TITLE: $XY-YX = J_4-I_4$
QUESTION [0 upvotes]: To find square matrices $X$ and $Y$ of order $4$ such that $$XY-YX = J_4-I_4,$$ where $J_4$ is the matrix of all ones and $I$ is the identity matrix.

There is an article "A Note on Matrix Solutions to A = XY - YX" by Charles R. Johnson saying that such matrices exist when $\mathrm{trace}(A) = 0$. Here the trace condition is satisfied, but how to find matrices $X$ and $Y$... there is no explicit method.
Can someone shed some light? Thank You.

REPLY [2 votes]: This works already for certain diagonal matrices $X$, e.g., set $X=\operatorname{diag}(0,1,2,3)$. Then $XY-YX=J_4-I_4$ is equivalent to a system of linear equations in the coefficients of $Y$. All solutions for $Y$ are given by
$$
Y=\frac{1}{6}\begin{pmatrix} a & -6 & -3 & -2 \cr
6 & b & -6 & -3 \cr
3 & 6 & c & -6 \cr
2 & 3 & 6 & d
 \end{pmatrix}
$$
for arbitrary $a,b,c,d\in K$.
You, or someone else already asked this for $3\times 3$-matrices, see below. Now it should be clear how to generalize to all $n$.
Reference:
Finding matrices $X,Y$ such that $XY - YX = \left[\begin{smallmatrix}0&1&1\\1&0&1\\1&1&0\end{smallmatrix}\right]$


TITLE: Does every simplicial polytope have a topology-preserving contractible edge?
QUESTION [7 upvotes]: An edge of a triangulated manifold is said to be contractible if it may be contracted to a vertex without modifying the topological type of the underlying manifold. Otherwise, the edge is noncontractible.
Not every edge of a triangulated manifold is contractible. For example, on a triangular bipyramid, the edges gluing the two pyramids together are noncontractible, while all other edges are contractible (yielding a simplex). On a $d$-simplex, no edge is contractible.
Dey, Edelsbrunner, Guha, and Nekyahev provide exact conditions for when an edge of a 2- or 3-manifold is contractible, noting that an edge $ab$ between vertices $a$ and $b$ is contractible iff  $link(ab) = link(a) \cap link(b)$.  I have found little in the literature about conditions for the contractibility of edges on higher-dimensional objects.
My questions concern the existence of contractible edges on polytopal $d$-spheres, ie, simplicial polytopes.
1) Does every simplicial polytope other than the simplex have a contractible edge?
2) Does contraction of an edge of a simplicial polytope always yield another polytope?
3) Are these questions any easier to answer if we restrict the question to 4-polytopes?

REPLY [6 votes]: I brought this thread to the attention of Eran Nevo, who gave some more references, and pointed out that the answer to the converse of your question (2) is "no".
In his thesis, which I gave the arXiv link to above, and in the paper version of this "Higher minors and Van Kampen's obstruction", Nevo defines a strongly edge decomposable sphere to be a sphere which can be reduced to the simplex by edge contractions.  He shows that in any PL-manifold (in any dimension), an edge is contractible if and only if it satisfies the link condition.
Then Satoshi Murai in "Algebraic shifting of strongly edge decomposable spheres" shows that any squeezed sphere is strongly edge decomposable.  (Murai also has an earlier paper which is related, "Generic initial ideals and squeezed spheres".)  Since there are more squeezed $d$-spheres than polytopes for $d \geq 5$, there are non-polytopal spheres where an edge contraction leaves a polytopal sphere.  Though I didn't see $d=4$ settled explicitly one way or the other in the papers I looked at, I suspect that there are non-polytopal squeezed 4-spheres as well (and probably this is known).  Every squeezed $3$-sphere is polytopal.
Another paper which is somewhat relevant is Babson and Nevo "Lefschetz properties and basic constructions on simplicial spheres".
I also thought a bit about the forward direction of your question (2).  If $vw$ is a contractible edge in the boundary of the convex hull of $V$, it looks plausible to me that the edge contraction should correspond to passing to (boundary of) the convex hull of $V \setminus \lbrace v,w \rbrace \cup x$, where $x$ is a point on $vw$.  For as you 'slide' $v$ and $w$ together, the link condition seems to prevent new faces from being created.  (Obviously, there is a lot to be checked with this idea!)


TITLE: Why does $\mathop {\lim }\limits_{n \to \infty } \left\| {T{x_n} - \lambda {x_n}} \right\| = 0$?
QUESTION [0 upvotes]: Let $T \in B(X)$ and $\lambda  \in \sigma (T)$, why is there a sequence like $\left\{ {{x_n}} \right\}$  in $X$ with $\left\| {{x_n}} \right\| = 1$ such that $\mathop {\lim }\limits_{n \to \infty } \left\| {T{x_n} - \lambda {x_n}} \right\| = 0$?

REPLY [1 votes]: The problem as stated is false. For example, consider the unilateral shift $S$ on $\ell^{2}$ defined by
$$
    (x_0,x_1,x_2,\cdots) \mapsto (0,x_0,x_1,x_2,\cdots)
$$
Then $\|Sx\|=\|x\|$ for all $x\in\ell^{2}$, and $0\in\sigma(S)$ because the range of $S$ is of co-dimension $1$ in $\ell^{2}$. However, if $\{ x_n \}$ is a sequence of unit vectors, then $\|(S-0I)x_n\|=\|Sx_n\|=\|x_n\|=1$.


text:
\begin{document}

\keywords{$p$-evolution equations, well-posedness in Sobolev spaces,
pseudo-differential operators}
\subjclass[2000]{Primary 35G10; Secondary 35A27}

\begin{abstract}
We consider $p$-evolution equations, for $p\geq2$, with complex valued
coefficients. We prove that a necessary condition for $H^\infty$ well-posedness
of
the associated Cauchy problem is that
the imaginary part of the coefficient of the subprincipal part
(in the sense of Petrowski) satisfies a decay estimate as $|x|\to+\infty$.
\end{abstract}

\maketitle
\markboth{\sc A necessary condition for
$H^\infty$ well-posedness \ldots}{\sc A.~Ascanelli, C.~Boiti, L.~Zanghirati}

\section{Introduction and main result}

Given an integer $p\geq 2$, we consider in $[0,T]\times\R$ the
linear partial differential operator $P$ of the form
\begin{equation}\label{op}
P(t,x,D_t,D_x)=D_t+a_p(t)D_x^p+\sum_{j=0}^{p-1}
a_j(t,x)D_x^j\,,
\end{equation}
with $D=\frac 1i \partial$, $a_p\in C([0,T];\R)$ and
$a_j\in C([0,T];{\mathcal B}^\infty)$ for $0\leq j\leq p-1$, (here
${\mathcal B}^\infty={\mathcal B}^\infty(\R_x)$ is the space of complex
valued functions which are bounded on $\R_x$ together with all their
$x$-derivatives). We are dealing with a non-kowalewskian evolution operator;
anisotropic evolution operators of the form (\ref{op}) are usually called
$p-$evolution operators. The condition that $a_p$ is real valued means that
the principal symbol
(in the sense of Petrowski)
of $P$ has the real characteristic $\tau=-a_p(t)\xi^p$; by the Lax-Mizohata
theorem (cf. \cite{M}), this is a necessary condition to have a unique
solution, in Sobolev spaces, of the Cauchy problem
\beqs
\label{CP1}
\begin{cases}
P(t,x,D_t,D_x)u(t,x)=f(t,x) & (t,x)\in[0,T]\times\R\cr
u(0,x)=g(x) & x\in\R,
\end{cases}
\eeqs
in a neighborhood of $t=0$.
We notice that for
$p=2$ the operator is of Schr\"odinger type, for $p=3$ we have the
same principal part as the Korteweg-De Vries equation.
Many
results of well-posedness
in Sobolev spaces of \eqref{CP1}
are available under the assumption that all the coefficients
$a_j$ of \eqref{op} are real (see, for instance,
\cite{Ag1}, \cite{Ag2}, \cite{AZ},
\cite{AC}, \cite{CC2}, \cite{CHR}). On the contrary, when the coefficients
$a_j(t,x)$ for $1\leq j\leq p-1$
are not real, the theory is well developed only in the case $p=2$: we
know from the pioneering papers
\cite{I1}, \cite{I2} that a decay condition as $|x|\to+\infty$ on
$\Im a_1$ is necessary and sufficient for well-posedness of the
Cauchy problem \eqref{CP1} in $H^\infty$.
Sufficient conditions for well-posedness in $H^\infty$ and/or Gevrey classes
for 2 or
$3-$evolution equations have been given by many authors (see, for instance,
\cite{D}, \cite{T}, \cite{B}, \cite{KB}, \cite{ACC}, \cite{CR1},
\cite{CR2}, \cite{CC1}).
The general case $p\geq2$ has been recently considered in \cite{ABZ1}, proving
$H^\infty$ well-posedness of the Cauchy problem \eqref{CP1} under suitable
decay conditions, as $|x|\to+\infty$, on $\Im D_x^\beta a_j$, for
$j\leq p-1$ and $[\beta/2]\leq j-1$.
These results have been extended to the case of weighted Sobolev spaces in
\cite{ACpp}, to the case of first order systems of pseudo-differential
operators in \cite{AB1}, to the case of higher order equations
in \cite{AB2}, and to semi-linear
3-evolution equations in \cite{ABZ2}.

As far as we know, there are no results available about necessary
conditions for $H^\infty$ well-posedness for $p$-evolution equations,
$p\geq 3$.

In this paper we give a necessary condition
for well-posedness of the
Cauchy problem \eqref{CP1} in $H^\infty$, generalizing to the case
$p\geq2$ the necessary condition given by Ichinose in \cite{I1} for
$p=2$.
More precisely, in \cite{I1} Ichinose considered, for $x\in\R^n$,
the operator
\begin{equation}\label{pichi}
P=D_t-a_2\triangle_x+\sum_{j=1}^na_1^{(j)}(x)D_{x_j}+c(x),
\end{equation}
with $a_2\in(0,1]$ and $a_1^{(j)},c\in{\mathcal B}^\infty(\R^n)$.
He proved that a necessary condition for $H^\infty$ well-posedness
of the associated Cauchy problem
is the existence of non-negative constants $M,\ N$ such that
\begin{equation}
\label{ichi}
\sup_{x\in\R^n\!,\,\omega\in S^{n-1}}\left\vert\sum_{j=1}^n\int_0^\varrho
\Im a_1^{(j)}(x+2a_2\theta\omega)\omega_jd\theta
\right\vert\leq M\log(1+\varrho)+N
\qquad\forall\varrho>0,
\end{equation}
where $S^{n-1}$ is the unit sphere in $\R^{n}$.
The same condition is also sufficient (cf. \cite{I2}) only in the
case of space dimension $n=1$.

In this paper we assume
that there exists a constant $m>0$ such that
\begin{equation}\label{a3}
|a_p(t)|\geq m\qquad \forall t\in [0,T]
\end{equation}
and prove the following:
\begin{Th}
\label{th1}
Let $P$ be the operator in \eqref{op} with $a_p\in C([0,T];\R)$
satisfying \eqref{a3} and $a_j\in C([0,T];\B^\infty)$ for $0\leq j\leq p-1$.
A necessary
condition for the Cauchy problem \eqref{CP1} to be well-posed in $H^\infty$ is
the existence of constants $M,N>0$ such that:
\begin{equation}
\label{CN2}
\sup_{x\in\R}\min_{0\leq\tau\leq t\leq T}\int_{-\varrho}^\varrho
\Im a_{p-1}(t,x+p a_p(\tau)\theta)d\theta\leq M\log(1+\varrho)+N,\qquad
\forall \varrho>0.
\end{equation}
\end{Th}

\begin{Rem}
\begin{em}
If the coefficient $a_p(t)$ vanishes at some point of the interval $[0,T]$,
the well-posedness in $H^\infty$ of the Cauchy problem \eqref{CP1}
may fail to be true also if the necessary condition
\eqref{CN2} is satisfied (see \cite{CR3}, \cite{ABZ1}).
In \cite{ABZ1}, for instance,
a Levi-type condition of the form
\beqs
\label{levi}
|\Im a_{p-1}(t,x)|\leq Ca_p(t)/\langle x\rangle
\eeqs
is needed to
weaken \eqref{a3} into
$a_p(t)\geq0$ for proving $H^\infty$ well-posedness of \eqref{CP1}.
Notice that condition \eqref{levi} is consistent with \eqref{CN2}.
In the present paper we focus
on the fact
that the dependence on space of the coefficient of the subprincipal part is
allowed only accompanied by a decay condition at infinity.
\end{em}
\end{Rem}

\section{Idea of the proof and auxiliary tools.}
\label{sec2}
We prove Theorem \ref{th1} by contradiction, taking $f\equiv 0$
without any loss of generality.

We assume the Cauchy problem \eqref{CP1} to be well-posed, so that
for every $g\in H^\infty$, there exists a unique
$u\in C([0,T];H^\infty)$
solution of \eqref{CP1} and there exists $q\in\N_0:=\N\cup\{0\}$
and $C>0$ such that
\begin{equation}\label{q}
\|u(t,\cdot)\|_0\leq C\|g\|_q\qquad\forall t\in [0,T],
\end{equation}
where $\|\cdot\|_s$ stands for the norm in the Sobolev space $H^s$
(we shall write $\|\cdot\|:=\|\cdot\|_0$ for simplicity).

Then we assume, by contradiction, that \eqref{CN2} does not hold.
This implies that, for every $M>0$ and $k\in\N$ there exist a sequence
of points $x_k\in\R$ and a sequence $\varrho_k\to+\infty$ such that
\beqs
\label{-roro}
\int_{-\varrho_k}^{\varrho_k}\Im a_{p-1}(t,x_k+pa_p(\tau)\theta)d\theta\geq
M\log(1+\varrho_k)+k\qquad\forall 0\leq\tau\leq t\leq T.
\eeqs
We can then construct a sequence of initial data $g_k$ localized at high
frequency $n_k:=\varrho_k^a$,
for suitable $a>0$, so
obtaining a sequence $u_k$ of solutions of the corresponding Cauchy
problem. Further localizing these solutions
in the phase space along the trajectory of
the hamiltonian $a_p(t)\xi^p$, we produce a sequence of functions
$v_k^{\alpha,\beta}$ (for $\alpha,\beta\in\N_0$) satisfying some energy
estimates, because of \eqref{q}.

Taking, finally, a suitable linear combination $\sigma_k(t)$ of the
$L^2$-norms $\|v_k^{\alpha,\beta}(t,\cdot)\|$, we obtain, in Section~\ref{sec3},
that \eqref{-roro} implies an estimate from below of $\sigma_k(t)$;
this estimate will contradict an estimate from above
for $\sigma_k(t)$ which is stated and proved in Section~\ref{sec4}.

In this section we discuss condition \eqref{CN2}, construct the
sequence $\{v_k^{\alpha,\beta}\}$ and collect some estimates that will be crucial
in the proofs of the contradictory estimates from below
and from above of $\sigma_k(t)$.

The next section is completely devoted to the proof of the estimate from
below \eqref{formulabelow}.

In Section~\ref{sec4} we give the estimate from above
\eqref{formulaabove}, and finally prove Theorem \ref{th1}.

Let us start by remarking that if condition \eqref{CN2} does not hold, then
at least one of the following two conditions does not hold:
\beqs
\label{A}
\sup_{x\in\R}\min_{0\leq \tau\leq t\leq T}\int_0^\varrho
\Im a_{p-1}(t,x+p a_p(\tau)\theta)d\theta\leq M\log(1+\varrho)+N,\qquad
\forall \varrho>0,
\eeqs
or
\beqs
\label{B}
\sup_{x\in\R}\min_{0\leq \tau\leq t\leq T}\int_{-\varrho}^0
\Im a_{p-1}(t,x+p a_p(\tau)\theta)d\theta\leq M\log(1+\varrho)+N,\qquad
\forall \varrho>0.
\eeqs
Since
\beqsn
\int_{-\varrho}^0
\Im a_{p-1}(t,x+p a_p(\tau)\theta)d\theta=
\int_0^\varrho\Im a_{p-1}(t,x-p a_p(\tau)\theta)d\theta,
\eeqsn
we can assume, without any loss of generality,
that \eqref{A} does not hold and obtain then a contradiction
(if \eqref{B} does not hold we argue in the same way taking $-a_p$ instead
of $a_p$).

The following lemma will be the key to obtain the desired estimate from below
\eqref{formulabelow}:
\begin{Lemma}
\label{lemma1}
If \eqref{A} does not hold, then for every $M>0$ and $k\in\N$ there
esist $x_k\in\R$ and $\varrho_k>0$ such that:
\begin{itemize}
\item[(i)]
$\varrho_k\to+\infty$;
\item[(ii)]
$\ds\int_0^{\varrho_k}\Im a_{p-1}(t,x_k+p a_p(\tau)\theta)d\theta\geq
M\log(1+\varrho_k)+k\qquad\forall 0\leq\tau\leq t\leq T$;
\item[(iii)]
$\ds\int_0^\varrho\Im a_{p-1}(t,x_k+p a_p(\tau)\theta)d\theta\geq0\qquad\forall
\varrho\in[0,\varrho_k],\ 0\leq\tau\leq t\leq T$.
\end{itemize}
\end{Lemma}
\begin{proof}
If \eqref{A} fails to be true, then for every $M>0$,
$k\in\N$  there exist
$y_k\in\R$ and $\delta_k>0$ such that for all $0\leq\tau\leq t\leq T$
\beqs
\label{deltak}
\int_0^{\delta_k}\Im a_{p-1}(t,y_k+p\theta a_p(\tau))
d\theta\geq M\log(1+\delta_k)+k.
\eeqs
Let us set, for $s\in[0,\delta_k]$,
\beqsn
F_k(s):=\int_0^s\Im a_{p-1}(t,y_k+p\theta a_p(\tau))d\theta
\eeqsn
and let $s_k$ be point of minimum of $F_k$ on $[0,\delta_k]$. Define then
\beqsn
x_k:=&&y_k+ps_ka_p(\tau)\\
\varrho_k:=&&\delta_k-s_k.
\eeqsn
Remark that all $y_k,\delta_k,s_k,x_k,\varrho_k$ depend also on $M$.

