob to be defined' do
      source = "class ApplicationJob < ActiveJob::Base\nend"
      inspect_source(cop, source)
      expect(cop.offenses).to be_empty
    end

    it 'corrects jobs that subclass ActiveJob::Base' do
      source = "class MyJob < ActiveJob::Base\nend"
      inspect_source(cop, source)
      expect(cop.messages).to eq(msgs)
      expect(cop.highlights).to eq(['ActiveJob::Base'])
      expect(autocorrect_source(cop, source))
        .to eq("class MyJob < ApplicationJob\nend")
    end

    it 'corrects single-line class definitions' do
      source = 'class MyJob < ActiveJob::Base; end'
      inspect_source(cop, source)
      expect(cop.messages).to eq(msgs)
      expect(cop.highlights).to eq(['ActiveJob::Base'])
      expect(autocorrect_source(cop, source))
        .to eq('class MyJob < ApplicationJob; end')
    end

    it 'corrects namespaced jobs that subclass ActiveJob::Base' do
      source = "module Nested\n  class MyJob < ActiveJob::Base\n  end\nend"
      inspect_source(cop, source)
      expect(cop.messages).to eq(msgs)
      expect(cop.highlights).to eq(['ActiveJob::Base'])
      expect(autocorrect_source(cop, source))
        .to eq("module Nested\n  class MyJob < ApplicationJob\n  end\nend")
    end

    it 'corrects jobs defined using nested constants' do
      source = "class Nested::MyJob < ActiveJob::Base\nend"
      inspect_source(cop, source)
      expect(cop.messages).to eq(msgs)
      expect(cop.highlights).to eq(['ActiveJob::Base'])
      expect(autocorrect_source(cop, source))
        .to eq("class Nested::MyJob < ApplicationJob\nend")
    end

    it 'corrects jobs defined using Class.new' do
      source = 'MyJob = Class.new(ActiveJob::Base)'
      inspect_source(cop, source)
      expect(cop.messages).to eq(msgs)
      expect(cop.highlights).to eq(['ActiveJob::Base'])
      expect(autocorrect_source(cop, source))
        .to eq('MyJob = Class.new(ApplicationJob)')
    end

    it 'corrects nested jobs defined using Class.new' do
      source = 'Nested::MyJob = Class.new(ActiveJob::Base)'
      inspect_source(cop, source)
      expect(cop.messages).to eq(msgs)
      expect(cop.highlights).to eq(['ActiveJob::Base'])
      expect(autocorrect_source(cop, source))
        .to eq('Nested::MyJob = Class.new(ApplicationJob)')
    end

    it 'corrects anonymous jobs' do
      source = 'Class.new(ActiveJob::Base) {}'
      inspect_source(cop, source)
      expect(cop.messages).to eq(msgs)
      expect(cop.highlights).to eq(['ActiveJob::Base'])
      expect(autocorrect_source(cop, source))
        .to eq('Class.new(ApplicationJob) {}')
    end

    it 'allows ApplicationJob defined using Class.new' do
      source = 'ApplicationJob = Class.new(ActiveJob::Base)'
      inspect_source(cop, source)
      expect(cop.offenses).to be_empty
    end
  end
end

redpajama_set_name: RedPajamaGithub


Discussion in 'The Doctor's Clinic' started by NightFlyer, Aug 6, 2010.
Ok so I am new to meta game. I just need help with movesets, and EV/IV. I know how to train them i just need to know where to put them.
With that being said these are the pokemon I want to use.
Movesets, EV/IV placement, Natures and subs would be greatly apreciated.
I think this rule speaks for itself. A lot of the Pokemon in your team can fulfil a wide range of different roles, so chucking random movesets at you isn't going to help here. You think about your movesets first, post your ideas and what you feel you want most out of each Pokemon, then you will get advice.
Gravedigging with bad advice...and as you were informed Pokefin, that right there was your last chance.


