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To appear, Journal of Fractal Geometry

Constrained quantization for the Cantor distribution

1Megha Pandey  and  2Mrinal Kanti Roychowdhury 1Department of Mathematical Sciences
Indian Institute of Technology (Banaras Hindu University)
Varanasi, 221005, India.
2School of Mathematical and Statistical Sciences
University of Texas Rio Grande Valley
1201 West University Drive
Edinburg, TX 78539-2999, USA.
1[email protected], 2[email protected]
Abstract.

The theory of constrained quantization has been recently introduced by Pandey and Roychowdhury. In this paper, they have further generalized their previous definition of constrained quantization and studied the constrained quantization for the classical Cantor distribution. Toward this, they have calculated the optimal sets of nn-points, nnth constrained quantization errors, the constrained quantization dimensions, and the constrained quantization coefficients, taking different families of constraints for all nn\in\mathbb{N}. The results in this paper show that both the constrained quantization dimension and the constrained quantization coefficient for the Cantor distribution depend on the underlying constraints. It also shows that the constrained quantization coefficient for the Cantor distribution can exist and be equal to the constrained quantization dimension. These facts are not true in the unconstrained quantization for the Cantor distribution.

Key words and phrases:
Cantor distribution, constrained quantization error, optimal sets of nn-points, constrained quantization dimension, constrained quantization coefficient
2010 Mathematics Subject Classification:
Primary 28A80; Secondary 94A34, 60Exx.

1. Introduction

Real-life problems, such as information theory, data compression, signal processing, etc., consist of a large number of data that is not easy to handle. In order to deal with such a data set, the theory of quantization comes into play (see, for instance, [1, 2, 3, 4, 5, 6, 7]). Quantization is a process of discretization, in other words, to represent a set with a large number of elements, discrete or continuous, by a set with a smaller number of elements. Several mathematical theories have been introduced in the literature concerning the process of quantization. Graf and Luschgy gave the rigorous mathematical treatment in [2]. In [8], Graf and Luschgy studied the quantization problem for the canonical probability measure on the classical Cantor set.

Recently, in [9], the authors introduced the concept of constrained quantization. A quantization without a constraint is referred to as an unconstrained quantization, which traditionally in the literature is referred to as a quantization, as mentioned in the previous paragraph. The theory of constrained quantization is a fascinating area of research, and it invites a lot of new areas to work with a number of applications. With the introduction of constrained quantization, quantization now has two classifications: constrained quantization and unconstrained quantization. In this paper, the authors have further generalized the definition of constrained quantization given in [9], and study the concept of constrained quantization for the canonical probability measure on the classical Cantor set.

Definition 1.1.

Let PP be a Borel probability measure on space k\mathbb{R}^{k} equipped with a metric dd induced by a norm \lVert\cdot\rVert on k\mathbb{R}^{k}, and r(0,)r\in(0,\infty). Let {Sjk:j}\{S_{j}\subseteq\mathbb{R}^{k}:j\in\mathbb{N}\}, where \mathbb{N} denotes the set of all natural numbers, be a family of closed sets with S1S_{1} nonempty. Then, for nn\in\mathbb{N}, the nnth constrained quantization error for PP, of order rr with respect to the family of constraints {Sjk:j}\{S_{j}\subseteq\mathbb{R}^{k}:j\in\mathbb{N}\}, is defined by

Vn,r:=Vn,r(P)=inf{minaαd(x,a)rdP(x):αj=1nSj,1card(α)n},V_{n,r}:=V_{n,r}(P)=\inf\Big{\{}\int\mathop{\min}\limits_{a\in\alpha}d(x,a)^{r}dP(x):\alpha\subseteq\bigcup_{j=1}^{n}S_{j},~{}1\leq\text{card}(\alpha)\leq n\Big{\}}, (1)

where card(A)\text{card}(A) represents the cardinality of the set AA.

The number

Vr(P;α):=minaαd(x,a)rdP(x)V_{r}(P;\alpha):=\int\mathop{\min}\limits_{a\in\alpha}d(x,a)^{r}dP(x)

is called the distortion error for PP, of order rr, with respect to a set αk\alpha\subseteq\mathbb{R}^{k}. The sets SjS_{j} are the constraints in the constrained quantization error. We assume that d(x,0)r𝑑P(x)<\int d(x,0)^{r}dP(x)<\infty to make sure that the infimum in (1) exists (see [9]). A set αj=1nSj\alpha\subseteq\mathop{\bigcup}\limits_{j=1}^{n}S_{j} for which the infimum in (1) exists and does not contain more than nn elements is called an optimal set of nn-points for PP. Elements of an optimal set are called optimal elements.

Remark 1.2.

In Definition 1.1 of constrained quantization error, if all SjS_{j} for jj\in\mathbb{N} are identical, then it reduces to the definition of constrained quantization error introduced by Pandey and Roychowdhury in [9]. Furthermore, if Sj=kS_{j}=\mathbb{R}^{k} for all jj\in\mathbb{N}, then it reduces to the definition of nnth unconstrained quantization error, which traditionally in the literature is referred to as the nnth quantization error (see [2]). For some recent work in the direction of unconstrained quantization, one can see [2, 8, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20].

Let Vn,r(P)V_{n,r}(P) be a strictly decreasing sequence, and write V,r(P):=limnVn,r(P)V_{\infty,r}(P):=\mathop{\lim}\limits_{n\to\infty}V_{n,r}(P). Then, the number Dr(P)D_{r}(P) defined by

Dr(P):=limnrlognlog(Vn,r(P)V,r(P)),D_{r}(P):=\mathop{\lim}\limits_{n\to\infty}\frac{r\log n}{-\log(V_{n,r}(P)-V_{\infty,r}(P))}, (2)

if it exists, is called the constrained quantization dimension of PP of order rr. The constrained quantization dimension measures the speed at which the specified measure of the constrained quantization error converges as nn tends to infinity. For any κ>0\kappa>0, the number

limnnrκ(Vn,r(P)V,r(P)),\lim_{n\to\infty}n^{\frac{r}{\kappa}}(V_{n,r}(P)-V_{\infty,r}(P)), (3)

if it exists, is called the κ\kappa-dimensional constrained quantization coefficient for PP of order rr.

Remark 1.3.

In unconstrained quantization, V,r(P):=limnVn,r(P)=0V_{\infty,r}(P):=\mathop{\lim}\limits_{n\to\infty}V_{n,r}(P)=0. Hence, in unconstrained quantization, i.e., when V,r(P)=0V_{\infty,r}(P)=0, the definitions of constrained quantization dimension and the κ\kappa-dimensional constrained quantization coefficient defined by (2) and (3), respectively, reduce to the corresponding definitions in unconstrained scenario (see [2]).

This paper deals with r=2r=2 and k=2k=2, and the metric on 2\mathbb{R}^{2} as the Euclidean metric induced by the Euclidean norm \lVert\cdot\rVert. Instead of writing Vr(P;α)V_{r}(P;\alpha) and Vn,r:=Vn,r(P)V_{n,r}:=V_{n,r}(P) we will write them as V(P;α)V(P;\alpha) and Vn:=Vn(P)V_{n}:=V_{n}(P), i.e., rr is omitted in the subscript as r=2r=2 throughout the paper. Let us take the family {Sj:j}\{S_{j}:j\in\mathbb{N}\}, that occurs in Definition 1.1 as follows:

Sj={(x,y):0x1 and y=1j} for all j.S_{j}=\{(x,y):0\leq x\leq 1\text{ and }y=\frac{1}{j}\}\text{ for all }j\in\mathbb{N}. (4)

Let T1,T2:T_{1},T_{2}:\mathbb{R}\to\mathbb{R} be two contractive similarity mappings such that T1(x)=13xT_{1}(x)=\frac{1}{3}x and T2(x)=13x+23T_{2}(x)=\frac{1}{3}x+\frac{2}{3}. Then, there exists a unique Borel probability measure PP on \mathbb{R} such that P=12PT11+12PT21P=\frac{1}{2}P\circ T_{1}^{-1}+\frac{1}{2}P\circ T_{2}^{-1}, where PTi1P\circ T_{i}^{-1} denote the image measures of PP with respect to TiT_{i} for i=1,2i=1,2 (see [21]). If kk\in\mathbb{N}, and σ:=σ1σ2σk{1,2}k\sigma:=\sigma_{1}\sigma_{2}\cdots\sigma_{k}\in\{1,2\}^{k}, then we call σ\sigma a word of length kk over the alphabet I:={1,2}I:=\{1,2\}, and denote it by |σ|:=k|\sigma|:=k. By II^{\ast}, we denote the set of all words, including the empty word \emptyset. Notice that the empty word has a length zero. For any word σ:=σ1σ2σkI\sigma:=\sigma_{1}\sigma_{2}\cdots\sigma_{k}\in I^{\ast}, we write

Tσ:=Tσ1Tσk and Jσ:=Tσ([0,1]).T_{\sigma}:=T_{\sigma_{1}}\circ\cdots\circ T_{\sigma_{k}}\text{ and }J_{\sigma}:=T_{\sigma}([0,1]).

Then, the set C:=kσ{1,2}kJσC:=\bigcap_{k\in\mathbb{N}}\bigcup_{\sigma\in\{1,2\}^{k}}J_{\sigma} is known as the Cantor set generated by the two mappings T1T_{1} and T2T_{2}, and equals the support of the probability measure PP, where PP can be written as

P=σ{1,2}k12kPTσ1.P=\sum_{\sigma\in\{1,2\}^{k}}\frac{1}{2^{k}}P\circ T_{\sigma}^{-1}.

For this probability measure PP, Graf and Luschgy determined the optimal sets of nn-means and the nnth quantization errors for all nn\in\mathbb{N} (see [8]). They also showed that the unconstrained quantization dimension of the measure PP exists and equals log2log3\frac{\log 2}{\log 3}, which is the Hausdorff dimension of the Cantor set CC, and the unconstrained quantization coefficient does not exist. In fact, in [8], they showed that the lower and the upper unconstrained quantization coefficients exist as finite positive numbers.

1.4. Delineation

In this paper, first, we have determined the optimal sets of nn-points and the nnth constrained quantization errors for all nn\in\mathbb{N} for the Borel probability measure PP with support the Cantor set CC. Then, we have calculated the constrained quantization dimension and the constrained quantization coefficient. We have shown that both the constrained quantization dimension D(P)D(P) and the D(P)D(P)-dimensional constrained quantization coefficient exist and are equal to one, i.e., they coincide. Then, in the last section, taking different families of constraints for all nn\in\mathbb{N}, we investigate the optimal sets of nn-points, nnth constrained quantization errors, the constrained quantization dimensions, and the constrained quantization coefficients. From work in this paper, it can be seen that the constrained quantization dimension of the Cantor distribution depends on the family of constraints {Sj:j}\{S_{j}:j\in\mathbb{N}\}, i.e., the constrained quantization dimension is not always equal to the Hausdorff dimension of the Cantor set as it occurs in the case of unconstrained quantization (see [8]). In the unconstrained quantization, the D(P)D(P)-dimensional quantization coefficient does not exist (see [8]). But from work in the last section, we see that the D(P)D(P)-dimensional constrained quantization coefficient also depends on the constraints; it may or may not exist.

