text:
\section{Introduction}

Let us consider the following integrals of the form:
\begin{equation}\label{eqn:1.1}
Z_f(\varphi)(s)=\int_{\mathbb{R}^2}|f(x,y)|^s \varphi(x,y)dxdy
\quad\quad \mbox{for $s\in {\mathbb C}$,}
\end{equation}
where
$f,\varphi$ are real-valued ($C^{\infty}$) smooth functions
defined on an open neighborhood $U$ of the origin
in $\mathbb{R}^2$
and the support of $\varphi$ is contained in $U$.
Since the integrals
$Z_f(\varphi)(s)$
converge locally uniformly
on the region: ${\rm Re}(s)>0$,
they become holomorphic functions there.
It has been known in many cases that
they can be holomorphically continued to wider regions.
After this process, these integrals become holomorphic functions
on domains containing the region: ${\rm Re}(s)>0$, which
are sometimes called {\it local zeta functions}.
In this paper, we are interested in the case when
$f$ satisfies the condition:
\begin{equation}\label{eqn:1.2}
f(0,0)=0, \quad \nabla f(0,0)=(0,0)
\end{equation}
and hereafter we always assume this condition.
(Our issues in this paper are easy
unless (\ref{eqn:1.2}) is satisfied.)

In order to see the region where $Z_f(\varphi)$ can be holomorphically
continued, the following index plays important roles:
\begin{equation}\label{eqn:1.3}
c_0(f):=\sup
\left\{\mu>0:
\begin{array}{l}
\mbox{there exists an open neighborhood $V$ of} \\
\mbox{the origin in $U$ such that $|f|^{-\mu}\in L^1(V)$}
\end{array}
\right\}.
\end{equation}
This  $c_0(f)$ is called the {\it critical integrability index} of $f$
(it is also called {\it log canonical threshold} or
{\it singularity exponent}).
The determination of the value of $c_0(f)$
is an important issue in the singularity theory and
there have been many interesting works from many points of view
(this problem will be discussed soon later).
In order to see a clear relationship between the index
$c_0(f)$ and the region where $Z_f(\varphi)(s)$ becomes holomorphic,
we assume in the Introduction
that $\varphi$ satisfies the condition:
\begin{equation}\label{eqn:1.4}
\varphi(0,0)>0, \quad \varphi(x,y)\geq 0 \quad \mbox{ on $U$}.
\end{equation}
Indeed, under this condition (\ref{eqn:1.4}),
the relationship between the convergence of
the integrals (\ref{eqn:1.1}) and their holomorphy
implies the equality:
\begin{equation}\label{eqn:1.5}
c_0(f)=\sup
\left\{\rho>0:
\begin{array}{l}
\mbox{The domain where $Z_f(\varphi)$ can be holomorphically} \\
\mbox{continued contains the region: ${\rm Re}(s)>-\rho$}
\end{array}
\right\}
\end{equation}
Without the condition (\ref{eqn:1.4}),
the right handside of (\ref{eqn:1.5})
may be greater than $c_0(f)$
(see \cite{KaN16tams}, etc.).

The above equality (\ref{eqn:1.5}) means that
$Z_f(\varphi)$ is holomorphic
on the region: ${\rm Re}(s)>-c_0(f)$ and, moreover, that
$Z_f(\varphi)$ has some singularities on the vertical line:
${\rm Re}(s)=-c_0(f)$.
More exactly,
the following lemma implies that
$Z_f(\varphi)$ must have a singularity at $s=-c_0(f)$.
\begin{lemma}\label{lemma:1.1}
$Z_f(\varphi)$ cannot be holomorphically continued
to any open neighborhood of $s=-c_0(f)$.
\end{lemma}
The proof of the above lemma will be given in Section~6.
The purpose of this paper is to consider the following question:
\begin{question}
What kind of singularity does $Z_f(\varphi)$ have at $s=-c_0(f)$?
\end{question}

Under certain assumptions of $f$,
the integrals $Z_f(\varphi)(s)$
have meromorphic continuation to the whole complex plane and,
in particular, they have a pole at $s=-c_0(f)$.
A brief history of the studies done on
this phenomena
is as follows.
For a meanwile,
the general dimensional cases are treated.
In 1954 in an invited talk at ICM Amsterdam,
I. M. Gel'fand conjectured that
if $f$ is a polynomial and the support of $\varphi$ is
sufficiently small,
then
the integrals $Z_f(\varphi)(s)$
can be analytically continued as meromorphic functions
to the whole complex plane.
The case when $f$ are monomials is investigated in \cite{GeS64}.
Gel'fand's conjecture was affirmatively solved as
a stronger form by
Bernstein, S. I. Gel'fand \cite{BeG69}
and  Atiyah \cite{Ati70} independently.
They showed that
if $f$ is real analytic
and
the support of $\varphi$ is sufficiently small,
then
$Z_f(\varphi)$ can be analytically
continued as a meromorphic function
to the whole complex plane and
their poles belong to the union of finite number of arithmetic progressions
which consist of negative rational numbers.
Their proofs use Hironaka's resolution of singularities
\cite{Hir64}.
The authors \cite{KaN16jmsut}
generalize these results
when $f$ belongs to some class of smooth functions.

More precisely, let us consider an issue about
the determination of the value of $c_0(f)$.
We are also interested in more detailed local behavior of
$Z_f(\varphi)$ near $s=-c_0(f)$.
In the real analytic case,
this issue is to decide the location and order of
the leading pole for $Z_f(\varphi)$.
In the seminal work of Varchenko \cite{Var76},
when $f$ is real analytic and satisfies some conditions,
$c_0(f)$ can be expressed by using
the {\it Newton polyhedron} of $f$ as
\begin{equation}\label{eqn:1.6}
c_0(f)=1/d(f),
\end{equation}
where $d(f)$ is the {\it Newton distance} of $f$
(see \cite{Var76}, \cite{AGV88}) and the order of pole
at $s=-1/d(f)$ depends on some topological
information of the Newton polyhedron of $f$.
He uses the theory of toric varieties
based on the geometry of Newton polyhedra.
More detailed situation of meromorphic continuation
of $Z_f(\varphi)$ is investigated in
\cite{DS89}, \cite{DS92}, \cite{DNS05}, \cite{OkT13}, etc.
A recent interesting work \cite{CGP13} treating the equality
(\ref{eqn:1.6}) is from another approach.
In the same paper \cite{Var76},
Varchenko more deeply investigated the two-dimensional case.
Indeed, without any assumption, he shows that
the equality (\ref{eqn:1.6}) holds for real analytic $f$
on {\it adapted coordinates}.
Here adapted coordinates are important coordinates
in the study of oscillatory integrals
and their existence is shown in two-dimensions
in \cite{Var76}, \cite{PSS99}, \cite{IkM11tams}, etc.
More generally, let us consider the smooth case.
In the above cited paper \cite{KaN16jmsut},
the authors show that Varchenko's result can be naturally
generalized in a certain restricted class of smooth functions.
On the other hand,
M. Greenblatt \cite{Gre06} obtains a sharp result
which generalizes the Varchenko's two-dimensional result.
\begin{theorem}[Greenblatt \cite{Gre06}]
When $f$ is a smooth function defined on $U$ in ${\mathbb R}^2$,
the equality (\ref{eqn:1.6}) holds on adapted coordinates.
\end{theorem}
In more detail, in his same paper \cite{Gre06},
Greenblatt explains the delicate situation
about the local integrability of $|f|^{-\mu}$
around $\mu=c_0(f)$ in the smooth case
by using the specific function:
\begin{equation}\label{eqn:1.7}
f(x,y)=x^a y^b+x^a y^{b-2} e^{-1/|x|^{1/(2b)}},
\end{equation}
where $a,b$ are nonnegative integers satisfying
$a<b$ and $2\leq b$
(see Remark~3.2 in this paper for his result).
Note that the second term in (\ref{eqn:1.7}) is a
(non-real analytic) flat function.
The purpose of this paper is to investigate
a slight generalization of the above example more deeply
and to understand detailed situation
of analytic continuation of the
respective local zeta functions.

This paper is organized as follows.
In Section~2,
we state a main theorem showing the failure
of meromorphy of some local zeta functions.
In order to show the main theorem,
we substantially investigate similar integrals $Z(\sigma)$
in Section~3.
The main theorem is shown in Section~4
by using the results in Section~3.
In Section~5, we give some property of domains of
convergence of the integrals $Z_f(\varphi)(s)$,
which is analogous to Landau's theorem on the Dirichlet series
with positive coefficients.
This result implies Lemma~1.1.
Our computations in this paper are very specific
and it is hoped to give good observation for future studies about
properties of local zeta functions in the general smooth case.
From our results,
many elementary (but probably not so easy) questions
are naturally raised,
some of which are listed in Section~6.

In this paper, we use
$C, C_1, C_2, \epsilon, \delta$ for
various kinds of constants without further comments.

\section{Main results}

In this section, we consider
the integrals $Z_f(\varphi)(s)$
with
smooth functions $f$ of the following form:
\begin{equation}\label{eqn:2.1}
f(x,y)=x^a y^b+x^a y^{b-q} e^{-1/|x|^p},
\end{equation}
where
$a,b,p,q$ satisfy
\begin{itemize}
\item $a,b,q$ are nonnegative integers satisfying
$a<b$, $2\leq b$, $1\leq q\leq b$;
\item $p$ is a positive real number.
\end{itemize}

We remark that $e^{-1/|x|^p}$ is regarded as
a smooth function defined on ${\mathbb R}$
by considering that its value at $0$ takes $0$.
As mentioned in the Introduction,
the above function $f$ slightly
generalizes the function (\ref{eqn:1.7}) which
is investigated by Greenblatt \cite{Gre06}
(he considers the case when $q=2$ and $p=1/(2b)$).
Note that the coordinate $(x,y)$ satisfies
the adapted coordinates conditions in \cite{IkM11tams}
and that the Newton distance
of $f$ is $b$ (i.e., $d(f)=b$).

Let us consider the case when
the second term, which is flat,
does not appear in (\ref{eqn:2.1})
(i.e., $f(x,y)=x^a y^b$).
From \cite{GeS64}, it is easy to see that
$Z_f(\varphi)$ can be regarded as a meromorphic function
on ${\mathbb C}$ and the poles of $Z_f(\varphi)$ are contained
in the set $\{-j/a,-k/b:j,k\in{\mathbb N}\}$.
When $\varphi$ satisfies (\ref{eqn:1.4}),
the leading pole exists
at $s=-1/b$, whose order is one.

Now, when $f$ is as in (\ref{eqn:2.1}),
it follows from
Theorem~1.2 and Lemma~1.1 in the Introduction
that $Z_f(\varphi)$ is holomorphic on
the region: ${\rm Re}(s)>-1/b$ and
that $Z_f(\varphi)$
has a singularity at $s=-1/b$.
(Note that $c_0(f)=1/b$.)
More precisely,
we see the behavior at $-1/b$ of the restriction of
$Z_f(\varphi)$ to the real axis as follows.
In this paper, we use the symbol $s=\sigma+i t$
with $\sigma, t\in{\mathbb R}$ which is
traditionally used in the analysis of the Riemann
zeta function.
\begin{theorem}\label{thm:1.1}
Let $f$ be as in \eqref{eqn:2.1}
and $\varphi$ as in \eqref{eqn:1.1}.
We assume that $q$ is even.
Then
the following hold:
\begin{enumerate}
\item
If $p>1-a/b$,
then
\begin{equation}\label{eqn:2.2}
\lim_{\sigma\to-1/b+0}
(b\sigma+1)^{1-\frac{1-a/b}{p}}\cdot Z_f(\varphi)(\sigma)
=4A\cdot\varphi(0,0)
\end{equation}
where $A$ is the positive constant defined by
\begin{equation}\label{eqn:A}
A=\int_0^{\infty}x^{-a/b}(1-e^{-1/(qx^p)})dx.
\end{equation}
Note that the above improper integral converges.
\item
If $p=1-a/b$,
then
\begin{equation}\label{eqn:2.3}
\lim_{\sigma\to-1/b+0}
|\log (b\sigma+1)|^{-1}\cdot Z_f(\varphi)(\sigma)
=\frac{4}{pq}\cdot\varphi(0,0).
\end{equation}
\item
If $0<p< 1-a/b$, then
there exists a constant $B(\varphi)$,
which depends on $a,b,p,q,\varphi$ but is independent
of $\sigma$, such that
\begin{equation}\label{eqn:2.4}
\lim_{\sigma\to-1/b+0}Z_f(\varphi)(\sigma)=B(\varphi),
\end{equation}
where $B(\varphi)$ is positive
if $\varphi$ satisfies the condition (\ref{eqn:1.4}).
\end{enumerate}
\end{theorem}
Of course, if $Z_f(\varphi)$ had a pole of order $m$ at $s=-1/b$,
then
$
\lim_{\sigma\to-1/b+0}
(b\sigma+1)^{m} \cdot Z_f(\varphi)(\sigma)
$ must be a positive value.
Noticing that $0<1-\frac{1-a/b}{p}<1$,
we can see the following from the above theorem
with Lemma~1.1 in the Introduction.
\begin{corollary
Under the assumption in Theorem~2.1
with the condition (\ref{eqn:1.4}) on $\varphi$,
$Z_f(\varphi)$ cannot be meromorphically
continued to any open neighborhood of $s=-1/b$ in ${\mathbb C}$.
In other words,
the singularity of $Z_f(\varphi)(s)$ at $s=-1/b$
is different from a pole.
\end{corollary}

If ${\rm Re}(s)>-c_0(f)$, then
$|f|^s$ can be regarded as a {\it distribution}
by considering the map from
$C_0^{\infty}(U)$ to ${\mathbb C}$ defined by
\begin{equation*}
\varphi
\longmapsto
\left\langle
|f|^s,\varphi
\right\rangle=
\int_{{\mathbb R}^2}|f|^s\varphi dxdy=Z_f(\varphi)(s).
\end{equation*}
Furthermore, the equalities (\ref{eqn:2.2}), (\ref{eqn:2.3})
in (i), (ii) can be interpreted as
in the following.
\begin{equation}\label{eqn:2.5}
\begin{split}
&\lim_{\sigma\to-1/b+0}
(b\sigma+1)^{1-\frac{1-a/b}{p}}|f|^{\sigma}=4A\delta,\\
&\lim_{\sigma\to-1/b+0}
|\log(b\sigma+1)|^{-1}|f|^{\sigma}=\frac{1}{pq}\delta,\\
\end{split}
\end{equation}
where $\delta\in{\mathcal D}'(U)$ is Dirac's delta function.
The limits in the left-hand sides of (\ref{eqn:2.5}) are
taken in the topology of ${\mathcal D}'(U)$.
On the other hand,
the map $B$ from $C_0^{\infty}(U)$ to ${\mathbb R}$,
defined by
$\varphi\mapsto B(\varphi)
(=\left\langle B,\varphi \right\rangle)$ from (iii),
can also be interpreted as a distribution:
\begin{equation*
\lim_{\sigma\to-1/b+0} |f|^{\sigma}=B.
\end{equation*}
The continuity of the above map $B$ will be explained
in Remark~4.1.

