\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 270, pp. 1--19.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/270\hfil Schr\"odinger Strichartz estimate]
{Remarks on the sharp constant for the Schr\"odinger Strichartz
estimate and applications}

\author[A. Selvitella \hfil EJDE-2015/270\hfilneg]
{Alessandro Selvitella}

\address{Alessandro Selvitella \newline
Department of Mathematics \& Statistics,
McMaster University,
Hamilton Hall, Office 401HH,
1280 Main Street West,
Hamilton, Ontario,
Canada L8S 4K1}
\email{aselvite@math.mcmaster.ca}

\thanks{Submitted October 14, 2014. Published October 21, 2015.}
\subjclass[2010]{35Q41, 35A23}
\keywords{Strichartz estimate; optimal constant;
 Schr\"odinger equation; \hfill\break\indent restriction inequality}

\begin{abstract}
 In this article, we compute the sharp constant for the homogeneous
 Schr\"{o}dinger Strichartz inequality, and for the Fourier restriction
 inequality  on the paraboloid in any dimension under the condition conjectured
 (and proved for dimensions 1 and 2) that the maximizers are Gaussians.
 We observe also how this would imply a far from optimal, but ``cheap''
 and sufficient, criterion of the global wellposedness in the $L^2$-critical case
 $p=1+4/n$.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{definition}[theorem]{Definition}
\allowdisplaybreaks


\section{Introduction}

Consider the nonlinear Schr\"{o}dinger equation (NLS for short)
 \begin{equation}\label{NLS0}
i\partial_tu(t,x)+\Delta u(t,x) +\mu |u|^{p-1}u(t,x)=0 \quad
(t,x) \in (0,\infty) \times \mathbb{R}^n,
\end{equation}
 with initial datum $u(0,x)=u_0(x), \quad x \in \mathbb{R}^n$.
 Here the space dimension $n \geq 1$, the nonlinearity has $p \geq 1$ and
 $\mu=-1,0,1$ in which cases the equation is said to be \emph{defocusing,
 linear} and \emph{focusing} respectively.

 A lot of research has been done to prove the global wellposedness of the above
problem in the scale of Hilbert Spaces
 $H^s(\mathbb{R}^n)$ (see Section \ref{Preliminaries} for a precise definition).
In the case of regular solutions $s >n/2$, the
algebra property of the space $H^s(\mathbb{R}^n)$ makes the proof simpler,
while in the case $s \leq n/2$ one needs
 Strichartz estimates to close the argument (see again Section \ref{Preliminaries}).
 We refer to \cite{Tao1} for more  details and references.

Strichartz estimates were originally proved by Strichartz \cite{Strichartz}
in the non end-point case and much  later for the end-point case by
Keel and Tao \cite{KeelTao} in the homogeneous case and by Foschi \cite{Foschi2}
 in the inhomogeneous case, following Keel and Tao's approach.
After Strichartz's work, a  research field opened and Strichartz estimates
 were proved for a lot of different equations.
See \cite{Tao1} and the references therein, for a more complete discussion
on Strichartz estimates.

 Several mathematicians have then been interested in the problem of the sharpness
of Strichartz Inequalities.
As far as we know, the first one addressing this problem has been
Kunze \cite{Kunze2}, who proved the existence of a maximizing function for the
 estimate
\[
\|e^{it\partial^2_x}u\|_{L^6_{t,x}(\mathbb{R}^2)}
\leq S_h(1) \|u\|_{L^2(\mathbb{R})}
\]
(case of dimension $n=1$), by means of the concentration compactness principle
used in the Fourier space and by means of multilinear
estimates due to Bourgain \cite{Bourgain1}. This method has been first
developed by him in relation to a
variational problem from nonlinear fiber optics on Strichartz-type estimates
 \cite{Kunze1}. The first author to give explicit values
of the sharp Strichartz constants and characterize the maximizers has been
Foschi \cite{Foschi1}, who proved that
in dimensions $n=1$ the sharp constant is $S_h(1)={12}^{-1/12}$,
while in dimension $n=2$ the sharp constant is
$S_h(2)=2^{-1/2}$. He also proved that the maximizer is the Gaussian function
$f(x)=e^{-|x|^2}$ (up to symmetries) in both dimensions $n=1$ and $n=2$
(see Section \ref{Preliminaries} below). He moreover conjectured
(Conjecture 1.10) that Gaussians are maximizers
in every dimension $n \geq 1$. Independently, this result has been reached
also by Hundertmark and Zharnitsky in \cite{HZ} that gave also a conjecture
on the value of the Strichartz Constant (Conjecture 1.7).
An extension of these results can be found in \cite{Carneiro}.
A step towards proving Foschi's conjecture has been done by Christ
and Quilod\'{a}n \cite{Christ1}, who demonstrated that
Gaussians are critical points in any dimension $n \geq 1$.
They do not give any conjecture on the explicit value of the sharp Strichartz
constant $S_h(n)$ for general dimension $n$. Duyckaerts, Merle and Roudenko
in \cite{DMR} give an estimate
of $S_h(n)$ and also precise asymptotics in the small data regime,
but not the explicit value. Here, assuming that Gaussians are actually maximizers,
as it is conjectured, and not just critical points, we compute the Strichartz
Constant in a setting a little more general than the one of the conjecture
of Hundertmark and Zharnitsky \cite{HZ} and this is the main contribution of
the paper.

\begin{theorem} \label{StrichartzConstant}
Suppose Gaussians maximize Strichartz estimates for any $n \geq 1$.
Then, for any $n \geq 1$ and  $(q,r)$ admissible pair
(see Section \ref{Preliminaries} below), the
\emph{sharp homogeneous Strichartz constant}  $S_h(n,q,r)=S_h(n,r)$
defined by
\begin{equation}
S_h(n,r):= \sup \Big\{ \frac{\|u\|_{L^q_tL^r_x(\mathbb{R} \times
\mathbb{R}^n)}}{\|u\|_{L_x^2(\mathbb{R}^n)}} :
u \in L_x^2(\mathbb{R}^n), u \neq 0 \Big\},
\end{equation}
is given by
\begin{equation}\label{CostanteGenerale}
S_h(n,r)= 2^{\frac{n}{4}-\frac{n(r-2)}{2r}}r^{-\frac{n}{2r}}.
\end{equation}
Moreover, if we define $S_h(n):=S_h(n,2+4/n,2+4/n)$ by
\begin{equation}
S_h(n)= \sup \Big\{ \frac{\|u\|_{L_{t,x}^{2+4/n}(\mathbb{R}
\times \mathbb{R}^n)}}{\|u\|_{L_x^2(\mathbb{R}^n)}} :
u \in L^2(\mathbb{R}^n), u \neq 0 \Big\},
\end{equation}
then for every $n \geq 1$ we have that
 \begin{equation}
S_h(n)=\Big( \frac{1}{2}(1+\frac{2}{n})^{-n/2}\Big)^{\frac{1}{2+4/n}};
\end{equation}
$S_h(n)$ is a decreasing function of $n$ and
$$
S_h(n)\to \frac{1}{(2e)^{1/2}}, \quad n \to +\infty.
$$
For any $n \geq 1$ and  $(\tilde{q},\tilde{r})$ admissible pair,
the \emph{sharp dual homogeneous Strichartz constant}  $S_d(q,r,n)=S_d(n,r)$ is
defined by
\begin{equation}
S_d(n,r):= \sup \Big\{ \|\int_{\mathbb{R}} e^{is\Delta} F(s)ds\|_{L^2_x}
{\|F\|_{L^{\tilde{q}'}_tL^{\tilde{r}'}_x}} : F \in L^{\tilde{q}'}_tL_x^{\tilde{r}'}
(\mathbb{R} \times \mathbb{R}^n), F \neq 0
\Big\},
\end{equation}
We have that $S_h(n,r)=S_d(n,r)$.
\end{theorem}

\begin{remark} \label{rmk1.2} \rm
We notice that $q$ and $r$ are not independent
since they are an admissible pair. For this reason,  $q$ appears in $S(n,r)$
just as a function of $r$. One could have also
expressed the sharp constant as a function of $q$ by
$$
S_h(n,q)= 2^{-\frac{1}{q}} \Big( 1-\frac{4}{qn} \Big)^{-1/q+n/4},
$$
since $r=\frac{2qn}{nq-4}$ (just plug this expression inside $S_h(n,r)$).
\end{remark}

