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Fixed point theorems for cyclic self-maps involving weaker Meir-Keeler functions in complete metric spaces and applications

Fixed Point Theory and Applications20132013:224

https://doi.org/10.1186/1687-1812-2013-224

  • Received: 7 May 2013
  • Accepted: 5 August 2013
  • Published:

Abstract

We obtain fixed point theorems for cyclic self-maps on complete metric spaces involving Meir-Keeler and weaker Meir-Keeler functions, respectively. In this way, we extend several well-known fixed point theorems and, in particular, improve some very recent results on weaker Meir-Keeler functions. Fixed point results for well-posed property and for limit shadowing property are also deduced. Finally, an application to the study of existence and uniqueness of solutions for a class of nonlinear integral equations is presented.

MSC:47H10, 54H25, 54E50, 45G10.

Keywords

  • fixed point
  • cyclic map
  • weaker Meir-Keeler function
  • complete metric space
  • integral equation

1 Introduction

In their paper [1], Kirk, Srinavasan and Veeramani started the fixed point theory for cyclic self-maps on (complete) metric spaces. In particular, they obtained, among others, cyclic versions of the Banach contraction principle [2], of the Boyd and Wong fixed point theorem [3] and of the Caristi fixed point theorem [4]. From then, several authors have contributed to the study of fixed point theorems and best proximity points for cyclic contractions (see, e.g., [513]). Very recently, Chen [14] (see also [15]) introduced the notion of a weaker Meir-Keeler function and obtained some fixed point theorems for cyclic contractions involving weaker Meir-Keeler functions.

In this paper we obtain a fixed point theorem for cyclic self-maps on complete metric spaces involving Meir-Keeler functions and deduce a variant of it for weaker Meir-Keeler functions. In this way, we extend in several directions and improve, among others, the main fixed point theorem of Chen’s paper [[14], Theorem 3]. Some consequences are given after the main results. Fixed point results for well-posedness property and for limit shadowing property in complete metric spaces are also given. Finally, an application to the study of existence and uniqueness of solution for a class of nonlinear integral equations is presented.

We recall that a self-map f of a (non-empty) set X is called a cyclic map if there exists m N such that X = i = 1 m A i , with A i non-empty and f ( A i ) A i + 1 , i = 1 , , m , where A m + 1 = A 1 .

In this case, we say that X = i = 1 m A i is a cyclic representation of X with respect to f.

2 Fixed point results

In the sequel, the letters , R + and will denote the set of real numbers, the set of non-negative real numbers and the set of positive integer numbers, respectively.

Meir and Keeler proved in [16] that if f is a self-map of a complete metric space ( X , d ) satisfying the condition that for each ε > 0 there is δ > 0 such that, for any x , y X , with ε d ( x , y ) < ε + δ , we have d ( f x , f y ) < ε , then f has a unique fixed point z X and f n x z for all x X .

This important result suggests the notion of a Meir-Keeler function:

A function ϕ : R + R + is said to be a Meir-Keeler function if for each ε > 0 , there exists δ > 0 such that for t > 0 with ε t < ε + δ , we have ϕ ( t ) < ε .

Remark 1 It is obvious that if ϕ is a Meir-Keeler function, then ϕ ( t ) < t for all t > 0 .

In [14], Chen introduced the following interesting generalization of the notion of a Meir-Keeler function.

Definition 1 [[14], Definition 3]

A function ϕ : R + R + is called a weaker Meir-Keeler function if for each ε > 0 , there exists δ > 0 such that for t > 0 with ε t < ε + δ , there exists n 0 N such that ϕ n 0 ( t ) < ε .

Now let ϕ , φ : R + R + . According to Chen [[14], Section 2], consider the following conditions for ϕ and φ, respectively.

  • ( ϕ 1 ) ϕ ( t ) = 0 t = 0 ;

  • ( ϕ 2 ) for all t > 0 , the sequence { ϕ n ( t ) } n N is decreasing;

  • ( ϕ 3 ) for t n > 0 ,
    1. (a)

      if lim n t n = γ > 0 , then lim n ϕ ( t n ) < γ , and

       
    2. (b)

      if lim n t n = 0 , then lim n ϕ ( t n ) = 0 ;

       
  • ( φ 1 ) φ is non-decreasing and continuous with φ ( t ) = 0 t = 0 ;

  • ( φ 2 ) φ is subadditive, that is, for every t 1 , t 2 R + , φ ( t 1 + t 2 ) φ ( t 1 ) + φ ( t 2 ) ;

  • ( φ 3 ) for t n > 0 , lim n t n = 0 if and only if lim n φ ( t n ) = 0 .

Definition 2 [[14], Definition 4]

Let ( X , d ) be a metric space. A self-map f of X is called a cyclic weaker ( ϕ φ ) -contraction if there exist m N , for which X = i = 1 m A i (each A i a non-empty closed set), and two functions ϕ , φ : R + R + satisfying conditions ( ϕ i ), i = 1 , 2 , 3 , and ( φ i ), i = 1 , 2 , 3 , respectively, with ϕ a weaker Meir-Keeler function such that
  1. (1)

    X = i = 1 m A i is a cyclic representation of X with respect to f;

     
  2. (2)
    for any x A i , y A i + 1 , i = 1 , 2 , , m ,
    φ ( d ( f x , f y ) ) ϕ ( φ ( d ( x , y ) ) ) ,
     

where A m + 1 = A 1 .

