TITLE: Using modular arithmetic to find$ [2^{16}]_8$...
QUESTION [2 upvotes]: This is a fairly simple problem. Just wanted to check if there's another method that I should of used. Here goes---
The problem:

Find $2^{16} \mod 8$

So I figured this one out pretty quickly, I'm just not sure if this was done as it needed to be done.
I first re-wrote our problem as an equivalence class $[2^{16}]_8$. This is what followed
$$[2^{16}]_8 = [2^{15}\cdot 2]_8 = [2^{15}]_8[2]_8 = [(2^3)^5]_8[2]_8 = [0]_8[2]_8 = [0]_8$$
$$\implies [2^{16}]_8 = [0]_8$$
$$\implies 2^{16} \equiv 0 \mod 8$$
So my overall question is: was what I did sufficient or is there another method that is generally advisable? I feel like it's sufficient...

REPLY [5 votes]: It is certainly sufficient, but an even easier way would be as follows:
$2^{16} = 2^3 \cdot 2^{13} = 8 \cdot 2^{13}$.
Hence the remainder when $8$ divides $2^{16}$ is $0$ and we get the desired result.

REPLY [3 votes]: This looks good, although you could have done this quicker by noting that
$$
2^{16} = 2^3 \cdot 2^{13} \equiv 0 \cdot 2^{13} \equiv 0 \pmod{8}
$$


text:
\begin{document}

\title{Tightness, weak compactness of nonlinear expectations and application to CLT }
\author{Shige PENG\thanks{Partially supported by National Basic Research Program of
China (973 Program) (No. 2007CB814906), and NSF of China (No. 10921101). The
author thanks to some discussions with Zengjing Chen, Fuqing Gao, Mingshang
Hu, Xinpeng Li, Zhengyan Lin, Qiman Shao and Liming Wu. It is those
discussions which motivated the author to the realization of the present
paper.}\\School of Mathematics, Shandong University\\250100, Jinan, China}
\date{version: June 10, 2010}

\maketitle

\medskip\medskip\medskip

\noindent\textbf{Abstract.} In this paper we introduce a notion of
tightness for a family of nonlinear expectations and show that the
tightness can be applied to obtain weak compactness in a framework
of nonlinear expectation space. This criterion is very useful for
obtaining the weak convergence for a sequence of nonlinear
expectations, which is a equivalent to the so-called convergence in
distribution, or in law for a sequence of random variables in a
nonlinear expectation space. We use the above result to give a new
proof to the central limit theorem under a sublinear expectation
space. The method can be also applied to prove the convergence of
some numerical schemes for degenerate fully nonlinear PDEs.

\medskip\medskip\medskip

\section{Introduction}

In this paper we introduce a notion of tightness for a family of
nonlinear expectations and show that the tightness can be applied to
obtain weak compactness in a framework of nonlinear expectation
space. This criterion is very useful for obtaining the weak
convergence for a sequence of nonlinear expectations, which is a
equivalent to the so-called convergence in distribution, or in law
for a sequence of random variables in a nonlinear expectation space.
Our results have nontrivially generalized those in the classical
situation of weak convergence, or convergence in law in classical
probability theory.

As an application of those results, we provide a new and probabilistic proof
for central limit theorem under sublinear expectation. Such new type of CLT
was obtained in [Peng2007b], [Peng2008] (see also [Peng2010] for a more
general situation). But the proof given in the above papers depend heavily on
a very deep estimate of fully nonlinear parabolic PDE of parabolic types. Our
new proof also provides a probabilistic method to prove the existence of fully
nonlinear parabolic PDEs. The method can be also applied to prove the
convergence of some numerical schemes for degenerate nonlinear PDEs.

This paper is organized as follows, in the next section we give a
preliminary introduction for the notion of sublinear and nonlinear
expectations. New notions of tightness and weak convergence of
nonlinear expectations and their relations will be introduced in
Section 3. In Section 4 we will use the result of the previous
section to give a new and probabilistic proof of the central limit
theorem under sublinear expectations.

\section{Preliminaries: sublinear expectation and upper expectation}

We present some basic notions and results of nonlinear and sublinear
expectations. For more details we refer to [Peng2007a], [Denis,Hu,Peng2008],
[Peng2010] Let $\Omega$ be a given set and $\mathcal{H}$ be a linear space of
real functions on $\Omega$ such that $1\in \mathcal{H}$ and $|\xi
|\in \mathcal{H}$ if $\xi \in \mathcal{H}$. A functional $\mathbb{E}
:\mathcal{H}\mapsto \mathbb{R}$ is called a nonlinear expectation if it
satisfies $\mathbb{E}[c]=c$, $c\in \mathbb{R}$ and $\mathbb{E}[X]\geq
\mathbb{E}[Y]$, for $X$, $Y\in \mathcal{H}$ such that $X\geq Y$. If
$\mathbb{E}$ satisfies furthermore $\mathbb{E}[X+Y]-\mathbb{E}[X]-\mathbb{E}
[Y]\leq0$ (resp. $=0$) and $\mathbb{E}[\lambda X]=\lambda \mathbb{E}[X]$, for
$X$, $Y\in \mathcal{H}$ and $\lambda \geq0$, then we call it a sublinear (resp.
linear) expectation.

Recently a new notion of coherent risk measures in finance caused much
attention to the study of such type of sublinear expectations and applications
to risk controls, see the seminal paper of Artzner, Delbaen, Eber and Heath
(1999) as well as in F\"{o}llmer and Schied (2004).

The following result is well-known as representation theorem. It is a direct
consequence of Hahn-Banach theorem (see Delbaen (2002), F\"{o}llmer and Schied (2004)).

\begin{theorem}
\bigskip Let {{$\mathbb{\hat{E}}$}} be a sublinear expectation defined on
$(\Omega,\mathcal{H})$. Then there exists a family of linear expectations
$\{E_{\theta}:\theta \in \Theta \}$ on $(\Omega,\mathcal{H})$ such that
\[
{{\mathbb{\hat{E}}}}[X]=\max_{\theta \in \Theta}E_{\theta}[X].
\]

\end{theorem}

A sublinear expectation $\mathbb{\hat{E}}$ on $(\Omega,\mathcal{H})$ is said
to be regular if for each sequence $\{X_{n}\}_{n=1}^{\infty}\subset
\mathcal{H}$ such that $X_{n}(\omega)\downarrow0$, for $\omega$, we have
${{\mathbb{\hat{E}}}}[X_{n}]\downarrow0$. If $\mathbb{\hat{E}}$ is regular
then from the above representation we have $E_{\theta}[X_{n}]\downarrow0$ for
each $\theta \in \Theta$. It follows from Daniell-Stone theorem that there
exists a unique ($\sigma$-additive) probability measure $P_{\theta}$ defined
on $(\Omega,\sigma(\mathcal{H}))$ such that
\[
E_{\theta}[X]=\int_{\Omega}X(\omega)dP_{\theta}(\omega),\  \ X\in
\mathcal{H}\text{.}
\]

The above representation theorem of sublinear expectation tells us that to use
a sublinear expectation for a risky loss $X$ is equivalent to take the upper
expectation of $\{E_{\theta}:\theta \in \Theta \}$. The corresponding model
uncertainty of probabilities, or ambiguity, is the subset $\{P_{\theta}
:\theta \in \Theta \}$. The corresponding uncertainty of distributions for an
$n$-dimensional random variable $X$ in $\mathcal{H}$ is
\[
\{F_{X}(\theta,A):=P_{\theta}(X\in A):A\in \mathcal{B}(\mathbb{R}^{n})\}.
\]

We consider the case where $\Omega$ is a complete separable metric space. Let
$\mathcal{B}(\Omega)$ be the Borel $\sigma$-algebra of $\Omega$ and
$\mathcal{M}$ the collection of all probability measures on $(\Omega
,\mathcal{B}(\Omega))$.

\begin{itemize}
\item $L^{0}(\Omega)$: the space of all $\mathcal{B}(\Omega)$-measurable real functions;

\item $B_{b}(\Omega)$: all bounded functions in $L^{0}(\Omega)$;

\item $C_{b}(\Omega)$: all continuous functions in $B_{b}(\Omega)$.
\end{itemize}

We are given a subset $\mathcal{P}\subseteq \mathcal{M}$. The upper expectation
of $\mathcal{P}$ is defined as follows: for each $X\in L^{0}(\Omega)$ such
that $E_{P}[X]$ exists for each $P\in \mathcal{P}$,
\[
\mathbb{E}[X]=\mathbb{E}^{\mathcal{P}}[X]:=\sup_{P\in \mathcal{P}}E_{P}[X].
\]

It is easy to verify that the upper expectation $\mathbb{E}[\cdot]$
of the family $\mathcal{P}$ is a sublinear expectation on
$(\Omega,B_{b}(\Omega))$ as well as on $(\Omega,C_{b}(\Omega)$).

Moreover, it is also easy to check that

\begin{theorem}
\begin{enumerate}
\item Let $\mathbb{E}[X_{n}]$ and $\mathbb{E}[\sum_{n=1}^{\infty}X_{n}]$ be
finite. Then $\mathbb{E}[\sum_{n=1}^{\infty}X_{n}] \leq \sum_{n=1}^{\infty}
\mathbb{E}[X_{n}].$

\item Let $X_{n} \uparrow X$ and $\mathbb{E}[X_{n}]$, $\mathbb{E}[X]$ be
finite. Then $\mathbb{E}[X_{n}] \uparrow \mathbb{E}[X]$.
\end{enumerate}
\end{theorem}

\begin{definition}
The functional $\mathbb{E}[\cdot]$ is said to be \textbf{regular} if for each
$\{X_{n}\}_{n=1}^{\infty}$ in $C_{b}(\Omega)$ such that $X_{n}\downarrow0$
\ on $\Omega$, we have $\mathbb{E}[X_{n}]\downarrow0$.
\end{definition}

We have:

\begin{theorem}
\label{Thm2}$\mathbb{\mathbb{E}}[\cdot]$ is regular if and only if
$\mathcal{P}$ is relatively compact.\
\end{theorem}

We denote
\[
c(A):=\sup_{P\in \mathcal{P}} P(A),\  \  \  \ A\in \mathcal{B}(\Omega).
\]

One can easily verify the following theorem.

$c(\cdot)$ is a Choquet capacity, i.e. (see [Choquet1953]),

\begin{theorem}
\begin{enumerate}
\item $0\leq c(A)\leq1,\  \  \forall A\subset \Omega$.

\item If $A\subset B$, then $c(A)\leq c(B)$.

\item If $(A_{n})_{n=1}^{\infty}$ is a sequence in $\mathcal{B}(\Omega)$, then
$c(\cup A_{n})\leq \sum c(A_{n})$.

