R E S E A R C H
Open Access
On best proximity points of upper
semicontinuous multivalued mappings
Ghada AlNemer
1,2, Jack Markin
3and Naseer Shahzad
4**Correspondence: [email protected]
4Operator Theory and Applications Research Group, Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah, 21589, Saudi Arabia Full list of author information is available at the end of the article
Abstract
In this paper we study the existence of best proximity points of a nonself upper semicontinuous multivalued mappingT:A→2Bin a strictly convex Banach space. This multivalued mapping commutes with affine, noncyclic, and relatively
u-continuous single-valued mappingf:A∪B→A∪B. Also, we study the case when
Tcommutes with a family of commuting, affine, noncyclic, and relatively
u-continuous single-valued mappings onA∪B. Moreover, we present some examples to illustrate our results.
MSC: 47H10; 54H25
Keywords: best proximity point; multivalued mapping; fixed point; upper semicontinuous mapping; relativelyu-continuous mapping
1 Introduction
LetA,Bbe nonempty subsets of a metric space (X,d) andT:A→B, where Bis the family of all nonempty subsets ofB. IfA∩B=∅, the operator inclusionx∈T(x) has no solution. In this case, it is logical to look for a pointx∈Asuch thatdist(x,T(x)) is mini-mum. Becausedist(x,T(x)) is at leastdist(A,B), the pointxis the solution of the equation dist(x,T(x)) =dist(A,B) =inf{d(x,y) :x∈A,y∈B}. This point is called the best proximity point ofT. Indeed, best proximity point theorems examine the existence of such optimal approximate solutions of the operator inclusionx∈T(x) when there is no exact solution. IfA∩B=∅, the best proximity point is the fixed point ofT.
For multivalued mappings, the existence of best proximity points was established by many authors, e.g., Abkar and Gabeleh in [] and [], Al-Thagafi and Shahzad in [], Amini-Harandi in [], De la Sen in [], Kirket al.in [] and Włodarczyket al. in []. Best proximity point theorems for relatively nonexpansive single-valued mapping were studied in [] in . Since then there has been a lot of activity in this area and a num-ber of results appeared by various authors. Best proximity point theorems for relatively u-continuous mapping were proved in [] and []. For other related results, we refer the reader to [–] and []. In this paper, we study the existence of best proximity points for an upper semicontinuous multivalued mapping with nonempty, compact, and convex val-uesT:A→Bwhich commutes with an affine and relativelyu-continuous single-valued mappingf :A∪B:→A∪Bsuch that f(A)⊆Aandf(B)⊆B(noncyclic). In addition, we present some support examples for our results and we also give an example showing
that the condition ‘T(x)∩B=∅for eachx∈A’ is necessary. Moreover, we add a similar
theorem for a multivalued mapping which commutes with a family of commuting, affine, noncyclic, and relativelyu-continuous single-valued mappings onA∪B.
2 Preliminaries
Definition .[] LetA,Bbe nonempty subsets of a metric spaceX. A mappingf :A∪ B→A∪Bis said to be relativelyu-continuous if for each> there existsδ> such that d(f(x),f(y)) <+dist(A,B) wheneverd(x,y) <δ+dist(A,B) for eachx∈A,y∈B.
Definition .[] LetA,Bbe nonempty subsets of a metric spaceX. A mappingf :A∪ B→A∪Bis called relatively nonexpansive ifd(f(x),f(y))≤d(x,y) for eachx∈A,y∈B.
Every relatively nonexpansive mapping is relativelyu-continuous. However, the con-verse is not true (see []).
Definition .[] LetA,Bbe nonempty subsets of a metric spaceXandT:A→Ba multivalued mapping. A pointx∈Ais called a (i) fixed point ofTifx∈T(x) and (ii) best proximity point ofT ifdist(x,T(x)) =dist(A,B). Note that ifdist(A,B) = , then we get a fixed point ofT.
Definition . LetA,Bbe nonempty subsets of a metric spaceX. A multivalued mapping T:A→Bis called upper semicontinuous ifT–(C) ={x∈A:T(x)∩C=∅}is closed in
AwheneverCis closed inB.
Proposition .[] Let X be a strictly convex Banach space,A a nonempty,compact, and convex subset of X,and B a nonempty closed subset of X.Let{xn}be a sequence in A and y∈B.Ifxn–y →dist(A,B),then xn→PA(y).