\begin{figure}[htb]
\begin{center}
\setlength{\unitlength}{1mm} \scriptsize
\begin{picture}(65,45)(0,0)
\put(5,5){\vector(1,0){60}}
\put(5,5){\vector(0,1){40}}
\put(63,7){$n$}
\put(7,43){$S_h(n)$}
\multiput(17,4)(12,0){4}{\line(0,1){2}}
\put(16,1){20}
\put(28,1){40}
\put(40,1){60}
\put(52,1){80}
\multiput(4,12)(0,13){3}{\line(1,0){2}}
\put(0,11.5){0.5}
\put(0,24.5){1.0}
\put(0,37.5){1.5}
\qbezier(6,23)(7,15)(13,14)
\qbezier(13,14)(36,13.5)(60,13)
\put(13,20){$S_h(n)=\big(\frac{1}{2}(1+\frac{2}{n})^{-n/2}\big)^{\frac{1}{2+4/n}}$}
\end{picture}
\end{center}
\caption{Homogeneous Strichartz constant in the
case $q=r=2+4/n$, $n\geq 1$.}
\label{fig1}
\end{figure}

\begin{remark} \label{rmk1.3} \rm
We can see that, for $n=1$ and $n=2$, we recover the values of $S_h(n)$ found by
 Foschi in \cite{Foschi1}.
\end{remark}

\begin{remark} \label{rmk1.4} \rm
 The asymptotic behavior of $S_h(n)$ basically says that in the non compact
 case of $\mathbb{R}^n$, the increase of the spatial dimension $n$ allows more
dispersion, but  the rate of dispersion, measured by the homogeneous Strichartz estimate, does not increase indefinitely. We believe that a similar phenomenon
should appear in the case of the Schr\"{o}dinger equation on the hyperbolic space.
 We think that it might not be the case for manifolds which become more and more
negatively curved with the increase of the dimension, in which case we might
observe an indefinitely growing dispersion rate.
\end{remark}

The knowledge of the Optimal Strichartz Constant gives a more precise upper bound
on the size of the $L^2$-norm for which the ``cheapest argument''
(standard Duhamel Principle) gives global wellposedness for \eqref{NLS0} in
the $L^2$-critical case $p=1+4/n$.
From now on we will concentrate on the case $s=0$ (note $0<n/2$ for every $n >0$),
namely we will consider just the case in which the initial datum
$u_0(x) \in  L^2(\mathbb{R}^n)$ and just the case of not \emph{supercritical}
nonlinearities $1<p \leq 1+4/n$. In the \emph{subcritical} case $1< p <1+4/n$,
 Tsustsumi \cite{Tsutsumi1} proved local wellposedness and also global
wellposedness due to the fact that the local time
 of existence given by his strategy depends just on the $L^2$-norm of the
initial datum and that the NLS have a
 conservation law at the $L^2$-regularity
($T_{\rm loc}=T_{\rm loc}(\|u_0\|_{L^2(\mathbb{R}^n)}))$.
Also in the \emph{critical}  case,
Tsutsumi proved local wellposedness,  thanks to the global bound of the
$L^{2(n+2)/n}_{t,x}$ Strichartz Norm (see Section \ref{Preliminaries}),
but now the conservation law could not lead to global existence because the
local existence time depends on the profile of the solution
($T_{\rm loc}=T_{\rm loc}(u_0)$). The problem of global
wellposedness for the NLS, in the $L^2$-critical case in  any dimension,
has been solved just recently in a series of
papers by Dodson (see \cite{Dodson1}, \cite{Dodson2}, \cite{Dodson3}).
However if the initial datum is ``sufficiently small'' in $L^2_x$ then
one can get global existence with the argument developed in \cite{Tsutsumi1},
namely by a straight contraction mapping argument.
Here, we give a more precise estimate of this ``sufficiently small''
 and so we have the following theorem.

\begin{theorem}\label{GWP}
Consider equation \eqref{NLS0} with initial datum
$u_0(x) \in L^2_x(\mathbb{R}^n)$ satisfying the following bound
\begin{equation} \label{UpperBound}
\|u_0(x)\|_{L^2_x}<\frac{1}{S_h(n,r) \alpha}
\Big(\frac{1}{ S_i(n,r)}-\frac{1}{S_i(n,r)\alpha} \Big)^{n/4}
\end{equation}
with $\alpha=2$  if $n \geq 4$ and  $\alpha=1+n/4$ for $1 \leq n \leq 4$.
Here $S_h(n,r)$ and $S_i(n,r)$ are, respectively, the sharp homogeneous and
inhomogeneous Strichartz constants. Then, there is a unique global solution
$u(t,x) \in L^2_x(\mathbb{R}^n)$ for every $t \geq 0$.
\end{theorem}

\begin{remark} \label{rmk1.6} \rm
This result reminds a bit what happens in the focusing case, in which there
is an upper bound on the size of the $L^2$-norm of the initial datum for
 which one can get global well-posedness and condition \eqref{UpperBound}
reminds the Gagliardo-Nirenberg Inequality (see \cite{Weinstein1} and \cite{Tao1}).
 Anyways, we want to make clear that condition \eqref{UpperBound} is in some
sense fictitious and it is not a threshold, since, for example, the results
of Dodson \cite{Dodson1,Dodson2,Dodson3}.
\end{remark}

Strichartz inequalities can be set in the more general framework of
 Fourier restriction inequalities in harmonic analysis. This connection has
been made already clear in the original paper of Strichartz \cite{Strichartz}.
Therefore, Theorem \ref{StrichartzConstant} can be rephrased in this framework.

\begin{theorem} \label{RestrictionTheorem}
Fix $n \geq 1$ and consider the paraboloid $(\mathbb{P}^n,dP^{n})$ defined
in \eqref{paraboloid} and \eqref{measure} below. Suppose Gaussians maximize
the Fourier restriction inequality
\begin{equation}\label{FRI}
 \|\widehat{fdP^{n}}\|_{L_{t,x}^{\frac{2(n+2)}{n}}(\mathbb{R}^{n+1})}\leq S_h(n) \|f\|_{L^2(\mathbb{P}^n,dP^{n})}
\end{equation}
Then, the sharp constant $S_h(n)$ is
$$
S_h(n)=\Big( \frac{1}{2}\Big(1+\frac{2}{n}\Big)^{-n/2}\Big )^{\frac{1}{2+4/n}}.
$$
\end{theorem}

The remaining part of the paper is organized as follows.
In Section \ref{Preliminaries}, we fix some notation and
collect some preliminary results, about the Fourier transform and the
fundamental solution for the linear Schr\"{o}dinger equation, about
the Strichartz estimates and their symmetries and the main results in the
literature about maximizers for the Strichartz inequality and about the sharp
Strichartz constant.
In Section \ref{proofThm1}, we prove Theorem \ref{StrichartzConstant}, while,
in Section \ref{proofThm2}, we prove
Theorem \ref{GWP}. In Section \ref{FourierRestriction}, we discuss the connection
between Strichartz and restriction inequalities, proving
Theorem \ref{RestrictionTheorem} in Subsection \ref{ProofOfRestrictionTheorem}.
In the appendix, we give some further comments on the inhomogeneous Strichartz
estimate and on the wave equation.

\section{Preliminaries}\label{Preliminaries}

By Schwartz functions we mean functions belonging to the  function space
$$
\mathcal{S} (\mathbb{R}^n) := \{ f \in C^\infty(\mathbb{R}^n) :
  \|f\|_{\alpha,\beta} < \infty\quad \forall \alpha, \beta  \},
 $$
with $\alpha$ and $\beta$  multi-indices, endowed with the following norm
$$
\|f\|_{\alpha,\beta}=\sup_{x\in\mathbb{R}^n} |x^\alpha D^\beta f(x) |.
$$
Let $(X, \Sigma, \mu)$ be a measure space. For $1 \leq p \leq +\infty$,
we define the space $L^p(X)$ of all
measurable functions from $f:X \to \mathbb{C}$ such that
$$
\|f\|_{L^p(X)} := \Big({\int_X |f|^p\;d\mu}\Big)^{1/p}<\infty.
$$
Consider $f:\mathbb{R}^n \to \mathbb{C}$ a Schwartz function in space and
$F(t,x):\mathbb{R} \times \mathbb{R}^n \to \mathbb{C}$ a Schwartz function
in space and time. We will use the following notation (and constants)
for the space Fourier transform
$$
\hat{f}(t,\xi)=\frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n}e^{-ix\cdot \xi}f(x)dx
$$
and for the Inverse space Fourier transform
$$
f(x):=\frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n}e^{ix\cdot \xi}\hat{f}(\xi) d\xi,
$$
and the following for the space-time Fourier transform
$$
\mathcal{F}(F)(\tau, \xi):=\frac{1}{(2\pi)^{\frac{n+1}{2}}}
\int_{\mathbb{R}^n}e^{-it\tau-ix\cdot \xi}f(t,x)\,dx\,dt
$$
and the Inverse space-time Fourier transform
$$
F(t,x):=\frac{1}{(2\pi)^{\frac{n+1}{2}}} \int_{\mathbb{R}^{n+1}}
e^{it \tau + ix\cdot \xi}\mathcal{F}(\tau, \xi) d\xi d\tau.
$$
By means of the Fourier transform, we can finally define $H^s$-spaces as the
set of functions such that
$$
\|u\|_{H^s(\mathbb{R}^n)}:=
 \Big( \int_{\mathbb{R}^n} |\hat{u}(\xi)|^2(1+|\xi|)^{2s} \Big)^{1/2}
< +\infty.
$$

