Quantization for a Condensation System

Abstract

For a given $r\in (0,+\infty )$, the quantization dimension of order r, if it exists, denoted by $D_r(\mu )$, represents the rate at which the nth quantization error of order r approaches zero as the number of elements n in an optimal set of n-means for $\mu$ tends to infinity. If $D_r(\mu )$ does not exist, we define ${\underline{D}}_r(\mu )$ and ${\overline{D}}_r(\mu )$ as the lower and the upper quantization dimensions of $\mu$ of order r, respectively. In this paper, we investigate the quantization dimension of the condensation measure $\mu$ associated with a condensation system $({\left\{S_j\right\}}_{j=1}^N, {(p_j)}_{j=0}^N,\nu ).$ We provide two examples: one where $\nu$ is an infinite discrete distribution on $\mathbb{R}$, and one where $\nu$ is a uniform distribution on $\mathbb{R}$. For both the discrete and uniform distributions $\nu$, we determine the optimal sets of n-means, calculate the quantization dimensions of condensation measures $\mu$, and show that the $D_r(\mu )$-dimensional quantization coefficients do not exist. Moreover, we demonstrate that the lower and upper quantization coefficients are finite and positive.

1. Introduction

Various types of dimensions, including Hausdorff and packing dimensions or the lower and the upper box-counting dimensions, are important to characterize the complexity of highly irregular sets. In recent years, paralleling methods have been adopted to study the corresponding dimensions of measures (see [1]). In this paper, we investigate the quantization dimensions of the condensation measures. The quantization problem consists of studying the quantization error induced by the approximation of a given probability measure with discrete probability measures of finite supports. This problem originated in information theory and some engineering technology. For rigorous mathematical foundation of this theory, one can see Graf-Luschgy’s book [2]. Further theoretical results and promising applications are contained in [3,4,5,6,7,8]. Two important objects in the quantization theory are the quantization coefficient and the quantization dimension.

Let $\mu$ be a Borel probability measure on a d-dimensional normed space ${\mathbb{R}}^d$ equipped with a metric $\rho$ induced by the norm $\parallel ·\parallel$. Let $r\in (0,+\infty )$ and $n\in \mathbb{N}$, where $\mathbb{N}$ is the set of natural numbers, and $\alpha \subseteq {\mathbb{R}}^d$ be a locally finite (i.e., intersection of $\alpha$ with any bounded subset of ${\mathbb{R}}^d$ is finite) subset of ${\mathbb{R}}^d$. This implies that $\alpha$ is countable and closed. Then, the nth quantization error of order r for $\mu$ is defined by

$$V_{n,r}(\mu ):=\inf \{\mathit{\int }\rho {(x,\alpha )}^{r}d\mu (x):\alpha \subseteq {\mathbb{R}}^d,\mathrm{card}(\alpha )\le n\},$$

where $\rho (x,\alpha )$ denotes the distance from the element x to the set $\alpha$ with respect to a given norm $\parallel ·\parallel$ on ${\mathbb{R}}^d$. If ${\int \parallel x\parallel }^{r}d\mu (x)<\infty$, then there is some set $\alpha$ for which the infimum is achieved (see [2,3,5,6]). Such a set $\alpha$ is called an optimal set of n-means, or optimal set of n-quantizers. For some recent work in this direction one can see [2,6,8,9,10,11,12,13,14,15,16,17,18,19,20,21]. An optimal set $\alpha$ of n-means can then be used to give a best approximation of $\mu$ by a discrete probability supported on a set with no more than n elements. This can be achieved by giving each element $a\in \alpha$ a mass corresponding to $\mu (M(a|\alpha ))$, where $M(a|\alpha )$ is the set of elements $x\in {\mathbb{R}}^d$, which are nearest to a, i.e., $\rho (x,\alpha )=\rho (x,a)$. So, the set $\{M(a|\alpha ):a\in \alpha \}$ is the Voronoi diagram or Voronoi tessellation of ${\mathbb{R}}^d$ with respect to $\alpha$. Of course, the idea of ‘best approximation’ is, in general, dependent on the choice of r. Such a set $\alpha$ for which the infimum occurs and contains no more than n elements is called an optimal set of n-means, or optimal set of n-quantizers (of order r). An optimal set of n-means for a probability distribution $\mu$ is also referred to as $V_{n,r}(\mu )$-optimal set. The collection of all optimal sets of n-means for a probability distribution $\mu$ is denoted by ${\mathcal{C}}_{n,r}(\mu )$. It can be shown that for a continuous Borel probability measure $\mu$, an optimal set of n-means always has exactly n elements (see [2]). The numbers

$${\underline{D}}_r(\mu ):=\underset{n\to \infty }{lim\; inf}{\displaystyle \frac{rlogn}{-log{V}_{n,r}(\mu )}},\mathrm{and}{\overline{D}}_r(\mu ):=\underset{n\to \infty }{lim\; sup}{\displaystyle \frac{rlogn}{-log{V}_{n,r}(\mu )}}$$

are, respectively, called the lower and the upper quantization dimensions of $\mu$ of order r. If ${\overline{D}}_r(\mu )={\underline{D}}_r(\mu )$, the common value is called the quantization dimension of $\mu$ of order r, and is denoted by $D_r(\mu )$. Quantization dimension measures the speed at which the specified measure of the error approaches zero as n tends to infinity. For any ${\kappa }_r>0$, the two numbers

$${\underline{Q}}_r^{{\kappa }_r}(\mu ):=\underset{n}{lim\; inf}{n}^{r/{\kappa }_r}{V}_{n,r}(\mu )\mathrm{and}{\overline{Q}}_r^{{\kappa }_r}(\mu ):=\underset{n}{lim\; sup}{n}^{r/{\kappa }_r}{V}_{n,r}(\mu )$$

are, respectively, called the ${\kappa }_r$-dimensional lower and the upper quantization coefficients for $\mu$ (of order r). The quantization coefficients provide us with more accurate information about the asymptotics of the quantization error than the quantization dimension. Compared to the calculation of quantization dimension, it is usually much more difficult to determine whether the lower and the upper quantization coefficients are finite and positive. Let $\nu$ be a Borel probability measure on ${\mathbb{R}}^d$ with compact support C, and $(p_0,p_1,p_2,\cdots ,p_N)$ be a probability vector with $p_j>0$ for all $1\le j\le N$, where N is a positive integer and $N\ge 2$. Let $S_1,S_2,\cdots ,S_N$ be a family of contractive mappings on ${\mathbb{R}}^d$. Then, there exist a unique Borel probability measure $\mu$ on ${\mathbb{R}}^d$ and a unique nonempty compact set K satisfying

$$\mu =\sum _{j=1}^N{p}_j\mu \circ S_j^{-1}+p_0\nu \mathrm{and}K=\underset{j=1}{\overset{N}{\cup }}{S}_j(K)\cup C.$$

Following [22,23], we call $({\left\{S_j\right\}}_{j=1}^N,{(p_j)}_{j=0}^N,\nu )$ a condensation system. The measure $\mu$ is called the attracting measure or the condensation measure for $({\left\{S_j\right\}}_{j=1}^N,{(p_j)}_{j=0}^N,\nu )$, and the set K, which is the support of the measure $\mu$, is called the attractor or invariant set for the system. If each $S_j$ is contractive similitude, then such a measure $\mu$ is also termed as an inhomogeneous self-similar measure (see [24]). There are several works that have been conducted on the condensation system, for example, $L^q$ spectra and Rényi dimensions of inhomogeneous self-similar measures were studied by Olsen-Snigireva, and then by Liszka (see [25,26]). Let $\mu$ be a condensation measure on ${\mathbb{R}}^d$ associated with a condensation system $({\left\{S_j\right\}}_{j=1}^N,{(p_j)}_{j=0}^N,\nu )$, where $\nu$ is any Borel probability measure on ${\mathbb{R}}^d$ with a compact support. Let K be the attractor of the condensation system, and C be the support of $\nu$. We say that $\{S_1(K),S_2(K),\cdots ,S_N(K),C\}$ satisfies the strong separation condition (SSC) if $S_1(K),S_2(K),\cdots ,S_N(K),C$ are pairwise disjointed. On the other hand, $\{S_1(K),S_2(K),\cdots ,S_N(K), C\}$ satisfies the inhomogeneous open set condition (IOSC), which is the modified version of the inhomogeneous open set condition proposed in [24], if there exists a bounded nonempty open set $U\subset {\mathbb{R}}^d$ such that the following conditions are satisfied: $(i)$$\underset{j=1}{\overset{N}{\cup }}{S}_j(U)\subset U$ and $S_i(U)\cap S_j(U)=\varnothing$ for $1\le i\ne j\le N$; $(ii)$$K\cap U\ne \varnothing$ and $C\subset U$; $(iii)$$P(\partial (U))=0$ and $C\cap S_j(\mathrm{cl}(U))=\varnothing$ for all $1\le j\le N$. Notice that if $\{S_1(K),S_2(K),\cdots ,S_N(K),C\}$ satisfies the SSC, then it also satisfies the IOSC. Take ${\left\{S_j\right\}}_{j=1}^N$ as a family of bi-Lipschitz mappings on ${\mathbb{R}}^d$, i.e., there exist $0<a_j\le b_j<1,$ such that $a_j\rho (x,y)\le \rho (S_j(x),S_j(y))\le b_j\rho (x,y)$ for all $1\le j\le N,$ and for all $x,y\in {\mathbb{R}}^d,$ we call $a_j$ and $b_j$ the lower and the upper bi-Lipschitz constants, respectively. For $r\in (0,+\infty )$, let $l_r,{\kappa }_r\in (0,+\infty )$ be the unique numbers such that ${\sum }_{j=1}^N{(p_j{a}_j^r)}^{{\displaystyle \frac{l_r}{r+l_r}}}=1$ and ${\sum }_{j=1}^N{(p_j{b}_j^r)}^{{\displaystyle \frac{{\kappa }_r}{r+{\kappa }_r}}}=1$, respectively. In [27], Priyadarshi et al. showed that, under the strong separation condition, $max\{l_r,{\underline{D}}_r(\nu )\}\le {\underline{D}}_r(\mu )$. Further, let $\{{\mathbb{R}}^d;g_1,g_2,\cdots ,g_M\}$ be an IFS with invariant set $C,$ satisfying the strong open set condition such that $\rho (g_i(x),g_i(y))\le c_i\rho (x,y)$ for all $x,y\in {\mathbb{R}}^d,$ where $0<c_i<1$ and $1\le i\le M.$ For a given probability vector $(q_1,q_2,\cdots ,q_M),$ let $\nu$ be the invariant measure associated with the IFS $\{{\mathbb{R}}^d;g_1,g_2,\cdots ,g_M\}.$ Then, ${\overline{D}}_r(\mu )\le max\{{\kappa }_r,d_r\},$ where $d_r$ is uniquely determined by the relation ${\sum }_{i=1}^M{(q_i{c}_i^r)}^{{\displaystyle \frac{d_r}{r+d_r}}}=1.$ In our recent work [28], by considering the Borel probability measure $\nu$ as the image measure of an ergodic measure with bounded distortion on the symbolic space ${\{1,2,\cdots ,M\}}^{\mathbb{N}}$ with support of a conformal set $C,$, which is a generalization of [27], we have shown that

$${\overline{D}}_r(\mu )\le max\{{\kappa }_r,D_r(\nu )\}.$$

The following problem remains open (also see Remark 1).

Open 1.

Let ν be any Borel probability measure and μ be the condensation measure associated with the condensation system $({\left\{S_i\right\}}_{i=1}^N,{(p_i)}_{i=0}^N,\nu ).$ Then, whether under the strong separation condition,

$$max\{l_r,{\underline{D}}_r(\nu )\}\le {\underline{D}}_r(\mu )\le {\overline{D}}_r(\mu )\le max\{{\kappa }_r,{\overline{D}}_r(\nu )\},$$

is not known yet.

Delineation 1.

In this paper, we estimate the quantization dimension of a condensation system given by $({\left\{S_j\right\}}_{j=1}^N,{(p_j)}_{j=0}^N,\nu ).$ In this regard, in Section 3 and Section 4, we give two examples of condensation measures μ associated with a condensation system $({\left\{S_j\right\}}_{j=1}^N,{(p_j)}_{j=0}^N,\nu ),$ consisting of contractive similarity mappings $S_j,$ with similarity ratios $s_j$ for $1\le j\le N$ such that one takes ν as an infinite discrete distribution on $\mathbb{R}$, and one takes ν as a uniform distribution on $\mathbb{R}$, which is the ‘main result’ of the paper. Notice that in the two examples, we fix $r=2$, and the underlying norm is the Euclidean distance. When $r=2$, and ${\kappa }_r$ is determined by the relation ${\sum }_{j=1}^N{(p_j{s}_j)}^{{\displaystyle \frac{{\kappa }_r}{r+{\kappa }_r}}},$ we denote ${\kappa }_r$, $D_r(\mu )$, and $D_r(\nu )$, respectively, by κ, $D(\mu )$, and $D(\nu )$. In Section 3, with respect to the squared Euclidean distance, we first determine the optimal sets of n-means, the nth quantization error, and the quantization dimension of an infinite discrete distribution ν with a compact support. For the discrete distribution ν, we see that $D(\nu )=0$. Then, in Section 3, we also show that $D(\mu )=\kappa =max\{\kappa ,D(\nu )\}$, the $D(\mu )$-dimensional quantization coefficient does not exist, and the lower and the upper quantization coefficients are finite and positive. In Section 4, we take ν as a uniform distribution with a compact support, determine the optimal sets of n-means, the nth quantization error, and show that $D(\mu )=1=D(\nu )=max\{\kappa ,D(\nu )\}$, the $D(\mu )$-dimensional quantization coefficient does not exist, and the lower and the upper quantization coefficients are finite and positive.

2. Preliminaries

First, we introduce some basic definitions, and then state and prove some lemmas and propositions. By a word $\omega$ of length k over the alphabet $I:=\{1,2,\cdots ,N\}$, we mean $\omega :={\omega }_1{\omega }_2\cdots {\omega }_k\in I^k$. A word of length zero is called the empty word and is denoted by ∅. The length of a word $\omega$ is denoted by $\left|\omega \right|$. By $I^{*}$, it is meant that the set of all words over the alphabet I includes the empty word ∅. For any two words $\omega :={\omega }_1{\omega }_2\cdots {\omega }_{\left|\omega \right|}$ and $\tau :={\tau }_1{\tau }_2\cdots {\tau }_{\left|\tau \right|}$ in $I^{*}$, by $\omega \tau$, the concatenation of the words $\omega$ and $\tau$ is meant, i.e., $\omega \tau :={\omega }_1{\omega }_2·{\omega }_{\left|\omega \right|}{\tau }_1{\tau }_2\cdots {\tau }_{\left|\tau \right|}$. Let $S_j$ be the bi-Lipschitz mappings with the lower and upper bi-Lipschitz constants $a_j$ and $b_j$, respectively, for $1\le j\le N$, let $(p_0,p_1,\cdots ,p_N)$ be the probability vector, and let $\nu$ be the Borel probability measure on ${\mathbb{R}}^d$ with a compact support C in the condensation system. Let K be the attractor of the condensation system as defined before. For $\omega ={\omega }_1{\omega }_2\cdots {\omega }_k\in I^k$, we define

$$S_{\omega }:=S_{{\omega }_1}\circ \cdots \circ S_{{\omega }_k},p_{\omega }:=\prod _{j=1}^k{p}_{{\omega }_j},\mathrm{and}{K}_{\omega }:=S_{\omega }(K).$$

For the empty word ∅ in $I^{*}$, by $S_{\varnothing }$ the identity mapping on $\mathbb{R}$ is meant. For every $n\ge 1$, by iterating, one easily gets,

$$\begin{array}{c}\hfill K=\left(\bigcup _{\left|\omega \right|=n}{S}_{\omega }(K)\right)\cup \left(\bigcup _{k=0}^{n-1}\bigcup _{\left|\omega \right|=k}{S}_{\omega }(C)\right),\mathrm{and}\mu =\sum _{\left|\omega \right|=n}{p}_{\omega }\mu \circ S_{\omega }^{-1}+p_0\sum _{k=0}^{n-1}\sum _{\left|\omega \right|=k}{p}_{\omega }\nu \circ S_{\omega }^{-1}.\end{array}$$

(1)

We call $\Gamma \subset I^{*}$ a finite maximal antichain if $\Gamma$ is a finite set of words in $I^{*}$, such that every sequence in ${\{1,2,\cdots ,N\}}^{\mathbb{N}}$ is an extension of some word in $\Gamma$, but no word of $\Gamma$ is an extension of another word in $\Gamma$.

Proposition 1.

Let $r\in (0,+\infty )$. Further, let $l_r\in (0,+\infty )$ be defined by ${\sum }_{j=1}^N{(p_j{a}_j^r)}^{{\displaystyle \frac{l_r}{r+l_r}}}=1$. Then,

$$max\{l_r,{\underline{D}}_r(\nu )\}\le {\underline{D}}_r(\mu )\mathit{and}max\{l_r,{\overline{D}}_r(\nu )\}\le {\overline{D}}_r(\mu ).$$

Lemma 1

(See, [27] Lemma 3.1). Let $r\in (0,+\infty )$, and let μ be the condensation measure associated with the condensation system $({\left\{S_j\right\}}_{j=1}^N,{(p_j)}_{j=0}^N,\nu )$. Then,

$${\underline{D}}_r(\nu )\le {\underline{D}}_r(\mu )\mathit{and}{\overline{D}}_r(\nu )\le {\overline{D}}_r(\mu ).$$

Lemma 2

(See, [27], Proposition 3.2). Assume that the sets $S_{1}K,\cdots ,S_{N}K,C$ satisfy the strong separation condition. Let $r\in (0,+\infty )$. Let $l_r\in (0,+\infty )$ be defined by ${\sum }_{j=1}^N{(p_j{a}_j^r)}^{{\displaystyle \frac{l_r}{r+l_r}}}=1$. Let μ be the condensation measure associated with the condensation system $({\left\{S_j\right\}}_{j=1}^N,{(p_j)}_{j=0}^N,\nu )$, and $e_{n,r}(\mu ):=V_{n,r}^{1/r}(\mu )$. Then, it follows that

$$\underset{n\to \infty }{lim\; inf}n{e}_{n,r}^{l_r}(\mu )>0.$$

Corollary 1.

Lemma 2 implies that $l_r\le {\underline{D}}_r(\mu )$.

Proof of Proposition 1.

Combining Lemma 1 and Corollary 1, we obtain the proof of the proposition. □

In the condensation system $({\left\{S_j\right\}}_{j=1}^N,{(p_j)}_{j=0}^N,\nu ),\mathrm{if}\nu$ is an image measure of an ergodic measure with bounded distortion on the symbolic space ${\{1,2,\cdots ,M\}}^{\mathbb{N}}$ with the support a conformal set $C,$ then under strong separation condition, ${\underline{D}}_r(\mu )\le {\overline{D}}_r(\mu )\le max\{{\kappa }_r,D_r(\nu )\}.$ For more details see, [28].

Proposition 2

(See, [28], Theorem 1.4). Assume that the sets $S_{1}K,\dots ,S_{N}K,C$ satisfy the strong separation condition. Let $r\in (0,+\infty )$. Let ${\kappa }_r\in (0,+\infty )$ be defined by ${\sum }_{j=1}^N{(p_j{b}_j^r)}^{{\displaystyle \frac{{\kappa }_r}{r+{\kappa }_r}}}=1$. Let C be the conformal set generated by a conformal IFS $({\mathbb{R}}^d;{\left\{g_i\right\}}_{i=1}^M)$ satisfying the strong separation condition, let ν be the image measure of an ergodic measure with bounded distortion on the code space ${\{1,2,\cdots ,M\}}^{\mathbb{N}}$ with support $C,$ let μ be the condensation measure associated with the condensation system $({\left\{S_j\right\}}_{j=1}^N,{(p_j)}_{j=0}^N,\nu )$, and let $e_{n,r}(\mu ):=V_{n,r}^{1/r}(\mu )$. Then, it follows that ${\underline{D}}_r(\mu )\le {\overline{D}}_r(\mu )\le max\{{\kappa }_r,D_r(\nu )\}.$

Proposition 2 leads us to the following corollory.

Corollary 2.

Let $s_j:=a_j=b_j$ in the above theorem, i.e., if $S_j$ are the similarity mappings for $1\le j\le N$ in the condensation system $({\left\{S_j\right\}}_{j=1}^N,{(p_j)}_{j=0}^N,\nu )$ and ${\kappa }_r$ is uniquely determined by the relation ${\sum }_{j=1}^N{(p_j{s}_j^r)}^{{\displaystyle \frac{{\kappa }_r}{r+{\kappa }_r}}}=1$, then $D_r(\mu )=max\{{\kappa }_r,D_r(\nu )\}.$

In the following remark, we examine whether the inequalities

$${\underline{D}}_r(\mu )\le max\{{\kappa }_r,{\underline{D}}_r(\nu )\}\mathrm{and}{\overline{D}}_r(\mu )\le max\{{\kappa }_r,{\overline{D}}_r(\nu )\}$$

hold in general for a condensation system $({\left\{S_j\right\}}_{j=1}^N,{(p_j)}_{j=0}^N,\nu )$, when $\nu$ is any Borel probability measure not necessarily obtained by some IFS or an image measure of an ergodic measure. However, we cannot make any conclusive statement regarding their validity. In Section 3 and Section 4, we give two examples in which for $r=2$, both $D_r(\nu )$ and $D_r(\mu )$ exist and satisfy the following relations:

$${\underline{D}}_r(\mu )=max\{{\kappa }_r,{\underline{D}}_r(\nu )\}\mathrm{and}{\overline{D}}_r(\mu )=max\{{\kappa }_r,{\overline{D}}_r(\nu )\}.$$

Remark 1.

Proceeding in a similar way as ([2], Lemma 4.14), we have

$$V_{n,r}(\mu )\left\{\begin{array}{c}\ge p_0{V}_{n,r}(\nu )+\sum _{j=1}^N{p}_j{s}_j^r{V}_{n,r}(\mu ),\hfill \\ \le p_0{V}_{\lfloor {\displaystyle \frac{n}{N+1}}\rfloor ,r}(\mu )+\sum _{j=1}^N{p}_j{s}_j^r{V}_{\lfloor {\displaystyle \frac{n}{N+1}}\rfloor ,r}(\mu ),\hfill \end{array}\right.$$

where $\lfloor x\rfloor$ represents the greatest integer not exceeding x for any real x. Thus, for any $s>0$, we have

$$\begin{array}{cc}\hfill & (1-\sum _{j=1}^N{p}_j{s}_j^r)\underset{n}{lim\; inf}{n}^{r/s}{V}_{n,r}(\nu )\le \underset{n}{lim\; inf}{n}^{r/s}{V}_{n,r}(\mu )\le \underset{n}{lim\; sup}{n}^{r/s}{V}_{n,r}(\mu )\hfill \\ \hfill & \le p_0{2}^{r/s}{(N+1)}^{r/s}\underset{n}{lim\; sup}{n}^{r/s}{V}_{n,r}(\nu )+\sum _{j=1}^N{p}_j{s}_j^r{2}^{r/{\kappa }_r}{(N+1)}^{r/{\kappa }_r}\underset{n}{lim\; sup}{n}^{r/s}{V}_{n,r}(\mu ).\hfill \end{array}$$

Let $s=max\{{\kappa }_r,{\underline{D}}_r(\nu )\}$, or $s=max\{{\kappa }_r,{\overline{D}}_r(\nu )\}$. Then, from the last inequalities, we see that ${lim\; inf}_n{n}^{r/s}{V}_{n,r}(\mu )$ and ${lim\; sup}_n{n}^{r/s}{V}_{n,r}(\mu )$ can either or both be zero, positive, or infinity. Thus, we can not draw any useful conclusion whether the following inequalities are correct:

$${\underline{D}}_r(\mu )\le max\{{\kappa }_r,{\underline{D}}_r(\nu )\},\mathit{and}{\overline{D}}_r(\mu )\le max\{{\kappa }_r,{\overline{D}}_r(\nu )\}.$$

In the sequel, we will denote the condensation measure as P. Take $r=2$, and consider the condensation system as $(\{S_1,S_2\},({\displaystyle \frac{1}{3}},{\displaystyle \frac{1}{3}},{\displaystyle \frac{1}{3}}),\nu )$ on the Euclidean space $\mathbb{R}$, i.e., the condensation measure P is given by $P:={\displaystyle \frac{1}{3}}P\circ S_1^{-1}+{\displaystyle \frac{1}{3}}P\circ S_2^{-1}+{\displaystyle \frac{1}{3}}\nu$, where $\nu$ is a Borel probability measure on $\mathbb{R}$ with a compact support. We define the two contractive similarity mappings $S_1$ and $S_2$ on $\mathbb{R}$ as follows: $S_1(x)={\displaystyle \frac{1}{5}}x$ and $S_2(x)={\displaystyle \frac{1}{5}}x+{\displaystyle \frac{4}{5}}$ for all $x\in \mathbb{R}$. Then, for $n\in \mathbb{N}$, by (1), we have

$$P={\displaystyle \frac{1}{3^n}}\sum _{\left|\omega \right|=n}P\circ S_{\omega }^{-1}+\sum _{k=0}^{n-1}{\displaystyle \frac{1}{3^{k+1}}}\sum _{\left|\omega \right|=k}\nu \circ S_{\omega }^{-1}.$$

(2)

Write $J:=[0,1]$. For $\omega ={\omega }_1{\omega }_2\cdots {\omega }_k\in I^k$, set $J_{\omega }:=S_{\omega }(J)$, and $C_{\omega }:=S_{\omega }(C)$.

The following lemma can easily be proved.

Lemma 3.

Let K be the support of the condensation measure. Then, for any $n\ge 1$,

$$K\subset (\underset{\omega \in I^n}{\cup }{J}_{\omega })\bigcup \left(\underset{k=0}{\overset{n-1}{\cup }}(\underset{\omega \in I^k}{\cup }{C}_{\omega })\right)\subset J.$$

By Equation (2), we can deduce the following lemma.

Lemma 4.

Let $g:\mathbb{R}\to {\mathbb{R}}^{+}$ be Borel measurable and $n\in \mathbb{N}$. Then,

$$\int g(x)dP(x)={\displaystyle \frac{1}{3^n}}\sum _{\left|\omega \right|=n}\int (g\circ S_{\omega })(x)dP(x)+\sum _{k=0}^{n-1}{\displaystyle \frac{1}{3^{k+1}}}\sum _{\left|\omega \right|=k}\int (g\circ S_{\omega })(x)d\nu (x).$$

The following lemma that appears in [29] is also true.

Lemma 5

(See [29], Lemma 3.7). Let α be an optimal set of n-means for the condensation measure P. Then, for any $\omega \in I^{*}$, the set $S_{\omega }(\alpha ):=\{S_{\omega }(a):a\in \alpha \}$ is an optimal set of n-means for the image measure $P\circ S_{\omega }^{-1}$. Conversely, if β is an optimal set of n-means for the image measure $P\circ S_{\omega }^{-1}$, then $S_{\omega }^{-1}(\beta )$ is an optimal set of n-means for P.

In the following two sections, we investigate the quantization for two condensation measures P: in Section 3, P is associated with an infinite discrete distribution $\nu$ borrowed from our recent paper [28]; and in Section 4, P is associated with a uniform distribution.

3. Quantization for a Condensation Measure Associated with a Discrete Distribution

Let $\nu$ be an infinite discrete distribution with support $C:=\{a_j:j\in \mathbb{N}\}$, where $a_j:={\displaystyle \frac{2}{5}}+{\displaystyle \frac{1}{5}}{\displaystyle \frac{2^{j-1}-1}{2^{j-1}}}$. Notice that $C=\left\{{\displaystyle \frac{2}{5}},{\displaystyle \frac{1}{2}},{\displaystyle \frac{11}{20}},{\displaystyle \frac{23}{40}},{\displaystyle \frac{47}{80}},{\displaystyle \frac{19}{32}},{\displaystyle \frac{191}{320}},{\displaystyle \frac{383}{640}},\cdots \right\}\subset [{\displaystyle \frac{2}{5}},{\displaystyle \frac{3}{5}}]\subset \mathbb{R}$. Let the probability mass function of $\nu$ be given by f such that

$$f(x)=\left\{\begin{array}{cc}{\displaystyle \frac{1}{2^j}}& \mathrm{if}x=a_j\mathrm{for}j\in \mathbb{N},\\ 0& \mathrm{otherwise}.\end{array}\right.$$

Set $A_k:=\{a_k,a_{k+1},a_{k+2},\cdots \}$, and $b_k=E(X:X\in A_k)$, where $k\in \mathbb{N}$. Then, by the definition of conditional expectation, we see that

$$b_k={\displaystyle \frac{{\sum }_{j=k}^{\infty }{a}_j\frac{1}{2^j}}{{\sum }_{j=k}^{\infty }\frac{1}{2^j}}}.$$

For any Borel measurable function g on $\mathbb{R}$, by $\int gd\nu$, it is meant that

$$\int gd\nu =\sum _{j=1}^{\infty }g(a_j)\nu (a_j)=\sum _{j=1}^{\infty }g(a_j){\displaystyle \frac{1}{2^j}}.$$

Then, $E(\nu )={\displaystyle \frac{7}{15}},\mathrm{and}V(\nu )={\displaystyle \frac{8}{1575}},$ where $E(\nu )$ and $V(\nu )$ represent the expected value of a random variable with distribution $\nu ,$ and its variance, respectively. The quantization dimension $D(\nu )$ of the measure $\nu$ exists and equals zero; for more details one can see [28]. In order to compute the quantization dimension of condensation measure P associated with discrete distribution $\nu ,$ first we give the following lemmas and proposition.

Lemma 6.

Let $E(P)$ represent the expected value and $V:=V(P)$ represent the variance of the condensation measure P. Then, $E(P)={\displaystyle \frac{19}{39}},$$V={\displaystyle \frac{\mathrm{86,696}}{\mathrm{777,231}}}$, and for any $x_0\in \mathbb{R}$, $\int {(x-x_0)}^{2}dP={(x_0-{\displaystyle \frac{19}{39}})}^2+V(P)$.

Proof. 

By Lemmas 4 and 5, we can see that $E(P)=\int xdP={\displaystyle \frac{19}{39}}$. Again, $\int x^{2}d\nu ={\sum }_{j=1}^{\infty }{a}_j^2{\displaystyle \frac{1}{2^j}}={\displaystyle \frac{39}{175}}.$ We have

$$\begin{array}{cc}\hfill & \int x^{2}dP={\displaystyle \frac{1}{3}}\int {(S_1(x))}^{2}dP+{\displaystyle \frac{1}{3}}\int {(S_2(x))}^{2}dP+{\displaystyle \frac{1}{3}}\int x^{2}d\nu \hfill \\ \hfill & ={\displaystyle \frac{1}{3}}\int {\left({\displaystyle \frac{1}{5}}x\right)}^{2}dP+{\displaystyle \frac{1}{3}}\int {\left({\displaystyle \frac{1}{5}}x+{\displaystyle \frac{4}{5}}\right)}^{2}dP+{\displaystyle \frac{1}{3}}{\displaystyle \frac{39}{175}}\hfill \\ \hfill & ={\displaystyle \frac{1}{75}}\int x^{2}dP+{\displaystyle \frac{1}{75}}\int x^{2}dP+{\displaystyle \frac{8}{75}}\int xdP+{\displaystyle \frac{16}{75}}+{\displaystyle \frac{13}{175}},\hfill \end{array}$$

which implies $\int x^{2}dP={\displaystyle \frac{6953}{19,929}}$, and hence, $V(P)=\int x^{2}dP-{(\int xdP)}^2={\displaystyle \frac{6953}{19,929}}-{\left({\displaystyle \frac{19}{39}}\right)}^2={\displaystyle \frac{86,696}{777,231}}$. For any $x_0\in \mathbb{R}$, $\int {(x-x_0)}^{2}dP={(x_0-{\displaystyle \frac{19}{39}})}^2+V(P)$ follows from the standard theory of probability. Thus, the proof of the lemma is complete. □

We now give the following proposition.

