text:
\begin{document}

\title{How long does the surplus stay close to its historical  high?}
\author{ Bo Li, Yun Hua and Xiaowen Zhou}

\AtEndDocument{\bigskip{\footnotesize
  \textsc{School of Mathematics and LPMC, Nankai University, Tianjin, P.R.China} \par
  \textit{E-mail address}, Bo Li: \texttt{libo@nankai.edu.cn} \par
  \addvspace{\medskipamount}
  \textsc{School of Mathematics and LPMC, Nankai University, Tianjin, P.R.China} \par
  \textit{E-mail address}, Yun Hua: \texttt{Huayun@mail.nankai.edu.cn} \par
  \addvspace{\medskipamount}
  \textsc{Department of Mathematics and Statistics, Concordia University} \par
  \textit{E-mail address}, Xiaowen Zhou: \texttt{xiaowen.zhou@concordia.ca}
}}

\date{}

\maketitle

\begin{abstract}
In this paper we find the Laplace transforms of the weighted occupation times for a spectrally negative L\'evy surplus process to spend below its running maximum up to the first exit times. The results are expressed in terms of generalized  scale functions. For step weight functions, the Laplace transforms can be further expressed in terms of scale functions.
\end{abstract}

\section{Introduction}\label{intro}

In risk theory, in addition to ruin behaviors, the surplus process's behaviors before ruin are also of interest.  Intuitively, it is an ideal situation if the surplus has typically small downward fluctuations from its historical   high.   It is thus interesting to know how long the surplus stays near its running maximum up to certain times. The amount of such time can serve  as  a criterion  for performance of the surplus process. We are not aware of any previous work along this line, which
 motivates our study of occupation times associated to the running maximum of a spectrally negative L\'evy process.

Since the work of \cite{Landriault2011:occupationtime:levy}, the occupation times  have been studied with different approaches in a series of papers for spectrally negative L\'evy processes and related processes such as the reflected or refracted spectrally negative L\'evy processes; see \cite{ZhouBo:localtime} for a summary of these results. So far all of these results concern the occupation times over deterministic intervals.
It is therefore also interesting, from the point of view of occupation time theory, to study the occupation time  over an interval with random boundaries. To our best knowledge, such occupation times have not been studied for spectrally negative L\'evy processes before, and our work represents the first attempt in this direction.

In this paper we consider the Laplace transforms of weighted occupation times the spectrally negative L\'evy process spends near its running maximum.  By considering the time it spends below the running maximum,
we can relate the problem with a problem on reflected L\'evy process. To this end we modify the approach of
\cite{BoLi:sub1}.  Taking use of Feynman-Kac type equations, see e.g. III.19 of \cite{Rogers:book:v1}, we express  the desired Laplace transforms using a generalized scale function, which is the unique solution to an integral equation involving the scale function and the weight function. Similar arguments to obtain Laplace transforms of weighted occupation times for  refracted spectrally negative L\'evy processes can be found in \cite{LiZh18}.
  When the weight function is a step function, the generalized scale function can be further expressed in terms of iterated integrals of the scale functions for spectrally negative L\'evy processes.

This paper is structured as follows. After the introduction in Section \ref{intro}, the main results and examples are presented in Section \ref{main}. Proofs of the main results are deferred to Section \ref{proof}.

\section{Main results}\label{main}

Let $X$ be a spectrally negative L\'{e}vy process,
$S:=\{S_{t}:=\sup_{s\in[0,t]}X_{s}, t\geq0\}$ be the running maximum process of $X$,
and $Y:=S-X=\{S_{t}-X_{t}, t\geq0\}$ be the reflected process from the running  maximum.
In this paper, we are interested in the
Laplace transforms of weighted occupation times of $Y$ up to the first passage times of $X$.

More precisely,
let $\omega$ be a nonnegative, locally bounded measurable function on $\mathbb{R}$,
we want to identify expressions of Laplace transforms
\[
\bE_{x}\big(e^{-L(\tau_{b}^{+})}; \tau_{b}^{+}<\tau_{0}^{-}\big)
\quad\text{and}\quad
 \bE_{x}\big(e^{-L(\tau_{0}^{-})}; \tau_{0}^{-}<\tau_{b}^{+}\big),
\]
where $\D L(t):=\int_{0}^{t}\omega(Y_{s})\,ds$ denotes the $\omega$-weighted occupation time near $S$,
and where the first passage times of $X$ are defined by
\begin{equation}
\tau_{x}^{+}:=\inf\{t>0, X_{t}>x\}
\quad\text{and}\quad
\tau_{x}^{-}:=\inf\{t>0, X_{t}<x\}
\end{equation}
with the convention $\inf\emptyset:=\infty$.
Notice that, for the case of $\omega(z)=\mathbf{1}(z\in[a,b))$ for some $b>a>0$,
$$\D L(t)=\int_{0}^{t}\mathbf{1}(S_{r}-b<X_{r}\leq S_{r}-a)\,dr$$
 is the occupation time process $X$ spends in a random interval below its maximum process $S$, which is also studied in examples at the end of this section.

In the fluctuation theory for L\'evy processes, the interested quantities are often expressed using scale functions $W^{(q)}$ and $Z^{(q)}$, where for $q\geq 0$, $W^{(q)}$ is a nonnegative and increasing function satisfying $W^{(q)}(x)=0$ for $x<0$ and
\[\int_0^\infty W^{(q)}(x)e^{-\lambda x}dx=\frac{1}{\psi(\lambda)-q}, \,\, \lambda>\Phi(q), \]

and
\[\D Z^{(q)}(x):=1+ q\int_{0}^{x}W^{(q)}(y)\,dy.\]
Here $\psi(\lambda):=\log \bE e^{\lambda X_1}$ for $\lambda\geq 0$ denotes the Laplace exponent for $X$ and  $\Phi(q):=\inf\{\lambda>0, \psi(\lambda)>q\}$ denotes the right inverse function of $\psi$. In addition,
\[\psi(\lambda)=\mu\lambda+\frac{1}{2}\sigma^2\lambda^2+\int_{(0,\infty)}(e^{-\lambda x}-1+\lambda x\mathbf{1}_{(x<1)})\Pi(dx),\]
where the $\sigma$-finite L\'evy measure $\Pi$ on $(0, \infty)$ satisfies $\int_{(0, \infty)} 1\wedge x^2 \Pi(dx)<\infty$.
Write $W$ and $Z$ for $W^{(0)}$ and $Z^{(0)}$, respectively. We refer to \cite{Kyprianou2014:book:levy} for more detailed introduction on scale functions.

To express our results, we need the so called $\omega$-scale function first introduced in \cite{BoLi:sub1},
which is defined as the unique locally bounded function satisfying the following equation
\begin{equation}\label{eqn:defn:w:1}
\wo(x,y)=W(x-y)+ \int_{y}^{x}W(x-z)\omega(z)\wo(z,y)\,dz, \,\, x, y\in \mathbb{R}.
\end{equation}
The $\omega$-scale function is further studied in \cite{ZhouBo:localtime} and shown to satisfy a dual version of the above equation.
\begin{equation}\label{eqn:defn:w:2}
\wo(x,y)=W(x-y)+\int_{y}^{x}\wo(x,z)\omega(z)W(z-y)\,dz;
\end{equation}
see Lemma 2 of \cite{ZhouBo:localtime}.

Throughout this paper we always assume that the derivative $W^{\prime}(x)$ is continuous  for $x\in (0, \infty)$, which holds if process $X$ has a Brownian component or the L\'evy measure allows a density.
We refer the readers to \cite{chan2011:scalefunction:smooth} for more detailed discussion on the smoothness of scale functions.
For convenience we also assume that function $\omega$ is  right continuous with left limit.

For $x>y$ denote by
\[
\begin{split}
\wo_{1}(x,y)
:=& \lim_{z\to x+}\frac{\wo(z,y)-\wo(x,y)}{z-x}\\
\text{and}\quad
\wo_{2}(x,y)
:=& \lim_{z\to y+}\frac{\wo(x,z)-\wo(x,y)}{z-y}
\end{split}
\]
the right partial derivatives of $\wo(x,y)$ on  $x$ and $y$, respectively.
One can check that
\begin{align}
\wo_{1}(x,y)=&\ W'(x-y)+ \int_{y}^{x}W'(x-z)\omega(z)\wo(z,y)\,dz+
W(0)\omega(x)\wo(x,y),\non\\
\label{eqn:wy}
\wo_{2}(x,y)=&\ -W'(x-y)- \int_{y-}^{x}\wo(x,z)\omega(z)W(dz-y)
\end{align}
where $W(dz)$ is the Stieltjes measure induced by $W$ with $W(\{0\})=W(0)$ and it is known that $W(0)>0$ if process $X$ has sample paths of bounded variation.

Given $\wo$ we also need to introduce function
\begin{equation}\label{eqn:defn:zh}
\zoh(x,y):=1+\int_{y}^{x}\omega(z)\wo(z,y)\,dz
\end{equation}
for $x, y\in
\mathbb{R}$.
The right partial derivatives of $\zoh$ are given by
\[\zoh_{1}(x,y):= \lim_{z\rightarrow x+}\frac{\zoh(z,y)-\zoh(x,y)}{z-x} =\omega(x)\wo(x,y)\]
and
\[\zoh_{2}(x,y):=  \lim_{z\rightarrow y+}\frac{\zoh(x,z)-\zoh(x,y)}{z-y}
=\int_{y}^{x}\omega(z)\wo_{2}(z,y)\,dz-\omega(y)W(0).\]

\begin{rmk}
We remark that, from  equations \eqref{eqn:defn:w:1} and \eqref{eqn:defn:w:2},
$\wo$ is absolutely continuous with respect to Lebesgue measure in both $x$ and $y$.
But  it may fail to be differentiable for general weight function $\omega$.
However, since $\omega$ is assumed to be right-continuous with left limit, both
the right derivative and  left derivative of $\wo(x,\cdot)$  exist and
the integral equation \eqref{eqn:wy} can be proved directly for $x>y$.
Moreover, $\wo_{2}(x,\cdot)$ is continuous at $y_{0}$ if and only if $\omega$ is continuous at $y_{0}$,
and in such a case, $\wo(x,\cdot)$ is differentiable at $y_{0}$. The same is true for $\wo_{1}$, $\zo_{1}$ and $\zo_{2}$.
\end{rmk}

For simplicity we write $\wo(x)=\wo(x,0)$, $\zoh(x)=\zoh(x,0)$ and write
\[
\wo_{1}(x):=\wo_{1}(x,0)=W'(x)+ \int_{0}^{x}W'(x-z)\omega(z)\wo(z)\,dz+ W(0)\omega(x)\wo(x),
\]
\begin{equation}\label{wo_2}
\wo_{2}(x):=\ \wo_{2}(x,0)= -W'(x)- \int_{0-}^{x}\wo(x,z)\omega(z)W(dz),
\end{equation}
\[\zoh_{1}(x):=\ \zoh_{1}(x,0)=\omega(x)\wo(x),\]
\[\zoh_{2}(x):=\ \zoh_{2}(x,0)=\int_{0}^{x}\omega(z)\wo_{2}(z)\,dz-\omega(0)W(0).\]

Define for $u>0$
\begin{equation}
\label{defn:ho}
\ho(u):=\exp\Big(-\int_{1}^{u} \frac{\wo_{2}(z)}{\wo(z)}\,dz\Big),
\end{equation}
where for $0<u<1$ the integral $\int_1^u $ is understood as $-\int_u^1 $.

We first present a result on potential density. Note that the exit time involved is for process $X$.

\begin{prop}\label{lem:3}
For any $0\leq x<z<b$ and $z>y>0$, we have
\begin{align}\label{eqn:thm1:ro+}
 \ro(x;dz,dy)&:=
\int_{0}^{\infty} \bE_{x}\Big(e^{-L(t)}; S_{t}\in dz, Y_{t}\in dy, t<\tau_{b}^{+}\wedge\tau_{0}^{-}\Big)\,dt \non\\
&= \frac{\ho(x)}{\ho(z)}
\Big(\frac{\wo_{2}(z)}{\wo(z)}\wo(y)- \wo_{2}(y)\Big)\,dy\,dz
\end{align}
and $$\D\ro(x;dz,\{0\})= \frac{\ho(x)}{\ho(z)} W(0)\,dz.$$
\end{prop}

\begin{thm}\label{thm:1} For any $b>0$ and $x\in[0,b]$, we have
\begin{equation}\label{eqn:thm1:b+}
\bE_{x}\Big(\exp\Big(-\int_{0}^{\tau_{b}^{+}} \omega(Y_{t})\,dt\Big);
\tau_{b}^{+}<\tau_{0}^{-}\Big)=
\frac{\ho(x)}{\ho(b)},
\end{equation}
and
\begin{equation}\label{eqn:thm1:c-}
\begin{split}
&\ \bE_{x}\Big(\exp\Big(-\int_{0}^{\tau_{0}^{-}} \omega(Y_{t})\,dt\Big);
\tau_{0}^{-}<\tau_{b}^{+}\Big)\\
=&\ \zoh(x)- \frac{\ho(x)}{\ho(b)}\zoh(b)
+ \int_{x}^{b} \frac{\ho(x)}{\ho(z)}\big(\zoh_{1}(z)+\zoh_{2}(z)\big)\,dz.
\end{split}
\end{equation}
\end{thm}

\begin{rmk}
We remark that for a locally bounded measurable and nonnegative function $\omega$, the partial derivatives   of $\wo(\cdot,\cdot)$ and $\zoh(\cdot,\cdot) $  exist Lebesgue a.e. Then the results in Theorem \ref{thm:1} still hold with the right partial derivatives replaced with the respective partial derivatives.
\end{rmk}

If $\omega\equiv q$ for some $q>0$,
then $\wo(x,y)=W^{(q)}(x-y)$ and one can check that $\D \ho(u)=\frac{W^{(q)}(u)}{W^{(q)}(1)}$.
Therefore, expression \eqref{eqn:thm1:b+} reduces to the classical Laplace transform for the two-sided passage problem.
On the other hand,
\begin{align*}
&\ \zoh(x,y)= 1+q \int_{y}^{x}W^{(q)}(z-y)\,dz=Z^{(q)}(x-y),\\
&\ \zoh(x)= Z^{(q)}(x)\quad\text{and}\quad\zoh_{1}(x,y)+\zoh_{2}(x,y)=0,
\end{align*}
and \eqref{eqn:thm1:c-} also coincides with the corresponding classical result for the two-sided exit problem.

