# Optimal Approximate Solutions of Fixed Point Equations

Authors: S. Sadiq Basha; N. Shahzad; R. Jeyaraj
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

The main objective of this paper is to present some best proximity point theorems for K-cyclic mappings and C-cyclic mappings in the frameworks of metric spaces and uniformly convex Banach spaces, thereby furnishing an optimal approximate solution to the equations of the formTx=x where T is a non-self mapping.


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## Body

## 1. Introduction

Fixed point theorems delve into the existence of a solution to the equations of the formTx=x where T is a self-mapping. However, when T is a nonself-mapping, the equation Tx=x does not necessarily have a solution, in which case best approximation theorems explore the existence of an approximate solution whereas best proximity point theorems analyze the existence of an approximate solution that is optimal. Indeed, a classical and well-known best approximation theorem, due to Fan [1], contends that if K is a nonempty convex compact subset of a Hausdorff topological vector space E and T is a continuous non-self mapping from K to E, then there exists an element x in K such that d(x,Tx)=d(A,B). Subsequently, many authors, including Prolla [2], Reich [3], and Sehgal and Singh [4, 5], accomplished several appealing extensions and variants of the preceding best approximation theorem. Further, Vetrivel et al. [6] elicited a more generalized result that unifies and subsumes many such results. Despite the fact that best approximation theorems produce an approximate solution to the equation Tx=x, they may not render an approximate solution that is optimal. On the contrary, best proximity point theorems are intended to furnish an approximate solution x that is optimal in the sense that the error d(x,Tx) is minimum. Indeed, in light of the fact that d(x,Tx) is at least d(A,B), a best proximity point theorem guarantees the global minimization of d(x,Tx) by the requirement that an approximate solution x satisfies the condition d(x,Tx)=d(A,B). Such optimal approximate solutions are called best proximity points of the mapping T.Eldred et al. [7] have established interesting best proximity point theorems for relatively nonexpansive mappings. A Best proximity point theorem for contractive mapping has been explored in [8]. Best proximity point theorems for various types of contractions have been obtained in [9–13]. Best proximity point theorems for several types of set valued mappings have been derived in [14–25]. Moreover, common best proximity point theorems for pairs of contractions and for pairs of contractive mappings have been elicited in [26].The main objective of this article is to prove some best proximity point theorems for K-cyclic mappings and C-cyclic mappings in the frameworks of metric spaces and uniformly convex Banach spaces, thereby furnishing an optimal approximate solution to the equations of the formTx=x where T is a non-self-K-cyclic mapping or a non-self-C-cyclic mapping.


## 2. Preliminaries

The following notions will be used in the sequel.Definition 2.1.
A pair of mappingsT:A→B and S:B→A is said to form a K-Cyclic mapping between A and B if there exists a nonnegative real number k<1/2 such that
(2.1)d(Tx,Sy)≤k[d(x,Tx)+d(y,Sy)]+(1-2k)d(A,B),
for all x∈A and y∈B.Definition 2.2.
A pair of mappingsT:A→B and S:B→A is said to form a C-Cyclic mapping between A and B if there exists a non-negative real number k<1/2 such that
(2.2)d(Tx,Sy)≤k[d(x,Sy)+d(y,Tx)]+(1-2k)d(A,B),
for all x∈A and y∈B.Definition 2.3.
A subsetC of a metric space is said to be boundedly compact if every bounded sequence in C has a subsequence converging to some element in C.It is evident that every compact set is boundedly compact but the converse is not true.


