In 2020, Alfaqih et al. [25] provided \(F^{*}\)-weak contractions by introducing a new class of auxiliary functions that eliminate conditions \((F1)\), \((F3)\) and only satisfy one-way implication of condition \((F2)\).
Firstly, we suppose that \(\mathcal{F}\) is the set of all functions \(F:(0,\infty )\rightarrow \mathbb{R}\) satisfying the following condition:
- \((C_{1})\):
-
\(\inf_{t>\varepsilon}F(t)>-\infty \) for any \(\varepsilon >0\).
It follows from Lemma 2.3 [7] that condition \((C_{1})\) is equivalent to \((C_{1}')\) and \((C_{1}'')\) stated as follows:
- \((C_{1}')\):
-
\(\lim_{n\rightarrow \infty} F(t_{n})=-\infty \) implies \(\lim_{n \rightarrow \infty} t_{n}=0\);
- \((C_{1}'')\):
-
\(\liminf_{t\rightarrow \varepsilon +} F(t)>-\infty \) for any \(\varepsilon >0\).
Apparently, \(\mathcal{H}\subset \mathcal{F}\). Next, we will introduce the concepts of the dual \(F^{*}-\) weak contraction and triple \(F^{*}-\) weak contraction, which can be regarded as a generalization of F-contraction.
Definition 2.1
Let \((X,d)\) be a metric space. We say that a mapping \(\mathfrak{F}:X\mapsto X\) is a dual \(F^{*}\)-weak contraction of type-I if there exists a real number \(\tau >0\) and \(F_{1}, F_{2}\in \mathcal{F}\) such that for all \(x,y\in X\), we have
$$\begin{aligned} &d\bigl(\mathfrak{F}^{2}x,\mathfrak{F}^{2}y \bigr)>0 \\ &\quad \text{implies} \\ &\qquad \tau +\min \bigl\{ F_{2}\bigl(d\bigl( \mathfrak{F}^{2}x, \mathfrak{F}^{2}y\bigr) \bigr),F_{1}\bigl(d(\mathfrak{F}x,\mathfrak{F}y)\bigr)\bigr\} \leq \alpha _{2}F_{2}\bigl(d(x,y)\bigr)+\alpha _{1}F_{1}\bigl(d(x,y)\bigr), \end{aligned}$$
(14)
where
$$\textstyle\begin{cases} \alpha _{1}=0,\alpha _{2}=1, & \text{if } F_{2}(d( \mathfrak{F}^{2}x,\mathfrak{F}^{2}y))\leq F_{1}(d(\mathfrak{F}x, \mathfrak{F}y)), \\ \alpha _{1}=1,\alpha _{2}=0, & \text{if } F_{2}(d( \mathfrak{F}^{2}x,\mathfrak{F}^{2}y))> F_{1}(d(\mathfrak{F}x, \mathfrak{F}y)). \end{cases} $$
Remark 2.1
For some \(x,y\in X\), the conditions of Definition 2.1 yield either
$$ \tau +F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x, \mathfrak{F}^{2}y\bigr)\bigr)\leq F_{2}\bigl(d(x,y)\bigr) \quad (L1) $$
or
$$ \tau +F_{1}\bigl(d(\mathfrak{F}x,\mathfrak{F}y)\bigr) \leq F_{1}\bigl(d(x,y)\bigr).\quad (L2) $$
It is worth noting that Definition 2.1 can be viewed as a combination of \((L1)\) and \((L2)\), which could reduce to the form of Wardowski’s contraction while it includes inequality \((L2)\) only.
Definition 2.2
Let \((X,d)\) be a metric space. We say that a mapping \(\mathfrak{F}:X\mapsto X\) is a dual \(F^{*}\)-weak contraction of type-II if there exists a real number \(\tau >0\) and \(F_{1}, F_{2}\in \mathcal{F}\) such that for all \(x,y\in X\), we have
$$\begin{aligned} &d\bigl(\mathfrak{F}^{2}x,\mathfrak{F}^{2}y \bigr)>0 \\ &\quad \text{implies} \\ & \qquad \tau +\min \bigl\{ F_{2}\bigl(d\bigl( \mathfrak{F}^{2}x, \mathfrak{F}^{2}y\bigr) \bigr),F_{1}\bigl(d(\mathfrak{F}x,\mathfrak{F}y)\bigr)\bigr\} \leq \alpha _{2}F_{2}\bigl(d(x,y)\bigr)+\alpha _{1}F_{1}\bigl(d(x,y)\bigr), \end{aligned}$$
(15)
where either \(\alpha _{1}=0\) or \(\alpha _{2}=0 \) and \(\alpha _{1}+\alpha _{2}=1\).
Remark 2.2
For all \(x,y\in X\), the dual \(F^{*}\)-weak contraction of type-II deals with one of the following two cases \((R1)\), \((R2)\):
$$\begin{aligned}& \tau +\min \bigl\{ F_{2}\bigl(d\bigl( \mathfrak{F}^{2}x,\mathfrak{F}^{2}y\bigr) \bigr),F_{1}\bigl(d( \mathfrak{F}x,\mathfrak{F}y)\bigr)\bigr\} \leq F_{2}\bigl(d(x,y)\bigr), \quad (R1) \\& \tau +\min \bigl\{ F_{2}\bigl(d\bigl( \mathfrak{F}^{2}x,\mathfrak{F}^{2}y\bigr) \bigr),F_{1}\bigl(d( \mathfrak{F}x,\mathfrak{F}y)\bigr)\bigr\} \leq F_{1}\bigl(d(x,y)\bigr). \quad (R2) \end{aligned}$$
For some \(x,y\in X, (R1)\) yields either
$$ \tau +F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x, \mathfrak{F}^{2}y\bigr)\bigr)\leq F_{2}\bigl(d(x,y)\bigr) \quad (a) $$
or
$$ \tau +F_{1}\bigl(d(\mathfrak{F}x,\mathfrak{F}y)\bigr) \leq F_{2}\bigl(d(x,y)\bigr). \quad (d) $$
Similarly, for some \(x,y\in X, (R2)\) yields either
$$ \tau +F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x, \mathfrak{F}^{2}y\bigr)\bigr)\leq F_{1}\bigl(d(x,y)\bigr) \quad (c) $$
or
$$ \tau +F_{1}\bigl(d(\mathfrak{F}x,\mathfrak{F}y)\bigr) \leq F_{1}\bigl(d(x,y)\bigr). \quad (d) $$
Apparently, inequality \((R1)\) takes the form as a combination of subcases \((a)\) and \((d)\). If the inequality \((R1)\) yields subcase \((d)\) only, we have \(\tau +F_{1}(d(\mathfrak{F}x,\mathfrak{F}y))\leq F_{2}(d(x,y))\), which is equivalent to the form of Wardowski’s contraction by taking \(F_{2}=F_{1}\).
Definition 2.3
Let \((X,d)\) be a metric space. We say that a mapping \(\mathfrak{F}:X\mapsto X\) is a triple \(F^{*}\)-weak contraction if there exists a real number \(\tau >0\) and \(F, F_{1}, F_{2}\in \mathcal{F}\) such that for all \(x,y\in X\) we have
$$\begin{aligned} &d\bigl(\mathfrak{F}^{2}x,\mathfrak{F}^{2}y \bigr)>0 \\ &\quad \text{implies}\quad \tau +\min \bigl\{ F_{2}\bigl(d\bigl( \mathfrak{F}^{2}x, \mathfrak{F}^{2}y\bigr) \bigr),F_{1}\bigl(d(\mathfrak{F}x,\mathfrak{F}y)\bigr)\bigr\} \leq F \bigl(d(x,y)\bigr). \end{aligned}$$
(16)
Remark 2.3
One can easily observe that if \(\min \{{{F}_{2}}(d({{\mathfrak{F}}^{2}}x,{{\mathfrak{F}}^{2}}y)),{{F}_{1}}(d( \mathfrak{F}x,\mathfrak{F}y))\}={{F}_{1}}(d(\mathfrak{F}x, \mathfrak{F}y))\) and \(F (t )=F_{1} (t )\) for all \(x,y \in X\), \(t\in (0,\infty )\), then inequality (16) becomes the form of Wardowski’s contraction.
Example 2.1
Let \(F_{1}, F_{2}\in \mathcal{F}\) be given by \(F_{1}(\alpha )=\ln \alpha \), \(F_{2}(\alpha )=\ln k\alpha \), where \(\alpha , k>0\).
