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# Fixed points for cyclic φ-contractions in generalized metric spaces

## Abstract

In this paper, we obtain a fixed point theorem for mappings satisfying cyclic φ-contractive conditions in complete metric spaces, which gives a positive answer to the question raised by Radenović (Fixed Point Theory Appl. 2015:189, 2015). We also find that this result and the fixed point result satisfying cyclic weak ϕ-contractions given by Karapınar (Appl. Math. Lett. 24:822-825, 2011) are independent of each other. Furthermore, when the number of cyclic sets is odd, we obtain fixed point theorems satisfying cyclic weak ϕ-contractions and cyclic φ-contractions in the setting of generalized metric spaces.

## Introduction and preliminaries

The main purpose of this paper is to answer an open question raised by Radenović in [1]. In order to go further, we attempt to extend our result and the result established by Karapınar [2, 3] to the setting of generalized metric spaces. We show these results are valid in generalized metric spaces when the number of cyclic sets is odd.

Let us recall the definition of a comparison function.

### Definition 1.1

[4]

A function $$\varphi:[0,\infty)\to[0,\infty)$$ is called a comparison function if it satisfies:

(i) φ :

φ is increasing;

(ii) φ :

$$(\varphi^{n}(t))_{n\in\mathbb{{N}}}$$ converges to 0 as $$n\to\infty$$, for all $$t\in(0,\infty)$$.

If the condition (ii) φ is replaced by
(iii) φ :

$$\sum_{k=0}^{\infty}\varphi^{k}(t)<\infty$$, for all $$t\in(0,\infty)$$,

then φ is called a strong comparison function.

It is clear that a strong comparison function is a comparison function, but the converse is not true.

### Example 1.2

Let $$\varphi:[0,\infty)\to[0,\infty)$$ be defined by $$\varphi(t)=\frac {t}{1+t}$$. Then φ is a comparison function, but it is not a strong comparison function. In fact,

$$\varphi^{n}(t)=\frac{t}{1+nt},$$

for all $$t>0$$. Consequently, for every $$t>0$$, $$(\varphi^{n}(t))$$ converges to 0 as $$n\to\infty$$, but $$\sum_{k=0}^{\infty}\varphi^{k}(t)=\infty$$.

Many authors considered fixed point results about cyclic φ-contractions in setting of different type of spaces; see, for example, [111]. Particularly, in [1], Radenović obtained a fixed point theorem for non-cyclic φ-contraction, where φ is comparison function, and raised the following question.

### Question 1.3

Prove or disprove the following.

Let $$\{A_{i}\}_{i}^{p}$$ be nonempty closed subsets of a complete metric space, and suppose $$f: \bigcup_{i=1}^{p} A_{i}\to\bigcup_{i=1}^{p} A_{i}$$ satisfies the following conditions (where $$A_{p+1}=A_{1}$$):

1. (i)

$$f(A_{i})\subset A_{i+1}$$ for $$1\leq i\leq p$$;

2. (ii)

there exists a comparison function $$\varphi:[0,\infty)\to[0,\infty )$$ such that

$$d(fx,fy)\leq\varphi\bigl(d(x,y)\bigr),$$

for any $$x\in A_{i}$$, $$y\in A_{i+1}$$, $$1\leq i\leq p$$.

Then f has a unique fixed point $$x^{*}\in\bigcap_{i=1}^{p} A_{i}$$ and a Picard iteration $$\{x_{n}\}_{n\geq1}$$ given by $$x_{n}=fx_{n-1}$$ converging to $$x^{*}$$ for any starting point $$x_{0} \in\bigcup_{i=1}^{p} A_{i}$$.

In Section 2, we give an answer to Question 1.3. In Section 3, we obtain a fixed point theorem for a mapping satisfying cyclic weak ϕ-contractions and cyclic φ-contractions in complete generalized metric spaces, where the number of cyclic sets is odd.

We start this section by presenting the notion of cyclic φ-contraction.

### Definition 2.1

Let $$(X,d)$$ be a metric space, $$p\in\mathbb{N}$$, $$A_{1},\ldots,A_{p}$$ nonempty subsets of X, and $$Y:=\bigcup_{i=1}^{p} A_{i}$$. An operator $$f: Y\to Y$$ is called a cyclic φ-contraction if:

1. (i)

$$\bigcup_{i=1}^{p} A_{i}$$ is a cyclic representation of Y with respect to f;

2. (ii)

there exists a comparison function $$\varphi:[0,\infty)\to[0,\infty )$$ such that

$$d(fx,fy)\leq\varphi\bigl(d(x,y)\bigr),$$
(2.1)

for any $$x\in A_{i}$$, $$y\in A_{i+1}$$, where $$A_{p+1}=A_{1}$$.

