text:
\subsubsection*{1. A remark on a counterexample
to inhomogeneous Strichartz estimates for
the Schr\"{o}dinger equation and motivation}
Consider the Cauchy problem
for the inhomogeneous free Schr\"{o}dinger  equation with zero initial data
\begin{align}\label{shreq}
\imath \partial_{t}u+\Delta u\,=\, F(t,x),\qquad (t,x)\in (0,\infty)\times{\mathbb{R}}^{n},\qquad u(0,x)\,=\,0.
\end{align}
Space time estimates of the form
\begin{align}\label{est1}
||u||_{L^{q}_{t}\left(
\mathbb{R};L^{r}_{x}({\mathbb{R}}^{n})\right)}\;\lesssim\;
||F||_{L^{{\widetilde{q}}^{\prime}}_{t}
\left(\mathbb{R};L^{{\widetilde{r}}
^{\prime}}_{x}({\mathbb{R}}^{n})\right)},
\end{align}
have been known as inhomogeneous Strichartz estimates.
The results obtained so far (see \cite{damianoinhom,
Kato,keeltao,vilela,YoungwooKoh})
are not conclusive when it comes to determining  the optimal values of the Lebesue exponents $\,q$, $r$, $\tilde{q}\,$ and $\,\tilde{r}\,$ for which
the estimate (\ref{est1}) holds. Trying to further understand this problem, we \cite{ahmed1} found
new necessary conditions on these exponents values.
The counterexample in \cite{ahmed1}, like
Example 6.10 in \cite{damianoinhom},
contains an oscillatory factor with high frequency. More precisely, we used
a forcing term given by
\begin{align}\label{myphase}
 F(t,x)= e^{-\imath\, N^2\,t } \,\chi_{[0,\frac{\eta}{N}]}(t)\,
 \chi_{B\left(\frac{\eta}{N}\right)}{(x)}
\end{align}
where $\,\eta>0\,$ is a fixed small number,
$\,N>>1\,$ and $ B\left(\frac{\eta}{N}\right)$ is the ball with radius $\,\eta/N\,$ about the origin.
While in \cite{damianoinhom}
the stationary phase method is applied to
the inhomogeneity
\begin{align}\label{fphase}
 F (t,x)= e^{-2\imath\, N^2\,t^{2} }\,\chi_{[0,1]}(t)\,\chi_{B\left(\frac{\eta}{N}\right)}{(x)}.
\end{align}
When $\,t\in [2,3],\,$
both data in (\ref{myphase}) and (\ref{fphase})
force the corresponding solution $u(t,x)$ to concentrate in a spherical shell centered at the origin with radius about $N.$ This agrees with the dispersive nature
of the Schr\"{o}dinger operator.
The shell thickness is different in both cases
though. It is about $1$ in the case of the data
(\ref{myphase}) but about $N$ in the case
of (\ref{fphase}). The necessary conditions obtained
are respectively
\begin{align*}
\frac{1}{q}\geq\frac{n-1}{\widetilde{r}}-\frac{n}{r},
\qquad  \quad\frac{1}{\widetilde{q}}\geq\frac{n-1}{r}-
\frac{n}{\widetilde{r}}
\end{align*}
and
\begin{align}\label{necess1}
|\frac{1}{r}-\frac{1}{\widetilde{r}}|\leq
\frac{1}{n}.
\end{align}
Observe that the oscillatory function in (\ref{myphase})
has a linear phase and is applied for the short time period of length $\:1/\sqrt{\text{frequency}}.\,$
The oscillatory function in (\ref{fphase})
on the other hand has a quadratic phase and
the oscillation is put to work for a whole time unit.
We noticed that the phase in \cite{damianoinhom}
need not be quadratic and we can get
the necessary condition (\ref{necess1}) using the data
\begin{align}\label{fphase1}
 F_{l} (t,x)= e^{-\imath\, N^2\,t}\,
 \chi_{[0,1]}(t)\,\chi_{B\left(\frac{\eta}{N}\right)}{(x)}
\end{align}
where the phase in the oscillatory function is linear.
Before we show this, we recall the following
approximation of oscillatory integrals according to
the principle of stationary phase.
\begin{lem}\label{stationary}
(see \cite{stein}, Proposition 2 Chapter VIII
and Lemma 5.6 in \cite{damianorem})
Consider the oscillatory integral $\;I(\lambda)=\displaystyle \int_{a}^{b}e^{\imath \lambda \phi(s)}\psi(s)d s.\;$ Let the phase $\,\phi \in C^5([a,b])\,$ and the amplitude $\psi\in C^3([a,b])$
such that
\begin{description}
\vspace*{-0.22 cm}
\item[(i)] \hspace{0.1 cm} $\;\phi^{\prime} (z)=0\,$ for a point $\;z\,\in\, ]a+c, b-c[\;$ with $\,c\,$ a positive constant,
\vspace*{-0.22 cm}
\item[(ii)] \hspace{0.0001 cm} $\;|\phi^{\prime} (s)|\,\gtrsim\, 1,\;$
for all $\;s\,\in\, [a, a+c]\,\cup\, [b-c, b],$
\vspace*{-0.22 cm}
\item[(iii)] $\;|\phi^{\prime\prime} (s)|\,\gtrsim\, 1,$
\vspace*{-0.22 cm}
\item[(iv)] $\;\psi^{(j)}\,$ and $\,\phi^{(j+3)}\;$
are uniformly bounded on $[a,b]$ for all $j=0,1,2$.\vspace*{-0.22 cm}
\end{description}
 \begin{align*}
\hspace*{-1 cm}\mbox{Then}\qquad \qquad  I(\lambda)\,=\, {\,\sqrt{\frac{2 \pi}{\lambda|\phi^{\prime\prime} (z)|}} \,\psi(z)\,e^{\imath \,\lambda\, \phi(z)+\imath\,\mbox{\small sgn}\left(
\phi^{\prime\prime} (z)\right)\, \frac{\pi}{4}}}+\mathcal{O}\left(\frac{1}{\lambda}\right),
\vspace*{-0.22 cm}
\end{align*}
where the implicit constant in the $\mathcal{O}-$symbol is absolute.
\end{lem}
The norm of the inhomogeneous term $F_{l} $
in (\ref{fphase1}) has the estimate
\begin{align}\label{normf}
\parallel F_{l} \parallel_{L^{\tilde{q}^{\prime}}
([0,1];L^{\tilde{r}^{\prime}}(\mathbb{R}^{n}))}\,\approx\,
{\eta}^{n-\frac{n}{\tilde{r}}}\,
{N}^{-n}\,{N}^{\frac{n}{\tilde{r}}}.
\end{align}
For the solution of (\ref{shreq}), we have the explicit formula
\begin{align}\label{solshro}
 u(t,x) \,=\, (4 \pi)^{-\frac{n}{2}}\int_{0}^{t}
(t-s)^{-\frac{n}{2}}\int_{{\mathbb{R}}^{n}}
e^{\imath\frac{|x-y|^2}{4(t-s)}}\,F(s,y)\,dy\, ds.
\end{align}
Let us estimate the solution
$u_{l}(t,x)$ that corresponds to $F_{l}.$
We shall restrict our attention to the region
\begin{align*}
\Omega_{\eta,N}=\left\{
(t,x)\in [2,3] \times \mathbb{R}^{n}\!\!:\,
2(t-{3}/{4})N+\eta N^{-1}\,<|x|<\,2(t-{1}/{4})N-\eta N^{-1}\right\}.
\end{align*}
It will be momentarily  seen
that this is the region
where we can exploit Lemma \ref{stationary} to approximate $\, u_{l}(t,x).$
Substituting from (\ref{fphase1}) into (\ref{solshro})
then applying Fubini's theorem we get
\begin{align}\label{corsol}
 u_{l}(t,x)\:=\:
(4\pi)^{-\frac{n}{2}} \,\int_{B({\eta}/{N})}
  \,I_{N}(t,x,y)\, d y
\end{align}
where  $\,I_{N}(t,x,y)\,$ is the oscillatory integral
\begin{align}\label{inoscints}
I_{N}(t,x,y)\;=\;\int_{0}^{1}\,
 e^{\imath N^2\, \phi_{N}{(s,t,x,y)}}\,\psi{(s,t)}\, d s,
\end{align}
with the phase $\;
\displaystyle \phi_{N}{(s,t,x,y)} =\frac{ |x-y|^2}{4\,N^2}\frac{1}{t-s}-s\,$
 and amplitude $\, \psi{(s,t)}=(t-s)^{-\frac{n}{2}}.$
\vspace{0.1 cm} For simplicity, we write $\phi(.)$ and $\psi(.)$ in place of $\,\phi_{N}{(.,t,x,y)}\,$ and $\,\psi{(.,t)}\,$ respectively.
Next, we verify the conditions
(\textbf{i}) - (\textbf{iv}) for $\,\phi\,$ and $\,\psi.$
Let $\,(t,x)\in \Omega_{\eta,N}\,$ and
$\,y\in B(\eta/N).$ Observe then that
$\;\displaystyle t-{3}/{4}<{|x-y|}/{2N}<t-{1}/{4}\;$
and $\; t-s \in [1,3]. $ Therefore
\begin{description}
\item[(i)] If $\,z\,$ is such that $\,\phi^{\prime}(z)=0\,$
then $\,\displaystyle
 z =t-{|x-y|}/{2N}.\,$ Moreover,
$\displaystyle \,z\in\:]{1}/{4},{3}/{4} [.$
\vspace*{-0.15 cm}
\item[(ii)] $\phi^{\prime}$ is monotonically increaing
so $\;\displaystyle \min_{s\in[0,1]}{\phi^{\prime}(s)}
= \phi^{\prime}(0)=\frac{|x-y|^2}{4 N^2\, t^2}
> \left(1-\frac{3}{4t}\right)^{2}
\gtrsim 1.$
\vspace*{-0.25 cm}
\item[(iii)]$\,\displaystyle
\phi^{\prime\prime}(s)\,=\,
\frac{|x-y|^2}{2 N^2}\frac{1 }{(t- s)^3}\,\approx\,1.$
\vspace*{-0.2 cm}
\item[(iv)]  $\;\displaystyle \phi^{(j)}(s)\,=\,
\frac{|x-y|^2}{4N^2}\frac{ j!}{(t- s)^{(j+1)}} \,\approx\,1,\; j=3,4,5$,
$\;\;\;\psi(s)\,=\,
(t- s)^{-\frac{n}{2}}
\,\approx\,1$,\\[0.15 cm]
$\psi^{\prime}(s)\,=\,
\frac{n}{2}(t- s)^{-\frac{n}{2}-1}
\,\approx\,1$, $\qquad \psi^{\prime\prime}(s)\,=\,
\frac{n}{2}(\frac{n}{2}+1)(t- s)^{-\frac{n}{2}-2}\,\approx\,1.$
\end{description}
Now, applying Lemma \ref{stationary} to the oscillatory integral $\,I_{N}(t,x,y)\,$ in (\ref{inoscints}) yields
\begin{align}\label{noosciny0}
I_{N}(t,x,y)\,=\,
{\,\sqrt{\frac{2 \pi}{\phi_N^{\prime\prime}(z,t,x,y)}} \,\psi(z,t)\,
\frac{e^{ \frac{\pi}{4} \imath}}{N}
\,e^{\imath N^2 \phi_N(z,t,x,y)}}+
\mathcal{O}\left(\frac{1}{N^2}\right).
\end{align}
Since $\,\phi_N(z,t,x,y)+t=|x-y|/N\, $ and
since $\, N\left(|x-y|-|x|\right)= \mathcal{O}\left(\eta\right)\,$
whenever \\ $\,(t,x)\in \Omega_{\eta,N},\;y\in B(\eta/N).$
Then
$\,\,N^2\,\phi_N(z,t,x,y)+N^2\,t
=N\,|x|+\mathcal{O}\left(\eta\right).$ Hence
\begin{align}\label{noosciny}
e^{\imath N^2\,\phi_N(z,t,x,y)}=
e^{\imath\left(N\,|x|-N^2\,t\right)}
\,e^{\mathcal{O}\left(\eta\right)}
=e^{\imath\left(N\,|x|-N^2\,t\right)}\,
\left(1+\mathcal{O}\left(\eta\right)\right).
\end{align}
Inserting (\ref{noosciny}) into (\ref{noosciny0})
then returning to (\ref{corsol}), we discover
\begin{align*}
 u_{l}(t,x)\:=\:&
\frac{(4\pi)^{\frac{1-n}{2}} }{\sqrt{2}}\frac{e^{ \frac{\pi}{4} \imath}}{N} \,e^{\imath\left(N\,|x|-N^2\,t\right)}\,
\int_{B({\eta}/{N})}\,{
\,\frac{\psi(z,t)}{\sqrt{\phi_N^{\prime\prime}(z,
t,x,y)}} \,
\,\left(1+\mathcal{O}\left(\eta\right)\right)}
\,d y\\&\;+
\mathcal{O}\left(\frac{1}{N^2}\right)\,
\int_{B({\eta}/{N})}\,\,
d y.
\end{align*}
Recalling that $\,\psi,\:
\phi^{\prime\prime}\approx 1,\,$ we immediately deduce
the estimate
\begin{align*}
&| u_{l}(t,x)|\,\gtrsim\,\frac{|B(\eta/N)|}{N}
\,\approx\,\eta^{n}\,N^{-(1+n)},\quad
(t,x)\in \Omega_{\eta,N}.\quad \text{Thus, for all}\;\;
t\in [2,3],\\
&\hspace*{-1 cm}
||u_{l}(t,x)||_{L^{r}_{x}\left({\mathbb{R}}^{n}\right)}
\,\geq\,
\left( \int_{2(t-{3}/{4})N+\eta N^{-1}\,<\,|x|\,<\,2(t-{1}/{4})N-\eta N^{-1}}\,
| u_{l}(t,x)|^{r}\,dx\,\right)^{\frac{1}{r}}
\,\gtrsim\,\eta^{n}\,N^{-(1+n)+\frac{n}{r}}.
\end{align*}
Consequently
\begin{align}\label{normul}
||u_{l}||_{L^{q}_{t}\left(
\mathbb{R};L^{r}_{x}({\mathbb{R}}^{n})\right)}
\,\geq\,||u_{l}||_{L^{q}_{t}\left(
[2,3];L^{r}_{x}({\mathbb{R}}^{n})\right)}
\,\gtrsim\,\eta^{n}\,N^{-(1+n)+\frac{n}{r}}.
\end{align}
Lastly, it follows from (\ref{normf}) and
(\ref{normul}) that
\begin{align*}
||u_{l}||_{L^{q}_{t}\left(
\mathbb{R};L^{r}_{x}({\mathbb{R}}^{n})\right)}/
\parallel F_{l} \parallel_{L^{\tilde{q}^{\prime}}
([0,1];L^{\tilde{r}^{\prime}}(\mathbb{R}^{n}))}\;
\gtrsim\;
\eta^{\frac{n}{\tilde{r}}}\,
N^{\frac{n}{r}-\frac{n}{\tilde{r}}-1}
\end{align*}
which, for a fixed $\,\eta,\,$ blows up
as $\,N\rightarrow +\infty\,$
if $\,\displaystyle \frac{n}{r}-\frac{n}{\tilde{r}}>1.\,$
In the light of duality this implies
the necessary condition (\ref{necess1}).

