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# Periodic points for the weak contraction mappings in complete generalized metric spaces

Fixed Point Theory and Applications20122012:79

https://doi.org/10.1186/1687-1812-2012-79

• Received: 25 December 2011
• Accepted: 9 May 2012
• Published:

## Abstract

In this article, we introduce the notions of (ϕ - φ)-weak contraction mappings and (ψ - φ)-weak contraction mappings in complete generalized metric spaces and prove two theorems which assure the existence of a periodic point for these two types of weak contraction.

Mathematical Subject Classification: 47H10; 54C60; 54H25; 55M20.

## Keywords

• Periodic point
• Meir-Keeler function
• (ϕ - φ)-weak contraction mapping
• (ψ - φ)-weak contraction mapping

## 1 Introduction and preliminaries

Let (X, d) be a metric space, D 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,
$d\left(fx,fy\right)\le \alpha \cdot d\left(x,y\right).$
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

Generalization of the above Banach contraction principle has been a heavily investigated research branch. (see, e.g., [3, 4]).

In 2000, Branciari  introduced the following notion of a generalized metric space where the triangle inequality of a metric space had been replaced by an inequality involing three terms instead of two. Later, many authors worked on this interesting space (e.g. ).

Let (X, d) be a generalized metric space. For γ > 0 and x X, we define
${B}_{\gamma }\left(x\right):=\left\{y\in X|d\left(x,y\right)<\gamma \right\}.$

Branciari  also claimed that {B γ (x): γ > 0, x X} is a basis for a topology on X, d is continuous in each of the coordinates and a generalized metric space is a Hausdorff space. We recall some definitions of a generalized metric space, as follows:

Definition 1  Let X be a nonempty set and d : X × X → [0, ∞) be a mapping such that for all x, y X and for all distinct point u, v X each of them different from × and y, one has

(i) d(x, y) = 0 if and only if × = y;

(ii) d(x, y) = d(y, x);

(iii) d(x, y) ≤ d(x, u) + d(u, v) + d(v, y) (rectangular inequality).

Then (X, d) is called a generalized metric space (or shortly g.m.s).

We present an example to show that not every generalized metric on a set X is a metric on X.

Example 1 Let X = {t, 2t, 3t, 4t, 5t} with t > 0 is a constant, and we define d : X × X → [0, ∞) by
1. (1)

d(x, x) = 0, for all × X;

2. (2)

d(x, y) = d(y, x), for all x, y X;

3. (3)

d(t, 2t) = 3γ;

4. (4)

d(t, 3t) = d(2t, 3t) = γ;

5. (5)

d(t, 4t) = d(2t, 4t) = d(3t, 4t) = 2γ;

6. (6)

$d\left(t,\phantom{\rule{2.77695pt}{0ex}}5t\right)=d\left(2t,\phantom{\rule{2.77695pt}{0ex}}5t\right)=d\left(3t,\phantom{\rule{2.77695pt}{0ex}}5t\right)=\left(4t,\phantom{\rule{2.77695pt}{0ex}}5t\right)=\frac{3}{2}\gamma$,

where γ > 0 is a constant. Then (X, d) be a generalized metric space, but it is not a metric space, because
$d\left(t,\mathsf{\text{2}}t\right)=\mathsf{\text{3}}\gamma >d\left(t,\mathsf{\text{3}}t\right)+d\left(\mathsf{\text{3}}t,\mathsf{\text{2}}t\right)=\mathsf{\text{2}}\gamma \mathsf{\text{.}}$

Definition 2  Let (X, d) be a g.m.s, {x n } be a sequence in X and x X. We say that {x n } is g.m.s convergent to × if and only if d(x n , x) → 0 as n → ∞. We denote by x n x as n → ∞.

Definition 3  Let (X, d) be a g.m.s, {x n } be a sequence in X and x X. We say that {x n } is g.m.s Cauchy sequence if and only if for each ε > 0, there exists ${n}_{0}\in ℕ$ such that d(x m , x n ) < ε for all n > m > n0.

Definition 4  Let (X, d) be a g.m.s. Then X is called complete g.m.s if every g.m.s Cauchy sequence is g.m.s convergent in X.

