# Constrained quantization for probability distributions

1Megha Pandey  and  2Mrinal K. 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<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

In this paper, for a Borel probability measure $P$ on a normed space
$\mathbb{R}^{k}$, we extend the definitions of $n$th unconstrained
quantization error, unconstrained quantization dimension, and unconstrained
quantization coefficient, which traditionally in the literature are known as
$n$th quantization error, quantization dimension, and quantization
coefficient, to the definitions of $n$th constrained quantization error,
constrained quantization dimension, and constrained quantization coefficient.
The work in this paper extends the theory of quantization and opens a new area
of research. In unconstrained quantization, the elements in an optimal set are
the conditional expectations in their own Voronoi regions, and it is not true
in constrained quantization. In unconstrained quantization, if the support of
$P$ contains infinitely many elements, then an optimal set of $n$-means always
contains exactly $n$ elements, and it is not true in constrained quantization.
It is known that the unconstrained quantization dimension for an absolutely
continuous probability measure equals the Euclidean dimension of the
underlying space where the support of the probability measure is defined. In
this paper, we show that this fact is not true as well for the constrained
quantization dimension. It is known that the unconstrained quantization
coefficient for an absolutely continuous probability measure exists as a
unique finite positive number. From the work in this paper, it can be seen
that the constrained quantization coefficient for an absolutely continuous
probability measure can be any nonnegative number depending on the constraint
that occurs in the definition of $n$th constrained quantization error.

###### Key words and phrases:

Probability measure, constrained quantization error, optimal sets of
$n$-points, constrained quantization dimension, constrained quantization
coefficient.

###### 2010 Mathematics Subject Classification:

60Exx, 94A34.

##  1\. Introduction

The most common form of quantization is rounding-off. Its purpose is to reduce
the cardinality of the representation space, in particular, when the input
data is real-valued. It has broad applications in communications, information
theory, signal processing, and data compression (see [1, 2, 3, 4, 5, 6, 7]).
For $k\in\mathbb{N}$, where $\mathbb{N}$ is the set of natural numbers, let
$d$ be a metric induced by a norm $\|\cdot\|$ on $\mathbb{R}^{k}$. Let $P$ be
a Borel probability measure on $\mathbb{R}^{k}$ and $r\in(0,\infty)$. Let $S$
be a nonempty closed subset of $\mathbb{R}^{k}$. Then, the distortion error
for $P$, of order $r$, with respect to a set $\alpha\subseteq S$, denoted by
$V_{r}(P;\alpha)$, is defined as

$V_{r}(P;\alpha)=\int\mathop{\min}\limits_{a\in\alpha}d(x,a)^{r}dP(x).$

Then, for $n\in\mathbb{N}$, the $n$th constrained quantization error for $P$,
of order $r$, is defined as

$V_{n,r}:=V_{n,r}(P)=\inf\Big{\\{}V_{r}(P;\alpha):\alpha\subseteq
S,~{}1\leq\text{card}(\alpha)\leq n\Big{\\}},$ (1)

where $\text{card}(A)$ represents the cardinality of a set $A$. If in the
definition of $n$th constrained quantization error, the set $S$, known as
constraint, is chosen as the set $\mathbb{R}^{k}$ itself, then the $n$th
constrained quantization error is referred to as the $n$th unconstrained
quantization error, which traditionally in the literature is referred to as
the $n$th quantization error. For some recent work in the direction of
unconstrained quantization, one can see [2, 8, 9, 10, 11, 12, 13, 14, 15, 16,
17, 18]. For the probability measure $P$, we assume that

$\int d(x,0)^{r}dP(x)<\infty.$ (2)

A set $\alpha$ for which the infimum in (1) exists and does not contain more
than $n$ elements is called an optimal set of $n$-points for $P$. The
collection of all optimal sets of $n$-points for $P$ is denoted by
$\mathcal{C}_{n,r}(P)$.

######  Proposition 1.1.

Let the assumption (2) be true. Then, $V_{n,r}(P)$ exists and is a decreasing
sequence of finite nonnegative numbers.

###### Proof.

Under the assumption, we claim that $\int d(x,a)^{r}dP(x)<\infty$ for any
$a\in\mathbb{R}^{k}$. Since $d(x,a)\leq d(x,0)+d(0,a)\leq
2\max\\{d(x,0),d(0,a)\\}$, we have

$d(x,a)^{r}\leq 2^{r}\max\\{d(x,0)^{r},d(0,a)^{r}\\}\leq
2^{r}\Big{(}d(x,0)^{r}+d(0,a)^{r}\Big{)}$

implying

$\displaystyle\int d(x,a)^{r}\,dP(x)$ $\displaystyle\leq
2^{r}\Big{(}\int(d(x,0)^{r}+d(0,a)^{r})\,dP(x)\Big{)}=2^{r}\Big{(}\int
d(x,0)^{r}\,dP(x)+d(0,a)^{r}\Big{)}<\infty.$

Hence, for $0<r<\infty$, $\int d(x,0)^{r}dP(x)<\infty$ yields the fact that
$\int d(x,a)^{r}dP(x)<\infty$ for any $a\in\mathbb{R}^{k}$, which is the
claim. By the claim, for any finite subset $\alpha\subseteq
S\subseteq\mathbb{R}^{k}$, we have

$0\leq\int\min_{a\in\alpha}d(x,a)^{r}dP(x)\leq\min_{a\in\alpha}\int
d(x,a)^{r}dP(x)<\infty\text{ implying }0\leq V_{r}(P;\alpha)<\infty,$

for $n\in\mathbb{N}$ which yields

$0\leq\inf\Big{\\{}V_{r}(P;\alpha):\alpha\subseteq
S,~{}1\leq\text{card}(\alpha)\leq n\Big{\\}}<\infty,$

i.e., $V_{n,r}(P)$ exists as a finite nonnegative number. The statement that
$V_{n,r}(P)$ is decreasing directly follows from the definition. Thus, the
proof of the proposition is complete. ∎

For a finite set $\alpha\subset\mathbb{R}^{k}$, and $a\in\alpha$, by
$M(a|\alpha)$ we denote the set of all elements in $\mathbb{R}^{k}$ which are
nearest to $a$ among all the elements in $\alpha$, i.e.,
$M(a|\alpha)=\\{x\in\mathbb{R}^{k}:d(x,a)=\mathop{\min}\limits_{b\in\alpha}d(x,b)\\}.$
$M(a|\alpha)$ is called the Voronoi region in $\mathbb{R}^{k}$ generated by
$a\in\alpha$. On the other hand, the set $\\{M(a|\alpha):a\in\alpha\\}$ is
called the Voronoi diagram or Voronoi tessellation of $\mathbb{R}^{k}$ with
respect to the set $\alpha$.

The following two propositions reflects two important properties of
constrained quantization.

######  Proposition 1.2.

In constrained quantization for any Borel probability measure $P$, an optimal
set of one-point always exists, i.e., $\mathcal{C}_{1,r}(P)$ is nonempty.

###### Proof.

Let $0<r<\infty$. Define a function

$\psi_{r}:S\to\mathbb{R}^{+}:\psi_{r}(a)=\int d(x,a)^{r}\,dP(x).$

The function $\psi_{r}$ is obviously continuous. Then, for every
$c\in\mathbb{R}^{+}$ with $c>\mathop{\inf}\limits_{b\in S}\int
d(x,b)^{r}\,dP(x)$, the level sets

$\\{\psi_{r}\leq c\\}:=\\{a\in S:\int d(x,a)^{r}\,dP(x)\leq c\\},$

are closed subsets of $S$. Proceeding in the similar way, as shown in the
previous proposition, for any $0<r<\infty$, we have

$\displaystyle d(0,a)^{r}\leq 2^{r}\Big{(}d(x,a)^{r}+d(x,0)^{r}\Big{)}.$

Thus, for $a\in\\{\psi_{r}\leq c\\}$, we have

$\displaystyle d(0,a)^{r}=\int d(0,a)^{r}\,dP(x)$ $\displaystyle\leq
2^{r}\Big{(}\int(d(x,a)^{r}+d(x,0)^{r})\,dP(x)\Big{)}\leq
2^{r}\Big{(}c+E\|X\|^{r}\Big{)},$

i.e.,

$d(0,a)\leq 2\Big{(}c+E\|X\|^{r}\Big{)}^{\frac{1}{r}}\text{ yielding
}\\{\psi_{r}\leq c\\}\subseteq
B\Big{(}0,2\Big{(}c+E\|X\|^{r}\Big{)}^{\frac{1}{r}}\Big{)},$

where $E\|X\|^{r}=\int d(x,0)^{r}\,dP(x)$. Hence, the level sets are bounded.
As the level sets are both bounded and closed, they are compact. Let us now
consider a decreasing sequence $\\{c_{n}\\}$ of elements in $\mathbb{R}^{+}$
such that

$c_{n}>\mathop{\inf}\limits_{b\in S}\int d(x,b)^{r}\,dP(x)\text{ and
}c_{n}\to\mathop{\inf}\limits_{b\in S}\int d(x,b)^{r}\,dP(x).$ (3)

Then,

$\\{a\in S:\int d(x,a)^{r}\,dP(x)\leq c_{n+1}\\}\subseteq\\{a\in S:\int
d(x,a)^{r}\,dP(x)\leq c_{n}\\},\text{ i.e., }\\{\psi_{r}\leq
c_{n+1}\\}\subseteq\\{\psi_{r}\leq c_{n}\\}.$

Also, $\\{\psi_{r}\leq c_{n}\\}\neq\emptyset$ for all $n\in\mathbb{N}$. If
this is not true, then there will exist an element $c_{N}$ for some
$N\in\mathbb{N}$ such that for all $b\in S$, we have

$c_{N}<\int d(x,b)^{r}\,dP(x)\text{ yielding }c_{N}\leq\inf_{b\in S}\int
d(x,b)^{r}\,dP(x),$

which contradicts (3). Thus, we see that the level sets $\\{\psi_{r}\leq
c_{n}\\}$ form a nested sequence of nonempty compact sets, and hence,

$\bigcap_{n=1}^{\infty}\\{\psi_{r}\leq c_{n}\\}\neq\emptyset.$

Let $a\in\mathop{\bigcap}\limits_{n=1}^{\infty}\\{\psi_{r}\leq c_{n}\\}$.
Then,

$\mathop{\inf}\limits_{b\in S}\int d(x,b)^{r}\,dP(x)\leq\int
d(x,a)^{r}\,dP(x)<c_{n}\text{ for all }n\in\mathbb{N},$

which by squeeze theorem implies that

$\mathop{\inf}\limits_{b\in S}\int d(x,b)^{r}\,dP(x)=\int d(x,a)^{r}\,dP(x),$

i.e., $\\{a\\}$ forms an optimal set of one-point, i.e.,
$\\{a\\}\in\mathcal{C}_{1,r}$, i.e., $\mathcal{C}_{1,r}$ is nonempty. Thus,
the proof of the proposition is complete. ∎

######  Proposition 1.3.

In constrained quantization for a Borel probability measure $P$ with
constraint $S$, an optimal set $\alpha_{n}$ of $n$-points will contain exactly
$n$ elements if there exist at least $n$ elements in $S$ the Voronoi regions
of which have positive probability.

###### Proof.

By Proposition 1.2, an optimal set of one-point always exists. Now, we prove
the proposition by contrapositive. Assume that $\alpha_{n}$ is an optimal set
of $n$-points which does not contain exactly $n$ elements, i.e., $\alpha_{n}$
contains $k$ elements for some $k<n$. This implies that the constraint $S$
cannot have more than $k$ elements with positive probability. If the
constraint $S$ has more than $k$ elements then by adding at most $n-k$
elements, which have positive probability, in $\alpha_{n}$ we can strictly
reduce the distortion error, which contradicts the fact that $\alpha_{n}$ is
an optimal set of $n$-points. Thus, the proof of the proposition is complete.
∎

In the case of unconstrained quantization as described in [2], the elements in
an optimal set are the conditional expectations in their own Voronoi regions.
This fact is not true in the case of constrained quantization. Because of
that, in the case of constrained quantization, a set $\alpha$ for which the
infimum in (1) exists and contains no more than $n$ elements is called an
optimal set of $n$-points instead of calling it as an optimal set of
$n$-means. Elements of an optimal set are called optimal elements. In
unconstrained quantization, as described in [2], if the support of $P$
contains infinitely many elements, then an optimal set of $n$-means always
contains exactly $n$ elements. It is not true in constrained quantization. In
the case of constrained quantization, an optimal set of one-point always
contains exactly one element; on the other hand, if the support of $P$
contains infinitely many elements, then an optimal set of $n$-points for any
$n\geq 2$ does not necessarily contain exactly $n$ elements. Notice that
unconstrained quantization, as described in [2], is a special case of
constrained quantization. There are some properties in unconstrained
quantization that are not true in constrained quantization.

This paper deals with $r=2$ and $k=2$, and the metric on $\mathbb{R}^{2}$ as
the Euclidean metric induced by the Euclidean norm $\lVert\cdot\rVert$. Thus,
instead of writing $V_{r}(P;\alpha)$ and $V_{n,r}:=V_{n,r}(P)$ we will write
them as $V(P;\alpha)$ and $V_{n}:=V_{n}(P)$.

###  1.4. Delineation

In this paper, we have determined the optimal sets of $n$-points and the $n$th
constrained quantization errors for all $n\in\mathbb{N}$ for different uniform
distributions: in Section 3, the uniform distribution has support a closed
interval $[a,b]$ and the optimal elements lie on another line segment; in
Section 4, the uniform distribution has support a circle, and the optimal
elements lie on another circle; and in Section 5, the uniform distribution has
support a chord of a circle, and the optimal elements lie on that circle.
Finally, in Section 6, we give the definitions of the constrained quantization
dimension and the constrained quantization coefficient and, with different
examples, show the differences between the constrained and the unconstrained
quantization dimensions, and the constrained and the unconstrained
quantization coefficients.

##  2\. Preliminaries

In this section, we give some basic notations and definitions which we have
used throughout this paper. For any two elements $\tilde{a}:=(a_{1},a_{2})$
and $\tilde{b}:=(b_{1},b_{2})$ in $\mathbb{R}^{2}$, we write
$\rho(\tilde{a},\tilde{b}):=\|(a_{1},a_{2})-(b_{1},b_{2})\|^{2}=(a_{1}-b_{1})^{2}+(a_{2}-b_{2})^{2}$,
which gives the squared Euclidean distance between the two elements
$(a_{1},a_{2})$ and $(b_{1},b_{2})$. Let $p$ and $q$ be two elements that
belong to an optimal set of $n$-points for some positive integer $n$, and let
$e$ be an element on the boundary of the Voronoi regions of the elements $p$
and $q$. Since the boundary of the Voronoi regions of any two elements is the
perpendicular bisector of the line segment joining the elements, we have

$\|p-e\|=\|q-e\|\text{ yielding }\|p-e\|^{2}=\|q-e\|^{2},\text{ i.e.,
}\rho(p,e)-\rho(q,e)=0.$

We call such an equation a canonical equation.

