text:
\section{Introduction and main result} \label{sec:Intro_and_main_results}

In this paper we are concerned with the Galton-Watson processes
(GW processes, in short) with immigration. Below we outline the
setting and refer to the classical treatises
\cite{Asmussen+Hering:1983, Athreya+Ney:1972} for more details on
the GW processes with and without immigration.

Let $(X_{i,k})_{i\in\mathbb{N}, k\in\mathbb{N}_0}$ and $(J_k)_{k\in\mathbb{N}_0}$, where $\mathbb{N}_0:=\mathbb{N}\cup\{0\}$, be mutually independent families of independent and
identically distributed (i.i.d.)\ random objects, where each
$X_{i,k}:=(X_{i,k}(n))_{n\in\mathbb{N}_0}$ is a GW
process with $X_{i,k}(0)=1$ and $\E X_{i,k}(1)=\mu\in (0,\infty)$,
and each $J_k$ is a nonnegative integer-valued random variable with
$\Prob\{J_k=0\}<1$. The random
sequence $Y:=(Y_n)_{n\in\mathbb{N}_0}$ defined by
$$
Y_n:=\sum_{k=0}^n\sum_{i=1}^{J_k}X_{i,k}(n-k),\quad n\in\mathbb{N}_0
$$
is called a {\it Galton-Watson process with immigration}. The random variable $J_k$ represents the number of immigrants which arrived at time $k$, while the GW process $(X_{i,k}(n))_{n\in\mathbb{N}_0}$ represents the number of descendants of the $i$th immigrant which arrived at time $k$, for $1\leq i \leq J_k$.

Let $J$ denote a random variable with the same law as the $J_k$'s.
It is known that the asymptotic behavior of $Y_n$ depends heavily upon the finiteness of the logarithmic moment $\E \log^+ J$, where $\log^+ x = \max(\log x,0)$. In the supercritical case $\mu>1$, the a.s.\ limit $\lim_{n\to\infty} Y_n/\mu^n$ exists and is finite a.s. provided that $\E \log^+ J<\infty$, whereas  $\lim_{n\to\infty} Y_n/c^n=\infty$ a.s.\ for every $c>0$ if $\E \log^+ J = \infty$; see~\cite{Seneta:1970a}, \cite{Seneta:1970b}.
In the subcritical case $\mu<1$, $Y_n$ converges in distribution to a non-degenerate random variable provided that $\E \log^+ J<\infty$, whereas $Y_n$ diverges to $+\infty$ in probability if $\E \log^+ J = \infty$; see~\cite{Heathcote:1966}. These and more refined results can be found in~\cite{Asmussen+Hering:1983}, Theorems 6.1 and 6.4.

Let $D:=D[0,\infty)$ denote the Skorokhod space of
right-continuous functions defined on $[0,\infty)$ with finite
limits from the left at positive points.
We intend to prove functional limit theorems for the process $\log^+(Y_{[n\cdot]})$ in $D$ as $n\to\infty$ under the assumption
\begin{equation}\label{stand}
\Prob\{\log J>x\}\sim x^{-\alpha}\ell(x),\quad x\to\infty
\end{equation}
for some $\alpha\in (0,1]$ and some $\ell$ slowly varying at $\infty$, which justifies the term ``very active immigration''.

To state our results we need to introduce certain Poisson random measures which appear as limits for the extremal order statistics of the sequence $(\log^+ J_k)_{k\in\N_0}$. Denote by $M_p$ the set of point measures $\nu$ on $[0,\infty)\times (0,\infty]$ which satisfy
\begin{equation}\label{1}
\nu([0,T]\times [\delta,\infty])<\infty
\end{equation}
for all $T>0$ and all $\delta>0$.  For positive $a$ and $b$, let
$$
N^{(a, b)}:=\sum_{k=1}^{\infty}
\varepsilon_{(t_k^{(a,b)},\,j_k^{(a,b)})}
$$
be a Poisson
random measure on $[0,\infty)\times (0,\infty]$ with mean measure
$\mathbb{LEB}\times \mu_{a, b}$, where $\varepsilon_{(t,\,x)}$
is the probability measure concentrated at $(t,x)\in [0,\infty)\times (0,\infty]$, $\mathbb{LEB}$ is the Lebesgue
measure on $[0,\infty)$, and $\mu_{a,b}$ is a measure on
$(0,\infty]$ defined by
$$
\mu_{a, b}\big((x,\infty]\big)=ax^{-b}, \ \ x>0.
$$
Throughout the paper we
use $\Rightarrow$ to denote weak convergence on the Skorokhod
space $D$ equipped with the $J_1$-topology (see \cite{Billingsley:1968, Lindvall:1973} for the necessary background) and on $M_p$ endowed with the vague topology.

Theorem \ref{main1} treats the situation in which the behaviour in
mean of the GW processes $X_{i,k}$ affects the limit behavior of
$Y$ (except in the less interesting case $\mu=1$), whereas in the
situation of Theorem \ref{main2} any traces of the $X_{i,k}$
disappear in the limit. We stipulate hereafter that the supremum
over the empty set is equal to zero.
\begin{Theorem}\label{main1} Assume that for some $c>0$,
\begin{equation}\label{2}
\underset{x\to\infty}{\lim} x\Prob\{\log J>x\}=c.
\end{equation}
Then, as $n\to\infty$,
\begin{equation}\label{317}
\frac {\log^+ (Y_{[n\cdot]})}{n}\quad \Rightarrow \quad \underset{t_k^{(c,1)}\leq
\cdot}{\sup}\big(j_k^{(c,1)} + (\cdot-t_k^{(c,1)})\log\mu\big).
\end{equation}
\end{Theorem}
\begin{Rem}
Realizations of the limit processes, which are called extremal shot noise processes~\cite{Dombry:2012}, are shown on Figure~\ref{figure:extremal_shot_noise}. In the critical case $\mu=1$, the limit is the well-known extremal process; see~\cite{Resnick:2008}, Sections 4.3--4.4.
We shall see in the course of the proof that in the supercritical case $\mu>1$, we also have
\begin{equation}\label{31}
\frac {\log^+ (\mu^{-[n\cdot]}Y_{[n\cdot]})} {n}\quad\Rightarrow\quad \underset{t_k^{(c,1)}\leq
\cdot}{\sup}\big(j_k^{(c,1)} - t_k^{(c,1)} \log\mu\big),
\end{equation}
in which case the marginal distributions of the limit process have support equal to $[0,\infty)$.
\end{Rem}

\begin{figure}[t]
\begin{center}
\includegraphics[width=0.32\textwidth]{subcrit.pdf}
\includegraphics[width=0.32\textwidth]{crit.pdf}
\includegraphics[width=0.32\textwidth]{supercrit.pdf}
\end{center}
\caption
{\small The limit processes appearing in Theorem~\ref{main1}. Left: subcritical case $\mu<1$.  Middle: critical case $\mu=1$. Right: supercritical case $\mu>1$. }
\label{figure:extremal_shot_noise}
\end{figure}

\begin{Theorem}\label{main2}
Suppose that \eqref{stand} holds. If $\alpha$ in \eqref{stand} equals $1$, suppose additionally that
$\underset{x\to\infty}{\lim} \ell(x)=\infty$. Let $(b_n)_{n\in\N}$ be a sequence of positive numbers satisfying
$$
\underset{n\to\infty}{\lim} n\Prob\{\log J>b_n\}=1.
$$
Then, as $n\to\infty$,
\begin{equation}\label{35}
\frac {\log^+ (Y_{[n\cdot]})} {b_n}\quad\Rightarrow\quad \underset{t_k^{(1,\,\alpha)}\leq
\cdot}{\sup}\,j_k^{(1,\alpha)}.
\end{equation}
\end{Theorem}

\begin{Rem}
Let us derive closed formulae for the marginal distributions of the limit processes. We claim that, with $r,s>0$ and $u\geq 0$,
\begin{eqnarray}\label{38}
\Prob\left\{\underset{t_k^{(r,1)}\leq
u}{\sup}\left(j_k^{(r,1)} + s (t_k^{(r,1)}-u)\right)\leq x\right\}
&=&
\Prob\left\{\underset{t_k^{(r,1)}\leq
u}{\sup}\left(j_k^{(r,1)}-st_k^{(r,1)}\right)\leq x\right\}\notag\\&=&\left(\frac {x}
{x+su}\right)^{r/s}
\end{eqnarray}
for all $x\geq 0$.
We only provide details for the second probability. Since
$$
N:= N^{(r,1)}\big((t,y): 0\leq t\leq u,
y-st>x\big)
$$
is a Poisson random variable, we have $\Prob\big\{N=0\big\}= e^{-\E N}$ and it remains to note that
\begin{eqnarray*}
\E N=\int_0^u\int_{[0,\infty)}\1_{\{y-st>x\}}\mu_{r,\,1}({\rm d}y){\rm
d}t=r\int_0^u(x+st)^{-1}{\rm
d}t=\frac rs \log\frac {x+su}{x}.
\end{eqnarray*}
Similarly, for the marginals of the extremal process appearing in~\eqref{35} we obtain, with $a,b>0$ and $u\geq 0$,
\begin{equation*
\Prob\left\{\underset{t_k^{(a,\,b)}\leq
u}{\sup}\,j_k^{(a,\,b)}\leq
x\right\}=\Prob\left\{N^{(a, b)}\left((t,y): 0\leq t\leq u,
y>x\right)=0\right\}= e^{-uax^{-b}}
\end{equation*}
for all $x\geq 0$. Armed with these observations we conclude that
relations \eqref{317} and \eqref{31} include the
results obtained by Pakes \cite{Pakes:1979} concerning weak convergence
of the one-dimensional distributions (Theorem 2, Theorem 6 and
Theorem 12 for the subcritical $\mu<1$, supercritical $\mu>1$ and
critical
$\mu=1$ cases, respectively). Similarly, the one-dimensional version of our relation \eqref{35}
is equivalent to the limit relations of Theorem 3 (case $\mu<1$),
Theorem 7 (case $\mu>1$) and Theorem 12 (case $\mu=1$) of
\cite{Pakes:1979}. To be more precise, Pakes states in Theorems 3
and 7 that
\begin{equation}\label{pak}
\underset{n\to\infty}{\lim} \Prob\bigg\{{\log(1+Y_n)\over a_n}\leq
x\bigg\}=\exp(-(|\log\mu|)^{-1}x^{-\alpha}),\quad x\geq 0
\end{equation}
whenever
\begin{equation}\label{pak1}
1-\E(1-e^{-x})^J\sim x^{-\alpha}\ell(x)\quad x\to\infty
\end{equation}
with $a_n$ defined by $1-\E(1-e^{-a_n})^J \sim (|\log \mu|n)^{-1}$
as $n\to\infty$. Formula \eqref{pak} is misleading because it
contains $|\log \mu|$ thereby suggesting that the contribution of the
$X_{i,k}$ persists in the limit. However, the relation
\begin{equation}\label{ours}
\underset{n\to\infty}{\lim} \Prob\bigg\{{\log(1+Y_n)\over b_n}\leq x\bigg\}=\exp(- x^{-\alpha}),\quad x\geq 0
\end{equation}
which is a consequence of \eqref{35} shows this is not the case.
Having observed that \eqref{pak1} is equivalent to \eqref{stand}
we infer $b_n\sim (|\log \mu|)^{-1/\alpha}a_n$ which implies that
\eqref{pak} and \eqref{ours} are actually equivalent.