\section{Asymptotic limits of associated integrals}

For a set $U$ in ${\mathbb R}^2$,
let us define the integral of the form:
\begin{equation}\label{eqn:3.1}
\begin{split}
Z_U(\sigma):=\int_{U}
\left|x^ay^b+x^ay^{b-q}e^{-1/|x|^p}\right|^{\sigma}
dxdy
\quad\quad \mbox{ for $\sigma<0$,}
\end{split}
\end{equation}
where $a,b,p,q$ satisfy the conditions in the previous section.

In this section, we consider the case when
$U=U_+(r_1,r_2)$ with $r_1,r_2\in(0,1)$,
where
$$
U_+(r_1,r_2)=
\{(x,y)\in{\mathbb R}^2:0\leq x\leq r_1, 0\leq y\leq r_2\}
$$
and simply denote
\begin{equation}\label{eqn:3.2}
Z(\sigma):=Z_{U_+(r_1,r_2)}(\sigma).
\end{equation}
Let $e(x)$ be the smooth function defined by
\begin{equation}\label{eqn:3.3}
e(x):=\exp\left(\frac{-1}{qx^p}\right)
\quad\quad \mbox{ for $x>0$}
\end{equation}
and $e(0)=0$,
which frequently appears in the computation below.
Since the function $e$ is monotonously increasing,
we can define the function $\rho:[0,\infty)\to[0,r_1]$ by
\begin{equation}\label{eqn:rho}
\rho(y)=
\begin{cases}
e^{-1}(y)=\left(
\frac{-1}{q\log y}
\right)^{1/p} \quad\quad \mbox{ if $0\leq y <e(r_1)$,} \\
r_1 \quad\quad\quad\quad \mbox{ if $y\geq e(r_1)$.}
\end{cases}
\end{equation}

Since the Newton distance of $f$ equals $b$,
Theorem~1.2 due to Greenblatt implies that
the integral $Z(\sigma)$ converges if $\sigma>-1/b$
and diverges if $\sigma<-1/b$.
The purpose of this section is
to compute exact asymptotic limits
of $Z(\sigma)$ as $\sigma\to-1/b+0$.

\begin{theorem}\label{thm:5.1}
The integral $Z(\sigma)$ satisfies the following.
\begin{enumerate}
\item
If $p>1-a/b$,
then
\begin{equation*
\lim_{\sigma\to-1/b+0}
(b\sigma+1)^{1-\frac{1-a/b}{p}}\cdot Z(\sigma)
=A,
\end{equation*}
where $A$ is as in (\ref{eqn:A}).
\item
If $p=1-a/b$,
then
\begin{equation*
\lim_{\sigma\to-1/b+0}
|\log (b\sigma+1)|^{-1}\cdot Z(\sigma)
=\frac{1}{pq}.
\end{equation*}
\item
If $0<p< 1-a/b$, then the limit of
$Z(\sigma)$ as
$\sigma\to-1/b+0$ exists
and it satisfies
\begin{equation*}
\begin{split}
&\max\left\{
\frac{L(\lambda)}{(1+\lambda^q)^{1/b}}+
\frac{M(\lambda)}{(1+\lambda^{-q})^{1/b}}
:\lambda>0
\right\} \\
&\quad\quad\leq
\lim_{\sigma\to-1/b+0}Z(\sigma) \leq
\min\{L(\lambda)+M(\lambda):\lambda>0\},
\end{split}
\end{equation*}
where $L(\lambda)$, $M(\lambda)$ are
positive constants depending on $\lambda$ as in
(\ref{eqn:3.12}), (\ref{eqn:3.38}) in the proof below.
\end{enumerate}
\end{theorem}

\begin{remark}
In \cite{Gre06},
Greenblatt shows the
boundedness of $Z(\sigma)$
near $\sigma=-1/b$
in the case of $p=1/(2b)$ and $q=2$,
which is contained in the above case (iii).
\end{remark}

\subsection{Auxiliary function with a parameter}

Let $\lambda$ be a positive number.
The set $U_+(r_1,r_2)$ is decomposed as
$U_1(\lambda)\cup U_2(\lambda)$ with
\begin{equation*}
\begin{split}
&U_1(\lambda)=\{(x,y)\in U_+(r_1,r_2):
\lambda y\geq e(x)\},\\
&U_2(\lambda)=\{(x,y)\in U_+(r_1,r_2):
\lambda y < e(x))\}.
\end{split}
\end{equation*}
The integral $Z(\sigma)$ is expressed as
\begin{equation}\label{eqn:3.5}
Z(\sigma)=
Z_1^{(\lambda)}(\sigma)
+
Z_2^{(\lambda)}(\sigma),
\end{equation}
where
\begin{equation}\label{eqn:3.6}
\begin{split}
Z_j^{(\lambda)}(\sigma)
=\int_{U_j(\lambda)}
(x^a y^b +x^a y^{b-q} e^{-1/x^p})^{\sigma}dxdy
\quad\quad \mbox{for $j=1,2$.}
\end{split}
\end{equation}
Note that
\begin{equation}\label{eqn:3.7}
\begin{split}
&x^a y^b(1 + y^{-q} e^{-1/x^p})\leq
(1+\lambda^q) x^a y^b \quad \mbox{for $(x,y)\in U_1(\lambda)$,} \\
&x^a y^{b-q} e^{-1/x^p}(y^q e^{1/x^p}+1)\leq
(1+\lambda^{-q}) x^a y^{b-q} e^{-1/x^p}
\quad \mbox{for $(x,y)\in U_2(\lambda)$.}
\end{split}
\end{equation}
Thus each $Z_j^{(\lambda)}(\sigma)$ can be estimated
by using the following integrals.
\begin{eqnarray}
&&\tilde{Z}_1^{(\lambda)}(\sigma)=
\int_{U_1(\lambda)}x^{a\sigma}y^{b\sigma}dxdy,
\label{eqn:3.8}\\
&&\tilde{Z}_2^{(\lambda)}(\sigma)=
\int_{U_2(\lambda)}
x^{a\sigma}y^{(b-q)\sigma}e^{-\sigma/x^p}dxdy.
\label{eqn:3.9}
\end{eqnarray}
Indeed,
${Z}_1^{(\lambda)}(\sigma)$ and
${Z}_2^{(\lambda)}(\sigma)$ are convergent
if and only if
so are
$\tilde{Z}_1^{(\lambda)}(\sigma)$ and
$\tilde{Z}_2^{(\lambda)}(\sigma)$.
Moreover, they satisfy
\begin{equation}\label{eqn:3.10}
\begin{split}
&(1+\lambda^q)^{\sigma}
\tilde{Z}_1^{(\lambda)}(\sigma)<
Z_1^{(\lambda)}(\sigma)<
\tilde{Z}_1^{(\lambda)}(\sigma),\\
&(1+\lambda^{-q})^{\sigma}
\tilde{Z}_2^{(\lambda)}(\sigma)<
Z_2^{(\lambda)}(\sigma)
< \tilde{Z}_2^{(\lambda)}(\sigma),
\end{split}
\end{equation}
for $\sigma<0$.
In order to prove Theorem~3.1,
let us investigate the behaviors of
$\tilde{Z}_1^{(\lambda)}(\sigma)$ and
$\tilde{Z}_2^{(\lambda)}(\sigma)$
as $\sigma\to-1/b+0$.

\subsection{Preliminary lemma}

For $\alpha>0$,
let $\psi_{\alpha}$ be the smooth function
defined by
\begin{equation}\label{eqn:3.11}
\psi_{\alpha}(x)
:=\frac{1-\alpha^x}{x} \quad \mbox{ for $x>0$.}
\end{equation}
The following properties of $\psi_{\alpha}$ play important roles
in the computation below.

\begin{lemma}
The function $\psi_{\alpha}$ satisfies the following
properties.
\begin{enumerate}
\item There exist positive constants $C_1, C_2$ such that
\begin{equation*}
-\log\alpha-C_1x<\psi_{\alpha}(x)
<-\log\alpha -C_2x
\end{equation*}
for $x\in(0,1)$.
In particular, $\lim_{x\to +0}\psi_{\alpha}(x)=-\log\alpha$.
\item If $\alpha\in(0,1)$,
then $\lim_{x\to \infty}\psi_{\alpha}(x)=0$.
\item If $\alpha\in(0,1)$,
then $\psi_{\alpha}$ is monotonously decreasing,
in particular,
$0<\psi_{\alpha}(x)<-\log \alpha$ for $x>0$.
\end{enumerate}
\end{lemma}

\begin{proof}
The above properties of $\psi_{\alpha}$ can be
easily seen by using Taylor's formula.
\end{proof}

\begin{remark}
In the computation below,
the function $1-e(x)$ often appears.
This function can be expressed by using $\psi_{\alpha}$
with $\alpha=\exp(-q^{-1})$ as follows.
\begin{equation*}
\begin{split}
1-e(x)=
\frac{1-\exp(-q^{-1}x^{-p})}{x^{-p}}\cdot x^{-p}
=\psi_{\alpha}(x^{-p})x^{-p}.
\end{split}
\end{equation*}
From Lemma~3.3, we can see that
\begin{equation}\label{eqn:e2}
q^{-1}x^{-p}-C_1x^{-2p}\leq
1-e(x)\leq
q^{-1}x^{-p}-C_2x^{-2p} \mbox{ for $x\geq 1$,}
\end{equation}
where $C_1,C_2$ are positive constants.
\end{remark}

\subsection{Asymptotics of $\tilde{Z}_1^{(\lambda)}(\sigma)$}

Let us investigate
the behavior of $\tilde{Z}_1^{(\lambda)}(\sigma)$
which is essentially important.

\begin{lemma}
\begin{enumerate}
\item
If $p>1-a/b$,
then
\begin{equation*
\lim_{\sigma\to-1/b+0}(b\sigma+1)^{1-\frac{1-a/b}{p}}\cdot
\tilde{Z}_1^{(\lambda)}(\sigma)=A,
\end{equation*}
where $A$ is as in (\ref{eqn:A}).
\item
If $p=1-a/b$,
then
\begin{equation*
\lim_{\sigma\to-1/b+0}
|\log (b\sigma+1)|^{-1}\cdot \tilde{Z}_1^{(\lambda)}(\sigma)
=\frac{1}{pq}.
\end{equation*}
\item
If $0<p<1-a/b$, then
$\lim_{\sigma\to-1/b+0}
\tilde{Z}_1^{(\lambda)}(\sigma)=L(\lambda)$ with
\begin{equation}\label{eqn:3.12}
L(\lambda)=
\frac{\rho(\lambda r_2)^{1-a/b}}{1-a/b}
\cdot\log(\lambda r_2)
+
\frac{\rho(\lambda r_2)^{1-a/b-p}}{q(1-a/b-p)}.
\end{equation}
\end{enumerate}
\end{lemma}

\begin{proof}
In the proof, we introduce the variable:
\begin{equation*
X=b\sigma+1,
\end{equation*}
which is convenient for many kinds of limit processes later.
Note that
$$\sigma\to -1/b+0 \Longleftrightarrow X\to +0.$$

Now, applying an iterated integral to (\ref{eqn:3.8}),
we have
\begin{equation}\label{eqn:3.13}
\tilde{Z}_1^{(\lambda)}\left(\sigma\right)
=\frac{1}{\lambda^X}
\frac{1}{X}\int_0^{\rho(\tilde{r}_2)}
x^{a\sigma} (\tilde{r}_2^X-e(x)^X)dx,
\end{equation}
where $\tilde{r}_2:=\lambda r_2$.

