text:
\section{Introduction and results}\label{intro}
The Fatou set $F(f)$ of an entire function $f$ is defined as
the set of all $z\in\mathbb C$ where the iterates $f^n$ of $f$
form a normal
family. The  Julia set is the complement of $F(f)$
and denoted by $J(f)$. The escaping set $I(f)$ is the set of
all $z\in\mathbb C$ for which $f^n(z)\to\infty$ as $n\to\infty$.
We note that $J(f)=\partial I(f)$ by a result of Eremenko~\cite{Eremenko89}.
For an introduction to the iteration theory of transcendental
entire functions we refer to~\cite{Bergweiler93}.

Considerable attention has been paid to the dimensions of Julia sets
of entire functions; see~\cite{Stallard08} for a survey,
as well as~\cite{Baranski08,BKZ,BK,BKS,BRS1,Rempe09,Rempe10,Schubert07}
for some recent results not covered there.
Many
results in this area are concerned with the Eremenko-Lyubich class
$B$ consisting of all transcendental
entire functions for which the set of critical
and finite asymptotic values is bounded. By a result of Eremenko and
Lyubich~\cite[Theorem~1]{Eremenko92} we have $I(f)\subset J(f)$ for
$f\in B$. For a function in the Eremenko-Lyubich class, a lower
bound for the dimension of the Julia set can thus be obtained from
such a bound for the escaping set.
This played a key role
 already in McMullen's seminal paper~\cite{McMullen87},
and it  has been used in many subsequent papers.

We denote the Hausdorff dimension, packing dimension
and  upper box dimension
of a subset $A$ of the complex
plane $\mathbb C$ by $\operatorname{dim_H} A$, $\operatorname{dim_P} A$ and $\operatorname{\overline{dim}_B} A$,
respectively,
noting that the upper box dimension
is defined only for bounded sets~$A$.
We refer to the book by Falconer~\cite{Falconer}
for the definitions
and a thorough treatment
of these concepts. Here we only note that we always
have~\cite[p.~48]{Falconer}
\[
\operatorname{dim_H} A\leq \operatorname{dim_P} A\leq \operatorname{\overline{dim}_B} A.
\]
The exceptional set $E(f)$ of a
transcendental entire function $f$ consists of all
points in $\mathbb C$ with finite backward orbit. It is an immediate
consequence of Picard's theorem that $E(f)$ contains at most one
point.  The following result is one part of a theorem of Rippon and
Stallard~\cite[Theorem~1.2]{Rippon05}.
\begin{theorema}
Let $f$ be a transcendental entire function, $A$ a backward
invariant subset of $J(f)$ and $U$ a bounded open subset
of $\mathbb C$ whose closure
does not intersect $E(f)$.
 Then $\operatorname{\overline{dim}_B}(U\cap
A)=\operatorname{dim_P} A$. In particular,  $\operatorname{\overline{dim}_B}(U\cap J(f))=\operatorname{dim_P} J(f)$.
\end{theorema}
For functions in the Eremenko-Lyubich class  they obtained the
following result~\cite[Theorem~1.1]{Rippon05}.
\begin{theoremb}
Let $f\in B$. Then $\operatorname{dim_P} J(f)=2$.
\end{theoremb}
It follows from Theorem~A that Theorem~B is equivalent to the result
that $\operatorname{\overline{dim}_B}(U\cap J(f))=2$ for some bounded open set~$U$
satisfying $\overline{U}\cap  E(f)=\emptyset$. In order
to show this, Rippon and Stallard actually proved that $\operatorname{\overline{dim}_B}(U\cap
I(f))=2$ for such a set $U$
and then used the result of Eremenko and Lyubich quoted
above.

The main tool used by Eremenko and Lyubich
to prove this result is a logarithmic change of variable.
This method shows in particular
that if $f\in B$, then $f$ is bounded on a curve tending
to~$\infty$; see~\cite[p.~993]{Eremenko92}.
 The $\cos\pi\rho$-theorem (see~\cite[Chapter~5,
Theorem~3.4]{Goldberg08}
or~\cite[Section~6.2]{Hayman89})
now implies that
\begin{equation}\label{1a3}
\liminf_{r\to\infty}\frac{\log M(r,f)}{\sqrt{r}}>0,
\end{equation}
where
\[
M(r,f):=\max_{|z|=r}|f(z)|
\]
is the maximum modulus. (The observation that~\eqref{1a3} holds for
functions in $B$ seems to have been made first in~\cite[Proof of
Corollary~2]{Bergweiler95} and \cite[p.~1788]{Langley95};
 see also~\cite[Lemma~3.5]{Rippon05}.)

It follows from a result of Baker~\cite[Corollary to Theorem~3.1]{Baker84}
that all components of $F(f)$ are simply connected if $f$ is bounded
on a curve tending to $\infty$. In particular, if $f\in B$, then
$F(f)$ has no multiply connected components~\cite[Proposition~3]{Eremenko92}.

In view of these results
 the following theorem can be considered as a generalization
of Theorem~B.
\begin{theorem}\label{thm1}
Let $f$ be a transcendental entire function satisfying
\begin{equation}\label{1a}
\liminf_{r\to\infty}\frac{\log\log M(r,f)}{\log\log r}=\infty.
\end{equation}
If $F(f)$ has no multiply connected component,
then
\begin{equation}\label{1a4}
\operatorname{dim_P}(I(f)\cap J(f))=2.
\end{equation}
\end{theorem}
Since multiply connected components of $F(f)$
are contained in $I(f)$, we see that
$I(f)$ has interior points if $F(f)$ has such a
component. We conclude that $\operatorname{dim}_P I(f)=2$
for all entire functions satisfying~\eqref{1a}.

The following result is an immediate consequence of Theorem~\ref{thm1}
and the results stated before it.
\begin{cor}\label{cor1}
Let $f$ be a transcendental entire
 function which is bounded on a
curve tending to~$\infty$. Then $\operatorname{dim_P}(I(f)\cap J(f))=2$.
\end{cor}
More generally, we have the following result involving
the minimum modulus
\[
L(r,f):=\min_{|z|=r}|f(z)|.
\]
\begin{cor}\label{cor2}
Let $f$ be a transcendental entire function and suppose that
\begin{equation}\label{1x}
\limsup_{r\to\infty}\frac{\log L(r,f)}{\log M(r,f)}<1.
\end{equation}
Then $\operatorname{dim_P}(I(f)\cap J(f))=2$.
\end{cor}
To deduce Corollary~\ref{cor2} from Theorem~\ref{thm1}
we note that the
$\cos\pi\rho$-theorem  yields that
\begin{equation}\label{1yy}
\liminf_{r\to\infty}\frac{\log\log M(r,f)}{\log r}>0
\end{equation}
under the hypothesis~\eqref{1x}. Clearly~\eqref{1a}
follows from~\eqref{1yy}.

Zheng~\cite[Corollary~1]{Zheng06}
proved, as a corollary to the main result of his paper,
 that if $F(f)$ has a
multiply connected component, then
\begin{equation}\label{1z}
\limsup_{r\to\infty}\frac{\log L(r,f)}{\log M(r,f)}>0.
\end{equation}
A slight extension of his argument shows that his main result
actually yields that
\begin{equation}\label{1y}
\limsup_{r\to\infty}\frac{\log L(r,f)}{\log M(r,f)}=1
\end{equation}
if $F(f)$ has a multiply connected component;
see~Proposition~\ref{zhengmod}. Thus the hypotheses of
Theorem~\ref{thm1} are satisfied if~\eqref{1x} holds.
  Actually the
condition~\eqref{1x} can be further relaxed; cf. Remark~\ref{rem2}.

 Zheng~\cite[Corollary~5]{Zheng06} also showed that the Fatou set
 of a transcendental entire function $f$ has no multiply connected
 component if
\begin{equation}\label{1zh}
\log M(2r,f)\geq d \log M(r,f)
\end{equation}
for some $d>1$ and all large~$r$. It is easy to see that~\eqref{1zh}
implies~\eqref{1yy} and hence~\eqref{1a}. Thus we obtain the
following corollary to Theorem~\ref{thm1}.
\begin{cor}\label{cor4}
Let $f$ be a transcendental entire function satisfying~\eqref{1zh}.
 Then $\operatorname{dim_P}(I(f)\cap J(f))=2$.
\end{cor}
We mention that in Theorem B and in Theorem~\ref{thm1}, as well as
in the corollaries to Theorem~\ref{thm1}, the packing dimension
cannot be replaced by the Hausdorff dimension. In fact, it is shown
in~\cite[Corollary~1.4]{Rempe10} that there exists a function $f\in
B$ for which $\operatorname{dim_H} I(f)=1$ and in~\cite{Stallard90} that for every
$\varepsilon>0$ there exists a function $f\in B$ such that $\operatorname{dim_H}
J(f)<1+\varepsilon$, and
the functions considered in~\cite{Rempe10,Stallard90}
satisfy~\eqref{1zh} as well.
On the other hand, for every transcendental entire function $f$ the
Hausdorff dimension of $I(f)\cap J(f)$ is at least~$1$, since this
set contains continua; see~\cite[Theorem~5]{Rippon05a}
 and~\cite[Theorem~1.3]{Rippon10}.

Concerning the proof of Theorem~\ref{thm1}, we note that
in view of Theorem~A the conclusion of Theorem~\ref{thm1} is
equivalent to the statement that
\begin{equation}\label{1a2}
\operatorname{\overline{dim}_B}(U\cap I(f)\cap J(f))=2
\end{equation}
for some bounded open set~$U$ satisfying $\overline{U}\cap E(f)=\emptyset$,
which in turn is equivalent to~\eqref{1a2} holding for all
bounded open sets~$U$.
We shall show in our proof that~\eqref{1a2} holds for some bounded
open set~$U$
whose closure does not intersect $E(f)$.

The main tools used in the proof
are certain estimates of the logarithmic derivative
and a version of the Ahlfors islands theorem. A similar
technique was used in~\cite{BK} where it was shown that
under a suitable regularity condition on the growth
of $f$ we even have
$\operatorname{dim_H}(I(f)\cap J(f))=2$.

\section{Preliminary lemmas}\label{prelim}
We use the standard terminology of Nevanlinna theory and,
in particular, denote by $T(r,f)$ the Nevanlinna characteristic
of a meromorphic function~$f$;
see~\cite{Goldberg08,Hayman64}.
First we note that using
the well-known inequality
\begin{equation}\label{2a}
T(r,f)\leq \log^+ M(r,f)\leq \frac{R+r}{R-r} T(R,f)
\quad\text{for}\ 0<r<R
\end{equation}
we easily see that the growth condition~\eqref{1a}
is equivalent to
\begin{equation}\label{2b}
\liminf_{r\to\infty}\frac{\log T(r,f)}{\log\log r}=\infty.
\end{equation}

We shall need a number of lemmas and
begin with the following estimate of the logarithmic
derivative~\cite[p.~88]{Goldberg08}.
\begin{lemma}\label{goldberg}
Let $f$ be an entire function satisfying $f(0)=1$.
Then
\[
\left| \frac{f'(z)}{f(z)}\right|
\leq \frac{4s}{(s-|z|)^2} T(s,f)+
\sum_{|z_j|\leq s} \frac{2}{|z-z_j|},
\]
for $s>|z|$,
where $(z_j)$ is the sequence of zeros of~$f$.
\end{lemma}

In order to estimate the sum on the right hand side
we shall use the following result due to Fuchs and
Macintyre~\cite{Fuchs40}. Here and in the following
we denote by $D(a,r)$ the open disk
and by $\overline{D}(a,r)$ the closed disk
of radius $r$ around
a point~$a$.
\begin{lemma} \label{fuchs}
Let $z_1,z_2,\dots,z_n\in\mathbb C$ and let $H>0$.
\begin{itemize}
\item[(i)]
There exist $l\in\{1,2,\dots,n\}$,
$u_1,u_2,\dots,u_l\in\mathbb C$ and $s_1,s_2,\dots,s_l>0$ satisfying
\[
\sum_{k=1}^l s_k^2 \leq 4H^2
\]
such that
\[
\sum_{k=1}^m\frac{1}{|z-z_k|}\leq  \frac{2n}{H}
\quad\text{for}\
z
\notin \bigcup_{k=1}^l D(u_k,s_k).
\]
\item[(ii)]
There exist
$m\in\{1,2,\dots,n\}$,
$v_1,v_2,\dots,v_m\in\mathbb C$ and $t_1,t_2,\dots,t_m>0$ satisfying
\[
\sum_{k=1}^m t_k \leq 2H
\]
such that
\[
\sum_{k=1}^n\frac{1}{|z-z_k|}\leq  \frac{n(1+\log n)}{H}
\quad\text{for}\
z
\notin \bigcup_{k=1}^m D(v_k,t_k).
\]
\end{itemize}
\end{lemma}
Actually Fuchs and Macintyre write $An/H$ instead of $2n/H$ in
part~(i), with a constant~$A$, but their argument shows that one can
take $A=2$.
We note that the term $\log n$ in (ii) cannot be omitted;
cf.~\cite{AndEid}.