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. Generalization of the above function has been a heavily investigated research branch. Praticularly, in [13, 14], the authors proved the existence and uniqueness of fixed points for various Meir-Keeler type contractive functions. In this study, we introduce the below notions of the weaker Meir-Keeler function ϕ : [0, ∞) → [0, ∞) and stronger Meir-Keeler function ψ : [0, ∞) → [0, 1).

Definition 5 We call ϕ : [0, ∞) → [0, ∞) a weaker Meir-Keeler function if the function ϕ satisfies the following condition
$\forall \eta >0\phantom{\rule{1em}{0ex}}\exists \delta >0\phantom{\rule{1em}{0ex}}\forall t\in \left[0,\infty \right)\phantom{\rule{1em}{0ex}}\left(\eta \le t<\delta +\eta \phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}\exists {n}_{0}\in ℕ\phantom{\rule{1em}{0ex}}\varphi {\left(t\right)}^{{n}_{0}}<\eta \right).$

The following provides an example of a weaker Meir-Keeler function which is not a Meir-Keeler function.

Example 2 Let $\varphi :{ℝ}^{+}\to {ℝ}^{+}$ be defined by
$\varphi \left(t\right)=\left\{\begin{array}{cc}\hfill 0,\hfill & \hfill if\phantom{\rule{2.77695pt}{0ex}}t\le 1,\hfill \\ \hfill 3t,\hfill & \hfill if\phantom{\rule{2.77695pt}{0ex}}1

Then ϕ is a weaker Meir-Keeler function which is not a Meir-Keeler function.

Definition 6 We call ψ : [0, ∞) → [0, 1) a stronger Meir-Keeler function if the function ψ satisfies the following condition
$\forall \eta >0\phantom{\rule{2.77695pt}{0ex}}\exists \delta >0\phantom{\rule{2.77695pt}{0ex}}\exists {\gamma }_{\eta }\in \left[0,1\right)\phantom{\rule{1em}{0ex}}\forall t\in \left[0,\infty \right)\phantom{\rule{1em}{0ex}}\left(\eta \le t<\delta +\eta \phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}\psi \left(t\right)<{\gamma }_{\eta }\right).$

The following provides an example of a stronger Meir-Keeler function.

Example 3 Let $\psi :{ℝ}^{+}\to \left[0,1\right)$ be defined by
$\psi \left(d\left(x,y\right)\right)=\frac{2t}{3t+1}.$

Then ψ is a stronger Meir-Keeler function.

The following provides an example of a Meir-Keeler function which is not a stronger Meir-Keeler function.

Example 4 Let $\phi :{ℝ}^{+}\to {ℝ}^{+}$ be defined by
$\phi \left(t\right)=\left\{\begin{array}{cc}\hfill t-1,\hfill & \hfill if\phantom{\rule{2.77695pt}{0ex}}t>1;\hfill \\ \hfill 0,\hfill & \hfill if\phantom{\rule{2.77695pt}{0ex}}t\le 1.\hfill \end{array}\right\$

Then φ is a Meir-Keeler function which is not a stronger Meir-Keeler function.

## 2 Main results

In the sequel, we let the function ϕ : [0, ∞) → [0, ∞) satisfies the following conditions:

(ϕ1) ϕ : [0, ∞) → [0, ∞) is a weaker Meir-Keeler function;

(ϕ2) ϕ(t) > 0 for t > 0 and ϕ(0) = 0;

(ϕ3) for all t (0, ∞), ${\left\{{\varphi }^{n}\left(t\right)\right\}}_{n\in ℕ}$ is decreasing;

(ϕ4) 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.

Let the function ψ : [0, ∞) → [0, 1) satisfies the following conditions:

(ψ1) ψ : [0, ∞) → [0, 1) is a stronger Meir-Keeler function;

(ψ2) ψ(t) > 0 for t > 0 and ϕ(0) = 0.

And, we let the function φ : [0, ∞) → [0, ∞) satisfies the following conditions:

(φ1) for all t (0, ∞), limn→∞t n = 0 if and only if limn→∞φ(t n ) = 0;

(φ2) φ(t) > 0 for t > 0 and φ(0) = 0;

(φ3) φ is subadditive, that is, for every μ1, μ2 [0, ∞), φ(μ1 + μ2) ≤ φ(μ1) + φ(μ2).

