TITLE: Question about a comment on a different question. A rational root of a monic polynomial over $\Bbb Z[x]$ is an integer
QUESTION [0 upvotes]: This questions contains a comment that states that $\gcd(r,s)\ne 1$ (which leads to a contradiction) where our rational root is $r/s$.
Why would that be true?

REPLY [2 votes]: Multiplying by $s^n$ gives $$r^n + a_{n-1}r^{n-1}s + \dots+ a_1rs^{n-1} + a_0s^n = 0$$ and so $$r^n=-s(a_{n-1}r^{n-1}+\dots+a_1rs^{n-2}+a_0s^{n-1}).$$  Thus $s\mid r^n$, which implies any prime factor of $s$ is also a prime factor of $r$.  So, we conclude $\gcd(s,r)\neq 1$ unless $s$ has no prime factors.


TITLE: About primes in an interval
QUESTION [0 upvotes]: I find this result without a proof:
The prime number theorem implies that for any $\epsilon > 0$, there exists an integer $N$ such that for all $x > N$, there exists a prime between $x$ and $x(1 + \epsilon)$.
I am asking about a simple proof for this result.

REPLY [2 votes]: By prime number theorem, there are approximately $\frac{N}{\ln(N)}$ primes less than $N$, for sufficiently large $N$. Therefore, between $x$ and $x(1+\epsilon)$ there are approximately $$\frac{x(1+\epsilon)}{\ln(x(1+\epsilon))}-\frac{x}{\ln(x)}$$ primes. This expression is equivalent to $$\frac{x(1+\epsilon)\ln(x)-x\ln(x(1+\epsilon))}{\ln(x(1+\epsilon))\ln(x)}$$ which, by L'Hopital's rule, approaches $\infty$ as $x\to\infty$.
In particular, this means there is some $N$ such that for all $x>N$ we have at least one prime between $x$ and $x(1+\epsilon)$.


TITLE: Fundamental domain for subgroup of fuchsian Schottky group.
QUESTION [0 upvotes]: Let G be a Fuchsian Schottky group defined by a possibly infinite set of disjoint halfplanes {C_i}_i. Let F be the fundamental domain obtained by intersecting the complements of the C_i's. If H i a subgroup of G then:
Can a fundamental domain for H be found in terms of the domain F?
Is H of schottky type?
Do the above answers change if we require 1) H to be normal in G ? 2) G to be finitely generated ? 3) H to be finitely generated?
It seems quite easy to find a fundamental region for H in terms of F. Yet I run into trouble trying to construct it in such a way that it is actually a domain.
I will be greatful for some good references.

REPLY [1 votes]: Here is a concrete construction of a fundamental domain for $H$ that works regardless of normality and finite generation. The translates of the fundamental domain $F$ by the action of $G$ form a tiling of $\mathbb{H}^2$. Let $T$ denote the dual tree of this tiling, with one vertex for each translate of $F$, and with two vertices connected by an edge if the corresponding translates of $F$ intersect along a line. Note that each oriented edge $E$ of $T$ corresponds to a transversely oriented line $L_E$ of the tiling which subdivides $\mathbb{H}^2$ into two halfplanes, and the transverse orientation on $L_E$ points into one of those half-planes which I'll denote $C_E$.
The action of $G$ on $\mathbb{H}^2$ induces a properly discontinuous action of $G$ on $T$ by simplicial isomorphisms. Consider the restricted action of $H$ on $T$, also properly discontinuous. The quotient graph $T/H$ has fundamental group identified with $H$. Let $\tau$ be a maximal tree in $T/H$. Let $\tilde\tau \subset T$ be a homeomorphic lift of $\tau$. As $E \subset T$ varies over all oriented edges not contained in $\tilde\tau$ but with initial endpoint in $\tau$, the collection of half-planes $C_E$ demonstrates that $H$ is a Schottky group, and the complement of their union is a fundamental domain for $H$.


TITLE: expectation of log(1+x) if x is a gamma random variable
QUESTION [3 upvotes]: I would like to know if there is a closed form expression for the expectation of log(1+x) when x is a gamma random variable.
Thank you.

REPLY [0 votes]: I may be mistaken, but if you are making the change of variable $s = \lambda t$, shouldn't there be an extra factor of $\lambda$ outside the integral?


TITLE: find the orbits of the operation of conjugation of $S_3$ on the group of subgroups of $S_3$
QUESTION [2 upvotes]: $G = S_3$ I need to examine the operation of conjugation of $G$ on the group of the subgroups of $G$ (meaning $gH = gHg^{-1}$)
here I already got confused from the definition.
Now I need to :
a) find the group of orbits $O$ of this operation.
b) for each orbit $o \in O$ choose a representative $H \in o$ and calculate $\operatorname{Stab}_G(H)$.
c) check the Orbit-stabilizer theorem on this operation.
I'm really confused from the definitions here.
What I tried -  If I understand the question correctly - I have H' which is a group that contains 6 subgroups of $S_3$ then i calculate the orbits for each of these groups. I get one orbit for each group. So I guess here the representative (asked in (b) ) is this one orbit that I found. for each of the 6 orbits I get that
$$\operatorname{Stab}_G(O_i) = S_3$$
so in (c) i get that
$|H'| = \sum_1^6 [G:\operatorname{Stab}_G(x_i)] =  1+1+1+1+1+1 = 6$ which is correct but I'm sure i did not understand the question correctly . any help will be very appreciated!

REPLY [2 votes]: As you suspect, your answer is not correct.  The first thing you need to list all the subgroups of $S_3$.  Now for each subgroup $H \leq S_3$ and for each $g \in S_3$, you need to compute $gHg^{-1}$.  These conjugate subgroups are the elements of the orbit of $H$.
For example, take $H = \langle (1 \ 2) \rangle \leq S_3$. Now we need to loop over all the $g \in S_3$ and compute $gHg^{-1}$.  If we take $g = 1$ or $g = (1\ 2)$, we just get $gHg^{-1} = H$.  Now consider $g = (1 \ 3)$.  Since
$$
(1\ 3) (1\ 2) (1\ 3)^{-1} = (1\ 3) (1\ 2) (1\ 3) = (2\ 3)
$$
we see that $gHg^{-1} = \langle (2\ 3) \rangle$.  Thus $H = \langle (1\ 2) \rangle$ and $\langle (2\ 3) \rangle$ are both in $O_H$, the orbit of $H$.  What other subgroups can you get by conjugating $H$ by other $g \in S_3$?  What happens for the other subgroups?  Try $H = \langle (1\ 2\ 3) \rangle$ and see how this differs from $\langle (1\ 2) \rangle$.


TITLE: Multidimensional Gaussian integral with complex variance and linear term
QUESTION [1 upvotes]: A formula is given by Wiki (https://en.wikipedia.org/wiki/Gaussian_integral#n-dimensional_with_linear_term):
$$ \int_{\mathbb{R}^n} \mathrm{d}^n\mathbf{x}\ \exp(-\frac{1}{2}\mathbf{x}^T\mathbf{A}\mathbf{x}+\mathbf{j}^T\mathbf{x}) = \sqrt{\frac{(2\pi)^n}{\det\mathbf{A}}}\exp(\frac{1}{2}\mathbf{j}^T\mathbf{A}^{-1}\mathbf{j}) $$
where $\mathbf{j}$ is a vector. It remains unclear to me what are the constraints on $\mathbf{j},\mathbf{A}$? Clearly, if both are real-valued and $\mathbf{A}$ is positive definite the the identity holds. Does it hold for complex $\mathbf{j}$ and Hermitian positive definite $\mathbf{A}$ (while integrating over $\mathbb{R}^n$ only)?
A few numerical tests suggest so and I can't see why it would not.

REPLY [0 votes]: In this lecture note from Berkeley, they say at top of page 3 that $\mathbf{j}$ is allowed to be complex and that indeed is the transpose of $\mathbf{j}$ that is meant on the right hand side of this equation.
However, I have not found any source what happens when $\mathbf{A}$ is Hermitian positive definite.


TITLE: Showing that two cartesian products are equinumerous
QUESTION [0 upvotes]: Math 3345
Section 16 Exercise 9
Let A, B, C, D be sets such that A is equinumerous to C and B is equinumerous to D. Show that AxB is equinumerous to CxD.
I believe I have the correct answer for when all A, B, C, D are finite. I think it works but I'm not sure if it's correct for when the sets are infinite.
My answer:
I'm using the '||' symbols to denote "the cardinality of"/"the length of", and 'x' to represent Cartesian product (I apologize, I'm new to formatting on the site)
A equinumerous to C, therefore |A| = |C|
B equinumerous to D, therefore |B| = |D|
|AxB| = (|A|)(|B|) = (|C|)(|B|) = (|C|)(|D|) = |CxD|
Therefore, since |AxB| = |CxD|, then AxB is equinumerous to CxD.

REPLY [0 votes]: As Asa Karagila said your argument is not right since essentially is what you have to prove (at least in the infinite case)
Try the following:
As $|A|=|C|$ then there exists a bijection $f:A\rightarrow C$. Analogous there exist a bijection $g:B \rightarrow D$.
The function $f\times g :A\times B\rightarrow C\times D$ defined as $(f\times g) (x,y)=(f(x),g(y))$ is a bijection. Prove it.
Finally conclude $|A\times B|=|C\times D|$.


text:
\begin{document}

\title[Poincar\'e problem for weighted projective foliations]{Poincar\'e problem for weighted projective foliations}

\author{F. E. Brochero Mart\'inez}

\address{F. E. Brochero Mart\'inez \\
Departamento de Matem\'atica \\
Universidade Federal de Minas Gerais\\
Av. Ant\^onio Carlos 6627 \\
30123-970 Belo Horizonte MG, Brazil} \email{fbrocher@mat.ufmg.br}

\author{M.  Corr\^ea Jr. }
\thanks{Partially supported by CNPq grant number 300352/2012-3 and FAPEMIG grant number PPM-00169-13.}
\address{\noindent Maur\' \i cio Corr\^ea Jr\\
Departamento de Matem\'atica \\
Universidade Federal de Minas Gerais\\
Av. Ant\^onio Carlos 6627 \\
30123-970 Belo Horizonte MG, Brazil} \email{mauricio@mat.ufmg.br}

\author{A. M. Rodr\'iguez  }

\address{A. M. Rodr\'iguez  \\
Departamento de Matem\'atica \\
Universidade Federal de Minas Gerais\\
Av. Ant\^onio Carlos 6627 \\
30123-970 Belo Horizonte MG, Brazil} \email{miguel.rodriguez.mat@gmail.com}

\subjclass[2010]{Primary 32S65} \keywords{ Holomorphic foliations}

\begin{abstract}
We give a  bounding of  degree of  quasi-smooth hypersurfaces  which are invariant
by a one dimensional holomorphic foliation of a given degree on a
weighted projective space.
\end{abstract}
\maketitle

\section{Introduction}

Henri  Poincar\'e  studied in \cite{Po} the problem  to decide if a holomorphic
foliation $\F$ on the complex projective plane
$\mathbb{P}^2$ admits a rational first integral.
Poincar\'e  observed that in order to solve this problem is sufficient
to find a bound for the degree of the generic curve  invariant by $\F$. In
general, this is not possible, but doing some hypothesis we obtain
an affirmative answer for this problem, which nowadays is known as
$\emph{Poincar\'e  Problem}$. This problem was treated by D. Cerveau and A. Lins Neto \cite{C-LN}.
M. Brunella in \cite{B1} observed that obstruction  to the
positive solution to Poincar\' e problem is given by the GSV index. There exist several works about Poincar\' e problem and its generalizations, see for instance the papers:    M. Carnicer \cite{Ca}, J. V. Pereira  \cite{Pe} , M . Brunella and L.G. Mendes \cite{BM}, E. Esteves and S. Kleiman \cite{EK}.

M. Soares in  \cite{So} proved the following Theorem for smooth hypersurfaces invariant by foliations on $\mathbb{P}^ n$ .

\begin{theorem} \cite{So}
Let $\mathcal{F}$ be a holomorphic one dimensional  foliation   on
$\mathbb{P}^ n$  with
isolated singularities. If  $V \subset  \mathbb{P}^ n$ is  a
smooth hypersurface invariant by  $\mathcal{F}$, then
$$
\deg(V)\leq \deg(\mathcal{F})+1.
$$
\end{theorem}

In \cite{CS} M. Corr\^ea Jr and M. Soares studied the Poincar\'e problem for foliations on weighted projective planes.

\begin{theorem}\label{Mauricio-Marcio}
Let $\F$ be a foliation on  the weighted projective plane $\mathbb P(\omega_0,\omega_1,\omega_2)$ such that $Sing(\F)\cap
Sing(\mathbb P(\omega_0,\omega_1,\omega_2))=\emptyset$ . If  $S$ is a  quasi-smooth invariant  curve, then
$$\deg(S)\leq
\deg(\F)+\omega_0+\omega_1+\omega_2 -2.$$
\end{theorem}

In this work, we give a  bounding of  degree of  quasi-smooth hypersurfaces  which are invariant
by a one dimensional holomorphic foliation of a given degree on a
weighted projective space.
Throughout this paper, $\mathbb P(\omega)$ will denote the weighted projective space of dimension $n$
and weights $(\omega_0,\dots,\omega_n)$.

We prove the following theorem.

\begin{theorem} \label{Theopri}
Let $\mathcal{F}$ be a holomorphic one dimensional  foliation on
$\mathbb{P}(\omega)$ with isolated singularities, let $V \subset  \mathbb{P}(\omega)$ be a
quasi-smooth hypersurface invariant by $\mathcal{F}$.
\begin{enumerate}[(i)]
\item If $n=2$, then
 $$
\deg(V)\leq \deg(\mathcal{F})+\omega_0+\omega_1+\omega_2 -2;
$$
\item if
 $n\geq 3$ and $\deg(\mathcal{F}) \ge \omega_0+\cdots+\omega_n+1$, then
$$
\deg(V)< \deg(\mathcal{F})+\alpha_n(\omega_0+\cdots+\omega_n) -1,
$$
where $\alpha_n=\begin{cases}\text{the positive root of $R_n(x):=x(x+1)^n-2=0$}& \text{if $n$ is odd}\\
\alpha_{n-1}&\text{if $n$ is even}\end{cases}$
\end{enumerate}
\end{theorem}
This result say us that the  hypothesis $Sing(\F)\cap
Sing(\mathbb P(\omega))=\emptyset$ is not necessary. Therefore, the Theorem \ref{Mauricio-Marcio} is improved  in the case $n=2$ and generalized  whenever
 the condition $\deg(\mathcal{F}) \ge \omega_0+\cdots+\omega_n+1$ holds.

\begin{remark} By a direct computation,  we can calculate the first values of $\alpha_n$ with 4 decimal places:
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}\hline
$ n$&$3$&$5$&$7$&$9$&$11$&$13 $&$15$&$17$&$19$\\ \hline
$\alpha_n$&\scriptsize$0.5436$&\scriptsize$0.3880$&\scriptsize$0.3069$&\scriptsize$0.2563$&\scriptsize$0.2214$&\scriptsize$0.1957$&\scriptsize$0.1759$&\scriptsize$0.1601$&\scriptsize$0.1471$
\\ \hline
\end{tabular}
\end{center}

In addition, since
$$
R_n\left(\frac{\ln n -\ln\ln n}{n}\right)< \frac{\ln n-\ln\ln n}{n} \exp(\ln n-\ln\ln n)-2=\frac {\ln n-\ln\ln n}{\ln n}-2 <-1
$$
and for any constant $\epsilon>0$, and all $n\gg 0$
\begin{align*}
R_n\left(\frac{\ln n -(1-\epsilon)\ln\ln n}{n}\right)&\approx \frac{\ln n-(1-\epsilon)\ln\ln n}{n} \exp(\ln n-(1-\epsilon)\ln\ln n)-2\\
&=(\ln n)^\epsilon\left(1-\frac {\ln n-\ln\ln n}{\ln n}\right)-2 \gg 0.
\end{align*}
Using that $R_n(x)$ is an increasing function in $\R^+$, it follows that $$\frac{\ln n -\ln\ln n}{n}<\alpha_n<\frac{\ln n -(1-\epsilon)\ln\ln n}{n}\quad \text{ for all $n\gg 0$.}$$

 In general, from the fact that  $\frac 2{n+1}< \frac{\ln n -\ln\ln n}{n}$ for all $n\ge 21$, we have that $\max\left\{\frac {2}{n+1}, \frac{\ln n -\ln\ln n}{n}\right\}<\alpha_n<\frac {\ln 2n}n$ for all $n\ge 3$.
\end{remark}

Let us give a family of examples of  holomorphic foliations satisfying the conditions of Theorem \ref{Theopri}:

Let $a_0,b_0,a_1,\dots, a_n,b_n$ be positive integers, without common factor in pairs and such that
 $$\xi:= a_{0}+b_{0} =\dots= a_{n}+b_{n}
. $$
and consider the  well formed   weighted projective space   $\mathbb{P}^{2n+1}(a_{0},b_{0},\dots,a_{n},b_{n})$.

Let $\F$ be the  holomorphic foliation  on  $\mathbb{P}^{2n+1}(a_{0},b_{0},\dots,a_{n},b_{n})$, induced by the quasi-homogeneous vector field
$$Z=\sum_{k=0}^{n}\Big(\beta_{k}Y_{k}^{\beta_{k}-1}\frac{\partial}{\partial X_{k}}-\alpha_{k}X_{k}^{\alpha_{k}-1}\frac{\partial}{\partial Y_{k}}\Big),$$
where the $\alpha_{k}, \beta_{k} \in \mathbb{N}$ satisfy the following relation
$$\zeta=a_{k}\alpha_{k}=b_{k}\beta_{k}\qquad \text{for all}\  k=0,\dots,n.$$

A quasi-smooth hypersurface on $ \mathbb{P}^{2n+1}(a_{0},b_{0},\dots,a_{n},b_{n})$ of degree $\zeta$ given by
$$
V=\left\{\sum_{k=0}^{n}\big(X_{k}^{\alpha_{k}}+Y_{k}^{\beta_{k}}\big)=0\right\}.
$$
We can see that $V$ is invariant by $\F$  and $\deg (\F) = \zeta - \xi + 1$ . Moreover,
   since $a_i$ and $b_i$ divide $\zeta$, it follows that $\zeta\ge a_0b_0\cdots a_nb_n\gg (n+2)\xi$ and
$$\deg (\F) = \zeta - \xi + 1 \geq (n+1)\xi+1=1+\sum_{j=0}^n a_j+b_j.$$
So,  the hypothesis of Theorem \ref{Theopri} is satisfied.

