TITLE: Problem in wedge constraint
QUESTION [1 upvotes]: My teacher told all this (u can see in picture)
Here both accelerations have component in -j direction, then why do we say that they cancel out?
[X-axis is surface of wedge]
I have problem in photo 2

Nobody answered so I have to ask it again...

REPLY [1 votes]: The answer is quite simple. In the constrained motion problems given above, the main constraint is that both surfaces must be in contact with each other. This is possible only if the component of acceleration of the two bodies in the direction perpendicular to that of motion is equal both in magnitude and direction i.e. the relative acceleration of the two bodies in the direction perpendicular to motion must be zero.

Here both accelerations have component in -j direction, then why do we say that they cancel out? [X-axis is surface of wedge]

The discussion here is on the relative acceleration of the two bodies in the direction perpendicular to that of motion.
Relative acceleration is given by:-$$\vec{a_{2/1}}=\vec{a_{2}}-\vec{a_{1}}$$
So, if the component of acceleration of the two bodies in the direction perpendicular to motion is equal both in magnitude and direction, we can expect their difference to be zero which is exactly your relative acceleration of the two bodies in the direction perpendicular to motion. Of course, the two accelerations do not cancel out each other as they are in the same direction!!
Hope it helps you.


TITLE: Function with zero average
QUESTION [2 upvotes]: What does it mean to have zero average? Does it mean that the integral over the whole integration domain is zero or for any symmetrical interval around $0$?
I came across the following example where $\Omega$ is a function with zero average, $x'=x/|x|$ and $\phi$ is some Schwartz function. It then says the equality comes from the fact that $\Omega$ has zero average.
\begin{align*}
\lim\limits_{\epsilon \to 0}\int_{|x|>\epsilon}\frac{\Omega(x')}{|x|^n}\phi(x)\mathrm{d}x=\int_{|x|<1}\frac{\Omega(x')}{|x|^n}[\phi(x)-\phi(0)]\mathrm{d}x+\int_{|x|>1}\frac{\Omega(x')}{|x|^n}\phi(x)\mathrm{d}x\\
\end{align*}
But for this to be true the integral $$\int_{|x|<1}\frac{\Omega(x')}{|x|^n}[\phi(0)]\mathrm{d}x$$ should be equal to $0$, so does this mean it has zero average on $(-1,1)$?

REPLY [3 votes]: Notice $Ω = Ω(x')$ is a function defined on the unit sphere. So what it means is that $\Omega$ has zero average on the unit sphere, i.e.
$$0=\int_{|x'|=1} \Omega(x') \,\sigma(dx').$$
This helps when you use polar coordinates to rewrite the integral over  $\epsilon<|x|<1$,
$$\int_{\epsilon<|x|<1}\frac{Ω(x') }{|x|^n} \, dx = \int_{r=\epsilon}^1 \frac{1}{r^n}r^{n-1}\underbrace{\int_{|x'|=1}Ω(x') \, \sigma(dx')}_{=0} \ dr  = 0.$$


TITLE: density function of $\frac{1}{\left(X+Y\right)^{2}+1}$.
QUESTION [0 upvotes]: X and Y are independent continuous random variables with the same density function. Find the density function of $\frac{1}{\left(X+Y\right)^{2}+1}$.
I have tried getting the Jacobian where T maps (x,y) to (u,v) by
u = $\frac{1}{\left(X+Y\right)^{2}+1}$,  $\ v = x $
and $\ T^{-1}$ maps (u,v) to (x,y) by
x = v, $y=\sqrt{\frac{1}{u}-1}$
Am I on the right track?

REPLY [0 votes]: $\newcommand{\+}{^{\dagger}}%
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\fermi}{\,{\rm f}}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\half}{{1 \over 2}}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\ol}[1]{\overline{#1}}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\begin{align}
\pp\pars{Z}
&=
\int_{-\infty}^{\infty}\dd X\,{\rm P}\pars{X}\int_{-\infty}^{\infty}\dd Y
\,{\rm P}\pars{Y}\,\delta\pars{Z - {1 \over \pars{X + Y}^{2} + 1}}
\\[3mm]&=
\int_{-\infty}^{\infty}\dd X\,{\rm P}\pars{X}\int_{-\infty}^{\infty}\dd Y
\,{\rm P}\pars{Y}\times
\\[3mm]&\Theta\pars{Z}\Theta\pars{1 - Z}\braces{%
{\delta\pars{Y - \bracks{X - \root{\pars{1 - Z}/Z}}} \over 2Z\root{Z\pars{1 - Z}}}
+
{\delta\pars{Y - \bracks{X + \root{\pars{1 - Z}/Z}}} \over 2Z\root{Z\pars{1 - Z}}}}
\\[3mm]&=
{\Theta\pars{Z}\Theta\pars{1 - Z} \over 2Z\root{Z\pars{1 - Z}}}\braces{%
\int_{-\infty}^{\infty}{\rm P}\pars{X}\bracks{%
{\rm P}\pars{X - \root{1 - Z \over Z}}
+
{\rm P}\pars{X + \root{1 - Z \over Z}}}\,\dd X}
\end{align}


TITLE: Behaviour of conductor ideal
QUESTION [2 upvotes]: Let $f : A \to B$ be a homomorphism of finitely generated $k$-algebras, where $k$ is a field.  Let $J_A$ and $J_B$ denote the conductor ideals of $A$ and $B$ respectively for the corresponding normalizations in the quotient fields (assume that $A, B$ are domains).  Is it true that $f(J_A) \subseteq J_B$ ?  What is the relationship between $f^{-1}(J_B)$ and $J_A$ ?

REPLY [2 votes]: There is no strong relations between $J_A$ and $J_B$ if you don't have some condition on $A\to B$.
First, it is not true that $f(J_A)\subseteq J_B$: just take $A$ integrally closed and $B$ not integrally closed. Then $1\in J_A$ but $1=f(1)\notin J_B$.
Second, suppose $B$ is the integral closure of $A$ and $B\ne A$. Then $J_B=B$ and $J_B$ is not contained in $f(J_A)B$.


text:
\begin{document}

\title{Partitioning digraphs with outdegree at least 4}

\author{Guanwu Liu\thanks{Partially supported by the China Scholarship Council.
E-mail address: liuguanwu@hotmail.com}\\
School of Mathematical Sciences\\ Dalian University of Technology\\
 Dalian 116024, P.R. China\\
and \\
School of Mathematics\\  Georgia Institute of Technology\\ Atlanta, GA 30332-0160, USA\\
\bigskip \\
 Xingxing Yu\thanks{Partially supported by NSF grant DMS 1954134.  E-mail address: yu@math.gatech.edu} \\
School of Mathematics\\
Georgia Institute of Technology\\
Atlanta, GA 30332-0160, USA}

\date{}

\maketitle

\begin{abstract}
Scott asked the question of determining $c_d$ such that if $D$ is a digraph with $m$ arcs and minimum outdegree $d\ge 2$ then $V(D)$ has a partition $V_1, V_2$
such that $\min\left\{e(V_1,V_2),e(V_2, V_1)\right\}\geq c_dm$, where $e(V_1,V_2)$ (respectively, $e(V_2,V_1)$)
is the number of arcs from $V_1$ to $V_2$ (respectively, from $V_2$ to $V_1$). Lee, Loh, and Sudakov showed that $c_2=1/6+o(1)$ and $c_3=1/5+o(1)$, and conjectured
that $c_d= \frac{d-1}{2(2d-1)}+o(1)$ for $d\ge 4$. In this paper, we show $c_4=3/14+o(1)$ and prove some partial results for $d\ge 5$.

\bigskip

AMS Subject Classification: 05C70, 05C20, 05C35, 05D40

Keywords: Judicious partition, digraph, outdegree
\end{abstract}

\section{Introduction}

Judicious partitioning problems concern partitions of graphs and
hypergraphs that provide bounds for several parameters simultaneously,
while classical partitioning problems seek for partitions that optimize
a single parameter. For a graph $G$ and $A,B\subseteq V(G)$, we use $e(A,B)$ to denote the number of
edges in $G$ between $A$ and $B$, and we write $e(A):=e(A,A)$.
An example of a classical partitioning result is Edwards' theorem \cite{Edwards1973,Edwards1975} that
if $G$ is a graph with $m$ edges then $V(G)$ has a partition $V_1,V_2$
such that $e(V_1,V_2)\ge m/2+(\sqrt{2m+1/4}-1/2)/4$, and the
inequality is tight for complete graphs of odd order.
Bollob\'{a}s and Scott \cite{Bollobas1999} proved the following
judicious version of Edwards' result: The vertex set of any $m$-edge graph
has a bipartition $V_1, V_2$ such that
$e(V_1, V_2)\geq m/2+(\sqrt{2m+1/4}-1/2)/4$ and $\max\{e(V_1),e(V_2)\}\leq m/4+(\sqrt{2m+1/4}-1/2)/8$, and both bounds are tight for
complete graphs of odd order.

 Bollob\'{a}s and Scott \cite{Bollobas2002_1, Scott2005} initiated a systematic study of judicious partitioning problems, which has lead to
a large amount of research in this area, see, for instance,
\cite{Hou2020, Lee2013, Lee2014, MaJCTB, Ma2012, Ma2016, Xu2011,
  Xu2014, Xu2008, Zhu2009}.

Partitioning problems concerning digraphs (i.e., directed graphs)  may be more
difficult.
For a digraph $D$ and $A, B \subseteq V(D)$, we use $e(A, B)$ to denote the number of arcs in $D$ directed from $A$ to $B$ and write $e(A):=e(A,A)$.
Edwards' result above implies that every digraph $D$ with $m$
arcs has a vertex partition $V_1, V_2$ such that $e(V_1, V_2)\geq
m/4+(\sqrt{2m+1/4}-1/2)/8$, and the bound is tight for complete graphs of
odd order with an Eulerian orientation. On the other hand, Alon, Bollob\'{a}s, Gy\'{a}rf\'{a}s, Lehel, and Scott \cite{Alon2007} constructed digraphs
whose maximum directed cut  is $m/4 + O(m^{4/5})$.

A natural judicious version of Edwards' result is to bound both $e(V_1,V_2)$ and $e(V_2,V_1)$. Indeed,
 Scott \cite{Scott2005} asked the following question for digraphs
 without loops or parallel arcs in the same direction. (Throughout this paper, all digraphs
 have no loops or parallel arcs in the same direction.) Note that the {\it outdegree} of a vertex in a digraph is the number of
 arcs directed away from that vertex.

\begin{prob}[Scott \cite{Scott2005}] \label{scottprob}
What is the maximum constant $c_{d}$ such that every digraph $D$ with $m$ arcs and minimum outdegree $d\geq 2$ admits a bipartition $V(D)=V_1\cup V_2$
such that
\[
\min\left\{e(V_1, V_2), e(V_2, V_1)\right\}\geq c_{d}m?
\]
\end{prob}

The reason for  the requirement $d\ge 2$ in Problem~\ref{scottprob} is the following: Take the
star $K_{1,n-1}$ with $n\ge 4$, and add
a single edge between two vertices of degree 1. Orient the unique
triangle so that it becomes a directed cycle, and orient all other
edges so that they are directed towards the unique vertex of degree $n-1$.  This digraph has minimum outdegree 1, and  $e(V_1,V_2)\le 1$
for any bipartition $V_1,V_2$ of its vertex set with $V_1$
containing  the unique vertex of degree $n-1$.  Thus, $c_1=0$.

Lee, Loh, and Sudakov \cite{Lee2014} proved that  $c_2=1/6+o(1)$ and
$c_3=1/5+o(1)$, and they  made the following conjecture for
 $d\geq 4$.

\medskip

\begin{conj}[Lee, Loh, and Sudakov \cite{Lee2014}]\label{conj1}
Let $d$ be an integer satisfying $d\geq 4$. Every digraph $D$ with $m$ arcs and minimum outdegree at least $d$ admits a bipartition
$V(D)=V_1\cup V_2$ with
\[
\min\left\{e(V_1,V_2),e(V_2, V_1)\right\}\geq \left(\frac{d-1}{2(2d-1)}+o(1)\right)m.
\]
\end{conj}

The main term $\frac{d-1}{2(2d-1)}$ in Conjecture \ref{conj1} is best
possible,  because of examples constructed in \cite{Lee2014}
using copies of $K_{2d-1}$ and one copy of $K_{2d+1}$. Lee, Loh, and
Sudakov \cite{Lee2014} also noted that their tools for $d=2,3$ appear to be
insufficient for $d\ge 4$.  Hence, much effort has been devoted to studying variations of this problem, for instance,  by considering minimum
total degree conditions, see  \cite{hou, Hou2020, Hou2017, HWY2017}.  In this paper, we show that Conjecture~\ref{conj1} holds under certain natural conditions. In particular,
we prove Conjecture~\ref{conj1} for $d=4$.

\begin{thm}\label{thm1}
Every digraph $D$ with $m$ arcs and minimum outdegree at least $4$ admits a bipartition
$V(D)=V_1\cup V_2$ with
\[
\min\left\{e(V_1,V_2),e(V_2, V_1)\right\}\geq
\left(3/14+o(1)\right)m.
\]
\end{thm}

 In Section 2, we set up  notation  and list previous results needed in our proof of Theorem \ref{thm1}.
In Section 3, we describe and discuss our  approach for all $d$ and
obtain information in terms of ``huge''
vertices, vertices whose indegree and outdegree have a large gap. In Section 4, we show that Conjecture~\ref{conj1} holds
under some additional conditions on the number of huge vertices.
We complete the proof of Theorem \ref{thm1} in Section 5 and offer some concluding remarks in Section 6.

\section{Notation and lemmas}

We start with  notation and terminology that will be used in this paper.

Let $D$ be a digraph.  For $x\in V(D)$,
let $N_{D}^{+}(x)=\{y : xy\in E(D)\}$ and $N_{D}^{-}(x)=\{y : yx\in E(D)\}$. Then
$d_{D}^{+}(x):=|N_{D}^{+}(x)|$ and $d_{D}^{-}(x):=|N_{D}^{-}(x)|$ are the the $outdegree$ and $indegree$ of $x$,
respectively.

The $degree$ of $x\in V(D)$ is defined as $d_D(x)= d^{+}_D(x)+d^-_D(x)$. We use
$\Delta(D)= \max\{d_D(x) : x\in V(D)\}$ to
denote the $maximum \ degree$ of $D$.

For any $X\subseteq  V(D)$, the subgraph of $D$ induced by $X$ is denoted by $D[X]$.

We will often omit the subscript $D$ in the above notation when there is no
danger of confusion. It will be convenient to write $[k]$ for $\{1, \ldots,
k\}$, where $k$ is any positive integer.