Using the functions ϕ and φ, we first introduce the notion of the (ϕ-φ)-weak contraction mapping and prove a theorem which assures the existence of a periodic point for the (ϕ-φ)-weak contraction mapping.

Definition 7 Let (X, d) be a g.m.s, and let f : XX be a function satisfying
$\phi \left(d\left(fx,fy\right)\right)\le \varphi \left(\phi \left(d\left(x,y\right)\right)$
(1)

for all x, y X. Then f is said to be a (ϕ - φ)-weak contraction mapping.

Theorem 1 Let (X, d) be a Hausdorff and complete g.m.s, and let f be a (ϕ - φ)-weak contraction mapping. Then f has a periodic point μ in X, that is, there exists μ X such that $\mu ={f}^{p}\mu$ for some $p\in ℕ$.

Proof. Given x0 and define a sequence {x n } in X by
${{x}_{n}}_{+\mathsf{\text{1}}}=f{x}_{n}\phantom{\rule{1em}{0ex}}\mathsf{\text{for}}\phantom{\rule{1em}{0ex}}n\in ℕ\cup \left\{0\right\}.$
Step 1. We shall prove that
$\underset{n\to \infty }{\text{lim}}\phi \left(d\left({x}_{n},{{x}_{n}}_{+\mathsf{\text{1}}}\right)\right)=0,$
(2)
$\underset{n\to \infty }{\text{lim}}\phi \left(d\left({x}_{n},{{x}_{n}}_{+2}\right)\right)=0.$
(3)
Using the inequality (1), we have that for each $n\in ℕ$
$\begin{array}{c}\phi \left(d\left({x}_{n},{x}_{n}+\text{1}\right)\right)=\phi \left(d\left(f{x}_{n-1},f{x}_{n}\right)\right)\\ \le \varphi \left(\phi \left(d\left({x}_{n-1},{x}_{n}\right)\right),\end{array}$
and so
$\begin{array}{c}\phi \left(d\left({x}_{n},{x}_{n}+\text{1}\right)\right)\le \varphi \left(\phi \left(d\left({x}_{n-1},{x}_{n}\right)\right)\right)\\ \le \varphi \left(\varphi \left(\phi \left(d\left({x}_{n-2},{x}_{n-1}\right)\right)\right)={\varphi }^{2}\left(\phi \left(d\left({x}_{n-2},{x}_{n-1}\right)\right)\right)\\ \le \cdots \cdots \\ \le {\varphi }^{n}\left(\phi \left(d\left({x}_{0},{x}_{1}\right)\right)\right).\end{array}$
Since ${\left\{{\varphi }^{n}\left(\phi \left(d\left({x}_{0},{x}_{\mathsf{\text{1}}}\right)\right)\right)\right\}}_{n\in ℕ}$ is 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, corresponding to η use, there exists δ > 0 such that for x0, x1 X with ηφ(d(x0, x1)) < δ + η, there exists ${n}_{0}\in ℕ$ such that ${\varphi }^{{n}_{0}}\left(\phi \left(d\left({x}_{0},{x}_{\mathsf{\text{1}}}\right)\right)\right)<\eta$. Since limn→∞ϕ n (φ(d(x0, x1))) = η, there exists ${p}_{0}\in ℕ$ such that ηϕ p (φ(d(x0, x1))) < δ + η, for all pp0. Thus, we conclude that ${\varphi }^{{p}_{0}+{n}_{0}}\left(\phi \left(d\left({x}_{0},{x}_{\mathsf{\text{1}}}\right)\right)\right)<\eta$. So we get a contradiction. Therefore limn→∞ϕ n (φ(d(x0, x1))) = 0, that is,
$\underset{n\to \infty }{\text{lim}}\phi \left(d\left({x}_{n},{{x}_{n}}_{+\mathsf{\text{1}}}\right)\right)=0.$
Using the inequality (1), we also have that for each $n\in ℕ$
$\begin{array}{c}\phi \left(d\left({x}_{n},{x}_{n}+2\right)\right)=\phi \left(d\left(f{x}_{n-1},f{x}_{n+1}\right)\right)\\ \le \varphi \left(\phi \left(d\left({x}_{n-1},{x}_{n+1}\right)\right),\end{array}$
and so
$\begin{array}{c}\phi \left(d\left({x}_{n},{x}_{n}+2\right)\right)\le \varphi \left(\phi \left(d\left({x}_{n-1},{x}_{n+1}\right)\right)\right)\\ \le \varphi \left(\varphi \left(\phi \left(d\left({x}_{n-2},{x}_{n}\right)\right)\right)={\varphi }^{2}\left(\phi \left(d\left({x}_{n-2},{x}_{n}\right)\right)\right)\\ \le \cdots \cdots \\ \le {\varphi }^{n}\left(\phi \left(d\left({x}_{0},{x}_{1}\right)\right)\right).\end{array}$
Since ${\left\{{\phi }^{n}\left(d\left({x}_{0},{x}_{2}\right)\right)\right\}}_{n\in ℕ}$ is decreasing, by the same proof process, we also conclude
$\underset{n\to \infty }{\text{lim}}\phi \left(d\left({x}_{n},{{x}_{n}}_{+2}\right)\right)=0.$

