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Fixed point theorems of JSquasicontractions
Fixed Point Theory and Applications volumeÂ 2016, ArticleÂ number:Â 40 (2016)
Abstract
In this paper, we introduce the concept of JSquasicontraction and prove some fixed point results for JSquasicontractions in complete metric spaces under the assumption that the involving function is nondecreasing and continuous. These fixed point results extend and improve many existing results since some assumptions made there are removed or weakened. In addition, we present some examples showing the usability of our results.
1 Introductions
Recall the Banach contraction principle [1], which states that each Banach contraction \(T:X\rightarrow X\) (i.e., there exists \(k \in[0,1)\) such that \(d(Tx,Ty)\leq k d(x,y)\) for all \(x,y\in X\)) has a unique fixed point, provided that \((X,d)\) is a complete metric space. According to its importance and simplicity, this principle have been extended and generalized in various directions (see [2â€“17]). For example, the concepts of Ä†iriÄ‡ contraction [5], quasicontraction [6], JScontraction [7], and JSÄ†iriÄ‡ contraction [8] have been introduced, and many interesting generalizations of the Banach contraction principle are obtained.
Following Hussain et al. [8], we denote by Î¨ the set of all nondecreasing functions \(\psi:[0,+\infty)\rightarrow[1,+\infty)\) satisfying the following conditions:
 (Î¨1):

\(\psi(t)=1\) if and only if \(t=0\);
 (Î¨2):

for each sequence \(\{t_{n}\}\subset(0,+\infty)\), \(\lim_{n\rightarrow\infty}\psi(t_{n})=1\) if and only if \(\lim_{n\rightarrow\infty}t_{n}=0\);
 (Î¨3):

there exist \(r\in(0,1)\) and \(l\in(0,+\infty]\) such that \(\lim_{t\rightarrow0^{+}}\frac{\psi(t)1}{t^{r}}=l\);
 (Î¨4):

\(\psi(t+s)\leq\psi(t)\psi(s)\) for all \(t,s>0\).
For convenience, we set:
Example 1
Let \(f(t)=e^{te^{t}}\) for \(t\geq0\). Then \(f\in \Phi_{2}\cap\Phi_{3}\), but \(f\notin\Psi\cup\Phi_{1}\cup\Phi_{4}\) since \(\lim_{t\rightarrow0^{+}}\frac{e^{te^{t}}1}{t^{r}}=0\) for each \(r\in (0,1)\) and \(e^{(s+t)e^{s+t}}>e^{se^{s}}e^{te^{t}}\) for all \(s,t>0\).
Example 2
Let \(g(t)=e^{t^{a}}\) for \(t\geq0\), where \(a>0\). When \(a\in(0,1)\), \(g\in\Psi\cap\Phi_{1}\cap\Phi_{2}\cap\Phi_{3}\cap\Phi_{4}\). When \(a=1\), \(g\in \Phi_{2}\cap\Phi_{3}\cap\Phi_{4}\), but \(g\notin\Psi\cup\Phi_{1}\) since \(\lim_{t\rightarrow0^{+}}\frac{e^{t}1}{t^{r}}=0\) for each \(r\in(0,1)\). When \(a>1\), \(g\in\Phi_{2}\cap\Phi_{3}\), but \(g\notin\Psi\cup\Phi_{1}\cup\Phi_{4}\) since \(\lim_{t\rightarrow0^{+}}\frac{e^{t^{a}}1}{t^{r}}=0\) for each \(r\in(0,1)\) and \(e^{(t+s)^{a}}>e^{t^{a}}e^{s^{a}}\) for all \(s,t>0\).
Example 3
Let \(h(t)=1\) for \(t\in[0,a]\) and \(h(t)=e^{ta}\) for \(t>a\), where \(a>0\). Then \(h\in\Phi_{2}\), but \(h\notin\Psi\cup\Phi_{1}\cup\Phi_{3}\cup\Phi_{4}\) since neither (Î¨1) nor (Î¨2) is satisfied.
