Definition 4
([24])
Denote by \(\mathcal{F}\) the set of all functions \(F\colon (0,\infty )\to \mathbb{R}\) fulfilling the properties:
-
(1)
F is strictly increasing;
-
(2)
For any sequence of positive terms \(\{a_{n}\}\), \(\lim_{n\to \infty }a_{n}=0 \iff \lim_{n\to \infty }F(a_{n})=-\infty \);
-
(3)
There is k in \((0,1)\) such that \(\lim_{\alpha \to 0^{+}}\alpha ^{k}F(\alpha )=0\).
Example 4
The following functions \(F\colon (0,\infty )\to \mathbb{R}\) given for all \(\ell \in (0,\infty )\) as:
-
(i)
\(F(\ell )=\ln (\ell )\),
-
(ii)
\(F(\ell )=\ell +\ln (\ell )\),
-
(iii)
\(F(\ell )=\ln (\ell ^{2}+\ell )\),
-
(iv)
\(F(\ell )=-\frac{1}{\sqrt{\ell }}\),
belong to the set \(\mathcal{F}\).
In [24], Wardowski introduced a remarkable contraction known as F-contraction and presented a method to obtain fixed points of such contractions in complete metric spaces. The following theorem is the variant of Wardowski’s fixed point theorem [24] in the complete dualistic partial metric spaces. Theorem 1 is a useful generalization of the mentioned theorem.
Theorem 1
Let \((M,\mathcal{V})\) be a complete dualistic partial metric space, \(F\in \mathcal{F}\), and \(T\colon M\to M\) be a continuous mapping for which there exists \(\tau >0\) such that, for all j, k from M, the following implication holds:
$$ \mathcal{V}(Tj,Tk)\ne 0 \quad \Longrightarrow \quad \tau +F\bigl( \bigl\vert \mathcal{V}(Tj,Tk) \bigr\vert \bigr)\le F\bigl( \bigl\vert \mathcal{V}(j,k) \bigr\vert \bigr). $$
(3.1)
Then T possesses a unique fixed point.
Proof
Let \(j_{0}\) be an element from M and \(j_{n+1}=Tj_{n}\) for all \(n\in \mathbb{N}\). If there is \(n_{o}\in \mathbb{N}\) for which \(j_{n_{o}+1}=j_{n_{o}}\), then \(j_{n_{o}}\) is a fixed point of T. Therefore, we may assume that \(j_{n}\ne j_{n+1}\) for any \(n\in \mathbb{N}\). Given that \(\mathcal{V}(j_{n},j_{n+1})\ne 0\) for any \(n\in \mathbb{N}\). By the use of the contractive condition, we obtain, for \(n\ge 1\),
$$\begin{aligned} F\bigl( \bigl\vert \mathcal{V}(j_{n},j_{n+1}) \bigr\vert \bigr)\le F\bigl( \bigl\vert \mathcal{V}(j_{n-1},j_{n}) \bigr\vert \bigr)- \tau < F\bigl( \bigl\vert \mathcal{V}(j_{n-1},j_{n}) \bigr\vert \bigr), \end{aligned}$$
(3.2)
which, by taking advantage of the monotonicity of F, \(\{|\mathcal{V}(j_{n},j_{n+1})|\}\) is decreasing. Let \(r\ge 0\) be its limit. By contractive condition (3.1), we have
$$ F\bigl( \bigl\vert \mathcal{V}(j_{n+1},j_{n}) \bigr\vert \bigr)=F\bigl( \bigl\vert \mathcal{V}\bigl(T(j_{n}), T(j_{n-1})\bigr) \bigr\vert \bigr) \leq F\bigl( \bigl\vert \mathcal{V}(j_{n},j_{n-1}) \bigr\vert \bigr)-\tau . $$
Thus
$$ F\bigl( \bigl\vert \mathcal{V}(j_{n+1},j_{n}) \bigr\vert \bigr)\leq F\bigl( \bigl\vert \mathcal{V}(j_{n},j_{n-1}) \bigr\vert \bigr)- \tau . $$
(3.3)
This in turns implies that
$$ F\bigl( \bigl\vert \mathcal{V}(j_{n+1},j_{n}) \bigr\vert \bigr)\leq F\bigl( \bigl\vert \mathcal{V}(j_{1},j_{0}) \bigr\vert \bigr)-n \tau . $$
Letting \(n\rightarrow \infty \) and using \((F_{2})\), we get