## 3. K-Cyclic Mappings

This section is concerned with best proximity point theorems for K-cyclic non-self mappings.Lemma 3.1.
LetA and B be two non-empty subsets of a metric space. Suppose that the mappings T:A→B and S:B→A form a K-Cyclic map between A and B. For a fixed element x0 in A, let x2n+1=Tx2n and x2n=Sx2n-1. Then, d(xn,xn+1)→d(A,B).Proof.
AsT and S form a K-Cyclic map,
(3.1)d(x1,x2)=d(Tx0,Sx1)≤k[d(x0,Tx0)+d(x1,Sx1)]+(1-2k)d(A,B)=k[d(x0,x1)+d(x1,x2)]+(1-2k)d(A,B).
So, it follows that d(x1,x2)≤(k/(1-k))d(x0,x1)+[1-(k/(1-k))]d(A,B).
Similarly, it can be seen that(3.2)d(x2,x3)≤(k1-k)2d(x0,x1)+[1-(k1-k)2]d(A,B).
Hence, it follows by induction that
(3.3)d(xn,xn+1)≤(k1-k)nd(x0,x1)+[1-(k1-k)n]d(A,B).
Therefore, d(xn,xn+1)→d(A,B) because of the fact that k<1/2.Lemma 3.2.
LetA and B be non-empty closed subsets of a metric space. Let the mappings T:A→B and S:B→A form a K-Cyclic map between A and B. For a fixed element x0 in A, let x2n+1=Tx2n and x2n=Sx2n-1. Then, the sequence {xn} is bounded.Proof.
It follows from Lemma3.1 that d(x2n-1,x2n) is convergent and hence it is bounded. Further, since S and T form a K-cyclic mapping, it follows that
(3.4)d(x2n,Tx0)≤k[d(x2n-1,x2n)+d(x0,Tx0)]+(1-2k)d(A,B).
Therefore, the subsequence {x2n} is bounded. Similarly, it can be shown that {x2n+1} is also bounded.Lemma 3.3.
LetA and B be non-empty closed subsets of a metric space. Let the mappings T:A→B and S:B→A form a K-Cyclic map between A and B. For a fixed element x0 in A, let x2n+1=Tx2n and x2n=Sx2n-1. Suppose that the sequence {x2n} has a subsequence converging to some element x in A. Then, x is a best proximity point of T.Proof.
Suppose that a subsequence{x2nk} converges to x in A. It follows from Lemma 3.1 that d(x2nk-1,x2nk) converges to d(A,B). As S and T form a K-cyclic mapping, it follows that
(3.5)d(A,B)≤d(x2nk,Tx)≤k[d(x2nk-1,x2nk)+d(x,Tx)]+(1-2k)d(A,B).
Therefore, d(x,Tx)=d(A,B).The preceding two lemmas yield the following best proximity point theorem for K-cyclic mappings in the setting of metric spaces.Corollary 3.4.
LetA and B be two non-empty and closed subsets of a metric space. Let the mappings T:A→B and S:B→A form a K-Cyclic map between A and B. If A is boundedly compact, then T has a best proximity point.The following lemma, due to Eldred and Veeramani [10], will be required subsequently to establish the next best proximity point theorem of this section.Lemma 3.5.
LetA be a non-empty, closed, and convex subset and B be a non-empty and closed subset of a uniformly convex Banach space. Suppose that {xn} and {yn} are sequences in A and {zn} is a sequence in B satisfying the following conditions:(a)