2. Preliminaries

In this section, we give some basic notations and definitions which we have used throughout the paper. As defined in the previous section, let I:={1,2}I:=\{1,2\} be an alphabet. For any two words σ:=σ1σ2σk\sigma:=\sigma_{1}\sigma_{2}\cdots\sigma_{k} and τ:=τ1τ2τ\tau:=\tau_{1}\tau_{2}\cdots\tau_{\ell} in II^{*}, by στ:=σ1σkτ1τ\sigma\tau:=\sigma_{1}\cdots\sigma_{k}\tau_{1}\cdots\tau_{\ell}, we mean the word obtained from the concatenation of the two words σ\sigma and τ\tau. For σ,τI\sigma,\tau\in I^{\ast}, σ\sigma is called an extension of τ\tau if σ=τx\sigma=\tau x for some word xIx\in I^{\ast}. The mappings Ti:,i=1,2,T_{i}:\mathbb{R}\to\mathbb{R},\ i=1,2, such that T1(x)=13xT_{1}(x)=\frac{1}{3}x and T2(x)=13x+23T_{2}(x)=\frac{1}{3}x+\frac{2}{3} are the generating maps of the Cantor set CC, which is the support of the probability measure PP on \mathbb{R} given by P=12PT11+12PT21P=\frac{1}{2}P\circ T_{1}^{-1}+\frac{1}{2}P\circ T_{2}^{-1}. For σ:=σ1σ2σkIk\sigma:=\sigma_{1}\sigma_{2}\cdots\sigma_{k}\in I^{k}, write Jσ=Tσ[0,1]J_{\sigma}=T_{\sigma}[0,1], where Tσ:=Tσ1Tσ2TσkT_{\sigma}:=T_{\sigma_{1}}\circ T_{\sigma_{2}}\circ\cdots\circ T_{\sigma_{k}} is a composition mapping. Notice that J:=J=T[0,1]=[0,1]J:=J_{\emptyset}=T_{\emptyset}[0,1]=[0,1]. Then, for any kk\in\mathbb{N}, as mentioned before, we have

C=kσIkJσ and P=σIk12kPTσ1.C=\bigcap_{k\in\mathbb{N}}\bigcup_{\sigma\in I^{k}}J_{\sigma}\text{ and }P=\sum_{\sigma\in I^{k}}\frac{1}{2^{k}}P\circ T_{\sigma}^{-1}.

The elements of the set {Jσ:σIk}\{J_{\sigma}:\sigma\in I^{k}\} are the 2k2^{k} intervals in the kkth level in the construction of the Cantor set CC, and are known as the basic intervals at the kkth level. The intervals Jσ1J_{\sigma 1}, Jσ2J_{\sigma 2}, into which JσJ_{\sigma} is split up at the (k+1)(k+1)th level are called the children of JσJ_{\sigma}.

With respect to a finite set α2\alpha\subset\mathbb{R}^{2}, by the Voronoi region of an element aαa\in\alpha, it is meant the set of all elements in 2\mathbb{R}^{2} which are nearest to aa among all the elements in α\alpha, and is denoted by M(a|α)M(a|\alpha). For any two elements (a,b)(a,b) and (c,d)(c,d) in 2\mathbb{R}^{2}, we write

ρ((a,b),(c,d)):=(ac)2+(bd)2,\rho((a,b),(c,d)):=(a-c)^{2}+(b-d)^{2},

which gives the squared Euclidean distance between the two elements (a,b)(a,b) and (c,d)(c,d). Let pp and qq be two elements that belong to an optimal set of nn-points for some positive integer nn. Then, pp and qq are called adjacent elements if they have a common boundary in their own Voronoi regions. Let ee be an element on the common boundary of the Voronoi regions of the adjacent elements pp and qq. Since the common boundary of the Voronoi regions of any two adjacent elements is the perpendicular bisector of the line segment joining the elements, we have

ρ(p,e)ρ(q,e)=0.\rho(p,e)-\rho(q,e)=0.

We call such an equation a canonical equation. Notice that any element xx\in\mathbb{R} can be identified as an element (x,0)2(x,0)\in\mathbb{R}^{2}. Thus, the nonnegative real-valued function ρ\rho on ×2\mathbb{R}\times\mathbb{R}^{2} defined by

ρ:×2[0,) such that ρ(x,(a,b))=(xa)2+b2,\rho:\mathbb{R}\times\mathbb{R}^{2}\to[0,\infty)\text{ such that }\rho(x,(a,b))=(x-a)^{2}+b^{2},

represents the squared Euclidean distance between an element xx\in\mathbb{R} and an element (a,b)2(a,b)\in\mathbb{R}^{2}. Let π:2\pi:\mathbb{R}^{2}\to\mathbb{R} such that π(a,b)=a\pi(a,b)=a for any (a,b)2(a,b)\in\mathbb{R}^{2} denote the projection mapping. For a random variable XX with distribution PP, let E(X)E(X) represent the expected value, and V:=V(X)V:=V(X) represent the variance of XX.

The following lemmas are well-known (see [8]).

Lemma 2.1.

Let f:+f:\mathbb{R}\to\mathbb{R}^{+} be Borel measurable and kk\in\mathbb{N}. Then,

f𝑑P=σ{1,2}kpσfTσ𝑑P.\int fdP=\sum_{\sigma\in\{1,2\}^{k}}p_{\sigma}\int f\circ T_{\sigma}dP.
Lemma 2.2.

Let XX be a random variable with probability distribution PP. Then, E(X)=12 and V:=V(X)=EX122=E(X12)2=18E(X)=\frac{1}{2}\text{ and }V:=V(X)=E\lVert X-\frac{1}{2}\rVert^{2}=E(X-\frac{1}{2})^{2}=\frac{1}{8}. Moreover, for any x0x_{0}\in\mathbb{R}, we have

(xx0)2𝑑P(x)=V(X)+(x12)2.\int(x-x_{0})^{2}dP(x)=V(X)+(x-\frac{1}{2})^{2}.
Remark 2.3.

For words β,γ,,δ\beta,\gamma,\cdots,\delta in II^{\ast}, by a(β,γ,,δ)a(\beta,\gamma,\cdots,\delta) we mean the conditional expectation of the random variable XX given JβJγJδ,J_{\beta}\cup J_{\gamma}\cup\cdots\cup J_{\delta}, i.e.,

a(β,γ,,δ)=E(X:XJβJγJδ)=1P(JβJδ)JβJδxdP.a(\beta,\gamma,\cdots,\delta)=E(X:X\in J_{\beta}\cup J_{\gamma}\cup\cdots\cup J_{\delta})=\frac{1}{P(J_{\beta}\cup\cdots\cup J_{\delta})}\int_{J_{\beta}\cup\cdots\cup J_{\delta}}x\,dP.

Recall Lemma 2.1, for each σI\sigma\in I^{\ast}, since TσT_{\sigma} is a similarity mapping, we have

a(σ)\displaystyle a(\sigma) =E(X:XJσ)=1P(Jσ)JσxdP=Jσxd(PTσ1)=Tσ(x)dP\displaystyle=E(X:X\in J_{\sigma})=\frac{1}{P(J_{\sigma})}\int_{J_{\sigma}}x\,dP=\int_{J_{\sigma}}xd(P\circ T_{\sigma}^{-1})=\int T_{\sigma}(x)\,dP
=E(Tσ(X))=Tσ(E(X))=Tσ(12).\displaystyle=E(T_{\sigma}(X))=T_{\sigma}(E(X))=T_{\sigma}(\frac{1}{2}).
Definition 2.4.

For nn\in\mathbb{N} with n2n\geq 2, let (n)\ell(n) be the unique natural number with 2(n)n<2(n)+12^{\ell(n)}\leq n<2^{\ell(n)+1} and Sn={(x,1n):0x1}S_{n}=\{(x,\frac{1}{n}):0\leq x\leq 1\}. For I{1,2}(n)I\subset\{1,2\}^{\ell(n)} with card(I)=n2(n)(I)=n-2^{\ell(n)} let αn(I)Sn\alpha_{n}(I)\subseteq S_{n} be the set such that

αn(I)={(a(σ),1n):σ{1,2}(n)I}{(a(σ1),1n):σI}{(a(σ2),1n):σI}.\alpha_{n}(I)=\{(a(\sigma),\frac{1}{n}):\sigma\in\{1,2\}^{\ell(n)}\setminus I\}\cup\{(a(\sigma 1),\frac{1}{n}):\sigma\in I\}\cup\{(a(\sigma 2),\frac{1}{n}):\sigma\in I\}.
Proposition 2.5.

Let αn(I)\alpha_{n}(I) be the set given by Definition 2.4. Then, the number of such sets is Cn2(n)2(n){}^{2^{\ell(n)}}C_{n-2^{\ell(n)}}, and the corresponding distortion error is given by

V(P;αn(I))=minaαn(I)ρ(x,a)dP=118(n)V(2(n)+1n+19(n2(n)))+1n2,V(P;\alpha_{n}(I))=\int\mathop{\min}\limits_{a\in\alpha_{n}(I)}\rho(x,a)\,dP=\frac{1}{18^{\ell(n)}}V\Big{(}2^{\ell(n)+1}-n+\frac{1}{9}(n-2^{\ell(n)})\Big{)}+\frac{1}{n^{2}},

where VV is the variance as stated in Lemma 2.2.

Proof.

If 2(n)n<2(n)+12^{\ell(n)}\leq n<2^{\ell(n)+1}, then the subset II can be chosen in Cn2(n)2(n){}^{2^{\ell(n)}}C_{n-2^{\ell(n)}} different ways, and so, the number of such sets is given by Cn2(n)2(n){}^{2^{\ell(n)}}C_{n-2^{\ell(n)}}, and the corresponding distortion error is obtained as

V(P;αn(I))=\displaystyle V(P;\alpha_{n}(I))= minaαn(I)ρ(x,a)𝑑P\displaystyle\int\min_{a\in\alpha_{n}(I)}\rho(x,a)\,dP
=\displaystyle= σ{1,2}(n)IJσρ(x,(a(σ),1n))𝑑P\displaystyle\sum_{\sigma\in\{1,2\}^{\ell(n)}\setminus I}\int_{J_{\sigma}}\rho(x,(a(\sigma),\frac{1}{n}))\,dP
+σI(Jσ1ρ(x,(a(σ1),1n))𝑑P+Jσ2ρ(x,(a(σ2),1n))𝑑P)\displaystyle+\sum_{\sigma\in I}\Big{(}\int_{J_{\sigma 1}}\rho(x,(a(\sigma 1),\frac{1}{n}))\,dP+\int_{J_{\sigma 2}}\rho(x,(a(\sigma 2),\frac{1}{n}))\,dP\Big{)}
=\displaystyle= σ{1,2}(n)I12(n)ρ(Tσ(x),(a(σ),1n))𝑑P\displaystyle\sum_{\sigma\in\{1,2\}^{\ell(n)}\setminus I}\frac{1}{2^{\ell(n)}}\int\rho(T_{\sigma}(x),(a(\sigma),\frac{1}{n}))\,dP
+σI12(n)+1(ρ(Tσ1(x),(a(σ1),1n))dP\displaystyle+\sum_{\sigma\in I}\frac{1}{2^{\ell(n)+1}}\Big{(}\int\rho(T_{\sigma 1}(x),(a(\sigma 1),\frac{1}{n}))\,dP
+ρ(Tσ2(x),(a(σ2),1n))dP)\displaystyle+\int\rho(T_{\sigma 2}(x),(a(\sigma 2),\frac{1}{n}))\,dP\Big{)}
=\displaystyle= σ{1,2}(n)I12(n)(19(n)V+1n2)+σI12(n)+1(29(n)+1V+2n2)\displaystyle\sum_{\sigma\in\{1,2\}^{\ell(n)}\setminus I}\frac{1}{2^{\ell(n)}}\Big{(}\frac{1}{9^{\ell(n)}}V+\frac{1}{n^{2}})+\sum_{\sigma\in I}\frac{1}{2^{\ell(n)+1}}\Big{(}\frac{2}{9^{\ell(n)+1}}V+\frac{2}{n^{2}}\Big{)}
=\displaystyle= 118(n)V(2(n)+1n+19(n2(n)))+1n2.\displaystyle\frac{1}{18^{\ell(n)}}V\Big{(}2^{\ell(n)+1}-n+\frac{1}{9}(n-2^{\ell(n)})\Big{)}+\frac{1}{n^{2}}.

