text:
\begin{document}

\title {Extremes of Spherical Fractional Brownian Motion}
\author{Dan Cheng \thanks{Supported by NSF Grant DMS-1811632.} \\ Arizona State University
 \and Peng Liu \thanks{Supported by Swiss National Science Foundation Grant 200021-175752/1.} \\ University of Lausanne}

\maketitle

\begin{abstract}
Let $\{B_\beta (x), x \in \mathbb \S^N\}$ be a fractional Brownian motion on the $N$-dimensional unit sphere $\S^N$ with Hurst index $\beta$. We study the excursion probability $\pk{\sup_{x\in T} B_\beta(x) > u }$ and obtain the asymptotics as $u\to \infty$, where $T$ can be the entire sphere $\S^N$ or a geodesic disc on $\S^N$.
\end{abstract}

\noindent{\small{\bf Keywords}: Excursion probability, Pickands constant, Sphere, Piterbarg constant, SFBM, Fractional Brownian motion, Gaussian random fields, Asymptotics.}

\noindent{\small{\bf Mathematics Subject Classification}:\ 60G15, 60G70.}

\section{Introduction}
Let $\{X(t), t\in T \}$ be a real-valued Gaussian random field living on some parameter space $T$. The extremes, especially excursion probabilities $\P\{\sup_{t\in T} X(t)> u \}$, of the field have been extensively studied in the literature due to the importance in both probability theory \cite{Rice:1945,ChanL06,Debicki:2016,HP99,Liu12,Sun93} and statistical applications such as the p-value computation for controlling the family-wise error \cite{TaylorW07,TaylorW08}, nonparametric density estimation \cite{Bickel:1973,Huckemann:2016,Qiao:2017} and construction of confidence bands \cite{MaYang:2012,WangYang:2007}. We refer to the survey \cite{Adler00} and monographs \cite{Piterbarg:1996,Adler07,AzaisBook09} for the history, recent developments and more related applications on this subject.

Recently, the study of random fields on spheres is attracting more and more attention due to vast applications in astronomy \cite{MP11}, spatial statistics \cite{Gelfand:2010,Stein:1999}, geoscience \cite{OhLi04,Jeong:2017} and environmental sciences \cite{Stein:2007}. In particular, Istas \cite{Istas:2005,Istas:2006} introduced spherical fractional Brownian motion (abbreviated as SFBM throughout this paper) on spheres and studied the Karhunen-Lo\`eve expansion and other properties. As an important extension to the classical fractional Brownian motion on Euclidean space, it would be very useful and valuable to study the excursion probability of SFBM, which is the main purpose of this paper.

Let $o$ be a fixed point on the $N$-dimensional unit sphere $\mathbb{S}^N \subset \R^{N+1}$. The SFBM, denoted by $B_\beta= \{B_\beta (x), x \in \mathbb S^N\}$, is defined in Istas \cite{Istas:2005} as a centered real-valued Gaussian random field on $\S^N$ such that $B_\beta(o)=0$ and
\begin{equation}\label{eq:SFBM-var}
\E[B_\beta(x)-B_\beta(y)]^2= d^{2\beta}(x,y), \quad \forall x,y\in \mathbb{S}^N,
\end{equation}
where the index $\beta\in (0, 1/2]$ and  $d(\cdot,\cdot)$ is the spherical distance on $\mathbb{S}^N$, that is $d(x,y)= \arccos \l x, y \r$, $\forall x, y\in \mathbb{S}^N$. Here $\l \cdot, \cdot \r$ is the usual inner product in $\R^{N+1}$. It follows immediately that the covariance structure is given by
\begin{equation}\label{eq:SFBM-cov}
{\rm Cov}(B_\beta(x), B_\beta(y))=\frac{1}{2}\big(d^{2\beta}(x,o)+d^{2\beta}(y,o)-d^{2\beta}(x,y) \big).
\end{equation}

In this paper, we shall study the asymptotics of the excursion probability $\P\{\sup_{x\in T} B_\beta(x)> u \}$ as $u\to \infty$. Two cases for the parameter set $T$ are considered separately: (i) $T=\S^N$ and (ii) $T=T_a:=\{x\in\mathbb{S}^N: d(x,o)\le a\}$, where $a\in (0, \pi)$. In other words, $T_a$ is the geodesic disc on $\S^N$ of radius $a$ centered at $o$, so that $T_a=\S^N$ when $a=\pi$. Notice that, the maximum of the variance function of $B_\beta(x)$ over $T$ will be attained at a single point for case (i) and on the boundary set $\{x\in\mathbb{S}^N: d(x,o)= a\}$, which is in fact an $(N-1)$-dimensional sphere, for case (ii), respectively, making the latter case more challenging.

Since the sphere $\S^N$ is not an Euclidean space, it would be hard to apply directly the traditional double sum method over $\S^N$ to derive the asymptotics of the excursion probability. Instead, we shall apply the main technique in Cheng and Xiao \cite{ChengXiaoSphere} to consider the SFBM as a Gaussian random field on Euclidean space by using spherical coordinate transformation. In such way, we can study the local behaviors of the standard deviation and correlation functions of the field under spherical coordinates (see Lemmas \ref{lemA} and \ref{lemB} below), and then apply the results in Euclidean space (see Lemma \ref{Piterbarg} and Theorem \ref{extension} below) to derive the desired asymptotics of the excursion probabilities in Theorems \ref{Thm-Sphere} and \ref{ThA}. In particular, for case (ii), the maximum of variance is attained on a set of dimension at least one (when $N\ge 2$) and there is no known asymptotic result in the literature except for the two-dimensional case studied in \cite{Debicki:2016}. In order to obtain the asymptotics on a geodesic disc in Theorem \ref{ThA}, we establish an asymptotic result on Euclidean space in Theorem \ref{extension} which is valuable itself in extreme value theory and will have further applications in the future.

The paper is organized as follows. We first introduce the preliminaries, such as spherical coordinate transformation and the Pickands and Piterbarg constants, in Section \ref{sec:prel}; and then study the asymptotics of the excursion probabilities of $B_\beta(x)$ on the entire sphere $\S^N$ and on a geodesic disc in Sections \ref{sec:sphere} and \ref{sec:disc}, respectively. Finally, the Appendix contains some auxiliary results and the proof of Theorem \ref{extension}.

\COM{ Define next
	\begin{equation*}
	\begin{split}
	\sigma(x)&=\sqrt{{\rm Var}(B_\beta(x))}, \quad \tilde{\sigma}(\theta)=\sigma(x)\\
	r(x,y)&={\rm Cov}(B_\beta(x)/d^\beta(x,o), B_\beta(y)/d^\beta(y,o)), \quad \tilde{r}(\theta,\varphi)=r(x,y).\\
	\end{split}
	\end{equation*}
}

\section{Preliminaries}\label{sec:prel}
\subsection{Spherical Coordinates and Notations}\label{subsection:SphereCoor}
For $x=(x_1, \ldots, x_N, x_{N+1})\in \mathbb{S}^N$, its corresponding spherical coordinate $\theta=(\theta_1, \ldots, \theta_N)$ is defined by the following way.
\begin{equation}\label{Eq:spherical coordinate}
x_1 = \cos \theta_1, \quad \ldots, \quad x_{N} = \left(\prod_{i=1}^{N-1}\sin \theta_i \right) \cos \theta_{N}, \quad x_{N+1} = \prod_{i=1}^{N}\sin \theta_i,
\end{equation}
where $\theta\in \Theta:=[0, \pi]^{N-1}\times[0, 2 \pi)$.

Throughout this paper, for two points $x=(x_1, \ldots, x_{N+1})$ and $y=(y_1, \ldots, y_{N+1})$ on $\mathbb{S}^N$, we always denote by $\theta=(\theta_1, \ldots, \theta_N)$ and $\varphi=(\varphi_1, \ldots, \varphi_N)$ the spherical coordinates of $x$ and $y$, respectively. For functions $f(x)$ and $g(x,y)$, $x, y\in \S^N$, we denote by $\tilde{f}(\theta):=f(x)$ and $\tilde{g}(\theta,\varphi):=g(x,y)$ the corresponding functions of $f(x)$ and $g(x,y)$ under spherical coordinates, respectively.

Let $d(\cdot,\cdot)$ denote the spherical distance on $\mathbb{S}^N$ and let $\|\cdot\|$ be the Euclidean norm in $\R^{N+1}$ or in $\R^N$, which will be clear from the context. For a set $D\subset \R^N$, denote by ${\rm mes} (D)$ the measure (volume) of $D$. Denote by $\Psi(u)$ the tail probability of standard normal distribution, that is $\Psi(u)=(2\pi)^{-1/2}\int_u^\infty e^{-v^2/2}dv$. For any two real-valued functions $h_1(u)$ and $h_2(u)$, we say $h_1(u) \sim h_2(u)$ as $u\to u_0\in [-\infty, +\infty]$ if $\lim_{u\to u_0} h_1(u)/h_2(u) =1$.

\subsection{Pickands and Piterbarg Constants}
Let $\{\chi_H(t)$, $t\in \mathbb{R}^N\}$, $H\in (0,1]$, be a  Gaussian random field with mean function
$$\mathbb{E}\left(\chi_H(t)\right)=-\|t\|^{2H}, \quad t\in \R^N,$$
and covariance
$${\rm Cov}(\chi_H(t), \chi_H(s))=\|t\|^{2H}+\|s\|^{2H}-\|t-s\|^{2H}, \quad t,s\in \R^N.$$
Let $\mathcal{H}_{2H}(E)=\mathbb{E}\left\{ \sup_{t\in E }\exp[\chi_H(t)]\right\}$, where $E\subset \R^N$ is a compact set. The \textit{Pickands constant} \cite{Piterbarg:1996} is defined by
\begin{equation}\label{eq:Pickands-Const}
\mathcal{H}_{2H}^N:=\lim_{S\to\infty}\frac{\mathcal{H}_{2H}^N([0,S])}{S},\quad \text{where }  \mathcal{H}_{2H}^N([0,S])=\lim_{S_1\to\infty}\frac{\mathcal{H}_{2H}([0,S]\times[0,S_1]^{N-1})}{S_1^{N-1}}.
\end{equation}
The \textit{Piterbarg constant} \cite{Bai:2018, Piterbarg:1996} is defined by
\begin{equation}\label{eq:Piterbarg-Const}
\mathcal{P}_{2H}^g :=\lim_{S\to\infty}\mathcal{P}_{2H}^g([-S,S]^N), \quad \text{where }\mathcal{P}_{2H}^g(E)=\mathbb{E}\left\{ \sup_{t\in E }e^{\chi_H(t)-g(t)}\right\},
\end{equation}
$E\subset \R^N$ is a compact set and $g$ is a continuous function over $\mathbb{R}^N$. Moreover, let
\begin{equation}\label{eq:M-Const}
\begin{split}
\mathcal{M}_{2H}^g &:=\lim_{S\to\infty}\lim_{S_1\to\infty}\frac{\mathcal{P}_{2H}^g([-S,S]\times[0,S_1]^{N-1})}{S_1^{N-1}},\\
\widehat{\mathcal{M}}_{2H}^g &:=\lim_{S\to\infty}\mathcal{M}_{2H}^g([0,S]), \quad \text{where } \mathcal{M}_{2H}^g([0,S])=\lim_{S_1\to\infty} \frac{\mathcal{P}_{2H}^g([0,S]\times[0,S_1]^{N-1})}{S_1^{N-1}},
\end{split}
\end{equation}
if the limits above exist.

\COM{\subsection{Excursion Probability over the Whole Sphere}
	Once we find the local structure of $\tilde{\sigma}(\theta)$ and $\tilde{r}(\theta,\varphi)$ near $\theta_0$ [see (\ref{Eq:sd sphere}) and (\ref{Eq:corr sphere})], then we can apply Piterbarg's results to get the approximation for $\P\{\sup_{x\in \mathbb{S}^N} B_\beta(x) \ge u \}=\P\{\sup_{\theta\in [0, \pi]^{N-1}\times[0, 2 \pi)} \wt{B}_\beta(\theta) \ge u \}$.
}

\section{Excursion Probability on $\S^N$}\label{sec:sphere}
We first study the excursion probability of $B_\beta$ over the entire sphere. Recall the notations introduced in Section \ref{subsection:SphereCoor}, we have that $\pk{ \sup_{x\in \mathbb{S}^N} B_\beta(x)>u}$ is equivalent to $\P\{ \sup_{\theta\in \Theta} \widetilde{B}_\beta(\theta)>u\}$, where $\wt{B}_\beta(\theta) := B_\beta(x)$ and $\Theta=[0, \pi]^{N-1}\times[0, 2 \pi)$ is an $N$-dimensional rectangle on $\R^N$. Therefore, to establish the asymptotics for the excursion probability of $B_\beta$, we will study the properties of the standard deviation and correlation functions of $\widetilde{B}_\beta$, which is a Gaussian random field living on $\Theta$, and then apply results on extremes for Gaussian random fields on Euclidean space.

