# Quantization for the mixtures of overlap probability distributions

1Asha Barua  ,  2Angelina Chavera  ,  3Ivan Djordjevic  ,  4Valerie Manzano  ,
5Sergio Soto Quintero  ,  6Mrinal Kanti Roychowdhury  and  7Hilda Tejeda

###### Abstract.

Mixtures of probability distributions, also known as mixed distributions, are
an exciting new area for optimal quantization. In this paper, we have
considered a mixed distribution which is generated by overlap uniform
distributions. For this mixed distribution we determine the optimal sets of
$n$-means and the $n$th quantization errors for all positive integers $n$.

###### Key words and phrases:

Mixed distribution, uniform distribution, optimal sets of $n$-means,
quantization error

###### 2010 Mathematics Subject Classification:

60Exx, 94A34.

##  1\. Introduction

Let $\mathbb{R}^{d}$ denote the $d$-dimensional Euclidean space equipped with
a metric $\|\cdot\|$ compatible with the Euclidean topology. Let $P_{1}$ and
$P_{2}$ be two Borel probability measures on $\mathbb{R}^{d}$. Then, a Borel
probability measure $P$ on $\mathbb{R}^{d}$ is called a mixture, or a mixed
distribution generated by $P_{1}$ and $P_{2}$ associated with a probability
vector $(p,1-p)$ if $P:=pP_{1}+(1-p)P_{2}$, where $0<p<1$. The $n$th
quantization error for $P$, with respect to the squared Euclidean distance, is
defined by

$V_{n}:=V_{n}(P)=\inf\Big{\\{}V(P;\alpha):\alpha\subset\mathbb{R}^{d},1\leq\text{
card}(\alpha)\leq n\Big{\\}},$

where $V(P;\alpha)=\int\min_{a\in\alpha}\|x-a\|^{2}dP(x)$ represents the
distortion error due to the set $\alpha$ with respect to the probability
distribution $P$. A set $\alpha\subset\mathbb{R}^{d}$ is called an optimal set
of $n$-means for $P$ if $V_{n}(P)=V(P;\alpha)$. It is known that for a Borel
probability measure $P$ if its support contains infinitely many elements and
$\int\|x\|^{2}dP(x)$ is finite, then an optimal set of $n$-means always has
exactly $n$-elements [AW, GKL, GL1, GL2]. The elements of an optimal set of
$n$-means are called optimal quantizers. For some work in this direction, one
can see [CR, DR1, DR2, GL3, L1, R1, R2, R3, R4, R5, R6, RR1].

Let us now state the following proposition (see [GG, GL1]):

######  Proposition 1.1.

Let $\alpha$ be an optimal set of $n$-means for $P$, and $a\in\alpha$. Then,

$(i)$ $P(M(a|\alpha))>0$, $(ii)$ $P(\partial M(a|\alpha))=0$, $(iii)$
$a=E(X:X\in M(a|\alpha))$, where $M(a|\alpha)$ is the Voronoi region of
$a\in\alpha,$ i.e., $M(a|\alpha)$ is the set of all elements $x$ in
$\mathbb{R}^{d}$ which are closest to $a$ among all the elements in $\alpha$.

Let $P_{1}$ and $P_{2}$ be two uniform distributions, respectively, on the
intervals given by $[0,1]$, and $[\frac{1}{2},\frac{3}{2}]$. Notice that
$P_{1}$ and $P_{2}$ have supports overlapped over the interval
$[\frac{1}{2},1]$. In this paper for the mixed distribution
$P:=\frac{1}{2}P_{1}+\frac{1}{2}P_{2}$ we determine the optimal sets of
$n$-means and the $n$th quantization errors for all positive integers $n$. The
main significance of this paper is that using the similar technique, given in
this paper, one can investigate the optimal sets of $n$-means and the $n$th
quantization errors for all $n\in\mathbb{N}$ for any mixed distribution
generated by overlap probability distributions.

##  2\. basic preliminaries

Let $f_{1}$ and $f_{2}$ be the respective density functions for the uniform
distributions $P_{1}$ and $P_{2}$ defined on the closed intervals $[0,1]$ and
$[\frac{1}{2},\frac{3}{2}]$. Then,

$f_{1}(x)=\left\\{\begin{array}[]{cc}1&\text{ if }x\in[0,1],\\\
0&\text{otherwise};\end{array}\right.\text{ and
}f_{2}(x)=\left\\{\begin{array}[]{cc}1&\text{ if
}x\in[\frac{1}{2},\frac{3}{2}],\\\ 0&\text{otherwise}.\end{array}\right.$

Let us consider the mixed distribution $P:=pP_{1}+(1-p)P_{2}$, where $0<p<1$.
Notice that $P$ has support the closed interval $[0,\frac{3}{2}]$, and the
component probabilities $P_{1}$ and $P_{2}$ have overlaps on the interval
$[\frac{1}{2},1]$. Since $f_{1}$ and $f_{2}$ are the density functions for the
probability distributions $P_{1}$ and $P_{2}$, respectively, we have

$dP_{1}(x)=P_{1}(dx)=f_{1}(x)dx=dx,\text{ and
}dP_{2}(x)=P_{2}(dx)=f_{2}(x)dx=dx.$

Notice that if $x\in[0,\frac{1}{2}]$, then $dP(x)=p\,dP_{1}(x)=pdx$; if
$x\in[\frac{1}{2},1]$, then
$dP(x)=pdP_{1}(x)+(1-p)dP_{2}(x)=pf_{1}(x)dx+(1-p)f_{2}(x)dx=dx;$ on the other
hand, if $x\in[1,\frac{3}{2}]$, then $dP(x)=(1-p)dP_{2}(x)=(1-p)dx$. Let us
now define a function $f$ on the real line $\mathbb{R}$ such that

(5) $\displaystyle f(x)=\left\\{\begin{array}[]{cc}p&\text{ if
}x\in[0,\frac{1}{2}],\\\ 1&\text{ if }x\in[\frac{1}{2},1],\\\ (1-p)&\text{ if
}x\in[1,\frac{3}{2}],\\\ 0&\text{otherwise}.\end{array}\right.$

Notice that the function $f$ satisfies the following properties to be a
density function:

$f(x)\geq 0\text{ for all }x\in\mathbb{R},\text{ and
}\int_{-\infty}^{\infty}f(x)dx=1;$

in fact, the mixed distribution $P$ can now be identified as a probability
distribution on $\mathbb{R}$ with the density function $f$, i.e., for any
$x\in\mathbb{R}$, we have $dP(x)=P(dx)=f(x)dx$.

######  Lemma 2.1.

Let $P$ be a probability distribution with the density function $f$, and $X$
be a random variable with distribution $P$. Then, its expectation and variance
are respectively given by $E(X)=\frac{2-p}{2}$ and
$V(X)=\frac{1}{12}(-3p^{2}+3p+1)$.

###### Proof.

We have

$E(X)=\int
xdP=p\int_{0}^{\frac{1}{2}}x\,dx+\int_{\frac{1}{2}}^{1}x\,dx+(1-p)\int_{1}^{\frac{3}{2}}x\,dx=\frac{2-p}{2},$

and

$\displaystyle V(X)=\int(x-E(X))^{2}dP$
$\displaystyle=p\int_{0}^{\frac{1}{2}}(x-\frac{2-p}{2})^{2}\,dx+\int_{\frac{1}{2}}^{1}(x-\frac{2-p}{2})^{2}\,dx+(1-p)\int_{1}^{\frac{3}{2}}(x-\frac{2-p}{2})^{2}\,dx,$

implying $V(X)=\frac{1}{12}(-3p^{2}+3p+1)$, and thus, the lemma is yielded. ∎

######  Note 2.2.

Lemma 2.1 implies that the optimal set of one-mean is the set
$\\{\frac{2-p}{2}\\}$, and the corresponding quantization error is the
variance $V:=V(X)$ of a random variable with distribution $P$. For a subset
$J$ of $\mathbb{R}$ with $P(J)>0$, by $P(\cdot|_{J})$, we denote the
conditional probability given that $J$ is occurred, i.e.,
$P(\cdot|_{J})=P(\cdot\cap J)/P(J)$, in other words, for any Borel subset $B$
of $\mathbb{R}$ we have $P(B|_{J})=\frac{P(B\cap J)}{P(J)}$.

The following proposition that appears in [BFGLR] plays an important role in
this paper. For the completeness of the paper we also give the proof of it.

######  Proposition 2.3.

Let $P$ be a Borel probability measure on $\mathbb{R}$ such that $P$ is
uniformly distributed over a closed interval $[a,b]$ with a constant density
function $f$ such that $f(x)=t$ for all $x\in[a,b]$, where $t\in\mathbb{R}$.
Then, the optimal set $\alpha_{n}(P(\cdot|_{[a,b]}))$ of $n$-means and the
corresponding quantization error $V_{n}(P(\cdot|_{[a,b]}))$ of $n$-means for
the probability distribution $P(\cdot|_{[a,b]})$ are, respectively, given by

$\alpha_{n}(P(\cdot|_{[a,b]})):=\Big{\\{}a+\frac{(2j-1)(b-a)}{2n}:1\leq j\leq
n\Big{\\}},\text{ and }V_{n}(P(\cdot|_{[a,b]}))=\frac{(b-a)^{3}t}{12n^{2}}.$

###### Proof.

Let $\alpha_{n}:=\\{a_{1}<a_{2}<\cdots<a_{n}\\}$ be an optimal set of
$n$-means for the probability distribution $P$ with a constant density
function $f$ on $[a,b]$ such that $f(x)=t$ for all $x\in[a,b]$, where
$t\in\mathbb{R}$. Then, proceedings analogously as [RR2, Theorem 2.1.1], we
can show that $a_{j}=a+\frac{(2j-1)(b-a)}{2n}$ implying

$\alpha_{n}(P(\cdot|_{[a,b]}))=\Big{\\{}a+\frac{(2j-1)(b-a)}{2n}:1\leq j\leq
n\Big{\\}}.$

Notice that the probability density function is constant, and the Voronoi
regions of the points $a_{j}$ for $1\leq j\leq n$ are of equal lengths. This
yields the fact that the distortion errors due to each $a_{j}$ are equal.
Hence, the $n$th quantization error is given by

$\displaystyle V_{n}(P(\cdot|_{[a,b]}))$
$\displaystyle=\int\min_{a\in\alpha_{n}(P(\cdot|_{[a,b]}))}(x-a)^{2}dP=nt\int_{a}^{\frac{1}{2}(a_{1}+a_{2})}(x-a_{1})^{2}dx=nt\int_{a}^{a+\frac{b-a}{n}}(x-(a+\frac{b-a}{2n}))^{2}dx$

implying

$V_{n}(P(\cdot|_{[a,b]}))=\frac{(b-a)^{3}t}{12n^{2}}.$

Thus, the proof of the proposition is complete. ∎

The following proposition plays an important role in the paper.

######  Proposition 2.4.

Let $P$ be a Borel probability measure on $\mathbb{R}$ such that $P$ is
uniformly distributed over a closed interval $[a,b]$ with a constant density
function $f$ such that $f(x)=t$ for all $x\in[a,b]$, where $t\in\mathbb{R}$.
Let $\alpha_{n+1}(P(\cdot|_{[a,b]})):=\\{a_{1}<a_{2}<\cdots<a_{n}<b\\}$ be an
optimal set of $(n+1)$-means such that it always contains the endpoint $b$ of
the interval $[a,b]$. Then,

$a_{j}=\left\\{\begin{array}[]{cc}a+\frac{(2j-1)(b-a_{1})}{2n}&\text{ for
}1\leq j\leq n,\\\ b&\text{ if }j=n+1,\end{array}\right.$

and the corresponding quantization error is given by

$V_{n+1}(P(\cdot|_{[a,b]}))=\frac{t}{12n^{2}}(b-a_{1})^{3}+\frac{t}{3}(a_{1}-a)^{3}.$

###### Proof.

Let $\alpha_{n+1}(P(\cdot|_{[a,b]})):=\\{a_{1}<a_{2}<\cdots<a_{n}<b\\}$ be an
optimal set of $(n+1)$-means for the probability distribution
$P(\cdot|_{[a,b]})$ on the closed interval $[a,b]$. Let us first prove the
following claim.

Claim. $a_{1},a_{2},\cdots,a_{n}$ are uniformly distributed over the closed
interval $[a,\frac{1}{2}(a_{n}+b)]$.

