text:
\section{Introduction}

\subsection{Strichartz estimates on compact manifolds}
The classical Strichartz estimates for the Schr\"odinger equation set in the Euclidean space $\mathbb{R}^d$ read (see~\cite{KT})
$$
\| e^{it\Delta} f(x) \|_{L^p_t (-\infty,\infty,L^q_x(\mathbb{R}^d))} \lesssim \| f \|_{L^2} \qquad \mbox{for $\frac{2}{p}+\frac{d}{q} = \frac{d}{2}, \;\; p \geq 2, \;\; (p,q) \neq (2,\infty)$},
$$
implying in particular (by Sobolev embedding) that, for $p \geq \frac{2(d+2)}{d}$,
$$
\| e^{it\Delta} f(x) \|_{L^p_{t,x} ((-\infty,\infty) \times \mathbb{R}^d)} \lesssim N^{\frac{d}{2}-\frac{d+2}{p}} \| f \|_{L^2} \qquad \mbox{if $\operatorname{Supp} \widehat{f} \subset B(0,N)$}.
$$
Given a compact Riemannian manifold $M$, with Laplace-Beltrami operator $\Delta$, and associated Sobolev spaces $H^s$, it is natural to ask for similar estimates on the Schr\"odinger group $e^{it\Delta}$: what is the best constant $C(M,p,T,N)$ in
$$
\|e^{it\Delta} f\|_{L^{p}_{t,x}([0,T] \times M)} \leq C(M,p,T,N) \| f \|_{L^2}
$$
(under some spectral assumption generalizing the Fourier support condition $\operatorname{Supp} \widehat{f} \subset B(0,N)$)?

Little is known about this question for general manifolds, but an upper bound on $C(M,p,T,N)$ was derived by Burq, G\'erard and Tzvetkov~\cite{BGT,BGT2}, which turns out to be sharp in some range for the sphere $\mathbb{S}^d$, at least if $d=3$. It was then showed~\cite{Thomann,ACMT} that the presence of a stable closed geodesic leads to a behavior similar to that of the sphere.

For tori, this question was recently answered by Bourgain and Demeter~\cite{BD} for time intervals $T \leq 1$.  They proved that if $R$ is a rectangular torus
$$
R = [0,\ell_1] \times \dots \times [0,\ell_d]
$$
(with the usual metric), then the following inequality holds:
for $p \geq 1$, $N \geq 1$, and for $\epsilon > 0$
\begin{equation}
\label{strichartz1}
\|e^{it\Delta} f\|_{L^{p}_{t,x}([0,1] \times R)} \lesssim _{\varepsilon}N^{\varepsilon} \left(1 + N^{\frac{d}{2}-\frac{d+2}{p}}\right)\|f\|_{L^2}\qquad\textrm{if }\operatorname{Supp} \widehat{f} \subset B(0,N).
\end{equation}
The question of Strichartz estimates on tori was first addressed by Bourgain~\cite{B1}, and this was followed by a number of works improving its results~\cite{B2,B3,CW,GOW}.  Before Bourgain and Demeter's paper, however, the sharp estimate seemed out of reach.  Also, before this paper, the estimates known for the cubic torus were better than for irrational tori, because of some number theoretic facts which were used in the arguments.  In this paper, we will study what happens for long time intervals $T >> 1$.  Over long time intervals, we will see that the behavior on irrational tori is actually better than the behavior on rational tori.

\subsection{Generic tori and times $T \geq 1$} Since the linear Schr\"odinger equation conserves the $L^2$ norm, the above estimate implies immediately, for $p \geq 1$, $N \geq 1$, $T \geq 1$, and for $\epsilon > 0$
\begin{equation}
\label{strichartz2}
\|e^{it\Delta} f\|_{L^{p}_{t,x}([0,T] \times R)} \lesssim _{\varepsilon}N^{\varepsilon} \left(1 + N^{\frac{d}{2}-\frac{d+2}{p}}\right) T^{\frac{1}{p}} \|f\|_{L^2}\qquad\textrm{if }\operatorname{Supp} \widehat{f} \subset B(0,N).
\end{equation}
This estimate is clearly optimal on the square torus, where the linear Schr\"odinger flow is periodic; but on irrational tori it raises the following question:
\textbf{For a generic choice of the parameters $(\ell_i)$, what is (up to sub-polynomial factors) the best constant $C(p,N,T)$ such that}
$$
\| e^{it\Delta} f \|_{L^p ([0,T] \times R)} \leq C(\ell_i,p,N,T) \| f \|_{L^2}, \qquad \mbox{if $\operatorname{Supp} \widehat{f} \subset B(0,N)$}?
$$
We will answer this question for $p > 6$ and obtain some upper and lower bounds for other $p$.

\subsection{Reformulation} Here we perform a change of variables to transform the problem to the square torus $\mathbb{T}^d=[0,1]^d$ (with periodic boundary conditions). Define the quadratic form
$$
Q (n_1,\dots,n_d) = \beta_1n_1^2 + \beta_2 n_2^2 + \cdots + \beta_d n_d^2, \quad \text{where $\beta_i=\ell_i^{-2}$,}
$$
and the corresponding differential operator
$$
\Delta_\beta = \frac{1}{2\pi} Q (\partial_1,\dots,\partial_d) = \frac{1}{2\pi}\big(\beta_1\partial_1^2 + \beta_2 \partial_2^2 + \dots + \beta_d \partial_d^2\big).
$$
For a function $f$ defined on $R$, observe that
$$
(e^{it\Delta} f)(\ell_1x_1,\ell_2 x_2,\dots,\ell_d x_d) =
(e^{2\pi it\Delta_\beta} g)(x_1,\dots,x_d) \quad \mbox{where $g(x_1,\dots,x_d) = f(\ell_1x_1,\ell_2 x_2,\dots,\ell_d x_d)$}.
$$
Therefore, Conjecture \ref{longstr} is equivalent to the corresponding estimates for the quantity\[\left\| e^{it\Delta_\beta} f \right\|_{L^p ([0,T] \times \mathbb{T}^d)}\quad\text{instead of}\quad \left\|e^{it\Delta}f\right\|_{L^p([0,T]\times R)}.\]
Moreover, the transformation between $(\ell_i)$ and $(\beta_i)$ is a diffeomorphism with positive Jacobian (so it maps null sets to null sets), thus below we will focus on the study of the quantity $\| e^{it\Delta_\beta} f\|_{L^p ([0,T] \times \mathbb{T}^d)}$ with parameters $(\beta_i)$.

\subsection{Genericity} To fix ideas, we will assume in the rest of this article that
$$
\beta_i \in [1,2] \quad \forall i \in \{1,\dots,d\}.
$$
\begin{df}
We will call a property \textit{generic} in $(\beta_1,\dots,\beta_d)$ if it is true for all $(\beta_1,\dots,\beta_d)$ outside of a null set (set with measure zero) of $[1,2]^d$.
\end{df}
Genericity will often be for us a consequence of a classical result on Diophantine approximation: it is well-known (see \cite{Cassels}) that, generically in $(\beta_i)$, there exists $C$ such that
\begin{equation}
\tag{D1} \label{dioph1}
\left|k_1+\beta_2k_2+\cdots+\beta_dk_d\right|\geq C \frac{1}{(|k_1| + \dots + |k_d|)^{d-1} \log(|k_1| + \dots + |k_d|)^{2d}}.
\end{equation}
A sharper version of this inequality, also true generically in $(\beta_i)$, is
\begin{equation}
\tag{D2} \label{dioph2}
\left|k_1+\beta_2k_2+\cdots+\beta_dk_d\right|\geq C \prod_{i=2}^{d}(1+|k_i|)^{-1}(\log(2+|k_i|))^{-2}.
\end{equation}

\subsection{The conjecture} We propose the following conjecture.
\begin{conj}\label{longstr} Let $d\geq 2$. For generic $(\beta_1,\cdots,\beta_d)$, one has
\begin{equation}
\label{strichartzlong}\left\|e^{it\Delta_\beta}f\right\|_{L^p([0,T]\times \mathbb{T}^d)}\lesssim_{\varepsilon}N^{\varepsilon}\left(1 + N^{\frac{d}{2}-\frac{d+2}{p}} \right) \left[1+\left(\frac{T}{N^{\theta(p)}}\right)^{\frac{1}{p}}\right]\|f\|_{L^2} \qquad\textrm{if }\operatorname{Supp} \widehat{f} \subset B(0,N)
\end{equation}
for $N\geq 1$, $T\geq 1$ and arbitrarily small $\varepsilon>0$, where\begin{equation}\label{expo}\theta(p)=\left\{\begin{aligned}&0,&p\in&\left[1,\frac{2(d+2)}{d} \right),\\
&\frac{d}{2} \left( p - \frac{2(d+2)}{d} \right), &p\in&\left[\frac{2(d+2)}{d},6\right),\\
& 2d-2,&p\in&\left[ 6,\infty\right). \end{aligned}\right.\end{equation} Moreover, these estimates are sharp up to $N^{\varepsilon}$ losses for arbitrarily small $\varepsilon$.
\end{conj}

Notice first that for $T = 1$ or $p \leq \frac{2(d+2)}{d}$, it follows from the estimate of Bourgain an Demeter~\eqref{strichartz1}.

Let us now explain briefly why this conjecture is plausible. It can be written equivalently: for generic $\beta$, and for $T \geq 1$,
\begin{equation}
\label{conj2}
\left\|e^{it\Delta_\beta} f \right\|_{L^p([0,T]\times \mathbb{T}^d )}\lesssim_{\varepsilon}N^{\varepsilon} \left[ T^{\frac{1}{p}} + N^{\frac{d}{2}-\frac{d+2}{p}} + N^{\frac{d}{2}-\frac{3d}{p}} T^{ \frac{1}{p}} \right]\| f\|_{L^2} \qquad\textrm{if }\operatorname{Supp} \widehat{f} \subset B(0,N).
\end{equation}
We will show heuristically that two simple examples saturate, in different regimes, the different terms in the right-hand side of the above. These two simple examples are $f \equiv 1$ and a peaked function living on a scale $\sim \frac{1}{N}$, such as
$$
\psi(x) = \frac{1}{N^{d/2}} \sum_{n \in \mathbb{Z}^d} \chi \left( \frac{n_1}{N} \right) \dots \chi \left( \frac{n_d}{N} \right)  e^{2\pi i n \cdot x},
$$
where $\chi \in \mathcal{C}_0^\infty$.

\bigskip

\noindent \underline{An example such that $\left\|e^{it\Delta_\beta} f \right\|_{L^p([0,T]\times R)} \sim T^{\frac{1}{p}} \| f \|_{L^2}$.} It obviously suffices to choose $f \equiv 1$, or $f(x) = e^{2\pi i n \cdot x}$ for some $n \in \mathbb{Z}^d$.

\bigskip

\noindent \underline{An example such that $\left\| e^{it\Delta_\beta} f \right\|_{L^p([0,T]\times R)} \sim N^{\frac{d}{2} - \frac{d+2}{p}} \| f \|_{L^2}$.} It is classical that $f = \psi$ satisfies this requirement as soon as $p \geq \frac{2(d+2)}{d}$, $T \geq 1$ (actually, $T \geq \frac{1}{N}$ suffices). This example shows the optimality of the result of Bourgain and Demeter~\eqref{strichartz1}.

\bigskip

\noindent \underline{An example such that $\left\| e^{it\Delta_\beta} f \right\|_{L^p([0,T]\times R)} \sim N^{\frac{d}{2} - \frac{3d}{p}}  T^{\frac{1}{p}} \| f \|_{L^2}$}. We now argue \underline{heuristically} that this should be the case for $f = \psi$, $p> \frac{2(d+2)}{d}$, $T > N^{2d-2}$. The main idea is that a significant contribution is made to the $L^p$ norm around the times $t_i$ where $u$ "refocuses", which is to say $u(t_i) \sim \psi$. These times occur with a period $\sim N^{2d-2}$, and around each time $t_i$, the contribution is of order $\sim N^{\frac{d}{2} - \frac{d+2}{p}}$ (by the previous paragraph). Therefore the $L^p$ norm on $[0,T] \times \mathbb{T}^d$ is of order $\left( \frac{T}{N^{2d-2}} \right)^{\frac{1}{p}} N^{\frac{d}{2} - \frac{d+2}{p}} \sim N^{\frac{d}{2} - \frac{3d}{p}}$.

We now explain why the time needed for the wave $u(t) = e^{it \Delta_\beta} \psi(x)$ to refocus is of order $N^{2d-2}$.
Observe that $u$ can be written
\begin{equation}
\label{equ}
u(t) = e^{it \Delta_\beta}  \psi(x) = \frac{1}{N^{d/2}} \sum_{n \in \mathbb{Z}^d} \chi \left( \frac{n}{N} \right) e^{2\pi i (n \cdot x + tQ_\beta(k))}.
\end{equation}
At the initial time, $u(t=0) = \psi(x)$, which is a very peaked function living on a scale $\sim \frac{1}{N}$. How long does it take before the wave $u$ "refocuses"? Fixing $\eta>0$, we argue that there is a time $t = q \sim N^{2d-2+\eta}$ such that $u(q) \sim \psi$. Indeed, by classical (simultaneous) Diophantine approximation theory~\cite{Cassels}, there exists, for generic $(\beta_i)$, an integer $q \sim N^{2d-2+\eta}$ such that, for all $i \in \{ 2,\dots,d\}$,
$$
\beta_i = \frac{p_i}{q} + O\left( \frac{1}{q N^{2+\frac{\eta}{d-1}}} \right) \qquad \mbox{for all $i \in \{ 2,\dots,d\}$}.
$$
This implies that, for $|k| \lesssim N$,
$$
\left\| q Q_\beta(k) \right\| = \left\| q \sum_{i=1}^d \beta_i |k_i|^2 \right\| \lesssim \frac{N^2}{N^{2+\frac{\eta}{d-1}}} << 1,
$$
where, for a real number $x$, we denote $\| x \|$ for the distance from $x$ to the closest integer.
Coming back to~\eqref{equ}, this implies that $u(t=q) \sim \psi$.

