Some Existence and Convergence Theorems of Best.pdf

Int. Journal of Math. Analysis, Vol. 6, 2012, no. 52, 2563 - 2578

Some Existence and Convergence Theorems of Best

Proximity Point for New Nonlinear Cyclic Maps

Ing-Jer Lin 1 and Shu-Huei Yang

Department of MathematicsNational Kaohsiung Normal University

Kaohsiung 824, Taiwan

Abstract

In this paper, some new existence and convergence theorems of bestproximity point for new nonlinear cyclic maps are established. Theseresults generalize and improve some main results in [5, 20, 21] andreferences therein.

Mathematics Subject Classification: 54H25

Keywords: MT - function, cyclic map, best proximity point, convergencetheorem

1. Introduction

Let A and B be nonempty subsets of a nonempty set D. A map G : A ∪B → A ∪ B is called a cyclic map if G(A) ⊂ B and G(B) ⊂ A. Let (X, d)be a metric space and T : A ∪ B → A ∪ B a cyclic map. For any nonemptysubsets A and B of X, let

dist(A, B) = inf{d(x, y) : x ∈ A, y ∈ B}.

A point x ∈ A ∪ B is called to be a best proximity point for T if

d(x, Tx) = dist(A, B).

Throughout this paper. we denote by N and R the sets of positive integersand real numbers, respectively.

Recently, the existence, uniqueness and convergence of iterates to the bestproximity point were investigated by many authors; see [1-5, 13-37] and ref-erences therein. In 2006, Eldred and Veeramani [13] first gave the concept ofcyclic contraction as follows.

[email protected]

2564 Ing-Jer Lin and Shu-Huei Yang

Definition 1.1. [13] Let A and B be nonempty subset of a metric space(X, d). A map T : A ∪ B → A ∪ B is called a cyclic contraction if the followconditions hold:

(1) F (A) ⊂ B and F (B) ⊂ A;

(2) There exists k ∈ (0, 1) such that d(Fx, Fy) ≤ kd(x, y)+(1−k)dist(A, B)for all x ∈ A, y ∈ B.

In [13], Eldred and Veeramani first establishd the following interesting bestproximity point theorem.

Theorem 1.1. [13, Proposition 3.2] Let A and B be nonempty closedsubsets of a complete metric space X. Let T : A ∪ B → A ∪ B be a cycliccontraction map, x1 ∈ A and define xn+1 = Txn, n ∈ N. Suppose {xn−1} hasa convergent subsequence in A. The there exists x ∈ A such that d(x, Tx) =dist(A, B).

Very recently, Basha, Shahzad and Jeyaraj [5] first introduced the conceptof K-cyclic and C-cyclic mapping as follows.

Definition 1.2. [5] A pair of mappings T : A → B and S : B → A is saidto form a K-Cyclic mapping between A and B if there exists a nonnegativereal number k < 1

2such that

d(Tx, Sy) ≤ k[d(x, Tx) + d(y, Sy)] + (1 − 2k)d(A, B)

for all x ∈ A and y ∈ B.

Definition 1.3. [5] A pair of mappings T : A → B and S : B → A is saidto form a C-Cyclic mapping between A and B if there exists a nonnegativereal number k < 1

2such that

d(Tx, Sy) ≤ k[d(x, Sy) + d(y, Tx)] + (1 − 2k)d(A, B)

for all x ∈ A and y ∈ B.

In [5], Basha, Shahzad and Jeyaraj proved the following convergence the-orems.

Theorem 1.2. [5] Let A and B be two non-empty subsets of a metricspace. Suppose that the mappings F : A → B and G : B → A form a K-Cyclic map between A and B. For a fixed element x0 in A, let x2n+1 = Fx2n

and x2n = Gx2n−1. Then d(xn, xn+1) → d(A, B).

Existence and convergence theorems 2565

Theorem 1.3. [5] Let A and B be two non-empty subsets of a metricspace. Suppose that the mappings F : A → B and G : B → A form a C-Cyclic map between A and B. For a fixed element x0 in A, let x2n+1 = Fx2n


Motivated by the concepts of K-cyclic and C-cyclic mapping and MT -function, Lin, Lakzian and Chou [20, 21] first introduce the concept of MT -Kand MT -C conditions as follows.