For all $s\in[0,\varrho_k]\subseteq[0,\delta_k]$:
\beqs
\nonumber
\int_0^s\Im a_{p-1}(t,x_k+p a_p(\tau)\theta)d\theta
=&&\int_{s_k}^{s+s_k}\Im a_{p-1}(t,y_k+p a_p(\tau)\theta')d\theta'\\
\label{2star}
=&&F_k(s+s_k)-F_k(s_k)\geq0
\eeqs
by definition of $s_k$. This proves
$(iii)$.

Moreover, $F_k(s_k)\leq F_k(0)=0$ and hence, from \eqref{2star}
and \eqref{deltak}:
\beqsn
\int_0^{\varrho_k}\Im a_{p-1}(t,x_k+p a_p(\tau)\theta)d\theta=&&
\int_0^{\delta_k-s_k}\Im a_{p-1}(t,x_k+p a_p(\tau)\theta)d\theta\\
=&&F_k(\delta_k)-F_k(s_k)\geq F_k(\delta_k)\\
\geq&&M\log(1+\delta_k)+k\geq M\log(1+\varrho_k)+k,
\eeqsn
proving $(ii)$.

Finally, the last inequality implies, for $k\to+\infty$,
\beqsn
\int_0^{\varrho_k}\Im a_{p-1}(t,x_k+p a_p(\tau)\theta)d\theta\geq k\to+\infty
\eeqsn
and hence $\varrho_k\to+\infty$, because $a_{p-1}\in\B^\infty$.
\end{proof}

\subsection{Solutions with high frequency initial data}
Let us fix, here and throughout all the paper, a cut-off
function $h\in C^\infty(\R)$, such that
\begin{equation}\label{h}
h(y)=\left\{\begin{array}{ll}
1 & |y|\leq 1/4
\\
0 & |y|\geq 1/2,
\end{array}\right.
\end{equation}
and a rapidly decreasing function $\psi$ such that $\psi(0)=2$ and
$$\supp \hat\psi\subseteq\{\xi\in\R:\  h(\xi)=1\}.$$
Define then
\begin{equation}\label{gk}
g_k(x)=e^{i(x-x_k)n}\psi(x-x_k),
\end{equation}
where
\begin{equation}\label{n}
n:=\varrho_k^{a}
\end{equation}
for some $a>0$ to be chosen later on (see \eqref{sceltamu}),
and $x_k, \varrho_k$ as in Lemma \ref{lemma1}.
Note that
\beqs
\label{fgk}
\hat{g}_k(\xi)=e^{-ix_k\xi}\hat{\psi}(\xi-n),
\eeqs
so ${g}_k$ is localized in the phase space around the point $(x_k,n)$.

Denote by $u_k\in C([0,T]; H^\infty)$
the solution of the Cauchy problem
\beqs
\label{CP2}
\begin{cases}
P(t,x,D_t,D_x)u_k(t,x)=0 & (t,x)\in[0,T]\times\R\cr
u_k(0,x)=g_k(x) & x\in\R.
\end{cases}
\eeqs
Then, by \eqref{q} and \eqref{fgk} we have, for all $t\in [0,T]$:
\beqs\nonumber
 \|u_k(t,\cdot)\|\leq C\|g_k\|_q
&=&C(2\pi)^{-1/2}\left(\ds\int_{\R}
\langle\xi\rangle^{2q}|\hat g_k(\xi)|^2d\xi\right)^{1/2}
 \\
  \nonumber
 &\leq&C_q (2\pi)^{-1/2}\langle n\rangle^{q}\left(\ds\int_{\R}
\langle\theta\rangle^{2q}|\hat \psi(\theta)|^2d\theta\right)^{1/2}
\\
\label{C'T}
&\leq& C'_qn^q,
\eeqs
for some $C_q,C'_q>0$.

\subsection{A localizing operator}
In this subsection we define, by giving its symbol $w_{n,k}(t,x,\xi)$,
a pseudo-differential operator $W_{n,k}(t,x,D_x)$
which localizes the
solutions of \eqref{CP1} in the phase space along the trajectory of
the hamiltonian $a_p(t)\xi^p$.

Let $w_{n,k}(t,x,\xi)$ be the solution of the Hamilton's equation of motion
\begin{equation}
\label{eqw}
\begin{cases}
\partial_tw_{n,k}=\left\{w_{n,k},-a_p(t)\xi^p\right\}
\\
w_{n,k}(0,x,\xi)=w_{0,n,k}(x,\xi):=\varrho_k^{1/2}h
\left(\varrho_k(x-x_k)\right)h\left(\varrho_k^\mu(\xi/n-1)\right),
\end{cases}
\end{equation}
with $\mu>0$ to be chosen later (see \eqref{sceltamu}),
where $\{\cdot,\cdot\}$ denotes the Poisson brackets defined by
\[\{p(x,\xi),q(x,\xi)\}=\partial_xp(x,\xi)\partial_\xi
q(x,\xi)-\partial_\xi p(x,\xi)\partial_xq(x,\xi).\]
Computing the Poisson brackets, equation \eqref{eqw} reduces to
\begin{equation}
\label{215}
\begin{cases}
\left(\partial_t +pa_p(t)\xi^{p-1}\partial_x\right) w_{n,k}=0
\\
w_{n,k}(0,x,\xi)=w_{0,n,k}(x,\xi),
\end{cases}
\end{equation}
which admits the solution
\[
w_{n,k}(t,x,\xi)=w_{0,n,k}(x-pA_p(t)\xi^{p-1},\xi),\quad A_p(t)=
\int_0^t a_p(\tau)d\tau.
\]
We thus obtain
\begin{equation}
\label{defwnk}
w_{n,k}(t,x,\xi):=\varrho_k^{1/2}h
\left(\varrho_k(x-x_k-pA_p(t)\xi^{p-1})\right)
h\left(\varrho_k^\mu(\xi/n-1)\right).
\end{equation}
The following lemma shows that the symbol $w_{n,k}(t,x,\xi)$ is
supported in a neighborhood of
the solution $(x_k+pA_p(t)n^{p-1},n)$ of
the Hamilton's canonical equation with
initial data $(x_k,n)$; moreover it introduces the sequence of symbols
$w_{n,k}^{\alpha,\beta}$ which naturally appear in the computation of
$\partial_\xi^\alpha D_x^\beta w_{n,k}$.

\begin{Lemma}
\label{lemma2}
Let us define, for $\alpha,\beta\in\N_0$,
$\mu\geq2$ and $n$ as in \eqref{n},
the symbols
\begin{equation}
\label{wnkab}
w_{n,k}^{\alpha,\beta}(t,x,\xi):=\left.\varrho_k^{1/2}(\partial_x^\alpha h)(x)
(\partial_\xi^\beta h)(\xi)
\right\vert_{\afrac{x=\varrho_k(x-x_k-pA_p(t)\xi^{p-1})}{\hspace*{-12mm}\xi
=\varrho_k^\mu(\xi/n-1)}}.
\end{equation}
Then, for every
$t\in\left[0,\ds\frac{\varrho_k}{n^{p-1}}\right]$ we have that
\beqsn
\supp w_{n,k}^{\alpha,\beta}(t)\subseteq\left\{(x,\xi):\
\vert x-(x_k+pA_p(t)n^{p-1})\vert\leq \frac{c_p}{\varrho_k},\
\vert\xi/n-1\vert\leq \frac{1}{2\varrho_k^\mu}\right\},
\eeqsn
for $c_p=\max\{1,p2^{p-1}\sup_{[0,T]}|a_p|\}$, if $k$ is large enough.
\end{Lemma}
\begin{proof}
The estimate $\vert\xi/n-1\vert\leq 1/(2\varrho_k^\mu)$ trivially
follows by definition \eqref{wnkab} and by \eqref{h}.
Moreover, \eqref{h} implies, for $t\in[0,\varrho_k/n^{p-1}]$, $\mu\geq2$
and $(x,\xi)\in\supp w_{n,k}^{\alpha,\beta}$:
\beqsn
\vert x-(x_k+pA_p(t)n^{p-1})\vert&\leq&\vert x-(x_k+pA_p(t)\xi^{p-1})
\vert+p|A_p(t)|n^{p-1}\left\vert \left(\frac{\xi}{n}\right)^{p-1}-1\right\vert
\\
&\leq&\frac 1{2\varrho_k}+p\sup_{[0,T]}|a_p| t n^{p-1}
\cdot\left|\frac\xi n-1\right|\cdot\left|\left(\frac{\xi}{n}\right)^{p-2}
+\left(\frac{\xi}{n}\right)^{p-3}+\ldots+1\right|\\
&\leq&\frac 1{2\varrho_k}+p\sup_{[0,T]}|a_p| \frac {\varrho_k}{2\varrho_k^\mu}
\left[2^{p-2}
+2^{p-3}+\ldots+1\right]\\
&\leq&\frac 1{2\varrho_k}+p\sup_{[0,T]}|a_p| \frac 1{2\varrho_k^{\mu-1}}2^{p-1}
\leq \frac{c_p}{\varrho_k},
\eeqsn
for $c_p=\max\{1,p2^{p-1}\sup_{[0,T]}|a_p|\}$,
since $\xi/n\leq|\xi/n-1|+1\leq1/(2\varrho_k^\mu)+1\leq2$ for $k$
large enough.
\end{proof}

As a consequence of Lemma \ref{lemma2},
we localize, in the phase space, the solution of \eqref{CP2},
defining
\begin{equation}\label{vkab}
v_k^{\alpha,\beta}(t,x):=W_{n,k}^{\alpha,\beta}(t,x,D_x)u_k(t,x),
\end{equation}
where $W_{n,k}^{\alpha,\beta}(t,x,D_x)$ is the pseudo-differential
operator with symbol $w_{n,k}^{\alpha,\beta}(t,x,\xi)$.
We shall denote, throughout all the paper, $W_{n,k}:=W_{n,k}^{0,0}(t,x,D_x)$
and  $v_k:=v_k^{0,0}$ for simplicity.

\subsection{Useful estimates}
In the next sections we need
estimates of the $L^2$-norms of the functions
$v_k^{\alpha,\beta}$ and of both operators $W_{n,k}^{\alpha,\beta}$
and $[a_j,W_{n,k}]$ acting on $u_k$ . In this subsection we state and
prove all these estimates. Proofs are quite technical, and the main
tools for obtaining them, collected in
Appendix A, are the Calderon-Vaillantcourt's Theorem \ref{thC}
and a skillful use of the expansion formula of the symbol of the
product of two pseudo-differential operators (Theorems \ref{thA} and
\ref{thB}).
To avoid losing his train of thought, the reader can skip
these estimates at a first reading,
passing directly to Section~\ref{sec3} and coming back to the estimates at
the moment of their application.

To estimate the $L^2$-norms of $v_k$ and of
$v_k^{\alpha,\beta}$ we first need estimates of the semi-norms
$|\cdot|^{0}_{\ell,\ell}$ of the
symbols $w_{n,k}^{\alpha,\beta}\in S^0_{0,0}$, defined in formula \eqref{semi000}
of the Appendix.
\begin{Lemma}
\label{lemmaAB1}
Let $n=\varrho_k^a$ with $a\geq \mu\geq 2$, and
$t\in \left[0,\ds\frac{\varrho_k}{n^{p-1}}\right]$.
Then, for every $\alpha,\beta\in\N_0$ we have, for $k$ large enough:
\begin{itemize}
\item[(i)]
for every $\gamma,\sigma\in\N_0$ there exists a constant
$C_{\alpha,\beta,\gamma,\sigma}>0$ such that, for all
$(t,x,\xi)\in[0,\frac{\varrho_k}{n^{p-1}}]\times\R^2$:
\beqsn
|\partial_\xi^\gamma\partial_x^\sigma w_{n,k}^{\alpha,\beta}(t,x,\xi)|
\leq C_{\alpha,\beta,\gamma,\sigma}\varrho_k^{\frac12+\sigma}
\left(\frac{\varrho_k^\mu}{n}\right)^\gamma;
\eeqsn
\item[(ii)]
for every $\ell\in\N$ there exists $C_{\alpha,\beta,\ell}>0$ such that
\beqsn
\left\vert w_{n,k}^{\alpha,\beta}\right\vert_{\ell,\ell}^{0}\leq
C_{\alpha,\beta,\ell}\varrho_k^{\frac 12+\ell};
\eeqsn
\item[(iii)]
for every $h\in \N_0$ and $\nu,\ell\in\N$ there exists
$C_{\alpha,\beta,\nu,\ell}>0$ such that
\beqs\label{AB5}
|\xi^h\partial_\xi^\nu w_{n,k}^{\alpha,\beta}|_{\ell,\ell}^0
\leq C_{\alpha,\beta,\nu,\ell}\,n^h\varrho_k^{\frac12+\ell}
\left(\frac{\varrho_k^\mu}{n}\right)^\nu.
\eeqs
\end{itemize}
\end{Lemma}

\begin{proof}
Let us write
\beqs
\nonumber
\partial_\xi^\gamma\partial_x^\sigma w_{n,k}^{\alpha,\beta}(t,x,\xi)=&&
\varrho_k^\sigma\partial_\xi^\gamma w_{n,k}^{\alpha+\sigma,\beta}(t,x,\xi)\\
\label{CC1}
=&&
\varrho_k^{\sigma+\frac12}\sum_{\gamma_1+\gamma_2=\gamma}C_\gamma
\partial_\xi^{\gamma_1}h^{(\alpha+\sigma)}(\varrho_k(x-x_k-pA_p(t)\xi^{p-1}))\cdot
\partial_\xi^{\gamma_2}h^{(\beta)}(\varrho_k^\mu(\xi/n-1)).
\eeqs
Since $|\xi|\leq 2n$ on $\supp w_{n,k}^{\alpha,\beta}$, by Lemma \ref{lemma2} we
have that
\beqsn
\left\vert \partial_\xi^{\gamma_1}h^{(\alpha+\sigma)}
\left(\varrho_k(x-x_k-pA_p(t)\xi^{p-1})\right)\right\vert&\leq&
C_{\alpha,\sigma,\gamma_1}(A_p(t)|\xi|^{p-2}\varrho_k)^{\gamma_1}\\
&\leq& C_{\alpha,\sigma,\gamma_1}\left(\sup_{[0,T]}|a_p|\cdot t
\cdot|\xi|^{p-2}\varrho_k \right)^{\gamma_1}
\leq C'_{\alpha,\sigma,\gamma_1}\left(\frac{\varrho_k^2}{n}\right)^{\gamma_1}
\eeqsn
for $t\in[0,\varrho_k/n^{p-1}]$.
Moreover,
\beqsn
\left\vert \partial_\xi^{\gamma_2}h^{(\beta)}
\left(\varrho_k^\mu(\xi/n-1)\right)\right\vert&\leq& C_{\gamma_2,\beta}
\left(\frac{\varrho_k^\mu}{n}\right)^{\gamma_2}.
\eeqsn
Substituting in \eqref{CC1} we thus obtain $(i)$, since $\mu\geq2$.

From $(i)$ and $a\geq\mu$ we get
\beqsn
\left\vert w_{n,k}^{\alpha,\beta}\right\vert_{\ell,\ell}^{0}
=\sup_{\afrac{\gamma,\sigma\leq \ell}{x,\xi\in\R}}\left\vert
\partial_\xi^\gamma\partial_x^\sigma w_{n,k}^{\alpha,\beta}(t,x,\xi)\right\vert
\leq
\sup_{\afrac{\gamma,\sigma\leq \ell}{x,\xi\in\R}}C_{\alpha,\beta,\gamma,\sigma}\,
\varrho_k^{\frac 12+\sigma}\left(\frac{\varrho_k^\mu}{n}\right)^{\gamma}
\leq C_{\alpha,\beta,\ell}\,
\varrho_k^{\frac 12+\ell}
\eeqsn
i.e. also $(ii)$ is satisfied.