In this section, we assume without loss of generality that $B_\beta$ starts at $o=(0,0,\ldots,1,0)\in \R^{N+1}$ whose spherical coordinate is given by $(\pi/2,\ldots,\pi/2,0)\in \Theta \subset \R^N$ according to \eqref{Eq:spherical coordinate}. Denote by $\sigma(x)$ the standard deviation function of $B_\beta(x)$. By \eqref{eq:SFBM-var},
\bqn{ \label{sigmax0}
	\sigma(x)=d^\beta(x,o)= \arccos^{\beta} \l x, o \r = \arccos^{\beta} (x_N), \quad x\in \S^N,}
which attains its unique maximum $\pi^\beta$ at $p:=(0,0,\ldots,-1,0)\in \R^{N+1}$ whose spherical coordinate is given by $\theta_0:=(\pi/2,\ldots,\pi/2,\pi)$. Note that, by \eqref{sigmax0}, we have the following standard deviation function under spherical coordinates,
\bqn{ \label{sigmax}
	\tilde{\sigma}(\theta):=\sigma(x)=\arccos^{\beta} (x_N)
	= \arccos^{\beta} \left(\left(\prod_{i=1}^{N-1}\sin \theta_i\right) \cos \theta_{N}\right),
}
which attains its unique maximum at the interior point $\theta_0\in \Theta$ above. Additionally, it follows from \eqref{eq:SFBM-var} and \eqref{eq:SFBM-cov} that the correlation function of $B_\beta(x)$ becomes
\bqn{\label{corrT}
	r(x,y)= \frac{d^{2\beta}(x,o)+d^{2\beta}(y,o)-d^{2\beta}(x,y)}{2d^\beta(x,o)d^\beta(y,o)}, \quad x, y \in \S^N,
}
whose form under spherical coordinates, denoted by $\tilde{r}(\theta,\varphi)$, can be obtained accordingly.

\begin{remark}
	We choose the starting point of $B_\beta(x)$ at $o=(0,0,\ldots,1,0)\in \R^{N+1}$ to make sure that the maximum of the variance function of $\widetilde{B}_\beta(\theta)$ will be attained at an interior point in $\Theta$. This will simplify a lot the arguments on deriving the asymptotics for the excursion probability. Note that the choice of starting point $o$ does not affect our results since the asymptotics of the excursion probability is determined only by the behavior of the field around the points attaining the maximum of the variance function. \qed
\end{remark}

We first derive a result below showing the local behaviors of the standard deviation and correlation functions of $\widetilde{B}_\beta(\theta)$, the SFBM under spherical coordinates, around $\theta_0$.
\BEL\label{lemA}
Let $\theta_0=(\pi/2,\ldots,\pi/2,\pi)\in \Theta \subset \mathbb{R}^N$. Then
\begin{equation}\label{Eq:sd sphere}
\begin{split}
\tilde{\sigma}(\theta)= \pi^{\beta}-\beta \pi^{\beta-1}\|\theta-\theta_0\|(1+o(1)), \quad \text{\rm as } \|\theta-\theta_0\|\to 0;
\end{split}
\end{equation}
and
\begin{equation}\label{Eq:corr sphere}
\begin{split}
\tilde{r}(\theta,\varphi)&= 1-\frac{\|\varphi-\theta\|^{2\beta}}{2\pi^{2\beta}}(1+o(1)), \quad \text{\rm as } \|\theta-\theta_0\|\vee\|\varphi-\theta_0\|\to 0.
\end{split}
\end{equation}
\EEL
\begin{proof} \, Note that, as $x_N \downarrow -1$,
\begin{equation}\label{eq:arccos}
\begin{split}
\arccos (x_N)-\arccos(-1)&= \int_{-1}^{x_N} \frac{-1}{\sqrt{1-t^2}}dt=\int_{0}^{1+x_N} \frac{-1}{\sqrt{2t-t^2}}dt\\
&\sim -\int_{0}^{1+x_N} \frac{1}{\sqrt{2t}}dt =-\sqrt{2(1+x_N)}.
\end{split}
\end{equation}
It then follows from \eqref{eq:arccos} and Taylor's expansion that, as $\|\theta-\theta_0\|\to 0$,
\begin{equation*}
\begin{split}
\tilde{\sigma}(\theta)-\pi^{\beta}&=\arccos^{\beta} (\prod_{i=1}^{N-1}\sin \theta_i \cos \theta_{N})-\pi^{\beta}\sim \beta \pi^{\beta-1}\left(\arccos (\prod_{i=1}^{N-1}\sin \theta_i \cos \theta_{N})-\arccos(-1)\right)\\
&\sim -\sqrt{2}\beta \pi^{\beta-1}\sqrt{1+\prod_{i=1}^{N-1}\sin \theta_i \cos \theta_{N}}\sim -\beta \pi^{\beta-1}\sqrt{\sum_{i=1}^{N-1}|\theta_i-\pi/2|^2 + |\theta_N-\pi|^2},
\end{split}
\end{equation*}
yielding \eqref{Eq:sd sphere}.

We derive next the expansion for the correlation function. First note that
\begin{equation}\label{eq:1-corr}
1-r(x,y)=\frac{d^{2\beta}(x,y)-\left(d^{\beta}(x,o)-d^{\beta}(y,o)\right)^2}{2d^\beta(x,o)d^\beta(y,o)}.
\end{equation}
As $x, y \to p=(0,0,\ldots,-1,0)\in \R^{N+1}$, which is equivalent to $\theta, \varphi \to \theta_0$, by Taylor's formula,
\begin{equation*}
\left(d^{\beta}(x,o)-d^{\beta}(y,o)\right)^2\sim \beta^2\pi^{2(\beta-1)}(d(x,o)- d(y,o))^2.
\end{equation*}
Combining this with the triangle inequality such that $|d(x,o)- d(y,o)|\le d(x,y)$, we obtain that for $\beta\in (0,1/2]$,
\[
\left(d^{\beta}(x,o)-d^{\beta}(y,o)\right)^2 = o\left(d^{2\beta}(x,y)\right), \quad \text{as} \ x, y \to p,
\]
implying $1-r(x,y) \sim d^{2\beta}(x,y)/(2\pi^{2\beta})$ by \eqref{eq:1-corr}. Note also that by Lemma 2.1 in Cheng and Xiao \cite{ChengXiaoSphere}, as $x, y \to p$, $d(x,y)\sim \|\theta-\varphi\|$. Therefore,
\[
1-\tilde{r}(\theta,\varphi)=1-r(x,y) \sim \frac{d^{2\beta}(x,y)}{2\pi^{2\beta}} \sim \frac{\|\varphi-\theta\|^{2\beta}}{2\pi^{2\beta}},
\]
yielding \eqref{Eq:corr sphere}.
\end{proof}

For convenience, we present here a simpler version of Theorem 8.2 in Piterbarg \cite{Piterbarg:1996}.
Let $\{X(t), t\in E\}$, where $E\subset \mathbb{R}^N$ is a compact set, be a centered Gaussian random field with variance function  attaining its maximum at the unique point $t_0\in E$. Moreover, there exist non-degenerate $N\times N $ matrices $A$ and $C$, and constants $\eta>0$ and $\alpha\in (0,2]$ such that
\begin{equation}\label{Var}
\sqrt{{\rm Var}(X(t))}=1-\|A(t-t_0)\|^\eta(1+o(1)), \quad \|t-t_0\|\to 0,
\end{equation}
and
\begin{equation}\label{Corr}
{\rm Corr}(X(t), X(s))=1-\|C(t-s)\|^\alpha(1+o(1)), \quad t,s\to t_0.
\end{equation}
Additionally, there exist $\gamma>0$ and $G>0$ such that
\begin{equation}\label{Holder}
\mathbb{E}\left[X(t)-X(s)\right]^2\leq G\|t-s\|^\gamma, \quad s,t\in E.
\end{equation}

\BEL\label{Piterbarg} Let $\{X(t), t\in E\}$, where $E\subset \mathbb{R}^N$ is a compact set, be a centered Gaussian random field with variance function  attaining its maximum at the unique point $t_0\in E$. Assume further that $t_0$ is an inner point of $E$ and (\ref{Var})-(\ref{Holder}) are satisfied. \\
If $\alpha<\eta$, then
\begin{equation}\label{eq:Piterbarg-1}
\pk{\sup_{t\in E}X(t)>u}\sim \mathcal{H}_{\alpha}^N\int_{\mathbb{R}^N} e^{-\|AC^{-1}t\|^\eta}dt\,u^{\frac{2N}{\alpha}-\frac{2N}{\eta}}\Psi(u), \quad u\to\infty.
\end{equation}
If $\alpha=\eta$, then
\begin{equation}\label{eq:Piterbarg-2}
\pk{\sup_{t\in E}X(t)>u}\sim \mathcal{P}_{\alpha}^{\|AC^{-1}t\|^{\alpha}}\Psi(u),\quad u\to\infty.
\end{equation}
If $\alpha>\eta$, then
$$\pk{\sup_{t\in E}X(t)>u}\sim \Psi(u), \quad u\to\infty.$$
\EEL

We are now ready to derive one of our main results as follows.
\BT \label{Thm-Sphere} Let $\{B_\beta (x), x \in \mathbb S^N\}$ be a SFBM, where $\beta\in(0,1/2]$.

(i) If $\beta\in (0,1/2)$, then
\[
	\pk{ \sup_{x\in \mathbb{S}^N} B_\beta(x)>u}
	\sim
	\mathcal{H}_{2\beta}^N\frac{N!\pi^{(2\beta-1/2) N}}{2^{\frac{N}{2\beta}}\beta^{N}\Gamma(N/2 +1)} u^{\frac{(1-2\beta)N}{\beta}}\Psi(\pi^{-\beta}u), \quad u\to\infty,
\]
where $\mathcal{H}_{2\beta}^N$ is the Pickands constant defined in \eqref{eq:Pickands-Const}.

(ii) If $\beta=1/2$, then
\[
	\pk{ \sup_{x\in \mathbb{S}^N} B_\beta(x)>u}
	\sim \mathcal{P}_{{1}}^g \Psi(\pi^{-1/2}u),\quad u\to\infty.
\]
where $\mathcal{P}_{{1}}^g$ is the Piterbarg constant defined in \eqref{eq:Piterbarg-Const} and $g(t)=\sqrt{\sum_{i=1}^Nt_i^2}$, $t\in \R^N$.
\ET

\begin{proof} \, (i) Note that
	\BQNY
	\pk{ \sup_{x\in \mathbb{S}^N} B_\beta(x)>u}=\pk{ \sup_{\theta\in \Theta } \frac{\widetilde{B}_\beta(\theta)}{\pi^\beta}>\frac{u}{\pi^{\beta}}}.
	\EQNY
	It follows from Lemma \ref{lemA} that
	\BQNY
	\begin{split}
		1-\frac{\tilde{\sigma}(\theta)}{\pi^\beta}
		&= \frac{\beta}{\pi} \|\theta-\theta_0\|(1+o(1)), \quad \|\theta-\theta_0\|\to 0,\\
		1-\tilde{r}(\theta,\varphi)&=
		\frac{1}{2\pi^{2\beta}}\|\theta-\varphi\|^{2\beta}(1+o(1)), \quad \|\theta-\theta_0\|\vee\|\varphi-\theta_0\|\to 0.
	\end{split}
	\EQNY
	Applying the identity
	$$\|x-y\|=2\sin\left(\frac{d(x,y)}{2}\right), \quad \forall x,y\in \mathbb{S}^N, $$
	there exists a positive constant $C_1$ such that
	$$d^{2\beta}(x,y)\leq C_1\|x-y\|^{2\beta}, \quad \forall x,y\in \mathbb{S}^N.$$
	Combining this inequality with \eqref{eq:SFBM-var}, there exists a positive constant $C_2$ such that
	$$\E\left[\widetilde{B}_\beta(\theta)-\widetilde{B}_\beta(\varphi)\right]^2= d^{2\beta}(x,y)\leq C_1\|x-y\|^{2\beta}\leq C_2 \|\theta-\varphi\|^{2\beta}, \quad \forall \theta, \varphi\in \Theta.$$
	Therefore, for $\beta\in (0,1/2)$, applying Lemma \ref{Piterbarg} with $\eta=1$, $\alpha=2\beta$, $A=\beta\pi^{-1}I_N$ and $C=2^{-1/(2\beta)}\pi^{-1} I_N$, we obtain
	\BQNY
	\pk{ \sup_{x\in \mathbb{S}^N} B_\beta(x)>u}\sim \mathcal{H}_{2\beta}^N\int_{\mathbb{R}^N}e^{-2^{1/(2\beta)}\beta\|s\|}ds\, v^{\frac{(1-2\beta)N}{\beta}}\Psi(v),
	\EQNY
	where $v=\pi^{-\beta}u$ and $I_N$ is the $N\times N$ identity matrix. Note that $\int_0^\IF r^{N-1}e^{-r}dr=\Gamma(N)$ and
	\begin{equation}\label{eq:area-S}
	\int_{[0,\pi]^{N-2}\times[0,2\pi]} \sin^{N-2}\theta_1\sin^{N-3}\theta_2\dots\sin\theta_{N-2}d\theta_1\cdots d\theta_{N-1} = {\rm Area}(\S^{N-1})=\frac{2\pi^{N/2}}{\Gamma(N/2)},
	\end{equation}
	one can use the spherical coordinate transformation to obtain
	\BQNY\int_{\mathbb{R}^N}e^{-\|s\|}ds=\Gamma(N)\times \frac{2\pi^{N/2}}{\Gamma(N/2)}=\frac{N!\pi^{N/2}}{\Gamma(N/2+1)}.
	\EQNY
	Therefore, as $u\to \infty$,
	\BQNY
	\pk{ \sup_{x\in \mathbb{S}^N} B_\beta(x)>u}\sim\mathcal{H}_{2\beta}^N\frac{N!\pi^{(2\beta-1/2) N}}{2^{\frac{N}{2\beta}}\beta^{N}\Gamma(N/2 +1)} u^{\frac{(1-2\beta)N}{\beta}}\Psi(\pi^{-\beta}u).
	\EQNY

	(ii) For $\beta=1/2$, applying again Lemma \ref{Piterbarg}, we have that
	\BQNY
	\pk{ \sup_{x\in \mathbb{S}^N} B_\beta(x)>u}\sim \mathcal{P}_{1}^g\Psi(\pi^{-\beta}u), \quad u\to\infty,
	\EQNY
	where $g(t)=\|t\|$, $t\in \R^N$.
\end{proof}

\section{Excursion Probability on a Geodesic Disc}\label{sec:disc}
In this section, we will study the excursion probability of $B_\beta(x)$ over a geodesic disc on $\S^N$. Without loss of generality, assume that $B_\beta(x)$ starts at $o'=(1,0,\ldots,0)\in \R^{N+1}$ whose spherical coordinate is given by $(0,\ldots,0)\in \Theta \subset \R^N$ according to \eqref{Eq:spherical coordinate}. The standard deviation function of $\widetilde{B}_\beta(\theta)$ now becomes
\bqn{ \label{Eq:sd geodesic ball}
	\tilde{\sigma}(\theta)=\sigma(x)=\arccos^{\beta} \l x, o'\r=\arccos^{\beta} (x_1) =\theta_1^\beta, \quad \theta\in \Theta.}
The geodesic disc on $\S^N$ with radius $a>0$ and center at $o'$ is defined as
\[
T_a=\{x\in\mathbb{S}^N: d(x,o')\le a\}.
\]
Since $d(x,o')=\theta_1$, the set corresponding to $T_a$ under spherical coordinates becomes
$$\Theta_a=[0,a]\times[0, \pi]^{N-2}\times[0, 2 \pi).$$
It is straightforward to check that $\tilde{\sigma}(\theta)$ attains its maximum only at
\[
\{\theta\in \Theta_a: \theta_1=a\}=\{a\}\times[0, \pi]^{N-2}\times[0, 2 \pi),
\]
which is one of the $(N-1)$-dimensional faces of the $N$-dimensional rectangle $\Theta_a$.

\begin{remark}
	We choose the starting point of $B_\beta(x)$ at $o'=(1,0,\ldots,0)\in \R^{N+1}$ to make the variance function of $\widetilde{B}_\beta(\theta)$ have a simple form so that the set attaining the maximum of variance would be easier to handle. Again, the choice of starting point does not affect our results. \qed
\end{remark}

Similarly to Lemma \ref{lemA}, we have the following result describing the local behaviors of the standard deviation and correlation functions of $\widetilde{B}_\beta(\theta)$ around $\Theta_a$.

