# Fixed point theorems of JS-quasi-contractions

## Abstract

In this paper, we introduce the concept of JS-quasi-contraction and prove some fixed point results for JS-quasi-contractions 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], quasi-contraction [6], JS-contraction [7], and JS-Ä†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:

\begin{aligned}& \begin{aligned}[b] \Phi_{1}={}&\bigl\{ \psi:(0,+\infty)\rightarrow(1,+ \infty): \psi\mbox{ is a nondecreasing function satisfying} \\ &{}(\Psi2)\mbox{ and }(\Psi3)\bigr\} , \end{aligned} \\& \Phi_{2}=\bigl\{ \psi:(0,+\infty)\rightarrow(1,+\infty): \psi\mbox{ is a nondecreasing continuous function}\bigr\} , \\& \begin{aligned}[b] \Phi_{3}={}&\bigl\{ \psi:[0,+\infty)\rightarrow[1,+ \infty): \psi\mbox{ is a nondecreasing continuous function satisfying} \\ &{}(\Psi1)\bigr\} , \end{aligned} \\& \begin{aligned}[b] \Phi_{4}={}&\bigl\{ \psi:[0,+\infty)\rightarrow[1,+ \infty): \psi\mbox{ is a nondecreasing continuous function satisfying} \\ &{}(\Psi1)\mbox{ and }(\Psi4)\bigr\} . \end{aligned} \end{aligned}

### 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 $$h(t)=e^{t-a}$$ 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

1. (i)

Clearly, $$\Psi\subseteq\Phi_{1}$$ and $$\Phi _{4}\subseteq\Phi_{3}\subseteq\Phi_{2}$$. Moreover, from Examples 2-4 it follows that $$\Psi\subset\Phi_{1}$$ and $$\Phi_{4}\subset\Phi_{3}\subset\Phi_{2}$$.

2. (ii)

From Examples 1-4 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:

1. (i)

a Ä†iriÄ‡ contraction [5] if there exist nonnegative numbers q, r, s, t with $$q+r+s+2t<1$$ such that

$$d(Tx, Ty)\leq qd(x,y)+rd(x, Tx) + sd(y, Ty)+t\bigl[d(x, Ty) + d(y, Tx)\bigr], \quad \forall x,y\in X;$$
2. (ii)

a quasi-contraction [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 $$M_{d}(x,y)=\max\{d(x,y),d(x,Tx),d(y,Ty),\frac {d(x,Ty)+d(y,Tx)}{2}\}$$;

3. (iii)

a JS-contraction [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)
4. (iv)

a JS-Ä†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 well-known 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 quasi-contraction. Then T has a unique fixed point in X.

Recently, Jleli and Samet [7] proved the following fixed point result of JS-contractions, 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 JS-contraction 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 JS-contraction. 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,

$$d(Tx, Ty)\leq k d(x,y), \quad\forall x,y\in X,$$

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 JS-Ä†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

$$\sqrt{d(Tx,Ty)}\leq q\sqrt{d(x,y)}+r\sqrt{d(x,Tx)}+s\sqrt {d(y,Ty)}+t \sqrt{d(x,Ty)+d(y,Tx)}, \quad\forall x,y\in X,$$

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 1-4 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 JS-quasi-contraction if there exist a function $$\psi:(0,+\infty )\rightarrow(1,+\infty)$$ and $$\lambda\in(0,1)$$ such that

$$\psi\bigl(d(Tx, Ty)\bigr)\leq\psi\bigl(M_{d}(x,y) \bigr)^{\lambda},\quad \forall x,y\in X \mbox{ with } Tx\neq Ty.$$
(3)

### Remark 4

(i) Each quasi-contraction is a JS-quasi-contraction with $$\psi(t)=e^{t}$$.

(ii) Each JS-contraction is a JS-quasi-contraction whenever Ïˆ is nondecreasing.

