text:
\section{Introduction}

\subsection{The cubic NLS on $\mathbb R$} \leavevmode\\
The starting point of this article is the cubic nonlinear Schrödinger equation on $\mathbb R$:
\begin{equation}\label{eq}
i\partial_t u+\partial_x^2 u \pm  |u|^2u=0.
\tag{NLS}
\end{equation}
This equation has been largely studied from theoretical and applied points of view, and appears in several areas of physics such as optics and plasma. \par
In the following, all the results will be valid for both focussing and defocussing cases. For simplicity we shall consider the focussing case.\par
Let us first recall the local well-posedness results in Sobolev spaces. It has been proven in \cite{ginibre1979class} and \cite{cazenave1990cauchy} that the equation is well-posed in $H^s$ for any $s\geq 0$. However, it is no longer the case when $s<0$. For exemple, in \cite{christ2003asymptotics} a norm inflation phenomena is pointed out. \par
Then, the critical Sobolev space associated with the scaling invariance $u\lambda(t,x):=\lambda u(\lambda^2t,\lambda x)$ is $\dot H^{-\frac12}$. For $s\leq -\frac12$, it has been proven in \cite{carles2017norm} and \cite{kishimoto2009wellposed} that norm inflation happens along with a loss of regularity, and in \cite{oh2017remark} that we have in fact a norm inflating phenomena around any data. For $-\frac 12<s<0$, the control of Sobolev norms of Schwartz solutions on the torus and the line is proven in \cite{killip2018low} and \cite{koch2018conserved}. More recently, global well-posedness has been proven in \cite{harrop2020} for any $s>-\frac 12$.\par
On the other hand, the critical Fourier-Lebesgue space associated to this equation is $\mathcal F(L^\infty)$, i.e. Fourier transform in $L^\infty$. The well-posedness has been proven for initial data with Fourier transform in $L^p$ for $p<+\infty$ in \cite{vargas2001global},\cite{grunrock2005bi} and \cite{christ2007power}. \par
One would like to consider the initial value problem with data a sum of Dirac masses. Unfortunately, Dirac masses are in $\mathcal F (L^\infty)$ and borderline in $\dot H^{-\frac12}$,  and the problem is ill-posed for data $u_0=\alpha\delta_0$. Indeed,  in \cite{kenig2001ill} the authors proved, using Galilean invariance, that there is either no weak solution or more than one to that Cauchy problem. More precisely, supposing uniqueness, the solution for $t>0$ of \eqref{eq} is:
\begin{equation}\label{ua}
u_\alpha(t,x)= \alpha e^{- i\alpha^2\ln t} \quad \frac{e^{i\frac{x^2}{4t}}}{\sqrt t},
\end{equation}
but does not converges towards $u_0$ as $t$ goes to zero. \par
However, this issue can be bypassed by a change of phase, which shows that
\[
\psi_\alpha(t,x)= \alpha  \frac{e^{i\frac{x^2}{4t}}}{\sqrt t}
\]
 is the solution of the renormalized equation \eqref{schro} with initial condition $\alpha \delta_0$. As a mater of fact, similar problems have also been treated using renormalisation in context of Gibbs measures, such as in \cite{lebo1988} or \cite{bourgain94periodic}. Once this obstruction has been identified, Banica and Vega constructed in \cite{BV1} and \cite{banica2012scattering} solutions of \eqref{eq} that are smoother perturbations of $u_\alpha$.\par
Now, we shall recall the results related to a sum of several Dirac masses. Let $q>\frac12$, let $(\alpha_j)_{j\in \mathbb Z} \in l^{2,q} (\mathbb C)$ \footnote{we define the space $l^{2,q}$ with $\|\alpha_j\|_{l^{2,q}}^2 = \displaystyle \sum_{j\in\mathbb Z}(1+|j|)^{2q}|\alpha_j|^2 $.}, and set:
\begin{equation}
M=\displaystyle\sum_{j\in\mathbb Z}|\alpha_j|^2.
\end{equation}
It has been proven in Theorem 1.3 of \cite{BV5} that there exists $T>0$ and a unique solution of \eqref{eq} on $(0,T]$ of the form
\begin{equation} \label{solA}
u_{\{ \alpha_j\} } (t,x)=  e^{-2iM\ln t}\sum_{j\in\mathbb Z} A_j(t) \frac{e^{i\frac{(x-j)^2}{4t}}}{\sqrt t},
\end{equation}
where
\[
 A_j(t)=e^{- i|\alpha_j|^2 \ln t}(\alpha_j+{R_j}( t)),
 \]
with $(R_j)_{j\in\mathbb Z}$ a family of functions such that  for any $0<\gamma<1$:
\begin{equation}\label{borneR}
\displaystyle\sup_{0<\tau<T} \tau^{-\gamma}\|R_j(\tau)\|_{l^{2,q}}< C(T, \|\alpha_j\|_{l^{2,q}}).
\end{equation}
In this article, we construct much more solutions of \eqref{eq} as large perturbations of the particular one given by \eqref{solA}. The theorem involves the pseudo-conformal transformation given by:
\[
\mathcal T(f)(t,x)=\frac{e^{i\frac{x^2}{4t}}}{\sqrt t}\overline{f}\left(\frac1t,\frac xt\right),
\]
and the pseudo-conformal operator:
\begin{equation}\label{defJ}
J(f)(t,x)=\left(\frac x2+it\nabla\right)f(t,x).
\end{equation}
\begin{theo} \label{stab}
Let $s\in\mathbb N^*$, $(\alpha_j)\in l^{2,q}$ with $q-s>\frac12$, $ u_+ \in H^s\cap \dot H^{-2} \cap W^{1,s}$  such that
\begin{equation}
\forall p\in\mathbb Z\quad\forall k\leq s \quad  \frac{\widehat{u_+}(\cdot)}{\cdot + p/2} \in H^k(\mathbb R).
\label{fracu}
\end{equation}
Let
\[
u_1(t,x)= u_{\{\alpha_j\} }+e^{-2iM\ln t} \widehat{\overline{u_+}}\left(-\frac x2\right).
\]
Then, if $\|\alpha_j\|_{l^{2,q}}$ is small enough, there exists $T_1<T$ and $u$ a unique solution of \eqref{eq} on $(0,T_1]$, such that:
\[
u-u_{\{\alpha_j\} } -e^{-2iM\ln t} \mathcal T \left(e^{it\partial_x^2}u_+(x)\right)\in \mathcal C((0, T_1],L^2(\mathbb R)),
\]
with:
\begin{equation}\label{decayschro}
\forall k \in \llbracket 0,s \rrbracket, \, \left\|(-i)^kJ^k\left[ u(t,x)-u_{\{\alpha_j\} }(t,x)\right] -e^{-2iM\ln t} \mathcal T \left(e^{it\partial_x^2}\nabla^ku_+(x)\right) \right\|_{L^2}=\mathcal O (t^{\frac12^-}).
\end{equation}
Moreover, if $ \nabla^k u_+\in (x^4L^2)$ for all $k\in \llbracket 0,s\rrbracket$,  then:
\[
u-u_1\in \mathcal C((0, T_1],L^2(\mathbb R)),
\]
with:
\begin{equation}\label{decayl2}
\forall k\in \llbracket 0,s\rrbracket, \, \left\| J^k \left[u(t,x)-u_{\{\alpha_j\} }(t,x)\right] -e^{-2iM\ln t}\left(\frac x2\right)^k\widehat{\overline{u_+}}\left(-\frac x2\right)  \right\|_{L^2} = \mathcal O (t^{\frac12^-}).
\end{equation}
In particular, for $k=0$ we have:
\begin{equation}\label{decay0}
\| u-u_1\|_{L^2}=\mathcal O (t^{\frac12^-}).
\end{equation}
\end{theo}
The condition \eqref{fracu} is not very restrictive and is only about $\widehat{u_+}$ being small enough at equidistant points. The fact that those points are in $\mathbb Z/2$ is not surprising since bounds in terms of $ 1/d(x,\mathbb Z/2)$ already appear in proofs of \cite{BV5}. \par
The proof of Theorem \ref{stab} starts by performing, as in \cite{BV5}, a change a phase
\[
\psi(t,x)= e^{i2M\ln t} u(t,x),
\]
leading to the Wick renormalisation of equation \eqref{eq}:
\begin{equation}\label{schro}
i\partial_t \psi +\partial_x^2 \psi +(|\psi|^2-\frac {2M}t)\psi=0.
\end{equation}
Then we perform a pseudo-conformal transformation:
\begin{equation} \label{PCTran}
\psi(t,x)=\mathcal T(v)(t,x) = \frac{e^{i\frac{x^2}{4t}}}{\sqrt t}\overline{v}\left(\frac1t,\frac xt\right).
\end{equation}
The equation \eqref{eq} is then equivalent to
\begin{equation}
i\partial_tv+\partial_x^2 v + \frac 1 {2t} (|v|^2-2M)v=0 ,
\label{NLS}
\end{equation}
and Theorem \ref{stab} rephrases for \eqref{NLS} as following.
Denoting
\begin{equation}\label{defv1}
v_1(t,x) = \mathcal T (e^{2iM\ln t}u_{\{\alpha_j\}})+e^{it\partial_x^2}u_+ (x)= A(t,x) + e^{it\partial_x^2}u_+(x),
\end{equation}
where
\begin{equation}
A(t,x) =  \sum_{j\in\mathbb Z} e^{- i|\alpha_j|^2\ln t }(\overline{\alpha_j}+\overline{R_j}(\frac1t))e^{-i\frac{tj^2}4+i\frac{xj}2},
\label{eqA}
\end{equation}
we want to find $ t_0>0$ and $v$ solution of \eqref{NLS} such that $v-v_1\in \mathcal C([T_1,\infty),H^s(\mathbb R))$, with a decay given by
\begin{equation} \label{decaynls}
\forall k \in \llbracket 0,s \rrbracket, \quad \|\nabla^k(v-v_1)(t)\|_{L^2}=\mathcal O (t^{-\frac12^-}).
\end{equation}
This $v$ will write
\[
v(t) = \sum_{j\in\mathbb Z} e^{- i|\alpha_j|^2\ln t }(\overline{\alpha_j}+\overline{R_j}(\frac1t))e^{-i\frac{tj^2}4+i\frac{xj}2}+ e^{it\partial_x^2} u_+(x) + r(t,x).
\]
for some remainder term $r$ of decay given by \eqref{decaynls}.
Note that this will prove the decay \eqref{decayschro}, but not \eqref{decayl2} yet. This latter estimation is obtained by using Schrödinger linear evolution properties, as explained in section \ref{fin}.\par
Although a bit computational, the proof can now be sum up as a fixed point theorem in an appropriate Sobolev space, using the Duhamel's formula as the functional, as in \cite{BV1}. The expansion of the cubic non linearity will provide several terms that we will group in powers of $e^{it\partial_x^2}u_+$. The smallness hypothesis on $\|\alpha_j\|_{l^{2,s}}$ comes from the linear term. The term that will provide the weaker time decay is the quadratic one, with conjugated phases. Finally, we perform the reverse pseudo conformal transformation in section \ref{fin} to conclude with the proof.\par
Concerning the tools at our disposal, we will deal with oscillatory integrals, on which we often perform integration by parts after a Fourier transform, that gives sharper estimates than the use of Sobolev embeddings.
\subsection{Link with vortex dynamics singularities}\leavevmode\\
In addition to the study of low regularity solutions of the 1-D cubic Schrödinger equation, this article is also motivated by the study of dynamics of vortex filaments. More precisely, we are referring to a model derived from Euler equations by Da Rios in 1906 in \cite{darios1906motion} called the binormal flow:
\begin{equation} \tag{BF} \label{BF}
\chi_t = \chi_x \wedge \chi_{xx},
\end{equation}
where $\chi$ is an arc length parameterized curve in $\mathbb R^3$ and where the vortex is supposed to be located near $\chi(t)$.
For further informations and references about this equation, see for instance the introduction of \cite{BV5}.\par
If $T$ represents the tangent vector of a solution $\chi$, then $T$ solves the Schrödinger map with values in $\mathbb S^2$:
\begin{equation}
T_t=T\wedge T_{xx}.
\end{equation}
Moreover, Hasimoto constructed in \cite{hasimoto1972soliton} a correspondance between solutions of \eqref{eq} and solutions of \eqref{BF} using the Frenet frame, considering the curvature of $\chi(t,x)$ as the modulus of the NLS-solution and the torsion as the derivative of its phase. This transformation stands in the spirit of the Madelung transform. \par
In \cite{BV5}, Banica and Vega showed that the solution \eqref{solA} we are proving the stability depicts, via the Hasimoto transform, the evolution of a polygonal line through the binormal flow. In a few words, every Dirac mass correspond to the formation of a corner in $\chi(t)$ at arclength parameter $x=j\in\mathbb Z$, as the one developed by the self-similar solution of \eqref{BF} of curvature $\frac{\alpha_j}{\sqrt t}$.
Hence, the odds are high that the class of solutions exhibited in Theorem \ref{stab} could correspond to the evolution of a curve with several corners through binormal flow. With the convergence rate of $\mathcal O(t^{\frac12-})$ in Theorem \ref{stab}, the method should be the same as for the case of one corner done in \cite{banica13} and \cite{banica2015initial}. However, this method is quite intricate, and adding much more complicated terms when considering several Dirac masses makes this task dense enough to be the object of a future work.
\section{Scattering in Sobolev spaces : proof of Theorem \ref{stab} }\label{proof}
 The proof follows the same path as the proof of Theorem 1.4 in \cite{BV1} and is based on a scattering argument. To do so, we introduce an appropriate functional $\phi$ on wich we perform a fixed point argument. The analysis will be much more delicate here, due to the increased complexity of the function $A(t,x)$ defined by \eqref{eqA}, (in \cite{BV1}, only one term is considered).
\subsection{Fixed point functional}\leavevmode\\
To say that $v$ solution of \eqref{NLS} is close to $v_1$ at infinity tends to consider the functional:
\[
\phi:v(t,x) \mapsto v_1(t,x) + i\displaystyle\int_t^\infty e^{i(t-\tau)\partial_x^2}\left( -\frac{|v|^2-2M}{2 \tau}v -(i\partial_t+\partial_{xx})v_1 \right)d\tau .
\]
Since $A$ solves \eqref{NLS}, we have
\[
(i\partial_t+\partial_{xx})v_1 =(i\partial_t+\partial_{xx})A+(i\partial_t+\partial_{xx})e^{it\partial_x^2}u_+ =- \frac{|A|^2-2M}{2 t}A .
\]
We will control the non linear term in $v$ by comparing it to the nonlinearity of $v_1$:
\begin{align*}
\phi(v) -v_1= + i\displaystyle\int_t^\infty e^{i(t-\tau)\partial_x^2} &\left( -\frac{|v|^2-2M}{2 \tau}v + \frac{|v_1|^2-2M}{ 2\tau}v_1 \right.\\
&\left. \quad \quad - \frac{|v_1|^2-2M}{2 \tau}v_1 + \frac{|A|^2-2M}{2 \tau}A  \right)d\tau .
\end{align*}
We then develop the $v_1$-cubic term:
\begin{align*}
(|v_1|^2-M)v_1= &((A+ e^{it\partial_x^2}u_+)(\overline A +   e^{-it\partial_x^2}\overline{u_+})-2M)(A+ e^{it\partial_x^2}u_+)\\
=&(|A|^2-2M)A + 2|A|^2 e^{it\partial_x^2}u_+ -2M e^{it\partial_x^2}u_+ + A^2  e^{-it\partial_x^2}\overline{u_+} + 2A|e^{it\partial_x^2}u_+|^2\\
+&\overline A (e^{it\partial_x^2}u_+)^2+|e^{it\partial_x^2}u_+|^2 e^{it\partial_x^2}u_+ ,
\end{align*}
so we get
\begin{align*}
\phi(v) -v_1=& + i\displaystyle\int_t^\infty e^{i(t-\tau)\partial_x^2}\left( -\frac{|v|^2-2M}{2 \tau}v + \frac{|v_1|^2-2M}{2 \tau}v_1 \right) d\tau\\
+& i\displaystyle\int_t^\infty e^{i(t-\tau)\partial_x^2}  ((-2|A|^2   e^{i\tau\partial_x^2}u_+ +2 M   e^{i\tau\partial_x^2}u_+ - A^2   e^{-i\tau\partial_x^2}\overline{u_+} - 2A|e^{i\tau\partial_x^2}u_+|^2 \\
&\quad\quad\quad\quad\quad\quad- \overline A e^{2i\tau\partial_x^2}u_+^2-|e^{it\partial_x^2}u_+|^2   e^{i\tau\partial_x^2}u_+ )\frac 1{ \tau})d\tau .
\end{align*}
Finally, after expending $|A|^2$, the functional $\phi$ writes
\begin{align}
&(\phi(v)-v_1)(t,x) \\
=&+ i\displaystyle\int_{t}^\infty e^{i(t-\tau)\partial_x^2}\left(\frac{(|v|^2-2M)v}{2 \tau}- \frac{(|v_1|^2-2M)v_1}{2 \tau}\right)d\tau \label{I}\\
&-2i \displaystyle\int_{t}^\infty e^{i(t-\tau)\partial_x^2}  \sum_{p\neq j} e^{- i (|\alpha_p|^2-|\alpha_j|^2)\ln \tau}e^{i(p-j)\frac x2} \overline{\alpha_p}\alpha_j e^{i\frac{j^2-p^2}4 \tau}  e^{i\tau\partial_x^2}u_+\frac{d\tau}{\tau} \label{Ja}\\
&-2i \displaystyle\int_{t}^\infty e^{i(t-\tau)\partial_x^2}  \sum_{(p,j)\in \mathbb Z^2} e^{- i(|\alpha_p|^2-|\alpha_j|^2)\ln \tau+i(p-j)\frac x2} \\
&\quad \quad \quad \times (\overline{\alpha_p}R_j + \overline{R_p}\alpha_j +\overline{R_p}R_j)e^{i\frac{j^2-p^2}4 \tau}e^{i\tau\partial_x^2}u_+\frac{d\tau}{\tau} \label{Jb}\\
&- i  \displaystyle\int_{t}^\infty e^{i(t-\tau)\partial_x^2} A^2 e^{-i\tau\partial_x^2}\overline{u_+} \frac{d\tau}{\tau} \label{Jc}\\
&- i \displaystyle\int_{t}^\infty e^{i(t-\tau)\partial_x^2} (2 A|e^{i\tau\partial_x^2}u_+|^2 + \overline{A}(e^{i\tau\partial_x^2}u_+)^2 )\frac{d\tau}{\tau} \label{Jd}\\
&- i \displaystyle\int_{t}^\infty e^{i(t-\tau)\partial_x^2}  |e^{i\tau\partial_x^2}u_+|^2e^{i\tau \partial_x^2} u_+ \frac{d\tau}\tau \label{Je}\\
=&I(v) + J_a+J_b+J_c +J_d+J_e , \nonumber
\end{align}
where $I(v)$ is the first term and the $J_*$ are the following source terms. Since we will treat each integral's norm independently, we won't be regarding whether its sign is positive or negative. We won't either keep the $-$ in the phase as it doesn't affect the computation. \par
Thanks to their behavior, the  $R_j$ terms will be very easy to treat as well as $J_e$ where we can largely exploit dispersion of the Schrödinger operator:
\begin{equation}\label{disp}
\|e^{it\partial_x^2}u_+\|_{L^\infty_x } \leq \frac1{\sqrt t}\|u_+\|_{L^1}.
\end{equation}
Then, the fact that there is a phase depending on $j$ or $p$ in some of these integrals will produce a shift after an integration by parts and requires the hypothesis on $\widehat{u_+}/(\cdot + \mathbb Z/2)$.
Finally, the limiting $1/t^{\frac12}$ decay comes from using dispersion estimates for $J_d$.\\