Finally,  we note that unlike us, Pakes \cite{Pakes:1979} imposed a regular
variation assumption on the tail of $X_{1,1}(1)$ for the critical
case and used the Seneta-Heyde norming rather than
$\mu^{-n}$ in the supercritical case in \eqref{31}.
\end{Rem}

\begin{Rem}
One may expect that, under \eqref{stand}, $Y_n$ is well approximated by $Z_n:=\E(Y_n|(J_k)_{k\in\N_0})=\sum_{k=0}^n \mu^{n-k}J_k$ for large $n$. Although this turns out to be true, it is worth stressing that both the behavior in mean and the survival probability (especially in the subcritical case) of the underlying GW processes affect the asymptotics of $Y_n$.
The sequence $(Z_n)_{n\in\N_0}$ is a rather particular case\footnote{Just take $A_n=\mu$ and $B_n=J_n$, $n\in\N_0$.} of the much studied Markov chain $(X_n)_{n\in\N_0}$ defined by
$$X_0:=B_0\quad\text{and}\quad X_n=A_nX_{n-1}+B_n,\quad n\in\N,$$ where $(A_n, B_n)$ are i.i.d.\ $\R^2$-valued random vectors independent of $B_0$.
Functional limit theorems for $\log^+ (X_{[n\cdot]})$ were obtained in
\cite{Buraczewski+Iksanov:2015} under the assumption that
\eqref{stand} holds with $B_1$ replacing $J$ and that $\underset{n\to\infty}{\lim}
(A_1\cdot\ldots\cdot A_n) =0$ a.s.\ (this corresponds to the
subcritical GW processes).
\end{Rem}

\section{Preparatory results}

 We start with a lemma that might have been known. Recall that $(X_{1,1}(n))_{n\in\N_0}$ is a GW process with $X_{1,1}(0)=1$ and mean $\mu \in (0,\infty)$.
\begin{Lemma}\label{aux1}
If $\mu\leq 1$, then for all $\delta>0$,
\begin{equation}\label{subexp}
\underset{n\to\infty}{\lim} e^{\delta n}\mu^{-n}\Prob\{X_{1,1}(n)\geq 1\}=\infty.
\end{equation}
\end{Lemma}
\begin{proof}
Put $p_n:=\Prob\{X_{1,1}(n)\geq 1\}$, $n\in\mathbb{N}$. If
$\Prob\{X_{1,1}(1)=1\}=1$, then $\mu=1$ and $p_n=1$, and
\eqref{subexp} holds trivially. To deal with the remaining cases
$\mu=1$ and $\Prob\{X_{1,1}=1\}<1$ or $\mu<1$ in which $\underset{n\to\infty}{\lim}
p_n=0$ we set $f_n(s):=\E s^{X_{1,1}(n)}$, $n\in\mathbb{N}$, $s\in
[0,1]$. Then
$$\mu=\underset{s\to 1-}{\lim}{1-f_1(s)\over 1-s}=\underset{n\to\infty}{\lim}
{1-f_1(f_n(0))\over 1-f_n(0)}=\underset{n\to\infty}{\lim} {p_{n+1}\over p_n}$$ having
utilized $1-p_n=\Prob\{X_{1,1}(n)=0\}=f_n(0)\to 1-$ as
$n\to\infty$ for the second equality. This implies $\underset{n\to\infty}{\lim}
{\mu^{-(n+k)}p_{n+k}\over \mu^{-n} p_n}=1$ for each $k\in\mathbb{N}$ and
thereupon \eqref{subexp}.
\end{proof}

Let $(c_n)_{n\in\mathbb{N}}$ be a sequence of positive numbers satisfying $\underset{n\to\infty}{\lim} n^{-1}c_n=\infty$ or $c_n=n$ for all $n\in\mathbb{N}$. For each $0<\gamma<1$, set
\begin{equation}\label{eq:def_Y_proc_truncated}
Y_{[n\cdot]}^{(\leq \gamma)}:=\sum_{k=0}^{[n\cdot]}\1_{\{J_k\leq e^{\gamma c_n}\}}\sum_{i=1}^{J_k}X_{i,k}([n\cdot]-k).
\end{equation}
The next lemma shows that $Y_{[n\cdot]}^{\leq \gamma}$, the contribution coming from times in which immigration is not extremely active, is negligible as $n\to\infty$ and $\gamma\to 0+$. No assumptions on the tail of $J$ are imposed in this lemma.
\begin{Lemma}\label{aux3}
Fix $T>0$ and $\gamma>0$.  For every $\delta>0$,
$$
\sum_{n\geq 1}\Prob\bigg\{\underset{0\leq t\leq T}{\sup}\,\frac 1 {c_n} \log^+\bigg(m([nt])Y_{[nt]}^{(\leq \gamma)}\bigg)>\gamma+\delta\bigg\} < \infty,
$$
where $m(n):=\mu^{-n}\wedge 1$, $n\in\N_0$ when $c_n=n$ and $m(n)=1$, $n\in\N_0$ when $\underset{n\to\infty}{\lim} n^{-1}c_n=\infty$.
\end{Lemma}
\begin{proof}
Suppose first that $c_n=n$. For $r\in\N_0$, set $d(r):=(\mu^{-r}\wedge 1)\sum_{k=0}^r\mu^k$ and note that $d(r)=r+1$ if $\mu=1$ and $d(r)\leq (1-\mu\wedge\mu^{-1})^{-1}<\infty$ if $\mu\neq 1$. For all $\delta>0$,
\begin{eqnarray*}
&&\sum_{n\geq 1}\Prob\bigg\{\underset{0\leq t\leq T}{\sup}\,\frac 1n \log^+\bigg(\big(\mu^{-[nt]}\wedge 1\big)Y_{[nt]}^{(\leq \gamma)}\bigg)>\gamma+\delta\bigg\}\\&\leq &\sum_{n\geq 1}\sum_{r=0}^{[nT]}\Prob\bigg\{\big(\mu^{-r}\wedge 1\big)\sum_{k=0}^r\1_{\{J_k\leq e^{\gamma n}\}}\sum_{i=1}^{J_k}X_{i,k}(r-k)>e^{(\gamma+\delta)n}\bigg\}\\&\leq& \sum_{n\geq 1}\sum_{r=0}^{[nT]}\Prob\bigg\{\big(\mu^{-r}\wedge 1\big)\sum_{k=0}^r\sum_{i=1}^{[e^{\gamma n}]}X_{i,k}(r-k)>e^{(\gamma+\delta)n}\bigg\}\\&\leq & \sum_{n\geq 1} e^{-\delta n}\sum_{r=0}^{[nT]}
d(r)<\infty.
\end{eqnarray*}
We have used Boole's inequality for the second line and Markov's inequality in combination with $\E X_{i,k}(r-k)=\mu^{r-k}$ for the fourth line.

When $\underset{n\to\infty}{\lim} n^{-1}c_n=\infty$, the proof is similar: for all $\delta>0$,
\begin{eqnarray*}
&&\sum_{n\geq 1}\Prob\bigg\{\underset{0\leq t\leq T}{\sup}\,\frac 1 {c_n}\log^+\bigg(Y_{[nt]}^{(\leq \gamma)}\bigg)>\gamma+\delta\bigg\}\\&\leq &\sum_{n\geq 1}\sum_{r=0}^{[nT]}\Prob\bigg\{\sum_{k=0}^r\1_{\{J_k\leq e^{\gamma c_n}\}}\sum_{i=1}^{J_k}X_{i,k}(r-k)>e^{(\gamma+\delta)c_n}\bigg\}\\&\leq& \sum_{n\geq 1}\sum_{r=0}^{[nT]}\Prob\bigg\{\sum_{k=0}^r\sum_{i=1}^{[e^{\gamma c_n}]}X_{i,k}(r-k)>e^{(\gamma+\delta)c_n}\bigg\}\\&\leq & \sum_{n\geq 1} e^{-\delta c_n}([nT]+1)\sum_{k=0}^{[nT]}\mu^k<\infty,
\end{eqnarray*}
because $([nT]+1)\sum_{k=0}^{[nT]}\mu^k$ grows at most exponentially whereas $e^{-\delta c_n}$ decreases superexponentially in view of $\underset{n\to\infty}{\lim} n^{-1}c_n=\infty$.
\end{proof}

In the next lemma, which is needed to prove Theorem~\ref{main1}, we identify the response functions of the limit extremal shot noise process. Roughly speaking, this lemma states that a GW process with finite mean $\mu$ starting at time $0$ with approximately $e^{an+o(n)}$ particles has at time $nt$ approximately $\mu^{nt} e^{an +o(n)}$ particles. However, there is one exception: if the process is subcritical, then the population dies out approximately at time $n a/ |\log \mu|$, and the number of particles after this time is $0$.
\begin{Lemma}\label{aux2}
Let $(A_n)_{n\in\mathbb{N}}\subset \N$ be a sequence satisfying $\underset{n\to\infty}{\lim} n^{-1} \log A_n = a$ for some $a>0$.
Then, for every $T>0$ and every sequence $(k_n)_{n\in\N}\subset \N_0$,
\begin{equation}\label{limit1}
\underset{n\to\infty}{\lim}
\sup_{0\leq t\leq T} \left|\frac 1n \log^+\bigg(\sum_{i=1}^{A_n}X_{i,k_n}([nt])\bigg) - (a+t \log\mu)^+\right| = 0
\quad \text{{\rm a.s.}}
\end{equation}
\end{Lemma}
\begin{proof}
According to the Borel-Cantelli lemma in combination with
$$\underset{n\to\infty}{\lim} \underset{t\in [0,T]}{\sup}\, |[nt]/n-t|=0$$ it suffices to check that for all $\varepsilon\in (0,a)$,
$$
\sum_{n\geq 1}\Prob\bigg\{\underset{0\leq t\leq T}{\sup}\,\bigg|\log^+\bigg(\sum_{i=1}^{A_n}X_{i,0}([nt])\bigg)-(an+[nt]\log\mu)^+\bigg|>\varepsilon n\bigg\}<\infty.
$$
We write $X^\ast_{i,0}(l):=\mu^{-l}X_{i,0}(l)$, so that $\E X^\ast_{i,0}(l)=1$. By Boole's inequality the last probability is bounded from above by $I_1(n)+I_2(n)$ with
\begin{align*}
I_1(n) &:=
\sum_{r=0}^{[nT]}\Prob\bigg\{\sum_{i=1}^{A_n}X_{i,0}(r)> e^{(an+r\log\mu)^++\varepsilon n}\bigg\},\\
I_2(n) &:= \sum_{r=0}^{r_n}
\Prob\bigg\{\sum_{i=1}^{A_n}X^\ast_{i,0}(r)<e^{(a-\varepsilon)n}\bigg\},
\end{align*}
where  $r_n:=[nT]\wedge [nT_0]$ and
$$
T_0:=
\begin{cases}
(a-\varepsilon)/|\log\mu|,& \text{ if } \mu<1,\\
+\infty, &\text{ if } \mu\geq 1.
\end{cases}
$$

To prove that $\sum_{n\geq 1} I_1(n)$ is finite, note that $|\log A_n-an|\leq \frac \varepsilon 2n$ for sufficiently large $n$.
Using Markov's inequality yields
$$
\sum_{n\geq 1}I_1(n)\leq \sum_{n\geq 1}\sum_{r=0}^{[nT]} \frac{e^{\log A_n+r\log\mu}}{e^{ (an+r\log\mu)^++\varepsilon n}}
\leq
C+ \sum_{n\geq 1}([nT]+1)e^{-\frac 12 \varepsilon n}<\infty.$$

In the following, we prove that $\sum_{n\geq 1} I_2(n)$ is finite. Let $p_n:=\Prob\{X_{1,0}(n)\geq 1\}$, $n\in\N$, be the probability that a GW process starting with a single particle at time $0$  does not die out at time $n$.  We fix $u>0$ and use Markov's inequality in combination with the
fact that $(e^{-u X^\ast_{1,0}(l)})_{l\in\N_0}$ is a
submartingale w.r.t.\ the natural filtration to infer that for large enough $n$ and all $r\leq r_n$,
\begin{align}
\log\Prob \left\{\sum_{i=1}^{A_n} X^\ast_{i,0}(r)<e^{(a-\varepsilon)n}\right\}
&\leq
ue^{(a-\varepsilon)n} + A_n\log \E e^{-uX^\ast_{1,0}(r_n)} \label{import} \\
&\leq
ue^{(a-\varepsilon)n}+A_n\log(1-p_{r_n}+p_{r_n}e^{-\mu^{-r_n}u}) \notag\\
&\leq
ue^{(a-\varepsilon)n}-A_n(1-e^{-\mu^{-r_n}u})p_{r_n} \notag\\
&\leq ue^{(a-\varepsilon)n}-e^{(a-\frac 12 \varepsilon)n}(1-e^{-\mu^{-r_n}u})p_{r_n}.\label{import2}
\end{align}
Further, we consider the three cases separately.

\vspace*{2mm}
\noindent {\sc Supercritical case $\mu>1$}. It is well known (see, for instance, Theorem 5.1 on p.~83 together
with Corollary 5.3 on p.~85 in \cite{Asmussen+Hering:1983}) that
there exists a function $L$ slowly varying at $\infty$ with $\liminf_{x\to\infty} L(x)>0$
such that, as $n\to\infty$, $X^\ast_{1,0}(n)/L(\mu^n)$ converges
a.s.\ to a random variable $W$, say, which is positive with
positive probability\footnote{The sequence $(\mu^n L(\mu^n))$ is
known as the Seneta-Heyde norming.}. In particular, by the dominated convergence,
$$
c := \underset{n\to\infty}{\lim} \log \E e^{-X^\ast_{1,0}(n)/L(\mu^n)} = \log \E e^{-W}   < 0.
$$
With
$u=1/L(\mu^{[nT]})$, inequality \eqref{import} takes the form
$$
\log\Prob\left\{\sum_{i=1}^{A_n} X^\ast_{i,0}(r)<e^{(a-\varepsilon)n}\right\} < \frac{e^{(a-\varepsilon) n}}{L(\mu^{[nT]})} + (c+o(1))A_n.
$$
The right-hand side goes to $-\infty$ as $n\to\infty$ exponentially fast  because $A_n > e^{(a-\frac 12 \varepsilon)n}$ for large $n$ and $L$ is slowly varying, thereby proving $\sum_{n\geq 1}I_2(n)<\infty$.

\vspace*{2mm}
\noindent {\sc Critical case $\mu=1$}. With $u>0$ fixed, inequality \eqref{import2} takes the form
$$
\log\Prob\left\{\sum_{i=1}^{A_n} X^\ast_{i,0}(r)<e^{(a-\varepsilon)n}\right\}
\leq ue^{(a-\varepsilon)n} - e^{(a-\frac 12 \varepsilon) n}(1-e^{-u})p_{[nT]}.
$$
In view of \eqref{subexp} this goes to $-\infty$ as $n\to\infty$ exponentially fast, whence $\sum_{n\geq 1}I_2(n)<\infty$.

