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ON CHARACTERIZATIONS OF FIXED POINTS
ZEQING LIU, LILI ZHANG, and SHIN MIN KANG
(Received 16 January 2001 and in revised form 14 March 2001)
Abstract.We give some necessary and sufficient conditions for the existence of fixed points of a family of self mappings of a metric space and we establish an equivalent condi-tion for the existence of fixed points of a continuous compact mapping of a metric space.
2000 Mathematics Subject Classification. 54H25.
1. Introduction. Jungck [2] first gave a necessary and sufficient condition for the existence of fixed points of a continuous self mapping of a complete metric space. Afterwards, Park [8], Leader [4], and Khan and Fisher [3] established a few theorems similar to that of Jungck. Janos [1] and Park [9] proved fixed point theorems for com-pact self mappings of a metric space. Recently, Liu [5] established criteria for the existence of fixed points of a family of self mappings of a metric space. The aim of this paper is to offer some characterizations for the existence of fixed points of a family of self mappings and a continuous compact mappings of metric spaces, respectively. We also establish a fixed point theorem for two compact mappings, which extends properly the results of Janos [1] and Park [9].
Let ωand N denote the sets of nonnegative and positive integers, respectively. Suppose that(X, d)is a metric space. Forx, y∈X, define
Cf =g|g:X →Xandf g=gf,
Hf =g|g:X →Xandg∩n∈ωfnX⊆ ∩n∈ωfnX,
Hf(x)=hx|h∈Hf, Hf(x, y)=Hf(x)∪Hf(y),
O(x, f )=fnx|n∈ω, O(x, y, f )=O(x, f )∪O(y, f ).
(1.1)
Obviously,Cf ⊆Hf. LetΦbe a family of self mappings ofX. A pointx∈Xis said to be afixed pointofΦiff x=xfor allf∈Φ. LetF:X×X→[0,+∞)be continuous andF (x, y)=0 if and only ifx=y. ForA, B⊂X, define
δ(A, B)=supF (x, y)|x∈A, y∈B (1.2)
In order to prove our main results, we need the following lemmas.
Lemma1.1(see [6]). Letfbe a continuous compact self mapping of a metric space (X, d). IfA= ∩n∈ωfnX, then
(a1) Ais compact, (a2) f A=A= ∅,
(a3) d(fnX)→d(A)asn→ ∞, (a4) {fn|n∈ω} ⊆Hf.
Lemma1.2(see [7]). Letf be a continuous self mappings of a metric space(X, d) with the following properties:
(a5) fhas a unique fixed pointwinX,
(a6) for everyx∈X, the sequence of iterations{fnx}∞
n=0converges tow,
(a7) there exists an open neighborhoodUofwwith the property that given any open setVcontainingw, there existsk∈Nsuch thatn≥kimpliesfnU⊂V. Then for eachα∈ (0,1), there exists a metric d∈ M(X)relative to which f is a contraction with Lipschitz constantα.
2. Main results
Theorem2.1. LetΦbe a family of self mappings of a metric space(X, d). Then the following statements are equivalent:
(b1) Φhas a fixed point;
(b2) there existm, n∈Nand continuous compact self mappingsf,gof(X, d)such that eitherΦ⊆Cf orΦ⊆Cgand
Ffmx, gny< δHf(x), Hg(y), (2.1)
for allx, y∈Xwithfmx=gny;
(b3) there existm, n∈Nand continuous self mappingsf,gof(X, d)such thatf g is compact,f∈Cg,Φ∈Cf g, and
Ffmx, gny< δHf g(x, y), (2.2)
for allx, y∈Xwithfmx=gny,
(b4) there exists a continuous compact self mapping of(X, d)withΦ⊆Cf such that
F (f x, f y) <max
F (x, y), F (x, f x), F (y, f y),F (x, f x)F (y, f y) F (x, y) ,
F (f x, f y)F (x, f x) F (x, y) ,
F (x, f y)F (f x, y) F (x, y)
,
(2.3)
for allx, y∈Xwithx=y.
Moreover, if (b2) holds, thenf,g, andΦhave a unique common fixed point; if (b3) holds, thenf gandΦhave a unique common fixed point; if (b4) holds, thenf andΦ
Proof. Let (b1) hold andwbe a fixed point ofΦ. Definef , g:X→Xbyf x=gx= wfor allx∈X. It is easy to show that (b2), (b3), and (b4) hold.