Finally, $(iii)$ follows from $(i)$, since $|\xi|\leq2n$ on $\supp
w_{n,k}^{\alpha,\beta}$ and $a\geq\mu$.
\end{proof}

We are now ready to estimate $\|v_k\|$. By Calderon-Vaillalntcourt's
Theorem \ref{thC}, $(ii)$ of Lemma \ref{lemmaAB1}
and \eqref{C'T}, we have that for all $t\in[0,\varrho_k/n^{p-1}]$
\beqs\label{nomme}
\|v_k(t,\cdot)\|&=&\|W_{n,k}^{0,0}(t,\cdot,D_x)u_k(t,\cdot)\|
\leq C\left\vert w_{n,k}(t,x,\xi)\right\vert_{2,2}^{0}\|u_k(t,\cdot)\|
\leq C'\varrho_k^{\frac 12+2}n^q
\eeqs
for some $C,C'>0$; similarly, for every $\alpha,\beta\in\N_0$, it follows that
\beqs\label{normevkab}
\|v_k^{\alpha,\beta}(t,\cdot)\|\leq C_{\alpha,\beta}\varrho_k^{\frac 12+2}n^q
=C_{\alpha,\beta}\varrho_k^{\frac 12+2+aq}\qquad\forall t\in[0,\varrho_k/n^{p-1}]
\eeqs
for some $C_{\alpha,\beta}>0$.
To estimate also the derivatives of $v_k^{\alpha,\beta}$ we need the following:

\begin{Lemma}
\label{lemmaAB3}
Let $n=\varrho_k^a$  with $a\geq\mu$.
For every $\nu,r\in\N$ and $\alpha,\beta\in\N_0$ there exists
$C_{\alpha,\beta,r,\nu}>0$ such that for all $t\in[0,\varrho_k/n^{p-1}]$
with $k$ large enough:
\beqsn
\|D_x^rv_k^{\alpha,\beta}(t,\cdot)\|\leq c_1n^r\|v_k^{\alpha,\beta}\|
+C_{\alpha,\beta,r,\nu}\varrho_k^{4+\frac12}
\left(\frac{\varrho_k^{\mu+1}}{n}\right)^\nu
n^{r+q},
\eeqsn
for a fixed constant $c_1>0$.\end{Lemma}
\begin{proof}
We define the function
\beqs\label{chi1k}
\chi_{1,k}(\xi)=h\left(\frac{\varrho_k^\mu}3\left(\frac\xi n-1\right)\right).
\eeqs
By definition \eqref{h}, we have that
\beqs
\label{suppchi1k}
\supp \chi_{1,k}\subseteq \Big\{\xi\ :\
\Big\vert\frac \xi n-1\Big\vert\leq \frac3{2\varrho_k^\mu}\Big\}
\subseteq \left\{\xi\ :\ \vert\xi\vert\leq 3n\right\},
\eeqs
and
\beqs
\label{suppchi1}
\supp \left(1-\chi_{1,k}\right)\subseteq \Big\{\xi\ :\
\Big\vert\frac\xi n -1\Big\vert\geq \frac{3}{4\varrho_k^\mu}\Big\}.
\eeqs
This implies, by Lemma \ref{lemma2}, that
\beqs
\label{intvuota}
\supp (1-\chi_{1,k})\cap\supp w_{n,k}^{\alpha,\beta}=\emptyset.
\eeqs
Localizing now at frequency $n$
\beqs
\nonumber
D_x^rv_k^{\alpha,\beta}&=&\chi_{1,k}(D_x)D_x^rv_k^{\alpha,\beta}
+(1-\chi_{1,k}(D_x))D_x^rv_k^{\alpha,\beta}\\
\nonumber
&=&\chi_{1,k}(D_x)D_x^rv_k^{\alpha,\beta}+\sum_{j=0}^r\binom rj
(1-\chi_{1,k}(D_x))(D_x^jW_{n,k}^{\alpha,\beta})
D_x^{r-j}u_k,
\eeqs
and applying Calder\'on-Vaillancourt's Theorem \ref{thC}, we come to:
\beqs
\nonumber
\|D_x^rv_k^{\alpha,\beta}(t,\cdot)\|
\leq&&|\chi_{1,k}(\xi)\xi^r|_{2,2}^0\cdot\|v_k^{\alpha,\beta}\|\\
\label{C3}
&&+\sum_{j=0}^r\binom rj\varrho_k^j
\left |\sigma\left((1-\chi_{1,k}(D_x))W_{n,k}^{\alpha+j,\beta}
D_x^{r-j}\right)\right |_{2,2}^0\cdot\|u_k\|.
\eeqs
Note that $|\chi_{1,k}(\xi)\xi^r|_{2,2}^0\leq c_1n^r$ for some $c_1>0$,
because of \eqref{suppchi1k};
to estimate the second term of \eqref{C3}, by Theorem \ref{thA}
and \eqref{intvuota} we write, for every integer $\nu\geq1$:
\beqs
\nonumber
\hspace*{10mm}
&&\sigma\left((1-\chi_{1,k}(D_x))W_{n,k}^{\alpha+j,\beta}D_x^{r-j}\right)=
\sum_{0\leq\gamma\leq \nu-1}\frac{1}{\gamma!}\partial_\xi^\gamma(1-\chi_{1,k}(\xi))
D_x^\gamma(w_{n,k}^{\alpha+j,\beta}\xi^{r-j})\\
\nonumber
&&+\int_0^1\frac{(1-\theta)^{\nu-1}}{(\nu-1)!}\int\!\!\int e^{-iy\eta}
\partial_\xi^\nu(1-\chi_{1,k}(\xi+\theta\eta)D_x^\nu(w_{n,k}^{\alpha+j,\beta}
(t,x+y;\xi)
\xi^{r-j})dy\dbar\eta d\theta\\
\label{C2}
&&=\int_0^1\frac{(1-\theta)^{\nu-1}}{(\nu-1)!}
\Osc_\nu(t,\theta,x,\xi)
 d\theta,
\eeqs
where
\beqsn
\Osc_\nu(t,\theta,x,\xi):=\int\!\!\int e^{-iy\eta}
\partial_\xi^\nu(1-\chi_{1,k}(\xi+\theta\eta)D_x^\nu w_{n,k}^{\alpha+j,\beta}
(t,x+y;\xi)
\xi^{r-j}dy\dbar\eta.
\eeqsn

Writing
$
\xi^{r-j}=\sum_{h=0}^{r-j}\binom{r-j}{h}(\xi+\theta\eta)^h(-\theta\eta)^{r-j-h}
$
and $e^{-iy\eta}(-\eta)^{r-j-h}=D_y^{r-j-h}e^{-iy\eta}$, we have, integrating
by parts:
\beqsn
\Osc_\nu=&&-\sum_{h=0}^{r-j}\binom{r-j}{h}\theta^{r-j-h}\int\!\!\int
\partial_\xi^\nu\chi_{1,k}(\xi+\theta\eta)\cdot(\xi+\theta\eta)^h
D_x^\nu w_{n,k}^{\alpha+j,\beta}(t,x+y;\xi)\\
&&\cdot D_y^{r-j-h}e^{-iy\eta}dy\dbar\eta\\
=&&\sum_{h=0}^{r-j}(-1)^{r-j-h+1}\binom{r-j}{h}\theta^{r-j-h}\int\!\!\int
e^{-iy\eta}\partial_\xi^\nu\chi_{1,k}(\xi+\theta\eta)\cdot(\xi+\theta\eta)^h\\
&&\cdot
D_y^{\nu+r-j-h}w_{n,k}^{\alpha+j,\beta}(t,x+y;\xi)dy\dbar\eta\\
=&&\sum_{h=0}^{r-j}(-1)^{r-j-h+1}\binom{r-j}{h}\theta^{r-j-h}\varrho_h^{\nu+r-j-h}
\int\!\!\int
e^{-iy\eta}\partial_\xi^\nu\chi_{1,k}(\xi+\theta\eta)\cdot(\xi+\theta\eta)^h\\
&&\cdot
w_{n,k}^{\alpha+\nu+r-h,\beta}(t,x+y;\xi)dy\dbar\eta.
\eeqsn

By Theorem \ref{thB}, \eqref{chi1k}, \eqref{suppchi1k} and
Lemma \ref{lemmaAB1}, for $\theta
\in[0,1]$ we have that
\beqsn
|\Osc_\nu(t,\theta)|_{2,2}^0\leq&&\sum_{h=0}^{r-j}c_h
\varrho_k^{\nu+r-j-h}|\partial_\xi^\nu
\chi_{1,k}(\xi)\xi^h|_{4,4}^0\cdot|w_{n,k}^{\alpha+\nu+r-h,\beta}(t,x;\xi)|_{4,4}^0\\
\leq&&\sum_{h=0}^{r-j}C_{\alpha,\nu,r,h,\beta}\varrho_k^{\nu+r-j-h}
\left(\frac{\varrho_k^\mu}{n}\right)^\nu n^h
\varrho_k^{\frac12+4}\\
\leq
&&\sum_{h=0}^{r-j}C'_{\alpha,\beta,\nu,r,h}\left(\frac{\varrho_k^\mu}{n}\right)^\nu
\left(\frac{n}{\varrho_k}\right)^{r-j}\varrho_k^{\nu+r-j+\frac12+4}\\
=&&C_{\alpha,\beta,\nu,r,j}\,
\varrho_k^{\frac12+4}\left(\frac{\varrho_k^{\mu+1}}{n}\right)^\nu
n^{r-j}
\eeqsn
for some $c_h,C_{\alpha,\nu,r,h,\beta},C'_{\alpha,\beta,\nu,r,h},
C_{\alpha,\beta,\nu,r,j}>0$, since $(n/\varrho_k)^h\leq(n/\varrho_k)^{r-j}$
for $0\leq h\leq r-j$.

Substituting in \eqref{C2} and integrating with respect to $\theta$ we
thus have that
\beqs\label{AB6}
\left |\sigma\left((1-\chi_{1,k}(D_x))W_{n,k}^{\alpha+j,\beta}
D_x^{r-j}\right)\right |_{2,2}^0\leq|\Osc_\nu|_{2,2}^0
\leq C_{\alpha,\beta,\nu,r,j}\varrho_k^{4+\frac12}
\left(\frac{\varrho_k^{\mu+1}}{n}\right)^\nu n^{r-j}.
\eeqs

Substituting in \eqref{C3}, and taking into account \eqref{C'T}, we have that
\beqsn
\|D_x^rv_k^{\alpha,\beta}(t,\cdot)\|\leq&&
c_1n^r\|v_k^{\alpha,\beta}\|+\sum_{j=0}^r
C'_{\alpha,\beta,\nu,r,j}\varrho_k^{4+\frac12+j}
\left(\frac{\varrho_k^{\mu+1}}{n}\right)^\nu
n^{r-j+q}\\
\leq&&c_1n^r\|v_k^{\alpha,\beta}\|+
C_{\alpha,\beta,\nu,r}\varrho_k^{4+\frac12}
\left(\frac{\varrho_k^{\mu+1}}{n}\right)^\nu
n^{r+q}
\eeqsn
for some $C'_{\alpha,\beta,\nu,r,j},C_{\alpha,\beta,\nu,r}>0$,
since $\left(\frac{\varrho_k}{n}\right)^j\leq1$ for every $j$.
\end{proof}

The following two lemmas give estimates of some pseudo-differential operators
acting on the  functions $u_k$.

\begin{Lemma}
\label{lemmaAB2}
Let $n=\varrho_k^a$ with $a\geq\mu$.
Then for every $\sigma,\gamma,\lambda\in\N_0$
the operators
$W_{n,k}^{\sigma,\gamma}(t,x,D_x)$ satisfy
\beqs\label{ind}
W_{n,k}^{\sigma,\gamma}D_x^\lambda=\ds\sum_{j=0}^\lambda c_j
\varrho_k^j D_x^{\lambda-j}W_{n,k}^{\sigma+j,\gamma},
\eeqs
for some $c_0,\ldots,c_\lambda>0$.
Moreover, there are constants $C_\lambda>0$ and, for all $\nu\in\N_0$,
$C_{\sigma,\gamma,\lambda,\nu}>0$ such that for all $t\in[0,\varrho_k/n^{p-1}]$
with $k$ large enough:
\beqs
\label{2.28}
\ \ \|W_{n,k}^{\sigma,\gamma}(t,\cdot,D_x)D_x^\lambda u_k(t,\cdot)\|\leq
C_\lambda \sum_{j=0}^\lambda \varrho_k^jn^{\lambda-j}
\|v_k^{\sigma+j,\gamma}\|+
C_{\sigma,\gamma,\lambda,\nu}\varrho_k^{4+\frac12}\left(\frac{\varrho_k^{\mu+1}}{n}
\right)^\nu n^{\lambda+q}.
\eeqs
\end{Lemma}

\begin{proof}
Let us first prove \eqref{ind} by induction on $\lambda\in\N$.

For $\lambda=1$ we clearly have
$
W_{n,k}^{\sigma,\gamma}D_x=D_x W_{n,k}^{\sigma,\gamma}-\varrho_kW_{n,k}^{\sigma+1,\gamma}.
$

Let us assume \eqref{ind} to be true for every $\lambda'<\lambda$ and let us
prove it for $\lambda$. By Theorem \ref{thA}:
\beqsn
W_{n,k}^{\sigma,\gamma}D_x^\lambda=&&D_x^\lambda W_{n,k}^{\sigma,\gamma}
+[W_{n,k}^{\sigma,\gamma},
D_x^\lambda]\\
=&&D_x^\lambda W_{n,k}^{\sigma,\gamma}-\op\left(\sum_{\alpha=1}^\lambda
\frac{1}{\alpha!}\partial_\xi^\alpha\xi^\lambda\cdot
D_x^\alpha w_{n,k}^{\sigma,\gamma}\right)\\
=&&D_x^\lambda W_{n,k}^{\sigma,\gamma}-
\sum_{\alpha=1}^\lambda
\binom\lambda\alpha\varrho_k^\alpha\left(W_{n,k}^{\sigma+\alpha,\gamma}
D_x^{\lambda-\alpha}\right).
\eeqsn
By the inductive assumption, we thus have that
\beqsn
W_{n,k}^{\sigma,\gamma}D_x^\lambda
=&&D_x^\lambda W_{n,k}^{\sigma,\gamma}-\sum_{\alpha=1}^\lambda
\binom\lambda\alpha\varrho_k^\alpha\left(
\sum_{\ell=0}^{\lambda-\alpha}C_\ell\varrho_k^\ell D_x^{\lambda-\alpha-\ell}
W_{n,k}^{\sigma+\alpha+\ell,\gamma}\right)\\
=&&D_x^\lambda W_{n,k}^{\sigma,\gamma}-\sum_{\alpha=1}^\lambda\sum_{\ell=0}^{\lambda-\alpha}
C_{\alpha,\lambda,\ell}\varrho_k^{\alpha+\ell}D_x^{\lambda-\alpha-\ell}
W_{n,k}^{\sigma+\alpha+\ell,\gamma}\\
=&&\sum_{\alpha'=0}^\lambda C_{\alpha',\lambda}\varrho_k^{\alpha'}D_x^{\lambda-\alpha'}
W_{n,k}^{\sigma+\alpha',\gamma}.
\eeqsn
Therefore \eqref{ind} is proved and, applying
Lemma \ref{lemmaAB3} for $j\leq\lambda-1$, we have
that for every $\nu\in\N$:
\beqsn
\|W_{n,k}^{\sigma,\gamma}(t,\cdot,D_x)D_x^\lambda u_k(t,\cdot)\|
\leq&&\ds\sum_{j=0}^\lambda c_j\varrho_k^j \|D_x^{\lambda-j}v_k^{\sigma+j,\gamma}\|\\
\leq&&\sum_{j=0}^{\lambda-1}c_j\varrho_k^j\left(c_1n^{\lambda-j}
\|v_k^{\sigma+j,\gamma}\|+
C_{\sigma,j,\gamma,\lambda,\nu}\varrho_k^{4+\frac12}\left(\frac{\varrho_k^{\mu+1}}{n}
\right)^\nu n^{\lambda-j+q}\right)\\
&&+c_\lambda\varrho_k^\lambda\|v_k^{\sigma+\lambda,\gamma}\|\\
\leq&&C'_\lambda \sum_{j=0}^\lambda \varrho_k^j n^{\lambda-j}
\|v_k^{\sigma+j,\gamma}\|
+C_{\sigma,\gamma,\lambda,\nu}\varrho_k^{4+\frac12}\left(\frac{\varrho_k^{\mu+1}}{n}
\right)^\nu n^{\lambda+q},
\eeqsn
for some $C'_\lambda,C_{\sigma,\gamma,\lambda,\nu}>0$.
This proves \eqref{2.28}.
\end{proof}

\begin{Lemma}
\label{lemmaAB4}
Let $a_j=a_j(t,x)$, for $0\leq j\leq p-1$, be the coefficients of the
operator \eqref{op}, and let $n=\varrho_k^a$ with
$a\geq\mu+1\geq2$.
Then, for every $\nu\in\N$ there exists $C_\nu>0$ such that
\beqsn
\|[a_j,W_{n,k}]D_x^ju_k(t,\cdot)\|\leq&&
C_\nu\,n^j\sum_{1\leq\alpha_1+\alpha_2\leq (\nu-1)(p-1)+j}
\left(\frac{\varrho_k^\mu}{n}
\right)^{\alpha_1+\alpha_2}
\|v_k^{\alpha_1,\alpha_2}\|
+C_\nu\varrho_k^{4+\frac12}\left(\frac{\varrho_k^{\mu+1}}{n}\right)^\nu n^{q+j}
\eeqsn
for all $t\in[0,\varrho_k/n^{p-1}]$ with $k$ large enough.
\end{Lemma}

\begin{proof}
By Theorem \ref{thA}, for all $\nu\in\N$
\beqs
\nonumber
&&\sigma\left([a_j(t,x),W_{n,k}(t,x,D_x)]D_x^j\right)=
\sigma([a_j,W_{n,k}])\cdot\xi^j\\
\label{A5}
=&&-\left(\sum_{1\leq\alpha\leq \nu-1}\frac{1}{\alpha!}\partial_\xi^\alpha
w_{n,k}\cdot D_x^\alpha a_j\right)\xi^j
-\int_0^1\frac{(1-\theta)^{\nu-1}}{(\nu-1)!}
\tilde{\Osc}_\nu(t,\theta,x,\xi)
 d\theta,
\eeqs
where
\beqsn
\tilde{\Osc}_\nu(t,\theta,x,\xi):=
\int\!\!\int
e^{-iy\eta}\partial_\xi^\nu w_{n,k}(t,x;\xi+\theta\eta)D_x^\nu
a_j(t,x+y)\cdot\xi^jdy\dbar\eta.
\eeqsn

Arguing as in the proof of Lemma \ref{lemmaAB3} we can estimate,
by Theorem \ref{thB} and Lemma \ref{lemmaAB1}:
\beqs
\nonumber
|\tilde{\Osc}_\nu|_{2,2}^{0}=&&\left|
\sum_{h=0}^j\binom jh \theta^{j-h}\int\!\!\int
\partial_\xi^\nu w_{n,k}(t,x;\xi+\theta\eta)\cdot(\xi+\theta\eta)^h
D_x^\nu a_j(t,x+y)D_y^{j-h}e^{-iy\eta}dy\dbar\eta\right|_{2,2}^{0}\\
\nonumber
\leq&&\sum_{h=0}^j\binom jh\left|\int\!\!\int e^{-iy\eta}
\partial_\xi^\nu w_{n,k}(t,x;\xi+\theta\eta)\cdot(\xi+\theta\eta)^h
D_x^{\nu+j-h}a_j(t,x+y)dy\dbar\eta\right|_{2,2}^{0}\\
\nonumber
\leq&&\sum_{h=0}^jC_j\left|\xi^h\partial_\xi^\nu w_{n,k}(t,x;\xi)\right|_{4,4}^{0}
\cdot\left|D_x^{\nu+j-h}a_j(t,x)\right|_{4,4}^{0}\\
\label{A4}
\leq&&C_{\nu}n^j
\varrho_k^{4+\frac12}\left(\frac{\varrho_k^\mu}{n}\right)^\nu
\eeqs
for some $C_j,C_\nu>0$, since $a_j\in C([0,T];\B^\infty)$ for
$0\leq j\leq p-1$.

