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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.