\subsection{Fourier transform and fundamental solutions for
 linear Schr\"{o}dinger equations} \label{FourierSolution}

In this subsection we solve the  linear Schr\"odinger equation
 \begin{equation}\label{LS}
i\partial_tu(t,x)=\Delta u(t,x), \quad (t,x) \in (0,\infty) \times \mathbb{R}^n,
\end{equation}
with initial datum $u_0(x)=e^{-|x|^2} \in \mathcal{S} (\mathbb{R}^n) $.
These computations are well known, but we will rewrite them
 here in order to clarify what we will compute in the next sections.
Since $u_0(x) \in \mathcal{S} (\mathbb{R}^n)$, then also
 $\partial_tu(t,x) \in \mathcal{S} (\mathbb{R}^n)$ and
$\Delta u(t,x) \in \mathcal{S} (\mathbb{R}^n)$. So
 we can apply the Fourier transform to both sides of \eqref{LS} and get:
 $$
i\hat{u}_t=-|\xi|^2\hat{u},
$$
whose solution is
$$
\hat{u}(\xi,t)=e^{i|\xi|^2t}\hat{u}(\xi,0).
$$
So we just need to compute the Fourier transform of the initial datum
and then the inverse Fourier transform of $\hat{u}(t,\xi)$ to get the
explicit form of the solution.
\begin{align*}
\hat{u}(0,\xi)
&=\frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n}e^{-ix\cdot \xi}u(0,x)dx\\
&=\frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n}e^{-ix\cdot \xi} e^{-|x|^2}dx \\
&= \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n}e^{-(|x|^2+ix\cdot \xi-|\xi|^2/4)}
 e^{-|\xi|^2/4}dx \\
&=\frac{e^{-|\xi|^2/4}}{(2\pi)^{n/2}} \int_{\mathbb{R}^n}e^{-|x-i\xi/2|^2}dx,
\end{align*}
by using contour integrals. We notice that, with a simple change of variables,
we have:
$$
2^{n/2}\int_{\mathbb{R}^n}e^{-|x-i\xi/2|^2}dx
=2^{n/2}\int_{\mathbb{R}^n}e^{-|x|^2}dx
=\int_{\mathbb{R}^n}e^{-|x|^2/2}dx=(2\pi)^{n/2}.
$$
Hence
\[
 \hat{u}(0,\xi)= \frac{e^{-|\xi|^2/4}}{(2\pi)^{n/2}} \pi^{n/2}
=\frac{e^{-|\xi|^2/4}}{2^{n/2}}.
\]
With this we can conclude that

\begin{align*}
u(t, x)
&= \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n}e^{i|\xi|^2t+ ix\cdot \xi}
\frac{e^{-|\xi|^2/4}}{2^{n/2}} \\
&= \frac{1}{2^n}\frac{1}{\pi^{n/2}} \int_{\mathbb{R}^n}e^{-|\xi|^2(1/4-it)
 +ix \dot \xi}d \xi\\
&=  \frac{1}{2^n}\frac{1}{\pi^{n/2}} \int_{\mathbb{R}^n}e^{-(|\xi|^2(1/4-it)
 -ix \dot \xi -|x|^2/(1-4it))}e^{-|x|^2/(1-4it)}d \xi\\
&= \frac{1}{2^n}\frac{1}{\pi^{n/2}} e^{-|x|^2/(1-4it)}
 \int_{\mathbb{R}^n}e^{-|\xi \sqrt{1/4-it}+ix/(\sqrt{1-4it})|^2}d\xi.
\end{align*}
Now we make the change of variables $\eta=\xi \sqrt{1/4-it}+ix/(\sqrt{1-4it})$ to get
\begin{align*}
u(t,x)&= \frac{1}{2^n}\frac{1}{\pi^{n/2}} e^{-|x|^2/(1-4it)}
 \int_{\mathbb{R}^n}e^{-|\eta|^2}(1/4-it)^{-n/2}d \eta\\
&= \frac{1}{2^n}\frac{1}{\pi^{n/2}} e^{-|x|^2/(1-4it)} (1/4-it)^{-n/2}\pi^{n/2}\\
&=(1-4it)^{-n/2}e^{-\frac{|x|^2}{1-4it}}
\end{align*}
Hence
\begin{equation}\label{FundamentalSolution}
 u(t,x)=(1-4it)^{-n/2}e^{-\frac{|x|^2}{1-4it}}.
\end{equation}

\subsection*{Strichartz estimates and their symmetries}

In this subsection, we state the Strichartz estimates for the Schr\"odinger
 equation, since they are the main topic of
the present paper and it will help to clarify the statement of our main theorems.

\begin{definition}\label{admissible} \rm
Fix $n \geq 1$. We call a set of exponents $(q,r)$ admissible if
$2 \leq q,r \leq +\infty$ and
$$
\frac{2}{q}+\frac{n}{r}=\frac{n}{2}.
$$
\end{definition}


\begin{proposition}[\cite{KeelTao,Foschi2,Strichartz}]
\label{StrichartzEstimate}
 Suppose $n \geq 1$. Then, for every $(q,r)$ and $(\tilde{q}, \tilde{r})$
admissible and for every $u_0 \in L^2_x(\mathbb{R}^n)$ and
$F \in L^{\tilde{q}'}_tL^{\tilde{r}'}_x(\mathbb{R} \times \mathbb{R}^n)$,
the following hold:
\begin{itemize}
  \item the \emph{homogeneous Strichartz estimates}
$$
| |e^{-it\Delta}u_0||_{L^q_tL^r_x}  \leq S_h(n,q,r) \|u_0\|_{L^2_x};
$$

\item the \emph{dual homogeneous Strichartz estimates}
$$
\big\|\int_{\mathbb{R}} e^{is\Delta}F(s)ds\big\|_{L^2_x}
\leq S_d(n, q, r) \|F\|_{L^{\tilde{q}'}_tL^{\tilde{r}'}_x} ;
$$

 \item the \emph{Inhomogeneous Strichartz estimates}

$$ \big\|\int_{s<t} e^{-i(t-s)\Delta}F(s)ds\big\|_{L^q_tL^r_x}
\leq S_i(n, q,r, \tilde{q},\tilde{r})  \|F\|_{L^{\tilde{q}'}_tL^{\tilde{r}'}_x}.
$$
 \end{itemize}
\end{proposition}

As explained for example in \cite{Foschi1}, Strichartz estimates are invariant
 by the following set of symmetries.

\begin{lemma}[\cite{Foschi1}] \label{Symmetries}
 Let $\mathcal{G}$ be the group of transformations generated by:
  \begin{itemize}
  \item space-time translations:
    $u(t, x) \mapsto u(t + t_0, x + x_0)$,
    with $t_0 \in \mathbb{R}$, $x_0 \in \mathbb{R}^n$;
  \item parabolic dilations:
    $u(t, x) \mapsto u(\lambda^2 t, \lambda x)$,
    with $\lambda > 0$;
  \item change of scale:
    $u(t, x) \mapsto \mu u(t, x)$, with $\mu > 0$;
  \item space rotations:
      $u(t, x) \mapsto u(t, R x)$,
      with $R \in SO(n)$;
  \item phase shifts:
      $u(t, x) \mapsto e^{i \theta} u(t, x)$,
      with $\theta \in \mathbb{R}$;
  \item Galilean transformations:
    \begin{equation*}
      u(t, x) \mapsto
      e^{ \frac{i}{4} \Big (  |v|^2 t + 2 v \cdot x \Big )}
      u(t, x + t v),
    \end{equation*}
    with $v \in \mathbb{R}^n$.
  \end{itemize}
Then, if $u$ solves equation \eqref{LS} and $g \in \mathcal{G}$,
also $v = g \circ u$ solves equation \eqref{LS}.
Moreover, the constants $S_h(n,q,r)$, $S_d(n,q,r)$ and
$S_i(n, q,r, \tilde{q},\tilde{r})$ are left unchanged by the action
of $\mathcal{G}$.
\end{lemma}

\begin{remark} \label{rmk2.4} \rm
For Strichartz estimates for different equations and different regularities,
we refer to \cite{Tao1}.
\end{remark}