For the two-sided exit problems, one may also be interested in the time
$\tau_{c}^{-}$ instead of $\tau_{0}^{-}$  for some $c<b$.
Since $X$ is spatially homogenous, with a shift argument applied,
we have following results with general initial value.

\begin{cor} For any $x\in[c,b]$, we have
\begin{equation}
\bE_{x}\Big(\exp\big(-\int_{0}^{\tau_{b}^{+}} \omega(Y_{t})\,dt\big);
\tau_{b}^{+}<\tau_{c}^{-}\Big)=
\frac{\ho(x-c)}{\ho(b-c)},
\end{equation}
and
\begin{equation}
\begin{split}
&\ \bE_{x}\Big(\exp\big(-\int_{0}^{\tau_{c}^{-}} \omega(Y_{t})\,dt\big);
\tau_{c}^{-}<\tau_{b}^{+}\Big)\\
=&\ \zoh(x-c)- \frac{\ho(x-c)}{\ho(b-c)}\zoh(b-c)\\
&\quad + \int_{x}^{b} \frac{\ho(x-c)}{\ho(z-c)}\big(\zoh_{1}(z-c)+\zoh_{2}(z-c)\big)\,dz.
\end{split}
\end{equation}
\end{cor}

The Gerber-Shiu penalty function is of great interest in ruin theory,
which describes  the joint distribution of
the time of ruin, the surplus before ruin as well as the deficit at ruin.
It has been generalized to different forms.
With the occupation time near the running maximum taken into consideration,
we  have following version of Gerber-Shiu function.

\begin{prop}[Gerber-Shiu function]\label{prop:3}
For any $x,z\in[0,b]$ and $y>0$, we have
\begin{equation}\label{eqn:prop3:1}
\begin{split}
&\ \bE_{x}\big(e^{-\delta \tau_{0}^{-}-L(\tau_{0}^{-})};
X(\tau_{0}^{-}-)\in\,dz, |X(\tau_{0}^{-})|\in\,dy, S(\tau_{0}^{-})<b\big)\\
=&\ \left(\frac{\hoq(x)}{\hoq(b)}\woq(b-z)- \woq(x-z)\right) \Pi(dy+z)\,dz\\
&\quad -\Big(\int_{x}^{b} \frac{\hoq(x)}{\hoq(u)}
\big(\woq_{1}(u-z)+\woq_{2}(u-z)\big)\,du\Big) \Pi(dy+z)\,dz.
\end{split}
\end{equation}

\end{prop}
We can also find Laplace transform of the occupation time involving  the creeping event that occurs
when $X$ first exits  a lower level by hitting the level with positive probability,
 which is also called ruin caused by oscillation in ruin theory.
But more notations are needed and we omit it.

If $\omega$ is a step function,
a recursive expression for $\wo(x,y)$ in terms of scale function is given in (3.24) of \cite{BoLi:sub1};
 also see similar results in \cite{KuZh17} and \cite{LiZh18}.
\textcolor{red}{In this paper we present a similar result for  $\zoh(x,y)$ for $x,y\in\mathbb{R}$,
and an alternative recursive expression for $\wo(x,y)$.}
Note that $\zoh(x,y)=1$ for $x\leq y$ by definition.

Let $\{a_{k}\}_{k\geq1}$ with $a_{k+1}<a_{k}$ be a partition of $\mathbb{R}$,
and let $\{p_{k}\}_{k\geq0}$ be a sequence of nonnegative constants. Define $\omega_{0}(x):=p_{0}$ and for $n\geq1$
\begin{equation}\label{step_function}
\omega_{n}(x):= p_{0} \mathbf{1}(x\geq a_{1}) + \sum_{k=1}^{n-1} p_{k} \mathbf{1}(a_{k+1}\leq x<a_{k})
+ p_{n} \mathbf{1}(x< a_{n}).
\end{equation}
Denote by
$W_{n}(x,y):=W^{(\omega_{n})}(x,y)$ and
$\widehat{Z}_{n}(x,y):=\widehat{Z}^{(\omega_{n})}(x,y)$
the scale functions with respect to $\omega_{n}$.

\begin{prop}\label{prop:2}

$W_{0}(x,y)=W^{(p_{0})}(x-y)$ and for $n\geq 0$,
\[
W_{n+1}(x,y)=W_{n}(x,y)+ (p_{n+1}-p_{n})\int_{y}^{a_{n+1}}
W_{n}(x,z)W^{(p_{n+1})}(z-y)\,dz.
\]

$\widehat{Z}_{0}(x,y)=Z^{(p_{0})}(x-y)$ and for $n\geq 0$,

\[
\widehat{Z}_{n+1}(x,y)
=\widehat{Z}_{n}(x,y)+ (p_{n+1}-p_{n})\int_{y}^{a_{n+1}}
\widehat{Z}_{n}(x,z)W^{(p_{n+1})}(z-y)\,dz.
\]
\end{prop}
Noticing that for  $y>a_{n+1}$,
we have $\widehat{Z}_{n+1}(x,y)=\widehat{Z}_{n}(x,y)$ from the equation above,
since $W^{(p_{n+1})}(z-y)=0$ for $z\in(a_{n+1},y)$,
which can be observed from \eqref{eqn:prop:1} and used in our proofs.

Considering a special case that
\begin{equation}\label{step_function_a}
\omega(z)=p+ (q-p) \mathbf{1}(a_{2}\leq z<a_{1})
\end{equation}
 for some $q\geq p\geq 0$ and $0\leq a_{2}<a_{1}$,
which is the weight function first considered in \cite{Loeffen2014:occupationtime:levy} and \cite{Zhou2014:occupationtime:levy},
where the occupation time of intervals of spectrally negative L\'{e}vy process is studied and
 the following auxiliary functions are introduced.
\begin{align*}
W^{(p,q)}_{(a_{2})}(x):=&\ W^{(p)}(x)+(q-p)\int_{a_{2}}^{x}W^{(q)}(x-z)W^{(p)}(z)\,dz\\
=&\ W^{(q)}(x)- (q-p) \int_{0}^{a_{2}}W^{(q)}(x-z)W^{(p)}(z)\,dz,\\
W^{(p,q,p)}_{(a_{2},a_{1})}(x):=&\ W^{(p,q)}_{(a_{2})}(x)+ (p-q) \int_{a_{1}}^{x}W^{(p)}(x-z)W^{(p,q)}_{(a_{2})}(z)\,dz.
\end{align*}
It was pointed out  in \cite{BoLi:sub1} that
$W^{(p,q,p)}_{(a_{2},a_{1})}(x)=\wo(x,0)$ for the function $\omega$ defined above.

Note that $W^{(\omega_{u})}(u-y,u-x)=\wo(x,y)$ for every $u\in\mathbb{R}$ by Lemma \ref{lem:1},
where
\begin{equation*}
\begin{split}
\omega_{u}(z)&=\omega(u-z)=p+ (q-p) \mathbf{1}(a_{2}<u-z<a_{1})\\
&=p+ (q-p) \mathbf{1}(u-a_{1}<z<u-a_{2}).
\end{split}
\end{equation*}
Since
$\D \wo_{2}(x,y)=-W^{(\omega_{u})}_{1}(u-y,u-x)$,
taking $x=u=t$ and $y=0$ we have
\[
\frac{\wo_{2}(t)}{\wo(t)}
=\frac{\wo_{2}(t,0)}{\wo(t,0)}
=- \frac{W_{1}^{(\omega_{t})}(t,0)}{W^{(\omega_{t})}(t,0)}
=- \frac{W_{(t-a_{1},t-a_{2})}^{(p,q,p)\prime}(t)}{W_{(t-a_{1},t-a_{2})}^{(p,q,p)}(t)}.
\]
Therefore, we obtain the following alternative expression for the function $\omega$ given in (\ref{step_function_a}).
\[
\bE_{x}\Big(\exp\big(-\int_{0}^{\tau_{b}^{+}} \omega(Y_{t})\,dt\big);
\tau_{b}^{+}<\tau_{0}^{-}\Big)=
\exp\Big( -\int_{x}^{b}
\frac{W_{(t-a_{2},t-a_{1})}^{(p,q,p)\prime}(t)}{W_{(t-a_{2},t-a_{1})}^{(p,q,p)}(t)}\,dt
\Big).
\]
In addition,  by Proposition \ref{prop:2} function $\zoh\equiv\widehat{Z}^{(p,q,p)}_{(a_{2},a_{1})}$ for the above mentioned weight function $\omega$ is given by
\[\widehat{Z}^{(q,p)}_{(a_{1})}(x,y):=
Z^{(p)}(x-y)+ (q-p)\int_{y}^{a_{1}} Z^{(p)}(x-z)W^{(q)}(z-y)\,dz\]
 and
\[\widehat{Z}^{(p,q,p)}_{(a_{2},a_{1})}(x,y):=
\widehat{Z}^{(q,p)}_{(a_{1})}(x,y)+ (p-q)\int_{y}^{a_{2}} \widehat{Z}^{(q,p)}_{(a_{1})}(x,z)W^{(p)}(z-y)\,dz.\]
Then a more explicit expression for \eqref{eqn:thm1:c-} also follows.

To further simplify the expression,
let $a_{1}=a$ and $a_{2}=0$ for some $a>0$,
that is
\[\omega(z)=q\mathbf{1}(0\leq z<a)+p \mathbf{1}(z\geq a), \,\,\, z\geq0.\]
We have by definitions that for $t\geq0$,
\begin{align*}
W_{(t-a)}^{(p,q)}(t)
=&\ W^{(p)}(x)+ (q-p) \int_{t-a}^{t}W^{(q)}(t-z)W^{(p)}(z)\,dz\\
=&\ W^{(p)}(x)+ (q-p)  \int_{0}^{a} W^{(p)}(t-s)W^{(q)}(s)\,ds
=W_{(a)}^{(q,p)}(t)\\
=&\ W^{(q)}(x)+ (p-q) \int_{a}^{t} W^{(p)}(t-s)W^{(q)}(s)\,ds,\\
\widehat{Z}^{(q,p)}_{(a)}(t)=&\ Z^{(p)}(t)+ (q-p)\int_{0}^{a}Z^{(p)}(t-s)W^{(q)}(s)\,ds\\
=&\ Z^{(q)}(t)+ (p-q)\int_{a}^{t}Z^{(p)}(t-s)W^{(q)}(s)\,ds.
\end{align*}

\begin{exam}
Let $X=\mu t+\sigma B_{t}$ be a Brownian surplus process,
where $\mu$ and $\sigma>0$ are constants and $B_{t}$ is a standard Brownian motion.
It is known that for $x>0$
\[
\begin{aligned}
W^{(q)}(x)=&\ \frac{e^{\rho_{1}x}- e^{\rho_{2}x}}{D\cdot (\rho_{1}-\rho_{2})}
& \text{and}&&
W^{(p)}(x)=&\ \frac{e^{\eta_{1}x}- e^{\eta_{2}x}}{D\cdot (\eta_{1}-\eta_{2})},\\
Z^{(q)}(x)=&\ \frac{\rho_{2}e^{\rho_{1}x}-\rho_{1} e^{\rho_{2}x}}{\rho_{2}-\rho_{1}}
&\text{and}&&
Z^{(p)}(x)=&\ \frac{\eta_{2}e^{\eta_{1}x}-\eta_{1} e^{\eta_{2}x}}{\eta_{2}-\eta_{1}},
\end{aligned}
\]
where $D=\sigma^{2}/2$,
$\rho_{1}>\rho_{2}$ are the two roots to equation $D s^{2}+ \mu s=q$
and $\eta_{1}>\eta_{2}$ are the two roots to equation $D s^{2}+ \mu s=p$.
We thus have
\begin{align*}
&\ W_{(t-a)}^{(p,q)}(t)=W^{(q,p)}_{(a)}(t)
=W^{(q)}(t)+ (p-q)\int_{a}^{t}W^{(p)}(t-s)W^{(q)}(s)\,ds\\
=&\ \frac{e^{\rho_{1}t}- e^{\rho_{2}t}}{D\cdot (\rho_{1}-\rho_{2})}+
\frac{(p-q)\mathbf{1}(t>a)}{D^{2}(\rho_{1}-\rho_{2})(\eta_{1}-\eta_{2})}
\int_{a}^{t}
\big(e^{\eta_{1}(t-s)}-e^{\eta_{2}(t-s)}\big)
\big(e^{\rho_{1}s}-e^{\rho_{2}s}\big)\,ds\\
=&\ \frac{e^{\rho_{1}t}- e^{\rho_{2}t}}{D\cdot (\rho_{1}-\rho_{2})}+
\frac{(p-q)\mathbf{1}(t>a)}{D^{2}(\rho_{1}-\rho_{2})(\eta_{1}-\eta_{2})}
\Big(\frac{e^{\rho_{1}t}-e^{\rho_{1}a+\eta_{1}(t-a)}}{\rho_{1}-\eta_{1}}
+ \frac{e^{\rho_{2}t}-e^{\rho_{2}a+\eta_{1}(t-a)}}{\eta_{1}-\rho_{2}}\\
&\quad+ \frac{e^{\rho_{1}t}-e^{\rho_{1}a+\eta_{2}(t-a)}}{\eta_{2}-\rho_{1}}
+  \frac{e^{\rho_{2}t}-e^{\rho_{2}a+\eta_{2}(t-a)}}{\rho_{2}-\eta_{2}}\Big).
\end{align*}
Similarly, we have the following expression for $\zoh(t)$.
\begin{align*}
&\ \widehat{Z}_{(a)}^{(q,p)}(t)=
Z^{(q)}(t)+ (p-q)\int_{a}^{t} Z^{(p)}(t-s)W^{(q)}(s)\,ds\\
=&\  \frac{\rho_{2}e^{\rho_{1}t}-\rho_{1} e^{\rho_{2}t}}{\rho_{2}-\rho_{1}}
+ \frac{(p-q)\mathbf{1}(t>a)}{D(\eta_{2}-\eta_{1})(\rho_{1}-\rho_{2})}
\int_{a}^{t}
(\eta_{2}e^{\eta_{1}(t-s)}- \eta_{1} e^{\eta_{2}(t-s)})
(e^{\rho_{1}s}-e^{\rho_{2}s})\,ds\\
=&\ \frac{\rho_{2}e^{\rho_{1}t}-\rho_{1} e^{\rho_{2}t}}{\rho_{2}-\rho_{1}}
+ \frac{(p-q)\mathbf{1}(t>a)}{D(\eta_{2}-\eta_{1})(\rho_{1}-\rho_{2})}
\Big(\frac{\eta_{2}(e^{\rho_{1}t}-e^{\rho_{1}a+\eta_{1}(t-a)})}{\rho_{1}-\eta_{1}}
+ \frac{\eta_{2}(e^{\rho_{2}t}-e^{\rho_{2}a+\eta_{1}(t-a)})}{\eta_{1}-\rho_{2}}\\
&\quad+ \frac{\eta_{1}(e^{\rho_{1}t}-e^{\rho_{1}a+\eta_{2}(t-a)})}{\eta_{2}-\rho_{1}}
+  \frac{\eta_{1}(e^{\rho_{2}t}-e^{\rho_{2}a+\eta_{2}(t-a)})}{\rho_{2}-\eta_{2}}\Big).
\end{align*}
\end{exam}