By using the above concept, Chen established the following fixed point theorem.

Theorem 1 [[14], Theorem 3]

Let ( X , d ) be a complete metric space. Then every cyclic weaker ( ϕ φ ) -contraction f of X has a unique fixed point z. Moreover, z i = 1 m A i , where X = i = 1 m A i is the cyclic representation of X with respect to f of Definition  2.

We shall establish fixed point theorems which improve in several directions the preceding theorem. To this end, we start by obtaining a fixed point theorem for cyclic contractions involving Meir-Keeler functions.

Theorem 2 Let f be a self-map of a complete metric space ( X , d ) , and let X = i = 1 m A i be a cyclic representation of X with respect to f, with A i non-empty and closed, i = 1 , , m . If ϕ : R + R + is a Meir-Keeler function such that for any x A i , y A i + 1 , i = 1 , 2 , , m ,
d ( f x , f y ) ϕ ( d ( x , y ) ) ,

where A m + 1 = A 1 , then f has a unique fixed point z. Moreover, z i = 1 m A i .

Proof Let x 0 A m . For each n N { 0 } , put x n = f n x 0 . Note that x n m + i A i whenever n N { 0 } and i = 1 , 2 , , m .

If x n 0 = x n 0 + 1 for some n 0 , then x n is a fixed point of f. So, we assume that x n x n + 1 for all n N { 0 } . By Remark 1 and the contraction condition, it follows that { d ( x n , x n + 1 ) } n N is a strictly decreasing sequence in R + , so there exists r R + such that lim n d ( x n , x n + 1 ) = r . If r > 0 , there is n 0 N such that ϕ ( d ( x n , x n + 1 ) ) < r for all n n 0 by our assumption that ϕ is a Meir-Keeler function. Hence, d ( x n + 1 , x n + 2 ) < r for all n n 0 , a contradiction. Therefore lim n d ( x n , x n + 1 ) = 0 .

Next we prove that { x n } n N is a Cauchy sequence in ( X , d ) . Choose an arbitrary ε > 0 . Then, there is δ ( 0 , ε ) such that for t > 0 with ε t < ε + δ , we have ϕ ( t ) < ε . Let k 0 N be such that d ( x k , x k + 1 ) < δ / 2 , d ( x k , x k + m 1 ) < ε / 2 and d ( x k , x k + m + 1 ) < δ / 2 for all k k 0 .

Take any k > k 0 . Then k = n m + i for some n N and some i { 1 , 2 , , m } . By induction we shall show that d ( x n m + i , x ( n + j ) m + i + 1 ) < ε for all j N .

Indeed, for j = 1 , we have
d ( x n m + i , x n m + i + m + 1 ) = d ( x k , x k + m + 1 ) < δ 2 < ε .
Now, assume that d ( x n m + i , x ( n + j ) m + i + 1 ) < ε for some j N . Thus
d ( x n m + i 1 , x ( n + j + 1 ) m + i ) d ( x n m + i 1 , x n m + i ) + d ( x n m + i , x ( n + j ) m + i + 1 ) + d ( x ( n + j ) m + i + 1 , x ( n + j + 1 ) m + i ) < δ 2 + ε + δ 2 = δ + ε .
If ε d ( x n m + i 1 , x ( n + j + 1 ) m + i ) , then ϕ ( d ( x n m + i 1 , x ( n + j + 1 ) m + i ) ) < ε , and, by the contraction condition,
d ( x n m + i , x ( n + j + 1 ) m + i + 1 ) < ε .
If d ( x n m + i 1 , x ( n + j + 1 ) m + i ) < ε , we deduce
d ( x n m + i , x ( n + j + 1 ) m + i + 1 ) ϕ ( d ( x n m + i 1 , x ( n + j + 1 ) m + i ) ) < d ( x n m + i 1 , x ( n + j + 1 ) m + i ) < ε .

It immediately follows that { x n } n N is a Cauchy sequence in ( X , d ) . Hence, there exists z X such that x n z . Since each A i is closed, we deduce that z i = 1 m A i .

Moreover, z = f z . Indeed, let i 0 { 1 , , m } be such that f z A i 0 + 1 . Then
d ( z , f z ) d ( z , x n m + i 0 ) + d ( x n m + i 0 , f z ) d ( z , x n m + i 0 ) + ϕ ( d ( x n m + i 0 1 , z ) ) < d ( z , x n m + i 0 ) + d ( x n m + i 0 1 , z ) ,

for all n N . Since lim n d ( z , x n m + i 0 ) = lim n d ( z , x n m + i 0 1 ) = 0 , it follows that d ( z , f z ) = 0 , i.e., z = f z .

Finally, let u X with u = f u and u z . Since z i = 1 m A i , we have d ( f z , f u ) ϕ ( d ( z , u ) ) , so d ( z , u ) < d ( z , u ) , a contradiction. Hence u = z , and thus z is the unique fixed point of f. □

Next we analyze some relations between Chen’s conditions ( ϕ i ), i = 1 , 2 , 3 .

