# On the $UC$ and $UC^{*}$ properties and the existence of best proximity
points in metric spaces

Vasil Zhelinski1<EMAIL_ADDRESS>and  Boyan Zlatanov2
<EMAIL_ADDRESS>1 Faculty of Mathematics and Informatics, Plovdiv
University “Paisii Hilendarski”, 24 “Tsar Assen” str., 4000 Plovdiv, Bulgaria.

(Date: Received: 03 March 2023)

###### Abstract.

We investigate the connections between $UC$ and $UC^{*}$ properties for
ordered pairs of subsets $(A,B)$ in metric spaces, which are involved in the
study of existence and uniqueness of best proximity points. We show that the
$UC^{*}$ property is included into the $UC$ property. We introduce some new
notions: bounded $UC$ ($BUC$) property and uniformly convex set about a
function $\phi$. We prove that these new notions are generalizations of the
$UC$ property and that both of them are sufficient for to ensure existence and
uniqueness of best proximity points. We show that these two new notions are
different from a uniform convexity and even from a strict convexity. If we
consider the underlying space to be a Banach space we find a sufficient
condition which ensures that from the $UC$ property it follows the uniform
convexity of the underlying Banach space. We illustrate the new notions with
examples. We present an example of a cyclic contraction $T$ in a space, which
is not even strictly convex and the ordered pair $(A,B)$ has not the $UC$
property, but has the $BUC$ property and thus there is a unique best proximity
point of $T$ in $A$.

###### Key words and phrases:

cyclic map, best proximity points, uniformly convex Banach space, $UC$
property

  * •

We prove that the $UC^{*}$ property is a consequence of the $UC$ property

  * •

We introduce two new notions, which are weaker than $UC$ property, that ensure
the existence and uniqueness of best proximity points

  * •

We find a connection between $UC$ property and uniform convexity of the
underlying Banach space

  * •

We find a connection between $UC$ property and uniformly convex in a single
direction of the unit ball of the underlying Banach space

  * •

We illustrate the new notions and results with a sequence of examples

##  1\. Introduction

A fundamental result in the fixed point theory is the Banach contraction
principle in Banach spaces or in complete metric spaces. Fixed point theory is
an important tool for solving equations $Tx=x$ for mapping T defined on
subsets of metric or normed spaces.

One kind of a generalization of the Banach contraction principle is the notion
of cyclic maps, $T:A\to B$ and $T:B\to A$ [11]. Because a non-self mapping
$T:A\to B$ does not necessarily have a fixed point, one often attempts to find
an element $x$ which is in some sense ”closest“ to $Tx$. Thus we can alter the
fixed point problem into the optimization problem $\min\\{\|x-Tx\|:x\in A\cup
B\\}$. Best proximity point theorems, introduced in [7], are relevant in this
perspective. A sufficient condition for the existence and uniqueness of the
best proximity points in uniformly convex Banach spaces is given in [7]. The
uniform convexity of the underlying space ensures good geometric properties of
the space and is a key property in getting the results of existence and
uniqueness of best proximity points.

Naturally [13, 14], in solving the optimization problem $\min\\{\|x-Tx\|:x\in
A\cup B\\}$ for a cyclic map $T:A\to B$ and $T:B\to A$, where $A,B$ are
subsets of either a Banach space $(X,\|\cdot\|)$ or a metric space $(X,\rho)$,
it may be needed only some specific properties of the domain of $T$, i.e.
$A\cup B$, instead of the uniform convexity of the underlying Banach space
$(X,\|\cdot\|)$.

This idea to search for good properties of the ordered pair $(A,B)$ of sets,
which defines the domain of the cyclic map, have been firstly initiated in
[14], where the authors have investigated existence and uniqueness of best
proximity points in metric space over ordered pairs $(A,B)$ of sets in
complete metric spaces. The authors have introduced the notion for an ordered
pair $(A,B)$ of sets to satisfy the property $UC$. Some relations of the
property $UC$ for sets in a Banach space and the properties of uniform
convexity, uniform convexity in every direction and relatively compact sets
were presented in [14].

Later on a new notion for an ordered pair $(A,B)$ of sets to satisfy the
property $UC^{*}$ was put it, in order to investigate existence and uniqueness
of coupled best proximity points in complete metric spaces, rather than
uniformly convex Banach spaces [13].

Deep results about fixed points and the geometry of the underlying space can
be found in [2, 3, 10].

Some results about applications of coupled best proximity points for solving
of symmetric [9] and for solving of non-symmetric [16] systems of equations
have been presented. It is interesting to mention that the presented technique
in [9, 16] enables to find exact solutions in cases, where the classical fixed
point methods can find only approximations. Best proximity point results have
been used in searching of market equilibrium in duopoly markets, where the
cyclic maps have been replaced by semi-cyclic maps [1, 6]. The natural
underlying space in the market equilibrium theory is close to non convex
spaces, rather than convex spaces as pointed in [1], where results about
coupled best proximity points have been obtained in reflexive Banach spaces.

Thereafter it seem interesting to search for some conditions, different from
the uniform convexity of the underlying space, that will ensure some of the
properties, involved in the definitions of $UC$ or/and $UC^{*}$ and will lead
to positive conclusions on the existence and uniqueness of best proximity
points.

##  2\. Preliminaries

In what follows we will use the notations: $\mathbb{N}$ for the set of natural
numbers, $\mathbb{R}$ for the set of real numbers, $S_{X}$ and $B_{X}$ the
unit sphere and the unit ball, where $(X,\|\cdot\|)$ is a Banach space,
$B(x_{0},r)=\\{x\in X:\|x-x_{0}\|<r$} and $B[x_{0},r]=\\{x\in
X:\|x-x_{0}\|<r$} for the open and close balls with a center $x_{0}$ and
radius $r$, respectively. Let $(X,\rho)$ be a metric space and $A,B\subset X$.
We will denote by ${\rm dist}(A,B)=\inf\\{\rho(a,b):A\in A,b\in B\\}$ the
distance between the sets $A$ and $B$. Whenever the underlying space is a
Banach space $(X,\|\cdot\|)$ we consider the metric to be the one generated by
the norm, i.e. $\rho(x,y)=\|x-y\|$.

######  Definition 1.

([4, 8]) Let $(X,\|\cdot\|)$ be a Banach space. For every
$\varepsilon\in(0,2]$ we define the modulus of convexity of $\|\cdot\|$ by

$\delta_{\|\cdot\|}(\varepsilon)=\inf\left\\{1-\left\|\frac{x+y}{2}\right\|:x,y\in
B_{X},\|x-y\|\geq\varepsilon\right\\}.$

The norm is called uniformly convex if $\delta_{X}(\varepsilon)>0$ for all
$\varepsilon\in(0,2]$. The space $(X,\|\cdot\|)$ is then called uniformly
convex Banach space.

######  Definition 2.

([11]) Let $A$ and $B$ be two sets. A map $T:A\cup B\to A\cup B$ is called a
cyclic map if it satisfies $T:A\to B$ and $T:B\to A$.

######  Definition 3.

([7]) Let $(X,\rho)$ be a metric space, $A$ and $B$ be subsets of $X$ and
$T:A\cup B\to A\cup B$ be a cyclic map. We say that the point $x\in A$ is a
best proximity point of $T$ in $A$, if $\rho(x,Tx)={\rm dist}(A,B)$, where
${\rm dist}(A,B)=\inf\\{\rho(a,b):a\in A,b\in B\\}$.

######  Definition 4.

([7]) Let $(X,\rho)$ be a metric space, $A$ and $B$ be subsets of $X$. We say
that the map $T:A\cup B\to A\cup B$ be a cyclic contraction map if it is a
cyclic map and satisfies the inequality

$\rho(Tx,Ty)\leq\alpha\rho(x,y)+(1-\alpha){\rm dist}(A,B)$

for some $\alpha\in(0,1)$ and any $x\in A$, $y\in B$.

######  Theorem 5.

([7]) Let $A$ and $B$ be nonempty closed and convex subsets of a uniformly
convex Banach space $(X,\|\cdot\|)$. Suppose $T:A\cup B\to A\cup B$ is a
cyclic contraction map, then there exists a unique best proximity point $x$ of
$T$ in $A$.

It is also proven in [7], that for any initial guess $x_{0}\in A$, the
iterated sequence $x_{n}=T^{n}x_{0}$, for $n\in\mathbb{N}$, splits into two
sequences, such that $\left\\{x_{2n}\right\\}_{n=1}^{\infty}$ converges to the
best proximity point $x$ of $T$ in $A$ and
$\left\\{x_{2n-1}\right\\}_{n=1}^{\infty}$ converges to the best proximity
point $Tx$ of $T$ in $B$. The apriori and the aposteriori error estimates have
been found [15] of the iterated sequences $\\{x_{2n}\\}_{n=1}^{\infty}$ and
$\\{x_{2n-1}\\}_{n=1}^{\infty}$.

The next lemmas are crucial in getting of the results about best proximity
points in uniformly convex Banach spaces.

######  Lemma 6.

([7]) Let $(X,\|\cdot\|)$ be a uniformly convex Banach space. Let $A$ and $B$
be nonempty and closed subsets of $X$. Let $A$ be convex. Let
$\\{x_{n}\\}_{n=1}^{\infty}$ and $\\{z_{n}\\}_{n=1}^{\infty}$ be sequences in
$A$ and $\\{y_{n}\\}_{n=1}^{\infty}$ be a sequence in $B$ such that
$\displaystyle\lim_{{n\to\infty}}\|x_{n}-y_{n}\|=\lim_{{n\to\infty}}\|z_{n}-y_{n}\|={\rm
dist}(A,B)$. Then $\displaystyle\lim_{{n\to\infty}}\|x_{n}-z_{n}\|=0$.

######  Lemma 7.

([7]) Let $X$ be a uniformly convex Banach space. Let $A$ and $B$ be non empty
and closed subsets of $X$. Let $A$ be convex. Let $\\{x_{n}\\}_{n=1}^{\infty}$
and $\\{z_{n}\\}_{n=1}^{\infty}$ be sequences in $A$ and
$\\{y_{n}\\}_{n=1}^{\infty}$ be a sequence in $B$ such that:

  * •

$\lim_{n\to\infty}\|x_{n}-y_{n}\|={\rm dist}(A,B)$

  * •

for every $\varepsilon>0$ there is $N\in\mathbb{N}$ so that the inequality
$\|x_{m}-y_{n}\|\leq{\rm dist}(A,B)+\varepsilon$ holds for all $m>n\geq N$.

Then for every $\varepsilon>0$ there exists $N_{0}\in\mathbb{N}$ so that for
all $m>n\geq N_{0}$ there holds the inequality
$\|x_{m}-y_{n}\|\leq\varepsilon$.

######  Definition 8.

([8]) Let $(X,\|\cdot\|)$ be a Banach space. $X$ is called a strictly convex
Banach space if $\|x+y\|<2$ for all $x,y\in S_{X}$, such that $x\not=y$.

######  Lemma 9.

([7]) Let $A$, $B$ be closed subsets of a strictly convex normed space
$(X,\|\cdot\|)$, such that ${\rm dist}(A,B)>0$ and let $A$ be convex. If
$x,z\in A$ and $y\in B$ be such that $\|x-y\|=\|z-y\|={\rm dist}(A,B)$, then
$x=z$.

As pointed in [7], if the sets $A$ and $B$ satisfy some additional properties
$T:A\cup B\to A\cup B$ be a cyclic contraction map and either $A$ or $B$ is
boundedly compact, then there exists a best proximity point $x$ of $T$ in $A$.

The authors of [14] have found some properties of the sets $A$ and $B$ that
define the domain of the cyclic contraction map $T$, which ensure the
existence and uniqueness of the best proximity points, without assuming for
the underlying space to be a uniform convex Banach space.

######  Definition 10.

([14]) Let $A$ and $B$ be nonempty subsets of a metric space $(X,\rho)$. We
say that the ordered pair $(A,B)$ satisfies the property $UC$ if for any
sequences
$\left\\{x_{n}\right\\}_{n=1}^{\infty},\left\\{z_{n}\right\\}_{n=1}^{\infty}\subset
A$ and $\left\\{y_{n}\right\\}_{n=1}^{\infty}\subset B$, such that there holds
$\displaystyle\lim_{n\to\infty}\rho\left(x_{n},y_{n}\right)=\lim_{n\to\infty}\rho\left(z_{n},y_{n}\right)={\rm
dist}(A,B)$, then there holds
$\displaystyle\lim_{n\to\infty}\rho\left(x_{n},z_{n}\right)=0$.

