TITLE: Proof writing - $\lim_{x\to 1}{x^n}=1,n\in \mathbb N$
QUESTION [0 upvotes]: I was wondering how to write a formal proof of the following proposition,
$$\lim_{x\to 1}{x^n}=1,n\in \mathbb N$$
I tried using the $\epsilon-\delta$ definition but I am unable to solve the inequality
$$\lvert x^n-1\rvert<\epsilon$$
I am looking for a formal proof of the statement, and also could anyone please provide me with references to get better at proof writing and proving techniques?
Also, can we generalise it to be true for all $n\in \mathbb R$?
Thanks for any answers!!

REPLY [0 votes]: You could use the definition of continuity of $x^n$ at x=1. A bit more explicit, since $x^n$ is continuous in $x=1$ you have the following:
\begin{equation}
\forall \epsilon > 0 \exists \delta >0 \text{ s.t. } |1-x|< \delta \implies |x^n - 1^n| = |x^n-1| < \epsilon
\end{equation}


text:
\begin{document}

\begin{abstract}
In this article we study some aspects of dispersive and concentration
phenomena for the Schr\"odinger equation posed on hyperbolic space
$\mathbb{H}^n$, in order to see if the negative curvature of the manifold
gets the dynamics more stable than in the Euclidean case.
It is indeed the case for the dispersive properties : we prove that the dispersion inequality is
valid, in a stronger form than the one on $\mathbb{R}^n$.
However, the geometry does not have enough of an effect to avoid the
concentration phenomena and the picture is actually worse than expected. The critical nonlinearity power for blow-up turns out to be the same as in the euclidean case, and we prove that there are more explosive solutions for critical and supercritical nonlinearities.\\
Keywords : Nonlinear Schr\"odinger equations on manifolds,
hyperbolic space, representations of solutions, dispersion estimates,
blow-up.

\end{abstract}
\maketitle
\section{Introduction}
Let $(M,g)$ be a Riemannian manifold and let $\Delta_M$ be the
Laplace-Beltrami operator.
The study of the nonlinear Schr{\"o}dinger equation
$$\left\{\begin{array}{c}
i\DP{t}u+\Delta_M u=V'(\mo{u}^2)u,\\
u(0)=u_0,
\end{array}\right.$$
where $u$ is a space-time function with complex values, and $V$ is a real
function with controlled growth at infinity, was motivated by number of
problems coming from Physics.

It is known that the geometry influences the dynamics of the equation.
Instability phenomena appear, even in the defocusing case.\\
For instance, on the one hand, for the cubic defocusing Schr\"odinger equation on the sphere
$\mathbb{S}^2$
$$\left\{\begin{array}{c}
i\partial_t u+\Delta_{\mathbb{S}^2}u=\vert u\vert^2u,\\
u(0,x)\in H^s(\mathbb{S}^2)
\end{array}\right.,$$
the critical regularity index for having the uniform continuity of the flow
on the bounded sets of $H^s$ is $s=\frac{1}{4}$, as it was proved recently by
Burq-G\'erard-Tzvetkov (\cite{BGT12}, \cite{BGT32}, see also \cite{Val2}).\\
On the other hand Bourgain (\cite{Bo}) and Cazenave-Weissler (\cite{CW})
have proved that the Cauchy problem of the same equation,
considered on $\TT^2$ and
on $\RR^2$ respectively, is $H^\epsilon$ well-posed for all positive
$\epsilon$, meanwhile for negative
$s$ instability phenomena appear (\cite{BGT12},\cite{CCT}).
It follows that the critical regularity index for the flat torus and for
$\RR^2$ is zero.

Hence these results point out the importance played by the geometry of the
manifold in the dynamics of the equation.

It is expected that the positive curvature generates the
differences, since for having the instability result on the sphere the
dynamics of spherical harmonic concentrated on closed geodesics are studied.

In this article we study dispersive and concentration phenomena for the
Schr\"odinger equation posed on hyperbolic space $\mathbb{H}^n$,
manifold of negative curvature,
expecting that the dynamics are stabler than in the Euclidean case.\\

We shall define in \S 2 the hyperbolic space and the tools used on
it. For the moment, let us introduce some notations. We denote by $0$ the
origin of the hyperbolic space, $0=(1,0,..,0)$. In the sequel we shall
use the $L^p$ spaces on $\mathbb{H}^n$
$$L^p=L^p(\mathbb{H}^n)=L^p(d\Omega),$$
and weighted spaces, defined by
$$L^p(w)=L^p(w\,d\Omega),$$
where $d\Omega$ denotes the measure on hyperbolic space
$\mathbb{H}^n$ and $w$ is a function on $\mathbb{H}^n$.\\

First, we shall treat the linear equation. We obtain an explicit
representation of the solutions.

\begin{theorem}\label{exactsol}
The solution of the linear Schr\"odinger equation posed on
hyperbolic space $\mathbb{H}^n$
$$(SL)\left\{\begin{array}{c}
i\partial_t u+\Delta_{\mathbb{H}^n}u=0\\
u(0,x)=u_0
\end{array}\right.,$$
is, for $n\geq 3$ odd,
\begin{equation}\label{odd}
u(t,\Omega)=\frac{c}{|t|^\frac{1}{2}}e^{-it\frac{(n-1)^2}{4}}\int_{\mathbb{H}^n}u_o(\Omega')\left(\frac{\partial_\rho}{\sinh\rho}\right)^\frac{n-1}{2}e^{i\frac{\rho^2}{4t}}\,d\Omega',\end{equation}
and for $n\geq 2$ even
\begin{equation}\label{even}
u(t,\Omega)=\frac{c}{|t|^\frac{3}{2}}e^{-it\frac{(n-1)^2}{4}}\int_{\mathbb{H}^n}u_o(\Omega')\left(\frac{\partial_\rho}{\sinh
\rho}\right)^\frac{n-2}{2}\int_\rho^\infty\frac{e^{i\frac{s^2}{4t}}s}{\sqrt{\cosh
s-\cosh\rho}}\,ds\,d\Omega',\end{equation}
where by $\rho$ we denote the hyperbolic distance between the points
$\Omega$ and $\Omega'$.

\end{theorem}
As harmonic analysis can be done on $\mathbb{H}^n$,
the proof is based on the representation of the solution via the Fourier
transform and on calculus of oscillatory integrals. There are many
points in common with the proof of the inverse Fourier formula on hyperbolic space.

This explicit representation of the solution allows us to study the
dispersive properties and to obtain the following results.

\begin{theorem}\label{disp}

i) For all dimension $n\geq 2$, the solution satisfies the following local
dispersion inequality
\begin{equation}\label{disph}
|u(t,\Omega)|\leq
c\frac{1}{|t|^\frac{n}{2}}\int_{\mathbb{H}^n}|u_0(\Omega')|\left(\frac{\rho}{\sinh\rho}\right)^\frac{n-1}{2}\,d\Omega'.\end{equation}

For large times, the following dispersion estimate holds for all $n\geq 3$ odd,
\begin{equation}\label{disphtlarge}
|u(t,\Omega)|\leq
c\frac{1}{|t|^\frac{3}{2}}\int_{\mathbb{H}^n}|u_0(\Omega')|\left(\frac{\rho}{\sinh\rho}\right)^\frac{n-1}{2}\,d\Omega',\end{equation}
and for all $n\geq 2$ even
\begin{equation}\label{disphtlarge2}
|u(t,\Omega)|\leq
c\frac{1}{|t|^\frac{3}{2}}\int_{\mathbb{H}^n}|u_0(\Omega')|\left(\frac{\rho}{\sinh\rho}\right)^\frac{n-1}{2}\,\frac{1+\rho}{\sqrt{\rho}}\,d\Omega',\end{equation}

ii)
Moreover, for $n\geq 3$ and radial initial data, we have a weighted-space local dispersion inequality
\begin{equation}\label{dispw}
\|u(t)\|_{L^\infty(w)}\leq \frac{c}{|t|^\frac{n}{2}}
\|u_0\|_{L^1(w^{-1})},\end{equation}
where the weight is
$$w(\Omega)=\frac{\sinh d(0,\Omega)}{d(0,\Omega)}.$$
Finally, for $n\geq 3$, for a finite time $T$ and for radial initial data, we have the local Strichartz weighted estimates
\begin{equation}\label{Strw}
\|u\|_{L^p\left([0,T],\,
L^{q}\left(w^{q-2}\right)\right)}\leq
c\|u_0\|_{L^2},\end{equation}
for all pairs $(p,q)$ satisfying $\frac{2}{p}+\frac{n}{q}=\frac{n}{2}$, $(p,q,n)\neq (2,\infty,2)$ and $2\leq p,q$.
\end{theorem}

The new term that appears in (\ref{disph}),
$$\frac{\rho}{\sinh \rho},$$
is specific to hyperbolic space ; it comes from the Harish-Chandra
coefficient and from a kernel that involves the eigenfunctions of the Laplace-Beltrami operator.
This new term informs us that, apart from the classical decay in time,
we have a new one, in space.

\begin{remark}
In dimension $3$, all the dispersion estimates from Theorem
\ref{disp} are global in time, which is not the case for other dimensions.
\end{remark}

\begin{remark}\label{dispsupp}
For dimensions allowed in Theorem \ref{disp}, away from the support of the initial data, the
decay is stronger than in the $\mathbb{R}^n$ case. More precisely, for initial data with support a domain $A$ of $\mathbb{H}^n$, the
solution satisfies, for any point $\Omega$ not included in $A$, at small times,
$$|u(t,\Omega)|\left(
\frac{\sinh d(\Omega,A)}{d(\Omega,A)}\right)^\frac{n-1}{2}\leq
\frac{c}{|t|^\frac{n}{2}}\int_{\mathbb{H}^n}|u_0(\Omega')|
\left(\frac{d(\Omega',^cA)}{\sinh d(\Omega',^cA)}\right)^\frac{n-1}{2}\,d\Omega'.$$
\end{remark}

\begin{remark}\label{disptime}
In dimension $3$, for radial initial data in a more restrictive space, that is
$L^1(\widetilde{w}^{-1})$ with the weight
$$\widetilde{w}(\Omega)=\sinh d(\Omega,0),$$ we
obtain a local dispersion-type estimate stronger in time,
$$\|u(t)\|_{L^\infty(\widetilde{w})}\leq \frac{c}{|t|^{\frac{1}{2}}}
\|u_0\|_{L^1(\widetilde{w}^{-1})}.$$
\end{remark}

However, it is only in (\ref{dispw}), in the radial case, that we
obtained the dispersion stated in weighted spaces, independently of
the initial data.

Let us also notice that the improvements in the dispersive estimates
stated in Theorem \ref{disp} are better decays away from the
origin. Near the origin -where the Laplace-Beltrami operator on
hyperbolic space is almost the one on $\mathbb{R}^n$- we have the same
decays as in the euclidean case. The infinite speed of propagation of
the Schr\"odinger operator is not sufficient to impose on the
dispersion estimate at the origin an influence from the metric at infinity.

As a consequence of (\ref{disph}), we obtain the classical local dispersion
inequality
$$\|u(t)\|_{L^\infty}\leq \frac{c}{|t|^{\frac{n}{2}}}
\|u_0\|_{L^1}.$$\\
From the dispersion inequality, by using the TT* functional analysis
argument (\cite{To}), the local Strichartz estimates are obtained
too.
We recall here that for the wave equation on hyperbolic space, the Strichartz estimates have been proved recently by Tataru (\cite{Ta}).

Let us now remark that on the sphere, the local Strichartz-type
estimates are known to hold with a loss of $\frac{1}{p}$ derivatives,
that is the
$L^pL^q$ norms of the solutions are controlled by the
$H^\frac{1}{p}$ norm of the initial data, instead of the
$L^2$ norm (\cite{BGT0}). This shows that on hyperbolic space, the
dispersion estimates are stronger.

\begin{remark}  An important problem related to the linear Schr\"odinger equation is the problem of the potential. The radial Schr\"odinger equation perturbed with a rough time dependent potential on hyperbolic space was recently treated by Pierfelice in (\cite{Vi}). By using Theorem \ref{disp} ii) and fixed point arguments she obtains for the radial perturbed Cauchy problem the Strichartz estimates with the same weight, pointing out again the influence of the negative curvature on the dispersive properties. These estimates hold even in the case of more general nonlinearities.
\end{remark}

\begin{remark}
It is expected that the results obtained should hold on all symmetric spaces of
rank 1.

Moreover, let us notice that the dispersion kernel obtained is
close to the kernel of the heat operator $e^{t\Delta}$, which has
been studied intensively on hyperbolic space, usually for $t$ real,
but also for complex $t$ with $\Re
t>0$ (\cite{D}, \cite{An}, \cite{DM}). In view of this work, we
expect that the dispersion estimate for large time can be improved by
an additional decay, for all $n\geq 2$ :
$$|u(t,\Omega)|\leq
c\frac{1}{|t|^\frac{3}{2}}\int_{\mathbb{H}^n}|u_0(\Omega')|
\left(\frac{\rho}{\sinh\rho}\right)^\frac{n-1}{2}
\left(\frac{1}{1+\rho}\right)^\frac{n-3}{2}\,d\Omega'.$$
\end{remark}

\bigskip

Let us turn now our attention to the concentration phenomena for the
nonlinear equation, and more
precisely to blow-up of solutions in the sense of the explosion in
finite time of the $L^2$ norm of their gradient.

We consider the focusing Schr\"odinger equation with power nonlinearity
$$(S)\left\{\begin{array}{c}
i\partial_t u+\Delta_{\mathbb{H}^n}u+\vert u\vert^{p-1}u=0\\
u(0,x)=u_0\in H^1(\mathbb{H}^n)
\end{array}\right..$$

First, let us notice that, as for the Euclidean case, since integration by parts works on hyperbolic space, the mass of a solution of $(S)$
$$\int_{\mathbb{H}^n}\mo{u}^2d\Omega,$$
and its energy
$$E(u)=
\frac{1}{2}\int_{\mathbb{H}^n}\mo{\nabla u}^2d\Omega-
\frac{1}{p+1}\int_{\mathbb{H}^n}\mo{u}^{p+1}d\Omega,$$
are conserved in time.

By using the classical Strichartz estimates, derived from the
dispersion estimate (\ref{disph}) on the linear equation,
one can obtain, as in the Euclidean case (\cite{GV},\cite{Ya}), the $H^1$ local existence of solutions of the equation $(S)$ with $p\leq 1+\frac{4}{n-2}$.

The argument used on $\mathbb{R}^n$ in the most of the blow-up results is the scale invariance.
The construction of explicit blow-up solutions is done using the
pseudo-conformal invariance. Informations on the blow-up solutions are
obtained by studying the dilatations of the solution and by using the
virial identity (\cite{Me}).
The main difficulty when working on hyperbolic space is that these
techniques and notions do not have an obvious analogue.