We want to apply the fixed point theorem on $\phi: X_\delta\rightarrow X_\delta$, where
\[
X_\delta = \left\{ v\in \mathcal C([t_0,\infty),L^\infty(\mathbb R)) / \|v-v_1\|_S \leq \delta \right \}  ,
\]
with
\[
\|f\|_S=  \sum_{0\leq k\leq s} \sup_{t_0\leq t} t^\mu \|\nabla^k(f)(t)\|_{L^2}
\]
for $\mu$ and $\delta$ strictly positives, to be chosen later.\\
The first step is to verify that $\phi(X_\delta)\subset X_\delta$.

\subsection{Estimates on the fixed point argument terms}\leavevmode\\
In this subsection, we give estimations for each of the terms listed above.
\begin{lemma}[Estimation of $I(v)$] Let $I$ defined by \eqref{I}:
\[
I(v)=i\displaystyle\int_{t}^\infty e^{i(t-\tau)\partial_x^2}\left(\frac{(|v|^2-2M)v}{2 \tau}- \frac{(|v_1|^2-2M)v_1}{ 2\tau}\right)d\tau,
\]
then,
\[
 \|I(v)\|_S\lesssim \footnote{Where $\lesssim$ means "$\leq$ up to a constant".}\|v-v_1\|_S\left( \frac{\|\alpha_j\|_{l^{2,q}}^2}\mu +\frac{C(t_0,\|\alpha_j\|_{l^{2,s}})}{t_0^{2\gamma}(\mu+2\gamma)}+  \frac{\|u_+\|_{L^1}^2}{t_0(1+\mu)}  + \frac{ \|v-v_1\|^2_S }{t_0^{\mu} } \right) .
\]
\label{lemmi}
\end{lemma}
Note that we don't need any smallness hypothesis on $u_+$ if we choose $t_0$ large and $\|\alpha_j\|_{l^{2,q}}$ small.
\begin{proof}
The idea of the proof is to first apply Strichartz inhomogeneous inequality with the admissible couple $(\infty,2)$  and then use dispersion inequality \eqref{disp} and also \eqref{borneR} after usual upper-bounds:
\begin{align*}
&\sup_{t_0\leq t} t^\mu\left \|\displaystyle\int_{t}^\infty e^{i(t-\tau)\partial_x^2}\left(\frac{(|v|^2-2M)v}{ 2\tau}- \frac{(|v_1|^2-2M)v_1}{ 2 \tau}\right)d\tau\right \|_{L^2}\\
\lesssim &\sup_{t_0\leq t}t^\mu \displaystyle\int_{t}^\infty \| |v|^2v-|v_1|^2v_1 -2M(v-v_1)\|_{L^2} \frac{d\tau}\tau\\
\lesssim&\sup_{t_0\leq t}t^\mu \displaystyle\int_{t}^\infty (2M+\|v_1\|^2_{L^\infty}+\|v\|^2_{L^\infty})\|v-v_1\|_{L^2}\frac{d\tau}\tau\\
\lesssim& \|v-v_1\|_S \sup_{t_0\leq t}t^\mu \displaystyle\int_{t_0}^\infty (2M+\|v_1\|^2_{L^\infty}+\|v\|^2_{L^\infty})\frac{d\tau}{\tau^{1+\mu}}\\
\lesssim&\|v-v_1\|_S \left( \frac{ 2M}\mu +  \sup_{t_0\leq t}t^\mu \displaystyle\int_{t}^\infty  3\|v_1\|^2_{L^\infty} +2 \|v-v_1\|_{L^\infty}^2  \frac{d\tau}{\tau^{1+\mu}} \right)\\
\lesssim&\|v-v_1\|_S\left(\frac{2M}{\mu}+\frac{6   \|\alpha_j\|_{l^{2,q}}^2}\mu +\frac {C(t_0,\|\alpha_j\|_{l^{2,s}})}{t_0^{2\gamma}(\mu+2\gamma)} +\frac{6}{1+\mu}\frac{\|u_+\|_{L^1}^2}{t_0}  +2 \sup_{t_0\leq t}t^\mu  \displaystyle\int_{t}^\infty   \|v-v_1\|_{L^\infty}^2\frac{d\tau}{\tau^{1+\mu}} \right) .
\end{align*}
Since $q>\frac12$, we have also controlled in the last steps the $l^1$ norm with the $l^{2,s}$ norm of $(\alpha_j)_{j\in\mathbb Z}$, using Cauchy-Schwarz.
Finally, we apply Gagliardo-Nirenberg's interpolation
\begin{equation}\label{GN}
|f|^2\leq \|f\|_{L^2}\|f'\|_{L^2}
\end{equation}
on the last integral:
\begin{align*}
\sup_{t_0\leq t}t^\mu \displaystyle\int_{t}^\infty   \|v-v_1\|_{L^\infty}^2\frac{d\tau}{\tau^{1+\mu}} \leq & \sup_{t_0\leq t}t^\mu   \displaystyle\int_{t}^\infty  \|v-v_1\|_{L^2} \|\nabla(v-v_1)\|_{L^2}  \frac{d\tau}{\tau^{1+\mu}}   \\
 =& \frac{ \|v-v_1\|^2_S }{t_0^{\mu}} ,
\end{align*}
to conclude that
\[
 \|I(v)\|_S\lesssim \|v-v_1\|_S\left( \frac{\|\alpha_j\|_{l^{2,q}}^2}\mu +\frac{C(t_0,\|\alpha_j\|_{l^{2,s}})}{t_0^{2\gamma}(\mu+2\gamma)}+  \frac{\|u_+\|_{L^1}^2}{t_0(1+\mu)}  + \frac{ \|v-v_1\|^2_S }{t_0^{\mu} } \right) .
\]
\end{proof}

We now have to control the gradient of this integral with the following lemma.

