Fixed points via a generalized local commutativity

© Hindawi Publishing Corp.

FIXED POINTS VIA A GENERALIZED LOCAL COMMUTATIVITY

GERALD F. JUNGCK

(Received 24 May 2000)

Abstract.Letg:X→X. The concept of a semigroup of maps which is “nearly commu-tative atg” is introduced. We thereby obtain newfixed point theorems for functions with bounded orbit(s) which generalize a recent theorem by Huang and Hong, and results by Jachymski, Jungck, Ohta, and Nikaido, Rhoades and Watson, and others.

2000 Mathematics Subject Classification. Primary 47H10, 54H25.

1. Introduction. By asemi-group of maps we mean a familyH of self maps of a setXwhich is closed with respect to composition of maps (f◦g=f g) and includes the identity mapid(x)=x, forx∈X. We often associate with a functiong:X→X following semi-groups:

Og=gn_{|n∈N∪{0}, (1.1)}

whereNis the set of positive integers andg0_{=id, and}

Cg=f:X →X|f g=gf. (1.2)

A quick check confirms thatCgis a semi-group.

IfHis a semi-group of self maps of a setXanda∈X, H(a)= {h(a)|a∈H}.In particular, ifH=Og, Og(a)= {gn_{(a)|n∈N∪{0}}and is called the orbit ofgata.}

In general,Lemma 3.2and some theorems inSection 3will be stated in the context of semi-metric spaces. Asemi-metricon a setXis a functiond:X×X→[0,∞)such thatd(x,y)=d(y,x)forx,y∈Xandd(x,y)=0 if and only ifx=y. Asemi-metric space is a pair(X;d), whereX is a topological space andd is a semi-metric onX. The topologyt(d)onXis generated by the setsS(p,)= {x|d(x,p) < }with the requirement thatp is an interior point ofS(p,). A sequence{xn}inX converges in t(d) top ∈X (denoted as xn →p) if and only if d(xn,p)→0. We let t(d) be T2(Hausdorff) to ensure unique limits. Thus, a metric space(X,d)is a semi-metric space having the triangle inequality. For further details on semi-metric spaces, see, for example, [1,4,6].

2. Nearly commutative semi-groups. In [2], a semi-groupHof maps is said to be near-commutativeif and only if for each pairf ,g∈H, there existsh∈Hsuch that f g=gh. We generalize as follows.

Definition2.1. A semi-groupH of self maps of a setX is nearly commutative (n.c.) at g:X→X if and only if(f ∈H)implies that there existsh∈H such that f g=gh.

Of course,Og and Cg are n.c. at g. Observe also that a near-commutative semi-group H of self maps of a setX is n.c. ateachg∈H. The following provides for eacha∈(0,∞)an example of a semi-groupH=Sa of self maps which is not near-commutative but is n.c. at a particularg:X→X.

Example2.2. LetX=[0,∞)anda∈(0,∞). Letg(x)=axand define

Sa=am_xn_{|x∈[0,∞), n∈N, m∈N∪{0}, (2.1)}

whereSa is nearly commutative (n.c.) at g. For iff (x)=am_xn _{is a representative} element ofSa, thenf g(x)=f (g(x))=am_(ax)n_=am+n_xn_{. We w anth(x)=a}r_xs_∈ Sa such thatf g=gh. Now,g(h(x))=a(ar_xs_)=ar+1_xs_{, so we can lets=nand} r+1=m+n; that is,r=m+(n−1). Sincen∈Nand(n−1), m∈N∪{0},sandr so designated implyh∈Sa. Thus,(f∈H=Sa)implies that there existsh∈Hsuch thatf g=gh. Sinceid∈Sa,Sa is clearly a semi-group, and we are finished. On the other hand,Sa is not anear-commutativesemi-group. For example, letf (x)=a2_x2 andh(x)=a2_x3_{. We wantt(x)=a}r_xs _{such thatf h=ht. So we must have 3s=6} and(2+3r )=6. But thenr=4/3, andr∉ N∪{0}.