\begin{exam}
Let $X$ be a Cram\'er-Lundberg surplus process with exponentially distributed claim sizes,
that is, $\D \psi(s)=\mu s- \frac{\lambda s}{s+\beta}$ for some constants $\mu, \lambda, \beta>0$.
Then we have for $x>0$
\[
\begin{aligned}
W^{(q)}(x)=&\ \frac{\beta+\rho_{1}}{\mu(\rho_{1}-\rho_{2})}
e^{\rho_{1}x}+ \frac{\beta+\rho_{2}}{\mu(\rho_{2}-\rho_{1})} e^{\rho_{2}x},\\
W^{(p)}(x)=&\ \frac{\beta+\eta_{1}}{\mu(\eta_{1}-\eta_{2})}
e^{\eta_{1}x}+ \frac{\beta+\eta_{2}}{\mu(\eta_{2}-\eta_{1})} e^{\eta_{2}x},\\
Z^{(q)}(x)=&\ \frac{\rho_{2}(\beta+\rho_{1})}{\beta(\rho_{2}-\rho_{1})}
e^{\rho_{1}x}+ \frac{\rho_{1}(\beta+\rho_{2})}{\beta(\rho_{1}-\rho_{2})} e^{\rho_{2}x},\\
Z^{(p)}(x)=&\ \frac{\eta_{2}(\beta+\eta_{1})}{\beta(\eta_{2}-\eta_{1})}
e^{\eta_{1}x}+ \frac{\eta_{1}(\beta+\eta_{2})}{\beta(\eta_{1}-\eta_{2})} e^{\eta_{2}x},
\end{aligned}
\]
where $\rho_{1}>\rho_{2}$ are the roots to equation
$\mu s^{2}+ (\mu\beta-\lambda-q)s- q\beta=0$
and $\eta_{1}>\eta_{2}$ are the roots to equation
$\mu s^{2}+ (\mu\beta-\lambda-p)s- p\beta=0$.
We thus have
\begin{align*}
&\ W^{(q,p)}_{(a)}(t)=W^{(q)}(t)+
(p-q)\int_{a}^{t} W^{(p)}(t-s)W^{(q)}(s)\,ds\\
=&\ \frac{(\beta+\rho_{1})e^{\rho_{1}t}}{\mu(\rho_{1}-\rho_{2})}
+ \frac{(\beta+\rho_{2})e^{\rho_{2}t}}{\mu(\rho_{2}-\rho_{1})}
+ \frac{(p-q)\mathbf{1}(t>a)}{\mu^{2}(\rho_{1}-\rho_{2})(\eta_{1}-\eta_{2})}\\
&\quad\times
\Big(\frac{e^{\rho_{1}t}-e^{\rho_{1}a+\eta_{1}(t-a)}}{\rho_{1}-\eta_{1}}
(\beta+\eta_{1})(\beta+\rho_{1})
+ \frac{e^{\rho_{2}t}-e^{\rho_{2}a+\eta_{1}(t-a)}}{\eta_{1}-\rho_{2}}
(\beta+\eta_{1})(\beta+\rho_{2})\\
&\quad\quad+ \frac{e^{\rho_{1}t}-e^{\rho_{1}a+\eta_{2}(t-a)}}{\eta_{2}-\rho_{1}}
(\beta+\eta_{2})(\beta+\rho_{1})
+  \frac{e^{\rho_{2}t}-e^{\rho_{2}a+\eta_{2}(t-a)}}{\rho_{2}-\eta_{2}}
(\beta+\eta_{2})(\beta+\rho_{2})\Big).
\end{align*}
Similarly, we can also obtain the following expression for $\zoh(t)$.
\begin{align*}
&\ \widehat{Z}_{(a)}^{(q,p)}(t)=Z^{(q)}(t)+ (p-q)\int_{a}^{t}Z^{(p)}(t-s)W^{(q)}(s)\,ds\\
=&\ \frac{\rho_{2}(\beta+\rho_{1})}{\beta(\rho_{2}-\rho_{1})}
e^{\rho_{1}t}+ \frac{\rho_{1}(\beta+\rho_{2})}{\beta(\rho_{1}-\rho_{2})} e^{\rho_{2}t}
+ \frac{(q-p)\mathbf{1}(t>a)}{\mu\beta(\eta_{1}-\eta_{2})(\rho_{1}-\rho_{2})}\\
&\quad\times
\Big(\frac{e^{\rho_{1}t}-e^{\rho_{1}a+\eta_{1}(t-a)}}{\rho_{1}-\eta_{1}}
\eta_{2}(\beta+\eta_{1})(\beta+\rho_{1})
+ \frac{e^{\rho_{2}t}-e^{\rho_{2}a+\eta_{1}(t-a)}}{\eta_{1}-\rho_{2}}
\eta_{2}(\beta+\eta_{1})(\beta+\rho_{2})\\
&\quad\quad+ \frac{e^{\rho_{1}t}-e^{\rho_{1}a+\eta_{2}(t-a)}}{\eta_{2}-\rho_{1}}
\eta_{1}(\beta+\eta_{2})(\beta+\rho_{1})
+  \frac{e^{\rho_{2}t}-e^{\rho_{2}a+\eta_{2}(t-a)}}{\rho_{2}-\eta_{2}}
\eta_{1}(\beta+\eta_{2})(\beta+\rho_{2})\Big).
\end{align*}
\end{exam}

\section{Proof}\label{proof}
This section is dedicated to the proofs of our main results.
We first find an expression for
 the expected time spent by $(S,Y)$ until the ruin time for $X$ in the following lemma.

\begin{lem}\label{lem:2}
For $z> x, y\geq 0$, we have
\begin{equation*}
\begin{split}
R(x; dz,dy)&:=
\int_{0}^{\infty} \bP_{x}\big( S_{t}\in dz, Y_{t}\in dy; t<\tau_{0}^{-}\big)\,dt\\
&=\ \frac{W(x)}{W(z)} \Big(W(dy)- \frac{W'(z)}{W(z)} W(y)\,dy\Big) \,dz.
\end{split}
\end{equation*}
In particular,
\[R(x; dz, \{0\})= \frac{W(x)}{W(z)}W(0)\,dz.\]
\end{lem}
\begin{proof}[Proof of Lemma \ref{lem:2}]
Let $f$ be a nonnegative, continuous and bounded function on $[0,b]$. Then
\begin{equation*}
\begin{split}
 \int_{0}^{\infty} \bE_{x}\big(f(X_{t}); t<\tau_{0}^{-}, S_{t}<b\big)\,dt
&= \int_{0}^{\infty} \bE_{x}\big(f(X_{t}); t\leq \tau_{b}^{+}\wedge\tau_{0}^{-}\big)\,dt \\
&= \int_{0}^{b} f(y) \Big(\frac{W(x)}{W(b)}W(b-y)-W(x-y)\Big)\,dy.
\end{split}
\end{equation*}
Differentiating in $b$ on both sides of the above equation, we have for $z>x$ and $z>y\geq0$,
\begin{align*}
&\ \int_{0}^{\infty} \bP_{x}\big(S_{t}\in dz, X_{t}\in dy, t<\tau_{0}^{-}\big)\,dt\\
=&\ \frac{W(x)}{W(z)} \Big(W'(z-y)- \frac{W'(z)}{W(z)} W(z-y)\Big) \,dzdy
+ \frac{W(x)}{W(z)}W(0) \delta_{\{z\}}(dy)\,dz
\end{align*}
where $\delta_{\{z\}}(dy)$ denotes the Dirac measure at $\{z\}$.
The result of the lemma follows by change of variable.
\end{proof}
\begin{rmk}\label{rmk:cal:1} For $z>x\geq 0$ and $0<u\leq z$, by Lemma \ref{lem:2} we obtain the following density
\begin{align*}
&\ \frac{1}{dz}
\Big(\int_{0-}^{z} \omega(y) \wo(u,y) R(x;dz,dy)\Big) \\
=&\ \int_{0-}^{z}
\frac{W(x)}{W(z)} \wo(u,y)\omega(y)
\Big(W(dy)- \frac{W'(z)}{W(z)} W(y)\,dy\Big)\\
=&\ \frac{W(x)}{W(z)} \Big(-\wo_{2}(u)-W'(u)
- \frac{W'(z)}{W(z)} (\wo(u)-W(u))\Big),
\end{align*}
where we need identity \eqref{wo_2}  and
identity \eqref{eqn:defn:w:2} for the last equality.
In particular, for $u=z$
it becomes \[\D -\wo(z)\frac{W(x)}{W(z)}
\Big(\frac{W'(z)}{W(z)}+ \frac{\wo_{2}(z)}{\wo(z)} \Big).\]
\end{rmk}

To obtain the Feymann-Kac identity,
we need the following result from \cite{BoLi:sub1} and more properties of $\wo$.

\begin{prop}\label{prop:1}

For $x\in(c,b)$ we have
\[
\bE_{x}\Big(\exp\big(-\int_{0}^{\tau_{b}^{+}}
\omega(X_{s})\,ds\big); \tau_{b}^{+}<\tau_{c}^{-}\Big)
=\frac{\wo(x,c)}{\wo(b,c)}.
\]
In addition, the $\omega$-resolvent measure of $X$ killed at $\tau_{b}^{+}\wedge\tau_{c}^{-}$ is given by for $c<y<b$,
\begin{equation}
\begin{aligned}
&\ \int_{0}^{\infty}\bE_{x}\Big(
\exp\big(-\int_{0}^{t}\omega(X_{s})\,ds\big);  t\leq \tau_{b}^{+}\wedge\tau_{c}^{-}, X_{t}\in dy\Big)\,dt\\
&=\  \Big(\frac{\wo(x,c)}{\wo(b,c)}\wo(b,y)-\wo(x,y)\Big)\,dy.
\end{aligned}
\end{equation}
\end{prop}

\begin{lem}\label{lem:1}
Let $\omega$ be a nonnegative locally bounded measurable function,
$\wo$ and $W^{(\omega_{u})}$ be scale functions with respect to
$\omega$ and $\omega_{u}$, respectively,
where $\omega_{u}(z):=\omega(u-z)$ for some $u\in\mathbb{R}$,
that is
\begin{equation}\label{eqn:lem:wo}
W^{(\omega_{u})}(x,y)=W(x-y)
+\int_{y}^{x} W(x-z)\omega(u-z)W^{(\omega_{u})}(z,y)\,dz
\end{equation}
for $x,y\in\mathbb{R}$. Then \[W^{(\omega_{u})}(u-y,u-x)=\wo(x,y).\]
\end{lem}

\begin{proof}[Proof of Lemma \ref{lem:1}]
Denoting by $g(x,y):=W^{(\omega_{u})}(u-y,u-x)$
and by change of variable,
we have from \eqref{eqn:lem:wo} that
\begin{align*}
g(x,y)=&\ W(x-y)+
\int_{u-x}^{u-y}W(u-y-z)\omega(u-z) W^{(\omega_{u})}(z,u-x)\,dz\\
=&\ W(x-y)+
\int_{y}^{x}g(x,z) \omega(z)W(z-y)\,dz.
\end{align*}
It shows that $g$ satisfies the equation \eqref{eqn:defn:w:2}, which
 concludes the proof by the uniqueness of solution to \eqref{eqn:defn:w:2}.
\end{proof}

\begin{rmk}\label{rmk:2}
Combining  Lemma \ref{lem:1} and Proposition \ref{prop:1} gives
\[
\ \bE_{b-x}\Big(\exp\big(-\int_{0}^{\tau_{b}^{+}}\omega(b-X_{s})\,ds\big); \tau_{b}^{+}<\tau_{0}^{-}\Big)
=\frac{\wo(b,x)}{\wo(b,0)}\]
and
\begin{equation*}
\begin{split}
&\ \int_{0}^{\infty} \bE_{b-x}\Big(\exp\big(-\int_{0}^{t}\omega(b-X_{s})\,ds\big) f(b-X_{t});
t<\tau_{b}^{+}\wedge\tau_{0}^{-}\Big)\,dt\\
&=\ \int_{0}^{b}f(y)
\Big(\frac{\wo(b,x)}{\wo(b,0)}\wo(y,0)- \wo(y,x)\Big)\,dy,
\end{split}
\end{equation*}
which are the corresponding  results for the dual process of $X$.
\end{rmk}

Since both processes $X$ and $Y$ are involved in our problems,
we need to consider the joint probability law of processes $S$ and $Y$.

We slightly abuse notations in the following discussion.