It is easy to observe that the $UC$ property replaces Lemma 6 and that the
assumption in Lemma 6 for the sets $A$ and $B$ to be closed ones and $A$ to be
convex is not necessary, when we replace the uniform convexity of the
underlying Banach space with the $UC$ property. We would like to points out
that it may happen the ordered pair $(A,B)$ to satisfy $UC$, but the ordered
pair $(B,A)$ not to satisfy it.

Some property of the $UC$ ordered pairs $(A,B)$ of subsets are presented in
[14].

######  Proposition 11.

([14]) Let $A$ and $B$ be nonempty subsets of a uniformly convex Banach space
$(X,\|\cdot\|)$. If $A$ is convex, then the ordered pair $(A,B)$ has the
property $UC$.

######  Proposition 12.

([14]) Let $A$ and $B$ be nonempty subsets of a metric space $(X,\rho)$, such
that ${\rm dist}(A,B)=0$. Then the ordered pair $(A,B)$ satisfies the property
UC.

######  Proposition 13.

([14]) Let $A,A^{\prime},B,B^{\prime}$ be nonempty subsets of a metric space
$(X,\rho)$, such that $A\subseteq A^{\prime}$, $B\subseteq B^{\prime}$ and
${\rm dist}(A,B)={\rm dist}(A^{\prime},B^{\prime})$. If the ordered pair
$(A^{\prime},B^{\prime}$) satisfies the property UC, then the ordered pair
$(A,B)$ satisfies the property UC too.

If in the next result the underlying metric space $(X,\rho)$ is replaced by a
uniformly convex Banach space $(X,\|\cdot\|)$ and $A$ and $B$ be nonempty
closed and convex subsets then it generalizes Theorem 5.

######  Theorem 14.

([14]) Let $A$ and $B$ be nonempty closed subsets of a complete metric space
$(X,\rho)$, such that the ordered pairs $(A,B)$ satisfy the property $UC$. Let
$T:A\cup B\to A\cup B$ be a cyclic map and there exists $k\in(0,1)$, so that
the inequality

$\rho(Tx,Ty)\leq k\max\\{\rho(x,y),\rho(x,Tx),\rho(y,Ty)\\}+(1-k){\rm
dist}(A,B)$

holds for all $x\in A$ and $y\in B$. Then there is a unique best proximity
point $x$ of $T$ in $A$, the sequence of successive iterations
$\\{T^{2n}x_{0}\\}_{n=1}^{\infty}$ converges to $x$ for any initial guess
$x_{0}\in A$. There is at least one best proximity point $y\in B$ of $T$ in
$B$. Moreover best proximity point $y\in B$ of $T$ in $B$ is unique, provided
that the ordered pair $(B,A)$ has the $UC$ property.

Examples show that it is possible to have a Banach space $(X,\|\cdot\|)$ that
is not even strictly convex, but there are sets $A$ and $B$, so that the
ordered pair $(A,B)$ satisfies the property $UC$.

Figure 1. Example 15

######  Example 15.

Let us consider the spaces $X_{\infty}=(\mathbb{R}^{2},\|\cdot\|_{\infty})$,
$X_{1}=(\mathbb{R}^{2},\|\cdot\|_{1})$ and the sets $A,B\subset X_{1}$ and
$B,C\subset X_{\infty}$ (Figure 1).

It easy to see that ${\rm dist}(A,B)=\inf\\{\|x-y\|_{1}:x\in A,y\in B\\}=1$.

If the sequences $\\{a_{n}\\}_{n=1}^{\infty}\subset A$ and
$\\{b_{n}\\}_{n=1}^{\infty}\subset B$ satisfy
$\lim_{n\to\infty}\|a_{n}-b_{n}\|={\rm dist}(A,B)$, then
$\lim_{n\to\infty}a_{n}=(1,0)$ and $\lim_{n\to\infty}b_{n}=(2,0)$. Therefore
if there hold $\lim_{n\to\infty}\|x_{n}-y_{n}\|={\rm dist}(A,B)$ and
$\lim_{n\to\infty}\|z_{n}-y_{n}\|={\rm dist}(A,B)$ for
$\\{x_{n}\\}_{n=1}^{\infty},\\{z_{n}\\}_{n=1}^{\infty}\subset A$ and
$\\{y_{n}\\}_{n=1}^{\infty}\subset B$, then there holds
$\lim_{n\to\infty}\|x_{n}-z_{n}\|=0$. Thus the ordered pair of sets $(A,B)$
has the $UC$ property.

There holds ${\rm dist}(B,C)=\inf\\{\|x-y\|_{\infty}:x\in B,y\in C\\}=1$. Let
us consider the sequences $\\{c_{n}\\}_{n=1}^{\infty}\subset C$ and
$\\{b_{n}\\}_{n=1}^{\infty}\subset B$ satisfy
$\lim_{n\to\infty}\|b_{n}-c_{n}\|={\rm dist}(B,C)$, where
$c_{n}=(x_{n},y_{n})$ and $b_{n}=(u_{n},v_{n})$, be such that
$\lim_{n\to\infty}x_{n}=5$, $\lim_{n\to\infty}u_{n}=4$ and
$y_{n},v_{n}\in[-1/2,1/2]$ can be arbitrary. Therefore there exist sequences
$\\{x_{n}\\}_{n=1}^{\infty},\\{z_{n}\\}_{n=1}^{\infty}\subset B$ and
$\\{y_{n}\\}_{n=1}^{\infty}\subset C$, satisfying
$\lim_{n\to\infty}\|x_{n}-y_{n}\|={\rm dist}(B,C)$ and
$\lim_{n\to\infty}\|z_{n}-y_{n}\|={\rm dist}(B,C)$, so that the
$\lim_{n\to\infty}\|x_{n}-z_{n}\|$ does not exist. Consequently the ordered
pair of sets $(B,C)$ has not the $UC$ property.

If we consider the set $\mathbb{R}^{2}$ endowed with an uniformly convex norm
(for example the Hilbert norm $\|\cdot\|_{2}$, then the ordered pair of sets
$(B,C)$ has the $UC$ property according to Lemma 6.

Figure 2. Example 16

######  Example 16.

Let us consider the space $X_{\infty}=(\mathbb{R}^{2},\|\cdot\|_{\infty})$ and
the sets $A$, $B$ and $C$ (Figure 2).

It easy to see that ${\rm dist}(A,B)={\rm dist}(B,C)=1$.

If the sequences $\\{a_{n}\\}_{n=1}^{\infty}\subset A$ and
$\\{b_{n}\\}_{n=1}^{\infty}\subset B$ satisfy
$\lim_{n\to\infty}\|a_{n}-b_{n}\|={\rm dist}(A,B)$, then
$\lim_{n\to\infty}a_{n}=(1,0)$ and $\lim_{n\to\infty}b_{n}=(2,0)$. Therefore
if there hold $\lim_{n\to\infty}\|x_{n}-y_{n}\|={\rm dist}(A,B)$,
$\lim_{n\to\infty}\|z_{n}-y_{n}\|={\rm dist}(A,B)$ for the sequences
$\\{x_{n}\\}_{n=1}^{\infty},\\{z_{n}\\}_{n=1}^{\infty}\subset A$ and
$\\{y_{n}\\}_{n=1}^{\infty}\subset B$, then there holds
$\lim_{n\to\infty}\|x_{n}-z_{n}\|=0$. Thus the ordered pair of sets $(A,B)$
has the $UC$ property.

By similar arguments we get that the ordered pair of sets $(B,C)$ has the $UC$
property, never mind the geometry of the unit ball in $\mathbb{R}^{2}$.

By the observations, in the two examples, we see that the property $UC$ depend
on three conditions: the geometry of the unit ball of the underlying Banach
space, the geometry properties of the sets $A$ and $B$ and the positioning of
the sets into the space.

######  Definition 17.

([5]) Let $(X,\|\cdot\|)$ be a Banach space. For every $\varepsilon\in(0,2]$
and every $z\in X$ we define $\delta_{\|\cdot\|}(z,\varepsilon)$ by

$\delta_{\|\cdot\|}(z,\varepsilon)=\inf\left\\{1-\textstyle\left\|\frac{x+y}{2}\right\|:x,y\in
S_{X},\|x-y\|\geq\varepsilon,x-y=\lambda z\ \mbox{for some}\
\lambda\in\mathbb{R}\right\\}.$

We call $\delta_{\|\cdot\|}(z,\varepsilon)$ the modulus of convexity in the
direction $z\in X$.

The norm is called uniformly convex in every direction ($UCED$) if
$\delta_{X}(z,\varepsilon)>0$ for all $z\in X$ and $\varepsilon\in(0,2]$. The
space $(X,\|\cdot\|)$ is then called uniformly convex in every direction
Banach space.

######  Proposition 18.

([14]) Let $A$ and $B$ be nonempty subsets of a $UCED$ Banach space
$(X,\|\cdot\|)$. If $A$ is convex and relatively compact then the ordered pair
$(A,B)$ has the property $UC$.

A notion that replaces Lemma 7 have been introduced in [13] in order to get
existence and uniqueness results about coupled best proximity points in metric
spaces.

######  Definition 19.

([13]) Let $A$ and $B$ be nonempty subsets of the metric space $(X,\rho)$. We
say that the ordered pair $(A,B)$ satisfies the property $UC^{*}$ if $(A,B)$
satisfies the property $UC$ and there holds

  * •

for any sequences
$\\{x_{n}\\}_{n=1}^{\infty},\\{z_{n}\\}_{n=1}^{\infty}\subset A$ and
$\\{y_{n}\\}_{n=1}^{\infty}\subset B$, so that

$\displaystyle\lim_{{n\to\infty}}\rho(z_{n},y_{n})={\rm dist}(A,B)$

  * •

for any $\varepsilon>0$ there exists $N\in\mathbb{N}$, so that the inequality
$\rho(x_{m},y_{n})\leq{\rm dist}(A,B)+\varepsilon$ holds for all $m>n\geq N$.

Then for any $\varepsilon>0$ there is $N_{1}\in\mathbb{N}$ so that the
inequality $\rho(x_{m},z_{n})\leq\varepsilon$ holds for all $m>n\geq N_{1}$.

######  Proposition 20.

([13]) Every ordered pair $(A,B)$ of nonempty subsets of a metric space
$(X,\rho)$, so that ${\rm dist}(A,B)=0$, satisfies the property $UC^{*}$.

######  Proposition 21.

([13]) Every ordered pair $(A,B)$ of nonempty subsets of a uniformly convex
Banach space $(X,\|\cdot\|)$, such that $A$ is convex, satisfies the property
$UC^{*}$.

######  Definition 22.

([13]) Let $A$ and $B$ be nonempty subsets of a metric space $(X,\rho)$ and
$T:A\times A\to B$. A point $(x,y)\in A\times A$ is called a coupled best
proximity point of $T$ in $A\times A$ if $\rho(x,F(x,y))=\rho(y,F(y,x))={\rm
dist}(A,B)$.

######  Definition 23.

([13]) Let $A$ and $B$ be nonempty subsets of a metric space $(X,\rho)$. An
ordered pair of maps $(F,G)$ is called an ordered pair of cyclic maps (or for
short cyclic maps) if $F:A\times A\to B$ and $G:B\times B\to A$.

######  Definition 24.

([13]) Let $A$ and $B$ be nonempty subsets of a metric space $(X,\rho)$ an
ordered pair of cyclic maps $(F,G)$ is called an ordered pair of cyclic
contraction maps if there exists $\alpha,\beta\in(0,1)$ with $\alpha+\beta<1$
so that the inequality

$\rho(F(x,y),G(u,v))\leq\alpha\rho(x,u)+\beta\rho(y,v)+(1-(\alpha+\beta)){\rm
dist}(A,B)$ (1)

holds for any $(x,y)\in A\times A$ and $(u,v)\in B\times B$.