We will also require the following version of the Borel-Nevanlinna
growth lemma;
see~\cite[Chapter~3, Theorem~1.2]{Goldberg08}
or~\cite[Section~3.3]{Cherry01}.
 Here
a measurable subset $E$ of $(0,\infty)$ is said to be of finite
logarithmic measure if
\[
\int_E\frac{dt}{t}<\infty.
\]
\begin{lemma}\label{growth}
Let $F:[r_0,\infty)\to [t_0,\infty)$
and $\varphi:[t_0,\infty)\to(0,\infty)$
be non-decreasing functions,
with $r_0,t_0>0$.
Suppose that
\[
\int_{t_0}^\infty\frac{dt}{\varphi(t)}<\infty.
\]
Then there exists a set $E\subset [r_0,\infty)$
 of finite logarithmic measure such that
\[
F\left(r\left(1+\frac{1}{\varphi(F(r))}\right)\right) \leq F(r)+1
\quad \text{for}\ r\notin E.
\]
\end{lemma}
We shall also need a result from the Ahlfors theory of covering
surfaces; cf.~\cite[Theorem~6.2]{Hayman64} or~\cite{Bergweiler98}.
Here $\mathbb D:=D(0,1)$ is the unit disk.
\begin{lemma}\label{ahlfors}
Let $D_1,D_2,D_3$ be Jordan domains with pairwise disjoint closures.
Then there exists $\mu>0$ such that if
$h:\mathbb D\to\mathbb C$ is a holomorphic function satisfying
\[
\frac{|h'(0)|}{1+|h(0)|^2}\geq \mu,
\]
then $\mathbb D$ has a subdomain which is mapped bijectively
onto one of the domains~$D_\nu$ by~$h$.
\end{lemma}
Finally, we shall repeatedly   use the following result known as
the
 Koebe distortion theorem and
the Koebe one quarter  theorem.

\begin{lemma}\label{koebe}
Let $g:D(a,r)\to \mathbb C$ be univalent, $0< \lambda <1$ and $z\in
\overline{D}(a,\lambda r)$. Then
\[
\frac{1}{(1+\lambda)^2}|g'(a)|\leq \frac{|g(z)-g(a)|}{|z-a|}
\leq \frac{1}{(1-\lambda)^2}|g'(a)|
\]
and
\[
\frac{1-\lambda}{(1+\lambda)^3}|g'(a)|\leq |g'(z)|\leq
\frac{1+\lambda}{(1-\lambda)^3}|g'(a)|.
\]
Moreover,
\[
g(D(a,r))\supset D\left(g(a),\frac{1}{4}|g'(a)|r\right).
\]
\end{lemma}
Usually Koebe's theorems are stated only for the special case
that $a=0$, $r=1$, $g(0)=0$ and $g'(0)=1$, but the above
result easily follows from this special case.

\section{Proof of Theorem~\ref{thm1}}
Without loss of generality we may assume that $f(0)=1$.
We denote by $n(r,0)$ the number of zeros of $f$ in the
closed disk of radius $r$ around $0$ and put
\[
N(r,0):=\int_0^r\frac{n(t,0)}{t}dt.
\]
By Nevanlinna's first fundamental theorem we have
\begin{equation}\label{2d}
T(s,f)
\geq N(s,0)\geq\int_r^s\frac{n(t,0)}{t}dt\geq n(r,0) \int_r^s\frac{dt}{t}
= n(r,0) \log\left(\frac{s}{r}\right)
\end{equation}
for $s>r>0$. Applying Lemma~\ref{growth} with $\varphi(x)=x^2/6$ and
$F(r)=\log T(r,f)$ we obtain a set $E$ of finite logarithmic measure
such that
\begin{equation}\label{3a}
T\left(r\left(1+\frac{6}{[\log T(r,f)]^2}\right),f\right) \leq e T(r,f)
\quad \text{for}\ r\notin E.
\end{equation}
For $m\in\mathbb N$ we shall use the abbreviation
\[
R_m(r):=r\left(1+\frac{m}{[\log T(r,f)]^2}\right)
\]
so that~\eqref{3a} takes the form
\begin{equation}\label{3b}
T\left(R_6(r),f\right)\leq e T(r,f) \quad \text{for}\ r\notin E.
\end{equation}
Using~\eqref{2a} we find that if $r\notin E$ is sufficiently large,
then
\begin{equation}\label{3b1}
\log M(r,f)\leq \frac{R_6(r)+r}{R_6(r)-r}T(R_6(r),f)
\leq T(r,f)[\log T(r,f)]^2.
\end{equation}

For measurable $X\subset\mathbb R$ and $Y\subset \mathbb C$
we denote by $\operatorname{length} X$ the $1$-dimensional Lebesgue measure
of $X$ and by $\operatorname{area} Y$ the $2$-dimensional Lebesgue measure
of~$Y$.
\begin{lemma}\label{lemma31}
For sufficiently large $r\notin E$  there exists
a closed subset $F_r$ of $\left[R_1(r),R_3(r)\right]$
with
\begin{equation}\label{3c}
\operatorname{length} F_r\geq \frac{r}{[\log T(r,f)]^2}
\end{equation}
such that
\begin{equation}\label{3d}
\left|\frac{f'(z)}{f(z)}\right|
\leq \frac{T(r,f) [\log T(r,f)]^7}{r}
\quad\text{for}\ |z|\in F_r.
\end{equation}
Moreover, for $\eta>0$ there exist
an integer $l\in\left\{1,2,\dots, n\left(R_5(r),0\right)\right\}$,
points $u_1,u_2,\dots,u_l\in\mathbb C$ and $s_1,s_2,\dots,s_l>0$ satisfying
\begin{equation}\label{3d1}
\sum_{k=1}^l s_k^2 \leq \frac{r^2}{T(r,f)^\eta}
\end{equation}
such that
\begin{equation}\label{3e}
\left|\frac{f'(z)}{f(z)}\right|
\leq \frac{T(r,f)^{1+\eta}}{r}
\quad\text{if}\
r\leq |z|\leq R_4(r)
\ \text{and}\ z\notin \bigcup_{k=1}^l D(u_k,s_k).
\end{equation}
\end{lemma}
\begin{proof}
It follows from Lemma~\ref{goldberg} that if $z_1,z_2,\dots$ are
the zeros of $f$, then
\begin{equation}\label{3g}
\left| \frac{f'(z)}{f(z)}\right|
\leq \frac{4R_5(r)}{(R_5(r)-|z|)^2} T(R_5(r),f)+
\sum_{|z_j|\leq R_5(r)} \frac{2}{|z-z_j|}
\end{equation}
for $|z|< R_5(r)$. Using~\eqref{3b} and noting that $R_5(r)\leq 2r$
for large $r$ we see that if $|z|\leq R_4(r)$, then
\begin{equation}\label{3h}
\frac{4R_5(r)}{(R_5(r)-|z|)^2} T(R_5(r),f) \leq 8e
\frac{T(r,f)}{r}[\log T(r,f)]^4
\end{equation}
for large $r\notin E$.
To estimate the sum on the right hand side of~\eqref{3g} we apply
Lemma~\ref{fuchs}, part~(ii), with
\begin{equation}\label{3i}
H=\frac{r}{8[\log T(r,f)]^2}
\end{equation}
and conclude that
\begin{equation}\label{3j}
\sum_{|z_j|\leq R_5(r)} \frac{1}{|z-z_j|}
\leq \frac{n(R_5(r),0)(1+\log n(R_5(r),0))}{H}
\end{equation}
outside a union of disks whose sum of radii is at most $2H$. Let $P$
be the set of all $s>0$ such that $\{z\in\mathbb C:|z|=s\}$ intersects the
union of these disks. Then $\operatorname{length} P\leq 4H$. Thus
$F_r:=[R_1(r),R_3(r)]\setminus P$ satisfies~\eqref{3c},
and~\eqref{3j} holds for $|z|\in F_r$.

From~\eqref{2d} and~\eqref{3b} we can deduce that
\begin{equation}\label{3j1}
\begin{aligned}
 n(R_5(r),0)
&\leq
\frac{T(R_6(r),f)}{\log\left(R_6(r)/R_5(r)\right)}\\
&\leq 2 T(R_6(r),f) [\log T(r,f)]^2
\leq 2e T(r,f) [\log T(r,f)]^2
\end{aligned}
\end{equation}
for large $r\notin E$.
  Combining this with~\eqref{3i} and~\eqref{3j}
we obtain
\begin{equation}\label{3k}
\begin{aligned}
&\quad \sum_{|z_j|\leq R_5(r)} \frac{1}{|z-z_j|}\\
&\leq \frac{16 e T(r,f) [\log T(r,f)]^4\left(
1+\log\left(2e T(r,f) [\log T(r,f)]^2\right) \right)}{r}\\
&\leq \frac{T(r,f)[\log T(r,f)]^6}{r}
\end{aligned}
\end{equation}
if $|z|\in F_r$ and $r$ is large.
Now~\eqref{3d} follows from~\eqref{3g}, \eqref{3h} and~\eqref{3k}.

In order to prove~\eqref{3e} we use part~(i) of Lemma~\ref{fuchs}
with
\begin{equation}\label{3l}
H=\frac{r}{2 T(r,f)^{\eta/2}}.
\end{equation}
This yields
$l$ disks $D(u_k,s_k)$ satisfying~\eqref{3d1} such that
\begin{equation}\label{3l1}
\sum_{|z_j|\leq R_5(r)} \frac{1}{|z-z_j|}
\leq \frac{2 n(R_5(r),0)}{H}
\quad\text{for}\
z\notin \bigcup_{k=1}^l D(u_k,s_k),
\end{equation}
with $l\leq n(R_5(r),0)$.
Combining~\eqref{3j1}, \eqref{3l} and~\eqref{3l1} we find that
\[
\quad \sum_{|z_j|\leq R_5(r)} \frac{1}{|z-z_j|} \leq \frac{8e T(r,f)
[\log T(r,f)]^2 T(r,f)^{\eta/2}}{r}
\]
if $z\notin \bigcup_{k=1}^l D(u_k,s_k)$.
This, together with~\eqref{3g} and~\eqref{3h},
implies~\eqref{3e}.
\end{proof}
\begin{lemma}\label{lemma32}
Let $F_r$ be as in Lemma {\rm \ref{lemma31}} and let $\delta>0$. If
$r$ is sufficiently large, then for each $s\in F_r$ there exists a
closed subset $J_s$ of $[0,2\pi]$ with
\begin{equation}\label{3n1}
\operatorname{length} J_s \geq \frac{1}{T(r,f)^\delta}
\end{equation}
such that if $\theta\in J_s$, then
\begin{equation}\label{3n}
|f(se^{i\theta})|\geq \sqrt{M(r,f)}
\end{equation}
and
\begin{equation}\label{3o}
\left|\frac{f'(se^{i\theta})}{f(se^{i\theta})}\right|
\geq \frac{T(r,f)^{1-\delta}}{r}.
\end{equation}
\end{lemma}
\begin{proof}
First we consider the case that
\begin{equation}\label{p3h}
\log L(s,f)=\min_{|z|=s}\log|f(z)|\leq\frac12 \log M(s,f).
\end{equation}
Then there exists $z_1$ and $z_2$ with $|z_1|=|z_2|=s$
satisfying
\[
\log|f(z_1)|=\frac12 \log M(s,f)
\quad\text{and}\quad
\log|f(z_2)|=\log M(s,f)
\]
while
\[
\frac12 \log M(s,f)\leq\log|f(z)|\leq\log M(s,f)
\]
on one of the two arcs between $z_1$ and $z_2$.
With $z_1=se^{i\theta_1}$ and $z_2=se^{i\theta_2}$ we may assume
without loss of generality
that $\theta_1<\theta_2$ and
\[
\frac12 \log M(s,f)\leq\log|f(se^{i\theta})|\leq\log M(s,f)
\]
for $\theta_1\leq \theta\leq \theta_2$. Let $J_s$ be the set of all
$\theta\in[\theta_1,\theta_2]$ for which~\eqref{3o} holds.
Since~\eqref{3n} holds for all $\theta\in[\theta_1,\theta_2]$
and thus in particular for $\theta\in J_s$, we
only have to estimate the length of~$J_s$. By the choice of $z_1$
and $z_2$ we have
\begin{equation}\label{p3ww}
\begin{aligned}
\frac12 \log M(s,f)
&=\log|f(z_2)|-\log|f(z_1)|
=\operatorname{Re}\left(\int_{z_1}^{z_2}
\frac{f'(z)}{f(z)} dz \right)\\
&\leq
\int_{z_1}^{z_2}
\left|\frac{f'(z)}{f(z)}\right|
|dz|
= s\int_{\theta_1}^{\theta_2}
\left|\frac{f'(se^{i\theta})}{f(se^{i\theta})}\right|d\theta.
\end{aligned}
\end{equation}
Now
\begin{equation}\label{p3xx}
\begin{aligned}
s\int_{\theta_1}^{\theta_2}
\left|\frac{f'(se^{i\theta})}{f(se^{i\theta})}\right| d\theta
&
=
s \int_{J_s} \left|\frac{f'(se^{i\theta})}{f(se^{i\theta})}\right|
d\theta + s \int_{[\theta_1,\theta_2]\setminus J_s}
\left|\frac{f'(se^{i\theta})}{f(se^{i\theta})}\right| d\theta
\\
&\leq
s \frac{T(r,f) [\log T(r,f)]^7}{r}\operatorname{length} J_s +
2\pi s\frac{T(r,f)^{1-\delta}}{r}  \\
&\leq
2 T(r,f) [\log T(r,f)]^7\operatorname{length} J_s +
4\pi\, T(r,f)^{1-\delta},
\end{aligned}
\end{equation}
where we used $s\leq R_3(r)\leq 2r$ in the last inequality.
Since $T(r,f)\leq \log M(r,f)$
we have
\[
4\pi\, T(r,f)^{1-\delta}\leq\frac14 \log M(r,f)
\]
for large $r$
and this, together with~\eqref{p3ww} and~\eqref{p3xx},
yields
\[
\frac14 \log M(r,f)\leq
2 T(r,f) [\log T(r,f)]^7\operatorname{length} J_s.
\]
We deduce that
\[
\operatorname{length} J_s
\geq\frac{1}{8 [\log T(r,f)]^7}
\]
and thus obtain~\eqref{3n1}.

