text:
\begin{document}

\title[On the optimal Voronoi partitions for Ahlfors-David measures]{On the optimal Voronoi partitions for Ahlfors-David measures with respect to the geometric mean error}
\author{Sanguo Zhu, Youming Zhou}

\address{School of Mathematics and Physics, Jiangsu University
of Technology\\ Changzhou 213001, China.}
\email{sgzhu@jsut.edu.cn}

\subjclass[2000]{Primary 28A80, 28A78; Secondary 94A15}
\keywords{Ahlfors-David measures, geometric mean error, Voronoi partition.}

\begin{abstract}
Let $\mu$ be an Ahlfors-David probability measure on $\mathbb{R}^q$ with support $K_\mu$. For every $n\geq 1$, let $C_n(\mu)$ denote the collection of all the $n$-optimal sets for $\mu$ with respect to the geometric mean error. We prove that, there exist constants $d_1,d_2>0$, such that for each $n\geq 1$, every $\alpha_n\in C_n(\mu)$ and an arbitrary Voronoi partition $\{P_a(\alpha_n)\}_{a\in\alpha_n}$ with respect to $\alpha_n$, we have
\[
d_1n^{-1}\leq\min_{a\in\alpha_n}\mu(P_a(\alpha_n))\leq\max_{a\in\alpha_n}\mu(P_a(\alpha_n))\leq d_2n^{-1}.
\]
Moreover, we prove that for each $a\in\alpha_n$, the set $P_a(\alpha_n)$ contains a closed ball of radius $d_3|P_a(\alpha_n)\cap K_\mu|$ which is centered at $a$, where $d_3$ is a constant and $|B|$ denotes the diameter of a set $B\subset\mathbb{R}^q$. Some estimates for the measure and the geometrical size of the elements of a Voronoi partition with respect to an $n$-optimal set are established in a more general context. These estimates also allow us to drop an additional condition in our previous work on self-similar measures.
\end{abstract}

\maketitle

\section{Introduction}

Let $\nu$ be a Borel probability measure on $\mathbb{R}^q$. The quantization problem for $\nu$ is concerned with the approximation of $\nu$ by discrete measures of finite support in $L_r$-metrics. This problem has a deep background in information theory and engineering technology such as signal processing and pattern recognition \cite{BW:82,GN:98}. We refer to \cite{GL:00,GL:04} for rigorous mathematical theory of the quantization problem. In the past decades, this problem has attracted great interest of mathematicians (cf. \cite{GL:00,GL:04,GL:08,GL:12,GPM:02,KZ:15,KZ:16,Kr:13,PG:97,PK:01,PK:04}).
\subsection{Some definitions and basic facts}
Let $r\in [0,\infty)$ and $k\in\mathbb{N}$. Let $d$ denote the Euclidean metric on $\mathbb{R}^q$. For every $k\geq 1$, let $\mathcal{D}_k:=\{\alpha\subset\mathbb{R}^q:1\leq {\rm card}(\alpha)\leq k\}$. For $x\in\mathbb{R}^q$ and $\alpha\subset\mathbb{R}^q$, let $d(x,\alpha):=\inf_{a\in\alpha}d(x,a)$. The $k$th quantization error for $\nu$ of order $r$ can be defined by
\begin{eqnarray}\label{quanerror}
e_{k,r}(\nu)=\left\{\begin{array}{ll}\big(\inf\limits_{\alpha\in\mathcal{D}_k}\int d(x,\alpha)^{r}d\nu(x)\big)^{1/r}&r>0\\
\inf\limits_{\alpha\in\mathcal{D}_k}\exp\big(\int\log d(x,\alpha)d\nu(x)\big)&r=0\end{array}\right..
\end{eqnarray}
One may see \cite{GL:00} for some equivalent definitions and interpretations in various contexts. For $r\in[1,\infty)$, $e_{n,r}(\nu)$ is equal to the minimum error when approximating $\nu$ by discrete probability measures supported on at most $n$ points in the $L_r$-metrics.

A set $\alpha\in\mathcal{D}_k$ is called a $k$-optimal set for $\nu$ of order $r$, if the infimum in (\ref{quanerror}) is attained at $\alpha$. We call the points in such an $\alpha$ $k$-\emph{optimal points}. As in \cite{GL:00,GL:04}, we denote the collection of all the $k$-optimal sets for $\nu$ of order $r$ by $C_{k,r}(\nu)$ and simply write $C_k(\nu)$ for $C_{k,0}(\nu)$. For $r>0$, $C_{k,r}(\nu)\neq\emptyset$ if $\int |x|^rd\nu(x)<\infty$; and $C_k(\nu)$ is non-empty if the following condition is satisfied:
\[
\int_0^1 s^{-1}\sup_{x\in\mathbb{R}^q}\nu(B(x,s))ds<\infty.
\]
In particular, $C_k(\nu)\neq\emptyset$ if for some constants $C,t>0$, we have
\[
\sup_{x\in\mathbb{R}^q}\nu(B(x,\epsilon))\leq C\epsilon^t
 \]
for every $\epsilon>0$. Furthermore, whenever the support $K_\nu$ of $\nu$ is an infinite set, we have that $e_{n,r}(\nu)$ is strictly decreasing as $n$ increases and ${\rm card}(\alpha_n)=n$ for every $\alpha_n\in C_{n,r}(\nu)$. One can see Theorem 4.12 of \cite{GL:00} and Theorem 2.4 of \cite{GL:04} for more detailed information.

Let $\alpha$ be a non-empty finite subset of $\mathbb{R}^q$.  For each $a\in\alpha$, the Voronoi region  generated by $a$ with respect to $\alpha$ is defined by
\begin{equation}\label{vregion}
W(a|\alpha):=\{x\in\mathbb{R}^q:d(x,\alpha)=d(x,a)\}.
\end{equation}
A Voronoi partition (VP) with respect to  the set $\alpha$ is defined to be a Borel partition $\{P_a(\alpha)\}_{a\in\alpha}$ of $\mathbb{R}^q$ such that
\[
P_a(\alpha)\subset W(a|\alpha),\;a\in\alpha.
\]
Let us call a VP with respect to an $\alpha\in C_{k,r}(\nu)$ a $k$-\emph{optimal Voronoi partition}.

For a Borel set $A\subset\mathbb{R}^q$ and a non-empty finite subset $\alpha$ of $\mathbb{R}^q$, we define
\begin{eqnarray}
I_{\nu,r}(A,\alpha):=\left\{\begin{array}{ll}\int_{A}d(x,\alpha)^rd\nu(x)&r>0\\
\int_{A}\log d(x,\alpha)d\nu(x)&r=0\end{array}\right..
\end{eqnarray}
In the following, we simply write $I_\nu(A,\alpha)$ for $I_{\nu,0}(A,\alpha)$ for convenience.
\subsection{A significant concern about the Voronoi partition}
Let $\nu$ be an absolutely continuous probability measure on $\mathbb{R}^q$. In \cite{Ger:79}, Gersho conjectured that for $\alpha_n\in C_{n,r}(\nu)$ and an arbitrary VP $\{P_a(\alpha_n)\}_{a\in\alpha_n}$ with respect to $\alpha_n$, we have
\begin{equation}\label{gersho}
\lim_{n\to\infty}\frac{I_{\nu,r}(P_a(\alpha_n),\{a\})}{n^{-1}e_{n,r}^r(\nu)}=1,\;\;a\in\alpha_n.
\end{equation}
This conjecture is also significant for singular  Borel probability measures.

In \cite{GL:12}, Graf, Luschgy and Pag\`{e}s proved that, for a large class of absolutely continuous probability measures $\nu$, there exist constants $A_1,A_2>0$ such that
\begin{equation}\label{g5}
A_1n^{-1}e_{n,r}^r(\nu)\leq I_{\nu,r}(P_a(\alpha_n),\{a\})\leq A_2n^{-1}e_{n,r}^r(\nu),\;\;a\in\alpha_n.
\end{equation}
This is a weak version of (\ref{gersho}). One may see \cite{Kr:13,Zhu:20} for some other related work. We remark that for general probability measures, it is very difficult even to examine whether (\ref{g5}) holds.

It is known from \cite{GL:04} that $e_{n,r}(\nu)\to e_n(\nu)$ as $r$ decreases to zero. Thus, by letting $r\to 0$ in (\ref{g5}), it is natural to conjecture that, for a Borel probability measure $\nu$, there exist some constants $B_1,B_2$ such that, for an arbitrary $\alpha_n\in C_n(\nu)$ and an arbitrary VP $(P_a(\alpha_n))_{a\in\alpha_n}$, the following holds:
\begin{equation}\label{wgc}
B_1n^{-1}\leq\min_{a\in\alpha_n}\nu(P_a(\alpha_n))\leq\max_{a\in\alpha_n}\nu(P_a(\alpha_n))\leq B_2n^{-1},\;\;a\in\alpha_n.
\end{equation}
This can be regarded as a limiting case of the weak version (\ref{g5}).
\subsection{Statement of the main result}
A Borel measure $\mu$ $\mathbb{R}^q$ is called an $s_0$-dimensional Ahlfors-David measure if there exists some $\epsilon_0>0$ such that, for every $x\in{\rm supp}(\mu)$,
\begin{equation}\label{AD}
C_1\epsilon^{s_0}\leq\mu(B(x,\epsilon))\leq C_2\epsilon^{s_0},\;\epsilon\in(0,\epsilon_0).
\end{equation}
The asymptotics of the quantization errors for Ahlfors-David measures have been studied in detail by Graf and Luschgy (cf. \cite[Theorem 12.18]{GL:00}). One can also see \cite{GL:00,Mattila:95} for various examples of such measures.

In the remaining part of the paper, we always denote by $\mu$ a probability measure satisfying (\ref{AD}). In addition, by Lemma 12.3 of \cite{GL:00}, we assume that the second inequality in (\ref{AD}) holds for all $x\in\mathbb{R}^q$ and all $\epsilon>0$. For a set $B\subset\mathbb{R}^q$, we denote the diameter of $B$ by $|B|$. We will prove
\begin{theorem}\label{mthm}
Let $\mu$ be an Ahlfors-David probability measure on $\mathbb{R}^q$ with support $K_\mu$. There exist positive constants $d_1, d_2, d_3$, such that for each $n\geq 1$, every $\alpha_n\in C_n(\mu)$ and an arbitrary VP $\{P_a(\alpha_n)\}_{a\in\alpha_n}$, we have
\[
d_1n^{-1}\leq\min_{a\in\alpha_n}\mu(P_a(\alpha_n))\leq\max_{a\in\alpha_n}\mu(P_a(\alpha_n))\leq d_2n^{-1}.
\]
Moreover, for every $a\in\alpha_n$, $P_a(\alpha_n)$ contains a ball of radius $d_3|P_a(\alpha_n)\cap K_\mu|$ which is centered at $a$.
\end{theorem}

Our main idea for the proof of Theorem \ref{mthm} is to reduce the quantization problem for $\mu$ with respect to an arbitrarily large $n$ to that for some conditional measures of $\mu$ with respect to some bounded integers, and then we apply Theorem 2.4 of \cite{GL:04} which says that a subset $\beta$ of a $k$-optimal set is ${\rm card}(\beta)$-optimal for the corresponding conditional measure of $\mu$. In order to accomplish the above-mentioned reduction, we will select some auxiliary integers and establish a characterization for $n$-optimal sets for $\mu$ with respect to the geometric mean error. In order to complete the proof of Theorem \ref{mthm} by using \cite[Theorem 2.4]{GL:04}, we will prove some weak estimates for the measures and geometrical size of elements of an optimal Voronoi partition. These results will be given in a more general context and allow us to drop an additional condition in \cite{Zhu:13} that the considered measure vanishes on every hyperplane.

Unlike the $L_r$-quantization problem, in the study of the geometric mean error, the involved integrals are usually negative and the integrands are in logarithmic form. It turns out that those methods to characterize the optimal sets in the $L_r$-quantization problem are often not applicable. For instance, let $A\supset B$ be Borel sets and $\alpha$ a non-empty finite set, we have
$I_{\nu,r}(A,\alpha)\geq I_{\nu,r}(B,\alpha)$ for $r>0$; while for $r=0$, we usually have an inequality in the reverse direction:
$I_\nu(A,\alpha)\leq I_\nu(B,\alpha)$, because the integrands are usually negative. For this reason, the arguments in the present paper are substantially different from those in \cite{Zhu:20} which are for $L_r$-quantization for $\mu$.
\section{Preliminaries}
For a probability measure $\nu$ on $\mathbb{R}^q$, we always denote the support of $\nu$ by $K_\nu$.
Let $m$ be the smallest integer with $m>2(C_1^{-1}C_2)^{1/s_0}$. Let $k_0$ be the smallest integer such that $2m^{-k_0}<\epsilon_0$. Note that, for the Ahlfors-David probability measure $\mu$, $K_\mu$ is compact. Thus, for every $k\geq k_0$, we denote by $\phi_k$ the largest number of closed balls of radii $m^{-k}$ which are pairwise disjoint and centered in $K_\mu$. We fix such $\phi_k$ closed balls and denote them by $E_{k,i},1\leq i\leq\phi_k$. We define
\[
\Omega_k:=\{(k,i):1\leq i\leq \phi_k\}.
\]
By the definition of $m$, one can show that $\phi_k<\phi_{k+1}$ by using (\ref{AD}) and the arguments in the proof of \cite[Lemma 2.1]{Zhu:20}. For $\sigma\in\Omega_k$, we denote the center of $E_\sigma$ by $c_\sigma$ and define $A_\sigma:=B(c_\sigma,2m^{-k})$. Then
we have $K_\mu\subset\bigcup_{\sigma\in\Omega_k}A_\sigma$. The following lemma is a consequence of (\ref{AD}).
\begin{lemma}\label{lem5}
There exist constants $\eta_1,\eta_2>0$, such that for every $\sigma\in\Omega_k$,
\[
\eta_1\phi_k^{-1}\leq\mu(A_\sigma)\leq\eta_2\phi_k^{-1}.
\]
\end{lemma}
\begin{proof}
Note that $K_\mu\subset\bigcup_{\sigma\in\Omega_k}A_\sigma$. By (\ref{AD}), for every $\sigma\in\Omega_k$, we have
\begin{eqnarray*}
&&1\leq\sum_{\tau\in\Omega_k}\mu(A_\tau)\leq\phi_k C_2C_1^{-1}\mu(A_\sigma);
\\&&\mu(A_\sigma)\leq C_2C_1^{-1}2^{s_0}\mu(E_\sigma).
\end{eqnarray*}
It follows that $\mu(A_\sigma)\geq C_2^{-1}C_1\phi_k^{-1}$. Because the sets $E_\tau,\tau\in\Omega_k$, are pairwise disjoint, by (\ref{AD}), we have
\[
1\geq\sum_{\tau\in\Omega_k}\mu(E_\tau)\geq\phi_k C_1C_2^{-1}\mu(E_\sigma)\geq\phi_k C_1^2C_2^{-2}2^{-s_0}\mu(A_\sigma).
\]
Hence, we have $\mu(A_\sigma)\leq (C_1^{-1}C_2)^22^{s_0}\phi_k^{-1}$. It suffices to define
\[
\eta_1:=C_2^{-1}C_1\;{\rm and}\;\eta_2:=(C_1^{-1}C_2)^22^{s_0}.
\]
\end{proof}

Let $C_1, C_2$ be as given in (\ref{AD}). We define
\begin{eqnarray}
&&\delta:=\frac{1}{16}\big(C_1C_2^{-1}\big)^{\frac{1}{s_0}};\label{g3}\\
&&\mathcal{A}_\sigma:=\{\tau\in\Omega_k: (A_\tau)_{2\delta|A_\tau|}\cap(A_\sigma)_{2\delta|A_\sigma|}\neq\emptyset\};\label{z8}\\
&&M_\sigma:={\rm card}(\mathcal{A}_\sigma);\;A_\sigma^*:=\bigcup_{\tau\in\mathcal{A}_\sigma}A_\tau,\;\;\sigma\in\Omega_k.\label{z7}
\end{eqnarray}
\begin{remark}
The number $\delta$ is defined as above for two reasons. First, $\delta$ should be small enough so that the set $E_\omega\setminus B(x_0,2^{-1}\delta|A_\omega|)$ is large enough. Second, it will be convenient for us to estimate the $\mu$-measure of a ball $B(x_0,2^{-1}\delta|A_\omega|)$ by using (\ref{AD}). One may see Lemma \ref{lem2} below for more details.
\end{remark}

For $x\in\mathbb{R}$, let $[x]$ denote the largest integer not exceeding $x$. For $t>0$ and a set $A\subset\mathbb{R}^q$, we denote the closed $t$-neighborhood of $A$ by $(A)_t$.
 \begin{remark}\label{rem0}
 Let $L_0:=[2\delta^{-1}+10]$. By estimating the volumes,  we know that for each $\sigma\in\Omega_k$, the set $(A_\sigma)_{2\delta|A_\sigma|}$ can be covered by $L_0$ closed balls of radii $2^{-1}\delta|A_\sigma|$ which are centered in $(A_\sigma)_{2\delta|A_\sigma|}$. This can be seen as follows. First, we consider the largest number of pairwise disjoint closed balls of radii $4^{-1}\delta|A_\sigma|$ which are centered in $(A_\sigma)_{2\delta|A_\sigma|}$; and then we double the radii of the balls and get a cover for $(A_\sigma)_{2\delta|A_\sigma|}$. In the remaining part of the paper, we always denote by $B_\sigma$ the set of the centers of such $L_0$ closed balls.
 \end{remark}

 Using the next lemma, we collect some basic facts regarding $A_\sigma^*$. These facts will allow us to adjust the number of prospective optimal points in $(A_\sigma)_{\delta|A_\sigma|}$ without affecting the points in $K_\mu\setminus A_\sigma^*$ unfavorably. One may apply Lemma 8 of \cite{Zhu:08} to obtain an optional proof.
\begin{lemma}\label{lem9}
Let $\sigma\in\Omega_k$ and let $\emptyset\neq\beta\subset\mathbb{R}^q$ be a finite set. Then
\begin{enumerate}
\item[(a1)] there exists an integer $M_0$ such that $M_\sigma\leq M_0$.
\item[(a2)] for every $\tau\in\Omega_k\setminus\mathcal{A}_\sigma$ and every $x\in A_\tau$, we have
\[
d(x,(\beta\setminus (A_\sigma)_{\delta|A_\sigma|})\cup B_\sigma)\leq d(x,\beta).
\]
\end{enumerate}
\end{lemma}
\begin{proof}
(a1) Note that $A_\sigma^*\subset B(c_\sigma,2(1+2\delta)|A_\sigma|)$ and $E_\tau,\tau\in\mathcal{A}_\sigma$, are pairwise disjoint. By estimating the volumes, one can see that
\[
M_\sigma (4^{-1}|A_\sigma|)^q\leq(2(1+2\delta)|A_\sigma|)^q.
\]
Hence, it is sufficient to define $M_0:=[(8(1+2\delta))^q]+1$.

