text:
\section{Introduction}
Let~$\Omega \subset \mathbb R^2$ be a convex domain  and let us
denote by~$\gamma$ and~$dm_\gamma$ the standard Gaussian
function and measure in~$\mathbb{R}^{2}$ respectively, that is
\begin{equation*}
  \gamma(x,y):= \exp\left(-\frac{x^2+y^2}{2}\right)
  \qquad \mbox{and} \qquad
  dm_\gamma := \gamma(x,y) \, dx \, dy
  \,.
\end{equation*}
In this paper we consider the following Neumann eigenvalue problem
for the Hermite operator
\begin{equation}\label{problem}
\left\{
\begin{array}{ll}
  - \mathop{\mathrm{div}}\nolimits (\gamma \nabla u)
   = \mu \gamma u
  &  \mbox{in} \quad \Omega \,,
  \\ \\
  \dfrac{\partial u}{\partial \bf{n}} =0
   & \mbox{on} \quad \partial\Omega \,,
\end{array}
\right.
\end{equation}
where~$\bf{n}$ stands for the outward normal to $\partial \Omega$.
As usual, we understand~\eqref{problem} as a spectral problem for
the self-adjoint operator~$T$ in the Hilbert space
$L^2_\gamma(\Omega) := L^2(\Omega,dm_\gamma)$
associated with the quadratic form
$
  t[u] := \|\nabla u\|^2
$,
$
  \mathsf{D}(t) := H_\gamma^1(\Omega)
$.
Here $\|\cdot\|$ denotes the norm in $L^2_\gamma(\Omega)$ and
\begin{equation*}
  H_\gamma^1(\Omega) :=
  \{u \in L^2_\gamma(\Omega) \ | \
  \nabla u \in L^2_\gamma(\Omega) \}
\end{equation*}
is a weighted Sobolev space equipped with the norm
$\sqrt{\|\cdot\|^2+\|\nabla \cdot\|^2}$.
Since the embedding
$H_\gamma^1(\Omega) \hookrightarrow L^2_\gamma(\Omega)$
is compact (see Remark \ref{oss} below), the spectrum of~$T$ is purely discrete.
We arrange the eigenvalues of~$T$ in a non-decreasing sequence
$\{\mu_n(\Omega)\}_{n=0}^{+\infty}$ where each eigenvalue
is repeated according to its multiplicity.
The first eigenfunction of~\eqref{problem} is clearly a constant
with eigenvalue $\mu_{0}(\Omega)=0$ for any~$\Omega$.
We shall be interested in the first non-trivial eigenvalue
$\mu _{1}(\Omega)$ of~\eqref{problem},
which admits the following variational characterization
\begin{equation}\label{minimax.intro}
  \mu _{1}(\Omega ) =
  \min
  \left\{
  \dfrac{\displaystyle\int_{\Omega }|\nabla u|^{2}\,dm_\gamma}
  {\displaystyle\int_{\Omega } u^{2}\,dm_\gamma} \ : \
  u\in H_\gamma^{1}(\Omega)\setminus \{0\} \,, \
  \int_{\Omega } u \, dm_\gamma=0\right\}
  .
\end{equation}

A classical Poincar\'e-Wirtinger type inequality
which goes back to Hermite (see for example \cite[Chapter II, p.~91 ff]{CH1})
states that
\begin{equation}\label{poincare}
\mu_1(\mathbb R^2)=1
\end{equation}
and therefore
\begin{equation*}
  \int_{\mathbb R^2 }\left( u-\int_{\mathbb R^2}u \, dm_\gamma\right) ^{2}dm_\gamma
  \leq
  \int_{\mathbb R^2}|\nabla u|^{2} \, dm_\gamma
  \,,
  \qquad \forall u\in H_\gamma^{1}(\mathbb R^2)
  \,.
\end{equation*}
Very recently an inequality analogous to \eqref{poincare} raised up in connection with the proof of the ``gap conjecture'' for bounded sets (see \cite{AC}). In \cite{Andrews-Ni_2012} the authors prove that if $\Omega$ is a bounded, convex set then
\begin{equation}\label{an-ni}
\mu_1(\Omega) \ge \mu_1\left(-\frac{\mathrm{d}(\Omega)}{2},\frac{\mathrm{d}(\Omega)}{2}\right)
\end{equation}
where $\mathrm{d}(\Omega)$ is the diameter of $\Omega$ and, here and throughout, $\mu_1\left(a,b\right)$ will denote the first nontrivial eigenvalue of the Sturm-Liouville problem
\begin{equation}\label{N_1d}
\left\{
\begin{array}{ll}
-\left( \gamma_1v^{\prime }\right) ^{\prime }=\mu \gamma_1v &
\mbox{in}\ \left( a,b\right) \,, \\ \\
v^{\prime }\left( a\right) =v^{\prime }\left( b\right) =0 \,, &
\end{array}
\right.
\end{equation}
with $-\infty \le a <b \le +\infty$ and
\begin{equation*}
  \gamma_1(x):= \exp\left(-\frac{x^2}{2}\right).
\end{equation*}
Again, we understand~\eqref{N_1d}
as a spectral problem for a self-adjoint operator
with compact resolvent in $L^2_{\gamma_1}((a,b))$.
It is well-known that
\begin{equation}\label{l(R)}
\mu _{1}(a,b)\geq 1
\quad \text{ with } \quad \mu _{1}(a,b)=1
\text{ \ if and only if \ }(a,b)=\mathbb{R}.
\end{equation}

As first result of this paper we extend the validity of~\eqref{an-ni}
to any convex, possibly unbounded, planar domain (see \cite{BCHT} for the smooth case).
\begin{theorem}\label{main copy(1)}
Let $\Omega \subset \mathbb{R}^{2}$ be any convex domain. Then
\begin{equation}\label{mi}
\mu _{1}(\Omega )\geq \mu_1(\mathbb R)=1 \,.
\end{equation}
\end{theorem}

The result is sharp in the sense that
the equality in~\eqref{mi} is achieved
for~$\Omega$ being any two-dimensional strip.
It is natural to ask if the strips are the unique domains
for which the equality in~\eqref{mi} is achieved.

We provide a partial answer to the uniqueness question
via the following theorem, which is the main result of this paper.
\begin{theorem}\label{main}
Let $\Omega $ be a convex  subset of
$
  S_{y_{1},y_{2}} :=
  \left\{ (x,y)\in \mathbb{R}^{2}:y_{1}<y<y_{2}\right\}
$
for some $y_1,y_2\in \mathbb R$, $y_1 < y_2$.
If $\mu _{1}(\Omega )=1$, then $\Omega $ is a strip.
\end{theorem}

Inequality \eqref{an-ni} is a Payne-Weinberger type inequality
for the Hermite operator. We recall that the classical Payne-Weinberger inequality
states that the first nontrivial eigenvalue of the Neumann Laplacian
in a bounded convex set~$\Omega$, $\mu_1^{\Delta}(\Omega)$,
satisfies the following bound
\begin{equation}\label{PW}
\mu_1^{{\Delta}}(\Omega)\ge \frac{\pi^2}{\mathrm{d}(\Omega)^2},
\end{equation}
where $\pi^2/\mathrm{d}(\Omega)^{2}$  is the first
nontrivial Neumann eigenvalue of the one-dimensional Laplacian
in $\left( -\mathrm{d}(\Omega)/2,\mathrm{d}(\Omega)/2\right)$ (see \cite{PW}).
The above estimate is the best bound that can be given in terms of
the diameter alone in the sense that
$\mu_1^{\Delta}(\Omega){\mathrm{d}(\Omega)^2}$ tends to $\pi^2$ for a
parallepiped all but one of whose dimensions shrink to zero (see \cite{HW,V}).

Estimate \eqref{mi} is sharp, not only asymptotically,
since the equality sign is achieved when $\Omega$ is any strip~$S$.
Indeed, it is straight-forward to verify that $\mu_1(S)=\mu_1(\mathbb R)=1$ for any strip~$S$.
Hence the question faced in Theorem \ref{main} appears quite natural.

The paper is organized as follows.
Section~\ref{Sec.ext} contains the proof of Theorem~\ref{main copy(1)},
while Section~\ref{Sec.main} is devoted to the proof of Theorem~\ref{main}.
The latter consists in various steps.
We firstly deduce from~\eqref{mi} that any optimal set must be unbounded;
then we show that it is possible to split an optimal set~$\Omega$
getting two sets that are  still optimal and have Gaussian area $m_\gamma(\Omega)/2$.
Repeating  this procedure we obtain a sequence of thinner and thinner,
optimal sets~$\Omega_k$ and
we finally prove that there exists $a\in \overline \mathbb R$
such that $\mu_1(\Omega_k)$ converges as $k\to+\infty$ to $\mu_1(a,+\infty)$,
which is strictly greater than 1 unless $a=-\infty$.
This circumstance implies that $\Omega$ contains a straight-line,
and hence~$\Omega$ is a strip.

The convergence of $\mu_1(\Omega_k)$ to $\mu_1(a,+\infty)$
follows by a more general result established in Section~\ref{Sec.as},
where we actually prove a convergence of \emph{all} eigenvalues of~$T$
in thin domains to eigenvalues of a one-dimensional problem
(see ~Theorem~\ref{Thm.convergence}).
We also establish certain convergence of eigenfunctions.
We believe that the convergence results are of independent interest,
since our method of proof differs from known techniques
in the case of the Neumann Laplacian in thin domains
\cite{Arrieta_1995,Arrieta-Carvalho_2004,Jerison2,Schatzman_1996}.

For optimisation results related to the present work,
we refer the interested reader to \cite{BCT_p,ENT, FNT,BCM,CdB}.

\section{Proof of Theorem~\ref{main copy(1)}}\label{Sec.ext}
Repeating step by step the arguments contained in~\cite{BCHT},
Theorem~\ref{main copy(1)} is a consequence of the following extension result,
which we believe is interesting on its own.
\begin{theorem}\label{ext}
Let $\Omega\subset \mathbb R^2$ be a convex domain
and let $u \in H_\gamma^1(\Omega)$.
Then there exists a function $\tilde u \in H_\gamma^1(\mathbb R^2)$
which is an extension of $u$ to $\mathbb R^2$, that is
$$
\tilde u|_{\Omega}=u
$$
and
\begin{equation}\label{extension}
  \|\tilde u\|_{H_\gamma^1(\mathbb R^2)}\le C \, \|u\|_{H_\gamma^1(\Omega)}
  \,,
\end{equation}
where $C=C(\Omega)$.
\end{theorem}

\begin{proof}
We preliminarily observe that, if $\Omega$ is bounded,
the theorem can be immediately obtained from the classical result
for the unweighted case
(see for instance \cite[Thm.~4.4.1]{Evans-Gariepy}).
So, from now on, we assume that $\Omega$ is unbounded and we adapt
the arguments in \cite{Evans-Gariepy} to treat our case.
We distinguish two cases: $0 \in \Omega$ and $0 \notin \Omega$.

\medskip
\fbox{\emph{Case 1:} $0 \in \Omega$.}
The convexity of $\Omega$ ensures there exists a constant $L>0$ such that,
for every $(x_0,y_0)\in \partial\Omega$, up to a rotation,
there exist $r>0$ and an $L$-Lipschitz continuous function
$\beta:\mathbb R \to [0,+\infty)$ such that,
if we set $Q(x_0,y_0,r):=\{(x,y)\in \mathbb R^2:\> |x-x_0|<r,\> |y-y_0|<r\}$,
it holds
$$
\Omega \cap Q(x_0,y_0,r)=\{(x,y)\in \Omega: \> |x-x_0|<r,\> y_0-r<y<\beta(x)\}, \quad \max_{|x-x_0|<r}|\beta(x)-y_0|<\frac{r}{2}.
$$
In other words,
\begin{equation}\label{beta'}
|\beta'(x)|\le L \qquad \mbox{for a.e.} \ \, x\in (x_0-r,x_0+r),
\end{equation}
with $L$ independent from $x_0,y_0,r$.

Fix $(x_0,y_0)\in \partial\Omega$ and set
$\Omega^i:=Q(x_0,y_0,r)\cap \Omega$ and
$\Omega^e:=Q(x_0,y_0,r)\setminus \overline{\Omega}$.
Let $u \in C^1(\overline{\Omega})$ and suppose for the moment that the support of $u$ is contained in
$ Q(x_0,y_0,r)\cap \overline{\Omega}$.
Set
$$
u^e(x,y) := u(x,2\beta(x)-y)
\qquad \mathrm{if} \> (x,y)\in \overline{\Omega^e}\,.
$$
We get
\begin{eqnarray*}
\lefteqn{\int_{\Omega^e}{u^e(x,y)^2} \, \exp\left(-\frac{x^2+y^2}{2}\right)dxdy}
\\
&&=\int_{\Omega^e}u(x,2\beta(x)-y)^2 \, \exp\left(-\frac{x^2+y^2}{2}\right) dxdy
\\
&&= \int_{\Omega^i} u(s,t)^2 \, \exp\left(-\frac{s^2+(2\beta(s)-t)^2}{2}\right)dsdt
\\
&&=\int_{\Omega^i} u(s,t)^2 \, \exp\left(-\frac{s^2+(2\beta(s)-t)^2}{2}
+\frac{s^2+t^2}{2}\right) \, \exp\left(-\frac{s^2+t^2}{2}\right)dsdt.
\end{eqnarray*}
By elementary geometric considerations, taking into account the assumption $0\in \Omega$, it is easy to verify that
\begin{equation}
\exp\left(-\frac{s^2+(2\beta(s)-t)^2}{2}+\frac{s^2+t^2}{2}\right) \leq 1
  \,, \qquad
  \forall \>(s,t)\in \Omega^i.  \label{exp}
\end{equation}
Thus
\begin{equation}\label{normau}
\displaystyle\int_{\Omega^e}{u^e(x,y)}^{2}
\, \exp\left(-\frac{x^2+y^2}{2}\right)dxdy
\le \int_{\Omega^i} u(s,t)^2 \, \exp\left(-\frac{s^2+t^2}{2}\right)dsdt.
\end{equation}
On the other hand, by \eqref{beta'} and \eqref{exp} it holds
\begin{eqnarray}\label{normaDu}
\lefteqn{\int_{\Omega^e}|\nabla u^e(x,y)|^2
\,\exp\left(-\frac{x^2+y^2}{2}\right)dxdy}
\notag
\\
&&\le \int_{\Omega^i}
\left[\left(\partial_s u(s,t)+2 \partial_t u(s,t)
\beta'(s)\right)^2+\big(\partial_t u(s,t)\big)^2\right]
\exp\left(-\frac{s^2+(2\beta(s)-t)^2}{2}\right)dsdt
\notag
\\
&&\le C(L)\int_{\Omega^i} |\nabla u(s,t)|^2
\,\exp\left(-\frac{s^2+t^2}{2}\right)dsdt.
\end{eqnarray}
Define
$$
\tilde u:=\left\{
\begin{array}{ll}
u & \mathrm{on}\quad \overline{\Omega^i} \,,
\\
u^e & \mathrm{on}\quad \overline{\Omega^e} \,,
\\
0 & \mathrm{on} \quad \mathbb R^2 \setminus(\Omega^i\cup\Omega^e) \,.
\end{array}
\right.
$$
If~$\Omega $ contains the origin and the support of $u$ is contained in
$Q(x_0,y_0,r)\cap \overline{\Omega}$,
then~\eqref{normau} and~\eqref{normaDu} imply~\eqref{extension} with $C=C(L)$.