######  Fact 2.1.

Let $P$ be a Borel probability measure on $\mathbb{R}^{2}$ with support a
curve $C$ given by the parametric representations

$x=f(t)\text{ and }y=g(t),\text{ where }a\leq t\leq b.$

Let us fix a point on the curve $C$. Let $s$ be the distance of a point on the
curve tracing along the curve starting from the fixed point. Then,

$ds=\sqrt{(dx)^{2}+(dy)^{2}}=\sqrt{(f^{\prime}(t))^{2}+(g^{\prime}(t))^{2}}|dt|,$
(4)

where $|dt|=dt$ if $t$ increases, and $|dt|=-dt$ if $t$ decreases. Then,

$dP(s)=P(ds)=f(x,y)ds,$

where $f(x,y)$ is the probability density function for the probability measure
$P$, i.e., $f(x,y)$ is a real-valued function on $\mathbb{R}^{2}$ with the
following properties: $f(x,y)\geq 0$ for all $(x,y)\in\mathbb{R}^{2}$, and

$\int_{\mathbb{R}^{2}}f(x,y)\,dA(x,y)=\int_{\mathbb{R}^{2}\setminus
C}f(x,y)\,dA(x,y)+\int_{C}f(x,y)\,dA(x,y)=\int_{C}f(x,y)\,dA(x,y)=1,$

where for any $(x,y)\in\mathbb{R}^{2}$ by $dA(x,y)$, it is meant the
infinitesimal area $dxdy$, and if $(x,y)\in C$, then by $dA(x,y)$, it is meant
the infinitesimal length $ds$ given by (4).

##  3\. Constrained quantization when the support lies on a line segment and
the optimal elements lie on another line segment

Let $a,b\in\mathbb{R}$ with $a<b$, and $c,m\in\mathbb{R}$. Let $L$ be a line
given by $y=mx+c$, the parametric representation of which is

$L:=\\{(x,mx+c):x\in\mathbb{R}\\}.$

Let $P$ be a Borel probability measure on $\mathbb{R}^{2}$ such that $P$ is
uniform on its support $\\{(x,y)\in\mathbb{R}^{2}:a\leq x\leq b\text{ and
}y=0\\}$. Then, the probability density function $f$ for $P$ is given by

$f(x,y)=\left\\{\begin{array}[]{ccc}\frac{1}{b-a}&\text{ if }a\leq x\leq
b\text{ and }y=0,\\\ 0&\text{ otherwise}.\end{array}\right.$

Recall Fact 2.1. Here we have $dP(s)=P(ds)=P(dx)=dP(x)=f(x,0)dx$. In this
section, we determine the optimal sets of $n$-points and the $n$th constrained
quantization errors for the probability measure $P$ for all positive integers
$n$ so that the elements in the optimal sets lie on the line $L$ between the
two elements $(d,md+c)$ and $(e,me+c)$, where $d,e\in\mathbb{R}$ with $d<e$.

$A_{1}$$A_{2}$$A_{3}$$B_{1}$$(a,0)$$(b,0)$$B_{2}$$B_{3}$$A_{n}$$B_{n}$$O$$y=mx+c$$A_{4}$$B_{4}$$(d,md+c)$$(e,me+c)$
Figure 1. Support of the probability distribution $P$ is the closed interval
joining the points $(a,0)$ and $(b,0)$; $A_{i}(a_{i},ma_{i}+c)$ are the
elements in an optimal set of $n$-points lying on the line $y=mx+c$ between
the two points $(d,md+c)$ and $(e,me+c)$; $B_{i}((m^{2}+1)a_{i}+mc,0)$ are the
points where the perpendiculars through $A_{i}$ on the line $y=mx+c$ intersect
the support of $P$.

Let us now give the following theorem.

######  Theorem 3.1.

Let $P$ be a Borel probability measure on $\mathbb{R}^{2}$ such that $P$ is
uniform on its support $\\{(x,y)\in\mathbb{R}^{2}:a\leq x\leq b\text{ and
}y=0\\}$. For $n\in\mathbb{N}$ with $n\geq 2$, let
$\alpha_{n}:=\\{(a_{i},ma_{i}+c):1\leq i\leq n\\}$ be an optimal set of
$n$-points for $P$ so that the elements in the optimal sets lie on the line
$L$ between the two elements $(d,md+c)$ and $(e,me+c)$ (see Figure 1), where
$d,e\in\mathbb{R}$ with $d<e$. Assume that

$\max\\{a,(m^{2}+1)d+mc\\}=a\text{ and }\min\\{b,(m^{2}+1)e+mc\\}=b.$

Then, $a_{i}=\frac{2i-1}{2n(1+m^{2})}(b-a)+\frac{a-cm}{1+m^{2}}$ for $1\leq
i\leq n$ with constrained quantization error

$\displaystyle V_{n}=$
$\displaystyle\frac{1}{12\left(m^{2}+1\right)n^{3}}\Big{(}-48(a-b)^{2}m^{2}+(a-b)(a-b+72cm+8(11a-2b)m^{2})n$
$\displaystyle\qquad\qquad-12(a-b)m(5c+(4a+b)m)n^{2}+12(c+am)^{2}n^{3}\Big{)}.$

###### Proof.

For $n\geq 2$, let $\alpha_{n}:=\\{(a_{i},ma_{i}+c):1\leq i\leq n\\}$ be an
optimal set of $n$-points on $L$ such that $d\leq
a_{1}<a_{2}<\cdots<a_{n-1}<a_{n}\leq e$. Notice that the boundary of the
Voronoi region of the element $(a_{1},ma_{1}+c)$ intersects the support of $P$
at the elements $(a,0)$ and $((m^{2}+1)\frac{(a_{1}+a_{2})}{2}+mc,0)$, the
boundary of the Voronoi region of $(a_{n},ma_{n}+c)$ intersects the support of
$P$ at the elements $((m^{2}+1)\frac{(a_{n-1}+a_{n})}{2}+mc,0)$ and $(b,0)$.
On the other hand, the boundaries of the Voronoi regions of $(a_{i},ma_{i}+c)$
for $2\leq i\leq n-1$ intersect the support of $P$ at the elements
$((m^{2}+1)\frac{(a_{i-1}+a_{i})}{2}+mc,0)$ and
$((m^{2}+1)\frac{(a_{i}+a_{i+1})}{2}+mc,0)$. Since the Voronoi regions of the
elements in an optimal set must have positive probability, we have

$\displaystyle\max\\{a,(m^{2}+1)d+mc\\}$
$\displaystyle\leq(m^{2}+1)a_{1}+mc<(m^{2}+1)a_{2}+mc$
$\displaystyle<\cdots<(m^{2}+1)a_{n}+mc\leq\min\\{b,(m^{2}+1)e+mc\\}.$

Let us consider the following two cases:

Case 1: $n=2$.

In this case, the distortion error due to the set $\alpha_{2}$ is given by

$\displaystyle V(P;\alpha_{2})=$
$\displaystyle\int_{\mathbb{R}}\min_{a\in\alpha_{2}}\|(x,0)-a\|^{2}dP(x)=\frac{1}{b-a}\Big{(}\int_{a}^{(m^{2}+1)\frac{(a_{1}+a_{2})}{2}+mc}\rho((x,0),(a_{1},ma_{1}+c))\,dx$
$\displaystyle\qquad+\int_{(m^{2}+1)\frac{(a_{1}+a_{2})}{2}+mc}^{b}\rho((x,0),(a_{2},ma_{2}+c))\,dx\Big{)}.$

Notice that $V(P;\alpha_{2})$ is not always differentiable with respect to
$a_{1}$ and $a_{2}$. By the hypothesis, we have

$\max\\{a,(m^{2}+1)d+mc\\}=a\text{ and }\min\\{b,(m^{2}+1)e+mc\\}=b.$

This guarantees that $V(P;\alpha_{2})$ is differentiable with respect to
$a_{1}$ and $a_{2}$. Since $\frac{\partial}{\partial a_{1}}V(P;\alpha_{2})=0$
and $\frac{\partial}{\partial a_{2}}V(P;\alpha_{2})=0$, we deduce that

$-3a_{1}m^{2}+a_{2}m^{2}+2a-3a_{1}+a_{2}-2cm=0~{}\text{ and
}~{}a_{1}m^{2}-3a_{2}m^{2}+a_{1}-3a_{2}+2b-2cm=0$

implying

$a_{1}=\frac{1}{4(1+m^{2})}(b-a)+\frac{a-cm}{1+m^{2}}~{}\text{ and
}~{}a_{2}=\frac{3}{4(1+m^{2})}(b-a)+\frac{a-cm}{1+m^{2}}$

with quantization error

$V_{2}=\frac{a^{2}\left(16m^{2}+1\right)+2ab\left(8m^{2}-1\right)+48acm+b^{2}\left(16m^{2}+1\right)+48bcm+48c^{2}}{48\left(m^{2}+1\right)}.$

Case 2: $n\geq 3$.

In this case, the distortion error due to the set $\alpha_{n}$ is given by

$\displaystyle V(P;\alpha_{n})$
$\displaystyle=\int_{\mathbb{R}}\min_{a\in\alpha_{n}}\|(x,0)-a\|^{2}dP(x)=\frac{1}{b-a}\Big{(}\int_{a}^{(m^{2}+1)\frac{(a_{1}+a_{2})}{2}+mc}\rho((x,0),(a_{1},ma_{1}+c))\,dx$
$\displaystyle+\sum_{i=2}^{n-1}\int_{(m^{2}+1)\frac{(a_{i-1}+a_{i})}{2}+mc}^{(m^{2}+1)\frac{(a_{i}+a_{i+1})}{2}+mc}\rho((x,0),(a_{i},ma_{i}+c))\,dx$
$\displaystyle\qquad\qquad+\int_{(m^{2}+1)\frac{(a_{n-1}+a_{n})}{2}+mc}^{b}\rho((x,0),(a_{n},ma_{n}+c))\,dx\Big{)}.$

Since $V(P;\alpha_{n})$ gives the optimal error and is always differentiable
with respect to $a_{i}$ for $2\leq i\leq n-1$, we have
$\frac{\partial}{\partial a_{i}}V(P;\alpha_{n})=0$ yielding

$a_{i+1}-a_{i}=a_{i}-a_{i-1}\text{ for }2\leq i\leq n-1$

implying

$a_{2}-a_{1}=a_{3}-a_{2}=\cdots=a_{n}-a_{n-1}=k$ (5)

for some real $k$. Due to the same reasoning as given in 3, we have
$\frac{\partial}{\partial a_{1}}V(P;\alpha_{n})=0$ and
$\frac{\partial}{\partial a_{n}}V(P;\alpha_{n})=0$, i.e.,

$2(a-cm)-3a_{1}\left(m^{2}+1\right)+a_{2}\left(m^{2}+1\right)=0\text{ and
}a_{n-1}\left(m^{2}+1\right)-3a_{n}\left(m^{2}+1\right)+2(b-cm)=0$

implying

$a_{1}=\frac{a-cm}{1+m^{2}}+\frac{k}{2}~{}\text{ and
}~{}a_{n}=\frac{b-cm}{m^{2}+1}-\frac{k}{2}.$ (6)

Now we have

$\displaystyle b-a=$
$\displaystyle(a_{1}-a)+\sum_{i=2}^{n-1}(a_{i}-a_{i-1})+(b-a_{n})=\left(\frac{a-cm}{1+m^{2}}+\frac{k}{2}-a\right)+(n-1)k+\left(b-\frac{b-cm}{1+m^{2}}+\frac{k}{2}\right),$

which implies $k=\frac{b-a}{n(1+m^{2})}.$ Putting $k=\frac{b-a}{n(1+m^{2})}$,
by the expressions given in (5) and (6), we deduce that

$a_{i}=\frac{2i-1}{2n(1+m^{2})}(b-a)+\frac{a-cm}{1+m^{2}}\text{ for }1\leq
i\leq n.$

To obtain the quantization error $V_{n}$, we proceed as follows:

Since the probability distribution $P$ is uniform on its support, Equation (5)
helps us to deduce that the distortion errors contributed by
$a_{2},a_{3},\cdots,a_{n-1}$ in their own Voronoi regions are equal, i.e.,
each term in the sum

$\sum_{i=2}^{n-1}\int_{(m^{2}+1)\frac{(a_{i-1}+a_{i})}{2}+mc}^{(m^{2}+1)\frac{(a_{i}+a_{i+1})}{2}+mc}\rho((x,0),(a_{i},ma_{i}+c))\,dx$

has the same value. Now, putting the values of $a_{i}$ for $2\leq i\leq n$ in
terms of $a_{1}$ and $k$, we have

$\displaystyle V(P;\alpha_{n})=$
$\displaystyle\int_{\mathbb{R}}\min_{a\in\alpha_{n}}\|(x,0)-a\|^{2}dP(x)=\frac{1}{b-a}\Big{(}\int_{a}^{(m^{2}+1)\frac{(2a_{1}+k)}{2}+mc}\rho\Big{(}(x,0),(a_{1},ma_{1}+c)\Big{)}\,dx$
$\displaystyle+(n-2)\int_{(m^{2}+1)\frac{(2a_{1}+k)}{2}+mc}^{(m^{2}+1)\frac{(2a_{1}+3k)}{2}+mc}\rho\Big{(}(x,0),(a_{1}+k,m(a_{1}+k)+c)\Big{)}\,dx$
$\displaystyle\qquad+\int_{(m^{2}+1)\frac{(2a_{1}+k(2n-3))}{2}+mc}^{b}\rho\Big{(}(x,0),(a_{1}+k(n-1),m(a_{1}+k(n-1))+c)\Big{)}\,dx\Big{)}.$

Upon simplification, and putting
$a_{1}=\frac{b-a}{2\left(m^{2}+1\right)n}+\frac{a-cm}{m^{2}+1}$ and
$k=\frac{b-a}{\left(m^{2}+1\right)n}$ in the above expression, we have the
quantization error as

$\displaystyle V_{n}=$
$\displaystyle\frac{1}{12\left(m^{2}+1\right)n^{3}}\Big{(}-48(a-b)^{2}m^{2}+(a-b)(a-b+72cm+8(11a-2b)m^{2})n$
$\displaystyle\qquad\qquad-12(a-b)m(5c+(4a+b)m)n^{2}+12(c+am)^{2}n^{3}\Big{)}.$

Thus, the proof of the theorem is complete. ∎

######  Remark 3.2.

In Theorem 3.1, the assumptions

$\max\\{a,(m^{2}+1)d+mc\\}=a\text{ and }\min\\{b,(m^{2}+1)e+mc\\}=b$

are necessary to guarantee that the elements in the optimal sets of $n$-points
lie on the line segment joining the points $(d,md+c)$ and $(e,me+c)$. For more
details, please see Proposition 3.5.