\vspace{.5 em}

{\bf [The case (i) : $p>1-a/b$.]} \quad

\vspace{.2 em}

Changing the integral variable in (\ref{eqn:3.13}) by
\begin{equation*
x=X^{1/p} u
\Longleftrightarrow
u=X^{-1/p}x, \quad (dx=X^{1/p}du),
\end{equation*}
we have
\begin{equation}\label{eqn:3.14}
\tilde{Z}_1^{(\lambda)}(\sigma)=
\frac{1}{\lambda^X}
\frac{1}{X^{1-\frac{1+a\sigma}{p}}}
\int_0^{\rho(\tilde{r}_2)/X^{1/p}} u^{a\sigma}(\tilde{r}_2^X-e(u))du.
\end{equation}
We decompose the integral in (\ref{eqn:3.14})
as $G_1(\sigma)+G_2(\sigma)+G_3(\sigma)$
with
\begin{eqnarray}
&&G_1(\sigma)=
\int_0^{1} u^{a\sigma}(\tilde{r}_2^X-e(u))du,
\nonumber\\
&&G_2(\sigma)=
\int_1^{\rho(\tilde{r}_2)/X^{1/p}}
u^{a\sigma}(1-e(u))du,
\label{eqn:3.15} \\
&&G_3(\sigma)=
(\tilde{r}_2^X-1)
\int_1^{\rho(\tilde{r}_2)/X^{1/p}}
u^{a\sigma}du.
\nonumber
\end{eqnarray}

\vspace{.2 em}
\underline{The limit of $G_1(\sigma)$.}

\vspace{.2 em}

Since $\tilde{r}_2^X\leq \max\{1,\lambda\}$,
the integrand can be estimated as
\begin{equation*}
u^{a\sigma}(\tilde{r}_2^X-e(u))< u^{-a/b} \max\{1,\lambda\}
\end{equation*}
for $(X,u)\in (0,1]\times(0,1]$.
Since $-a/b>-1$,
the Lebesgue convergence theorem implies
\begin{equation}\label{eqn:3.16}
\lim_{\sigma\to-1/b+0} G_1(\sigma)
=\int_0^{1}u^{-a/b}(1-e(u))du.
\end{equation}

\vspace{.2 em}
\underline{The limit of $G_2(\sigma)$.}

\vspace{.2 em}

Let $\epsilon_0:=
a/b+p-1>0$, then
\begin{equation}\label{eqn:3.20}
a\sigma-p+1=\frac{a}{b}(b\sigma+1)-\frac{a}{b}-p+1
=\frac{a}{b}X-\epsilon_0.
\end{equation}
Since $u<\rho(\tilde{r}_2)/X^{1/p}$
($\Leftrightarrow X<\rho(\tilde{r}_2)^p u^{-p}$)
and $e^{x}\leq 1+(e-1)x$ for $x\in(0,1)$,
there exists $\delta>0$ such that
\begin{equation}\label{eqn:3.23}
\begin{split}
1<u^{\frac{a}{b}X}=
\exp\left(\frac{a}{b}X\cdot \log u\right)
< 1+ D u^{-p}\log u,
\end{split}
\end{equation}
for
$(u,X)\in [1,\rho(\tilde{r}_2)/X^{1/p})\times(0,\delta)$,
where $D:=(e-1)\frac{a}{b}\rho(\tilde{r}_2)^p$.
From (\ref{eqn:3.20}), (\ref{eqn:e2}), (\ref{eqn:3.23}),
we have
\begin{equation*}
\begin{split}
u^{a\sigma}(1-e(u))
&< q^{-1}u^{a\sigma-p}
= q^{-1}u^{\frac{a}{b}X-1-\epsilon_0} \\
&
< q^{-1}(1+Du^{-p}\log u)u^{-1-\epsilon_0}
\leq Cu^{-1-\epsilon_0}
\end{split}
\end{equation*}
for $(u,X)\in [1,\rho(\tilde{r}_2)/X^{1/p})\times(0,\delta)$,
where $C>0$ is a constant independent of $u$ and $\sigma$.
Thus, the Lebesgue convergence theorem implies
\begin{equation}\label{eqn:3.17}
\begin{split}
\lim_{\sigma\to-1/b+0}
G_2(\sigma)
=\int_1^{\infty}u^{-a/b}(1-e(u))du.
\end{split}
\end{equation}

\vspace{.2 em}
\underline{The limit of $G_3(\sigma)$.}

\vspace{.2 em}

Since $1-\tilde{r}_2^X=X\psi_{\tilde{r}_2}(X)$,
$G_3(\sigma)$ can be computed as follows.
\begin{equation*}
\begin{split}
G_3(\sigma)
=X^{1-\frac{a\sigma+1}{p}}
\psi_{\tilde{r}_2}(X)\cdot
\frac{\rho(\tilde{r}_2)^{a\sigma+1}-X^{(a\sigma+1)/p}}{a\sigma+1}.
\end{split}
\end{equation*}
From Lemma~3.3 (i),
we can see the following.
\begin{equation}\label{eqn:3.18}
\begin{split}
\lim_{\sigma\to -1/b} X^{-1+\frac{a\sigma+1}{p}}
G_3(\sigma)=
(-\log \tilde{r}_2) \cdot
\frac{\rho(\tilde{r}_2)^{1-a/b}}{1-a/b}.
\end{split}
\end{equation}

\vspace{.2 em}

Therefore, (\ref{eqn:3.16}), (\ref{eqn:3.17}), (\ref{eqn:3.18})
imply
\begin{equation*}
\begin{split}
\lim_{\sigma\to-1/b+0}
X^{1-\frac{1-a/b}{p}}
\tilde{Z}_1^{(\lambda)}(\sigma)
=
\lim_{\sigma\to-1/b+0} G_1(\sigma)+
\lim_{\sigma\to-1/b+0}G_2(\sigma)=A.
\end{split}
\end{equation*}
Here we used the fact:
$\lim_{X\to +0} X^{cX}=1$ ($c\in{\mathbb R}$).

\vspace{.5 em}
{\bf [The case (ii) : $p=1-a/b$.]} \quad

Since
the equalities (\ref{eqn:3.16}) and (\ref{eqn:3.18})
always hold for any $p>0$,
it suffices to consider the behavior of $G_2(\sigma)$
in the case of $p=1-a/b$.

\vspace{.2 em}

\underline{The limit of $G_2(\sigma)$.}

\vspace{.2 em}

Since $\epsilon_0=0$ in (\ref{eqn:3.20}),
$\a\sigma=p-1+\frac{a}{b}X$ holds.
Thus, the estimates
(\ref{eqn:e2}), (\ref{eqn:3.23}) imply that
there exist $C_1, C_2,\delta>0$ such that
\begin{equation}\label{eqn:e3}
q^{-1}u^{-1}-C_1 u^{-p-1}<
u^{a\sigma}(1-e(u))<
q^{-1}u^{-1}+C_2 u^{-p-1}\log u,
\end{equation}
for $(u,X)\in
[1,\rho(\tilde{r}_2)/X^{1/p})\times(0,\delta)$.

Now, we rewrite (\ref{eqn:3.15}) as follows.
\begin{equation*
\begin{split}
G_2(\sigma)&=
q^{-1}\int_1^{\rho(\tilde{r}_2)/X^{1/p}} \frac{1}{u}du-
\int_1^{\rho(\tilde{r}_2)/X^{1/p}}
\left(\frac{q^{-1}}{u}-u^{a\sigma}(1-e(u))\right)du \\
&=:H_1(\sigma)-H_2(\sigma).
\end{split}
\end{equation*}
A direct computation gives
\begin{equation}\label{eqn:3.24}
H_1(\sigma)=q^{-1}(\log\rho(\tilde{r}_2)-p^{-1}\log X).
\end{equation}
From (\ref{eqn:e3}),
there exist $\epsilon, C>0$
such that
\begin{equation}\label{eqn:3.25}
|H_2(\sigma)|
\leq C \int_1^{\rho(\tilde{r}_2)/X^{1/p}}
u^{-p+\epsilon-1}du
\leq \frac{C}{p-\epsilon}.
\end{equation}
From
(\ref{eqn:3.16}), (\ref{eqn:3.18}),
(\ref{eqn:3.24}), (\ref{eqn:3.25}),
we have
\begin{equation*}
\begin{split}
&\lim_{\sigma\to -1/b}
|\log X|^{-1} \tilde{Z}_1^{(\lambda)}(\sigma)
=
\lim_{\sigma\to -1/b}
|\log X|^{-1} G_2(\sigma)\\
&\quad\quad
=
\lim_{\sigma\to -1/b}
|\log X|^{-1} H_1(\sigma)=\frac{1}{pq}.
\end{split}
\end{equation*}

 \vspace{.5 em}
{\bf [The case (iii) : $0<p<1-a/b$.]} \quad

\vspace{.2 em}

From (\ref{eqn:3.13}),
$\tilde{Z}_1^{(\lambda)}(\sigma)$
can be decomposed as
$J_1(\sigma)+J_2(\sigma)$ with

\begin{equation*}
\begin{split}
&J_1(\sigma)=
\frac{1}{\lambda^X}
\int_0^{\rho(\tilde{r}_2)}
x^{a\sigma} \frac{1-e(x)^X}{X}dx, \\
&J_2(\sigma)=
\frac{1}{\lambda^X}\cdot
\frac{\tilde{r}_2^X-1}{X}
\int_0^{\rho(\tilde{r}_2)}
x^{a\sigma}dx.
\end{split}
\end{equation*}

\vspace{.2 em}

\underline{The limit of $J_1(\sigma)$.}

\vspace{.2 em}

Using the function $\psi_{\alpha}$ with
$\alpha:=\exp(-q^{-1})$ and Lemma~3.3 (i),
we have
\begin{equation}\label{eqn:3.26}
\begin{split}
\frac{1-e(x)^{X}}{X}
=\frac{\psi_{\alpha}(X x^{-p})}{x^p} \longrightarrow
q^{-1}x^{-p}
\quad \mbox{ as $\sigma\to-1/b+0$.}
\end{split}
\end{equation}
On the other hand,
Lemma~3.3 (iii) implies
\begin{equation*}
\left|
x^{a\sigma}\cdot\frac{1-e(x)^{X}}{X}
\right|
\leq C x^{-a/b-p}
\end{equation*}
for $(x,X)\in(0,\rho(\tilde{r}_2)]\times(0,1)$,
where $C>0$ is a constant depending only
on $\tilde{r}_2$.
Since $-a/b-p>-1$,
the Lebesgue convergence theorem
implies
\begin{equation}\label{eqn:3.27}
\begin{split}
\lim_{\sigma\to -1/b+0}
J_1(\sigma)=
\frac{1}{q}
\int_0^{\rho(\tilde{r}_2)} x^{-a/b-p}dx
=
\frac{\rho(\tilde{r}_2)^{1-a/b-p}}{q(1-a/b-p)}.
\end{split}
\end{equation}

\vspace{.2 em}

\underline{The limit of $J_2(\sigma)$.}

\vspace{.2 em}

A direct computation gives
$$
J_2(\sigma)=
-\frac{1}{\lambda^X}\cdot
\psi_{\tilde{r}_2}(X)\cdot
\frac{\rho(\tilde{r}_2)^{a\sigma+1}}{a\sigma+1}.
$$
Therefore we have
\begin{equation}\label{eqn:3.28}
\lim_{\sigma\to -1/b+0}
J_2(\sigma)=
\frac{\rho(\tilde{r}_2)^{1-a/b}}{1-a/b}
\cdot\log(\tilde{r}_2)
\end{equation}

From (\ref{eqn:3.27}), (\ref{eqn:3.28}),
we obtain the limit
of $\tilde{Z}_1^{(\lambda)}(\sigma)$ in (iii).

\end{proof}

\subsection{Asymptotics of $\tilde{Z}_2^{(\lambda)}(\sigma)$}

The behavior of $\tilde{Z}_2^{(\lambda)}(\sigma)$
can be easily seen by
a direct computation.

\begin{lemma}
$\lim_{\sigma\to -1/b}\tilde{Z}_2^{(\lambda)}(\sigma)=M(\lambda)$
with
\begin{equation}\label{eqn:3.38}
\begin{split}
M(\lambda)=
\frac{b^2}{q(b-a)}\cdot \frac{1}{\lambda^{q/b}}\cdot
\rho(\lambda r_2)^{1-a/b}
+\frac{b}{q}r_2^{q/b}\int_{\rho(\lambda r_2)}^{r_1}
x^{-a/b}\exp(1/(bx^p))dx.
\end{split}
\end{equation}
\end{lemma}

\begin{proof}
By decomposing the integral region
$U_2(\lambda)$ into the following two sets:
$$
\{(x,y)\in U_2(\lambda):x\leq \rho(\tilde{r}_2)\},\quad
\{(x,y)\in U_2(\lambda):x > \rho(\tilde{r}_2)\},
$$
the integral $\tilde{Z}_2^{(\lambda)}(\sigma)$ can be computed as
\begin{equation*}
\tilde{Z}_2^{(\lambda)}(\sigma)
=\frac{1}{X-q\sigma}\cdot\frac{1}{\lambda^{X-q\sigma}}
\int_0^{\rho(\tilde{r}_2)} x^{a\sigma} e(x)^{X}dx+
\frac{r_2^{X-q\sigma}}{X-q\sigma}
\int_{\rho(\tilde{r}_2)}^{r_1} x^{a\sigma}e^{-\sigma/x^p}dx.
\end{equation*}
It can be easily computed that
the limit of the right-hand side of the above equation as
$\sigma\to-1/b+0$ is $M(\lambda)$
in (\ref{eqn:3.38}).
\end{proof}

\subsection{Proof of Theorem~3.1}

From (\ref{eqn:3.5}), (\ref{eqn:3.10}),
when the integrals
$\tilde{Z}_1^{(\lambda)}(\sigma)$ and
$\tilde{Z}_2^{(\lambda)}(\sigma)$
converge, $Z(\sigma)$ can be estimated as
\begin{equation}\label{eqn:3.39}
\begin{split}
&(1+\lambda^q)^{\sigma}\cdot \tilde{Z}_1^{(\lambda)}(\sigma)+
(1+\lambda^{-q})^{\sigma}\cdot \tilde{Z}_2^{(\lambda)}(\sigma) \\
&\quad\quad\quad\quad\quad\quad\quad\quad <Z(\sigma)<
\tilde{Z}_1^{(\lambda)}(\sigma)+
\tilde{Z}_2^{(\lambda)}(\sigma).
\end{split}
\end{equation}

First, let us consider the case (i).
Since Lemmas~3.5 (i) and 3.7 imply
$$
\lim_{\sigma\to-1/b+0}X^{1-\frac{1-a/b}{p}}\cdot
\tilde{Z}_1^{(\lambda)}(\sigma)=A,
\quad
\lim_{\sigma\to-1/b+0}X^{1-\frac{1-a/b}{p}}\cdot
\tilde{Z}_2^{(\lambda)}(\sigma)=0,
$$
the estimates (\ref{eqn:3.39}) give
\begin{equation}\label{eqn:3.40}
\begin{split}
(1+\lambda^q)^{-1/b}\cdot A
&\leq
\varliminf_{\sigma\to-1/b+0}X^{1-\frac{1-a/b}{p}}\cdot Z(\sigma)\\
&\leq
\varlimsup_{\sigma\to-1/b+0}X^{1-\frac{1-a/b}{p}}\cdot Z(\sigma)
\leq A.
\end{split}
\end{equation}
Note that $Z(\sigma)$ is independent of $\lambda$.
Considering the limit  as $\lambda\to 0$ in (\ref{eqn:3.40}),
we obtain (i).
The case (ii) can be similarly shown.