\subsection*{Previous results on sharp Strichartz constant and maximizers}

Here we collect the results concerning the optimization of Strichartz
inequalities that we need for the next sections.
For a broader discussion, we refer to \cite{Tao2} and the references therein.

\begin{proposition}[\cite{Kunze2,Christ1,Foschi1}] \label{Known}
For any $n \geq 1$ and  $(q,r)$ admissible pair, we define
$S_h(n):=S_h(n,2+4/n,2+4/n)$ by
\begin{equation}
S_h(n):= \sup \Big\{ \frac{\|u\|_{L_{t,x}^{2+4/n}(\mathbb{R}
\times \mathbb{R}^n)}}{\|u\|_{L^2(\mathbb{R}^n)}} :
u \in L^2(\mathbb{R}^n), u \neq 0 \Big\}.
\end{equation}
Then we have the following results:
\begin{itemize}
 \item Radial Gaussians are critical points of the homogeneous Strichartz
inequality in any dimension $n \geq 1$ for all admissible pairs
$(q,r) \in (0, +\infty) \times (0, +\infty)$;

 \item The explicit sharp Strichartz constants $S_h(n)$ can be computed
explicitly in dimension
 $n=1$: $S_h(1)=12^{-1/12}$; and dimension $n=2$: $S_h(2)=2^{-1/2}$.
Moreover, in both the cases $n=1$ and $n=2$, the maximizers are Gaussians.
\end{itemize}
\end{proposition}

\section{Proof of Theorem \ref{StrichartzConstant}} \label{proofThm1}

We are now ready to prove Theorem \ref{StrichartzConstant}.
We assume, as conjectured, that radial Gaussians are mazimizers and not just
critical points as proved in \cite{Christ1}.
So we will take $u_0(x)=e^{-|x|^2}$. By Lemma \ref{Symmetries}, the choice of
the Gaussian is done without loss of generality. We start to compute the
$L^2$-norm of the initial datum and so of the solution:
\begin{align*}
\|u(t,x)\|_{L^2_x}
&= \|u_0(x)\|_{L^2_x}
=\Big ( \int_{\mathbb{R}^n} e^{-2|x|^2}dx \Big )^{1/2}\\
&=\Big ( \int_{\mathbb{R}^n} e^{-2|x|^2/4}2^{-n}dy  \Big )^{1/2}\\
&= 2^{-n/2}\Big ( \int_{\mathbb{R}^n} e^{-|x|^2/2}dy  \Big )^{1/2} \\
&= 2^{-n/2} (2\pi)^{n/4}=\Big(\frac{\pi}{2}\Big)^{n/4}
\end{align*}
by similar computations as in Subsection \ref{FourierSolution}.

Now we compute the $L^q_tL^r_x$-norm of the linear solution
$$
u(t,x)=(1-4it)^{n/2}e^{-\frac{|x|^2}{1-4it}}.
$$
First
\begin{align*}
|u(t,x)|^r
&=|1-4it|^{-rn/2}|e^{-\frac{|x|^2}{1-4it}} |^r\\
&=|1+16t^2|^{-rn/4}| e^{-\frac{(1+4it)|x|^2}{1+16t^2}} |^r\\
&=|1+16t^2|^{-rn/4}e^{-\frac{r|x|^2}{1+16t^2}}.
\end{align*}
Then
\[
\|u(t,x)\|^r_{L^r_x}=|1+16t^2|^{-rn/4} \int_{\mathbb{R}^n}
e^{-\frac{r|x|^2}{1+16t^2}}dx
\]
 By the change of variable $y=r^{1/2}(1+16t^2)^{-1/2} $
 and hence $dy=r^{n/2}x(1+16t^2)^{-n/2}dx$, we get
\[
\|u(t,x)\|^r_{L^r_x}=|1+16t^2|^{n/2-rn/4}r^{-n/2} \int_{\mathbb{R}^n} e^{-|y|^2}dy
=|1+16t^2|^{n/2-rn/4}r^{-n/2} \pi^{n/2},
\]
 which implies
$$
\|u(t,x)\|_{L^r_x}=|1+16t^2|^{n/(2r)-n/4}r^{-n/(2r)} \pi^{n/(2r)}.
$$
Now we have to take the $L^q_t$-norm of what we obtained:
$$
\|u(t,x)\|_{L^q_tL^r_x}=\Big ( \int_{\mathbb{R}^n} \|u(t,x)\|^q_{L^r_x} \Big )^{1/q}
$$
which means, since $(q,r)$ is an admissible pair (and so $q=4r/[n(r-2)]$),
that
$$
\|u(t,x)\|_{L^q_tL^r_x}=\Big ( \int_{\mathbb{R}^n}
\|u(t,x)\|^{\frac{4r}{n(r-2)}}_{L^r_x} \Big )^{\frac{n(r-2)}{4r}}
=\Big [ \int_{\mathbb{R}}|1+16t^2|^{-1} \Big ]^{\frac{n(r-2)}{4r}}
\Big(\frac{\pi}{r}\Big)^{n/(2r)},
$$
since $(n/(2r)-n/4)q=-1$. Now by a simple change of variable inside the
integral ($4t=s$) we get:
$$
\|u(t,x)\|_{L^q_tL^r_x}=\Big(\frac{\pi}{r} \Big)^{\frac{n}{2r}}
\Big(\frac{\pi}{4}\Big)^{\frac{n(r-2)}{4r}}.
$$
Putting everything together we get the equation:
\[
S(n,r)\Big ( \frac{\pi}{2} \Big )^{n/4}= \Big(\frac{\pi}{r}\Big)^{\frac{n}{2r}}
\Big(\frac{\pi}{4}\Big)^{\frac{n(r-2)}{4r}}
\]
and so
$$
S(n,r)=2^{\frac{n}{4}-\frac{n(r-2)}{2r}}r^{-\frac{n}{2r}}.
$$
In the case $q=r=2+4/n$ one gets
\[
\|u(t,x)\|^q_{L^q_{t,x}}=q^{-n/2} \pi^{n/2}\int_{\mathbb{R}}|1+16t^2|^{-1}
=\pi^{n/2}(2+4/n)^{-n/2}\frac{\pi}{4}.
\]
Putting all the information together we obtain
$$
2^{-2}\pi^{1+n/2}(2+4/n)^{-n/2}=S_h(n)^{2+4/n} (\pi/2)^{1+n/2}
$$
and solving for $S_h(n)$ one gets
\[
 S_h(n)=\Big( \frac{1}{2} \Big(1+\frac{2}{n} \Big)^{-n/2}\Big)^{\frac{1}{2+4/n}}
\]
Now we have to prove that $S_h(n)$ is a decreasing function of $n$,
namely we have to prove that
$$
\Big( \frac{1}{2} \Big(1+\frac{2}{n+1} \Big)^{-(n+1)/2}
\Big)^{\frac{1}{2+4/(n+1)}}
=S_h(n+1) \leq S_h(n)
=\Big( \frac{1}{2} \Big(1+\frac{2}{n} \Big)^{-n/2}\Big)^{\frac{1}{2+4/n}}.
$$
Taking the natural logarithm to both sides and using the fact that the
logarithm is  a monotone increasing function of his argument we obtain
\begin{align*}
&\frac{1}{2+4/(n+1)}\Big [ -\log(2) -\frac{n+1}{2}\log(1+2/(n+1))\Big]\\
&\leq \frac{1}{2+4/n}\Big [ -\log(2) -\frac{n}{2}\log(1+2/n)\Big ].
\end{align*}
 We can easily see that
$$
\frac{-\log(2)}{2+4/(n+1)} \leq  \frac{-\log(2)}{2+4/n},
$$
so it remains to prove that
 $$
\frac{1}{2+4/(n+1)}\Big [-\frac{n+1}{2}\log(1+2/(n+1))\Big ]
\leq \frac{1}{2+4/n}\Big [ -\frac{n}{2}\log(1+2/n)\Big ] .
$$
Changing variables to $x:=(n+1)/2$ and $y:=n/2$ leads to
 $$
\frac{x\log(1+1/x)}{1+1/x} \geq \frac{y\log(1+ 1/y)}{1+1/y}
$$
 and changing variables again $\alpha:=1+1/x>1$ and $\beta:=1+1/y>1$ we remain with
 $$
\frac{\log(\alpha)}{\alpha(\alpha-1)} \geq \frac{\log(\beta)}{\beta(\beta-1)}.
$$
 So now it remains to show that the function $f:\mathbb{R} \to \mathbb{R}$,
defined by
$$
f(t)=\frac{\log(t)}{t(t-1)},
$$
is decreasing in $t$ and this would lead to the conclusion since $\alpha < \beta$.
Computing its derivative $f'(t)$ one gets
 $$
f'(t)=\frac{t-1-\log(t)(2t-1)}{t^2(t-1)^2}.
$$
We have to verify the inequality just for $t \geq 1$.
We define then
$$
g(t)=\log(t)-\frac{t-1}{2t-1}
$$
 and compute its derivative:
 $$
g'(t)=\frac{(2t-1)^2-t}{t(2t-1)^2}
$$
and so we can see (remember $t \geq 1$) that $g'(t) \leq 0$ if and only if
$t\leq 1$,  and $g'(1)=0$, so $t=1$ is a minimum. $g(1)=0$ and then positive.
So, going backwards with the computations, the inequality $S_h(n+1)<S_h(n)$
is verified.