\BEL\label{lemB}
Let $\Theta_a=[0,a]\times[0, \pi]^{N-2}\times[0, 2 \pi)$. Then
\begin{equation}\label{eqvar}
\frac{\tilde{\sigma}(\theta)}{a^\beta}=1-\frac{\beta}{a} |a-\theta_1|(1+o(1)), \quad  \theta\in \Theta_a, \ \theta_1\to a;
\end{equation}
and
\begin{equation}\label{eqcorr}
\begin{split}
\tilde{r}(\theta,\varphi)&=1-(1+o(1))\frac{1}{2a^{2\beta}}\bigg[(\varphi_1-\theta_1)^2 + (\sin^2 a)(\varphi_2-\theta_2)^2+\cdots \\
&\quad + \bigg(\sin^2 a\prod_{i=2}^{N-1} \sin^2 \theta_i\bigg)(\varphi_N-\theta_N)^2\bigg]^\beta, \quad \theta,\varphi\in \Theta_a, \ \|\theta-\varphi\|\to 0, \ \theta_1\to a.
\end{split}
\end{equation}
\EEL
\begin{proof}\, Note that \eqref{eqvar} follows immediately from Taylor's formula. By similar arguments in the proof of Lemma \ref{lemA}, we obtain
\[
1-\tilde{r}(\theta,\varphi)=1-r(x,y) \sim \frac{d^{2\beta}(x,y)}{2a^{2\beta}}.
\]
Then \eqref{eqcorr} follows from Lemma 2.1 in Cheng and Xiao \cite{ChengXiaoSphere}.
\end{proof}

Here, we present a result extending both Theorems 7.1 and 8.2 in Piterbarg \cite{Piterbarg:1996}. It is not only useful to prove Theorem \ref{ThA} below, but valuable itself in extreme value theory. The proof is given in the Appendix.

Let $\{X(t), t\in E\}$, where $E=\prod_{i=1}^N[a_i, b_i]$, be a Gaussian random field with continuous trajectories. Its standard deviation function $\sigma_X(t)$ attains the maximum $1$ at the hyperspace $E_0=\{t_1^*\}\times \prod_{i=2}^N[a_i,  b_i]$, where $t_1^*\in [a_1, b_1]$, and satisfies
\BQN\label{var}
\lim_{|t_1-t_1^*|\to 0}\sup_{\substack{t_1\neq t_1^*\\ \hat{t}\in \prod_{i=2}^N[a_i,  b_i]} }\left|\frac{1-\sigma_X(t)}{h(\hat{t})|t_1-t_1^*|^{\gamma}}-1\right|=0,
\EQN
where $\gamma>0$ and $h(\hat{t})$, $\hat{t}\in \prod_{i=2}^N[a_i,  b_i]$, is a positive continuous function with $\hat{t}=(t_2,\dots, t_N)$.
Moreover,
\BQN\label{cor}
\lim_{\delta\to 0, \, u\to\infty}\sup_{\substack{s\neq t, \, s,t\in E_u\\ \|s-t\|\leq \delta} }\left|\frac{1-r(s,t)}{ \left(c_1(t_1-s_1)^2+\sum_{i=2}^Nc_i(\hat{t})(t_i-s_i)^2\right)^\beta}-1\right|=0,
\EQN
where $E_u=\left([t_1^*-((\log u)/u)^{2/\gamma}, t_1^*+((\log u)/u)^{2/\gamma}]\times \prod_{i=2}^N[a_i,  b_i] \right)\cap E$, $\delta>0$, $\beta\in (0,1)$, $c_1>0$ and $c_i(\hat{t})$, $2\leq i\leq N$, are positive and continuous functions over $\prod_{i=2}^N[a_i,  b_i]$.

Additionally, assume that
\BQN\label{cor1}
r(s,t)<1,\quad  s\neq t, \ s,t\in E.
\EQN

\BT\label{extension}
Let $\{X(t), t\in E\}$, where $E=\prod_{i=1}^N[a_i, b_i]$, be a Gaussian random field with continuous trajectories satisfying (\ref{var})-(\ref{cor1}) and let $t_1^*= a_1$ or $b_1$. Then we have, as $u\to\infty$,

(i) for $\beta<\gamma/2$,
\begin{equation}\label{eq:asym-1}
\pk{\sup_{t\in E}X(t)>u}\sim \sqrt{c_1}\Gamma(1/\gamma+1)\mathcal{H}_{2\beta}^N \int_{\hat{t}\in \prod_{i=2}^N[a_i,  b_i] }h^{-1/\gamma}(\hat{t})\prod_{i=2}^N\sqrt{ c_i(\hat{t})}d\hat{t} \, u^{\frac{N}{\beta}-\frac{2}{\gamma}}\Psi(u);
\end{equation}

(ii) for $\beta=\gamma/2$,
\begin{equation}\label{eq:asym-2}
\pk{\sup_{t\in E}X(t)>u}\sim \int_{\hat{t}\in \prod_{i=2}^N[a_i,  b_i] }\widehat{\mathcal{M}}_{2\beta}^{g(t)}\prod_{i=2}^N\sqrt{ c_i(\hat{t})}d\hat{t} \, u^{\frac{N-1}{\beta}}\Psi(u),
\end{equation}
where $g(t)=c_1^{-\beta}h(\hat{t})|t_1|^\gamma$, $t\in \R^N$;

(iii) for $\beta>\gamma/2$,
\begin{equation}\label{eq:asym-3}
\pk{\sup_{t\in E}X(t)>u}\sim \mathcal{H}_{2\beta}^{N-1}\int_{\hat{t}\in \prod_{i=2}^N[a_i,  b_i] }\prod_{i=2}^N\sqrt{ c_i(\hat{t})}d\hat{t} \, u^{\frac{N-1}{\beta}}\Psi(u).
\end{equation}
\ET
\begin{remark}
In Theorem \ref{extension} above, we consider the case when $t_1^*$ is the boundary of the interval $[a_1, b_1]$. If $t_1^*\in (a_1,b_1)$, the following results will be obtained by modifying the proof accordingly. (i) For $\beta<\gamma/2$, replace $\mathcal{H}_{2\beta}^N$ by $2\mathcal{H}_{2\beta}^N$ in the asymptotics in \eqref{eq:asym-1}; (ii) for $\beta=\gamma/2$, replace $\widehat{\mathcal{M}}_{2\beta}^{g(t)}$ by $\mathcal{M}_{2\beta}^{g(t)}$ in the asymptotics in \eqref{eq:asym-2}; (iii) for $\beta>\gamma/2$, the asymptotics in \eqref{eq:asym-2} still holds. \qed
\end{remark}

We formulate our next main result as following.
\BT \label{ThA} Let $\{B_\beta (x), x \in \mathbb S^N\}$ be a SFBM, where $\beta\in(0,1/2]$, and let $T_a=\{x\in\mathbb{S}^N: d(x,o')\le a\}$ with $a\in (0,\pi)$.

(i) If $\beta\in (0,1/2)$, then
\[
	\pk{\sup_{x\in T_a} B_\beta(x) > u }
	\sim \mathcal{H}_{2\beta}^N\frac{N\pi^{N/2}(\sin a)^{N-1}}{2^{\frac{N}{2\beta}}a^{2N-2\beta-1}\beta\Gamma(N/2+1)} u^{\frac{N}{\beta}-2}\Psi(a^{-\beta}u),
	\quad u\to\infty,
\]
where $\mathcal{H}_{2\beta}^N$ is the Pickands constant defined in \eqref{eq:Pickands-Const}.

(ii) If $\beta=1/2$, then
\[ \pk{\sup_{x\in T_a} B_\beta(x) > u }

	\sim \widehat{\mathcal{M}}_{{1}}^{g}\frac{N\pi^{N/2}(\sin a)^{N-1}}{2^{N-1}a^{2(N-1)}\Gamma(N/2+1)} u^{2(N-1)}\Psi(a^{-1/2}u),
	\quad u\to\infty,
\]
where $\widehat{\mathcal{M}}_{{1}}^{g}$ is defined in \eqref{eq:M-Const} and is finite by Lemma \ref{finiteness}, $g(t)=|t_1|$, $t=(t_1, \ldots, t_N)\in \R^N$.
\ET
\begin{proof}\,
 Note that
\BQNY
\pk{ \sup_{x\in T_a} B_\beta(x)>u}=\pk{ \sup_{\theta\in \Theta_a } \frac{\widetilde{B}_\beta(\theta)}{a^\beta}>\frac{u}{a^\beta}},
\EQNY
and we will focus on studying the excursion probability on the right hand side which turns out to be of Euclidean case. It is straightforward that for any $0<\ep<\pi/2$,
\begin{equation}\label{eq:bounds}
\begin{split}
\pk{ \sup_{\theta\in \Theta_a^\ep } \frac{\widetilde{B}_\beta(\theta)}{a^\beta}>\frac{u}{a^\beta}}&\leq \pk{ \sup_{\theta\in \Theta_a } \frac{\widetilde{B}_\beta(\theta)}{a^\beta}>\frac{u}{a^\beta}}\\
&\leq \pk{ \sup_{\theta\in \Theta_a^\ep } \frac{\widetilde{B}_\beta(\theta)}{a^\beta}>\frac{u}{a^\beta}}+\pk{ \sup_{\theta\in \Theta_a\setminus \Theta_a^\ep } \frac{\widetilde{B}_\beta(\theta)}{a^\beta}>\frac{u}{a^\beta}},
\end{split}
\end{equation}
where $$\Theta_a^\ep=[0,a]\times [\ep, \pi-\ep]^{N-2}\times[0, 2 \pi-\ep].$$
Applying Lemma \ref{lemB} and Theorem \ref{extension} with $\beta=\beta$, $\gamma=1$, $h(\hat{\theta})=\beta/a$, $c_1=(2a^{2\beta})^{-1/\beta}$, $c_2(\hat{\theta})=(2a^{2\beta})^{-1/\beta}\sin^2 a$ and $c_j(\hat{\theta})=(2a^{2\beta})^{-1/\beta}(\sin^2 a)\prod_{i=2}^{j-1} \sin^2 \theta_i$ for $3\le j\le N$, we have that for $\beta\in (0,1/2)$,
\begin{equation}\label{eq:asym-ep-1}
\begin{split}
&\pk{ \sup_{\theta\in \Theta_a^\ep } \frac{\widetilde{B}_\beta(\theta)}{a^\beta}>\frac{u}{a^\beta}}\\
&\quad \sim \mathcal{H}_{2\beta}^N \frac{a}{\beta} (2a^{2\beta})^{-N/(2\beta)}(\sin a)^{N-1}a^{2\beta-N}\int_{\hat{\theta}\in \hat{\Theta}_\ep}\prod_{i=2}^{N-1} (\sin\theta_i)^{N-i}d\hat{\theta} \, u^{N/\beta-2}\Psi(\frac{u}{a^\beta});
\end{split}
\end{equation}
and for $\beta=1/2$,
\begin{equation}\label{eq:asym-ep-2}
\begin{split}
&\pk{ \sup_{\theta\in \Theta_a^\ep } \frac{\widetilde{B}_\beta(\theta)}{a^\beta}>\frac{u}{a^\beta}}\\
&\quad \sim \widehat{\mathcal{M}}_{1}^{g} (2a)^{1-N}(\sin a)^{N-1} a^{1-N} \int_{\hat{\theta}\in \hat{\Theta}_\ep}\prod_{i=2}^{N-1} (\sin\theta_i)^{N-i}d\hat{\theta}\, u^{2(N-1)}\Psi(\frac{u}{a^{1/2}}),
\end{split}
\end{equation}
where $g(t)=|t_1|$, $t=(t_1, \ldots, t_N)\in \R^N$, and $$\hat{\Theta}_\ep=[\ep, \pi-\ep]^{N-2}\times[0, 2 \pi-\ep], \quad \hat{\theta}=(\theta_2,\dots, \theta_N).$$