Let us consider the case (iii).
Since the integral $Z(\sigma)$ is a monotone decreasing function
as $\sigma\in(-1/b,0)$, the boundedness
of $Z(\sigma)$ from (\ref{eqn:3.39}) implies
the existence of the limit $\lim_{\sigma\to-1/b}Z(\sigma)$.
Moreover, since $Z(\sigma)$ is independent of $\lambda$,
the following inequalities can be obtained
from (\ref{eqn:3.39}), (\ref{eqn:3.12}), (\ref{eqn:3.38}).
\begin{equation}\label{eqn:3.41}
\begin{split}
&\sup\left\{
\frac{L(\lambda)}{(1+\lambda^q)^{1/b}}+
\frac{M(\lambda)}{(1+\lambda^{-q})^{1/b}}
:\lambda>0
\right\} \\
&\quad\quad\leq
\lim_{\sigma\to-1/b+0}Z(\sigma) \leq
\inf\{L(\lambda)+M(\lambda):\lambda>0\}.
\end{split}
\end{equation}
Furthermore, the supremum and infimum in (\ref{eqn:3.41}) can be
respectively replaced by the maximum and minimum
by using the lemma below.
As a result, the inequalities in (iii)
in the theorem is obtained.

\begin{lemma}
\begin{equation*}
\begin{split}
&\lim_{\lambda\to 0}L(\lambda)
=0, \quad
\lim_{\lambda\to 0}M(\lambda)
=\infty, \quad
\lim_{\lambda\to \infty}L(\lambda)
=\infty, \quad
\lim_{\lambda\to \infty}M(\lambda)
=0, \\
&\lim_{\lambda\to 0}
(1+\lambda^{q})^{-1/b}\cdot L(\lambda)=0, \quad
\lim_{\lambda\to 0}
(1+\lambda^{-q})^{-1/b}\cdot M(\lambda)=0, \\
&\lim_{\lambda\to \infty}
(1+\lambda^{q})^{-1/b}\cdot L(\lambda)=0, \quad
\lim_{\lambda\to \infty}
(1+\lambda^{-q})^{-1/b}\cdot M(\lambda)=0.
\end{split}
\end{equation*}
\end{lemma}
The proof of the above lemma is easy,
so it is left to the readers.

\section{Proof of Theorem~2.1}

For $r_1,r_2\in(0,1)$, let
$$
U(r_1,r_2):=\{(x,y)\in{\mathbb R}^n:|x|<r_1,|y|<r_2\}.
$$
The behavior of $Z_f(\varphi)$ can be appropriately
approximated by that of a more simple function
$Z_{U(r_1,r_2)}(\sigma)$ (see (\ref{eqn:3.1})).
Furthermore,
under the assumption that $q$ is even,
$f(x,y)=f(|x|,|y|)$ for any $(x,y)\in{\mathbb R}^2$,
which implies
\begin{equation}\label{eqn:5.1}
Z_{U(r_1,r_2)}(\sigma)=4Z_{U_+(r_1,r_2)}(\sigma)=4Z(\sigma).
\end{equation}
Therefore, Theorem~2.1 can be proved by using
Theorem~3.1 as follows.

\vspace{.5 em}

{\bf The cases (i), (ii).}\quad
For any $\epsilon>0$,
there exist $r_1, r_2\in(0,1)$ such that
\begin{equation*}
\varphi(0,0)-\epsilon \leq
\varphi(x,y)\leq
\varphi(0,0)+\epsilon
\quad \mbox{ for $(x,y)\in U(r_1,r_2)$.}
\end{equation*}
These inequalities imply that
\begin{equation}\label{eqn:5.2}
\begin{split}
(\varphi(0,0)-\epsilon) \cdot
Z_{{U(r_1,r_2)}}(\sigma)
&\leq
Z_f(\varphi)(\sigma)-
\int_{U\setminus U(r_1,r_2)}|f(x,y)|^{\sigma}\varphi(x,y)dxdy\\
&\leq
(\varphi(0,0)+\epsilon) \cdot
Z_{U(r_1,r_2)}(\sigma).
\end{split}
\end{equation}
We remark that the integral in (\ref{eqn:5.2}) converges
and is bounded by a positive constant
which is independent of $\sigma$
since $f$ does not vanish on $U\setminus U(r_1,r_2)$.

As a result,
the inequalities (\ref{eqn:5.2}) and
Theorem~3.1 easily imply (i), (ii) in Theorem~2.1.

\vspace{.5 em}
{\bf The case (iii).} \quad
We define
\begin{equation*}
\varphi_+(x,y)
=\max\{\varphi(x,y),0\}
\,\,\, \mbox{ and  }\,\,\,
\varphi_-(x,y)
=\max\{-\varphi(x,y),0\}.
\end{equation*}
Of course,
$\varphi(x,y)=\varphi_+(x,y)-\varphi_-(x,y)$ holds.

Now,
let $R_1,R_2$ be positive constants such that
the support of $\varphi$ is contained in
$U(R_1,R_2)$.
Then, Theorem~3.1 (iii) implies that
there exist  $\delta>0$ and $C_{\pm}(R_1,R_2)>0$,
which depends on $R_1, R_2$ and is independent of $\sigma$,
such that
\begin{equation}\label{eqn:5.3}
\begin{split}
Z_f(\varphi_{\pm})(\sigma)
&\leq
\max_{(x,y)\in U} \varphi_{\pm}(x,y) \cdot
Z_{{\rm Supp}(\varphi_{\pm})}(\sigma) \\
&\leq
\max_{(x,y)\in U} \varphi_{\pm}(x,y)
\cdot Z_{U(R_1,R_2)}(\sigma) \\
&\leq
C_{\pm}(R_1,R_2)\cdot
\max_{(x,y)\in U} \varphi_{\pm}(x,y),
\end{split}
\end{equation}
for $\sigma\in(-1/b,-1/b+\delta)$.
Since
$Z_f(\varphi_{\pm})(\sigma)$ are monotone decreasing functions
of $\sigma\in(-1/b,0)$,
the above estimates easily imply that
there exist nonnegative constants
$B_+(\varphi)$ and $B_-(\varphi)$ such that
\begin{equation*}
\lim_{\sigma\to-1/b+0} Z_f(\varphi_{\pm})(\sigma)=B_{\pm}(\varphi).
\end{equation*}
Let $B(\varphi):=B_+(\varphi)-B_-(\varphi)$, then
we can get the limit in (iii).
Note that when $\varphi$ satisfies (\ref{eqn:1.4}),
$B(\varphi)=B(\varphi_+)$ and $B(\varphi_-)=0$ hold,
so $B(\varphi)$ is positive.

\begin{remark}
The inequalities in (\ref{eqn:5.3}) imply the continuous
property for the distribution defined by the map:
$\varphi\mapsto
B(\varphi)=\lim_{\sigma\to-1/b} Z_f(\varphi)(\sigma)$.
\end{remark}

\section{Landau type theorem for local zeta functions}

In this section, we deal with $Z_f(\varphi)(s)$
in the general dimensional case,
i.e.
\begin{equation}\label{eqn:6.1}
Z_f(\varphi)(s)=\int_{\mathbb{R}^n}|f(x)|^s \varphi(x)dx
\quad\quad \mbox{for $s\in {\mathbb C}$,}
\end{equation}
where
$f,\varphi$ are real-valued smooth functions
defined on an open neighborhood $U$ of the origin
in $\mathbb{R}^n$
and the support of $\varphi$ is contained in $U$.

Lemma~1.1  in the Introduction follows from
the following theorem.

\begin{theorem}\label{prop:3.1}
Suppose
that $f(0)=0$, $|f(x)|<1$ and $\varphi(x)\geq 0$ on $U$.
Let $\rho$ be nonpositive number such that
the integral $Z_f(\varphi)(s)$ in (\ref{eqn:6.1})
converges if ${\rm Re}(s)>\rho$.
If
$Z_f(\varphi)$ can be analytically continued
as a holomorphic function
to some open neighborhood of $s=\rho$,
then there exists a positive number $\delta$
such that the integral $Z_f(\varphi)(\rho-\delta)$
converges.
\end{theorem}

Indeed,
if $Z_f(\varphi)(s)$ can be holomorphically continued across
the point $s=-c_0(f)$,
then
the above theorem implies that
$Z_f(\varphi)(s)$ becomes a holomorphic function
on the region: ${\rm Re}(s)>-c_0(f)-\delta$
with some positive $\delta$, which
is a contradiction to (\ref{eqn:1.5}).

The above property of local zeta functions
itself is interesting and
is analogous to Landau's theorem on the Dirichlet series
$\sum_{n=1}^{\infty} a_n n^{-s}$
where $a_n$ are nonnegative numbers
(see \cite{Zag81}).

\begin{proof}
From the assumption,
$Z_f(\varphi)$ can be considered as a holomorphic function
on the region: ${\rm Re}(s)>\rho$ and, moreover, there exist
an open neighborhood $V$ of $s=\rho$ and
a holomorphic function $\tilde{Z}$ defined
on the set $V\cup\{s\in \mathbb{C}: {\rm Re}(s)>\rho\}$
such that
$\tilde{Z}=Z_f(\varphi)$ on the region: ${\rm Re}(s)>\rho$.

Now, there exists a positive number $\delta$
such that
the disc:
$$
D:=\{z\in{\mathbb C}:|z-(\rho+1)|<1+2\delta\}
$$
is contained in the region
$V\cup\{s\in \mathbb{C}: {\rm Re}(s)>\rho\}$.
Since $\tilde{Z}$ is holomorphic on $D$,
its Taylor series converges to the value of $\tilde{Z}(s)$
at any point of $D$,
i.e.,
\begin{equation}\label{eqn:6.2}
\tilde{Z}(s)=\sum_{j=0}^{\infty}
\frac{1}{j!}\frac{d^j\tilde{Z}}{ds^j}(\rho+1)(s-(\rho+1))^j
\quad \mbox{ for $s\in D$.}
\end{equation}
Since the point $s=\rho-\delta$ is contained
in $D$,
we have
\begin{equation}\label{eqn:6.3}
\tilde{Z}(\rho-\delta)=\sum_{j=0}^{\infty}
\frac{1}{j!}\frac{d^j \tilde{Z}}{ds^j}(\rho+1)(-\delta-1)^j
\end{equation}
and this series converges.
On the other hand, we have
\begin{eqnarray}\label{eqn:6.4}
\frac{d^j \tilde{Z}}{d s^j}(s)=
\int_{\mathbb{R}^n}|f(x)|^s (\log |f(x)|)^{j}
\varphi(x)dx \quad \mbox{for $j\in{\mathbb N}$}
\end{eqnarray}
if $s$ satisfies ${\rm Re}(s)>\rho$.
Indeed, it is easy to show the possibility of
the exchange of integral and derivatives.
Substituting (\ref{eqn:6.4}) to (\ref{eqn:6.3}),
we have
\begin{equation}\label{eqn:6.5}
\begin{split}
\tilde{Z}(\rho-\delta)
&=\sum_{j=0}^{\infty}
\frac{1}{j!}(-\delta-1)^j
\int_{\mathbb{R}^n}|f(x)|^{\rho+1} (\log |f(x)|)^{j}\varphi(x)dx\\
&=
\sum_{j=0}^{\infty}
\int_{\mathbb{R}^n}|f(x)|^{\rho+1}
\frac{1}{j!}
((-\delta-1)\log |f(x)|)^{j}\varphi(x)dx.
\end{split}
\end{equation}
Since
all terms of the series in \eqref{eqn:6.5} are positive
and
the series
converges,
the order of the summation and the integral
can be exchanged.
Therefore,
\begin{equation*}
\begin{split}
\tilde{Z}(\rho-\delta)
&=
\int_{\mathbb{R}^n}|f(x)|^{\rho+1}
\left(
\sum_{j=0}^{\infty}\frac{1}{j!}
((-\delta-1)\log |f(x)|)^{j}
\right)
\varphi(x)dx \\
&=
\int_{\mathbb{R}^n}|f(x)|^{\rho+1}e^{(-\delta-1)\log|f(x)|}\varphi(x)dx\\
&=
\int_{\mathbb{R}^n}|f(x)|^{\rho-\delta}\varphi(x)dx.
\end{split}
\end{equation*}
The above equalities imply that
the last integral converges.
\end{proof}

\section{Discussion and open questions}

\subsection{Singularities of $Z_f(\varphi)(s)$}

First,
let us consider the case when
$f$ is as in (\ref{eqn:2.1}) and $\varphi$ satisfies
the condition (\ref{eqn:1.4}).
As mentioned in Collorary 2.2,
the singularity of $Z_f(\varphi)(s)$ at $s=-1/b$
is different from a pole.
To be more elementary,
the following question is naturally raised.

\begin{question}
Is the singularity of $Z_f(\varphi)$ at $s=-1/b$ isolated or not?
\end{question}

If this singularity was isolated,
then it must be an essential singularity.
At present,
it seems impossible to answer this question
from the information from Theorem~2.1 only.

Next, let us consider
more global property of $Z_f(\varphi)(s)$.
It should be expected that
$Z_f(\varphi)(s)$ can be holomorphically extended
to a wider domain containing the region: ${\rm Re}(s)>-1/b$.
The following question is considered
as a first step to this problem.

\begin{question}
Does $Z_f(\varphi)$ have another singularity on
the vertical line: ${\rm Re}(s)=-c_0(f)$?
\end{question}

\subsection{Openness problem}

Let $f$ be a smooth function with the condition
(\ref{eqn:1.2}).
Let us consider the following subset in ${\mathbb R}$:
\begin{equation*
H(f):=
\left\{\mu>0:
\begin{array}{l}
\mbox{there exists an open neighborhood $V$ of} \\
\mbox{the origin in $U$ such that $|f|^{-\mu}\in L^1(V)$}
\end{array}
\right\}.
\end{equation*}
(Of course, $c_0(f)=\sup H(f)$.)
When $f$ is real analytic, the set $H(f)$ is {\it open} in ${\mathbb R}$
from the fact that $s=-c_0(f)$ is a pole for $Z_f(\varphi)$
for $\varphi$ satisfying (\ref{eqn:1.4}).
Without the real analyticity assumption,
our observation implies that
$H(f)$ is not always open.
More precisely, in the case of (\ref{eqn:2.1}),
the openness of $H(f)$ depends on
the relationship among the parameters $a,b,p$.
Generally, when $f$ is a smooth function,
the following question seems interesting.

\begin{question}
Which condition on $f$ gives the openness (or closedness)
of $H(f)$?
\end{question}

An analogous problem has been deeply investigated
in the case of complex variables from the viewpoint of
complex geometry.
The {\it openness conjecture},
raised by Demailly and Koll\'ar \cite{DeK01},
is the following:
``If $\phi$ is plurisubharmonic,
then the set $H(e^{-\phi})$ is always open.''
This conjecture has been affirmatively solved
in \cite{FaJ05jams},  \cite{FaJ05inv},  \cite{Ber13}.
Our observation indicates that the openness of $H(f)$ needs
some kind of good property of $f$.

