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Remotest Points and Best Proximity Points in Metric

Spaces

M.

Ahmadi.

Baseri

*

,

H.

Mazaheri

DepartmentofMathematics,YazdUniversity,Yazd,Iran

Received February 28, 2019; Revised April 12, 2019; Accepted April 21, 2019

Copyright c2019 by authors, all rights reserved. Authors agree that this article remains permanently open access under the terms of the Creative Commons Attribution License 4.0 International License

Abstract

In this paper, we give sufficient conditions for the existence of remotest points with recurrence relations. Then, we apply these relations to generalize recent best proximity point theorems. The findings of the present research contribute and enrich previous results reviewed in the literature.

Keywords

RecurrenceRelations,Remotestpoints,PropertyUCf,PropertyUC,BestProximityPoints

MSC(2010):41A65,41A52,46N10.

1

Introduction

LetAandBbe nonempty bounded subsets of a metric space(X, d).The setAis said to be remotest from a pointx∈X,if there exists a pointx0∈Asuch thatδ(x, A) = sup{d(x, y) : y∈A}=d(x, x0).The pointx0is called a remotest point ofAfrom

x.If there is a pair(x0, y0)∈A×Bfor whichd(x0, y0) =δ(A, B), thatδ(A, B)is remotest distance ofAandB, define by

δ(A, B) = sup{d(x, y) : x∈A, y∈B}.

Then the pair(x0, y0)is called a remotest pair forAandBand put

F(A, B) ={(x, y)∈A×B: d(x, y) =δ(A, B)}

as the set of all remotest pairs(A, B).

The problem of characterizing remotest sets has applications in the study of approximation theory and geometry of Banach spaces, see [5, 6, 13]. One could find more results about remotest points in [9, 11, 12].

In the following this research, in 2015, we introduced the concept of remotest points for a cyclic mapT :A∪B →A∪B

i.e.T(A)⊆B and T(B)⊆A.As follows.

Definition 1.1 [2] LetAandBbe nonempty bounded subsets of a metric spaceX andT :A∪B →A∪Bbe a cyclic map.

The pointx∈A∪Bis a remotest point of the mapT,ifd(x, T x) =δ(A, B).

On the other hand, the study of existence of best proximity points for cyclic contraction mappings is very interesting. More precisely, for two given nonempty subsetsAandBof a metric space(X, d), a pointx∈A∪Bis called a best proximity point of cyclic mapT :A∪B →A∪B,ifd(x, T x) =d(A, B)whered(A, B) = inf{d(a, b) :a∈A, b ∈B}.The first result in this area was reported by Kirk and et al [10]. Later, many authors continued investigation and more results have been obtained, such as, [1, 3, 4, 7, 8].

In this paper, we first present recurrence relation as follows,T :A∪B →A∪B is a continuous cyclic contraction map, assumingx0 ∈A,definexn+1 =T xn for everyn∈N∪ {0}and letdn =d(xn, xn+1).Then we find remotest points of the

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Definition 1.2 [14] LetAandBbe nonempty bounded subsets of a metric spaceX.Then(A, B)is said to satisfy the property

U Cif the following holds:

If{xn}and{zn}are sequences inAand{yn}is a sequence inBsuch thatlimn→∞d(xn, yn) =d(A, B)andlimn→∞d(zn, yn) =

d(A, B),thend(xn, zn)→0holds.

2

Remotest Points

We prove the following result which will be needed in what follows.

Proposition 2.1 LetAandBbe nonempty bounded subsets of a metric spaceXand

dn+k≥αkdn+ (1−αk)δ(A, B), (2.1)

whereα, kare constant such that0< α <1and k∈N.Thendn+k→δ(A, B)asn→ ∞.

Proof.Letm∈N.By inequality (2.1),

dkm+k−1≥αkmdk−1+ (1−αk)(Pmi=0−1αki)δ(A, B),

dkm+k−2≥αkmdk−2+ (1−αk)(P m−1

i=0 α

ki)δ(A, B),

dkm+k−3≥αkmdk−3+ (1−αk)(P m−1

i=0 α

ki)δ(A, B),

continuing this process,

dkm+1≥αkmd1+ (1−αk)(P m−1

i=0 α

ki)δ(A, B)and

dkm≥αkmd0+ (1−αk)(P m−1

i=0 α

ki)δ(A, B).

