Several fixed point theorems concerning -distance

CONCERNING

_τ

-DISTANCE

Received 21 October 2003 and in revised form 10 March 2004

Using the notion ofτ-distance, we prove several fixed point theorems, which are general-izations of fixed point theorems by Kannan, Meir-Keeler, Edelstein, and Nadler. We also discuss the properties ofτ-distance.

1. Introduction

In 1922, Banach proved the following famous fixed point theorem [1]. Let (X,d) be a complete metric space. LetTbe a contractive mapping onX, that is, there existsr∈[0, 1) satisfying

d(Tx,T y)≤rd(x,y) (1.1)

for allx,y∈X. Then there exists a unique fixed pointx0∈XofT. This theorem, called

the Banach contraction principle, is a forceful tool in nonlinear analysis. This princi-ple has many applications and is extended by several authors: Caristi [2], Edelstein [5], Ekeland [6,7], Meir and Keeler [14], Nadler [15], and others. These theorems are also extended; see [4,9,10, 13,23,25,26,27] and others. In [20], the author introduced the notion ofτ-distance and extended the Banach contraction principle, Caristi’s fixed point theorem, and Ekeland’sε-variational principle. In 1969, Kannan proved the follow-ing fixed point theorem [12]. Let (X,d) be a complete metric space. LetTbe a Kannan mapping onX, that is, there existsα∈[0, 1/2) such that

d(Tx,T y)≤αd(Tx,x) +d(T y,y) (1.2)

for allx,y∈X. Then there exists a unique fixed pointx0∈X ofT. We note that

Kan-nan’s fixed point theorem is not an extension of the Banach contraction principle. We also know that a metric spaceX is complete if and only if every Kannan mapping has a fixed point, while there exists a metric spaceXsuch thatX is not complete and every contractive mapping onXhas a fixed point; see [3,17].

In this paper, using the notion ofτ-distance, we prove several fixed point theorems, which are generalizations of fixed point theorems by Kannan, Meir-Keeler, Edelstein, and Nadler. We also discuss the properties ofτ-distance.

2.τ-distance

Throughout this paper, we denote byNthe set of all positive integers. In this section, we discuss some properties ofτ-distance. Let (X,d) be a metric space. Then a function p fromX×Xinto [0,∞) is called aτ-distanceonX[20] if there exists a functionηfrom X×[0,∞) into [0,∞) and the following are satisfied:

(τ1) p(x,z)≤p(x,y) +p(y,z) for allx,y,z∈X;

(τ2)η(x, 0)=0 andη(x,t)≥tfor allx∈Xandt∈[0,∞), andηis concave and con-tinuous in its second variable;

(τ3) limnxn = x and limnsup{η(zn,p(zn,xm)) : m ≥ n} = 0 imply p(w,x) ≤

lim infnp(w,xn) for allw∈X;

(τ4) limnsup{p(xn,ym) :m≥n} =0 and limnη(xn,tn)=0 imply limnη(yn,tn)=0;

(τ5) limnη(zn,p(zn,xn))=0 and limnη(zn,p(zn,yn))=0 imply limnd(xn,yn)=0.

We may replace (τ2) by the following (τ2)(see [20]):

(τ2)inf{η(x,t) :t >0} =0 for allx∈X, andηis nondecreasing in its second variable. The metricdis aτ-distance onX. Many useful examples are stated in [11,16,18,19,20,

21,22,24]. It is very meaningful that oneτ-distance generates otherτ-distances. In the sequel, we discuss this fact.

Proposition2.1. Let(X,d)be a metric space. Let pbe aτ-distance onX and letηbe a function satisfying (τ2), (τ3), (τ4), and (τ5). Letqbe a function fromX×Xinto[0,∞) satisfying (τ1)q. Suppose that

(i)there existsc >0such thatmin{p(x,y),c} ≤q(x,y)forx,y∈X,

(ii) limnxn=xandlimnsup{η(zn,q(zn,xm)) :m≥n}=0implyq(w,x)≤lim infnq(w,xn)

forw∈X.