\subsection{Oscillatory integrals}
Let us consider an oscillatory integral of the form:
\begin{equation*}
I_f(\varphi)(\tau):=
\int_{{\mathbb R}^2}e^{i \tau f(x,y)}\varphi(x,y)dxdy
\quad \mbox{for $\tau>0$},
\end{equation*}
where  $f$, $\varphi$ are as in (\ref{eqn:1.1})
and they satisfy the conditions
(\ref{eqn:1.2}), (\ref{eqn:1.4}).

It is known (see \cite{Var76}, \cite{AGV88}, etc.)
that there is a deep relationship
between the behavior of oscillatory integrals at infinity
and the distribution of poles of local zeta functions.
Indeed, the Mellin transformation gives a
clear relationship between oscillatory integrals
and some functions similar to local zeta functions.

First, let us consider the case when $f$ is real analytic.
As mentioned in the Introduction,
the integral $Z_f(\varphi)(s)$ can be
analytically continued as
a meromorphic function to the whole complex plane.
Furthermore, under some assumption,
its leading pole exists at $s=-1/d(f)$ and
its order is $m(f)$.
(Note that $m(f)$ is the positive integer
determined by some topological information
of the Newton polyhedron of $f$.)
By using this fact, we have
\begin{equation}\label{eqn:7.1}
\lim_{\tau\to+\infty}
\tau^{1/d(f)}(\log \tau)^{-m(f)+1}
I_f(\varphi)(\tau)=C_f(\varphi),
\end{equation}
where $C_f(\varphi)$ is a positive constant independent of
$\tau$.

Next, let us consider the smooth case.
Although the formula (\ref{eqn:7.1})
can be directly generalized in many smooth cases
(\cite{Gre09}, \cite{IkM11jfaa}, \cite{KaN16jmsut}),
there exist examples for which
the behavior of $I_f(\varphi)$ is different from (\ref{eqn:7.1}).
In \cite{KaN17},
the authors investigate the case when
\begin{equation}\label{eqn:7.2}
f(x,y)=y^b+e^{-1/|x|^p},
\end{equation}
where $p>0$ and
$b\in{\mathbb Z}$ with $b\geq 2$, and obtain the following:
\begin{equation*}
\lim_{\tau\to+\infty}\tau^{1/b}(\log \tau)^{1/p}\cdot
I_f(\varphi)(\tau)=C_b\varphi(0,0),
\end{equation*}
where $C_b$ is a nonzero constant depending only on
$b$ and is explicitly computed.
We remark that $d(f)=b$ and $m(f)=1$ in this case.
Since (\ref{eqn:7.2}) is a special case of (\ref{eqn:2.1})
($a=0$, $b=q$),
Theorem~2.1 implies that if $p>1$, then
\begin{equation*}
\lim_{\sigma\to -1/b+0}
(b\sigma+1)^{-1/p+1}\cdot Z_f(\varphi)(\sigma)=
C\varphi(0,0),
\end{equation*}
where $C$ is a positive constant.

Now, we are interested in how flat functions affect
the behavior of $I_f(\varphi)(\tau)$.
Observing the case when $f$ is real analytic or
$f$ is as in (\ref{eqn:7.2}),
one can easily recognize some correspondence between the bahaviors of
$Z_f(\varphi)(\sigma)$ and $I_f(\varphi)(\tau)$.
It will be valuable to affirmatively answer the following question.
\begin{question}
When a positive limit of
$(\sigma+c_0(f))^{\alpha}Z_f(\varphi)(\sigma)$
as $\sigma\to-c_0(f)+0$ exists,
does a nonvanishing limit of
$\tau^{c_0(f)}(\log \tau)^{-\alpha+1}\cdot I_f(\varphi)(\tau)$
as $\tau\to+\infty$ exist?

In particular,
let $f$ be as in (\ref{eqn:2.1}) with
$p>1-a/b$, then
does the following hold?
\begin{equation*}
\lim_{\tau\to +\infty}
\tau^{1/b}(\log \tau)^{\frac{1-a/b}{p}}\cdot I_f(\varphi)(\tau)=
C\varphi(0,0),
\end{equation*}
where $C$ is a positive constant
which is independent of $\tau$.
\end{question}

\vspace{1 em}

{\sc Acknowledgements.}\quad
The authors would like to express their sincere gratitude
to the referee for giving the authors valuable comments.
This work was partially supported by
JSPS KAKENHI Grant Numbers
JP15K04932, JP15K17565, JP15H02057.


text:
\section{Introduction}

The Friedland-Hayman inequality \cite{FH} is a sharp inequality
concerning the growth rates of homogeneous harmonic functions with Dirichlet
boundary conditions on complementary cones dividing Euclidean space into
two parts.  It plays a crucial role in the interior regularity theory
of two-phase free boundary problems, as developed by Alt, Caffarelli, and Friedman
\cite{ACF}.   In this paper, we prove an analogous inequality in which
one divides a convex cone into two parts, placing Neumann
conditions on the boundary of the convex cone, and Dirichlet conditions
on the interface.  This analogous inequality was already proved in \cite{BJR}
and leads to regularity of two-phase
free boundaries at points near a fixed boundary with Neumann conditions
in convex domains.  Here we present a new  proof of independent interest
that leads, in addition, to the characterization of the case of equality.
In keeping with the theory introduced in \cite{ACF} (see also \cite{CaS}),
such an improvement should ultimately yield further regularity properties of the free boundary.

The Friedland-Hayman inequality can be stated as follows.
\begin{thm} \label{thm:Friedland-Hayman} \emph{\cite{FH}} \ Let $u_1$ and $u_2$ be non-negative, H\"older continuous
functions defined on $\mathbb{R}^n$, with $u_1u_2\equiv0$, and harmonic where they are positive, that is,
$\Delta u_i(x) = 0$ whenever $u_i(x)>0$.  If $u_i$ is homogeneous of
degree $\alpha_i$, then
\[
\alpha_1  + \alpha_2 \ge 2.
\]
\end{thm}
Equality holds if and only if the two functions are (up to constant multiples)
the positive and negative parts of a linear function.  In other words,
after rigid motion,
\[
u_1(x) = c_1 x_1^+; \quad u_2(x) = c_2 x_1^-, \quad c_1 > 0 , \ c_2>0,
\]
with $x_1$ the first coordinate of $x$ and
$u^\pm (x) : = \max (0, \pm u(x))$.  This was proved in \cite{ACF} (Lemma 6.6)  in dimension 2.  Also, in Remark 6.1 of that paper, the authors show that the characterization
of equality is valid in all dimensions,  provided one knows that the case of equality in a rearrangment theorem on the sphere due to Sperner \cite{Spe} is acheived only for rotationally symmetric caps. This
case of equality in Sperner's rearrangement theorem was subsequently proved by Brothers and Ziemer \cite{BZ}.

To state our main theorem, consider a convex, open cone $\Gamma$,
and two nonempty, disjoint, open, connected, conic subsets $\Gamma_\pm$,
that is,
\[
\bar \Gamma_+ \cup \bar \Gamma_- \subset \bar \Gamma, \quad \Gamma_+ \cap \Gamma_- = \emptyset, \quad \Gamma_\pm \neq \emptyset.
\]
Define the Neumann and Dirichlet portions of the boundary, $\gamma_N^\pm$ and
$\gamma_D^\pm$,  by
\[
\gamma_N^\pm = \partial \Gamma \cap \partial \Gamma_\pm, \quad \mbox{and} \quad \gamma_D^\pm : = \Gamma \cap \partial \Gamma_\pm, \quad \mbox{so that} \quad \partial \Gamma_\pm = \gamma_N^\pm \cup \gamma_D^\pm.
\]
\begin{thm}  \label{thm:main} (Friedland-Hayman inequality for convex cones) Suppose that $\Gamma\subset\mathbb{R}^n$ is an open,
convex cone containing two disjoint, open, connected cones $\Gamma_\pm$ as above.
Let $u_\pm$ denote the unique (up to constant
multiples) positive, harmonic function
on $\Gamma_\pm$ that is  homogeneous of positive degree and
satisfies the mixed boundary conditions, $u_\pm = 0$ on $\gamma_D^\pm$ and $(\partial/\partial \nu) u_\pm = 0$  on $\gamma_N^\pm$ in the weak sense.   If the degree of $u_\pm$ is denoted $\alpha_\pm$, then
\[
\alpha_+ + \alpha_- \ge 2.
\]
Moreover, if equality holds, then after rotation, there is an open, convex
cone $\Gamma' \subset \mathbb{R}^{n-1}$ such that
\[
\Gamma = \mathbb{R}\times \Gamma', \quad \Gamma_\pm = \{\pm x_1 > 0\} \times \Gamma', \quad \mbox{and} \quad u_\pm(x) = c_\pm x_1^\pm
\]
for some constants $c_\pm >0$.
\end{thm}
The original Friedland-Hayman inequality is the case $\Gamma  = \mathbb{R}^n$
with $\gamma_N^\pm  = \emptyset$.   If $\Gamma$ is a half space, then
the result also follows from the original Friedland-Hayman inequality by an
argument using reflection.   The proof of the inequality stated in Theorem \ref{thm:main} in our joint work with
Sarah Raynor \cite{BJR} uses the L\'evy-Gromov isoperimetric
inequality on Ricci non-negative manifolds.  As we mentioned
earlier, the new proof here will permit us to characterize the case of equality.

The exponents $\alpha_\pm$ can be expressed in terms
of the lowest eigenvalues of a mixed boundary problem on the spherical
cross sections $\Gamma_\pm \cap \mathbb{S}^{n-1}$.  The relationship
is given by \eqref{eq:sphere-eigenvalue} below.
Those eigenvalues have a variational characterization which is
important to the applications to the free boundary regularity, that
is, to the monotonicity formulas of \cite{ACF} and \cite{BJR}.  Variational
characterizations are also important to this proof, as we shall see.

By rearrangement, the
Friedland-Hayman inequality reduces to a family of
one-dimensional problems parametrized by the dimension $n$.
On the other hand,
the inequality in dimension $n+1$ implies the inequality in dimension $n$,
as one can see by  considering the product of a cone with a line.
Thus the one-dimensional inequality gets harder to prove as $n$ increases.
Beckner, Kenig and Pipher \cite{BKP} gave a more conceptual proof of the
original Friedland-Hayman theorem that relies on this rearrangement, but
circumvents the Brothers-Ziemer result.     They took the limit as $n$ tends
to infinity and identified and fully analyzed the problem one obtains
in the limit, an extremal eigenvalue problem on the real line with a gaussian weight.
The limiting ``infinite-dimensional"
problem dominates all the finite dimensional ones, and its
extremal sets are the half lines $x_1>0$, $x_1 < 0$. See Section 12.2 in \cite{CaS} for the details of this approach.

Rearrangement cannot be used to solve the problem on convex domains.
But it is possible to take the dimension to infinity
by considering cones of
the form $\Gamma_\pm \times \mathbb{R}^N$ as $N\to \infty$.  Using this device we will
prove the following key proposition.

\begin{prop} \label{prop:main}  If $\alpha_+$ is the characteristic exponent associated with $\Gamma_+$ in
Theorem \ref{thm:main}, then
\[
\alpha_+ \ge
\inf_{f\in Y_+} \frac{\int_{\Gamma_+}\left| \nabla f(x) \right|^2 e^{-|x|^2/2} \, dx}{\int_{\Gamma_+} f(x)^2 e^{-|x|^2/2} \, dx} \, ,
\]
with $Y_+ = \{f \in C_0^\infty(\mathbb{R}^n): f(x) = 0 \ \mbox{for} \ x\in \gamma_D^+\}$.
Evidently, the same result holds with $+$ replaced by $-$.
\end{prop}

Note that the weight in the variational expression on the right side of the inequality
in Proposition \ref{prop:main} is the restriction to $\Gamma_+$ of
\[
e^{-|x|^2/2} e^{-F(x)} \, dx, \quad F(x) =
\begin{cases}
0  & \ x\in \Gamma \\
 \infty & \ x\in \mathbb{R}^n \setminus \Gamma.
\end{cases}
\]
Because $\Gamma$ is convex, $F$ is a generalized convex function, that is, a convex function allowing for the value $+\infty$.
Put another way, $e^{-F}$ is a generalized log-concave function.

Because our measure is ``more log concave" than the gaussian,
we will be able to invoke a variant of Caffarelli's contraction theorem
for the Brenier optimal transport mapping.  Recall that Brenier's mapping
can be characterized as follows.

\begin{thm}[Brenier \cite{Br}] \label{thm:Brenier}
Let $\mu$, $\nu$ be positive measures on $\mathbb{R}^{n}$ and with finite second order moments.  Suppose also that $\mu$ is  absolutely continuous with respect to the Lebesgue measure. Then, there exists a convex function $\varphi:\mathbb{R}^{n}\to\mathbb{R}$ such that $T = \nabla \varphi:\mathbb{R}^{n}\to\mathbb{R}^{n}$ transports $\mu$ onto $c\, \nu$. That is,  $\mu(T^{-1}(E)) = c\, \nu(E)$, with normalizing constant
$c = \mu(\mathbb{R}^n)/\nu(\mathbb{R}^n)$.
\end{thm}

The variant of Caffarelli's contraction theorem that we require can be stated
as follows.
\begin{thm} \label{thm:Caf}  Let $K$ be a convex subset of $\mathbb{R}^n$ and let
$F$ be a convex function on $\mathbb{R}^n$. Set
\[
\mu = e^{-|x|^2/2}\, dx, \quad \nu  = e^{-F} 1_K \mu.
\]
Then the Brenier mapping $T = \nabla \varphi$ is
a contraction:  $|T(x) - T(y)| \le |x-y|$.  Put another way,
$D^2\varphi$ has eigenvalues bounded above by $1$.
\end{thm}
Caffarelli's theorem is the case $K = \mathbb{R}^n$.  The Friedland-Hayman inequality
will follow from Proposition \ref{prop:main} and Theorem \ref{thm:Caf} with $K$ a convex cone and $F \equiv 0$ on $K$.