Now assume that $u \in C^1(\overline{\Omega})$
and drop the restriction on its support.
Clearly $\partial \Omega$ is not compact,
but we can cover $\partial \Omega$  with a countable family of squares
$\{Q_{2k-1}\}=\{Q(x_0^k,y_0^k,r'_k)\}_{k=1}^\infty$,
with $(x_0^k,y_0^k)\in\partial\Omega$ and $r'_k>0$
such that \eqref{beta'} holds true for each~$k$.
Analogously, we can cover the set
$\Omega\setminus \bigcup_{k=1}^{+\infty} Q_{2k-1}$
with a countable family of squares
$\{Q_{2k}\}=\{Q(x_1^k,y_1^k,r''_k)\}_{k=1}^\infty$
with $(x_1^k,y_1^k)\in\Omega$ and $r''_k>0$.
For the countable cover $\{Q_l\}_{l=1}^\infty$ of $\overline{\Omega}$
there is a partition of unity $\varphi_l$  subordinated to $Q_l$ with $\varphi_l$
smooth for each $l$ (see for instance~\cite[Thm. 3.14]{A}).
Define $\widetilde{\varphi_l u}$ as above when~$l$ is odd and set
$$
\tilde u := \sum_{l \>\mathrm{odd}}\widetilde{\varphi_l u}
+\sum_{l \>\mathrm{even}} \varphi_l u.
$$
Clearly $\tilde u$ satisfies \eqref{extension}.

Finally if $u \in H_\gamma^1(\Omega)$,
the claim follows by approximation arguments
and the proof of Case~1 is accomplished.

\medskip
\fbox{\emph{Case 2:} $0 \not\in \Omega$.}
Suppose now that $0\notin \Omega $ and denote
$d_{0} := \mathop{\mathrm{dist}}\nolimits(0,\partial \Omega )$.
Let us fix a vector $(\delta_1, \delta_2)$,
with $\sqrt{\delta_1^2+\delta_2^2}>d_0$, in such a way that the translation
$
  \Phi: \mathbb R^2 \to \mathbb R^2:
  \{(x,y) \mapsto (x-\delta_1,y-\delta_2) \}
$
maps $\Omega $ onto a set $\Phi(\Omega )$ containing the origin.
Defining
\begin{equation*}
v(x,y) := u(x+\delta_1,y+\delta_2)\exp \left( -
\frac{x\delta_1 }{2}-\frac{\delta_1 ^{2}}{4}\right)\exp \left( -
\frac{y\delta_2 }{2}-\frac{\delta_2 ^{2}}{4}\right)
,
\end{equation*}
for every $(x,y)\in \Phi(\Omega)$,
we have
\begin{equation*}
\int_{\Omega }u^{2} \, dm_\gamma = \int_{\Phi(\Omega )} v^{2} \, dm_\gamma
\,.
\end{equation*}
Since by construction $\Phi(\Omega )$ contains the origin,
there exists a function $\tilde{v}\in H_\gamma^{1}(\mathbb{R}^{2})$
such that
$\tilde{v}\left\vert _{\Phi(\Omega )}\right. =v$
and
\begin{equation*}
\|\tilde{v}\|_{H_\gamma^{1}(\mathbb{R}^{2})}
\leq C(L) \, \|v\|_{H_\gamma^{1}(\Phi(\Omega ))}
\,.
\end{equation*}
Letting
\begin{equation*}
\tilde{u}(x,y) :=\tilde{v}(x-\delta_1
,y-\delta_2)\exp \left( \frac{x\delta_1 }{2}-\frac{\delta_1 ^{2}}{4}\right)
\exp \left( \frac{y\delta_2 }{2}-\frac{\delta_2 ^{2}}{4}\right)
,
\end{equation*}
we finally get that $\tilde u|_\Omega=u$ and
\begin{equation*}
\|\tilde{u}\|_{H_\gamma^{1}(\mathbb{R}^{2})}
\leq C(L,d_{0}) \, \|u\|_{H_\gamma^{1}(\Omega)}.
\end{equation*}
This completes the proof of the theorem.
\end{proof}

\begin{remark}\label{oss}
Using the fact that $H^{1}_\gamma(\mathbb{R}^{2})$ is compactly
embedded into $L^{2}_\gamma(\mathbb{R}^{2})$
(see for example~\cite{D}) and the above extension theorem
one can easily deduce the compact embedding of
$H^{1}_\gamma(\Omega)$ into $L^{2}_\gamma(\Omega)$ (see also \cite{FP,BCHT}).
Therefore, by the classical spectral theory on compact self-adjoint
operators, $\mu _{1}(\Omega )$ satisfies the variational
characterization~\eqref{minimax.intro}.
\end{remark}

\section{Proof of Theorem~\ref{main}}\label{Sec.main}

The main ingredient in our proof of Theorem~\ref{main}
is the following lemma, which  tells us that
cutting the optimiser of~\eqref{mi} in two convex, unbounded sets with equal Gaussian area,
we again get two optimisers.
\begin{lemma}
Let $\Omega$ be a convex subset of $S_{y_1,y_2}$
 with $\mu_1(\Omega)=1$.
Let $\bar y\in (y_1,y_2)$ be such that the straight-line
$\{y=\bar y\}$ divides $\Omega$ into two convex subsets with equal Gaussian area $m_\gamma(\Omega)$.
Then
$$
\mu_1\big(\Omega \cap \{y<\bar y\}\big)
=\mu_1\big(\Omega \cap \{y>\bar y\}\big)=1.
$$
\end{lemma}
\begin{proof}
Let $u$ be an eigenfunction of
(\ref{problem}) corresponding to $\mu _{1}(\Omega ).$
By~\eqref{minimax.intro}, we know that $\int_{\Omega} u \, dm_\gamma=0$ and
\begin{equation*}
1=\dfrac{\int_{\Omega }|\nabla u|^{2}\,dm_\gamma}
{\int_{\Omega }u^{2}\,dm_\gamma}
\,.
\end{equation*}
For each $\alpha \in [0,2\pi]$ there is a unique straight-line $r_\alpha$
orthogonal to $(\cos \alpha,\sin \alpha)$
such that it divides $\Omega$ into two convex sets
$\Omega_\alpha',\Omega_\alpha''$ with equal Gaussian measure.
Let $I(\alpha):=\int_{\Omega_\alpha'}u \, dm_\gamma$.
Since $I(\alpha)=-I(\alpha+\pi)$,
by continuity there is $\bar \alpha $ such that $I(\bar \alpha)=0$.
Now we claim that $r_{\bar \alpha}$ is parallel to the $x$-axis.
Note firstly that $\Omega_{\bar \alpha}'$ and $\Omega_{\bar \alpha}''$
are obviously convex and by \eqref{an-ni}, \eqref{l(R)} and \eqref{mi} we have
\begin{equation}
\mu _{1}(\Omega_{\bar \alpha}')\geq 1 \,,
\qquad \mu_1(\Omega_{\bar \alpha}'')\ge 1 \,.
\label{mu+->=1}
\end{equation}
Moreover, it is immediate to verify that
\begin{equation*}
1=\mu_1(\Omega)
=\dfrac{\int_{\Omega_{\bar \alpha}'}|\nabla u|^{2}\,dm_\gamma
+\int_{\Omega_{\bar \alpha}''}|\nabla u|^{2}\,dm_\gamma}
{\int_{\Omega_{\bar \alpha}'}u^{2}\,dm_\gamma
+\int_{\Omega_{\bar \alpha}''}u^{2}\,dm_\gamma}
\geq \min \left\{ \dfrac{\int_{\Omega_{\bar \alpha}'}|\nabla u|^{2}\,dm_\gamma}
{\int_{\Omega_{\bar \alpha}'}u^{2}\,dm_\gamma},
\dfrac{\int_{\Omega_{\bar \alpha}''}|\nabla u|^{2}\,dm_\gamma}
{\int_{\Omega_{\bar \alpha}''}u^{2}\,dm_\gamma}\right\},
\end{equation*}
with equality holding if and only if
$$
\dfrac{\int_{\Omega_{\bar \alpha}'}|\nabla u|^{2}\,dm_\gamma}
{\int_{\Omega_{\bar \alpha}'}u^{2}\,dm_\gamma}
=\dfrac{\int_{\Omega_{\bar \alpha}''}|\nabla u|^{2} \, dm_\gamma}{
\int_{\Omega_{\bar \alpha}''}u^{2}\,dm_\gamma}.
$$
Without loss of generality we can assume that
\begin{equation*}
\min \left\{ \dfrac{\int_{\Omega_{\bar \alpha}'}|\nabla u|^{2}\,dm_\gamma}
{\int_{\Omega_{\bar \alpha}'}u^{2}\,dm_\gamma},
\dfrac{\int_{\Omega_{\bar \alpha}''}|\nabla u|^{2}\,dm_\gamma}
{\int_{\Omega_{\bar \alpha}''}u^{2}\,dm_\gamma}\right\}
=\dfrac{\int_{\Omega_{\bar \alpha}'}|\nabla u|^{2}\,dm_\gamma}
{\int_{\Omega_{\bar \alpha}'}u^{2}\,dm_\gamma}.
\end{equation*}
Finally, (\ref{mu+->=1}) ensures that
\begin{equation}
1=\mu _{1}(\Omega )=\mu _{1}(\Omega_{\bar \alpha}')
=\mu _{1}(\Omega_{\bar \alpha}'') \,.  \label{1=m}
\end{equation}

Now we want to show that both $\Omega_{\bar \alpha}'$ and $\Omega_{\bar \alpha}''$ are unbounded, and hence $r_{\bar \alpha}$ is parallel to the $x$-axis.
Suppose by contradiction that, for instance, $\Omega_{\bar \alpha}'$ is bounded. In
such a case \eqref{an-ni} yields
\begin{equation*}
\mu _{1}(\Omega_{\bar \alpha}')\geq \mu _{1}\left( -\frac{\mathrm{d}(\Omega_{\bar \alpha}')}{2},
\frac{\mathrm{d}(\Omega_{\bar \alpha}')}{2}\right) .
\end{equation*}
Taking into account~(\ref{1=m}) and (\ref{l(R)}),
we get that
\begin{equation*}
\mu_{1}\left( -\frac{\mathrm{d}(\Omega_{\bar \alpha}')}{2},\frac{\mathrm{d}
(\Omega_{\bar \alpha}')}{2}\right) =1
\end{equation*}
 that is $\mathrm{d}(\Omega_{\bar \alpha}')=+\infty $, which is a
contradiction.
\end{proof}

\begin{proof}[Proof of Theorem \ref{main}]

By contradiction, let us assume that $\Omega \subset S_{y_1,y_2}$
is a convex domain different from a strip
and $\mu_1(\Omega)=1$. Let us denote
$$
\Omega=\left\{(x,y)\in \mathbb R^2: \, y_1<y<y_2,\, p(y)<x\right\},
$$
where $p$ is a convex, non-trivial function.
From~\eqref{an-ni} and~\eqref{l(R)}
it follows that~$\Omega$ is necessarily unbounded.
By employing a separation of variables,
we also deduce from~\eqref{an-ni}  and~\eqref{l(R)}
that~$\Omega$ cannot be a semi-strip.
Finally, we may assume that
$
  \inf\{x: \exists\, y\in[y_1,y_2], \, (x,y)\in\Omega\}
$
is finite
(otherwise, we would have the finite supremum,
which can be transferred to our situation by a reflection
of the coordinate system).

Repeating the procedure described in the above lemma,
since at any step we are dividing into two convex subsets with equal Gaussian area,
we can obtain a sequence of unbounded convex domains
\begin{equation}\label{Omega.cut}
  \Omega_{\epsilon_k}
  := \left\{ (x,y)\in \mathbb{R}^{2}:\ y_0<y<d_k,p(y)<x \right\}
  = \left\{ (x,y)\in \Omega:\ y_0<y<d_k \right\}
\end{equation}
such that
$$
  \mu _{1}(\Omega_{\epsilon_k}) =1
  \,, \qquad
  \epsilon_{k} := d_{k}-y_0 \xrightarrow[k \to +\infty]{} 0
  \,.
$$
Here the point~$y_0$ is chosen in such a way that
$
  p^{\prime}(y_0) \not= 0
$,
which is always possible because the situation of semi-strips
has been excluded.
Without loss of generality
(reflecting again the coordinate system if necessary),
we may in fact assume
\begin{equation}\label{Ass.phi}
  p^{\prime}(y_0) > 0
  \,,
\end{equation}
so that~$\phi$ is increasing on $[y_0,d_k]$
whenever~$k$ is sufficiently large.
Applying now a more general convergence result for eigenvalues
in thin Neumann domains that we shall establish
in the following section (Theorem~\ref{Thm.convergence}),
we have
\begin{lemma}\label{Lem.convergence}
$
\displaystyle
  \lim_{k\to\infty}\mu_1(\Omega_{\epsilon_k})
  =\mu_1(p^{-1}(y_0),+\infty)
$.
\end{lemma}
\noindent
Since $\mu _{1}(\Omega_{\epsilon_k})$ equals~$1$ for every~$k$,
we conclude that
$$
  \mu_1\big(p^{-1}(y_0),+\infty\big) = 1
  \,.
$$
However, from \eqref{l(R)},
we then deduce that $p^{-1}(y_0)=-\infty$,
which contradicts our assumptions
from the beginning of the proof.
In other words, $\Omega$ contains a straight-line
and the theorem immediately follows.
\end{proof}

It thus remains to establish Lemma~\ref{Lem.convergence}.