Now we have to prove the asymptotic behavior and this is easy:
 $$
\lim_{n \to +\infty} S(n)
=\lim_{n \to +\infty} \Big ( \frac{1}{2}(1+\frac{2}{n})^{-n/2}
\Big )^{\frac{1}{2+4/n}}=\lim_{n \to +\infty}
 2^{-1/2}1/e^{\frac{1}{2+4/n}}= \frac{1}{\sqrt{2e}}.
$$
 It remains to prove the equivalence between the homogeneous and the dual constant.
It basically comes from a duality argument. Define $Tu:=e^{it\Delta}u$.
Then for every $f \in L^2_x$ an $F \in L^q_tL^r_x$ we have
 $$
|\langle f, T^* F \rangle |
= |\langle Tf,F\rangle|
\leq \|Tf\|_{L^q_tL^r_x} \|F\|_{L^{q'}_tL^{r'}_x}
\leq S_h \|f\|_{L^2_x} \|F\|_{L^{q'}_tL^{r'}_x}.
$$
So
 $$
\|T^*F\|_{L^2_x}:=\sup_{f \in L^2_x}
\frac{|\langle f, T^* F \rangle |}{\|f\|_{L^2_x}}
\leq S_h \|F\|_{L^{q'}_tL^{r'}_x},
$$
hence $S_d \leq S_h$.  Analogously,
 $$
|\langle Tf,  F \rangle |
= |\langle f,T^*F\rangle |
 \leq \|f\|_{L^2_x} \|T^*F\|_{L^2_x}
\leq S_d \|f\|_{L^2_x} \|F\|_{L^{q'}_tL^{r'}_x}.
$$
So
 $$
\|Tf\|_{L^{q}_tL^{r}_x}:=\sup_{F \in L^{q'}_tL^{r'}_x}
\frac{|\langle Tf,  F \rangle |}{\|F\|_{L^{q'}_tL^{r'}_x}}
\leq S_d \|f\|_{L^2_x},
$$
hence $S_h \leq S_d$ and so we get $S_h=S_d$.
This concludes the proof of the theorem.

\section{Proof of Theorem \ref{GWP}} \label{proofThm2}

Here we will give the proof of Theorem \ref{GWP}. We will skip some of
the details because standard in the theory of global wellposedness for the NLS.
 We refer to \cite{Tao1} for some of the details skipped. We consider
equation \eqref{NLS0}:
 \begin{equation}\label{NLS}
i\partial_tu(t,x)+\Delta u(t,x) +\mu |u|^{p-1}u(t,x)=0 \quad
 (t,x) \in (0,\infty) \times \mathbb{R}^n,
\end{equation}
 with initial datum $u(0,x)=u_0(x)$, space dimension is $n \geq 1$, $p \geq 1$
in both the focusing and defocusing case:
 $\mu=-1,1$, since we are dealing with a small data analysis.
By Duhamel Principle we define
\begin{equation}
 Lu:=\chi(t/T) e^{-it\Delta}u_0(x)-i\mu\chi(t/T)
\int_0^t e^{-i(t-s)\Delta}|u(s,x)|^{p-1}u(s,x)ds,
\end{equation}
where $T >0$ and $\chi(r)$ is a smooth cut-off function supported on
$-2 \leq r \leq 2$ and such that $\chi(r)=1$ on $-1 \leq r \leq 1$.
Using Duhamel formula, we take the $L^q_tL^r_x$-norm of $Lu$
(from now on, unless specified, $t \in [-T,T]$ in the definition of
$L^q_tL^r_x$-norm), and get
\begin{align*}
 \|Lu\|_{L^q_tL^r_x}
&\leq S_h(n,r) \|u\|_{L^2_x}+S_i(n,r)  \|u\|^p_{L^{\tilde{q}'p}_tL^r_x}\\
&\leq S_h(n,r) \|u\|_{L^2_x}+S_i(n,r)  T^{1/(\tilde{q}')-p/q}\|u\|^p_{L^{q}_tL^r_x}
\end{align*}
choosing $\tilde{r}'p=r$.

 Now we need to do  numerical considerations.
Since $(q,r)$ and $(\tilde{q}, \tilde{r})$ are admissible pairs:
 $2/q+n/r=n/2$, $2/\tilde{q}+n/\tilde{r}=n/2$. Moreover, since we are
in the $L^2$-critical case we can choose $\tilde{r}'p=r$ and $\tilde{q}'p=q$,
having still some freedom on the choice of $(q,r)$ as it can be seen by
the following lemma. The conditions on $(q,r)$ and $(\tilde{q}, \tilde{r})$
 can be rewritten as a system of linear equations in
$(1/q,1/\tilde{q},1/r,1/\tilde{r})$.

\begin{lemma}
There exist infinite many solutions to the system $Se=N$, where
\[
S=\begin{pmatrix}
2 & 0 & n & 0 \\
0 & 2 & 0 & n \\
0 & 0 & p & 1 \\
p & 1 & 0 & 0 \end{pmatrix},
\]
$E=(1/q,1/\tilde{q},1/r,1/\tilde{r})^T$ and $N=(n/2, n/2, 1, 1)^T$,
if and only if $p=1+4/n$. If $p \neq 1+4/n$ the system has no solutions.
\end{lemma}

\begin{remark} \label{rmk4.2} \rm
Basically this lemma implies that, using the estimates that we have used above
in the $H^s$-scale, we cannot remove a power of $T$ in front of the nonlinear
term in the \emph{subcrtical} (good) and \emph{supercritical} (bad) cases.
\end{remark}

\begin{proof}
We can see that $\det (S)=0$ and $\operatorname{rank} (S)=3$, because the upper-left
$3 \times 3$ matrix is not
singular for $p \neq 0$. If $p \neq 1+4/n$, then $\operatorname{rank} ([S,N])=4$,
so the system has no solutions, while for
$p=1+4/n$, $\operatorname{rank} ([S,N])=3$ and so the system has infinite solutions.
\end{proof}

\begin{remark} \label{rmk4.3} \rm
Similar computations can be done for any regularity $s$, and with nonlinear
exponent $p(s)=1+4/(n-2s)$. The \emph{critical} case $\tilde{q}'p=q$
is the interesting one for us, because in the \emph{subcritical} case
$\tilde{q}'p < q$ one can shrink the interval, since $T^{1/(\tilde{q}')-p/q}$
appear with a positive power, and so does not really need to do a small data theory.
\end{remark}

Now we will see how big the datum can be in order to have a ``cheap''
 contraction with only the estimates done above. Define
$R:=\alpha S_h(n,r)\|u_0\|_{L^2_x}$
and
$$
B_R:= \{ u \in L^{q}_tL^r_x : \|u\|_{L^{q}_tL^r_x} \leq R \}.
$$
Choose also $\beta>0$ such that
$$
S_i(n,r)R^{p-1}<1/\beta.
$$
 With these choices we get
\begin{align*}
 \|Lu\|_{L^q_tL^r_x}
&\leq S_h(n,r) \|u\|_{L^2_x}+S_i(n,r)  T^{1/(\tilde{q}')-p/q}\|u\|^p_{L^{q}_tL^r_x} \\
&\leq R(1/\alpha+1/\beta) \leq R
\end{align*}
for every $1/\alpha+1/\beta\leq 1$ and with $1/\alpha+1/\beta= 1$
in the less restrictive case. So the Duhamel operator $L$ sends the balls
$B_R$ into themselves if $\|u_0\|_{L^2_x}$ is small enough, more precisely when
$$
\|u_0\|_{L^2_x}=\frac{R}{S_h(n)\alpha}.
$$
This implies that
$$
S_i(n,r)\big(\alpha S_h(n,r)\|u\|_{L^2_x}\big)^{p-1}<1/\beta,
$$
which means
$$
\|u\|_{L^2_x}<\frac{1}{S_h(n,r) \alpha}
\Big(\frac{1}{\beta S_i(n,r)}\Big)^{1/(p-1)}.
$$
Using our hypotheses on $p, \alpha, \beta$ we obtain
\begin{equation}\label{condizione}
\|u\|_{L^2_x}<\frac{1}{S_h(n,r) \alpha}
\Big(\frac{1}{ S_i(n,r)}-\frac{1}{S_i(n,r)\alpha} \Big)^{n/4}.
\end{equation}
For now, the only restriction on $\alpha$ is $1/\alpha+1/\beta\leq 1$.