\subsection{Obtained result: optimality of the conjecture}
Our first result gives the optimality of the conjecture. We saw above very simple examples such that $\left\|e^{it\Delta_\beta} f \right\|_{L^p([0,T]\times R)} \sim T^{\frac{1}{p}} \| f \|_{L^2}$ and $\left\| e^{it\Delta_\beta} f \right\|_{L^p([0,T]\times R)} \sim N^{\frac{d}{2} - \frac{2(d+2)}{p}} \| f \|_{L^2}$. This implies that the two first terms on the right-hand side of~\eqref{conj2} are necessary. The third term, namely $N^{\frac{d}{2}-\frac{3d}{p}} T^{\frac{1}{p}}$, becomes dominant in the range $p>6$, and was justified heuristically above. We now provide a rigorous statement.

\begin{thm} For any $\eta>0$, and provided $\beta_2, \dots , \beta_d$ are generic, there exists $f$ such that $\operatorname{Supp} \widehat f \subset B(0,N)$, and, if $T > N^{2d-2+\eta}$
$$
\| e^{it \Delta_\beta} f \|_{L^p([0,T] \times \mathbb{T}^d)} \sim N^{\frac{d}{2} - \frac{3d}{p}} T^{\frac{1}{p}} \| f \|_{L^2(\mathbb{T}^d)}.
$$
\end{thm}
This theorem is proved in Section~\ref{smt0}.

\subsection{Obtained result: partial proof of the conjecture}

\begin{thm}\label{mainthmfin} Conjecture \ref{longstr} is true for $p>6$, and a weaker version holds for $p<6$. More precisely, the inequality~\eqref{strichartzlong}, holds, up to sub-polynomial factors, with $\theta(p)$ given by
\[
\mbox{if $d=2$}, \;\;
\theta(p)=\left\{\begin{aligned}&0,&p\in&\left[1,4\right),\\
&\frac{2(p-4)}{p+4},&p\in&\left[4,6\right),\\
&2,&p\in&\left[6,\infty\right),\end{aligned}\right.
\qquad \quad
\mbox{if $d=3$}, \;\;
\theta(p)=\left\{\begin{aligned}&0,&p\in&\left[1, \frac{10}{3} \right),\\
&\frac{4(3p-10)}{3p+14},&p\in&\left[ \frac{10}{3},6\right),\\
&4,&p\in&\left[6,\infty\right),\end{aligned}\right.
\]
and
\[
\mbox{if $d\geq 4$}, \qquad
\theta(p)=\left\{\begin{aligned}&0,&p\in&\left[1,\frac{2(d+2)}{d} \right),\\
&\frac{2(d-1)(pd-2d-4)}{pd+6d-4},&p\in&\left[\frac{2(d+2)}{d}, \frac{2d}{d-2} \right),\\
&\frac{(d-2)(pd-2d-4)}{4(d-1)},&p\in&\left[ \frac{2d}{d-2},6\right),\\
&2d-2,&p\in&\left[6,\infty\right),\end{aligned}\right.\]
\end{thm}
This theorem is the combination of theorems~\ref{mainthm1},~\ref{mainthm2} and~\ref{mainthm3}, which are proved in sections~\ref{smt1},~\ref{smt2} and~\ref{smt3} respectively.

\subsection{Notations}
\label{notations}
The Fourier transform of a function $f$ defined on $\mathbb{T}^d\times\mathbb{R}$
is
\[\widehat{f}(\tau,k)=\mathcal{F}_{x,t}f(\tau,k)=\int_{\mathbb{T}^d\times\mathbb{R}}e^{-2\pi i(k\cdot x+\tau t)}f(t,x)\,\mathrm{d}x\mathrm{d}t\] for $k\in\mathbb{Z}^d$ and $\tau\in\mathbb{R}$.

The function $\chi(z)$ is a smooth, even, nonnegative function that equals $1$ for $|z|\leq 1$ and equals $0$ for $|z|\geq 2$.

We write $A\lesssim B$ if $A\leq CB$ for some constant $C$; and $A\lesssim_a B$ if the constant $C$ depends on a parameter $a$: $A \leq C(a) B$. Finally, $A \sim B$ if $A \lesssim B$ and $B \lesssim A$

When we fix a scale $N$, we write $A\preceq B$ if $A\leq_\epsilon C_\epsilon N^{\varepsilon}B$.

For a real number $n$, we denote $\| n \|$ for the smallest distance from $n$ to an integer.

\section{Optimality for $p>6$}
\label{smt0}

\begin{thm} \label{mainthm0} Assume $p>6$, and that $\beta_2, \dots , \beta_d$ satisfy~\eqref{dioph2}. Define $f$ by its Fourier transform
$$
\widehat{f}(k) = \chi \left( \frac{k_1}{N} \right) \dots \chi \left( \frac{k_d}{N} \right)
$$
(so that in particular $\operatorname{Supp} \widehat{f} \subset B(0,N)$). Then for any $\eta>0$, and $N$ sufficiently big,
\begin{equation}
\label{oriole}
\| e^{it \Delta_\beta} f \|_{L^p([0,T] \times \mathbb{T}^d)} \sim N^{\frac{d}{2} - \frac{3d}{p}} T^{\frac{1}{p}} \| f \|_{L^2(\mathbb{T}^d)}
\qquad \mbox{if $T > N^{2d-2+\eta}$}.
\end{equation}
\end{thm}

\subsection*{Proof of Theorem~\ref{mainthm0}} First notice that it suffices to prove that~\eqref{oriole} holds for $p\geq 6$ an even integer, and for
\begin{equation}
\label{blackbird1}
f(x) = h(x_1) \dots h(x_d)
\end{equation}
if $h$ is a function on $\mathbb{T}$ such that
\begin{equation}
\label{blackbird2}
\operatorname{Supp} \widehat h \subset B(0,N) \qquad \mbox{and} \qquad \| e^{i (2\pi)^{-1} t \partial_{x}^2} f \|_{L^p([0,1] \times \mathbb{T})} \sim N^{\frac{1}{2} - \frac{3}{p}}.
\end{equation}
Indeed, $\sqrt N \chi(Nx)$ satisfies this latter condition for all $p \geq 6$. The statement of the theorem for all $p \geq 6$ follows by interpolation.

\bigskip

\noindent \underline{Step 1: the expression for the Strichartz norm.} Consider $f$ and $h$ as above (equations~\eqref{blackbird1} and~\eqref{blackbird2}); and normalize furthermore $\| h \|_{L^2} = 1$.

In order to take advantage of the tensorial definition of $f$, let $F(t) = \| e^{i(2\pi)^{-1} t \partial_x^2} h \|^p_{L^p(\mathbb{T})}$, which is a $1$-periodic function. Assuming that $T \in \mathbb{N}$, the Strichartz norm of $f$ can be written
\begin{align*}
\| e^{it \Delta_\beta} f \|_{L^p([0,T] \times \mathbb{T}^d)}^p & = \int_0^T F(t) F(\beta_2 t) \dots F(\beta_d t)\,\mathrm{d}t \\
& = T \int_0^1 F(t) \underbrace{ \frac{1}{T} \sum_{n=0}^{T-1} F(\beta_2 t +  \beta_2 n ) \dots F(\beta_d t + \beta_d n )}_{\displaystyle G(t)}\, \mathrm{d} t.
\end{align*}

\noindent \underline{Step 2: Fourier series expansion of $F$.} Expand $F$ in Fourier series:
$$
F(t) = \sum_{k \in \mathbb{Z}} a_k e^{2\pi i k t}.
$$
First,
$$
a_0 = \int_0^1 F(t)\,\mathrm{d}t = \| e^{i(2\pi)^{-1} t \partial_x^2} h \|^p_{L^p([0,1] \times \mathbb{T})} \sim N^{\frac{p}{2}-3}.
$$
Second, by H\"older's inequality,
\[|\partial_tF(t)|\leq p\int_{\mathbb{T}}\big|e^{i(2\pi)^{-1}t\partial_x^2}h(x)\big|^{p-1}\big|\partial_te^{i(2\pi)^{-1}t\partial_x^2}h(x)\big|\,\mathrm{d}x\lesssim F(t)^{\frac{p-1}{p}} \big\|\partial_x^2e^{i(2\pi)^{-1}t\partial_x^2}h(x)\big\|_{L^p}\lesssim N^2 F(t),\] since $\widehat{h}$ is supported at frequencies $\leq N$. Similar bounds can be obtained for higher order derivatives of $F$, leading to the estimate, for all $n \geq 0$,
\begin{equation}
\label{boundak}
|a_k| \lesssim_n a_0 \frac{N^{2n}}{|k|^n}.
\end{equation}

\noindent \underline{Step 3: convergence of $G$ to $a_0^{d-1}$.} Using the Fourier expansion of $F$, $G$ can be written
$$
G(t) = \frac{1}{T} \sum_{k_1 \dots k_d} a_{k_2}\dots a_{k_d} e^{2\pi i (k_2 \beta_2 + \dots + k_d \beta_d)t} \frac{1 - e^{2\pi i (T-1)(k_2 \beta_2 + \dots + k_d \beta_d )}}{1 - e^{2\pi i (k_2 \beta_2 + \dots + k_d \beta_d )}}
$$
so that
$$
|G(t) - a_0^{d-1}| \leq \frac{1}{T} \sum_{(k_2,\dots,k_d) \neq (0,\dots,0)} |a_{k_2}\dots a_{k_d}| \frac{1 }{\| k_2 \beta_2 + \dots + k_d \beta_d \|}.
$$
To bound the above right-hand side, we split it into two pieces: assume first that one of the $|k_i|$ is $> N^{100d}$, for instance $|k_2| > N^{100d}$ while $|k_3| + |k_4| + \dots + |k_d| < N^{100d}$. The corresponding contribution is bounded by
$$
\dots \lesssim \frac{a_0^{d-1}}{T} \sum_{|k_2| > N^{100 d}} \frac{N^{8d}}{|k_2|^{4d}}\left( \sum_{|k_3| < N^{100 d}} \frac{N^2}{|k_3|} \right)^{d-2} |k_2|^{d+1} \leq \frac{a_0^{d-1}}{T},
$$
where we used in the first inequality the bounds~(\ref{boundak}) as well as the Diophantine condition~\eqref{dioph1}.

We are left with the sum over $|k_2| + \dots + |k_d| \lesssim N^{100d}$, which, using once again the bound~(\ref{boundak}), is less than
$$
\frac{(a_0 N^2)^{d-1}}{T} \sum_{|k_2| + \dots + |k_d| \lesssim N^{100d}} \frac{1}{|k_2|\dots|k_d|} \frac{1}{\| \beta_2 k_2 + \dots + \beta_d k_d \|}.
$$

\bigskip

\noindent \underline{Step 4: Proof of $\sum_{|k_2| + \dots + |k_d| \lesssim N^{100d}} \frac{1}{|k_2|\dots|k_d|} \frac{1}{\| \beta_2 k_2 + \dots + \beta_d k_d \|} \preceq 1$.} For $i = (i_2,\dots,i_d) \in \mathbb{N}^d$, $j \in \mathbb{N}$, let
$$
E_{ij} = \{ (k_2,\dots,k_d) \;:\; |k_{r}| \sim 2^{i_r} \; \mbox{and} \; \| \beta_2 k_2 + \dots + \beta_d k_d \| \sim 2^{-j}.
$$By (\ref{dioph1}), we must have \[j\leq i_2+\cdots+i_d+O(\log \log N),\] thus we can decompose\[j=\sum_{r=2}^d j_r,\quad j_r\leq i_r+O(\log N \log N).\]
Proceeding as in Step 3 of Section 2 and using (\ref{dioph1}) again, we see that for any two elements $k$ and $k'$ of $E_{ij}$ we must have $|k_r-(k')_r| \succeq 2^{-j_r}$ for at least one $r$ (if $2^{i_r} < N^{100d}$). By decomposing the box $\prod_{r=2}^d[-2^{i_r},2^{i_r}]$ into small boxed of size $2^{j_2}\times\cdots 2^{j_d}$, we see that
$$
\# E_{ij} \preceq 2^{i_2 + \dots + i_{d} - j} \preceq 1.
$$
But then
$$
\sum_{|k_2| + \dots + |k_d| \lesssim N^{100d}} \frac{1}{|k_2|\dots|k_d|} \frac{1}{\| \beta_2 k_2 + \dots + \beta_d k_d \|} \leq \sum_{i,j} 2^{-i_2 -\dots - i_d + j} \# E_{ij} \preceq 1.
$$

\bigskip

\noindent \underline{Step 5: conclusion.} Gathering the previous estimates,
$$
|G(t) - a_0^{d-1} | \preceq \frac{N^{2d-2}}{T} a_0^{d-1}.
$$
Choosing $T > N^{2d-2+\epsilon}$, we obtain $|G(t) - a_0^{d-1} | < \frac{1}{10} a_0^{d-1}$. But then
$$
\| e^{it \Delta_\beta} f \|_{L^p([0,T] \times \mathbb{T}^d)}^p T = \int_0^1 F(t)  G(t)\, \mathrm{d} t \sim T a_0^{d-1} \int_0^1 F(t)\, \mathrm{d} t \sim T a_0^{d},
$$
which is the desired result.