(a2) Let $\tau\in\Omega_k\setminus\mathcal{A}_\sigma$ and $x\in A_\tau$. Then we have $(A_\sigma)_{2\delta|A_\sigma|}\cap(A_\tau)_{2\delta|A_\tau|}=\emptyset$. Therefore, $x\in\mathbb{R}^q\setminus(A_\sigma)_{2\delta|A_\sigma|}$. We have two cases:

 Case 1: $d(x,\beta)=d(x,\beta\setminus(A_\sigma)_{\delta|A_\sigma|})$, then (a2) is clearly true.

 Case 2: $d(x,\beta)=d(x,\beta\cap(A_\sigma)_{\delta|A_\sigma|})$. We denote the boundary of a set $B$ by $\partial B$. Note that $(A_\sigma)_{2\delta|A_\sigma|}$ is compact with non-empty interior. We may select a $z_0\in \partial(A_\sigma)_{2\delta|A_\sigma|}$ such that
 \[
 d(x,z_0)=d(x,(A_\sigma)_{2\delta|A_\sigma|})=d(x,\partial(A_\sigma)_{2\delta|A_\sigma|})
 \]
 By the definition of $B_\sigma$, there exists some $b\in B_\sigma$ such that $d(z_0,b)\leq 2^{-1}\delta|A_\sigma|$. For every $a\in \beta\cap(A_\sigma)_{\delta|A_\sigma|}$, we have
\begin{equation*}
d(x,a)\geq d(x,z_0)+\delta|A_\sigma|>d(x,z_0)+d(z_0,b)\geq d(x,b).
\end{equation*}
Hence, $d(x,\beta\cap(A_\sigma)_{\delta|A_\sigma|})\geq  d(x,b)$.
\begin{eqnarray*}
d(x,\beta)=d(x,\beta\cap(A_\sigma)_{\delta|A_\sigma|})\geq d(x,B_\sigma)\geq d(x,(\beta\setminus (A_\sigma)_{\delta|A_\sigma|})\cup B_\sigma)).
\end{eqnarray*}
This completes the proof of the lemma.
\end{proof}

\section{Weak estimates for measures and geometrical size of $P_a(\alpha_k)$}

Let $C,t>0$. We consider compactly supported measures $\nu$ satisfying
\begin{equation}\label{holder}
\sup_{x\in\mathbb{R}^q}\nu(B(x,\epsilon))\leq C\epsilon^t \;\;{\rm for\; every}\; \epsilon>0.
\end{equation}
Without loss of generality, we may assume that $C\geq1$. As in \cite{GL:04}, we write
\[
\hat{e}_k(\nu)=\log e_{k,0}(\nu)=\inf_{\alpha\in\mathcal{D}_k}I_\nu(\mathbb{R}^q,\alpha).
\]

The following lemma can be seen as an analogue of \cite[Lemma 2.1]{Zhu:20}.
\begin{lemma}\label{lem1}
Let $\nu$ be a Borel probability measure on $\mathbb{R}^q$ with compact support $K_\nu$. Assume that $|K_\nu|\leq 1$ and (\ref{holder}) is satisfied.
Then for every $k\geq 2$, there exists a real number $\zeta_k>0$, which depends on $C$ and $t$ such that
\[
\hat{e}_{k-1}(\nu)-\hat{e}_k(\nu)\geq\zeta_k.
\]
\end{lemma}
\begin{proof}
Let $\alpha_{k-1}=\{a_i\}_{i=1}^{k-1}\in C_{k-1}(\nu)$. We define
\begin{eqnarray*}
&&\delta_{k,1}:=(4C(k-1))^{-\frac{1}{t}};\;\delta_{k,2}:=(2C(k-1))^{-\frac{1}{t}};\\
&&\delta_k:=2^{-1}\min\{\delta_{k,1},\delta_{k,2}-\delta_{k,1}\}.
\end{eqnarray*}
By (\ref{holder}), we have $\nu(K_\nu\setminus\bigcup_{i=1}^{k-1}B(a_i,\delta_{k,2}))\geq 2^{-1}$.
Let $l_k:=[(2\delta_k^{-1}+2)^q]+1$. Note that $|K_\nu|\leq 1$. Hence, $K_\nu\setminus\bigcup_{i=1}^{k-1}B(a_i,\delta_{k,2})$ can be covered by $l_k$ closed balls $B_i (1\leq i\leq l_k$) of radii $\delta_k$ which are centered in the set $K_\nu\setminus\bigcup_{i=1}^{k-1}B(a_i,\delta_{k,2})$ (cf. Remark \ref{rem0}). Thus, there exists some ball $B_i$ such that $\nu(B_i)\geq (2l_k)^{-1}$. Let $b_i$ denote the center of $B_i$. Then
\begin{eqnarray*}
\hat{e}_{k-1}(\nu)-\hat{e}_k(\nu)&\geq& I_\nu(\mathbb{R}^q,\alpha_{k-1})-I_\nu(\mathbb{R}^q,\alpha_{k-1}\cup\{b_i\})\\
&\geq&I_\nu(B_i,\alpha_{k-1})-I_\nu(B_i,\alpha_{k-1}\cup\{b_i\})\\&\geq&I_\nu(B_i,\alpha_{k-1})-I_\nu(B_i,\{b_i\})\\&\geq&\nu(B_i)(\log\delta_{k,1}-\log\delta_k)
\\&\geq&(2l_k)^{-1}\log 2.
\end{eqnarray*}
The proof of the lemma is complete by defining $\zeta_k:=(2l_k)^{-1}\log 2$.
\end{proof}

Using the next lemma, we establish a lower bound for the $\nu$-measure of the elements of a VP with respect to a $k$-optimal set for $\nu$ of order zero.
\begin{lemma}\label{lem11}
 Assume that the hypothesis of Lemma \ref{lem1} is satisfied. For each $k\geq 2$, there exists a positive real number $\underline{d}_k$ such that, for every $\alpha_k\in C_k(\nu)$ and an arbitrary VP $\{P_a(\alpha_k)\}_{a\in\alpha_k}$ with respect to $\alpha_k$, we have
\[
\min_{a\in\alpha_k}\nu(P_a(\alpha_k))\geq \underline{d}_k.
\]
\end{lemma}
\begin{proof}
Let $\alpha_k\in C_k(\nu)$ and $\{P_a(\alpha_k)\}_{a\in\alpha_k}$ a VP. Let $a\in\alpha_k$ be fixed. By Theorem 2.4 of \cite{GL:04}, $\nu(P_a(\alpha_k))>0$. We define $\beta:=\alpha_k\setminus\{a\}$. Then for every $x\in\bigcup_{b\in\beta}P_b(\alpha_k)$, we have $d(x,\beta)=d(x,\alpha_k)$. Thus,
\begin{eqnarray}\label{s10}
\hat{e}_{k-1}(\nu)-\hat{e}_k(\nu)&\leq& I_\nu(\mathbb{R}^q,\beta)-I_\nu(\mathbb{R}^q,\alpha_k)\nonumber\\&=&I_\nu(P_a(\alpha_k),\beta)-I_\nu(P_a(\alpha_k),\{a\}).
\end{eqnarray}
Note that $\sup_{x\in K_\nu}d(x,\alpha_k)\leq 2|K_\nu|\leq 2$ (cf. \cite[Lemma 5.8]{GL:04}). Therefore for every $x\in P_a(\alpha_k)\cap K_\nu$, we have,
$d(x,\beta)\leq 3|K_\nu|\leq 3$. It follows that
\begin{equation}\label{s8}
I_\nu(P_a(\alpha_k),\beta)\leq\nu(P_a(\alpha_k))\log 3.
\end{equation}
Now by \cite[Lemma 3.4]{GL:04}, we have
\begin{equation}\label{s9}
I_\nu(P_a(\alpha_k),\{a\})\geq\frac{1}{t}\big(\nu(P_a(\alpha_k))\log\nu(P_a(\alpha_k))-C\nu(P_a(\alpha_k))\big).
\end{equation}
We define $h(x):=-x\log x$ for $x>0$. Then $h(x)\to 0$ as $x$ decreases to zero. Thus,
there exists some $\eta_k>0$ such that $0<x<\eta_k$ implies $-x\log x<2^{-1}t\zeta_k$. Therefore, if $\mu(P_a(\alpha_k))<\eta_k$, using Lemma \ref{lem1} and (\ref{s10})-(\ref{s9}), we deduce
\begin{eqnarray*}
\zeta_k\leq\hat{e}_{k-1}(\nu)-\hat{e}_k(\nu)\leq\nu(P_a(\alpha_k))\log 3+\frac{1}{2}\zeta_k+\frac{C}{t}\nu(P_a(\alpha_k)).
\end{eqnarray*}
Thus, we obtain $\nu(P_a(\alpha_k))\geq 2^{-1}(\log3+Ct^{-1})^{-1}\zeta_k$. It suffices to define
\[
\underline{d}_k:=\min\big\{\eta_k,2^{-1}(\log3+Ct^{-1})^{-1}\zeta_k\big\}.
\]
\end{proof}

Next, we establish an upper bound for $\nu(P_a(\alpha_k)),a\in\alpha_k$.
\begin{lemma}\label{l1}
 Assume that the hypothesis of Lemma \ref{lem1} is satisfied. For each $k\geq 1$, there exists a positive real number $\overline{d}_k$ such that, for every $\alpha_k\in C_k(\nu)$ and an arbitrary VP $\{P_a(\alpha_k)\}_{a\in\alpha_k}$ with respect to $\alpha_k$, we have
\[
\max_{a\in\alpha_k}\nu(P_a(\alpha_k))\leq \overline{d}_k.
\]
\end{lemma}
\begin{proof}
Let $\alpha_k\in C_k(\nu)$ and $a\in\alpha_k$. Set
\[
\delta_{k,3}:=(2C)^{-\frac{1}{t}}\nu(P_a(\alpha_k))^{\frac{1}{t}}.
\]
Then we have $\delta_{k,3}<1$, since $C\geq 1$. By (\ref{holder}), we deduce that
\[
\nu(B(a,\delta_{k,3}))\leq C\delta_{k,3}^t=2^{-1}\nu(P_a(\alpha_k)).
\]
It follows that
\begin{equation}\label{z1}
\nu(P_a(\alpha_k)\setminus B(a,\delta_{k,3}))\geq\nu(P_a(\alpha_k))-\nu(B(a,\delta_{k,3}))\geq\frac{1}{2}\nu(P_a(\alpha_k)).
\end{equation}
Let $N_k:=[(8\delta_{k,3}^{-1})^q]+3$. One can easily see that
\begin{equation}\label{zsz1}
N_k\leq (16\delta_{k,3}^{-1})^{q}\;\; {\rm and}\;\;N_k^{-1}\geq16^{-q}\delta_{k,3}^q.
\end{equation}
Note that $|P_a(\alpha_k)\cap K_\nu|\leq |K_\nu|\leq 1$. By estimating volumes, one can see that
\[
(P_a(\alpha_k)\cap K_\nu)\setminus B(a,\delta_{k,3})
 \]
 can be covered by $N_k$ closed balls $B_i (1\leq i\leq N_k)$ of radii $4^{-1}\delta_{k,3}$ which are centered in $(P_a(\alpha_k)\cap K_\nu)\setminus B(a,\delta_{k,3})$. Thus, by (\ref{z1}) and (\ref{zsz1}), there exists some ball $B_i$ such that
\begin{eqnarray}\label{z3}
\nu(B_i\cap P_a(\alpha_k))&\geq& \frac{1}{2N_k}\nu(P_a(\alpha_k)\geq32^{-q}\delta_{k,3}^q\nu(P_a(\alpha_k))\nonumber
\\&\geq&32^{-q}(2C)^{-\frac{q}{t}}\nu(P_a(\alpha_k))^{1+\frac{q}{t}}\nonumber\\
&=:&D_1\nu(P_a(\alpha_k))^{1+\frac{q}{t}}.
\end{eqnarray}
Now we define $\beta:=\alpha_k\cup\{b_i\}$. Then we have the following estimate:
\begin{eqnarray}\label{z2}
\hat{e}_k(\nu)-\hat{e}_{k+1}(\nu)&\geq& I_\nu(\mathbb{R}^q,\alpha_k)-I_\nu(\mathbb{R}^q,\beta)\nonumber\\
&\geq& I_\nu(B_i\cap P_a(\alpha_k),\alpha_k)-I_\nu(B_i\cap P_a(\alpha_k),\beta)\nonumber\\&\geq& I_\nu(B_i\cap P_a(\alpha_k),\{a\})-I_\nu(B_i\cap P_a(\alpha_k),\{b_i\})
\end{eqnarray}
By the definition of $B_i$, for every $x\in B_i$, we have
\begin{equation}\label{g4}
d(x,a)\geq \frac{1}{2}\delta_{k,3},\;\;d(x,b_i)\leq \frac{1}{4}\delta_{k,3}.
\end{equation}
Now by the proof of \cite[Lemma 5.8]{GL:04}, for every $n\geq 1$, we have
\[
\hat{e}_n(\nu)-\hat{e}_{n+1}(\nu)\leq\frac{1}{n+1}\log3+C^{\frac{1}{2}}\frac{2}{t}\big(\frac{1}{n+1}\big)^{1/2}=:\chi_n.
\]
Using  this and (\ref{z3})-(\ref{g4}), we deduce
\[
\chi_k\geq\hat{e}_k(\nu)-\hat{e}_{k+1}(\nu)\geq\nu(B_i\cap P_a(\alpha_k))\log2\geq D_1\log 2\;(\nu(P_a(\alpha_k))^{1+\frac{q}{t}}.
\]
The proof of the lemma is complete by defining $\overline{d}_k:=\big(\chi_k (D_1\log 2)^{-1}\big)^{\frac{t}{t+q}}$.