Recall that the points in an optimal set are the conditional expectations in
their own Voronoi regions. Thus,

$a_{1}=E(X:X\in[a,\frac{1}{2}(a_{1}+a_{2})])=\frac{1}{4}(2a+a_{1}+a_{2})\text{
implying }a_{2}-a_{1}=2(a_{1}-a).$

Set $a_{m+1}:=b$. Then, for $2\leq i\leq m$,

$a_{i}=E(X:X\in[\frac{1}{2}(a_{i-1}+a_{i}),\frac{1}{2}(a_{i}+a_{i+1})])=\frac{1}{4}(a_{i-1}+2a_{i}+a_{i+1}),$

yielding $a_{i}-a_{i-1}=a_{i+1}-a_{i}$. Hence,

$a_{2}-a_{1}=a_{3}-a_{2}=\cdots=b-a_{m}=2(a_{1}-a).$

Notice that $b-a_{m}=2(a_{1}-a)$ implies that
$\frac{1}{2}(a_{m}+b)-a_{m}=a_{1}-a$. Thus, we deduce that
$a_{1},a_{2},\cdots,a_{n}$ are uniformly distributed over the closed interval
$[a,\frac{1}{2}(a_{n}+b)]$, which is the claim. Due to the claim, using
Proposition 2.3, we have

$a_{j}=\\{a+\frac{(2j-1)}{2n}(\frac{1}{2}(a_{n}+b)-a)\text{ for }1\leq j\leq
n\\}$

yielding

(8) $\displaystyle
a_{j}=\left\\{\begin{array}[]{cc}a+\frac{(2j-1)}{2n}(\frac{1}{2}(a_{n}+b)-a)&\text{
for }1\leq j\leq n,\\\ b&\text{ if }j=n+1.\end{array}\right.$

Notice that $b-\frac{1}{2}(a_{n}+b)=a_{1}-a$ yielding
$\frac{1}{2}(a_{n}+b)-a=b-a_{1}.$ Hence, by (8), we have

$a_{j}=\left\\{\begin{array}[]{cc}a+\frac{(2j-1)(b-a_{1})}{2n}&\text{ for
}1\leq j\leq n,\\\ b&\text{ if }j=n+1.\end{array}\right.$

To find the quantization error we proceed as follows: Since the points
$a_{1},a_{2},\cdots,a_{n}$ are uniformly distributed over the closed interval
$[a,\frac{1}{2}(a_{n}+b)]$, by Proposition 2.3, the quantization error
contributed by the points $a_{1},a_{2},\cdots,a_{n}$ over the closed interval
$[a,\frac{1}{2}(a_{n}+b)]$ is given by

(9)
$\frac{(\frac{1}{2}(a_{n}+b)-a)^{3}t}{12n^{2}}=\frac{t}{12n^{2}}(b-a_{1})^{3}.$

The quantization error contributed by $b$ in the closed interval
$[\frac{1}{2}(a_{n}+b),b]$ is given by

(10)
$\int_{\frac{1}{2}(a_{n}+b)}^{b}(x-b)^{2}dP=t\int_{b+a-a_{1}}^{b}(x-b)^{2}\,dx=\frac{t}{3}(a_{1}-a)^{3}.$

By (9) and (10), we have

$V_{n+1}(P(\cdot|_{[a,b]}))=\frac{t}{12n^{2}}(b-a_{1})^{3}+\frac{t}{3}(a_{1}-a)^{3}.$

Thus, the proof of the proposition is complete. ∎

##  3\. Optimal quantization for the mixed distribution $P$ when
$p=\frac{1}{2}$

Recall that the mixed distribution $P$ is identified as a probability
distribution with density function $f$. If $p=\frac{1}{2}$, then
$E(X)=\frac{3}{4}$ and $V(X)=\frac{7}{48}$, i.e., the optimal set of one-mean
for $p=\frac{1}{2}$ is $\\{\frac{3}{4}\\}$ and the corresponding quantization
error is $V(X)=\frac{7}{48}$. For $p=\frac{1}{2}$, the density function $f$
for $P$ represented by (3) reduces to

$\displaystyle f(x)=\left\\{\begin{array}[]{cc}\frac{1}{2}&\text{ if
}x\in[0,\frac{1}{2}]\cup[1,\frac{3}{2}],\\\ 1&\text{ if
}x\in[\frac{1}{2},1],\\\ 0&\text{ otherwise}.\end{array}\right.$

Notice that the probability measure $P$ is ‘symmetric’ about the point
$\frac{3}{4}$, i.e., if two intervals of equal lengths are equidistant from
the point $\frac{3}{4}$, then they have the same $P$-measure (see Figure 1).

######  Remark 3.1.

Since the probability distribution is symmetric about the point $\frac{3}{4}$,
without any loss of generality we can always assume that if $\alpha_{n}$ is an
optimal set of $n$-means, then for an odd positive integer $n$ the point
$\frac{3}{4}\in\alpha_{n}$ and all other elements in $\alpha_{n}$ are equally
distributed on both sides of $\frac{3}{4}$; on the other hand, if $n$ is an
even positive integer, than all the elements in the optimal set will be
equally distributed on both sides of $\frac{3}{4}$. Thus, we see that $n$ is
even or odd, in any case, an optimal set $\alpha_{n}$ of $n$-means contains
equal number of points from both sides of the point $\frac{3}{4}$ (see Figure
2).

Figure 1. Density function $f$ for the nonuniform probability distribution
$P$.

######  Proposition 3.2.

The optimal set of two-means is $\\{\frac{7}{16},\frac{17}{16}\\}$ with
quantization error $V_{2}=\frac{37}{768}$.

###### Proof.

Let $\alpha:=\\{a_{1},a_{2}\\}$ be an optimal set of two means. Since the
points in an optimal set are the conditional expectations in their own Voronoi
region, we can assume that $0<a_{1}<a_{2}<\frac{3}{2}$. Again, due to symmetry
of the probability distribution $P$ about the point $\frac{3}{4}$, we can
assume that the boundary $\frac{1}{2}(a_{1}+a_{2})$ of the Voronoi regions of
$a_{1}$ and $a_{2}$ passes through the midpoint $\frac{3}{4}$ of the support
of $P$. Thus, we have

$a_{1}=E(X:X\in[0,\frac{3}{4}])=\frac{\frac{1}{2}\int_{0}^{\frac{1}{2}}x\,dx+\int_{\frac{1}{2}}^{\frac{3}{4}}x\,dx}{\frac{1}{2}\int_{0}^{\frac{1}{2}}1\,dx+\int_{\frac{1}{2}}^{\frac{3}{4}}1\,dx}=\frac{7}{16},$

and since $\frac{1}{2}(a_{1}+a_{2})=\frac{3}{4}$, we have
$a_{2}=\frac{17}{16}$. Again, due to symmetry, the quantization error for two-
means is given by

$V_{2}=2\Big{(}\frac{1}{2}\int_{0}^{\frac{1}{2}}(x-\frac{7}{16})^{2}\,dx+\int_{\frac{1}{2}}^{\frac{3}{4}}(x-\frac{7}{16})^{2}\,dx\Big{)}=\frac{37}{768}.$

Thus the proof of the proposition is complete (also see Figure 2). ∎

######  Proposition 3.3.

The optimal set of three-means is $\\{\frac{1}{4},\frac{3}{4},\frac{5}{4}\\}$
with quantization error $V_{3}=\frac{1}{48}$.

###### Proof.

Let $\alpha$ be an optimal set of three-means. As mentioned in Remark 3.1, we
can assume that $\frac{3}{4}\in\alpha$. Let the other two elements in $\alpha$
be $a_{1}$ and $a_{2}$ such that $0<a_{1}<\frac{3}{4}<a_{2}<\frac{3}{2}$. Now,
the boundary of the Voronoi regions of $a_{1}$ and $\frac{3}{4}$ is
$\frac{1}{2}(a_{1}+\frac{3}{4})$. The following two cases can arise:

Case 1. $\frac{1}{2}(a_{1}+\frac{3}{4})\leq\frac{1}{2}.$

In this case, due to symmetry the distortion error is given by

$\displaystyle\int\min_{a\in\alpha}(x-a)^{2}dP=2\Big{(}\frac{1}{2}\int_{0}^{\frac{1}{2}(a_{1}+\frac{3}{4})}(x-a_{1}){}^{2}\,dx+\frac{1}{2}\int_{\frac{1}{2}(a_{1}+\frac{3}{4})}^{\frac{1}{2}}(x-\frac{3}{4})^{2}\,dx+\int_{\frac{1}{2}}^{\frac{3}{4}}(x-\frac{3}{4})^{2}\,dx\Big{)}$
$\displaystyle=\frac{1}{768}\Big{(}192a_{1}^{3}+144a_{1}^{2}-108a_{1}+31\Big{)},$

the minimum value of which is $0.0208333$ and it occurs when
$a_{1}=\frac{1}{4}$.

Case 2. $\frac{1}{2}\leq\frac{1}{2}(a_{1}+\frac{3}{4}).$

In this case, due to symmetry the distortion error is given by

$\displaystyle\int\min_{a\in\alpha}(x-a)^{2}dP=2\Big{(}\frac{1}{2}\int_{0}^{\frac{1}{2}}(x-a_{1}){}^{2}\,dx+\int_{\frac{1}{2}}^{\frac{1}{2}(a_{1}+\frac{3}{4})}(x-a_{1}){}^{2}\,dx+\int_{\frac{1}{2}(a_{1}+\frac{3}{4})}^{\frac{3}{4}}(x-\frac{3}{4})^{2}\,dx\Big{)}$
$\displaystyle=\frac{1}{384}\Big{(}192a_{1}^{3}-48a_{1}^{2}-12a_{1}+11\Big{)},$

the minimum value of which is $0.0208333$ and it occurs when
$a_{1}=\frac{1}{4}$.

Thus, considering all the possible cases we see that the distortion error is
smallest when $a_{1}=\frac{1}{4}$, and since
$\frac{1}{2}(a_{1}+a_{2})=\frac{3}{4}$, we have $a_{2}=\frac{5}{4}$. Thus, the
optimal set of three-means is $\\{\frac{1}{4},\frac{3}{4},\frac{5}{4}\\}$ with
quantization error $V_{3}=\frac{1}{48}$ (also see Figure 2). ∎

######  Proposition 3.4.

The optimal set of four-means is $\\{0.198223,0.59467,0.90533,1.30178\\}$ with
quantization error $V_{4}=0.01057$.

###### Proof.

Let $\alpha:=\\{a_{1},a_{2},a_{3},a_{4}\\}$ be an optimal set of four-means.
Due to symmetry of the probability measure we can say that the points in the
optimal set will be symmetrically located on the line with respect to the
point $\frac{3}{4}$, i.e.,
$0<a_{1}<a_{2}<\frac{3}{4}<a_{3}<a_{4}<\frac{3}{2}$, and $\frac{3}{4}$ is the
midpoint of $a_{2}$ and $a_{3}$. The following cases can arise:

Case 1. $0<a_{1}<a_{2}\leq\frac{1}{2}$.

In this case, due to symmetry the distortion error is given by

$\displaystyle\int\min_{a\in\alpha}(x-a)^{2}dP=2\Big{(}\frac{1}{2}\int_{0}^{\frac{1}{2}(a_{1}+a_{2})}(x-a_{1}){}^{2}\,dx+\int_{\frac{1}{2}}^{\frac{3}{4}}(x-a_{2}){}^{2}\,dx+\frac{1}{2}\int_{\frac{1}{2}(a_{1}+a_{2})}^{\frac{1}{2}}(x-a_{2}){}^{2}\,dx\Big{)}$
$\displaystyle=\frac{1}{96}\Big{(}24a_{1}^{3}+24a_{2}a_{1}^{2}-24a_{2}^{2}a_{1}-24a_{2}^{3}+96a_{2}^{2}-84a_{2}+23\Big{)},$

the minimum value of which is $0.0150463$ and it occurs when $a_{1}=0.166667$
and $a_{2}=0.5$.

Case 2. $0<a_{1}\leq\frac{1}{2}<a_{2}<\frac{3}{4}$.

In this case, the following two subcases can occur.