Now we consider the case that~\eqref{p3h} does not hold.
Then~\eqref{3n} holds for all $\theta\in[0,2\pi]$.
Let $J_s$ be the subset of all $\theta\in[0,2\pi]$
for which~\eqref{3o} holds.
By the argument principle, we have
\[
n(s,0)=\frac{1}{2\pi i}\int_{|z|=s} \frac{f'(z)}{f(z)} dz
\leq
 \frac{s}{2\pi} \int_0^{2\pi} \left|\frac{f'(se^{i\theta})}{f(se^{i\theta})}\right|
d\theta.
\]
The same argument as in~\eqref{p3xx} now yields
\begin{equation}\label{p3l}
2\pi \, n(s,0)
\leq
2 T(r,f) [\log T(r,f)]^7\operatorname{length} J_s +
4\pi\, T(r,f)^{1-\delta}.
\end{equation}
Since we are assuming that~\eqref{p3h} does not hold,
we have $|f(re^{i\theta})|\geq 1$ for $0\leq\theta\leq 2\pi$
and thus $m(r,1/f)=0$, where $m(r,\cdot)$ denotes the Nevanlinna
proximity function.
Nevanlinna's first fundamental theorem, together with the
assumption that $f(0)=1$, yields
\[
T(r,f)= N(r,0)\leq N(1,0)+ n(r,0)\log r
\]
and thus
\begin{equation}\label{p3m}
2\pi\, n(s,0)\geq 2\pi\, n(r,0)\geq \frac{T(r,f)}{\log r}
\geq 2 T(r,f)^{1-\delta/2}
\end{equation}
for large $r$ by~\eqref{2b}.
Combining~\eqref{p3l} and~\eqref{p3m} we obtain
$$
T(r,f)^{1-\delta/2} \leq 2 T(r,f) [\log T(r,f)]^7\operatorname{length} J_s ,
$$
from which~\eqref{3n1} easily follows.
\end{proof}
For small $\delta>0$ and large $r\notin E$ we consider the set
\[
\begin{aligned}
A(r):=\bigg\{
z
\in\mathbb C
:
\
& R_1(r)\leq |z|\leq R_3(r),\;\\
&|f(z)|\geq \sqrt{M(r,f)},\;
\left|\frac{f'(z)}{f(z)}\right|
\geq \frac{T(r,f)^{1-\delta}}{r}
\bigg\}.
\end{aligned}
\]
With the notation of Lemmas~\ref{lemma31} and ~\ref{lemma32}
we have
\[
A(r)\supset
\left\{se^{i\theta}: s\in F_r,\;\theta\in J_s\right\}
\]
and thus we can deduce from these lemmas that
\begin{equation}\label{4b}
\begin{aligned}
\operatorname{area} A(r)
&\geq
\operatorname{length} (F_r)
\inf_{s\in F_r}( s\operatorname{length}(J_s))\\
&\geq
\frac{r^2}{[\log T(r,f)]^2 T(r,f)^\delta}
\geq
\frac{2r^2}{ T(r,f)^{2\delta}}
\end{aligned}
\end{equation}
for large $r$.
We put
$$
\rho(r):=\frac{r}{ T(r,f)^{1-2\delta}}
$$
and
\[
B(r):=A(r)\setminus \bigcup_{j=1}^l D\left(u_j,s_j+\rho(r)\right),
\]
where $l,u_1,\dots,u_l,s_1,\dots,s_l$ are chosen according to
Lemma~\ref{lemma31}, taking $\eta:=3\delta$ there.

We use the notation $\operatorname{ann}(r,R):=\{z\in\mathbb C:r<|z|<R\}$ for an annulus
with radii $r$ and~$R$.
 Given a constant $M>1$ we have $M\rho(r)\leq r/[\log
T(r,f)]^2$ for large $r$ and thus
\begin{equation}\label{p4xy}
D(b,M\rho(r))\subset \operatorname{ann}(r,R_4(r)) \quad\text{for} \ b\in A(r).
\end{equation}
Hence
\[
D(b,\rho(r))\subset \operatorname{ann}(r,R_4(r))\setminus
\bigcup_{j=1}^l D(u_j,s_j)
\]
and thus
\begin{equation}\label{3e1}
\left|\frac{f'(z)}{f(z)}\right|
\leq \frac{T(r,f)^{1+3\delta}}{r}
\quad\text{if}\
b\in B(r) \ \text{and}\
z\in D(b,\rho(r))
\end{equation}
by~\eqref{3e}.
In particular, $f$ has no zeros in $D(b,\rho(r))$.

We want to show that the area of $B(r)$ is not much smaller
than that of $A(r)$.
In order to do so we note that~\eqref{3d1}, \eqref{3j1} and
the definition of $\rho(r)$
yield
\[
\begin{aligned}
\operatorname{area} \left(\bigcup_{j=1}^l D\left(u_j,s_j+\rho(r)\right)\right)
&=\pi\sum_{j=1}^l\left(s_j+\rho(r)\right)^2\\
&\leq 2\pi\sum_{j=1}^l \left(s_j^2+\rho(r)^2\right)\\
&\leq 2\pi \frac{r^2}{ T(r,f)^{3\delta}}
+2\pi \rho(r)^2 n(R_5(r),0)\\
&\leq 2\pi\frac{r^2}{ T(r,f)^{3\delta}}
+4\pi e \frac{r^2[\log T(r,f)]^2}{T(r,f)^{1-4\delta}}\\
&\leq \frac{r^2}{ T(r,f)^{2\delta}}
\end{aligned}
\]
for large~$r$, provided $\delta<1/6$.
 Combining this with~\eqref{4b} we obtain
\begin{equation}\label{4e}
\operatorname{area} B(r)
\geq \operatorname{area} A(r)-
\operatorname{area} \left(\bigcup_{j=1}^l D\left(u_j,s_j+\rho(r)\right)\right)
\geq \frac{r^2}{ T(r,f)^{2\delta}}
\end{equation}
for large~$r$.

Now let $m(r)$ be the maximal number of pairwise disjoint
disks of radius $\rho(r)$ whose centers are in $B(r)$.
Then $B(r)$ is contained in a union of $m(r)$ disks
of radius $2\rho(r)$ and thus
\[
\operatorname{area} B(r)\leq 4\pi m(r) \rho(r)^2=4\pi m(r)\frac{r^2}{T(r,f)^{2-4\delta}}.
\]
for large~$r$. Together with~\eqref{4e} we obtain
\begin{equation}\label{4g}
m(r)\geq \frac{1}{4\pi} T(r,f)^{2-6\delta}
\geq T(r,f)^{2-7\delta}.
\end{equation}

Recall that if $b\in B(r)$, for some large $r\notin E$, then
$f$ has no zeros in $D(b,\rho(r))$.
Thus we can define a branch $\phi$
of the logarithm of $f$ in $D(b,\rho(r))$;
that is, there exists a holomorphic function
$\phi:D(b,\rho(r))\to \mathbb C$ such that $\exp \phi(z)=f(z)$ for
$z\in D(b,\rho(r))$.
Of course, the other branches of the logarithm of $f$ are then given
by $z\mapsto \phi(z)+2\pi i n$ where $n\in \mathbb Z$.

For $a\in\mathbb R$ we will consider the domain
\[
Q(a):=\{z\in\mathbb C: |\operatorname{Re} z-a|< 1,|\operatorname{Im} z|< 2\pi\}
\]
\begin{lemma}\label{lemma33}
Let $b\in B(r)$, with $r\notin E$ sufficiently large.
Then there exists a branch $\phi$ of the logarithm of $f$
defined in $D(b,\rho(r))$ and a subdomain $U$ of $D(b,\rho(r))$
such that
$\phi$ maps $U$ bijectively onto $Q(\log|f(b)|)$.
\end{lemma}
\begin{proof}
Let $\phi_0$ be a fixed branch of the logarithm of $f$ defined in
$D(b,\rho(r))$. Define $h:\mathbb D\to \mathbb C$,
$h(z)=\phi_0(b+\rho(r)z)-\phi_0(b)$. Then $h(0)=0$ and
\[
|h'(0)|=\rho(r)|\phi_0'(b)|=\rho(r)\left|\frac{f'(b)}{f(b)}\right|
\geq \rho(r)\frac{T(r,f)^{1-\delta}}{r} = T(r,f)^{\delta}
\]
by the definition of $\rho(r)$ and
$A(r)$ and since $B(r)\subset A(r)$.
For $k\in\{1,2,3\}$ we put
$D_k:=\{z\in\mathbb C:|\operatorname{Re} z|<1,|\operatorname{Im} z -12k\pi |<4\pi\}$.
Lemma~\ref{ahlfors} now implies that if $r\notin E$ is large
enough, then there exists a subdomain $V$ of $\mathbb D$ and
$k\in\{1,2,3\}$ such that $h$ maps $V$ bijectively
onto $D_k$.
This implies that $D(b,\rho(r))$ contains a subdomain $W$
which is mapped bijectively onto
$$\{z\in\mathbb C: \left|\operatorname{Re} z-\log|f(b)|\right|< 1,
|\operatorname{Im} z-\operatorname{Im} \phi_0(b)-12\pi k|< 4\pi\}$$ by $\phi_0$.
 Choosing $n\in \mathbb Z$
such that $|2\pi n-\operatorname{Im} \phi_0(b) -12k\pi |\leq \pi$ we find
that there exists a domain $U\subset W\subset D(b,\rho(r))$ which is
mapped bijectively onto
$$\{z\in\mathbb C: \left|\operatorname{Re} z-\log|f(b)|\right|< 1,|\operatorname{Im} z-2\pi n|< 2\pi\}$$
by $\phi_0$.
The branch $\phi$ of the logarithm given by $\phi(z)=\phi_0(z)-2\pi i n$
now has the required property.
\end{proof}
We note that if $U$ is as in Lemma~\ref{lemma33}, then
\begin{equation}\label{4f}
f(U)=\exp Q(\log|f(b)|)=\operatorname{ann}\left(\frac{1}{e}|f(b)|,e|f(b)|\right).
\end{equation}

Now we fix some large $r_0\notin E$, put $\rho_0:=\rho(r_0)$
and choose $b_0\in B(r_0)$.
Lemma~\ref{lemma33} yields a branch $\phi_0$ of the logarithm
of $f$ and a domain $U_0\subset D(b_0,\rho_0)$ which is mapped
bijectively onto $Q(\log|f(b_0)|)$ by $\phi_0$.
Since $|f(b_0)|\geq \sqrt{M(r_0,f)}$
we also have
$|f(b_0)|\geq 2r_0$
and the interval $\left[|f(b_0)|,2|f(b_0)|\right]$ contains
a point $r_1$ which does not belong to~$E$,
provided $r_0$ is chosen large enough.
Note that $r_1\geq 2r_0$.
We now choose $b_1\in B(r_1)$ and
with $\rho_1:=\rho(r_1)$
we find a domain $U_1\subset D(b_1,\rho_1)$
which is mapped
bijectively onto $Q(\log|f(b_1)|)$ by a branch $\phi_1$ of the logarithm
of $f$.