Lee, Loh, and Sudakov \cite{Lee2014} proved that certain partial partitions of a digraph  may be extended
to a good partition of the entire digraph.

\begin{lem}[Lee, Loh, and Sudakov, \cite{Lee2014}]\label{lem1}
Let $D$ be a digraph with  $m$ arcs. Let $p$ be a real satisfying $p\in[0,1]$,
and let $\varepsilon>0$. Suppose that a subset $X\subseteq V$
and its partition $X=X_{1}\cup X_{2}$ are given, and let $Y=V\setminus X$.
Further suppose that $\max_{y\in Y}d(y)\le \varepsilon^{2}m/{4}$.
Then there exists a bipartition $V(D)=V_{1}\cup V_{2}$ with $X_i\subseteq V_i$ for $i\in [2]$ such that
$$e(V_{1},V_{2})  \ge e(X_{1},X_{2})+(1-p)\cdot e(X_{1}, Y)+p\cdot e(Y, X_{2})+p(1-p)\cdot e(Y)-\varepsilon m,$$
$$e(V_{2},V_{1})  \ge e(X_{2},X_{1})+p\cdot e(X_{2}, Y)+(1-p)\cdot e(Y, X_{1})+p(1-p)\cdot e(Y)-\varepsilon m.$$
\end{lem}

By applying  Lemma \ref{lem1}  with $p=1/2$ and $X_1=X_2=\emptyset$ and by noting that $d_D(v)\le 2|V(D)|$, we obtain the following.

\begin{coro}[Lee, Loh, and Sudakov \cite{Lee2014}]\label{lem2}
Let $D$ be a digraph with $n$ vertices and $m$ arcs. For any $\varepsilon>0$, if $m\geq8n/\varepsilon^{2}$
or $\Delta(D)\leq\varepsilon^{2}m/4$, then $D$ admits a bipartition $V(D)=V_{1}\cup V_{2}$ with
$\min\{e(V_{1},V_{2}),e(V_{2},V_{1})\}\geq m/4-\varepsilon m$.
\end{coro}

From Corollary \ref{lem2} we see that if the maximum degree of a digraph $D$ is not too large,
then $V(D)$ admits a partition $V_1, V_2$ such that both $e(V_1,V_2)$ and $e(V_2,V_1)$ are close to $m/4$.
We will see that the vertices causing problems for obtaining the desired partition in Conjecture~\ref{conj1} are those
whose outdegree and indegree differ significantly. Hence, for $x\in V(D)$, let
 $$s^+(x):=d^+(x)-d^-(x), s^-(x):=d^-(x)-d^+(x), \mbox{ and }
 s(x):=\max\{s^+(x),s^-(x)\}.$$
Note that $d(x)-s(x)$ is an even integer, and we often write $$2b=\sum_{x\in X}\big(d(x)-s(x)\big).$$

To study those vertices $x$ with large $s(x)$, we need the concept of the gap of a partition.
Let $D$ be a digraph and let $X,Y$ be a partition of $V(D)$. For each partition $X_1,X_2$  of $X$,
the {\it gap} of $X_1,X_2$ is defined as
$$\theta(X_1,X_2)=  \Big(e(X_{1},Y)+e(Y,X_{2})\Big)-\Big(e(X_{2},Y)
+e(Y,X_{1})\Big).$$
The {\it huge} vertices of $D$ with respect to the partition $X,Y$ are
the vertices $x$ such that $$s(x) \ge \min\{|\theta(X_1,X_2)| :
X_1,X_2  \mbox{ is a partition of } X\}.$$

Let $D$ be a digraph, $X,Y$ a partition of $V(D)$, and
$X_1,X_2$ a partition of $X$.  For convenience, let
$m_f(X_1,X_2)=e(X_1,Y)+e(Y,X_2)$ and $m_b(X_1,X_2)=e(X_2,Y)+e(Y,X_1)$;
so
$$\theta(X_1,X_2)=m_f(X_1,X_2)-m_b(X_1,X_2).$$
Note that $$\theta(X_1,X_2)   =  \Big(e(X_{1},Y)-e(Y,X_1)\Big)-\Big(e(X_{2},Y) -e(Y,X_{2})\Big).$$
Thus, if $e(X)=0$ then
\begin{equation}\label{theta}
\begin{aligned}
\theta(X_1,X_2)   =   \left(\sum_{x\in X_1}s^+(x)\right)-\left( \sum_{x\in X_2}s^+(x)\right)
\end{aligned}
\end{equation}
For any $x\in X$, we say that $x$ is
\begin{itemize}
\item [] {\it $(X_1,X_2)$-forward}  if $x\in X_1$ and $s^+(x)>0$, or $x\in X_2$ and $s^-(x)>0$, and
\item [] {\it $(X_1,X_2)$-backward}  if  $x\in X_1$ and $s^-(x)>0$, or $x\in X_2$ and $s^+(x)>0$.
\end{itemize}
Let $X_f:=\{x\in X: x \mbox{ is $(X_1,X_2)$-forward}\}$ and $X_b:=\{x\in X: x \mbox{ is $(X_1,X_2)$-backward}\}$.
By (\ref{theta}), if $e(X)=0$ then
\begin{equation}\label{theta2}
\theta(X_1,X_2)  =\sum_{x\in X_f}s(x) - \sum_{x\in X_b}s(x).
\end{equation}

We will need the following result from \cite{HWY2017}.

\begin{lem}[Hou, Wu, and Yan \cite{HWY2017}]\label{lem5}
Let $D$ be a digraph and $V(D)=X\cup Y$ be a partition of $D$ with $e(X)=0$.
Let $X=X_{1}\cup X_{2}$ be a partition of $X$ that minimizes
$|\theta(X_1,X_2)|$ among all partitions of $X$. Then
\begin{itemize}
\item [$(1)$]  $|\theta(X_1,X_2)|\leq|Y|$,  and
\item [$(2)$] $g:=\sum_{\{v\in X : s(v)<|\theta(X_1,X_2)|\}}s(v) \leq |Y|-|\theta(X_1,X_2)|$.
\end{itemize}

\end{lem}

\section{Properties of partitions with minimum gap}

In this section, we explore  the probabilistic approach
used by Lee, Loh, and Sodakov \cite{Lee2013, Lee2014}.  In particular, we investigate digraph partitions whose gaps have minimum absolute value.
We will prove several properties about gaps and huge vertices, by
considering various ways to partition the set of huge vertices. Those properties may be useful for the
eventual  resolution of Conjecture~\ref{conj1}.

\begin{lem}\label{property1}
Let $D$ be a digraph with $m$ arcs and minimum outdegree $d\ge 4$, and
let $X,Y$ be a partition of $V(D)$ with $e(X)=0$. Let
$\theta=\min\{|\theta(X_1,X_2)|: X_1,X_2 \mbox{ is a partition of } X\}$, and let $X'=\{x\in X: s(x)\ge \theta\}$.
Let $\varepsilon >0$ such that $\max_{y\in Y}d(y) \le \varepsilon^2m/4$.
Then there exists a partition
         $V_1,V_2$ of $V(D)$ such that $\min\{e(V_1,V_2), e(V_2,V_1)\}\ge \left(
  \frac{d-1}{2(2d-1)}-\varepsilon\right) m$, or the following statements hold:
   \begin{itemize}
       \item [$(1)$]  $\theta >m/(2d-1)$;
       \item [$(2)$] $|X'|$ is an odd integer;
       \item [$(3)$] letting $g=\sum_{x\in X\setminus X'}s(x)$ and
         $X'=\{v_1, \ldots, v_{2k+1}\}$ such that $s(v_1)\ge s(v_2)\ge
         \ldots \ge s(v_{2k+1})$, we have
$\sum_{j=k+1}^{2k+1}s(v_j)-\sum_{j=1}^{k}s(v_j)\geq g+\theta.$
   \end{itemize}
\end{lem}

\begin{proof}
 Suppose, for any partition $V_1,V_2$ of $V(D)$, $\min \{e(V_1,V_2), e(V_2,V_1)\}< \left( \frac{d-1}{2(2d-1)}-\varepsilon\right) m$. We show that (1), (2), and (3)
hold.

First, we prove (1).   Let $m_1=e(X,Y)+e(Y,X)$ and $m_2=e(Y)$. Thus,
$m=m_1+m_2$,  as $e(X)=0$. Let
$(X_1,X_2)$ be a partition of $X$ such that  $\theta(X_1,X_2)=\theta$.
Applying Lemma \ref{lem1} with $p=1/2$,
there is a bipartition $V_{1},V_{2}$ of $V(D)$ such that
$X_{i}\subseteq  V_{i}$ for $i\in [2]$, and
\begin{align*}
& \min\{e(V_{1},V_{2}),e(V_{2},V_{1})\}
\\ \geq &\frac{1}{2}\min\{e(X_{1},Y)+e(Y,X_{2}),e(X_{2},Y) +e(Y,X_{1})\}+\frac{e(Y)}{4}-\varepsilon m
\\ = &\frac{m_{1}-\theta}{4}+\frac{m_{2}}{4}-\varepsilon m
\\ =&\frac{m-\theta}{4}-\varepsilon m.
\end{align*}
If $\theta \le  m/(2d-1)$ then $(m-\theta)/4\ge (d-1)m/\big(2(2d-1)\big)$; so $\min \{e(V_1,V_2), e(V_2,V_1)\}\ge \left( \frac{d-1}{2(2d-1)}-\varepsilon\right) m$, a contradiction. Thus, $\theta >  m/(2d-1)$, and (1) holds.

\medskip

Let $X=\{v_1, \ldots, v_{|X|}\}$ such that $s(v_1)\ge s(v_2)\ldots \ge s(v_{|X|})$.
To prove (2), let us assume $|X'|=2k$ for some non-negative integer $k$.  Then $X'=\{v_1, \ldots, v_{2k}\}$.

First, suppose $k=0$. Then $s(v_1)<\theta$.  Let $X_1^*,X_2^*$ be the partition of  $X$ such that $v_1,v_3,\cdots, v_{2p-1}$ are $(X_1^*,X_2^*)$-forward, where $p=\lceil |X|/2 \rceil$,  and all other vertices are $(X_1^*,X_2^*)$-backward.
If $|X|$ is even then, by \eqref{theta2},
\[
\left|\theta(X_1^*,X_2^*) \right|
                               = \left|s(v_1) -\sum_{i=1}^{p-1} \big(s(v_{2i})-s(v_{2i+1})\big) -s(v_{|X|})\right|
                                \le  s(v_1)
                                <\theta,
\]
a contradiction. If $|X|$ is odd then, by \eqref{theta2},
\[
   \left|\theta(X_1^*,X_2^*)  \right| = \left|s(v_1)-\sum_{i=1}^{p-1}\big(s(v_{2i}) -s(v_{2i+1})\big)\right| \le s(v_1)< \theta,
\]
a contradiction.

Now suppose $k>0$.  Then $s(v_{2k})\ge \theta$. Let $X_1^*,X_2^*$ be the partition of  $X$ such that
$v_1,v_3,\cdots, v_{2k-1}$ are $(X_1^*,X_2^*)$-forward, and all other vertices in $X$ are $(X_1^*,X_2^*)$-backward.
Then  by \eqref{theta2}, $\left|\theta(X_1^*,X_2^*)\right|=\left|\sum_{i=1}^ks(v_{2i-1}) -\big(\sum_{i=1}^ks(v_{2i})+g\big)\right|$.
Note that
\begin{align*}
     \sum_{i=1}^ks(v_{2i-1}) -\left(\sum_{i=1}^ks(v_{2i})+g\right)
& = s(v_1)-\sum_{i=1}^{k-1}\Big(s(v_{2i})-s(v_{2i+1})\Big) -s(v_{2k})-g\\
& \le s(v_1)-s(v_{2k})\\
& \le  |Y|-\theta,
\end{align*}
and, since $g\le |Y|-\theta$ (by Lemma~\ref{lem5}),
\begin{align*}
 \left(\sum_{i=1}^ks(v_{2i})+g\right)-  \sum_{i=1}^ks(v_{2i-1})
 & =  \sum_{i=1}^{k}\Big(s(v_{2i})-s(v_{2i-1})\Big)+g
  \le g  \le |Y|-\theta.
\end{align*}
Hence, $\left|\theta(X_1^*,X_2^*)\right|\le |Y|-\theta$.
Because $\theta >m/(2d-1)$ (by (1)) and $m\ge d|V(D)|$, we see that $\theta > |V(D)|/2\ge |Y|/2$. Thus, $\left|\theta(X_1^*,X_2^*)\right|\le |Y|-\theta<\theta$,
a contradiction. Thus, $|X'|$ must be odd, and we have (2).

\medskip

By (2), let $X':=\{v_1, \ldots, v_{2k+1}\}$ for some $k\ge 0$.
Recall that $d(x)-s(x)$ is an even integer for all $x\in X$, and we write $2b=\sum_{x\in X}\big(d(x)-s(x)\big)$.
To prove (3), we consider the partition $X_1^1,X_2^1$ of $X$ such that $\{v_{1},v_{3},\cdots,v_{2k-1}\}\cup (X\setminus X')$ is the set of $(X^1_1,X^1_2)$-forward
vertices, and $\{v_{2},v_{4},\cdots,v_{2k},v_{2k+1}\}$ is the set of $(X^1_1,X^2_2)$-backward vertices.
Then  $m_f(X^1_1,X^1_2)=\sum_{j=1}^{k}s(v_{2j-1})+g+b$ and $m_b(X^1_1,X^1_2)=\sum_{j=1}^{k}s(v_{2j})+s(v_{2k+1})+b$.
Note that
\begin{align*}
 \theta(X_1^1,X_2^1) =& m_f(X^1_1,X^1_2)-m_b(X_1^1,X_2^1)\\
 =&\sum_{j=1}^{k}s(v_{2j-1})+g-\sum_{j=1}^{k}s(v_{2j})-s(v_{2k+1})\\
   =  & \big(s(v_1)-s(v_{2k})-s(v_{2k+1})\big)+\sum_{j=1}^{k-1}\big(s(v_{2j+1})-s(v_{2j})\big)+g\\
   \le & s(v_1)-s(v_{2k})-s(v_{2k+1})+|Y|-\theta  \quad (\mbox{since $g\le |Y|-\theta$ by Lemma~\ref{lem5}})\\
   \le & s(v_1)+|Y|-3\theta\\
   \le  & 2|V(D)|-3\theta \\
  < & \theta \quad (\mbox{as $\theta >|V(D)|/2$ because $m\ge d|V(D)|$ and by (1)}).
\end{align*}
Thus, since $|m_f(X^1_1,X^1_2)-m_b(X^1_1,X^1_2)|=\left|\theta(X^1_1,X^1_2)\right|\ge \theta$, we have
$$m_b(X^1_1,X^1_2)-m_f(X^1_1,X^1_2)=\sum_{j=1}^{k}s(v_{2j})+s(v_{2k+1})-\sum_{j=1}^{k}s(v_{2j-1})-g\geq \theta.$$

Now consider the partition $X_1^2,X_2^2$ of $X$ such that $X_1^2=\left(X^1_1\setminus \{v_{2k-1}\}\right)\cup \{v_{2}\}$ and $X_2^2=\left(X_2^1\setminus \{v_2\}\right)\cup \{v_{2k-1}\}$. Then $m_f(X_1^2,X_2^2)=m_f(X_1^1,X_2^1)-s(v_{2k-1})+s(v_2)$ and $m_b(X_1^2,X_2^2)=m_b(X_1^1,X_2^1)-s(v_2)+s(v_{2k-1})$.
Hence, $$m_b(X_1^2,X_2^2)-m_f(X_1^2,X_2^2)=\Big(m_b(X_1^1,X_2^1)-m_f(X_1^1,X_2^1)\Big)-2\Big(s(v_{2})-s(v_{2k-1})\Big),$$
which implies that
$$m_b(X_1^2,X_2^2)-m_f(X_1^2,X_2^2)\ge \theta-2(|Y|-\theta)>\theta-2\theta=-\theta.$$

Therefore, since $\left|m_b(X_1^2,X_2^2)-m_f(X_1^2,X_2^2)\right|=|\theta(X_1^2,X_2^2)|\ge \theta$, we see that
$$m_b(X_1^2,X_2^2)-m_f(X_1^2,X_2^2)\ge \theta.$$

Repeating the same argument by exchanging the sides for $v_{2(k-i)+1}$ and $v_{2i}$, one step at a time in the order $i=2, \ldots, \lfloor k/2 \rfloor$, we arrive at the partition
$X_1^{\lfloor k/2 \rfloor},X_2^{\lfloor k/2 \rfloor}$ of $X$, such that $\{v_1, v_2, \ldots, v_k\}\cup (X\setminus X')$ is the set of $(X_1^{\lfloor k/2 \rfloor},X_2^{\lfloor k/2 \rfloor})$-forward vertices, $\{v_{k+1}, v_{k+2}, \ldots, v_{2k+1}\}$
is the set of $(X_1^{\lfloor k/2 \rfloor},X_2^{\lfloor k/2 \rfloor})$-backward vertices, and
 $$m_b(X_1^{\lfloor k/2 \rfloor},X_2^{\lfloor k/2 \rfloor})-m_f(X_1^{\lfloor k/2 \rfloor},X_2^{\lfloor k/2 \rfloor})\ge \theta.$$
On the other hand, we have, by \eqref{theta2}, that
\[
 m_b(X_1^{\lfloor k/2 \rfloor},X_2^{\lfloor k/2 \rfloor})-m_f(X_1^{\lfloor k/2 \rfloor},X_2^{\lfloor k/2 \rfloor}) =\sum_{j=k+1}^{2k+1}s(v_j)-\sum_{j=1}^{k}s(v_j)-g.
\]
Hence, (3) holds.
\end{proof}

\begin{lem}\label{property2}

Let $D$ be a digraph with $m$ arcs and minimum outdegree $d\ge 4$, and
let $X,Y$ be a partition of $V(D)$ with $e(X)=0$. Let
$\theta=\min\{|\theta(X_1,X_2)|: X_1, X_2 \mbox{ is a partition
  of } X\}$,  $X'=\{x\in X: s(x)\ge \theta\}$,
$g=\sum_{x\in X\setminus X'}s(x)$, and  $2b=\sum_{x\in
  X}\big(d(x)-s(x)\big)$.
Let $\varepsilon>0$ and assume that $\max_{y\in Y}d(y)\le \varepsilon^2m/4$.
Then there exists a partition $V_1,V_2$ of $V(D)$ such that
$\min\{e(V_1,V_2),e(V_2,V_1)\}\ge
\left( \frac{d-1}{2(2d-1)}-\varepsilon\right) m$;  or $|X'|$ is odd
and if we let $X'=\{v_1, \ldots, v_{2k+1}\}$ such that
$s(v_1)\ge s(v_2)\ge \ldots \ge s(v_{2k+1})$ and write $\Delta_j=s(v_j)$ for $j\in [2k+1]$  then
   \begin{itemize}
       \item [$(1)$]  $d\left(\sum_{j=1}^{k}\Delta_{j}+g\right)-(d-1)\sum_{j=k+1}^{2k+1}\Delta_{j} +b+e(Y)/2<0$,
       \item [$(2)$]  $b>  \frac{d^2+2d-1}{d-1}\sum_{j=k}^{2k-1}\Delta_{j}-d\left(\sum_{j=1}^{k-1}\Delta_j+
     \Delta_{2k}+\Delta_{2k+1}\right)+(d-1)g +\frac{d-1}{2d}e(Y)$,
       \item [$(3)$]  $b< \frac{2(2d-1)(k+1)}{3d-1}|V(D)|+\frac{d^2-5d+2}{3d-1}\left(\sum_{j=1}^{k-1}\Delta_j+\Delta_{2k}+\Delta_{2k+1}\right)
   -\frac{d^2+2d-1}{3d-1}\sum_{j=k}^{2k-1}\Delta_{j} +\frac{d(d-1)}{3d-1}g
  -\frac{(d-1)^2}{2d(3d-1)} e(Y)$, and
      \item [$(4)$] when $d=4$ and $k=1$, $2\Delta_2+2\Delta_3-3\Delta_1-3g-b+3e(Y)/14<0$, or both
        $6\Delta_1-3\Delta_2-3\Delta_3+2g-b+3e(Y)/14<0$ and  $6\Delta_3-3\Delta_1-3\Delta_2-3g+b/3+3e(Y)/14<0$.