Subcase 1. $\frac{1}{2}(a_{1}+a_{2})\leq\frac{1}{2}.$

In this subcase, due to symmetry the distortion error is given by

$\displaystyle\int\min_{a\in\alpha}(x-a)^{2}dP=2\Big{(}\frac{1}{2}\int_{0}^{\frac{1}{2}(a_{1}+a_{2})}(x-a_{1}){}^{2}\,dx+\int_{\frac{1}{2}}^{\frac{3}{4}}(x-a_{2}){}^{2}\,dx+\frac{1}{2}\int_{\frac{1}{2}(a_{1}+a_{2})}^{\frac{1}{2}}(x-a_{2}){}^{2}\,dx\Big{)}$
$\displaystyle=\frac{1}{96}\Big{(}24a_{1}^{3}+24a_{2}a_{1}^{2}-24a_{2}^{2}a_{1}-24a_{2}^{3}+96a_{2}^{2}-84a_{2}+23\Big{)},$

the minimum value of which is $0.01057$ and it occurs when $a_{1}=0.198223$
and $a_{2}=0.59467$.

Subcase 2. $\frac{1}{2}\leq\frac{1}{2}(a_{1}+a_{2}).$

In this case, due to symmetry the distortion error is given by

$\displaystyle\int\min_{a\in\alpha}(x-a)^{2}dP=2\Big{(}\frac{1}{2}\int_{0}^{\frac{1}{2}}(x-a_{1}){}^{2}\,dx+\int_{\frac{1}{2}}^{\frac{1}{2}(a_{1}+a_{2})}(x-a_{1}){}^{2}\,dx+\int_{\frac{1}{2}(a_{1}+a_{2})}^{\frac{3}{4}}(x-a_{2}){}^{2}\,dx\Big{)}$
$\displaystyle=\frac{1}{96}(48a_{1}^{3}+48(a_{2}-1)a_{1}^{2}+(24-48a_{2}^{2})a_{1}-48a_{2}^{3}+144a_{2}^{2}-108a_{2}+23),$

the minimum value of which is $0.0169271$ and it occurs when $a_{1}=0.3125$
and $a_{2}=0.6875$.

Case 3. $\frac{1}{2}\leq a_{1}<a_{2}<\frac{3}{4}$.

Notice that in this case we obtain

$\displaystyle\int\min_{a\in\alpha}(x-a)^{2}dP\geq\frac{2}{2}\int_{0}^{\frac{1}{2}}(x-\frac{1}{2})^{2}\,dx=\frac{1}{24}=0.0416667,$

which is larger than the distortion errors obtained in at least one of the
previous cases. So, this case cannot happen.

Thus, considering all the possible cases, we can deduce that the smallest
distortion error is $V_{4}=0.01057$, and it occurs when $a_{1}=0.198223$ and
$a_{2}=0.59467$. Since $\frac{1}{2}(a_{2}+a_{3})=\frac{3}{4}$ and
$\frac{1}{2}(a_{1}+a_{4})=\frac{3}{4}$, we have $a_{3}=0.90533$ and
$a_{4}=1.30178$. Thus, the optimal set of four-means is
$\\{0.198223,0.59467,0.90533,1.30178\\}$ with quantization error
$V_{4}=0.01057$, which is the proposition (also see Figure 2). ∎

######  Proposition 3.5.

The optimal set of five-means is
$\\{0.169821,0.509464,\frac{3}{4},0.990536,1.33018\\}$ with quantization error
$V_{5}=0.00721728$.

###### Proof.

Let $\alpha:=\\{a_{1}<a_{2}<a_{3}<a_{4}<a_{5}\\}$ be an optimal set of five-
means. As mentioned in Remark 3.1, we can assume that $a_{3}=\frac{3}{4}$. The
following cases can happen.

Case 1. $a_{2}\leq\frac{1}{2}$

In this case the following subcases can happen.

Subcase 1. $\frac{1}{2}(a_{2}+\frac{3}{4})\leq\frac{1}{2}$.

Due to symmetry the distortion error is given by

$\displaystyle\int\min_{a\in\alpha}(x-a)^{2}dP$
$\displaystyle=2\Big{(}\frac{1}{2}\int_{0}^{\frac{1}{2}(a_{1}+a_{2})}(x-a_{1}){}^{2}\,dx+\frac{1}{2}\int_{\frac{1}{2}(a_{2}+\frac{3}{4})}^{\frac{1}{2}}(x-\frac{3}{4})^{2}\,dx+\frac{1}{2}\int_{\frac{1}{2}(a_{1}+a_{2})}^{\frac{1}{2}(a_{2}+\frac{3}{4})}(x-a_{2}){}^{2}\,dx$
$\displaystyle\qquad\qquad\qquad+\int_{\frac{1}{2}}^{\frac{3}{4}}(x-\frac{3}{4})^{2}\,dx\Big{)}$
$\displaystyle=\frac{1}{768}\Big{(}192a_{1}^{3}+192a_{2}a_{1}^{2}-192a_{2}^{2}a_{1}+144a_{2}^{2}-108a_{2}+31\Big{)},$

the minimum value of which is $0.0162037$, and it occurs when
$a_{1}=0.0833333$ and $a_{2}=0.25$.

Subcase 2. $\frac{1}{2}\leq\frac{1}{2}(a_{2}+\frac{3}{4})$.

Due to symmetry the distortion error is given by

$\displaystyle\int\min_{a\in\alpha}(x-a)^{2}dP$
$\displaystyle=2\Big{(}\frac{1}{2}\int_{0}^{\frac{1}{2}(a_{1}+a_{2})}(x-a_{1}){}^{2}\,dx+\int_{\frac{1}{2}(a_{2}+\frac{3}{4})}^{\frac{3}{4}}(x-\frac{3}{4})^{2}\,dx+\int_{\frac{1}{2}}^{\frac{1}{2}(a_{2}+\frac{3}{4})}(x-a_{2}){}^{2}\,dx$
$\displaystyle\qquad\qquad\qquad+\frac{1}{2}\int_{\frac{1}{2}(a_{1}+a_{2})}^{\frac{1}{2}}(x-a_{2}){}^{2}\,dx\Big{)}$
$\displaystyle=\frac{1}{384}\Big{(}96a_{1}^{3}+96a_{2}a_{1}^{2}-96a_{2}^{2}a_{1}+96a_{2}^{3}-48a_{2}^{2}-12a_{2}+11\Big{)},$

the minimum value of which is $0.0162037$, and it occurs when
$a_{1}=0.0833333$ and $a_{2}=0.25$.

Case 2. $0<a_{1}\leq\frac{1}{2}<a_{2}<\frac{3}{4}$.

In this case the following subcases can happen.

Subcase 1. $\frac{1}{2}(a_{1}+a_{2})\leq\frac{1}{2}$.

Due to symmetry the distortion error is given by

$\displaystyle\int\min_{a\in\alpha}(x-a)^{2}dP$
$\displaystyle=2\Big{(}\int_{\frac{1}{2}(a_{2}+\frac{3}{4})}^{\frac{3}{4}}(x-\frac{3}{4})^{2}\,dx+\frac{1}{2}\int_{0}^{\frac{1}{2}(a_{1}+a_{2})}(x-a_{1}){}^{2}\,dx+\int_{\frac{1}{2}}^{\frac{1}{2}(a_{2}+\frac{3}{4})}(x-a_{2}){}^{2}\,dx$
$\displaystyle\qquad\qquad\qquad+\frac{1}{2}\int_{\frac{1}{2}(a_{1}+a_{2})}^{\frac{1}{2}}(x-a_{2}){}^{2}\,dx\Big{)}$
$\displaystyle=\frac{1}{384}\Big{(}96a_{1}^{3}+96a_{2}a_{1}^{2}-96a_{2}^{2}a_{1}+96a_{2}^{3}-48a_{2}^{2}-12a_{2}+11\Big{)},$

the minimum value of which is $0.00721728$, and it occurs when
$a_{1}=0.169821$ and $a_{2}=0.509464$.

Subcase 2. $\frac{1}{2}\leq\frac{1}{2}(a_{1}+a_{2})$.

Due to symmetry the distortion error is given by

$\displaystyle\int\min_{a\in\alpha}(x-a)^{2}dP$
$\displaystyle=2\Big{(}\int_{\frac{1}{2}(a_{2}+\frac{3}{4})}^{\frac{3}{4}}(x-\frac{3}{4})^{2}\,dx+\frac{1}{2}\int_{0}^{\frac{1}{2}}(x-a_{1}){}^{2}\,dx+\int_{\frac{1}{2}}^{\frac{1}{2}(a_{1}+a_{2})}(x-a_{1}){}^{2}\,dx$
$\displaystyle\qquad\qquad\qquad+\int_{\frac{1}{2}(a_{1}+a_{2})}^{\frac{1}{2}(a_{2}+\frac{3}{4})}(x-a_{2}){}^{2}\,dx\Big{)}$
$\displaystyle=\frac{1}{384}\Big{(}192a_{1}^{3}+192(a_{2}-1)a_{1}^{2}-96(2a_{2}^{2}-1)a_{1}+144a_{2}^{2}-108a_{2}+11\Big{)},$

the minimum value of which is $0.0167955$, and it occurs when $a_{1}=0.315741$
and $a_{2}=0.684259$.

Case 3. $\frac{1}{2}<a_{1}<a_{2}<\frac{3}{4}$.

Due to symmetry the distortion error is given by

$\displaystyle\int\min_{a\in\alpha}(x-a)^{2}dP>2\times\frac{1}{2}\int_{0}^{\frac{1}{2}}(x-\frac{1}{2})^{2}\,dx=0.0416667$

which is larger than the distortion error that arises in at least one of the
previous cases.

Taking into consideration all the above possible cases, we see that the
quantization error for optimal set of five-means is $V_{5}=0.00721728$, and it
occurs when $a_{1}=0.169821$ and $a_{2}=0.509464$. Due to symmetry, we have
$a_{4}=0.990536$ and $a_{5}=1.33018$. Thus, the proof of the proposition is
complete (also see Figure 2). ∎

Let us now prove the following lemma.

######  Lemma 3.6.

Let $\alpha_{n}$ be an optimal set of $n$-means for $n\geq 4$. Then,
$\alpha_{n}$ contains points from both the open intervals $(0,\frac{1}{2})$
and $(\frac{1}{2},\frac{3}{4})$.

###### Proof.

By Propositions 3.4 and Proposition 3.5, the lemma is true for $n=4$ and
$n=5$. Let us now prove the lemma for $n\geq 6$. We prove it by contradiction.
Recall Remark 3.1, and also recall that for $n\geq 6$, we have $V_{n}\leq
V_{6}<V_{5}$. For $n\geq 6$, if $\alpha_{n}$ does not contain any point from
the open interval $(\frac{1}{2},\frac{3}{4})$, then due to symmetry we have

$V_{n}\geq
2\int_{\frac{1}{2}}^{\frac{3}{4}}(x-\frac{1}{2})^{2}\,dx=\frac{1}{96}=0.0104167>V_{5},$

which leads to a contradiction. For $n\geq 6$, if $\alpha_{n}$ does not
contain any point from the open interval $(0,\frac{1}{2})$, then due to
symmetry we have

$V_{n}\geq\frac{2}{2}\int_{0}^{\frac{1}{2}}(x-\frac{1}{2})^{2}\,dx=\frac{1}{24}=0.0416667>V_{5},$

which is a contradiction. Hence, we can conclude that the lemma is also true
for $n\geq 6$. Thus, the proof of the lemma is complete. ∎

######  Proposition 3.7.

The optimal set of six-means is $\\{0.125,0.375,0.625,0.875,1.125,1.375\\}$
with quantization error $V_{6}=0.00520833$.

###### Proof.

Let $\alpha:=\\{a_{1},a_{2},a_{3},a_{4},a_{5},a_{6}\\}$ be an optimal set of
six-means. Due to symmetry of the probability measure we can say that the
points in the optimal set will be symmetrically located on the line with
respect to the point $\frac{3}{4}$, i.e.,
$0<a_{1}<a_{2}<a_{3}<\frac{3}{4}<a_{4}<a_{5}<a_{6}<\frac{3}{2}$, and
$\frac{3}{4}$ is the midpoint of $a_{2}$ and $a_{3}$. By Lemma 3.6, we can say
that $a_{1}<\frac{1}{2}$, and $\frac{1}{2}<a_{3}$.

The following cases can arise:

Case 1. $0<a_{1}<a_{2}\leq\frac{1}{2}$.

The following two subcases can occur.

Subcase 1.
$0<a_{1}<a_{2}<\frac{1}{2}(a_{2}+a_{3})\leq\frac{1}{2}<a_{3}<\frac{3}{4}$.