\vspace*{2mm}
\noindent {\sc Subcritical case $\mu<1$}. With $u=e^{-(a-\varepsilon)n}$, expression \eqref{import2} takes the form
$$
1-(1-e^{-\mu^{-[nT_0]} e^{-(a-\varepsilon)n}})p_{[nT_0]}e^{(a-\frac 12 \varepsilon)n}
\leq
1-(1-e^{-\mu}) p_{[nT_0]}\mu^{-{[nT_0]}}e^{\frac 12 \varepsilon n},
$$
because $\mu<\mu^{-{[nT_0]}}e^{-(a-\varepsilon)n}\leq 1$.  In view of \eqref{subexp} this goes to $-\infty$ as $n\to\infty$ exponentially fast, thus proving that $\sum_{n\geq 1}I_2(n)<\infty$ in this case, too.
\end{proof}
In the next lemma, we consider a GW process starting at time $0$ with $e^{(a + o(1)) c_n}$ particles, where, as before, $(c_n)_{n\in\N}$ is a sequence of positive numbers satisfying $\lim_{n\to\infty} n^{-1}c_n =\infty$. At time $nt$, the number of particles in such a process is approximately $\mu^{nt} e^{(a+o(1))c_n} = e^{(a+o(1))c_n}$, so that we do not see any changes on the logarithmic scale. The subcritical case plays no special role here, because the process is very unlikely to die out on the time scale $n$.

\begin{Lemma}\label{aux2a}
Let $(A_n)_{n\in\N}\subset \N$ be a sequence satisfying $\underset{n\to\infty}{\lim} c_n^{-1} \log A_n = a$ for some $a>0$.
Then, for every $T>0$ and every sequence $(k_n)_{n\in\N}\subset \N_0$,
\begin{equation}\label{limit2}
\underset{n\to\infty}{\lim} \sup_{0\leq t\leq T} \left| \frac 1 {c_n}\log^+\bigg(\sum_{i=1}^{A_n}X_{i,k_n}([nt])\bigg) - a\right| =0
\quad \text{{\rm a.s.}}
\end{equation}
\end{Lemma}
\begin{proof}
The proof  is similar to that of Lemma \ref{aux2}. Therefore we only give an outline. It suffices to prove that  for all $\varepsilon\in (0,a)$,
$$
\sum_{n\geq 1}\Prob\bigg\{\underset{0\leq t\leq T}{\sup}\,\bigg|\log^+\bigg(\sum_{i=1}^{A_n}X_{i,0}([nt])\bigg)-ac_n\bigg|>\varepsilon c_n\bigg\}<\infty.
$$
The last probability is bounded from above by $J_1(n)+J_2(n)$, where
\begin{align*}
J_1(n) &:= \sum_{r=0}^{[nT]}\Prob\bigg\{\sum_{i=1}^{A_n}X_{i,0}(r)>\exp((a+\varepsilon)c_n)\bigg\}\\
J_2(n) &:= \sum_{r=0}^{[nT]}
\Prob\bigg\{\sum_{i=1}^{A_n}X^\ast_{i,0}(r)<\mu^{-r}\exp((a-\varepsilon)c_n)\bigg\}.
\end{align*}
For sufficiently large $n$, we have $|\log A_n-ac_n|\leq \frac \varepsilon 2 c_n$.
We use Markov's inequality to obtain
$$
\sum_{n\geq 1}J_1(n)
\leq
\sum_{n\geq 1}  A_n e^{-(a+\varepsilon)c_n}  \sum_{r=0}^{[nT]} \mu^r
\leq C+
\sum_{n\geq 1}e^{-\frac12 \varepsilon c_n}\sum_{r=0}^{[nT]}\mu^r<\infty,
$$ where the finiteness follows from the fact that
$e^{-\frac 12 \varepsilon c_n}$ decreases superexponentially in view of $\underset{n\to\infty}{\lim} n^{-1}c_n=\infty$. While analyzing $J_2(n)$ we only treat the subcritical case. A counterpart of \eqref{import2} reads
$$
\log\Prob\bigg\{\sum_{i=1}^{A_n}X^\ast_{i,0}(r)<\mu^{-r}e^{(a-\varepsilon)c_n}\bigg\}\notag\leq u\mu^{-[nT]}e^{(a-\varepsilon)c_n}-e^{(a-\frac 12 \varepsilon)c_n} (1-e^{-\mu^{-[nT]}u})p_{[nT]}
$$
for large enough $n$, $r\leq [nT]$ and any $u>0$. On setting $u=\mu^{[nT]}$ the right-hand side takes the form
$$ e^{(a-\varepsilon)c_n}(1-(1-e^{-1}) e^{\frac 12 \varepsilon c_n} p_{[nT]}).$$ As $n\to\infty$, this goes to $-\infty$ superexponentially fast by \eqref{subexp}. The proof of Lemma \ref{aux2a} is complete.
\end{proof}

\begin{Lemma}\label{lem:log_+_subadd}
For all $x,y\geq 0$, we have
\begin{equation}\label{ineq}
\log^+x\leq \log^+(x+y)\leq \log^+ x+\log^+y+2\log 2.
\end{equation}
\end{Lemma}
\begin{proof}
While the left-hand inequality follows by monotonicity, the
right-hand inequality is a consequence of $$\log^+ x\leq \log
(1+x)\leq \log^+ x+\log 2,\quad x\geq 0$$ and the subadditivity of
$x\mapsto \log (1+x)$, namely,
\begin{eqnarray*}
\log^+(x+y)\leq \log (1+x+y)\leq \log (1+x)+\log
(1+y)\leq \log^+ x+\log^+y+2\log 2.
\end{eqnarray*}
\end{proof}

\section{Proofs of the main results}

\begin{proof}[Proof of Theorem \ref{main1}]
It is a standard fact of the extreme-value theory that
condition \eqref{2} entails the point processes convergence
\begin{equation*}
N_n := \sum_{k\geq 0}\1_{\{J_{k}\neq 0\}}\varepsilon_{(\frac kn,\,
\frac 1 n \log J_{k})}\quad \Rightarrow\quad
N^{(c,1)}
\end{equation*}
weakly on $M_p$, as $n\to\infty$; see, for instance, Corollary 4.19 (ii) on p.~210 in \cite{Resnick:2008}.

\vspace*{2mm}
\noindent
\textsc{Step 1: Passing to a.s.\ convergence.}
By the Skorokhod representation theorem there are versions $\widehat N_n$ and $\widehat N^{(c,1)}$ of $N_n$ and $N^{(c,1)}$ (defined on some new probability space) which converge
a.s.  That is, with probability $1$,
\begin{equation}\label{7}
\widehat{N}_n:=\sum_{k\geq 0} \1_{\{y_k^{(n)}\neq -\infty\}} \varepsilon_{(\frac kn,\,
y_k^{(n)})}\quad \to\quad
\widehat{N}^{(c,1)}=\sum_{m\geq 0}\varepsilon_{(\tau_m,\,y_m)},
\end{equation}
vaguely on $M_p$, as $n\to\infty$.
Extending, if necessary, the probability space on which $(\widehat{N}_n)_{n\in\N}$ and $\widehat{N}^{(c,1)}$ are defined,  we can independently construct GW processes $(\widehat{X}_{i,k})_{i\in \N, k\in\N_0}$ having the same law as $({X}_{i,k})_{i\in \N, k\in\N_0}$. Write
$$
\widehat{Z}_n(t):=\sum_{k=0}^{[nt]} \sum_{i=1}^{\exp(n y_k^{(n)})} \widehat{X}_{i,k}([nt]-k),\quad n\in\mathbb{N}_0,\quad  t\geq 0,
$$
so that for each $n\in\N$, the distributions of the processes $(\widehat Z_n(t))_{t\geq 0}$ and $(Y_{[nt]})_{t\geq 0}$ coincide. Fix some $T>0$ and let $d_T$ be the standard $J_1$-metric on the Skorokhod space $D[0,T]$. Our aim is to prove that with probability $1$,
\begin{equation}\label{eq:need}
\lim_{n\to\infty} d_T\left(\frac 1n \log^+ \widehat{Z}_n(\cdot),  \underset{\tau_k\leq
\cdot}{\sup}\big(y_k + (\cdot-\tau_k)\log\mu\big)\right) =0.
\end{equation}

\vspace*{2mm}
\noindent
\textsc{Step 2: Estimate for non-extremal order statistics.}
We shall decompose the process $(\widehat{Z}_n(t))_{t\geq 0}$ into the contribution coming from times with extremely active immigration, and the contribution of all the other times.
For a truncation parameter $0< \gamma < 1$, put
\begin{align*}
\widehat Z_{n}^{(\leq \gamma)}(t)&:=\sum_{k=0}^{[nt]}\1_{\{y_k^{(n)}\leq \gamma\}}\sum_{i=1}^{\exp(ny_k^{(n)})}\widehat X_{i,k}([nt]-k),\\
\widehat Z_{n}^{(>\gamma)}(t) &:=\sum_{k=0}^{[nt]}\1_{\{y_k^{(n)}>\gamma\}}\sum_{i=1}^{\exp(ny_k^{(n)})}\widehat X_{i,k}([nt]-k),
\end{align*}
so that
\begin{equation}\label{equ}
\widehat Z_{n}(t) = \widehat Z_{n}^{(\leq \gamma)}(t) + \widehat Z_{n}^{(>\gamma)}(t),\quad n\in\N_0, \; t\geq 0.
\end{equation}
Suppose for a moment that with probability $1$,
\begin{equation}\label{eq:need1}
\underset{\gamma\to 0+}{\lim}\,\underset{n\to\infty}{\lim\sup}\,d_T\bigg(\frac 1n\log^+ \widehat Z_n^{(>\gamma)}(\cdot),\,\underset{\tau_k\leq\cdot}{\sup}\,\big(y_k + (\cdot-\tau_k)\log\mu\big)\bigg)=0,
\end{equation}
where $\gamma$ is restricted to the set $\{1/m\colon m=2,3,\ldots\}$. Let us argue that~\eqref{eq:need1} implies~\eqref{eq:need}.

\vspace*{2mm}
\noindent
\textit{Proof in the case $\mu\leq 1$.} Using \eqref{equ} in combination with Lemma~\ref{lem:log_+_subadd} yields
\begin{equation*}
\log^+(\widehat Z_{n}^{(>\gamma)}(t))
\leq
\log^+(\widehat Z_{n}(t))
\leq
\log^+(\widehat Z_{n}^{(>\gamma)}(t)) + \log^+(\widehat Z_{n}^{(\leq \gamma)}(t))+2\log 2.
\end{equation*}
Since the processes $(\widehat Z_n^{(\leq \gamma)}(t))_{t\geq 0}$ and $(Y_{[nt]}^{(\leq \gamma)})_{t\geq 0}$ (c.f.~\eqref{eq:def_Y_proc_truncated}) have the same distribution, we can use Lemma~\ref{aux3} and the Borel-Cantelli lemma to conclude that with probability $1$,
$$
\underset{n\to\infty}{\lim\sup}\,\underset{0\leq t\leq T}{\sup}\, \frac 1n \log^+(\widehat Z_n^{(\leq \gamma)}(t)) \leq \gamma.
$$
It follows from the last two inequalities that with probability $1$,
$$
 \underset{n\to\infty}{\lim\sup}\,\underset{0\leq t\leq T}{\sup}\, \left|\frac 1n \log^+(\widehat Z_{n}(t)) - \frac 1n \log^+(\widehat Z_{n}^{(>\gamma)}(t))\right| \leq \gamma.
$$
The Skorokhod distance is majorized by the sup-distance, whence we conclude that with probability $1$,
$$
\underset{n\to\infty}{\lim\sup}\, d_T\left(\frac 1n \log^+ \widehat Z_{n}(\cdot), \frac 1n \log^+ \widehat Z_{n}^{(>\gamma)}(\cdot)\right) \leq \gamma.
$$
The triangle inequality entails  that~\eqref{eq:need1} implies~\eqref{eq:need}.

