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)