Finally,
 we have
$$\deg (V)-\deg(\F)= \xi -1 =\frac 1{n+1}\Bigl(\sum_{j=0}^n a_j+b_j \Bigr)- 1<\alpha_{2n+1}\Bigl(\sum_{j=0}^n a_j+b_j \Bigr) - 1.$$
We can construct a similar foliation on even dimensional  weighted projective spaces $\mathbb{P}^{2n+2}(a_{0},b_{0},\dots,a_{n},b_{n},a_{n+1})$  where $\xi=a_{k}+b_{k}\ \text{for all }\ k=0,\dots,n.$

Let suppose that $\zeta=a_{k}\alpha_{k}=b_{k}\beta_{k}=a_{n+1}\alpha_{n+1}$ for all $\, k=0,\dots,n$ and consider
the vector field $Z$ in the previous example. Thus, the quasi-smooth hypersurface on $ \mathbb{P}^{2n+2}(a_{0},b_{0},\dots,a_{n},b_{n},a_{n+1})$ of degree $\zeta$ given by
$$
V=\left\{\sum_{k=0}^{n}\big(X_{k}^{\alpha_{k}}+Y_{k}^{\beta_{k}}\big)+X_{n+1}^{\alpha_{n+1}}=0\right\}
$$
is invariant by $Z$ and therefore we obtain the same conclusions.

\section{Weighted projective foliations}

Let $\omega_0,\dots,\omega_n$ be integers $\geq 1$. Consider the
$\mathbb{C}^*$-action on $\mathbb{C}^{n+1}\backslash \{0\}$ given by
$$
\lambda\cdot(z_0,\dots,z_n)=(\lambda^{\omega_0}
z_0,\dots,\lambda^{\omega_n} z_n),
$$
where $\lambda\in \mathbb{C}^*$ and $(z_0,\dots,z_n)\in
\mathbb{C}^{n+1}\backslash \{0\}$. The \emph{weighted projective
space of type $(\omega_0,\dots,\omega_n)$} is the quotient space $
\mathbb{P}(\omega_0,\dots,\omega_n)=(\mathbb{C}^{n+1}\setminus
\{0\}/\sim) $, induced by the action above . We will abbreviate
$\mathbb{P}(\omega_0,\dots,\omega_n):=\mathbb{P}(\omega)$.

Consider the open $\mathcal{U}_i=\{[z_0:\dots:z_n]\in
\mathbb{P}(\omega_0,\dots,\omega_n);\ z_i\neq0 \}\subset
\mathbb{P}(\omega_0,\dots,\omega_n)$, with $i=0,1,\dots,n.$ Let
$\mu_{\omega_i}\subset \mathbb{C}^*$ be the subgroup of
$\omega_i$-th roots of unity. We can define the homeomorphisms
$\phi_i:\mathcal{U}_i\longrightarrow \mathbb{C}^{n}/\mu_{\omega_i}$,
by
$$
\phi_i([z_0:\dots:z_n])=\left(\frac{z_0}{z_i^{\omega_0/\omega_i,}},
\dots,\frac{\widehat{z_i}}{z_i},\dots,\frac{z_n}{z_i^{\omega_n/\omega_i,}}
\right)_{\omega_i},
$$
where the symbol $``\;\widehat{\;}\;"$ means omission and
$(\cdot)_{\omega_i}$ is a $\omega_i$-conjugacy class in
$\mathbb{C}^{n}/\mu_{\omega_i}$ with $\mu_{\omega_i}$ acting on
$\mathbb{C}^{n}$ by
$$
\lambda\cdot(z_0,\dots, \hat z_i,\dots,z_n)=(\lambda^{\omega_0} z_0,\dots
\widehat{z_i},\dots,\lambda^{\omega_n} z_n), \lambda \in
\mu_{\omega_i}.
$$
On $\phi_i(\mathcal{U}_i\cap\mathcal{U}_j)\subset
\mathbb{C}^{n}/\mu_{\omega_i}$ we have the transitions maps  $(j<i$)
$$
\phi_i\circ
\phi_j^{-1}((z_0,\dots,\hat z_i, \dots, z_n)_{\omega_i})=\left(\frac{z_0}{z_j^{\omega_0/\omega_j,}},\dots,\frac{\widehat{z_j}}{z_j},
\dots,\frac{1}{z_j^{\omega_i/\omega_j}},\dots,\frac{z_n}{z_j^{\omega_n/\omega_j}}
\right)_{\omega_j}.
$$

\subsection{Line bundles on
$\mathbb{P}(\omega)$ and quasi-homogeneous hypersurface}

Let $d \in \mathbb{Z}$. Consider the $\mathbb{C}^*$-action
$\zeta_{d}$ on $\mathbb{C}^{n+1}\backslash
\{0\}\times \mathbb{C}$ given by
$$
\begin{array}{ccc}
   \zeta_{d}:\mathbb{C}^*\times\mathbb{C}^{n+1}\backslash
   \{0\}\times \mathbb{C}& \longrightarrow & \mathbb{C}^{n+1}\backslash \{0\}\times \mathbb{C} \\
  (\lambda,(z_0,\dots,z_n),t) & \longmapsto & ((\lambda^{\omega_0}
z_0,\dots,\lambda^{\omega_n} z_n),\lambda^{d}t).
\end{array}
$$
We denote  the quotient space induced by the action
$\zeta_{d}$ by
$$\mathcal{O}_{\mathbb{P}(\omega)}(d):=(\mathbb{C}^{n+1}\backslash
\{0\}\times \mathbb{C})/\sim \zeta_{d}.$$ The
space $\mathcal{O}_{\mathbb{P}(\omega)}(d)$ is a line orbibundle
on $\mathbb{P}(\omega)$. It is possible to show that the
Picard group of $\mathbb{P}(\omega)$ is generated by
$\mathcal{O}_{\mathbb{P}(\omega)}(1)$, i.e.
$$
Pic(\mathbb{P}(\omega)) :=\mathbb{Z}\cdot
 \mathcal{O}_{\mathbb{P}(\omega)}(1).
$$
As usual we set $\mathcal{O}_{\mathbb{P}(\omega)}(1)^{\otimes d}:=\mathcal{O}_{\mathbb{P}(\omega)}(d)$ for $d \in \mathbb{Z}$. Moreover, we have the identification $(d\geq 0)$
$$
\mathrm{H}^0(\mathbb{P}(\omega),\mathcal{O}_{\mathbb{P}(\omega)}(d))=
\bigoplus_{\omega_0k_0+\cdots+\omega_nk_n=d}\mathbb{C}\cdot(z_0^{k_1}\cdots
z_n^{k_n}).
$$

Thus, the global sections of
$\mathcal{O}_{\mathbb{P}(\omega)}(d)$ can be identify, in
homogeneous coordinates, with quasi-homogeneous polynomials of
degree equal to $d$.

Thus, a quasi-homogeneous hypersurface  $V$ on $\mathbb{P}(\omega)$,
of  quasi-homogeneity  degree $d_0$, is given by $V=\{f=0\}$, where
$f\in
\mathrm{H}^0(\mathbb{P}(\omega),\mathcal{O}_{\mathbb{P}(\omega)}(d_0))$.
We say that  $V=\{f=0\}$ is quasi-smooth if its tangent cone
$\{f=0\}$ on $\mathbb{C}^{n+1}-\{0\}$ is smooth.

\subsection{Foliations on $\mathbb{P}(\omega)$ and quasi-homogeneous vector fields}

A singular  one dimensional holomorphic foliation on $\mathbb{P}(\omega)$, of degree $d$, is given by an element of $\mathbb{P}\mathrm{H}^0(\mathbb{P}(\omega),T\mathbb{P}(\omega)\otimes\mathcal{O}_{\omega}(d-1))$.

On  $\mathbb{P}(\omega)$ we have an Euler sequence
$$
0\longrightarrow
\mathcal{O}_{\mathbb{P}(\omega)} \stackrel{\varsigma}{\longrightarrow}\bigoplus_{i=0}^{n
}\mathcal{O}_{\mathbb{P}(\omega)}(\omega_i) \longrightarrow
T\mathbb{P}(\omega) \longrightarrow 0,
$$
where $\mathcal{O}_{\mathbb{P}(\omega)}$ is the trivial line orbibundle  and  $T\mathbb{P}(\omega) = \mathrm{Hom} (\Omega_{\mathbb{P}(\omega)}^ 1,  \mathcal{O}_{\mathbb{P}(\omega)} )$ is the tangent  orbibundle of
$\mathbb{P}(\omega)$. The map $\varsigma$ is given explicitly by $
 \varsigma(1)=(\omega_0z_0,\dots,\omega_nz_n)$ (see \cite{So}).

Now,  let $X$ be a quasi-homogeneous  vector field of type $(\omega_0,
\dots, \omega_n)$ and degree $d$ on $\C^{n+1}$, i.e.
$X =\sum\limits_{i=0}^n P_i (z) \frac{\partial}{\partial z_i}$
where each polynomial $P_i$ satisfies the ``weight-homogeneous'' relation
$$P_i (\lambda^{\omega_0}z_0, \dots, \lambda^{\omega_n}z_n) = \lambda^{d
+ \omega_i - 1}P_i (z_0,\dots, z_n), \ \ \forall i=1,\dots,n.$$
These
vector fields descend well to $\mathbb{P}(\omega)$. In fact,
tensoring the Euler sequence by $\OO_{\p_\omega}(d-1)$, we obtain
$$ 0 \longrightarrow
\OO_{\mathbb{P}(\omega)}(d-1) \longrightarrow
\bigoplus\limits_{i=0}^n \OO_{\mathbb{P}(\omega)}(d + \omega_i - 1)
\longrightarrow T \mathbb{P}(\omega) \otimes
\OO_{\mathbb{P}(\omega)}(d-1) \longrightarrow 0.
$$
It follows that a quasi-homogeneous vector field $X$ induces a
foliation $\mathcal{F}$ of $\mathbb{P}(\omega)$ and that $g\,R_\omega +
X$ define the same foliation as $X$, where $R_\omega$ is the adapted
radial vector field $R_\omega = \omega_0 z_0 \frac{\partial}{\partial z_0}
+\cdots+   + \omega_n z_n \frac{\partial}{\partial z_2}$, with $g$ a
quasi-homogeneous polynomial of  type $(\omega_0,\dots, \omega_n)$  and degree
$d-1$. Therefore,   a quasi-homogeneous  vector field of type $(\omega_0,
\dots, \omega_n)$ and degree $d$ on $\C^{n+1}$  induces a holomorphic foliation
on  $\mathbb{P}(\omega)$ given by a global section of
$\mathrm{H}^0(\mathbb{P}(\omega),T\mathbb{P}(\omega)\otimes\mathcal{O}_{\omega}(d-1)).$

We have the following condition on the degree of a foliation
$$
d>1-\max_{0 \leq i<j \leq n}\{\omega_{i}+\omega_{j} \}.
$$
In fact,  by  Bott's Formulae for weighted projective spaces
(see \cite{D}), we have that
\begin{center}$
\mathrm{H}^0(\mathbb{P}(\omega),T\mathbb{P}(\omega)\otimes\mathcal{O}_{\omega}(d-1))\simeq
\mathrm{H}^0(\mathbb{P}(\omega),
\Omega^{n-1}_{\mathbb{P}(\omega)}(\sum_{i=0}^{n} \omega_{i}+d-1))\neq
\emptyset$,
\end{center}
if and only if, $d-1>-\max\limits_{0 \leq i\neq j \leq n}\{\omega_{i}+\omega_{j} \}$.

An algebraic hypersurface $V\subset \mathbb{P}(\omega)$ in invariant by a foliation $\F$ if $T_p\F_p \subset T_pV$ for all $p\in V\backslash Sing(V)\cup Sing(\F)$.
\subsection{Orbifold Milnor numbers and Baum-Bott formula}

\begin{definition}\label{df14:definition 14}\rm
Let  $M$ be a complex orbifold and $\F$ a singular holomorphic
foliation on $M$. Let  $p \in M$ be and
$(\widetilde{U},G_{p},\varphi)$ an orbifold chart  $U$ of $p$, the
\emph{orbifold Minor number} of $\F$ on $p$ is the  rational number
$$\mu_{p}^{orb}(\F)=\frac{\mu_{\widetilde{p}}(\,\widetilde{\xi}\,)}{|G_p|},$$
where  $\mu_{\widetilde{p}}(\,\widetilde{\xi}\,)$ is the milnor
number of the local lift $\widetilde{\xi}$ on $\widetilde{p}$ of a
the vector field $\xi$ tangent to $\F$ on $\widetilde{U}/G_p.$
\end{definition}

We will use the following Baum-Bott theorem for orbifolds due to M.
Corr\^ea,  A. M. Rodr\'iguez, M. G. Soares \cite{CPM}.
\begin{theorem}\label{teor7}
Let  $M$ be a compact  complex orbifold, of dimension $n$, and $\F$
a singular holomorphic foliation on $M$ induced by a global section
of $TM\otimes L$, with isolated singularities. Then
$$\int_M^{orb} c_n(TM\otimes
L)=\sum_{p\in Sing(\mathcal{F})}\mu_{p}^{orb}(\F).$$

\end{theorem}

On Weighted projective spaces and quasi-homogeneous and quasi-smooth hypersurfaces we have the following.
\begin{corollary} \label{sing(F)}\cite{CPM} Let $\F$ be a foliation of degree $d$ on
$\mathbb{P}(\omega)$. Then
$$\sum_{p\in Sing(\mathcal{F})} \mu_{p}(\F)^{orb}=\frac{1}{\omega_{0}\cdot\cdot\cdot \omega_{n}}\sum_{j=0}^{n}(d-1)^{n-j}\sigma_{j}(\omega),$$
where $\sigma_k(\omega)$ denotes the $k$-th
elementary symmetric function.
\end{corollary}

\begin{corollary}\label{sing(V)} Let $V$ be a quasi-homogeneous and quasi-smooth hypersurface, of degree
$d_{0}$, invariant by a holomorphic foliations $\F$ of degree $d.$
Then
$$\sum_{p\in Sing(\mathcal{F})\cap V} \mu_{p}(\F)^{orb}=\frac{1}{\omega_{0}\cdots \omega_{n}}\sum_{j=0}^{n-1}\left[\sum_{k=0}^{j}(-1)^{k}\sigma_{j-k}(\omega)d_{0}^{\,k+1}\right](d-1)^{n-1-j},$$
where $\sigma_\ell(\omega)$ denotes the $\ell$-th
elementary symmetric function.
\end{corollary}
\begin{proof} It follows from theorem \ref{teor7} that
$$\sum_{p\in Sing(\mathcal{F})\cap V}
\mu_{p}(\F)^{orb}=\int_{V}^{orb} C_{n-1}(TV\otimes\mathcal{O}_{\omega}(d-1)|_{V}).$$
 In order to calculate this integral, consider the Euler sequence

$$0 \longrightarrow TV \longrightarrow T\mathbb{P}^{n}_{\omega}|_{V} \longrightarrow N_{V} \longrightarrow 0\, ,$$
where $N_V$ is the normal bundle. Then
$$C(\mathbb{P}^{n}_{\omega})=C(V)C(N_V)$$
and
\begin{eqnarray}\label{eq:1.10}
C_{j}(V)=C_{j}(\mathbb{P}^{n}_{\omega})-C_{j-1}(V)C_{1}(N_V)\,,\,\, 1
\leq j \leq n-1.
\end{eqnarray}
Moreover, by using the Euler formula  we get
\begin{eqnarray}\label{eq:1.11}
C_{j}(\mathbb{P}^{n}_{\omega})=P_{j}(\omega)C_{1}(\mathcal{O}_{\omega}(1))^j.
\end{eqnarray}
On the other hand, since
$$N_{V}=\mathcal{O}_{\mathbb{P}^{n}_{\omega}}(V)|_{V}=\mathcal{O}_{\omega}(d_0)|_{V},$$
we have that
\begin{eqnarray}\label{eq:1.12}
C_{1}(N_V)=d_{0}C_{1}(\mathcal{O}_{\omega}(1)|_{V}),
\end{eqnarray}
and replacing (\ref{eq:1.11}) and  (\ref{eq:1.12}) in (\ref{eq:1.10}), we obtain
$$C_{j}(V)=\left(\sum_{k=0}^{j}(-1)^{k}\sigma_{j}(\omega)d_{0}^{\,k}\right)\mathcal{O}_{\omega}(1)^j\,,\,\, 0 \leq j \leq n-1.$$
Therefore
\begin{align}
 &\ \hspace{-1.5cm} C_{n-1}(TV\otimes\mathcal{O}_{\omega}(d-1)|_{V}) \nonumber\\

&=\sum_{j=0}^{n-1}C_{n-1-j}(V)C_{1}(\mathcal{O}_{\omega}(d-1))^{j}\nonumber\\
&= \sum_{j=0}^{n-1}C_{n-1-j}(V)C_{1}(\mathcal{O}_{\omega}(1))^j(d-1)^{j}\nonumber\\

&=
\sum_{j=0}^{n-1}\left[\left(\sum_{k=0}^{j}(-1)^{k}\sigma_{j-k}(\omega)d_{0}^{\,k}\right)C_{1}(\mathcal{O}_{\omega}(1))^{n-1}\right](d-1)^{n-1-j}.\label{eq:13}
\end{align}
By Satake-Poincar\'e duality, since  $C_{1}(\mathcal{O}_{\omega}(d_0))$ is the  Satake-Poincar\'e dual of $V$, we obtain
\begin{align*}
\int_{V}^{orb} C_{1}(\mathcal{O}_{\omega}(1))^{n-1}
&=\int_{\mathbb{P}^{n}_{\omega}}^{orb} C_{1}(\mathcal{O}_{\omega}(1))^{n-1}\wedge C_{1}(\mathcal{O}_{\omega}(d_0))\\
&=d_{0}\int_{\mathbb{P}^{n}_{\omega}}^{orb} C_{1}(\mathcal{O}_{\omega}(1))^{n}=\frac{d_0}{\omega_{0}\cdots \omega_{n}}.\\
\end{align*}
The last integral is calculated in \cite{M} and \cite{CPM}. Finally, the result follows from  the equation (\ref{eq:13}), integrating on  the variety  $V$.
\end{proof}