In this subcase, due to symmetry the distortion error is given by

$\displaystyle\int\min_{a\in\alpha}(x-a)^{2}dP=2\Big{(}\frac{1}{2}\int_{0}^{\frac{1}{2}(a_{1}+a_{2})}(x-a_{1}){}^{2}\,dx+\frac{1}{2}\int_{\frac{1}{2}(a_{1}+a_{2})}^{\frac{1}{2}(a_{2}+a_{3})}(x-a_{2}){}^{2}\,dx+\int_{\frac{1}{2}}^{\frac{3}{4}}(x-a_{3}){}^{2}\,dx$
$\displaystyle\qquad\qquad\qquad+\frac{1}{2}\int_{\frac{1}{2}(a_{2}+a_{3})}^{\frac{1}{2}}(x-a_{3}){}^{2}\,dx\Big{)}$
$\displaystyle=\frac{1}{96}\Big{(}24a_{1}^{3}+24a_{2}a_{1}^{2}-24a_{2}^{2}a_{1}-24a_{3}^{3}-24(a_{2}-4)a_{3}^{2}+12(2a_{2}^{2}-7)a_{3}+23\Big{)},$

the minimum value of which is $0.00520833$ and it occurs when $a_{1}=0.125$,
$a_{2}=0.375$ and $a_{3}=0.625$.

Subcase 2.
$0<a_{1}<a_{2}<\frac{1}{2}\leq\frac{1}{2}(a_{2}+a_{3})<a_{3}<\frac{3}{4}$.

In this subcase, due to symmetry the distortion error is given by

$\displaystyle\int\min_{a\in\alpha}(x-a)^{2}dP=2\Big{(}\frac{1}{2}\int_{0}^{\frac{1}{2}(a_{1}+a_{2})}(x-a_{1}){}^{2}\,dx+\int_{\frac{1}{2}}^{\frac{1}{2}(a_{2}+a_{3})}(x-a_{2}){}^{2}\,dx+\frac{1}{2}\int_{\frac{1}{2}(a_{1}+a_{2})}^{\frac{1}{2}}(x-a_{2}){}^{2}\,dx$
$\displaystyle\qquad\qquad\qquad+\int_{\frac{1}{2}(a_{2}+a_{3})}^{\frac{3}{4}}(x-a_{3}){}^{2}\,dx\Big{)}$
$\displaystyle=\frac{1}{96}\Big{(}24a_{1}^{3}+24a_{2}a_{1}^{2}-24a_{2}^{2}a_{1}+24a_{2}^{3}-48a_{3}^{3}+144a_{3}^{2}+48a_{2}^{2}(a_{3}-1)-108a_{3}+a_{2}(24-48a_{3}^{2})+23\Big{)},$

the minimum value of which is $0.00520833$ and it occurs when $a_{1}=0.125$,
$a_{2}=0.375$ and $a_{3}=0.625$.

Case 2. $0<a_{1}<\frac{1}{2}<a_{2}<a_{3}<\frac{3}{4}$.

In this case, proceeding as before considering the two subcases:
$\frac{1}{2}\left(a_{1}+a_{2}\right)\leq\frac{1}{2}<a_{2}<a_{3}<\frac{3}{4}$,
and
$\frac{1}{2}\leq\frac{1}{2}\left(a_{1}+a_{2}\right)<a_{2}<a_{3}<\frac{3}{4}$,
it can be shown that the distortion error is larger than the distortion error
obtained in Case 1. Therefore, this case cannot happen.

Hence, the quantization error for six-means is $V_{6}=0.00520833$, and it
occurs when $a_{1}=0.125,\,a_{2}=0.375,\,a_{3}=0.625$. Due to symmetry, we
have $a_{4}=0.875,\,a_{5}=1.125$, and $a_{6}=1.375$. Thus, the proof of the
proposition is complete (also see Figure 2). ∎

Figure 2. Points in the optimal sets of $n$-means for $1\leq n\leq 9$.

In the following section we calculate the optimal sets of $n$-means and the
$n$th quantization errors for all $n\geq 5$.

##  4\. Optimal sets of $n$-means and the $n$th quantization errors for all
$n\geq 5$

Let $n\geq 5$ be a positive integer. By Remark 3.1, we know that if $n$ is
odd, an optimal set $\alpha_{n}$ of $n$-means always contains the point
$\frac{3}{4}$. Notice that whether $n$ is an even or an odd positive integer,
it is enough to find the elements in an optimal set which are to the left side
of $\frac{3}{4}$, i.e., which are belonged to the interval $(0,\frac{3}{4})$;
the remaining elements in $\alpha_{n}$ can be obtained by taking the
reflections with respect to the point $\frac{3}{4}$. By Lemma 3.6, an optimal
set $\alpha_{n}$ contains points from both the open intervals
$(0,\frac{1}{2})$ and $(\frac{1}{2},\frac{3}{4})$. Thus, there exist two
positive integers $k:=k(n)$ and $m:=m(n)$ such that

$\displaystyle\alpha_{n}\cap[0,\frac{1}{2}]$
$\displaystyle:=\\{a_{1}<a_{2}<\cdots<a_{k}\\},\text{ and }$
$\displaystyle\alpha_{n}\cap(\frac{1}{2},\frac{3}{4}]$
$\displaystyle:=\left\\{\begin{array}[]{cc}\\{b_{1}<b_{2}<\cdots<b_{m}\\}&\text{
if $n$ is even,}\\\ \\{b_{1}<b_{2}<\cdots<b_{m}<\frac{3}{4}\\}&\text{ if $n$
is odd.}\end{array}\right.$

Observe that in the above expression, if $n$ is even, then $2(k+m)=n$; and if
$n$ is odd, then $2(k+m)+1=n$. Notice that the following two cases can happen:
either $\frac{1}{2}(a_{k}+b_{1})\leq\frac{1}{2}$, or
$\frac{1}{2}\leq\frac{1}{2}(a_{k}+b_{1})$. Whether $n$ is even or odd, let
$V1(k,m)$ be the $n$th quantization error when
$\frac{a_{k}+b_{1}}{2}\leq\frac{1}{2}$, and $V2(k,m)$ be the $n$th
quantization error when $\frac{1}{2}\leq\frac{a_{k}+b_{1}}{2}$. The optimal
sets of $n$-means and the $n$th quantization errors for $n=1,2,3,4,5$ are
given in the previous sections. The following propositions will give the
optimal sets of $n$-means and the $n$th quantization errors for all $n\geq 6$.

######  Proposition 4.1.

Let $k\geq 2$ and $m=1$. Then, if $\frac{1}{2}(a_{k}+b_{1})\leq\frac{1}{2}$,
we have $a_{j}=\frac{(2j-1)(a_{k}+b_{1})}{4k}$ for $1\leq j\leq k$, and

$\displaystyle
b_{1}=\left\\{\begin{array}[]{cc}E(X:X\in[\frac{1}{2}(a_{k}+b_{1}),\frac{3}{4}])&\text{
if $n$ is even},\\\
E(X:X\in[\frac{1}{2}(a_{k}+b_{1}),\frac{1}{2}(b_{1}+\frac{3}{4})])&\text{ if
$n$ is odd};\end{array}\right.$

and if $\frac{1}{2}\leq\frac{1}{2}(a_{k}+b_{1})$, we have
$a_{j}=\frac{(2j-1)(a_{k-1}+a_{k})}{4(k-1)}$ for $1\leq j\leq(k-1)$,
$a_{k}=E(X:X\in[\frac{1}{2}(a_{k-1}+a_{k}),\frac{1}{2}(a_{k}+b_{1})])$, and

$\displaystyle
b_{1}=\left\\{\begin{array}[]{cc}E(X:X\in[\frac{1}{2}(a_{k}+b_{1}),\frac{3}{4}])&\text{
if $n$ is even},\\\
E(X:X\in[\frac{1}{2}(a_{k}+b_{1}),\frac{1}{2}(b_{1}+\frac{3}{4})])&\text{ if
$n$ is odd}.\end{array}\right.$

The quantization errors for $n$-means are given by

$\displaystyle
V1(k,1)=\left\\{\begin{array}[]{cc}2\Big{(}\frac{(a_{k}+b_{1}){}^{3}}{192k^{2}}+\frac{1}{2}\int_{\frac{1}{2}(a_{k}+b_{1})}^{\frac{1}{2}}(x-b_{1}){}^{2}\,dx+\int_{\frac{1}{2}}^{\frac{3}{4}}(x-b_{1}){}^{2}\,dx\Big{)}&\text{
if $n$ is even},\\\
2\Big{(}\frac{(a_{k}+b_{1}){}^{3}}{192k^{2}}+\frac{1}{2}\int_{\frac{1}{2}(a_{k}+b_{1})}^{\frac{1}{2}}(x-b_{1}){}^{2}\,dx+\int_{\frac{1}{2}}^{\frac{1}{2}(b_{1}+\frac{3}{4})}(x-b_{1}){}^{2}\,dx\\\
+\int_{\frac{1}{2}(b_{1}+\frac{3}{4})}^{\frac{3}{4}}(x-\frac{3}{4})^{2}\,dx\Big{)}&\text{
if $n$ is odd};\end{array}\right.$

and

$\displaystyle
V2(k,1)=\left\\{\begin{array}[]{cc}2\Big{(}\frac{(a_{k-1}+a_{k}){}^{3}}{192(k-1)^{2}}+\frac{1}{2}\int_{\frac{1}{2}(a_{k-1}+a_{k})}^{\frac{1}{2}}(x-a_{k}){}^{2}\,dx+\int_{\frac{1}{2}}^{\frac{1}{2}(a_{k}+b_{1})}(x-a_{k}){}^{2}\,dx\\\
+\int_{\frac{1}{2}(a_{k}+b_{1})}^{\frac{3}{4}}(x-b_{1}){}^{2}\,dx\Big{)}&\text{
if $n$ is even},\\\
2\Big{(}\frac{(a_{k-1}+a_{k}){}^{3}}{192(k-1)^{2}}+\frac{1}{2}\int_{\frac{1}{2}(a_{k-1}+a_{k})}^{\frac{1}{2}}(x-a_{k}){}^{2}\,dx+\int_{\frac{1}{2}}^{\frac{1}{2}(a_{k}+b_{1})}(x-a_{k}){}^{2}\,dx\\\
+\int_{\frac{1}{2}(a_{k}+b_{1})}^{\frac{1}{2}(b_{1}+\frac{3}{4})}(x-b_{1}){}^{2}\,dx+\int_{\frac{1}{2}(b_{1}+\frac{3}{4})}^{\frac{3}{4}}(x-\frac{3}{4})^{2}\,dx\Big{)}&\text{
if $n$ is odd}.\end{array}\right.$

###### Proof.

If $\frac{a_{k}+b_{1}}{2}\leq\frac{1}{2}$, then $a_{1},a_{2},\cdots,a_{k}$ are
uniformly distributed over the closed interval $[0,\frac{a_{k}+b_{1}}{2}]$; on
the other hand, if $\frac{1}{2}\leq\frac{a_{k}+b_{1}}{2}$, then
$a_{1},a_{2},\cdots,a_{k-1}$ are uniformly distributed over the closed
interval $[0,\frac{a_{k-1}+a_{k}}{2}]$. Thus, by Proposition 2.3 and
Proposition 1.1, the expressions for $a_{j}$ and $b_{j}$ can be obtained. With
the help of the formula given in Proposition 2.3, the quantization errors are
also obtained as routine. ∎

######  Proposition 4.2.

Let $k=1$ and $m\geq 2$. Then, if $\frac{1}{2}(a_{1}+b_{1})\leq\frac{1}{2}$,
we have $a_{1}=E(X:X\in[0,\frac{1}{2}(a_{1}+b_{1})]$,
$b_{1}=E(X:X\in[\frac{1}{2}(a_{1}+b_{1}),\frac{1}{2}(b_{1}+b_{2})])$, and

$\displaystyle
b_{1+j}=\left\\{\begin{array}[]{cc}\frac{1}{2}(b_{1}+b_{2})+\frac{(2j-1)}{2(m-1)}(\frac{3}{4}-\frac{1}{2}(b_{1}+b_{2}))\text{
for }1\leq j\leq m-1&\text{ if $n$ is even},\\\
\frac{1}{2}(b_{1}+b_{2})+\frac{(2j-1)}{2(m-1)}(\frac{3}{4}-b_{2})\text{ for
}1\leq j\leq m-1&\text{ if $n$ is odd};\end{array}\right.$

and if $\frac{1}{2}\leq\frac{1}{2}(a_{1}+b_{1})$, we have
$a_{1}=E(X:X\in[0,\frac{1}{2}(a_{1}+b_{1})]$, and

