text:
\begin{document}

\date{2020-6-30\\E-mail:  cuifangsun@163.com,\;\;  mengchixiong@126.com}

\maketitle

\begin{abstract}
For any positive integer $m$, let $\mathbb{Z}_{m}$ be the set of residue classes modulo $m$. For $A\subseteq \mathbb{Z}_{m}$ and $\overline{n}\in \mathbb{Z}_{m}$, let representation function $R_{A}(\overline{n})$ denote the number of solutions of the equation $\overline{n}=\overline{a}+\overline{a'}$ with ordered pairs $(\overline{a}, \overline{a'})\in A \times A$. In this paper, we determine all sets $A, B\subseteq \mathbb{Z}_{m}$ with $A\cup B=\mathbb{Z}_{m}$ and $|A\cap B|=2$ or $m-2$ such that  $R_{A}(\overline{n})=R_{B}(\overline{n})$ for all $\overline{n}\in \mathbb{Z}_{m}$. We also prove that if $m$ is a positive integer with $4|m$, then there exist two distinct sets $A, B\subseteq \mathbb{Z}_{m}$ with $A\cup B=\mathbb{Z}_{m}$ and $|A\cap B|=4$ or $m-4$, $B\neq A+\overline{\frac{m}{2}}$ such that $R_{A}(\overline{n})=R_{B}(\overline{n})$ for all $\overline{n}\in \mathbb{Z}_{m}$. If $m$ is a positive integer with $2\|m$, $A\cup B=\mathbb{Z}_{m}$ and $|A\cap B|=4$ or $m-4$, then $R_{A}(\overline{n})=R_{B}(\overline{n})$ for all $\overline{n}\in \mathbb{Z}_{m}$ if and only if $B=A+\overline{\frac{m}{2}}$.

\noindent{{\bf Keywords:}\hspace{2mm} Representation function, partition, residue class.}
\end{abstract}

\maketitle

\section{Introduction}
Let $\mathbb{N}$ be the set of all nonnegative integers. For $S\subseteq \mathbb{N}$ and $n\in S$, let the representation function $R'_{S}(n)$ denote the number of solutions of the equation $n=s+s'$ with $s, s'\in S$. S\'{a}rk\H{o}zy asked whether there exist two sets $A, B\subseteq \mathbb{N}$ with $|(A\cup B)\backslash (A\cap B)|=\infty$ such that $R'_{A}(n)=R'_{B}(n)$ for all sufficiently large integers $n$. In 2002, Dombi \cite{D} showed that the answer is negative. There are many other related results (see \cite{CB, K, KS, LT, TC, TL} and the references therein).

For a positive integer $m$, let $\mathbb{Z}_{m}$ be the set of residue classes modulo $m$. For any residue classes $\overline{a}, \overline{b}\in \mathbb{Z}_{m}$, there exist two integers $a', b'$ with $0\leq a', b'\leq m-1$ such that $\overline{a'}=\overline{a}$ and $\overline{b'}=\overline{b}$.  We define the ordering $\overline{a}\leq \overline{b}$ if $a'\leq b'$. For any $\overline{n}\in \mathbb{Z}_{m}$, without loss of generality, we may suppose that $0\leq n\leq m-1$. For $A\subseteq \mathbb{Z}_{m}$ and $\overline{n}\in \mathbb{Z}_{m}$, let $R_{A}(\overline{n})$ denote the number of solutions of $\overline{n}=\overline{a}+\overline{a'}$ with $\overline{a}, \overline{a'}\in A$. For $\overline{n}\in \mathbb{Z}_{m}$ and $A\subseteq \mathbb{Z}_{m}$, let $\overline{n}+A=\{\overline{n}+\overline{a}: \overline{a}\in A\}$. For $A, B\subseteq \mathbb{Z}_{m}$ and $\overline{n}\in \mathbb{Z}_{m}$, let $R_{A, B}(\overline{n})$ be the number of solutions of $\overline{n}=\overline{a}+\overline{b}$ with $\overline{a}\in A$ and $\overline{b}\in B$.
The characteristic function of $A\subseteq \mathbb{Z}_{m}$ is denoted by
$$\chi_{A}(n)=\begin{cases}1  &\overline{n}\in A,\\
0 & \overline{n}\not\in A.
\end{cases}$$

In 2012, Yang and Chen \cite{YC} determined all sets $A, B\subseteq \mathbb{Z}_{m}$ with $|(A\cup B)\backslash (A\cap B)|=m$ such that $R_{A}(\overline{n})=R_{B}(\overline{n})$ for all $\overline{n}\in \mathbb{Z}_{m}$. In 2017, Yang and Tang \cite{YT} determined all sets $A, B\subseteq \mathbb{Z}_{m}$ with $|(A\cup B)\backslash (A\cap B)|=2$ or $m-1$ such that  $R_{A}(\overline{n})=R_{B}(\overline{n})$ for all $\overline{n}\in \mathbb{Z}_{m}$. Yang and Tang \cite{YT} also posed the following problem for further research.

\begin{problem}\label{prob1}
 Given a positive even integer $m$ and an integer $k$ with $2\leq k\leq m-1$. Do there exist two distinct sets $A, B\subseteq \mathbb{Z}_{m}$ with $|A|=|B|=k$ and $B\neq A+\{\frac{m}{2}\}$ such that $R_{A}(\overline{n})=R_{B}(\overline{n})$ for all $\overline{n}\in \mathbb{Z}_{m}$?
\end{problem}

 In this paper, we consider for which positive even integers $m$ there exist two distinct sets $A, B\subseteq\mathbb{Z}_{m}$ with $A\cup B=\mathbb{Z}_{m}$ and $|A\cap B|=2, m-2, 4$ or $m-4$ such that $R_{A}(\overline{n})=R_{B}(\overline{n})$ for all $\overline{n}\in \mathbb{Z}_{m}$ and obtain the following results:

\begin{theorem}\label{thm1}
Let $m$ be a positive even integer. Let $A, B\subseteq \mathbb{Z}_{m}$ with $A\cup B=\mathbb{Z}_{m}$ and $|A\cap B|=2$ or $m-2$. Then $R_{A}(\overline{n})=R_{B}(\overline{n})$ for all $\overline{n}\in \mathbb{Z}_{m}$ if and only if $B=A+\overline{\frac{m}{2}}$.
\end{theorem}

\begin{theorem}\label{thm2}
Let $m$ be a positive integer with $4|m$. Then there exist two distinct sets $A, B\subseteq \mathbb{Z}_{m}$ with $A\cup B=\mathbb{Z}_{m}, B\neq A+\overline{\frac{m}{2}}$ and $|A\cap B|=4$ or $m-4$ such that $R_{A}(\overline{n})=R_{B}(\overline{n})$ for all $\overline{n}\in \mathbb{Z}_{m}$.
\end{theorem}

\begin{theorem}\label{thm3}
Let $m$ be a positive integer with $2\| m$. Let $A, B\subseteq \mathbb{Z}_{m}$ with $A\cup B=\mathbb{Z}_{m}$ and $|A\cap B|=4$ or $m-4$. Then $R_{A}(\overline{n})=R_{B}(\overline{n})$ for all $\overline{n}\in \mathbb{Z}_{m}$ if and only if $B=A+\overline{\frac{m}{2}}$.
\end{theorem}

\section{Lemmas}

\begin{lemma}\label{lem1}
Let $m$ be a positive even integer and $t$ be a positive integer with $t<\frac{m}{2}$. Let $A, B\subseteq \mathbb{Z}_{m}$ with $A\cup B=\mathbb{Z}_{m}$ and $|A\cap B|=2t$. If $R_{A}(\overline{n})=R_{B}(\overline{n})$ for all $\overline{n}\in \mathbb{Z}_{m}$, then $|A|=|B|=\frac{m}{2}+t$.
\end{lemma}

\begin{proof}
If $R_{A}(\overline{n})=R_{B}(\overline{n})$ for all $\overline{n}\in \mathbb{Z}_{m}$, then
$$|A|^{2}=\sum\limits_{\overline{n}\in \mathbb{Z}_{m}}R_{A}(\overline{n})=\sum\limits_{\overline{n}\in \mathbb{Z}_{m}}R_{B}(\overline{n})=|B|^{2}.$$
Thus $|A|=|B|$. Noting that
$$|A|+|B|=|A\cup B|+|A\cap B|=m+2t,$$
we have $|A|=|B|=\frac{m}{2}+t$.