$$ \lim_{n\rightarrow \infty }F\bigl( \bigl\vert \mathcal{V}(j_{n+1},j_{n}) \bigr\vert \bigr)=-\infty \quad \iff\quad \lim_{n\rightarrow \infty } \bigl\vert \mathcal{V}(j_{n+1},j_{n}) \bigr\vert =0. $$
This yields that
$$ \lim_{n\rightarrow \infty }\mathcal{V}(j_{n+1},j_{n})=0=r. $$
(3.4)
Focusing now on the self-distance, we have, for \(n\in \mathbb{N}\),
$$\begin{aligned} F\bigl( \bigl\vert \mathcal{V}(j_{n+1},j_{n+1}) \bigr\vert \bigr)\le F\bigl( \bigl\vert \mathcal{V}(j_{n},j_{n}) \bigr\vert \bigr)- \tau < F\bigl( \bigl\vert \mathcal{V}(j_{n},j_{n}) \bigr\vert \bigr), \end{aligned}$$
and so \(\{|\mathcal{V}(j_{n},j_{n})|\}\) is a sequence decreasing to some \(l\ge 0\). Arguing as above, we get that \(l=0\). By the third property of functions in \(\mathcal{F}\), there is \(\theta \in (0,1)\) such that \(\lim_{n\to \infty }|\mathcal{V}_{n}|^{\theta } F(|\mathcal{V}_{n}|)=0\).
Consider \(\mathcal{V}_{n}=\mathcal{V}(j_{n},j_{n+1})\) for \(n\in \mathbb{N}\). By applying repeatedly inequality (3.2), it follows that
$$ F\bigl( \vert \mathcal{V}_{n} \vert \bigr)\le F\bigl( \vert \mathcal{V}_{0} \vert \bigr)-n\tau $$
or
$$ \vert \mathcal{V}_{n} \vert ^{\theta }F\bigl( \vert \mathcal{V}_{n} \vert \bigr)\le \vert \mathcal{V}_{n} \vert ^{\theta }\bigl(F\bigl( \vert \mathcal{V}_{0} \vert \bigr)-n\tau \bigr),\quad n\in \mathbb{N}. $$
Letting \(n\to \infty \) and taking advantage of the properties of the function F, we get that \(n|\mathcal{V}_{n}|^{\theta }\to 0\) as \(n\to \infty \). There is \(N_{1}\in \mathbb{N}\) such that
$$ \vert \mathcal{V}_{n} \vert \le \frac{1}{{n^{\frac{1}{\theta }}}}, \quad n \ge N_{1}. $$
(3.5)
Denote by \(\mathcal{V}_{n}=\mathcal{V}(j_{n},j_{n})\) for \(n\in \mathbb{N}\). Similarly, there is \(N_{2}\in \mathbb{N}\) such that, for any \(n\ge N_{2}\),
$$ \vert \mathcal{V}_{n} \vert \le \frac{1}{{n^{\frac{1}{\theta }}}}. $$
(3.6)
Having in mind inequalities (3.5) and (3.6), we have, for \(m>n\ge \max \{N_{1},N_{2}\}\),
$$\begin{aligned} d^{*}(j_{n},j_{m}) \le &\sum _{\ell =0}^{m-n-1}d^{*}(j_{n+\ell },j_{n+ \ell +1}) \\ \le &\sum_{\ell =0}^{m-n-1} \bigl( \bigl\vert \mathcal{V}(j_{n+\ell },j_{n+\ell +1}) \bigr\vert + \bigl\vert \mathcal{V}(j_{n+\ell },j_{n+\ell }) \bigr\vert \bigr) \\ \le &2\sum_{\ell =0}^{m-n-1} \frac{1}{{\ell ^{\frac{1}{\theta }}}}. \end{aligned}$$
Taking the limit to ∞, it follows that \(\{d^{*}(j_{n},j_{m})\}\to 0\). Applying an analogous procedure, we get that \(\{d^{*}(j_{m},j_{n})\}\to 0\), hence \(d_{\mathcal{V}}^{s}(j_{n},j_{m})\to 0\), so \(\{j_{n}\}\) is a Cauchy sequence in the complete metric space \((M,d_{\mathcal{V}}^{s})\). Let j be its limit. Obviously,
$$ \lim_{n\to \infty }\mathcal{V}(j_{n},j)=\mathcal{V}(j,j)=0,\qquad \lim_{n,m\to \infty }\mathcal{V}(j_{n},j_{m})=0. $$
Since T is a continuous mapping, \(\{Tj_{n}\}\) converges to Tj. This implies that \(\mathcal{V}(Tj_{n},Tj_{n})\to \mathcal{V}(Tj,Tj)\). Since \(\mathcal{V}(Tj_{n},Tj_{n})\to \mathcal{V}(j,j)=0\), it follows that \(\mathcal{V}(Tj,Tj)=0\).
