SOME RESULTS ON COINCIDENCE AND FIXED POINT THEOREMS FOR GENERALIZED CONTRACTION TYPE MAPPINGS

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SOME RESULTS ON COINCIDENCE AND FIXED POINT THEOREMS FOR GENERALIZED CONTRACTION TYPE MAPPINGS

D. K. GANGULY and D. BANDYOPADHYAY (Received 24 April 1996 and in revised form 1 October 1996)

Abstract.Some coincidence and fixed point theorems are proved for certain generalized contraction type single-valued and set-valued compatible mappings.

2000 Mathematics Subject Classification. 47H10, 54H25.

1. Introduction. Jungck [1] generalized the Banach contraction principle using the commuting map concept, which is extended by Sessa [4] giving weakly commuting map concept; this again modified in [2] by compatibility condition. Several authors [3, 5,6] discussed various results on coincidence and fixed point theorem for compatible single-valued and multi-valued maps. Here we develop some coincidence and fixed point theorems for compatible single-valued and multi-valued maps satisfying some generalized contraction type condition. Henceforth, we denote byNandR+, the set of naturals and nonnegative reals, respectively, andω=N∪{0}and(X, d), a metric space, unless otherwise stated.

2. Preliminaries

Definition2.1(see [3]). Two mappingsf , g:X→Xare compatible if and only if d(f gxn, gf xn)→0 whenever{xn}is a sequence inXsuch thatf xn→t,gxn→t, t∈X.

LetC(X)=class of closed subsets ofX, CB(X)=class of closed bounded subsets ofX, co(K)=convex hull ofK⊂X. The Hausdorff metricH on CB(X)is defined as H(A, B)=max{supx∈AD(x, B),supX∈BD(x, A)}, for allA, B∈CB(X), whereD(x, A)= inf_y∈Ad(x, y).

Definition2.2(see [3]). The mapsf:X→XandT:X→CB(X)are compatible if and only iff T x∈CB(X)for allx∈XandH(f T xn, T f xn)→0 whenever{xn}is a sequence inXsuch thatT xn→M∈CB(X),f xn→t∈M, whereHis the Hausdorff metric onX.

We now recall the following lemmas.

Lemma2.3(see [7]). Leth:R+ →R+ be a nondecreasing upper semi-continuous (u.s.c.) function. Thenh(t) < tif and only ifhⁿ(t)→0for eacht >0wherehⁿdenotes the composition ofhwith itselfntimes.

Lemma2.4(see [3]). LetT:X→CB(X)andf:X→Xbe compatible. Iff z∈T zfor somez∈X, thenf T z=T f z.

3. Coincidence and fixed point theorems for single-valued maps

Theorem3.1. LetXbe any nonempty set and(Y , d) be a complete metric space.

Letf , g, T:X→Y satisfy (i) f (X), g(X)⊆T (X);

(ii) T (X)is closed inY; (iii) for allx, y∈X, d(f x, gy)≤ϕ

max

d(T x, T y), d(T x, f x), d(T x, gy), d(T y, f x), d(T y, gy) , (3.1) where h(t)=ϕ[max{t, t, at, bt, t}] < t, for eacht >0, a, b∈ {0,1,2} with a+b=2andϕ:R+→R+is nondecreasing u.s.c function. Thenf , g, T have a coincidence point inX.

Further if

(iv) f orgis injective, then the coincidence point is unique inX.

Proof. Choose anyx0∈X. From (i), we define an iterationy2n=f x2n=T x2n+1, y_2n+1=gx_2n+1=T x_2n+2. Letdn=d(T xn, T x_n+1). Then from (iii), we have

d_2n+1=d

T x_2n+1, T x_2n+2

=d

y2n, y_2n+1

=d

f x2n, gx_2n+1

≤ϕ

max d

T x2n, T x_2n+1 , d

T x2n, f x2n

,

d

T x2n, gx_2n+1 , d

T x_2n+1, f x2n

, d

T x_2n+1, gx_2n+1

≤ϕ

max

d2n, d2n,

d2n+d2n+1

,0, d2n .

(3.2)

If d_2n+1 > d2n then contradiction arises; so takingd_2n+1 ≤d2n, we have d_2n+1≤ h(d2n). Similarly,d2n+2≤d2n+1, d2n+2≤h(d2n+1). Hencedn+1≤dnanddn≤h(dn−1)≤

··· ≤hⁿ(d0),for alln∈ω.