Here, for some $u\in\mathbb{R}$ and $v\geq0$,
let $\D S_{t}:=u\vee \sup_{s\in[0,t]} X_{s}$ be the running maximum process of $X$,
and $Y:=S_{t}-X_{t}$ be the  process reflected from the running maximum.
The two-dimensional process
$Z\equiv\{Z_{t}, t\geq0\}:=
\{(S_{t},  Y_{t}),t\geq0\}$ is still a Markov process.
The law of $Z$ starting from $(u,v)$ is denoted by
$\bP_{u,v}:=\bP\big(\cdot\big|(S_{0}, Y_{0})=(u, v)\big)$
and the corresponding expectation is denoted by $\bE_{u,v}$.
We write $\bP_{x}:=\bP_{x,0}=\bP(\cdot|X_{0}=x)$ for $x\in\mathbb{R}$
which reduces to the  probability law of $X$ given $X_{0}=x$.

\begin{proof}[Proof of Proposition \ref{lem:3}]
 Define an operator on
the space of continuous and bounded functions
by
\[
\ro f(u,v):=
\int_{0}^{\infty} \bE\Big(e^{-L(t)} f(S_{t},Y_{t}); t\leq \tau_{b}^{+}\wedge\tau_{0}^{-}\Big|(S_{0},Y_{0})=(u,v)\Big)\,dt,
\]
for $(u,v)\in\mathbb{R}\times\mathbb{R}^{+}$.
Write
$$\ro f(u):=\ro f(u,0)\,\,\,\text{ and}\,\,\, Rf(u,v):=\ro f(u,v)\,\,\,\text{ for}\,\,\, \omega\equiv 0.$$
Then $Rf(u)=R^{(0)}f(u,0)$ whose expression can be found  in Lemma \ref{lem:2}.

Firstly, by the additivity of $L$, we have for every $t>0$,
\begin{equation}\label{eqn:feykac}
1- e^{-L(t)}=e^{-L(t)}\int_{0}^{t} e^{L(s)}\omega(Y_{s})\,ds
=\int_{0}^{\infty} \omega(Y_{s}) e^{-L(t-s)\circ \theta_{s}} \mathbf{1}_{(s<t)}\,ds,
\end{equation}
where $\theta_{\cdot}$ denotes the shift operator and $\mathbf{1}_{A}$ denotes the indicator function.
Plugging it into the following equation and
applying the Markov property at time $s>0$,
we have
\begin{equation}\label{eqn:ro:2}
\begin{split}
&Rf(x)-\ro f(x)\\
&=
\int_{0}^{\infty} \bE_{x,0}\Big(\big(1-e^{-L(t)}\big) f(S_{t},Y_{t}); t<\tau_{b}^{+}\wedge\tau_{0}^{-}\Big)\,dt\\
&=
\int_{0}^{\infty}\int_{0}^{\infty}
\bE_{x,0}\Big( \omega(Y_{s}) \big(e^{-L(t-s)} f(S_{t-s},Y_{t-s})\big)\circ\theta_{s}; s<t<\tau_{b}^{+}\wedge\tau_{0}^{-}\Big)\,ds\,dt\\
&=\ \int_{x}^{b} \int_{0-}^{z} R(x; dz,dy)\omega(y) \ro f(z,y)\\
&=: R(\omega \ro)f(x).
\end{split}
\end{equation}

On the other hand, under $\bP_{z,y}$ we have $X_{0}=z-y$.
The absence of positive jumps gives
$S_{\tau_{z}^{+}}=z$ on the set $\{\tau_{z}^{+}<\infty\}$
and $S_{t}=z$, $Y_{t}=z-X_{t}$ for $t<\tau_{z}^{+}$.
Conditioning on $\tau_{z}^{+}\wedge\tau_{0}^{-}$, we have
\begin{align*}
\ro&  f(z,y)
=\bE\Big(\exp\big(-\int_{0}^{\tau_{z}^{+}}\omega(z-X_{s})\,ds\big); \tau_{z}^{+}<\tau_{0}^{-}
\Big|X_{0}=z-y\Big)\cdot \ro f(z)\\
&\ + \int_{0}^{\infty}
\bE\Big(\exp\big(-\int_{0}^{t}\omega(z-X_{s})\,ds\big) f(z,z-X_{t}); t<\tau_{z}^{+}\wedge\tau_{0}^{-} \Big|X_{0}=z-y\Big)\,dt.
\end{align*}
By  the discussion in Remark \ref{rmk:2},  the identity above reduces to
\begin{equation}\label{eqn:ro:1}
\begin{aligned}
\ro f(z,y)=&\ \frac{\wo(z,y)}{\wo(z,0)}
\Big(\ro f(z)+ \int_{0}^{z} f(z,u) \wo(u,0)\,du\Big)\\
&\quad - \int_{0}^{z} f(z,u) \wo(u,y)\,du.
\end{aligned}
\end{equation}

Plugging it into the definition of $R(\omega \ro)f$  gives
\begin{align*}
R(\omega \ro)& f(x)=
 \int_{x}^{b} \int_{0-}^{z} R(x;dz,dy)\omega(y)
\Big(- \int_{0}^{z}f(z,u)\wo(u,y)\,du\\
&\quad+  \frac{\wo(z,y)}{\wo(z,0)}
\big(\ro f(z)+ \int_{0}^{z} f(z,u)\wo(u,0)\,du\big)\Big).
\end{align*}
By Remark \ref{rmk:cal:1}
and $\wo(z,0)=\wo(z)$, we further have
\begin{align*}
R&(\omega \ro) f(x)=
\int_{x}^{b}\int_{0}^{z} f(z,u) \frac{W(x)}{W(z)}
\Big(W'(u)- \frac{W'(z)}{W(z)}W(u)\Big)dudz\\
&\quad +
\int_{x}^{b}\int_{0}^{z} f(z,u) \frac{W(x)}{W(z)}
\Big(\wo_{2}(u)+ \frac{W'(z)}{W(z)}\wo(u)\Big)dudz\\
&\quad -\int_{x}^{b}
\Big(\ro f(z)+ \int_{0}^{z} f(z,u)\wo(u)\,du\Big)
\frac{W(x)}{W(z)}
\Big(\frac{\wo_{2}(z)}{\wo(z)}+ \frac{W'(z)}{W(z)}\Big)\,dz.
\end{align*}
Notice that  the resolvent density of $R(x;dz,du)$ for $u>0$ in Lemma \ref{lem:2} appears in the first integrand of the above equation.
Comparing it with the left hand side of \eqref{eqn:ro:2} and applying Lemma \ref{lem:2} gives the following integral equation on $\frac{\ro f(x)}{W(x)} $.
\begin{equation}\label{eqn:ro:3}
\begin{split}
 \frac{\ro f(x)}{W(x)}
&=  \frac{Rf(x)- R(\omega \ro)f(x)}{W(x)}\\
&= \int_{x}^{b}
\Big(\frac{W'(z)}{W(z)}+ \frac{\wo_{2}(z)}{\wo(z)} \Big)
\frac{\ro f(z)}{W(z)}\,dz
+ W(0) \int_{x}^{b}  \frac{f(z,0)}{W(z)} \,dz\\
&\quad+  \int_{x}^{b}\int_{0}^{z} \frac{f(z,u)}{W(z)}
 \Big( \frac{\wo_{2}(z)}{\wo(z)}\wo(u)-\wo_{2}(u) \Big)\,dudz.
\end{split}
\end{equation}

Recalling the definition of $\ho$ in \eqref{defn:ho},
one can check that
\[
\frac{\ho(x)}{W(x)}=
\frac{1}{W(1)}-\int_{1}^{x}\Big(\frac{W'(z)}{W(z)}+ \frac{\wo_{2}(z)}{\wo(z)} \Big)
\frac{\ho(z)}{W(z)}\,dz.
\]
Solving the equation \eqref{eqn:ro:3} for $\frac{\ro f(x)}{W(x)} $, we have that for $x\in[0,b]$,
\begin{align*}
\ro f(x)=&\ \int_{x}^{b}\int_{0}^{z} f(z,u)
\frac{\ho(x)}{\ho(z)}
\Big(\frac{\wo_{2}(z)}{\wo(z)}\wo(u)- \wo_{2}(u)\Big)\,dudz\\
&\quad + W(0) \int_{x}^{b} f(z,0) \frac{\ho(x)}{\ho(z)}\,dz,
\end{align*}
which gives the desired formula \eqref{eqn:thm1:ro+}.
\end{proof}

We are now ready to prove the  main results.

\begin{proof}[Proof of Theorem \ref{thm:1}]
We take use of the following version of the Feynman-Kac identity:
\begin{equation}
\label{eqn:feykac:2}
1- e^{-L(t)}=\int_{0}^{t} e^{-L(s)}\omega(Y_{s})\,ds
=\int_{0}^{\infty} \omega(Y_{s})  e^{-L(s)} \mathbf{1}_{(s<t)}\,ds.
\end{equation}
To simplify notations we denote  for $u\in[0,b]$ and $u\geq v\geq 0$
 $$
 \mathcal{A}^{(\omega)}(u,v):=
\bE\big(e^{-L(\tau_{b}^{+})}; \tau_{b}^{+}<\tau_{0}^{-}\big|(S_{0},Y_{0})=(u,v)\big),\\
$$
and write $\mathcal{A}^{(\omega)}(u,0):=\mathcal{A}^{(\omega)}(u)$.
It holds  that for $\omega\equiv0$
\begin{equation}\label{eqn:b+:1}
\mathcal{A}(u,v):=\mathcal{A}^{(0)}(u,v)=\bP_{u-v}\big(\tau_{b}^{+}<\tau_{0}^{-}\big)=\frac{W(u-v)}{W(b)}.
\end{equation}
Similar to \eqref{eqn:ro:2}, but with \eqref{eqn:feykac}  replaced by \eqref{eqn:feykac:2},
we have from the Markov property that
\begin{equation}\label{eqn:b+:2}
\mathcal{A}(x)- \mathcal{A}^{(\omega)}(x)
=\int_{x}^{b} \int_{0-}^{z} \ro(x;dz,dy) \omega(y) \mathcal{A}(z,y).
\end{equation}
Plugging \eqref{eqn:b+:1} and \eqref{eqn:thm1:ro+} into the equation gives
$$
\mathcal{A}^{(\omega)}(x)
=\frac{W(x)}{W(b)}+ \int_{x}^{b}
\frac{\ho(x)}{\ho(z)}
\Big( \frac{W'(z)}{W(b)}
+\frac{\wo_{2}(z)}{\wo(z)} \frac{W(z)}{W(b)}\Big)\,dz.
$$
Noticing that
$$\D \frac{\wo_{2}(z)}{\wo(z)}
=\frac{d}{dz}\Big(-\log \ho(z)\Big),$$
 we further have
$$
\mathcal{A}^{(\omega)}(x)=
 \frac{W(x)}{W(b)}+
\Big(\frac{\ho(x)}{\ho(z)} \frac{W(z)}{W(b)}\Big)\Big|_{z=x}^{z=b}
=\frac{\ho(x)}{\ho(b)},
$$
which proves the formula \eqref{eqn:thm1:b+}.

We proceed to show formula (\ref{eqn:thm1:c-}).
Observe first that

\begin{equation}
\label{eqn:c-:1}

\bE_{x,0}\left(\int_{0}^{\infty} e^{-L(t)}\omega(Y_{t}) \mathbf{1}_{\{ t<\tau_{b}^{+}\wedge\tau_{0}^{-}\}}\,dt\right)
=1- \bE_{x,0}\left(e^{-L(\tau_{b}^{+}\wedge\tau_{0}^{-})}\right).
\end{equation}
On the other hand,
by \eqref{eqn:thm1:ro+} and integration by parts  the left hand side of \eqref{eqn:c-:1} is equal to
\begin{equation*}
\begin{split}

 &
 \int_{x}^{b} \frac{\ho(x)}{\ho(z)}
 \Big(\frac{\wo_{2}(z)}{\wo(z)}
 \big(\zoh(z)-1\big)- \zoh_{2}(z) \Big)\,dz\\
&=\ \frac{\ho(x)}{\ho(z)}(\zoh(z)-1)\Big|_{x}^{b}
- \int_{x}^{b} \frac{\ho(x)}{\ho(z)}(\zoh_{1}(z)+\zoh_{2}(z))\,dz\\
&=\ 1- \frac{\ho(x)}{\ho(b)}- \Big(\zoh(x)- \frac{\ho(x)}{\ho(b)}\zoh(b)\Big)\\
&\quad\ - \int_{x}^{b} \frac{\ho(x)}{\ho(z)}\big(\zoh_{1}(z)+\zoh_{2}(z)\big)\,dz.
\end{split}
\end{equation*}
Comparing with (\ref{eqn:c-:1}) gives the formula (\ref{eqn:thm1:c-}).
We thus completes the proof of Theorem \ref{thm:1}.
\end{proof}

\begin{proof}[Proof of Proposition \ref{prop:3}]
Observing that
$$\D \delta t+ L(t)=\int_{0}^{t}(\delta+\omega(X_{s}))\,ds$$ for every $t>0$,
the result for general case of $\delta>0$
can be derived from that of the special case $\delta=0$,
with the weight function replaced by $\omega+\delta$.
We thus focus on the case $\delta=0$ in the proof.

