text:
\section{Introduction}
Recall that the catenoid is the minimal surface of revolution surface given by
\begin{equation*}
Cat=\set{x_1^2+x_2^2=\cosh^2 x_3}\subset \mathbb R^3.
\end{equation*}
The catenoid was discovered by Euler in 1744 and is one of the classic examples in the theory of minimal surfaces and, more broadly, in the calculus of variations.  For instance, a sequence of homothetic blow-downs of $Cat$ provides the simplest model of the failure of smooth convergence for a sequence of minimal surfaces.
  Due to the invariance of the minimal surface equation under rigid motions and homotheties of $\mathbb R^3$, $Cat$ sits within a six dimensional family of catenoids which we henceforth denote by $\mathcal{C}$.  In other words, $C\in \mathcal{C}$ if $C$ can be obtained from $Cat$ by a rigid motion composed with a homothety.

The catenoid (or rather $\mathcal{C}$) has been characterized in many ways.  We list some notable results:  In the spirit of Euler, O. Bonnet, in the mid-nineteenth century, showed that the catenoid and plane are the only minimal surfaces of revolution.  More recently, in \cite{SCH}, R. Schoen showed that the catenoid is the unique complete, embedded, minimal surface with finite total curvature and two ends.  In a similar vein, but by very different techniques, F. J. L\'opez and A. Ros showed that the catenoid and plane are the unique complete, embedded minimal surfaces of finite total curvature and genus-zero \cite{LopezRos}.  Additionally, building on work of D. Fischer-Colbrie \cite{FC}, L\'opez and Ros characterized the catenoid as the unique complete, embedded minimal surface in $\mathbb R^3$ of Morse index one -- see \cite{LopezRos2}.    Finally, we note that in \cite{Pyo} pieces of the catenoid are shown to be the only minimal annuli in a slab that meet the boundary of the
  slab in a constant contact angle -- a fact that will be relevent in this note.  It bears mentioning that work of P. Collin \cite{Co} and T.H. Colding and W. P. Minicozzi \cite{CY} allows one to replace the geometric assumption of finite total curvature in \cite{LopezRos} and \cite{SCH} with the weaker topological assumption that the surfaces are of finite topology -- that is, diffeomorphic to finitely punctured compact surfaces.

 In this note, we characterize the catenoid as the unique minimal surface which minimizes area within a geometrically natural class of minimal annuli.  This turns out to be a simple consequence of the proof due to R. Osserman and M. Schiffer \cite{OS} of the isoperimetric inequality for minimal annuli in $\mathbb R^3$.

Let us now describe in what sense the catenoid minimizes area.  Fix two parallel planes $P_-,P_+\subset \mathbb R^3$ with $P_-\neq P_+$ and denote by $\Omega$ the open slab between them.    We remark that for any plane $P\subset \Omega$, $P$ must be parallel to $P_-$.
Let us denote by $\mathcal{M}(\Omega)$ the class of smooth minimal surfaces spanning $P_-$ and $P_+$.  That is, $\Sigma\in \mathcal{M}(\Omega)$ if $\Sigma\subset \Omega$ may be parameterized by a conformal, harmonic immersion $\mathbf{F}: M\to \mathbb R^3$ so that $b\Sigma:=\overline{\Sigma}\backslash \Sigma \subset \partial \Omega=P_+\cup P_-$.    Here $M$ is an open orientable surface.  Notice that an element $\Sigma \in \mathcal{M}(\Omega)$ may have arbitrarily bad behavior as one approaches $\partial \Omega$; however, if $\Sigma$ has the structure of a surface with boundary then $\partial \Sigma=b\Sigma$.
The class $\mathcal{M}(\Omega)$ is too broad for our methods and so we will restrict attention to the subclass $\mathcal{A}(\Omega)\subset \mathcal{M}(\Omega)$ consisting of embedded annuli.  Precisely, $\Sigma \in \mathcal{A}(\Omega)$ if, in addition to lying in $\mathcal{M}(\Omega)$, $\Sigma$ may be parameterized by an annulus; i.e.  there is a smooth embedding $\mathbf{F}: (0,1)\times \mathbb{S} \to \mathbb R^3$ with image $\Sigma$.  It bears mentioning that in \cite{MeeksWhite}, by using global analysis techniques, W. Meeks and B. White have shown that the subset of $\mathcal{A}(\Omega)$ consisting of surfaces with $b\Sigma$ a pair of $C^{2,\alpha}$ (planar) convex curves has the structure of a contractible Banach manifold.

Recall that given a pair of connected, simple closed curves $\sigma_+\in P_+$ and $\sigma_- \in P_-$ there need not exist $\Sigma \in \mathcal{M}(\Omega)$ with $\partial \Sigma=\sigma_+ \cup \sigma_-$.  Indeed, for $\sigma_\pm$ of sufficiently small length relative to the size of the slab, the existence of such a surface would violate the isoperimetric inequality for minimal surfaces. In Theorem 4 of \cite{OS}, Osserman and Schiffer give a sharp condition on the lengths of the $\sigma_\pm$ that precludes the existence of a minimal annulus spanning the $\sigma_\pm$.  In Section \ref{NoSpanSec}, we refine their result and show that their condition actually precludes the existence of \emph{any} connected minimal surface spanning the curves and prove, in addition, an interesting rigidity result.

It is important to emphasize that our class of surfaces consists of \emph{minimal} surfaces.  If one were to consider classes of arbitrary surfaces spanning $P_-$ and $ P_+$ then the infimum of area would be zero -- as can be seen by considering thin cylinders. Similarly, the surfaces in $\mathcal{M}(\Omega)$ are, in general, not stationary for area with respect to variations moving their boundary.  In particular, we are not considering the free boundary problem as usually formulated for minimal surfaces.

We claim that for some $C\in \mathcal{C}$, $C\cap \Omega\in \mathcal{A}(\Omega)$ has less area than any other surface in $\mathcal{A}(\Omega)$.   More precisely, denote by:
\begin{equation*}
Cat_{MS}=\set{x_1^2+x_2^2=\cosh^2 x_3}\cap \set{1-x_3\tanh x_3> 0}\subset Cat
\end{equation*}
 the maximally symmetric marginally stable piece of the catenoid.  That is $Cat_{MS}$ is stable but any domain in $Cat$ strictly containing $Cat_{MS}$ is unstable.  Recall a minimal surface  is \emph{stable} if no compactly supported infinitesimal deformation decreases area; it is \emph{unstable} if there is a compactly supported infinitesimal deformation decreasing area.  Marginally stable surfaces are on the boundary between these two classes.  We show that $Cat_{MS}$ provides the model for least area surfaces in $\mathcal{A}(\Omega)$.
\begin{thm}\label{MainThm}
Let $P_-$ and $P_+$ be distinct parallel planes in $\mathbb R^3$ and let  $\Omega$ be the open slab between them.
Let $C_{MS}\in \mathcal{A}(\Omega)$ be the (unique up to translations parallel to $P_\pm$)  minimal surface in $\mathcal{A}(\Omega)$ obtained from rigid motions and homotheties of $Cat_{MS}$.
Then for any $\Sigma \in \mathcal{A}(\Omega)$:
\begin{equation}
\mathcal{H}^2(\Sigma)\geq \mathcal{H}^2(C_{MS})
\end{equation}
 with equality if and only if $\Sigma$ is a translate of $C_{MS}$.
\end{thm}
Here $\mathcal{H}^k$ denotes $k$-dimensional Hausdorff measure.  We point out that the two disks $D_\pm \subset P_\pm$ with $\partial D_+ \cup \partial D_- = \partial C_{MS}$ satisfy $\mathcal{H}^2(D_-\cup D_+)< \mathcal{H}^2(C_{MS})$.
That is $C_{MS}$ is not an area minimizer with respect to its boundary even though it does minimize area in the class of spanning minimal annuli $\mathcal{A}(\Omega)$.

The restriction to $\mathcal{A}(\Omega)$ in Theorem \ref{MainThm} rather than $\mathcal{M}(\Omega)$ is necessitated by our argument.  However, it seems reasonable to believe that an area minimizer, $\Sigma_0 \in\mathcal{M}(\Omega)$ should have $b\Sigma_0$ rather nice -- for instance consisting of convex planar curves.
 Hence, in light of the embeddedness results of T. Ekholm, B. White and D. Wienholtz \cite{EBW} and a long standing conjecture of W. Meeks on the non-existence of postive genus surfaces in $\mathcal{M}(\Omega)$ bounded by convex curves (see Conjecture 3.10 of \cite{MeeksConj}), it is natural to conjecture:
\begin{conj}
Let $P_-$ and $P_+$ be distinct parallel planes in $\mathbb R^3$ and let  $\Omega$ be the open slab between them.
Let $C_{MS}\in \mathcal{M}(\Omega)$ be the (unique up to translations parallel to $P_\pm$)  minimal surface in $\mathcal{M}(\Omega)$ obtained from rigid motions and homotheties of $Cat_{MS}$.
Then for any $\Sigma \in \mathcal{M}(\Omega)$:
\begin{equation}
\mathcal{H}^2(\Sigma)\geq \mathcal{H}^2(C_{MS})
\end{equation}
 with equality if and only if $\Sigma$ is a translate of $C_{MS}$.
\end{conj}

 Returning to the more restricted setting  of $\mathcal{A}(\Omega)$, we note that Theorem \ref{MainThm} is a simple consequence of a more general area minimization property of the catenoid.  Indeed, minimal annuli have a natural scale which may be computed as the length of the flux vector associated to the generator of the homology group -- we refer to Section \ref{FluxSec} for precise definitions.  Normalizing with respect to this scale gives an area lower bound:
\begin{thm}\label{SecondThm}
Let $P_-=\set{x_3=h_-}$ and $P_+=\set{x_3=h_+}$ be distinct parallel planes in $\mathbb R^3$ with $h_-<h_+$ and let $\Omega$ be the open slab between them.   Fix $\Sigma \in \mathcal{A}(\Omega)$.
Let $P_0=\set{x_3=h_0}\subset \bar{\Omega}$ denote the plane that satisfies:
\begin{equation*}
\mathcal{H}^1 (\Sigma \cap P_0)=\inf_{t\in (h_-,h_+)} \mathcal{H}^1(\Sigma_t).
\end{equation*}
Here $\Sigma_t=\Sigma\cap \set{x_3=t}$ and $\mathcal{H}^1 (\Sigma \cap P_+)$ is defined as $\liminf_{t\nearrow h_+ } \mathcal{H}^1 (\Sigma_t)$ and likewise for $\mathcal{H}^1 (\Sigma \cap P_-)$.
Let $F_3$ denote the vertical component of $Flux(\Sigma)$. If $C$ is the vertical catenoid with $Flux(C)=(0,0,F_3)$ and symmetric with respect to reflection through the plane $P_0$ then:
\begin{equation}
\mathcal{H}^2(\Sigma)\geq \mathcal{H}^2(C\cap  \Omega)
\end{equation}
 with equality if and only if $\Sigma$ is a translate of  $C\cap \Omega$.
\end{thm}

The proof of Theorem \ref{SecondThm} will comprise the bulk of this note.  Let us first use it to prove Theorem \ref{MainThm}:
\begin{proof}
By Theorem \ref{SecondThm}, the least area surface must be a piece of a catenoid.  Up to a rescaling and rigid motion we may take $\Omega=\set{-1<x_3<1}$ and so may restrict attention to subsets of vertical catenoids.   The space of these catenoids are parameterized by $\lambda>0$ and $t$ where
\begin{equation*}
Cat_{\lambda,t}=\lambda Cat+t\mathbf{e}_3.
\end{equation*}
Set
\begin{equation*}
A(\lambda,t)=\mathcal{H}^2(Cat_{\lambda,t}\cap \Omega).
\end{equation*}
We will check that this value is minimized at $\lambda_0, t_0$ chosen so $Cat_{\lambda_0, t_0} \cap \Omega=\lambda_0 Cat_{MS}$.