In order to estimate now the first term of \eqref{A5},
we previously compute, by the
Fa\`a di Bruno formula:
\beqsn
\partial_\xi^\alpha
w_{n,k}
=&&\sum_{\alpha_1+\alpha_2=\alpha}\frac{\alpha!}{\alpha_1!\alpha_2!}
\varrho_k^{1/2}
\cdot\partial_\xi^{\alpha_1}h\Big(\varrho_k(x-x_k-pA_p(t)\xi^{p-1})\Big)
\cdot\partial_\xi^{\alpha_2}h
\Big(\varrho_k^\mu\Big(\frac\xi n-1\Big)\Big)\\
&&=\varrho_k^{1/2}
h\Big(\varrho_k(x-x_k-pA_p(t)\xi^{p-1})\Big)
\cdot\partial_\xi^{\alpha}h
\Big(\varrho_k^\mu\Big(\frac\xi n-1\Big)\Big)\\
&&+\sum_{\afrac{\alpha_1+\alpha_2=\alpha}{\alpha_1\geq1}}
\frac{\alpha!}{\alpha_1!\alpha_2!}
\sum_{\afrac{r_1+\ldots+r_s=\alpha_1}{r_h\geq1}}C_{s,r}\varrho_k^{1/2}
h^{(s)}\Big(\varrho_k(x-x_k-pA_p(t)\xi^{p-1})\Big)\\
&&\cdot\partial_\xi^{r_1}\Big[\varrho_k(x-x_k-pA_p(t)\xi^{p-1})\Big]
\cdots\partial_\xi^{r_s}\Big[\varrho_k(x-x_k-pA_p(t)\xi^{p-1})\Big]\\
&&\cdot\Big(\frac{\varrho_k^\mu}{n}\Big)^{\alpha_2}
h^{(\alpha_2)}\Big(\varrho_k\Big(\frac\xi n-1\Big)\Big)\\
=&&\left(\frac{\varrho_k^\mu}{n}\right)^\alpha\varrho_k^{1/2}
h\Big(\varrho_k(x-x_k-pA_p(t)\xi^{p-1})\Big)
h^{(\alpha)}
\Big(\varrho_k^\mu\Big(\frac\xi n-1\Big)\Big)\\
&&+\sum_{\afrac{\alpha_1+\alpha_2=\alpha}{\alpha_1\geq1}}
\frac{\alpha!}{\alpha_1!\alpha_2!}
\sum_{\afrac{r_1+\ldots+r_s=\alpha_1}{1\leq r_h\leq p-1}}C'_{s,r}
\Big(\varrho_kA_p(t)\Big)^{\alpha_1}\cdot\xi^{s(p-1)-\alpha_1}\\
&&\cdot
\Big(\frac{\varrho_k^\mu}{n}\Big)^{\alpha_2}
\varrho_k^{1/2}
h^{(s)}\Big(\varrho_k(x-x_k-pA_p(t)\xi^{p-1})\Big)
h^{(\alpha_2)}\Big(\varrho_k\Big(\frac\xi n-1\Big)\Big)
\eeqsn
for some $C_{s,r},C'_{s,r}>0$.
Coming back to the first term of \eqref{A5} and taking into account
the definition \eqref{defwnk} of $w_{n,k}$:
\beqsn
&&\left(\sum_{1\leq\alpha\leq \nu-1}\frac{1}{\alpha!}\partial_\xi^\alpha
w_{n,k}\cdot D_x^\alpha a_j\right)\xi^j
\leq\sum_{1\leq\alpha\leq \nu-1}
\frac{D_x^\alpha a_j}{\alpha!}\left(\frac{\varrho_k^\mu}{n}\right)^{\alpha}
w_{n,k}^{0,\alpha}\cdot\xi^j\\
&&+\sum_{1\leq\alpha\leq \nu-1}\sum_{\afrac{\alpha_1+\alpha_2=\alpha}{\alpha_1\geq1}}
\frac{D_x^\alpha a_j}{\alpha_1!\alpha_2!}
\sum_{\afrac{r_1+\ldots+r_s=\alpha_1}{1\leq r_h\leq p-1}}C'_{s,r}
\varrho_k^{\alpha_1}A_p(t)^{\alpha_1}
\left(\frac{\varrho_k^\mu}{n}\right)^{\alpha_2}
w_{n,k}^{s,\alpha_2}\cdot\xi^{s(p-1)-\alpha_1+j}
\eeqsn
and hence
\beqsn
&&\Big\|\op\Big[\Big(\sum_{1\leq\alpha\leq \nu-1}\frac{1}{\alpha!}
\partial_\xi^\alpha
w_{n,k}\cdot D_x^\alpha a_j\Big)\xi^j\Big]u_k\Big\|\\
\nonumber
\leq&&\sum_{1\leq\alpha\leq \nu-1}C_{\alpha,j}
\left(\frac{\varrho_k^\mu}{n}\right)^{\alpha}
\left\|W_{n,k}^{0,\alpha}D_x^ju_k\right\|\\
\nonumber
&&+\sum_{1\leq\alpha\leq \nu-1}\sum_{\afrac{\alpha_1+\alpha_2=\alpha}{\alpha_1\geq1}}
C_{\alpha,j}
\sup_{[0,\varrho_k/n^{p-1}]}|A_p(t)|^{\alpha_1}
\varrho_k^{\alpha_1}\left(\frac{\varrho_k^\mu}{n}\right)^{\alpha_2}
\sum_{s=1}^{\alpha_1}\left\|W_{n,k}^{s,\alpha_2}D_x^{s(p-1)-\alpha_1+j}u_k\right\|.
\eeqsn

Applying \eqref{2.28}, since $s\leq\alpha_1$, $\mu\geq2$ and $0\leq j\leq p-1$,
we thus obtain, for $t\in[0,\varrho_k/n^{p-1}]$:
\beqs
\nonumber
\hspace*{12mm}
&&\Big\|\op\Big[\Big(\sum_{1\leq\alpha\leq \nu-1}\frac{1}{\alpha!}
\partial_\xi^\alpha
w_{n,k}\cdot D_x^\alpha a_j\Big)\xi^j\Big]u_k\Big\|\\
\nonumber
\leq&&
\sum_{1\leq\alpha\leq \nu-1}C'_{\alpha,j}
\left(\frac{\varrho_k^\mu}{n}\right)^{\alpha}
\bigg[\sum_{h=0}^j\varrho_k^h n^{j-h}
\|v_k^{h,\alpha}\|+C_{\alpha,j,\nu}\varrho_k^{4+\frac12}
\left(\frac{\varrho_k^{\mu+1}}{n}\right)^\nu n^{j+q}\bigg]\\
\nonumber
&&+\sum_{1\leq\alpha\leq \nu-1}\sum_{\afrac{\alpha_1+\alpha_2=\alpha}{\alpha_1\geq1}}
C'_{\alpha,j}\left(\frac{\varrho_k}{n^{p-1}}\right)^{\alpha_1}
\varrho_k^{\alpha_1}\left(\frac{\varrho_k^\mu}{n}\right)^{\alpha_2}\\
\nonumber
&&\cdot
\sum_{s=1}^{\alpha_1}\bigg[C_{s,\alpha,j}
\sum_{h=0}^{s(p-1)-\alpha_1+j}\varrho_k^h n^{s(p-1)-\alpha_1+j-h}
\|v_k^{s+h,\alpha_2}\|\\
\nonumber
&&
+C_{s,\alpha,j,\nu}\varrho_k^{4+\frac12}
\left(\frac{\varrho_k^{\mu+1}}{n}\right)^\nu
n^{s(p-1)-\alpha_1+j+q}\bigg]\\
\nonumber
\leq&&C_\nu n^j\sum_{1\leq\alpha\leq \nu-1}
\left(\frac{\varrho_k^\mu}{n}\right)^{\alpha}
\sum_{h=0}^j\left(\frac{\varrho_k}{n}\right)^h
\|v_k^{h,\alpha}\|
+C_\nu\varrho_k^{4+\frac12}
\left(\frac{\varrho_k^{\mu+1}}{n}\right)^\nu n^{j+q}\\
\nonumber
&&+C_\nu\sum_{1\leq\alpha\leq \nu-1}
\sum_{\afrac{\alpha_1+\alpha_2=\alpha}{\alpha_1\geq1}}
\left(\frac{\varrho_k^\mu}{n}\right)^{\alpha_1+\alpha_2}\frac{1}{n^{\alpha_1(p-2)}}
\\
\nonumber
&&\cdot\bigg[n^{\alpha_1(p-2)+j}\sum_{s=1}^{\alpha_1}
\sum_{h=0}^{s(p-1)-\alpha_1+j}\left(\frac{\varrho_k}{n}\right)^h
\|v_k^{s+h,\alpha_2}\|
+\varrho_k^{4+\frac12}
\left(\frac{\varrho_k^{\mu+1}}{n}\right)^\nu n^{\alpha_1(p-2)+j+q}\bigg]\\
\nonumber
\leq&&C_\nu n^j\sum_{1\leq\alpha_2\leq \nu-1}\sum_{h=0}^j
\left(\frac{\varrho_k^\mu}{n}\right)^{h+\alpha_2}\|v_k^{h,\alpha_2}\|
+C'_{\nu}\varrho_k^{4+\frac12}
\left(\frac{\varrho_k^{\mu+1}}{n}\right)^\nu n^{q+j}\\
\nonumber
&&+C_\nu n^j\sum_{\afrac{1\leq \alpha_1+\alpha_2\leq \nu-1}{\alpha_1\geq1}}
\sum_{s=1}^{\alpha_1}\sum_{h=0}^{s(p-1)-\alpha_1+j}
\left(\frac{\varrho_k^\mu}{n}\right)^{s+h+\alpha_2}
\|v_k^{s+h,\alpha_2}\|\\
\label{A6}
\leq&&C_\nu n^j\sum_{1\leq \alpha_1+\alpha_2\leq (\nu-1)(p-1)+j}
\left(\frac{\varrho_k^\mu}{n}\right)^{\alpha_1+\alpha_2}
\|v_k^{\alpha_1,\alpha_2}\|
+C'_{\nu}\varrho_k^{4+\frac12}
\left(\frac{\varrho_k^{\mu+1}}{n}\right)^\nu n^{q+j}
\eeqs
for some $C_{\alpha,j},C_{\alpha,j,\nu},C'_{\alpha,j},C'_{\alpha,j,\nu},
C_{s,\alpha,j},C_{s,\alpha,j,\nu},C_\nu,C'_{\nu}>0$.

By the Calder\'on-Vaillancourt's Theorem \ref{thC}, by
\eqref{A5}, \eqref{A4} and \eqref{A6} we get:
\beqsn
&&\|[a_j,W_{n,k}]D_x^ju_k\|\leq C|\tilde{\Osc}_\nu|_{2,2}^{0}\cdot\|u_k\|
+\Big\|\op\Big[\Big(\sum_{1\leq\alpha\leq \nu-1}
\frac{1}{\alpha!}\partial_\xi^\alpha w_{n,k}\cdot D_x^\alpha a_j\Big)
\xi^j\Big]u_k\Big\|\\
\leq&&C'_{\nu}\varrho_k^{4+\frac12}
\left(\frac{\varrho_k^{\mu+1}}{n}\right)^\nu n^{q+j}
+C_{\nu}n^j\sum_{1\leq \alpha_1+\alpha_2\leq (\nu-1)(p-1)+j}
\left(\frac{\varrho_k^\mu}{n}\right)^{\alpha_1+\alpha_2}
\|v_k^{\alpha_1,\alpha_2}\|\\
\eeqsn
for some $C,C_\nu,C'_\nu>0$.
\end{proof}

\section{Estimates from below}
\label{sec3}

In this section we want to produce estimates from below of the $L^2$-norms
of the functions $v_k$ and $v_k^{\alpha,\beta}$, and then of a
linear combination $\sigma_k(t)$ of the $L^2$-norms of $v_k^{\alpha,\beta}$,
$\alpha+\beta\geq0.$

We start with the estimate of $\|v_k(0,\cdot)\|$.
For $n$ as in \eqref{n} and $k$ large enough, from \eqref{fgk} we have that
\beqs
\label{hint}
\supp \hat g_k=\supp \hat\psi(\xi-n)
\subseteq\{\xi\in\R:\
h(\varrho_k^\mu(\xi/ n-1))=1\}.
\eeqs
Therefore
\beqsn
v_k(0,x)&=&W_{n,k}u_k(0,x)=\int e^{ix\xi}w_{n,k}(0,x,\xi)\widehat {g_k}(\xi)
\dbar\xi
\\
\nonumber
&=&\int e^{ix\xi} \varrho_k^{1/2}h\left(\varrho_k(x-x_k)\right)
\underbrace{h\left(\varrho_k^\mu(\xi/n-1)\right)}_1e^{-ix_k\xi}
\hat\psi(\xi-n)\dbar\xi
\\
\nonumber
&=& \varrho_k^{1/2}h\left(\varrho_k(x-x_k)\right)e^{i(x-x_k)n}\psi(x-x_k)
\eeqsn
and
\beqs\label{vk0}
\|v_k(0,\cdot)\|^2&=&\int \varrho_k|h\left(\varrho_k(x-x_k)\right)|^2
|\psi(x-x_k)|^2dx=\int |h(y)|^2|\psi(y/\varrho_k)|^2dy
\\
\nonumber
&\geq &\int |h(y)|^2dy=\|h\|^2>0
\eeqs
if $k$ is large enough, since $\psi(0)=2$ and $\varrho_k\to+\infty$.

Now, to produce an estimate from below of $\|v_k(t,\cdot)\|$, our
idea is to follow the energy method, producing a "reverse energy estimate".
To this aim, denoting by $\langle\cdot,\cdot\rangle$
the scalar product on $L^2$,
we consider
\beqs
\nonumber
\ds\frac{d}{dt}\|v_k(t,\cdot)\|^2=&&
2\Re \langle \partial_tv_k,v_k\rangle\\
\label{encontrary}
=&& 2\Re i \langle Pv_k,v_k\rangle-2\Re i
a_p(t)\langle D_x^pv_k,v_k\rangle
-2\Re i \sum_{j=0}^{p-1}\langle (a_j(t,x)D_x^jv_k,v_k\rangle.
\eeqs
We compute separately estimates from below of each term in formula
\eqref{encontrary}.
By definition of $v_k$ we have that
\beqsn
Pv_k&=&PW_{n,k}u_k=W_{n,k}Pu_k+[P, W_{n,k}]u_k=
\\
&=&0+[D_t+a_p(t)D_x^p, W_{n,k}]u_k+\sum_{j=0}^{p-1}[a_j(t,x)D_x^j, W_{n,k}]u_k,
\eeqsn
since $Pu_k=0$.

Developing the symbol of the commutator $[D_t+a_p(t)D_x^p, W_{n,k}]$ and using
the fact that $w_{n,k}$ is the solution of Hamilton's
equation \eqref{215} we obtain, by Theorem \ref{thA}:
\beqsn
\sigma\left(\left[D_t+a_p(t)D_x^p, W_{n,k}\right]\right)(t,x,\xi)=&&
D_tw_{n,k}+a_p(t)\sigma\left([D_x^p, W_{n,k}]\right)\\
=&&D_tw_{n,k}+a_p(t)\sum_{\alpha=1}^p\frac{1}{\alpha!}\partial_\xi^\alpha
\xi^p\cdot D_x^\alpha w_{n,k}\\
=&&(D_t+pa_p(t)\xi^{p-1}D_x)w_{n,k}+a_p(t)\sum_{\alpha=2}^p\binom p\alpha
\xi^{p-\alpha}D_x^\alpha w_{n,k}\\
=&&a_p(t)\sum_{\alpha=2}^p\binom p\alpha
\xi^{p-\alpha}D_x^\alpha w_{n,k}.
\eeqsn
Defining then
\beqs
\label{fk}
f_k:=\op\left(a_p(t)\sum_{\alpha=2}^p\binom p\alpha
\xi^{p-\alpha}D_x^\alpha w_{n,k}\right)u_k
+\sum_{j=0}^{p-1}[a_j(t,x)D_x^j,W_{n,k}]u_k,
\eeqs
we have that
\beqsn
Pv_k=f_k
\eeqsn
and hence from \eqref{encontrary} we get
\beqs
\nonumber
\ds\frac{d}{dt}\|v_k(t,\cdot)\|^2=&&2\Re i\langle f_k,v_k\rangle
-2\Re ia_p(t)\langle D_x^pv_k,v_k\rangle-\sum_{j=0}^{p-1}2\Re i
\langle a_jD_x^jv_k,v_k\rangle\\
\label{EB3}
=&&2\Re i\langle f_k,v_k\rangle
-\sum_{j=0}^{p-1}\langle (ia_jD_x^j+(ia_jD_x^j)^*)v_k,v_k\rangle
\eeqs
since $\Re i\langle D_x^p v_k,v_k\rangle=0$.
Now,
\beqsn
\sigma(ia_j(t,x)D_x^j)^*=\sum_{\alpha\geq0}\frac{1}{\alpha!}\partial_\xi^\alpha
D_x^\alpha(\overline{ia_j(t,x)\xi^j})
=\sum_{\alpha=0}^j\binom j\alpha D_x^\alpha(
-i\Re a_j-\Im a_j(t,x))\xi^{j-\alpha},
\eeqsn
and hence
\beqsn
\sum_{j=0}^{p-1}\sigma[(ia_jD_x^j)+(ia_jD_x^j)^*]=&&
\sum_{j=0}^{p-1}\left[-2\Im a_j\xi^j
+\sum_{\alpha=1}^j\binom j\alpha D_x^\alpha\left(-i\Re a_j-
 \Im a_j\right)\xi^{j-\alpha}\right]\\
=&&-2\sum_{j=0}^{p-1}\Im a_j\xi^j+
\sum_{h=0}^{p-2}\sum_{j=h+1}^{p-1}\binom jh D_x^{j-h}\left(-i\Re a_j-\Im a_j
\right)\xi^h\\
=&&-2\Im a_{p-1}\xi^{p-1}\\
&&+\sum_{h=0}^{p-2}\left[-2\Im a_h
+\sum_{j=h+1}^{p-1}\binom jh D_x^{j-h}\left(-i\Re a_j-\Im a_j\right)\right]\xi^h.
\eeqsn