\section{Proof of the conjecture for $p>6$}
\label{smt1}

\begin{thm} \label{mainthm1}
For $\beta_2, \dots, \beta_d$ satisfying~\eqref{dioph1}, there holds for $p>6$
$$
\| e^{i t \Delta_\beta} f \|_{L^p([0,T] \times \mathbb{R}^d)} \preceq N^{\frac{d}{2} - \frac{d+2}{p}} \left[ 1 + \left( \frac{T}{N^{2d-2}} \right)^{\frac{1}{p}} \right] \| f \|_{L^2} \qquad\textrm{if }\operatorname{Supp} \widehat{f} \subset B(0,N).
$$
\end{thm}

\subsection*{Proof of Theorem~\ref{mainthm1}} \noindent \underline{Step 0: preliminaries.}
It suffices to prove that
$$
\big\|e^{it\Delta_{\beta}} f\big\|_{L_{t,x}^{6}([0,N^{2d-2}]\times\mathbb{T}^d)}^6 \preceq N^{2d-2}\|f\|_{L^2}^6.
$$
Indeed, if $p=6$, the estimate for $T<N^{2d-2}$ follows immediately, while for $T>N^{2d-2}$ it suffices to add up the above estimate on intervals of size $\sim N^{2d-2}$. Finally, the result follows for $p>6$ by interpolation with the trivial $p=\infty$ case.

Assuming that $f$ is supported in Fourier on $B(0,N)$, expand in Fourier series
\begin{align*}
& f(x) = \sum_{k \in \mathbb{Z}^d} a_k e^{2\pi i k \cdot x} \quad \mbox{where $a_k = 0$ for $|k|>N$}\\
& e^{it\Delta_\beta} f(x) = \sum_{k \in \mathbb{Z}^d} a_k e^{2\pi i ( k \cdot x - tQ(k))}.
\end{align*}
We may normalize $f$ so that \[\sum_{|k|\leq N}|a_k|^2=1.\]Finally, we write
\begin{align*}
& \big\|e^{it\Delta_{\beta}} f\big\|_{L_{t,x}^{6}([0,N^{2d-2}]\times\mathbb{T}^d)}^6 \\
& \qquad = \sum_{\substack{k_i\in\mathbb{Z}^d,|k_i|\leq N;\\k_1+k_3+k_5=k_2+k_4+k_6}}a_{k_1}\overline{a_{k_2}}a_{k_3}\overline{a_{k_4}}a_{k_5}\overline{a_{k_6}}\int_0^{N^{2d-2}}e^{-2\pi i[Q(k_1)-Q(k_2)+Q(k_3)-Q(k_4)+Q(k_5)-Q(k_6)]t}\,\mathrm{d}t,
\end{align*}

\bigskip

\noindent \underline{Step 1: decomposition in $\Lambda_A$.}
For a dyadic number $A\in[N^{2-2d},100 N^{2}]$, define the set\[\begin{aligned}\Lambda_A=\big\{(k_1,\cdots,k_6)&\in(\mathbb{Z}^d)^6:|k_i|\leq N,k_1+k_3+k_5=k_2+k_4+k_6,\\& A \leq \big|Q(k_1)-Q(k_2)+Q(k_3)-Q(k_4)+Q(k_5)-Q(k_6)\big| <2 A\big\}.\end{aligned}\]
When $A$ is the smallest dyadic number larger than $N^{2-2d}$, the lower bound $A \leq$ above is removed. Note that for $(k_1,\cdots,k_6)\in \Lambda_A$ one has\[\bigg|\int_0^{N^2}e^{-2\pi i[Q(k_1)-Q(k_2)+Q(k_3)-Q(k_4)+Q(k_5)-Q(k_6)]t}\,\mathrm{d}t\bigg|\lesssim A^{-1},\]
thus we only need to show that
\begin{equation}\label{mainest4}\sum_{(k_1,\cdots,k_6)\in \Lambda_A}|a_{k_1}a_{k_2}a_{k_3}a_{k_4}a_{k_5}a_{k_6}|\preceq N^{2d-2} A.\end{equation}

\bigskip

\noindent \underline{Step 2: decomposition in $\Sigma_{X_1, \dots, X_d}$.} For $(X_1, \dots, X_d)$ in $\mathbb{Z}^d$, let
\[\begin{aligned}\Sigma_{X_1 ,\dots, X_d}=\big\{(k_1,\cdots,k_6)&\in(\mathbb{Z}^d)^6:|k_i|\leq N,k_1+k_3+k_5=k_2+k_4+k_6,\\&(k_1^i)^2-(k_2^i)^2+(k_3^i)^2-(k_4^i)^2+(k_5^i)^2-(k_6^i)^2=X_i \mbox{ for $1 \leq i \leq d$}.\big\}.\end{aligned}\]
Write then
$$
\sum_{(k_1,\cdots,k_6)\in \Lambda_A}|a_{k_1}a_{k_2}a_{k_3}a_{k_4}a_{k_5}a_{k_6}| = \sum_{\substack{(X_1, \dots , X_d)\in \mathbb{Z}^d, \, |X_i| \lesssim N^2 \\ |X_1 + \beta_2 X_2 + \dots + \beta_d X_d|\sim A }} \sum_{(k_1, \dots ,k_6) \in \Sigma_{X_1, \dots, X_d}} |a_{k_1}a_{k_2}a_{k_3}a_{k_4}a_{k_5}a_{k_6}|.
$$
Therefore, to prove~\eqref{mainest4}, it suffices to show that
\begin{subequations}
\begin{align}
\label{bound1}
& \# \{ (X_1,\dots,X_d) \in \mathbb{Z}^d \; : \; |X_i| \lesssim N^2 \mbox{ and } |X_1 + \beta_2 X_2 + \dots + \beta_d X_d| \sim A \} \preceq N^{2d-2} A \\
\label{bound2}
&\sum_{(k_1,\cdots,k_6)\in \Sigma_{X_1, \dots, X_d}}|a_{k_1}a_{k_2}a_{k_3}a_{k_4}a_{k_5}a_{k_6}|\preceq 1\quad \mbox{for fixed $(X_1,\dots,X_d)$}.
\end{align}
\end{subequations}

\bigskip
\noindent \underline{Step 3: proof of the bound~\eqref{bound1}} We will actually prove that, if $K^{d-1} A \gtrsim 1$,
\begin{equation}
\label{boundX}
\# \{ (X_1,\dots,X_d) \in \mathbb{Z}^d  \; : \; |X_i| < K \mbox{ and } |X_1 + \beta_2 X_2 + \dots + \beta_d X_d|<A\} \preceq K^{d-1} A.
\end{equation}
If $A>1$, this is trivial: one can choose freely $X_1 \dots X_{d-1}$, and then at most $\sim A$ choices for $X_d$ are allowed.

Assume now that $A<1$, and that $X_1,\dots ,X_d$ and $X_1',\dots, X_d'$ satisfy
\begin{align*}
& | X_1 + \beta_2 X_2 + \dots + \beta_d X_d | < A \\
& | X_1' + \beta_2 X_2' + \dots + \beta_d X_d' | < A.
\end{align*}
But then
$$
A \gtrsim | (X_1-X_1') + \beta_2 (X_2-X_2') + \dots + \beta_d (X_d-X_d') | \succeq \frac{1}{(|X_2 - X'_2| + \dots + |X_d - X_d'|)^{d-1}},
$$
where the last inequality follows from the Diophantine condition~\eqref{dioph1}. This means that
$$|(X_2\dots X_d) - (X'_2 \dots X'_d)| \succeq A^{-\frac{1}{d-1}},$$
or in other words that the density of admissible coordinates in $(X_2,\dots, X_d)$ is bounded by $A^{-1}$. Since $A<1$, $X_1$ is completely determined by $(X_2,\dots,X_d)$ and the desired bound~\eqref{boundX} follows.

\bigskip

\noindent \underline{Step 4: proof of the bound~\eqref{bound2}} By the Cauchy-Schwarz inequality,
$$
\sum_{(k_1,\cdots,k_6)\in \Sigma_{X_1 \dots X_d}}|a_{k_1}a_{k_2}a_{k_3}a_{k_4}a_{k_5}a_{k_6}| \lesssim \sum_{(k_1,\cdots,k_6)\in \Sigma_{X_1\dots X_d}}|a_{k_1}a_{k_3}a_{k_5}|^2+\sum_{(k_1,\cdots,k_6)\in \Sigma_{X_1 \dots X_d}}|a_{k_2}a_{k_4}a_{k_6}|^2.
$$
By symmetry it suffices to estimate the first sum, which is bounded by\[\bigg(\sum_{k\in\mathbb{Z}^d,|k|\leq N}|a_k|^2\bigg)^3\cdot\sup_{k_1,k_3,k_5}\#\big\{(k_2,k_4,k_6):(k_1,\cdots,k_6)\in\Sigma_{X_1 ,\cdots ,X_d}\big\},\] so that we only need to show, for fixed $X_1, \dots ,X_d$ and $k_1,k_3,k_5$, that there are $\preceq 1$ choices for $(k_2,k_4,k_6)$ such that $(k_1,\cdots k_6)\in\Sigma_{X_1,\dots,X_d}$.

Now, if $X_1 \dots X_d$ and $k_1,k_3,k_5$ are given, and $(k_1,\cdots k_6)\in\Sigma_{X_1,\dots,X_d}$, then \[k_2^1+k_4^1+k_6^1=M,\quad \mbox{and} \quad (k_2^1)^2+(k_4^1)^2+(k_6^1)^2=S\] are both fixed. Thus\[(3k_2^1-M)^2+(3k_4^1-M)^2+(3k_2^1-M)(3k_4^1-M)=\frac{9S-3M^2}{2}\] is also fixed (and is nonzero, unless $k_2^1=k_4^1=k_6^1=M/3$).

Denoting $a = 3k_2^1-M$ and $b = 3k_4^1-M$, this can be written
$$
\frac{9S - 3M^2}{2} = a^2 + b^2 + ab = (a - \omega b) (a - \overline{\omega} b), \quad \mbox{with $\omega = e^{\frac{2\pi i}{3}}$}.
$$
By the divisor estimate in $\mathbb{Z}[\omega]$, there are $\preceq 1$ choices for $(k_2^1,k_4^1,k_6^1)$. A similar argument works for $(k_2^i,k_4^i,k_6^i)$, with $2 \leq i \leq d$, so that the bound for the number of choices for $(k_2,k_4,k_6)$ is proved, which completes the proof.

\section{A first bound for $p>\frac{2(d+2)}{d}$}
\label{smt2}

\begin{thm}
\label{mainthm2} Assume $\beta_2, \dots, \beta_d$ are chosen generically. Then if $p> \frac{2(d+2)}{d}$, and $\operatorname{Supp} \widehat{f} \subset B(0,N)$,
$$
\| e^{it\Delta_\beta} f \|_{L^p([0,T] \times \mathbb{T}^d)} \preceq N^{\frac{d}{2}-\frac{d+2}{p}} \left( 1 + \left( \frac{T}{N^{\theta_1(p)}} \right)^{1/p} \right) \| f \|_{L^2}
\quad \mbox{with} \quad
\left\{
\begin{array}{l}
\displaystyle \theta_1(p) = 2(d-1) \frac{p-p^*}{p+8-p^*} \\ \displaystyle  p^* = \frac{2(d+2)}{d}.
\end{array}
\right.
$$
\end{thm}

\begin{rem}
Conjecture \ref{longstr} gives the exponent $\theta(p)=\frac{d}{2}(p-p^*)$ for $p^* < p < 6$.
\end{rem}