Lemma 1 If ϕ : R + R + satisfies ( ϕ 3 )(a), then ϕ is a Meir-Keeler function that satisfies conditions ( ϕ 2 ) and ( ϕ 3 )(b).

Proof Suppose that ϕ is not a Meir-Keeler function. Then there exists ε > 0 such that for each n N we can find a t n > 0 with ε t n < ε + 1 / n and ϕ ( t n ) ε . Then lim n t n = ε > 0 , but ϕ ( t n ) ε for all n, so condition ( ϕ 3 )(a) is not satisfied. We conclude that condition ( ϕ 3 )(a) implies that ϕ is a Meir-Keeler function. Hence, by Remark 1, ϕ ( t ) < t for all t > 0 , so the sequence { ϕ n ( t ) } n N is (strictly) decreasing for all t > 0 , and thus condition ( ϕ 2 ) is satisfied. Finally, if lim n t n = 0 , with t n > 0 , we deduce that lim n ϕ ( t n ) = 0 because ϕ ( t n ) < t n for all n, so condition ( ϕ 3 )(b) also holds. □

Proposition 1 Let φ : R + R + be a function satisfying conditions ( φ 1 ) and ( φ 2 ). If ( X , d ) is a metric space, then the function p : X × X R + , given by
p ( x , y ) = φ ( d ( x , y ) ) ,

is a metric on X. If, in addition, ( X , d ) is complete and φ satisfies condition ( φ 3 ), then the metric space ( X , p ) is complete.

Proof We first show that p is a metric on X. Let x , y , z X :

  • Suppose p ( x , y ) = 0 . Then φ ( d ( x , y ) ) = 0 , so d ( x , y ) = 0 by ( φ 1 ). Hence x = y .

  • Clearly, p ( x , y ) = p ( y , x ) .

  • Since d ( x , y ) d ( x , z ) + d ( z , y ) , and φ is non-decreasing and subadditive, we deduce that φ ( d ( x , y ) ) φ ( d ( x , z ) ) + φ ( d ( z , y ) ) , i.e., p ( x , y ) p ( x , z ) + p ( z , y ) .

Finally, suppose that ( X , d ) is complete with φ satisfying ( φ i ), i = 1 , 2 , 3 . Let { x n } n N be a Cauchy sequence in ( X , p ) . If { x n } n N is not a Cauchy sequence in ( X , d ) , there exist ε > 0 and sequences { n k } k N and { m k } k N in such that k < n k < m k < n k + 1 and d ( x n k , x m k ) ε for all k N . By ( φ 3 ), the sequence { p ( x n k , x m k ) } k N does not converge to zero, which contradicts the fact that { x n } n N is a Cauchy sequence in ( X , p ) . Consequently, { x n } n N is a Cauchy sequence in ( X , d ) , so it converges in ( X , d ) to some x X . From ( φ 3 ) we deduce that { x n } n N converges to x in ( X , p ) . Therefore ( X , p ) is a complete metric space. □

Remark 2 Note that the continuity of φ is not used in the preceding proposition.

Now we easily deduce the following improvement of Chen’s theorem.

Theorem 3 Let f be a self-map of a complete metric space ( X , d ) , and let X = i = 1 m A i be a cyclic representation of X with respect to f, with A i non-empty and closed, i = 1 , , m . If ϕ , φ : R + R + satisfy conditions ( ϕ 3 )(a) and ( φ i ), i = 1 , 2 , 3 , respectively, and for any x A i , y A i + 1 , i = 1 , 2 , , m , it follows
φ ( d ( f x , f y ) ) ϕ ( φ ( d ( x , y ) ) ) ,

where A m + 1 = A 1 , then f has a unique fixed point z. Moreover, z i = 1 m A i .

Proof Define p : X × X R + by p ( x , y ) = φ ( d ( x , y ) ) for all x , y X . By Proposition 1, ( X , p ) is a complete metric space. Moreover, from the condition
φ ( d ( f x , f y ) ) ϕ ( φ ( d ( x , y ) ) ) ,
for all x A i , y A i + 1 , i = 1 , , m , it follows that
p ( f x , f y ) = φ ( d ( f x , f y ) ) ϕ ( φ ( d ( x , y ) ) ) = ϕ ( p ( x , y ) )

for all x A i , y A i + 1 , i = 1 , , m .

Finally, since by Lemma 1 ϕ is a Meir-Keeler function, we can apply Theorem 2, so there exists z i = 1 m A i , which is the unique fixed point of f. □

Note that the continuity of φ can be omitted in Theorem 3. Moreover, the condition that ϕ is a weaker Meir-Keeler function turns out to be irrelevant by virtue of Lemma 1. This fact suggests the question of obtaining a fixed point theorem for cyclic contractions involving explicitly weaker Meir-Keeler functions. In particular, it is natural to wonder if Theorem 2 remains valid when we replace ‘Meir-Keeler function’ by ‘weaker Meir-Keeler function’. In the sequel we answer this question. First we give an easy example which shows that it has a negative answer in general, but the answer is positive whenever the weaker Meir-Keeler function is non-decreasing as Theorem 5 below shows.