(iii) Let $$T:X\rightarrow X$$ and $$\psi:[0,+\infty)\rightarrow[1,+\infty )$$ be 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\biggl( \frac{d(x, Ty)+d(y, Tx)}{2}\biggr)^{2t}, \\ &\quad\quad\forall x,y\in X, \end{aligned}
(4)

where q, r, s, t are nonnegative numbers with $$q+r+s+2t<1$$. Then T is a JS-quasi-contraction 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 JS-quasi-contraction with $$\lambda=p+r+s+2t$$. Therefore, a JS-Ä†iriÄ‡ contraction with $$\psi\in\Phi_{4}$$ or $$\psi \in\Psi$$ is certainly a JS-quasi-contraction.

### Theorem 5

Let $$(X,d)$$ be a complete metric space, and $$T:X\rightarrow X$$ a JS-quasi-contraction 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

$$d(x_{n},x_{n+1})>0,\quad \forall n.$$
(5)

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

\begin{aligned} &\max\bigl\{ d(x_{n-1},x_{n}),d(x_{n},x_{n+1}) \bigr\} \\ &\quad\leq\max\biggl\{ d(x_{n-1},x_{n}),d(x_{n},x_{n+1}), \frac {d(x_{n-1},x_{n+1})+d(x_{n},x_{n})}{2}\biggr\} \\ &\quad=M_{d}(x_{n-1},x_{n})\leq\max\biggl\{ d(x_{n-1},x_{n}),d(x_{n},x_{n+1}), \frac {d(x_{n-1},x_{n})+d(x_{n},x_{n+1})}{2}\biggr\} \\ &\quad=\max\bigl\{ d(x_{n-1},x_{n}),d(x_{n},x_{n+1}) \bigr\} , \quad \forall n. \end{aligned}

Then by (3) and (5) we get

$$\psi\bigl(d(x_{n},x_{n+1})\bigr)\leq\psi\bigl(M_{d}(x_{n-1},x_{n}) \bigr)=\psi\bigl(\max\bigl\{ d(x_{n-1},x_{n}),d(x_{n},x_{n+1}) \bigr\} \bigr)^{\lambda},\quad \forall n.$$
(6)

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

$$\psi\bigl(d(x_{m_{0}},x_{m_{0}+1})\bigr)\leq \psi \bigl(d(x_{m_{0}},x_{m_{0}+1})\bigr)^{\lambda}< \psi \bigl(d(x_{m_{0}},x_{m_{0}+1})\bigr),$$

$$d(x_{n},x_{n+1})\leq d(x_{n-1},x_{n}), \quad \forall n,$$

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

$$d(x_{n},x_{n+1})\geq\alpha.$$
(7)

Suppose that $$\alpha>0$$. By (6) and (7), since Ïˆ is nondecreasing, we get

$$1< \psi(\alpha)\leq\psi\bigl(d(x_{n},x_{n+1})\bigr)\leq\psi \bigl(d(x_{n-1},x_{n})\bigr)^{\lambda}\leq\cdots\leq\psi \bigl(d(x_{0},x_{1})\bigr)^{\lambda ^{n}}, \quad\forall n.$$
(8)

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,

$$\lim_{n\rightarrow\infty}d(x_{n},x_{n+1})=0.$$
(9)

We claim that

$$\lim_{m,n\rightarrow\infty}d(x_{n},x_{m})=0.$$
(10)

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

$$d(x_{n_{k}},x_{m_{k}})\geq\beta,$$
(11)

which indicates that

$$d(x_{n_{k}},x_{m_{k}-1})< \beta.$$
(12)

By (11), (12), and the triangle inequality we get

\begin{aligned} \beta&\leq d(x_{n_{k}},x_{m_{k}})\leq d(x_{n_{k}},x_{m_{k}-1})+d(x_{m_{k}-1},x_{m_{k}}) \\ &< \beta+d(x_{m_{k}-1},x_{m_{k}}), \quad\forall m_{k}>n_{k}>k. \end{aligned}