Let us parameterize a scale $\lambda$ vertical catenoid by
\begin{equation*}
\mathbf{F}_\lambda (h, \theta)=(\lambda \cosh \frac{h}{\lambda}\cos \theta , \lambda\cosh \frac{h}{\lambda} \sin \theta, h).
\end{equation*}
Then we have by the area formula:
\begin{align*}
A( \lambda,t)&=\lambda \int_{-1-t}^{1-t} \int_0^{2\pi}  \cosh^2 \frac{h}{\lambda} d\theta dh\\
   					&=2\pi \lambda^2 \left.\left( \frac{1}{4} \sinh \frac{2h}{\lambda}+\frac{h}             {2\lambda}\right)\right|_{h=-1-t}^{h=1-t} \\
   					&=\frac{\lambda^2 \pi}{2} \left(\sinh \frac{2(1-t)}{\lambda}-\sinh  \frac{2(-1-t)}{\lambda} \right) +2\pi\lambda\\
   					&={\lambda^2 \pi}\cosh \frac{2t}{\lambda}  \sinh \frac{2}{\lambda}+2\pi\lambda.
 \end{align*}
It is elementary to check that $A(\lambda,t)\to \infty $ as $|(\ln \lambda,t)|\to \infty$, so in order to find the least area catenoid we look for critical points of $A$.
 We expect these to occur for catenoids that are symmetric about the plane $\set{x_3=0}$. Indeed,
$\partial_t A=2\pi \lambda \sinh\frac{2t}{\lambda} \sinh \frac{2}{\lambda}$ and this equals zero if and only if $t=0$ and hence $t_0=0$.  Thus, we need only find $\lambda$ so
\begin{equation*}
0=\partial_\lambda A(\lambda,0)=2\pi \lambda \sinh \frac{2}{\lambda}- 2\pi\cosh \frac{2}{\lambda}+2\pi
\end{equation*}
Using $\cosh x=\sqrt{1+\sinh^2 x}$,  this equation is equivalent to solving:
\begin{equation*}
(\lambda^2-1) \sinh^2 \frac{2}{\lambda} +2\lambda \sinh \frac{2}{\lambda} =0
\end{equation*}
which has a unique solution $\lambda=\lambda_0\approx 0.833$ determined by
\begin{equation}\label{lambdanot}
\frac{2\lambda}{1-\lambda^2} = \sinh \frac{2}{\lambda}.
\end{equation}
Uniqueness follows from properties of the two functions in \eqref{lambdanot} on the domain $\lambda \in [0, 1)\cup (1, \infty)$.  First, observe that the right hand side is always positive on this domain while the left hand function is only positive on $[0,1)$.  Second, $\sinh \frac{2}{\lambda}$ is strictly decreasing on $[0,1)$ while $\frac{2\lambda}{1-\lambda^2}$ is strictly increasing.
Finally, $\lim_{\lambda\searrow 0}  \sinh \frac{2}{\lambda}-\frac{2\lambda}{1-\lambda^2}=\infty$ while  $\lim_{\lambda\nearrow 1}  \sinh \frac{2}{\lambda}-\frac{2\lambda}{1-\lambda^2}=-\infty$.

Lastly, we must verify that $Cat_{\lambda_0, t_0}\cap \Omega=\lambda_0 Cat_{MS}$.  To that end we note that
standard properties of hyperbolic functions give that
\begin{equation*}
\sinh \frac{1}{\lambda_0}=\frac{\lambda_0}{\sqrt{1-\lambda_0^2}}
\end{equation*}
and so
\begin{equation*}
\tanh \frac{1}{\lambda_0}=\lambda_0.
\end{equation*}
Hence, $1-\frac{1}{\lambda_0} \tanh \frac{1}{\lambda_0}=0$.  But this implies that $1-\frac{1}{\lambda_0} \tanh \frac{x_3}{\lambda_0}>0$ on $ Cat_{\lambda_0, t_0}\cap \Omega$.  That is, $\frac{1}{\lambda_0} (Cat_{\lambda_0, t_0}\cap \Omega)= Cat_{MS}$.
\end{proof}

\subsection*{Acknowledgments}
We would like to thank David Hoffman for his interest and his many helpful comments.  We
also thank Brian White for suggesting a simplification of the proof of Lemma \ref{AltLem}.
\section{Convexity of the length of level sets}
Before we proceed with the proof of Theorem \ref{SecondThm} we must recall some important definitions.  We also state the result of Osserman and Schiffer regarding the convexity of the length of certain families of curves in minimal annuli in $\mathbb R^3$.
\subsection{The Flux Vector} \label{FluxSec}
  For the purposes of this discussion we assume that $\Sigma\in \mathcal{A}(\Omega)$ for some $\Omega$.  Fix an orientation of $\Sigma$ and let $\gamma\subset \Sigma$ be a simple $C^1$ closed curve in $\Sigma$ on which we also fix an orientation.
Our choices of orientation give rise to a normal vector field in $\Sigma$ along $\gamma$ which we denote by $\nu$.  We always think of the vectors $\nu$ as vectors in $\mathbb R^3$. The \emph{flux} of $\gamma$ is defined to be the vector:
\begin{equation}
Flux(\gamma)=\int_{\gamma} \nu  \; d\mathcal{H}^1\in \mathbb R^3.
\end{equation}
 As $\nu\cdot \mathbf{e}_i =\nu\cdot \nabla_\Sigma x_i$ and on a minimal surface $\Delta_\Sigma x_i=0$, the divergence theorem implies that the flux of a curve depends only on its homology class. In particular, for a minimal annulus, $\Sigma$, we may associate a vector $Flux(\Sigma)$, by choosing $\gamma$ so that $[\gamma]$ is  a generator of $H_1(\Sigma)$ and setting $Flux(\Sigma)=Flux(\gamma)$;  up to a reflection through the origin, $Flux(\Sigma)$ is independent of the choice of orientation of $\Sigma$ and of $\gamma$.  In the sequel, we will consider
\begin{equation*}
F_3(\Sigma) = Flux(\Sigma) \cdot \mathbf{e}_3,
\end{equation*}
the vertical component of the flux of the minimal annulus $\Sigma$.  We always choose orientations so that $F_3 \geq 0$.

An important property of the flux is that it sets a natural scale for a minimal annulus.  Namely, suppose that $\Sigma$ is a minimal annulus and $\Sigma'=\lambda \Sigma$ is the annulus obtained by homothetically scaling $\Sigma$ by $\lambda>0$. Then one computes $Flux(\Sigma')=\lambda Flux(\Sigma)$.    In particular, the flux allows one to distinguish between catenoids of differing scales.
A more subtle property of the flux is that it also helps to set a natural conformal scale for  elements of $\mathcal{A}(\Omega)$.
More precisely consider $\Sigma\in \mathcal{A}(\Omega)$.  By the uniformization theorem there is a conformal diffeomorphism $\psi$ between $\Sigma$ and a flat open cylinder, $(h_-, h_+)\times \mu \mathbb{S}$ where here $(h_-, h_+)$ denotes a (possibly infinite) interval and $\mu \mathbb{S}$ denotes the circle of radius $\mu$.   Moreover, the ratio between $|h_+-h_-|$ and $\mu$ is determined by $\Sigma$.  We claim this ratio is actually determined only by $\Omega$ and $Flux(\Sigma)$.

In order to show this we first need the following fact:
\begin{lem}\label{TransverseLem}
Let $\Sigma\in \mathcal{A}(\Omega)$ and suppose that $\bar{\Sigma}$ has the structure of a surface with boundary and that $\partial\Sigma$ is smooth.  Then for any plane $P\subset \bar{\Omega}$, $P$ meets $\Sigma$ transversally.
\end{lem}
\begin{proof}
Using an ambient rigid motion and homothety, we take $\Omega=\set{-1<x_3<1}$.
As $\partial \Sigma$ is smooth, standard boundary regularity results imply that $\Sigma$ may be viewed as a smooth surface with boundary.
We first show that $\Sigma$ meets the planes $P_1=\set{x_3=1}$ and $ P_2=\set{x_3=-1}$ transversally.  To that end we note that as
$\partial \Sigma$ is smooth there is a uniform constant $r_0>0$ so that for any point $p\in \partial \Sigma$, the inner and outer
osculating circles have radius greater than $r_0$.  Moreover, there is a uniform bound on the ratio between intrinsic and extrinsic
distance between points of $\partial \Sigma$. Hence, there exists $r_1$ with $0<r_1<r_0$ so for any $p \in P_1\cap \partial \Sigma$ the following holds:
there are circles in $P_1$ denoted by $C_{in}(p)$ and $C_{out}(p)$, both of radius $r_1$, such that $C_{in}(p)$ lies within
$\partial \Sigma\cap P_1$ (thought of as a plane curve in $P_1$) while $C_{out}(p)$ lies outside $\partial \Sigma \cap P_1$.  Moreover, both circles $C_{in}(p), C_{out}(p)$
 meet $\partial \Sigma$ only at $p$. A similar result holds for
$p\in \partial \Sigma \cap P_2$.  Without loss of generality we consider only $p\in \partial \Sigma\cap P_1$.

Now let $Cat^+=Cat \cap \set{x_3\geq 0}$.  Denote by $Cat_{in}(p)$ the set obtained from $Cat^+$ by
translations and homotheties so that $\partial
Cat_{in}(p)=C_{in}(p)$ and let $Cat_{out}(p)$ be defined in an analogous manner;
notice that both $Cat_{in}(p)$ and $Cat_{out}(p)$ are disjoint from $\Omega$.
Denote by $Cat_{in}'(p)$ the surfaces obtained by scaling $Cat_{in}(p)$ by $\frac{1}{2}$ about the center of $C_{in}(p)$ and define
$Cat_{out}'(p)$ similarly.  By the strict maximum principle and the definition
of $\mathcal{A}(\Omega)$ we have that $\overline{\Sigma} \cap \partial \Omega=\partial \Sigma$.  Hence, there is a $\delta>0$ so that for all $p$ both
$Cat_{in}'(p)$ and $Cat_{out}'(p)$ can be translated along their axes by $\delta$ into $\Omega$ while
remaining disjoint from $\Sigma$ (that is translated in the direction $-\textbf{e}_3$).  Let us denote by $Cat_{in}''(p)$ and $Cat_{out}''(p)$ the surfaces resulting from this translation and let $\lambda Cat_{in}''(p)$ denote the result of scaling $Cat_{in}''(p)$ by
$\lambda>0$ about the center of $\partial Cat_{in}''(p)$ and similarly for $\lambda Cat_{out}''(p)$.  As
$\lambda \to 0$ both $\lambda Cat_{in}''(p)$ and $\lambda Cat_{out}''(p)$ converge to a plane $P''(p)\subset \Omega$
which must meet $\Sigma$ as otherwise $\partial \Sigma \cap P_1=\emptyset$ which is impossible by the convex hull
property and our definition of $\mathcal{A}(\Omega)$.   Hence, there are $0<\lambda_{in}
(p)<1$ and $0<\lambda_{out}(p)<1$ so that,  for $\lambda<\lambda_{in}(p)$, $\lambda  Cat_{in}''(p)$ meets $\Sigma$
but, for $\lambda>\lambda_{in}(p)$, $\lambda  Cat_{in}''(p)$ is disjoint from $\Sigma$; and the same for
$\lambda_{out}(p)$ with respect to $\lambda Cat_{out}''(p)$.   As a consequence, $\lambda_{in}(p)Cat_{in}''(p)$ is
disjoint from $\Sigma\backslash \partial \Sigma$ but meets $\partial \Sigma$ and it must do so precisely at $p$ and the same is true for $\lambda_{out}(p)Cat_{out}''(p)$.   By the boundary maximum principle we
then see that the normal to $\Sigma$ at $p$ cannot be orthogonal to $P_1$ and hence $P_1$ meets $\Sigma$ transversally as do all planes $\set{x_3=t}$ for $1-\epsilon<t\leq 1$. A similar result holds for $P_2$.  Thus, for all planes $P\subset \Omega$ near $P_1$ or $P_2$,  $P\cap \Sigma$ consists of a single smooth simple closed curve.

Now let $f=x_3$ be the function whose level sets are planes in $\Omega$.  As $\Sigma$ is minimal $f$ is a harmonic function on $\Sigma$ and so has no local maxima or minima. In particular, at any critical points of $f$ the vector field $\nabla f$  has negative index.  By our previous discussion $f$ has no critical points near $\partial \Sigma$ and, moreover, $\nabla f$ is transverse to $\partial \Sigma$.  As $\Sigma$ is an annulus, the Hopf index theorem then implies that $f$ has no critical points.
\end{proof}
As a consequence, $Flux(\Sigma)$ and $\Omega$ determine the cylinder with which $\Sigma$ is conformally equivalent:
\begin{cor} \label{DiffeoCor}
Suppose that $\Omega=\set{h_-<x_3<h_+}$ is an open slab.  For any $\Sigma \in \mathcal{A}(\Omega)$ there is a conformal diffeomorphism
\begin{equation*}
\psi:\Sigma \to (h_-,h_+)\times \mu \mathbb{S}
\end{equation*}
given by $\psi(p)=(x_3(p), x_3^*(p))$.  Moreover, $\mu=\frac{1}{2\pi} F_3(\Sigma )>0$.
\end{cor}
\begin{proof}
Let $J$ be the almost-complex structure on $\Sigma$ arising from the metric and some choice of orientation. As $x_3$ is a harmonic function,  $dx_3$ is a harmonic one form.  Moreover, $dx_3^*=dx_3\circ J$, the conjugate differential,  is also harmonic. In general, $dx_3^*$, while closed, will never be exact.  Indeed, for a general closed curve $\gamma$ in $\Sigma$:
\begin{equation*}
Flux(\gamma)\cdot \mathbf{e}_3=\int_{\gamma} dx_3^*.
\end{equation*}
Thus, integrating $dx_3^*$ gives a map $x_3^*: \Sigma \to \mathbb{R}/F_3\mathbb{Z}$ where $F_3=F_3(\Sigma)\geq 0$.   We will see that we must have $F_3> 0$ and so $ \mathbb{R}/F_3\mathbb{Z}=\mu \mathbb{S}$.
As $dx_3$ and $dx_3^*$ have the same length and are orthogonal, if we set $\psi=(x_3, x_3^*)$ then $\psi$ is a conformal map.

