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A new generalization of the Banach contraction principle

  • Mohamed Jleli
  • Bessem SametEmail author
Open Access
Research
Part of the following topical collections:
  1. International Conference Anatolian Communications in Nonlinear Analysis

Abstract

We present a new generalization of the Banach contraction principle in the setting of Branciari metric spaces.

Keywords

Banach contraction generalized metric fixed point 

1 Introduction

The fixed-point theorem, generally known as the Banach contraction principle, appeared in explicit form in Banach’s thesis in 1922 [1], where it was used to establish the existence of a solution to an integral equation. Since then, because of its simplicity and usefulness, it has become a very popular tool in solving existence problems in many branches of mathematical analysis. This principle states that, if ( X , d ) Open image in new window is a complete metric space and T : X X Open image in new window is a contraction map (i.e., d ( T x , T y ) λ d ( x , y ) Open image in new window for all x , y X Open image in new window, where λ ( 0 , 1 ) Open image in new window is a constant), then T has a unique fixed point.

The Banach contraction principle has been generalized in many ways over the years. In some generalizations, the contractive nature of the map is weakened; see [2, 3, 4, 5, 6, 7, 8, 9] and others. In other generalizations, the topology is weakened; see [10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23] and others. In [24], Nadler extended the Banach fixed-point theorem from single-valued maps to set-valued contractive maps. Other fixed point results for set-valued maps can be found in [25, 26, 27, 28, 29, 30] and references therein.

In 2000, Branciari [11] introduced the concept of generalized metric spaces, where the triangle inequality is replaced by the inequality d ( x , y ) d ( x , u ) + d ( u , v ) + d ( v , y ) Open image in new window for all pairwise distinct points x , y , u , v X Open image in new window. Various fixed point results were established on such spaces; see [10, 13, 17, 18, 19, 20, 22] and references therein.

In this paper, we introduce a new type of contractive maps and we establish a new fixed-point theorem for such maps on the setting of generalized metric spaces.

2 Main results

We denote by Θ the set of functions θ : ( 0 , ) ( 1 , ) Open image in new window satisfying the following conditions:

( Θ 1 Open image in new window) θ is non-decreasing;

( Θ 2 Open image in new window) for each sequence { t n } ( 0 , ) Open image in new window, lim n θ ( t n ) = 1 Open image in new window if and only if lim n t n = 0 + Open image in new window;

( Θ 3 Open image in new window) there exist r ( 0 , 1 ) Open image in new window and ( 0 , ] Open image in new window such that lim t 0 + θ ( t ) 1 t r = Open image in new window.

Before we prove the main results, we recall the following definitions introduced in [11].

Definition 2.1 Let X be a non-empty set and d : X × X [ 0 , ) Open image in new window be a mapping such that for all x , y X Open image in new window and for all distinct points u , v X Open image in new window, each of them different from x and y, one has
  1. (i)

    d ( x , y ) = 0 x = y Open image in new window;

     
  2. (ii)

    d ( x , y ) = d ( y , x ) Open image in new window;

     
  3. (iii)

    d ( x , y ) d ( x , u ) + d ( u , v ) + d ( v , y ) Open image in new window.

     

Then ( X , d ) Open image in new window is called a generalized metric space (or for short g.m.s.).

Definition 2.2 Let ( X , d ) Open image in new window be a g.m.s., { x n } Open image in new window be a sequence in X and x X Open image in new window. We say that { x n } Open image in new window is convergent to x if and only if d ( x n , x ) 0 Open image in new window as n Open image in new window. We denote this by x n x Open image in new window.

Definition 2.3 Let ( X , d ) Open image in new window be a g.m.s. and { x n } Open image in new window be a sequence in X. We say that { x n } Open image in new window is Cauchy if and only if d ( x n , x m ) 0 Open image in new window as n , m Open image in new window.

Definition 2.4 Let ( X , d ) Open image in new window be a g.m.s. We say that ( X , d ) Open image in new window is complete if and only if every Cauchy sequence in X converges to some element in X.

The following result was established in [31] (Lemma 1.10).

Lemma 2.1 Let ( X , d ) Open image in new windowbe a g.m.s., { x n } Open image in new windowbe a Cauchy sequence in ( X , d ) Open image in new window, and x , y X Open image in new window. Suppose that there exists a positive integer N such that
  1. (i)

    x n x m Open image in new window, for all n , m > N Open image in new window;

     
  2. (ii)

    x n Open image in new window and x are distinct points in X, for all n > N Open image in new window;

     
  3. (iii)

    x n Open image in new window and y are distinct points in X, for all n > N Open image in new window;

     
  4. (iv)

    lim n d ( x n , x ) = lim n d ( x n , y ) Open image in new window.