\section{Proof of Theorem}

Denote by $d_0=\deg(V)$ and $d=\deg(\F)$. It follows from Corollary \ref{sing(V)} that
\begin{equation}\label{a}
\sum_{j=0}^{n-1}\left[\sum_{k=0}^{j}(-1)^{k}\sigma_{j-k}(w)d_{0}^{\,k+1}\right] (d-1)^{n-1-j} = (w_{0}\cdots w_{n})\sum_{p\,\in V /\, \xi|_{V}(p)=0}\mathcal{I}_{p}(\xi|_{V}) \geq 0.
\end{equation}

Also, by  Corollary  \ref{sing(F)}, we have
\begin{equation}\label{b}
\sum_{k=0}^{n}(d-1)^{n-k}\sigma_{k}(w)-\sum_{j=0}^{n-1}\left[\sum_{k=0}^{j}(-1)^{k}\sigma_{j-k}(w)\,d_{0}^{\,k+1}\right] (d-1)^{n-1-j}
\end{equation}
\begin{equation*}
= (w_{0}\cdots w_{n})\left(\sum_{p\,\in \mathbb{P}^{n}_{w} /\, \xi(p)=0}\mathcal{I}_{p}(\xi)
-\sum_{p\,\in V /\, \xi|_{V}(p)=0}\mathcal{I}_{p}(\xi|_{V})\right)\geq 0.
\end{equation*}\\

Firstly, we prove for  $n=2$. It follows from equation (\ref{a}) that
$$-d_{0}^{2}+(\sigma_{1}(w)+d-1)d_0=(w_{0}w_{1}w_{2})\sum_{p\,\in V /\, \xi|_{V}(p)=0}\mathcal{I}_{p}(\xi|_{V})\,\geq 0.$$
Then $0<d_{0}\leq \sigma_{1}(w)+d-1$. We have that $V \cap Sing(\mathcal{F}) \neq \varnothing$.  In fact, suppose that $V \cap Sing(\mathcal{F}) =\varnothing$. In this case $V$
is a leaf of the foliation and we have the isomorphism of normal orbibundles $\mathcal{N}_{\F}|_V\simeq N_{V| \mathbb{P}^{2}_{\omega}}=\mathcal{O}_{\omega}(d_0)|_V$. Since the curve $V$ is a orbifold and $V \cap Sing(\mathcal{F}) =\varnothing$, it follows from \cite{CPM} that
$\deg(\mathcal{N}_{\F}|_V)=0$.  But this is absurd since
$$\deg(\mathcal{N}_{\F}|_V)=\deg(\mathcal{O}_{\omega}(d_0)|_V)= [\mathcal{O}_{\omega}(d_0)]\cdot V= [\mathcal{O}_{\omega}(d_0)]\cdot [\mathcal{O}_{\omega}(d_0)]= d_0^2\neq 0.$$
Observe that $V \cap Sing(\mathcal{F}) \neq \varnothing$ implied that $d_{0} \neq \sigma_{1}(w)+d-1$,
thus  $d_{0} \leq \sigma_{1}(w)+d-2$ and this proves the Theorem for $n=2.$\\

For $n\geq 3$, in order to prove that $d_0\leq d-1+\alpha_n\sigma_1(\omega),$
consider  the polynomial  $\Psi(t)\in\mathbb{Z}[t]$ defined by
$$\Psi(t)=\sum_{j=0}^{n-1}\left(\sum_{k=0}^{j}(-1)^{k}\sigma_{j-k}(\omega)\,t^{\,k+1} \right)(d-1)^{n-1-j},$$
and define $\Omega_n(t)$ as
$$
\Omega_{n}(t)= \begin{cases}
                 \frac{d}{d t}\left(\frac{\Psi(t)}{t}\right), &  \mbox{if $n$ is even}\\
             \\  \frac{d}{d t}\,\Psi(t)   ,                   & \mbox{if $n$ is odd}. \\
             \end{cases}
$$
We claim that $\Omega_n(t)\ne 0$ for all $t\ge 0$ and $d\gg 0$. In order  to prove that, for each $m$ positive integer, we define $$P_m(t)=\sum_{j=0}^m (-1)^j (m+1-j)t^{m-j}\quad\text{and}\quad
Q_m(t)=P_m(t)-P_{m-1}(t)
$$
where  $P_0(t):=1$.

\begin{lemma}\label{Q_positivo}  Let $m$ be a positive integer. If $m$ is even then the polynomial
$Q_{m}(t)$
 is positive for all  $t\ge 0$.
\end{lemma}

\proof
The result is equivalent to prove that the polynomial
$$F(t)=(t+1)^2Q_m(t)=(m+1)t^{m+2}+2t^{m+1}-(m+1)t^m+2$$
is positive for all $t\ge 0$. Observe that $F(0)=2$ and $F(t)\ge 4$ for all $t\ge 1$, then it is enough to prove that $F(t)$ is positive in $[0,1]$.
Since
$$F'(t)=(m+1)t^{m-1}((m+2)t^2+2t-m),$$
the  critical points of $Q(t)$ are $t=-1$, $t=0$ and $t=\dfrac m{m+2}$. In addition
\begin{align*}
F\left(\frac m{m+2}\right)&=\left(\frac m{m+2}\right)^m\left((m+1)\left(\frac m{m+2}\right)^2+2\left(\frac m{m+2}\right)-(m+1)\right)+2\\
&=\left(\frac m{m+2}\right)^m \left(\frac {-m^2-4m-4}{(m+2)^2}\right)+2\\
&>\left(\frac {-m^2-4m-4}{(m+2)^2}\right)+2=\frac {4(m+1)}{(m+2)^2}>0.
\end{align*}
Therefore
$Q_m(t)\ge \dfrac {4(m+1)}{(m+2)^2(t+1)^2}>0$ for all $t\ge 0$.

\begin{lemma}\label{desig_simetricas}  For all  $k\ge 1$ we have
$\dfrac {\sigma_{k+1}(\omega)}{\sigma_k(\omega)}< \dfrac {\sigma_1(\omega)}{k+1}.$
\end{lemma}
\proof Let  $\check\omega_I$ be  the tuple
$(\omega_1,\dots,\omega_n)$ omitting the coordinates $\omega_i$'s
with $i\in I$. Observe that

\begin{align*}
\sigma_k(\omega)\sigma_1(\omega)&=\sum_{i=1}^n \omega_i\sigma_k(\omega)=\sum_{i=1}^n \omega_i\left( \omega_i \sigma_{k-1}(\check \omega_i)+\sigma_k(\check\omega_i)\right)\\
&=\sum_{i=1}^n  \omega_i^2 \sigma_{k-1}(\check \omega_i)+\sum_{i=1}^n  \omega_i\sigma_k(\check \omega_i))\\
&=\sum_{i=1}^n  \omega_i^2 \sigma_{k-1}(\check\omega_i)+(k+1) \sigma_{k+1}(\omega)).\\
\end{align*}

 Thus, we get that
$$\frac {\sigma_{k+1}(\omega)}{\sigma_k(\omega)}= \frac {\sigma_1(\omega)}{k+1}- \frac {\sum\limits_{i=1}^n  \omega_i^2 \sigma_{k-1}(\check \omega_i)}{(k+1)\sigma_k(\omega)}
<   \frac {\sigma_1(\omega)}{k+1}.\qed$$

\begin{proposition}\label{positividade}
Suppose that $d\ge \sigma_1(\omega)+1$, then the polynomial
$(-1)^{n-1}\Omega_n(t)$ is positive for all $t\ge 0$.
\end{proposition}

\proof :
First,  we consider the case when $n$ is odd. We have  that
\begin{align*}
\Omega_n(t)&=\sum_{j=0}^{n-1}\left(\sum_{k=0}^{j}(-1)^{k}\sigma_{j-k}(\omega)(k+1)t^{k} \right)(d-1)^{n-1-j}\\
&=\sum_{l=0}^{n-1} \sigma_l(\omega)\left(\sum_{k=0}^{n-1-l} (-1)^k(k+1) t^k (d-1)^{n-1-k-l}\right)\\
&=\sum_{l=0}^{n-1} \sigma_l(\omega)(d-1)^{n-1-l}\left(\sum_{k=0}^{n-1-l} (-1)^k (k+1)s^k\right)\\
&=\sum_{l=0}^{n-1} \sigma_l(\omega)(d-1)^{n-1-l}(-1)^{n-1-l}P_{n-1-l}(s)\\
\end{align*}
where $s=\frac t{d-1}$.

We can write
$\Omega_n(t)-\sigma_{n-1}(\omega)$  as
$$= \sum_{j=0}^{\frac {n-3}2}(d-1)^{n-2-2j}\left(
\sigma_{2j}(\omega)(d-1)P_{n-1-2j}(s)-
\sigma_{2j+1}P_{n-2-2j}(s)\right).
$$
On the other hand, by Lemma \ref{desig_simetricas} and the
hypothesis,  we know that $$\sigma_{2l}(\omega)(d-1)\ge
\sigma_{2l}(\omega)\sigma_1(\omega)\ge \sigma_{2l+1}(\omega).$$ It
follows that
\begin{align*}
\sigma_{2j}(\omega)(d-1)P_{n-1-2j}(s)& -\sigma_{2j+1}P_{n-2-2j}(s)\ge \\
&\ge\sigma_{2j}(\omega)(d-1)P_{n-1-2j}(s)- \sigma_{2j+1}\left|P_{n-2-2j}(s)\right|\\
&\ge\sigma_{2j}(\omega)(d-1)\left(P_{n-1-2j}(s)-\left|P_{n-2-2j}(s)\right|\right)\\
&\ge\sigma_{2j}(\omega)(d-1)\min\left\{(n-2j)s^{n-1-2j}, Q_{n-1-2j}(s)\right\}\\
&>0.
\end{align*}
This last inequality follows from Lemma \ref{Q_positivo}. Therefore
$\Omega_n(t)> \sigma_{n-1}(\omega)$ for all $t\ge 0$.

In the case $n$ even,  similarly we obtain that
\begin{align*}
\Omega_n(t)& =\sum_{l=0}^{n-1} \sigma_l(\omega)(d-1)^{n-2-l}(-1)^{n-1-l}P_{n-2-l}(s)\\
&=
\sum_{j=0}^{\frac {n-2}2}(d-1)^{n-2-2j}\left(
-\sigma_{2j}(\omega)(d-1)P_{n-2-2j}(s)+
\sigma_{2j+1}P_{n-3-2j}(s)\right),
\end{align*}
and by the  previous argument, we have that  each term of this sum is negative, therefore $\Omega_n(t)< 0$ for all $t\ge 0$.
\qed \\

Now, in order to  finish the proof of Theorem, we have two case to consider:

\subsection{$n$ odd:}
It follows from  Proposition \ref{positividade} that
$\Psi'(t)=\Omega_{n}(t)> 0$ for all $ t \in \mathbb{R}^{+}$. Then,
$\Psi$ is a increasing function and by equation (\ref{b}), $\Psi(d_0)\leq\sum\limits_{l=0}^{n}(d-1)^{n-l}\sigma_{l}(w)$.
We claim that
$\Psi\left(d-1+\alpha_n\sigma_1(\omega)\right)>
\sum\limits_{l=0}^{n}(d-1)^{n-l}\sigma_{l}(w)$. In fact,  since
\begin{align*}
\Psi(t)& =\sum_{j=0}^{n-1}\left(\sum_{k=0}^{j}(-1)^{k}\sigma_{j-k}(\omega)\,t^{k+1} \right)(d-1)^{n-1-j}\\
&=\sum_{l=0}^{n-1} \sigma_l(\omega)(d-1)^{n-l} \sum_{k=0}^{n-1-l}(-1)^k\left(\frac t{d-1}\right)^{k+1}\\
&=\frac t{d-1+t} \sum_{l=0}^{n-1} \sigma_l(\omega)(d-1)^{n-l}\left(1-\left(\frac {-t}{d-1}\right)^{n-l}\right)\\
&=\frac t{d-1+t} \sum_{l=0}^{n-1} \sigma_l(\omega)((d-1)^{n-l}-(-t)^{n-l}),
\end{align*}
then, putting  $t=d-1+\alpha_n\sigma_1(\omega)$
 and using that $\sigma_{2k+1}(\omega)<\frac 1{2k}\cdot  \sigma_{2k}(\omega)\sigma_1(\omega)$ for each $k\ge 1$, we have that
\begin{align*}
-\Psi(t)+&\sum\limits_{l=0}^{n}\sigma_{l}(w)(d-1)^{n-l}=\\
&= \sigma_n(\omega) +\frac 1{d-1+t} \sum_{l=0}^{n-1} \sigma_l(\omega)((d-1)^{n-l+1}-(-t)^{n-l+1})\\
&=\sigma_n(\omega)+\sigma_{n-1}(\omega)\frac {(d-1)^2-t^2}{d-1+t}+\\
&\hspace{1cm}+\frac 1{d-1+t} \sum_{k=0}^{\frac{n-3}2} \left( \sigma_{2k}(\omega)((d-1)^{n-2k+1}-t^{n-2k+1}) \right.\\
&\hspace{55mm}\left.+
\sigma_{2k+1}(\omega)((d-1)^{n-2k}+t^{n-2k})\right)\\

&<\frac 1{d-1+t} \sum_{k=1}^{\frac{n-3}2} \sigma_{2k}(\omega)\left( (d-1)^{n-2k}\left(d-1+\frac 1{2k}\sigma_1(\omega)\right)-t^{n-2k}\left(t-\frac 1{2k}\sigma_1(\omega)\right)\right)\\
&\hspace{4cm}
+(d-1)^{n+1}-t^{n+1}+\sigma_{1}(\omega)(d-1)^{n}+\sigma_1(\omega)t^{n}.
\end{align*}
Then, in order to conclude the proof, it is enough to show that each term of this summation is less or equal  to zero, or equivalently,  putting $s=d-1$, $\sigma=\sigma_1(\omega)$, we have to prove that
\begin{enumerate}[(I)]
\item
$\left(\dfrac t{s}\right)^{n-2k}\ge \dfrac{s+\frac 1{2k}\sigma}{t-\frac 1{2k}\sigma}$
for each $k=1,\dots, \frac {n-3}2$  and
\item
$t^n(t-\sigma)\ge s^n(s+\sigma)$.
\end{enumerate}

For item (I), the case $n=3$ is empty, then we can suppose that $n\ge 5$. By Bernoulli's inequality, we have
$$\left(\dfrac t{s}\right)^{n-2k}=\left(1+\frac{\alpha_n \sigma}{s}\right)^{n-2k}\ge 1+\frac{(n-2k)\alpha_n \sigma}{s},$$
so, it is enough to prove that right side of this inequality  is greater that
$$\dfrac{s+\frac 1{2k}\sigma}{t-\frac 1{2k}\sigma}=1+\frac {(\frac 1k -\alpha_n)\sigma}{s+(\alpha_n-\frac 1{2k})\sigma}.$$
In fact, if $n\ge 7$, then $\alpha_n\ge \frac 2n$ and
\begin{align*}
(n-2k)\alpha_n&\left(s+\left(\alpha_n-\frac 1{2k}\right)\sigma\right)-\left(\frac 1k -\alpha_n\right)s\\
&>(n-2k)\alpha_n\left(s-\frac 1{2k}\sigma\right)-\left(\frac 1k -\alpha_n\right)s\\
&>\left((n-2k)\alpha_n\frac {2k-1}{2k}-\frac 1k +\alpha_n\right)s\\
&>\left(\frac n2\alpha_n-1\right)s>0,\\
\end{align*}
and when $n=5$, it follows that $k=1$ and
$$(n-2k)\alpha_n\left(s+\left(\alpha_n-\frac 1{2k}\right)\sigma\right)-\left(\frac 1k -\alpha_n\right)s>\left( \frac 52\alpha_5+3\alpha_5^2\right)s>0.$$

 Finally, we are going to prove (II). Making $U:=\frac s{\sigma}>1$, observe that
\begin{align*}
t^n(t-&\sigma)- s^n(s+\sigma)\\
&=(s+\alpha_n\sigma)^n(s+(\alpha_n-1)\sigma)-s^{n+1}-\sigma s^n\\
&=\sigma^{n+1}\left((U+\alpha_n)^n(U+\alpha_n-1)-U^{n+1}-U^n\right)\\
&=\sigma^{n+1}\left(((n+1)\alpha_n-2)U^n+\sum_{j=1}^{n+1} \left(\binom nj \alpha_n^j+\binom n{j-1} \alpha_n^{j-1}(\alpha_n-1)\right) U^{n-j+1}\right)\\
&=\sigma^{n+1}\left(((n+1)\alpha_n-2)U^n+\sum_{j=0}^{n+1} \left(\frac {n+1}j \alpha_n-1\right)  \binom n{j-1}\alpha_n^{j-1}U^{n-j+1}\right)
\end{align*}
Now, the polynomial $$F(X)=((n+1)\alpha_n-2)X^n+\sum\limits_{j=0}^{n+1} \left(\frac {n+1}j \alpha_n-1\right) \binom n{j-1} \alpha_n^{j-1}X^{n-j+1}$$ and its derivate $F'(X)$   satisfy  that
\begin{itemize}
\item  the  leading  coefficient  is positive,
\item the list of other coefficients is a decreasing sequence, and then it   only has one change of sign,
\item $F(0)<0$ and $F'(0)<0$,
\item $F(1)=(1+\alpha_n)^n\alpha_n-2=0$.
\end{itemize}
 Then, by Descartes' rule of signs, $F'(X)$ only has one positive root and that root  is in the interval $(0,1)$, thus $F(X)$ is an increasing function in $(1,\infty)$. Therefore
$$t^n(t-\sigma)- s^n(s+\sigma)=\sigma^{n+1}F(U)\ge \sigma^{n+1}F(1)=0,$$
as we want to prove. \qed

\subsection{$n$ even} By the equation (\ref{a}) we know that
$\Psi(d_0) \geq 0.$
Therefore, if we  define $\Phi (t) = \frac{\Psi(t)}{t}$, then $\Phi(d_0)\ge 0$ and from  Proposition \ref{positividade} we have that  $\Phi(t)$ is a decreasing function.

