Fixed Point Theory, Volume 9, No.

1, 2008, 267-273
http://www.math.ubbcluj.ro/∼ nodeacj/sfptcj.html

A NOTE ON RAKOTCH CONTRACTIONS

SIMEON REICH∗ AND ALEXANDER J. ZASLAVSKI∗∗

∗
Department of Mathematics
The Technion-Israel Institute of Technology
32000 Haifa, Israel
E-mail: sreich@tx.technion.ac.il
∗∗
Department of Mathematics
The Technion-Israel Institute of Technology
32000 Haifa, Israel
E-mail: ajzasl@tx.technion.ac.il

Abstract. We establish fixed point and convergence theorems for certain mappings of
contractive type which take a closed subset of a complete metric space X into X.
Key Words and Phrases: Complete metric space, contractive mapping, fixed point, infi-
nite product
2000 Mathematics Subject Classification: 47H09, 47H10, 54H25.

One of the most important results in fixed point theory is the Banach
fixed point theorem [1]. As far as we know, the first significant generaliza-
tion of Banach’s theorem was obtained in 1962 by Rakotch [4], who replaced
Banach’s strict contractions with contractive mappings, that is, with those
mappings which satisfy condition (1) below. Since then, such mappings, as
well as their numerous modifications, were studied and used by many authors
[3]. Recently, a renewed interest in contractive mappings has arisen [2]. See,
for example, [6, 7] where well-posedness and genericity results were estab-
lished. Another important topic in fixed point theory is the search for fixed
points of nonself-mappings. In the present paper, as in [5], we combine these
This paper was presented at the International Conference on Nonlinear Operators, Dif-
ferential Equations and Applications held in Cluj-Napoca (Romania) from July 4 to July 8,
2007.
267
268 SIMEON REICH AND ALEXANDER J. ZASLAVSKI

two themes by proving fixed point and convergence theorems for contractive
nonself-mappings.
In Theorem 1 we provide a new sufficient condition for the existence and
approximation of the unique fixed point of a contractive mapping which maps a
nonempty and closed subset of a complete metric space X into X. In Theorem
2 we present a new proof of the fixed point theorem established in [5, Theorem
1(A)]. This new proof is based on Theorem 1. In Theorem 3 we obtain a
convergence result for (unrestricted) infinite products [8] of mappings which
satisfy a weak form of condition (1). Its proof is analogous to the proof of
Theorem 1(B) in [5].
Let K be a nonempty and closed subset of a complete metric space (X, ρ).
For each x ∈ X and r > 0, set

B(x, r) = {y ∈ X : ρ(x, y) ≤ r}.

Theorem 1. Assume that T : K → X satisfies

ρ(T x, T y) ≤ φ(ρ(x, y))ρ(x, y) for all x, y ∈ K, (1)

where φ : [0, ∞) → [0, 1] is a monotonically decreasing function such that

φ(t) < 1 for all t > 0.
Assume that there exists a sequence {xn }∞
n=1 ⊂ K such that

lim ρ(xn , T xn ) = 0. (2)

n→∞

Then there exists a unique point x̄ ∈ K such that T x̄ = x̄.

Proof. The uniqueness of x̄ is obvious. To establish its existence, let  ∈ (0, 1)
and choose a positive number γ such that

γ < (1 − φ())/8. (3)

By (2), there is a natural number n0 such that

ρ(xn , T xn ) < γ for all integers n ≥ n0 . (4)

Suppose that the integers m, n ≥ n0 . We claim that ρ(xm , xn ) ≤ . Assume

the contrary. Then
ρ(xm , xn ) > . (5)
By (3), (1), (5), the monotonicity of φ, and (4),

ρ(xm , xn ) ≤ ρ(xm , T xm ) + ρ(T xm , T xn ) + ρ(T xn , xn )

 A NOTE ON RAKOTCH CONTRACTIONS 269

≤ 2γ + φ(ρ(xm , xn ))ρ(xm , xn ) ≤ 2γ + φ()ρ(xm , xn )

= ρ(xm , xn ) − (1 − φ())ρ(xm , xn ) + 2γ
< ρ(xm , xn ) − (1 − φ())ρ(xm , xn ) + (1 − φ())/4
< ρ(xm , xn ) − (1 − φ())ρ(xm , xn )(3/4)
= ρ(xm , xn )[(1/4) + (3/4)φ()] < ρ(xm , xn ),
a contradiction.
The contradiction we have reached proves that ρ(xm , xn ) ≤  for all integers
m, n ≥ n0 , as claimed.
Since  is an arbitrary number in (0, 1), we conclude that {xn }∞ n=1 is a
Cauchy sequence and there exists x̄ ∈ X such that limn→∞ xn = x̄. By (1),
for all integers n ≥ 1,

ρ(T x̄, x̄) ≤ ρ(T x̄, T xn ) + ρ(T xn , xn ) + ρ(xn , x̄)

≤ 2ρ(xn , x̄) + ρ(T xn , xn ) → 0 as n → ∞.