Definition .[] LetA,Bbe nonempty convex subsets of a Banach spaceX. A mapping f :A∪B→A∪Bis called affine iff(αx+βy) =αf(x) +βf(y) for allx,y∈Aorx,y∈Band
α,β∈[, ] withα+β= .
Lemma .[] If A is a nonempty,compact,and convex subset of a Banach space,and T:A→Acan be expressed as a composition of finitely many upper semicontinuous multi-valued mappings with nonempty,compact,and convex values,then T has a fixed point.
LetA,Bbe nonempty subsets of a Banach spaceX.f :A∪B→A∪Ba relatively nonex-pansive mapping such thatf(A)⊆A,f(B)⊆B,T:A→KC(B), whereKC(B) is the set of all nonempty, compact, and convex subsets ofB. The mappingf andTare said to commute if for eachx∈A,f(T(x))⊆T(f(x)). Define
A=
x∈A:x–y=dist(A,B) for somey∈B,
B=
y∈B:x–y=dist(A,B) for somex∈A.
Remark . Note that ifAandBare nonempty, compact, and convex sets, thenAand
Bare nonempty, compact, and convex sets withdist(A,B) =dist(A,B). For details see
Remark .[, ] LetAbe a nonempty subset of a normed spaceX. The metric projec-tion operator is defined byPA(x) ={y∈A:x–y=dist(x,A)}for eachx∈X. IfAis a nonempty, compact, and convex subset of a Banach spaceX, thenPAis upper semicon-tinuous with nonempty, compact, and convex values. Observe that whenAis a nonempty, compact, and convex subset of a strictly convex Banach space X, PA is a single-valued mapping fromXtoA.
Theorem .[] Let A,B be nonempty,compact,and convex subsets of a strictly convex Banach space X.If f :A∪B:→A∪B is relatively u-continuous such that f(A)⊆A and f(B)⊆B.Then there exists(x,y)∈A×B such that f(x) =x,f(y) =y,andx–y=
dist(A,B).
3 Main results
The following proposition is a noncyclic version of Proposition . in [].
Proposition . Let A,B be nonempty,compact,and convex subsets of a strictly convex Banach space X.Let f :A∪B→A∪B be a relatively u-continuous mapping such that f(A)⊆A and f(B)⊆B.P:A∪B→A∪B is a mapping defined by
P(x) =
PB(x) if x∈A, PA(x) if x∈B.
Then f(P(x)) =P(f(x))for each x∈A∪B,i.e.,PA(f(y)) =f(PA(y))for each y∈Band
PB(f(x)) =f(PB(x))for each x∈A.
Proof Letx∈A. Then there existsy∈Bsuch thatx–y=dist(A,B). So,y=PB(x) and x=PA(y). Then for eachδ> ,x–y<δ+dist(A,B). Sincef is relativelyu-continuous, for each > we have dist(A,B)≤ f(x) –f(y)< +dist(A,B). Thus,f(x) –f(y)= dist(A,B). So,f(x) =PA(f(y)) andf(y) =PB(f(x)). SinceA,Bare nonempty, compact, and convex subsets of a strictly convex Banach space, the metric projection is unique. Now, x=PA(y)⇒f(x) =f(PA(y))⇒PA(f(y)) =f(PA(y)) for eachy∈B. Also,y=PB(x)⇒ f(y) =f(PB(x))⇒PB(f(x)) =f(PB(x)) for eachx∈A. Hence,f(P(x)) =P(f(x)) for each
x∈A∪B.
A cyclic version of the following proposition can be found in [] (see the proof of The-orem . in []).
Proposition . Let A,B be nonempty,compact,and convex subsets of a strictly convex Banach space X.Let f :A∪B→A∪B be a relatively u-continuous mapping such that f(A)⊆A and f(B)⊆B.Then f is continuous on Aand B.
Proof Letx∈Aand{xn} ⊆Asuch thatxn→x. We want to show thatf(xn)→f(x).
Using the triangle inequality, we obtain
xn–PB(x)≤ xn–x+x–PB(x)
=xn–x+dist(A,B)
Then for eachδ> there existsN∈Nsuch that for eachn≥N, we have|xn–PB(x)–
dist(A,B)|<δ. So,n≥N⇒ xn–PB(x)<δ+dist(A,B). By relativeu-continuity off, f(xn) –f(PB(x))<+dist(A,B) for eachn≥N. Since{f(xn)} ⊆AandPB(f(x))∈B,
Proposition . gives
f(xn)→PA
fPB(x)
=fPA
PB(x)
=f(x).