Example 1 Let X = { 0 , 1 } and let d be the discrete metric on X, i.e., d ( 0 , 0 ) = d ( 1 , 1 ) = 0 and d ( x , y ) = 1 otherwise. Of course ( X , d ) is a complete metric space. Define f : X X by f 0 = 1 and f 1 = 0 , and consider the function ϕ : R + R + defined by ϕ ( t ) = t / 2 for all t [ 0 , 1 ) , ϕ ( 1 ) = 2 and ϕ ( t ) = 1 / 2 for all t > 1 . Clearly, ϕ is a weaker Meir-Keeler function (note, in particular, that ϕ 2 ( 1 ) = 1 / 2 < 1 ), but it is not a Meir-Keeler function because ϕ ( 1 ) > 1 . Finally, since d ( f 0 , f 1 ) = 1 and ϕ ( d ( 0 , 1 ) ) = 2 , we deduce that d ( f x , f y ) ϕ ( d ( x , y ) ) for all x , y X . However, f has no fixed point.

The function ϕ of the preceding example is not non-decreasing. This fact is not casual as Theorem 5 below shows.

Lemma 2 Let ϕ : R + R + be a non-decreasing weaker Meir-Keeler function. Then the following hold:
  1. (i)

    ϕ ( t ) < t for all t > 0 ;

     
  2. (ii)

    lim n ϕ n ( t ) = 0 for all t > 0 .

     

Proof (i) Suppose that there exists t 0 > 0 such that t 0 ϕ ( t 0 ) . Since ϕ is non-decreasing, we deduce that { ϕ n ( t 0 ) } n N { 0 } is a non-decreasing sequence in R + , so, in particular, t 0 ϕ n ( t 0 ) for all n N . Finally, since ϕ is a weaker Meir-Keeler function, there exists n 0 N such that ϕ n 0 ( t 0 ) < t 0 , which yields a contradiction.

(ii) Fix t > 0 . By (i) the sequence { ϕ n ( t ) } n N is (strictly) decreasing, so there exists r 0 such that r = lim n ϕ n ( t ) . If r > 0 , there is δ > 0 such that for s > 0 with r s < r + δ , there exists n s N with ϕ n s ( s ) < r . Let n r N be such that r < ϕ n ( t ) < r + δ for all n n r . Putting s = ϕ n r ( t ) , we deduce that ϕ n s ( s ) < r , i.e., ϕ n s + n r ( t ) < r , a contradiction. We conclude that lim n ϕ n ( t ) = 0 . □

Remark 3 Observe that, as a partial converse of the above lemma, if ϕ : R + R + satisfies lim n ϕ n ( t ) = 0 for all t > 0 , then ϕ is a weaker Meir-Keeler function. Indeed, otherwise, there exist ε > 0 and a sequence { t n } n N with t n ε for all n N , lim n t n = ε but ϕ k ( t n ) ε for all k , n N , a contradiction.

We also will use the following cyclic extension of the celebrated Matkowski fixed point theorem [[17], Theorem 1.2], where for a self-map f of a metric space ( X , d ) , we define
M d ( x , y ) = max { d ( x , y ) , d ( x , f x ) , d ( y , f y ) , 1 2 [ d ( x , f y ) + d ( f x , y ) ] }

for all x , y X .

Theorem 4 (cf. [[18], Corollary 2.14])

Let f be a self-map of a complete metric space ( X , d ) , and let X = i = 1 m A i be a cyclic representation of X with respect to f, with A i non-empty and closed, i = 1 , , m . If ϕ : R + R + is a non-decreasing function such that lim n ϕ n ( t ) = 0 for all t > 0 , and for any x A i , y A i + 1 , i = 1 , 2 , , m ,
d ( f x , f y ) ϕ ( M d ( x , y ) ) ,

where A m + 1 = A 1 , then f has a unique fixed point z. Moreover, z i = 1 m A i .

Then from Lemma 2 and Theorem 4 we immediately deduce the following theorem.

Theorem 5 Let f be a self-map of a complete metric space ( X , d ) , and let X = i = 1 m A i be a cyclic representation of X with respect to f, with A i non-empty and closed, i = 1 , , m . If ϕ : R + R + is a non-decreasing weaker Meir-Keeler function such that for any x A i , y A i + 1 , i = 1 , 2 , , m ,
d ( f x , f y ) ϕ ( M d ( x , y ) ) ,

where A m + 1 = A 1 , then f has a unique fixed point z. Moreover, z i = 1 m A i .

Corollary Let f be a self-map of a complete metric space ( X , d ) , and let X = i = 1 m A i be a cyclic representation of X with respect to f, with A i non-empty and closed, i = 1 , , m . If ϕ : R + R + is a non-decreasing weaker Meir-Keeler function such that for any x A i , y A i + 1 , i = 1 , 2 , , m ,
d ( f x , f y ) ϕ ( d ( x , y ) ) ,

where A m + 1 = A 1 , then f has a unique fixed point z. Moreover, z i = 1 m A i .

Proof Since ϕ is non-decreasing, we deduce that for each x , y X , ϕ ( d ( x , y ) ) ϕ ( M d ( x , y ) ) , so d ( f x , f y ) ϕ ( M d ( x , y ) ) . Hence, by Theorem 5, f has a unique fixed point z and z i = 1 m A i . □

Theorem 5 can be generalized according to the style of Chen’s theorem as follows.