One can expect that maybe there will be no blow-up solutions, due to the negative curvature of the manifold.
In fact it turns out that the geometry does not have enough of an effect to avoid the
concentration phenomena. Moreover, we have more types of initial data generating blow-up in finite time than in the Euclidean case.
\begin{theorem}\label{blup}

For $p<1+\frac{4}{n}$, the solutions of the equation $(S)$ are global
in $H^1(\mathbb{H}^n)$.
The global existence still holds for the power $p=1+\frac{4}{n}$ and
for initial data of mass smaller than a certain constant.\par
However, for $p\geq 1+\frac{4}{n}$, blow-up solutions exist. More
precisely, if the initial data is radial, of finite variance
$$\int_{\mathbb{H}^n}|u_0(\Omega)|^2d^2(0,\Omega)d\Omega<+\infty,$$
and its energy satisfies
$$E(u_0)< c_n\|u_0\|^2_2,$$
then the solution blows up in finite time.
Here $c_n$ is a geometric positive constant given by
$$c_n=\frac{\inf \Delta^2_{\mathbb{H}^n}d(0,\cdot)}{16}.$$

\end{theorem}
The proof of the global existence up to the critical power $p=1+\frac{4}{n}$ -which is also the one of the Euclidean space- involves
the Sobolev
embeddings on $\mathbb{H}^n$ (\cite{He}), together with the
conservation laws of the equation.
The existence of blow-up solutions is shown by
adapting the arguments of Glassey and Zakharov on $\mathbb{R}^n$
(\cite{Gl}, \cite{Za}) and of Kavian on a star-shaped domain of
$\mathbb{R}^n$ (\cite{Ka}), to the particular metric on $\mathbb{H}^n$.

\begin{remark}
We obtain a new virial identity adapted to hyperbolic space that
allows us to
conclude that solutions of null energy and finite virial blow up in
finite time.
Probably this new concentration phenomena is the consequence of the
improved dispersion of the
solutions. We recall that this is not the case on $\mathbb{R}^n$
for the critical power,
since the ground state, that is the unique positive solution of the
elliptic equation
$$\Delta_{\mathbb{R}^n} Q+Q^p=Q,$$
gives us the global solution $e^{it}Q$ of null energy.\\ Moreover, on
$\mathbb{R}^n$, the mass of the ground state is the critical mass for
having explosive solutions in critical power.
In our case, the Theorem \ref{blup} implies that the ground state on hyperbolic space has positive
energy and mass smaller than the expected critical mass for blow-up on
$\mathbb{H}^n$.
\end{remark}

It should be interesting to see what is the minimal blow-up mass, and to get informations about the blow-up speed and about the profiles of the explosive solutions.\\

This article is organized as follows. In the second section we give the
definition of hyperbolic spaces and of the tools used on it, namely the
Laplace-Beltrami operator and the Fourier transform.
In \S3.1 we recall the proof of the representation of the
solutions and the dispersion inequality in the
Euclidean case, then we give in \S3.2 the proof of
Theorem \ref{exactsol}.
In \S4 we prove the first point of Theorem \ref{disp}, concerning the
$L^1-L^\infty$ estimates.
The section \S5 concerns the weighted dispersion inequalities of the second
point of Theorem \ref{disp}, and of the Remarks \ref{dispsupp} and
\ref{disptime}.
The last section contains the proof of Theorem \ref{blup}. In Appendix A is proved a technical proposition, crucial for
obtaining the dispersion estimates in \S4 .\\

Acknowledgments : I am very grateful to V. Georgiev, to P. G\'erard and to F. Ricci for having encouraged my research in this direction. I thank T. Duyckaerts and V. Pierfelice for having carefully read some parts of the article.

\section{hyperbolic space}
\subsection{Definition}

We shall use here the model of hyperbolic space given by the upper branch of the hyperboloid.

We define the hyperbolic space as being the following surface of
$\RR^{n+1}$, given by the parametrization :
$$\mathbb{H}^n=\{\Omega=(t,x)\in\RR^{n+1}, (t,x)=(\cosh r,\sinh r\,\omega),
r\geq 0, \omega\in\mathbb{S}^{n-1}\}.$$

Let us introduce the inner product on $\RR^{n+1}$
\begin{equation}\label{prod}
[x,y]=x_0y_0-....-x_ny_n.\end{equation}

An alternative definition for hyperbolic space is
$$\mathbb{H}^n=\{x=(x_0,x_1,...,x_n)\in\mathbb{R}^{n+1} , [x,x]=1 , x_0>0\}.$$
This space is invariant under $SO(1,n)$, the group of Lorentz
transformations of $\RR^{n+1}$ that preserve this inner product.

One has
$$dt=\sinh r dr, dx=\cosh r\,\omega \,dr+\sinh r \,d\omega,$$
and the metric induced on $\mathbb{H}^n$ by the Lorenzian metric on $\RR^{n+1}$
$$dl^2=-dt^2+dx^2,$$
is
$$ds^2=dr^2+\sinh^2 r\,d\omega^2,$$
where $d\omega^2$ is the metric on the sphere $\mathbb{S}^{n-1}$.

The volume element is
$$\int_{\mathbb{H}^n} f(\Omega)\,d\Omega=\int_0^\infty\int_{\mathbb{S}^{n-1}}f(r,\omega)\sinh^{n-1}r\,dr\,d\omega.$$

The length of a curve
$$\gamma (t)=(\cosh r(t),\sinh r(t) \,\omega(t)),$$ with $t$ varying from $a$ to
$b$, is defined to be as usual,
$$L(\gamma)=\int_a^b\sqrt{|r'(t)|^2+|\sinh r(t)|^2|\omega'(t)|^2}dt.$$
The distance between two points of $X$ will be the infimum of the lengths of
the paths connecting the points.
By direct calculus, one has that the distance of a point to the origin of the hyperboloid $O=(1,0,...,0)$ is
$$d((\cosh r,\sinh r\,\omega), O)=r.$$
More generally, the distance between two arbitrary points is
\begin{equation}\label{dist}
d(\Omega,\Omega')=\cosh ^{-1}([\Omega,\Omega']).\end{equation}

Starting from the general definition of the Laplace-Beltrami
operator, one can find its expression on $\mathbb{H}^n$
$$\Delta_{\mathbb{H}^n}=\partial_r^2+(n-1)\frac{\cosh r}{\sinh r}\,\partial_r+
\frac{1}{\sinh^2r}\,\Delta_{\mathbb{S}^{n-1}}.$$
In Section $6$ we will denote $\Delta=\Delta_{\mathbb{H}^n}$.

\subsection{The Fourier transform}\par
$$ $$
\par
For $\theta\in\mathbb{S}^{n-1}$ and $\lambda$ a real number, the functions of the
type
\begin{equation}\label{fp}
h_{\lambda,\theta}(\Omega)=[\Omega,\Lambda(\theta)]^{i\lambda-\frac{n-1}{2}},
\end{equation}
are generalized eigenfunctions of the Laplacian-Beltrami operator.
Here we denoted by $\Lambda(\theta)$ the point of $\RR^{n+1}$ given
by $(1,\theta)$.
Indeed, we have
$$-\Delta_{\mathbb{H}^n}h_{\lambda,\theta}=
\left(\lambda^2+\frac{(n-1)^2}{4}\right)h_{\lambda,\theta}.$$
By analogy with the $\RR^n$ case, the definition of the Fourier transform is,
for a function on $\mathbb{H}^n$,
$$\widehat{f}(\lambda,\theta)=
\int_{\mathbb{H}^n}h_{\lambda,\theta}(\Omega)
f(\Omega)\,d\Omega.$$
It turns out that this is the good definition, and one has the Fourier inversion formula for function on $\mathbb{H}^n$
$$f(\Omega)=\int_{-\infty}^\infty
\int_{\mathbb{S}^{n-1}}\overline{h}_{\lambda,\theta}(\Omega)
\widehat{f}(\lambda,\theta)\frac{d\theta d\lambda}{|c(\lambda)|^2},$$
where $c(\lambda)$ is the Harish-Chandra coefficient
$$\frac{1}{|c(\lambda)|^2}=\frac{1}{2(2\pi)^n}
\frac{|\Gamma(i\lambda+\frac{n-1}{2})|^2}{|\Gamma(i\lambda)|^2}.$$

Fore more details on hyperbolic space, see Helgason (\cite{Hel})
and Terras (\cite{Ter}).

\section{The representation of the solutions}

\subsection{The Euclidean case}

$$ $$

For the linear equation on $\RR^n$
$$\left\{ \begin{array}{c}
i\DP{t}u+\Delta_{\R^n} u=0,\\
u(0)=u_0\in L^2(\R^n),
\end{array}\right.,$$
the solution can be written explicitly.
By applying the Fourier transform to the equation, one gets
$$\widehat{u}(\xi)=e^{-it|\xi|^2}\widehat{u_0}(\xi),$$
so the solution is
$$u(t,x)=
\frac{1}{(2\pi)^n}
\int_{\mathbb{R}^n}e^{-it|\xi|^2+ix\xi}\widehat{u_0}(\xi)d\xi.$$
By the Plancherel formula, the solution can also be written
$$u(t,x)=
\frac{1}{(4\pi it)^{\frac{n}{2}}}\int_{\mathbb{R}^n}
e^{i\frac{\mo{x-y}^2}{4t}}u_0(y)dy,$$
and therefore one obtains the following estimate
$$\|u(t)\|_\infty\leq \frac{c}{\mo{t}^\frac{n}{2}}\|u_0\|_1,$$
called dispersion estimate.

\subsection{Proof of Theorem \ref{exactsol}}
$$ $$
\par

For the hyperbolic space we will try the same approach, by using the Fourier transform introduced previously. By Fourier transforming on $\mathbb{H}^n$, one has

$$\left(i\partial_t-\left(\lambda^2+\frac{(n-1)^2}{4}\right)\right)
\widehat{u}(t,\lambda,\theta)=0,$$
so that
$$\widehat{u}(t,\lambda,\theta)=e^{-it\left(\lambda^2+\frac{(n-1)^2}{4}\right)}\widehat{u_0}(\lambda,\theta).$$
By applying the inverse Fourier transform,

$$u(t,\Omega)=\int_{-\infty}^\infty
\int_{\mathbb{S}^{n-1}}e^{-it(\lambda^2+\frac{(n-1)^2}{4})}
\overline{h}_{\lambda,\theta}(\Omega)
\widehat{u_0}(\lambda,\theta)\frac{d\theta d\lambda}{|c(\lambda)|^2}.$$
Finally, by making explicit the Fourier transform of the initial data,
\begin{equation}\label{kernel}
u(t,\Omega)=e^{-it\frac{(n-1)^2}{4}}\int_{\mathbb{H}^n}u_o(\Omega')\int_{-\infty}^\infty
e^{-it\lambda^2}L_\lambda(\Omega,\Omega') \frac{d\lambda}{|c(\lambda)|^2}d\Omega',\end{equation}
where
$$L_\lambda(\Omega,\Omega')=
\int_{\mathbb{S}^{n-1}}\overline{h}_{\lambda,\theta}(\Omega)h_{\lambda,\theta}(\Omega')d\theta.$$

\begin{lemma}\label{Lg}
For all $g\in SO(1,n)$,
$$L_\lambda(\Omega,\Omega')=L_\lambda(g\Omega,g\Omega').$$
\end{lemma}

\begin{proof}

Using the definition (\ref{fp}) of the functions $h_{\lambda,\theta}$,
and the fact that $g$ preserves the inner product, we have
\begin{equation}\label{identLg}
L_\lambda(\Omega,\Omega')=\int_{\mathbb{S}^{n-1}}[g\Omega,g\Lambda(\theta)]^{-i\lambda-\frac{n-1}{2}}
[g\Omega',g\Lambda(\theta)]^{i\lambda-\frac{n-1}{2}}d\theta.
\end{equation}

Consider now the map $F$ on $\mathbb{S}^{n-1}$,
$$F(\theta)=\tilde{\theta},$$
given by the relation
\begin{equation}\label{mu}
g\Lambda(\theta)=\mu(\theta)\Lambda(\tilde{\theta}),\end{equation}
for some real $\mu(\theta)$.
Then, one can get (\cite{Vl}, Prop. 8.4.1) that this application is unique, is a
diffeomorphism,
$$\mu(\theta)=\cosh r_g+\sinh r_g \,\theta_1,$$
for some constant $r_g$ depending only on $g$, and the relation between the volume elements $d\tilde{\theta}$ and
$d\theta$ on $\mathbb{S}^{n-1}$ is
\begin{equation}\label{vol}
d\tilde{\theta}=\frac{d\theta}{(\cosh r_g+\sinh r_g\, \theta_1)^{n-1}}=\frac{d\theta}{\left(\mu(\theta)\right)^{n-1}}.\end{equation}
Here we denoted by $\theta_1$ the first $\mathbb{R}^n$-component of $\theta$.\\

Then, by making in the right hand side of (\ref{identLg})
the change of variable $F$, and by using (\ref{mu}) and
(\ref{vol}), we obtain
$$L_\lambda(\Omega,\Omega')=\int_{\mathbb{S}^{n-1}}[g\Omega,\Lambda(\tilde{\theta})]^{-i\lambda-\frac{n-1}{2}}
[g\Omega',\Lambda(\tilde{\theta})]^{i\lambda-\frac{n-1}{2}}d\tilde{\theta}=
L_\lambda(g\Omega,g\Omega'),$$
so the Lemma is proved.
\end{proof}

Let us remark that Lemma \ref{Lg} can also be proved by using the classical
argument of the fundamental point pair invariant (\cite{D}, p.177).

\begin{lemma}\label{linkLegendre}
We have the following identity
$$L_\lambda(\Omega,\Omega')=c (\sinh \rho)^{-\frac{n-2}{2}}
P^{-\frac{n-2}{2}}_{-\frac{1}{2}+i\lambda}(\cosh \rho),$$
where $\rho=d(\Omega,\Omega')$, and by $P^\mu_\nu(z)$ we denoted the Legendre
function, solution of the equation
$$(1-z^2)\partial^2_z w-2z\partial_z
w+\left(\nu(\nu+1)-\frac{\mu^2}{1-z^2}\right) w=0.$$
\end{lemma}

\begin{proof}

Let us choose a transformation $g\in SO(1,n)$, which maps $\Omega$ on the
origin of $\mathbb{H}^n$.
Then $g$ will send $\Omega'$ in a point of $\mathbb{H}^n$ of
coordinates
$$\Omega^*=(\cosh \rho,\sinh \rho\,\gamma).$$
Since $g$ preserves the product, the radius of $\Omega^*$ can be calculated,
$$\rho=\cosh^{-1}[\Omega,\Omega']=d(\Omega,\Omega').$$

So, by using Lemma \ref{Lg},
$$L_\lambda(\Omega,\Omega')=L_\lambda(0,\Omega^*)=
\int_{\mathbb{S}^{n-1}}[\Omega^*,\Lambda(\theta)]^{i\lambda-\frac{n-1}{2}}d\theta,$$
therefore
\begin{equation}\label{L}
L_\lambda(\Omega,\Omega')=
\int_{\mathbb{S}^{n-1}}(\cosh \rho-\sinh \rho\,\gamma.\theta)^{i\lambda-\frac{n-1}{2}}d\theta.
\end{equation}

Consider now a transformation $T\in SO(n)$ such that
$$T(1,0,0...)=\gamma.$$
Then for $\theta\in\mathbb{S}^{n-1}$, we obtain an $\alpha\in[0,\pi[$ and a $\theta'\in\mathbb{S}^{n-2}$, well defined by the relation
$$\theta=T(\cos\alpha,\sin\alpha\,\theta').$$
Moreover, the angle between $\gamma$ and $\theta$ is $\alpha$,
$$\gamma.\theta=\cos \alpha.$$
The change of variable from $\theta$ to $\alpha$ and $\theta'$ is
$$d\theta=\sin^{n-2}\alpha \,d\alpha \,d\theta'.$$

Let us now apply this change of variables in (\ref{L}), and obtain
$$L_\lambda(\Omega,\Omega')=c\int_o^\pi (\cosh \rho-\sinh
\rho\,\cos\alpha)^{i\lambda-\frac{n-1}{2}}\sin^{n-2}\alpha \,d\alpha .$$
Therefore, in view of the integral form of $P^\mu_\nu(\cosh r)$
(\cite{B}), we obtain the identity of the Lemma.

\end{proof}

We shall now use the following Lemma, corresponding to Lemma 8.5.2 and
8.5.3 of \cite{Vl}.

\begin{lemma}\label{Legendre}
The function defined by
$$L_\lambda^n(\rho)=c (\sinh \rho)^{-\frac{n-2}{2}}
P^{-\frac{n-2}{2}}_{-\frac{1}{2}+i\lambda}(\cosh \rho),$$
is, for $\rho>0$ and for $n\geq 1$ odd,
\begin{equation}\label{Lodd}
L_\lambda^n(\rho)=c
\frac{|\Gamma(i\lambda)|^2}{|\Gamma(i\lambda+\frac{n-1}{2})|^2}\left(\frac{\partial_\rho}{\sinh\rho}\right)^\frac{n-1}{2}\cos\lambda\rho,
\end{equation}
and for $n\geq 2$ even,
\begin{equation}\label{Leven}
L_\lambda^n(\rho)=c
\frac{|\Gamma(i\lambda)|^2}{|\Gamma(i\lambda+\frac{n-1}{2})|^2}
\int_\rho^\infty\frac{\sinh s}{\sqrt{\cosh s-\cosh\rho}}
\left(\frac{\partial_s}{\sinh s}\right)^\frac{n}{2}\cos\lambda s\, ds.
\end{equation}
\end{lemma}

By using in the formula (\ref{kernel}) the expression of
$L_\lambda(\Omega,\Omega')$ given by the Lemma \ref{linkLegendre} and
Lemma \ref{Legendre}, we obtain that
$$u(t,\Omega)=ce^{-it\frac{(n-1)^2}{4}}\int_{\mathbb{H}^n}u_o(\Omega')K^n(t,d(\Omega,\Omega'))d\Omega',$$
where the kernel $K^n$ is, for$\rho>0$ and for $n\geq 3$ odd,

$$K^n(t,\rho)=\int_{-\infty}^\infty
e^{-it\lambda^2}
\left(\frac{\partial_\rho}{\sinh\rho}\right)^\frac{n-1}{2}\cos\lambda\rho\,
d\lambda,$$
and for $n\geq 2$ even,
\begin{equation}\label{Keven}
K^n(t,\rho)=\int_{-\infty}^\infty e^{-it\lambda^2}
\int_\rho^\infty\frac{\sinh s}{\sqrt{\cosh s-\cosh\rho}}
\left(\frac{\partial_s}{\sinh s}\right)^\frac{n}{2}\cos\lambda s\, ds\,
d\lambda.
\end{equation}

\par
In the case of odd dimensions, the kernel can be written
\begin{equation}\label{Kodd}
K^n(t,\rho)=\left(\frac{\partial_\rho}{\sinh\rho}\right)^\frac{n-1}{2}
\int_{-\infty}^\infty
e^{-it\lambda^2}\cos\lambda\rho\,
d\lambda=\frac{c}{|t|^\frac{1}{2}}\left(\frac{\partial_\rho}{\sinh\rho}\right)^\frac{n-1}{2}e^{i\frac{\rho^2}{4t}},\end{equation}
so the representation (\ref{odd}) of the Theorem \ref{exactsol} is obtained.