\begin{lemma}[Estimation of $\nabla^kI(v)$]
\label{gradi}
Let $I$ defined by \eqref{I}:
\[
I(v)=i\displaystyle\int_{t}^\infty e^{i(t-\tau)\partial_x^2}\left(\frac{(|v|^2-2M)v}{2 \tau}- \frac{(|v_1|^2-2M)v_1}{ 2\tau}\right)d\tau,
\]
then, for $k\in\llbracket 1, s\rrbracket$:
\[
\sup_{t_0\leq t } t^\mu \| \nabla^kI (t)\|_{L^2} \lesssim  \|v-v_1\|_S \left(\left(\frac{\|\alpha_j\|_{l^{2,q}}}{\mu}\right)^2+\left(\frac {C(t_0,\|\alpha_j\|_{l^{2,s}})}{(\mu+\gamma)t_0^\gamma}\right)^2+\left(\frac{\|u_+\|_{W^{1,s}}}{\sqrt t_0 (\frac12+\mu)}\right)^2+\left(\frac{\delta}{2\mu t_0^\mu}\right)^2\right) .
\]
\end{lemma}

\begin{proof}

To obtain this, we need the following computations.Using Gagliardo–Nirenberg interpolation \eqref{GN}, we have for $k\in \llbracket 0,s-1\rrbracket$:
\begin{equation}
|\nabla^k(v-v_1)(t)|\leq C \|\nabla^{k+1}(v-v_1)(t)\|^{\frac12}_{L^2}   \|\nabla^k(v-v_1)(t)\|^{\frac12}_{L^2} \lesssim \frac{\delta}{t^\mu}  .
\end{equation}
Since $\nabla^k$ commute with the Schrödinger operator, we have
\begin{equation}
|\nabla^k e^{it\partial_x^2} u_+ |\leq  \frac {\|\nabla^k u_+ \| _{L^1}}{\sqrt t}.
\end{equation}
Thus, we can control the $k^{th}$ gradient $v$, with  $k\in \llbracket 0,s-1\rrbracket$, introducing $v_1$:
\begin{align}
|\nabla^kv(t)| \leq C |\nabla^k v_1(t)|+ |\nabla^k (v-v_1)(t) | \lesssim & \|\nabla^kA\|_{L^\infty} +  \frac{\|\nabla^ku_+\|_{L^1}}{\sqrt t} + \frac{\delta}{t^\mu} \\
&\lesssim \|\alpha_j\|_{l^{2,q}} +\frac 1{t^\gamma}+  \frac{\|\nabla^ku_+\|_{L^1}}{\sqrt t} + \frac{\delta}{t^\mu},
\end{align}
using that for each $k$, $q-k>\frac12$ to control $\|\nabla^kA\|_{L^\infty}$ with the $l^{2,q}$ norm and \eqref{borneR}.
We then do the same with the $k^{th}$ gradient of $v^2$ for $k\in \llbracket 1,s-1\rrbracket$ using Leibniz formula:
\begin{align}
|\nabla^k v^2| \leq& \sum_{p=0}^k\binom kp   |\nabla^p v||\nabla^{k-p} v| \nonumber \\
\lesssim & \sum_{p=0}^k\binom kp   \left(\|\nabla^pA\|_{L^\infty} +  \frac{\|\nabla^pu_+\|_{L^1}}{\sqrt t} + \frac{\delta}{t^\mu} \right) \\
&\times \left( \|\nabla^{k-p}A\|_{L^\infty} +  \frac{\|\nabla^{k-p}u_+\|_{L^1}}{\sqrt t} + \frac{\delta}{t^\mu} \right).
\end{align}

Now, let $k\in \llbracket 1,s-1\rrbracket$, the first step is to use $L^\infty L^2$ Strichartz estimates on $I$:
\[
\sup_{t_0\leq t } t^\mu \| \nabla^kI (t)\|_{L^2} \leq \sup_{t_0\leq t } t^\mu  \int_t^\infty \| \nabla^k(|v|^2v - |v_1|^2v_1)\|_{L^2} + \|2M \nabla^k (v-v_1)\|_{L^2} \frac {d\tau}\tau .
\]
We compute the first term here:
\begin{align*}
&\nabla^k(|v|^2v - |v_1|^2v_1 ) \\
=& \sum_{p=0}^k \binom kp   (\nabla^p \overline v \nabla^{k-p}v^2 - \nabla^p \overline{v_1}\nabla^{k-p}v_1^2 + \nabla^p \overline{v_1}\nabla^{k-p}v^2 - \nabla^p \overline{v_1}\nabla^{k-p}v^2 )\\
=&\sum_{p=0}^k \binom kp   ( \nabla^p (\overline{v-v_1})\nabla^{k-p}v^2 + \nabla^{p}(v^2-v_1^2)\nabla^{k-p}\overline{v_1} )\\
=& \sum_{p=0}^k \binom kp   ( \nabla^p (\overline{v-v_1})\nabla^{k-p}v^2 +  \nabla^{p}[(v-v_1)(v+v_1)]\nabla^{k-p}\overline{v_1} )\\
=& \sum_{p=0}^k \binom kp   \left( \nabla^p (\overline{v-v_1})\nabla^{k-p}v^2 + \sum_{q=0}^p \binom pq  (\nabla^q(v-v_1)\nabla^{p-q}(v+v_1) ) \nabla^{k-p}\overline{v_1} \right).
\end{align*}
Therefore, using the preliminary computations:
\begin{align*}
&\sup_{t_0\leq t } t^\mu \| \nabla^kI (t)\|_{L^2} \\
\leq &  \frac{ 2M}{\mu} \|v-v_1\|_S + \sup_{t_0\leq t}t^\mu\int_t^\infty  \left( \sum_{p=0}^k  \binom kp  \|\nabla^p (v-v_1)\|_{L^2}\|\nabla^{k-p}v^2\|_{L^\infty} \right. \\
&+ \left. \sum_{p=0}^k \sum_{q=0}^p \binom kp  \binom pq  \|\nabla^q(v-v_1)\|_{L^2}  \|\nabla^{k-p}v_1\|_{L^\infty}(\|\nabla^{p-q}v\|_{L^\infty} + \|\nabla^{p-q} v_1\|_{L^\infty} ) \right) \frac {d\tau}\tau \\
\lesssim& \frac{2M}\mu \|v-v_1\|_S + \sup_{t_0\leq t}t^\mu \int_t^\infty\sum_{p=0}^k \frac{\|v-v_1\|_S }{\tau^{1+\mu}}  \sum_{j=0}^{k-p}\binom {k-p} j \left(\|\nabla^jA\|_{L^\infty} +  \frac{\|\nabla^ju_+\|_{L^1}}{\sqrt \tau} + \frac{\delta}{\tau^\mu} \right)\\
& {\left( \|\nabla^{k-p-j}A\|_{L^\infty} +  \frac{\|\nabla^{k-p-j}u_+\|_{L^1}}{\sqrt \tau } + \frac{\delta}{\tau^\mu} \right)}\\
& + \sup_{t_0\leq t}t^\mu \int_t^\infty \sum_{p=0}^k\sum_{q=0}^p \frac{\|v-v_1\|_S}{\tau^{1+\mu}} \left( \|\nabla^{k-p}A\|_{L^\infty} +  \frac{\|\nabla^{k-p}u_+\|_{L^1}}{\sqrt \tau} \right) \\
&\quad \quad \quad \left( \|\nabla^{p-q}A\|_{L^\infty} +  \frac{\|\nabla^{p-q}u_+\|_{L^1}}{\sqrt \tau} +\frac{\delta}{\tau^\mu} + \|\nabla^qA\|_{L^\infty} +  \frac{\|\nabla^qu_+\|_{L^1}}{\sqrt \tau}  \right) d\tau  .
\end{align*}
We are now left with integrations:
\begin{align*}
\sup_{t_0\leq t } t^\mu \| \nabla^kI (t)\|_{L^2} \leq & \|v-v_1\|_S \left( \frac{ 2M}\mu +\left(\frac{ \|\alpha_j \|_{l^{2,q}} ^2 }{\mu} + \frac {C(t_0,\|\alpha_j\|_{l^{2,s}})}{(\mu+2\gamma)t_0^{2\gamma}} +\frac {\|u_+\|_{W^{1,s}}}{\sqrt t_0 (\frac12+\mu)} + \frac{C\delta}{2\mu t_0^{2\mu}} \right)^2 \right.\\
& +\left(\frac{ \|\alpha_j \|_{l^{2,q}}  }{\mu} + \frac 1{(\mu+\gamma)t_0^\gamma} +\frac {\|u_+\|_{W^{1,s}}}{\sqrt {t_0} (\frac12+\mu)} + \frac{\delta}{2\mu t_0^{2\mu}} \right)\\
&\left. \quad \quad \quad  \left(\frac{ \|\alpha_j\|_{l^{2,q}}  }{\mu} + \frac {C(t_0,\|\alpha_j\|_{l^{2,s}})}{(\mu+\gamma)t_0^\gamma} +\frac {\|u_+\|_{W^{1,s}}}{\sqrt t_0 (\frac12+\mu)} \right)  \right)  ,
\end{align*}
wich implies the conclusion of the Lemma for $k\in \llbracket 1, s-1 \rrbracket  $.\par
For $k=s$, we proceed similarly but instead of estimating
\[
\sup_{t_0\leq t}t^\mu \int_t^\infty  \|v-v_1\|_{L^2}\|v\|_{L^\infty}\|\nabla^sv\|_{L^\infty}\frac{d\tau}\tau
\]
by using Gagliardo-Nirenberg \eqref{GN} on $\|\nabla^sv\|_{L^\infty}$, we rather write
\[
\sup_{t_0\leq t}t^\mu \int_t^\infty ( \|v-v_1\|_{L^\infty}\|v\|_{L^\infty}\|\nabla^s(v-v_1)\|_{L^2}+ \|v-v_1\|_{L^2}\|v\|_{L^\infty} \|\nabla^sv_1\|_{L^\infty})\frac{d\tau}\tau.
\]
Similarly, instead of estimating
\[
\sup_{t_0\leq t}t^\mu \int_t^\infty  \|v-v_1\|_{L^\infty}\|v_1\|_{L^2}\|\nabla^sv\|_{L^\infty}\frac{d\tau}\tau,
\]
we rather write
\[
\sup_{t_0\leq t}t^\mu \int_t^\infty ( \|v-v_1\|_{L^\infty}\|v_1\|_{L^\infty}\|\nabla^s(v-v_1)\|_{L^2}+ \|v-v_1\|_{L^2}\|v_1\|_{L^\infty} \|\nabla^sv_1\|_{L^\infty})\frac{d\tau}\tau,
\]
and obtain a similar bound.
\end{proof}