Now, letᏹn and ᏺn denote the set of alln×nreal matrices and the set of all nonsingularn×nreal matrices, respectively. Then, both setsᏹn andᏺnare semi-groups of linear transformationsA:Rn_→Rn_{relative to composition of maps (matrix} multiplication).

Example2.3. ᏺnis n.c. For ifA,B∈ᏺn, there existsC=B−1(AB)∈ᏺn such that AB=BC.

Example2.4. ᏹnis n.c. at anyB∈ᏺn, byExample 2.3. Butᏹnis not near commu-tative. For instance, ifn=2,B=1 1

0 0

, andA=1 2 3 4

, there exists no 2×2 matrixC such thatAB=BC.

Now, letg:X→X. Since any semi-group of self maps which commute withgis a subset ofCg, we might hope thatHg= {f:X→X|f g =ghfor someh:X→X} would be a maximal semi-group which is n.c. at g. However, Hg so defined need not be n.c. at g! For example, let X = [0,∞), g(x)=1/(x+1), and f (x)= x/2. Thenh(x)=2x+1 satisfiesf (g(x))=g(h(x))forx∈[0,∞). However, there ex-ists no k∈Hg such that h(g(x))=g(k(x)); that is, 2(x+1)−1_+1=(k(x)+1)−1 (note thatx, k(x)≥0).

Example2.5. LetXbe any set and letg:X→Xbe surjective. Then the family of all self mappings ofX,Ᏺ= {f|f:X→X}, is n.c. atg. For supposef∈Ᏺ; we need h∈Ᏺsuch thatf g(x)=gh(x)for all x∈X. So leta∈X. Sincegis onto, we can choosexa∈X such thatg(xa)=f (g(a)). Choose such anxa for each a∈X and defineh(a)=xa. Thenh:X→Xandg(h(a))=g(xa)=f (g(a))fora∈X; that is, f g=gh.

Proposition2.6. Suppose thatHis a semigroup of maps which is n.c. atg:X→X. Iff∈Handn∈N, there existshn∈Hsuch thatf gn_=gn_{hn(i.e.,His n.c. atg}n_).

Proof. Letf∈H. Since,His n.c. atg, there existsh1∈Hsuch thatf g=gh1. So suppose thatk∈Nsuch thatf gk_=gk_{hkfor somehk∈H. Then}

f gk+1_{=f g}k_g=gk_hkg=gk_{hkg. (2.2)} Sincehk∈H, there existshk+1∈Hsuch thathkg=ghk+1, and therefore (2.2) implies

f gk+1_=gk_(ghk

+1)=gk+1hk+1, as desired.

Throughout this paper, P denotes a function P : [0,∞)→ [0,∞) which is non-decreasing, and satisfies limn→∞Pn(t)=0 fort∈[0,∞). (For example, we could let P(t)=αtfor someα∈(0,1), ort/(t+1).) And throughout this paper, we appeal to the following lemma.

Lemma2.7. LetH be a semi-group of self maps of a setXand suppose thatH is nearly commutative atg:X→X. Letd:X×X→[0,∞). Suppose that for each pair

x,y∈Xthere exists a choicer=r ({x,y}), s=s({x,y})∈H, andu,v∈ {x,y}for which

d(gx,gy)≤Pd(r u,sv). (2.3)

Then, if n∈N, for each pair x,y ∈ X there exist rn,sn ∈ H and un,vn ∈ {x,y} such that

dgn_x,gn_y≤Pn_{drnun,snvn. (2.4)} Proof. By (2.3), inequality (2.4) holds forn=1, so suppose thatn∈Nfor which (2.4) is true. Then, ifx,y∈X,

dgn+1_x,gn+1_y=dggn_x,ggn_{y≤Pd(r u,sv), (2.5)} wherer ,s∈Handu,v∈ {gn_x,gn_{y}, by (2.3). Specifically,u=g}n_{c, v=g}n_{d, where} c,d∈ {x,y}. And since r ,s∈H, there existr_{,s∈H such that r g}n_=gn_{r and} sgn_=gn_{s,byProposition 2.6. So (2.4) implies that}

d(r u,sv)=dr gn_(c),sgn_(d)=dgn_rc,gn_sd≤Pn_{drnun,snvv, (2.6)}

wherern,sn∈Handun,vn∈ {r_{c,sd}. Thus,rnun∈ {(rnr)c,(rns)d}, wherernr} andrns_{are elements ofH, sinceHis a semi-group. Sornun=rn+1un+1, wherern+1∈} {rnr_{,rns} (i.e.,rn}