   \end{itemize}

\end{lem}

\begin{proof}
For convenience, we introduce two functions which we will use to compare $\min \{e(V_1,V_2)$, $e(V_2, V_1)\}$ with
$\left( \frac{d-1}{2(2d-1)}-\varepsilon\right) m$ for any partition
$V_1,V_2$ of $V(D)$.  For any partition
$X_1,X_2$ of $X$, let $z(X_1,X_2)=e(X_1,Y)$ and
$z'(X_1,X_2)=e(Y,X_2)$; so
$m_f(X_1,X_2)=z(X_1,X_2)+z'(X_1,X_2)$.
Let $m_1:=e(X,Y)+e(Y,X)$,
$m_2:=e(Y)$, and  $$\ell(p,X_1,X_2):= (d-1)\sum_{j=1}^{2k+1}\Delta_j+(d-1)g + (2d-2)b-\Big(2(2d-1)p(1-p)-(d-1)\Big)m_2.$$
Define
\begin{itemize}
   \item [] $f(p,X_1,X_2)= 2(1-p)(2d-1)z(X_1,X_2)+2p(2d-1)z'(X_1,X_2) -\ell(p,X_1,X_2)$, and
   \item [] $h(p,X_1,X_2)= 2p(2d-1)\Big(m_1-z(X_1,X_2)-z'(X_1,X_2)\Big)-\ell(p,X_1,X_2)$.
\end{itemize}

By Lemma \ref{lem1}, for any $0\leq p\leq 1$, there is a partition $V(D)=V_1\cup V_2$ such that
$X_i\subseteq V_i$ for $i\in [2]$, and
\begin{equation}\label{eq:1}
\left\{
\begin{aligned}
e(V_{1},V_{2})  \ge (1-p)\cdot e(X_1,Y)+p\cdot e(Y,X_2)+p(1-p)\cdot e(Y)-\varepsilon m, \\
e(V_{2},V_{1})  \ge p\cdot e(X_{2}, Y)+(1-p)\cdot e(Y,X_1)+p(1-p)\cdot e(Y)-\varepsilon m.
\end{aligned}
\right.
\end{equation}
Without loss of generality, we may assume $p\leq 1-p$; so $p\le
1/2$. Then, from \eqref{eq:1}, we have
\begin{equation}\label{eq6}
\left\{
\begin{aligned}
 e(V_{1},V_{2}) & \ge (1-p)z(X_1,X_2)+p z'(X_1,X_2)+p(1-p) m_2-\varepsilon m, \\
 e(V_{2},V_{1}) & \ge  p\big(m_1-z(X_1,X_2)-z'(X_1,X_2)\big)+p(1-p)m_2-\varepsilon m.
\end{aligned}
\right.
\end{equation}

Note that
\begin{equation}\label{eq5}
m=m_1+m_2=\sum_{j=1}^{2k+1}\Delta_{j}+g+2b+m_2.
\end{equation}
By \eqref{eq6} and \eqref{eq5}, we have
\begin{align*}
& e(V_{1},V_{2})-\left(\frac{d-1}{2(2d-1)}m-\varepsilon m\right)\\
\geq & (1-p)z(X_1,X_2)+p z'(X_1,X_2)+p(1-p) m_2-\frac{d-1}{2(2d-1)}\left(\sum_{j=1}^{2k+1}\Delta_{j}+g+2b+m_2\right)\\
= & \frac{1}{2(2d-1)} f(p,X_1,X_2),
\end{align*}
and
\begin{align*}
& e(V_{2},V_{1})-\left(\frac{d-1}{2(2d-1)}m-\varepsilon m\right)\\
\geq & p\Big(m_1-z(X_1,X_2)-z'(X_1,X_2)\Big)+p(1-p)m_2-\frac{d-1}{2(2d-1)}\left(\sum_{j=1}^{2k+1}\Delta_{j}+g+2b+m_2\right)\\
= & \frac{1}{2(2d-1)} h(p,X_1,X_2).
\end{align*}

If $f(p,X_1,X_2)\ge 0$ and $h(p,X_1,X_2)\ge 0$ for some choice of $p,X_1,X_2$, we see that the there is a partition $V_1, V_2$ of $V(D)$ such that
for $i\in [2]$, $X_i\subseteq V_i$ and
$e(V_i,V_{3-i})\ge \left( \frac{d-1}{2(2d-1)}-\varepsilon\right) m$.
Hence, we may assume that
\begin{equation}\label{eq:a}
f(p,X_1,X_2)<0 \mbox{ or } h(p,X_1,X_2)<0 \mbox{  for any choice of } p,X_1,X_2.
\end{equation}

To see (1), we consider the partition $X_1^1,X_2^1$ of $X$ such that $\{v_{1},v_{2},\ldots,v_{k}\}\cup (X\setminus X')$ is the set of $(X_1^1,X_2^1)$-forward vertices,
 and  $\{v_{k+1},v_{k+2},\ldots,v_{2k+1}\}$ is the set of $(X_1^1,X_2^1)$-backward vertices.  Then $z(X_1^1,X_2^1)+z'(X_1^1,X_2^1)=m_f(X_1^1,X_2^1)=\sum_{j=1}^{k}\Delta_{j}+g+b$ and
$m_b(X_1^1,X_2^1)=\sum_{j=k+1}^{2k+1}\Delta_{j}+b$.  Setting $p=1/2$, it follows from a simple calculation that
\begin{equation*}\label{eq42}
\left\{
\begin{aligned}
& f(1/2,X_1^1,X_2^1)=d\left(\sum_{j=1}^{k}\Delta_{j}+g\right)-(d-1)\sum_{j=k+1}^{2k+1}\Delta_{j}
+b+m_2/2,\\
& h(1/2,X_1^1,X_2^1)=d\sum_{j=k+1}^{2k+1}\Delta_{j}-(d-1)\left(\sum_{j=1}^{k}\Delta_{j}+g\right)+b+m_2/2.
\end{aligned}
\right.
\end{equation*}
By (3) of Lemma~\ref{property1}, we may assume $ h(1/2,X_1^1,X_2^1)>0$; so by \eqref{eq:a},  we have $f(1/2,X_1^1,X_2^1)<0$.
Thus, (1) holds.

\medskip

For  (2) and (3), we note that at least $k$ members of  $\{s^+(v_1),s^+(v_2),\cdots,s^+(v_{2k-1})\}$
have the same sign; so we may assume that
$s^+(v_{j_1})$, $s^+(v_{j_2})$, $\cdots$, $s^+(v_{j_k})$ are positive,
where $1\leq j_1<j_2<\cdots<j_{k}\leq 2k-1$.
\medskip

To prove (2), let $X_1^2,X_2^2$ be the partition of $X$ such that $\{v_{j_1},v_{j_2},\cdots,v_{j_k}\}\cup (X\setminus X')$ is the set of $(X_1^2,X_2^2)$-forward
vertices, and all other vertices in $X$ are $(X_1^2, X_2^2)$-backward.
Then  $$m_f(X_1^2,X_2^2)=\sum_{i=1}^{k}\Delta_{j_i}+g+b$$
and
$$m_b(X_1^2,X_2^2)=\sum_{j=1}^{2k+1}\Delta_j-\sum_{i=1}^{k}\Delta_{j_i}+b\geq \sum_{j=k+1}^{2k+1}\Delta_j+b.$$
Note that $z(X_1^2,X_2^2)=e(X_1^2,Y)\geq \sum_{i=1}^{k}\Delta_{j_i}$
and $z'(X_1^2,X_2^2)=e(Y,X_2^2)=m_f(X_1^2,X_2^2)-e(X_1^2,Y)$. Setting  $p=(d-1)/(2d)$, we see that
\begin{align*}
& f\left(\frac{d-1}{2d},X_1^2,X_2^2\right) \\
\ge & \frac{(d+1)(2d-1)}{d}\sum_{i=1}^{k}\Delta_{j_i}  +\frac{(d-1)(2d-1)}{d}(g+b)- (d-1)\sum_{j=1}^{2k+1}\Delta_j\\
       & -(d-1)g-(2d-2)b+\frac{(d-1)^2}{2d^2}m_2\\
 = & \frac{d-1}{d} \left( \frac{d^2+2d-1}{d-1}\sum_{i=1}^{k}\Delta_{j_i}-d\Big(\sum_{j=1}^{2k+1}\Delta_{j}-\sum_{i=1}^{k}\Delta_{j_i}\Big) +(d-1)g-b+\frac{d-1}{2d}m_2 \right) \\
 \geq & \frac{d-1}{d} \left(\frac{d^2+2d-1}{d-1}\sum_{j=k}^{2k-1}\Delta_{j}-d\Big(\sum_{j=1}^{k-1}\Delta_{j}
 +\Delta_{2k}+\Delta_{2k+1}\Big) +(d-1)g-b+\frac{d-1}{2d}m_2 \right),
\end{align*}
and
\begin{align*}
& h\left(\frac{d-1}{2d},X_1^2,X_2^2\right)\\
\geq & \frac{(d-1)(2d-1)}{d}\left(\sum_{j=k+1}^{2k+1}\Delta_j+b\right)-(d-1)\sum_{j=1}^{2k+1}\Delta_j-(d-1)g
  -(2d-2)b+\frac{(d-1)^2}{2d^2}m_2\\
 = & \frac{d-1}{d}\left((d-1)\sum_{j=k+1}^{2k+1}\Delta_{j}-d\Big(\sum_{j=1}^{k}\Delta_{j}+g\Big)-b+\frac{d-1}{2d}m_2\right).
\end{align*}
By (1),  $h\left(\frac{d-1}{2d},X_1^2,X_2^2\right)>0$. So  $f\left(\frac{d-1}{2d},X_1^2,X_2^2\right)<0$ by \eqref{eq:a}. Hence, (2) holds.

\medskip

To prove (3), consider the partition $X_1^3,X_2^3$ of $X$ such that $\{v_{j_1},v_{j_2},\cdots,v_{j_k}\}\cup (X\setminus X')\subseteq X_1^3$,
and the  vertices in $X'\setminus \{v_{j_1}, \ldots, v_{j_k}\}$ are $(X_1^3, X_2^3)$-backward.
Since $s^+(v_{j_i})>0$ for $i\in [k]$, the vertices  $v_{j_1},v_{j_2},\cdots,v_{j_k}$  are $(X_1^3, X_2^3)$-forward.
Hence,   $$m_f(X_1^3,X_2^3)\ge \sum_{i=1}^{k}\Delta_{j_i}+b$$ and
$$m_b(X_1^3,X_2^3)\ge \sum_{j=1}^{2k+1}\Delta_j-\sum_{i=1}^{k}\Delta_{j_i}+b\geq \sum_{j=k+1}^{2k+1}\Delta_j+b.$$
Let $n:=|V(D)|$. Note that $X_2^3\subseteq X'\setminus \{v_{j_1},
\ldots, v_{j_k}\}$; so
$$z'(X_1^3,X_2^3)=e(Y,X_2^3)\le (k+1)n- \Big(\sum_{j=1}^{2k+1}\Delta_{j}-\sum_{i=1}^{k}\Delta_{j_i}\Big).$$
Hence,
$$z(X_1^3,X_2^3)=e(X_1^3,Y)\geq \sum_{i=1}^{k}\Delta_{j_i}+b-\left((k+1)n-
\Big(\sum_{j=1}^{2k+1}\Delta_{j}-\sum_{i=1}^{k}\Delta_{j_i}\Big)\right).$$
Setting  $p=(d-1)/(2d)$, we have
\begin{align*}
 & f\left(\frac{d-1}{2d},X_1^3,X_2^3\right) \\
 \ge & \frac{(d+1)(2d-1)}{d}\left[\sum_{i=1}^{k}\Delta_{j_i}+b-\Big((k+1)n-
 \big(\sum_{j=1}^{2k+1}\Delta_{j}-\sum_{i=1}^{k}\Delta_{j_i}\big)\Big)\right]\\
 & +\frac{(d-1)(2d-1)}{d}\left((k+1)n-
 \Big(\sum_{j=1}^{2k+1}\Delta_{j}-\sum_{i=1}^{k}\Delta_{j_i}\Big)\right)-(d-1)\sum_{j=1}^{2k+1}\Delta_j\\
&  -(d-1)g-(2d-2)b+\frac{(d-1)^2}{2d^2}m_2\\
 = & \frac{3d-1}{d}\Big(\frac{d^2+2d-1}{3d-1}\sum_{i=1}^{k}\Delta_{j_i}-\frac{d^2-5d+2}{3d-1}\big(\sum_{j=1}^{2k+1}\Delta_{j}-
\sum_{i=1}^{k}\Delta_{j_i}\big)-\frac{2(2d-1)(k+1)}{3d-1}n \\
 & -\frac{d(d-1)}{3d-1}g+b+\frac{(d-1)^2}{2d(3d-1)}m_2\Big) \\
 \geq &\frac{3d-1}{d}\Big( \frac{d^2+2d-1}{3d-1}\sum_{j=k}^{2k-1}\Delta_{j}-\frac{d^2-5d+2}{3d-1}\big(\sum_{j=1}^{k-1}\Delta_{j}
 +\Delta_{2k}+\Delta_{2k+1}\big)-\frac{2(2d-1)(k+1)}{3d-1}n\\
 & -\frac{d(d-1)}{3d-1}g+b+\frac{(d-1)^2}{2d(3d-1)}m_2\Big),
\end{align*}
and
\begin{align*}
& h\left(\frac{d-1}{2d},X_1^3,X_2^3\right)\\
\geq & \frac{(d-1)(2d-1)}{d}\left(\sum_{j=k+1}^{2k+1}\Delta_j+b\right)-(d-1)\sum_{j=1}^{2k+1}\Delta_j-(d-1)g-(2d-2)b+\frac{(d-1)^2}{2d^2}m_2\\
 = & \frac{d-1}{d}\left((d-1)\sum_{j=k+1}^{2k+1}\Delta_{j}-d\Big(\sum_{j=1}^{k}\Delta_{j}+g\Big)-b+\frac{d-1}{2d}m_2\right).
\end{align*}
By (1), $h\left(\frac{d-1}{2d},X_1^3,X_2^3\right)>0$. So
$f(\frac{d-1}{2d},X_1^3,X_2^3)<0$ by \eqref{eq:a}. Hence, (3) holds.