\section{Eigenvalue asymptotics in thin strips}\label{Sec.as}
In this section we establish Lemma~\ref{Lem.convergence}
as a consequence of a general result about convergence
of \emph{all} eigenvalues of~$T$
in thin domains of the type~\eqref{Omega.cut}.

\subsection{The geometric setting}
Let $f:[0,+\infty) \to [0,+\infty)$ be a concave
non-decreasing continuous non-trivial function such that $f(0) = 0$
(the case $f(0)>0$ is actually much easier to deal with).
Given a positive number $\varepsilon < \sup f$,
we put
$$
  f_\varepsilon(x) := \min\{\varepsilon,f(x)\}
$$
and define an unbounded domain
$$
  \Omega_\varepsilon := \{(x,y)\in\mathbb{R}^2 :
  0 < x \,, \ 0<y<f_\varepsilon(x)\}
  \,.
$$
Clearly, \eqref{Omega.cut} can be cast into this form after
identifying $f=p^{-1}$ and a translation.
However, keeping in mind that the problem~\eqref{problem}
is not translation-invariant, we accordingly change
the definition of the Gaussian weight
throughout this section
$$
  \gamma(x,y) := \exp\left(-\frac{(x_0+x)^2+(y_0+y)^2}{2}\right)
  \,.
$$
Here~$y_0$ is primarily thought as the point from~\eqref{Omega.cut}
and~$x_0$ is then such that $(x_0,y_0) \in \Omega_{\epsilon_k}$.
For the results established in this section, however,
$x_0$ and $y_0$ can be thought as arbitrary real numbers.
For our method to work, it is only important to assume~\eqref{Ass.phi},
which accordingly transfers to
\begin{equation}\label{Ass}
  f'(0) < +\infty \,.
\end{equation}

\subsection{The analytic setting and main result}
Keeping the translation we have made in mind,
instead of~\eqref{problem} we equivalently consider the eigenvalue problem
\begin{equation}\label{bv}
\left\{
\begin{array}{ll}
  - \mathop{\mathrm{div}}\nolimits (\gamma \nabla u)
  = \mu \gamma u
  & \mbox{in} \quad \Omega_\varepsilon \,,
  \\ \\
  \dfrac{\partial u}{\partial \bf{n}}
  =0
  & \mbox{on} \quad \partial\Omega_\varepsilon \,.
\end{array}
\right.
\end{equation}
We understand~\eqref{bv} as a spectral problem for
the self-adjoint operator~$T_\varepsilon$ in the Hilbert space
$L^2_\gamma(\Omega_\varepsilon)$
associated with the quadratic form
$
  t_\varepsilon[u] := \|\nabla u\|_\varepsilon^2
$,
$
  \mathsf{D}(t_\varepsilon) := H_\gamma^1(\Omega_\varepsilon)
$.
Here $\|\cdot\|_\varepsilon$ denotes the norm in $L^2_\gamma(\Omega_\varepsilon)$.
We arrange the eigenvalues of~$T_\varepsilon$ in a non-decreasing sequence
$\{\mu_n(\Omega_\varepsilon)\}_{n \in \mathbb{N}}$ where each eigenvalue
is repeated according to its multiplicity.
In this paper we adopt the convention $0 \in \mathbb{N}$.
We are interested in the behaviour of the spectrum as $\varepsilon \to 0$,
particularly $\mu_1(\Omega_\varepsilon)$ because of Lemma~\ref{Lem.convergence}.

It is expectable that the eigenvalues will be determined
in the limit $\varepsilon \to 0$ by the one-dimensional problem
\begin{equation}\label{bv.1D}
\left\{
\begin{array}{ll}
  - (\gamma_0 \, u')'
  = \nu \gamma_0 u
  & \mbox{in} \quad (0,+\infty) \,,
  \\ \\
  u' (0)=0,
    &
\end{array}
\right.
\end{equation}
where
$$
  \gamma_0(x) := \gamma(x,0)
  = \exp\left(-\frac{(x_0+x)^2+y_0^2}{2}\right)
  \,.
$$
Again, we understand~\eqref{bv.1D} as a spectral problem for
the self-adjoint operator~$T_0$ in the Hilbert space
$L^2_{\gamma_0}((0,+\infty))$
associated with the quadratic form
$
  t_0[u] := \|\nabla u\|_0^2
$,
$
  \mathsf{D}(t_0) := H_{\gamma_0}^1((0,+\infty))
$,
where $\|\cdot\|_0$ denotes the norm in $L^2_{\gamma_0}((0,+\infty))$.
As above, we arrange the eigenvalues of~$T_0$ in a non-decreasing sequence
$\{\nu_n\}_{n \in \mathbb{N}}$ where each eigenvalue
is repeated according to its multiplicity. By construction,  for each $n\in\mathbb N$, $\nu_n$ coincides with the eigenvalue $\mu_n(x_0,+\infty)$ defined in \eqref{N_1d}.

In this section we prove the following convergence result.
\begin{theorem}\label{Thm.convergence}
Let $f:[0,+\infty) \to [0,+\infty)$ be a concave
non-decreasing continuous non-trivial function such that $f(0) = 0$.
Assume in addition~\eqref{Ass}.
Then
$$
  \forall n\in\mathbb{N} \,, \qquad
  \mu_n(\Omega_\varepsilon) \xrightarrow[\varepsilon \to 0]{} \nu_n
  \,.
$$
\end{theorem}

We shall also establish certain convergence
of eigenfunctions of~$T_\varepsilon$ to eigenfunctions of~$T_0$.

Clearly, Lemma~\ref{Lem.convergence} is the case $n=1$
of this general theorem.

The rest of this section is devoted to
a proof of Theorem~\ref{Thm.convergence}.

\subsection{From the moving to a fixed domain}
Our main strategy is to map~$\Omega_\varepsilon$ into a fixed strip~$\Omega$.
We introduce a refined mapping in order to effectively
deal with the singular situation $f(0)=0$.

Let
$$
  a_\varepsilon := \inf f_\varepsilon^{-1}(\{\varepsilon\})
  \,.
$$
By the definition of~$f_\varepsilon$ and since~$f$ is non-decreasing,
$a_\varepsilon \to 0$ as $\varepsilon \to 0$ and
$f_\varepsilon(x)=\varepsilon$ for all $x>a_\varepsilon$.
If $f(0)>0$, then there exists $\varepsilon_0>0$ such that
$a_\varepsilon=0$ for all $\varepsilon \leq \varepsilon_0$.
On the other hand, if $f(0)=0$,
then $a_\varepsilon>0$ for all $\varepsilon>0$.
The troublesome situation is the latter,
to which we have restricted from the beginning.
In this case, we introduce an auxiliary function
$$
  g_\varepsilon(s) :=
  \begin{cases}
    a_\varepsilon s + a_\varepsilon
    &\mbox{if} \quad s \in [-1,0) \,,
    \\
    s+a_\varepsilon
    &\mbox{if} \quad s \in [0,+\infty) \,.
  \end{cases}
$$
Since we are interested in the limit $\varepsilon \to 0$, we may henceforth assume
\begin{equation}\label{Ass.small}
  \varepsilon \leq 1
  \qquad \mbox{and} \qquad
  a_\varepsilon \leq 1
  \,.
\end{equation}

Define $\varepsilon$-independent sets
$$
  \Omega_-:=(-1,0)\times(0,1)
  \,, \qquad
  \Omega_+:=(0,+\infty)\times(0,1)
  \,, \qquad
  \Omega:=(-1,+\infty)\times(0,1)
  \,.
$$
The mapping
\begin{equation}\label{layer}
  \mathcal{L}_\varepsilon : \Omega \to \Omega_\varepsilon :
  \left\{
  (s,t) \mapsto
  \mathcal{L}_\varepsilon(s,t) :=
  \big(
  g_\varepsilon(s), f_\varepsilon(g_\varepsilon(s)) \, t
  \big)
  \right\}
\end{equation}
represents a $C^{0,1}$-diffeomorphism
between~$\Omega$ and~$\Omega_\varepsilon$
($f$~is differentiable almost everywhere,
as it is supposed to be concave).
In this way, {we obtain a convenient parameterisation of~$\Omega_\varepsilon$
via the coordinates $(s,t) \in \Omega$}
whose Jacobian is
\begin{equation}\label{Jacobian}
  j_\varepsilon(s,t)
  = g_\varepsilon'(s) f_\varepsilon(g_\varepsilon(s))
  \,.
\end{equation}
Note that the Jacobian is independent of~$t$
and singular at $s=-1$.
Now we reconsider~\eqref{bv} in~$\Omega$.
With the notation
$$
  \gamma_\varepsilon(s,t) := (\gamma \circ \mathcal{L}_\varepsilon)(s,t)
  = \exp\left(-\frac{[x_0+g_\varepsilon(s)]^2
  +[y_0+f_\varepsilon(g_\varepsilon(s))t]^2}{2}\right)
  \,,
$$
introduce the unitary transform
$$
  U_\varepsilon:
  L^2_\gamma(\Omega_\varepsilon) \to
  L^2_{\gamma_\varepsilon j_\varepsilon/\varepsilon}(\Omega) :
  \left\{
  u \mapsto \sqrt{\varepsilon} \, u \circ \mathcal{L}_\varepsilon
  \right\}
  \,.
$$
Here, in addition to the change of variables~\eqref{layer},
we also make an irrelevant scaling transform
(so that the renormalised Jacobian~$j_\varepsilon/\varepsilon$ is~$1$ in~$\Omega_+$).
The operators $H_\varepsilon := U_\varepsilon T_\varepsilon U_\varepsilon^{-1}$
and~$T_\varepsilon$ are isospectral.
By definition, $H_\varepsilon$~is associated with the quadratic form
$h_\varepsilon[\psi] := t_\varepsilon[U_\varepsilon^{-1}\psi]$,
$\mathsf{D}(h_\varepsilon) := U_\varepsilon \mathsf{D}(t_\varepsilon)$.
\begin{proposition}\label{Prop.form}
Assume~\eqref{Ass}.
Then
\begin{align}
  h_\varepsilon[\psi] &= \int_\Omega
  \left[
  \left(
  \frac{\partial_s \psi}{g_\varepsilon'}
  - \frac{f_\varepsilon'\circ g_\varepsilon}{f_\varepsilon \circ g_\varepsilon} \, t \, \partial_t \psi
  \right)^2
  +
  \frac{(\partial_t \psi)^2}{(f_\varepsilon\circ g_\varepsilon)^2}
  \right]
  \gamma_\varepsilon \, g_\varepsilon' \, \frac{f_\varepsilon \circ g_\varepsilon}{\varepsilon} \, ds \, dt
  \,, \qquad
  \label{form1}
  \\
  \mathsf{D}(h_\varepsilon) &\subset H_{\gamma_\varepsilon j_\varepsilon/\varepsilon}^1(\Omega)
  \,.
  \label{form2}
\end{align}
\end{proposition}
\noindent
Here we have started to simplify the notation
by suppressing arguments of the functions.
\begin{proof}
The space
$
  \mathcal{D}_\varepsilon := C_0^1(\mathbb{R}^2) \upharpoonright\Omega_\varepsilon
$
is a core of~$t_\varepsilon$.
The transformed space
$
  \mathcal{D} := U_\varepsilon \mathcal{D}_\varepsilon
$
is a subset of $C_0^0(\mathbb{R}^2) \upharpoonright \Omega$ consisting of
Lipschitz continuous functions on~$\Omega$ which belong to
$C^1(\overline{\Omega_-}) \oplus C^1(\overline{\Omega_+})$
(we do not have $C^1$ globally,
because~$g_\varepsilon$ and~$f_\varepsilon$ are not smooth).
For any $\psi \in \mathcal{D}$, it is easy to check~\eqref{form1};
this formula extends to all~$\psi$ from the domain
$$
  \mathsf{D}(h_\varepsilon) = \overline{\mathcal{D}}^{\|\cdot\|_{h_\varepsilon}}
  \,, \qquad
  \|\cdot\|_{h_\varepsilon} := \sqrt{h_\varepsilon[\cdot] + \|\cdot\|^2}
  \,,
$$
where $\|\cdot\|$ denotes the norm of
$L^2_{\gamma_\varepsilon j_\varepsilon/\varepsilon}(\Omega)$.
Let $\psi \in \mathcal{D}$.
Using elementary estimates, we easily check
\begin{equation}\label{bound.form}
  h_\varepsilon^-[\psi] \leq h_\varepsilon[\psi]
\end{equation}
where
\begin{align*}
  h_\varepsilon^-[\psi] &:=
  \delta \int_{\Omega}
  \left(
  \frac{\partial_s \psi}{g_\varepsilon'}
  \right)^2
  \gamma_\varepsilon \, g_\varepsilon' \, \frac{f_\varepsilon \circ g_\varepsilon}{\varepsilon} \, ds \, dt
  + \left(
  1 - \frac{\delta}{1-\delta} \, \|f_\varepsilon'\|_\infty^2
  \right)
  \int_{\Omega} \frac{(\partial_t \psi)^2}{(f_\varepsilon \circ g_\varepsilon)^2}
  \, \gamma_\varepsilon \, g_\varepsilon' \, \frac{f_\varepsilon \circ g_\varepsilon}{\varepsilon} \, ds \, dt
\end{align*}
with any $\delta \in (0,1)$.
Note that~$f_\varepsilon'$ is bounded under the assumption~\eqref{Ass}
and the concavity.
For any $\varepsilon > 0$,
we can choose~$\delta$ so small that $h_\varepsilon^-[\psi]$
is composed of a sum of two non-negative terms
($\delta$~can be made independent of~$\varepsilon$
if we restrict the latter to a fixed bounded interval,
say $(0,1]$, see \eqref{Ass.small}, because
$
  \|f_\varepsilon'\|_{L^\infty((0,1))}
  \leq \|f'\|_{L^\infty((0,1))}
$,
but this assumption is not needed for the property
we are proving).
Using that~$g_\varepsilon'$ is bounded for any fixed~$\varepsilon$
and the estimate $f_\varepsilon \circ g_\varepsilon \leq \varepsilon$,
we thus deduce from~\eqref{bound.form} that
there is a positive constant~$c_{\varepsilon,\delta}$
(again, this constant can be made independent of~$\varepsilon$
if $\varepsilon \leq 1$)
such that
$$
  c_{\varepsilon,\delta} \, \|\psi\|_{H_{\gamma_\varepsilon j_\varepsilon/\varepsilon}^1(\Omega)}^2
  \leq \|\psi\|_{h_\varepsilon}
  \,.
$$
This proves~\eqref{form2}
because $\mathcal{D}$ is dense in $H_{\gamma_\varepsilon j_\varepsilon/\varepsilon}^1(\Omega)$.
\end{proof}

