Vol. 23, No. 3 (2000) 205–210 S0161171200001599 © Hindawi Publishing Corp.
RELATED FIXED POINTS FOR SET VALUED MAPPINGS
ON TWO METRIC SPACES
BRIAN FISHER and DURAN TÜRKO¯GLU
(Received 30 July 1998)
Abstract.Some related fixed points theorems for set valued mappings on two complete and compact metric spaces are proved.
Keywords and phrases. Set valued mapping, complete metric space, compact metric space, fixed point.
2000 Mathematics Subject Classification. Primary 47H10; Secondary 54H25.
We let(X,d)be a complete metric space and letB(X)be the set of all nonempty sub-sets ofX. As in [1, 2], we define the functionδ(A,B)withAandBinB(X)byδ(A,B)=
sup{d(a,b):a∈A,b∈B}. IfAconsists of a single pointawe writeδ(A,B)=δ(a,B). IfB also consists of single pointb, we writeδ(A,B)=δ(a,B)=δ(a,b)=d(a,b). It follows immediately thatδ(A,B)=δ(B,A)≥0, andδ(A,B)≤δ(A,C)+δ(C,B)for all A,BinB(X).
Now if{An:n=1,2,...}is a sequence of sets inB(X), we say that it converges to the setAinB(X)if
(i) each point a∈A is the limit of some convergent sequence{an ∈An:n=
1,2,...},
(ii) for arbitrary >0, there exists an integerNsuch thatAn⊂Aforn > N, where Ais the union of all open spheres with centers inAand radius.
The setAis then said to be the limit of the sequence{An}. The following lemma was proved [1].
Lemma. If{An}and{Bn}are sequences of bounded subsets of a complete metric space (X,d)which converge to the bounded subsets AandB, respectively, then the sequence{δ(An,Bn)}converges toδ(A,B).
Now, letFbe a mapping ofXintoB(X). We say that the mappingFis continuous at a pointXif whenever{xn}is a sequence of points inXconverging tox, the sequence
{Fxn}inB(X)converges toFxinB(X). We say thatF is continuous mapping ofX intoB(X)ifFis continuous at each pointxinX. We say that a pointzinXis a fixed point ofF ifzis inFz. IfAis inB(X), we define the setFA=a∈AFa.
Theorem1. Let(X,d1)and(Y ,d2)be complete metrics spaces, letF be mapping ofXintoB(Y )and letGbe mapping ofY intoB(X)satisfying the inequalities
δ1GFx,GFx≤cmax
δ2FGy,FGy≤cmax
d2y,y,δ2(y,FGy),δ1y,FGy ,δ1Gy,Gy (2) for allx,x inXandy,y, where0≤c <1. IfF is continuous, thenGF has a unique fixed pointzinXandFGhas a unique fixed pointwinY.
Proof. Letx1be an arbitrary point inX. Define sequences{xn}and {yn}inX
andY, respectively, as follows. Choose a pointy1inFx1and then a pointx2inGy1.
In general, having chosenxninXandyninY choosexn+1inGynand thenyn+1in Fxn+1forn=1,2,.... Then,
d1xn+1,xn+2≤δ1GFxn,GFxn+1
≤cmaxd1xn,xn+1,δ1xn,GFxn, δ1xn+1,GFxn, δ1xn+1,GFxn+1,δ2Fxn,Fxn+1
≤cmaxδ1GFxn−1,GFxn,δ1GFxn,GFxn+1,δ2Fxn,Fxn+1 =cmaxδ1GFxn−1,GFxn,δ2Fxn,Fxn+1
(3) and, similarly,
d2yn+1,yn+2≤δ2FGyn,GFyn+1
≤cmaxδ2FGyn−1,FGyn,δ1Gyn,Gyn+1.
(4)
It follow that, forr=1,2,...,
d1xn+1,xn+r+1≤δ1GFxn,GFxn+r
≤δ1GFxn,GFxn+1+···+δ1GFxn+r−1,GFxn+r ≤cn+cn+1+···+cn+r−1δ
1x1,GFx1<
(5)
forngreater than some N, sincec <1. The sequence{xn}is, therefore, a Cauchy sequence in the complete metric space X and so has a limitz in X. Similarly, the sequence{yn}is a sequence in complete metric spaceY and so has a limitwinY.