\vspace*{2mm}
\noindent
\textit{Proof in the case $\mu > 1$.}
Our aim is to obtain an upper bound for $\log^+(\widehat Z_{n}(t))$. Since the processes $(\widehat Z_n^{(\leq \gamma)}(t))_{t\geq 0}$ and $(Y_{[nt]}^{(\leq \gamma)})_{t\geq 0}$ have the same distribution,  Lemma~\ref{aux3} in conjunction with the Borel-Cantelli lemma allows us to conclude that with probability $1$,
$$
\underset{n\to\infty}{\lim\sup}\, \underset{0\leq t\leq T}{\sup}\, \frac 1n \log^+ \left(\mu^{-[nt]} \widehat  Z_n^{(\leq \gamma)}(t)\right) \leq \gamma.
$$
Since $\log x\leq \log^+ x$ and $[nt]\leq nt$, it follows that with probability $1$, for sufficiently large $n$,
\begin{equation}\label{eq:est1}
\widehat  Z_n^{(\leq \gamma)}(t) \leq e^{n(t\log \mu+ 2\gamma)}, \quad 0\leq t \leq T.
\end{equation}
Noting that $e^{n(t\log \mu+ 2\gamma)}\geq e^{2n\gamma}>1$ for sufficiently large $n$ and all $t\geq 0$, we obtain the estimate
\begin{align*}
\log^+(\widehat Z_{n}(t))
&\leq
\log^+\left(\widehat Z_{n}^{(>\gamma)}(t)  + e^{n(t\log \mu + 2\gamma)}\right)\\
&\leq
\log\left(\widehat Z_{n}^{(>\gamma)}(t)  + e^{n(t\log \mu + 2\gamma)}\right)\\
&\leq
n(t\log \mu - 2\sqrt \gamma) + \log\left( e^{-n(t\log \mu - 2\sqrt \gamma)} \widehat Z_{n}^{(>\gamma)}(t) + e^{4n\sqrt \gamma}\right).
\end{align*}
Using again the inequality $\log x\leq \log^+ x$ and then Lemma~\ref{lem:log_+_subadd}, we arrive at
\begin{equation}\label{eq:222}
\log^+(\widehat Z_{n}(t))
\leq
n(t\log \mu - 2\sqrt \gamma)  + \log^+ \left(e^{-n(t\log \mu - 2\sqrt \gamma)} \widehat Z_{n}^{(>\gamma)}(t)\right) + 4n\sqrt \gamma +2\log 2.
\end{equation}
Below we shall prove that with probability $1$ there exist a random $0< \gamma_0 < 1$ and $n_1\in\N$ such that for all $0<\gamma<\gamma_0$ and $n>n_1$, we have
\begin{equation}\label{eq:333}
\widehat Z_{n}^{(>\gamma)}(t) \geq  e^{n(t\log \mu - 2\sqrt \gamma)}, \quad \sqrt \gamma\leq t\leq  T.
\end{equation}
Given~\eqref{eq:333}, we conclude that $\log^+$ on the right-hand side of~\eqref{eq:222} can be replaced by $\log$, thus yielding for all $\sqrt \gamma\leq t\leq  T$ the estimate
\begin{equation*}
\log^+(\widehat Z_{n}^{(>\gamma)}(t)) \leq \log^+(\widehat Z_{n}(t))
\leq
4n\sqrt \gamma + \log^+ (\widehat Z_{n}^{(>\gamma)}(t)) + 2\log 2,
\end{equation*}
where the first inequality is an immediate consequence of~\eqref{equ}. For $0\leq t \leq \sqrt \gamma$, Lemma~\ref{lem:log_+_subadd} and~\eqref{eq:est1} yield
$$
\log^+(\widehat Z_{n}^{(>\gamma)}(t)) \leq \log^+(\widehat Z_{n}(t))
\leq
n(\sqrt{\gamma} \log \mu + 2\gamma) + \log^+ (\widehat Z_{n}^{(>\gamma)}(t)) + 2\log 2.$$
From now on, we can argue as in the case $\mu\leq 1$ to conclude that~\eqref{eq:need1} implies~\eqref{eq:need}

\vspace*{2mm}
\noindent
\textit{Proof of~\eqref{eq:333}}.  First we prove that with probability $1$ there is a $\gamma_0>0$ such that for  all  $0<\gamma<\gamma_0$ at least one atom $(\tau, y)$ of the point process $\widehat N^{(c,1)}$ satisfies $0<\tau <\sqrt \gamma$ and $y>\gamma$. Indeed, the number of points of $\widehat N^{(c,1)}$ in the set $(0,1/\sqrt{2k})\times (1/k,\infty)$ is Poisson-distributed with parameter $c\sqrt{k/2}$, for all $k\in\N$. Since $\sum_{k\geq 1} e^{-c\sqrt{k/2}}$ is finite,  the Borel-Cantelli lemma implies that
with probability $1$ we can find $k_0\in\N$ such that for every $k\geq k_0$ at least one atom $(\tau, y)$ of $\widehat N^{(c,1)}$ satisfies $0<\tau < 1/\sqrt{2k}$ and $y>1/k$. If $1/(k+1)\leq \gamma \leq  1/k$, then it follows $0<\tau <\sqrt \gamma$ and $y>\gamma$, so that we can take $\gamma_0 = 1/k_0$.

Since with probability $1$, $\widehat N_n$ converges to $\widehat N^{(c,1)}$ vaguely on $M_p$, there is a random $n_0\in\N$ such that for all $n\geq n_0$ at least one atom, say $(k_n/n, z_n)$, of $\widehat N_n$ satisfies $k_n/n<\sqrt \gamma$ and $z_n>\gamma$. In the following, we condition on the $\sigma$-field generated by $\widehat N^{(c,1)}$ and $\widehat N_n$, so that we can view $\gamma_0$ and $n_0$ as deterministic quantities.

Recall that we consider the case $\mu>1$. As has already been mentioned in the proof of Lemma \ref{aux2} (case $\mu>1$),
there exists a function $L$ slowly varying at $\infty$ such that
$$
\frac{X_{1,0} (m)}{\mu^m L(\mu^m)} \to W \quad \text{a.s. as } m\to\infty,
$$
the limit random variable $W$ being a.s.\ positive on the survival event of the GW process $X_{1,0}$. It follows that there is $\varepsilon>0$ such that
$$
\Prob\left\{\inf_{m\in\N}\frac{\widehat X_{i,k} (m)}{\mu^m L(\mu^m)} > \varepsilon \right\} >\varepsilon, \quad i\in \N,~ k\in\N_0.
$$
Since $z_n>\gamma$ for $n\geq n_0$ we have
$$
\sum_{n\geq n_0} \Prob\left\{\inf_{m\in \N}\frac{\widehat X_{i,k_n} (m)}{\mu^m L(\mu^m)} \leq \varepsilon \text{ for all } i=1,\ldots, e^{nz_n}\right\}
<
\sum_{n\geq n_0} (1-\varepsilon)^{e^{nz_n}} < \infty.
$$
By the Borel-Cantelli lemma, for sufficiently large $n$, there is an immigrant $i_n$ arriving at time $k_n< n\sqrt \gamma$ whose offspring numbers satisfy $\widehat X_{i_n,k_n}(m)>\varepsilon \mu^m L(\mu^m)$ for all $m\in\N$. For all $\sqrt \gamma \leq t \leq T$ and sufficiently large $n\geq n_1$ we have
$$
\widehat Z_n^{(>\gamma)} (t) \geq \widehat X_{i_n,k_n}([nt] - k_n) >
\varepsilon \mu^{[nt] - k_n} L(\mu^{[nt] - k_n})
>
e^{n(t\log \mu - 2 \sqrt \gamma)},
$$
thereby completing the proof of~\eqref{eq:333}.

\vspace*{2mm}
\noindent
\textsc{Step 3: Enumerating the points.} In the following we prove~\eqref{eq:need1}.
Let $\mathcal{F}$ be the $\sigma$-field generated by $(y_k^{(n)})_{k\in\N_0, n\in\N}$ and $(\tau_k,\,
y_k)_{k\in\N_0}$. Until further notice we work conditionally on $\mathcal{F}$, so that all $\mathcal{F}$-measurable variables can be treated as deterministic constants. After discarding an event of probability $0$, we can assume that the points $(\tau_k,
y_k)$, $k\in\N_0$, have the following properties:

\vspace*{1mm}
\noindent {\sc Property 1}: $\tau_k \notin \{0, T\}$ and $y_k\notin \{1/2,1/3,\ldots,\infty\}$ for all $k\in\N_0$.

\vspace*{1mm}
\noindent {\sc Property 2}: $\sup_{\tau_k\leq t}\big(y_k+(t-\tau_k)\log\mu\big)\geq 0$ for all $t\geq 0$.

\vspace*{1mm}
\noindent {\sc Property 3}: $\tau_k\neq\tau_j$ a.s.\ for $k\neq j$.

\vspace*{1mm}
\noindent {\sc Property 4}: $(0,0)$ is an accumulation point of $(\tau_k,
y_k)$, $k\in\N_0$.

Relation \eqref{7} implies that  for large
enough $n$ and some $p\in \N$,
$$
\widehat{N}_n ([0,T]\times
(\gamma,\infty])=\widehat{N}^{(c,1)}([0,T]\times (\gamma,\infty])=p,
$$
where $p\neq 0$ if $\gamma$ is sufficiently small. Denote by
$(\bar{\tau}_j,\bar{y}_j)_{1\leq j\leq p}$ an enumeration of the
points of $\widehat{N}^{(c,1)}$ in $[0,T]\times (\gamma,\infty]$ with
$0<\bar{\tau}_1<\bar{\tau}_2<\ldots< \bar{\tau}_p<T$
and by $(\bar{\tau}_j^{(n)}, \bar{y}_j^{(n)})_{1\leq j\leq p}$
the analogous enumeration of the points of $\widehat{N}_n$ in $[0,T]\times (\gamma,\infty]$. Then, relation~\eqref{7} implies that
(possibly, after renumbering the points),
\begin{equation}\label{2.2}
\lim_{n\to\infty} \bar{\tau}^{(n)}_j = \bar{\tau}_j,
\quad
\lim_{n\to\infty} \bar{y}^{(n)}_j =\bar{y}_j,
\quad
j=1,\ldots,p.
\end{equation}
For sufficiently large $n\in\N$, $j=1,\ldots, p$, and $t\geq 0$, set
$$
U_{n,j}(t):=\sum_{i=1}^{\exp(n\bar{y}_j^{(n)})}\widehat X_{i,\,n\bar{\tau}_j^{(n)}}([nt]),
$$
so that $\widehat Z_n^{(>\gamma)}(t)=\sum_{\bar{\tau_j}^{(n)}\leq t}U_{n,j}(t-\bar{\tau}_j^{(n)})$. Put also
$$
Z_{n,j}(t):=n^{-1}\log^+U_{n,j}(t),
\quad
Z_j(t):=(\bar{y}_j+t \log\mu )^+.
$$
For later needs, we also define these functions to be zero for $t<0$.
We rewrite~\eqref{eq:need1} in the following form:
\begin{equation}\label{princ10}
\underset{\gamma\to 0+}{\lim}\,\underset{n\to\infty}{\lim\sup}\,d_T\bigg(\frac 1n \log^+\bigg(\sum_{\bar{\tau_j}^{(n)}\leq \cdot}U_{n,j}(\cdot-\bar{\tau}_j^{(n)})\bigg),\,\underset{\tau_k\leq\cdot}{\sup}\,\big(y_k+(\cdot-\tau_k)\log\mu \big)\bigg)=0.
\end{equation}

\vspace*{2mm}
\noindent
\textsc{Step 4: Proof of~\eqref{princ10}}.
By the triangle inequality, we have
\begin{multline}\label{tech1}
d_T\bigg(\frac 1n \log^+\bigg(\sum_{\bar{\tau}_j^{(n)}\leq \cdot}U_{n,j}(\cdot-\bar{\tau}_j^{(n)})\bigg),\underset{\tau_k\leq\cdot}{\sup}\,\big(y_k+(\cdot-\tau_k)\log\mu\big)\bigg) %
\\
\begin{aligned}
&\leq d_T\bigg(\frac 1n \log^+\bigg(\sum_{\bar{\tau}_j^{(n)}\leq \cdot}U_{n,j}(\cdot-\bar{\tau}_j^{(n)})\bigg), \underset{\bar{\tau}_j\leq \cdot}{\sup}\,Z_j(\cdot-\bar{\tau}_j)\bigg)
\\&+
\underset{0\leq t\leq T}{\sup}\,\big|\underset{\tau_k\leq t}{\sup}\,\big(y_k+(t-\tau_k)\log\mu\big)-\underset{\bar{\tau}_j\leq t}{\sup}\,Z_j(t-\bar{\tau}_j)\big|,
\end{aligned}
\end{multline}
where for the last term we have used the fact that the Skorokhod metric $d_T$ is dominated by the uniform metric on $[0,T]$. In the following, we estimate both terms on the right-hand side.