Substituting in \eqref{EB3},
we have that there exist
postive constants $A_1,c'$ such that
\beqs
\nonumber
\ds\frac{d}{dt}\|v_k(t,\cdot)\|^2\geq &&-2\|f_k\|\cdot\|v_k\|
+2\langle\Im a_{p-1} D_x^{p-1}v_k,v_k\rangle-A_1\|v_k\|^2\\
\nonumber
&&+\sum_{h=1}^{p-2}
\left[2\langle\Im a_hD_x^h v_k,v_k\rangle
+\sum_{j=h+1}^{p-1}
\binom jh
\langle(D_x^{j-h}(i\Re a_j+\Im a_j))D_x^hv_k,v_k\rangle\right]\\
\label{encontrary2}
\geq&&2\langle\Im a_{p-1} D_x^{p-1}v_k,v_k\rangle
-2\|f_k\|\cdot\|v_k\|-A_1\|v_k\|^2-
c'\frac{\,n^{p-1}}{\varrho_k}
\|v_k\|^2,
\eeqs
since
\beqsn
|\langle\Im a_hD_x^hv_k,v_k\rangle|
\leq c n^h\|v_k\|^2
\leq c n^{p-2}\|v_k\|^2
\leq c\frac{\,n^{p-1}}{\varrho_k}\|v_k\|^2
\eeqsn
because of the support of $w_{n,k}$,
and analogously
\beqsn
|\langle(D_x^{j-h}(i\Re a_j+\Im a_j))D_x^hv_k,v_k\rangle|
\leq c\frac{\,n^{p-1}}{\varrho_k}\|v_k\|^2.
\eeqsn

 Now we want to give
 estimates of the terms in \eqref{encontrary2}. This is done in the
following Propositions \ref{ima2} and \ref{lemma24}.

\begin{Prop}
\label{ima2}
Let $n=\varrho_k^a$ with $a\geq\mu\geq 2$.
Then, for all
$\nu\in\N$ there exists $C_\nu>0$ such that, for every
$t\in \left[0,\ds\frac{\varrho_k}{n^{p-1}}\right]$ with
$k$ large enough:
\beqs
\label{AB8}
\langle \Im a_{p-1}(t,x)D_x^{p-1}v_k,v_k\rangle&\geq&
\left( \Im a_{p-1}(t,x_k+pA_p(t)n^{p-1})n^{p-1}-
C\frac{\,n^{p-1}}{\varrho_k}\right)\|v_k\|^2
\\
\nonumber
&&-C_\nu{\varrho_k}^{4+\frac12}\left(\frac{{\varrho_k}^{\mu+1}}{n}\right)^\nu
n^{q+p-1}\|v_k\|,
\eeqs
for some fixed $C>0.$
\end{Prop}
\begin{proof}
We split
\beqs
\nonumber
\Im a_{p-1}(t,x)D_x^{p-1}&=&\Im a_{p-1}(t,x_k +pA_p(t)n^{p-1})n^{p-1}\\
\label{split}
&&+\Im a_{p-1}(t,x_k +pA_p(t)n^{p-1})(D_x^{p-1}-n^{p-1})\\
\nonumber
&&+\left(\Im a_{p-1}(t,x)-\Im a_{p-1}(t,x_k +pA_p(t)n^{p-1})\right)D_x^{p-1}
\eeqs
and set
\beqsn
&I_1:=\Im a_{p-1}(t,x_k +pA_p(t)n^{p-1})n^{p-1},&
\\
&I_2:=\Im a_{p-1}(t,x_k +pA_p(t)n^{p-1})(D_x^{p-1}-n^{p-1})&
\\
&I_3:= \left(\Im a_{p-1}(t,x)-\Im a_{p-1}(t,x_k +pA_p(t)n^{p-1})\right)D_x^{p-1}.&
\eeqsn
We have
\beqs\label{I1vk}
\langle I_1v_k,v_k\rangle=\Im a_{p-1}(t,x_k +pA_p(t)n^{p-1})n^{p-1}\|v_k\|^2.
\eeqs
To estimate $\langle I_2v_k,v_k\rangle$, we localize at frequency
$n$ by means of the
function $\chi_{1,k}$ defined in \eqref{chi1k}
and write
\beqsn
I_2v_k&=&\chi_{1,k}(D_x)I_2v_k+(1-\chi_{1,k}(D_x))I_2v_k
\\
&=&\Im a_{p-1}(t,x_k +pA_p(t)n^{p-1})[\chi_{1,k}(D_x)(D_x^{p-1}-n^{p-1})v_k\\
&&+\left(1-\chi_{1,k}(D_x)\right)(D_x^{p-1}-n^{p-1})v_k],
\eeqsn
so, denoting by
\beqs
&J_1:=\|\chi_{1,k}(D_x)(D_x^{p-1}-n^{p-1})v_k\|,&
\\
\label{j2}
&J_2:=\|\left(1-\chi_{1,k}(D_x)\right)(D_x^{p-1}-n^{p-1})v_k\|,&
\eeqs
we have
\beqs\label{normaI2vk}
\|I_2v_k\|\leq |\Im a_{p-1}(t,x_k +pA_p(t)n^{p-1})|(J_1+J_2).
\eeqs
By Calderon-Vaillantcourt's Theorem \ref{thC},
\beqs
\label{normaj1}
J_1\leq C\vert\chi_{1,k}(\xi)(\xi^{p-1}-n^{p-1})\vert_{2,2}^{0}\|v_k\|
\leq C'\frac{n^{p-1}}{\varrho_k^\mu}\|v_k\|
\eeqs
for some $C,C'>0$, since by \eqref{suppchi1k}:
\beqsn
|\chi_{1,k}(\xi)(\xi^{p-1}-n^{p-1})|=&&
|\chi_{1,k}(\xi)(\xi-n)(\xi^{p-2}+n\xi^{p-3}+n^2\xi^{p-4}+\ldots+
n^{p-2})|\\
\leq&& c\frac{n}{\varrho_k^\mu}(p-1)n^{p-2}=c'\frac{n^{p-1}}{\varrho_k^\mu},
\eeqsn
for some $c,c'>0$, and for all $\gamma=\gamma_1+\gamma_2$
with $|\gamma|\leq 2$ there are constants $C_{\gamma_1}$, $C_\gamma>0$ such that:
\beqsn
|\partial_\xi^{\gamma_1}\chi_{1,k}(\xi)
\partial_\xi^{\gamma_2}(\xi^{p-1}-n^{p-1})|\leq
\begin{cases}
 C_{\gamma_1} \ds\frac{n^{p-1}}{\varrho_k^\mu}& \gamma_2=0
\\
C_{\gamma}n^{p-1-\gamma_2}\leq  C_\gamma\ds\frac{n^{p-1}}{\varrho_k^\mu}&
\gamma_2\geq1.
\end{cases}
\eeqsn
As it concerns \eqref{j2}, by definition of $v_k$ we write
\beqs
\label{j22}
(D_x^{p-1}-n^{p-1})v_k=
\left(W_{n,k}(D_x^{p-1}-n^{p-1})+[D_x^{p-1}-n^{p-1},W_{n,k}]\right)u_k.
\eeqs
Since
$
\sigma([D_x^{p-1}-n^{p-1},W_{n,k}])=
\sum_{\alpha=1}^{p-1}\binom{p-1}{\alpha}\xi^{p-1-\alpha}
\varrho_k^\alpha w_{n,k}^{\alpha,0},
$
we have that
\beqs
\label{j222}
[D_x^{p-1}-n^{p-1},W_{n,k}]=
\sum_{\alpha=1}^{p-1}\binom{p-1}{\alpha}
\varrho_k^\alpha W_{n,k}^{\alpha,0}D_x^{p-1-\alpha}
\eeqs
and therefore, by \eqref{j2}, \eqref{j22}, \eqref{j222}, the
Calderon-Vaillantcourt's Theorem \ref{thC} and \eqref{AB6}, for every
$\nu\in\N$ there are constants $C,C'_\nu,C''_\nu>0$ such that:
\beqs
\label{normaj2}
\nonumber
J_2&\leq &\|\left(1-\chi_{1,k}(D_x)\right)W_{n,k}(D_x^{p-1}-n^{p-1})u_k\|\\
\nonumber
&&+\sum_{\alpha=1}^{p-1}\binom{p-1}{\alpha}\varrho_k^\alpha
\|(1-\chi_{1,k}(D_x))W_{n,k}^{\alpha,0}D_x^{p-1-\alpha}u_k\|
\\
\nonumber
&\leq & C\bigg(\left |\sigma\left(\left(1-\chi_{1,k}(D_x)\right)
W_{n,k}(D_x^{p-1}-n^{p-1})\right)\right |_{2,2}^{0}\\
\nonumber
&&+\sum_{\alpha=1}^{p-1}\varrho_k^\alpha
\left |\sigma\left(\left(1-\chi_{1,k}(D_x)\right)W_{n,k}^{\alpha,0}D_x^{p-1-\alpha}
\right)
\right |_{2,2}^{0}\bigg)\|u_k\|
\\
\nonumber
&\leq& C'_\nu\varrho_k^{4+\frac12}\left(\frac{\varrho_k^{\mu+1}}{n}\right)^\nu
(n^{p-1}+\varrho_k n^{p-2}+\ldots+\varrho_k^{p-1})n^q
\\
&\leq& C''_\nu\varrho_k^{4+\frac12}\left(\frac{\varrho_k^{\mu+1}}{n}\right)^\nu
n^{q+p-1}.
\eeqs
Substituting now \eqref{normaj1} and \eqref{normaj2}
in \eqref{normaI2vk} we come to
\beqs\label{3.7}
\|I_2v_k\|\leq C\ds\frac{n^{p-1}}{\varrho_k^\mu}\|v_k\|+
C_\nu\varrho_k^{4+\frac12}\left(\frac{\varrho_k^{\mu+1}}{n}\right)^\nu n^{q+p-1}
\eeqs
for some $C,C_\nu>0$, and hence
\beqs\label{I2vk}
\langle I_2v_k,v_k\rangle\geq -C\ds\frac{n^{p-1}}{\varrho_k^\mu}\|v_k\|^2
-C_\nu\varrho_k^{4+\frac12}\left(\frac{\varrho_k^{\mu+1}}{n}\right)^\nu n^{q+p-1}
\|v_k\|.
\eeqs
Finally, to estimate $\langle I_3v_k,v_k\rangle$, we localize in a
neighborhood of $x_k+pA_p(t)\xi^{p-1}$ by defining,
for $h$ as
in \eqref{h}, the function
\beqs\label{chi2k}
\chi_{2,k}(x):= h\left(\varrho_k\frac{x-x_k-pA_p(t)\xi^{p-1}}{4pc_p}\right),
\eeqs
where $c_p$ is the constant defined in Lemma \ref{lemma2}.
We have that
\beqs\label{suppchi2k}
\supp \chi_{2,k}\subseteq \left\{x\ :\ \vert x-x_k-pA_p(t)\xi^{p-1}\vert\leq
\frac{2pc_p}{\varrho_k}\right\}
\eeqs
and
\beqs\label{supp1-chi2k}
\supp \left(1-\chi_{2,k}\right)\subseteq \left\{x\ :\
\vert\ x-x_k-pA_p(t)\xi^{p-1}\vert\geq \frac{pc_p}{\varrho_k}\right\}.
\eeqs
We now claim that
\beqs\label{yuppi}
\supp (1-\chi_{2,k})\cap\supp W_{n,k}^{\alpha,\beta}=\emptyset
\qquad\forall t\in
\left[0,\frac{\varrho_k}{n^{p-1}}\right].
\eeqs
This holds true because on the support of
$w_{n,k}^{\alpha,\beta}$, given by Lemma \ref{lemma2}, we have that, for all
$t\in \left[0,\ds\frac{\varrho_k}{n^{p-1}}\right]$,
\beqsn
\vert x-x_k-pA_p(t)\xi^{p-1}\vert&\leq&\vert x-x_k-pA_p(t)n^{p-1}\vert
+p|A_p(t)|\vert \xi^{p-1}-n^{p-1}\vert
\\
&\leq& \frac{c_p}{\varrho_k}+p\sup_{[0,T]}|a_p|\cdot t\cdot |\xi-n|
\cdot|\xi^{p-2}+n\xi^{p-3}+\ldots+n^{p-2}|
\\
&\leq&  \frac{c_p}{\varrho_k}+c_p \frac{\varrho_k}{n^{p-1}}
\frac{n}{2\varrho_k^\mu}(p-1)n^{p-2}
\leq p\frac{c_p}{\varrho_k},
\eeqsn
by the definition of $c_p$.
Therefore \eqref{yuppi} is proved and
\beqsn
I_3v_k=(1-\chi_{2,k}(x))I_3v_k+\chi_{2,k}(x)I_3v_k
=\chi_{2,k}(x)I_3v_k.
\eeqsn

Then, by Lemma \ref{lemmaAB3}:
\beqsn
\|I_3v_k\|=&&\|\chi_{2,k}(x)I_3v_k\|=
|\Im a_{p-1}(t,x)-\Im a_{p-1}(t,x_k +pA_p(t)n^{p-1})|
\cdot\|\chi_{2,k}(x)D_x^{p-1}v_k\|
\\
\leq &&\Big(\sup_{[0,T]\times\R} |\Im \partial_xa_{p-1}(t,x)|\Big)
\cdot |x-x_k-pA_p(t)n^{p-1}|
\cdot\|\chi_{2,k}(x)D_x^{p-1}v_k\|\\
\leq&&\frac{c}{\varrho_k}\|D_x^{p-1}v_k\|
\leq\frac{c}{\varrho_k}\left(c_1n^{p-1}\|v_k\|+C_\nu\varrho_k^{4+\frac12}
\left(\frac{\varrho_k^{\mu+1}}{n}\right)^\nu n^{q+p-1}\right)
\\
\leq&& C'\frac{\,n^{p-1}}{\varrho_k}\|v_k\|+C'_\nu{\varrho_k}^{3+\frac12}
\left(\frac{{\varrho_k}^{\mu+1}}{n}\right)^\nu n^{q+p-1},
\eeqsn
for some $c,C',C'_\nu>0$,
and so
\beqs
\label{I3vk}
\langle I_3v_k,v_k\rangle\geq -C'\frac{n^{p-1}}{\varrho_k}\|v_k\|^2
-C'_\nu{\varrho_k}^{3+\frac12}\left(\frac{{\varrho_k}^{\mu+1}}{n}\right)^\nu
n^{q+p-1}\|v_k\|.
\eeqs
Summing up \eqref{I1vk}, \eqref{I2vk} and \eqref{I3vk}
we finally get the desired estimate \eqref{AB8}.
\end{proof}

\begin{Prop}
\label{lemma24}
Let $n=\varrho_k^{a}$ with $a>\mu+1$. Then for all $\nu\in\N$
there exists $C_\nu>0$ such that
the function $f_k$
defined in \eqref{fk} satisfies
\beqsn
\|f_k(t,\cdot)\|\leq&& C\varrho_k^2n^{p-2}\sum_{j=1}^p\|v_k^{j,0}\|+C_\nu
n^{p-1}\!\!
\sum_{1\leq\alpha_1+\alpha_2\leq \nu(p-1)}\!\left(\frac{\varrho_k^\mu}{n}
\right)^{\alpha_1+\alpha_2}\|v_k^{\alpha_1,\alpha_2}\|\\
&&+C_\nu\,n^{q+p-1}\varrho_k^{4+\frac12}
\left(\frac{\varrho_k^{\mu+1}}{n}\right)^{\nu}
\eeqsn
for some fixed $C>0$ and for every $t\in[0,\varrho_k/n^{p-1}]$ with
$k$ large enough.
\end{Prop}

\begin{proof}
Let us recall that
\beqs
\label{addendi}
f_k=&&\op\left(a_p(t)\sum_{\alpha=2}^p
\binom p\alpha \xi^{p-\alpha}D_x^\alpha w_{n,k}\right)u_k
+\sum_{j=0}^{p-1}[a_j(t,x)D_x^j,W_{n,k}]u_k,
\eeqs
and estimate the above terms  separately.
For $\alpha=p$
\beqsn
\op(a_p(t)D_x^pw_{n,k})u_k=&&\int e^{ix\xi}a_p(t)D_x^pw_{n,k}(t,x;\xi)
\hat{u}_k(t,\xi)\dbar\xi\\
=&&a_p(t)\varrho_k^p\int e^{ix\xi}w_{n,k}^{p,0}(t,x;\xi)\hat{u}_k(t,\xi)\dbar\xi\\
=&&a_p(t)\varrho_k^pW_{n,k}^{p,0}(t,x;D_x)u_k(t,x)
=a_p(t)\varrho_k^pv_k^{p,0}(t,x)
\eeqsn
and hence
\beqs
\label{A9}
\|\op(a_p(t)D_x^pw_{n,k})u_k\|\leq C\varrho_k^p\|v_k^{p,0}\|
\eeqs
for some $C>0$.

For $2\leq\alpha\leq p-1$, by \eqref{2.28} we have:
\beqs
\nonumber
\|\op(a_p(t)\xi^{p-\alpha} D_x^\alpha w_{n,k})u_k(t,\cdot)\|\leq
&&C\varrho_k^\alpha\|W_{n,k}^{\alpha,0}D_x^{p-\alpha}u_k\|\\
\nonumber
\leq&&C'\varrho_k^\alpha\left(n^{p-\alpha}\sum_{j=0}^{p-\alpha}\|v_k^{\alpha+j,0}\|
+C_\nu\varrho_k^{4+\frac12}
\left(\frac{\varrho_k^{\mu+1}}{n}\right)^\nu n^{q+p-\alpha}\right)\\
\label{A1}
\leq&& C''\varrho_k^2n^{p-2}\sum_{s=2}^p\|v_k^{s,0}\|
+C'_\nu\varrho_k^{4+\frac12}
\left(\frac{\varrho_k^{\mu+1}}{n}\right)^\nu n^{q+p-1}
\eeqs
for some $C,C',C'',C_\nu,C'_\nu>0$, since
$(\varrho_k/n)^\alpha\leq(\varrho_k/n)^2$ and
$\varrho_k^2/n^2\leq1/n=\varrho_k^{-a}$ for
$2\leq\alpha\leq p-1$ and $a\geq2$.