Hence,f(xn)→f(x). Sincex∈Awas arbitrary,f is continuous onA. Similarly,f is
continuous onB. Therefore,f is continuous onA∪B.
Theorem . Let A, B be nonempty, compact, and convex subsets in a strictly convex Banach space X.Suppose f :A∪B→A∪B is an affine relatively u-continuous mapping with f(A)⊆A,f(B)⊆B.Then there exists(x,y)∈A×B such that f(x) =x,f(y) =y
andx–y=dist(A,B).
In addition,if T:A→KC(B)is an upper semicontinuous multivalued mapping,f and T commute,and T(x)∩B=∅for each x∈A,then there exists a∈A such that f(a) =a
anddist(a,T(a)) =dist(A,B).
Proof Foru∈A, there is av∈Bsuch thatu–v=dist(A,B). Then by the relative
u-continuity off,f(u) –f(v)=dist(A,B), implying thatf(u)∈A. Therefore, the
com-pact convex setAis invariant under the continuous mappingf, and the Schauder fixed
point theorem implies the existence of a fixed pointx=f(x)∈A. Letybe the unique
closest point toxinB. Then by the relativeu-continuity off and the uniqueness of the
closest point projection ontoB,y=f(y) andx–y=dist(A,B).
Now, we will prove that there exists a∈Asuch thatdist(a,T(a)) =dist(A,B). Define Fix(f) ={x∈A∪B:f(x) =x},FixA(f) =Fix(f)∩A andFixB(f) =Fix(f)∩B. Clearly,
FixA(f) andFixB(f) are nonempty, becausex∈FixA(f) andy∈FixB(f). The setFixA(f) is closed. Indeed, let{xn} ⊆FixA(f) such thatxn→x. Since{xn} ⊆AandAis closed
by Remark ., we havex∈A⊆A. Using Proposition .,f(xn)→f(x). Butf(xn) =xn for eachn. Soxn→f(x). Consequentlyx=f(x). Thusx∈FixA(f). Therefore,FixA(f) is closed. Similarly,FixB(f) is closed. So,FixA(f) andFixB(f) are compact sets as they are closed subsets of the compact setsA,B. In addition,FixA(f) is a convex set. Indeed, let x,y∈FixA(f) andα,β∈[, ] withα+β= . Sincef is affine,f(αx+βy) =αf(x) +
βf(y) =αx+βy,i.e.,αx+βy∈Fix(f). Also,αx+βy∈AasAis convex andx,y∈A.
Consequently,αx+βy∈Fix(f)∩A=FixA(f). Similarly,FixB(f) is a convex set.
Assumex∈FixA(f) and choosev∈T(x). Sincef andTcommute,f(v)∈T(f(x)) =T(x), which implies thatT(x) is invariant underf. Then the invariance ofBunderf shows that
the compact convex setT(x)∩Bis invariant underf. Sincef is continuous onB, by
the Schauder fixed point theoremf has a fixed point inT(x)∩B, implying thatT(x)∩
FixB(f)=∅for eachx∈FixA(f).
Now, defineF:FixA(f)→FixB(f)byF(x) =T(x)∩FixB(f) for eachx∈FixA(f). ThenF is an upper semicontinuous multivalued mapping with nonempty, compact, and convex values. Note thatPA:FixB(f)→FixA(f). To see this, letx∈FixB(f)⊆B. Then there exists
y∈Asuch thatx–y=dist(A,B). So,y=PA(x) andx=PB(y). For eachδ> , we have
x–y<δ+dist(A,B). Using the relativeu-continuity for anyf,dist(A,B)≤ f(x) –f(y)<
andf(x) =PB(f(y)). Sincex∈FixB(f) andy=PA(x), we havef(y) =f(PA(x)) =PA(f(x)) = PA(x) and soPA(x)∈FixA(f)⊆A. Note thatPA◦F:FixA(f)→FixA(f). By Lemma ., there existsa∈FixA(f)⊆Asuch thata∈(PA◦F)(a),i.e.,a=f(a) anda∈PA(F(a)). So, there existsb∈F(a) =T(a)∩FixB(f)⊆Bsuch thata=PA(b)⊆FixA(f). Asa=PA(b),
a–b=dist(b,A). Sinceb∈F(a) =T(a)∩FixB(f)⊆B, thenb∈T(a) andb∈B. Since
b∈B, there existsa∈Asuch thata–b=dist(A,B). Sincea∈AandT(a)⊆B, we
have
dist(A,B)≤dista,T(a)
≤ a–b
=dist(b,A)
≤b–a
=dist(A,B).