Example 4
Let \(p(t)=e^{\sqrt{te^{t}}}\) for \(t\geq0\). Then \(p\in\Phi_{1}\cap\Phi_{2}\cap\Phi_{3}\), but \(p\notin\Psi\cup\Phi_{4}\) since \(e^{\sqrt {(t_{0}+s_{0})e^{(t_{0}+s_{0})}}}=e^{\sqrt{2}e}>e^{2\sqrt{e}}=e^{\sqrt{t_{0}e^{t_{0}}}}e^{\sqrt{s_{0}e^{s_{0}}}}\) whenever \(t_{0}=s_{0}=1\).
Remark 1

(i)
Clearly, \(\Psi\subseteq\Phi_{1}\) and \(\Phi _{4}\subseteq\Phi_{3}\subseteq\Phi_{2}\). Moreover, from Examples 24 it follows that \(\Psi\subset\Phi_{1}\) and \(\Phi_{4}\subset\Phi_{3}\subset\Phi_{2}\).

(ii)
From Examples 14 we can conclude that \(\Phi_{2}\not\subset\Phi _{1}\), \(\Phi_{4}\not\subset\Psi\), \(\Phi_{1}\cap\Phi_{2}\neq\varnothing\), and \(\Psi\cap\Phi_{4}\neq\varnothing\).
Definition 1
Let \((X,d)\) be a metric space. A mapping \(T:X\rightarrow X\) is said to be:

(i)
a Ä†iriÄ‡ contraction [5] if there exist nonnegative numbers q, r, s, t with \(q+r+s+2t<1\) such that
$$d(Tx, Ty)\leq qd(x,y)+rd(x, Tx) + sd(y, Ty)+t\bigl[d(x, Ty) + d(y, Tx)\bigr], \quad \forall x,y\in X; $$ 
(ii)
a quasicontraction [6] if there exists \(\lambda\in[0,1)\) such that
$$d(Tx, Ty)\leq\lambda M_{d}(x,y),\quad \forall x,y\in X, $$where \(M_{d}(x,y)=\max\{d(x,y),d(x,Tx),d(y,Ty),\frac {d(x,Ty)+d(y,Tx)}{2}\}\);

(iii)
a JScontraction [7] if there exist \(\psi\in\Phi_{1}\) and \(\lambda \in[0,1)\) such that
$$ \psi\bigl(d(Tx, Ty)\bigr)\leq\psi\bigl(d(x,y)\bigr)^{\lambda}, \quad\forall x,y\in X \mbox{ with } Tx\neq Ty; $$(1) 
(iv)
a JSÄ†iriÄ‡ contraction [8] if there exist \(\psi\in\Psi\) and nonnegative numbers q, r, s, t with \(q+r+s+2t<1\) such that
$$\begin{aligned} &\psi\bigl(d(Tx, Ty)\bigr)\leq\psi\bigl(d(x,y)\bigr)^{q}\psi\bigl(d(x, Tx)\bigr)^{r} \psi\bigl(d(y, Ty)\bigr)^{s}\psi\bigl(d(x, Ty) + d(y, Tx)\bigr)^{t}, \\ &\quad \forall x,y\in X. \end{aligned}$$(2)
In the 1970s, Ä†iriÄ‡ [5, 6] established the following two wellknown generalizations of the Banach contraction principle.
Theorem 1
(see [5])
Let \((X,d)\) be a complete metric space, and \(T:X\rightarrow X\) a Ä†iriÄ‡ contraction. Then T has a unique fixed point in X.
Theorem 2
(see [6])
Let \((X,d)\) be a complete metric space, and \(T:X\rightarrow X\) a quasicontraction. Then T has a unique fixed point in X.
Recently, Jleli and Samet [7] proved the following fixed point result of JScontractions, which is a real generalization of the Banach contraction principle.
Theorem 3
(see [7], CorollaryÂ 2.1)
Let \((X,d)\) be a complete metric space, and \(T:X\rightarrow X\) a JScontraction with \(\psi\in\Phi_{1}\). Then T has a unique fixed point in X.