\subsection{The eigenvalue equation}
Recall that we denote the eigenvalues of~$T_\varepsilon$ (and hence~$H_\varepsilon$)
by $\mu_n(\Omega_\varepsilon)$ with $n \in \mathbb{N}$ ($=\{0,1,\dots\}$).
The~$(n+1)^\mathrm{th}$ eigenvalue can be characterised
by the Rayleigh-Ritz variational formula
\begin{equation}\label{minimax.as}
  \mu_n(\Omega_\varepsilon) =
  \inf_{\stackrel{\dim \mathfrak{L}_n=n+1}
  {\mathfrak{L}_n \subset \mathsf{D}(h_\varepsilon)}}
  \sup_{\psi \in \mathfrak{L}_n}
  \frac{h_\varepsilon[\psi]}{\|\psi\|^2}
  \,.
\end{equation}
\begin{proposition}\label{Prop.bound}
For any~$n \in \mathbb{N}$,
there exists a positive constant~$C_n$ such that
for all $\varepsilon \leq 1$,
$$
  \mu_n(\Omega_\varepsilon) \leq C_n
  \,.
$$
\end{proposition}
\begin{proof}
Assuming $\varepsilon \leq 1$, we have the following two-sided
$\varepsilon$- and $t$-independent bound
\begin{equation}\label{2-bound}
  \gamma_-(s)\leq \gamma_\varepsilon(s,t) \leq \gamma_+(s)
\end{equation}
valid for every $(s,t) \in \Omega_+$ with
\begin{equation*}
  \gamma_-(s) :=
  \exp\left(-\frac{(|x_0|+s+1)^2
  +(|y_0|+1)^2}{2}\right)
  , \qquad
  \gamma_+(s) :=
  \exp\left(-\frac{(-|x_0|+s)^2-2|y_0|}{2}\right)
  .
\end{equation*}
Using in addition that
$g_\varepsilon'=1$ and $f_\varepsilon\circ g_\varepsilon = \varepsilon$ in~$\Omega_+$,
we obviously have
$$
  \forall \psi \in C_0^\infty((0,+\infty)) \otimes \{1\}
  \,, \qquad
  \frac{h_\varepsilon[\psi]}{\|\psi\|^2}
  \leq
  \frac{\displaystyle
  \int_{\Omega_+}
  (\partial_s \psi)^2 \, \gamma_+(s) \, ds \, dt}
  {\displaystyle \int_{\Omega_+} \psi^2 \, \gamma_-(s) \, ds \, dt}
  \,.
$$
It then follows from~\eqref{minimax.as} that
the inequality of the proposition holds with the numbers
$$
  C_n :=
  \inf_{\stackrel{\dim \mathfrak{L}_n=n+1}
  {\mathfrak{L}_n \subset C_0^\infty((0,+\infty))}}
  \sup_{\psi \in \mathfrak{L}_n}
  \frac{\displaystyle \int_{0}^{+\infty} \psi'(s)^2 \, \gamma_+(s) \, ds}
  {\displaystyle \int_{0}^{+\infty} \psi(s)^2 \, \gamma_-(s) \, ds}
  \,,
$$
which are actually eigenvalues of the one-dimensional operator
$-\gamma_-^{-1} \partial_s \gamma_+ \partial_s$ in
$L^2_{\gamma_-}((0,+\infty))$,
subject to Dirichlet boundary conditions.
\end{proof}

Let us now fix $n \in \mathbb{N}$ and abbreviate the~$(n+1)^\mathrm{th}$
eigenvalue of~$H_\varepsilon$ by $\mu_\varepsilon := \mu_n(\Omega_\varepsilon)$.
We denote an eigenfunction corresponding to~$\mu_\varepsilon$ by~$\psi_\varepsilon$
and normalise it to~$1$ in $L^2_{\gamma_\varepsilon j_\varepsilon/\varepsilon}(\Omega)$,
\emph{i.e.},
\begin{equation}\label{norm}
  \|\psi_\varepsilon\|=1
  \,.
\end{equation}
for every admissible $\varepsilon > 0$.

The weak formulation of the eigenvalue equation
$H_\varepsilon \psi_\varepsilon = \mu_\varepsilon \psi_\varepsilon$ reads
\begin{equation}\label{weak}
  \forall \phi \in \mathsf{D}(h_\varepsilon)
  \,, \qquad
  h_\varepsilon(\phi,\psi_\varepsilon) = \mu_\varepsilon \, (\phi,\psi_\varepsilon)
  \,,
\end{equation}
where $(\cdot,\cdot)$ stands for the inner product in
$L^2_{\gamma_\varepsilon j_\varepsilon/\varepsilon}(\Omega)$
{and $h_\varepsilon(\cdot,\cdot)$ denotes the sesquilinear form
corresponding to~$h_\varepsilon[\cdot]$}, that is $ \forall \phi \in \mathsf{D}(h_\varepsilon) $

\begin{align}
   \int_\Omega
  \left[
  \left(
  \frac{\partial_s \psi_{\varepsilon}}{g_\varepsilon'}
  - \frac{f_\varepsilon'\circ g_\varepsilon}{f_\varepsilon \circ g_\varepsilon} \, t \, \partial_t \psi_{\varepsilon}
  \right)\left(
  \frac{\partial_s \phi}{g_\varepsilon'}
  - \frac{f_\varepsilon'\circ g_\varepsilon}{f_\varepsilon \circ g_\varepsilon} \, t \, \partial_t \phi
  \right)
  +
  \frac{(\partial_t \psi_{\varepsilon})}{(f_\varepsilon\circ g_\varepsilon) }  \frac{(\partial_t \phi)}{(f_\varepsilon\circ g_\varepsilon) }
  \right]
  \gamma_\varepsilon \, g_\varepsilon' \, \frac{f_\varepsilon \circ g_\varepsilon}{\varepsilon} \, ds \, dt
   \qquad
  \label{weform}
  \\
  \notag
  = \mu_\varepsilon \int_\Omega \psi_{\varepsilon}\, \phi \, \gamma_\varepsilon \, g_\varepsilon' \, \frac{f_\varepsilon \circ g_\varepsilon}{\varepsilon} \, ds \, dt
  \,.
\end{align}

\subsection{What happens in \texorpdfstring{$\Omega_+$}{Omega+}}
Using $|t| \leq 1$,  we easily verify
\begin{equation}\label{2-bound.bis}
  \forall (s,t) \in \Omega_+
  \,, \qquad
  \gamma_\varepsilon(s,t) \geq \rho_\varepsilon(s) \gamma_0(s)
  \,,
\end{equation}
where the function
$$
  \rho_\varepsilon(s) := \exp\left(
  -\frac{a_\varepsilon^2+2|x_0| a_\varepsilon + \varepsilon^2+2|y_0|\varepsilon}{2}
  \right)
  \exp(-a_\varepsilon s)
$$
is converging pointwise to~$1$ as $\varepsilon \to 0$.

Choosing $\phi = \psi_\varepsilon$ as a test function in~\eqref{weak}
and using~\eqref{2-bound.bis}
together with Proposition~\ref{Prop.bound} and~\eqref{norm},
we obtain
\begin{equation}\label{bound.plus}
  \int_{\Omega_+} (\partial_s\psi_\varepsilon)^2 \, \rho_\varepsilon \, \gamma_0 \, ds \, dt
  + \int_{\Omega_+} \frac{(\partial_t\psi_\varepsilon)^2}{\varepsilon^2}
  \, \rho_\varepsilon \, \gamma_0 \, ds \, dt
  \leq h_\varepsilon[\psi_\varepsilon] = \mu_\varepsilon \|\psi_\varepsilon\|^2 \leq C
  \,.
\end{equation}
Here and in the sequel, we denote by~$C$ a generic
constant which is independent of~$\varepsilon$
and may change its value from line to line.
Writing
\begin{equation}\label{dec.plus}
  \psi_\varepsilon(s) = \varphi_\varepsilon(s) + \eta_\varepsilon(s,t)
  \,,
\end{equation}
where
\begin{equation}\label{orthogonal}
  \int_0^1 \eta_\varepsilon(s,t) \, dt = 0
  \qquad \mbox{for a.e.\ }
  s \in (0,+\infty)
  \,,
\end{equation}
we deduce from the second term on the left hand side of~\eqref{bound.plus}
\begin{equation}\label{eta.plus}
  \pi^2
  \int_{\Omega_+} \eta_\varepsilon^2 \, \rho_\varepsilon\, \gamma_0 \, ds \, dt
  \leq
  \int_{\Omega_+} (\partial_t\eta_\varepsilon)^2 \, \rho_\varepsilon \, \gamma_0 \, ds \, dt
  \leq C \varepsilon^2
  \,.
\end{equation}
Differentiating~\eqref{orthogonal} with respect to~$s$,
we may write
$$
  \int_{\Omega_+} (\partial_s\psi_\varepsilon)^2 \, \rho_\varepsilon \, \gamma_0 \, ds \, dt
  = \int_{\Omega_+} {\varphi_\varepsilon'}^2 \, \rho_\varepsilon \,\gamma_0 \, ds \, dt
  + \int_{\Omega_+} (\partial_s\eta_\varepsilon)^2 \, \rho_\varepsilon \, \gamma_0 \, ds \, dt
$$
and putting this decomposition into~\eqref{bound.plus},
we get from the first term on the left hand side
\begin{equation}\label{varphi'.plus}
  \int_0^{+\infty} {\varphi_\varepsilon'}^2 \, \rho_\varepsilon \, \gamma_0 \, ds
  \leq C
  \,, \qquad
  \int_0^{+\infty} (\partial_s\eta_\varepsilon)^2 \, \rho_\varepsilon \, \gamma_0 \, ds
  \leq C
  \,.
\end{equation}

At the same time, from~\eqref{norm} using~\eqref{2-bound.bis},
we obtain
\begin{equation}\label{bound.plus.norm}
  \int_{\Omega_+} \varphi_\varepsilon^2 \, \rho_\varepsilon \, \gamma_0 \, ds \, dt
  +\int_{\Omega_+} \eta_\varepsilon^2 \, \rho_\varepsilon \, \gamma_0 \, ds \, dt
  =\int_{\Omega_+} \psi_\varepsilon^2 \, \rho_\varepsilon \, \gamma_0 \, ds \, dt
  \leq \|\psi_\varepsilon\|^2 = 1
  \,,
\end{equation}
where the first equality employs~\eqref{orthogonal}.
Consequently,
\begin{equation}\label{varphi.plus}
  \int_0^{+\infty} \varphi_\varepsilon^2 \, \rho_\varepsilon \, \gamma_0 \, ds
  \leq 1
  \,.
\end{equation}

Finally, employing the first inequality from~\eqref{varphi'.plus}
and~\eqref{varphi.plus},
we get
\begin{equation}\label{varphi'.plus.bis}
  \int_0^{+\infty} {(\sqrt{\rho_\varepsilon}\varphi_\varepsilon)'}^2 \, \gamma_0 \, ds
  \leq C
  \,.
\end{equation}

From~\eqref{varphi.plus} and~\eqref{varphi'.plus.bis},
we see that $\{\sqrt{\rho_\varepsilon}\varphi_\varepsilon\}_{\varepsilon>0}$
is a bounded family in $H^1_{\gamma_0}((0,+\infty))$
and therefore precompact in the weak topology of this space.
Let~$\varphi_0$ be a weak limit point,
\emph{i.e.}\ for a decreasing sequence of positive numbers
$\{\varepsilon_i\}_{i \in \mathbb{N}}$
such that $\varepsilon_i \to 0$ as $i \to +\infty$,
\begin{equation}\label{weak.limit}
  \sqrt{\rho_{\varepsilon_i}} \varphi_{\varepsilon_i}
  \xrightarrow[i\to+\infty]{w} \varphi_0
  \qquad \mbox{in} \qquad H^1_{\gamma_0}((0,+\infty))
  \,.
\end{equation}
Since $H^1_{\gamma_0}((0,+\infty))$ is compactly embedded
in $L^2_{\gamma_0}((0,+\infty))$, we may assume
\begin{equation}\label{strong.limit}
  \sqrt{\rho_{\varepsilon_i}}\varphi_{\varepsilon_i} \xrightarrow[i\to+\infty]{s} \varphi_0
  \qquad \mbox{in} \qquad L^2_{\gamma_0}((0,+\infty))
  \,.
\end{equation}

\subsection{What happens in \texorpdfstring{$\Omega_-$}{Omega-}}
Here~$\gamma_\varepsilon$ can be estimated from below just by
an $\varepsilon$-independent positive number, \emph{e.g.},
\begin{equation}\label{2-bound.minus}
  \forall (s,t) \in \Omega_-
  \,, \qquad
  \gamma_\varepsilon(s,t) \geq
  \exp\left(
  -\frac{(|x_0| +1)^2 + |y_0| +1)^2}{2}
  \right)
  \,.
\end{equation}
On the other hand, we need a lower bound to~$f_\varepsilon$.
Employing that~$f$ is concave and non-decreasing,
we can use
\begin{equation}\label{f.bound}
  \forall s \in (-1,0)
  \,, \qquad
  f_\varepsilon(g_\varepsilon(s)) \geq \varepsilon \, (s+1)
  \,.
\end{equation}
Recall also that $g_\varepsilon'=a_\varepsilon$ on $(-1,0)$.