### Theorem 2.2

Let $$(X,d)$$ be a complete metric space, $$p\in\mathbb{N}$$, $$A_{1},\ldots ,A_{p}$$ nonempty closed subsets of X, and $$Y:=\bigcup_{i=1}^{p} A_{i}$$. Assume that $$f: Y\to Y$$ is a cyclic φ-contraction. Then f has a unique fixed point $$x^{*}\in\bigcap_{i=1}^{p} A_{i}$$ and a Picard iteration $$\{x_{n}\}_{n\geq1}$$ given by $$x_{n}=fx_{n-1}$$ converging to $$x^{*}$$ for any starting point $$x_{0} \in\bigcup_{i=1}^{p} A_{i}$$.

### Proof

Let $$x_{0}$$ be an arbitrary point in Y. Define the sequence $$\{x_{n}\}$$ in Y by $$x_{n}=fx_{n-1}$$, $$n=1,2,\ldots$$ . If there exists $$n_{0}$$ such that $$x_{n_{0}+1}=x_{n_{0}}$$ then $$fx_{n_{0}}=x_{n_{0}+1}=x_{n_{0}}$$ and the existence of the fixed point is proved. Consequently, we always assume that $$x_{n}\neq x_{n+1}$$ for all $$n\in\mathbb{N}$$.

Step 1. We will prove that

$$\lim_{n\to\infty}d(x_{n},x_{n+1})=0, \qquad \lim_{n\to\infty}d(x_{n},x_{n+2})=0,\qquad \ldots, \qquad \lim_{n\to\infty}d(x_{n},x_{n+p})=0.$$
(2.2)

Using (2.1), we have

$$d(x_{n},x_{n+1})=d(fx_{n-1},fx_{n}) \leq\varphi\bigl(d(x_{n-1},x_{n})\bigr),$$
(2.3)

for all $$n\in\mathbb{N}$$. From this, we deduce that

$$d(x_{n},x_{n+1})\leq\varphi\bigl(d(x_{n-1},x_{n}) \bigr)\leq\varphi ^{2}\bigl(d(x_{n-2},x_{n-1})\bigr) \leq\cdots\leq\varphi^{n}\bigl(d(x_{0},x_{1}) \bigr).$$

Using the definition of φ, we get

$$\lim_{n\to\infty}d(x_{n},x_{n+1})=0,$$
(2.4)

using the triangle inequality, we have

$$d(x_{n},x_{n+k})\leq d(x_{n},x_{n+1})+d(x_{n+1},x_{n+2})+ \cdots+d(x_{n+k-1},x_{n+k}),$$

for $$k=2,3,\ldots,p$$. Combining this and (2.4), we conclude that (2.2) holds.

Step 2. We will prove the following claim.

### Claim

For every $$\varepsilon>0$$, there exists $$N\in\mathbb{N}$$ such that if $$n>m>N$$ with $$n-m\equiv1 \operatorname{mod} p$$ then $$d(x_{n},x_{m})<\varepsilon$$.

In fact, if the claim is not true, then there exists $$\varepsilon_{0}>0$$ such that for any $$N\in\mathbb{N}$$ we can find $$n>m>N$$ with $$n-m\equiv1 \operatorname{mod} p$$ satisfying $$d(x_{n},x_{m})\geq\varepsilon_{0}$$. By (2.2), corresponding to this $$\varepsilon_{0}$$, there exists $$n_{0}$$ such that if $$n>n_{0}$$ then

$$d(x_{n},x_{n+1})< \varepsilon_{0}, \qquad d(x_{n},x_{n+2})< \varepsilon_{0}, \qquad \ldots,\qquad d(x_{n},x_{n+p})< \varepsilon_{0}.$$
(2.5)

Taking $$N=n_{0}$$, we can find that $$n_{1}'>m_{1}>n_{0}$$ with $$n_{1}'-m_{1}\equiv1 \operatorname{mod} p$$ such that $$d(x_{n_{1}'},x_{m_{1}})\geq\varepsilon_{0}$$. Due to (2.5), we can choose a $$n_{1}\in\{m_{1}+p+1,m_{1}+2p+1,\ldots,n_{1}'\}$$ in such a way that it is smallest integer satisfying $$d(x_{n_{1}},x_{m_{1}})\geq\varepsilon_{0}$$. Then we obtain

$$d(x_{n_{1}},x_{m_{1}})\geq\varepsilon_{0},\qquad d(x_{n_{1}-p},x_{m_{1}})< \varepsilon_{0}\quad \mbox{and} \quad n_{1}-m_{1}\equiv1 \operatorname{mod} p.$$

Taking $$N=n_{1}$$, we can find that $$n_{2}'>m_{2}>n_{1}$$ with $$n_{2}'-m_{2}\equiv1 \operatorname{mod} p$$ such that $$d(x_{n_{2}'},x_{m_{2}})\geq\varepsilon_{0}$$. Similar to the choice of $$n_{1}$$, we can get a $$n_{2}\in\{m_{2}+p+1,m_{2}+2p+1,\ldots,n_{2}'\}$$ such that

$$d(x_{n_{2}},x_{m_{2}})\geq\varepsilon_{0},\qquad d(x_{n_{2}-p},x_{m_{2}})< \varepsilon_{0}\quad \mbox{and} \quad n_{2}-m_{2}\equiv1 \operatorname{mod} p.$$