Letting $$k\rightarrow\infty$$ in this inequality, by (9) we obtain

$$\lim_{k\rightarrow\infty}d(x_{n_{k}},x_{m_{k}})= \beta.$$
(13)

Also by the triangle inequality we get

\begin{aligned} &d(x_{n_{k}},x_{m_{k}})-d(x_{n_{k}+1},x_{n_{k}}) -d(x_{m_{k}},x_{m_{k}+1}) \\ &\quad\leq d(x_{n_{k}+1},x_{m_{k}+1})\leq d(x_{n_{k}+1},x_{n_{k}}) +d(x_{n_{k}},x_{m_{k}})+d(x_{m_{k}},x_{m_{k}+1}), \quad\forall m_{k}>n_{k}>k. \end{aligned}

Letting $$k\rightarrow\infty$$ in this inequality, by (9) and (13) we obtain

$$\lim_{k\rightarrow\infty}d(x_{n_{k}+1},x_{m_{k}+1})= \beta.$$
(14)

In analogy to (14), by (9) and (13) we can prove that

$$\lim_{k\rightarrow\infty}d(x_{n_{k}},x_{m_{k}+1})=\lim _{k\rightarrow\infty}d(x_{n_{k}+1},x_{m_{k}})=\beta.$$
(15)

It follows (9), (13), and (14) that

$$\lim_{k\rightarrow\infty}M_{d}(x_{n_{k}},x_{m_{k}})= \beta,$$
(16)

where

\begin{aligned} \beta&\leq d(x_{n_{k}},x_{m_{k}})\leq M_{d}(x_{n_{k}},x_{m_{k}}) \\ &=\max\biggl\{ d(x_{n_{k}},x_{m_{k}}) d(x_{n_{k}},x_{n_{k}+1}), d(x_{m_{k}},x_{m_{k}+1}),\frac {d(x_{n_{k}},x_{m_{k}+1})+d(x_{n_{k}+1},x_{m_{k}})}{2}\biggr\} . \end{aligned}

Note that (14) and (16) implies that there exists a positive integer $$k_{0}$$ such that

$$d(x_{n_{k}+1},x_{m_{k}+1})>0 \quad\mbox{and} \quad M_{d}(x_{n_{k}},x_{m_{k}})>0,\quad \forall k\geq k_{0}.$$

Thus, by (3) we get

$$\psi\bigl(d(x_{n_{k}+1},x_{m_{k}+1})\bigr)=\psi\bigl(d(Tx_{n_{k}},Tx_{m_{k}}) \bigr)\leq\psi \bigl(M_{d}(x_{n_{k}},x_{m_{k}}) \bigr)^{\lambda}, \quad\forall m_{k}>n_{k}>k\geq k_{0}.$$

Letting $$k\rightarrow\infty$$ in this inequality, by (14), (16), and the continuity of Ïˆ we obtain

$$\psi(\beta)=\lim_{k\rightarrow\infty}\psi \bigl(d(x_{n_{k}+1},x_{m_{k}+1}) \bigr)\leq \lim_{k\rightarrow\infty}\psi\bigl(M_{d}(x_{n_{k}},x_{m_{k}}) \bigr)^{\lambda }=\psi(\beta)^{\lambda}< \psi(\beta),$$

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

$$\lim_{n\rightarrow\infty}d\bigl(x_{n},x^{*}\bigr)=0.$$
(17)

Suppose that $$d(x^{*},Tx^{*})>0$$. It follows from (9) and (17) that there exists a positive integer $$n_{0}$$ such that

$$d\bigl(x_{n},x^{*}\bigr)\leq d\bigl(x^{*},Tx^{*}\bigr)\quad\mbox{and}\quad d(x_{n},x_{n+1})\leq d\bigl(x^{*},Tx^{*}\bigr),\quad \forall n\geq n_{0}.$$
(18)

