## 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 [217]). For example, the concepts of Ćirić contraction [5], quasi-contraction [6], JS-contraction [7], and JS-Ćirić contraction [8] have been introduced, and many interesting generalizations of the Banach contraction principle are obtained.

Following Hussain et al. [8], we denote by Ψ the set of all nondecreasing functions $$\psi:[0,+\infty)\rightarrow[1,+\infty)$$ satisfying the following conditions:

(Ψ1):

$$\psi(t)=1$$ if and only if $$t=0$$;

(Ψ2):

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

(Ψ3):

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

(Ψ4):

$$\psi(t+s)\leq\psi(t)\psi(s)$$ 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 (Ψ1) nor (Ψ2) 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 Ćirić 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-Ćirić 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, Ćirić [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 Ćirić 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], Corollary 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 Theorem 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 Theorem 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 Theorem 3 with $$\psi (t)=e^{t}$$ since $$e^{t}\notin\Theta$$.

Recently, Hussain et al. [8] presented the following extension of Theorem 1 and Theorem 3.

### Theorem 4

(see [8], Theorem 2.3)

Let $$(X,d)$$ be a complete metric space, and $$T:X\rightarrow X$$ a continuous JS-Ćirić contraction with $$\psi\in\Psi$$. Then T has a unique fixed point in X.

### Remark 3

It is clear that Theorem 1 is not a particular case of Theorem 4 since, in Theorem 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 Ćirić contraction whenever $$qr+rs+st\neq0$$, and hence Theorem 1 cannot be derived from Theorem 4 by using the same method as in [7]. Therefore, Theorem 4 may not be a real generalization of Theorem 1.

In this paper, we generalize and improve Theorems 1-4 and remove or weaken the assumptions made on ψ 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 ψ 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 (Ψ1) is satisfied.

(iv) Let $$T:X\rightarrow X$$ and $$\psi:[0,+\infty)\rightarrow[1,+\infty )$$ be such that (2) is satisfied. Suppose that ψ is a nondecreasing function such that (Ψ4) 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 (Ψ1) is satisfied, then it follows from (iii) that T is a JS-quasi-contraction with $$\lambda=p+r+s+2t$$. Therefore, a JS-Ćirić 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 ψ 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 ψ 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 ψ 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. □

### Remark 5

In view of Example 2 and (i) of Remark 4, Theorem 2 is a particular case of Theorem 5 with $$\psi(t)=e^{t}\in\Phi_{2}$$. The following example shows that Theorem 5 is a real generalization of Theorem 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 Example 1. Then from Theorem 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, Theorem 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 Theorem 3 nor Theorem 4 is applicable in this situation since $$e^{te^{t}}\notin\Psi\cup\Phi_{1}$$ by Example 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 Example 4. Then from Theorem 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, Theorem 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, Theorem 4 is not applicable here since $$e^{\sqrt{te^{t}}}\notin\Psi$$ by Example 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 Remark 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 (Ψ1) 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 Theorem 5. The proof is completed. □

### Remark 6

Theorem 2 and Theorem 1 are respectively particular cases of Theorem 5 and Theorem 6 with $$\psi(t)=e^{t}$$, whereas they are not particular cases of Theorem 3 and Theorem 4 with $$\psi(t)=e^{t}$$ since $$e^{t}\in\Phi_{2}\cap\Phi_{3}$$ but $$e^{t}\notin\Psi\cup\Phi_{1}$$. Hence, Theorem 5 and Theorem 6 are new generalizations of Theorem 2 and Theorem 1.

In view of (ii) and (iv) of Remark 4, we have the following two corollaries of Theorem 5 and Theorem 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-Ćirić contraction with $$\psi\in\Phi_{4}$$. Then T has a unique fixed point in X.

### Remark 7

Conditions (Ψ2) and (Ψ3) assumed in Theorems 3 and 4 are removed from Corollaries 1 and 2 at the expense that ψ is continuous. Thus, Corollaries 1 and 2 partially improve Theorems 3 and 4.

Taking $$\psi(t)=e^{t^{a}}$$ ($$a>0$$) in Theorem 6, we have the following new generalization of Theorem 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

Theorem 1 is a particular case of Corollary 3 with $$a=1$$.

### Corollary 4

(see [8], Theorem 2.4 and Corollary 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 Corollary 3, T has a unique fixed point. The proof is completed. □

### Remark 9

Theorem 2.4 and Corollary 2.9 of [8] are consequences of Theorem 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 Ćirić contraction in $$(X,D)$$. Therefore, Theorem 2.4 and Corollary 2.9 of [8] immediately follow from Theorem 1. However, Corollary 3 cannot be derived from Theorem 1 by the previous method since the pair $$(X,D)$$ is not a metric space whenever $$a>1$$.