CONCERNING τ -DISTANCE

CONCERNING τ -DISTANCE

TOMONARI SUZUKI

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].

Copyright©2004 Hindawi Publishing Corporation Fixed Point Theory and Applications 2004:3 (2004) 195–209 2000 Mathematics Subject Classification: 54H25, 54E50 URL:http://dx.doi.org/10.1155/S168718200431003X

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) lim_nx_n = x and lim_nsup{η(z_n,p(z_n,x_m)) : m ≥ n} = 0 imply p(w,x) ≤ lim inf_np(w,xn) for allw∈X;

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

(τ5) lim_nη(z_n,p(z_n,x_n))=0 and lim_nη(z_n,p(z_n,y_n))=0 imply lim_nd(x_n,y_n)=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) lim_nxn=xandlim_nsup{η(zn,q(zn,xm)) :m≥n}=0implyq(w,x)≤lim inf_nq(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 lim_nsup{q(x_n,y_m) : m≥n} =0 and lim_nθ(xn,tn)=0. Then lim_nsup{p(xn,ym) :m≥n} =0 and lim_ntn= lim_nη(xn,tn)= 0 clearly hold. From (τ4), we have limnη(yn,tn)= 0 and hence lim_nθ(y_n,t_n)=0. Therefore, we have shown (τ4)_q,θ. We assume that lim_nθ(z_n,q(z_n,x_n))= 0 and lim_nθ(zn,q(zn,yn))=0. By the definition ofθ, we have limnη(zn,q(zn,xn))=0 and lim_nq(zn,xn)=0. So, by the assumption, lim_nη(zn,p(zn,xn))=0 holds. We can similarly prove lim_nη(z_n,p(z_n,y_n))=0. Therefore, from (τ5), lim_nd(x_n,y_n)=0. Hence, we have

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

As a direct consequence ofProposition 2.1, we obtain the following proposition.

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 inf_np(w,xn) for allw∈X,q(w,x)≤lim inf_nq(w,x_n)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)=maxα(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 {x_n} satisfies lim_nx_n=x and p(w,x)≤ lim inf_np(w,xn) for all w∈X. Then it is clear that q1(w,x)≤lim inf_nq1(w,xn) and q2(w,x)≤lim inf_nq2(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{x_n}in Xis calledp-Cauchy[20] if there exist a functionηfromX×[0,∞) into [0,∞) satisfying (τ2)–(τ5) and a sequence{zn}inXsuch that lim_nsup{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{x_n}is a p-Cauchy sequence, then {xn} is a Cauchy sequence. Moreover, if {yn} is a sequence satisfying lim_nsup{p(xn,ym) :m≥n} =0, then {yn} is also a p-Cauchy sequence and lim_nd(x_n,y_n)=0.

Lemma2.5 [20]. Let(X,d)be a metric space and letpbe aτ-distance onX. If a sequence {xn} inX satisfieslim_np(z,xn)=0 for somez∈X, then{xn}is a p-Cauchy sequence.

Moreover, if a sequence{y_n}inXalso satisfieslim_np(z,y_n)=0, thenlim_nd(x_n,y_n)=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 satisfieslim_nsup{p(xn,xm) :m > n} =0, then {xn} is a p-Cauchy se- quence. Moreover, if a sequence{y_n}inX satisfieslim_np(x_n,y_n)=0, then{y_n}is also a p-Cauchy sequence andlim_nd(x_n,y_n)=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,T²x)≤r p(x,Tx)for all x∈X. Assume that either of the following holds:

(i)if lim_nsup{p(xn,xm) :m > n} =0,lim_np(xn,Txn)=0, andlim_np(xn,y)=0, then T y=y;

(ii)if{x_n}and{Tx_n}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(T²x,Tx),p(Tx,T²x)} ≤rmax{p(Tx,x),p(x,Tx)}for allx∈X;

(b) p(T²x,Tx) +p(Tx,T²x)≤r p(Tx,x) +r p(x,Tx)for allx∈X.