This completes the proof of Lemma \ref{lem1}.
\end{proof}

\begin{lemma}\label{lem2}
Let $m$ be a positive even integer and $t$ be a positive integer with $t<\frac{m}{2}$. Let $A, B\subseteq \mathbb{Z}_{m}$ with $A\cup B=\mathbb{Z}_{m}$ and $A\cap B=\{\overline{r_{1}}, \overline{r_{2}}, \ldots, \overline{r_{2t}}\}$. If $R_{A}(\overline{n})=R_{B}(\overline{n})$ for all $\overline{n}\in \mathbb{Z}_{m}$, then
\begin{equation}\label{2.1}
\sum_{i=1}^{2t}\chi_{A}(n-r_{i})=t+\frac{1}{2}R_{\{\overline{r_{1}}, \overline{r_{2}}, \ldots, \overline{r_{2t}}\}}(\overline{n}).
\end{equation}
\end{lemma}

\begin{proof}
Noting that $B=(\mathbb{Z}_{m}\backslash A)\cup \{\overline{r_{1}}, \overline{r_{2}}, \ldots, \overline{r_{2t}}\}$, we have
\begin{eqnarray*}
R_{B}(\overline{n})&=& R_{\mathbb{Z}_{m}\backslash A}(\overline{n})+2R_{\mathbb{Z}_{m}\backslash A, \{\overline{r_{1}}, \overline{r_{2}}, \ldots, \overline{r_{2t}}\}}(\overline{n})+
 R_{\{\overline{r_{1}}, \overline{r_{2}}, \ldots, \overline{r_{2t}}\}}(\overline{n})\\
&=& |\{(a, a'): \overline{a}, \overline{a'}\in\mathbb{Z}_{m}\backslash A, 0\leq a, a'\leq m-1, a+a'=n \text{ or } a+a'=n+m\}|\\
&& \hspace{2mm}+2\sum\limits_{i=1}^{2t}(1-\chi_{A}(n-r_{i}))+R_{\{\overline{r_{1}}, \overline{r_{2}}, \ldots, \overline{r_{2t}}\}}(\overline{n})
\end{eqnarray*}
\begin{eqnarray*}
&=& \sum\limits_{0\leq i\leq \frac{n}{2}}(1-\chi_{A}(i))(1-\chi_{A}(n-i))+\sum\limits_{n+1\leq i\leq \frac{n+m}{2}}(1-\chi_{A}(i))(1-\chi_{A}(n-i))\\
&&\hspace{2mm} +\sum\limits_{0\leq i<\frac{n}{2}}(1-\chi_{A}(i))(1-\chi_{A}(n-i))+\sum\limits_{n+1\leq i<\frac{n+m}{2}}(1-\chi_{A}(i))(1-\chi_{A}(n-i))\\
&&\hspace{2mm} +2\sum\limits_{i=1}^{2t}(1-\chi_{A}(n-r_{i}))+R_{\{\overline{r_{1}}, \overline{r_{2}}, \ldots, \overline{r_{2t}}\}}(\overline{n})\\
&=& \sum\limits_{0\leq i\leq \frac{n}{2}}1-\sum\limits_{0\leq i\leq n}\chi_{A}(i)-\chi_{A}(\frac{n}{2})+\sum\limits_{0\leq i\leq \frac{n}{2}}\chi_{A}(i)\chi_{A}(n-i)\\
&&\hspace{2mm} +\sum\limits_{n+1\leq i\leq \frac{n+m}{2}}1-\sum\limits_{n+1\leq i\leq m-1}\chi_{A}(i)-\chi_{A}(\frac{n+m}{2})+\sum\limits_{n+1\leq i\leq \frac{n+m}{2}}\chi_{A}(i)\chi_{A}(n-i)\\
&&\hspace{2mm} +\sum\limits_{0\leq i<\frac{n}{2}}1-\sum\limits_{0\leq i\leq n}\chi_{A}(i)+\chi_{A}(\frac{n}{2})+\sum\limits_{0\leq i<\frac{n}{2}}\chi_{A}(i)\chi_{A}(n-i)\\
&&\hspace{2mm} +\sum\limits_{n+1\leq i<\frac{n+m}{2}}1-\sum\limits_{n+1\leq i\leq m-1}\chi_{A}(i)+\chi_{A}(\frac{n+m}{2})+\sum\limits_{n+1\leq i< \frac{n+m}{2}}\chi_{A}(i)\chi_{A}(n-i)\\
&&\hspace{2mm} +2\sum\limits_{i=1}^{2t}(1-\chi_{A}(n-r_{i}))+R_{\{\overline{r_{1}}, \overline{r_{2}}, \ldots, \overline{r_{2t}}\}}(\overline{n})\\
&=& \left[\frac{n}{2}\right]+1+\left[\frac{n+m}{2}\right]-n+\left[\frac{n-1}{2}\right]+1+\left[\frac{n+m-1}{2}\right]-n-2|A|
\\
&&\hspace{2mm}+R_{A}(\overline{n})+2\sum\limits_{i=1}^{2t}(1-\chi_{A}(n-r_{i}))+R_{\{\overline{r_{1}}, \overline{r_{2}}, \ldots, \overline{r_{2t}}\}}(\overline{n})\\
&=& -2t+R_{A}(\overline{n})+2\sum\limits_{i=1}^{2t}(1-\chi_{A}(n-r_{i}))+R_{\{\overline{r_{1}}, \overline{r_{2}}, \ldots, \overline{r_{2t}}\}}(\overline{n})\\
&=& 2t+R_{A}(\overline{n})-2\sum\limits_{i=1}^{2t}\chi_{A}(n-r_{i})+R_{\{\overline{r_{1}}, \overline{r_{2}}, \ldots, \overline{r_{2t}}\}}(\overline{n}).
\end{eqnarray*}
Since $R_{A}(\overline{n})=R_{B}(\overline{n})$ for all $\overline{n}\in \mathbb{Z}_{m}$, we have
$$ \sum\limits_{i=1}^{2t}\chi_{A}(n-r_{i})=t+\frac{1}{2}R_{\{\overline{r_{1}}, \overline{r_{2}}, \ldots, \overline{r_{2t}}\}}(\overline{n}).$$

This completes the proof of Lemma \ref{lem2}.
\end{proof}

\section{Proofs}

\begin{proof}[{\bf Proof of  Theorem 1.1}]
It is clear that $R_{A}(\overline{n})=R_{B}(\overline{n})$ for all $\overline{n}\in \mathbb{Z}_{m}$ if $B=A+\overline{\frac{m}{2}}$.
Now we suppose that $R_{A}(\overline{n})=R_{B}(\overline{n})$ for all $\overline{n}\in \mathbb{Z}_{m}$. Clearly, the result is true for $m=2$. Now we may assume that $m\geq 4$.

{\bf Case 1.} $|A\cap B|=2$. Let $A\cap B=\{\overline{r_{1}}, \overline{r_{2}}\}$ with $\overline{r_{1}}\neq \overline{r_{2}}$. By choosing $n=2r_{1}$ in (\ref{2.1}), we have
$$ 1+\chi_{A}(2r_{1}-r_{2})=\chi_{A}(r_{1})+\chi_{A}(2r_{1}-r_{2})=1+\frac{1}{2}R_{\{\overline{r_{1}}, \overline{r_{2}}\}}(\overline{2r_{1}})\geq \frac{3}{2}.$$
Then $\chi_{A}(2r_{1}-r_{2})=1$ and $\overline{r_{2}}=\overline{r_{1}}+\overline{\frac{m}{2}}$. Let $\overline{k}\in A$ with $\overline{k}\neq \overline{r_{1}}$ and $\overline{k}\neq \overline{r_{2}}$. By choosing $n=k+r_{1}$ in (\ref{2.1}), we have
$$ \chi_{A}(k)+\chi_{A}(k+\frac{m}{2})=1+\frac{1}{2}R_{\{\overline{r_{1}}, \overline{r_{2}}\}}(\overline{k+r_{1}})=1.$$
It follows that $B=A+\overline{\frac{m}{2}}$.