\begin{lemma}[Estimation of $\nabla^k J_a$]
Let $J_a$ defined by \eqref{Ja}:
\[
J_a(t,x)=-2i \displaystyle\int_{t}^\infty e^{i(t-\tau)\partial_x^2}  \sum_{p\neq j} e^{-i(|\alpha_p|^2-|\alpha_j|^2)\ln \tau}e^{i(p-j)\frac x2} \overline{\alpha_p}\alpha_j e^{i\frac{j^2-p^2}4 \tau}  e^{i\tau\partial_x^2}u_+\frac{d\tau}{\tau},
\]
then
\[
\sup_{t_0<t} t^\mu \|\nabla^kJ_a\|_{L^2} \leq C(\|\alpha_j\|_{l^{2,k}}) \sup_{t_0\leq t}\frac {t^\mu }t \left \| \frac{\widehat{u_+}}{\cdot+\mathbb Z/2} \right\|_{H^k} .
\]
\end{lemma}
\begin{proof}
The first step is to perform a Fourier transform on the space variable and set the domain exploit the linear phase in $x$:
\begin{align*}
\nabla^k J_a (t,x)
=&e^{it\partial_x^2} \int_t^\infty \int_{\mathbb R} (i\xi)^k e^{i\tau\xi^2} e^{ix\xi} \sum_{p\neq j} \overline{\alpha_p}\alpha_j e^{-i(|\alpha_p|^2-|\alpha_j|^2)\ln \tau} e^{-i\tau(\xi-\frac{j-p}2)^2}\\
&\times \widehat{u_+}(\xi-\frac{j-p}2)e^{i\frac{j^2-p^2}4\tau}d\xi \frac{d\tau}\tau \\
=&e^{it\partial_x^2} \sum_{p\neq j}\overline{\alpha_p}\alpha_j  \int_{\mathbb R} (i\xi)^k e^{ix\xi} \widehat{u_+}(\xi-\frac{j-p}2)  \int_t^\infty e^{i\tau (j-p)(\frac{j+p}4 + \xi -\frac {j-p}4 )} e^{-i(|\alpha_p|^2-|\alpha_j|^2)\ln \tau}   \frac{d\tau}\tau d\xi
\end{align*}
As $p\neq j$, for all $\xi$ except $\xi=\frac p2$ we now perform an integration by parts on the $\tau$ variable to gain integrability:
\begin{align*}
 &\int_t^\infty e^{i\tau (j-p)(\xi- \frac{j-p}2+\frac j2 )}  e^{-i(|\alpha_p|^2-|\alpha_j|^2)\ln \tau}   \frac{d\tau}\tau \\
 =&- \frac{ e^{it (j-p)(\xi- \frac{j-p}2 +\frac j2 )}  }{i (j-p)(\xi- \frac{j-p}2 +\frac j2 )} \frac{e^{-i(|\alpha_p|^2-|\alpha_j|^2)\ln t}   }t + \int_t^\infty\frac{ e^{i\tau (j-p)(\xi- \frac{j-p}2 +\frac j2 )}  }{i (j-p)(\xi- \frac{j-p}2 +\frac j2 )} \left( \frac{ e^{-i(|\alpha_p|^2-|\alpha_j|^2)\ln \tau}}\tau \right)_\tau  d\tau.\\
\end{align*}
Undoing the Fourier transform, we have:
\begin{align*}
\nabla^kJ_a(t,x)= &\frac {e^{it\partial_x^2}i}t \sum_{p\neq j} \frac{e^{-i(|\alpha_p|^2-|\alpha_j|^2)\ln t}}{j-p} \overline{\alpha_p}\alpha_j e^{i\frac{j^2-p^2}4 t}  \nabla^k [e^{i(p-j)\frac x2}e^{it\partial_x^2}u_+^j (x)]\\
&+ \frac{e^{it\partial_x^2}}i\int_t^\infty  \sum_{p\neq j} \left( \frac{e^{-i(|\alpha_p|^2-|\alpha_j|^2)\ln \tau}}{\tau}\right)_\tau \overline{\alpha_p}\alpha_j e^{i\frac{j^2-p^2}4 \tau}  \nabla^k [e^{i(p-j)\frac x2}e^{i\tau\partial_x^2}u_+^j (x)] d\tau,
\end{align*}
where
\begin{equation}\label{upj}
\widehat{u_+^j} (\xi) = \frac {\widehat{u_+}(\xi)}{\xi+\frac j2}.
\end{equation}
To conclude, we now take the $L^2$ norm and use the fact that $(\alpha_j)\in l^{2,k}$ to obtain that:
\[
\sup_{t_0<t} t^\mu \|\nabla^kJ_a\|_{L^2} \leq C(\|\alpha_j\|_{l^{2,k}}) \sup_{t_0\leq t}\frac {t^\mu }t \left\| \frac{\widehat{u_+}(\cdot) }{\cdot+\mathbb Z/2}\right\|_{H^k} .
\]
\end{proof}
\begin{lemma}[Estimation of $\nabla^k J_b$]
Let $J_b$ defined by \eqref{Jb}:
\begin{align*}
J_b=&-2i \displaystyle\int_{t}^\infty e^{i(t-\tau)\partial_x^2}  \sum_{(p,j)\in \mathbb Z^2} e^{-i(|\alpha_p|^2-|\alpha_j|^2)\ln \tau+i(p-j)\frac x2} (\overline{\alpha_p}R_j + \overline{R_p}\alpha_j +\overline{R_p}R_j)\\
&\times e^{i\frac{j^2-p^2}4 \tau}e^{i\tau\partial_x^2}u_+\frac{d\tau}{\tau},
\end{align*}
 then
\[
\sup_{t_0\leq t} t^\mu \|\nabla^kJ_b(t)\|_{L^2} \leq C( t_0,\|\alpha_j\|_{l^{2,k}} )  \sup_{t_0\leq t} \frac {t^\mu}{t^\gamma} \|u_+\|_{H^k}.
\]
\end{lemma}
\begin{proof}
This proof is straightforward: we exploit the integrability provided by the decay \eqref{borneR} of the $R_j$'s:
\begin{align*}
\| \nabla^kJ_b(t) \|_{L^2_x} \leq &\int_t^\infty   \sum_{(p,j)\in \mathbb Z^2} \| e^{-i(|\alpha_p|^2-|\alpha_j|^2)\ln \tau}(\overline{\alpha_p}R_j + \overline{R_p}\alpha_j +\overline{R_p}R_j)\\
&\times e^{i\frac{j^2-p^2}4 \tau}   \nabla^k[e^{i(p-j)\frac x2} e^{i\tau\partial_x^2} \widehat{u_+}(x)] \|_{L^2_x} \frac{d\tau}\tau\\
\leq & \int_t^\infty   \sum_{(p,j)\in \mathbb Z^2} C(t_0 \|\alpha_j\|_{l^{2,k}} ) \|u_+\|_{H^k} \frac{d\tau}{\tau^{1+\gamma}},
\end{align*}
so
\[
\sup_{t_0\leq t} t^\mu \|\nabla^kJ_b(t)\|_{L^2} \leq C(t_0, \|\alpha_j\|_{l^{2,k}}  )  \sup_{t_0\leq t} \frac {t^\mu}{t^\gamma} \|u_+\|_{H^k}.
\]
\end{proof}
\begin{lemma}[Estimation of $\nabla^k J_c$] Let $J_c$ defined by \eqref{Jc}:
\[
J_c= - i  \displaystyle\int_{t}^\infty e^{i(t-\tau)\partial_x^2} A^2 e^{-i\tau\partial_x^2}\overline{u_+} \frac{d\tau}{\tau},
\]
 then
\[
\sup_{t_0\leq t}t^\mu\|\nabla^kJ_c\|_{L^2}\leq C( t_0,\|\alpha_j\|_{l^{2,k}} ) \left(\sup_{t_0<t} \frac{t^\mu}t \left\| \frac{\widehat{u_+}(\cdot) }{\cdot+\mathbb Z/2} \right\|_{H^{k-1}}+ \sup_{t_0<t} \frac{t^\mu}{t^\gamma} \|u_+\|_{H^k}  \right).
\]
\end{lemma}
\begin{proof}
The proof is in two steps, according to the following decomposition:
\begin{align*}
J_c = &- \int_t^\infty e^{i(t-\tau)\partial_x^2}  \sum_{(j,p)\in\mathbb Z^2} \overline{\alpha_p}\overline{\alpha_j} e^{-i\tau\frac{p^2+j^2}4+ix\frac{p+j}2} e^{-i(|\alpha_j|^2-|\alpha_p|^2)\ln \tau} e^{-i\tau\partial_x^2}\overline{u_+}\frac {d\tau}\tau\\
&-  \int_t^\infty e^{i(t-\tau)\partial_x^2}    \sum_{(j,p)\in\mathbb Z^2}  \left(\overline{\alpha_p}\overline{R_j} + \overline{R_p}\overline{\alpha_j} +\overline{R_p}\overline{R_j}\right) e^{-i\tau\frac{p^2+j^2}4+ix\frac{p+j}2} e^{-i(|\alpha_j|^2+|\alpha_p|^2)\ln \tau} \\
&\times e^{-i\tau\partial_x^2}\overline{u_+}\frac {d\tau}\tau\\
=&-J_{c_1}- J_{c_2}
\end{align*}
The diagonal term of $J_{c_1}$ could be controlled with inhomogeneous Strichartz with the couple $(4,\infty)$ of dual $(\frac43,\infty)$, but we rather use IBP in Fourier space on the whole term to obtain a better time decay:
\begin{align*}
\nabla^k J_{c_1}=&\nabla^k\int_t^\infty e^{i(t-\tau)\partial_x^2}  \sum_{(j,p)\in\mathbb Z^2} \overline{\alpha_p}\alpha_j e^{-i\tau \frac{p^2+j^2}4+ix\frac{p+j}2} e^{-i(|\alpha_j|^2+|\alpha_p|^2)\ln \tau}  e^{-i\tau\partial_x^2}\overline{u_+}\frac {d\tau}\tau\\
=&e^{-it\partial_x^2}\int_t^\infty \int_{\mathbb R} (i\xi)^k e^{i\tau\xi^2} e^{ix\xi}   \sum_{(j,p)\in\mathbb Z^2} \overline{\alpha_p}\alpha_j e^{-i\tau\frac{p^2+j^2}4} e^{-i(|\alpha_j|^2+|\alpha_p|^2)\ln \tau} e^{i\tau(\xi + \frac{p+j}2)^2}\\
&\times \widehat{\overline{u_+}}(\xi + \frac{p+j}2) d\xi \frac {d\tau}\tau\\
= &e^{-it\partial_x^2}   \sum_{(j,p)\in\mathbb Z^2} \overline{\alpha_p}\alpha_j  \int_{\mathbb R} (i\xi)^k e^{ix\xi} \widehat{\overline{u_+}}(\xi + \frac{p+j}2)\\
&\times \int_t^\infty e^{-i\tau\frac{p^2+j^2}4-i(|\alpha_j|^2+|\alpha_p|^2)\ln \tau +i\tau \frac{(p+j)^2}4}e^{2 i\tau \xi^2} e^{i\tau\xi (p+j)}\frac {d\tau}\tau d\xi \\
= &e^{-it\partial_x^2}   \sum_{(j,p)\in\mathbb Z^2} \overline{\alpha_p}\alpha_j  \int_{\mathbb R} (i\xi)^k e^{ix\xi} \widehat{\overline{u_+}}(\xi + \frac{p+j}2)\\
&\times \int_t^\infty e^{-i(|\alpha_j|^2+|\alpha_p|^2)\ln \tau }e^{ i\tau(2 \xi(\xi+\frac{p+j}2) +\frac {pj}2)}\frac {d\tau}\tau d\xi
\end{align*}
We now perform for any $\xi$ except the countably set of roots of the phase an IBP on the time integral:
\begin{align*}
  &\int_t^\infty e^{-i(|\alpha_j|^2+|\alpha_p|^2)\ln \tau }e^{ i\tau (2\xi(\xi+\frac{p+j}2) +\frac {pj}2)} \frac {d\tau}\tau \\
=& -\frac{e^{-i(|\alpha_j|^2+|\alpha_p|^2)\ln t }e^{ it (2\xi(\xi+\frac{p+j}2) +\frac {pj}2)} }{(2 i \xi(\xi+\frac{p+j}2) +\frac {pj}2 )t}
+ \int_t^\infty \frac{e^{ i\tau( 2\xi(\xi+\frac{p+j}2) +\frac {pj}2)} }{2 \xi(\xi+\frac{p+j}2) +\frac {pj}2} \left( \frac{e^{-i(|\alpha_j|^2+|\alpha_p|^2)\ln \tau}}{\tau} \right)_\tau d\tau.
\end{align*}
Undoing the Fourier transform, we have:
\begin{align*}
\nabla^k J_{c_1} =& \frac{ie^{it\partial_x^2}}{2t} \sum_{(j,p)\in\mathbb Z^2} \overline{\alpha_j}\overline{\alpha_p} e^{-it\frac{p^2+j^2}4} e^{-i(|\alpha_j|^2-|\alpha_p|^2)\ln t} \nabla^{k-1}[e^{it\frac{(p+j)}2x} e^{-it\partial_x^2}\overline{u_+^{pj}} (x)] \\
&+\frac{e^{it\partial_x^2}}{2i} \sum_{(j,p)\in\mathbb Z^2} \int_t^\infty \overline{\alpha_j}\overline{\alpha_p} e^{-i\tau\frac{p^2+j^2}4}  \left( \frac{e^{-i(|\alpha_j|^2+|\alpha_p|^2)\ln \tau}}{\tau} \right)_\tau \nabla^{k-1}[e^{i\tau\frac{(p+j)}2x} e^{-i\tau\partial_x^2}\overline{u_+^{pj}} (x)] .
\end{align*}
where $u_+^{pj}$ has been defined by \eqref{upj}.
As for $J_a$, we conclude with this term by taking the $L_2$ norm on the space variable and use the fact that $(\alpha_j)\in l^{2,q}$:
\[
\sup_{t_0<t} t^\mu \|\nabla^kJ_{c_1}\|_{L^2_x} \leq C(\|\alpha_j\|_{l^{2,k}}) \sup_{t_0<t}\frac{ t^\mu}t  \left\| \frac{\widehat{u_+}(\cdot) }{\cdot+\mathbb Z/2} \right\|_{H^{k-1}}.
\]
Finally, as for $J_b$ we use \eqref{borneR}:
\begin{align*}
\sup_{t_0<t} t^\mu \|\nabla^kJ_{c_2}\|_{L^2_x} \leq &C(t_0, \|\alpha_j\|_{l^{2,k}}) \sup_{t_0<t} t^\mu \int_t^\infty \frac 1{\tau^\gamma} \left \| \nabla^k \overline{u_+}  \right \|_{L^2} \frac{d\tau }\tau \\
\leq &C( t_0,\|\alpha_j\|_{l^{2,k}}  ) \|u_+\|_{H^k} \sup_{t_0<t} \frac{t^\mu}{t^\gamma} ,
\end{align*}
hence:
\[
\sup_{t_0\leq t}t^\mu\|\nabla^kJ_c\|_{L^2}\leq C( t_0,\|R_j\|_{l^{2,k}} ) \left(\sup_{t_0<t} \frac{t^\mu}t \left\| \frac{\widehat{u_+}(\cdot) }{\cdot+\mathbb Z/2} \right\|_{H^{k-1}}+ \sup_{t_0<t} \frac{t^\mu}{t^\gamma} \|u_+\|_{H^k}  \right).
\]
\end{proof}

\begin{lemma}[Estimation of $\nabla^kJ_d$] Let $J_d$ defined by \eqref{Jd}:
\[
J_d=-i\displaystyle\int_{t}^\infty e^{i(t-\tau)\partial_x^2} (2 A|e^{i\tau\partial_x^2}u_+|^2 + \overline{A}(e^{i\tau\partial_x^2}u_+)^2 )\frac{d\tau}{\tau},
\]
then
\[
\sup_{t_0\leq t}  t^\mu \|\nabla^k J_{d}\|_{L^2} \leq  C(t_0,\|\alpha_j\|_{l^{2,k}} )\sup_{t_0\leq t} \frac {t^\mu}{t^{\frac12}} \|u_+\|_{L^1}\|u_+\|_{H^k}.
\]
\end{lemma}

