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.
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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
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
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
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
where
\(\varphi:\Bbb{R}^{+}\to\Bbb{R}^{+}\) 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:
where \(i\in\{1,2,3\}\), \(\psi\in\Phi_{3}\), \((\psi,\varphi)\in\Phi_{1}\times\Phi_{2}\), respectively,
and
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
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
Then \(\psi_{1}\in\Phi_{3}\) and
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
Assume that \(\{a_{n}\}_{n\in\Bbb{N}}\) is an arbitrary sequence in \(\Bbb{R}^{+}\) with
Suppose that \(a_{n_{0}}=0\) for some \(n_{0}\in\Bbb{N}\). It follows from (1.6), (1.8) and (1.9) that
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
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
which together with (1.6) and \((\psi,\varphi)\in\Phi_{1}\times\Phi_{2}\) means
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
which yields
and
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
which yields
and
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
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
Put \(d_{n}=d(y_{n},y_{n+1})\) for all \(n\in\Bbb{N}\).
Now we prove
Using (2.4) and (2.5), we derive
and
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
which is a contradiction. Hence
Similarly we infer
which together with (2.9) ensures
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
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
where \(2n(k)\) is the smallest index satisfying (2.10). It follows that
Taking advantage of (2.10), (2.11), and the triangle inequality, we get
and
Letting \(k\to\infty\) in (2.12) and (2.13) and using (2.6), we deduce
Note that (1.3) and (2.14) yield
In view of (2.4), (2.14), (2.15), \(\psi\in\Phi_{3}\), and Lemma 1.1, we gain
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
Suppose that \(Tv\ne z\). Note that (1.3) and (2.16) imply
which together with (2.4), \(\psi\in\Phi_{3}\), and Lemma 1.1 gives
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
which together with (2.4), \(\psi\in\Phi_{3}\), and Lemma 1.1 yields
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
and
which is a contradiction, and hence \(Tz=Sz\).
Suppose that \(Tz\ne z\). It follows from (1.3) that
which together with (2.4), \(\psi\in\Phi_{3}\), and Lemma 1.1 implies
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
and
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
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
and
Suppose that \(d_{2n_{0}-1}< d_{2n_{0}}\) for some \(n_{0}\in\Bbb{N}\). It follows that
that is,
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
which is a contradiction. Consequently, we deduce
Similarly we have
It follows from (2.19) and (2.20) that
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
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
By virtue of (2.14), (2.17), (2.21), \(\psi\in\Phi_{3}\), and Lemma 1.1, we infer
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
which together with (2.17), \(\psi\in\Phi_{3}\), and Lemma 1.1 gives
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
which together with (2.17), \(\psi\in\Phi_{3}\), and Lemma 1.1 yields
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
Taking account of (2.17), \(\psi\in\Phi_{3}\), and Lemma 1.1, we conclude
which is a contradiction, and hence \(Tz=Sz\).
Suppose that \(Tz\ne z\). It follows from (1.4) that
which together with (2.17), \(\psi\in\Phi_{3}\), and Lemma 1.1 means
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
which together with (2.17), \(\psi\in\Phi_{3}\), and Lemma 1.1 ensures
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
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
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
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
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
and
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
Case 2. \(x=\frac{1}{2}\). Clearly we have
It follows that
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
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
which implies
that is,
which is a contradiction.
Suppose that there exists \(\psi,\varphi\in\Phi_{1}\) satisfying
which yields
that is,
which is impossible.
Suppose that there exists \((\psi,\varphi)\in\Phi_{1}\times\Phi_{2}\) satisfying
which gives
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
and
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
Case 2. \(x=\frac{1}{32}\). It follows that
and
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
and
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
Case 2. \(x=1\). It follows that
and
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
which implies
which means
which is a contradiction.
Suppose that there exist \(\psi,\varphi\in\Phi_{1}\) satisfying
which yields
which gives
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
and
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
Case 2. \(y=1\). It follows that
and
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
which yields
which is a contradiction.
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Acknowledgements
The authors are very grateful to the referees for their valuable comments and suggestions. This research was supported by the Science Research Foundation of Educational Department of Liaoning Province (L2012380) and the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Science, ICT & Future Planning (2013R1A1A2057665).
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Liu, Z., Zhang, X., Ume, J.S. et al. Common fixed point theorems for four mappings satisfying ψ-weakly contractive conditions. Fixed Point Theory Appl 2015, 20 (2015). https://doi.org/10.1186/s13663-015-0271-z
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DOI: https://doi.org/10.1186/s13663-015-0271-z