Thus,dist(a,T(a)) =dist(A,B).
Remark . The conditionT(x)∩B=∅for eachx∈Ais necessary in Theorem .. For
example, in the real space ifA= [, ]×[–, ],B= [–,–]×[–, ]. Define
f :A∪B→A∪B by f(x,y) =
x,y+
and
T:A→KC(B) by T(x,y) =
–,– x
× {y}.
Clearly, T is upper semicontinuous and f is affine and relatively u-continuous. Also, f(A)⊆Aandf(B)⊆B. There are fixed points of f, x= (, )∈A,y= (–, )∈Bsuch
thatx–y=dist(A,B) = .. In addition,f andTcommute. Suppose that there exists
a∈Fix(f)∩Asuch thatdist(a,T(a)) = .. Thena= (z, ), for some ≤z≤. So,
dista,T(a)=dist
(z, ),
–,– z
× {} =(z, ) –
–
z, = ..
Consequently, z– .z+ = . So, z
= . + .i, z=
. – .i, which are not real numbers, and z =
–., which does not belong to [, ]. Note thatA={}×[–, ],B={–}×
[–, ]. Forx= (,y)∈A, we haveT(x) =T(,y) ={(–,y)}. So,T(x)∩B={(–,y)} ∩ {(–,y) : –≤y≤}=∅.
Corollary . Let A,B be nonempty,compact,and convex sets in a strictly convex Banach space X.If T :A→KC(B)is an upper semicontinuous multivalued mapping and T(x)∩ B=∅for each x∈A,then there exists a∈A such thatdist(a,T(a)) =dist(A,B).
Proof Takingf =I(the identity mapping onA∪B) in Theorem ., we obtain the desired
Corollary . Let A be a nonempty,compact,and convex set in a strictly convex Banach space.Suppose f :A→A is an affine continuous mapping.If T:A→KC(A)is an upper semicontinuous multivalued mapping and f,T commute,then there exists a∈A such that a∈Fix(f)∩Fix(T).
Proof Since any continuous mapping on a compact set is relativelyu-continuous on that set, takingA=Bin Theorem ., we see that there existsa∈A such thatf(a) =aand dist(a,T(a)) =dist(A,A) = ,i.e.,a∈T(a). So, f(a) =a∈T(a). Therefore, a∈Fix(f)∩
Fix(T).
Theorem . Let X be a strictly convex Banach space.Let A,B be nonempty,compact, and convex subsets of X and let f,g:A∪B→A∪B be commuting,affine,and relatively u-continuous mappings such that f(A)⊆A,f(B)⊆B and g(A)⊆A,g(B)⊆B.Then there exist points x∈A and y∈B such that x=f(x) =g(x),y=f(y) =g(y)andx–y=
dist(A,B).
Proof Foru∈A, there is av∈Bsuch thatu–v=dist(A,B). Then by the relative
u-continuity off,f(u) –f(v)=dist(A,B), implying thatf(u)∈A. Therefore, the
com-pact convex setAis invariant under the continuous mappingf, and the Schauder fixed
point theorem implies the existence of a fixed pointx=f(x)∈A. The set of fixed points
off inA(denoted byFixA(f)) is closed and convex sincef is continuous and affine. If x∈FixA(f), commutativity off andg impliesf(g(x)) =g(f(x)) =g(x). Therefore,FixA(f) is invariant underg, and sincegis continuous it has a fixed point inFixA(f). Letxbe a
common fixed point off andg inA, that is,x=f(x) =g(x), and letybe the unique
closest point toxinB. Then by the relativeu-continuity offandgand the uniqueness of
the closest point projection ontoB,y=f(y) =g(y) andx–y=dist(A,B).