In view that
$$ \mathcal{V}(j,Tj)\le \mathcal{V}(j,j_{n+1})+\mathcal{V}(j_{n+1},Tj)- \mathcal{V}(j_{n+1},j_{n+1}), \quad n\in \mathbb{N}, $$
and
$$ \mathcal{V}(j_{n+1},Tj)\le \mathcal{V}(j_{n+1},j)+ \mathcal{V}(j,Tj)- \mathcal{V}(j,j), \quad n\in \mathbb{N}, $$
by considering \(n\to \infty \) and using the continuity of T, we obtain \(\mathcal{V}(j,Tj)=0\).
Thus \(\mathcal{V}(Tj,Tj)=\mathcal{V}(j,j)=\mathcal{V}(j,Tj)\), so \(Tj=j\). Let us focus now on the uniqueness of the fixed point of T. Assume that j and k are two distinct fixed points of T. If \(\mathcal{V}(j,k)\ne 0\), then the following relations hold true:
$$ F\bigl( \bigl\vert \mathcal{V}(j,k) \bigr\vert \bigr)=F\bigl( \bigl\vert \mathcal{V}(Tj,Tk) \bigr\vert \bigr)\le F\bigl( \bigl\vert \mathcal{V}(j,k) \bigr\vert \bigr)- \tau , $$
which is a contradiction. Therefore, \(\mathcal{V}(j,k)=0\). Similarly, it can be proved that \(\mathcal{V}(j,j)=0\) and \(\mathcal{V}(k,k)=0\). It follows that \(j=k\), and so the fixed point is unique. □
Remark 3
By setting \(\mathcal{V}(j,j)=0\) for all \(j\in M\), we have \(\mathcal{V}(j,k)\geq 0\) for all \(j,k\in M\) and thus \(\mathcal{V}:M\times M\rightarrow \mathbb{R}\) is restricted to \(\mathcal{V}:M\times M\rightarrow \mathbb{R}^{+}\), which is equal to the induced metric \(d_{\mathcal{V}}^{s}\) on M. Hence Wardowski fixed point theorem [24] is a particular case of Theorem 1.
We continue with an example with respect to this theorem. Example 5 does not only explain Theorem 1, but also shows that the Wardowski fixed point theorem [24] is not applicable for \(\ell =\hbar \).
