Open Access

Fixed and coupled fixed points of a new type set-valued contractive mappings in complete metric spaces

Fixed Point Theory and Applications20122012:215

https://doi.org/10.1186/1687-1812-2012-215

Received: 20 June 2012

Accepted: 7 November 2012

Published: 26 November 2012

Abstract

In this paper, motivated by the recent work of Wardowski (Fixed Point Theory Appl. 2012:94, 2012), we introduce a new concept of set-valued contraction and prove a fixed point theorem which generalizes some well-known results in the literature. As an application, we derive a new coupled fixed point theorem. Some examples are also given to support our main results.

MSC:47H10.

Keywords

fixed pointcoupled fixed pointset-valued contractions

1 Introduction

In the literature, there are plenty of extensions of the famous Banach contraction principle [1], which states that every self-mapping T defined on a complete metric space ( X , d ) satisfying
d ( T x , T y ) k d ( x , y ) for each  x , y X ,
(1)

where k [ 0 ; 1 ) , has a unique fixed point, and for every x 0 X , the sequence { T n x 0 } n N is convergent to the fixed point. Some of the extensions weaken the right side of the inequality in the condition (1) by replacing k with a mapping; see, e.g., [24]. In other results, the underlying space is more general; see, e.g., [58]. In 1969, Nadler [9] extended the Banach contraction principle to set-valued mappings. For other extensions of the Banach contraction principle, see [1021] and the references therein.

Recently, Wardowski [22] introduced a new concept of contraction and proved a fixed point theorem which generalizes the Banach contraction principle in a different way than in the known results from the literature. In this paper, we present an improvement and generalization of the main result of Wardowski [22]. To set up our results, in the next section, we introduce some definitions and facts.

Let ( X , d ) be a metric space and let C B ( X ) denote the class of all nonempty bounded closed subsets of X. Let H be the Hausdorff metric with respect to d, that is,
H ( A , B ) = max { sup u A d ( u , B ) , sup v B d ( v , A ) }

for every A , B C B ( X ) , where d ( u , B ) = inf { d ( u , y ) : y B } .

Theorem 1.1 (Nadler [9])

Let ( X , d ) be a complete metric space and let T : X C B ( X ) be a set-valued map. Assume that there exists k [ 0 , 1 ) such that
H ( T x , T y ) k d ( x , y ) for each  x , y X .
(2)

Then T has a fixed point.

In 1989 Mizoguchi and Takahashi [13] proved the following generalization of Theorem 1.1.

Theorem 1.2 (Mizoguchi and Takahashi [13])

Let ( X , d ) be a complete metric space and let T : X C B ( X ) be a set-valued map satisfying
H ( T x , T y ) α ( d ( x , y ) ) d ( x , y ) for each  x , y X ,

where α : [ 0 , ) [ 0 , 1 ) satisfies lim sup t r + α ( t ) < 1 for each r [ 0 , ) . Then T has a fixed point.

2 Main results

Let F : ( 0 , ) R and θ : ( 0 , ) ( 0 , ) be two mappings. Throughout the paper, let Δ be the set of all pairs ( F , θ ) satisfying the following:

( δ 1 ) θ ( t n ) 0 for each strictly decreasing sequence { t n } ;

( δ 2 ) F is strictly increasing;

( δ 3 ) For each sequence { α n } n N of positive numbers, lim n α n = 0 if and only if lim n F ( α n ) = ;

( δ 4 ) If t n 0 and θ ( t n ) F ( t n ) F ( t n + 1 ) for each n N , then n = 1 t n < .

Example 2.1 Let θ 1 ( t ) = τ for each t ( 0 , ) , where τ > 0 is a constant, and let F 1 : ( 0 , ) R be a mapping satisfying lim x 0 + x k F ( x ) = 0 for some k ( 0 , 1 ) where F : ( 0 , ) R is strictly increasing. Then the proof of the main result in [22] shows that ( F 1 , θ 1 ) Δ . We give the details for completeness. Using ( δ 4 ), the following holds for every n N :
F ( t n ) F ( t n 1 ) τ F ( t n 2 ) 2 τ F ( t 0 ) n τ .
(3)
By (3), the following holds for every n N :
t n k F ( t n ) t n k F ( t 0 ) t n k ( F ( t 0 ) n τ ) t n k F ( t 0 ) = t n k n τ 0 .
(4)

Since lim n t n k F ( t n ) = 0 , letting n in (4), we obtain lim n n t n k = 0 . Then there exists n 1 N such that n t n k 1 for n n 1 . Consequently, we have t n 1 n 1 k for all n n 1 . Thus, n = 1 t n < (note that n = 1 1 n 1 k < ).