Thenqis also aτ-distance onX. Proof. We put

θ(x,t)=t+η(x,t) (2.1)

forx∈X andt∈[0,∞). Note that η(x,t)≤θ(x,t) for allx∈X andt∈[0,∞). Then, by the assumption, (τ1)q, (τ2)’θ, and (τ3)q,θ hold. We assume that limnsup{q(xn,ym) :

m≥n} =0 and limnθ(xn,tn)=0. Then limnsup{p(xn,ym) :m≥n} =0 and limntn=

limnη(xn,tn)= 0 clearly hold. From (τ4), we have limnη(yn,tn)= 0 and hence

limnθ(yn,tn)=0. Therefore, we have shown (τ4)q,θ. We assume that limnθ(zn,q(zn,xn))=

0 and limnθ(zn,q(zn,yn))=0. By the definition ofθ, we have limnη(zn,q(zn,xn))=0 and

limnq(zn,xn)=0. So, by the assumption, limnη(zn,p(zn,xn))=0 holds. We can similarly

prove limnη(zn,p(zn,yn))=0. Therefore, from (τ5), limnd(xn,yn)=0. Hence, we have

shown (τ5)q,θ. This completes the proof.

Proposition2.2. Letpbe aτ-distance on a metric spaceX. Letqbe a function fromX×X into[0,∞)satisfying (τ1)q. Suppose that

(i)there existsc >0such thatmin{p(x,y),c} ≤q(x,y)forx,y∈X,

(ii)for every convergent sequence{xn}with limitxsatisfyingp(w,x)≤lim infnp(w,xn)

for allw∈X,q(w,x)≤lim inf_nq(w,xn)holds for allw∈X.

Thenqis also aτ-distance onX.

Using the above proposition, we obtain the following one which is used in the proof of generalized Kannan’s fixed point theorem.

Proposition2.3. Letpbe aτ-distance on a metric spaceXand letαbe a function fromX into[0,∞). Then two functionsq1andq2fromX×Xinto[0,∞), defined by

(i)q1(x,y)=max{α(x),p(x,y)}forx,y∈X,

(ii)q2(x,y)=α(x) +p(x,y)forx,y∈X,

areτ-distances onX. Proof. We have

q1(x,z)=max

α(x),p(x,z)

≤maxα(x) +α(y),p(x,y) +p(y,z) ≤q1(x,y) +q1(y,z),

q2(x,z)=α(x) +p(x,z)

≤α(x) +α(y) +p(x,y) +p(y,z) =q2(x,y) +q2(y,z),

(2.2)

for allx,y,z∈X. We note that

p(x,y)≤q1(x,y)≤q2(x,y) (2.3)

for all x,y∈X. We assume that a sequence {xn} satisfies limnxn=x and p(w,x)≤

lim infnp(w,xn) for all w∈X. Then it is clear that q1(w,x)≤lim infnq1(w,xn) and

q2(w,x)≤lim infnq2(w,xn) for allw∈X. ByProposition 2.2,q1 andq2areτ-distances

onX. This completes the proof.

Let (X,d) be a metric space and letpbe aτ-distance onX. Then a sequence{xn}in

Xis calledp-Cauchy[20] if there exist a functionηfromX×[0,∞) into [0,∞) satisfying (τ2)–(τ5) and a sequence{zn}inXsuch that limnsup{η(zn,p(zn,xm)) :m≥n} =0. The

following lemmas are very useful in the proofs of fixed point theorems inSection 3.

Lemma2.4 [20]. Let (X,d)be a metric space and let pbe a τ-distance onX. If{xn}is

a p-Cauchy sequence, then {xn} is a Cauchy sequence. Moreover, if {yn} is a sequence

satisfying limnsup{p(xn,ym) :m≥n} =0, then {yn} is also a p-Cauchy sequence and

limnd(xn,yn)=0.

Moreover, if a sequence{yn}inXalso satisfieslimnp(z,yn)=0, thenlimnd(xn,yn)=0. In

particular, forx,y,z∈X,p(z,x)=0andp(z,y)=0implyx=y.

Lemma 2.6 [20]. Let (X,d) be a metric space and let p be a τ-distance on X. If a se-quence{xn} inX satisfieslimnsup{p(xn,xm) :m > n} =0, then {xn} is a p-Cauchy

se-quence. Moreover, if a sequence{yn}inX satisfieslimnp(xn,yn)=0, then{yn}is also a

p-Cauchy sequence andlim_nd(xn,yn)=0.

3. Fixed point theorems

In this section, we prove several fixed point theorems in complete metric spaces. In [20], the following theorem connected with Hicks-Rhoades theorem [8] was proved and used in the proofs of generalizations of the Banach contraction principle, Caristi’s fixed point theorem, and so on. In this paper, this theorem is used in the proof of a generalization of Kannan’s fixed point theorem.

