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\vspace*{1.5pc} \noindent \normalsize\textbf{\Large { Ergodic property for Galton-Watson processes in which individuals have variable lifetimes}} \hfill

\vspace{12pt} \hspace*{0.75pc}{\small\textrm{\uppercase{Jiangrui Tan
}}}\hspace{-2pt}$^{*}$, {\small\textit{Beijing Normal University }}

\hspace*{0.75pc}{\small\textrm{\uppercase{Junping Li}}}
\hspace{-2pt}$^{**}$, {\small\textit{Central South University }}

\par
\footnote{\hspace*{-0.75pc}$^{*}\,$Postal address:
 School of Mathematical Sciences, Laboratory of Mathematics and Complex Systems,
Beijing Normal University, Beijing, 100875, China. E-mail address:
670816453@qq.com }

\par
\footnote{\hspace*{-0.75pc}$^{**}\,$Postal address:
 School of Mathematics and Statistics, Central
South University, Changsha, 410083, China. E-mail address:
jpli@mail.csu.edu.cn }

\par
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\begin{abstract}
   This paper is concerned with an extended Galton-Watson process so as to allow individuals to live and reproduce for more than one unit time. We assume that each individual can live $k$ seasons (time-units) with probability $h_k$, and produce $m$ offspring with probability $p_m$ during each season. These can be seen as Galton-Watson processes with countably infinitely many types in which particles of type $i$ may only have offspring of type $i+1$ and type $1$. Let $\emph{\textbf{M}}$ be its mean progeny matrix and $\gamma$ be the convergence radius of the power series $\sum_{k\geq 0}r^k(\emph{\textbf{M}}^k)_{ij}$. We first derive formula of calculating $\gamma$ and show that $\gamma$, in supercritical case, is actually the extinction probability of a Galton-Watson process. Next, we give clear criteria for $\emph{\textbf{M}}$ to be $\gamma$-transient, $\gamma$-positive and $\gamma$-null recurrent from which the ergodic property of the process is discussed. The criteria for $\gamma$ and $\gamma$-recurrence of $\emph{\textbf{M}}$ rely on the properties of lifetime distribution which are easier to be verified than current results. Finally, we show the asymptotic behavior of the total population size of each type of individuals under certain conditions which illustrates the evolution of Galton-Watson process in which individuals have variable lifetimes.
\end{abstract}

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{\footnotesize {\bf Keywords:}

Galton-Watson process; variable lifetimes; convergence radius.

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\normalsize

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\noindent {\large \bf 1. Introduction and Preliminaries}
\vspace{3mm}
\par
  A Galton-Watson process in which individuals have variable lifetimes is a discrete time stochastic process $\{Z_n;n\geq 0\}$ on $\mathbb{Z}_+=\{0,1,\cdots\}$. Let $\{p_k;k\geq 0\}$ and $\{h_k;k\geq 0\}$ denote the offspring distribution function and the lifetime distribution function, and let $f(s)$ and $g(s)$ denote the p.g.f of $\{p_k;k\geq 0\}$ and $\{h_k;k\geq 0\}$ respectively, $m$ the offspring mean and $l$ the lifetime mean, that is,
   \begin{equation}\label{ml}
   f(s)=\sum_{k=0}^{\infty}p_ks^{k},~~~~g(s)=\sum_{k=0}^{\infty}h_ks^{k},~~~m=\sum_{k=0}^{\infty}kp_k~~~\text{and}~~~l=\sum_{k=0}^{\infty}kh_k.
   \end{equation}
  Further, we define
   \begin{equation}\label{e1.1}
       q_k=
\begin{cases}
\sum_{i=k}^{\infty}h_i/\sum_{i=k-1}^{\infty}h_i, \ \ \ if~ \sum_{i=k-1}^{\infty}h_k \neq 0\\
0, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  if~ \sum_{i=k-1}^{\infty}h_k=0,
\end{cases}
\end{equation}
provide $\prod_{i=1}^{0}=1$. It is easy to see that
\begin{equation}\label{e1.2}
       h_k=(1-q_{k+1})\prod_{i=1}^{k}q_i.
\end{equation}
If $L$ denotes the life length of an individual, then $h_k=\mathbb{P}(L=k)$ and $q_k=\mathbb{P}(L\geq k|L\geq k-1)$.
\par
 The process can be regarded as representing an evolving population of individuals. It starts at time $0$ with $Z_0$ newborn individuals, at the end of first season, each individual produces $k$ ($k\geq 0$) offspring and remains alive with probability $q_1p_k$, or dies with probability $1-q_1$. At the end of $n$th season, an individual of age $j<n$ (alive for $j$ seasons) has a probability $q_jp_k$ of producing $k$ ($k\geq0$) offspring and remains alive, or it has a probability $1-q_j$ of immediate death. When $h_0>0$, it means there are individuals that die at end of their first season without any descendants.
 \par
 Whittle \cite{Whittle} was first to consider Galton-Watson processes in which individuals have variable lifetimes and gave the following extinction criterion.
 \begin{proposition}\label{pro} The extinction probability $q$ of a Galton-Watson process in which individuals have lifetimes $\{Z_n;n\geq0\}$ is the smallest nonnegative root of $g(f(s))=s$ in $[0,1]$, where $f(s)$ and $g(s)$ are defined in (\ref{ml}). Moreover, $q=1$ if and only if $f'(1)g'(1)\leq1$.
 \end{proposition}
  \par
  In this paper, we look at the process $\{Z_n;n\geq0\}$ as a Galton-Watson process with countably infinitely many types (we use the abbreviation GWCIMT to represent such process).
 Consider a GWCIMT $\{\emph{\textbf{Z}}_{n}=(Z_n^{(1)},Z_n^{(2)},Z_n^{(3)},\cdots); n\geq 0\}$ defined on $\mathbb{Z}_+^{\infty}$, where $Z_{n}^{(t)}$ represents the number of individuals of type $t$ particle at the $n$th generation, and $t$ in the type-set $\chi=\{1,2,3,\cdots\}$. The transition function is
\begin{eqnarray*}\
   \mathbb{P}(\emph{\textbf{i}},\emph{\textbf{j}})&=& \mathbb{P}(\emph{\textbf{Z}}_{n+1}=\emph{\textbf{j}}|\emph{\textbf{Z}}_{n}=
   \emph{\textbf{i})}\ \ \ \ \ \ \ \ \ \ \ \ \ \emph{\textbf{i}},\emph{\textbf{j}} \in \mathbb{Z}_+^{\infty} \\
   &=& \text{the coefficient of }\textbf{s}^{\emph{\textbf{j}}}\text{ in } [\emph{\textbf{f}}(\emph{\textbf{s}})]^{\emph{\textbf{i}}},
\end{eqnarray*}
where $\emph{\textbf{f}}(\emph{\textbf{s}})=(\emph{f}^{(1)}(\emph{\textbf{s}}),\emph{f}^{(2)}(\emph{\textbf{s}}),\cdots)$ and
\begin{eqnarray*}\
\emph{f}^{(i)}(\emph{\textbf{s}})=\sum_{\emph{\textbf{j}}\in \mathbb{Z}_+^{\infty}}p_{\emph{\textbf{j}}}^{i}\prod_{k\in\chi}s_k^{j_k}.~~~~~~~~\textbf{s}\in [0,1]^{\chi}
\end{eqnarray*}
 $p_{j_{1},j_{2},j_{3}\cdots}^{(i)}$ is the probability that a type $i$ parent individual produces $j_{1}$ particles of type $1$, $j_{2}$ individuals of type $2$,$\cdots$, $j_{n}$ individuals of type $n$ and so on.
 \par
Naturally, if we think of an individual of age $i$ as type $i+1$ individual, then the only possible transition is from $k$th type to $1$th type and $(k+1)$th type for $k>0$. Hence we can write out the specific expression of $p_{\emph{\textbf{j}}}^{i}$, that is
\begin{equation}\label{tran}
  p_{j_{1},j_{2},j_{3}\cdots}^{(i)}=
\begin{cases}
  1-q_{i}, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ j_{1},j_{2},\cdots=0 \\
  q_{i}p_{k},\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ j_{1}=k,j_{i+1}=1,j_{2},\cdots,j_{i},j_{i+2},\cdots=0 \\
  0,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ otherwise.
\end{cases}
\end{equation}
We can also calculate the mean progeny matrix $\emph{\textbf{M}}=(m_{ij};i,j\geq 1)$ of $\{\emph{\textbf{Z}}_n;n\geq0\}$ with
\begin{equation}\label{mean}
m_{ij}=
\begin{cases}
mq_i,\ \ \ & \mbox{if}~~~j=1 \\
q_i,\ \ \  & \text{if}~~~j=i+1 \\
0, \ \ \ \ & otherwise,
\end{cases}
\end{equation}
where $m_{ij}=\mathbb{E}Z_{1j}^{(i)}$ represents the expected number of type $j$ offspring of a single type $i$ individual in one generation. We give a natural classification of Galton-Watson process in which individuals have variable lifetimes according to the extinction criterion of Proposition \ref{pro}.
\begin{definition}\
 A Galton-Watson process with countably infinitely many types is called a Galton-Watson process in which individuals have variable lifetimes if its transition function satisfies (\ref{tran}), where $\{q_k;k>0\}$ and $\{p_k;k\geq0\}$ are defined in (\ref{ml}) and (\ref{e1.1}). Furthermore, the process is called subcritical if $ml<1$, critical if $ml=1$, and supercritical if $ml>1$, where $m,l$ are defined in (\ref{ml}).

\end{definition}
\par

 Obviously, if $h_1=1$, then $\{Z^{(1)}_n; n\geq 0\}$ is actually a classical Galton-Watson process.
If the lifetimes variable of each individual is bounded (i.e., there exists $k>0$, $h_i=0$ for all $i\geq k$), then $\{\emph{\textbf{Z}}_{n}; n\geq0\}$ degenerates to a classical multi-type Galton-Watson process with a finite number of types. To avoid trivialities, we make the following basic assumptions throughout this paper:
\begin{flalign*}
&\emph{\bf \text{A1:}}~\emph{\textbf{Z}}_0=(1,0,0,\cdots).&\\
&\emph{\bf \text{A2:}}~q_i>0~~\text{for all}~~i>0.&\\
&\emph{\bf \text{A3:}}~ 0<m<\infty.&

\end{flalign*}
\par
    A matrix $\emph{\textbf{A}}$ is said to be irreducible if for any pair $(i,j)$ there exists $n\geq 1$ such that $(\emph{\textbf{A}}^{n})_{ij}>0$. A GWCIMT is said to be irreducible if and only if its mean progeny matrix is irreducible, then the mean matrix $\emph{\textbf{M}}$ of $\{\emph{\textbf{Z}}_n; n\geq 0\}$ is irreducible consequently by assumption A2 and (\ref{mean}).

\par
 When the number of types is finite and the mean matrix of the process is irreducible, it is well-known that the extinction probability vector $\emph{\textbf{q}}<\textbf{1}$ if and only if the Perron-Frobenius eigenvalue, or in another words the spectral radius of mean matix is strictly greater than $1$ (see \cite{Athreya1972Branching}). When the type is countably infinite, the analogue of Perron-Frobenius eigenvalue often refers to the convergence radius of the power series $\sum_{k\geq 0}r^k(\emph{\textbf{A}}^k)_{ij}$, where $\emph{\textbf{A}}$ is the mean progeny matrix of the process. For any irreducible matrix $\emph{\textbf{A}}$, Vere-Jones \cite{Jone} proves that there exists a positive number $\tau$,
 \begin{equation}\label{e4.1}
\lim\limits_{k\rightarrow\infty}\{(\emph{\textbf{A}}^k)_{ij}\}^{1/k}=1/\tau.
\end{equation}
 for any pair $(i,j)$.
  $\tau$ and $1/\tau$ is often referred to as the convergence radius and the convergence norm of $\emph{\textbf{A}}$ respectively. Note that the convergence norm is equivalent to spectral radius of $\emph{\textbf{A}}$ in the finite type case. Sagitov \cite[Definition 7]{Sagitov2013Linear} gives a natural classification for GWCIMT according to the asymptotic properties of its mean matrix $\emph{\textbf{A}}^{n}=(a_{ij}^{(n)})_{i,j=1}^{\infty}$ as $n\rightarrow\infty$.
\par
\begin{definition}\ \label{def}
An irreducible matrix $\emph{\textbf{A}}$ is called $\tau\emph{\text{-recurrent}}$ or $\tau\emph{\text{-transient}}$ depending on the divergence or the convergence of the series $\sum_{k\geq 0}\tau^k(\emph{\textbf{A}}^k)_{ij}$. A $\tau\emph{\text{-recurrent}}$ matrix $\emph{\textbf{A}}$ is called $\tau\emph{\text{-positive}}$ if $\lim\limits_{k\rightarrow\infty}\tau^k(\emph{\textbf{A}}^k)_{ij}>0$ for some pair $(i,j)$ and hence for all pairs $(i,j)$, and it is called $\tau\emph{\text{-null}}$ if this limit is zero. Furthermore, a Galton-Watson process with countably many types is said to be $\emph{\text{transient}}$ \{$\emph{\text{positive recurrent}}$, $\emph{\text{null recurrent}}$\} if its mean matrix is $\tau\emph{\text{-transient}}$ \{$\tau\emph{\text{-positive}}$, $\tau\emph{\text{-null}}$\}.
\end{definition}
\par
  For a general GWCMIT with offspring generating function $\emph{\textbf{g}}(\emph{\textbf{s}})$, there are two different types of extinction: global extinction $\emph{\textbf{q}}$ and partial extinction $\tilde{\emph{\textbf{q}}}$ whose entries contain the probability of global extinction and partial extinction respectively. $\emph{\textbf{q}}$ and $\tilde{\emph{\textbf{q}}}$ are both the solution of $\emph{\textbf{g}}(\emph{\textbf{s}})=(\emph{\textbf{s}})$ and it is possible for the population of each type to become extinct almost surely ($\tilde{\emph{\textbf{q}}}=\textbf{1}$) while the total population size explodes with positive probability ($\emph{\textbf{q}}<\textbf{1}$; see \cite{Hautphenne2013Extinction}). For the process $\{\emph{\textbf{Z}}_{n}; n\geq0\}$ considered in this paper, by (\ref{tran}), we can write $\emph{\textbf{f}}(\emph{\textbf{s}})=(\emph{\textbf{s}})$ componentwise as
 \begin{equation}\label{ite}
 1-q_i+q_is_{i+1}f(s_1)=s_i,~~~~~~(i\geq1).
 \end{equation}
 Iterating (\ref{ite}) to infinity and using (\ref{e1.2}) give $g(f(s_1))=s_1$. Then from Proposition \ref{pro} and (\ref{ite}), we know $\emph{\textbf{q}}\equiv\tilde{\emph{\textbf{q}}}$.