Inductively we thus obtain sequences $(r_k)$, $(\rho_k)$, $(b_k)$,
$(U_k)$ and $(\phi_k)$ satisfying
 $$r_{k}\in
\left[\,\left|f(b_{k-1})|,2|f(b_{k-1})\right|\,\right]\setminus E
 \subset [2r_{k-1},\infty)\setminus E,$$
 $\rho_k=\rho(r_k)$ and $b_k\in B(r_k)$
such that $U_k$ is a subdomain of $D(b_k,\rho_k)$  and $\phi_k$ is a
branch of the logarithm of $f$ which has the property that
 $\phi_k:U_k\to Q(\log|f(b_k)|)$
is bijective.

For large $r_0$ we have
$R_4(r_k)\leq 5 r_k/4 \leq 5|f(b_k)|/2$
for all $k$ and thus
\begin{equation} \label{A54}
\begin{aligned}
D(b_k,M \rho_k)
&\subset
\operatorname{ann}\left(r_k,R_4(r_k)\right)
\\ &
\subset
\operatorname{ann}\left(|f(b_{k-1})|,\frac52 |f(b_{k-1})|\right)
\subset
\exp Q(\log|f(b_{k-1})|)
\end{aligned}
\end{equation}
by~\eqref{p4xy} and~\eqref{4f}.
Hence there exists a branch $L$ of the logarithm
which maps $D(b_k,M \rho_k)$ into $Q(\log |f(b_{k-1})|)$.
Then $\psi_k:= \phi_{k-1}^{-1}\circ L$
is a branch of the inverse function of $f$ which maps
$D(b_k,M\rho_k)$ into $U_{k-1}$.

We put
\[
V_k:=\left(\psi_1\circ\psi_2\circ\ldots\circ\psi_k\right)\left(
\overline{D}(b_k,\rho_k)\right).
\]
Then $V_k$ is compact and $V_{k+1}\subset V_k$. Thus
$\bigcap_{k=1}^\infty V_k\neq\emptyset$. We will show that this
intersection contains only one point.

 In order to do so we note that since
$\psi_k:D(b_k,M\rho_k)\to U_{k-1}$ is univalent and $U_{k-1}\subset
D(b_{k-1},\rho_{k-1}) \subset D(\psi_k(b_k),2\rho_{k-1})$,
Koebe's one quarter
theorem implies that $2\rho_{k-1}\geq M\rho_k|\psi_k'(b_k)|/4$ and
thus $|\psi_k'(b_k)|\leq 8\rho_{k-1}/(M\rho_k)$.
The Koebe distortion theorem, applied with $\lambda=1/M$,
now yields
$$
\sup_{z\in \overline{D}(b_k,\rho_k)}|\psi_k'(z)|
\leq
\frac{1+\lambda}{(1-\lambda)^3} |\psi_k'(b_k)|
=\frac{M^2(M+1)}{(M-1)^3} |\psi_k'(b_k)|
\leq \frac{8M(M+1)}{(M-1)^3} \frac{\rho_{k-1}}{\rho_k}.
$$
Choosing $M=20$ we obtain
$$
\sup_{z\in \overline{D}(b_k,\rho_k)}|\psi_k'(z)|
\leq \frac12 \frac{\rho_{k-1}}{\rho_k}.
$$
If $K\subset \overline{D}(b_k,\rho_k)$ is compact, we thus have
$$
\operatorname{diam} \psi_k(K)
\leq \frac12 \frac{\rho_{k-1}}{\rho_k}
\operatorname{diam} K,
$$
where $\operatorname{diam} K$ denotes the diameter of~$K$.
Hence
$$
\frac{\operatorname{diam} \psi_k(K)}{\rho_{k-1}}\leq \frac12 \frac{\operatorname{diam}
K}{\rho_k}.
$$
Inductively we obtain
$$
 \frac{\operatorname{diam} V_k}{\rho_0}
 =\frac{\operatorname{diam}
\left( \psi_1\circ\psi_2\circ\ldots\circ\psi_k\right)
\left(\overline{D}(b_k,\rho_k)\right)}{\rho_{0}}
\leq\frac{1}{2^k} \frac{\operatorname{diam} \overline{D}(b_k,\rho_k)}{\rho_k}
= \frac{1}{2^{k-1}}
$$
and thus
\begin{equation}\label{4k}
\lim_{k\to\infty} \operatorname{diam} V_k = 0
\end{equation}
so that
$$
\bigcap_{k=1}^\infty V_k=\{z_0\}
$$
 for some~$z_0$.

It follows from the definition of $V_k$ and~\eqref{p4xy} that
$$
f^k(V_k)=\overline{D}(b_k,\rho_k)\subset \operatorname{ann}\left(r_k,R_4(r_k)\right)
$$
and hence that $z_0\in I(f)$.
Moreover,
$$
\begin{aligned}
f^{k+1}(V_k)
&= f\left(\overline{D}(b_k,\rho_k)\right)
\supset f(U_k) \\
&=
\operatorname{ann}\left(\frac{1}{e}|f(b_k)|, e|f(b_k)| \right)
\supset
\operatorname{ann}\left(\frac{1}{e}r_{k+1},\frac{e}{2} r_{k+1} \right)  .
\end{aligned}
$$
As $F(f)$ does not have multiply connected components,
$$
\operatorname{ann}\left(\frac{1}{e}r_{k+1},\frac{e}{2} r_{k+1} \right)
\cap
J(f)\neq\emptyset
$$
 for large~$k$. Since $J(f)$ is completely invariant, we
conclude that $V_k$ intersects $J(f)$, and
since $J(f)$ is  closed, this yields that $z_0\in I(f)\cap
J(f)$.

In order to estimate the upper box dimension of $I(f)
\cap J(f)$, we note that in
the above construction of the sequences $(r_k)$, $(\rho_k)$,
$(b_k)$, $(U_k)$ and $(\phi_k)$ we have $m(r_k)$ choices
for the point $b_k\in B(r_k)$ such that the disks of radius
$\rho_k$ around these points are pairwise disjoint, with $m(r_k)$
satisfying~\eqref{4g}.
We fix $k\in\mathbb N$, choose
$r_j$, $\rho_j$, $b_j$, $U_j$ and $\phi_j$ for
$0\leq j\leq k-1$  as before
and denote by
$b_k^\nu$
 choices of $b_k$ with the
above property,
with $1\leq \nu\leq m_k:=m(r_k)$

In other words, we take
$r_k\in [|f(b_{k-1})|,2|f(b_{k-1})|]\setminus E$
and $\rho_k:=\rho(r_k)$
as before and choose $b_k^\nu\in B(r_k)$, where $1\leq \nu\leq m_k$,
such that
\begin{equation}\label{pwd}
D(b_k^i,\rho_k)\cap D(b_k^j,\rho_k)=\emptyset
\quad\text{for}\ i\neq j.
\end{equation}
Then for $1\leq \nu\leq m_k$ there exists a
branch $\psi_k^\nu:D(b_k^\nu,2 \rho_k)\to U_{k-1}$ of the inverse
function of $f$
that is of the form $\psi_k^\nu=\phi_{k-1}^{-1}\circ L_\nu$ for
some branch  $L_\nu$ of the logarithm which maps $D(b_k^\nu,2 \rho_k)$
into $Q(\log|f(b_{k-1})|)$.
For $a\in \mathbb R$ we  put
\[
P(a):=
\left\{z\in\mathbb C:\, 0\leq \operatorname{Re} z-a\leq  \log \frac52,\,
|\operatorname{Im} z|\leq  \frac32 \pi\right\} .
\]
In view of~\eqref{A54}
we may actually assume that
$L_\nu$ maps $D(b_k^\nu,2 \rho_k)$ into
the compact subset
$P(\log|f(b_{k-1})|)$
of $Q(\log|f(b_{k-1})|)$.
Since $\phi_{k-1}^{-1}$ is univalent in $Q(\log|f(b_{k-1})|)$
we thus conclude that
there exists an absolute constant $\alpha>1$ such that
\begin{equation}\label{alphaphi}
\frac{1}{\alpha}\leq
\frac{|(\phi_{k-1}^{-1})'(\zeta)|}{|(\phi_{k-1}^{-1})'(z)|}
\leq \alpha
\quad\text{for}\
\zeta,z\in P(\log|f(b_{k-1})|).
\end{equation}
An explicit upper bound
 for $\alpha$ could be determined from the Koebe
distortion theorem, but we do not need  such an estimate.
Put
\[
\Lambda_k^\nu:=\psi_1\circ\psi_2\circ \ldots\circ \psi_{k-1}\circ\psi_k^\nu
=\psi_1\circ\psi_2\circ \ldots\circ \psi_{k-1}\circ\phi_{k-1}^{-1}\circ L_\nu.
\]
Since $\psi_1\circ\psi_2\circ \ldots\circ \psi_{k-1}$ is univalent in
$D(b_{k-1},2\rho_{k-1})$,
we deduce from~\eqref{alphaphi}  and the Koebe distortion theorem
that there exists $\beta>1$ such that
\begin{equation}\label{betaLambda}
\frac{1}{\beta}\leq
\frac{|(\Lambda_k^\nu)'(b_k^\nu)|}{|(\Lambda_k^1)'(b_k^1)|}
\leq \beta
\quad\text{for}\
1\leq \nu\leq m_k.
\end{equation}

We put
\[
V_k^\nu:=\Lambda_k^\nu\left(\overline{D}(b_k^\nu,\rho_k)\right)
\quad\text{and}\quad
v_k^\nu:=\Lambda_k^\nu(b_k^\nu).
\]
As above we see that each $V_k^\nu$
contains a point of
$I(f)\cap J(f)$.

Since $\Lambda_k^\nu$ is univalent in $D(b_k^\nu,2\rho_k)$ we
deduce fron the
Koebe distortion theorem (with $\lambda=1/2$) that
\begin{equation}\label{Vkl}
\overline{D}\left(v_k^\nu,\frac49 \sigma_k^\nu\right)
\subset
V_k^\nu\subset \overline{D}\left(v_k^\nu,4 \sigma_k^\nu\right)
\end{equation}
where
\[
\sigma_k^\nu:=\left|(\Lambda_k^\nu)'(b_k^\nu)\right|\rho_k
=\frac{\rho_k}{\left|(f^k)'(v_k^\nu)\right|}.
\]
We put $\sigma_k:=\sigma_k^1$.
It follows from~\eqref{4k} and~\eqref{Vkl} that
$$\lim_{k\to\infty}\sigma_k=0.$$
Using~\eqref{betaLambda}
and~\eqref{Vkl}
we obtain
\begin{equation}\label{sizeWkl}
\overline{D}\left(v_k^\nu,\frac{4}{9\beta} \sigma_k\right)
\subset
V_k^\nu\subset \overline{D}\left(v_k^\nu,4\beta \sigma_k\right)
\end{equation}

Fix a square of sidelength $\sigma_k$
centered at a point~$c$ and denote by
$N$ the cardinality of the set of all $\nu\in\{1,\dots,m_k\}$ for
which $V_k^\nu$ intersects this square.
It follows from~\eqref{sizeWkl} that if
if $V_k^\nu$ intersects  this square,
 then $V_k^\nu\subset D(c,(8\beta+1)\sigma_k)$.
On the other hand, \eqref{sizeWkl} also says
that $V_k^\nu$ contains a disk of
radius $4\sigma_k/(9\beta)$. Since the
$V_k^\nu$ have pairwise disjoint interior by~\eqref{pwd},
we obtain
\[
N\pi
\left(\frac{4\sigma_k}{9\beta} \right)^2
\leq
\pi
\left( (8\beta+1)\sigma_k\right)^2.
\]
Thus
$N\leq N_0:=\lfloor 81\beta^2(8\beta+1)^2/16\rfloor$.