Lettingm → ∞,limm→∞dkm+i ≥ δ(A, B), i = 0,1, ..., k−1.On the other hand, δ(A, B) ≥ limm→∞dkm+i, i =

0,1, ..., k−1.So

lim

m→∞dkm+i=δ(A, B), i= 0,1, ..., k−1.

Thereforedn+k→δ(A, B)asn→ ∞.

In following theorem give remotest point when one of the sets is boundedly compact.

Theorem 2.1 LetAand B be nonempty bounded subsets of a metric spaceX and relation (2.1) holds. If eitherA orB is

boundedly compact, then there existsx∈A∪Bwithd(x, T x) =δ(A, B).

Proof. Letkbe an even positive integer. Sox2n+k ∈ A.SupposeAis boundedly compact. Since the sequence{x2n+k}is

bounded andAis boundedly compact. Hence{x2n+k}has a convergent subsequence inA.Let{x2nl+k}be a subsequence of {x2n+k}withx2nl+k → x ∈ Aas l → ∞. SinceT is a continuous map andd(x2nl+k, T x2nl+k) → d(A, B)asl → ∞. Therefored(x, T x) =δ(A, B).The proof whenkis an odd positive integer andBis boundedly compact is similar.

Example 2.1 Given0< α <1, k ∈N.letAandBbe nonempty bounded subsets oflp, 1 ≤p < ∞wherepis odd number,

defined byA={((1−α2n)e

2n) :n∈N}and B={((1−α2m−1)e2m−1) :m∈N}.Suppose

T((1−α2n)e2n) = (1−α2n+1)e2n+1

and

T((1−α2m−1)e2m−1) = (1−α2m)e2m.

Then

dn+k≥αkdn+ (1−αk)δ(A, B),

whereα, kare constant such that0< α <1and k∈N.

Proof.Hereδ(A, B) = 21/p.

αk((1−α2m)p+ (1−α2n+1)p)1/p+ (1−αk)21/p = ((1 +αk−1−α2m+k)p

+ (1 +αk−1−α2n+1+k)p)1/p

+ (1−αk)21/p

≤ ((1−α2m+k)p+ (1−α2n+1+k)p)1/p

− 21/p(1−αk)

+ (1−αk)21/p

= ((1−α2m+k)p

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Which implies that

dn+k≥αkdn+ (1−αk)21/p.

Definition 2.1 LetAandBbe nonempty bounded subsets of a metric spaceX.Then(A, B)is said to satisfy the propertyU Cf

if the following holds:

If{xn}and{zn}are sequences inAand{yn}is a sequence inBsuch thatlimn→∞d(xn, yn) =δ(A, B)andlimn→∞d(zn, yn) =

δ(A, B),thend(xn, zn)→0holds.

Theorem 2.2 LetAand B be nonempty closed and bounded subsets of a complete metric spaceX such that(A, B) has a

propertyU Cfand relation (2.1) holds. ThenThas a remotest point.

Proof.We show that for each >0,there exists a positive integerN0such that for allm > n≥N0,

d(x2m+k, x2n+k+1)> δ(A, B)−. (2.2)

On the contrary, assume that there exists0>0such that for eachl≥1,there isml> nl≥lsatisfying

d(x2ml+k, x2nl+k+1)≤δ(A, B)−0 (2.3)

and

d(x2ml−2+k, x2nl+k+1)> δ(A, B)−0. (2.4)

It follows from (2.3) and (2.4) that

d(x2ml−2+k, x2nl+k+1)−d(x2ml+k, x2ml−2+k) ≤ d(x2ml+k, x2nl+k+1)

≤ δ(A, B)−0.

