R E S E A R C H
Open Access
The existence of best proximity points with
the weak
P
-property
Tomonari Suzuki
**Correspondence:
suzuki-t@mns.kyutech.ac.jp Department of Basic Sciences, Kyushu Institute of Technology, Tobata, Kitakyushu 804-8550, Japan
Abstract
We improve some existence theorem of best proximity points with the weak
P-property, which has been recently proved by Zhanget al. MSC: Primary 54H25; secondary 54E50
Keywords: best proximity point; the weakP-property; the Banach contraction principle; Kannan’s fixed point theorem; completion
1 Introduction
Let (A,B) be a pair of nonempty subsets of a metric space (X,d), and letT be a map-ping fromAintoB. Thenx∈Ais called abest proximity pointifd(x,Tx) =d(A,B), where
d(A,B) =inf{d(x,y) :x∈A,y∈B}. We have proved many existence theorems of best prox-imity points. See, for example, [–]. Very recently, Caballeroet al.[] proved a new type of existence theorem, and Zhanget al.[] generalized the theorem. The theorem proved in [] is Theorem with an additional assumption of the completeness ofB. The essence of the result in [] becomes very clear in [], however, we have not learned the essence completely.
Motivated by the fact above, in this paper, we improve the result in []. Also, in order to consider the discontinuous case, we give a Kannan version.
2 Preliminaries
In this section, we give some preliminaries.
Definition Let (A,B) be a pair of nonempty subsets of a metric space (X,d), and define
AandBby
A=
x∈A: there existsu∈Bsuch thatd(x,u) =d(A,B) () and
B=
u∈B: there existsx∈Asuch thatd(x,u) =d(A,B). () Then
• (Sankar Raj [])(A,B)is said to have theP-propertyifA=∅and the following holds:
x,y∈A,u,v∈B, d(x,u) =d(y,v) =d(A,B) ⇒ d(x,y) =d(u,v).
• (Zhanget al.[])(A,B)is said to have theweakP-propertyifA=∅and the following holds:
x,y∈A,u,v∈B, d(x,u) =d(y,v) =d(A,B) ⇒ d(x,y)≤d(u,v).
Proposition Let(A,B)be a pair of nonempty subsets of a metric space(X,d),and define Aand Bby()and().Assume that A=∅.Then the following are equivalent:
(i) (A,B)has the weakP-property. (ii) The conjunction of the following holds:
(ii-) For everyu∈B,there exists a uniquex∈Awithd(x,u) =d(A,B). (ii-) There exists a nonexpansive mappingQfromBintoAsuch that
d(Qu,u) =d(A,B)for everyu∈B.
Proof We note thatB=∅becauseA=∅. First, we assume (i). Letx,y∈Aandu∈B
satisfyd(x,u) =d(y,u) =d(A,B). Then from (i), we have
d(x,y)≤d(u,u) = ,
thus,x=y. So (ii-) holds. We putQu=x. Then from the definition of the weakP-property, we haved(Qu,Qv)≤d(u,v) foru,v∈B, that is,Qis nonexpansive. Conversely, we
as-sume (ii). Letx,y∈Aandu,v∈Bsatisfyd(x,u) =d(y,v) =d(A,B). Then from (ii-), we
haveQu=xandQv=y. Therefore,
d(x,y) =d(Qu,Qv)≤d(u,v)
holds.
Lemma Let(A,B)be a pair of subsets of a metric space(X,d),and define Aand Bby
()and().Assume that A=∅.Let T be a mapping from A into B,and let Q be a mapping
from Binto Asuch that d(Qu,u) =d(A,B)for every u∈B.Then the following holds:
{un} ⊂B, lim
n→∞un=w, T
lim
n→∞Qun
=w ⇒ w∈B. ()
Proof Let{un}be a sequence inBsuch that{un}converges tow∈X, andT(limnQun) =w.
We puty=limnQun. SinceTy=w, we havey∈Aandw∈B. Since
d(y,w) = lim
n→∞d(Qun,un) =d(A,B),
we havey∈Aandw∈B.
Lemma Let(X,d)be a metric space,let A,A,Bbe nonempty subsets such that A is
complete and A⊂A.Let T be a mapping from A into X such that T(A)⊂B,and let Q
be a nonexpansive mapping from Binto A.LetQ be the mapping whose graph¯ Gr(Q¯)is
the completion ofGr(Q).Assume().Then the following hold:
(i) Q¯ is well-defined and nonexpansive.