Example 5
Let \(\mathcal{A}= ( -\infty ,0 ] \). Define the mapping
$$ \mathcal{V}\colon \mathcal{A} \times \mathcal{A} \rightarrow \mathbb{R}, \mathcal{V}(\ell ,\hbar )=\textstyle\begin{cases} \vert \ell -\hbar \vert &\text{if }\ell \neq \hbar, \\ \ell \vee \hbar &\text{if } \ell =\hbar .\end{cases} $$
Then \((\mathcal{A},\mathcal{V})\) is a complete dualistic partial metric space. Let \(\tau >0\) and define the mapping:
$$ T\colon \mathcal{A}\rightarrow \mathcal{A}, T(\ell )=\textstyle\begin{cases} \frac{e^{-\tau }\ell }{2} & \text{if }\ell \in \mathbb{Q} ( \text{a set of rational numbers}), \\ \frac{e^{-\tau }\ell }{3} & \text{if }\ell \in \mathbb{Q}^{\prime } ( \text{a set of irrational numbers}). \end{cases} $$
Contractive condition (2.1) is satisfied for \(F(\ell )=\ln (\ell )\) for all \(\ell >0\). For this, consider the following:
Case 1. (a) \(\ell \neq \hbar \); if \(\ell ,\hbar \in \mathbb{Q}\), then
$$ \bigl\vert \mathcal{V}\bigl(T(\ell ),T(\hbar )\bigr) \bigr\vert = \frac{1}{2}e^{-\tau } \vert \ell -\hbar \vert < e^{-\tau } \bigl\vert \mathcal{V}(\ell ,\hbar ) \bigr\vert , $$
which implies \(\tau +F(|\mathcal{V}(T\ell ,T\hbar )|)\le F(|\mathcal{V}(\ell , \hbar )|)\).
(b) \(\ell \neq \hbar \); if \(\ell ,\hbar \in \mathbb{Q}^{\prime }\), then
$$ \bigl\vert \mathcal{V}\bigl(T(\ell ),T(\hbar )\bigr) \bigr\vert = \frac{1}{3}e^{-\tau } \vert \ell -\hbar \vert < e^{-\tau } \bigl\vert \mathcal{V}(\ell ,\hbar ) \bigr\vert , $$
which leads to \(\tau +F(|\mathcal{V}(T\ell ,T\hbar )|)\le F(|\mathcal{V}(\ell , \hbar )|)\).
Case 2. (a) \(\ell =\hbar \); if \(\ell \in \mathbb{Q}\), then
$$ \bigl\vert \mathcal{V}\bigl(T(\ell ),T(\hbar )\bigr) \bigr\vert = \biggl\vert \frac{1}{2}e^{- \tau }\ell \biggr\vert < e^{-\tau } \bigl\vert \mathcal{V}(\ell ,\hbar ) \bigr\vert . $$
(b) \(\ell =\hbar \); if \(\ell \in \mathbb{Q}^{\prime }\), then
$$ \bigl\vert \mathcal{V}\bigl(T(\ell ),T(\hbar )\bigr) \bigr\vert = \biggl\vert \frac{1}{3}e^{- \tau }\ell \biggr\vert < e^{-\tau } \bigl\vert \mathcal{V}(\ell ,\hbar ) \bigr\vert . $$
From (a) and (b), we infer that \(\tau +F(|\mathcal{V}(T\ell ,T\hbar )|)\le F(|\mathcal{V}(\ell , \hbar )|)\).
Case 3. If \(\ell \in \mathbb{Q}\), \(\hbar \in \mathbb{Q}^{\prime }\) and vice versa, then
$$ \bigl\vert \mathcal{V}\bigl(T(\ell ),T(\hbar )\bigr) \bigr\vert = \biggl\vert \frac{\ell }{2}- \frac{\hbar }{3} \biggr\vert e^{-\tau }\leq e^{-\tau } \bigl\vert \mathcal{V}(\ell , \hbar ) \bigr\vert . $$
Thus, \(\tau +F(|\mathcal{V}(T\ell ,T\hbar )|)\le F(|\mathcal{V}(\ell , \hbar )|)\). Hence, T satisfies all the conditions of Theorem 1. Note that \(\ell =0\) is the unique fixed point of T.
The following theorem is for Kannan type F-contraction in the dualistic partial metric spaces. It is a useful generalization of [40, Theorem 2].
Theorem 2
Let \((\mho ,\mathcal{V})\) be a complete dualistic partial metric space, \(F\in \mathcal{F}\), and \(T\colon \mho \to \mho \) be a continuous mapping for which there exists \(\tau >0\) such that, for all j and k in ℧, the following implication holds:
$$ \mathcal{V}(Tj,Tk)\ne 0 \quad \Longrightarrow\quad \tau +F\bigl( \bigl\vert \mathcal{V}(Tj,Tk) \bigr\vert \bigr)\le F \biggl( \frac{1}{2}\bigl( \bigl\vert \mathcal{V}(j,Tj) \bigr\vert + \bigl\vert \mathcal{V}(k,Tk) \bigr\vert \bigr) \biggr). $$
(3.7)
Then T admits a unique fixed point.