GWCIMTs have been investigated by several authors. Moyal \cite{J1962Multiplicative} considers a general type space and proves that the extinction probability is a solution of $\emph{\textbf{s}}=\emph{\textbf{g}}(\emph{\textbf{s}})$, where $\emph{\textbf{g}}(\cdot)$ is the progeny generating function of such processes. Hautphenne et al \cite{Hautphenne2013Extinction} give a sufficient condition for the appearance of the situation $\tilde{\emph{\textbf{q}}}=\textbf{1}$ and $\tilde{\emph{\textbf{q}}}<\textbf{1}$ in terms of truncated Galton-Watson processes and discuss the connection between the convergence norm of the mean progeny matrix and  the extinction criteria in the irreducible and reducible cases. Braunsteins and Hautphenne \cite{2017arXiv170602919B} consider a so-called lower Hessenberg branching processes with countably many types, which restrict individuals of type $i$ to give birth to type $j\leq i+1$ only. They prove the existence of a continuum of fixed points of the progeny generating function and show that the minimum of this continuum is $\emph{\textbf{q}}$ and the maximum is $\tilde{\emph{\textbf{q}}}$. Sagitov \cite{Sagitov2013Linear} considers a linear-fractional GWCIMT whose mean progeny matrix $\emph{\textbf{A}}$ has a form of
\[
\emph{\textbf{A}}=\emph{\textbf{H}}+a\emph{\textbf{H}}\emph{\textbf{1}}^{t}\emph{\textbf{g}},
\]
where $\emph{\textbf{H}}=(h_{ij})_{i,j=1}^{\infty}$ is a sub-stochastic matrix, $\emph{\textbf{g}}=(g_1,g_2,\cdots)$ is a proper probability distribution, $\emph{\textbf{1}}^t$ is the transpose of the row vector $\textbf{1}=(1,1,\cdots)$ and $a$ is a positive constant.
 For linear-fractional GWCIMT, Sagitov presents criteria for $\tau$-positive recurrence \cite[Theorem 8]{2017arXiv170602919B} which depend on the value of $d(s)=\sum_{n\geq1}d_ns^n$ in the point of its convergence radius, where
 \[
 d_n=\emph{\textbf{g}}\emph{\textbf{H}}^n\emph{\textbf{1}}^t.
 \]
 While it is not easy to know about the specific form of $d(s)$ to get the recurrence property of linear-fractional GWCIMT.
 \par
  In this paper, we first derive criteria for convergence radius $\gamma$ of $\emph{\textbf{M}}$ to be greater, smaller than and equal with 1. We give formula of calculating $\gamma$ and show that $\gamma$, in supercritical case, is actually the extinction probability of a Galton-Watson process, and the method of which are more efficient than Lemma \ref{L3.1} below. Next, we give clear criteria for $\emph{\textbf{M}}$ to be $\gamma$-transient, $\gamma$-positive and $\gamma$-null recurrent from which the ergodic property of the process is discussed. The criteria only rely on the properties of lifetime probability generating function $g(s)$ which are more clear and easier to be verified than the criterion of linear-fractional GWCIMT. Labeling each individual to translate the process to GWCIMT helps to know more about the evolution of Galton-Watson process in which individuals have variable lifetimes. We can see the exact geometric growth of total population size of $\{\emph{\textbf{Z}}_n;n\geq0\}$ in supercritical case due to $\gamma$. At last, an example of geometric law of lifetimes is given to illustrate our results.
\par
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\noindent {\large \bf 2. Results and Proofs}
 \vspace{3mm}
 \par
  \par
\par
 Denote convergence norm of $\emph{\textbf{M}}$ by $\rho$ then $\rho=\gamma^{-1}$, where $\emph{\textbf{M}}$ is defined in (\ref{mean}). Further, let vector $\emph{\textbf{v}}=\{v^{(i)}\}_{i>0}$ and $\emph{\textbf{u}}^{T}=\{u^{(i)}\}_{i>0}$ be the $\gamma$-invariant measure and vector of $\emph{\textbf{M}}$ respectively, with $\emph{\textbf{u}},\emph{\textbf{v}}>\textbf{0}$ and
\begin{eqnarray*}

\gamma\emph{\textbf{v}}\emph{\textbf{M}}=\emph{\textbf{v}},~~~~~ \gamma\emph{\textbf{M}}\emph{\textbf{u}}=\emph{\textbf{u}},
\end{eqnarray*}
where we write $\emph{\textbf{v}}>\textbf{0}$ to indicate that $v^{(i)}\geq 0$ for all $i>0$ but at least one component strictly greater.
Denote the $(k\times k)$ northwest corner truncation matrix of $\emph{\textbf{M}}$ by $\emph{\textbf{M}}^{(k)}$, the spectral radius of $\emph{\textbf{M}}^{(k)}$ by $\rho_k$.
\par
 It is known that $\tilde{\emph{\textbf{q}}}<1$ if $\rho>1$ and $\tilde{\emph{\textbf{q}}}=1$ if $\rho\leq1$ in the irreducible case (see \cite{Hautphenne2013Extinction}). Convergence norm $\rho$ plays a key role in investigating Galton-Watson processes with countably many types. However, it is not easy to evaluate $\rho$. In this paper we provide a more efficient method to evaluate the convergence norm of $\emph{\textbf{M}}$ (an unique solution of an equation which relies on the specific form of $\emph{\textbf{M}}$) than Lemma \ref{L3.1} below (approximating method), when $\emph{\textbf{M}}$ is the mean progeny matrix of a Galton-Watson process in which individuals have variable lifetimes. To do so, we introduce the following two Lemmas which correspond to Theorem 6.8 and 6.4 in Seneta \cite{Seneta2006Non}.
\begin{lemma}\label{L3.1}
If the matrix $\textbf{M}^{(k)}$ is irreducible for all $k>0$, then $\rho_k\uparrow\rho$. \hfill $\Box$
\end{lemma}
\begin{lemma}\label{L3.2} The matrix $\textbf{M}$ is $\gamma$-positive if and only if the $\gamma$-invariant measure and $\gamma$-invariant vector satisfy
\begin{eqnarray*}
\textbf{v}\textbf{u}=\sum_{i>0}u^{(i)}v^{(i)}<\infty.
\end{eqnarray*}\hfill $\Box$
\end{lemma}
Define $Q_k=\prod_{i=1}^{k}q_i$ if $k\geq 1$ and $Q_0=1$, $F(s)=m\sum_{j=1}^{\infty}Q_js^j$. Denote the radius of convergence of the power series $F(s)$ by $R$. Noting that $Q_k=P(L>k)$ by (\ref{qk}) below, then $F(1)=ml$. Now we state our main theorems.
\begin{theorem} \label{th3.3}
(1) If $ml<1$ and $F(R)\geq1$, then $\gamma=\rho^{-1}>1$ is the unique solution of $F(s)=1$ in $(1,R]$.\\
(2) If $ml<1$ and $F(R)<1$, then $\rho=R^{-1}$.\\
(3) If $ml=1$, then $\rho=1$.\\
(4) If $ml>1$, then $\rho>1$ and $\gamma=\rho^{-1}$ is the extinction probability of a Galton-Watson process whose offspring generating function is
\begin{displaymath}
B(s)=\frac{1+mg(s)}{1+m}.
\end{displaymath}

\end{theorem}
\begin{proof} The characteristic polynomial of $\emph{\textbf{M}}^{(k)}$ is
\begin{eqnarray*}
f^{(k)}(
\lambda)=|\lambda E-\emph{\textbf{M}}^{(k)}|&=&\left|\begin{array}{ccccc}
    \lambda-mq_1 &    -q_1    & 0       & \cdots  &0 \\
    -mq_2        &    \lambda & -q_2    & \cdots  &0 \\
    -mq_3        &    0       & \lambda & \cdots  &0 \\
    \vdots       &    \vdots  & \vdots  & \ddots  &0 \\
    -mq_{k-1}    &    0       & 0       & \cdots  &-q_{k-1}\\
    -mq_{k}      &    0       & 0       & \cdots  &\lambda \\
\end{array}\right| \\
                           &=&(-mq_k)(-1)^{k+1}(-q_1(-q_2)\cdots(-q_{k-1}))+\lambda(-1)^{2k}f^{(k-1)}(\lambda) \\
                           &=&\lambda f^{(k-1)}(\lambda)-mq_1q_2\cdots q_{k},

\end{eqnarray*}
\par
 where $E$ is the identity matrix. Let $s=\lambda^{-1}$, $\phi_k(s)=f^{(k)}(s^{-1})$. Hence the recursion becomes $s\phi_{k}(s)=\phi_{k-1}(s)-msQ_k$ for $k>1$, i.e., $s^j\phi_j(s)-s^{j-1}\phi_{j-1}(s)=-mQ_js^{j}$ valid for $j>1$. Now $\phi_1(s)=s^{-1}-mQ_1$, so summing gives
$$
s^k\phi_k(s)=1-m\sum_{j=1}^{k}Q_js^j,~~~~~~(k>1).
$$
Denote the subtracted term on right-hand side by $F_k(s)$.
\par
It follows that the positive zeros of $f^{(k)}(\lambda)$ are in one to one correspondence with the positive solutions of
\begin{equation}\label{equ}
F_k(s)=1.
\end{equation}
Clearly $F_k(0)=0$ and $F_k(s)$ is a polynomial which is convex increasing in $[0,\infty)$, hence (\ref{equ}) has exactly one positive root, denoted by $s_k$. It is easy to see that
\begin{equation}\label{qk}
Q_k=\sum_{j=k}^{\infty}h_j,
\end{equation}
which implies
$$
F_k(1)=m\sum_{j=1}^{k}Q_j=m\sum_{j\geq1}\sum_{i\geq j}h_i-r_k,~~~\mbox{where}~r_k:=m\sum_{j>k}\sum_{i\geq j}h_i.
$$
 So if $ml\leq 1$, then $F_k(1)\leq 1$ for all $k$ and hence (\ref{equ}) has a unique solution in $[1,\infty)$. The solution $s_k=1$ holds if and only if $ml=1$ and $r_k=0$ which contradicts assumption A2. Next, when $ml>1$, it is clear that $r_k\downarrow0$ as $k\uparrow\infty$. So there exists a positive number $N$, if $k>N$, $F_k(1)>1$ and then (\ref{equ}) has a unique solution $s_k\in(0,1)$. If $k<N$, then $F_k(1)<1$ and (\ref{equ}) has a unique solution $s_k\in(1,\infty)$.
\par
According to Lemma \ref{L3.1}, the convergence norm of $\emph{\textbf{M}}$ is $\rho=\lim_{k\rightarrow\infty}\rho_k=\lim_{k\rightarrow\infty}s_k^{-1}$. Hence we need to consider  $F(s)=\lim_{k\rightarrow\infty}F_k(s)$. Let $\hat{s}$ be the unique solution of
\begin{equation}\label{eq1}
F(s)=1,
\end{equation}
if there is a solution.
\par
Noting that $F(1-)=ml$, then there clearly is such a solution if $ml>1$ and $\hat{s}\in(0,1)$. Hence we first consider the case $ml>1$ and $s\in(0,1)$. It follows from the definition of $F_k(s)$ and (\ref{qk}) that
\[
F(s)=m\sum_{j\geq1}s^j\sum_{i\geq j}h_i=ms\frac{1-g(s)}{1-s},
\]
where $g(s)=\sum_{i\geq0}h_is^i$. Hence the limiting form of $(\ref{equ})$ becomes $1-s=ms(1-g(s))$, i.e.,
\begin{equation}\label{ps}
B(s):=\frac{1+msg(s)}{1+m}=s.
\end{equation}
\par
It can be verified that $B(s)$ is a probability generating function and hence (\ref{ps}) can be interpreted as the equation satisfied by the extinction probability of a Galton-Watson process whose offspring probability generating function is $B(s)$ and offspring mean is $m(1+l)/(1+m)$. This exceeds unity if and only if $ml>1$. We are assuming this condition at present, so $\hat{s}$ is this extinction probability (see \cite{Athreya1972Branching}). Also, $\hat{s}>B(0)=(1+m)^{-1}$, implying that $\rho\in(1,1+m)$. Then $\emph{(4)}$ of the theorem is proved.
\par
Now there is an important point to be made here because (\ref{ps}) also has the solution $s=1$ for any value of $ml$. This solution is an artefact of the multiplication by $1-s$ of (\ref{equ}) required to produce (\ref{ps}). This addition solution is irrelevant to the problem.
\par
If $ml=1$, then (\ref{ps}) has the unique solution $\hat{s}=1$. Hence $\rho=1$ in this case and $\emph{(3)}$ follows.
\par
In the case that $ml<1$ it is not clear whether (\ref{eq1}) has a solution. Recall that $F(1)=ml$ implies $R\geq1$. Hence if $R>1$ and $F(R-)\geq1$ (can be $\infty$), then it is clear that there is a unique solution $\hat{s}=\rho^{-1}\in(1,R]$ and $\emph{(1)}$ follows.
\par
However, if $R>1$ and $F(R-)<1$, or $R=1$, (\ref{eq1}) has no solution. In these cases, $\rho=\lim_{k\rightarrow\infty}\rho_k$ and $\rho_k^{-1}=s_k\downarrow R$ from the definition of $F_k(s)$, implying $\rho=1/R$. Then $\emph{(2)}$ is proved and the proof of the theorem is complete.
\hfill $\Box$
\end{proof}
 \begin{remark}\label{mar1}
In case (2) of the above theorem, (\ref{eq1}) has no solution and hence the $\gamma$-invariant vector and measure do not exist.
\end{remark}
\par
In the following theorem, we derive the ergodic property of $\{\emph{\textbf{Z}}_n;n\geq0\}$ due to Definition \ref{def}.
\begin{theorem}\label{th3.4}
(1) If $ml<1$ and $F(R)<1$, then $\{\textbf{Z}_n;n\geq0\}$ is transient. \\
(2) If $ml<1$ and $F(R)\geq1$, or $ml=1$ and $g''(1-)<\infty$, or $ml>1$, then $\{\textbf{Z}_n;n\geq0\}$ is positive recurrent.\\
(3) If $ml=1$ and $g''(1-)=\infty$, then $\{\textbf{Z}_n;n\geq0\}$ is null recurrent.
\end{theorem}
\begin{proof}\ From Seneta \cite[Theorem 6.2]{Seneta2006Non}, for a $\tau$-recurrent matrix $\emph{\textbf{A}}$, there always exists $\tau$-invariant measure. Then by Remark \ref{mar1} and Theorem \ref{th3.3}, $\emph{\textbf{M}}$ is $\gamma$-transient if $ml<1$, $R>1$ and $F(R-)\leq1$, or $ml<1$ and $R=1$. Hence $\{\emph{\textbf{Z}}_n; n\geq 0\}$ is transient in these cases. We assume the existence of $\hat{s}=\rho^{-1}$ where $\hat{s}$ solves (\ref{eq1}) in the rest of proof.
\par
Without loss of generality, let $u^{(1)}=1$, $v^{(1)}=1$. Then by $\emph{\textbf{M}}\emph{\textbf{u}}=\rho\emph{\textbf{u}}$ and $\emph{\textbf{v}}\emph{\textbf{M}}=\rho\emph{\textbf{v}}$, one can present that
\begin{eqnarray}\label{e3.10}
u^{(k)}=\cfrac{f^{(k-1)}(\rho)}{q_1q_2\cdots q_{k-1}},~~~~~~~~~~~~
v^{(k)}=\cfrac{q_1q_2\cdots q_{k-1}}{\rho^{k-1}}.
\end{eqnarray}
Noting that the derivation of the explicit form (\ref{e3.10}) shows that they are the unique outcomes starting from $u^{(1)}=v^{(1)}=1$, which implies that $\emph{\textbf{M}}$ is $\gamma$-recurrent by Kendall's Theorem (See \cite{Continuous-time Markov chains}).
\par
Next, consider the infinite sum
\begin{eqnarray}
\emph{\textbf{v}}\emph{\textbf{u}}&=&\sum_{i\geq1}v^{(i)}u^{(i)}=\sum_{i\geq1}\hat{s}^i\phi_i(\hat{s}) \nonumber \\
&=&\sum_{i\geq1}(1-m\sum_{j=1}^{i}Q_j\hat{s}^j)=m\sum_{i\geq1}\sum_{j>i}Q_j\hat{s}^j          \nonumber\\
&=&m\sum_{j=2}^{\infty}(j-1)Q_j\hat{s}^j=:S.\label{sum}
\end{eqnarray}
The first equality follows from (\ref{e3.10}) and the definition of $\hat{s}$ and $\phi_k(\cdot)$ and the second equality from (\ref{equ}). Then we only need to consider in which case $S$ is finite due to Lemma \ref{L3.2}.
\par
If $ml>1$, then $\hat{s}<1$ by Theorem \ref{th3.3} and hence the sum (\ref{sum}) is finite.
\par
From the definition of $Q_j$ and (\ref{sum}), we obtain
\begin{equation}\label{eee}
S=m\sum_{i=2}^{\infty}h_i\sum_{j=2}^{i}(j-2)\hat{s}^j.
\end{equation}
If $ml=1$, then $\hat{s}=1$ and clearly $S<\infty$ if and only if $g''(1-)<\infty$.\par
If $ml<1$ then $\hat{s}>1$ and reversing the order of summation in the inner sum of (\ref{eee}) it becomes
\[
S=m\sum_{i=2}^{\infty}ih_i\hat{s}^i\sum_{k=0}^{i-2}(1-(k+1)/i)\hat{s}^{-k}.
\]
The monotone convergence theorem implies that the inner sum converges to $\sum_{k\geq0}\hat{s}^{-k}<\infty$. It follows that $S<\infty$ if and only if $g'(\hat{s}-)<\infty$. Recall that $\hat{s}$ solves (\ref{eq1}), then
\begin{eqnarray}
1&=&m\sum_{i\geq1}Q_i\hat{s}^i=m\sum_{j\geq1}\hat{s}^j\sum_{i\geq j}h_i=m\sum_{i\geq1}h_i\sum_{j=1}^{i}\hat{s}^j \nonumber\\
&\geq&m\sum_{i\geq1}h_ii\hat{s}^{i-1}=mg'(\hat{s}).\label{ee2}
\end{eqnarray}
 The inequality follows from the fact that $\sum_{j=1}^is^j\geq is^{i-1}$ if $s\geq1$ for all $i>0$. Then the proof is complete according to the Lemma \ref{L3.2} and Definition \ref{def}.
\hfill $\Box$ \end{proof}
\par
In general, compared to the case that the mean matrix is $\tau$-transient or $\tau$-null in which the invariant vector and measure correspond to the convergence norm of mean matrix do not exist or can not be normalized, it is easier to discuss limit theorems of a GWCIMT when the mean matrix is $\tau$-positive. In this case, the mean matrix behaves like a finite matrix and there are several published results. Sagitov \cite[Propostion 9,10,11]{Sagitov2013Linear} presents basic asymptotic results for positive recurrent linear-fractional GWCIMT in supercritical, critical and subcritical case. Vatutin et al \cite[Theorem 3]{VV} extend mentioned Sagitov result in critical case by considering a general class of the reproduction generating function of individuals and without assuming the finiteness of the second moments for the offspring number of individuls. In supercritical and positive recurrent case, Moy \cite[Theorem 1]{1967Extensions} extends the theorem of Everret, Ulam and Harris to GWCIMT. It tells that if
\begin{equation}\label{last}
\sum_{i=1}^{\infty}v^{(i)}\mathbb{E}\big((\emph{\textbf{Z}}_1\emph{\textbf{u}})^2|\emph{\textbf{Z}}_0=\emph{\textbf{e}}_i\big)
\end{equation}
where $\emph{\textbf{e}}_i=(0,\cdots,0,1,0,\cdots)$ with 1 in the $i$th component, then there is a nonnegative real random variable $Y$ with a finite second moment such that for any $\emph{\textbf{w}}$ satisfying $\emph{\textbf{w}}\leq c\emph{\textbf{u}}^t$ for some positive constant $c$, the sequence $\{\rho^{-n}\emph{\textbf{Z}}_n\emph{\textbf{w}}^t\}$ converges in mean square to $Y\emph{\textbf{v}}\emph{\textbf{w}}^t$. Here $\emph{\textbf{v}}$ and \emph{\textbf{u}} are normalized $\tau$-invariant measure and vector with $\emph{\textbf{v}}\emph{\textbf{u}}=1$.
\par
From Theorem \ref{th3.3} and \ref{th3.4}, Galton-Watson process in which individuals have variable lifetimes covers all the three cases of GWCIMT. In the case of positive recurrence, one can easily apply results of Sagitov, Vatutin and Moy to get the corresponding properties for the process considered in this paper. The total population size of $\{\emph{\textbf{Z}}_n;n\geq0\}$ is actually the population size of $\{\emph{Z}_n;n\geq0\}$, hence constructing a Markovian GWCIMT helps understand the evolution of Galton-Watson process in which individuals have variable lifetimes. Here we only give the property in supercritical case. It is easy to know (\ref{last}) holds if and only if $f''(1)<\infty$, then we have the following proposition by applying Theorem $1$ of \cite{1967Extensions}.
\begin{proposition}
If $ml>1$ and $f''(1)<\infty$, then $\lim_{n\rightarrow\infty}\rho^{-n}\mathbb{E}(\textbf{1}\cdot\textbf{Z}_n)=(1+1/m)S^{-1}$, where $S$ is defined in (\ref{sum}).
\end{proposition}

