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Fixed point theory for the cyclic weaker Meir-Keeler function in complete metric spaces

Abstract

In this article, we introduce the notions of cyclic weaker ϕ φ-contractions and cyclic weaker (ϕ, φ)-contractions in complete metric spaces and prove two theorems which assure the existence and uniqueness of a fixed point for these two types of contractions. Our results generalize or improve many recent fixed point theorems in the literature.

MSC: 47H10; 54C60; 54H25; 55M20.

1 Introduction and preliminaries

Throughout this article, by +, we denote the sets of all nonnegative real numbers and all real numbers, respectively, while is the set of all natural numbers. Let (X, d) be a metric space, D be a subset of X and f: DX be a map. We say f is contractive if there exists α [0,1) such that for all x, y D,

d ( f x , f y ) α d ( x , y ) .

The well-known Banach's fixed point theorem asserts that if D = X, f is contractive and (X, d) is complete, then f has a unique fixed point in X. It is well known that the Banach contraction principle [1] is a very useful and classical tool in nonlinear analysis. In 1969, Boyd and Wong [2] introduced the notion of Φ-contraction. A mapping f: XX on a metric space is called Φ-contraction if there exists an upper semi-continuous function Φ: [0, ∞) → [0, ∞) such that

d ( f x , f y ) Φ ( d ( x , y ) ) for all x , y X .

Generalization of the above Banach contraction principle has been a heavily investigated branch research. (see, e.g., [3, 4]). In 2003, Kirk et al. [5] introduced the following notion of cyclic representation.

Definition 1 [5] Let X be a nonempty set, m and f: XX an operator. Then X= i = 1 m A i is called a cyclic representation of X with respect to f if

  1. (1)

    A i , i = 1, 2,..., m are nonempty subsets of X;

  2. (2)

    f (A1) A2, f (A2) A3,..., f (Am-1) A m , f (A m ) A1.

Kirk et al. [5] also proved the below theorem.

Theorem 1 [5] Let (X, d) be a complete metric space, m , A1, A2,..., A m , closed nonempty subsets of X and X= i = 1 m A i . Suppose that f satisfies the following condition.

d ( f x , f y ) ψ ( d ( x , y ) ) , f o r a l l x A i , y A i + 1 , i { 1 , 2 , . . . , m } ,

where ψ: [0, ∞) → [0, ∞) is upper semi-continuous from the right and 0 ≤ ψ(t) < t for t > 0. Then, f has a fixed point z i = 1 n A i .

Recently, the fixed theorems for an operator f: XX that defined on a metric space X with a cyclic representation of X with respect to f had appeared in the literature. (see, e.g., [610]). In 2010, Pǎcurar and Rus [7] introduced the following notion of cyclic weaker φ-contraction.

Definition 2 [7] Let (X, d) be a metric space, m , A1, A2,...,A m closed nonempty subsets of X and X= i = 1 m A i . An operator f: XX is called a cyclic weaker φ-contraction if

  1. (1)

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

  2. (2)

    there exists a continuous, non-decreasing function φ: [0, ∞) → [0, ∞) with φ(t) > 0 for t (0, ∞) and φ(0) = 0 such that

    d ( f x , f y ) d ( x , y ) - φ ( d ( x , y ) ) ,

for any x A i , y Ai+1, i = 1,2,...,m where Am+1= A1.

And, Pǎcurar and Rus [7] proved the below theorem.

Theorem 2 [7] Let (X, d) be a complete metric space, m , A1, A2,..., A m closed nonempty subsets of X and X= i = 1 m A i . Suppose that f is a cyclic weaker φ-contraction. Then, f has a fixed point z i = 1 n A i .

In this article, we also recall the notion of Meir-Keeler function (see [11]). A function ϕ: [0, ∞) → [0, ∞) is said to be a Meir-Keeler function if for each η > 0, there exists δ > 0 such that for t [0, ∞) with ηt < η + δ, we have ϕ (t) < η. We now introduce the notion of weaker Meir-Keeler function ϕ: [0, ∞) → [0,∞), as follows:

Definition 3 We call ϕ: [0, ∞) → [0, ∞) a weaker Meir-Keeler function if for each η > 0, there exists δ > 0 such that for t [0, ∞) with ηt < η + δ, there exists n0 such that ϕ n 0 ( t ) <η.