So deflation's not all bad after all then.
The average commuter will see the price of a season ticket fall by around 0.4% next year, after the annual rate of retail price index inflation came in at -1.4% in July. It might not sound like much of a price cut, but it's a damn sight better than the 6% raise most had to pay out this year.
"We need to do everything we can to avoid deflation!" It's the war cry of central bankers across the world. And so far, the Bank of England under Mervyn King has actually achieved this goal rather well compared to its contemporaries.
Yet again, British inflation data surprised on the upside. Our inflation data does this so often - 12 times out of the past 16 months, according to the Financial Times - that the most surprising thing is that anyone gets surprised when the figures inevitably roll in above analysts' estimates. You wonder why they bother even guessing at the figures at all given how bad they are at it.
The annual rate of increase in the consumer price index (CPI) was expected to fall to 1.5% in July. Instead it remained at 1.8%, the same as June's reading. As for the retail price index (RPI), it fell at a slower annual rate of 1.4%, compared to the 1.6% year-on-year decline seen in June.
Why the disparity? That's straightforward. RPI includes home loan costs. So by cutting interest rates, the Bank of England actually drives RPI down. That's why CPI is used as the Bank's target measure of inflation – because unlike RPI, it doesn't rise or fall alongside interest rates.
A trickier question is why Britain is proving so successful at avoiding deflation. After all, in Europe, CPI is now falling at an annual rate of 0.7%, In the States, it's falling at a whopping 2.1% a year. What's the secret of the Bank of England's success?
Well, on the one hand, it's because Britain's economy is in such a terrible mess. Sterling took a beating against the dollar last year partly because international investors took fright at the sheer scale of our debts. That pushed up import costs for retailers and everyone else. The sorts of goods that rose in price year-on-year in June included computer games and toys, while furniture prices fell less rapidly than they did at the same time last year.
Meanwhile, interest rate cuts mean that consumers who still have jobs and also have home loans, have been left with a lot more disposable income than they had at this time last year. With a large number of shops also already wiped out by the credit crunch (think of all the holes in the high street that used to be Woolworths branches for example – no wonder DVD prices are up), retailers seem to be unconcerned about at least trying to pass higher prices on to customers.
Plenty of people – most analysts in fact – reckon that inflation will, despite this stickiness, continue to decline over the next few months or even years. And there's a lot to recommend this point of view. There is little or no pressure on wages to rise. By some measures, there's a lot of spare capacity in the global economy, and that will continue to push prices downwards. And indeed, it's one reason why almost every other major economy is experiencing deflation.
But this all comes back to why the Bank has been so successful at avoiding deflation so far. The argument among inflationists is that if a central bank really wants to avoid deflation and encourage inflation – and doesn't care about other consequences - then it can, simply by destroying its currency. But of course, currencies are a relative game. You might want your currency to be weak, but if your neighbour is pursuing those policies too, then it becomes a straight fight as to which of you is deemed less investment-worthy than the other.
The 'good' news – depending on your perspective – is that the Bank of England is in an almost uniquely strong position to trash sterling if it really feels it needs to. The dollar has plenty of problems, but it's backed by what is still by far the most powerful country in the world. The euro has its troubles too, but monetary policy there is effectively set for Germany, and Germany doesn't like the idea of inflation. Sterling isn't a commodity currency either – we have no great swathes of resources (anymore) to back the pound.
So of all the world's major central banks, the Bank of England is probably the one which has the easiest job of driving its currency down and creating inflation to order if necessary. And its actions earlier this month – in asking for permission to pump an extra £50bn into the economy – suggest that it intends to err on the side of inflation for the foreseeable future.
So while the pound may have strengthened yesterday as the inflation data suggested an early rise in interest rates could be on the cards, we doubt that'll happen. As one of our Roundtable experts points out in the current issue of MoneyWeek (Twelve stocks that will weather the storm, if you're not already a subscriber, claim your first three issues free here), there's a general election coming up, and the party in power wants to see the economy looking as healthy as possible by then. If that means propping it up on steroids and worrying about the inevitable collapse once the ballot papers have been counted, then so be it.
What does this mean for you? Well, expect monetary policy to remain "accommodating", regardless of how frequently the inflation figures surprise on the upside. That suggests that – as we've been saying for a long time now – you should keep avoiding gilts. And it's an increasingly good idea to have exposure to non-sterling denominated assets outside the UK in your portfolio –we suggested a few US defensive stocks yesterday, while a few weeks ago in MoneyWeek we listed some of the 'emerging market' giants that are worth having exposure to. And of course, there's gold, the classic hedge against inflation or financial panic.
By John Stepek for Money Morning , the free daily investment email from MoneyWeek magazine .


text:
\subsection*{Acknowledgements}
The authors acknowledge discussions with  D. Abanin, G. Murthy, Z. Papic, I. Sodemann,  and M. Zaletel and the experimental assistance of Scott Hannahs.  Magnetocapacitance measurements were funded by the NSF under DMR-1654186. A portion of the nanofabrication and transport measurements were funded by ARO under proposal 69188PHH. AFY acknowledges the support of the David and Lucile Packard Foundation. EMS acknowledges the support of the Elings Fellowship. The research reported here made use of shared facilities of the UCSB MRSEC (NSF DMR 1720256), a member of the Materials Research Facilities Network (www.mrfn.org).  Measurements above 14T were performed at the National High Magnetic Field Laboratory, which is supported by National Science Foundation Cooperative Agreement No. DMR-1157490 and the State of Florida.  K.W. and T.T. acknowledge support from the Elemental Strategy Initiative conducted by the MEXT, Japan and JSPS KAKENHI grant number JP15K21722.