Thus, the proof of the proposition is complete. ∎

In the next sections, we give the main results of the paper.

3. Optimal sets of nn-points for all n1n\geq 1

In this section, we calculate the optimal sets of nn-points and the nnth constrained quantization errors for all nn\in\mathbb{N}. For jj\in\mathbb{N}, we have the constraints as

Sj={(x,y):0x1 and y=1j} for all j.S_{j}=\{(x,y):0\leq x\leq 1\text{ and }y=\frac{1}{j}\}\text{ for all }j\in\mathbb{N}.

For all jj\in\mathbb{N}, the perpendiculars on the constraints SjS_{j} passing through the points (a,1j)Sj(a,\frac{1}{j})\in S_{j} intersect JJ at the points aa, where 0a10\leq a\leq 1. Thus, for each jj\in\mathbb{N}, there exists a one-one correspondence between the element (a,1j)(a,\frac{1}{j}) on SjS_{j} and the element aa on JJ. Thus, for jj\in\mathbb{N}, there exist bijective functions

Uj:SjJ such that Uj(a,1j)=a.U_{j}:S_{j}\to J\text{ such that }U_{j}(a,\frac{1}{j})=a. (5)

Hence, the inverse functions Uj1U_{j}^{-1} are defined as

Uj1:JSj such that Uj1(x)=(x,1j).U_{j}^{-1}:J\to S_{j}\text{ such that }U_{j}^{-1}(x)=(x,\frac{1}{j}).
Remark 3.1.

For n2n\geq 2, let αn(I)\alpha_{n}(I) be the set given by Definition 2.4, and for each jj\in\mathbb{N}, let UjU_{j} be the bijective functions defined by (5). Then, Proposition 2.5 implies that

V(P;αn(I))=V(P;Un(αn(I)))+1n2.V(P;\alpha_{n}(I))=V(P;U_{n}(\alpha_{n}(I)))+\frac{1}{n^{2}}.
Proposition 3.2.

An optimal set of one-point is {(12,1)}\{(\frac{1}{2},1)\} with constrained quantization error V1=98V_{1}=\frac{9}{8}.

Proof.

Let α:={(a,b)}\alpha:=\{(a,b)\} be an optimal set of one-point. Since αS1\alpha\subseteq S_{1}, we have b=1b=1. Now, the distortion error for PP with respect to the set α\alpha is give by

V(P;α)=ρ(x,(a,1))𝑑P=a2a+118,V(P;\alpha)=\int\rho(x,(a,1))dP=a^{2}-a+\frac{11}{8},

the minimum value of which is 98\frac{9}{8} and it occurs when a=12a=\frac{1}{2}. Thus, the proof of the proposition is complete. ∎

The following lemma plays an important role in the paper.

Lemma 3.3.

Let αnj=1nSj\alpha_{n}\subseteq\mathop{\cup}\limits_{j=1}^{n}S_{j} be an optimal set of nn-points for PP such that

αn:={(aj,bj):1jn},\alpha_{n}:=\{(a_{j},b_{j}):1\leq j\leq n\},

where a1<a2<a3<<ana_{1}<a_{2}<a_{3}<\cdots<a_{n}. Then, aj=E(X:Xπ(M((aj,bj)|αn)))a_{j}=E(X:X\in\pi(M((a_{j},b_{j})|\alpha_{n}))) and bj=1nb_{j}=\frac{1}{n}, where M((aj,bj)|αn)M((a_{j},b_{j})|\alpha_{n}) are the Voronoi regions of the elements (aj,bj)(a_{j},b_{j}) with respect to the set αn\alpha_{n} for 1jn1\leq j\leq n.

Proof.

Let αn:={(aj,bj):1jn}\alpha_{n}:=\{(a_{j},b_{j}):1\leq j\leq n\}, as given in the statement of the lemma, be an optimal set of nn-points. Take any (aq,bq)αn(a_{q},b_{q})\in\alpha_{n}. Since αnj=1nSj\alpha_{n}\subseteq\mathop{\cup}\limits_{j=1}^{n}S_{j}, we can assume that (aq,bq)St(a_{q},b_{q})\in S_{t}, i.e., bq=1tb_{q}=\frac{1}{t} for some 1tn1\leq t\leq n. Since the Voronoi region of (aq,bq)(a_{q},b_{q}), i.e., M((aq,bq)|αn)M((a_{q},b_{q})|\alpha_{n}) has positive probability, M((aq,bq)|αn)M((a_{q},b_{q})|\alpha_{n}) contains some basic intervals from JJ that generates the Cantor set CC. Let {Jσ(j):jΛ}\{J_{\sigma^{(j)}}:j\in\Lambda\}, where Λ\Lambda is an index set, be the family of all basic intervals that are contained in M((aq,bq)|αn)M((a_{q},b_{q})|\alpha_{n}). Now, the distortion error contributed by (aq,bq)(a_{q},b_{q}) in its Voronoi region M((aq,bq)|αn)M((a_{q},b_{q})|\alpha_{n}) is given by

M((aq,bq)|αn)ρ(x,(aq,bq))𝑑P=jΛ12(σ(j))Jσ(j)ρ(x,(aq,bq))d(PTσ(j)1)\displaystyle\int_{M((a_{q},b_{q})|\alpha_{n})}\rho(x,(a_{q},b_{q}))\,dP=\sum_{j\in\Lambda}\frac{1}{2^{\ell(\sigma^{(j)})}}\int_{J_{\sigma^{(j)}}}\rho(x,(a_{q},b_{q}))\,d(P\circ T_{\sigma^{(j)}}^{-1})
=jΛ12(σ(j))19(σ(j))V+jΛ12(σ(j))ρ(Tσ(j)(12),(aq,1t))\displaystyle=\sum_{j\in\Lambda}\frac{1}{2^{\ell(\sigma^{(j)})}}\frac{1}{9^{\ell(\sigma^{(j)})}}V+\sum_{j\in\Lambda}\frac{1}{2^{\ell(\sigma^{(j)})}}\rho(T_{\sigma^{(j)}}(\frac{1}{2}),(a_{q},\frac{1}{t}))
=jΛ12(σ(j))19(σ(j))V+jΛ12(σ(j))((Tσ(j)(12)aq)2+1t2)\displaystyle=\sum_{j\in\Lambda}\frac{1}{2^{\ell(\sigma^{(j)})}}\frac{1}{9^{\ell(\sigma^{(j)})}}V+\sum_{j\in\Lambda}\frac{1}{2^{\ell(\sigma^{(j)})}}\Big{(}(T_{\sigma^{(j)}}(\frac{1}{2})-a_{q})^{2}+\frac{1}{t^{2}}\Big{)}
=jΛ12(σ(j))19(σ(j))V+jΛ12(σ(j))(Tσ(j)(12)aq)2+jΛ12(σ(j))1t2.\displaystyle=\sum_{j\in\Lambda}\frac{1}{2^{\ell(\sigma^{(j)})}}\frac{1}{9^{\ell(\sigma^{(j)})}}V+\sum_{j\in\Lambda}\frac{1}{2^{\ell(\sigma^{(j)})}}(T_{\sigma^{(j)}}(\frac{1}{2})-a_{q})^{2}+\sum_{j\in\Lambda}\frac{1}{2^{\ell(\sigma^{(j)})}}\frac{1}{t^{2}}.

The above expression is minimum if both jΛ12(σ(j))(Tσ(j)(12)aq)2\sum_{j\in\Lambda}\frac{1}{2^{\ell(\sigma^{(j)})}}(T_{\sigma^{(j)}}(\frac{1}{2})-a_{q})^{2} and
jΛ12(σ(j))1t2\sum_{j\in\Lambda}\frac{1}{2^{\ell(\sigma^{(j)})}}\frac{1}{t^{2}} are minimum, i.e., when

aq=jΛ12(σ(j))Tσ(j)(12)jΛ12(σ(j))=E(X:Xπ(M((aq,bq)|αn))) and bq=1t=1n.a_{q}=\frac{\sum_{j\in\Lambda}\frac{1}{2^{\ell(\sigma^{(j)})}}T_{\sigma^{(j)}}(\frac{1}{2})}{\sum_{j\in\Lambda}\frac{1}{2^{\ell(\sigma^{(j)})}}}=E(X:X\in\pi(M((a_{q},b_{q})|\alpha_{n})))\text{ and }b_{q}=\frac{1}{t}=\frac{1}{n}.

Since (aq,bq)αn(a_{q},b_{q})\in\alpha_{n} is chosen arbitrarily, the proof of the lemma is complete. ∎

Remark 3.4.

Let αnj=1nSj\alpha_{n}\subseteq\mathop{\cup}\limits_{j=1}^{n}S_{j} be an optimal set of nn-points for PP such that

αn:={(aj,bj):1jn},\alpha_{n}:=\{(a_{j},b_{j}):1\leq j\leq n\},

where a1<a2<a3<<ana_{1}<a_{2}<a_{3}<\cdots<a_{n}. Then, by Lemma 3.3, we can deduce that 0<a1<<an<10<a_{1}<\cdots<a_{n}<1.

Remark 3.5.

Lemma 3.3 implies that if αn\alpha_{n} is an optimal set of nn-points for PP, then αnSn\alpha_{n}\subseteq S_{n} for all nn\in\mathbb{N}.

Proposition 3.6.

The set {(16,12),(56,12)}\{(\frac{1}{6},\frac{1}{2}),(\frac{5}{6},\frac{1}{2})\} forms an optimal set of two-points with constrained quantization error V2=1972V_{2}=\frac{19}{72}.

Proof.

Due to symmetry, the distortion error due to the set β:={(16,12),(56,12)}\beta:=\{(\frac{1}{6},\frac{1}{2}),(\frac{5}{6},\frac{1}{2})\} is given by

V(P;β)=2J1ρ(x,(16,12))𝑑P=1972.V(P;\beta)=2\int_{J_{1}}\rho(x,(\frac{1}{6},\frac{1}{2}))\,dP=\frac{19}{72}.