Choosing $\phi = \psi_\varepsilon$ as a test function in~\eqref{weak}
and using~\eqref{2-bound.minus} and~\eqref{f.bound},
we obtain
\begin{equation}\label{bound.minus}
  \int_{\Omega_-}
  \left(
  \frac{\partial_s\psi_\varepsilon}{a_\varepsilon}
  - \frac{f_\varepsilon'\circ g_\varepsilon}{f_\varepsilon \circ g_\varepsilon} \, t \, \partial_t\psi_\varepsilon
  \right)^2
  a_\varepsilon \, (s+1) \, ds \, dt
  + \int_{\Omega_-} \frac{(\partial_t\psi_\varepsilon)^2}{(f_\varepsilon \circ g_\varepsilon)^2}
  \, a_\varepsilon \, (s+1) \, ds \, dt
  \leq C
  \,.
\end{equation}
Assume~\eqref{Ass}.
Using elementary estimates as in the proof of Proposition~\ref{Prop.form},
this inequality implies
\begin{equation}\label{bound.minus.bis}
  \delta \int_{\Omega_-}
  \left(
  \frac{\partial_s\psi_\varepsilon}{a_\varepsilon}
  \right)^2 a_\varepsilon \, (s+1) \, ds \, dt
  + \left(
  1 - \frac{\delta}{1-\delta} \, \|f_\varepsilon'\|_\infty^2
  \right)
  \int_{\Omega_-} \frac{(\partial_t\psi_\varepsilon)^2}{(f_\varepsilon \circ g_\varepsilon)^2}
  \, a_\varepsilon \, (s+1) \, ds \, dt
  \leq C
\end{equation}
with any $\delta \in (0,1)$.
We can choose~$\delta$ (independent of~$\varepsilon$ due to~\eqref{Ass.small})
so small that the left hand
side of~\eqref{bound.minus.bis} is composed of a sum of
two non-negative terms.
Using in addition $f_\varepsilon \circ g_\varepsilon \leq \varepsilon$,
we thus deduce from~\eqref{bound.minus.bis}
\begin{equation*}
  \frac{1}{a_\varepsilon}
  \int_{\Omega_-}
  (\partial_s\psi_\varepsilon)^2 \, (s+1) \, ds \, dt
  + \frac{a_\varepsilon}{\varepsilon^2}
  \int_{\Omega_-} (\partial_t\psi_\varepsilon)^2
  \, (s+1) \, ds \, dt
  \leq C
  \,.
\end{equation*}
Moreover, it follows from~\eqref{Ass}
and the convexity bound
\begin{equation}\label{convexity}
  \forall s \geq 0
  \,, \qquad
  f(s) \leq f'(0) s
\end{equation}
that
\begin{equation}\label{convexity.corol}
  \varepsilon \leq f'(0) \, a_\varepsilon
  \,.
\end{equation}
Hence
\begin{equation}\label{bound.minus.bis2}
  \int_{\Omega_-}
  |{\nabla }\psi_\varepsilon|^2 \, (s+1) \, ds \, dt
  \leq C a_\varepsilon
  \,.
\end{equation}

Now we write ($\varphi_\varepsilon$ is constant!)
\begin{equation}\label{dec.minus}
  \psi_\varepsilon(s,t) = \varphi_\varepsilon + \eta_\varepsilon(s,t)
  \,,
\end{equation}
where
\begin{equation}\label{orthogonal.minus}
  \int_{\Omega_-} \eta_\varepsilon(s,t) \, (s+1) \, ds\, dt = 0
  \,.
\end{equation}
Then we deduce from~\eqref{bound.minus.bis2}
\begin{equation}\label{eta.minus}
  \pi^2
  \int_{\Omega_-} \eta_\varepsilon^2 \, (s+1) \, ds \, dt
  \leq
  \int_{\Omega_-} |{\nabla }\eta_\varepsilon|^2 \, (s+1) \, ds \, dt
  \leq C a_\varepsilon
  \,.
\end{equation}
Note that~$\pi^2$ is indeed the minimum between the first
non-zero Neumann eigenvalue in the interval of unit length
and the first non-zero Neumann eigenvalue in the unit disk.

At the same time, from~\eqref{norm} using~\eqref{2-bound.minus}
and~\eqref{f.bound}, we obtain
\begin{equation}\label{bound.minus.norm}
  \int_{\Omega_-} \varphi_\varepsilon^2 \, a_\varepsilon \, (s+1) \, ds \, dt
  +\int_{\Omega_-} \eta_\varepsilon^2 \, a_\varepsilon \, (s+1) \, ds \, dt
  = \int_{\Omega_-} \psi_\varepsilon^2 \, a_\varepsilon \, (s+1) \, ds \, dt
  \leq C
  \,,
\end{equation}
where the first equality employs~\eqref{orthogonal.minus}.
Consequently, recalling that~$\varphi_\varepsilon$ is constant,
\begin{equation}\label{varphi.minus}
  \varphi_\varepsilon^2 \, a_\varepsilon
  \leq C
  \qquad \mbox{on} \quad \Omega_-
  \,.
\end{equation}

\subsection{The limiting eigenvalue equation in \texorpdfstring{$\Omega_+$}{Omega+}}
Now we consider~\eqref{weak}
for the sequence $\{\varepsilon_i\}_{i\in\mathbb{N}}$
and a test function $\phi(s,t) = \varphi(s)$,
where $\varphi \in C_0^\infty(\mathbb{R})$
is such that $\varphi'=0$ on $[-1,0]$,
and take the limit $i \to +\infty$.

We shall need a lower bound analogous
to the upper bound~\eqref{convexity.corol}.
From the fundamental theorem of calculus,
we deduce
\begin{equation}\label{Taylor}
  \forall s \in [0,a_\varepsilon] \,, \qquad
  f(s) \geq \big( \mathop{\mathrm{ess\;\!inf}}_{(0,a_\varepsilon)} f' \big) \, s
  \,.
\end{equation}
Note that the infimum cannot be zero unless~$f$ is trivial
(we assume from the beginning $\varepsilon < \sup f$
and that~$f$ is non-decreasing)
and that it converges to $f'(0)>0$ as $\varepsilon \to 0$.
Consequently, for all sufficiently small~$\varepsilon$, we have
\begin{equation}\label{convexity.corol.lower}
  \varepsilon \geq \frac{1}{2} \, f'(0) \, a_\varepsilon
  \,.
\end{equation}

At the same time, in analogy with~\eqref{2-bound.bis},
we have
\begin{equation}\label{2-bound.new}
  \forall (s,t) \in \Omega_+
  \,, \qquad
  \gamma_\varepsilon(s,t) \leq c_\varepsilon \rho_\varepsilon(s) \gamma_0(s)
  \,,
\end{equation}
where
$$
  c_\varepsilon := \exp\left(
  \frac{2 |x_0| a_\varepsilon + \varepsilon^2 + 2|y_0|\varepsilon}{2}
  \right)
$$
is converging to~$1$ as $\varepsilon \to 0$.

We first look at the right hand side of~\eqref{weak}.
Using the decompositions~\eqref{dec.plus} and~\eqref{dec.minus},
we have
\begin{multline*}
  (\varphi,\psi_{\varepsilon})
  = \int_{\Omega_-} \varphi \, \varphi_{\varepsilon} \,
  \gamma_\varepsilon \, a_\varepsilon \, \frac{f_\varepsilon \circ g_\varepsilon}{\varepsilon} \, ds \, dt
  + \int_{\Omega_-} \varphi \, \eta_\varepsilon \,
  \gamma_\varepsilon \, a_\varepsilon \, \frac{f_\varepsilon \circ g_\varepsilon}{\varepsilon} \, ds \, dt
  \\
  + \int_{\Omega_+} \varphi \, \varphi_\varepsilon \,
  \gamma_\varepsilon \, ds \, dt
  + \int_{\Omega_+} \varphi \, \eta_\varepsilon \,
  \gamma_\varepsilon \, ds \, dt
  \,.
\end{multline*}
Estimating $\gamma_\varepsilon \leq 1$ and using~\eqref{convexity}
and~\eqref{convexity.corol.lower},
we get
$$
  \left|
  \int_{\Omega_-} \varphi \, \eta_\varepsilon \,
  \gamma_\varepsilon \, a_\varepsilon \, \frac{f_\varepsilon \circ g_\varepsilon}{\varepsilon} \, ds \, dt
  \right|
  \leq
  \frac{a_\varepsilon^2}{\varepsilon} f'(0)
  \int_{\Omega_-} |\varphi| \, |\eta_\varepsilon| \,
  (s+1) \, ds \, dt
  \leq
  2 a_\varepsilon
  \int_{\Omega_-} |\varphi| \, |\eta_\varepsilon| \,
  (s+1) \, ds \, dt
  \,,
$$
where the right hand side tends to zero as $\varepsilon \to 0$
due to the Schwarz inequality and~\eqref{eta.minus}.
At the same time,
recalling that~$\varphi_\varepsilon$ is constant in~$\Omega_-$,
$$
  \left|
  \int_{\Omega_-} \varphi \, \varphi_\varepsilon \,
  \gamma_\varepsilon \, a_\varepsilon \, \frac{f_\varepsilon \circ g_\varepsilon}{\varepsilon} \, ds \, dt
  \right|
  \leq
  \frac{a_\varepsilon^2}{\varepsilon} f'(0)
  \int_{\Omega_-} |\varphi| \, |\varphi_\varepsilon| \,
  (s+1) \, ds \, dt
  \leq
  2 a_\varepsilon |\varphi_\varepsilon|
  \int_{\Omega_-} |\varphi| \,
  (s+1) \, ds \, dt
  \,,
$$
where the right hand side tends to zero as $\varepsilon \to 0$
due to~\eqref{varphi.minus}.
Using~\eqref{2-bound.new}, we also get
$$
  \left|
  \int_{\Omega_+} \varphi \, \eta_\varepsilon \,
  \gamma_\varepsilon \, ds \, dt
  \right|
  \leq
  c_\varepsilon
  \int_{\Omega_+} |\varphi| \, |\eta_\varepsilon| \,
  \rho_\varepsilon \, \gamma_0 \, ds \, dt
  \leq
  c_\varepsilon
  \sqrt{\int_{\Omega_+} \varphi^2 \,
  \gamma_0 \, ds \, dt}
  \sqrt{\int_{\Omega_+} \eta_\varepsilon^2 \,
  \rho_\varepsilon \, \gamma_0 \, ds \, dt}
  \,,
$$
where the right hand side tends to zero as $\varepsilon \to 0$
due to~\eqref{eta.plus}.
Finally, we write
$$
  \int_{\Omega_+} \varphi \, \varphi_{\varepsilon_i}
  \gamma_{\varepsilon_i} ds \, dt
  = \int_{\Omega_+} \varphi \, \varphi_{\varepsilon_i}
  \sqrt{\rho_{\varepsilon_i}} \gamma_{0} \, ds \, dt
  + \int_{\Omega_+} \varphi \, \varphi_{\varepsilon_i}
  \sqrt{\rho_{\varepsilon_i}} \gamma_{0} \,
  \left(\frac{\gamma_{\varepsilon_i}}{\sqrt{\rho_{\varepsilon_i}} \gamma_0}
  -1\right)
  \, ds \, dt
  \,.
$$
Here the first term on the right hand side converges to
$
  \int_{\Omega_+} \varphi \, \varphi_{0}
  \, \gamma_{0} \, ds \, dt
$
as $i \to +\infty$ due to~\eqref{weak.limit},
while the second term vanishes in the limit because of
\begin{multline*}
  \left|
  \int_{\Omega_+} \varphi \, \varphi_{\varepsilon_i}
  \sqrt{\rho_{\varepsilon_i}} \gamma_{0} \,
  \left(\frac{\gamma_{\varepsilon_i}}{\sqrt{\rho_{\varepsilon_i}} \gamma_0}
  -1\right)
  \, ds \, dt
  \right|
  \\
  \leq
  \sqrt{
  \int_{\Omega_+} \varphi^2
  \, \gamma_{0} \,
  \left(\frac{\gamma_{\varepsilon_i}}{\sqrt{\rho_{\varepsilon_i}} \gamma_0}
  -1\right)^2
  \, ds \, dt
  }
  \sqrt{
  \int_{\Omega_+} \varphi_{\varepsilon_i}^2
  \rho_{\varepsilon_i} \gamma_{0} \,
  \, ds \, dt
  }
  \,.
\end{multline*}
Indeed the second term on the right hand side is bounded
by~\eqref{bound.plus.norm},
while first term tends to zero as $i \to +\infty$
by the dominated convergence theorem.
Summing up,
\begin{equation}\label{lim.right}
  \lim_{i \to +\infty} (\varphi,\psi_{\varepsilon_i})
  = \int_{0}^{+\infty} \varphi \, \varphi_{0}
  \, \gamma_{0} \, ds
  \,.
\end{equation}

Employing that the test function~$\varphi$ is constant on $[-1,0]$
and the decomposition~\eqref{dec.plus},
we have
$$
  h_\varepsilon(\varphi,\psi_{\varepsilon})
  = \int_{\Omega_+} \varphi' \, \varphi_\varepsilon' \,
  \gamma_\varepsilon \, ds \, dt
  + \int_{\Omega_+} \varphi' \, \partial_s\eta_\varepsilon \,
  \gamma_\varepsilon \, ds \, dt
  \,.
$$
Here the first term on the right hand side can treated
in the same way as above with the conclusion
$$
  \int_{\Omega_+} \varphi' \, \varphi_{\varepsilon_i}' \,
  \gamma_{\varepsilon_i} \, ds \, dt
  \xrightarrow[i \to +\infty]{}
  \int_{\Omega_+} \varphi' \, \varphi_{0}' \,
  \gamma_{0} \, ds \, dt
  = \int_{0}^{+\infty} \varphi' \, \varphi_{0}' \,
  \gamma_{0} \, ds
  \,,
$$
while we integrate by parts to handle the second term,
$$
  \int_{\Omega_+} \varphi' \, \partial_s\eta_\varepsilon \,
  \gamma_\varepsilon \, ds \, dt
  = - \int_{\Omega_+} \varphi'' \, \eta_\varepsilon \,
  \gamma_\varepsilon \, ds \, dt
  - \int_{\Omega_+} \varphi' \, \eta_\varepsilon \,
  \partial_s\gamma_\varepsilon \, ds \, dt
  \,.
$$
Notice that the boundary terms vanish because~$\varphi$
has a compact support in~$\mathbb{R}$ and $\varphi'(0)=0$.
As above, the first term on the right hand side
vanishes as $\varepsilon \to 0$ due to~\eqref{eta.plus}.
Similarly,
$$
\begin{aligned}
  \left|
  \int_{\Omega_+} \varphi' \, \eta_\varepsilon \,
  \partial_s\gamma_\varepsilon \, ds \, dt
  \right|
  &\leq
  c_\varepsilon
  \int_{\Omega_+} |\varphi| \, |\eta_\varepsilon| \,
  \rho_\varepsilon \, \gamma_0 \, (x_0+s+a_\varepsilon)
  \, ds \, dt
  \\
  &\leq
  c_\varepsilon
  \sqrt{\int_{\Omega_+} \varphi^2 \,
  \gamma_0 \, (x_0+s+a_\varepsilon)^2 \, ds \, dt}
  \sqrt{\int_{\Omega_+} \eta_\varepsilon^2 \,
  \rho_\varepsilon \, \gamma_0 \, ds \, dt}
  \,,
\end{aligned}
$$
where the right hand side tends to zero as $\varepsilon \to 0$
due to~\eqref{eta.plus}.
Summing up,
\begin{equation}\label{lim.left}
  \lim_{i \to +\infty} h_{\varepsilon_i}(\varphi,\psi_{\varepsilon_i})
  = \int_{0}^{+\infty} \varphi' \, \varphi_{0}'
  \, \gamma_{0} \, ds
  \,.
\end{equation}