By Sard's theorem, for each $\epsilon>0$ there is an $0<\epsilon' <\epsilon$ so that both $\set{x_3=h_-+\epsilon'}$ and $\set{x_3=h_+-\epsilon'}$ meet $\Sigma$ transversally.  Moreover, as $b\Sigma \subset \partial \Omega$, each of the finitely many components of $\set{h_-+\epsilon'\leq x_3\leq h_--\epsilon'}\cap \Sigma$ has the structure of a surface with boundary.  By the convex hull property of minimal surfaces there is exactly one such component $\Sigma_{\epsilon'}$ and it is an annulus.
 Lemma \ref{TransverseLem} implies that for $h_-+\epsilon'<t<h_+-\epsilon'$, each plane $\set{x_3=t}$ meets $\Sigma_{\epsilon'}$, and hence $\Sigma$, transversally.  In particular, $dx_3$ does not vanish on  $\Sigma_{\epsilon'}$  and hence  $\psi$  restricted to $\Sigma_{\epsilon'}$ is a local diffeomorphism.    As each level set of $x_3=t$ for $h_-+\epsilon'\leq t\leq h_+-\epsilon'$ is connected and $dx_3^*$ does not vanish, $\psi$ is injective on these level sets and hence the restriction of $\psi$ to $\Sigma_{\epsilon'}$ is injective.  In addition, it is then clear that $F_3\neq 0$.  Taken together it follows that $\psi$ restricts to a conformal diffeomorphism between $\Sigma_{\epsilon'}$ and $ (h_-+\epsilon',h_+-\epsilon')\times \mu \mathbb{S}$.  As $\epsilon$ may be taken as small as we like, the result is shown.
\end{proof}

\subsection{Osserman and Schiffer's Result}
We now record the convexity result of Osserman and Schiffer from \cite{OS} that we will use. This result was a key step in their proof -- also in \cite{OS} -- of the sharp isoperimetric inequality for doubly connected minimal surfaces in $\mathbb R^3$.  We point out  that the restriction to $\mathbb R^3$ comes from their use of the Weierstrass representation in order to prove the convexity result.  Roughly speaking, Osserman and Schiffer show that when a minimal annulus $\Sigma\subset\mathbb R^3$ is conformally parametrized by an annulus $A$ in the complex plane, then the length of the images in $\Sigma$ of the circles foliating $A$ satisfy a convexity condition that is sharp on catenoids and planar annuli.  Precisely,
\begin{lem}\label{ShiffOssLem}
Let $A_{r,R}=\set{z: r<|z|<R}\subset \mathbb{C}$ and suppose that $\mathbf{F}: A_{r,R}\to \mathbb R^{3}$ is a conformal harmonic immersion (so in particular the image of $\mathbf{F}$ is a minimal surface).  If we let $\sigma_\rho$ be the image of $|z|=\rho$ under $\mathbf{F}$ and define:
\begin{equation}
L(t)=\mathcal{H}^1(\sigma_{e^t})
\end{equation}
then
\begin{equation}
L''(t)\geq L(t)
\end{equation}
with equality if and only if $\mathbf{F}$ maps into a planar annulus or into a piece of a catenoid bounded by coaxial circles in parallel planes.
\end{lem}
For the sake of completeness we sketch Osserman and Schiffer's proof in Appendix \ref{OSApp}.
Rather than using Lemma \ref{ShiffOssLem} directly we use the following corollary:
\begin{cor}\label{OSCor}
Suppose that $\Omega=\set{h_-<x_3<h_+}$ and $\Sigma \in \mathcal{A}(\Omega)$.  Set $\Sigma_t=\Sigma\cap \set{x_3=t}$. Then for $t\in (h_-,h_+)$:
\begin{equation*}
\frac{d^2}{dt^2} \mathcal{H}^1(\Sigma_t) \geq \frac{(2\pi)^2}{F_3(\Sigma)^2} \mathcal{H}^1(\Sigma_t)
\end{equation*}
with equality if and only if $\Sigma$ is a piece of a vertical catenoid $C$.
\end{cor}
\begin{proof}
By Corollary \ref{DiffeoCor}, $F_3\neq 0$ and there is a conformal diffeomorphism
\begin{equation*}
\psi :(h_-,h_+)\times \mu \mathbb{S}\to \mathbb R^3
\end{equation*}
with image $\Sigma$ and so that $\Sigma_t$ is the image of $(t, \cdot)$ under $\psi$.
Here $\mu=\frac{1}{2\pi} F_3$.
One verifies that the map:
\begin{align*}
G:(h_-,h_+)\times \mu \mathbb{S} &\to A_{R_-, R_+} \subset \mathbb{C}\\
 (h, \theta)&\mapsto e^{\frac{h}{\mu}+i \frac{\theta}{\mu}}
 \end{align*} is a conformal diffeomorphism.
 Here $A_{R_-, R_+}=\set{R_-<|z|<R_+}$ with $R_-=e^{\frac{h_-}{\mu}}$ and $R_+=e^{\frac{h_+}{\mu}}$.
As a consequence, we obtain a conformal diffeomorphism:
\begin{equation*}
\mathbf{F}=\psi \circ G^{-1}
\end{equation*}
as in Lemma \ref{ShiffOssLem}.
We check that $\Sigma_t=\mathbf{F}( |z|=e^{ \frac{t}{\mu}} )$ and so
\begin{equation*}
\mathcal{H}^1(\Sigma_t)=L\left( \frac{t}{\mu}\right).
\end{equation*}
The corollary then follows immediately from
Lemma \ref{ShiffOssLem} and the fact that $\frac{1}{\mu^2}=\frac{(2\pi)^2}{F_3^2}$.
\end{proof}
\begin{rem}
We give an alternate approach to Corollary \ref{OSCor} in Appendix \ref{OtherApp}.  While this approach avoids the use of the Weierstrass representation and gives a sharper conclusion, it requires a certain geometric estimate that is still conjectural.
\end{rem}

\section{The Area Bound}
In order to prove Theorem $\ref{SecondThm}$ we use Corollary \ref{OSCor} to obtain a bound for the lengths of level sets:
\begin{prop} \label{LengthCompProp}
Let $P_-=\set{x_3=h_-}$ and $P_+=\set{x_3=h_+}$ be distinct parallel planes in $\mathbb R^3$ with $h_-<h_+$ and let $\Omega$ be the open slab between them.   Fix $\Sigma \in \mathcal{A}(\Omega)$.
Let $P_0=\set{x_3=h_0}\subset \bar{\Omega}$ denote the plane that satisfies:
\begin{equation*}
\mathcal{H}^1 (\Sigma \cap P_0)=\inf_{t\in (h_-,h_+)} \mathcal{H}^1(\Sigma_t).
\end{equation*}
Here $\Sigma_t=\Sigma\cap \set{x_3=t}$ and $\mathcal{H}^1 (\Sigma \cap P_+)$ is defined as $\liminf_{t\nearrow h_+ } \mathcal{H}^1 (\Sigma_t)$ and likewise for $\mathcal{H}^1 (\Sigma \cap P_-)$.
Let $C$ denote the vertical catenoid with $Flux(C)=(0,0,F_3(\Sigma))$, symmetric with respect to reflection through the plane $P_0$.
If $C_t=C\cap \set{x_3=t}$ then for $t\in [h_-,h_+]$:
\begin{equation}
\mathcal{H}^1(\Sigma_t)\geq \mathcal{H}^1(C_t).
\end{equation}
 Equality can hold when $t\neq h_0$ if and only if  $\Sigma$ is a translate of $C\cap \Omega$.
\end{prop}
\begin{proof}
Set $L_\Sigma(t)=\mathcal{H}^1(\Sigma_t)$ for $t\in (h_-,h_+)$.  By Lemma \ref{TransverseLem}, $L_\Sigma(t)$ depends smoothly on $t$ and by Lemma \ref{OSCor} one has:
\begin{equation*}
\frac{d^2}{dt^2} L_{\Sigma} (t)\geq \frac{(2\pi)^2}{F_3(\Sigma)^2} L_\Sigma (t),
\end{equation*}
with equality if and only if $\Sigma$ is piece of a catenoid.
Notice that $L_\Sigma$ is a convex function on $(h_-,h_+)$.  By setting
$L_\Sigma(h_-)=\lim_{t\searrow h_- }L_\Sigma(t)$ and $L_\Sigma(h_+)=\lim_{t\nearrow h_+} L_\Sigma(t)$, we may think of $L_\Sigma$ as a function on $[h_-,h_+]$ but possibly taking the value $\infty$ at the end points.  The convexity ensures these limits exist.

For $C$ as in the statement of the theorem, $C\cap \Omega \in \mathcal{A}(\Omega)$.  Set $L_C(t)=\mathcal{H}^1(C_t)$.  As $Flux(C)\cdot \mathbf{e}_3=F_3(\Sigma)=F_3$ by assumption, Corollary \ref{OSCor} implies:
\begin{equation*}
\frac{d^2}{dt^2} L_{C} (t)= \frac{(2\pi)^2}{F_3^2} L_C (t).
\end{equation*}
Notice that $L_C$ is smooth on $[h_-,h_+]$ and the symmetry about $P_0=\set{x_3=h_0}$ implies $L'_C(h_0)=0$.

We claim that $L_\Sigma \geq L_C$ on $[h_-,h_+]$ with equality if and only if $\Sigma $ is a piece of a catenoid.  To see this we distinguish between when $h_0\in (h_-,h_+)$ and when $h_0$ is an endpoint.
For any $t\in (h_-,h_+)$
\begin{equation*}
F_3=\left|\int_{\Sigma_t} \mathbf{e}_3\cdot \mathbf{\nu} ds\right| \leq \int_{\Sigma_t} ds = L_\Sigma(t)
\end{equation*}
with equality if and only if $\Sigma_t$ is a geodesic in $\Sigma$.
Similarly, if $t=h_-$ then
\begin{equation*}
F_3=\lim_{t\searrow h_- } \left|\int_{\Sigma_t} \mathbf{e}_3\cdot \mathbf{\nu} ds\right|\leq \lim_{t\searrow h_- } \int_{\Sigma_t} ds = L_\Sigma(h_-)
\end{equation*}
 and the corresponding result holds when $t=h_+$.
As $C_{h_0}$ is a geodesic in $C$,  $F_3=F_3(C)=L_C(h_0)$.  Hence, $L_C(h_0)\leq L_\Sigma(h_0)$.

Now assume that $h_0\in (h_-,h_+)$.
For $t\in (h_-,h_+)$, the choice of $C$ and Corollary \ref{OSCor} ensure that  $L_C(h_0)\leq L_\Sigma(h_0)$, $L'_\Sigma(h_0)= 0=L'_C(h_0)$
and $\frac{d^2}{dt^2} \left( L_\Sigma(t)-L_C(t)\right) \geq \frac{(2\pi)^2}{F_3^2} \left( L_\Sigma(t)-L_C(t)\right)$. An ODE comparison
then implies that for all  $t\in [h_-,h_+]$,  $L_C(t)\leq L_\Sigma(t)$ with equality holding for any $t\neq h_0$ if and only if $\Sigma$ is a piece of a catenoid and $\Sigma_{h_0}$ is a geodesic in $\Sigma$.   Thus, equality holds for any $t \neq h_0$, if and only if $\Sigma$ is equal (up to a translation) to $C\cap \Omega$.

When $h_0=h_-$ we argue as follows: For $\epsilon>0$ small, set $L_{C,\epsilon}(t)=L_C(t-\epsilon)$ and restrict attention to $(h_-+\epsilon, h_+)$.  Clearly, $L_{C,\epsilon}''=\frac{(2\pi)^2}{F_3^2} L_{C,\epsilon}$, $L'_{C,\epsilon}(h_-+\epsilon)=0$ and  $L_\Sigma(h_-+\epsilon)>L_\Sigma (h_-) \geq L_{C,\epsilon}(h_-+\epsilon)$.  Moreover, as $L_\Sigma$ is convex and has its minimum at $h_-$, $L_\Sigma'(h_-+\epsilon)>0$.  Hence by an ODE comparison,  $L_\Sigma(t)>L_{C,\epsilon}(t)$ for $t\in [h_-+\epsilon, h_+)$.  Letting $\epsilon\to 0$ implies $L_\Sigma(t)\geq L_{C}(t)$ for $t\in [h_-,h_+]$.  Equality can hold if and only if $\Sigma$ is equal (up to a translation) to $C\cap \Omega$.  An identical argument applies when $h_0=h_+$.
\end{proof}
\begin{rem}
Proposition \ref{LengthCompProp} fails if $\Sigma$ were taken in the larger class of embedded elements of $\mathcal{M}(\Omega)$.
Indeed, normalizing as in the proposition, it can be verified that (a suitably modified) version of Corollary \ref{OSCor} continues to hold for $\Sigma$ at $t$ so that $\set{x_3=t}$ meets $\Sigma$ transversally.  In particular, a modified version of Proposition \ref{LengthCompProp} holds between critical values of $x_3$.  However, it can be verified that while the length of level sets is continuous across critical values of $x_3$, the rate of change of the length of these level sets becomes infinite at a critical value.  In particular, there is never convexity across critical levels.
\end{rem}

Let us now use Proposition \ref{LengthCompProp} to prove Theorem \ref{SecondThm}:
\begin{proof}
We have verified:
\begin{equation*}
\mathcal{H}^1( C_t)\leq \mathcal{H}^1( \Sigma_t)
\end{equation*}
with equality for all $t$ if and only if $\Sigma$ is equal (up to a translation) to $C\cap \Omega$. Fix $h_*\in(h_-,h_+)$ and define the function $A_{\Sigma,h_*}(t)$ on $[h_*, h_+) $ by
 \begin{equation*}
 A_{\Sigma,h_*} (t)=\mathcal{H}^2(\Sigma \cap \set{h_*\leq x_3 < t})
 \end{equation*}
 and the function $A_{C,h_*}$ similarly.