Proposition 3.

Let $\omega \in I^k$, $k\ge 0$, and let X be the random variable with probability distribution P. Then, $E(X:X\in J_{\omega })=S_{\omega }({\displaystyle \frac{19}{39}}),\mathit{and}E(X:X\in C_{\omega })=S_{\omega }({\displaystyle \frac{7}{15}}).$ Moreover, for any $x_0\in \mathbb{R}$ and any $\omega \in I^k$, $k\ge 0$, we have

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\int }_{J_{\omega }}{(x-x_0)}^{2}dP(x)={\displaystyle \frac{1}{3^k}}\left({\displaystyle \frac{1}{{25}^k}}V+{(S_{\omega }({\displaystyle \frac{19}{39}})-x_0)}^2\right),\\ {\int }_{C_{\omega }}{(x-x_0)}^{2}dP(x)={\displaystyle \frac{1}{3^{k+1}}}\left({\displaystyle \frac{1}{{25}^k}}V(\nu )+{(S_{\omega }({\displaystyle \frac{7}{15}})-x_0)}^2\right),\\ {\int }_{S_{\omega }(\{a_1,a_2\})}{(x-x_0)}^{2}dP(x)={\displaystyle \frac{1}{3^{k+1}}}\left({(S_{\omega }(a_1)-x_0)}^2{\displaystyle \frac{1}{2}}+{(S_{\omega }(a_2)-x_0)}^2{\displaystyle \frac{1}{2^2}}\right),\mathit{and}\\ {\int }_{S_{\omega }(A_3)}{(x-x_0)}^{2}dP(x)={\displaystyle \frac{1}{3^{k+1}}}\sum _{j=3}^{\infty }{(S_{\omega }(a_j)-x_0)}^2{\displaystyle \frac{1}{2^j}}.\end{array}\right.\end{array}$$

(3)

Proof. 

By Equation (2), we have $P(J_{\omega })={\displaystyle \frac{1}{3^k}}$ and $P(C_{\omega })={\displaystyle \frac{1}{3^{k+1}}}$. Then, by Lemma 4, the proof follows. □

Note 1.

By Lemma 6, it follows that the optimal set of one-mean for the condensation measure P consists of the expected value ${\displaystyle \frac{19}{39}}$ and the corresponding quantization error is the variance $V(P)$ of P, i.e., $V_1(P)=V(P)$.

3.1. Essential Lemmas and Propositions

In this subsection, we give some lemmas and propositions that we need to determine the optimal sets of n-means and the nth quantization errors for all $n\ge 2$. To determine the quantization error, we will frequently use the formulas given in Expression (3).

Proposition 4

(See [4,6]). Let α be an optimal set of n-means, $a\in \alpha$, and $M(a|\alpha )$ be the Voronoi region generated by $a\in \alpha$. Then, for every $a\in \alpha$, $(i)$$\mu (M(a|\alpha ))>0$, $(ii)$$\mu (\partial M(a|\alpha ))=0$, $(iii)$$a=E(X:X\in M(a|\alpha ))$, and $(iv)$μ-almost means surely the set $\{M(a|\alpha ):a\in \alpha \}$ forms a Voronoi partition of ${\mathbb{R}}^d$.

Proposition 5.

Let $\alpha :=\{c_1,c_2\}$ be an optimal set of two-means with $c_1<c_2$, and let $V_{\mathrm{2,1}}$ and $V_{\mathrm{2,2}}$, respectively, be the distortion errors contributed by the points $c_1$ and $c_2$. Then, $c_1={\displaystyle \frac{659}{2730}}$, $c_2={\displaystyle \frac{1621}{1950}}$, and the corresponding quantization error is $V_2=V_{2,1}+V_{2,2}=0.0269686$, where $V_{2,1}={\displaystyle \frac{8,469,163}{466,338,600}}\mathit{and}{V}_{2,2}={\displaystyle \frac{20,536,627}{2,331,693,000}}$, respectively, denote the distortion errors due to $c_1$ and $c_2$.

Proof. 

Consider the set of two points $\beta :=\{d_1,d_2\}$, where $d_1=E(X:X\in J_1\cup \{a_1,a_2\})$ and $d_2=E(X:X\in A_3\cup J_2)$, i.e.,

$$\begin{array}{cc}\hfill d_1& ={\displaystyle \frac{1}{P(J_1\cup \{a_1,a_2\})}}\left({\int }_{J_1}xdP+{\displaystyle \frac{1}{3}}(a_1{\displaystyle \frac{1}{2}}+a_2{\displaystyle \frac{1}{2^2}})\right)={\displaystyle \frac{659}{2730}},\hfill \end{array}$$

and

$$d_2={\displaystyle \frac{1}{P(A_3\cup J_2)}}\left({\displaystyle \frac{1}{3}}\sum _{j=3}^{\infty }{a}_j{\displaystyle \frac{1}{2^j}}+{\int }_{J_2}xdP\right)={\displaystyle \frac{1621}{1950}}.$$

The distortion error due to the set $\beta$ is given by

$$\begin{array}{cc}\hfill & \int \underset{d\in \beta }{min}{(x-d)}^{2}dP\hfill \\ \hfill & ={\int }_{J_1}{(x-{\displaystyle \frac{659}{2730}})}^{2}dP+{\displaystyle \frac{1}{3}}({(a_1-{\displaystyle \frac{659}{2730}})}^2{\displaystyle \frac{1}{2}}+{(a_2-{\displaystyle \frac{659}{2730}})}^2{\displaystyle \frac{1}{2^2}}+\sum _{j=3}^{\infty }{(a_j-{\displaystyle \frac{1621}{1950}})}^2{\displaystyle \frac{1}{2^j}}\hfill \\ \hfill & +{\int }_{J_2}{(x-{\displaystyle \frac{1621}{1950}})}^{2}dP={\displaystyle \frac{\mathrm{499,067}}{\mathrm{18,505,500}}}=0.0269686.\hfill \end{array}$$

Since $V_2$ is the quantization error for two-means, we have $V_2\le 0.0269686$. Let $\alpha :=\{c_1,c_2\}$ be an optimal set of two-means for P. Suppose that ${\displaystyle \frac{2}{5}}\le c_1$. Then,

$$V_2\ge {\int }_{J_1}{(x-{\displaystyle \frac{2}{5}})}^{2}dP={\displaystyle \frac{\mathrm{47,833}}{\mathrm{1,494,675}}}=0.0320023>V_2,$$

which leads to a contradiction. So, we can assume that $c_1<{\displaystyle \frac{2}{5}}$. Similarly, we can show that ${\displaystyle \frac{3}{5}}<c_2$. Since ${\displaystyle \frac{3}{10}}\le {\displaystyle \frac{1}{2}}(c_1+c_2)\le {\displaystyle \frac{7}{10}}$, the Voronoi region of $c_1$ cannot contain any point from $J_2$, and the Voronoi region of $c_2$ cannot contain any point from $J_1$. Suppose that the Voronoi region of $c_1$ does not contain any point from $A_1$. Then, $c_1=E(X:X\in J_1)={\displaystyle \frac{19}{195}}$, and $c_2=E(X:X\in A_2\cup J_2)={\displaystyle \frac{133}{195}}$, and then the distortion error is

$${\int }_{J_1}{(x-c_1)}^{2}dP+{\displaystyle \frac{1}{3}}\sum _{j=1}^{\infty }{(a_j-c_2)}^2{\displaystyle \frac{1}{2^j}}+{\int }_{J_2}{(x-c_2)}^{2}dP={\displaystyle \frac{\mathrm{76,848}}{\mathrm{2,158,975}}}=0.0355947>V_2,$$

which leads to a contradiction. Similarly, we can show that if the Voronoi region of $c_2$ does not contain any point from $A_1$ a contradiction will arise. Thus, we conclude that both the Voronoi regions of $c_1$ and $c_2$ contain at least one point from $A_1$. Let the Voronoi region of $c_1$ contain points $a_1,a_2,\cdots ,a_k$ from $A_1$, and the Voronoi region of $c_2$ contain $\{a_j:j>k\}$ for some positive integer $k\ge 1$. Let $F(k)$ be the corresponding distortion error. Then,

$$F(k)={\int }_{J_1}{(x-c_1)}^{2}dP+{\displaystyle \frac{1}{3}}\left(\sum _{j=1}^k{(a_j-c_1)}^2{\displaystyle \frac{1}{2^j}}+\sum _{j=k+1}^{\infty }{(a_j-c_2)}^2{\displaystyle \frac{1}{2^j}}\right)+{\int }_{J_2}{(x-c_2)}^{2}dP.$$

Using Calculus, we see that $min\{F(k):k\in \mathbb{N}\}={\displaystyle \frac{\mathrm{499,067}}{\mathrm{18,505,500}}}=0.0269686,$ which occurs when $k=2$. This yields the fact that $c_1=E(X:X\in J_1\cup \{a_1,a_2\})={\displaystyle \frac{659}{2730}}$, and $c_2=E(X:X\in A_3\cup J_2)={\displaystyle \frac{1621}{1950}}$, and the corresponding quantization error is $V_2=V_{\mathrm{2,1}}+V_{\mathrm{2,2}}={\displaystyle \frac{\mathrm{499,067}}{\mathrm{18,505,500}}}=0.0269686$, where $V_{\mathrm{2,1}}={\int }_{J_1}{(x-c_1)}^{2}dP+{\displaystyle \frac{1}{3}}{\sum }_{j=1}^2{(a_j-c_1)}^2{\displaystyle \frac{1}{2^j}}={\displaystyle \frac{\mathrm{8,469,163}}{\mathrm{466,338,600}}}$, and $V_{\mathrm{2,2}}={\displaystyle \frac{1}{3}}{\sum }_{j=3}^{\infty }{(a_j-c_2)}^2{\displaystyle \frac{1}{2^j}}+{\int }_{J_2}{(x-c_2)}^{2}dP={\displaystyle \frac{\mathrm{20,536,627}}{\mathrm{233,1693,000}}}$. Thus, the proof of the proposition is complete. □

Lemma 7.

Let $\omega \in I^k$ for $k\ge 0$. Then,

$${\int }_{J_{\omega 1}\cup S_{\omega }(\{a_1,a_2\})}{(x-S_{\omega }({\displaystyle \frac{659}{2730}}))}^{2}dP={\displaystyle \frac{1}{{75}^k}}{V}_{\mathrm{2,1}},\mathit{and}{\int }_{S_{\omega }(A_3)\cup J_{\omega 2}}{(x-S_{\omega }({\displaystyle \frac{1621}{1950}}))}^{2}dP={\displaystyle \frac{1}{{75}^k}}{V}_{\mathrm{2,2}}.$$

Proof. 

For any $\omega \in I^k$, $k\ge 0$, we have

$$\begin{array}{cc}\hfill & {\int }_{J_{\omega 1}\cup S_{\omega }(\{a_1,a_2\})}{(x-S_{\omega }({\displaystyle \frac{659}{2730}}))}^{2}dP={\displaystyle \frac{1}{3^k}}{\int }_{J_{\omega 1}\cup S_{\omega }(\{a_1,a_2\})}{(x-S_{\omega }({\displaystyle \frac{659}{2730}}))}^{2}d(P\circ S_{\omega }^{-1})\hfill \\ \hfill & ={\displaystyle \frac{1}{3^k}}{\int }_{J_1\cup \{a_1,a_2\}}{(S_{\omega }(x)-S_{\omega }({\displaystyle \frac{659}{2730}}))}^{2}dP={\displaystyle \frac{1}{3^k}}{\displaystyle \frac{1}{{25}^k}}{\int }_{J_1\cup \{a_1,a_2\}}{(x-{\displaystyle \frac{659}{2730}})}^{2}dP={\displaystyle \frac{1}{{75}^k}}{V}_{\mathrm{2,1}}.\hfill \end{array}$$

Similarly, ${\int }_{S_{\omega }(A_3)\cup J_{\omega 2}}{(x-S_{\omega }({\displaystyle \frac{1621}{1950}}))}^{2}dP={\displaystyle \frac{1}{{75}^k}}{V}_{\mathrm{2,2}}$. Thus, the proof of the lemma is complete. □

From the above lemma, the following corollary follows.

Corollary 3.

Let $\omega \in I^k$ for $k\ge 0$. Then, for any $a\in \mathbb{R}$,

$$\begin{array}{cc}\hfill & {\int }_{J_{\omega 1}\cup S_{\omega }(\{a_1,a_2\})}{(x-a)}^{2}dP={\displaystyle \frac{1}{3^k}}\left({\displaystyle \frac{1}{{25}^k}}{V}_{2,1}+{\displaystyle \frac{7}{12}}{(S_{\omega }({\displaystyle \frac{659}{2730}})-a)}^2\right),\mathit{and}\hfill \\ \hfill & {\int }_{S_{\omega }(A_3)\cup J_{\omega 2}}{(x-a)}^{2}dP={\displaystyle \frac{1}{3^k}}\left({\displaystyle \frac{1}{{25}^k}}{V}_{2,2}+{\displaystyle \frac{5}{12}}{(S_{\omega }({\displaystyle \frac{1621}{1950}})-a)}^2\right).\hfill \end{array}$$

Proposition 6.

Let $\alpha :=\{a_1,a_2,a_3\}$ be an optimal set of three-means with $a_1<a_2<a_3$. Then, $a_1=S_1({\displaystyle \frac{19}{39}})$, $a_2=E(\nu )={\displaystyle \frac{7}{15}}$, and $a_3=S_2({\displaystyle \frac{19}{39}})$. The corresponding quantization error is $V_3={\displaystyle \frac{\mathrm{30,232}}{\mathrm{6,476,925}}}=0.00466765$.

Proof. 

Let $\beta :=\{S_1({\displaystyle \frac{19}{39}}),{\displaystyle \frac{7}{15}},S_2({\displaystyle \frac{19}{39}})\}$. Then, using (3), we have

$$\begin{array}{cc}\hfill & \int \underset{a\in \beta }{min}{(x-a)}^{2}dP={\int }_{J_1}{(x-{\displaystyle \frac{1}{10}})}^{2}dP+{\int }_C{(x-{\displaystyle \frac{7}{15}})}^{2}dP+{\int }_{J_2}{(x-{\displaystyle \frac{9}{10}})}^{2}dP\hfill \\ \hfill & ={\displaystyle \frac{1}{3}}{\displaystyle \frac{1}{25}}V+{\displaystyle \frac{1}{3}}V(\nu )+{\displaystyle \frac{1}{3}}{\displaystyle \frac{1}{25}}V={\displaystyle \frac{\mathrm{30,232}}{\mathrm{6,476,925}}}=0.00466765.\hfill \end{array}$$

Since $V_3$ is the quantization error for three-means, we have $0.00466765\ge V_3$. Let $\alpha :=\{c_1,c_2,c_3\}$ be an optimal set of three-means with $c_1<c_2<c_3$. Since $c_1$, $c_2$, and $c_3$ are the expected values of their own Voronoi regions, we have $0<c_1<c_2<c_3<1$. Suppose that ${\displaystyle \frac{1}{5}}\le c_1$. Then,

$$V_3\ge {\int }_{J_1}{(x-{\displaystyle \frac{1}{5}})}^{2}dP={\displaystyle \frac{2488}{\mathrm{498,225}}}=0.00499373>V_3,$$

which is a contradiction. So, we can assume that $c_1<{\displaystyle \frac{1}{5}}$. We now show that the Voronoi region of $c_1$ does not contain any point from C. For the sake of contradiction, assume that the Voronoi region of $c_1$ contains at least one point, say $a_1$, from C. Then,

$$V_3\ge min\{{\int }_{J_1}{(x-c_1)}^{2}dP+{\displaystyle \frac{1}{3}}{(a_1-c_1)}^2{\displaystyle \frac{1}{2}}:0<c_1<{\displaystyle \frac{1}{5}}\}={\displaystyle \frac{\mathrm{2,038,879}}{\mathrm{174,876,975}}}=0.0116589>V_3,$$

which leads to a contradiction. Hence, the Voronoi region of $c_1$ does not contain any point from C. If $c_3\le ({\displaystyle \frac{4}{5}}-{\displaystyle \frac{1}{1000}})$, then

$$V_3\ge {\int }_{J_2}{(x-({\displaystyle \frac{4}{5}}-{\displaystyle \frac{1}{1000}}))}^{2}dP={\displaystyle \frac{\mathrm{94,007,843}}{\mathrm{19,929,000,000}}}=0.00471714>V_3,$$

which leads to a contradiction. Hence, $({\displaystyle \frac{4}{5}}-{\displaystyle \frac{1}{1000}})<c_3$. Suppose that $c_2\le {\displaystyle \frac{2}{5}}$. Then, $a_9<{\displaystyle \frac{1}{2}}\left({\displaystyle \frac{2}{5}}+({\displaystyle \frac{4}{5}}-{\displaystyle \frac{1}{1000}})\right)<a_{10}$, and so, we have

$$\begin{array}{cc}\hfill V_3& \ge {\int }_{J_1}\underset{c\in S_1({\alpha }_2)}{min}{(x-c)}^{2}dP+{\displaystyle \frac{1}{3}}\sum _{j=1}^9{(a_j-{\displaystyle \frac{2}{5}})}^2{\displaystyle \frac{1}{2^j}}+{\int }_{J_2}(x-S_2{({\displaystyle \frac{19}{39}})}^{2}dP\hfill \\ \hfill & ={\displaystyle \frac{1}{75}}(V_{2,1}+V_{2,2})+{\displaystyle \frac{1}{3}}\sum _{j=1}^9{(a_j-{\displaystyle \frac{2}{5}})}^2{\displaystyle \frac{1}{2^j}}+{\displaystyle \frac{1}{75}}V\hfill \\ \hfill & ={\displaystyle \frac{\mathrm{1,221,371,823,004,331}}{\mathrm{244,495,731,916,800,000}}}=0.00499547>V_3,\hfill \end{array}$$

which leads to a contradiction. Hence, we can assume that ${\displaystyle \frac{2}{5}}<c_2$. Then, as ${\displaystyle \frac{1}{5}}<{\displaystyle \frac{1}{2}}(c_1+c_2)$, the Voronoi region of $c_2$ does not contain any point from $J_1$. Recall that we already proved that the Voronoi region of $c_1$ does not contain any point from C. Assume that $({\displaystyle \frac{4}{5}}-{\displaystyle \frac{1}{1000}})<c_3\le {\displaystyle \frac{4}{5}}$. Then, ${\displaystyle \frac{1}{2}}(c_2+c_3)\le {\displaystyle \frac{3}{5}}$ implying $c_2\le {\displaystyle \frac{6}{5}}-c_3\le {\displaystyle \frac{6}{5}}-({\displaystyle \frac{4}{5}}-{\displaystyle \frac{1}{1000}})={\displaystyle \frac{401}{1000}}<a_2$, and so

$$V_3\ge {\displaystyle \frac{1}{3}}\sum _{j=2}^{\infty }{(a_j-0.401)}^2{\displaystyle \frac{1}{2^j}}+{\int }_{J_2}{(x-{\displaystyle \frac{4}{5}})}^{2}dP={\displaystyle \frac{\mathrm{310,181,843}}{\mathrm{39,858,000,000}}}=0.00778217>V_3,$$

which is a contradiction. So, we can assume that ${\displaystyle \frac{4}{5}}<c_3$. If the Voronoi region of $c_3$ contains points form C, we have ${\displaystyle \frac{1}{2}}(c_2+c_3)<{\displaystyle \frac{3}{5}}$, implying $c_2\le {\displaystyle \frac{6}{5}}-c_3\le {\displaystyle \frac{6}{5}}-{\displaystyle \frac{4}{5}}={\displaystyle \frac{2}{5}}$, which leads to a contradiction, as we have seen ${\displaystyle \frac{2}{5}}<c_2$. Hence, the Voronoi region of $c_3$ does not contain any point from C. If ${\displaystyle \frac{3}{5}}\le c_2$, then

$$\begin{array}{cc}\hfill V_3& \ge {\int }_{J_1}{(x-S_1({\displaystyle \frac{19}{39}}))}^{2}dP+{\displaystyle \frac{1}{3}}\sum _{j=1}^{\infty }{(a_j-{\displaystyle \frac{3}{5}})}^2{\displaystyle \frac{1}{2^j}}+{\int }_{J_2}\underset{c\in {\alpha }_2}{min}{(x-c)}^{2}dP\hfill \\ \hfill & ={\displaystyle \frac{1}{75}}V+{\displaystyle \frac{1}{3}}\sum _{j=1}^{\infty }{(a_j-{\displaystyle \frac{3}{5}})}^2{\displaystyle \frac{1}{2^j}}+{\displaystyle \frac{1}{75}}(V_{2,1}+V_{2,2})={\displaystyle \frac{\mathrm{275,894,407}}{\mathrm{29,146,162,500}}}=0.00946589>V_3,\hfill \end{array}$$

which gives a contradiction. So, we can assume that $c_2<{\displaystyle \frac{3}{5}}$. Then, notice that the Voronoi region of $c_2$ does not contain any point from $J_3$ as ${\displaystyle \frac{4}{5}}<c_3<1$. Hence, we have $c_1=S_1({\displaystyle \frac{19}{39}})$, $c_2=E(\nu )={\displaystyle \frac{7}{15}}$, and $c_3=S_2({\displaystyle \frac{19}{39}})$, and the corresponding quantization error is $V_3={\displaystyle \frac{\mathrm{30,232}}{\mathrm{6,476,925}}}=0.00466765$. Thus, the proof of the proposition is complete. □

Lemma 8.

Let α be an optimal set of four-means. Then, $\alpha \cap J_1\ne \varnothing$, $\alpha \cap J_2\ne \varnothing$, and $\mathit{card}(\alpha \cap C)=2$.

Proof. 

The distortion error due to the set $\beta :=\{S_1({\displaystyle \frac{19}{39}}),S_2({\displaystyle \frac{19}{39}})\}\cup {\alpha }_2(\nu )$ is given by

$$\int \underset{a\in \beta }{min}{(x-a)}^{2}dP={\displaystyle \frac{1}{75}}V+{\displaystyle \frac{1}{3}}\sum _{j=2}^{\infty }{(a_j-b_2)}^2{\displaystyle \frac{1}{2^j}}+{\displaystyle \frac{1}{75}}V={\displaystyle \frac{\mathrm{185,729}}{\mathrm{58,292,325}}}=0.00318617.$$

Since $V_4$ is the quantization error for four-means, we have $V_4\le 0.00318617$. Let an optimal set of four-means be given by $\alpha :=\{c_1,c_2,c_3,c_4\}$ such that $c_1<c_2<c_3<c_4$. Since $c_1,c_2,c_3$, and $c_4$ are the expected values of their own Voronoi regions, we have $0<c_1<c_2<c_3<c_4<1$. If $c_1\ge {\displaystyle \frac{1}{5}}$, then

$$V_4\ge {\int }_{J_1}{(x-{\displaystyle \frac{1}{5}})}^{2}dP={\displaystyle \frac{2488}{\mathrm{498,225}}}=0.00499373>V_4,$$

which leads to a contradiction. So, we can assume that $c_1<{\displaystyle \frac{1}{5}}$. If $c_4\le {\displaystyle \frac{4}{5}}$, then

$$V_4\ge {\int }_{J_2}{(x-{\displaystyle \frac{4}{5}})}^{2}dP={\displaystyle \frac{6953}{\mathrm{1,494,675}}}=0.00465185>V_4,$$

which is a contradiction. So, we can assume that ${\displaystyle \frac{4}{5}}<c_4$. Thus, we see that $\alpha \cap J_1\ne \varnothing$, and $\alpha \cap J_2\ne \varnothing$. Assume that $\alpha \cap C=\varnothing$. Then, the following cases can arise:

Case 1. $c_3<{\displaystyle \frac{2}{5}}$.

Then,

$$V_4\ge {\displaystyle \frac{1}{3}}\sum _{j=1}^{\infty }{(a_j-{\displaystyle \frac{2}{5}})}^2{\displaystyle \frac{1}{2^j}}+{\int }_{J_2}{(x-S_2({\displaystyle \frac{19}{39}}))}^{2}dP={\displaystyle \frac{\mathrm{271,751}}{\mathrm{58,292,325}}}=0.00466187>V_4,$$

which is a contradiction.

Case 2. $c_2<{\displaystyle \frac{2}{5}}<{\displaystyle \frac{3}{5}}<c_3$.

First, assume that ${\displaystyle \frac{7}{20}}\le c_2<{\displaystyle \frac{2}{5}}$, and ${\displaystyle \frac{3}{5}}<c_3\le {\displaystyle \frac{13}{20}}$. Then, ${\displaystyle \frac{1}{2}}(c_1+c_2)<{\displaystyle \frac{1}{5}}$ implying $c_1<{\displaystyle \frac{2}{5}}-{\displaystyle \frac{7}{20}}={\displaystyle \frac{1}{20}}$. Similarly, ${\displaystyle \frac{1}{2}}(c_3+c_4)>{\displaystyle \frac{4}{5}}$ implying ${\displaystyle \frac{19}{20}}<c_4$. Then,

$$\begin{array}{cc}\hfill V_4& \ge {\int }_{S_1(A_3)\cup J_{12}}{(x-{\displaystyle \frac{1}{20}})}^{2}dP+{\displaystyle \frac{1}{3}}\sum _{j=1}^2{(a_j-{\displaystyle \frac{2}{5}})}^2{\displaystyle \frac{1}{2^j}}+{\displaystyle \frac{1}{3}}\sum _{j=3}^{\infty }{(a_j-{\displaystyle \frac{3}{5}})}^2{\displaystyle \frac{1}{2^j}}\hfill \\ \hfill & +{\int }_{J_{21}\cup S_1(\{a_1,a_2\})}{(x-{\displaystyle \frac{19}{20}})}^{2}dP={\displaystyle \frac{\mathrm{143,519}}{\mathrm{27,594,000}}}=0.00520109>V_4,\hfill \end{array}$$

which gives a contradiction. Next, assume that ${\displaystyle \frac{7}{20}}\le c_2<{\displaystyle \frac{2}{5}}$, and ${\displaystyle \frac{13}{20}}\le c_3$. Then, $c_1<{\displaystyle \frac{1}{20}}$, and $a_2<{\displaystyle \frac{1}{2}}({\displaystyle \frac{2}{5}}+{\displaystyle \frac{13}{20}})<a_3$, and so

$$\begin{array}{cc}\hfill V_4& \ge {\int }_{S_1(A_3)\cup J_{12}}{(x-{\displaystyle \frac{1}{20}})}^{2}dP+{\displaystyle \frac{1}{3}}\sum _{j=1}^2{(a_j-{\displaystyle \frac{2}{5}})}^2{\displaystyle \frac{1}{2^j}}+{\displaystyle \frac{1}{3}}\sum _{j=3}^{\infty }{(a_j-{\displaystyle \frac{13}{20}})}^2{\displaystyle \frac{1}{2^j}}\hfill \\ \hfill & ={\displaystyle \frac{\mathrm{473,663}}{\mathrm{137,970,000}}}=0.00343309>V_4,\hfill \end{array}$$

which leads to a contradiction. Last, assume that $c_2<{\displaystyle \frac{7}{20}}$, and ${\displaystyle \frac{13}{20}}<c_3$. Then, as ${\displaystyle \frac{1}{2}}({\displaystyle \frac{7}{20}}+{\displaystyle \frac{13}{20}})={\displaystyle \frac{1}{2}}$, and ${\int }_{J_1}{min}_{c\in S_1({\alpha }_2)}{(x-c)}^{2}dP={\displaystyle \frac{1}{75}}(V_{2,1}+V_{2,2})$, we have

$$\begin{array}{cc}\hfill V_4& \ge {\displaystyle \frac{1}{75}}(V_{2,1}+V_{2,2})+{\displaystyle \frac{1}{3}}\sum _{j=1}^2{(a_j-{\displaystyle \frac{7}{20}})}^2{\displaystyle \frac{1}{2^j}}+{\displaystyle \frac{1}{3}}\sum _{j=3}^{\infty }{(a_j-{\displaystyle \frac{13}{20}})}^2{\displaystyle \frac{1}{2^j}}\hfill \\ \hfill & ={\displaystyle \frac{\mathrm{126,549,001}}{\mathrm{38,861,550,000}}}=0.00325641>V_4,\hfill \end{array}$$

which is a contradiction.

Case 3. ${\displaystyle \frac{3}{5}}<c_2$.

Since $c_1<{\displaystyle \frac{1}{5}}$, this is the reflection of Case 1 about the point ${\displaystyle \frac{1}{2}}$, thus a contradiction arises in this case as well.

Hence, we can conclude that $\alpha \cap C\ne \varnothing$. We now show that $\mathrm{card}(\alpha \cap C)=2$. For the sake of contradiction, assume that $\mathrm{card}(\alpha \cap C)=1$. Recall that $\alpha \cap J_1\ne \varnothing$, and $\alpha \cap J_2\ne \varnothing$. Thus, either $c_2\in C$, or $c_3\in C$. Without any loss of generality assume that $c_3\in C$. Then, $c_2<{\displaystyle \frac{2}{5}}$. The following cases can arise:

Case I. ${\displaystyle \frac{7}{20}}\le c_2<{\displaystyle \frac{2}{5}}$.

Then, ${\displaystyle \frac{1}{2}}(c_1+c_2)<{\displaystyle \frac{1}{5}}$ implying $c_1<{\displaystyle \frac{1}{20}}$, and so

$$V_4\ge {\int }_{S_1(A_3)\cup J_{12}}{(x-{\displaystyle \frac{1}{20}})}^{2}dP+{\displaystyle \frac{1}{3}}{V}_2(\nu )+{\displaystyle \frac{1}{75}}V={\displaystyle \frac{\mathrm{344,484,113}}{\mathrm{93,267,720,000}}}=0.0036935>V_4,$$

which leads to a contradiction.

Case II. ${\displaystyle \frac{3}{10}}\le c_2\le {\displaystyle \frac{7}{20}}$.

Then, ${\displaystyle \frac{1}{2}}(c_1+c_2)<{\displaystyle \frac{1}{5}}$ implying $c_1<{\displaystyle \frac{1}{10}}$. Moreover, $a_1<{\displaystyle \frac{1}{2}}({\displaystyle \frac{7}{20}}+b_2)<a_2$. First assume that ${\displaystyle \frac{2}{25}}\le c_1<{\displaystyle \frac{1}{10}}$. Then,

$$\begin{array}{cc}\hfill V_4& \ge {\int }_{J_{11}}{(x-{\displaystyle \frac{2}{25}})}^{2}dP+{\int }_{S_1(A_3)\cup J_{12}}{(x-{\displaystyle \frac{1}{10}})}^{2}dP+{\displaystyle \frac{1}{3}}{(a_1-{\displaystyle \frac{7}{20}})}^2{\displaystyle \frac{1}{2}}\hfill \\ \hfill & +{\displaystyle \frac{1}{3}}\sum _{j=2}^{\infty }{(a_j-b_2)}^2{\displaystyle \frac{1}{2^j}}+{\displaystyle \frac{1}{75}}V\hfill \\ \hfill & ={\displaystyle \frac{\mathrm{152,465,267}}{\mathrm{46,633,860,000}}}=0.00326941>V_4,\hfill \end{array}$$

which is a contradiction. Next, assume that $c_1\le {\displaystyle \frac{2}{25}}$. Then, $S_{12}({\displaystyle \frac{3}{5}})<{\displaystyle \frac{1}{2}}({\displaystyle \frac{2}{25}}+{\displaystyle \frac{3}{10}})<S_{122}(0)$, and so

$$\begin{array}{cc}\hfill V_4& \ge {\int }_{J_{11}}{(x-S_{11}({\displaystyle \frac{19}{39}}))}^{2}dP+{\int }_{S_1(C)}{(x-{\displaystyle \frac{2}{25}})}^{2}dP+{\int }_{J_{121}}{(x-{\displaystyle \frac{2}{25}})}^{2}dP+{\int }_{S_{12}(C)}{(x-{\displaystyle \frac{2}{25}})}^{2}dP\hfill \\ \hfill & +{\int }_{J_{122}}{(x-{\displaystyle \frac{3}{10}})}^{2}dP+{\displaystyle \frac{1}{3}}{(a_1-{\displaystyle \frac{7}{20}})}^2{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{3}}\sum _{j=2}^{\infty }{(a_j-b_2)}^2{\displaystyle \frac{1}{2^j}}+{\displaystyle \frac{1}{75}}V\hfill \\ \hfill & ={\displaystyle \frac{\mathrm{89,570,473}}{\mathrm{27,980,316,000}}}=0.0032012>V_4,\hfill \end{array}$$

which leads to a contradiction.