\vspace*{2mm}
\noindent
\textit{First term in~\eqref{tech1}.}
We intend to check that
\begin{equation}\label{inter}
\underset{n\to\infty}{\lim} d_T\bigg(\frac 1n\log^+\bigg(\sum_{\bar{\tau}_j^{(n)}\leq \cdot}U_{n,j}(\cdot-\bar{\tau}_j^{(n)})\bigg),\, \underset{\bar{\tau}_j\leq \cdot}{\sup}\,Z_j(\cdot-\bar{\tau}_j)\bigg)=0 \quad \text{a.s.}
\end{equation}
In view of
\begin{eqnarray*}
\underset{\bar{\tau}_j^{(n)}\leq t}{\sup}\,\log^+\big(U_{n,j}(t-\bar{\tau}_j^{(n)})\big)&\leq& \log^+\bigg(\sum_{\bar{\tau}_j^{(n)}\leq t}U_{n,j}(t-\bar{\tau}_j^{(n)})\bigg)\\&\leq& \log^+\#\{j: \bar{\tau}_j^{(n)}\leq t\}+\underset{\bar{\tau}_j^{(n)}\leq t}{\sup}\,\log^+\big(U_{n,j}(t-\bar{\tau}_j^{(n)})\big)\\&\leq& \log p+\underset{\bar{\tau}_j^{(n)}\leq t}{\sup}\,\log^+\big(U_{n,j}(t-\bar{\tau}_j^{(n)})\big) \end{eqnarray*}
for $t\in [0,T]$, Equation~\eqref{inter} is equivalent to
\begin{equation}\label{inter2}
\underset{n\to\infty}{\lim} d_T\bigg(\underset{\bar{\tau}_j^{(n)}\leq\cdot}{\sup}\,Z_{n,j}(\cdot-\bar{\tau}_j^{(n)}),\, \underset{\bar{\tau}_j\leq \cdot}{\sup}\,Z_j(\cdot-\bar{\tau}_j)\bigg)=0 \quad \text{a.s.}
\end{equation}
In view of \eqref{2.2}, an application of Lemma \ref{aux2} yields
\begin{equation}\label{imp2}
Z_{n,j}(t)\quad \to\quad Z_j(t),\quad j=1,\ldots, p
\end{equation}
a.s.\ uniformly on $[0,T]$ (recall that we work conditionally on $\mathcal{F}$).
Define $\lambda_n$ to be continuous and strictly increasing
functions on $[0,T]$ with $\lambda_n(0) =0$, $\lambda_n(T) =T$,
$\lambda_n(\bar{\tau}_j)=\bar{\tau}^{(n)}_j$ for
$j=1,\ldots,p$, and let $\lambda_n$ be linearly interpolated
elsewhere on $[0,T]$. It is easily seen that $\underset{n\to\infty}{\lim} \underset{0\leq t\leq T}{\sup}\,|\lambda_n(t)-t|=0$.
This implies that
\begin{equation}\label{imp1}
\underset{n\to\infty}{\lim} \underset{t\in[0,T]}{\sup}\,\big|Z_{n,j}\big(\lambda_n(t)-\bar{\tau}_j^{(n)}\big)-Z_j(t-\bar{\tau}_j)\big|=0,\quad j=1,\ldots,p
\end{equation}
a.s. Indeed, for $t\in [0,\bar{\tau}_j)$ we have $Z_{n,j}(\lambda_n(t)-\bar{\tau}_j^{(n)})=Z_j(t-\bar{\tau}_j)=0$.
Also, as a consequence of \eqref{2.2} and \eqref{imp2} we obtain the relation
$$
\underset{n\to\infty}{\lim}\underset{t\in [0,T-\bar{\tau}_j]}{\sup}\,|Z_{n,j}(\lambda_n(t+\bar{\tau}_j)-\bar{\tau}_j^{(n)})-Z_j(t)|=0,
$$
which proves \eqref{imp1}.
Now \eqref{inter2} follows from
\begin{eqnarray*}
&&\underset{t\in[0,T]}{\sup}\,\big|\underset{\bar{\tau}_j^{(n)}\leq \lambda_n(t)}{\sup}\,Z_{n,j}\big(\lambda_n(t)-\bar{\tau}_j^{(n)}\big)-\underset{\bar{\tau}_j\leq t}{\sup}\,Z_j(t-\bar{\tau}_j)\big|\\&=&
\underset{t\in[0,T]}{\sup}\,\big|\underset{\bar{\tau}_j\leq t}{\sup}\,Z_{n,j}\big(\lambda_n(t)-\bar{\tau}_j^{(n)}\big)-\underset{\bar{\tau}_j\leq t}{\sup}\,Z_j(t-\bar{\tau}_j)\big|\\&\leq&
\underset{t\in[0,T]}{\sup}\,\sum_{\bar{\tau}_j\leq t}\big|Z_{n,j}\big(\lambda_n(t)-\bar{\tau}_j^{(n)}\big)-Z_j(t-\bar{\tau}_j)\big|\\&\leq& \sum_{j=1}^p \underset{t\in[0,T]}{\sup}\,\big|Z_{n,j}\big(\lambda_n(t)-\bar{\tau}_j^{(n)}\big)-Z_j(t-\bar{\tau}_j)\big|
\end{eqnarray*}
because the right-hand side converges to zero a.s.\ by \eqref{imp1}.

\vspace*{2mm}
\noindent
\textit{Second term in~\eqref{tech1}.}
Left with proving that
\begin{equation}\label{imp3}
\underset{\gamma\to 0+}{\lim}\,\underset{0\leq t\leq T}{\sup}\,\big|\underset{\tau_k\leq t}{\sup}\,\big(y_k+(t-\tau_k)\log\mu\big)-\underset{\bar{\tau}_j\leq t}{\sup}\,\big(\bar{y}_j+(t-\bar{\tau}_j)\log\mu\big)^+\big|=0,
\end{equation}
we first recall that the points $(\bar{\tau}_1, \bar{y_1}), \ldots, (\bar{\tau}_p, \bar{y}_p)$ belong to the collection $(\tau_k,y_k)_{k\in\mathbb{N}_0}$. Hence, for $t\in [0,T]$,
\begin{eqnarray}\label{3}
\sup_{\tau_k\leq t}\big(y_k+(t-\tau_k)\log\mu\big)
\geq
\sup_{\bar{\tau}_j\leq t}\big(\bar{y}_j+(t-\bar{\tau}_j)\log\mu\big)^+,
\end{eqnarray}
where we also used that the supremum on the left-hand side is nonnegative by Property~2.
Pick now
$\tau_k\notin\{\bar{\tau}_1,\ldots, \bar{\tau}_p\}$ satisfying $\tau_k\leq t$.
Recall that all $y_k$ other than
$\bar{y}_1,\ldots, \bar{y}_p$ do not exceed $\gamma$.
If $\mu\leq 1$, we infer
\begin{equation*}
y_k+(t-\tau_k)\log\mu
\leq y_k
\leq \gamma
\leq \gamma + \sup_{\bar{\tau}_j\leq t}(\bar{y}_j+(t-\bar{\tau}_j)\log\mu)^+.
\end{equation*}
Together with \eqref{3} this proves~\eqref{imp3}.  In the following, let $\mu>1$. Fix some $\delta>0$. It suffices to show that for sufficiently small $\gamma>0$ we have
\begin{equation}\label{3_suff}
y_k+(t-\tau_k)\log\mu \leq \sup_{\bar{\tau}_j\leq t}\big(\bar{y}_j+(t-\bar{\tau}_j)\log\mu\big)^+ + \gamma+\delta
\end{equation}
for all $k\in\N_0$ such that $\tau_k \leq t$. If $y_k > \gamma$, then  $(\tau_k,y_k)$ is one of the points $(\bar{\tau}_1, \bar{y}_1),\ldots, (\bar{\tau}_p, \bar{y}_p)$, and~\eqref{3_suff} is evident. Let therefore $y_k\leq \gamma$. Then,
$$
y_k + (t-\tau_k) \log \mu \leq \gamma + t\log \mu.
$$
This immediately implies~\eqref{3_suff} if $t\leq \delta/\log \mu$. Therefore, let $t>\delta/\log \mu$.  If $\gamma>0$ is sufficiently small, then by Property~4 we can find a point $(\bar{\tau}_{j'}, \bar{y}_{j'})$ such that $\bar{\tau}_{j'} < \delta/\log \mu$. We infer
$$
\sup_{\bar{\tau}_j\leq t}\big(\bar{y}_j+(t-\bar{\tau}_j)\log\mu\big)^+
\geq
\bar{y}_{j'}+(t-\bar{\tau}_{j'})\log\mu
\geq t\log \mu -\delta.
$$
Taking the last two inequalities  together we arrive at~\eqref{3_suff}, which proves \eqref{3}. The proof of Theorem \ref{main1} is complete.
\end{proof}

The proof of Theorem \ref{main2} runs the same path as that of Theorem \ref{main1}. We note that the proof is essentially based on the convergence
\begin{equation*}
\sum_{k\geq 0}\1_{\{J_{k}\neq 0\}}\varepsilon_{\left(\frac kn,\,
\frac {\log J_{k}}{b_n}\right)}\quad \Rightarrow\quad
N^{(1,\alpha)},\quad n\to\infty
\end{equation*}
on $M_p$ and uses the corresponding part of Lemma \ref{aux3} together with Lemma \ref{aux2a} in which we take $c_n=b_n$. We refrain from discussing the details which are much simpler here.

\vspace{1cm}
\noindent   {\bf Acknowledgements}  \quad
\footnotesize
A part of this work was done while A.~Iksanov was visiting M\"{u}nster in
July 2015, 2016. He gratefully acknowledges hospitality and the financial support by DFG SFB 878 ``Geometry, Groups and Actions''.

\normalsize

redpajama_set_name: RedPajamaArXiv


Given your unique space, consider organizing your room to best fit the needs and traffic patterns of your children.
Ask yourself these questions when designing your space.
What is the purpose of each space in your room?
Where will quiet places in my room be? Where will louder activities happen?
Consider putting loud and quiet activity spaces in opposite parts of your room.
Where will children do projects that make a wet mess? Where are items that need to stay dry located?
Where will the children walk and move around the room?
How can I make cozy corners and big open areas for different group sizes?
Consider hanging fabric from the ceiling to define spaces.
What safety precautions do I need to consider?


100% Chardonnay. Enticing aromas of bright citrus and melon are layered with spice, vanilla and a touch of butter. Vibrant flavors of mango and guava meld with vanilla and soft spice on the creamy, rich palate. The finish is juicy and satisfying with refreshing acidity that makes you crave another sip. Our favorite Joy of Cooking® pairings for this wine include Tuna-Avocado Burgers, Lobster Rolls and Broiled Salmon with Snail Butter and Orange-Hazelnut Asparagus.
100% Chardonnay. Enticing aromas of vanilla, melon, papaya and mango are layered with baked apples, creme brulee, butter and a slight minerality. Those scents are echoed as flavors on the creamy, rich palate, along with peach, pie crust and baking spices. The long, smooth finish is marked by mouthwatering acidity. Our favorite Joy of Cooking® pairings* for this wine include Tuna-Avocado Burgers, Chicken Satay with Green Papaya Salad; Baked Stuffed Peaches.
95% Zinfandel, 5% Petit Sirah. Our classic California Zinfandel is dark purple-ruby, with a bright crimson hue. On the nose, concentrated aromas of blackberry and raspberry meld with clove and spice. Lush flavors of raspberry, blackberry jam, plums and vanilla flood the palate, followed by touches of white pepper in the background. Plush and round tannins and great length make for a fresh and enticing finish. Our favorite Joy of Cooking pairings* for this wine include Spring Lamb Stew; Chocolate bark clusters; Buttermilk Layer cake with Bittersweet Chocolate frosting and coconut.
88% Merlot, 12% Cabernet Sauvignon. Aromas of fresh, ripe plums, red currant and violets intertwine with subtle notes of cedar and bramble bush. This medium-bodied wine is ripe and plush with flavors of plums, cherries and figs at the forefront, leading to juicy red fruit on the midpalate. Delicious now, this wine's depth and structure provide for a long cellaring potential. Our favorite pairings for this wine include Seared porkchops with Mustard Green Chimichurri; Winter Squash Mac & Cheese; Beef Braciole.
87.5% Cabernet Sauvignon, 7% Petit Verdot, 3.5% Cabernet Franc, 2% Merlot. Vibrant purple-black hue with expressive aromas of fresh cherry pie, blackberry, currant and hints of spice. The wine is medium- to full-bodied with a bright, fresh palate brimming with black cherry, fresh baked pie crust, sweet vanilla and currant. Supple tannins carry through to a lingering finish with layers of delicious fruit.. Our favorite Joy of Cooking® pairings for this wine include sauteed mushrooms with garlic & sherry; Braised short ribs with mushrooms, Red Wine and Crispy Fried shallots; Braised lamb shanks.