These examples made us wonder
how exactly different are linear oscillations from
quadratic ones if we capture the cancellations in
Lebesgue spaces.
One way to see this is to
consider the operators
$\,T^{j,k}_{N}:L^{p}(B)\,
\rightarrow\,L^{q}([0,1])\,$ defined by
\begin{align}\label{intop}
T^{j,k}_{N}f(s):=\int_{B}
\,f(x)\,e^{\imath N{|x|}^{j}s^{k}}\,dx,
\qquad (j,k)\in\{1,2\}^{2},
\end{align}
where $\,B\,$ is the unit ball in ${\mathbb{R}}^{n},\,$
and compare the asymptotic behaviour as $\,N\rightarrow +\infty\,$ of their operator norms for all
$\,p,\,q\in [1,+\infty].\,$
Let $\,C_{j,k,n}:[0,1]^{2}\rightarrow \mathbb{R}\,$
be the functions defined by
\begin{align*}
C_{j,k,n}\left(\frac{1}{p},\frac{1}{q}\right)\,:=\,
\alpha \quad \text{if}\qquad
\parallel T^{j,k}_{N}\parallel_{L^{p}\left(B\right)
\rightarrow L^{q}([0,1])}
\;\approx\; N^{ - \alpha}.
\end{align*}
We discover that $\,C_{j,k,n}\,$ is a continuous function
with range $\,[0,{1}/{4}]\,$
when $n=1,$ $j=2$ and  $\,[0,{1}/{2}{k}]\,$
 otherwise (see the figure below).
We actually prove that
\begin{thm}\label{mainthm}
\begin{equation*} C_{j,k,n}\left(\frac{1}{p},\frac{1}{q}\right)\;=\;
\left\{
  \begin{array}{ll}
\frac{1}{4}\, \sigma\left(\frac{1}{p},\frac{1}{q}\right), & \hbox{$n=1,\;$ $j=2$;} \\\\
\frac{1}{2\,k}\, \sigma\left(\frac{1}{p},\frac{1}{q}\right), & \hbox{
$n\geq j$.}
  \end{array}
\right.
\end{equation*}
where
\begin{equation}\label{sgmab}
 \sigma(a,b):=\left\{
                  \begin{array}{ll}
                   2b , & \hbox{$\; 0\leq a \leq 1-b,\;\;
0\leq b \leq \frac{1}{2}$;} \\
                   2(1-a) , & \hbox{$\; \frac{1}{2}\leq a \leq 1,\;\;a+ b \geq 1$;} \\
                   1 , & \hbox{$\;0\leq a \leq \frac{1}{2},\;\;\frac{1}{2}\leq b \leq 1$.}
                  \end{array}
                \right.
\end{equation}
 \end{thm}
\begin{align*}
 \begin{tikzpicture} [scale=9]
     \draw[->] (0.0, 0) -- (0.6, 0) node[below] {$\frac{1}{p}$};
    \draw[->] (0,0.0) -- (0, 0.6)  node[left] {$\frac{1}{q}$};
    \draw (0.5, 0) node[below] {${1}$};
    \draw (0, 0.5) node[left] {${1}$};
        \draw (0.25, 0) node[below] {$\frac{1}{2}$};
    \draw (0, 0.25) node[left] {$\frac{1}{2}$};
    \draw
    (0, 0) -- (0.5, 0.0) -- (0.5, 0.5) -- (0.0, 0.5) -- cycle;
    \draw
    (0.5, 0.0) -- (0.25, 0.25)-- (0.25, 0.5);
   \draw (0.25, 0.25) -- (0.0, 0.25);
\draw [loosely dotted] (0.25, 0.0) -- (0.25, 0.25);
\draw (0.165,0.125) node{$\;\frac{1}{2}\frac{1}{q}$};
\draw (0.125,0.375) node{$\;\frac{1}{4}$};
\draw (0.375,0.3) node{$\;
\frac{1}{2}(1-\frac{1}{p})$};
\draw (0.3, -0.1) node[below] {$C_{2,k,1}$};
\end{tikzpicture}\qquad\qquad\quad
 \begin{tikzpicture} [scale=9]
     \draw[->] (0.0, 0) -- (0.6, 0) node[below] {$\frac{1}{p}$};
    \draw[->] (0,0.0) -- (0, 0.6)  node[left] {$\frac{1}{q}$};
    \draw (0.5, 0) node[below] {${1}$};
    \draw (0, 0.5) node[left] {${1}$};
        \draw (0.25, 0) node[below] {$\frac{1}{2}$};
    \draw (0, 0.25) node[left] {$\frac{1}{2}$};
    \draw
    (0, 0) -- (0.5, 0.0) -- (0.5, 0.5) -- (0.0, 0.5) -- cycle;
    \draw
    (0.5, 0.0) -- (0.25, 0.25)-- (0.25, 0.5);
   \draw (0.25, 0.25) -- (0.0, 0.25);
\draw [loosely dotted] (0.25, 0.0) -- (0.25, 0.25);
\draw (0.165,0.125) node{$\;\frac{1}{k}\frac{1}{q}$};
\draw (0.125,0.375) node{$\;\frac{1}{2}\frac{1}{k}$};
\draw (0.375,0.3) node{$\;
\frac{1}{k}(1-\frac{1}{p})$};
\draw (0.3, -0.1) node[below] {$C_{j,k,n}$};
\end{tikzpicture}
\end{align*}
\begin{rem}
For each $\,p,q\in [1,\infty],$
and all dimension $\,n>1,\,$
the asymptotic behaviour of
$\,\parallel T^{j,k}_{N}\parallel_{L^{p}\left(B\right)
\rightarrow L^{q}([0,1])}\,$
as $\,n\rightarrow +\infty\,$
is determined only by the linearity or quadraticity of the phase in $s$. The role
of the power $j$ of $x$
appears exclusively in the dimension $n=1.$
\end{rem}
\begin{rem}
There is nothing special about
neither the unit interval nor the unit ball
in defining the operators $T^{j,k}_{N}$.
Actually we shall make use of H\"{o}lder
inclusions of $L^{p}$ spaces
on measurable sets of finite measure
(see Lemma \ref{holder} below).
So we may take any suitable two such
sets provided their finite measures are
asymptotically equivalent to a constant
independent of $ N $ as $ N\rightarrow +\infty.$
\end{rem}
Foschi \cite{damianorem} studied
a discrete version of an operator a little simpler than the integral operator $\,T^{1,1}_{N}.\,$ He considered the operator $\,D_{N}:\ell^{p}(\mathbb{C}^{N})
\rightarrow L^{q}(-\pi,\pi)\,$
that assigns to each vector
$\,a=(a_{0},a_{1},...a_{N-1})\in {\mathbb{C}}^{N}\,$ the $\,2\pi$-periodic trigonometric polynomial $\,D_{N}a(t)=\sum_{m=0}^{N-1}a_{m}\,e^{\imath\,m\,t}\,$
and described the asymptotic behaviour
of $\, \displaystyle\sup_{a\in \mathbb{C}^{N}-\{0\}}
{{\parallel D_{N}a\parallel_{L^{q}([-\pi,\pi])} }/
{\parallel a\parallel_{\ell^{p}\left(\mathbb{C}^{N} \right)}}}$ as $N\rightarrow+\infty,$
 for all $ 1\leq p,\,q\leq+\infty.$
The norms there are defined by
\begin{align*}
&\parallel a \parallel_{\ell^{p}}=
\left( \sum_{m=0}^{N-1}|a_{m}|^{p}\right)^{\frac{1}{p}},
\quad
1\leq p <\infty, \qquad
\parallel a \parallel_{\ell^{\infty}}=
\max_{0\leq m\leq N-1}|a_{m}|,\\
&
 \parallel f \parallel_{L^{q}}=
\left(\frac{1}{2\pi} \int_{-\pi}^{\pi}|f(t)|^{q}dt\right)^{\frac{1}{q}},
\quad
1\leq q <\infty, \qquad
\parallel f \parallel_{L^{\infty}}=
\max_{|t|\leq \pi}|f(t)|.
\end{align*}
This was followed by
a similar investigation (see Section 5 in \cite{damianorem}) of a linear
integral operator with an oscillatory kernel
 $\, L_{N}: L^{p}([0,1])\rightarrow L^{q}([0,1])\,$
defined by
\begin{align*}
L_{N}f(t)\,:=\,\int_{0}^{1}\,
e^{\imath N/(1+t+s)}\,\frac{f(s)}{(1+t+s)^{\gamma}}\,ds,
\quad\text{\small for some fixed}\;\;  \gamma \geq 0.
\end{align*}
\subsubsection*{2.  Proof of Theorem \ref{mainthm}}
In order to show Theorem \ref{mainthm}, we
shall go through the following steps.\\
\textbf{\emph{\underline{Step 1}}}.\hspace{0.1 cm}
Find lower bounds
for $\,\parallel T^{j,k}_{N} \parallel_
{L^{p}(B)\rightarrow L^{q}([0,1])}\,$
for all $\,p,q\in [1,+\infty]\,$:\\
Test the ratio $ \parallel T^{j,k}_{N}f \parallel_{L^{q}([0,1])}/
\parallel f \parallel_{L^{p}(B)} $ for
functions $f \in {L^{p}(B)}$
that kill or at least slow down the oscillations in the integrals $T^{j,k}_{N}f.\,$ Of course this ratio is majorized by $\displaystyle  \parallel T^{j,k}_{N} \parallel_
{L^{p}(B)\rightarrow L^{q}([0,1])}=\sup_{f\in L^{p}(B)-\{0\}}
{{\parallel T^{j,k}_{N}f\parallel_{L^{q}([0,1])} }/{\parallel f\parallel_{L^{p}\left(B\right)}}}. $
But what is really interesting is the fact that such functions likely maximize the ratio as well.\\
\textbf{\emph{\underline{Step 2}}}.\hspace{0.1 cm}
We find upper bounds
for $\,\parallel T^{j,k}_{N} \parallel_
{L^{p}(B)\rightarrow L^{q}([0,1])}\,$
for all $\,p,q\in [1,+\infty].$
Thanks to interpolation and  H\"{o}lder's inequality,
we merely need an upper bound
for $\parallel T^{j,k}_{N} \parallel_
{L^{2}(B)\rightarrow L^{2}([0,1]).}$
\begin{lem}\label{holder}
Let $\,T^{j,k}_{N}:L^{p}(B)\,
\rightarrow\,L^{q}([0,1])\,$ be as in (\ref{intop}). Assume that
\begin{align}\label{en11}
 \parallel T^{j,k}_{N}f \parallel_{L^{2}([0,1])}
\,\leq\, c_{j,k,N}
\parallel f \parallel_{L^{2}(B)}.
\end{align}
Then
\begin{align}\label{consigma}
 \parallel T^{j,k}_{N} \parallel_{L^{p}(B)
\rightarrow L^{q}([0,1])}
\;\lesssim_{p,q,n}\; c^{\sigma\left(\frac{1}{p},\frac{1}{q}
\right)}_{j,k,N}
\end{align}
where $\,\sigma:[0,1]^{2}\rightarrow [0,1]\,$
is the continuous function
in (\ref{sgmab}).
\end{lem}
\begin{proof}
If we take
 absolute values of both sides of (\ref{intop}) we get
the trivial estimate \\
$\;\parallel T^{j,k}_{N}f\parallel_{L^{\infty}([0,1])}
\,\leq\,\parallel f\parallel_{L^{1}\left(B\right)}.$
Interpolating this with (\ref{en11})
using Riesz-Thorin theorem (\cite{loukas})
implies
\begin{align}\label{int1}
 \parallel T^{j,k}_{N}f \parallel_{L^{q}([0,1])}
\,\leq\, c^{2\left(1-\frac{1}{p}\right)}_{j,k,N} \parallel f \parallel_{L^{p}(B)},
\qquad \frac{1}{2}\leq\frac{1}{p}\leq 1,\;\;
\frac{1}{q}=1-\frac{1}{p}.
\end{align}
Since, by H\"{o}lder's inequality,
$\; \parallel T^{j,k}_{N}f \parallel_{L^{\bar{q}}([0,1])}
\,\leq\,\parallel T^{j,k}_{N}f \parallel_{L^{q}([0,1])}\;$ whenever \\$\,1\leq \bar{q}\leq q\leq \infty,$ then
\begin{align}\label{int2}
 \parallel T^{j,k}_{N}f \parallel_{L^{q}([0,1])}
\,\leq\, c^{2\left(1-\frac{1}{p}\right)}_{j,k,N} \parallel f \parallel_{L^{p}(B)},
\qquad \frac{1}{2}\leq\frac{1}{p}\leq 1,\;\;
1-\frac{1}{p}\leq\frac{1}{q}\leq 1.
\end{align}
Applying H\"{o}lder's inequality once more
we find that if $\;1\leq p\leq\bar{p}\leq \infty,\,$ then
\begin{align}
\nonumber&\hspace{-1 cm}\parallel f \parallel_{L^{p}(B)}
\,\leq\,|B|^{\frac{1}{p}-\frac{1}{\bar{p}}}\,
\parallel f \parallel_{L^{\bar{p}}(B)}.
\;\; \text{Therefore by}\;(\ref{int1})\;
\text{we have}\\
