Common fixed point theorems for four mappings satisfying ψ-weakly contractive conditions

Abstract

The existence and uniqueness of common fixed points for four mappings satisfying ψ- and $(\psi, \varphi)$-weakly contractive conditions in metric spaces are proved. Four examples are given to demonstrate that the results presented in this paper generalize indeed some well-known results in the literature.

1 Introduction and preliminaries

In 2001, Rhoades [1] introduced the concept of φ-weakly contractive mappings and proved the following fixed point theorem, which is a generalization of the Banach fixed point theorem.

Theorem 1.1

([1])

Let $(X, d)$ be a complete metric space, and let $T: X\to X$ be a mapping such that

$$d(Tx,Ty)\leq d(x,y)-\varphi\bigl(d(x,y)\bigr), \quad\forall x,y\in X, $$

where $\varphi:\Bbb{R}^{+}\to\Bbb{R}^{+}$ is continuous and nondecreasing, and $\varphi(t)=0$ if and only if $t=0$. Then T has a unique fixed point.

Afterwards, the researchers [2–8] continued the study of Rhoades by introducing a few φ- and $(\psi, \varphi)$-weakly contractive conditions relative to one, two or three mappings and discussed the existence of fixed and common fixed point for these mappings. In particular, Abbas and Dorić [2], Abbas and Khan [3], and Dutta and Choudhury [5] proved the following fixed and common fixed point theorems for the φ- and $(\psi, \varphi)$-weakly contractive mappings.

Theorem 1.2

([5])

Let $(X, d)$ be a complete metric space, and let $T: X\to X$ be a mapping satisfying the inequality

$$\psi\bigl(d(Tx,Ty)\bigr)\leq\psi\bigl(d(x,y)\bigr)-\varphi\bigl(d(x,y)\bigr),\quad \forall x,y\in X, $$

where $\psi,\varphi:\Bbb{R}^{+}\to\Bbb{R}^{+}$ are both continuous and monotone nondecreasing functions with $\psi(t)=\varphi(t)=0$ if and only if $t=0$. Then T has a unique fixed point.

Theorem 1.3

([3])

Let T, S be two self mappings in a metric space $(X, d)$ satisfying

$$\psi\bigl(d(Tx,Ty)\bigr)\leq\psi\bigl(d(Sx,Sy)\bigr)-\varphi\bigl(d(Sx,Sy) \bigr),\quad \forall x,y\in X, $$

where $\psi,\varphi:\Bbb{R}^{+}\to\Bbb{R}^{+}$ are both continuous and monotone nondecreasing functions with $\psi(t)=\varphi(t)=0$ if and only if $t=0$. If range of S contains the range of T and $S(X)$ is a complete subspace of X, then T and S have a unique point of coincidence in X. Moreover, if T and S are weakly compatible, then T and S have a unique common fixed point.

Theorem 1.4

([2])

Suppose that A, B, S, and T are selfmaps of a complete metric space $(X, d)$, $T(X)\subseteq B(X)$, $S(X)\subseteq A(X)$ and the pairs $\{A,T\}$ and $\{B,S\}$ are weakly compatible. If

$$\psi\bigl(d(Tx,Sy)\bigr)\leq\psi\bigl(M(x,y)\bigr)-\varphi\bigl(M(x,y)\bigr), \quad\forall x,y\in X, $$

where

$$M(x,y)=\max \biggl\{ d(Ax,By),d(Ax,Tx),d(By,Sy),\frac {1}{2} \bigl[d(Ax,Sy)+d(Tx,By)\bigr] \biggr\} , \quad\forall x,y\in X, $$

$\varphi:\Bbb{R}^{+}\to\Bbb{R}^{+}$ is lower semi-continuous, $\varphi(0)=0$, $\varphi(t)>0$ for all $t>0$, $\psi:\Bbb{R}^{+}\to\Bbb{R}^{+}$ is continuous and nondecreasing with $\psi(t)=0$ if and only if $t=0$, then A, B, S and T have a unique common fixed point in X provided one of the ranges of $A(X)$, $B(X)$, $S(X)$ and $T(X)$ is closed.

Motivated by the results in [1–9], in this paper, we introduce the concepts of ψ- and $(\psi, \varphi)$-weakly contractive conditions relative to four mappings A, B, S and T:

$$\begin{aligned}& d(Tx,Sy)\leq\psi\bigl(M_{i}(x,y)\bigr),\quad \forall x,y\in X, \end{aligned}$$

(1.1)

$$\begin{aligned}& \psi\bigl(d(Tx,Sy)\bigr)\leq\psi\bigl(M_{i}(x,y)\bigr)- \varphi\bigl(M_{i}(x,y)\bigr), \quad\forall x,y\in X, \end{aligned}$$

(1.2)

where $i\in\{1,2,3\}$, $\psi\in\Phi_{3}$, $(\psi,\varphi)\in\Phi_{1}\times\Phi_{2}$, respectively,

$$\begin{aligned}& M_{1}(x,y)=\max \biggl\{ d(Ax,By),d(Ax,Tx),d(By,Sy), \\& \hphantom{M_{1}(x,y)=}\frac{1}{2}\bigl[d(Ax,Sy)+d(Tx,By)\bigr],\frac {d(Ax,Sy)d(Tx,By)}{1+d(Ax,By)}, \\& \hphantom{M_{1}(x,y)=}\frac{d(Ax,Tx)d(By,Sy)}{1+d(Ax,By)}, \frac{1+d(Ax,Sy)+d(Tx,By)}{1+d(Ax,Tx)+d(By,Sy)}d(Ax,Tx) \biggr\} ,\quad \forall x,y\in X, \end{aligned}$$

(1.3)

$$\begin{aligned}& M_{2}(x,y)=\max \biggl\{ d(Ax,By),d(Ax,Tx),d(By,Sy), \frac{1}{2}\bigl[d(Ax,Sy)+d(Tx,By)\bigr], \\& \hphantom{M_{2}(x,y)=}\frac {1+d(Ax,Tx)}{1+d(Ax,By)}d(By,Sy),\frac{1+d(By,Sy)}{1+d(Ax,By)}d(Ax,Tx), \\& \hphantom{M_{2}(x,y)=}\frac {1+d(Ax,Sy)+d(Tx,By)}{1+d(Ax,Tx)+d(By,Sy)}d(By,Sy) \biggr\} ,\quad \forall x,y\in X \end{aligned}$$

(1.4)

and

$$\begin{aligned}& M_{3}(x,y)=\max \biggl\{ d(Ax,By),d(Ax,Tx),d(By,Sy), \frac {1}{2}\bigl[d(Ax,Sy)+d(Tx,By)\bigr] \biggr\} , \\ & \quad \forall x,y\in X \end{aligned}$$

(1.5)

and establish sufficient conditions which ensure the existence and uniqueness of common fixed points for the four mappings A, B, S and T satisfying ψ- and $(\psi, \varphi)$-weakly contractive conditions, respectively, in metric spaces. Our results extend, improve and unify the corresponding results in [1–5]. Four nontrivial examples are included.

Throughout this paper, ℕ denotes the set of all positive integers, $\Bbb{N}_{0}=\{0\}\cup\Bbb{N}$, $\Bbb{R}^{+}=[0,+\infty)$ and

$$\begin{aligned}& \Phi_{1} =\bigl\{ \psi:\psi:\Bbb{R}^{+}\to\Bbb{R}^{+} \text{ is continuous and nondecreasing,} \\& \hphantom{\Phi_{1} =}\text{and } \psi(t)=0 \text{ if and only if } t=0\bigr\} , \\& \Phi_{2} =\bigl\{ \varphi:\varphi:\Bbb{R}^{+}\to\Bbb{R}^{+} \text{ is lower semi-continuous, and } \varphi(t)=0 \text{ if and only if } t=0\bigr\} , \\& \Phi_{3} = \Bigl\{ \psi:\psi:\Bbb{R}^{+}\to\Bbb{R}^{+} \text{ is upper semi-continuous,} \\& \hphantom{\Phi_{3} =}\text{and } \lim_{n\to\infty}a_{n}=0\text{ for each sequence } \{a_{n}\}_{n\in\Bbb{N}}\subset\Bbb{R}^{+} \text{ with } a_{n+1}\le\psi(a_{n}), \forall n\in\Bbb{N} \Bigr\} . \end{aligned}$$

Definition 1.1

([10])

A pair of self mappings f and g in a metric space $(X, d)$ are said to be weakly compatible if for all $t\in X$ the equality $ft=gt$ implies $fgt=gft$.

Lemma 1.1

([9])

Let $\psi\in\Psi_{3}$. Then $\psi(0)=0$ and $\psi(t)< t$ for all $t>0$.