Let α:={(a1,12),(a2,12)}\alpha:=\{(a_{1},\frac{1}{2}),(a_{2},\frac{1}{2})\}, where 0<a1<a2<10<a_{1}<a_{2}<1, be an optimal set of two-points. Since V2V_{2} is the constrained quantization error for two-points, we have V2V(P;β)=1972V_{2}\leq V(P;\beta)=\frac{19}{72}. We first show that U2(α)J1U_{2}(\alpha)\cap J_{1}\neq\emptyset. Suppose that T21(1)=79a1T_{21}(1)=\frac{7}{9}\leq a_{1}. Then,

V2>J1ρ(x,(79,12))𝑑P=4131296>V2,V_{2}>\int_{J_{1}}\rho(x,(\frac{7}{9},\frac{1}{2}))\,dP=\frac{413}{1296}>V_{2},

which leads to a contradiction. Suppose that 23a1<79\frac{2}{3}\leq a_{1}<\frac{7}{9}. Then, the Voronoi region of (a1,12)(a_{1},\frac{1}{2}) does not contain any element from J22J_{22}. For the sake of contradiction, assume that the Voronoi region of (a1,12)(a_{1},\frac{1}{2}) contains elements from J22J_{22}. Then, we must have 12(a1+a2)>89\frac{1}{2}(a_{1}+a_{2})>\frac{8}{9} implying a2>169a1>16979=1a_{2}>\frac{16}{9}-a_{1}>\frac{16}{9}-\frac{7}{9}=1, which gives a contradiction. Thus, we see that J22J_{22} is contained in the Voronoi region of (a2,12)(a_{2},\frac{1}{2}). Hence,

V2>J1ρ(x,(23,12))𝑑P+J22ρ(x,(a(22),12))𝑑P=8292592>V2,V_{2}>\int_{J_{1}}\rho(x,(\frac{2}{3},\frac{1}{2}))\,dP+\int_{J_{22}}\rho(x,(a(22),\frac{1}{2}))\,dP=\frac{829}{2592}>V_{2},

which gives a contradiction. Assume that 13a1<23\frac{1}{3}\leq a_{1}<\frac{2}{3}. Then, the distortion error is obtained as

V2J1ρ(x,(13,12))𝑑P+J21ρ(x,(23,12))𝑑P+J22ρ(x,(a(22),12))𝑑P=3531296>V2,V_{2}\geq\int_{J_{1}}\rho(x,(\frac{1}{3},\frac{1}{2}))\,dP+\int_{J_{21}}\rho(x,(\frac{2}{3},\frac{1}{2}))\,dP+\int_{J_{22}}\rho(x,(a(22),\frac{1}{2}))\,dP=\frac{353}{1296}>V_{2},

which yields a contradiction. Hence, we can assume that a1<13a_{1}<\frac{1}{3}, i.e., U2(α)J1U_{2}(\alpha)\cap J_{1}\neq\emptyset. Similarly, we can show that 23<a2\frac{2}{3}<a_{2}, i.e., U2(α)J2U_{2}(\alpha)\cap J_{2}\neq\emptyset. Hence, U2(α)JjU_{2}(\alpha)\cap J_{j}\neq\emptyset for j=1,2j=1,2. Notice that then the Voronoi region of (a1,12)(a_{1},\frac{1}{2}) does not contain any element from J2J_{2}, and the Voronoi region of (a2,12)(a_{2},\frac{1}{2}) does not contain any element from J1J_{1} yielding

(a1,12)=(a(1),12)=(16,12), and (a2,12)=(a(2),12)=(56,12),(a_{1},\frac{1}{2})=(a(1),\frac{1}{2})=(\frac{1}{6},\frac{1}{2}),\text{ and }(a_{2},\frac{1}{2})=(a(2),\frac{1}{2})=(\frac{5}{6},\frac{1}{2}),

with constrained quantization error V2=1972V_{2}=\frac{19}{72}. Thus, the proof of the proposition is complete. ∎

Lemma 3.7.

Let α3\alpha_{3} be an optimal set of three-points. Then, U3(α3)JjU_{3}(\alpha_{3})\cap J_{j}\neq\emptyset for j=1,2j=1,2, and U3(α3)U_{3}(\alpha_{3}) does not contain any element from the open line segment joining (13,0)(\frac{1}{3},0) and (23,0)(\frac{2}{3},0).

Proof.

The distortion error due to the set β:={(a(11),13),(a(12),13),(a(2),13)}\beta:=\{(a(11),\frac{1}{3}),(a(12),\frac{1}{3}),(a(2),\frac{1}{3})\} is given by

V(P;β)=minaα3ρ(x,a)𝑑P=77648.V(P;\beta)=\int\min_{a\in\alpha_{3}}\rho(x,a)\,dP=\frac{77}{648}.

Let α3:={(a1,13),(a2,13),(a3,13)}\alpha_{3}:=\{(a_{1},\frac{1}{3}),(a_{2},\frac{1}{3}),(a_{3},\frac{1}{3})\} be an optimal set of three-points such that 0<a1<a2<a3<10<a_{1}<a_{2}<a_{3}<1. Since V3V_{3} is the constrained quantization error for three-points, we have V377648V_{3}\leq\frac{77}{648}. Let us first show that U3(α3)J1U_{3}(\alpha_{3})\cap J_{1}\neq\emptyset, i.e., 13<a1\frac{1}{3}<a_{1}. We prove it by contradiction. Notice that

J1ρ(x,(a,13))>77648,\int_{J_{1}}\rho(x,(a,\frac{1}{3}))>\frac{77}{648},

if a>136(146+6)a>\frac{1}{36}\big{(}\sqrt{146}+6\big{)}. Choose a number 211420>136(146+6)\frac{211}{420}>\frac{1}{36}\big{(}\sqrt{146}+6\big{)}, and consider the following cases:

Case 1. 211420a1\frac{211}{420}\leq a_{1}.

Then,

V3J1ρ(x,(211420,13))𝑑P=465939200>V3,V_{3}\geq\int_{J_{1}}\rho(x,(\frac{211}{420},\frac{1}{3}))\,dP=\frac{4659}{39200}>V_{3},

which gives a contradiction.

Case 2. 1127a1<211420\frac{11}{27}\leq a_{1}<\frac{211}{420}.

Then, we have 12(a1+a2)>23\frac{1}{2}(a_{1}+a_{2})>\frac{2}{3} implying a2>43a1>43211420=349420>T21(1)a_{2}>\frac{4}{3}-a_{1}>\frac{4}{3}-\frac{211}{420}=\frac{349}{420}>T_{21}(1). Hence,

V3J1ρ(x,(1127,13))𝑑P+J21ρ(x,(349420,13))𝑑P=700687157153600>V3,V_{3}\geq\int_{J_{1}}\rho(x,(\frac{11}{27},\frac{1}{3}))\,dP+\int_{J_{21}}\rho(x,(\frac{349}{420},\frac{1}{3}))\,dP=\frac{7006871}{57153600}>V_{3},

which leads to a contradiction.

Case 3. 13<a1<1127\frac{1}{3}<a_{1}<\frac{11}{27}.

Then, we have 12(a1+a2)>23\frac{1}{2}(a_{1}+a_{2})>\frac{2}{3} implying a2>43a1>431127=2527=T221(1)a_{2}>\frac{4}{3}-a_{1}>\frac{4}{3}-\frac{11}{27}=\frac{25}{27}=T_{221}(1). Hence,

V3J1ρ(x,(13,13))𝑑P+J21J221ρ(x,(2527,13))𝑑P=601346656>V3,V_{3}\geq\int_{J_{1}}\rho(x,(\frac{1}{3},\frac{1}{3}))\,dP+\int_{J_{21}\cup J_{221}}\rho(x,(\frac{25}{27},\frac{1}{3}))\,dP=\frac{6013}{46656}>V_{3},

which gives a contradiction.

Taking into account the above cases, we see that a contradiction arises. Hence, U3(α3)J1U_{3}(\alpha_{3})\cap J_{1}\neq\emptyset. Reflecting the above arguments with respect to the line x=12x=\frac{1}{2}, we can show that U3(α3)J2U_{3}(\alpha_{3})\cap J_{2}\neq\emptyset. Thus, U3(α3)JjU_{3}(\alpha_{3})\cap J_{j}\neq\emptyset for j=1,2j=1,2. We now show that U3(α3)U_{3}(\alpha_{3}) does not contain any element from the open line segment joining (13,0)(\frac{1}{3},0) and (23,0)(\frac{2}{3},0). For the sake of contradiction, assume that U3(α3)U_{3}(\alpha_{3}) contains an element from the open line segment joining (13,0)(\frac{1}{3},0) and (23,0)(\frac{2}{3},0). Since U3(α3)JjU_{3}(\alpha_{3})\cap J_{j}\neq\emptyset for j=1,2j=1,2, we must have 13<a2<23\frac{1}{3}<a_{2}<\frac{2}{3}. The following cases can happen:

Case I. 12a2<23\frac{1}{2}\leq a_{2}<\frac{2}{3}.

Then, we have 12(a1+a2)<13\frac{1}{2}(a_{1}+a_{2})<\frac{1}{3} implying a1<23a216a_{1}<\frac{2}{3}-a_{2}\leq\frac{1}{6}. Again, E(X:XJ1)=16E(X:X\in J_{1})=\frac{1}{6}. Hence,

V3\displaystyle V_{3} =J1mina{a1,a2}ρ(x,a)𝑑P+J21J22mina{a2,a3}ρ(x,a)𝑑P\displaystyle=\int_{J_{1}}\min_{a\in\{a_{1},a_{2}\}}\rho(x,a)\,dP+\int_{J_{21}\cup J_{22}}\min_{a\in\{a_{2},a_{3}\}}\rho(x,a)\,dP
=J1ρ(x,(16,13))𝑑P+J21ρ(x,(23,13))𝑑P+J22ρ(x,(a(22),13))𝑑P\displaystyle=\int_{J_{1}}\rho(x,(\frac{1}{6},\frac{1}{3}))\,dP+\int_{J_{21}}\rho(x,(\frac{2}{3},\frac{1}{3}))\,dP+\int_{J_{22}}\rho(x,(a(22),\frac{1}{3}))\,dP
=1551296>V3,\displaystyle=\frac{155}{1296}>V_{3},

which yields a contradiction.

Case II. 13<a2<12\frac{1}{3}<a_{2}<\frac{1}{2}.

This case is the reflection of Case I with respect to the line x=12x=\frac{1}{2}. Hence, a contradiction also arises in this case.

Considering Case I and Case II, we can deduce that U3(α3)U_{3}(\alpha_{3}) does not contain any element from the open line segment joining (13,0)(\frac{1}{3},0) and (23,0)(\frac{2}{3},0). Thus, the proof of the lemma is complete. ∎

Proposition 3.8.

Let αn\alpha_{n} be an optimal set of nn-points for all n2n\geq 2. Then, Un(αn)JjU_{n}(\alpha_{n})\cap J_{j}\neq\emptyset for j=1,2j=1,2, and Un(αn)U_{n}(\alpha_{n}) does not contain any element from the open line segment joining (13,0)(\frac{1}{3},0) and (23,0)(\frac{2}{3},0).

Proof.

For all n2n\geq 2, let us first prove that Un(αn)J1U_{n}(\alpha_{n})\cap J_{1}\neq\emptyset. By Proposition 3.6 and Lemma 3.7, it is true for n=2,3n=2,3. Using a similar technique as Lemma 3.7, we can also prove that it is true for any n4n\geq 4. However, here we give a general proof for all n16n\geq 16. The distortion error due to the set β:={(a(σ),116):σ{1,2}4}\beta:=\{(a(\sigma),\frac{1}{16}):\sigma\in\{1,2\}^{4}\} is given by

V(P;β)=minaβρ(x,a)𝑑P=65931679616.V(P;\beta)=\int\min_{a\in\beta}\rho(x,a)\,dP=\frac{6593}{1679616}.