+1∈H) andun+1∈ {c,d} ⊂ {x,y}. Similarly, snvn=sn+1vn+1, wheresn+1∈Handvn+1∈ {x,y}. Thus, (2.6) implies that

d(r u,sv)≤Pn_drn

ButPis nondecreasing, and therefore (2.7) and (2.5) yield

dgn+1_x,gn+1_y≤PPn_drn

+1un+1,sn+1vn+1 =Pn+1_drn

+1un+1,sn+1vn+1, (2.8) with rn+1,sn+1∈H and un+1,vn+1∈ {x,y}. So, (2.4) is true for allnby induction.

3. Fixed point theorems

Definition3.1. Let(X;d)be a semi-metric space and letHbe a semi-group of self maps ofX. A mapg:X→XisP-contractive relative toH if and only if (2.3) holds. (We will also say, “gis a P-contraction relative toH.”)

Lemma3.2. Let(X;d)be aT2semi-metric space and letHbe a semi-group of self maps ofXn.c. atg:X→X. Suppose thatgisP-contractive relative toHand thatM⊂X

such thatB= ∪{H(c)|c∈M}is bounded. Thend(gn_(x),gn_{(y))→0uniformly onB}

asn→ ∞. Specifically, if >0, there existsk∈Nsuch that

(n≥k) ⇒dgn_(x),gn_{(y)< ∀x,y∈B. (3.1)}

Proof. By hypothesisδ(B) <∞,Pn_{(δ(B))→0 asn→ ∞. Let >0. We can choose} k∈Nsuch that

Pn_{δ(B)< forn≥k. (3.2)}

Let x,y ∈B. Ifn∈ N, since g is P-contractive relative to H, Lemma 2.7 yields rn,sn∈Handun,vn∈ {x,y}(⊂B) such that

dgn_(x),gn_(y)≤Pn_{drnun,snvn. (3.3)}

Since un ∈B, there exist h ∈H and c ∈M such that un = h(c). Butrn,h∈H, so rnh∈H. Therefore, rnun =(rnh)(c)∈H(c)⊂B. Likewise,snvn∈B. But then d(rnun,snvn)≤δ(B)and therefore,

Pn_{drnun,snvn≤P}n_{δ(B) forn∈N, (3.4)}

sincePis nondecreasing andnis arbitrary. Formulae (3.2), (3.3), and (3.4) imply

dgn_(x),gn_{(y)< forn≥k. (3.5)}

Since the choice ofkin (3.2) was independent ofxandy, (3.5) holds for allx,y∈B.

Theorem3.3. Let(X;d)be aT2semi-metric space, and letHbe a semi-group of self maps ofXwhich is n.c. atg∈H. Suppose thatH(a)is bounded for somea∈X

andXisg-orbitally complete. Ifgis aP-contraction relative toH, thengn_(a)→cfor

somec∈X. Ifgis continuous atc,g(c)=c.

To this end, let >0. Since,H(a)is bounded,Lemma 3.2withB=H(a)implies that there existsk∈Nsuch that

n≥k ⇒dgn_(x),gn_{(y)< ∀x,y∈H(a). (3.6)}

Therefore, ifm > n≥k , m=n+r for somer∈N, and

dgn_(a),gm_(a)=dgn_(a),gn_gr_{(a)< , (3.7)}

sincea,gr_{(a)∈H(a). We conclude that{g}n_{(a)}is Cauchy, and there existsc∈X} such thatgn_(a)→c.

Now, ifgis continuous atc, limn→∞ g(gn(a))=g(c), since gn(a)→c. But then gn+1_{(a)→calso, sog(c)=csince(X;d)is aT}

2semi-metric space.

Definition3.4. LetX andY be topological spaces. A mapg:X→Y isclosed if and only ifg(M)is closed inY wheneverMis a closed subset ofX.