Theorem 6 Let f be a self-map of a complete metric space ( X , d ) , and let X = i = 1 m A i be a cyclic representation of X with respect to f, with A i non-empty and closed, i = 1 , , m . If ϕ : R + R + is a non-decreasing weaker Meir-Keeler function, φ : R + R + is a function satisfying conditions ( φ i ), i = 1 , 2 , 3 , and for any x A i , y A i + 1 , i = 1 , 2 , , m , it follows
φ ( d ( f x , f y ) ) ϕ ( φ ( M d ( x , y ) ) ) ,

where A m + 1 = A 1 , then f has a unique fixed point z. Moreover, z i = 1 m A i .

Proof Construct the complete metric space ( X , p ) of Proposition 1, and observe that from the well-known fact that for a i R + , i = 1 , , k , one has ϕ ( max i a i ) = max i ϕ ( a i ) , one has
M p ( x , y ) = φ ( M d ( x , y ) )
for all x , y X . Therefore, for any x A i , y A i + 1 , i = 1 , 2 , , m , we deduce that
p ( f x , f y ) ϕ ( M p ( x , y ) ) .

Theorem 5 concludes the proof. □

We finish this section with two examples illustrating Theorem 5 and its corollary.

Example 2 Let A = { n N : n  is even } { 0 } , B = { n N : n  is odd } { 0 } , X = A B = N , and let d be the complete metric on X defined by d ( x , x ) = 0 for all x X and d ( x , y ) = x + y otherwise. Since d induces the discrete topology on X, we deduce that A and B are closed subsets of ( X , d ) .

Let f be the self-map of X defined by f 0 = 0 and f x = x 1 otherwise. It is clear that X = A B is a cyclic representation of X with respect to f.

Now we define the function ϕ : R + R + by ϕ ( 0 ) = 0 , and ϕ ( t ) = n 1 if t ( n 1 , n ] , n N . It is immediate to check that ϕ is a non-decreasing weaker Meir-Keeler function which is not a Meir-Keeler function.

Furthermore, we have:

  • For x = 0 and y = 1 , d ( f x , f y ) = d ( 0 , 0 ) = 0 .

  • For x = 0 and y = n N { 1 } ,
    d ( f x , f y ) = d ( 0 , n 1 ) = n 1 = ϕ ( n ) = ϕ ( d ( x , y ) ) .
  • For x = n A { 0 } and y = m B { 0 } ,
    d ( f x , f y ) = d ( n 1 , m 1 ) = n + m 2 < n + m 1 = ϕ ( n + m ) = ϕ ( d ( x , y ) ) .

Consequently, the conditions of the corollary of Theorem 5 are verified; in fact, z = 0 A B is the unique fixed point of f.

Example 3 Let A = [ 0 , 1 / 2 ] { 1 } , B = [ 1 , 2 ] , X = A B and let d be the restriction to X of the Euclidean metric on . Obviously, ( X , d ) is a complete metric space (in fact, it is compact), with A and B closed subsets of ( X , d ) .

Let f be the self-map of X defined by f x = 2 x if x A , and f x = 1 if x B . It is clear that X = A B is a cyclic representation of X with respect to f.

Now we define the function ϕ : R + R + by ϕ ( t ) = t / 2 if t [ 0 , 1 ] , and ϕ ( t ) = 1 if t > 1 . (Notice that ϕ is a non-decreasing weaker Meir-Keeler function which is not a Meir-Keeler function.)

Furthermore, we have:

  • For x = 1 A and y B , d ( f x , f y ) = d ( 1 , 1 ) = 0 .

  • For x = 1 / 2 A and y B ,
    d ( f x , f y ) = d ( 3 / 2 , 1 ) = 1 / 2 = ϕ ( 1 ) = ϕ ( d ( x , f x ) ) .
  • For x A { 1 , 1 / 2 } and y B ,
    d ( f x , f y ) = d ( 2 x , 1 ) = 1 x 1 = ϕ ( 2 2 x ) = ϕ ( d ( x , f x ) ) .

Consequently, the conditions of Theorem 5 are verified; in fact, z = 1 A B is the unique fixed point of f.

Finally, observe that the corollary of Theorem 5 cannot be applied in this case because for x = 1 / 2 A and y = 1 B , we have
d ( f x , f y ) = 1 / 2 > ϕ ( 1 / 2 ) = ϕ ( d ( x , y ) ) .

3 Applications to well-posedness and limit shadowing property of a fixed point problem

The notion of well-posedness of a fixed point problem has evoked much interest to several mathematicians, for example, De Blasi and Myjak [19], Lahiri and Das [20], Popa [21, 22] and others.

Definition 3 [19]

Let f be a self-map of a metric space ( X , d ) . The fixed point problem of f is said to be well posed if:
  1. (i)

    f has a unique fixed point z X ;

     
  2. (ii)

    for any sequence { x n } n N in X such that lim n d ( f x n , x n ) = 0 , we have lim n d ( x n , z ) = 0 .

     

Definition 4 [22]

Let f be a self-map of a metric space ( X , d ) . The fixed point problem of f is said to have limit shadowing property in X if for any sequence { x n } n N in X satisfying lim n d ( f x n , x n ) = 0 , it follows that there exists z X such that lim n d ( f n z , x n ) = 0 .