Next, we claim that {x n } is g.m.s Cauchy. We claim that the following result holds:

Step 2. Claim that $\underset{n\to \infty }{\text{lim}}\phi \left(d\left({x}_{{p}_{n}},{x}_{{q}_{n}}\right)\right)=0$, that is, for every ε > 0, there exists $n\in ℕ$ such that if p, qn then φ(d(x p , x q )) < ε.

Suppose the above statement is false. Then there exists ε > 0 such that for any $n\in ℕ$, there are ${p}_{n},{q}_{n}\in ℕ$ with p n > q n n satisfying
$\phi \left(d\left({x}_{{q}_{n}},{x}_{{p}_{n}}\right)\right)\ge \epsilon .$
Further, corresponding to q n n, we can choose p n in such a way that it the smallest integer with p n > q n n and $\phi \left(d\left({x}_{{q}_{n}},{x}_{{p}_{n}}\right)\right)\ge \epsilon$. Therefore $\phi \left(d\left({x}_{{q}_{n}},{x}_{{p}_{n}-1}\right)\right)<\epsilon$. By the rectangular inequality and (2), (3), we have
$\begin{array}{c}\epsilon \le \phi \left(d\left({x}_{{p}_{n}},{x}_{{q}_{n}}\right)\right)\\ \le \phi \left(d\left({x}_{{p}_{n}},{x}_{{p}_{n}-2}\right)+d\left({x}_{{p}_{n}-2},{x}_{{p}_{n}-1}\right)+d\left({x}_{{p}_{n}-1},{x}_{{q}_{n}}\right)\right)\\ \le \phi \left(d\left({x}_{{p}_{n}},{x}_{{p}_{n}-2}\right)\right)+\phi \left(d\left({x}_{{p}_{n}-2},{x}_{{p}_{n}-1}\right)\right)+\epsilon .\end{array}$
Letting n → ∞. Then we get
$\underset{n\to \infty }{\text{lim}}\phi \left(d\left({x}_{{p}_{n}},{x}_{{q}_{n}}\right)\right)=\epsilon .$
On the other hand, we have
$\begin{array}{ll}\hfill \phi \left(d\left({x}_{{p}_{n}},{x}_{{q}_{n}}\right)\right)& \le \phi \left(d\left({x}_{{p}_{n}},{{x}_{{p}_{n}-}}_{\mathsf{\text{1}}}\right)+d\left({{x}_{{p}_{n}-}}_{\mathsf{\text{1}}},{{x}_{{q}_{n}-}}_{\mathsf{\text{1}}}\right)+d\left({{x}_{{q}_{n}-}}_{1},{x}_{{q}_{n}}\right)\right)\phantom{\rule{2em}{0ex}}\\ \le \phi \left(d\left({x}_{{p}_{n}},{x}_{{p}_{n}-1}\right)\right)+\phi \left(d\left({x}_{{p}_{n}-1},{x}_{{q}_{n}-1}\right)\right)+\phi \left(d\left({x}_{{q}_{n}-1},{x}_{{q}_{n}}\right)\right)\phantom{\rule{2em}{0ex}}\end{array}$
and
$\begin{array}{ll}\hfill \phi \left(d\left({x}_{{p}_{n}-1},{x}_{{q}_{n}-1}\right)\right)& \le \phi \left(d\left({x}_{{p}_{n}-1},{x}_{{p}_{n}}\right)+d\left({x}_{{p}_{n}},{x}_{{q}_{n}}\right)+d\left({x}_{{q}_{n}},{x}_{{q}_{n}-1}\right)\right)\phantom{\rule{2em}{0ex}}\\ \le \phi \left(d\left({x}_{{p}_{n}-1},{x}_{{p}_{n}}\right)\right)+\phi \left(d\left({x}_{{p}_{n}},{x}_{{q}_{n}}\right)\right)+\phi \left(d\left({x}_{{q}_{n}},{x}_{{q}_{n}-1}\right)\right).\phantom{\rule{2em}{0ex}}\end{array}$
Letting n → ∞. Then we get
$\underset{n\to \infty }{\text{lim}}\phi \left(d\left({x}_{{p}_{n}-1},{x}_{{q}_{n}-1}\right)\right)=\epsilon .$
Using the inequality (1), we have
Letting n → ∞, by the definitions of the functions ϕ and φ, we have
$\epsilon \le \underset{n\to \infty }{\text{lim}}\varphi \left(\phi \left(d\left({x}_{{p}_{n}-1},{x}_{{q}_{n}-1}\right)\right)\right)<\epsilon .$