Note that the conclusion ofLemma 3.2asserts thatd(gk_(xk),gk_{(yk))→0 for any} sequences{xk}and{yk}inB.

Theorem3.5. Let(X;d)be a bounded and completeT2semi-metric space, and let

Hbe a semi-group of maps n.c. atg∈H. Ifgis closed andP-contractive relative toH, (i) there existsp∈Xsuch that{p} = ∩{gn_(X)|n∈N},

(ii) pis the unique fixed point ofg, (iii) gn_{(x)→pfor allx∈X.}

Proof. Letx∈X. ByTheorem 3.3,{gn_{(x)}converges topfor somep∈X.} More-over,p∈ ∩{gn_{(X)|n∈N}. Otherwise, there existsk∈Nsuch thatp∉g}k_{(X). Since} gk_{(X)is closed, there exists >0 such thatS(p,)∩g}k_{(X)= ∅. Thus,d(g}n_(x),p)≥ forn≥ksincegn_{(X)is a subset ofg}k_{(X)forn≥k. This contradicts the fact that} gn_(x)→p.

In fact,{p} = ∩{gn_{(X)|n∈N}. For ifq∈ ∩{g}n_{(X)|n∈N}, for eachk∈Nwe can} choosexk,yk∈Xsuch thatgk_(xk)=pandgk_{(yk)=q. So}

d(p,q)=dgk_xk,gk_{yk →0, (3.8)}

byLemma 3.2withM=X.

Clearly, (i) implies thatpis a fixed point ofg, sinceg({p})⊂ {p}. Thus, ifx∈X, d(gn_(x),p)=d(gn_(x),gn_{(p))→0 asn→ ∞, so (iii) holds. Similarly, ifq is a fixed} point ofg, thend(p,q)=(gn_(p),gn_{(q))→0, so thatq=p. Thus,pis the only fixed} point ofg.

In the following we need the triangle inequality, so we require the underlying space to be a metric space.

Theorem3.6. Let(X,d)be a metric space and letHbe a semi-group of self maps ofXn.c. at someg∈H. Suppose thatXisg-orbitally complete and there existsk∈N such that for each pairx,y∈X, there existr ,s∈Handu,v∈ {x,y}for which

(i) If there exists a∈Xsuch thatH(a)is bounded, then there existsc∈X such thatlimn→∞gn(a)=c. Ifhis continuous for someh∈H, thenh(c)=c. (Specifically, g(c)=cifgis continuous atc.)

(ii) If H(x) is bounded for each x ∈X, there exists a unique c ∈ X such that

gn_{(x)→c for all x ∈X. If g is continuous atc, c is a unique common fixed point}

for allh∈H.

Proof. Suppose thatH(a)is bounded. SinceHis n.c. atg,Proposition 2.6says thatH is n.c. atgk_{. And Xis g}k_{-orbitally complete since Xis g-orbitally complete.} Therefore, (3.9) andTheorem 3.3imply that

lim m→∞

gkm_{(a)=c for somec∈X. (3.10)}

To see that limn→∞gn(a)=c, let >0. Then (3.10) andLemma 3.2(withB=H(a)) imply that there existsp∈Nsuch thatd((gk₎p_{(a),c) < /2 andd(g}kp_(x),gkp_{(y)) <} /2 forx,y∈B; that is,

dgkp_(a),c<

2, d

gkp_gi_(a),gkp_(a)<

2 ∀i∈N, (3.11) sinceg∈H⇒gi_{(a)∈H(a). So, ifn > k p,n=kp+ifor somei∈N,and}

dgn_(a),c≤dgn_(a),gkp_(a)+dgkp_{(a),c, (3.12)}

or

dgn_(a),c≤dgkp_gi_(a),gkp_(a)+dgkp_(a),c< 2+

2=, (3.13)

by (3.11). Consequently,gn_(a)→c.