\item If $(A_{n})_{n=1}^{\infty}$ is an increasing sequence in $\mathcal{B}
(\Omega)$: $A_{n}\uparrow A=\cup A_{n}$, then $c(\cup A_{n})=\lim
_{n\rightarrow \infty}c(A_{n})$.
\end{enumerate}
\end{theorem}

Furthermore, for each $A\in \mathcal{B}(\Omega)$, we have
\[
c(A)=\sup \{c(K):\ K\text{ compact,}\ K\subset A\}.
\]
Indeed,
\[
c(A)=\sup_{P\in \mathcal{P}}\sup_{\substack{K\, \text{compact}\\K\subset
A}}P(K)=\sup_{\substack{K\, \text{compact}\\K\subset A}}\sup_{P\in \mathcal{P}
}P(K)=\sup_{\substack{K\, \text{compact}\\K\subset A}}c(K).
\]
We use the standard capacity-related vocabulary: a set $A$ is \textbf{polar}
if $c(A)=0$ and a property holds \textquotedblleft \textbf{quasi-surely}
\textquotedblright \ (q.s.) if it holds outside a polar set. In other words,
$A\in \mathcal{B}(\Omega)$ is polar if and only if $P(A)=0$ for any
$P\in \mathcal{P}$. We also have in a trivial way a Borel-Cantelli Lemma: Let
$(A_{n})_{n\in \mathbb{N}}$ be a sequence of Borel sets such that
\[
\sum_{n=1}^{\infty}c(A_{n})<\infty.
\]
Then $c(\limsup_{n\rightarrow \infty}A_{n})=0$. To prove it we just need to
applying the Borel-Cantelli Lemma under each probability $P\in \mathcal{P}$.

The following theorem is Prohorov's theorem.

\begin{theorem}
$\mathcal{P}$ is relatively compact if and only if for each $\varepsilon>0$,
there exists a compact set $K$ such that $c(K^{c})<\varepsilon$.
\end{theorem}

\section{Tightness and weak convergence of nonlinear expectations}

Let $\Omega$ be a complete separable metric space and let $\{ \mathbb{E}
_{\alpha}\}_{\alpha \in \mathcal{A}}$ be a family of sublinear expectations on
$(\Omega,B_{b}(\Omega))$.

\begin{definition}
A sublinear expectation $\mathbb{\hat{E}}$ on $(\Omega,B_{b}(\Omega))$ is said
to be tight if for each $\varepsilon>0$, there exists a compact subset
$K\subset \Omega$ such that $\mathbb{\hat{E}}[1_{K^{c}}]<\varepsilon$. A family
of sublinear expectations $\{ \mathbb{E}_{\alpha}\}_{\alpha \in \mathcal{A}}$ on
$(\Omega,B_{b}(\Omega))$ is said to be tight if the sublinear expectation
defined by
\[
\mathbb{\hat{E}}[X]:=\sup_{\alpha \in \mathcal{A}}\mathbb{E}_{\alpha
}[X],\  \ X\in B_{b}(\Omega),
\]
is tight. A family of nonlinear expectations $\{ \mathbb{\tilde{E}}_{\alpha
}\}_{\alpha \in \mathcal{A}}$ on $(\Omega,C_{b}(\Omega))$ is said to be tight if
there exists a tight sublinear expectation $\mathbb{\hat{E}}$ on
$(\Omega,B_{b}(\Omega))$ such that, for each $\alpha \in \mathcal{A}$,
\[
\mathbb{\tilde{E}}_{\alpha}[X]-\mathbb{\tilde{E}}_{\alpha}[Y]\leq
\mathbb{\hat{E}}[X-Y],\  \ X,Y\in C_{b}(\Omega).
\]

\end{definition}

\begin{definition}
A sequence $\{ \mathbb{E}_{n}\}_{n=1}^{\infty}$ of nonlinear linear
expectations defined on $(\Omega,C_{b}(\Omega))$ is said to be weakly
convergent if, for each $\varphi \in C_{b}(\Omega)$, $\{ \mathbb{E}_{n}
[\varphi]\}_{n=1}^{\infty}$ is a Cauchy sequence. A family $\{ \mathbb{E}
_{\alpha}\}_{\alpha \in \mathcal{A}}$ of nonlinear linear expectations defined
on $(\Omega,C_{b}(\Omega))$ is said to be weakly compact if for each sequence
$\{ \mathbb{E}_{\alpha_{i}}\}_{i=1}^{\infty}$ there exists a weak convergent subsequence.
\end{definition}

\begin{theorem}
\label{Theorem1}Let a family of nonlinear expectations $\{ \mathbb{E}_{\alpha
}\}_{\alpha \in \mathcal{A}}$ be tight, namely there exists a tight sublinear
expectation $\mathbb{\hat{E}}$ on $(\Omega,B_{b}(\Omega)\mathbb{)}$ such that
\[
\mathbb{E}_{\alpha}[X]-\mathbb{E}_{\alpha}[Y]\leq \mathbb{\hat{E}
}[X-Y],\  \ X,Y\in C_{b}(\Omega).
\]
Then $\{ \mathbb{E}_{\alpha}\}_{\alpha \in \mathcal{A}}$ is weakly compact,
namely, for each sequence $\{ \mathbb{E}_{\alpha_{n}}\}_{n=1}^{\infty}$, there
exists a subsequence $\{ \mathbb{E}_{\alpha_{n_{i}}}\}_{i=1}^{\infty}$ such
that, for each $\varphi \in C_{b}(\Omega)$, $\{ \mathbb{E}_{\alpha_{n_{i}}
}[\varphi]\}_{i=1}^{\infty}$ is a Cauchy sequence.
\end{theorem}

\begin{proof}
Let $K_{i}$, $i=1,2,\cdots$, be a sequence of compact subsets in
$\Omega$ such that $K_{i}\subset K_{i+1}$ and
$\mathbb{\hat{E}}[\mathbf{1}_{(K_{i}})^c]\leq1/i$. Let $\{
\varphi_{j}\}_{j=1}^{\infty}$ constitute a linear subspace of
$C_{b}(\Omega)$ such that, for each $K_{i}$, $\{ \varphi_{j}(x)\}_{j=1}
^{\infty}|_{x\in K}$ is dense in $C_{b}(K_{i})$. Since $\{ \mathbb{E}
_{\alpha_{n}}[\varphi_{1}]\}_{n=1}^{\infty}$ is a bounded sequence, there
exists a subsequence $\{n_{i}^{1}\}_{i=1}^{\infty}$ of $n=1,2,\cdots$, such
that $\{ \mathbb{E}_{\alpha_{n_{i}^{1}}}[\varphi_{1}]\}_{i=1}^{\infty}$ is a
Cauchy sequence. Similarly, there exists a \ subsequence $\{n_{i}^{2}
\}_{i=1}^{\infty}$ of $\{n_{i}^{1}\}_{i=1}^{\infty}$ such that $\{
\mathbb{E}_{\alpha_{n_{i}^{2}}}[\varphi_{2}]\}_{i=1}^{\infty}$ is a Cauchy
sequence. Proceeding in this way, we can find, for each $j=1,2,\cdots$, a
subsequence $\{n_{i}^{j}\}_{i=1}^{\infty}$ of $\{n_{i}^{j-1}\}_{i=1}^{\infty}$
such tat $\{ \mathbb{E}_{\alpha_{n_{i}^{j}}}[\varphi_{j}]\}_{i=1}^{\infty}$ is
a Cauchy sequence. It follows that, for the diagonal subsequence $\{n_{i}
^{i}\}_{i=1}^{\infty}$ and for each fixed $j=1,2,\cdots$, $\{ \mathbb{E}
_{\alpha_{n_{i}^{i}}}[\varphi_{j}]\}_{i=1}^{\infty}$ is a Cauchy sequence.

We now prove that, for each $\varphi \in C_{b}(\Omega)$, $\{ \mathbb{E}
_{\alpha_{n_{i}^{i}}}[\varphi]\}_{i=1}^{\infty}$ is also a Cauchy
sequence. We denote $M=\sup_{\omega \in {\Omega}}|\varphi(\omega)|$.
For each $\varepsilon>0$, let $m$ be an integer larger than
$16M/\varepsilon$, thus
$\mathbb{\hat{E}}[\mathbf{1}_{(K_{m})^{c}}]\leq \varepsilon/16M$.
Let $\{
\varphi_{j_{l}}\}_{l=1}^{\infty}$ be a subsequence of $\{ \varphi_{j}
\}_{j=1}^{\infty}$ such that $\sup_{l}\sup_{\omega \in \Omega}|\varphi_{i_{l}
}(\omega)|\leq M$ and
\[
\lim_{l\rightarrow \infty}\sup_{\omega \in K_{m}}|\varphi_{i_{l}}(\omega
)-\varphi(\omega)|=0.
\]
Thus there exists an integer $l$ such that $\mathbb{\hat{E}}[|\varphi
-\varphi_{j_{l}}|\mathbf{1}_{K_{m}}]\leq \varepsilon/8$. It follows that
\begin{align*}
\mathbb{\hat{E}}[|\varphi-\varphi_{j}|]  &  \leq \mathbb{\hat{E}}
[|\varphi-\varphi_{j_{l}}|\mathbf{1}_{(K_{m})^{c}}]+\mathbb{\hat{E}}
[|\varphi-\varphi_{j_{l}}|\mathbf{1}_{K_{m}}]\\
&  \leq2M\varepsilon/16M+\mathbb{\hat{E}}[|\varphi-\varphi_{j_{l}}
|\mathbf{1}_{K_{m}}]\\
&  \leq \varepsilon/4.\  \
\end{align*}
For this fixed $l$, there exists an integer $i_{0}$, such that $|\mathbb{E}
_{\alpha_{i}^{i}}[\varphi_{j_{l}}]-\mathbb{E}_{\alpha_{j}^{j}}[\varphi_{j_{l}
}]|\leq \varepsilon/2$, for each $i,j\geq i_{0}$. We then have
\begin{align*}
|\mathbb{E}_{\alpha_{n_{i}^{i}}}[\varphi]-\mathbb{E}_{\alpha_{n_{j}^{j}}
}[\varphi]|  &  \leq \mathbb{\hat{E}}[|\varphi-\varphi_{j_{l}}|]+|\mathbb{E}
_{\alpha_{n_{i}^{i}}}[\varphi_{j_{l}}]-\mathbb{E}_{\alpha_{n_{j}^{j}}}
[\varphi_{j_{l}}]|+\mathbb{\hat{E}}[|\varphi-\varphi_{j_{l}}|]\\
&  \leq2\mathbb{\hat{E}}[|\varphi-\varphi_{j_{l}}|]+|\mathbb{E}_{\alpha
_{n_{i}^{i}}}[\varphi_{j_{l}}]-\mathbb{E}_{n_{j}^{j}}[\varphi_{j_{l}}
]|\leq \varepsilon.\  \  \
\end{align*}
Thus $\{ \mathbb{E}_{\alpha_{n_{i}^{i}}}[\varphi]\}_{i=1}^{\infty}$ is a
Cauchy sequence. The proof is complete.
\end{proof}

\section{Convergence in distribution}

In this section we apply Theorem \ref{Theorem1} to give a probabilistic proof
of the central limit theorem under a sublinear expectation space.