As announced, this is the limiting term for the time decay.
\begin{proof}
The proof is straightforward, we just use direct upper-bounds and dispersion estimate \eqref{disp} on the Schrödinger group.
\begin{align*}
\sup_{t_0\leq t} t^\mu \|\nabla^k J_{d}\|_{L^2} \leq& \sup_{t_0\leq t} t^\mu  \int_t^\infty \sum_{j\in \mathbb Z} (|\alpha_j| +|R_j | )\| \nabla^k [e^{i\frac{xj}2} |e^{i\tau\partial_x^2}u_+|^2 + e^{-i\frac{xj}2}(e^{i\tau\partial_x^2}u_+(x))^2 ] \|_{L^2} \frac{d\tau}\tau
\\
\leq & C(t_0,\|\alpha_j\|_{l^{2,k}} )\sup_{t_0\leq t} t^\mu  \int_t^\infty \|u_+ \|_{H^k} \| e^{i\tau\partial_x^2}u_+\|_{L^\infty} \frac{d\tau}{\tau} \\
\leq & C(t_0,\|\alpha_j\|_{l^{2,k}})\sup_{t_0\leq t} t^\mu  \int_t^\infty \|u_+ \|_{H^k} \frac{\| u_+\|_{L^1}}{\sqrt \tau}  \frac{d\tau}{\tau} \\
\leq & C(t_0,\|\alpha_j\|_{l^{2,k}} )\sup_{t_0\leq t} \frac {t^\mu}{t^{\frac12}} \|u_+\|_{L^1}\|u_+\|_{H^k}.
\end{align*}
\end{proof}
The  last term  $J_e$ is easier to control, using again Schrödinger dispersion inequality.
\begin{lemma}[Estimation of $\nabla^k J_e$] Let $J_e$ defined by \eqref{Je}:
\[
J_e=- i \displaystyle\int_{t}^\infty e^{i(t-\tau)\partial_x^2}  |e^{i\tau\partial_x^2}u_+|^2e^{i\tau \partial_x^2} u_+ \frac{d\tau}\tau,
\]
 then
\[
\sup_{t_0\leq t}t^\mu\|\nabla^kJ_e\|_{L^2} \lesssim \sup_{t_0\leq t} \frac {t^\mu}{t}   \|u_+\|^2_{L^1}\|u_+\|_{H^s}
\]
\end{lemma}
\begin{proof}
It is straightforward to write that:
\begin{align*}
\|J_e(t)\|_{H^s}\leq&  \int_t^\infty \| |e^{i\tau\partial_x^2}u_+|^2e^{i\tau\partial_x^2}u_+ \|_{H^s} \frac{d\tau}\tau\\
&\lesssim \int_t^\infty \|e^{i\tau\partial_x^2}u_+\|^2_{L^\infty} \|e^{i\tau\partial_x^2}u_+\|_{H^s} \frac{d\tau}\tau \lesssim \frac{\|u_+\|^2_{L^1}\|u_+\|_{H^s}}t .
\end{align*}
\end{proof}
\subsection{The stability}\leavevmode\\
Regarding the results of the previous part,
\[
\|\phi(v)-v_1\|_{S}\leq C(t_0, u_+,\|\alpha_j\|_{l^{2,q}}) \left(\|v-v_1\|_{S} +\sup_{t_0<t} \frac{t^\mu}t+\sup_{t_0<t} \frac{t^\mu}{t^\gamma}+\sup_{t_0<t} \frac{t^\mu}{t^{\frac12}}\right),
\]
so we have the stability inclusion $\phi(X_\delta) \subset X_\delta $ for $\|\alpha_j\|_{l^{2,q}}$ and $\delta$ small and $t_0$ large enough,  and choosing $\mu<\max(\frac12,\gamma)$.
\subsection{The contraction}\leavevmode\\
In order to apply the fixed point, we still have to prove that $\phi$ contracts. \\
Let $v$ and $\tilde v$ in $Y$.\\
From
\begin{equation*}
\phi(v) - \phi (\tilde v) = \int_t^\infty e^{i(t-\tau)\partial_x^2}\left(\frac{|v|^2-2M}{ 2\tau}v - \frac{|\tilde{v}| ^2-2M}{2 \tau}\tilde{v} \right)d\tau,
\end{equation*}
yields
\begin{align*}
&\sup_{t_0\leq t}t^\mu \|\phi(v) - \phi (\tilde v) \|_{L^2} \\
\lesssim& \sup_{t_0\leq t} t^\mu \int_t^\infty \| |v|^2v - |\tilde v|^2\tilde v - 2M (v - \tilde v) \|_{L^2} \frac{d\tau}\tau\\
\lesssim&  \sup_{t_0\leq t} t^\mu \int_t^\infty (2M + \|v\|^2_{L^\infty} + \|\tilde v\|^2_{L^\infty})\|v-\tilde v\|_{L^2} \frac{d\tau}\tau\\
\lesssim&\|v-\tilde v\|_2   \left( \frac {2M}{\mu} + \sup_{t_0\leq t}t^\mu \int_t^\infty ( \|v(\tau)\|^2_{L^\infty} + \|\tilde v(\tau)\|^2_{L^\infty}) \frac{d\tau}{\tau^{1+\mu}} \right) \\
\lesssim&\|v-\tilde v\|_2   \left( \frac {2M}{\mu} +  \sup_{t_0\leq t}t^\mu \int_t^\infty (2\|v-v_1\|^2_{L^\infty}+ 2\|\tilde v-v_1\|^2_{L^\infty}+ 4 \|v_1\|^2_{L^\infty}) \frac{d\tau}{\tau^{1+\mu}} \right) \\
\lesssim& \|v-\tilde v\|_2   \left( \frac {2M}{\mu} +  \sup_{t_0\leq t} t^\mu \int_t^\infty ( 8\|A\|_{L^{\infty}}^2+ 8\frac{\|u_+\|^2_{L^1}}\tau +4\frac{\delta^2}{\tau^{2\mu}}) \frac{d\tau}{\tau^{1+\mu}} \right)\\
\lesssim& \|v-\tilde v\|_2   \left( \frac {2M}{\mu} + \frac{8\|A\|_{L^{\infty}} }\mu + \frac{8\|u_+\|^2_{L^1}}{t_0(1+\mu)} + \frac{(2\delta)^2}{t_0^{2\mu}(1+2\mu)} \right).
\end{align*}
Moreover, for $k\in \llbracket 1, s-1\rrbracket $:
\begin{align*}
&\sup_{t_0\leq t } t^\mu \| \nabla^k(\phi(v)-\phi(\tilde v)) \|_{L^2}\\
 \leq &  \frac{ 2M}{\mu} \|v-\tilde v\|_S + \sup_{t_0\leq t}t^\mu\int_t^\infty  \left( \sum_{p=0}^k  \binom kp  \|\nabla^p (v-\tilde v)\|_{L^2}\|\nabla^{k-p}v^2\|_{L^\infty} \right. \\
&+ \left. \sum_{p=0}^k \sum_{q=0}^p \binom kp  \binom pq  \|\nabla^q(v-\tilde v)\|_{L^2}  \|\nabla^{k-p}\tilde v\|_{L^\infty}(\|\nabla^{p-q}v\|_{L^\infty} + \|\nabla^{p-q} \tilde v\|_{L^\infty} ) \right) \frac {d\tau}\tau \\
\lesssim &\|v-\tilde v\|_S \left(\frac{2M}\mu + \int_t^\infty \sum_{p=0}^k  \binom kp  \sum_{q=0}^{k-p} \binom {k-p}q \left(\|\nabla^qA\|_{L^\infty} + \frac{\|\nabla^q u_+\|}{\sqrt \tau} + \frac{\delta}{\tau^\mu} \right) \right. \\
&{ \left. \left(\|\nabla^{k-p-q}A\|_{L^\infty} + \frac{\|\nabla^{k-p-q} u_+\|}{\sqrt \tau} + \frac{\delta}{\tau^\mu} \right) \frac{d\tau}{\tau^{1+\mu}}\right)}\\
&+ \|v-\tilde v\|_S\int_t^\infty  \sum_{p=0}^k \sum_{q=0}^p \binom kp  \binom pq \left(\|\nabla^{k-p}A\|_{L^\infty} + \frac{\|\nabla^{k-p} u_+\|}{\sqrt \tau} + \frac{\delta}{\tau^\mu} \right) \\
&{2\left(\|\nabla^{p-q}A\|_{L^\infty} + \frac{\|\nabla^{p-q} u_+\|}{\sqrt \tau} + \frac{\delta}{\tau^\mu} \right) \frac{d\tau}\tau.}
\end{align*}
As for Lemma \ref{gradi}, we must avoid estimating
\[
\sup_{t_0\leq t}t^\mu\int_t^\infty  \| v-\tilde v\|_{L^2} \|w_1\|_{L^\infty} \|\nabla^s w_2\|_{L^\infty} \frac{d\tau}\tau,
\]
by using Gagliardo-Nirenberg \eqref{GN} on  $\|\nabla^s v\|_{L^\infty}$. Therefore we rather write
\[
\sup_{t_0\leq t}t^\mu\int_t^\infty  (\| v-\tilde v\|_{L^\infty} \| w_1\|_{L^\infty} \|\nabla^s (w_2-v_1)\|_{L^2} + \| v-\tilde v\|_{L^2} \| w_1\|_{L^\infty} \|\nabla^s v_1\|_{L^\infty} )\frac{d\tau}\tau ,
\]
for $(w_1,w_2)\in \{\tilde v,v\}^2 $.\\
To conclude, we finally use the Gagliardo-Niremberg interpolation inequality
\[
\|v-\tilde v\|_{L^\infty}\leq \|v-\tilde v\|_{L^2}^{\frac 12}\|\nabla (v-\tilde v)\|_{L^2}^{\frac 12} \leq \frac{\|v-\tilde v\|_S}{\tau^\mu}
\]
so that
\begin{align*}
&\sup_{t_0\leq t}t^\mu\int_t^\infty  (\| v-\tilde v\|_{L^\infty} \| w_1\|_{L^\infty} \|\nabla^s (w_2-v_1)\|_{L^2} + \| v-\tilde v\|_{L^2} \| w_1\|_{L^\infty} \|\nabla^s v_1\|_{L^\infty} )\frac{d\tau}\tau\\
\leq &\sup_{t_0\leq t}t^\mu \|v-\tilde v\|_S \int_t^\infty \|w_1\|_{L^\infty}\left(\frac{\|w_2\|_S}{\tau^\mu} +  \|\nabla^s A\|_{L^\infty} + \frac{\nabla^ku_+\|_{L^1}}{\sqrt \tau} \right)\frac{d\tau}{\tau^{1+\mu}}.
\end{align*}
Putting it together, we have that
\[
\|\phi(v)-\phi(\tilde v)\|_S\leq \|v-\tilde v\|_S C(t_0,\|\alpha_j\|_{l^{2,q}},u_+) ,
\]
therefore $\phi$ contracts for $\|\alpha_j\|_{l^{2,q}}$ and $\delta$ small and $t_0$ large enough, and solutions exists on $[t_0,\infty)$. This ends the first part of existence of solutions of Theorem \ref{stab}.

\subsection{End of the proof: estimates  \eqref{decayschro} to \eqref{decay0} }\label{fin}\leavevmode\\

Let $v$ the solution of \eqref{NLS} on $[t_0,\infty)$ constructed in Section \ref{proof}. Note that we also dispose of the following estimate \eqref{decaynls} on $v$:
\begin{equation}\label{decayv}
\forall k \in \llbracket 0,s \rrbracket , \quad \|\nabla^k(v-v_1)(t)\|_{L^2}=\mathcal O (t^{-\frac12^-}).
\end{equation}
First we set $ T_1 = \frac 1{t_0}$ and use the pseudo conformal transformation \eqref{PCTran} on $v$ to obtain $\psi$, a solution of \eqref{schro} on $(0,T_1]$ that satisfies:
\[
v(t,x) = \mathcal T (\psi)(t,x)= \frac{e^{i\frac{x^2}{4t}}}{\sqrt t}\overline{\psi}\left(\frac1t,\frac xt\right).
\]
Since $v_1$ is defined by \eqref{defv1}, we have that:
\[
v_1(t,x)=\mathcal T  ( e^{2iM \ln t} u_{\{\alpha_j\}} (t,x))  +e^{it\partial_x^2}u_+(x),
\]
and estimate \eqref{decayv} becomes:
\[
\forall k\in \llbracket 0,s\rrbracket,\quad \left \| \nabla^k \left[\mathcal T \left( \psi(t,x)-e^{2iM \ln t} u_{\{\alpha_j\}} (t,x) \right)- e^{it\partial_x^2}u_+(x) \right]\right\|_{L^2}=\mathcal O (t^{-\frac12^-}).
\]
Then we normalize back the solutions $\psi$ of \eqref{schro} to obtain an estimate on the solutions $u$ of \eqref{eq} on $(0,T_1]$. As $\mathcal T$ preserves the $L^2$ norm, we use the fact that, for $J$ defined in the introduction by \eqref{defJ} we have:
\[
\nabla^k \mathcal T = \mathcal T (-i)^k J^k,
\]
to obtain estimate \eqref{decayschro}:
 \[
\forall k\in \llbracket 0,s\rrbracket,\quad \left\| (-i)^kJ^k \left[u (t,x)-u{\{\alpha_j\}} (t,x)\right] - e^{-2iM\ln t} \mathcal T \left(e^{it\partial_x^2}\nabla^ku_+(x)\right) \right\|_{L^2}=\mathcal O (t^{\frac12^-}).
\]
Finally, when $\nabla^k u_+\in (x^4L^2)$ for all $k\in \llbracket 0,s\rrbracket$, the $L^2$ estimate \eqref{decayl2} comes from Schrödinger evolution properties applied to the perturbation and has already been done in section 2.2 of \cite{BV1}:
\begin{equation}
\left\| \mathcal T \left(e^{it\partial_x^2}\nabla^ku_+(x) \right)  - \left(-i\frac x2\right)^k \widehat{\overline {u_+}}\left( -\frac x2\right)  \right\|_{L^2}=\mathcal O(t ).
\end{equation}
The last estimate \eqref{decay0} is exactly estimate \eqref{decayl2} with $k=0$ and Theorem \ref{stab} is proven.
\section*{Acknowledgments}
This paper has been written during my PhD under the supervision of Valeria Banica and Nicolas Burq, I would like to thank them for their precious help.
\bibliographystyle{aaai-named}