\bibliographystyle{unsrt}


These high quality drum sticks are ideal for all types of drum kits and all levels of ability.
Made from top grade Hard Wearing Maple these sticks are designed to last.
Designed to be lightweight for fast playing with great flex and rebound. The sticks feature a rounded oval style tip for full tones and fast attack from around the drumkit and provide an all round natural feel and look.
You will receive 12 sticks in total that's 6 pairs for the price you would usually pay for 1.


text:
Image: Samsung
Samsung unveils new 870 EVO SSD with faster performance
by Lance Whitney in Hardware
on January 19, 2021, 8:49 AM PST
The 2.5-inch SATA solid-state drive aims for a speeder ride than its predecessor and offers storage options from 250GB to 4TB.
Samsung's 870 EVO solid-state drive
Samsung has introduced a new SSD that promises speed, reliability, and budget-friendly prices.
SEE: Quick glossary: Storage (Tech Pro Research)
Available this month, the new 870 EVO SSD is Samsung's follow-up to the 860 EVO, which debuted in 2018. Packaged with a 2.5-inch SATA configuration, the 870 EVO is built with Samsung's latest V-NAND controller to help it reach maximum SATA sequential read and write speeds of 560 and 530 megabytes per second, respectively.
With a large variable SLC buffer, the drive's TurboWrite technology keeps it at peak performance levels, according to Samsung.
The 870 EVO also kicks in a nearly 38% improvement in random read speed and a 30% gain in sustained performance over its 860 predecessor. With those boosts, Samsung touts the 870 EVO as offering a better experience whether you're multitasking, web browsing, or just booting up your PC.
Samsung is positioning the new drive as a solid choice for a variety of consumers, from general PC users to IT professionals. With its 2.5-inch SATA interface, the 870 is designed to support a wide array of computers, from laptops to desktops to NAS drives. The drive's power saving sleep mode also works with devices that use the standby function in Windows.
"Representing the culmination of our SATA SSD line, the new 870 EVO delivers a compelling mix of performance, reliability, and compatibility for casual laptop and desktop PC users, as well as Network Attached Storage (NAS) users," KyuYoung Lee, vice president of Memory Brand Product Biz at Samsung Electronics, said in a press release.
With availability starting in January 2021, the 870 EVO will be sold in five different storage options and prices: 250GB for $49.99, 500GB for $79.99, 1TB for $139.99, 2TB for $269.99, and 4TB for $529.99. Curious customers can learn more at Samsung's SSD website.
Lance Whitney
Published: January 19, 2021, 8:49 AM PST Modified: January 19, 2021, 7:40 AM PST See more Hardware
3D printing: A cheat sheet (TechRepublic)
Five ways to upgrade your Raspberry Pi (TechRepublic download)
Flash storage: A guide for IT pros (TechRepublic Premium)
How to securely erase hard drives (HDDs) and solid state drives (SSDs) (ZDNet)
Best 2-in-1 laptops, convertibles, and hybrid laptops for business 2018 (ZDNet)
Best cell phone trade-in options for iPhones and Android phones (CNET)
Clean out junk files in Windows 7, 8.1, and 10 (Download.com)
Raspberry Pi: More must-read coverage (TechRepublic on Flipboard)
Share: Samsung unveils new 870 EVO SSD with faster performance
By Lance Whitney
Lance Whitney is a freelance technology writer and trainer and a former IT professional. He's written for Time, CNET, PCMag, and several other publications. He's the author of two tech books--one on Windows and another on LinkedIn.
Contact Lance Whitney
See all of Lance's content

redpajama_set_name: RedPajamaCommonCrawl


It is the science dealing with the artificial replacement of external parts of the human body, in cases of amputations or congenital limb deficiencies. Prosthetics belong in the scientific field of Biomechanics, which includes the research and analysis of the human system (muscles, skeleton and nervous systems), in order to contribute to the field of biomedicine.
The Orthotics is the science that deals with the construction of aiding devices, which are applied on the outer surface of the body, exerting forces in order to keep the member parts in a properly aligned position.


text:
Jean Demangeot, né le  à Paris et mort le  à Neuilly-sur-Seine, est un géographe français qui fut le secrétaire général de l'association de géographes français de 1972 à 1982, puis son président de 1982 à 1986.