2 Fixed point theory for the cyclic weaker ϕ φ-contractions

The main purpose of this section is to present a generalization of Theorem 1. In the section, we let ϕ: [0, ∞) → [0, ∞) be a weaker Meir-Keeler function satisfying the following conditions:

  • (ϕ1) ϕ(t) > 0 for t > 0 and ϕ (0) = 0;

  • (ϕ2) for all t (0, ∞), {ϕn(t)}nis decreasing;

  • (ϕ3) for t n [0, ∞), we have that

    1. (a)

      if limn→∞t n = γ > 0, then limn→∞ϕ (t n ) < γ, and

    2. (b)

      if limn→∞t n = 0, then limn→∞ϕ (t n ) = 0.

And, let φ: [0, ∞) → [0, ∞) be a non-decreasing and continuous function satisfying

  • (φ1) φ(t) > 0 for t > 0 and φ(0) = 0;

  • (φ2) φ is subadditive, that is, for every μ1, μ2 [0, ∞), φ( μ1 + μ2) ≤ φ(μ1) + φ(μ2);

  • (φ3) for all t (0, ∞), limn→∞t n = 0 if and only if limn→∞φ(t n ) = 0.

We state the notion of cyclic weaker ϕ φ-contraction, as follows:

Definition 4 Let (X, d) be a metric space, m , A1, A2,..., A m nonempty subsets of X and X= i = 1 m A i . An operator f: XX is called a cyclic weaker ϕ φ-contraction if

  1. (i)

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

  2. (ii)

    for any x A i , y Ai+1, i = 1, 2,..., m,

    φ ( d ( f x , f y ) ) ϕ ( φ ( d ( x , y ) ) ) ,

where Am+1= A1.

Theorem 3 Let (X, d) be a complete metric space, m , A1, A2, ..., A m nonempty subsets of X and X= i = 1 m A i . Let f: XX be a cyclic weaker ϕ φ-contraction. Then, f has a unique fixed point z i = 1 m A i .

Proof. Given x0 and let xn+1= fx n = fn+1x0, for n {0}. If there exists n0 {0} such that x n 0 + 1 = x n 0 , then we finished the proof. Suppose that xn+1x n for any n {0}. Notice that, for any n > 0, there exists i n {1,2,...,m} such that x n - 1 A i n and x n A i n + 1 . Since f: XX is a cyclic weaker ϕ φ-contraction, we have that for all n

φ ( d ( x n , x n + 1 ) ) = φ ( d ( f x n - 1 , f x n ) ) ϕ ( φ ( d ( x n - 1 , x n ) ) ) ,

and so

φ ( d ( x n , x n + 1 ) ) ϕ ( φ ( d ( x n 1 , x n ) ) ) ϕ ( ϕ ( φ ( d ( x n 2 , x n 1 ) ) ) = ϕ 2 ( φ ( ( d ( x n 2 , x n 1 ) ) ) ϕ n ( φ ( d ( x 0 , x 1 ) ) ) .

Since {ϕn(φ(d(x0, x 1)))}nis decreasing, it must converge to some η ≥ 0. We claim that η = 0. On the contrary, assume that η > 0. Then by the definition of weaker Meir-Keeler function ϕ, there exists δ > 0 such that for x0, x1 X with ηφ(d(x0, x1)) < δ + η, there exists n0 such that ϕ n 0 ( φ ( d ( x 0 , x 1 ) ) ) <η. Since limn→∞ϕn(φ(d(x0, x1))) = η, there exists p0 such that ηϕp(φ(d(x0, x1)) < δ + η, for all pp0. Thus, we conclude that ϕ p 0 + n 0 ( φ ( d ( x 0 , x 1 ) ) ) <η. So we get a contradiction. Therefore limn→∞ϕn(φ(d(x0, x1))) = 0, that is,

lim n φ ( d ( x n , x n + 1 ) ) = 0 .