\begin{remark} \label{rmk4.4} \rm
 The coefficients $\alpha$ and $\beta$ are almost conjugate exponents,
suggesting an orthogonal decomposition
 of the solution on the linear flow and on the nonlinear one.
\end{remark}

Now we check that the operator $Lu$ is a contraction. Let
\begin{gather}
 u(t)=e^{-it\Delta}u_0-i\mu \int_0^t e^{-i(t-s)\Delta}|u(s)|^{p-1}u(s)ds, \\
 v(t)=e^{-it\Delta}u_0-i\mu \int_0^t e^{-i(t-s)\Delta}|v(s)|^{p-1}v(s)ds.
\end{gather}
be two solutions of \eqref{NLS}. Then
\begin{align*}
\|Lu-Lv\|_{L^q_tL^r_x}
&=  \big\|\int_0^t e^{-i(t-s)\Delta}
 \left ( |u(s)|^{p-1}u(s)-|v(s)|^{p-1}v(s) \right )ds\big\|_{L^q_tL^r_x} \\
&\leq S_i(n,r) \||u|^{p-1}u-|v|^{p-1}v\|_{L^{\tilde{q}'}_tL^{\tilde{r}'}_x}  \\
&\leq S_i(n,r) \Big (\|u\|^{p-1}_{L^{q}_tL^r_x}+\|v\|^{p-1}_{L^{q}_tL^r_x} \Big )
\|u-v\|_{L^q_tL^r_x}
\end{align*}
in the above choice of exponents $(q,r)$ and $(\tilde{q}, \tilde{r})$.
This implies:
$$
\|Lu-Lv\|_{L^q_tL^r_x} \leq 2S_i(n)R^{p-1}\|u-v\|_{L^q_tL^r_x}
< 2/\beta \|u-v\|_{L^q_tL^r_x},
$$
so we need $2/\beta \leq 1,$ namely $\beta \geq 2$ and so
$1 \leq \alpha \leq 2$, since $1/\alpha+1/\beta \leq 1$.
This is the last restriction on $\alpha$ that we need to apply to the
estimate \eqref{condizione}. We remark here that \eqref{condizione}
holds for every $1\leq \alpha \leq 2$ and so we are allowed to take the
maximum on both sides of \eqref{condizione}. Notice also that the left hand
side of \eqref{condizione} does not depend on $\alpha$.

\begin{remark} \label{rmk4.5} \rm
 To have a contraction the ball needs to be big enough, but not that much
namely $S_h(n,r)\|u\|_{L^2_x} \leq R \leq 2S_h(n,r)\|u\|_{L^2_x}$.
\end{remark}

Now we want to optimize on $\|u_0\|_{L^2_x}$, namely we want to take it as
big as possible, maintaining the property of $Lu$ of being a contraction.
In other words we have to find the maximum of the function
$$
F_n(\alpha)=\frac{1}{ \alpha}\Big(1-\frac{1}{\alpha} \Big)^{n/4},
$$
 when $\alpha \in [1, 2]$. Taking the derivative, we get
$$
F_n'(\alpha)=-\alpha^{-2-n/4}( \alpha -1 )^{n/4-1}
\big( -(1+n/4)(\alpha-1)+\alpha n/4\big).
$$
So $F_n'(\alpha) \geq 0$ if and only if
$$
1\leq \alpha \leq 1+n/4.
$$
In particular when $n \geq 4$, $\alpha_{\rm max}=2$ and when $n \leq 4$,
$\alpha_{\rm max}=1+n/4$. This concludes the proof of Theorem \ref{GWP}.

\begin{remark} \label{rmk4.6} \rm
 The coefficient $\alpha=2$ is not always the optimal one, as it is
usually used in every exposition on the topic. The optimal $\alpha$ depends
on the dimension $n$.
 We can compute explicitly the values of $F_n(\alpha_{\rm max})$ in any dimension:
for $n=1$ $F_n(\alpha_{\rm max})=F_1(5/4)=5^{-5/4}4$, for $n=2$,
$F_n(\alpha_{\rm max})=F_2(3/2)=3^{-3/2}2$, for $n=3$,
$F_n(\alpha_{\rm max})=F_3(7/4)=3^{3/4}7^{-7/4}4$ and for $n \geq 4$,
  $F_n(\alpha_{\rm max})=2^{-1-n/4}$.
\end{remark}


\section{Applications to Fourier restriction inequalities}\label{FourierRestriction}

Strichartz inequalities can be set in the more general framework of Fourier
restriction inequalities in Harmonic Analysis. This connection has
been made clear already in the original paper of Strichartz \cite{Strichartz}.
In this section we will highlight this relationship in the
Schr\"{o}dinger/paraboloid  case and we will see how to prove
Theorem \ref{RestrictionTheorem}.  For the case of different flows and
hypersurfaces, like the Wave/Cone or Helmholtz/Sphere cases, we refer
to \cite{Tao2} and the references therein for more details.

Consider a function $f \in L^1(\mathbb{R}^n)$, then its Fourier transform $\hat{f}$
is a bounded and continuous function on all $\mathbb{R}^n$ and it vanishes at infinity. So $\hat{f}|_{\mathcal{S}}$, the restriction of $\hat{f}$ to a set $\mathcal{S}$ is well defined even if $\mathcal{S}$ has measure zero, like, for example, if $\mathcal{S}$ is a hypersurface. It becomes then interesting to understand what happens if $f \in L^p(\mathbb{R}^n)$ for $1< p <2$. From Hausdorff-Young inequality, we can see that if $f \in L^p(\mathbb{R}^n)$ then $\hat{f} \in L^{p'}(\mathbb{R}^n)$ with $1/p+1/p'=1$,
so $\hat{f}$ can be naturally restricted to any set $\mathcal{A}$ of positive
 measure. It turns out that a big role is played by the geometry of the
set $\mathcal{S}$. Stein proved that if the set $\mathcal{S}$ is sufficiently
smooth and its curvature is big enough (in fact it is not true for hyperplanes),
then it makes sense to talk about $\hat{f}|_{\mathcal{S}}$ belonging to $L^p$-spaces.

\subsection*{Proof of Theorem \ref{RestrictionTheorem}}
\label{ProofOfRestrictionTheorem}

From now on we will focus on the case where the hypersurface is the paraboloid
$\mathcal{S}=\mathbb{P}^n$, where $\mathbb{P}^n$ is defined as
\begin{equation} \label{paraboloid}
\mathbb{P}^n:= \{(\tau, \xi) \in \mathbb{R} \times \mathbb{R}^n: -\tau=|\xi|^2 \}
\end{equation}
and is endowed with the measure $dP^{n}$ that is given by
\begin{equation}\label{measure}
 \int_{\mathbb{P}^n} h(\tau, \xi)dP^{n}=\int_{\mathbb{R}^n} h(-|\xi|^2, \xi)d \xi.
\end{equation}
(here $h$ is a Schwartz function) and induced by the embedding
$\mathbb{P}^n \hookrightarrow \mathbb{R}^{n+1}$. To prove the theorem,
we have just to show the equivalence of Restriction Inequalities
and Strichartz Inequalities.