Now, leth∈Hand suppose thathis continuous atc. Then, limn→∞h(gn(a))=h(c) and

dh(c),c=_nlim_→∞dhgn_(a),gn_(a)=lim n→∞d

hgkn_(a),gkn_{(a). (3.14)}

ButHis n.c. atgk_{, so forn∈Nthere existshn∈Hsuch thathg}kn_=gkn_{hn. Then,} by (3.14),

dh(c),c= lim n→∞d

gkn_hn(a),gkn_{(a)=0, (3.15)}

sincea,hn(a)∈H(a)andLemma 3.2holds forgk_{. Thus, (i) holds.}

To prove (ii), suppose thatH(x)is bounded for eachx∈X. Ifa,b∈X, gn_(a)→ca andgn_{(b)→cbfor someca,cb∈Xby (i). Butca=cb, sinceH(a)∪H(b)is bounded,} and therefore,Lemma 3.2applied togk_{implies thatd(ca,cb)=limn→∞d((g}k₎n_(a), (gk₎n_(b))=0.

Thus, there exists a uniquec∈Xsuch thatgn_{(x)→cfor allx∈X. We knowthat} g(c)=c by part (i), ifgis continuous atc. Sincegn_{(d)=dfor allnifdis a fixed} point ofg, and thereforegn_{(d)→d,c must be the only fixed point ofg. Moreover,} h(c)=c for allh∈H (even though hmay not be continuous). This follows, since

Proposition 2.6applied togk_{implies that for eachn∈N,}

dc,h(c)=dgkn_(c),hgkn_(c)=dgkn_(c),gkn_{hn(c) (3.16)}

Remark3.7. Theorem 3.3appreciably generalizes Theorem 2.1 in [5] andTheorem 3.6generalizes Corollary 2.3 in [5]—and hence Theorem 2 in [3] and the theorems of Rhoades and Watson [9]. Note that inTheorem 3.6(ii), the mappingsh∈H (h≠g) need not be continuous. Remember also thatCgandOg are special instances ofH.

The following example suggests that the requirement inTheorem 3.6(ii), thatH(x) be bounded for eachx∈X, is not as restrictive as may first appear.

Example 3.8. Let S = {continuous functionsf : [0,∞) → [0,∞) | there exists af ∈(0,∞)such thatf (x) < xforx > af}. (So, e.g.,{f|f (x)=mx+b, m∈[0,1) andb≥0} ⊂S, and ln(x+b)∈S forb≥1.) Then (1)S∪ {id}is a semi-group under composition of functions, and (2)Of(x)is bounded forf∈Sandx∈[0,∞).

First note that, we can letMf denote the maximum value off on[0,af]for each f∈S since eachf is continuous. To see that (1) is true, letf ,g∈S. We need only to showthat g◦f =gf ∈S. Clearly,gf is a continuous self map of [0,∞). So let agf =max{af,Mg}and suppose thatx > agf. We want gf (x) < x. Now,x > agf implies thatx > af so that (i)f (x) < x. Iff (x) > ag, theng(f (x)) < f (x) < xby (i) and the definition ofag. Iff (x)≤ag, g(f (x))≤Mg≤agf < x. So, in any event, (g◦f )(x) < xif x > agf, and thus,g◦f ∈S. (2) follows easily by using induction to showthat(f∈S)implies that (ifx∈[0,∞), fn_{(x)≤max{x,Mf}forn∈N). We} omit the details.

If we letP(t)=αtfor fixedα∈(0,1)andt∈[0,∞), we have the following corollary. Corollary3.9. Let(X,d)be a bounded complete metric space and letg:X→Xbe continuous. Suppose thatHis a semi-group of self maps ofXn.c. atgandg∈H. If there existsα∈(0,1)such that for any pairx,y∈Xthere existr ,s∈Handu,v∈ {x,y} for which

d(gx,gy)≤αd(r u,sv), (3.17)

then there exists a uniquec∈Xsuch thatgn_{(x)→cforx∈X, andc=gc=hcfor}

allh∈H.

4. Some consequences

Definition 4.1. Agauge function is an upper semicontinuous (u.s.c.) function φ:[0,∞)→[0,∞)such thatφ(0)=0 andφ(t) < tfor allt >0.