Let us now give the following corollary.

######  Corollary 3.3.

Let $P$ be a Borel probability measure on $\mathbb{R}^{2}$ such that $P$ is
uniform on its support $\\{(x,y)\in\mathbb{R}^{2}:0\leq x\leq 2\text{ and
}y=0\\}$. For $n\in\mathbb{N}$ with $n\geq 2$, let $\alpha_{n}$ be an optimal
set of $n$-points for $P$ such that the elements in the optimal set lie on the
line $y=\sqrt{3}x$ between the elements $(0,0)$ and $(2,2\sqrt{3})$. Then,

$\alpha_{n}=\Big{\\{}\Big{(}\frac{2i-1}{4n},\frac{2i-1}{4n}\sqrt{3}\Big{)}:1\leq
i\leq n\Big{\\}}\text{ and }V_{n}=\frac{144n^{2}+196n-576}{48n^{3}}.$

###### Proof.

Putting $a=0,\,b=2,\,m=\sqrt{3},\,c=0,\,d=0,\text{ and }e=2$ in Theorem 3.1,
we see that

$\max\\{a,(m^{2}+1)d+mc\\}=0=a\text{ and }\min\\{b,(m^{2}+1)e+mc\\}=2=b.$

Hence, by Theorem 3.1, we obtain the optimal sets $\alpha_{n}$ and the
corresponding quantization errors $V_{n}$ as follows:

$\alpha_{n}=\Big{\\{}\Big{(}\frac{2i-1}{4n},\frac{2i-1}{4n}\sqrt{3}\Big{)}:1\leq
i\leq n\Big{\\}}\text{ and }V_{n}=\frac{36n^{2}+49n-144}{12n^{3}}.$

Thus, the proof of the corollary is complete. ∎

######  Remark 3.4.

If $m=0$, $c=0$, $d=a$ and $e=b$, then by Theorem 3.1, the optimal set of
$n$-points is given by $\alpha_{n}:=\\{a+\frac{2i-1}{2n}(b-a):1\leq i\leq
n\\}$, and the corresponding quantization error is
$V_{n}:=V_{n}(P)=\frac{(a-b)^{2}}{12n^{2}},$ which is Theorem 2.1.1 in [19].
Thus, Theorem 3.1 generalizes Theorem 2.1.1 in [19].

The following proposition plays an important role in finding the optimal sets
of $n$-points.

######  Proposition 3.5.

Let $P$ be a Borel probability measure on $\mathbb{R}^{2}$ such that $P$ is
uniform on its support $\\{(x,y)\in\mathbb{R}^{2}:a\leq x\leq b\text{ and
}y=0\\}$. For $n\in\mathbb{N}$ with $n\geq 2$, let
$\alpha_{n}:=\\{(a_{i},ma_{i}+c):1\leq i\leq n\\}$ be an optimal set of
$n$-points for $P$ so that the elements in the optimal set lie on the line $L$
between the two elements $(d,md+c)$ and $(e,me+c)$, where $d,e\in\mathbb{R}$
with $d<e$. Then, $(i)$ if $(m^{2}+1)d+mc>a$ (or $(m^{2}+1)e+mc<b$), let $N$
be the largest positive integer such that

$d<\frac{1}{2N(1+m^{2})}(b-a)+\frac{a-cm}{1+m^{2}}\ \Big{(}\text{or
}\frac{2N-1}{2N(1+m^{2})}(b-a)+\frac{a-cm}{1+m^{2}}<e\Big{)}.$

Then, for all $n\geq N+1$, the optimal sets $\alpha_{n}$ always contain the
end element $(d,md+c)$ (or $(e,me+c))$. On the other hand, $(ii)$ if
$(m^{2}+1)d+mc>a$ and $(m^{2}+1)e+mc<b$, let $N:=\max\\{N_{1},N_{2}\\}$, where
$N_{1}$ and $N_{2}$ are the largest positive integers such that

$d<\frac{1}{2N_{1}(1+m^{2})}(b-a)+\frac{a-cm}{1+m^{2}}\text{ and
}\frac{2N_{2}-1}{2N_{2}(1+m^{2})}(b-a)+\frac{a-cm}{1+m^{2}}<e.$

Then, for all $n\geq N+1$, the optimal sets $\alpha_{n}$ always contain the
end elements $(d,md+c)$ and $(e,me+c)$.

###### Proof.

Let $\alpha_{n}:=\\{(a_{i},ma_{i}+c):1\leq i\leq n\\}$ be an optimal set of
$n$-points for $P$ so that the elements in the optimal sets lie on the line
$L$ between the two elements $(d,md+c)$ and $(e,me+c)$, where
$d,e\in\mathbb{R}$ with $d<e$. Also notice that the perpendiculars on the line
$L$ passing through the elements $(a_{i},ma_{i}+c)$ intersect the support of
$P$ at the elements $((m^{2}+1)a_{i}+mc,0)$, respectively, where $1\leq i\leq
n$. By Theorem 3.1, we know that

$a_{i}=\frac{2i-1}{2n(1+m^{2})}(b-a)+\frac{a-cm}{1+m^{2}}\text{ for }1\leq
i\leq n.$

Suppose that $(m^{2}+1)d+mc>a$. Let $n=N$ be the largest positive integer such
that

$(m^{2}+1)d+mc<(m^{2}+1)a_{1}+mc,\text{ i.e., }d<a_{1},\text{ i.e.,
}d<\frac{1}{2N(1+m^{2})}(b-a)+\frac{a-cm}{1+m^{2}}.$ (7)

Notice that the sequence
$\\{\frac{1}{2n(1+m^{2})}(b-a)+\frac{a-cm}{1+m^{2}}\\}$ is strictly
decreasing, and hence for all $n\geq N+1$, the optimal sets $\alpha_{n}$
always contain the end element $(d,md+c)$. Suppose that $(m^{2}+1)e+mc<b$. Let
$n=N$ be the largest positive integer such that

$(m^{2}+1)a_{N}+mc<(m^{2}+1)e+mc,\text{ i.e., }a_{N}<e,\text{ i.e.,
}\frac{2N-1}{2N(1+m^{2})}(b-a)+\frac{a-cm}{1+m^{2}}<e.$ (8)

Notice that the sequence
$\\{\frac{2n-1}{2n(1+m^{2})}(b-a)+\frac{a-cm}{1+m^{2}}\\}$ is strictly
increasing, and hence for all $n\geq N+1$, the optimal sets $\alpha_{n}$
always contain the end element $(e,me+c)$. Next, suppose that
$(m^{2}+1)d+mc>a$ and $(m^{2}+1)e+mc<b$. Choose $N_{1}$ and $N_{2}$ same as
$N$ described in (7) and (8), respectively. Let $N=\max\\{N_{1},N_{2}\\}$.
Then, for all $n\geq N+1$, the optimal sets $\alpha_{n}$ always contain the
end elements $(d,md+c)$ and $(e,me+c)$. Thus, the proof of the proposition is
complete. ∎

######  Note 3.6.

In the following, we state and prove two theorems, Theorem 3.7 and Theorem
3.9. To facilitate the proofs in both the theorems, Proposition 3.5 can be
used. However, in the proof of Theorem 3.7, we have not used Proposition 3.5;
on the other hand, in the proof of Theorem 3.9, we have used Proposition 3.5.

######  Theorem 3.7.

Let $P$ be a Borel probability measure on $\mathbb{R}^{2}$ such that $P$ is
uniform on its support $\\{(x,y)\in\mathbb{R}^{2}:0\leq x\leq 2\text{ and
}y=0\\}$. For $n\in\mathbb{N}$, let $\alpha_{n}:=\\{(a_{i},1):1\leq i\leq
n\\}$ be an optimal set of $n$-points for $P$ so that the elements in the
optimal sets lie on the line $y=1$ between the two elements $(\frac{1}{2},1)$
and $(\frac{3}{2},1)$. Then, $\alpha_{1}=\\{(1,1)\\}$,
$\alpha_{2}=\\{(\frac{1}{2},1),(\frac{3}{2},1)\\}$, and for $n\geq 3$, we have

$a_{i}=\left\\{\begin{array}[]{ccc}\frac{1}{2}&\text{ if }i=1,\\\
\frac{1}{2}+\frac{(i-1)}{(n-1)}&\text{ if }2\leq i\leq n-1,\\\
\frac{3}{2}&\text{ if }i=n,\end{array}\right.$

and the quantization error for $n$-points is given by
$V_{n}=\frac{25n^{2}-50n+26}{24(n-1)^{2}}$.

###### Proof.

The proofs of $\alpha_{1}=\\{(1,1)\\}$,
$\alpha_{2}=\\{(\frac{1}{2},1),(\frac{3}{2},1)\\}$ are routine. We just give
the proof for $n\geq 3$. Let $\alpha_{n}:=\\{(t,1):t=a_{i}\text{ for }1\leq
i\leq n\\}$ be an optimal set of $n$-points such that $\frac{1}{2}\leq
a_{1}<a_{2}<\cdots<a_{n-1}<a_{n}\leq\frac{3}{2}$. Notice that the boundary of
the Voronoi region of the element $(a_{1},1)$ intersects the support of $P$ at
the elements $(0,0)$ and $(\frac{1}{2}(a_{1}+a_{2}),0)$, the boundary of the
Voronoi region of $(a_{n},1)$ intersects the support of $P$ at the elements
$(\frac{1}{2}(a_{n-1}+a_{n}),0)$ and $(2,0)$. On the other hand, the
boundaries of the Voronoi regions of $(a_{i},1)$ for $2\leq i\leq n-1$
intersect the support of $P$ at the elements $(\frac{1}{2}(a_{i-1}+a_{i}),0)$
and $(\frac{1}{2}(a_{i}+a_{i+1}),0)$. Thus, the distortion error due to the
set $\alpha_{n}$ is given by

$\displaystyle
V(P;\alpha_{n})=\int_{\mathbb{R}}\min_{a\in\alpha_{n}}\|(x,0)-a\|^{2}dP(x)$
$\displaystyle=\int_{0}^{\frac{1}{2}\left(a_{1}+a_{2}\right)}\frac{1}{2}\left(\left(x-a_{1}\right){}^{2}+1\right)\,dx+\sum_{i=2}^{n-1}\int_{\frac{1}{2}\left(a_{i-1}+a_{i}\right)}^{\frac{1}{2}\left(a_{i+1}+a_{i}\right)}\frac{1}{2}\left(\left(x-a_{i}\right){}^{2}+1\right)\,dx$
$\displaystyle\qquad+\int_{\frac{1}{2}\left(a_{n-1}+a_{n}\right)}^{2}\frac{1}{2}\left(\left(x-a_{n}\right){}^{2}+1\right)\,dx.$

Since $V(P;\alpha_{n})$ gives the optimal error and is differentiable with
respect to $a_{i}$ for $2\leq i\leq n-1$, we have $\frac{\partial}{\partial
a_{i}}V(P;\alpha_{n})=0$ implying

$a_{i+1}-a_{i}=a_{i}-a_{i-1}\text{ for }2\leq i\leq n-1.$

This yields the fact that

$a_{2}-a_{1}=a_{3}-a_{2}=\cdots=a_{n}-a_{n-1}=k$ (9)

for some real number $0<k<1$. By the equations in $\eqref{eq0000}$, we see
that all terms in the sum
$\sum_{i=2}^{n-1}\int_{\frac{1}{2}\left(a_{i-1}+a_{i}\right)}^{\frac{1}{2}\left(a_{i+1}+a_{i}\right)}\frac{1}{2}\left(\left(x-a_{i}\right){}^{2}+1\right)\,dx$
have the same value. Again, by the equations in (9) we have

$a_{2}=k+a_{1},\,a_{3}=2k+a_{1},\cdots,a_{n}=(n-1)k+a_{1}.$

Hence,

$\displaystyle V(P;\alpha_{n})$
$\displaystyle=\int_{0}^{\frac{1}{2}\left(2a_{1}+k\right)}\frac{1}{2}\left(\left(x-a_{1}\right){}^{2}+1\right)\,dx+(n-2)\int_{\frac{1}{2}\left(2a_{1}+k\right)}^{\frac{1}{2}\left(2a_{1}+3k\right)}\frac{1}{2}\left(\left(x-\left(a_{1}+k\right)\right){}^{2}+1\right)\,dx$
$\displaystyle\qquad\qquad+\int_{\frac{1}{2}\left(2a_{1}+k(2n-3)\right)}^{2}\frac{1}{2}\left(\left(x-\left(a_{1}+k(n-1)\right)\right){}^{2}+1\right)\,dx,$

which upon simplification yields

$\displaystyle V(P;\alpha_{n})$
$\displaystyle=\frac{1}{24}\Big{(}-12a_{1}(k(n-1)-2)\left(a_{1}+k(n-1)-2\right)$
$\displaystyle\qquad\qquad-k(n-1)\left(4k^{2}n^{2}-8k(k+3)n+3(k+4)^{2}\right)+56\Big{)},$

which is minimum if $a_{1}=\frac{1}{2}$ and $k=\frac{1}{n-1}$, and the minimum
value is $\frac{25n^{2}-50n+26}{24(n-1)^{2}}$. As $k=\frac{1}{n-1}$ and
$a_{1}=\frac{1}{2}$, using the expression (9), we obtain

$a_{i}=\left\\{\begin{array}[]{ccc}\frac{1}{2}&\text{ if }i=1,\\\
\frac{1}{2}+\frac{(i-1)}{(n-1)}&\text{ if }2\leq i\leq n-1,\\\
\frac{3}{2}&\text{ if }i=n,\end{array}\right.$

with quantization error $V_{n}=\frac{25n^{2}-50n+26}{24(n-1)^{2}}$. Thus, the
proof of the theorem is complete. ∎

######  Remark 3.8.

Comparing Theorem 3.7 with Proposition 3.5, we have
$a=0,\,b=2,\,m=0,\,c=1,d=\frac{1}{2},\text{ and }e=\frac{3}{2}$, and so

$(m^{2}+1)d+mc=\frac{1}{2}>a\text{ and }(m^{2}+1)e+mc=\frac{3}{2}<b.$

Let $n=N_{1}$ be the largest positive integer such that

$d<\frac{1}{2N_{1}(1+m^{2})}(b-a)+\frac{a-cm}{1+m^{2}},$

which is true if $N_{1}<2$, i.e., $N_{1}=1$. Let $n=N_{2}$ be the largest
positive integer such that

$\frac{2N_{2}-1}{2N_{2}(1+m^{2})}(b-a)+\frac{a-cm}{1+m^{2}}<e,$

which is true if $N_{2}<2$, i.e., $N_{2}=1$. Take $N=\max\\{N_{1},N_{2}\\}$.
Then, $N=1$. By Proposition 3.5, we can conclude that for all $n\geq 2$, the
optimal sets $\alpha_{n}$ will contain the end elements $(\frac{1}{2},1)$ and
$(\frac{3}{2},1)$, which is clearly true by Theorem 3.7.