Biographie

Il a fait ses études de géographie à la Sorbonne. Il a été professeur émérite à l'université Paris X-Nanterre et Membre de l'Académie des sciences d'Outre-Mer.

Sélection de publications
 Les États-Unis, petite géographie et histoire, 1942, éditions de la France nouvelle, Grenoble.
 Les États-Unis, présentation géographique d'un grand pays, 1946, Bordas, avec R. Blanchard.
 L'économie de la Ruhr, 1947, P.U.F., avec Cl. Simon
 Géomorphologie des Abruzzes adriatiques, 1965, compte-rendu de la thèse sur Persée.
 Les milieux « naturels » du globe, Masson, 1984, compte-rendu de l'ouvrage sur Persée.

Références

Annexes

Bibliographie
 Christian Daudel, Jean Demangeot, géographe de terrain, L'Harmattan, 2008, 349 p.
 Gérard Joly, « Demangeot (Jean) », in Dictionnaire biographique de géographes français du , aujourd'hui disparus, PRODIG, Paris, hors-série Grafigéo, 2013,

Liens externes

Naissance en octobre 1916
Naissance dans le 14e arrondissement de Paris
Géographe français
Professeur à l'université Paris X Nanterre
Décès en février 2009
Décès à Neuilly-sur-Seine
Décès à 92 ans

redpajama_set_name: RedPajamaWikipedia


text:
Carlo Tresca : IWW Leader and Enemy of Big-Business-Owned, Yellow Unions
Revolt Library >> People >> Tresca, Carlo
Though he started as a Socialist, Tresca died an Anarchist. He edited a number of papers which stood up for workers' rights and denounced the hypocrisy and corruption of those in power. One of his favorite targets was the clergy who he attacked relentlessly.
Top Tags :
anarchist (244)
socialist (157)
From : Anarchy Archives
"The army is the most monstrous, immoral, degenerate organism of brutal force."
From : "Carlo Tresca: Portrait of a Rebel," by Nunzio Pernicone
"It is necessary to see these slaves as I have seen them. Then no one would repeat the lie that work ennobles; rather, as a reproach to capitalism, they would say that work brutalizes and kills."
"Come redeeming socialism, come. Only then will the mine cease to be what it is today, a rich tomb created for men by the cruel and blind improvidence of capitalism."
► Listen
On : of 0 Words (Requires Chrome)
OPDS
About Carlo Tresca
Carlo Tresca 1
Carlo Tresca was born in Italy and moved to the United States in 1904 to escape a jail sentence for libel associated with the paper he was editing at the time.
Though he started as a Socialist, Tresca died an Anarchist. He edited a number of papers which stood up for workers' rights and denounced the hypocrisy and corruption of those in power. One of his favorite targets was the clergy who he attacked relentlessly. Tresca was also a skilled labor agitator, leading strikes and urging workers to stand up for their rights.
Dorothy Gallagher's book All the Right Enemies--The Life and Murder of Carlo Tresca is an excellent history of Tresca's life, as is the more recent biography by Nunzio Pernicone, Carlo Tresca: Portrait of a Rebel. Gallagher says in her preface, "to thousands of Italian immigrants Tresca was a hero; to the FBI he was 'notorious'; to a number of American intellectuals and labor leaders he was a counsel; to American and Italian fascists, a serious adversary; to the Communist party of the 1930's a renegade and Trotskyite; to rival anarchists, a spy and traitor; to his friends a joy; to women, overpoweringly attractive; to the man who killed him, little more than a contract." Given this, there can be little doubt that Tresca was a man who was as respected as he was hated, and ultimately hated enough to be murdered.
March 09, 1879 :
Birth Day.
January 11, 1943 :
Death Day.
November 16, 2016 ; 4:52:31 PM (America/Los_Angeles) :
Added to http://www.RevoltLib.com.
April 21, 2019 ; 5:06:04 PM (America/Los_Angeles) :
Last Updated on http://www.RevoltLib.com.
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Vanzetti, Bartolomeo - After they returned the two became more active in the anarchist community. Vanzetti began reading about industrial society and revolt and both began distributing anarchist and revolutionary literature.
Voline - March of 1920 saw him taken to Moscow, where he would remain prisoner until October, when he and many other anarchists were released by virtue of a treaty between the Soviet Union and Makhno's army. Voline then returned to Kharkov, resuming his old activities...
Walker, James L. - James L. Walker (June 1845 – April 2, 1904), sometimes known by the pen name Tak Kak, was an American individualist anarchist of the Egoist school, born in Manchester.
Ward, Colin - ...lived with the title of Britain's most famous anarchist for nearly half a century, bemused by this ambivalent sobriquet. In Anarchy in Action (1973), he set out his belief that an anarchist society was not an end goal.
Warren, Josiah - Equally notable as an inventive genius, a social philosopher, and a peaceful revolutionist, Josiah Warren stands forth, by descent, by his practical, all-round talents, by the force of an earnest life's work, as an American of the sturdy pioneer type whose brawn and brains have formed the true foundation of the republic.
Wilson, Charlotte - Charlotte M. Wilson was an English Fabian and anarchist who co-founded Freedom newspaper in 1886 with Peter Kropotkin, and edited, published, and largely financed it during its first decade. She remained editor of Freedom until 1895.
Woodcock, George - ...sought to raise awareness of the revolution in Spain and of what was being achieved by the Spanish working class against great odds. He was a firm believer in the working class's ability to reorganize society along fundamentally democratic and egalitarian lines.