Lemma4.2. Let(X,d)be a metric space and letHbe a semi-group of self maps of

Xwhich is n.c. atg∈H. Suppose thatH(x,y)=H(x)∪H(y)is bounded forx,y∈X

and there exists a gauge functionφsuch that

d(gx,gy)≤φδH(x,y) forx,y∈X. (4.1)

Then, there exists a nondecreasing continuous function P:[0,∞)→[0,∞)such that

Pn_{(t)→0for allt >0and which satisfies the following condition: for any pairx,y∈X}

there existr=r (x,y), s=s(x,y)∈H, andu,v∈ {x,y}such that

Proof. Letx,y∈Xand suppose that (4.1) holds. Since,φis a gauge function, as is well known [2], there exists a nondecreasing continuous functionP:[0,∞)→[0,∞) such thatPn_{(t)→0 fort≥0, and}

φ(t) < P(t), P(t) < t ∀t∈(0,∞). (4.3)

SincePis continuous, (4.3) implies that for anyt >0, there existst∈(0,t)such that

t_{∈t−t,t+t ⇒φ(t) < Pt. (4.4)}

And sinceH(x,y)is bounded, the definition ofδimplies that there existr ,s∈H andu,v∈ {x,y}such that, witht=δ(H(x,y)),

t=δH(x,y)≥d(r u,sv) > δH(x,y)−t. (4.5)

So, witht_{=d(r u,sv), (4.4) and (4.5) imply that}

φδH(x,y)< Pd(r u,sv). (4.6)

Therefore, (4.1) implies thatd(gx,gy)≤P(d(r u,sv)).

The following theorem provides a generalization of Theorem 2.1 in [2].

Theorem4.3. Let(X,d)be a complete metric space and letHbe a semi-group of self maps ofXwhich is n.c. at someg∈H. Suppose that the following conditions are satisfied:

(i) H(x)is bounded for allx∈X, gis continuous, (ii) there exists a gauge functionφandk∈Nsuch that

dgk_x,gk_{y≤φδH(x,y) forx,y∈X. (4.7)}

Then

(a)Hhas a unique common fixed pointcandgn_{(x)→cforx∈X.}

(b)If for eachh∈H− {id}there existsk=kh∈Nsuch that (4.7) holds withg= h, then

hn_{(x)→c ∀x∈X, h∈H−id. (4.8)} Proof. Now, (i) implies thatH(x,y)=H(x)∪H(y)is bounded forx,y∈X. To see that (a) is true, note thatHis n.c. atgk_{byProposition 2.6and substituteg}k_forgin

Lemma 4.2to conclude that (3.9) holds. Consequently, we can appeal toTheorem 3.6(ii) to obtain ac∈Xsuch thatgn_{(x)→cforx∈X. And sincegis continuous,cis the} unique fixed point ofgand a fixed point for eachh∈H. Thus,cis the unique common fixed point ofH(remember,g∈H) and therefore (a) holds.

To prove (b) note that, by part (a), ifh∈H− {id}, h≠g, hn_{(c)=g(c)=c for} n∈N. ButTheorem 3.6applied tohyields a uniquec1∈Xsuch thathn(x)→c1for allx∈X. Sincehn_{(c)=cfor alln,c}

1=c.

Remark4.4. Theorem 4.3generalizes Theorem 2.1 in [2] in the following ways: (i) The semi-groupH is not required to be near-commutative (i.e., n.c. at each

(ii) gis the only member ofHrequired to be continuous,

(iii) in (b), (4.7) is required to hold only fork=kh, not for allk≥kh.

Theorem 4.3yields the following corollary, which generalizes the theorem of Ohta and Nikaido [8] by requiring only that the orbits off—but not all ofX—be bounded. Corollary4.5. Letfbe a continuous self mapping of a metric space(X,d)having bounded orbitsOf(x)for allx∈X. If there existc∈(0,1)andk∈Nsuch that

dfk_x,fk_y≤cδfi_{t|t∈ {x,y}, i∈N∪{0} (4.9)}

for allx,y∈X, thenfhas a unique fixed point.

Observe that Lemma 3.2does not require that g∈H, whereas the theorems in

Section 3do. The requirement thatg∈Hwas convenient in the proof, but the follow-ing proposition says that it is not necessary whenOg(a)is bounded. Moreover, this result is needed for the proof ofTheorem 4.7.