Since the set of functions $\varphi \in C_0^\infty(\mathbb{R})$
satisfying $\varphi'(0)=0$ is a core for the form
domain of the operator~$T_0$,
we conclude from~\eqref{lim.left} and~\eqref{lim.right}
that~$\varphi_0$ belongs to~$\mathsf{D}(T_0)$
and solves the one-dimensional problems
\begin{equation}\label{1Dproblems}
\begin{aligned}
  T_0\varphi_0 &= \mu_0^+ \varphi_0
  \,, \qquad &
  \mu_0^+ &:= \limsup_{i\to+\infty} \mu_{\varepsilon_i}
  \,,
  \\
  T_0\varphi_0 &= \mu_0^- \varphi_0
  \,, \qquad &
  \mu_0^- &:= \liminf_{i\to+\infty} \mu_{\varepsilon_i}
  \,.
\end{aligned}
\end{equation}
If $\varphi_0 \not=0$ on $(0,+\infty)$,
then~$\mu_0^\pm$ must coincide
with some eigenvalues of~$T_0$.
It remains to check that indeed $\varphi_0 \not=0$ on $(0,+\infty)$.

\subsection{The limiting problem in \texorpdfstring{$\Omega_-$}{Omega-}:
a crucial step}
Define
$$
  \Omega_-' := (-1/2,0)\times(0,1)
  \,, \qquad
  \Omega_+' := (0,1/2)\times(0,1)
  \,, \qquad
  \Omega' := (-1/2,1/2)\times(0,1)
  \,.
$$
From~\eqref{bound.minus.bis2} and~\eqref{bound.minus.norm},
we respectively have
\begin{equation}
  \int_{\Omega_-'} |{\nabla }\psi_\varepsilon|^2 \, ds \, dt \leq 2C a_\varepsilon
  \,, \qquad
  \int_{\Omega_-'} \psi_\varepsilon^2 \, ds \, dt \leq \frac{2C}{a_\varepsilon}
  \,.
\end{equation}
At the same time, denoting
$
  m_0 := \min_{[0,1/2]} \gamma_0
$
and assuming $\varepsilon \leq 1$,
from~\eqref{bound.plus} and~\eqref{bound.plus.norm},
we respectively get
\begin{equation}
  \int_{\Omega_+'} |{\nabla }\psi_\varepsilon|^2 \, ds \, dt
  \leq \frac{C}{m_0 \, \rho_\varepsilon(1/2)}
  \,, \qquad
  \int_{\Omega_+'} \psi_\varepsilon^2 \, ds \, dt
  \leq \frac{1}{m_0 \, \rho_\varepsilon(1/2)}
  \,.
\end{equation}
Consequently, $\psi_\varepsilon \in H^1(\Omega')$ for any $\varepsilon \leq 1$
(although, in principle,
$\|\psi_\varepsilon\|_{H^1(\Omega')}$ might not be uniformly bounded in~$\varepsilon$).

It follows that the boundary values $\psi_\varepsilon(0-,t)$ and $\psi_\varepsilon(0+,t)$
exist in the sense of traces in~$\Omega_-'$ and~$\Omega_+'$, respectively,
and they must be equal as functions of~$t$ in $L^2((0,1))$.
Using the decompositions~\eqref{dec.plus} and~\eqref{dec.minus},
we therefore have, for almost every $t \in (0,1)$,
\begin{multline*}
  [\varphi_\varepsilon(0-) - \varphi_\varepsilon(0+)]^2
  = \left[
  \eta_\varepsilon(0+,t) - \eta_\varepsilon(0-,t)
  \right]^2
  \leq 2\,[\eta_\varepsilon(0+,t)]^2 + 2\,[\eta_\varepsilon(0-,t)]^2
  \\
  \leq 2\,C \int_0^{1/2}
  \left([\eta_\varepsilon(s,t)]^2 + [\partial_s\eta_\varepsilon(s,t)]^2\right) ds
  +  2\,C \int_{-1/2}^0
  \left([\eta_\varepsilon(s,t)]^2 + [\partial_s\eta_\varepsilon(s,t)]^2\right) ds
  \,,
\end{multline*}
where~$C$ is a constant coming from the Sobolev embedding theorem.
Recall that~$\varphi_\varepsilon$ is constant on~$(-1,0)$
and $\varphi_\varepsilon \in H^1((0,1/2)) \hookrightarrow C^0([0,1/2])$;
more specifically, the first inequality of~\eqref{varphi'.plus}
and~\eqref{varphi.plus} respectively yield
\begin{equation}\label{varphi.H1}
  \int_0^{1/2} {\varphi_\varepsilon'}^2 \, ds
  \leq \frac{C}{m_0 \, \rho_\varepsilon(1/2)}
  \,, \qquad
  \int_0^{1/2} \varphi_\varepsilon^2 \, ds
  \leq \frac{1}{m_0 \, \rho_\varepsilon(1/2)}
  \,.
\end{equation}
Integrating with respect to~$t$ above, we deduce
$$
  [\varphi_\varepsilon(0-) - \varphi_\varepsilon(0+)]^2
  \leq 2\,C \int_{\Omega_+'}
  \left[\eta_\varepsilon^2 + (\partial_s\eta_\varepsilon)^2\right] ds \, dt
  + 2\,C \int_{\Omega_-'}
  \left[\eta_\varepsilon^2 + (\partial_s\eta_\varepsilon)^2\right] ds \, dt
  \,.
$$
Applying~\eqref{eta.plus},
the second inequality of~\eqref{varphi'.plus} and~\eqref{eta.minus},
we may write
\begin{equation}\label{trace.trick}
  [\varphi_\varepsilon(0-) - \varphi_\varepsilon(0+)]^2 \leq C
  \,,
\end{equation}
where~$C$ is a constant (different from the above) independent of~$\varepsilon$,
provided that~\eqref{Ass.small} holds.
Finally, applying~\eqref{varphi.H1} and the Sobolev embedding
$H^1((0,1/2)) \hookrightarrow C^0([0,1/2])$,
we deduce from~\eqref{trace.trick} the following improvement
upon~\eqref{varphi.minus}
\begin{equation}\label{varphi.minus.improve}
  \varphi_\varepsilon^2
  \leq C
  \qquad \mbox{on $\Omega_-$}
  \,.
\end{equation}

\subsection{As \texorpdfstring{$\varepsilon \to 0$}{epsilonto0} only
\texorpdfstring{$\Omega_+$}{Omega+} matters:
convergence of eigenvalues and eigenfunctions}
Estimate \eqref{varphi.minus.improve}~provides a crucial information
whose significance consists in that
what happens in~$\Omega_-$ is insignificant.
\begin{proposition}\label{Prop.norm}
One has
$$
  \|\psi_{\varepsilon_i}\| \xrightarrow[i \to +\infty]{}
  \|\varphi_0\|_{L^2_{\gamma_0}((0,+\infty))}
  \,.
$$
\end{proposition}
\begin{proof}
We have
\begin{align*}
  \|\psi_{\varepsilon}\|^2
  = & \int_{\Omega_-} \varphi_{\varepsilon}^2 \,
  \gamma_\varepsilon \, a_\varepsilon \, \frac{f_\varepsilon \circ g_\varepsilon}{\varepsilon} \, ds \, dt
  + \int_{\Omega_-} \eta_\varepsilon^2 \,
  \gamma_\varepsilon \, a_\varepsilon \, \frac{f_\varepsilon \circ g_\varepsilon}{\varepsilon} \, ds \, dt
  \\
  & + \int_{\Omega_-} 2 \, \varphi_{\varepsilon} \, \eta_\varepsilon \,
  \gamma_\varepsilon \, a_\varepsilon \, \frac{f_\varepsilon \circ g_\varepsilon}{\varepsilon} \, ds \, dt
  \\
  & + \int_{\Omega_+} \varphi_{\varepsilon}^2 \,
  \gamma_\varepsilon \, ds \, dt
  + \int_{\Omega_+} \eta_\varepsilon^2 \,
  \gamma_\varepsilon \, ds \, dt
  + \int_{\Omega_+} 2 \, \varphi_{\varepsilon} \, \eta_\varepsilon \,
  \gamma_\varepsilon \, ds \, dt
  \,.
\end{align*}
The right hand side of the first line
together with the mixed term on the second line
goes to zero as $\varepsilon \to 0$.
Indeed, recalling~\eqref{convexity}, \eqref{convexity.corol.lower}
and $\gamma_\varepsilon \leq 1$,
$$
  \int_{\Omega_-} \varphi_{\varepsilon}^2 \,
  \gamma_\varepsilon \, a_\varepsilon \, \frac{f_\varepsilon \circ g_\varepsilon}{\varepsilon} \, ds \, dt
  \leq
  2 \, a_\varepsilon \, \varphi_{\varepsilon}^2 \int_{\Omega_-} (s+1) \, ds \, dt
  \xrightarrow[\varepsilon \to 0]{} 0
$$
due to~\eqref{varphi.minus.improve};
$$
  \int_{\Omega_-} \eta_{\varepsilon}^2 \,
  \gamma_\varepsilon \, a_\varepsilon \, \frac{f_\varepsilon \circ g_\varepsilon}{\varepsilon} \, ds \, dt
  \leq
  2 \, a_\varepsilon \int_{\Omega_-} \eta_{\varepsilon}^2 \, (s+1) \, ds
  \xrightarrow[\varepsilon \to 0]{} 0
$$
due to~\eqref{eta.minus};
and the mixed term goes to zero by the Schwarz inequality.
Similarly, recalling~\eqref{2-bound.new},
$$
  \int_{\Omega_+} \eta_{\varepsilon}^2 \,
  \gamma_\varepsilon \, ds \, dt
  \leq
  c_\varepsilon \int_{\Omega_+} \eta_{\varepsilon}^2 \, \rho_\varepsilon \gamma_0 \, ds \, dt
  \xrightarrow[\varepsilon \to 0]{} 0
$$
due to~\eqref{eta.plus};
while the Schwarz inequality yields
$$
  \left|
  \int_{\Omega_+} 2\, \varphi_{\varepsilon} \, \eta_\varepsilon \,
  \gamma_\varepsilon \, ds \, dt
  \right|
  \leq
  2 c_\varepsilon
  \sqrt{
  \int_{\Omega_+} \eta_{\varepsilon}^2 \, \rho_\varepsilon \gamma_0 \, ds \, dt
  }
  \sqrt{
  \int_{\Omega_+} \varphi_{\varepsilon}^2 \, \rho_\varepsilon \gamma_0 \, ds \, dt
  }
  \xrightarrow[\varepsilon \to 0]{} 0
  \,,
$$
where the second square root is bounded in~$\varepsilon$
due to~\eqref{varphi.plus}.
Finally, we write
$$
  \int_{\Omega_+} \varphi_{\varepsilon_i}^2 \,
  \gamma_{\varepsilon_i} \, ds \, dt
  = \int_{\Omega_+} \varphi_{\varepsilon_i}^2 \,
  \rho_{\varepsilon_i} \gamma_0 \, ds \, dt
  + \int_{\Omega_+} \varphi_{\varepsilon_i}^2 \,
  (\gamma_{\varepsilon_i}-\rho_{\varepsilon_i} \gamma_0) \, ds \, dt
$$
and observe that the first term on the right hand side
tends to the desired result
$\|\varphi_0\|_{L^2_{\gamma_0}((0,+\infty))}^2$
as $i \to+ \infty$ by the strong convergence~\eqref{strong.limit},
while the second term vanishes in the limit.
In more detail,
\begin{multline*}
  \left|
  \int_{\Omega_+} \varphi_{\varepsilon_i}^2 \,
  (\gamma_{\varepsilon_i}-\rho_{\varepsilon_i} \gamma_0) \, ds \, dt
  \right|
  = \left|
  \int_{\Omega_+}
  \left(
  \varphi_{\varepsilon_i}^2 \, \rho_{\varepsilon_i}
  - \varphi_{0}^2 + \varphi_{0}^2
  \right)
  \left(\frac{\gamma_{\varepsilon_i}}{\rho_{\varepsilon_i}}-\gamma_0\right) \, ds \, dt
  \right|
  \\
  \leq
  \int_{\Omega_+}
  \left|
  \varphi_{\varepsilon_i}^2 \, \rho_{\varepsilon_i}
  - \varphi_{0}^2
  \right|
  \left(c_{\varepsilon_i}\gamma_0+\gamma_0\right) \, ds \, dt
  + \int_{\Omega_+}
  \varphi_{0}^2
  \left(\frac{\gamma_{\varepsilon_i}}{\rho_{\varepsilon_i}}-\gamma_0\right) \, ds \, dt
  \,,
\end{multline*}
where the the first term after the inequality tends to zero as $i \to +\infty$
by the strong convergence again, while the second term vanishes
by the dominated convergence theorem.
\end{proof}

It follows from Proposition~\ref{Prop.norm} that $\varphi_0 \not= 0$,
so that it is indeed an eigenfunction of~$T_0$
due to~\eqref{1Dproblems}.
In particular, $\mu_0^+ = \mu_0^-$.