Definition 1.4. (Lin, Lakzian and Chou [20]) Let A and B be non-empty subsets of a metric space (X, d) and T : A → B and S : B → A bemaps. We call the pair of maps T and S satisfy MT -K condition if thereexists an MT -function ϕ : [0,∞) → [0, 1) such that

d(Tx, Sy) ≤ 1

2ϕ(d(x, y)

)[d(x, Tx) + d(y, Sy)] +

(1 − ϕ

(d(x, y)

))d(A, B)

for all x ∈ A, y ∈ B.

Definition 1.5. (Lin, Lakzian and Chou [21]) Let A and B be non-empty subsets of a metric space (X, d) and T : A → B and S : B → A bemaps. We call the pair of maps T and S satisfy MT -C condition if thereexists an MT -function ϕ : [0,∞) → [0, 1) such that

d(Tx, Sy) ≤ 1

2ϕ(d(x, y)

)[d(x, Sy) + d(y, Tx)] +

(1 − ϕ

(d(x, y)

))d(A, B)

for all x ∈ A, y ∈ B.

Theorem 1.4. (Lin, Lakzian and Chou [20]) Let A and B be non-emptysubsets of a metric space (X, d) and T : A → B and S : B → A. If the pair ofmaps T and S satisfy MT − K condition, then there exists a sequence {xn}in X such that

limn→∞

d(xn, xn+1) = d(A, B).

Theorem 1.4. (Lin, Lakzian and Chou [21]) Let A and B be two non-empty subsets of a metric space (X, d) and T : A → B and S : B → A bemaps. If the pair of mapsT and S satisfy MT -C condition, then there existsa sequence xn in X such that

limn→∞

d(xn, xn+1) = d(A, B).


In this paper, we will establish some generalizations of the main results of[20, 21]. Our results also generalizes and improves some results in the literatureand references therein.

2. Preliminaries

Let f be a real-valued function defined on R. For c ∈ R, we recall that

lim supx→c

f(x) = infε>0

sup0<|x−c|<ε

f(x)

and

lim supx→c+

f(x) = infε>0

sup0<x−c<ε

f(x).

Definition 2.1. [6-12] A function ϕ : [0,∞) → [0, 1) is said to be anMT -function (or R-function) if limsups→t+ϕ(s) < 1 for all t ∈ [0,∞).

It is obvious that if ϕ : [0,∞) → [0, 1) is a nondecreasing function ora nonincreasing function, then ϕ is an MT -function. So the set of MT -functions is a quite rich class. But it is worth to mention that there existfunctions which are not MT -functions.

Example 1.1. [11] Let ϕ : [0,∞) → [0, 1) be defined by

ϕ(t) :=

{sin t

t, if t ∈ (0, π

2]

0 , otherwise.

Since lim sups→0+

ϕ(s) = 1, ϕ is not an MT -function.

Very recently, Du [11] first proved some characterizations of MT -functions.

Theorem D. [11] Let ϕ : [0,∞) → [0, 1) be a function. Then the followingstatements are equivalent.

(a) ϕ is an MT -function.

(b) For each t ∈ [0,∞), there exist r(1)t ∈ [0, 1) and ε

(1)t > 0 such that

ϕ(s) ≤ r(1)t for all s ∈ (t, t + ε

(1)t ).

(c) For each t ∈ [0,∞), there exist r(2)t ∈ [0, 1) and ε

(2)t > 0 such that

ϕ(s) ≤ r(2)t for all s ∈ [t, t + ε

(2)t ].


(d) For each t ∈ [0,∞), there exist r(3)t ∈ [0, 1) and ε

(3)t > 0 such that

ϕ(s) ≤ r(3)t for all s ∈ (t, t + ε

(3)t ].

(e) For each t ∈ [0,∞), there exist r(4)t ∈ [0, 1) and ε

(4)t > 0 such that

ϕ(s) ≤ r(4)t for all s ∈ [t, t + ε

(4)t ).