\label{int3}
&\hspace{-0.6 cm}\parallel T^{j,k}_{N}f \parallel_{L^{q}([0,1])}
\,\leq\, |B|^{1-\frac{1}{p}-\frac{1}{q}}\,
c^{2/q}_{j,k,N} \parallel f \parallel_{L^{p}(B)},
\quad 0\leq\frac{1}{q}\leq \frac{1}{2},\;\;
0\leq\frac{1}{p}\leq 1-\frac{1}{q}.
\end{align}
Moreover, since we know from (\ref{int2})
that
\begin{align}
\nonumber &\hspace*{-1 cm}\parallel T^{j,k}_{N}f \parallel_{L^{q}([0,1])}
\,\leq\, c_{j,k,N} \parallel f \parallel_{L^{2}(B)},
\quad \frac{1}{2}\leq\frac{1}{q}\leq 1,\quad
\text{then}\\
&\label{int4}
\parallel T^{j,k}_{N}f \parallel_{L^{q}([0,1])}
\,\leq\, |B|^{\frac{1}{2}-\frac{1}{p}}\,
c_{j,k,N} \parallel f \parallel_{L^{p}(B)},
\quad 0\leq\frac{1}{p}\leq \frac{1}{2},\;\;
\frac{1}{2}\leq\frac{1}{q}\leq 1.
\end{align}\vspace{-0.5 cm}
\end{proof}
If the constants in inequalities (\ref{int1}) - (\ref{int4}) were sharp, they would be precisely the values of the corresponding norms
$\,\parallel T^{j,k}_{N}\parallel_{L^{p}\left(B\right)
\rightarrow L^{q}([0,1])}.$
Unfortunately, we are not able to
compute the optimal constant $\,c_{j,k,N}\,$  in the energy estimate (\ref{en11}).
Nevertheless, the constants $\,c^{\sigma\left(\frac{1}{p},\frac{1}{q}
\right)}_{j,k,N}\,$ in (\ref{consigma})
 would be good enough
for our purpose if,
for each $p,q\in [1,+\infty],$ they were asymptotically equivalent, as $ N\rightarrow +\infty$, to the corresponding lower bounds
of $\,\parallel T^{j,k}_{N}\parallel_{L^{p}\left(B\right)
\rightarrow L^{q}([0,1])}\,$ that we compute in \emph{Step 1}.\\
\textbf{\emph{\underline{Step 1}}. }\\
(i) \textbf{Focusing data}\\
When $\,x\in B(\eta /N^{\frac{1}{j}})\,$ we have
$\;\displaystyle e^{\imath N{|x|}^{j}s^{k}}=
e^{\mathcal{O}\left(\eta\right)}=
1+\mathcal{O}\left(\eta\right), \;\;
\text{\small for all}\;s\in [0,1].$
Thus, if we
take $f_{j}$ to be the focusing functions
$\,f_{j}=\displaystyle \chi_{B(\eta /N^{\frac{1}{j}})}\,$ then $\; \displaystyle \parallel f \parallel_{L^{p}(B)}\,=\,
|B(\eta /N^{\frac{1}{j}})|^{\frac{1}{p}}\;$
and
\begin{align*}
T^{j,k}_{N}f_{j}(s)\,=\,\int_{B(\eta /N^{\frac{1}{j}})}
\,e^{\imath N{|x|}^{j}s^{k}}\,dx=
\int_{B(\eta /N^{\frac{1}{j}})}
\,\left(1+\mathcal{O}\left(\eta\right)\right)\,dx
\,\gtrsim \,|B(\eta /N^{\frac{1}{j}})|
\end{align*}
for all $ \;0\leq s\leq 1.\;$
Consequently, since $\eta$ is fixed,
\begin{align}\label{lb1}
\parallel T^{j,k}_{N}\parallel_{L^{p}\left(B\right)
\rightarrow L^{q}([0,1])}
\,\geq\,\frac{\parallel T^{j,k}_{N}f_{j} \parallel_{L^{q}([0,1])}}{
\parallel f_{j} \parallel_{L^{p}(B)}}
\;\gtrsim\;
N^{-\frac{n}{j}\left(1-\frac{1}{p}\right)}.
\end{align}
The figure below illustrates the one dimensional case.
\begin{align*}
&\hspace{-1 cm} \begin{tikzpicture} [scale=10]
\fill[fill = black!10] (0,0.35)--(0.048,0.35)--(0.048,0.0)--
(0,0)--cycle;
     \draw[->] (0, 0) -- (0.38, 0) node[below] {\small$x$};
    \draw[->] (0,0.0) -- (0, 0.4)  node[left] {\small$ f_{j}(x)$};
        \draw (0.35, 0) node[below] {\small ${1}$};
        \draw (0,0.35) node[left] {\small ${1}$};
\draw[thick] (0,0.35)--(0.05,0.35);
\draw[thick] (0.05,0.0001)--(0.35,0.0001);
\draw (0.07,0.0)node[below] { \footnotesize {$N^{-\frac{1}{j}}$}};
 \draw[dotted] (0.35, 0) -- (0.35,0.35)--(0,0.35);
\end{tikzpicture}\quad
  \begin{tikzpicture} [scale=3.5]
\draw[->] (0.0, 0) -- (1.1, 0) node[below] {\small $s$};
\draw[->] (0.0,0.0) -- (0.0,1.1);
 \draw (0.9,1.2) node[left] { \footnotesize  {$N^{1/j}$\text{Re}$\left\{T^{j,k}_{N}f_{j}(s)\right\}$}};
\draw (1,0) node[below] {\small${1}$};
\draw (1,-0.001) --(1,0.003);
\draw (0.0,0.95) node[left] {\small${1}$};
\draw [blue,samples=500,domain=0.0:1] plot
(\x, {cos(0.65*\x r)});
\draw [red,samples=500,domain=0.0:1] plot
(\x, {cos(0.65*\x*\x r)});
 \draw(0.7,1) node[right,thick] {\tiny {$k=2$}};
 \draw(0.6,0.8) node[right,thick] {\tiny {$k=1$}};
 \draw(0.0,-0.15) node {};
\end{tikzpicture}\qquad
  \begin{tikzpicture} [scale=3.5]
\draw[->] (0.0, 0) -- (1.1, 0) node[below] {\small $s$};
\draw[->] (0.0,0.0) -- (0.0,1.1);
 \draw (0.9,1.2) node[left] { \footnotesize  {$N^{1/j}$\text{Im}$\left\{T^{j,k}_{N}f_{j}(s)\right\}$}};
\draw (1,0) node[below] {\small${1}$};
\draw (1,-0.001) --(1,0.003);
\draw (-0.001,1) --(0.003,1);
\draw (0.0,0.95) node[left] {\small${1}$};
\draw [blue,samples=500,domain=0.0:1] plot
(\x, {sin(0.65*\x r)});
\draw [red,samples=500,domain=0.0:1] plot
(\x, {sin(0.65*\x*\x r)});
 \draw(0.7,0.3) node[right,thick] {\tiny {$k=2$}};
 \draw(0.6,0.6) node[right,thick] {\tiny {$k=1$}};
  \draw(0.0,-0.15) node {};
\end{tikzpicture}\\
&\text{\small \emph{Both real and imaginary parts of the functions}}
\;T^{1,k}_{N}f_{1}\; \text{\small\emph{and}}\;
T^{2,k}_{N}f_{2}\; \text{\small \emph{have the same profile}}.
\end{align*}
(ii) \textbf{Constant data}\\
Let $\,g(x)=1.\,$
Whenever $\,\displaystyle s \in [0,\eta/N^{\frac{1}{k}}]\,$ we have
$\;\imath N{|x|}^{j}s^{k}\,=\,\mathcal{O}\left(\eta\right)\;$
for all $\;x\in B\;$ and it follows that
$\;\displaystyle e^{\imath N{|x|}^{j}s^{k}}=
1+\mathcal{O}\left(\eta\right).\;$
Hence, when $\,\displaystyle s \in [0,\eta/N^{\frac{1}{k}}],\,$
\begin{align*}
T^{j,k}_{N}g(s)\,=\,
\int_{B}\,e^{\imath N{|x|}^{j}s^{k}}\,dx\,=\,
\int_{B}\,\left(1+\mathcal{O}\left(\eta\right)\right)\,dx
\,\gtrsim 1.
\end{align*}
Therefore, recalling that $\eta$ is fixed,
\begin{equation}\label{lb20}
\int_{0}^{1}|T^{j,k}_{N}g(s)|^{q}\,ds
\,\geq\,
\int_{0}^{\eta/N^{\frac{1}{k}}}|T^{j,k}_{N}g(s)|^{q}\,ds
\,\gtrsim\,
\int_{0}^{\eta/N^{\frac{1}{k}}}\,ds
\,\approx\,N^{-\frac{1}{k}}.
\end{equation}
In view of (\ref{lb20}), we deduce that
\begin{align}\label{lb2}
\parallel T^{j,k}_{N}\parallel_{L^{p}\left(B\right)
\rightarrow L^{q}([0,1])}
\,\geq\,\frac{\parallel T^{j,k}_{N} g \parallel_{L^{q}([0,1])}}{
\parallel g \parallel_{L^{p}(B)}}
\;\gtrsim\;
N^{-\frac{1}{k}\frac{1}{q}}.
\end{align}
By rescaling, it is easy to verify that the estimate (\ref{lb2}) follows for any complex-valued constant function $g$. The figure below shows the behaviour of $ T_{N}^{j,k}g $ on $[0,1]$ in the dimension $n=1.$
\begin{align*}
\hspace*{-0.5 cm}
\begin{tikzpicture} [scale=10]
\draw[->] (0.0, 0.0) -- (0.48,0.0) node[below right] {$s$};
\draw[->] (0.0,0.0) -- (0.0,0.52);
\draw(0.25,0.44)  node[above] {\scriptsize $
    \text{Re}\left\{T^{1,k}_{N}g(s)\right\}=2$};
    \draw(0.45,0.44)  node[above] {\large $ \frac{\sin{(Ns^{k})}}{N\,s^{k}}$};
\draw (0.0,0.49) node[left] {\scriptsize ${2}$};
\draw (0.1,0.2) node[left] {\scriptsize $k=1$};
\draw (0.175,0.3) node[left] {\scriptsize $k=2$};
\draw [very thin,blue,samples=500,domain=0.001:0.45] plot
(\x, {2*sin((250*\x) r)/((1000*\x) )});
\draw [very thin,red,samples=500,domain=0.0012:0.45] plot
(\x, {2*sin((250*\x*\x) r)/((1000*\x*\x) )});
\end{tikzpicture}\qquad\qquad\qquad
\begin{tikzpicture} [scale=10]
\draw(0.23,0.44)  node[above] {\scriptsize $
    \text{Im}\left\{T^{1,k}_{N}g(s)\right\}=4$};
    \draw(0.46,0.44)  node[above] {\large $ \frac{\sin^{2}{(Ns^{k}/2)}}{N\,s^{k}}$};
\draw (0.0,0.49) node[left] {\scriptsize ${2}$};
\draw (0.0,0.49) --(0.001,0.49);
\draw (0.09,0.38) node[left] {\scriptsize $k=1$};
\draw (0.2,0.38) node[left] {\scriptsize $k=2$};
\draw [very thin,red,samples=500,domain=0.001:0.45] plot
(\x, {4*sin((100*\x*\x) r)*sin((100*\x*\x) r)/((800*\x*\x) )});
\draw [very thin,blue,samples=500,domain=0.0001:0.45] plot
(\x, {4*sin((50*\x) r)*sin((50*\x) r)/((400*\x) )});
\draw (-0.084, -0.084)  node[below] {};
\draw[->] (0.0, 0.0) -- (0.48,0.0) node[below right] {$s$};
\draw[->] (0.0,0.0) -- (0.0,0.52);
\end{tikzpicture}
\end{align*}
\begin{align*}
&\hspace*{-1.58 cm}\begin{tabular}{c c}
\hspace{-4.75 cm}\vspace{0.25 cm}   \scriptsize $2$& \\
\hspace{-1.7 cm}   \scriptsize $ k=2$&\\
\hspace{-3.3 cm}   \scriptsize $ k=1$&\\
&\hspace{-1 cm} \scriptsize $ k=1\quad k=2$
\vspace{-2.5 cm}\\
\hspace{0.5 cm}\scriptsize{ \text{Re}$\left\{T^{2,k}_{N}g(s)\right\}$}
&\hspace{2 cm}\scriptsize
{\text{Im}$\left\{T^{2,k}_{N}g(s)
    \right\}$} \vspace{-1.5 cm}\\
\includegraphics[scale=0.32]{t1001.pdf}
&\qquad\quad\;
\includegraphics[scale=0.32]{t2001.pdf}\vspace{-1 cm}\\
  \hspace{5.5 cm} \small$ s$&   \hspace{7 cm}
 \small $ s$
\end{tabular}\\
&\hspace{-2.3 cm} \text{\small\emph{Functions}}\; \text{\small \emph{Re}}\small \{ T^{1,k}_{N}g(s)\}\;
\text{\small\emph{vanish and}}\;\text{\small \emph{Re}}\small \{T^{2,k}_{N}g(s)\}\;
\text{\small \emph{change monotonicity, for the first time, when }}\;s=\sqrt[k]{\pi/N}
\end{align*}
(iii) \textbf{Oscillatory data}\\
Consider the oscillatory function
$\,h(x)=e^{2\imath N \left(|x|^2-|x|\right)}.\,$
Using polar coordinates we can write
\begin{align*}
T^{j,k}_{N}h(s)\,=\,
\int_{S^{n-1}}
\int_{0}^{1}\,e^{\imath N
\,\left({\rho}^{j}s^{k}+2\rho^2-2\rho\right)}\,
\rho^{n-1}\,d\rho \,d\omega\,=\,
\omega_{n-1}
\:I^{j,k}_{N}(s)
\end{align*}
where $I^{j,k}_{N}(s) $ is the oscillatory integral
given by
\begin{align}\label{inoscint}