\medskip

Now we prove (4); so assume $d=4$ and $k=1$.
First, let $X_1^4,X_2^4$ be the partition of $X$ such that $\{v_1\}\cup (X\setminus X')$ is the set of $(X_1^4,X_2^4)$-forward
vertices, and $v_2, v_3$ are $(X_1^4, X_2^4)$-backward.
Then  $m_f(X_1^4,X_2^4)=\Delta_1+g+b$ and
$m_b(X_1^4,X_2^4)=\Delta_2+\Delta_3+b.$
Also, we have $e(X_1^4,Y)\geq \Delta_1$. Setting  $p=5/14$, we see that
\begin{align*}
  h\left(5/14,X_1^4,X_2^4\right)
 \ge & 5(\Delta_2+\Delta_3+b)-3(\Delta_1+\Delta_2+\Delta_3)-3g-6b+3m_2/14\\
  = & 2\Delta_2+2\Delta_3-3\Delta_1-3g-b+3m_2/14\\
\end{align*}
and
\begin{align*}
  f\left(5/14,X_1^4,X_2^4\right)
  \ge & 9\Delta_1+5(g+b)-3(\Delta_1+\Delta_2+\Delta_3)-3g-6b+3m_2/14\\
   = & 6\Delta_1-3\Delta_2-3\Delta_3+2g-b+3m_2/14.
 \end{align*}
Thus, we have  $2\Delta_2+2\Delta_3-3\Delta_1-3g-b+3m_2/14<0$ or
$6\Delta_1-3\Delta_2-3\Delta_3+2g-b+3m_2/14<0$.

Next, we choose some $i\in [3]$ such that the number of arcs from $v_i$ to $Y$
counted in $b$ is maximum. Consider the partition $X_1^5,X_2^5$ of $X$
such that $\{v_i\}\cup (X\setminus X')\subseteq X_1^5$, and the
vertices in $X'\setminus \{v_i\}$ are $(X_1^5, X_2^5)$-backward.
Then, clearly, $m_f(X_1^5,X_2^5)\geq \Delta_i+b\geq \Delta_3+b$ and
$m_b(X_1^5,X_2^5)\geq \Delta_2+\Delta_3+b.$
Also, we have $e(X_1^5,Y)\geq (\Delta_i+b)-2b/3\geq \Delta_3+b/3$. Setting  $p=5/14$, we see that
\begin{align*}
  h\left(5/14,X_1^5,X_2^5\right)
&\ge 5(\Delta_2+\Delta_3+b)-3(\Delta_1+\Delta_2+\Delta_3)-3g-6b+3m_2/14\\

&  \ge 2\Delta_2+2\Delta_3-3\Delta_1-3g-b+3m_2/14,
\end{align*}
and
\begin{align*}
  f\left(5/14,X_1^5,X_2^5\right)
  & \ge 5(\Delta_3+b)+4(\Delta_3+b/3)-3(\Delta_1+\Delta_2+\Delta_3)-3g-6b+3m_2/14\\
 & =  6\Delta_3-3\Delta_1-3\Delta_2-3g+b/3+3m_2/14.

\end{align*}
Thus,  $2\Delta_2+2\Delta_3-3\Delta_1-3g-b+3m_2/14<0$ or
$6\Delta_3-3\Delta_1-3\Delta_2-3g+b/3+3m_2/14<0$.   This
completes the proof of (4).
\end{proof}

\section{Huge vertices}

In this section, we show that if $V(D)$ has a partition $X,Y$ such that $e(X)=0$, $\max_{y\in Y}d(y)\le \varepsilon^2m/4$, and $X$ has at least $d$ huge vertices or a unique huge vertex then
Conjecture~\ref{conj1} holds.

\begin{prop} \label{00001}

Let $d\geq 4$ be an integer and $\varepsilon>0$ be a real.
Let $D$ be a digraph with $m$ arcs and minimum outdegree at least $d$. Let $X,Y$ be a partition of $V(D)$
with $e(X)=0$ and $\max_{y\in Y}d(y)\le \varepsilon^2m/4$. Let
$\theta=\min\{|\theta(X_1,X_2)|: X_1,X_2  \mbox{ is a partition
  of } X\}$ and  $X'=\{x\in X: s(x)\ge \theta\}$.
  Suppose $|X'|\ge d$. Then $V(D)$ admits a partition $V_1,V_2$ such that
$\min\{e(V_1,V_2),e(V_2, V_1)\}\geq \left(\frac{d-1}{2(2d-1)}-\varepsilon\right)m$.
\end{prop}

\begin{proof}  Suppose the desired partition $V_1, V_2$ does not exist. By (2) of Lemma~\ref{property1},
let $X'=\{v_1, \cdots, v_{2k+1}\}$. Then  $2k+1\ge d\ge 4$ by assumption. Let
$\Delta_i=s(v_i)$ for $i\in [2k+1]$ and assume, without loss of
generality,   $\Delta_1\ge \Delta _2\ge \ldots \ge \Delta_{2k+1}$.
Let $X_1,X_2$ be a partition of $X$ such that
$\theta(X_1,X_2)=\theta$. Then, by  (1) and (2) of Lemma~\ref{property2},
we have
\begin{align*}
0> & d\left(\sum_{j=1}^{k}\Delta_{j}+g\right)-(d-1)\sum_{j=k+1}^{2k+1}\Delta_{j}
+b+m_2/2\\
  > & d\left(\sum_{j=1}^{k-1}\Delta_{j}+\Delta_{2k}\right)-(d-1)\left(\sum_{j=k}^{2k-1}\Delta_{j}+\Delta_{2k+1}\right)
 +\frac{d^2+2d-1}{d-1}\sum_{j=k}^{2k-1}\Delta_{j}\\
& -d\left(\sum_{j=1}^{k-1}\Delta_j+\Delta_{2k}+\Delta_{2k+1}\right)\\
 = & \frac{4d-2}{d-1}\sum_{j=k}^{2k-1}\Delta_{j}-(2d-1)\Delta_{2k+1}\\
\geq & \frac{k(4d-2)}{d-1}\Delta_{2k+1}-(2d-1)\Delta_{2k+1} \\
= & \frac{(2d-1)(2k+1-d)}{d-1}\Delta_{2k+1},
\end{align*}
This  is a contradiction, as $2k+1\geq d$.
\end{proof}

\medskip

{\it Remark}.  The requirement $e(X)=0$ in Proposition \label{1} can be replaced by $e(X)=o(m)$.

\medskip

Next,  we show that if $V(D)$ admits a partition $X,Y$ such that $|X|=o(|V(D)|)$, $\max_{y\in Y}d(y)\le \varepsilon^2m/4$,  and $D$ has a unique huge vertex in $X$ then
the conclusion of Conjecture~\ref{conj1} holds. For this, we need
another concept  introduced by Lee, Loh, and Sudakov \cite{Lee2013},
and we use the result of  Lu, Wang, and Yu in \cite{lu2015} to give its definition. We say that a connected graph is {\it tight}
if all its blocks are odd cliques.
If a disconnected graph $G$ is the underlying graph of a
digraph $D$, the {\it tight components} of $D$ are the
components of $G$ that are tight. (The $underlying \ graph$ of $D$ is obtained from $D$ by ignoring arc orientations and
removing redundant parallel edges.)
 For a tight component $T$ of $D$, we say $T$ is $essential$ if
 $D[V(T)]$, the subgraph of $D$ induced by $V(T)$, does not contain any parallel arcs in opposite directions. Recently, Hou, Li, and Wu \cite{Hou2020} proved the following.

\begin{lem}[Hou, Li, and Wu \cite{Hou2020}]\label{lem4}
For any positive constants $C$ and $\varepsilon$, there exist $\gamma$, $n_{0}>0$ for which the following holds.
Let $D$ be a digraph with $n\geq n_{0}$ vertices and at most $Cn$ arcs. Suppose $X\subseteq V(D)$ is a set
of at most $\gamma n$ vertices and $X_1,X_2$ is a partition of $X$.
Let $Y=V(D)\setminus X$ and let $\tau$ be the number of essential tight components in $D[Y]$.
If every vertex in $Y$ has degree at most
$\gamma n$ in $D$, then there is a bipartition
$V(D)=V_{1}\cup V_{2}$ with $X_{i}\subseteq V_{i}$ for $i=1,2$ such that
$$e(V_{1},V_{2})\geq e(X_{1},X_{2})+\frac{e(X_{1},Y)+e(Y,X_{2})}{2}+\frac{e(Y)}{4}+\frac{n-\tau}{8}-\varepsilon n,$$
$$e(V_{2},V_{1})\geq e(X_{2},X_{1})+\frac{e(X_{2},Y)+e(Y,X_{1})}{2}+\frac{e(Y)}{4}+\frac{n-\tau}{8}-\varepsilon n.$$
\end{lem}

\begin{prop}\label{huge-1}
Let $d\ge 4$ be an integer and let $C, \varepsilon$ be positive reals. Let $D$ be a digraph with $n$ vertices, $m\le Cn$ arcs, and minimum outdegree at least $d$. Then there exists $\gamma$ with $0<\gamma<\varepsilon$ such that the following holds:
Let $X,Y$ be a partition of $V(D)$ with $|X|\le \gamma n$, $e(X)=0$,
and $\max_{y\in Y}d(y)\le \gamma n$. Let
$\theta=\min\{|\theta(X_1,X_2)|: X_1,X_2 \mbox{ is a partition
  of } X\}$,  $X'=\{x\in X: s(x)\ge \theta\}$, $g=\sum_{x\in
  X\setminus X'}s(x)$, and  $2b=\sum_{x\in X}\big(d(x)-s(x)\big)$.
Then the following statements hold.

\begin{itemize}
\item [$(1)$]  $\tau\le (n+2g+2b)/(2d-2|X'|+1)$,  where $\tau$ is the number of essential tight components
in $D[Y]$.

\item [$(2)$] Suppose $|X'|=1$. Then $V(D)$ admits a partition $V_1,V_2$ such that
$\min\{e(V_1,V_2),e(V_2, V_1)\}\geq \left(\frac{d-1}{2(2d-1)}-\varepsilon\right)m.$
\end{itemize}
\end{prop}

\begin{proof}

First, we prove (1). Let $\alpha:=|X'|$.  For $i=1, 3, \ldots, 2d-2\alpha-1$, let $\tau_i$ be the number of essential tight components of order $i$;  and let $\tau'$ be the
number of essential tight components of order at least $2d-2\alpha+1$. Then
\begin{equation}\label{eq:i}
\tau_1+3\tau_3+\cdots+(2d-2\alpha-1)\tau_{2d-2\alpha-1}+(2d-2\alpha+1)\tau'\leq n.
\end{equation}

For each essential tight
component $D_i$ of order $i$, we see that $e(D_i)\le i(i-1)/2$ and $e(D_i,X')\le \alpha i$. Thus, since the outdegree of $D$ is
at least $d$, we see that $e(D_i,X\setminus X')\ge d i -\alpha i -i(i-1)/2$. Viewing $ d i -\alpha i -i(i-1)/2$ as a function of $i$
over the interval $[1, 2d-2\alpha]$, we see
that it achieves its minimum at $i=1$ (as well as at $i=2d-2\alpha$). Hence, $e(D_i,X\setminus X')\ge d-\alpha$ for $i\in [2d-2\alpha]$.
Thus, $e(Y, X\setminus X')\ge \sum_{i=1}^{d-\alpha}(d-\alpha)\tau_{2i-1}$. On the other hand, we have
$e(Y, X\setminus X')\le g+b$. Hence,
\begin{equation}\label{eq:j}
\sum_{i=1}^{d-\alpha} (d-\alpha) \tau_{2i-1} \leq g+b.
\end{equation}
Multiplying \eqref{eq:j} by 2 and adding the resulting inequality  to
\eqref{eq:i}, we derive that
$(2d-2\alpha+1)\tau \leq n+2g+2b$, completing the proof of (1).

\medskip
To prove (2), let $X'=\{v_0\}$ and
let $\Delta=s(v_0)$.  Let $X_1, X_2$ be the partition of $X$ such that $v_0$ is the only $(X_1,X_2)$-forward vertex. Then  $m_f(X_1,X_2)=\Delta+b$ and $m_b(X_1,X_2)=g+b$.

By Lemma \ref{lem4} (with $p=1/2$), there is a bipartition $V_1,V_2$ of $V(D)$ such that $X_i\subseteq V_i$ for $i\in [2]$ and
\begin{align*}
  &  \min \{e(V_1,V_2), e(V_2,V_2)\}\\
  \ge & \frac 12 \min\left\{e(X_1,Y)+e(Y,X_2), e(X_2,Y)+e(Y,X_1)\right\}+e(Y)/4+(n-\tau)/8-\varepsilon n\\
  = & (m-\theta)/4+(n-\tau)/8-\varepsilon n.
\end{align*}

Hence,
\begin{align*}
&\min \{e(V_1,V_2), e(V_2,V_2)\}
  -\left(\frac{d-1}{2(2d-1)}-\varepsilon \right)m\\
\ge & (m-\theta)/4+(n-\tau)/8-\varepsilon
      n-\left(\frac{d-1}{2(2d-1)}-\varepsilon \right)m\\
\ge & \frac{1}{4}\left( \frac{m}{2d-1}+\frac{n}{2}+4(d-1)\varepsilon n-\theta-\frac{\tau}{2}\right) \quad (\mbox{since $m\ge dn$}).
\end{align*}
Since the minimum outdegree of $D$ at least $d$ and
$|Y|=n-|X|>n-\varepsilon n$, we have $$m\geq b+d|Y|\geq b+dn-d\varepsilon n.$$
Therefore,
\begin{align*}

    &\frac{m}{2d-1}+\frac{n}{2}+4(d-1)\varepsilon n-\theta-\frac{\tau}{2} \notag \\
\geq & \frac{b+dn-d\varepsilon n}{2d-1}+\frac{n}{2}+4(d-1)\varepsilon n-\theta-\frac{n+2g+2b}{2(2d-1)} \quad (\mbox{by (1)})\\
\geq & \frac{1}{2(2d-1)}\Big( 2b+2dn+(2d-1)n-(4d-2)\theta-n-2g-2b \Big)\\
= & \frac{1}{2(2d-1)}\Big((4d-2)n-(4d-2)\theta-2g\Big)\\
\ge &0 \quad (\mbox{since $n\ge |Y|\ge \theta+g$ by Lemma \ref{lem5}}).
\end{align*}
So $V_1,V_2$ gives the desired partition for (2).
\end{proof}

We now use Lemma~\ref{lem4} to refine (1) of Lemma~\ref{property2} for the case
when there are only three huge vertices.

\begin{lem}\label{3huge-vertices}

Let $D$ be a digraph with $m$ arcs and minimum outdegree $d\ge 4$, and
let $X,Y$ be a partition of $V(D)$ with $e(X)=0$.  Let
$\theta=\min\{|\theta(X_1,X_2)|: X_1,X_2 \mbox{ is a partition of } X\}$,  $X'=\{x\in X: s(x)\ge \theta\}$,
$g=\sum_{x\in X\setminus X'}s(x)$, $2b=\sum_{x\in X}\big(d(x)-s(x)\big)$, and $\tau$ the number of essential tight components in $D[Y]$.
Let $\varepsilon>0$ and assume that $\max_{y\in Y}d(y)\le \varepsilon^2m/4$.
Suppose $|X'|=3$. Then there exists a partition $V_1,V_2$ of $V(D)$
such that $\min\{e(V_1,V_2),e(V_2,V_1)\}\ge (3/14-\varepsilon) m$;  or
if we write  $X'=\{v_1,v_2,v_3\}$ with $s(v_1)\ge
s(v_2)\ge s(v_3)$ and $\Delta_j=s(v_j)$ for $j\in [3]$  then
  $$b <3 \Delta_2+3\Delta_3-4\Delta_1-4g-m_2/2-7(|V(D)|-\tau)/4.$$
\end{lem}

\begin{proof}  We consider the partition $X_1,X_2$ of $X$ such
  that $\{v_{1}\}\cup (X\setminus X')$ is the set of $(X_1,X_2)$-forward vertices,
 and  $\{v_{2},v_{3}\}$ is the set of
 $(X_1,X_2)$-backward vertices.  Then $m_f(X_1,X_2)=\Delta_{1}+g+b$ and
$m_b(X_1,X_2)=\Delta_{2}+\Delta_3+b$. Note that
\begin{equation}\label{eq-1}
m=m_1+m_2=\sum_{j=1}^{3}\Delta_{j}+g+2b+m_2.
\end{equation}

Let $n:=|V(D)|$. By applying Lemma \ref{lem4} with $p=1/2$, there is a partition $V(D)=V_1\cup V_2$ such that
$X_i\subseteq V_i$ for $i\in [2]$, and
\begin{equation}\label{eq-2}
\left\{
\begin{aligned}
 e(V_{1},V_{2}) & \ge (\Delta_{1}+g+b)/2+ m_2/4 +(n-\tau)/8-\varepsilon n, \\
 e(V_{2},V_{1}) & \ge  (\Delta_{2}+\Delta_3+b)/2+m_2/4+(n-\tau)/8-\varepsilon n.
\end{aligned}
\right.
\end{equation}

Define
\begin{itemize}
   \item [] $f(1/2,X_1,X_2)= 4\Delta_1-3\Delta_2-3\Delta_3+4g+b+m_2/2+7(n-\tau)/4$, and
   \item [] $h(1/2,X_1,X_2)= 4\Delta_2+4\Delta_3-3\Delta_1-3g+b+m_2/2+7(n-\tau)/4$.
\end{itemize}

By (3) of Lemma~\ref{property1}, we may assume
$\Delta_2+\Delta_3-\Delta_1\ge g+\theta$; for otherwise the desired
partition of $V(D)$ exists. Hence, $ h(1/2,X_1,X_2)>0$.