Lettingl→ ∞,propertyU Cf implies

lim

k→∞d(x2ml+k, x2nl+k+1) =δ(A, B)−0. (2.5)

By relation (2.1),

δ(A, B)−0 ≥ d(x2ml+2k, x2nl+2k+1)

≥ αkd(x2ml+k, x2nl+k+1) + (1−α

k)δ(A, B)

Lettingl→ ∞and using (2.5), we obtain

δ(A, B)−0 ≥ αk(δ(A, B)−0) + (1−αk)δ(A, B)

= δ(A, B)−αk0,

which is a contradiction. Therefore by (2.1) and propertyU Cf,{x2n+k}is a Cauchy sequence by propertyU Cf. Now the

completeness ofXand the closedness ofAimply thatlimn→∞x2n+k =z ∈A,wheneverkis a even number. (ifkis a odd

number,x2n+k → z ∈ B asn → ∞.) SinceT is a continuous map andd(x2n+k, x2n+k+1) → δ(A, B)asn → ∞,hence

d(z, T z) =δ(A, B).Therefore,Thas a remotest point.

3

Best Proximity Points

First, we give a useful approximation result.

Proposition 3.1 LetAandBbe nonempty subsets of a metric spaceXand

dn+k≤αkdn+ (1−αk)d(A, B), (3.1)

whereα, kare constant such that0< α <1, k∈N.Thendn+k →d(A, B)asn→ ∞.

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dkm+k−1≤αkmdk−1+ (1−αkm)d(A, B),

dkm+k−2≤αkmdk−2+ (1−αkm)d(A, B),

dkm+k−3≤αkmdk−3+ (1−αkm)d(A, B),

continuing this process,

dkm+1≤αkmd1+ (1−αkm)d(A, B)and

dkm≤αkmd0+ (1−αkm)d(A, B),

Lettingm→ ∞,

dkm+i→d(A, B), i= 0,1, ..., k−1.

Sodn+k→d(A, B)asn→ ∞.

Next, we are ready to state our main results in this section.

Proposition 3.2 LetAandB be nonempty subsets of a metric spaceXand relation (3.1) holds. Then the sequences{x2n+k}

and{x2n+k+1}are bounded.

Proof. Since the sequence {d(x2n+k, x2n+k+1)} converges tod(A, B)as n → ∞,it is enough to prove that {x2n+k+1}is

bounded. Suppose{x2n+k+1}is not bounded, then there exists aN0such that

d(x1+k, x2N0+k+1)> M

and

d(x1+k, x2N0+1)≤M,

whereM >max{ αk

1−αkd(x1, x1+k) +d(A, B), d(x1+k, x1+2k}.By relation (3.1),

M < d(x1+k, x2N0+k+1)

≤ αkd(x1, x2N0+1) + (1−α

k)d(A, B)

≤ αkd(x1, x1+k) +αkd(x1+k, x2N0+1) + (1−α

k)d(A, B)

≤ αkd(x1, x1+k) +αkM + (1−αk)d(A, B)

ThusM < 1αkαkd(x1, x1+k) +d(A, B),which is a contradiction. The following Theorem explain a result when one of the sets is boundedly compact.

Theorem 3.1 LetAandB be nonempty subsets of a metric spaceX and relation (3.1) holds. If eitherAorB is boundedly

compact, then there existsx∈A∪Bwithd(x, T x) =d(A, B).

Proof. Letkbe even number. So x2n+k ∈ A.SupposeAis boundedly compact. Since the sequence {x2n+k} is bounded

andAis boundedly compact. Hence{x2n+k}has a convergent subsequence inA.Let{x2nl+k}be a subsequence of{x2n+k} withx2nl+k → x ∈ Aas l → ∞.SinceT is a continuous map andd(x2nl+k, T x2nl+k) → d(A, B)asl → ∞.Therefore

d(x, T x) =d(A, B).The proof whenkis odd number andBis boundedly compact is similar.

Example 3.1 Given0< α <1, k∈N.letAandBbe nonempty subsets oflP, 1≤p≤ ∞defined byA={((1 +α2n)e2n) :

n∈N}and B={((1 +α2m−1)e2m−1) :m∈N}.Suppose

T((1 +α2n)e2n) = (1 +α2n+1)e2n+1

and

T((1 +α2m−1)e2m−1) = (1 +α2m)e2m.

Then relation (2.1) holds.

Proof.The case whenp=∞is easy to check, so we consider1≤p <∞.

((1 +α2m+k)p+ (1 +α2n+1+k)p)1/p = ((1 +αk−αk+α2m+k)p

+ (1 +αk−αk+α2n+1+k)p)1/p

≤ ((αk+α2m+k)p+ (αk+α2n+1+k)p)1/p

+ 21/p(1−αk)

≤ αk((1 +α2m)p+ (1 +α2n+1)p)1/p

+ 21/p(1−αk).