######  Theorem 25.

([13]) Let $A$ and $B$ be nonempty closed subsets of a complete metric space
$(X,\rho)$, such that the ordered pairs $(A,B)$ and $(B,A)$ satisfy the
property $UC^{*}$. Let $F:A\times A\to B$, $G:B\times B\to A$ and $(F,G)$ be a
cyclic contraction. Then there exits coupled best proximity point $(x,y)$ of
$F$ in $A\times A$ and a coupled best proximity point $(u,v)$ of $G$ in
$B\times B$ such that $\rho(x,u)+\rho(y,v)=2{\rm dist}(A,B)$.

It seems that the properties $UC$ and $UC^{*}$ overlap for a wide class of
sets with the key lemmas (Lemma 6 and Lemma 7) from [7].

By the imposed conditions in Theorem 14 and Theorem 25 it seems, at first
glimpse, that there may be a gap in the proof of Theorem 14, as far the
generalization of Lemma 7, in terms of a $UC^{*}$ property, is missing. We
will see in the next section, that it is not the case, but the property
$UC^{*}$ overlaps with the property $UC$, when the sequences
$\left\\{x_{n}\right\\}_{n=1}^{\infty},\left\\{z_{n}\right\\}_{n=1}^{\infty}\subset
A$ and $\left\\{y_{n}\right\\}_{n=1}^{\infty}\subset B$ are the iterated
sequence, generated by a cyclic contraction $T$. We will show that we can
remove the assumption $(A,B)$ to satisfy the property $UC^{*}$ in the
conditions of Theorem 25.

Actually the properties $UC$ and $UC^{*}$ can holds just for the two sets $A$
and $B$ that are the domain of the cyclic maps $T:A\cup B\to A\cup B$. We will
try to find conditions which will ensure that whenever any ordered pair
$(A,B)$ of subsets of a Banach space $(X,\|\cdot\|)$ satisfies the property
$UC$, then $(X,\|\cdot\|)$ will be uniformly convex Banach space.

We will try to introduce a generalization of the notions of convexity, which
will be different from $UC$, strict convexity or uniform convexity, but will
insure existence and uniqueness of best proximity points for classes of cyclic
maps $T:A\cup B\to A\cup B$.

##  3\. Connection between $UC$ and $UC^{*}$ properties

We will start with some comments on the introduced in [13, 14] properties $UC$
and $UC^{*}$. The notions introduced in [13, 14] search for some good
properties to be satisfied for an ordered pair of sets $(A,B)$. The $UC$
states that for three sequences
$\left\\{x_{n}\right\\}_{n=1}^{\infty},\left\\{z_{n}\right\\}_{n=1}^{\infty}\subset
A$, $\left\\{y\right\\}_{n=1}^{\infty}\subset B$, such that
$\displaystyle\lim_{n\to\infty}\rho(x_{n},y_{n})=\lim_{n\to\infty}\rho(z_{n},y_{n})={\rm
dist}(A,B)$, then $\displaystyle\lim_{n\to\infty}\rho(x_{n},z_{n})=0$.

The $UC^{*}$ requires, at first glimpse, a stronger property by insisting the
sequences
$\left\\{x_{n}\right\\}_{n=1}^{\infty},\left\\{z_{n}\right\\}_{n=1}^{\infty}\subset
A$ to verify that for any $\varepsilon>0$ there is $N\in\mathbb{N}$ so that
the inequality $\rho(x_{m},z_{n})\leq\varepsilon$ holds for all $m>n\geq N$.
Actually the property $UC^{*}$ is included in the property $UC$ as we will see
in Theorem 30, because of the additional requirement on the sequences
$\left\\{x_{n}\right\\}_{n=1}^{\infty}\subset A$,
$\left\\{y_{n}\right\\}_{n=1}^{\infty}\subset B$ to satisfy for any
$\varepsilon>0$ there exists $N\in\mathbb{N}$, so that the inequality
$\rho(x_{m},y_{n})\leq{\rm dist}(A,B)+\varepsilon$ holds for all $m>n\geq N$.

An auxiliary result in ([7], Proposition 3.3) is that for a cyclic contraction
map $T$, the iterated sequences $\left\\{T^{2n}x_{0}\right\\}_{n=1}^{\infty}$,
$\left\\{T^{2n-1}x_{0}\right\\}_{n=1}^{\infty}$, for any arbitrary chosen
initial guess point $x_{0}\in A\cup B$, are bounded ones. The authors in [7]
apply Lemma 7 only for bounded sequences
$\left\\{x_{n}\right\\}_{n=1}^{\infty},\\{z_{n}\\}_{n=1}^{\infty}\subset A$,
$\left\\{y_{n}\right\\}_{n=1}^{\infty}\subset B$ to show that the sequences
$\left\\{T^{2n}x_{0}\right\\}_{n=1}^{\infty}$,
$\left\\{T^{2n-1}x_{0}\right\\}_{n=1}^{\infty}$ are Cauchy ones.

A crucial lemma in [14] presents a condition for a sequence to be a Cauchy
one.

######  Lemma 26.

([14]) Let $A$ and $B$ be subsets of a metric space $(X,\rho)$. Assume that
the ordered pair $(A,B)$ has the property $UC$ and
$\left\\{x_{n}\right\\}_{n=1}^{\infty}\subset A$,
$\left\\{y_{n}\right\\}_{n=1}^{\infty}\subset B$, so that either one of
following holds

$\lim_{m\to\infty}\sup_{n\geq m}\rho(x_{m},y_{n})={\rm dist}(A,B)\ \mbox{or}\
\lim_{n\to\infty}\sup_{m\geq n}\rho(x_{m},y_{n})={\rm dist}(A,B).$

Then $\\{x_{n}\\}$ is a Cauchy sequence.

In the proof of the main result in [14], the authors show that

$\lim_{m\to\infty}\sup_{n\geq m}\rho\left(T^{2m}x,T^{2n+1}x\right)={\rm
dist}(A,B),$

which replaces the assumption the ordered pair of sets $(A,B)$ to satisfies
the $UC^{*}$ property.

As far as the application of Lemmas 6 and 7 or properties $UC$ or $UC^{*}$ are
for the iterated sequences $\left\\{T^{2n}x_{0}\right\\}_{n=1}^{\infty}$,
$\left\\{T^{2n-1}x_{0}\right\\}_{n=1}^{\infty}$, that are bounded ones, we
will introduce a new property for an ordered pair of sets $(A,B)$ just about
any bounded sequences
$\left\\{x_{n}\right\\}_{n=1}^{\infty},\left\\{z_{n}\right\\}_{n=1}^{\infty}\subset
A$.

######  Definition 27.

Let $A$ and $B$ be nonempty subsets of a metric space $(X,\rho)$. We say that
the ordered pair $(A,B)$ satisfies the bounded property $UC$ ($BUC$) if for
any bounded sequences
$\left\\{x_{n}\right\\}_{n=1}^{\infty},\left\\{z_{n}\right\\}_{n=1}^{\infty}\subset
A$ and an arbitrary sequence $\left\\{y_{n}\right\\}_{n=1}^{\infty}\subset B$,
such that, whenever there holds

$\lim_{n\to\infty}\rho(x_{n},y_{n})=\lim_{n\to\infty}\rho(z_{n},y_{n})={\rm
dist}(A,B),$ (2)

then there holds $\lim_{n\to\infty}\rho(x_{n},z_{n})=0$.

Let an ordered pair $(A,B)$ has the property $UC$. If equality (2) holds for
bounded sequences
$\left\\{x_{n}\right\\}_{n=1}^{\infty},\left\\{z_{n}\right\\}_{n=1}^{\infty}\subset
A$, then $(A,B)$ has the property $BUC$. It is possible that for the ordered
pair $(A,B)$ there are no any bounded sequences
$\left\\{x_{n}\right\\}_{n=1}^{\infty},\left\\{z_{n}\right\\}_{n=1}^{\infty}\subset
A$ to satisfy (2), i.e. the set of sequences that satisfy Definition 27 is the
empty set. In this case we agree to say that the ordered pair $(A,B)$ has the
$BUC$ property. If an ordered pair $(A,B)$ satisfies $BUC$, then it may happen
that there are unbounded sequences
$\left\\{x_{n}\right\\}_{n=1}^{\infty},\left\\{z_{n}\right\\}_{n=1}^{\infty}\subset
A$ so that the assumptions about property $UC$ are not satisfied and thus
$(A,B)$ will have the $BUC$ property, but will not have the $UC$ property.

######  Example 28.

Let us consider the sets

$A=\left\\{(x,y)\in\mathbb{R}^{2}:y\geq\frac{1}{x},x>0\right\\}\ \mbox{and}\
B=\left\\{(x,y)\in\mathbb{R}^{2}:y\geq\frac{1}{|x|},x<0\right\\}.$

  1. (1)

If we consider $A,B\subset(\mathbb{R}^{2},\|\cdot\|_{2})$. Then the ordered
pair $(A,B)$ has the $UC$ property

  2. (2)

If we consider $A,B\subset(\mathbb{R}^{2},\|\cdot\|_{\infty})$. Then the
ordered pair $(A,B)$ has not the $UC$ property.

In both cases of Example 28 the sequences that satisfy (2) are unbounded ones.
Thus it is an example of an ordered pair $(A,B)$, which has the $BUC$ property
by our agreement that if the set set of sequences that satisfy Definition 27
is the empty set, then the ordered pair has the $BUC$ property.

First we will show that the conditions imposed on the sequences in the
definition of the $UC^{*}$ property ensure that they are bounded sequences and
therefore the set of the sequences in Definition 27 is not the empty set for
any ordered set $(A,B)$, satisfying the $UC^{*}$ property.

######  Lemma 29.

Let $(X,\rho)$ be a metric space. If the sequences

$\left\\{x_{n}\right\\}_{n=1}^{\infty},\
\left\\{z_{n}\right\\}_{n=1}^{\infty},\
\left\\{y_{n}\right\\}_{n=1}^{\infty}\subset X$

satisfy

$\lim_{n\to\infty}\rho(x_{n},y_{n})=a<\infty$ (3)
$\lim_{n\to\infty}\sup_{n<k,m}\rho(z_{m},y_{k})=b<\infty,$ (4)

then $\left\\{x_{n}\right\\}_{n=1}^{\infty},\
\left\\{z_{n}\right\\}_{n=1}^{\infty},\ \left\\{y_{n}\right\\}_{n=1}^{\infty}$
are bounded sequences.

Proof. From (4) it follows that there exists $n_{0}\in\mathbb{N}$ such that
$\displaystyle\sup_{n_{0}<m}\rho(z_{m},y_{n_{0}})<\infty$ and consequently the
sequence $\\{z_{n}\\}_{n=1}^{\infty}$ is bounded. By similar arguments we get
that $\\{y_{n}\\}_{n=1}^{\infty}$ is a bounded sequence too, From the
boundedness of $\\{y_{n}\\}_{n=1}^{\infty}$ and (3) it follows that
$\\{x_{n}\\}_{n=1}^{\infty}$ is a bounded sequence too. $\Box$

######  Theorem 30.

Let $(X,\rho)$ be a metric space and $A,B\subset X$. If the ordered pair
$(A,B)$ satisfies the $BUC$ property, then the ordered pair $(A,B)$ satisfies
the $UC^{*}$ property.