\end{proof}

We end this section with an estimate for the geometrical size of the elements of a VP with respect to a $k$-optimal set $\alpha_k\in C_k(\nu)$.
\begin{lemma}\label{l2}
 Assume that the hypothesis of Lemma \ref{lem1} is satisfied. For each $k\geq 2$, there exists a number $g_k>0$ such that, for every $\alpha_k\in C_k(\nu)$ and every VP $\{P_a(\alpha_k)\}_{a\in\alpha_k}$ with respect to $\alpha_k$ and every $a\in\alpha_k$, we have, $P_a(\alpha_k)$ contains a closed ball of radius $g_k|P_a(\alpha_k)\cap K_\nu|$ which is centered at $a$.
\end{lemma}
\begin{proof}
Let $d_H$ denote the Hausdorff metric. We define $\phi: C_k(\nu)\mapsto\mathbb{R}$ by:
\[
\phi(\alpha_k):=\min_{a\in\alpha_k}\min_{b\in\alpha_k\setminus\{a\}}d(a,b).
\]
We first show that $\phi$ is continuous on $C_k(\nu)$. To see this, it is sufficient to consider an arbitrary accumulation point  (if any) $\alpha_k=\{a_i\}_{i=1}^k$ of $C_k(\nu)$. Assume that
$\beta_{k,n}=\{b_{i,n}\}_{i=1}^k\in C_k(\nu)$ and $d_H(\beta_{k,n},\alpha_k)\to 0$ as $n\to\infty$. Without loss of generality, we assume that $\phi(\alpha_k)=d(a_1,a_2)$. Let $\eta_0:=4^{-1}\phi(\alpha_k)$. Then for every $\epsilon\in (0,\eta_0)$, there exists some $N_0\geq 1$, such that for all $n\geq N_0$, we have $d_H(\beta_{k,n},\alpha_k)<\epsilon$. Thus, for every $1\leq i\leq k$, there exists a unique $1\leq j(i)\leq k$ such that $d(b_{j(i),n},a_i)<\epsilon$. Thus, we rewrite $\beta_{k,n}$ as $\{b_{j(i),n}\}_{i=1}^k$. For $1\leq i_1\neq i_2\leq k$, by the triangle inequality, we have
\begin{eqnarray}\label{ss1}
d(b_{j(i_1),n},b_{j(i_2),n})&\geq& d(a_{i_1},a_{i_2})-d(a_{i_1},b_{j(i_1),n})-d(a_{i_2},b_{j(i_2),n})\nonumber\\
&\geq&d(a_1,a_2)-2\epsilon.
\end{eqnarray}
It follows that $\phi(\beta_{k,n})\geq \phi(\alpha_k)-2\epsilon$ for every $n\geq N_0$. Also, we have
\begin{eqnarray}\label{ss2}
\phi(\beta_{k,n})&\leq& d(b_{j(1),n},b_{j(2),n})\nonumber\\&\leq& d(b_{j(1),n},a_1)+d(a_1,a_2)+d(a_2,b_{j(2),n})\nonumber\\&<&\phi(\alpha_k)+2\epsilon.
\end{eqnarray}
From (\ref{ss1}) and (\ref{ss2}), we obtain that $|\phi(\beta_{k,n})-\phi(\alpha_k)|<2\epsilon$ for every $n\geq N_0$. It follows that $\phi(\beta_{k,n})\to\phi(\alpha_k)$ as $n\to\infty$. Thus, $\phi$ is continuous on $C_k(\nu)$.

By \cite[Theorem 2.5]{GL:04}, $C_k(\nu)$ is $d_H$-compact. Thus, by the continuity of $\phi$, there exist some $\alpha_{k,1}\in C_k(\nu)$, such that
\[
\underline{\lambda}_k(\nu):=\min_{\alpha\in C_k(\nu)}\phi(\alpha)=\phi(\alpha_{k,1}).
\]
Clearly, we have $\underline{\lambda}_k(\nu)>0$. Now let $\alpha_k=\{a_i\}_{i=1}^k$ be an arbitrary $k$-optimal set for $\nu$ and $\{P_a(\alpha_k)\}_{a\in\alpha_k}$ an arbitrary VP with respect to $\alpha_k$. Then
\begin{eqnarray*}
B(a,3^{-1}\underline{\lambda}_k(\nu))\subset P_a(\alpha_k)\;{\rm and}\;|P_a(\alpha_k)\cap K_\nu|\leq|K_\nu|\leq 1.
\end{eqnarray*}

Let $\eta_k,\zeta_k$ be as defined in the preceding lemmas. Next, we establish a lower bound for $\underline{\lambda}_k(\nu)$ in terms of $\eta_k$ and $\zeta_k$ which depend only on $C,t,k,q$. Set
\[
B_k:=\min\big\{\frac{t\zeta_k}{4C},\eta_k\big\};\;\epsilon_k:=\min\big\{\big(\frac{B_k}{2C}\big)^{1/t},2^{-1}\big\};
\;s_k:=\frac{1}{2}(e^{\frac{\zeta_k}{4}}-1)\epsilon_k.
\]
Note that $\zeta_k<1$, we have that $s_k<\epsilon_k$. We are going to show that $\underline{\lambda}_k(\nu)\geq s_k$. Suppose that $\underline{\lambda}_k(\nu)=d(a_1,a_2)<s_k$, we deduce a contradiction. Write
\[
A_{\epsilon_k}:=B(a_1,\epsilon_k)\cap P_{a_1}(\alpha_k);\;\;\beta:=\alpha_k\setminus\{a_1\}.
\]
Then we have $\nu(A_{\epsilon_k})\leq C\epsilon_k^t<\min\{\eta_k,(4C)^{-1}t\zeta_k\}$. By the proof of Lemma \ref{lem11}, we know that $-\frac{1}{t}(\nu(A_{\epsilon_k})\log \nu(A_{\epsilon_k})<2^{-1}\zeta_k$. Further, one can easily see that for every $b\in\beta$ and $x\in P_b(\alpha_k)$, we have $d(x,\alpha_k)=d(x,b)=d(x,\beta)$. Thus,
\begin{eqnarray}\label{append1}
\hat{e}_{k-1}(\nu)-\hat{e}_k(\nu)&\leq& I_\nu(\mathbb{R}^q,\beta)-I_\nu(\mathbb{R}^q,\alpha_k)\nonumber\\&=&I_\nu(P_{a_1}(\alpha_k),\beta)-I_\nu(P_{a_1}(\alpha_k),\alpha_k).
\end{eqnarray}
Note that for $x\in A_{\epsilon_k}$, we have $d(x,\alpha_k)=d(x,a_1)$ and
\[
d(x,\beta)\leq d(x,a_2)\leq d(x,a_1)+d(a_1,a_2)<\epsilon_k+s_k<2\epsilon_k<1.
\]
Using this and \cite[Lemma 3.6]{GL:04}, we deduce
\begin{eqnarray*}
\Delta_1:&=&I_\nu(A_{\epsilon_k},\beta)-I_\nu(A_{\epsilon_k},\alpha_k)\\&\leq&I_\nu(A_{\epsilon_k},\{a_2\})-I_\nu(A_{\epsilon_k},\{a_1\})
\\&\leq&\nu(A_{\epsilon_k})\log(2\epsilon_k)-\frac{1}{t}\big(\nu(A_{\epsilon_k})\log \nu(A_{\epsilon_k})-C\nu(A_{\epsilon_k})\big)\\
&\leq&-\frac{1}{t}\nu(A_{\epsilon_k})\log \nu(A_{\epsilon_k})+\frac{C}{t}\nu(A_{\epsilon_k})\\
&<&2^{-1}\zeta_k+4^{-1}\zeta_k.
\end{eqnarray*}
For every $x\in P_{a_1}(\alpha_k)\setminus A_{\epsilon_k}=:B_{\epsilon_k}$, we have
\[
d(x,\alpha_k)=d(x,a_1)>\epsilon_k,\;d(x,\beta)\leq d(x,a_2)<d(x,a_1)+s_k.
\]
From this, we deduce that
\[
\frac{d(x,\beta)}{d(x,\alpha_k)}\leq\frac{d(x,a_2)}{d(x,a_1)}\leq \frac{d(x,a_1)+s_k}{d(x,a_1)}=1+\frac{s_k}{\epsilon_k}<e^{\frac{\zeta_k}{4}}.
\]
By the preceding inequality and the fact that $\nu(B_{\epsilon_k})<1$, we obtain
\begin{eqnarray*}
\Delta_2:=I_\nu(B_{\epsilon_k},\beta)-I_\nu(B_{\epsilon_k},\alpha_k)\leq\nu(B_{\epsilon_k})\log e^{\frac{\zeta_k}{4}}<\frac{\zeta_k}{4}.
\end{eqnarray*}
From this and (\ref{append1}), we deduce that $\hat{e}_{k-1}(\nu)-\hat{e}_k(\nu)\leq\Delta_1+\Delta_2<\zeta_k$, contradicting Lemma \ref{lem1}.
Thus, $\underline{\lambda}_k(\nu)\geq s_k$ and the proof of the lemma is complete by defining $g_k=3^{-1}s_k$.
\end{proof}
\section{Auxiliary measures and auxiliary integers}

\subsection{Some subsets of $A_\omega^*$ and auxiliary measures}
For a finite subset $\alpha$ of $\mathbb{R}^q$, let $W(a|\alpha),a\in\alpha$, be as defined in (\ref{vregion}).
Let $\delta$ be as defined in (\ref{g3}). Let $\omega,\sigma\in\Omega_k$ with $\sigma\neq\omega$. Let $x_0\in A_\omega\cap K_\mu$. The following three types of subsets of $A_\omega^*$ will be considered in the characterization for the optimal sets for $\mu$:
\begin{eqnarray*}
&&D_{\omega,1}^{(\alpha)}:=E_\omega\cup\bigg(\bigcup_{a\in\alpha\cap (A_\omega)_{\delta|A_\omega|}}(W(a|\alpha)\cap A_\omega^*)\bigg)\setminus B(x_0,2^{-1}\delta|A_\omega|);\\
&&D_{\omega,2}^{(\alpha)}(\sigma):=E_\omega\cup\bigg(\bigcup_{a\in\alpha\cap (A_\omega)_{\delta|A_\omega|}}(W(a|\alpha)\cap A_\omega^*)\bigg)\setminus E_\sigma;\\
&&D_{\omega,3}^{(\alpha)}:=E_\omega\cup\bigg(\bigcup_{a\in\alpha\cap (A_\omega)_{\delta|A_\omega|}}(W(a|\alpha)\cap A_\omega^*)\bigg).
\end{eqnarray*}
If no confusion arises, we write $D_{\omega,i}$ for $D_{\omega,i}^{(\alpha)}$ and write $D_{\omega,2}$ for $D_{\omega,2}^{(\alpha)}(\sigma)$.

For $\omega\in\Omega_k$, recall that $c_\omega$ is the center of $E_\omega$. We define
\[
D_{\omega,4}:=B(c_\omega,(2^{-1}-\delta)|E_\omega|)\subset E_\omega.
\]

For $1\leq i\leq 4$, let $\mu(\cdot|D_{\omega,i})$ denote the conditional measure of $\mu$ on $D_{\omega,i}$:
\begin{equation}\label{auxmeasure}
\mu(\cdot|D_{\omega,i})(A)=\frac{\mu(A\cap D_{\omega,i})}{\mu(D_{\omega,i})},\;A\;{\rm is\;a\; Borel\;set}.
\end{equation}
Let $f_{D_{\omega,i}}$ be a similarity mapping of similarity ratio $|D_{\omega,i}|$ and define
\begin{equation}
\nu_{D_{\omega,i}}:=\mu(\cdot|D_{\omega,i})\circ f_{D_{\omega,i}},\;K_{\nu_{D_{\omega,i}}}:={\rm supp}(\nu_{D_{\omega,i}}).
\end{equation}
Then $\nu_{D_{\omega,i}}$  is a probability measure satisfying $|K_{\nu_{D_{\omega,i}}}|\leq 1$.

In a similar manner, we define the measures $\nu_{E_\sigma},\sigma\in\Omega_k$. We have

\begin{lemma}\label{lem2}
There exists a constant $C$ such that, for $B\in \{D_{\omega,i}\}_{i=1}^4\cup\{E_\omega\}$, we have $\sup_{x\in\mathbb{R}^q}\nu_B(B(x,\epsilon))\leq C\epsilon^{s_0}$ for every $\epsilon>0$.
\end{lemma}
\begin{proof}
By the definition, $D_{\omega,i}\subset A_\omega^*$ and $A_\omega^*\subset B(c_\omega, 2(1+2\delta)|A_\omega|)$. Hence,
\begin{equation}\label{g6}
|D_{\omega,i}|\leq |A_\omega^*|\leq 4(1+2\delta)|A_\omega|,\;1\leq i\leq 3.
\end{equation}
Since the diameter of $B(x_0,2^{-1}\delta|A_\omega|)$ is equal to $\delta|A_\omega|$, we have
\begin{equation}\label{zz1}
(1-2\delta)|E_\omega|\leq\bigg|E_\omega\setminus B\big(x_0,2^{-1}\delta|A_\omega|\big)\bigg|\leq|D_{\omega,1}|\leq4(1+2\delta)|A_\omega|.
\end{equation}
It follows that $|E_\omega|=2^{-1}|A_\omega|\geq(8(1+2\delta))^{-1}|D_{\omega,1}|$. This and (\ref{AD}) yield that
\begin{eqnarray*}
\mu(D_{\omega,1})&\geq&\mu(E_\omega\setminus B\big(x_0,2^{-1}\delta|A_\omega|)\\&\geq&\mu(E_\omega)-\mu(B\big(x_0,2^{-1}\delta|A_\omega|))\\&\geq& C_1(2^{-1}|E_\omega|)^{s_0}-C_2(2^{-1}\delta|A_\omega|)^{s_0}\\&\geq&C_1(2^{-1}|E_\omega|)^{s_0}-C_1(16^{-1}|E_\omega|)^{s_0}
\\&=&C_1(2^{-s_0}-16^{-s_0})|E_\omega|^{s_0}\\&\geq&C_1(2^{-s_0}-16^{-s_0})(8(1+2\delta))^{-s_0}|D_{\omega,1}|^{s_0}.
\end{eqnarray*}
We write $\xi_1:=C_1(2^{-s_0}-16^{-s_0})(8(1+2\delta))^{-s_0}$. On the other hand, by (\ref{zz1}),
\begin{eqnarray*}
\mu(D_{\omega,1})&\leq&\mu(A_\omega^*)\leq C_2(2(1+2\delta)|A_\omega|)^{s_0}\\&\leq& C_2(4(1+2\delta))^{s_0}(1-2\delta)^{-s_0}|D_{\omega,1}|^{s_0}=:\xi_2|D_{\omega,1}|)^{s_0}.
\end{eqnarray*}
Note that for distinct words $\sigma,\omega\in\Omega_k$, we have $E_\sigma\cap E_\omega=\emptyset$. Thus, for $i=2,3$, we have $E_\omega\subset D_{\omega,i}\subset A_\omega^*$. Using these facts and (\ref{g6}), we deduce
\begin{eqnarray*}
\mu(D_{\omega,i})&\leq&\mu(A_\omega^*)\leq C_2(2(1+2\delta)|A_\omega|)^{s_0}\\&\leq& C_24^{s_0}(1+2\delta)^{s_0}|E_\omega|^{s_0}\leq C_24^{s_0}(1+2\delta)^{s_0}|D_{\omega,i}|^{s_0}=:\xi_3|D_{\omega,i}|^{s_0};\\
\mu(D_{\omega,i})&\geq&\mu(E_\omega)\geq C_1(2^{-1}|E_\omega|)^{s_0}\geq C_1 4^{-s_0}|A_\omega|^{s_0}\\
&\geq& C_1 4^{-s_0}(4(1+2\delta))^{-s_0}|D_{\omega,i}|^{s_0}=:\xi_4|D_{\omega,i}|^{s_0}.
\end{eqnarray*}
For every $\omega\in\Omega_k$, we have
\begin{eqnarray*}
C_1(2^{-1}|E_\omega|)^{s_0})\leq\mu(E_\omega)\leq C_2(2^{-1}|E_\omega|)^{s_0});\\
C_1(2^{-1}|D_{\omega,4}|)^{s_0})\leq\mu(D_{\omega,4})\leq C_2(2^{-1}|D_{\omega,4}|)^{s_0})
\end{eqnarray*}
We define $\xi:=\max\{\xi_1^{-1},\xi_4^{-1},\xi_2,\xi_3\}$. Then by the above analysis, we obtain
\begin{equation}\label{xi}
\xi^{-1}|B|^{s_0}\leq\mu(B)\leq \xi|B|^{s_0}.
\end{equation}
for $B\in\{D_{\omega,i}\}_{i=1}^4\cup\{E_\omega\}$. Thus, the lemma follows from \cite[Lemma 2.5]{Zhu:20}.
\end{proof}
\begin{remark}\label{rem2}
Let $\xi$ be as defined in (\ref{xi}). For $1\leq i\leq 3$, we have
\begin{eqnarray*}
\mu(D_{\omega,i})&\leq&\xi|D_{\omega,i}|^{s_0}\\&\leq& \xi|A_\omega^*|^{s_0}\leq \xi(4(1+2\delta)|A_\omega|)^{s_0}\\&\leq&
 \xi(8(1+2\delta))^{s_0}|E_\omega|^{s_0}\\&\leq&\xi(8(1+2\delta))^{s_0}C_1^{-1}(2^{-1}-\delta)^{-s_0}\min_{\sigma\in\Omega_k}\mu(D_{\sigma,4}).
\end{eqnarray*}
Let $\zeta:=\xi(8(1+2\delta))^{s_0}C_1^{-1}(2^{-1}-\delta)^{-s_0}$. Then for every $\sigma\in\Omega_k$, we have
\[
\max_{1\leq i\leq 3}\mu(D_{\omega,i})\leq\zeta\mu(D_{\sigma,4})\leq \zeta\mu(E_\sigma).
\]
\end{remark}