To this end, we first evaluate,
for a continuous $f\geq0$ on $[0,b]$,
\begin{equation}\label{eqn:p3:1}
\begin{split}
&\ \int_{0}^{\infty} \bE_{x}\big(e^{-L(t)}f(X_{t}); t<\tau_{b}^{+}\wedge \tau_{0}^{-}\big)\,dt\\
=&\ \int_{x}^{b}f(z-y) \,dz \int_{0}^{z}
\frac{\ho(x)}{\ho(z)} \Big(\frac{\wo_{2}(z)}{\wo(z)}\wo(y)-\wo_{2}(y)\Big)\,dy\\
&\quad + W(0)\int_{x}^{b}f(z)\frac{\ho(x)}{\ho(z)}\,dz
\end{split}
\end{equation}
with Proposition \ref{lem:3} applied.
Changing the order of integrals, the first term of the right hand side of \eqref{eqn:p3:1} equals to
\begin{align*}
&\ \int_{x}^{b}\,dz \int_{0}^{z} f(y)
\frac{\ho(x)}{\ho(z)} \Big(\frac{\wo_{2}(z)}{\wo(z)}\wo(z-y)-\wo_{2}(z-y)\Big)\,dy\\
=&\ \int_{0}^{b} f(y)\,dy \int_{x\vee y}^{b}
\frac{\ho(x)}{\ho(z)} \Big(\frac{\wo_{2}(z)}{\wo(z)}\wo(z-y)-\wo_{2}(z-y)\Big)\,dz.
\end{align*}
Noticing that $$\frac{d}{dx}\ho(x)=-\ho(x)\frac{\wo_{2}(x)}{\wo(x)},$$ we further have
\begin{align*}
&\ \int_{x\vee y}^{b}
\frac{\ho(x)}{\ho(z)} \Big(\frac{\wo_{2}(z)}{\wo(z)}\wo(z-y)-\wo_{2}(z-y)\Big)\,dz\\
=&\ \frac{\ho(x)}{\ho(z)}\wo(z-y)\Big|_{z=x\vee y}^{z=b}
- \int_{x\vee y}^{b} \frac{\ho(x)}{\ho(z)}\big(\wo_{1}(z-y)+\wo_{2}(z-y)\big)\,dz.
\end{align*}
Since $\wo(z-y)=0$ for $z<y$, we  have
\[
\frac{\ho(x)}{\ho(z)}\wo(z-y)\Big|_{z=x\vee y}^{z=b}
=\frac{\ho(x)}{\ho(b)}\wo(b-y)-\wo(x-y)- W(0)\frac{\ho(x)}{\ho(y)}\mathbf{1}(y>x).
\]
Plugging into \eqref{eqn:p3:1} and using the fact that $\wo_{1}(z-y)=\wo_{2}(z-y)=0$ for $z<y$,
we further have
\begin{equation}
\begin{split}
&\ \int_{0}^{\infty} \bE_{x}\big(e^{-L(t)}f(X_{t}); t<\tau_{b}^{+}\wedge \tau_{0}^{-}\big)\,dt\\
=&\ \int_{0}^{b}
\Big(\frac{\ho(x)}{\ho(b)}\wo(b-y)-\wo(x-y)\Big) f(y)\,dy\\
&\quad - \int_{0}^{b}f(y)
\Big(\int_{x}^{b} \frac{\ho(x)}{\ho(z)}\big(\wo_{1}(z-y)+\wo_{2}(x-y)\big)\,dz\Big)\,dy.
\end{split}
\end{equation}
The formula \eqref{eqn:prop3:1} then follows from the compensating formula.
\end{proof}

For the proof of Proposition \ref{prop:2},
we apply \eqref{eqn:defn:w:2} for $\wo$, and definition
\eqref{eqn:defn:zh} for $\zoh$. We also need  the following identity
which first appears in \cite{Loeffen2014:occupationtime:levy}.
\begin{equation}
\label{eqn:scaleiden}
W^{(q)}(x-y)=W(x-y)+ q \int_{y}^{x}W(x-z)W^{(q)}(z-y)\,dz.
\end{equation}
\begin{proof}[Proof of Proposition \ref{prop:2}]
Firstly, we have from \eqref{eqn:defn:zh} for $\zoh$ and the Fubini's theorem that
\begin{align}
&\ \int_{y}^{x}\zoh(x,u)\omega(u)W(u-y)\,du\non\\
=&\  \int_{y}^{x}\omega(u)W(u-y)\,du
+ \iint_{x>u>v>y}\omega(u)\wo(u,v)\omega(v)W(v-y)\,dudv\non\\
=&\  \int_{y}^{x}\omega(u)
\Big(W(u-y)+ \int_{y}^{u}\wo(u,v)\omega(v)W(v-y)\,dv\Big)\,du\non\\
=&\ \int_{y}^{x}\omega(u)\wo(u,y)\,du=\zoh(x,y)-1\label{eqn:prop:1}
\end{align}
where identity \eqref{eqn:defn:w:2} for $\wo$ is used for the second to the last equation.

On the other hand, for any $q\geq0$,
integrating on both sides of the equation above with respect to $W^{(q)}(y-c)\,dy$ over interval $(c,x)$,
by the scale function identity \eqref{eqn:scaleiden} we have
\begin{align}
&\ q\int_{c}^{x}\big(\zoh(x,y)-1\big)W^{(q)}(y-c)\,dy\non\\
=&\ q\int_{c}^{x} \big(\int_{y}^{x} \zoh(x,u)\omega(u)W(u-y)\,du\big) W^{(q)}(y-c)\,dy\non\\
=&\ q\int_{c}^{x} \zoh(x,u)\omega(u) \int_{c}^{u}W(u-y)W^{(q)}(y-c)\,dy\non\\
=&\ \int_{c}^{x}\zoh(x,u)\omega(u) (W^{(q)}(u-c)-W(u-c))\,du.\label{eqn:prop:2}
\end{align}
Combining \eqref{eqn:prop:1}, \eqref{eqn:prop:2}  and
\[Z^{(q)}(x-y)=1+ q \int_{y}^{x} W^{(q)}(z-y)\,dz,\]
 we have
\begin{align}
\zoh(x,y)-Z^{(q)}(x-y)=\int_{y}^{x}\zoh(x,z)(\omega(z)-q)W^{(q)}(z-y)\,dz.\label{eqn:prop:4}
\end{align}

We now consider the case that $\omega$ is a step function given by (\ref{step_function}).
Given $\widehat{Z}_{n}$ on $\mathbb{R}\times\mathbb{R}$,
 since $\omega_{n}(y)=\omega_{n+1}(y)$ for  $y\geq a_{n+1}$,
it follows from \eqref{eqn:prop:1} that  $\widehat{Z}_{n+1}(x,y)= \widehat{Z}_{n}(x,y)$.

For $y<a_{n+1}$, taking $q=p_{n+1}$ in the equation \eqref{eqn:prop:4},
and noticing that $\D \omega_{n+1}(z)-p_{n+1}=0$ for $z< a_{n+1}$
and $\D \omega_{n+1}(z)=\omega_{n}(z)$ for $z\geq a_{n+1}$,
we have

\begin{equation*}
\begin{split}
\widehat{Z}_{n+1} (x,y)
&=\ \widehat{Z}^{(p_{n+1})}(x-y)+ \int_{y}^{x} \widehat{Z}_{n+1}(x,z)(\omega_{n+1}(z)-p_{n+1})W^{(p_{n+1})}(z-y)\,dz\\
&=\ \widehat{Z}^{(p_{n+1})}(x-y)+\int_{a_{n+1}}^{x} \widehat{Z}_{n+1}(x,z)(\omega_{n+1}(z)-p_{n+1}) W^{(p_{n+1})}(z-y)\,dz\\
&=\ \widehat{Z}^{(p_{n+1})}(x-y)+\int_{a_{n+1}}^{x} \widehat{Z}_{n}(x,z)(\omega_{n}(z)-p_{n+1}) W^{(p_{n+1})}(z-y)\,dz\\
&=\ \widehat{Z}_{n}(x,y)- \int_{y}^{a_{n+1}} \widehat{Z}_{n}(x,z)(\omega_{n}(z)-p_{n+1})W^{(p_{n+1})}(z-y)\,dz\\
&=\ \widehat{Z}_{n}(x,y)+ (p_{n+1}-p_{n})\int_{y}^{a_{n+1}}\widehat{Z}_{n}(x,z)W^{(p_{n+1})}(z-y)\,dz,
\end{split}
\end{equation*}
where identity \eqref{eqn:prop:4} is used for the first and the fourth equation.
The proof is thus completed.

The recursive expression for $W_n(x,y)$ can be proved similarly.
\end{proof}

\medskip

\noindent
{\bf Acknowledgement} Yun Hua and Bo Li thank Concordia University where the first draft of this paper was completed during their visit in early 2017.

\bibliographystyle{apalike}

config: arxiv


TITLE: Fractional part of rational power arbitrary small
QUESTION [7 upvotes]: I think that $\{a^n\}$ (where $\{x\}$ is $x \pmod 1$), where $a$ is fixed rational greater than 1 and $n$ is positive integer, is dense in $[0,1]$ is unsolved. However what about $\{a^n\}$ is arbitrary small for some $n$ ($a$ is fixed rational as well).

REPLY [7 votes]: I think that the wolfram link cited the first result (Vijayaraghavan 1941) incorrectly where it reports that

$\{ (3/2)^n \}$ has infinitely many accumulation points in both $[0, 1/2)$ and $[1/2, 1]$.

The first result actually is

$\{ (3/2)^n \}$ has infinitely many accumulation points in $[0,1]$.

The second result cited by wolfram is by Flatto, Lagarias, Pollington (1995) which is available here: http://matwbn.icm.edu.pl/ksiazki/aa/aa70/aa7023.pdf. In the introduction of this paper by them, it reports that

Vijayaraghavan later remarked that it was striking that
  one could not even decide whether or not $(3/2)^n$ mod $1$ has infinitely
  many limit points in $[0, 1/2)$ or in $[1/2, 1)$.

This paper proved that

$$\limsup\{ (3/2)^n \} - \liminf \{(3/2)^n\} \geq \frac13. $$

A few later results are mentioned (Dubickas 2006, 2008) in Yann Bugeaud's book "Distribution Modulo One and Diophantine Approximation" (for example, p. xi, p. 67 and p. 68).

$\{ (3/2)^n \}$ has a limit point in $[0.238117 . . . , 0.761882 . . .]$ which has length $0.523764 . . .$.
$\{(3/2)^n\}$ has a limit point in $[0, 8/39] \cup
[18/39, 21/39] \cup [31/39, 1]$, of total length $19/39$.


TITLE: Do finite morphisms preserve dimensions?
QUESTION [5 upvotes]: If $f: Y\to Z$ is a finite, surjective morphism of normal integral schemes (of finite type over a field) and $y$ is a prime divisor of $Y$, is then also $z= f(y)$ of codimension 1?
We have an inclusion $\mathcal{O}_{Z,z} \to \mathcal{O}_{Y,y}$ and the latter is a DVR, i.e. it has dimension 1. Does the integrality of both local rings imply that $\operatorname{codim}z =1$?
More generally, is there a relationship between $\dim y, \dim f(z)$ and $\dim f^{-1}(z'), \dim z'$ for $z'\in Z$ arbitrary?

REPLY [5 votes]: Answer to your first question: We prove the more general statement that if $f: Y \to Z$ is a finite surjective morphism of integral schemes, then $\dim y = \dim f(y)$ (in the case that $Y,Z$ are also of finite type over a field this will also yield that $codim \; y = codim \; f(y)$).
Claim: $\dim Y = \dim Z$, the reason being that if $Spec B$ is any affine open of $Z$, then $f^{-1}(Spec B) = Spec A$ with $B \hookrightarrow A$ being an integral extension hence $dim Spec A = dim Spec B$, now cover $Y$ with affine opens and recall that a scheme has dimension $n$ if and only if it has an affine cover where each affine in the cover has dimension $\le n$ and equality is achieved for at least one, from taking preimages of such a cover of $Z$ we now see that $\dim Z = \dim Y$.
Now we have that the morphism $y \to f^{-1}(z) \to f(y) = z$ is a finite surjective morphism of integral schemes (being the composition of a closed embedding together with a projection from a base change of a finite morphism) hence it follows from the claim that $\dim y = \dim z$ and since codimension is the difference of dimensions for integral schemes of finite type over a field it then follows that $codim y = codim z$.
Answer to your second question:
If $f: Y \to Z$ is any morphism of locally Noetherian schemes, and $p \in Y$, $q = f(p)$ then we always have
$$(Eq \; 1) \; \; \; \; codim_Y p \le codim_Z q + codim_{f^{-1}(q)}p$$ (here $codim$ of a point is interpeted as the codimension of the closure in the appropriate space). Now for integral schemes of finite type over a field (and more generally for Catenary schemes if I am not mistaken?) this translates to
$$ (Eq \; 2) \; \; \; \; \dim Y - \dim \overline{\{p\}} \le \dim Z - \dim \overline{\{q\}} + codim_{f^{-1}(q)}p$$
If the morphism is flat we have equality, and of course if the morphism is finite then the second term on the right hand side can be removed.


TITLE: Can you write down a chain of derivabilities?
QUESTION [0 upvotes]: Suppose you have a sentence $A$, from which you can derive $B$, from which you can derive $C$. What is the best way to write this down? I would like to write
$$
A \vdash B \vdash C,
$$
but I'm not sure if it 'allowed' to use multiple turnstiles in a chain.
What is in this situation best to do style-wise?

REPLY [1 votes]: It looks like what you want to mark is not derivability but inference.  Inferences are the atomic steps from formula to formula, by means of which derivations are constructed.
One convention for marking steps of inference is to use the sign $\therefore$.  But this doesn't work very well for derivations involving successive inferences, since an inference may have more than one premise.  Another convention is to write the conclusion of the inference underneath of the premises.
On the other hand, an expression like $\phi\vdash\psi$ means "there is a derivation of $\psi$ from $\phi$."  So without some further convention, an expression like $\phi\vdash\psi\vdash \chi$ doesn't really make sense: it sounds a bit like "there is a road from Urumqi to there is a road from Kashgar to Khotan".
One might ask "so, why not add a convention that $\phi\vdash \psi\vdash \chi$ means '$\phi\vdash \psi$ and $\psi\vdash \chi$'"?  You could if you wanted, but it's not a great idea, for the following reason.
Let $\to$ express the material conditional, so that $\phi\to\psi$ roughly means "if $\phi$ then $\psi$".  Now, this automatically gives a meaning to $\phi\to \psi\to \chi$ (following e.g., the convention that conditionals associate to the right): it means roughly the same as "if $\phi$ then if $\psi$ then $\chi$".  So, the meanings of $\to$ and $\vdash$ are quite different: the meaning of one, but not of the other, automatically makes sense in the context $\phi\cdots \psi\cdots \chi$.  Writing  $\phi\vdash\psi\vdash \chi$ obscures this difference.


TITLE: How to show locally Lipschitz on $A \subset \mathbb{R}^n$ implies continuity on $A$?
QUESTION [2 upvotes]: I know this question has been answered before but the way that I am trying to show is different and uses the open balls for the proof. I need a clarification for some part of my proof.
Let $A \subset \mathbb{R}^n$ where $A$ is open and let $f: A \rightarrow \mathbb{R}^n$ be locally Lipschitz on $E$. Prove that $f$ is continuous on $E$.