\subsection{A generic Diophantine property}

\begin{lem}
For generic $\beta_2,\dots,\beta_d$ in $[1,2]$, there exists a constant $C_0$ such that, for all $a=(a_i)\in\mathbb{Z}^d$ and $b=(b_i)\in\mathbb{Z}^d$,
\begin{equation}
\prod_{i=2}^d\left|\beta_i - \frac{a_ib_1}{a_1b_i}\right|\geq C_0 \prod_{i=1}^{d}\left[(1+|a_i|)^{-1}(\log(2+|a_i|))^{-d}\cdot (1+|b_i|)^{-1}(\log(2+|b_i|))^{-d}\right],
\label{dioph3}
\end{equation} given that $a_1\neq 0$ and $b_i\neq 0$ for $2\leq i\leq d$.
\end{lem}

\begin{proof}
Since the left hand side of (\ref{dioph3}) is never zero given that all $\beta_i(2\leq i\leq d)$ are irrational, (\ref{dioph3}) will hold true if there is some dyadic number $M\geq 1$ such that
\begin{equation}(\beta_2,\cdots,\beta_{d})\not\in Q_M,\end{equation} where the set $Q_M$ is defined by
\begin{equation}\begin{split}Q_M:=\bigg\{(\beta_2,\cdots,\beta_d)\in[1,2]^{d-1}:\,\,&\text{(\ref{dioph2}) is false with $C_0$ replaced by $1$, for some}\\
&\text{$(a_i)$ and $(b_i)$ such that $a_1b_i\neq 0$ for all $2\leq i\leq d$,}\\
&\text{and $|a_i|\geq M$ or $|b_i|\geq M$ for at least one $1\leq i\leq d$}\bigg\}
\end{split}\end{equation}
Now we shall prove that $|Q_M|\to 0$ as $M\to\infty$, which, by the Borel-Cantelli theorem, clearly implies (\ref{dioph3}). In fact, by elementary calculus one has that
\begin{equation}\left|\left\{(\beta_2,\cdots,\beta_d)\in[1 ,2]^{d-1}:\prod_{i=2}^{d}|\beta_i-y_i|\leq\varepsilon\right\}\right|\lesssim \varepsilon(\log(1/\varepsilon))^{d-2} \end{equation} for any $y=(y_i)\in\mathbb{R}^{d-1}$ and $\varepsilon\leq 1/2$. This gives
\begin{equation}\label{summation}|Q_M|\lesssim\sum_{A\gtrsim M;\text{ dyadic}}(\log A)^{d-2}\sum_{(a_i),(b_i)}^*\prod_{i=1}^{d}\left[(1+|a_i|)^{-1}(\log(2+|a_i|))^{-d}\cdot (1+|b_i|)^{-1}(\log(2+|b_i|))^{-d}\right],\end{equation} where the second summation is restricted to the set where the maximum of $|a_i|$ and $|b_i|$ is $\sim A$. Evaluating the sum in (\ref{summation}), one gets \[|Q_M|\lesssim\sum_{A\gtrsim M;\text{ dyadic}}(\log A)^{-2}\lesssim(\log M)^{-1},\] as desired.
\end{proof}

\subsection{Bounds on the fundamental solution}

We will denote $K_N$ the fundamental solution of $i\partial_t + Q(D)$ smoothly truncated to frequencies $\lesssim N$. More precisely, set
$$
K_N(t,x) = \sum_{k \in \mathbb{Z}^d} e^{i 2\pi (x\cdot k - tQ(k))} \chi \left( \frac{k_1}{N} \right) \dots \chi \left( \frac{k_d}{N} \right)
$$
(recall that $\chi$ is a smooth, nonnegative function, supported on $B(0,2)$, and equal to $1$ on $B(0,1)$).

\begin{lem}[Dispersive bound] \label{dispersive}
If $|t| \lesssim \frac{1}{N}$,
$$
\forall y \in \mathbb{R}, \qquad \bigg|\sum_{k\in\mathbb{Z}}e^{2\pi i(yk+k^2t)}\chi\bigg(\frac{k}{N}\bigg)\bigg|\lesssim \min \left( N,\frac{1}{\sqrt{t} } \right).
$$
\end{lem}

\begin{lem}[Weyl bound] \label{weyl0}
If $a\in\mathbb{Z} \setminus \{0\}$, $q \in \{ 1,\dots, N \}$, $(a,q)=1$ and $\left|t-\frac{a}{q}\right|\leq \frac{4}{Nq}$:
\begin{equation}\label{weyl} \forall y \in \mathbb{R}, \qquad \bigg|\sum_{k\in\mathbb{Z}}e^{2\pi i(yk+k^2t)}\chi\bigg(\frac{k}{N}\bigg)\bigg|\lesssim\frac{N}{\sqrt{q}\left(1+N\bigg|t-\frac{a}{q}\bigg|^{1/2}\right)}.\end{equation}
\end{lem}
\begin{proof} See for instance Bourgain \cite{B1}, Lemma 3.18.\end{proof}

A consequence of the Weyl bound and of the genericity of the $(\beta_i)$ is the following pointwise bound on $K_N(t,x)$.

\begin{prop}\label{boundk} Assume that $\beta =(\beta_2,\cdots,\beta_d)$ is chosen generically. Then for $\frac{2}{N}< t < N^K$, for a constant $K$, we have
\begin{equation}\label{bound}|K_N(t,x)|\preceq N^{\frac{d+1}{2}}t^{\frac{1}{4}}.\end{equation}
\end{prop}
\begin{proof} \underline{The case d=2.} By Dirichlet's lemma, there exists $a,a' \in \mathbb{Z}$ and $q,q' \in \{ 1 \dots N \}$ such that $\delta = \left| t - \frac{a}{q} \right| \leq \frac{1}{Nq}$ and $\delta' = \left| \beta t - \frac{a'}{q'} \right| \leq \frac{1}{Nq'}$. By the Weyl bound above, $K_N$ can be bounded by
\begin{equation}
\label{pinguin}
\left| K_N(t,x) \right| \lesssim \frac{N^2}{\sqrt{qq'}\left(1 + N \left|t - \frac{a}{q}\right|^{1/2} \right) \left( 1 + N \left|\beta t - \frac{a'}{q'}\right|^{1/2} \right)}.
\end{equation}
The Diophantine condition~(\ref{dioph1}) implies that $\delta + \delta' \succeq \frac{1}{tq^2 (q')^2}$. Inserting this bound in the above gives
$$
\left| K_N(t,x) \right| \lesssim \frac{N^2}{\sqrt{qq'}(1 + N\sqrt{\delta})(1+ N \sqrt{\delta'} )} \preceq \frac{N^2}{\sqrt{qq'} \left( 1 + \frac{N}{\sqrt{t} q q' }\right)}
$$
Observe that the map $y \mapsto \sqrt{y} \left( 1 + \frac{\alpha}{y} \right)$ reaches its minimum, equal to $2 \sqrt{\alpha}$, when $y = \alpha$. This implies that the above right-hand side is maximum for $qq' = \frac{N}{\sqrt{t}}$, leading to the bound
$$
\left| K_N(t,x) \right| \preceq N^{3/2} t^{1/4}.
$$

\noindent \underline{The general case $d \geq 2$.} By Dirichlet's lemma, for each $1 \leq i \leq d$ there are integers $a_i \in \mathbb{Z}$, $q_i \in \{ 1,\dots, N\}$ such that $(a_i,q_i) = 1$ and
$$
\left| \beta_i t - \frac{a_i}{q_i} \right| \leq \frac{1}{Nq_i}.
$$ Since $|t|>2/N$ we have $a_i\neq 0$. Let $|q_i|\sim Q_i$ and $|\beta_it-a_i/q_i|\sim K_i$. Then by Lemma \ref{weyl0},
\[|K_N(t,x)|\lesssim(Q_1\cdots Q_d)^{-1/2}N^d\prod_{i\geq 1:K_i\geq 1/N^2}(N^{-1}K_i^{-1/2}).\]
We may assume without loss of generality that $K_1\sim\min_i K_i$, then\[\left|\beta_i-\frac{a_iq_1}{a_1q_i}\right|\lesssim\frac{K_i}{t}\] for $2\leq i\leq d$. Since $|a_i|\sim tQ_i$, by generic Diophantine condition (\ref{dioph3}) gives \[\prod_{i=1}^{d}Q_i\cdot\prod_{i=1}^d(TQ_i)\cdot\prod_{i=2}^d(t^{-1}K_i)\succeq 1.\] This gives \[\prod_{i=1}^dQ_i\succeq t^{-1/2}\prod_{i=2}^dK_i^{-1/2},\] and thus \[|K_N(t,x)|\preceq t^{1/4}N^d \prod_{i=2}^dK_i^{1/4}\prod_{i\geq 1:K_i\geq 1/N^2}(N^{-1}K_i^{-1/2}).\] This final expression is maximized when each $K_i\sim N^{-2}$, and gives\[|K_N(t,x)|\preceq N^{\frac{d+1}{2}}t^{\frac{1}{4}},\]
which concludes the proof.
\end{proof}

In the case when $|t|$ is extremely small, we use instead the following bound.
\begin{prop}\label{boundsm} For any $A>0$,
\begin{equation}\label{boundsm0}\bigg\|\mathcal{F}_{t,x}\bigg[\chi\bigg(\frac{t}{A}\bigg)K_N(t,x)\bigg]\bigg\|_{L^{\infty}}\lesssim A.\end{equation}
\end{prop}
\begin{proof} Note that \[\mathcal{F}_{t,x}\bigg[\chi\bigg(\frac{t}{A}\bigg)K_N(t,x)\bigg](k,\tau)=A\chi\bigg(\frac{k_1}{N}\bigg)\dots \chi\bigg(\frac{k_d}{N}\bigg)\widehat{\chi}(A(\tau+Q(k))),\] the result follows.
\end{proof}

\subsection{Proof of Theorem~\ref{mainthm2}}
To prove Theorem \ref{mainthm2}, we only need to show that \begin{equation}
\| e^{it\Delta_\beta} f \|_{L^p([0,N^{\theta_1(p)}] \times \mathbb{T}^d)} \preceq N^{\frac{d}{2}-\frac{d+2}{p}}\| f \|_{L^2}, \end{equation} since the theorem follows then by iterating on time intervals of length $N^{\theta_1(p)}$.

\bigskip
\noindent
\underline{Step 1: decomposition of the kernel} Let $\phi$ be a smooth, real, non-negative function supported on $B(0,2)$ such that $\phi > 1$ on $B(0,1)$ and $\widehat{\phi} \geq 0$. For a number $A \in (0,\frac{1}{N})$ to be fixed later, decompose $\phi \left(\frac{t}{T}\right)K_N(t,x) $ into
\begin{align*}
\phi \left(\frac{t}{T}\right) K_N(t,x) =
\underbrace{ \phi \left(\frac{t}{T}\right)\chi \left( \frac{t}{A} \right)K_N(t,x)}_{\displaystyle J_1(t,x)}
& + \underbrace{\phi \left(\frac{t}{T}\right) \chi(Nt) \left[ 1- \chi \left( \frac{t}{A}  \right)\right]K_N(t,x)}_{\displaystyle J_2(t,x)} \\
& + \underbrace{\phi \left(\frac{t}{T}\right) \left[ 1 - \chi(Nt) \right]K_N(t,x)}_{\displaystyle J_3(t,x)}.
\end{align*}
Using lemmas~\ref{dispersive}, \ref{boundk} and~\ref{boundsm}, we obtain
\begin{align*}
& \| \widehat{J_1} \|_{L^\infty} \lesssim A \\
& \| J_2 \|_{L^\infty} \lesssim \frac{1}{A^{\frac{d}{2}}} \\
& \| J_3 \|_{L^\infty} \preceq T^{\frac{1}{4}} N^{\frac{d+1}{2}}.
\end{align*}

\bigskip \noindent
\underline{Step 2: level set estimates.} We essentially follow the argument in Bourgain~\cite{B1}, which is a modification adapted to level set estimates of the Stein-Tomas argument~\cite{Tomas}. Start with $f \in L^2 (\mathbb{T}^d)$ supported in Fourier on $B(0,N)$ and of norm 1: $\|f\|_{L^2(\mathbb{T}^d)} = 1$. Setting $F = e^{it \Delta_\beta} f$, we want to estimate the size of
$$
E_\lambda = \{ (x,t) \in \mathbb{T}^d \times [-T,T] \; : \; |F(x,t)| > \lambda \},
$$
for a time $T \geq 1$ yet to be fixed. Setting $\widetilde{F} = \frac{F}{|F|} \mathbbm{1}_{E_\lambda}$, we can bound, using successively Plancherel's theorem, the Cauchy Schwarz inequality, Plancherel's theorem again, and finally $\|f\|_2 = 1$,
\begin{align*}
\lambda^2 &|E_\lambda|^2  \lesssim \left[ \int_{\mathbb{T}^2 \times \mathbb{R}} \widetilde{F}(x,t) \overline{F(x,t) \phi \left( \frac{t}{T} \right)} \,\mathrm{d}x\,\mathrm{d}t \right]^2 \\
& = \left[ \sum_{k} \int \widehat{ \widetilde{F}}(\tau,k) \overline{T \widehat{f}(k) \widehat{\phi}(T(\tau + Q(k)))} \chi \left( \frac{k_1}{N} \right)^{1/2} \dots \chi \left( \frac{k_d}{N} \right)^{1/2} \,\mathrm{d}\tau \right]^2 \\
& \leq \left[ \sum_{k \in \mathbb{Z}^d} \int_{\mathbb{R}} \left| \widehat{\widetilde{F}} (\tau,k) \right|^2 T \widehat{\phi}(T(\tau + Q(k))) \chi \left( \frac{k_1}{N} \right) \dots \chi \left( \frac{k_d}{N} \right)  \,\mathrm{d}\tau \right] \left[ \sum_k |\widehat{f}(k)|^2 \int T \widehat{\phi}(T(\tau + Q(k))) \,\mathrm{d}\tau \right] \\
& \lesssim \sum_k \int \left| \widehat{\widetilde{F}} (\tau,k) \right|^2 T \widehat{\phi} (T(\tau + Q(k)))\chi \left( \frac{k_1}{N} \right) \dots \chi \left( \frac{k_d}{N} \right) \,\mathrm{d}\tau.
\end{align*}
Applying once more Plancherel's theorem, the above gives
\begin{align*}
\lambda^2 |E_\lambda|^2 & \lesssim \sum_k \int \left| \widehat{\widetilde{F}} (\tau,k) \right|^2 T \widehat{\phi}(T(\tau + Q(k)))\chi \left( \frac{k_1}{N} \right) \dots \chi \left( \frac{k_d}{N} \right) \,\mathrm{d}\tau \\
& = \int \left[ \left( K_N \phi \left(\frac{.}{T}\right) \right) * \widetilde{F} \right] (t,x) \overline{\widetilde{F}(t,x)} \,\mathrm{d}x\,\mathrm{d}t.
\end{align*}
Now using the decomposition of Step 1,
\begin{align*}
\lambda^2 |E_\lambda|^2 & \lesssim \left< (J_1 + J _2 + J_3) * \widetilde{F}\,,\,\widetilde{F} \right> \\
& \lesssim \|\widehat{J_1}\|_{L^\infty} \|\widetilde{F}\|_{L^2}^2 + \left(\|J_2\|_{L^\infty}  + \|J_3\|_{L^\infty} \right) \|\widetilde{F}\|_{L^1}^2 \\
& \preceq A |E_\lambda| + \left( \frac{1}{A^{\frac{d}{2}}} + T^{\frac{1}{4}} N^{\frac{d+1}{2}} \right) |E_\lambda|^2
\end{align*}
Summarizing, we get if $A < \frac{1}{N}$
\begin{equation}
\label{levelsetestimate}
\lambda^2 |E_\lambda|^2 \preceq A |E_\lambda| + \left( \frac{1}{A^{\frac{d}{2}}} + T^{\frac{1}{4}} N^{\frac{d+1}{2}} \right) |E_\lambda|^2
\end{equation}