In the following we denote by $f_B$ the similarity mapping in the definition of the measure $\nu_B$. The subsequent two lemmas will be very important for the characterization for the optimal sets. One of them is a consequence of the definition of the auxiliary measures $\nu_B$, and the other is based on Lemma \ref{lem9}.
\begin{lemma}\label{l3}
Let $B\in\{D_{\omega,i}\}_{i=1}^4\cup\{E_\omega\}$. Let $\alpha$ be a non-empty finite subset of $\mathbb{R}^q$ with ${\rm card}(\alpha)=l_\alpha$. Then
$I_\mu(B,\alpha)\geq\mu(B)\log |B|+\mu(B)\hat{e}_{l_\alpha}(\nu_B)$, and equality holds if $f_B^{-1}(\alpha)\in C_{l_\alpha}(\nu_B)$.
\end{lemma}
\begin{proof}
By the definition of $\nu_B$ (cf. (\ref{auxmeasure})), we have
\begin{eqnarray*}
I_\mu(B,\alpha)&=&\int_{B}\log d(x,\alpha)d\mu(x)\\
&=&\mu(B)\int_{B}\log d(x,\alpha)d\mu(\cdot|B)(x)\\&=&\mu(B)\int_{B}\log d(x,\alpha)d\nu_B\circ f_B^{-1}(x)\\
&=&\mu(B)\log |B|+\mu(B)\int_{f_B^{-1}(B)}\log d(x,f^{-1}_B(\alpha))d\nu_B(x)\\&\geq&\mu(B)\log |B|+\mu(B)\hat{e}_{l_\alpha}(\nu_B).
\end{eqnarray*}
This completes the proof of the lemma.
\end{proof}

Let $\omega,\tau,\sigma\in\Omega_k$ with $\sigma\neq\omega$ and $\tau\in\Omega_k\setminus\mathcal{A}_\omega$. Let $D_{\omega,i},1\leq i\leq 3$, be as defined above. We write
\[
F_{\omega,i}=\left\{\begin{array}{ll}D_{\omega,1}\cup B(x_0,2^{-1}\delta|A_\omega|)&\;\;\;\;i=1\\
D_{\omega,2}\cup E_\sigma&\;\;\;\;i=2
\\D_{\omega,3}\cup D_{\tau,4}&\;\;\;\;i=3\end{array}\right..
\]
\begin{lemma}\label{compare2}
Let $\alpha,\gamma$ be non-empty finite subsets of $\mathbb{R}^q$. Let $B_\omega$ be as defined in Remark \ref{rem0}. We define
$\beta:=(\alpha\setminus(A_\omega)_{\delta|A_\omega|})\cup B_\omega\cup\gamma$. Then
\[
I_\mu(\mathbb{R}^q\setminus F_{\omega,i},\beta)\leq I_\mu(\mathbb{R}^q\setminus F_{\omega,i},\alpha),\;\;1\leq i\leq 3.
\]
\end{lemma}
\begin{proof}
Let $1\leq i\leq 3$ be fixed. By the definition of $D_{\omega,i}$ and $F_{\omega,i}$, we have
\[
\{x\in A_\omega^*: d(x,\alpha)=d(x,\alpha\cap(A_\sigma)_{\delta|A_\sigma|})\}\subset F_{\omega,i}.
\]
Therefore, for every $x\in \mathbb{R}^q\setminus F_{\omega,i}$, we have the following two cases:
\begin{enumerate}
\item[(b1)] $x\in A_\omega^*$ and $d(x,\alpha)=d(x,\alpha\setminus(A_\sigma)_{\delta|A_\sigma|})$; then clearly $d(x,\beta)\leq d(x,\alpha)$;
\item[(b2)] $x\in K_\mu\setminus A_\omega^*$. Note that $K_\mu\subset\bigcup_{\tau\in\Omega_k}A_\tau$. Thus, $x\in A_\tau$ for some $\tau\in\Omega_k\setminus\mathcal{A}_\omega$. By Lemma \ref{lem9} (a2), we also have, $d(x,\beta)\leq d(x,\alpha)$.
\end{enumerate}
Thus, $d(x,\beta)\leq d(x,\alpha)$ for every $x\in \mathbb{R}^q\setminus F_{\omega,i}$, which implies the lemma.
\end{proof}
\subsection{Selection of some auxiliary integers}

Let $L_0$ be as defined in Remark \ref{rem0}. We define
\begin{eqnarray*}
L_1:=[(2\delta^{-1}+1)^q]+1,\;\;L_2:=6^q;\;\;n_0:=[(4\delta^{-1}+1)^q]+1.
\end{eqnarray*}

\begin{remark}\label{rem4}
By estimating the volumes, one can see the following facts:
\begin{enumerate}
\item[(c1)] for each $\sigma\in\Omega_k$, the set $E_\sigma$ can be covered by $L_1$ closed balls of radii $2^{-1}\delta|E_\sigma|$ which are centered in $E_\sigma$. We denote by $\gamma_{E_\sigma}$ the set of the centers of such $L_1$ balls.

\item[(c2)] for $x\in A_\sigma\cap K_\mu$, the ball $B(x,2^{-1}\delta|A_\sigma|)$ can always be covered by $L_2$ closed balls of radii $4^{-1}\delta|A_\sigma|$ which are centered in $B(x,2^{-1}\delta|A_\sigma|)$. We will denote by $G_x$ the set of the centers of such $L_2$ closed balls.
\item[(c3)]$A_\sigma$ can be covered by $n_0$ closed balls of radii $4^{-1}\delta|A_\sigma|$ which are centered in $A_\sigma$. We denote by $H_\sigma$ the set of the centers of such $n_0$ balls.
\end{enumerate}
\end{remark}

With the above preparations, we are able to define an integer $n_1$ which will be used to give a lower estimate for the number of optimal points in $(A_\sigma)_{\delta|A_\sigma|}$.
\begin{lemma}\label{lem10}
Let $\zeta$ be as defined in Remark \ref{rem2}. There exists a smallest integer $n_1$ such that for every $\omega\in\Omega_k$ and $n\geq n_1$, we have
\[
\hat{e}_{n-L_0-L_2}(\nu_{D_{\omega,2}})-\hat{e}_{n+L_1}(\nu_{D_{\omega,2}})<\zeta^{-1}C_1C_2^{-1}\delta^{s_0}\log 2.
\]
\end{lemma}
\begin{proof}
By Lemma \ref{lem2} and \cite[Lemma 5.8]{GL:04}, we deduce,
\begin{eqnarray*}
&&\lim_{n\to\infty}(\hat{e}_{n-L_0-L_2}(\nu_{D_{\omega,2}})-\hat{e}_{n+L_1}(\nu_{D_{\omega,2}}))\\
&&\;\;\;\;\;\;\;\;=\sum_{h=-(L_0+L_2)}^{L_1-1}\lim_{n\to\infty}(\hat{e}_{n+h}(\nu_{D_{\omega,2}})-\hat{e}_{n+h+1}(\nu_{D_{\omega,2}}))=0.
\end{eqnarray*}
This implies the lemma.
\end{proof}

By \cite[Lemma 2.1]{Zhu:20}, there exists an integer $N$ such that $\phi_{k+1}\leq N\phi_k$. Next, we select three more integers $n_2,n_3, n_4$. These integers will be used to establish an upper bound for the number of optimal points in $(A_\sigma)_{\delta|A_\sigma|}$.

\begin{lemma}\label{lem6}
Let $\zeta$ and $M_0$ be as defined in Remark \ref{rem2} and Lemma \ref{lem9}. Then

 (d1) there exists a smallest integer $n_2>n_1+L_0+L_1$, such that for every $n\geq n_2$, $\sigma,\omega\in\Omega_k$, the following holds:
\[
\hat{e}_{n-L_0-n_1-L_1}(\nu_{D_{\omega,2}})-\hat{e}_{n+L_1}(\nu_{D_{\omega,2}})
<\zeta^{-1}\big(\hat{e}_{n_1+L_1-1}(\nu_{E_\sigma})-\hat{e}_{n_1+L_1}(\nu_{E_\sigma})\big);
\]

(d2) let $n_3:=(n_2+n_0)N$; there exists a smallest integer $n_4>M_0n_3+L_0+L_1$, such that for $n\geq n_4$ and every pair $\sigma,\omega\in\Omega_k$, the following holds:
\[
\hat{e}_{n-L_0-n_3-L_1}(\nu_{D_{\sigma,3}})-\hat{e}_{n+L_1}(\nu_{D_{\sigma,3}})
<\zeta^{-1}\big(\hat{e}_{n_3+L_1-1}(\nu_{D_{\omega,4}})-\hat{e}_{n_3+L_1}(\nu_{D_{\omega,4}})\big).
\]
\end{lemma}
\begin{proof}
This is a consequence of \cite[Lemma 5.8]{GL:04} and Lemmas \ref{lem1} and \ref{lem2}.
\end{proof}

\section{A characterization for the $n$-optimal sets}

Our first lemma in this section is analogous to \cite[Lemma 4.1]{Zhu:20}.
\begin{lemma}\label{lc}
We have $L_c:={\rm card}(\alpha_n\setminus\bigcup_{\sigma\in\Omega_k}(A_\sigma)_{\delta|A_\sigma|})\leq n_0\phi_k$.
\end{lemma}
\begin{proof}
Assume that $L_c>n_0\phi_k$. Let $H_\sigma$ be as defined in Remark \ref{rem4} (c3). Set
\[
\beta:=\bigg(\alpha_n\cap \bigcup_{\sigma\in\Omega_k}(A_\sigma)_{\delta|A_\sigma|}\bigg)\cup\bigg(\bigcup_{\sigma\in\Omega_k}H_\sigma\bigg).
\]
Then ${\rm card}(\beta)\leq n$. For $x\in K_\mu\subset\bigcup_{\sigma\in\Omega_k}A_\sigma$, we have $d(x,\beta)\leq d(x,\alpha_n)$. We choose an arbitrary $x\in K_\mu  $ with $d(x,\alpha_n)=d(x,\alpha_n\setminus\bigcup_{\sigma\in\Omega_k}(A_\sigma)_{\delta|A_\sigma|})$. Then we have, $d(x,\alpha_n)>\delta|A_\sigma|$ for some $\sigma\in\Omega_k$. Thus, for every $y\in B_x:=B(x,4^{-1}\delta|A_\sigma|)$, we have $d(y,\alpha_n)\geq \frac{3}{4}\delta|A_\sigma|$. Hence,
\begin{eqnarray*}
I_\mu(\mathbb{R}^q,\alpha_n)-I_\mu(\mathbb{R}^q,\beta)&\geq& I_\mu(B_x,\alpha_n)-I_\mu(B_x,\beta)\\&\geq&\mu(B_x)\big(\log(\frac{3}{4}\delta|A_\sigma|)-\log(\frac{1}{4}\delta|A_\sigma|)\big)\\&=&\mu(B_x)\log 3>0.
\end{eqnarray*}
It follows that $I_\mu(\mathbb{R}^q,\alpha_n)>I_\mu(\mathbb{R}^q,\beta)$, contradicting the optimality of $\alpha_n$.
\end{proof}

For every $n\geq (n_0+n_2)\phi_{k_0}$, there exists a unique $k\geq k_0$, such that
\begin{equation}\label{g2}
(n_0+n_2)\phi_k\leq n<(n_0+n_2)\phi_{k+1}.
\end{equation}
Recall that $n_3=(n_0+n_2)N$. By \cite[Lemma 2.1]{Zhu:20}, we have
\begin{equation}\label{ncomp}
(n_0+n_2)\phi_k\leq n<(n_0+n_2)N\phi_k=n_3\phi_k.
\end{equation}

From now on, we assume that $n,k$ satisfy (\ref{g2}). We fix an $\alpha_n\in C_n(\mu)$ and an arbitrary VP $\{P_a(\alpha_n)\}_{a\in\alpha_n}$. We write $D_{\omega,i}$ for $D_{\omega,i}^{(\alpha_n)}$. We define
\[
L_\sigma:={\rm card}(\alpha_n\cap(A_\sigma)_{\delta|A_\sigma|}),\;\sigma\in\Omega_k.
\]

Using the subsequent two lemmas, we establish a lower bound for $L_\sigma$.
\begin{lemma}\label{key1}
Let $\omega\in\Omega_k$ and $B\in\{D_{\omega,i}\}_{i=1}^3\cup\{E_\omega\}$. Then
\[
I_\mu(B,\alpha_n)\geq \mu(B)(\log|B|+\hat{e}_{L_\omega+L_1}(\nu_B)).
\]
\end{lemma}
\begin{proof}
We divide $B$ into two parts:
\[
B(1):=\{x\in B:d(x,\alpha_n)=d(x,\alpha_n\cap(A_\omega)_{\delta|A_\omega|)}\};\;B(2):=B\setminus B(1).
\]
By the definition, we have $B(2)\subset E_\omega$.  Let $\gamma_{E_\omega}$ be as defined in Remark \ref{rem4}. We define
$\gamma(\omega):=\big(\alpha_n\cap(A_\omega)_{\delta|A_\omega|}\big)\cup \gamma_{E_\omega}$.
Then ${\rm card}(\gamma(\omega))\leq L_\omega+L_1$ and clearly
$d(x,\alpha_n)\geq d(x,\gamma(\omega))$ for every $x\in B(1)$.
For $x\in B(2)$, we have
\[
d(x,\alpha_n))>\delta|A_\omega|=2\delta|E_\omega|>d(x,\gamma_{E_\omega})\geq d(x,\gamma(\omega)).
\]
Thus, for every $x\in B$, we have $d(x,\alpha_n)\geq d(x,\gamma(\omega))$. Thus, by Lemma \ref{l3},
\begin{eqnarray*}
I_\mu(B,\alpha_n)\geq I_\mu(B,\gamma(\omega))
\geq\mu(B)(\log|B|+\hat{e}_{L_\omega+L_1}(\nu_B)).
\end{eqnarray*}
This completes the proof of the lemma.
\end{proof}