Concerning the well-posedness and limit shadowing of the fixed point problem for a self-map of a complete metric space satisfying the conditions of Theorem 5, we have the following results.

Theorem 7 Let ( X , d ) be a complete metric space. If f is a self-map of X and ϕ : R + R + is a non-decreasing weaker Meir-Keeler function satisfying the conditions of Theorem  5, then the fixed point problem of f is well posed.

Proof Owing to Theorem 5, we know that f has a unique fixed point z X . Let { x n } be a sequence in X such that lim n d ( x n , f x n ) = 0 . Then
d ( x n , z ) d ( x n , f x n ) + d ( f x n , f z ) d ( x n , f x n ) + ϕ ( max { d ( x n , z ) , d ( x n , x n + 1 ) , d ( z , f z ) , d ( x n , f z ) + d ( z , x n + 1 ) 2 } ) .

Passing to the limit as n in the above inequality, it follows that lim n d ( x n , z ) = 0 . □

Theorem 8 Let ( X , d ) be a complete metric space. If f is a self-map of X and ϕ : R + R + is a non-decreasing weaker Meir-Keeler function satisfying the conditions of Theorem  5, then f has the limit shadowing property.

Proof From Theorem 5, we know that f has a unique fixed point z X . Let { x n } n N be a sequence in X such that lim n d ( x n , f x n ) = 0 . Then, as in the proof of the previous theorem,
d ( x n , z ) d ( x n , f x n ) + ϕ ( max { d ( x n , z ) , d ( x n , x n + 1 ) , d ( z , f z ) , d ( x n , f z ) + d ( z , x n + 1 ) 2 } ) .

Passing to the limit as n in the above inequality, it follows that lim n d ( x n , f n z ) = d ( x n , z ) = 0 . □

4 An application to integral equations

In this section we apply Theorem 5 to study the existence and uniqueness of solutions for a class of nonlinear integral equations.

We consider the nonlinear integral equation
u ( t ) = 0 T G ( t , s ) K ( s , u ( s ) ) d s for all  t [ 0 , T ] ,
(1)

where T > 0 , K : [ 0 , T ] × R + R + and G : [ 0 , T ] × [ 0 , T ] R + are continuous functions, and M : = max ( s , x ) [ 0 , T ] 2 K ( s , x ) .

We shall suppose that the following four conditions are satisfied:
  1. (I)

    0 T G ( t , s ) d s 1 for all t [ 0 , T ] .

     
  2. (II)
    K ( s , ) is a non-increasing function for any fixed s [ 0 , 1 ] , that is,
    x , y R + , x y K ( s , x ) K ( s , y ) .
     
  3. (III)
    There exists a Meir-Keeler function ψ : R + R + that is non-decreasing on [ 0 , 2 M ] and such that
    | K ( s , x ) K ( s , y ) | ψ ( | x y | )

    for all s , x [ 0 , T ] and y R + with | x y | 2 M .

     
  4. (IV)

    There exists a continuous function α : [ 0 , T ] [ 0 , T ] such that:

    For all t [ 0 , T ] , we have
    α ( t ) 0 T G ( t , s ) K ( s , T ) d s
    and
    T 0 T G ( t , s ) K ( s , α ( s ) ) d s .
     
Now denote by C ( [ 0 , T ] , R + ) the set of non-negative real continuous functions on [ 0 , T ] . We endow C ( [ 0 , T ] , R + ) with the supremum metric
d ( u , v ) = max t [ 0 , T ] | u ( t ) v ( t ) | , for all  u , v C ( [ 0 , T ] , R + ) .

It is well known that ( C ( [ 0 , T ] , R + ) , d ) is a complete metric space.

Consider the self-map f : C ( [ 0 , T ] , R + ) C ( [ 0 , T ] , R + ) defined by
f u ( t ) = 0 T G ( t , s ) K ( s , u ( s ) ) d s for all  t [ 0 , T ] .

Clearly, u is a solution of (1) if and only if u is a fixed point of f.

In order to prove the existence of a (unique) fixed point of f, we construct the closed subsets A 1 and A 2 of C ( [ 0 , T ] , R + ) as follows:
A 1 = { u C ( [ 0 , T ] , R + ) : u ( s ) T  for all  s [ 0 , T ] } ,
and
A 2 = { u C ( [ 0 , T ] , R + ) : u α } .
We shall prove that
f ( A 1 ) A 2 and f ( A 2 ) A 1 .
(2)
Let u A 1 , that is,
u ( s ) T for all  s [ 0 , T ] .
Since G ( t , s ) 0 for all t , s [ 0 , T ] , we deduce from (II) and (IV) that
0 T G ( t , s ) K ( s , u ( s ) ) d s 0 T G ( t , s ) K ( s , T ) d s α ( t )

for all t [ 0 , T ] . Then we have f u A 2 .

Similarly, let u A 2 , that is,
u ( s ) α ( s ) for all  s [ 0 , T ] .
Again, from (II) and (IV), we deduce that
0 T G ( t , s ) K ( s , u ( s ) ) d s 0 T G ( t , s ) K ( s , α ( s ) ) d s T

for all t [ 0 , T ] . Then we have f u A 1 . Thus, we have shown that (2) holds.

Hence, if X : = A 1 A 2 , we have that X is closed in C ( [ 0 , T ] , R + ) , so the metric space ( X , d ) is complete.