redpajama_set_name: RedPajamaArXiv


This week on the podcast I interview Nathan Michaud of Investors Underground.
Nate has been a day trader since 2003, and the stories from his first few years in the game are enough to make your head spin. After experiencing major highs and lows, both financially and psychologically (all covered during the interview), he's gone from strength to strength playing small-cap momentum stocks.
Over the next 60 minutes you'll learn about; Nate's approach to fast moving stocks, the additional risks linked with short-selling, a smart way to scale up, the power of compounding, and heaps more.
For educational material and on-going support from the Investors Underground community, go to chatwithtraders.com/IU for a discounted rate.
Nate shares the wild ups 'n downs of his early years as a trader; the infamous 'motorcycle story', getting stung by the SEC, and how he bounced back after losing close to $1M.
The areas to focus your energy as a less-experienced trader, and why it's okay to be 'inspired' as long as you're not 'influenced' – mould into the trader YOU need to be.
Insight to how Nate trades small-cap momentum stocks, his two most reliable setups, and why he naturally gravitates towards short-selling.
The challenges that come with scaling up, and a smart way to approach this. Plus, letting go of sloppy habits to reach your maximum potential.
InvestorsUndergound.com – To learn more about educational material and on-going support from the Investors Underground community, visit the website for a discounted rate.
Momo Traders by Brady Dahl – A book recently published, featuring interviews with ten momentum traders who started from the ground up – Nate is one of the traders interviewed.
Traders4ACause.org – The charity founded by Nate which raised over $200,000+ in 2015, hosting live events for traders and going from strength to strength.
Interview with @LX21 – This is the must-watch interview with Greg (better known as @LX21 on Twitter), by Cam of Investors Underground.
@InvestorsLive – Be sure to follow Nate on Twitter, and tell him what you thought of the interview!


European Commission chief Jean-Claude Juncker admitted on Tuesday (30 May) that he overlooked certain effects of Luxembourg's tax policy during his tenure as prime minister of the tiny member state.
Juncker was thrust into the world spotlight in late 2014 when his native EU duchy was singled out in the LuxLeaks scandal as a tax haven for dozens of multinationals, including Ikea and Pepsi.
European Commission President Jean-Claude Juncker admitted that he has been weakened by the revelations of controversial corporate tax avoidance schemes during his long tenure as prime minister of Luxembourg, his confirmation coinciding with a new batch of publications exposing the schemes.


This is one cool looking, and awesome playing guitar. Robin made some of the coolest guitars in last fifty years. Here is another NEW Robin fro our vaults. Truly a collectors piece...and a players rock machine!


Air conditioning services are essential for residential homes and commercial use as they provide great cooling and heating services to the whole building. There are different types of air conditioners and this depends on the owner of the house due to the purposes they intend for it to serve. It is important to consider the following factors when buying air conditioning products from the best sellers as you will be able to make a concrete decision.
It is important that you search for the companies that provide these services as this will help you engage the best company. You need to ensure that you get a company that will install the products to you after selling to ensure that you get the best services. You need to look for a company that deals with affordable sale and installation of air conditioners as this will help you work within your budget.
Ensure that when you are buying air conditioner you look for the quality as this will determine the kind of service it will offer to you. You need to ensure that you buy a product that is quite effective and fast in cooling or heating the surrounding to ensure that the place stays fresh. It is important that you choose a licensed and recognized company that will provide the best services to you.
You need to know if the company you buy these products from is willing to offer free maintenance and repair services to you whenever the need arises. When buying the products, look for a company that provides free delivery and installation services as this will leave you satisfied and save on your money. It is important to ask for the warranty of the goods to ensure that you are free to return them in case it gets damaged.
There are many benefits that will come your way when you purchase the products from a company that is reputable as I will explain below. You are assured that when you engage a company that offers affordable installation services for products. You can be satisfied knowing that a good company will ensure they provide the best quality of installation services that you need.
Clients are able to benefit from great and reliable services and this allows you to call on their services whenever you need them. You are able to rest knowing that a good company will deliver the products to you and offer the installation services as required of them. The best company will sell air conditioner and their parts and offer you with the warranty to ensure that in case there is defect on it, you are able to return to them.


Soros is very close to the Ukrainian president, who was put in power after a US-backed coup deposed the elected leader of Ukraine last year. In the email correspondence, Soros tells the Ukrainian leadership that the US should provide Ukraine "with same level of sophistication in defense weapons to match the level of opposing force." In other words, despite the February ceasefire, Soros is pushing behind the scenes to make sure Ukraine receives top-of-the-line lethal weapons from the United States. Of course it will be up to us to pay the bill because Ukraine is broke.
So George Soros will use his influence in the US government to put the American people on the hook for a bankrupt Ukraine – forcing us to pay for weapons, more military training, and Ukraine's crippling debt.
Who is thrilled with Soros' drawing the US government into more intervention in the region? The military-industrial complex for one is happy at the prospect of big weapons "sales" to Ukraine. The bankers are thrilled. Washington power-brokers are thrilled. There is something in this for everyone who is politically well-connected. The only losers are the people who will be forced to pay for it, the American taxpayers.
No one seems to ask why we are involved in Ukraine at all. Is it really any of our business if the east wants to break away from the west? Is it a vital US interest which flag the people wish to hang in Donetsk?
One thing we should be sure of is that Ukraine's debt will not be paid. As in other bailouts, much of it will be transferred to the US taxpayer through the IMF and the Federal Reserve. All of this is only possible because of the perception that the dollar is still the world's reserve currency. But this too is coming to an end. US military and financial interventionism worldwide are only speeding up the process.


text:
Олег Васильович Вітович ( 7 квітня 1967, Кременець, Тернопільська область —  22 жовтня 2011, Київ) — український політик, народний депутат України 2-го скликання.

Життєпис
Народився на Волині. Навчався і працював у Львові. Закінчив Львівський політехнічний інститут. Здобув дві вищі освіти, закінчив аспірантуру.

Громадсько-політична діяльність
В кінці 1980-х і на початку 1990-х був легендарною постаттю на Львівщині — сотні мітингів, пікетів, панахид і 18 кримінальних справ за розстрільними статтями (антидержавна агітація і пропаганда, заклики до повалення наявного державного ладу і т. д., врятувала лише вказівка Михайла Горбачова що всі статті, за якими судили дисидентів, не застосовувати на практиці).

Олег Вітович відомий численними публіцистичними колонками, редагував та був співредактором самвидаву, редагував самвидав «Голос націоналіста»

Олег Вітович одним з перших, хто висував націоналістичні гасла у своїх статтях і публічних виступах. На одному з львівських мітингів Олег оголосив, що гасло сьогодення: «Україна для українців», чим спантеличив львівських демократів з Руху. Дехто оголосив його провокатором, але відомо, що ідеї Олега суттєво радикалізували національний рух. Олег мав численних прихильників і симпатиків. Жінки надсилали йому листи зі своїми зізнаннями, чоловіки захоплювалися його ідеями. Талановитий оратор, цікавий публіцист, Олег став організатором львівських осередків практично всіх національно-визвольних організацій (УГС, «Руху», «Меморіалу», «Товариства української мови», страйкових комітетів (організував і розпочав перший львівський страйк з політичними вимогами).

Один із засновників Української націоналістичної спілки — закритої організації орденського типу, співголова і співзасновник Спілки незалежної української молоді, а дещо пізніше — націоналістичної фракції СНУМ. Вже на першому з'їзді СНУМу оголосив про створення цієї фракції. Один із організаторів УНА. З 1994 по 1999 — голова УНА-УНСО.

В першій половині 1990-х років по Львову ходили численні чутки про постійні напади бойовиків Вітовича на різні установи, зокрема приміщення Російського культурного центру у Львові.

З 1990 по 1994 рік — депутат Львівської обласної Ради, з 1994 по 1998 роки — Народний депутат України. Запам'ятався як депутат тим, що першим зайшов у сесійний зал ВРУ в камуфляжі, порванням російського прапора на трибуні Верховної Ради та численними законопроєктами щодо ядерного статусу України. На численні запитання про його правозахисну діяльність (був головою підкомітету Верховної Ради України з прав людини) завжди відповідав — хіба можна захистити права людини без ядерної зброї, чим викликав у представників іноземних делегацій шок.

Інші його дії — народний суд над головою сільради в себе на окрузі («як адвокат я не маю що доброго сказати, як прокурор я звинувачую в недбалому ставленні до односельців похилого віку, як суддя я вас засуджую до 20 ударів буком» (тобто ніжкою від дубового стола) — газети тоді гнівно розкритикували його дії, зате під час наступного приїзду на округ його очікувало кілька сотень прохачів з інших сіл, інших районів, інших округів і навіть із сусідньої області з проханням зробити аналогічні візити в їх населені пункти.

Двічі брав участь в бойових діях (в 1992—1993 роках) — в Придністров'ї та Абхазії. В середині 1990-х допомагав білоруській демократичній опозиції й паралельно з цим в 4-х державах був оголошений військовим злочинцем.

І разом з тим, Олег Вітович відомий як філософ, історик, публіцист і навіть співак (в 90-х на рок-концерті підмінив п'яного виконавця, свого друга, публіка була в захваті від депутата, який співав), товаришував практично зі всіма політв'язнями, багатьма відомими в 90-х роках журналістами. Олег мав і багато ворогів, які заздрили його популярності серед молоді.

Після депутатської каденції він відійшов від націоналістів УНА, у 1999 році несподівано для багатьох став функціонером «медведчуківської» СДПУ(о).

Похований на 79 полі Личаківського цвинтаря.

Родина
Дружина Христина та донька Ксенія.

Відзнаки
Нагороджений медаллю уряду ПМР та подякою уряду Грузії.

Примітки

Джерела
 Біографія на офіційному вебсайті Верховної Ради України
 Хто такий Вітович
 Помер один із засновників УНА

Народні депутати України 2-го скликання
Члени УНА-УНСО
Уродженці Кременця
Померли в Києві

