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 [2–17]). 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

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

a contradiction. Consequently, we obtain

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

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 \(a>1\).