Continuing the above process, by induction, we obtain two subsequences $$\{x_{m_{k}}\}$$ and $$\{x_{n_{k}}\}$$ of $$\{x_{n}\}$$ such that

$$d(x_{n_{k}},x_{m_{k}})\geq\varepsilon_{0}, \qquad d(x_{n_{k}-p},x_{m_{k}})< \varepsilon _{0} \quad \mbox{and}\quad n_{k}-m_{k}\equiv1 \operatorname{mod} p.$$
(2.6)

Now, using (2.6) and the triangle inequality, we have

\begin{aligned} \varepsilon_{0} \leq& d(x_{n_{k}},x_{m_{k}}) \leq d(x_{n_{k}},x_{n_{k}-p})+d(x_{n_{k}-p},x_{m_{k}}) \\ \leq& d(x_{n_{k}},x_{n_{k}-p})+\varepsilon_{0}. \end{aligned}

Letting $$k\to\infty$$ in the above inequality, using (2.2), we obtain

$$d(x_{n_{k}},x_{m_{k}})\to\varepsilon_{0} \quad \mbox{as } k\to\infty.$$
(2.7)

Using the triangle inequality, we get

$$d(x_{n_{k}-p+1},x_{m_{k}+1})\leq d(x_{n_{k}-p+1},x_{n_{k}})+d(x_{n_{k}},x_{m_{k}})+d(x_{m_{k}},x_{m_{k}+1})$$

and

$$d(x_{n_{k}-p+1},x_{m_{k}+1})\geq d(x_{n_{k}},x_{m_{k}})-d(x_{n_{k}},x_{n_{k}-p+1})-d(x_{m_{k}+1},x_{m_{k}}).$$

Letting $$k\to\infty$$ in the above two inequalities, using (2.2) and (2.7), we get

$$d(x_{n_{k}-p+1},x_{m_{k}+1})\to\varepsilon_{0} \quad \mbox{as } k\to\infty.$$
(2.8)

Now, using (2.1) and (2.6), we have

$$d(x_{n_{k}-p+1},x_{m_{k}+1})= d(fx_{n_{k}-p},fx_{m_{k}}) \leq\varphi \bigl(d(x_{n_{k}-p},x_{m_{k}})\bigr)\leq\varphi( \varepsilon_{0}).$$
(2.9)

Taking the limit in (2.9) as $$k\to\infty$$, from (2.8), we see

$$\varepsilon_{0}\leq\varphi(\varepsilon_{0}),$$

which is a contradiction with $$\varphi(\varepsilon_{0})<\varepsilon_{0}$$. Therefore our claim is proved.

Step 3. We will prove $$\{x_{n}\}$$ is a Cauchy sequence in X.

Let $$\varepsilon>0$$ be given. Using the claim, we find that $$N_{1}\in \mathbb{N}$$ such that if $$n>m>N_{1}$$ with $$n-m\equiv1 \operatorname{mod} p$$ then

$$d(x_{n},x_{m})< \frac{\varepsilon}{p}.$$

On the other hand, using (2.4), we also find $$N_{2}\in\mathbb{N}$$ such that, for any $$n>N_{2}$$,

$$d(x_{n},x_{n+1})< \frac{\varepsilon}{p}.$$

Let $$n,m>N=\max\{N_{1},N_{2}\}$$ with $$n>m$$. Then we can find a $$s\in\{ 0,1,2,\ldots,p-1\}$$ such that $$n-(m+s)\equiv1 \operatorname{mod} p$$. Using the triangle inequality, we obtain

\begin{aligned} d(x_{m},x_{n}) \leq& d(x_{m},x_{m+1})+d(x_{m+1},x_{m+2})+ \cdots +d(x_{m+s},x_{n}) \\ < & \frac{\varepsilon}{p}+\frac{\varepsilon}{p}+\cdots+\frac {\varepsilon}{p} \\ =& (s+1)\cdot\frac{\varepsilon}{p}\leq\varepsilon. \end{aligned}

This proves that $$\{x_{n}\}$$ is a Cauchy sequence.

Step 4. We will prove f has a unique fixed point $$x^{*}\in \bigcap_{i=1}^{p} A_{i}$$.

As X is a complete metric space, there exists $$x\in X$$ such that $$\lim_{n\to\infty} x_{n}=x$$. Using the cyclic character of f, there exists a subsequence of $$\{x_{n}\}$$ for which belongs to $$A_{i}$$ for $$i\in\{1,2,\ldots,p\}$$. Hence, from $$A_{i}$$ is closed, we see that $$x\in\bigcap_{i=1}^{p} A_{i}$$. Now, we consider the restriction $$f|_{\bigcap_{i=1}^{p} A_{i}}$$ of f on $$\bigcap_{i=1}^{p} A_{i}$$. Since $$\bigcap_{i=1}^{p} A_{i}$$ is also complete, by Theorem 2.3 in [1], we see that f has a unique fixed point $$x^{*}$$ in $$\bigcap_{i=1}^{p} A_{i}$$.