The co-area formula implies that
\begin{equation*}
\frac{d}{dt}A_{\Sigma,h_*}(t)= \int_{\set{x_3=t}\cap \Sigma} \frac{1}{|\nabla_\Sigma x_3|} .
\end{equation*}
Applying the Cauchy-Schwarz inequality yields:
\begin{equation*}
\frac{d}{dt}A_{\Sigma,h_*}(t)\geq \frac{\mathcal{H}^1 (\Sigma_t)^2}{\int_{\Sigma_t} |\nabla_\Sigma x_3| }=\frac{\mathcal{H}^1(\Sigma_t)^2}{F_3}.
\end{equation*}
Notice that one has equality if and only if $ \frac{1}{|\nabla_\Sigma x_3|}$ and $|\nabla_\Sigma x_3|$ are linearly dependent, in other words are both constant.  This is readily checked to be the case on $C$ and so using the above estimate for length:
\begin{equation*}
\frac{d}{dt}A_{\Sigma,h_*}(t)\geq \frac{\mathcal{H}^1( C_t)^2}{F_3 }=\frac{d}{dt}A_{C,h_*}(t).
\end{equation*}
Integrating implies $A_{\Sigma,h_*} (t)\geq A_{C, h_*}(t)$ for  $t\in [h_*,h_+)$ with equality if and only if $\Sigma$ is a piece of of $C$.  Letting $h_*\to h_-$ proves the theorem.
\end{proof}

\section{Sharp non-existence result}
\label{NoSpanSec}
As discussed in the introduction, given an open slab $\Omega$ with $\partial \Omega=P_+\cup P_-$ and connected, simple closed curves $\sigma_\pm \subset P_\pm$ there need not be a surface $\Sigma\in \mathcal{M}(\Omega)$ with $\partial \Sigma=\sigma_+\cup \sigma_-$.  For instance, if the curves $\sigma_\pm$ are too short relative to the height of the slab then there cannot be a connected minimal surface $\Sigma\in \mathcal{M}(\Omega)$ spanning $\sigma_\pm$.  Indeed, the monotonicity formula gives a lower bound on the area of such a $\Sigma$ in terms of the distance between the planes, while the isoperimetric inequality gives an upper bound in terms of the lengths of the curves  (for surfaces in $\mathcal{M}(\Omega)$ with two boundary components the isoperimetric inequality with sharp constant is known to hold -- see \cite{LiSchoenYau}). Alternatively, if the $\sigma_{\pm}$ are well separated, barrier arguments can be used to rule out the existence of such $\Sigma$.   Using Proposition \ref{LengthCompProp}, we are able to  give a sharp condition (see also Theorem 6 of \cite{OS} for a related result):
\begin{thm}\label{NoSpanThm}
Fix $\Omega$ an open slab with $\partial \Omega=P_-\cup P_+$ the union of two parallel planes.  Let $\sigma_\pm \subset P_\pm$ be a pair of connected simple closed curves. Let $C_{MS}$ be the unique (up to translations parallel to $P_\pm$) minimal surface in $\mathcal{A}(\Omega)$ obtained via rigid motions and homotheties from $Cat_{MS}$.  If we define $L_{crit}(\Omega):=\mathcal{H}^1(\partial C_{MS})$ and
\begin{equation*}
 \mathcal{H}^1(\sigma_+\cup \sigma_-)<L_{crit}(\Omega)
\end{equation*}
 then there is no surface $\Sigma\in \mathcal{M}(\Omega)$ with $\partial \Sigma=\sigma_+\cup \sigma_-$.  Moreover, if $\Sigma \in \mathcal{M}(\Omega)$ is a smooth minimal surface with $\partial \Sigma =\sigma_-\cup \sigma_+$ and
\begin{equation*}
 \mathcal{H}^1(\partial \Sigma)=L_{crit}(\Omega)
\end{equation*}
 then $\Sigma$ is a translate of $C_{MS}$.
\end{thm}

In order to prove this theorem, we first prove a more general result.  Namely, we will show the existence of a
$\Sigma \in \mathcal{M}(\Omega)$ with $\partial \Sigma = \sigma_-\cup \sigma_+$ is precluded if one boundary curve is
too short as determined by an explicit function of the length of the other boundary curve.  Roughly speaking, the existence of
such a $\Sigma$ relies on the existence of a vertical catenoid $C$ so that $\mathcal{H}^1(C\cap P_\pm)= \mathcal{H}^1(\sigma_+ \cup \sigma_-)$. We point out that Theorem 4 of \cite{OS} gives the same result when one considers only $\Sigma\in \mathcal{A}(\Omega)$. As in the case
for area bounds, a marginally stable piece of a catenoid will serve as the model.  However, here the marginally stable pieces are
generally not obtained from rigid motions and homotheties of $Cat_{MS}$.

We begin by describing the general class of marginally stable pieces of $Cat$ we will need.
First note that the
rotational symmetry and convexity of the function $\cosh t$ imply that for each point $p=z\mathbf{e}_3$ on the $x_3$-axis, there are unique cones over $p$ that intersect $Cat$ tangentially.  Precisely, there exist values $t_+=t_+(p)>0$ and $t_-=t_-(p)<0$
 with the following property: the cones $C_+(p)$ (resp. $C_-(p)$) over $p$ of $Cat_{t_+(p)}=Cat\cap\set{x_3=t_+(p)}$ (resp.
$Cat_{t_-(p)}=Cat\cap\set{x_3=t_-(p)}$) meet $Cat$ only at $Cat_{t_+(p)}$ (resp. $Cat_{t_-(p)}$).  We observe also that $t_{+}$ is an increasing and continuous
function of $z$ with range $(0,\infty)$; similarly, $t_-$ is increasing and continuous with range $(-\infty,0)$.  Notice that $Cat$  must be tangential to $C_\pm(p)$ at $Cat_{t_\pm(p)}$. We refer the reader to Figure \ref{MSCatFig}.

\begin{figure}
 \centering
\def\columnwidth{\columnwidth}
\input{MSCat2.tex}
\caption{The subset $Cat_{MS}(p)$ of $Cat$ is indicated as well as the cones $C_+(p)$ and $C_-(p)$. }
\label{MSCatFig}
\end{figure}

 Let $Cat_{MS}(p)$ be the bounded component of $Cat\backslash \left(C_+(p)\cup C_-(p)\right)$.
One verifies that $Cat_{MS}=Cat_{MS}(0)$ and that as $p\to (0,0,\infty)$, $Cat_{MS}(p)$
converges to $Cat\cap \set{x_3>0}$.  We claim that $Cat_{MS}(p)$ is marginally stable for each $p$.  This
follows from  the observation that for $\lambda>0$ the surfaces $Cat_{MS}^\lambda(p)=p+\lambda\left(Cat_{MS}(p)-p\right)$
give a foliation of the component $\bar{C}(p)$ of $\mathbb R^3\backslash \left(C_{+}(p)\cup C_-(p)\right)$ containing $Cat_{MS}(p)$.
 Moreover,  as $Cat$ meets $C(p)=\partial \bar{C}(p)$ tangentially,  the motion of $\partial Cat_{MS}^\lambda(p)$ at $\lambda=1$ is
tangential to $Cat_{MS}(p)$.  As a consequence, the normal variation at $\lambda=1$ gives a positive
Jacobi field on $Cat_{MS}(p)$ vanishing on $\partial Cat_{MS}(p)$.  One also verifies that if $C_0=Cat\cap \set{h_-<x_3<h_+}$ is
marginally stable then $C_0=Cat_{MS}(p)$ for exactly one value $p$.

 Using these marginally stable pieces, we now determine the explicit function of interest:
 \begin{lem} \label{MSExistLem}
Fix $\Omega_0=\set{-1<x_3<1}$ and $\partial \Omega_0=P_-\cup P_+$.  There exist two well-defined functions $C_{MS}:\mathbb R^+ \to \mathcal{A}(\Omega_0)$ and $F_{\Omega_0}:\mathbb R^+\to \mathbb R^+$ determined in the following way: For each $L_->0$ let $C_{MS}(L_-)$ denote the unique (up to translations parallel to $P_\pm$) marginally stable piece of a vertical catenoid with
\begin{equation*}
 \partial C_{MS}(L_-)=\sigma_-\cup \sigma_+\subset P_-\cup P_+
\end{equation*}
and $\mathcal{H}^1(\sigma_-) = L_-$. Given $C_{MS}(L_-)$, define
\begin{equation*}
 F_{\Omega_0}(L_-)=\mathcal{H}^1(\sigma_+).
\end{equation*}
Furthermore, $F_{\Omega_0}$ has the following properties:
\begin{enumerate}
\item \label{PropOne}If $L_+\geq F_{\Omega_0}(L_-)$ then there is a vertical catenoid $C=C(L_-,L_+)$ so that writing $\partial(C \cap \Omega_0)=\gamma_-\cup \gamma_+$ gives $\mathcal{H}^1(\gamma_\pm)=L_\pm$, while if $L_+<F_{\Omega_0}(L_-)$ no such vertical catenoid exists.
\item \label{PropTwo}If $C$ is a vertical catenoid with $\partial(C\cap \Omega_0)=\gamma_-\cup \gamma_+$ and $\mathcal{H}^1(\gamma_-)<L_-$ then $\mathcal{H}^1(\gamma_+)> F_{\Omega_0}(L_-)$.
\item \label{PropThree} If $C$ is a vertical catenoid with $\partial(C\cap \Omega_0)=\gamma_-\cup \gamma_+$, $\mathcal{H}^1(\gamma_-)=L_-$, and $\mathcal{H}^1(\gamma_+)=F_{\Omega_0}(L_-)$ then $C\cap \Omega_0$ is a translate of $C_{MS}(L_-)$.
\end{enumerate}
\end{lem}
\begin{rem}
We note that for other slabs $\Omega$ it is straightforward to determine $F_{\Omega}$ in terms of $F_{\Omega_0}$. Indeed, rigid motions leave the function invariant and $F_{\lambda \Omega}(L)=\lambda F_{\Omega}(\lambda^{-1} L)$.
\end{rem}
\begin{proof}
We claim that given any $L_->0$ there is a marginally stable piece of a vertical catenoid $C_{MS}(L_-)$ with $\partial C_{MS}(L_-)=\sigma_- \cup \sigma_+\subset P_-\cup P_+$ and $\mathcal{H}^1(\sigma_-)=L_-$.  Rather than prove this by direct computation, we use global arguments.  Set
\begin{equation*}
C=\frac{L_-}{2\pi} Cat -\eE_1.
\end{equation*}
Let $C_0=C\cap \Omega_0$ and denote $\partial C_0=\sigma_-^0\cup \sigma_+^0$ so that  $\mathcal{H}^{1} (\sigma_-^0)=L_-$.  By domain monotonicity for eigenvalues, $C_0$ is strictly stable because $C\cap \set{x_3\geq -1}$ is stable.
 Consider now the following smooth family of coaxial circles in $P_\pm$: for $t>0$ set  $\sigma_-^t=\sigma_-^0$ and  $\sigma_+^t=t(\sigma_+^0-\eE_1)+\eE_1$.  By \cite{MeeksWhite} and a barrier argument, there is a $1>T_{crit}>0$ so that for each $t\in (T_{crit},1]$ there are $C_t$, strictly stable minimal annuli smoothly depending on $t$, with $\partial C_t=\sigma_-^t\cup \sigma_+^t$ and for $t=T_{crit}$ there is a marginally stable annulus, $C_{T_{crit}}$, with $\partial C_{T_{crit}}=\sigma_-^{T_{crit}}\cup \sigma_+^{T_{crit}}$. As the boundaries consist of coaxial circles, the proof of \cite{SCH} implies that each $C_{t}$ is a piece of a catenoid.
 In fact, for all $t\in (T_{crit},\infty)$ there is a strictly stable minimal annulus $C_t$ with $\partial C_t=\sigma_-^t\cup \sigma_+^t$.  To verify the claim, it suffices to consider $t\in (1,\infty)$ and in this range the $C_t$ are obtained from appropriate rescalings and translations of subsets of $C$.