We claim that $\Phi(d-1+\alpha_n\sigma_1(\omega))<0$, and therefore $d_0< d-1+\alpha_n\sigma_1(\omega)$.
In fact, following the same procedure using previously, we have that
\begin{align*}
\Phi(t)&=\frac 1{d-1+t} \sum_{l=0}^{n-1} \sigma_l(\omega)((d-1)^{n-l}-(-t)^{n-l})\\
&=\frac 1{d-1+t} \sum_{k=0}^{\frac{n-2}2} \left( \sigma_{2k}(\omega)((d-1)^{n-2k+1}-t^{n-2k+1}) \right.\\
&\hspace{50mm}\left.+
\sigma_{2k+1}(\omega)((d-1)^{n-2k}+t^{n-2k})\right)
\end{align*}
From here, the same argument of the case odd works.\qed

\vskip 0.5cm

\vskip 0.5cm

\vskip 0.5cm

config: arxiv


TITLE: How to use Concentration Inequalitites to bound this probability from above?
QUESTION [4 upvotes]: I'm attempting a problem in Rohatgi & Saleh's Intro to Probability and Statistics, from the Concentration inequalities section. It's actually the first exercise in the section, so the fact that I can't get it means I'm missing something obvious:
We have a random variable X with PDF: $f(x;\lambda) = \frac{e^{-x} x^\lambda}{\lambda !}, x>0$ for some $\lambda \geq 0$ an integer. They ask that I show that $P\{0<X<2(\lambda + 1)\} > \frac{\lambda}{\lambda + 1}$ .
I'd like to use the concentration/Markov/Chebyshev inequalities to for this but all the inequalities are in the wrong direction. Is there a nifty way I'm not thinking of to get them to mirror this result? Hints are greatly appreciated. Also please don't go easy on me if this is something obvious. I need to start being better at probability.
Also here are the concentration inequalities I've been referring to:
$$\text{If $E[h(X)]$ exists,  } \quad  P[h(X)\geq M] \leq \frac{E[h(X)]}{M} \quad \text{for any } M>0,\\
\text{If $E[|X|^k]$ exists,  } \quad  P[|X|\geq M] \leq \frac{E[|X|^k]}{M^k} \quad \text{for any } M>0,$$

REPLY [3 votes]: We begin by noticing that
$$P(0 < X < 2(\lambda +1)) = P(-(\lambda +1) < X - (\lambda + 1) < \lambda +1) = P(|X - (\lambda + 1 )| < \lambda + 1)$$
Then, being good detectives and noticing that the numerator of our pdf correlates to $\Gamma(\lambda + 1) = \lambda!$, we have
$$ E(X) = \int_{0}^{\infty} \frac{e^{-x}x^{\lambda +1}}{\lambda!} = \frac{(\lambda +1)!}{\lambda!} = \lambda + 1$$
By a similar calculation, which I leave to you, we have that $E(X^2) = (\lambda +2)(\lambda + 1)$, and thus $$\sigma^2 = (\lambda + 1)[(\lambda +2) - (\lambda + 1)] = \lambda +1$$
So, letting $k = \sigma = \sqrt{\lambda +1}$, we have by Chebyshev's inequality
$$ P(|X - (\lambda + 1 )| < \lambda + 1) = 1 - P(|X - (\lambda + 1 )| \geq \lambda + 1) > \frac{\lambda}{\lambda + 1}$$ as required.


TITLE: Degree of extension is equal to linear combination of prime factor multiplicities with prime factor index coefficients in Dedekind domains
QUESTION [2 upvotes]: I'm working on the following problem...

Suppose that $A$ is a Dedekind domain with fraction field $K$.  $L/K$ is a finite separable extension of $A$ of degree $n$, and $B$ is the integral closure of $A$ in $L$ (so that $B$ is also a Dedekind domain).  Suppose that $\mathfrak{p}$ is a nonzero prime ideal of $A$, and $B\mathfrak{p}=\mathfrak{q}^{e_1}\cdots \mathfrak{q}_r^{e_r}$ is the prime factorization of $B\mathfrak{p}$.  Show that $$e_1f_1+\cdots +e_rf_r=n$$ where $f_i=[B/\mathfrak{q}_i:A/\mathfrak{p}]$.

My first question is, why does integrality over $A$ imply that $B$ is also a Dedekind domain?
My second question is, well, how to do the problem.  I need an outline so that I know what I'm looking for. What results will be needed here?

REPLY [2 votes]: This just follows from the definition of a Dedekind domain. Since it is the integral closure in its own field of fractions, you know that $B$ is normal. It is dimension $1$ since it is an integral extension of $A$ which is dimension $1$. But, since $L/K$ is separable finite, it follows (essentially by bounding $B$ between $A$ and $\frac{1}{d_{B/A}}A$) that $B$ is a actually a finite $B$-module, and so Noetherian. So, $B$ is dimension $1$, normal, and Noetherian, and so a Dedekind domain.
Here's a rough idea. On one hand, you have that $B/\mathfrak{p}B$ is an $n$-dimensional $A/\mathfrak{p}$-space (why?). But, you can factor $\mathfrak{p}B$ as $\mathfrak{q}_1^{e_1}\cdots\mathfrak{q}_n^{e_n}$. So,
$$B/\mathfrak{p}B\cong \prod_i B/\mathfrak{q}_i^{e_i}B$$
by CRT. Then check that $|B/\mathfrak{q}_i^{e_i}B|=|B/\mathfrak{q}_i B|^{e_i}$. You can find an outline of this here (be prepared for a rabbit hole). Finally, use this to check that $\dim_{A/\mathfrak{p}}B/\mathfrak{q}_i^{e_i}B=e_i f_i$. Then, conclude.


TITLE: Let $(X,Y)$ be a Gaussian random vector with given means, variances and correlation, determine $E[X\mid X+Y]$ and $E[X\mid Y]$
QUESTION [3 upvotes]: Let $(X,Y)$ be a Gaussian random vector, with $E[X]=E[Y]=0$, $var(X)=\sigma^2,\ var(Y)=\tau^2$, and $Cor(X,Y)=\rho$ (correlation coefficient). Find $E[X\mid X+Y]$ and $E[X\mid Y]$.
My attempt for $E[X\mid X+Y]$ is to first write
$$E[X\mid X+Y]+E[Y\mid X+Y]=E[X+Y\mid X+Y]=X+Y$$
Since $(X,Y)$ is a Gaussian vector, $X+Y$ is also Gaussian. We can compute that $E[X+Y]=0$ and $var(X+Y)=\tau^2+\sigma^2+2\sigma\tau\rho$. Therefore, $X+Y\sim N(0,\tau^2+\sigma^2+2\sigma\tau\rho)$. But after this I don't know how to continue.
Another problem is that this does not seem to work for $E[X\mid Y]$.
Any help is appreciated.

REPLY [1 votes]: $X|Y$ is also gaussian random variable.This is because Gaussian Conditional Property.This holds In general for any finite dimensional gaussian random vectors. Look up for derivations because they are a bit tedious and i am just stating $2D$ result here.You can find the general result here. MVN
$$\begin{pmatrix} X_1 \\ X_2 \end{pmatrix} \sim \mathcal{N} \left( \begin{pmatrix} \mu_1 \\ \mu_2 \end{pmatrix} , \begin{pmatrix} \sigma^2_1 & \rho \sigma_1 \sigma_2 \\ \rho \sigma_1 \sigma_2 & \sigma^2_2 \end{pmatrix} \right) $$
The conditional expectation of X1 given X2 is:
$$\operatorname{E}(X_1 \mid X_2=x_2) = \mu_1 + \rho \frac{\sigma_1}{\sigma_2}(x_2 - \mu_2) =$$
In your case it reduces to following
$$\operatorname{E}[X|Y=y] = \rho \frac{\sigma}{\tau}.y$$


text:
\section{Analysis on $\boldsymbol{PW_{-}}$}
\label{08/07/03/0:40}
We shall give the proofs of Theorem \ref{08/06/12/9:48}, Theorem \ref{08/04/21/9:28}, Proposition \ref{08/11/02/22:52} and Proposition \ref{10/01/26/14:49}.
\par
We begin with the proof of Theorem \ref{08/06/12/9:48}. We employ the idea of Nawa \cite{Nawa8}.  Here, the generalized virial identity below (see (\ref{08/03/29/19:05})) plays an important role:
\[
\begin{split}
&(W_{R}, |\psi(t)|^{2})
\\
&=(W_{R},|\psi_{0}|^{2})
+2t \Im{(\vec{w}_{R} \cdot \nabla \psi_{0},\psi_{0})}+2\int_{0}^{t}\int_{0}^{t'}\mathcal{K}(\psi(t''))\,dt''dt'
\\
& \qquad
-2\int_{0}^{t}\int_{0}^{t'}\mathcal{K}^{R}(\psi(t''))\,dt''dt'
-\frac{1}{2}\int_{0}^{t}\int_{0}^{t'}(\Delta ({\rm div}\,{\vec{w}_{R}}), |\psi(t'')|^{2})\,dt''dt'
.
\end{split}
\]

\begin{proof}[Proof of Theorem \ref{08/06/12/9:48}]
Since we have already proved (\ref{09/12/23/22:48}) (see Proposition \ref{08/05/26/10:57}), proofs of (\ref{09/05/14/13:32}) and (\ref{09/05/13/15:58}) remain.  For simplicity, we consider the forward time only. The problem for the backward time can be proved in a similar way.
\par
Take any $\psi_{0} \in PW_{-}$ and let $\psi$ be the corresponding solution to our equation (\ref{08/05/13/8:50}) with $\psi(0)=\psi_{0}$. When the maximal existence time $T_{\max}^{+}<\infty$, we have (\ref{09/05/14/13:32}) as mentioned in (\ref{10/01/27/11:26}). Therefore, it suffices to prove (\ref{09/05/13/15:58}).\par
We suppose the contrary that (\ref{09/05/13/15:58}) fails when $T_{\max}^{+}=\infty$, so that there exists $R_{0}>0$ such that
\begin{equation}\label{08/06/12/9:57}
M_{0}:=\sup_{t \in [0,\infty)} \int_{|x|\ge R_{0}}|\nabla \psi(x,t)|^{2}\,dx < \infty.
\end{equation}
Then, we shall derive a contradiction in three steps.
\\
{\it Step1}. We claim that: there exists a constant $m_{0}>0$ such that
\begin{equation}\label{08/06/13/04:35}
m_{0}< \inf\left\{
\int_{|x|\ge R}|v(x)|^{2}\,dx \left|
\begin{array}{l}
v \in H^{1}(\mathbb{R}^{d}),
\quad
\mathcal{K}^{R}(v) \le
-\frac{1}{4} \varepsilon_{0},
\\[6pt]
\|\nabla v \|_{L^{2}(|x|\ge R)}^{2} \le M_{0},
\quad  \|v\|_{L^{2}}\le \|\psi_{0}\|_{L^{2}}
\end{array}
\right.
\right\}
\
\mbox{for all $R>0$},
\end{equation}
where $\varepsilon_{0}=\mathcal{B}(\psi_{0})-\mathcal{H}(\psi_{0})>0$. Let us prove this claim. Take any $v \in H^{1}(\mathbb{R}^{d})$ with the following properties:
\begin{equation}\label{10/03/01/15:45}
\mathcal{K}^{R}(v) \le
-\frac{1}{4} \varepsilon_{0},
\quad
\|\nabla v \|_{L^{2}(|x|\ge R)}^{2} \le M_{0},
\quad  \|v\|_{L^{2}}\le \|\psi_{0}\|_{L^{2}}.
\end{equation}
Note here that the second and third properties in (\ref{10/03/01/15:45}) shows that
\begin{equation}\label{10/03/01/16:22}
\|v \|_{H^{1}(|x|\ge R)}^{2}
\le
\|\psi_{0}\|_{L^{2}}^{2}+M_{0}.
\end{equation}
We also have by the first property in (\ref{10/03/01/15:45}) and (\ref{08/04/20/12:50}) that
\begin{equation}\label{08/06/14/22:02}
\frac{1}{4}\varepsilon_{0}
\le
-\mathcal{K}^{R}(v)
\le
\int_{|x|\ge R}\rho_{3}(x)|v(x)|^{p+1}\, dx.
\end{equation}

Now, we define $p_{*}$ by
\begin{equation}\label{10/05/20/17:06}
p_{*}=\left\{
\begin{array}{ccc}
2p-1 &\mbox{if}& d=1,2,
\\
2^{*}-1 &\mbox{if }& d\ge 3.
\end{array}
\right.
\end{equation}
Then, the H\"older inequality and the Sobolev embedding show that
\begin{equation}\label{08/06/14/22:18}
\begin{split}
\int_{|x|\ge R}\rho_{3}(x)|v(x)|^{p+1}\, dx
&\le
\|\rho_{3}\|_{L^{\infty}}
\|v \|_{L^{2}(|x|\ge R)}^{p+1-\frac{(p-1)(p_{*}+1)}{p_{*}-1}}
\|v \|_{L^{p_{*}+1}(|x|\ge R)}^{\frac{(p-1)(p_{*}+1)}{p_{*}-1}}
\\
&\lesssim
 \|\rho_{3}\|_{L^{\infty}}
\|v \|_{L^{2}(|x|\ge R)}^{p+1-\frac{(p-1)(p_{*}+1)}{p_{*}-1}}
\|v\|_{H^{1}(|x|\ge R)}^{\frac{(p-1)(p_{*}+1)}{p_{*}-1}},
\end{split}
\end{equation}
where the implicit constant depends only on $d$ and $p$.
This estimate, together with (\ref{10/03/01/16:22}) and (\ref{08/06/14/22:02}), yields that
\begin{equation}\label{10/03/01/16:35}
\frac{\varepsilon_{0}}
{\|\rho_{3}\|_{L^{\infty}}
\sqrt{ \left\|\psi_{0}\right\|_{L^{2}}^{2}+M_{0} }^{\frac{(p-1)(p_{*}+1)}{p_{*}-1}}
}
\lesssim
\|v\|_{L^{2}(|x|\ge R)}^{p+1-\frac{(p-1)(p_{*}+1)}{p_{*}-1}},
\end{equation}
where the implicit constant depends only on $d$ and $p$. Since $\|\rho_{3}\|_{L^{\infty}}\lesssim 1$ (see (\ref{08/12/29/12:53})), the estimate (\ref{10/03/01/16:35}) gives us the desired result (\ref{08/06/13/04:35}).
\\[12pt]
{\it Step2}. Let $m_{0}$ be a constant found in (\ref{08/06/13/04:35}). Then, we prove that
\begin{equation}\label{10/03/01/17:26}
\sup_{0\le t <\infty}\int_{|x|\ge R}|\psi(x,t)|^{2}\,dx \le m_{0}
\end{equation}
for all $R$ satisfying the following properties:
\begin{align}
\label{10/03/01/17:21}
&R>R_{0},
\\[6pt]
\label{08/03/29/20:04}
&\frac{10d^{2}K}{R^{2}}\|\psi_{0}\|_{L^{2}}^{2}< \varepsilon_{0},
\\[6pt]
\label{08/03/29/20:05}
&\int_{|x|\ge R}|\psi_{0}(x)|^{2}\,dx < m_{0},
\\[6pt]
\label{08/03/29/20:06}
&\frac{1}{R^{2}}\left( 1+ \frac{2}{\varepsilon_{0}} \left\|\nabla \psi_{0}\right\|_{L^{2}}^{2} \right)(W_{R},|\psi_{0}|^{2})<m_{0},
\end{align}
where $K$ is the constant given in (\ref{09/09/26/11:16}).
We remark that Lemma \ref{10/03/02/17:47} shows that we can take $R$ satisfying (\ref{08/03/29/20:06}).
\par
Now, for $R>0$ satisfying (\ref{10/03/01/17:21})--(\ref{08/03/29/20:06}), we put\begin{equation}
\label{10/03/03/16:49}
T_{R}=\sup
\left\{ T>0 \biggm| \sup_{0\le t <T }\int_{|x|\ge R}|\psi(x,t)|^{2}\,dx \le m_{0} \right\}.
\end{equation}
Note here that since $\psi \in C(\mathbb{R};L^{2}(\mathbb{R}^{d}))$ and $\psi(0)=\psi_{0}$, we have by (\ref{08/03/29/20:05}) that $T_{R}>0$.
It is clear that $T_{R}=\infty$ shows (\ref{10/03/01/17:26}).
\par
We suppose the contrary that  $T_{R}<\infty$. Then, it follows from  $\psi \in C(\mathbb{R};L^{2}(\mathbb{R}^{d}))$ that
\begin{equation}\label{10/03/03/17:19}
\int_{|x|\ge R}|\psi(x,T_{R})|^{2}\,dx =m_{0}.
\end{equation}
Hence, the definition of $m_{0}$ (see (\ref{08/06/13/04:35})), together with (\ref{08/06/12/9:57}), (\ref{10/03/01/17:21}) and the mass conservation law (\ref{08/05/13/8:59}),
 leads us to that
\begin{equation}\label{10/03/01/19:19}
-\frac{1}{4}\varepsilon_{0}\le \mathcal{K}^{R}(\psi(T_{R})).
\end{equation}
Moreover, applying this inequality (\ref{10/03/01/19:19}) and (\ref{09/12/23/22:48}) (or (\ref{09/06/21/18:52})) to the generalized virial identity (\ref{08/03/29/19:05}), we obtain that
\begin{equation}\label{10/03/02/15:40}
\begin{split}
(W_{R}, |\psi(T_{R})|^{2})
&< (W_{R},|\psi_{0}|^{2})
+2T_{R} \Im{(\vec{w}_{R} \cdot \nabla \psi_{0},\psi_{0})}
-\varepsilon_{0}^{2}T_{R}^{2}
\\[6pt]
&\qquad
+\frac{1}{4}\varepsilon_{0}T_{R}^{2}
-\frac{1}{2}\int_{0}^{T_{R}}\int_{0}^{t'}(\Delta ({\rm div}\,{\vec{w}_{R}}), |\psi(t'')|^{2})\,dt''dt'
.
\end{split}
\end{equation}
Here, the estimate (\ref{08/04/20/16:54}), the mass conservation law (\ref{08/05/13/8:59}) and (\ref{08/03/29/20:04}) show that the last term on the right-hand side above is estimated as follows:
\begin{equation}\label{10/03/02/16:08}
\begin{split}
-\frac{1}{2}\int_{0}^{T_{R}}\int_{0}^{t'}(\Delta ({\rm div}\,{\vec{w}_{R}}), |\psi(t'')|^{2})\,dt''dt'
&\le \frac{1}{2}\int_{0}^{T_{R}}\int_{0}^{t'}\frac{10d^{2}K}{R^{2}} \left\|\psi(t'') \right\|_{L^{2}}^{2}\,dt''dt'
\\[6pt]
&\le
\frac{1}{4}\varepsilon_{0}T_{R}^{2}.
\end{split}
\end{equation}
Therefore, we have
\begin{equation}\label{09/09/24/13:05}
\begin{split}
(W_{R}, |\psi(T_{R})|^{2})
&< (W_{R},|\psi_{0}|^{2})
+2T_{R} \Im{(\vec{w}_{R} \cdot \nabla \psi_{0},\psi_{0})}-\frac{1}{2}T_{R}^{2}\varepsilon_{0}.
\\[6pt]
&=(W_{R},|\psi_{0}|^{2}) -\frac{1}{2}\varepsilon_{0}
\left\{ T_{R} -\frac{2}{\varepsilon_{0}}\Im{(\vec{w}_{R} \cdot \nabla \psi_{0},\psi_{0})} \right\}^{2}
+ \frac{2}{\varepsilon_{0}}\left| (\vec{w}_{R} \cdot \nabla \psi_{0},\psi_{0}) \right|^{2}
\\[6pt]
&\le (W_{R},|\psi_{0}|^{2}) + \frac{2}{\varepsilon_{0}}\left| (\vec{w}_{R} \cdot \nabla \psi_{0},\psi_{0}) \right|^{2}
\\[6pt]
&\le
(W_{R},|\psi_{0}|^{2}) + \frac{2}{\varepsilon_{0}}
\left\| \nabla \psi_{0} \right\|_{L^{2}}^{2}
\int_{\mathbb{R}^{d}} W_{R}(x)|\psi_{0}(x)|^{2}\,dx,
\end{split}
\end{equation}
where we have used the Schwarz inequality and (\ref{09/09/25/13:24})
to derive the last inequality. Combining this inequality (\ref{09/09/24/13:05}) with (\ref{08/03/29/20:06}), we see that
\begin{equation}\label{10/03/02/16:28}
(W_{R}, |\psi(T_{R})|^{2}) \le
\left( 1+ \frac{2}{\varepsilon_{0}}
 \left\|\nabla \psi_{0}\right\|_{L^{2}}^{2} \right)
 (W_{R},|\psi_{0}|^{2})<R^{2}m_{0}.
\end{equation}
On the other hand, since $W_{R}(x) \ge R^{2}$ for $|x|\ge R$ (see (\ref{10/02/27/22:19})), we have
\begin{equation}\label{09/09/24/16:30}
\int_{|x|\ge R } |\psi(x,T_{R})|^{2} \,dx
=
\frac{1}{R^{2}}\int_{|x|\ge R} R^{2} |\psi(x,T_{R})|^{2} \,dx
\le \frac{1}{R^{2}} (W_{R}, |\psi(T_{R})|^{2}).
\end{equation}
Thus, it follows form (\ref{10/03/02/16:28}) and (\ref{09/09/24/16:30}) that
\begin{equation}\label{10/03/06/23:56}
\int_{|x|\ge R } |\psi(x,T_{R})|^{2} \,dx  <m_{0},
\end{equation}
which contradicts (\ref{10/03/03/17:19}), so that  $T_{R}=\infty$ and (\ref{10/03/01/17:26}) hold.
\\
{\it Step3}. We complete the proof of Theorem \ref{08/06/12/9:48}. The definition of $m_{0}$, together with the mass conservation law (\ref{08/05/13/8:59}), (\ref{08/06/12/9:57}) and (\ref{10/03/01/17:26}), shows that
\begin{equation}\label{10/03/03/17:48}
-\frac{1}{4}\varepsilon_{0}\le \mathcal{K}^{R}(\psi(t))
\qquad
\mbox{for all $R>0$ satisfying (\ref{10/03/01/17:21})--(\ref{08/03/29/20:06}), and all $t\ge 0$}.
\end{equation}
Applying this estimate (\ref{10/03/03/17:48}) to the generalized virial identity (\ref{08/03/29/19:05}), we obtain the following estimate as well as {\it Step2} (see (\ref{10/03/02/15:40}) and (\ref{10/03/02/16:08})):
\begin{equation}\label{10/03/03/17:56}
(W_{R}, |\psi(t)|^{2})
\le (W_{R},|\psi_{0}|^{2})
+2t \Im{(\vec{w}_{R} \cdot \nabla \psi_{0},\psi_{0})}-\frac{1}{2}t^{2}\varepsilon_{0}
\quad
\mbox{for all $t\ge 0$}.
\end{equation}
This inequality means that $(W_{R}, |\psi(t)|^{2})$ becomes negative in a finite time, so that $T_{\max}^{+}$ must be finite. However, this contradicts $T_{\max}^{+}=\infty$. Hence, (\ref{08/06/12/9:57}) derives an absurd conclusion: Thus,  (\ref{09/05/13/15:58}) holds.
\end{proof}