So we get a contradiction. Therefore $\underset{n\to \infty }{\text{lim}}\phi \left(d\left({x}_{{p}_{n}},{x}_{{q}_{n}}\right)\right)=0$, by the condition (φ1), we have $\underset{n\to \infty }{\text{lim}}d\left({x}_{{p}_{n}},{x}_{{q}_{n}}\right)=0$. Therefore {x n } is g.m.s Cauchy.

Step 3. We claim that f has a periodic point in X.

Suppose, on contrary, f has no periodic point. Then {x n } is a sequence of distinct points, that is, x p x q for all $p,q\in ℕ$ with pq. By step 2, since X is complete g.m.s, there exists ν X such that x n ν. Using the inequality (1), we have
$\phi \left(d\left(f{x}_{n},f\nu \right)\right)\le \varphi \left(\phi \left(d\left({x}_{n}\mathsf{\text{,}}\nu \right)\right)\right)$
Letting n → ∞, we have
$\phi \left(d\left(f{x}_{n},f\nu \right)\right)\to 0,\phantom{\rule{1em}{0ex}}\mathsf{\text{as}}\phantom{\rule{1em}{0ex}}n\to \infty ,$
by the condition (φ1), we get
$d\left(f{x}_{n},f\nu \right)\to 0,\phantom{\rule{1em}{0ex}}\mathsf{\text{as}}\phantom{\rule{1em}{0ex}}n\to \infty ,$
that is,
${{x}_{n}}_{+\mathsf{\text{1}}}=f{x}_{n}\to f\nu ,\phantom{\rule{1em}{0ex}}\mathsf{\text{as}}\phantom{\rule{1em}{0ex}}n\to \infty .$

As (X, d) is Hausdorff, we have ν = , a contradiction with our assumption that f has no periodic point. Therefore, there exists ν X such that$v={f}^{p}\left(v\right)$ for some $p\in ℕ$. So f has a periodic point in X.   □

Using the functions ψ and φ, we next introduce the notion of the (ψ-φ)-weak contraction mapping and prove a theorem which assures the existence of a periodic point for the (ψ-φ)-weak contraction mapping.

Definition 8 Let (X, d) be a g.m.s, and let f : XX be a function satisfying
$\phi \left(d\left(fx,fy\right)\right)\le \psi \left(\phi \left(d\left(x,y\right)\right)\cdot \phi \left(d\left(x,y\right)$
(4)

for all x, y X. Then f is said to be a (ψ - φ)-weak contraction mapping.

Theorem 2 Let (X, d) be a Hausdorff and complete g.m.s, and let f be a (ψ - φ)-weak contraction mapping. Then f has a periodic point μ in X.