Case III. $c_2\le {\displaystyle \frac{3}{10}}$.

Then, $a_1<{\displaystyle \frac{1}{2}}({\displaystyle \frac{3}{10}}+b_2)<a_2$, and so

$$\begin{array}{cc}\hfill V_4& \ge {\displaystyle \frac{1}{3}}{(a_1-{\displaystyle \frac{3}{10}})}^2{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{3}}\sum _{j=2}^{\infty }{(a_j-b_2)}^2{\displaystyle \frac{1}{2^j}}+{\displaystyle \frac{1}{75}}V={\displaystyle \frac{\mathrm{104,633}}{\mathrm{31,089,240}}}=0.00336557>V_4,\hfill \end{array}$$

which is a contradiction.

Thus, $\mathrm{card}(\alpha \cap C)=1$ leads to a contradiction. Hence, we can assume that $\mathrm{card}(\alpha \cap C)=2$. Thus, the proof of the lemma is complete. □

From Lemma 8, the following proposition follows.

Proposition 7.

The set $\{S_1({\displaystyle \frac{19}{39}}),S_2({\displaystyle \frac{19}{39}})\}\cup {\alpha }_2(\nu )$ is an optimal set of four-means with quantization error $V_4={\displaystyle \frac{185,729}{58,292,325}}=0.00318617$.

We now prove the following proposition.

Proposition 8.

Let $n\in \mathbb{N}$ and $n\ge 3$, and let α be an optimal set of n-means. Then, $\alpha \cap J_1\ne \varnothing$, $\alpha \cap J_2\ne \varnothing$, and $\alpha \cap C\ne \varnothing$.

Proof. 

By Proposition 6 and Lemma 8, the proposition is true for $n=3,4$. We now prove that the proposition is true for $n\ge 5$. Consider the set of five points $\beta :=S_1({\alpha }_2)\cup {\alpha }_2(\nu )\cup \left\{S_2({\displaystyle \frac{19}{39}})\right\}$. The distortion error due to the set $\beta$ is given by

$$\int \underset{c\in \beta }{min}{(x-c)}^{2}dP={\displaystyle \frac{1}{75}}(V_{2,1}+V_{2,2})+{\displaystyle \frac{1}{3}}{V}_2(\nu )+{\displaystyle \frac{1}{75}}V={\displaystyle \frac{\mathrm{6,666,323}}{\mathrm{3,238,462,500}}}=0.00205848.$$

Since $V_n$ is the quantization error for n-means with $n\ge 5$, we have $V_n\le V_5\le 0.00205848$. Let $\alpha :=\{c_1<c_2<\cdots <c_n\}$ be an optimal set of n-means for P. In a similar way as shown in Lemma 8, we can show that $\alpha \cap J_1\ne \varnothing$ and $\alpha \cap J_2\ne \varnothing$. For the sake of contradiction, assume that $\alpha \cap C=\varnothing$. Let $j:=max\{i:c_i<{\displaystyle \frac{2}{5}}\}$. Then, $c_j<{\displaystyle \frac{2}{5}}$ and $c_{j+1}>{\displaystyle \frac{3}{5}}$. The following cases can arise:

Case 1. ${\displaystyle \frac{3}{10}}<c_j<{\displaystyle \frac{2}{5}}$ and ${\displaystyle \frac{3}{5}}<c_{j+1}<{\displaystyle \frac{7}{10}}$.

Then, ${\displaystyle \frac{1}{2}}(c_{j-1}+c_j)<{\displaystyle \frac{1}{5}}$ and ${\displaystyle \frac{1}{2}}(c_{j+1}+c_{j+2})>{\displaystyle \frac{4}{5}}$, yielding $c_{j-1}<{\displaystyle \frac{1}{10}}$ and $c_{j+1}>{\displaystyle \frac{9}{10}}$. Thus,

$$\begin{array}{cc}\hfill V_n& \ge {\int }_{S_1(A_3)\cup J_{12}}{(x-{\displaystyle \frac{1}{10}})}^{2}dP+{\displaystyle \frac{1}{3}}\sum _{j=1}^2{(a_j-{\displaystyle \frac{2}{5}})}^2{\displaystyle \frac{1}{2^j}}+{\displaystyle \frac{1}{3}}\sum _{j=3}^{\infty }{(a_j-{\displaystyle \frac{3}{5}})}^2{\displaystyle \frac{1}{2^j}}\hfill \\ \hfill & +{\int }_{J_{21}\cup S_1(\{a_1,a_2\})}{(x-{\displaystyle \frac{9}{10}})}^{2}dP={\displaystyle \frac{1123}{\mathrm{459,900}}}=0.00244184>V_n,\hfill \end{array}$$

which gives a contradiction.

Case 2. ${\displaystyle \frac{3}{10}}<c_j<{\displaystyle \frac{2}{5}}$ and ${\displaystyle \frac{7}{10}}\le c_{j+1}$. Then, $c_{j-1}<{\displaystyle \frac{1}{10}}$, and $a_2<{\displaystyle \frac{1}{2}}({\displaystyle \frac{2}{5}}+{\displaystyle \frac{7}{10}})=a_3$, and so

$$\begin{array}{cc}\hfill V_n& \ge {\int }_{S_1(A_3)\cup J_{12}}{(x-{\displaystyle \frac{1}{10}})}^{2}dP+{\displaystyle \frac{1}{3}}\sum _{j=1}^2{(a_j-{\displaystyle \frac{2}{5}})}^2{\displaystyle \frac{1}{2^j}}+{\displaystyle \frac{1}{3}}\sum _{j=3}^{\infty }{(a_j-{\displaystyle \frac{7}{10}})}^2{\displaystyle \frac{1}{2^j}}\hfill \\ \hfill & ={\displaystyle \frac{\mathrm{1,834,513}}{\mathrm{597,870,000}}}=0.00306841>V_n,\hfill \end{array}$$

which leads to a contradiction.

Case 3. $c_j\le {\displaystyle \frac{3}{10}}$ and ${\displaystyle \frac{7}{10}}\le c_{j+1}$.

Then, as ${\displaystyle \frac{1}{2}}({\displaystyle \frac{3}{10}}+{\displaystyle \frac{7}{10}})={\displaystyle \frac{1}{2}}$, we have

$$V_n\ge {\displaystyle \frac{1}{3}}{(a_1-{\displaystyle \frac{3}{10}})}^2{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{3}}\sum _{j=2}^{\infty }{(a_j-{\displaystyle \frac{7}{10}})}^2{\displaystyle \frac{1}{2^j}}={\displaystyle \frac{41}{6300}}=0.00650794>V_n,$$

which is a contradiction.

Thus, we see that $\alpha \cap C=\varnothing$ leads to a contradiction. Hence, we can assume that $\alpha \cap C\ne \varnothing$. Thus, the proof of the proposition is complete. □

Lemma 9.

Let α be an optimal set of five-means. Then, α does not contain any point from the open intervals $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$ and $({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$.

Proof. 

Let $\alpha :=\{c_1<c_2<c_3<c_4<c_5\}$ be an optimal set of five means. As shown in the proof of Proposition 8, we have $V_5\le 0.00205848$. First, prove the following claim.

$\mathrm{card}(\alpha \cap C)=2$.

Since $\alpha \cap J_1\ne \varnothing$ and $\alpha \cap J_2\ne \varnothing$, we have $\mathrm{card}(\alpha \cap C)\le 3$. If $\mathrm{card}(\alpha \cap C)=3$, then

$$V_5\ge {\int }_{J_1}{(x-S_1({\displaystyle \frac{19}{39}}))}^{2}dP+{\int }_{J_2}{(x-S_2({\displaystyle \frac{19}{39}}))}^{2}dP={\displaystyle \frac{2}{75}}V={\displaystyle \frac{\mathrm{173,392}}{\mathrm{58,292,325}}}=0.00297453>V_5,$$

which is a contradiction. So, we can assume that $\mathrm{card}(\alpha \cap C)\le 2$. Suppose that $\mathrm{card}(\alpha \cap C)=1$. Then, the following cases can arise:

Case 1. $c_4\in C$.

Then, $c_3<{\displaystyle \frac{2}{5}}$. Assume that ${\displaystyle \frac{3}{10}}<c_3<{\displaystyle \frac{2}{5}}$. Then, ${\displaystyle \frac{1}{2}}(c_2+c_3)<{\displaystyle \frac{1}{5}}$ implying $c_2<{\displaystyle \frac{1}{10}}$, and so

$$V_5\ge {\int }_{S_1(A_3)\cup J_{12}}{(x-{\displaystyle \frac{1}{10}})}^{2}dP+{\int }_{J_2}{(x-S_2({\displaystyle \frac{19}{39}}))}^{2}dP={\displaystyle \frac{\mathrm{25,816,591}}{\mathrm{11,658,465,000}}}=0.00221441>V_5,$$

which gives a contradiction. Next, assume that $c_3\le {\displaystyle \frac{3}{10}}$. Then, $a_1<{\displaystyle \frac{1}{2}}({\displaystyle \frac{3}{10}}+b_2)<a_2$, and so

$$\begin{array}{cc}\hfill V_5& \ge {\displaystyle \frac{1}{3}}{(a_1-{\displaystyle \frac{3}{10}})}^2{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{3}}\sum _{j=2}^{\infty }{(a_j-b_2)}^2{\displaystyle \frac{1}{2^j}}+{\int }_{J_2}{(x-S_2({\displaystyle \frac{19}{39}}))}^{2}dP\hfill \\ \hfill & ={\displaystyle \frac{\mathrm{104,633}}{\mathrm{31,089,240}}}=0.00336557>V_5,\hfill \end{array}$$

which is a contradiction. Thus, $c_4\in C$ leads to a contradiction.

Case 2. $c_3\in C$.

Then, $c_2<{\displaystyle \frac{2}{5}}$ and ${\displaystyle \frac{3}{5}}<c_4$. First, assume that ${\displaystyle \frac{1}{3}}<c_2<{\displaystyle \frac{2}{5}}$ and ${\displaystyle \frac{3}{5}}<c_4<{\displaystyle \frac{2}{3}}$. Then, ${\displaystyle \frac{1}{2}}(c_1+c_2)<{\displaystyle \frac{1}{5}}$ and ${\displaystyle \frac{1}{2}}(c_4+c_5)>{\displaystyle \frac{4}{5}}$ implying $c_1<{\displaystyle \frac{1}{15}}$ and ${\displaystyle \frac{14}{15}}<c_5$. Thus,

$$\begin{array}{cc}\hfill V_5& \ge {\int }_{S_1(A_3)\cup J_{12}}{(x-{\displaystyle \frac{1}{15}})}^{2}dP+{\int }_{J_{21}\cup S_1(\{a_1,a_2\})}{(x-{\displaystyle \frac{14}{15}})}^{2}dP={\displaystyle \frac{32531}{\mathrm{10,347,750}}}=0.00314378>V_5,\hfill \end{array}$$

which is a contradiction. Next, assume that ${\displaystyle \frac{3}{10}}\le c_2\le {\displaystyle \frac{1}{3}}$ and ${\displaystyle \frac{3}{5}}<c_4<{\displaystyle \frac{2}{3}}$. Then, ${\displaystyle \frac{1}{2}}(c_1+c_2)<{\displaystyle \frac{1}{5}}$ and ${\displaystyle \frac{1}{2}}(c_4+c_5)>{\displaystyle \frac{4}{5}}$ implying $c_1<{\displaystyle \frac{1}{10}}$ and ${\displaystyle \frac{14}{15}}<c_5$. Moreover, ${\displaystyle \frac{1}{2}}(c_2+c_3)\ge {\displaystyle \frac{2}{5}}$ yielding $c_3\ge {\displaystyle \frac{4}{5}}-c_2\ge {\displaystyle \frac{4}{5}}-{\displaystyle \frac{1}{3}}={\displaystyle \frac{7}{15}}$, and so,

$$\begin{array}{cc}\hfill V_5& \ge {\int }_{S_1(A_3)\cup J_{12}}{(x-{\displaystyle \frac{1}{10}})}^{2}dP+{\displaystyle \frac{1}{3}}{(a_1-{\displaystyle \frac{7}{15}})}^2{\displaystyle \frac{1}{2}}+{\int }_{J_{21}\cup S_1(\{a_1,a_2\})}{(x-{\displaystyle \frac{14}{15}})}^{2}dP\hfill \\ \hfill & ={\displaystyle \frac{\mathrm{3,354,097}}{\mathrm{1,076,166,000}}}=0.00311671>V_5,\hfill \end{array}$$

which leads to a contradiction. Now, assume that ${\displaystyle \frac{3}{10}}\le c_2\le {\displaystyle \frac{1}{3}}$ and ${\displaystyle \frac{2}{3}}\le c_4\le {\displaystyle \frac{7}{10}}$. Then, $c_1<{\displaystyle \frac{1}{10}}$ and $c_5>{\displaystyle \frac{9}{10}}$. Moreover, ${\displaystyle \frac{1}{2}}(c_2+c_3)\ge {\displaystyle \frac{2}{5}}$ and ${\displaystyle \frac{1}{2}}(c_3+c_4)\le {\displaystyle \frac{3}{5}}$ implying $a_1<{\displaystyle \frac{7}{15}}\le c_3\le {\displaystyle \frac{8}{15}}<a_3$. Thus, we have

$$\begin{array}{cc}\hfill V_5& \ge {\int }_{S_1(A_3)\cup J_{12}}{(x-{\displaystyle \frac{1}{10}})}^{2}dP+{\displaystyle \frac{1}{3}}\left({(a_1-{\displaystyle \frac{7}{15}})}^2{\displaystyle \frac{1}{2}}+\sum _{j=3}^{\infty }{(a_j-{\displaystyle \frac{8}{15}})}^2{\displaystyle \frac{1}{2^j}}\right)\hfill \\ \hfill & +{\int }_{J_{21}\cup S_1(\{a_1,a_2\})}{(x-{\displaystyle \frac{9}{10}})}^{2}dP={\displaystyle \frac{2593}{\mathrm{1,103,760}}}=0.00234924>V_5,\hfill \end{array}$$

which is a contradiction. Last, assume that $c_2\le {\displaystyle \frac{3}{10}}$ and ${\displaystyle \frac{7}{10}}\le c_4$. Then, as ${\displaystyle \frac{1}{2}}({\displaystyle \frac{3}{10}}+{\displaystyle \frac{7}{10}})={\displaystyle \frac{1}{2}}$, we have

$$V_5\ge {\displaystyle \frac{1}{3}}\left({(a_1-{\displaystyle \frac{3}{10}})}^2{\displaystyle \frac{1}{2}}+\sum _{j=3}^{\infty }{(a_j-{\displaystyle \frac{7}{10}})}^2{\displaystyle \frac{1}{2^j}}\right)={\displaystyle \frac{1}{315}}=0.0031746>V_5,$$

which gives a contradiction.

Case 3. $c_2\in C$.

This case is similar to Case 1, and similarly shows that a contradiction arises.

Thus, considering all the above possible cases, we can say that $\mathrm{card}(\alpha \cap C)=1$ leads to a contradiction. Hence, we can assume that $\mathrm{card}(\alpha \cap C)=2$, i.e., the claim is true.

We now show that $\alpha$ does not contain any point from $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})\cup ({\displaystyle \frac{2}{5}},{\displaystyle \frac{3}{5}})$. Recall that $\alpha \cap J_1\ne \varnothing$, $\alpha \cap J_2\ne \varnothing$, and $\mathrm{card}(\alpha \cap C)=2$. Suppose that $\alpha$ contains a point from $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$. Then, $\alpha$ contains $c_2$ from $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$. First, assume that ${\displaystyle \frac{3}{10}}<c_2<{\displaystyle \frac{2}{5}}$. Then, ${\displaystyle \frac{1}{2}}(c_1+c_2)<{\displaystyle \frac{1}{5}}$ implying $c_1<{\displaystyle \frac{1}{10}}$, and so

$$V_5\ge {\int }_{S_1(A_3)\cup J_{12}}{(x-{\displaystyle \frac{1}{10}})}^{2}dP+{\int }_{J_2}{(x-S_2({\displaystyle \frac{19}{39}}))}^{2}dP={\displaystyle \frac{\mathrm{25,816,591}}{\mathrm{11,658,465,000}}}=0.00221441>V_5,$$

which is a contradiction. Next, assume that $c_2\le {\displaystyle \frac{3}{10}}$. Then, ${\displaystyle \frac{1}{2}}(c_2+c_3)\ge {\displaystyle \frac{2}{5}}$ implying $c_3\ge {\displaystyle \frac{4}{5}}-c_2\ge {\displaystyle \frac{4}{5}}-{\displaystyle \frac{3}{10}}={\displaystyle \frac{1}{2}}=a_2$, and so

$$V_5\ge {\displaystyle \frac{1}{3}}{(a_1-{\displaystyle \frac{1}{2}})}^2{\displaystyle \frac{1}{2}}+{\int }_{J_2}{(x-S_2({\displaystyle \frac{19}{39}}))}^{2}dP={\displaystyle \frac{\mathrm{1,470,799}}{\mathrm{466,338,600}}}=0.00315393>V_5,$$

which is a contradiction. Hence, $\alpha$ cannot contain any point from $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$. Suppose that $\alpha$ contains a point from $({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$. Then, $\alpha$ contains $c_4$ from $({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$. First, assume that ${\displaystyle \frac{3}{5}}<c_4<{\displaystyle \frac{7}{10}}$. Then, ${\displaystyle \frac{1}{2}}(c_4+c_5)>{\displaystyle \frac{4}{5}}$ implying $c_5>{\displaystyle \frac{9}{10}}$, and so

$$\begin{array}{cc}\hfill V_5& \ge {\int }_{J_1}{(x-S_1({\displaystyle \frac{19}{39}}))}^{2}dP+{\int }_{J_{21}\cup S_1(\{a_1,a_2\})}{(x-{\displaystyle \frac{9}{10}})}^{2}dP\hfill \\ \hfill & ={\displaystyle \frac{\mathrm{26,226,559}}{\mathrm{11,658,465,000}}}=0.00224957>V_5,\hfill \end{array}$$

which gives a contradiction. Next, assume that ${\displaystyle \frac{7}{10}}\le c_4$. Then, ${\displaystyle \frac{1}{2}}(c_3+c_4)\le {\displaystyle \frac{3}{5}}$ implying $c_3\le {\displaystyle \frac{6}{5}}-c_4\le {\displaystyle \frac{6}{5}}-{\displaystyle \frac{7}{10}}={\displaystyle \frac{1}{2}}$, and so

$$\begin{array}{cc}\hfill V_5& \ge {\int }_{J_1}{(x-S_1({\displaystyle \frac{19}{39}}))}^{2}dP+{\displaystyle \frac{1}{3}}\sum _{j=3}^{\infty }{(a_j-{\displaystyle \frac{1}{2}})}^2{\displaystyle \frac{1}{2^j}}+{\int }_{J_2}\underset{c\in S_2({\alpha }_2)}{min}{(x-c)}^{2}dP\hfill \\ \hfill & ={\displaystyle \frac{1}{75}}V+{\displaystyle \frac{1}{3}}\sum _{j=3}^{\infty }{(a_j-{\displaystyle \frac{1}{2}})}^2{\displaystyle \frac{1}{2^j}}+{\displaystyle \frac{1}{75}}(V_{2,1}+V_{2,2})={\displaystyle \frac{\mathrm{130,788,689}}{\mathrm{58,292,325,000}}}=0.00224367>V_5,\hfill \end{array}$$

which is a contradiction. Hence, $\alpha$ cannot contain any point from $({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$. Thus, the proof of the lemma is complete. □

Lemma 10.

Let α be an optimal set of six-means. Then, α does not contain any point from the open intervals $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$ and $({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$.

Proof. 

Consider the set of six points $\beta :=S_1({\alpha }_2)\cup {\alpha }_2(\nu )\cup S_2({\alpha }_2)$. The distortion error due to set $\beta$ is given by

$$\int \underset{c\in \beta }{min}{(x-c)}^{2}dP={\displaystyle \frac{2}{75}}(V_{2,1}+V_{2,2})+{\displaystyle \frac{1}{3}}{V}_2(\nu )={\displaystyle \frac{\mathrm{13,564,657}}{\mathrm{14,573,081,250}}}=0.000930802.$$

Since $V_6$ is the quantization error for six-means, we have $V_6\le 0.000930802$. Let $\alpha :=\{c_1<c_2<\cdots <c_6\}$ be an optimal set of six-means.

First, prove the following claim.

$\mathrm{card}(\alpha \cap C)=2$.

Since $\alpha \cap J_1\ne \varnothing$ and $\alpha \cap J_2\ne \varnothing$, we have $\mathrm{card}(\alpha \cap C)\le 4$. If $\mathrm{card}(\alpha \cap C)=4$, then

$$V_6\ge {\int }_{J_1}{(x-S_1({\displaystyle \frac{19}{39}}))}^{2}dP+{\int }_{J_2}{(x-S_2({\displaystyle \frac{19}{39}}))}^{2}dP={\displaystyle \frac{2}{75}}V={\displaystyle \frac{\mathrm{173,392}}{\mathrm{58,292,325}}}=0.00297453>V_6,$$

which is a contradiction. Assume that $\mathrm{card}(\alpha \cap C)=3$. Recall that $\alpha \cap J_1\ne \varnothing$, and $\alpha \cap J_2\ne \varnothing$. First, assume that $\alpha$ does not contain any point from the open intervals $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$ and $({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$. Then, either $\alpha$ contains two points from $J_1$ and one point from $J_2$, or $\alpha$ contains one point from $J_1$ and two points from $J_2$. In any case, we have the distortion error as

$${\displaystyle \frac{1}{75}}(V_{2,1}+V_{2,2})+{\displaystyle \frac{1}{3}}{V}_3(\nu )+{\displaystyle \frac{1}{75}}V={\displaystyle \frac{\mathrm{109,198,939}}{\mathrm{58,292,325,000}}}=0.0018733>V_6,$$

which is a contradiction. Next, assume that $\alpha$ contains a point from the set $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})\cup ({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$. First assume that $\alpha$ contains the point from the open interval $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$. Then, $c_2\in ({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$. The following two cases can arise:

Case 1. ${\displaystyle \frac{3}{10}}\le c_2<{\displaystyle \frac{2}{5}}$.

Then, ${\displaystyle \frac{1}{2}}(c_1+c_2)<{\displaystyle \frac{1}{5}}$, implying $c_1<{\displaystyle \frac{1}{10}}$, and so

$$V_6\ge {\int }_{S_1(A_3)\cup J_{12}}{(x-{\displaystyle \frac{1}{10}})}^{2}dP+{\int }_{J_2}{(x-S_2({\displaystyle \frac{19}{39}}))}^{2}dP={\displaystyle \frac{\mathrm{25,816,591}}{\mathrm{11,658,465,000}}}=0.00221441>V_6,$$

which is a contradiction.

Case 2. ${\displaystyle \frac{1}{5}}<c_2\le {\displaystyle \frac{3}{10}}$.

Then, ${\displaystyle \frac{1}{2}}(c_2+c_3)>{\displaystyle \frac{2}{5}}$ implying $c_4>{\displaystyle \frac{4}{5}}-c_2\ge {\displaystyle \frac{4}{5}}-{\displaystyle \frac{3}{10}}={\displaystyle \frac{1}{2}}$, and so

$$V_6\ge {\displaystyle \frac{1}{3}}{(a_1-{\displaystyle \frac{1}{2}})}^2{\displaystyle \frac{1}{2}}+{\int }_{J_2}{(x-S_2({\displaystyle \frac{19}{39}}))}^{2}dP={\displaystyle \frac{\mathrm{1,470,799}}{\mathrm{466,338,600}}}=0.00315393>V_6,$$

which gives a contradiction.

Similarly, we can show that if $\alpha$ contains the point from the open interval $({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$, a contradiction arises. Hence, we can assume that $\mathrm{card}(\alpha \cap C)\le 2$. Suppose that $\mathrm{card}(\alpha \cap C)=1$. Then, the following cases can arise:

Case A. $\alpha$ does not contain any point from the set $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})\cup ({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$.

Then, $V_6\ge {\displaystyle \frac{1}{3}}{V}_1(\nu )={\displaystyle \frac{1}{3}}V(\nu )={\displaystyle \frac{8}{4725}}=0.00169312>V_6$, which leads to a contradiction.

Case B. $\alpha$ contains only one point from the set $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})\cup ({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$.

In this case, the following subcases can arise:

Subcase (i). $c_1,c_2,c_3\in J_1$.

Then, $c_4\in ({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$, $c_5\in C$, and $c_6\in J_2$; or $c_4\in C$, $c_5\in ({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$, and $c_6\in J_2$. First, assume that $c_4\in ({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$, $c_5\in C$, and $c_6\in J_2$. Suppose that ${\displaystyle \frac{3}{10}}<c_4<{\displaystyle \frac{2}{5}}$. Then, ${\displaystyle \frac{1}{2}}(c_3+c_4)<{\displaystyle \frac{1}{5}}$, implying $c_3<{\displaystyle \frac{2}{5}}-c_4\le {\displaystyle \frac{1}{10}}$, and hence,

$$V_6\ge {\int }_{S_1(A_3)\cup J_{12}}{(x-{\displaystyle \frac{1}{10}})}^{2}dP+{\int }_{J_2}{(x-S_2({\displaystyle \frac{19}{39}}))}^{2}dP={\displaystyle \frac{\mathrm{25,816,591}}{\mathrm{11,658,465,000}}}=0.00221441>V_6,$$

which is a contradiction. Suppose that ${\displaystyle \frac{1}{5}}<c_4\le {\displaystyle \frac{3}{10}}$. Then, ${\displaystyle \frac{1}{2}}(c_4+c_5)\ge {\displaystyle \frac{2}{5}}$, implying $c_5\ge {\displaystyle \frac{1}{2}}$, and so

$$V_6\ge {\displaystyle \frac{1}{3}}{(a_1-{\displaystyle \frac{1}{2}})}^2{\displaystyle \frac{1}{2}}+{\int }_{J_2}{(x-S_2({\displaystyle \frac{19}{39}}))}^{2}dP={\displaystyle \frac{\mathrm{1,470,799}}{\mathrm{466,338,600}}}=0.00315393>V_6,$$

which yields a contradiction. Similarly, we can show that if $c_4\in C$, $c_5\in ({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$, and $c_6\in J_2$, a contradiction arises.

Subcase (ii). $c_1,c_2,\in J_1$.

Then, $c_3\in ({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$, $c_4\in C$, $c_5,c_6\in J_2$; or $c_3\in C$, $c_4\in ({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$, and $c_5,c_6\in J_2$. First, assume that $c_3\in ({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$, $c_4\in C$, $c_5,c_6\in J_2$. Suppose that ${\displaystyle \frac{3}{10}}<c_3<{\displaystyle \frac{2}{5}}$. Then, ${\displaystyle \frac{1}{2}}(c_2+c_3)<{\displaystyle \frac{1}{5}}$, implying $c_2<{\displaystyle \frac{2}{5}}-c_3\le {\displaystyle \frac{1}{10}}$, and hence,

$$V_6\ge {\int }_{S_1(A_3)\cup J_{12}}{(x-{\displaystyle \frac{1}{10}})}^{2}dP+{\int }_{J_2}\underset{c\in S_2({\alpha }_2)}{min}{(x-c)}^{2}dP={\displaystyle \frac{\mathrm{7,038,641}}{\mathrm{6,476,925,000}}}=0.00108673>V_6,$$

which is a contradiction. Suppose that ${\displaystyle \frac{1}{5}}<c_3\le {\displaystyle \frac{3}{10}}$. Then, ${\displaystyle \frac{1}{2}}(c_3+c_4)\ge {\displaystyle \frac{2}{5}}$ implying $c_4\ge {\displaystyle \frac{1}{2}}$, and so

$$V_6\ge {\displaystyle \frac{1}{3}}{(a_1-{\displaystyle \frac{1}{2}})}^2{\displaystyle \frac{1}{2}}+{\int }_{J_2}\underset{c\in S_2({\alpha }_2)}{min}{(x-c)}^{2}dP={\displaystyle \frac{\mathrm{5,624,509}}{\mathrm{2,775,825,000}}}=0.00202625>V_6,$$

which yields a contradiction. Similarly, we can show that if $c_3\in C$, $c_4\in ({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$, and $c_5,c_6\in J_2$, a contradiction arises.

Subcase (iii). $c_1\in J_1$.

This case is the reflection of Subcase (i) about the point ${\displaystyle \frac{1}{2}}$, and thus a contradiction also arises in this case.

Case C. $\alpha$ contains two points from the set $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})\cup ({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$.

By Proposition 4, $\alpha$ contains one point from $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$ and one point from $({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$. Without any loss of generality, we can assume that $c_1,c_2\in J_1$, $c_3\in ({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$, $c_4\in C$, $c_5\in ({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$, and $c_6\in J_6$. Then, we can also show that it leads to a contradiction.

Thus, we can assume that $\mathrm{card}(\alpha \cap C)=2$, which is the claim.

We now show that $\alpha$ does not contain any point from $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})\cup ({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$. For the sake of contradiction, first assume that $\alpha$ contains a point from $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$, and does not contain any point from $({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$. Suppose that $c_1,c_2\in J_1$, $c_3\in ({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$, $c_4,c_5\in C$, and $c_6\in J_2$, and then

$$V_6\ge {\int }_{J_2}{(x-S_2({\displaystyle \frac{19}{39}}))}^{2}dP={\displaystyle \frac{\mathrm{86,696}}{\mathrm{58,292,325}}}=0.00148726>V_6,$$

which is a contradiction. Suppose that $c_1\in J_1$, $c_2\in ({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$, $c_3,c_4\in C$, and $c_5,c_6\in J_2$. Then, either ${\displaystyle \frac{3}{10}}\le c_2<{\displaystyle \frac{2}{5}}$, or ${\displaystyle \frac{1}{5}}<c_2\le {\displaystyle \frac{3}{10}}$. First, assume that ${\displaystyle \frac{3}{10}}\le c_2<{\displaystyle \frac{2}{5}}$. Then, ${\displaystyle \frac{1}{2}}(c_1+c_2)<{\displaystyle \frac{1}{5}}$ implies $c_1\le {\displaystyle \frac{1}{10}}$, and so

$$V_6\ge {\int }_{S_1(A_3)\cup J_{12}}{(x-{\displaystyle \frac{1}{10}})}^{2}dP+{\int }_{J_2}\underset{c\in S_2({\alpha }_2)}{min}{(x-c)}^{2}dP={\displaystyle \frac{\mathrm{7,038,641}}{\mathrm{6,476,925,000}}}=0.00108673>V_6,$$

which leads to a contradiction. Next, assume that ${\displaystyle \frac{1}{5}}<c_2\le {\displaystyle \frac{3}{10}}$. Then, ${\displaystyle \frac{1}{2}}(c_2+c_3)>{\displaystyle \frac{2}{5}}$, implying $c_3>{\displaystyle \frac{4}{5}}-c_2\ge {\displaystyle \frac{4}{5}}-{\displaystyle \frac{3}{10}}={\displaystyle \frac{1}{2}}>a_1$, and so

$$V_6\ge {\displaystyle \frac{1}{3}}{(a_1-{\displaystyle \frac{1}{3}})}^2{\displaystyle \frac{1}{2}}+{\int }_{J_2}\underset{c\in S_2({\alpha }_2)}{min}{(x-c)}^{2}dP={\displaystyle \frac{\mathrm{5,624,509}}{\mathrm{2,775,825,000}}}=0.00202625>V_6,$$

which is a contradiction. Likewise, we can show that if $\alpha$ contains a point from $({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$, and does not contain any point from $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$, a contradiction arises.

