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FIXED POINTS VIA A GENERALIZED LOCAL
COMMUTATIVITY: THE COMPACT CASE
GERALD F. JUNGCK
Received 19 July 2001 and in revised form 6 August 2002
In an earlier paper, the concept of semigroups of self-maps which are nearly com-mutative at a functiong:X→X was introduced. We now continue the investi-gation, but with emphasis on the compact case. Fixed-point theorems for such semigroups are obtained in the setting of semimetric and metric spaces. 2000 Mathematics Subject Classification: 47H10, 54H25.
1. Introduction. By a semigroup of maps we mean a familyHof self-maps of a setXwhich is closed with respect to the composition of maps and includes the identity map. In [6] we obtained fixed-point theorems for semigroups of maps by introducing the following concept.
Definition1.1[6]. A semigroupHof self-maps of a setXisnearly com-mutative (n.c.)atg:X→Xif and only iff∈Himplies that there existsh∈H
such thatf g=gh.
As in [6], we consider this concept in the context of generalized metric spaces, namely, semimetric spaces. A semimetric on a setXis a functiond:
X×X→[0,∞)such thatd(x, y)=0 if and only ifx=yandd(x, y)=d(y, x)
forx, y∈X. Forp∈Xand >0, we letS(p, )= {x∈X:d(x, p) < }. A semimetric space is a pair(X;d)in whichXis a topological space anddis a semimetric onX. The topology onX is the family t(d)= {U⊂X:p∈U⇒
S(p, )⊂Ufor some >0}. We require that the pointpbe an interior point of the setS(p, ); that is, there existsU∈t(d)such thatp∈U⊂S(p, ). Conse-quently, a sequence{xn}inXconverges int(d)top∈X, denoted asxn→p if and only ifd(xn, p)→0. And a function (map)g:X→X is continuous if and only iff xn→f xwheneverxn→x. To ensure unique limits, all spacesX will be assumed to be Hausdorff(T2). We will conclude with two examples of semimetrics, one of which is notT2. For further detail regarding semimetric spaces, see [1,2,6].
This paper is a continuation of [6] but with focus on the effects of com-pactness requirements. We obtain fixed-point theorems for semigroupsHof self-maps ofXwhich are n.c. at continuous mapsg:X→X. These theorems generalize known results involving, for example, the semigroupsCg= {f:X→
that whenever a semigroupHof self-maps of a setXis n.c. atg:X→X, then
His n.c. at the compositegnfor alln∈Z+. (Z+denotes the set of positive in-tegers,ω=Z+∪{0}, andRdenotes the real numbers.) Observe also that ifH is a semigroup of self-maps ofXanda, b∈X, thenH(a)= {h(a):h∈H}and
H(a, b)=H(a)∪H(b). Furthermore, ifA⊂X,δ(A)=sup{d(x, y):x, y∈A}
and cl(A)denotes the closure ofA.
2. Preliminaries. We now state and prove our first lemma. Note that most results stipulate that the semimetricdin(X;d)is upper semicontinuous (u.s.c.). Thus,d−1(−∞, a)is open fora∈R. Specifically, ifa
n→a, bn→bin(X;d) andd(an, bn)↓η, thenη≤d(a, b).
Lemma2.1. Let(X;d)be a semimetric space withdu.s.c. and letg, f:X→X. Suppose there is a nonempty compact subsetAofXsuch thatg(A)=f (A)=A and a semigroupHof self-maps ofX such thath(A)⊂Aforh∈H. If there existm, n∈ωsuch that
fmx≠gny ⇒dfmx, gny< δH(x, y), forx, y∈A, (2.1)
then there is a unique pointa∈Asuch thata=f a=ga=hafor allh∈H. In fact,A= {a}.
Proof. SinceAis compact,A×Ais compact. Therefore, there existsa, b∈ Asuch thatd(a, b)=δ(A), sincedis u.s.c. Butf (A)=g(A)=A=fm(A)=
gn(A), so there existsc, d∈Asuch thatfmc=aandgnd=b. Ifa≠b, (2.1) implies
δ(A)=d(a, b)=dfmc, gnd< δH(c, d). (2.2)
Butc, d∈Aso thathc, hd∈Afor allh∈Hby hypothesis; that is,H(c, d)⊂A. Thus, (2.2) yields the contradictionδ(A) < δ(H(c, d))≤δ(A). Consequently,
a=bso thatA= {a}. Therefore, by hypothesis,f ({a})=g({a})=h({a})= {a}for allh∈H; that is,a=f a=ga=hafor allh∈H. Clearly,ais the only point ofAwhich is a common fixed point off , g, andh∈H.