\begin{example} Consider a geometric law of lifetimes. Let $g(s)=(1+l-ls)^{-1}$, where $l$ is a fixed positive number, then $g'(1)=l$. Denote the offspring mean by $m$ and assume $ml>1$. It follows from (\ref{ps}) that $\rho=(1+m)r$, where $r:=l/(1+l)$.
\par
calculation yields $Q_j=r^j$, so
\[
\sum_{j=1}^{k}Q_js^j=rs\frac{1-(rs)^k}{1-rs},
\]
and hence
\[
s^k\phi_k(s)=1-mrs\frac{1-(rs)^k}{1-rs}.
\]
It follows that
\[
\hat{s}^k\phi_k(\hat{s})=(1+m)^{-k}
\]
where $\hat{s}$ is the unique solution of (\ref{eq1}). We thus conclude that $\emph{\textbf{v}}\emph{\textbf{u}}$ converges, hence the process is positive recurrent in this case.
\par
Further calculation yields
\[
\lim_{n\rightarrow\infty}\rho^{-n}\mathbb{E}(\emph{\textbf{1}}\cdot\emph{\textbf{Z}}_n)=1.
\]
\end{example}
\section*{Acknowledgement}
\par
 This work is substantially supported by the National Natural Sciences Foundations of China (No. 11771452, No. 11971486) and Natural Sciences Foundations of Hunan (No. 2020JJ4674).

config: arxiv


TITLE: Rewrite $\int_{0}^1 \int_{0}^{1-y^2} \int_{0}^{1-y} dz \ dx \ dy$ in the five other orders of integration
QUESTION [1 upvotes]: I'm struggling to find limits of integration when the outermost integral depends on $x$ and $z$.
For example, when we study the shadow of the region on the $xz$-plane, we find that it's a square. But its projection upwards (in the direction of the positive $y$ axis) depends on two different surfaces (the cylinder $1-y^2$ and the plane $1-y$), and I don't know how to express everything in terms of one variable.
How can I approach these kinds of exercises?
Edit: my textbook says we should verify the following equalities
$$z = 1 - y \iff y = 1- z \\ x = 1 - y^2 \ \  \underset\iff{y\geq0} \ \ y = \sqrt{1-x}$$
Therefore
$$x = z^2 - 2z \\ z=1-\sqrt{1-x}$$
But what are these surfaces bounds of? I'm really confused.

REPLY [3 votes]: Simplest (and easiest) way to approach this kind of problems is to look straight at the limits of integration.
The domain can be described by the following inequalities.
$$0\leq y \leq 1,\ 0\leq x \leq 1-y^2,\ 0\leq z\leq 1-y $$
Those are equivalent to the following
$$0\leq z\leq 1, \ 0\leq x\leq 1,\ y^2 \leq 1-x, \  0\leq y\leq 1-z $$
Well, the last two inequalities can be written as $0\leq y\leq \min (\sqrt{1-x},\ 1-z)$.
This leaves us with $$0\leq z\leq 1,\ 0\leq x\leq 1,\ 0\leq y\leq \min (\sqrt{1-x},\ 1-z) $$ and the integral can be written as $$\int_0^1 \int_0^1 \int_0^{\min (\sqrt{1-x},\ 1-z)} dydxdz. $$


TITLE: About the definition of divergence
QUESTION [0 upvotes]: The divergence is defined as:
$\nabla . \mathbf{A}=\lim \limits_{V \to 0}  \dfrac{ \unicode{x222F}_{\partial V} \mathbf{A}.d\mathbf{S}}{V}$
My question is of two parts:
$(1)$ If we are using Cartesian coordinate system, should $V$ be necessarily a cuboid?
$(2)$ If we are using another coordinate system, should $V$ be necessarily a cuboid?

REPLY [1 votes]: If $\bf A$ is $C^1$ and the border of $V$ is piecewise $C^1$, using the divergence theorem you can prove that this definition is equivalent to the alternative definition $\sum\bf\partial_i A_i$.


TITLE: Explicit example of Abelian group with non-zero ulm invariants for infinite ordinals
QUESTION [1 upvotes]: I have seen a few posts here about Ulm invariants of infinite Abelian groups, and I know that countable Abelian groups are classified up to isomorphism by their Ulm invariants (and I think any combination of Ulm invariants is possible?). If I recall correctly, the length of the group (supremum of ordinals such that Ulm invariants are non-zero) can be any countable ordinal. However, I don't have a sense of what a group with infinite Ulm invariants should look like, I haven't been able to puzzle it out on my own, and I haven't been able to find any source that gives an explicit example of one.
I would appreciate it if someone could provide me with one (or multiple) explicit constructions of a reduced Abelian p-group with elements of infinite height, and ideally a way to construct one of any given length.

REPLY [1 votes]: Any ordinal can be the length of a reduced abelian p-group. A construction by generators and relations of a simply presented group whose length is a successor ordinal  was given in E. A. Walker, The groups $P_\beta$, Symposia Math. (13), 482-488, (1974); the construction is presented on p. 356, L. Fuchs, Abelian Groups, (2015). For limit ordinals, just take the direct limit of a direct system of groups of smaller length. These examples probably occur in earlier editions of Fuchs, but I don't know where.
Here is an explicit example  due to Pruefer, of a countable p--group $G$ of length $\omega + 1$. Let $a_0, a_1, \dots$ be a countable set of generators with relations $pa_0=0,\ pa_1=a_0,\dots, p^na_n=a_0,\dots$. Then $a_0$ has order $p$ and height $\omega$ and for all $n$, $a_n$ has order $p^{n+1}$ and height $0$.


text:
\section{Conclusion}
\label{sec:conclu}

In this paper, we introduced SSNMF, a new variant of the NMF model combining the assumptions of separability and sparsity.
We presented \brassens{}, an algorithm able to solve exactly SSNMF, based on SNPA and an exact sparse NNLS solver.
We showed its efficiency for various setups and in the successful unmixing of a multispectral image.
The present work provides a new way to perform underdetermined blind source separation, under mild hypothesis, and a new way to regularize NMF.
It makes NMF identifiable even when atoms of $W$ are nonnegative linear combinations of other atoms (as long as these combinations have sufficiently many non-zero coefficients).
Further work includes the theoretical analysis of the proposed model and algorithm in the presence of noise.

\subsubsection*{Acknowledgments.}

The authors are grateful to the reviewers, whose insightful comments helped improve the paper.
NN and NG acknowledge the support by the European Research Council (ERC starting grant No 679515), and by the Fonds de la Recherche Scientifique - FNRS and the Fonds Wetenschappelijk Onderzoek - Vlanderen (FWO) under EOS project O005318F-RG47.

file: 2006.07553/arXiv SSNMF/70conclu.tex


TITLE: Step in Proof of Cardinality of Product of two Groups
QUESTION [3 upvotes]: Let $A,B < G$ be two subgroups of some group $G$, then I have a question on the proof of the following:
$$
 |AB| = \frac{|A||B|}{|A \cap B|}.
$$
Proof: Let $D = A \cap B$, arrange $A$ and $B$ in left cosets and right cosets regarding $D$ (which is also a subgroup)
$$
 A = D \cup s_2 D \cup s_3 D \cup \ldots \cup s_n D, \quad
 B = D \cup D t_2 \cup D t_3 \cup \ldots \cup D t_m
$$
Now the fact that $\frac{|A||B|}{|D|}$ are distinct results from the following equations:
If
$$
 s_{\alpha}D D t_{\beta} = s_{\alpha_1} D D t_{\beta_1}
$$
then
$$
 s_{\alpha_1}^{-1} s_{\alpha} D = D t_{\beta} t_{\beta}^{-1}
$$
As the first member of the last equation represents an element of $A$ while the second member represents an element of $B$, it results that the last equation implies $\alpha_1 = \alpha$ and $\beta_1 = \beta$.
The last step is not clear to me, why does $\alpha_1 = \alpha$ and $\beta_1 = \beta$ follows? Is it the case that if $aD = Db$ for $a \in A$ and $b \in B$ that $a=b=1$ follows? I don't believe this and cannot prove it either? Could someone please explain the last step?

REPLY [1 votes]: It is not true in general that $aD = Db$, $a \in A$, $b \in B$, implies $a = b$.  In fact $aD = Db$ is true for any choice of elements $a, b \in D$.  What's being used here is the fact that the $s_\alpha$ and $t_\beta$ are a set of representatives for the cosets.  This means if $s_\alpha \neq s_{\alpha_1}$ then $s_\alpha D \neq s_{\alpha_1}D$.  Turning this around, if $s_\alpha D = s_{\alpha_1}D$ then we must have $s_\alpha = s_{\alpha_1}$.
Now, you know that $s_{\alpha_1}, s_\alpha \in A$ and $D \subseteq A$ so $s_{\alpha_1}^{-1}s_\alpha D \subseteq A$.  Similarly, $Dt_\beta t_{\beta_1}^{-1} \subseteq B$.  As these are equal you get $s_{\alpha_1}^{-1}s_\alpha D \subseteq A \cap B = D$ and hence $s_{\alpha_1}^{-1}s_\alpha D = D$.  But this can be rewritten as $s_{\alpha_1}D = s_\alpha D$, therefore $s_{\alpha_1} = s_\alpha$.
A similar argument works for the $t_\beta$.