\par
In the case of even dimensions, the kernel can be expressed by
induction (\cite{Vl}, formula (8.5.23)) in terms of the kernel
of the dimension $2$
$$K^n(t,\rho)=\left(\frac{\partial_\rho}{\sinh
\rho}\right)^\frac{n-2}{2}K^2(t,\rho).$$
As $K^2$ can be written, after applying the Fubini theorem in
(\ref{Keven}), in the form
$$K^2(t,\rho)=\int_\rho^\infty\frac{1}{\sqrt{\cosh s-\cosh\rho}}
\int_{-\infty}^\infty e^{-it\lambda^2}\sin \lambda s\,\lambda \,d\lambda\,ds=
\frac{c}{|t|^\frac{3}{2}}\int_\rho^\infty\frac{e^{i\frac{s^2}{4t}}s}
{\sqrt{\cosh s-\cosh\rho}}ds,$$
we get that
\begin{equation}\label{kerpair}
K^n(t,\rho)=\frac{c}{|t|^\frac{3}{2}}\left(\frac{\partial_\rho}{\sinh
\rho}\right)^\frac{n-2}{2}\int_\rho^\infty\frac{e^{i\frac{s^2}{4t}}s}{\sqrt{\cosh
s-\cosh\rho}}ds,\end{equation}
so we obtain also the representation (\ref{even}) and the first point
of Theorem \ref{exactsol} is proved.

\section{The $L^1-L^\infty$ dispersion estimates}

We shall use the following Proposition, which shall be proved in the
Appendix A.
\begin{prop}\label{F}
For all $m\geq 1$ integer, the following identity holds
\begin{equation}\label{derexp}
\left(\frac{\partial_s}{\sinh s}\right)^me^{i\frac{s^2}{4t}}=
\sum_{k=1}^m \frac{e^{i\frac{s^2}{4t}}}{t^k}\,F_k^m(s),\end{equation}
where, modulo constants,
\begin{equation}\label{idFm}
F_m^m(s)=\left(\frac{s}{\sinh s}\right)^m,\end{equation}
and, for $1\leq k<m$, $F_k^m(s)$ can be written
\begin{equation}\label{idF}
F_k^m(s)=\sum_{r=0}^k\,\,\sum_{\left\{\begin{array}{c}i_1+\cdot\cdot\cdot+i_r=m-k\\i_j\geq 1\end{array}\right.}\left(\frac{s}{\sinh s}\right)^{k-r}
\left(\frac{\partial_s}{\sinh s}\right)^{i_1}
\frac{s}{\sinh s}...
\left(\frac{\partial_s}{\sinh s}\right)^{i_r}
\frac{s}{\sinh s}.\end{equation}
Moreover, for $\alpha\in\{0,1\}$ and $1\leq k\leq m$
\begin{equation}\label{estFk}
\left|\partial_s^\alpha\left(\frac{\sinh s}{s}\,F_k^m(s)\right)\right|\leq
c\,s^\alpha\left(\frac{s}{\sinh s}\right)^{m-1},\end{equation}
and the second derivative of $\frac{\sinh s}{s}\,F_k^m(s)$ is bounded for $m\geq 2$.

\end{prop}
\subsection{Dispersive estimates for odd dimensions}

From the expression (\ref{Kodd}), for odd dimensions, the kernel is for $\rho>0$
$$K^n(t,\rho)=\frac{c}{|t|^\frac{1}{2}}\left(\frac{\partial_\rho}{\sinh\rho}\right)^\frac{n-1}{2}e^{i\frac{\rho^2}{4t}},$$
so by (\ref{derexp}) it can be developed in powers of $t$ as follows
\begin{equation}\label{kerneldevp}
K^n(t,\rho)=\sum_{k=1}^{\frac{n-1}{2}} \frac{e^{i\frac{s^2}{4t}}}{t^{k+\frac{1}{2}}}F_k^{\frac{n-1}{2}}(\rho).
\end{equation}
Combining this with (\ref{estFk}) for $\alpha =0$ we obtain the dispersion estimates
(\ref{disph}) and (\ref{disphtlarge}) for all $n\geq 3$ odd.\\

\subsection{Local dispersion for even dimensions}
Between two kernels of consecutive order, there is a induction
relation, namely
$$K^n(t,\rho)=c\int_\rho^\infty \frac{\sinh s}{\sqrt{\cosh
s-\cosh\rho}}\,K^{n+1}(t,s)\,ds,$$
(see for example formula ($5.7.4$) in the book of Davies (\cite{D})).

Therefore, by using the development (\ref{kerneldevp}) of the kernels
of odd order, we can get the one for kernels of even order for $\rho>0$
\begin{equation}\label{kernel2devp}
K^n(t,\rho)=\sum_{k=1}^{\frac{n}{2}}\frac{1}{t^{k+\frac{1}{2}}}\int_\rho^\infty
\frac{e^{i\frac{s^2}{4t}}s}{\sqrt{\cosh s-\cosh \rho}}\frac{\sinh s}{s} F_k^\frac{n}{2}(s)\,ds.\end{equation}
Therefore the local dispersion (\ref{disph}) is obtained for $n=2$ by (\ref{idFm}) and the following Proposition. For all even dimensions $n\geq 4$, the estimates (\ref{mainoscintegral}) and (\ref{oscintegral}) of Corollary \ref{corol} give us
$$|K^n(t,\rho)|\leq \left(\frac{\rho}{\sinh \rho}\right)^\frac{n-1}{2}\left(\frac{1}{t^{\frac{n}{2}}}+\sum_{k=1}^{\frac{n-2}{2}}\frac{1}{t^{k+\frac{1}{2}}}\right),$$
and the local dispersion (\ref{disph}) follows.

\begin{prop}\label{decay}
The integral
$$I(t,\rho)=\int_\rho^\infty\frac{e^{i\frac{s^2}{4t}}s}{\sqrt{\cosh s-\cosh\rho}}ds$$
can be estimated, for small times, by
\begin{equation}\label{decayappendix}
|I(t,\rho)|\leq c\sqrt{t}\sqrt{\frac{\rho}{\sinh \rho}}.
\end{equation}
\end{prop}

\begin{proof}
In the following we shall use frequently the estimate
\begin{equation}\label{devlimite}
\frac{1}{\sqrt{\cosh s-\cosh\rho}}\leq
\frac{c}{(s-\rho)\sqrt{\cosh\rho}}\leq \frac{c}{s-\rho}
\end{equation}
for $s>\rho\geq 0$, and
\begin{equation}\label{devlimite2}
\frac{1}{\sqrt{\cosh s-\cosh\rho}}\leq \frac{c}{\sqrt{(s-\rho)\sinh\rho}}
\end{equation}
for $s>\rho>0$.\\

For small time, in order to get the decay (\ref{decayappendix}),
we have to get advantage of the imaginary phase. Let $t<\frac{1}{2}$.

{\underline{The case $\rho\geq\frac{\sqrt{t}}{2}$}}.\\

By doing in the integral $I(t,\rho)$ the change of variable
$s=\tau\frac{t}{\rho}+\rho$, we obtain
$$I(t,\rho)=\frac{te^{i\frac{\rho^2}{4t}}}{\rho}
\int_0^\infty
\frac{e^{i\tau^2\frac{t}{4\rho^2}+i\frac{\tau}{2}}(\tau\frac{t}{\rho}+\rho)}
{\sqrt{\cosh (\tau\frac{t}{\rho}+\rho)-\cosh\rho}}d\tau=$$
$$=2e^{i\frac{\rho^2}{4t}}\sqrt{t}\sqrt{\frac{\rho}{\sinh\rho}}
\int_0^\infty
\frac{e^{i\tau^2\frac{t}{4\rho^2}+i\frac{\tau}{2}}(\tau\frac{t}{2\rho^2}+\frac{1}{2})}
{\sqrt{\frac{\rho}{t\sinh\rho}\left(\cosh (\tau\frac{t}{\rho}+\rho)-\cosh\rho\right)}}d\tau.$$
We split the integral into two pieces
\begin{equation}\label{unifbound}
I(t,\rho)=2e^{i\frac{\rho^2}{4t}}\sqrt{t}\sqrt{\frac{\rho}{\sinh\rho}}
\left(\int_0^1 \, +\int_1^\infty\right)=2e^{i\frac{\rho^2}{4t}}\sqrt{t}\sqrt{\frac{\rho}{\sinh\rho}}
(J_1+J_2).
\end{equation}
We shall show that $J_1$ and $J_2$ are bounded independently of $t$ and
$\rho$.

By using (\ref{devlimite2}),
$$|J_1|\leq\frac{1}{2}\int_0^1\frac{\tau\frac{t}{\rho^2}+1}
{\sqrt{\frac{\rho}{t\sinh\rho}\tau\frac{t}{\rho}\sinh\rho}}=
\frac{1}{2}\int_0^1\frac{\tau\frac{t}{\rho^2}+1}{\sqrt{\tau}}.$$
Since we are in the case $\rho\geq\frac{\sqrt{t}}{2}$, it follows that
$$\frac{t}{\rho^2}\leq 4,$$
and therefore
$$|J_1|\leq\frac{1}{2}\int_0^1\frac{5}{\sqrt{\tau}}=5,$$
The second term can be written
$$J_2=\int_1^\infty e^{i\phi (\tau)}\phi'(\tau)\alpha(\tau)d\tau,$$
where
$$\phi(\tau)=\tau^2\frac{t}{4\rho^2}+\frac{\tau}{2},$$
and
$$\alpha(\tau)=\frac{1}{\sqrt{\frac{\rho}{t\sinh\rho}\left(\cosh (\tau\frac{t}{\rho}+\rho)-\cosh\rho\right)}}.$$
By integrating by parts in $J_2$ we get
$$J_2=\frac{e^{i\phi (\tau)}}{i}\alpha(\tau)|_1^\infty-
\int_1^\infty \frac{e^{i\phi (\tau)}}{i}\alpha'(\tau)d\tau,$$
so
\begin{equation}\label{I0}
|J_2|\leq 2\,\underset{1\leq\tau}{\sup}|\alpha|+\int_1^\infty |\alpha'(\tau)|d\tau.\end{equation}

Let us notice that by (\ref{devlimite2})
$$\alpha(1)\leq
\frac{1}{\sqrt{\frac{\rho}{t\sinh\rho}\frac{t}{\rho}\sinh\rho}}=1,$$
and that $\alpha$ is a decreasing function. Therefore,
$$|J_2|\leq 2+\int_1^\infty |\alpha'(\tau)|d\tau.$$
Since the derivative of $\alpha$ is negative,
$$|J_2|\leq 2+\left|\int_1^\infty \alpha'(\tau)d\tau\right|
=2+\left| \alpha(\tau)|_1^\infty\right|\leq 4.$$

Therefore we have obtained that $J_1$ and $J_2$ are bounded independently of $t$ and
$\rho$. In view of (\ref{unifbound}) the decay (\ref{decayappendix}) is obtained in the region
$\rho\geq\frac{\sqrt{t}}{2}$.

{\underline{The case $\rho<\frac{\sqrt{t}}{2}$}}.\\

Since $t<\frac{1}{2}$, then $\rho<\frac{1}{2\sqrt{2}}$, and by noticing that the quotient $\frac{\rho}{\sinh\rho}$ is bounded
near zero, it will be sufficient to prove that
\begin{equation}\label{small}
\int_\rho^\infty\frac{e^{i\frac{s^2}{4t}}s}{\sqrt{\cosh
s-\cosh\rho}}ds\leq c\sqrt{t}.
\end{equation}
Let us split this integral in three parts :
$$\int_\rho^\infty\frac{e^{i\frac{s^2}{4t}}s}{\sqrt{\cosh s-\cosh\rho}}ds=
\int_\rho^{2\rho}+\int_{2\rho}^{\sqrt{t}}+\int_{\sqrt{t}}^\infty=
I_1+I_2+I_3.$$

If $\rho=0$ then $I_1$ is zero, otherwise by (\ref{devlimite2})
$$|I_1|\leq \int_\rho^{2\rho}\frac{s}{\sqrt{\cosh s-\cosh\rho}}ds\leq
2\rho \int_\rho^{2\rho}\frac{1}{\sqrt{(s-\rho)\sinh \rho}}ds= $$
$$=4\rho\frac{\sqrt{\rho}}{\sqrt{\sinh \rho}}.$$
Since we are in the case $\rho<\frac{\sqrt{t}}{2}$ we get that
$$|I_1|\leq 2\sqrt{t}\sqrt{\frac{\rho}{\sinh \rho}}\leq c\sqrt{t}.$$

If $s\geq 2\rho$, which is the case in $I_2$, then by (\ref{devlimite})
$$\frac{s}{\sqrt{\cosh s-\cosh\rho}}\leq \frac{s}{s-\rho}\leq 3,$$
and we obtain
$$|I_2|\leq 3(\sqrt{t}-2\rho)\leq 3\sqrt{t}.$$

By performing in $I_3$ the change of variable
$s=\sqrt{t}\tau$,
$$I_3=\sqrt{t}\int_1^\infty e^{i\frac{\tau^2}{4}}\,a(\sqrt{t}\tau)d\tau,$$
where
$$a(s)=\frac{s}{\sqrt{\cosh s-\cosh\rho}}$$
for $s\geq\sqrt{t}$. By integrating by parts two times the integral $I_3$, we get
$$|I_3|\leq c\sqrt{t},$$
provided that the following lemma holds. (see
Theorem I.8.1 of \cite{AG}).

\begin{lemma}\label{ab}
The function $a(\sqrt{t}\tau)$, its first and second
derivative in $\tau$, are all bounded independently of $t$, for all
$\tau\geq 1$.
\end{lemma}

\begin{proof}
By using the same argument as in the estimate of $I_2$, we have that
$a(\sqrt{t}\tau)$ is upper-bounded by $3$ for $\tau\geq 1$.

The derivative of $a$ is
$$a'(s)=\frac{1}{\sqrt{\cosh s-\cosh\rho}}-
\frac{s\sinh s}{2(\cosh s-\cosh\rho)^{\frac{3}{2}}}.$$
Since we are in the region $s\geq\sqrt{t}$,
$\rho<\frac{\sqrt{t}}{2}$, the first term can be estimated by (\ref{devlimite})
$$\frac{1}{\sqrt{\cosh s-\cosh\rho}}\leq \frac{c}{s-\rho}\leq \frac{c}{\sqrt{t}-\frac{\sqrt{t}}{2}}
\leq \frac{c}{\sqrt{t}}.$$
The second term in the derivative of $a$
can be written as
$$-\frac{a(s)}{2}\frac{\sinh s}{\cosh s-\cosh\rho}.$$
Since $s\geq\sqrt{t}>\rho$, the second fraction is positive, and as $a$ was already proved to be bounded, we get
$$|a'(s)|\leq \frac{c}{\sqrt{t}}+c\,\underset{s\geq\sqrt{t}}{\sup}\,\,\frac{\sinh s}{\cosh s-\cosh\rho}.$$
We have
$$\left(\frac{\sinh s}{\cosh s-\cosh\rho}\right)'=\frac{1-\cosh s\cosh\rho}{\left(\cosh s-\cosh\rho\right)^2}\leq 0,$$
so it follows that
$$\underset{s\geq\sqrt{t}}{\sup}\,\,\frac{\sinh s}{\cosh s-\cosh\rho}=\frac{\sinh \sqrt{t}}{\cosh \sqrt{t}-\cosh\rho}.$$
As $t<\frac{1}{2}$ and $\rho<\frac{\sqrt{t}}{2}$,
\begin{equation}\label{decr}
\underset{s\geq\sqrt{t}}{\sup}\,\,\frac{\sinh s}{\cosh s-\cosh\rho}\leq c\frac{\sqrt{t}}{(\sqrt{t}-\rho)^2}\leq c\frac{\sqrt{t}}{(\sqrt{t}-\frac{\sqrt{t}}{2})^2}\leq \frac{c}{\sqrt{t}},\end{equation}
and we have obtained that for $s\geq\sqrt{t}$
$$|a'(s)|\leq \frac{c}{\sqrt{t}}.$$
It follows that
$$\partial_\tau a(\sqrt{t}\tau)=\sqrt{t}a'(\sqrt{t}\tau)$$
is bounded independently of $t$ for $\tau\geq 1$.