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Boston Children's Primary Care Alliance »
Our Providers & Staff
Wareham Pediatrics 508-295-8622
53 Marion Road, Suite 1
Wareham, Massachusetts 02571
MyChart Patient PortalNew Patient Info
What Others Say | Wareham Pediatrics
We love hearing from our patient families! Here is a collection of some things our patients say about us. We strive to provide our patients with the finest medical care in a friendly neighborhood environment and are honored by their appreciation.
The staff was amazing. My son is 1 month old now and I did not know what to expect from peds office. My husband and I were more than pleased with the office, staff and mostly Amanda!! We are so happy with our choice of office and feel very comfortable with our doc!
"We like best the caring and knowledgeable approach from Dr. Reynolds."
"I appreciate how thoughtful Dr. Mendes is with my children. He explains things to them and answers their questions with ease."
"We like the short wait and nurses and especially the doctors were very thorough."
"I really like working with a nurse practitioner just because she is able to give so much of her time and attention."
I called to make a last minute appointment and was in an exam within 30 minutes of my phone call. The staff was friendly and courteous, the waiting room was clean and child-friendly, and I was treated with respect and consideration. I was in the office for a minor issue (which in all honesty probably did not warrant an office visit), but my concerns were still listened to, taken seriously, and addressed. I was even given a clean diaper and wipes since I didn't have a diaper bag with me!
"The doctors and nurses always spend plenty of time with my children. I never feel rushed and they are always more than willing to answer all of the questions I may have."
"I like the amount of time spent with the doctor. I don't mind waiting a few minutes for our appointment because I know that Dr. Mendes is giving other patients the time they need for a thorough visit and that we will receive the same consideration."
"We like best the amount of time Dr Reynolds spends with his patients. I never feel rushed."
"Dr. Conway is the best!"
"I like the treatment of the nurse and the cleanliness of the office."
"I love the entire staff, I never have anything but a positive experience.
"I like the way the Dr. interacts with my infant son… great bed side manner."
"To us, the best thing is Dr. Russell's familiarity with my children."
"The nurse that administered the allergy shot was wonderful. The first time our son did not cry."
"We like the nurses!"
"We had no waiting time, got right in and right out."
I am very appreciative of the phone calls received back from the on-call nurses that work during non-office hours. There have been a few times in the past month where my son has had a fever and other symptoms that have been a concern, and they call back within the half-hour to answer any questions. This being my first child, I think it is wonderful that they even call back the next day to check in and see how the night went.
"I just switched pediatricians and I was very pleased with everything. The nurse was wonderful and the doctor gave us 110 percent of her attention. I was extremely happy when I left the office."
"I like that they are very gentle and patient with my child."
"We arrived a couple of minutes early for the appointment and the nurse was able to get the process started early as well."
"I felt valued and respected."
"I really like how the office is decorated and is more spacious than it used to be. More importantly, the staff, nurse and doctor are very knowledgeable, efficient and friendly."
As a parent it made me really feel like the doctors in the practice really care for the children and want whats best for them. I think it was great! My previous pediatrician office never did anything like that and the fact that it was you, a physician taking the time to send the [phone] message was even better.. It wasn't just a recording!! I couldn't be happier with my decision to change doctors and things like this make me realize it was the best decision for my children!! Keep up the good work :)
"It is always super fast to see Libby!"
"We like that we have an opportunity to have all questions answered. All my questions were answered and I was not rushed."
"Amanda is absolutely wonderful, and Marianne the nurse also wonderful."
"Love having Libby she is very nice & knowledgeable."
"I like that the staff especially the doctors treat you as if you are their only patient and don't rush the visits."
"Dr. Conway always takes the time to make my daughter feel comfortable before examining her. He is very kind and compassionate."
"We love Dr. Conway! Libby has also been very nice to work with."
"Dr. Russell is patient and very forthcoming with all types of helpful information."
"It was just a physical but he was very attentive and never made me feel like my questions were stupid. He was great with both kids!"