Next, we claim that {x n } is a Cauchy sequence. We claim that the following result holds:

Claim: for each ε > 0, there is n0(ε) such that for all p, qn0(ε),

φ ( d ( x p , x q ) ) < ε , ( * )

We shall prove (*) by contradiction. Suppose that (*) is false. Then there exists some ε > 0 such that for all n , there are p n , q n with p n > q n n satisfying:

  1. (i)

    φ ( d ( x p n , x q n ) ) ε, and

  2. (ii)

    p n is the smallest number greater than q n such that the condition (i) holds.

Since

ε φ ( d ( x p n , x q n ) ) φ ( d ( x p n , x p n - 1 ) + d ( x p n - 1 , x q n ) ) φ ( d ( x p n , x p n - 1 ) ) + φ ( d ( x p n - 1 , x q n ) ) φ ( d ( x p n , x p n - 1 ) ) + ε ,

hence we conclude lim p φ ( d ( x p n , x q n ) ) =ε. Since φ is subadditive and nondecreasing, we conclude

φ ( d ( x p n , x q n ) ) φ ( d ( x p n , x q n + 1 ) + d ( x p n + 1 , x q n ) ) φ ( d ( x p n , x q n + 1 ) ) + φ ( d ( x p n + 1 , x q n ) ) ,

and so

φ ( d ( x p n , x q n ) ) - φ ( d ( x p n , x p n + 1 ) ) φ ( d ( x p n + 1 , x q n ) ) φ ( d ( x p n , x p n + 1 ) + d ( x p n , x q n ) ) φ ( d ( x p n , x p n + 1 ) ) + φ ( d ( x p n , x q n ) ) .

Letting n → ∞, we also have

lim n φ ( d ( x p n + 1 , x q n ) ) =ε.

Thus, there exists i, 0 ≤ im - 1 such that p n - q n + i = 1 mod m for infinitely many n. If i = 0, then we have that for such n,

ε φ ( d ( x p n , x q n ) ) φ ( d ( x p n , x p n + 1 ) + d ( x p n + 1 , x q n + 1 ) + d ( x q n + 1 , x q n ) ) φ ( d ( x p n , x p n + 1 ) ) + φ ( d ( x p n + 1 , x q n + 1 ) ) + φ ( d ( x q n + 1 , x q n ) ) = φ ( d ( x p n , x p n + 1 ) ) + φ ( d ( f x p n , f x q n ) ) + φ ( d ( x q n + 1 , x q n ) ) φ ( d ( x p n , x p n + 1 ) ) + ϕ ( φ ( d ( x p n , x q n ) ) ) + φ ( d ( x q n + 1 , x q n ) ) .

Letting n → ∞. Then by, we have

ε 0 + lim n ϕ ( φ ( d ( x p n , x q n ) ) ) + 0 < ε ,

a contradiction. Therefore lim n φ ( d ( x p n , x q n ) ) =0, by the condition (φ3), we also have lim n d ( x p n , x q n ) =0. The case i ≠ 0 is similar. Thus, {x n } is a Cauchy sequence. Since X is complete, there exists ν i = 1 m A i such that limn→∞x n = ν. Now for all i = 0, 1, 2,..., m - 1, {fx mn-i } is a sequence in A i and also all converge to ν. Since A i is clsoed for all i = 1, 2,..., m, we conclude ν i = 1 m A i , and also we conclude that i = 1 m A i ϕ. Since

φ ( d ( ν , f ν ) ) = lim n φ ( d ( f x m n , f ν ) ) lim n ϕ ( φ ( d ( f x m n - 1 , ν ) ) ) = 0 ,

hence φ(d(ν, fν)) = 0, that is, d(ν, fν) = 0, ν is a fixed point of f.