By \eqref{eq-1} and \eqref{eq-2}, we have
\begin{eqnarray*}
& & e(V_{1},V_{2})-(3m/14-\varepsilon m)\\
& \geq &  (\Delta_{1}+g+b)/2+m_2/4+(n-\tau)/8-(3/14)\left(\sum_{j=1}^{3}\Delta_{j}+g+2b+m_2\right)\\
& = & (1/14) f(1/2,X_1,X_2),
\end{eqnarray*}
and
\begin{eqnarray*}
& & e(V_{2},V_{1})-(3m/14-\varepsilon m)\\
& \geq & (\Delta_{2}+\Delta_3+b)/2+m_2/4+(n-\tau)/8-(3/14)\left(\sum_{j=1}^{3}\Delta_{j}+g+2b+m_2\right)\\
& = & (1/14) h(1/2,X_1,X_2).
\end{eqnarray*}

Thus, if  $f(1/2,X_1,X_2)\ge 0$ then, since  $ h(1/2,X_1,X_2)>0$, the
desired partition of $V(D)$ exists. So, we may assume $f(1/2,X_1,X_2)<0$ which implies the desired inequality.
\end{proof}

\section{Proof of Theorem \ref{thm1}}

In this section, we prove Theorem~\ref{thm1}, by using Propositions
\ref{00001} and \ref{huge-1} and Lemma~\ref{3huge-vertices} and by choosing $X$ to consist of vertices
of degree at most $n^{3/4}$. Our proof is much simpler when applied to
the cases $d=2$ and $d=3$, and gives the results of Lee, Loh, and Sudakov \cite{Lee2014} that
Conjecture \ref{conj1} is true for these cases.

\medskip

\noindent {\it Proof of Theorem~\ref{thm1}}.  Let $D$ be a digraph with $n$ vertices and $m$ arcs, and assume that the minimum outdegree of $D$ is at least $4$.   We wish to find a partition $V(D)=V_1\cup V_2$, such
that $\min\{e(V_1,V_2),e(V_2,V_1)\}\geq \big(3/14+o(1)\big)m$.  We may assume  that $n$ is sufficiently large so that all
lemmas in the previous sections can be applied. We claim that

\begin{itemize}
\item [(1)] $m\ge 4n$ and we may assume $m< 128\cdot  7^2 n$.
\end{itemize}
Since $D$ has minimum outdegree at least $4$, we have $m\ge 4n$. Now suppose  $m\geq 128\cdot 7^2 n$.
Then applying Corollary \ref{lem2} with $\varepsilon=1/28$ we obtain a partition $V(D)=V_1\cup V_2$ such that
\[
min\{e(V_1,V_2),e(V_2,V_1)\}\geq \left(1/4-1/28\right)m= 3m/14.
\]
So we may assume $m< 128 \cdot 7^2 n$. $\Box$

\medskip

Consider the partition $X,Y$ of $V(D)$ such that  $X=\{v\in
V(D):d(v)\geq n^{3/4}\}$ and $Y=V(D)\setminus X$. Then, by (1),
\[
|X|\cdot n^{3/4}\leq \sum_{v\in X}d(v)\leq \sum_{v\in V(D)}d(v)=2m< 256\cdot 7^2 n.
\]
Hence,  $|X|=O(n^{1/4})$ and,  thus,  $e(X)\leq|X|^{2}=O(n^{1/2})=o(m)$. Therefore, we may assume
\begin{itemize}
\item [(2)] $e(X)=0$.
\end{itemize}

Let $\theta=\min\{|\theta(S,T)|: S,T \mbox{ is a partition  of } X\}$.
Let  $X_1,X_2$ be a partition of $X$ such that
$\theta(X_1,X_2)=\theta$, and let $X'=\{x\in X: s(x)\ge \theta\}$. If
$|X'|=1$ or $|X'|\ge 5$ then the desired partition exists by
Propositions~\ref{00001} and \ref{huge-1}. So

 let $X'=\{v_1,v_2,v_3\}$ and $\Delta_i=s(v_i)$ for $i\in [3]$ such
 that $\Delta_1\ge \Delta_2\ge \Delta_3$.
Since $|X|=o(n)$,
$$\Delta_1+\Delta_2+\Delta_3+g+b+m_2\ge \sum_{y\in Y}d^+(y) \ge
4|Y|=4n-o(n).$$
Hence,  writing $t=2\Delta_1-(\Delta_2+\Delta_3)$, we have
\begin{equation}\label{5.2}
b\ge 4n-\Delta_1-\Delta_2-\Delta_3-g-m_2-o(n)=4n- 3\Delta_1-g-m_2+t-o(n),
\end{equation}

Next, we will derive bounds on $m_2,\Delta_1,t$ and $b$ in terms of
$n$ and $g$, so that we can use (4) of Lemma~\ref{property2}.

By Lemma~\ref{3huge-vertices},
$$b<3\Delta_2+3\Delta_3-4\Delta_1-4g-m_2/2-7(n-\tau)/4=2\Delta_1-3t-4g-m_2/2-7(n-\tau)/4, $$
which, combined with $\tau\leq (n+2g+2b)/3$ (by (1) of Proposition~\ref{huge-1}), imlpies
\begin{equation}\label{5.1}
b>7n+3m_2+17g-12\Delta_1+18t.

\end{equation}

We may assume that (2) and
(3) of Lemma~\ref{property2} hold; for, otherwise, by
Lemma~\ref{property2}, the desired partition of $V(D)$ exists. Thus,
\begin{equation}\label{5.3}
b>3g+3m_2/8-\Delta_1/3+4t,

\end{equation}
and
\begin{equation}\label{5.4}
b< \frac{28}{11}n-\frac{27}{11}\Delta_1+\frac{12}{11}g-\frac{9}{88}m_2+\frac{2}{11}t.

\end{equation}

Combining \eqref{5.4} with \eqref{5.2}, \eqref{5.1}, \eqref{5.3},
respectively, we obtain (by eliminating $b$)

\begin{equation}\label{5.6}
\frac{79}{8}m_2+23g>16n-6\Delta_1+9t-o(n),
\end{equation}
\begin{equation}\label{5.5}
\frac{273}{8}m_2+175g<105\Delta_1-49n-196t.
\end{equation}

\begin{equation}\label{5.7}
\frac{63}{4}m_2+63g<84n-70\Delta_1-126t.
\end{equation}
Noting that $\Delta_1\le n-|X|$ (by (2)) and $n$ is large, we see from \eqref{5.6}
that $\frac{79}{8}m_2+23g> 9.99n$; so
\begin{equation}\label{000}
m_2+2.31g> n
\end{equation}

We now combine \eqref{5.6} with \eqref{5.5} and
\eqref{5.7}, respectively,  and we get (by eliminating $m_2$)
\begin{equation}\label{5.8}
1.806t+0.759g<\Delta_1-0.829n,
\end{equation}
\begin{equation}\label{5.9}
2.322t+0.435g<0.968n-\Delta_1.
\end{equation}
From \eqref{5.8}, we have
\begin{equation}\label{001}
\Delta_1>0.829n.
\end{equation}
 Combining \eqref{5.8} and \eqref{5.9} to eliminate $\Delta_1$, we have

\begin{equation}\label{002}
3t+g<0.117n.
\end{equation}

From \eqref{5.4}, we have
$$b
<\frac{28}{11}n-\frac{27}{11}\Delta_1-\left(\frac{9}{88}m_2+\frac{21}{88}g\right)+\left(\frac{2}{11}t+\frac{117}{88}g\right).$$
Hence, by \eqref{000}, \eqref{001} and \eqref{002}, we have
\begin{equation}\label{003}
b  <\frac{28}{11}n-\frac{27}{11}\times 0.829n-\frac{9}{88}n+\frac{117}{88}\times 0.117n <0.564n.
\end{equation}

We wish to use (4) of Lemma~\ref{property2}. Note that
$$  2\Delta_2+2\Delta_3-3\Delta_1-3g-b+3m_2/14=\Delta_1+(3m_2/14+g/2)-b-(2t+7g/2).$$
So by \eqref{000}, \eqref{001}, \eqref{002}, and \eqref{003}, we see that
\begin{align*}
  & 2\Delta_2+2\Delta_3-3\Delta_1-3g-b+3m_2/14\\
   > & 0.829n+3n/14-0.564n-(7/2)\times  0.117n\\
  > & 0.069n\\
  > & 0.
\end{align*}

Hence, by (4) of Lemma \ref{property2}, we have $
6\Delta_1-3\Delta_2-3\Delta_3+2g-b+3m_2/14<0$. Thus,
\begin{equation}\label{5.10}
b > 6\Delta_1-3\Delta_2-3\Delta_3+3m_2/14+2g=3m_2/14+2g+3t.
\end{equation}
Combining \eqref{5.4} and \eqref{5.10} (by eliminating $b$), we have

$$\frac{195}{56}m_2+10g<28n-27\Delta_1-31t,$$

which, combined with \eqref{5.6} (by eliminating $m_2$), gives
\begin{equation}\label{5.12}
1.373t+0.075g<0.899n-\Delta_1,
\end{equation}
which implies $\Delta_1<0.899n$; so

by \eqref{5.2}, we have
\begin{equation}\label{005}
g+b+m_2>4n-3\Delta_1+t-o(n)>4n-3\times 0.9n=1.3n.
\end{equation}
Combining \eqref{5.8} and \eqref{5.12} (to eliminate $\Delta_1$), we derive

\begin{equation}\label{006}
3t+g<0.084n.
\end{equation}

Again by (4) of Lemma~\ref{property2}, we have
$$0  >  6\Delta_3-3\Delta_1-3\Delta_2-3g+b/3+3m_2/14 =
 3(m_2+b+g)/14+5b/42-(6t+2g+17g/14)+9(\Delta_1-\Delta_2).$$
Hence, noting that $b>3n/14$ (by \eqref{000} and \eqref{5.10}) and by \eqref{005} and \eqref{006},
$$0 >(3/14)\times 1.3n+(5/42) \times (3n/14)-(2+17/14))\times 0.084n>0.034n.$$
This is a contradiction,  completing the proof of Theorem~\ref{thm1}.

\medskip

\section{Concluding remarks}
We studied partitions of digraphs with minimum outdegree $d\ge 4$ and proved Conjecture \ref{conj1} in the case when $d=4$.
We used a typical approach  for finding a partition $V_1,V_2$ in a digraph $D$ that bounds $e(V_1,V_2)$ and $e(V_2,V_1)$ simultaneously:
Start with a partition $X,Y$ of $V(D)$ such that $X$ consists of
large degree vertices;  partition $X$ by
considering the ``huge'' vertices in $X$, those vertices with large
gap between their outdegree and their indegree; and randomly partition the
vertices in $Y$. Huge vertices play an important role in the process
for  obtaining the desired partition. For instance,
 we showed that Conjecture~\ref{conj1}
holds when there exists a partition of $V(D)$ for which  the number of
huge vertices is at least $d$ or exactly 1.
We hope that our work would shed light on how the set $V(D)$ should be
partitioned into $X,Y$ and how the set $X$ should be
partitioned.

In \cite{Lee2014}, Lee, Loh, and Sudakov point out that one needs  to
combine both Lemma $\ref{lem1}$ and Lemma \ref{lem4}
to prove Conjecture \ref{conj1}. They also remarked that a naive
combination is not adequate for $d\geq 4$ because of the following
example.  Let $D'$ be the digraph obtained from $K_{5,n-5}$ (with
$n>9$) by  orienting the edges so that
one vertex, say $v_{1}$, has outdegree $n-5$ and  four
vertices, say  $v_{2}, v_{3}, v_{4}, v_{5}$, each have indegree $n-5$.
Let $D$ be obtained from $D'$ by adding an arc directed from $v_i$ to $v_j$ for
each ordered pair $(i,j)$ with $1\le i\ne j\le 5$. Then the minimum
outdegree of $D$ is   4 and the number of arcs in $D$ is $m=5n-5$.

Let $X=\{v_1, \ldots, v_5\}$. If we partition $V(D)$ to $X$ (consisting of large degree vertices)
and $Y=V(D)\setminus X$ (consisting of small degree vertices), then
$X$  is the set of huge vertices.  One can check  that
$X_1=\{v_2 \}$ and $X_2=\{v_1,v_3,v_4,v_5 \}$ form  a partition of $X$
with  minimum gap.  However, for any partition $Y_1,Y_2$ of $Y$, we see that
 $e(X_2\cup Y_{2},X_1\cup Y_{1})=n-5+4=n-1=m/5$, which is smaller than $3m/14$.
What this means is that one need to consider different partitions of the
set  of huge vertices. In this paper, we have managed to prove Conjecture~\ref{conj1} in the case when $d=4$ by
carefully partitioning the huge vertices.

For digraphs with minimum outdegree $d\ge 5$, new ideas seem needed (in addition to better
partitioning the huge vertices), as shown by the following example.
Let $D'$ be the digraph obtained from $K_{3,n-3}$ (with $n>900$) by
orienting all edges from the part $Y$ of
size $n-3$ to the part $X$ of size 3. Let $D$ be obtained from $D'$ by
adding
six arcs directed from each vertex in $X$ to  6 vertices in $Y$ (so
that no two arcs get directed towards the same vertex in $Y$),
and adding a 3-out-regular graph on $Y$.
Hence, the minimum outdegree of $D$ is  $6$, and $X$ is the set of
huge vertices with respect to the partition $X,Y$.     It is not difficult to verify that for any value $p$
and any partition $X_1,X_2$ of $X$, we have
$f(p,X_1,X_2)<0$ and $h(p,X_1,X_2)<0$ (see Section 3). Therefore,  one needs to better
partition $V(D)\setminus X$ in order to achieve the bound in Conjecture \ref{conj1}.

config: arxiv


TITLE: Change of variables inside conditional expectation
QUESTION [0 upvotes]: Let $X, Y$ be random variables, $g$ be some function on $\mathbb{R}$ that is invertible.
Suppose also that we know $\mathbb{E}[g(Y)|X]$, and the goal is to find $\mathbb{E}[Y|X]$. Is this possible in closed-form? I'm aware we could do a Taylor approximation here, but what about an exact solution?

REPLY [0 votes]: You can do a Taylor approximation here. Let's denote $Z=g(Y)$ then $Y = g^{-1}(Z)$. For simplicity's sake, denote also $f(Z)=g^{-1}(Z)$ and $a(X)=E[Z|X]$.
We have
$$f(Z)=f(E[Z|X]) + \sum_{i=1}^{+ \infty}(Z-E[Z|X])^n\frac{d^n}{dz^n}f(E[Z|X])$$
$$f(Z)=f(a(X)) + \sum_{i=1}^{+ \infty}(Z-E[Z|X])^n\frac{d^n}{dz^n}f(a(X))$$
This implies that
$$E[Y|X]=E[f(Z)|X] = f(a(X)) + \sum_{i=1}^{+ \infty}E[(Z-a(X))^n |X]\frac{d^n}{dz^n}f(a(X))$$
But if you know only $E[Z|X]$ and not the variable $Z$ (or $Y$) itself, the term $E[(Z-a(X))^n|X] = E[(g(Y)-a(X))^n|X]$ is undefined.
The conclusion: it is unable to determine $E[Y|X]$ from the sole information $E[g(Y)|X]$.


text:
\begin{document}

\title{A note on the zeroes of the Fredholm Series}

\author{Umberto  Zannier}
\thanks{Umberto Zannier, Scuola Normale Superiore, Piazza dei Cavalieri 7, 56126 Pisa, ITALY\\ umberto.zannier@sns.it}

\maketitle
\vspace{-0.8cm}\begin{center}With an appendix by F. Veneziano\end{center}\vspace{1cm}

\centerline{\it Dedicated to the memory of Edoardo Vesentini}

\bigskip

\bigskip

{\sc Abstract}.  { The issue had been raised whether the so-called  {\it Fredholm series} $z+z^2+\dotsb+z^{2^n}+\dotsb$ has infinitely many zeroes in the unit disk. We provide an affirmative answer, proving that in fact every complex number occurs as a value infinitely many times. }

\medskip

\section{Introduction}

Let $D=\{z\in \C:|z|<1\}$ denote the complex unit disk and let
\begin{equation}\label{E.F}
f(z)=z+z^2+z^4+\dotsb+ z^{2^n}+\dotsb , \qquad z\in D.
\end{equation}

This function  is proposed in several books  as a simple instance of a holomorphic function on $D$ which cannot be holomorphically extended to any connected open domain of $\C$ containing $D$ strictly  (see e.g. \cite{T}, p. 159, or \cite{WW}, p. 98).
It is often called the {\it Fredholm series}, though apparently Fredholm had considered a different series;  the terminology was introduced  by A. J. Kempner in the belief that Fredholm studied it, and then this was followed by several authors: see  J. Shallit's paper \cite{S}, especially pp.~161--162, for an accurate description. (Fredholm is also mentioned in

R. Remmert's book  \cite{Re}, p. 254  but concerning indeed a different series sharing the said property.)

We shall not change here this convention,  though perhaps the function should be called {\it Kempner series}.