Finally, the second derivative of $a$ is
$$a''(s)=-\frac{\sinh s}{2(\cosh s-\cosh\rho)^{\frac{3}{2}}}-\frac{s\cosh s}{(\cosh s-\cosh\rho)^{\frac{3}{2}}}+
\frac{3s\sinh^2s}{4(\cosh s-\cosh\rho)^{\frac{5}{2}}}.$$
By using (\ref{decr}) and then (\ref{devlimite}) we can treat the first term
$$\frac{\sinh s}{2(\cosh s-\cosh\rho)^{\frac{3}{2}}}\leq \frac{c}{\sqrt{t}}\frac{1}{\sqrt{\cosh s-\cosh\rho}}\leq \frac{c}{\sqrt{t}}\frac{c}{s-\rho}\leq \frac{c}{t}.$$
The second term in the derivative can be written
$$\frac{s\cosh s}{(\cosh s-\cosh\rho)^{\frac{3}{2}}}=a(s)\frac{\cosh s}{\cosh s-\cosh\rho}=a(s)\left(1+\frac{\cosh \rho}{\cosh s-\cosh\rho}\right).$$
As $a$ was proved to be bounded and as the fraction in the brackets is decreasing, for $s\geq \sqrt{t}$
$$\frac{s\cosh s}{(\cosh s-\cosh\rho)^{\frac{3}{2}}}\leq c\left(1+\frac{\cosh\rho}{(\sqrt{t}-\rho)^2}\right).$$
We are in the region $\rho<\frac{\sqrt{t}}{2}<\frac{1}{2\sqrt{2}}$, so the second term in the derivative can be estimated
$$\frac{s\cosh s}{(\cosh s-\cosh\rho)^{\frac{3}{2}}}\leq c\left(1+\frac{c}{t}\right)\leq \frac{c}{t}.$$
Finally, the third term in the derivative can be written as
$$\frac{3a(s)}{4}\left(\frac{\sinh s}{\cosh s-\cosh\rho}\right)^2,$$
and can be upper-bounded using (\ref{decr}) by $\frac{c}{t}$.
In conclusion, for $s\geq\sqrt{t}$
$$|a''(s)|\leq \frac{c}{t}.$$
So we get that
$$\partial_\tau ^2a(\sqrt{t}\tau)=ta''(\sqrt{t}\tau)$$
is bounded independently of $t$ for $\tau\geq 1$.
\end{proof}

In conclusion, we have obtained that the three integrals $I_1, I_2$
and $I_3$ are upper-bounded by $c\sqrt{t}$, and (\ref{small})
follows. Therefore the estimate (\ref{decayappendix}) is proved and
Proposition \ref{decay} follows.

\end{proof}

\begin{corollary}\label{corol}
For small times and $n\geq 4$ the following estimates hold
\begin{equation}\label{mainoscintegral}
\left|\int_\rho^\infty\frac{e^{i\frac{s^2}{4t}}s}{\sqrt{\cosh s-\cosh\rho}}
\frac{\sinh s}{s}F^{\frac{n}{2}}_{\frac{n}{2}}(s)\,ds\right|\leq
c\sqrt{t}\left(\frac{\rho}{\sinh \rho}\right)^{\frac{n-1}{2}},
\end{equation}
and for $1\leq k< \frac{n}{2}$
\begin{equation}\label{oscintegral}
\left|\int_\rho^\infty\frac{e^{i\frac{s^2}{4t}}s}{\sqrt{\cosh s-\cosh\rho}}
\frac{\sinh s}{s}F^{\frac{n}{2}}_k(s)\,ds\right|\leq
c\left(\frac{\rho}{\sinh \rho}\right)^{\frac{n-1}{2}}.\end{equation}
\end{corollary}

\begin{proof}
We redo the proof of Proposition \ref{decay}, which is the particular
case $n=2$ of its Corollary. The only delicate point will be the integral $J_2$ which will generate
the two different estimates (\ref{mainoscintegral}) and (\ref{oscintegral}).

By using the upper-bound (\ref{estFk}) for $\alpha =0$ and $1\leq k\leq\frac{n}{2}$, and the fact that the integration is done for $s\geq\rho$,
$$\left|\frac{\sinh s}{s}F^{\frac{n}{2}}_k(s)\right|\leq \left(\frac{s}{\sinh s}\right)^{\frac{n-2}{2}}\leq \left(\frac{\rho}{\sinh \rho}\right)^{\frac{n-2}{2}}.$$
Therefore everytime the imaginary phase is ignored, we find ourselves in the same situation as in Proposition \ref{decay}, and we obtain the desired estimates.

The imaginary phase has been taken in account only in the terms $J_2$
and $I_3$.

For estimating $I_3$ we have to prove Lemma \ref{ab} with
$$\widetilde{a}(s)=a(s)\frac{\sinh s}{s}F^{\frac{n}{2}}_k(s).$$
We had that $a$ is bounded by a constant, its first derivative bounded
by $ct^{-\frac{1}{2}}$, and its second derivative bounded by
$ct^{-1}$. We want the same for $\widetilde{a}$. As $t$ is small, it will be sufficient to prove that $\frac{\sinh s}{s}F^{\frac{n}{2}}_k(s)$ is
bounded, and also are its first and second derivative. Since we are in the case $n\geq 4$, Proposition \ref{F} tells us already that its second derivative is bounded. Moreover, for $1\leq k\leq\frac{n}{2}$, by (\ref{estFk}) for $\alpha =0$
$$\left|\frac{\sinh s}{s}F^{\frac{n}{2}}_k(s)\right|\leq \left(\frac{s}{\sinh
s}\right)^{\frac{n-2}{2}},$$
and by (\ref{estFk}) for $\alpha =1$
$$\left|\partial_s\frac{\sinh s}{s}F^{\frac{n}{2}}_k(s)\right|\leq s\left(\frac{s}{\sinh
s}\right)^{\frac{n-2}{2}}=\frac{s^2}{\sinh s}\left(\frac{s}{\sinh
s}\right)^{\frac{n-4}{2}}.$$
Again, since we are in the case $n\geq 4$, these two quantities are bounded, so Lemma \ref{ab} still holds in the context of the Corollary.

In conclusion, up to the term $J_2$, our integrals have the decay of Proposition \ref{decay}
$$\sqrt{t}\sqrt{\frac{\rho}{\sinh \rho}}$$
improved by the extra term
$$\left(\frac{\rho}{\sinh \rho}\right)^{\frac{n-2}{2}}.$$

Let us now first treat $J_2$ for obtaining (\ref{mainoscintegral}). By (\ref{idFm}),
$$\frac{\sinh s}{s}F^{\frac{n}{2}}_{\frac{n}{2}}(s)=\left(\frac{s}{\sinh
s}\right)^{\frac{n-2}{2}}.$$
Therefore the same arguments as in Proposition \ref{decay} can be performed in $J_2$, by replacing the
function $\alpha(\tau)$ by

$$\alpha(\tau)\left(\frac{\tau\frac{t}{\rho}+\rho}{\sinh\left(\tau\frac{t}{\rho}+\rho\right)}\right)^{\frac{n-2}{2}},$$
which is still a decreasing function, bounded at $1$ by
$\left(\frac{\rho}{\sinh\rho}\right)^{\frac{n-2}{2}}$. So $J_2$ will
be also upper-bounded by $\left(\frac{\rho}{\sinh\rho}\right)^{\frac{n-2}{2}}$ and (\ref{mainoscintegral}) is proved.

In view to obtain (\ref{oscintegral}), we restart the argument performed on $J_2$ in Proposition \ref{decay}, with $\alpha(\tau)$ replaced by
$$\alpha(\tau)\frac{\sinh s}{s}F^{\frac{n}{2}}_k(s),$$
where
$$s=\tau\frac{t}{\rho}+\rho.$$
When getting to (\ref{I0}) we need to estimate
$$I_4=\int_1^\infty\left|\partial_\tau\left(\alpha(\tau)\frac{\sinh s}{s}F^{\frac{n}{2}}_k(s)\right)\right|d\tau.$$
Now the function $\frac{\sinh s}{s}F^{\frac{n}{2}}_k(s)$ is not necessarily decreasing, so we are not able to get rid of the modulus in the integral, as easily as before. We will be able to estimate
\begin{equation}\label{I4}
I_4\leq\frac{1}{\sqrt{t}}\left(\frac{\rho}{\sinh\rho}\right)^{\frac{n-2}{2}},\end{equation}
that is with a loss of $\sqrt{t}$.\\
In the derivation of the product,
$$I_4=
\int_1^\infty\left|\alpha'(\tau)\frac{\sinh s}{s}F^{\frac{n}{2}}_k(s)+\alpha(\tau)\frac{t}{\rho}\partial_s\left(\frac{\sinh s}{s}F^{\frac{n}{2}}_k(s)\right)\right|d\tau,$$
we can use the estimate (\ref{estFk}) for $\alpha\in\{0,1\}$ and then
$$I_4\leq
\left(\frac{\rho}{\sinh\rho}\right)^{\frac{n-2}{2}}\int_1^\infty\left|\alpha'(\tau)\right|d\tau+
\left(\frac{\rho}{\sinh\rho}\right)^{\frac{n-2}{2}}\int_1^\infty\left|
\alpha(\tau)\frac{t}{\rho}\left(\tau\frac{t}{\rho}+\rho\right)\right|d\tau.$$
The first integral is exactly the one in Proposition \ref{decay}, so it has been proved that it is bounded independently of $t$ and $\rho$. For treating the second integral, we shall use that for all odd $N$,
$$\alpha(\tau)=\frac{1}{\sqrt{\cosh \left(\tau\frac{t}{\rho}+\rho\right)-\cosh \rho}}\leq
\frac{1}{\sqrt{\sinh\rho}}\left(\tau\frac{t}{\rho}\right)^{-\frac{N}{2}}.$$
Using this estimate for $N=5$ and for $N=3$,
$$\int_1^\infty\left|
\alpha(\tau)\frac{t}{\rho}\left(\tau\frac{t}{\rho}+\rho\right)\right|d\tau\leq
\frac{1}{\sqrt{\sinh\rho}}\int_1^\infty \left(\left(\tau\frac{t}{\rho}\right)^{-\frac{5}{2}}\tau\left(\frac{t}{\rho}\right)^2+\left(\tau\frac{t}{\rho}\right)^{-\frac{3}{2}}\frac{t}{\rho}\rho\right)d\tau.$$
The integrals in $\tau$ are bounded, so
$$\int_1^\infty\left|\alpha(\tau)\frac{t}{\rho}\left(\tau\frac{t}{\rho}+\rho\right)\right|d\tau\leq
\frac{1}{\sqrt{\sinh\rho}}\left(\frac{t}{\rho}\right)^{-\frac{1}{2}}(1+\rho)\leq \frac{c}{\sqrt{t}},$$
so estimate (\ref{I4}) follows and the (\ref{oscintegral}) is proved.
\end{proof}

\subsection{Large time dispersion for even dimensions}

By using the development (\ref{kernel2devp}) of the dispersion kernel and the estimates (\ref{estFk}) for $\alpha =0$ we obtain
\begin{equation}\label{kernel2devplarget}
|K^n(t,\rho)|\leq \sum_{k=1}^\frac{n}{2}\frac{1}{t^{k+\frac{1}{2}}}\left(\frac{\rho}{\sinh \rho}\right)^\frac{n-2}{2}\left|\int_\rho^\infty
\frac{s}{\sqrt{\cosh s-\cosh \rho}}\,ds\right|.\end{equation}
Let us notice that we have not taken into account the oscillatory phase in
the integral. We shall split the
remaining integral into two pieces
$$\left| \int_\rho^\infty\frac{s}{\sqrt{\cosh s-\cosh\rho}}ds \right|=
\left| \int_0^\infty\frac{s+\rho}{\sqrt{\cosh (s+\rho)-\cosh\rho}}ds \right|=
\int_0^1+\int_1^\infty.$$
For $\rho>0$, the first part is upper-bounded by
$$\int_0^1\frac{s+\rho}{\sqrt{s\sinh\rho}}ds\leq c\frac{1+\rho}{\sqrt{\sinh\rho}},$$
and the second, for a $N$ odd integer sufficiently large, by
$$\int_1^\infty\frac{s+\rho}{\sqrt{s^N\sinh\rho}}ds\leq c\frac{1+\rho}{\sqrt{\sinh\rho}}.$$
Now, by (\ref{kernel2devplarget}) we get for $\rho>0$
$$|K^n(t,\rho)|\leq \sum_{k=1}^\frac{n}{2}\frac{1}{t^{k+\frac{1}{2}}}\left(\frac{\rho}{\sinh \rho}\right)^\frac{n-2}{2}\frac{1+\rho}{\sqrt{\sinh\rho}}.$$
Therefore we have obtained the dispersion estimate (\ref{disphtlarge2}) for large times, and the last part
of Theorem \ref{disp} i) follows.

\section{The weighted dispersion estimates}
Let us prove now the weighted dispersion (\ref{dispw}). From
(\ref{disph}) we have that for small times and $n\geq 3$
\begin{equation}\label{ppp}
|u(t,\Omega)|\leq \frac{c}{|t|^\frac{n}{2}}
\int_{\mathbb{H}^n}|u_0(\Omega')|\left(\frac{\rho}{\sinh\rho}\right)^\frac{n-1}{2}\,d\Omega'\leq \frac{c}{|t|^\frac{n}{2}}
\int_{\mathbb{H}^n}|u_0(\Omega')|\frac{\rho}{\sinh\rho}\,d\Omega'.
\end{equation}
The initial data is considered here radial, that is
$$u_0(\Omega')=u_0(\cosh r', \sinh r'\omega')=u_0(r').$$
Also in hyperbolic coordinates we can write, using (\ref{prod}) and
(\ref{dist})
$$\rho=d(\Omega,\Omega')=\cosh^{-1}[\Omega,\Omega']=
\cosh ^{-1}(\cosh r\cosh r'-\sinh r\sinh r'\,\omega.\omega').$$
Therefore, by passing in hyperbolic coordinates in (\ref{ppp}),
\begin{equation}\label{qqq}
|u(t,\Omega)|=|u(t,r)|\leq \frac{c}{|t|^\frac{n}{2}}
\int_0^\infty |u_0(r')|K(t,r,r')\sinh^{n-1}r'\,dr',\end{equation}
where
$$K(t,r,r')=\int_{\mathbb{S}^{n-1}}
\frac{\cosh ^{-1}(\cosh r\cosh r'-\sinh r\sinh r'\,\omega.\omega')}
{\sinh \cosh ^{-1}(\cosh r\cosh r'-\sinh r\sinh
r'\,\omega.\omega')}\,d\omega'.$$
By doing a rotation as in the proof of the Lemma \ref{linkLegendre},
$$K(t,r,r')=c\int_0^\pi
\frac{\cosh ^{-1}(\cosh r\cosh r'-\sinh r\sinh r'\,\cos\alpha)}
{\sinh \cosh ^{-1}(\cosh r\cosh r'-\sinh r\sinh
r'\,\cos\alpha)}\sin^{n-2}\alpha\,d\alpha.$$
Now let us do the change of variable $\cos\alpha=x$ and get
$$K(t,r,r')=c\int_{-1}^1
\frac{\cosh ^{-1}(\cosh r\cosh r'-\sinh r\sinh r'\,x)}
{\sinh \cosh ^{-1}(\cosh r\cosh r'-\sinh r\sinh
r'\,x)}(1-x^2)^\frac{n-3}{2}dx.$$
Since here $n$ is larger or equal to $3$,
$$K(t,r,r')\leq c\int_{-1}^1
\frac{\cosh ^{-1}(\cosh r\cosh r'-\sinh r\sinh r'\,x)}
{\sinh \cosh ^{-1}(\cosh r\cosh r'-\sinh r\sinh
r'\,x)}dx.$$
Finally, let us do the change of variable
$$\cosh r\cosh r'-\sinh r\sinh r'\,x=\cosh y,$$
and then
$$K(t,r,r')\leq \frac{c}{\sinh r\sinh r'}\int_{|r-r'|}^{r+r'} y\,dy=
c\frac{r\,r'}{\sinh r\sinh r'}.$$
Using now (\ref{qqq}), we obtain that
$$|u(t,\Omega)|=|u(t,r)|\leq \frac{c}{|t|^\frac{n}{2}}
\int_0^\infty |u_0(r')|\frac{r\,r'}{\sinh r\sinh r'}\sinh^{n-1}r'\,dr'.$$
Remembering that $r=d(\Omega,0)$ we can go back to the
$\Omega$-coordinates and
$$|u(t,\Omega)|\,\frac{\sinh d(\Omega,0)}{d(\Omega,0)}\leq \frac{c}{|t|^\frac{n}{2}}
\int_{\mathbb{H}^n}|u_0(\Omega')|\,\frac{d(\Omega',0)}{\sinh
d(\Omega',0)}d\Omega',$$
so the weighted local dispersion estimate (\ref{dispw}) is proved.\\