TITLE: Paper plane between two fans - is this possible?
QUESTION [8 upvotes]: The setup: two fans facing each other, distance around 1m. Both are turned on. In between them, place a simple paper plane and according to this video, it will fly.
https://www.youtube.com/watch?v=0ca9rxQkEiA
Is this really true? If yes, what is the physics behind it?
The fan that is blowing from behind seems to run at less RPM.

REPLY [0 votes]: I disagree with @Dmitry's answer. Although in the border the air has more speed it still has velocity in the middle and when the two "cylinders" of air collide the would create turbulence that would make it impossible to create a toroid. And if you managed to get a toroid with positive pressure it should be the same above and below the plane (as it would be a cylinder it should be symmetric) and then there wouldn't be a net force to counter gravity.
Apart from that, the video has a low resolution, the plane always pivots around the same point (like hanging from a wire), it goes near one of the fans with a no perfect aligned axis which should make it shake a lot and "fly" away. So, in my opinion, the video is fake.


TITLE: Induction: How does the electrons in the coil know in which way the magnetic field is moving?
QUESTION [0 upvotes]: It is well known that a coil will react to a moving external magnetic field in such a way, that the electrons inside it will start moving to produce an internal magnetic field, which counteracts the movement of the external magnetic field.
One question has struck my mind though; in order for the electrons inside the coil to flow in the right direction, they will have to "know"

The alignment of the external field.
The direction of movement of the external field.

Since the external magnetic field extends into the coil, it is not so hard to imagine the electrons "knowing" the alignment of the external magnetic field. What I don't understand is how the electrons can "know" the direction of movement of the external magnetic field, unless the movement in itself generates a field of some sort?
I have been thinking quite a bit on this, and the only clue that has struck my mind is that as two magnetic fields (aligned in the same way) approach other, the individual field lines slowly rotate versus each other, and goes from being opposite, to being aligned.
I have not studied physics at any higher level, so I would appreciate if you try to keep your answers as non-mathematical as possible.

REPLY [1 votes]: First, magnetic fields don't move. Every point in space has a (possibly zero) magnetic field vector at that point. And that vector at that point can change in time.
Electrons and other charges react to the electric field vector at the location of the charge and to the magnetic field vector at the location of the charge. They do not react to vectors at other locations.
And the vectors don't move from one location to another. There is a vector at each location at each time. Multiple locations could have vectors of the same size and direction, and a location could even have a zero vector. That's all OK.
Now, about changing magnetic fields (such as those associated with moving magnets, a magnet is a device, a magnetic field is when you have a vector at each location). Time changing magnetic fields are associated with particular arrangements of electric fields.
It's not entirely wrong to think of the arrangement of the electric fields being the cause of the change in the magnetic field. But really the two happen together, so there isn't really any evidence for one causing the other.
On the other hand, unlike an object that can have a position and a velocity and the velocity tells you how it is going and which can be affected by forces which then indirectly change the location by changing the velocity and the velocity changes the location. Unlike that, when you have an arrangement of electric field vectors throughout space and an arrangement of magnetic field vectors and some electric current then the arrangement of electric field vectors forces the magnetic field vectors to change a certain way (so it's like the velocity of the magnetic field isn't free to be adjusted, it is determined by the electric field, but it isn't moving, the vector at each point can change or stay the same), so unless the electric fields are balanced in a perfectly static way, then the magnetic fields have to change in time (though that's different than moving).  And the arrangement of magnetic fields, unless it perfectly balances the electric current, it will result in the electric field changing (so it's like the velocity of the electric field isn't free to be adjusted, it is jointly determined by the arrangement of magnetic field vectors and the electric current but the electric field isn't moving, the vector at each point can change or stay the same).
So you can have an electric and magnetic field vector at each point. And unless the arrangement is in perfect static balance, the fields will change. But an imbalanced electric field (one with a net circulation) determines the rate the magnetic field changes. And an imbalanced magnetic field (one with a net circulation different than the electric current) determines the rate the electric field changes.


TITLE: Difference between emission spectrum and black body spectrum
QUESTION [2 upvotes]: Why is an emission spectrum dependent on the chemical composition of the body's material, but the black body spectrum is independent of the material and only based on the body's temperature?  Aren't both spectrums caused by photons being emitted due to electrons transitioning between states?

REPLY [0 votes]: What you call emission spectra come from "optically thin" gases or objects. i.e. They are transparent to their own radiation. The light emerges unmodified by interactions with other atoms or ions and exhibits the characteristics of whatever atom, ion or molecule that emits the radiation.
A blackbody is always "optically thick" at all wavelengths (by definition), since it absorbs all radiation incident upon it and is at a uniform temperature. The requirement of thermal equilibrium means there are just as many emission as absorption processes and the material is in equilibrium with the radiation field, and that radiation field is a Planck function of the same temperature, which is completely independent of any details of the material involved.


TITLE: Creating a formal system which corresponds to multiplication (of natural numbers)
QUESTION [1 upvotes]: I've begun studying mathematical logic with Hodel's 'An Introduction to Mathematical Logic', and have just learnt about formal systems and their components. In the textbook, Hodel gives as an example of a formal system the system 'ADD', whose theorems correspond to true statements about addition; it is defined as follows:

Symbols: $| \,,+,=$
Formulae: statements of the form $x+y=z$ where $x,y$, and $z$ are expressions in $|$
Axiom: $|+|=||$
Rules of Inference: $\frac{x+y=z}{x|+y=z|}$ (R1) and $\frac{x+y=z}{y+x=z}$ (R2)

As an exercise (exercise 3 of section 1.3 for those with the book), Hodel then asks the reader to create a formal system 'MULT', which has theorems corresponding to our standard notions of multiplication; he also asks the reader to prove within this system that $|| \times ||| = ||||||$.
I believe I have a solution to this, but I would be grateful if someone wouldn't mind checking it for me - I'm new to this area of maths so want to make sure my fundamentals are strong. My system is defined as follows:

Symbols: $|\, , \times, + , =$
Formulae: statements of the form $x\times y = z$ or $x+y=z$ where $x, y,$ and $z$ are expressions in $|$
Axioms: $|+|=||$ (A1) and $| \times x = x$ (A2)
Rules of inference: $\frac{x+y=z}{x|+y=z|}$ (R1), $\frac{x+y=z}{y+x=z}$ (R2), $\frac{x\times y=z}{x|\times y=z+y}$ (R3), and $\frac{x\times y=z}{y\times x=z}$ (R4)

and my proof that $\vdash_{\text{MULT}}|| \times ||| = ||||||$:

$|\times||=||$ (A2)
$||\times || = ||+||$ (R3)
$|+|=||$ (A1)
$||+| = |||$ (R1)
$|+|| = |||$ (R2)
$||+|| = |||| = || \times || $(R1 and step 2)
$||| \times || = ||||+||$ (R3)
$|| + || = ||||$ (step 6 and previous)
$||||+|| = |||||| = ||| \times ||$ (R1 twice and step 7)
$ || \times ||| = ||||||$ (R4) $\square$

What makes me doubt whether this is the simplest solution is that I have included all of ADD within my new system, however I suppose this does reflect how multiplication is defined in terms of iterated addition.
Improvements, corrections, and other solutions welcomed and greatly appreciated :)

REPLY [0 votes]: Here is my solution:

Symbols: $|, \cdot, =$
Formulas of the form $a\cdot b = c$
Axiom: $|\cdot|=|$
Rules of inference: $\frac{a\cdot b = c}{a|\cdot b = b c} $ (R1) and $ \frac{a\cdot b = c}{a\cdot b| = ac}$ (R2)

Here is the proof:

$|\cdot|=|$ (Axiom)
$||\cdot|=||$ (R1)
$||\cdot||=||||$ (R2)
$||\cdot|||=||||||$ (R2)

Q.E.D.


TITLE: Why learn to solve differential equations when computers can do it?
QUESTION [63 upvotes]: I'm getting started learning engineering math. I'm really interested in physics especially quantum mechanics, and I'm coming from a strong CS background.
One question is haunting me.
Why do I need to learn to do complex math operations on paper when most can be done automatically in software like Maple. For instance, as long as I learn the concept and application for how aspects of linear algebra and differential equations work, won't I be able to enter the appropriate info into such a software program and not have to manually do the calculations?
Is the point of math and math classes to learn the big-picture concepts of how to apply mathematical tools or is the point to learn the details to the ground level?
Just to clarify, I'm not trying to offend any mathematicians or to belittle the importance of math. From CS I recognize that knowing the deep details of an algorithm can be useful, but that is equally important to be able to work abstractly. Just trying to get some perspective on how to approach the next few years of study.

REPLY [3 votes]: One issue is that you have to communicate your problem to the computer, before it can "automatically" solve it. Even if the computer understands the communicated problem, the actual form of the reported solution or whether any solution at all is reported can depend on "irrelevant" minor details. Here is an excerpt from a question about my most recent issue of this form:

I "suspect" that $\mathcal{F}\{(1-r^2)^n\operatorname{circ}(r)\}(\rho)=\frac{n!J_{n+1}(2\pi\rho)}{\pi^n\rho^{n+1}}$.
...
I tried to tell Wolfram Alpha that I want to know the 2d radially symmetric Fourier transform of $\operatorname{circ}(r)$, but it neither understands "$\operatorname{circ}$" nor "2d Fourier transform". Finally I wrote

integral of 2 pi x besselj(0,2 pi r x) from 0 to 1
integral of 2 pi x (1-x^2) besselj(0,2 pi r x) from 0 to 1

The first query gave the correct answer $\frac{J_1(2\pi r)}{r}$, but the second gave a cryptic $\pi{}_0\tilde{F}_1(;3;-\pi^2r^2)$.  But

integral of x (1-x^2)^2 besselj(0,r x) from 0 to 1

where I omitted the $2\pi$ factors, gave the helpful answer $\frac{8J_3(r)}{r^3}$.

Note that the only answer to this question proposes a way to communicate this problem to the computer in a "more direct less preprocessed" way, with the result that the computer doesn't generate any answer at all.


TITLE: On the validity of a simplification made on a differential equation
QUESTION [0 upvotes]: I am given the differential equation
$$y' + \frac2t y = \frac{y^3}{t^2}, t > 0$$
and I am asked to show that the transformation $v = \frac1{y^2}$ makes this a linear equation.
I guessed I had to differentiate $v$ in order to make $y'$ appear and then substitute back into the original equation. So I did:
$$\frac{d}{dt}v = -\frac{2}{y^3}y'$$
And then I manipulated the original equation like so:
$$y' + \frac2t y = \frac{y^3}{t^2} \iff \frac{1}{y^3}\left(y' + \frac2t y \right) = \frac1{t^2}$$
which then makes everything look like something I'd want:
$$\frac{y'}{y^3} + \frac2t \frac{1}{y^2} = \frac1{t^2} \iff -\frac12 v' +\frac2tv = \frac1{t^2}$$
and of course this is an ordinary linear differential equation.
My problem is, can I actually divide through by $y^3$? How can I be sure that $y \neq 0$? I tried to look at the first equation and reason about it with the fact that $t > 0$ but I got nothing.

REPLY [0 votes]: You can treat the case $y=0$ separately. Look at the equation: $y=0$ is clearly a solution.
When $y\neq 0$, you define $v=\frac{1}{y^2}$ and what you've done is correct.


TITLE: Hermitian Matrices $AA^{*}=CC^{*}$,
QUESTION [0 upvotes]: Could any one help me proving or disproving this statement:
If $A,B,C$ are $n\times n$ invertible matrices, with $AA^{*}=CC^{*}$, and $B^{*}=B$, does it follow that
$$ ABA^{*}=CBC^{*} $$
thanks in advance!

REPLY [0 votes]: Here is one counter example with $2\times 2$ matrices.
$A=\begin{pmatrix} 1 & i \\ i & 1\end{pmatrix}\quad A^*=\begin{pmatrix} 1 & -i \\ -i & 1\end{pmatrix}\quad AA^*=\begin{pmatrix} 2 & 0 \\ 0 & 2\end{pmatrix}$
$C=\begin{pmatrix} 0 & i\sqrt{2} \\ i\sqrt{2} & 0\end{pmatrix}\quad C^*=\begin{pmatrix} 0 & -i\sqrt{2} \\ -i\sqrt{2} & 0\end{pmatrix}\quad CC^*=\begin{pmatrix} 2 & 0 \\ 0 & 2\end{pmatrix}$
$B=\begin{pmatrix} -1 & 0 \\ 0 & 1\end{pmatrix}$
Then we have
$ABA^*=\begin{pmatrix} 0 & 2i \\ -2i & 0\end{pmatrix}\quad CBC^*=\begin{pmatrix} 2 & 0 \\ 0 & -2\end{pmatrix}$