Remark 2
The Banach contraction principe follows immediately from TheoremÂ 3. Indeed, let \(T:X\rightarrow X\) be a JScontraction. Then, if we choose \(\psi(t)=e^{\sqrt{t}}\in\Phi_{1}\) and \(\lambda=\sqrt{k}\) in (1), then we get \(\sqrt{d(Tx,Ty)} \leq\sqrt{k}\sqrt{d(x,y)}\), that is,
which means that T is a Banach contraction. Remark that TheoremÂ 3 is a real generalization of the Banach contraction principle (see Example in [7]), but the Banach contraction principle is not a particular case of TheoremÂ 3 with \(\psi (t)=e^{t}\) since \(e^{t}\notin\Theta\).
Recently, Hussain et al. [8] presented the following extension of TheoremÂ 1 and TheoremÂ 3.
Theorem 4
(see [8], TheoremÂ 2.3)
Let \((X,d)\) be a complete metric space, and \(T:X\rightarrow X\) a continuous JSÄ†iriÄ‡ contraction with \(\psi\in\Psi\). Then T has a unique fixed point in X.
Remark 3
It is clear that TheoremÂ 1 is not a particular case of TheoremÂ 4 since, in TheoremÂ 1, a mapping T does not have to be continuous. In addition, letting \(\psi(t)=e^{\sqrt{t}}\) in (2), we can only obtain
which does not imply that T is a Ä†iriÄ‡ contraction whenever \(qr+rs+st\neq0\), and hence TheoremÂ 1 cannot be derived from TheoremÂ 4 by using the same method as in [7]. Therefore, TheoremÂ 4 may not be a real generalization of TheoremÂ 1.
In this paper, we generalize and improve Theorems 14 and remove or weaken the assumptions made on Ïˆ appearing in [7, 8].
2 Main results
Definition 2
Let \((X,d)\) be a metric space. A mapping \(T:X\rightarrow X\) is said to be a JSquasicontraction if there exist a function \(\psi:(0,+\infty )\rightarrow(1,+\infty)\) and \(\lambda\in(0,1)\) such that
Remark 4
(i) Each quasicontraction is a JSquasicontraction with \(\psi(t)=e^{t}\).
(ii) Each JScontraction is a JSquasicontraction whenever Ïˆ is nondecreasing.
(iii) Let \(T:X\rightarrow X\) and \(\psi:[0,+\infty)\rightarrow[1,+\infty )\) be such that
where q, r, s, t are nonnegative numbers with \(q+r+s+2t<1\). Then T is a JSquasicontraction with \(\lambda=p+r+s+2t\), provided that (Î¨1) is satisfied.
(iv) Let \(T:X\rightarrow X\) and \(\psi:[0,+\infty)\rightarrow[1,+\infty )\) be such that (2) is satisfied. Suppose that Ïˆ is a nondecreasing function such that (Î¨4) is satisfied. Then, \(\psi(d(x,Ty)+d(y,Tx))^{t} \leq\psi(\frac{d(x,Ty)+d(y,Tx)}{2})^{2t}\) for all \(x,y\in X\), and so (4) holds. Moreover, if (Î¨1) is satisfied, then it follows from (iii) that T is a JSquasicontraction with \(\lambda=p+r+s+2t\). Therefore, a JSÄ†iriÄ‡ contraction with \(\psi\in\Phi_{4}\) or \(\psi \in\Psi\) is certainly a JSquasicontraction.
Theorem 5
Let \((X,d)\) be a complete metric space, and \(T:X\rightarrow X\) a JSquasicontraction with \(\psi\in\Phi_{2}\). Then T has a unique fixed point in X.
Proof
Fix \(x_{0}\in X\) and let \(x_{n}=T^{n}x_{0}\) for each n.