The dual \(F^{*}\)-weak contraction of type-I will take the form
$$\begin{aligned} \tau +\min \bigl\{ \ln \bigl(kd\bigl(\mathfrak{F}^{2}x, \mathfrak{F}^{2}y\bigr)\bigr),\ln \bigl(d( \mathfrak{F}x,\mathfrak{F}y) \bigr)\bigr\} \leq \alpha _{2}\ln \bigl(kd(x,y)\bigr)+\alpha _{1} \ln \bigl(d(x,y)\bigr), \end{aligned}$$
which can also be rewritten as
$$\begin{aligned} \tau +\ln \bigl(\min \bigl\{ kd\bigl(\mathfrak{F}^{2}x, \mathfrak{F}^{2}y\bigr),d( \mathfrak{F}x,\mathfrak{F}y)\bigr\} \bigr) \leq \ln \bigl(k^{\alpha _{2}}d(x,y)\bigr)^{{ \alpha _{1}}+{\alpha _{2}}}. \end{aligned}$$
From the definition of dual \(F^{*}\)-weak contraction of type-I, we have \(\alpha _{1}+\alpha _{2}=1\) and
$$\begin{aligned} \min \bigl\{ kd\bigl(\mathfrak{F}^{2}x,\mathfrak{F}^{2}y \bigr),d(\mathfrak{F}x, \mathfrak{F}y)\bigr\} \leq e^{-\tau}k^{\alpha _{2}}d(x,y). \end{aligned}$$
It is worth noting that \(\mathfrak{F}\) may be a contraction provided by \(d(\mathfrak{F}x,\mathfrak{F}y)< kd(\mathfrak{F}^{2}x,\mathfrak{F}^{2}y)\) for all \(x,y\in X\), that is, \(d(\mathfrak{F}x,\mathfrak{F}y)\leq e^{-\tau}d(x,y)\) with \(\alpha _{2}=0\) or be neither a contraction nor an expansion provided by \(d(\mathfrak{F}x,\mathfrak{F}y)>kd(\mathfrak{F}^{2}x,\mathfrak{F}^{2}y)\) for all \(x,y\in X\), that is, \(d(\mathfrak{F}^{2}x,\mathfrak{F}^{2}y)\leq e^{-\tau}kd(x,y)\).
Example 2.2
Let \(F_{1}, F_{2}\in \mathcal{F}\) be given by \(F_{1}(\alpha )=\ln \alpha \), \(F_{2}(\alpha )=\ln k\alpha \), where \(\alpha , k>0\).
Then, by the definition of dual \(F^{*}\)-weak contraction of type-II, for all \(x,y\in X\), we have
$$\begin{aligned} \tau +\min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x, \mathfrak{F}^{2}y\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x, \mathfrak{F}y)\bigr)\bigr\} \leq \alpha _{2}F_{2}\bigl(d(x,y) \bigr)+\alpha _{1}F_{1}\bigl(d(x,y)\bigr). \end{aligned}$$
From \((R1)\), \((R2)\), we have
$$\begin{aligned}& \tau +\min \bigl\{ F_{2}\bigl(d\bigl( \mathfrak{F}^{2}x,\mathfrak{F}^{2}y\bigr) \bigr),F_{1}\bigl(d( \mathfrak{F}x,\mathfrak{F}y)\bigr)\bigr\} \leq F_{1}\bigl(d(x,y)\bigr), \end{aligned}$$
(17)
$$\begin{aligned}& \tau +\min \bigl\{ F_{2}\bigl(d\bigl( \mathfrak{F}^{2}x,\mathfrak{F}^{2}y\bigr) \bigr),F_{1}\bigl(d( \mathfrak{F}x,\mathfrak{F}y)\bigr)\bigr\} \leq F_{2}\bigl(d(x,y)\bigr). \end{aligned}$$
(18)
Together with the definition of \(F_{1}\), inequality (17) can be rewritten as
$$\begin{aligned} \tau +\ln \bigl(\min \bigl\{ kd\bigl(\mathfrak{F}^{2}x, \mathfrak{F}^{2}y\bigr),d( \mathfrak{F}x,\mathfrak{F}y)\bigr\} \bigr) \leq \ln \bigl(d(x,y)\bigr). \end{aligned}$$
Then we have
$$\begin{aligned} \min \bigl\{ kd\bigl(\mathfrak{F}^{2}x, \mathfrak{F}^{2}y\bigr),d(\mathfrak{F}x, \mathfrak{F}y)\bigr\} \leq e^{-\tau}d(x,y). \end{aligned}$$
(19)
Similarly, inequality (18) can be rewritten as
$$\begin{aligned} \min \bigl\{ kd\bigl(\mathfrak{F}^{2}x, \mathfrak{F}^{2}y\bigr),d(\mathfrak{F}x, \mathfrak{F}y)\bigr\} \leq e^{-\tau}kd(x,y). \end{aligned}$$
(20)
It is noted that inequalities (19) and (20) can imply that \(\mathfrak{F}\) is a Lipschitzian mapping provided by \(kd(\mathfrak{F}^{2}x,\mathfrak{F}^{2}y)>d(\mathfrak{F}x,\mathfrak{F}y)\) for all \(x,y\in X\) or a contraction provided by \(kd(\mathfrak{F}^{2}x,\mathfrak{F}^{2}y)< d(\mathfrak{F}x,\mathfrak{F}y)\) for all \(x,y\in X\).
Example 2.3
Define \(F, F_{1}, F_{2}\in \mathcal{F}\) by \(F(\alpha )=\ln k\alpha \), \(F_{1}(\alpha )=\ln k_{1}\alpha \), \(F_{2}(\alpha )=\ln k_{2}\alpha \), where \(k, k_{1}, k_{2}>0\).
Then, by the definition of triple \(F^{*}\)-weak contraction, for all \(x,y\in X\), we have
$$\begin{aligned} \tau +\min \bigl\{ \ln \bigl(k_{2}d\bigl(\mathfrak{F}^{2}x, \mathfrak{F}^{2}y\bigr)\bigr),\ln \bigl(k_{1}b( \mathfrak{F}x,\mathfrak{F}y)\bigr)\bigr\} \leq \ln \bigl(kd(x,y)\bigr), \end{aligned}$$
which can also be rewritten as
$$\begin{aligned} \min \bigl\{ k_{2}d\bigl(\mathfrak{F}^{2}x, \mathfrak{F}^{2}y\bigr),k_{1}b( \mathfrak{F}x,\mathfrak{F}y) \bigr\} \leq e^{-\tau}kd(x,y). \end{aligned}$$
That is, either
$$\begin{aligned} k_{2}d\bigl(\mathfrak{F}^{2}x,\mathfrak{F}^{2}y \bigr)\leq e^{-\tau}kd(x,y) \end{aligned}$$
or
$$\begin{aligned} k_{1}b(\mathfrak{F}x,\mathfrak{F}y)\leq e^{-\tau}kd(x,y). \end{aligned}$$
We can define \(k_{2}=2\alpha _{2}\), \(k_{1}=2\alpha _{1}\), \(k_{1}+k_{2}=2\), where \(\alpha _{1}+\alpha _{2}=1\). Then we have
$$\begin{aligned} \alpha _{2}d\bigl(\mathfrak{F}^{2}x,\mathfrak{F}^{2}y \bigr)\leq \frac{k}{2}e^{- \tau}d(x,y) \end{aligned}$$
or
$$\begin{aligned} \alpha _{1}d(\mathfrak{F}x,\mathfrak{F}y)\leq \frac{k}{2} e^{-\tau}d(x,y). \end{aligned}$$
Therefore, we have
$$\begin{aligned} \alpha _{2}\bigl(d\bigl(\mathfrak{F}^{2}x, \mathfrak{F}^{2}y\bigr)\bigr)+\alpha _{1}\bigl(d( \mathfrak{F}x,\mathfrak{F}y)\bigr)\leq ke^{-\tau}\bigl(d(x,y)\bigr), \end{aligned}$$
which shows that \(\mathfrak{F}\) is a mean Lipschitzian mapping [36].
Next, we present the existence of a unique fixed point for dual \(F^{*}\)-weak contractions and triple \(F^{*}\)-weak contraction as follows.
Theorem 2.1
Suppose that a mapping \(\mathfrak{F}:X\mapsto X\) is a dual \(F^{*}\)-weak contraction of type-I in a complete metric space \((X,d)\). If \(\mathfrak{F}\) is orbitally continuous or k-continuous for some \(k \in \mathbb{N}\) and for all \(t_{1}, t_{2}\in \mathbb{R}\mathbbm{_{+}}\), there exist \(\upsilon >0\), \(\tau >2\upsilon \) such that
$$\begin{aligned} &F_{2}(t_{2})< F_{1}(t_{1})\leq F_{2}(t_{2})+\upsilon \end{aligned}$$
(A)
or
$$\begin{aligned} &F_{1}(t_{1})< F_{2}(t_{2})\leq F_{1}(t_{1})+\upsilon . \end{aligned}$$
(B)
Then, for every \(x_{0}\in X\), the sequence \(\{\mathfrak{F}^{m}x_{0}\}_{m=1}^{+\infty}\) converges to the unique fixed point of \(\mathfrak{F}\).
Proof
Consider a sequence \(\{x_{m}\}\subseteq X\) such that, for all \(m\in \mathbb{N}_{0}\), \(x_{m+1}=\mathfrak{F}x_{m}=\mathfrak{F}^{m}x_{0}\), where \(x_{0}\) is an arbitrary point in X.
If there exists \(m\in \mathbb{N}_{0}\) such that \(d(x_{m},\mathfrak{F}x_{m})=0\), then \(\mathfrak{F}\) admits a fixed point. So, we assume that \(d(x_{m},\mathfrak{F}x_{m})=d(\mathfrak{F}x_{m-1},\mathfrak{F}x_{m})>0\) for all \(m\in \mathbb{N}\).
We will prove that \(\lim_{m\rightarrow +\infty} d(x_{m},\mathfrak{F}x_{m})=0\).