Further, assume that either of the following holds:

(i)if lim_nsup{p(x_n,x_m) :m > n} =0, lim_np(Tx_n,x_n)=0,lim_np(x_n,Tx_n)=0, and lim_np(xn,y)=0, thenT y=y;

(ii)if{x_n}and{Tx_n}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

qTx,T²x≤rq(x,Tx) (3.1)

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

pT²x,Tx≤αpT²x,Tx+αp(Tx,x), (3.3) we have

pT²x,Tx≤ α

1−αp(Tx,x)≤p(Tx,x) (3.4) forx∈X. Puttingr=2α∈[0, 1), we have

maxpT²x,Tx,pTx,T²x≤αpT²x,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(x_n,x_m) : m > n} =0, lim_np(Tx_n,x_n)= 0, lim_np(xn,Txn)=0, and lim_np(xn,y)=0. Then, byLemma 2.6,{xn}and{Txn}are p- Cauchy and

nlim→∞dx_n,y=lim

n→∞dTx_n,y=0. (3.6)

Now we have

p(T y,y)≤lim inf

n→∞ pT y,Tx_n

≤lim inf

n→∞

αp(T y,y) +αpTxn,xn

=αp(T y,y),

(3.7)

and hencep(T y,y)=0. Sincep(T²y,T y)≤p(T y,y)=0 andp(T²y,y)≤p(T²y,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,T²x)≤r p(Tx,x) andp(T²x,Tx)≤r p(x,Tx) for allx∈X. So,

pT²x,Tx+pTx,T²x≤r p(Tx,x) +r p(x,Tx) (3.9) for allx∈X. We assume lim_nsup{p(x_n,x_m) :m > n} =0, lim_np(Tx_n,x_n)=0, lim_np(x_n, Txn)=0, and lim_np(xn,y)=0. Then{xn}and{Txn}arep-Cauchy and limnd(xn,y)= lim_nd(Txn,y)=0. So we have

p(T y,y)≤lim inf

n→∞ pT 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,T²y)≤r p(T y,y)=0, we havey=T²ybyLemma 2.5. So,p(y,T y)=p(T²y,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,T²x)≤αp(x,Tx) + αp(T²x,Tx) andp(T²x,Tx)≤αp(Tx,T²x) +αp(Tx,x), we have

pT²x,Tx+pTx,T²x≤r p(Tx,x) +r p(x,Tx) (3.11)

for allx∈X. So, byTheorem 3.2, we prove the desired result. In the case of (d), we have p(Tx,T²x)≤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

nlim→∞pTⁿx,Tⁿy=0 ∀x,y∈X. (3.12) In fact,{p(Tⁿx,Tⁿy)}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(T^mx,T^my)< r+δ. So we havep(T^m+1x,T^m+1y)< r. This is a contradiction, and hence (3.12) holds. Letu∈Xand putun=Tⁿufor everyn∈N. From (3.12), we have lim_np(un,un+1)=0. We will show that

lim

n→∞sup

m>npun,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,u_m, (3.14)

there existsk∈Nwith < k < msuch that

pu,u_k< ε+δ≤pu,u_k+1

. (3.15)

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

≤pu,u+1

+pu+1,u_k+1

< δ+ε. (3.16)

This is a contradiction. Therefore,m > n≥n0impliesp(u_n,u_m)≤2ε, and hence (3.13) holds. ByLemma 2.6,{un}isp-Cauchy. So,{un}is also a Cauchy sequence byLemma 2.4.

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

n→∞ pun,Tx0

≤lim sup

n→∞ pun−1,x0

=lim sup

n→∞ pun,x0

≤lim sup

n→∞ lim inf

m→∞ pun,um

≤lim

n→∞sup

m>npun,um

=0.

(3.17)

ByLemma 2.6again,{u_n}converges toTx0, and henceTx0=x0. From (3.12), we obtain px0,x0

=lim

n→∞pTⁿx0,Tⁿx0

=0. (3.18)

Ifz=Tz, then

px0,z=lim

n→∞pTⁿx0,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

nlim→∞pTⁿx,Tⁿy=0 (3.20)

for allx,y∈X. Letx,y∈Xbe fixed. Then there existu0,u1,u2,...,u∈Xsuch thatu0= x,u=y, andp(u_i−1,u_i)<εfori=1, 2,...,. Sincep(u_i−1,u_i)< ε, we havep(Tu_i−1,Tu_i)≤ 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→∞ pTⁿx,Tⁿy≤lim sup

n→∞

i=1

pTⁿu_i−1,Tⁿu_i

≤lim

n→∞

i=1

rⁿpui−1,ui

=0.