For proving the weighted Strichartz estimates we will apply the general
lemma of Keel and Tao (Theorem 10.1 of \cite{KT}), in the case
$$H=B_0=L^2_{rad}\;\;,\;\;B_1=L^1_{rad}(w^{-1})\;\;,\;\;
\sigma=\frac{n}{2}\;\;,\;\;2\leq \frac{4}{n\theta}\;\;,\;\;
0\leq\theta\leq 1 \;\;,\;\;\left(\frac{4}{n\theta},\theta,\sigma\right)\neq (2,1,1).$$
We obtain that for radial $u$
$$\|u\|_{L^\frac{4}{n\theta}\left([0,T],\,
(L^2,L^1(w^{-1}))^*_{\theta,2}\right)}\leq
c\|u_0\|_{L^2},$$
where $(L^2,L^1(w^{-1}))^*_{\theta,2}$ is the dual of the real
interpolation space between $L^2$ and $L^1(w^{-1})$. The conditions on $\theta$ can be rewritten, for $n\geq 2$,
$$0\leq \theta\leq \frac{2}{n}\;\;,\;\;(\theta,n)\neq (1,2).$$

For any $r\leq 2$ we have
$$(L^2,L^1(w^{-1}))_{\theta,r}\subseteq (L^2,L^1(w^{-1}))_{\theta,2},$$
so that
\begin{equation}\label{rrr}
\|u\|_{L^\frac{4}{n\theta}\left([0,T],\,
(L^2,L^1(w^{-1}))^*_{\theta,r}\right)}\leq
c\|u_0\|_{L^2}.\end{equation}
By the Theorem 5.5.1. of \cite{BL}, for $0<\theta<1$,
$$(L^2,L^1(w^{-1}))_{\theta,r}=L^r(\widetilde{w}),$$
with
$$\frac{1}{r}=\frac{1-\theta}{2}+\frac{\theta}{1}\;\;\;\;,\;\;\;\;
\widetilde{w}=w^{-r\theta}.$$
So the value of $r$ has to be $\frac{2}{1+\theta}$, which implies indeed that $r\leq 2$, and we have
$$(L^2,L^1(w))_{\theta,r}=L^\frac{2}{1+\theta}(w^{-\frac{2\theta}{1+\theta}}).
$$
By using the classical formula
$$L^p(w)^*=L^{p'}(w^{-\frac{p'}{p}})$$
(see for example (7.4.15) of \cite{Vl} with $s=0$ and the definition of $L^p(w)$ corresponding to $L^p(w^\frac{1}{p})$ in our notations) we get that
$$(L^2,L^1(w))^*_{\theta,r}=L^\frac{2}{1-\theta}(w^{\frac{2\theta}{1-\theta}}).$$

Now the relation (\ref{rrr}) gives us the weighted Strichartz
estimates
$$\|u\|_{L^\frac{4}{n\theta}\left([0,T],\,
L^{\frac{2}{1-\theta}}(w^{\frac{2\theta}{1-\theta}})\right)}\leq
c\|u_0\|_{L^2},$$
for all $\theta$ in $]0,\frac{2}{n}]$ satisfying $(\theta,n)\neq (1,2)$. Let us notice that we are able to include the value $\theta=0$, since in this case the estimate corresponds to the mass conservation.

By denoting
$$p=\frac{4}{n\theta}\,\,\,,\,\,\,q=\frac{2}{1-\theta},$$
we have that
$$\frac{2\theta}{1-\theta}=-2+\frac{2}{1-\theta}=q-2.$$
Also, the couple $(p,q)$ satisfies $\frac{2}{p}+\frac{n}{q}=\frac{n}{2}$, $(p,q,n)\neq (2,\infty,2)$ and $2\leq p,q$, therefore the weighted Strichartz estimates (\ref{Strw}) are found and the Theorem \ref{disp} is completely proved.

\subsection{Proof of the Remark \ref{dispsupp}}
$$ $$
\par

Let us choose a point $\Omega$ outside the support $A$ of the initial
data, and a point $\Omega'$ in $A$.
Let us denote $M$ a point of the intersection of the boundary of $A$
with the part of geodesic relying
$\Omega$ and $\Omega'$. Then
$$d(\Omega,\Omega')=d(\Omega,M)+d(M,\Omega'),$$
and
$$d(\Omega,\Omega')\geq d(\Omega,^cA)+d(\Omega',A),$$
As the function $\frac{r}{\sinh r}$ is a decreasing function,
$$\frac{d(\Omega,\Omega')}{\sinh d(\Omega,\Omega')}\leq
\frac{d(\Omega,^cA)+d(\Omega',A)}{\sinh \left(d(\Omega,^cA)+d(\Omega',A)\right)}.$$
By using that
$$\frac{a+b}{\sinh (a+b)}\leq c
\frac{a}{\sinh a}\frac{b}{\sinh b},$$
we finally obtain
$$\frac{d(\Omega,\Omega')}{\sinh d(\Omega,\Omega')}\leq
c\,\frac{d(\Omega,^cA)}{\sinh d(\Omega,^cA)}\,
\frac{d(\Omega',A)}{\sinh d(\Omega',A)}.$$
By using now the inequality (\ref{disph}) we obtain the estimate of
the Remark \ref{dispsupp}.

\subsection{Proof of the Remark \ref{disptime}}
$$ $$
\par

We shall follow the same approach as in \S 3.2 for proving the
weighted dispersion estimate (\ref{dispw}).\\
In dimension $3$, by the explicit form (\ref{odd}), the relation
(\ref{ppp}) becomes
$$u(t,\Omega)=\frac{c}{|t|^\frac{3}{2}}
\int_{\mathbb{H}^n}u_0(\Omega')\frac{\rho}{\sinh\rho}
e^{i\frac{\rho^2}{4t}}\,d\Omega',$$
and the kernel $K$ given by
$$u(t,\Omega)=\frac{c}{|t|^\frac{3}{2}}
\int_0^\infty |u_0(r')|K(t,r,r')\sinh^{n-1}r'\,dr',$$
is exactly
$$K(t,r,r')=\frac{c}{\sinh r\sinh r'}\int_{|r-r'|}^{r+r'}e^{i\frac{y^2}{4t}} y\,dy=t\,\frac{c}{\sinh r\sinh r'}e^{i\frac{r^2+r'^2}{4t}}\cos\frac{rr'}{2t}.$$
Therefore
$$|u(t,\Omega)|\,\sinh d(\Omega,0)\leq
\frac{c}{|t|^\frac{1}{2}}
\int_{\mathbb{H}^n}|u_0(\Omega')|\,\frac{d\Omega'}{\sinh d(\Omega',0)},$$
so the dispersion estimate of the Remark
\ref{disptime} is proved.\\

\section{Global existence and blow-up solutions}

\subsection{Global existence}
$$ $$

The Sobolev embeddings have their analogue on hyperbolic space (\cite{He})
$$\|v\|_{2^*}\leq K(n,2) \|\nabla v\|_2-\omega_n^{-\frac{2}{n}}\|v\|_2,$$
where $K(n,2)$ is the best constant for the Sobolev embeddings on $\RR^n$,
and $\omega_n$ is the volume of the sphere $\mathbb{S}^n$.
By interpolation between the $L^2$ and the $L^{2^*}$ norms,
we get the Gagliardo-Nirenberg inequality for functions on $\mathbb{H}^n$
$$\|v\|_{p+1}^{p+1}\leq C_{p+1} \|v\|_2^{2+(p-1)\frac{2-n}{2}}
\|\nabla v\|_2^{(p-1)\frac{n}{2}}.$$
This inequality implies that the energy of the solution $u$ of the
equation $(S)$ is bounded from below by
$$\|\nabla u\|_2^2\left(\frac{1}{2}-
\frac{C_{p+1}}{p+1}\|u\|_2^{2+(p-1)\frac{2-n}{2}}
\|\nabla u\|_2^{(p-1)\frac{n}{2}-2}\right).$$
As a consequence, if $p<1+\frac{4}{n}$, since the mass is conserved,
the gradient of $u$ is controlled by the energy.
Therefore the solution does not blow up and global existence occurs.\\
In the case $p=1+\frac{4}{n}$, if the mass of the initial
condition is small enough so that
$$\|u\|_2^{\frac{4}{n}}<
\frac{2+\frac{4}{n}}{2C_{2+\frac{4}{n}}},$$
then the energy controls the gradient and again, the global existence is
proved for the equation $(S)$.\\

\begin{remark}
The best constant in the Gagliardo-Niremberg inequality can be proved to be
larger or equal to the one on $\RR^n$, but it is not obvious that it is exactly equal to it.
\end{remark}

\subsection{Blow-up solutions}
$$ $$

The power $p=1+\frac{4}{n}$ shall be proved to be the critical power,
in the sense that the
nonlinearity is strong enough to generate solutions blowing up in a
finite time. In the following we shall show the existence of blow-up
solutions, by analyzing an appropriate virial function on hyperbolic space.

Let $u$ be a radial solution of $(S)$ and let $h$ be a radial
$\mathcal{C}_0^\infty(\mathbb{H}^n)$ function.
Then, by using the fact that $u$ satisfies $(S)$, we obtain the first
derivative in time of a virial-type function
$$\partial_t\int_{\mathbb{H}^n}|u(t)|^2h \,d\Omega=
2\int_{\mathbb{H}^n} \Re \left(u(t)\overline{u}_t(t)\right)h \,d\Omega=
2\int_{\mathbb{H}^n} \Im \left(u(t)\Delta\overline{u}(t)\right)h \,d\Omega.$$
By integrating by parts
\begin{equation}\label{g'}
\partial_t\int_{\mathbb{H}^n}|u(t)|^2h \,d\Omega=
-2\int_{\mathbb{H}^n} \Im \left(u(t)\nabla\overline{u}(t)\right)\nabla h \,d\Omega.
\end{equation}
By using again the equation $(S)$ we obtain
$$\DP{t}^2\int_{\mathbb{H}^n}\mo{u}^2h=
-2\int_{\mathbb{H}^n} \Re \left((\Delta u+|u|^{p-1}u)\nabla\overline{u}\right)
\nabla h+
2\int_{\mathbb{H}^n}\Re \left(u\nabla(\Delta \overline{u}+|u|^2\overline{u})
\right)
\nabla h$$

$$=\int_{\mathbb{H}^n}
2|\nabla u|^2\Delta h
-\left(2-\frac{4}{p+1}\right)
|u|^{p+1}\Delta h-
4\Re\left(\Delta u\nabla\overline{u}\right)\nabla h-
|u|^2\Delta^2h.$$
Since $u$ and $h$ are radial functions, it follows that
$$\DP{t}^2\int_{\mathbb{H}^n}\mo{u}^2h=
\int_{\mathbb{H}^n}
4|\nabla u|^2\partial_r^2 h-
2\frac{p-1}{p+1}
|u|^{p+1}\Delta h-
|u|^2\Delta^2h.$$
By performing a density argument in the spirit of (\cite{Ca}, Lemmas 6.4.3-6), the
previous identities are valid for
$$h(\Omega)=h(\cosh r,\sinh r\omega)=r^2.$$
By direct computation we obtain
$$\partial_r^2 r^2=2,$$
and
$$\Delta r^2=2+2(n-1)\frac{\cosh r}{\sinh r}r.$$
Therefore we can display the energy in the second derivative of the
virial
\begin{equation}\label{vir}
\DP{t}^2\int_{\mathbb{H}^n}\mo{u}^2r^2=
16E(u)-\int_{\mathbb{H}^n}|u|^2\Delta^2r^2-
2\frac{p-1}{p+1}\int_{\mathbb{H}^n} |u|^{p+1}2(n-1)
\left(\frac{\cosh r}{\sinh r}r-1\right)-
\end{equation}

$$-\left(-\frac{16}{p+1}+
2\frac{p-1}{p+1}2n\right)\int_{\mathbb{H}^n} |u|^{p+1}.$$
We have
\begin{equation}\label{delta2}
\Delta^2 r^2=4(n-1)\frac{1}{\sinh^3r}(r\cosh r-\sinh r)+
2(n-1)^2\frac{\cosh r}{\sinh^3r}(\cosh r\sinh r-r),\end{equation}
nonnegative for all $r\geq 0$, since
$$\left\{\begin{array}{c}r\cosh r-\sinh r\geq 0\\
\cosh r\sinh r-r\geq 0\end{array}\right..$$
Therefore, since $p$ is considered larger than $1$, the second and the third
term in the right hand side of (\ref{vir}) are negative. The last one is also negative, provided that
$$-\frac{16}{p+1}+
4n\frac{p-1}{p+1}\geq 0,$$
that is exactly
$$p\geq 1+\frac{4}{n}.$$
Let us notice,
by (\ref{delta2}), that $\Delta^2 r^2$ is bounded between two
positive constants depending only on the dimension $n$,
$$0<k_n<\Delta^2 r^2<K_n.$$

In conclusion, if $u$ is a solution of $(S)$ of initial data satisfying
$$16E(u_0)<k_n\|u_0\|^2_2,$$
there exists a constant $C$ such that for any $t$
\begin{equation}\label{negder}
\DP{t}^2\int_{\mathbb{H}^n}\mo{u(t)}^2r^2<C<0.\end{equation}
It follows that there is a finite time $T$ for which
$$\lim_{t\tend T}\int_{\mathbb{H}^n}\mo{u(t)}^2r^2=0.$$
Then, using the uncertainty principle
$$\left(\int_{\mathbb{H}^n}\mo{u}^2\right)^2\leq c
\left(\int_{\mathbb{H}^n}\mo{u}^2r^2\right)
\left(\int_{\mathbb{H}^n}\mo{\nabla u}^2\right),$$
tells us that
$$\lim_{t\tend T}\int_{\mathbb{H}^n}\mo{\nabla u(t)}^2=+\infty,$$
so the Theorem \ref{blup} is proved.