\begin{definition}
Let $(\Omega,\mathcal{H},\mathbb{E})$ be a nonlinear expectation space and
$\left \{  \xi_{i}\right \}  _{i=1}^{\infty}$ be a sequence of $\mathbb{R}^{d}
$-valued random variables in this space. $\left \{  \xi_{i}\right \}
_{i=1}^{\infty}$ is said to converge in distribution, or in law, if, for each
$\varphi \in C_{b}(\mathbb{R}^{d})$, $\{ \mathbb{E}[\varphi(\xi_{i}
)]\}_{i=1}^{\infty}$ is a Cauchy sequence. We define
\[
\mathbb{F}_{\xi}[\varphi]=\lim_{i\rightarrow \infty}\mathbb{E}[\varphi(\xi
_{i})],\  \  \varphi \in C_{b}(\mathbb{R}^{d})\text{.}
\]
$\mathbb{F}:C_{b}(\mathbb{R}^{d})\mapsto \mathbb{R}$ forms a nonlinear
expectation on $(\mathbb{R}^{d},C_{b}(\mathbb{R}^{d}))$.
\end{definition}

\begin{remark}
If $\mathbb{E}$ is a sublinear (resp. linear) expectation then $\mathbb{F}$ is
also a sublinear (resp. linear) on $(\mathbb{R}^{d},C_{b}(\mathbb{R}^{d}))$.
\end{remark}

\begin{lemma}
Let $(\bar{\Omega},\mathcal{\bar{H}},\mathbb{\bar{E}})$ be a sublinear
expectation space and let $\bar{X}=(\bar{X}_{1},\cdots,\bar{X}_{d})$ be such
that $\bar{X}_{1}$,$\cdots,\bar{X}_{d}\in \mathcal{\bar{H}}$ and $\psi(\bar
{X})\in \mathcal{\bar{H}}$ for each $\psi \in L_{0}(\mathbb{R}^{d})$ such that
$|\psi(x)|\leq c(1+|x|^{l})$, where $l>0$ is fixed. Then the sublinear
distribution on $(\mathbb{R}^{d},B_{b}(\mathbb{R}^{d}))$ defined by
$\mathbb{\bar{F}}_{\bar{X}}[\varphi]:=\mathbb{\bar{E}}[\varphi(\bar{X}
)]:B_{b}(\mathbb{R}^{d})\mapsto \mathbb{R}$ is tight.
\end{lemma}

\begin{proof}
For each $\varepsilon>0$, we can choose an integer $N>0$ such that
$N^{l}>\mathbb{\bar{E}}[|\bar{X}|^{l}]/\varepsilon$ and a compact
set $K=\{x\in \mathbb{R}^{d}:|x|\leq N\}$. Since for each
\[
\mathbb{\bar{F}}_{{\bar X}}[\mathbf{1}_{K^{c}}]=\mathbb{\bar{E}}[\mathbf{1}
_{\{|{\bar X}|>N\}}]\leq \frac{1}{N^{l}}\mathbb{\bar{E}}[|{\bar
X}|^{l}]<\varepsilon.
\]
It follows that the sublinear distribution $\mathbb{\bar{F}}_{\bar{X}}$ of
$\bar{X}$ is tight. The proof is complete.
\end{proof}

\begin{theorem}
\label{Theorem2}Let $(\Omega,\mathcal{H},\mathbb{E})$ be a nonlinear
expectation space and let $\{X_{\alpha}\}_{\alpha \in \mathcal{A}}$ be a family
of $\mathbb{R}^{d}$-valued random variables such that
\[
\mathbb{E}[\varphi(X_{\alpha})]-\mathbb{E}[\psi(X_{\alpha})]\leq
\mathbb{\bar{E}}[\varphi(\bar{X})-\psi(\bar{X})],\  \  \forall \alpha
\in \mathcal{A}
\]
where $\bar{X}$ is as in the above lemma. Then the family of distributions $\{
\mathbb{F}_{X_{\alpha}}[\varphi]\}_{\alpha \in \mathcal{A}}$ on $(\mathbb{R}
^{d},C_{b}(\mathbb{R}^{d}))$ is weakly compact.
\end{theorem}

\begin{proof}
We denote by $L_{l}(\mathbb{R}^{d})$ the space of all $L_{0}(\mathbb{R}^{d}
)$-measurable functions such that $|\psi(x)|\leq c(1+|x|^{l})$. It is clear
that, for each $n=1,2,\cdots$, $\mathbb{F}_{\xi_{n}}[\varphi]:=\mathbb{E}
[\varphi(\xi_{n})]:L_{l}(\mathbb{R}^{d})\mapsto \mathbb{R}$ forms a nonlinear
distribution on $(\mathbb{R}^{d},L_{l}(\mathbb{R}^{n}))$. Moreover
\[
\mathbb{F}_{X_{\alpha}}[\varphi]-\mathbb{F}_{X_{\alpha}}[\psi]\leq
\mathbb{\bar{F}}_{\bar{X}}[\varphi-\psi],\  \  \varphi,\psi \in C_{b}
(\mathbb{R}^{d}),\  \alpha \in \mathcal{A}.
\]
Since by the above lemma $\mathbb{\bar{F}}_{\bar{X}}$ is tight. It follows
from Theorem \ref{Theorem1} and the above lemma that $\{ \mathbb{F}
_{X_{\alpha}}[\varphi]\}_{\alpha \in \mathcal{A}}$ is weakly compact. The proof
is complete.
\end{proof}

Let $S$ be a complete separable metric space. We denote by $C_{b}(S)$, the
linear space of all real valued bounded and continuous functions on $S$. We
can also apply Theorem \ref{Theorem1} to random variables on a nonlinear
expectation space with values on $S$. A typical situation is $S=C([0,\infty
),\mathbb{R}^{d})$.

\begin{definition}
Let $(\Omega,\mathcal{H},\mathbb{E})$ be a nonlinear expectation space. A
mapping $\xi:\Omega \mapsto S$ is said to be an $S$-valued random variable if
$f(\xi)\in \mathcal{H}$ for each $f\in C_{b}(S\mathcal{)}$.
\end{definition}

\begin{definition}
Let $(\Omega,\mathcal{H},\mathbb{E})$ be a nonlinear expectation space and
$\left \{  \xi_{i}\right \}  _{i=1}^{\infty}$ be a sequence of $S$-valued random
variables in this space. $\left \{  \xi_{i}\right \}  _{i=1}^{\infty}$ is said
to be convergent in distribution, or in law if for each $\varphi \in C_{b}(S)$,
the sequence $\{ \mathbb{E}[\varphi(\xi_{i})]\}_{i=1}^{\infty}$ is Cauchy. We
define
\[
\mathbb{F}[\varphi]=\lim_{n\rightarrow \infty}\mathbb{E}[\varphi(\xi
_{i})],\  \  \varphi \in C_{b}(S)\text{.}
\]
$\mathbb{F}:C_{b}(S)\mapsto \mathbb{R}$ forms a nonlinear expectation on
$(S,C_{b}(S))$.
\end{definition}

\section{Central limit theorem under sublinear expectations}

Let $(\Omega,\mathcal{H},\mathbb{\hat{E}})$ be a sublinear
expectation space. We consider an i.i.d. sequence $\left \{
X_{i}\right \} _{i=1}^{\infty}$
valued in $\mathbb{R}^{d}$ such that $\mathbb{\hat{E}}[X_{1}]=\mathbb{\hat{E}
}[-X_{1}]=0$, and $\mathbb{\hat{E}}[|X_{1}|^{2+\beta}]<\infty$ for some
constant $\beta>0$. The following result was obtained in [Peng2007] (see also
[Peng2008, 2010]) using a very deep result of partial differential equation.
We shall give a short and probabilistic proof.

\begin{theorem}
Let $S_{n}=X_{1}+\cdots+X_{n}$. The sequence $\left \{  n^{-1/2}S_{n}\right \}
_{n=1}^{\infty}$ converges in distribution to $\xi$ satisfying
\[
\mathbb{E}[\xi]=0,\  \  \mathbb{E}[\left \langle A\xi,\xi \right \rangle ]=\frac
{1}{2}G(A):=\mathbb{E}[\left \langle AX_{1},X_{1}\right \rangle ],\  \ A\in
\mathbb{S}(d).
\]
and
\[
a\xi+b\bar{\xi}\sim \sqrt{a^{2}+b^{2}}\xi,\  \  \ a,b\geq0.
\]
Namely, the limit $\xi$ is $G$-normal.
\end{theorem}

\begin{proof}
We set
\[
\mathbb{\bar{F}}[\varphi]=\sup_{n}\mathbb{\hat{E}}\left[  \varphi
(n^{-1/2}S_{n})\right]  ,\  \  \varphi \in C_{b}(\mathbb{R}^{d}).
\]
$\mathbb{\bar{F}}$ is also a sublinear expectation on $(\mathbb{R}^{d}
,C_{b}(\mathbb{R}^{d}))$ satisfying
\[
\mathbb{\hat{E}}\left[  \varphi(n^{-1/2}S_{n})\right]  -\mathbb{\hat{E}
}\left[  \psi(n^{-1/2}S_{n})\right]  \leq \mathbb{\bar{F}}[\varphi
-\psi],\  \varphi,\psi \in C_{b}(\mathbb{R}^{d}).\
\]
Since $\mathbb{\hat{E}}\left[  \left \vert n^{-1/2}S_{n}\right \vert
^{2}\right]  =\mathbb{\hat{E}}[|X_{1}|^{2}]$, it is clear that, for
$\varphi=x^{2}$,
\[
\mathbb{\bar{F}}[|x|^{2}]=\mathbb{\hat{E}}[|X_{1}|^{2}]<\infty.
\]
\[
.
\]
By Theorem \ref{Theorem1}, there exists a subsequence $\left \{  n_{i}
^{-1/2}S_{n_{i}}\right \}  _{i=1}^{\infty}$ of $\left \{  n^{-1/2}S_{n}\right \}
_{n=1}^{\infty}$ which converges in law to some $\xi_{1}$ in $(\Omega
,\mathcal{H},\mathbb{E}\mathbb{)}$. We also have
\[
\mathbb{E}[\xi_{1}]=\mathbb{E}[-\xi_{1}]=0,\  \  \frac{1}{2}\mathbb{E}
[\left \langle A\xi_{1},\xi_{1}\right \rangle ]=\frac{1}{2}\mathbb{E}
[\left \langle AX_{1},X_{1}\right \rangle ]=G(A).
\]
We will prove that $\xi_{1}$ is $G$-normal distributed for the above $G$.
Since this distribution is uniquely defined, thus we can prove that, in fact,
the sequence $\{ \mathbb{\hat{E}}\left[  \varphi(n^{-1/2}S_{n})\right]
\}_{n=1}^{\infty}$ converges in law to the $G$-normal distribution.