Further
δ1z,GFxn≤d1z,xm+1+δ1xm+1,GFxn
≤d1z,xm+1+δ1GFxm,GFxn, (6)
sincexm+1∈GFxm. Thus, on using inequality (5), we have
δ1z,GFxn≤d1z,xm+1+ (7)
form,n≥N. Lettingmtends to infinity it follows that
δ1z,GFxn< (8)
forn > Nand so
lim
n→∞GFxn= {z} (9)
sinceis arbitrary. Similarly, lim
sincexn+1is inGyn. Using the continuity ofF, we see that
lim
n→∞Fxn=Fz= {w}. (11)
Using inequality (1), we now have
δ1GFxn,GFz≤cmax
d1(xn,z),δ1xn,GFxn,δ1(z,GFz),δ2Fz,Fxn. (12)
Lettingntends to infinity and using (9) and (11), we have
δ1(z,GFz)≤cδ1(z,GFz). (13)
Sincec <1,δ1(z,GFz)=0 and, so, we must haveGFz= {z}, proving thatzis a fixed
point ofGF.
Further, using (11), we have
FGw=FGFz=Fz=w, (14)
proving thatwis a fixed point ofFG.
Now suppose thatGF has a second fixed pointz. Then using inequalities (1) and (2), we have
δ1z,GFz≤δ1GFz,GFz
≤cmaxd1z,z,δ1z,GFz,δ2Fz,Fz
=cδ2Fz,Fz≤cδ2Fz,FGFz≤cδ2FGFz,FGFz ≤c2maxδ
2Fz,Fz,δ2Fz,FGFz,δ1GFz,FGz =c2δ
2GFz,GFz
(15)
and soFz is a singleton andGFz = {z}, sincec <1. Thus, d1z,z=δ1GFz,GFz
≤cmaxd1z,z,δ1(z,GFz),δ1z,GFz,δ2Fz,Fz =cd2Fz,Fz.
(16)
But
d2Fz,Fz≤δ2FGFz,FGFz
≤cmaxδ2Fz,Fz,δ2(Fz,FGFz),δ2Fz,FGFz,δ1GFz,GFz =cmaxd2Fz,Fz,d2(Fz,Fz),d2Fz,Fz,d1z,z
=cd1z,z
(17)
and so
d1z,z≤c2d1z,z. (18)
Sincec <1, the uniqueness ofzfollows.
If we letFbe a single valued mappingTofXintoYandGbe a single valued mapping ofY intoX, we obtain the following result given in [3].
Corollary1. Let(X,d1)and(Y ,d2)be complete metric spaces. IfTis a continuous mapping ofXintoY, andSis a mapping ofY intoXsatisfying the inequalities
d1ST x,ST x≤cmax
d1x,x,d1(x,ST x),d1x,ST x,d2T x,T x, d2ST y,ST y≤cmax
d2y,y ,d2(y,T Sy),d2y,T Sy ,d1Sy,Sy
(19)
for allx,x inXandy,y inY, where0≤c <1, thenST has a unique fixed pointz inXandT S has a unique fixed pointwinY. FurtherT z=wandSw=z.
Theorem2. Let(X,d1)and(Y ,d2)be compact metric spaces. IfF is a continuous mapping ofXintoB(Y ), andGis a continuous mapping ofY intoB(X)satisfying the inequalities
δ1(GFx,GFx) <max
d1x,x,δ1(x,GFx),δ1x,GFx,δ2Fx,Fx, δ2FGy,FGy<max
d2y,y,δ2y,FGy,δ2y,FGy ,δ1Gy,Gy
(20)
for allx,x inX andy,y inY for which the right-hand sides of the inequalities are positive. ThenFGhas a unique fixed pointzinXandGF has a unique fixed pointw inY. FurtherFGz= {z}andGFw= {w}.