Since VnV_{n} is the constrained quantization error for nn-points with n16n\geq 16, and VnV_{n} is a decreasing sequence, we have VnV1665931679616V_{n}\leq V_{16}\leq\frac{6593}{1679616}. Let αn:={(aj,1n):1jn}\alpha_{n}:=\{(a_{j},\frac{1}{n}):1\leq j\leq n\} be an optimal set of nn-points such that 0<a1<a2<<an<10<a_{1}<a_{2}<\cdots<a_{n}<1. For the sake of contradiction, assume that Un(αn)J1=U_{n}(\alpha_{n})\cap J_{1}=\emptyset. Then, 13<a1\frac{1}{3}<a_{1}, and so,

Vn>J1ρ(x,(13,1n))𝑑P=12n2+148148>Vn,V_{n}>\int_{J_{1}}\rho(x,(\frac{1}{3},\frac{1}{n}))\,dP=\frac{1}{2n^{2}}+\frac{1}{48}\geq\frac{1}{48}>V_{n},

which is a contradiction. Hence, we can assume that Un(αn)J1U_{n}(\alpha_{n})\cap J_{1}\neq\emptyset. Similarly, we can show Un(αn)J2U_{n}(\alpha_{n})\cap J_{2}\neq\emptyset. Thus, the proof of the first part of the proposition is complete. We now show that Un(αn)U_{n}(\alpha_{n}) does not contain any element from the open line segment joining (13,0)(\frac{1}{3},0) and (23,0)(\frac{2}{3},0). For the sake of contradiction, assume that Un(αn)U_{n}(\alpha_{n}) contains an element from the open line segment joining (13,0)(\frac{1}{3},0) and (23,0)(\frac{2}{3},0). Let

k=min{j:aj>13 for all 1jn}.k=\min\{j:a_{j}>\frac{1}{3}\text{ for all }1\leq j\leq n\}.

Since Un(αn)J1U_{n}(\alpha_{n})\cap J_{1}\neq\emptyset, we have 2k2\leq k. Thus, we see that ak113<aka_{k-1}\leq\frac{1}{3}<a_{k}. Again, recall that the Voronoi regions of the elements in an optimal set of nn-points must have positive probability. Hence, we have ak113<ak<23ak+1a_{k-1}\leq\frac{1}{3}<a_{k}<\frac{2}{3}\leq a_{k+1}.

The following cases can happen:

Case 1. 59ak<23\frac{5}{9}\leq a_{k}<\frac{2}{3}.

Then, 12(ak1+ak)<13\frac{1}{2}(a_{k-1}+a_{k})<\frac{1}{3} implying ak1<23ak2359=19a_{k-1}<\frac{2}{3}-a_{k}\leq\frac{2}{3}-\frac{5}{9}=\frac{1}{9}. Hence,

VnJ12ρ(x,(19,1n))𝑑P=14n2+192592192592>Vn,V_{n}\geq\int_{J_{12}}\rho(x,(\frac{1}{9},\frac{1}{n}))\,dP=\frac{1}{4n^{2}}+\frac{19}{2592}\geq\frac{19}{2592}>V_{n},

which gives a contradiction.

Case 2. 49<a2<59\frac{4}{9}<a_{2}<\frac{5}{9}.

Then, 12(ak1+ak)<13\frac{1}{2}(a_{k-1}+a_{k})<\frac{1}{3} implying ak1<23ak2349=29=T12(0)a_{k-1}<\frac{2}{3}-a_{k}\leq\frac{2}{3}-\frac{4}{9}=\frac{2}{9}=T_{12}(0). Again, 12(ak+ak+1)>23\frac{1}{2}(a_{k}+a_{k+1})>\frac{2}{3} implying ak+1>43ak>4359=79=T21(1)a_{k+1}>\frac{4}{3}-a_{k}>\frac{4}{3}-\frac{5}{9}=\frac{7}{9}=T_{21}(1). Hence,

VnJ12ρ(x,(29,1n))𝑑P+J21ρ(x,(79,1n))𝑑P=12n2+111296111296>Vn,V_{n}\geq\int_{J_{12}}\rho(x,(\frac{2}{9},\frac{1}{n}))\,dP+\int_{J_{21}}\rho(x,(\frac{7}{9},\frac{1}{n}))\,dP=\frac{1}{2n^{2}}+\frac{11}{1296}\geq\frac{11}{1296}>V_{n},

which leads to a contradiction.

Case 3. 13<ak<49\frac{1}{3}<a_{k}<\frac{4}{9}.

Then, 12(ak+ak+1)>23\frac{1}{2}(a_{k}+a_{k+1})>\frac{2}{3} implying ak+1>43ak>4349=89a_{k+1}>\frac{4}{3}-a_{k}>\frac{4}{3}-\frac{4}{9}=\frac{8}{9}. Hence,

VnJ21ρ(x,(89,1n))𝑑P=14n2+192592192592>Vn,V_{n}\geq\int_{J_{21}}\rho(x,(\frac{8}{9},\frac{1}{n}))\,dP=\frac{1}{4n^{2}}+\frac{19}{2592}\geq\frac{19}{2592}>V_{n},

which gives a contradiction.

Considering the above all possible cases, we see that a contradiction arises. Hence, Un(αn)U_{n}(\alpha_{n}) does not contain any element from the open line segment joining (13,0)(\frac{1}{3},0) and (23,0)(\frac{2}{3},0). Thus, the proof of the proposition is complete. ∎

Corollary 3.9.

Let n2n\geq 2. Then, the Voronoi region of any element in αnUn1(J1)\alpha_{n}\cap U_{n}^{-1}(J_{1}) does not contain any element from J2J_{2}, and the Voronoi region of any element in αnUn1(J2)\alpha_{n}\cap U_{n}^{-1}(J_{2}) does not contain any element from J1J_{1}.

Proof.

Let αn:={(aj,1n):1jn}\alpha_{n}:=\{(a_{j},\frac{1}{n}):1\leq j\leq n\} be an optimal set of nn-points such that 0<a1<a2<<an<10<a_{1}<a_{2}<\cdots<a_{n}<1. By Proposition 3.8, we see that Un(αn)U_{n}(\alpha_{n}) contains elements from both J1J_{1} and J2J_{2}, and does not contain any element from the open line segment joining (13,0)(\frac{1}{3},0) and (23,0)(\frac{2}{3},0). Let

k=max{j:aj13 for all 1jn}.k=\max\{j:a_{j}\leq\frac{1}{3}\text{ for all }1\leq j\leq n\}.

Then, ak13<23ak+1a_{k}\leq\frac{1}{3}<\frac{2}{3}\leq a_{k+1}. For the sake of contradiction, assume that the Voronoi region (ak,1n)(a_{k},\frac{1}{n}) contains an element from J2J_{2}. Then, 12(ak+ak+1)23\frac{1}{2}(a_{k}+a_{k+1})\geq\frac{2}{3} yielding ak+143ak4313=1a_{k+1}\geq\frac{4}{3}-a_{k}\geq\frac{4}{3}-\frac{1}{3}=1, which is a contradiction. Similarly, we can show that if the Voronoi region (ak+1,1n)(a_{k+1},\frac{1}{n}) contains an element from J1J_{1}, then a contradiction arises. Thus, the proof of the corollary is complete. ∎

Lemma 3.10.

For n2n\geq 2 let αn\alpha_{n} be an optimal set of nn-points. Set β1:=Un(αn)J1\beta_{1}:=U_{n}(\alpha_{n})\cap J_{1}, β2:=Un(αn)J2\beta_{2}:=U_{n}(\alpha_{n})\cap J_{2}, and n1:=card(β1)n_{1}:=\text{card}(\beta_{1}). Then, Un11(T11(β1))U_{n_{1}}^{-1}(T_{1}^{-1}(\beta_{1})) is an optimal set of n1n_{1}-points, Unn11(T21(β2))U_{n-n_{1}}^{-1}(T_{2}^{-1}(\beta_{2})) is an optimal set of n2:=(nn1)n_{2}:=(n-n_{1})-points, and

Vn=118(Vn1+Vnn11n121(nn1)2)+1n2.V_{n}=\frac{1}{18}\Big{(}V_{n_{1}}+V_{n-n_{1}}-\frac{1}{n_{1}^{2}}-\frac{1}{(n-n_{1})^{2}}\Big{)}+\frac{1}{n^{2}}.
Proof.

By Proposition 3.8, we have Un(αn)=β1β2U_{n}(\alpha_{n})=\beta_{1}\cup\beta_{2}. Proceeding in a similar way as Lemma 3.3, we can show that

Vn(P)=minaαnρ(x,a)𝑑P=minaUn(αn)ρ(x,(a,0))𝑑P+1n2.V_{n}(P)=\int\min_{a\in\alpha_{n}}\rho(x,a)\,dP=\int\min_{a\in U_{n}(\alpha_{n})}\rho(x,(a,0))\,dP+\frac{1}{n^{2}}.

Hence,

Vn=\displaystyle V_{n}= J1minaβ1ρ(x,(a,0))𝑑P+J2minaβ2ρ(x,(a,0))𝑑P+1n2\displaystyle\int_{J_{1}}\min_{a\in\beta_{1}}\rho(x,(a,0))\,dP+\int_{J_{2}}\min_{a\in\beta_{2}}\rho(x,(a,0))\,dP+\frac{1}{n^{2}}
=\displaystyle= 12minaT11(β1)ρ(T1(x),(T1(a),0))𝑑P+12minaT21(β2)ρ(T2(x),(T2(a),0))𝑑P+1n2\displaystyle\frac{1}{2}\int\min_{a\in T_{1}^{-1}(\beta_{1})}\rho(T_{1}(x),(T_{1}(a),0))\,dP+\frac{1}{2}\int\min_{a\in T_{2}^{-1}(\beta_{2})}\rho(T_{2}(x),(T_{2}(a),0))\,dP+\frac{1}{n^{2}}
=\displaystyle= 118minaT11(β1)ρ(x,(a,0))𝑑P+118minaT21(β2)ρ(x,(a,0))𝑑P+1n2\displaystyle\frac{1}{18}\int\min_{a\in T_{1}^{-1}(\beta_{1})}\rho(x,(a,0))\,dP+\frac{1}{18}\int\min_{a\in T_{2}^{-1}(\beta_{2})}\rho(x,(a,0))\,dP+\frac{1}{n^{2}}
=\displaystyle= 118(minaT11(β1)ρ(x,(a,1n1))𝑑P+minaT21(β2)ρ(x,(a,1n2))𝑑P1n121n22)+1n2\displaystyle\frac{1}{18}\Big{(}\int\min_{a\in T_{1}^{-1}(\beta_{1})}\rho(x,(a,\frac{1}{n_{1}}))\,dP+\int\min_{a\in T_{2}^{-1}(\beta_{2})}\rho(x,(a,\frac{1}{n_{2}}))\,dP-\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\Big{)}+\frac{1}{n^{2}}

implying

Vn=118(minaUn11T11(β1)ρ(x,a)𝑑P+minaUn21T21(β2)ρ(x,a)𝑑P1n121n22)+1n2.V_{n}=\frac{1}{18}\Big{(}\int\min_{a\in U_{n_{1}}^{-1}T_{1}^{-1}(\beta_{1})}\rho(x,a)\,dP+\int\min_{a\in U_{n_{2}}^{-1}T_{2}^{-1}(\beta_{2})}\rho(x,a)\,dP-\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\Big{)}+\frac{1}{n^{2}}. (6)