For the above convergent subsequence $\left \{  n_{i}^{-1/2}S_{n_{i}}\right \}
_{i=1}^{\infty}$, it is clear that for an arbitrarily increasing integers of
$\{ \bar{n}_{i}\}_{i=1}^{\infty}$ such that $|\bar{n}_{i}-n_{i}|\leq1$, both
$\left \{  n_{i}^{-1/2}S_{n_{i}}\right \}  _{i=1}^{\infty}$ and $\left \{
\bar{n}_{i}^{-1/2}S_{\bar{n}_{i}}\right \}  _{i=1}^{\infty}$ converges in law
to a same limit. Thus, without loss of generality, we can assume that $n_{i}$,
$i=1,2,\cdots$, are all even numbers and decompose $n_{i}^{-1/2}S_{n_{i}}$
into two parts:
\[
n_{i}^{-1/2}S_{n_{i}}=\frac{1}{\sqrt{2}}(n_{i}/2)^{-1/2}S_{\frac{n_{i}}{2}
}+\frac{1}{\sqrt{2}}(n_{i}/2)^{-1/2}(S_{n_{i}}-S_{\frac{n_{i}}{2}})
\]
We then use the same argument to each part and prove that there
exists a subsequence of
$\{\frac{1}{\sqrt{2}}(n_{i}/2)^{-1/2}S_{\frac{n_{i}}{2}}\}_{i=1}^{\infty}$
(resp.
$\{\frac{1}{\sqrt{2}}(n_{i}/2)^{-1/2}(S_{n^{i}}-S_{\frac{n_{i}}{2}})\}_{i=1}^{\infty}$)
converging in law to $\xi_{1/2}$ such that $\xi_{1}-\xi_{1/2}$ is
identically distributed with respect to $\xi_{1/2}$ as well as
independent of $\xi_{1/2}$. We set $\xi_{0}=0$. It is easy to check
that
\begin{equation}
\mathbb{E}[\xi_{t}]=\mathbb{E}[-\xi_{t}]=0,\  \  \frac{1}{2}\mathbb{E}
[\left \langle A\xi_{t},\xi_{t}\right \rangle ]=G(A)t,\ A\in \mathbb{S}
(d),\  \label{3.1}
\end{equation}
for $t=0,\frac{1}{2},1$.

We can continue this procedure to prove that there exist random variables
$\xi_{t}$, $t=i/4$, $i=0,1,\cdots,4$, such that (\ref{3.1}) holds for each $t$
and each increment $\xi_{t}-\xi_{s}$ is identically distributed as $\xi_{t-s}$
and independent of $\xi_{t_{1}}$, $\xi_{t_{2}}$, for $t,s,t_{1},t_{2}=i/4$,
$i=0,1,\cdots,4$ and $t\geq s\geq t_{i}$, $i=1,2$. Here, using the technique
of the product of expectation spaces (see [Peng2010]), we always use the same
sublinear expectation space $(\Omega,\mathcal{H},\mathbb{E})$ like in the
classical situation.

We proceeding in this way to prove that we can find $\xi_{t}\in \mathcal{H}$,
$t\in \mathbf{Q}[0,1]$ ($\mathbf{Q}[0,1]$ denotes the set of all rational
numbers in $[0,1]$) such that $\xi_{t+h}-\xi_{t}$ is identically distributed
with respect to $\xi_{h}$ and is independent of $\xi_{t_{1}}$, $\xi_{t_{2}}
$,$\cdots,\xi_{t_{k}}$, for each $t$, $s$, $h$, $t_{i}$ in $[0,1]$ such that
$t\geq t_{i}$, $i=1,2,\cdots k$. $t+h\in \lbrack0,1]$. We also have (\ref{3.1})
for $t\in \mathbf{Q}[0,1]$.

It remains to prove that $\xi_{1}$ is $G$-normally distributed. The
method is very similar to that of Theorem 1.6 in [Peng2010, ChIII].
We only give a sketch of the proof.

We define, for a bounded and Lipschitz function $\varphi$,
\[
u(t,x)=\mathbb{E}[\varphi(x+\xi_{t})],\  \ x\in \mathbb{R}^{d},\  \ t\in
\mathbf{Q}[0,1].
\]
It is easy to check that $u(t,x)$ is a Lipschitz function of $x$ with the same
Lipschitz constant as $\varphi$ and, for each $t$, $s$ $\in \mathbf{Q}[0,1]$
such that $t+s\leq1$,
\begin{equation}
u(t+s,x)=\mathbb{E}[u(t,x+\xi_{s})],\  \ x\in \mathbb{R}^{d}.\  \  \label{3.2}
\end{equation}
From which we can prove that
\begin{align*}
|u(t+s,x)-u(t,x)|  &  =\mathbb{E}[|u(t,x+\xi_{s})-u(t,x)|]\\
&  \leq C\mathbb{E}[|\xi_{s}|]\leq \bar{C}\sqrt{s}.
\end{align*}
We then continuously extend the domain $u(t,x)$ from $\mathbf{Q}
[0,1]\times \mathbb{R}^{d}$ to $[0,1]\times \mathbb{R}^{d}$. Relation
(\ref{3.2}) still holds for $t\in \lbrack0,1)$, $s\in \mathbf{Q}[0,1]$ such that
$t+s\leq1$.

For each $\bar{x}\in \mathbb{R}^{d}$, $\bar{t}\in \lbrack0,1)$, if $\psi$ is a
smooth function on $[0,1]\times \mathbb{R}^{d}$ such that $\psi \geq u$ and
$\psi(\bar{t},\bar{x})=u(\bar{t},\bar{x})$. From (\ref{3.2}) we have, for a
sufficiently small rational number $s>0$,
\[
\partial \psi(\bar{t},\bar{x})s-\mathbb{E[}\left \langle D\psi(\bar{t},\bar
{x}),\xi_{s}\right \rangle +\frac{1}{2}\left \langle D^{2}\psi(\bar{t},\bar
{x})\xi_{s},\xi_{s}\right \rangle ]+o(s)\leq0\text{.}
\]
Thus
\[
\partial_{t}\psi(\bar{t},\bar{x})-G(D^{2}\psi(\bar{t},\bar{x}))\leq
0,\  \ u|_{t=0}=\varphi.
\]
This implies $u$ is a viscosity subsolution of the PDE
\[
\partial_{t}u-G(D^{2}u)=0.
\]
Similarly we can also prove that $u$ is a viscosity supersolution.
Thus $\xi_{1}$ is $G$-normally distributed. The proof is complete.
\end{proof}

\begin{remark}
The above probabilistic proof of central limit theorem can be also applied to
prove the following more general situation. We assume that $\{(X_{i}
,Y_{i})\}_{i=1}^{\infty}$ is an i.i.d sequence in the following sense: for
each $i$, $(X_{i+1},Y_{i+1})$ is identically distributed with respect to
$(X_{1},Y_{1})$ and is independent of $(X_{1},Y_{1})$, $\cdots,(X_{i},Y_{i})$,
we also assume that $\mathbb{E}[X_{1}]=\mathbb{E}[-X_{1}]=0$, and $|X_{i}
|^{3}$, $|Y_{i}|^{2}\in \mathcal{H}$. Then we have
\[
\sum_{i=1}^{n}(\frac{X_{i}}{\sqrt{n}}+\frac{Y_{i}}{n})\overset{\text{in law}
}{\rightarrow}(\xi,\eta)
\]
where $(\xi,\eta)$ is $G$-distributed with
\[
G(p,A):=\mathbb{E}[\left \langle p,Y_{1}\right \rangle +\frac{1}{2}\left \langle
AX_{1},X_{1}\right \rangle ].
\]
This implies that the function $u(t,x):=\mathbb{E}[\varphi(x+\sqrt{t}\xi
+t\eta)]$ is the unique solution of the PDE
\[
\partial_{t}u-G(Du,D^{2}u)=0,\  \ u|_{t=0}=\varphi.
\]

\end{remark}

\begin{remark}
The above result provides a new method to the proof of the existence of
solutions for a type of degenerate fully nonlinear and possibly degenerate
PDEs. It also indicates a numerical method to solve such type of PDE. This
approach is quite different from the well-known approach of Perron's method
for the existence. Perron's method is note constructive.
\end{remark}

\begin{remark}
The above method can be also applied to treat the solution existence of the
following type of SDE
\[
dX_{t}=b(X_{t})dt+\beta(X_{t})d\left \langle B\right \rangle _{t}+\sigma
(X_{t})dB_{t}
\]
for the case when $b$, $\beta$ and $\sigma$ are only continuous.
\end{remark}

config: arxiv


TITLE: A question about the socle of an algebra.
QUESTION [0 upvotes]: Let $A$ be a finite dimensional algebra over a field K. Suppose $e$ is an idempotent such that $Ae$ is the direct sum of all indecomposable injective-projective left $A$-modules. Is that $soc(_AA) \in add(soc(Ae))$ holds?

REPLY [0 votes]: No, this is not true. For example, an algebra need not have any projective-injective modules. Consider the path algebra of the quiver $1\to 2\leftarrow 3$. Then none of the projective modules is injective, hence $e=0$, but $soc(A)=3S_2$.
It is true for algebras of positive dominant dimension, i.e. if the algebra, considered as a left module over itself embeds into a projective-injective module. Then, since $\operatorname{soc}$ is a left exact functor, $\operatorname{soc}(A)\in\operatorname{add} \operatorname{soc}(I)\subseteq  \operatorname{add}\operatorname{soc}(Ae)$ for this projective-injective module $I$. This class of algebras for example contains Auslander algebras as well as selfinjective algebras.


TITLE: Invertible (isometric) sections of certain hom bundles over sphere
QUESTION [2 upvotes]: Assume  that we  have  a  vector  bundle $E$ over  $S^n$.

Is  there  a  continuous  family of invertible linear  maps  $T_x:E_x \to  E_{-x}$?

Here  continuity has  the  obvious  meaning as soon as  we  have  trivialization for  the  bundle  around $x$ and  $-x$
There is an obvious affirmative answer for the non orientable line bundle over $S^1$,  for the tangent bundle of $S^n$ and  for  the  $2$  dimensional real vector  bundle $E \to S^2 \simeq \mathbb{C}P^1$ where  $E$ is the  tautological  complex line  bundle  over  $\mathbb{C}P^1$.
If the  answer is  yes, what  about  if  we  consider  $E$  as  a  Riemannian  bundle  and  we  require that every  $T_x$  be an  isometric  linear  map?

REPLY [1 votes]: You are asking if a real vector bundle of rank $k$ over $S^n$ is isomorphic to its antipodal image.
If $n$ is odd, the answer is always yes because the antipodal map is homotopic to the identity of $S^n$.
If $k=2$, the answer is easily seen to be yes, because the oriented $2$-bundles are classified by an integer, the Euler class.
If $n$ is even, such a real vector bundle amounts to a free homotopy class of continuous maps $S^n\to BO(k)$. i.e. to a pair of elements $\alpha, \bar\alpha$ of the homotopy group
$\pi_n(BSO(k))$, where $BSO(k)$ is the Grassmannian space of the oriented $k$-planes in $R^\infty$, and where $\bar\alpha$ is the image of $\alpha$ by an orientation-reversing isometry. One has $\pi_n(BSO(k))\cong\pi_{n-1}(SO(k))$. Your question amounts to ask that $\alpha=\alpha^{-1}$ or $\alpha=\bar\alpha^{-1}$. I believe that the second always holds, but I'm not sure.


text:
{\bf Problem.} In rectangle $ABCD$, $P$ is a point on $BC$ so that $\angle APD=90^{\circ}$. $TS$ is perpendicular to $BC$ with $BP=PT$, as shown.  $PD$ intersects $TS$ at $Q$.  Point $R$ is on $CD$ such that $RA$ passes through $Q$.  In $\triangle PQA$, $PA=20$, $AQ=25$ and $QP=15$.  Find $SD$. (Express your answer as a common fraction.) [asy]
size(7cm);defaultpen(fontsize(9));
real sd = 7/9 * 12;
path extend(pair a, pair b) {return a--(10 * (b - a));}

// Rectangle
pair a = (0, 0); pair b = (0, 16); pair d = (24 + sd, 0); pair c = (d.x, b.y);
draw(a--b--c--d--cycle);
label("$A$", a, SW);label("$B$", b, NW);label("$C$", c, NE);label("$D$", d, SE);