Finally, let us mention that in the Euclidean case, one has
$$\DP{t}^2\int_{\mathbb{R}^n}\mo{u}^2|x|^2=16E(u),$$
so the argument used before does not work for any solutions of null energy. \\

\section{Appendix A}
In this Appendix the following Proposition shall be proved.\\

{\bf{Proposition \ref{F}}}
{\it{For all $m\geq 1$ integer, the following identity holds
$$(\ref{derexp})\,\,\,
\left(\frac{\partial_s}{\sinh s}\right)^me^{i\frac{s^2}{4t}}=
\sum_{k=1}^m \frac{e^{i\frac{s^2}{4t}}}{t^k}\,F_k^m(s),$$
where, modulo constants,
$$(\ref{idFm})\,\,\,
F_m^m(s)=\left(\frac{s}{\sinh s}\right)^m,$$
and, for $1\leq k<m$, $F_k^m(s)$ can be written
$$(\ref{idF})\,\,\,
F_k^m(s)=\sum_{r=0}^k\,\,\sum_{\left\{\begin{array}{c}i_1+\cdot\cdot\cdot+i_r=m-k\\i_j\geq 1\end{array}\right.}\left(\frac{s}{\sinh s}\right)^{k-r}
\left(\frac{\partial_s}{\sinh s}\right)^{i_1}
\frac{s}{\sinh s}...
\left(\frac{\partial_s}{\sinh s}\right)^{i_r}
\frac{s}{\sinh s}.$$
Moreover, for $\alpha\in\{0,1\}$ and $1\leq k\leq m$
$$(\ref{estFk})\,\,\,
\left|\partial_s^\alpha\left(\frac{\sinh s}{s}\,F_k^m(s)\right)\right|\leq
c\,s^\alpha\left(\frac{s}{\sinh s}\right)^{m-1},$$
and the second derivative of $\frac{\sinh s}{s}\,F_k^m(s)$ is bounded for $m\geq 2$.}}\\

\begin{proof}
Each time we differentiate the exponential $e^{i\frac{s^2}{4t}}$ we obtain a term with one negative power of $t$ times $\frac{s}{\sinh s}$.
Therefore the terms $F_k^m(s)$ are the ones corresponding to the derivation
of $e^{i\frac{s^2}{4t}}$ $k$ times, and they contain $k$ times
the function $\frac{s}{\sinh s}$ or its derivatives.\\

\noindent
We have denoted in (\ref{idF}) by $k-r$ the number of times
$\frac{s}{\sinh s}$ appears, where $0\leq r\leq k$ ~; the other $r$ terms are
derivatives of $\frac{s}{\sinh s}$.
Since only $k$ derivatives have fallen on the exponential term, there are
$m-k$ left derivatives which act on the remainder term.
That is the reason for which in (\ref{idF}) we have the sum
$$i_1+...+i_r=m-k.$$

For showing the estimates of $F_k^m(s)$ we shall need the following Lemma.

\begin{lemma}\label{Pk}
The following property
\begin{equation}\label{derex}\exists\; c_l>0 \;\;\mbox{ such that }\;\;
\left|\left(\frac{\partial_\rho}{\sinh\rho}\right)^le^{i\rho^2}\right|\leq
c_l\left(\frac{\rho}{\sinh\rho}\right)^l, \forall \rho\geq 0,\end{equation}
is valid for all positive integer $l$. From it follows also that
\begin{equation}\label{dersh} P(l) \,\,:\,\, \,\,\exists\;c_l>0 \;\;\mbox{ such that }\;\;
\left|\left(\frac{\partial_\rho}{\sinh\rho}\right)^l
\frac{\rho}{\sinh\rho}\right|\leq
c_l\left(\frac{\rho}{\sinh\rho}\right)^{l+1}, \forall \rho\geq 0.\end{equation}
\end{lemma}
\begin{proof}
We shall split the proof into two cases, one when $\rho$ is small, and one when it is large.\\

{\underline{The case $\rho<1$}}.\par

The function
$$G_l(\rho)=\left(\frac{\partial_\rho}{\sinh\rho}\right)^le^{i\rho^2}$$
is a $\mathcal{C}^\infty$ even function.
It is obvious for $l=0$, and for other values it can be proved by
induction as follows.
Supposing that the fact is true for $l$, we have the existence of a $\mathcal{C}^\infty$
odd function $H_l$ such that
$$G'_l(\rho)=H_l(\rho).$$
Therefore
$$G_{l+1}(\rho)=\frac{\partial_\rho}{\sinh\rho}G_l(\rho)=\frac{G'_l(\rho)}{\sinh \rho}=\frac{H_l(\rho)}{\sinh\rho},$$
and the induction argument is done.\par
As $\rho$ is small, and $G_l$ is a $\mathcal{C}^\infty$
even fuction, we obtain the existence of a constant $c$ such that
$$|G_l(\rho)|\leq c.$$
Again because $\rho$ is small, we have the quotient $\frac{\rho}{\sinh\rho}$ bounded, and
we can find for all integer $l$ a constant $c_l$ such that
$$|G_l(\rho)|\leq c_l\left(\frac{\rho}{\sinh\rho}\right)^l.$$
So in the case $\rho<1$ the property $(\ref{derex})$ is satisfied.\\

{\underline{The case $\rho\geq 1$}}.\par

Let us perform the change of variable $r=\cosh\rho$. Then
$$\frac{\partial_\rho}{\sinh\rho}=\partial_r,$$
and $\rho$ can be written
$$\rho=\log (r+\sqrt{r^2-1}).$$
With these notations, the property $(\ref{derex})$ becomes
$$\left|\partial_r^le^{i\left(\log (r+\sqrt{r^2-1})\right)^2}\right|\leq
c_l\left(\frac{\log (r+\sqrt{r^2-1})}{\sqrt{r^2-1}}\right)^l, \forall \rho\geq 1.$$

Let us denote
$$a(r)=r+\sqrt{r^2-1},$$
and $$b(r)=\log a(r).$$

Since $r$ is larger than $\cosh 1$, then for proving the property $(\ref{derex})$ it suffices to prove
that
\begin{equation}\label{expib}
\left|\partial_r^le^{ib^2(r)}\right|\leq
c_l\frac{b^l(r)}{r^l}, \forall \,r\geq 3.\end{equation}

In the following we shall use the following general formula
\begin{equation}\label{Faa}
\partial_r^lf(g(r))=\sum_{q=1}^lc_qf^{(q)}(g(r))\sum_{p_1+...+p_q=l}c_{p_1,..,p_q}\partial_r^{p_1}g(r)...\partial_r^{p_q}g(r).\end{equation}
This is a weaker form of the Faa di Bruno formula, in which the constants are described.

Let us notice first that
 \begin{equation}\label{a}
|\partial_r^pa(r)|\leq cr^{1-p},
\end{equation}
and that, since $r$ is large enough,
\begin{equation}\label{loga}
|\log^{(q)}(a(r))|\leq cr^{-q},\end{equation}
for all $q\geq 1$.
Then, by using (\ref{Faa}), for $l\geq 1$,
$$\partial_r^l\log(a(r))=\sum_{q=1}^lc_q\log^{(q)}(a(r))\sum_{p_1+...+p_q=l}c_{p_1,..,p_q}\partial_r^{p_1}a(r)...\partial_r^{p_q}a(r).$$
Now the estimates (\ref{a}) and (\ref{loga}) tell us that
$$|\partial_r^l\log(a(r))|\leq c\sum_{q=1}^lr^{-q}\sum_{p_1+...+p_q=l}r^{1-p_1}...r^{1-p_q}=cr^{-l}.$$
Therefore we have obtained that for $l\geq 1$,
\begin{equation}\label{b}
|\partial_r^l b(r)|\leq cr^{-l}.\end{equation}
Let us compute now the derivatives of $b^2$
$$\partial_r^l b^2(r)=\sum_{q=0}^l\partial_r^q b(r)\partial_r^{l-q} b(r)=2b(r)\partial_r^l b(r)+\sum_{q=1}^{l-1}\partial_r^q b(r)\partial_r^{l-q} b(r).$$
By using (\ref{b}) we get
$$|\partial_r^l b^2(r)|\leq c|b(r)|r^{-l}+\widetilde{c}\sum_{q=1}^{l-1}r^{-q}r^{-(l-q)}\leq (c|b(r)|+c)r^{-l}.$$
As the variable $r$ is large, it follows that
$$b(r)=\log (r+\sqrt{r^2-1})\geq c,$$
and therefore
\begin{equation}\label{b2}
|\partial_r^l b^2(r)|\leq c|b(r)|r^{-l}.\end{equation}

Finally, by using again the general formula (\ref{Faa}),
$$\left|\partial_r^le^{ib^2(r)}\right|\leq c\sum_{q=1}^l\sum_{p_1+...+p_q=l}\partial_r^{p_1}b^2(r)...\partial_r^{p_q}b^2(r),$$
so the estimates (\ref{b2}) allows us to write
$$\left|\partial_r^le^{ib^2(r)}\right|\leq c\sum_{q=1}^lb^q(r)r^{-l}.$$
We have already noticed that $b(r)$ is large, therefore
$$\left|\partial_r^le^{ib^2(r)}\right|\leq cb^l(r)r^{-l},$$
and we finally proved (\ref{expib}), which implies the wanted property $(\ref{derex})$.\\

Property (\ref{dersh}) can be proved by induction. $P(0)$ is obvious. Suppose that the properties $P(1)$,...,$P(l-1)$ are true. Then the development (\ref{derexp}) for $m=l+1$ will contain, by (\ref{idFm}) and (\ref{idF}) only one term of the type
$$\left(\frac{\partial_\rho}{\sinh\rho}\right)^{l}
\frac{\rho}{\sinh\rho},$$
all others being of order less than $l$. More precisely, we have, modulo constants,
$$\left(\frac{\partial_s}{\sinh s}\right)^{l+1}e^{i\frac{s^2}{4t}}=
\sum_{k=1}^{l+1} \frac{e^{i\frac{s^2}{4t}}}{t^k}\,F_k^{l+1}(s)=$$
$$=\frac{e^{i\frac{s^2}{4t}}}{t}\left(\frac{\partial_\rho}{\sinh\rho}\right)^{l}
\frac{\rho}{\sinh\rho}+\frac{e^{i\frac{s^2}{4t}}}{t^{l+1}}\left(\frac{s}{\sinh s}\right)^{l+1}+$$
$$+\sum_{k=2}^l\frac{e^{i\frac{s^2}{4t}}}{t^k}\sum_{r=0}^k\,\,\sum_{\left\{\begin{array}{c}i_1+\cdot\cdot\cdot+i_r=l+1-k\\i_j\geq 1\end{array}\right.}\left(\frac{s}{\sinh s}\right)^{k-r}
\left(\frac{\partial_s}{\sinh s}\right)^{i_1}
\frac{s}{\sinh s}...
\left(\frac{\partial_s}{\sinh s}\right)^{i_r}
\frac{s}{\sinh s}.$$
In the last term all index $i_j$ are less than $l$, and the properties $P(1)$,...,$P(l-1)$ tell us that the term is upper-bounded by
$$\left(\frac{s}{\sinh s}\right)^{l+1}.$$
Therefore, by using (\ref{derex}) we obtain that also the first term in the right-hand-side is bounded by the same quantity
$$\left|\left(\frac{\partial_\rho}{\sinh\rho}\right)^{l+1}
\frac{\rho}{\sinh\rho}\right|\leq
c_k\left(\frac{\rho}{\sinh\rho}\right)^{l+1},$$
so the property $P(l)$ is proven and the induction argument is complete.

\end{proof}

Let us return now to the estimates of Proposition \ref{F}. \\

For $k=m$, estimate (\ref{estFk}) when $\alpha =0$ is obvious using (\ref{idFm}). For $0\leq k<m$, by (\ref{idF}) and we get that
$$\left|\frac{\sinh s}{s}
F_k^m(s)\right|\leq\sum_{r=0}^k\,\,\sum_{\left\{\begin{array}{c}i_1+\cdot\cdot\cdot+i_r=m-k\\i_j\geq 1\end{array}\right.}\left(\frac{s}{\sinh s}\right)^{k-r-1}
\left(\frac{s}{\sinh s}\right)^{i_1+1}...
\left(\frac{s}{\sinh s}\right)^{i_r+1},$$
so estimate (\ref{estFk}) when $\alpha =0$ follows by using (\ref{dersh}).\\

We start now to look at the derivatives. One has
\begin{equation}\label{1ereder}\partial_s\left(\frac{\sinh s}{s}\,F_k^m(s)\right)=\frac{s\cosh s-\sinh s}{s^2}F_k^m(s)+\frac{\sinh^2 s}{s}\frac{\partial_s}{\sinh s}F_k^m(s)\end{equation}
The forms (\ref{idFm}) and (\ref{idF}) of $F_k^m(s)$, together with estimate (\ref{dersh}), imply
$$\left|\partial_s\left(\frac{\sinh s}{s}\,F_k^m(s)\right)\right|\leq \frac{s\cosh s-\sinh s}{s^2}\left(\frac{s}{\sinh s}\right)^m+\frac{\sinh^2 s}{s}\left(\frac{s}{\sinh s}\right)^{m+1}.$$
We deduce that
$$\left|\partial_s\left(\frac{\sinh s}{s}\,F_k^m(s)\right)\right|\leq s\left(\frac{s}{\sinh s}\right)^{m-1}\left(\frac{s\cosh s-\sinh s}{s^2\sinh s}+1\right),$$
and estimate (\ref{estFk}) for $\alpha =1$ follows.\\

Knowing (\ref{1ereder}) one can get the expression of the second derivative
$$\partial_s^2\left(\frac{\sinh s}{s}\,F_k^m(s)\right)=\frac{s^2\sinh s-2(s\cosh s-\sinh s)}{s^3}F_k^m(s)+$$
$$+\frac{3s\cosh s\sinh s-2\sinh^2 s}{s^2}\frac{\partial_s}{\sinh s}F_k^m(s)+
\frac{\sinh^3 s}{s}\left(\frac{\partial_s}{\sinh s}\right)^2F_k^m(s).$$
By using again the forms (\ref{idFm}) and (\ref{idF}) of $F_k^m(s)$ and the estimate (\ref{dersh})
$$\left|\partial_s^2\left(\frac{\sinh s}{s}\,F_k^m(s)\right)\right|\leq \left(\frac{s}{\sinh s}\right)^{m-2}\left(\frac{s^2\sinh s-2(s\cosh s-\sinh s)}{s\sinh^2 s}+\right.$$
$$\left.+s\frac{3s\cosh s-2\sinh s}{\sinh^2 s}+\frac{s^3}{\sinh s}\right).$$
As the term in the brackets of the right-hand-side is bounded, for $m\geq 2$ we obtain that the second derivative of $\frac{\sinh s}{s}\,F_k^m(s)$ is bounded, and Proposition \ref{F} is completely proved.
\end{proof}

config: arxiv


TITLE: Spectrum of Unitary Operators
QUESTION [0 upvotes]: Let $T_1$ and $T_2$ be two unitary operators. Is it true that the spectrum of $T_1+T_2$ is contained in the closed disc of radius 2?

REPLY [2 votes]: Yes, this follows from $\left\|T_{1}+T_{2}\right\| \le \left\|T_{1}\right\| +\left\|T_{2}\right\|$.
To wit, suppose $v$ is an eigenfunction of $T_{1}+T_{2}$ with eigenvalue $\lambda$. Then
\begin{align}
|\lambda|\left\|v\right\| = \left\|(T_{1}+T_{2})v\right\| \le  \left\|(T_{1}+T_{2})\right\| \left\|v\right\|  \le  (\left\|T_{1}\right\| + \left\|T_{2}\right\| ) \left\|v\right\|  = 2\left\|v \right\|
\end{align}
So $|\lambda| \le 2$.


TITLE: Proof of positivity of $~ x+\sqrt{x^2+1} ~$ of $~\operatorname{arsinh}(x)=\operatorname{arcsinh}(x)=\sinh^{-1}(x)= \ln \left( x+\sqrt{x^2+1}\right)$
QUESTION [4 upvotes]: Proof of positivity of $~ x+\sqrt{x^2+1}  ~$
I found this formula appears at $~ \operatorname{arsinh}(x)= \ln \left( x+\sqrt{x^2+1}\right)~$
So, of course this argument inside the natural log function should be positive.
I tried to prove the positivity using proof of contradiction.
$$ \color{fuchsia}{\begin{align}
&x+\sqrt{x^2+1} \leq 0
\\&\iff \sqrt{x^2+1} \leq -x
\\&\iff \underbrace{x^2+1 \geq x^2}_{~~\text{Able to be held}~~ }
\end{align}}  $$
$$ \color{red}{\begin{align}
&x+\sqrt{x^2+1} >0
\\&\iff \sqrt{x^2+1} >-x
\\&\iff \underbrace{x^2+1 < x^2}_{~~\text{Unable to be held}~~ }
\end{align}}  $$
What is going on?