{\bf Case 2.} $|A\cap B|=m-2$. Let $A\cap B=T, A=\{\overline{a}\}\cup T, B=\{\overline{b}\}\cup T$ with $\overline{a}\neq \overline{b}$ and $\overline{a}, \overline{b}\not\in T$. For any $\overline{n}\in \mathbb{Z}_{m}$, we have
\begin{eqnarray*}
R_{A}(\overline{n})&=& R_{T}(\overline{n})+2R_{T, \{\overline{a}\}}(\overline{n})+R_{\{\overline{a}\}}(\overline{n})= R_{T}(\overline{n})+2\chi_{T}(n-a)+R_{\{\overline{a}\}}(\overline{n});\\
R_{B}(\overline{n})&=& R_{T}(\overline{n})+2R_{T, \{\overline{b}\}}(\overline{n})+R_{\{\overline{b}\}}(\overline{n})= R_{T}(\overline{n})+2\chi_{T}(n-b))+R_{\{\overline{b}\}}(\overline{n}).
\end{eqnarray*}
Then
\begin{equation}\label{3.1}
 2\chi_{T}(n-a)+R_{\{\overline{a}\}}(\overline{n})=2\chi_{T}(n-b)+R_{\{\overline{b}\}}(\overline{n}).
\end{equation}
By choosing $n=2b$ in (\ref{3.1}), we have $$2\chi_{T}(2b-a)+R_{\{\overline{a}\}}(\overline{2b})=2\chi_{T}(b)+R_{\{\overline{b}\}}(\overline{2b})=1.$$
Then $2\chi_{T}(2b-a)=0$ and $R_{\{\overline{a}\}}(\overline{2b})=1$. Thus $\overline{b}=\overline{a}+\overline{\frac{m}{2}}$. It follows that $B=A+\overline{\frac{m}{2}}$.

This completes the proof of Theorem \ref{thm1}.
\end{proof}

\begin{proof}[{\bf Proof of  Theorem 1.2}]  Let $m=4k$ with $k$ a positive integer.

{\bf Case 1.} $|A\cap B|=4$. Noting that $A\neq B$, we have $k\geq 2$. Let
\begin{eqnarray*}
A&=&\{\overline{0}, \overline{1}, \ldots, \overline{k-1}, \overline{k}\}\cup \{\overline{2k}, \overline{2k+1}, \ldots, \overline{3k-1}, \overline{3k}\},\\
B&=&\{\overline{k}, \overline{k+1}, \ldots, \overline{2k-1}, \overline{2k}\}\cup \{\overline{3k}, \overline{3k+1}, \ldots, \overline{4k-1}, \overline{0}\}.
\end{eqnarray*}
It is clear that $A, B\subseteq \mathbb{Z}_{m}$ with $A\cup B=\mathbb{Z}_{m}, B\neq A+\overline{\frac{m}{2}}$ and $A\cap B=\{\overline{0}, \overline{k}, \overline{2k}, \overline{3k}\}$. Let $S=\{\overline{0}, \overline{1}, \ldots, \overline{k-1}, \overline{k}\}$. Then $A=S\cup (S+\overline{2k})$ and $B=(S+\overline{k})\cup (S+\overline{3k})$.
Noting that
\begin{eqnarray*}
R_{S}(\overline{n})&=& R_{S+\overline{2k}}(\overline{n})=R_{S+\overline{k}, S+\overline{3k}}(\overline{n}),\\
R_{S+\overline{k}}(\overline{n})&=& R_{S, S+\overline{2k}}(\overline{n})=R_{S+\overline{3k}}(\overline{n})
\end{eqnarray*}
for all $\overline{n}\in \mathbb{Z}_{m}$,  we have
\begin{eqnarray*}
R_{A}(\overline{n})&=& R_{S}(\overline{n})+2R_{S, S+\overline{2k}}(\overline{n})+R_{S+\overline{2k}}(\overline{n})\\
&=& R_{S+\overline{k}}(\overline{n})+2R_{S+\overline{k}, S+\overline{3k}}(\overline{n})+R_{S+\overline{3k}}(\overline{n})\\
&=& R_{B}(\overline{n}).
\end{eqnarray*}

{\bf Case 2.} $|A\cap B|=m-4$. If $k=1$, then by choosing $A=\{\overline{0}, \overline{2}\}, B=\{\overline{1}, \overline{3}\}$, we have $A\cup B=\mathbb{Z}_{4}, B\neq A+\overline{2}$ and $R_{A}(\overline{n})=R_{B}(\overline{n})$ for all $\overline{n}\in \mathbb{Z}_{4}$. Now let $k\geq 2$ and
\begin{eqnarray*}
A&=&\{\overline{0}, \overline{1}, \ldots, \overline{k-1}\}\cup \{\overline{k+1}, \overline{k+2}, \ldots, \overline{3k-1}\}\cup \{\overline{3k+1}, \overline{3k+2}, \ldots, \overline{4k-1}\},\\
B&=&\{\overline{1}, \overline{2}, \ldots, \overline{2k-1}\}\cup \{\overline{2k+1}, \overline{2k+2}, \ldots, \overline{4k-1}\}.
\end{eqnarray*}
It is clear that $A\cup B=\mathbb{Z}_{m}, B\neq A+\overline{\frac{m}{2}}$ and
$$A\cap B=\{\overline{1}, \ldots, \overline{k-1}\}\cup \{\overline{k+1}, \ldots, \overline{2k-1}\}\cup \{\overline{2k+1}, \ldots, \overline{3k-1}\}\cup \{\overline{3k+1}, \ldots, \overline{4k-1}\}.$$
Noting that $R_{A\cap B, \{\overline{0}, \overline{2k}\}}(\overline{n})=R_{A\cap B, \{\overline{k}, \overline{3k}\}}(\overline{n})$ and $R_{\{\overline{0}, \overline{2k}\}}(\overline{n})=R_{\{\overline{k}, \overline{3k}\}}(\overline{n})$ for all $\overline{n}\in \mathbb{Z}_{m}$,  we have
\begin{eqnarray*}
R_{A}(\overline{n})&=& R_{\{\overline{0}, \overline{2k}\}}(\overline{n})+2R_{A\cap B, \{\overline{0}, \overline{2k}\}}(\overline{n})+R_{A\cap B}(\overline{n})\\
&=& R_{\{\overline{k}, \overline{3k}\}}(\overline{n})+2R_{A\cap B, \{\overline{k}, \overline{3k}\}}(\overline{n})+R_{A\cap B}(\overline{n})\\
&=& R_{B}(\overline{n}).
\end{eqnarray*}

This completes the proof of Theorem \ref{thm2}.
\end{proof}

\begin{proof}[{\bf Proof of  Theorem 1.3}]
It is clear that $R_{A}(\overline{n})=R_{B}(\overline{n})$ for all $\overline{n}\in \mathbb{Z}_{m}$ if $B=A+\overline{\frac{m}{2}}$.
Now we suppose that $R_{A}(\overline{n})=R_{B}(\overline{n})$ for all $\overline{n}\in \mathbb{Z}_{m}$. By $A\neq B$ and $2\|m$, we have $m\geq 6$.

{\bf Case 1.} $|A\cap B|=4$. Let $A\cap B=\{\overline{r_{1}}, \overline{r_{2}}, \overline{r_{3}}, \overline{r_{4}}\}$. By Lemma \ref{lem2}, we have
\begin{equation}\label{3.2}
\chi_{A}(n-r_{1})+\chi_{A}(n-r_{2})+\chi_{A}(n-r_{3})+\chi_{A}(n-r_{4})=2+\frac{1}{2}R_{\{\overline{r_{1}}, \overline{r_{2}}, \overline{r_{3}}, \overline{r_{4}}\}}(\overline{n}).
\end{equation}
Then $2|R_{\{\overline{r_{1}}, \overline{r_{2}}, \overline{r_{3}}, \overline{r_{4}}\}}(\overline{n})$ for any $\overline{n}\in \mathbb{Z}_{m}$. Thus $2|R_{\{\overline{r_{1}}, \overline{r_{2}}, \overline{r_{3}}, \overline{r_{4}}\}}(\overline{2r_{i}})$ for $i\in\{1, 2, 3, 4\}$. Without loss of generality, we may assume that
$\overline{r_{3}}=\overline{r_{1}}+\overline{\frac{m}{2}}$ and $\overline{r_{4}}=\overline{r_{2}}+\overline{\frac{m}{2}}$. Clearly, the result is true for $m=6$. Now
let $m=2k$ with $2\nmid k$ and $k\geq 5$. Let $$l=\min\{i\geq 2: \{\overline{r_{2}}, \overline{r_{2}}+\overline{\frac{m}{2}}\}=\{\overline{ir_{1}-(i-1)r_{2}}, \overline{ir_{1}-(i-1)r_{2}}+\overline{\frac{m}{2}}\}\}.$$
By $2\|m$, we have $3\leq l\leq k$ and $l|k$. Now we discuss the following two subcases according to $l$.