Lemma 1.2

Let A, B, S and T be self mappings in a metric space $(X, d)$ satisfying (1.2), where $(\psi,\varphi)\in\Phi_{1}\times\Phi_{2}$ and $i\in\{1,2,3\}$. Assume that $I:\Bbb{R}^{+}\to\Bbb{R}^{+}$ is the identity mapping and

$$ \psi_{1}(t)=(\psi+I)^{-1}(\psi+I-\varphi) (t),\quad \forall t\in\Bbb{R}^{+}. $$

(1.6)

Then $\psi_{1}\in\Phi_{3}$ and

$$ d(Tx,Sy)\leq\psi_{1}\bigl(M_{i}(x,y)\bigr),\quad \forall x,y\in X. $$

(1.7)

Proof

It follows from $\psi\in\Phi_{1}$ that $\psi+I:\Bbb{R}^{+}\to\Bbb{R}^{+}$ is continuous and increasing and $(\psi+I)(t)=0$ if and only if $t=0$. So does $(\psi+I)^{-1}$. Obviously, $(\psi,\varphi)\in\Phi_{1}\times\Phi_{2}$ and (1.6) guarantee

$$ \psi_{1} \text{ is upper semi-continuous and } \psi_{1}(0)=0. $$

(1.8)

Assume that $\{a_{n}\}_{n\in\Bbb{N}}$ is an arbitrary sequence in $\Bbb{R}^{+}$ with

$$ a_{n+1}\le\psi_{1}(a_{n}),\quad \forall n \in\Bbb{N}. $$

(1.9)

Suppose that $a_{n_{0}}=0$ for some $n_{0}\in\Bbb{N}$. It follows from (1.6), (1.8) and (1.9) that

$$0\le a_{n_{0}+1}\le\psi_{1}(a_{n_{0}})= \psi_{1}(0)=0, $$

that is, $a_{n_{0}+1}=0$. Similarly we have $a_{n}=a_{n-1}=\cdots=a_{n_{0}}=0$ for each $n>n_{0}$, that is, $\lim_{n\to\infty}a_{n}=0$. Suppose that $a_{n}>0$ for all $n\in\Bbb{N}$. If $a_{k+1}\ge a_{k}$ for some $k\in\Bbb{N}$, it follows from (1.6), (1.9) and $(\psi,\varphi)\in\Phi_{1}\times\Phi_{2}$ that

$$\begin{aligned} \psi(a_{k})+a_{k} \le&\psi(a_{k+1})+a_{k+1}=( \psi+I) (a_{k+1}) \le(\psi+I)\psi_{1}(a_{k})=(\psi+I-\varphi) (a_{k}) \\ =&\psi(a_{k})+a_{k}-\varphi(a_{k})< \psi(a_{k})+a_{k}, \end{aligned}$$

which is a contradiction. Consequently, $\{a_{n}\}_{n\in\Bbb{N}}$ is positive and decreasing, which implies that $\{a_{n}\}_{n\in\Bbb{N}}$ converges to some $a\ge0$. Suppose that $a>0$. By means of (1.8) and (1.9), we find

$$0< a=\limsup_{n\to\infty}a_{n+1}\le\limsup _{n\to\infty}\psi_{1}(a_{n})\le \psi_{1}(a), $$

which together with (1.6) and $(\psi,\varphi)\in\Phi_{1}\times\Phi_{2}$ means

$$\psi(a)+a\le\psi(a)+a-\varphi(a)< \psi(a)+a, $$

which is a contradiction. Hence $a=0$. Consequently, $\psi_{1}\in \Phi_{3}$.

In order to prove (1.7), we have to consider two possible cases as follows:

Case 1. $M_{i}(x_{0},y_{0})=0$ for some $x_{0},y_{0}\in X$. It is easy to verify

$$d(Ax_{0},By_{0})=d(Ax_{0},Tx_{0})=d(By_{0},Sy_{0})=0, $$

which yields

$$Tx_{0}=Ax_{0}=By_{0}=Sy_{0}, $$

and

$$d(Tx_{0},Sy_{0})=0=\psi_{1}\bigl(M_{i}(x_{0},y_{0}) \bigr); $$

Case 2. $M_{i}(x,y)>0$ for all $x,y\in X$. It follows from (1.2), (1.6) and $(\psi,\varphi)\in\Phi_{1}\times\Phi_{2}$ that

$$\psi\bigl(d(Tx,Sy)\bigr)\le\psi\bigl(M_{i}(x,y)\bigr)-\varphi \bigl(M_{i}(x,y)\bigr)< \psi\bigl(M_{i}(x,y)\bigr), \quad\forall x,y \in X, $$

which yields

$$d(Tx,Sy)< M_{i}(x,y), \quad\forall x,y\in X $$

and

$$\begin{aligned} (\psi+I) \bigl(d(Tx,Sy)\bigr) =&\psi\bigl(d(Tx,Sy)\bigr)+d(Tx,Sy) < \psi\bigl(M_{i}(x,y)\bigr)-\varphi\bigl(M_{i}(x,y) \bigr)+M_{i}(x,y) \\ =&(\psi+I-\varphi) \bigl(M_{i}(x,y)\bigr),\quad \forall x,y\in X, \end{aligned}$$

which together with (1.6) gives (1.7). This completes the proof. □

Remark 1.1

It follows from Lemma 1.2 that the $(\psi, \varphi)$-weakly contractive conditions (1.2) relative to four mappings A, B, S and T implies the $\psi_{1}$-weakly contractive conditions (1.1) relative to four mappings A, B, S and T.

2 Common fixed point theorems

Our main results are as follows.

Theorem 2.1

Let A, B, S, and T be self mappings in a metric space $(X, d)$ such that

$$\begin{aligned}& \{A,T\}\textit{ and } \{B,S\}\textit{ are weakly compatible;} \end{aligned}$$

(2.1)

$$\begin{aligned}& T(X)\subseteq B(X) \textit{ and } S(X)\subseteq A(X); \end{aligned}$$

(2.2)

$$\begin{aligned}& \textit{one of } A(X), B(X), S(X),\textit{ and } T(X) \textit{ is complete}; \end{aligned}$$

(2.3)

$$\begin{aligned}& d(Tx,Sy)\leq\psi\bigl(M_{1}(x,y)\bigr), \quad\forall x,y\in X, \end{aligned}$$

(2.4)

where $\psi_{1}$ is in $\Phi_{3}$ and $M_{1}$ is defined by (1.3). Then A, B, S, and T have a unique common fixed point in X.

Proof

Let $x_{0}\in X$. It follows from (2.2) that there exist two sequences $\{y_{n}\}_{n\in\Bbb{N}}$ and $\{x_{n}\}_{n\in\Bbb{N}_{0}}$ in X such that

$$ y_{2n+1}:=Bx_{2n+1}=Tx_{2n},\quad\quad y_{2n+2}:=Ax_{2n+2}=Sx_{2n+1},\quad \forall n\in\Bbb{N}_{0}. $$

(2.5)

Put $d_{n}=d(y_{n},y_{n+1})$ for all $n\in\Bbb{N}$.

Now we prove

$$ \lim_{n\to\infty}d_{n}=0. $$

(2.6)

Using (2.4) and (2.5), we derive

$$ d_{2n} =d(Tx_{2n},Sx_{2n-1})\leq\psi \bigl(M_{1}(x_{2n},x_{2n-1})\bigr), \quad\forall n\in\Bbb{N} $$

(2.7)

and

$$\begin{aligned}& M_{1}(x_{2n},x_{2n-1}) \\& \quad=\max \biggl\{ d(Ax_{2n},Bx_{2n-1}),d(Ax_{2n},Tx_{2n}),d(Bx_{2n-1}, Sx_{2n-1}), \\& \quad\quad\frac{1}{2}\bigl[d(Ax_{2n},Sx_{2n-1})+d(Tx_{2n},Bx_{2n-1}) \bigr], \\& \quad\quad\frac{d(Ax_{2n},Sx_{2n-1})d(Tx_{2n},Bx_{2n-1})}{1+d(Ax_{2n},Bx_{2n-1})}, \frac {d(Ax_{2n},Tx_{2n})d(Bx_{2n-1},Sx_{2n-1})}{1+d(Ax_{2n},Bx_{2n-1})}, \\& \quad\quad \frac{1+d(Ax_{2n},Sx_{2n-1})+d(Tx_{2n},Bx_{2n-1})}{ 1+d(Ax_{2n},Tx_{2n})+d(Bx_{2n-1},Sx_{2n-1})}d(Ax_{2n},Tx_{2n}) \biggr\} \\& \quad=\max \biggl\{ d(y_{2n},y_{2n-1}),d(y_{2n},y_{2n+1}),d(y_{2n-1},y_{2n}), \frac{1}{2}\bigl[d(y_{2n},y_{2n})+d(y_{2n+1},y_{2n-1}) \bigr], \\& \quad\quad\frac{d(y_{2n},y_{2n})d(y_{2n+1},y_{2n-1})}{1+d(y_{2n},y_{2n-1})}, \frac{d(y_{2n},y_{2n+1})d(y_{2n-1},y_{2n})}{1+d(y_{2n},y_{2n-1})}, \\& \quad\quad \frac {1+d(y_{2n},y_{2n})+d(y_{2n+1},y_{2n-1})}{1+d(y_{2n},y_{2n+1})+d(y_{2n-1}, y_{2n})}d(y_{2n},y_{2n+1}) \biggr\} \\& \quad=\max \biggl\{ d_{2n-1},d_{2n},d_{2n-1}, \frac {1}{2}d(y_{2n+1},y_{2n-1}),0,\frac{d_{2n}d_{2n-1}}{1+d_{2n-1}}, \frac{1+d(y_{2n+1},y_{2n-1})}{1+d_{2n}+d_{2n-1}}d_{2n} \biggr\} \\& \quad=\max\{d_{2n-1},d_{2n}\},\quad \forall n\in\Bbb{N}. \end{aligned}$$

(2.8)

Suppose that $d_{2n_{0}-1}< d_{2n_{0}}$ for some $n_{0}\in\Bbb{N}$. It follows from (2.7), (2.8), $\psi\in\Phi_{3}$, and Lemma 1.1 that

$$\begin{aligned} d_{2n_{0}} \le\psi\bigl(M_{1}(x_{2n_{0}},x_{2n_{0}-1}) \bigr)=\psi\bigl(\max\{d_{2n_{0}-1},d_{2n_{0}}\}\bigr) = \psi(d_{2n_{0}})< d_{2n_{0}}, \end{aligned}$$

which is a contradiction. Hence

$$ d_{2n}\le d_{2n-1}=M_{1}(x_{2n},x_{2n-1}),\quad \forall n\in\Bbb{N}. $$

(2.9)

Similarly we infer

$$d_{2n+1}\leq d_{2n}=M_{1}(x_{2n},x_{2n+1}),\quad \forall n\in\Bbb{N}, $$

which together with (2.9) ensures

$$d_{n+1}\leq d_{n}, \quad\forall n\in\Bbb{N}, $$

which means that the sequence $\{d_{n}\}_{n\in\Bbb{N}}$ is nonincreasing and bounded. Consequently there exists $r\geq0$ with $\lim_{n\to\infty}d_{n}=r$. Suppose that $r>0$. It follows from (2.7), (2.9), $\psi\in\Phi_{3}$, and Lemma 1.1 that

$$\begin{aligned} r =&\limsup_{n\to\infty}d_{2n}\leq\limsup _{n\to\infty}\psi \bigl(M_{1}(x_{2n},x_{2n-1}) \bigr) \\ =&\limsup_{n\to\infty}\psi(d_{2n-1})\leq\psi(r)< r, \end{aligned}$$

which is a contradiction. Hence $r=0$, that is, (2.6) holds.