De Philippis and Figalli \cite{DPF} have characterized
the case of equality in the eigenvalue formulation of Caffarelli's theorem.
This result will permit us to characterize the case of equality in the Friedland-Hayman
inequality, as stated in Theorem \ref{thm:main}.
Their theorem is stated here in our generalized setting.
\begin{thm} \label{thm:DPF}
 Let $\varphi$ be as in Theorem \ref{thm:Caf}, and let
\[
0\leq \lambda_1(D^2\varphi(x)) \leq \cdots \leq \lambda_{n}(D^2\varphi(x)) \leq 1
\]
be the eigenvalues of $D^2\varphi(x)$. If for some $m$, $1\leq m \leq n$,
$\lambda_{n-m+1}(D^2\varphi(x)) = 1$ for all $x\in\mathbb{R}^n$, then,
after translation and rotation, there is a convex set $K'\subset \mathbb{R}^{n-m}$ and a convex function
$G$ on $K'$ such that
\[
K  = \mathbb{R}^m \times K'; \quad \nu = e^{-|x'|^2/2 -|y'|^2/2 -G(y')} 1_{K'}(y') \, dx' dy'
\]
with $x = (x',y') \in \mathbb{R}^m \times \mathbb{R}^{n-m}$.
\end{thm}

Our second application is to the case of equality of a Poincar\'e-Wirtinger type inequality: For a given convex domain $K\subset \mathbb{R}^n$, define $\mu_1(K)$ by
\begin{align} \label{eqn:PW}
\mu_1(K) = \inf_f  \left\{\displaystyle{\frac{\int_{K}\left|\nabla f(x)\right|^2 e^{-|x|^2/2} \, d x }{\int_{K}|f(x)|^2 e^{-|x|^2/2} \, dx}}\text{ }:\text{ } \int_{K} f(x) e^{-|x|^2/2} \, dx = 0 \right\}.
\end{align}
Here the infimum is taken over function for which the expressions are finite.
The value $\mu_1(K)$ can also be viewed as the first non-zero eigenvalue of
$$
\begin{cases}
\text{div}\left(e^{-|x|^2/2} \nabla u\right)  =  -\mu e^{-|x|^2/2} u  \text{ in } K, \\
\frac{\partial u}{\partial n}  =  0  \text{ on } \partial K.
\end{cases}
$$
In \cite{BCHT}, Brandolini, Chiacchio, Henrot, and Trombetti show that if $K\subset \mathbb{R}^n$ has $C^2$-smooth boundary, then
\begin{align} \label{eqn:PW1}
\mu_1(K) \geq  \mu_1(\mathbb{R}^n) = \mu_1(\mathbb{R}) = 1.
\end{align}
In \cite{BCKT}, this inequality is shown to hold for any convex planar domain, and if $K\subset \mathbb{R}^2$ is contained in a strip, then equality in \eqref{eqn:PW1} holds precisely when $K$ is itself a strip. Here, we use Theorems \ref{thm:Caf} and \ref{thm:DPF} to prove the theorem in the non-smooth case and describe fully the case of equality.
\begin{thm} \label{thm:PW}
For any convex domain $K\subset \mathbb{R}^n$, and $\mu_1(K)$ as in \eqref{eqn:PW}, we have
\begin{align*}
\mu_1(K) \geq \mu_1(\mathbb{R}) = 1.
\end{align*}
Moreover, $($up to a rotation$)$ equality holds precisely when $K$ is of the form $K = \mathbb{R}\times K'$, for a convex domain $K' \subset \mathbb{R}^{n-1}$.
\end{thm}
Let us make a few remarks about the existing literature.
Our proof of Theorem \ref{thm:Caf} will follow Caffarelli's proof, exploiting the fact that $\varphi$ satisfies a Monge-Amp\`ere equation and that the second difference of $\varphi(x)$ is well-behaved as $|x|$ tends to infinity. Other proofs, variants, and extensions of this theorem have also been given in \cite{FGP}, \cite{KM1}, \cite{Ko1}, and \cite{Va1}.  The theorem of De Philippis and Figalli, Theorem 1.2 \cite{DPF}
 identifying the case of equality in Caffarelli's original theorem, has an alternative proof
due to Cheng and Zhou.  That proof involves
the first non-zero eigenvalue of a Laplacian with drift on a complete smooth metric space with a lower bound on the Bakry-\'Emery Ricci curvature (see Theorem 2 in \cite{CZ}).

The rest of the paper is structured as follows. In Section \ref{sec:contraction}, we prove the version of Caffarelli's contraction theorem and the equality case, Theorems \ref{thm:Caf} and \ref{thm:DPF}.   We then show how our two applications follow from these theorems in Section \ref{sec:consequences}. We first prove the Poincar\'e-Wirtinger type inequality and case of equality, which is a direct consequence of Theorems \ref{thm:Caf} and \ref{thm:DPF}.  Our version of
the Friedland-Hayman inequality requires Proposition \ref{prop:main}.
This converts our problem to one about Gaussian eigenvalues of a convex domain to which Theorems \ref{thm:Caf} and \ref{thm:DPF} apply
 in $\mathbb{R}^{n}$.    We emphasize here that our argument is inspired by
 the argument of Beckner, Kenig and Pipher.   Moreover, because our method
 reduces the case of the convex cone to case of the entire Euclidean space, it
 depends on the original Friedland-Hayman theorem and does not
 replace it.  We end by discussing a possible, natural variant
 of Caffarelli's contraction theorem for geodesically convex subsets
 of spheres.  This variant would provide an alternative
path to the main theorem, Theorem \ref{thm:main}.

\section{Proof of the Caffarelli contraction theorem} \label{sec:contraction}

Recall from Theorem \ref{thm:Brenier}, that $T = \nabla\varphi$ transports the gaussian measure $\mu = e^{-|x|^2/2}\, dx$ onto the measure $c\,\nu$, where $\nu = e^{-F(x)}1_K\, \mu$ for a convex domain $K\subset \mathbb{R}^n$ and a convex function $F$ on $K$. Here $\varphi(x)$ is a convex function on $\mathbb{R}^n$ and the constant $c$ is chosen so that $\mu$ and $c\,\nu$ have the same total measure. To prove Theorem \ref{thm:Caf}, we need to show that the eigenvalues of $D^2\varphi(x)$ are bounded above by $1$.
\begin{proof}{Theorem \ref{thm:Caf}}
We follow the proof Caffarelli used to prove Theorem 11 in \cite{Ca1}. In particular, rather than studying $D^2\varphi(x)$ directly, we work with second differences of $\varphi(x)$. That is, we fix $h>0$, and for each unit direction $e$, $x\in \mathbb{R}^n$, we set
 \begin{align*}
\delta_{e} \varphi(x) = \varphi(x + he) + \varphi(x -he) - 2 \varphi(x).
\end{align*}
To prove the theorem, we need to show that
\begin{align} \label{eqn:Caf-1}
0\leq \delta_{e} \varphi(x) \leq h^2
\end{align}
for all $x$, $e$, and $h>0$, since then letting $h\to0$ gives the desired upper bound. To prove \eqref{eqn:Caf-1} it is sufficient to study $\delta_{e} \varphi(x)$ at a point where it achieves its maximum in both $x$ and $e$, together with its behavior as $|x|$ tends to infinity. However, in the case where the convex set $K$ is not smooth, strictly convex, and bounded, then the behavior of $\varphi(x)$ at infinity can be more complicated. Therefore, we form a sequence of smooth, strictly convex, and bounded sets $K_j\subset K$, which converges to $K$ in Hausdorff distance on compact sets  \cite{Sc1}. (By strictly convex, we mean that each tangent plane to $K_j$ touches $\partial K_j$ at a unique point.) We also obtain corresponding Brenier maps $T_j = \nabla \varphi_j$ transporting $\mu = e^{-|x|^2/2}\, dx$ onto $c_j\nu_j$, with $\nu_j = e^{-F(x)}1_{K_j}\, \mu$. Here the constant $c_j>0$ is chosen to ensure that $c_j\nu_j$ and $\mu$ have the same total measure.

As $F$ is a convex function on $\mathbb{R}^n$, it is in particular in $L^{\infty}(K\cap B_R)$ for all $R>0$. Since the sets $K$ and $K_j$ are convex, this ensures that $\varphi$ and $\varphi_j$ are $C^2$ (see Theorem 1 in \cite{CEF}). Therefore, $\varphi_j$ is a classical solution to the Monge-Amp\'ere equation
\begin{align} \label{eqn:MA}
\text{det}\left(D^2\varphi_j(x)\right) = \frac{e^{-|x|^2/2}}{\exp\left\{-|\nabla\varphi_j(x)|^2/2-F(\nabla \varphi_j(x))\right\}},
\end{align}
and $\varphi$ satisfies the analogous equation. Moreover, since $K_j$ converges to $K$ on compact sets, the measures $e^{-F(x)}1_{K_j}\, \mu$ converge strongly to $\nu$ as $j$ tends to infinity. Therefore, $\varphi_j(x)$ converges to $\varphi(x)$ uniformly on compact sets (see Theorem 3 in \cite{Ca1}, also \cite{Vi1}, 5.23).  To show \eqref{eqn:Caf-1} and complete the proof of the theorem, it is thus sufficient to establish
\begin{align} \label{eqn:Caf-2}
0\leq \delta_{e} \varphi_j(x) \leq h^2
\end{align}
for all $x$, $e$, and $h>0$, and $j$ fixed.

Note that the lower bound is guaranteed since $\varphi_j$ is convex. To prove the upper bound, suppose first that $\delta_e\varphi_j(x)$ attains its maximum in $x$ and $e$ at $x=x_0$ and $e=e_0$. Then, as shown in the proof of Theorem 11 in \cite{Ca1} (see also \cite{Ca2}), using the fact that $\varphi_j(x)$ satisfies the Monge-Amp\`ere equation in \eqref{eqn:MA}, it satisfies a maximum principle ensuring that $\delta_{e_0}\varphi_j(x_0) \leq h^2$.