Next, assume that $\alpha$ contains a point from $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$ and a point from $({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$. Then, we have $c_1\in J_1$, $c_2\in ({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$, $c_3,c_4\in C$, $c_5\in ({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$, and $c_6\in J_2$. Then, the following cases can arise:

Case I. ${\displaystyle \frac{3}{10}}\le c_2<{\displaystyle \frac{2}{5}}$ and ${\displaystyle \frac{3}{5}}<c_5\le {\displaystyle \frac{7}{10}}$.

Then, ${\displaystyle \frac{1}{2}}(c_1+c_2)<{\displaystyle \frac{1}{5}}$ and ${\displaystyle \frac{1}{2}}(c_5+c_6)>{\displaystyle \frac{4}{5}}$, implying $c_1<{\displaystyle \frac{1}{10}}$ and $c_6>{\displaystyle \frac{9}{10}}$. Thus,

$$V_6\ge {\int }_{S_1(A_3)\cup J_{12}}{(x-{\displaystyle \frac{1}{10}})}^{2}dP+{\int }_{J_{21}\cup S_2(\{a_1,a_2\})}{(x-{\displaystyle \frac{9}{10}})}^{2}dP={\displaystyle \frac{137}{\mathrm{91,980}}}=0.00148945>V_6,$$

which leads to a contradiction.

Case II. ${\displaystyle \frac{1}{5}}<c_2\le {\displaystyle \frac{3}{10}}$ and ${\displaystyle \frac{3}{5}}<c_5\le {\displaystyle \frac{7}{10}}$.

Then, ${\displaystyle \frac{1}{2}}(c_2+c_3)>{\displaystyle \frac{2}{5}}$, implying $c_3>{\displaystyle \frac{4}{5}}-c_2\ge {\displaystyle \frac{4}{5}}-{\displaystyle \frac{3}{10}}={\displaystyle \frac{1}{2}}>a_1$, and $c_6>{\displaystyle \frac{9}{10}}$. Thus,

$$V_6\ge {\displaystyle \frac{1}{3}}{(a_1-{\displaystyle \frac{1}{3}})}^2{\displaystyle \frac{1}{2}}+{\int }_{J_{21}\cup S_2(\{a_1,a_2\})}{(x-{\displaystyle \frac{9}{10}})}^{2}dP={\displaystyle \frac{\mathrm{363,053}}{\mathrm{149,467,500}}}=0.00242898>V_6,$$

which is a contradiction.

Case III. ${\displaystyle \frac{1}{5}}<c_2\le {\displaystyle \frac{3}{10}}$ and ${\displaystyle \frac{7}{10}}\le c_5<{\displaystyle \frac{4}{5}}$.

Then, ${\displaystyle \frac{1}{2}}(c_2+c_3)\ge {\displaystyle \frac{2}{5}}$ and ${\displaystyle \frac{1}{2}}(c_4+c_5)\le {\displaystyle \frac{3}{5}}$, implying $c_3\ge {\displaystyle \frac{1}{2}}$ and $c_4\le {\displaystyle \frac{1}{2}}$, which leads to a contradiction because $c_3<c_4$.

Thus, we can conclude that $\alpha$ does not contain any point from the open intervals $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$ and $({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$, which is the lemma. □

Proposition 9.

Let ${\alpha }_n$ be an optimal set of n-means for all $n\ge 3$. Then, ${\alpha }_n$ does not contain any point from the open intervals $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$ and $({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$.

Proof. 

By Proposition 6 and Lemmas 8–10, the proposition is true for $n=3,4,5,6$. Proceeding in the similar way as Lemma 10, we can show that the proposition is true for $n=7$. We now show that the proposition is true for all $n\ge 8$. Consider the set of eight points $\beta :=S_1({\alpha }_3)\cup {\alpha }_2(\nu )\cup S_2({\alpha }_3)$. The distortion error due to set $\beta$ is given by

$$\begin{array}{cc}\hfill & \int \underset{c\in \beta }{min}{(x-c)}^{2}dP={\displaystyle \frac{1}{{75}^2}}V+{\displaystyle \frac{1}{3}}{\displaystyle \frac{1}{75}}V(\nu )+{\displaystyle \frac{1}{{75}^2}}V+{\displaystyle \frac{1}{3}}{V}_2(\nu )+{\displaystyle \frac{1}{{75}^2}}V+{\displaystyle \frac{1}{3}}{\displaystyle \frac{1}{75}}V(\nu )+{\displaystyle \frac{1}{{75}^2}}V\hfill \\ \hfill & ={\displaystyle \frac{\mathrm{489,817}}{\mathrm{1,457,308,125}}}=0.000336111.\hfill \end{array}$$

Since $V_n$ is the quantization error for n-means with $n\ge 8$, we have $V_n\le V_8\le 0.000633457$. Let $\alpha :=\{c_1<c_2<\cdots <c_n\}$ be an optimal set of n-means for P. Suppose that $\alpha$ contains a point from the open interval $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$. Let $j:=max\{i:c_i<{\displaystyle \frac{2}{5}}\}$. Then, ${\displaystyle \frac{1}{5}}<c_j<{\displaystyle \frac{2}{5}}$. The following cases can arise:

Case 1. ${\displaystyle \frac{3}{10}}<c_j<{\displaystyle \frac{2}{5}}$.

Then, ${\displaystyle \frac{1}{2}}(c_{j-1}+c_j)<{\displaystyle \frac{1}{5}}$, implying $c_{j-1}<{\displaystyle \frac{2}{5}}-{\displaystyle \frac{3}{10}}={\displaystyle \frac{1}{10}}$, and so

$$V_n\ge {\int }_{S_1(A_3)\cup J_{12}}{(x-{\displaystyle \frac{1}{10}})}^{2}dP={\displaystyle \frac{\mathrm{217,369}}{\mathrm{298,935,000}}}=0.000727145>V_n,$$

which leads to a contradiction.

Case 2. ${\displaystyle \frac{1}{5}}\le c_j\le {\displaystyle \frac{3}{10}}$.

Then, ${\displaystyle \frac{1}{2}}(c_j+c_{j+1})>{\displaystyle \frac{2}{5}}$, implying $c_{j+1}>{\displaystyle \frac{4}{5}}-c_j\ge {\displaystyle \frac{4}{5}}-{\displaystyle \frac{3}{10}}={\displaystyle \frac{1}{2}}$, and so

$$V_n\ge {\displaystyle \frac{1}{3}}{(a_1-{\displaystyle \frac{1}{2}})}^2{\displaystyle \frac{1}{2}}={\displaystyle \frac{1}{600}}=0.00166667>V_n,$$

which gives a contradiction.

Hence, we can say that $\alpha$ does not contain any point from $({\displaystyle \frac{2}{5}},{\displaystyle \frac{3}{5}})$. Suppose that $\alpha$ contains a point from $({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$. Let $j:=min\{i:{\displaystyle \frac{3}{5}}<c_i\}$. Then, ${\displaystyle \frac{3}{5}}<c_j<{\displaystyle \frac{4}{5}}$. The following cases can arise:

Case I. ${\displaystyle \frac{3}{5}}<c_j<{\displaystyle \frac{7}{10}}$.

Then, ${\displaystyle \frac{1}{2}}(c_j+c_{j+1})>{\displaystyle \frac{4}{5}}$, implying $c_{j+1}>{\displaystyle \frac{8}{5}}-{\displaystyle \frac{7}{10}}={\displaystyle \frac{9}{10}}$, and so

$$V_n\ge {\int }_{J_{21}\cup S_2(\{a_1,a_2\})}{(x-{\displaystyle \frac{9}{10}})}^{2}dP={\displaystyle \frac{\mathrm{227,881}}{\mathrm{298,935,000}}}=0.00076231>V_n,$$

which leads to a contradiction.

Case II. ${\displaystyle \frac{7}{10}}\le c_j<{\displaystyle \frac{4}{5}}$.

Then, ${\displaystyle \frac{1}{2}}(c_{j-1}+c_j)<{\displaystyle \frac{3}{5}}$, implying $c_{j-1}<{\displaystyle \frac{6}{5}}-c_j\le {\displaystyle \frac{6}{5}}-{\displaystyle \frac{7}{10}}={\displaystyle \frac{1}{2}}<a_2$, and so

$$V_n\ge {\displaystyle \frac{1}{3}}\sum _{j=2}^{\infty }{(a_j-{\displaystyle \frac{1}{2}})}^2{\displaystyle \frac{1}{2^j}}={\displaystyle \frac{1}{2520}}=0.000396825>V_n,$$

which is a contradiction.

Hence, $\alpha$ does not contain any point from the open interval $({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$. This completes the proof of the proposition. □

3.2. Optimal Sets and the Quantization Error for a Given Sequence $F(n)$

In this subsection, we first define the two sequences ${\left\{a(n)\right\}}_{n\ge 1}$ and ${\left\{F(n)\right\}}_{n\ge 1}$. These two sequences play important roles in this section.

Definition 1.

Define the sequence ${\left\{a(n)\right\}}_{n\ge 1}$ such that $a(n)=2n$ for all $n\ge 1$. Define the sequence ${\left\{F(n)\right\}}_{n\ge 1}$ such that $F(n)=5·2^n-2n-4$, i.e.,

$${\left\{F(n)\right\}}_{n\ge 1}=\{4,12,30,68,146,304,622,1260,2538,5096,10,214,20,452,40,930,\cdots \}.$$

Note 2.

Let ${\left\{a(n)\right\}}_{n\ge 1}$ and ${\left\{F(n)\right\}}_{n\ge 1}$ be the sequences defined by Definition 1. Then, notice that $F(n+1)=a(n+1)+2F(n)$. For $n=1$, we set ${\alpha }_{F(1)}:={\alpha }_{a(1)}(\nu )\cup \{S_1({\displaystyle \frac{19}{39}}),S_2({\displaystyle \frac{19}{39}})\}$, and for $n\ge 2$, we set ${\alpha }_{F(n)}:={\alpha }_{a(n)}(\nu )\cup S_1({\alpha }_{F(n-1)})\cup S_2({\alpha }_{F(n-1)}).$ Then, inductively, we have

$$\begin{array}{cc}\hfill {\alpha }_{F(n)}& ={\alpha }_{a(n)}\cup \left({\cup }_{\omega \in I}{S}_{\omega }({\alpha }_{a(n-1)}(\nu ))\right)\cup \left({\cup }_{\omega \in I^2}{S}_{\omega }({\alpha }_{a(n-2)}(\nu ))\right)\hfill \\ \hfill & \cdots \cup \left({\cup }_{\omega \in I^{n-1}}{S}_{\omega }({\alpha }_{a(1)}(\nu ))\right)\cup \{S_{\omega }({\displaystyle \frac{19}{39}}):\omega \in I^n\}.\hfill \end{array}$$

(4)

Notice that the set ${\alpha }_{F(n)}$ satisfies ${\alpha }_{F(n+1)}=S_1({\alpha }_{F(n)})\cup {\alpha }_{a(n+1)}(\nu )\cup {\alpha }_{F(n)}$. We will show that ${\alpha }_{F(n)}$ is an optimal set of $F(n)$-means for P. For $n\in \mathbb{N}$, we identify the sequence of sets ${\alpha }_{a(n)}(\nu ),{\cup }_{\omega \in I}{S}_{\omega }({\alpha }_{a(n-1)}(\nu )),{\cup }_{\omega \in I^2}{S}_{\omega }({\alpha }_{a(n-2)}(\nu )),\cdots ,{\cup }_{\omega \in I^{n-2}}{S}_{\omega }({\alpha }_{a(2)}(\nu )),$${\cup }_{\omega \in I^{n-1}}{S}_{\omega }({\alpha }_{a(1)}(\nu ))$, and $\{S_{\omega }({\displaystyle \frac{19}{39}}):\omega \in I^n\}$, respectively, by $S(n)$, $S(n-1)$, $S(n-2)$, ⋯, $S(2)$, $S(1)$, and $S(n+1)$. For $1\le \ell \le n$, and we write

$$\begin{array}{cc}\hfill S^{(2)}(\ell )& :={\cup }_{\omega \in I^{n-\ell }}{S}_{\omega }({\alpha }_{a(\ell )+1}(\nu ))\mathit{and}\hfill \\ \hfill S^{(2)(2)}(\ell )& :={\cup }_{\omega \in I^{n-\ell }}{S}_{\omega }({\alpha }_{a(\ell )+2}(\nu ))={\cup }_{\omega \in I^{n-\ell }}{S}_{\omega }({\alpha }_{a(\ell +1)}(\nu )).\hfill \end{array}$$

Further, we write

$$\begin{array}{cc}\hfill S^{(2)}(n+1)& :={\cup }_{\omega \in I^n}{S}_{\omega }({\alpha }_2(P))={\cup }_{\omega \in I^n}\{S_{\omega }({\displaystyle \frac{659}{2730}}),S_{\omega }({\displaystyle \frac{1621}{1950}})\},\hfill \\ \hfill S^{(2)(2)}(n+1)& :={\cup }_{\omega \in I^n}{S}_{\omega }({\alpha }_1(\nu ))\cup \{S_{\omega }({\displaystyle \frac{19}{39}}):\omega \in I^{n+1}\},\mathit{and}\hfill \\ \hfill S^{(2)(2)(2)}(n+1)& :={\cup }_{\omega \in I^n}{S}_{\omega }({\alpha }_{a(1)}(\nu ))\cup \{S_{\omega }({\displaystyle \frac{19}{39}}):\omega \in I^{n+1}\}.\hfill \end{array}$$

Moreover, for any $\ell \in \mathbb{N}$, if $A:=S(\ell )$, we identify $S^{(2)}(\ell )$, $S^{(2)(2)}(\ell )$, and $S^{(2)(2)(2)}(\ell )$, respectively, by $A^{(2)}$, $A^{(2)(2)}$, and $A^{(2)(2)(2)}$. For $n\in \mathbb{N}$, set

$$SF(n):=\{S(n),S(n-1),S(n-2),\cdots ,S(1),S(n+1)\}.$$

In addition, write

$$\begin{array}{cc}\hfill S{F}^{*}(n):=& \{S(n),S(n-1),S(n-2),\cdots ,S(1),S(n+1),S^{(2)}(n),S^{(2)}(n-1),S^{2)}(n-2),\cdots ,\hfill \\ \hfill & S^{(2)}(1),S^{(2)}(n+1),S^{(2)(2)}(n),S^{(2)(2)}(n-1),\cdots ,S^{(2)(2)}(1),S^{(2)(2)}(n+1),\hfill \\ \hfill & S^{(2)(2)(2)}(n+1)\}.\hfill \end{array}$$

For any element $a\in A\in S{F}^{*}(n)$, by the Voronoi region of a it is meant the Voronoi region of a with respect to the set ${\cup }_{B\in S{F}^{*}(n)}B$. Similarly, for any $a\in A\in SF(n)$, by the Voronoi region of a it is meant the Voronoi region of a with respect to the set ${\cup }_{B\in SF(n)}B$. Notice that if $a,b\in A$, where $A\in SF(n)$ or $A\in S{F}^{*}(n)$ except the sets $S^{(2)}(n+1)$, $S^{(2)(2)}(n+1)$, and $S^{(2)(2)(2)}(n+1)$, the error contributed by a in the Voronoi region of a equals to the error contributed by b in the Voronoi region of b. On the other hand, each of $S^{(2)}(n+1)$, $S^{(2)(2)}(n+1)$, and $S^{(2)(2)(2)}(n+1)$ contains two different sets: the error in the Voronoi region of an element in one set is larger than the error in the Voronoi region of an element in the other set. Let us now define an order ≻ on the set $S{F}^{*}(n)$ as follows: for $A,B\in S{F}^{*}(n)$ by $A\succ B$, it is meant that the error contributed by any element $a\in A$ in the Voronoi region of a is greater or equal to the error contributed by any element $b\in B$ in the Voronoi region of b. Similarly, we define the order relation ≻ on the set $SF(n)$.

Remark 2.

Let ${\alpha }_{F(n)}$ be the set defined by (4). Then,

$${\alpha }_{F(n+1)}=S^{(2)(2)}(1)\cup S^{(2)(2)}(2)\cup S^{(2)(2)}(3)\cup \cdots \cup S^{(2)(2)}(n)\cup S^{(2)(2)(2)}(n+1).$$

Remark 3.

Take any $1\le \ell <n$. The distortion errors due to any element in the sets $S^{(2)(2)}(\ell +1)$ and $S^{(2)(2)}(\ell )$ are, respectively, ${\displaystyle \frac{1}{3}}{\displaystyle \frac{1}{{75}^{n-(\ell +1)}}}{V}_{a(\ell +2)}(\nu )={\displaystyle \frac{1}{3}}{\displaystyle \frac{1}{{75}^{n-(\ell +1)}}}{\displaystyle \frac{2^{-6-6\ell }}{1575}}$ and ${\displaystyle \frac{1}{3}}{\displaystyle \frac{1}{{75}^{n-\ell }}}{V}_{a(\ell +1)}(\nu )={\displaystyle \frac{1}{3}}{\displaystyle \frac{1}{{75}^{n-\ell }}}{\displaystyle \frac{2^{-6\ell }}{1575}}$. Notice that

$${\displaystyle \frac{1}{3}}{\displaystyle \frac{1}{{75}^{n-(\ell +1)}}}{\displaystyle \frac{2^{-6-6\ell }}{1575}}>{\displaystyle \frac{1}{3}}{\displaystyle \frac{1}{{75}^{n-\ell }}}{\displaystyle \frac{2^{-6\ell }}{1575}},$$

and so, $S^{(2)(2)}(n)\succ S^{(2)(2)}(n-1)\succ S^{(2)(2)}(n-2)\succ \cdots \succ S^{(2)(2)}(2)\succ S^{(2)(2)}(1)$. Similarly, $S^{(2)}(n)\succ S^{(2)}(n-1)\succ S^{(2)}(n-2)\succ \cdots \succ S^{(2)}(2)\succ S^{(2)}(1)$, and $S(n)\succ S(n-1)\succ S(n-2)\succ \cdots \succ S(2)\succ S(1)$.

Lemma 11.

Let n be a large positive integer, in fact, $n\ge 40$. Then, the following inequalities are true:

$(i)$ If $2<n-k<n-k+26\le n$, then $S(n-k)\succ S^{(2)}(n-k+13)\succ S^{(2)(2)}(n-k+26)\succ S(n-k-1)$,

$(ii)$$S(14)\succ S^{(2)}(27)\succ S^{(2)(2)}(40)\succ S(n+1)\succ S(13)$.

$(iii)$$S(5)\succ S^{(2)}(18)\succ S^{(2)(2)}(31)\succ S^{(2)}(n+1)\succ S(4)$.

$(iv)$$S^{(2)(2)}(13)\succ S^{(2)(2)}(n+1)\succ S^{(2)(2)(2)}(n+1)\succ S^{(2)(2)}(12)$.

Proof. 

Let $2<n-k<n-k+26\le n$. Then, $S(n-k)\succ S^{(2)}(n-k+13)\succ S^{(2)(2)}(n-k+26)\succ S(n-k-1)$ will be true if

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{3}}{\displaystyle \frac{1}{{75}^k}}{V}_{a(n-k)}(\nu )& >{\displaystyle \frac{1}{3}}{\displaystyle \frac{1}{{75}^{k-13}}}{V}_{a(n-k+13)+1}(\nu )>{\displaystyle \frac{1}{3}}{\displaystyle \frac{1}{{75}^{k-26}}}{V}_{a(n-k+26)+2}(\nu )\hfill \\ \hfill & >{\displaystyle \frac{1}{3}}{\displaystyle \frac{1}{{75}^{k+1}}}{V}_{a(n-k-1)}(\nu ),\hfill \end{array}$$

i.e., if

$$\begin{array}{c}\hfill V_{a(n-k)}(\nu )>{75}^{13}{V}_{a(n-k+13)+1}(\nu )>{75}^{26}{V}_{a(n-k+26)+2}(\nu )>{\displaystyle \frac{1}{75}}{V}_{a(n-k-1)}(\nu ).\end{array}$$

(5)

By putting the values of $V_{a(n-k)}(\nu )$, $V_{a(n-k+13)+1}(\nu )$, $V_{a(n-k+26)+2}(\nu )$, and $V_{a(n-k-1)}(\nu )$, we see that the inequalities in (5) are clearly true. Thus, the proof of $(i)$ is complete.

By $(i)$, we know that $S(14)\succ S^{(2)}(27)\succ S^{(2)(2)}(40)$. Thus, to prove $(ii)$ it is enough to prove that $S^{(2)(2)}(40)\succ S(n+1)\succ S(13)$, which is true if

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{3}}{\displaystyle \frac{1}{{75}^{n-40}}}{V}_{a(40)+2}(\nu )>{\displaystyle \frac{1}{{75}^n}}V>{\displaystyle \frac{1}{3}}{\displaystyle \frac{1}{{75}^{n-13}}}{V}_{a(13)}(\nu ).\end{array}$$

(6)

By putting the values of $V_{a(40)+2}(\nu )$, V, and $V_{a(13)}(\nu )$, we see that the inequalities in (6) are clearly true. Thus, the proof of $(ii)$ is complete. Similarly, we can prove the inequalities in $(iii)$ and $(iv)$. Thus, the lemma is yielded. □

Remark 4.

By Remark 3 and Lemma 11, for any large positive integer n, in fact, if $n\ge 40$, it follows that

$$\begin{array}{cc}& S(n)\succ S(n-1)\succ \cdots \succ S(n-13)\succ S^{(2)}(n)\succ S(n-14)\succ S^{(2)}(n-1)\hfill \\ \hfill & \succ S(n-15)\succ S^{(2)}(n-2)\succ S(n-16)\succ \cdots \hfill \\ \hfill & \succ S(n-26)\succ S^{(2)}(n-13)\succ S^{(2)(2)}(n)\succ S(n-27)\succ \cdots \hfill \\ \hfill & S(14)\succ S^{(2)}(27)\succ S^{(2)(2)}(40)\succ S(n+1)\succ S(13)\succ S^{(2)}(26)\succ S^{(2)(2)}(39)\succ S(12)\succ \cdots \hfill \\ \hfill & S(5)\succ S^{(2)}(18)\succ S^{(2)(2)}(31)\succ S^{(2)}(n+1)\succ S(4)\succ S^{(2)}(17)\succ S^{(2)(2)}(30)\succ S(3)\succ \cdots \hfill \\ \hfill & S(1)\succ S^{(2)}(14)\succ S^{(2)(2)}(27)\succ S^{(2)}(13)\succ S^{(2)(2)}(26)\succ S^{(2)}(12)\succ S^{(2)(2)}(25)\succ \cdots \hfill \\ \hfill & S^{(2)}(1)\succ S^{(2)(2)}(14)\succ S^{(2)(2)}(13)\succ S^{(2)(2)}(n+1)\succ S^{(2)(2)(2)}(n+1)\succ S^{(2)(2)}(12)\hfill \\ \hfill & \succ S^{(2)(2)}(11)\succ S^{(2)(2)}(10)\succ \cdots S^{(2)(2)}(1).\hfill \end{array}$$

Lemma 12.

For any two sets $A,B\in SF(n)$, let $A\succ B$. Then, the distortion error due to the set $(SF(n)\setminus A)\cup A^{(2)}\cup B$ is less than the distortion error due to the set $(SF(n)\setminus B)\cup B^{(2)}\cup A$.

Proof. 

Let $V({\alpha }_{F(n)})$ be the distortion error due to the set ${\alpha }_{F(n)}$ with respect to the condensation measure P. First, take $A=S(k)$ and $B=S(k^{\prime })$ for some $1\le k,k^{\prime }\le n$. Then, $A\succ B$ implies $k>k^{\prime }$. The distortion error due to the set $({\alpha }_{F(n)}\setminus A)\cup A^{(2)}\cup B$ is given by

$$\begin{array}{cc}\hfill & V({\alpha }_{F(n)})-{\displaystyle \frac{1}{3}}{({\displaystyle \frac{2}{75}})}^{n-k}{V}_{a(k)}(\nu )+{\displaystyle \frac{1}{3}}{({\displaystyle \frac{2}{75}})}^{n-k}{V}_{a(k)+1}(\nu )+{\displaystyle \frac{1}{3}}{({\displaystyle \frac{2}{75}})}^{n-k^{\prime }}{V}_{a(k^{\prime })}(\nu )\hfill \\ \hfill & =V({\alpha }_{F(n)})-{\displaystyle \frac{56}{3}}{({\displaystyle \frac{2}{75}})}^{n-k}{\displaystyle \frac{2^{-6k}}{1575}}+{\displaystyle \frac{64}{3}}{({\displaystyle \frac{2}{75}})}^{n-k^{\prime }}{\displaystyle \frac{2^{-6k^{\prime }}}{1575}}.\hfill \end{array}$$

(7)

Similarly, the distortion error due to the set $({\alpha }_{F(n)}\setminus B)\cup B^{(2)}\cup A$ is

$$\begin{array}{c}\hfill V({\alpha }_{F(n)})-{\displaystyle \frac{56}{3}}{({\displaystyle \frac{2}{75}})}^{n-k^{\prime }}{\displaystyle \frac{2^{-6k^{\prime }}}{1575}}+{\displaystyle \frac{64}{3}}{({\displaystyle \frac{2}{75}})}^{n-k}{\displaystyle \frac{2^{-6k}}{1575}}.\end{array}$$

(8)

Thus, (7) will be less than (8) if ${\left({\displaystyle \frac{75}{128}}\right)}^{k^{\prime }}<{\left({\displaystyle \frac{75}{128}}\right)}^k$, which is clearly true since $k>k^{\prime }$. As mentioned in Remark 4, assume that $S(14)\succ S(n+1)\succ S(13)$. Take $A=S(14)$, and $B=S(13)$. Here, $A\succ B$. Proceeding as before, it can also be seen that the distortion error due to the set $({\alpha }_{F(n)}\setminus A)\cup A^{(2)}\cup B$ is less than the distortion error due to the set $({\alpha }_{F(n)}\setminus B)\cup B^{(2)}\cup A$. Similarly, we can show it for any two sets $A,B\in SF(n)$ with $A\succ B$. Thus, the lemma is yielded. □

Let us now state the following lemma. The proof is similar to the proof of Lemma 12.

Lemma 13.

For any two sets $A,B\in S{F}^{*}(n)$, let $A\succ B$. Then, the distortion error due to the set $(S{F}^{*}(n)\setminus A)\cup A^{(2)}\cup B$ is less than the distortion error due to the set $(S{F}^{*}(n)\setminus B)\cup B^{(2)}\cup A$.

Proposition 10.

For $n\ge 1$, let ${\alpha }_{F(n)}$ be the set defined by Note 2. Then, ${\alpha }_{F(n)}$ forms an optimal set of $F(n)$-means for P and the corresponding quantization error is given by

$$V_{F(n)}={\displaystyle \frac{\mathrm{769,208}}{\mathrm{5,884,749}}}{({\displaystyle \frac{2}{75}})}^n-{\displaystyle \frac{64}{3339}}{({\displaystyle \frac{1}{64}})}^n$$

Proof. 

We see that ${\alpha }_{F(1)}={\alpha }_{a(1)}(\nu )\cup \{S_{\omega }({\displaystyle \frac{19}{39}}):\omega \in I\}$, which by Proposition 7 is an optimal set of four-means for P. Thus, the proposition is true for $n=1$. Let ${\alpha }_{F(n)}$ be an optimal set of $F(n)$-means for some $n\ge 1$. We now show that ${\alpha }_{F(n+1)}$ is an optimal set of $F(n+1)$-means. We have ${\alpha }_{F(n)}={\cup }_{A\in SF(n)}A$. Recall that, by Proposition 9, an optimal set of n-means for any $n\ge 3$ does not contain any point from the open intervals $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$ and $({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$. In the first step, let $A(1)\in SF(n)$ be such that $A(1)\succ B$ for any other $B\in SF(n)$. Then, by Lemma 12, the set $({\alpha }_{F(n)}\setminus A(1))\cup A^{(2)}(1)$ gives an optimal set of $F(n)-\mathrm{card}(A(1))+\mathrm{card}(A^{(2)}(1))$-means. In the second step, let $A(2)\in (SF(n)\setminus \left\{A(1)\right\})\cup \left\{A^{(2)}(1)\right\}$ be such that $A(2)\succ B$ for any other set $B\in (SF(n)\setminus \left\{A(1)\right\})\cup \left\{A^{(2)}(1)\right\}$. Then, using the similar technique as Lemma 12, we can show that the distortion error is due to the following set:

$$\left((({\alpha }_{F(n)}\setminus A(1))\cup A^{(2)}(1))\setminus A(2)\right)\cup A^{(2)}(2)$$

(9)

where cardinality $F(n)-\mathrm{card}(A(1))+\mathrm{card}(A^{(2)}(1))-\mathrm{card}(A(2))+\mathrm{card}(A^{(2)}(2))$ is smaller than the distortion error due to the set obtained by replacing $A(2)$ in the set (9) by the set B. In other words, $\left((({\alpha }_{F(n)}\setminus A(1))\cup A^{(2)}(1))\setminus A(2)\right)\cup A^{(2)}(2)$ forms an optimal set of $F(n)-\mathrm{card}(A(1))+\mathrm{card}(A^{(2)}(1))-\mathrm{card}(A(2))+\mathrm{card}(A^{(2)}(2))$-means. Proceeding inductively in this way, we can see that the set ${\alpha }_{F(n+1)}=S^{(2)(2)}(1)\cup S^{(2)(2)}(2)\cup S^{(2)(2)}(3)\cup \cdots \cup S^{(2)(2)}(n)\cup S^{(2)(2)(2)}(n+1)$ forms an optimal set of $F(n+1)$-means. Thus, by the induction principle, we can say that for any $n\ge 1$, the set ${\alpha }_{F(n)}$ forms an optimal set of $F(n)$-means for the condensation measure P. The corresponding quantization error is given by

$$\begin{array}{cc}\hfill & V_{F(n)}\hfill \\ \hfill & ={\displaystyle \frac{1}{3}}\left(V_{a(n)}(\nu )+({\displaystyle \frac{2}{75}})V_{a(n-1)}(\nu )+{({\displaystyle \frac{2}{75}})}^2{V}_{a(n-2)}(\nu )+\cdots +{({\displaystyle \frac{2}{75}})}^{n-1}{V}_{a(1)}(\nu )\right)+{({\displaystyle \frac{2}{75}})}^{n}V\hfill \\ \hfill & ={\displaystyle \frac{1}{3}}{\displaystyle \frac{1}{1575}}\left({\displaystyle \frac{1}{{64}^{n-1}}}+({\displaystyle \frac{2}{75}}){\displaystyle \frac{1}{{64}^{n-2}}}+{({\displaystyle \frac{2}{75}})}^2{\displaystyle \frac{1}{{64}^{n-3}}}+\cdots +{({\displaystyle \frac{2}{75}})}^{n-2}{\displaystyle \frac{1}{64}}+{({\displaystyle \frac{2}{75}})}^{n-1}\right)+{({\displaystyle \frac{2}{75}})}^{n}V,\hfill \end{array}$$

which after simplification implies that

$$V_{F(n)}={\displaystyle \frac{64}{3339}}\left({({\displaystyle \frac{2}{75}})}^n-{({\displaystyle \frac{1}{64}})}^n\right)+{({\displaystyle \frac{2}{75}})}^{n}V={\displaystyle \frac{\mathrm{769,208}}{\mathrm{5,884,749}}}{({\displaystyle \frac{2}{75}})}^n-{\displaystyle \frac{64}{3339}}{({\displaystyle \frac{1}{64}})}^n.$$

Thus, the proof of the proposition is complete. □

We now prove the following proposition.