This concludes the proof of Theorem 1. 
Next we use Theorem 1 to present a new proof of [5, Theorem 1(A)].
Theorem 2. Let T : K → X satisfy

ρ(T x, T y) ≤ φ(ρ(x, y))ρ(x, y) for all x, y ∈ K,

where φ : [0, ∞) → [0, 1] is a monotonically decreasing function such that

φ(t) < 1 for all t > 0.
Assume that K0 ⊂ K is a nonempty and bounded set with the following
property:
For each natural number n, there exists yn ∈ K0 such that T i yn is defined
for all i = 1, . . . , n.
Then the mapping T has a unique fixed point x̄ in K.
Proof. By Theorem 1, it is sufficient to show that for each  ∈ (0, 1), there
is x ∈ K such that ρ(x, T x) < . Indeed, let  ∈ (0, 1). There is M > 0 such
that
ρ(y0 , yi ) ≤ M, i = 1, 2, . . . . (6)
By (1) and (6), for each integer i ≥ 1,

ρ(yi , T yi ) ≤ ρ(yi , y0 ) + ρ(y0 , T y0 ) + ρ(T y0 , T yi ) ≤ 2M + ρ(y0 , T y0 ). (7)

270 SIMEON REICH AND ALEXANDER J. ZASLAVSKI

Choose a natural number q ≥ 4 such that

(q − 1)(1 − φ()) > 4M + 2ρ(y0 , T y0 ). (8)

Set T 0 z = z, z ∈ K.
We will show that ρ(T q−1 yq , T q yq ) < . Assume the contrary. Then by (1),

ρ(T i yq , T i+1 yq ) ≥ , i = 0, . . . , q − 1. (9)

In view of (1), (9) and the monotonicity of φ, we have for i = 0, . . . , q − 2,

ρ(T i+1 yq , T i+2 yq ) ≤ φ(ρ(T i yq , T i+1 yq ))ρ(T i yq , T i+1 yq ) ≤ φ()ρ(T i yq , T i+1 yq )

and

ρ(T i yq , T i+1 yq ) − ρ(T i+1 yq , T i+2 yq ) ≥ (1 − φ())ρ(T i yq , T i+1 yq ) ≥ (1 − φ()).

(10)
By (7) and (10),

2M + ρ(y0 , T y0 ) ≥ ρ(yq , T yq ) − ρ(T q−1 yq , T q yq )

q−2
X
≥ [ρ(T i yq , T i+1 yq ) − ρ(T i+1 yq , T i+2 yq )]
i=0

≥ (q − 1)(1 − φ())
and
2M + ρ(y0 , T y0 ) ≥ (q − 1)(1 − φ()).
This contradicts (8). The contradiction we have reached shows that

ρ(T q−1 yq , T q yq ) < ,

as required. Theorem 2 is proved. 

Now we establish a convergence result for (unrestricted) infinite products
of mappings which satisfy a weak form of condition (1).
Theorem 3. Let φ : [0, ∞) → [0, 1] be a monotonically decreasing function
such that φ(t) < 1 for all t > 0.
Let
x̄ ∈ K, Ti : K → X, i = 0, 1, . . . , Ti x̄ = x̄, i = 0, 1, . . . , (11)
and assume that

ρ(Ti x, x̄) ≤ φ(ρ(x, x̄))ρ(x, x̄) for each x ∈ K, i = 0, 1, . . . . (12)

 A NOTE ON RAKOTCH CONTRACTIONS 271

Then for each M,  > 0, there exist δ > 0 and a natural number k such that
for each integer n ≥ k, each mapping r : {0, 1, . . . , n − 1} → {0, 1, . . . }, and
each sequence {xi }ni=0 ⊂ K satisfying
ρ(x0 , x̄) ≤ M and ρ(xi+1 , Tr(i) xi ) ≤ δ, i = 0, . . . , n − 1,
we have
ρ(xi , x̄) ≤ , i = k, . . . , n. (13)