It makes sense to talk about a restriction, if $\hat{f}|_{\mathcal{S}}$
is not infinite almost everywhere and a \emph{restriction estimate} holds:
$$
\|\hat{f}|_{\mathbb{P}^n}\||_{L^q(\mathbb{P}^n,dP^n)}
\leq \|f\|_{L^p(\mathbb{R}^n)},
$$
for some $1 \leq q <\infty$ and for every Schwartz function $f$.
This last estimate is equivalent, by a duality argument and Parseval Identity, to
$$
\|\mathcal{F}^{-1}(\hat{F}dP^n)|_{\mathbb{P}^n}\|_{L^{p'}(\mathbb{R}^n)}
\leq \|f\|_{L^{q'}(\mathbb{P}^n,dP^n)},
$$
for all Schwartz functions $F$ on $\mathbb{P}^n$ and where
$$
\mathcal{F}^{-1}(\hat{F}dP^n)(t,x)
=\int_{\mathbb{P}^n}e^{ix\xi+it\tau}\hat{F}(\tau, \xi) d\tau dP^n(\tau,\xi)
$$
is the inverse space-time Fourier transform of the measure $\hat{F}dP^n$.
 The dual formulation connects directly to the fundamental solution
\eqref{FundamentalSolution}
$$
u(t,x)=(1-4it)^{-n/2}e^{-\frac{|x|^2}{1-4it}}
$$
of equation \eqref{LS}
\begin{equation*}
i\partial_tu(t,x)=\Delta u(t,x), \quad (t,x) \in (0,\infty) \times \mathbb{R}^n.
\end{equation*}
Since $u$ can be rewritten in the form
$$
u=\mathcal{F}^{-1}(\hat{u}_0dP^n).
$$
In this way the homogeneous Strichartz inequality
 $$
\|e^{it\Delta}u_0\|_{L^q_tL^r_x}  \leq S_h(n,q,r) \|u_0\|_{L^2_x},
$$
for $q=r=2+4/n$, as in this present case, can be rewritten as
\begin{equation}%\label{FRI}
 \|\widehat{fdP^{n}}\|_{L_{t,x}^{\frac{2(n+2)}{n}}(\mathbb{R}^{n+1})}
\leq S_h(n) \|f\|_{L^2(\mathbb{P}^n,dP^{n})}
\end{equation}
where
 $$
S_h(n)=\Big ( \frac{1}{2}\Big(1+\frac{2}{n}\Big)^{-n/2}\Big )^{\frac{1}{2+4/n}}.
$$
This proves Theorem \ref{RestrictionTheorem}.

\begin{remark} \label{rmk5.1} \rm
We notice that results for the paraboloid seem easier to obtain than
for example for the sphere. For example there is not yet the counterpart
 of \cite{Christ1} in the wave/sphere case  and we do not have a conjecture
on the sharp Strichartz constant in general dimension in the case of the
 wave equation.
\end{remark}

\begin{remark} \label{rmk5.2} \rm
As we said above, the connection between restriction theorems and PDE
links a much broader class of hypersurfaces and PDEs.
For more details on the more recent results, we refer to
\cite{Christ1,Christ2,Christ3,FVV1,FVV2,Tao2} for a survey on restriction theorems.
\end{remark}

\begin{remark} \label{rmk5.3} \rm
In some of the proof of the existence of maximizers for restriction
inequalities it has been crucial the Hilbert structure.
See for example \cite{FVV1} and \cite{Christ1}. Here we are in $L^2_x$
and so a Hilbert case, but our analysis is not touched by this problem,
because we are interested in the optimal constants and not on the extremizers.
\end{remark}

\section{Comments on the inhomogeneous case and the wave equation}
In this section, we want to share some comments and computations on
the inhomogeneous Strichartz estimate and on the case of the wave equation.
We will not prove any theorem, but we will highlight some difficulties
 and make some remarks.

\subsection*{Inhomogeneous Strichartz constant $S_i$}

By the $TT^*$ principle (take $Tu:=e^{it\Delta}$) and by duality,
the homogeneous Strichartz and the dual Strichartz inequality are equivalent.
 By the same principle one can prove that the operator
$TT^*: L^q_tL^r_x \to L^{\tilde{q}'}_rL^{\tilde{r}'}_x$ is bounded if and
only if the operator $T: L^2_x \to L^{q}_tL^{r}_x$ is bounded.
Unfortunately, the inhomogeneous Strichartz inequality cannot be seen as such
a composition because it involves the retarded operator.
This does not prevent the retarded operator to keep the boundedness properties
of $TT^*$ but it complicates a lot the computation of
$S_i(n,r,q,\tilde{r}, \tilde{q})$ and the proof of the existence of critical points,
that, as far as we know, has not been treated  yet in the literature.
In the following, we will outline how the integrals become not tractable
in the inhomogeneous case already in the case of a Gaussian and so a simple
direct computation seems not to be enough to calculate the best Strichartz Constant.
We will concentrate also here on the $L^2$-critical case. See \cite{Tao1}
or \cite{Klainerman1} for more details on the $TT^*$-method.
 We now test the inhomogeneous inequality with Gaussians for every dimensions.
It is not known yet in the literature if they are maximizers or not, but an
explicit computation would lead at least to a lower bound on the constant.
We recall that the solutions that we want to test are
$$
u(t,x)=(1-4it)^{-n/2}e^{-\frac{|x|^2}{1-4it}},
$$
while the inequality we need to test is
 $$
\big\|\int_{s<t} e^{-i(t-s)\Delta}F(s)ds\big\|_{L^q_tL^r_x}
\leq S_i(n, q,r \tilde{q},\tilde{r})  \|F\|_{L^{\tilde{q}'}_tL^{\tilde{r}'}_x} .
$$
with $F(t,x)=|u(t,x)|^{p-1}u(t,x)$.

