For \(\psi\in\Psi_{2}\) and \(t\in[0,+\infty)\), set \(\eta(t)=\ln(\psi (t))\). Then it is easy to check that \(\eta:[0,+\infty)\rightarrow[0,+\infty)\) has the following properties:
 (\(\eta_{1}\)):

η is nondecreasing, and \(\eta(t)=0\) if and only if \(t=0\);
 (\(\eta_{2}\)):

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

\(\eta(a+b)\leq\eta(a)+\eta(b)\) for all \(a,b>0\).
Since (\(\eta_{1}\)) and (\(\eta_{2}\)) are clear, we only show (\(\eta_{3}\)). We have
$$\eta(a+b)=\ln\bigl(\psi(a+b)\bigr)\leq\ln\bigl(\psi (a)\psi(b)\bigr)=\ln\bigl( \psi(a)\bigr)+\ln\bigl(\psi(b)\bigr)=\eta(a)+\eta(b). $$
Lemma 1
Let
\((X,d)\)
be a metric space, and
\(\psi\in\Psi_{2}\). Then
\((X,D)\)
is a metric space, where
\(D(x,y)=\eta(d(x,y))=\ln(\psi (d(x,y)))\).
Proof
For each \(x\in X\), we have \(D(x,x)=\eta(d(x,x))=0\) by (\(\eta_{1}\)). For all \(x,y\in X\) with \(D(x,y)=0\), we have \(\eta(d(x,y))=0\) and hence \(d(x,y)=0\) by (\(\eta_{1}\)). Hence, for all \(x,y\in X\), \(D(x,y)=0\) if and only if \(x=y\).
For all \(x,y\in X\), we have \(D(x,y)=\eta(d(x,y))=\eta(d(y,x))=D(y,x)\).
For all \(x,y,z\in X\) with \(z\neq x\) and \(z\neq y\), by (\(\eta_{1}\)) and (\(\eta_{3}\)) we have \(D(x,y)=\eta(d(x,y))\leq\eta(d(x,z)+d(z,y))\leq\eta(d(x,z))+\eta (d(z,y))=D(x,z)+D(z,y)\). For all \(x\in X\) and \(y=z\in X\), we have \(D(x,y)=D(x,z)=D(x,z)+D(y,z)\) by (\(\eta_{1}\)). For all \(x=z\in X\) and \(y\in X\), we have \(D(x,y)=D(z,y)=D(x,z)+D(z,y)\) by (\(\eta_{1}\)). For all \(x=y=z\in X\), we have \(D(x,y)=0=D(x,z)+D(y,z)\) by (\(\eta_{1}\)). Hence, for all \(x,y,z\in X\), we always have \(D(x,y)\leq D(x,z)+D(z,y)\). This shows that \((X,D)\) is a metric space. The proof is complete. □
Lemma 2
Let
\((X,d)\)
be a metric space, and
\(\psi\in\Psi_{2}\). Then
\((X,D)\)
is complete if and only if
\((X,d)\)
is complete, where
\(D(x,y)=\eta(d(x,y))=\ln(\psi(d(x,y)))\).
Proof
Suppose that \((X,d)\) is complete and \(\{x_{n}\}\) is a Cauchy sequence of \((X,D)\), that is, \(\lim_{m,n\rightarrow\infty}D(x_{n},x_{m})=0\). Then we have \(\lim_{m,n\rightarrow\infty} \eta(d(x_{n},x_{m}))=0\), and hence \(\lim_{m,n\rightarrow\infty}d(x_{n}, x_{m})=0\) by (\(\eta_{2}\)). Moreover, by the completeness of \((X,d)\) there exists \(x\in X\) such that \(\lim_{n\rightarrow\infty}d(x_{n},x)=0\), and so \(\lim_{n\rightarrow\infty}D(x_{n},x)=\lim_{n\rightarrow\infty} \eta(d(x_{n},x))=0\) by (\(\eta_{2}\)). Hence, \((X,D)\) is complete. Similarly, we can show that if \((X,D)\) is complete, then \((X,d)\) is complete. The proof is complete. □
Lemma 3
Let
\((X,d)\)
be a metric space, and
\(T:X\rightarrow X\)
be a JScontraction with
\(\psi\in\Psi_{2}\). Then
T
is a Ćirić contraction in
\((X,D)\), where
\(D(x,y)=\eta (d(x,y))=\ln(\psi(d(x,y)))\).
Proof
It follows from (2) that, for all \(x,y\in X\),
$$\begin{aligned} D(Tx,Ty) =&\eta\bigl(d(Tx,Ty)\bigr)=\ln\bigl(\psi\bigl(d(Tx, Ty)\bigr)\bigr) \\ \leq&\ln\bigl(\psi\bigl(d(x,y)\bigr)^{q}\psi\bigl(d(x, Tx) \bigr)^{r} \psi\bigl(d(y, Ty)\bigr)^{s}\psi \bigl(d(x, Ty) + d(y, Tx)\bigr)^{t}\bigr) \\ =&q\ln\bigl(\psi\bigl(d(x,y)\bigr)\bigr)+r\ln\bigl(\psi\bigl(d(x,Tx)\bigr) \bigr)+s\ln\bigl(\psi\bigl(d(y,Ty)\bigr)\bigr) \\ &{}+t\bigl[\ln\bigl(\psi\bigl(d(x,Ty)\bigr)\bigr)+\ln\bigl(\psi\bigl(d(y,Tx) \bigr)\bigr)\bigr] \\ =&qD(x,y)+rD(x, Tx) + sD(y, Ty)+t\bigl[D(x, Ty) + D(y, Tx)\bigr], \end{aligned}$$
that is, (1) is satisfied with respect to the metric D, and hence T is a Ćirić contraction in \((X,D)\). The proof is complete. □
Theorem 4
Let
\((X,d)\)
be a complete metric space, and
\(T:X\rightarrow X\)
be a JScontraction with
\(\psi\in\Psi_{2}\). Then
T
has a unique fixed point in
X.
Proof
Since \((X,d)\) is a complete metric space, \((X,D)\) is also a complete metric space by Lemma 2. Note that T is a Ćirić contraction in \((X,D)\) by Lemma 3. Therefore, T has a unique fixed point in X by Theorem 1. The proof is complete. □
Remark 3
In comparison with Theorem 3, assumption (\(\psi_{3}\)) and the continuity of T have been removed from Theorem 4. Hence, Theorem 4 indeed improves Theorem 3.
Theorem 5
Theorem
4
implies Theorem
1.
Proof
Let \(\psi(t)=e^{t}\) for \(t\geq0\). Clearly, \(e^{t}\in\Psi_{2}\) by Remark 1. By (2) we have
$$d(Tx,Ty)\leq qd(x,y)+rd(x, Tx) + sd(y, Ty)+t\bigl[d(x,Ty)+d(y, Tx)\bigr],\quad \forall x,y\in X, $$
which implies that a Ćirić contraction \(T:X\rightarrow X\) is certainly a JScontraction with \(\psi(t)=e^{t}\). Thus, Theorem 1 immediately follows from Theorem 4. The proof is complete. □
Remark 4
It follows from Theorem 5 and the proof of Theorem 4 that Theorem 1 is equivalent to Theorem 4.
Remark 5
It is clear that Theorems 2.32.8 and Corollary 2.9 are immediate consequences of Theorem 1 but the converse is not true by Remark 2, and hence they are not real generalizations of Theorem 1. Note that Hussain et al. [1] also considered sufficient conditions for the existence of a fixed point of a JScontraction in bcomplete bmetric spaces.