May 8, 1989 - A fixed point theorem of Fisher andSessa is generalized by replacing the .... d(Tw,Tlxn) rd(lw,12Xn + a max {d(lw,Tw), d(Tlxn,12xn) for n. N.
Internat. J. Math. & Math. Sci. VOL. 13 NO. 3 (1990) 497-500
497
ON A FIXED POINT THEOREM OF FISHER AND SESSA
GERALD JUNGCK Department of Mathematlcs Bradley University Peoria, Illinois 61625, U.S.A.
(Received May 8, 1989)
A fixed point theorem of Fisher and Sessa is generalized by replacing the
ABSTRACT.
requirements
of
commutativity
and
by compatibility and continuity
nonexpansiveness
respectively.
KEY WORDS AND PHRASES.
Common fixed points, commuting and compatible maps.
1980 AMS SUBJECT CLASSIFICATION CODE. 54H25. 1.
INTRODUCTION.
In [1], Fisher and Sessa proved the following generalization of a theorem of
Gregus [2]. THEOREM I.I.
([I]).
Let T and I be two weakly commuting mappings of a closed
convex subset C of a Banach Space X into itself satisfying the inequality
for all x,y in C, where a (0,1). If I is linear and nonexpansive in C and if T(C) E I(C), then T and I have a unique common fixed point.
Sessa defined (see [I]) self maps I and T of a metric space (X,d) to be weakly commuting iff
such
TXn,
self
Ix
d(ITx,TIx) < d(Ix,Tx) for x
maps
to
be
compatible
iff
X.
Subsequently, Jungck [3] defined two
whenever {x
n}
is
a
sequence in X such that
0. some t E X, then d(ITx ,TIx Clearly, commuting maps are n n n weakly commuting, and weakly commuting maps are compatible. [I] and [3] give examples t for
which show that neither implication is reversible.
The
purpose
of
this
note
is
to
show that
Theorem
1.1
can
be
appreciably
generalized by substituting compatibility for weak commutativity and continuity for
the nonexpansive requirement.
G. JUNGCK
498
RESULTS.
2.
The following fact from [3] will shorten the proof of our first result.
LEMMA 2.1 (Proposition 2.2, [3]).
I.
If f(t)
2.
Suppose that
g(t), then fg(t)
IImnf(Xn
limng(Xn t for at t, llmngf(x n)
some t in X.
(a)
If f is continuous
(b)
If f and g are continuous at t, then f(t)
REMARK 2.1.
(X,d) be compatible.
Let f,g:(X,d)
gf(t). f(t).
g(t) and fg(t)
gf(t).
We shall use N to denote the set of positive Integers and cl(S) to
denote the closure of a set S.
Let T and I be compatible self maps of a metric space (X,d) with
PROPOSITION 2.1.
I continuous.
Suppose there exist real numbers r
>
(0,I) such that for all
0 and a
X,
x,y
rd(Ix,ly) + a max {d(Tx,lx), d(Ty,ly)}.
d(Tx,Ty) Then Twffilw for some w K,
X: d(Tx,lx)
{x
n
PROOF. Tw E T(K
n)
0 {cI(T(K ): n n
X iff A
where
I/n}.
X. Then w Suppose that Twffilw for some w A. Tw for all n; i.e, cl(Tn(Kn))
A, for each n there
exists
Consequently, for each n we can and do choose x
implies
that d(Tw, Iw)
A FIXED POINT THEOREM
499
Tw. 0, and thus lw Let I and T be compatible self maps of a closed convex subset C of a I(CI. If there Suppose that I is continuous, linear, and that T(CI
We conclude that d(Tw,lw)
THEOREM 2.1. Banach space X.
(0,11 such that for all x,y
exists a
C
then I and T have a unique common fixed point in C.
pages
n without appeal to the weak commutatlvlty of I and T or the nonexpanslveness of
24-26
I can
that by appeal to the convexity of C, the llnearity of I, and property (2.1), we infer
that A
N {cI(T(K )):n E N} is n
a
si,
lton
and
therefore
Consequently, Proposition (2.1) tien assures us that lw=Tw for some w e C.
nonempty.
We now
show that Tw is a common fixed point of I and T. Since I and T are compatible and lw=Tw, Lemma 2.1. implies that ITwTlw.
have
T2w
Tlw
We then
ITw, so that by (2.1)
TTw ITw; i.e., Tw is a By the above we then have Tw common fixed point of T and I. That Tw is the only common fixed point of I and T follows Immediately from (2.1).
thus, TT,W=-Tw since a
E
(0,11.
In conclusion, we wish to present an example which shows that our Theorem 2.1 Is indeed a generalization of Theorem I.I. To effect this, the following result from [4] is helpful.
LEMMA 2.2.
(Corollary 2.6, [4]).
Suppose that f and g are continuous self maps If fx-gx implies x-fx, then f and g are
of a metric space and that f is proper. compatible.
-I Y is proper iff f (C) is compact in X when
We remind the reader that a map f:X C is compact in Y.
5x/2 and Tx EXAMPLE 2.1. Let X be the reals with the usual norm, C[0,-I, IX -I C C C. C where T(CI Now for x is convex and closed, and I,T: 11 C and I is linear and continuous. Moreover, it is easy to see that the maximum I(C) value of dT/dt on C is 2, so by the Mean Value Theorem we can say x + x(x 2 +
4
for all x,y E C so
that
condition
5
5
4
(2.11 is satisfied.
To see that I and T are
compatible, note that both are continuous and that I is clearly proper, so we can 2 0 0. But I(0) appeal to Lemma 2.2. Now Tx Ix implies that x(3x + I) 0, so x
500
G. JUNGCK
Iff x
O; thus, T and 1 are Compatible by Lemma 2.2.
and so I is not nonexpanslve.
Moreover,
llx-
ly
Ix
Yl
Finally, to see that I and T are not weakly commuting,
REFERENCE S
I.
FISHER, B. and SESSA, S., On a Fixed Point Theorem of Gregus, Internat. J. Math. Math. Scl. 9 (1986), 23-28.
2.
GREGUS, Jr., M., A Fixed Point Theorem in Banach Space, Boll. Un. Mat. Ital. 5 17-A (1980), 193-198.
3.
JUNGCK,
4.
JUNGCK, G., Common Fixed Points for Commuting and Compatible Maps on Compacta Proceedings of the American Mathematical Society 103 (1988), 977-983.
G., Compatible Mappings and Math. Scl. 9 (1986), 771-779.
Common Fixed Polnts,
Internat.
J.
Math.