Next, we give the proof of Theorem \ref{08/04/21/9:28}.
\begin{proof}[Proof of Theorem \ref{08/04/21/9:28}]
Let $\psi_{0}$ be a radially symmetric function in $PW_{-}$, and let $\psi$ be the solution to (\ref{08/05/13/8:50}) with $\psi(0)=\psi_{0}$.
\par
To handle the term containing $\mathcal{K}^{R}$ in the generalized virial identity (\ref{08/03/29/19:05}) for $\psi$,  we consider the following variational problem, as well as Lemma 3.3 in \cite{Nawa8}:
\begin{lemma}
\label{08/04/20/18:30}
Assume that $d\ge 3$, $2+\frac{4}{d}< p+1 < 2^{*}$, and $p\le 5$ if $d=2$. Then, there exists $R_{*}>0$ and $m_{*}>0$ such that
\[
m_{*} < \inf\left\{  \int_{|x|\ge R}|v(x)|^{2}\,dx \left|
\begin{array}{l}
v \in H^{1}_{rad}(\mathbb{R}^{d}),
\ \mathcal{K}^{R}(v) \le -\frac{1}{4}\varepsilon_{0},
\\[6pt]
\|v\|_{L^{2}}\le \|\psi_{0}\|_{L^{2}}
\end{array}
\right.
\right\}
\quad
\mbox{for all $R\ge R_{*}$},
\]
where $H^{1}_{rad}(\mathbb{R}^{d})$ is the set of radially symmetric functions in $H^{1}(\mathbb{R}^{d})$, and $\varepsilon_{0}=\mathcal{H}(\psi_{0})-\mathcal{B}(\psi_{0})>0$.
\end{lemma}

\begin{proof}[Proof of Lemma \ref{08/04/20/18:30}]
We take a function $v \in H_{rad}^{1}(\mathbb{R}^{d})$ with the following properties:
\begin{align}
\label{10/03/05/13:28}
&\mathcal{K}^{R}(v) \le -\frac{1}{4}\varepsilon_{0},
\\[6pt]
\label{10/03/05/13:38}
&\|v\|_{L^{2}}\le \|\psi_{0}\|_{L^{2}},
\end{align}
where $R$ is a sufficiently large constant to be chosen later (see (\ref{10/03/06/18:09})). Since $v \in H_{rad}^{1}(\mathbb{R}^{d})$,
 $\mathcal{K}^{R}(v)$ is written as follows (see Remark \ref{09/09/25/16:47}):
\begin{equation}\label{10/03/05/13:56}
\mathcal{K}^{R}(v)
=
\int_{\mathbb{R}^{d}}\rho_{0}(x)|\nabla v(x)|^{2}\,dx
-\int_{\mathbb{R}^{d}}\rho_{3}(x)|v(x)|^{p+1}\,dx.
\end{equation}
Hence, it follows from (\ref{10/03/05/13:28}) that
\begin{equation}\label{08/04/20/19:49}
\begin{split}
\frac{1}{4}\varepsilon_{0} +\int_{\mathbb{R}^{d}}\rho_{0}(x)|\nabla v(x)|^{2}\,dx
&\le -\mathcal{K}^{R}(v)+ \int_{\mathbb{R}^{d}}\rho_{0}(x)|\nabla v(x)|^{2}\,dx
\\[6pt]
&=\int_{\mathbb{R}^{d}}\rho_{3}(x)|v(x)|^{p+1}\,dx.
\end{split}
\end{equation}
To estimate the right-hand side of (\ref{08/04/20/19:49}), we employ the following inequality (see
 Lemma 6.5.11 in \cite{Cazenave}): Assume that $d\ge 1$. Let $\kappa$ be a non-negative and radially symmetric function in $C^{1}(\mathbb{R}^{d})$ with $|x|^{-(d-1)}\max\{-\frac{x\nabla \kappa }{|x|},\, 0\} \in L^{\infty}(\mathbb{R}^{d})$. Then, we have that
\begin{equation}\label{08/06/12/7:28}
\begin{split}
&
\|\kappa^{\frac{1}{2}} f\|_{L^{\infty}}
\\[6pt]
&\le
K_{5}
\|f\|_{L^{2}}^{\frac{1}{2}}
\left\{
\left\| |x|^{-(d-1)}\kappa \, \frac{x \nabla f}{|x|} \right\|_{L^{2}}^{\frac{1}{2}}
+
\left\| |x|^{-(d-1)}
\max \left\{-\frac{x \nabla \kappa}{|x|}, \ 0 \right\}
\right\|_{L^{\infty}}^{\frac{1}{2}}
\left\| f \right\|_{L^{2}}^{\frac{1}{2}}
\right\}
\\[6pt]
&\hspace{9cm}
\mbox{for all $f \in H_{rad}^{1}(\mathbb{R}^{d})$},
\end{split}
\end{equation}
where $K_{5}>0$ is some constant depending only on $d$.
\\
This inequality (\ref{08/06/12/7:28}), together with (\ref{08/12/29/12:53}), (\ref{08/04/20/13:06}) and  ${\rm supp}\, \rho_{3}=\{|x|\ge R\}$ (see (\ref{08/12/29/14:16})), yields the following estimate for the right-hand side of (\ref{08/04/20/19:49}):
\begin{equation}\label{08/07/02/14:50}
\begin{split}
&\int_{\mathbb{R}^{d}}\rho_{3}(x)|v(x)|^{p+1}\,dx
\\[6pt]
&\le
\|\rho_{3}^{\frac{1}{4}}v\|_{L^{\infty}}^{p-1}
\int_{\mathbb{R}^{d}}\rho_{3}^{\frac{5-p}{4}}(x)|v(x)|^{2}\,dx
\\[6pt]
&\le \frac{K_{5}^{p-1}}{R^{\frac{p-1}{2}(d-1)}}
\left\| v \right\|_{L^{2}(|x|\ge R)}^{\frac{p-1}{2}}
\left\{
\left\|\sqrt{\rho_{3}} \frac{x\nabla v}{|x|}\right\|_{L^{2}}^{\frac{1}{2}}
\!\!
+
\left\|\max\left\{ -\frac{x \nabla \sqrt{\rho_{3}}}{|x|},\ 0 \right\}
\right\|_{L^{\infty}}^{\frac{1}{2}}
\!\!
\left\| v \right\|_{L^{2}}^{\frac{1}{2}}
\right\}^{p-1}\hspace{-12pt}K_{3}^{\frac{5-p}{4}}\|v\|_{L^{2}}^{2}
\\[6pt]
&\le
\frac{K_{5}^{p-1}K_{3}^{\frac{5-p}{4}}}{R^{\frac{p-1}{2}(d-1)}}
\left\| v \right\|_{L^{2}(|x|\ge R)}^{\frac{p-1}{2}}
\left\{
\left\|\sqrt{\rho_{3}}\nabla v \right\|_{L^{2}}^{\frac{1}{2}}
+
\frac{(K_{3}')^{\frac{1}{2}}}{R^{\frac{1}{2}}}
\left\| v \right\|_{L^{2}}^{\frac{1}{2}}
\right\}^{p-1}\hspace{-12pt}\|v\|_{L^{2}}^{2}
\\[6pt]
&\le
\frac{K_{5}^{p-1}K_{3}^{\frac{5-p}{4}}}{R^{\frac{p-1}{2}(d-1)}}
\left\| v \right\|_{L^{2}(|x|\ge R)}^{\frac{p-1}{2}}
C_{p}\left\{
\left\|\sqrt{\rho_{3}}\nabla v \right\|_{L^{2}}^{\frac{p-1}{2}}
+
\frac{(K_{3}')^{\frac{p-1}{2}}}{R^{\frac{p-1}{2}}}
\left\| v \right\|_{L^{2}}^{\frac{p-1}{2}}
\right\} \|v\|_{L^{2}}^{2}
\end{split}
\end{equation}
for some constant $C_{p}>0$ depending only on $p$.
Moreover, using $p\le 5$ (so that $\frac{p-1}{2}\le 2$), $\|v\|_{L^{2}}\le \|\psi_{0}\|_{L^{2}}$, and the Young inequality ($ab \le \frac{a^{r}}{r}+\frac{b^{r'}}{r'}$ with $\frac{1}{r}+\frac{1}{r'}=1$), we obtain that
\begin{equation}\label{10/03/06/17:54}
\begin{split}
\mbox{R.H.S. of (\ref{08/07/02/14:50})}
&\le
\frac{p-1}{4}\left\| v \right\|_{L^{2}(|x|\ge R)}^{2}
\left\|\sqrt{\rho_{3}}\nabla v \right\|_{L^{2}}^{2}
+\frac{5-p}{4}
\left(
\frac{C_{p}\left\|\psi_{0}\right\|_{L^{2}}^{2}K_{5}^{p-1}K_{3}^{\frac{5-p}{4}}}{R^{\frac{p-1}{2}(d-1)}}\right)^{\frac{4}{5-p}}
\\[6pt]
&\qquad
+C_{p}\frac{\left\|\psi_{0}\right\|_{L^{2}}^{p+1}K_{5}^{p-1}K_{3}^{\frac{5-p}{4}}(K_{3}')^{\frac{p-1}{2}}}{R^{\frac{d(p-1)}{2}}}.
\end{split}
\end{equation}
If $R$ is so large that
\begin{equation}\label{10/03/06/18:09}
\frac{5-p}{4}
\left(
\frac{C_{p}\left\|\psi_{0}\right\|_{L^{2}}^{2}K_{5}^{p-1}K_{3}^{\frac{5-p}{4}}}{R^{\frac{p-1}{2}(d-1)}}\right)^{\frac{4}{5-p}}
+
\frac{C_{p}\left\|\psi_{0}\right\|_{L^{2}}^{p+1}K_{5}^{p-1}K_{3}^{\frac{5-p}{4}}(K_{3}')^{\frac{p-1}{2}}}{R^{\frac{d(p-1)}{2}}}
 \le \frac{1}{8}\varepsilon_{0},
\end{equation}
then (\ref{08/04/20/19:49}), together with  (\ref{08/07/02/14:50}) and (\ref{10/03/06/17:54}),  shows that
\begin{equation}\label{10/03/06/18:12}
\int_{\mathbb{R}^{d}}\left\{ \rho_{0}(x)- \frac{p-1}{4}\|v\|_{L^{2}(|x|\ge R)}^{2}\rho_{3}(x)\right\}|\nabla v(x)|^{2} \le -\frac{\varepsilon_{0}}{8}.
\end{equation}
Hence, it must hold that
\begin{equation}\label{10/03/06/21:04}
\inf_{|x|\ge R}{ \frac{\rho_{0}(x)}{\rho_{3}(x)}}<
\frac{p-1}{4}\|v\|_{L^{2}(|x| \ge R)}^{2},
\end{equation}
which, together with (\ref{08/04/20/16:07}), gives us the desired result.
\end{proof}