Remark2.2. The definition of a semimetric space(X;d)requires thatS(x, )
be a neighborhood ofx; that is, the topological interior ofS(x, )is a set in
t(d)which containsx. It is an easy matter to show that this fact assures us that any compact semimetric space is sequentially compact.
Lemma2.3. Let(X;d)be a semimetric space and letf , g:X→X. Suppose thatHis a semigroup of self-maps ofXwhich is n.c. atf and atg. Then the following hold, whereA= ∩{(gf )n(X):n∈Z+}(= ∩(gf )n(X)):
(1) His n.c. atgf;
(3) iff , g∈Handgf (A)=A, thenf (A)=g(A)=A;
(4) gf (A)=Awhen(X;d)is compact andgfis continuous.
Proof. SinceHis n.c. atgand atf for eachh∈H, there existsh1, h2∈H
such that
h(gf )=(hg)f=gh1f=gh1f=gf h2=(gf )h2. (2.3)
Thus (1) holds. But then, sinceH is n.c. atgn forn∈Z+ ifH is n.c. atg, (1) implies that for eachn∈Z+ and h∈H, there exists hn ∈H such that
h(gf )n=(gf )nhn. So
h(A)⊂ ∩h(gf )n(X)= ∩(gf )nhn(X)⊂ ∩(gf )n(X)=A. (2.4)
Consequently, (2) is true. Moreover, iff , g∈Handgf (A)=A, theng(A)=A
sincef∈Himplies thatf (A)⊂A. Therefore,
A=gf (A) ⇒A=gf (A)⊂g(A)⊂A (2.5)
sinceg∈H. To see thatf (A)=A, first note that sinceg∈H andH is n.c. atf, there existsh∈H such thatgf (A)=f h(A). Buth(A)⊂A(by (2)) and
f∈H, so we can write
A=gf (A) ⇒A=gf (A)=f h(A)⊂f (A)⊂A. (2.6)
Thusf (A)=A, and (3) holds.
To prove (4), first note thatgf (A)⊂Aby the definition ofA. To show that
A⊂gf (A), leta∈A. Thena∈(gf )n+1(X)⊂(gf )n(X)for eachn∈Z+, so we can choosexn∈(gf )n(X)such thatgf (xn)=a. Since(X;d)is compact (sequentially), there exists a subsequence{xin}of{xn}andp∈Xsuch that xin →p. Butin≥n by definition of subsequence, so for anyk∈Z+, xin ∈ (gf )k(X)for alln≥k. Therefore, for a fixedk, since(gf )k(X)is compact and thus closed (X isT2),p∈(gf )k(X). Hencep∈A. Butgf is continuous and therefore,a=gf (xin)→gf (p). Thus,gf (p)=a.
Remark2.4. If(X;d)is compact andgfis continuous, thenA=∩(gf )n(X) ≠∅and is compact by the finite intersection property.
3. Main results
Theorem3.1. Letfandgbe self-maps of a compact semimetric space(X;d) withdu.s.c. andgf continuous. LetHbe a semigroup of self-maps ofX con-tainingfandg, and suppose thatHis n.c. atfand atg. If there existm, n∈ω such that
then there is a unique pointa∈Xsuch thata=f a=ga=hafor allh∈H. Moreover,(i)δ((gf )n(X))→0and(ii)(gf )n(x)→auniformly onX.
Proof. LetA= ∩{(gf )n(X):n∈Z+}. ByLemma 2.3(4), gf (A)=A and therefore,A=f (A)=g(A)byLemma 2.3(3). Since(X;d)is compact andgf
is continuous,Ais compact and nonempty. Moreover,Lemma 2.3(2) tells us that h(A)⊂A for allh∈H. But then Lemma 2.1implies that A= {a} and
ais the desired common fixed point. The pointais clearly unique since any common fixed pointb off andg satisfies(gf )n(b)=b forn∈Z+ and is therefore inA.
To see that (i) holds, note thatgf is continuous and therefore (gf )n(X) is compact forn∈Z+. Thus for eachn∈Z+, there existsxn, yn∈(gf )n(X) such that
αn=d
xn, yn
=δ(gf )n(X) (3.2)
since the semimetricdis u.s.c. Clearly, 0≤αn+1≤αn and thereforeαn↓b for someb≥0. SinceXis compact, there exist subsequences{xkn}and{ykn} of{xn}and{yn}which converge tox,y, respectively, for somex, y∈X. As in the proof ofLemma 2.3(4),x, y∈A= {a}, sox=y=a. Then the upper semicontinuity of the semimetricdyields
b=lim n→∞d
xkn, ykn
≤d(x, y)=0. (3.3)
To verify (ii), let >0. By (i), we can choosek∈Z+such thatδ((gf )n(X)) < forn≥k. Therefore, ifx∈X,a, (gf )n(x)∈(gf )n(X), sod(a, (gf )n(x)) < forn≥k.