Denoting

$$M_{d}\bigl(x_{n},x^{*}\bigr)=\max\biggl\{ d \bigl(x_{n},x^{*}\bigr),d(x_{n},x_{n+1}),d \bigl(x^{*},Tx^{*}\bigr),\frac {d(x_{n},Tx^{*})+d(x^{*},x_{n+1})}{2}\biggr\}$$

for each n, by (18) we get

$$M_{d}\bigl(x_{n},x^{*}\bigr)=d\bigl(x^{*},Tx^{*}\bigr),\quad \forall n\geq n_{0}.$$
(19)

From the continuity of d it follows that

$$\lim_{n\rightarrow\infty}d\bigl(x_{n+1},Tx^{*}\bigr)=d\bigl(x^{*},Tx^{*} \bigr),$$
(20)

which implies that there exists a positive integer $$n_{1}$$ such that

$$d\bigl(x_{n+1},Tx^{*}\bigr)>0, \quad\forall n\geq n_{1}.$$
(21)

Thus, by (3) we get

$$\psi\bigl(d\bigl(x_{n+1},Tx^{*}\bigr)\bigr)=\psi\bigl(d \bigl(Tx_{n},Tx^{*}\bigr)\bigr)\leq\psi \bigl(M_{d} \bigl(x_{n},x^{*}\bigr)\bigr)^{\lambda},\quad \forall n\geq n_{1},$$

and so, by (19),

$$\psi\bigl(d\bigl(x_{n+1},Tx^{*}\bigr)\bigr)\leq\psi\bigl(d\bigl(x^{*},Tx^{*} \bigr)\bigr)^{\lambda}, \quad\forall n\geq \max\{n_{0},n_{1} \}.$$
(22)

Letting $$n\rightarrow\infty$$ in this inequality, by (20) and the continuity of Ïˆ we obtain

$$\psi\bigl(d\bigl(x^{*},Tx^{*}\bigr)\bigr)=\psi\bigl(d\bigl(x_{n+1},Tx^{*} \bigr)\bigr)\leq\psi\bigl(d\bigl(x^{*},Tx^{*}\bigr)\bigr)^{\lambda }< \psi\bigl(d \bigl(x^{*},Tx^{*}\bigr)\bigr),$$

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

$$\psi\bigl(d\bigl(x,x^{*}\bigr)\bigr)=\psi\bigl(d\bigl(Tx,Tx^{*}\bigr)\bigr)\leq\psi \bigl(M_{d}\bigl(x,x^{*}\bigr)\bigr)^{\lambda},$$

where $$M_{d}(x,x^{*})=\max\{d(x,x^{*}),\frac{d(x,x^{*})+d(x^{*},x)}{2}\}=d(x,x^{*})$$. Thus, we obtain

$$\psi\bigl(d\bigl(x,x^{*}\bigr)\bigr)\leq \psi\bigl(d\bigl(x,x^{*}\bigr) \bigr)^{\lambda}< \psi\bigl(d\bigl(x,x^{*}\bigr)\bigr),$$

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 $$d(x,y)=|x-y|$$, 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 $$T\tau_{n}=\tau_{n-1}$$ for each $$n\geq2$$.

We show that T is a JS-quasi-contraction 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 $$Tx\neq Ty$$,

$$\frac{d(Tx,Ty)e^{d(Tx,Ty)-M_{d}(x,y)}}{M_{d}(x,y)}\leq\lambda.$$

In the case of $$m>2$$ and $$n=1$$, we have $$d(T\tau_{1},T\tau_{m})=d(\tau _{1},\tau_{m-1})=\frac{(m-1)m(m+1)-6}{3}$$ and