Finally, to prove the uniqueness of the fixed point, let μ be another fixed point of f. By the cyclic character of f, we have μ,ν i = 1 n A i . Since f is a cyclic weaker ϕ φ-contraction, we have

φ ( d ( ν , μ ) ) = φ ( d ( ν , f μ ) ) = lim n φ ( d ( f x m n , f μ ) ) lim n ϕ ( φ ( d ( f x m n - 1 , μ ) ) ) < φ ( d ( ν , μ ) ) ,

and this is a contradiction unless φ(d(ν, μ)) = 0, that is, μ = ν. Thus ν is a unique fixed point of f.

Example 1 Let X = 3 and we define d: X × X → [0,∞) by d(x,y) = |x1-y1 |+| x2-y2 |+| x3-y3|, for x = (x1, x2, x3), y = (y1, y2, y3) X, and let A = {(x, 0,0):x }, B = {(0,y,0):y },C = {(0,0, z): z } be three subsets of X. Define f: A B CA B C by

f ( ( x , 0 , 0 ) ) = 0 , 1 4 x , 0 ; f o r a l l x ; f ( ( 0 , y , 0 ) ) = 0 , 0 , 1 4 y ; f o r a l l y ; f ( ( 0 , 0 , z ) ) = 1 4 z , 0 , 0 ; f o r a l l z .

We define φ: [0, ∞) → [0, ∞) by

ϕ ( t ) = 1 3 t f o r t [ 0 , ) ,

and φ: [0, ∞) → [0, ∞) by

φ ( t ) = 1 2 t f o r t [ 0 , ) .

Then f is a cyclic weaker ϕ φ-contraction and (0, 0, 0) is the unique fixed point.

3 Fixed point theory for the cyclic weaker (ϕ, φ-contractions

The main purpose of this section is to present a generalization of Theorem 2. In the section, we let ϕ: [0, ∞) → [0, ∞) be a weaker Meir-Keeler function satisfying the following conditions:

  • (ϕ1) ϕ (t) > 0 for t > 0 and ϕ(0) = 0;

  • (ϕ2) for all t (0, ∞), {ϕn(t)}nis decreasing;

  • (ϕ3) for t n [0, ∞), if limn→∞t n = γ, then limn→∞ϕ(t n ) ≤ γ.

And, let φ: [0, ∞) → [0, ∞) be a non-decreasing and continuous function satisfying φ(t) > 0 for t > 0 and φ(0) = 0.

We now state the notion of cyclic weaker (ϕ, φ)-contraction, as follows:

Definition 5 Let (X, d) be a metric space, m , A1, A2,..., A m nonempty subsets of X and X= i = 1 m A i . An operator f: XX is called a cyclic weaker (ϕ,φ)-contraction if

  1. (i)

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

  2. (ii)

    for any x A i , y Ai+1, i = 1, 2,..., m,

    d ( f x , f y ) ϕ ( d ( x , y ) ) - φ ( d ( x , y ) ) ,

where Am+ 1= A1.

Theorem 4 Let (X, d) be a complete metric space, m , A1, A2,..., A m nonempty subsets of X and X= i = 1 m A i . Let f: XX be a cyclic weaker (ϕ, φ)-contraction. Then f has a unique fixed point z i = 1 m A i .

Proof. Given x0 and let xn+1= fx n = fn+1x0, for n {0}. If there exists n {0} such that x n 0 + 1 = x n 0 , then we finished the proof. Suppose that xn+ 1x n for any n {0}. Notice that, for any n > 0, there exists i n {1,2,...,m} such that x n - 1 A i n and x n A i n + 1 . Since f: XX is a cyclic weaker (ϕ, φ)-contraction, we have that n

d ( x n , x n + 1 ) = d ( f x n - 1 , f x n ) ϕ ( d ( x n - 1 , x n ) ) - φ ( d ( x n - 1 , x n ) ) ϕ ( d ( x n - 1 , x n ) ) ,

and so

d ( x n , x n + 1 ) ϕ ( d ( x n 1 , x n ) ) ϕ ( ϕ ( d ( x n 2 , x n 1 ) ) = ϕ 2 ( d ( x n 2 , x n 1 ) ) ϕ n ( d ( x 0 , x 1 ) ) .