text:
\section{Ommited Proofs from \Cref{sec:known} \label{app:known_proofs}}

\subsection{Proof of \Cref{prop:covariance_lb} \label{sec:prop:covariance_lb}}

\begin{proof}[Proof of \Cref{prop:covariance_lb}]
Let $v \in \R^{\dimu}$, with $\|v\| = 1$, and let $u_s = \matY_s v$ for $s \in \{1-h,2-h,\dots,t\}$, and set $u_s = 0$ for $s \le t - h$ and $s > t$. From \Cref{fact:vector_norm_lb}, which shows that $\|v+w\|_2^2 \ge \frac{1}{2}\|v\|^2 - \|w\|^2$, we have
\begin{align*}
v^\top \sum_{s=1}^t \matH_s^\top \matH_s v &:=  \sum_{s=1}^t \|\matH_s v\|_2^2 = \sum_{s=1}^t \left\|\sum_{i=0}^h G^{[i]} \matY_{s-i} v\right\|_2^2\\
&=  \sum_{s=1}^t \left\|\sum_{i=0}^h G^{[i]} u_{s-i} \right\|_2^2\\
&\ge  \sum_{s=1-h}^{t+h} \left\|\sum_{i=0}^h G^{[i]} u_{s-i} \right\|_2^2 - 2h \radG^2\radY^2.\\
&\overset{\normtext{(\Cref{fact:vector_norm_lb})}}{\ge}  \frac{1}{2}\sum_{s=1-h}^{t+h} \left\|\sum_{i=0}^{\infty} G^{[i]} u_{s-i} \right\|_2^2 - \sum_{s=1-h}^{t+h}\left\|\sum_{i> h}^{\infty} G^{[i]} u_{s-i} \right\|_2^2
 - 2h \radG^2\radY^2\\
 &\ge  \frac{1}{2}\sum_{s=1-h}^{t+h} \left\|\sum_{i=0}^{\infty} G^{[i]} u_{s-i} \right\|_2^2 - \sum_{s=1-h}^{t+h}\psiG(h+1)^2\radY^2
 - 2h \radG^2\radY^2\\
  &=  \frac{1}{2}\sum_{s=1-h}^{t+h} \left\|\sum_{i=0}^{\infty} G^{[i]} u_{s-i} \right\|_2^2 - \underbrace{\left(t\psiG(h+1)^2 + 4h\radG^2\right)\radY^2}_{:=\gamma_{t;h}},
\end{align*}
where we use $\psiG(h+1) \le \psiG(0) = \radG^2$ in the last line. \newcommand{\utild}{\tilde{u}} Moreover, setting $\utild_s = u_{s-h}$,
\begin{align*}
\sum_{s=1-h}^{t+h} \left\|\sum_{i=0}^{\infty} G^{[i]} u_{s-i} \right\|_2^2 &= \sum_{s=1}^{t+2h} \left\|\sum_{i=0}^{\infty} G^{[i]} \utild_{s-i} \right\|_2^2 \\
&\overset{(i)}{=} \sum_{s=1}^{\infty} \left\|\sum_{i=0}^{s} G^{[i]}\utild \right\|_2^2\\
&\overset{(ii)}{\ge} \kappa_0 \sum_{s=1}^{\infty} \|\utild_s\|_2^2\\
&= \kappa_0 \sum_{s=1}^{\infty} \|u_{s-h}\|_2^2
\end{align*}
where $(i)$ uses that we have $\utild_s = 0$ for $s \le 0$ and for $s \ge t+2h$, and $(ii)$ invokes \Cref{defn:input_recov}. Combining the two displays, we have
\begin{align*}
v^\top \sum_{s=1}^t \matH_s^\top \matH_s v  &\ge  \frac{\kappa_0}{2} \sum_{s=1}^{\infty} \|u_{s-h}\|_2^2 - \gamma_{t;h}\\
&\ge  \frac{\kappa_0 }{2}\sum_{s=1}^{t+h} \|\matY_{s-h}v\|_2^2 - \gamma_{t;h}\\
&=  v^\top\left(\frac{\kappa_0}{2} \sum_{s=1-h}^{t} \matY_{s}^{\top}\matY_{s} - \gamma_{t;h} I\right)v,
\end{align*}
where the last line uses $\|v\| = 1$. Finally, defining $\cpsi[t]:= \max\{1, \frac{t\psiG(h+1)^2}{h\radG^2}\}$, we have $\gamma_{t;h} = \radY^2(t\psiG(h+1)^2 + 4h\radG^2) \le \radY^2(h\cpsi[t]\radG^2 + 4h\radG^2) \le  5h \radH^2 \cpsi[t]$, yielding the desired bound.
\end{proof}

\subsection{Proof of \Cref{lem:semions_reg} \label{app:proof:ssec:online_semi_newton}}
Let $\zst \in \argmin_{z \in \calC}\sum_{t=1}^T f_t(z)$. Following the standard analysis of Online Newton Step (e.g. \citet[Chapter 4]{hazan2019introduction} with $\gamma \leftarrow 1/\eta$), one has
\begin{align*}
\sum_{t=1}^T \nabla_t(\matz_t - \zst) &\le \frac{\eta}{2}\sum_{t=1}^{T} \nabla_t^\top \Lambda_t^{-1}  \nabla_t + \frac{1}{2\eta}\sum_{t=1}^T(\matz_t - \zst)^\top(\Lambda_t - \Lambda_{t-1})(\matz_t - \zst) + \frac{1}{2\eta}(\matz_1 - \zst)^\top \Lambda_0 (\matz_1 - \zst)
\end{align*}
The last term is at most $\frac{\lambda}{2\eta} \diamz^2$. Moreover, since $\Lambda_t - \Lambda_{t-1} = \matH_t\matH_t^\top$,
\begin{align*}
\sum_{t=1}^T \nabla_t(\matz_t - \zst) - \frac{1}{2\eta}\|\matH_t(\matz_t - \zst)\|_2^2 \le\lambda \diamz^2 +  \frac{\eta}{2}\sum_{t=1}^{T} \nabla_t^\top \Lambda_t^{-1}  \nabla_t.
\end{align*}
Finally, for $\eta \ge \frac{1}{\alpha}$, we recognize that $\nabla_t(\matz_t - \zst) - \frac{1}{2\eta}\|\matH_t(\matz_t - \zst)\|_2^2 \ge \nabla_t(\matz_t - \zst) - \frac{\alpha}{2}\|\matH_t(\matz_t - \zst)\|_2^2 \ge f_t(\matz_t) - f_t(\zst)$ by \Cref{lem:quad_lb}. Thus,
\begin{align*}
\sum_{t=1}^T f_t(\matz_t)  - f_t(\zst) \le\lambda \diamz^2 +  \frac{\eta}{2}\sum_{t=1}^{T} \nabla_t^\top \Lambda_t^{-1}  \nabla_t,
\end{align*}
as needed. \qed

\subsection{Proof of \Cref{lem:Ft_diff}}

	\newcommand{\matzbar}{\overline{\matz}}
	We  have $F_t(\matz_t,\dots,\matz_{t-h}) - f_t(\matz_t) = F_t(\matz_t,\dots,\matz_{t-h}) - F_t(\matz_t,\dots\matz_t)$. Therefore Taylor's theorem, there exists some $\mu \in [0,1]$ such that, for $\matzbar_{t-i} = \mu \matz_{t-i} + (1-\mu)\matzbar_t$,
	\begin{align*}
	F_t(\matz_t,\dots,\matz_{t-h}) - f_t(\matz_t) &= (\nabla F_t(\matzbar_t,\dots,\matzbar_{t-h}))^\top (0,\matz_{t-1}-\matz_t,\matz_{t-2}-\matz_t,\dots,\matz_{t-h}-\matz_t).
	\end{align*}
	By the Chain Rule, we then have
	\begin{align*}
	|F_t(\matz_t,\dots,\matz_{t-h}) - f_t(\matz_t)| &=\left| \nabla \ell( \matv_t + \sum_{i=0}^h G^{[i]}\matY_{t-i}\matzbar_t)^\top \left(\sum_{i=1}^h G^{[i]}\matY_{t-i}(\matz_{t-i} - \matz_t)\right)\right|\\
	&\le \|\nabla \ell( \matv_t + \sum_{i=0}^h G^{[i]}\matY_{t-i}\matzbar_t)\|_2 \cdot \radG \cdot \max_{i \in \{1,\dots,h\}}\|\matY_{t-i}(\matz_{t-i} - \matz_t)\|_2.
	\end{align*}
	Analogous to the \Cref{lem:ft_facts}, we have $\|\nabla \ell( \matv_t + \sum_{i=0}^h G^{[i]}\matY_{t-i}\matzbar_t)\|_2 \le \Leff$, concluding the first part of the proof. For the second display, we have
	\begin{align*}
	\sum_{t=1}^{T} F_t(\matz_t,\dots,\matz_{t-h}) - f_t(\matz_t) &\le \Leff \radG \sum_{t=1}^T \max_{i\in \{1,\dots,h\}} \|\matY_{t-i}(\matz_t - \matz_{t-i})\|_2\\
	&\le \Leff \radG \sum_{t=1}^T \sum_{i=1}^h\|\matY_{t-i}(\matz_t - \matz_{t-i})\|_2\\
	&\le \Leff \radG \sum_{t=1}^T \sum_{i=1}^h\sum_{j=1}^{i-1}\|\matY_{t-i}(\matz_{t-j+1} - \matz_{t-j})\|_2\\
	&= \Leff \radG \sum_{s=1-h}^T \sum_{i=1}^h \sum_{j=1}^{i-1}\|\matY_{s}(\matz_{s+i-j+1} - \matz_{s+i-j})\|_2\\
	&\le h \Leff \radG \sum_{s=1-h}^T \sum_{i=1}^{h-1} \|\matY_{s}(\matz_{s+i+1} - \matz_{s+i})\|_2 \cdot \I_{s+i+1 \le t}.
	\end{align*}
	Finally, since $\matz_t - \matz_{t-1} =0$ for $t \le 1$, the above indicator $\I_{s+i+1 \le t}$ can be replaced with $\I_{2 \le s+i+1 \le t} = \I_{1 \le s+i \le t-1}$, completing the proof.\qed

\subsection{Proof of \Cref{lem:Y_bound}}
	For $t \le 0$, $\|\matY_s (\matz_{t+1}-\matz_{t})\|_2  = 0$. Otherwise, we have
	\begin{align}
	\|\matY_s (z_{t}-z_{t-1})\|_2 &= \|\matY_s \Lambda_t^{-1/2}\Lambda_t^{1/2}(\matz_{t+1}-\matz_{t})\|_2 \nonumber\\
	&\le \|\matY_s \Lambda_t^{-1/2}\|_{\op} \cdot \|\Lambda_t^{1/2}(\matz_{t+1}-\matz_{t})\|_2 \nonumber\\
	&\overset{(i)}{\le} \|\matY_s \Lambda_t^{-1/2}\|_{\op} \cdot \|\Lambda_t^{1/2}(\widetilde{\matz}_{t+1}-\matz_{t}) \|_2 \nonumber\\
	&= \|\matY_s \Lambda_t^{-1/2}\|_{\op} \|\Lambda_t^{1/2}\cdot \eta \Lambda_{t}^{-1}\nabla_t\|_2 \nonumber\\
	&= \eta \sqrt{\|\matY_s \Lambda_t^{-1/2}\|_{\op}^2 \| \Lambda_{t}^{-1/2}\nabla_t\|_2^2} \label{eq:operator_norm_bound_movement},
	\end{align}
	where $(i)$ follows from the Pythagorean theorem, using that $\matz_{t+1}$ is projected in the $\Lambda_t$-norm.
	Finally, we can crudely bound $\|\matY_s \Lambda_t^{-1}\matY_s\|_{\op} \le \trace(\matY_s \Lambda_t^{-1}\matY_s)$. Since we consider indices $t \ge s$, we have $\trace(\matY_s \Lambda_t^{-1}\matY_s)\le \trace(\matY_s \Lambda_s^{-1}\matY_s)$, where we have the understanding that $\Lambda_s = \Lambda_1$ for $s \le 0$. Thus, we see that for $t > 0$,
	\begin{align*}
	\|\matY_s (\matz_{t+1}-\matz_{t})\|_2 \le \eta  \trace(\matY_s \Lambda_s^{-1}\matY_s)^{\nicehalf} \trace(\nabla_t^\top \Lambda_t^{-1}\nabla_t))^{\nicehalf}
	\end{align*}
	Thus, from \Cref{lem:Ft_diff} and by Cauchy Schwartz,
	\begin{align*}
		\MoveDiff_T &\le  h \Leff \radG \sum_{i=1}^{h-1}\sum_{s=1-h}^T  \|\matY_{s}(\matz_{s+i+1} - \matz_{s+i})\|_2 \I_{1 \le s+i \le t-1}\\
		&\le \eta h\Leff \radG \cdot \sum_{i=1}^{h-1} \sqrt{\sum_{s=1-h}^{T} \I_{1 \le s+i \le t-1} \cdot \trace(\matY_s \Lambda_s^{-1}\matY_s)}\sqrt{\sum_{s=1-h}^{T} \I_{1 \le s+i \le t-1} \cdot \trace(\nabla_{s+i }^\top \Lambda_{s+i }^{-1}\nabla_{s+i })}\\
		&\le \eta h^2\Leff \radG \cdot \sqrt{\sum_{s=1-h}^{T}  \trace(\matY_s \Lambda_h^{-1}\matY_s)}\sqrt{\sum_{s=1}^{T} \trace(\nabla_t^\top \Lambda_t^{-1}\nabla_t)},
		\end{align*}
		as needed.
	\qed


TITLE: A question about convex functions of minimum and maximum
QUESTION [0 upvotes]: A question about convex functions of minimum and maximum
Let $f_1, f_2$ are two convex functions $\mathbb{R^n}\to \mathbb{R}$
then I prove that their $\max\{f_1(x),f_2(x),\forall x\in \mathbb{R}^n$ is convex
Is their minimum is also convex?

REPLY [2 votes]: $f(x)=x$ and $g(x)=-x$ gives a counterexample.


text:
\begin{document}
\title{On cyclic CAT(0)  domains of discontinuity}

\author
{Eric Swenson }

\subjclass{54F15,54F05,20F65,20E08}

\email [Eric Swenson]{eric@math.byu.edu}
\address
[Eric Swenson] {Mathematics Department, Brigham Young University,
Provo UT 84602}

\begin{abstract} Let $X$ be a CAT(0) space, and $G$ a discrete  cyclic group of isometries of $X$.
We investigate the domain of discontinuity for the action of $G$ on the boundary $\bd X$.
\end{abstract}
\maketitle
\section {Introduction}
The idea of a domain of discontinuity was first investigated in the setting of Kleinian groups, discrete groups of isometries of hyperbolic $m$-space, $\HH^m$.  Using the Poincare ball model  of $\HH^m$, where $\HH^m$ is the interior of the open unit $m$-ball, we see that the boundary of $\HH^m$ is the unit sphere $\SSS^{m-1}$, and we see that isometries of $\HH^m$ extent to homeomorphsims of  the closed $m$-ball, $\HH^m \cup \SSS^{m-1}$.

Let $G$ be a discrete group of isometries of $\HH^m$, and $(g_i)$ a sequence of distinct elements of $G$.  Fixing a point $a \in \HH^m$, using compactness and passing to a subsequence we may assume that $g_i(a) \to p \in \SSS^{m-1}$ and $g^{-1}_i(a) \to n \in \SSS^{m-1}$. It can be shown that  $g_i(x) \to p$ uniformly on compact subsets of $\SSS^{m-1} -\{n\}$.
\begin{Def}
Let $G$ be a group of homeomorphsim of a compact hausdorf space $Z$, we say that $G$ acts on $Z$ as a discrete convergence group if for each sequence of distinct elements of $G$, there is a subsequence $(g_i) \subset G$ and points $n,p \in Z$ such that $g_i(x) \to p$ uniformly on compact subsets of $Z$.  As shown above,  Kleinian groups of dimension $m$ act as a discrete convergences group on $\SSS^{m-1}$.
\end{Def}
The set of all such $p$ is the limit set of the Kleinian group $G$, $$\Lambda (G) = \{ p \in \SSS^{m-1}: \exists (g_i) \subset G\text{ with
}g_i(x) \to p \text{ for some }x \in \HH^m\}$$ and the domain of discontinuity of $G$,
$\Omega G = \SSS^{m-1} - \Lambda G$. Clearly $\Lambda G$ is $G$-invariant and  closed, so $\Omega G$ is $G$-invariant and  open.  Let $p \in \Lambda G$ and $(g_i)\subset G$ with $g_i(x) \to p$ and $g_i^{-1}(x) \to n \in \Lambda G$ for some (any) $x \in \HH^m$.  Then
for any $a \in \Omega G$, $g_i(a) \to p$.  Thus if $\Omega G \neq \emptyset$ then $p$ is a limit point of $\Omega G$, and so $\Omega G$ is dense in $\SSS^{m-1}$.

We have the following result: If $G$ is a discrete group of isometries of hyperbolic $m$-space, $\HH^m$, then $\bd \HH^m = \SSS^{m-1}$ is a disjoint union of the limit set of $G$, $\Lambda G$ and the domain of discontinuity of $G$, $\Omega(G)$, where $\Omega G$ is open and either dense or empty.  Since $G$ acts as a convergence group on $\SSS^{m-1}$,  the action of $G$ on $\Omega G$ is properly discontinuous (hence the name domain of discontinuity).