The previous theorem can be extended to an arbitrary family of commuting affine and noncyclic mappings. The proof depends on the following common fixed point result for commuting affineu-continuous mappings in strictly convex Banach spaces. The proof of this result is adapted from Przebieracz ([], Theorem .) and is included for convenience of the reader.
Lemma .(Markov-Kakutani theorem) Let X be a strictly convex Banach space.Let A, B be nonempty,compact,and convex subsets of X and letFbe a family of commuting affine and relatively u-continuous mappings on A∪B such that f(A)⊆A and f(B)⊆B.Then there is an x∈Asuch that f(x) =xfor every f ∈F.There is a y∈Bsuch that f(y) =y
for every f ∈F.
Proof Notice that the mappings in the familyFare continuous onA∪B. LetFix(f) = {x∈A∪B:f(x) =x},FixA(f) =Fix(f)∩A,f ∈F. As shown in the proof of Theorem .,
FixA(f)=∅andFixA(f) is convex and compact. To prove that
f∈FFixA(f)=∅, consider any finite collection fromF, sayf, . . . ,fn. Assume that
C=
≤i≤n
For eachx∈Candk∈ {, . . . ,n},fkfn+(x) =fn+fk(x) =fn+(x), which implies thatfn+(x)∈C.
Therefore, the compact convex setCis invariant underfn+, implying thatFixA(fn+)∩C= ∅sincefn+is continuous onA. Since every finite collection of the setsFixA(f),f∈F, has a nonempty intersection, we havef∈FFixA(f)=∅. Similarly,
f∈FFixB(f)=∅.
Theorem . Let X be a strictly convex Banach space.Let A,B be nonempty,compact,and convex subsets of X and letFbe a family of commuting affine and relatively u-continuous mappings on A∪B such that f(A)⊆A and f(B)⊆B.Then there exist points x∈A and
y∈B such that x=f(x)and y=f(y),for all f ∈Fwherex–y=dist(A,B).
Proof By Lemma . the mappings in the familyFhave a common fixed pointx∈A, that
is,f(x) =xforf∈F. Lety∈Bbe the unique closest point toxinB. Then, for anyf ∈F, f(x) –y=dist(A,B), but by the relativeu-continuity off,f(x) –f(y)=dist(A,B).
By the uniqueness of the closest point,y=f(y) forf ∈F.
Theorem . Let A, B be nonempty,compact,and convex subsets of a strictly convex Banach space X and let Fbe a family of commuting,affine and relatively u-continuous mappings on A∪B with f(A)⊆A,f(B)⊆B for each f ∈F.Let T:A→KC(B)be an upper semicontinuous mapping such that T(x)∩B=∅for each x∈A.IfFand T commute,then
there exists a point a∈A such that f(a) =a for each f ∈Fanddist(a,T(a)) =dist(A,B).
Proof By Lemma .,f∈FFixA(f) and
f∈FFixB(f) are nonempty.
As in the proof of Theorem .,T(x) is invariant under eachf∈F, forx∈FixA(f). Since
f∈FFixA(f)=∅, forx∈
f∈FFixA(f),T(x) is invariant underF. Also,Bis invariant un-derF. Therefore as in the proof of Theorem ., sinceT(x)∩Bis a compact convex set,
T(x)∩(f∈FFixB(f))=∅. By the proof of Theorem .,FixA(f) andFixB(f) are compact and convex sets forf ∈F. Therefore,f∈FFixA(f) andf∈FFixB(f) are compact and con-vex.
Now defineF:f∈FFixA(f)→
f∈FFixB(f)byF(x) =T(x)∩(
f∈FFixB(f)) for eachx∈
f∈FFixA(f). Clearly,Fis an upper semicontinuous multivalued mapping with compact convex values. Now,PA:
f∈FFixB(f)→
f∈FFixA(f). To see this, letx∈
f∈FFixB(f). Thenx∈Bandf(x) =xfor eachf ∈F. So, there existsy∈Asuch thatx–y=dist(A,B).