(f) For any nonincreasing sequence {xn}n∈� in [0,∞), we have 0≤ supn∈�

ϕ(xn) ≤1.

(g) ϕ is a function of contractive factor [8]; that is, for any strictlydecreasing sequence {xn}n∈� in [0,∞), we have 0≤ sup

n∈�ϕ(xn) ≤ 1.

3. Main results

In this section, we first establish a convergence theorem which is one of themain results in this paper.

Theorem 3.1. Let A and B be non-empty subsets of a metric space (X, d)and F : A → B and G : B → A be maps. If there exist a nondecreasingfunction μ : [0,∞) → [0,∞) and an MT -function ϕ : [0,∞) → [0, 1) suchthat

d(Fx,Gy) ≤ 1

2ϕ(μ(d(x, y)

))[d(x, Fx) + d(y, Gy)] +

[1 − ϕ

(μ(d(x, y)

))]d(A, B)

for all x ∈ A, y ∈ B, then there exists a sequence {xn} in X such that

limn→∞

d(xn, xn+1) = d(A, B).

Proof. Let x0 ∈ A be given. Define x2n+1 = Fx2n and x2n = Gx2n−1 for

each n ∈ N ∪ {0}. Then x2n ∈ A and x2n+1 ∈ B for each n ∈ N ∪ {0}. By(3.1), we have

d(x1, x2) = d(Fx0, Gx1)

≤ 1

2ϕ(μ(d(x0, x1)

))[d(x0, Fx0) + d(x1, Gx1)

]+

[1 − ϕ

(μ(d(x0, x1)

))]d(A, B)

=1

2ϕ(μ(d(x0, x1)

))[d(x0, x1) + d(x1, x2)

]+

[1 − ϕ

(μ(d(x0, x1)

))]d(A, B),


which implies

[1 − 1

2ϕ(μ(d(x0, x1)

))]d(x1, x2) ≤

[1 − 1

2ϕ(μ(d(x0, x1)

))]d(x0, x1)

+[1 − ϕ

(μ(d(x0, x1)

))]d(A, B). (3.2)

By (3.2), we obtain

d(x1, x2) ≤ϕ(μ(d(x0, x1)

))

2 − ϕ(μ(d(x0, x1)

))d(x0, x1) +[1 −

ϕ(μ(d(x0, x1)

))

2 − ϕ(μ(d(x0, x1)

))]d(A, B),

and hence

d(x1, x2) − d(A, B) ≤ϕ(μ(d(x0, x1)

))

2 − ϕ(μ(d(x0, x1)

))(d(x0, x1) − d(A, B)

).

By (3.1) again, we have

d(x2, x3) = d(Gx1, Fx2)

= d(Fx2, Gx1)

≤ 1

2ϕ(μ(d(x2, x1)

))[d(x2, Fx2) + d(x1, Gx1)

]+

[1 − ϕ

(μ(d(x2, x1)

))]d(A, B)

=1

2ϕ(μ(d(x2, x1)

))[d(x2, x3) + d(x1, x2)

]+

[1 − ϕ

(μ(d(x2, x1)

))]d(A, B),

which implies

[1 − 1

2ϕ(μ(d(x2, x1)

))]d(x2, x3) ≤ 1

2ϕ(μ(d(x2, x1)

))d(x1, x2)

+[1 − ϕ

(μ(d(x2, x1)

))]d(A, B),

and hence

d(x2, x3) ≤ϕ(μ(d(x2, x1)

))

2 − ϕ(μ(d(x2, x1)

))d(x2, x1) +[1 −

ϕ(μ(d(x2, x1)

))

2 − ϕ(μ(d(x2, x1)

))]d(A, B)

or

d(x2, x3) − d(A, B) ≤ϕ(μ(d(x2, x1)

))

2 − ϕ(μ(d(x2, x1)

))[d(x2, x1) − d(A, B)

].


By induction, we have

d(xn, xn+1) − d(A, B) ≤ϕ(μ(d(xn−1, xn)

))

2 − ϕ(μ(d(xn−1, xn)

))[d(xn−1, xn) − d(A, B)

].