Proof. Let assume the contrary, i.e. $(A,B)$ satisfies the $BUC$, but does not
satisfies the $UC^{*}$ property. Then there exist sequences
$\\{x_{n}\\}_{n=1}^{\infty},\\{z_{n}\\}_{n=1}^{\infty}\in A$ and
$\\{y_{n}\\}_{n=1}^{\infty}\in B$, satisfying

$\lim_{n\to\infty}\rho(x_{n},y_{n})={\rm dist}(A,B)\ \mbox{and}\
\lim_{n\to\infty}\sup_{n<k,m}\rho(z_{m},y_{k})={\rm dist}(A,B)$ (5)

and $\displaystyle\lim_{n\to\infty}\sup_{n<k,m}\rho(z_{m},x_{k})\neq 0$.
Therefore there exist subsequences
$\\{x_{k_{i}}\\}_{i=1}^{\infty}\subset\\{x_{n}\\}_{n=1}^{\infty}$ and
$\\{z_{m_{i}}\\}_{i=1}^{\infty}\subset\\{z_{n}\\}_{n=1}^{\infty}$, so that

$\lim_{i\to\infty}\rho(x_{n_{i}},z_{m_{i}})\neq 0.$ (6)

Let us consider the sequences
$\\{x_{k_{i}}\\}_{i=1}^{\infty},\\{z_{m_{j}}\\}_{j=1}^{\infty}\subset A$ and
$\\{y_{k_{i}}\\}_{i=1}^{\infty}\subset B$. By (5) it follows that

$\lim_{n\to\infty}\rho(x_{k_{i}},y_{k_{i}})={\rm dist}(A,B)$ (7)

and for every $\varepsilon>0$, there is $N\in\mathbb{N}$, so that for all
$i,j\geq N$ there holds the inequality

$\rho(z_{m_{j}},y_{k_{i}})\leq{\rm dist}(A,B)+\varepsilon.$ (8)

From (8) we get that $\lim_{i\to\infty}\rho(z_{m_{i}},y_{k_{i}})={\rm
dist}(A,B)$. By Lemma 29, (7) and (8) it follows that the sequences are
bounded ones. From the assumption of the theorem that the ordered pair $(A,B)$
has the $BUC$ property and the sequences
$\\{x_{k_{i}}\\}_{i=1}^{\infty},\\{z_{m_{j}}\\}_{j=1}^{\infty}\subset A$ and
$\\{y_{k_{i}}\\}_{i=1}^{\infty}\subset B$ are bounded ones it follows that
$\lim_{i\to\infty}\rho(x_{n_{i}},z_{m_{i}})=0$, which is a contradiction with
(7). $\Box$

Thus having in mind that the conditions imposed on the sequences in the
definition of the $UC^{*}$ property ensure that they are bounded ones it
follows that from the $UC$ property it follows the $UC^{*}$ property, that is
$UC^{*}$ is included in the $UC$ property. Consequently we can replace in
Theorem 25 the assumption that $(A,B)$ satisfies the $UC^{*}$ by $(A,B)$
satisfies the $UC$.

From Theorem 30 and [7], where the authors have proven that the iterated
sequence $\\{T^{n}x\\}_{n=1}^{\infty}$, for any arbitrary chosen initial guess
$x\in A\cup B$, is a bounded one for the maps investigated in Theorem 5 it
follows that we can present a generalization of their result. We will
illustrate in the last section with an example that the next theorem is
actually a generalization of Theorem 5. We will present an ordered pair
$(A,B)$ and a cyclic map $T$, so that $(A,B)$ has the $BUC$ property, but has
not the $UC$ property and $T$ satisfies the contracive condition in Theorem 5,
i.e. (9).

######  Theorem 31.

Let $A$ and $B$ be nonempty closed subsets of a complete metric space
$(X,\rho)$, such that the ordered pairs $(A,B)$ satisfy the property $BUC$.
Let $T:A\cup B\to A\cup B$ be a cyclic map and there exists $k\in(0,1)$, so
that the inequality

$\rho(Tx,Ty)\leq k\rho(x,y)+(1-k){\rm dist}(A,B)$ (9)

holds for all $x\in A$ and $y\in B$. Then there is a unique best proximity
point $x$ of $T$ in $A$, the sequence of successive iterations
$\\{T^{2n}x_{0}\\}_{n=1}^{\infty}$ converges to $x$ for any initial guess
$x_{0}\in A$. There is at least one best proximity point in $B$ of $T$ and if
the ordered pair $(B,A)$ has the $BUC$ property, then this point is unique.

We will show that the iterated sequence $\\{T^{n}x\\}_{n=1}^{\infty}$, for any
arbitrary chosen initial guess $x\in A\cup B$ is a bounded one for the maps
investigated by [14] in Theorem 14.

######  Lemma 32.

Let $(X,\rho)$ be a metric space and let $A$ and $B$ be nonempty subsets of
$X$ and $T:A\cup B\to A\cup B$ be a cyclic map, such that there is
$\lambda\in[0,1)$ such that the inequality

$\rho(Tx,Ty)\leq\lambda\max\\{\rho(x,y),\rho(x,Tx),\rho(y,Ty)\\}+(1-\lambda){\rm
dist}(A,B)$ (10)

holds for every $x\in A$ and $y\in B$. Then the iterated sequence
$\\{T^{n}x\\}_{n=1}^{\infty}$ is a bounded one for every $x\in A\cup B$.

Proof. Let $x\in A\cup B$ be arbitrary chosen. Let us denote $x_{n}=T^{n}x$,
for $n=0,1,2,\dots$.

We will show that

$\sup_{n\in\mathbb{N}}\rho(x_{n},x_{n-1})=\rho(x_{1},x_{0})<\infty.$ (11)

For any $n\in\mathbb{N}$ by (10) we have the inequality

$\rho(x_{n},x_{n-1})\leq\lambda\max\\{\rho(x_{n-1},x_{n-2}),\rho(x_{n-1},x_{n}),\rho(x_{n-1},x_{n-2})\\}+(1-\lambda){\rm
dist}(A,B).$ (12)

The inequality (12) reduces to one of the inequalities, either

$\rho(x_{n},x_{n-1})\leq\lambda\rho(x_{n-1},x_{n})+(1-\lambda){\rm dist}(A,B)$
(13)

or

$\rho(x_{n},x_{n-1})\leq\lambda\rho(x_{n-1},x_{n-2})+(1-\lambda){\rm
dist}(A,B).$ (14)

If there holds (13), then it is fulfilled

$\rho(x_{n},x_{n-1})={\rm dist}(A,B).$ (15)

For any arbitrary chosen $n\in\mathbb{N}$ there are two cases: for any $m<n$
there holds either

$\rho(x_{m},x_{m-1})={\rm dist}(A,B).$ (16)

or

$\rho(x_{m},x_{m-1})\leq\lambda\rho(x_{m-1},x_{m-2})+(1-\lambda){\rm
dist}(A,B).$ (17)

If there holds (17) for any $m\leq n$ then we get the chain of inequalities

$\begin{array}[]{rl}\rho(x_{n},x_{n-1})\leq&\lambda\rho(x_{n-1},x_{n-2})+(1-\lambda){\rm
dist}(A,B)\\\ \leq&\lambda^{2}\rho(x_{n-2},x_{n-3})+(1-\lambda^{2}){\rm
dist}(A,B)\\\ \cdots&\cdots\\\
\leq&\lambda^{k}\rho(x_{n-k},x_{n-k+1})+(1-\lambda^{k}){\rm dist}(A,B)\\\
\cdots&\cdots\\\ \leq&\lambda^{n-2}\rho(x_{2},x_{1})+(1-\lambda^{n-2}){\rm
dist}(A,B)\\\ \leq&\lambda^{n-1}\rho(x_{1},x_{0})+(1-\lambda^{n-1}){\rm
dist}(A,B)\leq\rho(x_{1},x_{0})\end{array}$

If for some $m<n$ the case (16) is valid, then we get

$\begin{array}[]{rl}\rho(x_{n},x_{n-1})\leq&\lambda\rho(x_{n-1},x_{n-2})+(1-\lambda){\rm
dist}(A,B)\\\ \leq&\lambda^{2}\rho(x_{n-2},x_{n-3})+(1-\lambda^{2}){\rm
dist}(A,B)\\\ \cdots&\cdots\\\
\leq&\lambda^{m-n}\rho(x_{m},x_{m-1})+(1-\lambda^{m-n}){\rm dist}(A,B)\leq{\rm
dist}(A,B)\leq\rho(x_{1},x_{0}),\end{array}$

which finishes the proof of (11).

For every $n$ from (10) we have

$\rho(x_{2n},x_{1})\leq\lambda\max\\{\rho(x_{2n-1},x_{0}),\rho(x_{2n-1},x_{2n}),\rho(x_{1},x_{0})\\}+(1-\lambda){\rm
dist}(A,B).$

By (11) we get
$\rho(x_{2n},x_{1})\leq\lambda\max\\{\rho(x_{2n-1},x_{0}),\rho(x_{1},x_{0})\\}+(1-\lambda){\rm
dist}(A,B)$. From the triangle inequality and (10) it follows

$\rho(x_{2n-1},x_{0})\leq\rho(x_{2n-1},x_{2n})+\rho(x_{2n},x_{1})+\rho(x_{1},x_{0})\leq
2\rho(x_{1},x_{0})+\rho(x_{2n},x_{1}).$

Therefore we get
$\rho(x_{2n},x_{1})\leq(2\lambda+1)\rho(x_{1},x_{0})+\lambda\rho(x_{2n},x_{1})+(1-\lambda){\rm
dist}(A,B)$, i.e.

$\rho(x_{2n},x_{1})\leq\frac{2\lambda+1}{1-\lambda}\rho(x_{1},x_{0})+{\rm
dist}(A,B).$

and consequently $\left\\{x_{2n}\right\\}_{n=0}^{\infty}$ is a bounded
sequence.

By similar arguments we can prove that $\\{x_{2n+1}\\}_{n=0}^{\infty}$ is a
bounded sequence too. $\Box$

From Theorem 30 and Lemma 32 we can present a generalization of Theorem 14.

######  Theorem 33.

([14]) Let $A$ and $B$ be nonempty closed subsets of a complete metric space
$(X,\rho)$, such that the ordered pairs $(A,B)$ satisfy the property $BUC$.
Let $T:A\cup B\to A\cup B$ be a cyclic map and there exists $k\in(0,1)$, so
that the inequality

$\rho(Tx,Ty)\leq k\max\\{\rho(x,y),\rho(x,Tx),\rho(y,Ty)\\}+(1-k){\rm
dist}(A,B)$

holds for all $x\in A$ and $y\in B$. Then there is a unique best proximity
point $x$ of $T$ in $A$, the sequence of successive iterations
$\\{T^{2n}x_{0}\\}_{n=1}^{\infty}$ converges to $x$ for any initial guess
$x_{0}\in A$. There is at least one best proximity point in $B$ of $T$ and if
the ordered pair $(B,A)$ has the $BUC$ property, then this point is unique.

We will show that the iterated sequences generated by the maps in Theorem 25
are bounded ones. Following a smart idea, from [12], that connects coupled
fixed points and fixed points, we will try to apply the technique from [12] to
show that the maps involved in Theorem 25 can be considered as like as the
maps investigated in [7].

######  Lemma 34.

Let $(X,\rho)$ be a metric space and let $A$ and $B$ be nonempty subsets of
$X$. Let $F$ be a cyclic contraction, satisfying (1), i.e.

$\rho(F(x,x^{\prime}),F(y,y^{\prime}))\leq\alpha\rho(x,y)+\beta\rho(x^{\prime},y^{\prime})+(1-(\alpha+\beta)){\rm
dist}(A,B)$ (18)

for every $x,x^{\prime}\in A$, $y,y^{\prime}\in B$ and some $\alpha,\beta\geq
1$ and $\alpha+\beta\in[0,1)$. Then the iterated sequences
$\\{x_{n}\\}_{n=1}^{\infty}$ and $\\{y_{n}\\}_{n=1}^{\infty}$ are bounded
ones.

Proof. Let us consider the metric space $(X\times X,d)$, where
$d((x,y),(u,v))=\rho(x,u)+\rho(y,v)$. Let us define the map $T:X\times X\to
X\times X$, by $T(x,y)=(F(x,y),F(y,x))$. The map $T$ is a cyclic maps as
$T(A\times A)\subseteq B\times B$ and $T(B\times B)\subseteq A\times A$. There
holds

$\begin{array}[]{lll}{\rm dist}(A\times A,B\times
B)&=&\inf\\{d((a,a^{\prime}),(b,b^{\prime})):a,a^{\prime}\in A,b,b^{\prime}\in
B\\}\\\ &=&\inf\\{\rho(a,b)+\rho(a^{\prime},b^{\prime}):a,a^{\prime}\in
A,b,b^{\prime}\in B\\}\\\ &=&2\inf\\{\rho(a,b):a,\in A,b\in B\\}=2{\rm
dist}(A,B).\end{array}$ (19)

For any two arbitrary chosen $x=(x,x^{\prime})\in A\times A$ and
$y=(y,y^{\prime})\in B\times B$, by using (18) and (19) there holds the chain
of inequalities

$\begin{array}[]{lll}d(Tx,Ty)&=&d((F(x,x^{\prime}),(F(x^{\prime},x)),(F(y,y^{\prime}),(F(y^{\prime},y)))\\\
&=&\rho(F(x,x^{\prime}),F(y,y^{\prime}))+\rho(F(x^{\prime},x),F(y^{\prime},y))\\\
&\leq&\alpha\rho(x,y)+\beta\rho(x^{\prime},y^{\prime})+(1-(\alpha+\beta)){\rm
dist}(A,B)\\\
&&+\alpha\rho(x^{\prime},y^{\prime})+\beta\rho(x,y)+(1-(\alpha+\beta)){\rm
dist}(A,B)\\\
&=&(\alpha+\beta)(\rho(x,y)+\rho(x^{\prime},y^{\prime}))+2(1-(\alpha+\beta)){\rm
dist}(A,B)\\\
&=&(\alpha+\beta)d((x,x^{\prime}),(y,y^{\prime}))+(1-(\alpha+\beta)){\rm
dist}(A\times A,B\times B).\end{array}$

Consequently the cyclic map $T$ satisfies the conditions imposed in [7] and
according to [7] the iterated sequence

$u_{n}=(x_{n},y_{n})=Tu_{n-1}=T(x_{n-1},y_{n-1})=(F(x_{n-1},y_{n-1}),F(y_{n-1},x_{n-1}))$

for any arbitrary chosen $u_{0}=(x_{0},y_{0})\in A\times A$ is a bounded
sequence. $\Box$

From Theorem 30 and Lemma 34 we can present a generalization of Theorem 25.