Next we prove that $\{y_{n}\}_{n\in\Bbb{N}}$ is a Cauchy sequence. Because of (2.6) it is sufficient to verify that $\{y_{2n}\}_{n\in\Bbb{N}}$ is a Cauchy sequence. Suppose that $\{y_{2n}\}_{n\in\Bbb{N}}$ is not a Cauchy sequence. It follows that there exist $\varepsilon>0$ and two subsequences $\{y_{2m(k)}\}_{k\in\Bbb{N}}$ and $\{y_{2n(k)}\}_{k\in\Bbb{N}}$ of $\{y_{2n}\}_{n\in\Bbb{N}}$ such that

$$ 2n(k)>2m(k)>2k,\quad\quad d(y_{2m(k)},y_{2n(k)})\geq \varepsilon,\quad \forall k\in\Bbb{N}, $$

(2.10)

where $2n(k)$ is the smallest index satisfying (2.10). It follows that

$$ d(y_{2m(k)},y_{2n(k)-2})< \varepsilon, \quad\forall k\in\Bbb{N}. $$

(2.11)

Taking advantage of (2.10), (2.11), and the triangle inequality, we get

$$\begin{aligned} \varepsilon \leq& d(y_{2m(k)},y_{2n(k)}) \\ \leq& d(y_{2m(k)},y_{2n(k)-2})+d(y_{2n(k)-2},y_{2n(k)-1})+d(y_{2n(k)-1}, y_{2n(k)}) \\ < &\varepsilon+d_{2n(k)-2}+d_{2n(k)-1},\quad \forall k\in\Bbb{N} \end{aligned}$$

(2.12)

and

$$\begin{aligned}& \bigl|d(y_{2m(k)},y_{2n(k)-1})-d(y_{2m(k)},y_{2n(k)})\bigr| \leq d_{2n(k)-1},\quad \forall k\in\Bbb{N}; \\& \bigl|d(y_{2m(k)+1},y_{2n(k)})-d(y_{2m(k)},y_{2n(k)})\bigr|\leq d_{2m(k)}, \quad\forall k\in\Bbb{N}; \\& \bigl|d(y_{2m(k)+1},y_{2n(k)-1})-d(y_{2m(k)},y_{2n(k)-1})\bigr|\leq d_{2m(k)}, \quad\forall k\in\Bbb{N}. \end{aligned}$$

(2.13)

Letting $k\to\infty$ in (2.12) and (2.13) and using (2.6), we deduce

$$\begin{aligned} \lim_{k\to\infty}d(y_{2m(k)},y_{2n(k)}) =&\lim_{k\to\infty}d(y_{2m(k)},y_{2n(k)-1}) =\lim _{k\to\infty}d(y_{2m(k)+1},y_{2n(k)}) \\ =&\lim_{k\to\infty}d(y_{2m(k)+1},y_{2n(k)-1})=\varepsilon. \end{aligned}$$

(2.14)

Note that (1.3) and (2.14) yield

$$\begin{aligned}& M_{1}(x_{2m(k)},x_{2n(k)-1}) \\& \quad=\max \biggl\{ d(Ax_{2m(k)},Bx_{2n(k)-1}), d(Ax_{2m(k)},Tx_{2m(k)}),d(Bx_{2n(k)-1},Sx_{2n(k)-1}), \\& \quad\quad\frac {1}{2}\bigl[d(Ax_{2m(k)},Sx_{2n(k)-1})+d(Tx_{2m(k)},Bx_{2n(k)-1}) \bigr], \\& \quad\quad\frac {d(Ax_{2m(k)},Sx_{2n(k)-1})d(Tx_{2m(k)},Bx_{2n(k)-1})}{1+d(Ax_{2m(k)},Bx_{2n(k)-1})}, \\& \quad\quad\frac {d(Ax_{2m(k)},Tx_{2m(k)})d(Bx_{2n(k)-1},Sx_{2n(k)-1})}{1+d(Ax_{2m(k)},Bx_{2n(k)-1})}, \\& \quad\quad\frac{1+d(Ax_{2m(k)},Sx_{2n(k)-1})+d(Tx_{2m(k)},Bx_{2n(k)-1})}{ 1+d(Ax_{2m(k)},Tx_{2m(k)})+d(Bx_{2n(k)-1},Sx_{2n(k)-1})}d(Ax_{2m(k)},Tx_{2m(k)}) \biggr\} \\& \quad=\max \biggl\{ d(y_{2m(k)},y_{2n(k)-1}), d(y_{2m(k)},y_{2m(k)+1}), d(y_{2n(k)-1},y_{2n(k)}), \\& \quad\quad\frac{1}{2}\bigl[d(y_{2m(k)},y_{2n(k)})+d(y_{2m(k)+1},y_{2n(k)-1}) \bigr], \\& \quad\quad\frac {d(y_{2m(k)},y_{2n(k)})d(y_{2m(k)+1},y_{2n(k)-1})}{1+d(y_{2m(k)},y_{2n(k)-1})}, \frac {d(y_{2m(k)},y_{2m(k)+1})d(y_{2n(k)-1},y_{2n(k)})}{1+d(y_{2m(k)},y_{2n(k)-1})}, \\& \quad\quad\frac{1+d(y_{2m(k)},y_{2n(k)})+d(y_{2m(k)+1},y_{2n(k)-1})}{ 1+d(y_{2m(k)},y_{2m(k)+1})+d(y_{2n(k)-1},y_{2n(k)})}d(y_{2m(k)},y_{2m(k)+1}) \biggr\} \\& \quad\to\max \biggl\{ \varepsilon,0,0,\frac{1}{2}(\varepsilon+\varepsilon ), \frac{\varepsilon^{2}}{1+\varepsilon},0,0 \biggr\} \\& \quad=\varepsilon\quad \text{as } k\to\infty. \end{aligned}$$

(2.15)

In view of (2.4), (2.14), (2.15), $\psi\in\Phi_{3}$, and Lemma 1.1, we gain

$$\begin{aligned} \varepsilon =&\limsup_{k\to\infty}d(y_{2m(k)+1},y_{2n(k)})= \limsup_{k\to\infty }d(Tx_{2m(k)},Sx_{2n(k)-1}) \\ \leq&\limsup_{k\to\infty}\psi\bigl(M_{1}(x_{2m(k)},x_{2n(k)-1}) \bigr)\leq\psi (\varepsilon)< \varepsilon, \end{aligned}$$

which is a contradiction. Hence $\{y_{n}\}_{n\in\Bbb{N}}$ is a Cauchy sequence.

Assume that $A(X)$ is complete. Observe that $\{y_{2n}\}_{n\in\Bbb{N}}$ is a Cauchy sequence in $A(X)$. Consequently there exists $(z, v)\in A(X)\times X$ with $\lim_{n\to\infty}y_{2n}=z=Av$. It is easy to see

$$ z=\lim_{n\to\infty}y_{n}=\lim _{n\to\infty}Tx_{2n}=\lim_{n\to\infty}Bx_{2n+1}= \lim_{n\to\infty}Sx_{2n-1}=\lim_{n\to\infty}Ax_{2n}. $$

(2.16)

Suppose that $Tv\ne z$. Note that (1.3) and (2.16) imply

$$\begin{aligned}& M_{1}(v,x_{2n+1}) \\& \quad=\max \biggl\{ d(Av,Bx_{2n+1}),d(Av,Tv),d(Bx_{2n+1},Sx_{2n+1}),\\& \quad\quad \frac{1}{2}\bigl[d(Av,Sx_{2n+1})+d(Tv,Bx_{2n+1})\bigr], \\& \quad\quad\frac{d(Av,Sx_{2n+1})d(Tv,Bx_{2n+1})}{1+d(Av,Bx_{2n+1})}, \frac{d(Av,Tv)d(Bx_{2n+1},Sx_{2n+1})}{1+d(Av,Bx_{2n+1})}, \\& \quad\quad\frac {1+d(Av,Sx_{2n+1})+d(Tv,Bx_{2n+1})}{1+d(Av,Tv)+d(Bx_{2n+1},Sx_{2n+1})}d(Av,Tv) \biggr\} \\& \quad\to\max \biggl\{ d(Av,z),d(Av,Tv),d(z,z),\frac{1}{2}\bigl[d(Av,z)+d(Tv,z) \bigr], \\& \quad\quad\frac{d(Av,z)d(Tv,z)}{1+d(Av,z)},\frac{d(Av,Tv)d(z,z)}{1+d(Av,z)}, \frac{1+d(Av,z)+d(Tv,z)}{1+d(Av,Tv)+d(z,z)}d(Av,Tv) \biggr\} \\& \quad=\max \biggl\{ 0,d(z,Tv),0,\frac{1}{2}d(Tv,z),0,0,d(z,Tv) \biggr\} \\& \quad=d(Tv,z) \quad\text{as } n\to\infty, \end{aligned}$$

which together with (2.4), $\psi\in\Phi_{3}$, and Lemma 1.1 gives

$$\begin{aligned} d(Tv,z) =&\limsup_{n\to\infty}d(Tv,y_{2n+2})=\limsup _{n\to\infty }d(Tv,Sx_{2n+1}) \\ \le&\limsup_{n\to\infty}\psi\bigl(M_{1}(v,x_{2n+1}) \bigr)\leq\psi\bigl(d(Tv,z)\bigr)< d(Tv,z), \end{aligned}$$

which is a contradiction. Hence $Tv=z$. It follows from (2.2) that there exists a point $w\in X$ with $z=Bw=Tv$. Suppose that $Sw\ne z$. In light of (1.3) and (2.16), we deduce