We claim that $C_{T_{crit}}$ is the desired $C_{MS}(L_-)$ and gives $F_{\Omega_0}$ as outlined.
We first note that by the uniqueness of the sets $Cat_{MS}(p)$ there are $p,\lambda$ and $h$ so that $C_{T_{crit}}=\lambda (Cat_{MS}(p)-p)+p+h\mathbf{e}_3$. That is, $C_{T_{crit}}$ is, up to a vertical translation, a rescaling of one of the marginally stable pieces described previously.

For $C'$ a vertical catenoid, let $C_0'=C'\cap \Omega_0$. Suppose $C'$ is such that $\partial C_0'=\sigma_-'\cup \sigma_+'\subset P_-\cup
P_+$ and $\mathcal{H}^1(\sigma_-')=L_-$. For the same $p$ as before, we may write $C_0'=(\lambda' (Cat-p)+p +(h+\Delta h)\mathbf{e}_3)\cap \Omega_0$. We claim $\Delta h \leq 0$.  This follows
by noting that as $Cat_{MS}(p)$ meets $C(p)$ tangentially $\left(\lambda' (Cat-p)+p+h\mathbf{e}_3\right) \cap P_-$ has length greater than $L_-$. An
upward translation only increases the length further and so one must translate downward, i.e. take $\Delta h\leq 0$. By a similar reasoning,
one concludes that $\mathcal{H}^1(\sigma'_+)\geq \mathcal{H}^1(\sigma_+^{T_{crit}})$.  As $\sigma_\pm'$ are coaxial circles,
we have
$\sigma_-'=\sigma_-^{T_{crit}}$ and $\sigma_+'$ surrounding $\sigma_+^{T_{crit}}$. By \cite{MeeksWhite}, as $C_{T_{crit}}$ is
marginally stable, if $C_0'$ differs from $C_{T_{crit}}$ then $\partial C_0'$ differs from $\partial C_{T_{crit}}$; hence in this case
$\sigma_+'$ strictly surrounds $\sigma_+^{T_{crit}}$.  By the preceeding paragraph we then see that $\partial C_0'$ bounds a strictly stable
 annulus and so, by \cite{MeeksWhite}, $C_0'$ cannot be marginally stable.  This proves the claimed uniqueness.

 Clearly, \eqref{PropOne} is an immediate consequence of the preceeding argument.  Furthermore, if \eqref{PropTwo} failed to hold for a vertical catenoid $C'$ then it could be used as a barrier allowing one to construct a piece of a vertical catenoid  violating \eqref{PropOne} (see the proof of Proposition \ref{SharpCurvProp} for a detailed argument).  Finally, by \cite{MeeksWhite}, as $C_{T_{crit}}$ is marginally stable there is no other minimal surface $C_{T_{crit}}'$ with $\partial C_{T_{crit}}=\partial C_{T_{crit}}'$ which verifies \eqref{PropThree}.
\end{proof}

As a consequence we may prove the following general proposition giving sharp conditions for the non-existence of minimal surfaces spanning a given pair of curves:
\begin{prop} \label{SharpCurvProp}
Fix $\Omega$ an open slab with $\partial \Omega=P_-\cup P_+$ the union of two parallel planes and let $C_{MS}:\mathbb R^+ \to \mathcal{A}(\Omega), \;F_{\Omega}:\mathbb R^+\to \mathbb R^+$ be the functions given by Lemma \ref{MSExistLem}.  Let $\sigma_\pm \subset P_\pm$ be a pair of connected, simple closed curves. If
\begin{equation*}
\mathcal{H}^1(\sigma_+)<F_\Omega(\mathcal{H}^1(\sigma_-))
\end{equation*}
then there is no surface $\Sigma\in \mathcal{M}(\Omega)$ with $\partial \Sigma=\sigma_+\cup \sigma_-$.  Moreover, if $\Sigma\in \mathcal{M}(\Omega)$ is a smooth minimal surface with $\partial \Sigma =\sigma_-\cup \sigma_+$ such that
\begin{equation*}
\mathcal{H}^1(\sigma_+)=F_\Omega(\mathcal{H}^1(\sigma_-))
\end{equation*}
  then  $\Sigma$ is a translate of $C_{MS}(\mathcal{H}^1(\sigma_-))$.
\end{prop}
\begin{rem}  Let $\mathcal{M}_2(\Omega)\subset \mathcal{M}(\Omega)$ be the set of all $\Sigma \in \mathcal{M}(\Omega)$ with $b\Sigma = \partial \Sigma$ consisting of exactly two connected boundary components.
 If one considers $\Psi_\Omega: \mathcal{M}_2(\Omega)\to \mathbb R^2$ the map defined by
\begin{equation*}
\Psi(\Sigma)=\left( \mathcal{H}^1(\partial \Sigma\cap P_-), \mathcal{H}^1(\partial \Sigma\cap P_+)\right)
\end{equation*}
then the proposition says that the image of $\Psi$ is an unbounded region in the first quadrant of the plane whose boundary consists of the images of marginally stable pieces of catenoids and is explicitly given as the graph of the function $F_{\Omega}$.
\end{rem}

\begin{proof}
Up to a rescaling and rigid motion we may take $\Omega=\set{-1<x_3<1}$.
 Suppose that $\Sigma\in \mathcal{M}(\Omega)$ has the structure of a smooth manifold with boundary and that $\partial \Sigma$ is embedded and consists of two connected components $\sigma_\pm \subset P_\pm$. By assumption, the $\sigma_\pm$ are connected, simple closed curves in $P_+$ and $P_-$.  It will suffice to show that $\mathcal{H}^1 (\sigma_+)\geq F_{\Omega}(\mathcal{H}^1(\sigma_-))$.   Note that $\Sigma$ is allowed to be immersed and have arbitrary genus, however it may still be used as a barrier to construct an embedded annulus with the same boundary.  Indeed, while $\Omega\backslash \Sigma$ may have more than $2$ components, only one of these, $\Omega'$, is unbounded.  Clearly, $\sigma_+$ and $\sigma_-$ are homotopic in $\bar{\Omega}'$ but are not null homotopic in $\bar{\Omega'}$.  In particular, there is an annulus $A$ in $\Omega'$ with $\partial A=\sigma_+\cup \sigma_-$ but no disk $D$ in $\Omega'$ with $\partial D=\sigma_+$ or $\partial D=\sigma_-$   Finally, we point out that $\bar{\Omega}'$ is mean convex in the sense of Meeks and Yau \cite{MeeksYau2}.  As a consequence, by \cite{MeeksYau2} there is an embedded minimal annulus $\Gamma\subset \Omega'$ with $\partial \Gamma =\sigma_+\cup \sigma_-$.

By Proposition \ref{LengthCompProp}, there is a vertical catenoid $C$ so that if we write $\partial(C\cap \Omega)=\gamma_+\cup \gamma_-$ where $\gamma_\pm\subset P_\pm$ then $\mathcal{H}^1(\sigma_\pm)\geq \mathcal{H}^1(\gamma_\pm)$. Moreover the inequality is strict unless $C\cap \Omega$ and $\Gamma$ agree up to a translation parallel to $P_\pm$.  As $\mathcal{H}^1(\gamma_-)\leq \mathcal{H}^1(\sigma_-)$, by \eqref{PropTwo} of Lemma \ref{MSExistLem}
\begin{equation*}
\mathcal{H}^1(\sigma_+)\geq \mathcal{H}^1(\gamma_+)\geq F_{\Omega}
( \mathcal{H}^1(\sigma_-)).
\end{equation*}
Finally, by Lemma \ref{MSExistLem} and Proposition \ref{LengthCompProp} equality is only achieved if $\Gamma$ is a horizontal translate of $C_{MS}(\mathcal{H}^1(\sigma_-))$.  In this case,  $\sigma_+\cup \sigma_-=\partial \Gamma$ consist of coaxial circles in parallel planes.  Hence, the proof of \cite{SCH} implies the $\sigma_\pm$ can bound only pieces of a catenoid; that is, $\Sigma$ is a horizontal translate of $C_{MS}(\mathcal{H}^1(\sigma_-))$.
\end{proof}

We now prove Theorem \ref{NoSpanThm}:
\begin{proof}Up to a rescaling and rigid motion we may take $\Omega=\set{-1<x_3<1}$.
By Proposition \ref{SharpCurvProp} we need only verify the theorem for vertical catenoids.
  The space of vertical catenoids is parameterized by $\lambda>0$ and $t$ where
\begin{equation*}
Cat_{\lambda,t}=\lambda Cat+t\mathbf{e}_3.
\end{equation*}
Just as in the proof of Theorem \ref{MainThm} where we saw $C_{MS}$ minimized area among all vertical catenoid pieces in $\Omega$, we now show $C_{MS}$ minimizes boundary length in this same class. Setting $L(\lambda,t)=\mathcal{H}^1(\partial(\Omega\cap Cat_{\lambda,t}))$,
one computes
\begin{equation*}
L(\lambda,t)=2\pi \lambda \cosh \frac{1-t}{\lambda}+2\pi \lambda \cosh \frac{-1-t}{\lambda}.
\end{equation*}
As
$L(\lambda,t)\to \infty $ when $|(\ln \lambda,t)|\to \infty$, it suffices to find critical points of $L$.  First observe that $\frac{\partial}{\partial t} L=-2\pi \sinh \frac{1-t}{\lambda}-2\pi  \sinh \frac{-1-t}{\lambda}$, which is zero only when $t=0$.  Thus one must only minimize $L(\lambda,0)=4\pi \lambda \cosh \frac{1}{\lambda}$.  One verifies that the critical points of $L(\lambda,t)$ are of the form $(\lambda_0,0)$ where $\lambda_0$ satisfes $\lambda_0=\tanh \frac{1}{\lambda_0}$. Hence, as in Theorem \ref{MainThm}, $\lambda_0$ is unique and $Cat_{\lambda_0,0}=C_{MS}$.
\end{proof}

redpajama_set_name: RedPajamaArXiv


text:
package mf.org.apache.xerces.impl.xs.util;

import java.lang.reflect.Array;
import java.util.AbstractList;

import mf.org.apache.xerces.xs.LSInputList;
import mf.org.w3c.dom.ls.LSInput;

/**
 * Contains a list of LSInputs.
 *
 * @xerces.internal
 *
 * @author Michael Glavassevich, IBM
 *
 * @version $Id: LSInputListImpl.java 776326 2009-05-19 14:27:24Z mrglavas $
 */
public final class LSInputListImpl extends AbstractList implements LSInputList {

    /**
     * An immutable empty list.
     */
    public static final LSInputListImpl EMPTY_LIST = new LSInputListImpl(new LSInput[0], 0);

    // The array to hold all data
    private final LSInput[] fArray;
    // Number of elements in this list
    private final int fLength;

    /**
     * Construct an LSInputList implementation
     *
     * @param array     the data array
     * @param length    the number of elements
     */
    public LSInputListImpl(LSInput[] array, int length) {
        fArray = array;
        fLength = length;
    }

    /**
     * The number of LSInputs in the list. The range of valid
     * child object indices is 0 to length-1 inclusive.
     */
    public int getLength() {
        return fLength;
    }

    /**
     * Returns the indexth item in the collection or
     * null if index is greater than or equal to
     * the number of objects in the list. The index starts at 0.
     * @param index  index into the collection.
     * @return  The LSInput at the indexth
     *   position in the LSInputList, or null if
     *   the index specified is not valid.
     */
    public LSInput item(int index) {
        if (index < 0 || index >= fLength) {
            return null;
        }
        return fArray[index];
    }

    /*
     * List methods
     */

    public Object get(int index) {
        if (index >= 0 && index < fLength) {
            return fArray[index];
        }
        throw new IndexOutOfBoundsException("Index: " + index);
    }

    public int size() {
        return getLength();
    }

    public Object[] toArray() {
        Object[] a = new Object[fLength];
        toArray0(a);
        return a;
    }

    public Object[] toArray(Object[] a) {
        if (a.length < fLength) {
            Class arrayClass = a.getClass();
            Class componentType = arrayClass.getComponentType();
            a = (Object[]) Array.newInstance(componentType, fLength);
        }
        toArray0(a);
        if (a.length > fLength) {
            a[fLength] = null;
        }
        return a;
    }

    private void toArray0(Object[] a) {
        if (fLength > 0) {
            System.arraycopy(fArray, 0, a, 0, fLength);
        }
    }

} // LSInputListImpl

redpajama_set_name: RedPajamaGithub


In a previous post, we explored the concept of the 'meaning' of a measurement. What exactly is it we are measuring? How does this relate to the quantity we actually want to know?
This led to a discussion in the last post of 'direct' and 'indirect' measurements, and the conclusion that almost all measurements today are of the indirect type. This has importance when we look more closely at errors, as indirect measurement of a quantity involves more actual measurements, each of which has its associated errors. These errors propagate through the reduction process in interesting ways, leading to interesting errors in the derived values. But that discussion must come after we have discussed some other topics.
In the post where we considered the 'meaning' of a measurement, we considered the apparently simple situation of measuring the length of a table. Things got progressively more complex as we thought through exactly what we were measuring and how it related to what we wanted for the 'length' of the table. However, what use is the length of the table by itself?
In the previous post, direct and indirect measurements were measured, but these were not elaborated upon. Before we go any further, we should discuss these two concepts and why they are different.
When we use 'direct' measurement, we are measuring the desired quantity directly, i.e., we apply a measurement device directly to the particular quantity we wish to measure. For example, to measure the length of a table, as discussed in the previous post, we might place a tape measure on the tabletop, position the zero mark on the tape at one edge of the table, and read the distance on the tape at the other edge.
By contrast, 'indirect' measurement is where we measures something other than the desired quantity, then use a mathematical model to transform what we actually measured into an estimate of the desired quantity. Modern measurement technology has led to increasing amounts of indirect measurements, and direct measurement is becoming much less common.