If Un11T11(β1)U_{n_{1}}^{-1}T_{1}^{-1}(\beta_{1}) is not an optimal set of n1n_{1}-points for PP, then there exists a set γ1Sn1\gamma_{1}\subset S_{n_{1}} with card(γ1)=n1\text{card}(\gamma_{1})=n_{1} such that minaγ1ρ(x,a)𝑑P<minaUn11T11(β1)ρ(x,a)𝑑P\int\min_{a\in\gamma_{1}}\rho(x,a)\,dP<\int\min_{a\in U_{n_{1}}^{-1}T_{1}^{-1}(\beta_{1})}\rho(x,a)\,dP. But then, δ:=T1Un1(γ1)β2\delta:=T_{1}U_{n_{1}}(\gamma_{1})\cup\beta_{2} is a set of cardinality nn. VnV_{n} being the constrained quantization error for nn-points, we have

Vnminaδρ(x,(a,1n)dP=minaδρ(x,(a,0))dP+1n2.V_{n}\leq\int\min_{a\in\delta}\rho(x,(a,\frac{1}{n})\,dP=\int\min_{a\in\delta}\rho(x,(a,0))\,dP+\frac{1}{n^{2}}. (7)

Notice that

J1minaT1Un1(γ1)ρ(x,(a,0))𝑑P\displaystyle\int_{J_{1}}\min_{a\in T_{1}U_{n_{1}}(\gamma_{1})}\rho(x,(a,0))\,dP
=\displaystyle= 118minaUn1(γ1)ρ(x,(a,0))𝑑P=118(minaUn1(γ1)ρ(x,(a,1n1))𝑑P1n12)\displaystyle\frac{1}{18}\int\min_{a\in U_{n_{1}}(\gamma_{1})}\rho(x,(a,0))\,dP=\frac{1}{18}\Big{(}\int\min_{a\in U_{n_{1}}(\gamma_{1})}\rho(x,(a,\frac{1}{n_{1}}))\,dP-\frac{1}{n_{1}^{2}}\Big{)}
=\displaystyle= 118(minaγ1ρ(x,a)𝑑P1n12)<118(minaUn11T11(β1)ρ(x,a)𝑑P1n12).\displaystyle\frac{1}{18}\Big{(}\int\min_{a\in\gamma_{1}}\rho(x,a)\,dP-\frac{1}{n_{1}^{2}}\Big{)}<\frac{1}{18}\Big{(}\int\min_{a\in U_{n_{1}}^{-1}T_{1}^{-1}(\beta_{1})}\rho(x,a)\,dP-\frac{1}{n_{1}^{2}}\Big{)}. (8)

Hence, by (6), (7), and (3), we have

Vn\displaystyle V_{n} J1minaT1Un1(γ1)ρ(x,(a,0))𝑑P+J2minaβ2ρ(x,(a,0))𝑑P+1n2\displaystyle\leq\int_{J_{1}}\min_{a\in T_{1}U_{n_{1}}(\gamma_{1})}\rho(x,(a,0))\,dP+\int_{J_{2}}\min_{a\in\beta_{2}}\rho(x,(a,0))\,dP+\frac{1}{n^{2}}
<118(minaUn11T11(β1)ρ(x,a)𝑑P+minaUn21T21(β2)ρ(x,a)𝑑P1n121n22)+1n2\displaystyle<\frac{1}{18}\Big{(}\int\min_{a\in U_{n_{1}}^{-1}T_{1}^{-1}(\beta_{1})}\rho(x,a)\,dP+\int\min_{a\in U_{n_{2}}^{-1}T_{2}^{-1}(\beta_{2})}\rho(x,a)\,dP-\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\Big{)}+\frac{1}{n^{2}}
=Vn,\displaystyle=V_{n},

which leads to a contradiction. Hence, Un11(T11(β1))U_{n_{1}}^{-1}(T_{1}^{-1}(\beta_{1})) is an optimal set of n1n_{1}-points. Similarly, we see that Unn11(T21(β2))U_{n-n_{1}}^{-1}(T_{2}^{-1}(\beta_{2})) is an optimal set of n2:=(nn1)n_{2}:=(n-n_{1})-points, and hence,

Vn=118(Vn1+Vnn11n121(nn1)2)+1n2.V_{n}=\frac{1}{18}\Big{(}V_{n_{1}}+V_{n-n_{1}}-\frac{1}{n_{1}^{2}}-\frac{1}{(n-n_{1})^{2}}\Big{)}+\frac{1}{n^{2}}.

Thus, the proof of the lemma is complete. ∎

In view of Lemma 3.10, we give the following example.

Example 3.11.

Because of Proposition 3.8 and Corollary 3.9, we can show that if αn\alpha_{n} is an optimal set of nn-points with constrained quantization error VnV_{n}, then

α3=\displaystyle\alpha_{3}= {(a(11),13),(a(12),13),(a(2),13)} with V3=127864,\displaystyle\{(a(11),\frac{1}{3}),(a(12),\frac{1}{3}),(a(2),\frac{1}{3})\}\text{ with }V_{3}=\frac{127}{864},
α4=\displaystyle\alpha_{4}= {(a(11),14),(a(12),14),(a(21),14),(a(22),14)} with V4=831296,\displaystyle\{(a(11),\frac{1}{4}),(a(12),\frac{1}{4}),(a(21),\frac{1}{4}),(a(22),\frac{1}{4})\}\text{ with }V_{4}=\frac{83}{1296},
α7=\displaystyle\alpha_{7}= {(a(111),17),(a(112),17),(a(121),17),(a(122),17),(a(211),17),(a(212),17),\displaystyle\{(a(111),\frac{1}{7}),(a(112),\frac{1}{7}),(a(121),\frac{1}{7}),(a(122),\frac{1}{7}),(a(211),\frac{1}{7}),(a(212),\frac{1}{7}),
(a(22),17)},\displaystyle(a(22),\frac{1}{7})\},

with V7=199395256V_{7}=\frac{1993}{95256}. Here

β1\displaystyle\beta_{1} =U7(α7)J1={a(111),a(112),a(121),a(122)} and\displaystyle=U_{7}(\alpha_{7})\cap J_{1}=\{a(111),a(112),a(121),a(122)\}\text{ and }
β2\displaystyle\beta_{2} =U7(α7)J2={a(211),a(212),a(22)},\displaystyle=U_{7}(\alpha_{7})\cap J_{2}=\{a(211),a(212),a(22)\},

with card(β1)=4\text{card}(\beta_{1})=4 and card(β2)=3\text{card}(\beta_{2})=3. Notice that

U41(T11(β1))\displaystyle U_{4}^{-1}(T_{1}^{-1}(\beta_{1})) ={(a(11),14),(a(12),14),(a(21),14),(a(22),14)},\displaystyle=\{(a(11),\frac{1}{4}),(a(12),\frac{1}{4}),(a(21),\frac{1}{4}),(a(22),\frac{1}{4})\},
U31(T21(β2))\displaystyle U_{3}^{-1}(T_{2}^{-1}(\beta_{2})) ={(a(11),13),(a(12),13),(a(2),13)},and\displaystyle=\{(a(11),\frac{1}{3}),(a(12),\frac{1}{3}),(a(2),\frac{1}{3})\},and
V7\displaystyle V_{7} =118(V4+V3142132)+172.\displaystyle=\frac{1}{18}(V_{4}+V_{3}-\frac{1}{4^{2}}-\frac{1}{3^{2}})+\frac{1}{7^{2}}.

Let us state and prove the following theorem, which gives the optimal sets of nn-points for all n2n\geq 2.

Theorem 3.12.

Let P=12PT11+12PT21P=\frac{1}{2}P\circ T_{1}^{-1}+\frac{1}{2}P\circ T_{2}^{-1} be a unique Borel probability measure on \mathbb{R} with support the Cantor set CC generated by the two contractive similarity mappings T1(x)=13xT_{1}(x)=\frac{1}{3}x and T2(x)=13x+23T_{2}(x)=\frac{1}{3}x+\frac{2}{3} for all xx\in\mathbb{R}. Then, the set αn:=αn(I)\alpha_{n}:=\alpha_{n}(I) given by Definition 2.4 forms an optimal set of nn-points for PP with the corresponding constrained quantization error Vn:=V(P;αn(I))V_{n}:=V(P;\alpha_{n}(I)), where V(P;αn(I))V(P;\alpha_{n}(I)) is given by Proposition 2.5.

Proof.

We will proceed by induction on (n)\ell(n). If (n)=1\ell(n)=1, then the theorem is true by Proposition 3.6. Let us assume that the theorem is true for all (n)<m\ell(n)<m, where mm\in\mathbb{N} and m2m\geq 2. We now show that the theorem is true if (n)=m\ell(n)=m. Let αn:=αn(I)\alpha_{n}:=\alpha_{n}(I) be an optimal set of nn-points for PP such that 2mn<2m+12^{m}\leq n<2^{m+1}. Set β1:=Un(αn)J1\beta_{1}:=U_{n}(\alpha_{n})\cap J_{1}, β2:=Un(αn)J2\beta_{2}:=U_{n}(\alpha_{n})\cap J_{2}, n1:=card(β1)n_{1}:=\text{card}(\beta_{1}), and n2:=card(β2)n_{2}:=\text{card}(\beta_{2}). Then, by Lemma 3.10, we have

Vn=118(Vn1+Vn21n121n22)+1n2.V_{n}=\frac{1}{18}\Big{(}V_{n_{1}}+V_{n_{2}}-\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\Big{)}+\frac{1}{n^{2}}. (9)

Without any loss of generality, we can assume that n1n2n_{1}\geq n_{2}. Let p,qp,q\in\mathbb{N} be such that

2pn1<2p+1 and 2qn2<2q+1.2^{p}\leq n_{1}<2^{p+1}\text{ and }2^{q}\leq n_{2}<2^{q+1}. (10)

We will show that p=q=m1p=q=m-1. Since n1n2n_{1}\geq n_{2}, we have n12m1n_{1}\geq 2^{m-1} and n2<2mn_{2}<2^{m}. Hence, pm1p\geq m-1 and qm1q\leq m-1. If V~n\tilde{V}_{n} is the distortion error due to the set

{(a(σ),1n):σ{1,2}mI}{(a(σ1),1n):σI}{(a(σ2),1n):σI},\{(a(\sigma),\frac{1}{n}):\sigma\in\{1,2\}^{m}\setminus I\}\cup\{(a(\sigma 1),\frac{1}{n}):\sigma\in I\}\cup\{(a(\sigma 2),\frac{1}{n}):\sigma\in I\},

where I{1,2}mI\subset\{1,2\}^{m} with card(I)=n2m(I)=n-2^{m}, then by Proposition 2.5, we have

V~n=118mV(2m+1n+19(n2m))+1n2.\displaystyle\tilde{V}_{n}=\frac{1}{18^{m}}V\Big{(}2^{m+1}-n+\frac{1}{9}(n-2^{m})\Big{)}+\frac{1}{n^{2}}.