(ii) Qw¯ =zis equivalent to that there exists a sequence{un}inBsuch thatlimnun=w
(iii) The domain ofQ¯ isB¯,whereB¯is the completion ofB.
(iv) The range ofQ¯ is a subset ofA¯,whereA¯is the completion ofA. (v) T◦ ¯Qw=wimpliesT◦Qw=w.
(vi) Q¯ ◦Tz=zimpliesQ◦Tz=z. (vii) The range ofQ¯ is a subset ofA.
Proof We consider that the whole space is the completion ofX. SinceQis Lipschitz con-tinuous,Q¯ is well-defined. The rest of (i) and (ii)-(iv) are obvious. By using (), we can
easily prove (v) and (vi). From the completeness ofA, we obtain (vii).
3 Fixed point theorems
In this section, we give fixed point theorems, which are used in the proofs of the main results.
Theorem Let(X,d)be a metric space,let A,A,Bbe nonempty subsets such that A is
complete and A⊂A.Let T be a contraction from A into X such that T(A)⊂B,and let
Q be a nonexpansive mapping from Binto A.Assume().Then Q◦T has a unique fixed
point in A.
Proof We consider that the whole space is the completion ofX. Define a nonexpansive mappingQ¯ as in Lemma . SinceT is continuous,T(A¯) is a subset ofB¯. LetSbe the
restriction ofTtoA¯. ThenQ¯◦Sis a contraction onA¯. So the Banach contraction
prin-ciple yields that there exists a unique fixed pointzofQ¯ ◦SinA¯. SinceQ¯ ◦Tz=z, by
Lemma (vi),zis a fixed point ofQ◦T.
Remark
• IfX=A=A=BandQis the identity mapping onB, then Theorem becomes the Banach contraction principle [].
• We can prove Theorem with the mappingT◦ ¯Qas in the proof of Theorem .
We prove generalizations of Kannan’s fixed point theorem [].
Theorem Let(X,d)be a metric space,let Y be a complete subset of X,and let T be a mapping from Y into X.Assume that the following hold:
(i) There existsα∈[, /)such thatd(Tx,Ty)≤αd(x,Tx) +αd(y,Ty)for allx,y∈Y. (ii) There exists a nonempty subsetZofYsuch thatT(Z)⊂Z.
Then there exists a unique fixed point z,and for every x∈Z,{Tnx}converges to z.
Proof Fixx∈Z. Then from the proof in Kannan [], we obtain that{Tnx}converges to a
fixed point, and the fixed point is unique.
Remark IfX=Y=Z, then Theorem becomes Kannan’s fixed point theorem []. Using Theorem , we obtain the following.
Theorem Let(X,d)be a metric space,let A,A,Bbe nonempty subsets such that A is
complete and A⊂A.Let T be a mapping from A into X such that T(A)⊂B,and let Q
• There existα∈[, /)andμ∈[,∞)such that
d(Tx,Ty)≤αd(x,Tx) –μ+αd(y,Ty) –μ
forx,y∈Aandd(Qu,u)≤μfor allu∈B.
Then T◦Q has a unique fixed point in B.
Proof We consider that the whole space is the completion ofX. Define a nonexpansive mappingQ¯ as in Lemma . From the continuity ofd,d(Qu¯ ,u)≤μforu∈ ¯B. Foru,v∈ ¯B,
we have
d(T◦ ¯Qu,T◦ ¯Qv)
≤αd(Qu¯ ,T◦ ¯Qu) –μ+αd(Qv¯ ,T◦ ¯Qv) –μ
≤αd(Qu¯ ,u) +d(u,T◦ ¯Qu) –μ+αd(Qv¯ ,v) +d(v,T◦ ¯Qv) –μ
≤αd(u,T◦ ¯Qu) +αd(v,T◦ ¯Qv).
HenceT◦ ¯Qis a Kannan mapping fromB¯intoX.T◦ ¯Q(B) =T◦Q(B)⊂Bis obvious.
So by Theorem , there exists a unique fixed pointwofT◦ ¯QinB¯. By Lemma (v),w∈B
andwis a fixed point ofT◦Q.
