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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  is a very useful and classical tool in nonlinear analysis. In 1969, Boyd and Wong  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.  introduced the following notion of cyclic representation.

Definition 1  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.  also proved the below theorem.

Theorem 1  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., ). In 2010, Pǎcurar and Rus  introduced the following notion of cyclic weaker φ-contraction.

Definition 2  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  proved the below theorem.

Theorem 2  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 ). 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

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

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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## Acknowledgements

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

## Author information

Correspondence to Chi-Ming Chen.

### Competing interests

The authors declare that they have no competing interests.

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• #### DOI

https://doi.org/10.1186/1687-1812-2012-17

### Keywords

• fixed point theory
• weaker Meir-Keeler function
• cyclic weaker ϕ φ-contraction
• cyclic weaker (ϕ, φ)-contraction 