Next we show that the last term in \eqref{eq:bounds} is negligible. Denote by
$$E_0=[0,a-\ep]\times [0, \pi]^{N-2}\times[0, 2 \pi), \quad E_j=[a-\ep, a]\times F_j, \quad 1\leq j\leq n, $$
where  $F_j, 1\leq j\leq n$, is a collection of compact rectangles forming a partition of $ [0, \pi]^{N-2}\times[0, 2 \pi)\setminus \hat{\Theta}_\ep$. Moreover, assume that  $F_j$ and $F_{j'}$ have no common inner point for $j\neq j'$ and the largest edge of $F_j$  has length $L$. Then we have that
\BQNY
\pk{ \sup_{\theta\in \Theta_a\setminus \Theta_a^\ep } \frac{\widetilde{B}_\beta(\theta)}{a^\beta}>\frac{u}{a^\beta}}\leq \sum_{j=0}^n\pk{ \sup_{\theta\in  E_j } \frac{\widetilde{B}_\beta(\theta)}{a^\beta}>\frac{u}{a^\beta}}.
\EQNY
It follows from Lemma \ref{lemB} that there exists $0<\delta<1$ such that $\sup_{\theta\in E_0 } \widetilde{\sigma}(\theta)/a^\beta< 1-\delta$.
By the Borell-TIS inequality \cite{Adler07}, for $u$ sufficiently large,
\BQNY
\pk{ \sup_{\theta\in E_0 } \frac{\widetilde{B}_\beta(\theta)}{a^\beta}>\frac{u}{a^\beta}}\leq \exp\left\{-\frac{(u/a^\beta)^2}{2(1-\delta)^2}\right\}.
\EQNY
By (\ref{eqvar})-(\ref{eqcorr}) and the Slepain inequality, we have that for $\ep>0$ and $L>0$ sufficiently small
\BQNY
\begin{split}
	\pk{ \sup_{\theta\in E_j } \frac{\widetilde{B}_\beta(\theta)}{a^\beta}>\frac{u}{a^\beta}}&\leq \pk{ \sup_{\theta\in E_j } \frac{\widetilde{B}_\beta(\theta)/\tilde{\sigma}(\theta)}{1+\frac{\beta}{2a} |a-\theta_1|}>\frac{u}{a^\beta}}\leq \pk{ \sup_{\theta\in E_j } \frac{Y(C\theta)}{1+\frac{\beta}{2a} |a-\theta_1|}>\frac{u}{a^\beta}},
\end{split}
\EQNY
where $1\leq j\leq n$, $C>2^{-1/(2\beta)}a^{-1}$, and  $Y(t)$ is a centered homogeneous Gaussian random field with continuous trajectories, unit variance and correlation function satisfying
$${\rm Corr}(Y(s),Y(t))=e^{-\|s-t\|^{2\beta}}, \quad s,t\in \R^N.$$
In light of Theorem \ref{extension} with $\beta=\beta$, $\gamma=1$, $h(\hat{\theta})=\beta/(2a)$, $c_1=C^2$, $c_i(\hat{\theta})=C^2,~ 2\leq i\leq n$, we have for $1\leq j\leq n$,
\BQNY
\begin{split}
\pk{ \sup_{\theta\in E_j } \frac{Y(C\theta)}{1+\frac{\beta}{2a} |a-\theta_1|}>\frac{u}{a^\beta}}&\sim\mathcal{H}_{2\beta}^N C^N \frac{2a}{\beta} {\rm mes}(F_j) (\frac{u}{a^\beta})^{\frac{N}{\beta}-2}\Psi(\frac{u}{a^\beta}), \quad \beta<1/2,\\
\pk{ \sup_{\theta\in E_j } \frac{Y(C\theta)}{1+\frac{\beta}{2a} |a-\theta_1|}>\frac{u}{a^\beta}}&\sim\widehat{\mathcal{M}}_{1}^{(4aC)^{-1}|t_1|}   {\rm mes}(F_j) (\frac{u\sqrt{C}}{\sqrt{a}})^{2N-2}\Psi(\frac{u}{a^{1/2}}), \quad \beta=1/2.
\end{split}
\EQNY
Further,
\BQNY
\begin{split}
\sum_{j=1}^n\pk{ \sup_{\theta\in E_j } \frac{Y(C\theta)}{1+\frac{\beta}{2a} |a-\theta_1|}>\frac{u}{a^\beta}}&\sim\mathcal{H}_{2\beta}^N C^N \frac{2a}{\beta} \sum_{j=1}^n{\rm mes}(F_j) (\frac{u}{a^\beta})^{\frac{N}{\beta}-2}\Psi(\frac{u}{a^\beta}), \quad \beta<1/2,\\
\sum_{j=1}^n\pk{ \sup_{\theta\in E_j } \frac{Y(C\theta)}{1+\frac{\beta}{2a} |a-\theta_1|}>\frac{u}{a^\beta}}&\sim\widehat{\mathcal{M}}_{1}^{(4aC)^{-1}|t_1|}  \sum_{j=1}^n {\rm mes}(F_j) (\frac{u\sqrt{C}}{\sqrt{a}})^{2N-2}\Psi(\frac{u}{a^{1/2}}), \quad \beta=1/2.
\end{split}
\EQNY
Note that $\lim_{\ep\to 0} \sum_{j=1}^n {\rm mes}(F_j)=0$ implies that the last term in \eqref{eq:bounds} is negligible.

Applying \eqref{eq:area-S}, one has
$$\lim_{\ep\to 0}\int_{\hat{\theta}\in \hat{\Theta}_\ep}\prod_{i=2}^{N-1} (\sin\theta_i)^{N-i}d\hat{\theta}=\int_{\hat{\theta}\in [0, \pi]^{N-2}\times[0, 2 \pi)}\prod_{i=2}^{N-1} (\sin\theta_i)^{N-i}d\hat{\theta}=\frac{N\pi^{N/2}}{\Gamma(N/2+1)}. $$
Plugging this into \eqref{eq:asym-ep-1} and \eqref{eq:asym-ep-2}, together with \eqref{eq:bounds}, we obtain the desired asymptotic results by letting $\ep\to 0$.
\end{proof}

It is worth mentioning that, when $N=1$, the geodesic disc $T_a$ becomes a circular arc and the maximum of variance is attained at only the two boundary points of $T_a$. Recall Lemma \ref{Piterbarg} and note that if $N=1$ and $t_0$ is a boundary point instead of an interior point, then we can obtain the asymptotics by multiplying the original asymptotics in \eqref{eq:Piterbarg-1} by 1/2 for $\beta\in (0,1/2)$, and by replacing $\mathcal{P}_{\alpha}^{\|AC^{-1}t\|^{\alpha}}$ in \eqref{eq:Piterbarg-2} by $\widehat{\mathcal{M}}_{\alpha}^{\|AC^{-1}t\|^{\alpha}}$ for $\beta=1/2$. Applying these results, together with Lemma \ref{lemB}, similarly to the proof of Theorem \ref{Thm-Sphere}, we have that
\begin{equation}\label{eq:N1}
\begin{split}
\pk{ \sup_{x\in T_a} B_\beta(x)>u}&\sim
\mathcal{H}_{2\beta}^1\frac{a^{2\beta-1}}{2^{\frac{1}{2\beta}-1}\beta} u^{\frac{1}{\beta}-2}\Psi(a^{-\beta}u), \quad \beta\in (0,1/2),\\
\pk{ \sup_{x\in T_a} B_\beta(x)>u}
&\sim 2\widehat{\mathcal{M}}_{{1}}^{g} \Psi(a^{-1/2}u), \quad \beta=1/2,
\end{split}
\end{equation}
where $g(t)=|t|$, $t\in \R$. Then it is easy to check that the asymptotics in \eqref{eq:N1} are exactly the same as those in Theorem \ref{ThA} for $N=1$.

\section*{Acknowledgments} The authors thank Enkelejd Hashorva from University of Lausanne for his helpful discussions and suggestions, and thank anonymous referees for their careful reading and valuable comments.

\section{Appendix}
The following useful lemma can be shown by similar arguments in the proof of Lemma 7.1 in \cite{Piterbarg:1996}. The proof is omitted in this paper.
\BEL\label{lemma} Let $Y(t), t\in \mathbb{R}^N$ be a centered homogeneous Gaussian random field with continuous trajectories, unit variance and correlation function satisfying, with $\beta\in (0,1]$,
$$1-{\rm Corr}(Y(s),Y(t))=\|s-t\|^{2\beta}(1+o(1)), \quad \|s-t\| \to 0.$$
Denote by $\{u_\lambda, \lambda\in \Lambda\}$ a series of function of $u$ with the property that
$$\lim_{u\to\infty}\sup_{\lambda\in \Lambda}\left|\frac{u_\lambda}{u}-1\right|=0.$$
Then for all $b\geq 0$
$$\lim_{u\to\infty}\sup_{\lambda\in \Lambda}\left|\frac{\pk{\sup_{t\in [0, u^{-1/\beta}S]\times \prod_{i=2}^N[a_i,b_i]}\frac{Y(t)}{1+b|t_1|^{2\beta}}>u_\lambda}}{u^{\frac{N-1}{\beta}}\Psi(u_\lambda)}- \mathcal{M}_{2\beta}^{b|t_1|^{2\beta}}([0,S]) \prod_{i=2}^N(b_i-a_i)\right|=0,$$
\EEL
where
$$\mathcal{M}
_{2\beta}^{b|t_1|^{2\beta}}([0,S]) =\lim_{S_1\to\infty}\frac{\mathcal{P}_{2\beta}^{b|t_1|^{2\beta}}([0,S]\times[0,S_1]^{N-1})}{S_1^{N-1}}\in (0,\IF).$$

The following lemma shows the finiteness of the constant $\widehat{\mathcal{M}}_{{1}}^{g}$ in Theorem \ref{ThA}.
\BEL\label{finiteness} For any $\beta\in (0,1]$ and $b>0$,
$$
\widehat{\mathcal{M}}_{2\beta}^{b|t_1|^{2\beta}}:=\lim_{S\to\infty}\mathcal{M}_{2\beta}^{b|t_1|^{2\beta}}([0,S])\in (0,\infty).
$$
\EEL
\begin{proof}\, Let $Y(t)$ be as in Lemma \ref{lemma}. Note that for $0<S_1<\log u$ and $S>0$,
\BQN\label{main1}
A_0(u,S_1)\leq \pk{\sup_{t\in [0, u^{-1/\beta}\log u]\times [0,1]^{N-1}}\frac{Y(t)}{1+b|t_1|^{2\beta}}>u}\leq \sum_{k=0}^{\lfloor (\log u)/S \rfloor+1} A_k(u,S),
\EQN
where
$$A_k(u,S)=\pk{\sup_{t\in [ u^{-1/\beta}kS, u^{-1/\beta}(k+1)S]\times [0,1]^{N-1}}\frac{Y(t)}{1+b|t_1|^{2\beta}}>u}.$$
In light of Lemma \ref{lemma} and subadditivity of $\mathcal{M}_{2\beta}^0([0,S])$, we have
\BQNY
\begin{split}
	A_0(u,S)&\sim \mathcal{M}_{2\beta}^{b|t_1|^{2\beta}}([0,S])u^{\frac{N-1}{\beta}}\Psi(u),\\
	A_k(u,S)&\leq \pk{\sup_{t\in [ u^{-1/\beta}kS, u^{-1/\beta}(k+1)S]\times [0,1]^{N-1}}Y(t)>u(1+u^{-2}b|kS|^{2\beta})}\\
	&\sim \mathcal{M}_{2\beta}^0([0,S])u^{\frac{N-1}{\beta}}\Psi(u(1+u^{-2}b|kS|^{2\beta}))\\
	&\leq CSe^{-b|kS|^{2\beta}}u^{\frac{N-1}{\beta}}\Psi(u), \quad 1\leq k\leq \lfloor (\log u)/S \rfloor+1,
\end{split}
\EQNY
where $C>0$ is a fixed constant. Dividing (\ref{main1}) by $u^{\frac{N-1}{\beta}}\Psi(u)$ on both sides, we have that
$$\mathcal{M}_{2\beta}^{b|t_1|^{2\beta}}([0,S_1])\leq \mathcal{M}_{2\beta}^{b|t_1|^{2\beta}}([0,S]) +\sum_{k=1}^\infty CSe^{-b|kS|^{2\beta}}<\IF.$$
Letting $S_1\to\infty$ leads to
$$\lim_{S_1\to\infty} \mathcal{M}
_{2\beta}^{b|t_1|^{2\beta}}([0,S_1])<\IF,$$
completing the proof.
\end{proof}

\prooftheo{extension} Without loss of generality, we assume that $t_1^*=a_1$, implying $E_u=[a_1,a_1+((\log u)/u)^{2/\gamma}]\times \prod_{i=2}^N[a_i,  b_i]$. Denote by $$F_{1,\ep}=[a_1+\ep, b_1]\times \prod_{i=2}^N[a_i,  b_i], \quad F_{2,\ep}(u)=[a_1+((\log u)/u)^{2/\gamma}, a_1+\ep]\times \prod_{i=2}^N[a_i,  b_i].$$
Then it follows that
\BQN\label{main}
\begin{split}
	\pk{\sup_{t\in E_u}X(t)>u}&\leq \pk{\sup_{t\in E}X(t)>u}\\
	&\leq \pk{\sup_{t\in E_u}X(t)>u}+\pk{\sup_{t\in F_{1,\ep}}X(t)>u}+\pk{\sup_{t\in F_{2,\ep}(u)}X(t)>u}.
\end{split}
\EQN
By (\ref{var}), for $\ep>0$ sufficiently small, there exists a constant $\delta>0$ such that
$\sup_{t\in F_{1,\ep}}\sigma^2(t)<1-\delta.$
By the Borell-TIS inequality \cite{Adler07}, for $u$ large enough,
\BQN\label{upper1}
\pk{\sup_{t\in F_{1,\ep}}X(t)>u}\leq \exp\left\{-\frac{u^2}{2(1-\delta)}\right\}.
\EQN
Moreover, in light of (\ref{cor}), there exists $C>0$ such that for $u$ sufficiently large and $\ep>0$ sufficiently small,
\BQNY
\E\{\left(\overline{X}(t)-\overline{X}(s)\right)\}\leq C\|t-s\|^{2\beta}\leq NC\sum_{i=1}^N|t_i-s_i|^{2\beta}, \quad s,t\in F_{2,\ep}(u),
\EQNY
where $\overline{X}$ is the standardized field of $X$. Additionally, it follows from \eqref{var} that there exists $C_1>0$ such that
$$\sup_{t\in F_{2,\ep}(u)}\sigma^2(t)<1-C_1\left(\frac{\log u}{u}\right)^2.$$
By Theorem 8.1 in \cite{Piterbarg:1996}, we have that, for $u$ sufficiently large,
\BQN\label{upper2}
\pk{\sup_{t\in F_{2,\ep}(u)}X(t)>u}\leq C_2 u^{\frac{N}{\beta}}\Psi\left(\frac{u}{\sqrt{1-C_1\left(\frac{\log u}{u}\right)^2}}\right).
\EQN

We study next $\pk{\sup_{t\in E_u}X(t)>u}$ to derive the exact asymptotics and show that
\[
\pk{\sup_{t\in F_{1,\ep}}X(t)>u} \quad {\rm and} \quad  \pk{\sup_{t\in F_{2,\ep}(u)}X(t)>u}
\]
are negligible as $u\to\infty$. We distinguish three scenarios: $\beta<\gamma/2$, $\beta=\gamma/2$ and $\beta>\gamma/2$.