From the definition of dual F-contraction of type I, we have
$$\begin{aligned} &\tau +\min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{m-1},\mathfrak{F}x_{m})\bigr)\bigr\} \\ &\quad \leq \alpha _{2}F_{2}\bigl(d(x_{m-1},x_{m}) \bigr)+\alpha _{1}F_{1}\bigl(d(x_{m-1},x_{m}) \bigr), \end{aligned}$$
so
$$\begin{aligned} &\min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{m-1},\mathfrak{F}x_{m})\bigr)\bigr\} \\ &\quad \leq \alpha _{2}F_{2}\bigl(d(x_{m-1},x_{m}) \bigr)+\alpha _{1}F_{1}\bigl(d(x_{m-1},x_{m}) \bigr)- \tau . \end{aligned}$$
(23)
Now we will discuss the following two possible cases \((\mathit{I})\), \((\mathit{II})\):
$$\begin{aligned} &F_{1}\bigl(d(\mathfrak{F}x_{m-1},\mathfrak{F}x_{m}) \bigr)=\min \bigl\{ F_{2}\bigl(d\bigl( \mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{m-1}, \mathfrak{F}x_{m})\bigr)\bigr\} , \quad (I) \\ &F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)\bigr)=\min \bigl\{ F_{2}\bigl(d\bigl( \mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{m-1}, \mathfrak{F}x_{m})\bigr)\bigr\} . \quad (\mathit{II}) \end{aligned}$$
If \((I)\) holds, then inequality (21) will take the form
$$\begin{aligned} \min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{m-1},\mathfrak{F}x_{m})\bigr)\bigr\} \leq F_{1}\bigl(d(x_{m-1},x_{m})\bigr)- \tau . \end{aligned}$$
(24)
Moreover, we also have either
$$\begin{aligned} F_{1}\bigl(d(x_{m-1},x_{m})\bigr)= \min \bigl\{ F_{2}\bigl(d(\mathfrak{F}x_{m-1}, \mathfrak{F}x_{m})\bigr),F_{1}\bigl(d(x_{m-1},x_{m}) \bigr)\bigr\} \end{aligned}$$
(25)
or
$$\begin{aligned} F_{2}\bigl(d(\mathfrak{F}x_{m-1}, \mathfrak{F}x_{m})\bigr)=\min \bigl\{ F_{2}\bigl(d( \mathfrak{F}x_{m-1},\mathfrak{F}x_{m})\bigr),F_{1} \bigl(d(x_{m-1},x_{m})\bigr)\bigr\} . \end{aligned}$$
(26)
If (23) holds, then (22) yields
$$\begin{aligned} &\min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{m-1},\mathfrak{F}x_{m})\bigr)\bigr\} \\ &\quad \leq \min \bigl\{ F_{2}\bigl(d(\mathfrak{F}x_{m-1}, \mathfrak{F}x_{m})\bigr),F_{1}\bigl(d(x_{m-1},x_{m}) \bigr) \bigr\} -\tau . \end{aligned}$$
(27)
If (24) holds, we can write
$$\begin{aligned} F_{2}\bigl(d(\mathfrak{F}x_{m-1},\mathfrak{F}x_{m}) \bigr)< F_{1}\bigl(d(x_{m-1},x_{m})\bigr). \end{aligned}$$
It follows from condition \((A)\) that
$$\begin{aligned} F_{2}\bigl(d(\mathfrak{F}x_{m-1}, \mathfrak{F}x_{m})\bigr)< F_{1}\bigl(d(x_{m-1},x_{m}) \bigr) \leq F_{2}\bigl(d(\mathfrak{F}x_{m-1}, \mathfrak{F}x_{m})\bigr)+\upsilon . \end{aligned}$$
(28)
Applying (26) in (22), we have
$$\begin{aligned} &\min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{m-1},\mathfrak{F}x_{m})\bigr)\bigr\} \\ &\quad \leq F_{2}\bigl(d(\mathfrak{F}x_{m-1},\mathfrak{F}x_{m}) \bigr)+\upsilon -\tau . \end{aligned}$$
So, it follows from (24) that
$$\begin{aligned} &\min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{m-1},\mathfrak{F}x_{m})\bigr)\bigr\} \\ &\quad \leq \min \bigl\{ F_{2}\bigl(d(\mathfrak{F}x_{m-1}, \mathfrak{F}x_{m})\bigr),F_{1}\bigl(d(x_{m-1},x_{m}) \bigr) \bigr\} +\upsilon -\tau . \end{aligned}$$
(29)
Combining the both possible cases of (25) and (27), we have
$$\begin{aligned} &\min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{m-1},\mathfrak{F}x_{m})\bigr)\bigr\} \\ &\quad \leq \min \bigl\{ F_{2}\bigl(d(\mathfrak{F}x_{m-1}, \mathfrak{F}x_{m})\bigr),F_{1}\bigl(d(x_{m-1},x_{m}) \bigr) \bigr\} +\delta _{m}\upsilon -\tau , \end{aligned}$$
(30)
where
$$\delta _{m}= \textstyle\begin{cases} 0, & \text{if } F_{2}(t_{2})>F_{1}(t_{1}), \\ 1, & \text{if } F_{2}(t_{2})< F_{1}(t_{1}), \end{cases} $$
\(t_{1}, t_{2}\in \mathbb{R}\mathbbm{_{+}}\), \(t_{1}\neq t_{2}\).
Therefore, \((\mathit{I})\) implies (28).
Similarly, with the existence of \((\mathit{II})\), (21) takes the following form:
$$\begin{aligned} \min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{m-1},\mathfrak{F}x_{m})\bigr)\bigr\} \leq F_{2}\bigl(d(x_{m-1},x_{m})\bigr)- \tau . \end{aligned}$$
(31)
If \(F_{2}(d(x_{m-1},x_{m}))\leq F_{1}(d(x_{m-1},x_{m}))\), then (29) can be rewritten as
$$\begin{aligned} \min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{m-1},\mathfrak{F}x_{m})\bigr)\bigr\} \leq F_{1}\bigl(d(x_{m-1},x_{m})\bigr)- \tau . \end{aligned}$$
(32)
If \(F_{2}(d(x_{m-1},x_{m}))>F_{1}(d(x_{m-1},x_{m}))\), from condition \((B)\), we have
$$\begin{aligned} \min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{m-1},\mathfrak{F}x_{m})\bigr)\bigr\} \leq F_{1}\bigl(d(x_{m-1},x_{m})\bigr)+ \upsilon -\tau . \end{aligned}$$
(33)
Combining inequalities (30), (31), we have
$$\begin{aligned} &\min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{m-1},\mathfrak{F}x_{m})\bigr)\bigr\} \\ &\quad \leq F_{1}\bigl(d(x_{m-1},x_{m})\bigr)+ \eta _{m}\upsilon -\tau , \end{aligned}$$
(34)
where
$$\eta _{m}= \textstyle\begin{cases} 1, & \text{if } F_{2}(t)>F_{1}(t), \\ 0, & \text{if } F_{2}(t)< F_{1}(t), \end{cases} $$
\(t\in \mathbb{R}\mathbbm{_{+}}\).
Together with inequalities (22)–(28), inequality (32) will take the form
$$\begin{aligned} &\min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{m-1},\mathfrak{F}x_{m})\bigr)\bigr\} \\ &\quad \leq \min \bigl\{ F_{2}\bigl(d(\mathfrak{F}x_{m-1}, \mathfrak{F}x_{m})\bigr),F_{1}\bigl(d(x_{m-1},x_{m}) \bigr) \bigr\} +(\delta _{m}+\eta _{m})\upsilon -\tau . \end{aligned}$$
(35)
Therefore, \((\mathit{II})\) implies (33).
The existence of \((\mathit{I})\) and \((\mathit{II})\) implies the existence of (28) and (33) respectively. Combining inequalities (28) and (33), we have
$$\begin{aligned} &\min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{m-1},\mathfrak{F}x_{m})\bigr)\bigr\} \\ &\quad\leq \min \bigl\{ F_{2}\bigl(d(\mathfrak{F}x_{m-1}, \mathfrak{F}x_{m})\bigr),F_{1}\bigl(d(x_{m-1},x_{m}) \bigr) \bigr\} +(\delta _{m}+\varsigma _{m}\eta _{m})\upsilon -\tau \\ &\quad=\min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-2}, \mathfrak{F}^{2}x_{m-1}\bigr)\bigr),F_{1} \bigl(d(x_{m-2},x_{m-1})\bigr) \bigr\} +(\delta _{m}+ \varsigma _{m}\eta _{m})\upsilon -\tau , \end{aligned}$$
where \(\varsigma _{m}\) is either 0 or 1.