Assume that (b2) holds. LetA= ∩n∈ωfnX,B= ∩n∈ωgnX. Sincef andgare con-tinuous compact self mappings of(X, d), it follows from (a1) and (a2) thatAandB are compact andf A=A,gB =B. Consequently,fmA=A, gnB=B. Suppose that δ(A, B) >0. Then there exista∈A,b∈Bwithδ(A, B)=F (a, b)becauseFis continu-ous andA×Bis compact. SincefmA=A,gnB=B, there existx∈A,y∈Bsuch that fmx=a,gny=b. In view of (2.1), we have
δ(A, B)=F (a, b)=Ffmx, gny< δHf(x), Hg(y)< δ(A, B), (2.4)
which is impossible, and henceδ(A, B)=0. That is,A=B= {w}for somew∈Xso f w=gw=w. Ifvis another fixed point of f, thenv∈ ∩n∈ωfnX= {w}, that is,
v=w. Hencewis the only fixed point off. Similarly,wis also the only fixed point ofg.
Without loss of generality, we assume thatΦ⊆Cf. It follows from Cf ⊆Hf that hA⊆Afor allh∈Φ. That is,hw=wfor allh∈Φ. Thuswis the only common fixed point off,g, andΦ. Therefore (b1) holds.
Assume that (b3) holds. PutA= ∩n∈ω(f g)nX. ThenAis compact andf gA=A. Sincef is continuous andf∈Cg, we infer that
f A=f∩n∈ω(f g)nX⊆ ∩n∈ω(f g)nf X⊆ ∩n∈ω(f g)nX=A. (2.5)
Similarly, we have
gA⊆A. (2.6)
It follows fromf gA=A, (2.5), and (2.6) that
f A⊆A=f gA⊆f A. (2.7)
That is, f A=A. Similarly, we have gA=A. Suppose thatδ(A) >0. BecauseF is continuous andAis compact, then there exista, b∈Asuch thatδ(A)=F (a, b). Since fmA=gnA=A, there existx, y∈Awithfmx=a,gny=b. Using (2.2), we have
δ(A)=F (a, b)=fmx, gny< δHF G(x, y)≤δ(A), (2.8)
which is a contradiction. Henceδ(A)=0. That is,A= {w} for somew∈X. This implies thatf w=gw=f gw=w. As in the proof of above, we can prove thatwis the only fixed point off g, andwis the unique common fixed point off gandΦ. So (b1) holds.
Assume that (b4) holds. As above we infer thatA= ∩n∈ωfnXis compact andf A=
virtue of (2.3), we get
φ(f w)=F (f w, f f w)
<max
F (w, f w), F (w, f w), F (f w, f f w),F (w, f w)F (f w, f f w) F (w, f w) ,
F (f w, f f w)F (w, f w) F (w, f w) ,
F (w, f f w)F (f w, f w) F (w, f w)
=F (w, f w)
=φ(w).
(2.9)
This is a contradiction to the definition ofw. Sow is a fixed point off. Iff has a second distinct fixed pointv, by (2.3), we obtain that
F (w, v)=F (f w, f v)
<max
F (w, v), F (w, w), F (v, v),F (w, w)F (v, v) F (w, v) ,
F (w, v)F (w, w) F (w, v) ,
F (w, v)F (w, v) F (w, v)
=F (w, v),
(2.10)
which is a contradiction. Therefore,wis the only fixed point off. It is a simple matter to show thatwis the unique common fixed point off andΦ. Thus (b1) holds. This completes the proof.
Next, we give a theorem about the equivalent condition for the existence of fixed points of a continuous compact self mapping on a metric space.
Theorem2.2. Letsbe a continuous compact self mapping of a metric space(X, d).
Thens has a fixed point if and only if there exists a continuous self mappingf ofX such thatf∈Csand
F (f x, f y) <max
F (sx, sy), F (sx, f x), F (sy, f y),F (sx, f x)F (sy, f y) F (sx, sy) ,
F (f x, f y)F (sx, f x) F (sx, sy) ,
F (sx, f y)F (f x, sy) F (sx, sy)
,
(2.11)
for allx, y∈Xwithsx=sy. Indeed,f andshave a unique common fixed point.
Proof. To see that the stated conditions is necessary, suppose thatshas a fixed pointw∈X. Definef:X→Xbyf x=wfor allX∈X. Thenf sx=w=sw=sf x for allx∈X, that is,f∈Cs. Clearly, (2.11) holds.