Example 2.2 Let F 2 ( t ) = ln t and let θ 2 ( t ) = ln ( α ( t ) ) for each t ( 0 , ) , where α : ( 0 , ) [ 0 , 1 ) satisfying
lim sup t r + α ( t ) < 1 for each  r [ 0 , ) .
Now, we show that ( F 2 , θ 2 ) Δ . It is easy to see that F 2 and θ 2 satisfy ( δ 1 )-( δ 3 ). To show ( δ 4 ), assume that t n 0 and
ln ( α ( t n ) ) ln t n ln t n + 1 n N .

Then t n + 1 α ( t n ) t n for each n N . Since lim sup t 0 + α ( t ) < 1 , then there exist n 0 N and 0 < r < 1 such that α ( t n ) < r for n n 0 . Thus, t n + 1 r t n for each n n 0 , and so n = 1 t n < .

Example 2.3 Let F 3 ( t ) = ln t + t and let θ 3 ( t ) = τ for each t ( 0 , ) , where τ > 0 is a constant. Now, we show that ( F 3 , θ 3 ) Δ . We only show ( δ 4 ). Suppose that t n 0 and
τ ( ln t n + t n ) ( ln t n + 1 + t n + 1 ) n N .
Then
s n + 1 e τ s n n N ,

where s n = t n e t n . Since e τ < 1 , then from the above we get n = 1 s n < , and so n = 1 t n < (note that t n s n for each n N ).

Now, we state the main result of the paper.

Theorem 2.4 Let ( X , d ) be a complete metric spaces, let T : X C B ( X ) be a set-valued mapping and let ( F , θ 2 ) Δ . Assume that either T is compact valued or F is continuous from the right. Furthermore, assume that
θ ( d ( x , y ) ) + F ( H ( T x , T y ) ) F ( d ( x , y ) ) x , y X  with  T x T y .
(5)

Then T has a fixed point.

Proof Let x 0 X and x 1 T x 0 . If T x 0 = T x 1 , then x 1 T x 0 = T x 1 and x 1 is a fixed point of T. So, we may assume that T x 0 T x 1 . Since either T is compact valued or F is continuous from the right, x 1 T x 0 and F ( d ( x 1 , T x 1 ) ) < F ( H ( T x 0 , T x 1 ) ) + θ ( d ( x 0 , x 1 ) ) 2 then there exists x 2 T x 1 such that (note that F is increasing)
F ( d ( x 1 , x 2 ) ) F ( H ( T x 0 , T x 1 ) ) + θ ( d ( x 0 , x 1 ) ) 2 .
(6)
From (5) and (6), we have
and so
θ ( d ( x 0 , x 1 ) ) 2 + F ( d ( x 1 , x 2 ) ) F ( d ( x 0 , x 1 ) ) .
(7)
We may also assume that T x 1 T x 2 (otherwise, x 2 T x 2 = T x 1 ). Proceeding this manner, we can define a sequence { x n } in X satisfying
x n + 1 T x n , θ ( t n ) 2 F ( t n ) F ( t n + 1 ) , for each  n N ,
(8)
where t n = d ( x n , x n + 1 ) . Since θ ( t n ) > 0 then from (8), we have F ( t n ) > F ( t n + 1 ) for each n N . Since F is strictly increasing, then we deduce that { t n } is a nonnegative strictly decreasing sequence and so is convergent to some r 0 , lim n t n = r . Now we show that r = 0 . On the contrary, assume that r > 0 . From (8), we get
1 2 i = 1 n θ ( t i ) F ( t 1 ) F ( t n + 1 ) for each  n N .
(9)
Since { t n } is strictly decreasing, then from ( δ 1 ) we get θ ( t n ) 0 . Thus, i = 1 θ ( t i ) = , and then from (9), we have lim n F ( t n ) = . Then by ( δ 3 ), t n 0 , a contradiction. Hence,
lim n t n = 0 .
(10)
From (8), (10) and ( δ 4 ), we have
i = 1 t i = i = 1 d ( x i , x i + 1 ) < .

Then, by the triangle inequality, { x n } is a Cauchy sequence. From the completeness of X, there exists x X such that lim n x n = x . Now, we prove that x is a fixed point of T. To prove the claim, we may assume that T x n T x for sufficiently large n N . On the contrary, assume that T x n i = T x for each i N . Since Tx is closed, x n i + 1 T x n i = T x and x n i + 1 x , then x T x , and we are finished.

From (5), we have (note that x n + 1 T x n and T x n T x for n N )
F ( d ( x n + 1 , T x ) ) θ ( d ( x n , x ) ) + F ( d ( x n + 1 , T x ) ) θ ( d ( x n , x ) ) + F ( H ( T x n , T x ) ) F ( d ( x n , x ) ) .
(11)
Since d ( x n , x ) 0 , then (11) together with ( δ 3 ) imply that
d ( x , T x ) = lim n d ( x n + 1 , T x ) = 0 ,

and so d ( x , T x ) = 0 . Hence, x T x (note that Tx is closed). □

Remark 2.5 By Example 2.1, Theorem 2.4 is an extension and improvement of Theorem 2.1 of Wardowski [22]. From Example 2.2, we infer that Theorem 2.4 is a generalization of the above mentioned Theorem 1.2 of Mizoguchi and Takahashi.