In order to estimate the second addend of \eqref{addendi} we compute,
for $0\leq j\leq p-1$:
\beqsn
[a_jD_x^j,W_{n,k}]u_k
=&&a_j\sum_{h=0}^j\binom jh(D_x^{j-h}W_{n,k})D_x^hu_k
-W_{n,k}a_jD_x^ju_k\\
=&&a_j\sum_{h=0}^{j-1}\binom jh\varrho_k^{j-h}W_{n,k}^{j-h,0}D_x^hu_k
+[a_j,W_{n,k}]D_x^ju_k.
\eeqsn
Then, by Lemmas \ref{lemmaAB2} and \ref{lemmaAB4}, for $0\leq j\leq p-1$,
we have that:
\beqs
\nonumber
\|[a_jD_x^j,W_{n,k}]u_k\|\leq&&
C\sum_{h=0}^{j-1}\varrho_k^{j-h}\|W_{n,k}^{j-h,0}D_x^hu_k\|
+\|[a_j,W_{n,k}]D_x^ju_k\|\\
\nonumber
\leq&&\sum_{h=0}^{j-1}C_h\varrho_k^{j-h}n^h\sum_{s=0}^h\|v_k^{j-h+s,0}\|
+C_\nu \varrho_k^{4+\frac12}
\left(\frac{\varrho_k^{\mu+1}}{n}\right)^\nu n^{q+j}
\\
\nonumber
&&+C_\nu n^j\sum_{1\leq\alpha_1+\alpha_2\leq(\nu-1)(p-1)+j}
\left(\frac{\varrho_k^\mu}{n}\right)^{\alpha_1+\alpha_2}
\|v_k^{\alpha_1,\alpha_2}\|
\\
\label{A3}
\leq&&C\varrho_k n^{j-1}\sum_{s=1}^j\|v_k^{s,0}\|
+C_\nu \varrho_k^{4+\frac12}
\left(\frac{\varrho_k^{\mu+1}}{n}\right)^\nu n^{q+j}\\
\nonumber
&&+C_\nu n^j
\sum_{1\leq\alpha_1+\alpha_2\leq(\nu-1)(p-1)+j}
\left(\frac{\varrho_k^\mu}{n}\right)^{\alpha_1+\alpha_2}
\|v_k^{\alpha_1,\alpha_2}\|
\eeqs
for some $C,C_\nu>0$.

By \eqref{addendi}, \eqref{A9}, \eqref{A1} and \eqref{A3}:
\beqsn
\|f_k(t,\cdot)\|\leq&&
C\varrho_k^p\|v_k^{p,0}\|+C''\varrho_k^2n^{p-2}\sum_{s=2}^p\|v_k^{s,0}\|
+C'_\nu\varrho_k^{4+\frac12}
\left(\frac{\varrho_k^{\mu+1}}{n}\right)^\nu n^{q+p-1}\\
&&+\sum_{j=0}^{p-1}\bigg[C\varrho_kn^{j-1}\sum_{s=1}^j\|v_k^{s,0}\|
+C'_\nu n^j
\sum_{1\leq\alpha_1+\alpha_2\leq(\nu-1)(p-1)+j}
\left(\frac{\varrho_k^\mu}{n}\right)^{\alpha_1+\alpha_2}
\|v_k^{\alpha_1,\alpha_2}\|\\
&&+C'_\nu\varrho_k^{4+\frac12}
\left(\frac{\varrho_k^{\mu+1}}{n}\right)^\nu n^{q+j}\bigg]\\
\leq&&\tilde{C}\varrho_k^2n^{p-2}\sum_{s=1}^p\|v_k^{s,0}\|
+\tilde{C}_\nu n^{p-1}\sum_{1\leq\alpha_1+\alpha_2\leq\nu(p-1)}
\left(\frac{\varrho_k^\mu}{n}\right)^{\alpha_1+\alpha_2}
\|v_k^{\alpha_1,\alpha_2}\|\\
&&+\tilde{C}_\nu\varrho_k^{4+\frac12}
\left(\frac{\varrho_k^{\mu+1}}{n}\right)^\nu n^{q+p-1}
\eeqsn
for some $\tilde{C},\tilde{C}_\nu>0$.
\end{proof}

Summing up, from \eqref{encontrary2}, by Propositions \ref{ima2} and
\ref{lemma24}, for every $\nu\in\N$ we come to the estimate:
\beqs
\label{lemma3.1quaderno}
\nonumber
\frac12\frac{d}{dt}\|v_k(t,\cdot)\|^2&\geq&
\left( \Im a_{p-1}(t,x_k+pA_p(t)n^{p-1})n^{p-1}-
A\left(1+\frac{n^{p-1}}{\varrho_k}\right)\right)\|v_k\|^2
\\
&&-C_\nu{\varrho_k}^{4+\frac12}\left(\frac{{\varrho_k}^{\mu+1}}{n}\right)^\nu
n^{q+p-1}\|v_k\|
-C\varrho_k^2n^{p-2}\sum_{j=1}^p\|v_k^{j,0}\|\cdot\|v_k\|\\
\nonumber
&&-C_\nu n^{p-1}
\sum_{1\leq\alpha_1+\alpha_2\leq \nu(p-1)}\!\left(\frac{\varrho_k^\mu}{n}
\right)^{\alpha_1+\alpha_2}
\|v_k^{\alpha_1,\alpha_2}\|\cdot\|v_k\|
\eeqs
for some $A,C,C_\nu>0$.
Now, for $a>\mu+1$, it is possible to take $\nu\in\N$
sufficiently large so that
\beqs
\label{sup}
\sup_k {\varrho_k}^{4+\frac12}\left(\frac{{\varrho_k}^{\mu+1}}{n}\right)^\nu
n^{q+p-1}\leq M_\nu
\eeqs
for some $M_\nu>0$.
After substituting \eqref{sup} in \eqref{lemma3.1quaderno},
we finally choose $a$ and $\mu$ such that
\beqs\label{postsup}
\frac{d}{dt}\|v_k(t,\cdot)\|&\geq&
\left( \Im a_{p-1}(t,x_k+pA_p(t)n^{p-1})n^{p-1}-
A\left(1+\frac{n^{p-1}}{\varrho_k}\right)\right)\|v_k\|- M'_\nu
\\
\nonumber
&&-C'_\nu n^{p-1}
\sum_{1\leq\alpha_1+\alpha_2\leq \nu(p-1)}\left(\frac{\varrho_k^\mu}{n}
\right)^{\alpha_1+\alpha_2}
\|v_k^{\alpha_1,\alpha_2}\|,
\eeqs
for some $M'_\nu,C'_\nu>0$; this can be done for
\beqs
\label{sceltamu}
\begin{cases}
\mu>p+1\cr
\ds\mu+1<a\leq\frac{p\mu-2}{p-1}=\mu+1+\frac{\mu-p-1}{p-1},
\end{cases}
\eeqs
since
$\varrho_k^2 n^{p-2}\leq n^{p-1}\left(\frac{\varrho_k^\mu}{n}
\right)^j$ for all $1\leq j\leq p$ if
$2\leq p\mu-a(p-1)$, and this implies, together with $a>\mu+1$, that
we must take $\mu>p+1$.

Using now $\varrho_k\left(\frac{\varrho_k^\mu}{n}\right)^{\alpha_1+\alpha_2}
\leq \left(\frac{\varrho_k^{\mu+1}}{n}\right)^{\alpha_1+\alpha_2}$, we
come to
\beqsn
\frac{d}{dt}\|v_k(t,\cdot)\|&\geq&\left( \Im a_{p-1}(t,x_k+pA_p(t)n^{p-1})n^{p-1}
-A\left(1+\frac{n^{p-1}}{\varrho_k}\right)\right)\|v_k\|- M'
\\
&&-C'\frac{n^{p-1}}{\varrho_k}
\sum_{1\leq\alpha_1+\alpha_2\leq \nu(p-1)}\!\left(\frac{\varrho_k^{\mu+1}}{n}
\right)^{\alpha_1+\alpha_2}
\|v_k^{\alpha_1,\alpha_2}\|
\eeqsn
for some constants $M',C'>0$, since $\nu$ has been fixed in \eqref{sup}.

Arguing in the same way for the functions $v_k^{\alpha,\beta}$ instead of $v_k$,
we finally get:
\begin{Prop}
\label{prop34}
Let $n$ be as in \eqref{n}, $a,\mu$ as in \eqref{sceltamu}, $\nu\in\N$
sufficiently large so that \eqref{sup} is satisfied. Then, for every
$\alpha, \beta\in\N_0$ there exists $C_{\alpha,\beta}>0$ such that
for all $t\in[0,\varrho_k/n^{p-1}]$ with $k$ large enough:
\beqs
\label{belowderivatives}
\frac{d}{dt}\|v_k^{\alpha,\beta}(t,\cdot)\|&\geq&
\left( \Im a_{p-1}(t,x_k+pA_p(t)n^{p-1})n^{p-1}-
A\left(1+\frac{n^{p-1}}{\varrho_k}
\right)\right)\|v_k^{\alpha,\beta}\|- C_{\alpha,\beta}
\\
\nonumber
&&-C_{\alpha,\beta}\frac{n^{p-1}}{\varrho_k}
\sum_{1\leq\tilde\alpha+\tilde \beta\leq \nu(p-1)}\!\left(\frac{\varrho_k^{\mu+1}}{n}
\right)^{\tilde\alpha+\tilde\beta}
\|v_k^{\alpha+\tilde\alpha,\beta+\tilde\beta}\|.
\eeqs
\end{Prop}

From Proposition \ref{prop34} it follows that:
\beqs
\label{bohh}
\nonumber
\frac{d}{dt}\left(\left(\frac{\varrho_k^{\mu+1}}n\right)^{\alpha+\beta}\!\!
\|v_k^{\alpha,\beta}\|\right)&\geq&\bigg( \Im a_{p-1}(t,x_k+pA_p(t)n^{p-1})n^{p-1}
-A\left(1+\frac{n^{p-1}}{\varrho_k}\right)\bigg)\\
&&\cdot
\left(\frac{\varrho_k^{\mu+1}}n
\right)^{\alpha+\beta}\!\!\|v_k^{\alpha,\beta}\|-C_{\alpha,\beta}
\\
\nonumber
&&-C_{\alpha,\beta}\frac{n^{p-1}}{\varrho_k}
\sum_{1\leq\tilde\alpha+\tilde \beta\leq \nu(p-1)}\!\left(\frac{\varrho_k^{\mu+1}}{n}
\right)^{\alpha+\tilde\alpha+\beta+\tilde\beta}
\|v_k^{\alpha+\tilde\alpha,\beta+\tilde\beta}\|.
\eeqs
We now choose $s\in\N$ sufficiently large so that, for all
$\bar{\alpha}+\bar{\beta}\geq s+1$, using \eqref{normevkab}
and $a>\mu+1$, we have
\beqs
\label{costanti}
\frac{n^{p-1}}{\varrho_k}\left(\frac{\varrho_k^{\mu+1}}{n}
\right)^{\bar\alpha+\bar\beta}
\|v_k^{\bar\alpha,\bar\beta}\|\leq
c_s\frac{n^{p-1}}{\varrho_k}\left(\frac{\varrho_k^{\mu+1}}{n}
\right)^{s+1}\varrho_k^{\frac 12+2}n^q\leq c'_s
\eeqs
for some $c_s,c'_s>0$. In order to satisfy \eqref{costanti}
it's enough to take $s$ such that
\beqsn
a(q+p-1)+\frac 12+1+(s+1)(\mu+1-a)\leq 0,
\eeqsn
i.e.
\beqs\label{chooses}
s\geq\frac{a(q+p-2)+\mu+\frac 52}{a-\mu-1}.
\eeqs
With this choice of $s$ we define:
\beqs\label{sigmak}
\sigma_k(t):=\ds\sum_{0\leq\alpha+\beta\leq s}\left(\frac{\varrho_k^{\mu+1}}{n}
\right)^{\alpha+\beta}
\|v_k^{\alpha,\beta}\|.
\eeqs
From \eqref{bohh} we have that:
\beqsn
\frac{d}{dt}\sigma_k(t)&=&
\ds\sum_{0\leq\alpha+\beta\leq s}\frac{d}{dt}
\left[\left(\frac{\varrho_k^{\mu+1}}{n}
\right)^{\alpha+\beta}
\|v_k^{\alpha,\beta}\|\right]
\\
&\geq &\ds\sum_{0\leq\alpha+\beta\leq s}
\left( \Im a_{p-1}(t,x_k+pA_p(t)n^{p-1})n^{p-1}
-A\left(1+\frac{n^{p-1}}{\varrho_k}\right)\right)
\left(\frac{\varrho_k^{\mu+1}}n
\right)^{\alpha+\beta}\|v_k^{\alpha,\beta}\| \\
&&
-C_s\sum_{1\leq\bar\alpha+\bar\beta\leq s}\frac{n^{p-1}}{\varrho_k}
\left(\frac{\varrho_k^{\mu+1}}n\right)^{\bar\alpha+\bar\beta}
\|v_k^{\bar\alpha,\bar\beta}\|-C_s\\
&\geq &\ds\left( \Im a_{p-1}(t,x_k+pA_p(t)n^{p-1})n^{p-1}
-A_s\left(1+\frac{n^{p-1}}{\varrho_k}\right)\right)
\sigma_k(t)-C_s
\eeqsn
for some $C_s,A_s>0$, because of \eqref{costanti}.

We have thus obtained for the function $\sigma_k$ the following
differential inequality:
\beqsn
\sigma_k'(t)-B_k(t)\sigma_k(t)+C_s\geq 0\qquad
t\in\left[0,\ds\frac{\varrho_k}{n^{p-1}}\right],\ k\gg1,
\\
\nonumber
B_k(t):=\Im a_{p-1}(t,x_k+pA_p(t)n^{p-1})n^{p-1}-A_s
\left(1+\frac{n^{p-1}}{\varrho_k}\right),
\eeqsn
which clearly implies that
\beqsn
\sigma_k(t)\geq e^{\int_0^t B_k(\theta)d\theta}\left[\sigma_k(0)-C_s\ds
\int_0^t e^{-\int_0^\tau B_k(\theta)d\theta}d\tau\right]
\qquad
t\in\left[0,\ds\frac{\varrho_k}{n^{p-1}}\right], \ k\gg1.
\eeqsn
For $t=\varrho_k/n^{p-1}$ we have
\beqs\label{frombelow!!!}
\sigma_k\left(\frac{\varrho_k}{n^{p-1}}\right)\geq e^{\int_0^{\frac{\varrho_k}{n^{p-1}}}
B_k(\theta)d\theta}\left[\sigma_k(0)-C_s\ds\int_0^{\frac{\varrho_k}{n^{p-1}}}
e^{-\int_0^\tau B_k(\theta)d\theta}d\tau\right].
\eeqs
Let us focus on the term $\int_0^{\frac{\varrho_k}{n^{p-1}}} B_k(\theta)d\theta$;
the choice of $x_k,\varrho_k$ of Lemma \ref{lemma1} gives for it, by
the change of variables $\theta'=n^{p-1}\theta$ and for $k$ large enough, the
following estimate from below:

\beqs
\label{okbelow}
\nonumber
\int_0^{\frac{\varrho_k}{n^{p-1}}} B_k(\theta)d\theta&=&\int_0^{\frac{\varrho_k}{n^{p-1}}}
\Im a_{p-1}(\theta,x_k+pA_p(\theta)n^{p-1})n^{p-1}d\theta-
A_s\int_0^{\frac{\varrho_k}{n^{p-1}}}
\left(1+\frac{n^{p-1}}{\varrho_k}\right)d\theta
\\
\nonumber
&\geq &\int_0^{\varrho_k}\Im a_{p-1}\left(\frac{\theta'}{n^{p-1}},
x_k+pA_p\left(\frac{\theta'}{n^{p-1}}
\right)
n^{p-1}\right)d\theta'-2A_s
\\
\nonumber
 &= & \int_0^{\varrho_k}\Im a_{p-1}\left(\frac{\theta'}{n^{p-1}},
x_k+pa_p(\tau_k)\theta'\right)d\theta'-2A_s
\\
&\geq &M\log(1+\varrho_k)+k-2A_s,
\eeqs
for some $\tau_k\in[0,\theta'/n^{p-1}]$, since
$A_p(\theta'/n^{p-1})n^{p-1}=\theta'a_p(\tau_k)$
by the mean value theorem for integration.

Similarly it follows that for every $\tau\in [0,\frac{\varrho_k}{n^{p-1}}]$:
\beqs\label{okpos}
\int_0^{\tau} B_k(\theta)d\theta&\geq &
\int_0^{n^{p-1}\tau}
\Im a_{p-1}\left(\frac{\theta'}{n^{p-1}},
x_k+pa_p(\tau'_k)\theta'\right)d\theta'-2A_s\geq -2A_s
\eeqs
for some $\tau_k'\in[0,\theta'/n^{p-1}]$,
because of Lemma \ref{lemma1},
since $n^{p-1}\tau\leq n^{p-1}\frac{\varrho_k}{n^{p-1}}\leq
\varrho_k$.