I^{j,k}_{N}(s) =
\int_{0}^{1}\,e^{\imath N
\,\phi_{j,k}(\rho;s)}\,
\rho^{n-1}\,d\rho
\end{align}
with the phase $\displaystyle \phi_{j,k}(\rho;s) ={\rho}^{j}s^{k}+2\rho^2-2\rho. $\\
The quadratic function $\displaystyle \rho\rightarrow\phi_{j,k}(\rho;s)$, after a suitable translation along the vertical axis,
has a single nondegenerate stationary point
that happens to lie well inside $]\frac{1}{5},\frac{4}{5}[.$ Indeed, one can simply
write
\begin{align*}
\phi_{j,k}(\rho;s)=\left\{
                     \begin{array}{ll}
2\left(\rho-\frac{2-s^{k}}{4}\right)^{2}-
\frac{\left(2-s^{k}\right)^{2}}{8}, & \hbox{$j=1$;} \\
\left(2+s^{k}\right)\left(\rho-\frac{1}{2+s^{k}}\right)^{2}-
\frac{1}{\left(2+s^{k}\right)^{2}}, & \hbox{$j=2$.}
                     \end{array}
                   \right.
\end{align*}
Notice also that $\, \left(2-s^{k}\right)/4\in[\frac{1}{4},\frac{1}{2}]\,$
and $\,\left(2+s^{k}\right)^{-1}\in[\frac{1}{3},
\frac{1}{2}]\,$
when $\,s\in [0,1].$
In fact, this is what we were after when we
used the oscillatory function $h$ with its
particular quadratic phase. Let us see how we benefit from this. We shall work on the integral $\,I^{1,k}_{N}(s)\,$ and the applicability
 of the same procedure to the integral
$\,I^{2,k}_{N}(s)\,$ will be obvious.
For simplicity, let
$z$ denote $\,\left(2-s^{k}\right)/4.\,$
Then
 \begin{align}
\nonumber
e^{2\imath N\,z^2}\,I^{1,k}_{N}(s) =&
\,\int_{0}^{1}\,e^{2\imath N
\,\left(\rho-z\right)^{2}}\,
\rho^{n-1}\,d\rho\\
\label{feq}=&\,z^{n-1}\,\int_{0}^{1}\,e^{2\imath N
\,\left(\rho-z\right)^{2}}\,d\rho+
\int_{0}^{1}\,e^{2\imath N
\,\left(\rho-z\right)^{2}}\,
\left(\rho^{n-1}-z^{n-1}\right)\,d\rho.
\end{align}
We compute
\begin{align}\label{t1}
\hspace{-1 cm}
\int_{0}^{1}\,e^{2\imath N
\,\left(\rho-z\right)^{2}}\,d\rho=
\int_{-\infty}^{+\infty}\,e^{2\imath N
\,\left(\rho-z\right)^{2}}\,d\rho-
\int_{-\infty}^{0}\,e^{2\imath N
\,\left(\rho-z\right)^{2}}\,d\rho-
\int_{1}^{+\infty}\,e^{2\imath N
\,\left(\rho-z\right)^{2}}\,d\rho.
\end{align}
Using the identity (See Exercise 2.26 in \cite{taobook})
\begin{align*}
\int_{-\infty}^{+\infty}
\,e^{-ax^2}\,e^{bx}\,dx=\sqrt{\frac{\pi}{a}}
\,e^{b^2/4a},\quad a,b \in \mathbb{C},\;
 \textrm{Re}(a) >0 \qquad \text{we get}
\end{align*}
\begin{align}\label{11}
\hspace{-0.5 cm}\int_{-\infty}^{+\infty}\,e^{2\imath N
\,\left(\rho-z\right)^{2}}\,d\rho
=\sqrt{\frac{\pi}{2N}}\,e^{\frac{\pi}{4}\imath}.
\end{align}
And since
\begin{align*}
\hspace*{-1 cm}
\left|\;\int_{-\infty}^{0}\,e^{2\imath N
\,\left(\rho-z\right)^{2}}\,\partial_{\rho}
\left(\rho-z\right)^{-1}\,d\rho \right|\,\leq\,
 \frac{1}{z},\quad \left|\;
\int_{1}^{+\infty}\,e^{2\imath N
\,\left(\rho-z\right)^{2}}\,\partial_{\rho}
\left(\rho-z\right)^{-1}\,d\rho \right|\,\leq\,
\frac{1}{1-z},
\end{align*}
then integration by parts implies
\begin{align}
\label{22}&\int_{-\infty}^{0}\,e^{2\imath N
\,\left(\rho-z\right)^{2}}\,d\rho=
\frac{\imath\, e^{2\imath N z^{2}}}{4 N z}
+\mathcal{O}\left(\frac{1}{Nz}\right),\\
\label{33}&\int_{1}^{+\infty}\,e^{2\imath N
\,\left(\rho-z\right)^{2}}\,d\rho=
\frac{\imath\, e^{2\imath N\left(1-z\right)^{2}}}{4 N\left(1-z\right)}
+\mathcal{O}\left(\frac{1}{N\left(1-z\right)}\right).
\end{align}
Recalling that $\,\frac{1}{4}\leq z\leq \frac{1}{2}\;$
and using (\ref{11}), (\ref{22}), (\ref{33})
in (\ref{t1}) we obtain
\begin{align}\label{44}
\int_{0}^{1}\,e^{2\imath N
\,\left(\rho-z\right)^{2}}\,d\rho\,=\,
\sqrt{\frac{\pi}{2N}}\,e^{\frac{\pi}{4}\imath}
+\mathcal{O}\left(\frac{1}{N}\right).
\end{align}
This gives us an estimate for the first integral
on the right hand side of (\ref{feq}).
The second integral is $\;\mathcal{O}\left({1}/{N}\right).\;$ This
follows from integration by parts and the smoothness
of the polynomial $\;P(\rho;z):={\left(\rho^{n-1}-z^{n-1}\right)}/{\left(\rho-z\right)}=
\sum_{\ell=0}^{n-2}\,\rho^{n-2-\ell}\,z^{\ell}\;$
as we can write
\begin{align*}
\int_{0}^{1}\,e^{2\imath N
\,\left(\rho-z\right)^{2}}\,
\left(\rho^{n-1}-z^{n-1}\right)\,d\rho\,=\,
\frac{1}{4\imath N}
\int_{0}^{1}\,
P(\rho;z)\,
\partial_{\rho}\,e^{2\imath N
\,\left(\rho-z\right)^{2}}\,d\rho.
\end{align*}
Plugging (\ref{44}) together with the latter estimate into (\ref{feq}) we get that
\begin{align}\label{sm}
e^{2\imath N\,z^2}\,I^{1,k}_{N}(s)
\,=\,z^{n-1}\,
\sqrt{\frac{\pi}{2N}}\,e^{\frac{\pi}{4}\imath}
+\mathcal{O}\left(\frac{1}{N}\right).
\end{align}
From (\ref{sm}) follows the estimate
\begin{align*}
\left|I^{1,k}_{N}(s)\right|
\,\gtrsim\,N^{-1/2}.
\end{align*}
An explanation for the estimate above comes from the fact that the function
$\,\lambda_{N}(\rho;z)=
\cos{\left(2N\,\left(\rho-z\right)^{2}\right)}\,$
remains positive for $\;|\rho-z|<\sqrt{\left(\pi/4N\right)}\;$
and the further we move from the stationary point
$\rho=z$ it, unlike the slowly varying factor $\rho^{n-1}$, oscillates rapidly for large $N$
so that, when summing over $\rho$, integrals over neighbouring halfwaves where
$\lambda_{N}$  changes sign almost cancel.
See the figure below.
An identical estimate for
$\,I^{2,k}_{N}(s)\,$
follows applying the same argument above.
The approach adopted here is
standard. It represents the key idea
of the proof of the stationary phase method
illustrated by Lemma \ref{stationary}.
\begin{align*}
\begin{tikzpicture}[yscale=1.5]
\fill[fill = black!50] (3*pi/8,0) -- plot [domain=3*pi/8:11*pi/13] (\x,{cos(64*\x*\x )}) -- (11*pi/13,0) -- cycle;
\fill[fill = black!50] (-3*pi/8,0) -- plot [domain=-3*pi/8:-11*pi/13] (\x,{cos(64*\x*\x )}) -- (-11*pi/13,0) -- cycle;
\fill[fill = black!25] (11*pi/13,0) -- plot [domain=11*pi/13:235*pi/208] (\x,{cos(64*\x*\x )}) -- (235*pi/208,0) -- cycle;
\fill[fill = black!25] (-11*pi/13,0) -- plot [domain=-11*pi/13:-235*pi/208] (\x,{cos(64*\x*\x )}) -- (-235*pi/208,0) -- cycle;
\fill[fill = black!5] (235*pi/208,0) -- plot [domain=235*pi/208:141*pi/104] (\x,{cos(64*\x*\x )}) -- (141*pi/104,0) -- cycle;
\fill[fill = black!5] (-235*pi/208,0) -- plot [domain=-235*pi/208:-141*pi/104] (\x,{cos(64*\x*\x )}) -- (-141*pi/104,0) -- cycle;
\draw [ <->] (-6.5,0) -- (6.5,0);
\draw [help lines,dashed,<-] (0,1.3) -- (0,0);
\draw (0,0) node[below] {$\rho=z$};
\draw (-5,1.3) node[above]{$ \cos{\left(N\,\left(\rho-z\right)^{2}\right)}$}; ;
\draw [thick,samples=500,domain=-2*pi:2*pi] plot
(\x, {cos(64*\x*\x )});
\draw [ <-](-3*pi/8+0.01,-0.7)
--(-2*pi/8+0.18,-0.7);
\draw [ ->](2*pi/8-0.1,-0.7)
--(3*pi/8-0.01,-0.7);
\draw (6.5,0) node[right] {$\rho$};
\draw (0,-0.7) node {$\sqrt{{\pi}/{2N}}$};
\draw [help lines,dashed] (-3*pi/8,1) -- (-3*pi/8,-1);
\draw [help lines,dashed] (3*pi/8,1) -- (3*pi/8,0-1);
\end{tikzpicture}
\end{align*}
Finally, since
$\;\displaystyle \parallel h \parallel_{L^{p}(B)}\,=\, |B|^{{1}/{p}}\,\approx\,1,\;$ then
\begin{align}\label{lb3}
\parallel T^{j,k}_{N}\parallel_{L^{p}\left(B\right)
\rightarrow L^{q}([0,1])}
\,\geq\,\frac{\parallel T^{j,k}_{N} h \parallel_{L^{q}([0,1])}}{
\parallel h \parallel_{L^{p}(B)}}
\;\gtrsim\;
N^{-\frac{1}{2}}.
\end{align}
Putting (\ref{lb1}),  (\ref{lb2}) and
(\ref{lb3}) together we deduce
\begin{align*}
 \parallel T^{j,k}_{N} \parallel_{L^{p}(B)
\rightarrow L^{q}([0,1])}
\;\;\gtrsim\;
N^{-\min\left\{\frac{n}{j}\left(1-\frac{1}{p}\right),
\,\frac{1}{k}\frac{1}{q},\,\frac{1}{2}\right\}}
\,=\,N^{- C_{j,k,n}\left(\frac{1}{p},\frac{1}{q}\right)}.
\end{align*}
\textbf{\emph{\underline{Step 2}}.}
The $\,L^{2} - L^{2}\,$ estimate takes the form:
\begin{eqnarray}\label{energy}
\left.
             \begin{array}{ll}
\vspace{0.3 cm}
\parallel T^{j,k}_{N}f\parallel_{L^{2}([0,1])}
\;\lesssim \; N^{-1/2k}\,\parallel f \parallel_{L^{2}\left(B\right)}, & \hbox{$n\geq j$,} \\
               \parallel T^{2,k}_{N}f\parallel_{L^{2}([0,1])}
\;\lesssim \; N^{-n/2j}\,\parallel f \parallel_{L^{2}\left(B\right)}, & \hbox{$n=1$.}
             \end{array}
           \right \}
\end{eqnarray}
 Besides (\ref{lb2}), the estimate (\ref{energy}) demonstrates the difference between linear ($k=1$) and quadratic ($k=2$) oscillations.
Let $\,x\in {\mathbb{R}}^{n}-\{0\}.\,$
The phase $\;s \longrightarrow  {|x|}^{j}\,s^{k}\;$
of the oscillatory factor in (\ref{intop})
is non-stationary when $\,k=1.\,$ While
in the case $\,k=2,\,$
it is stationary with the nondegenerate
critical point $s=0.\,$ This is where
non-stationary and stationary phase methods
(see lemmas \ref{nonstationary} and
\ref{stationary0} below)
for estimating oscillatory integrals come into play.
As expected from (\ref{lb1}), the role of $j$
appears only in the dimension $n=1.$
Using the estimate (\ref{energy}) in Lemma \ref{holder}
we infer
\begin{align*}
 \parallel T^{j,k}_{N} \parallel_{L^{p}(B)
\rightarrow L^{q}([0,1])}
\;\;\lesssim\; N^{- C_{j,k,n}\left(\frac{1}{p},\frac{1}{q}\right)}.
\end{align*}
\subsubsection*{3. Proof of the energy estimate
 (\ref{energy})}
To prove the estimate (\ref{energy})
we need lemmas \ref{kernelsk}, \ref{kernelsq} and \ref{even} that we give below.