Repeating this process, we have
$$\begin{aligned} &\min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{m-1},\mathfrak{F}x_{m})\bigr)\bigr\} \\ &\quad\leq \min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-3}, \mathfrak{F}^{2}x_{m-2}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{m-3},\mathfrak{F}x_{m-2})\bigr)\bigr\} \\ &\quad\quad{}+(\delta _{m}+\varsigma _{m}\eta _{m})\upsilon +(\delta _{m-1}+ \varsigma _{m-1}\eta _{m-1})\upsilon -2\tau \\ &\quad\cdots \\ &\quad\leq \min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{1}, \mathfrak{F}^{2}x_{0}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{1},\mathfrak{F}x_{0})\bigr)\bigr\} +\upsilon \sum_{j=1}^{m}( \delta _{j}+ \varsigma _{j}\eta _{j})-m\tau . \end{aligned}$$
Since \(\tau >2\upsilon \) and \(\upsilon \sum_{j=1}^{m}(\delta _{j}+\varsigma _{j}\eta _{j})< m \tau \), we deduce
$$\begin{aligned} \lim_{m\rightarrow +\infty} \upsilon \sum_{j=1}^{m}( \delta _{j}+\varsigma _{j}\eta _{j})-m\tau =- \infty . \end{aligned}$$
Therefore,
$$\begin{aligned} \lim_{m\rightarrow +\infty} \min \bigl\{ F_{2}\bigl(d \bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m} \bigr)\bigr),F_{1}\bigl(d(\mathfrak{F}x_{m-1}, \mathfrak{F}x_{m})\bigr) \bigr\} =-\infty . \end{aligned}$$
(36)
Now, equation (34) further has two possible cases:
$$\begin{aligned} &\lim_{m\rightarrow +\infty} F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)\bigr)=-\infty , \end{aligned}$$
(E)
$$\begin{aligned} &\lim_{m\rightarrow +\infty} F_{1}\bigl(d(\mathfrak{F}x_{m-1}, \mathfrak{F}x_{m})\bigr)=-\infty . \end{aligned}$$
(F)
Condition \((C_{1})\) with case \((E)\) yields
$$\begin{aligned} \lim_{m\rightarrow +\infty} d\bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)=0 \end{aligned}$$
or equivalently,
$$\begin{aligned} \lim_{m\rightarrow +\infty} d\bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)=\lim_{m\rightarrow +\infty} d(x_{m+1}, \mathfrak{F}x_{m+1})=\lim_{m\rightarrow +\infty} d(x_{m}, \mathfrak{F}x_{m})=0. \end{aligned}$$
Condition \((C_{1})\) with case \((F)\) yields
$$\begin{aligned} \lim_{m\rightarrow +\infty} d(\mathfrak{F}x_{m-1}, \mathfrak{F}x_{m})=\lim_{m\rightarrow +\infty} d(x_{m}, \mathfrak{F}x_{m})=0. \end{aligned}$$
Therefore, from (34) we get
$$\begin{aligned} \lim_{m\rightarrow +\infty} d(x_{m}, \mathfrak{F}x_{m})=0. \end{aligned}$$
(39)
Now, in order to prove that the sequence \(\{x_{m}\}_{m=1}^{+\infty }\) is a Cauchy sequence, we suppose on the contrary that there exists \(\varepsilon >0\) and two subsequences \(\{ x_{g(m)}\}_{m=1}^{+\infty }\) and \(\{x_{h(m)}\}_{m=1}^{+\infty }\) of \(\{x_{n}\}\),
$$\begin{aligned} \lim_{m\rightarrow \infty} d(x_{g(m)+2},x_{h(m)+2})= \lim_{m\rightarrow \infty} d(x_{g(m)+1},x_{h(m)+1})= \lim _{m\rightarrow \infty} d(x_{g(m)},x_{h(m)})=\varepsilon _{+}. \end{aligned}$$
(40)
Applying (14) by taking \(x=x_{g(m)}\) and \(y=x_{h(m)}\), we get
$$\begin{aligned} &\tau +\min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{g(m)} \bigr),\mathfrak{F}^{2}x_{h(m)}\bigr),F_{1}\bigl(d( \mathfrak{F}x_{g(m)},\mathfrak{F}x_{h(m)})\bigr)\bigr\} \\ &\quad \leq \alpha _{2}F_{2}\bigl(d(x_{g(m)},x_{h(m)}) \bigr)+\alpha _{1}F_{1}\bigl(d(x_{g(m)},x_{h(m)}) \bigr), \end{aligned}$$
which concludes the following two cases:
$$\begin{aligned} \tau +F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{g(m)}, \mathfrak{F}^{2}x_{h(m)}\bigr)\bigr) \leq F_{2} \bigl(d(x_{g(m)},x_{h(m)})\bigr) \end{aligned}$$
or
$$\begin{aligned} \tau +F_{1}\bigl(d\bigl(\mathfrak{F}^{2}x_{g(m)}, \mathfrak{F}^{2}x_{h(m)}\bigr)\bigr) \leq F_{1} \bigl(d(x_{g(m)},x_{h(m)})\bigr). \end{aligned}$$
Taking limits in the above two inequalities as \(k\rightarrow \infty \), we have
$$\begin{aligned} \tau +\min \bigl\{ F_{2}(\varepsilon _{+}),F_{1}( \varepsilon _{+})\bigr\} \leq \alpha _{2}F_{2}( \varepsilon _{+})+\alpha _{1}F_{1}(\varepsilon _{+}), \end{aligned}$$
which yields either \(\tau +F_{2}(\varepsilon _{+})\leq F_{2}(\varepsilon _{+})\) or \(\tau +F_{1}(\varepsilon _{+})\leq F_{1}(\varepsilon _{+})\).
Both of the above inequalities are contradictions. Therefore, \(\{x_{m}\}_{m=1}^{+\infty}\) is a Cauchy sequence. The completeness of \((X,d)\) proves that \(\{x_{m}\}_{m=1}^{+\infty}\) converges to some point \(x^{*}\) in X.
Now, if \(\mathfrak{F}\) is orbitally continuous, then we have
$$\begin{aligned} d\bigl(\mathfrak{F}x^{*},x^{*}\bigr)=\lim _{m\rightarrow +\infty} d( \mathfrak{F}x_{m},x_{m})=\lim _{m\rightarrow +\infty} d(x_{m+1},x_{m})=d \bigl(x^{*},x^{*}\bigr)=0. \end{aligned}$$
If \(\mathfrak{F}\) is k-continuous for some \(k\in \mathbb{N}\), we have
$$\begin{aligned} d\bigl(\mathfrak{F}x^{*},x^{*}\bigr)=\lim _{m\rightarrow +\infty} d\bigl( \mathfrak{F}\bigl(\mathfrak{F}^{k-1}x_{m} \bigr),\mathfrak{F}^{k-1}x_{m}\bigr)= \lim _{m\rightarrow +\infty} d(x_{k+m},x_{k+m-1})=d \bigl(x^{*},x^{*}\bigr)=0. \end{aligned}$$
Therefore, \(\mathfrak{F}\) has a fixed point \(x^{*}\).
Furthermore, from Proposition (1.1), it follows that \(x\mapsto d(x,\mathfrak{F}x)\) is \(\mathfrak{F}\)-orbitally lower semi-continuous, then we have
$$\begin{aligned} d\bigl(x^{*},\mathfrak{F}x^{*}\bigr)\leq \liminf _{m\rightarrow \infty} d(x_{m}, \mathfrak{F}x_{m})=0, \end{aligned}$$
which implies that \(\mathfrak{F}x^{*}=x^{*}\) and \(x^{*}\) is a fixed point of \(\mathfrak{F}\).
Now, for the uniqueness, let us suppose that \(\mathfrak{F}\) has more than one fixed point, that is, there exist two distinct \(x,y\in X\) such that \(\mathfrak{F}x=x\neq y=\mathfrak{F}y\).
Therefore, \(d(x,y)=d(\mathfrak{F}x,\mathfrak{F}y)=d(\mathfrak{F}^{2}x, \mathfrak{F}^{2}y)>0\). From (14), we have either
$$\begin{aligned} F_{1}\bigl(d(x,y)\bigr)=F_{1}\bigl(d( \mathfrak{F}x,\mathfrak{F}y )\bigr)< \tau +F_{1}\bigl(d( \mathfrak{F}x, \mathfrak{F}y)\bigr)\leq F_{1}\bigl(d(x,y)\bigr) \end{aligned}$$
(41)
or
$$\begin{aligned} F_{2}\bigl(d(x,y)\bigr)=F_{2}\bigl(d\bigl( \mathfrak{F}^{2}x,\mathfrak{F}^{2}y \bigr)\bigr)< \tau +F_{2}\bigl(d\bigl( \mathfrak{F}^{2}x,\mathfrak{F}^{2}y \bigr)\bigr)\leq F_{2}\bigl(d(x,y)\bigr). \end{aligned}$$
(42)
Both (37), (38) are two contradictions that prove the existence of a unique fixed point. □
Theorem 2.2
Let \((X,d)\) be a complete metric space. Suppose that a mapping \(\mathfrak{F}:X\mapsto X\) is a dual \(F^{*}\)-weak contraction of type-II and is orbitally continuous or k-continuous for some \(k\in \mathbb{N}\). Moreover, for all \(t_{1}, t_{2}\in \mathbb{R}\mathbbm{_{+}}\), there exist \(\upsilon >0\), \(\tau >2\upsilon \) such that
$$\begin{aligned} &F_{2}(t_{2})< F_{1}(t_{1})\leq F_{2}(t_{2})+\upsilon \end{aligned}$$
(A)
or
$$\begin{aligned} &F_{1}(t_{1})< F_{2}(t_{2})\leq F_{1}(t_{1})+\upsilon . \end{aligned}$$
(B)
Then, for every \(x_{0}\in X\), the sequence \(\{\mathfrak{F}^{m}x_{0}\}_{m=1}^{+\infty}\) converges to the unique fixed point of \(\mathfrak{F}\).