Next, we give a lower bound for $\min\limits_{\sigma\in\Omega_k}L_\sigma$.
\begin{lemma}\label{lem7}
For every $\sigma\in\Omega_k$, we have $L_\sigma\geq n_1$.
\end{lemma}
\begin{proof}
Assume that $L_\sigma<n_1$ for some $\sigma\in\Omega_k$. We deduce a contradiction. By the assumption and Lemma \ref{lc}, we obtain
\[
\sum_{\tau\in\Omega_k\setminus\{\sigma\}}L_\tau>n-L_c-n_1\geq(n_2+n_0)\phi_k-n_0\phi_k-n_1>(\phi_k-1)n_2.
\]
Hence, there exists some $\omega\in\Omega_k\setminus\{\sigma\}$ such that $L_\omega>n_2$. We consider
\[
D_{\omega,2}=E_\omega\cup\bigg(\bigcup_{a\in\alpha_n\cap (A_\omega)_{\delta|A_\omega|}}(W(a|\alpha_n)\cap A_\omega^*)\bigg)\setminus E_\sigma.
\]
Note that it is possible that $E_\sigma\cap (A_\omega)_{\delta|A_\omega|}=\emptyset$. Let
\begin{eqnarray*}
&&\gamma_{L_\omega-L_0-n_1-L_1}(D_{\omega,2})\in C_{L_\omega-L_0-n_1-L_1}(\nu_{D_{\omega,2}}),\;\gamma_{n_1+L_1}(E_\sigma)\in C_{n_1+L_1}(\nu_{E_\sigma});
\\&&\beta:=\big(\alpha_n\setminus (A_\omega)_{\delta|A_\omega|}\big)\cup B_\omega\cup f_{D_{\omega,2}}(\gamma_{L_\omega-L_0-n_1-L_1}(D_{\omega,2}))\cup f_{E_\sigma}(\gamma_{n_1+L_1}(E_\sigma)).
\end{eqnarray*}
Then by applying Lemma \ref{compare2} to $F_{\omega,2}=D_{\omega,2}\cup E_\sigma$, we obtain
\begin{equation}\label{s4}
I_\mu(\mathbb{R}^q\setminus F_{\omega,2},\beta)\leq I_\mu(\mathbb{R}^q\setminus F_{\omega,2},\alpha_n).
\end{equation}
Next, we focus on the sets $D_{\omega,2}$ and $E_\sigma$. By the assumption, we have $L_\sigma\leq n_1-1$. Hence, by applying Lemmas \ref{l3} and \ref{key1} with $B=E_\sigma$, we deduce
\begin{eqnarray}
\Delta_{E_\sigma}:&=&I_\mu(E_\sigma,\alpha_n)-I_\mu(E_\sigma,\beta)\nonumber\\
&\geq&I_\mu(E_\sigma,\alpha_n)-I_\mu(E_\sigma,f_{E_\sigma}(\gamma_{n_1+L_1}(E_\sigma)))\nonumber\\
&\geq&\mu(E_\sigma)\big(\hat{e}_{n_1+L_1-1}(\nu_{E_\sigma})-\hat{e}_{n_1+L_1}(\nu_{E_\sigma})\big).
\end{eqnarray}
On the other hand, we apply Lemmas \ref{l3} and \ref{key1} with $B=D_{\omega,2}$, we have
\begin{eqnarray}
\Delta_{D_{\omega,2}}:&=&I_\mu(D_{\omega,2},\beta)-I_\mu(D_{\omega,2},\alpha_n)\nonumber\\
&\leq&I_\mu(D_{\omega,2},f_{D_{\omega,2}}(\gamma_{L_\omega-L_0-n_1-L_1}(D_{\omega,2})))-I_\mu(D_{\omega,2},\alpha_n)\nonumber\\
&\leq&\mu(D_{\omega,2})\big(\hat{e}_{L_\omega-L_0-n_1-L_1}(\nu_{D_{\omega,2}})-\hat{e}_{L_\omega+L_1}(\nu_{D_{\omega,2}})\big).
\end{eqnarray}
Note that $L_\omega>n_2$. By Lemma \ref{lem6} (d1) and Remark \ref{rem2}, we obtain that $\Delta_{E_\sigma}>\Delta_{D_{\omega,2}}$. This and (\ref{s4}), yield that $I_\mu(\mathbb{R}^q,\alpha_n)>I_\mu(\mathbb{R}^q,\beta)$, contradicting the optimality of $\alpha_n$.
\end{proof}

Our next lemma is very helpful for us to characterize the geometrical structure of an optimal VP.
\begin{lemma}\label{lem8}
For every $\sigma\in\Omega_k$, the following holds:
\[
\sup_{x\in A_\sigma\cap K_\mu}d(x,\alpha_n)\leq \delta|A_\sigma|.
\]
\end{lemma}
\begin{proof}
Assume that for some $\sigma\in\Omega_k$ and some $x_0\in A_\sigma\cap K_\mu$, we have, $d(x_0,\alpha_n)>\delta|A_\sigma|$. We will deduce a contradiction. By the assumption, for $x\in B(x_0,2^{-1}\delta|A_\sigma|)$, we have $d(x,\alpha_n)>2^{-1}\delta|A_\sigma|$. Let $G_{x_0}$ be as defined in Remark \ref{rem4} (c2). We consider
\[
D_{\sigma,1}=E_\sigma\cup\bigg(\bigcup_{b\in\alpha_n\cap(A_\sigma)_{\delta|A_\sigma|}}(W_b(\alpha_n)\cap A_\sigma^*)\bigg)\setminus B(x_0,2^{-1}\delta|A_\sigma|).
\]
Let $\gamma_{L_\sigma-L_0-L_2}(D_{\sigma,1})\in C_{L_\sigma-L_0-L_2}(\nu_{D_{\sigma,1}})$. We define
\begin{eqnarray*}
\gamma:=\big(\alpha_n\setminus(A_\sigma)_{\delta|A_\sigma|}\big)\cup G_{x_0}\cup B_\sigma\cup f_{D_{\sigma,1}}(\gamma_{L_\sigma-L_0-L_2}(D_{\sigma,1})).
\end{eqnarray*}
Then by applying Lemma \ref{compare2} to $F_{\sigma,1}=D_{\sigma,1}\cup B(x_0,2^{-1}\delta|A_\sigma|)$, we obtain
\begin{equation}\label{s1}
I_\mu(\mathbb{R}^q\setminus F_{\sigma,1},\gamma)\leq I_\mu(\mathbb{R}^q\setminus F_{\sigma,1},\alpha_n).
\end{equation}
For every $x\in B(x_0,2^{-1}\delta|A_\sigma|)$, we have $d(x,\gamma)\leq 4^{-1}\delta|A_\sigma|$. It follows that
\begin{eqnarray}\label{s2}
\Delta_{x_0}:&=&I_\mu(B(x_0,2^{-1}\delta|A_\sigma|),\alpha_n)-I_\mu(B(x_0,2^{-1}\delta|A_\sigma|),\gamma)\nonumber\\&\geq&\mu(B(x_0,2^{-1}\delta|A_\sigma|))\log 2\nonumber\\&\geq&C_1(2^{-1}\delta|A_\sigma|)^{s_0}\log 2\nonumber\\&\geq& C_1C_2^{-1} \delta^{s_0}\mu(A_\sigma)\log2.
\end{eqnarray}
By applying Lemmas \ref{l3} and \ref{key1} with $B=D_{\sigma,1}$, we have
\begin{eqnarray}\label{s3}
\Delta_{D_{\sigma,1}}:&=&I_\mu(D_{\sigma,1},\gamma)-I_\mu(D_{\sigma,1},\alpha_n)
\nonumber\\&\leq&I_\mu(D_{\sigma,1},f_{D_{\sigma,1}}(\gamma_{L_\sigma-L_0-L_2}(D_{\sigma,1})))-I_\mu(D_{\sigma,1},\alpha_n)\nonumber\\&\leq&
\mu(D_{\sigma,1})\big(\hat{e}_{L_\sigma-L_0-L_2})(\nu_{D_{\sigma,1}})-\hat{e}_{L_\sigma+L_1}(\nu_{D_{\sigma,1}})\big).
\end{eqnarray}
From Lemma \ref{lem7}, we know that $L_\sigma\geq n_1$. Thus, by Lemmas \ref{lem2}, \ref{lem10} and Remark \ref{rem2}, we obtain that $\Delta_{x_0}>\Delta_{D_{\sigma,1}}$. This and (\ref{s1}) imply that $I_\mu(\mathbb{R}^q,\gamma)<I_\mu(\mathbb{R}^q,\alpha_n)$, which contradicts the optimality of $\alpha_n$.
\end{proof}

\begin{remark}\label{rem5}
By Lemma \ref{lem8}, we obtain that, whenever $n\geq(n_0+n_2)\phi_k$, we have
$L_c={\rm card}(\alpha_n\setminus\bigcup_{\sigma\in\Omega_k}(A_\sigma)_{\delta|A_\sigma|})=0$. Therefore, we have
\[
\alpha_n\subset\bigcup_{\sigma\in\Omega_k}(A_\sigma)_{\delta|A_\sigma|}.
\]
\end{remark}

\begin{lemma}\label{lemma1}
Let $\emptyset\neq\beta\subset\mathbb{R}^q$ be a finite set and $l_\beta(\omega):={\rm card}(\beta\cap E_\omega)$ for $\omega\in\Omega_k$. Then the following estimate holds:
\[
I_\mu(D_{\omega,4},\beta)\geq\mu(D_{\omega,4})(\log|D_{\omega,4}|+\hat{e}_{l_\beta(\omega)+L_1}).
\]
\end{lemma}
\begin{proof}
Let $\gamma_{E_\omega}$ be as defined in Remark \ref{rem4} (c1). Since $D_{\omega,4}\subset E_\omega$, we have $d(x,\gamma_{E_\omega})\leq2^{-1}\delta|E_\omega|$ for every $x\in D_{\omega,4}$. We define
\[
\gamma(\omega):=(\beta\cap E_\omega)\cup\gamma_{E_\omega}.
 \]
Then ${\rm card}(\gamma(\omega))\leq l_\beta(\omega)+L_1$. Let $x\in D_{\omega,4}$. If $d(x,\beta)=d(x,\beta\cap E_\omega)$, then it is clear that $d(x,\gamma(\omega))\leq d(x,\beta)$. Otherwise, we have
\[
d(x,\beta)=d(x,\beta\setminus E_\omega)\geq\delta|E_\omega|>d(x,\gamma_{E_\omega})\geq d(x,\gamma).
\]
Thus, $I_\mu(D_{\omega,4},\beta)\geq I_\mu(D_{\omega,4},\gamma(\omega))$. The lemma follows by Lemma \ref{l3}.
\end{proof}

Now we are able to give an upper bound for $\max_{\sigma\in\Omega_k}L_\sigma$.
\begin{lemma}\label{lem4}
For every $\sigma\in\Omega_k$, we have $L_\sigma\leq n_4$.
\end{lemma}
\begin{proof}
Assume that, for some $\sigma\in\Omega_k$, we have $L_\sigma>n_4(>M_0n_3)$. Next, we deduce a contradiction. By the assumption and (\ref{ncomp}), we deduce
\[
{\rm card}(\alpha_n\setminus (A_\sigma)_{\delta|A_\sigma|})\leq n-n_4<n_3\phi_k-M_0n_3\leq(\phi_k-M_0)n_3.
\]
 By Lemma \ref{lem9}, we have ${\rm card}(\Omega_k\setminus\mathcal{A}_\sigma)\geq \phi_k-M_0$. Note that $E_\rho,\rho\in\Omega_k$, are pairwise disjoint. There exists some $\omega\in\Omega_k\setminus\mathcal{A}_\sigma$ such that
${\rm card}(\alpha_n\cap E_\omega)<n_3$. We consider
\[
D_{\sigma,3}=E_\sigma\cup\bigg(\bigcup_{a\in\alpha_n\cap(A_\sigma)_{\delta|A_\sigma|}}(W_a(\alpha_n)\cap A_\sigma^*)\bigg).
\]
Using Lemma \ref{lem8} and the triangle inequality, for every $x\in E_\omega$, we have
\begin{eqnarray*}
d(x,\alpha_n)\leq \frac{1}{2}|E_\omega|+\delta|A_\omega|<d(x,(A_\sigma)_{\delta|A_\sigma|}).
\end{eqnarray*}
It follows that $E_\omega\cap D_{\sigma,3}=\emptyset$. We define
\begin{eqnarray*}
&&\gamma_{n_3+L_1}(D_{\omega,4})\in C_{n_3+L_1}(\nu_{D_{\omega,4}}),\;\gamma_{L_\sigma-L_0-n_3-L_1}(D_{\sigma,3})\in C_{L_\sigma-L_0-n_3-L_1}(\nu_{D_{\sigma,3}});\\
&&\beta:=\big(\alpha_n\setminus(A_\sigma)_{\delta|A_\sigma|}\big)\cup B_\sigma\cup f_{D_{\sigma,3}}(\gamma_{L_\sigma-L_0-n_3-L_1}(D_{\sigma,3}))\cup f_{D_{\omega,4}}(\gamma_{n_3+L_1}(D_{\omega,4})).
\end{eqnarray*}
Then ${\rm card}(\beta)\leq n$. By applying Lemma \ref{compare2} to $F_{\sigma,3}=D_{\sigma,3}\cup D_{\omega,4}$, we obtain
 \begin{eqnarray}\label{s6}
I_\mu(\mathbb{R}^q\setminus F_{\sigma,3},\beta)\leq I_\mu(\mathbb{R}^q\setminus F_{\sigma,3},\alpha_n).
\end{eqnarray}
This allows us to focus on integrals over the sets $D_{\sigma,3}$ and $D_{\omega,4}$. Note that for every $x\in D_{\omega,4}$, we have $d(x,\beta)\leq d(x,\gamma_{n_3+L_1}(D_{\omega,4}))$. Applying Lemma \ref{l3} with $B=D_{\omega,4}$  and Lemma \ref{lemma1} , we have
\begin{eqnarray*}
\Delta_{D_{\omega,4}}:&=&I_\mu(D_{\omega,4},\alpha_n)-I_\mu(D_{\omega,4},\beta)\nonumber\\
&\geq&I_\mu(D_{\omega,4},\alpha_n)-I_\mu(D_{\omega,4},f_{D_{\omega,4}}(\gamma_{n_3+L_1}(D_{\omega,4})))\nonumber
\\&=&\mu(D_{\omega,4})\big(\hat{e}_{n_3-1+L_1}(\nu_{D_{\omega,4}})-\hat{e}_{n_3+L_1}(\nu_{D_{\omega,4}})\big).
\end{eqnarray*}
Similarly, for every $x\in D_{\sigma,3}$, we have $d(x,\beta)\leq d(x,\gamma_{L_\sigma-L_0-n_3-L_1}(D_{\sigma,3}))$. Thus, we apply Lemmas \ref{l3} and \ref{key1} with $B=D_{\sigma,3}$ and obtain
\begin{eqnarray*}
\Delta_{D_{\sigma,3}}:&=&I_\mu(D_{\sigma,3},\beta)-I_\mu(D_{\sigma,3},\alpha_n)\nonumber\\
&\leq&I_\mu(D_{\sigma,3},f_{D_{\sigma,3}}(\gamma_{L_\sigma-L_0-n_3-L_1}(D_{\sigma,3})))-I_\mu(D_{\sigma,3},\alpha_n)\nonumber
\\&=&\mu(D_{\sigma,3})\big(\hat{e}_{L_\sigma-L_0-n_3-L_1}(\nu_{D_{\sigma,3}})-\hat{e}_{L_\sigma+L_1}(\nu_{D_{\sigma,3}})\big).
\end{eqnarray*}
By the assumption, we have $L_\sigma>n_4$. Thus, from Lemmas \ref{lem2}, \ref{lem6} (d2) and Remark \ref{rem2}, we deduce that $\Delta_{D_{\omega,4}}>\Delta_{D_{\sigma,3}}$. Combining this with
(\ref{s6}), we obtain that $I_\mu(\mathbb{R}^q,\beta)<I_\mu(\mathbb{R}^q,\alpha_n)$, contradicting the optimality of $\alpha_n$.
\end{proof}

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

Let $a\in\alpha_n$. By Remark \ref{rem5}, we have $a\in (A_\sigma)_{\delta|A_\sigma|}$ for some $\sigma\in\Omega_k$. Fix an arbitrary word $\tau_0\in\mathcal{A}_\sigma$. We define
\begin{eqnarray*}
&&\Gamma(\tau):=\alpha_n\cap (A_\tau)_{\delta|A_\tau|},\; \tau\in\mathcal{A}_\sigma;\\
&&G(a):=A_{\tau_0}\cup\bigcup_{\tau\in\mathcal{A}_\sigma}\bigcup_{b\in\Gamma(\tau)}(P_b(\alpha_n)\cap K_\mu);\\&&H(a):=\bigcup_{\tau\in\mathcal{A}_\sigma}\Gamma(\tau);\;T_a:={\rm card}(H(a)).
\end{eqnarray*}

Let $f_a$ be a similarity mapping of similarity ratio $|G(a)|$. We define
\[
\nu_{G(a)}=\mu(\cdot|G(a))\circ f_a=\mu\bigg(\cdot\bigg|\bigcup_{\tau\in\mathcal{A}_\sigma}\bigcup_{b\in\Gamma(\tau)}(P_b(\alpha_n)\cap K_\mu)\bigg)\circ f_a.
\]
\begin{lemma}\label{lem3}
Let $G(a)$ and $\nu_{G(a)}$ be as defined above. Then we have
\begin{enumerate}
\item[(i)]$P_a(\alpha_n)\cap K_\mu\subset G(a)$ and $n_1\leq T_a\leq M_0n_4=:n_5$;
\item[(ii)] there exists some constant $C$, such that
\[
\sup_{x\in\mathbb{R}^q}\nu_{G(a)}(B(x,\epsilon))\leq C\epsilon^{s_0}\;{\rm for\; every}\;\epsilon>0.
\]
\end{enumerate}
\end{lemma}
\begin{proof}
The first part of (i) is an easy consequence of the definition of $G(a)$. By Lemma \ref{lem9}, ${\rm card}(\mathcal{A}_\sigma)\leq M_0$. Further, for every $\tau\in\mathcal{A}_\sigma$, by Lemmas \ref{lem7} and \ref{lem4}, we have, $n_1\leq{\rm card}(\Gamma(\tau))\leq n_4$ for every $\tau\in\mathcal{A}_\sigma$. Hence, $n_1\leq T_a\leq n_5$.