Step 5. We prove that the Picard iteration converges to $$x^{*}$$ for any initial point $$x_{0}\in\bigcup_{i=1}^{p} A_{i}$$.

Using (2.1), we have

$$d\bigl(x_{n},x^{*}\bigr)=d\bigl(fx_{n-1},fx^{*}\bigr)\leq\varphi \bigl(d\bigl(x_{n-1},x^{*}\bigr)\bigr).$$

From this, we see that

$$d\bigl(x_{n},x^{*}\bigr)\leq\varphi\bigl(d\bigl(x_{n-1},x^{*} \bigr)\bigr)\leq\varphi^{2}\bigl(d\bigl(x_{n-2},x^{*}\bigr)\bigr) \leq \cdots\leq\varphi^{n}\bigl(d\bigl(x_{0},x^{*}\bigr)\bigr).$$

Using the definition of φ, we conclude that $$x_{n}\to x^{*}$$ as $$n\to\infty$$.

This completes the proof.  □

### Remark 2.3

From Theorem 2.2, we see that the open question raised by Radenović (that is, Question 1.3) has been answered.

### Remark 2.4

Following the idea of Radenović in [1], we see that Theorem 2.3 in [1] and Theorem 2.2 are equivalent.

## Cyclic weak ϕ-contractions and cyclic φ-contractions in generalized metric spaces

In 2000, Branciari [12] introduced the notion of generalized metric space and proved the Banach fixed point theorem in such spaces. For more information, the reader can refer to [1317]. For some notions and facts about generalized metric spaces, one may wish to see [12].

In [3], Karapınar gave a fixed point results satisfying cyclic weak ϕ-contractions. For convenience, we rewrite his theorem (i.e., [3], Theorem 2) as the following equivalent statement.

### Theorem 3.1

Let $$(X,d)$$ be a complete metric space, $$p\in\mathbb{N}$$, $$A_{1},\ldots ,A_{p}$$ closed nonempty subsets of X, $$Y:=\bigcup_{i=1}^{p} A_{i}$$ and $$f: Y\to Y$$ an operator. Assume that:

1. (i)

$$\bigcup_{i=1}^{p} A_{i}$$ is a cyclic representation of Y with respect to f;

2. (ii)

there exists a function $$\phi:[0,\infty)\to[0,\infty)$$ with $$\phi (t)< t$$ and $$t-\phi(t)$$ is nondecreasing for $$t\in(0,\infty)$$ and $$\phi(0)=0$$ such that

$$d(fx,fy)\leq\phi\bigl(d(x,y)\bigr),$$

for any $$x\in A_{i}$$, $$y\in A_{i+1}$$, where $$A_{p+1}=A_{1}$$.

Then f has a unique fixed point $$x^{*}\in\bigcap_{i=1}^{p} A_{i}$$.

Based on the concept of cyclic weak ϕ-contraction, we can introduce the following notion.

### Definition 3.2

A function $$\phi:[0,\infty)\to[0,\infty)$$ is called a (w)-comparison function if it satisfies:

(i) ϕ :

$$\phi(0)=0$$;

(ii) ϕ :

$$\phi(t)< t$$, for all $$t\in(0,\infty)$$;

(iii) ϕ :

the function $$\psi(t):=t-\phi(t)$$ is increasing, i.e., $$t_{1}\leq t_{2}$$ implies $$\psi(t_{1})\leq\psi(t_{2})$$, for $$t_{1},t_{2}\in [0,\infty)$$.

### Lemma 3.3

If $$\phi:[0,\infty)\to[0,\infty)$$ is a $$(w)$$-comparison function, then the following hold:

1. (1)

$$\phi(t)\leq t$$, for any $$t\in[0,\infty)$$;

2. (2)

for $$k\geq1$$, $$\phi^{k}(t)< t$$, for any $$t\in(0,\infty)$$;

3. (3)

$$(\phi^{n}(t))_{n\in\mathbb{{N}}}$$ converges to 0 as $$n\to\infty$$, for all $$t\in(0,\infty)$$.

### Proof

From the definition of ϕ, it is easy to verify that (1), (2), and (3) hold. Now, we only prove that (3) holds. Let $$t\in (0,\infty)$$. Then we have

$$\phi^{n}(t)=\phi\bigl(\phi^{n-1}(t)\bigr)\leq \phi^{n-1}(t),\quad \mbox{for all }n\in \mathbb{N}.$$

This means that $$(\phi^{n}(t))_{n\in\mathbb{{N}}}$$ is a decreasing sequence of non-negative real numbers. Therefore, there exists $$r\geq0$$ such that $$\lim_{n\to\infty}\phi^{n}(t)=r$$. Suppose that $$r>0$$. Then $$\phi(r)< r$$ and $$r-\phi(r)>0$$. Since $$r=\inf\{ \phi^{n}(t): n\in\mathbb{N}\}$$, $$0< r\leq\phi^{n}(t)$$, for all $$n\in\mathbb{N}$$. By the definition of ϕ, we get

$$r-\phi(r)\leq\phi^{n}(t)-\phi\bigl(\phi^{n}(t)\bigr),$$

for all $$n\in\mathbb{N}$$. Letting $$n\to\infty$$ in the above inequality, we obtain $$r-\phi(r)\leq r-r=0$$ and this contradicts $$r-\phi(r)>0$$. □

The next are two basic examples of the comparison function and the $$(w)$$-comparison function.