The following lemma tells us that the solution $\psi$ is well localized:
\begin{lemma}[cf. Lemma 3.4 in \cite{Nawa8}]
\label{08/03/30/15:12}
Assume that $d\ge 1$, $2+\frac{4}{d}<p+1<2^{*}$, and $p\le 5$ if $d=2$.
Let $R_{*}$ and $m_{*}$ be constants found in Lemma \ref{08/04/20/18:30}.
Then, we have
\[
T_{\max}^{+}=\sup{\left\{ T>0 \biggm|  \sup_{0\le t <T} \int_{|x|\ge  R}|\psi(x,t)|^{2}\,dx < m_{*}  \right\}}
\]
for all $R$ with the following properties:
\begin{align}
\label{10/03/07/0:10}
&R>R_{*},
\\[6pt]
\label{10/03/07/0:11}
&\frac{10d^{2}K}{R^{2}}\|\psi_{0}\|_{L^{2}}^{2}< \varepsilon_{0},
\\[6pt]
\label{10/03/07/0:12}
&\int_{|x|\ge R}|\psi_{0}(x)|^{2}\,dx < m_{*},
\\[6pt]
\label{10/03/07/0:13}
&\frac{1}{R^{2}}\left( 1+ \frac{2}{\varepsilon_{0}} \left\|\nabla \psi_{0}\right\|_{L^{2}}^{2} \right)(W_{R},|\psi_{0}|^{2})<m_{*}.
\end{align}
\end{lemma}
\begin{proof}[Proof of Lemma \ref{08/03/30/15:12}] For $R>0$ satisfying (\ref{10/03/07/0:10})--(\ref{10/03/07/0:13}), we put
\[
T_{R}:=\sup{\left\{ T>0 \biggm|  \sup_{0\le t \le T}\int_{|x|\ge  R}|\psi(x,t)|^{2}\,dx < m_{*} \right\}}.
\]
We suppose the contrary that $T_{R}<T_{\max}^{+}$, so that
it follows from $\psi \in C([0,T_{\max}^{+});L^{2}(\mathbb{R}^{d}))$ that
\begin{equation}\label{08/03/30/16:09}
\int_{|x|\ge R}|\psi(x,T_{R})|^{2}\,dx = m_{*}.
\end{equation}
Hence, we obtain by Lemma \ref{08/04/20/18:30} that
\begin{equation}\label{10/03/06/23:50}
-\frac{1}{4}\varepsilon_{0}< \mathcal{K}^{R}(\psi(T_{R})).
\end{equation}
Then, the same computation as (\ref{10/03/02/15:40})--(\ref{10/03/06/23:56})
 yields that
\begin{equation}\label{10/03/06/23:58}
\int_{|x|\ge R } |\psi(T_{R},x)|^{2} \,dx <m_{*},
\end{equation}
which contradicts (\ref{08/03/30/16:09}): Thus, $T_{R}=T_{\max}^{+}$.
\end{proof}
Now, we are in a position to complete the proof of Theorem \ref{08/04/21/9:28}.
\\
{\it Proof of Theorem \ref{08/04/21/9:28} (continued)} We take $R>0$ satisfying  (\ref{10/03/07/0:10})--(\ref{10/03/07/0:13}). Then, Lemma \ref{08/03/30/15:12} gives us that
\begin{equation}\label{10/03/07/0:19}
\int_{|x|\ge R}|\psi(t,x)|^{2}\,dx < m_{*}
\quad
\mbox{for all $t \in [0,T_{\max}^{+})$}.
\end{equation}
Moreover, Lemma \ref{08/04/20/18:30}, together with (\ref{10/03/07/0:19}), shows that
\begin{equation}\label{10/03/07/0:22}
-\frac{\varepsilon_{0}}{4}< \mathcal{K}^{R}(\psi(t))
\quad
\mbox{for all $t \in [0,T_{\max}^{+})$}.
\end{equation}
Applying (\ref{10/03/07/0:22}) and (\ref{09/12/23/22:48}) to the generalized virial identity (\ref{08/03/29/19:05}), we obtain that
\begin{equation}\label{10/03/07/0:29}
\begin{split}
(W_{R}, |\psi(t)|^{2})
&< (W_{R},|\psi_{0}|^{2})
+2t \Im{(\vec{w}_{R} \cdot \nabla \psi_{0},\psi_{0})}
-\varepsilon_{0}^{2}t^{2}
\\[6pt]
&\qquad
+\frac{1}{4}\varepsilon_{0}t^{2}
-\frac{1}{2}\int_{0}^{t}\int_{0}^{t'}(\Delta ({\rm div}\,{\vec{w}_{R}}), |\psi(t'')|^{2})\,dt''dt'
.
\end{split}
\end{equation}
Here, it follows from (\ref{08/04/20/16:54}), the mass conservation law (\ref{08/05/13/8:59}) and (\ref{10/03/07/0:11}) that the last term on the right-hand side above is estimated as follows :
\begin{equation}\label{10/03/07/0:31}
\begin{split}
-\frac{1}{2}\int_{0}^{t}\int_{0}^{t'}(\Delta ({\rm div}\,{\vec{w}_{R}}), |\psi(t'')|^{2})\,dt''dt'
&\le \frac{1}{2}\int_{0}^{t}\int_{0}^{t'}\frac{10d^{2}K}{R^{2}} \left\|\psi(t'') \right\|_{L^{2}}^{2}\,dt''dt'
\\[6pt]
&\le
\frac{1}{4}\varepsilon_{0}t^{2}.
\end{split}
\end{equation}
Combining (\ref{10/03/07/0:29}) with (\ref{10/03/07/0:31}), we obtain that
\begin{equation}\label{10/03/07/0:30}
(W_{R}, |\psi(t)|^{2})
< (W_{R},|\psi_{0}|^{2})
+2t \Im{(\vec{w}_{R} \cdot \nabla \psi_{0},\psi_{0})}-\frac{1}{2}t^{2}\varepsilon_{0}.
\end{equation}
This inequality means that $(W_{R}, |\psi(t)|^{2})$ becomes negative in a finite time, so that $T_{\max}^{+}<\infty$: This fact, together with (\ref{10/01/27/11:26}), shows  (\ref{10/02/22/20:33}).
\par
Now, we shall show the claims (i)--(iii) in Theorem \ref{08/04/21/9:28}.
\par
We can find that Lemma \ref{08/03/30/15:12} is still valid if we replace $m_{*}$ with any $m \in (0,m_{*})$. Thus, we see that (i) holds.
\par
We next consider (ii). Since we have (\ref{08/04/20/12:51}), in order to prove (\ref{08/06/10/10:21}), it suffices to show that
\begin{equation}\label{10/03/07/16:52}
\int_{0}^{T_{\max}^{+}}(T_{\max}^{+}-t)
\left(
 \int_{\mathbb{R}^{d}}\rho_{0}(x)|\nabla \psi(x,t)|^{2}\,dx
\right) dt < \infty
\quad
\mbox{for all sufficiently large $R>0$}.
\end{equation}
Let us prove this. Integrating by parts, we find that
\begin{equation}\label{10/03/07/17:54}
\begin{split}
&\int_{0}^{t} (t-t')\left( \int_{\mathbb {R}^{d}}\rho_{0}(x)|\nabla \psi(x,t')|^{2}\,dx
\right)dt'
\\[6pt]
&=
\int_{0}^{t}\int_{0}^{t'}
\left(
\int_{\mathbb{R}^{d}}\rho_{0}(x)|\nabla \psi(x,t'')|^{2}\,dx \right) dt''\,dt'
\quad
\mbox{for all $t\ge 0$}.
\end{split}
\end{equation}
This formula  (\ref{10/03/07/17:54}) and the generalized virial identity (\ref{08/03/29/19:05}) lead us to the following estimate:
\begin{equation}\label{10/03/07/17:48}
\begin{split}
&2 \int_{0}^{t} (t-t')\left( \int_{\mathbb {R}^{d}}\rho_{0}(x)|\nabla \psi(x,t')|^{2}\,dx
\right)dt'
\\[6pt]
&\le
(W_{R},|\psi_{0}|^{2})
+2t \Im{(\vec{w}_{R} \cdot \nabla \psi_{0},\psi_{0})}+2\int_{0}^{t}\int_{0}^{t'}\mathcal{K}(\psi(t''))\,dt''dt'
\\[6pt]
& \quad
+2\int_{0}^{t}(t-t') \left(
\int_{\mathbb{R}^{d}} \rho_{3}(x)|\psi(x,t')|^{p+1}\,dx
\right)dt'
-\frac{1}{2}\int_{0}^{t}\int_{0}^{t'}(\Delta ({\rm div}\,{\vec{w}_{R}}), |\psi(t'')|^{2})\,dt''dt' .
\end{split}
\end{equation}
Here, applying the estimate (\ref{10/03/07/0:31}) to the right-hand side of (\ref{10/03/07/17:48}) (abbreviated to R.H.S. of (\ref{10/03/07/17:48})), we have that
\begin{equation}\label{10/03/07/18:21}
\begin{split}
\mbox{R.H.S. of (\ref{10/03/07/17:48})}
&\le
(W_{R},|\psi_{0}|^{2})
+2t \Im{(\vec{w}_{R} \cdot \nabla \psi_{0},\psi_{0})}-\varepsilon_{0}t^{2}
\\[6pt]
&\quad +2\int_{0}^{t}(t-t') \left(
\int_{\mathbb{R}^{d}} \rho_{3}(x)|\psi(x,t')|^{p+1}\,dx
\right)dt'
+\frac{\varepsilon_{0}}{4}t^{2}.
\end{split}
\end{equation}
Moreover, taking $R$ so large that (\ref{10/03/06/18:09}) holds,
 we find by the same computation as (\ref{08/07/02/14:50})--(\ref{10/03/06/17:54}) that
\begin{equation}\label{10/03/07/19:03}
\int_{\mathbb{R}^{d}}\rho_{3}(x)|\psi(x,t)|^{p+1}\,dx
\le
\frac{p-1}{4}\left\|\psi(t) \right\|_{L^{2}(|x|\ge R)}^{2}
\int_{\mathbb{R}^{d}}\rho_{3}(x)|\nabla \psi(x,t)|^{2}\,dx + \frac{\varepsilon_{0}}{8}.
\end{equation}
Combining (\ref{10/03/07/17:48}), (\ref{10/03/07/18:21}) and (\ref{10/03/07/19:03}), we obtain that
\begin{equation}\label{10/03/07/19:11}
\begin{split}
&2\int_{0}^{t}
(t-t')
\left( \int_{\mathbb{R}^{d}} \left\{ \rho_{0}(x)-
\frac{p-1}{4}\left\|\psi(t) \right\|_{L^{2}(|x|\ge R)}^{2}
\rho_{3}(x)
\right\}
|\nabla \psi(x,t)|^{2}\,dx
\right)\,dt'
\\[6pt]
&\le
(W_{R},|\psi_{0}|^{2})
+2t \Im{(\vec{w}_{R} \cdot \nabla \psi_{0},\psi_{0})}
-\frac{\varepsilon_{0}}{2}t^{2}
\\[6pt]
&\hspace{5cm}
\mbox{for all $R>0$ satisfying (\ref{10/03/06/18:09}) and (\ref{10/03/07/0:10})--(\ref{10/03/07/0:13})}.
\end{split}
\end{equation}
Here, using the result of (i) (the formula (\ref{08/06/12/8:54})), we can take
 a constant $R_{1}>0$ such that
\begin{equation}\label{10/03/07/16:47}
\frac{p-1}{4}\left\| \psi(t) \right\|_{L^{2}(|x|\ge R_{1})}^{2}\frac{1}{K_{4}}
\le
\frac{1}{2}
\quad
\mbox{for all $t \in I_{\max}$},
\end{equation}
where $K_{4}$ is some positive constant found in (\ref{08/04/20/16:07}). Therefore, (\ref{10/03/07/19:11}), together with (\ref{10/03/07/16:47}) and (\ref{08/04/20/16:07}), shows that
\begin{equation}\label{10/03/07/22:23}
\begin{split}
&
\int_{0}^{t}(t-t')\left(
\int_{\mathbb{R}^{d}}\rho_{0}(x)|\nabla \psi(x,t)|^{2}\,dx
\right)dt'
\\[6pt]
&\le (W_{R},|\psi_{0}|^{2})
+2t \Im{(\vec{w}_{R} \cdot \nabla \psi_{0},\psi_{0})}
-\frac{\varepsilon_{0}}{2}t^{2}
\quad
\mbox{for all sufficiently large $R$ and $t \in [0,T_{\max}^{+})$}.
\end{split}
\end{equation}
This, together with (\ref{08/04/20/12:50}) and (\ref{08/04/20/12:51}), gives (\ref{08/06/10/10:21}).
\par
Once (\ref{08/06/10/10:21}) is proved, we can easily obtain (\ref{08/06/10/10:22}). Indeed, (\ref{10/03/07/19:03}), together with (\ref{08/06/10/10:21}) and (\ref{10/03/01/15:06}), immediately yields (\ref{08/06/10/10:22}). Thus, we have proved (ii).
\par
Finally, we prove (iii). We take $R>0$ so large that
 (\ref{08/06/10/10:21}) holds valid.
\par
To prove (\ref{08/06/12/6:56}), we employ the radial interpolation inequality by Strauss \cite{Strauss}:
\begin{equation}\label{10/05/30/14:01}
\left\|f \right\|_{L^{\infty}}^{2}
\lesssim
\frac{1}{R^{d-1}}
\left\|f \right\|_{L^{2}}
\left\| \nabla f \right\|_{L^{2}}
\quad
\mbox{for all $f \in H_{rad}^{1}(\mathbb{R}^{d})$},
\end{equation}
where the implicit constant depends only on $d$.
\\
This inequality (\ref{10/05/30/14:01}), together with (\ref{08/06/10/10:21})
 and the mass conservation law (\ref{08/05/13/8:59}), gives us (\ref{08/06/12/6:56}).
\par
The estimate (\ref{08/06/12/6:57}) follows from (\ref{08/06/12/6:56}) and the inequality
\[
\|\psi(t)\|_{L^{p+1}(|x|>R)}^{p+1}\le \|\psi_{0}\|_{L^{2}}^{2}\|\psi(t)\|_{L^{\infty}(|x|>R)}^{p-1}.
\]
We obtain (\ref{08/06/12/6:58}) and (\ref{08/06/12/6:59}) from (\ref{08/06/12/6:56}) and (\ref{08/06/12/6:57}), respectively.
\end{proof}

We also have variants of (\ref{08/06/12/6:56})--(\ref{08/06/12/6:59}). For example: in their proofs, replacing $R$ by $R(T_{\max}^{+}-t)^{-\frac{1}{2(d-1)}}$, we obtain
\[
\int_{0}^{T_{\max}^{+}}\|\psi(t)\|_{L^{\infty}(|x|>R(T_{\max}^{+}-t)^{-\frac{1}{2(d-1)}})}^{4}\,dt < \infty
\]
and
\[
\int_{0}^{T_{\max}^{+}}\|\psi(t)\|_{L^{p+1}(|x|>R(T_{\max}^{+}-t)^{-\frac{1}{2(d-1)}})}^{\frac{4(p+1)}{p-1}}\,dt < \infty .
\]
Moreover, replacing $R$ by $R(T_{\max}^{+}-t)^{-\frac{1}{d-1}}$ in (\ref{08/06/12/6:56}) and (\ref{08/06/12/6:57}), we  obtain
\[
\liminf_{t \to T_{\max}^{+}} \|\psi(t)\|_{L^{\infty}\left(|x|>R(T_{\max}^{+}-t)^{-\frac{1}{d-1}}\right)}^{2}=0
\]
and
\[
\liminf_{t \to T_{\max}^{+}} \|\psi(t)\|_{L^{p+1}\left(|x|>R(T_{\max}^{+}-t)^{-\frac{1}{d-1}}\right)}^{p+1} =0 ,
\]
respectively.
\\
\par
Now, we shall give the proof of Propositions \ref{08/11/02/22:52}.