$$\begin{aligned}& M_{1}(x_{2n},w) \\& \quad=\max \biggl\{ d(Ax_{2n},Bw),d(Ax_{2n},Tx_{2n}),d(Bw,Sw), \frac{1}{2}\bigl[d(Ax_{2n},Sw)+d(Tx_{2n},Bw)\bigr], \\& \quad\quad \frac {d(Ax_{2n},Sw)d(Tx_{2n},Bw)}{1+d(Ax_{2n},Bw)}, \frac{d(Ax_{2n},Tx_{2n})d(Bw,Sw)}{1+d(Ax_{2n},Bw)}, \\& \quad\quad\frac {1+d(Ax_{2n},Sw)+d(Tx_{2n},Bw)}{1+d(Ax_{2n},Tx_{2n})+d(Bw,Sw)}d(Ax_{2n},Tx_{2n}) \biggr\} \\& \quad\to\max \biggl\{ d(z,Bw),d(z,z),d(Bw,Sw),\frac{1}{2}\bigl[d(z,Sw)+d(z,Bw) \bigr], \\& \quad\quad \frac{d(z,Sw)d(z,Bw)}{1+d(z,Bw)},\frac{d(z,z)d(Bw,Sw)}{1+d(z,Bw)}, \frac{1+d(z,Sw)+d(z,Bw)}{1+d(z,z)+d(Bw,Sw)}d(z,z) \biggr\} \\& \quad=\max \biggl\{ 0,0,d(z,Sw),\frac{1}{2}d(z,Sw),0,0,0 \biggr\} \\& \quad=d(z,Sw) \quad\text{as } n\to\infty, \end{aligned}$$

which together with (2.4), $\psi\in\Phi_{3}$, and Lemma 1.1 yields

$$\begin{aligned} d(z,Sw) =&\limsup_{n\to\infty}d(y_{2n+1},Sw)=\limsup _{n\to\infty}d(Tx_{2n},Sw) \\ \le&\limsup_{n\to\infty}\psi\bigl(M_{1}(x_{2n},w) \bigr)\le\psi\bigl(d(z,Sw)\bigr)< (d(z,Sw), \end{aligned}$$

which is impossible, and hence $Sw=z$. Thus (2.1) means $Az=ATv=TAv=Tz$ and $Bz=BSw=SBw=Sz$. Suppose that $Tz\ne Sz$. It follows from (1.3), (2.4), $\psi\in\Phi_{3}$, and Lemma 1.1 that

$$\begin{aligned}& M_{1}(z,z) \\& \quad=\max \biggl\{ d(Az,Bz),d(Az,Tz),d(Bz,Sz),\frac {1}{2} \bigl[d(Az,Sz)+d(Tz,Bz)\bigr], \\& \quad\quad\frac{d(Az,Sz)d(Tz,Bz)}{1+d(Az,Bz)},\frac {d(Az,Tz)d(Bz,Sz)}{1+d(Az,Bz)},\\& \quad\quad \frac{1+d(Az,Sz)+d(Tz,Bz)}{1+d(Az,Tz)+d(Bz,Sz)}d(Az,Tz) \biggr\} \\& \quad=\max \biggl\{ d(Tz,Sz),0,0,\frac{1}{2}\bigl[d(Tz,Sz)+d(Tz,Sz)\bigr], \frac {d^{2}(Tz,Sz)}{1+d(Tz,Sz)},0,0 \biggr\} \\& \quad=d(Tz,Sz) \end{aligned}$$

and

$$\begin{aligned} d(Tz,Sz)\leq\psi\bigl(M_{1}(z,z)\bigr)=\psi\bigl(d(Tz,Sz) \bigr)< d(Tz,Sz), \end{aligned}$$

which is a contradiction, and hence $Tz=Sz$.

Suppose that $Tz\ne z$. It follows from (1.3) that

$$\begin{aligned}& M_{1}(z,w) \\& \quad=\max \biggl\{ d(Az,Bw),d(Az,Tz),d(Bw,Sw),\frac {1}{2} \bigl[d(Az,Sw)+d(Tz,Bw)\bigr], \\& \quad\quad \frac{d(Az,Sw)d(Tz,Bw)}{1+d(Az,Bw)},\frac {d(Az,Tz)d(Bw,Sw)}{1+d(Az,Bw)}, \\& \quad\quad \frac{1+d(Az,Sw)+d(Tz,Bw)}{1+d(Az,Tz)+d(Bw,Sw)}d(Az,Tz) \biggr\} \\& \quad=\max \biggl\{ d(Tz,z),0,0,\frac{1}{2}\bigl[d(Tz,z)+d(Tz,z)\bigr], \frac {d^{2}(Tz,z)}{1+d(Tz,z)},0,0 \biggr\} \\& \quad=d(Tz,z), \end{aligned}$$

which together with (2.4), $\psi\in\Phi_{3}$, and Lemma 1.1 implies

$$d(Tz,z)=d(Tz,Sw)\leq\psi\bigl(M_{1}(z,w)\bigr)=\psi\bigl(d(Tz,z) \bigr)< d(Tz,z), $$

which is impossible and hence $Tz=z$, that is, z is a common fixed point of A, B, S, and T.

Suppose that A, B, S, and T have another common fixed point $u\in X\setminus\{z\}$. It follows from (1.3), (2.4), $\psi\in\Phi_{3}$, and Lemma 1.1 that

$$\begin{aligned}& M_{1}(u,z) \\& \quad=\max \biggl\{ d(Au,Bz),d(Au,Tu),d(Bz,Sz),\frac {1}{2} \bigl[d(Au,Sz)+d(Tu,Bz)\bigr], \\& \quad\quad\frac{d(Au,Sz)d(Tu,Bz)}{1+d(Au,Bz)},\frac {d(Au,Tu)d(Bz,Sz)}{1+d(Au,Bz)}, \\& \quad\quad \frac{1+d(Au,Sz)+d(Tu,Bz)}{1+d(Au,Tu)+d(Bz,Sz)}d(Au,Tu) \biggr\} \\& \quad=\max \biggl\{ d(u,z),0,0,\frac{1}{2}\bigl[d(u,z)+d(u,z)\bigr], \frac {d^{2}(u,z)}{1+d(u,z)},0,0 \biggr\} \\& \quad=d(u,z) \end{aligned}$$

and

$$d(u,z)=d(Tu,Sz)\leq\psi\bigl(M_{1}(u,z)\bigr)=\psi\bigl(d(u,z) \bigr)< d(u,z), $$

which is a contradiction and hence z is a unique common fixed point of A, B, S, and T in X.

Similarly we conclude that A, B, S, and T have a unique common fixed point in X if one of $B(X)$, $S(X)$, and $T(X)$ is complete. This completes the proof. □

Theorem 2.2

Let A, B, S, and T be self mappings in a metric space $(X, d)$ satisfying (2.1)-(2.3) and

$$ d(Tx,Sy)\leq\psi\bigl(M_{2}(x,y)\bigr),\quad \forall x,y\in X, $$

(2.17)

where ψ is in $\Phi_{3}$ and $M_{2}$ is defined by (1.4). Then A, B, S, and T have a unique common fixed point in X.

Proof

Let $x_{0}\in X$. It follows from (2.2) that there exist two sequences $\{y_{n}\}_{n\in\Bbb{N}}$ and $\{x_{n}\}_{n\in\Bbb{N}_{0}}$ in X satisfying (2.5). Put $d_{n}=d(y_{n},y_{n+1})$ for all $n\in\Bbb{N}$.

Now we prove that (2.6) holds. In view of (1.4) and (2.17), we deduce

$$ d_{2n} =d(Tx_{2n},Sx_{2n-1})\leq\psi \bigl(M_{2}(x_{2n},x_{2n-1})\bigr),\quad \forall n\in\Bbb{N} $$

(2.18)

and

$$\begin{aligned}& M_{2}(x_{2n},x_{2n-1}) \\& \quad=\max \biggl\{ d(Ax_{2n},Bx_{2n-1}),d(Ax_{2n},Tx_{2n}),d(Bx_{2n-1},Sx_{2n-1}), \\& \quad\quad\frac{1}{2}\bigl[d(Ax_{2n},Sx_{2n-1})+d(Tx_{2n},Bx_{2n-1} \bigr], \\& \quad\quad\frac {1+d(Ax_{2n},Tx_{2n})}{1+d(Ax_{2n},Bx_{2n-1})}d(Bx_{2n-1},Sx_{2n-1}), \frac {1+d(Bx_{2n-1},Sx_{2n-1})}{1+d(Ax_{2n},Bx_{2n-1})}d(Ax_{2n},Tx_{2n}), \\& \quad\quad\frac{1+d(Ax_{2n},Sx_{2n-1})+d(Tx_{2n},Bx_{2n-1})}{ 1+d(Ax_{2n},Tx_{2n})+d(Bx_{2n-1},Sx_{2n-1})}d(Bx_{2n-1},Sx_{2n-1}) \biggr\} \\& \quad=\max \biggl\{ d(y_{2n},y_{2n-1}),d(y_{2n},y_{2n+1}),d(y_{2n-1},y_{2n}), \\& \quad\quad\frac{1}{2}\bigl[d(y_{2n},y_{2n})+d(y_{2n+1},y_{2n-1}) \bigr], \frac{1+d(y_{2n},y_{2n+1})}{1+d(y_{2n},y_{2n-1})}d(y_{2n-1},y_{2n}), \\& \quad\quad\frac{1+d(y_{2n-1},y_{2n})}{1+d(y_{2n},y_{2n-1})}d(y_{2n},y_{2n+1}), \frac {1+d(y_{2n},y_{2n})+d(y_{2n+1},y_{2n-1})}{1+d(y_{2n},y_{2n+1})+d(y_{2n-1},y_{2n})}d(y_{2n-1},y_{2n}) \biggr\} \\& \quad=\max \biggl\{ d_{2n-1},d_{2n},d_{2n-1}, \frac {1}{2}d(y_{2n+1},y_{2n-1}),\frac{1+d_{2n}}{1+d_{2n-1}}d_{2n-1},d_{2n}, \\& \quad\quad \frac{1+d(y_{2n+1},y_{2n-1})}{1+d_{2n}+d_{2n-1}}d_{2n-1} \biggr\} \\& \quad=\max \biggl\{ d_{2n-1},d_{2n},\frac{1+d_{2n}}{1+d_{2n-1}}d_{2n-1} \biggr\} ,\quad \forall n\in\Bbb{N}. \end{aligned}$$

