NONLINEAR CONTRACTIONS

NONLINEAR CONTRACTIONS

YONG-ZHUO CHEN

Received 29 August 2004 and in revised form 12 October 2004

Recently, W. A. Kirk proved an asymptotic fixed point theorem for nonlinear contractions by using ultrafilter methods. In this paper, we prove his theorem under weaker assump- tions. Furthermore, our proof does not use ultrafilter methods.

1. Introduction

There are many papers in the literature that discuss the asymptotic fixed point theory, in which the existence of the fixed points is deduced from the assumption on the iterates of an operator (e.g., [1,6] and the references therein). Recently, Kirk [5] studied an asymp- totic fixed point theorem concerning nonlinear contractions. He proved the following theorem [5, Theorem 2.1] by appealing to ultrafilter methods.

Theorem1.1. Let(M,d)be a complete metric space. LetT:M→Mbe a continuous map- ping such that

dTⁿx,Tⁿy≤φn

d(x,y) (1.1)

for allx,y∈M, whereφn: [0,∞]→[0,∞]andlim_n→∞φn=φuniformly on the range ofd.

Suppose thatφand allφ_nare continuous, andφ(t)< tfort >0. If there existsx0∈Mwhich has a bounded orbitO(x0)= {x0,Tx0,T²x0,...}, thenT has a unique fixed pointx∗∈M such thatlim_n→∞Tⁿx=x∗for allx∈M.

In this paper, we prove Theorem 1.1under weaker assumptions without the use of ultrafilter methods.

2. Main results

We need the following recursive inequality (cf. [2, Lemmas 2.1 and 3.1], [3, Lemmas 2.1 and 2.4], and [4, Lemma 1]).

Lemma2.1. Letφ:R+→R+be upper semicontinuous, that is,lim sup_t→t0φ(t)≤φ(t0)for allt0∈R+, andφ(t)< tfort >0. Suppose that there exist two sequences of nonnegative real

Copyright©2005 Hindawi Publishing Corporation Fixed Point Theory and Applications 2005:2 (2005) 213–217 DOI:10.1155/FPTA.2005.213

numbers{u_n}and{_n}such that

u2n≤φun

+n, (2.1)

wheren→0asn→ ∞. Then eithersupun= ∞orlim infun=0.

Proof. Suppose thatb=sup{un}<∞. Assume that lim infun=0. Then there existm >0 andN1>0 such thatun> mfor alln > N1.

Sinceφis upper semicontinuous,φ(t)/tis upper semicontinuous on [m,b] and so that Lm=max{φ(t)/t, t∈[m,b]}<1 due to the facts thatφ(t)< tfort >0 and thatφ(t)/t achieves its maximum on [m,b].

Let>0. By (2.1), there existsN2> N1such that u2n≤φun

+≤Lmun+ (2.2)

for alln > N2. Note that the contraction mappingf(x)=Lmx+has a unique fixed point /(1−Lm) and lim_n→∞fⁿ(x)=/(1−Lm) for any real numberx. Now for anyn > N2,

u2²n≤φu2n

+≤Lmu2n+≤Lmfun

+= f²un

. (2.3)

By induction,u2^kn≤f^k(un) for allk, so thatm≤ f^k(un). Lettingk→ ∞,m≤/(1−Lm).

This is impossible since>0 can be arbitrarily chosen.

Theorem2.2. Let(M,d)be a complete metric space. LetT:M→Mbe such that dTⁿx,Tⁿy≤φn

d(x,y) (2.4)

for allx,y∈M, whereφ_n: [0,∞]→[0,∞]andlim_n→∞φ_n=φuniformly on any bounded interval[0,b]. Suppose thatφis upper semicontinuous andφ(t)< tfort >0. Furthermore, suppose there exists a positive integern∗such thatφn∗is upper semicontinuous andφn∗(0)= 0. If there existsx0∈Mwhich has a bounded orbitO(x0)= {x0,Tx0,T²x0,...}, thenThas a unique fixed pointx∗∈Msuch thatlim_n→∞Tⁿx=x∗for allx∈M.

Proof. First we establish the uniqueness of the fixed point. Assume thatThas two differ- ent fixed pointsz1andz2. Thend(z1,z2)=d(Tⁿz1,Tⁿz2)≤φn(d(z1,z2)). Lettingn→ ∞, d(z1,z2)≤φ(d(z1,z2))< d(z1,z2). This is a contradiction.

Without loss of generality, we setφn(0)=0 andφ(0)=0. Let b be the diameter of clos{O(x0)}. For a givenx∈clos{O(x0)}, denotean=d(Tⁿ⁺¹x,Tⁿx). Then

a2n=dT²ⁿ⁺¹x,T²ⁿx

≤φ_ndTⁿ⁺¹x,Tⁿx by (2.4)

=φan +φn

an

−φan .

(2.5)

Let _n=φ_n(a_n)−φ(a_n). Since φ_n→φ uniformly on [0,b], _n→0. By Lemma 2.1, lim inf_n→∞a_n=0.

Assume that lim_n→∞a_n=0 is not true. Since lim inf_n→∞a_n=0, there existsn0>0 such thata_n0<lim sup_n→∞a_n. We choose a sequencen0< n1< n3<··· witha_n0< a_ni for all i=1, 2,...as can be done by choosing lim_i→∞ani=lim sup_n→∞an. Then,

an0< ani=dTⁿⁱ⁺¹x,Tⁿⁱx

≤φ_ni−n⁰

dTⁿ⁰⁺¹x,Tⁿ⁰x by (2.4)

=φan0

+φni−n0

an0

−φan0

.