text:
AO Exam Yr 4 Term 3 AmblesideOnline.org
AmblesideOnline Exam Questions Year 4 Term 3
Student's Name _________________________
Date of Exam ____________________________
1. Tell the story of Samson.
2. Tell a story about Samuel, OR one of the adventures of the ark of the Lord.
3. What is "the birthday of the church"? What happened on that day?
4. Tell about the first martyr, OR about Philip and the Ethiopian.
Write 2-4 lines of a poem that you memorized this term.
(Assigned passage is in the Exam Key.)
1. Tell the story of Cupid and Psyche, or Echo and Narcissus, or Niobe.
2. Describe a scene from The Incredible Journey.
1. Underline the Subjects (noun phrase) and circle the Predicates (verb phrase) in the following:
a. Sunday quickly became her favorite day of the week.
b. She and Abigail were always good friends.
c. The people must have access to the truth.
d. Abigail's trip to Washington was not a happy one.
2. Tell whether each bolded word is a noun or a verb:
Mis-shapes the beauteous forms of things:--
3. Fix the punctuation and capitalization in the following sentence: what are we to do now I asked
1. What do you know about Paul Revere's Ride, the Battle of Bunker Hill, or Valley Forge?
2. Tell about the Declaration of Independence.
3. "I haven't begun to fight yet." Tell the whole story. Or, discuss the relationship between Thomas Jefferson, and John and Abigail Adams.
1. Tell about the fall of the Bastille.
2. What do you know about Frederick the Great, Catherine the Great, or Dr. Jenner?
3. Tell about the Reign of Terror or Napoleon in Egypt.
1. "The Upper Mississippi made the Lower." Explain this. Or, Discuss some boats that changed the Mississippi River.
2. What is the difference between a fast flood and a slow one?
3. Describe a river from where it begins to where it ends.
1. Explain how Madame How's great ice-plough has helped make the ground on which we live.
2. Tell what you know about pollen, and why plants need insects.
3. Choose one question matched to your term's Nature Study theme from here.
Question will be taken from this term's Plutarch. *
Reading Skill
Father to choose an unseen passage, giving marks for enunciation.
1. If a toy car and a toy truck on a scale weighed 1 pound together, and the truck weighs 10 ounces, how much does the car weigh?
2. At breakfast the temperature outside was 66 degrees F. At lunchtime it had gone up 8 degrees F, and then by dinnertime it had gone down 6 degrees F. What was the temperature at dinnertime?
3. If 3 + n = 20, what is n?
1. Make sentences using the second person singular, present tense ("you hear...")
2. Optional; question will be taken from this term's Composer Study. *

redpajama_set_name: RedPajamaCommonCrawl


text:
Tadworth
Oxshott
Holland Park, Holland Park, London, W11 3RS, London, England, United Kingdom For Sale
Build Size: 9,542 ft2
The area is famous for its grand white stucco villas, tree lined avenues, private communal gardens and the public open space
of Holland Park. The park itself is located moments from the property and is home to The Belvedere Restaurant, the famous Japanese Kyoto Garden and the Walled Spring Tulip Garden.
The wider areas of Holland Park and Notting Hill are well known for being culturally vibrant and have a strong artistic heritage,
being home to many fashionable shops and a diverse offering of cafes and restaurants.
The property is well served by public transport, within walking distance are Holland Park and Notting Hill Gate Underground
stations, providing Central, Circle and District Line services, approximately 350 metres and 850 metres from the property
Situated toward the north western end of Holland Park, the Property comprises of a substantial 5 storey mansion, extending to approximately 9,542 sq ft.
Benefitting from a south facing garden, backing on to Holland Park Mews, the property has been an embassy for approximately 43 years, but has not been actively used for this purpose since 2011.
The property is now in need of full refurbishment, but subject to the necessary consents could be converted and restored as an elegant single family dwelling.
Notting Hill Notting Hill
I am interested in Holland Park, Holland Park, London, W11 3RS and would like more information
London Properties
Priory Gardens, Highgate, London, N6 5QS
Cloncurry Street, London, SW6 6DR
Asking Price £1,350 per week
Thorncote, Edgehill Road, Ealing, London
Guide Price £14,950,000
Asking Price £550 per week
Langdon Park Road, Highgate, London, N6 5QG
Ackmar Road, London, SW6 4UP
Trebovir Road, Earl's Court, London, SW5
Gloucester Street, Pimlico, London, SW1V 4EA
Chiddingstone Street, Parsons Green, London, SW6
Cumberland Terrace, Regent's Park, London, NW1 4HJ
Ormonde Place, Belgravia, London, SW1W 8HX
St. Quintin Avenue, North Kensington, London, W10 6NZ
Netherhall Gardens, Hampstead, London, NW3 5RL
Montagu Square, Marylebone, London, W1H
Asking Price £9,250,000
Flat 10 & 11
See More London Listings
Holland Park, Holland Park, London, W11 3RS, London, England, United Kingdom is a 9,542ft2 London luxury Single Family Home listed for sale Price on Application. This high end London Single Family Home is comprised of 9 bedrooms and N/A baths. Find more luxury properties in London or search for luxury properties for sale in London.


Us, on the other hand...well...we can only figure that Image Collections has had, for a time [24 years now ;o] a place in making some history of its own, and if you were a party to it, we hope it ended up for the best.
---Kermit the Frog--MUPPETS, now an all-ages COMIC BOOK!
"Batman: Better put 5 cents in the meter.
Batman: This money goes to building better roads. We all must do our part"
---Old 1960's BATMAN TV show...Now continuing again as a DC COMIC!!!
"You get what anyone gets,....you get a LIFETIME..."
. .. .. .. . . ..
... .. .. ... . .. .
.. ... . .. .. ..
. .. .. . . .
... .. .. .. .
. ..Way before the movie took many by surprise [...and went on to be the most successful of 2014!...], MARVEL COMICS just knew it had a hit on its hands, as the franchise had proven itself back in 2008-2010 with a series starring the same crew [Rocket Raccoon, Groot,...] which became a dark-horse critical smash, paving the way to the big screen...so it made sense to launch an ALL-NEW ONGOING title called 'GUARDIANS OF THE GALAXY'!
.. . . . . . ..


text:
Mark Donaldson
The Boy King
Moonlight Painting
Dogs And Kids
Norman Rockwell (American, 1894-1978) The Night Before Christmas (Santa Peering over Chair at Sleeping Child), Literary Digest magazine cover, December 22, 1923
Norman Rockwell Art
Olympia Le Tan
Norman Rockwell - Pipe and Bowl Sign Painter (Colonial Sign Painter), 1926 Oil on canvas, 27 x 21.88 in.
American Modern
"The Bookworm" cover illustration by Norman Rockwell (1926). 'The Saturday Evening Post' (August 14, 1926). Oil on canvas. 81.3 x 66 cm
Georgia Okeefe
Robert Motherwell
Norman Rockwell - "End of the Working Day" (Commuter Rush), 1920 oil on canvas, 76.8 x 63.5 cm. (30x 25 in.). Appeared on the cover of Literary Digest on November 6, 1920.
Most Famous Paintings
Mystic Messenger Memes
Pin Up Art
Artist Art
Norman Rockwell (American painter and illustrator) 1894 - 1978 Tender Years: Mowing the Lawn, 1957 oil on canvas 45.7 x 45.7 cm. (18 x 18 in.)
Freedom From Want
Four Freedoms
Lion Sculpture
Norman Rockwell (American painter and illustrator) 1894 - 1978; "Freedom of Worship (Study)", 1943 oil and pencil on board 27.3 x 26.7 cm. (10.75 x 10.5 in.) private collection, © photo Sotheby's
Very Cute Dogs
Newspaper Cover
Animal Medicine
Norman Rockwell - "The Veterinarian", 1961 oil on canvas, 53.3 x 50.8 cm. (21 x 20 in.)
Best Artist
Norman Rockwell - "The Organ Grinder", 1920 oil on canvasboard, 57.2 x 57.2 cm. (22.5 x 22.5 in.)
Little Girl Dancing
Online Painting
Norman Rockwell - "Jeff Raleigh's Piano Solo", The Saturday Evening Post, May 27, 1939.
Peintures Norman Rockwell
Framed Artwork
Norman Rockwell: "The Land of Enchantment", 1934 oil on canvas, 94 x 193 cm. (37 x 76 in.) New Rochelle Public Library, New York
Rockwell Paintings
Live Model
Renaissance Paintings
Rosie The Riveter
Painting Of Girl
"CHEERLEADERS" (LOSING THE GAME) - Norman Rockwell's "Cheerleader" appeared on the cover of The Saturday Evening Post on February 16, 1952.
Grand Art
Architecture Tattoo
Art Database
Old Master
Norman Rockwell - The Lineman, 1949.
American Illustration
Battle Of Borodino
Battle Of Moscow
Academic Art
Impressionist Artists
"Bright Future for Banking" by Norman Rockwell, circa 1955
Wall Art For Sale
LIBERTY GIRL (ROSIE TO THE RESCUE), Saturday Evening Post cover, 1943 Norman Rockwell hired a professional model for the Post cover Liberty Girl (Rosie to the Rescue), honoring the mythical embodiment of all the new roles that now belonged to the American woman during wartime. As a symbol of the ability of Americans to mobilize and transform themselves during the war effort, Rockwell chose to paint her as a girl-next-door, accentuating her authenticity. - Norman Rockwell Museum
Jc Leyendecker
Types Of Art
1956 Pan American ad by Norman Rockwell
Norman Rockwell (1894-1978) - Study for Dress Rehearsal, s.d. pencil on paper, 38.1 x 34.9 cm. (15 x 13.75 in.) signed Norman Rockwell (lower right); private collection. (The work is a study for Dress Rehearsal, which Rockwell created as part of a series of Christmas cards for Hallmark Cards, Inc., between 1946 and 1957.)
Norman Rockwell (American, 1894-1978) - "Homecoming Marine" (1945)
Dante Gabriel Rossetti
John William Waterhouse
Mary Cassatt
Norman Rockwell - "Overheard Lovers", 1936 oil on canvas, 76.5 x 61.3 cm. (30.13 x 24.13 in.)
Me And My Dog
Office Prints
Mermaid Art
Mermaid Poster
"Mermaid" or "Lobsterman" - Saturday Evening Post Cover, August 20,1955, by Norman Rockwell
Art Vintage
Norman Rockwell - "The Runaway" 1958 (The Saturday Evening Post)
Strange Art
Trumpet Practice, 1950 by Norman Rockwell
Paint Set
The Art Critic, 1972 by Norman Rockwell
Art Critique
Best Guacamole Recipe
Art Competitions
Live For Yourself
Illustration Art Nouveau
Art Tumblr
Norman Rockwell, "Murder in Mississippi, 1965" © Norman Rockwell Family Agency. Courtesy of the Norman Rockwell Museum and the New York Historical Society Museum & Library.
Norman Rockwell Christmas
Norman Rockwell (American,1894-1978) - "Christmas: Knight Looking In Stained Glass Window" (1930). Oil on canvas, 112.4 x 87 cm (44 1/4 x 34 1/4 in).