Proposition 11.

Let $n\in \mathbb{N}$ be large. Then, the set

$$\left({\alpha }_{F(n)}\setminus \left(\underset{k=18}{\overset{n}{\cup }}S(k)\cup S(n+1)\right)\right)\cup \left(\underset{k=31}{\overset{n}{\cup }}{S}^{(2)(2)}(k)\right)\cup \left(\underset{k=18}{\overset{30}{\cup }}{S}^{(2)}(k)\right)\cup S^{(2)}(n+1)$$

(10)

forms an optimal set of $\left(2^n\left(6+2^{-30}(2^{13}+1)\right)-2n-6\right)$-means for P and the corresponding quantization error is given by

$$\begin{array}{cc}& V_{2^n(6+2^{-30}(2^{13}+1))-2n-6}\hfill \\ \hfill & =V_{F(n)}-{\displaystyle \frac{128}{3975}}{({\displaystyle \frac{2}{75}})}^n\left({({\displaystyle \frac{75}{128}})}^{31}-{({\displaystyle \frac{75}{128}})}^{n+1}\right)-{\displaystyle \frac{1024}{\mathrm{35,775}}}{({\displaystyle \frac{2}{75}})}^n\left({({\displaystyle \frac{75}{128}})}^{18}-{({\displaystyle \frac{75}{128}})}^{31}\right)\hfill \\ \hfill & -{\displaystyle \frac{\mathrm{450,241}}{\mathrm{5,323,500}}}{({\displaystyle \frac{2}{75}})}^n.\hfill \end{array}$$

Proof. 

Due to Remark 4, Lemma 13, using the similar technique as Proposition 10, we can show that the set given by (10) is an optimal set. Recall the definitions of $S(k)$, $S^{(2)}(k)$, $S^{(2)(2)}(k)$, $S(n+1)$, and $S^{(2)}(n+1)$. Thus, for $31\le k\le n$, in a set when $S(k)$ is replaced by $S^{(2)(2)}(k)$, we are removing $2^{n-k}·a(k)$ elements, and adding $2^{n-k}·(a(k)+2)$ elements to the set leading to the number of elements increased by $2^{n-k}·2$, and the quantization error is decreased by ${\displaystyle \frac{1}{3}}{({\displaystyle \frac{2}{75}})}^{n-k}(V_{a(k)}(\nu )-V_{a(k)+2}(\nu ))$. For $18\le k\le 30$, when $S(k)$ is replaced by $S^{(2)}(k)$ in a set, we are removing $2^{n-k}a(k)$ elements, and adding $2^{n-k}·(a(k)+1)$ elements to the set, leading to the number of elements in the set being increased by $2^{n-k}$, and the quantization error being decreased by ${\displaystyle \frac{1}{3}}{({\displaystyle \frac{2}{75}})}^{n-k}(V_{a(k)}(\nu )-V_{a(k)+1}(\nu ))$. When $S(n+1)$ is replaced by $S^{(2)}(n+1)$, the number of elements in the set is increased by $2^n$, and the quantization error is decreased by $(V-(V_{2,1}+V_{2,2}))$. Thus, the cardinality of the set (10) is given by

$$\begin{array}{cc}\hfill & F(n)+2\sum _{k=31}^n{2}^{n-k}+\sum _{k=18}^{30}{2}^{n-k}+2^n=F(n)+2(2^{n-30}-1)+2^{n-30}(2^{13}-1)+2^n\hfill \\ \hfill & =(5·2^n-2n-4)+2^{n-30}(2^{13}+1)+2^n-2=2^n\left(6+2^{-30}(2^{13}+1)\right)-2n-6.\hfill \end{array}$$

The corresponding quantization error is

$$\begin{array}{cc}\hfill & V_{F(n)}-{\displaystyle \frac{1}{3}}\sum _{k=31}^n{({\displaystyle \frac{2}{75}})}^{n-k}(V_{a(k)}(\nu )-V_{a(k)+2}(\nu ))-{\displaystyle \frac{1}{3}}\sum _{k=18}^{30}{({\displaystyle \frac{2}{75}})}^{n-k}(V_{a(k)}(\nu )-V_{a(k)+1}(\nu ))\hfill \\ \hfill & -{({\displaystyle \frac{2}{75}})}^n(V-(V_{2,1}+V_{2,2}))\hfill \\ \hfill & =V_{F(n)}-{\displaystyle \frac{1}{75}}{({\displaystyle \frac{2}{75}})}^n\sum _{k=31}^n{({\displaystyle \frac{75}{128}})}^k-{\displaystyle \frac{8}{675}}{({\displaystyle \frac{2}{75}})}^n\sum _{k=18}^{30}{({\displaystyle \frac{75}{128}})}^k-{\displaystyle \frac{\mathrm{450,241}}{\mathrm{5,323,500}}}{({\displaystyle \frac{2}{75}})}^n\hfill \\ \hfill & =V_{F(n)}-{\displaystyle \frac{128}{3975}}{({\displaystyle \frac{2}{75}})}^n\left({({\displaystyle \frac{75}{128}})}^{31}-{({\displaystyle \frac{75}{128}})}^{n+1}\right)-{\displaystyle \frac{1024}{\mathrm{35,775}}}{({\displaystyle \frac{2}{75}})}^n\left({({\displaystyle \frac{75}{128}})}^{18}-{({\displaystyle \frac{75}{128}})}^{31}\right)\hfill \\ \hfill & -{\displaystyle \frac{\mathrm{450,241}}{\mathrm{5,323,500}}}{({\displaystyle \frac{2}{75}})}^n.\hfill \end{array}$$

□

3.3. Asymptotics for the nth Quantization Error $V_n(P)$

In this subsection, we show that the quantization dimension $D(P)$ of the condensation measure P exists, the quantization coefficient does not exist, and the $D(P)$-dimensional lower and upper quantization coefficients are finite and positive.

Theorem 1.

Let P be the condensation measure associated with the self-similar measure ν and κ be the unique number given by ${({\displaystyle \frac{1}{3}}{({\displaystyle \frac{1}{5}})}^2)}^{{\displaystyle \frac{\kappa }{2+\kappa }}}+{({\displaystyle \frac{1}{3}}{({\displaystyle \frac{1}{5}})}^2)}^{{\displaystyle \frac{\kappa }{2+\kappa }}}=1$. Then, ${lim}_{n\to \infty }{\displaystyle \frac{2logn}{-log{V}_n(P)}}=\kappa$, i.e., the quantization dimension $D(P)$ of the measure P exists and equals κ.

Proof.

${({\displaystyle \frac{1}{3}}{({\displaystyle \frac{1}{5}})}^2)}^{{\displaystyle \frac{\kappa }{2+\kappa }}}+{({\displaystyle \frac{1}{3}}{({\displaystyle \frac{1}{5}})}^2)}^{{\displaystyle \frac{\kappa }{2+\kappa }}}=1$ implies $\kappa ={\displaystyle \frac{2log2}{log75-log2}}$. For $n\in \mathbb{N}$, let $\ell (n)$ be the least positive integer such that $F(\ell (n))\le n<F(\ell (n)+1)$. Then, $V_{F(\ell (n)+1)}<V_n\le V_{F(\ell (n))}$. Thus, we have

$$\begin{array}{c}\hfill {\displaystyle \frac{2log\left(F(\ell (n))\right)}{-log\left(V_{F(\ell (n)+1)}\right)}}<{\displaystyle \frac{2logn}{-log{V}_n}}<{\displaystyle \frac{2log\left(F(\ell (n)+1)\right)}{-log\left(V_{F(\ell (n))}\right)}}.\end{array}$$

Notice that when $n\to \infty$, then $\ell (n)\to \infty$. Recall that $F(\ell (n))=5·2^{\ell (n)}-2\ell (n)-4$, and so we have

$$\begin{array}{cc}\hfill & \underset{\ell (n)\to \infty }{lim}{\displaystyle \frac{2log\left(F(\ell (n))\right)}{-log\left(V_{F(\ell (n)+1)}\right)}}=2\underset{\ell (n)\to \infty }{lim}{\displaystyle \frac{log\left(5·2^{\ell (n)}-2\ell (n)-4\right)}{-log\left(\frac{769,208}{5,884,749}{(\frac{2}{75})}^n-\frac{64}{3339}{(\frac{1}{64})}^n\right)}}\hfill \\ \hfill & ={\displaystyle \frac{2log2}{-log\frac{2}{75}}}={\displaystyle \frac{2log2}{log75-log2}}=\kappa .\hfill \end{array}$$

Similarly, $\underset{\ell (n)\to \infty }{lim}{\displaystyle \frac{2log\left(F(\ell (n)+1)\right)}{-log\left(V_{F(\ell (n))}\right)}}=\kappa$. Thus, $\kappa \le {lim\; inf}_n{\displaystyle \frac{2logn}{-log{V}_n}}\le {lim\; sup}_n{\displaystyle \frac{2logn}{-log{V}_n}}\le \kappa$, implying the fact that the quantization dimension of the measure P exists and equals $\kappa$. □

Theorem 2.

The $D(P)$-dimensional quantization coefficient for the condensation measure P does not exist, and the $D(P)$-dimensional lower and upper quantization coefficients for P are finite and positive.

Proof. 

We have $D(P)=\kappa ={\displaystyle \frac{2log2}{log75-log2}}$, and so for any $n\in \mathbb{N}$, $2^{2n/\kappa }={({\displaystyle \frac{75}{2}})}^n$. Hence, by Proposition 10, we have

$$\begin{array}{cc}\hfill & \underset{n\to \infty }{lim}{(F(n))}^{2/\kappa }{V}_{F(n)}(P)=\underset{n\to \infty }{lim}{(5·2^n-2n-4)}^{2/\kappa }\left({\displaystyle \frac{769,208}{5,884,749}}{({\displaystyle \frac{2}{75}})}^n-{\displaystyle \frac{64}{3339}}{({\displaystyle \frac{1}{64}})}^n\right)\hfill \\ \hfill & =\underset{n\to \infty }{lim}{({\displaystyle \frac{75}{2}})}^n{(5-{\displaystyle \frac{2n}{2^n}}-{\displaystyle \frac{4}{2^n}})}^{2/\kappa }\left({\displaystyle \frac{769,208}{5,884,749}}{({\displaystyle \frac{2}{75}})}^n-{\displaystyle \frac{64}{3339}}{({\displaystyle \frac{1}{64}})}^n\right)=5^{2/\kappa }{\displaystyle \frac{\mathrm{769,208}}{\mathrm{5,884,749}}}.\hfill \end{array}$$

Again, by Proposition 11, we have

$$\begin{array}{cc}\hfill & \underset{n\to \infty }{lim}{(2^n(6+2^{-30}(2^{13}+1))-2n-6)}^{2/\kappa }{V}_{2^n(6+2^{-30}(2^{13}+1))-2n-6}\hfill \\ \hfill & =\underset{n\to \infty }{lim}{(2^n(6+2^{-30}(2^{13}+1))-2n-6)}^{2/\kappa }(V_{F(n)}-{\displaystyle \frac{128}{3975}}{({\displaystyle \frac{2}{75}})}^n\left({({\displaystyle \frac{75}{128}})}^{31}-{({\displaystyle \frac{75}{128}})}^{n+1}\right)\hfill \\ \hfill & -{\displaystyle \frac{1024}{\mathrm{35,775}}}{({\displaystyle \frac{2}{75}})}^n\left({({\displaystyle \frac{75}{128}})}^{18}-{({\displaystyle \frac{75}{128}})}^{31}\right)-{\displaystyle \frac{\mathrm{450,241}}{\mathrm{5,323,500}}}{({\displaystyle \frac{2}{75}})}^n)\hfill \\ \hfill & =\underset{n\to \infty }{lim}{({\displaystyle \frac{75}{2}})}^n{(6+2^{-30}(2^{13}+1)-{\displaystyle \frac{2n}{2^n}}-{\displaystyle \frac{6}{2^n}})}^{2/\kappa }({\displaystyle \frac{\mathrm{769,208}}{\mathrm{5,884,749}}}{({\displaystyle \frac{2}{75}})}^n-{\displaystyle \frac{64}{3339}}{({\displaystyle \frac{1}{64}})}^n\hfill \\ \hfill & -{\displaystyle \frac{128}{3975}}{({\displaystyle \frac{2}{75}})}^n\left({({\displaystyle \frac{75}{128}})}^{31}-{({\displaystyle \frac{75}{128}})}^{n+1}\right)-{\displaystyle \frac{1024}{\mathrm{35,775}}}{({\displaystyle \frac{2}{75}})}^n\left({({\displaystyle \frac{75}{128}})}^{18}-{({\displaystyle \frac{75}{128}})}^{31}\right)\hfill \\ \hfill & -{\displaystyle \frac{\mathrm{450,241}}{\mathrm{5,323,500}}}{({\displaystyle \frac{2}{75}})}^n)\hfill \\ \hfill & =6^{2/\kappa }\left({\displaystyle \frac{769,208}{5,884,749}}-{\displaystyle \frac{128}{3975}}{({\displaystyle \frac{75}{128}})}^{31}-{\displaystyle \frac{1024}{35,775}}\left({({\displaystyle \frac{75}{128}})}^{18}-{({\displaystyle \frac{75}{128}})}^{31}\right)-{\displaystyle \frac{450,241}{5,323,500}}\right).\hfill \end{array}$$

Since ${(F(n))}^{2/\kappa }{V}_{F(n)}(P)$, and ${(2^n(6+2^{-30}(2^{13}+1))-2n-6)}^{2/\kappa }{V}_{2^n(6+2^{-30}(2^{13}+1))-2n-6}(P)$ are two subsequences of ${(n^{2/\kappa }{V}_n(P))}_{n\in \mathbb{N}}$ having two different limits, we can say that the sequence ${(n^{2/\kappa }{V}_n(P))}_{n\in \mathbb{N}}$ does not converge, in other words, the $D(P)$-dimensional quantization coefficient for P does not exist.

To show that the $D(P)$-dimensional lower and upper quantization coefficients for P are finite and positive, for $n\in \mathbb{N}$, let $\ell (n)$ be the least positive integer such that $F(\ell (n))\le n<F(\ell (n)+1)$. Then, $V_{F(\ell (n)+1)}<V_n\le V_{F(\ell (n))}$ implying ${(F(\ell (n)))}^{2/\kappa }{V}_{F(\ell (n)+1)}<n^{2/\kappa }{V}_n<{(F(\ell (n)+1))}^{2/\kappa }{V}_{F(\ell (n))}$. As $\ell (n)\to \infty$ whenever $n\to \infty$, we have

$$\underset{n\to \infty }{lim}{\displaystyle \frac{F(\ell (n))}{F(\ell (n)+1)}}=\underset{n\to \infty }{lim}{\displaystyle \frac{5·2^{\ell (n)}-2\ell (n)-4}{5·2^{\ell (n)+1}-2(\ell (n)+1)-4}}={\displaystyle \frac{1}{2}},$$

and so,

$$\begin{array}{cc}\hfill & \underset{n\to \infty }{lim}{(F(\ell (n)))}^{2/\kappa }{V}_{F(\ell (n)+1)}={\displaystyle \frac{1}{4^{1/\kappa }}}\underset{n\to \infty }{lim}{(F(\ell (n)+1))}^{2/\kappa }{V}_{F(\ell (n)+1)}\hfill \\ \hfill & ={\displaystyle \frac{1}{4^{1/\kappa }}}\underset{n\to \infty }{lim}{(5·2^{\ell (n)+1}-2(\ell (n)+1)-4)}^{2/\kappa }\left({\displaystyle \frac{769,208}{5,884,749}}{({\displaystyle \frac{2}{75}})}^{\ell (n)+1}-{\displaystyle \frac{64}{3339}}{({\displaystyle \frac{1}{64}})}^{\ell (n)+1}\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{4^{1/\kappa }}}{5}^{2/\kappa }{\displaystyle \frac{\mathrm{769,208}}{\mathrm{5,884,749}}}={\left({\displaystyle \frac{25}{4}}\right)}^{1/\kappa }{\displaystyle \frac{\mathrm{769,208}}{\mathrm{5,884,749}}},\hfill \end{array}$$

and similarly,

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}{(F(\ell (n)+1))}^{2/\kappa }{V}_{F(\ell (n))}=4^{1/\kappa }\underset{n\to \infty }{lim}{(F(\ell (n)))}^{2/\kappa }{V}_{F(\ell (n))}={(100)}^{1/\kappa }{\displaystyle \frac{\mathrm{769,208}}{\mathrm{5,884,749}}}\end{array}$$

yielding the fact that ${({\displaystyle \frac{25}{4}})}^{1/\kappa }{\displaystyle \frac{\mathrm{769,208}}{\mathrm{5,884,749}}}\le \underset{n\to \infty }{lim\; inf}{n}^{2/\kappa }{V}_n(P)\le \underset{n\to \infty }{lim\; sup}{n}^{2/\kappa }{V}_n(P)\le {(100)}^{1/\kappa }{\displaystyle \frac{\mathrm{769,208}}{\mathrm{5,884,749}}}$, i.e., the $D(P)$-dimensional lower and upper quantization coefficients for the condensation measure P are finite and positive. Thus, the proof of the theorem is complete. □

Remark 5.

Notice that ${({\displaystyle \frac{1}{3}}{({\displaystyle \frac{1}{5}})}^2)}^{{\displaystyle \frac{\kappa }{2+\kappa }}}+{({\displaystyle \frac{1}{3}}{({\displaystyle \frac{1}{5}})}^2)}^{{\displaystyle \frac{\kappa }{2+\kappa }}}=1$ implies $\kappa ={\displaystyle \frac{2log2}{log75-log2}}\approx 0.382496>0=D(\nu )$. In Theorem 1, we have proved that $D(P)=\kappa =max\{\kappa ,D(\nu )\}.$

4. Quantization for a Condensation Measure Associated with a Uniform Distribution

In this section, we investigate the optimal quantization for the condensation measure P when $\nu$ is the uniform distribution on the closed interval $[{\displaystyle \frac{2}{5}},{\displaystyle \frac{3}{5}}]$.

Lemma 14.

Let X and Y be the random variables with the probability distributions P and ν, respectively. Let $E(X)$, $E(Y)$ represent the expected values, and $V:=V(X)$, $W:=V(Y)$ represent the variances of X and Y, respectively. Then,

$$E(X)=E(Y)={\displaystyle \frac{1}{2}},V={\displaystyle \frac{97}{876}},\mathit{and}W={\displaystyle \frac{1}{300}}.$$

Proof. 

Using Lemma 4, we have $E(X)=\int xdP(x)={\displaystyle \frac{1}{3}}\int {\displaystyle \frac{1}{5}}xdP(x)+{\displaystyle \frac{1}{3}}\int ({\displaystyle \frac{1}{5}}x+{\displaystyle \frac{4}{5}})dP(x)$ $+{\displaystyle \frac{1}{3}}\int xd\nu (x)={\displaystyle \frac{1}{15}}E(X)+{\displaystyle \frac{1}{15}}E(X)+{\displaystyle \frac{4}{15}}+{\displaystyle \frac{1}{6}},$ which implies $E(X)={\displaystyle \frac{1}{2}}$. Again, $E(X^2)=\int x^{2}dP(x)={\displaystyle \frac{1}{3}}\int x^{2}d(P\circ S_1^{-1})(x)$ $+{\displaystyle \frac{1}{3}}\int x^{2}d(P\circ S_2^{-1})(x)+{\displaystyle \frac{1}{3}}\int x^{2}d\nu ={\displaystyle \frac{1}{3}}\int {({\displaystyle \frac{1}{5}}x)}^{2}dP(x)+$ ${\displaystyle \frac{1}{3}}\int {({\displaystyle \frac{1}{5}}x+{\displaystyle \frac{4}{5}})}^{2}dP(x)+$ ${\displaystyle \frac{1}{3}}{\int }_{{\displaystyle \frac{2}{5}}}^{{\displaystyle \frac{3}{5}}}5x^{2}dx={\displaystyle \frac{1}{75}}E(X^2)+{\displaystyle \frac{1}{75}}E(X^2)+{\displaystyle \frac{8}{75}}E(X)+{\displaystyle \frac{16}{75}}+{\displaystyle \frac{19}{225}}$, which implies $E(X^2)={\displaystyle \frac{79}{219}}$, and hence, $V(X)=E{(X-E(X))}^2=E(X^2)-{\left(E(X)\right)}^2={\displaystyle \frac{79}{219}}-{\displaystyle \frac{1}{4}}={\displaystyle \frac{97}{876}}$. On the other hand, $E(Y)=\int yd\nu ={\int }_{{\displaystyle \frac{2}{5}}}^{{\displaystyle \frac{3}{5}}}5ydy={\displaystyle \frac{1}{2}}$, and $V(Y)=E{(Y-{\displaystyle \frac{1}{2}})}^2={\int }_{{\displaystyle \frac{2}{5}}}^{{\displaystyle \frac{3}{5}}}5{(y-{\displaystyle \frac{1}{2}})}^{2}dy={\displaystyle \frac{1}{300}}$. Thus, the proof of the lemma is complete. □

Lemma 15.

Let P be the condensation measure and ν be the uniform distribution that occurs in P. Then, for any $x_0\in \mathbb{R}$, we have

$$\int {(x-x_0)}^{2}dP(x)={\displaystyle \frac{97}{876}}+{(x_0-{\displaystyle \frac{1}{2}})}^2,\mathit{and}\int {(x-x_0)}^{2}d\nu (x)={\displaystyle \frac{1}{300}}+{(x_0-{\displaystyle \frac{1}{2}})}^2.$$

Proof. 

Let X and Y be the random variables as defined in Lemma 14. Then, following the standard rule of probability, we have $\int {(x-x_0)}^{2}dP(x)=V(X)+{(x_0-{\displaystyle \frac{1}{2}})}^2$, and $\int {(x-x_0)}^{2}d\nu (x)=V(Y)+{(x_0-{\displaystyle \frac{1}{2}})}^2$, which is the lemma. □

Note 3.

Let $\omega \in I^k$, $k\ge 0$. Then, by Equation (2), it follows that $P(J_{\omega })={\displaystyle \frac{1}{3^k}}$, $P(C_{\omega })={\displaystyle \frac{1}{3^{k+1}}}$,

$$\begin{array}{cc}& E(X:X\in J_{\omega })={\displaystyle \frac{1}{P(J_{\omega })}}{\int }_{J_{\omega }}xdP=\int xd(P\circ S_{\omega }^{-1})=\int S_{\omega }(x)dP=S_{\omega }({\displaystyle \frac{1}{2}}),\mathit{and}\hfill \\ \hfill & E(X:X\in C_{\omega })={\displaystyle \frac{1}{P(C_{\omega })}}{\int }_{C_{\omega }}xdP=\int xd(\nu \circ S_{\omega }^{-1})=\int S_{\omega }(x)d\nu =S_{\omega }({\displaystyle \frac{1}{2}}).\hfill \end{array}$$

For any $x_0\in \mathbb{R}$,

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\int }_{J_{\omega }}{(x-x_0)}^{2}dP(x)={\displaystyle \frac{1}{3^k}}\left({\displaystyle \frac{1}{{25}^k}}V+{(S_{\omega }({\displaystyle \frac{1}{2}})-x_0)}^2\right),\\ {\int }_{C_{\omega }}{(x-x_0)}^{2}d\nu (x)={\displaystyle \frac{1}{3^{k+1}}}\left({\displaystyle \frac{1}{{25}^k}}W+{(S_{\omega }({\displaystyle \frac{1}{2}})-x_0)}^2\right).\end{array}\right.\end{array}$$

(11)

Remark 6.

By Lemma 15, it follows that the optimal set of one-mean for the condensation measure P consists of the expected value ${\displaystyle \frac{1}{2}}$ and the corresponding quantization error is the variance $V(X)$ of the random variable X.

The following proposition is useful.

Proposition 12.

Let ν be the uniform distribution on the interval $[{\displaystyle \frac{2}{5}},{\displaystyle \frac{3}{5}}]$. Then, for $n\ge 1$, the set $\{{\displaystyle \frac{2}{5}}+{\displaystyle \frac{2i-1}{10n}}:1\le i\le n\}$ is a unique optimal set of n-means for ν with quantization error $V_n(\nu )={\displaystyle \frac{W}{n^2}}$. Moreover, the quantization dimension $D(\nu )$ of ν is one.

Proof. 

Since $\nu$ is uniformly distributed on $[{\displaystyle \frac{2}{5}},{\displaystyle \frac{3}{5}}]$, the boundaries of the Voronoi regions of an optimal set of n-means will divide the interval $[{\displaystyle \frac{2}{5}},{\displaystyle \frac{3}{5}}]$ into n equal subintervals, i.e., the boundaries of the Voronoi regions are given by

$$\{{\displaystyle \frac{2}{5}},{\displaystyle \frac{2}{5}}+{\displaystyle \frac{1}{5n}},{\displaystyle \frac{2}{5}}+{\displaystyle \frac{2}{5n}},\cdots ,{\displaystyle \frac{2}{5}}+{\displaystyle \frac{n-1}{5n}},{\displaystyle \frac{3}{5}}\}.$$

This implies that an optimal set of n-means for $\nu$ is unique, and it consists of the midpoints of the boundaries of the Voronoi regions, i.e., the optimal set of n-means for $\nu$ is given by $\{{\displaystyle \frac{2}{5}}+{\displaystyle \frac{2i-1}{10n}}:1\le i\le n\}$ for any $n\ge 1$. Then, the nth quantization error for $\nu$ is

$$\begin{array}{cc}\hfill V_n(\nu )& =n\times (\mathrm{the}\mathrm{quantization}\mathrm{error}\mathrm{in}\mathrm{each}\mathrm{Voronoi}\mathrm{region})\hfill \\ \hfill & =n\left({\int }_{{\displaystyle \frac{2}{5}}}^{{\displaystyle \frac{2}{5}}+{\displaystyle \frac{1}{5n}}}5{\left(x-({\displaystyle \frac{2}{5}}+{\displaystyle \frac{1}{10n}})\right)}^{2}dx\right),\hfill \end{array}$$

which after simplification implies $V_n(\nu )={\displaystyle \frac{1}{300n^2}}={\displaystyle \frac{W}{n^2}}$. Hence, $D(\nu )=\underset{n\to \infty }{lim}{\displaystyle \frac{2log(n)}{-log(\frac{1}{300n^2})}}=1.$ Thus, the proof of the proposition is complete. □

Remark 7.

By Proposition 12, it follows that $\underset{n\to \infty }{lim}{n}^2{V}_n(\nu )={\displaystyle \frac{1}{300}}$, i.e., the one-dimensional quantization coefficient for ν is finite and strictly positive. In fact, for any probability measure μ on ${\mathbb{R}}^d$ with non-vanishing absolutely continuous part $\underset{n\to \infty }{lim}{n}^{2/d}{V}_n(\mu )$ is finite and strictly positive; in other words, the quantization dimension of a probability measure with non-vanishing absolutely continuous part is equal to the dimension d of the underlying space (see [30]).

Lemma 16.

Let ${\alpha }_n(\nu )$ be an optimal set of n-means for ν. Then, for any $\omega \in I^k$, $k\ge 0$, $S_{\omega }({\alpha }_n(\nu ))$ is an optimal set of n-means for the image measure $\nu \circ S_{\omega }^{-1}$. Moreover,

$${\int }_{C_{\omega }}\underset{a\in S_{\omega }({\alpha }_n(\nu ))}{min}{(x-a)}^{2}dP={\displaystyle \frac{1}{3}}·{\displaystyle \frac{1}{{75}^k}}·{\displaystyle \frac{W}{n^2}}.$$

Proof. 

Let ${\alpha }_n(\nu )$ be an optimal set of n-means for $\nu$. Then, $S_{\omega }({\alpha }_n(\nu ))$ is an optimal set of n-means for the image measure $\nu \circ S_{\omega }^{-1}$ follows equivalently from Lemma 5. Now, using (2) and Proposition 12, we have

$$\begin{array}{cc}\hfill & {\int }_{C_{\omega }}\underset{a\in S_{\omega }({\alpha }_n(\nu ))}{min}{(x-a)}^{2}dP={\displaystyle \frac{1}{3^{k+1}}}{\int }_{C_{\omega }}\underset{a\in S_{\omega }({\alpha }_n(\nu ))}{min}{(x-a)}^{2}d(\nu \circ S_{\omega }^{-1})\hfill \\ \hfill & ={\displaystyle \frac{1}{3^{k+1}}}{\int }_C\underset{a\in S_{\omega }({\alpha }_n(\nu ))}{min}{(S_{\omega }(x)-a)}^{2}d\nu ={\displaystyle \frac{1}{3^{k+1}}}·{\displaystyle \frac{1}{{25}^k}}{\int }_C\underset{a\in {\alpha }_n(\nu )}{min}{(x-a)}^{2}d\nu ={\displaystyle \frac{1}{3}}·{\displaystyle \frac{1}{{75}^k}}·{\displaystyle \frac{W}{n^2}}.\hfill \end{array}$$

Thus, the proof of the lemma is complete. □

4.1. Essential Lemmas and Propositions

In this subsection, we give some lemmas and propositions that we need to determine the optimal sets of n-means and the nth quantization errors for all $n\ge 2$. To determine the quantization error, we will frequently use the formulas given in the expression (11).

Let us now state the following proposition. The proof of this proposition follows similarly as the proof of Proposition 5.

Proposition 13.

Let $\alpha :=\{a_1,a_2\}$ be an optimal set of two-means with $a_1<a_2$. Then, $a_1={\displaystyle \frac{13}{60}}$, $a_2={\displaystyle \frac{47}{60}}$, and the corresponding quantization error is $V_2={\displaystyle \frac{8003}{262,800}}=0.0304528$.

Lemma 17.

Let $\omega \in I^k$ for any $k\ge 0$. Then,

$${\int }_{J_{\omega 1}\cup S_{\omega }[{\displaystyle \frac{2}{5}},{\displaystyle \frac{1}{2}}]}{(x-S_{\omega }({\displaystyle \frac{13}{60}}))}^{2}dP={\displaystyle \frac{1}{{75}^k}}·{\displaystyle \frac{1}{2}}{V}_2={\int }_{J_{\omega 2}\cup S_{\omega }[{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{5}}]}{(x-S_{\omega }({\displaystyle \frac{47}{60}}))}^{2}dP.$$

Corollary 4.