Lemma \ref{kernelsk} is based on the assertions
of lemmas \ref{nonstationary} and
\ref{stationary0}.
\begin{lem}\label{nonstationary}
(\cite{stein}, Proposition 1 Chapter VIII)
Let $\,\psi \in C^{\infty}_{c}\left(\mathbb{R}
\right)\,$ and let $\displaystyle\; I(\lambda)=
\int_{\mathbb{R}}\,\psi(s)\,e^{\imath \,\lambda\,s}\,ds. \,$ Then $\;\displaystyle |I(\lambda)|\;\lesssim\;
\min{\left\{
\frac{1}{1+|\lambda|},
\frac{1}{1+\lambda^{2}}\right\}}.$
\end{lem}
Observing that $\;\displaystyle \int_{0}^{1}\,e^{\imath \,\lambda\,s^{2}}\,ds\,=\,
\frac{1}{2}\int_{-1}^{1}\,e^{\imath \,\lambda\,s^{2}}\,ds\;$ and arguing as
in (\ref{t1})-(\ref{44}) implies the estimate
in Lemma \ref{stationary0}.
\begin{lem}\label{stationary0}
\begin{align*}
\left|\int_{0}^{1}\,e^{\imath \,\lambda\,s^{2}}\,ds\right|\;\lesssim \; \max{\left\{\frac{1}{1+\sqrt{|\lambda|}},
\frac{1}{1+|\lambda|}\right\}}.
\end{align*}
\end{lem}
\begin{lem}\label{kernelsk}
Let $\,\psi \in C^{\infty}_{c}\left(\mathbb{R}
\right)\,$ and let
$\;K_{N}^{j,k}:{\mathbb{R}}^{n}\times{\mathbb{R}}^{n}
\longrightarrow {\mathbb{C}}\;$ be defined by
\begin{align*}
K_{N}^{j,k}(x,y):=
\left\{
  \begin{array}{ll}
 \displaystyle   \int_{\mathbb{R}}\,\psi(s)\,e^{\imath N \left({|x|}^{j}-{|y|}^{j}\right)s}\,ds, & \hbox{$k=1$;} \\\\
 \displaystyle  \int_{0}^{1}\,\,e^{\imath N \left({|x|}^{j}-{|y|}^{j}\right)s^{2}}\,ds  , & \hbox{$k=2$.}
  \end{array}
\right.
\end{align*}
Then
\begin{align}
\label{kernelsk1}
&\hspace*{-1 cm}
|K_{N}^{j,1}(x,y)|\;\lesssim\;
\min{\left\{
\left(1+N\,\left|{|x|}^{j}-{|y|}^{j}\right|\right)^{-1},
\left(1+N^{2} \, \left({|x|}^{j}-{|y|}^{j}\right)^{2}\right)^{-1}
\right\}},
\\ \label{kernelsk2}
&\hspace*{-1 cm} |K_{N}^{j,2}(x,y)|\;\lesssim\;
\max{\left\{\left( 1+\sqrt{N}\, \sqrt{\left|{|x|}^{j}-{|y|}^{j}\right|}\right)^{-1},
\left(1+N\, \left|{|x|}^{j}-{|y|}^{j}\right|\right)^{-1}\right\}}.
\end{align}
\end{lem}
The next lemma is mainly a consequence of
Young's inequality.
\begin{lem}\label{kernelsq}
Let $\,p,q,r\geq 1\,$ and $\,1/p+1/q+1/r =2.\,$
Let $\,f\in L^{p}(B),\,$ $\,g\in L^{q}(B)\,$
and $\,h\in L^{r}([0,1]).\,$  Then
\begin{align*}
 \left|\,\int_{{B}}\,\int_{{B}}\,
f(x)\,f(y)\,h(|x|^{m}-|y|^{m})\,dx\,dy\,\right|\;\lesssim
\;\parallel f \parallel_{L^{p}(B)}\,
\parallel g \parallel_{L^{q}(B)}\,
\parallel h \parallel_{L^{r}([0,1])}
\end{align*}
provided $\,m\leq n$.
\end{lem}
\begin{proof}
Switching to polar coordinates by setting
$\,x=r_{1}\theta_{1}\,$ and $\,y=r_{2}\theta_{2}\,$
then applying Fubini's theorem gives
\begin{align}\label{newlemma1}
 \left|\,\int_{{B}}\,\int_{{B}}\,
f(x)\,f(y)\,h(|x|^{m}-|y|^{m})\,dx\,dy\,\right|
\,\leq\,\int_{S^{n-1}}\,\int_{S^{n-1}}\,
|Q(\theta_{1},\theta_{2})|
\,d\theta_{1}\,d\theta_{2}
\end{align}
where
\begin{align*}
Q(\theta_{1},\theta_{2})\,=\,
\int_{0}^{1}\,\int_{0}^{1}\,
f(r_{1}\theta_{1})\,g(r_{2}\theta_{2})
\,h\left({r_{1}}^{m}-{r_{2}}^{m}\right)
\,r_{1}^{n-1}\,r_{2}^{n-1}\,dr_{1}\,dr_{2}.
\end{align*}
Changing variables
$\:r_{i}^{m}\,\longrightarrow\, \rho_{i}\:$
then using Young's inequality we get
\begin{align*}
\hspace*{-1 cm}
|Q(\theta_{1},\theta_{2})|\,\lesssim\,
\left(\int_{0}^{1}\left|f(\sqrt[m]{\rho_{1}}\,\theta_{1})
\right|^{p}\,\rho_{1}^{p\frac{n-m}{m}}\,d\rho_{1}\right)
^{\frac{1}{p}}
\left(\int_{0}^{1}\left|g(\sqrt[m]{\rho_{2}}\,\theta_{2})
\right|^{q}\,\rho_{2}^{q\frac{n-m}{m}}\,d\rho_{2}\right)
^{\frac{1}{q}}
\parallel h \parallel_{L^{r}([0,1])}.
\end{align*}
Reversing the variables change
in the first two integrals on the right-hand side of the latter estimate we obtain
\begin{align}\label{newlemma2}
\hspace*{-1 cm}
\nonumber |Q(\theta_{1},\theta_{2})|\,\lesssim&\,
\left(\int_{0}^{1}\left|f({r_{1}}\,\theta_{1})
\right|^{p}\,r_{1}^{(p-1)(n-m)}\,
r_{1}^{n-1}\,dr_{1}\right)
^{\frac{1}{p}}\\&\;\nonumber
\left(\int_{0}^{1}\left|g({r_{2}}\,\theta_{2})
\right|^{q}\,r_{2}^{(p-1)(n-m)}\,
r_{2}^{n-1}\,dr_{2}\right)
^{\frac{1}{q}}\,
\parallel h \parallel_{L^{r}([0,1])}\\
\leq&
\left(\int_{0}^{1}\left|f({r_{1}}\,\theta_{1})
\right|^{p}\,r_{1}^{n-1}\,dr_{1}\right)
^{\frac{1}{p}}
\left(\int_{0}^{1}\left|g({r_{2}}\,\theta_{2})
\right|^{q}\,r_{2}^{n-1}\,dr_{2}\right)
^{\frac{1}{q}}
\,\parallel h \parallel_{L^{r}([0,1])}
\end{align}
as long as $\,m\leq n.$ Invoking H\"{o}lder's inequality it follows that
\begin{align}
\nonumber  &\int_{S^{n-1}}\,
\left(\int_{0}^{1}\left|f({r_{1}}\,\theta_{1})
\right|^{p}\,
r_{1}^{n-1}\,dr_{1}\right)^{\frac{1}{p}}\,d\theta_{1}\\
\label{newlemma3} &\hspace{0.8 cm}\leq\;\omega_{n-1}^{1-\frac{1}{p}}\;
 \left( \int_{S^{n-1}}\,\int_{0}^{1}
\left|f({r_{1}}\,\theta_{1})
\right|^{p}\,r_{1}^{n-1}\,dr_{1}\,d
\theta_{1}\right)^{\frac{1}{p}}\;=
\;\omega_{n-1}^{1-\frac{1}{p}}\;
\parallel f \parallel_{L^{p}(B)},\\
\nonumber  &
\int_{S^{n-1}}\,
\left(\int_{0}^{1}\left|g({r_{2}}\,\theta_{2})
\right|^{q}\,
r_{2}^{n-1}\,dr_{2}\right)^{\frac{1}{q}}\,d\theta_{2}\\
\label{newlemma4} &
\hspace{0.8 cm}\leq\;\omega_{n-1}^{1-\frac{1}{q}}\;
 \left( \int_{S^{n-1}}\,\int_{0}^{1}
\left|g({r_{2}}\,\theta_{2})
\right|^{q}\,r_{2}^{n-1}\,dr_{2}\,d
\theta_{2}\right)^{\frac{1}{q}}\;=
\;\omega_{n-1}^{1-\frac{1}{q}}\;
\parallel g \parallel_{L^{q}(B)}.
\end{align}
Returning to (\ref{newlemma1})
with the estimates (\ref{newlemma2}),
(\ref{newlemma3}) and (\ref{newlemma4}) concludes the proof.
\end{proof}
Remark \ref{even0} together with Lemma \ref{homogeneous} are needed to show Lemma \ref{even}.
\begin{rem}\label{even0}
Suppose that the integral
\begin{align*}
 J\,=\,
\int_{-b_{1}}^{b_{1}}...\int_{-b_{m}}^{b_{m}}
\,K(t_{1},...,t_{m})\,
f_{1}(t_{1})...f_{m}(t_{m})\,dt_{1}...dt_{m}
\end{align*}
exists. If $\,K\,$
is even in all its variables then
\begin{align*}
 J\,=\,
\int_{0}^{b_{1}}...
\int_{0}^{b_{m}}
\,K(t_{1},...,t_{m})\,
\prod_{i=1}^{m}\left(f_{i}(t_{i})+f_{i}(-t_{i})\right)
\,dt_{1}...dt_{m}.
\end{align*}
This follows easily from
the fact that the integrand
in the second expression for $\,J\,$
is even in all variables.
\end{rem}
Lemma \ref{homogeneous} discusses the boundedness
of a bilinear form with a homogeneous kernel.
\begin{lem}\label{homogeneous}
Let $\,f\in L^{p}([0,1])\,$ and
$\,g\in L^{q}([0,1])\,$
with $\,1\leq p \leq +\infty\,$ and $\,1/p\, +\, 1/q=1.\,$
Assume that $\,K:{[0,1]}\times
{[0,1]}\longrightarrow {\mathbb{R}}\,$
is homogeneous of degree $-1,\,$
that is, $\,K(\lambda x, \lambda y)=
\lambda^{-1} K(x,y),\,$ for $\,\lambda>0.\,$
Assume also that
\begin{align*}
 \int_{0}^{+\infty}
\,\left|K(x,1)\right|\,{x}^{-\frac{1}{p}}\,dx
\,\lesssim\,1 \qquad \text{or } \qquad
 \int_{0}^{+\infty}
\,\left|K(1,y)\right|\,{y}^{-\frac{1}{q}}\,dy
\,\lesssim\,1.
\end{align*}
Then
\begin{align*}
\left|\int_{0}^{1}\int_{0}^{1}\,
K(x,y)\,f(x)\,g(y)\,dx\,dy\right|\;\lesssim\;
\parallel f \parallel_{L^{p}([0,1])}\,
\parallel g \parallel_{L^{q}([0,1])}.
\end{align*}
 \end{lem}
In \cite{hardy}, one can find a proof
for the case when the integrals
that define the bilinear form are taken over $\,[0,+\infty[.\,$ We treat this slightly trickier
case of finite range without using the result in
\cite{hardy}.
\begin{proof}
Let $\,\displaystyle Q(f,g)\,=\,
\int_{0}^{1}\int_{0}^{1}\,
K(x,y)\,f(x)\,g(y)\,dx\,dy.\,$
Using a change of variables,
$\,x\rightarrow y.u,\,$ and exploiting the homogeneity
of the kernel we have
\begin{align*}
\hspace*{-0.8 cm} Q(f,g)\,=\,
\int_{0}^{1}y\,g(y)\int_{0}^{\frac{1}{y}}
K(y.u,y)\,f(y.u)\,du\,dy\,=\,
\int_{0}^{1}g(y)\int_{0}^{\frac{1}{y}}
K(u,1)\,f(y.u)\,du\,dy.
\end{align*}
By Fubini's theorem we may write
 \begin{align}\label{qfg}
\hspace*{-1 cm}
Q(f,g)=\int_{0}^{1} K(u,1) \int_{0}^{1} f(y.u)\,g(y)\,dy\,du+
\int_{1}^{+\infty} K(u,1) \int_{0}^{\frac{1}{u}} f(y.u)\,g(y)\,dy\,du.
\end{align}
But by  H\"{o}lder's inequality we have
\begin{align*}
\hspace*{-0.8 cm}
\left|\int_{0}^{1}\,f(y.u)\,g(y)\,dy\right|
\;\leq&\;\left(\int_{0}^{1}\,|f(y.u)|^{p}\,dy\right)
^{\frac{1}{p}}
\left(\int_{0}^{1}\,|g(y)|^{q}\,dy\right)^{\frac{1}{q}}\\
\,&\hspace*{-2 cm}=u^{-\frac{1}{p}}\,
\left(\int_{0}^{u}\,|f(x)|^{p}\,dx\right)
^{\frac{1}{p}}\,
\parallel g \parallel_{L^{q}([0,1])}
\;\leq\;u^{-\frac{1}{p}}\,\parallel f \parallel_{L^{q}([0,1])}\,
\parallel g \parallel_{L^{q}([0,1])}
\end{align*}
for all $\,0 < u < 1.\,$ Similarly
\begin{align*}
\hspace*{-0.8 cm}
\left|\int_{0}^{\frac{1}{u}}\,f(y.u)\,g(y)\,dy\right|
\;\leq&\;\left(\int_{0}^{\frac{1}{u}}
\,|f(y.u)|^{p}\,dy\right)
^{\frac{1}{p}}\,
\left(\int_{0}^{\frac{1}{u}}\,|g(y)|^{q}
\,dy\right)^{\frac{1}{q}}\\
\,&\hspace*{-3.4 cm}=u^{-\frac{1}{p}}\,
\left(\int_{0}^{1}\,|f(x)|^{p}\,dx\right)
^{\frac{1}{p}}\,
\left(\int_{0}^{\frac{1}{u}}\,|g(y)|^{q}
\,dy\right)^{\frac{1}{q}}
\;\leq\;u^{-\frac{1}{p}}\,\parallel f \parallel_{L^{q}([0,1])}\,
\parallel g \parallel_{L^{q}([0,1])}
\end{align*}
for all $\,1< u < +\infty.\,$ Using the last two inequalities together
with the triangle inequality  in (\ref{qfg})
we get