{\it \textbf{(i) Case $\bm{\beta<\gamma/2}$}}. We first introduce some notation for further analysis. Let
\BQN\label{IM}
I_{k}(u)=[a_1+ku^{-1/\beta}S, a_1+(k+1)u^{-1/\beta}S],\quad  M_u^{\pm}=\left[\frac{((\log u)/u)^{2/\gamma}}{u^{-1/\beta}S}\right]\pm 1.
\EQN
Split $\prod_{i=2}^N[a_i,  b_i]$ into $n^{N-1}$ rectangles with the form $\prod_{i=2}^N\left[a_i+\frac{k_i(b_i-a_i)}{n}, a_i+\frac{(k_i+1)(b_i-a_i)}{n} \right]$ with $n, k_i\in \mathbb{N}$, denoted by  $\{D_j, 1\leq j\leq n^{N-1}\}$. We assume that $D_j$ and $D_{j'}$ have no common inner points for $j\neq j'$. Let
\BQN\label{CJ}
I_{k,j}(u)&=&I_{k}(u)\times D_j, \quad \Lambda^{\pm}=\{(k,j): 0\leq k\leq M_u^{\pm}, 1\leq j\leq n^{N-1}\},\nonumber\\
u_{k,j,\ep}&=&u\left(1+(1-\ep)h_j\inf_{t_1\in I_k(u)}|t_1-t_1^*|^{\gamma}\right),\quad h_j=\inf_{\hat{t}\in D_j }h(\hat{t}),\\ c(j)&=&((c_1+\ep)^{1/2}, (c_{2,j}+\ep)^{1/2}, \dots, (c_{N,j}+\ep)^{1/2}),\quad c_{k,j}=\sup_{\hat{t}\in D_j}c_k(\hat{t}), 2\leq k\leq N.\nonumber
\EQN
Moreover, let $Y(t)$ be a centered homogeneous Gaussian random fields with continuous trajectories, unit variance and correlation function satisfying ${\rm Corr}(Y(s),Y(t))=e^{-\|s-t\|^{2\beta}}$ with $\beta\in (0,1]$.
It follows straightforwardly that
\BQN\label{Bonferroni}
\pi^-(u)-\Sigma(u)\leq \pk{\sup_{t\in E_u}X(t)>u}\leq \pi^+(u),
\EQN
where
$$\pi^\pm(u)=\sum_{(k,j)\in \Lambda^{\pm}}\pk{\sup_{t\in I_{k,j}(u)}X(t)>u},$$
$$\Sigma(u)=\sum_{(k,j)\neq (k',j'), k\leq k',(k,j), (k',j')\in \Lambda^-}\pk{\sup_{t\in I_{k,j}(u)}X(t)>u, \sup_{t\in I_{k',j'}(u)}X(t)>u}.$$
{\it \underline{Asymptotics for $\pi^\pm(u)$}}.
To derive the upper bound, in light of  Slepian inequlaity we have
\BQNY
\begin{split}
	\pk{\sup_{t\in I_{k,j}(u)}X(t)>u}&\leq\pk{\sup_{t\in I_{k,j}(u)}\overline{X}(t)>u_{k,j,\ep}}\leq\pk{\sup_{t\in I_{k,j}(u)}Y(c(j)t)>u_{k,j,\ep}}\\
	&=\pk{\sup_{t\in c(j)I_{0,1}(u)}Y(t)>u_{k,j,\ep}},
\end{split}
\EQNY
where for any $D\subset R^N$, $c(j)D=\{((c_1+\ep)^{1/2}t_1, (c_{2,j}+\ep)^{1/2}t_2, \dots, (c_{N,j}+\ep)^{1/2}t_N): t\in D\}$.
In light of Lemma \ref{lemma}, we have
\BQN\label{pkj}
\begin{split}
&\pk{\sup_{t\in I_{k,j}(u)}X(t)>u}\\
&\quad \leq  \mathcal{H}_{2\beta}^N([0, (c_1+\ep)^{1/2}S])\prod_{i=2}^N(c_{i,j}+\ep)^{1/2}{\rm mes}(D_j)u^{\frac{N-1}{\beta}}\Psi(u_{k,j,\ep})(1+o(1)),
\end{split}
\EQN
as $u\to\infty,$ uniformly with respect to $(k,j)\in\Lambda^+$. Hence, as $u\to\infty$,
\BQNY
\begin{split}
	\sum_{k=0}^{M_u^+}\pk{\sup_{t\in I_{k,j}(u)}X(t)>u}
	\leq  \mathcal{H}_{2\beta}^N([0, (c_1+\ep)^{1/2}S])\prod_{i=2}^N(c_{i,j}+\ep)^{1/2}{\rm mes}(D_j)u^{\frac{N-1}{\beta}}\sum_{k=0}^{M_u^+}\Psi(u_{k,j,\ep}).
\end{split}
\EQNY
Noting that, as $u\to\infty$,
\BQNY
\begin{split}
\sum_{k=0}^{M_u^+}\Psi(u_{k,j,\ep})&\leq \Psi(u)\sum_{k=0}^{M_u^+}
	e^{-(1-\ep)h_ju^2|ku^{-1/\beta}S|^{\gamma}}\\
&\leq \Psi(u)
	\left((1-\ep)^{1/\gamma}h_j^{1/\gamma}u^{2/\gamma-1/\beta}S\right)^{-1}\\
	&\quad \times\sum_{k=0}^{M_u^+}
	e^{-|k(1-\ep)^{1/\gamma}h_j^{1/\gamma}u^{2/\gamma-1/\beta}S|^{\gamma}}\times (1-\ep)^{1/\gamma}h_j^{1/\gamma}u^{2/\gamma-1/\beta}S\\
&\leq \Psi(u)
	\left((1-\ep)^{1/\gamma}h_j^{1/\gamma}u^{2/\gamma-1/\beta}S\right)^{-1}\int_0^\infty e^{-|t|^\gamma} dt,
\end{split}
\EQNY
we have
\BQNY
\begin{split}
	&\quad \sum_{k=0}^{M_u^+}\pk{\sup_{t\in I_{k,j}(u)}X(t)>u}\\
	&\leq \frac{\mathcal{H}_{2\beta}^N([0, (c_1+\ep)^{1/2}S])}{S}\Gamma(1/\gamma+1)(1-\ep)^{-1/\gamma}h_j^{-1/\gamma}\prod_{i=2}^N(c_{i,j}+\ep)^{1/2}{\rm mes}(D_j)
	u^{\frac{N}{\beta}-\frac{2}{\gamma}}\Psi(u)\\
	&\leq \mathcal{H}_{2\beta}^N\Gamma(1/\gamma+1)(c_1+\ep)^{1/2}(1-\ep)^{-1/\gamma}h_j^{-1/\gamma}\prod_{i=2}^N(c_{i,j}+\ep)^{1/2}{\rm mes}(D_j)
	u^{\frac{N}{\beta}-\frac{2}{\gamma}}\Psi(u),
\end{split}
\EQNY

as $u\to\infty$ and $S\to\infty$.
Furthermore,
\begin{equation}\label{pi+}
\begin{split}
\pi^+(u)&=\sum_{j=1}^{n^{N-1}}\sum_{k=0}^{M_u^+}\pk{\sup_{t\in I_{k,j}(u)}X(t)>u}\\
&\leq  \mathcal{H}_{2\beta}^N\frac{\Gamma(1/\gamma+1)(c_1+\ep)^{1/2}}{(1-\ep)^{1/\gamma}}u^{\frac{N}{\beta}-\frac{2}{\gamma}}\Psi(u)\sum_{j=0}^{n^{N-1}}\left(h_j^{-1/\gamma}\prod_{i=2}^N(c_{i,j}+\ep)^{1/2}{\rm mes}(D_j)\right)\\
&\sim  \mathcal{H}_{2\beta}^N \Gamma(1/\gamma+1)c_1^{1/2}u^{\frac{N}{\beta}-\frac{2}{\gamma}}\Psi(u)\int_{\hat{t}\in \prod_{i=2}^N [a_i,b_i] }h^{-1/\gamma}(\hat{t})\prod_{i=2}^N c_i^{1/2}(\hat{t})d\hat{t},
\end{split}
\end{equation}
as $u\to\infty, n\to\infty, \ep\to 0$. Analogously,
we can show that
\BQN\label{pi-}
\pi^-(u)\sim \mathcal{H}_{2\beta}^N \Gamma(1/\gamma+1)c_1^{1/2}u^{\frac{N}{\beta}-\frac{2}{\gamma}}\Psi(u)\int_{\hat{t}\in \prod_{i=2}^N [a_i,b_i] }h^{-1/\gamma}(\hat{t})\prod_{i=2}^N c_i^{1/2}(\hat{t})d\hat{t},
\EQN
as $u\to\infty, n\to\infty$. Next we show that $\Sigma(u)$ is negligible compared with $\pi^-(u)$ as $u\to\infty$. For this, denote by
\BQNY
\begin{split}
	\Lambda^-_1&=\{(k,j, k',j'): (k,j), (k',j')\in \Lambda^-, k\leq k', D_j\cap D_{j'}=\emptyset\},\\
	\Lambda^-_2&=\{(k,j, k',j'):(k,j), (k',j')\in \Lambda^-,  k\leq k'\leq k+1, D_j\cap D_{j'}\neq \emptyset, j\neq j'\},\\
	\Lambda^-_3&=\{(k,j, k',j'): (k,j), (k',j')\in \Lambda^-,   k+1<k',  D_j\cap D_{j'}\neq \emptyset\},\\
	\Lambda^-_4&=\{(k,j, k',j'): (k,j), (k',j')\in \Lambda^-,  k'=k+1, j=j' \}.
\end{split}
\EQNY
Then it follows that $\Sigma(u)\leq \sum_{i=1}^4\Sigma_i(u)$, where
$$\Sigma_i(u)=\sum_{(k,j,k',j')\in \Lambda^-_i}\pk{\sup_{t\in I_{k,j}(u)}X(t)>u, \sup_{t\in I_{k',j'}(u)}X(t)>u}.$$
{\it\underline{Upper bound for $\Sigma_1(u)$}}. Note that
\BQNY
\Sigma_1(u)\leq\sum_{(k,j,k',j')\in \Lambda^-_1}\pk{\sup_{s\in I_{k,j}(u), t\in I_{k,j}(u)}X(s)+X(t)>2u},
\EQNY
and by (\ref{var}) and (\ref{cor1}), there exists $0<\delta<1$ such that
\BQNY
{\rm Var}\left(X(s)+X(t)\right)= \sigma^2(s)+\sigma^2(t)+2\sigma(s)2\sigma(t)r(s,t)<4-\delta.
\EQNY
It follows from the Borell-TIS inequality \cite{Adler07} that, as $u\to\infty$,
\BQN\label{sigma1}
\Sigma_1(u)\leq\sum_{(k,j,k',j')\in \Lambda^-_1} e^{-\frac{\left(2u-\E\left(\sup_{t\in E}X(t)\right)\right)^2}{2(4-\delta)}}\leq (n^{N-1} M_u^+)^2e^{-\frac{\left(2u-\E\left(\sup_{t\in E}X(t)\right)\right)^2}{2(4-\delta)}}=o(\pi^-(u)).
\EQN
{\it\underline{Upper bound for $\Sigma_2(u)$}}.   For $(k,j,k',j')\in \Lambda^-_2$, without loss of generality, we assume that
\BQNY
\begin{split}
	D_j&=\prod_{i=2}^N\left[a_i+\frac{k_i(b_i-a_i)}{n}, a_i+\frac{(k_i+1)(b_i-a_i)}{n} \right],\\
	D_{j'}&=\left[a_2+\frac{(k_2+1)(b_2-a_2)}{n}, a_2+\frac{(k_2+2)(b_2-a_2)}{n} \right]\\
	&\quad \times\prod_{i=3}^N\left[a_i+\frac{k_i(b_i-a_i)}{n}, a_i+\frac{(k_i+1)(b_i-a_i)}{n} \right].
\end{split}
\EQNY
Split $D_{j'}$ into two parts:
\BQNY
\begin{split}
	D_{j'}^{(1)}&=\left[a_2+\frac{(k_2+1)(b_2-a_2)}{n}, a_2+\frac{(k_2+1)(b_2-a_2)}{n}+\frac{b_2-a_2}{n^{2}} \right]\\
	&\quad \times\prod_{i=3}^N\left[a_i+\frac{k_i(b_i-a_i)}{n}, a_i+\frac{(k_i+1)(b_i-a_i)}{n} \right],\\
	D_{j'}^{(2)}&=\left[a_2+\frac{(k_2+1)(b_2-a_2)}{n}+\frac{b_2-a_2}{n^{2}} , a_2+\frac{(k_2+2)(b_2-a_2)}{n} \right]\\
	&\quad \times\prod_{i=3}^N\left[a_i+\frac{k_i(b_i-a_i)}{n}, a_i+\frac{(k_i+1)(b_i-a_i)}{n} \right].
\end{split}
\EQNY
Then it follows that
\BQNY
\begin{split}
&\pk{\sup_{t\in I_{k,j}(u)}X(t)>u, \sup_{t\in I_{k',j'}(u)}X(t)>u}\\
&\quad \leq \pk{ \sup_{t\in I_{k',j'}^{(1)}(u)}X(t)>u}+\pk{\sup_{t\in I_{k,j}(u)}X(t)>u, \sup_{t\in I_{k',j'}^{(2)}(u)}X(t)>u}
\end{split}
\EQNY
with $I_{k'j'}^{(l)}(u)=I_{k'}(u)\times D_{j'}^{(l)}, l=1,2$.
By Lemma \ref{lemma} and (\ref{pkj}), we have as $u\to\infty$,
\BQNY
\pk{ \sup_{t\in I_{k',j'}^{(1)}(u)}X(t)>u}\leq C\frac{{\rm mes}(D_{j'}^{(1)})}{{\rm mes}(D_{j'})}\pk{ \sup_{t\in I_{k',j'}(u)}X(t)>u},
\EQNY
where $C>0$ is a constant independent of $k'$ and $j'$.
Using the fact that $D_{j'}$ has at most $3^{N-1}$ neighbors and
$$\lim_{n\to\infty}\sup_{1\leq j'\leq n^{N-1}}\frac{{\rm mes}(D_{j'}^{(1)})}{{\rm mes}(D_{j'})}=0,$$
we have
\BQNY
\begin{split}
	\sum_{(k,j,k',j')\in \Lambda^-_2}\pk{ \sup_{t\in I_{k',j'}^{(1)}(u)}X(t)>u}&\leq C\sum_{(k,j,k',j')\in \Lambda^-_2}\frac{{\rm mes}(D_{j'}^{(1)})}{{\rm mes}(D_{j'})}\pk{ \sup_{t\in I_{k',j'}(u)}X(t)>u}\\
	&\leq 3^{N-1}C\sum_{(k',j')\in \Lambda^-}\frac{{\rm mes}(D_{j'}^{(1)})}{{\rm mes}(D_{j'})}\pk{ \sup_{t\in I_{k',j'}(u)}X(t)>u}\\
	&=o(\pi^-(u)),\quad u\to\infty, n\to\infty.
\end{split}
\EQNY
Using the same argument as in (\ref{sigma1}), we have
\BQNY
\sum_{(k,j,k',j')\in \Lambda^-_2}\pk{\sup_{t\in I_{k,j}(u)}X(t)>u,  \sup_{t\in I_{k',j'}^{(2)}(u)}X(t)>u}=o(\pi^-(u)), \quad u\to\infty.
\EQNY
Hence,
$$\Sigma_2(u)=o(\pi^-(u)), \quad u\to\infty.$$
{\it \underline{Upper bound for $\Sigma_3(u)$}}.
Let