To complete the proof we therefore need to study the behavior of $\delta_e\varphi_j(x)$ as $|x|$ tends to infinity. This part of the proof is the reason for using the smooth, strictly convex approximating sets $K_j$, and also why we work with the second difference rather than the second derivative directly. Since the sets $K_j$ will not be balls centered at the origin when $K$ is a proper subset of $\mathbb{R}^n$, this part of the proof requires a small modification of Caffarelli's proof, and so we write it out in detail.
\begin{lem} \label{lem:infinity}
Suppose that $|x|$ tends to infinity, with $\tfrac{x}{|x|}$ converging to a direction ${\textbf{n}}$. Then,
\begin{align*}
\nabla\varphi_j(x) \to  {y}_{{\textbf{n}}},
\end{align*}
with uniform convergence in the direction $\textbf{n}$. Here ${y}_{{\textbf{n}}}$ is the unique point on $\partial K_j$ with outward unit normal pointing in the direction ${\textbf{n}}$.
\end{lem}
\begin{proof}{Lemma \ref{lem:infinity}}
The uniqueness of the point $y_{\textbf{n}}$ follows immediately from the strict convexity and smoothness of $K_j$. Given ${x}\in\mathbb{R}^{n}$, let ${y} = \nabla \varphi_j ({x}) \in K_j$. Define $\Gamma_{y}(\theta)$ to be the cone of vertex ${y}$ pointing in the direction of ${x}$, with angle $\theta$ (here $\theta$ is fixed, with $0<\theta<\tfrac{\pi}{2}$), so that
\begin{align*}
\Gamma_{{y}}(\theta) = \{{y}'\in\mathbb{R}^{n}: \text{angle}(x,{y}'-{y}) \leq \theta\}.
\end{align*}
By the cyclical monotonicity of the optimal transport mapping, if ${y}' = \nabla \varphi_j({x}')$, then the inner product of ${x}'-{x}$ and ${y}' -{y}$ is nonnegative. Therefore, the pre-image of $\Gamma_{y}(\theta)$ under $\nabla \varphi_j$ is contained in the cone
\begin{align*}
\Gamma_{x}(\theta) = \{{x}'\in\mathbb{R}^{n}: \text{angle}(x,{x}'-{x}) \leq \theta + \tfrac{\pi}{2}\}.
\end{align*}
There exists a constant $a_{\theta}>0$ such that the complement of $\Gamma_{x}(\theta)$ contains the ball of radius $a_{\theta}|x|$ centred at the origin. In particular, as $|x|$ tends to infinity, for $\theta$ fixed, the Gaussian measure of $\Gamma_{x}(\theta)$ tends to zero. Since the map $\nabla \varphi_j$ is measure preserving, and the measure $\nu_j$ is bounded from below on $K_j$, this means that the Lebesgue measure of $\Gamma_{{y}}(\theta)\cap K_j$ tends to $0$. Therefore, given $\eta>0$ and $0 < \theta < \tfrac{\pi}{2}$, there exists $M_{\theta}>0$ such that if $|x|>M_{\theta}$, then the Lebesgue measure of $\Gamma_{{y}}(\theta)\cap K_j$ is less than $\eta$. This in particular ensures dist$(y,\partial K_j)<C_1\eta$, where $C_1$ is a constant depending only on the Lipschitz bound for the convex set $K_j$. As $\theta$ tends to $\tfrac{\pi}{2}$, the cone $\Gamma_y(\theta)$ approaches the half-plane passing through $y$ in the direction $\tfrac{x}{|x|}$, and this forces $y$ to approach $y_n$, the unique point on $\partial K_j$ with outward unit normal $\textbf{n}$. More precisely, for any $\eta>0$, if $|\textbf{n}-\textbf{m}| <\eta$, then $|y_{\textbf{n}}-y_{\textbf{m}}| < C_2\eta$, for a constant $C_2$ depending only on the strict convexity of $K_j$ (but not $\textbf{n}$). Therefore, given $\eta>0$, we can choose $\delta>0$ and $\theta^* < \tfrac{\pi}{2}$ such that if $\left|\tfrac{x}{|x|} - \textbf{n}\right| < \delta$, with $|x|>M_{\theta^*}$, then $|y-{y}_\textbf{n}| < \epsilon$. This proves that $y$ converges to $y_{\textbf{n}}$, uniformly in the direction $\textbf{n}$.
\end{proof}
To conclude the proof of the theorem, we note that
\begin{align*}
0\leq h^{-2}\delta_e\varphi_j(x) \leq \frac{\nabla \varphi_j({x} + h{e})\cdot {e} - \nabla \varphi_j({x}-h{e})\cdot {e}}{2h},
\end{align*}
and for $h>0$, ${e}$ fixed, $\left|\tfrac{{x} + h{e}}{|{x} + h{e}|} - \tfrac{{x} - h{e}}{|{x} - h{e}|}\right|$ tends to $0$ as $|{x}|$ tends to infinity. Therefore, Lemma \ref{lem:infinity} implies that $\delta_e\varphi_j(x)$ tends to $0$ as $|x|$ tends to infinity, and  hence $\delta_e\varphi_j(x)\leq h^2$ for all ${x}$, ${e}$ and $h>0$ as required.
\end{proof}
\begin{proof}{Theorem \ref{thm:DPF}}
Now that we have proved Theorem \ref{thm:Caf}, after ordering the eigenvalues of $D^2\varphi(x)$, we can ensure that
\begin{align*}
0\leq \lambda_1(D^2\varphi(x)) \leq \cdots \leq \lambda_{n}(D^2\varphi(x)) \leq 1.
\end{align*}
To deal with the case of equality, where for some $k\geq1$, $\lambda_{n-k+1}(D^2\varphi(x)) = 1$ for all $x\in\mathbb{R}^n$, the proof of De Philippis and Figalli in Theorem 1.2 in \cite{DPF} (used to deal with the case of stability in Caffarelli's original contraction theorem) still applies and so we just briefly summarize their proof:  Defining the convex function $\Psi(x) = \tfrac{1}{2}|x|^2 - \varphi(x)$, the assumption of the theorem implies that det$(D^2\Psi)(x) = 0$. Subtracting a linear function from $\Psi(x)$ (which only translates $\nu$), we can assume that $\Psi(x) \geq\Psi(0) = 0$. Combining det$(D^2\Psi)(x) = 0$ with an Alexandrov estimate, de Philippis and Figalli show that the set $\Sigma = \{\Psi = 0\}$ does not have an exposed point. Therefore, the set $\Sigma$ must contain a line (see \cite{Bar}, Lemma 3.5 in Chapter 2). Rotating so that this line is $\mathbb{R}{e}_1$, and combining this with the convexity of $\Psi$ implies that $\partial_{{e}_1}\Psi({x}) \equiv 0$. Therefore, we can view $\Psi$ as a function of the variables $x' = (x_2,\ldots,x_n)$, and write the mapping $T = \nabla \varphi$ as $T(x) = \left(x_1,x' - \nabla \Psi(x')\right)$. This means that the measure $\nu$ can be written as $\nu = e^{-x_1^2/2}\, dx_1\otimes \nu_1$, where $T_1(x') = x' - \nabla \Psi(x')$ transports the $(n-1)$-dimensional gaussian measure $e^{-|x'|^2/2}$ onto $c\nu_1$.
Moreover, we can write $\nu_1$ as $e^{-|x'|^2/2-G_1(x')}1_{K_1} \, dx'$, for a convex set $K_1$ and convex function $G_1$, since these properties are preserved under taking marginals (see Theorem 4.3 in \cite{BL}). This proves the theorem for $k=1$, and by recursively applying this argument the theorem holds.
\end{proof}

\section{Consequences of the contraction theorem} \label{sec:consequences}

In this section we use Theorems \ref{thm:Caf} and \ref{thm:DPF} to study the Friedland-Hayman and Poincar\'e-Wirtinger type inequalities discussed in the introduction. The Poincar\'e-Wirtinger result follows as a direct application,
so we will prove it first.
\subsection{A Poincar\'e-Wirtinger inequality}
\begin{proof}{Theorem \ref{thm:PW}}
Let $u(x)$ be the eigenfunction corresponding to minimizing the quantity $\mu_1(K)$ given in \eqref{eqn:PW}.  Applying Theorem \ref{thm:Caf} with $F(x) \equiv 0$ (so that $\nu = e^{-|x|^2/2}1_K\, dx$), we obtain a transport map $T = \nabla \varphi$ from $\mu = e^{-|x|^2/2} \,dx$ to $c\,\nu$, such that $T$ is Lipschitz, with Lipschitz constant bounded by $1$. We define $v(x)$ by $v(x) = u(T(x))$, and since $T$ is a transport map, we have
\begin{align*}
\int_{\mathbb{R}^n} v(x) e^{-|x|^2/2} \, dx = \int_{\mathbb{R}^n}u(T(x)) e^{-|x|^2/2} \, dx = c\int_{K}u(x) \, d\nu = 0.
\end{align*}
Therefore, $v(x)$ is an admissible test function for $\mu_1(\mathbb{R}^n)$, and so
\begin{align} \label{eqn:PWproof1}
\mu_1(\mathbb{R}) = \mu_1(\mathbb{R}^n) \leq \displaystyle{\frac{\int_{\mathbb{R}^n}\left|\nabla v(x)\right|^2e^{-|x|^2/2} \, dx }{\int_{\mathbb{R}^n}|v(x)|^2e^{-|x|^2/2}\, dx}}.
\end{align}
Moreover, since the Lipschitz constant of $T$ is bounded by $1$, we have
\begin{align} \label{eqn:PWproof2}
\left|\nabla v(x)\right| \leq \left|(\nabla u)(T(x))\right|
\end{align}
for all $x\in \mathbb{R}^n$. Combining this with the fact that $T$ is a transport map, we can use \eqref{eqn:PWproof1} to obtain
\begin{align*}
\mu_1(\mathbb{R}) \leq \displaystyle{\frac{\int_{K}\left|\nabla u(x)\right|^2e^{-|x|^2/2} \, dx }{\int_{K}|u(x)|^2e^{-|x|^2/2}\, dx}} = \mu_1(K).
\end{align*}
To deal with the equality case $\mu_1(K) = \mu_1(\mathbb{R}) = 1$ we use Theorem \ref{thm:DPF}. If $K$ contains a line (say $\mathbb{R}_{e_1}$), then we can set $u(x) = x_1$ to obtain $\mu_1(K) = 1$. Now suppose that  $K$ does not contain a line and again let $u(x)$ be the eigenfunction corresponding to $\mu_1(K)$. Then by Theorem \ref{thm:DPF}, there exist $\epsilon>0$ and a set $U\subset \mathbb{R}^n$ of positive measure for which $\lambda_j(D^2\varphi(x))\leq1-\epsilon$ for all $x\in U$, $1\leq j \leq n$. The image of $U$ under $T$ also has positive measure. Setting $v(x) = u(T(x))$, we thus have
\begin{align*}
\int_{U}\left|\nabla v(x)\right|^2e^{-|x|^2/2} \, dx  \leq (1-\epsilon) \int_{U}\left|\nabla u(T(x))\right|^2e^{-|x|^2/2} \, dx.
\end{align*}
Using this inequality in the above argument in place of \eqref{eqn:PWproof2}, we obtain $\mu_1(K)>1$ and so we cannot have equality unless $K$ contains a line.
\end{proof}

\subsection{The Friedland-Hayman inequality}

In this section we prove Theorem \ref{thm:main}.  Recall that $u_+$ is
a harmonic function on the cone $\Gamma_+$ of homogeneous
degree $\alpha_+$ satisfying
boundary conditions $u_+ = 0$ on $\gamma_D^+$ and
$(\partial/\partial\nu) u_+ = 0$ (in the weak sense)
on $\gamma^+_N$. The first step is to prove Proposition \ref{prop:main}, a lower bound
on the characteristic exponent $\alpha_+$ in
 terms of the lowest Gaussian eigenvalue on the cone with the same boundary
conditions.  The convexity of the cone $\Gamma$ is not used in this proof.

Fix an integer $N\ge 0$, and extend the function $u_+$ to
the product cone $\Gamma_+ \times \mathbb{R}^N$ to be constant in the extra variable:
\[
u_N(x,y) = u_+(x), \quad (x,y) \in \Gamma_+\times \mathbb{R}^N.
\]
Denote the spherical cross-section of the cone $\Gamma_+ \times \mathbb{R}^N\subset \mathbb{R}^{m}$ by
\[
\Sigma_m = \{(x,y) \in \Gamma_+\times \mathbb{R}^N: |x|^2 + |y|^2 = 1\},
\]
where $m = n+N$. Let $d\sigma_m$, $\nabla_m$, $\Delta_m$ denote the spherical
measure, gradient and Laplace-Beltrami operator on the unit $(m-1)$ sphere in
$\mathbb{R}^m$.
Then, setting $\alpha = \alpha_+$, $u_N$ is homogeneous of degree
$\alpha$ in the product cone in $\mathbb{R}^m$, and so by separation of variables,
\[
\Delta_m u_N = -\alpha(\alpha+ m-2) u_N \quad \mbox{on}
\quad  \Sigma_m.
\]
Integrating by parts, using the boundary conditions, we get
\begin{equation}\label{eq:sphere-eigenvalue}
\int_{\Sigma_m} |\nabla_m u_N|^2 \, d\sigma_m = \alpha(\alpha+m-2)\int_{\Sigma_m} u_N^2 \, d\sigma_m \ .
\end{equation}
We rewrite this as
\[
\alpha(\alpha+ N+ n-2) =
\frac{\int_{\Sigma_m} |\nabla_m u_N|^2 \, d\sigma_m}
{\int_{\Sigma_m} u_N^2 \, d\sigma_m}
\]
The integrand in the numerator, on $|x|^2 + |y|^2=1$, is given by
\[
|\nabla_m u_N|^2
= |\nabla_{x,y} u_N (x,y)|^2 - |(x,y)\cdot \nabla_{x,y}u_N(x,y)|^2
= |\nabla  u_+(x))|^2 - |x\cdot \nabla u_+(x)|^2 =
|\nabla  u_+(x)|^2 - (\alpha \, u_+(x))^2,
\]
and the formula for the integral of a function $f$ on $\Sigma_m$ that depends
only on $x$ is
\[
\int_{\Sigma_m} f(x) \, d\sigma_m(x,y) = c_N \int_{\Gamma_+ \cap \{|x|<1\}} f(x) (1-|x|^2)^{N/2 - 1} dx.
\]
Therefore,
\[
\frac{\alpha(\alpha+ N+n-2)}{N} =
\frac{\frac1N  \int_{\Gamma_+ \cap \{|x|<1\}}
[|\nabla  u_+(x))|^2 - (\alpha u_+(x))^2] (1-|x|^2)^{N/2 - 1} dx}
 {\int_{\Gamma_+ \cap \{|x|<1\}} u_+(x)^2 (1-|x|^2)^{N/2 - 1} dx} \ .
 \]
Notice that the left hand side is $\alpha + O(1/N)$ as $N\to \infty$.

Next, change variables by setting $f_N(z) = u_+(z/\sqrt{N})$, $|z| < \sqrt{N}$,
to obtain
\begin{align*}
\alpha \
& = \
\frac{\int_{\Gamma_+ \cap \{|z|<\sqrt{N}\}}
[|\nabla  f_N(z)|^2 - \frac1N |\alpha \, f_N(z)|^2] \left(1-\frac{|z|^2}N\right)^{N/2 - 1} dz}
{\int_{\Gamma_+ \cap \{|z|<\sqrt{N}\}}
f_N(z)^2 \left(1-\frac{|z|^2}N\right)^{N/2 - 1}
dz} \ + \ O(1/N) \\
& = \
\frac{\int_{\Gamma_+ \cap \{|z|<\sqrt{N}\}}
|\nabla  f_N(z)|^2 \left(1-\frac{|z|^2}N\right)^{N/2 - 1} dz}
{\int_{\Gamma_+ \cap \{|z|<\sqrt{N}\}}
f_N(z)^2 \left(1-\frac{|z|^2}N\right)^{N/2 - 1}
dz} \ + \ O(1/N)
\end{align*}
Because
\[
\left(1-\frac{|z|^2}N\right)^{N/2 - 1}  \to e^{-|z|^2/2} \ \mbox{as} \ N\to \infty,
\]
this nearly completes the proof.  The additional property we need
to check is that as $N \to \infty$,
\begin{equation}\label{eq:cut-off}
\frac{\int_{\Gamma_+ \cap \{\sqrt{N} - 1 <|z|<\sqrt{N}\}}
[f_N(z)^2 + |\nabla f_N(z)|^2 ] \left(1-\frac{|z|^2}N\right)^{N/2 - 1} dz}
{\int_{\Gamma_+ \cap \{|z|<\sqrt{N}\}}f_N(z)^2 \left(1-\frac{|z|^2}N\right)^{N/2 - 1} dz}
\to 0.
\end{equation}
An estimate like \eqref{eq:cut-off} is required because $f_N$ is not a suitable
test function.  Although $f_N$ does vanish on $\gamma^+_D$ as required, it
does not vanish on the outer boundary $|z| = \sqrt{N}$.
The simplest truncation is by a radial cut-off function of slope $1$
on a band of unit width
$\sqrt{N} - 1 < |z| < \sqrt{N}$, which gives a legitimate test function and proves our proposition, assuming \eqref{eq:cut-off} holds.

