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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 → X a mapping such that, for each x,y ∈ X,
d(f x,f y)
0 ϕ(t)dt ≤ c
d(x,y)
0 ϕ(t)dt, (1)
where ϕ : R+→ R+is a Lebesgue-integrable mapping which is summable, non- negative, and such that, for each > 0,
0ϕ(t)dt > 0. Then f has a unique fixed point z ∈ X such that, for each x ∈ X, limnfnx = 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 → X a mapping such that, for each x,y ∈ X,
d(f x,f y)
0 ϕ(t)dt ≤ k
m(x,y)
0 ϕ(t)dt, (3)
where ϕ : R+→ R+is a Lebesgue-integrable mapping which is summable, non- negative, and such that
0ϕ(t)dt > 0 for each > 0. (4) Then f has a unique fixed point z ∈ X and, for each x ∈ X, limnfnx = z.
Proof. Let x ∈ X and, for brevity, define xn= fnx. For each integer n ≥ 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,xn+1),d(xn−1,xn)}
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), as n → ∞, gives
limn
d(xn,xn+1)
0 ϕ(t)dt = 0, (10)
which, from (4), implies that
limn d
xn,xn+1
= 0. (11)
TWO FIXED-POINT THEOREMS FOR MAPPINGS ...
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(xm(p)−1,xm(p))
0 ϕ(t)dt = lim
p
d(xn(p)−1,xn(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 limit z.
From (2),
d(f z,xn+1)
0 ϕ(t)dt ≤ k
m(z,xn)
0 ϕ(t)dt
= kmaxd(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) as n → ∞, 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 that d(z,f z) = 0 or z = f z.
Suppose that z and w are fixed points of f . 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, or z = 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)
TWO FIXED-POINT THEOREMS FOR MAPPINGS ...
But this is not possible since, as the following example shows, one must assume that the orbits are bounded.
Example3. Let f : N → N be defined by f (n) = n + 1 and φ,ϕ : [0,∞) → [0,∞), where φ(t) := (t +1)t+1−1, and ϕ(t) = φ(t).
Then, for n > m,
M(n,m) = max{n−m,1,n−m −1,n−m +1}
= n−m+1 = t +1, (27)
where t := n−m.
Note that, for any t ∈ N,
(t +2)t+2−1 = (t +1+1)t+2−1 ≥ (t +1)t+2+1t+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)
Since ϕ(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 and f has no fixed points.
Theorem 1is clearly a special case ofTheorem 2. With ϕ equal to the con- stant function 1,Theorem 2reduces to [2, Theorem 2.5]
It is possible to prove a weaker theorem involving condition (25).
Let O(x,n) := {x,f x,f2x,...,fnx}. Then O(x,n) is called the nth orbit of x. For any set A, δ(A) will denote the diameter of A.
Theorem4. Let (X,d) be a complete metric space, k ∈ [0,1), f : X → X a mapping such that, for each x,y ∈ X, (25) is satisfied, where ϕ : R+→ R+ is a Lebesgue-integrable mapping which is summable, nonnegative, and satisfies (4). If there exists a point x ∈ X with bounded orbit, then f has a unique fixed point z ∈ X.
Proof. From the definition of O(x,n), there exist integers i, j satisfying 0≤ i < j ≤ n such that δ(O(x,n)) = d(fix,fjx).
Claim5. For some integer k satisfying 0 < k ≤ n, δ(O(x,n)) = d(x,fkx).
Proof ofClaim5. We may assume that δ(O(x,n)) > 0 for each n, since, if there exists an n for which δ(O(x,n)) = 0, then f has a fixed point.
Suppose that δ(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 since δ(O(x,n)) > 0. Therefore i = 0.
Pick an x ∈ X with bounded orbit. Let m and n be 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(xn−1,xk1+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,xk2+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 as m,n → ∞ gives, since the orbit of x is 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 limit z. 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,xn+1)}
0 ϕ(t)dt.
(35)
TWO FIXED-POINT THEOREMS FOR MAPPINGS ...
Taking the limit of both sides, as n → ∞, gives
d(z,f z)
0 ϕ(t)dt ≤ k
d(z,f z)
0 ϕ(t)dt, (36)
which implies that d(z,f z) = 0, which, from (4), implies that z = f z.
Suppose that z and w are fixed points of f . From (25),
d(z,w)
0 ϕ(t)dt ≤ k
d(z,w)
0 ϕ(t)dt, (37)
which implies that z = w, and the fixed point is unique.
The following example shows that (2) is indeed a proper extension of (1).
Example6. Let X := {1/n : n ∈ Z, |n| ≥ 2} ∪ {0} endowed with the Eu- clidean metric. Define f : X → X by
f
1 n
:=
1
n+1, n > 1 and odd, 1
n, n > 0 and even or n < −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]
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