{\bf Subcase 1.1 } $l=k$. Then
$$\mathbb{Z}_{m}=\bigcup\limits_{i=1}^{k}\{\overline{ir_{1}-(i-1)r_{2}}, \overline{ir_{1}-(i-1)r_{2}}+\overline{\frac{m}{2}}\}.$$
By choosing $n-r_{2}=ir_{1}-(i-1)r_{2}$ for $i=1, 2, \ldots, k$ in (\ref{3.2}), we have
\begin{eqnarray*}
& &\chi_{A}(r_{2})+\chi_{A}(r_{2}+\frac{m}{2})+\chi_{A}(r_{1})+\chi_{A}(r_{1}+\frac{m}{2})=4,\\
& &\chi_{A}(r_{1})+\chi_{A}(r_{1}+\frac{m}{2})+\chi_{A}(2r_{1}-r_{2})+\chi_{A}(2r_{1}-r_{2}+\frac{m}{2})=3,\\
& &\chi_{A}(2r_{1}-r_{2})+\chi_{A}(2r_{1}-r_{2}+\frac{m}{2})+\chi_{A}(3r_{1}-2r_{2})+\chi_{A}(3r_{1}-2r_{2}+\frac{m}{2})=2,\\
&&\cdots\\
&& \chi_{A}((k-2)r_{1}-(k-3)r_{2})+\chi_{A}((k-2)r_{1}-(k-3)r_{2}+\frac{m}{2})\\
&&\hspace{10mm} +\chi_{A}((k-1)r_{1}-(k-2)r_{2})+\chi_{A}((k-1)r_{1}-(k-2)r_{2}+\frac{m}{2})=2,\\
&& \chi_{A}((k-1)r_{1}-(k-2)r_{2})+\chi_{A}((k-1)r_{1}-(k-2)r_{2}+\frac{m}{2})+\chi_{A}(r_{2})+\chi_{A}(r_{2}+\frac{m}{2})=3.
\end{eqnarray*}
Noting that
$$ \chi_{A}(r_{2})+\chi_{A}(r_{2}+\frac{m}{2})=\chi_{A}(r_{1})+\chi_{A}(r_{1}+\frac{m}{2})=2,$$
we have $$ \chi_{A}(ir_{1}-(i-1)r_{2})+\chi_{A}(ir_{1}-(i-1)r_{2}+\frac{m}{2})=1$$
for $i=2, 3, \ldots, k-1$. It follows that $B=A+\overline{\frac{m}{2}}$.

{\bf Subcase 1.2 } $3\leq l<k$. Then $k=ls$ with $s\geq 3$ and $2\nmid s$. Thus

$$\mathbb{Z}_{m}=\bigcup\limits_{i=1}^{l}\bigcup\limits_{j=0}^{s-1}\{\overline{ir_{1}-(i-1)r_{2}+j}, \overline{ir_{1}-(i-1)r_{2}+j}+\overline{\frac{m}{2}}\}.$$
By choosing $n-r_{2}=ir_{1}-(i-1)r_{2}$ for $i=1, 2, \ldots, l$ in (\ref{3.2}), we have
\begin{eqnarray*}
& &\chi_{A}(r_{2})+\chi_{A}(r_{2}+\frac{m}{2})+\chi_{A}(r_{1})+\chi_{A}(r_{1}+\frac{m}{2})=4,\\
& &\chi_{A}(r_{1})+\chi_{A}(r_{1}+\frac{m}{2})+\chi_{A}(2r_{1}-r_{2})+\chi_{A}(2r_{1}-r_{2}+\frac{m}{2})=3, \\
& &\chi_{A}(2r_{1}-r_{2})+\chi_{A}(2r_{1}-r_{2}+\frac{m}{2})+\chi_{A}(3r_{1}-2r_{2})+\chi_{A}(3r_{1}-2r_{2}+\frac{m}{2})\\
& & = 2+\frac{1}{2}R_{\{\overline{r_{1}}, \overline{r_{1}+\frac{m}{2}}, \overline{r_{2}}, \overline{r_{2}+\frac{m}{2}}\}}(\overline{3r_{1}-r_{2}}),\\
& &\cdots \\
& &\chi_{A}((l-1)r_{1}-(l-2)r_{2})+\chi_{A}((l-1)r_{1}-(l-2)r_{2}+\frac{m}{2})+\chi_{A}(r_{2})+\chi_{A}(r_{2}+\frac{m}{2})=3.
\end{eqnarray*}
Noting that
$$ \chi_{A}(r_{2})+\chi_{A}(r_{2}+\frac{m}{2})=\chi_{A}(r_{1})+\chi_{A}(r_{1}+\frac{m}{2})=2,$$
we have $$ \chi_{A}(ir_{1}-(i-1)r_{2})+\chi_{A}(ir_{1}-(i-1)r_{2}+\frac{m}{2})=1$$
for $i=2, \ldots, l-1$.

For any $j\in\{1, 2, \ldots, s-1\}$, by choosing $n-r_{2}=ir_{1}-(i-1)r_{2}+j$ for $i=1, 2, \ldots, l$ in (\ref{3.2}), we have
\begin{eqnarray*}
& &\chi_{A}(r_{2}+j)+\chi_{A}(r_{2}+j+\frac{m}{2})+\chi_{A}(r_{1}+j)+\chi_{A}(r_{1}+j+\frac{m}{2})=2,\\
& &\chi_{A}(r_{1}+j)+\chi_{A}(r_{1}+j+\frac{m}{2})+\chi_{A}(2r_{1}-r_{2}+j)+\chi_{A}(2r_{1}-r_{2}+j+\frac{m}{2})=2,\\
& &\chi_{A}(2r_{1}-r_{2}+j)+\chi_{A}(2r_{1}-r_{2}+j+\frac{m}{2})+\chi_{A}(3r_{1}-2r_{2}+j)\\
 & &\hspace{10mm}+\chi_{A}(3r_{1}-2r_{2}+j+\frac{m}{2})=2,\\
& &\cdots\\
& & \chi_{A}((l-1)r_{1}-(l-2)r_{2}+j)+\chi_{A}((l-1)r_{1}-(l-2)r_{2}+j+\frac{m}{2})\\
& & \hspace{10mm} +\chi_{A}(r_{2}+j)+\chi_{A}(r_{2}+j+\frac{m}{2})=2
\end{eqnarray*}
Noting that $2\nmid l$, we have
$$ \chi_{A}(ir_{1}-(i-1)r_{2}+j)+\chi_{A}(ir_{1}-(i-1)r_{2}+j+\frac{m}{2})=1$$
for $i=1, 2, \ldots, l-1$. It follows that $B=A+\overline{\frac{m}{2}}$.

{\bf Case 2.} $|A\cap B|=m-4$. Let $A\cap B=T, A=\{\overline{a_{1}}, \overline{a_{2}}\}\cup T, B=\{\overline{b_{1}}, \overline{b_{2}}\}\cup T$ with $\{\overline{a_{1}}, \overline{a_{2}}\}\cap \{\overline{b_{1}}, \overline{b_{1}}\}=\varnothing$ and $\{\overline{a_{1}}, \overline{a_{2}}, \overline{b_{1}}, \overline{b_{2}}\}\cap T=\varnothing$. For any $\overline{n}\in \mathbb{Z}_{m}$, we have
\begin{eqnarray*}
R_{A}(\overline{n})&=& R_{T}(\overline{n})+2R_{T, \{\overline{a_{1}}, \overline{a_{2}}\}}(\overline{n})+R_{\{\overline{a_{1}}, \overline{a_{2}}\}}(\overline{n})\\
&=& R_{T}(\overline{n})+2\chi_{T}(n-a_{1})+2\chi_{T}(n-a_{2})+R_{\{\overline{a_{1}}, \overline{a_{2}}\}}(\overline{n});\\
R_{B}(\overline{n})&=& R_{T}(\overline{n})+2R_{T, \{\overline{b_{1}}, \overline{b_{2}}\}}(\overline{n})+R_{\{\overline{b_{1}}, \overline{b_{2}}\}}(\overline{n})\\
&=& R_{T}(\overline{n})+2\chi_{T}(n-b_{1})+2\chi_{T}(n-b_{2})+R_{\{\overline{b_{1}}, \overline{b_{2}}\}}(\overline{n}).
\end{eqnarray*}
Then
\begin{equation}\label{3.3}
2\chi_{T}(n-a_{1})+2\chi_{T}(n-a_{2})+R_{\{\overline{a_{1}}, \overline{a_{2}}\}}(\overline{n})=2\chi_{T}(n-b_{1})+2\chi_{T}(n-b_{2})+R_{\{\overline{b_{1}}, \overline{b_{2}}\}}(\overline{n}).
\end{equation}