We now put a grid of sidelength $\sigma_k$ over $U_0$.
Then each square of this grid can intersect at most
$N_0$ of the $m_k$ domains $V_k^\nu$.
Recalling that each of the domains $V_k^\nu$ contains
a point of $I(f)\cap J(f)$
we see that at least $m_k/N_0$ squares of our grid
intersect $I(f)\cap J(f)$.
We conclude that
\begin{equation}\label{dimmk}
\operatorname{\overline{dim}_B}(U_0\cap I(f)\cap J(f))
\geq
\limsup_{k\to\infty}
\frac{\log (m_k/N_0)}{-\log \sigma_k}.
\end{equation}
By~\eqref{4g} we have
\begin{equation}\label{boundmk}
m_k=m(r_k)\geq T(r_k,f)^{2-7\delta}.
\end{equation}
It remains to estimate~$\sigma_k$.
In order to
do so
we note that
\begin{equation}\label{4l}
(f^k)'(v_k^1)=\prod_{j=0}^{k-1}f'(f^j(v_k^1)).
\end{equation}
Since
\[
f^j(v_k^1)\subset U_j\subset D(b_j,\rho_j)
\]
it follows from~\eqref{3e1}  that
\[
\left|\frac{f'(f^j(v_k^1))}{f(f^j(v_k^1))}\right|
\leq \frac{T(r_j,f)^{1+3\delta}}{r_j}.
\]
This yields
\begin{equation}\label{4m}
\left|
f'(f^j(v_k^1))\right|
\leq \frac{T(r_j,f)^{1+3\delta}}{r_j}\left|
f^{j+1}(v_k^1)\right|
\leq e \frac{T(r_j,f)^{1+3\delta}}{r_j} r_{j+1}.
\end{equation}
We deduce from~\eqref{4l} and ~\eqref{4m} that
\[
\left|(f^k)'(v_k^1)\right| \leq
 \prod_{j=0}^{k-1}
e \frac{T(r_j,f)^{1+3\delta}}{r_j} r_{j+1}
=e^k  \frac{r_{k}}{r_0}
\left(\prod_{j=0}^{k-1} T(r_j,f)\right)^{1+3\delta}
\]
Using the definition of $\rho_k$ and $\sigma_k$ we thus have
\begin{equation}\label{4n}
\sigma_k\geq
\frac{r_0}{e^k T(r_k,f)^{1-2\delta} \left( \prod_{j=0}^{k-1}
T(r_j,f)\right)^{1+3\delta}}
\end{equation}
By construction, we have
\[
\log r_{j+1}\geq \log|f(b_j)|\geq \frac12\log M(r_j,f)
\]
for $0\leq j\leq k-1$.
Given a large positive number $q$, we deduce from~\eqref{1a} that
if $r_0$ is sufficiently large, then
\[
\log M(r_{j+1},f)\geq \left(2\log r_{j+1}\right)^{q+1}
\]
for $0\leq j\leq k-1$.
Combining the last two estimates with~\eqref{3b1}
we find that
\[
T(r_j,f)\leq \log M(r_j,f)\leq 2\log r_{j+1}
\leq (\log M(r_{j+1},f))^{1/(q+1)}
\leq T(r_{j+1},f)^{1/q}
\]
for large $r_0$.
We conclude that
\begin{equation}\label{4o}
\prod_{j=0}^{k-1} T(r_j,f)
\leq T(r_k,f)^{\tau},
\end{equation}
where
\[
\tau:=\sum_{j=1}^k\left(\frac{1}{q}\right)^j
\leq \sum_{j=1}^\infty \left(\frac{1}{q}\right)^j
=\frac{1}{q-1}.
\]
For large $q$ we have $\tau(1+3\delta)\leq\delta$ and
thus
\[
\left( \prod_{j=0}^{k-1}
T(r_j,f)\right)^{1+3\delta}
\leq T(r_k,f)^\delta.
\]
We can also deduce from~\eqref{4o} that
\[
T(r_k,f)^\delta
\geq e^k
\]
if $r_0$ is chosen large enough. Combining the last two estimates
with~\eqref{4n}, and assuming that $r_0\geq 1$, we conclude that
 $\sigma_k\geq 1/T(r_k,f)$.

Together with~\eqref{dimmk} and~\eqref{boundmk} we thus find
that
\[
\operatorname{\overline{dim}_B}(U_0\cap I(f)\cap J(f))
\geq\limsup_{k\to\infty}\frac{(2-7\delta)\log T(r_k,f)-\log N_0}
{\log T(r_k,f)}
=2-7\delta.
\]
Since $\delta>0$ was arbitrary, we obtain~\eqref{1a2} for $U=U_0$.
We may assume that the closure of
$U_0$ does not intersect the exceptional set.
As mentioned in the introduction, Theorem~\ref{thm1} now follows.

\begin{remark}
Theorem~B has been extended to meromorphic functions with finitely
many poles~\cite{Rippon06} and in fact to meromorphic functions with a
logarithmic tract~\cite[Theorem~1.4]{BRS1}. It is conceivable that
our result admits similar extensions.
\end{remark}

\section{The minimum modulus of entire functions with multiply connected
Fatou components}
Zheng~\cite{Zheng06} proved that if  the Fatou set of
a transcendental entire
function $f$ has a multiply connected component~$U$,
then there exist sequences $(r_k)$ and $(R_k)$
satisfying
$\lim_{k\to\infty}r_k=\lim_{k\to\infty}R_k/r_k= \infty$
such that $\operatorname{ann}(r_k,R_k)\subset f^k(U)$
 for large $k$.
It is shown in~\cite{BRS} that
one can actually take $R_k=r_k^c$ for some $c>1$ and this is then
used to show that if $F(f)$ has a multiply connected component,
then there exists $C >0$ such that
\begin{equation}\label{5x}
\log L(r,f)\geq \left(1-\frac{C}{\log r}\right)\log M(r,f)
\end{equation}
on some unbounded sequence of $r$-values.

Using Zheng's result~\cite{Zheng06} instead of~\cite{BRS}  yields
the following proposition referred to in the introduction.
 We include its short proof for completeness.

\begin{proposition}\label{zhengmod}
Let $f$ be a transcendental entire function for which $F(f)$
has a multiply connected component. Then~\eqref{1y} holds.
\end{proposition}
\begin{proof}
Zheng~\cite[Corollary~1]{Zheng06} used hyperbolic geometry to prove
that the hypothesis of the proposition implies~\eqref{1z}.
 We will use the same idea and denote the hyperbolic
 distance of two points $a,b$ in a hyperbolic domain $V$ by
 $\lambda_V(a,b)$; see, e.g., \cite[Section 2.2]{McM94} for
the properties of the hyperbolic metric that are used.

Let $U$ be a multiply connected component of $F(f)$ and let $(r_k)$
and $(R_k)$ be as in Zheng's result mentioned above.
  Put $s_k=\sqrt{R_kr_k}$ and $U_k=f^k(U)$.
Choose $|a_k|=|b_k|=s_k$ such that
$|f(a_k)|=L(s_k,f)$ and $|f(b_k)|=M(s_k,f)$. Since
$R_k/r_k\to\infty$ we easily see that $\lambda_{U_k}(a_k,b_k)\leq
\lambda_{\operatorname{ann}(r_k,R_k)}(a_k,b_k)\to 0$. Thus
$\lambda_{U_{k+1}}(f(a_k),f(b_k))\to 0$. Since $0,1\notin U_{k+1}$
for large $k$
 we obtain
\begin{equation}\label{5a}
\lambda_{\mathbb C\setminus\{0,1\}}(f(a_k),f(b_k))\to 0
\end{equation}
as $k\to\infty$. As the
 density $\rho_{\mathbb C\setminus\{0,1\}}(z)$ of the hyperbolic metric in
$\mathbb C\setminus\{0,1\}$ satisfies $\rho_{\mathbb C\setminus\{0,1\}}(z)\geq
 c/(|z|\log|z|)$ for some $c>0$ and large~$|z|$,
 we obtain
\begin{equation}\label{5b}
\lambda_{\mathbb C\setminus\{0,1\}}(f(a_k),f(b_k))
\geq c\int_{|f(a_k)|}^{|f(b_k)|}\frac{dt}{t\log t}
 =c\log\frac{\log M(s_k,f)}{\log L(s_k,f)}.
\end{equation}
Now~\eqref{1y} follows from~\eqref{5a} and~\eqref{5b}.
\end{proof}

\begin{remark}
An alternative way to deduce Proposition~\ref{zhengmod} from Zheng's
result is via Harnack's inequality~\cite[p.~14]{Ransford}.
 This method was used by Hinkkanen~\cite[Lemma~2]{Hinkkanen94}
 and Rippon and Stallard~\cite[Lemma~5]{RS-slowescape},
 and it is also used in~\cite{BRS}.

With the notation as in the above proof, put $t_k=
\log s_k=\log \sqrt{R_k/r_k}$ and define
$$u_k:\{z\in\mathbb C: |\operatorname{Re} z|<t_k\}\to\mathbb R, \quad u_k(z)=\log|f(s_ke^z)|.
$$
We may assume that $|f(z)|>1$ for $z\in \operatorname{ann}(r_k,R_k)\subset f^k(U)$
so that $u_k$ is a positive harmonic function.
 Choose $y_1,y_2\in\mathbb R$ with
$|y_1-y_2|\leq\pi$ such that $u(iy_1)=\log L(s_k,f)$ and
$u(iy_2)=\log L(s_k,f)$.
 By Harnack's inequality we have
$$
u(iy_2)\leq \frac{t_k+\pi}{t_k-\pi} u(iy_1)=(1+o(1))u(iy_1)
$$
as $k\to\infty$, and~\eqref{1y} follows.
\end{remark}
\begin{remark}\label{rem2}
It follows from a result of  Fenton (\cite{Fenton80}, see
also~\cite{Chyz}) that if
\begin{equation}\label{6a}
p:=\liminf_{r\to\infty}\frac{\log\log M(r,f)}{\log\log r}<\infty
\end{equation}
and $\varepsilon>0$, then
\begin{equation}\label{6x}
\log L(r,f)\geq \log M(r,f) -  (\log r)^{p-2+\varepsilon}
\end{equation}
on some sequence of $r$-values tending to $\infty$. Hence
\begin{equation}\label{6c}
\log L(r,f)\geq \log M(r,f) - \frac{\log M(r,f)}{(\log r)^{2-2\varepsilon}}
\end{equation}
on such a sequence. Choosing $\varepsilon<1/2$ we see that if
\begin{equation}\label{6b}
\lim_{r\to\infty}\left(1-\frac{\log L(r,f)}{\log M(r,f)}\right)\log
r= \infty,
\end{equation}
 then~\eqref{6c} and hence~\eqref{6a} cannot hold and thus~\eqref{1a} holds.
The result of~\cite{BRS} quoted before Proposition~\ref{zhengmod}
shows that if~\eqref{6b} holds, then $F(f)$ has no multiply
connected components.
Thus we obtain the following consequence of Theorem~\ref{thm1}.
\begin{cor}\label{cor3}
Let $f$ be a transcendental entire function satisfying~\eqref{6b}.
Then  $\operatorname{dim_P}(I(f)\cap J(f))=2$.
\end{cor}
Finally we mention that it was actually shown
in~\cite{BRS} and~\cite{Fenton80} that~\eqref{5x}
and~\eqref{6x} hold on sets or $r$-values of a
certain size. This could be used to further strengthen
the statement of Corollary~\ref{cor3}.
\end{remark}

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Nerd. Artist. CrossPlayer. Friend. Gamer.
Drawing, Reading, Writing, Crossplaying, Swimming, Voice Acting, Dancing Silly, Quoting Everything.
If you want to know about me, get to know me, then you'll see...I'm a nerd, an artist, someone who likes laughing, someone who likes watching Disney movies constantly, someone who cries at every movie, someone who dies whenever Sasuke speaks. I am the voice in the clouds, the unoffical voice of Naruto. I am the reader who will read without putting the book down, will yell at you for dog-earring my books and then hug you. I say I love you to everyone because I do. I am a free spirit, I love the First Amendment, I enjoy playing video games and crying if I lose. Don't Judge. What's up with you?


text:
Proton Exchange Membrane Fuel Cell – PEMFC
Polymer Electrolyte Fuel Cell – PEFC
Commercial module NEXATM by Ballard (1.2 kW)
They use as electrolyte a polymeric membrane, based on perfluorinated sulfonic acids, which must be kept completely hydrated during the operation to favor the conduction of the protons. This requirement limits the operating temperature below 100 ºC and is critical to obtain good efficiency. As water is the only liquid used, corrosion problems are minimal. The commonly used catalyst is platinum supported on carbon, or bifunctional metallic electrocatilizadores with base platinum and a metallic element like ruthenium, that makes the electrode more resistant to the poisoning by carbon monoxide when the fuel used is not pure hydrogen. They are capable of working at high current densities, with a rapid response to variable power demands, which makes them very suitable for transport applications; in fact, its higher level of development is due to the interest aroused in the automobile industry.
They require hydrogen of high purity (greater than 99.9999%) at the anode and oxygen or air at the cathode. Any other fuel (ethanol, natural gas, gasoline and derivatives … etc.) Has to go through a previous process of reforming it to convert it into hydrogen, except methanol that can be used directly in the so-called direct methanol fuel cells ( DMFC, «Direct Methanol Fuel Cell»), which are also polymeric electrolyte. The current challenges are to reduce the cost and increase efficiency, which translates into reducing the platinum load, reducing the thickness of the catalytic layer and optimizing the dispersion of the catalyst, improving the performance of the polymer membrane (increase the ionic conductivity and water retention capacity) and find an alternative material to the graphite for the bipolar plate of high electrical and thermal conductivity, resistant to corrosion, lighter and cheaper. In direct methanol fuel cells (DMFC) the challenges are replacing the membrane with another that allows working at a temperature above 130 ºC and that does not present a crossover problem (passage of the anodic reactant to the cathode compartment through the membrane) and find a more active anodic catalyst for the direct oxidation of methanol.
PEMFC del modelo NECAR 5 de Daimler-Chrysler
In 1962 General Electric (GE), under the direction of Thomas Grubb and Leonard Niedrach developed the first polymer membrane electrolyte, which is the first milestone in this type of fuel cells. The GE stacks replace the batteries used in the previous Mercury Project in the Gemini Project. A new polymer formulation developed by Grot in 1972 improved the characteristics and durability of the electrolyte and, later, in 1986, Raistrick made major improvements in manufacturing processes. In 1981 Ballard was founded, which began the development of its models in 1983 and in 1989 produced the first conceptual demonstration units. Between 1992 and 1994, they built functional prototypes and in 2001 launched its first NexaTM commercial module.
Thomas Grubb (left) and Leonard Niedrach (right) next to a polymer membrane fuel cell.
Roy Mushrush, at the time, director of the Department of Energy Conversion of GE, showing part of the 96 single cells of the stack of the image. The Gemini Protecto ships incorporated three of these modules.