Since ϕ(t) < 1 for all t ∈ [0,∞), we have ϕ(t)2−ϕ(t)

< 1 for all t ∈ [0,∞). By (3.3),we get

d(xn, xn+1) − d(A, B) < d(xn−1, xn) − d(A, B),

which implies d(xn, xn+1) < d(xn−1, xn) for all n ∈ N. So, the sequence{d(xn, xn+1)

}is strictly decreasing in [0,∞). Since ϕ is an MT -function,

by Theorem D, we obtain

0 ≤ supn∈�

ϕ(μ(d(xn, xn+1)

))< 1

Let α = supn∈�

ϕ(μ(d(xn, xn+1)

)). So 0 ≤ α < 1. Since

ϕ(μ(d(xn, xn+1)

))≤ α,

we have

2 − ϕ(μ(d(xn, xn+1)

))≥ 2 − α, for all n ∈ N.

So, by (3.4) and (3.5), we get

ϕ(μ(d(xn, xn+1)

))

2 − ϕ(μ(d(xn, xn+1)

)) ≤ α

2 − αfor all n ∈ N.

By (3.6),

0 ≤ supn∈�

ϕ(μ(d(xn, xn+1)

))

2 − ϕ(μ(d(xn, xn+1)

)) ≤ α

2 − α< 1

Let

β = supn∈�

ϕ(μ(d(xn, xn+1

)))

2 − ϕ(μ(d(xn, xn+1)

)) .


Then β ∈ [0, 1). It follows from (3.3) that


))

2 − ϕ(μ(d(xn1, xn)

))(d(xn−1, xn) − d(A, B)

)

≤ β[d(xn−1, xn) − d(A, B)

]≤ β2

[d(xn−2, xn−1) − d(A, B)

]≤ . . .

≤ βn[d(x0, x1) − d(A, B)

]

Since β ∈ [0, 1), we have limn→∞

βn = 0. So, the last inequality implies

limn→∞

d(xn, xn+1) = d(A, B).

The proof is completed. �

The following result is immediate from Theorem 3.1.

Corollary 3.1. (Lin, Lakzian and Chou [20]) Let A and B be non-empty subsets of a metric space (X, d) and T : A → B and S : B → A. If thepair of maps T and S satisfy MT −K condition, then there exists a sequence{xn} in X such that

limn→∞

d(xn, xn+1) = d(A, B).

Corollary 3.2. [5] Let A and B be two non-empty subsets of a metricspace. Suppose that the mappings F : A → B and G : B → A form a K-Cyclic map between A and B. For a fixed element x0 in A, let x2n+1 = Fx2n


Theorem 3.2. Let A and B be non-empty subsets of a metric space (X, d)and F : A → B and G : B → A be maps. Suppose that there exist anondecreasing function μ : [0,∞) → [0,∞) and an MT -function ϕ : [0,∞) →[0, 1) such that

d(Fx,Gy) ≤ 1

2ϕ(μ(d(x, y)

))[d(x, Fx) + d(y, Gy)] +

[1 − ϕ

(μ(d(x, y)

))]d(A, B)

for all x ∈ A, y ∈ B. If x0 ∈ A, define x2n+1 = Fx2n and x2n = Gx2n−1, n ∈ N,then the sequence {xn} is bounded.


Proof. By Theorem 3.1, we have

limn→∞

d(xn, xn+1) = d(A, B).

Since {d(x2n+1, x2n)} is a subsequence of {d(xn, xn+1)}, we have

limn→∞

d(x2n−1, x2n) = d(A, B).

Hence {d(x2n−1, x2n)} is bounded. So there exists γ > 0 such that

d(x2n−1, x2n) ≤ γ for all n ∈ N.

For each n ∈ N, we have

d(x2n, Fx0) = d(Gx2n−1, Fx0)

≤ 1

2ϕ(μ(d(x2n−1, x0)

))[d(x2n−1, Gx2n−1) + d(x0, Fx0)

]+

[1 − ϕ

((d(x2n−1, x0)

))]d(A, B)

<1

2[d(x2n−1, x2n) + d(x0, Fx0)] + d(A, B)

≤ 1

2[γ + d(x0, Fx0)] + d(A, B).