Proof
Let \(x_{0}\in X\) be an arbitrary point and define a sequence \(\{x_{m}\}\subseteq X\) by \(x_{m+1}=\mathfrak{F}x_{m}=\mathfrak{F}^{m}x_{0}\) for all \(m\in \mathbb{N}_{0}\).
If there exists some \(m\in \mathbb{N}_{0}\) such that \(d(x_{m},\mathfrak{F}x_{m})=0\), then \(\mathfrak{F}\) admits a fixed point. So, we assume that \(d(x_{m},\mathfrak{F}x_{m})=d(\mathfrak{F}x_{m-1},\mathfrak{F}x_{m})>0\) for all \(m\in \mathbb{N}\).
We first take into account case \((R2)\). Analysis similar to the procedure of obtaining inequalities (22) to (28) in the proof of Theorem 2.1 shows that
$$\begin{aligned} &\min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{m-1},\mathfrak{F}x_{m})\bigr)\bigr\} \\ &\quad \leq \min \bigl\{ F_{2}\bigl(d(\mathfrak{F}x_{m-1}, \mathfrak{F}x_{m})\bigr),F_{1}\bigl(d(x_{m-1},x_{m}) \bigr) \bigr\} +\delta _{m}\upsilon -\tau , \end{aligned}$$
where
$$\delta _{m}= \textstyle\begin{cases} 0 & \text{if } F_{2}(t_{2})>F_{1}(t_{1}), \\ 1 & \text{if } F_{2}(t_{2})< F_{1}(t_{1}), \end{cases} $$
\(t_{1}, t_{2}\in \mathbb{R}\mathbbm{_{+}}\), \(t_{1}\neq t_{2}\).
The above relation takes the following form:
$$\begin{aligned} &\min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{m-1},\mathfrak{F}x_{m})\bigr)\bigr\} \\ &\quad \leq \min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-2}, \mathfrak{F}^{2}x_{m-1}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{m-2},\mathfrak{F}x_{m-1})\bigr)\bigr\} +\delta _{m}\upsilon - \tau . \end{aligned}$$
By continuing this process, we can write
$$\begin{aligned} &\min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{m-1},\mathfrak{F}x_{m})\bigr)\bigr\} \\ &\quad\leq \min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-3}, \mathfrak{F}^{2}x_{m-2}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{m-3},\mathfrak{F}x_{m-2})\bigr)\bigr\} \\ &\quad\quad{}+\delta _{m}\upsilon +\delta _{m-1}\upsilon -2\tau \\ &\quad\leq \cdots \\ &\quad\leq \min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{1}, \mathfrak{F}^{2}x_{0}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{1},\mathfrak{F}x_{0})\bigr)\bigr\} + \sum _{j=1}^{m} \delta _{j}\upsilon -m \tau \end{aligned}$$
or
$$\begin{aligned} &\min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{m-1},\mathfrak{F}x_{m})\bigr)\bigr\} \\ &\quad \leq F_{1}\bigl(d(x_{1},x_{0})\bigr)+ \sum _{j=1}^{m}\delta _{j} \upsilon -(m+1)\tau . \end{aligned}$$
Since \(\tau >2\upsilon \) and \(\sum_{j=1}^{m}\delta _{j}<2(m+1)\), therefore
$$\begin{aligned} \lim_{m\rightarrow +\infty} \sum_{j=1}^{m} \delta _{j} \upsilon -(m+1)\tau =-\infty . \end{aligned}$$
Therefore,
$$\begin{aligned} \lim_{m\rightarrow +\infty} \min \bigl\{ F_{2}\bigl(d\bigl( \mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m} \bigr)\bigr),F_{1}\bigl(d(\mathfrak{F}x_{m-1}, \mathfrak{F}x_{m})\bigr) \bigr\} =-\infty . \end{aligned}$$
Similar arguments apply to case \((R1)\). According to the procedure of inequalities (29) to (32) in the proof of Theorem 2.1, we have
$$\begin{aligned} &\min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{m-1},\mathfrak{F}x_{m})\bigr)\bigr\} \\ &\quad \leq F_{1}\bigl(d(x_{m-1},x_{m})\bigr)+\eta _{m}\upsilon -\tau , \end{aligned}$$
where
$$\eta _{m}= \textstyle\begin{cases} 1 & \text{if } F_{2}(t)>F_{1}(t), \\ 0 & \text{if } F_{2}(t)< F_{1}(t), \end{cases} $$
\(t\in \mathbb{R}\mathbbm{_{+}}\).
The above inequality will take the form
$$\begin{aligned} &\min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{m-1},\mathfrak{F}x_{m})\bigr)\bigr\} \\ &\quad \leq \min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-2}, \mathfrak{F}^{2}x_{m-1}\bigr)\bigr),F_{1} \bigl(d(x_{m-2},x_{m-1})\bigr) \bigr\} +(\delta _{m}+ \eta _{m})\upsilon -\tau . \end{aligned}$$
Repeating this process, we have
$$\begin{aligned} &\min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{m-1},\mathfrak{F}x_{m})\bigr)\bigr\} \\ &\quad\leq \min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-3}, \mathfrak{F}^{2}x_{m-2}\bigr)\bigr),F_{1} \bigl(d(x_{m-3},x_{m-2})\bigr) \bigr\} \\ &\quad\quad{}+(\delta _{m}+\eta _{m})\upsilon +(\delta _{m-1}+\eta _{m-1}) \upsilon -2\tau \\ &\quad\leq \cdots \\ &\quad\leq \min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{1}, \mathfrak{F}^{2}x_{0}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{1},\mathfrak{F}x_{0})\bigr)\bigr\} + \sum _{j=1}^{m}( \delta _{j}+\eta _{j})\upsilon -m\tau . \end{aligned}$$
Since \(\tau >2\upsilon \) and \(\sum_{j=1}^{m}(\delta _{j}+\eta _{j})\upsilon < m\tau \), therefore
$$\begin{aligned} \lim_{m\rightarrow +\infty} \sum_{j=1}^{m}( \delta _{j}+ \eta _{j})\upsilon -m\tau =-\infty . \end{aligned}$$
So that
$$\begin{aligned} \lim_{m\rightarrow +\infty} \min \bigl\{ F_{2}\bigl(d\bigl( \mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m} \bigr)\bigr),F_{1}\bigl(d(\mathfrak{F}x_{m-1}, \mathfrak{F}x_{m})\bigr) \bigr\} =-\infty . \end{aligned}$$
Hence, for both cases \((R1)\), \((R2)\), it follows from condition \((C_{1})\) that
$$\begin{aligned} \lim_{m\rightarrow +\infty} d(x_{m}, \mathfrak{F}x_{m})=0. \end{aligned}$$
(45)
Now, in order to prove that the sequence \(\{x_{m}\}_{m=1}^{+\infty }\) is a Cauchy sequence, we suppose on the contrary that there exist \(\varepsilon >0\) and two subsequences \(\{ x_{g(m)}\}_{m=1}^{+\infty }\) and \(\{x_{h(m)}\}_{m=1}^{+\infty }\) of \(\{x_{n}\}\),
$$\begin{aligned} \lim_{m\rightarrow \infty} d(x_{g(m)+2},x_{h(m)+2})= \lim_{m\rightarrow \infty} d(x_{g(m)+1},x_{h(m)+1})= \lim _{m\rightarrow \infty} d(x_{g(m)},x_{h(m)})=\varepsilon _{+}. \end{aligned}$$
(46)
Applying (15) by taking \(x=x_{g(m)}\) and \(y=x_{h(m)}\), we get
$$\begin{aligned} &\tau +\min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{g(m)}, \mathfrak{F}^{2}x_{h(m)}\bigr)\bigr),F_{1}\bigl(d \bigl( \mathfrak{F}^{2}x_{g(m)},\mathfrak{F}^{2}x_{h(m)} \bigr)\bigr)\bigr\} \\ &\quad \leq \alpha _{2}F_{2}\bigl(d(x_{g(m)},x_{h(m)}) \bigr)+\alpha _{1}F_{1}\bigl(d(x_{g(m)},x_{h(m)}) \bigr). \end{aligned}$$
Taking limits in the above inequalities as \(m\rightarrow \infty \), we have
$$\begin{aligned} \tau +\min \bigl\{ F_{2}(\varepsilon _{+}),F_{1}( \varepsilon _{+})\bigr\} \leq \alpha _{2}F_{2}( \varepsilon _{+})+\alpha _{1}F_{1}(\varepsilon _{+}). \end{aligned}$$
For both cases \((R1)\), \((R2)\), the above inequality yields the following possibilities:
$$\begin{aligned}& \tau +F_{2}(\varepsilon _{+})\leq F_{2}(\varepsilon _{+}), \end{aligned}$$
(47)
$$\begin{aligned}& \tau +F_{1}(\varepsilon _{+})\leq F_{1}(\varepsilon _{+}), \end{aligned}$$
(48)
$$\begin{aligned}& \tau +F_{2}(\varepsilon _{+})\leq F_{1}(\varepsilon _{+}), \end{aligned}$$
(49)
$$\begin{aligned}& \tau +F_{1}(\varepsilon _{+})\leq F_{2}(\varepsilon _{+}). \end{aligned}$$
(50)
As \(\tau >0\), inequalities (41) and (42) are the contradictions.