Theorem3.1 [20]. LetXbe a complete metric space and letTbe a mapping onX. Suppose that there exist aτ-distanceponXandr∈[0, 1)such thatp(Tx,T2_{x)≤r p(x,Tx)for all}

x∈X. Assume that either of the following holds:

(i)if limnsup{p(xn,xm) :m > n} =0,limnp(xn,Txn)=0, andlimnp(xn,y)=0, then

T y=y;

(ii)if{xn}and{Txn}converge toy, thenT y=y;

(iii)Tis continuous.

Then there existsx0∈Xsuch thatTx0=x0. Moreover, ifTz=z, thenp(z,z)=0.

As a direct consequence, we obtain the following theorem.

Theorem3.2. LetXbe a complete metric space and letpbe aτ-distance onX. LetTbe a mapping onX. Suppose that there existsr∈[0, 1)such that either (a) or (b) holds:

(a) max{p(T2_x,Tx),p(Tx,T2_{x)} ≤rmax{p(Tx,x),p(x,Tx)}for allx∈X;}

(b) p(T2_{x,Tx) +p(Tx,T}2_{x)≤r p(Tx,x) +r p(x,Tx)for allx∈X.}

Further, assume that either of the following holds:

(i)if lim_nsup{p(xn,xm) :m > n} =0, limnp(Txn,xn)=0,limnp(xn,Txn)=0, and

limnp(xn,y)=0, thenT y=y;

(ii)if{xn}and{Txn}converge toy, thenT y=y;

(iii)Tis continuous.

Then there existsx0∈Xsuch thatTx0=x0. Moreover, ifTz=z, thenp(z,z)=0.

Proof. In the case of (a), we define a functionqbyq(x,y)=max{p(Tx,x),p(x,y)}. In the case of (b), we define a functionqbyq(x,y)=p(Tx,x) +p(x,y). ByProposition 2.3, qis aτ-distance onX. In both cases, we have

for allx∈X. Conditions (ii) and (iii) are not connected withτ-distancep. In the case of (i), since

p(x,y)≤q(x,y), p(Tx,x)≤q(x,Tx), (3.2)

for allx,y∈X,Thas a fixed point inXbyTheorem 3.1. IfTz=z, thenq(z,z)=0, and

hencep(z,z)=0. This completes the proof.

We now prove a generalization of Kannan’s fixed point theorem [12]. LetXbe a metric space, letpbe aτ-distance onX, and letTbe a mapping onX. ThenTis called aKannan mappingwith respect topif there existsα∈[0, 1/2) such that either (a) or (b) holds:

(a) p(Tx,T y)≤αp(Tx,x) +αp(T y,y) for allx,y∈X; (b) p(Tx,T y)≤αp(Tx,x) +αp(y,T y) for allx,y∈X.

Theorem3.3. Let(X,d)be a complete metric space, letpbe aτ-distance onX, and letTbe a Kannan mapping onXwith respect top. ThenThas a unique fixed pointx0∈X. Further,

suchx0satisfiesp(x0,x0)=0.

Proof. In the case of (a), there exists α∈[0, 1/2) such that p(Tx,T y)≤αp(Tx,x) + αp(T y,y) forx,y∈X. Since

pT2_x,Tx≤αpT2_{x,Tx+αp(Tx,x), (3.3)}

we have

pT2_x,Tx≤ α

1−αp(Tx,x)≤p(Tx,x) (3.4)

forx∈X. Puttingr=2α∈[0, 1), we have

maxpT2_x,Tx,pTx,T2_x≤αpT2_{x,Tx+αp(Tx,x)}

≤r p(Tx,x)

≤rmaxp(Tx,x),p(x,Tx)

(3.5)

for all x ∈ X. We assume lim_nsup{p(xn,xm) : m > n} =0, limnp(Txn,xn)= 0,

limnp(xn,Txn)=0, and limnp(xn,y)=0. Then, byLemma 2.6,{xn}and{Txn}are p

-Cauchy and

lim

n→∞d

xn,y

=lim

n→∞d

Txn,y

=0. (3.6)