Finally, from \eqref{sigmak} and \eqref{vk0} we have
$\|\sigma_k(0)\|\geq \|v_k(0)\|\geq \|h\|>0$; therefore,
substituiting the estimates \eqref{okbelow} and \eqref{okpos} into
\eqref{frombelow!!!}, we have proved the following
desired estimate from below for the function $\sigma_k(t)$:
\begin{Prop}
\label{prop35}
For every $M>0$ and $k\in\N$ let $x_k,\varrho_k$  be as in Lemma~\ref{lemma1}.
Taking $\mu\geq 2$ in \eqref{eqw} and $n$ as in \eqref{n} with $a,\mu$
satisfying \eqref{sceltamu},
it is possible to construct the functions $v_k^{\alpha,\beta}$ in
\eqref{vkab} and then to choose $s$
great enough (see \eqref{chooses}) such that the function $\sigma_k(t)$
defined in \eqref{sigmak} satisfies the following estimate from below:
\beqs\label{formulabelow}
\sigma_k\left(\frac{\varrho_k}{n^{p-1}}\right)\geq c (1+\varrho_k)^{M},
\qquad k\gg1,
\eeqs
for some $c>0$.
\end{Prop}

\section{Estimate from above and proof of the main Theorem. }
\label{sec4}

The estimate from above is now quite simple to be obtained and it is
shown in the following:
\begin{Prop}
\label{prop41}
For every $M>0$ and $k\in\N$ let $x_k,\varrho_k$  be as in Lemma~\ref{lemma1}.
Taking $\mu\geq 2$ in \eqref{eqw} and $n$ as in \eqref{n} with $a,\mu$
satisfying \eqref{sceltamu},
it is possible to construct the functions $v_k^{\alpha,\beta}$ in
\eqref{vkab} and then to choose $s$
great enough (see \eqref{chooses}) such that the function $\sigma_k(t)$
defined in \eqref{sigmak} satisfies the following estimate from above
for all $t\in [0,\frac{\varrho_k}{n^{p-1}}]$:
\beqs\label{formulaabove}
\sigma_k(t)\leq C \varrho_k^{\frac 12+2+aq},\qquad k\gg1,
\eeqs
for some $C>0$.
\end{Prop}
\begin{proof}
The estimate \eqref{normevkab} obtained in Section \ref{sec2}
and definition \eqref{sigmak} immediately give:
\beqsn
\sigma_k(t)\leq \ds\sum_{0\leq\alpha+\beta\leq s}C_{\alpha,\beta}
\left(\frac{\varrho_k^{\mu+1}}{n}
\right)^{\alpha+\beta}\varrho_k^{\frac 12+2+aq}\leq C\varrho_k^{\frac 12+2+aq}
\eeqsn
for some $C>0$, since $s$ has been fixed in \eqref{chooses}.
\end{proof}

We are now ready to give the proof of Theorem \ref{th1}.
\begin{proof}[Proof of Theorem \ref{th1}]
Let us assume, by contradiction, that the Cauchy problem \eqref{CP1} is
well-posed in $H^\infty$ but \eqref{CN2} does not hold true.
Then at least one of the two conditions \eqref{A} or \eqref{B}
does not hold true.
As we remarked in Section~\ref{sec2}, we can assume,
without loss of generality, that \eqref{A} does not hold and apply
Lemma \ref{lemma1}. By Propositions \ref{prop35} and \ref{prop41} we
come to the estimate:
\[c (1+\varrho_k)^{M}\leq \sigma_k\left(\frac{\varrho_k}{n^{p-1}}\right)
\leq C\varrho_k^{\frac 12+2+aq},\]
for positive constants $c,C$ not depending on $k$,
giving rise to a contradiction for $k$ large enough, if we choose
\beqsn
M>\frac 12+2+aq.
\eeqsn

Therefore condition \eqref{CN2} must be satisfied and the proof is complete.
\end{proof}

\appendix
\section{}
The localized pseudo-differential operators $W_{n,k}^{\alpha,\beta}(t,x,D_x)$
of the present paper have symbols $w_{n,k}^{\alpha,\beta}(t,x,\xi)$ depending
on the parameter $t$ and belonging to the class $S^{0}_{0,0}$
of all functions $p(x,\xi)\in C^\infty(\R^2)$ such that for every
$\alpha,\beta\geq 0$
\beqs
\label{classe000}
\vert D_x^\beta\partial_\xi^\alpha p(x,\xi)\vert\leq C_{\alpha,\beta};
\eeqs
$S^{0}_{0,0}$ is a Fr\'echet space with semi-norms
\beqs
\label{semi000}
|p|_{\ell,\ell'}^{0}:=\max_{\alpha\leq\ell,\beta\leq \ell'}\sup_{x,\xi\in\R}\vert
\partial_\xi^\alpha D_x^\beta p(x,\xi)\vert.
\eeqs
The class $S^0_{0,0}$ corresponds to the classical class $S^m_{\varrho,\delta}$
(defined by $\vert D_x^\beta\partial_\xi^\alpha p(x,\xi)\vert
\leq C_{\alpha,\beta}\langle\xi\rangle^{m-\varrho\alpha+\delta\beta}$
instead of \eqref{classe000}; see \cite{KG})
with
$m=\varrho=\delta=0$. In the
$S^m_{0,0}$ classes
the usual asymptotic expansion formula
\[p(x,\xi)\sim\ds\sum_{\alpha\geq 0}\frac1{\alpha!}\partial_\xi^\alpha
p_1(x,\xi)D_x^\beta p_2(x,\xi)\]
fails to be true, and we need to use the expansion formula with a
remainder, as in \cite[Thm. 3.1, Chap. 2]{KG} (see also \cite[Thm. A]{I1}):
\begin{Th}
\label{thA}
Let $P_j(x,D_x)$ be pseudo-differential operators with symbols
$p_j(x,\xi)\in S^{m_j}_{0,0}$, $j=1,2$.
Then the operator $P(x,D_x)=P_1(x,D_x)\circ P_2(x,D_x)$ has symbol
given by the oscillatory integral
$$p(x,\xi)=\ds\int\!\!\int e^{-iy\eta}p_1(x,\xi+\eta)
p_2(x+y,\xi)dy\dbar\eta\in S^{m_1+m_2}_{0,0},$$
where $\dbar\eta=(2\pi)^{-1}d\eta$.

Moreover, the following expansion formula holds for every $\nu\in\N$:
\beqsn
p(x,\xi)=\ds\sum_{\alpha\leq\nu-1}\frac1{\alpha!}
\partial_\xi^\alpha p_1(x,\xi)D_x^\beta p_2(x,\xi)+\ds
\int_0^1\frac{(1-\theta)^{\nu-1}}{(\nu-1)!}r_{\theta,\nu}(x,\xi)d\theta,
\eeqsn
where
\beqsn
r_{\theta,\nu}(x,\xi):=\ds\int\!\!\int e^{-iy\eta}\partial_\xi^\nu
p_1(x,\xi+\theta\eta)D_x^\nu p_2(x+y,\xi)dy\dbar\eta\in S^{m_1+m_2}_{0,0}.
\eeqsn
\end{Th}

We recall from \cite[Lemm 2.2, Chap. 7]{KG}, (see also \cite[Thm. B]{I1}):
\begin{Th}
\label{thB}
Let $p_j(x,\xi)\in S^0_{0,0}$ for $j=1,2$ and define
\beqsn
p_\theta(x,\xi):=\ds\int\!\!\int e^{-iy\eta}p_1(x,\xi+\theta\eta)
p_2(x+y,\xi)dy\dbar\eta.
\eeqsn
Then for every $\ell\in\N_0$ there exists a constant $C_\ell>0$
such that
\beqsn
|p_\theta|_{\ell,\ell}^{0}\leq C_\ell|p_1|_{\ell+2,\ell+2}^{0}
|p_2|_{\ell+2,\ell+2}^{0}
\eeqsn
for all $\theta\in[0,1]$.
\end{Th}

We conclude the appendix with the statement of the
Calder\'on-Vaillancourt's Theorem about continuity of
pseudo-differential operators with symbols in the class $S^0_{0,0}$
acting on  $L^2$ (see \cite{CV} or \cite[Thm. C]{I2}):
\begin{Th}
\label{thC}
Let $p(x,D_x)$ be a pseudo-differential operator with symbol
$p(x,\xi)\in S^0_{0,0}$. Then:
\beqsn
\|p(x,D_x)u\|\leq C \vert p\vert^0_{2,2}\ \|u\|
\eeqsn
for all $u\in L^2$, with a positive constant $C$ independent of $p$ and $u$.
\end{Th}

{\bf Aknowledgment.}
The first two authors are members of the Gruppo Nazionale per
l'Analisi Matematica, la Probabilit\`a e le loro Applicazioni (GNAMPA)
of the Istituto Nazionale di Alta Matematica (INdAM).

config: arxiv


TITLE: A coin tossing game which leads to a contradiction
QUESTION [2 upvotes]: The question is:

Person $A$ and $B$ are going to play a coin tossing game. There is an initial score $0$, and whenever a head/tail appears, the score $+1$/$-1$. Repeating the coin tossing until one wins, that is, when the score reaches $+2$/$-2$, $A$/$B$ wins the game. There is also an initial stake \$$1$ for the game and person A has the option to double the stake every time before a coin toss. When one person wins the game, the other player needs to pay the amount of dollars on the stake to the winner. The question is: if you are person $A$, what is your strategy and what is your highest expected payoff of the game?

My idea is as follows:
We first denote the initial stake as $S$ ($S>0$), and denote the expected payoff $A$ can get when using some non-random strategy $g$ as $\mathop{\mathbb{E}}_g(S)$. Then no matter what the strategy $g$ is, we always have that
$$\mathop{\mathbb{E}}_g(S)=\frac{1}{2^{x_1}}\times (2^{y_1}\times (-1)^{z_1} \times S) + \frac{1}{2^{x_2}}\times (2^{y_2}\times (-1)^{z_2}\times S) + \dots = k_gS$$
where $x_i$ stands for some possible path leading to $A$'s winning or losing, $y_i$ stands for the possible stake-doubling actions, and $z_i$ stands for $A$'s winning or losing.
We denote the best strategy $A$ can use as $G$, and we have $\mathop{\mathbb{E}}_G(S)=k_GS$. It must hold that $k_G\ge0$ because if $A$ never doubles the stake, $A$'s expected payoff is $0$. Therefore under the best strategy, the payoff must be not less than $0$.
Now I claim that $A$ should always double the stake when the score reaches $+1$. The reason is that when you double the stake when your stake is $S'$ and the score is $+1$, you can earn an extra money of
$$\frac{1}{2} S'+\frac{1}{2}\mathop{\mathbb{E}}_{G'}(2S')-\frac{1}{2}\mathop{\mathbb{E}}_{G'}(S')=\frac{1}{2}S'+\frac{1}{2}k_{G'}S>0$$
And I also claim that $A$ should double the stake when the score reaches $0$. The reason is that since $A$'s expected payoff is $k_GS$ and $k_G\ge0$, if in strategy $G$, $A$ doesn't double the stake on score $0$, then after we double the stake on $0$, our expected payoff becomes $2k_GS\ge k_GS$, and thus we can at least break even comparing to not doubling the stake on $0$.
But the strange thing is, no matter we double the stake on $-1$ or not, there always remains a contradiction as follows:
Case $1$ is that we don't double the stake on $-1$. Then we can list an equation:
$$\mathop{\mathbb{E}}_G(S)=\frac{1}{4}\times 4S + \frac{1}{4}\mathop{\mathbb{E}}_G(4S)+\frac{1}{4}\mathop{\mathbb{E}}_G(2S)-\frac{1}{4}\times 2S$$
and if we substitute $\mathop{\mathbb{E}}_G(S)=k_GS$ into the equation, we can get that $k_G=-1$, which contradicts with $k_G\ge 0$.
Case $2$ is that we double the stake on $-1$. This indicates that
$$\frac{1}{2}\mathop{\mathbb{E}}_G(2S)-\frac{1}{2}\mathop{\mathbb{E}}_G(S)-\frac{1}{2}S\ge 0$$
(this is the expected extra payoff we can get when we double the stake on $-1$, and this should be $\ge0$ otherwise we won't double the stake on $-1$), which leads to $k_G\ge 1$. But when we list the similar equation as case $1$:
$$\mathop{\mathbb{E}}_G(S)=\frac{1}{4}\times 4S + \frac{1}{4}\mathop{\mathbb{E}}_G(4S)+\frac{1}{4}\mathop{\mathbb{E}}_G(4S)-\frac{1}{4}\times 4S$$
and substitute $\mathop{\mathbb{E}}_G(S)=k_GS$ into the equation, we can get that $k_G=0$, which contradicts with $k_G\ge 1$.
What's wrong with my reasoning and what's your thought to the original question? Thanks!
UPDATE:
Thanks to all the users' answers, and I think the contradiction comes from the fact that under some strategies, the expected payoff doesn't exist. Thus I cannot write $\mathbb{E}_g(S)=k_gS$ in the first place for all strategies. And a strategy without a existent expectation should not be considered as a "good" strategy (not to mention the optimal one) because it cannot ensure you a positive expected payoff. However, I have thought of a new strategy that can result in arbitrarily large expected payoffs: double on all $+1$'s, double on the first $k$ $0$'s, and never double on $-1$'s. First we calculate the probability that you end in $+2$ after $2j$ coin tosses: $$\sum_{i=0}^{j-1}{j-1\choose i}\times \frac{1}{2^{2j}} = \frac{1}{2^{j+1}}$$ and the expected payoff if you double on all $+1$'s and $0$'s and ends in $+2$ after $2j$ steps:  $$\sum_{i=0}^{j-1}{j-1\choose i}\times \frac{1}{2^{2j}}\times 2^{i+1} \times S = \frac{2}{3}\times (\frac{3}{4})^j\times S$$. Similarly when you end in $-2$ it is: $$\sum_{i=0}^{j-1}{j-1\choose i}\times \frac{1}{2^{2j}}\times 2^{i} \times (-S) = \frac{1}{3}\times (\frac{3}{4})^j\times (-S)$$. Thus if you add up these two expectations you can get the expected payoff if you end in $+2/-2$, and it is $$\frac{1}{3}\times (\frac{3}{4})^j\times S$$. Then the expected payoff of my new strategy can be calculated as: $$\sum_{j=1}^{k} \frac{1}{3}\times (\frac{3}{4})^j\times 2^j \times S + \sum_{j=k+1}^{+\infty} \frac{1}{3}\times (\frac{3}{4})^j\times 2^k \times S = (2\times (\frac{3}{2})^k-1)S$$, which can be arbitrarily large if we choose large enough $k$.

REPLY [2 votes]: Let's call the players Alice and Bob.
It always makes sense for Alice to double the bet when the score is $+1$:  You can think of the doubling as placing a side bet on the outcome of a "new" game in which Alice has a head start. In other words, Alice should expect no profit, on average, from the initial bet (because Bob has an equal probability of winning it), but can expect a net profit from any side bets she forces Bob to accept when she has the advantage.
Because Alice now has a strategy (double on $+1$) that gives her a positive expected profit, she would like the game to start with as large a stake as possible.  Thus anytime the score returns to $0$, Alice should also double the bet.
Should Alice ever double the bet when the score is $-1$? Ask Bob. He would like to see a side bet made on the "new" game in which has the head start.  If doubling on $-1$ ever offers Alice an advantage, then it would always offer her an advantage, so she would always double the bet no matter what the score. But this would be like not consulting either Alice or Bob on strategy, at which point the game back to a symmetric draw. So Alice should not double the bet when the score is $-1$.
A mathematical drawback to this optimal strategy (especially from Bob's point of view) is that it results in an expected value (to Alice) of $\infty$.  One way to see this is to imagine Alice is only allowed a fixed number of doublings, $n$, when the score is $0$ (but is always allowed to double on $+1$).  For $n=0$, the expected value satisfies
$$E_0={1\over4}(2)+{1\over4}(2E_0)+{1\over4}(E_0)-{1\over4}(1)$$
That is, after two coin tosses, Alice either doubles once on $+1$ and then wins the total bet of $2$, or she doubles once $+1$ and then drops back to $0$ with the stakes now doubled, or she drops to $-1$ and then gets back to $0$ with the original bet, or loses her (undoubled) bet of $1$. It's easy to see that this solves to $E_0=1$.
Recursion now says
$$E_n={1\over4}(2)+{1\over4}(4E_{n-1})+{1\over4}(2E_{n-1})-{1\over4}(1)$$
for $n\ge1$, which simplifies to
$$E_n={3\over2}E_{n-1}+{1\over4}$$
and it's now clear that $E_n\to\infty$ as $n\to\infty$. If you like, you can derive the direct formula
$$E_n=\left(3\over2\right)^{n+1}-{1\over2}$$
It might be more interesting to consider Alice's expected value if she is limited to a finite number of doublings altogether (and not just on doublings when the score is $0$), i.e., if the maximum amount that can be wagered in a single game is $2^n$ for some $n\ge0$.  Let's let $\mathcal{E}_n^*$ denote the expected value from an optimal strategy. The live question is whether (and when) an optimal strategy has her double on $0$.
It's obvious that $\mathcal{E}_0^*=0$.  For $n\gt0$, Alice needs to choose between a strategy $\mathcal{E}_n$, in which she waits for the score to reach $+1$ before her first doubling, and a strategy $\mathcal{E}_n'$, in which she doesn't wait (i.e., doubles if the score returns from $-1$ to $0$). The first possibility satisfies
$$\mathcal{E}_n={1\over4}(2)+{1\over4}(2\mathcal{E}_{n-1}^*)+{1\over4}(\mathcal{E}_n)-{1\over4}(1)$$
which solves to
$$\mathcal{E}_n={2\mathcal{E}_{n-1}^*+1\over3}$$
The second possibility satisfies
$$\mathcal{E}_n'={1\over4}(2)+{1\over4}(2\mathcal{E}_{n-1}^*)+{1\over4}(2\mathcal{E}_{n-1}^*)-{1\over4}(1)=\mathcal{E}_{n-1}^*+{1\over4}$$
The expected value for the optimal strategy is thus
$$\mathcal{E}_n^*=\max\{\mathcal{E}_n,\mathcal{E}_n'\}=\max\left\{{2\over3}\mathcal{E}_{n-1}^*+{1\over3},\mathcal{E}_{n-1}^*+{1\over4} \right\}$$
We get
$$\begin{align}
\mathcal{E}_1^*&=\max\left\{{1\over3},{1\over4} \right\}={1\over3}\\
\mathcal{E}_2^*&=\max\left\{{5\over9},{7\over12} \right\}={7\over12}\\
\mathcal{E}_3^*&=\max\left\{{13\over18},{5\over6} \right\}={5\over6}
\end{align}$$
etc. It's easy to see that the $\mathcal{E}_n'$ strategy is the best choice for $n\gt1$ and that
$$\mathcal{E}_n^*={1\over3}+{n-1\over4}\quad\text{for }n\ge1$$
In other words, Alice should double on $0$ until she gets down to her last doubling, at which point she should wait for the score to reach $+1$. Note, we still have $\mathcal{E}_n^*\to\infty$ as $n\to\infty$, but at a much slower rate than $E_n$, when the doublings on $+1$ didn't count against her limit.