(3.22)

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,T^mx≤

m−1 k=n

pT^kx,T^k+1x

≤

m−1 k=n

r^k−n0pTⁿ⁰x,Tⁿ⁰⁺¹x

≤rⁿ⁻ⁿ⁰

1−rpTⁿ⁰x,Tⁿ⁰⁺¹x.

(3.24)

Hence, lim_nsup{p(Tⁿx,T^mx) :m > n} =0. ByLemma 2.6,{Tⁿx}isp-Cauchy. ByLemma 2.4,{Tⁿx}is a Cauchy sequence. So,{Tⁿx}converges to somex0∈X. Since

lim sup

n→∞ pTⁿx,x0

≤lim sup

n→∞ lim inf

m→∞ pTⁿx,T^mx

≤lim

n→∞sup

m>npTⁿx,T^mx=0, (3.25) we have

lim sup

n→∞ pTⁿx,Tx0

≤lim

n→∞r pTⁿ⁻¹x,x0

=0. (3.26)

ByLemma 2.6, we obtainTx0=x0. Ifzis a fixed point ofT, then we have px0,z=lim

n→∞pTⁿx0,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].

Theorem3.7. Let(X,d)be a complete metric space and letpbe aτ-distance onX. LetT be a set-valued p-contractive mapping fromXinto itself such that for anyx∈X,Txis a nonempty closed subset ofX. Then there existsx0∈Xsuch thatx0∈Tx0andp(x0,x0)=0.

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{v_n}inT ysatisfying lim_np(u,v_n)=0.

ByLemma 2.5,{v_n}isp-Cauchy, and hence{v_n}is Cauchy. SinceXis complete andT y is closed,{vn}converges to some pointv∈T y. Then we have

p(u,v)≤lim

n→∞pu,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{u_n}inXsuch thatu_n+1∈Tu_nandp(u_n,u_n+1)≤ r p(u_n−1,u_n) 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

+···+r^m−1pu0,u1

≤ rⁿ

1−rpu0,u1

.

(3.32)

ByLemma 2.6,{u_n}is a p-Cauchy sequence. Hence, byLemma 2.4,{u_n}is a Cauchy sequence. So,{un}converges to some pointv0∈X. Forn∈N, from (τ3), we have

pun,v0

≤lim inf

m→∞ pun,um

≤ rⁿ

1−rpu0,u1

. (3.33)

By hypothesis, we also havew_n∈Tv0such thatp(u_n,w_n)≤r p(u_n−1,v0) forn∈N. So we have

lim sup

n→∞ pu_n,w_n≤lim sup

n→∞ r pu_n−1,v0

≤lim

n→∞

rⁿ

1−rpu0,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 thatv_n+1∈Tv_nandp(v0,v_n+1)≤r p(v0,v_n) for alln∈N. So we have

pv0,vn

≤r pv0,vn−1

≤ ··· ≤rⁿpv0,v0

. (3.35)

Hence

lim sup

n→∞ pun,vn

≤lim

n→∞

pun,v0

+pv0,vn

=0. (3.36)

ByLemma 2.6again,{v_n}is ap-Cauchy sequence and converges tov0. So we have pv0,v0

≤lim

n→∞pv0,v_n=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{p_n}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) forx,y∈X, is aτ-distance onX.

(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∈Np_n(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

p_n(x,y) (4.9)

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

Proof. Let{η_n}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)≤sup

n∈N

pn(x,y) +pn(y,z)≤sup

n∈Npn(x,y) + sup

n∈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⁻ⁿminη_n(x,t), 1 (4.11) forx∈Xandt∈[0,∞). Fixx∈X. For anyε >0, we choosek1∈Nwith 1/k1+ 2^1−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⁻ⁿminηn x,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 lim_nxn=xand lim_nsup{θ(zn,q5(zn,xm)) :m≥n} =0. Then, for anyk∈N, we have

lim sup

n→∞ sup

m≥nminηk zn,pk

zn,xm , 1

≤lim sup

n→∞ sup

m≥nminηk zn,q5

zn,xm , 1

≤lim

n→∞sup

m≥n2^k−1θzn,q5

zn,xm

=0,

(4.14)