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.

1 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, [1–11]. 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 $A_{p+1}=A_{1}$):

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

• (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.

2 Answer of Question 1.3

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:

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

• (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.

3 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 [13–17]. 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:

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

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

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

• (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:

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

• (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.