\bigskip

\noindent
\underline{Step 3: from level set estimates to $L^p$ bounds.}
Recall first that $\|F\|_{L^\infty(\mathbb{R} \times \mathbb{T}^d)} \lesssim N^{\frac{d}{2}}$ by the Sobolev embedding theorem.
Choose next $\delta>0$. When estimating $|E_\lambda|$, two cases have to be distinguished:
\begin{itemize}
\item If $\lambda^2 > T^{\frac{1}{4}} N^{\frac{d+1}{2}+\delta}$, then we choose $A = \frac{N^\delta}{ \lambda^{\frac{4}{d}}}$ (notice that $A < \frac{1}{N}$). The bound~\eqref{levelsetestimate} becomes then
$$
|E_\lambda| \preceq N^\delta \lambda^{-\frac{2(d+2)}{d}}.
$$
\item If $\lambda^2 < T^{\frac{1}{4}} N^{\frac{d+1}{2}+\delta}$, we rely on the Chebyshev inequality and the estimate $\|F\|_{L^{\frac{2(d+2)}{d}}([-T,T] \times \mathbb{T}^d)} \lesssim T^{\frac{d}{2(d+2)}}$ (which follows from the $L^{\frac{2(d+2)}{d}}$ bound of Bourgain-Demeter~\cite{BD}) to obtain
$$
|E_\lambda| \lesssim T \lambda^{-\frac{2(d+2)}{d}}.
$$
\end{itemize}
All in all, this gives for $p>\frac{2(d+2)}{d}$
\begin{align*}
\|F\|^p_{L^p([-T,T] \times \mathbb{T}^d)}
& = p \int_{0}^{N^{\frac{d}{2}}} \lambda^{p-1} |E_\lambda|\,d\lambda \\
& \preceq \int_0^{T^{\frac{1}{8}} N^{\frac{d+1}{4}+\frac{\delta}{2} }} T \lambda^{-\frac{2(d+2)}{d} + p-1}\,d\lambda
+ \int_{T^{\frac{1}{8}} N^{\frac{d+1}{4}+\frac{\delta}{2}}}^{N^{\frac{d}{2}}} N^\delta \lambda^{-\frac{2(d+2)}{d} + p-1}\,d\lambda  \\
& \preceq T \left( T^{\frac{1}{8}} N^{\frac{d+1}{4}+\frac{\delta}{2}} \right)^{-\frac{2(d+2)}{d} + p} + N^{\frac{pd}{2}-(d+2)+\delta}.
\end{align*}
Since the above is true for any $\delta>0$, we get upon choosing $T = N^{\theta_1(p)}$
$$
\|F\|_{L^p([0,N^{\theta_1(p)}] \times \mathbb{T}^d)} \preceq N^{\frac{d}{2}-\frac{d+2}{p}},
$$
from which the desired bound follows immediately.

\section{An improvement if $p>\frac{2(d+2)}{d}$ for $d \geq 4$}
\label{smt3}
\begin{thm}\label{mainthm3}  Assume $\beta_2, \dots, \beta_d$ are chosen generically. Then if $p> \frac{2(d+2)}{d}$, and $\operatorname{Supp} \widehat{f} \subset B(0,N)$,
$$
\| e^{it\Delta_\beta} f \|_{L^p([-T,T] \times \mathbb{T}^d)} \preceq \| f \|_{L^2} N^{\frac{d}{2}-\frac{d+2}{p}} \left( 1 + \left( \frac{T}{N^{\theta_2(p)}} \right)^{1/p} \right)
\quad \mbox{with} \quad
\left\{
\begin{array}{l}
\displaystyle \theta_2(p) = \frac{d^2-2d}{4d-4}(p-p^*) \\ \displaystyle p^* = \frac{2(d+2)}{d}.
\end{array}
\right.
$$
\end{thm}
\begin{rem} Observe that $\theta_2(p)>\theta_1(p)$ if and only if $p>\frac{2d}{d-2}$. Thus Theorem \ref{mainthm2} gives a better bound if $d\geq 4$ and $6>p>2d/(d-2)$.
\end{rem}

The proof will rely on a decomposition into major and minor arcs. Define to that effect, for $Q$ a power of 2,
$$
\Lambda_Q(t) = \sum_{\substack{q \in \mathbb{N}^* \\ Q \leq q < 2Q}} \sum_{\substack{a \in \mathbb{N}^* \\ (a,q) = 1}} \chi \left( NQ\left( t-\frac{a}{q} \right)\right)
$$
and
$$
\rho(t) = 1 - \sum_{\substack{Q \in 2^\mathbb{N} \\ Q < c_0 N}} \Lambda_Q(t),
$$
where $c_0$ is taken sufficiently small, in particular to ensure that the supports of the $\Lambda_Q(t)$ are disjoint for $Q < c_0 N$.

\subsection{A genericity condition on $\beta_2,\dots ,\beta_d$}

Notice that the support of $\Lambda_Q(t)$ has density $\sim \frac{Q}{N}$; therefore, it is natural to expect that for generic $\beta_2, \dots, \beta_d$, the support of $\Lambda_{Q_1}(t) \Lambda_{Q_2}(\beta_2 t) \dots \Lambda_{Q_k}(\beta_d t)$ has density $\sim \frac{Q_1 \dots Q_k}{N^k}$. This is the content of the next lemma.
\begin{lem}
\label{generic}
Generic $\beta_2, \dots, \beta_d \in [1,2]^d$ are such that, for $1 \leq k \leq d$, and $T \geq 1$,
$$
\int_0^T \Lambda_{Q_1}(t) \Lambda_{Q_2}(\beta_2 t) \dots \Lambda_{Q_k}(\beta_k t)\,\mathrm{d}t \preceq \frac{Q_1 \dots Q_k}{N^k} T.
$$
\end{lem}

\begin{proof} Using the Borel-Cantelli lemma, it suffices to prove that
$$
\int_{1}^{2} \dots \int_{1}^{2} \int_0^T \Lambda_{Q_1}(t) \Lambda_{Q_2}(\beta_2 t) \dots \Lambda_{Q_k}(\beta_k t)\,\mathrm{d}t \,\mathrm{d}\beta_2 \dots \mathrm{d}\beta_k \lesssim \frac{Q_1 \dots Q_k}{N^k} T.
$$
Observe first that, due to the definition of $\Lambda_Q$,
$$
\int_{1}^{2} \Lambda_Q(\beta t) \,\mathrm{d}\beta = \frac{1}{t} \int_{t/C_0}^{C_0t} \Lambda_Q(y)\,\mathrm{d}y \lesssim \frac{Q}{N}.
$$
But then by Fubini's theorem
\begin{align*}
& \int_{1}^{2} \dots \int_{1}^{2} \int_0^T \Lambda_{Q_1}(t) \Lambda_{Q_2}(\beta_2 t) \dots \Lambda_{Q_k}(\beta_k t)\,\mathrm{d}t \,\mathrm{d}\beta_2 \dots \mathrm{d}\beta_k \\
& \qquad \qquad = \int_0^T \Lambda_{Q_1}(t) \left[ \int_{1}^{2} \Lambda_{Q_2}(\beta_2 t) \,\mathrm{d}\beta_2 \right] \dots \left[ \int_{1}^{2} \Lambda_{Q_k}(\beta_2 t) \,\mathrm{d}\beta_k \right]\,\mathrm{d}t \\
& \qquad \qquad \lesssim \int_0^t \Lambda_{Q_1}(t) \frac{Q_2 \dots Q_k}{N^{k-1}}\,\mathrm{d}t \lesssim \frac{Q_1 \dots Q_k}{N^k} T.
\end{align*}
\end{proof}

\subsection{Kernel bounds}

Proceeding as in Section~\ref{smt2}, first decompose $K_N$ as follows:
\begin{align*}
\phi\left(\frac t T \right) K_N(t,x) = \underbrace{\phi\left(\frac t T \right) \chi \left(\frac{t}{A} \right) K_N(t,x)}_{\displaystyle J_1(t,x)} & + \underbrace{\phi\left(\frac t T \right) \chi(Nt) \left( 1- \chi \left(\frac{t}{A} \right)\right) K_N(t,x)}_{\displaystyle J_2(t,x)} \\
& + \underbrace{\phi\left(\frac t T \right)(1 - \chi(Nt)) K_N(t,x)}_{\displaystyle J_3(t,x)}.
\end{align*}
Turning to $J_3$,
\begin{align*}
& J_3(t,x) = J_3(t,x) \prod_{j=1}^{d} \left[ \sum_{Q < c_0 N} \Lambda_Q(\beta_j t) + \rho(\beta_j t) \right] \\
& = \sum_{k = 1}^d \sum_{Q_{1}, \dots, Q_{k}}
\underbrace{
\sum_{\substack{\{i_1 ,\dots, i_{d-k}\} \cup \{ j_1 ,\dots, j_k \} = \{1, \dots, d \} \\ \{i_1 ,\dots ,i_{d-k}\} \cap \{ j_1, \dots, j_k \}=\emptyset}}
J_3(t,x) \rho(\beta_{i_1} t) \dots \rho(\beta_{i_{d-k}} t)
\Lambda_{Q_{1}}(\beta_{j_1}(t)) \dots \Lambda_{Q_{k}}(\beta_{j_k}(t))
}_{\displaystyle J_3^{Q_1, \dots, Q_k}(t,x)}
\end{align*}
From Dirichlet's lemma as well as lemmas~\ref{weyl0},~\ref{dispersive} and~\ref{generic}, we conclude that, for generic $\beta_2 \dots \beta_d$, and $A < \frac{1}{N}$,
\begin{align*}
& \| \widehat{J_1} \|_{L^\infty} \lesssim A \\
& \| J_2 \|_{L^\infty} \lesssim \frac{1}{A^{\frac{d}{2}}} \\
& \| J_3^{Q_1 ,\dots, Q_k} \|_{L^\infty} \lesssim \frac{N^{\frac{k}{2} + \frac{d}{2}}}{\sqrt{Q_1 \dots Q_k}} \\
& \| \widehat{J_3^{Q_1 ,\dots, Q_k}} \|_{L^\infty} \lesssim \frac{Q_1 ,\dots, Q_k}{N^k}T
\end{align*}