\begin{proof}[Proof of Proposition \ref{08/11/02/22:52}]
 Suppose that $\psi$ is a solution to the equation (\ref{08/05/13/8:50}) satisfying that
\begin{equation}\label{10/01/26/12:50}
\limsup_{t\to T_{\max}^{+}}\left\|\nabla \psi(t)  \right\|_{L^{2}}=
\limsup_{t\to T_{\max}^{+}}\left\| \psi(t) \right\|_{L^{p+1}}=\infty.
\end{equation}
Then, there exists a sequence $\{t_{n}\}_{n\in \mathbb{N}}$ in
 $[0,T_{\max}^{+})$ such that
\begin{align}
\label{10/03/08/17:57}
&\lim_{n\to \infty}t_{n} = T_{\max}^{+},
\\[6pt]
\label{10/01/26/12:53}
&\left\| \psi(t_{n})\right\|_{L^{p+1}}=\sup_{t\in [0,t_{n})}
\left\| \psi(t)\right\|_{L^{p+1}}.
\end{align}
Using such a sequence $\{t_{n}\}_{n\in \mathbb{N}}$, we define a number $\lambda_{n}$ by
\begin{equation}\label{10/01/26/12:58}
\lambda_{n}=\left\| \psi(t_{n})\right\|_{L^{p+1}}^{-\frac{(p-1)(p+1)}{d+2-(d-2)p}},
\quad
n \in \mathbb{N}.
\end{equation}
It is easy to see that
\begin{equation}\label{10/03/08/17:59}
\lim_{n\to \infty}\lambda_{n} = 0.
\end{equation}
We consider the scaled functions $\psi_{n}$ defined by
\begin{equation}\label{10/01/26/13:14}
\psi_{n}(x,t):=\lambda_{n}^{\frac{2}{p-1}}\overline{\psi(\lambda_{n}x,t_{n}-\lambda_{n}^{2}t)},
\quad
(x,t)\in \mathbb{R}^{d}\times
\left(-\frac{T_{\max}^{+}-t_{n}}{\lambda_{n}^{2}}, \frac{t_{n}}{\lambda_{n}^{2}}\right],
\quad n \in \mathbb{N}.
\end{equation}
We can easily verify that
\begin{align}
\label{10/03/08/15:29}
&2i\displaystyle{\frac{\partial \psi_{n}}{\partial t}} +\Delta \psi_{n}+|\psi_{n}|^{p-1}\psi_{n}=0  \quad \mbox{in} \ \mathbb{R}^{d}\times
\left(-\frac{T_{\max}^{+}-t_{n}}{\lambda_{n}^{2}}, \frac{t_{n}}{\lambda_{n}^{2}}\right],
\\[6pt]
\label{10/03/08/15:33}
&\left\| \psi_{n}(t)\right\|_{L^{2}}
=
\lambda_{n}^{-\frac{d(p-1)-4}{2(p-1)}}\left\| \psi_{0} \right\|_{L^{2}}
\quad
\mbox{for all $t\in \left(-\frac{T_{\max}^{+}-t_{n}}{\lambda_{n}^{2}}, \frac{t_{n}}{\lambda_{n}^{2}}\right]$},
\\[6pt]
\label{10/03/08/15:34}
&\mathcal{H}(\psi_{n}(t))=\lambda_{n}^{\frac{2(p+1)}{p-1}-d}\mathcal{H}(\psi_{0})
\quad
\mbox{for all $t\in \left(-\frac{T_{\max}^{+}-t_{n}}{\lambda_{n}^{2}}, \frac{t_{n}}{\lambda_{n}^{2}}\right]$},
\\[6pt]
\label{08/11/16/16:22}
&\sup_{t\in \big[0,\ \frac{t_{n}}{\lambda_{n}^{2}}\big]}\left\| \psi_{n}(t)\right\|_{L^{p+1}}=1,
\end{align}
where we put $\psi_{0}=\psi(0)$. Besides, we have that
\begin{equation}\label{08/11/16/16:23}
\sup_{t\in \big[0,\frac{t_{n}}{\lambda_{n}^{2}}\big]}\left\| \nabla \psi_{n}(t)\right\|_{L^{2}}^{2} \le 1
\quad
\mbox{for all sufficiently large $n \in \mathbb{N}$}.
\end{equation}
Indeed, (\ref{10/03/08/15:34}) and (\ref{08/11/16/16:22}) lead us to that
\begin{equation}\label{10/03/08/15:52}
\begin{split}
\left\| \nabla \psi_{n}(t)\right\|_{L^{2}}^{2}
&=
\mathcal{H}(\psi_{n}(t))+\
\frac{2}{p+1}\left\| \psi_{n}(t) \right\|_{L^{p+1}}^{p+1}
\\
&\le \lambda_{n}^{\frac{2(p+1)}{p-1}-d}\mathcal{H}(\psi_{0})+\frac{2}{p+1}
\quad
\mbox{for all $t\in \big[0, \frac{t_{n}}{\lambda_{n}^{2}}\big]$},
\end{split}
\end{equation}
which, together with (\ref{10/03/08/17:59}), immediately yields (\ref{08/11/16/16:23}).
\par
Now, we suppose that
\begin{equation}\label{10/01/27/12:38}
\left\| \psi \right\|_{L^{\infty}([0,T_{\max}^{+});L^{\frac{d}{2}(p-1)})}<\infty,
\end{equation}
so that $\{\psi_{n}\}$ satisfies that
\begin{equation}\label{10/03/10/18:08}
\sup_{n\in \mathbb{N}}\left\| \psi_{n} \right\|_{L^{\infty}([0,\frac{t_{n}}{\lambda_{n}^{2}}];L^{\frac{d}{2}(p-1)})}
\le
\left\| \psi \right\|_{L^{\infty}([0,T_{\max}^{+});L^{\frac{d}{2}(p-1)})}<\infty
.
\end{equation}
Take any $T>0$. Then, extraction of some subsequence of $\{\psi_{n}\}$ allows us to assume that:
\begin{equation}\label{10/04/12/18:23}
\sup_{t \in [0,T]}
\left\| \psi_{n}(t) \right\|_{L^{\frac{d}{2}(p-1)}}\lesssim 1,
\
 \sup_{t \in [0,T]}
\left\| \psi_{n}(t) \right\|_{L^{p+1}}=1,
\
\sup_{t \in [0,T]}
\left\| \nabla \psi_{n}(t) \right\|_{L^{2}}\le 1
\quad  \mbox{for all $n \in \mathbb{N}$},
\end{equation}
where the implicit constant is independent of $n$. The condition (\ref{10/04/12/18:23}) enable us to apply Lemmata \ref{08/10/25/23:47} and \ref{08/10/25/23:48}, so that: There exist a constant $\delta>0$ and a sequence $\{y_{n}\}_{n\in \mathbb{N}}$ in $\mathbb{R}^{d}$ such that, putting $\widetilde{\psi}_{n}(x,t)=\psi_{n}(x+y_{n},t)$,
 we have
\begin{equation}\label{10/03/10/18:02}
\sup_{t \in [0,T]}\mathcal{L}^{d}\left(
\left[
\left| \widetilde{\psi}_{n}(\cdot +y_{n}, t)\right| \ge \delta
\right]\cap B_{1}(0)
\right)
\gtrsim 1
\quad
\mbox{for all $n \in \mathbb{N}$},
\end{equation}
where $B_{1}(0)=\{x \in \mathbb{R}^{d} | |x|< 1\}$ and the implicit constant is independent of $n$. Besides, we have by (\ref{10/03/08/15:29}) and
(\ref{10/04/12/18:23}) that
\begin{align}
\label{10/03/10/18:31}
&2i\displaystyle{\frac{\partial \widetilde{\psi}_{n}}{\partial t}} +\Delta \widetilde{\psi}_{n}+|\widetilde{\psi}_{n}|^{p-1}\widetilde{\psi}_{n}=0  \quad \mbox{in} \ \mathbb{R}^{d}\times [0,T],
\\[6pt]
\label{10/03/14/17:57}
&\widetilde{\psi}_{n} \in C([0,T];\dot{H}^{1}(\mathbb {R}^{d})\cap L^{p+1}(\mathbb{R}^{d}))
\quad
\mbox{for all $n\in \mathbb{N}$},
\\[6pt]
\label{10/03/10/18:33}
&
\sup_{t \in [0,T]}\left\| \widetilde{\psi}_{n}(t) \right\|_{L^{p+1}}= 1
,
\quad
\sup_{t \in [0,T]}\left\| \nabla \widetilde{\psi}_{n}(t) \right\|_{L^{2}}\le 1
\quad
\mbox{for all $n \in \mathbb{N}$}.
\end{align}
We shall prove that $\{\widetilde{\psi}_{n} \}$ is an  equicontinuous sequence in $C([0,T];L^{2}(\Omega))$ for any compact set $\Omega \subset \mathbb{R}^{d}$.
 Let $\chi$ be a function in $C_{c}^{\infty}(\mathbb{R}^{d})$ such that $\chi\equiv 1$ on $\Omega$. Then, we can verify that
\begin{equation}\label{10/03/09/14:13}
\begin{split}
&\left\| \chi \widetilde{\psi}_{n}(t)- \chi \widetilde{\psi}_{n}(s)
\right\|_{L^{2}}^{2}
\\[6pt]
&=\int_{s}^{t}\frac{d}{dt'} \left\|\chi\widetilde{\psi}_{n}(t')
-\chi\widetilde{\psi}_{n}(s) \right\|_{L^{2}}^{2}dt'
\\[6pt]
&=2\Re{
\int_{s}^{t}
\!\!
\int_{\mathbb{R}^{d}}
\!
\overline{\chi(x)(\partial_{t}\widetilde{\psi}_{n})(x,t')}
\left\{ \chi(x)\widetilde{\psi}_{n}(x,t')-\chi(x)\widetilde{\psi}_{n}(x,s) \right\}dxdt'}
\quad
\mbox{for all $s,t \in [0,T]$}.
\end{split}
\end{equation}
This identity (\ref{10/03/09/14:13}), together with (\ref{10/03/10/18:31}), yields that
\begin{equation}\label{08/12/19/15:09}
\begin{split}
&\left\| \widetilde{\psi}_{n}(t)-\widetilde{\psi}_{n}(s)\right\|_{L^{2}(\Omega)}^{2}
\\[6pt]
&\le \left\| \chi \widetilde{\psi}_{n}(t)- \chi \widetilde{\psi}_{n}(s) \right\|_{L^{2}}^{2}
\\[6pt]
&\le 2
|t-s| \left\| \chi \partial_{t} \widetilde{\psi}_{n} \right\|_{L^{\infty}([0,T];H^{-1})}
\left\| \chi \widetilde{\psi}_{n} \right\|_{L^{\infty}([0,T];H^{1})}
\\[6pt]
&\le 2
|t-s| \left\{ \left\| \chi \Delta \widetilde{\psi}_{n} \right\|_{L^{\infty}([0,T];H^{-1})}+
\left\| \chi (|\widetilde{\psi}_{n}|^{p-1}\widetilde{\psi}_{n}) \right\|_{L^{\infty}([0,T];H^{-1})}
\right\}
\\[6pt]
&\qquad  \times
 \left\{ \left\| \chi \widetilde{\psi}_{n}\right\|_{L^{\infty}([0,T];L^{2})}
+
\left\| (\nabla \chi) \widetilde{\psi}_{n }\right\|_{L^{\infty}([0,T];L^{2})}
+
\left\| \chi  \nabla \widetilde{\psi}_{n}\right\|_{L^{\infty}([0,T];L^{2})}
\right\}
\\[6pt]
&\hspace{264pt}\mbox{for all $s,t \in [0,T]$}.
\end{split}
\end{equation}
We consider the terms in the first parentheses on the right-hand side of (\ref{08/12/19/15:09}). They are estimated by the Sobolev embedding and the H\"older inequality as follows:
\begin{equation}\label{10/03/09/14:30}
\begin{split}
&\left\| \chi \Delta \widetilde{\psi}_{n} \right\|_{L^{\infty}([0,T];H^{-1})}
+
\left\|
\chi (|\widetilde{\psi}_{n}|^{p-1}\widetilde{\psi}_{n})
\right\|_{L^{\infty}([0,T];H^{-1})}
\\[6pt]
&=
\left\| \Delta (\chi  \widetilde{\psi}_{n})-(\Delta \chi)\widetilde{\psi}_{n} -2\nabla \chi \nabla \widetilde{\psi}_{n} \right\|_{L^{\infty}([0,T];H^{-1})}
+
\left\|
\chi (|\widetilde{\psi}_{n}|^{p-1}\widetilde{\psi}_{n})
\right\|_{L^{\infty}([0,T];H^{-1})}
\\[6pt]
&\lesssim  \left\| \nabla (\chi \widetilde{\psi}_{n}) \right\|_{L^{\infty}([0,T];L^{2})}
+ \left\| (\Delta \chi)\widetilde{\psi}_{n} \right\|_{L^{\infty}([0,T];L^{\frac{p+1}{p}})}
+  \left\| \nabla \chi \nabla \widetilde{\psi}_{n} \right\|_{L^{\infty}([0,T];L^{2})}
\\[6pt]
&\qquad  \qquad
+\left\|
\chi (|\psi_{n}|^{p-1}\widetilde{\psi}_{n})
\right\|_{L^{\infty}([0,T];L^{\frac{p+1}{p}})}
\\[6pt]
&\le
\left\| \nabla \chi \right\|_{L^{\frac{2(p+1)}{p-1}}}
\left\| \widetilde{\psi}_{n} \right\|_{L^{\infty}([0,T];L^{p+1})}
+
\left\| \chi \right\|_{L^{\infty}}
\left\| \nabla \widetilde{\psi}_{n} \right\|_{L^{\infty}([0,T];L^{2})}
\\[6pt]
&\qquad +
\left\| \Delta \chi \right\|_{L^{\frac{p+1}{p-1}}}
\left\| \widetilde{\psi}_{n} \right\|_{L^{\infty}([0,T];L^{p+1})}
+ \left\| \nabla \chi \right\|_{L^{\infty}}
\left\| \nabla \widetilde{\psi}_{n} \right\|_{L^{\infty}([0,T];L^{2})}
\\[6pt]
& \qquad \qquad +
\left\|
\chi \right\|_{L^{\infty}}\left\|\widetilde{\psi}_{n}
\right\|_{L^{\infty}([0,T];L^{p+1})}^{p+1},
\end{split}
\end{equation}
where the implicit constant depends only on $d$ and $p$.
 On the other hand, the terms in the second parentheses on the right-hand side of (\ref{08/12/19/15:09}) are estimated by the H\"older inequality as follows:
\begin{equation}\label{10/03/09/16:08}
\begin{split}
&
\left\| \chi \widetilde{\psi}_{n}\right\|_{L^{\infty}([0,T];L^{2})}
+
\left\| (\nabla \chi) \widetilde{\psi}_{n }\right\|_{L^{\infty}([0,T];L^{2})}
+
\left\| \chi  \nabla \widetilde{\psi}_{n}\right\|_{L^{\infty}([0,T];L^{2})}
\\[6pt]
&\le 2
\left\| \chi \right\|_{W^{1,\frac{2(p+1)}{p-1}}}
\left\| \widetilde{\psi}_{n} \right\|_{L^{\infty}([0,T];L^{p+1})}
+
\left\|\chi \right\|_{L^{\infty}}
\left\| \nabla \widetilde{\psi}_{n} \right\|_{L^{\infty}([0,T];L^{2})}.
\end{split}
\end{equation}
Hence, combining these estimates (\ref{08/12/19/15:09})--(\ref{10/03/09/16:08})  with (\ref{10/03/10/18:33}), we obtain that
\begin{equation}\label{10/01/26/18:23}
\left\| \widetilde{\psi}_{n}(t)-\widetilde{\psi}_{n}(s)\right\|_{L^{2}(\Omega)}^{2}
\lesssim |t-s|,
\end{equation}
where the implicit constant depends only on $d$, $p$ and $\Omega$ ($\chi$ is determined by $\Omega$). Thus, we see that $\{\widetilde{\psi}_{n}\}$ is equicontinuous in $C([0,T];L^{2}(\Omega))$ for all compact set $\Omega \subset \mathbb{R}^{d}$.
\par
The estimate (\ref{10/01/26/18:23}), with the help of the  Gagliardo-Nirenberg inequality and  (\ref{10/03/10/18:33}), also shows that $\{\widetilde{\psi}_{n} \}$ is an equicontinuous sequence in $C([0,T];L^{q}(\Omega))$ for all compact set $\Omega \subset \mathbb{R}^{d}$ and $q \in [2,2^{*})$. Hence, (\ref{10/03/10/18:33}) and the Ascoli-Arz\'era theorem, together with (\ref{10/03/10/18:02}) and (\ref{10/03/10/18:31}), yield that: There exist a subsequence of $\{\widetilde{\psi}_{n}\}$ (still denoted by the same symbol) and a nontrivial function $\psi_{\infty} \in L^{\infty}([0,\infty);\dot{H}^{1}(\mathbb{R}^{d})\cap L^{p+1}(\mathbb{R}^{d}))$
such that
\begin{align}
\label{10/01/27/12:25}
&\lim_{n\to \infty}\widetilde{\psi}_{n}= \psi_{\infty} \quad \mbox{ strongly in $L^{\infty}([0,T];L_{loc}^{q}(\mathbb{R}^{d}))$} \quad \mbox{for all $q \in [2, 2^{*})$},
\\[6pt]
\label{10/03/09/16:53}
&\lim_{n\to \infty}\nabla \widetilde{\psi}_{n} = \nabla \psi_{\infty}
\quad
\mbox{weakly* in $L^{\infty}([0,T];L^{2}(\mathbb{R}^{d}))$},
\\[6pt]
\label{10/03/09/18:11}
&2i\frac{\partial \psi_{\infty}}{\partial t}+\Delta \psi_{\infty}+|\psi_{\infty}|^{p-1}\psi_{\infty}=0
\quad
\mbox{in $\mathcal{D}'([0,\infty);\dot{H}^{-1}(\mathbb{R}^{d})+ L^{\frac{p+1}{p}}(\mathbb{R}^{d}))$}.
\end{align}

Finally, we hall prove the formula (\ref{10/03/10/18:47}). The Lebesgue dominated convergence theorem leads to that for all $\varepsilon>0$, there exists $R>0$ such that
\begin{equation}\label{10/04/16/18:52}
\int_{|x|\le R}|\psi_{\infty}(0)|^{\frac{d(p-1)}{2}}\,dx
\ge
(1-\varepsilon)
\int_{\mathbb{R}^{d}}|\psi_{\infty}(0)|^{\frac{d(p-1)}{2}}\,dx.
\end{equation}
Combining this estimate with the strong convergence (\ref{10/01/27/12:25}), we have that
\begin{equation}\label{10/04/16/19:00}
\lim_{n\to \infty}
\int_{|x|\le R}
|\widetilde{\psi}_{n}(x,0)|^{\frac{d(p-1)}{2}}\,dx
\ge
(1-\varepsilon)
\int_{\mathbb{R}^{d}}|\psi_{\infty}(0)|^{\frac{d(p-1)}{2}}\,dx.
\end{equation}
Hence, it follows from  (\ref{10/04/16/19:00}) and the relation
\begin{equation}\label{10/04/16/16:33}
\int_{|x|\le R}|\widetilde{\psi}_{n}(x,0)|^{\frac{d(p-1)}{2}}\,dx
=
\int_{|x-\lambda_{n}y_{n}|\le \lambda_{n}R}|\psi(x,t_{n})|^{\frac{d(p-1)}{2}}\,dx
\end{equation}
that the desired result holds.
\end{proof}

Finally, we shall give the proof of Proposition \ref{10/01/26/14:49}.