Let $\omega \in I^k$ for $k\ge 0$. Then, for any $a\in \mathbb{R}$,

$$\begin{array}{cc}\hfill & {\int }_{J_{\omega 1}\cup S_{\omega }([{\displaystyle \frac{2}{5}},{\displaystyle \frac{1}{2}}])}{(x-a)}^{2}dP={\displaystyle \frac{1}{3^k}}\left({\displaystyle \frac{1}{{25}^k}}{\displaystyle \frac{1}{2}}{V}_2+{\displaystyle \frac{1}{2}}{(S_{\omega }({\displaystyle \frac{13}{60}})-a)}^2\right),\mathit{and}\hfill \\ \hfill & {\int }_{S_{\omega }([{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{5}}])\cup J_{\omega 2}}{(x-a)}^{2}dP={\displaystyle \frac{1}{3^k}}\left({\displaystyle \frac{1}{{25}^k}}{\displaystyle \frac{1}{2}}{V}_2+{\displaystyle \frac{1}{2}}({\displaystyle \frac{47}{60}}){-a)}^2\right).\hfill \end{array}$$

Proposition 14.

Let $\alpha :=\{a_1,a_2,a_3\}$ be an optimal set of three-means with $a_1<a_2<a_3$. Then, $a_1=S_1({\displaystyle \frac{1}{2}})={\displaystyle \frac{1}{10}}$, $a_2={\displaystyle \frac{1}{2}}$, and $a_3=S_2({\displaystyle \frac{1}{2}})={\displaystyle \frac{9}{10}}$. The corresponding quantization error is $V_3={\displaystyle \frac{89}{21,900}}=0.00406393$.

Proof. 

Let $\beta :=\{{\displaystyle \frac{1}{10}},{\displaystyle \frac{1}{2}},{\displaystyle \frac{9}{10}}\}$. Then, using (11), we have

$$\begin{array}{cc}\hfill & \int \underset{a\in \beta }{min}{(x-a)}^{2}dP={\int }_{J_1}{(x-{\displaystyle \frac{1}{10}})}^{2}dP+{\int }_C{(x-{\displaystyle \frac{1}{2}})}^{2}dP+{\int }_{J_2}{(x-{\displaystyle \frac{9}{10}})}^{2}dP\hfill \\ \hfill & ={\displaystyle \frac{1}{3}}\left({\displaystyle \frac{1}{25}}V+{\displaystyle \frac{1}{3}}W+{\displaystyle \frac{1}{25}}V\right)={\displaystyle \frac{89}{\mathrm{21,900}}}=0.00406393.\hfill \end{array}$$

Since $V_3$ is the quantization error for three-means, we have $0.00406393\ge V_3$. Let $\alpha :=\{a_1,a_2,a_3\}$ be an optimal set of three-means with $a_1<a_2<a_3$. Since $a_1,a_2$, and $a_3$ are the centroids of their own Voronoi regions, we have $0<a_1<a_2<a_3<1$. Suppose that ${\displaystyle \frac{1}{5}}\le a_1$. Then,

$$V_3\ge {\int }_{J_1}{(x-{\displaystyle \frac{1}{5}})}^{2}dP={\displaystyle \frac{79}{\mathrm{16,425}}}=0.00480974>V_3,$$

which is a contradiction. So, we can assume that $a_1<{\displaystyle \frac{1}{5}}$. Similarly, we have ${\displaystyle \frac{4}{5}}<a_3$. Suppose that $a_2\le {\displaystyle \frac{2}{5}}$. Then,

$$V_3\ge {\int }_C{(x-{\displaystyle \frac{2}{5}})}^{2}dP={\displaystyle \frac{1}{3}}(W+{({\displaystyle \frac{1}{2}}-{\displaystyle \frac{2}{5}})}^2)={\displaystyle \frac{1}{225}}=0.00444444>V_3,$$

which is a contradiction. So, we can assume that ${\displaystyle \frac{2}{5}}<a_2$. Similarly, $a_2<{\displaystyle \frac{3}{5}}$. Thus, we have ${\displaystyle \frac{2}{5}}<a_2<{\displaystyle \frac{3}{5}}$. We now show that the Voronoi region of $a_1$ does not contain any point from C. Suppose that ${\displaystyle \frac{1}{2}}(a_1+a_2)\ge {\displaystyle \frac{2}{5}}$. Then, $a_2\ge {\displaystyle \frac{4}{5}}-a_1>{\displaystyle \frac{4}{5}}-{\displaystyle \frac{1}{5}}={\displaystyle \frac{3}{5}}$, which is a contradiction, and so the Voronoi region of $a_1$ does not contain any point from C. Similarly, the Voronoi region of $a_3$ does not contain any point from C. In a similar fashion, we can show that the Voronoi region of $a_2$ does not contain any point from $J_1$ and $J_2$. Hence,

$$a_1=S_1({\displaystyle \frac{1}{2}})={\displaystyle \frac{1}{10}},a_2=E(X:X\in C)={\displaystyle \frac{1}{2}},\mathrm{and}{a}_3=S_2({\displaystyle \frac{1}{2}})={\displaystyle \frac{9}{10}},$$

and the corresponding quantization error is $V_3={\displaystyle \frac{89}{\mathrm{21,900}}}=0.00406393$. Thus, the proof of the lemma is complete. □

Lemma 18.

Let α be an optimal set of four-means. Then, $\alpha \cap J_1\ne \varnothing$, $\alpha \cap J_2\ne \varnothing$, and $\alpha \cap C\ne \varnothing$. Moreover, α does not contain any point from the open intervals $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$ and $({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$.

Proof. 

Let us first consider a set of four points $\beta :=\{S_1({\displaystyle \frac{13}{60}}),S_1({\displaystyle \frac{47}{60}}),{\displaystyle \frac{1}{2}},S_2({\displaystyle \frac{1}{2}})\}$. Then, using Lemma 17,

$$\begin{array}{cc}\hfill & \int \underset{a\in \beta }{min}{(x-a)}^{2}dP=2{\int }_{J_{11}\cup S_1[{\displaystyle \frac{2}{5}},{\displaystyle \frac{1}{2}}]}{(x-S_1({\displaystyle \frac{13}{60}}))}^{2}dP+{\int }_C{(x-{\displaystyle \frac{1}{2}})}^{2}dP+{\int }_{J_2}{(x-S_2({\displaystyle \frac{1}{2}}))}^{2}dP\hfill \\ \hfill & ={\displaystyle \frac{1}{75}}{V}_2+{\displaystyle \frac{1}{3}}W+{\displaystyle \frac{1}{75}}V={\displaystyle \frac{\mathrm{59,003}}{\mathrm{19,710,000}}}=0.00299356.\hfill \end{array}$$

Since $V_4$ is the quantization error for four-means, we have $V_4\le 0.00299356$. Let $\alpha :=\{a_1<a_2<a_3<a_4\}$ be an optimal set of four-means. Proceeding in the similar way as shown in the proof of Proposition 14, we have $0<a_1<{\displaystyle \frac{1}{5}}$ and ${\displaystyle \frac{4}{5}}<a_4<1$, implying $\alpha \cap J_1\ne \varnothing$ and $\alpha \cap J_2\ne \varnothing$. We now show that $\alpha \cap C\ne \varnothing$. For the sake of contradiction, assume that $\alpha \cap C=\varnothing$. Then, if $a_3<{\displaystyle \frac{2}{5}}$, we have

$$\begin{array}{c}\hfill V_4\ge {\int }_C{(x-{\displaystyle \frac{2}{5}})}^{2}dP+{\int }_{J_2}{(x-S_2({\displaystyle \frac{1}{2}}))}^{2}dP={\displaystyle \frac{1}{225}}+{\displaystyle \frac{1}{75}}V={\displaystyle \frac{389}{\mathrm{65,700}}}=0.00592085>V_4,\end{array}$$

which leads to a contradiction. Hence, ${\displaystyle \frac{3}{5}}<a_3$. Suppose that $a_2\le {\displaystyle \frac{7}{20}}$. Then, due to symmetry, we can assume that ${\displaystyle \frac{13}{20}}\le a_3$, and so

$$\begin{array}{cc}\hfill V_4& \ge {\int }_{[{\displaystyle \frac{2}{5}},{\displaystyle \frac{1}{2}}]}{(x-{\displaystyle \frac{7}{20}})}^{2}dP+{\int }_{[{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{5}}]}{(x-{\displaystyle \frac{13}{20}})}^{2}dP={\displaystyle \frac{2}{3}}{\int }_{[{\displaystyle \frac{2}{5}},{\displaystyle \frac{1}{2}}]}{(x-{\displaystyle \frac{7}{20}})}^{2}d\nu ={\displaystyle \frac{13}{3600}}\hfill \\ \hfill & =0.00361111>V_4,\hfill \end{array}$$

which gives a contradiction. So, we can assume that ${\displaystyle \frac{7}{20}}<a_2<{\displaystyle \frac{2}{5}}<{\displaystyle \frac{3}{5}}<a_3<{\displaystyle \frac{13}{20}}$. Then, ${\displaystyle \frac{1}{2}}(a_1+a_2)<{\displaystyle \frac{1}{5}}$ and ${\displaystyle \frac{1}{2}}(a_3+a_4)>{\displaystyle \frac{4}{5}}$, yielding $a_1<{\displaystyle \frac{1}{20}}$ and $a_4>{\displaystyle \frac{19}{20}}$; otherwise, the quantization error can strictly be reduced by moving $a_2$ to ${\displaystyle \frac{2}{5}}$ and $a_3$ to ${\displaystyle \frac{3}{5}}$ contradicting the fact that $\alpha$ is an optimal set of four-means. Then,

$$\begin{array}{cc}\hfill V_4& \ge 2\left({\int }_{J_{12}}{(x-{\displaystyle \frac{1}{20}})}^{2}dP+{\int }_{[{\displaystyle \frac{2}{5}},{\displaystyle \frac{1}{2}}]}{(x-{\displaystyle \frac{2}{5}})}^{2}dP\right)={\displaystyle \frac{\mathrm{48,349}}{\mathrm{9,855,000}}}=0.00490604>V_4,\hfill \end{array}$$

which is a contraction. Hence, $\alpha \cap C\ne \varnothing$. Thus, we see that $\alpha \cap J_1\ne \varnothing$, $\alpha \cap J_2\ne \varnothing$, and $\alpha \cap C\ne \varnothing$. To complete the rest of the proof, for the sake of contradiction, assume that $\alpha$ contains a point from the open interval $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$. Then, $\alpha$ can not contain any point from $({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$ because $\mathrm{card}(\alpha )=4$, and $\alpha \cap J_1\ne \varnothing$, $\alpha \cap J_2\ne \varnothing$, and $\alpha \cap C\ne \varnothing$. Assume that $\alpha$ contains $a_2$ from $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$. Then, $0<a_1<{\displaystyle \frac{1}{5}}<a_2<{\displaystyle \frac{2}{5}}<a_3<{\displaystyle \frac{3}{5}}<{\displaystyle \frac{4}{5}}<a_4<1$, and so, the Voronoi region of $a_4$ does not contain any point from C, and the Voronoi region of $a_3$ does not contain any point from $J_2$. Under the assumption $a_2\in ({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$ the following three cases can arise:

Case 1: ${\displaystyle \frac{1}{3}}\le a_2<{\displaystyle \frac{2}{5}}$.

Then,

$$\begin{array}{cc}\hfill & \underset{a\in \{a_2,a_3\}}{min}\left\{{\int }_C{(x-a)}^{2}dP:{\displaystyle \frac{1}{3}}\le a_2<{\displaystyle \frac{2}{5}}\mathrm{and}{\displaystyle \frac{2}{5}}\le a_3\le {\displaystyle \frac{3}{5}}\right\}\hfill \\ \hfill & =\underset{a\in \{a_2,a_3\}}{min}\left\{{\displaystyle \frac{1}{3}}\left({\int }_{{\displaystyle \frac{2}{5}}}^{{\displaystyle \frac{a_2+a_3}{2}}}5{(x-a_2)}^{2}dx+{\int }_{{\displaystyle \frac{a_2+a_3}{2}}}^{{\displaystyle \frac{3}{5}}}5{(x-a_3)}^{2}dx\right):{\displaystyle \frac{1}{3}}\le a_2<{\displaystyle \frac{2}{5}},{\displaystyle \frac{2}{5}}\le a_3\le {\displaystyle \frac{3}{5}}\right\}\hfill \\ \hfill & ={\displaystyle \frac{1}{2025}}(\mathrm{when}{a}_2={\displaystyle \frac{2}{5}}\mathrm{and}{a}_3={\displaystyle \frac{8}{15}}).\hfill \end{array}$$

(12)

Again, ${\displaystyle \frac{1}{2}}(a_1+a_2)<{\displaystyle \frac{1}{5}}$ implies $a_1<{\displaystyle \frac{2}{5}}-a_2\le {\displaystyle \frac{2}{5}}-{\displaystyle \frac{1}{3}}={\displaystyle \frac{1}{15}}<{\displaystyle \frac{1}{10}}=S_1({\displaystyle \frac{1}{2}})$, and so

$$V_4\ge {\int }_{S_1[{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{5}}]\cup J_{12}}(x-{\displaystyle \frac{1}{15}})dP+{\displaystyle \frac{1}{2025}}+{\int }_{J_2}{(x-S_2({\displaystyle \frac{1}{2}}))}^{2}dP={\displaystyle \frac{\mathrm{20,833}}{\mathrm{5,913,000}}}=0.00352325>V_4,$$

which leads to a contradiction.

Case 2: ${\displaystyle \frac{4}{15}}\le a_2\le {\displaystyle \frac{1}{3}}$.

Then, proceeding as (12), we have

$$\begin{array}{c}\hfill \underset{a\in \{a_2,a_3\}}{min}\left\{{\int }_C{(x-a)}^{2}dP:{\displaystyle \frac{4}{15}}\le a_2\le {\displaystyle \frac{1}{3}},{\displaystyle \frac{2}{5}}\le a_3\le {\displaystyle \frac{3}{5}},\mathrm{and}{\displaystyle \frac{2}{5}}\le {\displaystyle \frac{a_2+a_3}{2}}\right\}={\displaystyle \frac{11}{\mathrm{10,935}}},\end{array}$$

which occurs when $a_2={\displaystyle \frac{1}{3}}$ and $a_3={\displaystyle \frac{23}{45}}$. Moreover, ${\displaystyle \frac{1}{2}}(a_1+a_2)<{\displaystyle \frac{1}{5}}$ implying $a_1<{\displaystyle \frac{2}{5}}-a_2\le {\displaystyle \frac{2}{5}}-{\displaystyle \frac{4}{15}}={\displaystyle \frac{2}{15}}$. First, assume that ${\displaystyle \frac{1}{15}}\le a_1<{\displaystyle \frac{2}{15}}$. Then,

$$\begin{array}{cc}\hfill V_4& \ge {\int }_{J_{11}}{(x-{\displaystyle \frac{1}{15}})}^{2}dP+{\int }_{C_1}{(x-S_1({\displaystyle \frac{1}{2}}))}^{2}dP+{\int }_{J_{12}}{(x-{\displaystyle \frac{2}{15}})}^{2}dP+{\displaystyle \frac{11}{\mathrm{10,935}}}+{\displaystyle \frac{1}{75}}V\hfill \\ \hfill & ={\displaystyle \frac{\mathrm{608,693}}{\mathrm{199,563,750}}},\hfill \end{array}$$

implying $V_4\ge {\displaystyle \frac{\mathrm{608,693}}{\mathrm{199,563,750}}}=0.00305012>V_4$, which yields a contradiction. Next, assume that $0<a_1<{\displaystyle \frac{1}{15}}$. Then, for any $x\in J_{12}$, we have

$$\underset{a\in \{a_1,a_2\}}{min}{(x-a)}^2\ge {(x-{\displaystyle \frac{1}{30}})}^2,$$

because $a_2-x\ge {\displaystyle \frac{4}{15}}-x\ge {\displaystyle \frac{1}{15}}\ge x-{\displaystyle \frac{1}{30}}\ge {\displaystyle \frac{2}{75}}>0$, and $x-a_1\ge x-{\displaystyle \frac{1}{15}}\ge {\displaystyle \frac{7}{75}}>x-{\displaystyle \frac{1}{30}}\ge {\displaystyle \frac{2}{75}}>0$. Thus, we have

$$\begin{array}{c}\hfill V_4\ge {\int }_{J_{12}}{(x-{\displaystyle \frac{1}{30}})}^{2}dP+{\displaystyle \frac{11}{\mathrm{10,935}}}+{\displaystyle \frac{1}{75}}V={\displaystyle \frac{\mathrm{488,149}}{\mathrm{99,781,875}}}=0.00489216>V_4,\end{array}$$

which leads to another contradiction. Thus, Case 2 gives a contradiction.

Case 3: ${\displaystyle \frac{1}{5}}<a_2\le {\displaystyle \frac{4}{15}}$.

Then, proceeding as (12), we have

$$\begin{array}{c}\hfill \underset{a\in \{a_2,a_3\}}{min}\left\{{\int }_C{(x-a)}^{2}dP:{\displaystyle \frac{1}{5}}<a_2\le {\displaystyle \frac{4}{15}},{\displaystyle \frac{2}{5}}\le a_3\le {\displaystyle \frac{3}{5}},\mathrm{and}{\displaystyle \frac{2}{5}}\le {\displaystyle \frac{a_2+a_3}{2}}\right\}={\displaystyle \frac{1}{675}},\end{array}$$

which occurs when $a_2={\displaystyle \frac{4}{15}}$ and $a_3={\displaystyle \frac{8}{15}}$. Moreover,

$$\begin{array}{cc}& \underset{a\in \{a_1,a_2\}}{min}\left\{{\int }_{J_1}{(x-a)}^{2}dP:0<a_1<{\displaystyle \frac{1}{5}}<a_2<{\displaystyle \frac{2}{5}}\right\}\ge {\int }_{J_1}\underset{a\in S_1({\alpha }_2)}{min}{(x-a)}^{2}dP={\displaystyle \frac{1}{75}}{V}_2.\hfill \end{array}$$

Thus, we see that

$$\begin{array}{c}\hfill V_4\ge {\displaystyle \frac{1}{75}}{V}_2+{\displaystyle \frac{1}{675}}+{\displaystyle \frac{1}{75}}V={\displaystyle \frac{7367}{\mathrm{2,190,000}}}=0.00336393>V_4,\end{array}$$

which is a contradiction.

Therefore, we can assume that $\alpha$ does not contain any point from $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$. Reflecting the situation with respect to the point ${\displaystyle \frac{1}{2}}$, we can also say that $\alpha$ does not contain any point from the open interval $({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$. Thus, the proof of the lemma is complete. □

Remark 8.

By Lemma 18, it is easy to see that the sets $\{S_1({\displaystyle \frac{13}{60}}),S_1({\displaystyle \frac{47}{60}}),{\displaystyle \frac{1}{2}},S_2({\displaystyle \frac{1}{2}})\}$ and $\{S_1({\displaystyle \frac{1}{2}}),{\displaystyle \frac{1}{2}},S_2({\displaystyle \frac{13}{60}}),S_2({\displaystyle \frac{47}{60}})\}$ form two different optimal sets of four-means.

Proposition 15.

Let α be an optimal set of n-means with $n\ge 3$. Then, $\alpha \cap J_1\ne \varnothing$, $\alpha \cap J_2\ne \varnothing$, and $\alpha \cap C\ne \varnothing$.

Proof. 

Due to Proposition 14 and Lemma 18, the proposition is true for $n=3$ and $n=4$. Let $\alpha :=\{a_1,a_2,\cdots ,a_n\}$ be an optimal set of n-means for $n\ge 5$, where $0<a_1<a_2<\cdots <a_n<1$. Proceeding in the similar way as shown in the proof of Proposition 14, we see that $0<a_1<{\displaystyle \frac{1}{5}}$, and ${\displaystyle \frac{4}{5}}<a_n<1$, yielding $\alpha \cap J_1\ne \varnothing$ and $\alpha \cap J_2\ne \varnothing$. Consider a set of five points $\beta :=S_1({\alpha }_2)\cup \left\{{\displaystyle \frac{1}{2}}\right\}\cup S_2({\alpha }_2)$, where ${\alpha }_2$ is an optimal set of two-means for P. Then, using Lemma 17, and the symmetry of P, we obtain the distortion error for the set $\beta$ as

$$\begin{array}{cc}\hfill & \int \underset{a\in \beta }{min}{(x-a)}^{2}dP=2{\int }_{J_{11}\cup S_1[{\displaystyle \frac{2}{5}},{\displaystyle \frac{1}{2}}]}{(x-S_1({\displaystyle \frac{13}{60}}))}^{2}dP+{\int }_C{(x-{\displaystyle \frac{1}{2}})}^{2}dP\hfill \\ \hfill & ={\displaystyle \frac{1}{75}}{V}_2+{\displaystyle \frac{1}{3}}W={\displaystyle \frac{\mathrm{29,903}}{\mathrm{19,710,000}}}=0.00151715.\hfill \end{array}$$

(13)

Since $V_n$ is the quantization error for n-means with $n\ge 5$, we have $V_n\le 0.00151715$. Notice that for any $a\in \alpha$, since $P(M(a|\alpha )>0$, there can not be more than one point from $\alpha$ in any of the open intervals $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$ and $({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$. We now show that $\alpha \cap C\ne \varnothing$. On the contrary, assume that $\alpha \cap C=\varnothing$. Let $j=max\{i:a_i<{\displaystyle \frac{2}{5}}\mathrm{and}1\le i\le n\}$. Then, $a_j<{\displaystyle \frac{2}{5}}$ and ${\displaystyle \frac{3}{5}}<a_{j+1}$. If $a_j\le {\displaystyle \frac{1}{2}}({\displaystyle \frac{1}{5}}+{\displaystyle \frac{2}{5}})={\displaystyle \frac{3}{10}}$, then

$$\begin{array}{cc}\hfill & \int \underset{a\in \alpha }{min}{(x-a)}^{2}dP\ge {\int }_C\underset{a\in \alpha }{min}{(x-a)}^{2}dP={\displaystyle \frac{1}{3}}{\int }_C\underset{a\in \{a_j,a_{j+1}\}}{min}{(x-a)}^{2}d\nu \hfill \\ \hfill & \ge {\displaystyle \frac{1}{3}}{\int }_{[{\displaystyle \frac{2}{5}},{\displaystyle \frac{9}{20}}]}{(x-{\displaystyle \frac{3}{10}})}^{2}d\nu +{\displaystyle \frac{1}{3}}{\int }_{[{\displaystyle \frac{9}{20}},{\displaystyle \frac{3}{5}}]}{(x-{\displaystyle \frac{3}{5}})}^{2}d\nu ={\displaystyle \frac{23}{7200}}=0.00319444>V_n,\hfill \end{array}$$

which yields a contradiction. Similarly, if $a_{j+1}\ge {\displaystyle \frac{7}{10}}$ a contradiction will arise. So, we can assume that ${\displaystyle \frac{3}{10}}<a_j<{\displaystyle \frac{2}{5}}<{\displaystyle \frac{3}{5}}<a_{j+1}<{\displaystyle \frac{7}{10}}$. Then, ${\displaystyle \frac{1}{2}}(a_{j-1}+a_j)<{\displaystyle \frac{1}{5}}$, implying $a_{j-1}<{\displaystyle \frac{2}{5}}-a_j\le {\displaystyle \frac{2}{5}}-{\displaystyle \frac{3}{10}}={\displaystyle \frac{1}{10}}$, and similarly, ${\displaystyle \frac{1}{2}}(a_{j+1}+a_{j+2})>{\displaystyle \frac{4}{5}}$ implies $a_{j+2}\ge {\displaystyle \frac{9}{10}}$. Thus, we see that

$$\begin{array}{cc}\hfill & V_n\ge {\int }_{J_{12}}{(x-{\displaystyle \frac{1}{10}})}^{2}dP+{\int }_{[{\displaystyle \frac{2}{5}},{\displaystyle \frac{1}{2}}]}{(x-{\displaystyle \frac{2}{5}})}^{2}dP+{\int }_{[{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{5}}]}{(x-{\displaystyle \frac{3}{5}})}^{2}dP+{\int }_{J_{21}}{(x-{\displaystyle \frac{9}{10}})}^{2}dP,\hfill \end{array}$$

which after simplification yields $V_n\ge {\displaystyle \frac{\mathrm{12,677}}{\mathrm{4,927,500}}}=0.0025727>V_n$, which is a contradiction. All these contradictions arise due to our assumption $\alpha \cap C=\varnothing$. Hence, we can conclude that $\alpha \cap C\ne \varnothing$. This completes the proof of the proposition. □

Lemma 19.

Let α be an optimal set of five-means. Then, α does not contain any point from $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})\cup ({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$.

Proof. 

Let $\alpha :=\{a_1<a_2<a_3<a_4<a_5\}$ be an optimal set of five-means. By Expression (13), we have $V_5\le 0.00151715$. Proposition 15 says that $\alpha$ contains points from $J_1$, C and $J_2$. Suppose that $\alpha$ contains a point from $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$. Then, two cases can arise:

Case 1. ${\displaystyle \frac{3}{10}}\le a_2<{\displaystyle \frac{2}{5}}$.

Due to symmetry of P about ${\displaystyle \frac{1}{2}}$, we can assume that $a_3={\displaystyle \frac{1}{2}}$ and ${\displaystyle \frac{3}{5}}<a_4\le {\displaystyle \frac{7}{10}}$. Then, ${\displaystyle \frac{1}{2}}(a_1+a_2)<{\displaystyle \frac{1}{5}}$ and ${\displaystyle \frac{4}{5}}<{\displaystyle \frac{1}{2}}(a_4+a_5)$, yielding $a_1<{\displaystyle \frac{1}{10}}$ and ${\displaystyle \frac{9}{10}}<a_5$, and so, we have

$$\begin{array}{cc}\hfill V_5& \ge 2({\int }_{J_{11}\cup S_1[{\displaystyle \frac{2}{5}},{\displaystyle \frac{1}{2}}]}{(x-S_1({\displaystyle \frac{13}{60}}))}^{2}dP+{\int }_{S_1[{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{5}}]\cup J_{12}}{(x-{\displaystyle \frac{1}{10}})}^{2}dP+{\int }_{[{\displaystyle \frac{2}{5}},{\displaystyle \frac{9}{20}}]}{(x-{\displaystyle \frac{2}{5}})}^{2}dP\hfill \\ \hfill & +{\int }_{[{\displaystyle \frac{9}{20}},{\displaystyle \frac{1}{2}}]}{(x-{\displaystyle \frac{1}{2}})}^{2}dP)={\displaystyle \frac{\mathrm{21,289}}{\mathrm{9,855,000}}}=0.00216022>V_5,\hfill \end{array}$$

which is a contradiction.

Case 2. ${\displaystyle \frac{1}{5}}<a_2\le {\displaystyle \frac{3}{10}}$.

Then, ${\displaystyle \frac{1}{2}}(a_2+a_3)>{\displaystyle \frac{2}{5}}$, implying $a_3>{\displaystyle \frac{4}{5}}-a_2\ge {\displaystyle \frac{4}{5}}-{\displaystyle \frac{3}{10}}={\displaystyle \frac{1}{2}}$, which leads to a contradiction, because by our assumption, $a_3={\displaystyle \frac{1}{2}}$. Therefore, $\alpha$ does not contain any point from $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$. Similarly, $\alpha$ does not contain any point from $({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$, which completes the proof of the lemma. □

Remark 9.

By Lemma 19 and due to the symmetry of P, it can be seen that the set $S_1({\alpha }_2)\cup \left\{{\displaystyle \frac{1}{2}}\right\}\cup S_2({\alpha }_2)$, where ${\alpha }_2$ is an optimal set of two-means for P, forms an optimal set of five-means.

Lemma 20.

Let α be an optimal set of six-means. Then, α does not contain any point from $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})\cup ({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$.

Proof. 

Let $\alpha :=\{a_1<a_2<a_3<a_4<a_6\}$ be an optimal set of six-means. Suppose that $\alpha$ contains a point from $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$. Proposition 15 says that $\alpha$ contains points from $J_1$, C and $J_2$. Consider the set of six points $\beta :=S_1({\alpha }_2)\cup S_2({\alpha }_2)\cup {\alpha }_2(\nu )$, where ${\alpha }_2\in {\mathcal{C}}_2(P)$, and ${\alpha }_2(\nu )\in {\mathcal{C}}_2(\nu )$. Then,

$$\begin{array}{cc}\hfill \int \underset{a\in \beta }{min}{(x-a)}^{2}dP& =4{\int }_{J_{11}\cup S_1[{\displaystyle \frac{2}{5}},{\displaystyle \frac{1}{2}}]}{(x-S_1({\displaystyle \frac{13}{60}}))}^{2}dP+{\displaystyle \frac{1}{3}}{V}_2(\nu )={\displaystyle \frac{\mathrm{21,481}}{\mathrm{19,710,000}}}=0.00108985.\hfill \end{array}$$

Since $V_6$ is an optimal set of six-means, $V_6\le 0.00108985$. Since $\alpha \cap C\ne \varnothing$, $\mathrm{card}(\alpha )$ is even, and P is symmetric about ${\displaystyle \frac{1}{2}}$, we can assume that $\alpha$ contains an equal number of points from both sides of the point ${\displaystyle \frac{1}{2}}$, implying $\mathrm{card}(\alpha \cap C)=2$, and $a_3,a_4\in C$. Since $\alpha$ contains a point from $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$, the following two cases can arise:

Case 1. ${\displaystyle \frac{3}{10}}\le a_2<{\displaystyle \frac{2}{5}}$.

Proceeding as Case 1 of Lemma 19 in this case, we have

$$\begin{array}{cc}\hfill V_6& \ge 2\left({\int }_{J_{11}\cup S_1[{\displaystyle \frac{2}{5}},{\displaystyle \frac{1}{2}}]}{(x-S_1({\displaystyle \frac{13}{60}}))}^{2}dP+{\int }_{S_1[{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{5}}]\cup J_{12}}{(x-{\displaystyle \frac{1}{10}})}^{2}dP\right)=0.00188245>V_6,\hfill \end{array}$$

which is a contradiction.

Case 2. ${\displaystyle \frac{1}{5}}<a_2\le {\displaystyle \frac{3}{10}}$.

Due to symmetry, we can assume that ${\displaystyle \frac{7}{10}}\le a_5<{\displaystyle \frac{4}{5}}$. Then, ${\displaystyle \frac{1}{2}}(a_2+a_3)>{\displaystyle \frac{2}{5}}$, implying $a_3>{\displaystyle \frac{4}{5}}-a_2\ge {\displaystyle \frac{4}{5}}-{\displaystyle \frac{3}{10}}={\displaystyle \frac{1}{2}}$. Similarly, ${\displaystyle \frac{1}{2}}(a_4+a_5)<{\displaystyle \frac{3}{5}}$, implying $a_4<{\displaystyle \frac{1}{2}}$, which leads to a contradiction as $a_3<a_4$.

Therefore, $\alpha$ does not contain any point from $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$. Likewise, $\alpha$ does not contain any point from $({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$, which completes the proof of the lemma. □

Remark 10.

By Lemma 20, it is easy to see that if ${\alpha }_6$ is an optimal set of six-means for P, then ${\alpha }_6=S_1({\alpha }_2)\cup {\alpha }_2(\nu )\cup S_2({\alpha }_2)$, where ${\alpha }_2$ is an optimal set of two-means for P, and the corresponding quantization error is given by $V_6={\displaystyle \frac{21,481}{19,710,000}}=0.00108985$.

Lemma 21.

Let α be an optimal set of seven-means. Then, α does not contain any point from $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})\cup ({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$.

Proof. 