Definition(locally Lipschitz on $A$): $f$ is locally Lipschitz on $A \subset \mathbb{R}^n$  if and only if $\forall x_0 \in A$ there exist $\delta >0$ such that $B_{\delta}(x_0) \subset A$ and f is Lipschitz on $B_{\delta}(x_0)$.

Now we need definition of Lipschitz on $B_{\delta}(x_0)$ as follows

Definition(Lipschitz on $B_{\delta}(x_0)$):
$f$ is Lipschitz on $A \subset \mathbb{R}^n$  if and only if there exist k>0 such that $\forall x,y \in A$
$$
\|f(x)-f(y)\|\leq k\|x-y\|
$$

Following the definitions makes it super easy to come up with:
$f$ is locally Lipschitz on $A$  then $\forall x_0 \in A$ there exist $\delta >0$ such that $B_{\delta}(x_0) \subset A$ and f is Lipschitz on $B_{\delta}(x_0)$. Therefore there exist a $k$ such that
$$
\|f(x)-f(x_0)\|\leq k\|x-x_0\| \,\,\,\,\, \forall x \in B_{\delta}(x_0)
$$
Then, since $\|x-x_0\| < \delta$, $\|f(x)-f(x_0)\|\leq k\delta=\epsilon \,\,\,\,\, \forall x \in B_{\delta}(x_0)$. $Done!$
But I do not understand two things:

1- Why at this stage we can conclude that $f$ is continuous on $A$? In other words why we can go backward and say $\forall \epsilon>0$ there exist $\delta' > 0 $ such that if $B_{\delta'}(x_0)$ then $B_{\epsilon}(f(x_0))$.
2- Why when we know $f$ is Lipschitz on $B_{\delta}(x_0)$ we do not use $x,y \in B_{\delta}(x_0)$ and we use $x_0$ as one of our point? In other words, can we prove this using $x,y \in B_{\delta}(x_0)$?

REPLY [1 votes]: Continuity is a local property on $A$. For $f$ to be continuous on $A$, we must show that $f$ is continuous at every fixed point $x_0 \in A$. Moreover, the continuity of $f$ at $x_0$ depends only on the behaviour of $f$ near $x_0$. Put otherwise, when discussing the continuity of $f$ at a fixed point $x_0$, we only care about how $f$ behaves in a neighbourhood of $x_0$. Therefore, $f$ being continuous on a ball $B(x_0,\delta) \subset A$ immediately gives the continuity of $f$ at the point $x_0$.
As for your second question, say we fix $x_0 \in A$ and choose a small ball $B(x_0,\delta) \subset A$ on which $f$ is Lipschitz. Because $f$ is only assumed to be locally Lipschitz, we cannot claim that the inequality
$$
||f(x)-f(y)|| \leq k ||x-y||
$$
holds for all $x,y \in A$. However, we can certainly claim that it holds for every $x,y \in B(x_0,\delta)$. This implies that $f$ is uniformly continuous on the small ball $B(x_0,\delta) \subset A$, which is stronger than simply being continuous there.

Edit:.
Fix $x_0 \in A$ and let $B(x_0,\delta) \subset A$ be an open ball on which $f$ is Lipschitz. Then, we can find $L > 0$ such that
$$
||f(x)-f(y)||\leq L||x-y||, \quad \forall x,y \in B(x_0, \delta).
$$
In particular, because $x_0 \in B(x_0,\delta)$, we have
$$
||f(x)-f(x_0)||\leq L||x-x_0||, \quad \forall x\in B(x_0, \delta).
$$
Using this, we will show that $f$ is continuous at $x_0$. Let $\epsilon > 0$ and choose $\delta^\prime > 0$ such that
$$
0 < \delta^\prime < \frac{\epsilon}{L} \quad \text{and} \quad \delta^\prime < \delta.
$$
Then, if $|x-x_0| < \delta^\prime$ one also has $x \in B(x_0,\delta)$ so that
\begin{align*}
||f(x)-f(x_0)|| \leq L||x-x_0|| < L\delta^\prime < \epsilon.
\end{align*}
Or, rather, $f$ sends $B(x_0,\delta^\prime)$ into the ball $B(f(x_0),\epsilon)$. This means that $f$ is continuous at $x_0$. By our remarks above the edit, $f$ is also continuous on $A$.


TITLE: Using definition of open set determine if A is open set
QUESTION [0 upvotes]: Let $x_{0} = (2,0,1,1)$ and determine whether $A = \{(x,y,z,t)\in \mathbb{R^4} : x^2+y^2=4, z >0, t \geq 0 \}$ is open or not.

$A$ is said to be open iff $\forall x \in A, \; \exists \varepsilon >0$ such that $\mathcal{B}_{\varepsilon}(x) \subset A$. It is easy to see $x_{0}\in A$ since $2^2+0^2=0,\; 1>0 ,\; 1\geq 0$ all properties of $A$ are satisfied. But I couldn't use the definition. Thanks in advance!

REPLY [0 votes]: A is said to be open iff $\forall x \in \mathbb{A}, \; \exists \varepsilon > 0 \;$ such that $\mathcal{B_{\varepsilon}(x)}\subset \mathbb{A}$. For $x_{0}$ it is easy to see $x_{0} \in \mathbb{A}$. Let's build a ball around $x_{0}$:
$$\mathcal{B_{\varepsilon}(x)}=\{(x,y,z,t)\in \mathbb{R^4}: \|(x,y,z,t)-(2,0,1,0)\| <\varepsilon\}$$
$$\mathcal{B_{\varepsilon}(x)}=\{(x,y,z,t)\in \mathbb{R^4}:\sqrt{(x-2)^2+y^2+(z-1)^2+t^2}\}$$
Let $y=(2,-\frac{\varepsilon}{2},1,0) \in \mathcal{B_{\varepsilon}(x)}:$
$$\sqrt{(2-2)^2+\frac{\varepsilon^2}{4}+(1-1)^2+0^2}$$
$$\frac{\varepsilon}{2}<\varepsilon$$
But $y \in \mathbb{A}$ since $-\frac{\varepsilon}{2}$ not $z > 0$. So $\mathbb{A}$ is closed.


TITLE: On a finite nilpotent group with a cyclic normal subgroup
QUESTION [3 upvotes]: I'm reading Dummit & Foote, Sec. 6.1.
My question is the following.
If $G$ is a finite nilpotent group with a cyclic normal subgroup $N$ such that $G/N$ is also cyclic, when is $G$ abelian?
I know that dihedral groups are not abelian, and I think the question is equivalent to every Sylow subgroup being abelian.
EDIT: So, the real question is about finding all metacyclic finite p-groups that are not abelian.
Thanks.

REPLY [2 votes]: A finite nilpotent group is abelian if and only if all of its Sylow subgroups are abelian.  This is just because a finite nilpotent group is the direct product of its Sylows.
Groups G with normal subgroups N such that both N and G/N are cyclic are called metacyclic groups.
Metacyclic p-groups need not be abelian: take the Sylow p-subgroup of the group of affine functions on $R=\mathbb{Z}/p^2\mathbb{Z}$, that is all functions $x \mapsto ax+b$ where a, b are in R and where the multiplicative order of a is a power of p.  This is a group of order p3 called the extra special group of exponent p2 and order p3.  For p=2 it is also known as the dihedral group of order 8.
Metacyclic p-groups were classified quite a few times.  My favorites are King (1973) and Liedahl (1996). Magma has a slightly different classification available as a database, and it was easy to write code for King and Liedahl in GAP.

King, Bruce W.
"Presentations of metacyclic groups."
Bull. Austral. Math. Soc. 8 (1973), 103–131.
MR323893
DOI:10.1017/S0004972700045500
Liedahl, Steven.
"Enumeration of metacyclic p-groups."
J. Algebra 186 (1996), no. 2, 436–446.
MR1423270
DOI:10.1006/jabr.1996.0381


TITLE: How to prove the limit of form $0^{\infty}$ equals to 0 without restriction?
QUESTION [0 upvotes]: This is a question from Calculus 7e by Stewart:

It's easy to prove by calculating the limit of $e^{ln(f(x))\cdot g(x)}$ if $f(x)$ is restricted to a positive function. However, seems that if we remove this restriction the conclusion still holds. But then we cannot use $ln(f(x))$ since $f(x)\leq 0$ is possible. So how to prove the general form of this limit?

REPLY [1 votes]: In calculus context for function of real variables, the function
$$[f(x)]^{g(x)}$$
is restricted and defined only for $f(x)>0$.
Refer also to the related

Finding domain of $ f(x) ^ {g(x)} $?

REPLY [0 votes]: Let $\varepsilon>0$. Then $\min\left\{\varepsilon,\frac12\right\}>0$ and therefore there is a $\delta_1>0$ such that $$0<\lvert x-a\rvert<\delta_1\implies\bigl\lvert f(x)\bigr\rvert<\min\left\{\varepsilon,\frac12\right\}.$$And there is a $\delta_2>0$ such that$$0<\lvert x-a\rvert<\delta_2\implies g(x)>1.$$Therefore, if $\delta=\min\{\delta_1,\delta_2\}$,$$0<\lvert x-a\rvert<\delta\implies\left\lvert f(x)^{g(x)}\right\rvert=\bigl\lvert f(x)\bigr\rvert^{g(x)}<\min\left\{\varepsilon,\frac12\right\}\leqslant\varepsilon.$$


TITLE: Fourier transform of $F(x)$
QUESTION [1 upvotes]: $$F(x)=\int^{\infty}_{-\infty} e^{-i \xi x}f(x)\ dx$$
Fourier transform of $F$:
$$\int^{\infty}_{-\infty}\left[\int^{\infty}_{-\infty} f(\xi)e^{-x^2\xi^2}\ d\xi\right]\ dx$$
Is that the way to proceed? If yes, how do I continue and if No, show me how I should proceed.

REPLY [0 votes]: $=2 \pi [\int^{\infty}_{-\infty} F(x) e^{ix(-\xi)}$ $dx]=2\pi$ $f(-\xi)$


TITLE: Continuous variables: Probability of outcome in a quadrature measurement
QUESTION [0 upvotes]: In the phase-space formulation of QM over continuous variables, how can I determine the probability of obtaining a particular measurement outcome $m$ in the following setting.
Given a quantum state $\rho$ described by the Wigner function $W_{\rho}(q,p)$, one measures the observable $\hat{p}$. What is the probability of obtaining a particular outcome $m\in\mathbb{R}$? I know that in standard QM I need to calculate
$$\text{Tr }(\rho \; |m\rangle _p\langle m |)$$
and I have found that such expectation values, in phase space formulation, are usually computed via
$$\text{Tr }(\rho A)=\int dq dp W_{\rho}(q,p)A(q,p)$$
where $\hat{A}$ is some observable expressed as a combintation of $\hat{q}$ and $\hat{p}$.
Once this easy quadrature measurement is clarified, I would also like to understand partial measurements, i.e. let $W(q_1,p_1,q_2,p_2)$ be the input state, then what is probability of outcome $m$ when measuring $\hat{p}$ only on the first mode.
Note, I am new to the phase-space formulation. Thus, I would also appreciate a good reference where I could simply read up on this.

REPLY [1 votes]: The Wigner map image of your momentum projection operator $|m\rangle \langle m|$ is
$$
A(q,p)= 2 \int dy ~ e^{2ipy/\hbar} \langle q+y| m\rangle \langle m|q-y\rangle =
\frac {2} { 2\pi} \int dy ~ e^{2ipy/\hbar} e^{-2imy/\hbar} = \delta (p-m),
$$
by virtue of $\langle x|p\rangle= \exp (ixp/\hbar)~/\sqrt{2\pi}$, etc.
Consequently, the expectation you presumably arrived at in your intro is
$$
\int\! dq ~ W_\rho (q,m).
$$
Mutatis mutandis, collapsing the momentum of $\hat p _1$ to m, yields
$$\int \! dq_1 dq_2 dp_2 ~ W(q_1,m,q_2,p_2).$$
Here is a concise review  with exercises to practice on.


TITLE: Fourier transform inversion formula for $f\in L_1(\mathbb{R}^n)$ and Dini condition
QUESTION [4 upvotes]: Let us define the Dini condition for a function $f\in L_1(-\infty,\infty)$, i.e. Lebesgue summable on $\mathbb{R}$, as

Given an $x\in\mathbb{R}$ there is a $\delta>0$ such that the Lebesgue integral $\int_{[-\delta,\delta]}|\frac{f(x+t)-f(x)}{t}|d\mu_t$ exists.

