TWO FIXED-POINT THEOREMS FOR MAPPINGS SATISFYING A GENERAL CONTRACTIVE

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TWO FIXED-POINT THEOREMS FOR MAPPINGS SATISFYING A GENERAL CONTRACTIVE

CONDITION OF INTEGRAL TYPE

B. E. RHOADES Received 6 August 2002

We establish two fixed-point theorems for mappings satisfying a general contrac- tive inequality of integral type. These results substantially extend the theorem of Branciari (2002).

2000 Mathematics Subject Classification: 47H10.

In a recent paper [1], Branciari established the following theorem.

Theorem1. Let(X,d)be a complete metric space,c∈[0,1),f:X→Xa mapping such that, for eachx,y∈X,

_{d(f x,f y)}

0 ϕ(t)dt≤c _d(x,y)

0 ϕ(t)dt, (1)

whereϕ:R⁺→R⁺is a Lebesgue-integrable mapping which is summable, non- negative, and such that, for each >0,

0ϕ(t)dt >0. Thenf has a unique fixed pointz∈Xsuch that, for eachx∈X,lim_nfⁿx=z.

In [1], it was mentioned that (1) could be extended to more general contrac- tive conditions. It is the purpose of this paper to make such an extension to two of the most general contractive conditions.

Define

m(x,y)=max

d(x,y),d(x,f x),d(y,f y),

d(x,f y)+d(y,f x) 2

. (2)

Our first result is the following theorem.

Theorem2. Let(X,d)be a complete metric space,k∈[0,1),f :X→Xa mapping such that, for eachx,y∈X,

_{d(f x,f y)}

0 ϕ(t)dt≤k _m(x,y)

0 ϕ(t)dt, (3)

whereϕ:R⁺→R⁺is a Lebesgue-integrable mapping which is summable, non- negative, and such that

0ϕ(t)dt >0 for each >0. (4) Thenf has a unique fixed pointz∈Xand, for eachx∈X,lim_nfⁿx=z. Proof. Letx∈Xand, for brevity, definexn=fⁿx. For each integern≥1, from (3),

_d(xn,xn+1)

0 ϕ(t)dt≤k

_m(xn−1,xn)

0 ϕ(t)dt. (5)

Using (2),

m

xn−1,xn

=max

d

xn−1,xn ,d

xn,xn+1 ,d

xn−1,xn+1

2

. (6) But

d

xn−1,xn+1

2 ≤d

xn−1,xn +d

xn,xn+1 2

≤max d

xn−1,xn

,d

xn,xn+1 .

(7)

Therefore, m

xn−1,xn

=max d

xn−1,xn ,d

xn,xn+1

. (8)

Substituting into (5), one obtains _d(xn,xn+1)

0 ϕ(t)dt≤k

max{d(xn,x_n+1),d(x_n−1,x_n)}

0 ϕ(t)dt

=kmax

_d(xn,xn+1)

0 ϕ(t)dt,

_d(xn−1,xn)

0 ϕ(t)dt

=k

_d(xn−1,xn)

0 ϕ(t)dt≤ ··· ≤kⁿ

_d(x0,x1)

0 ϕ(t)dt.

(9)

Taking the limit of (9), asn→ ∞, gives

limn

d(x_n,x_n+1)

0 ϕ(t)dt=0, (10)

which, from (4), implies that limn d

xn,xn+1

=0. (11)

...

We now show that{xn}is Cauchy. Suppose that it is not. Then there exists an >0 and subsequences{m(p)}and {n(p)} such that m(p) < n(p) <

m(p+1)with d

xm(p),xn(p)

≥, d

xm(p),xn(p)−1

< . (12)

From (2), n

xm(p)−1,xn(p)−1

=max

d

xm(p)−1,xn(p)−1 ,d

xn(p)−1,xm(p) ,d

xn(p)−1,xn(p) , d

xm(p)−1,xn(p) +d

xn(p)−1,xm(p) 2

.