Now we have

p(T y,y)≤lim inf

n→∞ p

T y,Txn

≤lim inf

n→∞

αp(T y,y) +αpTxn,xn

=αp(T y,y),

and hencep(T y,y)=0. Sincep(T2_{y,T y)≤p(T y,y)=0 andp(T}2_y,y)≤p(T2_{y,T y) +}

p(T y,y)=0, we have T y=y by Lemma 2.5. Therefore, by Theorem 3.2, there exists x0∈Xsuch thatTx0=x0andp(x0,x0)=0. Further, a fixed point ofTis unique. In fact,

ifTz=z, thenp(z,z)=0 byTheorem 3.2. So we have

px0,z

=pTx0,Tz

≤αpTx0,x0

+αp(Tz,z) =αpx0,x0

+αp(z,z)=0. (3.8)

ByLemma 2.5again, we havex0=z. In the case of (b), there existsα∈[0, 1/2) such that

p(Tx,T y)≤αp(Tx,x) +αp(y,T y) forx,y∈X. Then, puttingr=α/(1−α)∈[0, 1), we havep(Tx,T2_{x)≤r p(Tx,x) andp(T}2_{x,Tx)≤r p(x,Tx) for allx∈X. So,}

pT2_x,Tx+pTx,T2_{x≤r p(Tx,x) +r p(x,Tx) (3.9)}

for allx∈X. We assume lim_nsup{p(xn,xm) :m > n} =0, limnp(Txn,xn)=0, limnp(xn,

Txn)=0, and limnp(xn,y)=0. Then{xn}and{Txn}arep-Cauchy and limnd(xn,y)=

limnd(Txn,y)=0. So we have

p(T y,y)≤lim inf

n→∞ p

T y,Txn

≤lim inf

n→∞

αp(T y,y) +αpxn,Txn

=αp(T y,y),

(3.10)

and hencep(T y,y)=0. Since p(T y,T2_{y)≤r p(T y,y)=0, we havey=T}2_ybyLemma 2.5. So,p(y,T y)=p(T2_{y,T y)≤r p(y,T y), and hencep(y,T y)=0. We also havep(y,y)≤}

p(y,T y) +p(T y,y)=0. So we haveT y=ybyLemma 2.5. Therefore, byTheorem 3.2, there existsx0∈Xsuch thatTx0=x0and p(x0,x0)=0. As in the case of (a), we obtain

that a fixed point ofTis unique.

In general,τ-distance pdoes not satisfy p(x,y)=p(y,x). So conditions (a) and (b) in the definition of Kannan mappings differ from conditions (c) and (d) in the following theorem. Indeed, there exists a mappingT on a complete metric spaceX such that (c) and (d) hold, andThas no fixed points; see [19]. However, under the assumption thatT is continuous,Thas a fixed point.

Theorem3.4. LetXbe a complete metric space and letT be a continuous mapping onX. Suppose that there exist aτ-distance p onX andα∈[0, 1/2)such that either (c) or (d) holds:

(c)p(Tx,T y)≤αp(x,Tx) +αp(T y,y)for allx,y∈X; (d) p(Tx,T y)≤αp(x,Tx) +αp(y,T y)for allx,y∈X.

Then there exists a unique fixed pointx0∈XofT. Moreover, suchx0satisfiesp(x0,x0)=0.

Proof. In the case of (c), puttingr=α/(1−α)∈[0, 1), from p(Tx,T2_{x)≤αp(x,Tx) +}

αp(T2_{x,Tx) andp(T}2_{x,Tx)≤αp(Tx,T}2_{x) +αp(Tx,x), we have}

for allx∈X. So, byTheorem 3.2, we prove the desired result. In the case of (d), we have p(Tx,T2_{x)≤r p(x,Tx) for allx∈X. Therefore, byTheorem 3.1, we prove the desired}

result. This completes the proof.

We next prove a generalization of Meir and Keeler’s fixed point theorem [14].

Theorem3.5. LetXbe a complete metric space, let pbe aτ-distance onX, and letT be a mapping onX. Suppose that for anyε >0, there existsδ >0such that for everyx,y∈X, p(x,y)< ε+δimpliesp(Tx,T y)< ε. ThenThas a unique fixed pointx0inX. Further, such

x0satisfiesp(x0,x0)=0.