$\displaystyle
b_{j}=\left\\{\begin{array}[]{cc}\frac{1}{2}(a_{1}+b_{1})+\frac{(2j-1)}{2m}(\frac{3}{4}-\frac{1}{2}(a_{1}+b_{1}))$
for $1\leq j\leq m&\text{ if $n$ is even},\\\
\frac{1}{2}(a_{1}+b_{1})+\frac{(2j-1)}{2m}(\frac{3}{4}-b_{1})\text{ for }1\leq
j\leq m&\text{ if $n$ is odd};\end{array}\right.$

The quantization errors for $n$-means are given by

$\displaystyle
V1(1,m)=\left\\{\begin{array}[]{cc}2\Big{(}\frac{1}{2}\int_{0}^{\frac{1}{2}(a_{1}+b_{1})}(x-a_{1}){}^{2}\,dx+\frac{1}{2}\int_{\frac{1}{2}(a_{1}+b_{1})}^{\frac{1}{2}}(x-b_{1}){}^{2}\,dx\\\
+\int_{\frac{1}{2}}^{\frac{1}{2}(b_{1}+b_{2})}(x-b_{1}){}^{2}\,dx+\frac{(3-2(b_{1}+b_{2}))^{3}}{768(m-1)^{2}}\Big{)}&\text{
if $n$ is even},\\\
2\Big{(}\frac{1}{2}\int_{0}^{\frac{1}{2}(a_{1}+b_{1})}(x-a_{1}){}^{2}\,dx+\frac{1}{2}\int_{\frac{1}{2}(a_{1}+b_{1})}^{\frac{1}{2}}(x-b_{1}){}^{2}\,dx\\\
+\int_{\frac{1}{2}}^{\frac{1}{2}(b_{1}+b_{2})}(x-b_{1}){}^{2}\,dx+\frac{(3-4b_{2}){}^{3}}{768(m-1)^{2}}+\frac{1}{24}(b_{2}-b_{1}){}^{3}\Big{)}&\text{
if $n$ is odd};\end{array}\right.$

and

$\displaystyle
V2(1,m)=\left\\{\begin{array}[]{cc}2\Big{(}\frac{1}{2}\int_{0}^{\frac{1}{2}}\left(x-a_{1}\right){}^{2}\,dx+\int_{\frac{1}{2}}^{\frac{1}{2}\left(a_{1}+b_{1}\right)}\left(x-a_{1}\right){}^{2}\,dx+\frac{\left(3-2(a_{1}+b_{1})\right){}^{3}}{768m^{2}}\Big{)}&\text{
if $n$ is even},\\\
2\Big{(}\frac{1}{2}\int_{0}^{\frac{1}{2}}\left(x-a_{1}\right){}^{2}\,dx+\int_{\frac{1}{2}}^{\frac{1}{2}\left(a_{1}+b_{1}\right)}\left(x-a_{1}\right){}^{2}\,dx\\\
+\frac{(3-4b_{1}){}^{3}}{768m^{2}}+\frac{1}{24}(b_{1}-a_{1}){}^{3}\Big{)}&\text{
if $n$ is odd}.\end{array}\right.$

###### Proof.

If $\frac{1}{2}(a_{1}+b_{1})\leq\frac{1}{2}$ and $n$ is even, then
$b_{2},b_{3},\cdots,b_{m}$ are uniformly distributed over the closed interval
$[\frac{1}{2}(b_{1}+b_{2}),\frac{3}{4}]$, and so by Proposition 2.3, the
expressions for $b_{i+j}$, where $1\leq j\leq(m-1)$, and the corresponding
quantization error can be obtained. On the other hand, if
$\frac{1}{2}(a_{1}+b_{1})\leq\frac{1}{2}$ and $n$ is odd, then as
$\\{b_{2}<b_{3}<\cdots<b_{m}<\frac{3}{4}\\}$ is the set of optimal quantizers
with respect to the probability distribution
$P(\cdot|_{[\frac{1}{2}(b_{1}+b_{2}),\frac{3}{4}]})$ with constant density
$f(x)=1$ for all $x\in[\frac{1}{2}(b_{1}+b_{2}),\frac{3}{4}]$, the expressions
for $b_{j}$, and the corresponding quantization error can be obtained using
Proposition 2.4. Likewise, if $\frac{1}{2}\leq\frac{1}{2}(a_{1}+b_{1})$, using
Proposition 2.3 and Proposition 2.4, we get the expressions for the optimal
quantizers and the corresponding quantization error. ∎

######  Proposition 4.3.

Let $k\geq 2$ and $m\geq 2$. Then, if
$\frac{1}{2}(a_{k}+b_{1})\leq\frac{1}{2}$, we have
$a_{j}=\frac{(2j-1)(a_{k}+b_{1})}{4k}$ for $1\leq j\leq k$,
$b_{1}=E(X:X\in[\frac{1}{2}(a_{k}+b_{1}),\frac{1}{2}(b_{1}+b_{2})])$, and

$\displaystyle
b_{1+j}=\left\\{\begin{array}[]{cc}\frac{1}{2}(b_{1}+b_{2})+\frac{(2j-1)}{2(m-1)}(\frac{3}{4}-\frac{1}{2}(b_{1}+b_{2}))\text{
for }1\leq j\leq m-1&\text{ if $n$ is even},\\\
\frac{1}{2}(b_{1}+b_{2})+\frac{(2j-1)}{2(m-1)}(\frac{3}{4}-b_{2})\text{ for
}1\leq j\leq m-1&\text{ if $n$ is odd};\end{array}\right.$

and if $\frac{1}{2}\leq\frac{1}{2}(a_{k}+b_{1})$, we have
$a_{j}=\frac{(2j-1)(a_{k-1}+a_{k})}{4(k-1)}$ for $1\leq j\leq(k-1)$,
$a_{k}=E(X:X\in[\frac{1}{2}(a_{k-1}+a_{k}),\frac{1}{2}(a_{k}+b_{1})])$, and

$\displaystyle
b_{j}=\left\\{\begin{array}[]{cc}\frac{1}{2}(a_{k}+b_{1})+\frac{(2j-1)}{2m}(\frac{3}{4}-\frac{1}{2}(a_{k}+b_{1}))$
for $1\leq j\leq m&\text{ if $n$ is even},\\\
\frac{1}{2}(a_{k}+b_{1})+\frac{(2j-1)}{2m}(\frac{3}{4}-b_{1})\text{ for }1\leq
j\leq m&\text{ if $n$ is odd};\end{array}\right.$

The quantization errors for $n$-means are given by

$\displaystyle
V1(k,m)=\left\\{\begin{array}[]{cc}2\Big{(}\frac{(a_{k}+b_{1}){}^{3}}{192k^{2}}+\frac{1}{2}\int_{\frac{1}{2}(a_{k}+b_{1})}^{\frac{1}{2}}(x-b_{1}){}^{2}\,dx+\int_{\frac{1}{2}}^{\frac{1}{2}(b_{1}+b_{2})}(x-b_{1}){}^{2}\,dx\\\
+\frac{(3-2(b_{1}+b_{2}))^{3}}{768(m-1)^{2}}\Big{)}&\text{ if $n$ is even},\\\
2\Big{(}\frac{(a_{k}+b_{1}){}^{3}}{192k^{2}}+\frac{1}{2}\int_{\frac{1}{2}(a_{k}+b_{1})}^{\frac{1}{2}}(x-b_{1}){}^{2}\,dx\\\
+\int_{\frac{1}{2}}^{\frac{1}{2}(b_{1}+b_{2})}(x-b_{1}){}^{2}\,dx+\frac{(3-4b_{2}){}^{3}}{768(m-1)^{2}}+\frac{1}{24}(b_{2}-b_{1}){}^{3}\Big{)}&\text{
if $n$ is odd};\end{array}\right.$

and

$\displaystyle
V2(k,m)=\left\\{\begin{array}[]{cc}2\Big{(}\frac{(a_{k-1}+a_{k}){}^{3}}{192(k-1)^{2}}+\frac{1}{2}\int_{\frac{1}{2}(a_{k-1}+a_{k})}^{\frac{1}{2}}(x-a_{k}){}^{2}\,dx+\int_{\frac{1}{2}}^{\frac{1}{2}(a_{k}+b_{1})}(x-a_{k}){}^{2}\,dx\\\
+\frac{\left(3-2(a_{k}+b_{1})\right){}^{3}}{768m^{2}}\Big{)}&\text{ if $n$ is
even},\\\
2\Big{(}\frac{(a_{k-1}+a_{k}){}^{3}}{192(k-1)^{2}}+\frac{1}{2}\int_{\frac{1}{2}(a_{k-1}+a_{k})}^{\frac{1}{2}}(x-a_{k}){}^{2}\,dx+\int_{\frac{1}{2}}^{\frac{1}{2}(a_{k}+b_{1})}(x-a_{k}){}^{2}\,dx\\\
+\frac{(3-4b_{1}){}^{3}}{768m^{2}}+\frac{1}{24}(b_{1}-a_{k}){}^{3}\Big{)}&\text{
if $n$ is odd}.\end{array}\right.$

###### Proof.

Notice that Proposition 4.3 is a mixture of Proposition 4.1 and Proposition
4.2, and thus the proof follows in the similar lines. ∎

######  Lemma 4.4.

For any even positive integer $n\geq 4$, let $\alpha_{n}$ be an optimal set of
$n$-means for $P$. Assume that
$\text{card}(\alpha_{n}\cap[0,\frac{1}{2}])=k:=k(n)$ and
$\text{card}(\alpha_{n}\cap(\frac{1}{2},\frac{3}{4}))=m:=m(n)$ for some
positive integers $k$ and $m$. Then, either
$\text{card}(\alpha_{n+2}\cap[0,\frac{1}{2}])=k+1$ and
$\text{card}(\alpha_{n+2}\cap(\frac{1}{2},\frac{3}{4}))=m$, or
$\text{card}(\alpha_{n+2}\cap[0,\frac{1}{2}])=k$ and
$\text{card}(\alpha_{n+2}\cap(\frac{1}{2},\frac{3}{4}))=m+1$.

###### Proof.

For any even positive integer $n\geq 4$, let
$\text{card}(\alpha_{n}\cap[0,\frac{1}{2}])=k:=k(n)$ and
$\text{card}(\alpha_{n}\cap(\frac{1}{2},\frac{3}{4}))=m:=m(n)$ for some
positive integers $k$ and $m$. Let $V(k(n),m(n))$ be the corresponding
distortion error. By Proposition 3.4 and Proposition 3.7, we know that
$\text{card}(\alpha_{4}\cap[0,\frac{1}{2}])=1$,
$\text{card}(\alpha_{4}\cap(\frac{1}{2},\frac{3}{4}))=1$,
$\text{card}(\alpha_{6}\cap[0,\frac{1}{2}])=2$, and
$\text{card}(\alpha_{6}\cap(\frac{1}{2},\frac{3}{4}))=1$ Thus, the lemma is
true for $n=4$. Let the lemma be true for $n=N$ for some even positive integer
$N\geq 4$. Then, $\text{card}(\alpha_{N}\cap[0,\frac{1}{2}])=k(N)\text{ and
}\text{card}(\alpha_{N}\cap(\frac{1}{2},\frac{3}{4}))=m(N)$ imply that either
$\text{card}(\alpha_{N+2}\cap[0,\frac{1}{2}])=k(N)+1$ and
$\text{card}(\alpha_{N+2}\cap(\frac{1}{2},\frac{3}{4}))=m(N)$, or
$\text{card}(\alpha_{N+2}\cap[0,\frac{1}{2}])=k(N)$ and
$\text{card}(\alpha_{N+2}\cap(\frac{1}{2},\frac{3}{4}))=m(N)+1$. Suppose that
$\text{card}(\alpha_{N+2}\cap[0,\frac{1}{2}])=k(N)+1$ and
$\text{card}(\alpha_{N+2}\cap(\frac{1}{2},\frac{3}{4}))=m(N)$ hold. Now, for
the given $N$, by calculating the distortion errors $V(\ell,N+4-\ell)$ for all
$1\leq\ell\leq N+3$, we see that the distortion error is smallest if
$\text{card}(\alpha_{N+4}\cap[0,\frac{1}{2}])=k(N)+2\text{ and
}\text{card}(\alpha_{N+2}\cap(\frac{1}{2},\frac{3}{4}))=m(N),\text{ or if
}\text{card}(\alpha_{N+4}\cap[0,\frac{1}{2}])=k(N)+1\text{ and
}\text{card}(\alpha_{N+4}\cap(\frac{1}{2},\frac{3}{4}))=m(N)+1$, i.e., the
lemma is true for $n=N+2$ whenever it is true for $n=N$. Similarly, we can
show that the lemma is true for $n=N+2$ if
$\text{card}(\alpha_{N+2}\cap[0,\frac{1}{2}])=k(N)\text{ and
}\text{card}(\alpha_{N+2}\cap(\frac{1}{2},\frac{3}{4}))=m(N)+1$ hold. Thus, by
the induction principle, the proof of the lemma is complete. ∎

Proceeding in the similar lines as Lemma 4.4, the following lemma can be
proved.