REPLY [3 votes]: The mistake I made above my question-post is that I flipped the symbol of inequality even signs of both endpoints of inequality is different.
$$\begin{align}
\text{negative real value}<\text{positive real value}
\end{align}$$
At this sitch, one can't flip $~ <~$  to $~ > ~$ as both terms are squared . I've completely missed it.
$$\begin{align}
&0\leq x+\left| x \right|= x + \sqrt{x^2} \leq x+ \sqrt{x^2+1}~~\text{for}~~x\in\mathbb{R}_{<0}\\
&\therefore~~0\leq x+ \sqrt{x^2+1}
\end{align}$$
$~ \sqrt{x^2}\leq \sqrt{x^2+1} ~$ can be held since a root function is a  monotonous increasing function.


text:
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Function and mapping are synonyms. However according to
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that this index may have internal structure.
For instance, considering the algebra $A$ we enumerate coordinates of
$a\in A$ relative to basis $\Basis e$ by an index $i$.
This means that $a$ is a vector. However, if $a$
is matrix, then we need two indexes, one enumerates
rows, another enumerates columns. In the case, when index has
structure, we begin the index from symbol $\cdot$ in
the corresponding position.
For instance, if I consider the matrix $a^i_j$ as an element of a vector
space, then I can write the element of matrix as $a^{\cdot}{}^i_j$.
\qed
\end{convention}
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of dimension $1$. According to this statement, we can explore not only
homomorphisms of division ring $D_1$ into division ring $D_2$,
but also linear maps of division rings.
This means that map is multiplicative over
maximum possible field. In particular, linear map
of division ring $D$ is multiplicative over center $Z(D)$. This statement
does not contradict with
definition of linear map of field because for field $F$ is true
$Z(F)=F$.
When field $F$ is different from
maximum possible, I explicit tell about this in text.
\qed
\end{convention}
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For given field $F$, unless otherwise stated,
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In spite of noncommutativity of product a lot of statements
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left representation or right vector space by left
vector space.
To keep this symmetry in statements of theorems
I use symmetric notation.
For instance, I consider \Ds vector space
and \sD vector space.
We can read notation \Ds vector space
as either D\Hyph star\Hyph vector space or
left vector space.
We can read notation \Ds linear dependent vectors
as either D\Hyph star\Hyph linear dependent vectors or
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\begin{convention}
Let $A$ be free finite
dimensional algebra.
Considering expansion of element of algebra $A$ relative basis $\Basis e$
we use the same root letter to denote this element and its coordinates.
However we do not use vector notation in algebra.
In expression $a^2$, it is not clear whether this is component
of expansion of element
$a$ relative basis, or this is operation $a^2=aa$.
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of algebra. For instance,
\ShowEq{Expansion relative basis in algebra}
\qed
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If free finite dimensional algebra has unit, then we identify
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Although the algebra is a free module over some
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matrix of coordinates of element of algebra, I will use vector
notation to write corresponding element.
In order to avoid ambiguity when I use conjugation,
I denote $a^*$ element conjugated to element $a$.
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In \citeBib{Cohn: Universal Algebra},
an arbitrary operation of algebra is denoted by letter $\omega$,
and $\Omega$ is the set of operations of some universal algebra.
Correspondingly, the universal algebra with the set of operations
$\Omega$ is denoted as $\Omega$\Hyph algebra.
Similar notations we see in
\citeBib{Burris Sankappanavar} with small difference
that an operation in the algebra is denoted by letter $f$
and $\mathcal F$ is the set of operations.
I preferred first case of notations because in this case it is
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\begin{convention}
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$(A_1,...,A_n)$ of $\Omega_i$\Hyph algebras $A_i$, $i=1$, ..., $n$
as $\Vector A$. We denote the set of representations
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We
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In relation to the set $\Vector A$,
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that we discussed
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For instance, we
\fi
use the symbol $\Vector A_{[1]}$ to denote the
set of $\Vector{\Omega}$\Hyph algebras $(A_2,...,A_n)$.
In the corresponding notation $(\VX A,\Vector f)$ of tower
of representation, we assume that $\Vector f=(f_{2,3},...,f_{n-1,n})$.
\qed
\end{convention}

\begin{convention}
Since we use vector notation for elements of the
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\begin{remark}
I believe that diagrams of maps are an important tool.
However, sometimes I want
to see the diagram as three dimensional figure
and I expect that this would increase its expressive
power. Who knows what surprises the future holds.
In 1992, at a conference in Kazan, I have described to my colleagues
what advantages the computer preparation of papers has.
8 years later I learned from the letter from Kazan that now we can
prepare paper using LaTeX.
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Without a doubt, the reader may have questions,
comments, objections. I will appreciate any response.
\fi


text:
\section{Example: Exponential decay function}\label{app:exponential_kernel}

Here we give the analysis for the score function and the Fisher information under exponential decay function $\varphi_{ij}(\Delta t) = \beta e^{-\beta \Delta t}$.
The score function is,
$$
S_i(\vec\alpha_i^*) = \ \int_0^T\frac{\int_0^t\beta e^{-\beta(t-\tau)}d\vec N_\tau}{\mu_i + (\vec{\alpha}_i^*)^\intercal\int_0^t\beta e^{-\beta(t-\tau)}d\vec N_\tau}(d N_t^i - \lambda_i^*(t)dt),
$$
where $d\vec N_t = (dN_t^1,\cdots dN_t^D)^\intercal$.

We show that $S_i(\vec\alpha_i^*)$ is small by giving an upper bound of its covariance matrix, which is $T$ times the Fisher information.

Assume the Hawkes process with parameter $\vec{\alpha}_i,\mu_i,i=1,\cdots,D,\beta$ is stationary, we have
\begin{align*}
I_i^*
=&\ \E\left[\frac{\left(\int_{-\infty}^t\beta e^{-\beta(t-\tau)}d\vec N_\tau\right)\left(\int_{-\infty}^t\beta e^{-\beta(t-\tau)}d\vec N_\tau\right)^\intercal}{\mu_i + (\vec{\alpha}_i^*)^\intercal\int_{-\infty}^t\beta e^{-\beta(t-\tau)}d\vec N_\tau}\right]

\end{align*}
Since $
\vec{\alpha}_i^T\int_{-\infty}^t\beta e^{-\beta(t-\tau)}dN_\tau \geq 0
$, we have
$$
I_i^*\preceq \mu_i^{-1}\underbrace{\E\left[\left(\int_{-\infty}^t\beta e^{-\beta(t-\tau)}d\vec N_\tau\right)\left(\int_{-\infty}^t\beta e^{-\beta(t-\tau)}d\vec N_\tau\right)^\intercal\right]}_{W},
$$
where $W$ has a close-form expression for Hawkes processes with exponential influence function, derived from \cite{bacry2016first} and \cite{li2017detecting}.

\begin{lem}
\[
W = \Lambda\Lambda^\intercal + \frac{\beta}{2}\Sigma + \frac{\beta}{4}A(\mathbb I-A)^{-1}\Sigma + \frac{\beta}{4}\Sigma A^\intercal(\mathbb I-A^\intercal)^{-1},
\]
where $\mathbb I$ is the identity matrix, $A = (\vec\alpha_1^*,\cdots,\vec\alpha_D^*)^\intercal,$ $\Lambda = (\mathbb I-A)^{-1}\vec \mu$ is the expected intensity, and $\Sigma = \text{diag}(\Lambda)$.
\end{lem}
\begin{proof}
By Lemma 2 and 3 in \cite{li2017detecting}, we have
\[
\E[d\vec N_t] = \Lambda dt,
\]
and
\[
\text{Cov}[d\vec N_t,d\vec N_{t'}^\intercal] = c(t-t')dtdt',
\]
where
\[
c(\tau) = \begin{cases}
\beta e^{-\beta(\mathbb I - A)\tau} A\left(\mathbb I + \frac{1}{2}(\mathbb I-A)^{-1}A\right)\Sigma, & \tau > 0;\\
\Sigma\delta(\tau),&\tau = 0;\\
c(-\tau)^\intercal,&\tau<0,
\end{cases}
\]
where $\delta(\cdot)$ is the Dirac delta  function. Then
\begingroup
\allowdisplaybreaks
\begin{align*}
W = &\ \E\left[\left(\int_{-\infty}^t\beta e^{-\beta(t-\tau)}d\vec N_\tau\right)\left(\int_{-\infty}^t\beta e^{-\beta(t-\tau)}d\vec N_\tau\right)^\intercal\right]\\
=&\ \E\left[\int_{-\infty}^0\int_{-\infty}^0\beta^2 e^{\beta (t+t')}d\vec N_td\vec N_{t'}^\intercal\right]\\
=&\ \int_{-\infty}^0\int_{-\infty}^0 \beta^2 e^{\beta(t+t')}\E[d\vec N_td\vec N_{t'}]\\
=&\ \Lambda\Lambda^\intercal + \int_{-\infty}^0\int_{-\infty}^0 \beta^2 e^{\beta(t+t')}\text{Cov}[d\vec N_t,d\vec N_{t'}]\\
=&\ \Lambda\Lambda^\intercal + \int_{-\infty}^0 \beta^2 e^{2\beta t}\Sigma dt + \iint_{t\leq 0,\tau\in (0,-t]}\beta^2 e^{\beta(2t+\tau)}(c(\tau)+c(-\tau))dtd\tau\\
=&\ \Lambda\Lambda^\intercal + \frac{\beta}{2}\Sigma + \int_0^\infty (c(\tau)+c(-\tau))d\tau\int_{-\infty}^{-\tau} \beta^2 e^{\beta(2t+\tau)}dt\\
=&\ \Lambda\Lambda^\intercal + \frac{\beta}{2}\Sigma + \int_0^\infty (c(\tau)+c(-\tau))\frac{\beta}{2}e^{-\beta\tau}d\tau\\
=&\ \Lambda\Lambda^\intercal + \frac{\beta}{2}\Sigma + \int_0^\infty \frac{\beta^2}{2}\Bigg[e^{-\beta(2\mathbb I-A)\tau}A(\mathbb I+\frac{1}{2}(\mathbb I-A)^{-1}A)\Sigma \\
& \hspace{170pt}+\left(e^{-\beta(2\mathbb I-A)\tau}A(\mathbb I+\frac{1}{2}(\mathbb I-A)^{-1}A)\Sigma\right)^\intercal\Bigg]d\tau\\
=&\ \Lambda\Lambda^\intercal + \frac{\beta}{2}\Sigma + \frac{\beta}{2}\Bigg[(2\mathbb I-A)^{-1}A(\mathbb I+\frac{1}{2}(\mathbb I-A)^{-1}A)\Sigma \\
& \hspace{170pt} + \left((2\mathbb I-A)^{-1}A(\mathbb I+\frac{1}{2}(\mathbb I-A)^{-1}A)\Sigma\right)^\intercal\Bigg].
\end{align*}
\endgroup
We notice that
\[
(2\mathbb I-A)^{-1}A(\mathbb I+\frac{1}{2}(\mathbb I-A)^{-1}A) = (2\mathbb I-A)^{-1}A(\mathbb I-A)^{-1}(\mathbb I-A/2) = \frac{1}{2}A(\mathbb I-A)^{-1}.
\]
Together, we prove the lemma.
\end{proof}

\begin{lem}
For any vector $\vec z$,
$$
P \left(\vec z^\intercal S_i(\vec\alpha_i^*)
\geq \varepsilon\sqrt{T}\right)\leq \frac{\mu_i^{-1}\vec z^\intercal W\vec z}{\varepsilon^2}.
$$
\end{lem}
\begin{proof}
\[
\text{Var}\left[\vec z^\intercal S_i(\vec\alpha_i^*)\right] = T\vec z^\intercal I_i^* \vec z\leq \mu_i^{-1} T\vec z^\intercal W\vec z,
\]
by Markov's inequality on the random variable $(\vec z^\intercal S_i(\vec\alpha_i^*))^2$, we proof the lemma.

\end{proof}

\section{Proofs}

The proof of Theorem\,\ref{thm:concentration} and Theorem\,\ref{prop:concentration} is an immediate results of the following two lemmas.
\begin{lem}\label{lem:intrinsic_variance}
For any measurable random process $(\vec z(t)\in \R^D)_{t\in[0,T]}$ adapted to the same filtration $(\mathcal H_{t})_{t\in [0,T]}$ with the Hawkes process, let the intrinsic variance of $\int_0^t \vec z^\intercal (\tau) dS_{i,\tau}(\vec\alpha_i^*)$ (denoted by $V_{i,t}(\vec z)$) be a random process also adapted to $(\mathcal H_{t})_{t\in [0,T]}$, such that there exists a supermartingale $(M_t(\vec z))_{t\in[0,T]}$ with respect to $(\mathcal H_t)_{t\in[0,T]}$,
\[
\exp\left(\int_0^t\vec z^\intercal (\tau) d S_{i,\tau}(\vec\alpha_i^*) - V_{i,t}(\vec z)\right)\leq M_t(\vec z)
\]
almost surely. Then $\forall \varepsilon\in(0,1),$
\[
\Pr\left(\int_0^T\vec z^\intercal(t) dS_{i,t}(\vec\alpha_i^*) - V_{i,T}(\vec z)\geq \ln(\E[M_0(\vec z)]/\varepsilon)\right)\leq \varepsilon.
\]
\end{lem}
\begin{proof}
By the property of a supermartingale, we have
\[
\E\left[\exp\left(\int_0^T\vec z^\intercal(t) dS_{i,t}(\vec\alpha_i^*) - V_{i,T}(\vec z)\right)\right]\leq \E[M_T(\vec z)]\leq \E[ M_0(\vec z)],
\]
and by Markov's inequality,
\begin{align*}
&\ \Pr\left[\int_0^T\vec z^\intercal(t) dS_{i,t}(\vec\alpha_i^*) - V_{i,T}(\vec z)\geq \ln(\E[M_0(\vec z)]/\varepsilon)\right] \\
=&\ \Pr\left[\exp\left(\int_0^T\vec z^\intercal(t) dS_{i,t}(\vec\alpha_i^*) - V_{i,T}(\vec z)\right)\geq \E[M_0(\vec z)]/\varepsilon\right]\\
\leq&\ \frac{ \E\left[\exp\left(\int_0^T\vec z^\intercal(t) dS_{i,t}(\vec\alpha_i^*) - V_{i,T}(\vec z)\right)\right]}{\E[M_0(\vec z)]/\varepsilon}
\leq \varepsilon.
\end{align*}
\end{proof}
Moreover, the intrinsic variance can be characterized explicitly by the following result.

\begin{lem}\label{fact:v}
Let \begin{equation}
V_{i,t}(\vec z) = \int_0^{t}\left(\lambda_i^*(\tau)\exp(\lambda_i^{*-1}(\tau)\vec z^\intercal(\tau) \vec\eta_i(\tau)) - \vec z^\intercal(\tau) \vec\eta_i(\tau) - \lambda_i^*(\tau)\right)d\tau.\label{eq:define_v}
\end{equation}
\[
M_t(\vec z) =\exp\left( \int_0^t\vec z^\intercal(\tau) dS_{i,\tau}(\vec\alpha_i^*) - V_{i,t}(\vec z)\right)
\]
is a supermartingale, with $M_0(\vec z) = 1$ almost surely.
\end{lem}
\begin{proof}
For any $t$, since $V_{i,t}$ is continuous and the right derivative exists,
\begin{align*}
&\ \lim_{\Delta t\to 0^{+}}\frac{\log\E\left[M_{t+\Delta t}(\vec z)/M_t(\vec z)|\mathcal H_t\right]}{\Delta t} \\
= &\ \lim_{\Delta t \to 0^+}\frac{\log\E\left[\exp\left(\int_t^{t+\Delta t} \vec z^\intercal (\tau) dS_{i,\tau}(\vec{\alpha}_i^*)-\Delta V_{i,t}(\vec z)\right)|\mathcal H_{t}\right]}{\Delta t}\\
= &\ \lim_{\Delta t \to 0^+}\frac{\log\E\left[\exp\left(\vec z^\intercal(t)\vec\eta_i(t)(\lambda_i^{*-1}\Delta N_t^i -\Delta t\right)|\mathcal H_{t}\right]}{\Delta t}-(\lambda_i^*\exp(\lambda_i^{*-1}\vec z^\intercal (t)\vec\eta_i(t))-\vec z^\intercal (t) \vec\eta_i(t)-\lambda_i^*(t))\\
=&\ \lim_{\Delta t\to 0^+}\frac{\log\left(\lambda_i^*(t)\Delta t\exp(\lambda_i^{*-1}\vec z^\intercal (t) \vec\eta_i(t)) + (1-\lambda_i^*(t)\Delta t)\exp(-\vec z^\intercal (t)\vec\eta_i(t)\Delta t)\right)}{\Delta t}\\
&\ -(\lambda_i^*\exp(\lambda_i^{*-1}\vec z^\intercal (t) \vec\eta_i(t))-\vec z^\intercal(t) \vec\eta_i(t)-\lambda_i^*(t)) \\
=&\ 0.
\end{align*}
From this we can see that $M_t$ is actually a martingale.
\end{proof}

\begin{proof}[Proof of Theorem\,\ref{thm:concentration}, Theorem\,\ref{prop:concentration} and Corollary\,\ref{corollary:cs_adapted_z}]
From Lemma\,\ref{lem:intrinsic_variance} and Lemma \ref{fact:v}, we have immediately
\begin{equation}
\Pr\left[ \int_0^T \vec z^\intercal(t) dS_{i,t}(\vec\alpha_i^*) - V_{i,T}(\vec z)\geq \ln(1/\varepsilon)\right] \leq \varepsilon,\ \forall \vec z \in \R^D,\forall \varepsilon\in(0,1),
\label{eq:simple_prop_bound}
\end{equation}
where $V_{i,T}(\vec z)$ is chosen as (\ref{eq:define_v}).
Moreover, we can also choose multiple $\vec z$ to bound $S_{i}(\vec\alpha_i^*)$ in all directions. By simple union bound, it holds that
\[
\Pr\left[\exists k\in [K], \int_0^T\vec z_k^\intercal (t) dS_{i,t}(\vec\alpha_i^*) - V_{i,T}(\vec z_k)\geq \ln(K/\varepsilon) \right]\leq K\varepsilon/K = \varepsilon, \ \forall \vec z_1,\cdots,\vec z_K\in\R^D,\forall \varepsilon\in(0,1).
\]
We define continuous process $V_{i,t}$ for any $\vec\alpha_i$ as
\eqref{eq:v_alpha}, $\vec\alpha_i^*$ falls into the confidence set $\mathcal C_{i,\varepsilon}$ with probability at least $1-\varepsilon$.
\end{proof}