\begin{equation*}
\begin{split}
J_{l}(u)&=\prod_{i=2}^N[l_iu^{-1/\beta}S, (l_i+1)u^{-1/\beta}S], \quad \text{with}\quad l=(l_2,\dots,l_N), \quad \Xi_j=\{l: D_j\bigcap J_{l}(u)\neq \emptyset\},\\
J_{k,l}(u)&=I_k(u)\times J_l(u).
\end{split}
\end{equation*}
Then
\BQNY
\begin{split}
	&\pk{\sup_{t\in I_{k,j}(u)}X(t)>u, \sup_{t\in I_{k',j'}(u)}X(t)>u}\\
	&\quad \leq \pk{\sup_{t\in I_{k,j}(u)}\overline{X}(t)>u_{k,j,\ep}, \sup_{t\in I_{k',j'}(u)}\overline{X}(t)>u_{k',j',\ep}}\\
	&\quad \leq \sum_{l\in \Xi_j, l'\in \Xi_{j'}}\pk{\sup_{t\in J_{k,l}(u)}\overline{X}(t)>u_{k,j,\ep}, \sup_{t\in J_{k',l'}(u)}\overline{X}(t)>u_{k',j',\ep}}\\
	&\quad =   \sum_{l\in \Xi_j, l'\in \Xi_{j'}}\pk{\sup_{t\in J_{k,l}(1)}\overline{X}(u^{-1/\beta}t)>u_{k,j,\ep}, \sup_{t\in J_{k',l'}(1)}\overline{X}(u^{-1/\beta}t)>u_{k',j',\ep}}.
\end{split}
\EQNY
In view of \eqref{cor}, there exist $C_1, C_2>0$ such that for $u$ and $n$ sufficiently large
\BQNY
\begin{split}
	C_1\sum_{i=1}^N|s_i-t_i|^{2\beta}\leq u^2(1-{\rm Corr}(\overline{X}(u^{-1/\beta}s),\overline{X}(u^{-1/\beta}t)))\leq C_2\sum_{i=1}^N|s_i-t_i|^{2\beta},
\end{split}
\EQNY
and
$${\rm Corr}(\overline{X}(u^{-1/\beta}s),\overline{X}(u^{-1/\beta}t))\geq 1/2,$$
hold for all $s,t\in [a_1u^{1/\beta},a_1u^{1/\beta}+((\log u)/u)^{2/\gamma}u^{1/\beta}]\times \bigcup_{D_j\cap D_{j'}\neq \emptyset}u^{1/\beta}D_{j'}, 1\leq j\leq n^{N-1}$ with $u^{1/\beta}D_{j'}=\{u^{1/\beta}\hat{t}: \hat{t}\in D_{j'}  \}$.
Thus in light of Corollary 3.1 in \cite{Debicki:2017}, there exist $\mathcal{C}, \mathcal{C}_1>0$ such that for $u$ and $n$ sufficiently large, $l\in \Xi_j, l'\in \Xi_{j'}$, $D_j\cap D_{j'}\neq \emptyset$, $0\leq k,k'\leq M_u^+$ and $|k'-k-1|\geq 1$,
\BQN\label{new}
&\pk{\sup_{t\in J_{k,l}(1)}\overline{X}(u^{-1/\beta}t)>u_{k,j,\ep}, \sup_{t\in J_{k',l'}(1)}\overline{X}(u^{-1/\beta}t)>u_{k',j',\ep}}\nonumber\\
&\quad \leq \mathcal{C} S^{2N}e^{-\mathcal{C}_1S^{2\beta}(|k'-k-1|^{2\beta}+\|l-l'\|^{2\beta})}\Psi(u_{k,k',j,j'\ep}),
\EQN
where $$u_{k,k',j,j'\ep}=\min(u_{k,j,\ep}, u_{k',j',\ep}).$$
We have
\BQNY
\begin{split}
	\Sigma_3(u)&\leq \sum_{(k,j,k',j')\in \Lambda^-_3}\pk{\sup_{t\in I_{k,j}(u)}X(t)>u, \sup_{t\in I_{k',j'}(u)}X(t)>u}\\
	&\leq \sum_{ (k,j)\in \Lambda^-, |k-k'-1|\geq 1, D_j\bigcap D_{j'}\neq \emptyset}\pk{\sup_{t\in I_{k,j}(u)}X(t)>u, \sup_{t\in I_{k',j'}(u)}X(t)>u}\\
&\leq\sum_{ (k,j)\in \Lambda^-, |k-k'-1|\geq 1, D_j\bigcap D_{j'}\neq \emptyset}\sum_{l\in \Xi_j, l'\in \Xi_{j'}}\\
	&\quad \pk{\sup_{t\in J_{k,l}(1)}\overline{X}(u^{-1/\beta}t)>u_{k,j,\ep}, \sup_{t\in J_{k',l'}(1)}\overline{X}(u^{-1/\beta}t)>u_{k',j',\ep}}.
\end{split}
\EQNY
For  $(k,j)\in \Lambda^-$, it follows from (\ref{new}) that
\BQNY
\begin{split}
 &\sum_{|k-k'-1|\geq 1, D_j\bigcap D_{j'}\neq \emptyset}\sum_{l\in \Xi_j, l'\in \Xi_{j'}}
	 \pk{\sup_{t\in J_{k,l}(1)}\overline{X}(u^{-1/\beta}t)>u_{k,j,\ep}, \sup_{t\in J_{k',l'}(1)}\overline{X}(u^{-1/\beta}t)>u_{k',j',\ep}}\\
	&\leq \sum_{ |k-k'-1|\geq 1, D_j\bigcap D_{j'}\neq \emptyset}\sum_{l\in \Xi_j, l'\in \Xi_{j'}}\mathcal{C} S^{2N}e^{-\mathcal{C}_1S^{2\beta}(|k'-k-1|^{2\beta}+\|l-l'\|^{2\beta})}\Psi(u_{k,k',j,j',\ep})\\
	&\leq \sum_{l\in \Xi_j}C_3 S^{2N}e^{-C_4S^{2\beta}}\Psi(u_{k,j,\ep})\\
	&\leq C_3 S^{2N-1}u^{\frac{N-1}{\beta}}e^{-C_4S^{2\beta}}\Psi(u_{k,j,\ep}),
	\quad u\to\infty,
\end{split}
\EQNY
where $C_3$ and $C_4$ are two positive constants.
Hence
\BQNY
\begin{split}
\Sigma_3(u)&\leq \sum_{ (k,j)\in \Lambda^-}C_3 S^{2N-1}u^{\frac{N-1}{\beta}}e^{-C_4S^{2\beta}}\Psi(u_{k,j,\ep})\\
	&= C_3 S^{2N-1}e^{-C_4S^{2\beta}} \sum_{ (k,j)\in \Lambda^-}u^{\frac{N-1}{\beta}}\Psi(u_{k,j,\ep})=o(\pi^-(u)), \quad u\to\infty, S\to\infty.
\end{split}
\EQNY
{\it \underline{Upper bound for $\Sigma_4(u)$}}.
Observe that
\BQNY
\begin{split}
	&\pk{\sup_{t\in I_{k,j}(u)}X(t)>u, \sup_{t\in I_{k+1,j}(u)}X(t)>u}\\
	&\quad=\pk{\sup_{t\in I_{k,j}(u)\bigcup I_{k+1,j}(u)}X(t)>u}-\pk{\sup_{t\in I_{k,j}(u)}X(t)>u}-\pk{\sup_{t\in I_{k+1,j}(u)}X(t)>u}.
\end{split}
\EQNY
Thus in light of  (\ref{pi+}) and (\ref{pi-}), it follows that
\BQNY
\begin{split}
	\Sigma_4(u)&\leq \sum_{(k,j)\in \Lambda^-}\pk{\sup_{t\in I_{k,j}(u)}X(t)>u, \sup_{t\in I_{k+1,j}(u)}X(t)>u}\\
	&\leq \sum_{(k,j)\in \Lambda^-} \pk{\sup_{t\in I_{k,j}(u)\bigcup I_{k+1,j}(u)}X(t)>u}-\sum_{(k,j)\in \Lambda^-} \pk{\sup_{t\in I_{k,j}(u)}X(t)>u}\\
	&\quad -\sum_{(k,j)\in \Lambda^-} \pk{\sup_{t\in I_{k+1,j}(u)}X(t)>u}\\
	&=2\pi^+(u)(1+o(1))-2\pi^-(u)(1+o(1))=o(\pi^-(u)), \quad u\to\infty, S\to\infty.
\end{split}
\EQNY
Therefore we conclude that
\BQNY
\Sigma(u)=o(\pi^-(u)), \quad u\to\infty, S\to\infty,
\EQNY
together with (\ref{Bonferroni}), (\ref{pi+}) and (\ref{pi-}), yielding that as $u\to\infty$,
$$\pk{\sup_{t\in E_u}X(t)>u}\sim \mathcal{H}_{2\beta}^N \Gamma(1/\gamma+1)c_1^{1/2}u^{\frac{N}{\beta}-\frac{2}{\gamma}}\Psi(u)\int_{\hat{t}\in \prod_{i=2}^N [a_i,b_i] }h^{-1/\gamma}(\hat{t})\prod_{i=2}^N c_i^{1/2}(\hat{t})d\hat{t}.$$
Inserting the above asymptotics, (\ref{upper1}) and (\ref{upper2}) into (\ref{main}) establishes the claim.

{\it \textbf{(ii) Case $\bm{\beta=\gamma/2}$}}. It follows that
\BQN\label{Bonferroni1}
\pi_1^-(u)-\Sigma_5(u)\leq \pk{\sup_{t\in E_u}X(t)>u}\leq \pi_1^-(u)+ \pi_1^+(u),
\EQN
where
$$\pi_1^-(u)=\sum_{j=1}^{n^{N-1}}\pk{\sup_{t\in I_{0,j}(u)}X(t)>u},\quad \pi_1^+(u)=\sum_{k=1}^{M_u^+}\sum_{j=1}^{n^{N-1}}\pk{\sup_{t\in I_{k,j}(u)}X(t)>u},$$
$$\Sigma_5(u)=\sum_{1\leq j<j'\leq n^{N-1}}\pk{\sup_{t\in I_{0,j}(u)}X(t)>u, \sup_{t\in I_{0,j'}(u)}X(t)>u},$$
with $I_{k,j}$ and $M_u^{\pm}$ being defined in (\ref{IM}) and (\ref{CJ}).\\
{\it \underline{Asymptotics of $\pi_1^-(u)$}}. By (\ref{var}) and Slepain inequality,
\BQNY
\begin{split}
	\pi_1^-(u)&\leq \sum_{j=1}^{n^{N-1}}\pk{\sup_{t\in I_{0,j}(u)}\frac{\overline{X}(t)}{1+h_j|t_1-t_1^*|^{\gamma}}>u}\leq \sum_{j=1}^{n^{N-1}}\pk{\sup_{t\in I_{0,j}(u)}\frac{Y(c(j)t)}{1+h_j|t_1-t_1^*|^{\gamma}}>u}\\
	&\leq \sum_{j=1}^{n^{N-1}}\pk{\sup_{t\in c(j)I_{0,j}(u)}\frac{Y(t)}{1+h_j(c_1+\ep)^{-\gamma/2}|t_1-t_1^*|^{\gamma}}>u},
\end{split}
\EQNY
where $h_j$ and $c(j)$ are given in (\ref{CJ}).
In light of Lemma \ref{lemma} and Lemma \ref{finiteness}, we have that
\BQN\label{pi1-}
\begin{split}
	\pi_1^-(u)&\leq \sum_{j=1}^{n^{N-1}} \mathcal{M}_{2\beta}^{h_j(c_1+\ep)^{-\gamma/2}|t_1|^\gamma}[0,(c_1+\ep)^{1/2}S]\prod_{i=2}^N(c_{i,j}+\ep)^{1/2}{\rm mes}(D_j)u^{\frac{N-1}{\beta}}\Psi(u)\\
	&\sim \int_{\hat{t}\in \prod_{i=2}^N[a_i,  b_i] }\widehat{\mathcal{M}}_{2\beta}^{c_1^{-\beta}h(\hat{t})|t_1|^\gamma}\prod_{i=2}^N\sqrt{ c_i(\hat{t})}d\hat{t} u^{\frac{N-1}{\beta}}\Psi(u),
\end{split}
\EQN
as $u\to\infty, n\to\infty, \ep\to 0$.
Similarly, we can show that
\BQNY
\pi_1^-(u)\geq \int_{\hat{t}\in \prod_{i=2}^N[a_i,  b_i] }\widehat{\mathcal{M}}_{2\beta}^{c_1^{-\beta}h(\hat{t})|t_1|^\gamma}\prod_{i=2}^N\sqrt{ c_i(\hat{t})}d\hat{t} u^{\frac{N-1}{\beta}}\Psi(u)(1+o(1)), \quad u\to\infty, S\to\infty.
\EQNY
Hence
\BQNY
\pi_1^-(u)\sim\int_{\hat{t}\in \prod_{i=2}^N[a_i,  b_i] }\widehat{\mathcal{M}}_{2\beta}^{c_1^{-\beta}h(\hat{t})|t_1|^\gamma}\prod_{i=2}^N\sqrt{ c_i(\hat{t})}d\hat{t} u^{\frac{N-1}{\beta}}\Psi(u),\quad u\to\infty, S\to\infty.
\EQNY
{\it \underline{Upper bound for $\pi_1^+(u)$}}. In view of (\ref{pkj}), we have, as $u\to\infty$,
\BQNY
\begin{split}
	\pi_1^+(u)&\leq \sum_{k=1}^{M_u^+}\sum_{j=1}^{n^{N-1}}\pk{\sup_{t\in I_{k,j}(u)}\overline{X}(t)>u_{k,j,\ep}}\\
	&\leq \sum_{k=1}^{M_u^+}\sum_{j=1}^{n^{N-1}}\mathcal{H}_{2\beta}^N([0, (c_1+\ep)^{1/2}S])\prod_{i=2}^N(c_{i,j}+\ep)^{1/2}{\rm mes}(D_j)u^{\frac{N-1}{\beta}}\Psi(u_{k,j,\ep}).
\end{split}
\EQNY
Note that as $u\to\infty$,
\BQNY
\begin{split}
	\sum_{k=1}^{M_u^+}\Psi(u_{k,j,\ep})&\sim \Psi(u)\sum_{k=1}^{M_u^+}e^{-(1-\ep)h_j|kS|^{\gamma}}\leq \Psi(u)e^{-CS^{\gamma}},
\end{split}
\EQNY
where $C>0$ is a positive constant. It follows that
\BQN\label{pi1+}
\begin{split}
	\pi_1^+(u)
	&\leq Se^{-CS^{\gamma}}\sum_{j=1}^{n^{N-1}}\frac{\mathcal{H}_{2\beta}^N([0, (c_1+\ep)^{1/2}S])}{S}\prod_{i=2}^N(c_{i,j}+\ep)^{1/2}{\rm mes}(D_j)u^{\frac{N-1}{\beta}}\Psi(u)\\
	&=o(\pi_1^-(u)), \quad u\to\infty, S\to\infty.
\end{split}
\EQN
{\it \underline{Upper bound for $\Sigma_5(u)$}}.
Let
\begin{equation*}
\begin{split}
\Lambda_5&=\{(j,j'): 1\leq j<j'\leq n^{N-1}, D_j\bigcap D_{j'}=\emptyset\},\\
\Lambda_6&=\{(j,j'): 1\leq j<j'\leq n^{N-1}, D_j\bigcap D_{j'}\neq\emptyset\}.
\end{split}
\end{equation*}
Then
\BQNY
\Sigma_5(u)\leq \Sigma_6(u)+\Sigma_7(u),
\EQNY
with
$$\Sigma_i(u)=\sum_{(j,j')\in \Lambda_{i-1}}\pk{\sup_{t\in I_{0,j}(u)}X(t)>u, \sup_{t\in I_{0,j'}(u)}X(t)>u}, \quad i=6,7.$$
Using same arguments as in those to get the upper bounds of $\Sigma_1(u)$ and $\Sigma_2(u)$, we can show that  $\Sigma_i(u)=o(\pi_1^-(u)), i=6,7$, as $u\to\infty$ and $n\to\infty$.
Hence
$$\pk{\sup_{t\in E_u}X(t)>u}\sim \int_{\hat{t}\in \prod_{i=2}^N[a_i,  b_i] }\widehat{\mathcal{M}}_{2\beta}^{c_1^{-\beta}h(\hat{t})|t_1|}\prod_{i=2}^N\sqrt{ c_i(\hat{t})}d\hat{t} u^{\frac{N-1}{\beta}}\Psi(u), \quad u\to\infty,$$
together with (\ref{main})-(\ref{upper2}), establishing the claim.