redpajama_set_name: RedPajamaWikipedia


text:
About IWCF
Susan Smith Endowment
Join IWCF
Information for Grant Seekers
Above and Beyond Interest Area
Grants Historical Data
IWCF Calendar of Events
Share your photos!
Members Page – login landing spot
Idaho Womens Charitable Foundation
Positive Change Through Thoughtful Giving
What is Idaho Women's Charitable Foundation (IWCF)?
Idaho Women's Charitable Foundation is a membership organization created to expand the number of women involved in philanthropy in Idaho and to commit its members' resources to significantly impact the needs within the greater Treasure Valley.
Why was IWCF created?
IWCF was founded in 2001 to inspire women to become leaders in the area of philanthropy and to demonstrate the impact that pooled monies can have in bringing positive change to our communities. Anyone who is willing to commit to an annual membership contribution of $1,125 (the new membership amount starting in July 2019) for three years may become a member of IWCF. Half of your membership contribution goes into a pooled fund; the other half you may designate directly to the nonprofit entity of your choice, in your name, and in increments of $250 or $500. The remaining $125 goes to administrative expenses such as postage, rent, etc. The entire amount is tax deductible.
How is IWCF administered?
The majority of IWCF's workload is administered through a base of dedicated member volunteers who are supported by part-time office staff. Our aim is to keep the philanthropy dollars in the community and to be sustainable at the same time. We pride ourselves on keeping our operation costs very low. We try to do it as simply as possible and with much thought to financial responsibility.
Who manages funds?
The funds are managed by the Treasurer under direction and guidance of the Finance Committee and the Board. All accounts are insured through FDIC.
Is IWCF funding limited to Boise?
Our grants are focused in the Treasure Valley, including Ada, Canyon, Elmore, Owyhee, Boise and Valley counties. The founding members are from Treasure Valley, and are most familiar with charitable needs and trends in this area. Additionally, we are able to monitor the accountability of our grants closer to this base.
What kind of participation opportunities will be available for members?
Sharing your time and talents are a great way for you to make a personal impact and to really get to know this organization. Members find that when they are involved it greatly enhances their enjoyment of IWCF. There are many small ways to participate if or when you choose:
Attending educational programs throughout the year focusing on the areas in the community to which IWCF makes grants, philanthropy and personal leadership development.
Participating in social events to get to know other IWCF members and to introduce potential new members to the organization.
Being an on-call volunteer to assist as needed with office-related projects.
Serving on a committee which is welcomed and rewarding.
Committees include:
Education Committee: plans events on informed giving and our community's charitable needs.
Events Committee: coordinates logistics of IWCF events.
Finance Committee: advises and plans for the organization's financial stability and sustainability.
Governance Committee: provides guidance to the Board to foster best management practices on an as-needed basis.
Grants Committee: researches and nominates the nonprofit finalists for the pooled-fund grant ballot.
Grants Assessment Committee: evaluates grant awardees' project progress.
Leadership Development Committee: slates the Board of Directors for election and helps members get involved in IWCF opportunities that match their interests and skill sets.
Marketing & Communications Committee: develops strategies for increasing public awareness and keeping members informed.
Membership Committee: plans activities for attracting new members and meeting the needs of current members.
Symposium: plans the theme, program and logistics of the biennial symposium.
How much is annual membership? Why should I consider becoming a member if I already give to the community?
Membership is $1,125 per year (starting July 2019), and the entire amount is tax deductible.
$500 is credited to the pooled-fund for the grants process.
$500 is directed to the nonprofit of the member's choosing either in a single donation of $500 or two donations of $250 each as Individual Grant Designations (IGD).
$125 goes toward IWCF operating expenses
The collaborative giving concept leverages your membership contribution and those of other members to make a more significant impact, and the educational component of IWCF allows you to become a more informed charitable donor. Additionally, your knowledge of the community can be shared with other members. You can continue to give to your own personal interests with the $500 as your individual designation. This payment will be made for you, per your direction, to the organization(s) you designate as soon as you join or renew.
Will I be asked to participate in fundraising activities/events?
IWCF does not conduct fundraising events. We may seek underwriting/sponsorships from businesses or corporations for specific purposes. The organization's grant funding comes strictly from membership contributions. Members do have the additional opportunity to contribute to our endowment fund to support our long-term sustainability.
Can men become members of IWCF?
While IWCF was initially formed as a group of women, we certainly do not discriminate against gender! Men are encouraged and welcome to join if they are committed to the goals and objectives of the organization.
Can I transfer my membership, especially if an organization pays for or reimburses me for it?
IWCF membership is held by individual persons only and is non-transferable, regardless of the source of payment for any member's annual donation (e.g., a person other than the registered member, an organization, or other entity making payment on behalf of the member). The individual member is the specific person who is identified as such in the completed member registration form. IWCF does not have an institutional or organization membership category.
Pooled-Fund Grants
What is the process for determining the pooled-fund grants?
IWCF members contribute donations each year to a pooled fund from which we collectively award grants to nonprofit organizations, educational institutions and tax-exempt religious entities serving southwestern Idaho. The grants cycle begins in January and ends in May and is coordinated by the Grants Committee, open to all IWCF members. The committee researches proposals in five interest areas: cultural arts, education, environment, financial stability and health. After evaluations in each interest area, three to four proposals are selected for site visits, and projects are then narrowed down to two proposals for the member ballot. Ballots are then sent to the entire IWCF membership each spring for a vote on final funding decisions. The number of grants awarded and the total amount awarded depend on the size of IWCF's pooled fund and the amounts sought by the finalist proposals.
How will I receive my ballot?
Members may access an online ballot through their personal email or our Members Only website. Hard copy ballots are also available by request or at Q & A events during the voting period.
Are votes confidential?
Yes, votes are confidential. Only office staff will have access to the names attached to voting ballots.
Am I able to redirect my portion of the pooled fund if I am not happy with the chosen grantees?
Once the individual membership donation is paid, the amount the member contributes to the pooled fund for a given year will be awarded in accordance with the majority membership vote process for pooled grants. Each member only has the ability to control the IGD portion of their membership donation, directing it as either one $500 gift or two gifts of $250.
How does IWCF handle potential conflicts of interest on Grants Committees and subcommittees?
All members of the Grants Committee are required to disclose relationships they have with community organizations so that the committee can make a best-practice, informed decision regarding the member's level of involvement in grants selections.
Individual Grant Designations (IGDs)
What is an IGD?
Each member is requested to make their Individual Grant Designation (IGD) of $500 in a distribution of either two grants for $250 each or one grant for $500. IWCF will make contributions in each member's name to those organizations qualified by the IRS as having 501(c)(3) tax-exempt status, government entities such as schools, public libraries and park departments (as long as the grant furthers a public purpose), and religious organizations.
You may also designate to the IWCF Pooled Fund, Administrative Fund, or the Susan Smith Idaho Women's Charitable Foundation Endowment Fund.
When will organizations I've selected receive my donation?
This payment will be made for you, per your direction, to the organization(s) you designate as soon as you join or renew, or at the date of your specification. To make sure we get your money back to the community quickly, members have until March 31 to make their IGDs. If you choose to make your IGD to the current year's pooled fund, please make your designation by January 31. Individual grant designations not assigned by March 31 will be transferred to the pooled fund.
Is there a geographic restriction on IGDs?
There is no geographic restriction on these individual grants.
Idaho Women's Charitable Foundation fosters educated philanthropy through collaboration, pooled resources and individual giving to positively impact the community.
Address: 9454 W. Fairview Ave., Suite 120, Boise ID 83704
Phone: (208) 343- IWCF (4923)
information@IWCFBoise.org
Tax ID#: 82-0527986
© IWCF | website by CENTERLYNE

redpajama_set_name: RedPajamaCommonCrawl


"Dave Elman's book Hypnotherapy (the GreenBook) has been published by Westwood for over 30 years, and is a hypnotherapy classic. In a beautifully bound format designed for Gil Boyne to last the hypnotist a lifetime, this book is an essential for every practitioner of the hypnotic arts".
Dave Elman, Milton Erickson and Gil Boyne have been repeatedly voted as the three most influential persons in the history of modern hypnotherapy. Elman's book contains some of the best information that can be found from someone who was a genuine expert and master of his art. I consider it to be one of the most significant hypnotherapy books ever written.


text:
Olmedillo de Roa es un municipio español situado en la provincia de Burgos, en la comunidad autónoma de Castilla y León, que forma parte de la comarca de la Ribera del Duero.

Es un pueblo eminentemente agrícola, en sus campos se alternan el grano y la uva, combinados con hermosos pinares que coronan los cotarros. Con un clima continental, una altitud media de 800 metros y un suelo de aluvión y arcilloso, rico y fértil, han creado el hábitat apropiado para la uva Tinta del País, variedad principal de la zona – procedente del mismo tronco que la uva Tempranillo -, que otorga color, aroma y cuerpo a sus excelentes tintos muy cubiertos de color, con irisaciones moradas, aroma muy delicado y penetrante y suavemente ácidos.

En esta localidad está situada la Bodega Pagos del Rey, perteneciente al grupo  Félix Solís Avantis, que es la mayor elaboradora de toda la Denominación de Origen Ribera del Duero.

Demografía
El municipio, que tiene una superficie de 26,03 km², cuenta según el padrón municipal para  del INE con  habitantes y una densidad de  hab./km².

Economía

Evolución de la deuda viva
El concepto de deuda viva contempla sólo las deudas con cajas y bancos relativas a créditos financieros, valores de renta fija y préstamos o créditos transferidos a terceros, excluyéndose, por tanto, la deuda comercial.

La deuda viva municipal por habitante en 2014 ascendía a 0 €.

Patrimonio

Entre sus encantos destaca la Iglesia de Nuestra Señora de la Asunción, cuyo edificio fue construido, con la monumentalidad espacial propia del  y principios del XVII, por el cantero Juan de Navaceda. Es una construcción que consta de tres naves, muros de piedra y sillería de aparejo regular. Posee una cabecera con ventanas góticas, interior con yesos de las techumbres de las naves, bóveda estrellada en falso crucero, grandes pilastras que proporcionan esbeltez y dos arcos góticos. Tiene un gran salón de doce tramos de bóveda que se apean sobre pilares de base poligonal y sección circular en su alzado, con cubierta de falsa bóveda de ladrillo con decoración de yesones -excepto en el cuarto tramo que se aboveda con crucería reticular y de combados-.

En su interior alberga un retablo de Nuestra Señora de Basardilla, de mediados del XVII con la imagen de la Virgen de Basardilla que antiguamente se ubicaba en la Ermita del pueblo, también se puede encontrar un receptáculo del , realizado por Manuel Sandoval y un tabernáculo del  de Pedro Munar.

En la plaza del pueblo se conserva una casa del , cuya fachada posee un juego de ladrillo y piedra y dos bustos, uno de mujer y otro de hombre, encima de las dos ventanas del edificio. Existen numerosos ejemplos de este estilo de construcción, en el que van variando los bustos, en otros pueblos de la ribera (La Horra, Quinatanamanvirgo, Guzmán, Boada de Roa … )

No lejos del pueblo, sobre un cerro próximo se localiza la ermita barroca de la Virgen de la Basardilla. Externamente destaca su espadaña y sus grandes pilastras.

Iglesia parroquial de la Asunción de Nuestra Señora
Iglesia católica dependiente de la parroquia de Tórtoles de Esgueva en el Arciprestazgo de Roa, diócesis de Burgos.

Rutas

Olmedillo de Roa está situado en la ruta más corta entre los lugares santos de Caravaca de la Cruz y Santo Toribio de Liébana.

Cultura

Fiestas y costumbres
Fiestas patronales:La Semana Santa, la Festividad de la Santísima Trinidad y Natividad de Nuestra señora de Basardilla.

La fiesta más importante de la localidad es la Natividad de Nuestra Señora de Basardilla que se celebra el día 8 de septiembre, y al día siguiente conmemoran la fiesta de los Mártires de Cardeña. El día sábado de la fiesta, el pueblo realiza la Romería a la Ermita de la Virgen de Basardilla, después de celebrar misa en la ermita la virgen es llevada a la Iglesia del pueblo, durante todo el recorrido de vuelta al pueblo los jóvenes acompañan danzando sin parar en honor a la Virgen.

En Semana Santa, lo más llamativo son los cantos tradicionales, como el del Reloj -en el que se va narrando hora tras hora lo que va ocurriendo durante la pasión de Cristo-, que se entonan el día de Jueves Santo y Viernes Santo durante "la carrera" (procesión nocturna por las calles del pueblo", y es también importante el canto que se hace durante el lavatorio de pies que se realiza a los niños del pueblo en la iglesia el jueves santo, representando a los 12 apóstoles.

La Santísima Trinidad se celebra 40 días después de Semana Santa, la fiesta dura 2 días durante los que se organizan diferentes actos festivos. La procesión se desarrolla antes de la misa, y adquiere vistosidad ya que se danza durante el trayecto.

Historia
Las noticias relacionadas directamente con Olmedillo son escasas, por lo que el pasado de este municipio se reconstruye indirectamente. Del primer periodo de la edad de los metales hay que destacar una modalidad de cerámica que se conoce como estilo "Ciempozuelos" que en la Meseta tuvo gran desarrollo mil años antes de Cristo Se han encontrado posibles yacimientos de este periodo en diferentes municipios y en Olmedillo concretamente en los siguientes términos: El Calvario, El Carril, Las Eras, Busiles y La Serna.

Durante los inicios de la Edad de Bronce (1800- 750 a. C.) hubo gran diversidad de tipos de asentamientos de corta duración propios de una actividad económica itinerante,. La estructura de habitación permanente aparece en la denominada cultura del Soto. Periodo que destaca por la regionalización y porque el ritual funerario constituye la inhumación en enterramientos simples o colectivos. Se han encontrado yacimientos posibles de este periodo solo en Olmedillo de Roa: En la Basardilla, la Plaza y Santimia.

La Edad del Hierro, etapa de intercambios con el exterior y de asentamientos de población estables situados junto a los valles fluviales. Se han encontrado poblaciones de esta época cerca de nuestros pueblos.

La celtiberización, muy importante en estos pueblos ribereños, trajo una serie de cambios culturales y grandes novedades técnicas y económicas. El periodo de romanización, tras la conquista militar del pueblo vacceo, se construyó una importante red de caminos como la calzada de primera categoría que enlazaba Astorga con Zaragoza y otras de carácter secundario. En Olmedillo se conservan restos de estas vías, ejemplo de ello es el puente romano en las inmediaciones de la ermita. Los visigodos a su llegada a tierras del Duero ocuparon los asentamientos romanos, lo cual lo constatan los yacimientos encontrados en Olmedillo.

Hay que destacar, en el año 912, la ocupación de la fortaleza de Roa por parte del Conde Munio Nuñez y de ella pasaron a depender un conjunto de aldeas y núcleos de población entre ellos Olmedillo. Un conjunto de pequeñas fortalezas defienden estas tierras de los ataques islámicos a los largo del . Se dan sucesivos intentos de reorganizar y poblar el territorio según informan crónicas de 939 y 983 probablemente bajo la dirección de los cristianos. Esto se consigue en 1011 cuando se recuperan las fortalezas más significativas de la cuenca del Duero. Un hecho significativo es la reconstrucción del obispado de Osma al que Olmedillo pertenece desde 1136. Desde la entrada en vigor del fuero e Roa en 1143 cuando se funda la Comunidad de Villa y tierra de Roa, "Olmediello" tendrá la capacidad de poblar y ocupar las tierras yermas entre el Arlanza y la sierra.

En el  se vive una etapa de prosperidad y en la Edad Moderna el entramado urbano y el templo parroquial, recibirán una profunda remodelación adquiriendo en lo esencial el aspecto que ahora podemos contemplar.