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Consider this to be a resurrection day! The death of Jesus is the saddest part God's story. But what comes next? The resurrection! The happiest part! Make today a happy day! Choose to be happy, happy, happy! Send a message to 10 people today reminding them that God is good or that God loves them. Lift some spirits today!
Pray that each of us will have a renewed desire to pursue spiritual habits in their lives. Pray for our nurses.
Using a notebook, journal your thoughts about today's study. What insights did you take away from our time together? What does the resurrection of Jesus mean to you? Have you been raised up with Jesus in baptism? Will you raise up with Jesus on the last day?


text:
Jane's Urgent Christmas Appeal
My name is Jane O'Sullivan, and I am writing with a request that's very dear to my heart. I would also like to take the chance to tell you about my husband, Donie.
I met Donie in 1966 in Australia, where I'm from. We met in a café late one night after a dance and got married in Melbourne a year later. Donie said we'd just go back to Ireland to start a farm and see if I liked it or not. It's been 47 years, so I must have liked it all right!
Since the day I met him, Donie and I have had an incredibly strong love and companionship. Donie loved his family and his farm, and could learn just about any trade he needed. He built our first house on his own. As he'd tell you himself, he's a chancer.
Then a few years ago, things started to change. Donie seemed to be living only in the past – like the present wasn't of great importance to him. The signs were small, but the children and I noticed. Our GP suggested Donie go for tests.
I'll never forget the day Donie was diagnosed.
After he'd had an MRI, he walked straight in and said to the doctor, 'Do I have Alzheimer's?' It was the first time I'd ever heard him use the word. The doctor said, 'Yes, you have.' It was an absolutely devastating moment. With that sentence, we knew our lives had changed unbelievably.
We'd found a problem Donie couldn't fix, no matter how hard he tried. You know things have changed for the person with Alzheimer's, but your own life changes unbelievably, too. The whole implication of being a wife changes completely. We were retired, and we had plans of what we were going to do in our retirement. That all fell apart.
We live in Kerry and our children all now live far away, some even as far as Australia. They are always in touch and come home as often as they possibly can but it is difficult. In rural areas like ours there is very little support or help available for people living with dementia.
Without the help of The Alzheimer Society of Ireland (ASI), I don't know how Donie or I could have coped.
It's been five years since Donie's diagnosis now, and I've been his main carer during that time. This can be extremely challenging, as anyone who's cared for someone with dementia knows.He has good and bad days, and I try to let him be as independent as he can, but he needs constant supervision. He can be disorientated, emotional and on edge. Sometimes Donie's aware of it, and he'll say 'I don't know what's happening. I don't know why I'm like this.' It's been a horrible experience.
I sometimes feel as though I'm living two lives, mine and his both.
One of the main struggles has been loneliness. I don't think anything can prepare you for the isolation. We live a kilometre from our nearest neighbour, and over the years, as our situation has become more complicated, friends have fallen away.
Donie will sit in his chair, and I'll sit in mine beside him, and there'll be no conversation between us. All of that is gone. I feel that I've lost my husband, and I miss him.
As Christmas comes around this year, I find myself dreading it. Donie used to love Christmas. He was always in charge of bringing home a tree, but his work on the farm was so constant that he'd leave it to the last minute. I remember one year, he brought home a patchy one, and the kids all complained.
Donie said, 'I'll fix it!' and started cutting branches off from the thicker places and tying them round the bits that were spare, and soon we had a tree everybody was happy with and we all started decorating.
I don't know what is going to happen this Christmas. We can't have a big family Christmas at our house this year because Donie just can't cope. My daughter Grainne will have it at hers instead. I hope myself and Donie will be able to go. But I'm not sure. It might depend on how he is on the day.
I first contacted the Alzheimer's Helpline immediately after Donie was diagnosed, and it was the best thing I could have done.
My head was in such a whirl, thinking 'What does this mean? How am I going to deal with this? How can I get help?' It was so important to be able to talk to someone who understood and could help.
The ASI has provided carers to come into our home so I can go out to the shop or just take some time for myself. They have become more than friends – they're part of the family now. They've promised that they'll be with us for the long haul, and that means so much. Without them, I would never get past the door.
The ASI has so many resources for families like mine, and over the years, they've made me feel like a weight has been lifted from my shoulders. From Dementia Advisers to the helpline, from community events like coffee mornings to the practical and emotional support of the carers network, everything they do makes you feel less isolated and alone.
I was recently able to take a carers' course offered by The ASI, and it was just wonderful to talk to people who are going through similar situations.
The ASI also made it possible for Donie to attend eight weeks of cognitive stimulation therapy. This was so beneficial for everyone. During this hour, all of the carers got together for a coffee and chat, and it was something we all looked forward to.
The support we have received has made such a difference. That's why, in September of this year, my son Dominic came home for Australia and together we walked the Camino to help raise funds for The ASI.
But I know that there are still many other families who need help.
That's why this Christmas Appeal is so vital for The Alzheimer Society of Ireland and for families like mine all over Ireland.
Living in rural Ireland is especially difficult for families caring for someone living with dementia. That's why The ASI is so wonderful. Because the vital funds that you help to raise are used to support services in local communities around Ireland.
Amidst everything, there have still been some good memories. Donie was able to attend both of his son's weddings, which we weren't sure would be possible. The most recent one, last October, was such an undertaking that when I got him into the church, I just collapsed into a pew and cried. But when our son John saw that we'd made it, he had the biggest smile. A few weeks ago, Donie also got to meet his two-week-old grandson in Cork. These moments have been so joyful.
It's difficult to imagine what the next years will look like – but I know we will manage with the invaluable support of The ASI, made possible by the generosity of people like you. We will keep taking one day at a time.
By giving what you can today, you are making every single one of these days better – or even possible.
This is a true gift, and it's one that I'm grateful for every day.
Thank you so much for taking the time to read this letter. And thank you for everything you have done for The Alzheimer Society of Ireland and for families like mine.
And may I wish you and your family a very happy Christmas.
From Donie and me.
Jane O'Sullivan
P.S. The support my husband and I have received from The Alzheimer Society of Ireland has been crucial. Please consider a gift this Christmas to other families to provide vital care and support along their journey.
You can support The ASI with a donation HERE
Alzheimer National Helpline
The Alzheimer Society of Ireland National Helpline is open six days a week Monday to Friday 10am–5pm and Saturday 10am–4pm on 1800 341 341 or helpline@alzheimer.ie
Helpline Christmas Hours
Monday, 23rd Dec – 10am to 4pm
Tuesday, 24th Dec – closed
Wednesday, 25th Dec – closed
Thursday, 26th Dec – closed
Friday, 27th Dec – 10am to 4pm
Saturday, 28th Dec – closed
Monday, 30th Dec – 10am – 5pm
Tuesday, 31st Dec – 10am – 5pm
Wednesday, 1st Jan – closed
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The 8-9 Abbeville Wildcats snatched the lead late in the game in an 11-10 victory over Kaplan on Tuesday.
Abbeville was behind 7-1 in the sixth inning.
The game was tied at ten with Abbeville batting in the top of the ninth when Kyle Choate singled on the first pitch of the at bat, scoring Josiah Chambers for the win.
Hunter Duhon led Abbeville to victory by driving in four runs. He went 3-for-4 at the plate. Duhon drove in runs on a grand slam in the sixth inning.
After Abbeville scored one run in the top of the fourth inning, Kaplan answered with one of their own.
Abbeville scored when Josiah Chambers singled on the first pitch of the at bat, scoring one run. Kaplan then answered when Brayden Bass hit a solo homer.
Abbeville Wildcats put up nine runs in the sixth inning. The offensive onslaught came from singles by Chambers, Hogan Landry, Jakarri Grogan, and Channing Spears, a walk by Choate, and a home run by Duhon.
Kevin Linton pitched for Abbevile. He allowed nine hits and 10 runs over eight innings, striking out eight.
Abbeville scattered 17 hits in the game. Chambers, Duhon, Grogan, Spears, Garrett Hebert, and Choate each managed hits for Abbeville.
Grogan, Duhon, and Chambers each managed three hits to lead Abbeville.
Hitters for Kaplan were Lane Marceaux, Noah Gaspard, Brayden Bass, Kolby Bernard, Nathan Sistrunk and Zack Stelly.
Drew Winch and Bernard combined to throw nine innings for KHS.


Anhui, Gansu, Guangxi, Hebei, Heilongjiang, Henan, Hubei, Hunan, Jiangsu, Jiangxi, Liaoning, Nei Mongol, Qinghai, ?Shaanxi, Shandong, Shanxi, Taiwan, ?Zhejiang [Japan, Korea, Mongolia, Russia; Europe].
Flora of China Vol. 24: 226 in eFloras.org, Missouri Botanical Garden. Accessed Nov 12, 2008.
Rhizome terete, 5--14 mm thick. Stem arching, 20--50(--100) cm, glabrous, angled. Leaves 7--12, alternate; petiole short; leaf blade abaxially glaucous, elliptic to ovate-oblong, 5--12(--20) × 3--6(--8) cm, often smooth, sometimes papillose-scabrous on veins, apex acuminate and obtuse tipped. Inflorescences 1--4(--8)-flowered; peduncle usually 1--1.5 cm; bracts small or absent. Flowers pendulous; pedicel 5--10(--20) mm. Perianth yellowish green to white, cylindric to campanulate-cylindric, 1.3--2(--2.5) cm; lobes ca. 3 mm. Filaments filiform, smooth or verruculose; anthers ca. 4 mm. Ovary 3--4 mm. Style 1--1.4 cm, included. Berries blue-black, 7--10(--12) mm in diam., 7--9-seeded. Fl. May--Jun, fr. Jul--Sep. 2 n = (18*), 20*, (21, 22*, 23, 26, 28, 29, 30, 40*).
Convallaria odorata Miller, Gard. Dict., ed. 8, Convalla-ria no. 4. 1768; C. polygonatum Linnaeus; Polygonatum hondoense Nakai ex Koidzumi; P. japonicum C. Morren & Decaisne; P. langyaense D. C. Zhang & J. Z. Shao; P. maximowiczii F. Schmidt; P. odoratum f. ovalifolium Y. C. Chu et al.; P. officinale Allioni; P. officinale var. papillosum Franchet; P. planifilum Kitagawa & Hir. Takahashi; P. quelpaertense Ohwi; P. simizui Kitagawa; P. thunbergii C. Morren & Decaisne; P. vulgare Desfontaines.
Forests, shaded slopes; 500--3000 m.