Thus, by the induction hypothesis, we have V~nVn\tilde{V}_{n}\geq V_{n}, and then Equation (9) implies that

118mV(2m+1n+19(n2m))+1n2118(Vn1+Vnn11n121(nn1)2)+1n2,\displaystyle\frac{1}{18^{m}}V\Big{(}2^{m+1}-n+\frac{1}{9}(n-2^{m})\Big{)}+\frac{1}{n^{2}}\geq\frac{1}{18}\Big{(}V_{n_{1}}+V_{n-n_{1}}-\frac{1}{n_{1}^{2}}-\frac{1}{(n-n_{1})^{2}}\Big{)}+\frac{1}{n^{2}},

i.e.,

118mV(2m+1n+19(n2m))\displaystyle\frac{1}{18^{m}}V\Big{(}2^{m+1}-n+\frac{1}{9}(n-2^{m})\Big{)}\geq 118m+1(V(2p+1n1+19(n12p))\displaystyle\frac{1}{18^{m+1}}\Big{(}V\Big{(}2^{p+1}-n_{1}+\frac{1}{9}(n_{1}-2^{p})\Big{)}
+(2q+1n2+19(n22q))),\displaystyle+\Big{(}2^{q+1}-n_{2}+\frac{1}{9}(n_{2}-2^{q})\Big{)}\Big{)},

which is the same as Equation 5.9 in [8]. Thus, proceeding in a similar way as [8], we have p=q=m1p=q=m-1. By Lemma 3.10, Un11(T11(β1))U_{n_{1}}^{-1}(T_{1}^{-1}(\beta_{1})) is an optimal set of n1n_{1}-points, Unn11(T21(β2))U_{n-n_{1}}^{-1}(T_{2}^{-1}(\beta_{2})) is an optimal set of n2:=(nn1)n_{2}:=(n-n_{1})-points. Moreover, we have proved 2m1n1<2m2^{m-1}\leq n_{1}<2^{m}, and 2m1n2<2m2^{m-1}\leq n_{2}<2^{m}. Hence, by the induction hypothesis,

Un11(T11(β1))\displaystyle U_{n_{1}}^{-1}(T_{1}^{-1}(\beta_{1}))
=\displaystyle= {(a(σ),1n):σ{1,2}m1I1}{(a(σ1),1n):σI1}{(a(σ2),1n):σI1},\displaystyle\{(a(\sigma),\frac{1}{n}):\sigma\in\{1,2\}^{m-1}\setminus I_{1}\}\cup\{(a(\sigma 1),\frac{1}{n}):\sigma\in I_{1}\}\cup\{(a(\sigma 2),\frac{1}{n}):\sigma\in I_{1}\},

where I1{1,2}m1I_{1}\subset\{1,2\}^{m-1} with card(I1)=n12m1(I_{1})=n_{1}-2^{m-1}; and

Un21(T21(β2))\displaystyle U_{n_{2}}^{-1}(T_{2}^{-1}(\beta_{2}))
=\displaystyle= {(a(σ),1n):σ{1,2}m1I2}{(a(σ1),1n):σI2}{(a(σ2),1n):σI2},\displaystyle\{(a(\sigma),\frac{1}{n}):\sigma\in\{1,2\}^{m-1}\setminus I_{2}\}\cup\{(a(\sigma 1),\frac{1}{n}):\sigma\in I_{2}\}\cup\{(a(\sigma 2),\frac{1}{n}):\sigma\in I_{2}\},

where I2{1,2}m1I_{2}\subset\{1,2\}^{m-1} with card(I2)=n22m1(I_{2})=n_{2}-2^{m-1}. Then, notice that

β1={a(1σ):σ{1,2}m1I1}{a(1σ1):σI1}{a(1σ2):σI1},\beta_{1}=\{a(1\sigma):\sigma\in\{1,2\}^{m-1}\setminus I_{1}\}\cup\{a(1\sigma 1):\sigma\in I_{1}\}\cup\{a(1\sigma 2):\sigma\in I_{1}\},

and

β2={a(2σ):σ{1,2}m1I2}{a(2σ1):σI2}{a(2σ2):σI2}.\beta_{2}=\{a(2\sigma):\sigma\in\{1,2\}^{m-1}\setminus I_{2}\}\cup\{a(2\sigma 1):\sigma\in I_{2}\}\cup\{a(2\sigma 2):\sigma\in I_{2}\}.

Take I:=I1I2I:=I_{1}\cup I_{2}, and then

card(I)=card(I1)+card(I2)=n12m1+n22m1=n2m.\text{card}(I)=\text{card}(I_{1})+\text{card}(I_{2})=n_{1}-2^{m-1}+n_{2}-2^{m-1}=n-2^{m}.

Hence,

Un(αn)=β1β2={a(σ):σ{1,2}mI}{a(σ1):σI}{a(σ2):σI}.\displaystyle U_{n}(\alpha_{n})=\beta_{1}\cup\beta_{2}=\{a(\sigma):\sigma\in\{1,2\}^{m}\setminus I\}\cup\{a(\sigma 1):\sigma\in I\}\cup\{a(\sigma 2):\sigma\in I\}.

Thus, we have

αn:=αn(I)={(a(σ),1n):σ{1,2}mI}{(a(σ1),1n):σI}{(a(σ2),1n):σI},\alpha_{n}:=\alpha_{n}(I)=\{(a(\sigma),\frac{1}{n}):\sigma\in\{1,2\}^{m}\setminus I\}\cup\{(a(\sigma 1),\frac{1}{n}):\sigma\in I\}\cup\{(a(\sigma 2),\frac{1}{n}):\sigma\in I\},

and using Equation (9), we have the constrained quantization error as

Vn\displaystyle V_{n} =118(Vn1+Vn21n121n22)+1n2\displaystyle=\frac{1}{18}\Big{(}V_{n_{1}}+V_{n_{2}}-\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\Big{)}+\frac{1}{n^{2}}
=118(118m1(2mn1+19(n12m1))+118m1(2mn2+19(n22m1)))+1n2\displaystyle=\frac{1}{18}\Big{(}\frac{1}{18^{m-1}}\Big{(}2^{m}-n_{1}+\frac{1}{9}(n_{1}-2^{m-1})\Big{)}+\frac{1}{18^{m-1}}\Big{(}2^{m}-n_{2}+\frac{1}{9}(n_{2}-2^{m-1})\Big{)}\Big{)}+\frac{1}{n^{2}}
=118m(2m+1n+19(n2m))+1n2.\displaystyle=\frac{1}{18^{m}}\Big{(}2^{m+1}-n+\frac{1}{9}(n-2^{m})\Big{)}+\frac{1}{n^{2}}.

Thus, the theorem is true if (n)=m\ell(n)=m. Hence, by the induction principle, the proof of the theorem is complete. ∎

4. Constrained quantization dimension and constrained quantization coefficient

Let P=12PT11+12PT21P=\frac{1}{2}P\circ T_{1}^{-1}+\frac{1}{2}P\circ T_{2}^{-1} be a unique Borel probability measure on \mathbb{R} with support the Cantor set CC generated by the two contractive similarity mappings T1(x)=13xT_{1}(x)=\frac{1}{3}x and T2(x)=13x+23T_{2}(x)=\frac{1}{3}x+\frac{2}{3} for all xx\in\mathbb{R}. Since the Cantor set CC under investigation satisfies the strong separation condition, with each TjT_{j} having a contracting factor of 13\frac{1}{3}, the Hausdorff dimension of the Cantor set is equal to the similarity dimension. Hence, from the equation 2(13)β=12(\frac{1}{3})^{\beta}=1, we have dimH(C)=β=log2log3\dim_{\text{H}}(C)=\beta=\frac{\log 2}{\log 3}. It is known that the unconstrained quantization dimension of the probability measure PP exists and equals the Hausdorff dimension of the Cantor set (see [8]). The work in this section shows that it is not true in the constrained case, i.e., the constrained quantization dimension D(P)D(P) of the probability measure PP, though it exists, is not necessarily equal to the Hausdorff dimension of the Cantor set. In this section, we show that the constrained quantization dimension D(P)D(P) exists and equals one. We further show that the D(P)D(P)-dimensional constrained quantization coefficient exists as a finite positive number and equals D(P)D(P), which is also not true in the unconstrained quantization for the Cantor distribution.

Theorem 4.1.

The constrained quantization dimension D(P)D(P) of the probability measure PP exists, and D(P)=1D(P)=1.

Proof.

For nn\in\mathbb{N} with n2n\geq 2, let (n)\ell(n) be the unique natural number such that 2(n)n<2(n)+12^{\ell(n)}\leq n<2^{\ell(n)+1}. Then, V2(n)+1VnV2(n)V_{2^{\ell(n)+1}}\leq V_{n}\leq V_{2^{\ell(n)}}. By Theorem 3.12, we see that V2(n)+10V_{2^{\ell(n)+1}}\to 0 and V2(n)0V_{2^{\ell(n)}}\to 0 as nn\to\infty, and so Vn0V_{n}\to 0 as nn\to\infty, i.e., V=0V_{\infty}=0. We can take nn large enough so that V2(n)V<1V_{2^{\ell(n)}}-V_{\infty}<1. Then,

0<log(V2(n)V)log(VnV)log(V2(n)+1V)0<-\log(V_{2^{\ell(n)}}-V_{\infty})\leq-\log(V_{n}-V_{\infty})\leq-\log(V_{2^{\ell(n)+1}}-V_{\infty})

yielding

2(n)log2log(V2(n)+1V)2lognlog(VnV)2((n)+1)log2log(V2(n)V).\frac{2\ell(n)\log 2}{-\log(V_{2^{\ell(n)+1}}-V_{\infty})}\leq\frac{2\log n}{-\log(V_{n}-V_{\infty})}\leq\frac{2(\ell(n)+1)\log 2}{-\log(V_{2^{\ell(n)}}-V_{\infty})}.

Notice that

limn2(n)log2log(V2(n)+1V)\displaystyle\lim_{n\to\infty}\frac{2\ell(n)\log 2}{-\log(V_{2^{\ell(n)+1}}-V_{\infty})} =limn2(n)log2log(V9(n)+1+14(n)+1)=1, and\displaystyle=\lim_{n\to\infty}\frac{2\ell(n)\log 2}{-\log(\frac{V}{9^{\ell(n)+1}}+\frac{1}{4^{\ell(n)+1}})}=1,\text{ and }
limn2((n)+1)log2log(V2(n)V)\displaystyle\lim_{n\to\infty}\frac{2(\ell(n)+1)\log 2}{-\log(V_{2^{\ell(n)}}-V_{\infty})} =limn2((n)+1)log2log(V9(n)+14(n))=1.\displaystyle=\lim_{n\to\infty}\frac{2(\ell(n)+1)\log 2}{-\log(\frac{V}{9^{\ell(n)}}+\frac{1}{4^{\ell(n)}})}=1.

Hence, limn2lognlog(VnV)=1\lim_{n\to\infty}\frac{2\log n}{-\log(V_{n}-V_{\infty})}=1, i.e., the constrained quantization dimension D(P)D(P) of the probability measure PP exists and D(P)=1D(P)=1. Thus, the proof of the theorem is complete. ∎

Theorem 4.2.

The D(P)D(P)-dimensional constrained quantization coefficient for PP is a finite positive number and equals the constrained quantization dimension D(P)D(P).

Proof.