Now, we illustrate our main result by the following example.

Example 2.6 Consider the complete metric space ( X = { 0 , 1 , 2 , 3 , } , d ) , where d is defined as
d ( x , y ) = { 0 , x = y , x + y , x y .
Let T : X C B ( X ) be defined as
T x = { { 0 , 1 , 2 , 3 , } , x = 0 , { x 1 , x , x + 1 , } , x > 0 .
Let f : X X be given by
f x = { 0 , x = 0 , x 1 , x > 0 .
Now, we show that T satisfies (5), where θ ( t ) = 1 for each t ( 0 , ) and F ( x ) = ln x + x for each x ( 0 , ) . To show the claim, notice first that H ( T x 1 , T x 2 ) = d ( f x 1 , f x 2 ) for each x 1 , x 2 X . Now let x 1 , x 2 X with f x 1 f x 2 . Since d ( f x 1 , f x 2 ) d ( x 1 , x 2 ) 1 , then we have
d ( f x 1 , f x 2 ) d ( x 1 , x 2 ) e d ( f x 1 , f x 2 ) d ( x 1 , x 2 ) e 1 , for each  x 1 , x 2 X  with  f x 1 f x 2 .
Thus, from the above, we have
1 [ ln d ( x 1 , x 2 ) + d ( x 1 , x 2 ) ] [ ln d ( f x 1 , f x 2 ) + d ( f x 1 , f x 2 ) ] = F ( d ( x 1 , x 2 ) ) F ( d ( f x 1 , f x 2 ) ) .
Therefore, (note that H ( T x 1 , T x 2 ) = d ( f x 1 , f x 2 ) )
1 F ( d ( x 1 , x 2 ) ) F ( H ( T x 1 , T x 2 ) ) .

Then, by Theorem 2.4, T has a fixed point.

Now, we show that T does not satisfy the condition of Nadler’s theorem. On the contrary, assume that there exists a function k [ 0 , 1 ) such that
H ( T x 1 , T x 2 ) k d ( x 1 , x 2 )
for all x 1 , x 2 X . Then
d ( f x 1 , f x 2 ) k d ( x 1 , x 2 ) .
Then, for each x 1 > 1 and x 2 = x 1 + 1 , we have
2 x 1 1 k ( 2 x 1 + 1 ) , for each  x 1 > 1 .
Hence,
1 = lim x 1 2 x 1 1 2 x 1 + 1 k ,

a contradiction.

Example 2.7 For each t ( 0 , ) , let F 4 ( t ) = 1 t and let
θ 4 ( t ) = { ln t t , 0 < t < 1 , 1 , 1 t .

Then it is easy to see that ( F 4 , θ 4 ) Δ , but F 4 does not satisfy the condition ( F 3 ) of the definition of F-contraction in [22].

Now, by using the technique in [23], we present a new coupled fixed point result. For more details on coupled fixed point theory, see [2325] and the references therein.

Corollary 2.8 Let ( M , ρ ) be a complete metric space and let ( F , θ 2 ) Δ . Let f : M × M M be a mapping satisfying
(12)

for each x , y , u , v M . Then f has a coupled fixed point ( x 0 , y 0 ) , that is, f ( x 0 , y 0 ) = x 0 and f ( y 0 , x 0 ) = y 0 .

Proof Let X = M × M and let d be the metric on M which is defined by
d ( ( x , y ) , ( u , v ) ) = ρ ( x , u ) + ρ ( y , v ) .
Then it is straightforward to show that ( X , d ) is a complete metric space. Let T : X X be defined by T ( x , y ) = ( f ( x , y ) , f ( y , x ) ) . From (12), we get
θ ( d ( ( x , y ) , ( u , v ) ) ) + F ( d ( T ( x , y ) , T ( u , v ) ) ) F ( d ( ( x , y ) , ( u , v ) ) )

for each ( x , y ) , ( u , v ) X . Then from Theorem 2.4 we deduce that T has a fixed point u 0 = ( x 0 , y 0 ) . Then ( x 0 , y 0 ) is a coupled fixed point of f. □

Declarations

Acknowledgements

The author is grateful to the referees for their helpful comments leading to improvement of the presentation of the work. This work was supported by the University of Shahrekord and by the Center of Excellence for Mathematics, University of Shahrekord, Iran. This research was also in part supported by a grant from IPM (No. 91470412). This research is partially carried out in the IPM-Isfahan Branch.

Authors’ Affiliations

(1)
Department of Pure Mathematics, University of Shahrekord
(2)
School of Mathematics, Institute for Research in Fundamental Sciences (IPM)

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© Amini-Harandi; licensee Springer 2012

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