"We do not wait long in the waiting room, which is a godsend with a newborn who is frequently breastfeeding."
"Dr.Conway has an amazing bedside manner. It's always a pleasure to have my children examined by him. He puts everyone at ease."
"I appreciated the feedback from the Doctor related to the concerns that I had about a pending vaccination."
"Everyone is very caring!"
"We think the staff is best!"
"Everyone in the office is dedicated to making your appointment workable for your time frame. And they all go out of theie way to make the office visit pleasant."
"We love Dr. Reynolds! He always makes us feel like he has plenty of time for us and listens to our questions and concerns and is very knowledgeable. And he never makes us feel rushed. He also makes a point to ask about how our whole family is doing."
"Dr. Conway is the Best Pediatrician I have ever met in my life. He is the most kindest, attentive, knowledgeable, wise and compassionate doctor."
"My fourteen year old son felt comfortable enough with the male doctor to ask him questions, knowing all conversations are kept confidential."
"I enjoy the people working here."
"They all seem to know and care about my children."
"I really enjoy Dr. Conway and how he treats my children and the encouragement he gives me as a mother!"
"I like how promptly my son was seen on a Saturday."
"Dr. Mendes is very nice & knowledgeable."
"I love Dr. Reynolds and the nursing staff."
"Wareham is the absolute best (most pleasant, friendly and thorough) we have ever been to."
"I love your new facilities."
"Dr. Reynolds is the most amazing pediatrician we've had. He always spends as much time as it takes to make sure we understand what (if anything) we need to know."
"Michelle Anderson has a wonderful bedside manner with our children and has great communication with parents as well."
"The doctors are great and i recommend them to everyone I know."
"I really enjoy the attention that my child receives by all those that see him on a single visit."
"Dr. Reynolds makes our children feel very comfortable in a place that they don't always look forward to visiting."
"The great and friendly staff from the front desk to providers are awesome."
"Doctors and nurses are great. You guys are the best!"
"The nurse was wonderful – very friendly and approachable. She was very professional."
"It was my first trip to the office, [the nurse] took her time to ask my son questions and to get weight and temp. She didn't rush in and out. Great Job!"
"Dr. Russell was wonderful as well. She was easy to talk to and approach. She explained things in a common-sense way and was available for questions."
"Dr. Russell is calm, sincere and patient. She is great a what she does."
"My kids really like Dr. Reynolds and he never rushes with them."
"Dr. Reynolds is great! Taking the time to really listen and address our questions and concerns."
"The doctors and nurse practitioners have always been fantastic. The office staff have become much more pleasant in recent visits."
"The amount of time dedicated for my children's physicals. Never seems rushed."
"Dr. Conway is so kind and friendly."
"The doctor was very patient, caring, and really listened."
"All office staff, nurses, and doctors are always a pleasure. thank you!"
"I like that the doctor made sure to address all my questions no matter how silly they may have seemed."
"Amanda was fabulous with my daughter. I have never been so impressed by anyone as much as i am with her. Her bedside manner is very good and she made my frightened, sick daughter feel comfortable."
"I trust [Amanda] and I was completely satisfied with her professionalism and knowledge."
"I can call that day and get my child in to see a doctor the same day."
"ALL doctors at this office that we have seen have been wonderful."
"Short wait, questions answered."
"I love the atmosphere and the way I am treated as the mother. All my questions were taken seriously. She is very pleasant."
"I love this office, I travel a little to get here."
"We love going to Wareham Pediatrics. Whenever we walk through the door we are made to feel very welcome. The entire staff knows our son by name. We do not even have to tell them who we are when we arrive, they already know us."
"The time Dr. Conway spent with us provided a very thorough visit."
"We love your office. We live in Dartmouth but stayed with you because of the wonderful care my children receive. All the staff is very intelligent and caring."
"I love the staff. The nurses, and the receptionists are all great people. (and I love the HOURS)"
"[I like] the friendly nature of all of the staff- my son is only 3 weeks old and I feel like we have been coming there for years! Also, Michelle the lactation consultant is outstanding!!!"
"Dr. Russell always takes the time with my kids and answers all of my questions, we really think she is wonderful."
"The doctors and nurses are amazing!"
"I absolutely love the nurses and the doctors!"
"Very friendly staff."
53 Marion Road
Wareham, MA 02571


