Abstract
In this paper, we obtain a new convergence theorem for fixed points of weak contractions in Branciari type generalized metric spaces under weaker conditions. The proof process of the theorem is new and different from that of other authors. An illustrative example of this theorem is to show how the new conditions extend known results.
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1 Introduction
The concept of generalized metric spaces, which is a generalization of metric spaces, was first defined by Branciari (see [1]) in 2000. The generalization is via the fact that the triangle inequality is replaced by the rectangular inequality \(d(x,y)\leq d(x,u)+d(u,v)+d(v,y)\) for all pairwise distinct points x, y, u, v of X. Afterwards, many authors studied and extended the existence of old fixed point results in such spaces (see [1–18]). In this paper, we are to generalize well-known related fixed points theorems. For this, we recall some basic definitions and conclusions.
Definition 1.1
([1])
Let X be a nonempty set and \(d: X\times X\rightarrow [0,+\infty )\) be a mapping such that for all \(x,y\in X\) and for all distinct points \(u,v\in X\) each of them different from x and y satisfying the following conditions:
\((d_{1})\) \(d(x,y)=0\) if and only if \(x=y\);
\((d_{2})\) \(d(x,y)=d(y,x)\);
\((d_{3})\) \(d(x,y)\leq d(x,u)+d(u,v)+d(v,y)\) (the rectangular inequality).
Then \((X,d)\) is called a Branciari type generalized metric space.
Every metric space is a Branciari type generalized metric space, but the converse is not true (see [2]).
Definition 1.2
([1])
Let \((X,d)\) be a Branciari type generalized metric space and \(\{x_{n}\}\) be a sequence in X and \(x\in X\). We call that
(i) \(\{x_{n}\}\) is convergent to x if and only if \(d(x_{n},x)\rightarrow 0\) as \(n\rightarrow \infty \) (denoted by \(x_{n}\rightarrow x\)).
(ii) \(\{x_{n}\}\) is a Cauchy sequence if and only if for each \(\epsilon >0\) there exists a natural number N such that \(d(x_{n},x_{m})<\epsilon \) for all \(m,n>N\).
(iii) X is complete if and only if every Cauchy sequence is convergent in X.
In 2012, Lakzian and Samet [4] obtained a fixed point theorem of the generalized metric spaces.
Theorem 1.1
([4])
Let \((X,d)\) be a Hausdorff and complete generalized metric space, and let \(T: X\rightarrow X\) be a self-mapping satisfying
for all \(x, y\in X\), where (i) \(\psi : [0, +\infty )\rightarrow [0, +\infty )\) is a continuous and monotone nondecreasing function with \(\psi (t)=0\) if and only if \(t=0\); (ii) \(\varphi : [0, +\infty )\rightarrow [0, +\infty )\) is a continuous function with \(\varphi (t)=0\) if and only if \(t=0\). Then T has a unique fixed point.
In 2013, Liu and Chai [8] gave a generalization of Theorem 1.1.
Theorem 1.2
([5])
Let \((X,d)\) be a Hausdorff and complete generalized metric space, and let \(T: X \to X\) be a self-mapping satisfying
for all \(x, y \in X\), where (i) \(\psi : [0, +\infty )\rightarrow [0, +\infty )\) is a continuous and monotone nondecreasing function with \(\psi (t)=0\) if and only if \(t=0\); (ii) \(\theta : [0, +\infty )\rightarrow [0, +\infty )\) satisfies \(\lim_{t\rightarrow r}\theta (t)>0\) for \(r>0\) and \(\lim_{t\rightarrow r}\theta (t)=0\) if and only if \(r=0\); (iii) \(a_{i}\geq 0(i=1,2,3)\) with \(a_{1}+a_{2}+a_{3}\leq 1\). Then T has a unique fixed point.
It is worth mentioning that \(\psi : [0, +\infty )\rightarrow [0, +\infty )\) is both continuous and monotone nondecreasing function, but we cannot obtain that \(t_{1}\leq t_{2}\) if \(\psi (t_{1})\leq \psi (t_{2})\). In fact, the erroneous conclusion has been widely applied in proofs of the above theorems. This paper is to provide the correct results related to the above theorems and to weaken the conditions of theorems.