Proof
For \(j_{0}\) from ℧, set \(j_{n+1}=Tj_{n}\) for \(n\in \mathbb{N}\). If there is \(s\in \mathbb{N}\) for which \(j_{s+1}=j_{s}\), then \(j_{s}\) is a fixed point of T. We presume that \(j_{n}\ne j_{n+1}\) for any \(n\in \mathbb{N}\). Given \(\mathcal{V}(j_{n},j_{n+1})\ne 0\) for any \(n\in \mathbb{N}\). By the use of the contractive condition, we obtain, for \(n\ge 1\),
$$\begin{aligned} F\bigl( \bigl\vert \mathcal{V}(j_{n},j_{n+1}) \bigr\vert \bigr) \le& F \biggl( \frac{1}{2}\bigl( \bigl\vert d(j_{n-1},j_{n}) \bigr\vert + \bigl\vert \mathcal{V}(j_{n},j_{n+1}) \bigr\vert \bigr) \biggr)-\tau \\ < & F \biggl( \frac{1}{2}\bigl( \bigl\vert d(j_{n-1},j_{n}) \bigr\vert + \bigl\vert \mathcal{V}(j_{n},j_{n+1}) \bigr\vert \bigr) \biggr), \end{aligned}$$
(3.8)
from which, by taking advantage of the monotony of F, we have that \(\{|\mathcal{V}(j_{n},j_{n+1})|\}\) is a decreasing sequence. Let \(r\ge 0\) be its limit.
By applying inequality (3.8) and keeping in mind also that \(\{|\mathcal{V}(j_{n},j_{n+1})|\}\) is a decreasing sequence, we get, for \(n\ge 1\),
$$ F\bigl( \bigl\vert \mathcal{V}(j_{n},j_{n+1}) \bigr\vert \bigr)\le F\bigl( \bigl\vert \mathcal{V}(j_{n-1},j_{n}) \bigr\vert \bigr)- \tau \le F\bigl( \bigl\vert \mathcal{V}(j_{0},j_{1}) \bigr\vert \bigr)-n\tau .$$
(3.9)
It follows that \(\lim_{n\to \infty }F(|\mathcal{V}(j_{n},j_{n+1})|)=-\infty \). Having in mind \(F\in \mathcal{F}\), we obtain that \(\lim_{n\to \infty }|\mathcal{V}(j_{n},j_{n+1})|=0\).
Let us now analyze the self-distance. Presume that there exists \(n_{0}\in \mathbb{N}\) such that \(\mathcal{V}(j_{n},j_{n})\ne 0\) for any \(n\ge n_{0}\), where \(n_{0}\in \mathbb{N}\). The contractive condition imposes
$$ F\bigl( \bigl\vert \mathcal{V}(j_{n+1},j_{n+1}) \bigr\vert \bigr)\le F\bigl( \bigl\vert d(j_{n},j_{n+1}) \bigr\vert \bigr)-\tau . $$
From (3.9), it follows that
$$ F\bigl( \bigl\vert \mathcal{V}(j_{n+1},j_{n+1}) \bigr\vert \bigr)\le F\bigl( \bigl\vert \mathcal{V}(j_{n_{0}},j_{n_{0}+1}) \bigr\vert \bigr)-(n-n_{0}) \tau . $$
Letting \(n\to \infty \) and using the properties of the function F, we get that \(\lim_{n\to \infty }\mathcal{V}(j_{n},j_{n})=0\).
As in the proof of Theorem 1, we obtain that \(\{j_{n}\}\) is a Cauchy sequence and so it is convergent to some j. From the preceding lines, \(\mathcal{V}(j,j)=0\). Since T is continuous, \(\{Tj_{n}\}\) converges to Tj, and therefore \(\mathcal{V}(Tj_{n},Tj_{n})\to \mathcal{V}(Tj,Tj)\). On the other hand, \(\mathcal{V}(Tj_{n},Tj_{n})\to \mathcal{V}(j,j)=0\). We have obtained that \(\mathcal{V}(Tj,Tj)=0\).