 \begin{align*}
\hspace*{-1 cm}
|Q(f,g)|\leq& \,
\,\parallel f \parallel_{L^{q}([0,1])}\,
\parallel g \parallel_{L^{q}([0,1])}\,
\left(\int_{0}^{1} |K(u,1)| \,u^{-\frac{1}{p}}\,du+
\int_{1}^{+\infty} |K(u,1)|\,u^{-\frac{1}{p}}\,du
\right)\\
 \lesssim&\;
\parallel f \parallel_{L^{q}([0,1])}\,
\parallel g \parallel_{L^{q}([0,1])},
\qquad \text{when} \quad
\int_{0}^{+\infty} |K(x,1)|\,x^{-\frac{1}{p}}\,dx
\,\lesssim\,1.
\end{align*}
When $\;\displaystyle \int_{0}^{+\infty}\left|K(1,y)\right|
{y}^{-\frac{1}{q}}dy\lesssim 1\;$
the assertion follows analogously.
\end{proof}
\begin{rem}
If $\,K(x,y)=\left( x + y \right)^{-1}\,$
in Lemma \ref{homogeneous} we get Hilbert's
inequality.
\end{rem}
\begin{lem}\label{even}
Let $\,f,g \in L^2([-1,1]).$ Then
\begin{align}
\label{even1}&\int_{-1}^{1}\,\int_{-1}^{1}\,
\frac{|f(x)||g(y)|}{1+N\,
\left|x^2-y^2\right|}
\,dx\,dy \;\lesssim\; \frac{1}{\sqrt{N}}\,
\parallel f \parallel_{L^{2}([-1,1])}\,\parallel g \parallel_{L^{2}([-1,1])}, \\
\label{even2} &\int_{-1}^{1}\,\int_{-1}^{1}\,
\frac{|f(x)|\,|g(y)|}{
\sqrt{\left|{x}^{2}-{y}^{2}\right|}}
\,dx\,dy \;\lesssim\; \parallel f \parallel_{L^{2}([-1,1])}\,\parallel g \parallel_{L^{2}([-1,1])}.
\end{align}
\end{lem}
\begin{proof}
Beginning with the estimate (\ref{even1}),
Remark \ref{even0} suggests estimating\\\\
$\; \displaystyle \int_{0}^{1}\,\int_{0}^{1}\,
\frac{|f(\pm x)||g(\pm y)|}{1+N\,
\left|x^2-y^2\right|}
\,dx\,dy.\;$ Let $\;\displaystyle W_{N}(f,g):=
\int_{0}^{1}\,\int_{0}^{1}\,
\frac{|f(x)||g(y)|}{1+N\,
\left|x^2-y^2\right|}
\,dx\,dy.$\\\\
If $\,x,y \geq 0\,$ and
$\,|x-y|>>1/\sqrt{N}\,$ then we also have
$\,x+y>>1/\sqrt{N}\,$ and consequently $\,N\left|x^2-y^2\right|>>1.\,$
Therefore
\begin{align}
\hspace{-0.8 cm}
\nonumber W_{N}(f,g)&\approx
\int\,\int_{
\substack{0\leq x,y\leq1,\\
|x-y|\lesssim\; 1/\sqrt{N}}}
\frac{|f(x)||g(y)|}{1+N\,
\left|x^2-y^2\right|}
\,dx\,dy+
\int\,\int_{
\substack{0\leq x,y\leq1,\\
|x-y|>> 1/\sqrt{N}}}
\frac{|f(x)||g(y)|}{1+N\,
\left|x^2-y^2\right|}
\,dx\,dy\\
\nonumber &\lesssim
\int\,\int_{
\substack{0\leq x,y\leq1,\\
|x-y|\lesssim\; 1/\sqrt{N}}}
{|f(x)||g(y)|}\,dx\,dy+
\frac{1}{N}\,\int\,\int_{
\substack{0\leq x,y\leq1,\\
|x-y|>> 1/\sqrt{N}}}
\frac{|f(x)||g(y)|}{\left|x^2-y^2\right|}
\,dx\,dy\\
\label{h1} &\lesssim
\int_{0}^{1}\,\int_{0}^{1}\,
\chi_{N}{\left(|x-y|\right)}{|f(x)||g(y)|}\,dx\,dy+
\frac{1}{\sqrt{N}}\,\int_{0}^{1}\,\int_{0}^{1}\,
\frac{|f(x)||g(y)|}{x+y}
\,dx\,dy
\end{align}
where $\,\chi_{N}\,$ is the characteristic function
of the interval $\,[0,1/\sqrt{N}\,].$
By Young's inequality we have
\begin{align}\label{h2}
\int_{0}^{1}\,\int_{0}^{1}\,
\chi_{N}{\left(|x-y|\right)}{|f(x)||g(y)|}
\,dx\,dy\,\leq\,
\frac{1}{\sqrt{N}}\,
\parallel f \parallel_{L^{2}([0,1])}\,\parallel g \parallel_{L^{2}([0,1])}.
\end{align}
And by Hilbert's inequality
\begin{align}\label{h3}
\int_{0}^{1}\,\int_{0}^{1}\,
\frac{|f(x)||g(y)|}{x+y}
\,dx\,dy\,\lesssim \;\parallel f \parallel_{L^{2}([0,1])}\,\parallel g \parallel_{L^{2}([0,1])}.
\end{align}
Using (\ref{h2}) together with
(\ref{h3}) in (\ref{h1}) we obtain
\begin{align*}
\int_{0}^{1}\,\int_{0}^{1}\,
\frac{|f(x)||g(y)|}{1+N\,
\left|x^2-y^2\right|}
\,dx\,dy\,\lesssim\,\frac{1}{\sqrt{N}}\,
\parallel f \parallel_{L^{2}([0,1])}\,\parallel g \parallel_{L^{2}([0,1])}.
\end{align*}
In obtaining (\ref{h1}), we worked only on the kernel of $W_{N}.$ It is therefore easy to see that replacing the function $\,x\rightarrow f(x)\,$
by the function $\,x\rightarrow f(-x)\,$
or $\,y\rightarrow g(y)\,$
by $\,y\rightarrow g(-y)\,$
then repeating the routine above
eventually leads to the estimate
\begin{align*}
\int_{0}^{1}\,\int_{0}^{1}\,
\frac{|f(\pm x)||g(\pm y)|}{1+N\,
\left|x^2-y^2\right|}
\,dx\,dy\,\lesssim\,\frac{1}{\sqrt{N}}\,
\parallel f \parallel_{L^{2}([-1,1])}\,\parallel g \parallel_{L^{2}([-1,1])}.
\end{align*}
This proves (\ref{even1}). Taking advantage of Remark \ref{even0} again and arguing like before,
it suffices to\\\\ estimate
$\displaystyle
 V(f,g)=\int_{0}^{1}\,\int_{0}^{1}\,
\frac{|f(x)|\,|g(y)|}{\sqrt{\left|{x}^{2}-{y}^{2}\right|}}
\,dx\,dy.\;$  Since $\; \displaystyle \int_{0}^{+\infty}\frac{dz}{\sqrt{z}\,\sqrt{|1-z^2|}}
\,\approx\,1,$\\\\
a direct application of Lemma \ref{homogeneous} then
gives $\,V(f,g)\,\lesssim\,\parallel f \parallel_{L^{2}([0,1])}\,\parallel g \parallel_{L^{2}([0,1])}.$
\end{proof}
We are now ready to prove (\ref{energy}).
We do this for each of the cases
$k=1$ and $k=2$ separately.\\
\textbf{The phase is linear in $\textbf{s}\,$ $\,(k=1)$}:\\
Let $\psi$ be a nonnegative smooth cutoff function
such that $\,{supp}\:\psi \subset\;]-1,2[\,$
and $\,\psi(s)=1\,$ on $\,[0,1]$.
Since $\,|T^{j,1}_{N} f |^2 \,=\, T^{j,1}_{N} f\;\;\overline{T^{j,1}_{N} f}.\,$ Then
\begin{align*}
&\hspace{-1 cm}\parallel T^{j,1}_{N} f \parallel^{2}_{L^{2}([0,1])}
\,=
\int_{0}^{1}\,|T^{j,1}_{N} f(s)|^{2}\,ds
\,\leq\,\int_{\mathbb{R}}\psi(s)\,|T^{j,1}_{N} f(s)|^{2}\,ds\\
&\hspace{-1 cm}=\;\int_{\mathbb{R}}\psi(s)\, T^{j,1}_{N} f(s)\;\overline{T^{j,1}_{N} f(s)}\,ds\,
=\;
\int_{\mathbb{R}}\psi(s)\,  \int_{{B}}\,\int_{{B}}\,e^{\imath N \left({|x|}^{j}-{|y|}^{j}\right)s}\,
f(x)\,\overline{f(y)}\,dx\,dy\,ds.
\end{align*}
Let $f\in L^{2}(B)$. Applying Fubini's theorem we get
\begin{align}\label{energy01}
\parallel T^{j,1}_{N} f \parallel^{2}_{L^{2}([0,1])}\;\leq\;
 \int_{{B}}\,\int_{{B}}\,K_{N}^{j,1}(x,y)\,
f(x)\,\overline{f(y)}\,dx\,dy.
\end{align}
In the light of the estimate (\ref{kernelsk1})
of Lemma \ref{kernelsk}, it follows that
\begin{align}\label{energy11}
\parallel T^{j,1}_{N} f \parallel^{2}_{L^{2}([0,1])}\;\lesssim\;
 \int_{{B}}\,\int_{{B}}\,
\frac{|f(x)|\,|f(y)|}{1+N^{2} \, \left({|x|}^{j}-{|y|}^{j}\right)^{2}}
\,dx\,dy.
\end{align}
Since $\displaystyle
\int_{0}^{1}\frac{dz}{1+N^2 z^2}\approx \frac{1}{N},\,$
then, applying Lemma \ref{kernelsq} with
$\,h(z)=\left(1+N^2 z^2\right)^{-1}\,$ to
the\\\\ estimate  (\ref{energy11}), we obtain
\begin{align}\label{e1}
\parallel T^{j,1}_{N} f \parallel_{L^{2}([0,1])}\;\lesssim\;
\frac{1}{\sqrt{N}}\,\parallel  f \parallel_{L^{2}(B)},
\qquad\text{for all dimensions}\;\;n\geq j.
\end{align}
To finish this case, it remains
to estimate $\,T^{2,1}f\,$ in the dimension
$\,n=1.$ In view of  (\ref{kernelsk1}) and (\ref{energy01}), we have
\begin{align*}
\hspace{-1 cm}
\parallel T^{2,1}_{N} f \parallel^{2}_{L^{2}([0,1])}\:\lesssim\,
 \int_{-1}^{1}\,\int_{-1}^{1}\,
\frac{|f(x)|\,|f(y)|}{1+N\,
\left|x^2-y^2\right|}
\,dx\,dy.
\end{align*}
Hence, by (\ref{even1}) of Lemma \ref{even},
\begin{align}\label{e2}
\parallel T^{2,1}_{N} f \parallel_{L^{2}([0,1])}\:
\lesssim\,\frac{1}{N^{1/4}}\,
\parallel f \parallel_{L^{2}([-1,1])}.
\end{align}
\textbf{The phase is quadratic in $\textbf{s}\,$ $\,(k=2)$}:\\
For $f\in L^{2}(B)$, using Fubini's theorem
then employing the estimate
(\ref{kernelsk2}) implies
\begin{align}\label{energy21}
\hspace{-0.6 cm}
\parallel T^{j,2}_{N} f \parallel^{2}_{L^{2}([0,1])}\:=\,
 \int_{{B}}\,\int_{{B}}\,K_{N}^{j,2}(x,y)\,
f(x)\,\overline{f(y)}\,dx\,dy
\,\lesssim\, G^{j}_{N}(f)+H^{j}_{N}(f)
\end{align}
where
\begin{align*}
G^{j}_{N}(f)\,=&\, \int_{{B}}\,\int_{{B}}\,
\frac{|f(x)|\,|f(y)|}{1+\sqrt{N}\, \sqrt{\left|{|x|}^{j}-{|y|}^{j}\right|}}
\,dx\,dy,\\
H^{j}_{N}(f)\,=&\, \int_{{B}}\,\int_{{B}}\,
\frac{|f(x)|\,|f(y)|}{1+N\, \left|{|x|}^{j}-{|y|}^{j}\right|}
\,dx\,dy.
\end{align*}
Since $\;\displaystyle  \int_{0}^{1}\,\frac{dz}{1+\sqrt{N}\,\sqrt{z}}
\,\approx\, \frac{1}{\sqrt{N}},\quad
\int_{0}^{1}\,\frac{dz}{1+N\,z}
\,=\,
\text{\large o}\left(\frac{1}{\sqrt{N}}\right),
\quad \text{as}\;\;\; N\longrightarrow+\infty,
$\\\\
then applying Lemma \ref{kernelsq}
to both $\,G^{j}_{N}(f)\,$ and $\,H^{j}_{N}(f)\,$ gives the estimate
\begin{align}\label{energy22}
G^{j}_{N}(f)+
H^{j}_{N}(f)\;\lesssim\; \frac{1}{\sqrt{N}}
\parallel f \parallel^{2}_{L^{2}(B)},
\qquad n\geq j.
\end{align}
It remains to control $\:G^{2}_{N}(f)\,$ and $\,
H^{2}_{N}(f)\:$ in the dimension $\,n=1.\,$
But when $\,n=1,$
\begin{align*}
\hspace*{-0.4 cm}
G^{2}_{N}(f)\,=&\, \int_{-1}^{1}\,\int_{-1}^{1}\,
\frac{|f(x)|\,|f(y)|}{1+\sqrt{N}\, \sqrt{\left|{x}^{2}-{y}^{2}\right|}}
\,dx\,dy\\ \leq&\,\frac{1}{\sqrt{N}}\,
\int_{-1}^{1}\,\int_{-1}^{1}\,
\frac{|f(x)|\,|f(y)|}{
\sqrt{\left|{x}^{2}-{y}^{2}\right|}}
\,dx\,dy
\,\lesssim\,
\frac{1}{\sqrt{N}}\,\parallel f \parallel^{2}_{L^{2}([-1,1])}\quad
\text{by}\;\; (\ref{even2})\; \text{of}\;
\text{Lemma} \;\ref{even}.
\end{align*}
An identical estimate holds for $H^{2}_{N}(f)$ in the dimension $n=1$ because of (\ref{even1}).
Combining this with (\ref{energy22})
and using them in (\ref{energy21}) yields
\begin{align}\label{e3}
\parallel T^{j,2}_{N} f \parallel_{L^{2}([0,1])}
\;\lesssim\;\frac{1}{{N}^{1/4}}
\parallel f \parallel_{L^{2}(B)}.
\end{align}
Finally, bringing the estimates (\ref{e1}),  (\ref{e2}) and (\ref{e3}) together results in (\ref{energy}).
\section*{References}