{\it \textbf{(iii) Case $\bm{\beta>\gamma/2}$}}. Observe that
\BQNY
\pi_2(u)\leq \pk{\sup_{t\in E_u}X(t)>u}\leq \pi_1^-(u)+\pi_1^+(u),
\EQNY
with $\pi_1^-(u)$ and $\pi_1^+(u)$ defined in (\ref{Bonferroni1}) and
$$\pi_2(u)=\pk{\sup_{\hat{t}\in \prod_{i=2}^N [a_i,b_i] }X(t_1^*,\hat{t})>u}.$$
\textit{\underline{Upper bound of $\pi_1^\pm(u)$}}. By (\ref{var}) and (\ref{pi1-}) , it follows that for any $q, \ep>0$  and sufficiently large $S_1$, as $u\to\infty, n\to\infty$,
	\BQNY
	\begin{split}
		\pi_1^-(u)&\leq \sum_{j=1}^{n^{N-1}}\pk{\sup_{t\in I_{0,j}(u)}\frac{\overline{X}(t)}{1+q|t_1-t_1^*|^{2\beta}}>u}\\
		&\sim \mathcal{M}_{2\beta}^{c_1^{-\beta}q|t_1|^{2\beta}}[0,S]\int_{\hat{t}\in \prod_{i=2}^N[a_i,  b_i] }\prod_{i=2}^N\sqrt{ c_i(\hat{t})}d\hat{t} u^{\frac{N-1}{\beta}}\Psi(u)\\
		&\leq (1+\ep)\frac{\mathcal{P}_{2\beta}^{c_1^{-\beta}q|t_1|^{2\beta}}([0,S]\times[0,S_1]^{N-1})}{S_1^{N-1}}\int_{\hat{t}\in \prod_{i=2}^N[a_i,  b_i] }\prod_{i=2}^N\sqrt{ c_i(\hat{t})}d\hat{t} u^{\frac{N-1}{\beta}}\Psi(u).
	\end{split}
	\EQNY
	By the fact that
	$$\lim_{S_1\to\infty}\lim_{q\to\infty}\frac{\mathcal{P}_{2\beta}^{c_1^{-\beta}q|t_1|^{2\beta}}([0,S]\times[0,S_1]^{N-1})}{S_1^{N-1}}=
	\lim_{S_1\to\infty}\frac{\mathcal{H}_{2\beta}^N(\{0\}\times[0,S_1]^{N-1})}{S_1^{N-1}}=\mathcal{H}_{2\beta}^{N-1},$$
	we have that as $u\to\infty$,  $n\to\infty$, $q\to\infty$, $S_1\to\infty$ and $\ep\to 0$,
	\BQNY
	\begin{split}
		\pi_1^-(u)
		\leq \mathcal{H}_{2\beta}^{N-1}\int_{\hat{t}\in \prod_{i=2}^N[a_i,  b_i] }\prod_{i=2}^N\sqrt{ c_i(\hat{t})}d\hat{t} u^{\frac{N-1}{\beta}}\Psi(u).
	\end{split}
	\EQNY
	Using the same argument as in  (\ref{pi1+}), we have that
	$$\pi_1^+(u)=o(\pi_1^-), \quad u\to\infty, S\to\infty.$$
	\textit{\underline{Asymptotics of $\pi_2(u)$}}. Note that $X(t_1^*,\hat{t})$ is a Gaussian random field with unit variance and correlation function satisfying
	\BQNY
	\lim_{\delta\to 0}\sup_{\hat{s}\neq \hat{t}, \hat{s},\hat{t}\in \prod_{i=2}^N[a_i,  b_i], |\hat{t}-\hat{s}|\leq \delta }\left|\frac{1-{\rm Corr}(X(t_1^*,\hat{t}), X(t_1^*,\hat{s}))}{ \left(\sum_{i=2}^Nc_i(\hat{t})(t_i-s_i)^2\right)^\beta}-1\right|=0,
	\EQNY
	and $${\rm Corr}(X(t_1^*,\hat{t}), X(t_1^*,\hat{s}))<1, \quad \hat{t}\neq \hat{s}, \hat{t}, \hat{s}\in  \prod_{i=2}^N[a_i,  b_i].$$
	By Theorem 7.1 in \cite{Piterbarg:1996}, we have
	$$\pi_2(u)\sim\mathcal{H}_{2\beta}^{N-1}\int_{\hat{t}\in \prod_{i=2}^N[a_i,  b_i] }\prod_{i=2}^N\sqrt{ c_i(\hat{t})}d\hat{t} u^{\frac{N-1}{\beta}}\Psi(u), \quad u\to\infty.$$
	Therefore, we conclude that
	$$\pk{\sup_{t\in E_u}X(t)>u}\sim \mathcal{H}_{2\beta}^{N-1}\int_{\hat{t}\in \prod_{i=2}^N[a_i,  b_i] }\prod_{i=2}^N\sqrt{ c_i(\hat{t})}d\hat{t} u^{\frac{N-1}{\beta}}\Psi(u), \quad u\to\infty,$$
	together with  (\ref{main})-(\ref{upper2}), establishing the claim. This completes the proof. \QED

\begin{small}


TITLE: integrals of vector fields that yield vectors, not scalars
QUESTION [6 upvotes]: When I tried to think of how I'd answer this question, I realized that never in my undergraduate curriculum was I asked to compute the surface or line integral of a vector field. I don't mean I've never been asked to compute the flux or circulation of a field (meaning the field dotted with the surface normal, in the case of a surface integral, or the field dotted with a tangent vector, in the case of a line integral). I mean I've never been asked to compute things like $$\int_Sfd\mathbf{S}, \int_S \mathbf{f}dS, \int_c\mathbf{f}\times d\mathbf{r}$$
(where vectors are bolded and scalars aren't). I have two questions:

In what contexts do such integrals -- integrals of vector fields that yield a vector rather than a scalar -- arise?
Why are integrals like these pretty much never encountered in a standard course in undergraduate vector calculus? (I cannot think of a textbook where I could find them. If you can, please mention it.)

NOTE: I'm not asking how to do these integrals; I realize you can just compute them component-wise.

REPLY [1 votes]: These multidimensional scenarios arise in pdes, take for instance the navier stokes equations:
$\rho(\frac{\partial v}{\partial t}+v\cdot\nabla v)-\mu\Delta v+\nabla p=f(x)$
This system of equations in defined in $\Bbb R^n$, and it's derivation relies on the integration of vector fields.


TITLE: Can you prove that Average(f(x)) is not equal to f(average(x)) for non-linear f in more than one variable
QUESTION [2 upvotes]: I am seeking a general mathematical proof & a reference for the proof for something I know intuitively to be true, and can demonstrate by example, but would like to prove. Assume a function with 6 inputs [ f(z,y,x,w,v,u) ] that is non-linear with respect to all six variables. I do not get the same answer averaging the outputs vs using the average input and and then averaging the outputs. I need to know if there is a way to prove that this will always be the case, or to identify the minimum set of assumptions required to assert that it will not be true, or alternatively to identify the set of assumptions that MUST be true in order to get the same result (such that I might infer that if those assumptions are not true, then the results will not be equal).
This is not for school, it is for work. I am at a loss for where to find a proof for something so simple, but I need it to convince a co-worker who does not seem to believe my examples.
thanks

REPLY [6 votes]: Counterexample:
Let $\lbrace e_i \rbrace_{i\in I}$ be a Hamel basis, a basis for $\mathbb R$ over $\mathbb Q$, which means every real can be written uniquely as a finite linear combination of the basis elements with rational coefficients. The set of indices $I$ is uncountable. Choose a particular basis element $e_0$ (which might be taken to be $1$). Let $f$ assign to $(z,y,x,w,v,u)$ the coefficient of $e_0$ in the expression of $z$ as a rational linear combination of the Hamel basis. Then $f$ is discontinuous (it only takes rational values) and therefore not linear over the reals, but it satisfies $\operatorname{Avg}(f) = f(\operatorname{Avg})$ because it is $\mathbb Q$-linear.
There are foundational issues. I can't write down a Hamel basis, but one exists if you assume the Axiom of Choice. See this question.
If you assume that $f$ is continuous, or just measurable then it must actually be linear.


TITLE: Vertical Diameter of Convex Domains
QUESTION [2 upvotes]: Suppose we have a bounded, strictly convex domain $D\subset \mathbb{R}^2$ with smooth boundary with strictly positive curvature. Suppose further that the projection of $D$ onto the horizontal coordinate axis is given by the interval $[0,2]$. Now, for any $x\in [0,2]$ we can consider the vertical diameter $d_D(x)$ which is the length of the intersection of $D$ with a vertical ray through $x$.
For instance for $D$ the unit disc with midpoint $(1,0)^\top$ we have $d_D(x)= 2\sqrt{2x-x^2}$
for $x\in [0,2]$. Let us make the definition $d_0(x):=2\sqrt{2x-x^2}$.
For a convex set $D$ as above consider the quotient
$$\delta_D(x):= \frac{d_D(x)}{d_0(x)}.$$
It can be checked that $\delta_D$ is bounded from above and below for any convex $D$. My question is whether it can be shown that $ \delta_D $ is actually $C^\infty$ on $[0,2]$, that is with all derivatives
bounded as we approach the endpoints?

REPLY [8 votes]: Yes. Incidentally, just recently I had to write down a proof of a similar fact in one of my papers. It is quite technical.
Let us work at the endpoint $x=0$. We have to prove that the function $x\mapsto d_D(x)/\sqrt x$ is $C^\infty$. We need the following well-known facts about $C^\infty$ functions $f$ defined in a neighborhood of 0:

If $f(0)=0$, then $f(x)=xg(x)$ for some $g\in C^\infty$.
If $f(0)=f'(0)=0$, then $f(x)=x^2g(x)$ for some $g\in C^\infty$.
If $f(x)=f(-x)$ for all $x$, then $f(x)=g(x^2)$ for some $g\in C^\infty$.

(See this and this MO questions.)
We may assume that the boundary of $D$ contains the origin $(0,0)$. Then the boundary of $D$ near the origin is a graph $x=f(y)$ of a function $f\in C^\infty$ satisfying $f(0)=f'(0)=0$ and $f''(0)>0$. By the 2nd item above, we can write $f(y)=y^2g(y)$ where $g\in C^\infty$ and $g(0)=\frac12 f''(0)>0$. Let $h(y)=y \sqrt{g(y)}$, then $f(y)=h(y)^2$. Observe that $h(0)=0$ and $h'(0)>0$, so $h$ is invertible near 0. Denote $\varphi=h^{-1}$.
We can write $d_D(x)=d^+(x)-d^-(x)$ where $d^+(x)$ and $d^-(x)$ are the $y$-coordinates of the highest and lowest intersection point of $D$ and the vertical line through $(x,0)$. The values $d^\pm(x)$ are the solutions of the equation $f(y)=x$ (in the variable $y$), or, equivalently, $h(y)=\pm\sqrt x$, so
$$
 d^\pm(x) = \varphi(\pm\sqrt x) .
$$
It remains to prove that the function
$$
 x \mapsto \frac{\varphi(\sqrt x)-\varphi(-\sqrt x)}{\sqrt x}
$$
is $C^\infty$ on $[0,\varepsilon)$. Define $\psi(x)=\phi(x)-\phi(-x)$. The function $\psi$ is smooth and odd (i.e. $\psi(-x)=-\psi(x)$), therefore, by 1 and 3 above, it can be written in the form $\psi(x)=x \lambda(x^2)$ where $\lambda\in C^\infty$. Now we have
$$
 \lambda(x) = \frac{\psi(\sqrt x)}{\sqrt x}
= \frac{\varphi(\sqrt x)-\varphi(-\sqrt x)}{\sqrt x}
$$
for all $x\ge 0$, and $\lambda\in C^\infty$, q.e.d.