######  Theorem 3.9.

Let $P$ be a Borel probability measure on $\mathbb{R}^{2}$ such that $P$ is
uniform on its support $\\{(x,y)\in\mathbb{R}^{2}:0\leq x\leq 2\text{ and
}y=0\\}$. For $n\in\mathbb{N}$, let $\alpha_{n}:=\\{(a_{i},1):1\leq i\leq
n\\}$ be an optimal set of $n$-points for $P$ so that the elements in the
optimal set lie on the line $y=1$ between the two elements $(0,1)$ and
$(\frac{28}{15},1)$, i.e., $0\leq a_{1}<a_{2}<\cdots<a_{n}\leq\frac{28}{15}$.
Then, $\alpha_{1}=\\{(1,1)\\}$, and for $1\leq n\leq 7$,

$\alpha_{n}=\Big{\\{}\Big{(}\frac{2i-1}{n},1\Big{)}:1\leq i\leq n\Big{\\}}.$

On the other hand, for $n\geq 8$, we obtain

$a_{i}=\left\\{\begin{array}[]{ccc}\frac{28(2i-1)}{15(2n-1)}&\text{ if }1\leq
i\leq n-1,\\\ \frac{28}{15}&\text{ if }i=n,\end{array}\right.$

and the quantization error for $n$-points is given by
$V_{n}=\frac{7(5788(n-1)n+3015)}{10125(1-2n)^{2}}$.

###### Proof.

Let $\alpha_{n}:=\\{(t,1):t=a_{i}\text{ for }1\leq i\leq n\\}$ be an optimal
set of $n$-points such that $0\leq
a_{1}<a_{2}<\cdots<a_{n-1}<a_{n}\leq\frac{28}{15}$ for all $n\in\mathbb{N}$.
Using Proposition 3.5, it can be proved that for all $n\geq 8$, the optimal
sets always contain the end element $\frac{28}{15}$, i.e.,
$a_{n}=\frac{28}{15}$ for all $n\geq 8$. The proofs of
$\alpha_{1}=\\{(1,1)\\}$, and for $1\leq n\leq 7$,

$\alpha_{n}=\Big{\\{}\Big{(}\frac{2i-1}{n},1\Big{)}:1\leq i\leq n\Big{\\}},$

are routine. Here we prove the optimal sets of $n$-points for all $n\geq 8$.
Proceeding in the similar lines as given in the proof of Theorem 3.7, we have

$a_{2}-a_{1}=a_{3}-a_{2}=\cdots=a_{n}-a_{n-1}=k$

for some real $k$, which implies

$a_{1}=a_{n}-(n-1)k,\,a_{2}=a_{n}-(n-2)k,\cdots,a_{n-1}=a_{n}-k.$

Also, by using $\frac{\partial}{\partial a_{1}}V(P;\alpha_{n})=0$, we get
$3a_{1}-a_{2}=0$, which implies that $a_{1}=\frac{k}{2}$. Now we have

$\frac{k}{2}=a_{1}=a_{n}-(n-1)k=\frac{28}{15}-(n-1)k,$

this yields $k=\frac{56}{15(2n-1)}$. Using $a_{n}=\frac{28}{15}$ and
$k=\frac{56}{15(2n-1)}$, we get $a_{i}=\frac{28(2i-1)}{15(2n-1)}$ for $1\leq
i\leq n-1$ with quantization error

$\displaystyle V(P;\alpha_{n})=$
$\displaystyle\frac{1}{2}\left(\int_{0}^{\frac{1}{2}\left(2a_{n}-k(2n-3)\right)}\left(\left(x-\left(a_{n}-k(n-1)\right)\right)^{2}+1\right)\,dx\right.$
$\displaystyle+\left.(n-2)\int_{\frac{1}{2}\left(2a_{n}-3k\right)}^{\frac{1}{2}\left(2a_{n}-k\right)}\left(\left(x-\left(a_{n}-k\right)\right)^{2}+1\right)\,dx+\int_{\frac{1}{2}\left(2a_{n}-k\right)}^{2}\left(\left(x-a_{n}\right)^{2}+1\right)\,dx\right)$
$\displaystyle=$
$\displaystyle\frac{1}{24}\Big{(}12k^{2}n^{2}a_{n}-24k^{2}na_{n}+12k^{2}a_{n}-12kna_{n}^{2}+12ka_{n}^{2}+24a_{n}^{2}-48a_{n}-4k^{3}n^{3}$
$\displaystyle+12k^{3}n^{2}-11k^{3}n+3k^{3}+56\Big{)}=\frac{7(5788(n-1)n+3015)}{10125(1-2n)^{2}}.$

This completes the proof. ∎

##  4\. Constrained quantization when the support lies on a circle and the
optimal elements lie on another circle

Let $O(0,0)$ be the origin of the Cartesian plane. Let $C$ be the unit circle
given by the parametric equations:

$C:=\\{(x,y):x=\cos t,\,y=\sin t\text{ for }0\leq t\leq 2\pi\\}.$

Let the positive direction of the $x$-axis cut the circle at the element
$A_{0}$, i.e., $A_{0}$ is represented by the parametric value $t=0$. Let $s$
be the distance of an element on $C$ along the arc starting from the element
$A_{0}$ in the counterclockwise direction. Then,

$ds=\sqrt{\Big{(}\frac{dx}{dt}\Big{)}^{2}+\Big{(}\frac{dy}{dt}\Big{)}^{2}}\,dt=dt.$

Let $P$ be a uniform distribution with support the unit circle $C$. Then, the
probability density function $f(x,y)$ for $P$ is given by

$f(x,y)=\left\\{\begin{array}[]{ccc}\frac{1}{2\pi}&\text{ if }(x,y)\in C,\\\
0&\text{ otherwise}.\end{array}\right.$

Thus, we have $dP(s)=P(ds)=f(x,y)ds=\frac{1}{2\pi}dt$. Moreover, we know that
if $\hat{t}$ radians is the central angle subtended by an arc of length $S$ of
the unit circle, then $S=\hat{t}$, and

$P(S)=\int_{S}dP(s)=\frac{1}{2\pi}\int_{S}dt=\frac{\hat{t}}{2\pi}.$

Let $L$ be a concentric circle with $C$, and $L$ has radius $a$, i.e., the
parametric representation of the circle $L$ is given by

$L:=\\{(x,y):x=a\cos\theta,\,y=a\sin\theta\text{ for }0\leq\theta\leq
2\pi\\}.$

In this section, we determine the optimal sets of $n$-points and the $n$th
constrained quantization errors for the uniform distribution $P$ on $C$ under
the condition that the elements in an optimal set lie on the circle $L$. Let
the line $OA_{0}$ cut the circle $L$ at the element $B_{0}$, i.e., $B_{0}$ is
represented on the circle $L$ by the parameter $\theta=0$.

######  Proposition 4.1.

Any element on the circle $L$ forms an optimal set of one-point with
quantization error $V_{1}=1+a^{2}.$

###### Proof.

Let $\alpha:=\\{(a\cos\theta,a\sin\theta)\\}$, where $0\leq\theta\leq 2\pi$,
forms an optimal set of one-point. Then, the distortion error $V(P;\alpha)$ is
given by

$V(P;\alpha)=\int_{C}\frac{1}{2\pi}\rho((\cos t,\sin
t),(a\cos\theta,a\sin\theta))\,dt=1+a^{2},$

which does not depend on $\theta$ for any $0\leq\theta\leq 2\pi$. Hence, any
element on the circle $L$ forms an optimal set of one-point, and the
quantization error for one-point is given by $V_{1}=1+a^{2}$. ∎

######  Proposition 4.2.

A set of the form
$\\{(a\cos\theta,a\sin\theta),(-a\cos\theta,-a\sin\theta)\\}$, where
$0\leq\theta\leq 2\pi$, forms an optimal set of two-points with quantization
error $V_{2}=1+a^{2}-\frac{4a}{\pi}$.

###### Proof.

Let
$\alpha:=\\{(a\cos\theta_{1},a\sin\theta_{1}),(a\cos\theta_{2},a\sin\theta_{2})\\}$,
where $0\leq\theta_{1}<\theta_{2}\leq 2\pi$, form an optimal set of two-
points. Notice that the boundary of the Voronoi regions of the two elements in
the optimal set is the line joining the two points given by the parameters
$\theta=\frac{\theta_{1}+\theta_{2}}{2}$ and
$\theta=\pi+\frac{\theta_{1}+\theta_{2}}{2}$. Then, the distortion error is
given by

$\displaystyle V(P;\alpha)$
$\displaystyle=\frac{1}{2\pi}\Big{(}\int_{-\pi+\frac{\theta_{1}+\theta_{2}}{2}}^{\frac{\theta_{1}+\theta_{2}}{2}}\rho\Big{(}(\cos
t,\sin t),(a\cos\theta_{1},a\sin\theta_{1})\Big{)}\,dt$
$\displaystyle\qquad\qquad+\int_{\frac{\theta_{1}+\theta_{2}}{2}}^{\pi+\frac{\theta_{1}+\theta_{2}}{2}}\rho\Big{(}(\cos
t,\sin t),(a\cos\theta_{2},a\sin\theta_{2})\Big{)}\,dt\Big{)}$
$\displaystyle=\frac{1}{2\pi}\Big{(}\int_{-\pi+\frac{\theta_{1}+\theta_{2}}{2}}^{\frac{\theta_{1}+\theta_{2}}{2}}\Big{(}1+a^{2}-2a\cos(t-\theta_{1})\Big{)}\,dt+\int_{\frac{\theta_{1}+\theta_{2}}{2}}^{\pi+\frac{\theta_{1}+\theta_{2}}{2}}\Big{(}1+a^{2}-2a\cos(\theta-\theta_{2})\Big{)}\,dt\Big{)},$

which upon simplification yields that

$V(P;\alpha)=\frac{1}{2\pi}\Big{(}(1+a^{2})2\pi-8a\sin\frac{\theta_{2}-\theta_{1}}{2}\Big{)}.$

Since $0<\frac{\theta_{2}-\theta_{1}}{2}<\pi$, we can say that $V(P;\alpha)$
is minimum if $\theta_{2}-\theta_{1}=\pi$. Thus, an optimal set of two-points
is given by $\\{(a\cos\theta,a\sin\theta),(-a\cos\theta,-a\sin\theta)\\}$ for
$0\leq\theta\leq 2\pi$ with constrained quantization error
$V_{2}=1+a^{2}-\frac{4a}{\pi}$, which yields the proposition. ∎

######  Theorem 4.3.

Let $\alpha_{n}$ be an optimal set of $n$-points for the uniform distribution
$P$ on the unit circle $C$ for $n\in\mathbb{N}$ with $n\geq 3$. Then,

$\alpha_{n}=\Big{\\{}\Big{(}a\cos\frac{(2i-1)\pi}{n},a\sin\frac{(2i-1)\pi}{n}\Big{)}:i=1,2,\cdots,n\Big{\\}}$

and the corresponding quantization error is given by
$V_{n}=a^{2}+1-\frac{2an}{\pi}\sin\frac{\pi}{n}.$

###### Proof.

Let $\alpha_{n}:=\\{a_{1},a_{2},\cdots,a_{n}\\}$, where
$a_{i}=(a\cos\theta_{i},a\sin\theta_{i})$, be an optimal set of $n$-points for
$P$ with $n\geq 3$ such that the elements in the optimal set lie on the circle
$L$. Let the boundary of the Voronoi regions of $a_{i}$ cut the circle $L$, in
fact also the circle $C$, at the elements given by the parameters
$\theta_{i-1}$ and $\theta_{i}$, where $1\leq i\leq n$. Since the circles have
rotational symmetry, without any loss of generality, we can assume that
$\theta_{0}=0$, and $\theta_{n}=2\pi$. Then, each $a_{i}$ on $L$ has the
parametric representation $\frac{1}{2}(\theta_{i-1}+\theta_{i})$ for $1\leq
i\leq n$. Then, the quantization error for $n$-points is given by

$\displaystyle V(P;\alpha_{n})$
$\displaystyle=\int_{C}\min_{u\in\alpha_{n}}\rho((\cos t,\sin t)-u)\,dP(s)$
$\displaystyle=\sum_{i=1}^{n}\int_{\theta_{i-1}}^{\theta_{i}}\frac{1}{2\pi}\rho\Big{(}((\cos
t,\sin
t),(\frac{1}{2}\cos\frac{\theta_{i-1}+\theta_{i}}{2},\frac{1}{2}\sin\frac{\theta_{i-1}+\theta_{i}}{2}))\Big{)}\,dt$
$\displaystyle=\sum_{i=1}^{n}\int_{\theta_{i-1}}^{\theta_{i}}\frac{1}{2\pi}\Big{(}a^{2}-2a\cos(-\frac{\theta_{i-1}}{2}-\frac{\theta_{i}}{2}+t)+1\Big{)}\,d\theta$
$\displaystyle=\sum_{i=1}^{n}\frac{1}{2\pi}\Big{(}(a^{2}+1)(\theta_{i}-\theta_{i-1})-4a\sin\frac{\theta_{i}-\theta_{i-1}}{2}\Big{)},$

upon simplification, which yields

$V(P;\alpha_{n})=a^{2}+1-\frac{2a}{\pi}\sum_{i=1}^{n}\sin\frac{\theta_{i}-\theta_{i-1}}{2}.$
(10)

Since $V(P;\alpha_{n})$ gives the optimal error and is differentiable with
respect to $\theta_{i}$ for all $1\leq i\leq n-1$, we have
$\frac{\partial}{\partial\theta_{i}}V(P;\alpha)=0$. For $1\leq i\leq n-1$, the
equations $\frac{\partial}{\partial\theta_{i}}V(P;\alpha)=0$ implies that

$\cos\frac{\theta_{i}-\theta_{i-1}}{2}=\cos\frac{\theta_{i+1}-\theta_{i}}{2}\text{
yielding
}\frac{\theta_{i}-\theta_{i-1}}{2}=\frac{\theta_{i+1}-\theta_{i}}{2},\text{ or
}\frac{\theta_{i}-\theta_{i-1}}{2}=2\pi-\frac{\theta_{i+1}-\theta_{i}}{2}.$

Without any loss of generality, for $1\leq i\leq n-1$ we can take
$\frac{\theta_{i}-\theta_{i-1}}{2}=\frac{\theta_{i+1}-\theta_{i}}{2}$. This
yields the fact that

$\theta_{1}-\theta_{0}=\theta_{2}-\theta_{1}=\theta_{3}-\theta_{2}=\cdots=\theta_{n}-\theta_{n-1}=\frac{2\pi}{n}.$

Thus, we have $\theta_{i}=\frac{2\pi i}{n}$ for $i=1,2,\cdots,n$. Hence, if
$\alpha_{n}:=\\{a_{1},a_{2},\cdots,a_{n}\\}$ is an optimal set of $n$-points,
then

$a_{i}=\Big{(}a\cos\frac{(2i-1)\pi}{n},a\sin\frac{(2i-1)\pi}{n}\Big{)}\text{
for }i=1,2,\cdots,n,$

and the quantization error for $n$-points, by (10), is given by

$\displaystyle
V_{n}=V(P;\alpha_{n})=a^{2}+1-\frac{2an}{\pi}\sin\frac{\pi}{n}.$

Thus, the proof of the theorem is complete. ∎

##  5\. Constrained quantization when the support lies on a chord of a circle
and the optimal elements lie on the circle

Let $C$ be a circle with center $(0,0)$ and radius one, i.e., the equation of
the circle is $x^{2}+y^{2}=1,$ whose parametric representations are
$x=\cos\theta$ and $y=\sin\theta$, where $0\leq\theta\leq 2\pi$. Thus, if
$(\cos\theta,\sin\theta)$ is an element on the circle, then we will represent
it by $\theta$. Let $P$ be a Borel probability measure on $\mathbb{R}^{2}$
such that $P$ has support a chord of the circle, and $P$ is uniform on its
support. We now investigate the optimal sets of $n$-points and the $n$th
constrained quantization errors for all $n\in\mathbb{N}$ so that the optimal
elements lie on the circle. The two cases can happen as described in the
following two subsections.