redpajama_set_name: RedPajamaArXiv


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Styx Tour Dates & Concert Tickets
Styx Tour Dates and Concert Tickets
Styx (pronounced: /stɪks/) is an American rock band. Their hit songs have included "Come Sail Away", "Lady", "Mr. Roboto", "Renegade", "Babe", "Blue Collar Man" and "The Best of Times". The eight consecutive albums certified mu... read more
Styx Tickets
Other Tour Dates (24)
Jan 17 Styx Silver Legacy Casino Reno, Nevada
Jan 18 David Victor Presents Boston/Styx - Tribute Sandia Resort & Casino Albuquerque, New Mexico
Jan 19 Styx Pearl Concert Theater at Palms Las Vegas, Nevada
Jan 30 Styx Ruth Eckerd Hall Clearwater, Florida
Jan 31 Styx Suncoast Credit Union Arena at FSW Fort Myers, Florida
Feb 1 Styx Port Saint Lucie Port Saint Lucie, Florida
Feb 20 Styx Paradise Cove Tulsa, Oklahoma
Feb 21 Styx The Riverdome Bossier City, Louisiana
Mar 13 Styx Parx Casino Bensalem, Pennsylvania
Mar 14 STYX Penns Peak Jim Thorpe, Pennsylvania
Mar 15 Styx Beacon Theatre New York, New York
Mar 17 Styx Warner Theatre Washington, District of Columbia
Mar 19 Styx The Avalon Ballroom Theatre At Niagara Fallsview Casino Resort Niagara Falls, CAN
Mar 20 Styx Tropicana Casino and Resort Atlantic City, New Jersey
Mar 21 Styx The Grand Theater at Foxwoods Resort Mashantucket, Connecticut
(10 of 24 events shown) See More
More About Styx
Recommendations Similar to Styx
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MORE INFO ABOUT Styx
Styx (pronounced: /stɪks/) is an American rock band. Their hit songs have included "Come Sail Away", "Lady", "Mr. Roboto", "Renegade", "Babe", "Blue Collar Man" and "The Best of Times". The eight consecutive albums certified multi-platinum by the RIAA.
Styx formed in Chicago. Twin brothers Chuck and John Panozzo first got together with their neighbor Dennis DeYoung in 1961 in the Roseland section of the south side of Chicago, eventually taking the band name "The Tradewinds". Chuck Panozzo left to attend seminary school for a year but returned to the group by 1964. By this time, the others had brought in guitarist Tom Nardini to teach Chuck the guitar. Chuck decided to rejoin the others as bassist. Brother John was the drummer, while Dennis had switched from accordion to organ and piano. In 1965, the name "Tradewinds" was changed to TW4 after another band called The Trade Winds broke through nationally. By 1966, the Panozzo brothers had joined DeYoung at Chicago State University and kept the group together doing gigs at high schools and frat parties while studying to be teachers. In 1969, they added a college buddy, John Curulewski, on guitar after Nardini departed. Guitarist James "J.Y." Young came aboard in 1970 making TW4 a quintet.
In 1972, the band members decided to choose a new name when they signed to Wooden Nickel Records; several suggestions were made and, says DeYoung, Styx was chosen because it was "the only one that none of us hated".
The Grand Illusion (1977) became the group's breakthrough album, reaching Triple Platinum certification and spawning a top-ten hit and AOR radio staple in the DeYoung-written "Come Sail Away,", which reached #8 in 1978, as well as a second radio hit, Shaw's "Fooling Yourself (The Angry Young Man)", which reached #29 the same year.
Through the late 1970s, the band enjoyed its greatest success. The album Pieces of Eight (1978) found the group moving in a more straight-ahead pop-rock direction and spawned the Shaw-written hit singles "Renegade,", #16 in the U.S., and "Blue Collar Man (Long Nights)," #21 in the U.S., plus a minor hit in the release "Sing for the Day" that stopped just short of the Top Forty at #41.[2]
Cornerstone (1979) yielded the group's first number one hit, the DeYoung ballad "Babe" [3]. By early 1980, "Babe" had become the band's biggest international hit and first million-selling single, reaching number six in the United Kingdom [4]. The album also included the #26 DeYoung hit "Why Me", and "Borrowed Time" which was co-written with Shaw, plus Shaw's "Boat On The River". The popularity of the album helped win the band a People's Choice Award for Best New Song in 1980. Styx was nominated for the Twenty-Second Grammy Awards[5] for Best Rock Vocal Performance by a Duo or Group. Cornerstone's engineers Gary Loizzo and Rob Kingsland were nominated for a Grammy[6] at the 22nd Grammy Awards for Best Engineered Recording.
(From Wikipedia)
http://www.myspace.com/styxtheband
Categories: Music | Rock | Pop
http://www.styxworld.com/ | http://www.myspace.com/styxworldmusic | Styx Tickets

redpajama_set_name: RedPajamaCommonCrawl


MANAGING YOUR FOOTBALL CLUB'S ACCOUNTS.
It's important for Clubs to raise funds to cover essential expenditure such as Club Affiliation fees, League membership fees, pitch hire charges, kit and equipment.
All Clubs should operate a separate bank account for the transaction of Club business, which should be registered under the Club name.
Any cheques or withdrawals should require a minimum of two signatures from committee members. When a new Club official with the authority to sign cheques is appointed, the bank must be notified immediately and a new bank mandate must be completed.
Clubs should think about setting up a subscription rate for players and members to meet your costs for the season.
Any petty cash should be balanced against receipts. A cash book and all vouchers must be retained by the Club for at least two seasons preceding the current season to be available to Birmingham FA whenever required.
Every member of the Club is likely to be a joint owner of any assets of the Club, but also jointly liable for all or part of any debts. Clubs should therefore consider suitable insurance.
It is also important to make provision for how any cash, kit, or other assets will be dealt with if the Club ceases to exist.