TITLE: Definition of complex derivative in polar form
QUESTION [2 upvotes]: I have this defintion
$$\dfrac{d}{dz}f(z)=\lim\limits_{\Delta z\to0}\dfrac{f(z+\Delta z)-f(z)}{\Delta z}$$
and I know how to rewrite it in terms of the Cartesian representation of complex numbers as
$$\dfrac{d}{dz}f(z)=\lim\limits_{\substack{\Delta x\to0\\\Delta y\to0}}\dfrac{f\big((x+iy)+(\Delta x+i\Delta y)\big)-f\big((x+iy)\big)}{\Delta x+i\Delta y}$$
My question is, what is the corresponding expression for $z=re^{i\theta}$?  Do I just write $\Delta z$ as $\Delta re^{i\Delta\theta}$?

REPLY [0 votes]: In the Cartesian case $z$ was a function of two variables $z(x,y)=x+iy$, and the partial derivatives are $\frac{\partial z}{\partial x}=1,\frac{\partial z}{\partial y}=i$, that is the gradient of $z(x,y)$ is constant and equal to the vector $(1,i)$ so that applying ([1]) $dz$ to $(\Delta x,\Delta y)$ gives $\Delta x+ i \Delta y$.
The same reasoning will hold in the case of polar coordinates. Here:
$$z(r,\theta)=re^{i\theta}$$
so
$$\frac{\partial z}{\partial \theta}=rie^{i\theta},\quad \frac{\partial z}{\partial r}=e^{i\theta}$$
hence, the result of applying $dz$ in these coordinates to $(\Delta r,\Delta\theta)$ is:
$$e^{i\theta}\Delta r + rie^{i\theta}\Delta\theta$$
[1]("Applying" the gradient to a vector means taking the inner-product of the gradient with that vector)


TITLE: Fast evaluation of quadratic matrix product when changed slightly
QUESTION [0 upvotes]: Suppose I know that $x^T A x = b$, where $x$ is a vector with $x_i\in\{0,1\}$, $A$ is square with real numbers as elements.
Now suppose I flip two of the "bits" of $x$ i.e. change one of $x_i$ from $0$ to $1$ and another $x_i$ from $1$ to $0$.
Is there a fast way of calculating (for very large sized $A$) the new matrix product $x^{*T}Ax^*=c$ using the previous result?

REPLY [1 votes]: You're looking for the result of $y^TAy$, where $y$ (your $x^*$) is a vector such that $y-x$ has exactly two non-zero entries (one of which is $1$ and the other $-1$).
The value of $x^TAx$ alone isn't really enough information to do anything useful, but if you also stored the result $(A + A^T)x$, then you could use the following fact. Let $\delta = y-x$. We have
$$
y^TAy = (x+\delta)^TA(x + \delta) = x^TAx + [Ax + A^Tx]^T\delta + \delta^TA\delta.
$$
Because $\delta$ is sparse, the products $[(A + A^T)x]^T\delta$ and $\delta^TA\delta$ can be quickly computed (in particular by adding/subtracting 6 values).


TITLE: Counting poker hands: full house, straight, $4$ of a kind, straight flush
QUESTION [0 upvotes]: A deck of cards contains $52$ cards. There are $13$ different kinds of cards, with four cards of each kind. (Among the terms commonly used instead of “kind” are “rank,” “face value,” “denomination,” and “value.”) These kinds are twos, threes, fours, fives, sixes, sevens, eights, nines, tens, jacks, queens, kings, and aces. There are also four suits: spades, clubs, hearts, and diamonds, each containing $13$ cards, with one card of each kind in a suit. A hand consists of five cards.
How many possibilities are there for the following poker hands?:
(1) Full house = $3$ of one kind and $2$ of another kind
(2) Straight = $5$ consecutive kinds (note that Ace can be high or low)
(3) $4$ of a kind
(4) Straight flush = five cards of the same suit of consecutive kinds
I calculate answers for each above, but I am not totally sure about my answer. Can someone take a look? Thanks in advance!
For (1), $C(13, 2) \cdot 2 \cdot 4^5$ since we choose $2$ kinds from $13$ kinds, and $3$ cards can have one of the kind, while the other $2$ cards can have the other kind. Since there are two choices for which kind to be for the three kinds, we multiply by $2$. Each card can be one of four suits, so we multiply $4^5$.
For (2), there are $10$ consecutive kinds. Each cards can be one of $4$ suits. Thus, $4^5 \cdot 10$.
For (3), I think there are $13$ kinds in total, so the answer is $13$.
For (4), there are $10$ consecutive kinds and four suits totally. So $10 \cdot 4 = 40$.

REPLY [0 votes]: Full house: There are $\binom{13}{1}$ ways to choose the rank from which three cards are drawn.  There are $\binom{4}{3}$ ways to choose three of the four suits for that rank.  There are $\binom{12}{1}$ ways to choose the rank from which two cards are drawn from the remaining ranks.  There are $\binom{4}{2}$ ways to choose two of the four suits of that rank.
$$\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{2}$$
Your answer is incorrect since you do not have four choices for the rank of each card since choosing a card of a particular rank reduces the number of choices you have for selecting another rank of that card.  Suppose you select three jacks and two fives. There only four ways to select three of the four jacks since one of those four ranks must be omitted

$\color{red}{J\diamondsuit}, \color{red}{J\heartsuit}, J\clubsuit$
$\color{red}{J\diamondsuit}, \color{red}{J\heartsuit}, J\spadesuit$
$\color{red}{J\diamondsuit}, J\clubsuit, J\spadesuit$
$\color{red}{J\heartsuit}, J\clubsuit, J\spadesuit$

There are only $\binom{4}{2} = 6$ ways to select two of the four fives.

$\color{red}{5\diamondsuit}, \color{red}{5\heartsuit}$
$\color{red}{5\diamondsuit}, 5\clubsuit$
$\color{red}{5\diamondsuit}, 5\spadesuit$
$\color{red}{5\heartsuit}, 5\clubsuit$
$\color{red}{5\heartsuit}, 5\spadesuit$
$5\clubsuit, 5\spadesuit$

Straight:  In poker, a straight is usually defined as five cards of consecutive ranks, not all of the same suit, where an ace can be high or low.  However, we will use the author's definition.
Choosing the rank for the low card determines which five ranks appear in the straight.  The low card in a straight can be selected in ten ways (A, 2, 3, 4, 5, 6, 7, 8, 9, 10).  The suit for each rank can be selected in four ways.
$$\binom{10}{1}\binom{4}{1}^5$$
so your answer is correct.
Under the usual definition, we would have to exclude straight flushes, that is, a straight in which all cards are of the same suit.
$$\binom{10}{1}\left[\binom{4}{1}^5 - 4\right]$$
Four of a kind:  There are $\binom{13}{1}$ ways to choose the rank from which four cards are drawn.  We must select all four cards of that rank.  There are $\binom{12}{1}$ ways to choose the rank from which the other card is drawn.  There are $\binom{4}{1}$ ways to choose the suit of that card.
$$\binom{13}{1}\binom{4}{4}\binom{12}{1}\binom{4}{1}$$
Your answer is incorrect because you forgot to account for the rank and suit of the other card.
Straight flush: There are $\binom{10}{1}$ ways to choose the low card and $\binom{4}{1}$ ways to choose the suit.
$$\binom{10}{1}\binom{4}{1}$$
so your answer is correct.


TITLE: If $A^n = B$ and I know $B$, can I find $A$?
QUESTION [6 upvotes]: Suppose that $A$ and $B$ are invertible, $p \times p$ matrices. If $A^n = B$ and I know all of the entries in $B$, can I find an $A$ for some or all integers $n \ge 0$? How many solutions for $A$ exist? If I'm thinking correctly, then $A = B * (A^{-1})^{n-1},$ but this is sort of self referential. Thanks!

REPLY [2 votes]: We can use diagonalisation of $B$ to constract a solution of this equation, that mean if $B=PDP^{-1}$ and we suppose that $D=Diag(\lambda_1,\lambda_2,\dots,\lambda_k)$ if you put $A=PRP^{-1}$ where
$$
R=Diag(\sqrt[n]{\lambda_1},\sqrt[n]{\lambda_2},\dots,\sqrt[n]{\lambda_k})
$$
So you can verify that $A^n=B$


TITLE: Triangle perimeter and area
QUESTION [0 upvotes]: If you take the square root of the perimeter of a triangle and double it, you get the area of the triangle. The sides have integer lengths which are mutually different from each other. What are the lengths of its sides?

REPLY [0 votes]: Without the restriction that the integers are different, there are two other solutions: 2, 9, 9 and 4, 4, 4.


TITLE: Convergence in $L^2$ norm
QUESTION [1 upvotes]: Given a sequence of measurable functions $f_n: \mathbb{R} \rightarrow \mathbb{R}$ such that $\sup_n \|f_n\|_{L^2} < \infty$ where $f_n \rightarrow f$ almost everywhere, is it true that $$\int_{\mathbb{R}}^{ } |f_n(x)-f(x)|^2 dx \rightarrow 0?$$
In other words, does $f_n \rightarrow f$ in $L^2$ norm?

REPLY [1 votes]: This is not true, as was pointed out in the comments. David Mitra gave perhaps the simplest counter-example $f_n = \chi_{[n,n+1]}$. Another class of counter-examples (also with $f=0$) can be had be had by choosing $|f_n(x)|^2$ to be a nascent delta function; for example, $f_n = \sqrt n \chi_{[0,1/n]}$. These show that even replacing the domain of the functions with a compact interval won't make the statement true.


TITLE: The tensor product of monads.
QUESTION [1 upvotes]: It is known that the tensor product of endofunctors End(C) over a given category C is given by composition and the category of monads Mon(C) over a given category is cartesian. That cartesian product is roughly described in Monad Compositions I (Manes and Mulry) for a given set-indexed family of monads $\mathbf{T_i}=(T_i,\mu_i,\eta_i)$ as $\mathbf{T}=(T,\mu,\eta)$ by the rules:
$\hspace{15em}{TX}=\prod{{T_i}X}$
   $\hspace{15em}{\pi_j\eta_X}=\eta_{jX}$
 $\hspace{15em}{\pi_j{(X\rightarrow{TY})}}={(TX\rightarrow{T_jX}\rightarrow{T_jY})}$
 $\hspace{15em}{\pi_j{(TTX\rightarrow{TX})}}={(TTX\rightarrow{T_jT_jX}\rightarrow{T_jX})}$
where I've omitted the name of the arrows.
Is there any reason for this product not to be a tensor product and hence the composition induced by the category End(C) where Mon(C) lives?
Remark: In my version of the paper the third rule was stated as
$\hspace{15em}{\pi_j{(X\rightarrow{TY})}}={(TX\rightarrow{TX_j}\rightarrow{TY_j})}$
which I guess is wrong.

REPLY [3 votes]: $\mathrm{Mon}(C)$ does not have composition as a tensor. The composite of two monads is not a monad in any natural way unless there is a distributive law between them. Lots of research goes into getting around this fact, e.g. the paper you cite.
For example, for any set $A$ there is a functor $\Delta_A: \mathsf{Set} \to \mathsf{Set}$ which is constant at $A$: it sends all objects to $A$ and all morphisms to $\mathrm{id}_A$. If $A$ has more than one element, then $\Delta_A$ does not admit a monad structure (this is a fun exercise). (It's also true that if $A$ has less than one element, then $\Delta_A$ does not admit a monad structure!)
Now let $T$ be a monad on $\mathsf{Set}$ such that $T(1) \neq 1$. Consider also the monad $\Delta_1$ which is constant at 1. We can compose the underlying functors $T\circ \Delta_1 = \Delta_{T(1)}$ to get the constant functor at the set $T(1)$. But this composite functor does not admit a monad structure.


TITLE: Is the image of this operator on $\ell^2$ closed?
QUESTION [2 upvotes]: Suppose that $\ell^2 = \biggl\{(x_n)_n \in \mathbb{K}^{\mathbb{N}_0} \biggm|
\sum_{n=1}^{\infty}|{x_n}^2| < +\infty \biggr\}$ is a Hilbert-space with the inproduct $\langle\cdot,\cdot\rangle_2: \ell^2 \to \ell^2: (x,y) \mapsto \sum_{n=1}^\infty \overline{x_n}y_n$.
Consider the operator $f: \ell^2 \to \ell^2: (x_0, x_1, \ldots) \mapsto (x_0, 0, x_1, 0, \ldots)$.
I'm supposed to check whether the image of $f$ is closed or not. First of all, I don't think the image can be the whole $\ell^2$-space, because there's no element that's mapped on $\left(\frac{1}{n}\right)_n$. I tried to reason on like this:
If one should show that $\operatorname{Im}(f)$ is closed, it's sufficient that $\overline{\operatorname{Im}(f)} \subset \operatorname{Im}(f)$ (I'm having the habit to use $\overline{\cdot}$ for closure). Therefore, choose an element $(x_n)_n$ in $\overline{\operatorname{Im}(f)}$. As the element lies in the closure of $\operatorname{Im}(f)$, there should be a sequence $(x_n^k)_n$ that converges to $(x_n)_n$. That means that $\forall n: \lim_{k\to\infty}x_n^k=x_n$. I'm stuck at this point, because I don't know what the image of $f$ is.
Is this a good start or is there an easier way?

REPLY [2 votes]: Your notation seems off, but it seems you have the basic idea down. Try this:
Take $x$ in the closure of the image of $f$ and choose a sequence $(x_n)$ from the image of $f$ such that $x_n\rightarrow x$ in norm. Note the even coordinates of any $x_n$ are all zero.
If one denotes the $i$'th coordinate of a vector $y$ by $y(i)$, then for each $i$, since convergence in $\ell_2$ implies coordinatewise convergence, we have $\lim\limits_{n\rightarrow\infty} x_n(i)=x(i)$.
In particular, $x(i)=0$ if $i$ is even.  Let $z=(x_1,x_3,\ldots)$. Then $f(z)=x$; which shows $x$ is in the image of $f$.
Thus the image of $f$ is closed.
Alternatively, you could use the fact that $f$ is an isomorphism (isometry in fact):
Take $(y_n)$  in $f(\ell_2)$ with $y_n\rightarrow y$ for some $y\in\ell_2$. Then $(f^{-1}(y_n))$ is Cauchy in $\ell_2$ and thus converges to some $x\in\ell_2$. Then since $f(f^{-1}(y_n))=y_n$ and since $f$ is continuous, we have $y_n\rightarrow f(x)$. It follows that $f(x)=y$. It follows that $f(\ell_2)$ is closed.
You could also note (prove!) that the image of $f$ is simply the set of all vectors $x$ in $\ell_2$ such that $x(i)=0$ whenever $i $ is even. This set is closed (argue as in the first part of this answer).