This result holds whenever $G$ acts as a discrete convergence group on a compact Hausdorff space, in particular when $G$ is a discrete group of isometries of a proper $\delta $-hyperbolic space.

In this paper we consider the case where the negative curvature ($\delta$-hyperbolic) condition is relaxed to a non-positive curvature (CAT(0)) condition.  Considering the action of $\Z^m$ on $\R^m$
we see that often the action on the boundary may be trivial and so the domain of discontinuity must be empty whenever the group is infinite. Thus the boundary will not in general be the union of the limit set and the domain of discontinuity.

The conjecture seems to be that if a cyclic subgroup acts ``nicely" on a CAT(0) space and the action on the boundary is not virtually trivial, then there should be an open dense subset of the boundary on which the  cyclic subgroup acts properly discontinuously.

This conjecture is realized when the Tits diameter of the boundary is large, but still open in the other cases.  The case where the fixed points of the cyclic subgroup are spherical suspension points of the boundary will not be addressed in this note.

The author wishes to thank G. Levitt and K. Ruane for helpful conversations.

\section{Definitions and Basic results}
We refer the reader to \cite{BRI-HAE} or \cite{BAL} for more details of the following.
\begin{Def} For $X$ a geodesic metric space and $\Delta(a,b,c)$ a geodesic triangle in $X$ with vertices $a,b,c \in X$ there is a {\em comparison }triangle $\bar \Delta=\Delta(\bar a,\bar b, \bar c) \subset \E^2$ with
$d(a,b) =d(\bar a, \bar b)$, $d(a,c) = d(\bar a, \bar c)$ and $d(b,c)=d(\bar b, \bar c)$.
We define the comparison angle $\bar \angle_a(b,c) =\angle_{bar a}(\bar b,\bar c)$.

Each point $z \in \Delta(a,b,c)$ as a unique comparison point, $\bar z \in \bar \Delta$ on the edge corresponding to the edge of $z$ and in the corresponding location on that edge.  We say that the triangle $\Delta(a,b,c)$ is CAT(0) if for any $y, z \in \Delta(a,b,c)$ with comparison points $\bar y, \bar z \in \bar \Delta$, $d(y,z) \le d(\bar y,\bar z)$. The space $X$ is said to be CAT(0) if every geodesic triangle in $X$ is CAT(0).
\end{Def}
If $X$ is CAT(0), notice that for any geodesics $\alpha:[0, r] \to X$ and $\beta:[0,s] \to X$ with $\alpha(0)=\beta(0)=a$, the function $\theta(r,s) =\bar \angle_{r(0)}(\alpha(r),\beta(s))$ is an increasing function of  $r,s$.  Thus $\lim\limits_{r,s \to 0} \theta(r,s)$ exists and we call this limit $\angle_a(\alpha(r),\beta(s))$.  It follows that for any $a,b,c \in X$, a CAT(0) space, $\angle_a(b,c) \le \bar \angle_a(b,c)$.

The
(visual) boundary, $\bd X$, is the set of equivalence classes of
rays, where rays  are equivalent if they fellow travel.  Given a
ray $R$ and a point $x \in X$ there is a ray $S$ emanating from
$x$ with $R \sim S$.  Fixing a base point $\mathbf 0 \in X$ we
defined a Topology on $\bar X= X \cup \bd X$ by taking the basic
open sets of $ x \in X$ to be the open metric balls about $x$. For
$y \in \bd X$, and $R$ a ray from $\mathbf 0$ representing $y$, we
construct  basic open sets $U(R,n,\epsilon)$ where $n,\epsilon>0$.
We say $z \in U(R,n,\epsilon)$ if the unit speed geodesic,
$S:[0,d(\mathbf 0,z)] \to \bar X$, from $\mathbf 0$ to $z$
satisfies $d(R(n),S(n)) <\epsilon$.  These sets form a basis for a regular
topology on $\bar X$ and $\bd X$. For any $x \in X$ and $u,v \in \bd X$ we can define
$\angle_x(u,v) $ and $\bar \angle_x(u,v)$ by parameterizing the rays $[x,u)$ and $[x,v)$ by
$t\in [0,\infty)$ and taking the limit as $t\to 0$ and $t \to \iy$ respectively.

 for $u,v \in \bd X$, we define $\angle(u,v) =
\sup\limits_{p \in X} \angle_p(u,v)$.
 It follows from \cite{BRI-HAE} that $\angle(u,v) =
\overline{\angle}_p(u,v)$ for any $p \in X$.
 Notice that isometries of $X$ preserve the angle between points of $\bd X$.
 The angle defines a path metric, $d_T$ on the set $\bd X$, called the Tits
metric, whose topology is finer than the given topology of $\bd
X$.  Also $\angle(a,b)$ and $ d_T(a,b)$ are equal whenever either
of them are less than $\pi$.

  The set $\bd X$ with the Tits metric is called the Tits  boundary of $X$,
denoted  $TX$.  Isometries of $X$ extend to isometries of $TX$

  The identity function $TX \to \bd X$ is continuous, but the identity
function
 $\bd X \to TX$ is only lower semi-continuous.  That is for any sequences
$(u_n), (v_n)  \subset \bd X$ with $u_n \to u$ and $v_n \to v$ in
$\bd X$, then $$\varliminf d_T(u_n,v_n) \ge d_T(u,v)$$
\begin{Lem} \label{L:horo} Let $v \in \bd X$  and $H \subset X$ a horoball centered at $v$,
then $\bar H \cap \bd X \subset \bar B_T(v, \frac  \pi 2)$ the closed Tits ball  about $v$.
\end{Lem}
\begin{proof} Suppose that $w \in \bar H \cap \bd X$ with $\angle(v,w) > \frac \pi 2 $.
Choose a point $y \in X$ with $\angle_y(v,w) > \frac \pi 2 $.

Let $R:[0,\infty)\to X$ be the geodesic ray from $y$ to $v$, and $b_R:X \to \R$ be the Buseman function associated to $R$, $b_R(x) = \lim\limits_{t \to \infty} [d(x,R(t))-t]$.
Note the $b_R(y) = 0$ and $b_R(R(t)) = -t$ for $t>0$.
By \cite[Exercise II 8.23(1)]{BRI-HAE} any horoball based at $v$ is within finite distance of any other horoball based at $v$.  It follows that all horoballs based at $v$ will have the same limit points in the boundary.  Thus we may assume that $H = b_R^{-1}([-1,-\iy))$.

Since $w \in \bar H$, there exists $\hat w \in H$ such that $\angle_y(v,\hat w) >\frac \pi 2$.

We recall \cite[Exercise II 8.23(4)]{BRI-HAE}. Choose a geodesic line $\bar R:\R \to \E^2$.
For each $x\in X$ and $n \in \N$ consider a  comparison triangle $\bar \Delta(x, R(0), R(n))$ with $\bar R(i)$ the comparison point for $R(i)$ ($i=0,n$) and $x_n\in \E^2$ the comparison point for $x$.  Choose $r_n \in \R$ with $\bar R(r_n) = \pi_{\bar R}(x_n)$ (where $\pi_{\bar R}$ is orthogonal projection to the line $\bar R$).  Then $r_n \to r$ where $b_R(x)=-r $.  Since $b_R(\hat w) \le -1$,  for $n>>0$, $r_n >0$.  It follows that $\angle_{\bar R(0)}(\bar R(n), \hat w_n) < \frac \pi 2$.  However $$\frac \pi 2 < \angle_y(v,\hat w) =\angle_{R(0)}(R(n), \hat w) \le \bar \angle_{R(0)}(R(n), \hat w) = \angle_{\bar R(0)}(\bar R(n), \hat w_n) < \frac \pi 2$$ which is a contradiction.
\end{proof}

\section{Large Tits radius}
Recall that for any $u \in \bd X$, $B_T(u, \epsilon) = \{v \in \bd X : d_T(u,v) < \epsilon\}$
and $\bar B_T(u, \epsilon) = \{v \in \bd X : d_T(u,v) \le \epsilon\}$, where $d_T$ is the Tits metric.
For $A \subset \bd X$, we define
$$\radius_T (A)= \inf\{r: A \subset B_T(u, r) \text{ for some } u \in \bd X\}.$$

If $g$ is an infinite order isometry of $X$, and $\langle g\rangle$ is proper, then $g$ is either hyperbolic or parabolic.  When $g$ is hyperbolic it acts by translation on a line (called an axis of $g$) in $X$ with endpoints $g^+$ (in the direction of translation) and $g^-$ (see \cite{BRI-HAE}).

Recall that a hyperbolic isometry $h$ of $X$ is called {\em rank }1, if $h$ has an axis $L$ which doesn't bound a half flat.
From \cite{BAL}  we see that $d_T(h^+, \alpha) = \iy$ for all $\alpha \neq h^+$, so if $X$ has a rank 1 isometry then $\radius_T (\bd X)=\iy$

Recall a result from \cite{PA-SW}
\begin{Thm}\label{T:pi} \cite{PA-SW}
Let $X$ be a complete  CAT(0) space and $(g_i) $ a sequence of isometries of $X$
with the
property that $g_i(x) \to p \in \bd X$ and $g_i^{-1}(x) \to n \in
\bd X$ for any $x \in X$.
Then for any $\theta \in [0, \pi]$ and  any compact set $K \subset \bd X
-\overline{B}_T(n,\theta)$, $g_n(K) \to \overline{B}_T(p,  \pi
-\theta)$ (in the sense that for any open $U \supset
\overline{B}_T(p, \pi -\theta)$, $g_n(K) \subset U$ for all $n$
sufficiently large).
\end{Thm}
This Theorem is stated in \cite{PA-SW} with stronger hypothesis, but they were not used in the proof.
We will refer to this result as $\pi$-convergence.
For a given hyperbolic element $h$ with axis $L$, we set $g_i= h^i$, $n = L(-\iy)= h^-$ and $p = L(\iy)=h^+$.

The following theorem is due to Ballmann.
\begin{Thm}[Ballmann]\label{T:Ball}  If $h$ is a rank 1 isometry of the complete CAT(0) space $X$, then the group generated by $h$, $\langle h\rangle$ act properly discontinuously on the open set $\bd X - \{h^{\pm }\}$ which is dense if it is non-empty.
\end{Thm}
\begin{proof}  Let $K$ be a compact subset of $\bd X - \{h^{\pm }\}$.  The Tits distance from $K$ to $n =h^-$ will be infinite.  Thus by $\pi$-convergence $h^i(K) \to p= h^+$ and $h^{-i}(K) \to n = h^{-}$.  Thus $\{i \in \Z: h^i(K ) \cap K \neq \emptyset\}$ is finite and the action of $\langle h\rangle$ on $\bd X-\{h^{\pm}\}$ is properly discontinuous.  Clearly $\bd X-\{h^{\pm}\}$ is an open subset and for any $a \in \bd X-\{h^{\pm}\}$, $h^i(a) \to h^+$ and $h^{-i} (a) \to h^{-}$ so $\bd X-\{h^{\pm}\}$ is dense in $\bd X$.
\end{proof}

\begin{Def} For $X$ a complete CAT(0) space we define the limit set
$\Lambda X \subset \bd X$ to be the set $\{p \in \bd X: \exists (g_i)$ isometries of $X$ with $g_i(x) \to p$ and $g_i^{-1}(x) \to n$ for some $n \in \bd X$ and  for all $x \in X \}$
\end{Def}

The following is a slight generalization of a result of Karlsson \cite{KAR}.
\begin{Thm} Let $X$ be a complete CAT(0) space with $\radius_T(\bd X) > 3\pi$ and $|\bd X|>2$. If $h$ is a hyperbolic isometry of $X$ then $\langle h\rangle$ acts properly discontinuously on the open subset $\Omega= \bd X -[\bar B_T(h^+, \frac \pi 2) \cup \bar B_T(h^{-}, \frac \pi 2)]$. If $[\bar B_T(h^+, \frac \pi 2) \cup \bar B_T(h^{-}, \frac \pi 2)]\subset \Lambda X$ then $\Omega $ is dense in $\bd X$.
\end{Thm}
\begin{proof} By Theorem \ref{T:Ball} we may assume $h$ is not rank 1. Thus some axis of $h$ bounds a half-flat, which corresponds to a Tits geodesic of length $\pi$ from $h^+$ to $h^-$ so $d_T(h^+,h^-) =\pi$.

If there is any point of $w \in \bd X$ which is isolated in the Tits metric, ($d_T(w, v)= \iy, \, \forall v $) then by $\pi$-convergence the orbit of $w$ under $\langle h \rangle$ is infinite and each element of the orbit will be isolated as well.

By the triangle inequality, for any $q\in \bd X$, there exists $u,v \in \bd X$ such that
$d_T(q,u),d_T(q,v)\ge \pi$ and $ d_T(u,v) >2\pi$. If follows from $\pi$-convergence that for any $w \in \Lambda X$ and any neighborhood $W$ of $w$ in $\bd X$, the Tits diameter $\diam_T(W) > 2\pi$.

Since $\diam_T \left[ \bar B_T(h^+, \frac \pi 2) \cup \bar B_T(h^-, \frac \pi 2)\right]\le 2\pi$, $ W \not \subset \bar B_T(h^+, \frac \pi 2) \cup \bar B_T(h^-, \frac \pi 2)$ so
$W\cap \Omega \neq \emptyset$.  Thus $\Omega$ is dense in $\Lambda X$.. Since closed Tits balls are closed in $\bd X$, $\Omega$ is open.

For any compact $K \subset \Omega$, by $\pi$-convergence
$h^i(K) \to \bar B_T(h^+, \frac \pi 2)$ and $h^{-i}(K) \to \bar B_T(h^{-}, \frac \pi 2)$.  Using this, we can show that  $\{i \in \Z: h^i(K ) \cap K \neq \emptyset\}$ is finite and so the action of $\langle h\rangle$ on  $\Omega$ is properly discontinuous.
\end{proof}
\begin{Ex}
Notice that
$\Omega$ need not be dense in $\bd X$ if we remove the condition that $\bar B_T(h^\iy, \frac \pi 2) \cup  \bar B_T(h^{-\iy}, \frac \pi 2) \subset \Lambda X$.  You start with the half plane $\{(x,y) : y \ge 0\}$ where $h$ is unit translation in the 1st coordinate.  Now you attach a line at the origin and let $g$ act by translation in that line.  You construct the CAT(0) space $X$ by translating this picture by $g$ and $h$.  This gives you an action of $<g,h>$ on the CAT(0) space $X$.  The isometry $g$ is a rank 1 hyperbolic element, so $X$ is rank 1, but $\Omega= \bd X - [\bar B_T(h^\iy, \frac  \pi 2) \cup  \bar B_T(h^{-\iy}, \frac \pi 2)]$ is not dense in the boundary, in fact its closure hits $ [\bar B_T(h^\iy, \frac \pi 2) \cup  \bar B_T(h^{-\iy}, \frac \pi 2)]$ only in $h^{\pm \iy}$.
\end{Ex}
\begin{Thm} Let $X$ be a proper CAT(0) space with $\radius_T (\bd X)>3\pi$, and $h$ a parabolic isometry of $X$. There exists $m\in \bd X$ fixed point of $h$ such that $\langle h \rangle$ acts properly discontinuously on the open dense subset $\Omega = \bd X - \bar B_T(m,\pi)$ of $\bd X$.
If $\bar B_T(m, \pi) \subset \Lambda X$ then $\Omega $ is dense in $\bd X$.
\end{Thm}
\begin{proof}
Clearly $\Omega$ is non-empty and open.  By \cite[II 8.25]{BRI-HAE}, there exists $m \in \bd X$ such that $h$ leaves invariant each horoball centered at $m$ which implies that $h$ fixes $m$.  Suppose that  $\langle h \rangle$  doesn't act properly discontinuously on $\Omega$.  Then there exists $K $, a compact subset of $\Omega$, and a strictly increasing sequence $(i_j)\subset \N$ such that $K \cap h^{i_j} (K)\neq \emptyset$ for all $j \in \N$.