Let

ρ =1

2[γ + d(x0, Fx0)] + d(A, B).

So, x2n ∈ B(Fx0, ρ) for all n ∈ N. For each n ∈ N, since

d(x2n+1, Fx0) ≤ d(x2n, x2n+1) + d(x2n, Fx0) ≤ γ + ρ,

we get x2n+1 ∈ B(Fx0, γ + ρ) for all n ∈ N. On the other hand, since

x2n ∈ B(Fx0, ρ) ⊆ B(Fx0, γ + ρ) for all n ∈ N,

we also have x2n ∈ B(Fx0, ρ) for all n ∈ N. By above, we have

xn ∈ B(Fx0, γ + ρ) for all n ∈ N.

Therefore {xn} is bounded. The proof is completed. �

The following result is a special case of Theorem 3.2.

Corollary 3.3. (Lin, Lakzian and Chou [20]) Let A and B be two non-empty closed subsets of a metric space (X, d) and T : A → B and S : B


→ A be maps. If the pair of maps T and S satisfies MT -K condition. Ifx0 ∈ A, define x2n+1 = Tx2n and x2n = Sx2n−1, n ∈ N then the sequence xn isbounded.

Corollary 3.4. [5] Let A and B be two non-empty closed subsets of ametric space. Suppose that the mappings F : A → B and G : B → A form aK-Cyclic map between A and B. For a fixed element x0 in A, let x2n+1 = Fx2n

and x2n = Gx2n−1. Then the sequence xn is bounded.

Theorem 3.3. Let A and B be non-empty subsets of a metric space (X, d)and F : A → B and G : B → A be map. If there exist a nondecreasingfunction μ : [0,∞) → [0,∞) and an MT -function ϕ : [0,∞) → [0, 1) suchthat

d(Fx,Gy) ≤ 1

2ϕ(μ(d(x, y)

))[d(x, Gy) + d(y, Fx)] +

[1 − ϕ

(μ(d(x, y)

))]d(A, B)

(3.7)

for all x ∈ A, y ∈ B, then there exists a sequence {xn} in X such that

limn→∞

d(xn, xn+1) = d(A, B).

Proof. Let x0 ∈ A be given. Define x2n+1 = Fx2n and x2n = Gx2n−1 for

each n ∈ N ∪ {0}. Then x2n ∈ A and x2n+1 ∈ B for each n ∈ N ∪ {0}. By(3.5), we have

d(x1, x2) = d(Fx0, Gx1)

≤ 1

2ϕ(μ(d(x0, x1)

))[d(x0, Gx1) + d(x1, Fx0)

]+

[1 − ϕ

(μ(d(x0, x1)

))]d(A, B)

=1

2ϕ(μ(d(x0, x1)

))[d(x0, x2) + d(x1, x1)

]+

[1 − ϕ

(μ(d(x0, x1)

))]d(A, B)

≤ 1

2ϕ(μ(d(x0, x1)

))[d(x0, x1) + d(x1, x2)

]+

[1 − ϕ

(μ(d(x0, x1)

))]d(A, B),

which implies[1 − 1

2ϕ(μ(d(x0, x1)

))]d(x1, x2) ≤ 1

2ϕ(μ(d(x0, x1)

))d(x0, x1)

+[1 − ϕ

(μ(d(x0, x1)

))]d(A, B). (3.8)

By (3.8), we obtain

d(x1, x2) ≤ϕ(μ(d(x0, x1)

))

2 − ϕ(μ(d(x0, x1)

))d(x0, x1) +[1 −

ϕ(μ(d(x0, x1)

))

2 − ϕ(μ(d(x0, x1)

))]d(A, B).


and hence

d(x1, x2) − d(A, B) ≤ϕ(μ(d(x0, x1)

))

2 − ϕ(μ(d(x0, x1)

))(d(x0, x1) − d(A, B)

).