### Example 3.4

Let $$\phi:[0,\infty)\to[0,\infty)$$ be defined by

$$\phi(t)=\left \{ \textstyle\begin{array}{l@{\quad}l} \frac{1}{4}t, & 0\leq t< 1, \\ \frac{5}{4}t-1, & 1\leq t< 2, \\ t-\frac{1}{2}, & t\geq2. \end{array}\displaystyle \right .$$

Then ϕ is a comparison function. But ϕ is not a $$(w)$$-comparison function because $$t-\phi(t)$$ is not increasing.

### Example 3.5

Let $$\phi:[0,\infty)\to[0,\infty)$$ be defined by

$$\phi(t)=\left \{ \textstyle\begin{array}{l@{\quad}l} \frac{3}{4}t, & 0\leq t< 1 , \\ 1-\frac{1}{4}t, & 1\leq t< 2, \\ \frac{1}{4}t, & t\geq2. \end{array}\displaystyle \right .$$

Then ϕ is a $$(w)$$-comparison function. But ϕ is not a comparison function because $$\phi(t)$$ is not increasing.

### Remark 3.6

From Example 3.4 and Example 3.5, we see that the comparison function and the $$(w)$$-comparison function do not imply each other. Consequently, Theorem 2 in [3] and Theorem 2.2 are independent of each other.

Now we carry over the concept of cyclic weak ϕ-contraction to generalized metric space.

### Definition 3.7

Let $$(X,d)$$ be a generalized metric space, $$p\in\mathbb{N}$$, $$A_{1},\ldots ,A_{p}$$ nonempty subsets of X and $$Y:=\bigcup_{i=1}^{p} A_{i}$$. An operator $$f: Y\to Y$$ is called a cyclic weak ϕ-contraction if:

1. (i)

$$\bigcup_{i=1}^{p} A_{i}$$ is a cyclic representation of Y with respect to f;

2. (ii)

there exists a $$(w)$$-comparison function $$\phi:[0,\infty)\to[0,\infty )$$ such that

$$d(fx,fy)\leq\phi\bigl(d(x,y)\bigr),$$
(3.1)

for any $$x\in A_{i}$$, $$y\in A_{i+1}$$, where $$A_{p+1}=A_{1}$$.

### Theorem 3.8

Let $$(X,d)$$ be a complete generalized metric space, p an odd number, $$A_{1},\ldots,A_{p}$$ nonempty closed subsets of X and $$Y:=\bigcup_{i=1}^{p} A_{i}$$. Assume that $$f: Y\to Y$$ is a cyclic weak ϕ-contraction. Then f has a unique fixed point $$x^{*}\in\bigcap_{i=1}^{p} A_{i}$$ and a Picard iteration $$\{x_{n}\}_{n\geq1}$$ given by $$x_{n}=fx_{n-1}$$ converging to $$x^{*}$$ for any starting point $$x_{0} \in\bigcup_{i=1}^{p} A_{i}$$.

### Proof

Let $$x_{0}\in Y$$, and $$x_{n}=fx_{n-1}$$, $$n=1,2,\ldots$$ . If there exists $$n_{0}$$ such that $$x_{n_{0}+1}=x_{n_{0}}$$ then $$fx_{n_{0}}=x_{n_{0}+1}=x_{n_{0}}$$ and the existence of the fixed point is proved. Consequently, we will assume that $$x_{n}\neq x_{n+1}$$ for all $$n\in\mathbb{N}$$.

Step 1. We will prove that $$x_{n}\neq x_{m}$$ for all $$n\neq m$$.

Suppose that $$x_{n}= x_{m}$$ for some $$n\neq m$$. Without loss of generality, we may assume that $$n>m+1$$. Due to the property of ϕ, we see that

\begin{aligned} d(x_{m},x_{m+1}) =& d(x_{m},fx_{m})=d(x_{n},fx_{n}) \\ =& d(fx_{n-1},fx_{n}) \\ \leq& \phi\bigl(d(x_{n-1},x_{n})\bigr) \\ \leq& \cdots \\ \leq& \phi^{n-m}\bigl(d(x_{m},x_{m+1})\bigr). \end{aligned}

By Lemma 3.3(2), we get $$\phi ^{n-m}(d(x_{m},x_{m+1}))< d(x_{m},x_{m+1})$$, which is a contradiction.