In view that
$$ \mathcal{V}(j,Tj)\le \mathcal{V}(j,j_{n+1})+\mathcal{V}(j_{n+1},Tj)- \mathcal{V}(j_{n+1},j_{n+1}),\quad n\in \mathbb{N}, $$
and also that
$$ \mathcal{V}(j_{n+1},Tj)\le \mathcal{V}(j_{n+1},j)+ \mathcal{V}(j,Tj)- \mathcal{V}(j,j), \quad n\in \mathbb{N}, $$
by considering \(n\to \infty \) and using the continuity of T, we obtain \(\mathcal{V}(j,Tj)=0\). We conclude that \(Tj=j\), and so T has a fixed point.
Assume now that the set \(\{n\in \mathbb{N},\mathcal{V}(j_{n},j_{n})=0\}\) is not bounded. In this case, there exists a subsequence of \(\{j_{n}\}\), \(\{j_{n_{k}}\}\) for which \(\mathcal{V}(j_{n_{k}},j_{n_{k}})=0\), \(k\in \mathbb{N}\). We repeat the same lines as in the proof of Theorem 1 for this subsequence \(\{j_{n_{k}}\}\), and we obtain that \(\{j_{n_{k}}\}\) is convergent to \(j\in \mho \) with \(\mathcal{V}(j,j)=0\). The equality \(j=Tj\) follows as in the case when \(\{n\in \mathbb{N},\mathcal{V}(j_{n},j_{n})=0\}\) is bounded.
Summing up, we have proved that T has a fixed point.
Let us focus now on the uniqueness of the fixed point of T. Assume that j and k are two distinct fixed points of T. Now suppose that \(\mathcal{V}(j,j)\ne 0\). Then the following holds:
$$ F\bigl( \bigl\vert \mathcal{V}(j,j) \bigr\vert \bigr)=F\bigl( \mathcal{V}(Tj,Tj)\bigr)\le F\bigl( \bigl\vert \mathcal{V}(j,Tj) \bigr\vert \bigr)- \tau =F\bigl( \bigl\vert \mathcal{V}(j,j) \bigr\vert \bigr)-\tau , $$
which is impossible. Hence, \(\mathcal{V}(j,j)=0\). Similarly, it can be proved that \(\mathcal{V}(k,k)=0\).
If \(\mathcal{V}(j,k)\ne 0\), then the following holds:
$$ F\bigl( \bigl\vert \mathcal{V}(j,k) \bigr\vert \bigr)=F\bigl( \bigl\vert \mathcal{V}(Tj,Tk) \bigr\vert \bigr)\le F \biggl( \frac{1}{2}\bigl( \bigl\vert \mathcal{V}(j,Tj) \bigr\vert + \bigl\vert \mathcal{V}(k,Tk) \bigr\vert \bigr) \biggr)-\tau , $$
and it is not well defined. Therefore, \(\mathcal{V}(j,k)=0\).
It follows that \(j=k\), and so the fixed point is unique. □
To support our results, we give the next example.
Example 6
Let \(\mho =(-\infty ,0]\) and define the function \(\mathcal{V}\colon \mho \times \mho \to \mathbb{R}\) by
$$ \mathcal{V}(j,k)=\textstyle\begin{cases} \vert j-k \vert +1,& \text{if }j\neq k, \\ \max \{j,k\},&\text{if }j= k. \end{cases} $$
According to [21], the pair \((\mho ,\mathcal{V})\) is a complete dualistic partial metric space. Consider the mapping \(T\colon \mho \to \mho \) given as
$$ Tj=\textstyle\begin{cases} -1,&j\in (-\infty , -4], \\ -\frac{1}{2},&j\in (-4,0]. \end{cases} $$
We claim that T fulfills the conditions of Theorem 2 for \(\tau =\ln \frac{3}{2}\) and \(F(t)=\ln t\).
We have to analyze several cases.
Case I. \(j\neq k\) with \(j,k\in (-4,0]\).