InTheorem 3.1, (ii) states that the pointais a uniformly contractive point ofgf, or attractsX(see [7]). The following theorem tells us thatfandgneed not be members ofHiff g=gf.
Theorem3.2. Letf andgbe commuting maps of a compact semimetric space(X;d)withdu.s.c. LetHbe a semigroup of self-maps ofXwhich is n.c. at gf. Ifgfis continuous and (3.1) holds, then there is a unique pointa∈Xsuch that for allh∈X,a=f a=ha. Moreover,Theorem 3.1(i) and (ii) hold.
Proof. As above, letA= ∩{(gf )n(X):n∈Z+}. ByRemark 2.4, we know thatAis nonempty and compact. Moreover, since (X;d)is compact andgf
is continuous, the proof of Lemma 2.3(4) tells us thatgf (A)=A. And the argument in the proof ofLemma 2.3(2) is valid under our hypothesis, soh(A)⊂
Aforh∈H. Finally, sincef andgcommute, they commute with(gf )nfor
n∈Z+, so g(A)⊂A and f (A)⊂A (substitute g(f )for hand h
n in (2.4)). Consequently, A=gf (A)⊂g(A)⊂A and thusA=g(A). Therefore, since
forh∈H. And the proof of (i) and (ii) given in the proof ofTheorem 3.1holds under the present hypothesis.
Remark 3.3. Theorem 3.2 generalizes [5, Theorem 4.2] by requiring that the underlying space be a semimetric space(X;d)with u.s.c. semimetric d
instead of being a metric space(X, d), in which cased is uniformly contin-uous (see [4]). Moreover, it substitutes the more general semigroup H for
Cgf. Note also that [6, Theorem 5.1] is a consequence ofTheorem 3.2. In fact,
Theorem 3.2extends [6, Theorem 5.1] since the underlying space in Theorem 5.1 is a metric space, and Theorem 5.1 does not include results (i) and (ii) of
Theorem 3.2.
We now lift the requirement that the space(X;d)be compact by demanding thatH(a)be relatively compact (i.e., cl(H(a))is compact) for somea∈X. And if there existsx∈Xsuch thathx=xfor allh∈H, we say that the semigroup
Hhas afixed point.
Theorem3.4. Let(X, d)be a semimetric space withdu.s.c. and letHbe a semigroup of continuous self-maps ofXwhich is n.c. atf , g∈H. SupposeH(a) is relatively compact for somea∈X. If (2.1) holds forx, y∈cl(H(a)), thenH has a fixed point incl(H(a)).
Proof. Nowh(H(a))⊂H(a)forh∈HsinceHis a semigroup. Moreover, since eachh∈His continuous,h(cl(H(a)))⊂cl(h(H(a)))⊂cl(H(a)). Thus, cl(H(a))is a nonempty compacth-invariant subset ofX forh∈H. We can therefore applyTheorem 3.1to the compact semimetric space(M;d)withM=
cl(H(a))to obtain our conclusion.
Since any closed and bounded subset ofRnis compact, we have the following result. Recall that a semigroupHis n.c. atgnif it is n.c. atg.
Corollary3.5. SupposeX⊂RnandHis a semigroup of continuous
self-maps ofX n.c. atg∈H. IfH(a)is bounded for somea∈Xand there exists m, n∈ωsuch that
gm(x)≠gn(y)⇒dgmx, gny< δH(x, y) forx, y∈clH(a), (3.4)
thenHhas a fixed point.
The following example demonstrates the generality ofCorollary 3.5in that the semigroupHconstructed is n.c. at the functiongbutHis not n.c. (i.e., n.c. at eachh∈H, see [3,6]).
Remark3.6. Suppose thatgi:Xi→Xiforiin some indexing setλand that
Hiis a semigroup of mapshi:Xi→Xiwhich is n.c. atgifori∈λ. Letg=(gi)λ, where forx =(xi)λ∈X={Xi :i∈λ}, g(x)=g((xi)λ)=(gi(xi))λ, and
then it is immediate thatHis a semigroup of self-maps ofXwhich is n.c. at
g:X→X.
Example3.7. LetX=X1×X2⊂R2with the usual metricd, whereX1=X2= [0,∞), and letb∈(0,1). Defineg:X→X, whereg=(g1, g2)byg1(x1)=bx1
andg2(x2)=x21/2=√x2. LetH1andH2be semigroups defined by
H1=h1:h1x1=bmxn
1 :m∈ω, n∈Z+
,
H2=h2:h2x2=x(21/k), k∈Z+
. (3.5)
ThenH1is not n.c. but is n.c. atg1(see [6, Example 2.2]), andH2is n.c. atg2
since any member ofH2commutes withg2. Thus,H=H1×H2is n.c. atgby
Remark 3.6. Moreover,
H 1 2, 1 2
= bm 1 2n,
1
2 1/k
:m∈ω; n, k∈Z+
⊂0,1
2
×1
2,1
,
(3.6)
and is thus bounded. Furthermore, ifx1, y1∈[0,1/2],x2, y2∈[1/2,1], and
(x1, x2)≠(y1, y2), then
dgx1, x2, gy1, y2=dg1x1, g2x2,g1y1, g2y2
=bx1−y12+√x2−y22
<x1−y12+x2−y22.