Proof. Given x0 and define a sequence {x n } in X by
${{x}_{n}}_{+\mathsf{\text{1}}}=f{x}_{n}\phantom{\rule{1em}{0ex}}\mathsf{\text{for}}\phantom{\rule{1em}{0ex}}n\in ℕ\cup \left\{0\right\}.$
Step 1. We shall prove that
$\underset{n\to \infty }{\text{lim}}\phi \left(d\left({x}_{n},{x}_{n+1}\right)\right)=0,$
(5)
$\underset{n\to \infty }{\text{lim}}\phi \left(d\left({x}_{n},{x}_{n+2}\right)\right)=0.$
(6)
Taking into account (4) and the definition of stronger Meir-Keeler function ψ, we have that for each $n\in ℕ$
$\begin{array}{c}\phi \left(d\left({x}_{n},{x}_{n}+1\right)\right)=\phi \left(d\left(f{x}_{n-1},f{x}_{n}\right)\right)\\ \le \psi \left(\phi \left(d\left({x}_{n-1},{x}_{n}\right)\right)\cdot \phi \left(d\left({x}_{n-1},{x}_{n}\right)\\ <\phi \left(d\left({x}_{n-1},{x}_{n}\right).\end{array}$
Thus the sequence {φ(d(x n , xn+1))} is descreasing and bounded below and hence it is con-vergent. Let limn → ∞φ(d(x n , xn+1)) = η ≥ 0. Then there exists ${n}_{0}\in ℕ$ and δ > 0 such that for all $n\in ℕ$ with nn0
$\eta \le \phi \left(d\left({x}_{n},{x}_{n+1}\right)\right)<\eta +\delta .$
(7)
Taking into account (7) and the definition of stronger Meir-Keeler function ψ, corresponding to η use, there exists γ η [0, 1) such that
Thus, we can deduce that for each $n\in ℕ$ with nn0 + 1
$\begin{array}{c}\phi \left(d\left({x}_{n},{x}_{n}+1\right)\right)=\phi \left(d\left(f{x}_{n-1},f{x}_{n}\right)\right)\\ \le \psi \left(\phi \left(d\left({x}_{n-1},{x}_{n}\right)\right)\cdot \phi \left(d\left({x}_{n-1},{x}_{n}\right)\\ <{\gamma }_{\eta }\cdot \phi \left(d\left({x}_{n-1},{x}_{n}\right)\right),\end{array}$
and so
$\begin{array}{ll}\hfill \phi \left(d\left({x}_{n},{x}_{n+1}\right)\right)& \le {\gamma }_{\eta }\cdot \phi \left(d\left({x}_{n-1},{x}_{n}\right)\right)\phantom{\rule{2em}{0ex}}\\ \le {\gamma }_{\eta }^{2}\cdot \phi \left(d\left({x}_{n-2},{x}_{{n}_{0}-1}\right)\right)\phantom{\rule{2em}{0ex}}\\ \le \cdots \phantom{\rule{2em}{0ex}}\\ \le {\gamma }_{\eta }^{n-{n}_{0}}\cdot \phi \left(d\left({x}_{{n}_{0}},{x}_{{n}_{0}+1}\right)\right).\phantom{\rule{2em}{0ex}}\end{array}$
Since γ η [0, 1), we get
$\underset{n\to \infty }{\text{lim}}\phi \left(d\left({x}_{n},{x}_{n+1}\right)\right)=0.$
Taking into account (4) and the definition of stronger Meir-Keeler function ψ, we have that for each $n\in ℕ$
$\begin{array}{c}\phi \left(d\left({x}_{n},{x}_{n+2}\right)\right)=\phi \left(d\left(f{x}_{n-1},f{x}_{n+1}\right)\right)\\ \le \psi \left(\phi \left(d\left({x}_{n-1},{x}_{n+1}\right)\right)\cdot \phi \left(d\left({x}_{n-1},{x}_{n+1}\right)\\ <\phi \left(d\left({x}_{n-1},{x}_{n+1}\right).\end{array}$
Thus the sequence {φ(d(x n , xn+2))} is descreasing and bounded below and hence it is convergent. By the same proof process, we also conclude
$\underset{n\to \infty }{\text{lim}}\phi \left(d\left({x}_{n},{x}_{n+2}\right)\right)=0.$

Next, we claim that {x n } is g.m.s Cauchy.