Now, let~$\hat{\psi}_\varepsilon$ be a normalised eigenfunction
corresponding to possibly another eigenvalue
$\hat{\mu}_\varepsilon := \mu_m(\varepsilon)$.
Again, we use the decompositions~\eqref{dec.plus} and~\eqref{dec.minus}
and distinguish the individual components by tilde.
In the same way as we proved Proposition~\ref{Prop.norm},
we can establish
\begin{proposition}\label{Prop.inner}
One has
$$
  (\psi_{\varepsilon_i},\hat{\psi}_{\hat\varepsilon_j})
  \xrightarrow[i,j \to +\infty]{}
  (\varphi_0,\hat{\varphi}_0)_{L^2_{\gamma_0}((0,+\infty))}
  \,.
$$
\end{proposition}

If $m \not= n$, then $(\psi_{\varepsilon_i},\hat{\psi}_{\hat\varepsilon_j})=0$
and thus $(\varphi_0,\hat{\varphi}_0)_{L^2_{\gamma_0}((0,+\infty))}=0$.
Hence~$\varphi_0$ and~$\hat{\varphi}_0$ correspond to
distinct eigenvalues of~$T_0$.
In particular, $\varphi_0$~is an eigenfunction
corresponding to the $(n+1)^\mathrm{th}$ eigenvalue~$\nu_n$ of~$T_0$.
Since we get this result for \emph{any} weak limit
point of $\{\varphi_\varepsilon\}_{\varepsilon>0}$,
we have the convergence results actually in $\varepsilon \to 0$
(no need to pass to subsequences).

This completes the proof of Theorem~\ref{Thm.convergence}.

\vskip 1cm
\noindent \textsc{Acknowledgements}.
This paper was partially supported by the grants PRIN 2012
``Elliptic and parabolic partial differential equations:
geometric aspects, related inequalities, and applications'',
FIRB 2013  ``Geometrical and qualitative aspects of PDE's'',
and  STAR 2013 ``Sobolev-Poincar\'e inequalities: embedding constants,
stability issues, nonlinear eigenvalues'' (SInECoSINE).
{D.K.\ was partially supported
by the project RVO61389005 and the GACR grant No.\ 14-06818S.}

redpajama_set_name: RedPajamaArXiv


text:
import { Plugin, toRoute } from "../Plugin";
import { Registry } from "../Registry";
import { MockServer } from "../__mocks__/server";

import { Server } from "hapi";
import { T, identity } from "ramda";

let plugin: Plugin;
let server: Server;
let action: any;

beforeEach(() => {
  server = new MockServer() as any as Server;
  plugin = new Plugin(server, {
    root: "/test",
    secure: false
  });

  action = {
    method: "POST",
    path: "/users",
    handle: jest.fn().mockReturnValue({
      id: 1337
    }),
    decorate: jest.fn().mockImplementation(identity)
  };
});

it("registers a root action", () => {
  let onPreStartFn = (server.ext as jest.Mock<any>).mock.calls[0][1];
  onPreStartFn(server, () => { });

  expect(server.route).toHaveBeenCalledWith({
    handler: jasmine.any(Function),
    method: "GET",
    path: "/test",
    options: {}
  });
});

describe("when `options.secure` is not set to false", () => {
  beforeEach(() => {
    plugin = new Plugin(server, {
      root: "/test"
    });
  });

  describe("when `prism-security` plugin has not been registered", () => {
    it("throws an error when server starts", () => {
      let fn = () => server.start();
      expect(fn).toThrowError("Secure mode enabled but `prism-security` plugin has not been registered");
    });
  });

  it("mutates the the root action to contain `optional` auth mode", () => {
    server.plugins["prism-security"] = "mockSecurityPlugin" as any;
    server.start();

    expect(server.route).toHaveBeenCalledWith({
      handler: jasmine.any(Function),
      method: "GET",
      path: "/test",
      options: {
        auth: {
          mode: "optional"
        }
      }
    });
  });
});

describe("#registerAction()", () => {
  it("adds `root` suffix to Action `path` property", () => {
    plugin.registerAction(action);
    expect(action.path).toBe("/test/users");
  });

  it("registers route with server", () => {
    plugin.registerAction(action);
    expect(server.route).toHaveBeenCalledWith({
      handler: jasmine.any(Function),
      method: "POST",
      path: "/test/users"
    });
  });
});

describe("toRoute()", () => {
  let route: any;

  beforeEach(() => {
    route = toRoute(action);
  });

  it("strips URI template placeholders from path", () => {
    action.path = "/test/users{?where,page,order}";
    let route = toRoute(action);

    expect(route.path).toBe("/test/users");
  });

  describe(".handler()", () => {
    let request: any;

    beforeEach(() => {
      request = {
        auth: {},
        params: {
          method: "GET"
        },
        query: {
          where: "owner,2"
        }
      };
    });

    it("calls `action.handle()` with merged URI template parameters", () => {
      route.handler(request, T);

      expect(action.handle).toHaveBeenCalledWith({
        method: "GET",
        where: {
          owner: "2"
        }
      }, request);
    });

    it("creates a document with result of `action.handle()` and renders it", async () => {
      const response = await route.handler(request);
      expect(response).toEqual({
        _links: {
          self: {
            href: "/users"
          }
        },
        id: 1337
      });
    });
  });
});

redpajama_set_name: RedPajamaGithub


text:
D3 Swimming and Diving
A site dedicated to covering NCAA Division III Swimming and Diving.
'19 Conference Championship Central
2019 NCAA DIII Cut Times
PREVIEW: Centennial Conference
D3SwimDive	Conference Preview	 October 22, 2018 October 22, 2018 6 Minutes
For our first conference preview, we travel to the eastern part of the US and specifically eastern Pennsylvania and Maryland to look at the Centennial Conference. Last year, 11 total swimmers represented the conference at the NCAA Division III Championships, and eight of those swimmers return for the 2018-19 season with hopes of more joining them in Greensboro.
In 2018, the Ursinus women rolled to their fifth-straight league title by more than 300 points while Gettysburg reclaimed the men's title with a 215-point victory.
Here are our projected standings with previews:
CENTENNIAL WOMEN:
Ursinus – The margin won't be the 326 points over Swarthmore that it was last year, as the graduation of 2018 Most Outstanding Performer Sydney Gundersen and fellow HM All-Americans Meggie Leitz and NCAA B-cut flyer Katherine Porter all amount to a significant dent and loss of 195 individual points and 8 "A" relay swims. However, Ursinus does return the conference's lone individual All-American from 2018 in Clara Baker (HM, 100 Fly) and fellow relay HM All-Americans Peyton Lyons and Sophie May ('18 Outstanding Rookie). Add in the strongest recruiting class in the league – both Sophia Lear (23.72/52.43/1:55.24 FR, 2:11.82 2IM, 1:07.6 1BR) and Katie Schultz (2:28.9 2BR, 2:11.1/4:33.5 IM, 55.4/1:57.2/5:03.9 FR) could win conference titles and the league's Outstanding Rookie Award while Liv Rachael (24.38/53.14 FR) could be a big relay piece – and the Bears have little to worry about in 2019.
Swarthmore – The Garnet lose much less than Ursinus, but Maggie Eberts is a big loss, and her 50 individual points in the Breast/IM events and spots on four A relays will need to be filled. However, sophomore Hannah Kloetzer returns off her NCAA selection in the 1650 from last year, as do two other 40+ point scorers in junior Clare Cushing and senior Scout Clark. Anna Lyn (24.77/52.52/1:54.1, 58.82 1FL) may be the best of a strong recruiting class and will have a shot at titles in the 100 and 200 Free, while Liz Gallagher (24.63/54.32 FR, 58.53/2:10.2 FL) and Anna Kottakis (24.53/53.03/1:57.4, 58.33/2:07.17 BA) also should reach several A finals.
Gettysburg – The conference coaches selected Gettysburg ahead of Swarthmore in their preseason poll, but I just don't see it happening. The Bullets bring in a slightly deeper recruiting class than the Garnets but also lost a lot more, as Savannah Llewellyn, Sadie Stiles, Nicole Farina, Kate Hegwood, and Allie Mueller combined for 137 individual points and 13 total relay (6A/7B) slots. Katie Cooper (24.55/53.3/1:56.8, 2:13.9/4:44 IM), Katie Jenkins (54.5 100FR, 58.7/2:10.9 FL, 2:12.2 IM), Cate Kosko (25.2/54.08/1:56.7), and Megan Wojnar (54.8/1:55.3/5:04/18:01, 2:07.2 2BA, 2:11/4:36 IM) all make major impacts for the team.
Franklin & Marshall – F&M is a pretty good step behind the top three programs in the league, but they do seem to be trending in the right direction. The Diplomats lose just 52 individual points and two A relay slots from the 2018 Centennial Championships and return their top two individual scorers in junior Katherine Hirsch and Grace Pereles. However, their real strength may be among their newcomers. Tara Kupsky (2:08/4:38 IM, 54.29/1:58.6/18:17 FR, 57.96 1FL) should be one of the league's top freshmen and immediately could content for conference crowns in the IM events, while Amelia Bothwell (54.7 100FR, 17:40 1650, 2:12.4 2BA, 2:10.7 2FL) and Skye Howes (24.85/54.58 FR) also should be major contributors.
Dickinson – Distance standout Olivia Lyman and classmate Katie Schmidt return to lead the Red Devils for their senior seasons, and Dickinson returns all but 2 individual points and a single A relay slot. Their eight-woman recruiting class isn't particularly strong, but Sasha Arefyev (25.31/55.04/1:57.8/5:12), Brooke Kelly (54.8 100FR, 4:53 4IM), and Kelly Hester (2:33 2BR) should all be able to score right away.
Washington – Washington finished ahead of Dickinson both at last year's championships and in this year's preseason coaches' poll, but the Shorewomen graduated breaststroke standout Jenni Clune, Katie Gorman, and three other seniors, totaling 72 individual points and nine A-relay slots. Seniors Julia Portmann and Anna Schmittle lead the returners, but, from our research, only Teresa Draves (2:21 2FL, 4:50 4IM) and Anna Hall (1:00/2:17 FL) are expected to make much of an impact this year from the team's nine newcomers.
McDaniel – The Green Terror lost a lot – seven of the 12 swimmers on the team's roster from 2017-18 are not returning, counting for a total of 61 individual points, 10 A-relay slots, and eight B-relay slots. Samantha Smith (40 pts, 4 A-relays) is the biggest lost by far, but Blair Boyle and Sara Hegna do return after combining for 60 individual points last year. The Green Terror will need their five freshmen to contribute immediately – Cambrie Cooke (1:00/2:11 BA, 1:00.7 FL, 4:43 4IM) is best set to make an impact.
Bryn Mawr – Bryn Mawr's 2018-19 roster is still not on their website. The Owls didn't have a single senior on last year's roster, but new HC Pat McDevitt has a lot of work in front of him, as senior Lillian Oyen-Ustad was one of just three swimmers to score individual points and the only one to break double figures (23). Just a few strong newcomers could give the Owls enough to challenge McDaniel, but until we see the roster, we have to leave Bryn Mawr here.
CENTENNIAL MEN
Gettysburg – Yes, the Bullets lose 2x Centennial Most Outstanding Performer and 8x All-American Ty Dougherty, and that's a loss that can't be understated, but Dougherty is the only one of the team's top-12 scorers from last year's championships not returning. Aedan Collins, Josh Parker, Adam Dias, And Jack Edelson all scored 40 points or more at the 2018 league meet, and a strong four-man recruiting class should offset Dougherty's departure. Jack Freeman (1:45/4:45 FR, 52.4/1:57.5 FL, 1:56.8/4:24 IM) returns after not being on the roster last year, and CJ Hewitt (16:33 1650, 52.8/1:58.9 FL, 4:16.8 4IM), Sam Nonemaker (21.6/46.7/1:43 FR, 1:57.6/4:24.6 IM), and Oliver Pickering (17:23 1650, 52.22/1:54.78 FL) all should make significant impacts.
Swarthmore – The Garnet return their top four scorers from the 2018 Championships, including the three who earned trips to NCAAs in Alec Menzer ('18 Outstanding Rookie), Jeffrey Tse, and Michael Lutzker. Chris Smith and Charles Yang also return from relays that achieved B-Cuts in 2018. Swarthmore does lose 74.5 points and two A-relay slots (14 B-relay slots) from last year, but they bring in six recruits, all of whom are capable of scoring immediately. Jeremy Rockaway (51.27/1:52.69 FL, 1:54.6/4:05.3 IM, 2:05.89 2BR) should contend for multiple league titles right away, and Ethan Bergmann (21.6/47.0/1:42.6, 52.77 1FL), Horace Shew (51.4/1:55.5 BA, 59.4/2:10.7 BR, 52.9 FL, 1:56 2IM), and Tom Ward (1:44.8/4:42 FR, 55.3/2:03.5 BA, 53.0 1FL) should all score big points. Gettysburg's 215-point win over the Garnet in 215 was exaggerated (Swarthmore DQd 3 relays (2A, 1B), and Swarthmore will close the gap to under 50 points, but I think they ultimately come up just short, despite perhaps having more NCAA selections once again.
Washington – Washington loses 34 ind. points and seven A-relay slots from 2018 but does return its top seven individual scorers and has no seniors on its 2018-19 roster. Jake Brown (42 pts, 4 A-relays), Andrew Feeley (40 pts, 2/2 relays), and Aidan Sullivan (40 pts, 4 A-relays) are among the team's top returners, and Leo Johnson, Teague Sauter and Brett Edwards all reached multiple A-finals. The Shoremen have just two newcomers, but Joseph Bozzi (2:12.5 2BR, 50.8/1:55.4 FL, 1:55.2/4:26 IM) may be one of the league's top rookies, and Nate Braddock (55.1 1BA, 4:25.98 4IM) should be able to contribute as well. The league's coaches have Washington behind both F&M and Ursinus in their preseason poll, but all three squads lost similar amounts to what they are projected to have brought in, so we are leaving Washington here for now… though 3-5 will be VERY close.
Franklin & Marshall – Christopher Schiavone returns after finishing as conference runner-up in three events last year and reaching the NCAA Championships in the 50 free, but standout flyer Eric Dietrich-Peterson transferred to D1 Penn. Alec Wahl and Timothy Gould also return for the Diplomats, but, including Dietrich-Peterson, F&M lost 136 individual points, nine A-relay slots, and 12 B-relay slots to graduation or other departures. They did bring in the deepest recruiting class in the conference. Thomas Graminski (56.72/2:03.67 BR, 51.11 1FL, 1:52.3/4:16.94) could win multiple conference titles immediately, and Adam Moskus (21.27/46.42/1:42.92 FR) and Brendan Cline (21.49/47.2/1:43.5/17:06 FR, 52.06 1FL) should both be major factors.
Ursinus – The Bears return eight of their top nine scorers and lose just two scorers (42 ind., 3/2 relays) from last year's league championships. Jacob Menzer (48 pts, 2/2 relays) leads the way and will look to make it to nationals in the 500 after notching a B-Cut last year, while Casey Lear, Michael Prior, and James Callaghan all scored 31 or more points. Ryan Carkuff (21.14/47.2 FR, 56.29 1BA) is the best of a seven-man recruiting class, while Daniel Mueser (21.80 50, 52.24 1FL) should also be a significant factor.
Dickinson – The Red Devils return four of their top five scorers, but Dickinson does lose four seniors, including Mitch Reynolds, who combined for 66 individual points and 12 relay slots (6/6). Will Freeman, conference champ and B-cut in the 400 IM, leads the way for Dickinson while Graham Novitch and Kiernan Jordan are also key returners. Ben Should (50.23 1FL, 56.36 1BA) is the best of the team's seven newcomers, while Logan Cort (1:44.3/4:44.4/17:08) and Damon St. Prix (47.44/1:43.6 FR) should impact their events.
McDaniel – McDaniel was a distant seventh in 2018 and that shouldn't change this year. They lost two swimmers to graduation who combined for just seven individual points (6 A-relay slots), but neither of their recruits shows up on CollegeSwimming.com, so we aren't expecting much there. Paul Van Nevel, the Green Terror's only individual A-final swimmer from last year, will look to have a strong senior season after reaching seven A-finals over his first three years.
Projected Award Winners:
Women's Swimmer of the Year: Peyton Lyons, Ursinus
Women's Newcomer of the Year: Sophia Lear, Ursinus
Men's Swimmer of the Year: Alec Menzer, Swarthmore
Men's Newcomers of the Year: Thomas Graminski, F&M/Jeremy Rockaway, Swarthmore
conferencepreview
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redpajama_set_name: RedPajamaCommonCrawl