By choosing $n=a_{1}+a_{2}$ in (\ref{3.3}), we have
$$2\chi_{T}(a_{1}+a_{2}-b_{1})+2\chi_{T}(a_{1}+a_{2}-b_{2})+R_{\{\overline{b_{1}}, \overline{b_{2}}\}}(\overline{a_{1}}+\overline{a_{2}})=2.$$
If $\chi_{T}(a_{1}+a_{2}-b_{1})=1$, then
$$\chi_{T}(a_{1}+a_{2}-b_{2})=R_{\{\overline{b_{1}}, \overline{b_{2}}\}}(\overline{a_{1}}+\overline{a_{2}})=0.$$
Thus $\overline{a_{1}}+\overline{a_{2}}-\overline{b_{2}}\in \{\overline{b_{1}}, \overline{b_{2}}\}$ and $\overline{a_{1}}+\overline{a_{2}}\not\in \{\overline{2b_{1}}, \overline{b_{1}}+\overline{b_{2}}, \overline{2b_{2}}\}$, which is a contradiction. It follows that $\chi_{T}(a_{1}+a_{2}-b_{1})=0$. Similarly, we can get $\chi_{T}(a_{1}+a_{2}-b_{2})=0$. Therefore $R_{\{\overline{b_{1}}, \overline{b_{2}}\}}(\overline{a_{1}}+\overline{a_{2}})=2$. It means that $\overline{a_{1}}+\overline{a_{2}}=\overline{2b_{1}}=\overline{2b_{2}}$ or $\overline{a_{1}}+\overline{a_{2}}=\overline{b_{1}}+\overline{b_{2}}$.

By choosing $n=b_{1}+b_{2}$ in (\ref{3.3}), we have
$$2\chi_{T}(b_{1}+b_{2}-a_{1})+2\chi_{T}(b_{1}+b_{2}-a_{2})+R_{\{\overline{a_{1}}, \overline{a_{2}}\}}(\overline{b_{1}}+\overline{b_{2}})=2.$$
If $\chi_{T}(b_{1}+b_{2}-a_{1})=1$, then
$$\chi_{T}(b_{1}+b_{2}-a_{2})=R_{\{\overline{a_{1}}, \overline{a_{2}}\}}(\overline{b_{1}}+\overline{b_{2}})=0.$$
Thus $\overline{b_{1}}+\overline{b_{2}}-\overline{a_{2}}\in \{\overline{a_{1}}, \overline{a_{2}}\}$ and $\overline{b_{1}}+\overline{b_{2}}\not\in \{\overline{2a_{1}}, \overline{a_{1}}+\overline{a_{2}}, \overline{2a_{2}}\}$, which is a contradiction. It follows that $\chi_{T}(b_{1}+b_{2}-a_{1})=0$. Similarly, we can get $\chi_{T}(b_{1}+b_{2}-a_{2})=0$. Therefore $R_{\{\overline{a_{1}}, \overline{a_{2}}\}}(\overline{b_{1}}+\overline{b_{2}})=2$. It means that $\overline{b_{1}}+\overline{b_{2}}=\overline{2a_{1}}=\overline{2a_{2}}$ or $\overline{b_{1}}+\overline{b_{2}}=\overline{a_{1}}+\overline{a_{2}}$.

If $\overline{a_{1}}+\overline{a_{2}}=\overline{2b_{1}}=\overline{2b_{2}}$, then $\overline{b_{1}}+\overline{b_{2}}=\overline{2a_{1}}=\overline{2a_{2}}$. Thus
$\overline{a_{2}}=\overline{a_{1}}+\overline{\frac{m}{2}}, \overline{b_{2}}=\overline{b_{1}}+\overline{\frac{m}{2}}$ and $\overline{2a_{1}}+\overline{\frac{m}{2}}=\overline{2b_{1}}$, which contradicts $2\| m$. It follows that $\overline{a_{1}}+\overline{a_{2}}=\overline{b_{1}}+\overline{b_{2}}$. By choosing $n=2a_{1}$ in (\ref{3.3}), we have
\begin{equation}\label{3.4}
2\chi_{T}(2a_{1}-a_{2})+R_{\{\overline{a_{1}}, \overline{a_{2}}\}}(\overline{2a_{1}})=2\chi_{T}(2a_{1}-b_{1})+2\chi_{T}(2a_{1}-b_{2})+R_{\{\overline{b_{1}}, \overline{b_{2}}\}}(\overline{2a_{1}}).
\end{equation}

If $R_{\{\overline{a_{1}}, \overline{a_{2}}\}}(\overline{2a_{1}})=2$, then $\overline{2a_{1}}=\overline{2a_{2}}$ and $\chi_{T}(2a_{1}-a_{2})=\chi_{T}(a_{2})=0$.
Thus $R_{\{\overline{b_{1}}, \overline{b_{2}}\}}(\overline{2a_{1}})=0$ and $\chi_{T}(2a_{1}-b_{1})=\chi_{T}(2a_{1}-b_{2})=1$, which contradicts (\ref{3.4}).

If $R_{\{\overline{a_{1}}, \overline{a_{2}}\}}(\overline{2a_{1}})=1$, then $\overline{2a_{1}}\neq \overline{2a_{2}}$. By (\ref{3.4}), we have $R_{\{\overline{b_{1}}, \overline{b_{2}}\}}(\overline{2a_{1}})=1$. Then $\overline{2b_{1}}\neq\overline{2b_{2}}$ and $\overline{2a_{1}}\in\{\overline{2b_{1}}, \overline{2b_{2}}\}$.
Without loss of generality, we may assume that $\overline{2a_{1}}=\overline{2b_{1}}$. Then $\overline{b_{1}}=\overline{a_{1}}+\overline{\frac{m}{2}}$ and $\overline{b_{2}}=\overline{a_{2}}+\overline{\frac{m}{2}}$. It follows that $B=A+\overline{\frac{m}{2}}$.

This completes the proof of Theorem \ref{thm3}.
\end{proof}


TITLE: Does extreme cold make **everything** extremely brittle?
QUESTION [24 upvotes]: First of all, I'm genuinely sorry if this question isn't "serious" enough for this forum!
A common cliche in movies and tv is that a very tough object (eg the villain) is frozen, and then hit with something, shattering into a million pieces.
I've seen a demo of a flower being put into liquid nitrogen, then being crumbled, but a flower is a very delicate object to start off with.  If I take a leg of lamb (for example) out of the freezer, I don't feel like it's in any danger of shattering into a million bits (unlike my foot if i were to drop it).
So, is the whole "cold = brittle" thing just movie bullcrap?  Or is there anything to it?  Sticking with the leg of lamb example:  is there a temperature to which a leg of lamb could be dropped that would make the leg of lamb prone to shattering?
EDIT - i just realised that the question title could be read as "Is there anything which is rendered extremely brittle by extreme cold?".  Obviously there are some things, eg flowers.  Hence the title change.

REPLY [2 votes]: Great answer found here:

Ductile and Brittle Failure of Materials
When a stress is applied to any object, it deforms, i.e., changes shape and/or size. This deformation is called elastic if the object returns to its original shape after the applied stress has been removed. Deformation that is permanent is called plastic deformation.
All materials can undergo only a limited amount of elastic deformation, after which either plastic deformation sets in or the material fractures.
Materials that fracture without any plastic deformation are called brittle materials. Examples include glass and most other ceramic materials.
Ductile materials undergo plastic deformation before fracture. Examples include aluminum, copper, steel and many metals, as well as polyethylene, nylon and many other polymers.
A number of factors determine whether a material is going to behave in a ductile or brittle manner. Among these factors are
• The structure and composition of the material, i.e., what are the atoms that make up the material, how are they bonded to each other, are there impurities, etc.
• The rate at which the material is deformed
• The temperature at which the material is deformed
Generally high rates of deformation and low temperatures promote brittle fracture.
Brittle failure usually occurs very rapidly and can be catastrophic. Many materials which are ductile at high temperatures become brittle when cooled below a critical temperature. This temperature is called the ductile to brittle transition temperature (DBTT) for metals and the glass transition temperature (Tg) for polymers.


TITLE: Age of a black hole
QUESTION [1 upvotes]: Is there a way to measure the age of a black hole by find Hawking radiation or calculating the stable orbits around the black hole?

REPLY [0 votes]: No. The Hawking radiation could tell us the black hole's remaining lifetime, or equivalently how much mass it has. But that doesn't tell us anything of the following (which are all equivalent): how much mass it's shed; how much it once had; how long it's existed for. The argument for orbital details is analogous.