Since {ϕn(d(x0, x1))}nis decreasing, it must converge to some η ≥ 0. We claim that η = 0. On the contrary, assume that η > 0. Then by the definition of weaker Meir-Keeler function ϕ, there exists δ > 0 such that for x0, x1 X with ηd(x0, x1) < δ + η, there exists n0 such that ϕ n 0 ( d ( x 0 , x 1 ) ) < η . Since limn→∞, ϕn(d(x0, x1)) = η, there exists p0 such that ηϕp(d(x0, x1)) < δ + η, for all pp0. Thus, we conclude that ϕ p 0 + n 0 ( d ( x 0 , x 1 ) ) <η. So we get a contradiction. Therefore limn→∞ϕn(d(x0, x1)) = 0, that is,

lim n d ( x n , x n + 1 ) = 0 .

Next, we claim that {x n } is a Cauchy sequence. We claim that the following result holds:

Claim: For every ε > 0, there exists n such that if p, qn with p-q = 1 mod m, then d(x p , x q ) < ε.

Suppose the above statement is false. Then there exists ϵ > 0 such that for any n , there are p n , q n with p n > q n n with p n - q n = 1 mod m satisfying

d ( x q n , x p n ) ε .

Now, we let n > 2m. Then corresponding to q n n use, we can choose p n in such a way, that it is the smallest integer with p n > q n n satisfying p n - q n = 1 mod m and d ( x q n , x p n ) ε. Therefore d ( x q n , x p n - m ) ε and

ε d ( x q n , x p n ) d ( x q n , x p n - m ) + i = 1 m d ( x p n - i , x p n - i + 1 ) < ε + i = 1 m d ( x p n - i , x p n - i + 1 ) .

Letting n → ∞ , we obtain that

lim n d ( x q n , x p n ) = ε .

On the other hand, we can conclude that

ε d ( x q n , x p n ) d ( x q n , x q n + 1 ) + d ( x q n + 1 , x p n + 1 ) + d ( x x p n + 1 , p n ) d ( x q n , x q n + 1 ) + d ( x q n + 1 , x q n ) + d ( x q n , x p n ) + d ( x p n , x p n + 1 ) + d ( x x p n + 1 , p n ) .

Letting n → ∞, we obtain that

lim n d ( x q n + 1 , x p n + 1 ) = ε .

Since x q n and x p n lie in different adjacently labeled sets A i and Ai+1for certain 1 ≤ im, by using the fact that f is a cyclic weaker (ϕ, φ)-contraction, we have

d ( x q n + 1 , x p n + 1 ) = d ( f x q n , f x p n ) ϕ ( d ( x q n , x p n ) ) - φ ( d ( x q n , x p n ) ) .

Letting n → ∞, by using the condition ϕ3 of the function ϕ, we obtain that

ε ε - φ ( ε ) ,

and consequently, φ (ϵ) = 0. By the definition of the function φ, we get ϵ = 0 which is contraction. Therefore, our claim is proved.

In the sequel, we shall show that {x n } is a Cauchy sequence. Let ε > 0 be given. By our claim, there exists n1 such that if p, qn1 with p - q = 1 mod m, then

d ( x p , x q ) ε 2 .

Since limn→∞d(x n , xn+1) = 0, there exists n2 such that

d ( x n , x n + 1 ) ε 2 m ,

for any nn2.

Let p, q ≥ max{n 1, n2} and p > q. Then there exists k {1, 2,..., m} such that p -q = k mod m. Therefore, p - q + j = 1 mod m for j = m - k + 1, and so we have

d ( x q , x p ) d ( x q , x p + j ) + d ( x p + j , x p + j - 1 ) + + d ( x p + 1 , x p ) ε 2 + j × ε 2 m ε 2 + m × ε 2 m = ε .