Step 2. We will prove that

$$\lim_{n\to\infty}d(x_{n},x_{n+1})=0, \qquad \lim_{n\to\infty }d(x_{n},x_{n+2})=0,\qquad \ldots,\qquad \lim_{n\to\infty}d(x_{n},x_{n+p})=0.$$
(3.2)

Using (3.1), we get

$$d(x_{n},x_{n+1})=d(fx_{n-1},fx_{n})\leq\phi \bigl(d(x_{n-1},x_{n})\bigr),$$

for all $$n\in\mathbb{N}$$. Using the definition of ϕ, we see that

$$d(x_{n},x_{n+1})< d(x_{n-1},x_{n}).$$
(3.3)

This implies the sequence $$\{d(x_{n},x_{n+1})\}$$ is decreasing and bounded below. Consequently, $$d(x_{n},x_{n+1})\to r$$ for some $$r\geq0$$. Suppose that $$r>0$$. Then $$\phi(r)< r$$. Using the definition of ϕ and $$d(x_{n},x_{n+1})\geq r$$, we get

$$r-\phi(r)\leq d(x_{n},x_{n+1})-\phi\bigl(d(x_{n},x_{n+1}) \bigr),$$

for all $$n\in\mathbb{N}$$. From $$d(x_{n+1},x_{n+2})\leq\phi(d(x_{n},x_{n-1}))$$, we see that

$$r-\phi(r)\leq d(x_{n},x_{n+1})-d(x_{n+1},x_{n+2}),$$

for all $$n\in\mathbb{N}$$. Letting $$n\to\infty$$ in the above inequality, we get $$r-\phi(r)\leq0$$, which is a contradiction with $$\phi(r)< r$$. Thus, we conclude that

$$\lim_{n\to\infty}d(x_{n},x_{n+1})=0.$$
(3.4)

Using the rectangular inequality, we get

$$d(x_{n},x_{n+3}) \leq d(x_{n},x_{n+1})+d(x_{n+1},x_{n+2})+d(x_{n+2},x_{n+3}) .$$

From (3.4), we see that $$d(x_{n},x_{n+3})\to0$$ as $$n\to\infty$$. By induction, we deduce that

$$\lim_{n\to\infty}d(x_{n},x_{n+k})=0, \quad \mbox{for all } k\in\{1,3,5,\ldots ,p\}.$$
(3.5)

Now, we prove

$$\lim_{n\to\infty}d(x_{n},x_{n+p-1})=0.$$
(3.6)

Since $$x_{n}$$ and $$x_{n+p-1}$$ lie in different adjacently labeled sets $$A_{i}$$ and $$A_{i+1}$$ for certain $$i\in\{1,2,\ldots,p\}$$, from (3.1) we get

$$d(x_{n},x_{n+p-1}) = d(fx_{n-1},fx_{n+p-2}) \leq \phi\bigl(d(x_{n-1},x_{n+p-2})\bigr).$$

Similar to the proof of the conclusion (3.4), we can deduce that $$\{d(x_{n},x_{n+p-1})\}$$ is decreasing and converges to 0. This means that (3.6) holds.

For $$k=2,4,\ldots,p-3$$, using the rectangular inequality, we have

$$d(x_{n},x_{n+k}) \leq d(x_{n},x_{n+p-1})+d(x_{n+p-1},x_{n+p})+d(x_{n+p},x_{n+k}).$$
(3.7)

Since $$p-k$$ is odd, from (3.5) we get

$$\lim_{n\to\infty}d(x_{n+p},x_{n+k})= \lim_{n\to\infty}d(x_{n+p-k},x_{n})=0.$$
(3.8)

Therefore, from (3.5), (3.6), (3.7), and (3.8) we conclude that

$$\lim_{n\to\infty}d(x_{n},x_{n+k})=0, \quad \mbox{for all } k\in\{2,4,\ldots ,p-1\}.$$
(3.9)

Combining (3.4) and (3.9), we see (3.2) is proved.

Step 3. We will prove the following claim.

### Claim

For every $$\varepsilon>0$$, there exists $$N\in\mathbb{N}$$ such that if $$n>m>N$$ with $$n-m\equiv1 \operatorname{mod} p$$ then $$d(x_{n},x_{m})<\varepsilon$$.

In fact, if this is not true, then there exists $$\varepsilon_{0}>0$$ such that for any $$N\in\mathbb{N}$$ we can find $$n>m>N$$ with $$n-m\equiv1 \operatorname{mod} p$$ satisfying $$d(x_{n},x_{m})\geq\varepsilon_{0}$$. By the definition of ϕ, we get

$$\varepsilon_{0}-\phi(\varepsilon_{0})\leq d(x_{n},x_{m})-\phi\bigl(d(x_{n},x_{m}) \bigr).$$
(3.10)

Using (3.1), we get

$$d(x_{n+1},x_{m+1})\leq\phi\bigl(d(x_{n},x_{m}) \bigr).$$
(3.11)