TITLE: Stability of these fixed points
QUESTION [1 upvotes]: Say we have the set of nonlinear equations, where $\alpha>0$:
$$\begin{matrix}
\frac{dx}{dt}=x[1-\alpha x-y]\\
\frac{dy}{dt}=y[1-x-\alpha y]
\end{matrix}$$
I have determined that the fixed points of this system when $\alpha \neq 1$ are $$(0,0), (0,\frac{1}{\alpha}), (\frac{1}{\alpha},0), (\frac{\alpha (1-\alpha)}{1-\alpha ^{2}},\frac{1-\alpha}{1-\alpha^{2}})$$
And the Jacobian is $$J= \begin{pmatrix}
1-2\alpha x-y& -x\\
-y & 1-x-2\alpha y
\end{pmatrix}$$
So, the fixed point $(0,0)$ has eigenvalues $\lambda=1,1$, so this fixed point is unstable.
But what about the stability of the other fixed points? For example, for the fixed point $(0,\frac{1}{\alpha})$ the Jacobian becomes
$$J= \begin{pmatrix}
\frac{\alpha -1}{\alpha}& 0\\
-\frac{1}{\alpha} & -1
\end{pmatrix}$$
which gives eigenvalues of $\lambda=\frac{\alpha -1}{\alpha},-1?$ But I am struggling at this point because the first eigenvalue could be positive or negative depending on the value of $\alpha$, correct? Have I made a mistake somewhere or am I missing something?
Thanks.

REPLY [0 votes]: There is no problem with your answer. Just do it on a case by case basis depending on $\alpha$. For example:
$$\bigg(0,\frac 1\alpha\bigg) \text{ is a }\begin{cases} \text{saddle} & \alpha<0 \\ \text{stable node} & 0<\alpha <\frac 12 \\ \text{stable inflected node } & \alpha = \frac 12 \\ \text{stable node} & \frac 12<\alpha < 1 \\ \text{saddle} & \alpha > 1\end{cases}$$
where of course, a saddle is unstable by definition.
When $\alpha = 1$, you get a zero eigenvalue, so you will have to expand out the center manifold in order to determine the stability in this case.

EDIT:
Have you been taught how to find the eigenvalues for a general matrix?
For any matrix $A$, $\lambda$ is an eigenvalue of $A$ if and only if $\lambda$ satisfies $\det(A-\lambda I)=0$
In particular, the determinant of a $2 \times 2$ matrix is
$$ \det \begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad-bc$$
Here, the Jacobian is
$$J = \begin{pmatrix}
-\frac{\alpha}{\alpha+1} & -\frac{1}{\alpha+1} \\
-\frac{1}{\alpha+1} & -\frac{\alpha}{\alpha+1}
\end{pmatrix}
$$
So we want to find $\lambda$ such that $\det(J-\lambda I)=0$.
\begin{align}
0 & =\det(J-\lambda I) \\
& = \det \begin{pmatrix} -\frac{\alpha}{\alpha+1}-\lambda & -\frac{1}{\alpha+1} \\
-\frac{1}{\alpha+1} & -\frac{\alpha}{\alpha+1}-\lambda
\end{pmatrix} \\
& = \bigg(-\frac{\alpha}{\alpha+1}-\lambda\bigg)\bigg(-\frac{\alpha}{\alpha+1}-\lambda\bigg) - \bigg(-\frac{1}{\alpha+1}\bigg)\bigg(-\frac{1}{\alpha+1}\bigg) \\
& = \bigg(\lambda + \frac{\alpha}{\alpha+1}\bigg)^2 - \bigg(\frac{1}{\alpha+1}\bigg)^2 \\
\end{align}
Solving for $\lambda$, we find that
$$\lambda = \color{red}{-}1 \qquad \text{ or } \qquad \lambda = \color{red}{-}\frac{\alpha-1}{\alpha+1}$$
I leave it up to you to do the analysis.


TITLE: Collection: Results on stopping times for Brownian motion (with drift)
QUESTION [42 upvotes]: The aim of this question is to collect results on stopping times of Brownian motion (possibly with drift), with a focus on distributional properties:

distributions of stopping times (Laplace transform, moments,..)
distributional properties of the stopped process (computation/finiteness of moments, ...)

Many of the results, which I have in mind, are typical homework problems.

What is the motivation for such a collection?

There is a number of "classical" stopping times for Brownian motion, but unfortunately these stopping times don't have a specific name (apart from "exit time", "hitting time", ... - which is also not very specific), and this makes it hard to find results here on StackExchange. Sometimes, when I'm looking for a result, I know that it is somewhere here on MSE but I'm simply not able to find it. For other questions, which are asked very frequently in MSE, it is often difficult to find a good "old" answer.
In any case, I believe that it would be a benefit to make the knowledge easier to access - both for students (who are trying to solve their homework problems) as for the "teachers" (who are answering questions on MSE).
To make this list a helpful tool (e.g. for answering questions) please make sure to give a short but concise description of each result which you list in your answer.

REPLY [31 votes]: Below, $(X_t)_{t \geq 0}$ is either a Brownian motion (BM, for short) or a Brownian motion with drift. For each of the items in my list I will indicate for which process the corresponding result was obtained.

$\tau = \tau_a:=\inf\{t \geq 0; X_t = a\}$ for $a>0$.

Note: We have $\tau=\inf\{t \geq 0; X_t \geq a\}$ a.s. if $(X_t)_{t \geq 0}$ is a BM, see this answer.

(BM) $\tau_a = \tfrac{1}{c} \tau_{a \sqrt{c}}$ in distribution
(BM) $\mathbb{P}(\tau<\infty)=1$ (via martingale methods)
(BM with non-negative drift) $\mathbb{P}(\tau<\infty)=1$ (via law of iterated logarithm)
(BM with negative drift) Computation of $\mathbb{P}(\tau<\infty)$
(BM) $\mathbb{E}\tau = \infty$
(BM with positive drift) $\mathbb{E}(\tau^n)<\infty$ for all $n \in \mathbb{N}$ and computation of these moments
(BM with drift) density of the distribution of $\tau$
(BM) Laplace transform of $\tau$
(BM with positive drift) Laplace transform of $\tau$

$\tau= \inf\{t \geq 0; X_t \notin [a,b]\}$

(BM) $\mathbb{P}(\tau<\infty)=1$ (for $d$-dimensional BM)
(BM) Distribution of $X_{\tau}$
(BM with drift) Distribution of $X_{\tau}$ (i.e. computation of $\mathbb{P}(X_{\tau}=a)$ and $\mathbb{P}(X_{\tau}=b)$.)
(BM) Computation of $\mathbb{E}(\tau)$ (see this question for BM started at $X_0=x$)
(BM) Computation of $\mathbb{E}(\tau^2)$
(BM) $\mathbb{E}(\tau^p) < \infty$ for all $p \geq 1$
(BM; $a=b$) Laplace transform of $\tau$ (see also this question)
(BM) distribution of $\tau$

Hitting times for some curves

(BM) $\mathbb{E}\tau=\infty$ for $\tau := \inf\{t; B_t^2 \geq 1+t\}$ (see also this question)
(BM) $\mathbb{E}(\tau) =\tfrac{1}{2}$ for $\tau = \inf\{t; B_t^2 = 1-t\}$
(BM) $\mathbb{E}\tau<\infty$ for $\tau:=\inf\{t; |B_t| = \tfrac{1}{2}(1+\sqrt{1+t})\}$
(BM) $\mathbb{E}(\tau)=\infty$ for $\tau=\inf\{t; B_t \geq e^{-\lambda t}\}$

Random variables which are not stopping times

(BM) $\tau=\inf\{t; X_t = \max_{s \in [0,1]} X_s\}$ is not a stopping time (see also this question)
(BM) hitting time of an open set is not a stopping time with respect to the canonical filtration

Miscellaneous

(BM) $\mathbb{E}(\tau)$ for $\tau=\inf\{t; |X_t-x| \geq r\}$ where $X$ is a multi-dimensional Brownian motion
(BM) $\mathbb{E}(\tau)$ for $\tau=\inf\{t; \sup_{s \leq t} X_s \geq b, \inf_{s \leq t} X_s \leq a\}$ with $a<0<b$
(BM) Proof of Wald's identity (see also this question and this version for $\mathbb{E}(\sqrt{\tau})<\infty$)


TITLE: Is the topological index of a self-adjoint operator always zero?
QUESTION [5 upvotes]: By the Atiyah-Singer index theorem, the index of a self-adjoint opeartor D (e.g., Hamiltonian) is given by
Index(D) = dim Ker(D) − dim Ker(D*),
where D* is the adjoint operator of D. Since D is self-adjoint, D=D*, we conclude that Index(D)=0. Is this conclusion right? Can we define a non-zero index for the self-adjoint operator?