We start by computing the norm on the right hand side of this inequality.
By the choice of the exponents and the criticality of the problem $\tilde{r}'p=r$
and $\tilde{q}'p=q$. So we  get
$$
\|F\|_{L^{\tilde{q}'}_tL^{\tilde{r}'}_x}
= \||u|^p\|_{L^{q/p}_tL^{r/p}_x}= \|u\|^p_{L^{q}_tL^{r}_x} .
$$
By the computations done in Section \ref{proofThm1}, we then obtain
$$
\|F\|_{L^{\tilde{q}'}_tL^{\tilde{r}'}_x}
=\Big( \frac{\pi}{4} \Big)^{\frac{np(r-2)}{4r}}
\Big( \frac{\pi}{r} \Big)^{\frac{pn}{2r}}.
$$
Now we have to compute the left hand side of the inhomogeneous
Strichartz inequality:
$$
\big\|\int_{s<t} e^{-i(t-s)\Delta}F(s)ds \big\|_{L^q_tL^r_x} .
$$
We start computing explicitly $e^{-i(t-s)\Delta}F(s)$.
By definition of $e^{-i(t-s)\Delta}$, we have
\begin{align*}
e^{-i(t-s)\Delta}F(s)
=\frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n}e^{ix\cdot \xi}
\widehat{e^{-i(t-s)\Delta}F} d\xi
=\frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n}e^{ix \cdot \xi
+ i (t-s)    |\xi|^2 }\hat{F} d\xi.
\end{align*}
So we have now to compute $\hat{F}(s, \xi)$:
\begin{align*}
&\hat{F}(s, \xi)\\
&= \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n}e^{-ix\cdot \xi}F(s,x)dx\\
&=\frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n}e^{-ix\cdot \xi}|u(s,x)|^{p-1}u(s,x)dx \\
& =  (2 \pi)^{-n/2} |1+16s^2 |^{-(p-1)n/4}(1-4is)^{-n/2}\int_{\mathbb{R}^n}
 e^{-ix\cdot \xi}e^{-\frac{|x|^2}{1-4is}-\frac{(p-1)|x|^2}{1+16s^2}} \\
& =  (2 \pi)^{-n/2} |1+16s^2|^{-(p-1)n/4-n/2}(1+4is)^{n/2}\int_{\mathbb{R}^n}
 e^{-ix\cdot \xi} e^{-\frac{(p+4is)|x|^2}{1+16s^2}} \\
& = (2 \pi)^{-n/2}|1+16s^2 |^{-(p-1)n/4-n/2}(1+4is)^{n/2} \times \\
&\quad  \times   \int_{\mathbb{R}^n}e^{-\frac{(p+4is)|x|^2}{1+16s^2}
 -ix\cdot \xi+\frac{(1+16s^2)|\xi|^2}{4(p+4is)}}e^{-\frac{(1+16s^2)
 |\xi|^2}{4(p+4is)}} \\
& = (2 \pi)^{-n/2} |1+16s^2|^{-(p-1)n/4-n/2}(1+4is)^{n/2} \times \\
&\quad \times   \int_{\mathbb{R}^n}e^{-|\frac{x(p+4is)^{1/2}}{(1+16s^2)^{1/2}}
 +i\frac{\xi (1+16s^2)^{1/2}}{2(p+4is)^{1/2}}|^2}
e^{-\frac{(1+16s^2)|\xi|^2}{4(p+4is)}} \\
& = \frac{e^{-\frac{(1+16s^2)|\xi|^2}{4(p+4is)}}}{(2 \pi)^{n/2}}
|1+16s^2 |^{-(p-1)n/4-n/2}(1+4is)^{n/2} (1+16s^2)^{n/2}(p+4is)^{-n/2}\pi^{n/2} \\
& = 2^{-n/2} |1+16s^2 |^{-(p-1)n/4}\Big(\frac{1+4is}{p+4is} \Big)^{n/2}
 e^{-\frac{(1+16s^2)|\xi|^2}{4(p+4is)}}
\end{align*}
by completing the square and changing integration variables to
$$
y=\frac{x(p+4is)^{1/2}}{(1+16s^2)^{1/2}}+i\frac{\xi (1+16s^2)^{1/2}}{2(p+4is)^{1/2}},
$$
similarly to the computations  done in Section \ref{Preliminaries}. So
\begin{align*}
\hat{F}(s, \xi) = 2^{-n/2} |1+16s^2 |^{-(p-1)n/4}
\Big(\frac{1+4is}{p+4is} \Big)^{n/2}e^{-\frac{(1+16s^2)|\xi|^2}{4(p+4is)}}.
\end{align*}
Notice that this is consistent with what we got in Section \ref{Preliminaries}
in the case $s=0$ and $p=1$.
Now, putting everything together, we obtain
\begin{align*}
&e^{-i(t-s)\Delta}F(s)\\
&=\frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n}e^{ix \cdot \xi
 + i( t-s)    |\xi|^2 }\hat{F} d\xi \\
&= \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n} e^{ix \cdot \xi + i (t-s)
|\xi|^2 } 2^{-n/2}|1+16s^2 |^{-(p-1)n/4}
 \Big(\frac{1+4is}{p+4is} \Big)^{n/2}e^{-\frac{(1+16s^2)|\xi|^2}{4(p+4is)}}\\
&= \frac{1}{(2\pi)^{n/2}} 2^{-n/2}|1+16s^2|^{-(p-1)n/4}
 \Big(\frac{1+4is}{p+4is} \Big)^{n/2} \int_{\mathbb{R}^n}
e^{-\frac{(1+16s^2)|\xi|^2}{4(p+4is)}}e^{ix \cdot \xi + i (t-s) |\xi|^2 } \\
&= \frac{1}{(2\pi)^{n/2}} 2^{-n/2}|1+16s^2 |^{-(p-1)n/4}
\Big(\frac{1+4is}{p+4is} \Big)^{n/2}  \\
&\quad \times \int_{\mathbb{R}^n} e^{-|\xi|^2 \big[
\frac{(1+16s^2)}{4(p+4is)} -i(t-s)\big]+ix\xi
+\frac{|x|^2}{4\big[ \frac{(1+16s^2)}{4(p+4is)} -i(t-s)\big]}}
e^{-\frac{|x|^2}{4\big[
\frac{(1+16s^2)}{4(p+4is)} -i(t-s) \big]}} \\
&=  \frac{1}{(2\pi)^{n/2}} 2^{-n/2}|1+16s^2 |^{-(p-1)n/4}
\Big(\frac{1+4is}{p+4is} \Big)^{n/2}  \\
&\quad \times \int_{\mathbb{R}^n} e^{-|\xi
[\frac{(1+16s^2)}{4(p+4is)} -i(t-s) ]^{1/2}-\frac{ix}{2\big[
\frac{(1+16s^2)}{4(p+4is)} -i(t-s) \big]^{1/2}}|^2}
e^{-\frac{|x|^2}{4\big[
\frac{(1+16s^2)}{4(p+4is)} -i(t-s)\big]}}
\end{align*}
which by the change of variable
\[
\eta=\xi [\frac{(1+16s^2)}{4(p+4is)} -i(t-s)]^{1/2}
 -\frac{ix}{2[\frac{(1+16s^2)}{4(p+4is)} -i(t-s) ]^{1/2}},
\]
 becomes
\begin{align*}
 e^{-i(t-s)\Delta}F(s)
&= \frac{1}{(2\pi)^{n/2}} 2^{-n/2}|1+16s^2 |^{-(p-1)n/4}
\Big(\frac{1+4is}{p+4is} \Big)^{n/2}  \\
&\quad \times \int_{\mathbb{R}^n} e^{-|\eta|^2}
e^{-\frac{|x|^2}{4[\frac{(1+16s^2)}{4(p+4is)} -i(t-s) ]}}
\big[\frac{(1+16s^2)}{4(p+4is)} -i(t-s)\big]^{-n/2}.
\end{align*}
In conclusion,
\begin{align*}
&e^{-i(t-s)\Delta}F(s) \\
&= \frac{1}{(2\pi)^{n/2}} 2^{-n/2}|1+16s^2|^{-(p-1)n/4}
\Big(\frac{1+4is}{p+4is} \Big)^{n/2}\big[
\frac{(1+16s^2)}{4(p+4is)} -i(t-s) \big]^{-n/2}
\\
&\quad \times \pi^{n/2}
e^{-\frac{|x|^2}{4[\frac{(1+16s^2)}{4(p+4is)} -i(t-s) ]}} \\
&=  |1+16s^2 |^{-(p-1)n/4} \Big(\frac{1+4is}{p+4is} \Big)^{n/2}
\big[\frac{(1+16s^2)}{p+4is} -4i(t-s) \big]^{-n/2} e^{-\frac{|x|^2}
{4[\frac{(1+16s^2)}{4(p+4is)} -i(t-s) ]}}\\
& =  |1+16s^2 |^{-(p-1)n/4} \big[ \frac{1+4is}{1-4ip(t-s)+16ts}\big]^{n/2}
\end{align*}
Again, this is consistent with what we got in Section \ref{Preliminaries}
in the case $s=t=0$ and $p=1$. At this point the approach of the direct
computation seems not good enough anymore, because one should integrate
in the variable $s$ and this does not seem to have an explicit expression
with elementary functions. We refer to \cite{Selvitella11} for more
details on a possible numerical approach to the problem.

\begin{remark} \label{rmk} \rm
If one would be able to compute explicitly $S_i(n,r)$, one could
use Theorem \ref{GWP} also as a stability-type result for the solutions of
the NLS, in a similar spirit of the stability of solitons in the focusing case.
This connection links, in some sense, optimizers and stability, also when the
functionals involve both space and time.
\end{remark}

\subsection*{The wave equation case}

For completeness, we want to mention here that similar studies have been done
for several others homogeneous Strichartz estimates, like the wave equation.
The complete characterization of critical points done by \cite{Christ1} in
the case of the Schr\"{o}dinger
Equation is still not available in the case of the wave equation.
We believe that an argument completely similar to the
one that we have given in Section \ref{proofThm1} would lead to the computation
of the possible best homogeneous wave Strichartz constant $W(n)$ for the wave
equation, once a complete characterization of the maximizers would be available.
For more details on the case of the wave equation we refer to
\cite{Bulut,Carneiro,Foschi1}.

\begin{remark} \label{rmk6.2} \rm
There are well known transformations that send solutions to the
Schr\"{o}dinger equation to solutions of the wave equation,
see for example \cite{Tao1}. So one strategy here could be also to
transform the maximizers of $S_h(n,r)$ into solutions of the corresponding
wave equation and hope that the known transformation sends maximizers
to maximizers. Unfortunately, to our knowledge, no known transformation
does this job. This technique could be very helpful also for other equations.
\end{remark}

\begin{remark} \label{rmk6.3} \rm
Note that the functions that optimize the wave Strichartz inequality
(see \cite{Foschi1}), optimize also the Sobolev embeddings (see \cite{Talenti1}
and \cite{Aubin1}).
Let $1<p<n$ and $p^*=\frac{np}{n-p}$, then
$$
\|u\|_{L^{p^*}(\mathbb{R}^n)} \leq C(n,p) \|\nabla u\|_{L^{p}(\mathbb{R}^n)}
$$
with optimal constant $C(n,p)$ given by
$$
C(n,p)=\pi^{1/2}n^{-1/p}\Big( \frac{p-1}{n-p} \Big)^{1-1/p}
\Big( \frac{\Gamma(1+n/2) \Gamma(n)}{\Gamma(n/p) \Gamma(1+n-n/p)} \Big)
$$
and maximizers given by
$$
u(x)=(a+b |x|^{\frac{p}{p-1}})^{-\frac{n-p}{p}},
$$
with $a,b>0$. We notice that with $p=2$
and substituting $n$ with $n+1$ in the above optimizers, we recover the
optimizers given in \cite{Foschi1}. The correspondence
between the constants seems more involved.
\end{remark}

\subsection*{Acknowledgments}
This article is in memory of Zia Amelia.
I want to thank my family and Victoria Ban for their constant support.
I thank also my supervisor Prof. Narayanaswamy Balakrishnan for his help
and guidance.


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\end{document}