By (3.10), (3.11), and the rectangular inequality, we obtain

\begin{aligned} \varepsilon_{0}-\phi(\varepsilon_{0}) \leq& d(x_{n},x_{m})-d(x_{n+1},x_{m+1}) \\ \leq& d(x_{n},x_{n+1})+d(x_{n+1},x_{m+1})+d(x_{m+1},x_{m})-d(x_{n+1},x_{m+1}) \\ =& d(x_{n},x_{n+1}) +d(x_{m+1},x_{m}). \end{aligned}

From (3.3), it follows that

$$\varepsilon_{0}-\phi(\varepsilon_{0})\leq2d(x_{m+1},x_{m}) \quad \mbox{and}\quad d(x_{m+1},x_{m})\geq\frac{\varepsilon_{0}-\phi(\varepsilon_{0})}{2}>0.$$

Therefore, $$\{d(x_{m+1},x_{m})\}$$ does not converge to 0 as $$m\to\infty$$, which contradicts (3.4).

Step 4. We will prove $$\{x_{n}\}$$ is a Cauchy sequence in X.

Let $$\varepsilon>0$$ be given. Using the claim, we find that $$N_{1}\in \mathbb{N}$$ such that if $$n>m>N_{1}$$ with $$n-m\equiv1 \operatorname{mod} p$$ then

$$d(x_{n},x_{m})< \frac{\varepsilon}{3}.$$

On the other hand, using (3.2), we also find $$N_{2}\in\mathbb{N}$$ such that, for any $$n>N_{2}$$,

$$d(x_{n},x_{n+1})< \frac{\varepsilon}{3},\qquad d(x_{n},x_{n+2})< \frac{\varepsilon }{3}, \qquad \ldots,\qquad d(x_{n},x_{n+p})< \frac{\varepsilon}{3}.$$

Let $$n,m>N=\max\{N_{1},N_{2}\}+1$$ with $$n>m$$. Then we can find $$s\in\{ 0,1,2,\ldots,p-1\}$$ such that $$n-(m+s)\equiv1 \operatorname{mod} p$$.

In the case where $$s=0$$, we have

$$d(x_{n},x_{m})< \frac{\varepsilon}{3}< \varepsilon.$$

In the other case where $$s\geq1$$, using the rectangular inequality we have

\begin{aligned} d(x_{m},x_{n}) \leq& d(x_{m},x_{m-1})+d(x_{m-1},x_{m+s})+d(x_{m+s},x_{n}) \\ < & \frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3} \\ =& \varepsilon. \end{aligned}

This proves that $$\{x_{n}\}$$ is a Cauchy sequence.

Step 5. We will prove f has a unique fixed point $$x^{*}\in \bigcap_{i=1}^{p} A_{i}$$ and the Picard iteration $$\{x_{n}\}$$ converges to $$x^{*}$$.

Since X is a complete generalized metric space, there exists $$x^{*}\in X$$ such that $$\lim_{n\to\infty} x_{n}=x^{*}$$. Using the cyclic character of f, there exists a subsequence of $$\{x_{n}\}$$ for which belongs to $$A_{i}$$ for $$i\in\{1,2,\ldots,p\}$$. Hence, from $$A_{i}$$ is closed for $$i\in\{1,2,\ldots,p\}$$, we see that $$x^{*}\in\bigcap_{i=1}^{p} A_{i}$$. Now, we will prove $$d(x_{n},fx^{*})\to0$$ as $$n\to\infty$$. In fact, using (3.1), we have

$$d\bigl(x_{n},fx^{*}\bigr)=d\bigl(fx_{n-1},fx^{*}\bigr)\leq\phi \bigl(d\bigl(x_{n-1},x^{*}\bigr)\bigr)\leq d\bigl(x_{n-1},x^{*}\bigr) \to0\quad \mbox{as } n\to\infty,$$

which implies $$d(x_{n},fx^{*})\to0$$ as $$n\to\infty$$. Using Proposition 3 of [18], we deduce that $$fx^{*}=x^{*}$$, i.e., $$x^{*}$$ is a fixed point of f.

In order to prove that the uniqueness of the fixed point, we take $$y,z\in Y$$ such that y and z are fixed points of f. The cyclic character of f implies that $$y,z\in\bigcap_{i=1}^{p}A_{i}$$. Using (3.1),

$$d(y,z)=d(fy,fz)\leq\phi\bigl(d(y,z)\bigr)\leq d(y,z).$$

This means $$\phi(d(y,z))=d(y,z)$$. Since $$\phi(t)>0$$ for $$t>0$$, we get $$d(y,z)=0$$ and $$y=z$$. This finishes the proof.  □

### Theorem 3.9

Let $$(X,d)$$ be a complete generalized metric space, p an odd number, $$A_{1},\ldots,A_{p}$$ nonempty closed subsets of X and $$Y:=\bigcup_{i=1}^{p} A_{i}$$. Assume that $$f: Y\to Y$$ is a cyclic φ-contraction. Then f has a unique fixed point $$x^{*}\in\bigcap_{i=1}^{p} A_{i}$$ and a Picard iteration $$\{x_{n}\}_{n\geq1}$$ given by $$x_{n}=fx_{n-1}$$ converging to $$x^{*}$$ for any starting point $$x_{0} \in\bigcup_{i=1}^{p} A_{i}$$.