Suppose that $d_{2n_{0}-1}< d_{2n_{0}}$ for some $n_{0}\in\Bbb{N}$. It follows that

$$d_{2n_{0}}(1+d_{2n_{0}-1})=d_{2n_{0}}+d_{2n_{0}}d_{2n_{0}-1}>d_{2n_{0}-1}+d_{2n_{0}}d_{2n_{0}-1}=d_{2n_{0}-1}(1+d_{2n_{0}}), $$

that is,

$$d_{2n_{0}}>\frac{1+d_{2n_{0}}}{1+d_{2n_{0}-1}}d_{2n_{0}-1}, $$

which implies $M_{2}(x_{2n_{0}},x_{2n_{0}-1})=d_{2n_{0}}$. By means of (2.18), $\psi\in\Phi_{3}$, and Lemma 1.1, we conclude

$$\begin{aligned} d_{2n_{0}}\le\psi\bigl(M_{2}(x_{2n_{0}},x_{2n_{0}-1}) \bigr)=\psi(d_{2n_{0}})< d_{2n_{0}}, \end{aligned}$$

which is a contradiction. Consequently, we deduce

$$ d_{2n}\le d_{2n-1}=M_{2}(x_{2n},x_{2n-1}),\quad \forall n\in\Bbb{N}. $$

(2.19)

Similarly we have

$$ d_{2n+1}\leq d_{2n}=M_{2}(x_{2n},x_{2n+1}),\quad \forall n\in\Bbb{N}. $$

(2.20)

It follows from (2.19) and (2.20) that

$$d_{n+1}\leq d_{n}, \quad\forall n\in\Bbb{N}, $$

which means that the sequence $\{d_{n}\}_{n\in\Bbb{N}}$ is nonincreasing and bounded. Consequently there exists $r\geq0$ with $\lim_{n\to\infty}d_{n}=r$. Suppose that $r>0$. It follows from (2.18), (2.19), $\psi\in\Phi_{3}$, and Lemma 1.1 that

$$\begin{aligned} r =&\limsup_{n\to\infty}d_{2n}\leq\limsup _{n\to\infty}\psi \bigl(M_{2}(x_{2n},x_{2n-1}) \bigr) \\ =&\limsup_{n\to\infty}\psi(d_{2n-1})\leq\psi(r)< r, \end{aligned}$$

which is a contradiction. Hence $r=0$, that is, (2.6) holds.

In order to prove that $\{y_{n}\}_{n\in\Bbb{N}}$ is a Cauchy sequence, we need only to show that $\{y_{2n}\}_{n\in\Bbb{N}}$ is a Cauchy sequence. Suppose that $\{y_{2n}\}_{n\in\Bbb{N}}$ is not a Cauchy sequence. It follows that there exist $\varepsilon>0$ and two subsequences $\{y_{2m(k)}\}_{k\in\Bbb{N}}$ and $\{y_{2n(k)}\}_{k\in\Bbb{N}}$ of $\{y_{2n}\}_{n\in\Bbb{N}}$ satisfying (2.10)-(2.14) and

$$\begin{aligned}& M_{2}(x_{2m(k)},x_{2n(k)-1}) \\& \quad=\max \biggl\{ d(Ax_{2m(k)},Bx_{2n(k)-1}),d(Ax_{2m(k)},Tx_{2m(k)}),d(Bx_{2n(k)-1},Sx_{2n(k)-1}), \\& \quad\quad\frac {1}{2}\bigl[d(Ax_{2m(k)},Sx_{2n(k)-1})+d(Tx_{2m(k)},Bx_{2n(k)-1}) \bigr], \\& \quad\quad \frac {1+d(Ax_{2m(k)},Tx_{2m(k)})}{1+d(Ax_{2m(k)},Bx_{2n(k)-1})}d(Bx_{2n(k)-1},Sx_{2n(k)-1}), \\& \quad\quad\frac {1+d(Bx_{2n(k)-1},Sx_{2n(k)-1})}{1+d(Ax_{2m(k)},Bx_{2n(k)-1})}d(Ax_{2m(k)},Tx_{2m(k)}), \\& \quad\quad\frac{1+d(Ax_{2m(k)},Sx_{2n(k)-1})+d(Tx_{2m(k)},Bx_{2n(k)-1})}{ 1+d(Ax_{2m(k)},Tx_{2m(k)})+d(Bx_{2n(k)-1},Sx_{2n(k)-1})}d(Bx_{2n(k)-1},Sx_{2n(k)-1}) \biggr\} \\& \quad=\max \biggl\{ d(y_{2m(k)},y_{2n(k)-1}),d(y_{2m(k)},y_{2m(k)+1}),d(y_{2n(k)-1},y_{2n(k)}), \\& \quad\quad\frac{1}{2}\bigl[d(y_{2m(k)},y_{2n(k)})+d(y_{2m(k)+1},y_{2n(k)-1}) \bigr], \\& \quad\quad\frac {1+d(y_{2m(k)},y_{2m(k)+1})}{1+d(y_{2m(k)},y_{2n(k)-1})}d(y_{2n(k)-1},y_{2n(k)}), \\& \quad\quad \frac {1+d(y_{2n(k)-1},y_{2n(k)})}{1+d(y_{2m(k)},y_{2n(k)-1})}d(y_{2m(k)},y_{2m(k)+1}), \\& \quad\quad\frac{1+d(y_{2m(k)},y_{2n(k)})+d(y_{2m(k)+1},y_{2n(k)-1})}{1+d(y_{2m(k)}, y_{2m(k)+1})+d(y_{2n(k)-1},y_{2n(k)})}d(y_{2n(k)-1},y_{2n(k)}) \biggr\} \\& \quad\to\max \biggl\{ \varepsilon,0,0,\frac{1}{2}(\varepsilon+\varepsilon ),0,0,0 \biggr\} \\& \quad=\varepsilon\quad \text{as } k\to\infty. \end{aligned}$$

(2.21)

By virtue of (2.14), (2.17), (2.21), $\psi\in\Phi_{3}$, and Lemma 1.1, we infer

$$\begin{aligned} \varepsilon =&\limsup_{k\to\infty}d(y_{2m(k)+1},y_{2n(k)})= \limsup_{k\to\infty }d(Tx_{2m(k)},Sx_{2n(k)-1}) \\ \leq&\limsup_{k\to\infty}\psi\bigl(M_{2}(x_{2m(k)},x_{2n(k)-1}) \bigr)\leq\psi (\varepsilon)< \varepsilon, \end{aligned}$$

which is impossible. Hence $\{y_{n}\}_{n\in\Bbb{N}}$ is a Cauchy sequence.

Assume that $A(X)$ is complete. Observe that $\{y_{2n}\}_{n\in\Bbb{N}}\subseteq A(X)$ is a Cauchy sequence. It follows that there exists $(z, v)\in A(X)\times X$ with $\lim_{n\to\infty}y_{2n}=z=Av$. It is easy to show that (2.16) holds.

Suppose that $Tv\ne z$. Note that (1.4), (2.16), (2.17), and $\psi\in\Phi_{3}$ imply

$$\begin{aligned}& M_{2}(v,x_{2n+1}) \\& \quad=\max \biggl\{ d(Av,Bx_{2n+1}),d(Av,Tv),d(Bx_{2n+1},Sx_{2n+1}), \\& \quad\quad\frac{1}{2}\bigl[d(Av,Sx_{2n+1})+d(Tv,Bx_{2n+1})\bigr], \\& \quad\quad \frac{1+d(Av,Tv)}{1+d(Av,Bx_{2n+1})}d(Bx_{2n+1},Sx_{2n+1}),\frac {1+d(Bx_{2n+1},Sx_{2n+1})}{1+d(Av,Bx_{2n+1})}d(Av,Tv), \\& \quad\quad\frac {1+d(Av,Sx_{2n+1})+d(Tv,Bx_{2n+1})}{1+d(Av,Tv)+d(Bx_{2n+1},Sx_{2n+1})}d(Bx_{2n+1},Sx_{2n+1}) \biggr\} \\& \quad\to\max \biggl\{ d(Av,z),d(Av,Tv),d(z,z),\frac{1}{2}\bigl[d(Av,z)+d(Tv,z) \bigr], \\& \quad\quad \frac{1+d(Av,Tv)}{1+d(Av,z)}d(z,z),\frac{1+d(z,z)}{1+d(Av,z)}d(Av,Tv), \frac{1+d(Av,z)+d(Tv,z)}{1+d(Av,Tv)+d(z,z)}d(z,z) \biggr\} \\& \quad=\max \biggl\{ 0,d(z,Tv),0,\frac{1}{2}d(Tv,z),0,d(z,Tv),0 \biggr\} \\& \quad=d(Tv,z) \quad\text{as } n\to\infty, \end{aligned}$$

which together with (2.17), $\psi\in\Phi_{3}$, and Lemma 1.1 gives

$$\begin{aligned} d(Tv,z) =&\limsup_{n\to\infty}d(Tv,y_{2n+2})=\limsup _{n\to\infty }d(Tv,Sx_{2n+1}) \\ \le&\limsup_{n\to\infty}\psi\bigl(M_{2}(v,x_{2n+1}) \bigr)\leq\psi\bigl(d(Tv,z)\bigr)< d(Tv,z), \end{aligned}$$

which is a contradiction. Hence $Tv=z$.