\subsection{Proof of Theorem~\ref{mainthm3}} By the same argument as in Section~\ref{smt2}, we find
$$
\lambda^2 |E_\lambda|^2 \lesssim A |E_\lambda| + \frac{1}{A^{\frac{d}{2}}} |E_\lambda|^2 + \sum_k \sum_{Q_1, \dots, Q_k} \left< J_3^{Q_1, \dots, Q_k} * \widetilde{F}\,,\, \widetilde{F} \right>.
$$
For $\lambda > C_1 N^{\frac{d}{4}}$ (for a sufficiently big constant $C_1$) we can choose $A = \frac{C_2}{\lambda^{\frac{4}{d}}}$ (for a sufficiently big constant $C_2$), leading to
$$
\lambda^2 |E_\lambda|^2 \lesssim \frac{1}{\lambda^{\frac{4}{d}}} |E_\lambda| + \sum_k \sum_{Q_1 ,\dots, Q_k} \left< J_3^{Q_1, \dots, Q_k} * \widetilde{F}\,,\, \widetilde{F} \right>.
$$
This implies that
$$
\lambda^2 |E_\lambda|^2 \preceq \frac{1}{\lambda^{\frac{4}{d}}} |E_\lambda| + \max_{k,Q_1,\dots ,Q_k} \left< J_3^{Q_1 ,\dots, Q_k} * \widetilde{F}\,,\, \widetilde{F} \right>.
$$
We now fix $k,Q_1, \dots ,Q_k$ for which the $\max$ above is realized, as well as $\delta>0$. Two cases need to be considered
\begin{itemize}
\item If $\frac{N^{\frac{k}{2}+ \frac{d}{2}}}{\sqrt{Q_1 \dots Q_k}}< N^{-\delta} \lambda^2$, we use the $L^\infty$ bound on $J^{Q_1, \dots, Q_k}_3$ and obtain
$$
\lambda^2 |E_\lambda|^2 \preceq \frac{1}{\lambda^{\frac{4}{d}}} |E_\lambda| + N^{-\delta} \lambda^2 |E_\lambda|^2
$$
which implies
$$
|E_\lambda| \preceq \frac{1}{\lambda^{\frac{2(d+2)}{d}}}.
$$
\item Otherwise, $\frac{N^{\frac{k}{2}+ \frac{d}{2}}}{\sqrt{Q_1, \dots, Q_k}} >\lambda^2 N^{-\delta}$ and the $L^\infty$ bound on $\widehat{J^{Q_1 \dots Q_k}_3}$ gives
$$
\lambda^2 |E_\lambda|^2 \preceq \frac{1}{\lambda^{\frac{4}{d}}} |E_\lambda| + \frac{Q_1 \dots Q_k}{N^k} T |E_\lambda| \preceq \frac{1}{\lambda^{\frac{4}{d}}}|E_\lambda| + \frac{N^{d+2\delta} T}{\lambda^4} |E_\lambda|,
$$
which implies
$$
|E_\lambda| \preceq  \frac{1}{\lambda^{ \frac{2(d+2)}{d}}} + \frac{N^{d+2\delta} T}{\lambda^6}.
$$
\end{itemize}
On the one hand, we deduce, since these inequalities hold for any $\delta>0$, that, for any $\lambda> N^{\frac{d}{4}}$,
$$
|E_\lambda| \preceq \frac{1}{\lambda^{ \frac{2(d+2)}{d}}} + \frac{N^{d} T}{\lambda^6} \lesssim
\left\{
\begin{array}{ll}
\frac{1}{\lambda^{\frac{2(d+2)}{d}}} & \mbox{for $\lambda > N^{\frac{d^2}{4d-4}} T^{\frac{d}{4d-4}}$} \\
\frac{N^{d} T}{\lambda^6} & \mbox{for $\lambda < N^{\frac{d^2}{4d-4}} T^{\frac{d}{4d-4}}$}
\end{array}
\right.
$$
On the other hand, we can still resort to the bound of Bourgain-Demeter
$$
|E_\lambda| \lesssim \frac{T}{\lambda^{\frac{2(d+2)}{d}}}.
$$
Combining these bounds, we obtain the desired estimate: for $\frac{2(d+2)}{d} < p < 6$,
\begin{align*}
& \|F\|_{L^p([-T,T] \times \mathbb{T}^d)} = p \int_0^{N^{\frac{d}{2}}} \lambda^{p-1} |E_\lambda|\,d\lambda \\
& \quad \preceq \int_0^{N^{\frac{d^2}{4d-4}}} T \lambda^{-\frac{2(d+2)}{d}+p-1}\,d\lambda + \int_{N^{\frac{d^2}{4d-4}}}^{N^{\frac{d^2}{4d-4}}T^{\frac{d}{4d-4}}} T N^d \lambda^{p-7} \,d\lambda + \int_{N^{\frac{d^2}{4d-4}}T^{\frac{d}{4d-4}}}^{N^\frac{d}{2}} \lambda^{-\frac{2(d+2)}{d}+p-1}\,d\lambda \\
& \quad \preceq N^{\frac{d}{2} (p - \frac{2(d+2)}{d})} + T N^{\frac{d^2}{4d-4}(p-2\frac{d+2}{d})}.
\end{align*}

\bigskip \bigskip

\noindent \textbf{Acknowledgements}  PG was supported by the National Science Foundation grant DMS-1301380.

redpajama_set_name: RedPajamaArXiv


The fashion industry is one of the hardest to break into due to its sheer size and high level of competition – that's no secret. Getting your foot through the door isn't an obvious task, and according to emerging fashion stylist Anita-Hosanna Kangabe, there's no set path to "making" it. However, after working as a freelance fashion stylist for just over a year, this Ottawa-based artist was recently nominated for Faces Magazine's "Favourite Fashion Stylist" Award. She has styled celebrities, including the likes of Mia Martina, and worked alongside well-renowned fashion moguls including Ty Hunter, Beyonce's stylist. We caught up with Anita before she heads off to London Fashion Week to hear five things she's learned along her journey that might help you too.
The first thing I learned was that I would never be completely ready. No amount of overthinking would prepare me for every problem, and no amount of talking would win everyone's support. It takes a leap of faith and that means trusting that success is possible, no matter the setbacks, so envision your success. Set objectives and follow them with daily action. Look for that job, start that blog, volunteer for that magazine and go to that networking event.
Utilize anything and everything at your disposal to start curating a portfolio. This is what I call "doing everything within your power". Photo shoots and websites can be costly, so you may need to be innovative about how you get things done. Try opting for a makeshift studio, for instance.
I did this after landing my first job at a fashion showroom downtown. I didn't have the money for photographers, models, wardrobe or studio space, but I did have beautiful, talented friends, the showroom itself, and access to all the wardrobe in the showroom. I didn't have everything I wanted but I definitely had what I needed.
Doing "everything within your power" doesn't mean doing what you're comfortable with. The things you're comfortable with represent potential that you've already met. The things that scare and excite you, on the other hand, represent unmet potential. The prospect of meeting that potential is exciting, but the prospect of falling short is terrifying.
Everything you do to achieve your success should scare and excite you at least a little. Travelling is a great way of doing this. This takes me back to a fast-paced and disorienting experience in 2017 when I travelled to New York and Connecticut, meeting Claire Sulmers of Fashion Bomb Daily. She helped me see fashion through a different lens and gave me tips on how to pursue my career. I learned so much and I thank God that I went through with it.
This one I learned from Claire. Opportunity doesn't always come knocking at your door, especially at first. If you can't get a seat at the table, create your own table. Sometimes you'll need to look for opportunity in bold directions. This might mean offering your services for free in order to get experience and exposure, for example.
After launching my portfolio website, I applied to dress at Toronto Fashion Week. I saw the opportunity to style for the Toronto International Film Festival as well and thought, "Why not?" I cold-contacted an agent and offered my styling services free of charge. Not only did he answer, but he'd give me the chance to style not one but three of his top clients.
Taking the leap of faith isn't something you do once. It's something you probably have to do a few times every day. Every setback is a chance to overthink or to lose faith; to tackle problems outside of your power or give up on problems within it. And every distraction is a chance to lose sight of your success entirely. The truth is you can't stay focused alone.
Sometimes you will get discouraged and distracted. This is why you need people who have faith in you; people whose courage can encourage you, whose can be focused on you, and whose decisiveness can give you direction when you falter. And this is why you need role models; people whose faith in themselves you can emulate. Mentors are a must.
SHIFTER | home of the creators and influencers that shift culture.
Copyright © 2019 AMOVEO Media. All rights reserved.


A tuck away table presents a good solution for small living spaces. Even those with big houses can prefer to have the small end table at their homes. These are ideal for placement against the sofa. The sofa is often the place where you relax and enjoy watching TV or playing video games. You must have a small table by its side to keep the TV remote and the games controller. For the right tuck away table, look for a suitable online store. The online store of tuc away table must maintain a wide inventory covering different designs. Whether you need an antique design or a modern style table, the store must have all the options. You must be able to browse through the various categories in the site selecting the most suitable solution.
Besides the usual place beside a sofa, it can also be a good bedside solution. You can keep your wallet, mobile phone and small trinkets on the table. It is also a nice place for the alarm clock. While trying to decide on the right table, look at the design aspect. Think about the décor of your home and find a table that complements it. It also reflects your mood and style. An antique table can be an amazing value addition, attributing to the style quotient. Antique tables have the ability to blend with a modern décor effortlessly. However, you can also look for urban, sharp designs for your home. Find the site that can provide you with all the options.
Besides the design, consider the utility aspect of the table. Decide whether you want a drawer with the table. Tuck away tables do not usually have drawers. However, permanent end tables do have them. See whether the site presents you with both the options. Deciding on the color of the furniture is also crucial. Most tables are available in black and deep brown. You can also find light brown and white pieces, but they are less common. Black is a wonderful choice to blend in with every surrounding.
See whether the online store of tuc away table has an effective filter system in place. You may want to filter the furniture based on price. See whether that option is available. You may also want to filter based on size. The store must be able to help you find the perfect table for your home. You must also enquire about the return and refund policy at the store. In case, the tuck away mechanism is not working properly, you may want replacement.


Peaceful lake/mountain district. Restaurants, Irish music & dancing 1km. Horse riding, Fishing & Golf Courses 1km. Dingle & Ring of Kerry 2km. Breakfast Menu. Off R562. Historical sights. Walkers and climbers paradise. All tours arranged.


Find great deals on eBay for pinball machines for sale and pinball machines used.A free inside look at Coin Wrap salary trends. 9 salaries for 4 jobs at Coin Wrap.
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text:
Tex (1982): Tim Hunter's Promising Directing Debut, Starring Matt Dillon
December 12, 2018 by EmanuelLevy
Tim Hunter made a promising directing debut in 1992 with Tex, scripted by Charles S. Haas from the novel of the same title by S. E. Hinton. walks out on them.
A coming-of-age adventure about two brothers, Tex and Mason McCormick (Matt Dillon and Jim Metzler) struggling to make it on their own when their mother dies and their father leaves them in their Oklahoma home.
the studio behind the film, Disney, got credit for dealing with a mature content in a more realistic manner.

redpajama_set_name: RedPajamaCommonCrawl


Skyforge continues its tradition of rich storytelling by building on previous expansions and releasing New Horizons. New Horizons introduces an entirely new planet named Terra. With a new planet comes new features, such as the growth of the planet being entirely up to the players' actions. Immortals will need to develop the land, eliminate threats, and collect data when their ships arrive on Terra.
Skyforge's largest map to date comes with New Horizons, and as players unravel the mysteries of Terra, launch programs to change the world around them, and reinforce their holdings, they will see change and growth in the land. Programs are a new mechanic that will let players manage the development of Terra and also unlock new rewards. Skyforge releases today on PC and hits Playstation 4 on April 11th, April 12th for the Xbox One.
Largest In-Game Map To Date: Terra is a vast new world, full of ruined, overgrown cities, lush jungles and lifeless wastelands, providing new game mechanics, dangerous wildlife to endure and even a new race to discover.
New Draconid Race: The Draconids are strong and dangerous creatures that can interfere with the development of the new planet. According to intel, the only Draconids present on Terra are army units. For the time being, they are collecting resources and conducting reconnaissance. It appears as if the Draconids have no intention of conquering the planet, but why did they show up in force?
New interface: Returning players will find it easier to get back into the action as the interface makes searching for relevant activities easier. A new consolidated overview of missions and tasks allowing players to achieve the maximum rewards available on a selected day.
Changes to scaling and divine specializations: A revamped scaling and divine specialization system minimizing the daunting task for newcomers and returning players to catch up with their friends. To help them reach the current content while still enjoying the journey, scaling and divine specializations have been enhanced. The 'New Horizons' update streamlines the number of divine specializations by bringing them all as part of the Temple of Deeds.
Find out more and start your journey!
Is there success in the Free-to-Play market?