text:
BRIGHTBURN arrives on Digital 8/6 and Blu-ray & DVD 8/20
The Walking Dead S9 arrives on Blu-ray, DVD and Digital 8/20
Lost in Translation 211: Cortex/Cortex Plus
In previous examinations of works adapted to gaming, Lost in Translation focused on just one original work being turned into a game. This time around, multiple original works are being adapted into just one game system, the Cortex and its successor, Cortex Plus.
The ideal works to adapt as tabletop role-playing games provide a larger setting, one where the original provides for a larger setting than what the main characters there experience. This is the case in three previous game adaptations examined, Star Trek, Star Wars, and 007, where players can take on similar roles as the main characters in both franchises. Even the Buffy RPG could delve into both past and future, allowing players to take up the mantle of the Slayer in a different time. At the same time, if players want, they can still take the roles of the existing characters. RPGs need to keep that flexibility and allow for new characters with similar capabilities as existing ones.
The Cortex system debuted in its early form with the Sovereign Stone Game System, which was based on Margaret Weis and Tracy Hickman's Sovereign Stone novels. Since then, the Cortex system was refined and was used as the base for a number of licensed games published by Margaret Weis Productions – Serenity, based on the movie*; Battlestar Galactica, based on the rebooted series; Demon Hunters, based on videos by Dead Gentlemen Productions; and Supernatural, based on the CW TV series. A stand-alone version of the Cortex rules, the Cortex System Role Playing Game, came out in 2008, after Serenity, Battlestar Galactica, and Demon Hunters had been released.
The works adapted share some common features. Both Serenity and Battlestar Galactica are science fiction set far from Earth and have characters who spend much of their time in spaceships. Demon Hunters and Supernatural both deal with supernatural threats. All four original works have a devoted fan base, one that is likely to play RPGs. At the same time, each of the original works has its own tone. Serenity, the movie, was based around a government cover-up that affected one of the core characters. Battlestar Galactica showed the last of humanity escaping a relentless enemy intent on exterminating every last human. Demon Hunters is a comedy. Supernatural is the story of the Winchester brothers fighting against destiny, Hell, and Heaven.
Fortunately, game mechanics don't always set the tone. While there are games where the mechanics were written to support the tone of the game, a more generic system can be adapted to the desired result. The Serenity RPG presents the game in terms of the players being the crew of their own small ship trying to make a living while staying true to themselves. Battlestar Galactica focused on survival in an environment that is inherently deadly, with the push to find the lost planet, Earth. Demon Hunters focused on comedy, with additional writing from Dead Gentlemen contributors and bonus orientation DVD. Supernatural placed the focus on the players being a small group of supernatural hunters that banded together. Each release added rules needed for the setting. Serenity and Battlestar Galactica had rules for starships, with Galactica having expanded rules for dogfighting in space. Demon Hunters added rules for creating demons and spending plot points to summon the Purple Ninja. Supernatural also had rules for creating the supernatural and exploiting their weaknesses.
What helped Cortex be flexible enough for the range of adapted works is its simplicity. The core mechanic involves using dice – the full range of regular polyhedrons – for attributes, skills, traits, and complications. Players roll the dice from appropriate attribute, skill, and, if any, trait against a difficulty number set by the GM, modified by dice from the character's applicable complications, again, if any. Skill lists can be modified by setting; where Pilot would be a given for a Colonial Warrior to have in Battlestar Galactica, a group of hunters in Supernatural might just have the one character who can fly a small plane. Adding setting-specific rules, such as details about the Cylons, builds on top of the existing core rules, allowing for specialization.
Cortex worked well for settings that focused on action. However, not all settings focus on action. Gaming has seen a movement to expand towards a more narrative-driven focus, moving away from the hobby's wargaming background. The intent is to tell stories, not chop down opponents. While Cortex might not have been able to take advantage, its successor, Cortex Plus, was developed to do just that.
Cortex Plus developed into three streams – Cortex Plus Drama, focusing on the relationships between characters; Cortex Plus Action, looked at what the characters did; and Cortex Plus Heroic, which combined the Drama and the Action. The core die mechanic remained but was heavily modified as needed. The game also added more details for characters, including character distinctions that could help or hinder depending on the circumstance. Traditional hit points fell by the wayside. Instead, characters could suffer from stress or complications imposed on them. If either got too high, the character would be forced out of the scene, either because of injuries, exhaustion, or escaping.
The first Cortex Plus game released was the Smallville Roleplaying Game, licensed from the Smallville TV series. The RPG used the Cortex Plus Drama system, reflecting how what the characters on the show were driven by their relationships with each other. Instead of rolling attribute and skill, players rolled value and relationship, adding in relevant resources and assets, and keeping the best two dice rolled. Players wouldn't always be rolling against the GM; sometimes, two characters could work at cross purposes, coming into conflict. Character creation was a group effort; relationships between player and non-player characters were set up in the first session. The main drawback with the game was that a large number of players meant that the relationship map grew complex, It also meant that all the players had to show up for the first session, something not all groups can handle. That said, the system reflected the show; the values, replacing attributes, went to what motivated the characters. Clark Kent's highest rated value, Justice, and lowest, Power, were true to his appearance on the show.
The first Cortex Plus Action game released was based on the TV series, Leverage, which was about a group of five con artists and thieves using their skills to help the average person who is being run over roughshod by the powerful and corrupt. Leverage used the same attributes that were in the Cortex games above, but instead of skills, characters had roles. The roles came directly from the TV series – Mastermind, Grifter, Hitter, Hacker, and Thief. Every character would have a key role, a minor role, and roles that would make things interesting for them. Each attribute and role had a die type assigned, which would be added to the player's dice pool when it came time to raise the stakes. Characters also had distinction, which could work for or against them in the dice pool. For example, if Parker, the thief of the Leverage crew, needed to break into an office on the 34th floor of an office tower, she could decide to repel down the side of the building from the roof. To do this, Parker's player would her Strength die, her Thief die, and could add her "Crazy" distinction and her "No – Really Crazy", giving her four dice, though only the highest two would be added together. Parker's player could decide that "Crazy" merited the usual d8 for the distinction, but then say that, because she's going down facing the ground, "No – Really Crazy" would add just a d4 and provide a plot point.
The Leverage RPG added rules to reflect the show's episode structure, which included flashbacks to show the characters setting up twists to defeat the Mark. The game allows for players to set up their own flashbacks, allowing them to get an advantage when needed. This allows the crew's thief to place a convenient smartphone in a drawer where the crew's hacker can grab it later without the players spending two hours of gaming working out all the contingencies before committing their heist. The Leverage RPG works to keep the story flowing; there's no real need to spend hours on a plan when the players just need to work out a general idea of what they want to do.
The next Cortex Plus game out was also the first to use the Heroic approach. The Marvel Heroic Roleplaying Game featured the characters from Marvel Comics and was the fourth licensed game based on the Marvel Universe. The game pulled in ideas from both Smallville and Leverage, adapted to reflect the needs of the new setting. Instead of Smallville's values and Leverage's attributes, Marvel Heroic used affiliations, or how well a character worked with others. There were three types of affiliation, each getting a different die type – Solo, for characters like Daredevil and Wolverine, who worked best on their own; Buddy, for characters like Spider-Man who often works with another hero; and Team, for the likes of Captain America and Cyclops, who work most often with a large group, such as the Avengers or the X-Men.
Marvel Heroic also uses the stress system from Smallville, again, modified. Instead of taking six different types of stress, Marvel Heroic characters only have three sources – physical, mental, and emotional. It is possible for Spider-Man to keep up his constant wisecracking to force an opponent to break down and give up, a result that the previous three Marvel-based games didn't have a mechanic for. Skills were replaced by specialties, though their use is similar. The mechanics remain the same; a player builds a dice pool and takes the best two results and add them together.
The most recent game from Margaret Weis Productions was the Firefly RPG. The license for the Serenity RPG had expired, but the company had worked with Fox to get the license for the TV series*. The Firefly RPG uses Cortex Plus Action, modified from its first incarnation as the Leverage RPG and taking some ideas from Marvel Heroic. Firefly has attributes, but only three; Mental, Physical, and Social. The skill list is shorter than Serenity's, but are broader. The game introduced rules for spaceships, since the TV series was set on one. Players could not only create their characters but their own ship, turning the vessel from a stat block into something that the players and their characters could care about. The game also modified the distinctions. Instead of just giving a bonus die for the player's dice pool, distinction had some extra mechanical bits that helped players distinguish their characters. The rule book also uses scenes from the TV series to illustrate how the mechanics work, giving players a way to follow the action.
With four licensed games using Cortex and another four using Cortex Plus, how did the adapting fare? At its core, Cortex is a simple, flexible system, in the same vein as the Cinematic Unisystem rules used by the Buffy RPG. This allowed the developers to tailor the mechanics to the adapted setting by changing skill lists and adding and removing talents and complications. It is possible for characters from one of the published games to be used in another; it would be odd to see a member of the Brotherhood of the Celestial Torch on board the Rising Star, but less so if that same member met the Winchesters.
With Cortex Plus and its different streams, adapting the mechanics to the setting was a design goal. This means that characters from the different games wouldn't interact as easily – Spider-Man has no relationship ties to anyone from Smallville, for example – but the games reflect the TV series they're meant to portray. Smallville is a super-powered soap opera while Leverage is a series of heist mini-movies, and their games reflect those realities. The key is to choose the correct Cortex Plus stream to reflect the core of a work. So far, the developers have been able to do just that.
Speaking of the developers, Cam Banks has licensed Cortex and Cortex Plus from Margaret Weis Productions. The end goal is to create a game, Cortex Prime, that takes in all the prior work mentioned above and produce not just the rules but settings that aren't necessarily licensed works. His studio, Magic Vacuum Design Studio is running a Kickstarter, with stretch goals that will include a number of pre-made settings with the new game.
* The Serenity and Firefly licenses had an inherent problem – different studios held the licensing rights. Fox has the rights for Firefly, but Universal had the rights for Serenity. This split meant that information in one work could not appear in the licensed game of the other. Players, however, aren't restricted and can pull in characters and ideas from both works, but any work needed to stat up something not covered by the game fell to the GM.
This article was originally published at Seventh Sanctum.
Thanks to our friends at Seventh Sanctum for letting us share this content.
Seventh Sanctum is a partner in Crossroads Alpha along with Psycho Drive-In.
Battlestar GalacticaCortexCortex PlusFireflyLeverageMarvel Heroic Roleplaying GameScott DelahuntSerenitySmallvilleSupernatural
Early-Bird Review: Wonder Woman (2017)
American Gods 1.05 "Lemon Scented You"
By day, Scott Delahunt is an IT analyst, fixing problems and explaining operating systems for end users. By night, he takes his degree in Computer Science, his love of movies, his vast knowledge of tabletop gaming, his curiosity into how things work and becomes a geek! Although he has nothing published professionally, Scott has written fanfiction, scripted an anime music video, play tested role-playing games, and applied his love of bad movies to Lost In Translation. He has also helped put on an anime convention and organize bus trips to Anime North. In his spare time, he raises two cats to become Internet icons and maintains a personal blog, The Chaos Beast.


text:
// Copyright 2018 The Chromium Authors. All rights reserved.
// Use of this source code is governed by a BSD-style license that can be
// found in the LICENSE file.

#ifndef CHROME_BROWSER_NET_CHROME_REPORT_SENDER_H_
#define CHROME_BROWSER_NET_CHROME_REPORT_SENDER_H_

#include "base/callback_forward.h"
#include "base/memory/ref_counted.h"
#include "base/strings/string_piece.h"
#include "net/traffic_annotation/network_traffic_annotation.h"

class GURL;

namespace network {
class SharedURLLoaderFactory;
}

// Similar to net::ReportSender but uses network::SimpleURLLoader under the
// hood. Reports sent with this function do not save or send credentials.
void SendReport(
    scoped_refptr<network::SharedURLLoaderFactory> url_loader_factory,
    net::NetworkTrafficAnnotationTag traffic_annotation,
    const GURL& report_uri,
    const std::string& content_type,
    const std::string& report,
    base::OnceClosure success_callback,
    base::OnceCallback<void(int /* net_error */, int /* http_response_code */)>
        error_callback);

#endif  // CHROME_BROWSER_NET_CHROME_REPORT_SENDER_H_

redpajama_set_name: RedPajamaGithub


text:
AMC offering $5 movie tickets on Tuesdays
Must be a member of AMC Stubs program to take advantage
KMBC 9 News Staff
Here's a deal for Kansas City area movie-goers.AMC Theatres – such as AMC Ward Parkway, AMC Barrywoods 24, AMC Legends 14, AMC DINE-IN Studio 28 and AMC Town Center 20 – has a special deal for movie-goers on Tuesdays. AMC announced earlier this week all movies – outside of IMAX, Dolby Cinema and 3-D movies – will be $5 on Tuesdays. AMC said the concession stand will also have a $5 combo, offering a small drink and a small popcorn. There is one catch, though. You must be a member of the AMC Stubs program to take advantage of the offers. It doesn't cost to sign up for the Stubs program. The theater's $5 Ticket Tuesday deal was supposed to last for only a limited time, but AMC extended it.
Here's a deal for Kansas City area movie-goers.
AMC Theatres – such as AMC Ward Parkway, AMC Barrywoods 24, AMC Legends 14, AMC DINE-IN Studio 28 and AMC Town Center 20 – has a special deal for movie-goers on Tuesdays.
AMC announced earlier this week all movies – outside of IMAX, Dolby Cinema and 3-D movies – will be $5 on Tuesdays. AMC said the concession stand will also have a $5 combo, offering a small drink and a small popcorn.
There is one catch, though.
You must be a member of the AMC Stubs program to take advantage of the offers. It doesn't cost to sign up for the Stubs program. The theater's $5 Ticket Tuesday deal was supposed to last for only a limited time, but AMC extended it.


text:
DJ Marquis
By Khary Kimani Turner on Wed, May 5, 1999 at 12:00 am
DJ-producer-artist Marquis is a quiet fixture in Detroit's hip-hop and club history. He is also the first DJ to bring Latin freestyle dance music to radio – via WDRQ – where he currently spins. Indeed, Marquis hails from a school that includes Gary Chandler, Stacey "Hotwaxx" Hale, DJ Fingers and Jeff "The Wizard" Mills. He owns two record labels – Nu Flava, a hip-hop label, and Mixx Flava, a dance label – and released "Mega Mix," a club favorite, on the latter.