Proposition 4.6. If H is a semi-group of self maps n.c. at g and g∉ H, then

Hg = {gn_{h|n∈N∪ {0}andh∈H}is a semi-group which is n.c. atg. Moreover,} g∈HgandH⊂Hg.

Proof. Hgis a semi-group. For ifgn_h

1,gmh2∈Hg, sinceHis n.c. atg, we have

gn_h

1gmh2=gn(h1gm)h2=gn(gmh3)h2=gn+m(h4), whereh4=h3h2∈H.

Hgis n.c. atg, since (Hn.c. atg) implies that there existsh2∈Hsuch that(gnh)g=

gn_(hg)=gn_(gh

2)=g(gnh2).

It is clear that ifg:X→X is aP-contraction relative toH, then it is certainly a P-contraction relative toHgsinceH⊂Hg. We use this fact in the proof ofTheorem 4.7. Theorem4.7. LetCbe a compact subset of a normed linear spaceXwhich is star-shaped with respect toq∈C. LetT:C→Cbe continuous and letHbe a semi-group of affine mapsI:C→C n.c. atT such thatI(q)=q. If for each pairx,y∈Cthere exist

I,J∈Handu,v∈ {x,y}for which

_{T x−T y≤Iu−Jv, (4.10)}

then there existsa∈Csuch thata=T aanda=Iafor all continuousI∈H.

Proof. Choose a sequence{kn}in(0,1)such thatkn→1, and for eachn∈N, let

Tn(x)=knT x+1−knq. (4.11)

SinceCis star-shaped with respect toq,Tn:C→C forn∈N. Moreover, ifI∈H, there existsJ∈Hsuch that

ITnx=IknT x+1−knq=knI(T x)+1−knIq

=knT (Jx)+1−knq=TnJx, (4.12)

sinceIis affine,His n.c. atT, andIq=q. Thus, for eachn∈N,His a semi-group of affine maps which is n.c. atTn. Then, byProposition 4.6,HTn is a semi-group of self

Nowfix n. By hypothesis, for each pairx,y∈C there existI,J∈H(⊂HTn)and u,v∈ {x,y}such that

_{T x−T y≤Iu−Jv, (4.13)}

so

_{Tnx−Tny≤knIu−Jv, (4.14)}

by (4.11). Therefore, sinceTn is continuous andkn∈(0,1),Corollary 3.9applied to Tn andHTn (C compact implies thatCis bounded and complete) implies that there

exists a uniquexn∈Csuch that

xn=Tnxn=Ixn ∀I∈HTn. (4.15)

Thus we have a sequence{xn}inC which satisfies (4.15). SinceCis compact,{xn} has a subsequence{xin}which converges to somea∈C. Equations (4.11) and (4.15)

thus imply that

a=lim

n→∞xin=nlim→∞kinT xin+nlim→∞

1−kin

q= lim

n→∞Ixin. (4.16)

ButT is continuous, so (4.16) implies thata=T a, anda=Iafor all continuousI.

Remark4.8. We see thatTheorem 4.7does indeed extend Theorem 3 in [7] if w e observe that the familyᏲin Theorem 3 [7]. is a family of sets which is a subset ofCg. We can let

H=mapsh:C →C|his affine,h∈Cg. (4.17) ThenHis a semi-group andᏲ⊂H.

5. Conclusion. We conclude with further evidence of the generality and applicabil-ity of the concept of being nearly commutative at a functiong. The theorem below generalizes Theorem 4.2 in [5] by replacing the semi-groupCgf with a more general semi-groupH.

Theorem5.1. Letfandgbe commuting self maps of a compact metric space(X,d)

such thatgf is continuous. IfHis a semi-group of self maps ofXwhich is n.c. atgf, and

f x≠gy ⇒d(f x,gy) < δH(x,y), (5.1) then there exists a unique pointa∈Xsuch thata=f a=ga=hafor allh∈H.

We leave the proof ofTheorem 5.1to the interested reader.

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Gerald F. Jungck: Department of Mathematics, Bradley University, Peoria, IL61625, USA