In all the post the integrals are to be intended as Lebesgue integrals. I know (p. 423, here, of Kolmogorov-Fomin's Элементы теории функций и функционального анализа) that, if $f\in L_1(-\infty,\infty)$ satisfies the Dini condition in $x\in\mathbb{R}$, then the inversion formula for the Fourier transform holds: $$f(x)=\frac{1}{2\pi}\lim_{N\to+\infty}\int_{[-N,N]}\Bigg(\int_{\mathbb{R}}f(t)e^{-i\lambda t)}d\mu_t\Bigg)e^{i\lambda x} d\mu_\lambda.$$
The same famous text proves (p. 437-438) a similar theorem for functions $f\in L_1(\mathbb{R}^n)$, under conditions that I do not fully understand, stated in the following way:

Let function $f(x_1,x_2,\ldots,x_n)$ ne integrable on the whole space $\mathbb{R}^n$ and let it satisfy the conditions:
  $$|f(x_1+t_1,x_2,\ldots,x_n)-f(x_1,x_2,\ldots,x_n)|\leq C|t_1|^a,$$ $$|f(x_1,x_2+t_2,\ldots,x_n)-f(x_1,x_2,\ldots,x_n)|\leq C(x_1)|t_2|^a,$$$$\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots$$$$|f(x_1,x_2,\ldots,x_n+t_n)-f(x_1,x_2,\ldots,x_n)|\leq C(x_1,x_2,\ldots,x_{n-
> 1})|t_n|^a,$$where $0\le a\le1,\quad\int_{\mathbb{R}}C(x_1)d\mu_{x_1}<\infty,\ldots,\quad\int_{\mathbb{R}}\ldots\int_{\mathbb{R}}C(x_1,\ldots,x_{n-1})d\mu_{x_1}\ldots d\mu_{x_{n-1}}<\infty.$
  Then $$(2\pi)^nf(x_1,x_2,\ldots x_n)=$$$$=\lim_{N_1\to\infty}\int_{-N_1}^{N_1}\Big(\ldots\lim_{N_{n-1}\to\infty}\int_{-N_{n-1}}^{N_{n-1}}\Big(\lim_{N_n\to\infty}\int_{-N_n}^{N_n}g(\lambda_1,\ldots,\lambda_n)e^{ix_n\lambda_n }d\mu_{\lambda_n}\Big)e^{ix_{n-1}\lambda_{n-1}}d\mu_{\lambda_{n-1}}...\ldots\Big)e^{ix_1\lambda_1}d\mu_{\lambda_1} $$

where the integrals are Lebesgue integrals and $g(\lambda_1,\ldots,\lambda_n):=\int_{\mathbb{R}}\ldots\int_{\mathbb{R}}f(x_1,\ldots,x_n)e^{-i\sum_{k=1}^nx_k\lambda_k}d\mu_{x_1}\ldots d\mu_{x_n}$.
The theorem is precisely stated in this way in the book (I've only changed some notation to clearly show the integrals are Lebesgue integrals), but I am not sure to understand what are $a$ and the $t_k$ in this wording. Does anybody passing by know the theorem and what the inequalities mean: must they hold for some $a\in[0,1]$ or for all $a\in[0,1]$? As to the $t_k$, $k=1,\ldots ,n$ must they satisfy the inequality for some $\delta>0$ such that $t_1,\ldots,t_n\le\delta$?
The proof seems to use the fact that the function $f(-,x_2,\ldots ,x_n)$ as a function of $x_1$ satisfies the above defined Dini condition: does it? Why can we see that? In particular, I am not convinced that $a=0$ could guarantee the Dini condition ($\int_{0}^1 x^{-1}dx=+\infty$).
I think that my greatest problem is that I do not understand the conditions in the theorem's statement... Thank you so much for any explanation!

REPLY [1 votes]: The Riemann-Lebesgue lemma states that, for any $f \in L^{1}(\mathbb{R})$,
$$
                 \lim_{R\rightarrow\pm\infty}\int_{-\infty}^{\infty}f(t)e^{-iRt}\,dt = 0.
$$
This becomes the basis for a lot of basic pointwise convergence theorems. If $\mathcal{F}$ denotes the Fourier transform on $\mathbb{R}$, and $\mathcal{F}^{-1}$ its inverse transform, then
$$
             \mathcal{F}^{-1}(\chi_{[-R,R]}\mathcal{F}f)|_{x}-L=\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{f(t)-L}{t-x}\sin(R(t-x))\,dt.
$$
The function $\frac{f(t)-L}{t-x}$ is absolutely integrable for $t \in (-\infty,x-\delta]\cup[x+\delta,\infty)$ whenever $\delta > 0$. If it is also integrable on $[x-\delta,x+\delta]$, then the Riemann-Lebesgue lemma gives
$$
           L = \lim_{R\rightarrow\infty}\mathcal{F}^{-1}(\chi_{[-R,R]}\mathcal{F}f)|_{x}.
$$
The Dini condition is nothing more than the additional assumption required to guarantee that $f \in L^{1}$ or $f\in L^{2}$ gives rise to a function $\frac{f(t)-L}{t-x}$ which is absolutely integrable in $t$ for a fixed $x$, where $L=f(x)$. One way to guarantee this condition is by assuming $f$ is $\alpha$-Holder continuous at $x$ with exponent $\alpha \in (0,1]$:
$$
                          |f(t)-f(x)| \le C|t-x|^{\alpha}.
$$
This condition is sufficient in order the Dini condition to hold, but not necessary, but Holder continuity is convenient, especially for Sobolev spaces.
The conditions stated in your theorem allow you to evaluate the full inverse Fourier transform on $\mathbb{R}^{n}$ by evaluating with iterated integrals. That's why the conditions stated are "staggered" in some sense. But they're set up so that the inversion integral can be evaluated using iterated integrals and limits. That is, if $B_{N}$ is the box in $\mathbb{R}^{n}$ described by $\{ x : |x_{k}| \le N_{k},\; 1 \le k \le n\}$, then
$$
        \mathcal{F}^{-1}(\chi_{B_{N}}(x)\mathcal{F}f)
           = \mathcal{F_{n}}^{-1}\chi_{[-N_n,N_n]}\mathcal{F_{n-1}}^{-1}\chi_{[-N_{n-1},N_{n-1}]}\cdots \mathcal{F_{1}}^{-1}\chi_{[-N_1,N_1]}\mathcal{F}f,
$$
where $\mathcal{F_{k}}^{-1}$ is the inverse Fourier transform in the $k$-th coordinate variable only.
Knowing that the inversion integral exists on the whole space allows you to evaluate by iterated limits, and the imposed conditions allow you to move the limits inside the transforms to the location where they can be evaluated using the Riemann-Lebesgue/Dini procedure described above for a single variable, one at a time.


TITLE: The concept of gradient, related to lagrange multipliers, surface areas, tangent hyper planes
QUESTION [3 upvotes]: As we all know, gradient is always perpendicular to the level curve. On the other hand, $\nabla f(a,b) \dot\ h$ where $h=(x-a\     \ \ \ \ \ y-b)^T$, give a  tangent hyper plane which is tangent to the point on a surface, you can find in many texts that such hyper plane is in general, $z=f(a,b)+\nabla f(a,b) \dot\ h$ at some point $(a,b,f(a,b))$.  For instance, when we are calculating the surface area, we would denote $T_s\triangle s$ and $T_t\triangle t$ as the vector which is spanning a paralleogram to approximate the rectangle $ \triangle s \times \triangle t$ where $T_s =\frac{\partial F}{\partial s} $ and F is the surface. In lagrange multiplier, we know f has extremum point if $\nabla f= \lambda\nabla g$, from this expression, we know the grandient f should be perpendicular to the level set g, where it is the domain of f. From the proof, the idea is if $\nabla f $ is not parallel to $\nabla g$ then there is change in f, but here come the question, why if $\nabla f $ is not parallel to
$\nabla g$, then there will be a change in f either increase or decrease. Any explaination to gradients would be appreciate.d

REPLY [1 votes]: Okay, some background on the gradient. Since $\gamma$ carves out the level set $g(x,y)=c$, we clearly must have $g\circ\gamma=c$ identically. The derivative of a constant is zero, so we obtain
$$\frac{d}{dt}g\circ\gamma=(\nabla g\circ\gamma)\cdot\gamma\,'=0.$$
As above, when two vectors' dot product is zero they are perpendicular, thus $\nabla g$ is perpendicular to the tangent of the curve at any point on the curve. This is all in $\Bbb R^2$, two dimensions. Now go to $\Bbb R^3$.
Recall that a plane's equation is determined by its normal vector and one point. Indeed, if $n$ is the normal and $p$ a point on a plane $\pi$, then $\pi-p$ (the plane translated by $-p$) is a parallel plane (so it has the same normal) containing the origin, and therefore has equation $n\cdot x=0$ (the plane is defined to be all points perpendicular to given normal). Therefore $\pi$ has equation $n\cdot(x-p)=0$.
Now consider the graph of $g:\Bbb R^2\to\Bbb R$ as a surface in $\Bbb R^3$. The function is given by
$$h:(x,y)\mapsto(x,y,g(x,y)).$$
Two vectors tangent to the surface can be obtained partial differentiation
$$\frac{\partial h}{\partial x}=\left(1,0,\frac{\partial g}{\partial x}\right), \qquad \frac{\partial h}{\partial y}=\left(0,1,\frac{\partial g}{\partial y}\right).$$
A vector is normal to the plane iff it is normal to both of the above, and we can check that
$$n=\left(\frac{\partial g}{\partial x},\frac{\partial g}{\partial y},-1\right)$$
satisfies this. Our equation of a tangent plane is therefore
$$\left(\frac{\partial g}{\partial x},\frac{\partial g}{\partial y},-1\right)\cdot\big((x,y,z)-(a,b,g(a,b))\big)=0.$$
Rearranging, we obtain
$$z=g(a,b)+\nabla g(a,b)\cdot\big((x,y)-(a,b)\big),$$
where now the gradient and dot product above are two dimensional.


TITLE: Solving Euler equation
QUESTION [1 upvotes]: Let $t^2 x''-3tx'+3x = 0$.
We want to find the general solution of this equation using the substitution $t = e^s$.
Then, if we name $y(s):=x(e^s)=x(t)$ we have that $y''-3y'+3y=0$.
To find the solutions of this equation we would find the zeros of the associated polynomial, which would be $\lambda = \frac{3\pm \sqrt{9-12}}{2} $.
But that freaks me out because the solutions of the original Euler equation actually are $t$ and $t^3$, so I would have expected $\lambda \in \{1, 3\}$.
What is going on? Why do I get this weird result?

REPLY [2 votes]: We have $ y(s)=x(e^s)$ so when we differentiate we get
\begin{eqnarray*}
y'(s)=e^s x'(e^s) =tx'\\
y''(s)=e^{2s} x''(e^s)+\color{red}{e^s x'(s)}= t^2 x''+\color{red}{tx'} \\
\end{eqnarray*}
& when you substitute this into $t^2 x''-3tx'+3x = 0$ you get $y''-\color{red}{4}y'+3y=0$.


TITLE: A family $\mathscr{F}\subset \operatorname{Aut}(G)$ is compact
QUESTION [1 upvotes]: Let $G\subset\mathbb{C}$ be any bounded region, and let $w\in G$. Let $\mathscr{F}$ be the set $\{f\in \operatorname{Aut}(G):f(w)=w\}$ ; where $\operatorname{Aut}(G)$ is the group of all biholomorphic maps on G.
Then we have to show that $\mathscr{F}$ is compact.
My idea :
Let $\{f_n\}_{n=1}^{\infty}$ be a sequence in $\mathscr{F}$.
Since G is bounded region, $\overline{G}$ is compact in $\mathbb{C}$, and the diameter of $G$ is finite.. Thus there exists $M>0$ such that $\lvert f(z) \rvert \leq M$ for every $z\in G$ and for every $f\in \mathscr{F}$. Hence $\mathscr{F}$ is a locally uniformly bounded family. So, by Montel's theorem, there exists a subsequence $\{f_{n_k}\}_{k=1}^{\infty}$ of $\{f_n\}_{n=1}^{\infty}$ which converges uniformly to $f$ on every compact subset of $G$. Since $\mathscr{H}(G)$, the family of all holomorphic functions on $G$, is a complete metric space, we say that $f$ is also holomorphic on $G$. It is also clear that $f(w)=w$.
Now I am stuck. How do I show that $f$ is infact a member of $\mathscr{F}$ ? If this thing can be shown then we can say, by sequential compactness, that the family $\mathscr{F}$ is indeed compact.

REPLY [0 votes]: To simplify notation, we will denote by $f_n$ the sequence of functions in $\mathcal F$ that converges to $f$ unifmorly on compact subsets. By passing to a further subsequence, we can also assume that the sequence $h_n:=f_n^{-1}$ converges uniformly on compact subsets, as well, using the same Montel argument. Let us denote the limit of $h_n$ by $h$.
First we claim that $f(G) \subset G$ (this is not clear: a priori we only have $f(G) \subset \overline{G}$).
So let $p \in \Bbb C \setminus G$. Now consider the sequence of functions $g_n:G \to \Bbb C, z \mapsto f_n(z)-p$. This converges uniformly on compact subsets to $g_n:G \to \Bbb C,z \mapsto f(z)-p$. Note that for each $n$, the function $g_n$ has no zeroes on $G$ as $f_n(G) \subset G$. By Hurwitz' theorem, this implies that either $g$ has no zeroes on $G$ or $g \equiv 0$. But clearly we have $g \not\equiv 0$, because $g(w)=w-p \neq 0$. This shows that $g$ has no zeroes on $G$ which means that $f(z) \neq p$ for all $z \in G$. As $p \in \Bbb C \setminus G$ was arbitrary, this implies that $f(G) \subset G$. This argument shows that $h(G) \subset G$ as well.
Now as we have established that $f$ and $h$ are holomorphic functions $G \to G$, it makes sense to consider the compositions $f \circ h$ and $h \circ f$.
We can do this part in more generality:

Let $U \subset \Bbb C$ be a open. Let $f_n,h_n$ be sequences of functions $U \to U$ that converge uniformly on compact subsets to $f$ and $h$, respectively and assume that $f(U) \subset U$ and $h(U) \subset U$. Then the sequence $f_n \circ h_n$ converges uniformly on compact subsets to $f \circ h$

Proof Let $K$ be a compact subset of $U$. Let $z \in K$.
Then we have for every $n \in \Bbb N$:
$$|f(h(z))-f_n(h_n(z))| \leq |f(h(z)) - f_n(h(z))|+|f_n(h(z))-f_n(h_n(z))|$$
For the first term, note that $h(K)$ is a compact subset of $U$, thus $|f(h(z)) - f_n(h(z))| \to 0$ uniformly in $z \in K$.
We now claim the following: there is some compact subset $L \subset U$ such that for all $n$ sufficiently large, we have $h_n(K) \subset L$ and also $h(K) \subset L$. Because $h(K)$ is compact and $U \subset \Bbb C$ is open, there is some $\varepsilon >0$ such that $L:=\{z \in \Bbb C \mid \exists \zeta \in h(K): |z-\zeta| \leq \varepsilon\}$ is contained in $U$. $L$ is a closed and bounded subset of $\Bbb C$, so it is compact by Heine-Borel. We get that $h_n(K) \subset L$ for $n$ sufficiently large, because $h_n$ converges to $h$ uniformly on $K$.
Having found such an $L$, we can take care of the second term. We need to show that $|f_n(h(z)) - f_n(h_n(z))| \to 0$ uniformly in $z \in K$.
Using Cauchy's inequality and compactness, the set $\{|f'_n(z)|, \in \Bbb N, z \in L\}$ is bounded. Thus we can find a common Lipschitz constant $C > 0$ for $f_n$ on $L$ that does not depend on $n$. Thus we get for all $n$ sufficiently large:
$$|f_n(h(z)) - f_n(h_n(z))| \leq C |h(z)-h_n(z)|$$ which converges to $0$ uniformly on $K$ because $h_n \to h$ uniformly on $K$.
Now back to our situation: by what we have established so far, we get that $f_n \circ h_n$ converges uniformly on compact subsets to $f \circ h$. But for all $n$, we have $f_n \circ h_n=\mathrm{id}_G$. Thus $f \circ h=\mathrm{id}_G$. By the same argument, $h \circ f = \mathrm{id}_G$.
This shows that $f \in \mathrm{Aut}(G)$. Also clearly $f(w)=w$, so $f \in \mathcal F$.
Note: More generally, the group of isometries of a Riemannian manifold has compact stabilizers. For a reference, see theorem II.1.1 in Kobayashi's Transformation Groups in Differential Geometry. As we can alway put a Riemannian metric on a Riemann surface such that the group of orientation-preserving isometries is the group of holomorphic automorphisms and the group of orientation preserving isometries is closed in the group of all isometries, we get that the group of holomorphic automorphism of any Riemannian surface has compact stabilizers.