Now we consider inequality (43). Since \(\tau >2\upsilon \), we have
$$\begin{aligned} 2\upsilon +F_{2}(\varepsilon )\leq F_{1}(\varepsilon ). \end{aligned}$$
That is,
$$\begin{aligned} \upsilon +\bigl(\upsilon +F_{2}(\varepsilon )\bigr)< F_{1}( \varepsilon ). \end{aligned}$$
Moreover, from condition \((A)\) we also have
$$\begin{aligned} \upsilon +F_{1}(\varepsilon )< F_{1}(\varepsilon ). \end{aligned}$$
Therefore, inequality (43) yields a contradiction.
Similarly, \(\tau +F_{1}(\varepsilon )\leq F_{2}(\varepsilon )\) implies a contradiction \(\upsilon +F_{2}(\varepsilon )< F_{2}(\varepsilon )\). Contradictions of inequalities (41)–(44) prove that \(\{x_{m}\}_{m=1}^{+\infty }\) is a Cauchy sequence. Since \((X,d)\) is a complete metric space, the sequence \(\{x_{m}\}_{m=1}^{\infty }\) is convergent in X and \(x^{*}\) is the point of convergence. The completeness of \((X,d)\) proves that \(\{x_{m}\}_{m=1}^{\infty }\) converges to some point \(x^{*}\) in X.
Now, if \(\mathfrak{F}\) is orbitally continuous, then we have
$$\begin{aligned} d\bigl(\mathfrak{F}x^{*},x^{*}\bigr)=\lim _{m\rightarrow +\infty} d( \mathfrak{F}x_{m},x_{m})=\lim _{m\rightarrow +\infty} d(x_{m+1},x_{m})=d \bigl(x^{*},x^{*}\bigr)=0. \end{aligned}$$
If \(\mathfrak{F}\) is k-continuous for some \(k\in \mathbb{N}\), we have
$$\begin{aligned} d\bigl(\mathfrak{F}x^{*},x^{*}\bigr)=\lim _{m\rightarrow +\infty} d( \mathfrak{F}\bigl(\mathfrak{F}^{k-1}x_{m}, \mathfrak{F}^{k-1}x_{m}\bigr)= \lim_{m\rightarrow +\infty} d(x_{k+m},x_{k+m-1})=d\bigl(x^{*},x^{*} \bigr)=0. \end{aligned}$$
Therefore, \(\mathfrak{F}\) has a fixed point \(x^{*}\).
Furthermore, from Proposition 1.1 it follows that \(x\mapsto d(x,\mathfrak{F}x)\) is \(\mathfrak{F}\)-orbitally lower semi-continuous, then we have
$$\begin{aligned} d\bigl(x^{*},\mathfrak{F}x^{*}\bigr)\leq \liminf _{m\rightarrow \infty} d(x_{m}, \mathfrak{F}x_{m})=0, \end{aligned}$$
which implies that \(\mathfrak{F}x^{*}=x^{*}\) and \(x^{*}\) is a fixed point of \(\mathfrak{F}\).
Now, in order to prove the uniqueness, we suppose that there exist two distinct \(x,y\in X\) such that \(\mathfrak{F}x=x\neq y=\mathfrak{F}y\). Therefore, \(d(x,y)=d(\mathfrak{F}x,\mathfrak{F}y)=d(\mathfrak{F}^{2}x, \mathfrak{F}^{2}y)>0\). From the assumptions of the theorem, we may have the following four possibilities:
$$\begin{aligned}& F_{1}\bigl(d(x,y)\bigr)=F_{1}\bigl(d( \mathfrak{F}x,\mathfrak{F}y)\bigr)< \tau +F_{1}\bigl(d( \mathfrak{F}x, \mathfrak{F}y)\bigr)\leq F_{1}\bigl(d(x,y)\bigr), \end{aligned}$$
(51)
$$\begin{aligned}& F_{2}\bigl(d(x,y)\bigr)=F_{2}\bigl(d\bigl( \mathfrak{F}^{2}x,\mathfrak{F}^{2}y\bigr)\bigr)< \tau +F_{2}\bigl(d\bigl( \mathfrak{F}^{2}x,\mathfrak{F}^{2}y \bigr)\bigr)\leq F_{2}\bigl(d(x,y)\bigr), \end{aligned}$$
(52)
$$\begin{aligned}& F_{2}\bigl(d(x,y)\bigr)=F_{2}\bigl(d\bigl( \mathfrak{F}^{2}x,\mathfrak{F}^{2}y\bigr)\bigr)< \tau +F_{2}\bigl(d\bigl( \mathfrak{F}^{2}x,\mathfrak{F}^{2}y \bigr)\bigr)\leq F_{1}\bigl(d(x,y)\bigr), \end{aligned}$$
(53)
$$\begin{aligned}& F_{1}\bigl(d(x,y)\bigr)=F_{1}\bigl(d( \mathfrak{F}x,\mathfrak{F}y)\bigr)< \tau +F_{1}\bigl(d( \mathfrak{F}x, \mathfrak{F}y)\bigr)\leq F_{2}\bigl(d(x,y)\bigr). \end{aligned}$$
(54)
Inequalities (45) and (46) are both contradictions. Inequality (47) with condition \((B)\) yields a contradiction as follows:
$$\begin{aligned} \tau +F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x, \mathfrak{F}^{2}y\bigr)\bigr)\leq F_{1}\bigl(d(x,y)\bigr). \end{aligned}$$
Since \(\tau >2\upsilon \), we have
$$\begin{aligned} 2\upsilon +F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x, \mathfrak{F}^{2}y\bigr)\bigr)\leq F_{1}\bigl(d(x,y)\bigr) \end{aligned}$$
or
$$\begin{aligned} \upsilon +\bigl(\upsilon +F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x, \mathfrak{F}^{2}y\bigr)\bigr)\bigr) \leq F_{1}\bigl(d(x,y) \bigr), \end{aligned}$$
or
$$\begin{aligned} \upsilon +F_{1}\bigl(d\bigl(\mathfrak{F}^{2}x, \mathfrak{F}^{2}y\bigr)\bigr))\leq F_{1}\bigl(d(x,y)\bigr). \end{aligned}$$
As \(\mathfrak{F}^{2}x=x\), \(\mathfrak{F}^{2}y=y\), we have
$$\begin{aligned} \upsilon +F_{1}\bigl(d(x,y)\bigr))\leq F_{1}\bigl(d(x,y) \bigr), \end{aligned}$$
which is a contradiction.
Similarly, inequality (48) with condition \((A)\) implies a contradiction as follows:
$$\begin{aligned} \tau +F_{1}\bigl(d(\mathfrak{F}x,\mathfrak{F}y)\bigr)\leq F_{2}\bigl(d(x,y)\bigr). \end{aligned}$$
Since \(\tau >2\upsilon \), we can write
$$\begin{aligned} 2\upsilon +F_{1}\bigl(d(\mathfrak{F}x,\mathfrak{F}y)\bigr)\leq F_{2}\bigl(d(x,y)\bigr) \end{aligned}$$
or
$$\begin{aligned} \upsilon +\bigl(\upsilon +F_{1}\bigl(d(\mathfrak{F}x,\mathfrak{F}y) \bigr)\bigr)\leq F_{2}\bigl(d(x,y)\bigr), \end{aligned}$$
or
$$\begin{aligned} \upsilon +F_{2}\bigl(d(\mathfrak{F}x,\mathfrak{F}y)\bigr))\leq F_{2}\bigl(d(x,y)\bigr). \end{aligned}$$
As \(\mathfrak{F}x=x\), \(\mathfrak{F}y=y\), we have
$$\begin{aligned} \upsilon +F_{2}\bigl(d(x,y)\bigr))\leq F_{2}\bigl(d(x,y) \bigr), \end{aligned}$$
which is a contradiction. That proves the existence of a unique fixed point. □
Theorem 2.3
Let \((X,d)\) be a complete metric space. Suppose that a mapping \(\mathfrak{F}:X\mapsto X\) is a triple \(F^{*}\)-weak contraction and is orbitally continuous or k-continuous for some \(k\in \mathbb{N}\). Moreover, for all \(t, t_{1}, t_{2}\in \mathbb{R}\mathbbm{_{+}}\), there exist \(\upsilon >0\), \(\tau >2\upsilon \) such that
$$\begin{aligned} &F_{2}(t_{2})< F_{1}(t_{1})\leq F_{2}(t_{2})+\upsilon \end{aligned}$$
(A)
or
$$\begin{aligned} &F_{1}(t_{1})< F_{2}(t_{2})\leq F_{1}(t_{1})+\upsilon , \end{aligned}$$
(B)
or
$$\begin{aligned} &{{F}_{i}} ({{t}_{1}} )< F ( {{t}_{2}} )\le {{F}_{i}} ({{t}_{1}} )+\upsilon ,\quad i=1,2. \end{aligned}$$
(C)
Then, for every \(x_{0}\in X\), the sequence \(\{\mathfrak{F}^{m}x_{0}\}_{m=1}^{+\infty}\) converges to the unique fixed point of \(\mathfrak{F}\).