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Bosom Buddies Trust > News and Research
It can be tough to stay on top of all the latest news and research on breast cancer, so we want to share some of the most interesting stories we've been reading recently.
We want to hear from you as well, so if you find a gem somewhere, let us know (email link) and we'll try to pass it on.
We choose these links carefully when we publish them, however, Bosom Buddies are not responsible for the content of external websites.
The inclusion of a link to an external website from bosombuddiestrust.org should not be understood to be an endorsement of that website or the site's owners (or their products/services).
National Breast Reconstruction Audit Published
The fourth annual report of the National Mastectomy and Breast Reconstruction
Audit is available to download from the NHS website.
The audit is a comprehensive study, covering all women over the age of
16 who undergo mastectomies or reconstructive surgery following previous
mastectomies in both public and private sectors.
It delivers wide-reaching conclusions, recording women's responses to
surgery in terms of medical effectiveness, comfort and self-image. This leads
to recommendations, particularly to do with the facts and details given to
women before surgery, and their ability to make informed choices about their
own treatments. For example, 42 percent of women reported that a lack of
information led to them delaying reconstructive surgery. However, the overall
picture is encouraging, with 90 percent of women considering the car they
received to be excellent or very good.
The report also breaks down the data by individual hospital.
Get Active!
Daily exercise can reduce the risk of recurrence or mortality by up to 40%, a
Macmillan study has found.
Chief Executive Ciaran Devane is leading calls for an activity programme and
support to be provided to all recovering patients, and his appeal has been
backed by some famous faces! In this video nurse-turned commedienne Jo
Brand is joined by actors Joanna Scanlan and Peter Capaldi to promote this
excellent cause.
We'd love to hear how you go about keeping active after treatment. This
article suggests things as simple as a jog to the shop to buy milk, a bit of
a dance at home or even a spot of gardening – so perfect timing to take
advantage of the late summer weather and get active! Tweet us your best
exercise ideas @BosomBuddiesT.
Tamoxifen: Long-term benefits
Tamoxifen has been in the news recently. In March, WebMD reported on a
study which showed that a full five year course of the drug had a better chance
of preventing future cancers than a shorter two year course.
Recently the Pharmaceutical Journal has reported on research published in The
Lancet which gives the encouraging news that women with oestrogen receptor-
positive breast cancers who have used Tamoxifen continue to experience
benefits for a decade after they stop taking it. The study suggests long-term
mortality rates could be cut by as much as a third.
New Gel Treatment
A new approach to treating breast tumours is in the early stages of
clinical trials in America. Afimoxifene is a a gel treatment being tested at
Northwestern University near Chicago after an encouraging study in France.
The new trial will look at six weeks' use before breast cancer surgery. The
scientists hope this gel will reduce the side effects of some other treatments,
as up to nine times less of the drug will be absorbed into the bloodstream.
Read more about Afimoxifene here.
At Bosom Buddies we believe that sharing first hand experience is one of
the very best ways to help other people out, and BBC Cumbria reporter Val
Armstrong has been keeping a diary of her experiences since being diagnosed
with breast cancer in July. We wish Val all the very best and hope that you
find her thoughts helpful!
If anyone else is keeping a diary or blog of their experiences with the illness
and would like to share it with our readers, please drop us a line to let us
know!
Herceptin and Heart Risks
A first study in the journal Annals of Oncology has found that the drug
Herceptin may increase the likelihood of heart problems, particularly in
people with diabetes or existing heart conditions. The report stresses that
it's important to be cautious when looking at a relatively small sample, but
that the study will continue to look for the patient groups who may be most
vulnerable to this side-effect.
You can read the full article at Medical News Today.
Drug Trial Begins in Oxford
A new drug is being trialed at the Oxford Experimental Cancer Medicine Centre,
which doctors hope will help in the fight against certain genetic cancers,
announces Cancer Research, who are funding the trial. The Centre hopes to help
women with faults in the BRAC1 and BRAC2 genes, which increase the chances
of developing breast cancer.
The drug, known as 6MP, has shown very encouraging results in laboratory
tests, and now moves onto the clinical trial phase.
New Role for Ibuprofen?
Scientists at the University of Colorado are investigating whether readily
available anti-inflammatory drugs like Ibuprofen could reduce the spread of
post-partum breast cancers. Recent animal studies have shown encouraging
results, although research into the possibility of using the treatment for people
is only just beginning.