// Extra points and lines
pair q = (24, 7); pair s = (q.x, 0); pair t = (q.x, b.y);
pair r = IP(c--d, extend(a, q));
pair p = (12, b.y);
draw(q--a--p--d--r--cycle);draw(t--s);
label("$R$", r, E); label("$P$", p, N);label("$Q$", q, 1.2 * NE + 0.2 * N);label("$S$", s, S); label("$T$", t, N);

// Right angles and tick marks
markscalefactor = 0.1;
draw(rightanglemark(a, b, p)); draw(rightanglemark(p, t, s)); draw(rightanglemark(q, s, d));draw(rightanglemark(a, p, q));
add(pathticks(b--p, 2, spacing=3.4, s=10));add(pathticks(p--t, 2, spacing=3.5, s=10));

// Number labels
label("$16$", midpoint(a--b), W); label("$20$", midpoint(a--p), NW); label("$15$", midpoint(p--q), NE);
label("$25$", midpoint(a--q), 0.8 * S + E);
[/asy]
{\bf Level.} Level 4
{\bf Type.} Geometry
{\bf Solution.} We have $\angle B = \angle PTQ$ and $\angle TPQ = 180^\circ - \angle QPA - \angle APB = 90^\circ - \angle APB = \angle BAP$.  Therefore, $\triangle BAP \sim \triangle TPQ$.  Since $AB/AP = 4/5$, triangles $BAP$ and $PTQ$ are $\{3,4,5\}$ right triangles, and we have $TQ = \frac35(15) = 9$ and $TP = \frac45(15)=12$.  Since $ABCD$ is a rectangle and $TS$ is perpendicular to $BC$, then $ABTS$ is also a rectangle. Thus, $TS=BA=16$ and $QS=TS-QT=16-9=7$.

In triangles $PQT$ and $DQS$, $\angle PTQ=\angle DSQ=90^{\circ}$. Also, $\angle PQT$ and $\angle DQS$ are vertically opposite angles and are therefore equal. Therefore, $\triangle PQT$ and $\triangle DQS$ are similar triangles. Since $\triangle PQT$ and $\triangle DQS$ are similar triangles, the ratios of corresponding side lengths in these two triangles are equal.

That is, $\dfrac{SD}{TP}=\dfrac{QS}{QT}$ or $\dfrac{SD}{12}=\dfrac{7}{9}$ or $SD=12\times\dfrac{7}{9}=\boxed{\dfrac{28}{3}}$.

set_name: MATH


TITLE: Formal limit definition when x tends to infinity
QUESTION [1 upvotes]: Using the limit definition, show that:
$\lim\limits_{x\to\infty} \frac{x+7}{3x^2+2}=0$
I get blocked when I use the equation:
Definition $f(x)$ tends to $L$ as $x$ tends to $ \infty$ if and only if $$ \forall \epsilon >0, \exists \delta>0 \ \mbox{ such that} \ \forall x \ \mbox{where} \ x>\delta, \\ |f(x)-L|<\epsilon$$
if I apply the equation would look like this:
$\frac{x+7}{3x^2+2}-0 < ε$
but I do not know how to continue

REPLY [2 votes]: Here's one way to do it. Let's simplify the fraction a bit first with some inequalities. First, the denominator:
$$f(x) = \frac{x+7}{3x^2+2} < \frac{x+7}{3x^2}.$$
Now, for the numerator, notice that $x+7 < x+7x$ provided that $x>1$. So, continuing from above, we have
$$f(x) < \frac{x+7}{3x^2} < \frac{x+7x}{3x^2} = \frac{8x}{3x^2} = \frac8{3x}.$$
We want to guarantee that $f(x) < \epsilon$, so it is enough to guarantee that $\frac8{3x} < \epsilon$. Solving this last inequality for $x$, we have
$$x > \frac8{3\epsilon}.$$
Therefore we should choose any $\delta$ such that $$\delta > \max\left\{\frac8{3\epsilon},1\right\}.$$
(Recall that we need to make sure that $x>1$, which is why we're taking the maximum of $\frac8{3\epsilon}$ and $1$.)


TITLE: Hermitian Matrix eigenvalues
QUESTION [0 upvotes]: I am trying to show the following sentence: Let H be hermitian with $\sigma(H) \subseteq \lbrace-r,r\rbrace$. Show that H ∘ H = $r^2$* I.
Since H is hermitian, we know that we can decompose the matrix into H=SDS* with S unitary and D diagonal matrix. H ∘ H = $(SDS^*) ∘ (SDS^*) = SD(S^* S)DS^* = SD^2S^*$. I know since the eigenvalues are $\subseteq \lbrace -r,r\rbrace, D^2 = r^2*I $ but that's not the fact that I needed to show. Any ideas where I am wrong?
($\sigma(H)$ is the set of all eigenvalues of H)

REPLY [0 votes]: $S$ is unitary so $SS^\ast = I$. Thus
$$H^2 = SD^2S^\ast = S(r^2 I)S^\ast  = r^2 I SS^\ast = r^2I$$


TITLE: Is the fixpoint $x=0$ stable, attracting, repelling for $f(x)= x\sin(1/x)$?
QUESTION [1 upvotes]: I'm working on exercise 1.5.8 from Goodson's Chaotic Dynamics.
We have the function $f(x) = x\sin(\frac{1}{x})$ for $x\neq 0$ and $f(0)=0$. I have already proven that $x=0$ is not an isolated point. Now I need to prove if $x=0$ is a stable, attracting or repelling point.
I know I have to use the definition
$\forall\epsilon>0, \exists\delta>0, |x-0|<\delta\implies |f^n(x)-0|<\epsilon\forall n\in\mathbb{N}$.
I think $x=0$ is an unstable point, so I would have to find a counterexample to the definition given above. However I've got no idea how to solve this problem. Can somebody help me?

REPLY [1 votes]: You have to be careful here with which type of stability you are considering.
From scholarpedia, we have Lyapunov stability

Lyapunov stability:
$x^e$ is a stable equilibrium if for every neighborhood U of $x^e$ there is a neighborhood V ⊆ U such that every solution x(t) starting in V remains in U for all t≥0. Notice that x(t) need not approach $x^e$.

and asymptotic stability (among others).

Asymptotic stability:
An equilibrium $x^e$ is asymptotically stable if it is Lyapunov stable and additionally V can be chosen so that |x(t)−$x^e$|→0 as t→∞ for all x(0) ∈ V.

The definition you have stated is Lyapunov stability. For this system, $x=0$ is Lyapunov stable, but it is not asymptotically stable.
Lyapunov stability can be established by noticing that $|f(x)| \leq |x|$ for all $x$, and taking $\delta < \epsilon$. Then, you can directly see that $|f^n(x)-0|  \leq |x| < \delta < \epsilon$ for all n.
Asymptotic stability can be disproven by noticing that $f(x)$ has a fixed point or 2-cycle whenever $\sin\left(\frac{1}{x}\right) = \pm 1$. As this occurs whenever $x = \frac{2}{k\pi}$ for $k$ odd, there are infinitely many fixed points arbitrarily close to $0$ (they are counterexamples that cannot converge to $0$).


TITLE: How to calculate statistical "difference" of two samples in 0-100 range?
QUESTION [2 upvotes]: There are two groups of people, target and neutral
There are two group of events X and others
We're making assumption that people from target group react on event X very different to others.
We have some sets of reaction numeric values on different type of events by those two groups. And after running Student's T-Test in excel I get 0.032 for target group and 0.55 for others so I can tell that target group react on X different but is it possible to measure how much different in 0-100 range?
example of data set (not real data set) for both groups looks like
30, 20, 45, 15, 26 ... - reaction on X event
10, 14, 22, 8, 13 ... - reactions on other events

What can I use to measure how much reaction X different (alike in T-Test) to reaction on other events in 0-100 range and is it possible in general?

REPLY [1 votes]: If the data are approximately normal then use a Welch 2-sample t test. This test does not assume
that the two groups are sampled from populations with equal variances.
When you mention the 0-100 range, perhaps you mean the P-value which
gives the probability of a more extreme difference between groups than
observed in your data, if the two groups are really equal (null hypothesis).
A common standard is to reject the null hypothesis that the groups are
equal if the P-value is less than 5%. Most statistical software computes
the P-value of the test. Here is one description of the Welch test in Excel.


TITLE: Sheaf cohomology in non-commutative setup
QUESTION [3 upvotes]: Let $X$ be a topological space and $A$ a sheaf of noncommutative associative algebras over a fixed field $k$. My questions are:
1) Does the category of modules over A have enough injective?
2) If we have the affirmative answer of question (1), then suppose $M$ is a quasi-coherent sheaf over $A$ then the first cohomology of M: $H^1(X,M)$ is $0$?
I learn the cohomology of module over a scheme $X$ in Hartshorne book. But when we deform the structure sheaf $O_X$ of $X$, then does the cohomology theory still hold for quasi-coherent sheaves? Would you please tell me some books/papers about cohomology theory of module over a sheaf of (noncommutative) rings?
Many thanks!!!

REPLY [8 votes]: For the first question, the answer is yes, since the category of sheaves of $A$-modules is a Grothendieck Abelian Category. The least obvious condition one has to check is that this category has a generator - to construct it, just use
$\bigoplus_{U\subseteq X} (j_U)_!(A|_U)$,
where $U$ ranges over all open sets of $X$, $A$ is considered as sheaf of (left or right) $A$-modules, $j_U:U\rightarrow X$ is the natural inclusion, and $(j_U)_!$ is extension by zero.
For a nice introduction to Grothendieck Abelian Categories, check out section 14 of http://stacks.math.columbia.edu/download/injectives.pdf.
For the second question, if we take $X$ to be an elliptic curve, take $A = O_X$, and $M=A$, then we already have $H^1(X,M) \ne 0$.


TITLE: Does the common matrix expression $A \cdot B \cdot A'$ have a name?
QUESTION [1 upvotes]: In an algorithm I'm working with, the same type of expression comes up multiple times:
$X = A \cdot B \cdot A'$
Where $A$ is a row-vector, $B$ is a square matrix, and then (naturally) $A'$ is a column vector, so the result $X$ is a scalar value.
Since this comes up so often, I thought the expression might have a name.
But it has been difficult to google for this, and I have no idea what it might be called.
Is there a common name for this operation to take a vector and matrix to produce a scalar?

REPLY [3 votes]: This construction define a map $V\to\mathbb R$, where $V$ is a vector space
of dimension $n$ and $B$ is a $n\times n$ matrix. That is, vectors $A$ in $V$ are transformed to the scalar is a map
$$A\mapsto ABA^{\top},$$
that is dubbed quadratic form.


TITLE: How to show that $a_{n+1} = \sqrt{12+4a_n}$ is upper bounded?
QUESTION [1 upvotes]: I am trying to prove that the sequence given by the recursive relation $a_{n+1} = \sqrt{12+4a_n}$ is convergent, $a_1 = 1$.
I managed to prove using induction that the sequence is stricly increasing.
I'm using the theorem that states that

If a sequence is monotonous and bounded then it is convergent.

Therefore, I must now prove there exists an upper bound

My attempt
I will use induction again.