Proof
Consider a sequence \(\{x_{m}\}\subseteq X\) such that, for all \(m\in \mathbb{N}_{0}\), \(x_{m+1}=\mathfrak{F}x_{m}=\mathfrak{F}^{m}x_{0}\), where \(x_{0}\) is an arbitrary point in X. If there exists some \(m\in \mathbb{N}_{0}\) such that \(d(x_{m},\mathfrak{F}x_{m})=0\), then \(\mathfrak{F}\) admits a fixed point. So, we assume that \(d(x_{m},\mathfrak{F}x_{m})=d(\mathfrak{F}x_{m-1},\mathfrak{F}x_{m})>0\) for all \(m\in \mathbb{N}\).
We will show that \(\lim_{m\rightarrow +\infty} d(x_{m},\mathfrak{F}x_{m})=0\). From the definition of triple \(F^{*}\)-weak contraction, we can write
$$\begin{aligned} \min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{m-1},\mathfrak{F}x_{m})\bigr)\bigr\} \leq F \bigl(d(x_{m-1},x_{m})\bigr)- \tau . \end{aligned}$$
(58)
Using condition \((C)\), we can rewrite inequality (49) as follows:
$$\begin{aligned} &\min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{m-1},\mathfrak{F}x_{m})\bigr)\bigr\} \\ &\quad \leq F_{1}\bigl(d(x_{m-1},x_{m})\bigr)- \tau . \end{aligned}$$
(59)
Then we have either
$$\begin{aligned} F_{1}\bigl(d(x_{m-1},x_{m})\bigr)= \min \bigl\{ F_{2}\bigl(d(\mathfrak{F}x_{m-1}, \mathfrak{F}x_{m})\bigr),F_{1}\bigl(d(x_{m-1},x_{m}) \bigr)\bigr\} \end{aligned}$$
(60)
or
$$\begin{aligned} F_{2}\bigl(d(\mathfrak{F}x_{m-1}, \mathfrak{F}x_{m})\bigr)=\min \bigl\{ F_{2}\bigl(d( \mathfrak{F}x_{m-1},\mathfrak{F}x_{m})\bigr),F_{1} \bigl(d(x_{m-1},x_{m})\bigr)\bigr\} . \end{aligned}$$
(61)
If relation (51) exists, then (50) can be written as
$$\begin{aligned} &\min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{m-1},\mathfrak{F}x_{m})\bigr)\bigr\} \\ &\quad \leq \min \bigl\{ F_{2}\bigl(d(\mathfrak{F}x_{m-1}, \mathfrak{F}x_{m})\bigr),F_{1}\bigl(d(x_{m-1},x_{m}) \bigr) \bigr\} -\tau . \end{aligned}$$
(62)
If inequality (52) holds, we can write \(F_{2}(d(\mathfrak{F}x_{m-1},\mathfrak{F}x_{m}))< F_{1}(d(x_{m-1},x_{m}))\).
Using \((A)\), we can write
$$\begin{aligned} F_{2}\bigl(d(\mathfrak{F}x_{m-1}, \mathfrak{F}x_{m})\bigr)< F_{1}\bigl(d(x_{m-1},x_{m}) \bigr) \leq F_{2}\bigl(d(\mathfrak{F}x_{m-1}, \mathfrak{F}x_{m})\bigr)+\upsilon . \end{aligned}$$
(63)
Using inequality (54) in (50), we have
$$\begin{aligned} \min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{m-1},\mathfrak{F}x_{m})\bigr)\bigr\} \leq F_{2}\bigl(d(\mathfrak{F}x_{m-1}, \mathfrak{F}x_{m}) \bigr)+\upsilon -\tau . \end{aligned}$$
Moreover, from (52), we have
$$\begin{aligned} &\min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{m-1},\mathfrak{F}x_{m})\bigr)\bigr\} \\ &\quad\leq \min \bigl\{ F_{2}\bigl(d(\mathfrak{F}x_{m-1}, \mathfrak{F}x_{m})\bigr),F_{1}\bigl(d(x_{m-1},x_{m}) \bigr) \bigr\} +\upsilon -\tau . \end{aligned}$$
(64)
Combining both inequalities (53) and (55), we have
$$\begin{aligned} &\min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{m-1},\mathfrak{F}x_{m})\bigr)\bigr\} \\ &\quad\leq\min \bigl\{ F_{2}\bigl(d(\mathfrak{F}x_{m-1}, \mathfrak{F}x_{m})\bigr),F_{1}\bigl(d(x_{m-1},x_{m}) \bigr) \bigr\} +\delta _{m}\upsilon -\tau , \end{aligned}$$
(65)
where
$$\delta _{m}= \textstyle\begin{cases} 0 & \text{if } F_{2}(t_{2})>F_{1}(t_{1}), \\ 1 & \text{if } F_{2}(t_{2})\leq F_{1}(t_{1}), \end{cases} $$
\(t_{1}, t_{2}\in \mathbb{R}\mathbbm{_{+}}\), \(t_{1}\neq t_{2}\).
The above relation is equivalent to
$$\begin{aligned} &\min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{m-1},\mathfrak{F}x_{m})\bigr)\bigr\} \\ &\quad \leq \min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-2}, \mathfrak{F}^{2}x_{m-1}\bigr)\bigr),F_{1} \bigl(d(x_{m-2},x_{m-1})\bigr) \bigr\} +\delta _{m} \upsilon -\tau . \end{aligned}$$
Repeating this process, we have
$$\begin{aligned} &\min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{m-1},\mathfrak{F}x_{m})\bigr)\bigr\} \\ &\quad\leq \min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-3}, \mathfrak{F}^{2}x_{m-2}\bigr)\bigr),F_{1} \bigl(d(x_{m-3},x_{m-2})\bigr) \bigr\} \\ &\quad\quad{}+\delta _{m}\upsilon +\delta _{m-1}\upsilon -2\tau \\ &\quad\leq \cdots \\ &\quad\leq \min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{1}, \mathfrak{F}^{2}x_{0}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{1},\mathfrak{F}x_{0})\bigr)\bigr\} + \sum _{j=1}^{m} \delta _{j}\upsilon -m \tau \end{aligned}$$
or
$$\begin{aligned} &\min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)\bigr),F_{1}\bigl(d( \mathfrak{F}x_{m-1},\mathfrak{F}x_{m})\bigr)\bigr\} \\ &\quad \leq F_{1}\bigl(d(x_{1},x_{0})\bigr)+ \sum _{j=1}^{m}\delta _{j} \upsilon -(m+1)\tau . \end{aligned}$$
Since \(\tau >2\upsilon \) and \(\sum_{j=1}^{m}\delta _{j}< m+1\), we have
$$\begin{aligned} \lim_{m\rightarrow +\infty} \sum_{j=1}^{m} \delta _{j} \upsilon -(m+1)\tau =-\infty . \end{aligned}$$
So that we have
$$\begin{aligned} \lim_{m\rightarrow +\infty} \min \bigl\{ F_{2}\bigl(d \bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m} \bigr)\bigr),F_{1}\bigl(d(\mathfrak{F}x_{m-1}, \mathfrak{F}x_{m})\bigr) \bigr\} =-\infty . \end{aligned}$$
(66)
Now, equation (57) further has two possible cases:
$$\begin{aligned} &\lim_{m\rightarrow +\infty} F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)\bigr)=-\infty , \end{aligned}$$
(G)
$$\begin{aligned} &\lim_{m\rightarrow +\infty} F_{1}\bigl(d(\mathfrak{F}x_{m-1}, \mathfrak{F}x_{m})\bigr)=-\infty . \end{aligned}$$
(H)
Condition \((C_{1})\) with case \((G)\) yields
$$\begin{aligned} \lim_{m\rightarrow +\infty} d\bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)=0, \end{aligned}$$
or equivalently,
$$\begin{aligned} \lim_{m\rightarrow +\infty} d\bigl(\mathfrak{F}^{2}x_{m-1}, \mathfrak{F}^{2}x_{m}\bigr)=\lim_{m\rightarrow +\infty} d(x_{m+1}, \mathfrak{F}x_{m+1})=\lim_{m\rightarrow +\infty} d(x_{m}, \mathfrak{F}x_{m})=0. \end{aligned}$$
Condition \((C_{1})\) with case \((H)\) yields
$$\begin{aligned} \lim_{m\rightarrow +\infty} d(\mathfrak{F}x_{m-1}, \mathfrak{F}x_{m})=\lim_{m\rightarrow +\infty} d(x_{m}, \mathfrak{F}x_{m})=0. \end{aligned}$$
Therefore, from (57), we get
$$\begin{aligned} \lim_{m\rightarrow +\infty} d(x_{m}, \mathfrak{F}x_{m})=0. \end{aligned}$$
(69)
Now, in order to prove that the sequence \(\{x_{m}\}_{m=1}^{+\infty }\) is a Cauchy sequence, we suppose on the contrary that there exist \(\varepsilon >0\) and two subsequences \(\{ x_{g(m)}\}_{m=1}^{+\infty }\) and \(\{x_{h(m)}\}_{m=1}^{+\infty }\) of \(\{x_{n}\}\),
$$\begin{aligned} \lim_{m\rightarrow \infty} d(x_{g(m)+2},x_{h(m)+2})= \lim_{m\rightarrow \infty} d(x_{g(m)+1},x_{h(m)+1})= \lim _{m\rightarrow \infty} d(x_{g(m)},x_{h(m)})=\varepsilon _{+}. \end{aligned}$$
(70)
Applying (16) by taking \(x=x_{g(m)}\) and \(y=x_{h(m)}\), we get
$$\begin{aligned} &\tau +\min \bigl\{ F_{2}\bigl(d\bigl(\mathfrak{F}^{2}x_{g(m)}, \mathfrak{F}^{2}x_{h(m)}\bigr)\bigr),F_{1}\bigl(d \bigl( \mathfrak{F}^{2}x_{g(m)},\mathfrak{F}^{2}x_{h(m)} \bigr)\bigr)\bigr\} \\ &\quad \leq F\bigl(d(x_{g(m)},x_{h(m)})\bigr). \end{aligned}$$
Taking limits in the above inequalities as \(m\rightarrow \infty \), we have
$$\begin{aligned} \tau +\min \bigl\{ F_{2}(\varepsilon _{+}),F_{1}( \varepsilon _{+})\bigr\} \leq F_{1}( \varepsilon _{+}), \end{aligned}$$
which yields either \(\tau +F_{2}(\varepsilon _{+})\leq F(\varepsilon _{+})\) or \(\tau +F_{1}(\varepsilon _{+})\leq F(\varepsilon _{+})\).