(13)

Using (11), limp

d(x_m(p)−1,x_m(p))

0 ϕ(t)dt=lim

p

d(x_n(p)−1,x_n(p))

0 ϕ(t)dt=0. (14)

Using the triangular inequality and (12), d

xm(p)−1,xn(p)−1

≤d

xm(p)−1,xm(p)

+d

xm(p),xn(p)−1

< d

xm(p)−1,xm(p)

+. (15)

Hence,

limp

_{d(xm(p)−1,xn(p)−1)}

0 ϕ(t)dt≤

0ϕ(t)dt. (16)

Using the triangular inequality and (12), v(m,n):=d

xm(p)−1,xn(p)

+d

xn(p)−1,xm(p)

2

≤d

xm(p)−1,xm(p) +2d

xm(p),xn(p)−1 +d

xn(p)−1,xn(p) 2

<d

xm(p)−1,xm(p)

+d

xn(p)−1,xn(p)

2 +.

(17)

Therefore, using (11), limp

_v(m,n)

0 ϕ(t)dt≤

0ϕ(t)dt. (18)

Using (3), (12), (13), (14), (16), and (18), it then follows that

0ϕ(t)dt≤

_{d(xm(p),xn(p))}

0 ϕ(t)dt

≤k

_{m(xm(p)−1,xn(p)−1)}

0 ϕ(t)dt≤k

0ϕ(t)dt,

(19)

which is a contradiction. Therefore, {xn} is Cauchy, hence convergent. Call the limitz.

From (2), _{d(f z,xn+1)}

0 ϕ(t)dt≤k

_m(z,xn)

0 ϕ(t)dt

=kmax

d(z,xn)

0 ϕ(t)dt, _{d(z,f z)}

0 ϕ(t)dt, _d(xn,xn+1)

0 ϕ(t)dt,

_d(z,xn+1)

0 ϕ(t)dt,

_{d(xn,f z)}

0 ϕ(t)dt

.

(20)

Taking the limit of (20) asn→ ∞, one obtains _{d(f z,z)}

0 ϕ(t)dt≤k _{d(f z,z)}

0 ϕ(t)dt, (21)

which implies that

_{d(f z,z)}

0 ϕ(t)dt=0, (22)

which, from (4), implies thatd(z,f z)=0 orz=f z. Suppose thatzandware fixed points off. Then, from (2),

d(z,w)

0 ϕ(t)dt=

d(f z,f w)

0 ϕ(t)dt≤k m(z,w)

0 ϕ(t)dt

=kmax

_d(z,w)

0 ϕ(t)dt,0

=k _d(z,w)

0 ϕ(t)dt, (23)

which implies that

_d(z,w)

0 ϕ(t)dt=0, (24)

which, from (4), implies that d(z,w)=0, orz =w, and the fixed point is unique.

One would like to be able to replace (2) with the integral form of ´Ciri´c’s condition [3], that is,

_{d(f x,f y)}

0 ϕ(t)dt≤k _M(x,y)

0 ϕ(t)dt, (25)

where

M(x,y):=max d(x,y),d(x,f x),d(y,f y),d(x,f y),d(y,f x)

. (26)

...

But this is not possible since, as the following example shows, one must assume that the orbits are bounded.

Example3. Letf:N→Nbe defined byf (n)=n+1 andφ,ϕ:[0,∞)→ [0,∞), whereφ(t):=(t+1)^t+1−1, andϕ(t)=φ(t).

Then, forn > m,

M(n,m)=max{n−m,1,n−m−1,n−m+1}

=n−m+1=t+1, (27) wheret:=n−m.

Note that, for anyt∈N,

(t+2)^t+2−1=(t+1+1)^t+2−1≥(t+1)^t+2+1^t+2−1

=(t+1)^t+1(t+1)≥2(t+1)^t+1

≥2(t+1)^t+1−2=2

(t+1)^t+1−1 .

(28)

Sinceϕ(t)=φ(t), it follows from (28) that t

0ϕ(t)dt≤1 2

_t+1

0 ϕ(t)dt (29)

or, equivalently,

_{d(f n,f m)}

0 ϕ(t)dt≤1 2

_M(n,m)

0 ϕ(t)dt, (30)

and (25) is satisfied. However, the orbits are not bounded andf has no fixed points.

Theorem 1is clearly a special case ofTheorem 2. Withϕequal to the con- stant function 1,Theorem 2reduces to [2, Theorem 2.5]

It is possible to prove a weaker theorem involving condition (25).

LetO(x,n):= {x,f x,f²x,...,fⁿx}. ThenO(x,n)is called thenth orbit of x. For any setA,δ(A)will denote the diameter ofA.