\begin{aligned} M_{d}(\tau_{1},\tau_{m})&=\max\biggl\{ d( \tau_{1},\tau_{m}),d(\tau_{1},\tau _{1}),d(\tau_{m},\tau_{m-1}), \frac{d(\tau_{1},\tau_{m-1})+d(\tau_{m},\tau_{1})}{3} \biggr\} \\ &=d(\tau_{1},\tau_{m})=\frac{m(m+1)(m+2)-6}{3}, \end{aligned}

and so

$$\frac{d(T\tau_{1},T\tau_{m})e^{d(T\tau_{1},T\tau_{m})-M_{d}(\tau_{1},\tau _{m})}}{M_{d}(\tau_{1},\tau_{m})}= \frac{(m-1)m(m+1)-6}{m(m+1)(m+2)-6}e^{-m(m+1)}< e^{-6}.$$

In the case $$m>n>1$$, we have

$$d(T\tau_{n},T\tau_{m})=d(\tau_{n-1}, \tau_{m-1})=\frac{(m-n)(m^{2}+n^{2}+mn-1)}{3}$$

and

\begin{aligned} M_{d}(\tau_{n},\tau_{m})&=\max\biggl\{ d( \tau_{n},\tau_{m}),d(\tau_{n},\tau _{n-1}),d(\tau_{m},\tau_{m-1}), \frac{d(\tau_{n},\tau_{m-1})+d(\tau_{m},\tau_{n-1})}{2} \biggr\} \\ &=\max\biggl\{ \tau_{m}-\tau_{n},\frac{\tau_{m}+\tau_{m-1}-\tau_{n}-\tau_{n-1}}{2}\biggr\} =\tau_{m}-\tau_{n} \\ &=\frac{(m-n)(m^{2}+n^{2}+mn+3(m+n)+2)}{3}, \end{aligned}

and so

\begin{aligned} \frac{d(T\tau_{n},T\tau_{m})e^{d(T\tau_{n},T\tau_{m})-M_{d}(\tau_{n},\tau _{m})}}{M_{d}(\tau_{n},\tau_{m})}&=\frac {m^{2}+n^{2}+mn-1}{m^{2}+n^{2}+mn+3(m+n)+2}e^{(n-m)(m+n+1)} \\ &\leq e^{6(n-m)}\leq e^{-6}. \end{aligned}

This shows that T is a JS-quasi-contraction 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

$$\lim_{m\rightarrow\infty}\frac{d(T\tau_{1},T\tau_{m})}{M_{d}(\tau _{1},\tau_{m})}= \lim_{m\rightarrow\infty} \frac{(m-1)m(m+1)-6}{m(m+1)(m+2)-6}=1,$$

which implies that T is not a quasi-contraction. 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 $$d(x,y)=|x-y|$$. 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 JS-quasi-contraction 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 $$Tx\neq Ty$$,

$$\frac{d(Tx,Ty)e^{d(Tx,Ty)-M_{d}(x,y)}}{M_{d}(x,y)}\leq\lambda^{2}.$$

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

$$\frac{d(T1,T3)e^{d(T1,T3)-M_{d}(1,3)}}{M_{d}(1,3)}=\frac {d(T2,T3)e^{d(T2,T3)-M_{d}(2,3)}}{M_{d}(2,3)}=\frac{e^{-1}}{2},$$

which implies that T is a JS-quasi-contraction 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 JS-contraction 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 JS-quasi-contraction 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 JS-contraction 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 JS-Ä†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

\begin{aligned} &d(Tx,Ty)^{a}\leq qd(x,y)^{a}+rd(x,Tx)^{a}+sd(y,Ty)^{a}+2t \biggl(\frac {d(x,Ty)+d(y,Tx)}{2} \biggr)^{a}, \\ &\quad\forall x,y\in X. \end{aligned}
(23)

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

$$d(Tx,Ty)^{a}\leq qd(x,y)^{a}+rd(x,Tx)^{a}+sd(y,Ty)^{a}+t \bigl(d(x,Ty)+d(y,Tx)\bigr)^{a},\quad \forall x,y\in X,$$
(24)

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

$$D(Tx,Ty)\leq qD(x,y)+rD(x,Tx)+sD(y,Ty)+t\bigl(D(x,Ty)+D(y,Tx)\bigr), \quad \forall x,y\in X,$$

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 $$a>1$$.