Proof. We first showp(Tx,T y)≤p(x,y) for allx,y∈X. For an arbitraryλ >0, there exists δ >0 such that for everyz,w∈X, p(z,w)< p(x,y) +λ+δ implies p(Tz,Tw)< p(x,y) +λ. Since p(x,y)< p(x,y) +λ+δ, we have p(Tx,T y)< p(x,y) +λ. Sinceλ >0 is arbitrary, we obtainp(Tx,T y)≤p(x,y). We next show

lim

n→∞p

Tnx,Tny=0 ∀x,y∈X. (3.12)

In fact,{p(Tn_x,Tn_{y)}is nonincreasing and hence converges to some real numberr. We}

assumer >0. Then there existsδ >0 such that for everyz,w∈X,p(z,w)< r+δimplies p(Tz,Tw)< r. For suchδ, we can choosem∈Nsuch that p(Tm_x,Tm_{y)< r+δ. So we}

havep(Tm+1_x,Tm+1_{y)< r. This is a contradiction, and hence (3.12) holds. Letu∈Xand}

putun=Tnufor everyn∈N. From (3.12), we have limnp(un,un+1)=0. We will show

that

lim

n→∞supm>np

un,um

=0. (3.13)

Letε >0 be arbitrary. Then, without loss of generality, there existsδ∈(0,ε) such that for everyz,w∈X,p(z,w)< ε+δimpliesp(Tz,Tw)< ε. For suchδ, there existsn0∈N

such that p(un,un+1)< δfor everyn≥n0. Assume that there existsm > ≥n0such that

p(u,um)>2ε. Since

pu,u+1

< ε+δ < pu,um

, (3.14)

there existsk∈Nwith < k < msuch that

pu,uk

< ε+δ≤pu,uk+1

. (3.15)

Then, sincep(u,uk)< ε+δ, we havep(u+1,uk+1)< ε. On the other hand, we have

pu,uk+1

≤pu,u+1

+pu+1,uk+1

< δ+ε. (3.16)

This is a contradiction. Therefore,m > n≥n0impliesp(un,um)≤2ε, and hence (3.13)

Hence there existsx0∈Xsuch that{un}converges tox0. From (τ3), we have

lim sup

n→∞ p

un,Tx0

≤lim sup

n→∞ p

un−1,x0

=lim sup

n→∞ p

un,x0

≤lim sup

n→∞ lim infm→∞ p

un,um

≤lim

n→∞supm>np

un,um

=0.

(3.17)

ByLemma 2.6again,{un}converges toTx0, and henceTx0=x0. From (3.12), we obtain

px0,x0

=lim

n→∞p

Tn_x

0,Tnx0

=0. (3.18)

Ifz=Tz, then

px0,z=lim

n→∞p

Tnx0,Tnz=0. (3.19)

So, fromLemma 2.5,x0=z. Therefore, a fixed point ofTis unique. This completes the

proof.

LetX be a metric space and let p be a τ-distance onX. For ε∈(0,∞], X is called ε-chainable with respect to p if, for each (x,y)∈X×X, there exists a finite sequence {u0,u1,u2,...,u}inXsuch thatu0=x,u=y, and p(ui−1,ui)< εfori=1, 2,...,. We

will prove a generalization of Edelstein’s fixed point theorem [5].

Theorem3.6. LetXbe a complete metric space. Suppose thatXisε-chainable with respect topfor someε∈(0,∞]and for someτ-distanceponX. LetTbe a mapping onX. Suppose that there existsr∈[0, 1)such thatp(Tx,T y)≤r p(x,y)for allx,y∈Xwithp(x,y)< ε. Then there exists a unique fixed pointx0∈XofT. Further, suchx0satisfiesp(x0,x0)=0.

Proof. We first show

lim

n→∞p

Tnx,Tny=0 (3.20)

for allx,y∈X. Letx,y∈Xbe fixed. Then there existu0,u1,u2,...,u∈Xsuch thatu0=

x,u=y, andp(ui−1,ui)<εfori=1, 2,...,. Sincep(ui−1,ui)< ε, we havep(Tui−1,Tui)≤

r p(ui−1,ui)< ε. Thus

pTnui−1,Tnui

≤r pTn−1ui−1,Tn−1ui

≤ ··· ≤rnpui−1,ui

. (3.21)

Therefore

lim sup

n→∞ p

Tn_x,Tn_{y≤lim sup} n→∞

i=1

pTn_u i−1,Tnui

≤lim

n→∞

i=1

rnpui−1,ui=0.