\begin{proof}[Proof of Proposition \ref{10/01/26/14:49}]
We use the same assumptions, definitions and notation in Proposition \ref{08/11/02/22:52}, except the condition (\ref{10/01/26/15:36}).
\par
We find from the proof of Proposition \ref{08/11/02/22:52} that: For all $T>0$, there exists a subsequence of $\{\psi_{n}\}$ in $C([0,T];H^{1}(\mathbb{R}^{d}))$ (still denoted by the same symbol) with the following properties:
\begin{align}
\label{10/03/11/15:27}
&\sup_{t\in [0,T]}\left\|\psi_{n}(t) \right\|_{L^{p+1}}=1
\quad \mbox{for all $n \in \mathbb{N}$},
\\[6pt]
&\label{10/03/11/15:39}
\sup_{t \in [0,T]}\left\| \nabla \psi_{n}(t) \right\|_{L^{2}}\le 1
\quad
\mbox{for all $n \in \mathbb{N}$},
\\[6pt]
&\label{10/03/11/15:36}
2i\frac{\partial \psi_{n}}{\partial t}+\Delta \psi_{n}+|\psi_{n}|^{p-1}\psi_{n}=0
\quad
\mbox{in $\mathbb{R}^{d}\times [0,T]$}.
\end{align}
For such a subsequence $\{\psi_{n}\}$, we define renormalized functions $\Phi_{n}^{RN}$ by \begin{equation}\label{10/01/26/13:15}
\Phi_{n}^{RN}(x,t)=\psi_{n}(x,t)-e^{\frac{i}{2}t\Delta}\psi_{n}(x,0),
\quad n \in \mathbb{N}.
\end{equation}
Here, it is worth while noting that
\begin{align}
\label{10/03/15/18:18}
&\Phi^{RN}_{n} \in C([0,T];H^{1}(\mathbb{R}^{d}))
\quad
\mbox{for all $n \in \mathbb{N}$},
\\[6pt]
\label{10/03/11/16:00}
&\Phi_{n}^{RN}(t)
=\frac{i}{2}\int_{0}^{t}e^{\frac{i}{2}(t-t')\Delta}|\psi_{n}(t')|^{p-1}\psi_{n}(t')\,dt'
\quad
\mbox{for all $n \in \mathbb{N}$}.
\end{align}
We shall show that
\begin{equation}
\label{08/11/21/17:55}
\sup_{n\in \mathbb{N}}\left\|\Phi^{RN}_{n}\right\|_{L^{\infty}([0,T];H^{1})}
\le C_{T}
\end{equation}
for some constant $C_{T}>0$ depending only on $d$, $p$ and $T$.
Applying the Strichartz estimate to the formula (\ref{10/03/11/16:00}), and using (\ref{10/03/11/15:27}), we obtain the following two estimates:
\begin{equation}\label{10/03/11/16:22}
\begin{split}
\left\|\Phi^{RN}_{n}\right\|_{L^{\infty}([0,T];L^{2})}
&\lesssim
\left\||\psi_{n}|^{p-1}\psi_{n} \right\|_{L^{\frac{4(p+1)}{4(p+1)-d(p-1)}}([0,T];L^{\frac{p+1}{p}})}
\\[6pt]
& \le T^{1-\frac{d(p-1)}{4(p+1)}}\left\| \psi_{n} \right\|_{L^{\infty}([0,T];L^{p+1})}^{p}
\\[6pt]
&\le T^{1-\frac{d(p-1)}{4(p+1)}},
\end{split}
\end{equation}
\begin{equation}\label{10/03/11/17:29}
\begin{split}
\left\| \nabla \Phi^{RN}_{n} \right\|_{L^{\infty}([0,T];L^{2})}
&\lesssim
\left\| \nabla \left( |\psi_{n}|^{p-1}\psi_{n}\right) \right\|_{L^{\frac{4(p+1)}{4(p+1)-d(p-1)}}([0,T];L^{\frac{p+1}{p}})}
\\
& \le T^{1-\frac{d(p-1)}{2(p+1)}}\left\| \psi_{n}\right\|_{L^{\infty}([0,T];L^{p+1})}^{p-1}\left\| \nabla \psi_{n}\right\|_{L^{\frac{4(p+1)}{d(p-1)}}([0,T];L^{p+1})}
\\
&\le T^{1-\frac{d(p-1)}{2(p+1)}}\left\| \nabla \psi_{n}\right\|_{L^{\frac{4(p+1)}{d(p-1)}}([0,T];L^{p+1})},
\end{split}
\end{equation}
where the implicit constants depend only on $d$ and $p$. Therefore, for the desired estimate (\ref{08/11/21/17:55}), it suffices to show that
\begin{equation}\label{08/11/19/23:00}
\sup_{n\in \mathbb{N}}
\left\| \nabla \psi_{n} \right\|_{L^{\frac{4(p+1)}{d(p-1)}}([0,T];L^{p+1})}
\le D_{T}
\end{equation}
for some constant $D_{T}>0$ depending only on $d$, $p$ and $T$. Here, note that  the pair $(p+1,\ \frac{4(p+1)}{d(p-1)})$ is admissible. In order to prove
 (\ref{08/11/19/23:00}), we introduce an admissible pair $(q,r)$ with $q=p+2$ if $d=1,2$ and $q=\frac{1}{2}(p+1+2^{*})$ if $d\ge 3$, so that $p+1<q<2^{*}$.
 Then, it follows from the integral equation for $\psi_{n}$ and the Strichartz estimate that
\begin{equation}\label{10/04/16/17:45}
\begin{split}
&\left\|\nabla \psi_{n} \right\|_{L^{r}\left([0,T];L^{q}\right)}
\\[6pt]
&\lesssim
\left\| \nabla \psi_{n}(0)\right\|_{L^{2}}
+ \left\| \nabla \left( |\psi_{n}|^{p-1} \psi_{n}\right) \right\|_{L^{\frac{4(p+1)}{4(p+1)-d(p-1)}}([0,T]; L^{\frac{p+1}{p}})}
\\[6pt]
&\lesssim
\left\| \nabla \psi_{n}(0)\right\|_{L^{2}}
+ \left\| \psi_{n} \right\|_{ L^{\frac{2(p+1)(p-1)}{d+2-(d-2)p}}([0,T]; L^{p+1})}^{p-1}
\left\| \nabla \psi_{n} \right\|_{L^{\frac{4(p+1)}{d(p-1)}}([0,T];L^{p+1})}
\\[6pt]
&\le
\left\| \nabla \psi_{n}(0)\right\|_{L^{2}} +
T^{\frac{d+2-(d-2)p}{2(p+1)}}\left\| \psi_{n} \right\|_{L^{\infty}([0,T];L^{p+1})}^{p-1}
\left\|\nabla \psi_{n} \right\|_{L^{\infty}([0,T];L^{2})}^{1-\frac{q(p-1)}{(q-2)(p+1)}}
\left\| \nabla \psi_{n} \right\|_{L^{r}([0,T];L^{q})}^{\frac{q(p-1)}{(q-2)(p+1)}},
\end{split}
\end{equation}
where the implicit constants depend only on $d$ and $p$.
 Combining (\ref{10/04/16/17:45}) with (\ref{10/03/11/15:27}) and (\ref{10/03/11/15:39}), we obtain that
\begin{equation}\label{10/03/11/17:56}
\left\|\nabla \psi_{n} \right\|_{L^{r}\left([0,T];L^{q}\right)}
\lesssim
 1 +
T^{\frac{d+2-(d-2)p}{2(p+1)}}\left\| \nabla \psi_{n} \right\|_{L^{r}([0,T];L^{q})}^{\frac{q(p-1)}{(q-2)(p+1)}},
\end{equation}
where the implicit constant depends only on $d$ and $p$. Since $0<\frac{q(p-1)}{(q-2)(p+1)}<1$, this estimate, together with the Young inequality ($ab \le
 \frac{1}{\gamma}a^{\gamma}+\frac{1}{\gamma'}b^{\gamma'}$ for $\gamma,\gamma'>1$ with $\frac{1}{\gamma}+\frac{1}{\gamma'}=1$), yields that
\begin{equation}\label{08/11/19/22:29}
\left\| \nabla \psi_{n} \right\|_{L^{r}([0,T];L^{q})}
\lesssim  1+T^{\frac{(q-2)\{d+2-(d-2)p\}}{4\{ q-(p+1)\}}}
,
\end{equation}
where the implicit constant depends only on $d$ and $p$. Hence, interpolating
 (\ref{10/03/11/15:39}) and  (\ref{08/11/19/22:29}), we obtain (\ref{08/11/19/23:00}), so that (\ref{08/11/21/17:55}) holds.
\par
Next, we shall show that $\{\Phi^{RN}_{n}\}$ is an equicontinuous sequence in
 $C([0,T];L^{q}(\mathbb{R}^{d}))$ for all $q \in [2,2^{*})$.
Differentiating the both sides of (\ref{10/03/11/16:00}), we obtain that
\begin{equation}\label{10/03/15/16:24}
\begin{split}
\partial_{t}\Phi^{RN}_{n}(t)
&= \frac{i}{2}|\psi_{n}(t)|^{p-1}\psi_{n}(t)
-\frac{1}{4}\Delta \int_{0}^{t} e^{\frac{i}{2}(t-t')\Delta}|\psi_{n}(t')|^{p-1}\psi_{n}(t')\,dt'
\\[6pt]
&=\frac{i}{2}|\psi_{n}(t)|^{p-1}\psi_{n}(t)+\frac{i}{2}\Delta \Phi^{RN}_{n}(t).
\end{split}
\end{equation}
This formula (\ref{10/03/15/16:24}) and the H\"older inequality show that
\begin{equation}\label{10/03/15/16:36}
\begin{split}
&\left\|\Phi^{RN}_{n}(t)-\Phi^{RN}_{n}(s)\right\|_{L^{2}}^{2}
\\[6pt]
&
=\int_{s}^{t}\frac{d}{dt'} \left\|\Phi^{RN}_{n}(t')-\Phi^{RN}_{n}(s) \right\|_{L^{2}}^{2}dt'
\\[6pt]
&=
2\Re \int_{s}^{t}\int_{\mathbb{R}^{d}} \overline{\partial_{t}\Phi^{RN}_{n}(x,t')}\left\{ \Phi^{RN}_{n}(x,t')-\Phi^{RN}_{n}(x,s)\right\}dxdt'
\\[6pt]
&\lesssim
|t-s|\left\|\psi_{n} \right\|_{L^{\infty}([0,T];L^{p+1})}^{p}
\left\| \Phi^{RN}_{n} \right\|_{L^{\infty}([0,T];L^{p+1})}
+ |t-s|\left\| \nabla \Phi^{RN}_{n}\right\|_{L^{\infty}([0,T];L^{2})}^{2}
,
\end{split}
\end{equation}
where the implicit constant depends only on $d$ and $q$. Combining this estimate with (\ref{10/03/11/15:27}) and (\ref{08/11/21/17:55}),  we obtain
\begin{equation}\label{10/03/15/16:54}
\left\|\Phi^{RN}_{n}(t)-\Phi^{RN}_{n}(s)\right\|_{L^{2}}^{2}
\lesssim |t-s|
\quad
\mbox{for all $s,t \in [0,T]$},
\end{equation}
where the implicit constant depends only on $d$, $q$ and $T$. Moreover, the Gagliardo-Nirenberg inequality, together with (\ref{08/11/21/17:55}) and (\ref{10/03/15/16:54}), shows that $\{\Phi^{RN}_{n}\}$ is an equicontinuous sequence in
 $C([0,T];L^{q}(\mathbb{R}^{d}))$ for all $q \in [2,2^{*})$.
\par
Now, we are in a position to prove the main property of $\{\Phi^{RN}_{n}\}$, the alternatives (i) and (ii). We first suppose that
\begin{equation}\label{10/03/15/17:02}
\lim_{n\to \infty}\sup_{t\in [0,T]}\left\| \Phi_{n}^{RN}(t)\right\|_{L^{\frac{d}{2}(p-1)}}=0.
\end{equation}
Then, a direct calculation immediately yields (\ref{08/11/20/17:56}).
\par
On the other hand, when
\begin{equation}\label{10/03/15/17:04}
A:=\limsup_{n\to \infty}\sup_{t\in [0,T]}\left\| \Phi_{n}^{RN}(t) \right\|_{L^{p+1}}>0,
\end{equation}
we can take a subsequence $\{\Phi^{RN}_{n}\}$ (still denoted by the same symbol) such that
\begin{equation}\label{08/11/21/18:29}
\sup_{t \in [0,T]}\left\|\Phi^{RN}_{n}(t) \right\|_{L^{p+1}}
 \ge \frac{A}{2}
\qquad
\mbox{for all  $n \in \mathbb{N}$}.
\end{equation}
This inequality (\ref{08/11/21/18:29}) and the uniform bound (\ref{08/11/21/17:55}) enable us to apply Lemmata \ref{08/10/25/23:47} and \ref{08/10/25/23:48}. Thus, we find that there exist a constant $\delta>0$ and a sequence $\{y_{n}\}_{n\in \mathbb{N}}$ in $\mathbb{R}^{d}$ such that, putting $\widetilde{\Phi}^{RN}_{n}(x,t)=\Phi^{RN}_{n}(x+y_{n},t)$, we have
\begin{equation}\label{10/03/15/17:57}
\sup_{t \in [0,T]}\mathcal{L}^{d}\left(
\left[ \left|\widetilde{\Phi}^{RN}_{n}(t)\right| >
\frac{\delta}{2} \right]
\cap B_{1}(0)
\right)>C
\end{equation}
for some constant $C>0$ being independent of $n$.
Besides, we find by (\ref{08/11/21/17:55}) and (\ref{10/03/15/16:54}) that:
 $\{\widetilde{\Phi}^{RN}_{n}\}$ is
\begin{align}
\label{10/04/16/18:13}
&\mbox{a uniformly bounded sequence in $C([0,T];H^{1}(\mathbb{R}^{d}))$, and}
\\[6pt]
&\mbox{an equicontinuous sequence in $C([0,T];L^{q}(\mathbb{R}^{d}))$ for all $q\in [2,2^{*})$}.
\end{align}
Hence, the Ascoli-Arzel\'a theorem, together with (\ref{10/03/15/17:57}), gives us that: There exist a subsequence of $\{\widetilde{\Phi}^{RN}_{n}\}$ (still denoted by the same symbol) and a nontrivial function $\Phi \in L^{\infty}([0,\infty);H^{1}(\mathbb{R}^{d}))$ such that
\begin{align}
\label{08/12/18/16:24}
&\lim_{n\to \infty}\widetilde{\Phi}^{RN}_{n} = \Phi
\quad \mbox{weakly* in $L^{\infty}([0,T];H^{1}(\mathbb{R}^{d}))$},
\\[6pt]
\label{08/12/17/18:26}
&
\lim_{n\to \infty}\widetilde{\Phi}^{RN}_{n}= \Phi
\quad \mbox{strongly in $C([0,T];L_{loc}^{q}(\mathbb{R}^{d}))$
\quad
for all  $q\in[2,2^{*})$}.
\end{align}
It remains to prove (\ref{10/03/15/19:52})--(\ref{08/11/20/17:57}). We find by (\ref{10/03/11/15:27}) that: There exists a function
$F \in L^{\infty}([0,\infty);L^{\frac{p+1}{p}}(\mathbb{R}^{d}))$ such that
\begin{equation}\label{08/12/18/16:23}
\lim_{n\to \infty}|\psi_{n}|^{p-1}\psi_{n} = F
\quad
\mbox{weakly* in $L^{\infty}([0,T];L^{\frac{p+1}{p}}(\mathbb{R}^{d}))$}.
\end{equation}
Then, it follows from the equation (\ref{10/03/15/16:24}) and (\ref{08/12/18/16:24}) that
\begin{equation}\label{08/12/18/16:25}
2i\frac{\partial \Phi}{\partial t} +\Delta \Phi + F=0.
\end{equation}
Here, if $F$ were trivial, then $\Phi$ is so since $\displaystyle{\Phi(0)=\lim_{n\to \infty}\Phi^{RN}_{n}(0)=0}$. Therefore, $F$ is nontrivial.
\par
We can prove the formula (\ref{08/11/20/17:57}) in a way similar to the proof of (\ref{10/03/10/18:47}) (cf. the estimates (\ref{10/04/16/18:52})--
 (\ref{10/04/16/16:33})).
\end{proof}


TITLE: Is this function differentiable at $\pi/2$?
QUESTION [0 upvotes]: Consider the function $f(x)$ defined by
\begin{equation*}
f(x) = \begin{cases}
\tan(x/2 + \pi/4) - (\pi/4)/(1-2 x/\pi) & \text{if $x \ne \pi/2$,} \\
0 & \text{if $x = \pi/2$}.
\end{cases}
\end{equation*}
Combining fractions and then using L'Hopital's rule (twice), it can be seen that this function is continuous at $\pi/2$.
Since the only discontinutites of this function are at multiples of $\pi/2$, can we conclude, without any further calculations, that the function is also differentiable at $\pi/2$?

REPLY [0 votes]: make a change of variable $$\begin{align}x = \pi/2 + 2h, f(\pi/2 + 2h) &= \tan (\pi/2 + h) - (\pi/4)\left(1 - \frac{2}{\pi}(\pi/2 + 2h) \right)\\
&= -\tan h + h  = -\frac 13 h^3+\cdots\end{align}$$ therefore $f$ is differentiable  at $x = \pi/2$ and its derivative is zero.


TITLE: Zig-Zag traversal of list
QUESTION [0 upvotes]: Suppose that each node of the list has the following structs:

string
num
next

It is given a string $w$. Suppose that $w$ is in the list at a node $p$, of which the struct num has the value n.
We are looking for the value of the struct string of the node that precedes $p$ for $n$ positions in the list.

So, do we have to traverse the whole list, till we find a node with struct num equal to $n$ and then with a while loop find the desired value or is there also a better way?
EDIT:That's what I have tried:
Algorithm(pointer L, string w){
    pointer R=L;
    int times=0;
    if (R==NULL) return;
    while (R!=NULL and R->string!=w){
          times++;
          R=R->next;
    }
    n=R->num;
    if (times>n){
        int i=0;
        p=L;
        while (i<times-n){
               p=p->next;
               i++;
        }
        return p->string;
    }
    }

Is it right or have I done something wrong?

REPLY [1 votes]: You have a pointer $L$ to the beginning of the list. First, create a new pointer $R$ that also points to the beginning of the list. Now, traverse the list with $R$, looking for the struct with string $w$. While $R$ is traversing the list, you should also keep a count $p$ equal to how many elements $R$ has iterated over. When you find your the struct with string $w$, simply compute $p-n$ and iterate this many times from $L$ to find what you're looking for.
In C++, the algorithm may look something this:
int spaces_to_move(LL* R, std::string word_to_find)
{
    int i = 0;
    while(R->w != word_to_find){
        R = R->next;
        i++;

        if(!R){
            std::cout << "Word not found\n";
            exit(0);
        }
    }

    return i-R->n;
}


text:
{\bf Problem.} The red parabola shown is the graph of the equation $x = ay^2 + by + c$.  Find $a+b+c$.

[asy]
size(150);
real ticklen=3;
real tickspace=2;

real ticklength=0.1cm;
real axisarrowsize=0.14cm;
pen axispen=black+1.3bp;
real vectorarrowsize=0.2cm;
real tickdown=-0.5;
real tickdownlength=-0.15inch;
real tickdownbase=0.3;
real wholetickdown=tickdown;
void rr_cartesian_axes(real xleft, real xright, real ybottom, real ytop, real xstep=1, real ystep=1, bool

useticks=false, bool complexplane=false, bool usegrid=true) {

import graph;

real i;

if(complexplane) {

label("$\textnormal{Re}$",(xright,0),SE);

label("$\textnormal{Im}$",(0,ytop),NW);

} else {

label("$x$",(xright+0.4,-0.5));

label("$y$",(-0.5,ytop+0.2));

}

ylimits(ybottom,ytop);

xlimits( xleft, xright);

real[] TicksArrx,TicksArry;

for(i=xleft+xstep; i<xright; i+=xstep) {

if(abs(i) >0.1) {

TicksArrx.push(i);

}

}

for(i=ybottom+ystep; i<ytop; i+=ystep) {

if(abs(i) >0.1) {

TicksArry.push(i);

}

}

if(usegrid) {

xaxis(BottomTop(extend=false), Ticks("%", TicksArrx ,pTick=gray

(0.22),extend=true),p=invisible);//,above=true);

yaxis(LeftRight(extend=false),Ticks("%", TicksArry ,pTick=gray(0.22),extend=true),

p=invisible);//,Arrows);

}

if(useticks) {

xequals(0, ymin=ybottom, ymax=ytop, p=axispen, Ticks("%",TicksArry ,

pTick=black+0.8bp,Size=ticklength), above=true, Arrows(size=axisarrowsize));

yequals(0, xmin=xleft, xmax=xright, p=axispen, Ticks("%",TicksArrx ,

pTick=black+0.8bp,Size=ticklength), above=true, Arrows(size=axisarrowsize));

} else {

xequals(0, ymin=ybottom, ymax=ytop, p=axispen, above=true, Arrows(size=axisarrowsize));

yequals(0, xmin=xleft, xmax=xright, p=axispen, above=true, Arrows(size=axisarrowsize));

}
};
real lowerx, upperx, lowery, uppery;
real f(real x) {return -(x+4)*(x+4)/2+5;}
lowery = -9;
uppery = 1;
rr_cartesian_axes(-8,7,lowery,uppery);
draw(reflect((0,0),(1,1))*(graph(f,lowery,uppery,operator ..)), red);
[/asy]

Each tick mark on the graph is one unit.
{\bf Level.} Level 5
{\bf Type.} Algebra
{\bf Solution.} The vertex of the parabola is $(5,-4)$, so the equation of the parabola is of the form \[x = a(y + 4)^2 + 5.\] The parabola passes through the point $(3,-2)$.  Substituting these values into the equation above, we get \[3 = a(-2 + 4)^2 + 5.\] Solving for $a$, we find $a = -1/2$.  Hence, the equation of the parabola is given by \[x = -\frac{1}{2} (y + 4)^2 + 5 = -\frac{1}{2} (y^2 + 8y + 16) + 5 = -\frac{1}{2} y^2 - 4y - 3.\] The answer is $-1/2 - 4 - 3 = \boxed{-\frac{15}{2}}$.

set_name: MATH