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Jay! Tomlinson
Podcasters & Radio Producers: We've Been Excluding Millions from Our Conversations
I just figured out a way to make listener participation far more inclusive and I did it by realizing how exclusive it had been.
Voicemail lines for listener feedback has been a staple of podcasts since the beginning and call-in lines for radio have been with us for decades. It's time for an inclusive upgrade.
The Problem: Who are we NOT hearing from?
My anecdotal data is far from scientific, but it is illustrative of clearly-recognized, wider trends.
Fewer Calls Overall:
I've been producing my podcast since 2006 and I've had a listener voicemail line since 2009. Back then, it was normal to receive a dozen or more calls each week. Ten years later, receiving only two voicemails in a week is more normal, even as my download numbers increased.
Gender Disparity:
Here's another trend, way more men call the show than women. Seriously, WAY more. And before you jump to conclusions, my latest demographics survey says the audience is statistically even.
During 2020, I've received voicemails from 29 men and 11 women, nearly a 3 to 1 ratio. But when you consider that many people call in multiple times, it gets worse. 29 men have made 109 calls while 11 women made only 18.
Age disparity:
This one is harder to measure, for obvious reasons. That said, it is clear that voicemails are left, on average, by those in GenX and older. As an elder Millennial myself, it's pretty rare to hear from people younger than me. And again, my demographics survey says that about 50% of my audience is 35 and under.
So, how much of an open conversation with the audience am I really having?
Generational Phone Phobia, the Socialization of Men and Women, and Other Communities We've Ignored
This article isn't going to go into the details of the general downward trend of phone use, phone phobia/anxiety which disproportionately impacts younger generations or the tendency for women to not feel comfortable speaking their opinions in a variety of scenarios.
Nor do I have the expertise to detail the many other groups of people who you won't hear from on a call-in show or voicemail line, such as those with hearing and speech impairment, those learning English as a second language not yet confident in their new tongue, those with accents that make them hesitant to speak up or even just those who need to write things down to get their thoughts straight.
So let's take it as granted that these massive structural trends have something to do with the pattern we're seeing in voicemail messages, make a point to no longer ignore communities of people who simply don't or can't communicate in a traditional audio context, and move on to a solution.
We Can Do Better
As podcasters and radio producers seeking to make a connection with our listeners (and those reading our transcripts), and helping them make connections with each other, if we stick exclusively with traditional voicemail and call-in lines then the biggest loss will be the perspectives we never hear and don't even know we've missed out on.
But wait, why not just read emails from listeners? Because listening to someone read emails is boring. It's a solid rule-of-thumb to do this as little as possible and if you're going to do it, the best way is to hand off to a co-host who reads the message that you can then respond to. This helps break up the auditory monotony and signals to listeners that what they're hearing is from someone else. It's not ideal in any case, doubly so if you don't have a co-host.
VoicedMail for the Otherwise Unheard
My proposal is that we give voice to the far more numerous and diverse perspectives we receive through emails and comments and encourage even more people to send in their messages. Huge portions of our audiences have things to say but just aren't comfortable calling in to a live show, making an audio recording or think it's even worth the effort to write in a message that won't be shared.
"VoicedMail" is the term I have coined to describe the creation of audio messages from text sources using human-sounding text-to-speech voice technology and playing them on a radio show or podcast. Making this an additional communication option for listeners means that you will be able to share with your audience the perspectives of a range of people who never would have been heard from otherwise.
Speaking of giving options, you can also give people an option of what voice they would like to have represent their messages.
Now, I'm not here to write an ad for any particular company, this is not a sponsored post and these are not affiliate links so please use whatever software works for you. But if you're interested, I use Descript and their feature "Stock Voices" to make my VoicedMails.
The Response
After a month of offering VoicedMail, the response has been overwhelmingly positive. Regular callers are excited that we're expanding the conversation (no, they don't mind the robot voices) and the VoicedMails have been rolling in, plus the promise of more to come:
"I love the voiced mail section. I've mentioned before that I'm autistic and can't call, but would love to contribute. I'm so glad to have an option." — Listener email
Not everyone who sends a VoicedMail tells me why they are choosing to do it, but the "why" really doesn't matter — all that matters is they now have a way to be heard.
Unforeseen Benefits
Here's the bottom line: accessible design benefits not just a few people, it benefits everyone. From closed captioning, to curb cuts, to automatic sliding doors, to text messaging — these are all modifications and communication tools that were originally introduced to accommodate people with disabilities. Today, these accessible features are so seamlessly integrated into our daily routines that they become taken for granted. When we design universally, all of society is rewarded in sometimes-unforeseeable ways. As custodians of what is known as an "intimate" medium where deep and thoughtful conversations are possible, let's make sure that everyone is invited to the discussion. We are all sure to be better off for it.
Jay! Tomlinson hosts and produces Best of the Left, an award-winning, expertly-(and human!)-curated podcast of progressive politics and culture. Est. 2006. Contact: Jay@BestOfTheLeft.com
Producer and host of the Best of the Left podcast since 2006. Punctuator of my name with an exclamation point since 1999.
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