We first show that T has a fixed point. We may assume that
Otherwise, there exists some positive integer p such that \(x_{p}=x_{p+1}\) and so \(x_{p}\) is a fixed point of T, and the proof is complete. Note that
If there exists some \(m_{0}\) such that \(d(x_{m_{0}},x_{m_{0}+1})>d(x_{m_{0}1},x_{m_{0}})\), then by (6) we get
a contradiction. Consequently, we obtain
which implies that \(\{d(x_{n},x_{n+1})\}\) is a decreasing sequence of nonnegative reals, and so there exists \(\alpha\geq0\) such that \(\lim_{n\rightarrow\infty }d(x_{n},x_{n+1})=\alpha\) and
Suppose that \(\alpha>0\). By (6) and (7), since Ïˆ is nondecreasing, we get
Letting \(n\rightarrow\infty\) in this inequality, we get \(\psi(\alpha)=1\), which contradicts the assumption that \(\psi(t)>1\) for each \(t>0\). Consequently, we have \(\alpha=0\), that is,
We claim that
Otherwise, there exist \(\beta>0\) and two subsequences \(\{x_{m_{k}}\}\) and \(\{x_{n_{k}}\}\) of \(\{x_{n}\}\) such that \(m_{k}\) is the smallest index with \(m_{k}>n_{k}>k\) for which
which indicates that
By (11), (12), and the triangle inequality we get
Letting \(k\rightarrow\infty\) in this inequality, by (9) we obtain
Also by the triangle inequality we get
Letting \(k\rightarrow\infty\) in this inequality, by (9) and (13) we obtain
In analogy to (14), by (9) and (13) we can prove that
It follows (9), (13), and (14) that
where
Note that (14) and (16) implies that there exists a positive integer \(k_{0}\) such that
Thus, by (3) we get
Letting \(k\rightarrow\infty\) in this inequality, by (14), (16), and the continuity of Ïˆ we obtain
a contradiction. Consequently, (10) holds, that is, \(\{x_{n}\}\) is a Cauchy sequence in X. Moreover, by the completeness of \((X,d)\) there exists \(x^{*}\in X\) such that
Suppose that \(d(x^{*},Tx^{*})>0\). It follows from (9) and (17) that there exists a positive integer \(n_{0}\) such that
Denoting
for each n, by (18) we get
From the continuity of d it follows that
which implies that there exists a positive integer \(n_{1}\) such that
Thus, by (3) we get
and so, by (19),
Letting \(n\rightarrow\infty\) in this inequality, by (20) and the continuity of Ïˆ we obtain
a contradiction. Consequently, \(d(x^{*},Tx^{*})=0\), that is, \(x^{*}=Tx^{*}\).
Let x be another fixed point of T. Suppose that \(d(x,x^{*})>0\). Then by (3) we get
where \(M_{d}(x,x^{*})=\max\{d(x,x^{*}),\frac{d(x,x^{*})+d(x^{*},x)}{2}\}=d(x,x^{*})\). Thus, we obtain
a contradiction. Consequently, we have \(x=x^{*}\). This shows that \(x^{*}\) is the unique fixed point of T. The proof is completed.â€ƒâ–¡
Remark 5
In view of ExampleÂ 2 and (i) of RemarkÂ 4, TheoremÂ 2 is a particular case of TheoremÂ 5 with \(\psi(t)=e^{t}\in\Phi_{2}\). The following example shows that TheoremÂ 5 is a real generalization of TheoremÂ 2.
Example 5
Let \(X=\{\tau_{n}\}\) and \(d(x,y)=xy\), where \(\tau_{n}=\frac{n(n+1)(n+2)}{3}\) for each n. Clearly, \((X,d)\) is a complete metric space. Define the mapping \(T:X\rightarrow X\) by \(T\tau_{1}=\tau_{1}\) and \(T\tau_{n}=\tau_{n1}\) for each \(n\geq2\).
We show that T is a JSquasicontraction with \(\psi(t)=e^{te^{t}}\). In fact, it suffices to show that there exists \(\lambda\in(0,1)\) such that, for all \(x,y\in X\) with \(Tx\neq Ty\),
In the case of \(m>2\) and \(n=1\), we have \(d(T\tau_{1},T\tau_{m})=d(\tau _{1},\tau_{m1})=\frac{(m1)m(m+1)6}{3}\) and
and so
In the case \(m>n>1\), we have
and
and so
This shows that T is a JSquasicontraction with \(\psi(t)=e^{te^{t}}\) and \(\lambda\in[e^{6},1)\). Note that \(e^{te^{t}}\in\Phi_{2}\) by ExampleÂ 1. Then from TheoremÂ 5 we know that T has a unique fixed point \(\tau_{1}\).