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Paintin' P Town Brown - An Evening of Ween - 10/12
Maine's Funniest Comedian Competition Finals - 8/11
Steel Panther - Heavy Metal Rules Tour - 10/20	On sale 7/19
Nick Kroll: Middle-Aged Boy Tour - 9/27	On sale 7/19
Alice Phoebe Lou - 10/18	On sale 7/19
Jon Stickley Trio - 9/10	On sale 7/19
The Masterstroke Queen Experience - 10/31
Ronnie Baker Brooks - 8/14
The Movement - 11/17
Superhero Lady Armwrestlers of Portland - Welcome Back To Camp Wannabrawl - 8/17
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Thompson's Point
Thompsons Point Listing
Port City Listing
Port City Calendar
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State Theatre Portland
Boston.com – February 26, 2011 – "part of the core of Portland's Arts District."
The Portland Press Herald – December 26, 2010 – "an impressive lineup of national acts"
Maine Magazine – November 11, 2010 – "This is a beautiful, old, historic theater. Its got great bones, great history, and amazing character."
The Bowdoin Orient – October 22, 2010 – "…epitomizes traditional theatrical elegance without compromising the casual aesthetic—something that has been lacking in Portland's music venues."
The Bangor Daily News – October 21, 2010 – "The pictures of the renovations at this historic Portland venue look absolutely gorgeous…"
Relix – October 18, 2010 – "The acoustics of the intimate venue (between 1450 and 1680 capacity) enhanced the warm, immediate sound coming from the speakers…" – opening night with My Morning Jacket, October 15, 2010
Portland, Maine Daily Photo – October 81, 2010 – "The venue is truly a gem…"
Dangerous Events Inc – October 18, 2010 – "splendid, elegant…"
Brooklyn Vegan – October 18, 2010 – "The encore was a thing of perfection for me, but hell the whole show was perfect!"
The Theatre Historical Society of America – October 18, 2010
The Daily Swarm – October 18, 2010
The Portland Press Herald – October 13, 2010 – "glimmering"
Nylon Magazine – October 1, 2010 – "majestic"
The New York Times – August 8, 2010 – "a live music anchor…"
The Portland Press Herald – July 28, 2010 – "It's exciting to be part of a project that will eventually help make a unique and amazing connection between artists and musicians and their audience, and to be able to witness this kind of connection happen at such a historic and beautiful venue makes it even more exciting."
The New York Times – May 27, 2010
Angioplasty – May 26, 2010 – "a jewel in the Portland-area music scene and a legacy in New England."
Bangor Daily News – May 26, 2010
Brooklyn Vegan – May 26, 2010
Hilly Town – May 26, 2010 – "a critical spot for a lot of the bigger names in music"
Live Work Portland – May 26, 2010 – "…we fell in love with the [Portland] and saw the potential in the theatre that will appeal to Portland's creative community."
WCYY – May 26, 2010
The Portland Phoenix – May 26, 2010
The Sun Journal – May 26, 2010
The Portland Press Herald – May 26, 2010 – "It's about time. The State is back."
Original Portland – May 26 2010
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**TONIGHT JULY 17** Silversun Pickups with I Dont Know How but They Found Me. DOORS 7PM, SHOW 8PM. Tickets availabl… https://t.co/qOm94cSN4W about 1 hour ago
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The bright crescent of Enceladus slides past distant Rhea in this mutual event, or occultation, movie from Cassini.
The movie was created from 40 original images taken over the course of about 20 minutes as the spacecraft's narrow angle camera remained pointed toward Enceladus. Additional frames were inserted between the 40 Cassini images in order to smooth the appearance of the moons' movement - a scheme called interpolation.
As Enceladus (504 kilometers, 313 miles across) approaches Rhea (1,528 kilometers, 949 miles across), the shadow of Saturn can be seen, projected onto the unlit side of its incredible rings.
The images were taken in visible light with the Cassini spacecraft narrow-angle camera on March 2, 2006 at a distance of approximately 2.8 million kilometers (1.7 million miles) from Rhea and 2 million kilometers (1.2 million miles) from Enceladus. Image scale is 17 kilometers (11 miles) per pixel on Rhea and 12 kilometers (7 miles) per pixel on Enceladus.


Take a moment to browse through our gallery of custom candle labels for our candles and expressive word labels! You can personalize these labels with fun quotes, names and more in font varieties and styles of your choosing! You can find more of our label designs in our store.


Shire is known as one of the world's fastest growing specialty pharmaceutical companies. Today, it has value around US $5billion and is the third largest pharma company in the UK. It is the market leader in attention deficit and hyperactivity disorder (ADHD) in the United States, as well as having a growing European network and an exciting set of products being developed for launch and marketing in the future. But only 27 years ago was it a tiny business above a shop. How did it become so successful?
In the beginning, Shire was a tiny company called AimCane, founded and funded in 1986 in the UK by four entrepreneurs. These four entrepreneurs were Dennis Stephens, Harry Stratford, Peter Moriarty, and Geoff Hall. Each one had considerable experience either in or directly serving the pharmaceutical industry.
AimCane was situated in a shop in Hampshire and when the company came to change its name, it chose Shire, after the name of the building that housed the company in those early days.
Initially, AimCane launched and sold a range of calcium products designed to help with the bone disease, osteoporosis. One of these products, called Calcichew, is still being successfully marketed today.
Then, under its new name Shire, the company expanded its specialism to include hormone replacement therapy (HRT) and Alzheimer's disease. It began trading fully in 1987.
In 1995, Shire made its first acquisition of another Hampshire-based business called Imperial Pharmaceutical Services. This acquisition brought to Shire a range of product licenses as well as capital that enabled the company to embark on expansion and a series of further acquisitions.
The same year, Shire entered into a global co-development agreement with Johnson and Johnson Pharmaceutical Research and Development to develop Reminyl (galantamine) for the treatment of Alzheimer's disease in Japan. Today, this product has made millions for Shire.
The company was floated on the London Stock Exchange in February 1996. This was a significant moment in Shire's history as it reflected its ambitions to invest, to expand and to become an international force in the industry while still retaining a point of difference from the other big pharmaceutical giants.
Shire then, in 1997, acquired two companies. The first was Pharmavene, a company with a broad portfolio of drug delivery and screening technologies. The second was Richwood, which ultimatelylead to the launch of ADHD treatments, Adderall (amphetamine and dextroamphetamine) and Adderall XR. Together, these two acquisitions led to the creation of US-based Shire Laboratories and facilitated Shire's access to the US market.
By 1999, Shire had effectively doubled in size, following the merger with Roberts. With this merger, Shire gained the anti-inflammatory medication, Pentasa, for ulcerative colitis, as well as ProAmatine for low blood pressure and Agrylin for low blood platelet levels. Shire then opened its marketing capabilities in Europe even further, by acquiring Fuisz EU.
2001 saw Shire merge with Canadian pharma company, BioChem Pharma. In 2003, Shire granted its strategic alliance partner, Bayer Yakuhin Ltd., an exclusive license to develop and market chronic kidney disease drug Fosrenol (Lanthanum carbonate) in Japan. This drug was launched into the Japanese market in 2009.
Shire began broadening into rare diseases with the acquisition of orphan drug company, TKT, which later became Shire Human Genetic Therapies. Two years previously, TKT signed an exclusive licensing agreement with Genzyme Corporation to market Elaprase (idursulfase), which is a drug used to treat Hunter syndrome (also called MPS-II), in Asian markets and South Africa.
Shire acquired New River, gaining control of Vyvanse, in 2007 – a move which also broadened its products pipeline and technology platform. As did the acquisition of German pharma company, Jerini AG, a year later, which brought Firazyr into its portfolio. The same year, 2008, Shire opened an office in Japan.
In order to enter the ADHD market outside of the US, Shire acquired the worldwide rights to Equasym from UCB, excluding the US, Canada and Barbados. This was in 2009.
By 2010, Shire had added to its gastrointestinal portfolio through the acquisition of Movetis NV. This gave the company constipation treatment, Resolor (prucalopride).
The year 2011 saw the acquisition of Advanced BioHealing, which brought Dermagraft, a treatment for diabetic foot ulcers, into Shire's portfolio, as well as the creation of a new strategic platform. This platform was subsequently renamed as Shire Regenerative Medicine. The same year, Shire entered into a partnership with Shionogi & Co to co-develop Vyvanse and Intuniv in Japan.
Shire acquired Ferrokin BioSciences, in 2012, adding an iron chelator therapy in development for the treatment of chronic iron overload following numerous blood transfusions, to Shire's Specialty Pharmaceuticals pipeline. The company also acquired substantially all the assets of Pervasis Therapeutics including certain vascular repair technologies.
This year, Shire has already acquired three companies. The first was privately held biotech Lotus Tissue Repair, which is developing the first and only protein replacement therapy currently being investigated for the treatment of dystrophic epidermolysis bullosa (DEB). The second, was Premacure AB, allowing Shire's HGT business to continue the ongoing development a protein replacement therapy for the treatment of retinopathy of prematurity (ROP). The third and final acquisition of 2013 has been with SARcode Bioscience, expanding Shire's presence in Ophthalmology.
Through all these acquisitions over the past two decades, Shire has had one aim: "to achieve leading market positions in each of [its] target therapeutic areas while imagining and leading the future of healthcare, enabling people with life-altering conditions to lead better lives, thus creating value for society".
Chief executive in the crucial early 2000s period was Matthew Emmens, who steered the company through growth from 2003 until 2008, handing over to former CFO Angus Russell, and moving to the chairman of the Board role until 2014.
Angus Russell left the company in 2012, just as key drug, ADHD treatment Adderall XR was facing generic competition.
Succeeding him was Flemming Ornskov took on the position of chief executive, joining the company from Bayer, where he had headed up the Germany company's general and speciality medicine marketing.
Ornskov set about aggressive expansion through a string of M&A deals.
Under Emmens and Ornskov's guidance, Shire continued to focus on the following four areas: Neuroscience, Rare Diseases, Gastrointestinal and Regenerative Medicine, as well as early development projects in additional therapeutic areas. In Neuroscience.


text:
Pauk () is a town in Pauk Township, Pakokku District, Magway Region, in north-west Myanmar. It is the administrative center for Pauk Township. Pauk is located on the western bank of the Kyaw River just above its intersection with the Yaw River.

The current town was founded in 1840.  The local airport ICAO designation is VYPK (IATA code: PAU).

Notes

Further reading
 Winston, W. R. (1892) Four years in Upper Burma C.H. Kelly, London, pages 241-242,

External links
 "Pauk
 Map — Satellite Im
 ages of Pauk" Maplandia

Township capitals of Myanmar
Populated places in Magway Region

redpajama_set_name: RedPajamaWikipedia


Did you know that what you eat can affect your medications? Kelly Groth, Manager of Memorial's Anti-Coagulation Management Services (Coumadin Clinic) discusses the effects of Coumadin on the blood and how diet can affect it. Groth also discussed the relocation of the Coumadin Clinic from the hospital to the lower level of the Cornerstone Medical building at 402 S. 12th Ave in Yakima.
This entry was posted in News and tagged blood, coumadin. Bookmark the permalink.


Jeff Passan of Yahoo Sports reports that the Braves have agreed to terms on a four-year contract with free agent outfielder Nick Markakis. Dan Connolly of the Baltimore Sun says the deal is for $44 million.
Markakis has a lot of name recognition and he put together a few nice seasons with the Orioles, but the 31-year-old owns a troubling .274/.335/.371 batting line in 1,410 plate appearances since the beginning of the 2013 campaign. He won the 2014 American League Gold Glove Award for right field, but his defensive metrics are far from sparkling.
What exactly are the Braves doing?