######  Theorem 35.

([14]) Let $A$ and $B$ be nonempty closed subsets of a complete metric space
$(X,\rho)$, such that the ordered pairs $(A,B)$ and $(B,A)$ satisfy the
property $BUC$. Let $F:A\times A\to B$, $G:B\times B\to A$ and $(F,G)$ be a
cyclic contraction. Then there exits coupled best proximity point $(x,y)$ of
$F$ in $A\times A$ and a coupled best proximity point $(u,v)$ of $G$ in
$B\times B$ such that $\rho(x,u)+\rho(y,v)=2{\rm dist}(A,B)$.

##  4\. Characterization of the $UC$ property

We have seen in the examples, that it is possible to have the underlying
Banach space $(X,\|\cdot\|)$ not to be uniformly convex, but some particularly
chosen ordered pair of subsets $(A,B)$ to be either $UC$ one or $BUC$ one. We
will try to find some sufficient conditions that will ensure an ordered pair
of subsets $(A,B)$ to be a $UC$ one in an arbitrary Banach space.

######  Definition 36.

Let $(X,\|\cdot\|)$ is a Banach space and $A\subset X$. We say that $A$ is
uniformly convex set if for every $\epsilon>0$ there exists
$\eta(\epsilon)>0$, such that for every $x,y\in A$, satisfying the inequality
$\|x-y\|\geq\epsilon$ there holds
$B\left(\frac{x+y}{2},\eta(\epsilon)\right)\subset A$.

From Definition 1 and Definition 36 it is easy to observe that for any
$\varepsilon>0$ and for any $x,y\in B_{X}$, where $B_{X}$ is the unit ball in
a uniformly convex Banach space $(X,\|\cdot\|)$, such that
$\|x-y\|\geq\varepsilon$, there exists $\eta(\varepsilon)>0$, so that
$B\left(\frac{x+y}{2},\eta(\epsilon)\right)\subset B_{X}$. We can choose
$\eta(\varepsilon)=\delta_{X}(\varepsilon)$. Thus the unit ball in a uniformly
convex Banach space satisfies Definition 36.

We have seen in Example 16, that if the sets $A$ or $B$ have some good
geometric properties then the ordered pair $(A,B)$ will be an $UC$ one.

######  Theorem 37.

Let $(X,\|\cdot\|)$ be a Banach space, $A,B\subset X$ and $A$ is a uniformly
convex set. Then the ordered pair $(A,B)$ has the $UC$ property.

Proof. Let us assume the contrary, i.e. there are sequences
$\\{x_{n}\\}_{n=1}^{\infty},\\{z_{n}\\}_{n=1}^{\infty}\subset A$ and
$\\{y_{n}\\}_{n=1}^{\infty}\subset B$, such that
$\lim_{n\to\infty}\|x_{n}-y_{n}\|=\lim_{n\to\infty}\|z_{n}-y_{n}\|={\rm
dist}(A,B)$ but the sequence $\\{\|x_{n}-z_{n}\|\\}_{n=1}^{\infty}$ does not
converge to zero. Then there exists $\varepsilon_{0}>0$, such that for any
$N\in\mathbb{N}$ there is $n>N$ so that the inequality
$\left\|x_{n}-z_{n}\right\|>\varepsilon_{0}$ holds true. From the assumption,
that $A$ is a uniformly convex set it follows that for $\varepsilon_{0}$,
there is $\eta(\varepsilon_{0})$ so that the inclusion
$B\left(\frac{x_{n}+z_{n}}{2},\eta(\varepsilon_{0})\right)\subset A$ holds.
Consequently we can write the chain of inequalities

$\begin{array}[]{lll}\eta(\epsilon_{0})+{\rm dist}(A,B)&\leq&{\rm
dist}\left(\frac{x_{n}+z_{n}}{2},B\right)\leq\left\|\displaystyle\frac{x_{n}+z_{n}}{2}-y_{n}\right\|\\\\[10.0pt]
&\leq&\displaystyle\frac{\|x_{n}-y_{n}\|+\|z_{n}-y_{n}\|}{2}.\end{array}$ (20)

By the choice of the sequences $\\{x_{n}\\}_{n=1}^{\infty}$,
$\\{z_{n}\\}_{n=1}^{\infty}$ and $\\{y_{n}\\}_{n=1}^{\infty}$ it follows that
there exists $N_{1}(\varepsilon_{0})\in\mathbb{N}$, such that for every
$n>N_{1}$ there holds the inequality

$\frac{\|x_{n}-y_{n}\|+\|z_{n}-y_{n}\|}{2}<\eta(\varepsilon_{0})+{\rm
dist}(A,B),$ (21)

which is a contradiction with (20). Thus $\lim_{n\to\infty}\|x_{n}-z_{n}\|=0$
and we get that the ordered pair $(A,B)$ has the $UC$ property. $\Box$

By the results up to now it follows that if $A$ is a uniformly convex set,
then any ordered pair of sets $(A,B)$ has the $BUC$ property and the $UC$
property. It seems that the assumption $A$ to be a uniformly convex set is too
restrictive. We will try to find a weaker property, which will ensure that the
ordered pair of sets $(A,B)$ has the $BUC$ property, without being $UC$
ordered pair.

######  Definition 38.

Let $(X,\|\cdot\|)$ is a Banach space and $A\subset X$. We say that a function
$\phi$ has the positive property about the set $A$ if
$\phi:A\times\mathbb{R}^{+}\to\mathbb{R}^{+}$ be such that for every bounded
subset $A^{\prime}\subset A$ and and every $\epsilon_{0}>0$ there holds the
inequality

$\displaystyle\inf\\{\phi(x,\epsilon):x\in
A^{\prime},\epsilon\geq\epsilon_{0}\\}>0.$

We will illustrate Definition 38 by an example.

######  Example 39.

Let us consider the space $X=(\mathbb{R}^{2},\|\cdot\|_{\infty})$, the set

$\displaystyle A=\left\\{(x,y)\in\mathbb{R}^{2}:\frac{1}{x}\leq
y,x>0\right\\}$

and the function
$\displaystyle\phi:A\times\mathbb{R}^{+}\rightarrow\mathbb{R}^{+}$ be defined
by
$\displaystyle\phi(x,\varepsilon)=\frac{\varepsilon^{2}}{320+5\varepsilon^{2}+5\|x\|^{3}}$.
Then $\phi$ has positive property about $A$.

Let us define the function
$\phi_{1}:\mathbb{R}^{+}\times\mathbb{R}^{+}\rightarrow\mathbb{R}^{+}$ by
$\displaystyle\phi_{1}(r,\varepsilon)=\frac{\varepsilon^{2}}{320+5\varepsilon^{2}+5r^{3}}$.
Let us consider the function

$\phi(x,\varepsilon)=\phi_{1}(\|x\|,\varepsilon).$ (22)

From the boundedness of $A^{\prime}\in A$ it follows the existence of
$r_{1}=r_{1}(A^{\prime})$, so that

$\displaystyle\sup\\{\|x\|:x\in A^{\prime}\\}=r_{1}<\infty.$

For every $\varepsilon_{1}>0$ there holds

$\displaystyle\inf\\{\phi_{1}(r,\varepsilon):0\leq r\leq
r_{1},0<\varepsilon_{1}\leq\varepsilon\\}=\phi_{1}(r_{1},\varepsilon_{1})>0.$

By (22) and the definition of $r_{1}$ it follows that

$\displaystyle\inf\\{\phi(x,\varepsilon):x\in
A^{\prime},0<\varepsilon_{1}\leq\varepsilon\\}\geq\phi_{1}(\sup\\{\|x\|:x\in
A^{\prime}\\},\varepsilon_{1})>0.$

Consequently $\phi$ has positive property about $A$.

######  Definition 40.

Let $(X,\|\cdot\|)$ be a Banach space, $A\subset X$ and
$\phi:A\times\mathbb{R}^{+}\to\mathbb{R}^{+}$ has the positive property about
$A$. We say that $A$ is uniformly convex set about $\phi$, if for any
$\varepsilon>0$ and every $x,y\in A$, satisfying $\|x-y\|\geq\varepsilon>0$,
there holds
$B\left(\frac{x+y}{2},\phi(\frac{x+y}{2},\varepsilon)\right)\subset A$.

We will show that the set in Example 39 is uniformly convex about $\phi$.

######  Example 41.

Let us consider the space $X=(\mathbb{R}^{2},\|\cdot\|_{\infty})$, the set

$\displaystyle A=\left\\{(x,y)\in\mathbb{R}^{2}:\frac{1}{x}\leq
y,x>0\right\\}$

and the function $\phi:A\times\mathbb{R}^{+}\to\mathbb{R}^{+}$ be defined by
$\displaystyle\phi(x,\varepsilon)=\frac{\varepsilon^{2}}{320+5\varepsilon^{2}+5\|x\|^{3}}$.
Then $A$ is uniformly convex about $\phi$.

We have proven in Example 39 that $\phi$ has positive property about $A$. It
remains to show that Then $A$ is uniformly convex about $\phi$.

Let $\varepsilon>0$ be arbitrary chosen and let $p_{1}=(x_{1},y_{1})\in A$,
$p_{2}=(x_{2},y_{2})\in A$ satisfy $\|p_{1}-p_{2}\|=\varepsilon$. Let us put
$p_{3}=(x,y)=\frac{p_{1}+p_{2}}{2}=\left(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2}\right)$.

There are two cases: either 1) $|x_{1}-x_{2}|=\varepsilon$ or 2)
$|y_{1}-y_{2}|=\varepsilon$ is fulfilled.

Let us consider first the case 1) $|x_{1}-x_{2}|=\varepsilon$.

Without loss generality we can assume that $x_{1}<x_{2}$. Then for $x$ we
obtain the equalities

$\begin{array}[]{lll}x_{2}&=&x+\frac{\varepsilon}{2}\\\
x_{1}&=&x-\frac{\varepsilon}{2}.\end{array}$ (23)

From $p_{1}\in A$ it follows that $x_{1}>0$ and therefore $\displaystyle
0<x-\frac{\varepsilon}{2}$. Thus $\varepsilon<2x$.

From the inequalities $\displaystyle y_{1}\geq\frac{1}{x_{1}}$, $\displaystyle
y_{2}\geq\frac{1}{x_{2}}$ and by using (23) we get that there hold

$\displaystyle y_{1}\geq\frac{1}{x-\varepsilon},\ \displaystyle
y_{2}\geq\frac{1}{x+\varepsilon}\ \mbox{and}\ \displaystyle
y\geq\frac{x}{x^{2}-\frac{\varepsilon^{2}}{4}}.$

Thus if $\displaystyle p_{3}=(x,y)$, then
$y\geq\frac{x}{x^{2}-\frac{\varepsilon^{2}}{4}}$ and $0<\varepsilon<2x$.

If there holds case 2) $|y_{1}-y_{2}|=\varepsilon$, by similar arguments, we
get that there hold $x\geq\frac{y}{y^{2}-\frac{\varepsilon^{2}}{4}}$ and
$0<\varepsilon<2y$, provided that $\displaystyle p_{3}=(x,y)$.