For $n = 1: a_1 = 5 < M\in\mathbb{R}$
For $n = k: a_k <  M\in\mathbb{R}$
For $n = k+1: a_{k+1} = \sqrt{12+4a_k} = 2\sqrt{3+a_k} < M \iff a_k < \frac{M^2}{2} -3 < \frac{M^2}{2} < M^2  $

I've shown that $a_{k+1}$ is less that $M^2$ whereas in the induction step I stated that $a_{k+1}$ is less than $M$. The square kinda confuses me, and I am not sure if I indeed proved boundness here, thus I ask this question

REPLY [3 votes]: The way to approach this sort of problems is usually as follows.
Imagine you have already proven that the sequence converges... so $\lim_{n\to\infty}a_n=a\in\mathbb R$. Wouldn't you be interested to find what's $a$? The way to do it is: in the equation $a_{n+1}=\sqrt{12+4a_n}$ you calculate limits of the left- and right-hand side when $n\to\infty$. You get:
$$a=\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}\sqrt{12+4a_n}=\sqrt{12+\lim_{n\to\infty}a_n}=\sqrt{12+4a}$$
so $a=\sqrt{12+4a}$ which implies $a=6$.
So what you have proven is that, if $a_n$ converges, it has to converge to $6$ and no other number. Also you know it converges (as you wouldn't be asked to prove it if it didn't!) so knowing that it is monotonically increasing, you immediately see that $a_n\lt 6$, approaching $6$ "from below", and actually $6=\sup\{a_n:n\in\mathbb N\}$.
Thus, maybe it pays off to now try to forget all that we said until this point, and prove that $a_n\lt 6$, which will immediately mean that your sequence is monotonically increasing and bounded - hence convergent.
And, indeed (proof by induction), $a_1=5\lt 6$ and if $a_n\lt 6$, then $a_{n+1}=\sqrt{12+4a_n}\lt\sqrt{12+4\cdot 6}=6$.


TITLE: Finding the characteristic polynomial
QUESTION [0 upvotes]: let $A$ be \begin{pmatrix}
    7 & -5 & -4  \\
    -1 & -2 & 1\\
    9 & -5 & -6
\end{pmatrix}
find the characteristic polynomial, eigenvalues and eigenvectors

So we need to find $|x\cdot I-A|=0$ or in this case:
\begin{vmatrix}
    x-7 & 5 & 4  \\
    1 & x+2 & -1\\
    -9 & 5 & x+6
\end{vmatrix}
Developing by the first row we get:

 $(x-7)\cdot \begin{vmatrix}
      x+2 & -1\\
    5 & x+6
\end{vmatrix}-5\cdot \begin{vmatrix}
    1 & -1\\
    -9 & x+6
\end{vmatrix}+4\cdot \begin{vmatrix}
    1 & x+2 \\
    -9 & 5 &
\end{vmatrix}$
$(x-7)[(x+2)(x+6)+5]-5(x+6-9)+4(5-9x-18)=-156-80x+x^2+x^3$
How should I approach from here? can apply I row/columns operations on $|x\cdot I-A|$?

REPLY [1 votes]: You can do elementary operations on lines and columns of determinants, and here, adding 3rd column to the 1st one, then substracting the 1st row to the 3rd will greatly simplify the computations:
\begin{align*}
\begin{vmatrix}
    x-7 & 5 & 4  \\
    1 & x+2 & -1\\
    -9 & 5 & x+6
\end{vmatrix}&=
\begin{vmatrix}
    x-3 & 5 & 4  \\
    0 & x+2 & -1\\
    x-3 & 5 & x+6
\end{vmatrix}=
(x-3)\begin{vmatrix}
    1 & 5 & 4  \\
    0 & x+2 & -1\\
    1 & 5 & x+6
\end{vmatrix}\\
&=
(x-3)\begin{vmatrix}
    1 & 5 & 4  \\
    0 & x+2 & -1\\
    0 & 0 & x+2
\end{vmatrix}=(x-3)(x+2)^2.
\end{align*}
You should find $E_3=\langle(1,0,-1)\rangle$ and $E_{-2}=\langle(1,1,1)\rangle$. As $\dim E_{-2}<2$, $A$ is not diagonalisable.


TITLE: Existence of a continuous surjective map from the surface of a sphere to a circle?
QUESTION [5 upvotes]: Does there exist a continuous surjective mapping  $f : S^2 \to S^1$ ? Intuitively, I think not, because every time in my head I try to demonstrate such a mapping, I always end up with a "tearing" in $S^2$. But I'm not sure how to prove that there doesn't exist such a mapping.

The first approach I thought of follows something like this outline:

Pick a path $\gamma$ in $S^1$ that is not homotopic to the constant path.
Artificially "lift it" to a path in $S^2$ (in some similar manner to how you life paths to covering spaces)
Show that the lifted path is homotopic to the constant path, and hence the initial path is homotopic to the constant path. Contradiction!

It doesn't seem like this way is fruitful however, since it isn't apparent how to "artificially lift the path" since $f$ isn't given as a covering map.
I feel like this comes down to a more basic fact however, and so I attempted to make the problem easier by generalising:

If $X$ is simply connected and $f : X \to Y$ is a continuous surjective map, then $Y$ is simply connected

because that seemed to be at the core of my thinking. But I couldn't prove this either, and I couldn't find mention of a theorem like this in my textbook.

REPLY [2 votes]: Just to make the map given in @Arthur's answer explicit, consider the map $f \colon S^2 \to S^1$ given by $$f(x,y,z) = e^{2\pi i z}.$$
Here, we view $S^2 \hookrightarrow \mathbb{R}^3$ and $S^1 \hookrightarrow \mathbb{C}$ in the natural way (precisely, as the unit spheres centered at the origin in $\mathbb{R}^3$ and $\mathbb{C} \cong \mathbb{R}^2$, respectively).
This is a continuous map since it is the restriction of the continuous map $\mathbb{R}^3 \to S^1$ given by the same formula. It is not difficult to see that this map is also surjective.
(It is in fact also a smooth surjection from $S^2$ to $S^1$, if we view $S^2$ and $S^1$ as smooth manifolds.)


TITLE: $L \otimes \bar{K}$ is never a field.
QUESTION [1 upvotes]: Let $K$ be a field and $L$ be a simple extension $(i.e.\  L=K(\alpha))$. Let $\bar K$ denotes the algebraic closure of $K$. Then prove that  $L \otimes \bar{K}$ is never a field.

If $f(x)$ be the minimal polynomial of $\alpha$ then  $$K[x]/(f(x))\cong K(\alpha) \implies K[x]/(f)\otimes_{K} \bar{K}  \cong K(\alpha)\otimes_K \bar{K}. $$
If, I can show that $$ K[x]/(f)\otimes_{K} \bar{K} \cong\bar{K}[x]/(f) ,$$ then I can prove the result. So help me in proving the above claim.

REPLY [1 votes]: USe the map $$ \phi : \bar{K}[x]\to  K[x]/(f)\otimes_{K} \bar{K} $$ $$ \sum_{i=0}^na_i x^i\mapsto \sum_{i=0}^n\bar{x}^i\otimes a_i $$

Show that $\phi$ is a surjective homomorphism.
Ker$(\phi)=(f)$

Conclude your claim.


TITLE: Norm from a scalar product only for $p = 2$
QUESTION [0 upvotes]: How to prove that the following norm, defined on $C([0,1],\mathbb{C})$ :
$||f||_p = (\int_0^1 |f(t)|^p dt)^{1/p}$ (for $p \in [1,+\infty[$) is "from" a scalar product only if $p=2$ ?
I don't know how to show it at all, I need food for thought... Thank you in advance !

REPLY [3 votes]: Since your name is French, I assume with scalar product (produit scalaire) that you're talking about the inner product.
You could use that the parallelogram law holds if and only if the norm is induced by an inner product. So you only need to check that $$2\|f\|_p^2+2\|g\|_p^2=\|f+g\|_p^2+\|f-g\|_p^2 $$
holds for all $f,g\in C([0,1],\mathbb{C})$ only for $p=2$.
This post contains more details about the proof of the parallelogram law.


TITLE: A question about the proof of Hahn-banach theorem
QUESTION [1 upvotes]: I am reading the proof of Hahn-banach on Bass(https://www.math.wustl.edu/~victor/classes/ma5051/rags100514.pdf) Theorem 18.5. When proving that the functional can be extended for 1 more dimension, why "we need to verify that the norm of f_1 is less than or equal to
1" instead of equal to 1, given that the statement of Theorem 18.5 says the extended functional has the same operator norm?

REPLY [1 votes]: The Hahn-Banach Theorem as presented in your attached document:

If $M$ is a subspace of a normed linear space $X$ and $f$ is a bounded real linear functional on $M$, then $f$ can be extended to a bounded linear functional $F$ on $X$ such that $\|F\| = \|f\|$.

At the relevant point in the proof, you are assuming that $\|f\|=1$. $M_1$ is defined to be the linear span of $M$ and $\{x_0\}$, and the extension $f_1$ to $M_1$ is defined by $f_1(x+\lambda x_0):=f(x)+\lambda\alpha$.
So for each $\epsilon>0$ there exists $x\in X$ with $\|x\|\leq 1$ and $|f(x)|>1-\epsilon$. Then $x\in M_1$, $\|x\|\leq 1$ and $|f_1(x)|=|f(x)|>1-\epsilon$ gives that, so $\|f_1\|>1-\epsilon$. Since this holds for arbitrary $\epsilon>0$, we have that $\|f_1\|\geq 1$. Now it suffices to prove that $\|f_1\|\leq 1$.


TITLE: Dirac equation without $i$
QUESTION [5 upvotes]: In Witten's review paper "Fermion path integrals and topological phases", the Dirac equation (Eq(2.2)) is
$$(\gamma^{\mu}D_{\mu}-m)\psi=0$$
which appears very strange to me. Initially I thought this is the equation in Euclidean metric, but this is not true. My argument is the following: Assuming we Wick rotate from the Minkowski metric $(-,+,+)$, then both $\gamma^0$ and $\partial_0$ will contribute an $i$, which will cancel each other out. Then overall we do not get an extra $i$ to cancel the $i$ in $i\gamma^{\mu}\partial_{\mu}$.
Another attempt: maybe we can extract a factor of $i$ out of the gamma matrices in a particular basis. This however, doesn't work. The reason is that after factoring out an $i$ from each of the gamma matrices, the new set of "gamma matrices" will no longer satisfy the Clifford algebra because $\{\gamma^{\mu}/i,\gamma^{\nu}/i\}=-2\eta^{\mu\nu}$.
Any help would be greatly appreciated.
========= Edit with detailed answer=============
Based on Madmax's answer, I elaborate on what is happening here. In short, whether there is an $i$ or not depends on whether the Minkowski metric used is $(+,-,-,-,...)$ or $(-,+,+,+,...)$. Apparently, people call these west coast convention and east coast convention\footnote{\url{https://www.math.columbia.edu/~woit/wordpress/?p=7773&cpage=1}}.
The basic idea of Dirac equation is to factorize the Klein Gordon (KG)equation, and the crucial thing here is that the KG equation takes slightly different forms in the two different conventions. In the many-plus metric, the KG eq takes the form $(-\partial^2+m^2)\psi=0$, whereas in the many-minus metric, it takes the form $(\partial^2+m^2)\psi=0$.
The we can easily see that in the many-minus metric, the Dirac equation should be $(i\gamma^{\mu}\partial_{\mu}-m)\psi=0$. Simply act $(-i\gamma^{\mu}\partial_{\mu}-m)$ on the left to obtain the KG eq in the many-minus metric. Very similarly, we can check that the Dirac equation has to take the form $(\gamma^{\mu}\partial_{\mu}-m)\psi=0$ in order to recover the KG equation in the many-plus convention.