For each \(m>2\), we have
which implies that T is not a quasicontraction. Hence, TheoremÂ 2 is not applicable here.
On the other hand, it is not hard to check that there exists \(\lambda \in(0,1)\) (resp. nonnegative numbers q, r, s, t with \(q + r + s + 2t < 1\)) such that (1) (resp. (2)) is satisfied with \(\psi(t)=e^{te^{t}}\). But neither TheoremÂ 3 nor TheoremÂ 4 is applicable in this situation since \(e^{te^{t}}\notin\Psi\cup\Phi_{1}\) by ExampleÂ 1.
Example 6
Let \(X=\{1,2,3\}\) and \(d(x,y)=xy\). Clearly, \((X,d)\) is a complete metric space. Define the mapping \(T:X\rightarrow X\) by \(T1=T2=1\) and \(T3=2\).
We show that T is a JSquasicontraction with \(\psi(t)=e^{\sqrt {te^{t}}}\). In fact, it suffices to show that there exists \(\lambda\in(0,1)\) such that, for all \(x,y\in X\) with \(Tx\neq Ty\),
Then, we only need to consider the cases \(x=1\), \(y=3\) and \(x=2\), \(y=3\). For both cases, we have \(d(T1,T3)=d(T2,T3)=1\) and \(M_{d}(1,3)=M_{d}(2,3)=2\), and so
which implies that T is a JSquasicontraction with \(\psi(t)=e^{\sqrt {te^{t}}}\) and \(\lambda\in[\sqrt{\frac{e^{1}}{2}},1)\). Note that \(e^{te^{t}}\in\Phi_{2}\) by ExampleÂ 4. Then from TheoremÂ 5 we know T has a unique fixed point \(x=1\).
When \(x=2\) and \(y=3\), we have \(d(T2,T3)=d(2,3)=1\) and hence \(\frac {d(T2,T3)e^{d(T2,T3)d(2,3)}}{d(2,3)}=1\), which implies that T is not a JScontraction with \(\psi(t)=e^{\sqrt{te^{t}}}\). Therefore, TheoremÂ 3 is not applicable here.
In addition, it is not hard to check that there exist nonnegative numbers q, r, s, t with \(q + r + s + 2t < 1\) such that (2) is satisfied with \(\psi(t)=e^{\sqrt {te^{t}}}\). However, TheoremÂ 4 is not applicable here since \(e^{\sqrt{te^{t}}}\notin\Psi\) by ExampleÂ 4.
Theorem 6
Let \((X,d)\) be a complete metric space, and \(T:X\rightarrow X\). Assume that there exist \(\psi\in\Phi_{3}\) and nonnegative numbers q, r, s, t with \(q+r+s+2t<1\) such that (4) is satisfied. Then T has a unique fixed point in X.
Proof
In view of (iii) of RemarkÂ 4, T is a JSquasicontraction with \(\lambda=q+r+s+2t\). In the case where \(q+r+s+2t=0\), by (3) we have \(\psi(d(Tx,Ty))=1\) for all \(x,y\in X\). Moreover, by (Î¨1) we get \(d(Tx,Ty)=0\) for all \(x,y\in X\). This shows that \(y=Tx\) is a fixed point of T. Let z be another fixed point of T. Then \(d(y,z)=d(Ty,Tz)=0\), and hence \(y=z\), that is, T has a unique fixed point. In the case where \(0< q+r+s+2t<1\), the conclusion immediately follows from TheoremÂ 5. The proof is completed.â€ƒâ–¡
Remark 6
TheoremÂ 2 and TheoremÂ 1 are respectively particular cases of TheoremÂ 5 and TheoremÂ 6 with \(\psi(t)=e^{t}\), whereas they are not particular cases of TheoremÂ 3 and TheoremÂ 4 with \(\psi(t)=e^{t}\) since \(e^{t}\in\Phi_{2}\cap\Phi_{3}\) but \(e^{t}\notin\Psi\cup\Phi_{1}\). Hence, TheoremÂ 5 and TheoremÂ 6 are new generalizations of TheoremÂ 2 and TheoremÂ 1.