2 The main results
In this section, we denote by Ψ the set of functions \(\psi : [0, +\infty )\rightarrow [0, +\infty )\) satisfying the following conditions:
\((a_{1})\) ψ is monotone nondecreasing;
\((a_{2})\) \(\lim_{t\rightarrow r}\psi (t)>0\) for \(r>0\) and \(\lim_{t\rightarrow 0^{+}}\psi (t)=0\);
\((a_{3})\) \(\psi (t)=0\) if and only if \(t=0\).
We denote by Φ the set of functions \(\varphi : [0, +\infty )\rightarrow [0, +\infty )\) satisfying the following conditions:
\((b_{1})\) \(\lim_{t\rightarrow r}\inf \varphi (t)>0\) for each \(r>0\);
\((b_{2})\) \(\varphi (t)\rightarrow 0\) implies that \(t\rightarrow 0\);
\((b_{3})\) \(\varphi (t)=0\) if and only if \(t=0\).
Theorem 2.1
Let \((X,d)\) be a Branciari type complete generalized metric space, and let \(T: X\to X\) be a self-mapping satisfying
for all \(x, y \in X\), where \(\psi \in \Psi \), \(\varphi \in \Phi \) and \(a_{i}\geq 0(i=1,2,3)\) with \(a_{1}+a_{2}+a_{3}\leq 1\). Then T has a unique fixed point.
Proof
In order to prove our conclusion, we divide the proof into the following steps.
Step 1. If there exists a fixed point of T, the fixed point is unique.
Suppose that there exist two fixed points p, q with \(Tp=p\neq q=Tq\), it means that \(d(p,q)\neq 0\). Taking \(x=p\) and \(y=q\) in (2.1), we have
If \(a_{1}=0\), then \(\psi (d(p, q))=0\), i.e., \(d(p, q)=0\), which contradicts \(d(p, q)\neq 0\). If \(a_{1}>0\), we deduce from (2.2) that
is a contradiction and \(p=q\). Hence the fixed point is unique.
Step 2. T has the fixed point in X.
Let \(x_{0} \in X\) and construct a sequence \(\{x_{n}\}\) in X by \(x_{n+1}=Tx_{n}\) for all \(n\geq 0\).
Case 1. T has a periodic point.
Case 1-1. If \(x_{n+1}=x_{n}\) for some n, then \(x_{n}\) is a fixed point of T.
For the rest, we assume that \(d(x_{n+1},x_{n})\neq 0\) for all n.
Case 1-2. If \(x_{n+2}=x_{n}\) for some n, then \(Tx_{n}\) is a fixed point of T. Suppose it is not true, then \(Tx_{n}\neq T^{2}x_{n}\), i.e., \(d(Tx_{n},T^{2}x_{n})>0\), which implies that \(\varphi (d(Tx_{n},T^{2}x_{n}))>0\). By (2.1), we have
i.e., \(\varphi ((a_{1}+a_{2}+a_{3})d(x_{n},x_{n+1}))=0\). If \(\sum_{i=1}^{3}a_{i}\neq 0\), we get that \(d(x_{n}, x_{n+1})=0\), a contradiction. If \(\sum_{i=1}^{3}a_{i}=0\), we get from (2.3) that \(\psi (d(x_{n}, x_{n+1}))=0\), i.e., \(d(x_{n}, x_{n+1})=0\) is a contradiction to the assumption, and so \(Tx_{n}\) is a fixed point of T.
Case 1-3. If there are two natural numbers m, n such that \(x_{m}=x_{n}\) with \(m-n>2\) and \(x_{i}\neq x_{j}\) for all \(n\leq i\neq j< m\), we claim that \(T^{m-n-1}x_{n}\) is a fixed point of T. Suppose that it does not hold, then
it implies that
Again using (2.1), we get
If \(d(x_{m},x_{m-1})< d(x_{m+1},x_{m})\), then
i.e.,
that is,
which implies that \(a_{1}=a_{2}=a_{3}=0\). From (2.4), we obtain that \(\psi (d(x_{m+1},x_{m}))=0\Leftrightarrow d(x_{m+1},x_{m})=0\), which is a contradiction and so \(d(x_{m+1},x_{m})\leq d(x_{m},x_{m-1})\). Then we have
then \(a_{1}d(x_{m},x_{m-1})+a_{2}d(x_{m},x_{m+1})+a_{3}d(x_{m-1},x_{m})>0\). Otherwise, \(a_{1}=a_{2}=a_{3}=0\), we obtain a contradiction. Therefore, (2.6) turns into
a contradiction. Hence, the conclusion holds.