The proof of Lemma \ref{lem:3.1} relies on the following lemma:

\begin{lem}[Ogata \cite{ogata1978asymptotic}, Lemma 2]
If $\xi_t$ is a stationary predictable process, then
\[
\frac{1}{T}\int_0^T\xi_tdt \to \E\left[\xi\right]
\]
with probability 1. In addition, if $\xi_t$ has finite second moment, then
\[
\frac{1}{T}\int_0^T\xi_t\frac{dN_t}{\lambda(t)} \to \E\left[\xi\right]
\]
with probability 1.
\label{lem:ogata_convergence}
\end{lem}

\begin{proof}[Proof of Lemma\,\ref{lem:3.1}]
Denote
$$
\Delta \vec\alpha_i = \vec\alpha_i - \hat{\vec\alpha}_i.
$$
Using Taylor expansion and based on the mean value theorem, there exists $\tilde{\vec{\alpha}}_i$ between $\vec{\alpha}_i$ and $\hat{\vec{\alpha}}_i$, such that
\[
\frac{1}{T}S_i(\vec\alpha_i)=\frac{1}{T} H_i(\hat{\vec{\alpha}}_i) \Delta\vec\alpha_i +\frac{1}{2T} \sum_{k,l\in[D]}\frac{\partial^2 S_i(\tilde{\vec\alpha}_i)}{\partial\alpha_{ik}\partial\alpha_{il}}\Delta\alpha_{ik}\Delta\alpha_{il}.
\]
For any $j,k,l\in [D]$, the $j$-th entry of $\frac{\partial^2 S_i(\tilde{\vec\alpha}_i)}{\partial\alpha_{ik}\partial\alpha_{il}}$ is
\[
\int_0^T \frac{2\eta_{ij}(t)\eta_{ik}(t)\eta_{il}(t)}{\tilde\lambda_i^3(t)}dN_t^i.
\]
Under the assumption that the moment generating function of $\vec\eta_i$ exists, and $\tilde\lambda_i\geq \mu_i>0$, by Lemma \ref{lem:ogata_convergence}, as $T\to\infty$,
\[
\frac{1}{T}\int_0^T \frac{2\eta_{ij}(t)\eta_{ik}(t)\eta_{il}(t)}{\tilde\lambda_i^3(t)}dN_t^i\leq \frac{1}{T\mu_i^3}\int_0^T2\eta_{ij}(t)\eta_{ik}(t)\eta_{il}(t)dN_t^i
\]
is uniformly bounded for any $\tilde{\vec\alpha}_i\geq 0$ with probability 1. Now we have for any $\vec\alpha_i$,
\[
\|S_i(\vec\alpha_i) - H_i(\hat{\vec\alpha}_i)\Delta\vec\alpha_i \| \leq  O(T)\|\Delta\vec\alpha_i\|^2.
\]
Since \[
\frac{1}{T}H_i(\hat{\vec\alpha}_i)\to I_i^*
\]
with probability 1, we have
\[
\|S_i(\vec\alpha_i) - TI_i^*\Delta\vec\alpha_i\| \leq O(T)\|\Delta\vec\alpha_i\|^2 + \|(H_i(\hat{\vec\alpha}_i)-TI_i^*)\Delta\vec\alpha_i\| \leq  O(T)\|\Delta\vec\alpha_i\|^2 + o(T)\|\Delta\vec\alpha_i\|,
\]
which is \eqref{eq:S_approx}.

For \eqref{eq:V_approx}, similarly we use the Taylor expansion at $\vec z = 0$ and $\hat{\vec\alpha}_i$, by the mean value theorem,
\begin{align*}
V_i(\vec z,\vec\alpha_i) = &\ V_i(\vec 0,\vec\alpha_i) + \frac{\partial V_i(\vec 0,\vec\alpha_i)}{\partial \vec z^\intercal}\vec z + \frac{1}{2}\vec z^\intercal \frac{\partial^2 V_i(\vec 0,\vec\alpha_i)}{\partial \vec z\partial\vec z^\intercal}\vec z +\frac{1}{6}\sum_{j,k,l\in[D]}\frac{\partial^3V_i(\tilde{\vec z},\vec\alpha_i)}{\partial z_j\partial z_k\partial z_l}z_jz_kz_l\\
=&\ \frac{1}{2}\vec z^\intercal \frac{\partial^2 V_i(\vec 0,\vec\alpha_i)}{\partial \vec z\partial\vec z^\intercal}\vec z+\frac{1}{6}\sum_{j,k,l\in[D]}\frac{\partial^3V_i(\tilde{\vec z},\vec\alpha_i)}{\partial z_j\partial z_k\partial z_l}z_jz_kz_l\\
=&\ \frac{1}{2}\vec z^\intercal \frac{\partial^2 V_i(\vec 0,\hat{\vec\alpha}_i)}{\partial \vec z\partial\vec z^\intercal}\vec z+\frac{1}{2}\sum_{j,k,l\in[D]}\frac{\partial^3 V_i(\vec 0,\tilde{\vec\alpha}_i)}{\partial z_j\partial z_k\partial \alpha_{il}}z_jz_k\Delta\alpha_{il}+\frac{1}{6}\sum_{j,k,l\in[D]}\frac{\partial^3V_i(\tilde{\vec z},\vec\alpha_i)}{\partial z_j\partial z_k\partial z_l}z_jz_kz_l,
\end{align*}
for some $\tilde{\vec z}$ between $\vec 0,\vec z$, some $\tilde{\vec\alpha}_i$ between $\hat{\vec\alpha}_i,\vec\alpha_i$. By the assumption that the moment generating function of $\vec\eta_i$ exists and by Lemma \ref{lem:ogata_convergence}, for the first term
\[
\frac{1}{2T}\vec z^\intercal \frac{\partial^2 V_i(\vec 0,\hat{\vec\alpha}_i)}{\partial \vec z\partial\vec z^\intercal}\vec z =\frac{1}{2T} \vec z^\intercal\int_0^T\frac{\vec\eta_i(t)\vec\eta_i(t)^\intercal}{\hat\lambda_i(t)}dt\vec z\to \frac{\vec z^\intercal I_i^*\vec z}{2}
\]
with probability 1. For the second term, for any $j,k,l\in [D]$,
\[
\left|\frac{1}{T}\frac{\partial^3 V_i(\vec 0,\tilde{\vec\alpha}_i)}{\partial z_j\partial z_k\partial \alpha_{il}}\right| = \left|\frac{1}{T}\int_0^T\frac{\eta_{ij}(t)\eta_{ik}(t)\eta_{il}(t)}{\tilde\lambda_i^2(t)}dt\right|\leq \left|\frac{1}{T}\int_0^T\frac{\eta_{ij}(t)\eta_{ik}(t)\eta_{il}(t)}{\mu_i^2}dt\right|.
\]
is uniformly bounded for any $\tilde{\vec\alpha}_i\geq 0$ with probability 1. For the third term, for any $i,j,k\in [D]$,
\begin{align*}
\left|\frac{1}{T}\frac{\partial^3V_i(\tilde{\vec z},\vec\alpha_i)}{\partial z_j\partial z_k\partial z_l}\right| =&\ \left|\frac{1}{T}\int_0^T \frac{\eta_{ij}(t)\eta_{ik}(t)\eta_{il}(t)}{\lambda_i^{2}(t)}\exp\left(\lambda_i^{-1}\vec\eta_i^\intercal(t)\tilde{\vec z}\right)dt\right|\\
\leq &\ \left|\frac{1}{T}\int_0^T\frac{\eta_{ij}(t)\eta_{ik}(t)\eta_{il}(t)}{\mu_i^{2}}\min\left\{\exp\left(\mu_i^{-1}\vec\eta_i^\intercal(t)\tilde{\vec z}\right),1\right\}dt\right|,
\end{align*}
is convex in $\vec z$. There exists a neighborhood $U$ of $\vec 0$ such that the expectation of the term above for any $\vec z\in U$ is finite, and by its convexity, it is uniformly bounded in $U$ with probability 1.

Together, we have

\[
\left|V_i(\vec z,\vec\alpha_i) - \frac{T}{2} \vec z^\intercal I_i^*\vec z\right| \leq o(T)\|\vec z\|^2 + O(T)\|\Delta\vec\alpha_i\|\|\vec z\|^2 + O(T)\|\vec z\|^3.
\]
\end{proof}

\begin{proof}[Proof of Proposition\,\ref{prop:width}]
We prove a slightly weaker version: for any neighborhood $U_1$ of $\vec\alpha_i^*$, such that the diameter of $U_1$ is $o(1)$, the width of $\mathcal C_{i,\varepsilon}\cap U_1$ in $\alpha_{ij}$ converges to $2\sqrt{2\ln(K/\varepsilon)\sigma_{ij}^2/T}$ with probability 1.

Before proving the proposition, we explain why we choose $\vec z_1,\cdots,\vec z_K$ this way. Let $\vec z_1,\cdots,\vec z_{2D}$ be $\pm c_j I_i^{*-1}\vec e_j$, $c_j>0$, $j=1,\cdots,D$. By Lemma\,\ref{lem:3.1}, we have
\begin{align*}
(\pm c_jI_i^{*-1}\vec e_j)^\intercal S_i(\vec\alpha_i) = (\pm c_jI_i^{*-1}\vec e_j)^\intercal TI_i^*(\vec\alpha_i - \hat{\vec\alpha}_i) + c_j\left( O(T) \|\vec\alpha_i - \hat{\vec\alpha}_i\|^2+ o(T)\|\vec\alpha_i - \hat{\vec\alpha}_i\|\right),
\end{align*}
for any $\vec \alpha_i\geq 0$.
Note that
$$(\pm c_jI_i^{*-1}\vec e_j)^\intercal T  I_i^*(\vec\alpha_i - \hat{\vec\alpha}_i) = \pm c_jT(\alpha_{ij}-\hat\alpha_{ij}).$$
By \eqref{eq:V_approx}, we have
\begin{align*}
V_{i}(\pm c_jI_i^{*-1}\vec e_j,\vec\alpha_i) =   \frac{c^2T}{2}\vec e_j^\intercal I_i^{*-1}\vec e_j   + c^2\left(o(T) + O(T)\|\vec\alpha_i-\hat{\vec\alpha}_i\|\right) + O(T)c^3.
\end{align*}
The constraints
\[
\vec z_k^\intercal S_i(\vec\alpha_i) - V_i(\vec z_k,\vec\alpha_i)\leq \ln(K/\varepsilon),\quad k=1,\cdots,2D
\]
becomes
\begin{align*}
c_jT|\alpha_{ij}-\hat\alpha_{ij}| - \frac{c_j^2T}{2}\sigma_{ij}^2 +&\ o(T)c_j^2+ O( T)c_j^3+(O(T)c_j^2+o(T)c_j)\|\vec\alpha_i-\hat{\vec\alpha}_i\|  + O(T)c_j\|\vec\alpha_i-\hat{\vec\alpha}_i\|^2\\
\leq&\ \ln(K/\varepsilon),\quad j=1,\cdots,D.
\end{align*}
If all the $o(\cdot),O(\cdot)$ terms are negligible when $T\to \infty$, the width of $\mathcal C_{i,\varepsilon}$ in $\alpha_{ij}$
is \[
2\left(\frac{\ln(K/\varepsilon)}{c_jT} + \frac{c_j\sigma_{ij}^2}{2}\right),
\]
and is minimized when
\[
c_j = \sqrt{\frac{2\ln(K/\varepsilon)}{T\sigma_{ij}^2}}.
\]
The $o(\cdot),O(\cdot)$ terms are indeed negligible with this choice of $c_j$, because the constraints now becomes
\begin{equation}
\sqrt{2T\ln(K/\varepsilon)/\sigma_{ij}^2}|\alpha_{ij}-\hat\alpha_{ij}|   +o(T^{1/2})\|\vec\alpha_i-\hat{\vec\alpha}_i\|  + O(T^{1/2})\|\vec\alpha_i-\hat{\vec\alpha}_i\|^2
\leq 2\ln(K/\varepsilon)+ o(1),
\label{eq:constraints}
\end{equation}
$j=1,\cdots,D$. Let $U_1$ be any neighborhood of $\vec\alpha_i^*$ with diameter $o(1)$. For any $\vec\alpha_i\in\mathcal C_{i,\varepsilon}\cap U_1$, we choose  \[
j' =\mathop{\arg\max}_{j\in [D]}
|\alpha_{ij}-\hat\alpha_{ij}|/\sigma_{ij}.
\]
By the way we choose $j'$, $\|\vec\alpha_i-\hat{\vec\alpha}_i\|$ can be upper bounded by $|\alpha_{ij'}-\hat\alpha_{ij'}|$ up to some constant scale, and $|\alpha_{ij'}-\hat\alpha_{ij'}| = o(1)$. There is
\[
\sqrt{2T\ln(K/\varepsilon)/\sigma_{ij'}^2}|\alpha_{ij'}-\hat\alpha_{ij'}|   +o(T^{1/2})|\alpha_{ij'}-\hat\alpha_{ij'}|
\leq 2\ln(K/\varepsilon)+ o(1),
\]
and
\[
\frac{|\alpha_{ij'}-\hat\alpha_{ij'}|}{\sigma_{ij'}}\leq \sqrt{\frac{2\ln (K/\varepsilon)}{T}}(1+o(1)).
\]
Again by the way we choose $j'$, this inequality holds for any $j\in[D]$. So the width of $\mathcal C_{i,\varepsilon}$ in $\alpha_{ij}$ is upper bounded by $2\sqrt{2\ln (K/\varepsilon)\sigma_{ij}^2/T}(1+o(1))$ with high probability. It is easy to see from \eqref{eq:constraints} that there exists $\vec\alpha_i\in\mathcal C_{i,\varepsilon}$ with $\alpha_{ij} = \hat\alpha_{ij}\pm \sqrt{2\ln (K/\varepsilon)\sigma_{ij}^2/T}(1-o(1))$. Together, we know that the width of $\mathcal C_{i,\varepsilon}$ converges to $2\sqrt{2\ln (K/\varepsilon)\sigma_{ij}^2/T}$ with probability 1.
\end{proof}


TITLE: The Four Fours puzzle using only $+$, $-$, $\times$, $\div$, $\sqrt{}$, and concatenation. Can we make $19$?
QUESTION [1 upvotes]: I know the classic Four Fours puzzle, which asks to create whole numbers using exactly four "$4$"s combined via certain operations (arithmetic, concatenation, grouping, factorials, etc.), the exact roster of which can vary.
I have tried this specific version of it:
You can use only the basic arithmetic operations ($\times \div + - $), square root ($\sqrt{}$), and concatenation (eg, "$44$" is allowed).

How far you can go? I cannot go further than $18$. I am stuck with $19$ and does not seem that there is a solution for it.

REPLY [3 votes]: Let $\#$ denote concatenation. That is, $4\#3 = 43$. Then
$$(4 \# \sqrt 4) / \sqrt 4 - \sqrt4 = 42 / 2 - 2 = 19.$$
Not sure if you can use parentheses, but you would just have to perform concatenation first above all else.