Next, we show (ii). By the definitions of $G(a),\mathcal{A}_\rho$ and  $A_\rho,\rho\in\Omega_k$ and Lemma \ref{lem8}, we have
\begin{equation}\label{g1}
A_{\tau_0}\subset G(a)\subset\bigcup_{\tau\in\mathcal{A}_\sigma}\bigcup_{\rho\in\mathcal{A}_\tau}A_\rho\subset B\big(c_\sigma,(8\delta+\frac{5}{2})|A_{\tau_0}|\big)
\end{equation}
Thus, we have the following estimate:
\begin{equation}\label{z4}
|A_{\tau_0}|\leq|G(a)|\leq (5+16\delta)|A_{\tau_0}|.
\end{equation}
Let $\eta_3:= C_2(3+8\delta)^{s_0}$ and $\eta_4:=C_12^{-s_0}$. By (\ref{g1}), (\ref{z4}) and (\ref{AD}),
\begin{eqnarray}\label{compare1}
&&\mu(G(a))\leq C_2(3+8\delta)^{s_0}|A_{\tau_0}|^{s_0}\leq\eta_3|G(a)|^{s_0};\label{compare11}\\
&&\mu(G(a))\geq C_12^{-s_0}|A_{\tau_0}|^{s_0}\geq \eta_4(5+16\delta)^{-s_0}|G(a)|^{s_0}.\nonumber
\end{eqnarray}
Thus, from \cite[Lemma 2.5]{Zhu:20}, we obtain (ii).
\end{proof}

\emph{Proof of Theorem \ref{mthm}}

By (\ref{AD}), Lemmas \ref{lem11} and \ref{l1}, it is sufficient to consider $n\geq(n_0+n_2)\phi_{k_0}$.
Let $a\in\alpha_n$ and let $G(a),H(a),\nu_{G(a)}$ be as defined above. By Theorem 2.4 and Lemma 2.3 of \cite{GL:04} and the similarity of $f_a$, we know that $f_a^{-1}(H(a))\in C_{T_a}(\nu_{G(a)})$. From Lemma \ref{lem3} (i), we have that $n_1\leq T_a\leq n_5$. Because of Lemma \ref{lem3} (ii), we may apply Lemmas \ref{lem11}, \ref{l1} to the measure $\nu_{G(a)}$. We define
\[
\underline{d}:=\min_{2\leq h \leq n_5}\underline{d}_h,\;\overline{d}:=\max_{2\leq h \leq n_5}\overline{d}_h,\;d_3:=\min_{2\leq h \leq n_5}g_h.
\]
Thus, using the similarity of $f_a$ and Lemmas \ref{lem11} and \ref{l1}, we obtain
\[
\mu(G(a))\underline{d}\leq\mu(P_a(\alpha_n))=\mu(G(a))\nu_{G(a)}\big(P_{f_a^{-1}(a)}(f_a^{-1}(H(a)))\big)\leq \mu(G(a))\overline{d}.
\]
By Lemma \ref{lem5}, (\ref{compare11}) and (\ref{g2}), we have
\begin{eqnarray*}
&&\mu(G(a))\leq \eta_3|A_{\tau_0}|^{s_0}\leq\eta_3C_1^{-1}2^{s_0}\eta_2\phi_k^{-1}\leq n_3\eta_2\eta_3\eta_4^{-1}n^{-1};\\
&&\mu(G(a))\geq \mu(A_{\tau_0})\geq\eta_1\phi_k^{-1}\geq (n_0+n_2)\eta_1n^{-1}.
\end{eqnarray*}
It suffices to define $d_1:=\underline{d}\eta_1(n_0+n_2)$ and $d_2:=\overline{d}n_3\eta_2\eta_3\eta_4^{-1}$.

By Lemma \ref{l2}, we know that for every $b\in H(a)\setminus\{a\}$,
\[
d(f_a^{-1}(b),f_a^{-1}(a))\geq 3d_3|P_{f^{-1}(a)}(f_a^{-1}(H(a)))\cap K_{\nu_{G(a)}}|.
 \]
Thus, using the similarity of $f_a$, we obtain that
\[
\min_{b\in H(a)\setminus\{a\}}d(a,b)\geq 3d_3|P_a(\alpha_n)\cap K_\mu|.
\]
By Lemma \ref{lem8}, we know that $|P_a(\alpha_n)\cap K_\mu|\leq 2\delta|A_\sigma|$.
On the other hand, for every $b\in\alpha_n\setminus H(a)$, there exists some $\tau\in\Omega_k\setminus\mathcal{A}_\sigma$ such that $b\in (A_\tau)_{\delta|A_\tau|}$. Note that $(A_\tau)_{2\delta|A_\tau|}\cap(A_\sigma)_{2\delta|A_\sigma|}=\emptyset$ and $a\in (A_\sigma)_{\delta|A_\sigma|}$, we deduce
\[
d(b,a)\geq2\delta|A_\tau|=2\delta|A_\sigma|\geq |P_a(\alpha_n)\cap K_\mu|.
\]
Note that $3d_3<2^{-1}$. It follows that $d(b,a)\geq 3d_3|P_a(\alpha_n)\cap K_\mu|$ for every $b\in\alpha_n\setminus\{a\}$. Thus, the set $P_a(\alpha_n)$ contains a closed ball of radius $d_3|P_a(\alpha_n)\cap K_\mu|$ which is centered at $a$.


TITLE: Why expected value from a tries example $n=3$ is equal to expected value $n=654$. Why the population mean is still the same?
QUESTION [0 upvotes]: From a sample we have the expected value:
$E[x] = μ$.
And if we have differents trials $n=654$ the mean still is
$E[x] = μ$.
What is the reason of this relation? What causes this relationship to be true?

REPLY [3 votes]: It depends what you mean when you say from the different trials the mean is the same. Presumably you're describing the statistic that corresponds to the mean of the trials (since I can make any function of the trials and have it not equal the mean of a single trial!)
Consider $S_n = \frac{1}{n} \sum_{j=1}^n X_j$, where the $X_j$ are the results of your trial, each independent and with mean $E(X_j) = \mu$. Then one can show (by linearity of the expectation operator):
$$ E(S_n) = E \left(\frac{1}{n} \sum_{j=1}^n X_j \right) = \frac{1}{n} \sum_{j=1}^n E(X_j) = \frac{1}{n} \sum_{j=1}^n \mu = \mu. $$


text:
\section{Gibbardian collapse without Left Logical Equivalence}\label{sec:khoo}

\citet[489]{khoo2019triviality} prove Gibbard's collapse result using Reasoning by Cases. They do not use Left Logical Equivalence (as in our reconstruction of Gibbard's original proof) and explicitly refer to principles (i)--(iii) only (i.e., Import-Export, Stronger-than-Material and Supraclassicality). However, like Gibbard, they actually make use of more assumptions, in particular (v): the classicality of $\supset$. Their proof can be formalized thus:

\begin{thm}
Let $L$ be a reflexive, monotonic, and transitive consequence relation, with $\vee$ satisfying Reasoning by Cases. Then if (i), (ii), (iii) and (v) hold in $L$, $\supset$ entails $\to$, that is, for any $A, B \in {\sf For}(L)$, $A \supset B \models_L A \to B$.
\end{thm}

1. $\neg A \wedge A \models_{\sf CL} B$, classical logic

2. $\models_L (\neg A \wedge A)\to B$, by 1 and (iii)

3. $\models_L \neg A \to (A \to B)$, by 2 and (i)

4. $\neg A \to (A\to B) \models_L \neg A \supset (A \to B)$, by (ii)

5. $\models_L \neg A \supset (A \to B)$, by 3, 4 and Transitivity

6. $\neg A \models_L \neg A \supset (A \to B)$, by 5 and Monotonicity

7. $\neg A \models_L \neg A$, by Reflexivity

8. $\neg A \models_L A\to B$, by 6, 7 and (v), using (meta) Modus Ponens for $\supset$

9. $B \wedge A \models_{\sf CL} B$, classical logic

10. $\models_L (B \wedge A) \to B$, by 9 and (iii)

11. $\models_L B \to (A \to B)$, by 10 and (i)

12. $B \to (A \to B) \models_L B \supset (A \to B)$, by (ii)

13. $\models_L B \supset (A \to B)$, by 11, 12, and Transitivity

14. $B \models_L B \supset (A \to B)$, by 13 and Monotonicity

15. $B \models_L B$ by Reflexivity

16. $B \models_L A\to B$, by 14, 15, (v), using (meta) Modus Ponens

17. $\neg A \vee B\models_L A\to B$, by 8, 16 and Reasoning by Cases

18. $A\supset B\models_L A\to B$, by 17 and (v)\bigskip

This version does not use the replacement principle (iv) of Gibbard's original proof, making it particularly interesting, in particular in relation to {\sf DF/TT}. Indeed, Reasoning by Cases is valid in {\sf DF/TT} and {\sf CC/TT}, as are structural assumptions on logical consequence. We know that {\sf CC/TT} fails Supraclassicality and so step 2 and 10 of the proof are blocked. Interestingly, however, all steps of the proof here are \emph{sound in {\sf DF/TT}}. Although principle (v) does not hold of $\supset$ in full generality in {\sf DF/TT}, all instances of (v) are sound in this case, unlike in Gibbard's original proof. Readers may observe that the proof of $A \supset B \vdash_{\sf DF/TT} A \rightarrow B$ produced in the sequent-system of Appendix \ref{sec:technical} also mirrors Reasoning by Cases (see Lemma \ref{lemma:seq}): on the third line from the root of the tree, the left branch of the derivation tree actually establishes that $\neg A \vdash_{\sf DF/TT} A \rightarrow B$, while the right branch establishes that $B \vdash_{\sf DF/TT} A \rightarrow B$.

config: arxiv


TITLE: Proof for Unique Factorization Domain
QUESTION [6 upvotes]: Prove that the quotient ring $\mathbb{C}[x,y]/(x^2+y^2-1)$ is a unique factorization domain.

I am trying to prove first it is a principal ideal domain. However I am really stuck on this problem

REPLY [4 votes]: Show that the ring is isomorphic to the ring of complex trigonometric polynomials $\mathbb{C}[e^{i\theta},e^{-i\theta}]$.  This is a localization of $\mathbb{C}[t]$ so is a PID.


TITLE: Approximation of the solution of an IVP
QUESTION [1 upvotes]: Consider the initial value problem
$$\frac{dy}{dx} = x^2 + y^2, \\
y(0) = 0$$
on D = {|x| <= 1, |y| <= 1}
Find the third approximation to the solution
If someone could maybe walk me through the steps that would be most appreciated

REPLY [2 votes]: Take, as suggested, $D:|x|,|y|\le 1$, where $f(x,y)=x^2+y^2\le 2=M.$ Then according to Picard-Lindelof your IVP enjoys existence and uniqueness when
$$
\lvert x\rvert\le\min\left\{a,b/M\right\}=\min\left\{1,1/2\right\}=1/2.
$$
Approximations (Picard's iteration)
$$
y_0(x)=0,\quad y _1(x)=0+\int_0^x \big(s^2+y_0(s)\big)\,ds=\frac{x^3}{3}, \quad \\
y _2(x)=0+\int_0^x \big(s^2+y_1(s)\big)\,ds=\int_0^x\left(s^2+\frac{s^6}{9}\right)\,ds=\frac{x^3}{3}+\frac{x^{7}}{63}, \\
y _3(x)=0+\!\!\int_0^x\! \big(s^2+y_2(s)\big)\,ds=\!\int_0^x\!\left(s^2+\frac{s^6}{9}+\frac{2s^{10}}{189}+\frac{s^{14}}{63^2}\right)\,ds=\frac{x^3}{3}+\frac{x^{7}}{63}+\frac{2x^{11}}{2079}+\cdots
$$


TITLE: Concerning the proof of an elementary theorem regarding $s = \sup A $.
QUESTION [2 upvotes]: Is my argument to the proposition below correct?

Proposition. Assume $s\in\mathbf{R}$ is an upper bound for a set $A\subseteq R$. Then, $s = \sup A$ if and only if, for every choice of
  $\epsilon > 0$, there exists an element $a\in A$ satisfying $s − \epsilon < a$.

Proof. $(\Rightarrow)$. Assume $s = \sup A$ and let $\epsilon>0$, by definition $s$ is the smallest upper bound for $A$, so $s-\epsilon$ cannot be an upper bound for $A$ consequently for some $a\in A$ we have $s-\epsilon<a$.
$(\Rightarrow).$ For the converse assume on the contrary that $s-\epsilon<a,\forall\epsilon>0$ but $s\neq\sup A$, then $s>b$ where $b$ is some upper bound of $A$ but this implies that $s-b>0$ and thus for $\epsilon =s-b$ we have $s-\epsilon = b<a$ for some $a\in A$ contradicting the fact that $b$ is an upperbound for $A$.
$\blacksquare$

REPLY [1 votes]: It is correct, but I have two remarks:

The second $\Rightarrow$ should be a $\Leftarrow$ (this is a minor remark, of course).
There's a quantifier missing in the second proof. It should be $(\forall\varepsilon>0)\color{red}{(\exists a\in A)}:s-\varepsilon<a$.


TITLE: Is it possible to determine whether distant galaxies are gravitationally bound
QUESTION [9 upvotes]: In a previous question, one issue was related to the potential energy
of cosmic structures. This raised in particular the question of
whether these structures are gravitationally bound.
If you consider a group of say, 50 galaxies, I guess one can decide
whether they are gravitationally bound by comparing their positions
and velocities, even if the galaxies considered do not constitute the
whole group thus bound.
Considering a group of very remote galaxies, I presume (is it
correct?) that all one can measure only angular distance, and their
redshift which is due to their peculiar velocity and their distance
(through universe expansion).
Given a set of such measurements for each member of a group of
galaxies, is it possible on that limited basis to determine whether these
galaxies are gravitationally bound, and to infer from that more
precise information about these galaxies?
Are there other measurements that can be made to help that
determination?

REPLY [6 votes]: You're right that determining gravitational boundedness at large distances is difficult. Quick recap of the information on hand regarding the positions and velocities:

Angular separation between any two galaxies
Redshift

Which is really not a whole lot to work with. You can't get a handle on the (projected) physical separation without knowing the distance. And you can't get the peculiar velocity (line of sight component) without the distance so that you can use Hubble's law, and you can't get an accurate-ish distance from Hubble's law because you don't know the peculiar velocity...
But there are some tricks astronomers use to try and do better. It helps if you have a redshift independent measure of the distance to the galaxies, but beyond the local Universe only SNIa luminosity tends to be any use, and you'd need to be uncommonly lucky to get SNIa measurements in the two galaxies you happen to be interested in. There are also the Tully-Fisher and Faber-Jackson relations (thanks Chris White for the reminder) that can help with distances, but the scatter in these relations is large enough that it's essentially useless for determining boundedness (this is actually also true of the SNIa measurements, usually).
About the best we can do is come up with bound pair/group candidates and then try and reinforce or invalidate that hypothesis. Coming up with candidates is pretty easy - you look for galaxies with small angular separations, and that have similar redshifts within some tolerance. There's always a chance you could have closer galaxy with positive peculiar velocity and a more distant galaxy with a negative peculiar velocity conspire and have a similar redshift, so this doesn't confirm that the two galaxies have similar distances, but it can refute it. If the two galaxies have a velocity offset (velocity in the sense of peculiar + recessional) of say $15,000\,{\rm km}/{\rm s}$, you can be quite certain that they are not bound.
Another thing people often look for is morphological disturbance. If a pair of galaxies are interacting enough to pull big tidal tails off of each other, then they are almost certainly gravitationally bound. A particularly striking example:

No one argues that this pair is gravitationally bound! For a more technical overview, here is a recent paper that uses morphological disturbances to look for mergers/interacting pairs.
There is another technique which works reasonable well, but only in the case of larger groups and clusters of galaxies. A galaxy colour magnitude diagram shows the colour (difference in luminosity in two broadband filters) as a function of the magnitude in some broadband filter. The textbook example is as a function of absolute magnitude, and in this case most galaxies tend to land either along a line called the red sequence or in a region called the blue cloud. Here's an example from the MSc thesis of a friend of mine (the blue cloud is not visible here because there is a well known deficit of blue galaxies in clusters, and this sample specifically targets a cluster; also note that the x-axis is flipped relative to the schematic in the Wikipedia article I linked):

A more distant galaxy is on average further right (apparent magnitude gets fainter) and up (colour redder from redshifting) on this diagram. The nearly horizontal bands, one of which is highlighted, are the red sequences of a series of bound structures at different distances along roughly the same line of sight. The dots between the lines are candidates for membership in a single galaxy cluster. Of course there's a very obvious chance that some of those belong to the scatter in the distributions of some of those other red sequences, but a majority of the dots between the lines are from a single gravitationally bound galaxy cluster.
Finally, I'll shamelessly plug some of my own MSc work, the gist of which is that given only the two observables (bullet points) I started this answer with, you can get the probability for a given galaxy being a member of a cluster that it's close to in projected position and line of sight velocity (the probabilities are all calculated from simulations, where you know everything you could care to about the kinematics of the dark matter haloes that host galaxies). The catch is that you need to know some properties of the cluster to begin with, but in practice for large clusters it's easy to get a sufficiently accurate estimate of what you need to get the method to work. The cluster members occupy the red portion of this diagram, while the non-members ("interlopers") occupy mostly the blue region:


TITLE: If $f$ is Lipschitz function, prove that $\frac{f(x)}{(1+||f(x)||)}$ is Lipschitz function
QUESTION [1 upvotes]: What is the best way to go about this? Any help is appreciated.