Case 2. T has no periodic point, i.e., \(x_{m}\neq x_{n}\) for all \(m\neq n\).
Step 2-1. Show that \(\lim_{n\rightarrow \infty }d(x_{n+1},x_{n})=0\). Taking \(x=x_{n}\), \(y=x_{n-1}\) in (2.1), we have
If \(d(x_{n}, x_{n-1})< d(x_{n+1}, x_{n})\), then
it implies that
then \(a_{1}=a_{2}=a_{3}\). Thus
a contradiction. Hence
for all n. Since ψ is monotone nondecreasing, then
There exist nonnegative numbers r and \(r^{*}\) such that
If \(r>0\), we get
then
By (2.8), we have
Letting \(n\rightarrow \infty \) in (2.13), taking lower limits on each side of the above inequality, we deduce that
a contradiction, and so \(\lim_{n\rightarrow \infty }d(x_{n+1},x_{n})=0\).
Step 2-2. Show that \(\lim_{n\rightarrow \infty }d(x_{n+2},x_{n})=0\). Again taking \(x=x_{n+1}\), \(y=x_{n-1}\) in (2.1), then
If \(\sum_{i=1}^{3}a_{i}=0\), then \(a_{i}=0\) for \(i=1,2,3\). Thus, \(\psi (d(x_{n+2},x_{n}))=0\), a contradiction. If \(\sum_{i=1}^{3}a_{i}\neq 0\), we consider the following cases.
Case 2-2-1. If there exists a infinite subsequence \(\{x_{n(i)}\}\) of \(\{x_{n}\}\) such that \(d(x_{n(i)},x_{n(i)-1})< d(x_{n(i)+1},x_{n(i)-1})\) for all i. Without loss of generality, we have
for all \(i\geq 1\). Again by (2.1), we get
for all i. If \(\sum_{i=1}^{3}a_{i}=0\), then \(a_{i}=0\) for \(i=1,2,3\). Thus, we have \(\psi (d(x_{n(i)+1},x_{n(i)-1}))=0\), i.e., \(d(x_{n(i)+1},x_{n(i)-1})=0\) is a contradiction. If \(\sum_{i=1}^{3}a_{i}\neq 0\), then we get from (2.16) and (2.15) that
It follows from (2.17) and the result of Step 2-1 that
as \(i\rightarrow \infty \), that is,
And we also obtain from (2.17) that
which implies that
so the sequence \(\{d(x_{n(i)+1},x_{n(i)-1})\}\) is monotone decreasing and bounded below, there exists \(R\geq 0\) such that
If \(R>0\), then
contradicts (2.18). Thus \(d(x_{n(i)+1},x_{n(i)-1})\rightarrow 0\) as \(i\rightarrow \infty \).
Case 2-2-2. If there exists an infinite subsequence \(\{x_{n(j)}\}\) of \(\{x_{n}\}\) such that
then \(d(x_{n(j)+1},x_{n(j)-1})\rightarrow 0\) as \(j\rightarrow \infty \).
Therefore, in two cases we proved that \(\lim_{n\rightarrow \infty }d(x_{n+2},x_{n})=0\).
Step 2-3. Show that \(\{x_{n}\}\) is a Cauchy sequence. Suppose, on the contrary, that there exists \(\epsilon >0\) for which we can find subsequences \(\{x_{m(k)}\}\) and \(\{x_{n(k)}\}\) of \(\{x_{n}\}\) such that
for \(n(k)>m(k)>k\) with \(n(k)\) is the smallest index, and so we have
for all k. Applying the rectangular inequality, we obtain that
then \(d(x_{m(k)},x_{n(k)})\rightarrow \epsilon \) as \(k\rightarrow \infty \). Similarly,
then \(d(x_{m(k)-1},x_{n(k)-1})\rightarrow \epsilon \) as \(k\rightarrow \infty \). Furthermore, there exists K such that
for \(m(k),n(k)>K\). Again using (2.1), then
Taking the lower limit as \(n\rightarrow \infty \) in the above inequality, (2.19) yields
On the other hand,
If \(a_{1}=0\), then we obtain from (2.19) that
as \(k\rightarrow \infty \), i.e., \(\lim_{k\rightarrow \infty }\psi (d(x_{m(k)},x_{n(k)}))=0\), a contradiction. If \(a_{1}\neq 0\), then (2.20) implies that
which is a contradiction. Therefore, \(\{x_{n}\}\) is a Cauchy sequence. Since \((X,d)\) is complete, there exists \(q\in X\) such that \(\lim_{n\rightarrow \infty }x_{n}=q\).