## References

1. Banach, S: Sur les opÃ©rations dans les ensembles abstraits et leur application aux Ã©quations intÃ©grales. Fundam. Math. 3, 133-181 (1922) (in French)

2. Kannan, R: Some results on fixed points. Bull. Calcutta Math. Soc. 60, 71-76 (1968)

3. Chatterjea, SK: Fixed point theorems. C. R. Acad. Bulgare Sci. 25, 727-730 (1972)

4. Reich, S: Some remarks concerning contraction mappings. Can. Math. Bull. 14, 121-124 (1971)

5. Ä†iriÄ‡, L: Generalized contractions and fixed-point theorems. Publ. Inst. Math. (Belgr.) 12(26), 19-26 (1971)

6. Ä†iriÄ‡, L: A generalization of Banachâ€™s contraction principle. Proc. Am. Math. Soc. 45(2), 267-273 (1974)

7. Jleli, M, Samet, B: A new generalization of the Banach contraction principle. J. Inequal. Appl. 2014, 38 (2014)

8. Hussain, N, Parvaneh, V, Samet, B, Vetro, C: Some fixed point theorems for generalized contractive mappings in complete metric spaces. Fixed Point Theory Appl. 2015, 185 (2015)

9. Jleli, M, Samet, B: A generalized metric space and related fixed point theorems. Fixed Point Theory Appl. 2015, 61 (2015)

10. Argoubi, H, Samet, B, Vetro, C: Nonlinear contractions involving simulation functions in a metric space with a partial order. J. Nonlinear Sci. Appl. 8, 1082-1094 (2015)

11. Samet, B: Fixed points for Î±-Ïˆ contractive mappings with an application to quadratic integral equations. Electron. J.Â Differ. Equ. 2014, 152 (2014)

12. Argoubi, H, Samet, B, Turinici, M: Fixed point results on a metric space endowed with a finite number of graphs and applications. Czechoslov. Math. J. 64, 241-250 (2014)

13. Aydi, H, Jellali, M, KarapÄ±nar, E: Common fixed points for generalized Î±-implicit contractions in partial metric spaces: consequences and application. Rev. R. Acad. Cienc. Exactas FÃ­s. Nat., Ser. A Mat. 109, 367-384 (2015)

14. KarapÄ±nar, E, Lakzian, H: $$(\alpha,\psi)$$-Contractive mappings on generalized quasi-metric spaces. J. Funct. Spaces 2014, 914398 (2014)

15. KarapÄ±nar, E: A discussion on â€˜Î±-Ïˆ-Geraghty contraction type mappingsâ€™. Filomat 28, 761-766 (2014)

16. KarapÄ±nar, E, Samet, B: Generalized $$(\alpha,\psi)$$ contractive type mappings and related fixed point theorems with applications. Abstr. Appl. Anal. 2012, 793486 (2012)

17. KarapÄ±nar, E, Aydi, H, Samet, B: Fixed points for generalized $$(\alpha,\psi)$$-contractions on generalized metric spaces. J.Â Inequal. Appl. 2014, 229 (2014)

## 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).

## Author information

Authors

### Corresponding author

Correspondence to Shujun Jiang.

### Competing interests

The authors declare that they have no competing interests.

### Authorsâ€™ contributions

The authors have contributed in this work on an equal basis. All authors read and approved the final manuscript.

## Rights and permissions

Reprints and permissions

Li, Z., Jiang, S. Fixed point theorems of JS-quasi-contractions. Fixed Point Theory Appl 2016, 40 (2016). https://doi.org/10.1186/s13663-016-0526-3