Consider the set $\beta :=S_1({\alpha }_2)\cup S_2({\alpha }_3)\cup {\alpha }_2(\nu )$, where ${\alpha }_2\in {\mathcal{C}}_2(P)$, ${\alpha }_3\in {\mathcal{C}}_3(P)$, and ${\alpha }_2(\nu )\in {\mathcal{C}}_2(\nu )$. Then,

$$\begin{array}{cc}\hfill & \int \underset{a\in \beta }{min}{(x-a)}^{2}dP={\int }_{J_1}\underset{a\in S_1({\alpha }_2)}{min}{(x-a)}^{2}dP+{\int }_{J_2}\underset{a\in S_2({\alpha }_3)}{min}{(x-a)}^{2}dP+{\displaystyle \frac{1}{3}}{V}_2(\nu )\hfill \\ \hfill & ={\displaystyle \frac{1}{75}}{V}_2+{\displaystyle \frac{1}{75}}\left({\displaystyle \frac{1}{75}}V+{\displaystyle \frac{1}{3}}W+{\displaystyle \frac{1}{75}}V\right)+{\displaystyle \frac{1}{3}}{\displaystyle \frac{1}{1200}}={\displaystyle \frac{7273}{\mathrm{9,855,000}}}=0.000738001.\hfill \end{array}$$

Since $V_7$ is the quantization error for seven-means, we have $V_7\le 0.000738001$. Let $\alpha :=\{a_1<a_2<a_3<a_4<a_6<a_7\}$ be an optimal set of seven-means. Proposition 15 says that $\alpha$ contains points from $J_1$, C and $J_2$. Suppose that $\alpha$ contains a point from $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$. We now prove the following claim:

Claim 1. $\mathit{card}(\alpha \cap J_1)\ge 2$ and $\mathit{card}(\alpha \cap J_2)\ge 2$.

Suppose that $\mathrm{card}(\alpha \cap J_1)=1$. Then, two cases can arise:

Case 1. ${\displaystyle \frac{3}{10}}\le a_2<{\displaystyle \frac{1}{5}}$.

Then, ${\displaystyle \frac{1}{2}}(a_1+a_2)<{\displaystyle \frac{1}{5}}$ implying $a_1<{\displaystyle \frac{1}{10}}$, and so

$$\begin{array}{cc}\hfill V_7& \ge {\int }_{J_{11}\cup S_1[{\displaystyle \frac{2}{5}},{\displaystyle \frac{1}{2}}]}{(x-S_1({\displaystyle \frac{13}{60}}))}^{2}dP+{\int }_{S_1[{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{5}}]\cup J_{12}}{(x-{\displaystyle \frac{1}{10}})}^{2}dP\hfill \\ \hfill & ={\displaystyle \frac{\mathrm{37,103}}{\mathrm{39,420,000}}}=0.000941223>V_7,\hfill \end{array}$$

which is a contradiction.

Case 2. $a_2\le {\displaystyle \frac{3}{10}}$.

Then, ${\displaystyle \frac{1}{2}}(a_2+a_3)>{\displaystyle \frac{2}{5}}$ yielding $a_3>{\displaystyle \frac{1}{2}}$, and so

$$\begin{array}{cc}\hfill V_7& \ge {\int }_{J_1}\underset{a\in \{S_1({\displaystyle \frac{13}{60}}),S_1({\displaystyle \frac{47}{60}})\}}{min}{(x-a)}^{2}dP+{\int }_{[{\displaystyle \frac{2}{5}},{\displaystyle \frac{1}{2}}]}{(x-{\displaystyle \frac{1}{2}})}^{2}dP={\displaystyle \frac{1}{75}}{V}_2+{\displaystyle \frac{1}{3}}{\int }_{[{\displaystyle \frac{2}{5}},{\displaystyle \frac{1}{2}}]}{(x-{\displaystyle \frac{1}{2}})}^{2}d\nu ,\hfill \end{array}$$

which implies $V_7\ge {\displaystyle \frac{\mathrm{18,953}}{\mathrm{19,710,000}}}=0.000961593>V_7$, and so a contradiction arises.

Thus, we can assume that $\mathrm{card}(\alpha \cap J_1)\ge 2$. Similarly, we can show that $\mathrm{card}(\alpha \cap J_2)\ge 2$. Thus, the claim is true.

Next, we prove the following claim:

Claim 2. $\mathit{card}(\alpha \cap C)=2$.

By the hypothesis, $\alpha$ contains a point from $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$; by Claim 1, $\mathrm{card}(\alpha \cap J_1)\ge 2$ and $\mathrm{card}(\alpha \cap J_2)\ge 2$. So, we can assume that $card(\alpha \cap C)\le 2$. Suppose that $card(\alpha \cap C)=1$. Let $a_j=max\{a_i:a_i<{\displaystyle \frac{2}{5}}\mathrm{and}1\le i\le 7\}$. Then, $a_j<{\displaystyle \frac{2}{5}}$. Consider the following two cases:

Case A. ${\displaystyle \frac{3}{10}}\le a_j<{\displaystyle \frac{2}{5}}$.

Then, ${\displaystyle \frac{1}{2}}(a_{j-1}+a_j)<{\displaystyle \frac{1}{5}}$ implying $a_{j-1}<{\displaystyle \frac{1}{10}}$. First, suppose that $\alpha$ does not contain any point from $({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$. Proceeding as (12), we have

$$\begin{array}{c}\hfill \underset{a\in \{a_j,a_{j+1}\}}{min}\left\{{\int }_C{(x-a)}^{2}dP:{\displaystyle \frac{3}{10}}<a_j\le {\displaystyle \frac{2}{5}},{\displaystyle \frac{2}{5}}\le a_{j+1}\le {\displaystyle \frac{3}{5}},\mathrm{and}{\displaystyle \frac{2}{5}}\le {\displaystyle \frac{a_j+a_{j+1}}{2}}\right\}={\displaystyle \frac{1}{2025}},\end{array}$$

which occurs when $a_j={\displaystyle \frac{2}{5}}$ and $a_{j+1}={\displaystyle \frac{8}{15}}$. Then,

$$\begin{array}{c}\hfill V_7\ge {\int }_{S_1[{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{5}}]\cup J_{12}}{(x-{\displaystyle \frac{1}{10}})}^{2}dP+{\displaystyle \frac{1}{2025}}={\displaystyle \frac{1457}{\mathrm{1,182,600}}}=0.00123203>V_7,\end{array}$$

which gives a contradiction. Next, suppose that $\alpha$ contains a point from $({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$, i.e., $a_{j+2}\in ({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$. Then,

$$\begin{array}{cc}\hfill \underset{a\in \{a_j,a_{j+1},a_{j+2}\}}{min}& \{{\int }_C{(x-a)}^{2}dP:a_j\in [{\displaystyle \frac{3}{10}},{\displaystyle \frac{2}{5}}),a_{j+1}\in [{\displaystyle \frac{2}{5}},{\displaystyle \frac{3}{5}}],a_{j+2}\in ({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}}),\hfill \\ \hfill & \mathrm{and}{\displaystyle \frac{2}{5}}\le {\displaystyle \frac{a_j+a_{j+1}}{2}}<{\displaystyle \frac{a_{j+1}+a_{j+2}}{2}}\le {\displaystyle \frac{3}{5}}\}={\displaystyle \frac{1}{3600}},\hfill \end{array}$$

which occurs when $a_j={\displaystyle \frac{2}{5}},a_{j+1}={\displaystyle \frac{1}{2}}$ and $a_{j+2}={\displaystyle \frac{3}{5}}$, and so

$$\begin{array}{c}\hfill V_7\ge {\int }_{S_1[{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{5}}]\cup J_{12}}{(x-{\displaystyle \frac{1}{10}})}^{2}dP+{\displaystyle \frac{1}{3600}}={\displaystyle \frac{89}{\mathrm{87,600}}}=0.00101598>V_7,\end{array}$$

which leads to a contradiction. Thus, a contradiction arises in Case 3.

Case B. ${\displaystyle \frac{1}{5}}<a_j\le {\displaystyle \frac{3}{10}}$.

Suppose that $\alpha$ does not contain any point from $({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$. Then,

$$\begin{array}{c}\hfill \underset{a\in \{a_j,a_{j+1}\}}{min}\left\{{\int }_C{(x-a)}^{2}dP:{\displaystyle \frac{1}{5}}<a_j\le {\displaystyle \frac{3}{10}},{\displaystyle \frac{2}{5}}\le a_{j+1}\le {\displaystyle \frac{3}{5}},\mathrm{and}{\displaystyle \frac{2}{5}}\le {\displaystyle \frac{a_j+a_{j+1}}{2}}\right\}={\displaystyle \frac{1}{900}},\end{array}$$

which occurs when $a_j={\displaystyle \frac{3}{10}}$ and $a_{j+1}={\displaystyle \frac{1}{2}}$ yielding $V_7\ge {\displaystyle \frac{1}{900}}>0.00111111>V_7$, which is a contradiction. Next, suppose that $\alpha$ contains a point from $({\displaystyle \frac{3}{5}},{\displaystyle \frac{7}{10}}]$. This case can be considered as a reflection of the last part of Case A. Thus, if $\alpha$ contains a point from $({\displaystyle \frac{3}{5}},{\displaystyle \frac{7}{10}}]$, a contradiction arises. Now, suppose that $\alpha$ contains a point from $[{\displaystyle \frac{7}{10}},{\displaystyle \frac{4}{5}})$, i.e., $a_{j+2}\in [{\displaystyle \frac{7}{10}},{\displaystyle \frac{4}{5}})$. Then,

$$\begin{array}{cc}\hfill \underset{a\in \{a_j,a_{j+1},a_{j+2}\}}{min}& \{{\int }_C{(x-a)}^{2}dP:a_j\in ({\displaystyle \frac{1}{5}},{\displaystyle \frac{3}{10}}],a_{j+1}\in [{\displaystyle \frac{2}{5}},{\displaystyle \frac{3}{5}}],a_{j+2}\in [{\displaystyle \frac{7}{10}},{\displaystyle \frac{4}{5}}),\hfill \\ \hfill & \mathrm{and}{\displaystyle \frac{2}{5}}\le {\displaystyle \frac{a_j+a_{j+1}}{2}}<{\displaystyle \frac{a_{j+1}+a_{j+2}}{2}}\le {\displaystyle \frac{3}{5}}\}={\displaystyle \frac{1}{900}},\hfill \end{array}$$

which occurs when $a_j={\displaystyle \frac{3}{10}},a_{j+1}={\displaystyle \frac{1}{2}}$ and $a_{j+2}={\displaystyle \frac{7}{10}}$. Then,

$$\begin{array}{c}\hfill V_7\ge {\displaystyle \frac{1}{900}}=0.00111111>V_7\end{array}$$

which yields a contradiction.

Hence, $\mathrm{card}(\alpha \cap C)=2$, i.e., Claim 2 is true.

By the hypothesis $\alpha$ contains a point from $({\displaystyle \frac{1}{5}},{\displaystyle \frac{3}{5}})$; by Claim 1 and Claim 2, we have $\mathrm{card}(\alpha \cap J_1)\ge 2$, $\mathrm{card}(\alpha \cap J_2)\ge 2$, and $\mathrm{card}(\alpha \cap C)=2$. So, we can assume that $\mathrm{card}(\alpha \cap J_1)=\mathrm{card}(\alpha \cap J_2)=2$, $\mathrm{card}(\alpha \cap C)=2$, and $\alpha$ does not contain any point from $({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$. Then, the following two cases arise:

Case 5. ${\displaystyle \frac{3}{10}}\le a_3<{\displaystyle \frac{2}{5}}$.

Then,

$$\begin{array}{cc}\hfill V_7& \ge {\int }_{S_1[{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{5}}]\cup J_{12}}{(x-{\displaystyle \frac{1}{10}})}^{2}dP+{\int }_{J_2}\underset{a\in S_2({\alpha }_2)}{min}{(x-a)}^{2}dP=0.00114424>V_7,\hfill \end{array}$$

which is a contradiction.

Case 6. ${\displaystyle \frac{1}{5}}<a_3\le {\displaystyle \frac{3}{10}}$.

Then,

$$\begin{array}{cc}\hfill \underset{a\in \{a_3,a_4,a_5\}}{min}& \left\{{\int }_C{(x-a)}^{2}dP:{\displaystyle \frac{1}{5}}<a_3\le {\displaystyle \frac{3}{10}},{\displaystyle \frac{2}{5}}\le a_4<a_5\le {\displaystyle \frac{3}{5}},\mathrm{and}{\displaystyle \frac{2}{5}}\le {\displaystyle \frac{a_3+a_4}{2}}\right\}={\displaystyle \frac{1}{1620}},\hfill \end{array}$$

which occurs when $a_3={\displaystyle \frac{3}{10}},a_4={\displaystyle \frac{1}{2}}$, and $a_5={\displaystyle \frac{17}{30}}$. Thus, in this case,

$$V_7\ge {\int }_C\underset{a\in \{a_3,a_4,a_5\}}{min}{(x-a)}^{2}dP+{\int }_{J_2}\underset{a\in S_2({\alpha }_2)}{min}{(x-a)}^{2}dP={\displaystyle \frac{1}{1620}}+{\displaystyle \frac{1}{75}}{V}_2={\displaystyle \frac{\mathrm{60,509}}{\mathrm{59,130,000}}},$$

i.e., $V_7\ge {\displaystyle \frac{\mathrm{60,509}}{\mathrm{59,130,000}}}=0.00102332>V_7$, which gives a contradiction.

Therefore, we can assume that $\alpha$ does not contain any point from the interval $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$. Reflecting the situation with respect to the point ${\displaystyle \frac{1}{2}}$, we can also assume that $\alpha$ does not contain any point from the interval $({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$. Hence, the proof of the lemma is complete. □

Remark 11.

In Lemma 21, we have proved that if α is an optimal set of seven-means, then $\mathit{card}(\alpha \cap J_1)\ge 2$, $\mathit{card}(\alpha \cap J_2)\ge 2$, and $\mathit{card}(\alpha \cap C)=2$. Moreover, α does not contain any point from $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})\cup ({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$. Therefore, we can assume that if α is an optimal set of seven-means, then $\alpha =S_1({\alpha }_3)\cup S_2({\alpha }_2)\cup {\alpha }_2(\nu )$, or $\alpha =S_1({\alpha }_2)\cup S_2({\alpha }_3)\cup {\alpha }_2(\nu )$, where ${\alpha }_2\in {\mathcal{C}}_2(P)$, ${\alpha }_3\in {\mathcal{C}}_3(P)$, and ${\alpha }_2(\nu )\in {\mathcal{C}}_2(\nu )$, and the corresponding quantization error is $V_7={\displaystyle \frac{7273}{9,855,000}}=0.000738001$.

Proposition 16.

Let α be an optimal set of n-means with $n\ge 3$. Then, α does not contain any point from $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})\cup ({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$.

Proof. 

By Proposition 14 and Lemmas 18–21, the proposition is true for $3\le n\le 7$. We now prove that the proposition is true for all $n\ge 8$. Let $\alpha :=\{a_1<a_2<\cdots <a_n\}$ be on optimal set of n means for $n\ge 8$. Consider the set $\beta =S_1({\alpha }_3)\cup S_2({\alpha }_3)\cup {\alpha }_2(\nu )$, where ${\alpha }_3\in {\mathcal{C}}_3(P)$ and ${\alpha }_2(\nu )\in {\mathcal{C}}_2(\nu )$. Then,

$$\begin{array}{cc}\hfill & \int \underset{a\in \beta }{min}{(x-a)}^{2}dP\hfill \\ \hfill & =2\left({\int }_{J_{11}}{(x-S_{11}({\displaystyle \frac{1}{2}}))}^2+{\int }_{C_1}{(x-S_1({\displaystyle \frac{1}{2}}))}^{2}dP+{\int }_{J_{12}}{(x-S_{12}({\displaystyle \frac{1}{2}}))}^2\right)+{\int }_C\underset{a\in {\alpha }_2(\nu )}{min}{(x-a)}^{2}dP\hfill \\ \hfill & =2\left({\displaystyle \frac{1}{{75}^2}}V+{\displaystyle \frac{1}{75}}{\displaystyle \frac{1}{3}}W+{\displaystyle \frac{1}{{75}^2}}V\right)+{\displaystyle \frac{1}{3}}{\displaystyle \frac{1}{1200}}={\displaystyle \frac{2537}{\mathrm{6,570,000}}}=0.000386149,\hfill \end{array}$$

Since $V_n$ is the quantization error for n-means with $n\ge 8$, we have $V_n\le V_8\le 0.000386149$. For the sake of contradiction, assume that $\alpha$ contains a point from the open interval $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$. Recall that there can not be more than one point from $\alpha$ in any of the open intervals $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$ and $({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$. Suppose that $a_j\in ({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$, where $j=max\{i:a_i<{\displaystyle \frac{2}{5}}\mathrm{and}1\le i\le n\}$. Then, following cases can arise:

Case 1. ${\displaystyle \frac{3}{10}}\le a_j<{\displaystyle \frac{2}{5}}$.

Then, $a_{j-1}<{\displaystyle \frac{1}{10}}$ yielding

$$V_n\ge {\int }_{S_1[{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{5}}]\cup J_{12}}{(x-{\displaystyle \frac{1}{10}})}^{2}dP={\displaystyle \frac{97}{\mathrm{131,400}}}=0.000738204>V_n,$$

which is a contradiction.

Case 2. ${\displaystyle \frac{1}{5}}<a_j\le {\displaystyle \frac{3}{10}}$.

Then, ${\displaystyle \frac{1}{2}}(a_j+a_{j+1})>{\displaystyle \frac{2}{5}}$ implies $a_{j+1}>{\displaystyle \frac{4}{5}}-a_j\ge {\displaystyle \frac{4}{5}}-{\displaystyle \frac{3}{10}}={\displaystyle \frac{1}{2}}$. Thus,

$$\begin{array}{cc}\hfill V_n& \ge {\int }_{[{\displaystyle \frac{2}{5}},{\displaystyle \frac{1}{2}}]}\underset{a\in \{a_j,a_{j+1}\}}{min}{(x-a)}^{2}dP\ge {\int }_{[{\displaystyle \frac{2}{5}},{\displaystyle \frac{1}{2}}]}{(x-{\displaystyle \frac{1}{2}})}^{2}dP={\displaystyle \frac{1}{1800}}=0.000555556>V_n,\hfill \end{array}$$

which leads to a contradiction.

Similarly, we can show that for any $a\in \alpha$ if $a\in ({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$, then a contradiction arises. Thus, the proof of the proposition is complete. □

4.2. Optimal Sets and the Quantization Error for a Given Sequence $F(n)$

In this subsection we first define the two sequences ${\left\{a(n)\right\}}_{n\ge 0}$ and ${\left\{F(n)\right\}}_{n\ge 0}$. These two sequences play important role in the rest of the paper.

Definition 2.

Define the sequence ${\left\{a(n)\right\}}_{n\ge 0}$ such that $a(0)=0$, and $a(n)=2n-1$ for all $n\ge 1$. Define the sequence ${\left\{F(n)\right\}}_{n\ge 0}$ such that $F(n)=4^n+2^{n+1}$, i.e.,

$${\left\{F(n)\right\}}_{n\ge 0}=\{3,8,24,80,288,1088,4224,16640,66048,263168,1050624,\cdots \}.$$

Lemma 22.

Let ${\left\{a(n)\right\}}_{n\ge 0}$ and ${\left\{F(n)\right\}}_{n\ge 0}$ be the sequences defined by Definition 2. Then, $F(n+1)=2^{a(n+1)}+2F(n)$.

Proof. 

We have, $2^{a(n+1)}+2F(n)=2^{2n+1}+2(4^n+2^{n+1})=4^{n+1}+2^{n+2}=F(n+1)$, and thus the lemma follows. □

For $n\in \mathbb{N}$, we identify the sequence of sets ${\alpha }_{2^{a(n)}}(\nu ),$ ${\cup }_{\omega \in I}{S}_{\omega }({\alpha }_{2^{a(n-1)}}(\nu )),$ ${\cup }_{\omega \in I^2}{S}_{\omega }({\alpha }_{2^{a(n-2)}}(\nu )),$ $\cdots ,{\cup }_{\omega \in I^{n-2}}{S}_{\omega }({\alpha }_{2^{a(2)}}(\nu )),$ ${\cup }_{\omega \in I^{n-1}}{S}_{\omega }({\alpha }_{2^{a(1)}}(\nu )),{\cup }_{\omega \in I^n}{S}_{\omega }({\alpha }_{2^{a(0)}}(\nu ))$, and $\{S_{\omega }({\displaystyle \frac{1}{2}}):\omega \in I^{n+1}\}$, respectively, by $S(n)$, $S(n-1)$, $S(n-2)$, ⋯, $S(2)$, $S(1)$, $S(0)$, and $S(n+1)$. For $0\le \ell \le n$, we write

$$S^{(2)}(\ell ):={\cup }_{\omega \in I^{n-\ell }}{S}_{\omega }({\alpha }_{2^{a(\ell )+1}}(\nu ))\mathrm{and}{S}^{(2)(2)}(\ell ):={\cup }_{\omega \in I^{n-\ell }}{S}_{\omega }({\alpha }_{2^{a(\ell )+2}}(\nu )).$$

Further, we write $S^{(2)}(n+1):=\{S_{\omega }({\alpha }_2(P)):\omega \in I^{n+1}\}=\{S_{\omega }({\displaystyle \frac{13}{60}}),S_{\omega }({\displaystyle \frac{47}{60}}):\omega \in I^{n+1}\}$, and $S^{(2)(2)}(n+1):={\cup }_{\omega \in I^{n+1}}{S}_{\omega }({\alpha }_{2^{a(0)}}(\nu ))\cup \{S_{\omega }({\displaystyle \frac{1}{2}}):\omega \in I^{n+2}\}.$ Moreover, for any $\ell \in \mathbb{N}\cup \left\{0\right\}$, if $A:=S(i)$, we identify $S^{(2)}(i)$ and $S^{(2)(2)}(i)$, respectively, by $A^{(2)}$ and $A^{(2)(2)}$. For $n\in \mathbb{N}\cup \left\{0\right\}$, set

$${\alpha }_{F(n)}:=S(n)\cup S(n-1)\cup S(n-2)\cdots S(1)\cup S(0)\cup S(n+1),$$

(14)

and

$$SF(n):=\{S(n),S(n-1),S(n-2),\cdots ,S(1),S(0),S(n+1)\}.$$

In addition, write

$$\begin{array}{c}\hfill S{F}^{*}(n):=\{S(n),S(n-1),\cdots ,S(0),S(n+1),S^{(2)}(n),S^{(2)}(n-1),\cdots ,S^{(2)}(1),S^{(2)}(n+1)\}.\end{array}$$

For any element $a\in A\in S{F}^{*}(n)$, by the Voronoi region of a it is meant the Voronoi region of a with respect to the set ${\cup }_{B\in S{F}^{*}(n)}B$. Similarly, for any $a\in A\in SF(n)$, by the Voronoi region of a, the Voronoi region of a is meant with respect to the set ${\cup }_{B\in SF(n)}B$. Notice that if $a,b\in A$, where $A\in SF(n)$ or $A\in S{F}^{*}(n)$, the error contributed by a in the Voronoi region of a is equal to the error contributed by b in the Voronoi region of b. Let us now define an order ≻ on the set $S{F}^{*}(n)$ as follows: for $A,B\in S{F}^{*}(n)$ by $A\succ B$, it is meant that the error contributed by any element $a\in A$ in the Voronoi region of a is larger than the error contributed by any element $b\in B$ in the Voronoi region of b. Similarly, we define the order relation ≻ on the set $SF(n)$.

Remark 12.

By Definition 2, we have

$${\alpha }_{F(n)}=S_1({\alpha }_{F(n-1)})\cup {\alpha }_{2^{a(n)}}(\nu )\cup S_2({\alpha }_{F(n-1)}).$$

Lemma 23.

Let ≻ be the order relation on $S{F}^{*}(n)$. Then,

$(i)$$S(n)\succ S(n-1)\succ \cdots \succ S(2)\succ S(1)$ for all $n\ge 2$, and $S^{(2)}(n)\succ S^{(2)}(n-1)\succ \cdots \succ S^{(2)}(2)\succ S^{(2)}(1)$ for all $n\ge 3$.

$(ii)$$S(1)\succ S^{(2)}(n)$ for all $0\le n\le 14$.

$(iii)$$S(n-13)\succ S^{(2)}(n)\succ S(n-14)$ for all $n\ge 14$.

$(iv)$$S^{(2)}(2)\succ S(n+1)$ for all $n\ge 2$, and $S(1)\succ S(n+1)\succ S^{(2)}(1)\succ S(0)\succ S^{(2)}(n+1)\succ S^{(2)}(0)$ for all $n\ge 1$. On the other hand, $S(1)\succ S(0)\succ S^{(2)}(1)\succ S^{(2)}(0)$ for $n=0$.

Proof.

$(i)$ Let $1\le p<q\le n$. The distortion error due to any element in the set $S(q):={\cup }_{\omega \in I^{n-q}}({\alpha }_{2^{a(q)}}(\nu ))$ is given by ${\displaystyle \frac{1}{{75}^{n-q}}}{\displaystyle \frac{1}{3}}{\displaystyle \frac{W}{2^{3(a(q))}}}={\displaystyle \frac{1}{{75}^{n-q}}}{\displaystyle \frac{1}{3}}{\displaystyle \frac{W}{2^{6q-1}}}$. Similarly, the distortion error due to any element in the set $S(p)$ is given by ${\displaystyle \frac{1}{{75}^{n-p}}}{\displaystyle \frac{1}{3}}{\displaystyle \frac{W}{2^{6p-1}}}$. Thus, $S(q)\succ S(p)$ is true if ${\displaystyle \frac{1}{{75}^{n-q}}}{\displaystyle \frac{1}{3}}{\displaystyle \frac{W}{2^{6q-1}}}>{\displaystyle \frac{1}{{75}^{n-p}}}{\displaystyle \frac{1}{3}}{\displaystyle \frac{W}{2^{6p-1}}}$, i.e., if ${\left({\displaystyle \frac{75}{64}}\right)}^{q-p}>1$, which is clearly true since $q>p$. Hence, $S(n)\succ S(n-1)\succ \cdots \succ S(2)\succ S(1)$ for all $n\ge 2$, and similarly, we can prove that $S^{(2)}(n)\succ S^{(2)}(n-1)\succ \cdots \succ S^{(2)}(2)\succ S^{(2)}(1)$ for all $n\ge 3$.

$(ii)$$S(1)\succ S^{(2)}(n)$ is true if ${\displaystyle \frac{1}{{75}^{n-1}}}{\displaystyle \frac{1}{3}}{\displaystyle \frac{W}{2^{3a(1)}}}>{\displaystyle \frac{1}{3}}{\displaystyle \frac{W}{2^{3(a(n)+1)}}}$, i.e., if ${\displaystyle \frac{75}{8}}{\left({\displaystyle \frac{64}{75}}\right)}^n>1$, which is true for $0\le n\le 14$.

$(iii)$ For $0\le n\le 14$, we see that $S(n-13)\succ S^{(2)}(n)\succ S(n-14)$ is true if ${\displaystyle \frac{1}{{75}^{13}}}{\displaystyle \frac{1}{3}}{\displaystyle \frac{W}{2^{3a(n-13)}}}>{\displaystyle \frac{1}{3}}{\displaystyle \frac{W}{2^{3(a(n)+1)}}}>{\displaystyle \frac{1}{{75}^{14}}}{\displaystyle \frac{1}{3}}{\displaystyle \frac{W}{2^{3a(n-14)}}}$, i.e., if ${\displaystyle \frac{1}{{75}^{13}}}{\displaystyle \frac{1}{2^{6n-81}}}>{\displaystyle \frac{1}{2^{6n}}}>{\displaystyle \frac{1}{{75}^{14}}}{\displaystyle \frac{1}{2^{6n-87}}}$, i.e., if ${\displaystyle \frac{2^{81}}{{75}^{13}}}>1>{\displaystyle \frac{2^{87}}{{75}^{14}}}$, which is obviously true.

$(iv)$ For $n\ge 2$, the relation $S^{(2)}(2)\succ S(n+1)$ is true if ${\displaystyle \frac{1}{{75}^{n-2}}}{\displaystyle \frac{1}{3}}{\displaystyle \frac{W}{2^{3(a(2)+1)}}}>{\displaystyle \frac{1}{{75}^{n+1}}}V$, i.e., if ${\displaystyle \frac{{75}^3}{3}}{\displaystyle \frac{W}{2^{12}}}>V$, which is obvious. Similarly, we can prove the rest of the inequalities. □

We now give the following proposition.

Proposition 17.

Let ≻ be the order relation on $S{F}^{*}(n)$. Then,

$(i)$$S(n)\succ S(n-1)\succ S(n-2)\succ \cdots \succ S(2)$$\succ S(1)\succ S^{(2)}(n)$$\succ S^{(2)}(n-1)\succ S^{(2)}(n-2)\succ \cdots \succ$$S^{(2)}(2)\succ S(n+1)\succ S^{(2)}(1)\succ S(0)\succ$$S^{(2)}(n+1)\succ S^{(2)}(0)$ for all $2\le n\le 14$.

$(ii)$$S(n)\succ S(n-1)\succ S(n-2)\succ \cdots \succ S(n-13)\succ S^{(2)}(n)\succ S(n-14)\succ$$S^{(2)}(n-1)\succ S(n-15)\succ S^{(2)}(n-2)\succ S(n-16)\succ$$\cdots \succ S^{(2)}(17)$$\succ S(3)\succ S^{(2)}(16)\succ S(2)\succ$$S^{(2)}(15)\succ S(1)\succ$$S^{(2)}(14)\succ S^{(2)}(13)$$\succ \cdots \succ S^{(2)}(2)\succ S(n+1)\succ S^{(2)}(1)$$\succ S(0)\succ S^{(2)}(n+1)\succ$$S^{(2)}(0)$ for all $n\ge 15$.

Proof. 

The proof of the proposition follows by combining the inequalities in Lemma 23. □

Lemma 24.

Let ${\alpha }_{F(n)}$ and $SF(n)$ be the sets as defined before. Then,

$${\alpha }_{F(n+1)}=\left(\underset{A\in (SF(n)\setminus S(0))}{\cup }{A}^{(2)(2)}\right)\cup S^{(2)}(0).$$

Proof. 

For $S(n),S(n-1),\cdots ,S(2),S(0),S(n+1)\in SF(n)$, we have

$$\begin{array}{cc}\hfill S^{(2)(2)}(n)& ={\alpha }_{2^{a(n)+2}}(\nu )={\alpha }_{2^{a(n+1)}}(\nu ),\hfill \\ \hfill S^{(2)(2)}(n-1)& ={\cup }_{\omega \in I}{S}_{\omega }({\alpha }_{2^{a(n-1)+2}}(\nu ))={\cup }_{\omega \in I}{S}_{\omega }({\alpha }_{2^{a(n)}}(\nu )),\hfill \\ \hfill & ⋮\hfill \\ \hfill S^{(2)(2)}(1)& ={\cup }_{\omega \in I^{n-1}}{S}_{\omega }({\alpha }_{2^{a(1)+2}}(\nu ))={\cup }_{\omega \in I^{n-1}}{S}_{\omega }({\alpha }_{2^{a(2)}}(\nu )),\hfill \\ \hfill S^{(2)}(0)& ={\cup }_{\omega \in I^n}{S}_{\omega }({\alpha }_{2^{a(0)+1}}(\nu ))={\cup }_{\omega \in I^n}{S}_{\omega }({\alpha }_{2^{a(1)}}(\nu )),\hfill \\ \hfill S^{(2)(2)}(0)& ={\cup }_{\omega \in I^{n+1}}{S}_{\omega }({\alpha }_{2^{a(0)}}(\nu ))\cup \{S_{\omega }({\displaystyle \frac{1}{2}}):\omega \in I^{n+2}\}\hfill \\ \hfill & ={\cup }_{\omega \in I^{n+1}}{S}_{\omega }({\alpha }_{2^{a(0)}}(\nu ))\cup \{S_{\omega }({\displaystyle \frac{1}{2}}):\omega \in I^{n+2}\}.\hfill \end{array}$$

Thus, by the expression (14), the proof of the lemma follows. □

We now prove the following lemma.