This impliesx=PB(y) andy=PA(x). For eachδ> , we havex–y<δ+dist(A,B). Us-ing the relative u-continuity for anyf ∈F, dist(A,B)≤ f(x) –f(y)<+dist(A,B) for each> . Thus,f(x) –f(y)=dist(A,B). Therefore,f(y) =PA(f(x)) andf(x) =PB(f(y)) for each f ∈F. Now,y=PA(x)⇒f(y) =f(PA(x))⇒PA(x) =f(PA(x)) for each f ∈F. Hence,PA(x)∈
f∈FFixA(f) for eachx∈
f∈FFixB(f). Note thatPA◦F:
f∈FFixA(f)→
f∈FFixA(f). By Lemma .,P
A◦Fhas a fixed point. So, there existsa∈f∈FFixA(f) such thata∈(PA◦F)(a). So,f(a) =afor eachf ∈Fanda∈PA(F(a)),i.e., there existsb∈F(a) such thata=PA(b). Sinceb∈F(a),b∈T(a)∩(
f∈FFixB(f)). So,b∈T(a),b∈B, and f(b) =bfor eachf ∈F.a=PA(b) impliesa–b=dist(b,A). Sinceb∈B, there exists
a∈Asuch thata–b=dist(A,B). Sincea∈AandT(a)⊆B, we have
dist(A,B)≤dista,T(a)
≤ a–b
≤b–a
=dist(A,B).
Thus,dist(a,T(a)) =dist(A,B).
Corollary . Let A be a nonempty,compact,and convex subset of a strictly convex
Ba-nach space X and letFbe a family of commuting,affine and continuous self-mappings of A. Let T:A→KC(A)be an upper semicontinuous mapping.IfFand T commute,then there exists a point a∈A such that a=f(a)∈T(a)for each f ∈F.
4 Examples
Examples . to . are related to Theorem .. On other hand, the last two examples are related to Theorem . (and Theorem .).
Example . LetX=R with the usual metric. The setsA={(x,y) : ≤x≤, ≤y≤
},B={(x, ) : ≤x≤}are nonempty, compact, and convex withdist(A,B) = . Define f :A∪B→A∪Bbyf(x,y) = (x+,y) andT:A→KC(B) byT(x,y) = [x, ]× {}. Then T is upper semicontinuous andf is relativelyu-continuous and affine withf(A)⊆Aand f(B)⊆B. AsFix(f) ={(,y) : ≤y≤ ory= }, we getx= (, )∈Fix(f)∩A,y= (, )∈
Fix(f)∩Bwithx–y= . In addition,f andTcommute. Indeed,f(T(x,y)) =f([x, ]× {}) ={z+:z∈[x, ]}×{}andT(f(x,y)) =T(x+,y) = [x+, ]×{}. Forz∈[x, ],z+ ∈ [x+, ]⊆[x+, ]. Thus,f(T(x,y))⊆T(f(x,y)) for each (x,y)∈A. Also,T(x)∩B=∅for
eachx∈AsinceA={(x, ) : ≤x≤}andB=B. For (, )∈A, we havef(a) =aand
dist(a,T(a)) =dist(A,B) = .
Example . LetX=Rwith the usual metric. The setsA={(,a) : ≤a≤},B={(x,y) :
≤x≤, ≤y≤}are nonempty, compact, and convex withdist(A,B) = . Definef : A∪B→A∪Bbyf(x,y) = (x,y+ ) andT :A→KC(B) byT(,a) = [,a]× {}. ThenT is upper semicontinuous andf is relativelyu-continuous and affine withf(A)⊆Aand f(B)⊆B. AsFix(f) ={(x, ) :x= or ≤x≤}, we getx= (, )∈Fix(f)∩A,y= (, )∈
Fix(f)∩Bwithx–y= . In addition,fandTcommute. Indeed,f(T(,a)) =f([,a]× {}) = [,a]× {} andT(f(,a)) =T(,a+ ) = [,a+ ]× {}. Fora∈[, ], a+ ≥a,i.e., [,a]⊆[,a+ ]. Thus,f(T(,a))⊆T(f(,a)) for each (,a)∈A. Also,T(x)∩B=∅for
eachx∈AsinceA=AandB={(,y) : ≤y≤}. Fora= (, )∈A, we havef(a) =a
anddist(a,T(a)) =dist(A,B) = .