REPLY [1 votes]: Your conclusion is right. The index defined in the way you did, of course, can not be non-zero for self-adjoint operator. One can try to define some other index, if one wants, but I am afraid that it would have nothing to do with the Atiyah-Singer theorem.


TITLE: Suppose B is a compact set in Real numbers and A is a subset of B. Prove that if A is an infinite set, then A has a cluster point.
QUESTION [0 upvotes]: So far I have that
Proof:
Suppose $B$ is a compact set in real numbers and $A\subseteq B$.
I will do proof by contradiction.
Assume $A$ is an infinite set and $A$ does not have a cluster point in K.
I’m not sure the next step I know I want to show a contradiction but I’m not sure where it will come from yet. Is it sufficient to use proof by contradiction.

REPLY [0 votes]: I think this statement is not true. For example $B=[0,1]$ but $A=\{1,1/2,...,1/n,...\}$.


TITLE: limit along ultrafilter is multiplicative
QUESTION [0 upvotes]: I am trying to prove that the limit of a sequence along a non-principal ultrafilter on $\mathbb{N}$ is multiplicative and I am getting stuck. This is what I have until now.
Let $U$ be a non-principal ultrafilter on $\mathbb{N}$ and let $\boldsymbol{x}=(x_n)_n$ and $\boldsymbol{y}=(y_n)_n$ be two elements in $\ell^{\infty}$. Define a map $f$ such that $f(\boldsymbol{x})=U-\lim\limits_{n}x_n$. Then
$$
f(\boldsymbol{x})\cdot f(\boldsymbol{y})=U-\lim\limits_{n}x_n \cdot U-\lim\limits_{n}y_n=L_1 L_2,
$$
where $L_1,L_2$ are such that for every $\varepsilon >0$ $\{n\in \mathbb{N} : |x_n-L_1|< \varepsilon  \}\subset U$ and $\{n\in \mathbb{N} : |y_n-L_2|< \varepsilon  \}\subset U$.
Now,
$$
f(\boldsymbol{x}\boldsymbol{y})=U-\lim\limits_{n}x_ny_n=L,
$$
where $L$ is such that for every $\varepsilon >0$ $\{n\in \mathbb{N} : |x_ny_n-L|< \varepsilon  \}\subset U$.
How do I conclude that $L_1 L_2 =L$?

REPLY [1 votes]: HINT: Fix $\epsilon>0$. You know by the continuity of multiplication that there is a $\delta>0$ such that $|x_ny_n-L_1L_2|<\epsilon$ whenever $|x_n-L_1|<\delta$ and $|y_n-L_2|<\delta$. Let
$$u_x=\{n\in\Bbb N:|x_n-L_1|<\delta\}$$
and
$$u_y=\{n\in\Bbb N:|y_n-L_2|<\delta\}\,,$$
and consider $u=u_x\cap u_y$.


TITLE: Most general definition of Borel and parabolic Lie algebras?
QUESTION [6 upvotes]: Let $\mathbb{K}$ be an arbitrary field, and $G$ an algebraic group (=group variety?) over this field. A Borel subgroup of $G$ is a connected solvable subgroup variety $B$ of $G$ such that $G/B$ is complete. A parabolic subgroup of $G$ is a subgroup variety $P$ such that $G/P$ is complete.
Given a group $G$ with lie algebra $\mathfrak{g}$, I would expect that a parabolic Lie subalgebra $\mathfrak{p} \leq \mathfrak{g}$ should be a Lie subalgebra which is the Lie algebra of a parabolic subgroup, and likewise for Borel subalgebras $\mathfrak{b} \leq \mathfrak{g}$.
Since distinct groups can obviously have isomorphic Lie algebras, I would expect that there should be a characterization of Borel and parabolic subalgebras in terms of Lie algebra properties alone, without reference to groups.
However, it seems that in every source I've found that discusses Borel and parabolic subalgebras, that they are defined only for finite-dimensional semisimple and complex lie algebras $\mathfrak{g}$. A Borel subalgebra of a complex finite-dimensional semisimple Lie algebra $\mathfrak{g}$ is usually defined to be a maximal solvable subalgebra $\mathfrak{b} \leq \mathfrak{g}$, and a parabolic subalgebra $\mathfrak{p} \leq \mathfrak{g}$ is any subalgebra containing a Borel subalgebra.
Is there a reason for restricting to complex semisimple Lie algebras? Does the definition not work for general Lie algebras?
More generally, what is the relationship between parabolic subgroups and subalgebras? Does the relationship depend at all on the field $\mathbb{K}$ we are working with (characteristic, whether or not it if it is algebraically closed etc.)? Does it depend on the properties of the group (to connected, simply connected, etc.)?
Unfortunately my algebraic geometry knowledge is very minimal. I'm working on stuff in parabolic differential geometry, so I'm hoping to get a better understanding of parabolic subgroups and algebras, especially in the real case.

REPLY [1 votes]: Bourbaki, Lie Groups and Algebras, ch. VIII §3 nos. 3--5, uses the following definition. The setting is: $\mathfrak g$ is a reductive Lie algebra over a field $k$ of characteristic $\mathbf{0}$.
A subalgebra $\mathfrak a \subseteq g$ is called Borel / parabolic if, for an algebraic closure $\bar k \vert k$, the subalgebra $\mathfrak a \otimes_k \bar k \subseteq \mathfrak g \otimes_k \bar k$ is Borel / parabolic.
(Where it is used that over an algebraically closed field $\bar k$, Borels and parabolics are defined in one of the well-known ways, equivalently w.r.t. some choice of Cartan subalgebra and appropriate subsets of the corresponding root system.)
So basically a Borel or parabolic is something which after scalar extension to an algebraic closure becomes a Borel or parabolic. A nice side effect of this is that for any scalar extension $K \vert k$, one gets that $\mathfrak a \subseteq \mathfrak g$ is Borel / parabolic if and only if the scalar extension $\mathfrak a_K$ is Borel / parabolic in the scalar extension $\mathfrak g_K$.
Of course what happens now is that many $\mathfrak g$ contain no Borel subalgebras at all ($\mathfrak g$ itself is always a parabolic, but in some cases it's the only parabolic). But that actually becomes part of classifications: Semisimple Lie algebras without proper parabolics are anisotropic (over the reals, "compact"); semisimple Lie algebras which contain some Borel are quasi-split.
While excluding the notoriously troublesome case of positive characteristic, the generality of this definition notably applies to semisimple Lie algebras over the reals, $p$-adics and numbers fields, and should be compared to other definitions for real semisimple ones in other answers. I used it freely in my thesis about semisimple Lie algebras over characteristic $0$ fields, focussing more on the $p$-adic case.


TITLE: Probability question involving 4 coins (gold and silver) in 3 boxes each. Is my explanation correct?
QUESTION [0 upvotes]: Key:
g = Gold coin.
s = Silver coin.
Box A)
g1
g2
g3
g4
Box B)
g5
g6
s5
s6
Box C)
s1
s2
s3
s4
Waz chooses a box at random and picks up a coin from it. It turns out to be a gold coin.
( Pick 1 from Box ? results in: g being picked )
Question 1:
Which box do you think g (pick 1) is more likely to have been picked from?
My explanation (answer 1):
Box C is impossible. Box A has more g than Box B, so Box A is more likely.
Box A: 4/8 or 1/2
Box B: 2/8 or 1/4
Question 2: For pick 2 from that same Box ?, which do you think is more likely to be picked by Waz: g or s?
My explanation (answer 2):
Pick 1 establishes Box C is impossible leaving 8 possibilities. However, since one g has been taken out of the equation, this now leaves only 7 possibilities.
5/7 = g and 2/7 = s.
Hence pick 2 is more likely to be g than s.
Is my probability calculation correct here? (bolded parts)
A1: Box A: 1/2 , Box B: 1/4
A2: I think it's either 5/7 = g or 4/6 = g and 2/7 = s or 2/6 = s.
I'm not sure.
Are there only 7 possibilities or 6 remaining for pick 2? Or something else?