REPLY [1 votes]: Please have a look at eq. (5.5.43) of Weinberg's monograph Theory of Quantum fields where Dirac equation appears without an i. The gamma matrices appear in (5.4.13) and (5.4.14). His spacetime metric is (1,1,1,-1) -- see equation (2.3.3). This is apparent, for example, from the equation that follows (2.5.27) and many other. Equation (5.4.17) gives the Dirac matrices explicitly.  His Dirac matrices are just -i x the usual in the Weyl representation (modulo a sign, related, I think to his choice of spacetime metric.


TITLE: Question about mathematical logic ∀x ∈ S, ∀z ∈ S, ∃y ∈ C
QUESTION [0 upvotes]: ∀x ∈ S, ∀z ∈ S, ∃y ∈ C,(x != z) ⇒ ¬(T(x, y) ∧ T(z, y))
I'm trying to express this in English, but I can't use the variables x or y in my sentence.
Basically it means for elements x in S, and all elements z in S, There exists a y in C, such that x being different than z implies.....
How can I say this without using the variables?  Cheers.

REPLY [1 votes]: "For any two different elements of the set $S$, there is an element of set $C$ which is not in the $T$ relation with both of them."


TITLE: I think I've found an invariant distribution for a transient discrete Markov chain - Where is my mistake?
QUESTION [0 upvotes]: Let

$E$ be an at most countable set equipped with the discrete topology $\mathcal E$
$X=(X_n)_{n\in\mathbb N_0}$ be a discrete Markov chain with values in $(E,\mathcal E)$, distributions $(\operatorname P_x)_{x\in E}$ and transition matrix $$p=\left(p(x,y)\right)_{x,y\in E}:=\left(\operatorname P_x\left[X_1=y\right]\right)\;.$$
$\tau_x^1:=\inf\left\{n\in\mathbb N:X_n=x\right\}$ and $$\varrho(x,y):=\operatorname P_x\left[\tau_y^1<\infty\right]$$

A measure $\mu$ on $(E,\mathcal E)$ is called invariant $:\Leftrightarrow$ $$\mu p=\mu\;,\tag 1$$ where $$\mu p\left(\left\{x\right\}\right):=\sum_{y\in E}\mu\left(\left\{y\right\}\right)p(y,x)\;\;\;\text{for }y\in E\;.$$ If $\mu$ is a probability measure with $(1)$, it's called an invariant distribution. Moreover, $x\in E$ is called transient $:\Leftrightarrow$ $$\varrho(x,x)<1\;.$$

At the German Wikipedia page, they state, that if $X$ is transient, i.e. all states are transient, then there exists no invariant distribution.
However, I think, that I've found a counterexample:

Consider the random walk on $\mathbb Z$ with transition probabilities $$p(x,x+1)=r\;\text{and}\;p(x,x-1)=1-r\;\;\;\text{for }x\in\mathbb Z$$ for some $r\in (0,1)$
Let $$\mu_1\left(\left\{x\right\}\right):=1\;\text{and}\;\mu_2\left(\left\{x\right\}\right):=\left(\frac r{1-r}\right)^x\;\;\;\text{for }x\in\mathbb Z$$ and $\mu_1(\emptyset):=\mu_2(\emptyset):=0$

It's easy to see, that $\mu_1$ and $\mu_2$ both satisfy $(1)$. Moreover, if $r\ne 1/2$, the random walk is transient and $\mu_1\ne \mu_2$. In addition, each non-negative linear combination of $\mu_1$ and $\mu_2$ is an invariant measure, too.

While these combinations are no invariant distribution in general, at least $\mu_1$ should be one. So, I think I've found an invariant distribution of a transient discrete Markov chain. Is this the world's end or did I made a mistake?

REPLY [2 votes]: You made more than one mistake. You mistranslated the German Wikipedia article – it doesn't talk about measures (Maß) but about distributions (Verteilung). You also seem to be equivocating between measures, probability measures and distributions in your arguments in English. The counting measure is indeed invariant under these transitions, but it's neither a probability measure nor a distribution.


TITLE: Integration against divergence free vector fields
QUESTION [4 upvotes]: Let $\chi:\Omega\to \mathbb{R}^n$ be a vector field on a bounded, smooth domain $\Omega \subset \mathbb{R}^n$. Assume that for any divergence free vector field $\eta:\Omega \to \mathbb{R}^n$ we have
$
\int_\Omega \eta \cdot \chi =0
$
Does this imply that $\chi = 0$ ?

REPLY [4 votes]: This does not imply that $\chi = 0$. What can be said is the following:
Let $\chi \colon \overline{\Omega} \rightarrow \mathbb{R}^n$ be a vector field on a bounded, smooth domain $\overline{\Omega} \subset \mathbb{R}^n$. Then,
$$ \int_{\Omega} \eta \cdot \chi \, d\Omega = 0 $$
for all smooth divergence free vector fields $\eta \colon \overline{\Omega} \rightarrow \mathbb{R}^n$ if and only if there exist a smooth function $\phi \colon \overline{\Omega} \rightarrow \mathbb{R}$ such that $\chi = \nabla\phi$ and $\left. \phi \right|_{\partial \Omega} = 0$.
Proof: Assume that $\chi = \nabla \phi$ with $\left. \phi \right|_{\partial \Omega} = 0$. Then, using Green's identity for integration by parts we have
$$ \int_{\Omega} (\nabla \phi) \cdot \eta \, d\Omega = \int_{\partial \Omega} \phi (\eta \cdot \hat{N}) \, dS - \int_{\Omega} \phi (\nabla \cdot \eta) \, d\Omega = 0 $$
for all divergence free vector fields $\eta$. For the other direction, given $\chi$ we can solve the Poisson problem $\Delta \phi = \nabla \cdot \chi$ on $\Omega$ with boundary condition $\left. \phi \right|_{\partial \Omega} = 0$. Such a solution satisfies $\Delta \phi = \nabla \cdot (\nabla \phi) = \nabla \cdot \chi$ which implies that $\nabla \cdot (\nabla \phi - \chi) = 0$. Thus, we can write $\nabla \phi = \chi + F$ with $F = \nabla \phi - \chi$ and $\nabla \cdot F = 0$. Taking $\eta = F$, we see that
$$ 0 = \int_{\Omega} \eta \cdot \chi \, d\Omega = \int_{\Omega} F \cdot  (\nabla \phi - F) \, d\Omega = \int_{\Omega} \nabla \phi \cdot F \, d\Omega - \int_{\Omega} F \cdot F \, d \Omega = -\int_{\Omega} F \cdot F \, d\Omega.$$
Thus, $F \equiv 0$ and $\nabla \phi = \chi$ with $\left. \phi \right|_{\partial \Omega} = 0$.

Remark: This shows that the orthogonal complement to the subspace of smooth divergent free vector fields on $\overline{\Omega}$ is given by the space of "grounded gradients" (gradients of smooth functions that vanish on the boundary of $\Omega$) where the inner product on the space of vector fields is given by the formula
$$ \left< X, Y \right> = \int_{\Omega} X \cdot Y \, d\Omega. $$


text:
\begin{document}

\bibliographystyle{hplain}

\title{Maps on graphs can be deformed to be coincidence free\thanks{MSC2000: 54H25, 55M20}}
\author{P. Christopher Staecker\thanks{
Address: Department of Mathematics and Computer Science, Fairfield
University, Fairfield CT, USA}
\thanks{Email: cstaecker@fairfield.edu}
\thanks{Keywords: Nielsen theory, coincidence theory}
}

\maketitle
\begin{abstract}
We give a construction to remove coincidence points of continuous maps
on graphs ($1$-complexes) by changing the maps by homotopies. When the
codomain is not homeomorphic to the circle, we show that any pair of
maps can be changed by homotopies to be coincidence free.
This means that there can
be no nontrivial coincidence index, Nielsen coincidence number, or coincidence
Reidemeister trace in this setting, and the results of our
previous paper ``A formula for the coincidence Reidemeister trace of
selfmaps on bouquets of circles" are invalid.
\end{abstract}

\section{Introduction}
Let $X$ and $Y$ be graphs, which are always assumed to be
nontrivial. Throughout, we will
consider continuous maps $f,g:X \to Y$ (continuous maps of
$X$ and $Y$ as dimension $1$ CW-complexes)
and examine the coincidence set
\[ \Coin(f,g) = \{ x \mid f(x) = g(x) \}. \]

The paper \cite{stae09a} attempts, in the special case of bouquets of
circles, to study coincidence points of $f$ and $g$ by computing the
Reidemeister trace, which would then allow the computation of the
Nielsen number of the pair $(f,g)$. This Nielsen number would be a
lower bound on the minimal number of coincidence points achievable by
deforming $f$ and $g$.

A serious error in \cite{stae09a} renders the approach fundamentally
misguided. The approach makes heavy use of the coincidence index,
which is not well-behaved for bouquets of circles. Our main result
(Theorem \ref{mccor}) is that maps $f,g:X \to Y$ of graphs
with $Y$ not homeomorphic to the circle can always be changed by
homotopy to be coincidence free.
Thus any coincidence index in this setting must always
be zero, and so any Nielsen number or Reidemeister
trace which were being computed in \cite{stae09a} must have the
value zero.

In Section \ref{intsection} we give our main result. We conclude in
Section \ref{errorsection} with a note on
the specific errors in \cite{stae09a}.

We would like to thank Robert F. Brown for many helpful suggestions on
the organization of the paper, and the referee for suggestions which
substantially simplified the paper.

\section{Removing coincidences by homotopy}\label{intsection}

Our strategy for removing coincidences can be intuitively described
using a road traffic analogy. Consider a coincidence point which
occurs on the interior of an edge of the
domain space. Then we parameterize this edge ($1$-cell) as the time interval
$[0,1]$, and we can view the maps $f$ and $g$ as being represented
by a pair of points which travel around the space $Y$.

Let us imagine
that these points represent cars traveling on a network of single-lane
roads (so that the cars may not pass one another), and a coincidence
point of the maps will represent a collision of the cars.
A removal of a coincidence point by a homotopy will consist of a
strategy for letting the two cars pass one another without colliding.

Avoiding a collision is possible provided that there is a fork in the
network of roads where at least three roads meet: When two cars are
about to collide, one of them reverses direction until the fork is
reached. At this point, the car which reversed direction moves onto
the third road and allows the other to pass. The cars can now proceed
back to their original meeting point, this time with their positions
reversed. Repeating this process before each imminent collision allows
the cars to complete their trips without colliding.