It satisfies the functional equation $f(z)=z+f(z^2)$.

The function became better known also through the work of K. Mahler \cite{Mah}, who considered its values at (nonzero) algebraic points in $D$ and proved their transcendency (together with  similar results for a much wider class of functions). In fact, concerning the function itself, the transcendency of the values at points $1/n$, $n\in\N^+$,  had been already established by Kempner \cite{K}; these values are interesting and related to iterated paperfolding (see \cite{S}, pp.  162--163).
All of this gave rise to many-sided generalisations and several works, both of arithmetical and functional nature, by a number of authors. See
also D. Masser's book \cite{M} for an account of Mahler's method in particular, and for some  references,  and see \cite{Mah2} p. 132 for Mahler's reminiscences on this topic.

\medskip

In the book \cite{M} the issue is raised (see Ex. 3.15) {\it whether the function has infinitely many zeroes in $D$}, and it is remarked that Mahler found 16 zeroes (apparently using a pocket calculator). Actually already Mahler enquired about the location of the zeroes in the paper \cite{Mah3}, commenting that there are probably zeroes ``{\it in every neighborhood of the unit circle}''.  It is easy to locate the  (unique)  real nontrivial zero $-0.658626...$ on looking at signs, but otherwise the location of zeroes seems not obvious to detect, especially due to the fairly complicated oscillatory nature of $f(z)$ near the boundary of $D$ (which is related to exponential sums with widely growing arguments - see \cite{Mah3} and see  Remarks \ref{R.R}, \ref{R.R2} below). Of course in any disk $rD$  of fixed radius $r<1$ there are only finitely many zeroes of $f$, so Mahler's expectation literally amounts to the existence of an infinity of zeroes.

Note that several `general' methods for studying the zeroes (and the values)  of holomorphic functions (e.g. Picard Theorems, Nevanlinna Theory)  usually apply to functions with singularities, or  defined on the whole complex plane,  and do not seem to be immediately helpful  here since $f(z)$ is regular on the whole $D$ and has the unit circle as a natural boundary. (See  however W. Hayman's book \cite{Hay}, especially Chs. 5 and  6, and also W. Rudin's book \cite{R}, Chs. 15, 17, for some results about the distribution of zeroes of functions holomorphic in $D$. See Remark  \ref{R.R} below for some comments on this.)

\medskip

It is the main purpose of this note to prove that indeed there are infinitely many zeroes of $f(z)$ in $D$, confirming the expectation of Mahler. Actually we shall prove a sharpening of this result,  namely that the function attains every complex number as a value infinitely often, and even arbitrarily near $1$.

\medskip

By the results of Mahler any of the zeroes, apart from $0$,  is a transcendental number. Masser asked if for instance one can prove sharper results, e.g. of algebraic independence, or if one can study more accurately  the distribution of zeroes. We have no answer to the former issue; the latter maybe admits some kind of answer through the present method, but we haven't carried out any analysis in this direction (see the final remarks for a few comments).

\medskip

This  note contains also an  appendix by Francesco Veneziano, who kindly followed some  requests of mine, on performing  several computations concerning this problem; in the Appendix he lists some approximate zeroes of the function and locates a zero of a related function which appears in the proof.  Also, he represented  in Figure 1 below some zeroes of a truncated series, rescaled with respect to the Poincar\'e distance in $D$, from the origin.

\bigskip

Before stating our results, we introduce some notation. We denote as usual $e(w)=\exp(2\pi iw)$ and we let $\H=\{w\in \C:\Im w>0\}$ denote  the upper-half plane, with which we shall work in place of $D$. Indeed,  for $w\in \H$ we set

 \begin{equation}\label{E.Fw}
F(w):=f(e(w))=e(w)+e(2w)+e(4w)+\ldots , \qquad  w\in \H.
\end{equation}

This function is holomorphic in $\H$ and satisfies the functional equations
\begin{equation}\label{E.fe}
F(w+1)=F(w),\qquad F(w)=e(w)+F(2w).
\end{equation}

Consider now the function
\begin{equation}\label{E.G}
G(w):=\sum_{l=1}^{\infty} e\left(-\frac{1}{2^l}\right)\left(e\left(\frac{w}{2^l}\right)-1\right),\qquad w\in\H.
\end{equation}
 Note that for $w$ in a given compact set $K\subset\C$ we have    $|e(w/2^l)-1|\ll_K 2^{-l}$, hence  the series defining $G$  converges uniformly and boundedly on compact sets of $\C$, so in fact $G$ is in particular holomorphic on the whole $\C$.

 We further define
\begin{equation}\label{E.S}
S(w)=F(w)+G(w),\qquad w\in \H.
\end{equation}

 We have

 \begin{prop} \label{P.P}  Let $V=S(\H)\subset \C$ be the image  of $S$ on $\H$. Then, for every open disk $U$ centered at $1$ and for every $v\in V$,  the Fredholm function $f(z)$ assumes on $D\cap U$  infinitely many times the  value $v$.
 \end{prop}

 The following result also shows that the set $V$ of Proposition \ref{P.P} is stable by certain translations, thus is in a sense `large'.

  \begin{prop} \label{P.P2}  Let $V=S(\H)\subset \C$ be as above  the image  of $S$ on $\H$. Then  $V+\Z=V$ and moreover for each $m\in\Z$ we have $V+c_m\subset V$, where $c_m$ is as in \eqref{E.cm} below.
 \end{prop}

 A study of the numbers $c_m$ carried out in \S \ref{SS.cm} will in fact imply the following sharpening of the proposition (which essentially then becomes a lemma):

 \begin{thm} \label{T.V} We have $V=\C$, that is, the function $S(w)$ attains every complex value.
 \end{thm}

  In particular, combining these results we obtain the following consequence,  which contains as a corollary the main motivation for this note:

 \begin{thm}\label{T.T}
 For every open disk $U$ centered at $1$, the Fredholm series attains on $D\cap U$ every complex number as a value  infinitely many times.
 \end{thm}

So  in particular $f(z)$  has infinitely many zeroes in $D$ and even in $D\cap U$. It may be that the    methods  allow to replace $U$ with any disk centered on  the unit circle but we have not verified this.

\bigskip

{\bf Acknowledgements}.  I thank David Masser for raising several issues and for remarks and references; similarly I thank Pietro Corvaja. I thank Francesco Veneziano for some computer calculations, especially  helpful when writing the paper, and for providing the Appendix. I further thank Jeffrey  Shallit for pointing out to me a correct historical account of the terminology, and some references.

\section{Some  formulae}
We now let $\theta\in\R$ be real and $n>0$ be a positive integer, and seek an approximation formula for $F(w2^{-n}+\theta)$.  On iterating \eqref{E.fe} we have

\begin{equation}\label{E.app}
\begin{split}
F\left({w\over 2^n}+\theta\right) & = e\left({w\over 2^n}+\theta\right)+e\left({w\over 2^{n-1}}+2\theta\right)+\dotsb +e\left({w\over 2}+2^{n-1}\theta\right)+F(w+2^n\theta)
\\
& =e\left({w\over 2^n}\right)e(\theta)+e\left({w\over 2^{n-1}}\right)e(2\theta)+\dotsb +e\left({w\over 2}\right)e(2^{n-1}\theta)+F(w+2^n\theta).
\end{split}
\end{equation}

\begin{comment}

Using the expansion
\begin{equation*}
e({w\over 2^m})=\sum_{r=0}^\infty{1\over r!}\left({2\pi i w\over 2^m}\right)^r
\end{equation*}
for the terms on the bottom of \eqref{E.app} we obtain

\begin{equation*}
\begin{split}
e({w\over 2^n})e(\theta)+e({w\over 2^{n-1}})e(2\theta)+\ldots +e({w\over 2})e(2^{n-1}\theta) &
=\sum_{l=0}^{n-1}e(2^l\theta)\sum_{r=0}^\infty{2^{lr}\over r!}({2\pi i w\over 2^n})^r\\
& = \sum_{r=0}^\infty{1\over r!}(2\pi i w)^r {S_r(n,\theta)\over 2^{nr}},
\end{split}
\end{equation*}
where we have used the notation
\begin{equation}\label{E.Sr}
S_r(n,\theta)=\sum_{l=0}^{n-1}2^{lr}e(2^l\theta).
\end{equation}
In conclusion, we obtain the identity

\begin{equation}\label{E.ident}
F({w\over 2^n}+\theta)= \sum_{r=0}^\infty{(2\pi i w)^r\over r!} \cdot {S_r(n,\theta)\over 2^{nr}}+F(w+2^n\theta).
\end{equation}

\end{comment}

We shall use this formula, with suitable choices for the involved quantities, to approximate $F$  with convenient  functions.

\medskip

\subsubsection{A Ramanujan sum and related ones} As remarked by Mahler in \cite{Mah3} the behaviour of $F(w)$ (for rational $\Re  w$) is linked  with certain Gaussian periods. Here we adopt a related but different viewpoint, constructing vanishing {\it Ramanujan sums}.

We let $q=3^k$, where $k\ge 2$ is an integer.  Note  the Euler function value $\phi(q)=2\cdot 3^{k-1}$.

Since $2$ is a primitive root modulo $3$ \footnote{In what follows this choice is immaterial and there is wide possibility of changing it.} and since $2^{2}\not\equiv 1\pmod{9}$, it follows easily (e.g. by induction on $k$)  that $2$ is a primitive root modulo $q$, namely the powers $2^m$, $0\le m\le \phi(q)-1$ form a complete set of residues modulo $q$ and coprime to $q$.  Since $q$ is not squarefree, it follows  (see e.g. \cite{HW}, \S 16.6, especially Thm. 271) that, for any integer $s$ coprime with $q$,  the {\it Ramanujan sum},  i.e.  of the  primitive $q$-th roots of unity, vanishes, which amounts to:
\begin{equation}\label{E.ram}
\sum_{m=0}^{\phi(q)-1}e\left({2^ms\over q}\right)=0,  \qquad \gcd(s,q)=1,
\end{equation}
the same equation actually holding when the summation for $m$  runs through a complete set of residues modulo $\phi(q)$.

\medskip

We now choose  an integer $a\ge 1$ and   $k=2^a$, a power of $2$. We put
\begin{equation*}
\theta_0={1\over q},\qquad n_0=\phi(q)=2\cdot 3^{2^a-1}.
\end{equation*}

We also note that $q\equiv 1\pmod {2^{a+2}}$, as follows easily by induction on $a\ge 1$. Hence for $l\le a+2$ the ratio $(1-q)/2^l$ is an integer. Also, we have  for any integer $n\ge l$,
 \begin{equation*}
2^{n-l}\theta_0\equiv2^n{(1-q)\over 2^l}\theta_0\pmod \Z, \qquad 0\le l\le a.
\end{equation*}
In particular, let $n=n_0$, so $2^{n}\equiv 1\pmod q$. Then  for $l\le a\le n_0$, on writing $2^{n_0}=1+q\ \cdot\ $integer,  this congruence yields
 \begin{equation}\label{E.modZ}
2^{n_0-l}\theta_0\equiv {(1-q)\over 2^l q}={\theta_0\over 2^l}-{1\over 2^l}\pmod \Z, \qquad 0\le l\le a.
\end{equation}

\begin{comment}

Now, we have, on writing $n-l$ in place of $l$ in the defining formula \eqref{E.Sr} for $S_r$,
\begin{equation*}
{S_r(n_0,\theta_0)\over 2^{n_0r}} =\sum_{l=1}^{n_0}{e(2^{n_0-l}\theta_0)\over 2^{lr}},
\end{equation*}
which, after using \eqref{E.modZ} gives

\begin{equation}\label{E.Sr2}
{S_r(n_0,\theta_0)\over 2^{n_0r}} =\sum_{l=1}^{n_0}e(-{1\over 2^l}){e({\theta_0\over 2^l})\over 2^{lr}}.
\end{equation}

 $\theta=\theta_0$, $n=n_0$ into \eqref{E.ident} and  using  \eqref{E.Sr2} we obtain, taking into account \eqref{E.S0} and  that  $2^{n_0}\theta_0\equiv\theta_0\pmod\Z$ and that $F$ is $\Z$-periodic,

 \begin{equation}\label{E.app2}
 \begin{split}
 F({w\over 2^{n_0}}+\theta_0) &
 =\sum_{l=1}^{n_0} e\left({\theta_0-1\over 2^l}\right)\sum_{r=1}^\infty{(2\pi i w)^r\over r!} \cdot {1\over 2^{lr}}+F(w+\theta_0)\\
 &= \sum_{l=1}^{n_0} e\left({\theta_0-1\over 2^l}\right)(e({w\over 2^l})-1)+F(w+\theta_0).
 \end{split}
 \end{equation}

\end{comment}

\medskip

 \subsubsection{Preparing an  approximate formula}  Setting  $s=1$ into \eqref{E.ram} we obtain
  \begin{equation*}
 \sum_{l=0}^{n_0-1}e(2^l\theta_0)=0,
 \end{equation*}
 whence, setting   $\theta=\theta_0$, $n=n_0$ into \eqref{E.app} we derive,  taking into account  that  $2^{n_0}\theta_0\equiv\theta_0\pmod\Z$ and that $F$ is $\Z$-periodic,

 \begin{equation*}
 F\left({w\over 2^{n_0}}+\theta_0\right) =\sum_{l=0}^{n_0-1} e(2^l\theta_0)\left(e\left({w\over 2^{n_0-l}}\right)-1\right)+F(w+\theta_0).
 \end{equation*}

 Further, on  writing $n_0-l$ in place of $l$ in the sum and using \eqref{E.modZ}  for $1\le l\le a$, this becomes

  \begin{equation}\label{E.app2}
 F\left({w\over 2^{n_0}}+\theta_0\right) =\sum_{l=1}^{a} e\left({\theta_0 -1\over 2^l}\right)\left(e\left({w\over 2^{l}}\right)-1\right)+\sum_{l=a+1}^{n_0} e(2^{n_0-l}\theta_0)\left(e\left({w\over 2^{l}}\right)-1\right)+F(w+\theta_0).
 \end{equation}

 \medskip

\section{Proofs of main assertions}

 \begin{proof}[Proof of Proposition \ref{P.P}]

Let   $K\subset \H$ be a given compact subset of $\H$, to which we shall refer by a subscript to indicate dependence of the implicit constants which shall appear. For $w\in K$ we have
\begin{equation*}
|F(w+\theta_0)-F(w)|\ll_K {1\over q},\quad \left|e\left({w\over 2^{l}}\right)-1\right|\ll_K{1\over 2^l},\quad  \left| e\left({\theta_0 -1\over 2^l}\right)-e\left(-{1\over 2^l}\right)\right|\ll {1\over q}\quad (l\ge 1),
\end{equation*}
whence, by \eqref{E.app2},  again for $w\in K$,
\begin{equation}\label{E.app3}
\left|F\left({w\over 2^{n_0}}+\theta_0\right) -S(w)\right|\ll_K\sum_{l=a+1}^\infty\left|e\left({w\over 2^{l}}\right)-1\right|+{1\over q}\ll_K{1\over 2^a}.
\end{equation}

The proof is now easily completed by a quite standard argument. Let  $v\in V$ be any  value of $S$ on $\H$, say $v=S(w_0)$ for some $w_0\in\H$.  Let $C$ be a circle inside $\H$ and centred at  $w_0$ such that $S(w)\neq v$ for $w\in C$, and choose  $K$ to be a closed disk inside $\H$,  centred at $w_0$,  and of radius larger than that of $C$.