Fix $ x \in X$. Passing to a subsequence we may assume that $h^{i_j} (x) \to p\in \bd X$ and $h^{-i_j}(x) \to n\in \bd X$.  Since $h$ leaves the horosphere $S$ centered at $m$ passing through $x$ invariant, then $n,p \in \bd S \subset \bar B_T(m, \frac \pi 2)$ by Lemma \ref{L:horo}.  Since $\bd X$ is Hausdorf, there exists  $U\subset \bd X$ open with $\bar B_T(m, \pi) \subset U$ and $U \cap K= \emptyset$.  Since $\bar B_T(n, \frac \pi 2), \bar B_T(p, \frac \pi 2) \subset \bar B_T(m, \pi) \subset U$, by $\pi$-convergence $h^{i_j}(K) \subset U$ for all $j>>0$ which implies $h^{i_j}(K) \cap K = \emptyset$ for all $j>>0$ contradicting the choice of $(i_j)$. Thus $\langle h\rangle $ acts properly on $\Omega$.

By the triangle inequality, for any $q\in \bd X$, there exists $u,v \in \bd X$ such that
$d_T(q,u),d_T(q,v)\ge \pi$ and $ d_T(u,v) >2\pi$. If follows from $\pi$-convergence that for any $w \in \Lambda X$ and any neighborhood $W$ of $w$ in $\bd X$, the Tits diameter $\diam_T(W) > 2\pi$.

Since $\diam_T (\bar B_T(m, \pi))\le 2\pi$, $ W \not \subset\bar B_T(m, \pi)$ so
$W\cap \Omega \neq \emptyset$.  Thus $\Omega$ is dense in $\Lambda X$.
\end{proof}

\section{Small Tits Radius}
\begin{Def}Recall that a metric space is {\em proper} if closed metric balls are compact, and a metric space is {\em cocompact} if the quotient of the space by the isometry group is compact.
\end{Def}
\subsection*{We now assume that $X$ is a proper CAT(0) space.}

\begin{Def}We say that $a,b \in \bd X$ are antipodes if $d_T(a,b) \ge \pi$.    The suspension of the antipodes $a$ and $b$,   $S_a^b$, is the union of $a$ and $b$ together with of all Tits geodesics from $a$ to $b$ of length $\pi$ (if any). Notice that $S_a^b =S_b^a$.
\end{Def}
 Let $a \in \bd X$ and $(g_i)$ a sequence of isometries of a CAT(0) space $X$.  We say that $(g_i)$ {\em pulls from} $n$ if there is some unit speed geodesic ray $R:[0,\iy) \to X$ representing $n$, and a sequence  $(s_i) \subset [0,\iy)$ with $s_i \to \iy$ such that  the sequence $(g_i(R(s_i)))$ is bounded.
Clearly this is independent of the ray chosen.   Passing to a subsequence, we may assume that $g_i(R(s_i)) \to b$, and that
the sequence of rays $(g_i(R))$  (each reparametrized) converges uniformly on compact subsets to a geodesic line $L$ with $g_i(-i) \to L(-\iy)$, and $L(0)= b$.  Notice that for some (and so any) $x \in X$,  $g_i(x) \to L(-\iy)$ and
$g_i^{-1}(x) \to n$.

 For any $x \in \bd X$, passing to a subsequence, we may assume that  $g_i(x) \to\hat x \in \bd X$.   Thus for any compact set $C \subset TX$ (where $TX$ is $\bd X$ with the topology of the Tits metric)   we can define $f:C \to \bd X$ by $f(x)= \hat x$.  In fact we can define $f$ whenever $C$ has a countable dense subset as a subset of $TX$.   By \cite{BRI-HAE} $f:C \to TX$ is Lipschitz with constant one.  Notice that $f(n) = L(-\iy)$.

  \begin{Lem}\label{L:f} In the above setting  for any $a \in \bar B_T(n,\pi)$, $f:[n,a]\to TX$ is an isometric embeding of $[n,a]$ into $S_{L(-\iy)}^{L(\iy)}$.
\end{Lem}
\begin{proof}
Because $f$ is Lipschitz with constant one, if $d_T(n,a) = d_T(f(n),f(a))$ then $f:[n,a] \to TX$ is an isometry.   Since $f(n) = L(-\iy)$, it suffices to show that $d_T(f(n), f(a)) + d_T(f(a), L(\iy)) = \pi$.

 Let $\theta= d_T(n,a)$, so by $\pi$-convergence $\pi-\theta\ge d_T(f(a), L(\iy))$.
Since $f$ is Lipschitz with constant one, $d_T(f(n),f(a)) \le \theta$.  However
\begin{align*}\pi= \theta + (\pi-\theta) \ge d_T(f(n),f(a))+ d_T(f(a), L(\iy)) \ge  d_T(L(-\iy), L(\iy)) \ge \pi \end{align*}
\end{proof}
\begin{Ex}
The function $f$ defined above need not be an embedding on $B_T(n,\pi)$.
Consider the half flat $Y =\R \times [0,\iy)$.  For each $n \in \Z$   glue the quarter flat $X_n=[n,\iy) \times [0,\iy)$  by identifying $(a,b) \in Y$ with $(c,d)\in X_n$ if $a=c$, $b=d$, and $\ln (a+1-n) \ge b$.
Let $Z$ be the resulting space.   $\bd Z$ is  broom attached at $\iy$ in the $\R$ factor of $Y$.  Notice that $\Z$ acts on $Z$  ($1 \in \Z$ sends $X_n$ to $X_{n+1}$). Letting $g_i$ be subtraction by $i$, we have that $f(\bd Z) =\bd Y$.
\end{Ex}
\begin{Def} Let $g$ be a hyperbolic isometry of the CAT(0) space $X$.  We define
$S_g = S^{g^+}_{g^{-}}$, that is the suspension of the endpoints of $g$.
\end{Def}

\begin{Lem}\label{L:discrete} The group $\langle g\rangle $ acts discretely on the open set $\Omega =\bd X- S_g$.
\end{Lem}
\begin{proof} It suffices to show that for any point $a \in \bd X$, the limit points of the $\{g^i(a): \, i\in \N \}$ are in $S_g$. Let $b$ be a limit point of $\{g^i(a): \, i\in \N \}$.  Choose an increasing sequence $(i_j) \subset \N$ such that $g^{i_j}(a) \to b$.

  Let $L$ be an axis of $g$, then the sequence $(g^{i_j})$ pulls from $n = L(-\iy)= g^-$, leaving $L$ invariant.  If $d_T(a,n) \ge \pi$ then by $\pi$-convergence $b = L(\iy)=g^+$.  If $d_T(a,n) < \pi$ then by
  we construct  the function $f:[n,a] \to TX$ as above and so $f(a) = b$.

 By  Lemma \ref{L:f}  $f$ embeds $[n,a]$ into $S_{L(-\iy)}^{L(\iy)}= S_g$.  Thus $b \in S_g$.
\end{proof}

\begin{Def} An bounded metric space $X$ is almost a product($K$) of $Y$ and
$Z$ if there is a convex subset $W\cong Y \times Z$ and $X \subset \nb(W, K)$.
\end{Def}

\begin{Lem}\label{L:limprod} Let $X$ be a  proper cocompact CAT(0) space  $X$.  If
there is a convex subset $W$ of $X$ where $W = Y\times Z$, and a non-empty open subset $V \subset \bd X$ with $ V \subset \bd W$ then $X$ is almost a product of $\hat Y$ and $\hat Z$, convex subsets of $X$.  Additionally if $Y$ is cocompact  then  $\hat Y \cong Y$.
\end{Lem}
\begin{proof} By \cite{GEA-ONT} $X$ has almost extendable($C$) geodesics for some $C$.
That is for any $x,y \in X$ there exists $z \in B(y,C)$ such that the geodesic segment $[x,z]$ extends to a ray from $x$.

We may assume that $Y$ and $Z$ are (convex) subsets of $W$.  Fix a base point $w \in W$.
Let $\alpha \in V$.  Let $R:[0,\iy) \to W$ be a geodesic ray with $R(0) =w$ and $R(\iy) = \alpha$.  There exists $N \in \N$, $\epsilon >0$ such that $U(\alpha, N,\epsilon)\subset V$.

Using cocompactness, we find $(g_i)\subset G$ and $t_i\to \iy$ such that
$g_i(R_{t_i}) \to L$ uniformly on compact subset  where $R_{t_i}$ is the ray $R$ reparametrize to have domain $[-t_i, \iy)$.  Using the fact that $X$ is proper and passing to a subsequence, there are unbounded sequences $(y_i) \subset Y$, $(z_i) \subset Z$ so that $g_i(w_i) \to \hat w \in X$ where $w_i=(y_i,z_i)$.  Using properness and passing to a subsequence we have that $g_i(W) \to \hat W$, a convex subset of $X$.   Let $Y_i$ be the copy of $Y$ in $W$ which contains $z_i$, and $Z_i$ the copy of $Z$ in $W$ which contains $y_i$, so that $Y_i \cap Z_i = \{w_i\}$.  Passing to a subsequence, we may assume  that $g_i(Y_i) \to \hat Y$, a convex subset of $\hat W$ and $g_i(Z_i) \to \hat Z$, a convex subset of $\hat W$.   Now let $\hat a, \hat b \in \hat W$.  Thus there are sequences $(a_i), (b_i) \subset W$ with $g_i(a_i) \to \hat a$ and $g_i(b_i) \to \hat b$.
For each $i$, let  $a_i^Y= \pi_{Y_i}(a_i)$, the projection of $a_i$ in $Y_i$. Define $b_i^Y$, $a_i^Z$ and $b_i^Z$ similarly.  For each $i$,  $W$ is a product of $Y_i$ and $Z_i$ so $d(a_i,b_i)^2 = d(a_i^Y,b_i^Y)^2+d(a_i^Z,b_i^Z)^2$.  The sequence $(g_i(a_i^Y))$ will clearly converge to the to the closest point  projection $\pi_{\hat Y}(\hat a)$. Similarly $g_i(a_i^Z) \to \pi_{\hat Z}(\hat a)$,
$g_i(b_i^Y) \to \pi_{\hat Y}(\hat b)$, and $g_i(b_i^Z) \to \pi_{\hat Z}(\hat b)$.
It follows that
$$d\left(\hat a,\hat b\right)^2 = d\left(\pi_{\hat Y}(\hat a), \pi_{\hat Y}(\hat b)\right)^2+
d\left(\pi_{\hat Z}(\hat a),\pi_{\hat Z}(\hat b)\right)^2$$

 so  $\hat W = \hat Y\times \hat Z$.

When $Y$ admits a cocompact action we can arrange it so that for some $y \in Y$, and $h_i$ isometry of $Y$, $g_i(h_i(y))$ is bounded and so there is an isometric embedding $h:Y \to X$ such that $h(Y) = \hat Y$, and similarly for $Z$.

 Now we claim that $X \subset \nb(\hat W, C)  $.  Suppose not, then let $x \in X$ with $d(x,\hat W)> C$.
 Notice that $g^{-1}_i(\bar B(x, C)) \to \alpha$.  Thus for $i >>0$ $g^{-1}_i(\bar B(x, C)) \subset \tilde U(\alpha ,N,\epsilon)$.

 By almost extendable geodesics($C$) there is a point $v_i \in g^{-1}_i(\bar B(x, C))$ such that $v_i$ is on a ray $S:[0,\iy) \to X$ from our base point $ w$.  Notice that for $i >>0$,  $d(x, g_i(W)) >C$, so in that case $d(v_i, W) > 0$, so $S([0,\iy))\not \subset W$.

  Since the (unit speed geodesic) rays based at $ w$ give unique representatives of $\bd X$ and every element of $\bd  W$ is represented by a ray in $W$ from $w$, this implies that $S(\iy) \not \in \bd W$, but $S(\iy) \in U(\alpha ,N,\epsilon)\subset \bd W$ which is a contradiction.

\end{proof}

\begin{Cor}
Let $X$ be a proper  cocompact CAT(0) space and $W$  a convex subset  with cocompact stabilizer in the isometry group of $X$.  If for some nonempty open $V \subset \bd X$, $V \subset \bd W$, then $X$ lies in a uniform neighborhood of $W$.
\end{Cor}
\begin{proof} In the proof of Lemma \ref{L:limprod}, we may choose the sequence $(g_i)$ in the stabilizer of $W$, so $\hat W=W$.
\end{proof}
Unfortunately, the previous result raises more questions than it answers
\begin{Qu} If $X$ and $Y$ are cocompact proper CAT(0) spaces and $U\subset \bd X$, $V\subset \bd Y$ non-empty
with $U \cong V$, is $\bd X \cong \bd Y$?
(where $\cong$ is a homeomorphism in cone topology and an isometry in the Tits metric)

What if additionally we have a group $G$ acting geometrically on both $X$ and $Y$?
The answer is clearly yes  if $X$ or $Y$ is rank 1 and the isomorphism $U \cong V$ is $G$-equivariant.
\end{Qu}
See \cite{CRO-KLE} and \cite{MOO} for related questions.

We now generalize part of a result of Lytchak to our setting.

\begin{Thm} \cite{LYT}  Let $Z$ be a geodesically complete  finite dimensonal CAT(1) space.
Then $Z$ has a unique decomposition $Z = \SSS^n \ast G_1 \ast \dots \ast G_k \ast Y_1 \ast \dots \ast Y_m$
where $G_j$ is a thick irreducible building  and $Y_j$ is an irreducible (via spherical join) non-building.
\end{Thm}
In our setting, the Tits boundary $T X$ is not geodesically complete.  Nonetheless, we obtain the following result: If $Z$ is a finite dimensional CAT(1) space then there is a unique decomposition
$Z = S^n \ast Y$ where $Y$ doesn't have a sphere as a spherical join factor.

\begin{Lem}\label{L:TG} If $Y$ and $Z$ are CAT(0) spaces and $F$ is a flat sector in $Y \times Z$, then
$\pi_Y(F)$ is a flat sector in $A$.
\end{Lem}
\begin{proof}
Since a flat sector is the nested union of flat triangles (and this is sufficiant).
It suffices to show that the projection of a flat triangle is a flat triangle.