Step 2. Claim that $\underset{n\to \infty }{\text{lim}}\phi \left(d\left({x}_{{p}_{n}},{x}_{{q}_{n}}\right)\right)=0$, that is, for every ε > 0, corresponding to above n0 use, there exists $n\in ℕ$ with nn0 +1 such that if p, qn then φ(d(x p , x q )) < ε.

Suppose the above statement is false. Then there exists ε > 0 such that for any $n\in ℕ$, there are ${p}_{n},{q}_{n}\in ℕ$ with p n > q n nn0 + 1 satisfying
$\phi \left(d\left({x}_{{q}_{n}},{x}_{{p}_{n}}\right)\right)\ge \epsilon .$
Following from Theorem 1, we have that
$\underset{n\to \infty }{\text{lim}}\phi \left(d\left({x}_{{p}_{n}},{x}_{{q}_{n}}\right)\right)=\epsilon .$
and
$\underset{n\to \infty }{\text{lim}}\phi \left(d\left({x}_{{p}_{n}-1},{x}_{{q}_{n}-1}\right)\right)=\epsilon .$
Using the inequality (4), we have
Letting n → ∞, by the definitions of the functions ψ and φ, we have
$\epsilon <\underset{n\to \infty }{\text{lim}}{\gamma }_{\eta }\cdot \phi \left(d\left({x}_{{p}_{n}-1},{x}_{{q}_{n}-1}\right)\right)<{\gamma }_{\eta }\cdot \epsilon <\epsilon .$

So we get a contradiction. Therefore $\underset{n\to \infty }{\text{lim}}\phi \left(d\left({x}_{{p}_{n}},{x}_{{q}_{n}}\right)\right)=0$, by the condition (φ1), we have $\underset{n\to \infty }{\text{lim}}d\left({x}_{{p}_{n}},{x}_{{q}_{n}}\right)=0$. Therefore {x n } is g.m.s Cauchy.

Step 3. We claim that f has a periodic point in X.

Suppose, on contrary, f has no periodic point. Then {x n } is a sequence of distinct points, that is, x p x q for all $p,q\in ℕ$ with pq. By step 2, since X is complete g.m.s, there exists ν X such that x n ν. Using the inequality (4), we have
$\phi \left(d\left(f{x}_{n},f\nu \right)\right)\le \psi \left(\phi \left(d\left({x}_{n},\nu \right)\right)\right)\cdot \phi \left(d\left({x}_{n},\nu \right)\right)$
Letting n → ∞, we have
$\phi \left(d\left(f{x}_{n},f\nu \right)\right)\to 0,\phantom{\rule{1em}{0ex}}\mathsf{\text{as}}\phantom{\rule{1em}{0ex}}n\to \infty ,$
by the condition (φ1), we get
$d\left(f{x}_{n},f\nu \right)\to 0,\phantom{\rule{1em}{0ex}}\mathsf{\text{as}}\phantom{\rule{1em}{0ex}}n\to \infty ,$
that is,
${{x}_{n}}_{+\mathsf{\text{1}}}=f{x}_{n}\to f\nu ,\phantom{\rule{1em}{0ex}}\mathsf{\text{as}}\phantom{\rule{1em}{0ex}}n\to \infty .$

As (X, d) is Hausdorff, we have ν = , a contradiction with our assumption that f has no periodic point. Therefore, there exists ν X such that $v={f}^{p}\left(v\right)$ for some $p\in ℕ$. So f has a periodic point in X.   □

In conclusion, by using the new concepts of (ϕ-φ)-weak contraction mappings and (ψ - φ)-weak contraction mappings, we obtain two theorems (Theorems 1 and 2) which assure the existence of a periodic point for these two types of weak contraction in complete generalized metric spaces. Our results generalize or improve many recent fixed point theorems in the literature.

## Declarations

### Acknowledgements

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

## Authors’ Affiliations

(1)
Department of Applied Mathematics, National Hsinchu University of Education, Hsin-Chu, Taiwan
(2)
Department of Applied Mathematics, Chung Yuan Christian University, Chungli City, Taiwan

## References

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