∥yn-zn∥→d(A,B),(b)
for everyϵ>0, ∥xm-zn∥≤d(A,B)+ϵ, for sufficiently large values of m and n.
Then, for everyϵ>0, ∥xm-yn∥≤ϵ for sufficiently large values of m and n.The following best proximity point theorem is for K-cyclic mappings in the setting of uniformly convex Banach spaces.Theorem 3.6.
LetA and B be non-empty, closed, and convex subsets of a uniformly convex Banach space. If the mappings T:A→B and S:B→A form a K-Cyclic map between A and B, then there exist a unique element x∈A and a unique element y∈B such that
(3.6)d(x,Tx)=d(A,B),d(y,Sy)=d(A,B),d(x,y)=d(A,B).
Further, if x0 is any fixed element in A, x2n+1=Tx2n and x2n=Sx2n-1, then the sequences {x2n} and {x2n+1} converge to the best proximity points x and y, respectively.Proof.
It follows from Lemma3.1 that
(3.7)‖x2m-1-x2m‖⟶d(A,B),‖x2n-x2n+1‖⟶d(A,B).
Therefore, for every ϵ>0,
(3.8)∥x2m-1-x2m∥<d(A,B)+ϵ2k,∥x2n-x2n+1∥<d(A,B)+ϵ2k,
for sufficiently large values of m and n. As T and S form a K-cyclic mapping,
(3.9)∥x2m-x2n+1∥≤k[∥x2m-1-x2m∥+∥x2n-x2n+1∥]+(1-2k)d(A,B)<d(A,B)+ϵ,
for sufficiently large values of m and n. Thus, {x2n} is a Cauchy sequence by Lemma 3.5. Since the space is complete, {x2n} converges to some element x∈A, which becomes a best proximity point of the mapping T by Lemma 3.3. Similarly, {x2n+1} converges to some element y∈B, which is a best proximity point of the mapping S. Further, d(Tx,Sy)≤k[d(x,Tx)+d(y,Sy)]+(1-2k)d(A,B)=d(A,B). Therefore, d(Tx,Sy)=d(A,B). By strict convexity of the space, Tx and y should be identical, and Sy and x should be identical. Consequently, d(x,y)=d(A,B). To prove the uniqueness, let us suppose that there exists another element x* such that
(3.10)∥x*-Tx*∥=d(A,B).
Then, ∥Tx*-STx*∥≤k[∥x*-Tx*∥+∥Tx*-STx*∥]+(1-2k)d(A,B). Consequently, ∥Tx*-STx*∥=d(A,B). By strict convexity of the space, STx*=x*. Moreover,
(3.11)∥Tx-x*∥=∥Tx-STx*∥≤k[∥x-Tx∥+∥Tx*-STx*∥]+(1-2k)d(A,B)=d(A,B).
Therefore, ∥Tx-x*∥=d(A,B). By strict convexity of the space, x and x* are identical. This completes the proof of the theorem.The following example illustrates Lemma3.3. Further, it shows that uniqueness of best proximity point is not feasible.Example 3.7.
Consider the nonuniformly convex Banach spaceR2 with the norm ∥(x,y)∥=max{|x|,|y|}.
Let(3.12)A∶={(x,0):0≤x≤1},B∶={(x,1):0≤x≤1}.
Then, d(A,B)=1 and d(u,v)=1 for all u in A and v in B. Let T:A→B and S:B→A be defined as
(3.13)T((x,0))=(x,1),S((x,1))=(x,0).
For any positive number k,
(3.14)‖T((x1,0))-S((x2,1))‖=1=k[∥(x1,0)-T((x1,0))∥+∥(x2,1)-S((x2,1))∥]+(1-2k)d(A,B).
So, the mappings S and T form a K-cyclic mapping. Further, it can be observed that every element of A is a best proximity point of the mapping T.