Finally, \eqref{eq:cut-off}  follows from the fact that $f_N$ and $\nabla f_N$
grow like powers of $|z|$, whereas the weight resembles $e^{-|z|^2/2}$.   In detail, set
\[
A = \int_{\Sigma_n} u_+^2 \, d\sigma_n .
\]
Then, since $u_+$ is homogeneous of degree $\alpha$,
\[
\int_{\Gamma_+ \cap \{|z|<\sqrt{N}\}}f_N(z)^2 w(|z|)\, dz
=  A \int_0^{\sqrt{N}} \frac{r^{2\alpha}}{N^\alpha} w(r) r^{n-1} \, dr
\]
By equation \eqref{eq:sphere-eigenvalue} for $m=n$, $N=0$, we have
\[
\int_{\Sigma_n} |\nabla u_+|^2 \, d\sigma_n = \alpha(\alpha  + n-2) A.
\]
Thus,  since $\nabla u_+$ is homogeneous of degree $\alpha-1$,
\[
\int_{\Gamma_+ \cap \{|z|<\sqrt{N}\}}  |\nabla f_N(z)|^2   w(|z|) dz
=
\alpha(\alpha  + n-2) A \int_0^{\sqrt{N}} \frac{r^{2\alpha-2}}{N^\alpha} w(r) r^{n-1} \, dr.
\]
Note that the denominator $N^\alpha$ is the same in both formulas because
of the choice of change of variable $z/\sqrt{N}$.

With these formulas for the numerator and denominator,
one sees that \eqref{eq:cut-off} is valid.  Indeed, setting $w(r) = \left(1-r^2/N\right)^{N/2-1}$,
\[
\int_{\sqrt{N}-1}^{\sqrt{N}} ( r^{2\alpha} + r^{2\alpha-2}) (1-r^2/N)^{N/2-1} r^{n-1}\, dr
\]
tends to zero very fast since $(1-r^2/N)^{N/2-1} \approx e^{-r^2/2} \approx e^{-N/2}$
in the range $\sqrt{N} - 1 < r < \sqrt{N}$, whereas
\[
\int_{0}^{\sqrt{N}} r^{2\alpha}  (1-r^2/N)^{N/2-1} r^{n-1}\, dr
\]
tends to a positive limit.  This concludes the proof of Proposition \ref{prop:main}.

We are now ready to prove the Friedland-Hayman inequality in the
original case with $\Gamma = \mathbb{R}^{n}$.
\begin{defn} \label{def:full}
For $\Omega\subset\mathbb{R}^{n}$ open, with Gaussian measure satisfying $\int_{\Omega}e^{-|x|^2/2}\,\mathrm{d} x <1$, define $\lambda_g(\Omega)$ by
\begin{align*}
\lambda_g(\Omega) =  \inf_{w\in H_0^1(\Omega)} \frac{\int_{\Omega}\left| \nabla w(x) \right|^2 e^{-|x|^2/2} \, dx}{\int_{\Omega} w(x)^2 e^{-|x|^2/2} \, dx}.
\end{align*}
\end{defn}
Alternatively, $\lambda_g(\Omega)$ is the first eigenvalue of
$$
\begin{array}{rll}
-(\partial/\partial x_i) \left(e^{-|x|^2/2} \partial u/\partial x_i\right)  = &
\lambda \, e^{-|x|^2/2}u(x)  & \emph{ in } \Omega \\
u(x)  = & 0 &  \emph{ on } \partial\Omega.
\end{array}
$$
Via Gaussian symmetrization, the following theorem holds:
\begin{thm}[Erhard, Proposition 2.3 in \cite{Er} and Carlen, Kerce, Theorem 3 in \cite{CK}] \label{thm:BCF}
Let $\Omega$ and $\lambda_g$ be as in \emph{Definition \ref{def:full}}. Let $H$ be a half-space with the same Gaussian measure as $\Omega$, then
\begin{align*}
\lambda_g(\Omega) \geq \lambda_g(H),
\end{align*}
with equality if and only if $\Omega$ is equal to $H$, up to a rotation.
\end{thm}
For half spaces, the corresponding eigenfunction is a function of a single variable. Therefore, we can apply the results of Beckner, Kenig, and Pipher.
\begin{thm}[Beckner, Kenig, Pipher \cite{BKP}; in Section 12.2 of \cite{CaS}]
\label{thm:BKP}
Let $H_\pm$ be complementary half-spaces. Then,
\begin{align*}
\lambda_g(H_+) + \lambda_g(H_-) \geq 2,
\end{align*}
with equality if and only if $\int_{H_+}e^{-|x|^2/2} \, dx = \int_{H_-}e^{-|x|^2/2} \, dx$.
\end{thm}
In particular, combining these two theorems with Proposition \ref{prop:main} establishes the original Friedland-Hayman inequality and classifies the case of equality. For the general case where $\Gamma$ is a proper, convex subset of $\mathbb{R}^n$, we will use Theorems \ref{thm:Caf} and \ref{thm:DPF} to compare the Gaussian eigenvalues on $\Gamma_{\pm}$ to those on complementary subsets of $\mathbb{R}^{n}$. In particular, the proof is now similar to that of Theorem \ref{thm:PW} with some small modifications coming from the fact that we have to consider eigenvalues on complementary subsets.

We first apply Theorem \ref{thm:Caf} with $K = \Gamma$ and $F(x)\equiv0$. This gives a transport map $T = \nabla \varphi$ from $\mu = e^{-|x|^2/2}\, dx$ to $c_{n,K}\nu$, with $\nu = e^{-|x|^2/2}1_\Gamma \, dx$ (and $c_{n,K}$ a constant ensuring $c_{n,K}\nu$ and $\mu$ have the same total measure), such that $T$ is Lipschitz with Lipschitz constant bounded above by $1$. Let $w_{\pm}$, defined on $\Gamma_{\pm}$ respectively, be admissible test functions for the infimum given in Proposition \ref{prop:main}. Extending $w_{\pm}$ by zero so that they are defined on $\Gamma$, we set $v_+(x) = w_+(T(x))$, $v_-(x) = w_-(T(x))$. Since $T$ is a transport map, we have
\begin{align} \label{eqn:transport1}
\int_{\mathbb{R}^n} v_{\pm}(x)^2  e^{-|x|^2/2}\, dx = \int_{\mathbb{R}^n}w_{\pm}(T(x))^2 e^{-|x|^2/2} \, dx = c\int_{\Gamma} w_i(x)^2 \,d \nu.
\end{align}
$T$ is Lipschitz, with Lipschitz constant bounded by $1$, and so
\begin{align} \label{eqn:transport2}
\left|\nabla v_{\pm}(x)\right| \leq \left|\left(\nabla w_{\pm}\right)(T(x))\right| \left|\nabla T(x)\right| \leq  \left|\left(\nabla w_{\pm}\right)(T(x))\right|.
\end{align}
Combining \eqref{eqn:transport1} and \eqref{eqn:transport2} gives
\begin{align} \label{eqn:transport3}
\frac{\int_{\Gamma_{+}}\left| \nabla w_+(x) \right|^2 e^{-|x|^2/2} \, dx}{\int_{\Gamma_{+}} w_+(x)^2  e^{-|x|^2/2} \, dx} \geq \frac{\int_{\mathbb{R}^{n}}\left| \nabla v_+(x) \right|^2  e^{-|x|^2/2} \, dx}{\int_{\mathbb{R}^{n}} v_+(x)^2 e^{-|x|^2/2} \,\mathrm{d} x},
\end{align}
and the same for $\Gamma_{-}$, $w_-$ and $v_-$. Since $w_1$ and $w_2$ have disjoint supports in the cone $\Gamma$, so do the functions $v_1$ and $v_2$ in $\mathbb{R}^{n}$. Applying Theorems \ref{thm:BCF} and \ref{thm:BKP} thus implies that
\begin{align*}
\frac{\int_{\mathbb{R}^{n}}\left| \nabla v_+(x) \right|^2  e^{-|x|^2/2} \, dx}{\int_{\mathbb{R}^{n}} v_+(x)^2 e^{-|x|^2/2} \, dx} + \frac{\int_{\mathbb{R}^{n}}\left| \nabla v_-(x) \right|^2  e^{-|x|^2/2} \,\mathrm{d} x}{\int_{\mathbb{R}^{n}} v_-(x)^2  e^{-|x|^2/2} \, dx} \geq 2,
\end{align*}
and by Proposition \ref{prop:main}, this proves the desired inequality in Theorem \ref{thm:main}.

We now turn to the case of equality. Suppose that
the cone $\Gamma$ does not contain a line.  Then by Theorem \ref{thm:DPF}, there exist $\epsilon>0$ and a set $U_1\subset\mathbb{R}^{n}$ of positive measure for which $\lambda_{j}(D^2\varphi(x))\leq1-\epsilon$ for all $x\in U_1$, $1\leq j \leq n$. The image of $U_1$ under $T$ is also of positive measure, and so without loss of generality, assume that $T(U_1)\cap \Gamma_{+}$ has positive measure. Letting $w_+$ be the minimizer for the infimum in Proposition \ref{prop:main} for $\Gamma_{+}$, we therefore have
\begin{align*}
\left|\nabla v_+(x)\right| \leq (1-\epsilon) \left|\left(\nabla w_+\right)(T(x))\right| \quad \text{on } T^{-1}\left(T(U_1)\cap \Gamma^{+}\right).
\end{align*}
Inserting this strict inequality into the argument above ensures that we cannot have the equality $\alpha_{+}+\alpha_{-} = 2$.

Now suppose that we have equality. Then we can write the cone $\Gamma$ in the form $\Gamma = \mathbb{R}^{k}\times \Gamma'$, where $\Gamma'$ does not contain a line. In particular, by Theorem \ref{thm:DPF}, we have $T(x) = (x_1-p_1,\ldots,x_k-p_k,T'(x_{k+1},\ldots,x_{n+1}))$ for some fixed $(p_1,\ldots,p_k)\in\mathbb{R}^k$ and with  $|\nabla T'|<1$ on a set of positive measure $U_2$ in $\mathbb{R}^{n-k}$. Assuming that $T'(U_2)\cap \Pi'_{n-k}\Gamma_{+}$ has positive measure in $\mathbb{R}^{n-k}$ (where $\Pi'_{n-k}\Gamma_+ = \{x'\in \mathbb{R}^{n-k}:(x_1,\ldots,x_k,x')\in \Gamma_+$ for some $(x_1,\ldots,x_k)\in\mathbb{R}^k\}$), in order to have equality, the minimizer $w_{+}$ for $\Gamma_{+}$ in Proposition
\ref{prop:main} must depend only on the variables $x_1,\ldots,x_k$. Therefore, $v_{+}(x) = w_{+}(x-p)$ for some $p\in R^n$, and by Theorem \ref{thm:BCF}, $w_{+}$ is an explicit ODE solution depending only on one variable and $\Gamma_{+}$ equals  the intersection of $\Gamma$ with a half-space. That is we can write $\Gamma_{+} = H\times \Gamma'$, $\Gamma_{-} = \left(\mathbb{R}^k\backslash H\right)\times \Gamma'$ for a half-space $H\subset \mathbb{R}^k$. This in particular ensures that $T'(U_2)\cap\Pi'_{n-k} \Gamma_{-}$ also has positive measure in $\mathbb{R}^{n-k}$, and so for equality $w_{-}$ depends only on the variables $x_1,\ldots,x_k$. We have therefore reduced to the setting of Theorem \ref{thm:BKP} in $\mathbb{R}^k$, and so for equality, after a rotation, we have $\Gamma_{+} = \{x_1>0\}\cap \Gamma$ and $\Gamma_{-} = \{x_1<0\}\cap \Gamma$.

\section{Remarks on transport maps on the sphere $\mathbb{S}^{n-1}$}

To prove the Friedland-Hayman type inequality, Theorem \ref{thm:main}, we used Proposition \ref{prop:main} to obtain a lower bound on the characteristic exponents $\alpha_\pm $ in terms of Gaussian eigenvalues. This then allowed us to apply the Caffarelli contraction theorem, Theorem \ref{thm:Caf}. A possible alternative to this is to work directly on the sphere $\mathbb{S}^{n-1}$ and prove an analogous version of the Caffarelli contraction theorem on the sphere. In \cite{Mc1}, McCann shows the following: Let $(M,g)$ be a smooth, compact manifold without boundary, with distance function $d(x,y)$, and let $\mu_1$, $\mu_2$ be probability measures on $M$. There exists a unique mapping $T$ transporting $\mu_1$ onto $\mu_2$ minimizing the total cost with respect to the cost function $c(x,y) = d(x,y)^2/2$. The mapping $T(x) = \exp_{x}\left[-\nabla \psi(x)\right]$ for a $c$-concave function $\psi$, where one defines a function $\psi$ to be $c$-concave if $\psi = \left(\psi^c\right)^c$, for
\begin{align*}
\psi^c(y) = \inf_{x\in M} c(x,y) - \psi(x).
\end{align*}
Now let $\mu$ be the uniformly distributed probability measure on a hemisphere
$\mathbb{S}^{n-1}_{+}$ in $\mathbb{S}^{n-1}$, the appropriate multiple of the volume form
of the round metric. Define the measure $\nu$ by
\begin{align*}
\nu = e^{-f}\mu\big|_{W}.
\end{align*}
Here $W$ is a (geodesically) convex subset of $\mathbb{S}^{n-1}_+$ and $f$ is a convex function on $W$, such that $\nu$ is a probability measure. It is then natural to pose the following questions, for which the tools of Fathi et al. \cite{FGP} seem appropriate:
\\
\\
1) Does there exist a transport map $T(x) = \exp_{x}\left[-\nabla \psi(x)\right]$ from $\mu$ to $\nu$ for a $c$-concave function $\psi$ such that $T$ is Lipschitz on $\mathbb{S}^{n-1}_{+}$ with Lipschitz constant bounded by $1$?
\\

\noindent
2) If, at almost every point of $\mathbb{S}^{n-1}_+$, there is a direction in which $T$ is not a strict contraction, then is it true that, after a rotation, $W$ contains the antipodal points $(\pm 1,0,\ldots,0)$ in
$\mathbb{S}^{n-1}_+$ and that $f$ is independent of the $x_1$-variable? \\

\noindent
3)  If there is a pair of points $x$ and $y$ for which $d(x,y) = d(T(x), T(y))$,
then does $W$ contain the full geodesic through $x$ and $y$,
all the way to the antipodal points? And does $T$ split in that direction as in
Question (2)?  Is there an infinitesimal version of this phenomenon at
a single point?

redpajama_set_name: RedPajamaArXiv