###  5.1. Chord is a diameter of the circle

Without any loss of generality, let us consider the horizontal diameter as the
support of $P$, i.e., the support of $P$ is the closed interval
$\\{(x,y):-1\leq x\leq 1,\,y=0\\}$. Then, the probability density function is
given by

$f(x,y)=\left\\{\begin{array}[]{ccc}\frac{1}{2}&\text{ if }-1\leq x\leq
1\text{ and }y=0,\\\ 0&\text{ otherwise}.\end{array}\right.$

Recall Fact 2.1. Here we have $dP(s)=P(ds)=P(dx)=dP(x)=f(x,0)dx$. We know that
an optimal set of one-point always exists. Let $\alpha_{n}$ be an optimal set
of $n$-points for any $n\geq 2$. Since the boundary of the Voronoi regions of
any two optimal elements, in this case, passes through the center of the
circle, from the geometry, we see that among $n$ Voronoi regions, only two
Voronoi regions contain elements from the support of $P$. Hence, an optimal
set of $n$-points for any $n\geq 2$ contains exactly two elements. We now
calculate the optimal sets of one-point and the two-points in the following
propositions:

######  Proposition 5.1.1.

Any element on the circle forms an optimal set of one-point with constrained
quantization error $V_{1}=\frac{4}{3}$.

###### Proof.

Let $(\cos\theta,\sin\theta)$ be an element on the circle. Then, the
distortion error for $P$ with respect to this element is given by

$V(P;\\{(\cos\theta,\sin\theta)\\})=\int_{-1}^{1}\rho((x,0),(\cos\theta,\sin\theta))\,dP(x)=\frac{1}{2}\int_{-1}^{1}\rho((x,0),(\cos\theta,\sin\theta))\,dx=\frac{4}{3},$

which does not depend on $\theta$. Hence, any element on the circle forms an
optimal set of one-point with constrained quantization error
$V_{1}=\frac{4}{3}$. ∎

######  Proposition 5.1.2.

The set $\\{(-1,0),(1,0)\\}$ forms an optimal set of two-points with
constrained quantization error $V_{2}=\frac{1}{3}$.

###### Proof.

From the geometry, we see that the boundary of any two elements on the circle
passes through the center of the circle. Thus, in an optimal set of two-
points, one Voronoi region will contain the left half, and the other Voronoi
region will contain the right half of the support of $P$. Hence, by the
routine calculation, we can show that $\\{(-1,0),(1,0)\\}$ forms an optimal
set of two-points with constrained quantization error

$V_{2}=\frac{1}{2}\Big{(}\int_{-1}^{0}\rho((x,0),(-1,0))dx+\int_{0}^{1}\rho((x,0),(1,0))dx\Big{)}=\frac{1}{3}.$

Thus, the proof of the proposition is complete. ∎

###  5.2. Chord is not a diameter of the circle

In this case, for definiteness sake, we investigate the optimal sets of
$n$-points and the $n$th constrained quantization errors for a Borel
probability measure $P$ on $\mathbb{R}^{2}$ such that $P$ has support the
chord $y=-\frac{1}{2}$ for $-\frac{\sqrt{3}}{2}\leq x\leq\frac{\sqrt{3}}{2}$,
and $P$ is uniform there. Then, the probability density function for $P$ is
given by

$f(x,y)=\left\\{\begin{array}[]{ccc}\frac{1}{\sqrt{3}}&\text{ if
}-\frac{\sqrt{3}}{2}\leq x\leq\frac{\sqrt{3}}{2}\text{ and }y=-\frac{1}{2},\\\
0&\text{ otherwise}.\end{array}\right.$

Recall that the circle has rotational symmetry. Thus, for any other chord, the
technique of finding the optimal sets of $n$-points and the $n$th constrained
quantization errors will be similar. Here we have
$dP(s)=P(ds)=P(dx)=dP(x)=f(x,-\frac{1}{2})dx$, where $x$ varies over the line
$y=-\frac{1}{2}$. The arc of the circle subtended by the chord is represented
by $\theta$ for $\frac{7\pi}{6}\leq\theta\leq\frac{11\pi}{6}$. Moreover, the
circle is geometrically symmetric with respect to the line $y=0$, and also the
probability measure is symmetric with respect to the line $y=0$, i.e., if two
intervals of the same length lie on the support of $P$ and are equidistant
from the line $y=0$, then they have the same probability. In proving the
results, we can use this symmetry of the circle.

Figure 2. Optimal configuration of $n$ elements for $1\leq n\leq 9$.

######  Proposition 5.2.1.

The set $\\{(0,-1)\\}$ forms an optimal set of one-point with quantization
error $V_{1}=\frac{1}{2}$.

###### Proof.

Let us consider an element $(\cos\theta,\sin\theta)$ on the circle. The
distortion error for $P$ with respect to the set
$\\{(\cos\theta,\sin\theta)\\}$ is given by

$V(P;\\{(\cos\theta,\sin\theta)\\})=\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}\frac{1}{\sqrt{3}}\rho((x,-\frac{1}{2}),(\cos\theta,\sin\theta))\,dx=\sin\theta+\frac{3}{2},$

the minimum value of which is $\frac{1}{2}$ and it occurs when
$\theta=\frac{3\pi}{2}$ (see Figure 2). Thus, the proof of the proposition is
yielded. ∎

######  Proposition 5.2.2.

The optimal set of two-points is given by

$\Big{\\{}2\pi-2\tan^{-1}\Big{(}\frac{\sqrt{3}}{2}+\frac{\sqrt{7}}{2}\Big{)},\pi+2\tan^{-1}\Big{(}\frac{\sqrt{3}}{2}+\frac{\sqrt{7}}{2}\Big{)}\Big{\\}}$

with quantization error $V_{2}=\frac{1}{2}\left(3-\sqrt{7}\right)$.

###### Proof.

Since the probability measure is symmetric with respect to the line $y=0$, we
can assume that in an optimal set of two-points, the Voronoi region of one
element will contain the left half of the chord, and the Voronoi region of the
other element will contain the right half of the chord, i.e., the boundary of
the two Voronoi regions is the $y$-axis. Let the left element is
$(\cos\theta,\sin\theta)$. Then, due to symmetry, the distortion error for the
two elements is given by

$2\int_{-\frac{\sqrt{3}}{2}}^{0}\frac{1}{\sqrt{3}}\rho((x,-\frac{1}{2}),(\cos\theta,\sin\theta))\,dx=\sin\theta+\frac{1}{2}\sqrt{3}\cos\theta+\frac{3}{2},$

which is minimum if
$\theta=2\pi-2\tan^{-1}(\frac{\sqrt{3}}{2}+\frac{\sqrt{7}}{2})$, and the
minimum value is $\frac{1}{2}\left(3-\sqrt{7}\right)$. Thus, the one element
is represented by
$\theta=2\pi-2\tan^{-1}(\frac{\sqrt{3}}{2}+\frac{\sqrt{7}}{2})$, and due to
symmetry the other element is represented by
$\theta=\pi+2\tan^{-1}(\frac{\sqrt{3}}{2}+\frac{\sqrt{7}}{2})$ with
quantization error for two-points $V_{2}=\frac{1}{2}\left(3-\sqrt{7}\right)$
(see Figure 2). Thus, the proof of the proposition is complete. ∎

######  Remark 5.2.3.

Due to the symmetry of the probability measure $P$ and the geometrical
symmetry of the circle, we can assume that in an optimal set of $n$-points,
where $n\geq 3$, if $n$ is even, then there are $\frac{n}{2}$ elements to the
left of the $y$-axis and $\frac{n}{2}$ elements to the right of the $y$-axis.
On the other hand, if $n$ is odd, then there are $\frac{n-1}{2}$ elements to
the left of the $y$-axis and $\frac{n-1}{2}$ elements to the right of the
$y$-axis, and the remaining one element will be the element $(-1,0)$.
Moreover, whether $n$ is even or odd, the set of elements on the left side and
the set of elements on the right side are reflections of each other with
respect to the $y$-axis. Due to this fact, in the sequel of this section, we
calculate the optimal sets of $n$-points for $n=8$ and $n=9$. Following the
similar technique, whether $n$ is even or odd, one can calculate the locations
of the elements for any positive integer $n\geq 3$.

######  Proposition 5.2.4.

The optimal set of eight-points is given by

$\displaystyle\\{(-0.821938,-0.569577),(-0.680768,-0.732499),(-0.4608,-0.887504),$
$\displaystyle(-0.164598,-0.986361),(0.821938,-0.569577),(0.680768,-0.732499),$
$\displaystyle(0.4608,-0.887504),(0.164598,-0.986361)\\}$

with quantization error $V_{8}=0.12327$.

###### Proof.

Let $\alpha_{8}:=\\{\theta_{1},\theta_{2},\cdots\theta_{8}\\}$ be an optimal
set of eight-points. Without any loss of generality, we can assume that
$\theta_{1}<\theta_{2}<\cdots<\theta_{8}$. Due to symmetry as mentioned in
Remark 5.2.3, the boundary of the Voronoi regions of $\theta_{4}$ and
$\theta_{5}$ is the $y$-axis, and the elements on the right side of $y$-axis
are the reflections of the elements in the left side of $y$-axis with respect
to the $y$-axis. Thus, it is enough to calculate the first four elements
$\theta_{1},\theta_{2},\theta_{3},\theta_{4}$. Let the boundaries of the
Voronoi regions of $\theta_{i}$ and $\theta_{i+1}$ intersects the support of
$P$ at the element $(a_{i},-\frac{1}{2})$, where $1\leq i\leq 3$. Because of
the symmetry, the distortion error is given by

$\displaystyle V(P;\alpha_{8})=$ $\displaystyle
2\Big{(}\int_{-\frac{\sqrt{3}}{2}}^{a_{1}}\rho((x,0),(\cos\theta_{1},\sin\theta_{1}))\,dP(x)$
$\displaystyle\qquad+\sum_{i=1}^{2}\int_{a_{i}}^{a_{i+1}}\rho((x,0),(\cos\theta_{i+1},\sin\theta_{i+1}))\,dP(x)$
(11)
$\displaystyle\qquad+\int_{a_{3}}^{0}\rho((x,0),(\cos\theta_{4},\sin\theta_{4}))\,dP(x)\Big{)}.$

The canonical equations are

$\rho((a_{i},-\frac{1}{2}),(\cos\theta_{i},\sin\theta_{i}))-\rho((a_{i},-\frac{1}{2}),(\cos\theta_{i+1},\sin\theta_{i+1}))=0\text{
for }i=1,2,3.$

Solving the canonical equations, we have

$a_{1}=\frac{\sin\theta_{1}-\sin\theta_{2}}{2\left(\cos\theta_{1}-\cos\theta_{2}\right)},a_{2}=\frac{\sin\theta_{2}-\sin\theta_{3}}{2\left(\cos\theta_{2}-\cos\theta_{3}\right)},a_{3}=\frac{\sin\theta_{3}-\sin\theta_{4}}{2\left(\cos\theta_{3}-\cos\theta_{4}\right)}.$

Putting the values of $a_{1},a_{2},a_{3}$ in (5.2), we see that
$V(P;\alpha_{8})$ is a function of $\theta_{i}$ for $i=1,2,3,4$. Since
$V(P;\alpha_{8})$ is optimal we have

$\frac{\partial}{\partial\theta_{i}}V(P;\alpha_{8})=0\text{ for }i=1,2,3,4.$

Solving the above four equations, we obtain the values of $\theta_{i}$ for
which $V(P;\alpha_{8})$ is minimum as

$\theta_{1}=3.74758,\theta_{2}=3.96358,\theta_{3}=4.23349,\theta_{4}=4.54704.$

Due to symmetry $\theta_{5},\theta_{6},\theta_{7},\theta_{8}$ can also be
obtained. Recall that $\theta_{i}$ represents the element
$(\cos\theta_{i},\sin\theta_{i})$. Thus, we obtain the optimal set of eight-
points as mentioned in the proposition with quantization error $V_{8}=0.12327$
(see Figure 2). Thus, the proof of the proposition is complete. ∎

######  Proposition 5.2.5.

The optimal set of nine-points is given by

$\displaystyle\\{(-0.827126,-0.562016),(-0.708531,-0.70568),(-0.529525,-0.848294),(-0.286494,-0.958082),$
$\displaystyle(0.,-1),(0.827126,-0.562016),(0.708531,-0.70568),(0.529525,-0.848294),(0.286494,-0.958082)\\}$

with quantization error $V_{9}=0.122546$.