=====Arthrodesis===== Also known as artificial ankylosis or syndesis, is the artificial induction of joint ossification between two bones via surgery. This is done to relieve intractable pain in a joint which cannot be managed by pain medication, splints, or other normally-indicated treatments. The typical causes of such pain are fractures which disrupt the joint, and arthritis. It is most commonly performed on joints in the spine, hand, ankle, and foot. Historically, knee and hip arthrodeses were also performed as pain relieving procedures, however with the great successes achieved in hip and knee arthroplasty, arthrodesis of these large joints has fallen out of favour as a primary procedure, and now are only used as procedures of last-resort in some failed arthroplasties. see [[Spine arthrodesis]]. ====Indications==== Instability Examples pediatric [[occipitocervical instability]].


text:
Q: How to create cells as dropdown list with more than 700 records using EpPlus package in .Net Core In my .Net Core 5 API project there is a requirement to have particular column cells as dropdown list i.e., when the user clicks on any cell of that particular column (which will vary in different excels) it should show a dropdown list with options (the list can have more than 700 options), hence we have to also make it searchable. Our project is little old & its using EPPlus package (version 5+) to generate the excel sheet.
How to implement this ? Thanks in advance!!!

redpajama_set_name: RedPajamaStackExchange


text:
using System.ComponentModel.DataAnnotations;

namespace OdeToFood.Entities
{
    public enum CuisineType
    {
        None,
        Italian,
        Franch,
        Japanese,
        American
    }

    public class Restaurant
    {
        public int Id { get; set; }

        [Required, MaxLength(80)]
        [Display(Name = "Restaurant name")]
        public string Name { get; set; }

        [Display(Name = ("Type of cuisine"))]
        public CuisineType Cuisine { get; set; }
    }
}

redpajama_set_name: RedPajamaGithub


text:
Review: e.l.f. Essential 14-Piece Nail Cube - Jewel Tones
For a nail polish collector, the e.l.f. Essential 14-Piece Nail Cube is like being in an ice cream parlor and to be so lucky to be allowed to taste a spoonful of many different flavors of ice cream. I still haven't experienced tasting ice cream with the iconic pink spoon of Baskin Robbins yet, but it should be that feeling. I have never bought an e.l.f. nail polish before and it is great that I get to try nail polishes with different kinds of finishes, glitters, and colors.
There are two e.l.f. Nail Cubes: the Brights and the Jewel Tones. On the e.l.f. Netherlands website, only the Jewel Tones Nail Cube (14 euro) is available at the moment, but on www.eyeslipsface.com, you can find both Nail Cubes ($14) . The e.l.f. Bright Nail Cube has two glitter top coats in common with the Jewel Tones Nail Cube, namely the Gold Star and Chic Confetti nail polishes, but besides that, it is the total opposite. The Brights Nail Cube has many bright and colorful nail polishes, whereas the Jewel Tones Nail Cube has neutrals and darker nail polishes with shimmers in them.
The e.l.f. Nail Cube - Jewel Tones*
Pictures from the e.l.f. website
The nail polish bottles
The e.l.f. Nail Cube contains 14 e.l.f nail polishes in a smaller bottle than the regular e.l.f. nail polishes. These only contain 6 ml. (0.20 fl oz.) of nail polish instead of 10 ml. (0.34 fl. oz.).
Neutrals and a very Christmas like red and green.
Shimmery nail polishes.
The glitter top coats.
Ingredients of the e.l.f. Essential 14-Piece Nail Cube nail polishes
Nail swatches on a nail polish display wheel
All of these nail polishes were opaque with just one layer, except for Cherry Bomb, which still stays a little bit sheer after two layers.
Because I like the shimmery nail polishes a lot, I have two more pictures of those with slightly different angles and lighting.
These two glitter top coats look very similar to two other nail polishes that I own: the Peripera nail polish YE905 has the same muted gold glitters as the Gold Star and the Etude House nail polish PWH901 also has multicolor glitters like the Chic Confetti.
The two glitter top coats
I was quite interested in the Sea Escape and Violet Velvet nail polishes, but these appear to be a case of looking gorgeous in the bottle, but disappointing on the nails. The Violet Velvet looks like a satiny purple in the bottle, but on the nail, it looks like a deep, dark purple and you really need to get the light directed in the right way for the shimmers to show. I had the same thing with the Sea Escape nail polish. In the bottle, it looked like a dark blue with beautiful light blue shimmers, but on the nail, it almost looks pitch black and you really need to search for the shimmers.
Nail polish swatches of the Blushing Beauty and the Dazzle Diva
Of all the nail polishes, I like the Blushing Beauty and the Dazzle Diva nail polishes the most. That's why, I have taken these two out to wear on my nails first. During application of the nail polish, the nail polish smell can be a bit strong, but it is doable in a well-aired room. The two glitter top coats have the strongest smell. After drying, the nail polishes look glossy and feel smooth. The smell coming from the nail polishes after drying is similar to the smell of a whole bunch of Crocs shoes (imagine: standing in a narrow storage room filled with plastic shoes). This goes away after about a day, so it is not a big problem. I also didn't experience any chipping for a couple of days, since I used a base and a top coat.
Blushing Beauty nail polish
Dazzle Diva nail polish
Labels: e.l.f. nail polish nail polish review Nail products nail swatch received

redpajama_set_name: RedPajamaCommonCrawl


text:
Q: Printing Two Dimensional Associative Array in Bash Script I have created associative array but when I try to print it in for loop, I don't get all the values. I tried to look up for an sample but there's no example which prints the two dimensional associative array. Here's my script:
#!/bin/bash
declare -A links
links["A",0]="linkA1"
links["A",1]="linkA2"
links["B",0]="linkB1"
links["B",1]="linkB2"
links["C",0]="linkC1"
links["C",1]="linkC2"
links["C",2]="linkC2"

#echo "${links["A",0]}"
#echo "${links["A",1]}"
#echo "${links["B",0]}"

for i in ${#links[@]} ; do
  echo "key: $i"
  j=0
  while [ "${j}" -le 2 ]; do
    echo "value: ${links["${i}","${j}"]}"
    j=$(( $j + 1 ))
  done
done


text:
Politics / Ademola Adeleke / Dotun Babayemi / Osun state News
Supreme Court Affirms Ademola Adeleke As PDP Governorship Candidate
by Prince Admin · Published September 30, 2022 · Updated November 3, 2022
Osun State governor-elect, Ademola Adeleke
The Nigerian Supreme Court dismissed an appeal filed by Prince Oyedotun Babayemi seeking to be declared the Peoples Democratic Party (PDP) candidate for the now-rested July 16 Governorship election in Osun State on Thursday.
In Suit No: SC/CV/914/2022: (Prince Oyedotun Babayemi vs. PDP, INEC, and Ademola Adeleke), the Supreme Court upheld Senator Ademola Adeleke as the validly elected PDP candidate for the said election.
Babayemi filed his Notice of Appeal on time, but the second one was filed on Day 15, in violation of Section 285. He dropped the first and built his Brief around the second 'incompetent' one. The Justices observed that the Notice of Appeal was ineffective and dismissed it without cost.
The panel led by Justice Amina Augie ruled that the case was statute-barred having been filed outside the 14 days Constitutional period set out for Appeal.
Prince Babayemi and Senator Adeleke had locked horns in a legal battle over who was validly elected after parallel primaries were conducted in different locations of the State Capital, Osogbo.
On March 8, Babayemi was declared winner of the primary election held at the Women and Children Development Initiative Foundation (WOCDIF) in Osogbo.
However, on the same day, the Independent National Electoral Commission (INEC) monitored and recognized the NWC-backed Primary held at the Osogbo City Stadium, under the supervision of Deputy Governor of Bayelsa State, Lawrence Ewhrudjakpo, as Chairman with Senator Adeleke announced winner.
In a list of candidates later released by INEC, Adeleke was recognised as duly elected, a decision which set a template for a flurry of litigations.
It was for this reason that Babayemi filed a suit at the Federal High Court to challenge the nomination of Adeleke by the PDP to INEC and the validity of the primaries conducted at the Stadium against the one conducted at WOCDIF.
In his judgement on the matter on May 18, Justice Nathaniel Ayo-Emmanuel of the Federal High Court, sitting in Osogbo, dismissed the suit and validated the governorship primary that produced Adeleke.
The Judge, held that the Court can not dabble into the internal affairs of a political party. Justice Emmanuel said the primary that produced Babayemi did not follow provisions of Sections 82 (1) and (5) of the Electoral Act, 2022.
As a corollary to the above, the learned Judge added that Section 84(5) of the Electoral Act 2022 vests only the NEC/NWC of the Party (not the State Chapter or any faction) with the powers to conduct the primaries of that Party.
Not happy with the ruling, Prince Babayemi approached the penultimate Court in Akure, challenging the judgment of the Federal High Court and seeking to be declared the PDP Governorship Candidate.
The Appeal Court, on July 20, accordingly upheld the decisions of the trial Court and dismissed the suit. Unsurprisingly dissatisfied with the judgement, Babayemi filed an appeal at the apex Court.
Adeleke and the PDP went on to defeat incumbent, Adegboyega Oyetola (APC) in the Guber Election.
Next story Federal High Court sacks Osun state governor, Gboyega Oyetola and his deputy as APC candidates in the July 16 governorship election
Previous story Osun PDP Expels Dotun Babayemi After Supreme Court Ruling
Ned Nwoko Wins PDP Candidate For Delta North Senatorial District
by Prince Admin · Published May 24, 2022 · Last modified November 3, 2022
Soludo will answer to Igbo deities — Ohanaeze Ndi Igbo
I Will Restructure Nigeria, Revamp The Economy – NNPP presidential candidate Kwankwaso


text:
BBC Children in Need funding
Mental Health Resource has been awarded funding by BBC Children in Need to part-fund our Reachout Youth project and help make a difference to the lives of disadvantaged young people in West Kent.
Reachout Youth's café style sessions are aimed at young people aged 13-19, who are experiencing emotional or mental health difficulties. The groups, which take place after school twice a week, bring young people together to reduce stigma, support the development of coping strategies and provide a safe space for the young people to make friends.
BBC Children in Need funding will mean that young people can continue to access a safe space to receive 1-2-1 support to help them work through issues such as the effects of social media, bullying and relationships. Funding will also enable us to establish a new Reachout Youth group so that even more young people can attend our groups.
BBC Children in Need is currently supporting over 2,800 projects in communities across the UK that are helping children and young people facing a range of disadvantages such as living in poverty, being disabled or ill, or experiencing distress, neglect or trauma.
And groups like our Reachout Youth project are needed now more than ever. In Kent, 50% of lifetime mental illness begins by the age of 14. Through providing these facilitated groups, young people will be able to share stories meet other young people who've had similar experiences.
"We could not continue to reach out to the young people in Tunbridge Wells without the support of BBC Children in Need. This grant will go on to change the lives of so many young people and give them the support they need." Rachel Corry, Charity Manager.
Fundraising For Reachout Youth
We are delighted to win BBC Children in Need funding, but we still need to raise £5000. Want to help? You can donate today!
£5 will pay for art materials for one session
£20 will pay for a Support Worker to run one Reachout Youth session.
£35 will cover the room hire for one group for one session.
Set up a regular payment Donate
Author: Jenny Williams
TW Boorman run a pop up Hair Salon at the Hub!
Samantha and Tansy from TW Boorman Funeral Services recently came to the Hub to give free hair cuts to our clients, as part of their pop-up hair salon!
Sandersons UK raise £1975
Sandersons UK have raised an incredible £1975 for Mental Health Resource by completing a gruelling 12km Commando Run, complete with 48 obstacles!
Mental health training workshops
Mental Health Resource delivers Mental Health Awareness training workshops and is accredited to deliver Mental Health First Aid training. If you would like to receive mental health training organisations at your company, society or school, get in touch.
Mental Health First Aid Training
Mental Health Resource is delighted to offer on-site Mental Health First Aid (MHFA) classes, delivered by our approved Instructor who has been accredited by the Royal Society for Public Health, helping you to take a proactive approach to mental health issues.
Sandersons Commando Run
We are fundraising for Mental Health Resource because their small charity makes a big difference to people who live with complex, long-term mental health conditions throughout West Kent.
Mental Health Resource runs 'The Hub', our vibrant wellbeing centre in the heart of Tunbridge Wells, Kent, where people can access wellbeing activities and a safe space to talk for as long as they need.
Example - Support us
Push already-interested visitors over the line with stories, stats and ideas. Make next steps clear
Example - How we're funded
State your case for the support you need, while showing supporters can trust you (be transparent)
When a person is detained under the Mental Health Act it can be distressing. Sometimes patients who are being detained feel unable to ask questions or get information. The support of a trained Independent Mental Health Advocate (IMHA) can help.
Support for my Bipolar
Mental Health Resource ex-Service User and now part time employee, Chris Smith, talks about the positive impact of our charity's work.
Mental Health Resource run a "Serenity" Crisis Cafe for anyone experiencing a mental health crisis to prevent escalation. It is funded by the Kent Police and Crime Commissioner's 'Mental Health & Policing Fund'.
Example - Our history
Tell the relevant, inspiring parts of how you started, where you are today and any major milestones
Mental Health Reachout Youth Young people


text:
Thank you for "40 Years of Fiber"
FORTA Corporation announces the retirement of Daniel Biddle
GROVE CITY, PA, April 30, 2020 /24-7PressRelease/ — FORTA Corporation would like to announce the retirement of Daniel Biddle, Vice President of Sales. Dan has been a staple in not only the concrete industry, but also our hearts for the past 40 years. His impact on the use of fiber reinforcement in the industry is legendary and his presence will be sorely missed.
Dan started his career as the first salesman of synthetic fiber for concrete not only for FORTA, but for the Unites States. In the Fall of 1979, he was hired by Rodger Lindh, FORTA Founder. Selling an unknown product to a huge construction industry had its challenges. Dan kept faith in the products and made it his life's mission to reinforce the future of concrete.
Dan has been on more job sites and testing facilities than we can count. His involvement in the dozens of organizations, associations, and committees over the years introduced fiber reinforcement to standard-setting industry experts. Undoubtedly, his determination in bringing fibers to the forefront has changed the face of concrete for all future generations.
Above all, Dan attributes his personal success to his family, and his professional success to Rodger Lindh, a man he views as the father of synthetic fiber reinforcement. Rodger Lindh, Chairman and Founder of FORTA states,
"Dan's commitment to FORTA, and to his sales team, was always unparalleled. His drive to make FORTA an industry leader was ever present, and he accomplished just that with his passion for the products, his relationships within the industry, and his unwavering loyalty to FORTA through thick and thin. We wouldn't be where we are today without Dan's vision and contributions. I can't express how much he will be missed!"
Dan's dedication is apparent and although his title currently reads Vice President of Sales, he is truly an Ambassador of Fiber to us. Dan will be irreplaceable in our daily lives and to the industry he loves. Had it not been for his dedication and hard work, the fiber reinforcement would not be what it is today.
Since 1978, the products, services, and people of FORTA have played an important role in the design, construction, and preservation of STRONGER LASTING™ concrete around the globe. Based in Grove City, Pennsylvania, FORTA Corporation holds more patents than any other company in the industry and is recognized as a worldwide leader in synthetic fiber reinforcement. Through an ongoing commitment to research and development, FORTA will continue to envision and construct the future of reinforcement fibers.
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Grégory Labrousse: "How is legal open data organized in France?"