(3.7)
Note that for x1≠y1, |√x2− √y2|<|x2−y2|since x2, y2∈[1/2,1]. Now the identity function id (id(xi) =xi) is a member of Hi (i= 1,2), and if
h2(x2)= √x2, then h2∈ H2. Therefore, (3.4) is satisfied with m =n =1 anda=(1/2,1/2), andHhas a fixed point byCorollary 3.5. In fact, for any
h=(h1, h2)∈H, the compositeshp1(x1)→0 andh
p
2(x2)→1, asp→ ∞for x1∈[0,1/2]andx2∈[1/2,1]. Thus,(0,1)is a fixed point ofH.
In the above example, withM=[0,1/2]×[1/2,1],g(M)⊂Mandd(g(p), g(q)) < d(p, q)when p≠q onM; that is, gwas a contraction on M. Thus ghas a fixed point by the classic theorem of Edelstein. We now provide an example in whichgis not a contraction but satisfies the hypothesis ofCorollary 3.5.
Example3.8. LetX=[0,∞)and letg(x)=(15/4)x3−(59/4)x2+15x+1
forx ∈X. IfH =Og = {gn: n∈ω}, then g satisfies condition (3.4) with
m=n=1, and hasa=2 as a fixed point. Forx < y,
d(gx, gy) <
did(x),max{gx, gy}
, x <2,
Clearly, g is not a contraction onX. However, on the small interval J =
[1.87,2], g is a contraction and gnx ↑2 onJ. So suppose we let gbe the continuous piecewise linear function g:X →X given by g(x)=3x+1 on
[0,1],g(x)= −2x+6 on[1,2], andg(x)=2x−2 forx >2. Thengsatisfies (3.8), has 2 as a fixed point, andgso defined is a contraction on no subinterval ofX.
4. Retrospect. The arguments given in the proofs of Lemma 2.1 and
Theorem 3.1appear to be “standard” in that they are similar in form to those given in the metric case (see, e.g., [5, Theorem 4.2]). However, these proofs demonstrate that much can be done in spaces not having all of the attributes of metric spaces such as the triangle inequality or the continuity of the dis-tance function. Moreover, by using semigroupsH of maps n.c. at a function
g instead of Cg, the results are appreciably generalized. See [6] for further examples of semigroups n.c. at a functiong.
We now give two examples of nonmetric semimetrics. The first produces a topology which is notT2and therefore does not yield a semimetric space.
Example 4.1. Let X = R, and define d(x, y) =1/|x−y| if x ≠y and
d(x, x)=0, forx, y∈R. Clearly,dis a semimetric onX. To see that(X;d)is a semimetric space, note that for anya∈Xand >0,S(a, )= {a}∪(−∞, a−
1/)∪(a+1/,∞). It is then an easy matter to show thatS(a, )∈t(d); that is, S(a, )is “open.” Moreover, it is clear that S(a, )∩S(b, δ)≠∅for any
a, b∈Xand, δ >0 and therefore the definition oft(d)assures us that any two nonempty members oft(d)have a nonempty intersection.
The next semimetric induces a topology which does not produce a semimet-ric space sincepis not always an interior point ofS(p, ).
Example4.2. LetX=[0,1]and define
d(x, y)= |x−y|, ifx, y∈X, x, y >0,
d(y,0)=d(0, y)=
y, ify∈X, yis rational,
2, ify∈X, yis irrational.
(4.1)
Let∈(0,2). Then 0 is not an interior point ofS(0, ). Suppose there exists
U∈t(d)such that
0∈U⊂S(0, ). (4.2)
SinceU∈t(d), we can choose a positive rationalr such that
Nowris rational, so thatr /2∈S(0, r )and thereforer /2∈U. SinceU∈t(d)
by definition oft(d), there existsα∈(0, r /2)such that
S
r
2, α
⊂U⊂S(0, ). (4.4)
Thus, the definition of the functiondimplies
I=
r
2−α,
r
2+α
⊂U⊂S(0, ). (4.5)
But since 0< <2,S(0, )contains no irrationals whereas the intervalIdoes. This contradiction assures us that 0 is not an interior point ofS(0, ).
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Gerald F. Jungck: Department of Mathematics, Bradley University, Peoria, IL 61625, USA