Thus, {x n } is a Cauchy sequence. Since X is complete, there exists ν i = 1 m A i such that limn→∞x n = ν. Since X= i = 1 m A i is a cyclic representation of X with respect to f, the sequence {x n } has infinite terms in each A i for i {1,2,...,m}. Now for all i = 1,2,...,m, we may take a subsequence { x n k } of {x n } with x n k A i - 1 and also all converge to ν. Since

d ( x n k + 1 , f ν ) = d ( f x n k , f ν ) ϕ ( d ( x n k , ν ) ) - φ ( d ( x n k , ν ) ) ϕ ( d ( x n k , ν ) ) .

Letting k → ∞ , we have

d ( ν , f ν ) 0 ,

and so ν = fν.

Finally, to prove the uniqueness of the fixed point, let μ be the another fixed point of f. By the cyclic character of f, we have μ,ν i = 1 n A i . Since f is a cyclic weaker (ϕ, φ)-contraction, we have

d ( ν , μ ) = d ( ν , f μ ) = lim n d ( x n k + 1 , f μ ) = lim n d ( f x n k , f μ ) lim n [ ϕ ( d ( x n k , μ ) ) - φ ( d ( x n k , μ ) ) ] d ( ν , μ ) - φ ( d ( ν , μ ) ) ,

and we can conclude that

φ ( d ( ν , μ ) ) = 0 .

So we have μ = ν. We complete the proof.

Example 2 Let X = [-1,1] with the usual metric. Suppose that A1 = [-1,0] = A3 and A2 = [0,1] = A4. Define f: XX by f ( x ) = - x 6 for all x X, and let ϕ, φ: [0,∞) → [0, ∞) be ϕ ( t ) = 1 2 ,φ ( t ) = t 4 . Then f is a cyclic weaker (ϕ, φ)-contraction and 0 is the unique fixed point.

Example 3 Let X = + with the metric d:X × X+ given by

d ( x , y ) = max { x , y } , f o r x , y X .

Let A1 = A2 = ... = A m = +. Define f: XX by

f ( x ) = x 2 77 f o r x X ,

and let ϕ, φ: [0, ∞) → [0,∞) be φ ( t ) = t 3 2 ( t + 2 ) and

ϕ ( t ) = 2 t 3 3 t + 8 , i f t 1 ; t 2 2 , i f t < 1 .

Then f is a cyclic weaker (ϕ, φ)-contraction and 0 is the unique fixed point.

Example 4 Let X = 3 and we define d: X × X → [0, ∞) by

d ( x , y ) = max x 1 - y 1 , x 2 - y 2 , x 3 - y 3 ,

for x = (x1,x2,x3), y = (y1, y2, y3) X, and let A = {(x,0,0): x [0,1]}, B = {(0,y,0): y [0,1]}, C = {(0,0, z): z [0,1]} be three subsets of X.

Define f: A B CA B C by

f ( ( x , 0 , 0 ) ) = 0 , 1 8 x 2 , 0 ; f o r a l l x [ 0 , 1 ] ; f ( ( 0 , y , 0 ) ) = 0 , 0 , 1 8 y 2 ; f o r a l l y [ 0 , 1 ] ; f ( ( 0 , 0 , z ) ) = 1 8 z 2 , 0 , 0 ; f o r a l l z [ 0 , 1 ] .

We define φ: [0, ∞) → [0,∞) by

ϕ ( t ) = t 2 t + 1 f o r t [ 0 , ) ,

and φ: [0, ∞) → [0,∞) by

φ ( t ) = t 2 t + 2 f o r t [ 0 , ) .

Then f is a cyclic weaker (ϕ, φ)-contraction and (0,0,0) is the unique fixed point.

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The authors would like to thank referee(s) for many useful comments and suggestions for the improvement of the article.

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Chen, CM. Fixed point theory for the cyclic weaker Meir-Keeler function in complete metric spaces. Fixed Point Theory Appl 2012, 17 (2012). https://doi.org/10.1186/1687-1812-2012-17

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