     

Then we have x = y Open image in new window.

We observe easily that if one of the conditions (ii) or (iii) is not satisfied, then the result of Lemma 2.1 is still valid.

Now, we are ready to state and prove our main result.

Theorem 2.1 Let ( X , d ) Open image in new windowbe a complete g.m.s. and T : X X Open image in new windowbe a given map. Suppose that there exist θ Θ Open image in new windowand k ( 0 , 1 ) Open image in new windowsuch that
x , y X , d ( T x , T y ) 0 θ ( d ( T x , T y ) ) [ θ ( d ( x , y ) ) ] k . Open image in new window
(1)

Then T has a unique fixed point.

Proof Let x X Open image in new window be an arbitrary point in X. If for some p N Open image in new window, we have T p x = T p + 1 x Open image in new window, then T p x Open image in new window will be a fixed point of T. So, without restriction of the generality, we can suppose that d ( T n x , T n + 1 x ) > 0 Open image in new window for all n N Open image in new window. Now, from (1), for all n N Open image in new window, we have
θ ( d ( T n x , T n + 1 x ) ) [ θ ( d ( T n 1 x , T n x ) ) ] k [ θ ( d ( T n 2 x , T n 1 x ) ) ] k 2 [ θ ( d ( x , T x ) ) ] k n . Open image in new window
Thus, we have
1 θ ( d ( T n x , T n + 1 x ) ) [ θ ( d ( x , T x ) ) ] k n , for all  n N . Open image in new window
(2)
Letting n Open image in new window in (2), we obtain
θ ( d ( T n x , T n + 1 x ) ) 1 as  n , Open image in new window
which implies from ( Θ 2 Open image in new window) that
lim n d ( T n x , T n + 1 x ) = 0 . Open image in new window
(3)
From condition ( Θ 3 Open image in new window), there exist r ( 0 , 1 ) Open image in new window and ( 0 , ] Open image in new window such that
lim n θ ( d ( T n x , T n + 1 x ) ) 1 [ d ( T n x , T n + 1 x ) ] r = . Open image in new window
Suppose that < Open image in new window. In this case, let B = / 2 > 0 Open image in new window. From the definition of the limit, there exists n 0 N Open image in new window such that
| θ ( d ( T n x , T n + 1 x ) ) 1 [ d ( T n x , T n + 1 x ) ] r | B , for all  n n 0 . Open image in new window
This implies that
θ ( d ( T n x , T n + 1 x ) ) 1 [ d ( T n x , T n + 1 x ) ] r B = B , for all  n n 0 . Open image in new window
Then
n [ d ( T n x , T n + 1 x ) ] r A n [ θ ( d ( T n x , T n + 1 x ) ) 1 ] , for all  n n 0 , Open image in new window

where A = 1 / B Open image in new window.

Suppose now that = Open image in new window. Let B > 0 Open image in new window be an arbitrary positive number. From the definition of the limit, there exists n 0 N Open image in new window such that
θ ( d ( T n x , T n + 1 x ) ) 1 [ d ( T n x , T n + 1 x ) ] r B , for all  n n 0 . Open image in new window
This implies that
n [ d ( T n x , T n + 1 x ) ] r A n [ θ ( d ( T n x , T n + 1 x ) ) 1 ] , for all  n n 0 , Open image in new window

where A = 1 / B Open image in new window.

Thus, in all cases, there exist A > 0 Open image in new window and n 0 N Open image in new window such that
n [ d ( T n x , T n + 1 x ) ] r A n [ θ ( d ( T n x , T n + 1 x ) ) 1 ] , for all  n n 0 . Open image in new window
Using (2), we obtain
n [ d ( T n x , T n + 1 x ) ] r A n ( [ θ ( d ( x , T x ) ) ] k n 1 ) , for all  n n 0 . Open image in new window
Letting n Open image in new window in the above inequality, we obtain
lim n n [ d ( T n x , T n + 1 x ) ] r = 0 . Open image in new window
Thus, there exists n 1 N Open image in new window such that
d ( T n x , T n + 1 x ) 1 n 1 / r , for all  n n 1 . Open image in new window
(4)
Now, we shall prove that T has a periodic point. Suppose that it is not the case, then T n x T m x Open image in new window for every n , m N Open image in new window such that n m Open image in new window. Using (1), we obtain
θ ( d ( T n x , T n + 2 x ) ) [ θ ( d ( T n 1 x , T n + 1 x ) ) ] k [ θ ( d ( T n 2 x , T n x ) ) ] k 2 [ θ ( d ( x , T 2 x ) ) ] k n . Open image in new window
Letting n Open image in new window in the above inequality and using ( Θ 2 Open image in new window), we obtain
lim n d ( T n x , T n + 2 x ) = 0 . Open image in new window
(5)
Similarly, from condition ( Θ 3 Open image in new window), there exists n 2 N Open image in new window such that
d ( T n x , T n + 2 x ) 1 n 1 / r , for all  n n 2 . Open image in new window
(6)