This yields, byLemma 2.3, limndn=0=limnd(yn, y_n+1). Now, the sequence{yn} is a Cauchy sequence inf (X), which can be proved using the same technique as used in [6, Theorem 2.1] so from (ii),∃u∈Xlimnyn=T u, that is, limnT xn=T uand limnf x2n=T u=limngx_2n+1. Suppose thatf u≠T u≠gu. Then

d(f u, T u)≤d

f u, gx_2n+1 +d

gx_2n+1, T u

≤ϕ

max d

T u, T x_2n+1

, d(T u, f u), d

T u, gx_2n+1 , d

T x_2n+1, f u ,

d

T x_2n+1, gx_2n+1 +d

gx_2n+1, T u

⇒d(f u, T u)

≤ϕ

max

0, d(T u, f u),0, d(T u, f u),0 ,

(3.3) asn→ ∞; henced(f u, T u) < d(f u, T u)which is absurd. Hencef u=T u. Similarly, gu=T u. Thus,f u=T u=guand uniqueness ofufollows from (iii) and (iv).

Lemma3.2. Letf , g:X→Xbe compatible. Iff z=gzfor somez∈X, thenf gz=gf z.

Proof. The proof is similar to that of Kaneko and Sessa [3].

Theorem3.3. Let(X, d, δ)be a bimetric space such thatXis complete with respect toδ. Letf , g, T:X→Xsatisfy conditions (i)–(iii) ofTheorem 3.1with respect tod, and

(v) (f , T )and(g, T )are compatible pairs;

(vi) δ(x, y)≤k(d(x, y))for allx, y∈X,

wherek:R+→R+ is continuous withk(0)=0. Thenf , g, T have a unique common fixed point inX.

Proof. ByTheorem 3.1,{T xn}is Cauchy with respect todand hence from (vi) it is Cauchy with respect toδ. SinceXis complete with respect toδ, fromTheorem 3.1(ii), there exists z∈Xf z=T z=gz. Thus, by Lemma 3.2and (v), T f z=f T z and gT z=T gz. SoT T z=T f z=f T z=f f z=f gz=gf z=ggz=gT z=T gz. Now, fromTheorem 3.1(iii) it is easy to show thatf z=gf z. Thus,f z=gf z=T f z= f f zis a common fixed point off , T andginX. The uniqueness part follows from Theorem 3.1(iii).

Corollary 3.4. Let(X, d) be a complete metric space f , g, T :X→X satisfying (i)–(iv) ofTheorem 3.1and (v) ofTheorem 3.3. Thenf , g, andThave a unique common fixed point inX.

Corollary3.5. Let(X, d)be a complete metric space and letbe a family of self maps ofX. If there is a mapTinsuch that for each pairf , ginsatisfying (i)–(iv) of Theorem 3.1and (v) ofTheorem 3.3, then each member ofhas a unique fixed point inXwhich is a unique common fixed point of the family.

Theorem3.6. Let(X, d)be a complete metric space. Thenf , g, T:X→Xsatisfying Theorem 3.1(iii) have a unique common fixed point if and only if there isu∈Xsuch thatf u=gu=T uandf²u=g²u=T²u.

Proof. The necessary part is trivial. To prove the sufficient part, let there be a u∈X(a)f u=gu=T u, (b)f²u=g²u=T²u. Lety=f u=gu=T u. Then from Theorem 3.1(iii) and (b), we can show thaty=f y=T y=gy, that is,yis a common fixed point of f , g, T in X. Further, from (iii) ofTheorem 3.1, the uniqueness of y follows at once.

Theorem3.7. LetXbe a set andY a Banach space. Letf , g:X→Y be such that (i) co(f (X))⊂g(X);

(ii) g(X)is closed inY;

(iii) f x−f y ≤ϕ[max{gx−gy,gx−f x,gy−f y}]for all x, y ∈X whereϕ:R+→R+is nondecreasingu.s.c.function withϕ(qt) < t,1≤q≤2.

Then there is au∈Xsuch thatf u=gu. Further, iff orgis injective, thenu is unique.