###### Proof.

Recall Remark 5.2.3. We can assume that the optimal set of nine-points is
$\alpha_{9}=\\{\theta_{i}:1\leq i\leq 9\\}$ such that
$\theta_{i}<\theta_{i+1}$ for $1\leq i\leq 8$, where
$\theta_{5}=\frac{3\pi}{2}$. Because of the same reasoning as given in the
proof of Proposition 5.2.4, we have the distortion error as

$\displaystyle V(P;\alpha_{9})$
$\displaystyle=2\Big{(}\int_{-\frac{\sqrt{3}}{2}}^{a_{1}}\rho((x,0),(\cos\theta_{1},\sin\theta_{1}))\,dP(x)+\sum_{i=1}^{3}\int_{a_{i}}^{a_{i+1}}\rho((x,0),(\cos\theta_{i+1},\sin\theta_{i+1}))\,dP(x)$
(12) $\displaystyle\qquad+\int_{a_{4}}^{0}\rho((x,0),(0,-1))\,dP(x)\Big{)}.$

The canonical equations are

$\rho((a_{i},-\frac{1}{2}),(\cos\theta_{i},\sin\theta_{i}))-\rho((a_{i},-\frac{1}{2}),(\cos\theta_{i+1},\sin\theta_{i+1}))=0\text{
for }i=1,2,3,4.$

Solving the canonical equations, we obtain the values of $a_{i}$ for $1\leq
i\leq 4$. Putting the values of $a_{i}$ in (12), we see that $V(P;\alpha_{9})$
is a function of $\theta_{i}$ for $i=1,2,3,4$. Since $V(P;\alpha_{9})$ is
optimal we have

$\frac{\partial}{\partial\theta_{i}}V(P;\alpha_{9})=0\text{ for }i=1,2,3,4.$

Solving the above four equations, we obtain the values of $\theta_{i}$ for
which $V(P;\alpha_{9})$ is minimum as

$\theta_{1}=3.73841,\theta_{2}=3.92497,\theta_{3}=4.15435,\theta_{4}=4.42182,$

Due to symmetry $\theta_{6},\theta_{7},\theta_{8},\theta_{9}$ can also be
obtained. Recall that $\theta_{i}$ represents the element
$(\cos\theta_{i},\sin\theta_{i})$. Hence, we obtain the optimal set of nine-
points as mentioned in the proposition with quantization error
$V_{9}=0.122546$ (see Figure 2). Thus, the proof of the proposition is
complete. ∎

##  6\. Quantization dimensions and quantization coefficients

Let $P$ be a Borel probability measure on $\mathbb{R}^{k}$ equipped with a
metric, and let $r\in(0,\infty)$. A quantization without any constraint is
called an unconstrained quantization. In unconstrained quantization (see [2]),
the numbers

$\underline{D}_{r}(P):=\liminf_{n\to\infty}\frac{r\log n}{-\log
V_{n,r}(P)}\text{ and }\overline{D}_{r}(P):=\limsup_{n\to\infty}\frac{r\log
n}{-\log V_{n,r}(P)},$ (13)

are called the lower and the upper quantization dimensions of the probability
measure $P$ of order $r$, respectively. If
$\underline{D}_{r}(P)=\overline{D}_{r}(P)$, the common value is called the
quantization dimension of $P$ of order $r$ and is denoted by $D_{r}(P)$. In
unconstrained quantization (see [2]) for any $\kappa>0$, the two numbers
$\liminf_{n}n^{\frac{r}{\kappa}}V_{n,r}(P)$ and
$\limsup_{n}n^{\frac{r}{\kappa}}V_{n,r}(P)$ are, respectively, called the
$\kappa$-dimensional lower and upper quantization coefficients for $P$. The
quantization coefficients provide us with more accurate information about the
asymptotics of the quantization error than the quantization dimension. In
unconstrained case, it is known that for an absolutely continuous probability
measure, the quantization dimension always exists and equals the Euclidean
dimension of the support of $P$, and the quantization coefficient exists as a
finite positive number (see [20]). If the $\kappa$-dimensional lower and upper
quantization coefficients for $P$ are finite and positive, then $\kappa$
equals the quantization dimension of $P$.

Unconstrained quantization error $V_{n,r}(P)$ goes to zero as $n$ tends to
infinity (see [2]). This is not true in the case of constrained quantization.
Constrained quantization error $V_{n,r}(P)$ can approach to any nonnegative
number as $n$ tends to infinity, and it depends on the constraint $S$ that
occurs in the definition of constrained quantization error as given in (1). In
this regard, we give the following examples.

Let $P$ be a Borel probability measure on $\mathbb{R}^{2}$ such that $P$ is
uniform on its support $\\{(x,y)\in\mathbb{R}^{2}:0\leq x\leq 2\text{ and
}y=0\\}$. Let $V_{n}(P):=V_{n,2}(P)$ be its constrained quantization error. If
the elements in the optimal sets lie on the line $y=1$ between the two
elements $(\frac{1}{2},1)$ and $(\frac{3}{2},1)$, then by Theorem 3.7, for
$n\geq 3$,

$V_{n}(P)=\frac{25n^{2}-50n+26}{24(n-1)^{2}}\text{ implying
}\lim_{n\to\infty}V_{n}(P)=\frac{25}{24}.$ (14)

If the elements in the optimal sets lie on the line $y=1$ between the two
elements $(0,1)$ and $(\frac{28}{15},1)$, then by Theorem 3.9, for $n\geq 8$,

$V_{n}(P)=\frac{7(5788(n-1)n+3015)}{10125(1-2n)^{2}}\text{ implying
}\lim_{n\to\infty}V_{n}(P)=\frac{10129}{10125}.$ (15)

On the other hand, if the elements in the optimal sets lie on the line
$y=\sqrt{3}x$ between the two elements $(0,0)$ and $(2,2\sqrt{3})$, then by
Corollary 3.3, for $n\geq 2$,

$V_{n}(P)=\frac{36n^{2}+49n-144}{12n^{3}}\text{ implying
}\lim_{n\to\infty}V_{n}(P)=0.$ (16)

Moreover, notice that if $P$ is a uniform distribution on a unit circle, and
if the elements in an optimal set of $n$-points lie on a concentric circle
with radius $a$, then by Theorem 4.3, for $n\geq 3$,

$V_{n}(P)=a^{2}+1-\frac{2an}{\pi}\sin\frac{\pi}{n}\text{ implying
}\lim_{n\to\infty}V_{n}(P)=(a-1)^{2},$ (17)

which is a nonnegative constant depending on the values of $a$.

Let us now give the following definition.

######  Definition 6.1.

Let $P$ be a Borel probability measure on $\mathbb{R}^{k}$ equipped with a
metric $d$, and let $r\in(0,\infty)$. Let $V_{n,r}(P)$ be the $n$th
constrained quantization error of order $r$ for a given $S$ that occurs in
(1). Then, the $n$th constrained quantization error $V_{n,r}(P)$ is a
decreasing sequence and converges to its exact lower bound, which is a
nonnegative constant. Set

$V_{\infty,r}(P):=\lim_{n\to\infty}V_{n,r}(P).$

Then, $(V_{n,r}(P)-V_{\infty,r}(P))$ is a decreasing sequence of nonnegative
real numbers such that

$\mathop{\lim}\limits_{n\to\infty}(V_{n,r}(P)-V_{\infty,r}(P))=0.$

Write

$\displaystyle\left\\{\begin{array}[]{ll}\underline{D}_{r}(P):=\mathop{\liminf}\limits_{n\to\infty}\frac{r\log
n}{-\log(V_{n,r}(P)-V_{\infty,r}(P))},\text{ and }\\\
\overline{D}_{r}(P):=\mathop{\limsup}\limits_{n\to\infty}\frac{r\log
n}{-\log(V_{n,r}(P)-V_{\infty,r}(P))}.\end{array}\right.$ (20)

$\underline{D}_{r}(P)$ and $\overline{D}_{r}(P)$ are called the lower and the
upper constrained quantization dimensions of the probability measure $P$ of
order $r$, respectively. If $\underline{D}_{r}(P)=\overline{D}_{r}(P)$, the
common value is called the constrained quantization dimension of $P$ of order
$r$ and is denoted by $D_{r}(P)$. The constrained quantization dimension
measures the speed at which the specified measure of the constrained
quantization error converges as $n$ tends to infinity. For any $\kappa>0$, the
two numbers

$\liminf_{n}n^{\frac{r}{\kappa}}(V_{n,r}(P)-V_{\infty,r}(P))\text{ and
}\limsup_{n}n^{\frac{r}{\kappa}}(V_{n,r}(P)-V_{\infty,r}(P))$

are, respectively, called the $\kappa$-dimensional lower and upper constrained
quantization coefficients for $P$. If the $\kappa$-dimensional lower and upper
constrained quantization coefficients for $P$ exist and are equal, then we
call it the $\kappa$-dimensional constrained quantization coefficient for $P$.
If the $\kappa$-dimensional lower and upper constrained quantization
coefficients for $P$ are finite and positive, then $\kappa$ equals the
constrained quantization dimension of $P$.

Let $V_{n,2}(P)$ be the $n$th constrained quantization error of order $2$.
Then,
(14) implies that

$\lim_{n\to\infty}\frac{2\log n}{-\log(V_{n,2}(P)-V_{\infty,2}(P))}=1\text{
and }\lim_{n\to\infty}n^{2}(V_{n,2}(P)-V_{\infty,2}(P))=\frac{1}{24},$ (21)

(15) implies that

$\lim_{n\to\infty}\frac{2\log n}{-\log(V_{n,2}(P)-V_{\infty,2}(P))}=1\text{
and }\lim_{n\to\infty}n^{2}(V_{n,2}(P)-V_{\infty,2}(P))=\frac{2744}{10125},$
(22)

(16) implies that

$\lim_{n\to\infty}\frac{2\log n}{-\log(V_{n,2}(P)-V_{\infty,2}(P))}=2\text{
and }\lim_{n\to\infty}n(V_{n,2}(P)-V_{\infty,2}(P))=3,$ (23)

and (17) implies that

$\lim_{n\to\infty}\frac{2\log n}{-\log(V_{n,2}(P)-V_{\infty,2}(P))}=1\text{
and }\lim_{n\to\infty}n^{2}(V_{n,2}(P)-V_{\infty,2}(P))=\frac{\pi^{2}a}{3}.$
(24)

###  6.2. Observations and Conclusions

  1. (1)

In unconstrained quantization, the elements in an optimal set are the
conditional expectations in their own Voronoi regions. It is not true in
constrained quantization, for example, for the probability measure $P$,
defined in Corollary 3.3, the optimal set of two-points is obtained as
$\\{(\frac{1}{8},\frac{1}{8}\sqrt{3}),(\frac{3}{8},\frac{3}{8}\sqrt{3})\\}$,
and the set of conditional expectations of the Voronoi regions is
$\\{(\frac{1}{2},0),(\frac{3}{2},0)\\}$, i.e., the two sets are different.

  2. (2)

In unconstrained quantization if the support of $P$ contains infinitely many
elements, then an optimal set of $n$-points contains exactly $n$ elements.
This is not true in constrained quantization. For example, from Subsection
5.1, we see that if a Borel probability measure $P$ on $\mathbb{R}^{2}$ has
support the diameter of a circle and the constraint $S$ is the circle, then
the optimal sets of $n$-points for all $n\geq 2$ always contain exactly two-
elements although the support has infinitely many elements.

  3. (3)

In unconstrained quantization, the quantization dimension for an absolutely
continuous probability measure exists and equals the Euclidean dimension of
the support of $P$. This fact is not true in constrained quantization, as can
be seen from the expressions (21), (22), and (23). Each of the probability
measures has support the closed interval $[0,2]$ on a line, but the
quantization dimensions are different, i.e., the quantization dimension in
constrained quantization depends on the constraint $S$ that occurs in the
definition of constrained quantization. The quantization dimension, in the
case of unconstrained quantization, if it exists, measures the speed at which
the specified measure of the error goes to zero as $n$ tends to infinity, on
the other hand, in the case of constrained quantization, if it exists,
measures the speed at which the specified measure of the error converges as
$n$ tends to infinity.

  4. (4)

In unconstrained quantization, the quantization coefficient for an absolutely
continuous probability measure exists as a unique finite positive number. In
constrained quantization, the quantization coefficient for an absolutely
continuous probability measure also exists, but it is not unique, and can be
any nonnegative number as can be seen from the expressions of quantization
coefficients in (21), (22), (23), and (24), i.e., the quantization coefficient
in constrained quantization depends on the constraint $S$ that occurs in the
definition of constrained quantization.

### Acknowledgement

The authors are grateful to Professor Carl P. Dettmann of the University of
Bristol, UK, for his valuable comments and suggestions in preparing this
manuscript. The first author is grateful to her supervisor Professor Tanmoy
Som of the IIT(BHU), Varanasi, India, for his 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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Jian Cui , Peter Danchev and Danya-Jin

# Rings whose Nil-Clean and Clean Elements are
Uniquely Nil-Clean

Jian Cui  School of Mathematics and Statistics, Anhui Normal University
Wuhu, Anhui 241002, China<EMAIL_ADDRESS>,  Peter Danchev  Institute of
Mathematics and Informatics, Bulgarian Academy of Sciences
”Acad. G. Bonchev” str., bl. 8, 1113 Sofia, Bulgaria<EMAIL_ADDRESS><EMAIL_ADDRESS>and  Danya Jin  School of Mathematics and Statistics,
Anhui Normal University
Wuhu, Anhui 241002, China<EMAIL_ADDRESS>
###### Abstract.

We consider and study those rings in which each nil-clean or clean element is
uniquely nil-clean. We establish that, for abelian rings, these rings have a
satisfactory description and even it is shown that the classes of abelian
rings and the rings in which nil-clean elements are uniquely nil-clean do
coincide. Moreover, we prove that the rings in which clean elements are
uniquely nil-clean coincide with the subclass of abelian rings consisting of
only unipotent units and, in particular, that in the semipotent case we have a
complete characterization only in terms of the former ring and its divisions.
Likewise, some extension properties and group rings for such kinds of rings
are also considered.

###### Key words and phrases:

Nil-clean elements; clean elements; uniquely nil-clean elements; strongly nil-
clean elements; UU-rings

###### 2020 Mathematics Subject Classification:

16U60; 16U99; 16S34

##  1\. Introduction and Conventions

All rings into consideration in the current work are associative and contain
an identity element. Let $R$ be a ring. The symbols $U(R)$, $J(R)$, ${\rm
nil}(R)$ and ${\rm idem}(R)$ stand for the group of all units, the Jacobson
radical, the set of all nilpotents of $R$ and the set of all idempotents of
$R$, respectively. For an arbitrary $n\in\mathbb{N}$, the ring consisting of
all $n\times n$ matrices over $R$ and the ring consisting of all $n\times n$
upper triangular matrices over $R$ are, respectively, denoted by
$\mathbb{M}_{n}(R)$ and $\mathbb{T}_{n}(R)$. Let $\mathbb{Z}_{n}$ be the ring
of integers modulo $n$.