Since $T(X)\subseteq B(X)$, it follows that there exists a point $w\in X$ such that $z=Bw=Tv$. Suppose that $Sw\ne z$. In light of (1.4) and (2.16), we obtain

$$\begin{aligned}& M_{2}(x_{2n},w) \\& \quad=\max \biggl\{ d(Ax_{2n},Bw),d(Ax_{2n},Tx_{2n}),d(Bw,Sw), \frac{1}{2}\bigl[d(Ax_{2n},Sw)+d(Tx_{2n},Bw)\bigr], \\& \quad\quad\frac{1+d(Ax_{2n},Tx_{2n})}{1+d(Ax_{2n},Bw)}d(Bw,Sw),\frac {1+d(Bw,Sw)}{1+d(Ax_{2n},Bw)}d(Ax_{2n},Tx_{2n}), \\& \quad\quad \frac {1+d(Ax_{2n},Sw)+d(Tx_{2n},Bw)}{1+d(Ax_{2n},Tx_{2n})+d(Bw,Sw)}d(Bw,Sw) \biggr\} \\& \quad\to\max \biggl\{ d(z,z),d(z,z),d(z,Sw),\frac{1}{2}\bigl[d(z,Sw)+d(z,Bw) \bigr], \\& \quad\quad\frac{1+d(z,z)}{1+d(z,z)}d(z,Sw),\frac {1+d(z,Sw)}{1+d(z,z)}d(z,z), \\& \quad\quad\frac{1+d(z,Sw)+d(z,z)}{1+d(z,z)+d(z,Sw)}d(z,Sw) \biggr\} \\& \quad=\max \biggl\{ 0,0,d(z,Sw),\frac{1}{2}d(z,Sw),d(z,Sw),0,d(z,Sw) \biggr\} \\& \quad=d(z,Sw)\quad \text{as } n\to\infty, \end{aligned}$$

which together with (2.17), $\psi\in\Phi_{3}$, and Lemma 1.1 yields

$$\begin{aligned} d(z,Sw) =&\limsup_{n\to\infty}d(y_{2n+1},Sw)=\limsup _{n\to\infty}d(Tx_{2n},Sw) \\ \le&\limsup_{n\to\infty}\psi\bigl(M_{2}(x_{2n},w) \bigr)\le\psi\bigl(d(z,Sw)\bigr)< d(z,Sw), \end{aligned}$$

which is impossible, and hence $Sw=z$. Clearly, (2.1) yields $Az=ATv=TAv=Tz$ and $Bz=BSw=SBw=Sz$. Suppose that $Tz\ne Sz$. It follows from (1.4) that

$$\begin{aligned} M_{2}(z,z) =&\max \biggl\{ d(Az,Bz),d(Az,Tz),d(Bz,Sz),\frac {1}{2} \bigl[d(Az,Sz)+d(Tz,Bz)\bigr], \\ &\frac{1+d(Az,Tz)}{1+d(Az,Bz)}d(Bz,Sz),\frac {1+d(Bz,Sz)}{1+d(Az,Bz)}d(Az,Tz), \\ & \frac{1+d(Az,Sz)+d(Tz,Bz)}{1+d(Az,Tz)+d(Bz,Sz)}d(Bz,Sz) \biggr\} \\ =&\max \biggl\{ d(Tz,Sz),0,0,\frac{1}{2}\bigl[d(Tz,Sz)+d(Tz,Sz) \bigr],0,0,0 \biggr\} \\ =&d(Tz,Sz). \end{aligned}$$

Taking account of (2.17), $\psi\in\Phi_{3}$, and Lemma 1.1, we conclude

$$d(Tz,Sz)\leq\psi\bigl(M_{2}(z,z)\bigr)=\psi\bigl(d(Tz,Sz) \bigr)< d(Tz,Sz), $$

which is a contradiction, and hence $Tz=Sz$.

Suppose that $Tz\ne z$. It follows from (1.4) that

$$\begin{aligned} M_{2}(z,w) =&\max \biggl\{ d(Az,Bw),d(Az,Tz),d(Bw,Sw),\frac {1}{2} \bigl[d(Az,Sw)+d(Tz,Bw)\bigr], \\ &\frac{1+d(Az,Tz)}{1+d(Az,Bw)}d(Bw,Sw),\frac {1+d(Bw,Sw)}{1+d(Az,Bw)}d(Az,Tz), \\ &\frac{1+d(Az,Sw)+d(Tz,Bw)}{1+d(Az,Tz)+d(Bw,Sw)}d(Bw,Sw) \biggr\} \\ =&\max \biggl\{ d(Tz,z),0,0,\frac{1}{2}\bigl[d(Tz,z)+d(Tz,z)\bigr],0,0,0 \biggr\} \\ =&d(Tz,z), \end{aligned}$$

which together with (2.17), $\psi\in\Phi_{3}$, and Lemma 1.1 means

$$d(Tz,z) =d(Tz,Sw)\leq\psi\bigl(M_{2}(z,w)\bigr)=\psi\bigl(d(Tz,z) \bigr)< d(Tz,z), $$

which is impossible, and hence $Tz=z$, that is, z is a common fixed point of A, B, S, and T.

Suppose that A, B, S, and T have another common fixed point $u\in X\setminus\{z\}$. It follows from (1.4) that

$$\begin{aligned} M_{2}(u,z) =&\max \biggl\{ d(Au,Bz),d(Au,Tu),d(Bz,Sz),\frac {1}{2} \bigl[d(Au,Sz)+d(Tu,Bz)\bigr], \\ & \frac{1+d(Au,Tu)}{1+d(Au,Bz)}d(Bz,Sz),\frac {1+d(Bz,Sz)}{1+d(Au,Bz)}d(Au,Tu), \\ & \frac{1+d(Au,Sz)+d(Tu,Bz)}{1+d(Au,Tu)+d(Bz,Sz)}d(Bz,Sz) \biggr\} \\ =&\max \biggl\{ d(u,z),0,0,\frac{1}{2}\bigl[d(u,z)+d(u,z)\bigr],0,0,0 \biggr\} \\ =&d(u,z), \end{aligned}$$

which together with (2.17), $\psi\in\Phi_{3}$, and Lemma 1.1 ensures

$$d(u,z)=d(Tu,Sz)\leq\psi\bigl(M_{2}(u,z)\bigr)=\psi\bigl(d(u,z) \bigr)< d(u,z), $$

which is a contradiction, and hence z is a unique common fixed point of A, B, S, and T in X.

Similarly we conclude that A, B, S, and T have a unique common fixed point in X if one of $B(X)$, $S(X)$, and $T(X)$ is complete. This completes the proof. □

Similar to the proofs of Theorems 2.1 and 2.2, we have the following result and omit its proof.

Theorem 2.3

Let A, B, S, and T be self mappings in a metric space $(X, d)$ satisfying (2.1)-(2.3) and

$$ d(Tx,Sy)\leq\psi\bigl(M_{3}(x,y)\bigr),\quad \forall x,y\in X, $$

(2.22)

where ψ is in $\Phi_{3}$ and $M_{3}$ is defined by (1.5). Then A, B, S, and T have a unique common fixed point in X.

Utilizing Theorems 2.1-2.3, Lemma 1.2, and Remark 1.1, we get the following results.

Theorem 2.4

Let A, B, S, and T be self mappings in a metric space $(X, d)$ satisfying (2.1)-(2.3) and

$$ \psi\bigl(d(Tx,Sy)\bigr)\leq\psi\bigl(M_{1}(x,y)\bigr)- \varphi\bigl(M_{1}(x,y)\bigr), \quad\forall x,y\in X, $$

(2.23)

where $(\psi,\varphi)$ is in $\Phi_{1}\times\Phi_{2}$ and $M_{1}$ is defined by (1.3). Then A, B, S, and T have a unique common fixed point in X.

Theorem 2.5

Let A, B, S, and T be self mappings in a metric space $(X, d)$ satisfying (2.1)-(2.3) and

$$ \psi\bigl(d(Tx,Sy)\bigr)\leq\psi\bigl(M_{2}(x,y)\bigr)- \varphi\bigl(M_{2}(x,y)\bigr), \quad\forall x,y\in X, $$

(2.24)

where $(\psi,\varphi)$ is in $\Phi_{1}\times\Phi_{2}$ and $M_{2}$ is defined by (1.4). Then A, B, S, and T have a unique common fixed point in X.

Theorem 2.6

Let A, B, S, and T be self mappings in a metric space $(X, d)$ satisfying (2.1)-(2.3) and (2.22), where $(\psi,\varphi)$ is in $\Phi_{1}\times\Phi_{2}$ and $M_{3}$ is defined by (1.5). Then A, B, S, and T have a unique common fixed point in X.

Remark 2.1

Condition (2.3) in Theorem 2.6 is weaker than the conditions of $(X,d)$ is complete and one of the ranges of the four mappings A, B, S, and T is closed in Theorem 2.1 in [2]. Hence Theorem 2.6 is a slight generalizations of Theorem 2.1 in [2]. Note that Theorem 2.4 generalizes Theorems 2.1 and 2.2 in [4]. Example 2.1 below shows that Theorem 2.6 is a substantial generalization of Theorem 2.1 in [2] and Theorems 2.1 and 2.2 in [4].

Example 2.1

Let $X=(-1,1)$ be endowed with the Euclidean metric $d(x,y)=|x-y|$ for all $x,y\in X$. Let $A,B,S,T:X\to X$ be defined by

$$ Ax=x^{2},\quad\quad Bx=x,\quad\quad Sx=0,\quad \forall x\in X, \quad\quad Tx= \begin{cases} 0,&\forall x\in X\setminus \{\frac{1}{2} \},\\ -\frac{1}{4},&x=\frac{1}{2}. \end{cases} $$

Since the metric space $(X,d)$ is not complete, it follows that Theorem 2.1 in [2] is useless in proving the existence of common fixed points of A, B, S, and T in X and Theorems 2.1 and 2.2 in [4] are unapplicable in proving the existence of common fixed points of S and T and fixed points of T, respectively.