Theorem4. Let(X,d)be a complete metric space,k∈[0,1),f :X→Xa mapping such that, for eachx,y∈X, (25) is satisfied, whereϕ:R⁺→R⁺ is a Lebesgue-integrable mapping which is summable, nonnegative, and satisfies (4). If there exists a pointx∈Xwith bounded orbit, thenf has a unique fixed pointz∈X.

Proof. From the definition ofO(x,n), there exist integersi,j satisfying 0≤i < j≤nsuch thatδ(O(x,n))=d(fⁱx,f^jx).

Claim5. For some integerksatisfying0< k≤n,δ(O(x,n))=d(x,f^kx).

Proof ofClaim5. We may assume thatδ(O(x,n)) >0 for eachn, since, if there exists annfor whichδ(O(x,n))=0, thenf has a fixed point.

Suppose thatδ(O(x,n))=d(xi,xj), where 0< i < j≤n. Then, from (25), _δ(O(x,n))

0 ϕ(t)dt=

_d(xi,xj)

0 ϕ(t)dt≤k

_{M(xi−1,xj−1)}

0 ϕ(t)dt

≤k

δ(O(x,n))

0 ϕ(t)dt,

(31)

which is a contradiction sinceδ(O(x,n)) >0. Thereforei=0.

Pick anx∈X with bounded orbit. Letmand nbe integers with m > n. Then, from (25),

_d(xn,xm)

0 ϕ(t)dt

≤k

_{M(xn−1,xm−1)}

0 ϕ(t)dt≤k

_{δ(O(xn−1,m−n+1))}

0 ϕ(t)dt

=k

d(x_n−1,x_k1+n−1)

0 ϕ(t)dt for some 0< k1≤m−n+1

≤k²

_{δ(O(xn−2,k1+n−1))}

0 ϕ(t)dt

=k²

_d(xn−2,xk

2+n−2)

0 ϕ(t)dt for some 0< k²≤m−n+2 ...

≤kⁿ

_δ(O(x,m))

0 ϕ(t)dt.

(32)

Taking the limit asm,n→ ∞gives, since the orbit ofxis bounded,

limm,n

_d(xn,xm)

0 ϕ(t)dt=0, (33)

which, from (4), implies that limm,nd

xn,xm

=0. (34)

Thus{xn}is Cauchy, hence convergent. Call the limitz. From (25), _{d(xn+1,f z)}

0 ϕ(t)dt≤k

_M(xn,z)

0 ϕ(t)dt

=k

_{max{d(xn,z),d(xn,xn+1}),d(z,f z),d(xn,f z),d(z,x_n+1)}

0 ϕ(t)dt.

(35)

...

Taking the limit of both sides, asn→ ∞, gives _{d(z,f z)}

0 ϕ(t)dt≤k _{d(z,f z)}

0 ϕ(t)dt, (36)

which implies thatd(z,f z)=0, which, from (4), implies thatz=f z. Suppose thatzandware fixed points off. From (25),

_d(z,w)

0 ϕ(t)dt≤k _d(z,w)

0 ϕ(t)dt, (37)

which implies thatz=w, and the fixed point is unique.

The following example shows that (2) is indeed a proper extension of (1).

Example6. LetX:= {1/n:n∈Z, |n| ≥2} ∪ {0} endowed with the Eu- clidean metric. Definef:X→Xby

f 1

n

:=

























 1

n+1, n >1 and odd, 1

n, n >0 and even orn <−1 and odd, 1

n+1, n <0 and even, 0, n= ∞.

(38)

Acknowledgment. The author wishes to thank the referee for careful reading of the original manuscript and for providing Examples3and6.

References

[1] A. Branciari,A fixed point theorem for mappings satisfying a general contractive condition of integral type, Int. J. Math. Math. Sci.29(2002), no. 9, 531–536.

[2] Lj. B. ´Ciri´c,Generalized contractions and fixed-point theorems, Publ. Inst. Math.

(Beograd) (N.S.)12(26)(1971), 19–26.

[3] ,A generalization of Banach’s contraction principle, Proc. Amer. Math. Soc.

45(1974), 267–273.

B. E. Rhoades: Department of Mathematics, Indiana University, Bloomington, IN 47405-7106, USA

E-mail address:[email protected]