Mimicking [14], an element $r$ of a ring $R$ is called clean, provided
$r=u+e$, where $u\in U(R)$ and $e\in{\rm idem}(R)$. If, in addition, $ue=eu$,
the element $r$ is said to be strongly clean (see [15]). Likewise, if in the
record of $r$ there exist a unique unit $u$ and an idempotent $e$ with
$r=u+e$, the element $r$ is then termed uniquely clean (see [16] and also [2]
in the case of group rings). Some other valuable sources in this topic are
[3], [10] and [11], as well. Moreover, when each element of a ring $R$ is
equipped with one of these properties, the ring $R$ also retains the same
name. It was proved in [16] that a ring $R$ is uniquely clean if, and only if,
$R$ is abelian (i.e., all idempotents of $R$ are central), $R/J(R)$ is boolean
and all idempotents of $R$ lift modulo $J(R)$.

On the other side, imitating these notions, in [8] were defined the kinds of
nil-clean, strongly nil-clean and uniquely nil-clean elements of rings as
follows: An element $r$ of a ring $R$ is called nil-clean, provided $r=q+e$,
where $q\in{\rm nil}(R)$ and $e\in{\rm idem}(R)$. If, in addition, $qe=eq$,
the element $r$ is said to be strongly nil-clean. Likewise, if in the record
of $r$ there exist a unique nilpotent $q$ and an idempotent $e$ with $r=q+e$,
the element $r$ is then termed uniquely nil-clean. It is worth mentioning that
the uniqueness of the nil-clean record in the commutative case is also
confirmed in [7]. When each element of a ring $R$ is equipped with one of
these properties, the ring $R$ also retains the same name. On this vein, it is
also worth noticing that arbitrary strongly nil-clean rings were independently
characterized in [6] and [12] as those rings $R$ for which $J(R)$ is nil and
$R/J(R)$ is boolean.

It is well known that, and also easy to check that, nil-clean rings are always
clean, while the converse fails. However, for the element-wise case, the
situation is totally different and unexpected. For a more detailed information
in this way, we refer to [18]. Nevertheless, one can say that if $2\in{\rm
nil}(R)$, then each nil-clean element is necessarily clean.

Incidentally, while studying rings with unipotent units (that are rings $R$
for which $U(R)=1+{\rm nil}(R)$ and, for shortness, hereafter abbreviating
them as UU-rings), it was shown in [6, Theorem 3.7] that in UU-rings any
element is (strongly) clean if, and only if, it is (strongly) nil-clean,
respectively (see Corollary 3.11 in [6] too). On that vein, recently, in [1],
it was provided a comprehensive exploration of those rings whose clean
elements are uniquely clean, calling them CUC-rings. They are a proper
generalization of uniquely clean rings. Specifically, the authors considered
certain crucial properties of such rings including their natural extensions
and group rings.

Our motivating tool to investigate such rings is that the uniqueness of nil-
clean elements is definitely not so good studied as that of clean elements,
bearing in mind the discussion presented above. Thus, our further plan is to
examine those rings in which every nil-clean or clean element is uniquely nil-
clean, calling them NCUNC-rings and CUNC-rings, respectively, and thus
extending the class of uniquely nil-clean rings. Concretely, in that aspect,
our main establishments are structured thus: In Section 2, we prove the
unexpected fact that the rings in which nil-clean elements possess an unique
record are exactly the abelian rings (see Theorem 2.2). In Section 3, we
describe rings in which clean elements have an unique record as nil-clean
elements and show that these are precisely the abelian rings whose units are
unipotents (i.e., a sum of the identity and a nilpotent) (see Theorem 3.1). In
the case of semipotent rings, we give an other valuable classification and
also provide a close relation with the aforementioned CUC-rings (see Theorem
3.7 and Proposition 3.9). In the final fourth section, we study group rings of
CUNC-rings. For a ring $R$ and a group $G$, we write $RG$ for the group ring
of $G$ over $R$. We establish some necessary and sufficient conditions by
proving that $RG$ is a CUNC-ring if, and only if, $R$ is a CUNC-ring and $G$
is a $2$-group, whenever $G$ is a locally finite group (see Theorem 4.4).

##  2\. Nil-clean elements are uniquely nil-clean

We begin here with the following useful technicality.

######  Proposition 2.1.

Let $R$ be a ring. Then the following are equivalent$:$

$(1)$ Every nilpotent of $R$ is uniquely nil-clean.

$(2)$ $({\rm nil}(R)+{\rm nil}(R))\cap{\rm idem}(R)=\\{0\\}$.

###### Proof.

“(1) $\Rightarrow$ (2)”. Let $e\in({\rm nil}(R)+{\rm nil}(R))\cap{\rm
idem}(R)$. Then, $e=b+c$ for some $b,c\in{\rm nil}(R)$. So $c=e+(-b)=0+c$ are
two nil-clean decomposition of $c$. By hypothesis, $e=0$, and hence $({\rm
nil}(R)+{\rm nil}(R))\cap{\rm idem}(R)=\\{0\\}$, as asserted.

“(2) $\Rightarrow$ (1)”. For any $b\in{\rm nil}(R)$, it is clear that $b=0+b$
is a nil-clean representation. Assume, for a moment, that there exists another
nil-clean decomposition, say $b=f+c$, where $f\in{\rm idem}(R)$ and $c\in{\rm
nil}(R)$. Then $f=b-c\in({\rm nil}(R)+{\rm nil}(R))\cap{\rm idem}(R)=\\{0\\}$,
and so $b$ is uniquely nil-clean, as claimed. ∎

A ring $R$ is said to be a _NCUNC-ring_ if every nil-clean element of $R$ is
uniquely nil-clean. Recall once again that a ring $R$ is _abelian_ if all
idempotents of $R$ are central. Besides, for shortness, we denote by
$C_{k}^{n}$ the binomial coefficient $\binom{n}{k}$, where $k\geq n\geq 1$ are
integers.

We now arrive at our first basic result, which is rather curious.

######  Theorem 2.2.

Let $R$ be a ring. Then the following are equivalent$:$

$(1)$ Every nil-clean element of $R$ is uniquely nil-clean, that is, $R$ is a
NCUNC-ring.

$(2)$ Every strongly nil-clean element of $R$ is uniquely nil-clean.

$(3)$ Every idempotent of $R$ is uniquely nil-clean.

$(4)$ $R$ is an abelian ring.

$(5)$ Every idempotent of $R$ commutes with all nilpotents and $({\rm
nil}(R)+{\rm nil}(R))\cap{\rm idem}(R)=\\{0\\}$.

$(6)$ $({\rm idem}(R)-{\rm idem}(R))\cap{\rm nil}(R)=\\{0\\}$.

###### Proof.

“(1) $\Rightarrow$ (2)”. This implication is obvious.

“(2) $\Rightarrow$ (3)”. Clearly, each idempotent is strongly nil-clean.
Therefore, (2) follows by hypothesis.

“(3) $\Rightarrow$ (4)”. Let $e^{2}=e\in R$. Thus, for any $r\in R$, we obtain
that

$e=(e-(1-e)re)+(1-e)re=(e-er(1-e))+er(1-e)$

are two nil-clean decompositions of $e$. So, we have $er(1-e)=(1-e)re$, which
implies that $er=ere=re$. Thus, $R$ is abelian.

“(4) $\Rightarrow$ (1)”. Suppose that $e+b=f+c$ for some $e,f\in{\rm idem}(R)$
and $b,c\in{\rm nil}(R)$. Then, $e-f=c-b$. As $R$ is abelian, we know that
$e,f,b$ and $c$ commute one another. So,

$(e-f)^{2}=(c-b)(e-f)\in{\rm idem}(R)\cap{\rm nil}(R)=\\{0\\}.$

It now follows that $e=(2e-1)f$, whence

$e=(2e-1)e=(2e-1)^{2}f=f.$

This ensures that $R$ is a NCUNC-ring, as stated.

“(4) $\Rightarrow$ (5)”. It is enough to show that $({\rm nil}(R)+{\rm
nil}(R))\cap{\rm idem}(R)=\\{0\\}$. To that goal, for any $e\in({\rm
nil}(R)+{\rm nil}(R))\cap{\rm idem}(R)$, one has that $e=b-c$ for some
$b,c\in{\rm nil}(R)$. We may let $b^{k}=c^{k}=0$ with $k\geq 1$. Then, we
deduce that

$\displaystyle(e+c)^{k}$
$\displaystyle=e+C_{k}^{1}ec+\cdots+C_{k}^{k-1}ec^{k-1}+C_{k}^{k}c^{k}$
$\displaystyle=e(1+C_{k}^{1}c+\cdots+C_{k}^{k-1}c^{k-1})$
$\displaystyle=b^{k}=0.$

Note that $C_{k}^{1}c+\cdots+C_{k}^{k-1}c^{k-1}\in{\rm nil}(R).$ So, we can
easily verify that

$1+C_{k}^{1}c+\cdots+C_{k}^{k-1}c^{k-1}\in U(R),$

which assures $e=0$, as required.

“(5) $\Rightarrow$ (6)”. Take

$x\in({\rm idem}(R)-{\rm idem}(R))\cap{\rm nil}(R).$

Then, there exist $e,f\in{\rm idem}(R)$ satisfying $x=e-f$. By hypothesis, we
have

$e(1-f)=x-xf\in({\rm nil}(R)+{\rm nil}(R))\cap{\rm idem}(R)=\\{0\\}.$

It now follows that

$-x=-xf=(f-e)f=(1-e)f$

must be zero, as needed.

“(6) $\Rightarrow$ (3)”. Given $e^{2}=e\in R$. Clearly, $e$ is nil-clean.
Write $e=f+b$, where $f\in{\rm idem}(R)$ and $b\in{\rm nil}(R)$. So, one has

$e-f=b\in({\rm idem}(R)-{\rm idem}(R))\cap{\rm nil}(R)=\\{0\\}.$

Thus, $e=f$, which proves that $e$ is uniquely nil-clean, as formulated. ∎

Now, according to Theorem 2.2, one sees that commutative rings, reduced rings
and rings with trivial idempotents are all NCUNC-rings.

######  Remark 2.3.

We claim that the condition “$({\rm nil}(R)+{\rm nil}(R))\cap{\rm
idem}(R)=\\{0\\}$” in Theorem 2.2(5) does not imply that $R$ is abelian. For
example, set $R=\mathbb{T}_{2}(\mathbb{Z}_{2})$. Then, one inspects by simple
matrix calculations that

$({\rm nil}(R)+{\rm nil}(R))\cap{\rm idem}(R)=\\{0\\},$

but $R$ is manifestly not abelian.

In view of “(3) $\Leftrightarrow$ (4)” in Theorem 2.2, we derive the following
result immediately.

######  Corollary 2.4.

An idempotent $e$ of a ring $R$ is uniquely nil-clean if, and only if, $e$ is
central.

Next, owing to Theorem 2.2, we can derive a similar result to that from [1].

######  Corollary 2.5.

Let $I$ be a nil-ideal of a ring $R$. Then the following are equivalent$:$

$(1)$ $R$ is an abelian ring.

$(2)$ $R/I$ is an abelian ring and idempotents lift uniquely modulo $I$.

###### Proof.

The proof surely can be extracted from [1, Lemma 2.8(1)], but we give an
independent proof from the view point of the nil-cleanness and uniquely nil-
cleanness.

“(1) $\Rightarrow$ (2)”. Let $\overline{e}^{2}=\overline{e}\in R/I$. As $I$ is
nil, idempotents lift modulo $I$ in the traditional manner. So, we may let
$e\in{\rm idem}(R)$. Notice that $e$ is central, and so $\overline{e}$ is
central too. Now, to prove that idempotent lift uniquely modulo $I$, we assume
that there exists $f\in{\rm idem}(R)$ such that $e-f\in I$. Then, $e=f+b$ for
some $b\in I\subseteq{\rm nil}(R)$. However, in view of Corollary 2.4, we
obtain that $e=f$, as desired.

“(2) $\Rightarrow$ (1)”. Given $e\in{\rm idem}(R)$. Let $e=f+b$ be a nil-clean
decomposition with $e\in{\rm idem}(R)$ and $b\in{\rm nil}(R)$. Then,
$\overline{e}=\overline{f}+\overline{b}$ is a nil-clean decomposition of
$\overline{e}\in R/I$. Since the quotient-ring $R/I$ is abelian, it follows
that $\overline{e}=\overline{f}$ by applying Theorem 2.2. Hence, $e=f$, and
thus $e$ is uniquely nil-clean in $R$. Therefore, $R$ is abelian, as wanted. ∎

##  3\. Clean elements are uniquely nil-clean

It is well known that nil-clean rings are clean (see also [8]), but there
exist nil-clean elements which are not clean ([18]). In this section, we study
rings whose clean elements are uniquely nil-clean. We say that a ring $R$ is a
_CUNC-ring_ if every clean element of $R$ is uniquely nil-clean. Recall that a
ring $R$ is known to be a _UU-ring_ (see, e.g., [6]) if $U(R)=1+{\rm nil}(R)$.

We are now able to prove the following major assertion.

######  Theorem 3.1.

Let $R$ be a ring. Then the following are equivalent$:$

$(1)$ $R$ is a CUNC-ring.

$(2)$ $R$ is an abelian UU-ring.

###### Proof.

“(1) $\Rightarrow$ (2)”. Notice the well-known fact that all idempotents are
clean. By hypothesis, all idempotents are uniquely nil-clean. Furthermore, in
view of Theorem 2.2, we conclude that $R$ is abelian. Now, given $u\in U(R)$,
we elementarily see that $u$ is nil-clean since it is clean. Thus, we write
that $u=e+b$, where $e\in{\rm idem}(R)$ and $b\in{\rm nil}(R)$. From $R$ is
abelian, one obtains that $e=u-b\in U(R)$, and hence $e=1$. So, $R$ is $UU$,
as asserted.

“(2) $\Rightarrow$ (1)”. Suppose that $a\in R$ is clean. Write $a=(1-e)+u$
with $e\in{\rm idem}(R)$ and $u\in U(R)$. As $R$ is a UU-ring, it must be that
$u=1+b$ for some $b\in{\rm nil}(R)$ and $2\in{\rm nil}(R)$. It now follows
that

$a=(1-e)+u=e+(1-2e+u)=e+(2-2e+b),$

where $n_{1}:=2-2e+b$ is nilpotent since $R$ is abelian. This proves that
$a=e+n_{1}$ is nil-clean. If, however, there is another nil-clean
decomposition $a=f+n_{2}$, where $f\in{\rm idem}(R)$ and $n_{1}\in{\rm
nil}(R)$, then we get that $e-f=n_{2}-n_{1}$. By application of [17, Corollary
2.4], ${\rm nil}(R)$ is a subring of $R$ as $R$ is a $UU$-ring. So, we may
write that

$e-f=(e-f)^{3}=(n_{2}-n_{1})^{3}\in{\rm nil}(R),$

and then it follows at once that $e=f$. Therefore, $a$ is uniquely nil-clean,
as claimed. ∎

The next statement is a direct consequence of a simple combination of Theorems
2.2 and 3.1.