Now we use Theorem 2.6 to prove the existence of common fixed points of A, B, S, and T in X. Define $\psi,\varphi:\Bbb{R}^{+}\to\Bbb{R}^{+}$ by

$$\psi(t)= \begin{cases}\sqrt{t},&\forall t\in [0,\frac{1}{2} ),\\ \frac{\sqrt{2}}{2},&\forall t\in [\frac{1}{2},+\infty ) \end{cases} $$

and

$$\varphi(t)= \begin{cases} t^{3},&\forall t\in [0,\frac{1}{2} ),\\ \frac{1}{16},&\forall t\in [\frac{1}{2},+\infty ). \end{cases} $$

It is easy to verify that (2.1)-(2.3) holds, $(\psi,\varphi)\in\Phi_{1}\times\Phi_{2}$, $\psi(t)\geq\varphi(t)$ for each $t\in\Bbb{R}^{+}$. Put $x,y\in X$. In order to verify (2.22), we consider two cases as follows:

Case 1. $x\in X\setminus \{\frac{1}{2} \}$. It is clear that

$$\psi\bigl(d(Tx,Sy)\bigr)=\psi(0)=0\le\psi\bigl(M_{3}(x,y)\bigr)- \varphi\bigl(M_{3}(x,y)\bigr); $$

Case 2. $x=\frac{1}{2}$. Clearly we have

$$\begin{aligned} M_{3}(x,y) =&\max \biggl\{ d(Ax,By),d(Ax,Tx),d(By,Sy),\frac {1}{2} \bigl[d(Ax,Sy)+d(Tx,By)\bigr] \biggr\} \\ \ge& d(Ax,Tx)=d \biggl(\frac{1}{4},-\frac{1}{4} \biggr)= \frac{1}{2}. \end{aligned}$$

It follows that

$$\begin{aligned} \psi\bigl(d(Tx,Sy)\bigr) =&\psi \biggl(d \biggl(-\frac{1}{4},0 \biggr) \biggr)=\psi \biggl(\frac{1}{4} \biggr)=\frac{1}{2} \\ \le&\frac{\sqrt{2}}{2}-\frac{1}{16}=\psi\bigl(M_{3}(x,y)\bigr)- \varphi\bigl(M_{3}(x,y)\bigr). \end{aligned}$$

That is, (2.22) holds. Hence the conditions of Theorem 2.6 are satisfied. It follows from Theorem 2.6 that A, B, S, and T in X possess a unique common fixed point $0\in X$.

Remark 2.2

Theorems 2.4-2.6 extend, improve and unify Theorem 2.1 in [3], Theorem 2.1 in [5] and Theorem 1 in [1]. Note that Examples 2.2-2.4 below deal with the existence of common fixed points of four mappings A, B, S, and T, but Theorem 2.1 in [3], Theorem 2.1 in [5] and Theorem 1 in [1] deal with the existence of fixed and common fixed points of at most three mappings, therefore the results in [1, 3, 5] are useless in proving the existence of common fixed points of four mappings A, B, S, and T. That is, Theorems 2.4-2.6 extend indeed Theorem 2.1 in [3], Theorem 2.1 in [5] and Theorem 1 in [1].

Example 2.2

Let $X=\Bbb{R}^{+}$ be endowed with the Euclidean metric $d(x,y)=|x-y|$ for all $x,y\in X$. Let $B,T:X\to X$ be defined by

$$Bx=x^{2},\quad \forall x\in X \quad\text{and}\quad Tx= \begin{cases} 1,&\forall x\in\Bbb{R}^{+}- \{\frac{1}{32} \},\\ \frac{15}{16},&x= \{\frac{1}{32} \}. \end{cases} $$

Firstly we claim that Theorem 2.1 in [5] and Theorem 1 in [1] and Theorem 2.1 in [3] cannot be used to prove the existence of fixed and common fixed points for the mapping T and the mappings B and T, respectively, in the complete metric space X.

Suppose that there exist $\varphi\in\Phi_{1}$ satisfying

$$d(Tx,Ty)\leq d(x,y)-\varphi\bigl(d(x,y)\bigr),\quad \forall x,y\in X, $$

which implies

$$\begin{aligned} \frac{1}{16} =&d \biggl(1,\frac{15}{16} \biggr)=d \biggl(T0,T \frac{1}{32} \biggr) \le d \biggl(0,\frac{1}{32} \biggr)-\varphi \biggl(d \biggl(0,\frac{1}{32} \biggr) \biggr) \\ =&\frac{1}{32}-\varphi \biggl(\frac{1}{32} \biggr), \end{aligned}$$

that is,

$$0< \varphi \biggl(\frac{1}{32} \biggr)\le\frac{1}{32}- \frac{1}{16}=-\frac{1}{32}, $$

which is a contradiction.

Suppose that there exists $\psi,\varphi\in\Phi_{1}$ satisfying

$$\psi\bigl(d(Tx,Ty)\bigr)\leq\psi\bigl(d(x,y)\bigr)-\varphi\bigl(d(x,y)\bigr), \quad\forall x,y\in X, $$

which yields

$$\begin{aligned} \psi \biggl(\frac{1}{16} \biggr) =&\psi \biggl(1-\frac{15}{16} \biggr) =\psi \biggl(d \biggl(T\frac{3}{32},T\frac{1}{32} \biggr) \biggr) \\ \le&\psi \biggl(d \biggl(\frac{3}{32},\frac{1}{32} \biggr) \biggr) - \varphi \biggl(d \biggl(\frac{3}{32},\frac{1}{32} \biggr) \biggr) = \psi \biggl(\frac{1}{16} \biggr)-\varphi \biggl(\frac{1}{16} \biggr), \end{aligned}$$

that is,

$$0< \varphi \biggl(\frac{1}{16} \biggr)\le\psi \biggl(\frac{1}{16} \biggr)-\psi \biggl(\frac{1}{16} \biggr)=0, $$

which is impossible.

Suppose that there exists $(\psi,\varphi)\in\Phi_{1}\times\Phi_{2}$ satisfying

$$\psi\bigl(d(Tx,Ty)\bigr)\leq\psi\bigl(d(Bx,By)\bigr)-\varphi\bigl(d(Bx,By) \bigr), \quad\forall x,y\in X, $$

which gives

$$\begin{aligned} \psi \biggl(\frac{1}{16} \biggr) =&\psi \biggl(1-\frac{15}{16} \biggr) =\psi \biggl(d \biggl(T\frac{3}{64},T\frac{1}{32} \biggr) \biggr) \\ \leq&\psi \biggl(d \biggl(B\frac{3}{64},B\frac{1}{32} \biggr) \biggr) -\varphi \biggl(d \biggl(B\frac{3}{64},B\frac{1}{32} \biggr) \biggr) \\ =&\psi \biggl( \biggl(\frac{3}{64} \biggr)^{2}- \biggl( \frac{1}{32} \biggr)^{2} \biggr) -\varphi \biggl( \biggl( \frac{3}{64} \biggr)^{2}- \biggl(\frac{1}{32} \biggr)^{2} \biggr) \\ =&\psi \biggl(\frac{5}{4{,}096} \biggr)-\varphi \biggl(\frac{5}{4{,}096} \biggr) \\ < &\psi \biggl(\frac{5}{4{,}096} \biggr)\le\psi \biggl(\frac{5}{4{,}095} \biggr)=\psi \biggl(\frac{1}{819} \biggr) \\ \le&\psi \biggl(\frac{1}{16} \biggr), \end{aligned}$$

which is impossible.

Secondly we claim that the mappings A, B, S, and T satisfy the conditions of Theorem 2.6, where $A,S:X\to X$ and $\psi,\varphi:\Bbb{R}^{+}\to\Bbb{R}^{+}$ are defined by

$$Ax=x^{3}, \quad\quad Sx=1, \quad\forall x\in X $$

and

$$\psi(t) = \begin{cases} 16t,&\forall t\in[0, \frac{1}{16}),\\ 512t^{2}-1,&\forall t\in[\frac{1}{16}, +\infty), \end{cases} \quad\quad \varphi(t) = \begin{cases} t^{2},&\forall t\in[0, \frac{1}{16}),\\ \frac{t^{2}}{1+8t^{2}},&\forall t\in[\frac{1}{16}, +\infty). \end{cases} $$

Clearly, (2.1)-(2.3) hold, $(\psi,\varphi)\in\Phi_{1}\times\Phi_{2}$, $\psi(t)\geq\varphi(t)$ for any $t\in\Bbb{R}^{+}$, and $\varphi(\Bbb{R}^{+})\subset [0, \frac{1}{8} )$. Put $x,y\in X$. In order to verify (2.22), we have to consider the following two possible cases:

Case 1. $x\in X\setminus\{\frac{1}{32}\}$. It follows that

$$\psi\bigl(d(Tx,Sy)\bigr)=\psi(0)=0\leq\psi\bigl(M_{3}(x,y)\bigr)- \varphi\bigl(M_{3}(x,y)\bigr); $$

Case 2. $x=\frac{1}{32}$. It follows that

$$\begin{aligned} M_{3} \biggl(\frac{1}{32}, y \biggr) =&\max \biggl\{ \biggl| \frac {1}{32^{3}}-y^{2} \biggr|, \biggl|\frac{1}{32^{3}}-\frac{15}{16} \biggr|, \bigl|1-y^{2} \bigr|, \frac{1}{2} \biggl( \biggl|\frac{1}{32^{3}}-1 \biggr|+ \biggl| \frac {15}{16}-y^{2} \biggr| \biggr) \biggr\} \\ \ge&\frac{15}{16}-\frac{1}{32^{3}}=\frac{30{,}719}{32{,}768}>\frac{1}{16} \end{aligned}$$

and

$$\begin{aligned} \psi \biggl(d \biggl(T\frac{1}{32},Sy \biggr) \biggr) =&\psi \biggl(d \biggl(\frac {15}{16},1 \biggr) \biggr) =\psi \biggl(\frac{1}{16} \biggr)=1 \\ < &400<\frac{941{,}559{,}809}{2{,}097{,}152}-\frac{1}{8} \\ <&\psi \biggl(\frac{30{,}719}{32{,}768} \biggr)-\varphi \biggl(M \biggl( \frac {1}{32}, y \biggr) \biggr) \\ \le&\psi \biggl(M_{3} \biggl(\frac{1}{32}, y \biggr) \biggr)- \varphi \biggl(M_{3} \biggl(\frac{1}{32}, y \biggr) \biggr). \end{aligned}$$

That is, (2.22) holds. Thus the conditions of Theorem 2.6 are satisfied. It follows from Theorem 2.6 that the mappings A, B, S, and T have a unique common fixed point $1\in X$.