On the other hand, suppose that there exists a continuous self mappingfofXsuch thatf∈Cs and (2.11) holds. LetA= ∩n∈ωSnX. FromLemma 1.1, we infer thatAis compact andsA=A. Sincefis continuous ands∈Cf, we have
Define the function φ(x)by φ(x)=F (sx, f x)for all x∈A. It is clear that φ(x) is continuous on A and attains its minimum value at somew ∈A. We claim that sw=f w. If not, from (2.12), there existsp∈Asatisfyingf w=sp. Using (2.11), we conclude that
φ(p)=F (sp, f p)=F (f w, f p)
<max
F (sw, sp), F (sw, f w), F (sp, f p),F (sw, f w)F (sp, f p) F (sw, sp) ,
F (f w, f p)F (sw, f w) F (sw, sp) ,
F (sw, f p)F (f w, sp) F (sw, sp)
=maxF (sw, f w), F (sp, f p)
=F (sw, f w)
=φ(w),
(2.13)
which is a contradiction to the choice ofw. Sosw=f w. By virtue off∈Cs, we have
f sw=sf w=ssw. (2.14)
Now suppose thatssw=sw. By (2.11) and (2.14), we get
F (ssw, sw)=F (f sw, f w)
<max
F (ssw, sw), F (ssw, f sw), F (sw, f w),F (ssw, f sw)F (sw, f w) F (ssw, sw) ,
F (f sw, f w)F (ssw, f sw) F (ssw, sw) ,
F (ssw, f w)F (f sw, sw) F (ssw, sw)
=F (ssw, sw),
(2.15)
which is a contradiction. Thusssw=sw, that is,swis a fixed point ofs. Therefore, the set M of fixed points of s is not empty. Now s is continuous, soM is closed. SinceM ⊆Aand Ais compact, M is compact. Moreover, since f and s commute, f (M)⊆M. Note also that (2.11) restricted toM reduces to (2.3). We can therefore applyTheorem 2.1(b4) tofM:M→Mto obtain a unique common fixed pointuoff ands inM. SinceM contains all the fixed points ofs, uis a unique common fixed pointf ands. This completes the proof.
Theorem2.3. Letf,gbe continuous compact self mappings of a metric space(X, d)
satisfying(2.1). Thenfandghave a unique fixed point, respectively, and furthermore, for anyα∈(0,1), there exist metrics dand d∈M(X)relative to whichf andg satisfy, respectively,
d(f x, f y)≤αd(x, y), d(gx, gy)≤αd(x, y), (2.16)
for allx, y∈X.
Proof. LetA= ∩n∈ωfnX,B= ∩n∈ωgnX, andU=X. As in the proof ofTheorem
d(fnX), d(gnX)→0 asn→ ∞. ThusfnXandgnXsqueeze into any neighborhood of w. That is (a7) is fulfilled. ThusTheorem 2.3follows fromLemma 1.2. This completes the proof.
Corollary2.4. Letfbe a continuous compact self mapping of a metric space(X, d)
satisfying
F (f x, f y) < δHf(x), Hf(y), (2.17)
for allx, y∈Xwithx=y. Thenf has a unique fixed point.
Furthermore, for anyα∈(0,1), there exists a metricd∈M(X)relative to whichf satisfies
d(f x, f y)≤αd(x, y), (2.18)
for allx, y∈X.
The following simple example reveals that Corollary 2.4 extends properly Theo-rem 1.1 of Janos [1] and Theorem 1 of Park [9].
Example2.5. LetX= {0,2,4,6,9}with the usual metric. Define a mappingf:X→ X byf0=f4=f6=6,f2=0, andf9=2. Then f is a continuous compact self mapping ofX. It is easy to check that
d(f x, f y)≤6<9=δHf(x), Hf(y), (2.19)
for allx, y∈Xwithx=y. So the conditions ofCorollary 2.4are satisfied. But Theo-rem 1.1 of Janos [1] and Theorem 1 of Park [9] are not applicable since
d(f2, f4)=6>2=12d(2, f2)+d(4, f4),
d(f2, f4)=6=δO(2,4, f ).
(2.20)
Acknowledgement. The authors express their thank to the referee for his help-ful suggestions, and the third author was supported by Korea Research Foundation Grant (KRF-99-005-D00003).
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Zeqing Liu: Department of Mathematics, Liaoning Normal University, Dalian, Liaoning116029, China
E-mail address:zeqingliu@sina.com.cn
Lili Zhang: Department of Mathematics, Liaoning Normal University, Dalian, Liaoning116029, China
Shin Min Kang: Department of Mathematics, Gyeongsang National University,
Chinju660-701, Korea