Moreover, X : = A 1 A 2 is a cyclic representation of the restriction of f with respect to X, which will be also denoted by f.

Now construct the function ϕ : R + R + given by
ϕ ( t ) = ψ ( t ) if  t [ 0 , 2 M ] ,
and
ϕ ( t ) = 2 M if  t > 2 M .

Since ψ is a Meir-Keeler function that is non-decreasing on [ 0 , 2 M ] , it immediately follows that ϕ is a non-decreasing weaker Meir-Keeler function. Note also that ϕ is not continuous at t = 2 M (in fact, it is not a Meir-Keeler function).

Finally we shall show that for each u A 1 and v A 2 , one has d ( f u , f v ) ϕ ( d ( u , v ) ) .

To this end, let u A 1 and v A 2 . Since u ( s ) [ 0 , T ] for each s [ 0 , T ] , we have that
f u ( t ) = 0 T G ( t , s ) K ( s , u ( s ) ) d s M 0 T G ( t , s ) d s M

for all t [ 0 , T ] .

Similarly, since v α and α ( s ) [ 0 , T ] for each s [ 0 , T ] , we deduce that
f v ( t ) 0 T G ( t , s ) K ( s , α ( s ) ) d s M
for all t [ 0 , T ] . Therefore
| f u ( t ) f ( v ( t ) | f u ( t ) + f v ( t ) 2 M
for all t [ 0 , T ] . So,
d ( f u , f v ) 2 M .
If d ( u , v ) > 2 M , we have ϕ ( d ( u , v ) ) = 2 M , so
d ( f u , f v ) ϕ ( d ( u , v ) ) .
If d ( u , v ) 2 M , then | u ( s ) v ( s ) | 2 M for all s [ 0 , T ] , so by (III), we deduce that for each t [ 0 , T ] ,
| f u ( t ) f ( v ( t ) | 0 T G ( t , s ) | K ( s , u ( s ) ) K ( s , v ( s ) ) | d s 0 T G ( t , s ) ψ ( | u ( s ) v ( s ) | ) d s ψ ( d ( u , v ) ) 0 T G ( t , s ) d s ψ ( d ( u , v ) ) = ϕ ( d ( u , v ) ) .

Consequently, by the corollary of Theorem 5, f has a unique fixed point u A 1 A 2 , that is, u C is the unique solution to (1) in A 1 A 2 .

Remark 4 The first author studied in [[9], Section 3] a variant of the problem discussed above for the case that ψ is the non-decreasing Meir-Keeler function given by ψ ( t ) = ( ln ( t 2 + 1 ) ) 1 / 2 for all t R + .

The next example illustrates the preceding development.

Example 4 Consider the integral equation
u ( t ) = 0 T G ( t , s ) K ( s , u ( s ) ) d s for all  t [ 0 , T ] ,
where T = 1 , G ( t , s ) = t for all t , s [ 0 , 1 ] , and
K ( s , x ) = cos s 1 + x

for all s [ 0 , 1 ] and x 0 .

Hence, M = max ( s , x ) [ 0 , 1 ] 2 K ( s , x ) = K ( 0 , 0 ) = 1 .

Furthermore, it is obvious that G satisfies condition (I), whereas K satisfies condition (II).

Now construct a Meir-Keeler function ψ : R + R + as
ψ ( t ) = t / ( 1 + t ) if  t [ 0 , 2 ] ,
and
ψ ( t ) = 0 if  t > 2 .

Note that ψ is non-decreasing on [ 0 , 2 ] and not continuous at t = 2 .

Moreover, for each s , x [ 0 , 1 ] and each y R + with | x y | 2 , we have
| K ( s , x ) K ( s , y ) | = cos s | 1 1 + x 1 1 + y | | x y | 1 + | x y | = ψ ( | x y | ) ,

so condition (III) is also satisfied.

Finally, define α : [ 0 , 1 ] [ 0 , 1 ] as α ( t ) = t / 3 for all t [ 0 , 1 ] . It is not hard to check that α verifies condition (IV), and consequently the integral equation has a unique solution u in A 1 A 2 , where A 1 = { u C ( [ 0 , 1 ] , R + ) : u ( s ) 1  for all  s [ 0 , 1 ] } and A 2 = { u C ( [ 0 , 1 ] , R + ) : u ( s ) s / 3  for all  s [ 0 , 1 ] } . In fact u A 1 A 2 , i.e., t / 3 u ( t ) 1 for all t [ 0 , 1 ] .

Note that, according to our constructions, for each pair u , v C ( [ 0 , 1 ] , R + ) with u 1 and v α , we have d ( f u , f v ) ϕ ( d ( u , v ) ) , where ϕ is the non-decreasing weaker Meir-Keeler function defined as ϕ ( t ) = t / ( t + 1 ) if t [ 0 , 2 ] and ϕ ( t ) = 2 if t > 2 .