REPLY [0 votes]: $\large|\frac{f(t)}{1+\|f(t)\|}-\frac{f(s)}{1+\|f(s)\|}|=|\frac{(f(t)-f(s))+f(t)\|f(s)\|-f(s)\|f(t)\|}{(1+\|f(t)\|)(1+\|f(s)\|)}|$
$\large\le L\frac{|t-s|}{(1+\|f(t)\|)(1+\|f(s)\|)}+|\frac{f(t)\|f(s)\|-f(s)\|f(t)\|}{(1+\|f(t)\|)(1+\|f(s)\|)}|$
$\large=L\frac{|t-s|}{(1+\|f(t)\|)(1+\|f(s)\|)}+|\frac{f(t)\|f(s)\|-f(s)\|f(s)\|+f(s)\|f(s)\|-f(s)\|f(t)\|}{(1+\|f(t)\|)(1+\|f(s)\|)}|$
$\large\le L\frac{|t-s|}{(1+\|f(t)\|)(1+\|f(s)\|)}+\|f(s)\|(\frac{|f(t)-f(s)|}{(1+\|f(t)\|)(1+\|f(s)\|)}+\frac{|\|f(s)\|-\|f(t)\||}{(1+\|f(t)\|)(1+\|f(s)\|)})$
$\large\le L(\frac{(1+\|f(s)\|)|t-s|}{(1+\|f(t)\|)(1+\|f(s)\|)})+\|f(s)\|(\frac{|f(s)-f(t)|}{(1+\|f(t)\|)(1+\|f(s)\|)})$
$\large\le L(\frac{(1+2\|f(s)\|)|t-s|}{(1+\|f(t)\|)(1+\|f(s)\|)})\le L(1+2\|f\|_\infty)|t-s|$
To obtain this I just made alot of manipulations and used the triangle inequality and th reverse triangle inequality.
A stronger bound would be:
$\large L\frac{(1+2\|f\|_\infty)|t-s|}{(1+\inf\|f\|)^2}$


TITLE: Is this matrix multiplication correct
QUESTION [0 upvotes]: Having two diagonal matrices $A$ and $D$ where ${\bf a}_i$ is a $M\times 1$ vector for $i\in[1:n]$ while $d_i$ is a scalar.
$$A  = \begin{bmatrix}
    {\bf a_1} & 0 & 0 & \dots  &0 \\
    0 & {\bf a}_{2} & 0 & \dots  & 0 \\
    \vdots & \vdots & \vdots & \ddots & \vdots \\
    0 &0 & 0 & \dots  & {\bf a}_{n}
\end{bmatrix}$$
$$D  = \begin{bmatrix}
    { d_1} & 0 & 0 & \dots  &0 \\
    0 & {d}_{2} & 0 & \dots  & 0 \\
    \vdots & \vdots & \vdots & \ddots & \vdots \\
    0 &0 & 0 & \dots  & { d}_{n}
\end{bmatrix}$$
Can I say that $$AD = \sum_i^{n} a_i d_i$$
Thanks.

REPLY [0 votes]: No, $AD$ is not equal to $$\sum_{i=1}^n a_id_i$$
For every $i$, the value $a_id_i$ is a vector (of dimension $M\times 1$). Therefore, the sum of such vectors is also a $M\times 1$ vector.
On the other hand, since $A$ is a $M\times n$ matrix and $D$ is a $n\times n$ matrix, the matrix $AD$ is also a $M\times n$ matrix.


TITLE: Show that Lie algebra $L$ is solvable (nilpotent) if and only if $ad(L)$ is solvable (nilpotent)
QUESTION [1 upvotes]: Show that Lie algebra $L$ is solvable (nilpotent) if and only if $ad(L)$ is solvable (nilpotent).
If $L$ is solvable, then $ad(L)$ is the image of $L$ under a lie algebra morphism and so will be solvable. I would appreciate some help with the $(\leftarrow)$ as well as the nilpotent cases... I've been scratching my head for awhile on this one.. Thanks!

REPLY [1 votes]: Hints:

If $$0 \rightarrow L_1 \rightarrow L_2 \rightarrow L_3 \rightarrow 0$$ is an exact sequence of Lie algebras, what do you know about solvability / nilpotency of $L_2$ related to solvability / nilpotency of $L_1$ and $L_3$?

Added: Every introductory treatment on solvable / nilpotent Lie algebras should contain the (easily proven) lemmata I am alluding to here, maybe in different language. Wikipedia phrases it like this, referencing standard textbooks by Serre and Knapp:

Given a Lie algebra $\mathfrak g$ and an ideal $\mathfrak h$ in it,
$\mathfrak{g}$ is solvable if and only if both $\mathfrak h$ and $\mathfrak{g}/\mathfrak h$ are solvable.
The analogous statement is true for nilpotent Lie algebras provided $\mathfrak h$ is contained in the center. Thus, an extension of a solvable algebra by a solvable algebra is solvable, while a central extension of a nilpotent algebra by a nilpotent algebra is nilpotent.

What is $\ker(ad)$?


TITLE: Upper bound for the Norm of an algebraic number
QUESTION [0 upvotes]: Let $L/\mathbb{Q} $ be a Galois extension of degree $n$ let $\mathbb{Q}(\alpha)$ be an intermidiate field extension of $L/\mathbb{Q} $ where $\alpha$ is an algebraic numbers with minimal polynomial $P$.let $\mathcal{N}_{L/\mathbb{Q}}(\alpha)$ be the norm of $\alpha $ and $H(P)$ is  the absolute maximum of the coefficiencts of $P$ .
I am looking for an upper bound for
$|\mathcal{N}_{L/\mathbb{Q}}(\alpha)|$
In terms of $H(P)$ and $n$ .

REPLY [1 votes]: The norm $\mathcal N_{L/\mathbb Q}(\alpha)$ is the product over all $\sigma \in \mathrm{Gal}(L/\mathbb Q)$ of $\sigma(\alpha)$.  If $m$ is the degree of $\alpha$ over $\mathbb Q$, then each of the $m$ algebraic conjugates of $\alpha$ appears exactly $n/m$ times in this product.
If $P(x) = x^m + c_{m-1} x^{m-1} + \cdots + c_1 x + c_0$, then the product of all the conjugates of $\alpha$ is $(-1)^m c_0$.  It follows that we have an exact formula for the norm in terms of $P$:
$$ \mathcal N_{L/\mathbb Q}(\alpha) = \left( (-1)^m c_0 \right)^{n/m} = (-1)^n c_0^{n/m}. $$
This is bounded in absolute value by $H(P)^{n/m}$, or by $H(P)^n$ if you really want to avoid dependence on $m$.


text:
\clearpage
\subsubsection{Tables for advection tests}
\label{A:AdvTestTables}

\begin{table}[h!]\small
	\centering

	\caption{Solid body rotation, cosine bell, grid R2B04, $T=1036800[s]$}
        \scalebox{0.5}{
	 \begin{tabular}{c c c c c c c c}
			\hline
\textbf{Error norms} &\textbf{ICON-FFSL,}  &\textbf{Standard adjoint,} &\textbf{Art. source adjoint,} &\textbf{ICON-FFSL,}                                                    &\textbf{Art. source adjoint,}                                          &\textbf{ICON-FFSL,}                                              &\textbf{Art. source adjoint,} \\
                     &\textbf{no limiter} &\textbf{no limiter}       &\textbf{no limiter}          &\textbf{limiter \cite{zalesak1979fully}, \cite{schar1996synchronous}} &\textbf{limiter \cite{zalesak1979fully}, \cite{schar1996synchronous}} &\textbf{limiter \cite{zalesak1979fully}, \cite{harris2011flux}} &\textbf{limiter \cite{zalesak1979fully}, \cite{harris2011flux}} \\
\hline
\hline
$l_{1,rel}$      & 2.230227E-02  & 4.617754E-02  & 2.230230E-02  & 1.399967E-02  & 1.399982E-02  & 1.358807E-02  & 1.358786E-02 \\
$l_{2,rel}$      & 1.264162E-02  & 2.890380E-02  & 1.264118E-02  & 1.073500E-02  & 1.073535E-02  & 1.297653E-02  & 1.297682E-02 \\
$l_{\infty,rel}$ & 1.145905E-02  & 2.296593E-02  & 1.145946E-02  & 1.516152E-02  & 1.516201E-02  & 2.414110E-02  & 2.414264E-02 \\
$l_{1,abs}$      & 3.737112      & 7.697128      & 3.737118      & 2.330792      & 2.330815      & 2.247145      & 2.247110     \\
$l_{2,abs}$      & 1.245850E-01  & 2.838235E-01  & 1.245807E-01  & 1.055754E-01  & 1.055789E-01  & 1.269004E-01  & 1.269032E-01 \\
$l_{\infty,abs}$ & 1.134216E-02  & 2.273165E-02  & 1.134256E-02  & 1.500686E-02  & 1.500734E-02  & 2.389484E-02  & 2.389635E-02 \\
\hline
Undershoot       & 9993          & 10007         & 9993          &  0            & 0             &  0            & 89           \\
Minimum          & -1.081394E-02 & -1.164300E-02 & -1.081451E-02 &  0            & 0             &  0            & -5.023336E-11\\
Overshoot        & 0             & 2             & 0             &  4            & 4             &  4            & 4            \\
Maximum          & 9.897216E-01  & 9.952338E-01  & 9.897270E-01  &  9.909216E-01 & 9.909271E-01  &  9.905530E-01 & 9.905582E-01 \\
\hline
\hline
		\end{tabular}}

	\label{tab:SolidBodyRotationcos}
\end{table}

\vfill

\begin{table}[h!]
	\centering

	\caption{Solid body rotation, slotted cylinder, grid R2B04, $T=1036800[s]$}
        \scalebox{0.5}{
	 \begin{tabular}{c c c c c c c c}
			\hline
\textbf{Error norms} &\textbf{ICON-FFSL,}  &\textbf{Standard adjoint,} &\textbf{Art. source adjoint,} &\textbf{ICON-FFSL,}                                                    &\textbf{Art. source adjoint,}                                          &\textbf{ICON-FFSL,}                                              &\textbf{Art. source adjoint,} \\
                     &\textbf{no limiter} &\textbf{no limiter}       &\textbf{no limiter}          &\textbf{limiter \cite{zalesak1979fully}, \cite{schar1996synchronous}} &\textbf{limiter \cite{zalesak1979fully}, \cite{schar1996synchronous}} &\textbf{limiter \cite{zalesak1979fully}, \cite{harris2011flux}} &\textbf{limiter \cite{zalesak1979fully}, \cite{harris2011flux}} \\
\hline
\hline
$l_{1,rel}$       & 3.069264E-01  & 3.171004E-01  & 3.069262E-01  & 2.522601E-01  & 2.522586E-01 & 2.599566E-01  & 2.599549E-01 \\
$l_{2,rel}$       & 2.673495E-01  & 2.693426E-01  & 2.673488E-01  & 2.676666E-01  & 2.676659E-01 & 2.736425E-01  & 2.736418E-01 \\
$l_{\infty,rel}$  & 7.106978E-01  & 6.999076E-01  & 7.107034E-01  & 8.149918E-01  & 8.150041E-01 & 8.057034E-01  & 8.057114E-01 \\
$l_{1,abs}$       & 3.157555E+02  & 3.263074E+02  & 3.157552E+02  & 2.593519E+02  & 2.593502E+02 & 2.669324E+02  & 2.669305E+02 \\
$l_{2,abs}$       & 8.582671      & 8.649268      & 8.582646      & 8.586382      & 8.586358     & 8.773138      & 8.773115 \\
$l_{\infty,abs}$  & 7.106978E-01  & 6.999076E-01  & 7.107034E-01  & 8.149918E-01  & 8.150041E-01 & 8.057034E-01  & 8.057114E-01 \\
\hline
Undershoot      & 9666          & 9649          &  9666         &  0            & 0            & 0             & 117    \\
Minimum           & -2.736402E-01 & -2.657430E-01 & -2.736436E-01 &  0            & 0            & 0             & -8.511675E-10 \\
Overshoot      & 435           & 431           & 435           &  415          & 415          & 414           & 414    \\
Maximum           & 1.2450956     & 1.246481      & 1.245110      & 1.284667      & 1.284693     & 1.327834      & 1.327850  \\
\hline
\hline
		\end{tabular}}

	\label{tab:SolidBodyRotationcyl}
\end{table}

\vfill

\begin{table}[h!]
	\centering

	\caption{Deformational flow, cosine bells, non-divergent velocity vector, grid R2B04, $T=1036800[s]$}
        \scalebox{0.5}{
	 \begin{tabular}{c c c c c c c c}
			\hline
\textbf{Error norms} &\textbf{ICON-FFSL,}  &\textbf{Standard adjoint,} &\textbf{Art. source adjoint,} &\textbf{ICON-FFSL,}                                                    &\textbf{Art. source adjoint,}                                          &\textbf{ICON-FFSL,}                                              &\textbf{Art. source adjoint,} \\
                     &\textbf{no limiter} &\textbf{no limiter}       &\textbf{no limiter}          &\textbf{limiter \cite{zalesak1979fully}, \cite{schar1996synchronous}} &\textbf{limiter \cite{zalesak1979fully}, \cite{schar1996synchronous}} &\textbf{limiter \cite{zalesak1979fully}, \cite{harris2011flux}} &\textbf{limiter \cite{zalesak1979fully}, \cite{harris2011flux}} \\
\hline
\hline
$l_{1,rel}$      & 3.256092E-02  & 4.011542E-02  & 3.256965E-02  & 2.342655E-02  & 2.341489E-02 & 1.895446E-02 & 1.894835E-02 \\
$l_{2,rel}$      & 2.464795E-02  & 2.847823E-02  & 2.463155E-02  & 2.138599E-02  & 2.136528E-02 & 1.901537E-02 & 1.899423E-02 \\
$l_{\infty,rel}$ & 3.967162E-02  & 4.035984E-02  & 3.958678E-02  & 3.179198E-02  & 3.170231E-02 & 2.966709E-02 & 2.957616E-02 \\
$l_{1,abs}$      & 2.438983E+01  & 2.999117E+01  & 2.439604E+01  & 1.759734E+01  & 1.758865E+01 & 1.427368E+01 & 1.426911E+01 \\
$l_{2,abs}$      & 5.131036E-01  & 5.930658E-01  & 5.127594E-01  & 4.469183E-01  & 4.464925E-01 & 3.979740E-01 & 3.975392E-01 \\
$l_{\infty,abs}$ & 3.966944E-02  & 4.035763E-01  & 3.958461E-02  & 3.179024E-02  & 3.170058E-02 & 2.966546E-02 & 2.957454E-02 \\
\hline
Undershoot       & 9201          & 9209          & 9208          & 0             & 0            & 0            & 1167          \\
Minimum          & -2.355757E-02 & -2.397034E-02 & -2.360643E-02 & 0             & 0            & 0            & -8.922257E-10 \\
Overshoot        & 0             & 1             & 0             & 0             & 0            & 0            & 0             \\
Maximum          & 9.9958314E-01 & 1.004633      & 9.994996E-01  & 9.988715E-01  & 9.987885E-01 & 9.991847E-01 & 9.991020E-01  \\
\hline
\hline