After boarding Celebrity Equinox in Rome on Monday afternoon we settled ourselves into our staterooms. The boarding process all went very smoothly and by the time we sailed away at 5.30pm we were sorted. I'm sharing with Scarlett and the boys are 2 doors down. Having never cruised before it was all a bit strange. Everyone else seemed to know what they were doing and we just staggered around, jaws dropping that something so big and beautiful could float. The next day was a full day at sea. The kids spent the morning at the kids club which they loved. The afternoon saw us all by the pool which reminded me of a human soup! That evening was the first formal dinner so they went back to the kids club while we enjoyed date night on the high seas. They are loving cruising so far but they are exhausted. So much to do and see! We arrived at Santorini about 2.00pm the next day. After some consideration and discussion we decided to leave the kids aboard. (Read.....they begged to stay on board). Sounds strange I know but they were adamant they did not want to come and once they have made up their minds they would have been, shall we say, quite vocal with their protests all afternoon. We only had a few hours there so with everyone happy Nick and I set off for a Greek lunch. I had forgotten how many steps there are around Oia but we were silently high fiveing ourselves for leaving the kids on board to have fun. It just would not have been enjoyable for them. For us however.......magic. Santorini is, well, Santorini! Simply amazing. We took the local bus from Fira to Oia for €1.60 each and spent a few hours eating, drinking and just BEING in Santorini. One of my favourite places in the world. Back on board we watched the sun go down with the kids on Celebrity's signature real grass roof, sipping wine and nibbling on cheese. To full for dinner we retired to our rooms where we had our own private room party, ordering room service and watching....what else...but Mamma Mia! Phew. What a day. A day at sea today as we make our way to Istanbul, Turkey.
Where to begin. I think I will let the pictures speak the volumes of our Summer adventure do far. I am currently typing this offline in the middle of the Med aboard Celebrity Equinox. Will post when I can find some WiFi tomorrow when we arrive in Santorini. Forgive me regular Vietnam readers as this is more keeping in touch with family in Aus and hopefully bringing a smile to their faces. Thinking of you constantly.


TUGUEGARAO CITY, Cagayan, Jan. 16 (PIA) -- The fisheries bureau in Cagayan Valley has started the implementation of its Mangrove Plantation Project which is part of a nationwide program to plant around 11 million mangrove seedlings to cover 3,667 hectares mangrove areas in the entire country.
The project aims to restore mangrove forests, mitigate climate change and provide additional source of source of livelihood for the coastal folks.
Max Prudencio, Bureau of Fisheries and Aquatic Resources (BFAR) information officer said that in Region 2, around 45,000 seedlings have been planted particularly in the towns of Sta. Teresita, Buguey and Aparri in Cagayan and Palanan, Isabela. This exceeded the bureau's target of 300,000 mangrove seedlings to be planted in the two provinces.
According to Venchito Villarao, officer-in-charge of the Coastal Resource Management Section and Cagayan Provincial Fishery Action Officer of BFAR RO2, the mangrove rehabilitation project currently being implemented is different from the previous practice of simply planting then leaving the delicate seedling to survive on its own.
"We will pay our fisherfolk associations who will provide and plant the seedlings, and render the necessary care to ensure that the seedlings will survive and grow," Villarao said.
The CRM expert further said the BFAR will inspect planted areas after one year to see if the seedlings planted have survived. "We will again provide additional payment to the planters based on the number of surviving plants and eventually the project will be turned-over to the association as additional source of livelihood," Villarao added.
Seedlings will consist mostly of nipa, due to its short maturity period of only three years and the number of economic benefits that can be derived from the plant. Nipa leaves are used to make shingles, which in turn is used as roofing or as partition material. Sap is also extracted from its fruit stem which is processed to make vinegar and wine.
Villarao said mangroves provide a number of economic and ecological benefits such as source of fishery products of both estuarine and coastal fishes estimated at 20 to 30 tons of fish per year per square kilometer; shoreline protection and as first defense during typhoon surges and tsunami; carbon sink and sequestration; sediment trapping; as nursery and habitat for various species; source of forest products like timber and firewood; and for aesthetic purposes.
In addition to the mangrove restoration project, the BFAR 2 has also forged last January 9, a Memorandum of Agreement with the school heads of Isabela State University – Roxas, Cagayan State University – Aparri, Abulug School of Fisheries, and Sabtang National School of Fisheries for the establishment of a community-based multi-species hatchery in their respective schools.


text:
Home » Suburbs » Ang Mo Kio
By mailer_diablo (Self-taken (Unmodified)) [GFDL or CC-BY-SA-3.0], via Wikimedia Commons
Ang Mo Kio is one of the seven new towns of Singapore located in the northeast of the country and is a reputable tourist destination.
Originally, Ang Mo Kio was a settlement of rubber plantation workers who were later relocated for construction of the new town. Currently, the centre includes seven districts, mega malls, hawker centres and two MRT stations.
Ang Mo Kio Singapore, Swimming Pool, Park & Shopping Centre
Amongst the most notable tourist attractions is the new Bishan-Ang Mo Kio Park. What started as a rehabilitation project is now a green park with water-edged walkways and playgrounds for kids. Moreover, there is a huge array of diverse wildlife in the park and man-made water features.
if you are looking for a place to take the kids to cool down during the hot Singaporean days, then Ang Mo Kio is perfect for you. Stepping stones, river bridges and other water features make the park unique and more than appealing.
Any visitor to this town must also pay a visit the AMK hub which is a fantastic shopping complex. But AMK is so much more than just a shopping venue as it has plenty of entertainment and learning centres here. The hub has more than 200 shops that sell a wide range of products and services. You can also access banking and health care services here. For entertainment, there are arcade machines, a cinema complex and a club.
When it comes to transport, Ang Mo Kio has a well-developed networks of highly organised road infrastructure; two mass transit stations (Ang Mo Kio and Yio Chu Kang) and a bus interchange. The town is connected to the country's transport system by the Central expressway.
If you are looking for hotel accommodation in Ang Mo Kio, click the link below to find the best deals.
Ang Mo Kio Images & Photos
Ang Mo Kio, Singapore Current Weather Now & 7 Day Forecast
Ang Mo Kio WEATHER
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Book Now & Pay Later on Most Rooms
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Safe & Secure Booking System
Book Online Now & Save on Accommodation Prices
Ang Mo Kio Area, Street & Hotel Map, Singapore


To the extent that you attempt to assert any such claim, you hereby expressly agree to present such claim only through binding arbitration to occur in Maui, HI. You further agree to and do hereby waive any right to class arbitration and agree, instead, to conduct an arbitration related solely to any individual claims you and/or any entity related to you asserts against the Company. To the fullest extent permissible by law, you further agree that you shall be responsible for all costs associated with initiating the arbitration and for the administration of the arbitration.


Adam Smith, Category Specialist at NEPO, explains more about the ambitions of NEPRO³ , and describes the progress to date that has resulted in this significant procurement opportunity.
NEPRO³ is the replacement solution for the current Specialist Professional Services framework contract (NEPRO²), which expires on 30 November 2019.
Since its inception, NEPRO has developed into a market-leading solution and recognised brand for appointing specialist professional services via a neutral vendor route within the UK public sector. The solution has seen continued and sustained growth year on year. There are currently around 225 contracting authorities accessing the solution. The solution has provided significant opportunities to the supply chain, with SME suppliers accounting for 60% of project spend.
NEPRO³ will build on this success, providing the public sector with a neutral vendor managed service for the delivery of specialist professional services and consultancy.
How will NEPRO³ differ from previous solutions?
NEPRO³ will continue to address the existing core requirements and objectives that were identified back in 2012 under the first iteration; namely to offer greater choice, value, control and speed to public sector buyers appointing specialist professional services.
In addition, NEPRO³ will bring a greater focus to the buyer and supply chain community. The new solution will allow participating users to communicate, share best practice, solve common problems and access a central repository of project specifications as a community within the specialist professional services category. The participating supply chain will also be encouraged to collaborate and work together, where appropriate, to access and deliver against strategic and complex requirements.
Almost eight years of experience with this model has enabled NEPO to build a wealth of knowledge within the category. This experience and knowledge will be translated into the design of NEPRO³ to ensure challenges previously faced are avoided and the solution can continue to achieve its objectives as originally outlined back in 2012.
NEPO has undertaken significant market consultation to ensure that both buyers' and suppliers' feedback has been considered.
In October 2018 interested suppliers were identified via a Prior Information Notice (PIN) and invited to attend market consultation sessions. This was followed by a soft market testing questionnaire issued via the NEPO portal. The insight and feedback gained through engaging with the marketplace was extremely valuable and has significantly supported the design of NEPRO³ .
Over the course of the last six months, substantial stakeholder engagement has taken place with various public sector colleagues across the UK. Workshops were held in Durham, Newcastle, Manchester and London to ensure the future solution is fit for purpose and meets public sector requirements. A national working group has been established to support the procurement process, comprising representatives from local government, central government, police and further education.
What are the key dates for the procurement process?
*Dates are indicative and are correct at the time of writing, please get in touch to find out the latest position.


Many people affected by cancer struggle with how to communicate about what is happening. There are so many questions. When is the right time? How do you do it? Do you tell everybody or just a few? How will they respond?
If you or someone close to you has cancer, then the dilemma of telling other people or not will be a familiar one. The same goes for other life-changing or life-shortening illnesses and traumatic events.
Why do some struggle with talking about cancer diagnosis, and others don't give the decision a second thought?
A cancer diagnosis changes everything. It can leave people feeling stunted, unable to comprehend what has just happened and what is about to happen. For some, talking about cancer happens automatically. It is a release; a search for help and safe ground, when all else has turned into nothingness.
Others are too shocked to speak about it. They need time for the new reality to settle in. It is not uncommon for people to decide to keep the diagnosis to themselves or only tell a selected few.
Being judged: will people label me and reduce me to my illness?
Being pitied: will people feel uncomfortable and sorry for me?
Being rejected: will people turn the other way, and friends or family leave me?
Being a burden: will others be able to cope with the news (eg children, aging parents)?
Being disadvantaged: will an employer or colleagues think I can no longer cope with my job?
Becoming the centre of attention: I don't want others to make a fuss about me.
Becoming weak: will people think I am weak and cannot cope?
Becoming drained: I need to keep my focus and do not want to be distracted by the emotions of others.
Loneliness: there might not be anybody we can talk to.
None of this is right or wrong. We are what we are. We do what we think best.
I told everyone that needed to know, including my neighbours. The one group of people I did not tell were my therapy clients. I have been regretting that ever since. Second time around I did it differently.
Most importantly, I told those around me what I needed from them – business as usual, no pity, no worry, and no change.
Telling my ageing parents, who live abroad, was the most difficult decision and conversation of all. Should I tell them after the tests and before the diagnosis, so they have time to prepare, but may worry unnecessarily? Or do I tell only if the diagnosis is positive and then cause them a massive shock? I opted for the former.
Whichever way you turn it, telling those close to you about a cancer diagnosis is not easy.
This time around it was different. Perhaps because first time I had less understanding of what I was going to be in for.
This time around I know much more about cancer, treatment options and its physical, emotional, mental, financial and social side effects.
I talked about my cancer diagnosis to others in stages.And this time I made a point in including my therapy and counselling clients. My previous therapeutic and ethical concern about self-disclosure no longer applied.
When my own therapist stopped working I found out by email. I also found out about his death by email. I am sure there were good reasons for it all. But it caused me anger, a lot of grief and anxiety. And I think I might have caused the same in the way I handled (not) telling my clients first time around. We have to learn our lessons.
This time around I spoke with my clients about may diagnosis, face-to-face in our sessions; about the impact this (and any) difficult news can have. We talked openly about my need to take a break from work, and how I would keep everyone posted on my future plans, once all the scan results are back in. We all agreed this was the right way to do it. And I am glad I felt well enough (physically and mentally) to do it.
Why did I tell others about my diagnosis?
In all my uncertainty, this was one area where I could exercise control, express a view and make an immediate difference to my life.
I did not want to second-guess who knew and who did not.
I also did not want to explain, again and again.
I wanted to be left to get on with things.
Keeping a secret of the magnitude of a highly distressing (and potentially terminal) illness and treatment experience comes at a cost.
Some of us are expert stoics, and you would never have a clue of what is really going on inside.On the downside, stoics' emotional self-reliance often leaves them isolated and lacking essential assistance from others.
Lying, pretending, thinking we need to protect others, and being frightened of what others may do as a result of knowing the truth about our health – all of that takes a lot of energy. This in turn increases the already substantial stress and anxiety we will find ourselves in.
Undergoing cancer treatment, facing up to a terminal diagnosis, and living with or beyond cancer is not easy. It can take single-minded determination to deal with treatment and its side effects, as well as life-long uncertainty or the certainty of death. It is a roller coaster of unexpected mixed emotions.
Having lost certainty of health and life can lead to an overwhelming sense of helplessness and loss of control. How we decide to share our news and talk about cancer with others, or not, can be an important moment in taking charge of what is happening.
It is our prerogative to talk about cancer, or not, to share or withhold, to decide when the time is right, and who needs to know what.
The key is that the choice we make is based on our explicit decision of what is best for us and not others, and that whatever we choose to do, we're not motivated by a fear of what might happen if we tell. That's not always easy, I know, especially when others depend on us.
Telling the truth can be tough at the outset but in the long-run less damaging for all concerned (including children, who otherwise may feel left excluded and abandoned).
Standing up for ourselves, making an informed choice about whether to tell others or not, may feel difficult. But it is our prerogative. No one else can give us this permission. It is ours to take and to insist on.