text:
August 11, 2015 3.01 pm This story is over 76 months old
Historians puzzled over mysterious double-edged sword found in Lincoln
Help needed: Historians are appealing for the public's help in deciphering an inscription on a double-edged sword found in Lincolnshire in the 19th century.
Help is needed to decipher the inscription on this sword found in Lincoln in the 19th century. Photo: British Museum
Historians are appealing for people's help in deciphering an inscription on a double-edged sword found in Lincolnshire in the 19th century.
The mysterious artefact was found in the River Witham in July 1825, and was presented to the Royal Archaeological Institute by the registrar to the Bishop of Lincoln.
The 13th century sword, believed to have been manufactured in Germany, features an inscription found along one of its edges and inlaid in gold wire.
It has been speculated that this is a religious invocation, since the language is unknown.
The inscription appears to read: +NDXOXCHWDRGHDXORVI+
The inscription on the sword. Photo: British Museum
The sword is on loan to the British Library from the British Museum as part of the Magna Carta: Law, Liberty, Legacy exhibition which runs until September 1.
Weighing 1.2 kg (2lb 10oz) and measuring 964 mm (38in) in length and 165 mm (6½in) across the hilt, it has been claimed that the sword could have easily sliced a man's head in two if struck with sufficient force.
Leasehold Home & Giftware Business for Sale


Alexander Sondheimer, banker and merchant, recognized as one of the representative and progressive business men of Muskogee, was born in St. Louis, Missouri, May 11, 1873, and is a son of Joseph and Johanna Sondheimer. His youthful days were spent in St. Louis and in Muskogee, for he was but six years of age when brought by his parents to this city. His father had first visited Muskogee in 1867 and, pleased with the city and its prospects, he brought his family here in 1879, so that Alexander Sondheimer pursued a part of his education in the schools of Muskogee. He finished his studies in the high school of St. Louis, where he also pursued a business course.
When his textbooks were put aside he worked for the Cotton Belt Railroad at Tyler, Texas, where he continued for a year and then came to Muskogee, where he became court reporter, being one of the first court reporters of the state. He also reported the first democratic convention in the state, held at McAlester in 1892. He continued to engage in court reporting for two years and afterward became associated with his father in the buying of hides, wool and furs, the business being carried on under the firm name of the Joseph Sondheimer Sons Company. The business has been developed to notable proportions, being now the largest of the kind in the southwest, and since the death of the father Alexander Sondheimer has continued in the business, having one of the oldest houses of the kind in Muskogee. He possesses sound judgment and unfaltering enterprise and his close application, thoroughness and industry have been. most potent forces in the attainment of his present-day success.
He also figures prominently in financial circles, being one of the stockholders and a director of the First National Bank of Muskogee, also of the Citizens National Bank of Okmulgee and the First National Bank of Stigler, Oklahoma. He likewise owns a large amount of stock in the Muskogee Wholesale Grocery Company and in the Ratcliff-Sanders Wholesale Grocery Company of Tulsa. He also has important farming interests and oil leases and is today regarded as one of the foremost business men of his section of the state.
On the 1st of October, 1895, Mr. Sondeimer was united in marriage to Miss Eudora. L. Cobb, a daughter of John O, Cobb, a native of Tennessee, who came to the Indian Territory in the late 1860 and who was a friend of Mr. Sondheimer. Mr.Cobb engaged in merchandising here and at the time of his death was a druggist of Muskogee. He served as an officer in the Union army during the Civil war and had a brother who served in the Confederate army.
In religion Mr. Sondheimer holds to the faith of his fathers. He belongs to the Benevolent Protective Order of Elks, of which he is a life member, and he also has membership connection with the United Commercial Travelers. His political endorsement is given to the Democratic Party, but the honors and emoluments of office have never had attraction for him. He has always preferred to concentrate his energies and attention upon his business affairs and the thoroughness with which he has directed his interests has made him one of the most prosperous merchants and financiers of this section of the state.
Benedict, John Downing. Muskogee and Northeastern Oklahoma : including the counties of Muskogee, McIntosh, Wagoner, Cherokee, Sequoyah, Adair, Delaware, Mayes, Rogers, Washington, Nowata, Craig, and Ottawa. Chicago: S.J. Clarke Pub. Co., 1922.


text:
Renault Zoe Slows, But Remains #1 In Europe (April Electric Car Sales Report)
The European passenger plug-in market had some 17,500 registrations last month, up only 6% over the same month last year. Despite this slowdown, the market is up 24% this year, with the electric vehicle (EV) share now standing at 1.4%.
Jose Pontes
Originally published on CleanTechnica.
Renault Zoe Hiccups, But Remains #1
Despite some supply chain issues that will only be completely cleared in June, the Renault Zoe remained Master in Command both in April and for the year to date (YTD). Meanwhile, the BMW i3 and Mitsubishi Outlander PHEV recovered some ground over the ageing Nissan Leaf.
Looking at the monthly model ranking:
#1 Renault Zoe – Due to some quality glitches, the deliveries of some 2,000 Zoe were delayed, hurting April's performance of the French hatchback, which last month only landed 1,627 registrations, its worst result of the year. But that was still enough to win Europe's monthly best seller status for the fourth month in a row. Due to these production hiccups, the model's performance suffered across Europe, from its home market (939 registrations, down 5%) to Austria (96, down 2%). Nevertheless, it somehow managed to grow in the all-important German market (280 registrations, up 27%).
#2 BMW i3 – The German hot hatch had its worst-selling month in Europe since the massive arrival of the 33 kWh battery, with 1,562 cars being delivered. Despite this, deliveries were up 27% year over year (YoY), with good numbers especially in Norway (426 registrations), all of them being fully electrics(!), but also in its home Germany (357), the United Kingdom (195), and Austria (93). Looking ahead, expect deliveries to slowly drop throughout the summer as people anticipate the revised version, coming later in the year.
#3 Mitsubishi Outlander PHEV – After a tight race with the BMW i3, the Japanese SUV ended March in 3rd place (1,507 registrations, down 15% YoY). Despite reaching interesting sales levels in important markets, like Germany (294 units) or Norway (222 units), Mitsubishi will have a hard time finding a replacement for UK sales — incentives for PHEVs have now become much less appealing there.
#4 Nissan Leaf – Thanks to deep discounting, the ageing hatch is resisting a wipeout from the newer offerings, reaching 4th place despite registrations going down 31% YoY to 1,326 units. With steady sales in its stronghold markets — like Norway (287 units), the United Kingdom (350), and France (198) — and a surprising performance in Sweden (189 units, a new record), the veteran model is keeping momentum by appealing to a new demographic still untapped by EVs: Bargain Hunters.
#5 Mercedes GLC350e – The (not so big) surprise of the month came from the Mercedes GLC350e. This trendy Mercedes model is in the intersection of two hot trends on the Old Continent: SUVs and plug-ins, all served in a Euro-friendlier size than the yacht-like models above it (Tesla Model X, Volvo XC90 …). The GLC350e landed #5 in April with 961 registrations. The largest markets for the Mercedes SUV were Belgium (204 units, new record), Norway (179 units), and Germany (126). The three-pointed star has hit a homerun with this model, especially considering that it doesn't have direct competition among the premium brands. (Well, perhaps not a homerun in Europe — maybe a hat trick.)
YTD Ranking – Zoe Steady in #1, Nissan Leaf climbs to #2.
Looking at the YTD ranking, the top positions remained stable* (see disclaimer), with the highest position change occurring only in the 7th spot, where the Mercedes GLC350e reached its best position to date and is now looking to reach the rear of the Tesla Model S and VW Passat GTE.
Other models on the rise are the BMW 330e, now #8, and its 225xe Active Tourer relative, #10, putting three BMWs in the top 10.
Underlining the positive momentum of pure electric cars, we salute the return to the top 20 of the VW e-Golf, with the German hatchback delivering 669 units in April, its best result in 12 months, thanks to the massive delivery of the larger 36 kWh battery. With the Opel Ampera-e (Chevrolet Bolt in Euro-spec) volume delivery now underway in Norway (more than 150 units delivered there in May) and elsewhere, expect the Opel EV to reach the top 20 during the summer.
Looking at the manufacturer ranking, BMW (20%, up 3%) is seated on the top spot, followed by Renault (13%, down 1%) in 2nd place, while 3rd place is being toughly fought by Nissan, Mercedes, and Volkswagen — all with 10% share. With the Wolfsburg brand now making the restyled Golf plug-ins at full speed, expect it to beat the competition in the next couple of months and race Renault for the #2 position.
Europe April YTD
Renault Zoe 1,627 10,676
Nissan Leaf 1,326 8,009
BMW i3 1,562 7,031
Mitsubishi Outlander PHEV 1,507 6,174
Tesla Model S 559 4,276
VW Passat GTE 872 4,252
Mercedes GLC350e 961 3,503
BMW 330e 777 3,442
Tesla Model X 481 3,386
BMW 225xe Act. Tourer 711 3,084
Volvo XC90 T8 PHEV 443 2,931
Mercedes C350e 677 2,821
Audi A3 e-Tron 681 2,287
BMW X5 40e 485 2,051
Hyundai Ioniq Electric 433 1,602
VW e-Golf 669 1,559
Kia Soul EV 343 1,540
VW Golf GTE 613 1,538
Audi Q7 e-Tron 330 1,295
Volvo V60 Plug-In 282 1,213
Others 2,174 8,574
TOTAL 17,513 81,244
*Disclaimer: Due to some data adjustments in the March numbers, the models positioning were subject to changes regarding the March article.
Also published on our EV sales page, EV Obsession, and the EV Sales blogspot.
Reprinted with permission.
In this article:Europe, Europe EV sales
Written By Jose Pontes
Always interested in the auto industry, particularly in electric cars, Jose has been overviewing the sales evolution of plug-ins through the EV Sales blogsince 2012, allowing him to gain an expert view on where EVs are right now, and where they are headed in the future. The EV Sales blog has become a go-to source for people interested in electric car sales around the world. Extending that work and expertise, Jose is now a partner in EV-Volumes.
Big EV Spreadsheet Compares Actual Energy Consumption Of Electric Vehicles, Cybertruck updates, Electric buses to debut in Zduńska Wola


text:
package com.example;

public interface Command {
    void execute();
    void undo();
}


text:
Q: Return a 2D array from a function with just the array size as the arguments i am new to c++ and i am trying to create an Array from within a function with just the array size, [row][col], as the argument. I know you cannot return an array from a function in c++. I would allocate memory in these situations if i were to code in C. However, i am not sure what is the C++ way (:
I tried,
    int** get2Darray(int row, int col){
         int** randInts = new int[row][col]; //Invalid. Constant required.
         return randInts;
    }

Thanks.
EDIT: vector class does the trick, however, my program is heavily modular. I dont have access to the main program. My function should be called and a fresh array should be created and returned. Implementation should only be done inside this function

A: Either change int** randInts = new int[row][col]; into int* randInts = new int[row * col]; or define int** randInts as follows:
int** randInts = new int*[row];
for (int i = 0; i < row; ++i) {
    randInts[i] = new int[col];
}

It's a C way, not a C++ way. In C++ you should use containers (e.g. a vector as suggested in the comments).

A: I have shown this using vector <vector<T> >, but you can also use Eigen matrix or boost::ublas::matrix
std::vector< std::vector<int> > foo(const int rows, const int cols)
{
    std::vector< std::vector<int> > v;
    v.resize(rows);
    for (auto i = 0; i < rows; i++)
    {
        v[i].resize(cols);
    }
    return v;
}

This will return a matrix of desired dimensions, and all elements will be initialized to 0.
The std::move part invokes a behavior called move semantics. You can read up on it. It's basically a better way of returning objects across a function boundary.
For compatibility's sake, if you must have a native array, then do this :
std::vector< vector<int> > temp = foo(rows, cols);
int *v = &temp[0][0];

Now, you can implement the gymnastics to get the correct index inside of int *v. You don't need multiple indirection.
P.S. This implementation assumes that your compiler supports C++ 11.

redpajama_set_name: RedPajamaStackExchange