Example . LetX=Rwith the usual metric. The setsA={(x,y) : –≤x≤–., –≤
y≤}, B={(x,y) : ≤x≤, –≤y≤} are nonempty, compact, and convex with dist(A,B) = .. Define f :A∪B→A∪Bby f(x,y) = (x,y+ ) andT :A→KC(B) by
T(x,y) = [,x]× {y}. ThenT is upper semicontinuous andf is relativelyu-continuous and affine withf(A)⊆Aandf(B)⊆B. AsFix(f) ={(x, ) : –≤x≤–. or ≤x≤}, we getx= (–., )∈Fix(f)∩A,y= (, )∈Fix(f)∩Bwithx–y= .. In addition,
f andTcommute. Also,T(x)∩B=∅for eachx∈AsinceA={(–.,y) : –≤y≤}
andB={(,y) : –≤y≤}. Fora= (–., )∈A, we havef(a) =aanddist(a,T(a)) =
Example . LetX=R with the usual metric. The setsA={(x,y) : –≤x≤, –≤
y≤–.},B={(x,y) : –≤x≤, ≤y≤}are nonempty, compact, and convex with dist(A,B) = .. Definef :A∪B→A∪Bbyf(x,y) = (x,y) andT:A→KC(B) byT(x,y) =
{x} ×[,y]. ThenT is upper semicontinuous andf is relativelyu-continuous and affine
with f(A)⊆Aandf(B)⊆B. AsFix(f) ={(,y) : ≤y≤ or – ≤y≤–.}, we get x= (, –.)∈Fix(f)∩A,y= (, )∈Fix(f)∩Bwithx–y= .. In addition,fand
T commute. Also,T(x)∩B=∅for eachx∈AsinceA={(x, –.) : –≤x≤}and
B={(x, ) : –≤x≤}. Fora= (, –.), we havef(a) =aanddist(a,T(a)) =dist(A,B) =
..
Example . LetX=Rwith the usual metric. The setsA={(x,y) : ≤x≤,y= –},
B={(x,y) : –≤x≤,y= }are nonempty, compact, and convex subsets of a strictly convex Banach space withdist(A,B) = . Definef,g:A∪B→A∪Bby f(x,y) = (x,y) andg(x,y) = (x,y). Thenf,gare relativelyu-continuous and affine withf(A)⊆A,f(B)⊆ B,g(A)⊆A, andg(B)⊆B. Alsof,gcommute. Now, defineT :A→KC(B) byT(x,y) = [–, –x]× {y}. ThenT is upper semicontinuous with nonempty, compact, and convex
values. In addition,T commutes withf andg. Clearly,A={(, –)}, B={(, )}, and
(, )∈T(, –) = [–, ]× {}. Fora= (, –)∈Aandb= (, )∈B, we havef(a) =g(a) = a,f(b) =g(b) =b, anda–b=dist(A,B) = . Moreover,dist(a,T(a)) =dist(A,B).
Example . LetX=R with the usual metric. The setsA={(x,y) : –≤x≤–, –≤
y≤},B={(x,y) : ≤x≤, –≤y≤}are nonempty, compact, and convex subsets of a strictly convex Banach space withdist(A,B) = . Definef,g:A∪B→A∪Bbyf(x,y) = (x,y) andg(x,y) = (x,y). Thenf,g are relativelyu-continuous and affine withf(A)⊆A, f(B)⊆B, g(A)⊆A, andg(B)⊆B. Alsof,g commute. Now, defineT :A→KC(B) by T(x,y) = [, –x]× {y}. ThenT is upper semicontinuous with nonempty, compact, and convex values. In addition,T commutes withf andg. Clearly,A={(–,y) : –≤y≤},
B={(,y) : –≤y≤}. So, (,y)∈T(–,y)∩B= ([, ]× {y})∩Bfor each (–,y)∈A.
Fora= (–, )∈Aandb= (, )∈B, we havef(a) =g(a) =a,f(b) =g(b) =b, anda–b= dist(A,B) = . Moreover,dist(a,T(a)) =dist(A,B).
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.
Author details
1Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah, 21589, Saudi Arabia. 2Department of Mathematics, Princess Nourah bint Abdulrahman University, P.O. Box 105862, Riyadh, 11656, Saudi Arabia. 3528 Rover Boulevard, Los Alamos, NM 87544, USA.4Operator Theory and Applications Research Group, Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah, 21589, Saudi Arabia.
Acknowledgements
This article was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah. The authors, therefore, acknowledge with thanks DSR for technical and financial support. The authors are grateful to three anonymous referees for their useful suggestions and comments.
Received: 29 June 2015 Accepted: 11 December 2015
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