In particular, we can deduce the following approximation to the value of u ( t ) for each t [ 0 , 1 ] :
| u ( t ) sin 1 2 t | = | u ( t ) 0 1 t cos s 2 d s | = | f u ( t ) 0 1 G ( t , s ) K ( s , 1 ) d s | ϕ ( d ( u , 1 ) ) = max t [ 0 , 1 ] ( 1 u ( t ) ) 1 + max t [ 0 , 1 ] ( 1 u ( t ) ) = 1 min t [ 0 , 1 ] u ( t ) 2 min t [ 0 , 1 ] u ( t ) 1 2 .
Note also that the contraction inequality d ( f u , f v ) ϕ ( d ( u , v ) ) does not follow when the weaker Meir-Keeler function ϕ is replaced by our initial Meir-Keeler function ψ: Take, for instance, the constant functions u = 0 and v = 3 ; then u 1 , v α , and
ψ ( d ( u , v ) ) = ψ ( 3 ) = 0 < d ( f u , f v ) .

Remark 5 In Example 4 above, the inequality | K ( s , x ) K ( s , y ) | ψ ( | x y | ) is not globally satisfied, i.e., there exist s , x [ 0 , 1 ] and y R + such that | K ( s , x ) K ( s , y ) | > ψ ( | x y | ) . In fact, this happens for all x , y R + with y > x + 2 . However, it is clear that for each s [ 0 , 1 ] , and x , y R + , one has | K ( s , x ) K ( s , y ) | ψ 1 ( | x y | ) for all s [ 0 , 1 ] , and x , y R + , where ψ 1 ( t ) = t / ( t + 1 ) for all t R + .

We conclude the paper with an example where conditions (I)-(IV) also hold (in particular, (III) for the function ψ 1 defined above) but the inequality | K ( s , x ) K ( s , y ) | ψ 1 ( | x y | ) is not globally satisfied.

Example 5 We modify Example 4 as follows. Consider the integral equation
u ( t ) = 0 T G ( t , s ) K ( s , u ( s ) ) d s for all  t [ 0 , T ] ,
where T = 2 , G ( t , s ) = t / 2 for all t , s [ 0 , 2 ] , and
K ( s , x ) = e s / ( 1 + x ) if  s [ 0 , 2 ] , x [ 0 , 1 ] ; K ( s , x ) = e s / ( 1 + x 1 / 2 ) if  s [ 0 , 2 ] , x ( 1 , 4 ] ; K ( s , x ) = e s / ( 4 x 13 ) if  s [ 0 , 2 ] , x > 4 .

Clearly K is continuous on [ 0 , 2 ] × R + . Moreover, M = 1 , and G and K satisfy conditions (I) and (II), respectively.

Now, construct a Meir-Keeler function ψ 1 : R + R + as ψ 1 ( t ) = t / ( 1 + t ) for all t R + .

By discussing the different cases, it is routine to show that for each s , x [ 0 , 2 ] and each y R + with | x y | 2 , we have
| K ( s , x ) K ( s , y ) | ψ 1 ( | x y | ) ,

so condition (III) is also satisfied.

Finally, define α : [ 0 , 2 ] [ 0 , 2 ] as α ( t ) = 6 t / 35 for all t [ 0 , 2 ] . Then, for each t [ 0 , 2 ] , we have
0 2 G ( t , s ) K ( s , 2 ) d s = t 2 0 2 e s 1 + 2 d s = t ( 1 e 2 ) 2 ( 1 + 2 ) > 6 t / 7 5 = α ( t ) .
Now observe that α ( s ) < 1 for all s [ 0 , 2 ] , so K ( s , α ( s ) ) = e s / ( 1 + α ( s ) ) . Hence, for each t [ 0 , 2 ] ,
0 2 G ( t , s ) K ( s , α ( s ) ) d s = t 2 0 2 e s 1 + ( 6 s / 35 ) d s = t 2 0 2 35 e s 35 + 6 s d s t 2 0 2 d s = t 2 .

Therefore α verifies condition (IV), and consequently the integral equation has a unique solution u in A 1 A 2 , where A 1 = { u C ( [ 0 , 1 ] , R + ) : u ( s ) 2  for all  s [ 0 , 2 ] } and A 2 = { u C ( [ 0 , 1 ] , R + ) : u ( s ) 6 s / 35  for all  s [ 0 , 2 ] } . In fact u A 1 A 2 , i.e., 6 t / 35 u ( t ) 2 for all t [ 0 , 2 ] .

It is interesting to observe that the Meir-Keeler function ψ 1 is continuous on R + but condition (III) is not globally satisfied: Indeed, take x = 0 and y > 14 / 3 . Then, for each s [ 0 , 1 ] , we obtain
K ( s , x ) K ( s , y ) = e s ( 1 1 4 y 13 ) > e s y 1 + y .

Hence, K ( 0 , 0 ) K ( 0 , y ) > ψ 1 ( y ) .

Declarations

Acknowledgements

The second author thanks for the support of the Ministry of Economy and Competitiveness of Spain under grant MTM2012-37894-C02-01, and the Universitat Politècnica de València, grant PAID-06-12-SP20120471.

Authors’ Affiliations

(1)
Department of Mathematics, Disha Institute of Management and Technology, Satya Vihar, Vidhansabha-Chandrakhuri Marg, Mandir Hasaud, Raipur, Chhattisgarh, 492101, India
(2)
Instituto Universitario de Matemática Pura y Aplicada, Universitat Politècnica de València, Camí de Vera s/n, Valencia, 46022, Spain

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© Nashine and Romaguera; licensee Springer. 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.

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