Consequently for any $\varepsilon>0$ and any $\displaystyle p_{1}\in A$,
$p_{2}\in A$, such that $\|p_{1}-p_{2}\|=\varepsilon$ we get that
$p_{3}=\frac{p_{1}+p_{2}}{2}\in P(\varepsilon)=C(\varepsilon)\cup
D(\varepsilon)$, where

$C(\varepsilon)=\left\\{(x,y)\in\mathbb{R}^{2}:x\geq\frac{y}{y^{2}-\frac{\varepsilon^{2}}{4}},0<\varepsilon<2y\right\\}$

and

$D(\varepsilon)=\left\\{(x,y)\in\mathbb{R}^{2}:y\geq\frac{x}{x^{2}-\frac{\varepsilon^{2}}{4}},0<\varepsilon<2x\right\\}.$

Thus to show that $A$ is uniformly convex about $\phi$ it is enough to
estimate $\phi(p,\varepsilon)$, for $p\in P(\varepsilon)$.

For any $\varepsilon>0$ and $p\in P(\varepsilon)$ there holds the inequality
$\displaystyle\phi(p,\varepsilon)\leq{\rm dist}\left(p,\overline{A}\right)$,
where $\overline{A}=\left\\{(x,y)\in\mathbb{R}^{2}:\frac{1}{x}=y,x>0\right\\}$
and consequently $B(p,\phi(p,\varepsilon))\in A$.

######  Theorem 42.

Let $(X,\|\cdot\|)$ be a Banach space, $A,B\subset X$,
$\phi:A\times\mathbb{R}^{+}\rightarrow\mathbb{R}^{+}$ has the positive
property about $A$ and $A$ be uniformly convex set about $\phi$. Then the
ordered pair $(A,B)$ satisfies the $BUC$ property.

Proof. We will prove the theorem by assuming the contrary, i.e. let the
ordered pair $(A,B)$ does not satisfy the $BUC$ property and $A$ be uniformly
convex set about $\phi$. Then there are three bounded sequences
$\left\\{x_{n}\right\\}_{n=1}^{\infty},\left\\{z_{n}\right\\}_{n=1}^{\infty}\subset
A$ and $\left\\{y_{n}\right\\}_{n=1}^{\infty}\subset B$ such that

$\lim_{n\to\infty}\|x_{n}-y_{n}\|=\lim_{n\to\infty}\|z_{n}-y_{n}\|={\rm
dist}(A,B),$ (24)

but the sequence $\\{\|x_{n}-z_{n}\|\\}_{n=1}^{\infty}$ does not converge to
zero.

Therefore there exists $\varepsilon_{0}>0$ such that for any $N\in\mathbb{N}$
there is $n>N$ so that the inequality
$\left\|x_{n}-z_{n}\right\|\geq\epsilon_{0}$ holds true. Thus there is a
subsequence $\\{\|x_{n_{k}}-z_{n_{k}}\|\\}_{k=1}^{\infty}$ so that
$\|x_{n_{k}}-z_{n_{k}}\|\geq\varepsilon_{0}$ for all $k\in\mathbb{N}$. The
sequence $\left\\{\frac{x_{n_{k}}+z_{n_{k}}}{2}\right\\}_{k=1}^{\infty}\in A$
is a bounded sequence too. By the assumption that $\phi$ has the positive
property about $A$ and that
$\left\\{\frac{x_{n_{k}}+z_{n_{k}}}{2}\right\\}_{k=1}^{\infty}\in A$ is a
bounded sequence it follows that there exists $\delta_{0}$, such that the
inequality

$\inf\left\\{\phi(a,\varepsilon):a\in\left\\{\frac{x_{n}+z_{n}}{2}\right\\}_{n=1}^{\infty},\varepsilon\geq\varepsilon_{0}\right\\}=\delta_{0}>0$
(25)

holds true.

From the assumption that $A$ is uniformly convex set about $\phi$ and the fact
that $\left\\{\frac{x_{n_{k}}+z_{n_{k}}}{2}\right\\}_{k=1}^{\infty}\in A$ is a
bounded sequence, satisfying the inequality
$\|x_{n_{k}}-z_{n_{k}}\|\geq\varepsilon_{0}$ we get the inclusions

$B\left(\frac{x_{n}+z_{n}}{2},\delta\right)\subseteq
B\left(\frac{x_{n}+z_{n}}{2},\phi\left(\frac{x_{n}+z_{n}}{2},\varepsilon_{0}\right)\right)\subset
A.$

By (24) it follows that for $\delta_{0}$ there exists $N_{1}\in\mathbb{N}$
such that for every $k\geq N_{1}$ the inequality

$\frac{\|x_{n_{k}}-y_{n_{k}}\|+\|z_{n_{k}}-y_{n_{k}}\|}{2}<\delta_{0}+{\rm
dist}(A,B)$ (26)

holds true. Consequently for all $k\geq N_{1}$ there holds the chain of
inequalities

$\begin{array}[]{lll}\delta_{0}+{\rm dist}(A,B)&\leq&{\rm
dist}\displaystyle\left(\frac{x_{n_{k}}+z_{n_{k}}}{2},B\right)\leq\left\|\frac{x_{n_{k}}+z_{n_{k}}}{2}-y_{n}\right\|\\\\[10.0pt]
&\leq&\displaystyle\frac{\|x_{n_{k}}-y_{n_{k}}\|+\|z_{n_{k}}-y_{n_{k}}\|}{2},\end{array}$

which is a contradiction and thus the sequence
$\\{\|x_{n}-z_{n}\|\\}_{n=1}^{\infty}$ converges to zero. $\Box$

######  Example 43.

Let us consider the space $X=(\mathbb{R}^{2},\|\cdot\|_{\infty})$ and the sets
$\displaystyle A=\left\\{(x,y)\in\mathbb{R}^{2}:\frac{1}{x}\leq
y,x>0\right\\}$, $\displaystyle
B=\left\\{(x,y)\in\mathbb{R}^{2}:\frac{1}{x+1}-1\geq y,x>-1\right\\}$, then
the ordered pair $(A,B)$ have $BUC$ property, but not have $UC$ property

We have proven in Examples 39 and 41 that there exists a function $\phi$, so
that $\phi$ has positive property about $A$ and $A$ is uniformly convex about
$\phi$. By Theorem 42 it follows that $(A,B)$ has the $BUC$ property, i.e. for
any two bounded sequences
$\\{x_{n}\\}_{n=1}^{\infty},\\{z_{n}\\}_{n=1}^{\infty}\subset A$ and a
sequence $\\{y_{n}\\}_{n=1}^{\infty}\subset B$, such that
$\lim_{n\to\infty}\|x_{n}-y_{n}\|=\lim_{n\to\infty}\|z_{n}-y_{n}\|={\rm
dist}(A,B)$ and the sequence $\\{\|x_{n}-z_{n}\|\\}_{n=1}^{\infty}$ converges
to zero.

It remains to show that $(A,B)$ has not the $UC$ property, i.e. we will show
that there are three unbounded sequences
$\\{x_{n}\\}_{n=1}^{\infty},\\{z_{n}\\}_{n=1}^{\infty}\subset A$ and a
sequence $\\{y_{n}\\}_{n=1}^{\infty}\subset B$, such that
$\lim_{n\to\infty}\|x_{n}-y_{n}\|=\lim_{n\to\infty}\|z_{n}-y_{n}\|={\rm
dist}(A,B)$, but the sequence $\\{\|x_{n}-z_{n}\|\\}_{n=1}^{\infty}$ does not
converge to zero.

Let consider the sequences $\\{a_{n}\\}_{n=0}^{\infty},\
\\{b_{n}\\}_{n=0}^{\infty}$ and $\\{c_{n}\\}_{n=0}^{\infty}$, defined as
follows

$\textstyle a_{n}=\left(n\ ,\ \frac{1}{n}\right)\ ,\ b_{n}=\left(n+1\ ,\
\frac{1}{n+1}\right)\in A\ ,\ c_{n}=\left(n\ ,\ \frac{1}{n+1}-1\right)\in B,\
\mbox{for}\ n\in\mathbb{N}.$

From $\displaystyle\|a_{n}-c_{n}\|_{\infty}=\left\|\left(0\ ,\
\frac{1}{n}-\frac{1}{n+1}+1\right)\right\|_{\infty}=\frac{1}{n}-\frac{1}{n+1}+1$
and $\displaystyle\|b_{n}-c_{n}\|=\left\|\left(1\ ,\
1\right)\right\|_{\infty}=1$ it follows that
$\displaystyle\lim_{n\to\infty}\|a_{n}-c_{n}\|_{\infty}=1$ and
$\displaystyle\lim_{n\to\infty}\|b_{n}-c_{n}\|_{\infty}=1$.

By $\displaystyle\|b_{n}-a_{n}\|_{\infty}=\left\|\left(1\ ,\
\frac{1}{n+1}-\frac{1}{n}\right)\right\|_{\infty}\geq 1$ we get that
$\displaystyle\lim_{n\to\infty}\|b_{n}-a_{n}\|_{\infty}=0\neq 0$.

From ${\rm dist}(A,B)=1$ it follows that
$\displaystyle\lim_{n\to\infty}\|a_{n}-c_{n}\|_{\infty}={\rm dist}(A,B)$,
$\displaystyle\lim_{n\to\infty}\|b_{n}-c_{n}\|_{\infty}={\rm dist}(A,B)$ and
$\displaystyle\lim_{n\to\infty}\|b_{n}-a_{n}\|_{\infty}\neq 0$ and
consequently the ordered pair $(A,B)$ has not the $UC$ property.

##  5\. The property $UC$ and uniform convexity of the underlying space

We will show that in some cases the validity of the $UC$ property in a Banach
space $(X,\|\cdot\|)$ leads to a conclusion that $(X,\|\cdot\|)$ is a
uniformly convex Banach space. We will start with two technical lemmas.

######  Lemma 44.

Let $(X,\|\cdot\|)$ be a Banach space. Let
$\\{x_{n}\\}_{n=1}^{\infty},\\{y_{n}\\}_{n=1}^{\infty}\in X$ be sequences,
such that for any $n\in\mathbb{N}$ there hold $\|x_{n}\|\leq a$,
$\|y_{n}\|\leq b$ and $\lim_{n\rightarrow\infty}\|x_{n}+y_{n}\|=a+b$, where
$a,b\in[0,+\infty)$. Then $\lim_{n\rightarrow\infty}\|x_{n}\|=a$ and
$\lim_{n\rightarrow\infty}\|y_{n}\|=b$.

Proof. From the chain of inequalities

$a=\lim_{n\to\infty}(\|x_{n}+y_{n}\|-b)\leq\lim_{n\to\infty}(\|x_{n}+y_{n}\|-\|y_{n}\|)\leq\lim_{n\to\infty}\|x_{n}\|\leq
a$

it follows that $\lim_{n\rightarrow\infty}\|x_{n}\|=a$.

By similar arguments we can prove that $\lim_{n\rightarrow\infty}\|y_{n}\|=b$.
$\Box$

######  Lemma 45.

Let $(X,\|.\|)$ be a Banach space and let $B=\\{x\in X:\|x\|\geq 2\\}$. If the
ordered pair $(B_{X},B)$ satisfies the UC property and the sequences
$\\{x_{n}\\}_{n=1}^{\infty},\\{z_{n}\\}_{n=1}^{\infty}\subset A$ be such that
$\lim_{n\rightarrow\infty}\left\|\frac{x_{n}+z_{n}}{2}\right\|=1$, then
$\lim_{n\rightarrow\infty}\|x_{n}-z_{n}\|=0$.