ViaWindowsLive is a showcase site!
Missing a ":" in your syndicated feeds in DNN 4.6?
WPF ClickOnce – Can i have my cake and eat it too?
Geek Girl Blogs Site Released!
You ever get sick of having mulitple liveids and different passwods and trying to remember which one is which (is is it just me) ? A great new feature has been released that means you can link your liveids, login with just one password and switch between your ids using a combo box..pretty cool hey….
Does Raid Striping make your HDD fail?
Today we expand our team and welcome Phillip Suh. Based in Brisbane, Phillip has been in the IT industry for 7 years working for a range of companies such as Qcom, Praxa and most recently Technology One. As his meego gives away, Phillip also loves playing base guitar.
A couple of weeks ago the session videos for Remix Australia were released. If you missed any of the sessions they can be found: http://blogs.msdn.com/postmixau/ John helped in Neil's sesson on Windows Live Mashups by showing what you can do with Virtual Earth in 10 Mins. I've that section of video below.


Hibs assistant Garry Parker believes they are going BACKWARDS and that the players need to turn things around or they'll be replaced.
The Hibees haven't won seven ahead of today's trip to play Hamilton – the last side they were able to win against back in October.
It's an alarming slide when compared to last season and Parker, who was again deputising on media duties for manager Neil Lennon – was just as bullish as he was following Wednesday's draw with St Mirren at Easter Road.
"We came in here, got us out of the Champinoship, finished fourth last year - but we've gone backwards," he said.
"WE're just not performing .We're not doing it. We did at the start of the season, now we've gone off the boil.
'We've not got the results, we're not playing well - at Kilmarnock last weekend we were terrible.
New faces in January are a certainty – as to how many and in which areas is still be to determined.
"We know what sort of players that we need to bring to play the type of football that we want to play. At the moment we don't have the players to play the pressing game like we have been last year and we need to get back to that," Parker explained.
"We may be talking about a few new players, we'll need to speak to the board and hopefully get backed because we do need to improve and we do need a better squad.
"In the mean time the guys here need to respond because we're not winning games. They've got to do it.
"Of course they're playing for their futures. If you're not performing, you don't play.
"And, if we have to bring players in, they'll be under pressure. Like we are.
Flo Kamberi came in for criticism in midweek following recent displays, but Parker was keen to stress the need for collective responsibility.
He added: "You could say the say about both Flo and Jamie Maclaren. Last year they did brilliant for us, we brought them back so everybody was happy and buzzing and we started off alright.
"But at the moment they are going through a bad patch so we need to turn that round.
"It's not just about them though, there's a few of them who's not doing it so we need to turn it round quickly.
"Everybody has to put in 100 percent. We can see a difference, he's trained ok today, we've spoken to him and some of the players have spoken to him.
"But it's not just him, it's everybody - from defence all the way through. We're a team and we just need to turn it around.
"Flo's an important player to us. When he's up front and he's on it, putting himself about, he's a threat and he needs to get back to doing that.


text:
Informed authorities on Zahran's call to kill all non-Muslims - Moulavi Sahlan
Authorities were notified of Zahran Hashim's extremist teachings that called to kill every non-Muslim, says Moulavi Sahlan of Sufi Muslims' organization in Kattankudy.
Testifying before the Special Parliament Select Committee yesterday (18), Moulavi Sahlan said, however, no authority had taken any action on the matter.
Pointing out that his faction hails from traditional Muslims who follow Sufism, Moulavi Sahlan said Zahran Hashim had left their organization following the clashes.
Zahran had established the National Thowheed Jama'ath organization in 2011 and later started issuing a magazine named 'Thowheed' on a monthly basis in 2013.
From 2013 to 2014, nearly 11 complaints were filed against Zahran attacking the Sufi Muslims through this monthly magazine, Moulavi Sahlan further commented.
The Select Committee member MP Field Marshal Sarath Fonseka went on to ask whom Zahran worked for during the previous presidential election. To this, Moulavi Sahlan responded that Zahran worked against Opposition Leader Mahinda Rajapaksa's election campaign.
Responding to Parliamentarian M.A. Sumanthiran's question, Moulavi Sahlan stated that Shafi Salley, Shibly Farook, M.L.A.M. Hizbullah, A.L.M. Ruby, Abdul Rahman had signed agreements with Zahran Hashim. He added that this agreement was against Sufi Muslims. "Back in 2016, Zahran spoke up against Catholic people and Christmas festival," Moulavi Sahlan continued.
During three meetings that Zahran had conducted from 2016 and 2017, he had spoken up against Sri Lanka as well, while stressing that non-Muslims should be killed, Moulavi Sahlan revealed.
Zahran had preached that hoisting the national flag was a threat to the Islamic State and that Sri Lanka should be overtaken by Muslims, Moulavi Sahlan stated testifying before the Parliament Select Committee.
"We had submitted a letter on the matter to the Presidential Secretariat, the Office of the Prime Minister, then Minister of Justice Wijeyadasa Rajapakshe, State Minister of Defence Ruwan Wijewardene as well as Minister Sagala Ratnayake. We even submitted videos in this regard. We also complained to former DIG Nalaka De Silva," Moulavi Sahlan said.
The Prime Minister's Office had responded that the letter would be directed to PM Ranil Wickremesinghe while the Attorney General's Department had responded that necessary measures would be taken, Moulavi Sahlan said. However, no action concerning the complaints had been taken by the authorities, he added.


text:
Download Fan Made FNAF Map | Season 3 Mod Latest Version for Minecraft 1.16.5 / 1.12.2
Download Fan Made FNAF Map | Season 3 Latest Version
Why make a FNAF map? That is, "Why?" That may be the more accurate question. Because this type of fan made map is so popular and available that people are getting in on the action. Here are some of the benefits to creating your own FNAF map.
A lot of people who like the show have become fans. In fact, many have taken to writing fan fiction about their favorite characters, and including their own custom made maps. This is actually a good thing. Because a map maker has the experience necessary to be able to know what will appeal to the target audience, it means a higher likelihood of success.
Creating a fan made map for a show can be a fun way to share your love of the show with others. A lot of the fan made maps are actually quite good. They capture the spirit of the show quite well. Once you've gotten a good map, you may even find that you'll want to start making your own. After all, you can always collaborate with the creator of the map, or even hire a freelancer to do the work for you if you aren't quite sure what to do.
But what exactly does a FNAF map creator do? A FNAF map maker has the experience necessary to know what will appeal to your target audience. This means they understand how to put together a cohesive story that makes it easy to enjoy. A good map designer will also have experience creating custom maps for similar television shows or movies. And of course, a good map maker understands the basics of computer graphics as well.
You should know that not all FNAF map downloads are created equal. There are some which are poorly constructed, don't contain many details, or contain copyrighted materials. If you want the best quality map, you should be looking for a FNAF map creator who has experience in creating the kind of maps you want. If you can find one of these creators, they will be able to help make your FNAF fan art a reality.
So, if you want your FNAF art to look as good as possible, look for a good FNAF map maker. You'll be very happy with the results. After all, a FNAF fan map is a great way to share your love of this classic animation with others. After all, isn't that what really keeps a fan going?
Get Fan Made FNAF Map | Season 3 Download Options
Download Links of Fan Made FNAF Map | Season 3 Mod for Minecraft 1.16.5 / 1.12.2
Terra Restore 2
Minecraft version: 1.16.5 / 1.9 , Adventure, Survival
Recently, in a recent video interview, Jens Podgulsky, creative director of Minecraft, stated that the new version of the game will have many new features. One of those features is Terra Map, which is said to be the overhaul of the vanilla game. Now, did you know that before the...
Project Zearth
Minecraft version: 1.16.5 / 1.14.4 , Adventure, Creation, Game Map, Puzzle, Survival
Project Earth is a new mod for Minecraft. It brings a great feature that the vanilla game does not have, the option to build your own farm and city. You can build it with different types of blocks such as cobblestone, stone, gravel, brick and more. These blocks have several...
Lucky Blocks Race Vanilla (No Mods)
Minecraft version: 1.16.5 / 1.11.2 , Game Map, Parkour
Lucky Blocks Race 1.16.5 is a mod for vanilla Minecraft. I won't go into details about it because it's more or less similar to the more famous mods. It adds several new blocks to your world and rearranges them. These changes are not always perfect but it can be fun...
Find The Button: Christmas Edition (30+ Levels)
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This Christmas, find the Button: Christmas Edition is coming to Switch, Wii and Nintendo DS this December. The fun little game promises to bring the romance back into Christmas with its cute, yet engaging characters and story. Children of all ages will have fun interacting with their beloved characters, while...
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Minecraft version: 1.16.5 / 1.17.1 , Adventure, Creation, Game Map, Parkour, Survival
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Siren Head Map
Minecraft version: 1.16.5 / 1.12.2 , Adventure, Game Map, Modded World, Survival
Siren Head Map 1.16.5 has been created by the creators of Minecraft and modifies the gameplay in a very unique way. Rather than using a block system similar to many games in the past, it utilizes a crafting system based on raw materials instead. If you are familiar with the...
The Library (A Community Dropper Collaboration)
Minecraft version: 1.16.5 / 1.4.2 , Game Map
The Library is a place to share ideas, stories, culture and history. A Community Contributor, or a Library User, is someone who gives to the community and helps to preserve the past for future generations. If you have always wanted to help preserve the past for future generations then now...
Protosky map
Minecraft version: 1.16.5 / 1.16.1 , Game Map, Modded World, Survival
Welcome to the review of the new and popular Minecraft Mod, Protosky Map 1.16.5 Maker. I am sincerely happy that after spending almost a month testing this mod, I have finally got round to writing up my honest review. My mission was to see if it lived up to all...
Portal jumps
Minecraft version: 1.16.5 / 1.16.4 , Adventure, Parkour, Puzzle
If you want to experience what it's like to fly then the Minecraft Mod is a perfect choice for you! This allows you to create your own virtual world. It allows you to play with other players as well. You can build, dig and mine anything you want in this...
Minecraft version: 1.16.5 / 1.15.2 , Adventure
Terminal Velocity 1.16.5 Minecraft Mod is a new version of the popular mod, Tile Aquarium. It has many exciting features and changes, and also improves on the tile aquarium's basic functions. This mod is aimed to make things easier for the experienced modders who already use the mod, but also...
Most Installed Mod: JourneyMap, Just Enough Items (JEI), LaggRemover, Easy Villagers, All The Tweaks, CodeChicken Core 1.8.+, Dungeon Crawl, Better Badlands, Parkour, Jumbo Furnace, Villager Names, Thermal Dynamics, Animal Crops, BuildCraft|Robotics, Silk Spawners, Smooth Boot (Forge), The Escapists MC Map Series, Masonry, FallingTree (Forge&Fabric), The Lost Cities, All in One [Modded One Block], Isla Unga Bunga by Forge Labs, Reliquary v1.3, ICBM – Classic, Blood Arsenal, World of Dragons – Start City, Dynmap, Multiverse-Portals, The Outer End, Elytra Fall, Tree Assist, Lockette, Chickens, Macaw's Bridges,


text:
Southwest Ends Operations at Branson, Key West and Jackson-Evers
By Richard Maslen
Posted 6 December 2013 17:50
US low-cost carrier Southwest Airlines has announced it will close its operations at Branson Airport (BKG), Key West International Airport (EYW) and Jackson-Evers International Airport (JAN) from June 7, 2014 due to low passenger demand.
"Unfortunately, the level of local demand no longer allows Southwest to profitably serve these markets," said Bob Jordan, executive vice president and chief commercial officer, Southwest Airlines. "Southwest takes pride in becoming not only a great choice for air travel in the cities we serve, but we also become a member of the community. These decisions are never easy."
The airline first introduced flights to Jackson-Evers International in 1997 and over the past ten years has been offering flights to Baltimore Washington International, Chicago Midway, Houston Hobby, Orlando International firmly establishing itself as one of the largest operators at the Mississippi facility. According to MIDT data it carried an estimated 371,000 bi-directional O&D passengers flew on its flights to and from the airport in 2012, a 29.8 per cent share of the total traffic. Although it had no direct competition on its route pairs at Jackson-Evers International, air services to the wider Chicago, Houston and Washington markets were served by feeder affiliates of the US majors.
The markets of Branson and Key West International are relative newcomers to the Southwest network and were only introduced in 2012 as part of its integration with AirTran Airways. The airline currently serves Chicago Midway, Dallas Love Field, Houston Hobby, Orlando International from Branson and New Orleans, Orlando International and Tampa International from Key West International.
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Editor, Routesonline
Richard provided insight to the Routesonline community delivering content every day via our Airlineroute schedules blog and through regular features and analysis on latest developments and trends within the sector. With almost 20 years in the industry working for a number of aviation publications he brought the team a wealth of experience and insight.
Jackson-Evers Intl, Mississippi
Key West International
Branson Airport, LLC
Southwest further expands Florida service Posted 13 November 2020 12:13
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