We have shown (3.20). Letx∈Xbe fixed. From (3.20), there existsn0∈Nsuch that

pTnx,Tn+1x< ε (3.23)

forn≥n0. Then, form > n≥n0, we have

pTnx,Tmx≤

m−1

k=n

pTkx,Tk+1x

≤

m−1

k=n

rk−n0_pTn0_x,Tn0+1_x

≤rn−n0 1−rp

Tn0_x,Tn0+1_x.

(3.24)

Hence, limnsup{p(Tnx,Tmx) :m > n} =0. ByLemma 2.6,{Tnx}isp-Cauchy. ByLemma

2.4,{Tn_{x}is a Cauchy sequence. So,{T}n_{x}converges to somex0∈X. Since}

lim sup

n→∞ p

Tn_x,x

0

≤lim sup

n→∞ lim infm→∞ p

Tn_x,Tm_x

≤lim

n→∞supm>np

Tn_x,Tm_x=0, (3.25)

we have

lim sup

n→∞ p

Tnx,Tx0

≤lim

n→∞r p

Tn−1x,x0

=0. (3.26)

ByLemma 2.6, we obtainTx0=x0. Ifzis a fixed point ofT, then we have

px0,z=lim

n→∞p

Tnx0,Tnz=0 (3.27)

from (3.20). We also havep(x0,x0)=0. Therefore,z=x0byLemma 2.5. This completes

the proof.

LetXbe a metric space and let pbe aτ-distance onX. Then, a set-valued mappingT fromXinto itself is calledp-contractiveifTxis nonempty for eachx∈Xand there exists r∈[0, 1) such that

Q(Tx,T y)≤r p(x,y) (3.28)

for allx,y∈X, where

Q(A,B)=sup

a∈A

inf

b∈Bp

(a,b). (3.29)

The following theorem is a generalization of Nadler’s fixed point theorem [15].

Remark 3.8. z∈Tzdoes not necessarily implyp(z,z)=0; seeExample 3.9.

Proof. By the assumption, there existsr∈[0, 1) such thatQ(Tx,T y)≤rp(x,y) for all x,y∈X. Put r=(1 +r)/2∈[0, 1) and fix x,y∈X andu∈Tx. Then, in the case of p(x,y)>0, there is v∈T y satisfying p(u,v)≤r p(x,y). In the case of p(x,y)=0, we haveQ(Tx,T y)=0. Then there exists a sequence{vn}inT ysatisfying limnp(u,vn)=0.

ByLemma 2.5,{vn}isp-Cauchy, and hence{vn}is Cauchy. SinceXis complete andT y

is closed,{vn}converges to some pointv∈T y. Then we have

p(u,v)≤lim

n→∞p

u,vn

=0=r p(x,y). (3.30)

Hence, we have shown that for anyx,y∈Xandu∈Tx, there isv∈T ywithp(u,v)≤ r p(x,y). Fixu0∈X and u1∈Tu0. Then there exists u2∈Tu1 such that p(u1,u2)≤

r p(u0,u1). Thus, we have a sequence{un}inXsuch thatun+1∈Tunandp(un,un+1)≤

r p(un−1,un) for alln∈N. For anyn∈N, we have

pun,un+1≤r pun−1,un≤r2pun−2,un−1≤ ··· ≤rnpu0,u1, (3.31)

and hence, for anym,n∈Nwithm > n,

pun,um

≤pun,un+1

+pun+1,un+2

+···+pum−1,um

≤rnpu0,u1

+rn+1pu0,u1

+···+rm−1pu0,u1

≤ rn 1−rp

u0,u1

.

(3.32)

ByLemma 2.6,{un}is a p-Cauchy sequence. Hence, byLemma 2.4,{un}is a Cauchy

sequence. So,{un}converges to some pointv0∈X. Forn∈N, from (τ3), we have

pun,v0

≤lim inf

m→∞ p

un,um

≤ rn 1−rp

u0,u1

. (3.33)

By hypothesis, we also havewn∈Tv0such thatp(un,wn)≤r p(un−1,v0) forn∈N. So we

have

lim sup

n→∞ p

un,wn

≤lim sup

n→∞ r p

un−1,v0

≤lim

n→∞ rn

1−rp

u0,u1

=0.