This is a stunning, ceramic ornament of a leopard named 'Jayden'. This is a magnificent ornament that would look amazing in any type of home, especially a modern or contemporary home. It's got a stunning design which is sure to stand out whilst also trying the room together nicely.


text:
Jomo Sono caught offside with a steep R40m tax bill
By NGWAKO MALATJI - 03 March 2019 - 10:06
Jomo Cosmos boss Jomo Sono in hot water with Sars.
Image: Samuel Shivambu/BackpagePix
The taxman has obtained a judgment at the South Gauteng High Court to attach assets belonging to legendary football icon Jomo Sono and his football club Jomo Cosmos.
The SA Receiver of Revenue (Sars) took Sono to court after he allegedly failed to pay a whopping R40m tax bill demanded of him since last year.
The judgment was obtained on February 21, after Sono allegedly failed to honour an agreement he entered into with Sars to pay the millions to get his books in order.
According to court papers seen by Sunday World, Sars had previously obtained a writ of execution to attach Sono and his club's assets after he allegedly failed to pay the staggering tax amount.
They also previously obtained civil judgment against him and his National First Division outfit on October 9, 2017, and blacklisted him and the club with TransUnion credit bureau.
The papers show that Sars reversed the punitive measures, apparently after initially entering into an agreement with "The Black Prince".
"With reference to all the previous correspondences and various meetings held during the course of 2018 at our Megawatt Park offices: This office has reached an agreement to suspend collection, based on the following understanding: all outstanding returns/declarations relating to the above entity would be submitted, followed by a complete and accurate application in terms of Section 200 of the TA Act.
To date Sars has assisted by stopping our collection process by terminating the warrant of execution issued, withdrawing the civil judgment issued on 09/10/2017 and assisting with the removal of the civil judgment on 06/04/2018 taken with TransUnion.
"However, according to current records, the above entity remains non-compliant as the following returns reflect outstanding and the Sars has not received the relevant compromise application," reads the papers.
Siyabonga Ngwekazi's hefty tax bill headache
SARS obtains court judgment against TV star.
The papers further state that Sono allegedly failed to submit all his pay-as-you-earn returns for 2017 and 2018.
He was also accused of failing to submit VAT returns for certain months of 2018 and also allegedly failed to submit returns for the skills development levy he received from Sector Education and Training Authority.
"In light of the above, the Sars will provide you with seven (7) calendar days in order to rectify the current situation, however should same not be done, the Sars will proceed to initiate our collection process without delay, which includes the immediate re-instatement of the civil judgement previously issued," reads the letter sent to Sono on November 19, last year.
It looks like Sono cocked a snook at the letter sent to him, leaving Sars with no option but to obtain the latest judgment against him.
According to the documents, Sono allegedly owes more than R22m in unpaid VAT, almost R13m in unpaid PAYE/ UIF and skills development levy as well as more than R4m in assessed tax and penalties. Sars also warned penalties would be charged on his outstanding amount if he failed to pay up.
Spokesperson Sandile Memela said: "Sars is bound by Chapter 6 of the Tax Administration Act, 2011, dealing with confidentiality of information, therefore it is not in a position to divulge specific details on the affairs of taxpayers."
Sono said he was shocked that Sars obtained a judgement against him and the team because they reached a settlement agreement on how to pay them in December.
He said Sars could have obtained the judgement after his auditors failed to submit certain returns they wanted when they reached the agreement.
"When you are in business, these are the things you go through my brother. Every businessman goes through that, that's why I'm not even ashamed to admit that yes I owe them and I'm paying." he said.
'Lucky escape' for veteran sports presenter Cebo Manyaapelo
DJ Tira on dancer's rape case: Let the law take its course
SA's celeb teenage hubby quits school
Top DJs may earn less as SABC mulls hourly rate cut
ANC presidential campaign debt haunts Lindiwe Sisulu
Durban taxi driver to serve 17 years for kidnapping and robbing passenger
New curator appointed to take control of Mpisanes' assets
Lucky R232m Powerball winner steps forward


text:
Рајнафелден () град је у њемачкој савезној држави Баден-Виртемберг. Једно је од 35 општинских средишта округа Лерах. Према процјени из 2010. у граду је живјело 32.397 становника. Посједује регионалну шифру (-{AGS}-) 8336069.

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Рајнафелден се налази у савезној држави Баден-Виртемберг у округу Лерах. Град се налази на надморској висини од 280 метара. Површина општине износи 62,8 -{km²}-. У самом граду је, према процјени из 2010. године, живјело 32.397 становника. Просјечна густина становништва износи 516 становника/-{km²}-.

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redpajama_set_name: RedPajamaWikipedia


It is a clear and cloudless Sunday morn.
August heat has yet to show its true form.
Only sound heard is the bees as they swarm.
The ubiquitous green begins to wane.
Queen of flowers, the rose resigned her reign.
Floating down each petal falls to its doom.
Though sad, there should be no grievous pain.
It is not death, merely taking a rest.
Needed for their strength to regain.
Soon to return in full beauty and zest.


ServerMenu is a mutator that adds a info menu (press F12) with up to 10 tabs.
I tried this yesterday and it worked offline (SP) but as soon as I tried it online (MP) on my server it crashed instantly when the server was launched. The .log didnt give any clue of what went wrong.
I inactivated, put an ; infront of ServerColor and removed it from the server mutators but that didnt help.
Checked files the its .ini file, which was in and saved it again in Unicode, no didnt help.
Tried everything removed all color codes from the example file, hopeless.
Then I inactived another mutator from ProAsm.com, the ServerLogo4a (not ever had it working), which anyway wasnt active as it was not added in mutator on the server.
Well, that did it, then the ServerMenu worked online, great, I even added a custom logo, it worked.
Today I tried the ServerColor program which I have in folder and cut and paste my current info menus from ServerMenu and added in it the color codes. Then copied the result and pasted it in ServerMenu .ini file and restarted UT2004.
No, it crashes server instantly when trying to launch it.
The problem now is that even when I copy back them it dont work today.
I had to remove it to get server up again.
Have you tried using a very basic colour like white for all the text?
Yes, I removed all color codes still didnt work.
Yesterday, when it worked I cut and paste the example color codes, and they worked, well maybe they dont work today as they still are there?
Ill try to remove all color codes again.
No, it didnt help, restarted 3 times, with and without ServerColor, ServerLogo4a.
PS the gold colors are the default colors without any color codes.
I tried many more times but still crashing server.
I don't really know what to suggest other then trial and error. Completely remove all 3 mods, add ServerMenuMutator with the default .ini file and then add ServerColour, test, add ServerLogo, test, change one setting at a time in ServerMenuMutatorV1c.ini, test.
If you don't get it to work beforehand, i can try on Sunday, but it would be helpful if you could send me your ServerMenuMutatorV1c.ini file.
Thanks, there are alot of mods, I like to try out diffrent stuff, there are some nice ones.
You and others are Welcome to try it out.
It might be down from time to time.
I even removed some lines that I had added ; infront of, if ServerMenu has some other way to comment out stuff, but it didnt help, to rule out stupid stuff It was a long time ago when I did some programming and then the code didnt work as expected. The usual , ; and blank spaces has wasted me alot of time.
I tried the one at a time aproach but it didnt help, its strange since it worked yesterday and I backup those files and using them today, not working.
I logged-out/Login WinXP but it didnt help, maybe a restart of computer is needed.
Well, I had desided to do a offline defrag of system files (PerfectDisk defragger), while restarting computer, it went was all good until 74% where it got stuck and its commandpromt stopped blinking, dead. Powered off and then tried to start windows in "last config that worked", it freeze.
The motherboard got 2 bios; 1 normal and 1 fail-safe, the fail-safe had kicked in and in that the Boot order diffrent. Anyway win8.1 started and did ran in the background a raid rebuild, another 4 hours, which has no interface to see what it is doing, Then I did the chkdsk of winXP and all other disks, all fine.
So now 24 hours later im on the Win81, which got UT2004 installed on the SSD, but it 6 months since I ran win81 and it was broken then (the System Restore didnt work, create Restore Points showed error message, and trying to restore didnt work, thats why Im on the WinXP thats works). I used to get random (graphics driver had stopped working) and then BSODs.
I have to get back on this thread later since I cant run UT2004 on win81 anymore.
I will have win81 running for some time now since im tired of raid rebuilds.
Uhh.. hope you can save all your data.
Hi, all data is safe, non lost.
I tried to google how to fix win81 but it looks like a bigger problem so I restarted and choose winXP but had to save it in bios 3 times before it acutally booted winXP and ofc the raid lost a drive but it started and is now rebuilding the raid, two more hours.
Think I have to do a remove pagefile in advanded to get rid of it and start clean again (but that would require another boot and another 4 hour raid rebuild, maybe tomorrow. If I dont reboot or shut down computer there are no problems at all with the raid for months maybe ½ year.
So its winXP again... later I can try to see if ServerMenu is working or do some troubleshooting.
PS my backup program is in Win81, TrueImage, but since that is broken no more "complete" backups is possible.


Learn to fly with the best flying school in the South West.
If you have always dreamt of flying, either as an aviation enthusiast or an aspiring commercial pilot, then our training courses can help you get your passion airborne and put you within reach of the skies. Our flight school is united by our enthusiasm and passion for flying, allowing us to provide a high quality, but still affordable training programme to all our students.
We are surrounded by some of the most beautiful scenery in the UK, offering a unique, busy and efficient GA friendly airfield open to all manner of fixed wing and rotary aircraft. We provide a fast and convenient departure service allowing you to complete your training on time and on budget.
Whether you are looking to gain a Pilot's Licence, develop your flying skills or simply enjoy one of our flight experiences, we have plenty to offer at our friendly and professional flying school.
Our quality team of dedicated instructors and operations staff can help you step by step over the course of your training in all aspects of flying and theoretical knowledge. Their combined background experience in many different fields of aviation is a true asset and will help you quickly gain the confidence and expertise that is needed in the aviation environment.
We operate a safe and reliable fleet of aircraft, maintained by our professional and CAA certified engineers. Our Five Cessna 152's, Three Cessna 172's, a PA-28, a Citabria Tailwheel, and Scheibe Motor Glider provide, what we believe to be, the perfect balance between pleasurable and professional flying.
We pride ourselves in providing the highest quality of training, so why not give us a call and decide for yourself by arranging your very first flight.
Some of the most stunning views anywhere in the UK.
Dunkeswell Aerodrome is situated in the heart of the Blackdown Hills, approximately 6 miles North of Honiton. Once a small quiet parish village, Dunkeswell is now a busy and vibrant Aero hub for one of the largest General Aviation communities in South West England. Initially established as an American Naval base during the peak of WW2; it now supports a network of aviation businesses and commercial developments.
Our location upon the beautiful Blackdown Hills provides us with, what we believe to be, some of the most stunning views anywhere in the UK. Quaint secluded villages, beautiful rolling countryside and rugged coastal views are a sight not worth missing.
We are situated at an outstanding 839 feet above Sea Level, making us the highest licenced airfield in the United Kingdom.
Fly in and see what we have to offer.
Dunkeswell Aerodrome operates on an Air/Ground radio frequency, which means we cannot provide any form of clearances, permissions or instructions to arriving, departing and maneuvering traffic. We operate 7 days a week on the frequency of 123.475, although there may be times when the frequency is not manned. Your first call on the frequency should be addressed to 'Dunkeswell Radio', where we will provide you with the current airfield information (Runway in use, QNH, QFE etc.) and any other relevant information. Any other calls made thereafter are to be made to 'Dunkeswell Traffic' announcing your position and intentions.
Dunkeswell is an active parachuting airfield. Parachuting normally takes place on Wednesdays, Fridays, Saturdays and Sundays although, throughout the year there are occasions where they will operate 7 days a week (See SkyDiveUK for details). Please assume parachuting is active unless told otherwise. All fixed wing aircraft are to join on downwind or base legs, unless otherwise advised. Fixed wing aircraft may continue whilst parachuting is in progress but should ensure that they do not fly through the deadside/overhead, this including crosswind joins as they bring you extremely close to the drop zone. Helicopter Pilot's must ensure that their rotors are not turning whilst parachuting is in progress within the ATZ. A call for rotor start-up must be made.
Dunkeswell operates daily on a RFFS Category 1 service, although RFFS Category 2 is available upon request to the Operations Staff. RFFS is available from 08.30-18.00 during the summer, and 08.30-17.00 during the winter. On request it may be possible for us to arrange RFFS outside these times, however this must be done by prior arrangement with our operations staff.
Contact should be established by 10nm or 10 minutes out (which ever is sooner) with Dunkeswell Radio on 123.475.
Joining is by downwind or base leg only. Dunkeswell is an active parachuting airfield (up to FL150); therefore overhead/deadside joins are only permitted when agreement is made via the ATSU. Circuits, for all active runways, are made at 800ft on Dunkeswell QFE. Refer to the ICAO Aerodrome Chart and Textual data for further information. Out of respect to our neighbours and to keep Dunkeswell Airfield safe and enjoyable for everybody, please avoid flying overhead Dunkeswell village and obey all correct RT procedures for the airfield.
Helicopters are to join downwind or base. Circuit height 800ft. Under no circumstances should rotors be turning when parachute canopies are in the air. Helicopters must call the ATSU before rotor/engine start up. Refer to the ICAO Aerodrome Chart and Textual data for further information. Out of respect to our neighbours and to keep Dunkeswell Airfield safe and enjoyable for everybody, please avoid flying overhead Dunkeswell village and obey all correct RT procedures for the airfield.
Dunkeswell offers fuel and a variety of oils at competitive prices.
Fuel available includes: Avgas 100LL (£ inc. VAT) and Avtur Jet A1 (£ inc.VAT).
Oil available includes: Aeroshell 80, 100, W80, W100 and 15W50.
Note: For the latest fuel prices contact the operations staff on 01404 891643.
General Declaration Forms must now be completed and submitted online via the AOPA website.
Contact the operations desk if you need a flight plan filed into the system at least one hour (Three hours for IFR) before departure. Flight plans will be activated upon departure from here and closed upon arrival here.
*Prices are updated regularly and are subject to change. If you have any queries please feel free to call the ops staff to confirm the current prices.
The Aviator is the meeting hub for all the pilots and visitors taking part in the various aviation activities at the airfield. Situated at the main entrance of the airfield, adjacent to the flight school, the Aviator is open seven days a week and offers Tea, Coffee, Hot/Cold food from sandwiches to the hearty English breakfast, lunches and evening meals. The bar sells various local brews not to mention the usual favourites.
They can also cater for weddings, functions, meetings, Christmas parties and all manner of other celebrations at a very competitive price. Booking is advisable on 01404 890009. See further details on their website.
Dunkeswell Aerodrome is owned and operated by DSFT, so we aim to keep our prices at a very competitive rate. All flying is done on a pay as you go basis, allowing you to fit your training around your personal life and on a budget that suits you.
PPL hire and dual rates in our aircraft include unlimited, free, landings at the airfield, and are inclusive of all the usual extras from VAT to fuel.
At DSFT we offer two pay as you go rates for instructional and self-hire flights.
The 'Dual Instruction' rate is inclusive of the aircraft hire (wet) and the cost of the instructor required for your training per hour.
The 'Self Hire' rate is inclusive of only the aircraft hire (wet).
Both rates are inclusive of V.A.T at 20%, and assume membership (See below) for the flying club has been paid in advance. No charges are made for the use of credit cards.
All examinations for the PPL/LAPL course can be carried out at Devon & Somerset Flight Training at the following prices.
Membership for Devon & Somerset Flight training can be purchased at any time during the year and will apply for the following time periods.
All prices quoted include V.A.T at 20%; no extra charge is made for the use of credit cards. Prices are subject to change.
The local area surrounding Dunkeswell Airfield has plenty to offer in the form of Hotels, Bed & Breakfast and country pubs.
The Deer Park Hotel in Honiton is situated amid 80 acres of unspoilt Devonshire countryside, just a short stroll away from the River Otter. The Hotel boasts 16 gorgeous bedrooms, in addition to a well established bar & restaurant. The Hotel caters for all manner of events, as well providing the perfect venue for a Devon wedding. Open to non-residents for lunch, afternoon tea and dinner.
The Lakeview Manor is situated within 45 acres of parkland, with panoramic views overlooking their secluded & tranquil lakeside. The Hotel offers an outstanding catering service, with a range of local dishes and Otter Ales. The Lakeview Manor is an ideal place to stay for any weary Pilot.
Situated in the heart of the Blackdown Hills, the Old Kennels offer beautiful 4 star Bed & Breakfast accommodation. They believe in providing an environmentally friendly service of home grown organic foods, supporting the small local Devonshire producers. At costs as little as £36 per night, they offer an unrivaled, comfortable and quaint stay.
Ample free car parking is available at the airfield, situated near to the entrance of the flight school. Further parking is available to those participating in skydiving outside the main centre of SkyDiveUK.
Note: All vehicles and contents are parked on the airfield at the owners own risk.
Dunkeswell is the perfect location for you to base your aircraft, offering fantastic airfield facilities in addition to a safe and well maintained hangarage & parking service.
Most Importantly... A friendly & helpful atmosphere!
Airfield operations staff, including our Rescue & Fire Fighting Service and Refueling services , are only available between 08.30-17.00 in the winter, to 08.30-18.00 in the summer (Or by prior arrangement). However, aircraft based at Dunkeswell are more than welcome to fly before/after hours on lovely summer mornings and evenings. See our Airfield Information for further details.
Should you require any maintenance on your aircraft, we can offer you two well established engineering teams to carry out your work. See our maintenance section for further options and details.
Whether you have just begun your training, or are looking to progress further in your flying, we have plenty of options available at our flight school. We can provide complete training for all the courses we offer, in addition to all the required ground examinations. We can also cater for a wide range of revalidation, renewals and differences training.
For further information, and quotes on hangarage/parking, contact Air Westward Ltd. on 01404 891271.
Our airfield is home to FlyMoore Aircraft Engineering, a well established business providing high quality maintenance solutions. They maintain our fleet of five aircraft, as well as providing an honest and reliable service to non home-base aircraft. The on-site engineering team is able to provide a range of services including airframe repairs, interior refurbishment, re-sprays, avionics maintenance in addition to the regular 50 hour, 150 hour and annual checks.
FlyMoore Aircraft Engineering offers a full aircraft refurbishment service using environmentally friendly dry and wet stripping methods. The on-site engineering team have specially developed facilities and equipment, designed to provide you with the highest quality painting service.
Please feel free to contact them on the following lines to discuss your requirements.
Opening times are subject to weather conditions. We may close earlier than our usual published times under extreme circumstances. Call for further details.