Proof. Choosex∈X. From (i) ofTheorem 3.7, we define{xn}inXasf xn=gxn+1, for alln∈ω. Writingdn= f xn−f x_n+1and using (iii) ofTheorem 3.7, we get

dn< dn−1, dn≤ϕ dn−1

≤ ··· ≤ϕⁿ d0

, ∀n∈ω. (3.4)

Now, for eachp∈N,

f xn−f x_n+p≤

p−1 i=1

f x_n+i−f xn+1+i

≤

p−1

i=0

ϕⁿ⁺ⁱ d0

=ϕⁿ d0

. ϕ^p

d0

−1 ϕ

d0

−1

→0

(3.5)

asn→ ∞byLemma 2.3implies{f xn}is Cauchy inY and by assumption, limnf xn

exists finitely inY. From (i), definegyn=af xn+(1−a)f x_n+1, 0≤a≤1 ing(X). We havef yn−gyn≤af xn−f yn+(1−a)f xn+1−f yn

≤aϕ

maxgxn−gyn,gxn−f xn,gyn−f yn

+(1−a)ϕ

maxgxn+1−gyn,gxn+1−f xn+1,gyn−f yn

. (3.6) Also,

gxn−gyn≤f xn−1f xn+(1−a)f xn−1−f xn+1

≤ϕⁿ⁻¹ d0

+(1−a)ϕⁿ d0

≤(2−a)ϕⁿ⁻¹

d0 usingϕ d0

< d0

, gxn+1−gyn=(1−a)f xn−f xn+1≤(1−a)ϕⁿ

d0 .

(3.7) Thus, from (3.4) and (3.7), (3.6) reduces to

f yn−gyn≤aϕ

max

(2−a)ϕⁿ⁻¹ d0

, ϕⁿ⁻¹ d0

,f yn−gyn

+(1−a)ϕ max

(1−a)ϕⁿ d0

, ϕⁿ d0

,f yn−gyn

≤aϕ

max

(2−a)ϕⁿ⁻¹ d0

,f yn−gyn

+(1−a)ϕ

max

(2−a)ϕⁿ⁻¹ d0

,f yn−gyn

, asϕ

d0

< d0,1≤2−a≤2

≤ϕ

max

(2−a)ϕⁿ⁻¹ d0

,f yn−gyn

≤ϕ

(2−a)ϕⁿ⁻¹ d0

< ϕⁿ⁻¹ d0

,

(3.8)

otherwise, iff yn−gynis maximum then a contradiction arises.

Now, for anyp∈N, writingKp=(ϕ^p(d0)−1)/(ϕ(d0)−1)we get gyn−gyn+p≤af xn−f xn+p+(1−a)f x_n+1−f x_n+1+p

≤ aϕⁿ

d0

+(1−a)ϕⁿ⁻¹ d0

Kp →0 asn → ∞ ⇒ gyn

(3.9)

is Cauchy ing(X)⊂Y, and from (ii) ofTheorem 3.7there existsu∈Xlimngyn= gu. So, from (3.4), (3.7), and (3.8) we have, f xn−f yn ≤ f xn−gxn + gxn− gyn + gyn−f yn → 0 as n → ∞. Hence, limnf xn = limnf yn = limngyn = limngxn=gu.

Now, let f u = gu. Then from (iii) of Theorem 3.7, we have f u−f xn ≤ ϕ[max{gu−gxn, gu−f u, gxn−f xn}]; taking limit as n→ ∞, we have f u−gu ≤ϕ[max{0,f u−gu,0}] <f u−guwhich is a contradiction. Hence f u=gu. The second part follows from (iii) of Theorem 3.7and injectiveness off org.

4. Coincidence point for multivalued mappings

Theorem4.1. LetXbe a Banach space; and letS, T:X→CB(X)andf:X→Xbe such that

(i) S(X)U T (X)⊆f (X)∈C(X),

(ii) for all x, y ∈ X, H(Sx, T y) ≤ ϕ{f x− f y, D(f x, Sx), D(f y, T y), D(f x, T y), D(f y, Sx)} where ϕ: R⁵₊ → R+ is u.s.c. and nondecreasing in each coordinate variable withγ(t)=max[ϕ(t, t, t, at, bt):a+b=2, a, b∈ {0,1,2}]≤qt,0≤q <1,t >0. Thenf , S andT have a coincidence point inX.