Proof. Let us denote the right-hand side of inequalities (20) by h(x,x) and k(y,y ), respectively. First of all suppose thath(x,x )=0 for all x,x ∈X and k(y,y )=0 for ally,y ∈Y. Define the real-valued functionf(x,x)onX2by
fx,x=δ1
GFx,GFx
hx,x . (21)
Then if (xn,xn)is an arbitrary sequence in X2 converning to (x,x), it follows
from the lemma and the continuity ofF andGthe sequence{f(xn,xn)}converges to f(x,x). The functionf is therefore a continuous function defined on the compact metric spaceX2and so achieves its maximum valuec
1. Because of inequality (9),c1<1
and so
δ1GFx,GFx≤c1max
d1x,x,δ(x,GFx),δ1x,GFx,δ2Fx,Fx (22)
for allx,x inX.
Similarly, there existsc2<1 such that δ2FGy,FGy≤c2max
d2y,y,δ2(y,FGy),δ2y,FGy ,δ1Gy,Gy (23)
for ally,y inY. It follows that the conditions of Theorem 2 are satisfied withc=
max{c1,c2}and, so, once again there existzinXandwinYsuch thatGFz= {z}and FGw= {w}.
Now, suppose thath(x,x )=0 for somex,x inX. ThenGFx=GFx = {x} = {x}
GFandGFz= {z}. Further,
FGw=FGFz=Fz= {w} (24)
and sowis a fixed point ofFG.
It follows similarly that ifk(y,y )=0 for somey,y inY, then againGFhas a fixed pointzandFGhas a fixed pointw.
Now let us suppose thatGF has a second fixed pointz inXso thatz is inGFz. Then on using inequalities (20), we have, on assuming thatδ2(Fz,Fz)=0,
δ1z,GFz≤δ1GFz,GFz
<maxd1z,z,δ1z,GFz,δ2Fz,Fz
=δ2Fz,Fz≤δ2Fz,FGFz≤δ2FGFz,FGFz <maxδ2Fz,Fz,δ2Fz,FGFz,δ1GFz,FGz =c2δ
2GFz,GFz
(25)
a contradiction and soFz is a singleton andGFz = {z}. Thus, ifz=z d1z,z=δ1GFz,GFz
<maxd1z,z,δ1(z,GFz),δ1z,GFz,δ2Fz,Fz =d2Fz,Fz.
(26)
But ifFz=Fz, we have
d2Fz,Fz≤δ2FGFz,FGFz
<maxδ2Fz,Fz,δ2(Fz,FGFz),δ2Fz,FGFz,δ1GFz,GFz =maxδ2Fz,Fz,d2(Fz,Fz),d2Fz,Fz,d1z,z
=d1z,z
(27)
and so
d1z,z< d1z,z, (28)
a contradiction. The uniqueness ofzfollows.
Similarly,wis the unique fixed point ofFG. This completes the proof of the theo-rem.
If we letFbe a single valued mappingTofXintoYandGbe a single valued mapping ofY intoX, we obtain the following result given in [3].
Corollary2. Let(X,d1)and(Y ,d2)be compact metric spaces. IfT is a contin-uous mapping ofY intoX, andS is a continuous mapping ofY intoXsatisfying the inequalities
d1ST x,ST x<max
d1x,x,d1(x,ST x),d1x ,ST x,d2T x,T x , d2T Sy,T Sy <max
d2y,y,d2(y,T Sy),d2y,T Sy ,d1Sy,Sy
for allx,x inX andy,y inY for which the right-hand sides of the inequalities are positive, thenST has a fixed point zinX andT S has a unique fixed pointw inY. Further,T z=wandSw=z.
References
[1] B. Fisher,Common fixed points of mappings and set-valued mappings, Rostock. Math. Kol-loq. (1981), no. 18, 69–77. MR 83e:54041. Zbl 479.54025.
[2] ,Set-valued mappings on metric spaces, Fund. Math.112(1981), no. 2, 141–145. MR 82h:54074. Zbl 456.54009.
[3] ,Related fixed points on two metric spaces, Math. Sem. Notes Kobe Univ.10(1982), no. 1, 17–26. MR 83k:54050. Zbl 501.54032.
Fisher: Department of Mathematics and Computer Science, University of Leicester, Leicester, LE1 7RH, England
E-mail address:fbr@le.ac.uk
Türko¯glu: Department of Mathematics, Kirikkale University, 71450 Yah¸sihan, Kirikkale, Turkey