In today's society, beautifying a recently purchased house has turned out to be a priority for some. Individuals are particularly keen on getting the kitchen and bathroom space looking tip-top. The Kitchen has its instant matching units that effortlessly fit together, yet bathrooms? They are to a greater extent a blend and match where you need to work harder to get the ideal bathroom or cloakroom space. Lots of different types of hardware including shower cubicle, bathroom suite and heating products need to be brought together to create a harmonious and peaceful space.
The walk in shower spaces add genuine magnificence to a shower tray and your bathroom. There are numerous sorts of shower cubicles that can give a cutting edge look to your home. There are numerous online decorators and bathroom decorators who offer good rates for fitting. One ought to be ready and mindful to seek at least 3 competitive written quotes as opposed to simply going with the first option. You should be clear on what kind of shower unit you might want to have in your home, get more details straight from the source.
There are a couple focuses which you ought to have a personality a main priority before you go in for a stroll in shower area. The materials in which it is made from matter and subsequently you ought to examine all the accessible materials to offer you some assistance with choosing the best choice for you. There are likewise units accessible in diverse sizes relying upon how much accessible space you have in your lavatory. Before you go in for such a stroll in shower work space, you ought to ensure that you have the required estimations for your washroom area.


text:
Home News Pablo Ferreira named Soft2Bet chief marketing officer
Pablo Ferreira named Soft2Bet chief marketing officer
Pablo Ferreira has stated that it's a "real privilege" and a "fantastic step", after being named as the chief marketing officer of casino and sportsbook platform provider Soft2Bet.
In the position, his remit will include creating the promotional strategy for the company's flagship 2021 sportsbook, lauded as well as "a highly anticipated brand launch scheduled for the coming months.
Boris Chaikin, CEO at Soft2Bet, stated: "We are thrilled to name Pablo Ferreira as chief marketing officer and have no doubt that his appointment will bring continued success to the business.
"We're gearing up for a busy year here at Soft2Bet, but with such talented employees as Pablo on board we will continue to exceed the requirements of partners across the globe."
Following a successful spell as Soft2Bet's head of country managers, Ferreira's new position will involve overseeing the marketing and promotion of Soft2Bet's more than 30 strong operator portfolio, including the SBC award-winning igaming entity Wazamba.
The platform provider's new CMO brings a wealth of industry experience to the role, following a multi-year stint as head of marketing in LatAm and Spain for the then GVC Holdings, and a managerial tenure at Sportingbet.
"Becoming chief marketing officer at Soft2Bet is a fantastic step for me personally, and managing such a talented and hardworking team will be a real privilege," added Pablo Ferreira, CMO at Soft2Bet.
"We've made tremendous progress in recent months, winning awards, releasing new products, and extending our ever-growing commercial reach even further. Looking to the year ahead, I have no doubt that that success will continue."
The group currently holds Malta Gaming Authority, Swedish Gambling Authority and Curacao licences, with further certification from the EU set to follow.
Pablo Ferreira
soft2bet
Previous articleLondoner Macao to launch in first phase on February 8
Next articleOhio becomes WynnBet's eighth state of access


text:
Vermonter of the Year: Hal Colston
Aki Soga
Free Press Reader Engagement Editor
Hal Colston is the founder of Good News Garage.
He launched NeighborKeepers.
He was a leader in confronting racial profiling.
He the director of Partnership for Change.
You might call Hal Colston a serial do-gooder.
The Winooski resident has laid down a track record of work to improve the lives of people in the community during his 25-year tenure in Vermont.
Colston is best known as the founder of Good News Garage, the nonprofit that fixes donated vehicles for low-income people.
He also is known for launching NeighborKeepers, an initiative to help families in need build relationships with those who can provide the support they need.
He was instrumental in forcing Vermont to confront the issue of racial profiling by the police.
Colston says he sees every day as an opportunity to make a difference.
"We get them one at a time with no guarantee for tomorrow," he said. "May all of us spend our days wisely to improve the common good."
Colston's understated demeanor often belies his passion. He works to give voice to those who are unable to be heard on their own. He has shown the courage to tackle difficult issues.
Colston's quiet devotion to bringing the people together and looking out for those who find themselves in unfortunate circumstances especially stand out in a year that saw the streets of this country fill with protest — both peaceful and violent — against injustice.
For his years of service to the community in the name of social and economic justice, the Burlington Free Press editorial board names Hal Colston 2014 Vermonter of the Year.
In nominating Colston for Vermonter of the Year, Laban Hill of Winooski writes, "Hal has spent his life devoted to our community and making a difference in both small and large ways."
Colston's work with Good News Garage has earned him national recognition, including an appearance on the Oprah show. The idea is simple. For many, one of the bigger barriers to economic independence is the lack of reliable transportation. Good News Garage tackles that problem in the most direct way possible by awarding reliable cars to those in need.
Colston founded the organization in 1996 after meeting a Lutheran minister he had heard in Philadelphia shortly before moving to Vermont. That chance encounter led to developing an idea and seed funding from what he calls a pan-Lutheran organization. He launched the nonprofit under the wing of the Lutheran Social Services New England.
Nearly two decades later, Good News Garage has awarded more than 4,000 vehicles.
Second career
His career in community service also includes a stint as associate director of Community Action in Burlington, now Champlain Valley Office of Economic Opportunity, as executive director of the Vermont Commission on National and Community Service and as diversity coordinator for the social services nonprofit HowardCenter.
Colston also spoke out clearly and firmly when African-Americans in the community charged that local police were using racial profiling in stops.
In an April 2007 My Turn piece in the Free Press, Colston wrote, "I believe that racial profiling in Vermont is an epidemic," going on to recount his own experience with "driving while black."
Colston did more than complain. He worked to open a dialogue within the community, including the police, that allowed people to talk openly about issues surrounding race.
"We're building trust," he said. "I don't believe we will ever eradicate racism, but how do we heal from the wounds?"
All this was a major shift for a man who had made a career as a chef and restauranteur in Philadelphia, and arrived in Vermont in 1989 to become the director of catering at the New England Culinary Institute in Essex.
Colston called his career change a midlife crisis, but said the work put him in touch with his core values, "truth and justice."
Today, Colston is engaged in what is perhaps his biggest challenge to date. He serves as director of Partnership for Change, a collaboration between the Burlington and Winooski school districts. The mission is to re-imagine public education to better prepare students from diverse social, economic and cultural backgrounds to succeed in school, in their careers and as members of their communities.
"The reason I love Vermont is it's got its challenges. But it's really on a human scale. You can have a conversation," he said.
For his steady faith in the ability of each person to make a difference, and putting that faith to work in the service of his community, Hal Colston is the 2014 Vermonter of the Year.
Join the conversation online at BurlingtonFreePress.com or send a letter to the editor to letters@freepressmedia.com. Contact Aki Soga at asoga@freepressmedia.com. Follow him on Twitter at @asoga.
A nominating letter
I would like to nominate Hal Colston for Vermonter of the Year.
Hal is director of Partnership for Change, which is remodeling Burlington and Winooski school districts by establishing a student-centered learning system that enables all learners to develop skills, knowledge, and relationships necessary to become confident, motivated, and self-sufficient learners who are successful in college and careers and are engaged in their communities.
Hal and his team are changing the way education is delivered in our communities so that it is more rigorous and more equitable.
Over the years Hal has been an integral and essential part of our community. He founded Good News Garage, which is one of the first nonprofit social enterprise car donation programs in the U.S. He also founded NeighborKeepers, which was a community nonprofit that helped the generational poor become financially secure.
Hal has spent his life devoted to our community and making a difference in both small and large ways. He seems like the ideal candidate for the Burlington Free Press's Vermonter of the Year. In addition, there has never been a person of color who has been recognized as Vermonter of the Year. It's about time.
There are so many people of color in Vermont who are making important contributions to our community. Now is the time to begin recognizing them.
LABAN HILL,Winooski
2013: Lois McClure, philanthropist


text:
Woman arrested after man dies in serious collision in Bacup
The 19-year-old woman remains in police custody
Chantelle HeedsLive News Reporter
Market Street in Bacup was closed for three hours due to investigation work (Image: SS Digital Images)
A 19-year-old woman has been arrested after a man died in a Bacup collision.
The collision involved a white Suzuki Alto and a male pedestrian.
It took place on Market Street, adjacent to the Texaco service station, at 1am on Monday (June 17).
The pedestrian, a 24-year-old local man, died at the scene.
The driver of the vehicle, a 19-year-old woman from Bacup, was arrested on suspicion of causing death by careless driving when under the influence of drink or drugs.
She is currently in police custody.
Lancashire Police attended the scene and closed the road for around three hours due to collision investigation work.
Shocking moment spurned lover had violent meltdown in Southport milkshake bar
Sergeant Andy Ainsworth said: "An investigation is underway following a fatal collision in Bacup.
"A man has died and our thoughts are with his family at this difficult time.
"A Family Liaison Officer has been appointed to support the man's family.
"We are appealing to anyone who witnessed the collision and has not yet spoken with police to come forward.
"I am particularly interested in speaking to a group of males who stopped at the scene – they may have information useful to our investigation."
Anyone with information is asked to contact police on 101 quoting log reference 69 of June 17th.
Alternatively, information can be reported anonymously via Crimestoppers UK on 0800 555 111.
Young thug who robbed taxi drivers of cars could be classed a 'dangerous offender'
Nine-day Rossendale Heritage Festival promises something for everyone


text:
Treason - 50 Years and Still Going Strong!
While on vacation last week, I had the opportunity to finish a book I started over Christmas vacation - Treason: Liberal Treachery from the Cold War to the War on Terror by Ann Coulter. In it, she meticulously (and I mean meticulously - there are 877 footnotes throughout 292 printed pages) details how the left, liberals, Democrats (pick your favorite label); these folks have been wrong about almost every foreign policy decision they've made over the last 50 years. (Today, another revelation occurred that, to me, shows that it hasn't stopped yet - but more on that later.)
The book starts with a look at Alger Hiss and Senator Joseph McCarthy. McCarthy is probably one of the most unfairly maligned people in United States history - especially since, within the last decade, declassified Soviet cables (intercepted by the Venona Project) have proved his allegations true. First, she makes the excellent point that Senator Joe McCarthy had nothing to do with the House Unamerican Activities Committee (HUAC). Instead, his modest job was to identify loyalty risks serving in sensitive positions. He was very good at his job, and identified several people that needed to be moved - not fired, not tried for treason and hanged, just moved out of the sensitive positions where they were currently.
Over the past 50+ years, the left has consistently been for containment, appeasement, and has been more than willing to cede portions of our national security to keep us from having to fight wars. I can't recall who said it, but the gist of it was "If these folks were just stupid, the laws of chance would dictate that they'd occasionally be right!" One paragraph (on page 154 of the paperback edition) summarizes just how duped they were in the Cold War, while debunking the claim that Truman's policies actually won the Cold War, it just took until Reagan's administration for them to work.
To review the record, as part of Truman's yeoman work on the Cold War, he cooperated with the Soviets at the Nuremberg Trials, white-washing their joint aggression with Hitler under the Nazi-Soviet Pact. He looked the other way when the Soviet Union murdered three million Russian prisoners of war returned home by the Allies. On his watch, the Soviet army consolidated its control over nine countries, China became a Communist dictatorship, and tens of millions of people were murdered under Communist tyrannies. Truman defended Communist spy Alger Hiss as a patriot who was framed by Republicans, and he tried to indict Whittaker Chambers for perjury. He refused to remove members of his administration identified to him by J. Edgar Hoover and others as Communist agents, including Harry Dexter White, whom Truman appointed to U. S. representative of the IMF. Among the Soviet spies advising Truman on China was Frank Coe, who refused to answer the question: "Are you a Soviet agent, Mr. Coe?" Soon thereafter Coe fled to Communist China, where he became a top policy-maker to Chairman Mao, helping the Chinese murder tens of millions of their own people. Truman was considered such a dupe of the Communists that the Army refused to tell him about the Venona Project. And that's how Truman won the Cold War!
Ann continues in other chapters to discuss the Bay of Pigs invasion, where we went back on our word by not supporting the Communist resitance there when Castro had very little power; Vietnam, where we were starting to make progress until a Democrat-controlled Congress used Watergate as an excuse to leave before winning; Grenada, where we went in and defeated Communism over a weekend; the Cold War, where an arms race did prove to be the way to bankrupt the Soviet economy; Iraq, where the same arguments once used against Vietnam and Grenada were once again being brought up, but yet we prevailed (not by appeasement, or inspections, but through force); and North Korea, where recent adminstrations have dropped the ball, but the current one is taking a strong stand.
While those who know Ann Coulter know that she simply delights in shredding the left to pieces, the content of this book is much more important that her amusing delivery method. What this book shows is that, using the 20/20 benefits of hindsight, the left's approach to defending our country only makes us weaker. When the decision was made to not prosecute Jane Fonda for her treasonous actions during the Vietnam War, the message was sent that this country wasn't interested in prosecuting anyone for these types of actions. So, Hollywood and the media have been more than willing to participate in this as well - which brings us to today's happenings.
Last week, Newsweek magazine ran a story that claimed that, as an interrogation technique, the Army had flushed a copy of the Koran down a toilet. This caused riots in Afghanistan, where 15 people have been killed and over 100 injured. Today, they ran another article where they said that their source for that piece of information is now unsure that that's what happened. (Separately, they have claimed that this is not a retraction of the story.) So, what we've got here is an international news magazine including uncorroborated hearsay from a single source in a story, claiming it as fact. I have several problems with this.
First, didn't Newsweek learn anything from the CBS News / Bush ANG document fiasco? The public deserves better, and our troops on the ground certainly deserve better fact-checking before they have to contain riots for no good reason.
Second, why was Newsweek so eager to run this part of the story? I believe it's the same reason that CBS News ran with the Bush ANG story - they wanted terribly for it to be true. For all their accusations, it's the left in this country that doesn't seem to realize that the "War on Terror" is not a "War on Islam". They want us to disrespect Islam, so then they can say they were right.
Third, this bad decision has placed our military in harm's way. Iraq is not the only place we've got a military presence, training locals to take over their own security. Afghanistan is farther along than Iraq, but we're still there. With claims like this, even if they are true, publicizing them as some scandalous new revelation gives aid and comfort to our enemies. All they need is one story like this to fire them up for days - and now, not only have they given the opposition the morale boost, 15 people are dead because of it. There's no way that either the reporter or the editors will be tried for either treason or murder - but I believe a case could be made for either of those charges.
Finally, what does it matter if they did do it? I'm not a student of Islam, but as a Christian, I believe the Bible is the holy, inspired Word of God. Any feeble attempt by man to eliminate it is simple futility - others have attempted to destroy God's Word, and have had little success. Flushing paper and ink down a toilet does not come anywhere close to destroying it - only that copy. I would think that Muslims would feel the same way about their holy book - but, I may be wrong. I don't see what it would hurt, though - these terrorists already call us the Great Satan!
If you have the opportunity, pick up Ann's book at the store or library, and read through it. It's a great analysis of conflicting foreign policy over the last 50 years - and, it's fun to read to boot!
Categorized Books Entertainment Liberal Moonbats National Defense Politics War on Terror
Tagged alger hiss ang ann coulter bay of pigs communism george w bush iraq joseph mccarthy koran newsweek spies toilet treason venona project
"No Nukes Is Good Nukes?"
"Terri Schiavo"


This research, however, is concerned with the identification and analysis of regional imbalances in agricultural productivity pattern in South Bihar Plain. Quantitatively, it analyses the environmetal, technological and social factors as the causes. This provides an appropriate conceptual base for increasing regional disparities in agricultural productivity and suggests methods for increasing productivity. Although this part of Bihar is characterised by high diversity of population, backward agriculture, small fields, uneconomic holdings, undue utilization of ground and surface water resources and object poverty.
The book analyses the environmental, technological and social factors as the causes for increasing regional disparities in agricultural productivity and suggests methods to increase it. It examines agricultural productivity at micro and macro level, etc.


text:
Q: Put a GeoJSON on a Leaflet map: Invalid GeoJSON object. throw new Error('Invalid GeoJSON object.'); I've some trouble with a Leaflet code that build a GeoJSON and put it on the map.
The GeoJSON is builded from an XML response of a service.
The error is
Invalid GeoJSON object. throw new Error('Invalid GeoJSON object.');
My code is ...