		\end{tabular}}

	\label{tab:DeformationalFlow1cos}
\end{table}

\vfill

\begin{table}[h!]
	\centering

	\caption{Deformational flow, slotted cylinders, non-divergent velocity vector, grid R2B04, $T=1036800[s]$}
        \scalebox{0.5}{
	 \begin{tabular}{c c c c c c c c}
			\hline
\textbf{Error norms} &\textbf{ICON-FFSL,}  &\textbf{Standard adjoint,} &\textbf{Art. source adjoint,} &\textbf{ICON-FFSL,}                                                    &\textbf{Art. source adjoint,}                                          &\textbf{ICON-FFSL,}                                              &\textbf{Art. source adjoint,} \\
                     &\textbf{no limiter} &\textbf{no limiter}       &\textbf{no limiter}          &\textbf{limiter \cite{zalesak1979fully}, \cite{schar1996synchronous}} &\textbf{limiter \cite{zalesak1979fully}, \cite{schar1996synchronous}} &\textbf{limiter \cite{zalesak1979fully}, \cite{harris2011flux}} &\textbf{limiter \cite{zalesak1979fully}, \cite{harris2011flux}} \\
\hline
\hline
$l_{1,rel}$      & 2.890181E-01   & 2.888711E-01  & 2.889820E-01  & 2.780968E-01  & 2.780884E-01  & 2.738582E-01  & 2.738496E-01 \\
$l_{2,rel}$      & 2.940940E-01   & 2.942568E-01  & 2.940659E-01  & 3.042498E-01  & 3.042283E-01  & 3.015322E-01  & 3.015094E-01 \\
$l_{\infty,rel}$ & 9.425960E-01   & 9.190518E-01  & 9.412042E-01  & 9.509492E-01  & 9.504477E-01  & 9.468321E-01  & 9.464949E-01 \\
$l_{1,abs}$      & 6.834999E+02   & 6.832123E+02  & 6.834093E+02  & 6.576075E+02  & 6.575934E+02  & 6.475465E+02  & 6.475301E+02 \\
$l_{2,abs}$      & 1.430822E+01   & 1.431595E+01  & 1.430670E+01  & 1.480470E+01  & 1.480354E+01  & 1.467501E+01  & 1.467376E+01 \\
$l_{\infty,abs}$ & 9.425960E-01   & 9.190518E-01  & 9.412042E-01  & 9.509492E-01  & 9.504477E-01  & 9.468321E-01  & 9.464949E-01 \\
\hline
Undershoot       & 8886           & 8879          & 8890          &  0            & 0             & 0             & 181           \\
Minimum          & -9.581544E-02  & -1.741285E-01 & -9.600165E-02 &  0            & 0             & 0             & -1.739492E-09 \\
Overshoot        & 910            & 910           & 909           & 796           & 796           & 837           & 837            \\
Maximum          & 1.193710       & 1.202044      & 1.193843      & 1.169024      & 1.169178      & 1.183066      & 1.183258       \\
\hline
\hline
		\end{tabular}}

	\label{tab:DeformationalFlow1cyl}
\end{table}

\vfill

\begin{table}[h!]
	\centering

	\caption{Deformational flow, cosine bells, divergent velocity vector, grid R2B04, $T=1036800[s]$}
        \scalebox{0.5}{
		\begin{tabular}{c c c c c}
			\hline
			\textbf{Error norms} &\textbf{Standard adjoint,} &\textbf{Art. source adjoint,} &\textbf{Art. source adjoint,}                                          &\textbf{Art. source adjoint,} \\
			                     &\textbf{no limiter}       &\textbf{no limiter}          &\textbf{limiter \cite{zalesak1979fully}, \cite{schar1996synchronous}} &\textbf{limiter \cite{zalesak1979fully}, \cite{harris2011flux}} \\
			\hline
			\hline
			$l_{1,rel}$      & 4.029138E-002  & 3.311189E-002  & 1.753717E-002  & 1.627544E-002 \\
			$l_{2,rel}$      & 3.351494E-002  & 2.831085E-002  & 1.925266E-002  & 1.848417E-002 \\
			$l_{\infty,rel}$ & 5.717197E-002  & 4.793155E-002  & 3.574307E-002  & 3.388867      \\
			$l_{1,abs}$      & 3.013273E+001  & 2.479282E+001  & 1.305616E+001  & 1.210955E+001 \\
			$l_{2,abs}$      & 6.964130E-001  & 5.885914E-001  & 3.994988E-001  & 3.835561E-001 \\
			$l_{\infty,abs}$ & 5.710458E-002  & 4.787505E-002  & 3.570094E-002  & 3.384873E-001\\
			\hline
			Undershoot       & 9187           & 9189           & 0              & 7135 \\
			Minimum          & -3.020335E-002 & -2.661911E-002 & 0              & -1.572671E-008 \\
			Overshoot        & 0              & 0              & 0              & 0   \\
			Maximum          & 9.955241E-001  & 9.937946E-001  & 9.933643E-001  & 9.933694E-001 \\
			\hline
			\hline
		\end{tabular}}

	\label{tab:DeformationalFlow2cos}
\end{table}

\vfill

\begin{table}[h!]
	\centering

	\caption{Deformational flow, slotted cylinders, divergent velocity vector, grid R2B04, $T=1036800[s]$}
        \scalebox{0.5}{
		\begin{tabular}{c c c c c}
			\hline
			\textbf{Error norms} &\textbf{Standard adjoint,} &\textbf{Art. source adjoint,} &\textbf{Art. source adjoint,}                                          &\textbf{Art. source adjoint,} \\
			                     &\textbf{no limiter}       &\textbf{no limiter}          &\textbf{limiter \cite{zalesak1979fully}, \cite{schar1996synchronous}} &\textbf{limiter \cite{zalesak1979fully}, \cite{harris2011flux}} \\
			\hline
			\hline
			$l_{1,rel}$      & 3.201877E-001  & 3.072660E-01  & 2.959957E-01  & 2.911967E-01 \\
			$l_{2,rel}$      & 3.004045E-001  & 2.950234E-01  & 3.033951E-01  & 3.008232E-01 \\
			$l_{\infty,rel}$ & 8.202367E-001  & 8.689133E-01  & 8.527525E-01  & 8.511831E-01 \\
			$l_{1,abs}$      & 6.792025E+002  & 6.511061E+02  & 6.267315E+02  & 6.163679E+02 \\
			$l_{2,abs}$      & 1.383153E+001  & 1.357617E+01  & 1.396507E+01  & 1.384493E+01 \\
			$l_{\infty,abs}$ & 8.202369E-001  & 8.689133E-01  & 8.527525E-01  & 8.511831E-01 \\
			\hline
			Undershoot       & 8935           & 8867           & 14             & 7445 \\
			Minimum          & -1.334238E-001 & -1.173093E-01  & -4.796964E-15  & -4.019092E-09\\
                        Overshoot        & 858            & 839            & 791            & 801\\
			Maximum          & 1.200032       & 1.185067       & 1.177634       & 1.179568  \\
			\hline
			\hline
		\end{tabular}}

	\label{tab:DeformationalFlow2cyl}
\end{table}

\vfill

\begin{table}[h!]
	\centering

	 \caption{Moving vortices, grid R2B04, $T=1036800[s]$}
         \scalebox{0.5}{
            \begin{tabular}{c c c c c}
	      \hline
	      \textbf{Error norms} & \textbf{Standard adjoint,} &\textbf{Art. source adjoint,} & \textbf{Art. source adjoint,} & \textbf{Art. source adjoint,} \\
	                           & \textbf{no limiter}        & \textbf{no limiter}  & \textbf{limiter \cite{zalesak1979fully}, \cite{schar1996synchronous}} & \textbf{limiter \cite{zalesak1979fully}, \cite{harris2011flux}} \\
	      \hline
	      \hline
	      $l_{1,rel}$       & 2.412546E-02 & 1.411150E-03 & 1.411459E-03 & 1.411150E-03 \\
	      $l_{2,rel}$       & 4.083858E-02 & 3.874989E-03 & 3.875515E-03 & 3.874989E-03 \\
 	      $l_{\infty,rel}$  & 2.048717E-01 & 3.017727E-02 & 3.017801E-02 & 3.017727E-02 \\
              $l_{1,abs}$       & 5.084637E+02 & 2.881504E+01 & 2.882144E+01 & 2.881504E+01 \\
	      $l_{2,abs}$       & 6.294867     & 5.789490     & 5.790304     & 5.789490E-01 \\
	      $l_{\infty,abs}$  & 3.148948E-01 & 4.638326E-02 & 4.638440E-02 & 4.638326E-02 \\
	      \hline
	      Undershoot        & 44          & 1            & 1            & 1            \\
	      Minimum           & 4.483181E-01 & 4.629725E-01 & 4.629725E-01 & 4.629725E-01 \\
	      Overshoot         & 259          & 1            & 1            & 1            \\
	      Maximum           & 1.613474     & 1.537027     & 1.537027     & 1.537027     \\
	      \hline
	      \hline
            \end{tabular}}

	\label{tab:MovingVortices}
   \end{table}

\vfill


TITLE: differentiating covariance
QUESTION [1 upvotes]: $$\sum_{i=1}^n x_{i}^2C_{ii} + \sum_{i=1}^n \sum_{j=1 \atop{j\neq i}}^n x_{i}x_{j}C_{ij}$$
I am trying to differentiate the above expression with respect to $x_i$. I did partial differentiation to get $$2x_i C_{ii} + \sum_{j=1 \atop{j\neq i}}^n x_{j}C_{ij}$$. However the answer is given as $$2x_i C_{ii} + 2\sum_{j=1 \atop{j\neq i}}^n x_{j}C_{ij}$$.
Could I get assistance with the regards to the extra 2 before the summation and why it should be there?

REPLY [0 votes]: Note that
$$\frac{\partial x_p}{\partial x_i} = \delta_{ip} = \begin{cases}1, & i=p\\0,& i \neq p \end{cases},$$
and by relabeling the first index,
$$S= \underset{j \neq i}{\sum_{i=1}^n \sum_{j=1 }^n} x_ix_{j}C_{ij} = \underset{j \neq k}{\sum_{k=1}^n \sum_{j=1 }^n} x_kx_{j}C_{kj}$$
Hence,
$$\frac{\partial S}{\partial x_i} =\frac{\partial}{\partial x_i}\underset{j \neq k}{\sum_{k=1}^n \sum_{j=1 }^n} x_kx_{j}C_{kj} = \underset{j \neq k}{\sum_{k=1}^n \sum_{j=1 }^n} \frac{\partial (x_k x_{j })}{\partial x_i}C_{kj} \\ = \underset{j \neq k}{\sum_{k=1}^n \sum_{j=1 }^n} x_k\frac{\partial  x_{j }}{\partial x_i}C_{kj}+  \underset{j \neq k}{\sum_{k=1}^n \sum_{j=1 }^n} x_j\frac{\partial  x_{k }}{\partial x_i}C_{kj} = \underset{j \neq k}{\sum_{k=1}^n \sum_{j=1 }^n} x_k\delta_{ij}C_{kj}+  \underset{j \neq k}{\sum_{k=1}^n \sum_{j=1 }^n} x_j\delta_{ik}C_{kj} $$
For the second sum on the RHS, since $\delta_{ik} = 0$ if $k \neq i$, we have
$$\underset{j \neq k}{\sum_{k=1}^n \sum_{j=1 }^n} x_j\delta_{ik}C_{kj} = \underset{j \neq i}{\sum_{j=1 }^n}x_jC_{ij} $$
For the first sum on the RHS, since $\delta_{ij} = 0$ if $j \neq i$, we have
$$\underset{j \neq k}{\sum_{k=1}^n \sum_{j=1 }^n} x_k\delta_{ij}C_{kj}= \underset{k \neq j}{\sum_{j=1}^n \sum_{k=1 }^n} x_k\delta_{ij}C_{kj} = \underset{k \neq i}{\sum_{k=1}^n } x_kC_{ki} =\underset{j \neq i}{\sum_{j=1}^n } x_jC_{ji}$$
Thus,
$$\tag{*}\frac{\partial S}{\partial x_i} = \underset{j \neq i}{\sum_{j=1 }^n}x_jC_{ij} + \underset{j \neq i}{\sum_{j=1}^n } x_jC_{ji}$$
A covariance matrix is symmetric.  Thus $C_{ij} = C_{ji}$ and (*) becomes
$$\frac{\partial S}{\partial x_i} = 2\underset{j \neq i}{\sum_{j=1 }^n}x_jC_{ij}$$


TITLE: Simulate discrete state space model with a for loop
QUESTION [0 upvotes]: I need some help with simulating a discrete state space model. I have succeeded to simulate a time continuous state space model via matlab/octave code.
$$\dot{x}(t) = Ax(t)+ Bu(t)\\ y(t) = Cx(t)+Du(t) \\ x(t) = x(t) + \dot{x}(t) \Delta t(t) $$
But how would the update of state $x(t)$ and input $u(t)$ be if the for loop is discrete?
First of all, I have the discrete A and B matrices.

REPLY [2 votes]: Hint:
Continuous state-space representation:
$$\dot{x}(t) = Ax(t)+ Bu(t)\\ y(t) = Cx(t)+Du(t)$$
Discrete state-space representation:
$$x(k+1) = A_dx(k)+ B_du(k)\\ y(k) = C_dx(k)+D_du(k)$$
It should be straightforward to calculate your state vector by using your initial conditions.


TITLE: Equivalent forms of weak Nullstellensatz
QUESTION [0 upvotes]: I read that the following proposition, that is basically a corollary of Zariski's lemma, is a form of the weak Nullstellensatz:

Let $k$ be a field, $A$ a finitely generated $k$-algebra. Let $m$ be a maximal
ideal of $A$. Then the residue field $A/m$ is a finite algebraic extension of $k$. In particular,
if $k$ is algebraically closed, $A/m \cong k$.

My question is: if $k$ is algebraically closed, saying that $A/m \cong k$ is actually equivalent to saying that for every proper ideal $I\subset k[x_1,\dots,x_n]$, there are $a_1,\dots ,a_n\in k$ such that $f(a_1,\dots ,a_n)=0$ for every $f\in I$? I suppose yes, but I really can't see how. Thanks in advance

REPLY [1 votes]: Well every such ideal $I$ will be contained in a maximal ideal $\mathfrak{m}$ so that $A/\mathfrak{m}\simeq k$.
Now take $a_i\in k$ such that $x_i$ corresponds to $a_i$ under this isomorphism. So  $\mathfrak{m}=(x_1-a_2,\dots, x_n-a_n).$
Now $I \subset \mathfrak{m}=(x_1-a_1,\dots, x_n-a_n)$ says that for all $f\in I$, $f(a_1,\dots, a_n)=0$.


TITLE: Show that there are infinitely many $k$-consecutive positive integers s.t. Möbius function takes the same value.
QUESTION [3 upvotes]: That is to say $\forall k \in \mathbb{N}$, there exist infinitely many $n\in \mathbb{N}$ s.t. $\mu \left ( n+1 \right )=\mu \left ( n+2 \right )=\cdots=\mu \left ( n+k \right )$, where $\mu$ is the Möbius function.
I saw this problem in a number theory textbook and it appears in the exercises for the section talking about the Möbius inversion formula. I cannot come up with an effective idea to tackle it.
This is not a homework problem.
Many thanks for your help!

REPLY [6 votes]: Let $p_1,p_2,\cdots,p_k$ be any distinct prime numbers.
Then by the Chinese Remainder Theorem, the system of congruences
$$\begin{align}
  n + 1 & \equiv 0 \mod p_1^2 \\
  n + 2 & \equiv 0 \mod p_2^2 \\
  n + 3 & \equiv 0 \mod p_3^2 \\
  \vdots \\
  n + k & \equiv 0 \mod p_k^2
\end{align}$$
has infinitely many solutions $n$. For any such solution $n$, we have that
$n + m$ is divisible by $p_m^2$ for $m=1,2,3,\cdots,k$, and so we have that $\mu(n+m)=0$ for $m=1,2,3\cdots,k$
We see that there are infinitely many natural numbers $n$ such that
$$ \mu(n+1)=\mu(n+2)=\mu(n+3)=\cdots=\mu(n+k)=0$$

edit
In fact, if $k \geq 4$ and for some $n$ we have that
$$ \mu(n+1)=\mu(n+2)=\mu(n+3)=\cdots=\mu(n+k)$$
then it must in fact be the case that
$$ \mu(n+1)=\mu(n+2)=\mu(n+3)=\cdots=\mu(n+k)=0$$
This is because amongst the set of numbers
$$\lbrace n+1, n+2, n+3, n+4\rbrace$$
there will be a multiple of $4$. The Möbius function evaluated at this multiple of $4$ will be $0$.