######  Corollary 3.2.

Let $R$ be a ring. Then the following are equivalent$:$

$(1)$ $R$ is a CUNC-ring.

$(2)$ $R$ is abelian and every clean element is nil-clean.

$(3)$ $R$ is a NCUNC-ring and every unit is nil-clean.

We now concentrate on the following reduction assertion.

######  Corollary 3.3.

Let $R$ be a ring. Then the following are equivalent$:$

$(1)$ $R$ is a CUNC-ring.

$(2)$ $J(R)$ is nil, $R/J(R)$ is a CUNC-ring and idempotents lift uniquely
modulo $J(R)$.

###### Proof.

“(1) $\Rightarrow$ (2)”. Utilizing Theorem 3.1, $R$ is an abelian UU-ring. So,
$R/J(R)$ is UU, and $J(R)$ is nil by consulting with [6, Theorem 2.4(2)].
Furthermore, employing Corollary 2.5, it follows that idempotents lift
uniquely modulo $J(R)$ and $R/J(R)$ is abelian. Now, applying Theorem 3.1,
$R/J(R)$ is a CUNC-ring.

“(2) $\Rightarrow$ (1)”. Since $R/J(R)$ is a CUNC-ring, $R/J(R)$ is an abelian
UU-ring. By hypothesis and with Corollary 2.5 at hand, $R$ has to be abelian.
Now, as $R/J(R)$ is a UU-ring and $J(R)$ is nil, [6, Theorem 2.4(2)] applies
to conclude that $R$ is a UU-ring, as expected. ∎

It is worthwhile to note the following: one readily verifies that the ring
$\mathbb{T}_{2}(\mathbb{Z}_{2})$ is a UU-ring, but not a CUNC-ring as it is
not abelian and even more, for any ring $R$ and any integer $n\geq 2$, both
rings $\mathbb{M}_{n}(R)$ and $\mathbb{T}_{n}(R)$ are not CUNC.

######  Corollary 3.4.

If $R$ is a CUNC-ring, then the corner $eRe$ is a CUNC-ring for any
$e^{2}=e\in R$.

###### Proof.

It is clear that the subring of an abelian ring is still abelian. By
assumption, $eRe$ is abelian. In view of Corollary 3.2, it suffices to show
that every clean element of $eRe$ is also nil-clean. To that goal, assume
$a\in eRe$ is clean. Then, there exist $f^{2}=f\in eRe$ and $w\in U(eRe)$ such
that $a=f+w$. So, one has

$a+(1-e)=f+(w+(1-e))$

is a clean decomposition in $R$. It thus follows that $a+(1-e)$ is nil-clean
in $R$. Hence, we write that

$a+(1-e)=g+b$

with $g^{2}=g\in R$ and $b\in{\rm nil}(R)$. It now follows that $a=ege+ebe$ is
a nil-clean element in the corner ring $eRe$, because $e$ is central, as
required. ∎

The next statement is helpful and shows the inheritance by subrings and direct
products of the property being a CUNC-ring. Note that our subrings need not be
unital (herein, a subring of a given ring is unital whenever it contains the
same identity element).

######  Proposition 3.5.

$(1)$ A subring $S$ of a CUNC-ring $R$ is again a CUNC-ring.

$(2)$ A direct product $\prod R_{i}$ is a CUNC-ring if, and only if, so is
$R_{i}$ for each $i$.

###### Proof.

(1) One easily checks that any subring of an abelian ring is abelian as well.
However, in view of Theorem 3.1, it suffices to show only that $S$ is a UU-
ring. For convenience, we use $1_{R}$ and $1_{S}$ to denote the identity of
$R$ and $S$, respectively. To that goal, let $v\in U(S)$. Then,
$v+(1_{R}-1_{S})\in U(R)$ with its inverse $v^{-1}+(1_{R}-1_{S})$. By
assumption, we have $v+(1_{R}-1_{S})=1_{R}+b$ for some $b\in{\rm nil}(R)$,
which in turn implies that $v=1_{S}+b$ and

$b=v-1_{S}\in S\cap{\rm nil}(R)={\rm nil}(S).$

Thus, $S$ is UU, as pursued.

(2) For the sufficiency, it is clear that $\prod R_{i}$ is abelian as all
$R_{i}$ are abelian. Let $(a_{i})\in\prod R_{i}$ be a clean element. Then, by
plain element-wise arguments, it is evident that all vector’s components
$a_{i}$ are clean. Referring to Corollary 3.2, one gets that $a_{i}$ is nil-
clean for each $i$, which in turn insures that $(a_{i})$ is nil-clean in
$\prod R_{i}$. Hence, $\prod R_{i}$ a CUNC-ring, as formulated. The necessity
follows from (1). ∎

The next result contrasts to [1, Corollary 2.13(1)] and somewhat refines [1,
Corollary 2.14(1)].

######  Proposition 3.6.

$(1)$ For any ring $R$, the power series ring $R[[x]]$ is not CUNC.

$(2)$ A polynomial ring $R[x]$ over a commutative ring $R$ is CUNC if, and
only if, so is $R$.

###### Proof.

(1) Note the principal fact that the Jacobson radical of $R[[x]]$ is not nil.
Thus, in view of Corollary 3.3, $R[[x]]$ is really not a CUNC-ring.

(2) It is a well-known fact that ${\rm nil}(R[x])={\rm nil}(R)[x]$ (see, for
instance, [13]) whenever $R$ is commutative. First, assume that $R$ is a CUNC-
ring. Let $f(x)=\sum_{i=0}^{n}a_{i}x^{i}\in U(R[x])$. Then, as it is known in
the existing literature (see, e.g., [13]), it follows that $a_{0}\in U(R)$ and
$a_{i}\in{\rm nil}(R)$ for all $1\leq i\leq n$. By hypothesis, we can write
$a_{0}=1+b_{0}$ for some $b_{0}\in{\rm nil}(R)$. It thus follows that

$f(x)=1+b_{0}+\sum_{i=1}^{n}a_{i}x^{i}\in 1+{\rm nil}(R[x]).$

Finally, Theorem 3.1 applies to get that $R[x]$ is, indeed, a CUNC-ring, as
claimed.

The converse claim follows with the aid of Proposition 3.5(1). ∎

A ring $R$ is known to be _semipotent_ if every one-sided ideal not contained
in $J(R)$ contains a non-zero idempotent. It is a well-known fact that clean
rings are semipotent (see, e.g., [14]).

We now manage to prove the following chief result.

######  Theorem 3.7.

Let $R$ be a ring. Then the following are equivalent$:$

$(1)$ $R$ is a uniquely nil-clean ring.

$(2)$ $R$ is a semipotent CUNC-ring.

$(3)$ $R$ is abelian, $J(R)$ is nil and $R/J(R)$ is boolean.

###### Proof.

“(1) $\Rightarrow$ (2)”. It is clear that $R$ is a CUNC-ring. Since uniquely
nil-clean rings are nil-clean and so clean, we derive that $R$ is semipotent.

“(2) $\Rightarrow$ (3)”. By virtue of Theorem 3.1 and Corollary 3.3, it
suffices to show that $R/J(R)$ is boolean. To this purpose, assume on the
contrary. Then, there exists $a\in R$ such that $a-a^{2}\not\in J(R)$. Since
$R$ is semipotent, there is a non-zero $e\in{\rm idem}(R)$ satisfying
$e\in(a-a^{2})R$. Therefore, $e=(a-a^{2})t$ for some $t\in R$. As $e$ is
central, we may write

$e=(ea)(e(1-a))(et),$

which yields that both $ea$ and $e(1-a)$ are units of the corner ring $eRe$.
Writing now $v=ea+(1-e)$, we then obtain $v\in U(R)$. But since $R$ is a UU-
ring, it must be that

$v-1=-e(1-a)\in{\rm nil}(R),$

which is the wanted contradiction. Hence, $R/J(R)$ is Boolean.

“(3) $\Rightarrow$ (1)”. It is clear that Boolean rings are always uniquely
nil-clean. By assumption, $R/J(R)$ is a CUNC-ring. In accordance with
Corollary 3.3, we have $R$ is a CUNC-ring. Now, to prove that $R$ is uniquely
nil-clean, we only need to show that each element of $R$ is clean. Indeed, the
quotient $R/J(R)$ is a clean ring, and idempotents lift modulo $J(R)$.
Consequently, [9, Proposition 6] allows us to deduce that $R$ is clean, as
required. ∎

Combing Theorem 3.7 with the corresponding results from [14], we obtain the
following result immediately.

######  Corollary 3.8.

If $R$ is a CUNC-ring, then the following are equivalent$:$

$(1)$ $R$ is a uniquely nil-clean ring.

$(2)$ $R$ is a nil-clean ring.

$(3)$ $R$ is a clean ring.

$(4)$ $R$ is an exchange ring.

$(5)$ $R$ is a semipotent ring.

We proceed by proving the next claim, which gives a relationship between CUC-
rings from [1] and our CUNC-rings in the case of semi-potentness.

######  Proposition 3.9.

Suppose that $R$ is a semipotent ring. Then $R$ is a CUNC-ring if, and only
if, $R$ is a CUC-ring and $J(R)$ is nil.

###### Proof.

Combining [1, Theorem 3.2] with Theorems 3.1 and 3.7, we are done. ∎

##  4\. CUNC group rings

Let $R$ be a ring, $G$ a group, and, as usual, we denote by $RG$ the group
ring of $G$ over $R$.

In this section, we intend to establish a criterion for a group ring to be
CUNC under some minimal restrictions on the former group and ring objects.
Specifically, we are able to achieve this, provided the group is locally
finite. Recall that a group $G$ is a _$2$ -group_ if every its element has
order that is a power of $2$. Also, we recollect that a group is _locally
finite_ if each finitely generated subgroup is finite.

The next two technicalities are helpful to us for proving the main result
stated below.

######  Lemma 4.1.

Suppose that $R$ is a CUNC-ring. Then the following hold$:$

$(1)$ $2\in{\rm nil}(R)$.

$(2)$ If $I$ is a nil-ideal of $R$, then $R/I$ is a CUNC-ring.

###### Proof.

$(1)$ follows by Theorem 3.1 and [6, Theorem 2.6(1)].

$(2)$ Notice that, consulting with Theorem 3.1, $R$ is both abelian and UU. As
$I$ is nil, by a similar argument to that in the implication “(1)
$\Rightarrow$ (2)” from Corollary 2.5, the factor-ring $R/I$ is abelian.
Further, in view of [6, Theorem 2.4(1)], we deduce that the quotient $R/I$ a
UU-ring. Thus, $R/I$ is a CUNC-ring by the usage of Theorem 3.1, as stated. ∎

######  Lemma 4.2.

Let $R$ be a ring and let $G$ be a group. If $RG$ is a CUNC-ring, then $R$ is
a CUNC-ring and $G$ is a $2$-group.

###### Proof.

As $R$ can be viewed as a subring of $RG$, we refer to Proposition 3.5(1) to
get that $R$ is a CUNC-ring. Now, Lemma 4.1 applies to infer that $2$ is
nilpotent in both $RG$ and $R$. which implies that $2RG$ is a nil-ideal of
$RG$, and so it follows that $(R/2R)G\cong RG/2RG$ is a CUNC-ring.
Furthermore, we may assume $2=0\in R$. Given any $g\in G$, we know that $g\in
U(RG)$. By assumption, $1-g$ is nilpotent in $RG$, say $(1-g)^{2^{m}}=0$ for
some $m\geq 1$. This guarantees that

$g^{2^{m}}-1=(g-1)^{2^{m}}=0,$

and, therefore, $g^{2^{m}}=1$. Thus, $G$ is a $2$-group, as required. ∎

We recall now that the map $f:RG\rightarrow R$, defined by $f(\sum_{g\in
G}a_{g}g)=\sum_{g\in G}a_{g}$, is a surjective ring homomorphism. The kernel
of $f$ is called the _augmentation ideal_ of $RG$ and, standardly, is denoted
by $\omega(RG)$. It is well known that $\omega(RG)$ is an ideal of $RG$
generated by the set $\\{1-g:g\in G\\}$.

Our next technical claim asserts the following.

######  Lemma 4.3.

If $R$ is a CUNC-ring and $G$ is a locally finite $2$-group, then $RG$ is a
CUNC-ring.

###### Proof.

According to Lemma 4.1, we have $2\in{\rm nil}(R)$, whence $2\in J(R)$. Hence,
$R$ is abelian with $2\in J(R)$. Now, [2, Lemma 11] enables us that every
idempotent of $RG$ lies in $R$ since $G$ is a locally finite $2$-group, which
means that $RG$ is abelian. To prove now that $RG$ is a CUNC-ring, as pursued,
we only need to show that $RG$ is a UU-ring. Indeed, as $G$ is a locally
finite $2$-group and $2\in{\rm nil}(R)$, the ideal $\omega(RG)$ is nil owing
to [4, Proposition 16(ii)]. However, we observe that $R\cong RG/\omega(RG)$ is
a CUNC-ring. Finally, [6, Theorem 2.4(1)] allows us to conclude that $RG$ is a
UU-ring, as desired. ∎

Now, combining Lemma 4.2 with Lemma 4.3, we can extract the following chief
result of this section.

######  Theorem 4.4.

Let $R$ be a ring and let $G$ be a locally finite group. Then $RG$ is a CUNC-
ring if, and only if, $R$ is a CUNC-ring and $G$ is a $2$-group.

It is worthwhile noticing that a valuable non-trivial necessary and sufficient
condition for a group ring to be UU was found in [5]. Notice also that Theorem
3.1 is a guarantor that CUNC-rings are always abelian UU-rings, and vice
versa.

We close our investigation with the following two intriguing and non-trivial
questions, which are closely related to our presentation given above.

Problem 1. Examine those rings whose nil-clean elements are either uniquely
clean or uniquely strongly clean.

Problem 2. Examine those rings whose strongly clean elements are either
uniquely clean or uniquely strongly clean.

FUNDING

The first-named author of this research paper was supported by Anhui
Provincial Natural Science Foundation (No. 2008085MA06), Key Laboratory of
Financial Mathematics of Fujian Province University (Putian University) (No.
JR202203) and the Key project of Anhui Education Committee (No.
gxyqZD2019009).

The second-named author of this research paper was partially supported by the
Bulgarian National Science Fund under Grant KP-06 No. 32/1 of December 07,
2019, as well as by the Junta de Andalucía, Grant FQM 264, and by the BIDEB
2221 of TÜBÍTAK.

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