(3.34)

ByLemma 2.6,{wn}converges tov0. SinceTv0is closed, we havev0∈Tv0. For suchv0,

there existsv1∈Tv0such thatp(v0,v1)≤r p(v0,v0). Thus, we also have a sequence{vn}

inXsuch thatvn+1∈Tvnandp(v0,vn+1)≤r p(v0,vn) for alln∈N. So we have

pv0,vn≤r pv0,vn−1

≤ ··· ≤rn_pv

0,v0

. (3.35)

Hence

lim sup

n→∞ p

un,vn

≤lim

n→∞

pun,v0

+pv0,vn

ByLemma 2.6again,{vn}is ap-Cauchy sequence and converges tov0. So we have

pv0,v0

≤lim

n→∞p

v0,vn

=0. (3.37)

This completes the proof.

Example 3.9. PutX= {0, 1}and define aτ-distanceponXbyp(x,y)=yfor allx,y∈X, and a set-valuedp-contractive mappingT fromX into itself byT(x)=Xfor allx∈X. Then 1∈Xis a fixed point ofTandp(1, 1)=0.

4. Other examples ofτ-distances

In this section, we give other examples ofτ-distances generated by either someτ-distance por a family ofτ-distances.

Proposition4.1. Let pbe aτ-distance on a metric spaceX. Fixc >0. Define a functionq fromX×Xinto[0,∞)by

q(x,y)=minp(x,y),c (4.1)

forx,y∈X. Thenqis also aτ-distance onX.

Proof. Fixx,y,z∈X. In the case ofp(x,y)< candp(y,z)< c, we have

q(x,z)≤p(x,z)≤p(x,y) +p(y,z)=q(x,y) +q(y,z). (4.2)

In the case ofp(x,y)≥corp(y,z)≥c, we have

q(x,z)≤c≤q(x,y) +q(y,z). (4.3)

Therefore, we have shown (τ1)q. So, by Proposition 2.2, we obtain the desired result.

Proposition4.2. Let(X,d)be a metric space. Let{pn}be a sequence ofτ-distances onX.

Then the following hold. (i)A functionq1, defined by

q1(x,y)=maxp1(x,y),p2(x,y) (4.4)

forx,y∈X, is aτ-distance onX. (ii)A functionq2, defined by

q2(x,y)=p1(x,y) +p2(x,y) (4.5)

forx,y∈X, is aτ-distance onX. (iii)For eachc >0, a functionq3, defined by

q3(x,y)=min

sup

n∈Npn(x,y),c

(4.6)

(iv)For eachc >0, a functionq4, defined by

q4(x,y)=min _∞

n=1

pn(x,y),c (4.7)

forx,y∈X, is aτ-distance onX. (v)If a functionq5, defined by

q5(x,y)=sup

n∈Npn(x,y) (4.8)

forx,y∈X, is a real-valued function, thenq5is aτ-distance onX.

(vi)If a functionq6, defined by

q6(x,y)= ∞

n=1

pn(x,y) (4.9)

forx,y∈X, is a real-valued function, thenq6is aτ-distance onX.

Proof. Let{ηn}be a sequence of functions satisfying (τ2)pn,ηn, (τ3)pn,ηn, (τ4)pn,ηn, and (τ5)pn,ηnforn∈N. We first prove thatq5is aτ-distance onX. Since

sup

n∈Npn(x,z)≤supn∈N

pn(x,y) +pn(y,z)

≤sup

n∈Npn(x,y) + supn∈Npn(y,z), (4.10)

we haveq5(x,z)≤q5(x,y) +q5(y,z) forx,y,z∈X. Define a functionθ fromX×[0,∞)

into [0,∞) by

θ(x,t)=t+

∞

n=1

21−n_minη n(x,t), 1

(4.11)

forx∈Xandt∈[0,∞). Fixx∈X. For anyε >0, we choosek1∈Nwith 1/k1+ 21−k1< ε/2.