Let N = max { n 0 , n 1 } Open image in new window. We consider two cases.

Case 1. If m > 2 Open image in new window is odd, then writing m = 2 L + 1 Open image in new window, L 1 Open image in new window, using (4), for all n N Open image in new window, we obtain
d ( T n x , T n + m x ) d ( T n x , T n + 1 x ) + d ( T n + 1 x , T n + 2 x ) + + d ( T n + 2 L x , T n + 2 L + 1 x ) 1 n 1 / r + 1 ( n + 1 ) 1 / r + + 1 ( n + 2 L ) 1 / r i = n 1 i 1 / r . Open image in new window
Case 2. If m > 2 Open image in new window is even, then writing m = 2 L Open image in new window, L 2 Open image in new window, using (4) and (6), for all n N Open image in new window, we obtain
d ( T n x , T n + m x ) d ( T n x , T n + 2 x ) + d ( T n + 2 x , T n + 3 x ) + + d ( T n + 2 L 1 x , T n + 2 L x ) 1 n 1 / r + 1 ( n + 2 ) 1 / r + + 1 ( n + 2 L 1 ) 1 / r i = n 1 i 1 / r . Open image in new window
Thus, combining all the cases we have
d ( T n x , T n + m x ) i = n 1 i 1 / r , for all  n N , m N . Open image in new window
From the convergence of the series i 1 i 1 / r Open image in new window (since 1 / r > 1 Open image in new window), we deduce that { T n x } Open image in new window is a Cauchy sequence. Since ( X , d ) Open image in new window is complete, there is z X Open image in new window such that T n x z Open image in new window. On the other hand, observe that T is continuous, indeed, if T x T y Open image in new window, then we have from (1)
ln [ θ ( d ( T x , T y ) ) ] k ln [ θ ( d ( x , y ) ) ] ln [ θ ( d ( x , y ) ) ] , Open image in new window
which implies from ( Θ 1 Open image in new window) that
d ( T x , T y ) d ( x , y ) , for all  x , y X . Open image in new window
From this observation, for all n N Open image in new window, we have
d ( T n + 1 x , T z ) d ( T n x , z ) . Open image in new window
Letting n Open image in new window in the above inequality, we get T n + 1 x T z Open image in new window. From Lemma 2.1, we obtain z = T z Open image in new window, which is a contradiction with the assumption: T does not have a periodic point. Thus T has a periodic point, say z, of period q. Suppose that the set of fixed points of T is empty. Then we have
q > 1 and d ( z , T z ) > 0 . Open image in new window
Using (1), we obtain
θ ( d ( z , T z ) ) = θ ( d ( T n z , T n + 1 z ) ) [ θ ( d ( z , T z ) ) ] k n < θ ( d ( z , T z ) ) , Open image in new window
which is a contradiction. Thus, the set of fixed points of T is non-empty, that is, T has at least one fixed point. Now, suppose that z , u X Open image in new window are two fixed points of T such that d ( z , u ) = d ( T z , T u ) > 0 Open image in new window. Using (1), we obtain
θ ( d ( z , u ) ) = θ ( d ( T z , T u ) ) [ θ ( d ( z , u ) ) ] k < θ ( d ( z , u ) ) , Open image in new window

which is a contradiction. Then we have one and only one fixed point. □

Since a metric space is a g.m.s., from Theorem 2.1, we deduce immediately the following result.

Corollary 2.1 Let ( X , d ) Open image in new windowbe a complete metric space and T : X X Open image in new windowbe a given map. Suppose that there exist θ Θ Open image in new windowand k ( 0 , 1 ) Open image in new windowsuch that
x , y X , d ( T x , T y ) 0 θ ( d ( T x , T y ) ) [ θ ( d ( x , y ) ) ] k . Open image in new window

Then T has a unique fixed point.