Which implies that

dn+k≤αkdn+ (1−αk)21/p,

whered(A, B) = 21/p.

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Theorem 3.2 LetAandBbe nonempty closed subsets of a complete metric spaceXsuch that(A, B)has a propertyU Cand relation (3.1) holds. ThenT has a best proximity point.

Proof.We show that for each >0,there exists a positive integerN0such that for allm > n≥N0,

d(x2m+k, x2n+k+1)< d(A, B) +. (3.2)

On the contrary, assume that there exists a0>0such that for eachl≥1,there isml> nl≥lsatisfying

d(x2ml+k, x2nl+k+1)≥d(A, B) +0 (3.3)

and

d(x2ml−2+k, x2nl+k+1)< d(A, B) +0. (3.4)

It follows from (3.3) and (3.4) that

d(A, B) +0 ≤ d(x2ml+k, x2nl+k+1)

≤ d(x2ml+k, x2ml−2+k) +d(x2ml−2+k, x2nl+k+1)

< d(x2ml+k, x2ml−2+k) +d(A, B) +0.

Lettingl→ ∞,and with property UC implies

lim

k→∞d(x2ml+k, x2nl+k+1) =d(A, B) +0. (3.5)

By relation (3.1),

d(A, B) +0 ≤ d(x2ml+2k, x2nl+2k+1)

≤ αkd(x2ml+k, x2nl+k+1) + (1−α

k)d(A, B)

Lettingl→ ∞and using (3.5), we obtain

d(A, B) +0 ≤ αk(d(A, B) +0) + (1−αk)d(A, B)

= d(A, B) +αk0

< d(A, B) +0,

which is a contradiction. Therefore by (3.1) and propertyU Cf, {x2n+k} is a Cauchy sequence by property U C. Now the

completeness ofXand the closedness ofAimply thatlimn→∞x2n+k =z ∈A,wheneverkis a even number. (ifkis a odd

number,x2n+k → z ∈ Bas n → ∞.) SinceT is a continuous map andd(x2n+k, x2n+k+1) → d(A, B)asn → ∞,hence

d(z, T z) =d(A, B).Therefore,T has a best proximity point.

REFERENCES

[1] A. Abkar, M. Gabeleh, Best proximity points of non-self mappings, Top 21, (2) (2013) 287-295.

[2] M. Ahmadi Baseri and H. Mazaheri, Remotest points and approximate remotest points in metric spaces, Iranian Journal of Science and Technology,Transactions A: Science, (2015) pp.4.

[3] M. A. Al-Thagafi, N. Shahzad, Convergence and existence result for best proximity points, Nonliner Analysis, Theory, Methods and Applications 70, (10) (2009) 3665-3671.

[4] A. Amini-Harandi, Common best proximity points theorems in metric spaces, Optimization Letters, (2014) 581-589. [5] M. Baronti, A note on remotal sets in Banach spaces,Publication deL,InstituteMath. (53) (1993) 95-98.

[6] M. Baronti, P. L. Papini, Remotal sets revisted,Taiwanese J. Math. (5) (2001) 367-373.

[7] A. A. Eldred, P. Veeramani, Existence and convergence of best proximity points, J. Math. Anal. Appl 323, (2) (2006) 1001-1006.

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[9] R. Khalil, Sh. Al-Sharif, Remotal sets in vector valued function spaces, Scientiae Mathematicae Japonicae 63, (3) (2006) 433-442.

[10] W. A. Kirk, P. S. Serinivasan and P. Veeramani, Fixed points for mappings satisfying cyclical contractive conditions, Fixed Point Theory 4, (1) (2003) 79-89.

[11] M. Mart´ın, T. S. S. R. K. Rao, On remotality for convex sets in Banach spaces, J. Approx. Theory. (162) (2010) 392-396. [12] M. Sababheh, R. Khalil, Remotality of closed bounded convex sets in reflexive spaces, Numer Funct. Anal. Optim, (29)

(2008) 1166-1170.

[13] T. D. Narang, A study of farthest points, Nieuw Arch. Wisk. 25 (1977) 54-79.

References

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