By (3.7) again, we have

d(x2, x3) = d(Gx1, Fx2)

= d(Fx2, Gx1)

≤ 1

2ϕ(μ(d(x2, x1)

))[d(x2, Gx1) + d(x1, Fx2)

]+

[1 − ϕ

(μ(d(x2, x1)

))]d(A, B)

=1

2ϕ(μ(d(x2, x1)

))[d(x2, x2) + d(x1, x3)

]+

[1 − ϕ

(μ(d(x2, x1)

))]d(A, B)

≤ 1

2ϕ(μ(d(x2, x1)

))[d(x1, x2) + d(x2, x3)

]+

[1 − ϕ

(μ(d(x2, x1)

))]d(A, B),

which implies[1 − 1

2ϕ(μ(d(x2, x1)

))]d(x2, x3) ≤ 1

2ϕ(μ(d(x2, x1)

))d(x1, x2)

+[1 − ϕ

(μ(d(x2, x1)

))]d(A, B),

and hence

d(x2, x3) ≤ϕ(μ(d(x2, x1)

))

2 − ϕ(μ(d(x2, x1)

))d(x2, x1) +[1 −

ϕ(μ(d(x2, x1)

))

2 − ϕ(μ(d(x2, x1)

))]d(A, B),

or

d(x2, x3) − d(A, B) ≤ϕ(μ(d(x2, x1)

))

2 − ϕ(μ(d(x2, x1)

))[d(x2, x1) − d(A, B)

].

By induction, we have


))

2 − ϕ(μ(d(xn−1, xn)

))[d(xn−1, xn) − d(A, B)

].

(3.9)

Since ϕ(t) < 1 for all t ∈ [0,∞), we have ϕ(t)2−ϕ(t)

< 1 for all t ∈ [0,∞). By (3.9),we get

d(xn, xn+1) − d(A, B) < d(xn−1, xn) − d(A, B),


which implies d(xn, xn+1) < d(xn−1, xn) for all n ∈ N. So, the sequence{d(xn, xn+1)

}is a strictly decreasing in [0,∞). Since ϕ is an MT -function,

by Theorem D, we obtain

0 ≤ supn∈�

ϕ(μ(d(xn, xn+1)

))< 1

Let α = supn∈�

ϕ(μ(d(xn, xn+1)

)). So 0 ≤ α < 1. Since

ϕ(μ(d(xn, xn+1)

))≤ α, (3.10)

we have

2 − ϕ(μ(d(xn, xn+1)

))≥ 2 − α, for all n ∈ N. (3.11)

So, by (3.10) and (3.11), we get

ϕ(μ(d(xn, xn+1)

))

2 − ϕ(μ(d(xn, xn+1)

)) ≤ α

2 − αfor all n ∈ N. (3.12)

By, (3.12),

0 ≤ supn∈�

ϕ(μ(d(xn, xn+1)

))

2 − ϕ(μ(d(xn, xn+1)

)) ≤ α

2 − α< 1

Let

β = supn∈�

ϕ(μ(d(xn, xn+1)

))

2 − ϕ(μ(d(xn, xn+1)

)) .

Then β ∈ [0, 1). It follows from (3.9) that


))

2 − ϕ(μ(d(xn1, xn)

))(d(xn−1, xn) − d(A, B)

)

≤ β[d(xn−1, xn) − d(A, B)

]≤ β2

[d(xn−2, xn−1) − d(A, B)

]≤ . . .

≤ βn[d(x0, x1) − d(A, B)

]


Since β ∈ [0, 1), we have limn→∞

βn = 0. So, the last inequality implies

limn→∞

d(xn, xn+1) = d(A, B).

Then proof is completed. �

Corollary 3.5. (Lin, Lakzian and Chou [21]) Let A and B be two non-empty subsets of a metric space (X, d) and T : A → B and S : B → A bemaps. If the pair of mapsT and S satisfy MT -C condition, then there existsa sequence xn in X such that

limn→∞

d(xn, xn+1) = d(A, B).

Corollary 3.6. [5] Let A and B be two non-empty subsets of a metricspace. Suppose that the mappings F : A → B and G : B → A form a C-Cyclic map between A and B. For a fixed element x0 in A, let x2n+1 = Fx2n


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Received: May, 2012

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