Step 2-4. Let us show that q is a fixed point of T. Suppose that it is not the case, then q is not a fixed point of T, i.e., \(d(q,Tq)>0\). Since
then
Thus,
By (2.1), we get
If \(a_{2}=0\), then (2.23) yields
as \(n\rightarrow \infty \), i.e., \(\lim_{n\rightarrow \infty }\psi (d(Tq,x_{n+1}))=0\), a contradiction. If \(a_{2}\neq 0\), then we have
And we get from (2.23) that
Since
then
Taking lower limits as \(n\rightarrow \infty \) on either side of inequality (2.25), then
which is a contradiction, and hence \(q=Tq\). □
Remark 2.1
If \(a_{1}=1\), \(a_{2}=a_{3}=0\) and ψ, φ are all continuous in Theorem 2.1, then we obtain Theorem 2.1 of [4]. If ψ is continuous and φ has limit in Theorem 2.1, then we obtain the main results of [8].
Remark 2.2
In the proofs of the main theorems of [4, 19], and [8], there exists a common problem for monotonicity of function ψ, which is unreasonable that \(\psi (t_{1})\leq \psi (t_{2})\) implies that \(t_{1}\leq t_{2}\). It is as follows:
(i) In page 3 of [19], (2.2) implies (2.3).
(ii) In line 5 to line 3 from the bottom of page 903 of [4] and in line 7 to 10 from the top of page 904 of [4].
(iii) The above problems still exist in the proof of the main theorem in [8].
In addition, the proof process of Theorem 2.1 is different from that of [4, 19], and [8].
As a corollary of Theorem 2.1, taking \(a_{1}=1\), \(a_{2}=a_{3}=0\), we obtain the following result.
Corollary 2.1
Let \((X,d)\) be a Branciari type complete generalized metric space, and let \(T: X \to X\) be a self-mapping satisfying
for all \(x,y \in X\), where ψ and φ are defined as in Theorem 2.1. Then T has a unique fixed point.
Similar results are obtained from Theorem 2.1 putting \(a_{1}=a_{3}=0\), \(a_{2}=1\) or \(a_{1}=a_{2}=0\), \(a_{3}=1\).
Corollary 2.2
Let \((X,d)\) be a Branciari type complete generalized metric space, and let \(T: X \to X\) be a self-mapping satisfying
for all \(x,y \in X\), where ψ and φ are defined as in Theorem 2.1. Then T has a unique fixed point.
Further, we obtain the following result which includes Corollary 5 of [7].
Corollary 2.3
Let \((X,d)\) be a Branciari type complete generalized metric space, and let \(T: X \to X\) be a self-mapping satisfying
for all \(x,y \in X\), where ψ and φ are defined as in Theorem 2.1. Then T has a unique fixed point.
Finally, we introduce a simple example [7] that supports the result of our Theorem 2.1.
Example 2.1
([7])
Let \(X=A\cup B\), where \(A=\{\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}\}\) and \(B=[1,2]\). Define the generalized metric on X as follows:
Then \((X,d)\) is a Branciari type generalized metric space, but it is not a metric space. In fact,
Let \(T: X\rightarrow X\) be defined by
Define \(\psi (t)=t\), \(\varphi (t)=\frac{t}{5}\), \(t\in [0, +\infty )\). Then T satisfies
for all \(x,y\in X\), where \(a_{1}=0.4\), \(a_{2}=0.4\), \(a_{3}=0.2\) and T has a unique fixed point \(x=\frac{1}{4}\).
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The authors are grateful to the anonymous referees for their good comments and valuable suggestions which helped to obtain the manuscript.
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This work was supported by Science and Technology Research of Higher Education in Hebei province (ZD2019047).
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Xue, Z., Lv, G. A fixed point theorem for generalized \((\psi ,\varphi )\)-weak contractions in Branciari type generalized metric spaces. Fixed Point Theory Algorithms Sci Eng 2021, 1 (2021). https://doi.org/10.1186/s13663-021-00688-2
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DOI: https://doi.org/10.1186/s13663-021-00688-2