TITLE: Hassle with Absolute Value and Square Root
QUESTION [1 upvotes]: Question 1:
By definition absolute value gives just no of units and does not indicate any direction neither positive nor negative then why in practice we use +ve direction like $\left|4\right|=+4$ it should be just 4 not +4
Question 2:
We know that  $$\sqrt{{x}^2} = \pm x = \left|x\right| $$  Then why   $\left|4\right|=+4$   and not   $\left|4\right| = \pm 4$  Also     $\sqrt{{4}^2} = +4$   and not   $\sqrt{{4}^2} = \pm 4$
Is this all have do with involvement of a variable only (when variable involved use $\pm$ and when constants involved don't use $\pm$)   ............ or there are some other rules !

REPLY [3 votes]: The plus does nothing, it's $+4=4$. It's there to draw attention to the fact that the absolute value left a positive value unchanged -- they wanted to stress the difference between $-$ and $+$ case.
Also, $\sqrt{x^2}=\pm x$ is wrong. It's $\sqrt{x^2}=|x|$. The result of square root is always positive by definition. You can't get two values from a function. If you want to solve a quadratic equation, for instance, $y^2=x$, then you have to explicitly put $\pm$ in front of the square root to specify both solutions: $y=\pm \sqrt{x}$.


TITLE: Structure of the proof of the fundamental theorem of arithmetics
QUESTION [1 upvotes]: Fundamental theorem of arithmetics: A principal ideal domain is factorial. i.e: Any non zero element of a principal ideal domain can be decomposed in a unique product of irreductibles.
The structure of the proof goes like this::

Proof of the existence of a decomposition: In this part, it is shown any non-zero and non-invertible elements of a principal ideal domain can be decomposed in a product of irrreductibles.
Proof of the uniqueness

My problem: In the first part they show that any non-zero and non-invertible elements of a principal ideal domain can be decomposed in a product of irrreductibles. What about non-zero but invertible elements of a principal ideal domain?

REPLY [1 votes]: In a commutative ring $R$, say that $a\sim b$ if there exists a unit $u$ such that $a=bu$. ($a$ is said to be 'an associate to $b$'.) This turns out to be an equivalence relation.
In the theory of unique factorization, when you say a particular factorziaton is a "unique product," it is subject to the caveat that things are unique up to being associates.
Now you can verify that the set of units of the ring forms an equivalence class in this relation, so that every unit is equivalent to $1$. Therefore, it's pointless to talk about factorizations of units because all units are "the same as $1$."


TITLE: Let $f(z)$ be continuous on a region A and holomorphic on $A\setminus \{z_0\}$ for a point $\{z_0\}$. Show that f is holomorphism on $A$
QUESTION [1 upvotes]: Let $f(z)$ be continuous on a region A and holomorphic on $A\setminus \{z_0\}$ for a point $\{z_0\}$. Show that $f$ is holomorphism on $A$
i really did't get any idea..can any help please

REPLY [2 votes]: This is just a particular case of Riemann's theorem on removable singularities.
It's not hard to prove it. Let, for each $z\in A$, $g(z)=(z-z_0)f(z)$. Then $g$ is holomorphic, since$$g'(z)=\begin{cases}f(z)+(z-z_0)f'(z)&\text{ if }z\neq z_0\\f(z_0)&\text{ if }z=z_0.\end{cases}$$Besides, $g(z_0)=0$.
Since $g$ is holomorphic, it is analytic and, since $g(z_0)=0$, its Taylor series around $z_0$ is of the type$$g(z)=a_1(z-z_0)+a_2(z-z_0)^2+\cdots$$and therefore$$f(z)=a_1+a_2(z-z_0)+\cdots$$So, $f$ is analytic and therefore holomorphic.


TITLE: Why do all magnetic field of a bar magnet do not start and end only at pole?
QUESTION [1 upvotes]: Iam very confused that why not all field lines emanate out of a pole? Why few field lines emanate and in middle of a bar magnet as shown in the following figure:

REPLY [4 votes]: The magnetic field of a bar magnet is made up from the individual magnetic dipoles of the outermost electrons in the iron atoms. The magnetic field of an individual electron is tiny, but since there are gazillions of electrons in a typical bar magnet their fields combine to generate a substantial overall field.
If you looked at the field from a single electron it would look a bit like:

(image from Wikipedia)
so the field lines do splay out sideways as well as forwards and backwards. However in a collection of electrons the field from any given electron interacts with the electrons in front of it and behind it, and the result is to channel the field lines:

I fear my drawing skils don't do the subject justice, but hopefully the general idea is clear. When you have a bar magnet stuffed with electrons that are all lined up with each other then the field lines tend to run along the lines of the electrons so the only emerge at the ends of the bar magnet where the lines of electrons stop.


TITLE: Lagrangian saddle-point set is compact?
QUESTION [0 upvotes]: Let $f: \mathbb R^n \to \mathbb R$ be a convex, continuously differentiable function. Let $\mathcal X \subset \mathbb R^n$ denote the set of solutions to the problem
$$ \min_{x \in \mathbb R^n} f(x) \quad \text{s.t.} \quad g(x) = 0, $$
where $g: \mathbb R^n \rightarrow \mathbb R^p$ is a linear map with $p \leq n$. Assume that $\mathcal X$ is non-empty and compact.
Define $L: \mathbb R^n \times \mathbb R^p \rightarrow \mathbb R$ to be $L(x, \lambda) = f(x) + \lambda^T g(x)$, and denote the set of min-max saddle points of $L$ by $$\mathcal S_L = \{(x^*, \lambda^*) \in \mathbb R^n \times \mathbb R^p: \; L(x^*, \lambda) \leq L(x^*, \lambda^*) \leq L(x, \lambda^*) \; \forall (x, \lambda) \in \mathbb R^n \times \mathbb R^p\}.$$
Under what condition(s) is $\mathcal S_{L}$ compact?

REPLY [1 votes]: Note that since $g$ is linear, $g(x) = Ax$ for some $A$, and $Dg(x) = A$.
Note that $x \mapsto L(x, \lambda)$ is convex, hence $x'$ is a minimiser of
$x \mapsto L(x, \lambda)$ iff $Df(x')+\lambda^T A = 0$.
Note that if $n^T A  = 0$ then $L(x,\lambda) = L(x,\lambda+tn)$ for all $t$.
Suppose $(x^*,\lambda^*) \in S_L$, then since $L(x^*,\lambda) \le L(x^*,\lambda^*)$
for all $\lambda$ we must have $g(x^*) = 0$.
Since $f(x^*)=L(x^*,\lambda^*) \le L(x,\lambda^*)$ for all $x$, we see that if $g(x) = 0$ then $f(x^*) \le L(x,\lambda^*) = f(x)$
and so $x^* \in {\cal X}$.
We also see that $x^*$ is a minimiser of $x \mapsto L(x,\lambda^*)$ and so
$Df(x^*)+ {\lambda^*}^T A = 0$.
If $\ker A^T = \{0\}$ then we see that $\lambda^* = -(AA^T)^{-1}A Df(x^*)^T$, in particular, since $x \mapsto (x, -(AA^T)^{-1}ADf(x)^T)$ is continuous, we see that $S_L$ is compact since ${\cal X}$ is.
If $A^Tn = 0$ for some $n \neq 0$, we see that $(x^*,\lambda^*+tn) \in S_L$ for all $t$, hence $S_L$ is unbounded.


TITLE: If $\int_0^T F(x_n(t)) dt- \int_0^T x_n^2(t) dt \to c$, how to prove that $\int_0^T F(x_n(t)) dt$ is bounded?
QUESTION [0 upvotes]: Let $T>0$ and $F$ be a real function. Let $(x_n)_n$ be a sequence of Lipschtizian functions over $[0, T]$ such that
$$\int_0^T F(x_n(t)) dt- \int_0^T x_n^2(t) dt \to c\quad\mbox{ as } n\to +\infty,$$
with $c\in\mathbb{R}$.
My question is: it is true that $\int_0^T F(x_n(t)) dt\le M$, for a suitable constant $M>0$. About me the answer is yes since we have the convergence to $c$, but I don't know how to formally prove it.
Could someone please help me?
Thank you in advance!

REPLY [2 votes]: The answer is positive, if you have the additional assumption that $(x_n)_n$ is bounded. Then
$$\left|\int_0^T F(x_n(t))dt\right|\leq \left|\int_0^T F(x_n(t))-x_n^2(t)dt\right|+ \left|\int_0^T x_n^2(t)dt\right|,$$
where the first term on the right hand side is bounded because of the convergence and the second term because $x_n$ is bounded and $[0,T]$ is compact. For a counterexample just take $x_n(t)=nt$ and $F$ the identity, then the statement is false.


TITLE: Estimating of the size of the Milky Way from the Sun's motion
QUESTION [6 upvotes]: I am trying to run a simple first-principles calculation to estimate the diameter of the Milky Way from what we know about the motion of the Sun around its center. In particular, from various online sources and books, we can roughly say that:

The period of the Sun's orbit around the center of the Milky Way is $T_{sun}\approx 200$ million years.
The speed of the Sun in its orbit is $v_{sun}\approx 200$ km/s.
The sun is located about halfway between the Milky Way's center and its edge.

This is a purely kinematic problem. I'll assume that the Sun's orbit is circular, so that we may write:
$$
\omega_{sun}r_{sun} = v_{sun},
$$
where $\omega_{sun}=2\pi/T_{sun}$ is the angular velocity of the Sun around the Milky Way center. Hence:
$$
r_{sun} = \frac{T_{sun}}{2\pi}v_{sun}.
$$
The radius of the Milky Way is twice that (the Sun is halfway out to the edge), and the diameter is twice that again. So the overall diameter of the Milky Way, in light years (denoting the speed of light by $c$):
$$
d_{MilkyWay} = \frac{2T_{sun}v_{sun}}{c\pi}.
$$
We know all the quantities on the right hand side. A short Python code evaluates the solution.
T_sun = (200e6)*365.25*24*3600 # [s]
v_sun = 200e3 # [m/s]
c = 2.99792458e8 # [m/s]
pi = 3.14159
d_MilkyWay = 2*T_sun*v_sun/(c*pi) # [ly]

print "d_MilkyWay = %.2e [light years]"%(d_MilkyWay)

The answer: $d_{MilkyWay} = 2.68\cdot 10^{12}$ light years. This is much, much larger than the current best estimate given by scientists: about $10^5$ light years.
How come there is such a huge discrepancy between the measured size, and my rough estimate? Is the assumption that the Sun's orbit is circular wrong?

REPLY [3 votes]: Problem: Wrong units

$$
 d_{MilkyWay} = \frac{2T_{sun}v_{sun}}{c\pi}
 $$

And that's your first problem, right here. You cannot disregard units and expect to get any correct or even well-defined result from your calculations.
The left-hand side is a distance (e.g. in meters, light-years or light-seconds), the right-hand side is a duration. This cannot work, and you cannot use this equation anywhere.
Light-years are not years, just like light-seconds are not seconds.
Solution: Remove $c$
The correct formula is simply:
$$
d_{MilkyWay} = \frac{2T_{sun}v_{sun}}{\pi}
$$
Order of magnitude
Before using any tool in order to calculate the exact result, you should first try to get some idea about the correct order of magnitude (see Fermi problem).
The goal should be to simplify the calculation until it becomes easy to do mentally, without being too wrong.

So $3 \approx 1$ and $4 \approx 10$ are perfectly acceptable, but $10^5\approx10^{12}$ isn't.
You could notice that 200 km/s is approximately $\frac{c}{1000}$.
Here's the order-of-magnitude estimate:
\begin{align}
d_{MilkyWay} & \approx \frac{2v_{sun}T_{sun}}{\pi} \\
 & \approx \frac{2*200\mathrm{km/s}*200\thinspace000\thinspace000\mathrm{y}}{\pi} \\
 & \approx \frac{2*\frac{c}{1\thinspace000}*200\thinspace000\thinspace000\mathrm{y}}{\pi} \\
 & \approx \frac{2*c*200\thinspace000\mathrm{y}}{\pi} \\
 & \approx \frac{400\thinspace000}{\pi}*c*\mathrm{y} \\
 & \approx 100\thinspace000 \mathrm{ly}
\end{align}
This is not the exact answer. But whichever tool you use, you should expect to get a result not too far away from $10^{5}\mathrm{ly}$.
Result
With Python
If you want to stick with Python, you could use @PålGD's excellent answer.
With Wolfram Alpha
Wolfram Alpha does a good job at converting units:
2*200km/s*200million years/pi to lightyears
You get many different units as a bonus, and even a direct comparison to the correct galactic diameter:

With Qalculate
Qalculate! is an excellent desktop calculator, and it supports many units:


TITLE: Extreme points of a modified unit ball $\ell^p$
QUESTION [0 upvotes]: Consider the space $\ell^p$ over the reals. Denote
$$
A = \lbrace x \in \ell^p; \lVert x \rVert_p \leq 1 \ \& \ \lVert x \rVert_\infty \leq \alpha \rbrace.
$$
For $p \in (1, \infty)$, $\alpha > 0$, find the set of extreme points of $A$.

I think the set of extreme points is
$$
B = \lbrace (x_n)_n \in S_{\ell^p}; \exists j: |x_j| = \alpha \rbrace
$$
where $S_{\ell^p}$ is the unit sphere. From $\lVert x \rVert \leq 1$ we have that $\alpha \leq 1$.
$\text{ext}A \supset B$
Pick $x \in B$, $y,z \in A$ and assume $x = \frac{1}{2}(y+z)$. Then
$$
1 = \lVert x \rVert = \frac{1}{2}\lVert y+z \rVert \leq \frac{1}{2} \left( \lVert y \rVert + \lVert z \rVert \right) \leq 1
$$
hence we have equalities.
For $j$ such that $|x_j| = \alpha$ we have
$$
\alpha = |x_j| = \frac{1}{2} |y_j + z_j| \leq \frac{1}{2}(|y_j| + |z_j|) \leq \alpha
$$
Again, we have equalities and $x=z=y$.
$\text{ext}A \subset B$
Pick $x \in A \setminus B$. Then $\forall j : |x_j| \neq \alpha$ and $ \lVert x \rVert_p < 1$, hence $\forall n: x_n \in (-1,1)$. We can find $\varepsilon > 0$ such that $[x_n - \varepsilon, x_n + \varepsilon] \subset [-1,1]$. Now define $y,z$ as $y_n = x_n - \varepsilon, z_n = x_n + \varepsilon$. Then $y,z \in A, y \neq z$ but $\frac{1}{2}(y+z) = x$. Hence $x \not\in \text{ext}A$.
Is this correct? Thank you.

REPLY [1 votes]: The set of extreme points of $A$ is equal
$$E:=\{x\in \ell^p\,:\, \|x\|_p=1, \ |x_n|\le a\}$$
Indeed every element $x$ in $E$ is an extreme point of the unit ball in $\ell^p.$ As $A$ is a subset of the unit ball, the element $x$ is an extreme point of $A.$
It remains to show that every extreme point of $A$ belongs to $E.$
To this end we will show that if
$(x_n)$ belongs to $A$ and $\|x\|_p<1,$ then $x$ is not an extreme point of $A.$ Indeed, let $F=\{n\in \mathbb{N}\,:|x_n|=a\}.$
As $x_n\to 0$ we have
$$m:=\inf_{n\notin F}(a-|x_n|) >0$$
Hence $$|x_n|\le a-m,\quad n\notin F$$ For $L\ge 1$
let $$y_n=\begin{cases}0 & n\in F \\
\displaystyle {m\over 2^{n/p}L} & n\notin F\end{cases}$$
Then $x+y$ and $x-y$ belong to $A.$ Indeed for $n\notin F$ we have
$$|x_n\pm y_n|\le |x_n|+y_n\le (a-m)+m=a$$
$$\|x\pm y\|_p\le \|x\|_p+\|y\|_p\le \|x\|_p+\left (\sum_{n=1}^\infty {m^p\over 2^{n}L^p} \right )^{1/p}=\|x\|_p+{m\over L} $$
For $L$ large enough we obtain $\|x\pm y\|_p\le 1.$
Eventually we get
$$x={1\over 2}(x+y)+{1\over 2}(x-y)$$ thus $x$ is not an extreme point of $A.$


TITLE: What is the order of discontinuity of this function?
QUESTION [0 upvotes]: Consider the function f(x) such that f(x) = 0 for all rational x and f(x) = 1 for all irrational x. It would seem that the number of 'jumps' up is uncountably infinite and the number of 'jumps' down is countably infinite; or is the other way around? Shouldn't the number of jumps up have the same order as the number of jumps down? Would the total number of jumps up or down have the same cardinality as the set of all real numbers? Is there a formal 'measure' for the order of discontinuity of a function? Thanks in advance for any guidance you can provide.

REPLY [1 votes]: Use the following definition and see what happens
One can instead require that for any sequence $(x_n)_{n\in\mathbb{N}}$ of points in the domain which converges to c, the corresponding sequence $\left(f(x_n)\right)_{n\in \mathbb{N}}$ converges to f(c). In mathematical notation, $\forall (x_n)_{n\in\mathbb{N}} \subset I:\lim_{n\to\infty} x_n=c \Rightarrow \lim_{n\to\infty} f(x_n)=f(c)\,.$
Pick up a sequence $a(n) \in Q $ with a limit point $ a \in Q' $ and apply the above definition.


TITLE: Finding domain of function and writing in interval notation
QUESTION [2 upvotes]: I think I have the answer but it always takes me a long time to find it and wanted to check it.  Equation is;$$ f(x) =\frac{\sqrt {x+4}}{x-4}$$ Interval notation then is $$ [-4, 4) \cup (4, \infty)$$

REPLY [1 votes]: Note that $\sqrt{x+4}$ yields $x\ge -4$ and $\frac{1}{x-4}$ yields $x\ne 4$, so $x$ can be anything greater than $-4$ excluding $4$. Therefore we can write $x \in [-4,+\infty) \setminus \{ 4\}$ or equivalently $x \in [-4,4)\cup (4, +\infty)$.