Then there existst1∈(0,ε/2) satisfying

k1

n=1

21−nηn

x,t1

< 1

k1. (4.12)

Hence

θx,t1< t1+ 1

k1+ ∞

n=k1+1

21−n_minη

nx,t1, 1≤ ε

2+ 1 k1+ 2

1−k1_{< ε. (4.13)}

Therefore,θ(x,·) is continuous at 0. Hence, (τ2)θ is shown. We suppose limnxn=xand

limnsup{θ(zn,q5(zn,xm)) :m≥n} =0. Then, for anyk∈N, we have

lim sup

n→∞ supm≥nmin

ηkzn,pkzn,xm, 1

≤lim sup

n→∞ supm≥nmin

ηkzn,q5zn,xm, 1

≤lim

n→∞supm≥n2 k−1_θz

n,q5

zn,xm=0,

and hence

lim

n→∞supm≥nηk

zn,pk

zn,xm

=0. (4.15)

From (τ3)pk,ηk,

pk(w,x)≤lim inf_n→∞ pk

w,xn

(4.16)

for allw∈X. Therefore, we have

sup

k∈Npk(w,x)≤supk∈Nlim infn→∞ pk

w,xn

≤lim inf

n→∞ supk∈Npk

w,xn,

(4.17)

and henceq5(w,x)≤lim infnq5(w,xn) for allw∈X. We have shown (τ3)q5,θ. We prove

(τ4)q5,θ. We assume that limnsup{q5(xn,ym) :m≥n} =0 and limnθ(xn,tn)=0. Then we

have limnsup{pk(xn,ym) :m≥n} =0 and limnηk(xn,tn)=0 for allk∈N. From (τ4)pk,ηk, we have limnηk(yn,tn)=0 for all k∈N. For any ε >0, we choosek2∈Nwith 1/k2+

21−k2_{< ε/2. Then there existsn}

2∈Nsatisfying

k2

k=1

21−kηk

yn,tn

< 1

k2 (4.18)

andtn< ε/2 forn≥n2. We now have

θyn,tn

≤tn+ 1

k2+ 2

1−k2_{< ε (4.19)}

for n≥n2. This implies limnθ(yn,tn)=0. We prove (τ5)q5,θ. We assume limnθ(zn,

q5(zn,xn))=0 and limnθ(zn,q5(zn,yn))=0. Then we have

lim sup

n→∞ min

η1zn,p1zn,xn, 1≤lim sup n→∞ θ

zn,p1zn,xn

≤lim

n→∞θ

zn,q5zn,xn=0,

(4.20)

and hence lim_nη1(zn,p1(zn,xn))=0. We can similarly prove limnη1(zn,p1(zn,yn))=0.

Therefore, we obtain limnd(xn,yn)=0. We have shown thatq5is aτ-distance onX. We

next prove thatq6is aτ-distance onX. Since

∞

n=1

pn(x,z)≤

∞

n=1

pn(x,y) +pn(y,z)=

∞

n=1

pn(x,y) +

∞

n=1

pn(y,z), (4.21)

we haveq6(x,z)≤q6(x,y) +q6(y,z) for x,y,z∈X. We note thatq5(x,y)≤q6(x,y) for

have limnsup{θ(zn,q5(zn,xm)) :m≥n} =0. In such case, we have already shown that

pk(w,x)≤lim infnpk(w,xn) forw∈Xandk∈N. Fixλwith

λ < ∞

k=1

lim inf

n→∞ pk

w,xn. (4.22)

Then there existk3,n3∈Nsuch thatλ <_kk₌31pk(w,xn) forn≥n3. Hence

λ≤lim inf

n→∞

k3

k=1

pk

w,xn

≤lim inf

n→∞ ∞

k=1

pk

w,xn

. (4.23)

Therefore, we have

∞

k=1

pk(w,x)≤

∞

k=1

lim inf

n→∞ pk

w,xn

≤lim inf

n→∞ ∞

k=1

pk

w,xn

, (4.24)

and henceq6(w,x)≤lim infnq6(w,xn) forw∈X. ByProposition 2.1,q6 is aτ-distance

onX. Since

q1(x,y)=sup

p1(x,y),p2(x,y),p2(x,y),p2(x,y),...

,

q2(x,y)=p1(x,y) +1

2p2(x,y) + 1

4p2(x,y) + 1

8p2(x,y) +···,

(4.25)

q1andq2areτ-distances onX. Since

q3(x,y)=sup

n∈Nmin

pn(x,y),c

,

q4(x,y)=sup

n∈Nmin _n

k=1

pk(x,y),c ,

(4.26)

q3andq4areτ-distances onX. This completes the proof.

Acknowledgment

The author wishes to express his sincere thanks to the referee for giving many suggestions concerning English expressions.

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Tomonari Suzuki: Department of Mathematics, Physics, and Computer-Aided Science, Kyushu Institute of Technology, 1-1 Sensuicho, Tobataku, Kitakyushu 804-8550, Japan