### Proof

Let $$x_{0}\in Y$$, and $$x_{n}=fx_{n-1}$$, $$n=1,2,\ldots$$ .

Similar to the Step 1 and Step 2 in the proof of Theorem 3.8, we can prove

$$x_{n}\neq x_{m},\quad \mbox{for all }n\neq m$$

and

$$\lim_{n\to\infty}d(x_{n},x_{n+1})=0, \qquad \lim_{n\to\infty }d(x_{n},x_{n+2})=0,\qquad \ldots,\qquad \lim_{n\to\infty}d(x_{n},x_{n+p})=0.$$
(3.12)

Now, we will prove the following claim.

### Claim

For every $$\varepsilon>0$$, there exists $$N\in\mathbb{N}$$ such that if $$n>m>N$$ with $$n-m\equiv1 \operatorname{mod} p$$ then $$d(x_{n},x_{m})<\varepsilon$$.

In fact, in the opposite case, similar to the Step 2 in the proof of Theorem 2.2, we can find that $$\varepsilon_{0}>0$$ and two subsequences $$\{x_{m_{k}}\}$$ and $$\{x_{n_{k}}\}$$ of $$\{x_{n}\}$$ such that

$$d(x_{n_{k}},x_{m_{k}})\geq\varepsilon_{0}, \qquad d(x_{n_{k}-p},x_{m_{k}})< \varepsilon _{0} \quad \mbox{and} \quad n_{k}-m_{k}\equiv1 \operatorname{mod} p.$$
(3.13)

Next, we only prove $$d(x_{n_{k}},x_{m_{k}})\to\varepsilon_{0}$$ as $$k\to \infty$$ because the other proof is the same as in the Step 2 of Theorem 2.2. In fact, using (3.13) and the rectangular inequality, we have

\begin{aligned} \varepsilon_{0} \leq& d(x_{n_{k}},x_{m_{k}}) \leq d(x_{n_{k}},x_{n_{k}-p+1})+d(x_{n_{k}-p+1},x_{n_{k}-p})+d(x_{n_{k}-p},x_{m_{k}}) \\ \leq& d(x_{n_{k}},x_{n_{k}-p+1})+d(x_{n_{k}-p+1},x_{n_{k}-p})+ \varepsilon_{0}. \end{aligned}

Letting $$k\to\infty$$ in the above inequality, using (3.12), we obtain

$$d(x_{n_{k}},x_{m_{k}})\to\varepsilon_{0}\quad \mbox{as } k\to\infty.$$

Similar to Step 4 and Step 5 in the proof of Theorem 3.8, we can finish the proof.  □

### Example 3.10

Let $$X=\{1,2,3,4,5\}$$. Define $$d:X\times X\to\infty$$ by

$$d(x,y)= \left \{ \textstyle\begin{array}{l@{\quad}l} 0, & x=y, \\ 1, & |x-y|=1, \\ 4, & x=3,y=4 \mbox{ or }x=4,y=3, \\ 2, & \mbox{otherwise}. \end{array}\displaystyle \right .$$

Then $$(X,d)$$ is a generalized metric space. But d is not a metric on X because

$$d(3,4)=4>3=d(3,2)+d(2,4).$$

Now, consider $$A_{1}=\{1,2,3\}$$, $$A_{2}=\{3\}$$, $$A_{3}=\{3,4,5\}$$, and $$T : X\to X$$ to be defined by

$$T1=T2=T3=3\quad \mbox{and} \quad T4=T5=2.$$

It is easy to prove that T satisfies all the conditions of Theorem 3.8 and Theorem 3.9 with $$\varphi(t)=\phi(t)=\frac{1}{2}t$$. Using Theorem 3.8 or Theorem 3.9, we see that T has a unique fixed point. In fact, 3 is the unique fixed point of f. But we do not apply Theorem 2.2 or Theorem 2 in [3] because d is not a metric on X.

Finally, a natural question arises.

### Question 3.11

If the number of cyclic sets is even, then we may ask whether Theorem 3.8 or Theorem 3.9 is valid or not.

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## Acknowledgements

The first author is supported by the National Natural Science Foundation of China (11471236, 11570049). The second author is supported by the National Natural Science Foundation of China (11371185).

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Correspondence to Fei He.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors contributed equally and significantly in writing this paper. All authors read and approved the final manuscript.

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He, F., Chen, A. Fixed points for cyclic φ-contractions in generalized metric spaces. Fixed Point Theory Appl 2016, 67 (2016). https://doi.org/10.1186/s13663-016-0558-8

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### Keywords

• fixed point
• comparison function
• generalized metric space
• cyclic φ-contraction
• cyclic weak ϕ-contraction