 The function $S(w)-v$ is holomorphic in $K$ and never zero on $C$,  hence its absolute value attains on $C$ a minimum $\delta>0$.

Put now, for any integer $a\ge 1$,  $F_a(w)=F(2^{-n_0}w+\theta_0)$, where $n_0,\theta_0$ are as above, so by \eqref{E.app3}, $|F_a(w)-S(w)|\le c_K2^{-a}$ for $w\in K$, where $c_K$ is a constant, i.e. independent of $a$. Now  suppose that $a$ is so large that  $2c_K<\delta 2^a$. Then on $C$ we have
\begin{equation*}
|(F_a(w)-v)-(S(w)-v)|<{\delta\over 2}< \min_{w\in C}|S(w)-v|.
\end{equation*}

By Rouch\'e's theorem (see e.g. \cite{T}) the functions $S(w)-v$, $F_a(w)-v$ have the same number of zeroes inside $C$, so since the former vanishes at $w_0$ (which is the centre of  $C$) the latter has at least a zero $\zeta_a$  inside $C$.

This means that $F(2^{-n_0}\zeta_a+\theta_0)=v$. Now, the points $2^{-n_0}\zeta_a+\theta_0\in\H$, for varying $a$,
 tend to $0$ (because $\zeta_a\in K$), so the corresponding images through $e(w)$ tend to $1$ (but are different from $1$ as they lie in $D$).
 This proves that, if $U\subset \C$ is any neighbourhood of $1$,  $f(z)$ attains the value $v$ at infinitely many points of $D\cap U$, concluding the argument. \end{proof}

\medskip

\subsection{About the function \texorpdfstring{$S(w)$}{S(w)}}  In this subsection we develop some properties of the function $S(w)$, in particular of its image on $\H$, above denoted $V$.

It will be equivalent and  notationally convenient to work with the function $S_1(w):=S(w+1)$. By definition we have
\begin{equation}\label{E.S1}
S_1(w)=F(w+1)+G(w+1)=F(w)+\sum_{l=1}^\infty\left(e\left({w\over 2^l}\right)-e\left(-{1\over 2^l}\right)\right).
\end{equation}

 For integers $m\in\Z$,  also define constants $c_m$ by
 \begin{equation}\label{E.cm}
 c_m:=\sum_{l=1}^\infty\left(e\left({m\over 2^l}\right)-1\right).
 \end{equation}
 Hence, setting $H(w):=\sum_{l\ge 1}\left(e\left(w/2^l\right)-1\right)$,  we may rewrite \eqref{E.S1} as
 \begin{equation}\label{E.S1H}
S_1(w)=F(w)+\sum_{l=1}^\infty\left(e\left({w\over 2^l}\right)-1\right)+\sum_{l=1}^\infty\left(1-e\left(-{1\over 2^l}\right)\right)=F(w)+H(w)-c_{-1}.
\end{equation}
Note that $H$ is holomorphic in the whole $\C$ and $c_m=H(m)$.

As Masser pointed out to me, the function $F(w)+H(w)$ appears already in S. Ramanujan's work (see \cite{H}, p. 39); it also appears (as a function on $D$) in  Exercise 3.11, p. 44 of Masser's book \cite{M} and in the previous article of lectures by  Masser in \cite{M2} (see Lecture 3).  It is observed that it satisfies a functional equation when (in the present notation) $w$ is changed into $2w$. Indeed, we easily find that
\begin{equation}\label{E.S1fe}
S_1(2w)=S_1(w)-1,
\end{equation}
 This already shows that  $V+\Z=V$. Note  also that this yields (Masser's observation)  that the function $S_1(w)+(\log w/\log 2)$ (which is well-defined in $\H$ on agreeing for instance on the value $\log i=\pi/2$) is holomorphic on $\H$ and invariant under  $w\to 2w$, as pointed out in the above quoted sources.

 Let us now seek  other functional equations.

 Setting $w+m2^s$ in place of $w$, for integers $m\in\Z$, $s\ge 0$, we obtain
 \begin{equation}\label{E.S1fe2}
 \begin{split}
S_1(w+m2^s) &=S_1(w)+\sum_{l=s+1}^\infty e\left({w\over 2^l}\right)\left(e\left({m\over 2^{l-s}}\right)-1\right)\\
& = S_1(w)+ \sum_{l=1}^\infty \left(e\left({w\over 2^{l+s}}\right) -1\right)  \left(e\left({m\over 2^l}\right)-1\right) +c_m\\
& = S_1(w)+\Delta_{m,s}(w)+c_m,
\end{split}
\end{equation}
say.  Note that for $w$ in a compact subset $K$ of $\H$, we have
\begin{equation}\label{E.D}
\Delta_{m,s}(w):=\sum_{l=1}^\infty \left(e\left({w\over 2^{l+s}}\right) -1\right)  \left(e\left({m\over 2^l}\right)-1\right) =O_K\left({1\over 2^s}\right).
\end{equation}

  \begin{proof}[Proof of Proposition \ref{P.P2}]   We have already noted above that, in view of \eqref{E.S1fe}, the image of $S$ is invariant under translation by integers, which yields the first claim of the proposition.

  Let now $K$ be any compact disk inside $\H$ and consider  formula  \eqref{E.S1fe2}.    Note that by \eqref{E.D}, for $s\to\infty$ (and fixed $m$ for instance)  the functions $\Delta_{m,s}$ converge uniformly and boundedly to $0$ for $w\in K$. Then, exactly by the same argument as in the proof of Proposition \ref{P.P},  if $v=S_1(w_0)$ is a value of $S_1$ at a point $w_0$ in the interior of $K$, then for large $s$  (and  fixed $m$) the function $S_1(w+m2^s)-c_m$ assumes as well that value.   Hence $S_1$ (and therefore $S$) assumes the value $v+c_m$, proving the contention.
  \end{proof}

  \medskip

  Before going ahead, we pause to give a proof of the special case of Theorem \ref{T.T} in which $\C$ is replaced  by $\R$ as a set of possible values. This proof is a bit simpler than the general case and does not require the considerations of \S \ref{SS.cm}.

  \begin{proof}[Proof of Theorem \ref{T.T} for real values]   By Proposition \ref{P.P} it suffices to prove that $\R\subset  V$ (i.e. that $S_1$  attains all real numbers as values  on $\H$). Also, by Proposition \ref{P.P2}  applied with $m=-1$, it suffices to show that $\R-c_{-1}\in V$;  namely, using \eqref{E.S1H}, it suffices to show that  $F(w)+H(w)$ assumes any real value in $\H$.   Let us look at the values for purely imaginary $w=it$, $t\in\R$.  These values are real and by the formula $S_1(2w)=S_1(w)-1$ this set of real values is invariant by any integer translation. Thus it suffices to prove that it contains an interval of length $1$; on the other hand this is clear  by the same formula  $S_1(2w)=S_1(w)-1$, since the imaginary line is connected.
   \end{proof}

  \subsection{About the constants \texorpdfstring{$c_m$}{cm}} \label{SS.cm}
  Note that $c_{2m}=c_m$ and that     $\overline{c_m}=c_{-m}$.
  As remarked above, $c_m$ is just the value of $H(z)$ at $z=m$. These numbers seem interesting on their own and we wonder whether they have irrationality of transcendency properties. We cannot prove any neat assertion of this type, but some small information in this direction will appear below.

    Proposition \ref{P.P2}  says that $V$ is stable under translation by the semigroup generated by $\Z$ and the $c_m$. Let $\sigma_m:=c_m+c_{-m}$ noting it  is real, and  in particular we find that $V$ is stable under  translation by the real semigroup generated by $\Z$ and the $\sigma_m$.  We shall study a bit  this semigroup, denoted here $\Gamma$.

    \medskip

    Now, it turns out that a related sequence has better approximation properties, useful for our purposes.

    Define $k_l:=e(1/2^l)-1$. We have  $k_0=0$, $k_1=-2$, $k_2=i-1$, $k_3=(\sqrt 2-1+i\sqrt 2)/2$ and $k_l=k_{l+1}(k_{l+1}+2)$. Setting $\mu_l:=|k_l|^2$ we also have $\mu_l=2-2\cos(\pi/2^{l-1})$  and $\mu_{l+1}=2-\sqrt{4-\mu_l}$.

    Further,  $e(m/2^l)=(k_l+1)^m$ and $c_m=\sum_{l=1}^\infty ((k_l+1)^m-1)$. For $m\in\Z$ we now set
    \begin{equation}\label{E.c(m)}
    c(m)=\sum_{l=1}^\infty k_l^m.
    \end{equation}

    Note that the $k_l$ tend to $0$ exponentially so this formula yields  good approximations of the $c(m)$   (by  linear recurrence sequences) on taking  truncations of the series.

    These numbers are linear combinations of the $c_m$ with integer coefficients and conversely. In fact
    $c(m)=\sum_{l=1}^\infty (e(m/2^l)-{m\choose 1}e((m-1)/2^l)+...)=c_m-{m\choose 1}c_{m-1}+\ldots +(-1)^{m-1}c_1$. Similarly, $c_m=\sum_{l=1}^\infty (k_l^m+{m\choose 1}k_l^{m-1}+\ldots +{m\choose 1}k_l)=c(m)+{m\choose 1}c(m-1)+\ldots +{m\choose 1}c(1)$.

    \medskip

    For the sake of curiosity, we also note the recurrence $c(m)=\sum_{l=0}^\infty k_{l+1}^m(k_{l+1}+2)^m=c(2m)+2{m\choose 1}c(2m-1)+4{m\choose 2}c(2m-2)+\ldots + 2^{m-1}{m\choose 1}c(m+1)+2^mc(m)$, whence
     \begin{equation*}
    (1-2^m)c(m)=\sum_{h=1}^m2^{m-h}{m\choose h}c(m+h),
    \end{equation*}
 so the  $c(m)$ satisfy several linear relations with rational coefficients. These correspond in fact to the more evident relations $c_{2m}=c_m$. We wonder whether there are other such linear relations holding among them.

    \medskip

    Now, for our purposes it is more useful to consider the numbers
  \begin{equation*}
  \sigma(m):=\sum_{l=1}^\infty(k_l+\bar k_l)^m=\sum_{l=1}^\infty \left(e\left({1\over 2^l}\right)+e\left(-{1\over 2^l}\right)-2\right)^m.
  \end{equation*}

  Observe that, for $2h\neq m$, $k_l^h\bar k_l^{m-h}+k_l^{m-h}\bar k_l^h$ equals an integer linear combination of terms $e(s/2^l)+e(-s/2^l)-2=\sigma_s$: this is seen on expanding $k_l^h=(e(1/2^l)-1)^h$ with the binomial theorem (and similarly for the complex conjugate). The same holds for a term $k_l^h\bar k_l^h$ in case $2h=m$. On summing over $l$ we deduce that $\sigma(m)$ is an integer linear combination of $\sigma_1,\ldots ,\sigma_m$.    (One can also write down an explicit formula in terms of Chebyshev polynomials.)

    Now, $(-1)^m\sigma(m)$ is of the shape $4^m+2^m+O((2-\sqrt 2)^m)$.  We conclude that  for growing $m$, either one among $\sigma_1,\ldots,\sigma_m$ is irrational, or the least common denominator of these numbers grows to infinity.  In fact, if all $\sigma_j$ would be rationals with a bounded denominator, the said relation would imply the same for $\sigma(m)$, whence, since $0<2-\sqrt 2<1$,  for large $m$ we would have  $\sigma(m)=(k_1+\bar k_1)^m+(k_2+\bar k_2)^m$, which is plainly false.

We in turn deduce that the semigroup $\Gamma$ generated by $\Z$ and the $\sigma_m$ is dense in $\R$. In fact, if some $\sigma_m$ is irrational this follows from well-known (easy)  theorems, and otherwise it follows from the deduction just obtained.

\subsection{Concluding arguments}  We can now use the results just obtained  to complete the proof of Theorems \ref{T.V} and  \ref{T.T}.

\begin{proof}[Proof of Theorem \ref{T.V} and Theorem  \ref{T.T}]  In view of Proposition \ref{P.P}, it suffices to prove Theorem \ref{T.V}. Let  $z_0\in V$ and let $A$ be an open disk with centre $z_0$, entirely contained in $V$ (which exists since $S(w)$ is holomorphic).
Recall that $V$ is stable under translation by the semigroup $\Gamma\subset \R$, so  in particular $A+\Gamma\subset V$.
Since $\Gamma$ is dense in $\R$ we deduce that $V$ contains a whole horizontal strip around the line $\Im w=\Im z_0$.
 In particular, $V$ contains the line $\Im w=\Im z_0$ itself.  Hence $V$ contains any line $\Im w=t_0 $ as soon as it contains a point in such a line. But since $V$ is invariant under translation by $c_{\pm 1}\neq 0$, thus by any multiple  $sc_{\pm 1}$ (any integer $s>0$, any sign),  the imaginary parts of elements of $V$ are unbounded both from above and from below. But the set of such imaginary parts is connected, hence is the whole $\R$. Combining this with the above deduction, we conclude that $V=\C$, as required.
\end{proof}

 \bigskip

 \begin{rem} \label{R.R}  Concerning the distribution of zeroes of $f(z)$, one could  also think  for instance of using criteria such as Theorem 15.23 of \cite{R}, which would predict a convergent sum $\sum(1-|\alpha_i|)$ for the zeros $\alpha_i$, {\it provided}  $\int_{0}^1\log^+|f(re(\theta))|\d\theta$ remained  bounded as $r\to 1^-$.

 However one may show that  this last condition does not hold  for the present function, that is, $f$ does not belong to the class denoted $N$ (Nevanlinna) in \cite{R}. We give a very brief sketch for this claim.  One first may expand  $f(re(\theta))$ for $r=e^{-c/2^n}$  ($1/2<c\le 1$), using the first $n$ terms and bounding the remainder by a constant (see e.g. \eqref{E.app} above).  Approximating  in turn $r^m=1+O(m/2^n)$,
 $m=1,2,\ldots ,2^{n-1}$,  for each of these $n$ terms, one is reduced to estimate the exponential sums $s(\theta)=s_n(\theta)=e(\theta)+e(2\theta)+\ldots +e(2^{n-1}\theta)$ in absolute value.  (See also \cite{Mah3} for such approximate formulae.)

 In conclusion,  if the said integral would be bounded then it is easy to see that,  in particular,  the measure of the set
  $\Aa=\Aa_n=\{\theta\in [0,1]: |s(\theta)|>\sqrt n/2\}$ would tend to $0$, hence would be
  $<\epsilon$, for any  prescribed $\epsilon >0$  and large enough $n$.
  Now, however, we easily find  the moments $\int_{0}^1|s(\theta)|^2\d\theta=n$, $\int_0^1|s(\theta)|^4\d\theta=2n^2-n$.  But on the one hand $\int_{\Aa}|s|^2\ge n-(\sqrt n/2)^2=3n/4$, on the other hand $(\int_{\Aa}|s|^2)^2\le \epsilon \int_{\Aa}|s^4|\le 2\epsilon n^2$. This is false for  $\epsilon<9/32$, proving the claim.

  We do not know if nevertheless  the series $\sum (1-|\alpha_i|)$ converges (though we are inclined to believe it does not).
  \end{rem}

 \begin{rem} \label{R.R2} Quite possibly there are other methods to prove  the results of the paper (and maybe more). For instance, one could apply Ahlfors' theory (see especially \cite{Hay}, Ch. 5). Also, on approximating $f(z)$ as indicated in the previous remark (or see \cite{Mah3}), one should obtain that for $r=e^{-c/4^n}$,  ($1/2<c\le 1$),  $f(re(\theta))$ is approximated by $s_{2n}(\theta)$ up to a bounded function. In turn, we have $s_{2n}(\theta)=s_n(\theta)+s_n(2^n\theta)$, and on changing $\theta$ to $\theta_a:=\theta+(a/2^n)$ for an arbitrary integer $a$, we get $s_{2n}(\theta_a)=s_n(\theta_a)+s_n(2^n\theta)$. On varying $a$, the first term essentially varies along the whole set of values of $s_n$, whereas the second term remains constant. So one  nearly reduces to study $s_n(u)+s_n(v)$ for {\it independent } variables $u,v$, and one can also iterate (and compare with a compatible probabilistic distribution).  This possibly leads to a better understanding of the image of $f(re(\theta))$ for fixed $r$ approaching $1$, and in turn to an understanding of the distribution of the values attained by the function.
 \end{rem}

\appendix
\setcounter{secnumdepth}{0}
\section{Appendix by Francesco Veneziano}

\subsection{Finding a zero of \texorpdfstring{$S(w)$}{S(w)}}
Here we will prove computationally that the holomorphic function $S(w)$ defined in~\eqref{E.S} and expressed by the series
\begin{equation*}\label{S.serie}
 S(w)=\sum_{n=0}^{\infty}\left( e(w 2^n)+e\left(\frac{w-1}{2^{n+1}}\right)-e\left(-\frac{1}{2^{n+1}}\right)\right)
\end{equation*}
has a zero in the upper-half plane. This fact, together with Proposition~\ref{P.P}, provides an alternative proof of the main assertion that the Fredholm series has infinitely many zeroes, independent from the study of the quantities $c_m$ carried out in Section~\ref{SS.cm}.