This strategy is formalized as follows:
\begin{thm}\label{remcoin}
Let $f,g:X \to Y$ be maps of connected graphs with $Y$ not a manifold,
and let $x\in \Coin(f,g)$ be a coincidence point in the interior of
some edge. Then there is an arbitrarily small neighborhood $U$ of
$x$ on which $f$ and $g$ can be changed by homotopy to be coincidence free.
\end{thm}
\begin{proof}
Let $\sigma$ be the edge ($1$-cell) containing $x$, which we identify with its
attaching map $\sigma:[0,1]\to X$. Let $x = \sigma(t_0)$, and let $U =
\sigma([t_0-\epsilon,t_0+\epsilon])$ for some small $\epsilon >0$. We
may assume that $f(x)=g(x)$ is a point on the interior of some
$1$-cell $\rho:[0,1]\to Y$. We can parameterize $\sigma$ and $\rho$
so that $f(x)=g(x)=\rho(1/2)$, and that $f$ and $g$ behave according
to the graph in Figure \ref{intgraph}. (We may perhaps have to
interchange the roles of $f$ and $g$.)

\begin{figure}
\newcommand{\xwidth}{60}
\[ \begin{xy}

(0,0)="ll";
(0,\xwidth) = "ul";
(\xwidth,0) = "lr";
(\xwidth,\xwidth) = "ur";

"ul";"ll"; **@{-};
"lr";"ll";**@{-};

(10,1);(10,-1);**@{-};
(30,1);(30,-1);**@{-};
(50,1);(50,-1);**@{-};
(-1,0);(0,0);**@{-};
(-1,10);(1,10);**@{-};
(-1,30);(1,30);**@{-};
(-1,50);(1,50);**@{-};
(-1,60);(1,60);**@{-};

(10,-3)*{\sigma(t_0-\epsilon)};
(30,-3)*{\sigma(t_0)};
(50,-3)*{\sigma(t_0+\epsilon)};
(-5,0)*{\rho(0)};
(-5,10)*{\rho(\frac14)};
(-5,30)*{\rho(\frac12)};
(-5,50)*{\rho(\frac34)};
(-5,60)*{\rho(1)};
(65,0)*{\sigma(t)};
(0,65)*{\rho(s)};

(10,50);(50,10); **@{-}; ?(0)*+!L{f};
(10,10);(50,50); **@{-}; ?(0)*+!L{g};
(30,30)*{\circ}
\end{xy} \]
\caption{Behavior of $f$ and $g$ on $U$.\label{intgraph}}
\end{figure}

The assumption that $Y$ is not a manifold means that we may choose the
CW-complex structure on $Y$ so that the vertex $\sigma(0)$ meets two
other $1$-cells $\gamma,\lambda:[0,1]\to Y$ with
$\gamma(0)=\lambda(0)=\sigma(0)$. This vertex is the ``fork in the road''.
Now we change $f$ and $g$ by homotopy on $U$ to maps $f'$ and $g'$
according to Figure \ref{intgraphremoved}. Informally, the two maps
retreat to the fork point, use the fork to pass one another
without colliding, and return to their original positions at time $t_0+\epsilon$.
\begin{figure}
\newcommand{\xwidth}{60}
\[ \begin{xy}

(0,0)="ll";
(0,\xwidth) = "ul";
(\xwidth,0) = "lr";
(\xwidth,\xwidth) = "ur";

"ul";"ll"; **@{-};
"lr";"ll";**@{-};

(10,1);(10,-1);**@{-};
(30,1);(30,-1);**@{-};
(50,1);(50,-1);**@{-};
(-1,0);(0,0);**@{-};
(-1,10);(1,10);**@{-};
(-1,30);(1,30);**@{-};
(-1,50);(1,50);**@{-};
(-1,60);(1,60);**@{-};

(10,-3)*{\sigma(t_0-\epsilon)};
(30,-3)*{\sigma(t_0)};
(50,-3)*{\sigma(t_0+\epsilon)};
(-3,0)*{0};
(-3,10)*{\frac14};
(-3,30)*{\frac12};
(-3,50)*{\frac34};
(-3,60)*{1};
(65,0)*{\sigma(t)};
(0,65)*{s};

(50,50);(36,0);**@{-}; ?(1)*+!L{g'};
(36,0);(24,25);**@{--};
(24,25);(16,0);**@{--};
(16,0);(10,10);**@{-};

(50,10);(44,0);**@{-}; ?(1)*+!L{f'};
(44,0);(36,25);**@{.};
(36,25);(24,0);**@{.};
(24,0);(10,50);**@{-};
\end{xy} \]
\caption{Behavior of $f'$ and $g'$ on $U$. Solid line indicates values
  in $\rho(s)$, dotted line indicates values in $\gamma(s)$, and dashed
  line indicates values in $\lambda(s)$.\label{intgraphremoved}}
\end{figure}
The maps $f'$ and $g'$ are free of coincidences on $U$, and the theorem is proved.
\end{proof}

The above theorem implies our main result, with the help of a lemma which is true for much more general spaces, though we only require it for complexes. Its proof is an exercise.
\begin{lem}
Let $f,g:X \to Y$ where $X$ and $Y$ are connected complexes, and let $x\in \Coin(f,g)$. Then for any neighborhood $U$ of $x$, we may change $f$ and $g$ by homotopy on $U$ so that $x$ is no longer a coincidence point.
\end{lem}

The lemma above means that we may assume that every coincidence of our maps occurs on the interior of an edge, and then Theorem \ref{remcoin} can be applied repeatedly to remove them. Thus we obtain:

\begin{thm}\label{mccor}
If $f,g:X \to Y$ are maps on connected graphs with $Y$ not homeomorphic to
the circle, then $f$ and $g$ can be changed by homotopy to be
coincidence free.
\end{thm}
\begin{proof}
If $Y$ is homeomorphic to the interval $[0,1]$, then $f$ and $g$ are trivially nullhomotopic and thus can be made to be coincidence free by deforming them into different constant maps. Thus we may assume that $Y$ is not a manifold, and we may freely use Theorem \ref{remcoin}.

First, we may change our maps by homotopy to be ``linear'' as in
\cite{stae09a} so that $\Coin(f,g)$ is a finite set. Furthermore by
the lemma we may assume that all coincidences occur at interior points
of edges. Then repeated application of Theorem \ref{remcoin} will
remove all coincidences.
\end{proof}

See the end of Section \ref{errorsection} for a note on the case where
$Y$ is the circle.

The above theorem highlights the fact that coincidence theory on
graphs is not a generalization of fixed point theory. It is
certainly possible for a selfmap $f$ on e.g.\ a bouquet of 2 circles to have
fixed points which cannot be removed by homotopy, even though (by
Theorem \ref{mccor}) any
coincidences of $f$ with the identity map \emph{can} be removed. This
occurs because our removal construction changes the
second map by homotopy as well as the first. This distinction does
not occur between fixed point and coincidence theory on manifolds and
some other spaces, as demonstrated by Brooks in \cite{broo71}, but
Brooks' result does not apply to complexes in general.

\section{The error of \cite{stae09a}, and the coincidence index}\label{errorsection}
The formula given for the Reidemeister trace in \cite{stae09a} uses essentially two ingredients: the computation of the Reidemeister class for each coincidence point, and the computation of the coincidence index for each coincidence point. The material concerning the Reidemeister class is essentially correct, and the material concerning the index is incorrect.

The error specifically arises on page 43 of \cite{stae09a}: ``Near any
point $x$ other than $x_0$, the space $X$ is an orientable
differentiable manifold, and we define the coincidence index as usual
for that setting.'' This formulation of the coincidence index is not
well-behaved under homotopy. If, over the course of the homotopy, the
coincidence value (the common value of $f(x)$ and $g(x)$) travels
through the wedge point $y_0$, this ``index'' will change unpredictably.

In fact, two fundamental properties of the coincidence index
are that it is invariant under
homotopies of $f$ and $g$, and that the index is zero when $f$ and $g$
are coincidence-free on $U$. Since (by Theorem \ref{mccor}) the
coincidence set for maps of graphs can always be made empty by
homotopies, any ``coincidence index'' in this setting must always give
the value zero.

Some nontrivial indices can be defined by restricting the
structure of either the domain or the codomain spaces. Gon\c{c}alves
in \cite{gonc99} gives a coincidence index for maps from a complex into a
manifold of the same dimension, which suffices to address the
exceptional case from Section \ref{intsection}, the case where $Y$ is
the circle. In this case Gon\c{c}alves's index does provide a
nontrivial coincidence index which generalizes the fixed point index.

Thus there are many
examples of maps $f,g:X \to S^1$ for which $\MC(f,g)$, the minimal
number of coincidence points when $f$ and $g$ are changed by homotopy,
is nonzero (this
will occur whenever Gon\c{c}alves's index is nonzero). In particular
when $X$ is also $S^1$,
it is known that $\MC(f,g)$ is the Nielsen number $N(f,g) = |\deg(f) -
\deg(g)|$, which is easily made nonzero.


TITLE: Is comparison lemme required for Hanson-Wright Inequality?
QUESTION [3 upvotes]: I am reading Chapter 6 of Vershynin’s book on High Dimensional probability. He proves the following theorem `Hanson-Wright’ in Theorem 6.2.1. (I am modifying it slightly to make my point succinct.)

$\textbf{Theorem}$ Let $X=(X_1, \ldots, X_n)$ be a random vector with independent mean $0$ sub-Gaussian coordinates. Let $A$ be an $n\times n$ matrix with $A_{ii}=0$ . Then, for every $t>0,$ we have $$P(|X^TAX|\ge t)\le 2\exp\left(-c\min\left(\frac{t^2}{K^4||A||_F^2}, \frac{t}{K^2||A||}\right)\right),$$
where $K=\max_i ||X_i||_{\psi_2}.$

The proof basically follows by controlling the moment generating function of $X^TAX.$ To this end, one uses a decoupling argument that allows us to replace $X^TAX$ by $X^TAX’$ where $X’$ is an independent copy of $X.$ This is the step that requires each coordinate of $X_i$ To be independent. More precisely, we get
$$\mathbb{E}(\exp(\lambda X^TAX))\le \mathbb{E}\exp(4\lambda X^TAX’).$$
The next step is to compare the moment generating function of $X^TAX$ With $g^TAg’$ Where $g, g’$ are independent $N(0, I_n)$ random vector. Using the comparison we obtain $$\mathbb{E}(4\lambda X^TAX’)\le \mathbb{E}(\exp(C\lambda g^TAg’)).$$
At this point, we compute the moment generating function (it is easier in Gaussian case) and do the usual optimization over $\lambda$ To get the desired inequality.
Now my question is the following. If I have $X$ and $X’$ Independent, using that $X$ is subgaussian with norm $K,$ I can show that $\mathbb{E}(\exp(\lambda X^TAX’))\le \mathbb{E}\exp(C\lambda^2 K^2 ||AX’||_2^2)$ for some $C.$ Now I use the fact that $||AX’||_2^2\le ||A||^2||X’||^2.$ Using that coordinate of $X’$ Are independent, I can show that $||X’||^2$ is subexponential and therefore I can bound the moment generating function of $||X’||_2^2.$ I did not work out the exact constants, but it seems to me that I am getting the same inequality.
Can anyone tell me if this approach is fine? Does it give the same inequality or not? My point is to understand if comparison lemma is just a tool to better organize the proofs or it really adds something that I do not get otherwise?

REPLY [0 votes]: It seems to me that you considerably overestimate with $\|AX\|\leqslant \|A\|\|X\|$. This makes the bound depend on the dimension of X. You dont want that, in fact matrix and vector are never decoupled like that in more accurate proofs.