Observe that the Banach contraction principle follows immediately from Corollary 2.1. Indeed, if T is a Banach contraction, i.e., there exists λ ( 0 , 1 ) Open image in new window such that
d ( T x , T y ) λ d ( x , y ) , for all  x , y X , Open image in new window
then we have
e d ( T x , T y ) [ e d ( x , y ) ] k , for all  x , y X . Open image in new window

Clearly the function θ : ( 0 , ) ( 1 , ) Open image in new window defined by θ ( t ) : = e t Open image in new window belongs to Θ. So, the existence and uniqueness of the fixed point follows from Corollary 2.1. In the following example (inspired by [9]), we show that Corollary 2.1 is a real generalization of the Banach contraction principle.

Example Let X be the set defined by
X : = { τ n : n N } , Open image in new window
where
τ n : = n ( n + 1 ) 2 , for all  n N . Open image in new window
We endow X with the metric d given by d ( x , y ) : = | x y | Open image in new window for all x , y X Open image in new window. It is not difficult to show that ( X , d ) Open image in new window is a complete metric space. Let T : X X Open image in new window be the map defined by
T τ 1 = τ 1 , T τ n = τ n 1 , for all  n 2 . Open image in new window
Clearly, the Banach contraction is not satisfied. In fact, we can check easily that
lim n d ( T τ n , T τ 1 ) d ( τ n , τ 1 ) = 1 . Open image in new window
Now, consider the function θ : ( 0 , ) ( 1 , ) Open image in new window defined by
θ ( t ) : = e t e t . Open image in new window
It is not difficult to show that θ Θ Open image in new window. We shall prove that T satisfies the condition (1), that is,
d ( T τ n , T τ m ) 0 e d ( T τ n , T τ m ) e d ( T τ n , T τ m ) e k d ( τ n , τ m ) e d ( τ n , τ m ) , Open image in new window
for some k ( 0 , 1 ) Open image in new window. The above condition is equivalent to
d ( T τ n , T τ m ) 0 d ( T τ n , T τ m ) e d ( T τ n , T τ m ) k 2 d ( τ n , τ m ) e d ( τ n , τ m ) . Open image in new window
So, we have to check that
d ( T τ n , T τ m ) 0 d ( T τ n , T τ m ) e d ( T τ n , T τ m ) d ( τ n , τ m ) d ( τ n , τ m ) k 2 , Open image in new window
(7)

for some k ( 0 , 1 ) Open image in new window. We consider two cases.

Case 1. n = 1 Open image in new window and m > 2 Open image in new window. In this case, we have
d ( T τ 1 , T τ m ) e d ( T τ 1 , T τ m ) d ( τ 1 , τ m ) d ( τ 1 , τ m ) = m 2 m 2 m 2 + m 2 e m e 1 . Open image in new window
Case 2. m > n > 1 Open image in new window. In this case, we have
d ( T τ m , T τ n ) e d ( T τ m , T τ n ) d ( τ m , τ n ) d ( τ m , τ n ) = m + n 1 m + n + 1 e n m e 1 . Open image in new window

Thus, the inequality (7) is satisfied with k = e 1 / 2 Open image in new window. Theorem 2.1 (or Corollary 2.1) implies that T has a unique fixed point. In this example τ 1 Open image in new window is the unique fixed point of T.

Note that Θ contains a large class of functions. For example, for
θ ( t ) : = 2 2 π arctan ( 1 t α ) , 0 < α < 1 , t > 0 , Open image in new window

we obtain from Theorem 2.1 the following result.

Corollary 2.2 Let ( X , d ) Open image in new windowbe a complete g.m.s. and T : X X Open image in new windowbe a given map. Suppose that there exist α , k ( 0 , 1 ) Open image in new windowsuch that
2 2 π arctan ( 1 [ d ( T x , T y ) ] α ) [ 2 2 π arctan ( 1 [ d ( x , y ) ] α ) ] k , for all  x , y X , T x T y . Open image in new window

Then T has a unique fixed point.

Notes

Acknowledgements

This project was supported by King Saud University, Deanship of Scientific Research, College of Science Research Center.

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© Jleli and Samet; licensee BioMed Central Ltd. 2013

This article is published under license to BioMed Central Ltd. This is an open access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Authors and Affiliations

  1. 1.Department of Mathematics, College of ScienceKing Saud UniversityRiyadhSaudi Arabia

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