We note that \(\mathcal{V}(Tj,Tk)=\mathcal{V}(-\frac{1}{2},-\frac{1}{2})=- \frac{1}{2}\). Also,
$$\begin{aligned} \ln \frac{3}{2}+\ln \bigl\vert \mathcal{V}(Tj,Tk) \bigr\vert =&\ln \frac{3}{2}+\ln \biggl\vert -\frac{1}{2} \biggr\vert \\ =&\ln \frac{3}{4} \\ \le &\ln \frac{{ \vert \mathcal{V} (j,-\frac{1}{2} ) \vert + \vert \mathcal{V} (k,-\frac{1}{2} ) \vert }}{2} \\ =&\ln \frac{{ \vert \mathcal{V}(j,Tj) \vert + \vert \mathcal{V}(k,Tk) \vert }}{2}, \end{aligned}$$
and so it behaves accordingly.
Case II. \(j\neq k\) with \(j,k\in (-\infty , -4]\).
Here, we have
$$\begin{aligned} \ln \frac{3}{2}+\ln \bigl\vert \mathcal{V}(Tj,Tk) \bigr\vert =&\ln \frac{3}{2}+\ln \bigl\vert \mathcal{V}(-1,-1) \bigr\vert \\ =&\ln \frac{3}{2} \\ \leq &\ln \frac{{ \vert \mathcal{V}(j,-1) \vert + \vert \mathcal{V}(k,-1) \vert } }{2} \\ =&\ln \frac{{ \vert \mathcal{V}(j,Tj) \vert + \vert \mathcal{V}(k,Tk) \vert }}{2}, \end{aligned}$$
and the relation is also checked.
Case III. \(j\neq k\) with \(j\in (-\infty , -4]\), \(k\in (-4,0]\) and vice versa.
In this case, we have \(\mathcal{V}(Tj,Tk)=\mathcal{V} (-1,-\frac{1}{2} )= \frac{3}{2}\). Also,
$$\begin{aligned} \ln \frac{3}{2}+\ln \bigl\vert \mathcal{V}(Tj,Tk) \bigr\vert =&\ln \frac{3}{2}+\ln \biggl\vert \frac{3}{2} \biggr\vert \\ =&\ln \frac{9}{4} \\ \le &\ln \frac{{ \vert \mathcal{V}(j,-1) \vert + \vert \mathcal{V} (k, \frac{-1}{2} ) \vert }}{2} \\ =&\ln \frac{{ \vert \mathcal{V}(j,Tj) \vert + \vert \mathcal{V}(k,Tk) \vert }}{2}. \end{aligned}$$
Case IV. \(j=k\in (-4,0]\).
We have
$$\begin{aligned} \ln \frac{3}{2}+\ln \bigl\vert \mathcal{V}(Tj,Tk) \bigr\vert =&\ln \frac{3}{2}+\ln \biggl\vert -\frac{1}{2} \biggr\vert \\ =&\ln \frac{3}{4} \\ \le &\ln \frac{{ \vert \mathcal{V} (j,-\frac{1}{2} ) \vert + \vert \mathcal{V} (k,-\frac{1}{2} ) \vert }}{2} \\ =&\ln \frac{{ \vert \mathcal{V}(j,Tj) \vert + \vert \mathcal{V}(k,Tk) \vert }}{2}, \end{aligned}$$
so, this case is also true.
Case V. \(j=k\in (-\infty , -4]\).
In this case, one writes
$$\begin{aligned} \ln \frac{3}{2}+\ln \bigl\vert \mathcal{V}(Tj,Tk) \bigr\vert =&\ln \frac{3}{2}+\ln \bigl\vert \mathcal{V}(-1,-1) \bigr\vert \\ =&\ln \frac{3}{2} \\ \le &\ln \frac{{ \vert \mathcal{V}(j,-1) \vert + \vert \mathcal{V}(k,-1) \vert } }{2} \\ =&\ln \frac{{ \vert \mathcal{V}(j,Tj) \vert + \vert \mathcal{V}(k,Tk) \vert }}{2}. \end{aligned}$$
The inequality in Theorem 2 is fulfilled. By applying this result, we get that T has a unique fixed point, which is \(j=-\frac{1}{2}\).
On the other hand, it can be easily verified that the inequality in Theorem 1 is not satisfied in the case \(j=k=0\).