Thanks to uniform convergence on compacts, we can write
\begin{equation*}\label{S.derivate}
 S^{(k)}(w)=\sum_{n=0}^{\infty}\left((2^{n+1}\pi i)^k  e(w 2^n)+\left(\frac{\pi i}{2^{n}}\right)^k e\left(\frac{w-1}{2^{n+1}}\right)\right).
\end{equation*}
Writing $t=\Im w$ we have
\begin{equation}\label{S.derivate.abs}
 \abs{S^{(k)}(w)}\leq(2\pi)^k\sum_{n=0}^{\infty}2^{k n} e^{-2^{n+1}\pi t}+\pi^k\sum_{n=0}^{\infty}2^{-k n} e^{-\frac{\pi t}{2^n}}.
\end{equation}

Set now $a=\log_2\frac{k}{2\pi t}$, and let $n$ be an integer, $n\geq \ceiling{a}+1$.

By  standard calculus we find
 \begin{equation}\label{stima1}
  2^{k n}e^{-2^{n+1}\pi t}=e^{kn\log2-2^{n+1}\pi t}\leq e^{-\frac{k}{2}}e^{-\frac{(\log2)^2}{2}kn}.
 \end{equation}
Splitting the first sum of \eqref{S.derivate.abs} at the summand $\ceiling{a}$ we get
\begin{equation*}
 \abs{S^{(k)}(w)}\leq(2\pi)^k\sum_{n=0}^{\ceiling{a}}2^{k n} e^{-2^{n+1}\pi t}+(2\pi)^k\sum_{n=\ceiling{a}+1}^{\infty}2^{k n} e^{-2^{n+1}\pi t}+\pi^k\sum_{n=0}^{\infty}2^{-k n} e^{-\frac{\pi t}{2^n}}.
 \end{equation*}
 Estimating the first and third terms with geometrical sums, putting these terms together, and  using \eqref{stima1} to estimate the middle term  we get, for $k\ge 2$,

  \begin{equation*}
 \abs{S^{(k)}(w)}  \leq \frac{(2\pi)^k}{2^k-1}2^{k(\ceiling{a}+1)}+3^{k+1}\leq \frac{4}{3}(\pi 2^{a+2})^k+3^{k+1}.
\end{equation*}
From the definition of $a$ and the elementary inequality $k!\leq \frac{k^k}{e^{k-1}}$

we eventually  obtain
\begin{equation}\label{stima2}
 \abs{S^{(k)}(w)}\leq\left(\frac{2e}{t}\right)^k k! +3^{k+1}.
\end{equation}
Consider now a point $w_0$ on the upper-half plane, with imaginary part $t_0$. For a point $w$ at distance $\rho$ from $w_0$ using \eqref{stima2} we have that
\begin{align*}
 \abs{S(w)}&=\abs{\sum_{k=0}^\infty \frac{S^{(k)}(w_0)}{k!}(w-w_0)^k}\geq \abs{S'(w_0)}\rho-\abs{S(w_0)}-\sum_{k=2}^{\infty}\frac{\abs{S^{(k)}(w_0)}}{k!}\rho^k\\
 &\geq \abs{S'(w_0)}\rho-\abs{S(w_0)}-\sum_{k=2}^{\infty}\left(\frac{2e\rho}{t_0}\right)^k-3\sum_{k=2}^{\infty}\frac{(3\rho)^k}{k!}\\
 &=\abs{S'(w_0)}\rho-\abs{S(w_0)}-\frac{4e^2\rho^2}{t_0(t_0-2e\rho)}-3e^{3\rho}+9\rho+3.
\end{align*}
Now we can finally set $w_0=-0.177323882+0.144626388i$ and $\rho=0.002$ and check numerically that $\abs{S(w)}>\abs{S(w_0)}$ on the circle of centre $w_0$ and radius $\rho$, thus proving that within that disk lies a zero of $S(w)$.

\subsection{Some zeroes of the Fredholm series}
As shown, the Fredholm series has infinitely many zeroes. We list here the approximate values of a few of them, other than the zero at $z=0$ and the only other real zero at $z=-0.65862675\dotso$
\begin{footnotesize}
\begin{align*}
 -0.93753706&\pm0.33482855 i, & -0.92536435&\pm0.35488765 i, & -0.82600557&\pm0.55115255 i,\\
-0.68520628&\pm0.67053411 i, & -0.55421602&\pm0.82701155 i, & -0.36921064&\pm0.92165325 i,\\
0.12031484&\pm0.93460594 i, & 0.14635991&\pm0.98490932 i, & 0.18459117&\pm0.95835110 i,\\
0.39186275&\pm0.89825762 i, & 0.74745993&\pm0.65768729 i, & 0.77543377&\pm0.61813413 i.
\end{align*}
\end{footnotesize}
Of course the zeroes tend to accumulate towards the unit circle. Figure~\ref{zeri} shows the zeroes of the 13th partial sum which lie in the interior of the unit circle (there are 1126 of them); in order to visualize them better, we have rescaled their distances from the origin according to the hyperbolic metric in the Poincar{\'e} disk model.
In other words, each zero $\rho e^{i\theta}$ in the unit circle has been mapped in the picture to the point $\log\left(\frac{1+\rho}{1-\rho}\right) e^{i\theta}$.

\begin{figure}[h!]
 \includegraphics[scale=0.9]{Zeri_riscalati_13SommaParziale_1126pt_pt2.pdf}
 \caption{}\label{zeri}
\end{figure}

\vfill

Francesco Veneziano

Universit{\`a} di Genova

Via Dodecaneso 35, 16146 Genova, ITALY

e-mail: veneziano@dima.unige.it

\FloatBarrier


TITLE: Probability help (Conditional probability) - very basic
QUESTION [2 upvotes]: first of all I want to say hey to all! I have been a long time lurker and follower of these forums and have been known to find a lot of your answers helping me out throughout my college life.
I also know that you just don't like to spit out the answers for those who don't put in the effort so here it goes:
This is a very basic conditional probability question, but I can't seem to find a trick in solving these questions and they always get me. Any help would be greatly appreciated. The question is as follows:
The air rescue service of the armed forces divided the flight path of a downed plane into search sectors. In one mission a search plane will overfly every square kilometer of the designated sector. From past experience, one knows however that in one mission there is only a 50% chance of spotting a plane down in a sector of tundra. There is a 40% chance of spotting a plane down in a forested sector and a 15% chance of spotting a plane down in a lake sector. (a) A small plane is down in an area containing two sectors of tundra, one lake sector and one forested sector and a priori the plane has an equal chance of being down in any one of the fours sectors. The mission director decides to use the first mission to search the two sectors of tundra. What is the probability of spotting the downed plane?
(b) Given that we did not find the plane in the tundra sections, what is the probability that the plane is (i) in the tundra sections? (ii) in the forested section? (iii) in the lake sector?
Ok, so for a) what I did was basically the plane is going over the two sectors of Tundra which has a probability of 50% of containing the downed plane. So P(T) . P(T) would be 0.5*0.5 = 0.25 is the probability of finding the plane in the two sectors of Tundra, right?
for b) (i) I said P(Finding Plane | Tundra) = P(Finding Plane intersect Tundra) / P(Finding Plane) (ii) P(Finding Plane | Forest) = P(Finding Plane intersect Forest) / P(Finding Plane) (iii) P(Finding Plane | Lake) = P(Finding Plane intersect Lake) / P(Finding Plane)
P(Finding Plane) would be the total probability of finding the plane in each sector so given that it has an equal chance of being in either sector 2/4 * P(Tundra) + 1/4 * P(Forest) + 1/4*P(Lake)
I am not sure about my answers nor my approach to this problem. Any tricks or tips to solve this problem or conditional probability in general?

REPLY [1 votes]: If you are finding this kind of problem difficult or just want to check your results, you could try this technique.
Suppose this happened $400$ times (you don't have to use this number, but there are four equally likely sectors and you are dealing with percentages).  Then the plane would be in the tundra an expected $200$ times, in the lake $100$ times and in the forest $100$ times.  It would be found in the tundra an expected $100$ times.  $\dfrac{100}{400} = 0.25$ so you have the right result for (a).
So exclude the expected $100$ times it is found in the tundra on the first mission, leaving $300$, of which $100$ are in the tundra, $100$ in the lake and $100$ in the forest.  So for (b) each of these has a  $\dfrac{100}{300} \approx 0.333$ probability of being true.
If you want to do this formally then for (a) you have
$\Pr(\text{found in tundra|down in tundra})\Pr(\text{down in tundra}) = \dfrac{50}{100} \times \dfrac{2}{4} = \dfrac{1}{4}.$
For (b), one of the calculations is
$\Pr(\text{down in lake|not found in tundra}) =\dfrac{\Pr(\text{down in lake})\Pr(\text{not found in tundra|down in lake})}{\Pr(\text{not found in tundra})} $ $= \dfrac{\frac{1}{4} \times 1}{1 - \frac{1}{4}} = \dfrac{1}{3}.$


TITLE: Square integrable and related limit
QUESTION [1 upvotes]: Let a continuous function $x:[0,\infty)\rightarrow\mathbb{R}$. Does $x\in\mathcal{L}_2[0,\infty)$ (square integrable i.e. $\lim_{t\rightarrow\infty}\int_0^t{x^2(s)ds=c<\infty}$) implies $\lim_{t\rightarrow\infty}\int_0^t{e^{-\lambda(t-s)}x(s)ds}=0$ for every $\lambda>0$?
I can prove this  if $x$ is bounded but does it also hold true for unbounded x?
Note that $\int_0^t{e^{-\lambda(t-s)}x(s)ds}$ is a bounded also square integrable function if $x$ is square integrable and no boundedness assumption on $x$ is needed for this.

REPLY [0 votes]: I will elaborate on this. Following Ma's suggestion, from the Cauchy-Schwarz inequality we have
\begin{align}
\bigg|\int_{t}^{t+\delta}{e^{- \lambda(t+\delta-s)}x(s)ds}\bigg| &\leq \int_{t}^{t+\delta}{\big|e^{- \lambda(t+\delta-s)}x(s)\big|ds}\\
&\leq \sqrt{\int_{t}^{t+\delta}{e^{-2\lambda (t+\delta-s)}ds}\int_{t}^{t+\delta}{x^2(s)ds}}\\
&= \sqrt{\frac{1-e^{-2\lambda \delta}}{2\lambda}}\sqrt{\int_{t}^{t+\delta}{x^2(s)ds}}.
\end{align}
For a square integrable $x$ it holds true that $\lim_{t\rightarrow\infty}\int_t^{t+\delta}{x^2(s)ds}=0$.
Using this fact and taking the limit $t\rightarrow\infty$ in the inequality above we obtain $\lim_{t\rightarrow\infty}\int_{t}^{t+\delta}{e^{- \lambda(t+\delta-s)}x(s)ds}=0$.
We can write $\int_0^{t+\delta}{e^{-\lambda(t+\delta-s)}x(s)ds}=e^{-\lambda\delta}\int_0^t{e^{-\lambda (t-s)}x(s)ds}+\int_{t}^{t+\delta}{e^{- \lambda(t+\delta-s)}x(s)ds}$.
If we now define $S:=\lim_{t\rightarrow\infty}\int_0^t{e^{-\lambda (t-s)}x(s)ds}$ and take the limit in the identity above  we have $S=e^{-\lambda\delta}S$ that yields the desired $S=0$. However, one has to prove that the limit $S$ actually exists.


TITLE: What is the 'super' parameter of superconductivity and what is the role of Cooper pair?
QUESTION [2 upvotes]: Only thing I know about superconductors is that here the electrical current face zero resistance. My first question is what is 'super' (physical or mathematical entity) about a superconductor. Or more precisely if I start from the Hamiltonian for a system, which parameter I should try find out to see whether the system is superconducting or not?
I read about the mechanism involving the formation of Cooper pair. I understand how they are formed, but don't understand why the pair formation is required for superconductors. Or in other words, how does the pair formation lead to a dissipationless current?

REPLY [3 votes]: Historically, european were saying supra-conductors, not super-conductor (it is still supraleitung in german, supraconductivité in french, and something similar in russian, but I don't know cyrillic, have fun checking this issue using Wikipedia and changing the language). Only americans were saying super-conductors at the begining (I guess, since I do not really know why the name changed along time, could be interesting to ask on this website if someone knows...). Supra means beyond or something like that : english speakers still use the preffix supra in supra-natural for instance. One usually opposes supra-/super-conductors to the normal ones. So supra-/super-conductivity means the conductivity of the material is not the natural/normal one: supra-/super-conductors do not follow the Ohm's law $j=\sigma E$ relating the current $j$ to the electric field $E$ through the conductivity tensor $\sigma$, as we expect in normal materials. Historically, normal was refering to everything known at that time.
The conceptual reason behind the distinction between super- and normal- conductors, is that a perfect metal should not in principle follow the Ohm's law neither. A perfect/theoretical metal can conduct current perfectly, without resistance. But what was discovered was not only the absence of resistance of a metallic slab at low temperature, but the possibility for persistent current, even large ones. See the website of the Lorentz Institute for some historical perspectives about the discovery by Kammerling Onnes of the disappearance of the resistance and the persistent current. As already suggested in comments, people started to understand the super-conductivity phenomenology only when the Meissner-Ochsenfeld effect was discovered. Superconductors are not only conductors without resistance (as any perfect metal would be as well), they are perfect diamagnetic (they repulse the magentic field from the inside, whereas the perfect metal allow any magnetic field to penetrate the material).
The difference between a normal metal (one should say normal-conductor if we were smart enough to simplify our life) and a superconductor is indeed the charge carrier (I'll come back latter on the reason why I put indeed in italics). In normal metals, the electric current is identified as current of electrons, in superconductors it is a current of Cooper pairs. The Cooper pairs are formed at low temperatures. Nevertheless, one may argue that Cooper pairs are not intrinsically the origin of the superconductivity. Some people prefer to say that the Cooper instability is more important: the fact that the Fermi surface is unstable at low temperature, resulting in a gap at the Fermi energy. This is more or less a semantic problem, since the Cooper instability goes along with the creation of the Cooper pairs as far as we know today. So somehow the Cooper pairs are not required for superconductivity, but come with the Cooper instability. You can read this CERN courier by S. Weinberg about this topic for instance.
Now the subbtle point explaining why the charge carriers are indeed different in normal- and super-conductors. There is in fact no charge carrier in both... more details in a paper by Kivelson and Rohksar from 1990 - PRB 41:11693. But this is far beyond your question I guess.


TITLE: What are all of the possible orders of $K$? Justify your answer.
QUESTION [0 upvotes]: Let $K\le A_5$.  Assume that $K$ is cyclic.  What are all of the possible orders of $K$?  Justify your answer.
So I know that $|K|\in\{1,2,3,5\}$ but I'm not sure how to justify it.

REPLY [0 votes]: Suppose $K\leqslant A_5$; with $K$ cyclic. Then $K$ is generated by one element, $\sigma\in A_5$, so $K=\langle \sigma\rangle$. Now, it remains to determine the order of $\sigma$. Now, what are the possible cycle types of $\sigma$? Why does this give you the answer? For example, $\sigma$ can be the identity, so $|K|=1$. It can also be a product of disjoint transpositions, and $|K|=2$.
SPOILER

 Remember that the cycle type of an element in $S_5$ can be $(1),(12),(123),(1234)(12345)$,$(123)(45),(12)(34)$ but only $(1);(123),(12345),(12)(34)$ are in $A_5$. Thus $\sigma$ has cycle type $(1),(123),(12345),(12)(34)$. These elements have order $1,3,5,2$ respectively, and those are the possible orders.