Let $abc$ be a triangle in a CAT(0) space.  Consider the following  condition: For $t \in (0,1)$ if $e \in [a,b]$ and $f \in [a,c]$ with $d(a,e) = t d(a,b)$ and $ d(a,f) =t d(a,c)$ then $d(e,f) = t d(b,c)$.
Notice that if $abc$ is flat then this condition is satisfied by similar triangles.  If on the the other hand this condition is satisfied, then by similar triangles, the Euclidian comparison angle $\bar \angle_a(b,c) =\bar \angle_a(e,f)$. It follows that $\angle_a(b,c) =\bar \angle_a(b,c)$ and so the triangle $abc$ is flat.

Thus we may assume that $abc$ is a triangle in satisfying this condition.  For $t \in (0,1)$ choose $e$ and $f$ as above.  Let $a_1,b_1,c_1,e_1,f_1$ be the projections of $a,b,c,e,f$ respectively into $Y$ and
$a_2,b_2, c_2,e_2,f_2$ the projections of $a,b,c,e,f$ respectively into $Z$.
By similar Euclidean triangles  $d(a_i, e_i) =t d(a_i,b_i)$ and $d(a_i,f_i)= td(a_i,c_i)$ for $i =1,2$.

By hypothesis $d(e,f) = t d(b,c)$ so $d(e,f)^2 = t^2 d(b,c) ^2$.  Thus
$$d(e_1,f_1)^2 +d(e_2,f_2)^2 = t^2d(b_1,c_1) ^2 + t^2d(b_2,c_2)^2$$

By the CAT(0) inequality applied to the triangles $a_ib_ic_i$, $i=1,2$.
$d(e_i,f_i) \le t d(b_i,c_i)$ for $i=1,2$.  It follows that $d(e_i,f_i) = t d(b_i,c_i)$ for $i=1,2$.
and so the triangle $a_ib_ic_i$ are flat for $i=1,2$.
\end{proof}
\begin{Def}
\cite[I 5.11]{BRI-HAE} For $Y$ and $Z$ metric spaces with metrics bounded by $\pi$, we define
$Y\ast Z$  to be the quotient of the space $Y\times Z \times [0, \frac \pi 2]$ by the identifications
$(y,z,0)= (y,\hat z,0)$ and $(y,z,\frac \pi 2) = (\hat y,z, \frac \pi 2)$.   We denote the class
$(y,z,\theta)$ by $y \cos \theta + z \sin \theta$ so $(y,z,0) = y$ and $(y,z,\frac \pi 2) = z$, and so we have
$Y,Z \subset Y\ast Z$. For $u=y\cos \theta + z \sin \theta, u'=y'\cos \theta' + z' \sin \theta' \in Y\ast Z$ we define $d(u,u')$ by
$$\cos(d(u,u')) = \cos \theta \cos \theta ' \cos (d(y,y')) +\sin \theta \sin \theta' \cos (d(z,z'))$$
Clearly for $\theta \neq 0 , \frac \pi 2$,  $d(u,u') = \pi$ if and only if $d(y, y') =\pi = d(z,z')$ and $\theta + \theta ' = \frac \pi 2$.

\end{Def}
\begin{Lem}Let $\alpha$ be a geodesic in $A\ast B$ where $A$ and $B$ are CAT(1) spaces. Notice that
$A$ and $B$ are $\pi$-convex subsets of $A\ast B$.
If  the image of $\alpha$ misses $B$, then the projection of $\alpha$ to $A$ is a geodesic.
\end{Lem}
\begin{proof}
Let $Y = C_0(A)$ and $Z= C_0(B)$ be the Euclidian cones on $A$ and $B$ respectively.  By \cite{BRI-HAE} $C_0(A\ast B) = X \cong Y \times Z$ and so $A\ast B = \bd (Y\times Z)$.
By \cite[I 5.15]{BRI-HAE} It suffices to show that the projection of $\alpha $ to $A$ is a local Tits geodesic.
 There is a flat sector $F \subset X$ based at the cone point
$0$ of $X$ with $\bd F = \alpha$.  We can think of $Y$ and $Z$ as being convex subsets of $X$.  By Lemma \ref{L:TG}, the projection $F_Y$ of $F$ into $Y$ is a Euclidean sector based at the cone point of $Y$ (which is $0$ of course).  By the proof of \cite[I 5.15]{BRI-HAE} the $\bd F_Y$ will be the projection of $\alpha$ to $A$.  It follows that this projection is a Tits geodesic in $A$.
\end{proof}

 \begin{Thm} \label{T:decomp}If $Z$ is a finite dimensional complete CAT(1) space, then there is a unique decomposition
$Z = S \ast D$ where $S\cong \SSS^n$ and $D$ doesn't have a non-empty sphere as a spherical join factor.
\end{Thm}
\begin{proof} Suppose that $Z = \SSS^k\ast U$ and $Z = \{a,b\} \ast A$, where $\{a,b\} \not \subset \SSS^k$.  It follows that $a$ is the unique antipode to $b$ in $Z$ and visa versa (so $\{a,b\}\cap \SSS^k= \emptyset$).  By the metric on $Z = \SSS^k \ast U$, we have $a_1,b_1 \in \SSS^k$,
$a_2,b_2 \in U$ and $\theta \in  (0,\pi/2]$ with $a = a_1\cos \theta +a_2\sin \theta$ and $ b = b_1\sin \theta + b_2\cos \theta$.  If follows that $a_2$ is the unique antipode of $b_2$ in $U$ and visa versa (and similarly for $a_1$ and $b_1$ in $\SSS^k$).

Since $Z =  \{a,b\} \ast A$ then $Z = S_a^b$, and every point of $Z$ is  on a Tits geodesic from $a$ to $b$.  Fix $u \in U \subset \SSS^k \ast U$;  the point $u$ is on a Tits geodesic from $a$ to $b$  which misses $\SSS^k$.  By Lemma \ref{L:TG}, the projection of this Tits geodesic into $U$  is a Tits geodesic from $a_2$ to $b_2$ passing through $u$.   It follows that $U$ is the union of all Tits geodesics in $U$ from $a_2$ to $b_2$.  This implies by \cite[4.1]{LYT} that $U = \{a_2,b_2\}\ast W$ where $W$ is the set of all points of $U$ at Tits distance $\pi/2$ from both $a_2$ and $b_2$.  Thus by \cite[I 5.15]{BRI-HAE}, $Z= \SSS^k\ast(\{a_2,b_2\} \ast W) = (\SSS^k \ast \{a_2,b_2\}) \ast W= \SSS^{k+1} \ast W$.  Since $Z$ is finite dimensional this process must terminate, and so for some $n$, $Z = \SSS^n \ast  Y$ where $Y$ doesn't have a non-empty sphere as a spherical join factor.  Notice by our construction that $$S= \{
a \in Z:\, Z \text{ is a spherical suspension with suspension point }a \}$$  Thus this decomposition is canonical.
\end{proof}
\begin{Cor}\label{C:sus}  Under the hypothesis of Lemma \ref{L:limprod},  if $Y \cong \E^n$ for some $n$ then $\bd Y \subset S$, the set of suspension points of $\bd X$. \end{Cor}
\begin{proof} By \cite{SWE} $\bd X$ is finite dimension, so $TX$ is finite dimensional,  and $\bd X = S \ast D$ as in Theorem \ref{T:decomp}.
  Since $\E^n $ admits a cocompact isometric group action, $\hat Y \cong Y \cong \E^n$.
$\bd X = \bd (\hat Y \times\hat Z) = ( \bd Y)\ast (\bd \hat Z)$.  Since $\bd \hat Y \cong \bd \E^n= S^{n-1}$ then $\bd \hat Y \subset S$.  Let $\alpha \in \bd Y - S$, so $d_T(\alpha, D) < \pi/2$.
We may assume  $g_i(\alpha) \to \beta$ preforce with $\beta \in \bd \hat Y \subset S$.  However $g_i(D) = D$ for all $i$, so $d_T(\beta, D) \le d_T(\alpha, D) < \pi /2$, but that is a contradiction since every point of $S$ is distance $\pi /2 $ form every point of $D$.  Thus $\bd Y \subset S$.
\end{proof}
\begin{Thm} Let $g$ be a hyperbolic isometry of the cocompact CAT(0) space $X$.  If $\Omega = \bd X - S_g \neq \emptyset$, then $\Omega$ is a dense open subset of $\bd X$, and
$\langle g \rangle $ acts discretely on $\Omega$. \end{Thm}
\begin{proof} By Lemma \ref{L:discrete}, $\langle g \rangle $ acts discretely on the open set  $\Omega$.  We must show that $\Omega$ is dense whenever it is non-empty.

 Suppose that $\Omega $ is not dense. Then there is an open subset $V$ of $\bd X$ with $V \subset \bd S_g$.  Recall that $S_g \cong \R \times Z$ for some $Z$, where the endpoints of $\R$ are $g^\pm$. Thus by Corollary \ref{C:sus}, $g^\pm \in S$, the set of suspension points of $\bd X$.  It follows that $S_g = \bd X$ as required.
\end{proof}


TITLE: How to solve the given multivariable integral using recursion relation?
QUESTION [0 upvotes]: This integral appeared in Statistical Mechanics while calculating the number of microstates for an ideal gas.
$B_N=\int_{0}^{\infty}dy_1y_1^2\cdot\cdot\cdot\cdot\int_{0}^{\infty}dy_N y_N^2\Theta(1-\sum_{r=1}^{N}y_r)$ where $\Theta$ denotes the unit step function.
The book gives the hint that because of the step function, the sum $y_2+y_3+\cdot\cdot\cdot +y_N$ cannot exceed the value $1-y_1$. For a fixed value of $y_1$, integration over all variables from $y_2$ and $y_N$ then yields $B_{N-1}(1-y_1)^{3(N-1)}$, and we can write
$B_N=\int_{0}^{1}dy_1y_1^2B_{N-1}(1-y_1)^{3(N-1)}$. Now $B_{N-1}$ can be taken out of the integral and this recursion relation can be used to obtain $B_N=\frac{2^N}{(3N)!}$.
I tried integrating over all variables except for $y_1$ to obtain the recursion relation. But I doubt it. Integrating first over $y_2$ gives cubic term and then higher powers with other variables. If the book had really followed this way then it would not have been much difficult to solve the last integral i.e over $y_1$ and no need to go for the recursion relation. There must be a way to rearrange the variables or substitute something to get $B_{N-1}(1-y_1)^{3(N-1)}$ without doing any integration.

REPLY [1 votes]: The whole point is to proceed recursively, not to try to do explicit iterated integration.
Notice that for $a>0$, if you change variables with $z_i = y_i/a$, then the condition that $\sum y_i \le a$ is equivalent to the condition that $\sum z_i \le 1$. Now when you do the integral, note that
$$y_i^2\,dy_i = a^3 z_i^2\, dz_i,$$
and so you'll pick up a factor of $a^3$ for each $y_i$. Note also that $\Theta(a-\sum y_i) = \Theta(1-\sum z_i)$. (The limits of integration remain $0$ to $\infty$, of course.) When you have $N-1$ of them and $a=1-y_1$, you get the factor $(1-y_1)^{3(N-1)}$, as desired.


TITLE: Confusion about the proof of Menger's Theorem in "Introduction to Graph Theory" by Douglas West
QUESTION [3 upvotes]: The proof of Menger's Theorem in the book "Introduction to Graph Theory" by Douglas West (2nd Edition; Page 167) has been divided into two cases.
The second case assumes that

"Every minimum $x,y$-cut is $N(x)$ or $N(y)$",

where $N(x)$ denote the set of neighbors of $x$.
However, it seems that the graph for Case 2 (see below) in the illustration does not satisfy this assumption. What is going on here?

REPLY [1 votes]: The illustrations will match up with the cases, if we change the descriptions of the cases to:
Case 1'. $G$ has a minimum $x,y$-cut $S$ not contained in $N(x) \cup N(y)$.
Case 2'. Every minimum $x,y$-cut is contained in $N(x) \cup N(y)$.
If we follow these descriptions, then the proof still works, because the only way we use the assumption in case 2 is to say that every vertex outside $\{x\} \cup N(x) \cup N(y) \cup \{y\}$ is in no minimum $x,y$-cut, and this is still true in case 2'. (Since case 1' is a subcase of case 1, there is nothing to worry about there.)
In general, whenever $G$ falls under both case 1 and case 2' (that is, every minimum $x,y$-cut is contained in $N(x) \cup N(y)$, but there is some minimum $x,y$-cut $S$ not equal to $N(x)$ or $N(y)$) then we can handle $G$ by the argument from either case, which is where this flexibility comes from.

Pedagogical note: when I taught this proof last year, I began by considering the case where $N(x) \cap N(y) = \varnothing$ and $V(G) = \{x\} \cup N(x) \cup N(y) \cup \{y\}$, which falls under case 2' and is the case where we can apply König-Egerváry. Then I dealt with the three possibilities below:

$v \in N(x) \cap N(y)$, which is handled in the case 2 proof. (Delete $v$, reducing $\kappa(x,y)$ by $1$.)
$v \notin \{x\} \cup N(x) \cup N(y) \cup \{y\}$, but $v$ is not part of any minimum $x,y$-cut, which is also handled in the case 2 proof. (Delete $v$, not changing $\kappa(x,y)$.)
There is a minimum $x,y$-cut $S$ not contained in $N(x) \cup N(y)$. (This is case 1', and we can apply the case 1 proof.)

In some sense, this is the logical progression: we apply König-Egerváry for some cases, and then show that all other cases can be reduced to smaller ones.


TITLE: Showing some sequence of function is delta converging.
QUESTION [0 upvotes]: Show that
$$f_n(x) = \frac{\sin nx}{\pi x}$$
converges to $\delta$ in the sense of distribution. I know the same question have been asked here. But I found some hint from Arfken Mathematical Methods for physicists which is

Show that
  $$\lim_{n\to\infty}\int_{-\infty}^{\infty}f(x)\frac{\sin nx}{\pi x}dx=0$$
  Assume that $f(x)$ is continuous at $x = 0$ and vanishes as $x → ±∞$.
Hint. Replace $x$ by $y/n$ and take $\lim\ n → ∞$ before integrating.

with this hint I found the following solution(simple to grasp)

from this site. Everything in the above solution is a pretty much convincing to me except while substitution the denominator should be $\pi y/n$. But having that "$n$" the limit become "ugly" and the result would be different. So, how can we show $\lim_{n\to\infty}\int_{-\infty}^{\infty}f(x)\frac{\sin nx}{\pi x}dx=0$ using the hint.

REPLY [1 votes]: Let $y = nx$, then $x = y/n$ and $dx = dy/n$, hence
$$\int_{-\infty}^{\infty}f(x)\frac{\sin nx}{\pi x}dx=\int_{-\infty}^{\infty}f(y/n)\frac{\sin y}{\pi (y/n)}d(y/n)=\int_{-\infty}^{\infty}f(y/n)\frac{\sin y}{\pi y}dy,$$
which allows you to pass to the limit.
Does this answer your question?