for the fragile cats to dine.
that made the lights twinkle so bright.
she heard a ghostly clatter.
when your dark clouds swirl.
as you have for theirs.
and I will drink your tears.
and I am here for you."
"for you and only you."


Beat eggs. Add milk and sugar. Beat together. Slice peaches. Add to egg mixture and stir together. Pour into pie crust.
Bake in 450° F oven for 10 minutes. Turn oven down to 375° F and continue to bake for another 45-55 minutes. To test doneness, insert knife. If it comes out clean, the pie is done.


text:
Bay Area Houston Magazine
Doug Peterson, left, kicked off his campaign for Houston City Council At Large Position 3 Sept. 8 with former Congressman Nick Lampson on hand to wish him well. After years in NASA External Affairs, Peterson promised to work for all Houston, from Clear Lake up to Kingwood and from the east side to the west side.
Crime spree has deadly end
A 30-year-old League City man, Bryan Winslow Phillips, is dead after a crime spree that ended in a hostage situation in the Walgreen's in Kemah.
League City Police said Phillips took his own life after freeing his last hostage, the Walgreen's pharmacist, dying of a self inflicted gunshot wound at approximately 9:45 a.m. Tuesday, Sept. 8. Throughout the incident, emergency responders were able to view the suspect's actions through live-feed cameras at the drugstore. No citizens or first responders were injured.
Phillips, a 2003 Clear Creek High graduate, led League City Police officers on the chase that ended in Kemah after stealing a car at gunpoint at 8:35 a.m. Tuesday in the 2500 block of South Shore Boulevard. Police said Phillips shot multiple rounds once inside Walgreens.
He has also been identified as the suspect in a robbery that occurred at 2:45 a.m. that Tuesday in League City at a local emergency room, and reports later surfaced that Phillips shot and killed Derek Zagone, after an argument over a gun Monday in Texas City. Both men reportedly pulled guns on each other and Zagone, 32, was shot in the chest and later died at a hospital.
Sister Rosanne Popp MD, original event co-creator and Duncan Finlayson, kickoff the 11th Annual Nun Run on Aug. 12 at Stubbs Harley-Davidson.
11th Nun Run on tap Oct. 17
Dozens are already making plan to participate in the 11th annual Nun Run, a charity motorcycle ride from Houston to Galveston, with nuns riding in the sidecar Saturday, Oct. 17.
The 50-mile run, escorted by the Blue Knights, benefits the Christus Foundation for HealthCare, a nonprofit that provides healthcare to children and families in need who may otherwise go without. It begins at 9 a.m. near the St. Austin Center, 2002 S. Wayside Drive in Houston and ends in San Leon at the Top Water Grill, located at 815 Avenue O.
Last year, 15,000 children received medical care from one of the many Christus Foundation for HealthCare programs and clinics that have the mission to Keep Children Healthy.
For information and sponsorships, go online to christusfoundation.org
Nassau Bay gets new solar lights
Nassau Bay has begun installing the first 50 solar lights that will eventually replace all current decorative street lights in the city.
Most light locations will remain the same, but some will be moved, depending on the results of a lighting analysis that factors in direct sunlight availability and overall neighborhood lighting levels.
Suspect arrested in man's murder
Derrick Dwayne Fields has been charged with murder in the death of a 24-year-old League City man, Antonio Martins, by the Galveston County District Attorney's Office. He was taken into custody Sept. 4 in the 11800 block of Dickinson Road, Houston. Bond is set at $750,000.
Martins vanished Thursday, July 30. His white Chevy pickup was later found at the intersection of Carefree Drive and Candlewood Drive in League City.
His family said about 11:30 p.m. Thursday he told them that he was going to meet a man. No one ever heard from him again. His body, wearing only a tattered shirt, washed onto the Texas City Levee Aug. 9 with a head wound. His death was ruled a homicide. The 6-foot-3 Martins was engaged to be married and had two small daughters.
The League City Police Department attributes the successful apprehension of Fields to the relentless investigation of its detectives, the support by Martins' family, and the collaborative efforts of multiple agencies.
Man held in arson at Memorial SE
Houston Fire Department Arson investigators have arrested David Williams, 33, for arson involving Memorial Hermann Southeast Hospital. Williams arrived at the hospital's emergency room for treatment of an injury. He became agitated and left the emergency room and proceeded to the professional building, which was unoccupied during the early morning hours.
He then broke into the building and set two fires, one on the first floor and one on the second floor, Houston Fire Department officials said. "Memorial Hermann security responded and confronted Williams, who at this point broke out a bottom floor window. Williams ignored commands to exit the building and proceeded to destroy the contents of the building," they added.
When Houston Police arrived, Williams exited the building but was detained without incident. Due to the fire, smoke, water and vandalism Williams caused, estimated damages to the hospital may exceed $1 million.
Posted on Thursday, October 1st, 2015 at 7:00 am | Category News | Tags: Antonio Martins, Derrick Dwayne Fields has been charged with murder in the death of a 24-year-old League City man, Kemah Walgreens shooting, Man held in arson at Memorial SE, Nassau Bay gets new solar lights|	Comment |
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Timothy K. Lewis (born 1954 in Pittsburgh, Pennsylvania), is a former United States circuit judge of the United States Court of Appeals for the Third Circuit and is currently  an attorney at the law firm of Schnader, Harrison, Segal & Lewis LLP, where he serves as the co-chair of the firm's alternative dispute resolution group. He also serves as a mediator, arbitrator, settlement counselor. Lewis is African American.

Education and career

Lewis graduated from Tufts University with a Bachelor of Arts degree in 1976 and from Duquesne University School of Law with a Juris Doctor in 1980. He worked as an assistant district attorney for Allegheny County, Pennsylvania from 1980 to 1983, and as an Assistant United States Attorney for the Western District of Pennsylvania from 1983 to 1991. He is the great-great-great-grandson of Lewis Woodson, believed to be the grandson of Thomas Jefferson and his alleged mixed-raced slave, Sally Hemings, with DNA evidence in 1998 suggesting this was true.

Federal judicial service

Lewis' judicial career began on April 25, 1991, when President George H. W. Bush nominated him to the bench of the United States District Court for the Western District of Pennsylvania, to fill the vacancy left by Paul A. Simmons.  He was confirmed by the Senate on June 14, 1991, and received his commission on June 18, 1991. His service terminated on October 23, 1992, due to elevation to the Third Circuit.

President Bush chose to elevate Lewis again when he nominated him for a newly created seat on the United States Court of Appeals for the Third Circuit on September 17, 1992.  He was confirmed to this seat by the Senate on October 8, 1992. He received his commission on October 9, 1992, making him the final George H. W. Bush appeals-court nominee confirmed by the Senate. At the time of both appointments, he was the youngest federal judge in the United States. On June 30, 1999, Lewis resigned his seat to return to private practice.

Later career

Since resigning from the bench, Lewis has worked at Schnader, Harrison, Segal & Lewis LLP, where he serves as co-chair of the firm's alternative dispute resolution group and often serves as a mediator, arbitrator, and settlement counselor. Lewis also serves as counsel on trial and appellate matters and is a past co-chair of Schnader's appellate practice group.

Lewis has formerly held several positions with the American Arbitration Association, including as a member of the Judicial Settlement Conference Service Panel, a member of the National Task Force on Issues Related to the Arbitration of Consumer Debt Issues, and as a board and executive committee member. At the International Institute for Conflict Prevention and Resolution (CPR), he served on the Task Force that created the Employment-Related Mass Claims Protocol for arbitrators and mediators, is co-chair of the Diversity in ADR Task Force, and is a former member of the board of directors. Lewis is a member of the American Law Institute, an honorary fellow of the American Academy of Appellate Lawyers, a fellow of the College of Commercial Arbitrators, an active member of the Pennsylvania Interbranch Commission for Gender, Racial, and Ethnic Fairness, and serves on the Board of Advisors of the Georgetown Supreme Court Institute. He is a former member of the board of directors of the National Jazz Museum in Harlem.

See also
 List of African-American federal judges
 List of African-American jurists

References

External links

Attorney bio from Schnader, Harrison, Segal & Lewis

1954 births
Living people
20th-century American judges
African-American judges
African-American lawyers
American prosecutors
Assistant United States Attorneys
Constitution Project
Duquesne University alumni
Judges of the United States Court of Appeals for the Third Circuit
Judges of the United States District Court for the Western District of Pennsylvania
Lawyers from Pittsburgh
Pennsylvania lawyers
Tufts University alumni
United States court of appeals judges appointed by George H. W. Bush
United States district court judges appointed by George H. W. Bush


Jonas Jerebko secured a win for the Warriors on Friday.
The 31-year-old forward, who signed a one-year contract with Golden State in the offseason, tipped in a missed shot from Kevin Durant with time expiring to lift his team to a 124-123 victory over the Jazz at Vivint Smart Home Arena.
Warriors coach Steve Kerr discussed Jerebko's game-winning play with reporters after the game.
"That's the moment where he's one of us," Kerr said (via ESPN). "Any time you're a new guy with a team, you're trying to make your mark. You do it in practice by going hard. You do it with your work ethic.
Golden State trailed by 16 points with 5:36 to play in the third quarter, but came back for the victory behind 38 points from Durant and 31 from Stephen Curry.
Jerebko, who averaged 5.8 points and 3.3 rebounds in 74 games for the Jazz last season, said it "felt great" to beat his old team Friday.
The Jazz shot 41.3 percent from 3-point range and scored 81 first-half points in the loss.