Proof. From the assumptions of the lemma we have that

${\rm dist}(A,B)=1,\ \|x_{n}\|\leq 1,\ \|z_{n}\|\leq 1\ \mbox{and}\
\lim_{n\rightarrow\infty}\|x_{n}+z_{n}\|=2.$ (27)

Therefore by lemma 44 it follows

$\lim_{n\rightarrow\infty}\|x_{n}\|=1,\ \lim_{n\rightarrow\infty}\|z_{n}\|=1.$
(28)

Let us define the sequence $\\{y_{n}\\}_{n=1}^{\infty}$ by
$y_{n}=2\frac{x_{n}+z_{n}}{\|x_{n}+z_{n}\|}$. Then $\|y_{n}\|=2$ for every
$n\in\mathbb{N}$ and consequently $\\{y_{n}\\}_{n=1}^{\infty}\subset B$. Using
the chain of inequalities

$\begin{array}[]{lll}\|y_{n}-x_{n}\|&=&\left\|2\frac{x_{n}+z_{n}}{\|x_{n}+z_{n}\|}-x_{n}\right\|=\left\|2\frac{x_{n}+z_{n}}{\|x_{n}+z_{n}\|}-(x_{n}+z_{n})+z_{n}\right\|\\\\[8.0pt]
&=&\left\|\left(\frac{2}{\|x_{n}+z_{n}\|}-1\right)(x_{n}+z_{n})+z_{n}\right\|\leq\left(\frac{2}{\|x_{n}+z_{n}\|}-1\right)\|x_{n}+z_{n}\|+\|z_{n}\|.\end{array}$

we conclude that $\lim_{n\to\infty}\|y_{n}-x_{n}\|\leq 1$

By (27) and the assumption that $\\{x_{n}\\}_{n=1}^{\infty}\subset A$ we get
the inequality $\|y_{n}-x_{n}\|\geq 1$. Consequently there holds
$\lim_{n\rightarrow\infty}\|y_{n}-x_{n}\|=1={\rm dist}(A,B)$.

By similar arguments we can prove that
$\lim_{n\rightarrow\infty}\|y_{n}-z_{n}\|=1={\rm dist}(A,B)$.

From the assumption that the ordered pair $(B_{X},B)$ satisfies the $UC$
property it follows that $\lim_{n\rightarrow\infty}\|x_{n}-z_{n}\|=0$. $\Box$

######  Theorem 46.

Let $(X,\|\cdot\|)$ be a Banach space and let $B=\\{x\in X:\|x\|\geq 2\\}$. If
the ordered pair $(B_{X},B)$ satisfies the UC property, then $(X,\|\cdot\|)$
is a uniformly convex Banach space.

Proof. Let us assume the contrary, i.e. there exists $\varepsilon>0$ such that
for every $\delta>0$ there are $x(\delta),z(\delta)\in A$, so that the
inequalities $\|x(\delta)-z(\delta)\|\geq\varepsilon$ and
$1-\delta\leq\left\|\frac{x(\delta)+z(\delta)}{2}\right\|\leq 1$ hold true.
Thus we can choose sequences
$\\{x_{n}\\}_{n=1}^{\infty},\\{z_{n}\\}_{n=1}^{\infty}\in A$, satisfying
$\|x_{n}-z_{n}\|\geq\varepsilon$ and
$1-\frac{1}{n}\leq\left\|\frac{x(\delta)+z(\delta)}{2}\right\|\leq 1$.
Therefore
$\displaystyle\lim_{n\to\infty}\left\|\frac{x_{n}+z_{n}}{2}\right\|=1$, which
contradicts with Lemma 45. $\Box$

######  Corollary 47.

Let $(X,\|\cdot\|)$ be a Banach space. If any ordered pair of subsets $(A,B)$
has the $UC$ property, where $A$ is a convex, then $(X,\|\cdot\|)$ is a
uniformly convex Banach space.

##  6\. The property $UC$ and $UCED$ of the underlying space

By Proposition 18 it follows that in $UCED$ Banach spaces any ordered pair
$(A,B)$, such that $A$ is convex and relatively compact set, satisfies the
property $UC$. Unfortunately the sets
$A=\\{(x,y)\in(\mathbb{R}^{2},\|\cdot\|):y\geq x^{2}\\}$, where $\|\cdot\|$ is
a $UCED$ norm or $A=B_{X}$, where $(X,\|\cdot\|)$ is a $UCED$ Banach space are
not relatively compact and therefore for an arbitrary set $B\subset X$ the
ordered pair $(A,B)$ does not satisfy the assumption of Proposition 18. The
next lemma presents a different condition on the sets of the ordered pair
$(A,B)$, by removing the too restrictive assumption the set $A$ to be
relatively compact.

######  Lemma 48.

Let $(X,\|\cdot\|)$ be a Banach space and let $B\subset X/B_{X}$ such that
${\rm dist}(B,B_{X})\geq 1$. If the ordered pair $(B_{X},B)$ satisfies the
$UC$ property, there exists $p\in B$ so that ${\rm dist}(p,B_{X})={\rm
dist}(B,B_{X})$ and there exist two sequences
$\left\\{x_{n}\right\\}_{n=1}^{\infty},\left\\{z_{n}\right\\}_{n=1}^{\infty}\subset
B_{X}$, satisfying $\lim_{n\to\infty}\left\|\frac{x_{n}+z_{n}}{2}\right\|=1$
and $x_{n}+z_{n}=|\lambda_{n}|p$, then
$\lim_{n\rightarrow\infty}\|x_{n}-z_{n}\|=0$.

Proof. From the assumptions we have that $\|x_{n}\|\leq 1$ , $\|z_{n}\|\leq 1$
and $\lim_{n\rightarrow\infty}\|x_{n}+z_{n}\|=2$. Therefore from Lemma 44 it
follows that

$\lim_{n\rightarrow\infty}\|x_{n}\|=1\ ,\
\lim_{n\rightarrow\infty}\|z_{n}\|=1\ \mbox{and}\
\lim_{n\rightarrow\infty}\|x_{n}+z_{n}\|=2.$ (29)

Let us define the sequence $\\{y_{n}\\}_{n=1}^{\infty}$ by $y_{n}=p$ for every
$n\in\mathbb{N}$.

From the assumption $x_{n}+z_{n}=|\lambda_{n}|p$ it follows that
$\|p-(x_{n}+z_{n})\|=|\|p\|-\|x_{n}+z_{n}\||$.

After using the inequalities $\|x_{n}+z_{n}\|\leq\|x_{n}\|+\|z_{n}\|\leq 2$,
${\rm dist}(B,B_{X})+1\geq 2$, the equality $\|p\|={\rm dist}(p,B_{X})+1={\rm
dist}(B,B_{X})+1$ and (29) we can write the chain of inequalities

$\begin{array}[]{lll}{\rm
dist}(B,B_{X})&\leq&\lim_{n\to\infty}\|y_{n}-x_{n}\|=\displaystyle\lim_{n\to\infty}\|p-x_{n}\|\\\
&=&\displaystyle\lim_{n\to\infty}\|p-(x_{n}+z_{n})+z_{n}\|\\\
&\leq&\displaystyle\lim_{n\to\infty}\left(\|p-(x_{n}+z_{n})\|+\|z_{n}\|\right)\\\
&=&\displaystyle\lim_{n\to\infty}\left(|\|p\|-\|x_{n}+z_{n}\||+\|z_{n}\|\right)\\\
&=&\displaystyle\lim_{n\to\infty}\left(|{\rm
dist}(B,B_{X})+1-\|x_{n}+z_{n}\||+\|z_{n}\|\right)\\\ &\leq&{\rm
dist}(B,B_{X})+1+\displaystyle\lim_{n\to\infty}\left(\|z_{n}\|-\|x_{n}+z_{n}\|\right)={\rm
dist}(B,B_{X}).\end{array}$

Thus $\lim_{n\to\infty}\|y_{n}-x_{n}\|={\rm dist}(B_{X},B)$.

By similar arguments we can prove that $\lim_{n\to\infty}\|y_{n}-z_{n}\|={\rm
dist}(B_{X},B)$.

From the assumption that the ordered pair $(B_{X},B)$ satisfies the $UC$
property it follows that $\lim_{n\to\infty}\|x_{n}-z_{n}\|=0$. $\Box$

######  Definition 49.

We say that a Banach space $(X,\|\cdot\|)$ is uniformly convex in the
direction $z$ if $\delta_{\|\cdot\|}(z,\varepsilon)>0$ for every
$\varepsilon\in(0,2]$.

######  Theorem 50.

Let $(X,\|\cdot\|)$ be a Banach space and let $B\subset X/B_{X}$. If there
holds ${\rm dist}(B,B_{X})\geq 1$, the ordered pair $(B_{X},B)$ satisfies the
$UC$ property and there is $p\in B$ so that ${\rm dist}(p,B_{X})={\rm
dist}(B,B_{X})$, then $(X,\|\cdot\|)$ is a uniformly convex in the direction
$\frac{p}{\|p\|}$.

Proof. Let us assume the contrary, i.e. there exists $\varepsilon>0$ such that
for every $\delta>0$ there are $x(\delta),z(\delta)\in B_{X}$, satisfying
$\|x(\delta)-z(\delta)\|\geq\varepsilon$,
$1-\delta\leq\left\|\frac{x(\delta)+z(\delta)}{2}\right\|\leq 1$ and
$x(\delta)+z(\delta)=\lambda(\delta)p$. Thus we can choose sequences
$\\{x_{n}\\}_{n=1}^{\infty},\\{z_{n}\\}_{n=1}^{\infty}\in B_{X}$ so that
$\|x_{n}-z_{n}\|\geq\varepsilon$,
$\lim_{n\to\infty}\left\|\frac{x_{n}+z_{n}}{2}\right\|=1$ and
$x_{n}+z_{n}=\lambda_{n}p$.

Let us construct the sequences
$\\{x^{\prime}_{n}\\}_{n=1}^{\infty}$,$\\{z^{\prime}_{n}\\}_{n=1}^{\infty}\subset
B_{X}$ as follows: $x^{\prime}_{n}=x_{n}$, $z^{\prime}_{n}=z_{n}$ if
$\lambda_{n}\geq 0$ and $x^{\prime}_{n}=-x_{n}$, $z^{\prime}_{n}=-z_{n}$ if
$\lambda_{n}<0$. Then
$\\{x^{\prime}_{n}\\}_{n=1}^{\infty}\,\\{z^{\prime}_{n}\\}_{n=1}^{\infty}\in
B_{X}$, $\|x^{\prime}_{n}-z^{\prime}_{n}\|\geq\varepsilon$,
$\lim_{n\to\infty}\left\|\frac{x^{\prime}_{n}+z^{\prime}_{n}}{2}\right\|=1$
and $x^{\prime}_{n}+z^{\prime}_{n}=|\lambda_{n}|p$ which contradicts with
Lemma 48 and consequently $(X,\|\cdot\|)$ is a uniformly convex in the
direction $\frac{p}{\|p\|}$. $\Box$

##  7\. Examples and applications

We will finish with an example of a cyclic map, which satisfies the conditions
of Theorem 14, but the ordered pair of sets $(A,B)$, that are the domain of
the map $T$, are just $BUC$ but not $UC$. Therefore we can not conclude the
existence of best proximity points using the result of [14].

######  Example 51.

Let us consider the space $X=(\mathbb{R}^{2},\|\cdot\|_{\infty})$, the sets

$\displaystyle A=\left\\{(x,y):\frac{1}{x}\leq y,x>0\right\\}$

and

$B=B_{1}\cup B_{2},$

where

$\displaystyle B_{1}=\left\\{(x,y):\frac{1}{x+1}-1\geq y\ and\ x>-1\right\\}\
\mbox{and}\ \displaystyle
B_{2}=\left\\{(x,y):y\in\mathbb{R},x\leq-1\right\\}.$

Let us denote

$\displaystyle\overline{A}=\left\\{(x,y):\frac{1}{x}=y,x>0\right\\},$

and

$\displaystyle\overline{B}=\left\\{(x,y):\frac{1}{x+1}-1=y\ and\
x>-1\right\\},$

Let the map $T:A\cup B\rightarrow A\cup B$ be defined by

$Tx=\left\\{\begin{array}[]{lcr}\displaystyle\left(-\frac{{\rm
dist}(x,\overline{A})}{2},-\frac{{\rm dist}(x,\overline{A})}{2}\right)&,&x\in
A\\\\[16.0pt] \displaystyle\left(1+\frac{{\rm
dist}(x,\overline{B})}{2},1+\frac{{\rm dist}(x,\overline{B})}{2}\right)&,&x\in
B.\end{array}\right.$ (30)

Then $T$ is a cyclic map on $A\cup B$ and satisfies the inequality

$\rho(Tx,Ty)\leq\frac{1}{2}\rho(x,y)+\frac{1}{2}{\rm dist}(A,B)$ (31)

for every $x\in A$ , $y\in B$, i.e. Theorems 31 with $k=1/2$.