######  Lemma 4.5.

For any odd positive integer $n\geq 5$, let $\alpha_{n}$ be an optimal set of
$n$-means for $P$. Assume that
$\text{card}(\alpha_{n}\cap[0,\frac{1}{2}])=k:=k(n)$ and
$\text{card}(\alpha_{n}\cap(\frac{1}{2},\frac{3}{4}))=m:=m(n)$ for some
positive integers $k$ and $m$. Then, either
$\text{card}(\alpha_{n+2}\cap[0,\frac{1}{2}])=k+1$ and
$\text{card}(\alpha_{n+2}\cap(\frac{1}{2},\frac{3}{4}))=m$, or
$\text{card}(\alpha_{n+2}\cap[0,\frac{1}{2}])=k$ and
$\text{card}(\alpha_{n+2}\cap(\frac{1}{2},\frac{3}{4}))=m+1$.

######  Definition 4.6.

Define a real valued function $F(k,m)$ on the domain
$\mathbb{N}\times\mathbb{N}\setminus\\{(1,1)\\}$ such that

$F(k,m)=\min\\{V1(k,m),V2(k,m)\\},$

where $V1(k,m)$ and $V2(k,m)$ are the distortion errors as defined before.

######  Definition 4.7.

Define the sequence $\\{a(n)\\}$ such that

$a(n):=\left\\{\begin{array}[]{cc}0&\text{ if }n=1,\\\
\Big{\lfloor}\frac{\lfloor\frac{n}{2}\rfloor}{\sum_{k=1}^{\lfloor\frac{n}{2}\rfloor}\frac{1}{k^{2}}}\Big{\rfloor}&\text{
if }n\geq 2.\end{array}\right.$

i.e.,

$\displaystyle\\{a(n)\\}_{n=1}^{\infty}=$
$\displaystyle\\{0,1,1,1,1,2,2,2,2,3,3,4,4,4,4,5,5,5,5,6,6,7,7,7,7,8,8,8,8,9,9,$
$\displaystyle
10,10,10,10,11,11,11,11,12,12,13,13,13,13,14,14,14,14,15\cdots\\},$

where $\lfloor x\rfloor$ represents the greatest integer not exceeding $x$.

######  Remark 4.8.

Let $n\geq 5$. Let $k$ and $m:=\lfloor\frac{n}{2}\rfloor-k$ be the positive
integers such that $\text{card}(\alpha_{n}\cap[0,\frac{1}{2}])=k$ and
$\text{card}(\alpha_{n}\cap(\frac{1}{2},\frac{3}{4}))=m$. Thus, for a given
$n\geq 5$, if we know the exact values of $k$ and $m$, then by Proposition 4.1
through Proposition 4.3, we can easily determine the optimal sets $\alpha_{n}$
and the corresponding quantization error.

The following algorithm helps us to calculate the exact value of $k$ and so,
$m=\lfloor\frac{n}{2}\rfloor-k$.

###  4.9. Algorithm.

Let $n\geq 5$ and $F(k,m)$ be the function defined by Definition 4.6, and let
$\\{a(n)\\}$ be the sequence defined by Definition 4.7. Then, the algorithm
runs as follows:

$(i)$ Write $k:=a(n)$ and calculate $F(k,\lfloor\frac{n}{2}\rfloor-k)$.

$(ii)$ If
$F(k-1,\lfloor\frac{n}{2}\rfloor-k+1)<F(k,\lfloor\frac{n}{2}\rfloor-k)$
replace $k$ by $k-1$ and return, else step $(iii)$.

$(iii)$ If
$F(k+1,\lfloor\frac{n}{2}\rfloor-k-1)<F(k,\lfloor\frac{n}{2}\rfloor-k)$
replace $k$ by $k+1$ and return, else step $(iv)$.

$(iv)$ End.

When the algorithm ends, then the value of $k$, obtained, is the exact value
of $k$ that $\alpha_{n}$ contains from the closed interval $[0,\frac{1}{2}]$.

#### Optimal sets of $n$-means and the $n$th quantization errors for all
positive integers $n\geq 5$.

If $n=5$, then $a(n)=1$, and by the algorithm, we also obtain $k=1$, i.e., an
optimal set $\alpha_{5}$ of five-means contains one element from the closed
interval $[0,\frac{1}{2}]$, which is clearly true by Proposition 3.5. If
$n=6$, then $a(n)=2$, and by the algorithm, we also obtain $k=2$, which is
also true by Proposition 3.7; if $n=9$, then $a(n)=2$, and by the algorithm,
we obtain $k=3$; if $n=50$, then $a(n)=15$, and by the algorithm, we also
obtain $k=15$. If $n=1001$, then $a(n)=304$, and by the algorithm, we obtain
$k=307$. Thus, we see that with the help of the sequence and the algorithm, we
can easily determine the exact values of $k$ and
$m:=\lfloor\frac{n}{2}\rfloor-k$ for any given positive integer $n\geq 5$.
Hence, as mentioned in Remark 4.8, we can obtain the optimal sets of $n$-means
and the corresponding quantization errors for all positive integers $n\geq 5$
(also see Figure 2).

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#  Existence of group nonexpansive retractions and ergodic theorems in
topological groups

Ebrahim Soori

###### Abstract.

Suppose that $G$ is a topological group and $C$ a compact subset of $G$. In
this paper we define group nonexpansive mappings and then we consider
$\mathcal{S}=\\{T_{i}:i\in I\\}$ as a family of the group nonexpansive
mappings on $C$. Also we study the existence of group nonexpansive retractions
$P_{i}$ from $C$ onto $\text{Fix}(\mathcal{S})$ such that
$P_{i}T_{i}=T_{i}P_{i}=P_{i}$.

   * Corresponding author
2010 Mathematics Subject Classification: 47H09,47H10
E-mail addresses<EMAIL_ADDRESS><EMAIL_ADDRESS>(E. Soori)

Department of Mathematics, Lorestan University, P. O. Box 465, Khoramabad,
Lorestan, Iran.

Keywords: Fixed point; Group nonexpansive mapping; Topologial group;
Retraction.

##  1\.  Introduction

A topological group $G$ is a set endowed with two structures, a group
structure and a topological structure. Specifically, $G$ is both an abstract
group and a topological space such that the two maps

$\displaystyle G\times G\rightarrow G:(x,y)\mapsto xy$ $\displaystyle
G\rightarrow G:x\mapsto x^{-1}$

are assumed to be continuous. Also, the Hausdorff condition will be imposed.

Let $D$ be a subset of $B$ where $B$ is a subset of a topological group $G$. A
mapping $P$ is called a retraction of $B$ onto $D$, if $Px=x$ for each $x\in
D$.

The first nonlinear ergodic theorem for nonexpansive mappings in a Hilbert
space was established by Baillon [1]: Let $C$ be a nonempty closed convex
subset of a Hilbert space $H$ and let $T$ be a nonexpansive mapping of $C$
into itself. If the set $\text{Fix}(T)$ of fixed points of $T$ is nonempty,
then for each $x\in C$, the Cesaro means
$S_{n}x=\frac{1}{n}\sum_{k=1}^{n}T^{k}x$ converge weakly to some
$y\in\text{Fix}(T)$. In Baillon’s theorem, putting $y=Px$ for each $x\in C$,
$P$ is a nonexpansive retraction of $C$ onto $\text{Fix}(T)$ such that
$PT^{n}=T^{n}P=P$ for all positive integers $n$ and
$Px\in\overline{co}\\{T^{n}x:n=1,2,...\\}$ for each $x\in C$. Takahashi [13]
proved the existence of such retractions, “ergodic retractions”, for non-
commutative semigroups of nonexpansive mappings in a Hilbert space: If $S$ is
an amenable semigroup, $C$ is a closed, convex subset of a Hilbert space $H$
and $\mathcal{S}=\\{T_{s}:s\in S\\}$ is a nonexpansive semigroup on $C$ such
that $\text{Fix}(\mathcal{S})\neq\emptyset$, then there exists a nonexpansive
retraction $P$ from $C$ onto $\text{Fix}(\mathcal{S})$ such that
$PT_{t}=T_{t}P=P$ for each $t\in S$ and $Px\in\overline{co}\,\\{T_{t}x:t\in
S\\}$ for each $x\in C$. These results were extended to uniformly convex
Banach spaces for commutative semigroups in [4] and for non-commutative
amenable semigroups in [5, 6] and recently for a family of $Q$-nonexpansive
mappings in locally convex spaces in [3]. For other results we refer the
reader to [11, 9, 10, 12].

In this paper, first we define group nonexpansive mappings. Then we establish
some ergodic retractions for topological groups, based on the definition.
Indeed we find a family of desired retractions by removing the important
condition ”convexity” in the above mentioned theorems, in the topological
groups.

##  2\. preliminaries

In this section, we introduce our definition and give some examples:

######  Definition 2.1.

Suppose that $G$ is a topological group and $C\subset G$. A mapping
$T:C\rightarrow C$ is said to be group nonexpansive if for each $x,y\in C$ and
each closed neighborhood $U\in\mathfrak{B}_{e}$ (where $\mathfrak{B}_{e}$ is a
local base in $e$ (identity element)) that $xy^{-1}\in U$ then we have
$Tx(Ty)^{-1}\in U$.

######  Example 2.2.

Our group nonexpansiveness is more general than nonexpansiveness i.e. every
nonexpansive mapping is a group nonexpansive mapping if the topological group
is a normed vector space. Indeed, let $E$ be a normed vector space and
$C\subset E$. Consider the closed neighborhood $U=\\{z\in E:\|z\|\leq
r\\}\in\mathfrak{B}_{0}$ (where $\mathfrak{B}_{0}$ is a local base in $0$)
that $x-y\in U$ for a positive number $r$. If $T$ is a nonexpansive mapping on
$C$ then we have

$\|Tx-Ty\|\leq\|x-y\|,$

and hence $Tx-Ty\in U$. Thus $T$ is a group nonexpansive.

In the following example we consider a case that some group nonexpansive
mappings are nonexpansive mapping.

######  Example 2.3.

Let $G$ be metric topological group with a right invariant metric (that is,
$d(yx,zx)=d(y,z)$ for all $x,y,z\in G$). Let $T$ be a group nonexpansive
mapping from $G$ into $G$. Consider the neighbourhoods
$N_{d(xy^{-1},e)+\frac{1}{n}}(e)$ of $e$ with $xy^{-1}\in
N_{d(xy^{-1},e)+\frac{1}{n}}(e)$ for each $n\in\mathbb{N}$. Then we have
$Tx(Ty)^{-1}\in N_{d(xy^{-1},e)+\frac{1}{n}}(e)$ for each $n\in\mathbb{N}$,
and hence

$d(Tx(Ty)^{-1},e)\leq d(xy^{-1},e)+\frac{1}{n},$

for each $n\in\mathbb{N}$. Therefore

$d(Tx(Ty)^{-1},e)\leq d(xy^{-1},e),$

and since $d$ is right invariant, we have

$d(Tx,Ty)\leq d(x,y).$

Then we conclude that $T$ is a nonexpansive mappings in the sense of
nonexpansive mappings in metric spaces.

##  3\. Ergodic retractions for families of group nonexpansive mappings on
topological groups

Let $G$ be a topological group. In this section, we study the existence of
group nonexpansive retractions onto the set of common fixed points of a family
of group nonexpansive mappings that commute with the mappings. A group
nonexpansive retraction that commutes with the mappings is usually called an
ergodic retraction.

First, we prove the following theorem (motivated in part by ideas in [11])
which is the main result of this section and will be essential in the sequel.

######  Theorem 3.1.

Let $G$ be a topological group and let $C$ be a compact subset of $G$. Suppose
that $\mathcal{S}=\\{T_{i}:i\in I\\}$ is a family of the group nonexpansive
mappings on $C$ such that $\text{Fix}(\mathcal{S})\neq\emptyset$. Also suppose
for every nonempty compact $\mathcal{S}$-invariant subset $K$ of $C$,
$K\cap\text{Fix}(\mathcal{S})\neq\emptyset$. Then, for each $i\in I$, there
exists a group nonexpansive retraction $P_{i}$ from $C$ onto
$\text{Fix}(\mathcal{S})$, such that $P_{i}T_{i}=T_{i}P_{i}=P_{i}$ and every
closed $\mathcal{S}$-invariant subset of $C$ is also $P_{i}$-invariant.