Example 2.3

Let $X=[0, 1]$ be endowed with the Euclidean metric $d(x,y)=|x-y|$ for all $x,y\in X$. Let $A,B,S,T:X\to X$ and $\psi,\varphi:\Bbb{R}^{+}\to\Bbb{R}^{+}$ be defined by

$$Ax=x^{2},\quad\quad Bx=\frac{1}{2}x^{2},\quad\quad Sx=0,\quad \forall x\in X, \quad\quad Tx = \begin{cases} 0,&\forall x\in[0, 1),\\ \frac{1}{4},& x=1 \end{cases} $$

and

$$\psi(t) = \begin{cases} 16t^{2},&\forall t\in[0, \frac{1}{4}),\\ 8t-1,&\forall t\in[\frac{1}{4}, +\infty), \end{cases} \quad\quad \varphi(t) = \begin{cases} 4t^{2},&\forall t\in[0, \frac{1}{4}),\\ \frac{1}{6+2\sqrt{t}},&\forall t\in[\frac{1}{4}, +\infty). \end{cases} $$

It is easy to see that (2.1)-(2.3) hold, $(\psi,\varphi)\in\Phi_{1}\times\Phi_{2}$, $\psi(t)\geq\varphi(t)$ for each $t\in\Bbb{R}^{+}$ and $\varphi(\Bbb{R}^{+})\subset [0,\frac{1}{4} )$. Let $x,y\in X$. In order to verify (2.23), we have to consider two possible cases as follows:

Case 1. $x\in X\setminus\{1\}$. It is clear that

$$\psi\bigl(d(Tx,Sy)\bigr)=\psi(0)=0\leq\psi\bigl(M_{1}(x,y)\bigr)- \varphi\bigl(M_{1}(x,y)\bigr); $$

Case 2. $x=1$. It follows that

$$\begin{aligned} M_{1}(1, y) =&\max \biggl\{ \biggl|1-\frac{y^{2}}{2} \biggr|, \frac{3}{4}, \frac{y^{2}}{2}, \frac{1}{2} \biggl(1+ \biggl| \frac{1}{4}-\frac{y^{2}}{2} \biggr| \biggr), \frac{ |\frac{1}{4}-\frac{y^{2}}{2} |}{1+ |1-\frac{y^{2}}{2} |}, \\ & \frac{\frac{3}{4}\cdot\frac{y^{2}}{2}}{1+ |1-\frac{y^{2}}{2} |}, \frac{1+1+ |\frac{1}{4}-\frac{y^{2}}{2} |}{1+\frac{3}{4}+\frac {y^{2}}{2}}\cdot\frac{3}{4} \biggr\} \\ \geq&\frac{3}{4} \end{aligned}$$

and

$$\begin{aligned} \psi\bigl(d(T1,Sy)\bigr) =&\psi \biggl(\frac{1}{4} \biggr)=1< 5- \frac{1}{4} \le\psi \biggl(\frac{3}{4} \biggr)-\varphi \bigl(M_{1}(1,y)\bigr) \\ \leq&\psi\bigl(M_{1}(1,y)\bigr)-\varphi\bigl(M_{1}(1,y) \bigr). \end{aligned}$$

That is, (2.23) holds. It follows from Theorem 2.4 that the mappings A, B, S, and T have a unique common fixed point $0\in X$. However, we neither use Theorem 1 in [1] nor employ Theorem 2.1 in [5] to show the existence of fixed points of the mapping T in X.

Suppose that there exists $\varphi\in\Phi_{1}$ satisfying

$$d(Tx,Ty)\leq d(x,y)-\varphi\bigl(d(x,y)\bigr), \quad\forall x,y\in X, $$

which implies

$$\frac{1}{4}=d \biggl(\frac{1}{4},0 \biggr)=d \biggl(T1,T \frac{7}{8} \biggr) \le d \biggl(1,\frac{7}{8} \biggr)-\varphi \biggl(d \biggl(1,\frac{7}{8} \biggr) \biggr) = \frac{1}{8}-\varphi \biggl(\frac{1}{8} \biggr), $$

which means

$$0< \varphi \biggl(\frac{1}{8} \biggr)\le\frac{1}{8}- \frac{1}{4}=-\frac{1}{8}, $$

which is a contradiction.

Suppose that there exist $\psi,\varphi\in\Phi_{1}$ satisfying

$$\psi\bigl(d(Tx,Ty)\bigr)\leq\psi\bigl(d(x,y)\bigr)-\varphi\bigl(d(x,y)\bigr), \quad\forall x,y\in X, $$

which yields

$$\begin{aligned} \psi \biggl(\frac{1}{4} \biggr) =&\psi \biggl(d \biggl(0, \frac{1}{4} \biggr) \biggr)=\psi\bigl(d(Tx,T1)\bigr) \le\psi\bigl(d(x,1) \bigr)-\varphi\bigl(d(x,1)\bigr) \\ =&\psi(1-x)-\varphi(1-x), \quad\forall x\in X\setminus\{1\}, \end{aligned}$$

which gives

$$\begin{aligned} 0 < &\psi \biggl(\frac{1}{4} \biggr)\le\limsup_{x\to1} \bigl[\psi(1-x)-\varphi (1-x)\bigr] \\ \le&\limsup_{x\to1}\psi(1-x)-\liminf_{x\to1} \varphi(1-x)\le\psi (0)-\varphi(0) = 0, \end{aligned}$$

which is impossible.

Example 2.4

Let $X=[-1, 1]$ be endowed with the Euclidean metric $d(x,y)=|x-y|$ for all $x,y\in X$. Let $A,B,S,T:X\to X$ and $\psi,\varphi:\Bbb{R}^{+}\to\Bbb{R}^{+}$ be defined by

$$\begin{aligned}& Ax=\frac{x^{2}}{2},\quad\quad Tx=0,\quad \forall x\in X, \\& Bx= \begin{cases} 0,&\forall x\in[-1, 1),\\ \frac{1}{2},& x=1, \end{cases} \quad\quad Sx = \begin{cases} 0,&\forall x\in[-1, 1),\\ \frac{1}{8},& x=1, \end{cases} \end{aligned}$$

and

$$\psi(t)= \begin{cases} 64t^{3},&\forall t\in[0, \frac{1}{4}),\\ 32t^{2}-1,&\forall t\in[\frac{1}{4}, +\infty), \end{cases} \quad\quad \varphi(t)= \begin{cases} 128t^{4},&\forall t\in[0, \frac{1}{4}),\\ \frac{1}{2}\sin\frac{\pi}{2+8t},&\forall t\in[\frac{1}{4}, +\infty). \end{cases} $$

Clearly, (2.1)-(2.3) holds, $(\psi,\varphi)\in\Phi_{1}\times\Phi_{2}$, $\psi(t)\geq\varphi(t)$ for each $t\in\Bbb{R}^{+}$ and $\varphi(t)\leq\frac{\sqrt{2}}{4}<\frac{1}{2}$ for all $t\in[\frac{1}{4}, +\infty)$. Let $x,y\in X$. In order to verify (2.24), we have to consider two possible cases as follows:

Case 1. $y\in X\setminus\{1\}$. Obviously

$$\psi\bigl(d(Tx,Sy)\bigr)=\psi(0)=0\leq\psi\bigl(M_{2}(x,y)\bigr)- \varphi\bigl(M_{2}(x,y)\bigr); $$

Case 2. $y=1$. It follows that

$$\begin{aligned} M_{2}(x, 1) =&\max \biggl\{ \frac{1-x^{2}}{2}, \frac{x^{2}}{2}, \frac{1}{2}-\frac{1}{8}, \frac{1}{2} \biggl( \biggl| \frac{x^{2}}{2}-\frac {1}{8} \biggr|+\frac{1}{2} \biggr), \frac{1+\frac{x^{2}}{2}}{1+\frac{1-x^{2}}{2}}\cdot\frac{3}{8}, \\ & \frac{1+\frac{3}{8}}{1+\frac{1-x^{2}}{2}}\cdot\frac{x^{2}}{2}, \frac{1+ |\frac{x^{2}}{2}-\frac{1}{8} |+\frac{1}{2}}{1+\frac{x^{2}}{2} +\frac{3}{8}}\cdot \frac{3}{8} \biggr\} \\ \geq&\frac{3}{8} \end{aligned}$$

and

$$\begin{aligned} \psi\bigl(d(Tx,S1)\bigr) =&\psi \biggl(\frac{1}{8} \biggr)=64\times \frac{1}{8^{3}} =\frac{1}{8}< 3=32\times \biggl(\frac{3}{8} \biggr)^{2}-1-\frac{1}{2} \\ <&\psi\bigl(M_{2}(x, 1)\bigr)-\varphi\bigl(M_{2}(x, 1) \bigr). \end{aligned}$$

That is, (2.24) holds. Consequently, Theorem 2.5 guarantees that the mappings A, B, S, and T have a unique common fixed point $0\in X$. However, we do not invoke that Theorem 2.1 in [5] proves the existence of fixed points of the mapping S in X. Otherwise there exist $\psi,\varphi\in\Phi_{1}$ satisfying

$$\psi\bigl(d(Sx,Sy)\bigr)\leq\psi\bigl(d(x,y)\bigr)-\varphi\bigl(d(x,y)\bigr), \quad\forall x,y\in X, $$

which yields

$$\begin{aligned} \psi \biggl(\frac{1}{8} \biggr) =&\psi \biggl(d \biggl(S \frac{31}{32},S1 \biggr) \biggr) \le\psi \biggl(d \biggl( \frac{31}{32},1 \biggr) \biggr)-\varphi \biggl(d \biggl( \frac{31}{32},1 \biggr) \biggr) \\ =&\psi \biggl(\frac{1}{32} \biggr)-\varphi \biggl(\frac{1}{32} \biggr) < \psi \biggl(\frac{1}{32} \biggr) \\ \le&\psi \biggl(\frac{1}{8} \biggr), \end{aligned}$$

which is a contradiction.