(2.6)

Letting i→ ∞, we havean0≤φ(an0)< an0, which is a contradiction. We conclude that lim_n→∞a_n=0.

We next show that{Tⁿx}is a Cauchy sequence. For if not, there exist pkandqksuch thatpk> qkfor eachk, and

klim→∞dT^pkx,T^qkx=δ >0. (2.7) Without loss of generality, assume that

dT^pkx,T^qkx>δ

2 ^∀k. (2.8)

Sinces−φ(s) is lower semicontinuous, there exists0>0 such that s−φ(s)>0 ∀s∈

δ 2,b

(2.9) due to the fact thatφ(s)< sfors >0 ands−φ(s) achieves its minimum on [δ/2,b].

Since lim_n→∞φn=φuniformly on [δ/2,b], there existsm0 such thatφm0(s)< φ(s) + 0< sfor alls∈[δ/2,b]. Now

dT^pkx,T^qkx

≤dT^pkx,T^pk+1x+dT^pk+1x,T^pk+2x+···+dT^pk+m0−1x,T^pk+m0x +dT^pk+m0x,T^qk+m0x+dT^qk+m0x,T^qk+m0−1x+···+dT^qk+1x,T^qkx

≤apk+apk+1+···+apk+m0−1+φm0

dT^pkx,T^qkx+aqk+m0−1+···+aqk

< apk+apk+1+···+apk+m0−1+φdT^pkx,T^qkx+0+aqk+m0−1+···+aqk. (2.10) Lettingk→ ∞and using (2.7), lim_n→∞an=0, and the upper semicontinuity ofφ,

δ≤φ(δ) +0< δ. (2.11)

This is a contradiction. Hence{Tⁿx}is a Cauchy sequence, there existsx∗∈Msuch that lim_n→∞Tⁿx=x∗.

Now for eachn >0,d(T^n∗+nx,T^n∗x∗)≤φn∗(d(Tⁿx,x∗)). Since lim sup_n→∞φn∗(d(Tⁿx, x∗))≤φ_n∗(0)=0, we have lim_n→∞Tⁿx=T^n∗x∗, so thatT^n∗x∗=x∗. Note thatT^n∗(Tx∗)

=T(T^n∗x∗)=Tx∗. By the uniqueness of the fixed point ofT^n∗,Tx∗=x∗.

For anyy0∈M\ {x0}, dTⁿy0,Tⁿx0

≤φn

dx0,y0

−→φdx0,y0

< dx0,y0

(2.12)

asn→ ∞. Hence,O(y0) is also bounded. By the previous argument, lim_n→∞Tⁿy0=x∗

due to the uniqueness of the fixed point.

Remark 2.3. Kirk’s paper [5] assumes the continuity forφand allφ_n. We only assume the upper semicontinuity ofφand one of theφn’s, which is weaker and easier to check.

If we have lim sup_t→∞(φ(t)/t)<1, then the assumption of the existence of a bounded orbit inTheorem 2.2can be removed. This observation is formulated as the following corollary.

Corollary2.4. Let(M,d)be a complete metric space. LetT:M→Mbe such that dTⁿx,Tⁿy≤φn

d(x,y) (2.13)

for allx,y∈M, whereφn: [0,∞]→[0,∞]andlim_n→∞φn=φuniformly on any bounded interval[0,b]. Suppose thatφis upper semicontinuous,φ(t)< t fort >0, andlim sup_t→∞

(φ(t)/t)<1. If there exists a positive integern∗such thatφn∗ is upper semicontinuous and φn∗(0)=0, then T has a unique fixed point x∗∈M such that lim_n→∞Tⁿx=x∗for all x∈M.

Proof. Examining the proofs of Lemma 2.1 and Theorem 2.2, one can find that the boundedness of the orbitO(x) is only used to guarantee that

sup φ(t)

t :t∈[m,b]

<1, inf t−φ(t) :t∈[m,b]>0

(2.14)

for someb >0 and allm >0 satisfying 0< m < b <∞. If we have lim sup_t→∞(φ(t)/t)<1, then there existsb >0 such that sup_t∈[b,∞](φ(t)/t)<1. For otherwise, there will bet_n→

∞with lim_t→∞(φ(tn)/tn)=1, which implies that lim sup_t→∞(φ(t)/t)=1 and leads to a contradiction. Hence, lim sup_t→∞(φ(t)/t)<1 combined with the upper semicontinuity ofφcan guarantee (2.14) for allm >0 with 0< m < b <∞. Acknowledgment

The author would like to thank the referees for their valuable comments and suggestions which helped to improve this paper.

References

[1] F. E. Browder,Asymptotic fixed point theorems, Math. Ann.185(1970), 38–60.

[2] Y.-Z. Chen,Inhomogeneous iterates of contraction mappings and nonlinear ergodic theorems, Nonlinear Anal. Ser. A: Theory Methods39(2000), no. 1, 1–10.

[3] , Path stability and nonlinear weak ergodic theorems, Trans. Amer. Math. Soc. 352 (2000), no. 11, 5279–5292.

[4] J. R. Jachymski,An extension of A. Ostrowski’s theorem on the round-offstability of iterations, Aequationes Math.53(1997), no. 3, 242–253.

[5] W. A. Kirk,Fixed points of asymptotic contractions, J. Math. Anal. Appl.277(2003), no. 2, 645–

650.

[6] R. D. Nussbaum,Some asymptotic fixed point theorems, Trans. Amer. Math. Soc.171(1972), 349–375.

Yong-Zhuo Chen: Department of Mathematics, Computer Science and Engineering, University of Pittsburgh at Bradford, Bradford, PA 16701, USA

E-mail address:[email protected]