TITLE: Is there a group that can act as a foundation of mathematics?
QUESTION [0 upvotes]: Implication is not associative. This results in quite a bit of structure that can be used in set theory with things like ZFC, but I’m curious if it is necessary, as the parentheses can be annoying to deal with. While prefix and postfix forms are options, they often aren’t very readable.
Are there any sets of axioms that seem to have enough structure to act as a mathematical foundation while having only associative operators?
Because multiple associative operations can be composed to simulate implication (and thus not be associative), I think I’m looking for a single associative operator that is sufficient to simulate ZFC. If we are going that route, we might as well require inverse and identity as well so we can use group theory. This results in the formal question (thanks to @NoahSchweber):

Is there a finitely generated computably presentable group $G$ on generator set $A$ and a computable function $f$ from first-order formulas to words on $A$ such that $\mathsf{ZFC}\vdash\sigma\leftrightarrow\tau$ iff $f(\sigma)$ and $f(\tau)$ represent the same element in $G$?

I could imagine that question being answered in a non-constructive way, so a constructive answer is preferred. In words, I’m asking for an infinite group that has enough structure to correspond to the standard kinds of things discussed in mathematics. Unlike the classification of finite simple groups, we still don’t have a good handle on all infinite groups, and there are plenty of infinite groups with undecidable decision problems on them (for example), so it’s conceivable that there are some that are expressive enough to work.
I’ve been looking through alternate metamath axiom sets, but all of them so far seem to require some form of implication or other operator that isn’t associative.

REPLY [3 votes]: EDIT: The followup question below/above was asked and answered affirmatively at mathoverflow.

In classical logic, implication doesn't need to be included as a primitive operation: we can define it in terms of the operations of negation and disjunction, which are respectively unary and associative, as $$A\rightarrow B\quad:\equiv\quad (\neg A)\vee B.$$ Indeed, the set of Boolean operations $\{\neg,\vee\}$ is functionally complete. So we can rewrite (say) all the $\mathsf{ZF}$ axioms using only disjunction and negation.
(You may also be interested in cylindrical algebras, which provide an algebraic semantics for first-order logic whose basic operations are all either unary or associative.)
Of course, taht doesn't actually solve the notation problem you mention: we still need parentheses to distinguish between e.g. $(\neg A)\vee B$ and $\neg(A\vee B)$. Basically, the problem is that associative operations can be combined to produce non-associative operations. If what we actually want is to avoid parentheses, then instead of changing our basic logical connectives we should instead (as Asaf Karagila comments) adopt something like Polish notation. For instance, the parentheses-laden expression $$p\rightarrow\neg(q\rightarrow r)$$ would be written as $$CpNCqr.$$ This, though, has the drawback of (at least in my experience) generally being less readable for long expressions, especially when quantifiers enter the picture.

You separately ask, however, a question about structures, e.g. whether there is

a certain infinite group that has enough structure to correspond to the standard kinds of things discussed in mathematics.

It's unclear to me exactly what this means, but there are certainly some observations we can make which seem relevant. Most obviously, we have on the one hand the set of first-order formulas modulo $\mathsf{ZFC}$-provable equivalence, and on the other hand - given a computably presentable group $G$ - the set of words on the generators of $G$ modulo equality given the presenting relations of $G$, and we could ask for an appropriate correspondence. The coarsest question is:

Is there a finitely generated computably presentable group $G$ whose word problem is as complicated as the problem of telling whether two sentences are $\mathsf{ZFC}$-provably equivalent?

The answer is yes: we can in fact find a finitely presentable group $G$ such that the halting problem is many-one reducible to the word problem for $G$, and $\mathsf{ZFC}$-provable equivalence is many-one equivalent to the halting problem.
On a more algebraic note, we could ask:

Is there a finitely generated computably presentable group $G$ on generator set $A$ and a computable function $f$ from first-order formulas to words on $A$ such that $\mathsf{ZFC}\vdash\sigma\leftrightarrow\tau$ iff $f(\sigma)$ and $f(\tau)$ represent the same element in $G$?

I believe that the answer is yes - and in fact that we can use the same $G$ as before - but I don't know.
I added the "finitely generated" bit since without it the question is trivial: take a generator $a_\sigma$ for each formula $\sigma$ and consider an appropriate set of relations making $a_\sigma=a_\tau$ iff $\mathsf{ZFC}\vdash\sigma\leftrightarrow\tau$.


TITLE: Perfect Groups with faithful complex irreps
QUESTION [5 upvotes]: Is it true that a perfect finite group $ G $ has a faithful complex irrep if and only if the center of $ G $ is cyclic?
The corresponding fact is true for quasisimple finite $ G $
Does every quasisimple finite group have a faithful complex irrep?
And certainly it is necessary to have cyclic center for a finite group to have a faithful complex irrep.
So the question is really: Does every perfect group with cyclic center have a faithful complex irrep?
An example of a non perfect group with cyclic center which has no faithful complex irrep is given here
https://mathoverflow.net/q/57129/387190

REPLY [4 votes]: The criterion  of Gaschütz for (a finite group) $G$ having a faithful irreducible complex representation described in the linked mathoverflow post can be used to show theoretically that examples like the one of ahulpke must exist.
The criterion for the existence is that the socle of $G$ should be generated by a single conjugacy class of $G$.
Consider groups $G$ with a normal elementary abelian subgroup $N$ of order $p^n$ such that $S = G/N$ is nonabelian simple. Then conjugacy classes in $G$ have size at most $|S|$, so if $n > |S|$ then there can be no such representation.
For all such $S$ and all primes $p$ we can find examples of the type, by letting $M$ be any nontrivial irreducible module for $S$ over ${\mathbb F}_p$, and setting $G = M^k \rtimes S$ for any $k>0$, with the induced module action of $S$ on $M^k$.
The example of ahulpke is of this type with $n=12$ and $M$ of dimension $4$.


TITLE: Accurate model of a laptop when it is not connected to any external device
QUESTION [2 upvotes]: You have a laptop with a fixed amount of memory and hard disk space and no external
storage devices connected (CD, USB drives, . . . ). Which of the following is the most
accurate formal model of your laptop?
(a) Turing machine
(b) Linear bounded automaton
(c) Pushdown automaton
(d) Finite state automaton
First of all I don't mind saying that I could not interpret well enough. I understand it conveys that the memory is limited but how much that affect when it is connected to any external device is blur to me.Help appreciated.
N.B- This is not a homework question.Its from a pg admission exam.
Source- http://www.cmi.ac.in/admissions/sample-qp/pgcs2011.pdf

REPLY [2 votes]: They just put the bit about external devices in so that you would be sure that you couldn't, for example, plug in a drive to make the storage bigger.
A Turing machine has an unlimited tape, so your finite-memory computer can't satisfy that. Similarly a pushdown automaton has a potentially-unlimited stack.
However, a computer is fully programmable, while a finite state machine is not. So, while strictly your computer has a finite number of states and well-defined transitions between them (since its memory and CPU together have only finitely many possible states), I would say it is more like a linear bounded automaton.


TITLE: Showing that a function $u:[a,b]\to \mathbb{R}$ is measurable
QUESTION [0 upvotes]: How does one show if one function $u:[a,b]\to \mathbb{R}$ is measurable?

Using Schilling's methodology, I understand this as to show that $u$ is $\mathcal{A}/\mathcal{B}$ measurable (see Chapter 8), where $\mathcal{A}$ is a $\sigma$-algebra on $[a,b]$. To do this, one may show that, for example, $\{u<\alpha\}\in \mathcal{A}$ for all $\alpha\in \mathbb{R}$.

Is this automatically true, if $u$ were continuous on $[a,b]$?

Here is what I did,
$\{u<\alpha\}=\{x\in [a,b]:u(x)<\alpha\}=u^{-1}((-\infty,\alpha))$
Then the right hand side is open, because of continuity. But does it belong to $\mathcal{A}$?

Another question is, no sure if it is relevant, what if $u$ were not continuous on $[a,b]$?

REPLY [0 votes]: Consider 2 examples:
(1) $\mathcal{A}=\sigma\{[a,(a+b)/2),[(a+b)/2,b]\}$ and $u(x)=x$ ($u$ is not $(\mathcal{A},\mathcal{B}_{\mathbb{R}})$-measurable).
(2) $\mathcal{A}$ is the Borel $\sigma$-algebra (i.e. generated from open sets) and $u$ is continuous ($u$ is $(\mathcal{A},\mathcal{B}_{\mathbb{R}})$-measurable according to your argument).
Regarding your second question, there are discontinuous functions that are measurable w.r.t. $\mathcal{A}$ in (1), e.g. $u(x)=1_{[a,(a+b)/2)}(x)$, and there exist functions that are not measurable w.r.t. $\mathcal{A}$ in (2) (which is much larger), e.g. $u(x)=1_N(x)$, where $N$ is a non-Borel set in $[a,b]$.


TITLE: Proof in ruin player problem
QUESTION [1 upvotes]: Let $M_i$ the average number of matches until the player, or lose all,
  or wins the capital $N$ as it began with the capital $i$. Show that
  $$M_i=i(N-1);p=\frac{1}{2}$$
  $$M_i=\frac{i}{1-2p}-\frac{N}{1-2p}\frac{1-(\frac{1-p}{p})^i}{1-(\frac{1-p}{p})^N};p\neq\frac{1}{2}$$

I'm litle lost but I believe we need to assume that
$$P_{WIN}=P_{LOSE}=\frac{1}{2}$$
in each match. Let $X_n$ the Markov Chain to denote the amount of capital in after $n$ matches, then since they start with $i$
$$P(X_{n+1}=i+1|X_n=i)=\frac{1}{2}$$
$$P(X_{n+1}=i-1|X_n=i)=\frac{1}{2}$$
I need to calculate the expectation here?

REPLY [2 votes]: You do not assume that the probability of losing is equal to the probability of winning; the first line you have given is the formula for a fair game, i.e. when the probability of winning is equal to the probability of winning, i.e. $p=\frac{1}{2}$ as written after the semicolon, whereas the second line is for when $p \neq \frac{1}{2}$.
(The player is betting 1 unit at a time, from what I gather.)
First off, I believe that for a fair game, the solution should in fact be $M_i = i(N-i)$. (Check the boundary condition when the player starts off with capital $N$; then $M_i$ should be zero.)
Moving on, we can denote the time before the player hits $0$ or $N$, starting from $i$, in one random realisation of the game, by $\tau_i$. Then $M_i = \mathbb E [\tau_i]$, and use a first-step decomposition:
$$M_i = 1 + (1-p) M_{i-1} + p M_{i+1}$$
This is a difference equation with boundary conditions $M_0=M_N=0$.
For the fair case when $p = \frac{1}{2}$, $$M_i=A+Bi-i^2$$ is the general solution (you should check this!); by the BCs, $A=0$ and $B=N$, which gives you the first required solution.
For when $p\neq \frac{1}{2}$, the general solution is $$M_i = \frac{i}{1-2p} + A + B \left( \frac {1-p}{p}\right)^i$$
and this should lead you to the second required solution after considering the BCs.
Alternatively: as this is a 'show' problem, it is sufficient for you to just check that the given expressions for $M_i$ satisfy the difference equation (i.e. plug it in), which is probably what you are expected to do!


TITLE: Infinite residue field extensions and algebraic closure of residue fields
QUESTION [1 upvotes]: Let $X$ be a $K$-scheme of finite type over a field $K$, let $L$ be an extension field of $K$, let $X_L := L \times_K X$, and let $p:X_L \rightarrow X$ be the projection. For each $x \in X_L$ we get an extension $k(x) \supset k(p(x))$. Can I choose $L$ such that $k(x)$ contains an algebraic closure of $k(p(x))$? What about $L$ being the algebraic closure of $K$?
All this holds obviously in case $x$ and $p(x)$ are closed, but I don't know about the general situation.

REPLY [1 votes]: I think the answer depends on what you mean by "contains".
Suppose $K = \mathbb{Q}$, $X = \text{Spec }\mathbb{Q}[t]$. For any extension field $L$, $X_L = \text{Spec }L[t]$. Let $x\in X_L$ be the generic point, which must map to the generic point of $X$. Here, $k(x) = L(t)$, and $k(p(x)) = K(t)$. So of course if $L = \overline{K(t)}$, then you could say that $k(x)$ contains $k(p(x))$, but this containment isn't compatible with the natural extension of residue fields induced from the map $p$.
If you allow all these containments, then you can even pick a single $L$ (at least in the case of char 0) that will work for all $x$ (assuming $K$ isn't too big). Namely, set $L = \mathbb{C}$, and simply note that $\mathbb{C}$ contains every field of characteristic zero who's transcendence degree over $\mathbb{Q}$ is at most uncountable.
The example above easily extends to the general affine case, and since your question is local, it extends to all schemes of finite type over $K$.


text:
\begin{definition}[Definition:Triangular Bipyramid]
A '''triangular bipyramid''' is a [[Definition:Polyhedron|polyhedron]] formed by taking two [[Definition:Regular Tetrahedron|regular tetrahedra]] and placing their [[Definition:Base of Tetrahedron|bases]] together:
:[[File:TriangularBipyramid.png|300px]]
The '''triangular bipyramid''' is an example of a [[Definition:Deltahedron|deltahedron]].
\end{definition}

config: wiki