<title>Prova caricamento WMS Indirizzi del PCN</title>
<meta charset="utf-8" />
<meta name="viewport" content="width=device-width, initial-scale=1.0">
    <link rel="stylesheet" href="http://cdn.leafletjs.com/leaflet-0.7.2/leaflet.css"/>

You can copy/paste it and shoud work on your desktop/laptop.
Note that if you uncomment the row ....
//var addressPCN = {"type": "FeatureCollection","features":[{"type": "Feature", "properties": { "popupContent": "Via Pegolo 9", "show_on_map": true }, "geometry": { "type": "Point", "coordinates": [7.719490,44.849197]}},{"type": "Feature", "properties": { "popupContent": "Via Pegolo 8", "show_on_map": true }, "geometry": { "type": "Point", "coordinates": [7.719490,44.849197]}},]};

.... you've the GeoJSON I'm building (you can match it with the alert message ....).
In this case all it's working right.
NOTE: you've to use Firefox or Chrome at the momenti, my code still doesn't work foi IE.
Any suggestion is appreciated!!
Thank you in advance!!
Cesare

A: The problem was about I tried to use a string as a JSON object.
The solutions is to use jQuery.parseJSON function to convert my string with my JSON code (addressPCN1) to a Javascript JSON object.
Here you are code that works ...
<!DOCTYPE html>

  <title>Prova caricamento WMS Indirizzi del PCN</title>
  <meta charset="utf-8" />
  <meta name="viewport" content="width=device-width, initial-scale=1.0">
  <link rel="stylesheet" href="http://cdn.leafletjs.com/leaflet-0.7.2/leaflet.css"/>

A: First, make sure the Geo JSON data is valid data with http://geojsonlint.com/ or any other service. Fix the errors if there are any.
And then parse your data which is in the text format with JSON.parse():
L.geoJSON(JSON.parse(data)).addTo(this.map);

redpajama_set_name: RedPajamaStackExchange


Joining this community will CONNECT you with thousands of local supportive women, MOTIVATE you to reach new heights of success & create new OPPORTUNITES for you to grow your empires!
Australia's most supportive community for women in business and women who dream of building their empires. We hold regular SELL OUT events to connect like minded local women all over Australia! You will leave our events with your passion re-ignited and the support and motivation to take your business to the next level. The Girls in Business Movement is for ALL women – in business or not!
All for less than a cup of coffee a week $15 Per Month (ex GST). No Contracts, Cancellation Periods or Hidden Fees! What are you waiting for? Join your SUPPORTIVE SISTERS today!
Just starting out with a new business venture? We will support you through those beginning stages, share tips and offer advice on how to grow your business empires. Join your FREE local facebook support group below!
Asha Peck – Wife, Mother, Friend, Motivator and the brains behind the fastest growing women's empowerment movement for women in business in Australia!
Want to be the first to know what's happening?
Sign up for special offers & event information! We promise not to send you junk, just the good stuff!
Amanda Clare – Plush Content Co.
This group is full of some of the most amazing, talented and downright gorgeous women I have ever had the pleasure of meeting!
My business is still small, but since joining GCGIB a couple of months ago, business has BOOMED!! I attribute this 100% to the support, advice and encouragement I receive from these ladies on a daily basis. They encourage and inspire me to take leaps I wouldn't have dreamed of taking in the past. Women supporting women is the most uplifting movement and I am so proud to be able to call myself a GCGIB member!
IT'S HERE! The GCGIB Gala Awards Night video thanks to the highly talented @paigehudsonproductions 💖🥰 Still pinching myself how amazing this event was!


text:
Атом Виллард (настоящее имя Адам Дэвид Виллард, ; 15 августа 1973) — американский барабанщик. Он начал свою карьеру в 1990 году, когда присоединился к группе Rocket from the Crypt, где играл до 2000 года. Впоследствии он был членом групп The Special Goodness и The Offspring, а в 2005 стал одним из основоположников группы Angels & Airwaves. В 2007 году Виллард заявил о своем уходе из группы The Offspring ради того, чтобы больше времени уделять работе в Angels & Airwaves.
В сентябре 2011 Атом ушёл из AvA.

Дискография
В этом разделе представлен список альбомов и EP, на которых Виллард выступал.

Примечания

Родившиеся в 1973 году
Панк-рок-музыканты США
Ударники США
Википедия:Персоналии, не категоризованные по месту рождения
Музыканты Angels & Airwaves

redpajama_set_name: RedPajamaWikipedia


The General Assembly comprises all members present when it meets during the annual congress. It has the right to make fundamental decisions on the life of IAML, discussing the plans, the current work and the problems of the Association. It votes on the budget for the next three financial years.
Available General Assembly Minutes are listed below.


Every so often parents and/or skaters will insist on purchasing skates that are outside our recommendations. It is important that staff provide information on getting a good fit and an appropriate model for that skaters height, weight and ability. Should the customer insist on going against North Edge staff recommendations, it is of critical importance the customer read and sign the Skate Fitting Waiver. Following this process waivers must be placed in the Skate Fitting binders for future reference. A note on the customers account must also be made.


text:
Diversions Front Page Money Reviews U.S.SURFIN' SAFARI
Facebook welcomes investors, pedophiles … huh?
Exclusive: Andrea Shea King covers bombshell news, hottest trends in technology
By Andrea Shea King
WND's stunning four-part expose of the dark underbelly of Facebook hit the front page on Sunday night, just ahead of the Internet social media giant's initial public offering expected to value the company as high as $100 billion.
The introduction to Chelsea Schilling's shocking expose that should be read by every parent offers the following warning: "This report contains graphic details of sexual abuse of children as it has appeared in numerous locations on Facebook. WND immediately reported images of child pornography and child sexual abuse to the FBI. Censored screenshots published are among the mildest of those found."
As mentioned, Facebook will begin pitching its initial public stock offering to investors and has posted the presentation so anyone can see it.
Meanwhile, Facebook is looking at up to a $95-billion-plus valuation. The IPO is predicted to be in the $28 to $35 range, but most are ignoring the lower figure. According to a regulatory filing, Facebook hopes to raise as much as $11.8 billion in its IPO, the highest ever for a Web company.
Irrelevant or spam?
TRENDING: Trump blasts 'power-hungry lunatics': 'We are not going to take it anymore'
"This comment seems irrelevant or inappropriate and can't be posted. To avoid having comments blocked, please make sure they contribute to the post in a positive way." – Facebook
Is Facebook determining what is inappropriate or irrelevant to an FB conversation? Apparently so, according to Robert Scoble, tech start-up enthusiast, who attempted to post a comment on a Facebook post written by a Carnegie Mellon student about the tech blogging scene and received this odd error message.
But according to Facebook, it's not censorship, it's spam filters.
Cyber complaint desk
The more people become acquainted with social media, the more effective it will be to tell companies and the government exactly what they like and don't like about their service or products.
Last week we told you about Gripevine, a new site developed by the musician whose guitar was smashed by careless baggage handlers with United Airlines.
As I wrote, Gripevine is a David-versus-Goliath story about how a guy with a little knowledge of YouTube, some determination and a gripe created a company that provides resolution for customers with a complaint.
This week, we've got news of a new app that gives you the ability to complain about your TSA screening. It's called the "FlyRights" free mobile application, and it was launched by a Sikh advocacy group to give travelers a way to complain immediately to the government if they think they've been unfairly treated by airport TSA screeners.
Complainers simply input relevant information, and click on the "report" button. They can also click on the "submit" and "share" buttons to post the complaint on social media.
In a related matter, Facebook users brought public opinion to bear on Spirit Airlines which took a gigantic PR hit last week when the carrier's CEO refused to back off the airline's policy of non-refundable tickets. The public relations nightmare began when Spirit refused to refund a 76-year-old Vietnam veteran's $197 ticket when he was unable to fly due to a diagnosis of terminal cancer. The story went viral on the "Boycott Spirit Airlines" Facebook page, which attracted 21,000 "likes," thus pressuring the airline to refund the man's ticket.
All electronic devices must be turned off ...
Some time ago I flew from Orlando to New York City beside an off-duty airline pilot. We talked about how the pilots in the cockpit are now using iPads instead of the traditional Jeppesen paper aviation charts. I asked him why pilots at the controls can use the same electronic devices that passengers are told to turn off during flight, lest they interfere with aircraft avionics.
He explained in confidential tones that despite FAA regulations preventing it, the electronic devices do not affect the aircraft, and he surmised that the rule is in place to ensure that passengers pay attention during take-off emergency instructions (they don't, generally. Look around the next time you fly. The flight attendants are playing to an empty room.) The friendly pilot confided that the rule also was in place to prevent phone conversations from disturbing other passengers.
Last week, a flyer with an iPad got a reprimand letter from the FAA for videotaping birds striking his Delta aircraft as it took off from New York's John F. Kennedy International Airport.
"If these electronics are actually dangerous, the FAA doesn't need to send me a letter. They should ban them from the airlines," the flyer said.
Apparently the FAA does not. The offending passenger won't be fined – but the letter "will be made a matter of record for a period of two years."
"A record with whom?" asked the frequent flier who is concerned that he's on some no-fly list. He said the letter "has a lot of Big Brother in it."
Just the federal government looking out for you and me.
Blogger Doug Ross of DirectorBlue.blogspot was on my radio program last week, telling listeners about a news aggregator he's designed that scans some 700 center-right blogs and publications for headlines of interest to the conservative moment. It's called Bad Blue, and the feed is real-time, with constant refreshings.
Doug says it's the news that the old media doesn't want you to see, and he's right. Add Bad Blue to your bookmarked toolbar tabs and visit it – in fact, when you're tempted to take a peek at what Drudge has up on his page, click on Bad Blue too.
Avast, mateys!
A computer technician who builds, rebuilds and repairs computers says he installs it on every computer he works on. What is it? A free anti-virus software program called Avast. It scans hard drives, e-mails, exe. files and more, blocking viruses and spyware, and best of all it's free. Of course, more comprehensive versions are available for a fee. But he swears by the free version.
Teachers no longer allowed to "friend" students
A new set of guidelines issued by the New York Education Department prohibits public school teachers from contacting students through personal pages on social media sites such as Twitter and Facebook. Teachers should communicate to their students only through pages set up for classroom use or on professional accounts.
BRAD: Berkeley Ridiculously Automated Dorm. A campus trend? Probably
A computerized system that uses remote appliance control modules, iPad and iPhone apps and voice recognition software controls lighting, music and curtains in a Berkeley college dorm room. Watch this video to see the technology in action, even in "party mode," with strobe and laser lighting! Even an emergency party button!
Family kidnapped by ninjas – need $ for karate lessons
Thinking of doing an on-line start-up? If so, understand these concepts before taking the plunge.
Andrea Shea King is a talk-radio host who also writes at The Radio Patriot website and is known as Central Florida's "First Lady of Space Coast Conservatism."
Super Bowl babies? Not a myth – watch them sing
Girl has unusual reaction to news she'll meet Trump
Lighter side of blizzard of the century
What is the most successful 1970s rock band?
Is this the world's most massive snowplow?


Burnshield 10cm x 10cm burns dressings are a highly effective emergency first aid burn dressings. The unique versatile foam dressing of the 10cm Burnshield dressing provides essential immediate treatment and protection following a burn. The 10cm burnshield dressing can be used on 1st, 2nd or 3rd degree burns - including facial burns and scalds.