We will prove first that $T$ is a cyclic map, i.e. $T(A)\subseteq B$ and
$T(B)\subseteq A$.

Let first $z=(x,y)\in A$. Then $\displaystyle
Tz=\displaystyle\left(-\frac{{\rm dist}(z,\overline{A})}{2},-\frac{{\rm{\rm
dist}}(z,\overline{A})}{2}\right)$.

By $\displaystyle-\frac{{\rm dist}(z,\overline{A})}{2}\leq 0$ it follows that
for any $z\in A$, if $\displaystyle-\frac{{\rm
dist}(z,\overline{A})}{2}\leq-1$ there holds the inclusion
$\left(-\displaystyle\frac{{\rm dist}(z,\overline{A})}{2},-\frac{{\rm
dist}(z,\overline{A})}{2}\right)\in B_{2}$. If $-\frac{{\rm
dist}(z,\overline{A})}{2}\in(-1,0]$, then $\left(-\displaystyle\frac{{\rm
dist}(z,\overline{A})}{2},-\frac{{\rm dist}(z,\overline{A})}{2}\right)\in
B_{1}$. Thus $Tz\in B$ for any $z\in A$.

Let $z=(x,y)\in B$. Then $\displaystyle Tz=\left(1+\frac{{\rm
dist}(z,\overline{B})}{2},1+\frac{{\rm dist}(z,\overline{B})}{2}\right)$.

From $\displaystyle 1+\frac{{\rm dist}(z,\overline{B})}{2}\geq 1$ and
$\displaystyle\frac{1}{x}\leq 1$ for $x\geq 1$ it follows

$\displaystyle\left(1+\frac{{\rm dist}(z,\overline{B})}{2},1+\frac{{\rm
dist}(z,\overline{B})}{2}\right)\in A,$

i.e. $Tz\in A$.

It remains to show that the map $T$ satisfies the inequality (31).

It is easy to calculate that

${\rm dist}(A,B)=\inf\\{\|x-y\|_{\infty}:x\in A,y\in B\\}=1$

and

${\rm dist}(a,\overline{A})+{\rm dist}(A,B)+{\rm
dist}(b,\overline{B})\leq\|a-b\|_{\infty}.$

Let $a=(x_{a},y_{a})\in A$ and $b=(x_{b},y_{b})\in B$, then

$\begin{array}[]{lll}\textstyle\|Ta-
Tb\|_{\infty}&=&\textstyle\left\|\left(1+\textstyle\frac{{\rm
dist}(b,\overline{B})}{2},1+\frac{{\rm
dist}(b,\overline{B})}{2}\right)-\left(-\textstyle\frac{{\rm
dist}(a,\overline{A})}{2},-\frac{{\rm
dist}(a,\overline{A})}{2}\right)\right\|_{\infty}\\\ &=&1+\frac{{\rm
dist}(b,\overline{B})}{2}+\frac{{\rm dist}(a,\overline{A})}{2}\\\
&=&\frac{1}{2}\left({\rm dist}(A,B)+{\rm dist}(b,\overline{B})+{\rm
dist}(a,\overline{A})\right)+\frac{1}{2}{\rm dist}(A,B)\\\\[8.0pt]
&\leq&\frac{1}{2}\|a-b\|_{\infty}+\frac{1}{2}{\rm dist}(A,B).\end{array}$

The the ordered pair of sets $(A,B)$ has the $BUC$ property, but not the $UC$
property by Example 43 therefore by Theorem 33 it follows that there is a
unique best proximity point of $T$ in $A$. As far as $(A,B)$ has not the $UC$
property we can not use Theorem 14 to conclude that there is a best proximity
point of $T$ in $A$.

We have seen that in order the key lemma from [7] to hold true we need either
to impose good properties on the unit ball $B_{X}$ of the underlying Banach
space $(X,\|\cdot\|)$ or to impose good properties on the set $A$, from the
ordered pair $(A,B)$. The set $A$ in Example 51 is a strictly convex set, i.e.
for any $x,y\in A$, there is $\varepsilon=\varepsilon(x,y)>0$ so that
$B\left(\frac{x+y}{2},\varepsilon\right)\subset A$ and we have proven that any
ordered pair $(A,B)$ has the $BUC$ property.

We will show in the next example, where the underlying space is strictly
convex without being uniformly convex. Thereafter we can not apply Lemma 6. We
will show that the ordered pair $(B_{X},\\{x\in X:\|x\|\geq 2\\})$ has not
either $BUC$ nor $UC$.

######  Example 52.

Let us denote $E_{n}=(R^{2},\|\cdot\|_{n})$, where $\|(x,y)\|_{n}=\root{n}
\of{|x|^{n}+|y|^{n}}$. Any of the spaces $E_{n}$ is a uniformly convex Banach
space. Let us consider the space
$X=\left(\prod_{n=2}^{\infty}E_{n},\|\cdot\|\right)$, where
$\|\\{x_{n}\\}\|=\sqrt{\sum_{n=2}^{\infty}\|x_{n}\|_{n}^{2}}$ and $x_{n}\in
E_{n}$. The space $X$ is a strictly convex Banach space, which is not
uniformly convex, i.e. its unit ball $B_{X}$ is strictly convex set.

Let us denote $B=\\{x\in X:\|x\|\geq 2\\}$. We will construct three sequences
$\left\\{x_{n}\right\\}_{n=1}^{\infty},\\{z_{n}\\}_{n=1}^{\infty}\in B_{x}$
and $\left\\{y_{n}\right\\}_{n=1}^{\infty}\in B$ such that
$\displaystyle\lim_{n\rightarrow\infty}\|x_{n}-y_{n}\|={\rm dist}(B_{X},B)$,
$\displaystyle\lim_{n\rightarrow\infty}\|z_{n}-y_{n}\|={\rm dist}(B_{X},B)$
and $\lim_{n\rightarrow\infty}\|x_{n}-z_{n}\|>0$, i.e. the ordered pair
$(B_{X},B)$ has not the $UC$ property.

Let $r_{x},r_{z}:\mathbb{N}\rightarrow\mathbb{R}^{2}$ be defined by and
$r_{x}(n)=\left(\frac{1}{\sqrt[n]{2}},\frac{1}{\sqrt[n]{2}}\right)$,
$r_{z}(n)=\left(\frac{1}{\sqrt[n]{2}},-\frac{1}{\sqrt[n]{2}}\right)$.

We can see that $\|r_{x}(n)\|_{n}=\|r_{z}(n)\|_{n}=1$.

Let

$x_{n}=(\underbrace{(0,0),(0,0),...,(0,0)}_{n-1},r_{x}(n+1),(0,0),...),$
$z_{n}=(\underbrace{(0,0),(0,0),...,(0,0)}_{n-1},r_{z}(n+1),(0,0),...)$

and

$y_{n}=(\underbrace{(0,0),(0,0),...,(0,0)}_{n-1},(2,0),(0,0),...).$

From $\|x_{n}\|=\|r_{x}(n+1)\|_{n+1}=1$, $\|z_{n}\|=\|r_{z}(n+1)\|_{n+1}=1$
and $\|y_{n}\|=2$ it follows that
$\left\\{x_{n}\right\\}_{n=1}^{\infty},\left\\{z_{n}\right\\}_{n=1}^{\infty}\subset
B_{X}$ and $\left\\{y_{n}\right\\}_{n=1}^{\infty}\subset B$

For any $n\in\mathbb{N}$ there holds

$\begin{array}[]{lll}\|x_{n}-y_{n}\|&=&\|r_{x}(n+1)-(2,0)\|_{n+1}=\left\|\left(\frac{1}{\sqrt[n+1]{2}}-2,\frac{1}{\sqrt[n+1]{2}}\right)\right\|_{n+1}\\\
&=&\sqrt[n+1]{\left(2-\frac{1}{\sqrt[n+1]{2}}\right)^{n+1}+\frac{1}{2}}.\end{array}$

Using the inequalities

$\begin{array}[]{lll}2-\frac{1}{\sqrt[n+1]{2}}&=&\sqrt[n+1]{\left(2-\frac{1}{\sqrt[n+1]{2}}\right)^{n+1}}\leq\sqrt[n+1]{\left(2-\frac{1}{\sqrt[n+1]{2}}\right)^{n+1}+\frac{1}{2}}\\\
&<&\sqrt[n+1]{2\left(2-\frac{1}{\sqrt[n+1]{2}}\right)^{n+1}}=2\sqrt[n+1]{2}-1\end{array}$

for any $n\in\mathbb{N}$, we get $\lim_{n\to\infty}\|x_{n}-y_{n}\|=1$.

We can get that $\lim_{n\to\infty}\|z_{n}-y_{n}\|=1$ in a similar fashion.

From
$\|x_{n}-z_{n}\|=\|r_{x}(n+1)-r_{z}(n+1)\|_{n+1}=\left\|\left(0,\frac{2}{\sqrt[n+1]{2}}\right)\right\|_{n+1}=\frac{2}{\sqrt[n+1]{2}}$
for any $n\in\mathbb{N}$ it follows that
$\displaystyle\lim_{n\rightarrow\infty}\|x_{n}-z_{n}\|=2$.

Thus $(B_{X},B)$ has not the $UC$ property. The sequences in the last example
are bounded ones, and therefore ordered pair $(B_{X},B)$ has not the $BUC$
property, too.

In view of Theorem 42, $B_{X}$ is not uniformly convex about any function
$\phi$. As far as $B_{X}$ is a strictly convex set it follows that there is a
difference between a strict convexity of a set and a uniform convexity of a
set about a function $\varphi$.

## References

  * [1] L. Ajeti, A Ilchev, and B Zlatanov.  On coupled best proximity points in reflexive banach spaces.  Mathematics, 10:1304, 2022.
  * [2] V. Berinde.  Iterative Approximation of Fixed Points.  Springer, 2007.
  * [3] C. Chidume.  Geometric Properties of Banach Spaces and Nonlinear Iterations.  Springer-Verlag, 2009.
  * [4] J.A. Clarkson.  Uniformly convex spaces.  Trans. Amer. Math. Soc., 40:396–414, 2022.
  * [5] M.M. Day, R.C. James, and S. Swaminathan.  Normed linear spaces that are uniformly convex in every direction.  Canadian J. Math., 83:1051–1059, 1972.
  * [6] Y. Dzhabarova, S. Kabaivanov, M. Ruseva, and B. Zlatano.  Existence, uniqueness and stability of market equilibrium in oligopoly markets.  Administrative Sciences, 10:70, 2020.
  * [7] A Eldred and P. Veeramani.  Existence and convergence of best proximity points.  Journal of Mathematical Analysis and Applications, 323:1001–1006, 2006.
  * [8] M. Fabian, P. Habala, P. Hájek, V. Montesinos, and V. Zizler.  Banach Space Theory - The Basis for Linear and Nonlinear Analysis.  Springer, 2010.
  * [9] A. Ilchev and B. Zlatanov.  Error estimates for approximation of coupled best proximity points for cyclic contractive maps.  Applied Mathematics and Computation, 290:412–425, 2016.
  * [10] M. Khamsi and W. Kozlowski.  Fixed Point Theory in Modular Function Spaces.  Birkhäuser, 2015.
  * [11] W. Kirk, P. Srinivasan, and P. Veeramani.  Fixed points for mappings satisfying cyclical contractive conditions.  Fixed Point Theory, 4:79–89, 2003.
  * [12] A. Petruşel.  Fixed points vs. coupled fixed points.  Journal of Fixed Point Theory and Applications, 20:150, 2018.
  * [13] W. Sintunavarat and P. Kumam.  Coupled best proximity point theorem in metric spaces.  Fixed Point Theory and Applications, 2012:93, 2012.
  * [14] T. Suzuki, M. Kikkawa, and C. Vetro.  The existence of best proximity points in metric spaces with the property ${UC}$.  Nonlinear Analysis, Theory, Methods and Applications, 71:2918–2926, 2009.
  * [15] B. Zlatanov.  Error estimates for approximating best proximity points for cyclic contractive maps.  Carpathian Journal of Mathematics, 32:265–270, 2016.
  * [16] B. Zlatanov.  Coupled best proximity points for cyclic contractive maps and their applications.  Fixed Point Theory, 22:431–452, 2021.