Proof. Choosea∈(0,1) such that q^1−a <1. Let x0∈X. Form (i), we define a sequence{xn}inXasf x_2n+1∈Sx2n,f x_2n+2∈T x_2n+1such that

f x_2n+1−f x_2n+2< q^−aH

Sx2n, T x_2n+1 , f x_2n+2−f x_2n+3< q^−aH

T x_2n+1, Sx_2n+2

, (4.1)

for all n∈ω, writing dn= f xn−f x_n+1, we have from (ii) by routine calcula- tions that d_2n+1≤d2n and d_2n+1≤q^1−ad2n. Similarly,d_2n+2≤d_2n+1 andd_2n+2≤ q^1−ad2n+1. Thus, combining these we can write

d_n+1≤dn, dn≤q^1−ad_n−1≤ ··· ≤q^(1−a)nd0, ∀n∈ω,0≤q^1−a<1. (4.2) This shows that{f xn}is Cauchy in f (X)and from (i) ofTheorem 4.1, there exists z∈Xlimf xn=f z,

D(f z, Sz)≤f z−f x_2n+2+D

f x_2n+2, Sz

≤f z−f x_2n+2+H

Sz, T x_2n+1

≤ϕf z−f x_2n+1, D(f z, Sz), D

f x_2n+1, T x_2n+1 , D

f z.T x_2n+1 , D

f x2n+1, Sz +f z−f x2n+2

≤ϕf z−f x2n+1, D(f z, Sz),f x2n+1−f x2n+2,f z−f x2n+2, f x_2n+1−f z+D(f z, Sz)

+f z−f x_2n+2.

(4.3)

Asn→ ∞, we haveD(f z, Sz)≤ϕ{0, D(f z, Sz),0,0, D(f z, Sz)} ≤ϕ{t, t, t, t, t} ≤qt (wheret=D(f z, Sz))which implies thatf z∈Sz=Sz. Similarlyf z∈T z.

Hencezis a coincidence point off , S andT inX.

In [3, Theorem 2] the continuity of the involved maps are taken; but inTheorem 4.1 instead of the continuity condition of the maps we take only f (X)∈C(X)for the existence of a coincidence point; to support this we give the following example.

Example4.2. LetX=[0,1]. DefineS, T:X→CB(X)andf:X→Xas follows:

Sx=









{0}, 0≤x≤1 2, 1

4

, 1

2< x≤1,

T x=









{0}, 0≤x≤1 2, 1

4

, 1

2< x≤1,

f x=



















0, 0≤x≤1 2, 1

4, 1

2< x <1, 2

3, x=1.

(4.4) ThenSX= {0,1/4} =T X,f X= {0,1/4,2/3} ∈C(X);S, T, andf are discontinuous.

Let ϕ:R⁵₊ →R+ be given byϕ(t1, t2, t3, t4, t5)=

t1/2, ti>0; then γ(t)= t1/2.

ClearlyS, T , f and ϕ,γ satisfy all the conditions ofTheorem 4.1with q=1/2 and 0=f0∈S0=T0, that is, 0 is a coincidence point ofS, T, andf.

Theorem4.3. LetXbe a Banach space andf:X→X,S, T:X→C(X)satisfy (i)–(ii) ofTheorem 4.1and (iii)(f , S)and(f , T ) are compatible pairs. Then there is a point z∈X such thatf z∈Sz∩T z. Suppose that{zn=fⁿz}is a sequence of iterate inX forzand{Sn},{Tn}are sequences of multifunctions onXwhereSnz=Sfⁿ⁻¹z,

Tnz=T fⁿ⁻¹z, fⁿz∈Snz∩Tnz, ∀n∈N. (4.5) Ifzn→zand{Sn},{Tn}converge, respectively, toSandT onXpointwise, thenzis a common fixed point ofSandT.

Proof. From Theorem 4.1, there is z∈ X f z ∈ Sz∩T z. Again from (ii) of Theorem 4.1, it is easy to show thatSz=T z. Again, from (iii) ofTheorem 4.3and Lemma 2.4, we havef z ∈Sz =T z ⇒f²z∈f Sz= Sf z, f²z∈ f T z =T f z, and Sf z=T f z. Continuing this process, we getSnz=Sfⁿ⁻¹z=T fⁿ⁻¹z=Tnz where zn=fⁿz∈Sfⁿ⁻¹z=T fⁿ⁻¹z. By hypothesis,Snz→SzandTnz→T z. Then

D(z, Sz)≤z−zn+D zn, Sz

≤z−zn+H

Snz, Sz

→0 asn → ∞, which implies thatz∈Sz=Sz.

(4.6) AsSz=T z, hencezis a common fixed point ofSandT inX.

Acknowledgement. The authors convey their gratitudes to the referee for his valuable suggestions.

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D. K. Ganguly and D. Bandyopadhyay: Department of Pure Mathematics, University of Calcutta,35, B.C. Road, Calcutta-700019, India