REPLY [3 votes]: Your general reasoning/intuition is correct, but your calculations are not.
Let $A$, $B$, $C$ represent the events that the box selected was $A$, $B$, or $C$ respectively.  Let $G$ represent the event that the first drawn coin was gold.
Then we are asked to compute and compare $\Pr[A \mid G]$, $\Pr[B \mid G]$, and $\Pr[C \mid G]$; these are the conditional probabilities that the selected box was $A$, $B$, or $C$ given the drawn coin was gold.  We have from Bayes' rule
$$\Pr[A \mid G] = \frac{\Pr[G \mid A]\Pr[A]}{\Pr[G]}.$$
Since before drawing a coin from the box, each box is equally likely to have been selected, we know $\Pr[A] = \Pr[B] = \Pr[C] = 1/3$.  We also know that $\Pr[G \mid A] = 1$ because every coin in box $A$ is gold.  The only unresolved quantity is $\Pr[G]$, the unconditional probability of drawing a gold coin.  Our intuition suggests that since among all three boxes, there are an equal number of gold and silver coins, that we should have $\Pr[G] = 1/2$.  Indeed, this is true as the law of total probability demonstrates:
$$\begin{align}
\Pr[G] &= \Pr[G \mid A]\Pr[A] + \Pr[G \mid B]\Pr[B] + \Pr[G \mid C]\Pr[C] \\
&= (1)(1/3) + (1/2)(1/3) + (0)(1/3) \\
&= 1/2.
\end{align}$$
So the desired probability is $$\Pr[A \mid G] = \frac{1/3}{1/2} = \frac{2}{3}.$$  Similarly,
$$\Pr[B \mid G] = \frac{\Pr[G \mid B]\Pr[B]}{\Pr[G]} = \frac{(1/2)(1/3)}{1/2} = \frac{1}{3},$$
and of course
$$\Pr[C \mid G] = \frac{\Pr[G \mid C]\Pr[C]}{\Pr[G]} = \frac{(0)(1/3)}{1/2} = 0.$$
This formally proves that the most likely box from which the gold coin was drawn is box $A$, but the probability is $2/3$, not $1/2$ as you asserted.
For the second coin drawn from the same box, again intuition suggests that this coin should also be gold, since $A$ was the more likely box (twice as likely as $B$) that was selected.  If we let $G'$ be the event that the second coin drawn (without replacement) is gold, then we are asked to compute $\Pr[G' \mid G]$, the probability of the second coin being gold given that the first coin was gold.  Thus,
$$\Pr[G' \mid G] = \Pr[G' \mid A, G]\Pr[A \mid G] + \Pr[G' \mid B, G]\Pr[B \mid G].$$  We omit box $C$ since $\Pr[C \mid G] = 0$.  Then
$$\Pr[G' \mid A, G] = 1,$$
since if the first gold coin was drawn from box $A$, the second coin will necessarily be gold.  But
$$\Pr[G' \mid B, G] = \frac{1}{3},$$
since if the first gold coin was drawn from box $B$, there remains only $1$ gold coin out of $3$ in that box.  Therefore,
$$\Pr[G' \mid G] = (1)(2/3) + (1/3)(1/3) = \frac{7}{9}.$$
Note how the calculation must match the sampling process.  Your calculation is incorrect because you seem to be allowing the second coin to be selected from a different box as the first, when the problem clearly states that once the first coin is drawn, the second draw is from the same box.


TITLE: A question about uniform convergent series
QUESTION [2 upvotes]: $X$ is a Banach space and $a_n:X\rightarrow \mathbb{R}$ is a sequence of (continuous) functions.
Suppose $\Sigma_{n=1}^\infty|a_n(x)|$ converges uniformly on $x\in B_X$, unit ball of $X$ (so we can find $N$ so that $\Sigma_{n=N}^\infty|a_n(x)|<\epsilon$ regardless of $x\in B_X$).
Is it true that for any sequence $(x_n)$ of $B_X$, the series $\Sigma_{n=1}^\infty|a_n(x_n)|$ converges?

REPLY [2 votes]: No, and you don't need Banach spaces for a counterexample - $\mathbb{R}$ will do. Let $a_n$ be a piecewise linear function that is equal to $1/n$ at $2^{-n}$ and is zero outside the interval $(2^{-n-1}, 2^{-n+1})$. Then
$\sum a_n$ converges uniformly, but $\sum a_n(2^{-n})$ diverges.
This is essentially asking whether the Weierstrass M-test is necessary for uniform convergence. It is sufficient but not necessary.


text:
\begin{document}
 \maketitle

 \begin{abstract}
 \noindent
 We consider a singular parabolic equation of form
 \[
  u_t = u_{xx} + \tfrac{\alpha}{2}(\sgn u_x)_x
 \]
 with periodic boundary conditions. Solutions to this kind of equations exhibit competition between smoothing due to one-dimensional Laplace operator and tendency to create flat facets due to strongly nonlinear operator $(\sgn u_x)_x$ coming from the total variation flow. We present results concerning analysis of qualitative behaviour and regularity of the solutions. Our main result states that locally (between moments when facets merge), the evolution is described by a system of free boundary problems for $u$ in intervals between facets coupled with equations of evolution of facets. In particular, we provide a proper law governing evolution of endpoints of facets. This leads to local smoothness of the motion of endpoints and the unfaceted part of the solution.
\end{abstract}
\smallskip
\begin{center}
MSC: 35K67, 35B65, 35R35
\end{center}
\smallskip
Keywords: very singular parabolic equation, total variation flow, regularity, free boundary, Stefan problems, crystal growth models, image processing
 \section{Introduction}
 \input{intro}

 \section{Basic properties of solutions}
 \input{basic}

 \section{Characterisation of regular evolutions}
 \input{regular}

 \section*{Acknowledgements}
 The author wishes to express his gratitude towards Piotr Mucha for suggesting the problem, encouragement and helpful discussions. Special thanks are also due to Jose Maz{\'o}n for his informative comments.

 The work has been supported by the grant no.\;2014/13/N/ST1/02622 of the National Science Centre, Poland.

 \bibliographystyle{plain}
 \bibliography{aniso}
\end{document}

config: arxiv


TITLE: Riemann integrability of a piecewise function over $[1, 7]$
QUESTION [2 upvotes]: I have the following function
$f(x) = \begin{cases} 2 & \text{if } 1 \leq x \leq 2, \\ 3 & \text{if } 2 < x \leq 4,\\ 1 & \text{if } 4 < x \leq 7. \end{cases}$
How do I determine if the function is Riemann integrable within $[1, 7]$?
I think I have to check if the lower and upper integrals of $f(x)$ match but I’m unsure how to approach.

REPLY [0 votes]: hint
Let $\epsilon>0$ given and consider the partition $\sigma$ definef by
$$\Bigl(1,2-\frac{\epsilon}{7},2+\frac{\epsilon}{7} ,4-\frac{\epsilon}{7}, 4+\frac{\epsilon}{7},7\Bigr)$$
then
$$U(f,\sigma)-L(\sigma,f)=$$
$$(3-2)\frac{2\epsilon}{7}+(3-1)\frac{2\epsilon}{7}=\frac{6\epsilon}{7}<\epsilon$$

REPLY [0 votes]: Since $\int_a^b f(x) dx = \int_a^cf(x) dx + \int_c^b f(x)dx$ you can split this integral into three parts, as $\int_1^7 f(x) dx = \int_1^2 f(x) dx + \int_2^4 f(x)dx + \int_4^7 f(x) dx$. Since $f(x)$ is constant on each of these intervals they are easily integrated.


TITLE: Why doesn't $S^n$ embed into $R^n$?
QUESTION [5 upvotes]: It seems obviously true, but how does one actually show this? Or what tools does one use? I only know the basics of homotopy theory and homology.
Can I use invariance of domain somehow? If $S^n$ embeds, then so does a neighborhood of it in $R^{n+1}$?

REPLY [4 votes]: To add to Ted Shifrin's hint, one can actually show a stronger result via the invariance of domain theorem:

If $M$ is a compact $n$-manifold and $N$ is a connected $n$-manifold, then an embedding $h : M \to N$ must be surjective, hence a homeomorphism.

This is corollary 2B.4 in Hatcher's Algebraic Topology. To prove it, note that $h(M)$ must be closed since $M$ is compact and $N$ Hausdorff. $h(M)$ is also open by the invariance of domain. Since $N$ is connected, it follows that $h(M) = N$.

Your question now follows since $S^n$ and $\Bbb R^n$ are not homeomorphic.


TITLE: Expected value of infinite sum
QUESTION [16 upvotes]: $x_1, x_2,
\dots, x_n, \dots$ - independent random variables.
Is it true that
$$
\sum_{i = 1}^{\infty}Ex_i = E(\sum_{i = 1}^{\infty} x_i)
$$
?

REPLY [10 votes]: Here it says:
It can be shown that linearity of expectation also holds for countably infinite summations in certain cases. For example, it holds that:
$E\left[\sum\limits_{i=1}^{\infty}X_i\right]=\sum\limits_{i=1}^{\infty}E[X_i]$
if $\sum\limits_{i=1}^{\infty}E[|X_i|]$ converges.
Also This Book says the same thing.