###### Proof.

Let $C^{C}$ be the product space with the product topology induced by the
relative topology on $C$. Now for a fixed $\alpha\in I$, consider the
following set

$\mathfrak{R}=\\{T\in C^{C}:T\;\text{is group nonexpansive},T\circ
T_{\alpha}=T\\\ \text{ }\text{ and every closed $\mathcal{S}$-invariant subset
of $C$ is also $T$ -invariant}\\}$.
From the fact that $G$ is Hausdorff, for each $z\in\text{Fix}(\mathcal{S})$,
the singleton set $\\{z\\}$ is a closed $\mathcal{S}$-invariant subset of $C$,
and then for each $T\in\mathfrak{R}$, $Tz=z$. Fix
$z_{0}\in\text{Fix}(\mathcal{S})$ and let for each $x\in C$,

$\displaystyle C_{x}:=$ $\displaystyle\\{y\in C:\text{for each closed
neighborhood }U\;\text{of}\;e\;\text{that}\;xz_{0}^{-1}\in U$
$\displaystyle\text{then}\;yz_{0}^{-1}\in U\\}.$

For all $x\in C$ and $T\in\mathfrak{R}$, we have that $T(x)\in C_{x}$. Since
$T$ is group nonexpansive for a closed neighborhood $U$ of $e$, if
$xz_{0}^{-1}\in U$ then $T(x)z_{0}^{-1}=T(x)(T(z_{0}))^{-1}\in U$. Hence
$\mathfrak{R}\subseteq\prod_{x\in C}C_{x}$, where $\prod_{x\in C}C_{x}$ is the
Cartesian product of sets $C_{x}$ for all $x\in C$. Let $\\{y_{\alpha}\\}$ be
a net in $C_{x}$ such that $y_{\alpha}\rightarrow y$. Consider a closed
neighborhood $U$ of $e$ such that $xz_{0}^{-1}\in U$. Then we have
$y_{\alpha}z_{0}^{-1}\in U$. From the fact that the mapping $(x,y)\mapsto xy$
is continuous we have $y_{\alpha}z_{0}^{-1}\rightarrow yz_{0}^{-1}$ and since
$U$ is closed we conclude that $yz_{0}^{-1}\in U$, and therefore $C_{x}$ is
closed and since $C$ is compact we conclude that $C_{x}$ is compact. By
Tychonoff’s theorem, we know that when $C_{x}$ is given the relative topology
and $\prod_{x\in C}C_{x}$ is given the corresponding product topology,
$\prod_{x\in C}C_{x}$ is compact. Next we prove that $\mathfrak{R}$ is closed
in $\prod_{x\in C}C_{x}$. Let $\\{T_{\lambda}:\lambda\in\Lambda\\}$ be a net
in $\mathfrak{R}$ which converges to $T_{0}$ in $\prod_{x\in C}C_{x}$. Hence
if $z\in\text{Fix}(\mathcal{S})$, then we have $T_{\lambda}z=z$ for each
$\lambda\in\Lambda$ (because $T_{\lambda}\in\mathfrak{R})$ and
$T_{0}z=\displaystyle\lim_{\lambda}T_{\lambda}(z)=z$. From the fact that the
mapping $(x,y)\mapsto xy$ and $x\mapsto x^{-1}$ are continuous, if we consider
a closed neighborhood $U$ of $e$ that $xy^{-1}\in U$ then we have
$T_{0}x(T_{0}y)^{-1}=\displaystyle\lim_{\lambda}T_{\lambda}x(T_{\lambda}y)^{-1}\in
U$. Hence, $T$ is group nonexpansive. Obviously, we have $T_{0}\circ
T_{\alpha}=T_{0}$ and every closed $\mathcal{S}$-invariant subset of $C$ is
also $T_{0}$-invariant. Therefore, $T_{0}\in\mathfrak{R}$. Then $\mathfrak{R}$
is closed in $\prod_{x\in C}C_{x}$. Since $\prod_{x\in C}C_{x}$ is compact,
hence $\mathfrak{R}$ is compact. Next, we show that
$\mathfrak{R}\neq\emptyset$. Consider the mappings
$S_{n}=T^{n-1}_{\alpha}\in\prod_{x\in C}C_{x}$ for each $n\in\mathbb{N}$. Then
from the fact that $\prod_{x\in C}C_{x}$ is compact, it has a convergent
subnet $\\{S_{n_{\eta}}\\}$. Define for each $x\in C$,
$T(x)=\lim_{\eta}S_{n_{\eta}}x$. We now check that $T\in\mathfrak{R}$. Note
that, from the continuity of the mapping $(x,y)\mapsto xy$ and $x\mapsto
x^{-1}$ and the group nonexpansiveness of $S_{n_{\eta}}$ and closedness of
$U$, $T$ is group nonexpansive. Indeed, if $xy^{-1}\in U$ for any closed
neighbourhood $U$ of $e$ and $x,y\in C$, then we have $\displaystyle
Tx(Ty)^{-1}=\lim_{\eta}S_{n_{\eta}}x(\lim_{\eta}S_{n_{\eta}}y)^{-1}=\lim_{\eta}S_{n_{\eta}}x(S_{n_{\eta}}y)^{-1}\in
U$. Moreover,
$T(T_{\alpha}x)=\lim_{\eta}S_{n_{\eta}}(T_{\alpha}x)=\lim_{\eta}S_{n_{\eta}}(x)=T(x)$.
Finally, if $D$ is a closed $\mathcal{S}$-invariant subset of $C$, it is clear
that $D$ is $S_{n_{\eta}}$-invariant and thus from the closedness of $D$, is
$T$-invariant. Therefore, we have shown that $T\in\mathfrak{R}\neq\emptyset$.

Now define a preorder $\preceq$ in $\mathfrak{R}$ by $T_{1}\preceq T_{2}$ if
for each $U\in\mathfrak{B}_{e}$ that $T_{2}x(T_{2}y)^{-1}\in U$ we have
$T_{1}x(T_{1}y)^{-1}\in U$, and by using a method similar to Bruck’s method
[2], we find a minimal element $T_{min}$ in $\mathfrak{R}$. Indeed, using
Zorn’s Lemma, it is enough that we show that each linearly ordered subset of
$\mathfrak{R}$ has a lower bound in $\mathfrak{R}$. Let $\\{A_{\lambda}\\}$ be
a linearly ordered subset of $\mathfrak{R}$. Then the family of sets
$\\{T\in\mathfrak{R}:T\preceq A_{\lambda}\\}$ is a linearly ordered subset of
$\mathfrak{R}$ by inclusion. Taking into account the closeness proof of
$\mathfrak{R}$ in $\prod_{x\in C}C_{x}$, these sets are closed in
$\mathfrak{R}$, and hence compact. Then from the finite intersection property,
there exists $R\in\bigcap_{\lambda}\\{T\in\mathfrak{R}:T\preceq
A_{\lambda}\\}$ with $R\preceq A_{\lambda}$ for all $\lambda$. Then each
linearly ordered subset of $\mathfrak{R}$ has a lower bound in $\mathfrak{R}$.
We have shown that there exist a minimal element $P_{\alpha}$ in the following
sense:

if $T\in\mathfrak{R}\;\text{and }\text{for each }U\in\mathfrak{B}_{e}\text{
that }P_{\alpha}x(P_{\alpha}y)^{-1}\in U\;\text{then }Tx(Ty)^{-1}\in U,\\\
\text{ }\text{ then }\text{for each }U^{\prime}\in\mathfrak{B}_{e}\text{ that
}Tx(Ty)^{-1}\in U^{\prime}\text{ we have }P_{\alpha}x(P_{\alpha}y)^{-1}\in
U^{\prime}$. ($*$)

Next we prove that $P_{\alpha}x\in\text{Fix}(\mathcal{S})$ for every $x\in C$.
For a given $x\in C$, consider $K:=\\{T(P_{\alpha}x):T\in\mathfrak{R}\\}$.
From the fact that $\mathfrak{R}$ is compact, from Proposition 3.3.18 and
Definition 3.3.19 in [8], we conclude that $K$ is a nonempty compact subset of
$C$. Now we have $S(K)\subset K$ for each $S\in\mathcal{S}$, because
$STT_{\alpha}=ST$ for each $T\in\mathfrak{R}$ hence $ST\in\mathfrak{R}$ i.e,
$K$ is $\mathcal{S}$-invariant.

From our assumption $K\cap\text{Fix}(\mathcal{S})\neq\emptyset$. Then there
exists $L\in\mathfrak{R}$ such that
$L(P_{\alpha}x)\in\text{Fix}(\mathcal{S})$. Suppose that $y=L(P_{\alpha}x)$.
Since $P_{\alpha},L\in\mathfrak{R}$ and the set $\\{y\\}$ is
$\mathcal{S}$-invariant, we have $P_{\alpha}(y)=L(y)=y$, and since $L$ is
group nonexpansive, $P_{\alpha}$ is minimal and
$L(P_{\alpha}x)(L(P_{\alpha}y))^{-1}=L(P_{\alpha}x)y^{-1}=yy^{-1}=e\in U$, for
each $U\in\mathfrak{B}_{e}$ and then we have
$(P_{\alpha}x)y^{-1}=P_{\alpha}x(P_{\alpha}y)^{-1}\in U$ for each
$U\in\mathfrak{B}_{e}$ (to see this consider $LP_{\alpha}$ instead of $T$ in
($*$)) and by (vi) of Corollary 1.11 in [7], $(P_{\alpha}x)y^{-1}=e$, hence
$P_{\alpha}x=y\in\text{Fix}(\mathcal{S})$ and this holds for each $x\in C$.

Since $P_{\alpha}\in\mathfrak{R}$, $T_{\alpha}\in\mathcal{S}$ and
$\\{P_{\alpha}x\\}$ is $\mathcal{S}$-invariant for each $x\in C$, hence, it
must be $P_{\alpha}$-invariant for each $x\in C$. Then we conclude that
$P_{\alpha}^{2}=P_{\alpha}$ and
$P_{\alpha}T_{\alpha}=T_{\alpha}P_{\alpha}=P_{\alpha}$.

∎

As a consequence of Theorem 3.1, we establish an ergodic retraction by a group
nonexpansive retraction.

######  Theorem 3.2.

Let $G$ be a topological group and let $C$ be a compact subset of $G$. Suppose
that $\mathcal{S}=\\{T_{i}:i\in I\\}$ is a family of group nonexpansive
mappings on $C$ such that $\text{Fix}(\mathcal{S})\neq\emptyset$. Also suppose
for every nonempty compact $\mathcal{S}$-invariant subset $K$ of $C$,
$K\cap\text{Fix}(\mathcal{S})\neq\emptyset$. If there is a group nonexpansive
retraction $R$ from $C$ onto $\text{Fix}(\mathcal{S})$, then for each $i\in
I$, there exists a group nonexpansive retraction $P_{i}$ from $C$ onto
$\text{Fix}(\mathcal{S})$, such that $P_{i}T_{i}=T_{i}P_{i}=P_{i}$, and every
closed $\mathcal{S}\cup\\{R\\}$-invariant subset of $C$ is also
$P_{i}$-invariant.