In view of (ii) and (iv) of RemarkÂ 4, we have the following two corollaries of TheoremÂ 5 and TheoremÂ 6.
Corollary 1
Let \((X,d)\) be a complete metric space, and \(T:X\rightarrow X\) a JScontraction with \(\psi\in\Phi_{2}\). Then T has a unique fixed point in X.
Corollary 2
Let \((X,d)\) be a complete metric space, and \(T:X\rightarrow X\) a JSÄ†iriÄ‡ contraction with \(\psi\in\Phi_{4}\). Then T has a unique fixed point in X.
Remark 7
Conditions (Î¨2) and (Î¨3) assumed in Theorems 3 and 4 are removed from Corollaries 1 and 2 at the expense that Ïˆ is continuous. Thus, Corollaries 1 and 2 partially improve Theorems 3 and 4.
Taking \(\psi(t)=e^{t^{a}}\) (\(a>0\)) in TheoremÂ 6, we have the following new generalization of TheoremÂ 1.
Corollary 3
Let \((X,d)\) be a complete metric space, and \(T:X\rightarrow X\). Assume that there exist \(a>0\) and nonnegative numbers q, r, s, t with \(q+r+s+2t<1\) such that
Then T has a unique fixed point in X.
Remark 8
TheoremÂ 1 is a particular case of CorollaryÂ 3 with \(a=1\).
Corollary 4
(see [8], TheoremÂ 2.4 and CorollaryÂ 2.9)
Let \((X,d)\) be a complete metric space, and \(T:X\rightarrow X\). Assume that there exist nonnegative numbers q, r, s, t with \(q+r+s+2t<1\) such that
where \(a=\frac{1}{2}\) or \(a=\frac{1}{n}\). Then T has a unique fixed point in X.
Proof
For each \(a\in(0,1]\), we have \((d(x,Ty)+d(y,Tx))^{a}\leq 2(\frac{d(x,Ty)+d(y,Tx)}{2})^{a}\), and so (23) immediately follows from (24). Thus, by CorollaryÂ 3, T has a unique fixed point. The proof is completed.â€ƒâ–¡
Remark 9
TheoremÂ 2.4 and CorollaryÂ 2.9 of [8] are consequences of TheoremÂ 1. In fact, let \(a\in(0,1]\) and \(D(x,y)=d(x,y)^{a}\) for all \(x,y\in X\). Then \((X,D)\) is a complete metric space by the completeness of \((X,d)\). Note that \((d(x,Ty)+d(y,Tx))^{a}\leq d(x,Ty)^{a}+d(y,Tx)^{a}\) for all \(x,y\in X\). Then (24) implies
that is, T is a Ä†iriÄ‡ contraction in \((X,D)\). Therefore, TheoremÂ 2.4 and CorollaryÂ 2.9 of [8] immediately follow from TheoremÂ 1. However, CorollaryÂ 3 cannot be derived from TheoremÂ 1 by the previous method since the pair \((X,D)\) is not a metric space whenever \(a>1\).
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Acknowledgements
The work was supported by the Natural Science Foundation of China (11161022, 11561026, 71462015), the Natural Science Foundation of Jiangxi Province (20142BCB23013, 20143ACB21012, 20151BAB201003, 20151BAB201023), the Natural Science Foundation of Jiangxi Provincial Education Department (KJLD14034, GJJ150479).
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Li, Z., Jiang, S. Fixed point theorems of JSquasicontractions. Fixed Point Theory Appl 2016, 40 (2016). https://doi.org/10.1186/s1366301605263
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DOI: https://doi.org/10.1186/s1366301605263