Remark
• SinceTis not necessarily continuous, the range ofT◦ ¯Qis not necessarily included by
¯
B. Because of the same reason, we cannot prove Theorem with the mappingQ¯ ◦T. • It is interesting that we do not need the completeness of any set related toBdirectly.
Of course, we need the completeness ofA.
4 Main results
In this section, we give the main results.
Theorem (Zhanget al.[]) Let(A,B)be a pair of subsets of a metric space(X,d),and define Aand B by()and().Let T be a contraction from A into B.Assume that the
following hold:
(i) (A,B)has the weakP-property. (ii) Ais complete.
(iii) T(A)⊂B.
Then there exists a unique z∈A such that d(z,Tz) =d(A,B).
Proof By Proposition (ii-), there exists a nonexpansive mapping Qfrom B into A
such thatd(Qu,u) =d(A,B) for everyu∈B. Then by Lemma , all the assumptions in
Theorem hold. So there exists a unique fixed pointzofQ◦T inA. This implies that
d(z,Tz) =d(A,B). Letx∈Asatisfyd(x,Tx) =d(A,B). Then from Proposition (ii-),x∈A,
Tx∈BandQ◦Tx=xhold. SinceQ◦T has a unique fixed point, we obtainx=z. Hence
Remark
• If we weaken (i) to the conjunction ofA=∅and (ii-) in Proposition , we obtain only the existence of best proximity points.
• In [], we assume the completeness ofB.
• Exactly speaking, in [], we obtained a theorem connected with Geraghty’s fixed point theorem []. However, in this paper, the difference between the two fixed point theorems is not essential. This means that we can easily modify Theorem to be connected with Geraghty’s theorem.
Theorem Let(A,B)be a pair of subsets of a metric space(X,d),and define Aand B
by()and().Let T be a mapping from A into B.Assume that(i)-(iii)in Theoremand the following hold:
(iv) There existsα∈[, /)such that
d(Tx,Ty)≤αd(x,Tx) –d(A,B)+αd(y,Ty) –d(A,B) forx,y∈A.
Then there exists a unique z∈A such that d(z,Tz) =d(A,B).
Proof By Proposition (ii-), there exists a nonexpansive mappingQfromBintoAsuch
thatd(Qu,u) =d(A,B) for everyu∈B. Then by Theorem , there exists a unique fixed
pointwofT◦QinB. This implies thatd(z,Tz) =d(A,B), wherez=Qw. Letx∈Asatisfy
d(x,Tx) =d(A,B). Then from Proposition (ii-),x∈A,Tx∈BandQ◦Tx=xhold. Since
T◦Q◦Tx=Tx, we haveTx=w, and hencex=Q◦Tx=Qw=z. Therefore,zis unique.
Remark If we weaken (i) to the conjunction of A=∅ and (ii-) in Proposition , we
obtain only the existence of best proximity points.
5 Additional result
In this section, we give a proposition similar to Proposition .
Proposition Let(A,B)be a pair of nonempty subsets of a metric space(X,d),and define Aand Bby()and().Assume that A=∅.Then the following are equivalent:
(i) (A,B)has theP-property.
(ii) The conjunction of the following holds:
(ii-) For everyu∈B,there exists a uniquex∈Awithd(x,u) =d(A,B). (ii-) There exists an isometryQfromBontoAsuch thatd(Qu,u) =d(A,B)for
everyu∈B.
Proof We note B=∅. First, we assume (i). Letx,y∈A andu∈B satisfy d(x,u) =
d(y,u) =d(A,B). Then from (i), we haved(x,y) =d(u,u) = , thus,x=y. So (ii-) holds. We putQu=x. Then it is obvious thatQis isometric. For everyx∈A, there existsu∈B
withd(x,u) =d(A,B). From (ii-),Qu=xobviously holds, and henceQis surjective. Con-versely, we assume (ii). Letx,y∈Aandu,v∈Bsatisfyd(x,u) =d(y,v) =d(A,B). Then
Competing interests
The author declares that he has no competing interests.
Acknowledgements
The author is supported in part by the Grant-in-Aid for Scientific Research from the Japan Society for the Promotion of Science.
Received: 24 June 2013 Accepted: 18 September 2013 Published:07 Nov 2013 References
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