## 4. C-Cyclic Mappings

This section is concerned with best proximity point theorems for C-cyclic non-self mappings.Lemma 4.1.
LetA and B be two non-empty subsets of a metric space. Suppose that the mappings T:A→B and S:B→A form a C-cyclic mapping between A and B. For a fixed element x0 in A, let x2n+1=Tx2n and x2n=Sx2n-1. Then, d(xn,xn+1)→d(A,B).Proof.
SinceT and S form a C-cyclic mapping,
(4.1)d(x1,x2)=d(Tx0,Sx1)≤k[d(x1,Tx0)+d(x0,Sx1)]+(1-2k)d(A,B)=kd(x0,x2)+(1-2k)d(A,B)≤k[d(x0,x1)+d(x1,x2)]+(1-2k)d(A,B).
So, it follows that d(x1,x2)≤(k/(1-k))d(x0,x1)+[1-(k/(1-k))]d(A,B).
Similarly,d(x2,x3)≤(k/(1-k))2d(x0,x1)+[1-(k/(1-k))2]d(A,B).
It can be shown by induction that(4.2)d(xn,xn+1)≤(k1-k)nd(x0,x1)+[1-(k1-k)n]d(A,B).
Therefore, d(xn,xn+1)→d(A,B) because of the fact that k<1/2.Lemma 4.2.
LetA and B be non-empty closed subsets of a metric space. Let the mappings T:A→B and S:B→A form a C-cyclic map between A and B. For a fixed element x0 in A, let x2n+1=Tx2n and x2n=Sx2n-1. Suppose that the sequence {x2n} has a subsequence converging to some element x in A. Then, x is a best proximity point of T.Proof.
Suppose that a subsequence{x2nk} converges to x in A.  Then, it follows from Lemma 4.1 that d(x2nk-1,x2nk)→d(A,B). Further, we have
(4.3)d(x2nk,Tx)=d(Sx2nk-1,Tx)≤k[d(x2nk-1,Tx)+d(x,Sx2nk-1)]+(1-2k)d(A,B)≤k[d(x2nk-1,x2nk)+d(x2nk,Tx)+d(x,x2nk)]+(1-2k)d(A,B).
So, it follows that
(4.4)d(A,B)≤d(x2nk,Tx)≤(k1-k)[d(x2nk-1,x2nk)+d(x,x2nk)]+[1-(k1-k)]d(A,B).
Letting k→∞, d(x,Tx)=d(A,B). This completes the proof of the Lemma.Lemma 4.3.
LetA and B be non-empty closed subsets of a metric space. Let the mappings T:A→B and S:B→A form a C-cyclic map between A and B. For a fixed element x0 in A, let x2n+1=Tx2n and x2n=Sx2n-1. Then, the sequence {xn} is bounded.Proof.
By Lemma4.1, d(x2n-1,x2n) is convergent and hence it is bounded. Further, we have
(4.5)d(x2n,Tx0)=d(Sx2n-1,Tx0)≤k[d(x2n-1,Tx0)+d(x0,Sx2n-1)]+(1-2k)d(A,B)≤k[d(x2n-1,x2n)+2d(x2n,Tx0)+d(x0,Tx0)+(1-2k)d(A,B).
Therefore, d(x2n,Tx0)≤(k/(1-2k))[d(x2n-1,x2n)+d(x0,Tx0)]+d(A,B).
Therefore, the subsequence{x2n} is bounded. Similarly, it can be shown that {x2n+1} is also bounded.The preceding two lemmas give rise to the following best proximity point theorem for C-cyclic mappings in the setting of metric spaces.Corollary 4.4.
LetA and B be two non-empty and closed subsets of a metric space. Let the mappings T:A→B and S:B→A form a C-cyclic map between A and B. If A is boundedly compact, then T has a best proximity point.The following best proximity point theorem is for C-cyclic mappings in the setting of uniformly convex Banach spaces.Theorem 4.5.
LetA and B be non-empty, closed, and convex subsets of a uniformly convex Banach space. Let the mappings T:A→B and S:B→A form a C-Cyclic map between A and B. If x0 is any fixed element in A, x2n+1=Tx2n and x2n=Sx2n-1, then the sequence {x2n} converges to a best proximity x of T and the sequence {x2n+1} converges to a best proximity point y of S such that d(x,y)=d(A,B).Proof.
It follows from Lemma4.1 that
(4.6)∥x2m-1-x2m∥⟶d(A,B),∥x2n-x2n+1∥⟶d(A,B).
Therefore, for every ϵ>0,
(4.7)∥x2m-1-x2m∥<d(A,B)+ϵ(1-2k)2k,∥x2n-x2n+1∥<d(A,B)+ϵ(1-2k)2k,
for sufficiently large values of m and n. As T and S form a K-cyclic mapping,
(4.8)∥x2m-x2n+1∥=∥Sx2m-1-Tx2n∥≤k[∥x2m-1-Tx2n∥+∥x2n-Sx2m-1∥]+(1-2k)d(A,B)≤k[∥x2m-1-x2m∥+∥x2m-x2n+1∥+∥x2n-x2n+1∥+∥x2n+1-x2m∥]+(1-2k)d(A,B).
Thus, it follows that
(4.9)∥x2m-x2n+1∥≤(k1-2k)[∥x2m-1-x2m∥+∥x2n-x2n+1∥]+d(A,B).
Therefore, it can be concluded that
(4.10)∥x2m-x2n+1∥<d(A,B)+ϵ,
for sufficiently large values of m and n. Thus, {x2n} is a Cauchy sequence by Lemma 3.5. Since the space is complete, {x2n} converges to some element x∈A, which becomes a best proximity point of the mapping T by Lemma 4.2. Similarly, {x2n+1} converges to some element y∈B, which is a best proximity point of the mapping S. Further, d(x2n,x2n+1)→d(x,y). However, by Lemma 4.1, d(x2n,x2n+1)→d(A,B). Consequently, d(x,y)=d(A,B). This completes the proof of the theorem.



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