Lemma 25.

For any two sets $A,B\in SF(n)$, let $A\succ B$. Then, the distortion error due to the set $(SF(n)\setminus A)\cup A^{(2)}\cup B$ is less than the distortion error due to the set $(SF(n)\setminus B)\cup B^{(2)}\cup A$.

Proof. 

By Lemma 23 and Proposition 17, for all $n\ge 0$, we have

$$S(n)\succ S(n-1)\succ S(n-2)\succ \cdots \succ S(1)\succ S(n+1)\succ S(0).$$

Let $V({\alpha }_{F(n)})$ be the distortion error due to the set ${\alpha }_{F(n)}$ with respect to the condensation measure P. First, take $A=S(a(k))$ and $B=S(a(k^{\prime }))$ for some $1\le k^{\prime }<k\le n$. Then, the distortion error due to the set $({\alpha }_{F(n)}\setminus A)\cup A^{(2)}\cup B$ is given by

$$\begin{array}{cc}\hfill & V({\alpha }_{F(n)})-{({\displaystyle \frac{2}{75}})}^{n-k}{\displaystyle \frac{1}{3}}{\displaystyle \frac{W}{2^{2a(k)}}}+{({\displaystyle \frac{2}{75}})}^{n-k}{\displaystyle \frac{1}{3}}{\displaystyle \frac{W}{2^{2(a(k)+1)}}}+{({\displaystyle \frac{2}{75}})}^{n-k^{\prime }}{\displaystyle \frac{1}{3}}{\displaystyle \frac{W}{2^{2a(k^{\prime })}}}\hfill \\ \hfill & =V({\alpha }_{F(n)})-{({\displaystyle \frac{2}{75}})}^{n-k}{\displaystyle \frac{1}{3}}{\displaystyle \frac{3W}{2^{4k}}}+{({\displaystyle \frac{2}{75}})}^{n-k^{\prime }}{\displaystyle \frac{1}{3}}{\displaystyle \frac{W}{2^{4k^{\prime }}}}\hfill \end{array}$$

(15)

Similarly, the distortion error due to the set $({\alpha }_{F(n)}\setminus B)\cup B^{(2)}\cup A$ is

$$\begin{array}{c}\hfill V({\alpha }_{F(n)})-{({\displaystyle \frac{2}{75}})}^{n-k^{\prime }}{\displaystyle \frac{1}{3}}{\displaystyle \frac{3W}{2^{4k^{\prime }}}}+{({\displaystyle \frac{2}{75}})}^{n-k}{\displaystyle \frac{1}{3}}{\displaystyle \frac{W}{2^{4k}}}.\end{array}$$

(16)

Thus, (15) will be less than (16) if ${({\displaystyle \frac{2}{75}})}^{n-k}{\displaystyle \frac{1}{3}}{\displaystyle \frac{7W}{2^{4k}}}>{({\displaystyle \frac{2}{75}})}^{n-k^{\prime }}{\displaystyle \frac{1}{3}}{\displaystyle \frac{7W}{2^{4k^{\prime }}}}$, i.e., if ${({\displaystyle \frac{75}{32}})}^{k-k^{\prime }}>1$, which is clearly true since $k^{\prime }<k$. Now, take $A=S(k)$ for $1\le k\le n$, and $B=S(n+1)$. Then, the distortion error due to the set $({\alpha }_{F(n)}\setminus A)\cup A^{(2)}\cup B$ is less than the distortion error due to the set $({\alpha }_{F(n)}\setminus B)\cup B^{(2)}\cup A$ if

$$V_{F(n)}-{({\displaystyle \frac{2}{75}})}^{n-k}{\displaystyle \frac{1}{3}}{\displaystyle \frac{3W}{2^{4k}}}+{({\displaystyle \frac{2}{75}})}^{n+1}V<V_{F(n)}-{({\displaystyle \frac{2}{75}})}^{n+1}V+{({\displaystyle \frac{2}{75}})}^{n+1}{\displaystyle \frac{1}{2}}{V}_2+{({\displaystyle \frac{2}{75}})}^{n-k}{\displaystyle \frac{1}{3}}{\displaystyle \frac{4W}{2^{4k}}},$$

i.e., if ${({\displaystyle \frac{2}{75}})}^{n-k}{\displaystyle \frac{1}{3}}{\displaystyle \frac{7W}{2^{4k}}}>{({\displaystyle \frac{2}{75}})}^{n+1}(2V-{\displaystyle \frac{1}{2}}{V}_2)$, i.e., if ${({\displaystyle \frac{75}{32}})}^k{\displaystyle \frac{7W}{3}}>{\displaystyle \frac{2}{75}}(2V-{\displaystyle \frac{1}{2}}{V}_2)$, which is obviously true for $k\ge 1$. Similarly, if $A=S(n+1)$ and $B=S(0)$, we can show that the distortion error due to the set $({\alpha }_{F(n)}\setminus A)\cup A^{(2)}\cup B$ is less than the distortion error due to the set $({\alpha }_{F(n)}\setminus B)\cup B^{(2)}\cup A$. Thus, the proof of the lemma is complete. □

Using the similar technique as Lemma 25, the following lemma can be proved.

Lemma 26.

For any two sets $A,B\in S{F}^{*}(n)$, let $A\succ B$. Then, the distortion error due to the set $(S{F}^{*}(n)\setminus A)\cup A^{(2)}\cup B$ is less than the distortion error due to the set $(S{F}^{*}(n)\setminus B)\cup B^{(2)}\cup A$.

The following lemma is useful.

Lemma 27.

Let α be an optimal set of n-means with $\alpha \cap ({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})=\varnothing =\alpha \cap ({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$, and let $i=1,2$. Set ${\beta }_i:=\alpha \cap J_i$, $n_i:=\mathit{card}({\beta }_i)$, ${\beta }_c:=\alpha \cap C$, and $n_c:=\mathit{card}({\beta }_c)$. Then, $S_i^{-1}({\alpha }_i)$ is an optimal set of $n_i$-means, and ${\beta }_c$ is an optimal set of $n_c$-means for ν. Moreover, $V_n(P)={\displaystyle \frac{1}{75}}(V_{n_1}(P)+V_{n_2}(P))+{\displaystyle \frac{1}{3}}{V}_{n_c}(\nu ).$

Proof. 

Proceeding in the similar way as ([29], Lemma 3.6), we can show that $S_i^{-1}({\beta }_i)$ is an optimal set of $n_i$-mean for $i=1,2$. Proof of ${\beta }_c$ is an optimal set of $n_c$-means for $\nu$ is also similar with a minor modification. $\alpha \cap ({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})=\varnothing =\alpha \cap ({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$ implies $n=n_1+n_2+n_c$ and $\alpha ={\beta }_1\cup {\beta }_2\cup {\beta }_c$. For $i=1,2$, we have

$${\int }_{J_i}\underset{a\in {\beta }_i}{min}{(x-a)}^{2}dP={\displaystyle \frac{1}{3}}{\int }_{J_i}\underset{a\in {\beta }_i}{min}{(x-a)}^{2}d(P\circ S_i^{-1})={\displaystyle \frac{1}{75}}\int \underset{a\in S_i^{-1}({\beta }_i)}{min}{(x-a)}^{2}dP={\displaystyle \frac{1}{75}}{V}_{n_i}(P).$$

Moreover,

$$V_{n_c}(\nu )={\int }_{{\beta }_c}\underset{a\in {\beta }_c}{min}{(x-a)}^{2}d\nu .$$

Thus, we have

$$V_n(P)=\sum _{i=i}^2{\int }_{J_i}\underset{a\in {\beta }_i}{min}{(x-a)}^{2}dP+{\int }_{{\beta }_c}\underset{a\in {\beta }_c}{min}{(x-a)}^{2}dP={\displaystyle \frac{1}{75}}(V_{n_1}(P)+V_{n_2}(P))+{\displaystyle \frac{1}{3}}{V}_{n_c}(\nu ).$$

This completes the proof of the lemma. □

Remark 13.

By Remark 12 and Lemma 27, we see that if ${\alpha }_{F(n)}$ is an optimal set of $F(n)$-means for any $n\ge 1$, then ${\alpha }_{F(n)}$ contains $F(n-1)$ elements from each of $J_1$ and $J_2$, and $2^{a(n)}$ elements from C, yielding the fact that $V_{F(n)}(P)={\displaystyle \frac{2}{75}}{V}_{F(n-1)}(P)+{\displaystyle \frac{1}{3}}{V}_{2^{a(n)}}(\nu )$.

Proposition 18.

For any $n\ge 0$, the set ${\alpha }_{F(n)}$ is an optimal set of $F(n)$-means for the condensation measure P with quantization error given by

$$V_{F(n)}:=V_{F(n)}(P)=\left\{\begin{array}{cc}{\displaystyle \frac{89}{21,900}}\hfill & \mathit{if}n=0,\hfill \\ {\displaystyle \frac{2537}{6,570,000}}\hfill & \mathit{if}n=1,\hfill \\ {\displaystyle \frac{1}{129}}{\displaystyle \frac{1}{{16}^n}}-{\displaystyle \frac{3473}{941,700}}{({\displaystyle \frac{2}{75}})}^n\hfill & \mathit{if}n\ge 2.\hfill \end{array}\right.$$

Proof. 

By Proposition 14, we know that ${\alpha }_{F(0)}$ is an optimal set of $F(0)$-means with quantization error ${\displaystyle \frac{89}{21,900}}$. Using Lemma 17 and the expression (11), the distortion error due to the set ${\alpha }_{F(1)}$ is obtained as

$$\begin{array}{cc}\hfill & \int \underset{a\in {\alpha }_{F(1)}}{min}{(x-a)}^{2}dP={\displaystyle \frac{1}{3}}{V}_{2^{2a(1)}}(\nu )+\sum _{\omega \in I}{\int }_{C_{\omega }}\underset{a\in S_{\omega }({\alpha }_{2^{a(0)}}(\nu ))}{min}{(x-a)}^{2}dP+\sum _{\omega \in I^2}{\int }_{J_{\omega }}{(x-a)}^{2}dP\hfill \\ \hfill & ={\displaystyle \frac{1}{3}}{\displaystyle \frac{W}{2^2}}+{\displaystyle \frac{1}{3}}{\displaystyle \frac{2}{75}}{\displaystyle \frac{W}{2^{2a(0)}}}+{({\displaystyle \frac{2}{75}})}^{2}V={\displaystyle \frac{2537}{\mathrm{6,570,000}}}=0.000386149.\hfill \end{array}$$

First, we show that ${\alpha }_{F(1)}$ is an optimal set of $F(1)$-means. Let $\beta$ be an optimal set of $F(1)$-means and $V_{F(1)}$ is the corresponding quantization error. Then, from the above calculation, we have $V_{F(1)}\le 0.000386149$. Recall that $\beta$ contains points from $J_1,J_2$ and C, and $\beta$ does not contain any point from the open intervals $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$ and $({\displaystyle \frac{2}{5}},{\displaystyle \frac{3}{5}})$. Due to symmetry of P, we can assume that $\beta$ contains an equal number of elements from each of $J_1$ and $J_2$, and the rest of the elements of $\beta$ are equally spaced over the set C with respect to the uniform distribution $\nu$. Thus, $\mathrm{card}(\beta \cap J_1)=\mathrm{card}(\beta \cap J_2)\le 3$. Suppose that $\mathrm{card}(\beta \cap J_1)=\mathrm{card}(\beta \cap J_2)=1$, then

$$V_{F(1)}\ge 2{\int }_{J_1}\underset{a\in \beta \cap J_1}{min}{(x-a)}^{2}dP={\displaystyle \frac{2}{75}}V=0.00295282>V_{F(1)},$$

which is a contradiction. Next, suppose that $\mathrm{card}(\beta \cap J_1)=\mathrm{card}(\beta \cap J_2)=2$, then,

$$V_{F(1)}\ge 2{\int }_{J_1}\underset{a\in \beta \cap J_1}{min}{(x-a)}^{2}dP={\displaystyle \frac{2}{75}}{V}_2=0.000812075>V_{F(1)},$$

which leads to another contradiction. So, we can assume that $\mathrm{card}(\beta \cap J_1)=\mathrm{card}(\beta \cap J_2)=3$. Then, by Lemma 27, we have $S_1^{-1}(\beta \cap J_1)=S_2^{-1}(\beta \cap J_2)={\alpha }_{F(0)}$. Hence, by Remark 12, $\beta =S_1({\alpha }_{F(0)})\cup S_2({\alpha }_{F(0)})\cup {\alpha }_2(\nu )={\alpha }_{F(1)}$, i.e., ${\alpha }_{F(1)}$ is an optimal set of $F(1)$-means, and the corresponding quantization error is given by $V_{F(1)}={\displaystyle \frac{2537}{\mathrm{6,570,000}}}$. The distortion error due to the set ${\alpha }_{F(n)}$ for any $n\ge 2$ is given by

$$\begin{array}{cc}\hfill & \int \underset{a\in {\alpha }_{F(n)}}{min}{(x-a)}^{2}dP\hfill \\ \hfill & ={\displaystyle \frac{1}{3}}{V}_{2^{2a(n)}}(\nu )+\sum _{k=0}^{n-1}\sum _{\omega \in I^{n-k}}{\int }_{C_{\omega }}\underset{a\in S_{\omega }({\alpha }_{2^{a(k)}}(\nu ))}{min}{(x-a)}^{2}dP+\sum _{\omega \in I^{n+1}}{\int }_{J_{\omega }}{(x-a)}^{2}dP\hfill \\ \hfill & ={\displaystyle \frac{1}{3}}{\displaystyle \frac{W}{2^{2a(n)}}}+\sum _{k=1}^{n-1}\sum _{\omega \in I^{n-k}}{\displaystyle \frac{1}{3}}{\displaystyle \frac{1}{{75}^{n-k}}}{\displaystyle \frac{W}{2^{2a(k)}}}+\sum _{\omega \in I^n}{\displaystyle \frac{1}{3}}{\displaystyle \frac{1}{{75}^n}}{\displaystyle \frac{W}{2^{2a(0)}}}+{({\displaystyle \frac{2}{75}})}^{n+1}V\hfill \\ \hfill & ={\displaystyle \frac{1}{3}}{\displaystyle \frac{4W}{{16}^n}}+\sum _{k=1}^{n-1}{\displaystyle \frac{1}{3}}{({\displaystyle \frac{2}{75}})}^{n-k}{\displaystyle \frac{4W}{{16}^k}}+{\displaystyle \frac{1}{3}}{({\displaystyle \frac{2}{75}})}^{n}W+{({\displaystyle \frac{2}{75}})}^{n+1}V\hfill \\ \hfill & ={\displaystyle \frac{1}{3}}{\displaystyle \frac{4W}{{16}^n}}+{({\displaystyle \frac{2}{75}})}^n{\displaystyle \frac{4W}{3}}\sum _{k=1}^{n-1}{({\displaystyle \frac{75}{32}})}^k+{\displaystyle \frac{1}{3}}{({\displaystyle \frac{2}{75}})}^{n}W+{({\displaystyle \frac{2}{75}})}^{n+1}V\hfill \\ \hfill & ={\displaystyle \frac{1}{3}}{\displaystyle \frac{4W}{{16}^n}}+{({\displaystyle \frac{2}{75}})}^n\left({\displaystyle \frac{W}{3}}-{\displaystyle \frac{4W}{3}}{\displaystyle \frac{75}{43}}\left(1-{({\displaystyle \frac{75}{32}})}^{n-1}\right)+{\displaystyle \frac{2}{75}}V\right),\hfill \end{array}$$

yielding

$$\int \underset{a\in {\alpha }_{F(n)}}{min}{(x-a)}^{2}dP={\displaystyle \frac{1}{129}}{\displaystyle \frac{1}{{16}^n}}-{\displaystyle \frac{3473}{\mathrm{941,700}}}{({\displaystyle \frac{2}{75}})}^n.$$

(17)

Now, we show that ${\alpha }_{F(2)}$ is an optimal set of $F(2)$-means. Let $\gamma$ be an optimal set of $F(2)$-means, and $V_{F(2)}$ is the corresponding quantization error. By (17), we have $V_{F(2)}\le 0.0000276584$. We show that $\mathrm{card}(\gamma \cap J_1)=\mathrm{card}(\gamma \cap J_2)=8$. Due to symmetry of P we can assume that $\gamma$ contains equal number of elements from each of $J_1$ and $J_2$, and the rest of the elements of $\beta$ are equally spaced over the set C with respect to the uniform distribution $\nu$. Suppose that $\mathrm{card}(\gamma \cap J_1)=\mathrm{card}(\gamma \cap J_2)\le 6$. Then, if ${\alpha }_6$ is an optimal set of six-means, by Remark 10, we have

$$V_{F(2)}\ge 2{\int }_{J_1}\underset{a\in \gamma \cap J_1}{min}{(x-a)}^{2}dP\ge 2{\int }_{J_1}\underset{a\in S_1({\alpha }_6)}{min}{(x-a)}^{2}dP={\displaystyle \frac{2}{75}}{\displaystyle \frac{\mathrm{21,481}}{\mathrm{19,710,000},}}$$

i.e., $V_{F(2)}\ge 0.0000290627>V_{F(2)}$, which is a contradiction. Suppose that $\mathrm{card}(\gamma \cap J_1)=\mathrm{card}(\gamma \cap J_2)=7$. Then, by Remark 11, we have

$$V_{F(2)}\ge 2{\int }_{J_1}\underset{a\in \gamma \cap J_1}{min}{(x-a)}^{2}dP+{\displaystyle \frac{1}{3}}{V}_{10}(\nu )={\displaystyle \frac{2}{75}}{\displaystyle \frac{7273}{\mathrm{9,855,000}}}+{\displaystyle \frac{1}{3}}{\displaystyle \frac{W}{{10}^2}}=0.0000307911>V_{F(2)},$$

which leads a contradiction. Similarly, we can show that if $\mathrm{card}(\gamma \cap J_1)=\mathrm{card}(\gamma \cap J_2)\ge 9$, a contradiction arises, i.e., $\mathrm{card}(\gamma \cap J_1)=\mathrm{card}(\gamma \cap J_2)=8$, and so by Remark 12, $\gamma =S_1({\alpha }_{F(1)})\cup S_2({\alpha }_{F(1)})\cup {\alpha }_{2^{a(2)}}(\nu )={\alpha }_{F(2)}$, i.e., ${\alpha }_{F(2)}$ is an optimal set of $F(2)$-means, and the corresponding quantization error is given by $V_{F(2)}=0.0000276584$.

Let ${\alpha }_{F(n)}$ be an optimal set of $F(n)$-means for some $n\ge 2$. We show that ${\alpha }_{F(n+1)}$ is an optimal set of $F(n+1)$-means. We have ${\alpha }_{F(n)}={\cup }_{A\in SF(n)}A$. Recall that, by Proposition 16, an optimal set of n-means for any $n\ge 3$ does not contain any point from the open intervals $({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}})$ and $({\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}})$. In the first step, let $A(1)\in SF(n)$ be such that $A(1)\succ B$ for any other $B\in SF(n)$. Then, by Lemma 25, the set $({\alpha }_{F(n)}\setminus A(1))\cup A^{(2)}(1)$ gives an optimal set of $F(n)-\mathrm{card}(A(1))+\mathrm{card}(A^{(2)}(1))$-means. In the second step, let $A(2)\in (SF(n)\setminus \left\{A(1)\right\})\cup \left\{A^{(2)}(1)\right\}$ be such that $A(2)\succ B$ for any other set $B\in (SF(n)\setminus \left\{A(1)\right\})\cup \left\{A^{(2)}(1)\right\}$. Then, using the similar technique as Lemma 25, we can show the distortion error due to the following set:

$$\left((({\alpha }_{F(n)}\setminus A(1))\cup A^{(2)}(1))\setminus A(2)\right)\cup A^{(2)}(2)$$

(18)

where cardinality $F(n)-\mathrm{card}(A(1))+\mathrm{card}(A^{(2)}(1))-\mathrm{card}(A(2))+\mathrm{card}(A^{(2)}(2))$ is smaller than the distortion error due to the set obtained by replacing $A(2)$ in the set (18) by the set B. In other words, $\left((({\alpha }_{F(n)}\setminus A(1))\cup A^{(2)}(1))\setminus A(2)\right)\cup A^{(2)}(2)$ forms an optimal set of $F(n)-\mathrm{card}(A(1))+\mathrm{card}(A^{(2)}(1))-\mathrm{card}(A(2))+\mathrm{card}(A^{(2)}(2))$-means. Proceeding inductively in this way, up to $(2n+3)$ steps, we can see that ${\alpha }_{F(n+1)}=\left(\underset{A\in (SF(n)\setminus S(0))}{\cup }{A}^{(2)(2)}\right)\cup S^{(2)}(0)$ forms an optimal set of $F(n+1)$-means. Thus, by the induction principle, we can say that for any $n\ge 0$, the set ${\alpha }_{F(n)}$ forms an optimal set of $F(n)$-means with quantization error $V_{F(n)}$ as given in the hypothesis. Thus, the proof of the proposition is complete. □

4.3. Asymptotics for the nth Quantization Error $V_n(P)$

In this subsection, we show that the quantization dimension of the condensation measure P exists and equals to the quantization dimension of the uniform distribution $\nu$. In addition, we show that the $D(P)$-dimensional quantization coefficient for the condensation measure P does not exist, and the lower and upper quantization coefficients for P are finite and positive.

Theorem 3.

Let P be the condensation measure associated with the uniform distribution ν. Then, ${lim}_{n\to \infty }{\displaystyle \frac{2logn}{log{V}_n(P)}}=1$, i.e., the quantization dimension $D(P)$ of the measure P exists and equals to the quantization dimension $D(\nu )$ of ν.

Proof. 

For $n\in \mathbb{N}$, $n\ge 4$, let $\ell (n)$ be the least positive integer such that $F(\ell (n))\le n<F(\ell (n)+1)$. Then, $V_{F(\ell (n)+1)}<V_n\le V_{F(\ell (n))}$. Thus, we have

$$\begin{array}{c}\hfill {\displaystyle \frac{2log\left(F(\ell (n))\right)}{-log\left(V_{F(\ell (n)+1)}\right)}}<{\displaystyle \frac{2logn}{-log{V}_n}}<{\displaystyle \frac{2log\left(F(\ell (n)+1)\right)}{-log\left(V_{F(\ell (n))}\right)}}.\end{array}$$

Notice that when $n\to \infty$, then $\ell (n)\to \infty$. By Proposition 18, we have

$$\begin{array}{cc}\hfill & \underset{\ell (n)\to \infty }{lim}{\displaystyle \frac{2log\left(F(\ell (n))\right)}{-log\left(V_{F(\ell (n)+1)}\right)}}=2\underset{\ell (n)\to \infty }{lim}{\displaystyle \frac{log(4^{\ell (n)}+2^{\ell (n)+1})}{-log\left(\frac{1}{129}\frac{1}{{16}^{\ell (n)+1}}-\frac{3473}{941,700}{(\frac{2}{75})}^{\ell (n)+1}\right)}}({\displaystyle \frac{\infty }{\infty }}\mathrm{form})\hfill \\ \hfill & =2\underset{\ell (n)\to \infty }{lim}\left({\displaystyle \frac{\frac{1}{129}\frac{1}{{16}^{\ell (n)+1}}-\frac{3473}{941,700}{(\frac{2}{75})}^{\ell (n)+1}}{4^{\ell (n)}+2^{\ell (n)+1}}}\right)\left({\displaystyle \frac{4^{\ell (n)}log4+2^{\ell (n)+1}log2}{\frac{1}{129}\frac{1}{{16}^{\ell (n)+1}}log16+\frac{3473}{941,700}{(\frac{2}{75})}^{\ell (n)+1}log\frac{2}{75}}}\right)\hfill \\ \hfill & ={\displaystyle \frac{2log4}{log16}}=1.\hfill \end{array}$$

Similarly, $\underset{\ell (n)\to \infty }{lim}{\displaystyle \frac{2log\left(F(\ell (n)+1)\right)}{-log\left(V_{F(\ell (n))}\right)}}=1$. Thus, $1\le {lim\; inf}_n{\displaystyle \frac{2logn}{-log{V}_n}}\le {lim\; sup}_n{\displaystyle \frac{2logn}{-log{V}_n}}\le 1$, implying the fact that the quantization dimension of the measure P exists and equals to the quantization dimension $D(\nu )$ of $\nu$, which is the theorem. □

Theorem 4.

The $D(P)$-dimensional quantization coefficient for the condensation measure P does not exist.

Proof. 

For any $n\in \mathbb{N}$, by Proposition 17 and Lemma 25, we see that $({\alpha }_{F(n)}\setminus S(n))\cup S^{(2)}(n)$ is an optimal set of $F(n)+2^{a(n)}$-means with quantization error

$$\begin{array}{cc}\hfill & V_{F(n)+2^{a(n)}}=V_{F(n)}-{\displaystyle \frac{1}{3}}{\displaystyle \frac{W}{2^{2a(n)}}}+{\displaystyle \frac{1}{3}}{\displaystyle \frac{W}{2^{2(a(n)+1)}}}=V_{F(n)}-{\displaystyle \frac{W}{{16}^n}}=({\displaystyle \frac{1}{129}}-W){\displaystyle \frac{1}{{16}^n}}-{\displaystyle \frac{3473}{\mathrm{941,700}}}{({\displaystyle \frac{2}{75}})}^n\hfill \\ \hfill & ={\displaystyle \frac{19}{4300}}{\displaystyle \frac{1}{{16}^n}}-{\displaystyle \frac{3473}{\mathrm{941,700}}}{({\displaystyle \frac{2}{75}})}^n.\hfill \end{array}$$

Since $F(n)+2^{a(n)}=4^n+2^{n+1}+2^{2n-1}={\displaystyle \frac{3}{2}}{4}^n+2^{n+1}$, we have

$$\underset{n\to \infty }{lim}{(F(n)+2^{a(n)})}^2{V}_{F(n)+2^{a(n)}}=\underset{n\to \infty }{lim}{({\displaystyle \frac{3}{2}}{4}^n+2^{n+1})}^2({\displaystyle \frac{19}{4300}}{\displaystyle \frac{1}{{16}^n}}-{\displaystyle \frac{3473}{941,700}}{({\displaystyle \frac{2}{75}})}^n)={\displaystyle \frac{171}{\mathrm{17,200}}}.$$

(19)

Again, ${(F(n))}^2={(4^n+2^{n+1})}^2={16}^n+2^{3n+2}+4^{n+1}={16}^n(1+{\displaystyle \frac{4}{2^n}}+{\displaystyle \frac{4}{4^n}})$, and

$$V_{F(n)}(P)={\displaystyle \frac{1}{129}}{\displaystyle \frac{1}{{16}^n}}-{\displaystyle \frac{3473}{\mathrm{941,700}}}{({\displaystyle \frac{2}{75}})}^n={\displaystyle \frac{1}{{16}^n}}\left({\displaystyle \frac{1}{129}}-{\displaystyle \frac{3473}{941,700}}{({\displaystyle \frac{32}{75}})}^n\right).$$

Thus,

$$\underset{n\to \infty }{lim}{(F(n))}^2{V}_{F(n)}={\displaystyle \frac{1}{129}}.$$

(20)

By Equations (19) and (20), we see that ${lim\; inf}_n{n}^2{V}_n(P)\le {\displaystyle \frac{1}{129}}<{\displaystyle \frac{171}{\mathrm{17,200}}}\le {lim\; sup}_n{n}^2{V}_n(P)$; in other words, ${lim}_{n\to \infty }{n}^2{V}_n(P)$ does not exist, i.e., the $D(P)$-dimensional quantization coefficient for the condensation measure P does not exist. □

Theorem 5.

The $D(P)$-dimensional lower and upper quantization coefficients for the condensation measure P are finite and positive.

Proof. 

For $n\in \mathbb{N}$, $n\ge 4$, let $\ell (n)$ be the least positive integer such that $F(\ell (n))\le n<F(\ell (n)+1)$. Then, $V_{F(\ell (n)+1)}<V_n\le V_{F(\ell (n))}$, implying ${(F(\ell (n)))}^2{V}_{F(\ell (n)+1)}<n^2{V}_n<{(F(\ell (n)+1))}^2{V}_{F(\ell (n))}$. As $\ell (n)\to \infty$ whenever $n\to \infty$, we have

$$\underset{n\to \infty }{lim}{\displaystyle \frac{{(F(\ell (n)))}^2}{{(F(\ell (n)+1))}^2}}=\underset{n\to \infty }{lim}{\left({\displaystyle \frac{4^{\ell (n)}+2^{\ell (n)+1}}{4^{\ell (n)+1}+2^{\ell (n)+2}}}\right)}^2={\displaystyle \frac{1}{16}}.$$

Again, ${(F(\ell (n)))}^2={(4^{\ell (n)}+2^{\ell (n)+1})}^2={16}^{\ell (n)}+2^{3\ell (n)+2}+4^{\ell (n)+1}={16}^{\ell (n)}(1+{\displaystyle \frac{4}{2^{\ell (n)}}}+{\displaystyle \frac{4}{4^{\ell (n)}}})$, and

$$V_{F(\ell (n))}(P)={\displaystyle \frac{1}{129}}{\displaystyle \frac{1}{{16}^{\ell (n)}}}-{\displaystyle \frac{3473}{\mathrm{941,700}}}{({\displaystyle \frac{2}{75}})}^{\ell (n)}={\displaystyle \frac{1}{{16}^{\ell (n)}}}\left({\displaystyle \frac{1}{129}}-{\displaystyle \frac{3473}{941,700}}{({\displaystyle \frac{32}{75}})}^{\ell (n)}\right).$$

Thus,

$$\underset{n\to \infty }{lim}{(F(\ell (n)))}^2{V}_{F(\ell (n)+1)}={\displaystyle \frac{1}{16}}\underset{n\to \infty }{lim}{(F(\ell (n)+1))}^2{V}_{F(\ell (n)+1)}={\displaystyle \frac{1}{16}}{\displaystyle \frac{1}{129}},\mathrm{and}\mathrm{similarly},$$

$$\underset{n\to \infty }{lim}{(F(\ell (n)+1))}^2{V}_{F(\ell (n))}={\displaystyle \frac{16}{129}},$$

yielding the fact that ${\displaystyle \frac{1}{16}}{\displaystyle \frac{1}{129}}\le \underset{n\to \infty }{lim\; inf}{n}^2{V}_n(P)\le \underset{n\to \infty }{lim\; sup}{n}^2{V}_n(P)\le {\displaystyle \frac{16}{129}}$, i.e., the $D(P)$-dimensional lower and quantization coefficients for the condensation measure P are finite and positive, which is the theorem. □

Remark 14.

Notice that ${({\displaystyle \frac{1}{3}}{({\displaystyle \frac{1}{5}})}^2)}^{{\displaystyle \frac{\kappa }{2+\kappa }}}+{({\displaystyle \frac{1}{3}}{({\displaystyle \frac{1}{5}})}^2)}^{{\displaystyle \frac{\kappa }{2+\kappa }}}=1$ implies $\kappa ={\displaystyle \frac{2log2}{log75-log2}}\approx 0.382496<1=D(\nu )=max\{\kappa ,D(\nu )\}$. In Theorem 3, we have proved that $D(P)=D(\nu ).$

We conclude this paper with the following remark.

Remark 15.

Using the similar techniques or by giving a major overhaul to our techniques given in the last two sections, Section 3 and Section 4, one can investigate optimal quantization for more general condensation measures.