Proof. Choose δ0 > 0 such that

δ0 < M (1 − φ(M/2))/4. (14)
Assume that
y ∈ K ∩ B(x̄, M ), i ∈ {0, 1, . . . }, z ∈ X and ρ(z, Ti y) ≤ δ0 . (15)
By (15) and (12),
ρ(x̄, z) ≤ ρ(x̄, Ti y) + ρ(Ti , z) ≤ φ(ρ(x̄, y))ρ(x̄, y) + δ0 . (16)
There are two cases:
ρ(y, x̄) ≤ M/2 (17)
and
ρ(y, x̄) > M/2. (18)
Assume that (17) holds. Then by (16), (17) and (14),
ρ(x̄, z) ≤ ρ(x̄, y) + δ0 ≤ M/2 + δ0 < M. (19)
If (18) holds, then by (16), (15), (14) and the monotonicity of φ,
ρ(x̄, z) ≤ δ0 + φ(M/2)ρ(x̄, y) ≤ δ0 + φ(M/2)M
< (M/4)(1 − φ(M/2)) + φ(M/2)M ≤ M.
Thus ρ(x̄, z) ≤ M in both cases.
We have shown that
if y ∈ K ∩ B(x̄, M ), i ∈ {0, 1, . . . }, z ∈ X, ρ(z, Ti y) ≤ δ0 , then ρ(x̄, z) ≤ M.
(20)
Since M is any positive number, we conclude that there is δ1 > 0 such that
if y ∈ K ∩ B(x̄, ), i ∈ {0, 1, . . . }, z ∈ X, ρ(z, Ti y) ≤ δ1 , then ρ(x̄, z) ≤ .
(21)
272 SIMEON REICH AND ALEXANDER J. ZASLAVSKI

Now choose a positive number δ such that

δ < min{δ0 , δ1 , (1 − φ())4−1 } (22)

and a natural number k such that

k > 4(M + 1)((1 − φ()))−1 + 4. (23)

Let n ≥ k be a natural number. Assume that r : {0, . . . , n − 1} → {0, 1, . . . }

and that
{xi }ni=0 ⊂ K
satisfies

ρ(x0 , x̄) ≤ M and ρ(xi+1 , Tr(i) xi ) ≤ δ, i = 0, . . . , n − 1. (24)

We claim that (13) holds. By (20), (24) and the inequality δ < δ0 ,

{xi }ni=0 ⊂ B(x̄, M ). (25)

Assume to the contrary that (13) does not hold. Then there is an integer j
such that
j ∈ {k, . . . , n} and ρ(xj , x̄) > . (26)
By (26) and (12),
ρ(xi , x̄) > , i = 0, . . . , j. (27)
Let i ∈ {0, . . . , j − 1}. By (24), (12) and the monotonicity of φ,

ρ(xi+1 , x̄) ≤ ρ(xi+1 , Tr(i) xi ) + ρ(Tr(i) xi , x̄) ≤ δ + φ(ρ(xi , x̄))ρ(xi , x̄)

≤ δ + φ()ρ(xi , x̄).
When combined with (22) and (27), this implies that

ρ(xi+1 , x̄) − ρ(xi , x̄) ≤ δ − (1 − φ())ρ(xi , x̄) ≤ δ − (1 − φ()) < −(1 − φ())/2.
(28)
Finally, by (24), (28) and (26),

−M ≤ −ρ(x0 , x̄) ≤ ρ(xj , x̄) − ρ(x0 , x̄)

j−1
X
= [ρ(xi+1 , x̄) − ρ(xi , x̄)] ≤ −j(1 − φ())/2 ≤ −k(1 − φ())/2.
i=0
This contradicts (23). The contradiction we have reached proves (13) and
Theorem 3 itself. 
 A NOTE ON RAKOTCH CONTRACTIONS 273

Acknowledgments. This research was supported by the Israel Science

Foundation (Grant No. 647/07), the Fund for the Promotion of Research at
the Technion and by the Technion President’s Research Fund.

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[4] E. Rakotch, A note on contractive mappings, Proc. Amer. Math. Soc., 13(1962), 459-
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Received: November 15, 2007; Accepted: December 21, 2007.