###### Proof.

Set $\mathcal{S}^{{}^{\prime}}:=\mathcal{S}\cup\\{R\\}$ and
$\mathfrak{R^{{}^{\prime}}}=\\{T\in C^{C}:T\;\text{is group
nonexpansive},T\circ T_{\alpha}=T\\\ \text{ }\text{ and every closed
$\mathcal{S}^{{}^{\prime}}$-invariant subset of $C$ is also $T$
-invariant}\\}$,
and we get that
$\text{Fix}(\mathcal{S}^{{}^{\prime}})=\text{Fix}(\mathcal{S})$ and by
replacing $\mathcal{S}$ with $\mathcal{S}^{{}^{\prime}}$ and $\mathfrak{R}$
with $\mathfrak{R^{{}^{\prime}}}$ in the proof of Theorem 3.1, we find a
minimal element $P_{\alpha}$ in the sense of ($*$). Now we have $R\circ
T\in\mathfrak{R^{{}^{\prime}}}$ for each $T\in\mathfrak{R^{{}^{\prime}}}$.
Indeed, $R\circ T\circ T_{\alpha}=R\circ T$ for each
$T\in\mathfrak{R^{{}^{\prime}}}$ and because $R\in\mathcal{S}^{{}^{\prime}}$,
we have that every closed $\mathcal{S}^{{}^{\prime}}$-invariant subset of $C$
is also $R$ -invariant, and therefore is $R\circ T$-invariant for each
$T\in\mathfrak{R^{{}^{\prime}}}$. Hence for each $x\in C$, the set
$K=\\{T(P_{\alpha}x):T\in\mathfrak{R^{{}^{\prime}}}\\}$ is an $R$-invariant
subset of $C$ for each $T\in\mathfrak{R^{{}^{\prime}}}$. Therefore from the
fact that $R(K)\subset K\cap R(C)=K\cap\text{Fix}(\mathcal{S})$, we have
$K\cap\text{Fix}(\mathcal{S}^{{}^{\prime}})=K\cap\text{Fix}(\mathcal{S})\neq\emptyset$.
Now by repeating the reasoning used in Theorem 3.1, we get the desired result.

∎

As an application of Theorem 3.2, we have the following result:

######  Theorem 3.3.

Let $G$ be a topological group with the topology $\tau$ and let $C$ be a
compact subset of $G$. Suppose that $\mathcal{S}=\\{T_{i}:i\in I\\}$ is a
family of group nonexpansive mappings on $C$ such that
$\text{Fix}(\mathcal{S})\neq\emptyset$. Consider the following assumptions:

  1. (a)

Suppose for every nonempty compact $\mathcal{S}$-invariant subset $K$ of $C$,
$K\cap\text{Fix}(\mathcal{S})\neq\emptyset$,

  2. (b)

there exists a group nonexpansive retraction $R$ from $C$ onto
$\text{Fix}(\mathcal{S})$.

Let $\\{P_{i}\\}_{i\in I}$ be the family of retractions obtained in the above
Theorem. Then for each $x\in C$,

$\overline{\\{T_{i}^{n}x:i\in
I,n\in\mathbb{N}\\}}^{\tau}\cap\text{Fix}(\mathcal{S})\subseteq\overline{\\{P_{i}(x):i\in
I\\}}^{\tau}.$

###### Proof.

Let $g\in\overline{\\{T_{i}^{n}x:i\in
I,n\in\mathbb{N}\\}}^{\tau}\cap\text{Fix}(\mathcal{S})$. Then for each
$U\in\mathfrak{B}_{e}$, there exists $i\in I$ and $n\in\mathbb{N}$ such that
$T_{i}^{n}xg^{-1}\in U$. From our assumptions and using Theorems 3.1 and 3.2,
there exists a group nonexpansive retraction $P_{i}$ such that
$P_{i}=P_{i}T_{i}$ and since from Theorems 3.1 and 3.2 every closed
$\mathcal{S}$-invariant or $\mathcal{S}\cup\\{R\\}$-invariant subset of $C$ is
also $P_{i}$-invariant then we have $P_{i}g=g$ for each $i\in I$. Hence from
the fact that $P_{i}$ is group nonexpansive and since $T_{i}^{n}xg^{-1}\in U$
then we have,

$(P_{i}x)g^{-1}=(P_{i}T_{i}^{n}x)(P_{i}T_{i}^{n}g)^{-1}\in U,$

and then we conclude $g\in\overline{\\{P_{i}(x):i\in I\\}}^{\tau}$. ∎

##  4\. Examples and applications

Recall every locally convex space is a topological group by the topology
generated by a family of seminorms. Theorem 3.1 extends and generalizes
Theorem 4.1(a) in [3], by removing the ”convex” and ”separated” conditions as
follows:

######  Corollary 4.1.

Suppose that $Q$ is a family of seminorms on a locally convex space $E$ which
determines the topology of $E$. Let $C$ be a compact subset of $E$. Suppose
that $\mathcal{S}=\\{T_{i}:i\in I\\}$ is a family of $Q$-nonexpansive mappings
on $C$ such that $\text{Fix}(\mathcal{S})\neq\emptyset$. If for every nonempty
compact $\mathcal{S}$-invariant subset $K$ of $C$,
$K\cap\text{Fix}(\mathcal{S})\neq\emptyset$, then, for each $i\in I$, there
exists a group nonexpansive retraction $P_{i}$ from $C$ onto
$\text{Fix}(\mathcal{S})$, such that $P_{i}T_{i}=T_{i}P_{i}=P_{i}$ and every
closed $\mathcal{S}$-invariant subset of $C$ is also $P_{i}$-invariant.

Theorem 3.1 extends and generalizes Theorem 2.1(a) in [11], by removing the
”convex” condition as follows:

######  Corollary 4.2.

Let $C$ be a compact subset of a Banach space $E$. Suppose that
$\mathcal{S}=\\{T_{i}:i\in I\\}$ is a family of nonexpansive mappings on $C$
such that $\text{Fix}(\mathcal{S})\neq\emptyset$. If for every nonempty
compact $\mathcal{S}$-invariant subset $K$ of $C$,
$K\cap\text{Fix}(\mathcal{S})\neq\emptyset$, then, for each $i\in I$, there
exists a group nonexpansive retraction $P_{i}$ from $C$ onto
$\text{Fix}(\mathcal{S})$, such that $P_{i}T_{i}=T_{i}P_{i}=P_{i}$ and every
closed $\mathcal{S}$-invariant subset of $C$ is also $P_{i}$-invariant.

In the following example for Theorem 3.1, we find a family of retractions
without the convexity condition on $K$.

######  Example 4.3.

Let $G=\mathbb{R}$ with the general topology, $C=[0,2]$. Suppose that
$\mathcal{S}=\\{T_{n}:n=2,3,4,\cdots\\}$ is a family of the group nonexpansive
mappings on $C$ such that

$T_{n}(x)=\left\\{\begin{array}[]{ll}x,&x\in[0,\frac{2n}{2n-1}];\\\
(\frac{1}{n}-1)x+2,&x\in(\frac{2n}{2n-1},2].\end{array}\right.$

So, $\text{Fix}(\mathcal{S})=[0,1]$. Also for every nonempty compact
$\mathcal{S}$-invariant subset $K$ of $C$,
$K\cap\text{Fix}(\mathcal{S})\neq\emptyset$. Indeed, since $K$ is
$\mathcal{S}$-invariant, then for each $x\in K\cap[1,2]$, there exists
$n_{1}\in\mathbb{N}$ such that $x\in[\frac{2n_{1}}{2n_{1}-1},2]$ and
$T_{n}(x)=(\frac{1}{n}-1)x+2\in K$ for each $n\geq n_{1}$, taking limit on
$n$, from the fact that $K$ is closed, $-x+2\in K$, but we know that
$-x+2\in[0,\frac{2n_{1}-2}{2n_{1}-1}]\subset[0,1]$, hence
$K\cap\text{Fix}(\mathcal{S})=K\cap[0,1]\neq\emptyset$. We show that $1\in K$.
Indeed, $-x+2\in K$, for each $x\in(1,2]$, because if $x\in(1,2]$ then there
exists an integer $n_{2}\in\mathbb{N}$ such that
$x\in[\frac{2n_{2}}{2n_{2}-1},2]$ therefore for each $n\geq n_{2}$,
$T_{n}(x)=(\frac{1}{n}-1)x+2\in K$, then taking limit we have $-x+2\in K$, for
each $x\in(1,2]$. So by putting $x=0.1,0.01,0.001,\ldots$ we get
$0.9,0.99,0.999,\ldots\in K$, hence from the fact that $K$ is closed, $1\in
K$. Now, let to define $P_{n}$:

$P_{n}(x)=\left\\{\begin{array}[]{ll}x,&x\in[0,1];\\\
1,&x\in(1,\frac{n}{n-1});\\\
(\frac{1}{n}-1)x+2,&x\in[\frac{n}{n-1},2].\end{array}\right.$

First, we show that $P_{n}$ is a retraction for each $n=2,3,4,\ldots$. Let
$x\in[0,1]$ then $P_{n}^{2}x=P_{n}(P_{n}x)=P_{n}x$. Next, let
$x\in(1,\frac{n}{n-1})$ then $P_{n}^{2}x=P_{n}(P_{n}x)=P_{n}(1)=1=P_{n}x$.
Finally, if $x\in[\frac{n}{n-1},2]$, then we have $0\leq(\frac{1}{n}-1)x+2\leq
1$, hence

$P_{n}^{2}x=P_{n}(P_{n}x)=P_{n}((\frac{1}{n}-1)x+2)=(\frac{1}{n}-1)x+2=P_{n}x,$

so $P_{n}^{2}=P_{n}$.

Now, we show that for each $n=2,3,\cdots$, $P_{n}$ is a group nonexpansive
retraction from $C$ onto $\text{Fix}(\mathcal{S})$. Indeed, in the case that
$x\in[\frac{n}{n-1},2]$ and $y\in(1,\frac{n}{n-1})$, we have
$\frac{x}{n}\geq\frac{1}{n-1}$ then $y-\frac{x}{n}\leq 1$, so
$x-\frac{x}{n}-1\leq x-y$ hence we get $(\frac{1}{n}-1)x+1\leq x-y$ then

$|P_{n}(x)-P_{n}(y)|\leq|x-y|.$

Now, consider another case that $x\in[\frac{n}{n-1},2]$ and $y\in[0,1]$, then
we have $2(y-1)\leq\frac{1}{n}x$ then $-\frac{1}{n}x\leq-2(y-1)$, so
$-\frac{1}{n}x-2\leq-2y$ hence

$-2(x-y)\leq(\frac{1}{n}-2)x+2\leq 0,$

therefore since $x-y\geq 0$, we have

$\displaystyle|P_{n}(x)-P_{n}(y)|$
$\displaystyle=|(\frac{1}{n}-1)x+2-y|=|x-y+(\frac{1}{n}-2)x+2|$
$\displaystyle\leq|x-y-2(x-y)|=|x-y|,$

another cases is obvious. Hence $P_{n}$ is a group nonexpansive mapping. Now,
to show $P_{n}T_{n}=T_{n}P_{n}=P_{n}$, first we show that $T_{n}P_{n}=P_{n}$:
the case that $x\in[0,1]$ is obvious. Let $x\in(1,\frac{n}{n-1})$, then we
have $T_{n}P_{n}x=T_{n}(1)=1=P_{n}x$. Finally, let $x\in[\frac{n}{n-1},2]$, so
we have $\frac{2}{n}\leq(\frac{1}{n}-1)x+2\leq 1$ then

$T_{n}P_{n}x=T_{n}((\frac{1}{n}-1)x+2)=(\frac{1}{n}-1)x+2=P_{n}x.$

Next, we show that $P_{n}T_{n}=P_{n}$: let $x\in(0,\frac{2n}{2n-1})$,
obviously we have $P_{n}T_{n}x=P_{n}x$. Finally, let
$x\in(\frac{2n}{2n-1},2]$, then we have

$P_{n}T_{n}x=P_{n}((\frac{1}{n}-1)x+2)=P_{n}x$

because we have two cases; (a): if $x\in(\frac{2n}{2n-1},\frac{n}{n-1}]$, then
we have

$1\leq(\frac{1}{n}-1)x+2\leq\frac{2n}{2n-1}<\frac{n}{n-1},$

then

$P_{n}T_{n}x=P_{n}((\frac{1}{n}-1)x+2)=1=P_{n}x;$

(b): if $x\in(\frac{n}{n-1},2]$, then we have $0\leq(\frac{1}{n}-1)x+2\leq 1$,
then

$P_{n}T_{n}x=P_{n}((\frac{1}{n}-1)x+2)=(\frac{1}{n}-1)x+2=P_{n}x.$

Now, we show that every closed $\mathcal{S}$-invariant subset $K$ of $C$ is
also $P_{n}$-invariant for each $n=2,3,\cdots$. First, let $x\in K\cap[0,1]$
then $P_{n}x=x\in K$. Next, let $x\in K\cap(1,\frac{n}{n-1})$ then
$P_{n}x=1\in K$. Finally, if $x\in K\cap[\frac{n}{n-1},2]$, then $x\in
K\cap[\frac{2n}{2n-1},2]$, hence from the fact that
$P_{n}x=(\frac{1}{n}-1)x+2=T_{n}x\in K$, we have $P_{n}x\in K$. Therefore, $K$
is also $P_{n}$-invariant for each $n=2,3,\cdots$.

Acknowledgements

The second author is grateful to the University of Lorestan for their support.

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