which is better to buy a baby yellow nape or a 7yr old for my family?
Baby or 7 year old Amazon for family?
Amazon's live for many years, so age is not important as far as how long it will live with you.
What IS important is if you have researched all can about Amazon parrots. Did you know they may go through a hormonal stage where you can not handle the bird for possibly years? Have seen the damage they can do to people they get upset with? Is the 7 year old Amazon you are considering tame and does it allow you and your family members to hold it?
Amazons are GREAT parrots, I am not trying to discourage you from purchasing one. I just want to make sure you have done your research and know both the positive and the negative things about Amazons.
So, if the 7yr old Amazon is hand tame and you have done your research on Amazons, then I vote for the older bird that probably will have a harder time finding a home.
I suggest you read the Amazon parrot stories that are at the bottom of our Amazon parrot information page to learn more about what it is like to live with an Amazon parrot.


text:
WikiLeaks Quo Vadis?
Gunnar Gällmo
WikiLeaks has just been nominated for the Nobel Peace Prize. It is true that, as an organization and a collective, they are doing a good and valuable job.
Let's hope, however, that the Norwegian Nobel Prize Committee - and WikiLleaks themselves - have the good sence not to mix up the work of WikiLeaks with the private behaviour of Julian Assange and Israel Shamir.
These two have been acting in a very blameworthy way, not to say a stupid one.
Julian Assange has misbehaved sexually. (I won't call his behaviour "rape", as he isn't judged yet. Perhaps he is guilty of rape, perhaps he is not; perhaps he is guilty according to Swedish law but not according to British, or vice versa. Perhaps the ladies who are accusing him are secret agents of the CIA, perhaps they are not. Perhaps the Swedish prosecutor has acted wrongly, perhaps - less likely, I think - she has not.) We should wish him a fair trial - nothing more, nothing less.
Israel Shamir has expressed both his personal antisemitism, and his personal support for the Belarusan dictator Alexander Lukashenko.
WikiLeaks' enemies will, of course, try to confuse the merits of the organization with the private behaviour of people like Assange and Shamir.
A natural defence of WikiLeaks would be to insist on not confusing the two. They could, for example, stress the fact that the accusations against Assange in Sweden have, according to the Swedish prosecutor, nothing to do with WikiLeaks; it is about his sexual behaviour towards two Swedish ladies, and whether he is guilty or not is a question separated from WikiLeaks.
The alleged rape also has nothing to do with the accusations against Assange in the USA, by Hillary Clinton and others. These accusations are not only extremely stupid, they also show a totalitarian tendency that many do not think is characteristic of the USA (but perhaps it is).
WikiLeaks could also stress that they, as an organization, have nothing to do with Shamir's antisemitism or his support for Lukashenko. After all, human history is full of examples of a brilliant genius with a doubtful personal morality.
Unfortunately, WikiLeaks does neither. They ask for support for WikiLeaks, and for support for Julian Assange, and don't make clear that these are two different things, that is it perfectly possible, and perfectly legitimate, to support one but not the other.
I've got a certain impression that WikiLeaks may at present be led by a group of highly specialized people who know a lot within a limited field, but who are less knowledgeable when they are forced to leave that field.
It might also seem as if WikiLeaks is not content to show us the ex-secret documents, but would also like to tell us how these documents should be interpreted. If they do so, they don't understand how serious journalism works. They may also have an exaggerated view of their own competence. I think they are good whistle-blowers, but not very good analysts.
Let's hope that the members of WikiLeaks will be wiser than their leaders, and put them in place. Let's hope that WikiLeaks will transfer responsibility to more responsible persons.
Or, if that turns out to be impossible, that the good work initiated by WikiLeaks will be continued by other organizations with a better judgment. (there are already some WikiLeak clones around.)
Internet was once initiated by Pentagon. The appearance of WikiLeaks showed that Pentagon at the time didn't understand what they were starting.
Perhaps WikiLeaks, too, don't understand what they have started.
Let's hope the process will continue, even if WikiLeaks themselves are not competent to lead it.
And quite regardless whether they do receive the Nobel Prize or (more probably, I'm afraid) not.


Family Dollar is an American variety store chain. With over 8,000 locations in all states except Alaska, Hawaii, Oregon and Washington, it is the second largest retailer of its type in the United States. You will get information about Family Dollar Hours Today, Sunday, What time does Family Dollar Open/ closed. You can also find out the Family Dollar Near My Locations and Holiday hours of Family Dollar.
Family Dollar was founded in 1959 by Leon Levine, a 21-year-old entrepreneur. In November of that year, the company's first store was opened, in Charlotte, North Carolina. In 1961, their first store in South Carolina opened, followed by stores in Georgia and Virginia, which were opened in 1962 and 1965, respectively. If you need to find out the Family Dollar Hours Of Operation, Holiday Hours & Family Dollar hours Locations Near Me Must read this post till the end. Family Dollar hours.
What Time Does Family Dollar Open and close?
Is Family Dollar Closed On Holidays?
Is Family Dollar Open On Christmas?
Is Family Dollar Open On?
location, click on the map shown below and nearest Family Dollar location around you will automatically pop up on your map. Also, use this store locator tool to find Family Dollar near me.
You can also get the Family Dollar Near Me Now through Embed Map / Family Dollar Locator. If you use Family Dollar locator to find the Family Dollar Locations. Just enter your areas zip code & find Family Dollar Near My Location.
Get ready for Black Friday shopping 2018 by finding the Family Dollar locations nearest you. Check out Black Friday store hours, scope out the best parking spots and check the store out ahead of time to get a feel for the layout. If you are traveling for your Thanksgiving celebrations and have asked yourself "Is there going to be a Family Dollar near me?" our handy store finder can help.
Hopefully, We helped you to get some genuine records for Family Dollar Hours information.


If the Library of Congress Chronicling America: Historic American Newspapers website is not already part of your go-to primary sources toolkit, put your coffee down and bookmark it already. Chronicling America is a huge collection of digitized newspapers from around the United States published between 1836 and 1925. There are a variety of ways to search the database but no matter how you search, you get visual reproductions of complete newspaper pages that contain your search terms.
Say you want your students to compare and contrast contemporary accounts of the Battle of Gettysburg. Head over to the Chronicling America site, adjust the dates to limit the search to 1863, and type in your keywords of "battle gettysburg." Over 800 results appear.
I've been on a current events kick lately. A recent newsletter from social studies guru Mike Hasley reminded me of another awesome news resource called Newseum. And apparently I've never really posted anything about Newseum here at History Tech.
Not sure how I've never gotten around to that. The Newseum is a very cool, actual museum located in Washington DC with a powerful online presence. Their mission is to "champion the five freedoms of the First Amendment through exhibits, public programs and education." And I know that you've got one or two other museum choices in DC but if you're in the area, the Newseum is a very fun place to spend some time. Last time I visited, they had an awesome exhibit highlighting Pulitzer Prize winner photographs and the stories behind them. Amazing.
Thousands of historical newspapers from all over the country? Yup. And over 7,892,470 actual newspaper pages? Let that sink in for just a moment. Yup. But where, you ask, can I find such an incredible research tool? The very useful Chronicling America site from the Library of Congress, of course.
You'd think I'd be happy with almost eight million pages to play with. I mean, it's 7,892,470 pages. Which is . . . you know, a lot. The 7,892,470+ pages cover newspapers from almost all 50 states and the District of Columbia from 1836 to 1922.
But once you get in the collection, it's easy to get a little greedy. Wouldn't it be nice to have some coverage from the Civil War? The Great Depression? Prohibition? WWII? Vietnam? Hippies? 9/11?
Google News for finding primary sources. Sweet!
While it is possible, I suppose, to teach social studies and history without primary sources, it's probably not a good idea. And while there are more and more places to find primary sources, it can still be difficult to track down stuff that you can use.
It's can be especially difficult finding newspapers.
Google News is already an awesome tool for finding resources for current events around the world. But if you know where to look, Google News is also great for finding old newspapers for use in your instruction.
For as long as I can remember, I've been a newspaper reader. I started small. The Garden City Telegram was an afternoon delivery and just couple of sections but it got me into a routine . . . comics, sports, opinion page.
Went off to college and had the best of both worlds – the Hillsboro Free Press, a weekly local paper with a focus on crop reports, small town events and high school basketball scores and the New York Times, delivered a day late to the college library.
Moves to Wichita and eventually back to Hillsboro hooked me into the Wichita Eagle-Beacon, later just the Eagle. Delivered everyday. To my driveway. In the morning. Comics, sports, opinion page.
The Wichita Eagle has decided that there are not enough subscribers in the county to justify home delivery. They cut me off a couple of weeks ago.
I'm past the denial stage and have moved into bargaining. I'll probably end up getting the Hutchinson News. (It's not bad but, seriously, the Hutch News? How creative, a newspaper called the News.) And I do have the online version of the Eagle, so . . .
My "friends" haven't been a lot of help.
You could print out the website on flimsy paper with smudgy ink.
I dropped my morning statewide paper, because the entire thing is available in their iPad app. The cost is less too.
I stopped getting the paper years ago. I get all the news on my laptop/iPad.
You could join the rest of us in the 21st century.
Yeah. I get it. I have an iPad. Two, actually. There are 13 internet devices in my house, not including the Wii, Apple TV and Netflix-enabled Blueray player. I have a website called History Tech. Another called Social Studies Central. I've got a Twitter and a Plurk account. I'm on Facebook and use a password app to help me log in to all the different sites I visit. So I am well aware of this whole new world of the Interwebs. But I still need a newspaper.
If you can't tell, I'm a little bitter.
I'll probably get over it.
But the whole thing has gotten me thinking about why I like paper so much. And not just newspaper but magazines and books. Especially books. I love books more than I love newspapers.
The smell of books. Especially old books. I walked into the Queen Anne Bookstore in Seattle several weeks ago and immediately fell in love. It smelled like books. And coffee.
I like the different feel and heft of books. A quality, hard copy is great by the fire. The paperbook version is perfect for the hammock.
It's hard to lend someone an eBook. More importantly, it's hard for someone else to loan me an ebook.
While it's possible to know how far along you are in an eBook, a percentage usually of the remaining total, I like dog-earring a page and seeing what I got left.
No one ever comments on or starts a conversation about the eBook I'm reading at the coffee shop.
But I also started thinking about why I'm falling in love with digital media.
Small size. I can load up 100s of books, magazines and newspapers on my iPad / iPhone.
I like being able to create my own books and articles using the ePub format, PDFs and apps like Book Creator. The opportunity for educators to customize content and resources for their own classes is huge.
I love the fact that digital newspapers, magazines and books can be incredibly deep and multimedia rich. Video, audio, photos, additional information, supplementary materials and collaboration can all be built into digital stuff.
The problem with Queen Anne Books? Limited selection. I could easily spend a whole day in there but it's still gonna be just a few thousand books. iBooks, Kindle, digital libraries have hundreds of thousands.
I guess what I'm trying to say is that it's both/and not either/or. I'm still ticked that the bean counters at the Wichita Eagle cut me off. But I do understand that eBooks and digital media have their own advantages. And as educators, we need to be willing to use both.
Yes, still a little bitter.


Through M-Corps, SecondMuse incubates a core cohort of startups, but we also create workshops and events for the wider founder community. Explore our upcoming invites, as well as past events, and level your startup.
Come learn about best practices in crafting your Request For Quotation (RFQ), from preparation and processing to vetting and selection. What should you consider when choosing a partner? To how many vendors should you send your RFQ? How do you compare the quality of your bids?
This event will be held at PowerBridgeNY, at Columbia University.
All 11 M-Corps startups will present on a technical issue that they're having and receive feedback from a panel of mentors and their fellow startups.
Looking to build a supply chain that's closer to home? SecondMuse and NextCorps are leading a tour of six Rochester-area manufacturers who specialize in circuitboards, injection molding, industrial-grade 3D printing and precision milling. Join us for this two-day trip! Space is limited, so RSVP now.