For nn\in\mathbb{N} with n2n\geq 2, let (n)\ell(n) be the unique natural number such that 2(n)n<2(n)+12^{\ell(n)}\leq n<2^{\ell(n)+1}. Then, 02(n)+1n<2(n)0\leq 2^{\ell(n)+1}-n<2^{\ell(n)} and 019(n2(n))<2(n)0\leq\frac{1}{9}(n-2^{\ell(n)})<2^{\ell(n)}. Hence,

02(n)+1n+19(n2(n))<2(n)+1,0\leq 2^{\ell(n)+1}-n+\frac{1}{9}(n-2^{\ell(n)})<2^{\ell(n)+1},

which implies

0118(n)V(2(n)+1n+19(n2(n)))<2(n)+118(n)\displaystyle 0\leq\frac{1}{18^{\ell(n)}}V\Big{(}2^{\ell(n)+1}-n+\frac{1}{9}(n-2^{\ell(n)})\Big{)}<\frac{2^{\ell(n)+1}}{18^{\ell(n)}}

yielding

0n2(118(n)V(2(n)+1n+19(n2(n))))<n22(n)+118(n)<22(n)+22(n)+118(n)=8(49)(n).0\leq{n^{2}}\Big{(}\frac{1}{18^{\ell(n)}}V\Big{(}2^{\ell(n)+1}-n+\frac{1}{9}(n-2^{\ell(n)})\Big{)}\Big{)}<n^{2}\frac{2^{\ell(n)+1}}{18^{\ell(n)}}<\frac{2^{2\ell(n)+2}2^{\ell(n)+1}}{18^{\ell(n)}}=8\Big{(}\frac{4}{9}\Big{)}^{\ell(n)}.

Hence, by the squeeze theorem, we have

limnn2(118(n)V(2(n)+1n+19(n2(n))))=0.\lim_{n\to\infty}{n^{2}}\Big{(}\frac{1}{18^{\ell(n)}}V\Big{(}2^{\ell(n)+1}-n+\frac{1}{9}(n-2^{\ell(n)})\Big{)}\Big{)}=0.

Again, as shown in the proof of Theorem 4.1, we have V=limnVn=0V_{\infty}=\lim_{n\to\infty}V_{n}=0. Thus, we deduce that

limnn2(VnV)=1,\lim_{n\to\infty}n^{2}(V_{n}-V_{\infty})=1,

i.e., the D(P)D(P)-dimensional constrained quantization coefficient for PP exists as a finite positive number and equals the constrained quantization dimension D(P)D(P), which is the theorem. ∎

5. Constrained quantization with some other families of constraints

As defined in the previous sections, let P=12PT11+12PT21P=\frac{1}{2}P\circ T_{1}^{-1}+\frac{1}{2}P\circ T_{2}^{-1} be the unique Borel probability measure on \mathbb{R} with support the Cantor set CC generated by the two contractive similarity mappings T1(x)=13xT_{1}(x)=\frac{1}{3}x and T2(x)=13x+23T_{2}(x)=\frac{1}{3}x+\frac{2}{3} for all xx\in\mathbb{R}. In this section, in the following subsections, we give the optimal sets of nn-points and the nnth constrained quantization errors for different families of constraints. Then, for each family, we investigate the constrained quantization dimension and the constrained quantization coefficient.

5.1. Constrained quantization when the family is Sj={(x,y):0x1 and y=1+1j}S_{j}=\{(x,y):0\leq x\leq 1\text{ and }y=1+\frac{1}{j}\} for jj\in\mathbb{N}.

Using the similar arguments as Lemma 3.3, it can be shown that if αn\alpha_{n} is an optimal set of nn-points for PP, then αnSn\alpha_{n}\subseteq S_{n} for all nn\in\mathbb{N}. Let us first state the following theorem, the proof of which is similar to Theorem 3.12.

Theorem 5.1.1.

For nn\in\mathbb{N} with n2n\geq 2 let (n)\ell(n) be the unique natural number with 2(n)n<2(n)+12^{\ell(n)}\leq n<2^{\ell(n)+1}. For I{1,2}(n)I\subset\{1,2\}^{\ell(n)} with card(I)=n2(n)(I)=n-2^{\ell(n)} let αn(I)Sn\alpha_{n}(I)\subseteq S_{n} be the set such that

αn(I)={(a(σ),1+1n):σ{1,2}(n)I}{(a(σ1),1+1n):σI}{(a(σ2),1+1n):σI}.\alpha_{n}(I)=\{(a(\sigma),1+\frac{1}{n}):\sigma\in\{1,2\}^{\ell(n)}\setminus I\}\cup\{(a(\sigma 1),1+\frac{1}{n}):\sigma\in I\}\cup\{(a(\sigma 2),1+\frac{1}{n}):\sigma\in I\}.

Then, αn:=αn(I)\alpha_{n}:=\alpha_{n}(I) forms an optimal set of nn-points for PP with constrained quantization error

Vn=118(n)V(2(n)+1n+19(n2(n)))+(1+1n)2,V_{n}=\frac{1}{18^{\ell(n)}}V\Big{(}2^{\ell(n)+1}-n+\frac{1}{9}(n-2^{\ell(n)})\Big{)}+(1+\frac{1}{n})^{2},

where VV is the variance.

Remark 5.1.2.

Notice that here V=limnVn=1V_{\infty}=\mathop{\lim}\limits_{n\to\infty}V_{n}=1. Thus, proceeding in the similar way as Theorem 4.1 and Theorem 4.2, it can be seen that

limn2lognlog(VnV)=2 and limnn2(VnV)=,\lim_{n\to\infty}\frac{2\log n}{-\log(V_{n}-V_{\infty})}=2\text{ and }\lim_{n\to\infty}n^{2}(V_{n}-V_{\infty})=\infty,

i.e., the constrained quantization dimension D(P)D(P) exists and equals 22, but the constrained quantization coefficient does not exist.

5.2. Constrained quantization when the family is Sj={(x,y):0x1 and y=1}S_{j}=\{(x,y):0\leq x\leq 1\text{ and }y=1\} for jj\in\mathbb{N}.

Proceeding in a similar way as Theorem 3.12, we can show that the following theorem is true.

Theorem 5.2.1.

For nn\in\mathbb{N} with n2n\geq 2 let (n)\ell(n) be the unique natural number with 2(n)n<2(n)+12^{\ell(n)}\leq n<2^{\ell(n)+1}. For I{1,2}(n)I\subset\{1,2\}^{\ell(n)} with card(I)=n2(n)(I)=n-2^{\ell(n)} let αn(I)Sn\alpha_{n}(I)\subseteq S_{n} be the set such that

αn(I)={(a(σ),1):σ{1,2}(n)I}{(a(σ1),1):σI}{(a(σ2),1):σI}.\alpha_{n}(I)=\{(a(\sigma),1):\sigma\in\{1,2\}^{\ell(n)}\setminus I\}\cup\{(a(\sigma 1),1):\sigma\in I\}\cup\{(a(\sigma 2),1):\sigma\in I\}.

Then, αn:=αn(I)\alpha_{n}:=\alpha_{n}(I) forms an optimal set of nn-points for PP with constrained quantization error

Vn=118(n)V(2(n)+1n+19(n2(n)))+1,V_{n}=\frac{1}{18^{\ell(n)}}V\Big{(}2^{\ell(n)+1}-n+\frac{1}{9}(n-2^{\ell(n)})\Big{)}+1,

where VV is the variance.

Notice that here V=limnVn=1V_{\infty}=\mathop{\lim}\limits_{n\to\infty}V_{n}=1. If β\beta is the Hausdorff dimension of the Cantor set 𝒞\mathcal{C}, then, β=log2log3\beta=\frac{\log 2}{\log 3}. Then, the following lemma and theorems are equivalent to the lemma and theorems that appear in the last section in [8]. For the proofs, one can consult [8].

Theorem 5.2.2.

The set of accumulation points of the sequence (n2β(VnV))n\Big{(}n^{\frac{2}{\beta}}(V_{n}-V_{\infty})\Big{)}_{n\in\mathbb{N}} equals

[V,f(178+4β)],\Big{[}V,f\Big{(}\frac{17}{8+4\beta}\Big{)}\Big{]},

where f:[1,2]f:[1,2]\to\mathbb{R} is such that f(x)=172x2β(178x)f(x)=\frac{1}{72}x^{\frac{2}{\beta}}(17-8x).

Lemma 5.2.3.

Let nn\in\mathbb{N}. Then,

172n2β(VnV)98.\frac{1}{72}\leq n^{\frac{2}{\beta}}(V_{n}-V_{\infty})\leq\frac{9}{8}.
Theorem 5.2.4.

The constrained quantization dimension of PP equals the Hausdorff dimension β:=log2log3\beta:=\frac{\log 2}{\log 3} of the Cantor set, i.e.,

D(P)=limn2lognlog(VnV)=β.D(P)=\lim_{n\to\infty}\frac{2\log n}{-\log(V_{n}-V_{\infty})}=\beta.
Remark 5.2.5.

Thus, in this case, we see that the constrained quantization dimension exists and equals the Hausdorff dimension β\beta of the Cantor set, but the constrained quantization coefficient does not exist.

5.3. Constrained quantization when the family is Sj={(x,y):0x1 and y=11j}S_{j}=\{(x,y):0\leq x\leq 1\text{ and }y=1-\frac{1}{j}\} for jj\in\mathbb{N}.

Using the similar arguments as Lemma 3.3, it can be shown that if αn\alpha_{n} is an optimal set of nn-points for PP, then αnS1\alpha_{n}\subseteq S_{1} for all nn\in\mathbb{N}. Let us first state the following theorem, the proof of which is similar to Theorem 3.12.

Theorem 5.3.1.

For nn\in\mathbb{N} with n2n\geq 2 let (n)\ell(n) be the unique natural number with 2(n)n<2(n)+12^{\ell(n)}\leq n<2^{\ell(n)+1}. For I{1,2}(n)I\subset\{1,2\}^{\ell(n)} with card(I)=n2(n)(I)=n-2^{\ell(n)} let αn(I)S1\alpha_{n}(I)\subseteq S_{1} be the set such that

αn(I)={(a(σ),0):σ{1,2}(n)I}{(a(σ1),0):σI}{(a(σ2),0):σI}.\alpha_{n}(I)=\{(a(\sigma),0):\sigma\in\{1,2\}^{\ell(n)}\setminus I\}\cup\{(a(\sigma 1),0):\sigma\in I\}\cup\{(a(\sigma 2),0):\sigma\in I\}.

Then, αn:=αn(I)\alpha_{n}:=\alpha_{n}(I) forms an optimal set of nn-points for PP with constrained quantization error

Vn=118(n)V(2(n)+1n+19(n2(n))),V_{n}=\frac{1}{18^{\ell(n)}}V\Big{(}2^{\ell(n)+1}-n+\frac{1}{9}(n-2^{\ell(n)})\Big{)},

where VV is the variance.

Remark 5.3.2.

By Theorem 5.3.1, we see that for the family Sj={(x,y):0x1 and y=11j}S_{j}=\{(x,y):0\leq x\leq 1\text{ and }y=1-\frac{1}{j}\}, where jj\in\mathbb{N}, the optimal sets of nn-points and the corresponding constrained quantization error VnV_{n} coincide with the optimal sets of nn-means and the corresponding quantization error for the Cantor distribution PP that occurs in [8]. Thus, all the results in [8] are also true here.

Acknowledgement

The first author is grateful to her supervisor Professor Tanmoy Som of the IIT(BHU), Varanasi, India, for his constant support in preparing this manuscript.

Declaration

Conflicts of interest. We do not have any conflict of interest.

Data availability: No data were used to support this study.

Code availability: Not applicable

Authors’ contributions: Each author contributed equally to this manuscript.

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