Villa rodeada de despoblados pertenecientes a las 33 aldeas que en el año 1.143 conformaron el territorio de la Comunidad de Villa y Tierra de Roa. Constituye una de las zonas más densamente pobladas de la Edad Media con una tradición de población que hunde sus raíces desde el Calcolítico.

En el Censo de Vecindarios de la Corona de Castilla  realizado en 1591 se denominada Durón y Olmedillo , pertenecía a la Tierra de Roa , incluida en la provincia de Burgos . La comunidad contaba con 1.569 vecinos pecheros, correspondiendo 563 a la capital.

Villa , conocida entonces como Olmedillo , perteneciente a la Tierra de Roa con jurisdicción, de señorío  ejercida por el Conde de Siruela  quien nombraba su alcalde ordinario.

Olmedilllo de Roa posee un conjunto de arquitectura nobiliar y civil con sorprendentes inscripciones en algunas de sus sillerías e interesantes construcciones del . Por otro lado, conserva su estructura urbana circular, heredada de su pasado medieval.

A la caída del Antiguo Régimen queda constituido en ayuntamiento constitucional   en el partido de Roa   perteneciente a la región de Castilla la Vieja que en el Censo de la matrícula catastral contaba con 163 hogares y 652 vecinos.

Siglo XIX

Así se describe a Olmedillo de Roa en la página 251 del tomo XII del Diccionario geográfico-estadístico-histórico de España y sus posesiones de Ultramar, obra impulsada por Pascual Madoz a mediados del :

Véase también

Referencias

Enlaces externos

 Ayuntamiento de Olmedillo de Roa
 Informe sobre exhumación de cadáveres de la Guerra Civil española

Municipios de la provincia de Burgos
Localidades de la provincia de Burgos
Tierra de Roa
Ribera del Duero


text:
Laughing matters: Yoga heals by splitting sides
WEST BRIDGEWATER – Wander past the people sitting in a circle in the library's back room and you'll quickly come to a conclusion: These people are deranged.
No one is speaking, but for the better part of 30 minutes, they cackle and crack up, giggle and guffaw and hoot and holler.
Welcome to "Laughter Yoga."
You've discovered a free weekly "exercise" program open to the public led by West Bridgewater Public Library director Ellen Snoeyenbos.
It's pretty simple: Sit down, start laughing and leave happy.
"You feel odd at first, but once you get into it, it makes you feel so good that you want to keep going and share it with everyone," said Middleboro's Denise Goodspeed. "You feel like you're happy for the rest of the day and you want to make everyone else around you happy, too."
It's not a joke — seriously. Laughter Yoga is a directed program that requires training to lead and has been around for nearly 30 years after a doctor in India believed there were medical benefits to, well, giggling.
Snoeyenbos said there are over 6,000 "laughter clubs" worldwide now — she even visited one recently in France. But it's not a joke-telling club.
And so they just laugh. They laugh each vowel, they laugh like farm animals, and they laugh as they act out silly scenarios: "Flossing" their brains, tickling themselves with "laughter cream" and laughing into a sneeze.
They even laugh act out emotions: Guffawing at the pain of opening a credit card bill, rolling in the aisles as they pretend they're on a roller coaster and screaming with delight while pretending an ice cube slides down their back.
The class is one part laughter, one part charades and altogether loony — and that is exactly the point.
Pat Williams said she's brought the program to the West Bridgewater Food Pantry, where she often volunteers. But all of this sidesplitting action has a very serious effect.
"This becomes a part of yourself; it's so awesome," said Williams. "I've been through so much and been so stressed but this is the best medicine ever."


Senior Norwegian politicians and business figures have rejected Norway-plus, the increasingly touted British cross-party plan for the UK to leave the EU but join Norway in a free trade trade area inside the EU single market.
They attacked the idea as "neither in Norway nor the UK's interest". The UK would need Norway's permission to join its Efta club.
The rejection is a blow to an influential cross-party group led by the Tory MP Nick Boles, with private cabinet support, that is looking for a plan B if, as expected, MPs reject Theresa May's deal next Tuesday.
The former Labour foreign secretary David Miliband and the Tory former universities minister Jo Johnson also condemned Norway-plus as throwing away a key advantage of current membership "in the form of our vote, voice and veto around the table".
The former ministers' joint assault on Norway-plus, in a pamphlet written by the People's Vote campaign, is in some senses confirmation that the Boles plan could become a credible rival to a second referendum as a MPs search for a way out of a potential Commons deadlock.
The two men insist Norway-plus "would not be easy to negotiate, would not mean reduced payments to the EU and would not allow the UK to end freedom of movement for migrants from the EU".
Speaking on BBC Radio 4's Today programme, Johnson said: "Norway-plus does not deliver on the referendum since it would reduce the UK to a nation of lobbyists camped outside the European parliament in Brussels." Describing such an outcome as undignified and the worst of all worlds, he said it would only add to a sense of deceit for voters who had backed Brexit.
People's Vote claims that at a time when trust in politics is at low point, it is better to confront voters with the true choices on offer, rather than "repeating the deeply misleading promises made for Brexit two years ago", Johnson said.
But Stephen Kinnock, one of the Labour advocates for Norway-plus, said the Efta court could diverge from the European court of justice, and that the Efta treaty allowed for an emergency brake on migration in exceptional circumstances. He claimed as many as 10 cabinet ministers backed a pivot to Norway-plus after the expected Commons defeat next week, and urged May to take up this option.
Any attempt to push through May's deal or a second referendum was deeply risky, he said.
Norway-plus, the softest form of Brexit, requires the UK to seek to apply to join the European Free Trade Area grouping, consisting of Norway, Iceland, Switzerland and Liechtenstein. The scheme, as revised by Boles, would require the UK also to remain in the EU customs union indefinitely, or at least until a solution to the border in Ireland could be found.
But the plan was rejected by Heidi Nordby Lunde, an MP in Norway's governing Conservative party, and leader of Norway's European movement. She said her views reflected those of the governing party even though the Norwegian prime minister, Erna Solberg, has been more diplomatic by saying Norway would examine a UK application.
Explaining Norway's fear of the UK joining the Efta club, she said: "The three countries in Efta have to agree on all the regulations coming from the EU, so if one country vetoes something we all have to veto, which means that if the UK enters the Efta platform and starts to veto regulations that we want, this will affect not just the UK but also us as well. Part of the success we have had with this EEA agreement is for the last 25 years is that we do accept the rules and regulations that do come out of the EU, mostly because it is in our interest.
A member of the parliament's economic affairs committee, she said "it is not in my country's interests to have the UK aboard, and I cannot see how possibly an EEA/Efta agreement could be in the interests of the UK".
She said the only politicians in Norway who wanted the UK to join Efta were the Eurosceptic party that wanted to destroy Norway's relationship with the EU.
Her fears have been echoed by Ole Erik Almlid, of the Confederation of Norwegian Enterprise business association, who has also questioned whether the UK would be willing to go from rule maker to rule taker. He said Norway would suffer and parts of the Efta agreement with the EU would have to be suspended if an Efta member such as the UK refused to abide by the obligations in the agreement.
• This article was amended on 12 December 2018 to include reference to Switzerland, the fourth member of EFTA.


The new collection of exclusive Chapkas in Aviator-Look by ANDREA JANKE made in Germany - finest nappa-leather with fantastical trimmings in precious furs, such as in Silver-, Shadow- and Bluefrost Fox, Finn Raccoon, Canadian and Russian Barguzinsky Sable amongst others, refined with soft quilted black inner-linings.
Chapkas' by ANDREA JANKE - To Fall In Love!
A Chapka 'to fall in love', - these hand-manufactured Aviator-headpieces by ANDREA JANKE are elaborated with precious furs, and combined with ultra-soft nappa-leather; inner-linings are black and softly quilted. The size is adjustable.
Eine Chapka zum Verlieben, - diese Fliegermützen sind aufwendigst verarbeitet mit hochwertigen Rauchwaren, ultrasoften Nappaleder und weich-gesteppten, schwarzen Innenfutter. Die Grösse ist durch einen Lederzug verstellbar.
This Chapka asks "Do you love me?" And, I'm sure you will!
start-up for the Style Lab!
The most important question for all women - "Which shoes should I wear?" - I selected between René Caovilla and CASADEI Swarovski sandals - The CASADEI'S!
... and, if you'd like, - the San Blas Swarovski Clutch.
"A Riviera Dream ... Early evening at the Hotel du Cap Eden-Roc, Cap d'Antibes: the sun on the waves below; the models tread the long, tree-lined path leading to the "Château". Their silhouettes are simple, their diamonds sparkle with the last rays of the sun, the luster of their pearls glows against their garments. An ethereal dream of an ultra chic Cruise show, where mischievous tomboys meet femmes fatales holding sway over submissive Gatsbys. Karl Lagerfeld transports us to the legendary era of the Riviera, an era of rivers of diamonds, of Russian princes, of stylish aristocrats who came to sample the delights of the seaside in magnificent beachfront villas."


The Coalition's plan to cut foreign aid by $4.5 billion over 4 years is a kick in the teeth to the 1.2 billion people living on less than $2 a day.
Shadow Treasurer Joe Hockey's announcement comes on top of the $5.8 billion worth of aid cuts already signalled by the Labor Government.
Both major parties are failing to live up to the promises that all politicians made at the last election – to increase foreign aid spending to 0.5% of gross national income by 2015-16.
"Australian aid is saving people's lives and giving opportunities for those who have next to nothing," said Samah Hadid, Australian Director of The Global Poverty Project. "It is disgraceful that a rich country like Australia is shirking its moral responsibility to help those in need".
Thousands of Australians from all walks of life have come together to demonstrate their support for ending extreme poverty in the run-up to the federal election.
More than 85,000 Australians have signed the Movement to End Poverty petition calling for all politicians to keep their promises on foreign aid funding.
Only last Saturday 3,000 of Tony Abbott's own constituents came together to send him a message that extreme poverty should not be forgotten, whoever forms the next Government.
"Most Australians are generous people. It's a shame that their main political leaders are proving to be so stingy", added Ms Hadid.


text:
President hails Orban for 'tremendous job' defending Christians
David Charter, Washington
May 14 2019, 12:01am, The Times
Viktor Orban, left, was welcomed back to the White House yesterday for the first time in twenty yearsMARK WILSON/GETTY IMAGES
President Trump has praised the "tremendous job" in defence of "Christian communities" done by Viktor Orban, the leader of Hungary known for his hard line on immigration and the creation of a self-professed illiberal democracy.
In his first bilateral meeting with Mr Trump, ten days before European parliament elections, Mr Orban returned the compliment by noting that the pair shared a tough approach to illegal immigration.
Mr Orban, 55, was welcomed back to the White House yesterday for the first time in two decades. He had been frozen out by President Obama when he took increasingly tough measures against immigration, the media and the judiciary during his second stint as prime minister from 2010.
Then he become the first EU leader to voice support for…


text:
import math

import torch.nn.functional as F

from fairseq import metrics, utils
from fairseq.criterions import FairseqCriterion, register_criterion

@register_criterion('cross_entropy')
class CrossEntropyCriterion(FairseqCriterion):

    def __init__(self, task, sentence_avg):
        super().__init__(task)
        self.sentence_avg = sentence_avg

    def forward(self, model, sample, reduce=True):
        """Compute the loss for the given sample.

        Returns a tuple with three elements:
        1) the loss
        2) the sample size, which is used as the denominator for the gradient
        3) logging outputs to display while training
        """
        net_output = model(**sample['net_input'])
        loss, _ = self.compute_loss(model, net_output, sample, reduce=reduce)
        sample_size = sample['target'].size(0) if self.sentence_avg else sample['ntokens']
        logging_output = {
            'loss': loss.data,
            'ntokens': sample['ntokens'],
            'nsentences': sample['target'].size(0),
            'sample_size': sample_size,
        }
        return loss, sample_size, logging_output

    def compute_loss(self, model, net_output, sample, reduce=True):
        lprobs = model.get_normalized_probs(net_output, log_probs=True)
        lprobs = lprobs.view(-1, lprobs.size(-1))
        target = model.get_targets(sample, net_output).view(-1)
        loss = F.nll_loss(
            lprobs,
            target,
            ignore_index=self.padding_idx,
            reduction='sum' if reduce else 'none',
        )
        return loss, loss

    @staticmethod
    def reduce_metrics(logging_outputs) -> None:
        """Aggregate logging outputs from data parallel training."""
        loss_sum = sum(log.get('loss', 0) for log in logging_outputs)
        ntokens = sum(log.get('ntokens', 0) for log in logging_outputs)
        sample_size = sum(log.get('sample_size', 0) for log in logging_outputs)

        metrics.log_scalar('loss', loss_sum / sample_size / math.log(2), sample_size, round=3)
        if sample_size != ntokens:
            metrics.log_scalar('nll_loss', loss_sum / ntokens / math.log(2), ntokens, round=3)
            metrics.log_derived('ppl', lambda meters: utils.get_perplexity(meters['nll_loss'].avg))
        else:
            metrics.log_derived('ppl', lambda meters: utils.get_perplexity(meters['loss'].avg))

    @staticmethod
    def logging_outputs_can_be_summed() -> bool:
        """
        Whether the logging outputs returned by `forward` can be summed
        across workers prior to calling `reduce_metrics`. Setting this
        to True will improves distributed training speed.
        """
        return True

redpajama_set_name: RedPajamaGithub