If \(\tau +{{F}_{2}} (\varepsilon _{+} )\le F ( \varepsilon _{+} )\), condition \((C)\) allows us to write \(\tau +{{F}_{2}} (\varepsilon _{+} )\le {{F}_{2}} ( \varepsilon _{+} )+\upsilon \). That yields a contradiction as \(\tau >2\upsilon \). Likewise, \(\tau +{{F}_{1}} (\varepsilon _{+} )\le F ( \varepsilon _{+} )\) gives a contradiction \(\tau +{{F}_{2}} (\varepsilon _{+} )\le {{F}_{2}} ( \varepsilon _{+} )+\upsilon \).
The completeness of \((X,d)\) proves that \(\{x_{m}\}_{m=1}^{+\infty}\) converges to some point \(x^{*}\) in X.
The rest of the proof runs as the proof of Theorem 2.1. For brevity, we omit it. □
Example 2.4
Consider a closed unit ball \(\mathcal{B}\) in the \(\ell _{1}\) space of all absolutely summable sequences, \(u =(u _{1},u_{2},\ldots )\) with a metric inherited from the standard norm \(\|u\|=\sum_{i=1}^{+\infty}|u_{i}|\). Let \(h:[-1,1]\mapsto [-1,1]\) be the function given by
$$h(w)= \textstyle\begin{cases} 1+2w,& \text{if } -1\leq w\leq -\frac{1}{2}, \\ 0,& \text{if } -\frac{1}{2}< w\leq \frac{1}{2}, \\ -1+2w,& \text{if } \frac{1}{2}< w\leq 1. \end{cases} $$
Observe that for all \(s, w\in [-1,1]\), we have \(|h(s)-h(w)|\leq 2|s-w|\) and \(|h(w)|\leq |w|\).
Define a continuous mapping \(\mathfrak{F}:\mathcal{B}\mapsto \mathcal{B}\) by
$$\begin{aligned} \mathfrak{F}u=\mathfrak{F}(u_{1},u_{2},\ldots )=e^{-\kappa}\biggl(h(u_{2}), \frac{2}{3}u_{3},u_{4},u_{5}, \ldots \biggr), \end{aligned}$$
where \(\kappa >0\) is a real number. Then we have
$$\begin{aligned} \mathfrak{F}^{2}u=e^{-\kappa}\biggl(h\biggl( \frac{2}{3}u_{3}\biggr),\frac{2}{3}u_{4},u_{5},u_{6}, \ldots \biggr). \end{aligned}$$
For each \(u= (u_{1},u_{2}, \ldots )\) and \(v=(v_{1},v_{2},\ldots )\) in \(\mathcal{B}\), we have
$$\begin{aligned} e^{\kappa} \Vert \mathfrak{F}u-\mathfrak{F}v \Vert &= \bigl\vert h(u_{2})-h(v_{2}) \bigr\vert + \frac{2}{3} \vert u_{3}-v_{3} \vert +\sum_{k=4}^{+\infty } \vert u_{k}-v_{k} \vert \\ & \leq 2 \vert u_{2}-v_{2} \vert +\frac{2}{3} \vert u_{3}-v_{3} \vert + \sum _{k=4}^{+ \infty} \vert u_{k}-v_{k} \vert \\ & \leq 2 \Vert u- v \Vert . \end{aligned}$$
That is,
$$\begin{aligned} \frac{e^{\kappa}}{2} \Vert \mathfrak{F}u-\mathfrak{F}v \Vert \leq \Vert u-v \Vert . \end{aligned}$$
(71)
The above inequality can be written as
$$\begin{aligned} \kappa +\ln \biggl(\frac{1}{2} \Vert \mathfrak{F}u- \mathfrak{F}v \Vert \biggr)\leq \ln \Vert u-v \Vert . \end{aligned}$$
(72)
Likewise,
$$\begin{aligned} e^{\kappa} \bigl\Vert \mathfrak{F}^{2}u-\mathfrak{F}^{2}v \bigr\Vert &= \biggl\vert h\biggl(\frac{2}{3}u_{3}\biggr)-h \biggl( \frac{2}{3}v_{3}\biggr) \biggr\vert + \frac{2}{3} \vert u_{3}-v_{3} \vert \\ &\quad{}+ \sum_{k=5}^{+\infty} \vert u_{k}-v_{k} \vert \\ & \leq \frac{4}{3} \vert u_{3}-v_{3} \vert + \frac{2}{3} \vert u_{3}-v_{3} \vert +\sum _{k=}^{+\infty} \vert u_{k}-v_{k} \vert \\ & \leq \frac{4}{3} \Vert u-v \Vert . \end{aligned}$$
That is,
$$\begin{aligned} \frac{3e^{\kappa}}{4} \bigl\Vert \mathfrak{F}^{2}u- \mathfrak{F}^{2}v \bigr\Vert \leq \Vert u-v \Vert . \end{aligned}$$
The above inequality can be written as
$$\begin{aligned} \kappa + \ln \biggl(\frac{3}{4} \bigl\Vert \mathfrak{F}^{2}u-\mathfrak{F}^{2}v \bigr\Vert \biggr) \leq \ln \Vert u-v \Vert . \end{aligned}$$
(73)
Now, we define \(F_{1}(t)=\ln (\frac{t}{2})\), \(F_{2}(t)=\ln (\frac{3t}{4})\), \(F(t)=\ln (t)\). So, inequalities (61), (62) will take the form
$$\begin{aligned} \kappa +\min \bigl\{ F_{1}\bigl( \Vert \mathfrak{F}u- \mathfrak{F}v \Vert \bigr),F_{2}\bigl( \bigl\Vert \mathfrak{F}^{2}u-\mathfrak{F}^{2}v \bigr\Vert \bigr)\bigr\} \leq \min \{F\bigl( \Vert u-v \Vert \bigr). \end{aligned}$$
(74)
Further, the definitions of \(F_{1}\), \(F_{2}\), and F yield \(F_{1}(t)< F_{2}(t)\). So that we have \(\ln \frac{t}{2}<\ln \frac{3t}{4}\). Now, we can define \(0<\upsilon \leq \ln \frac{3}{2}\) such that
$$\begin{aligned} \ln \frac{t}{2}< \ln \frac{3t}{4}\leq \ln \frac{t}{2}+ \upsilon . \end{aligned}$$
That is, \(F_{1}(t)< F_{2}(t)\leq F_{1}(t)+\upsilon \).
Therefore, we conclude from inequality (63) that \(\mathfrak{F}:\mathcal{B}\mapsto \mathcal{B}\) represents the triple \(F^{*}\)-weak mapping.
Next we will show that for every \(x_{0} \in \mathcal{B}\), the sequence \(\{\mathfrak{F}^{m}x_{0}\}_{m=1}^{+\infty }\) converges to a unique fixed point \(x^{*}=(0, 0,\ldots )\).
For some fixed \(i\in \mathbb{N}\), consider the absolutely summable sequences \(a=(a_{1},a_{2},\ldots ,a_{i},0,0, \ldots )\in \mathcal{B}\) with a metric inherited from the standard norm. As \(\mathcal{B}\) is a closed unit ball, we have \(\sum_{k=1}^{i}{|a_{k}|\leq 1}\).
Now,
$$\begin{aligned} \mathfrak{F}a=e^{-\kappa}\biggl(\tau (a_{2}), \frac{2}{3}a_{3},\ldots ,a_{i},0,0, \ldots \biggr) \end{aligned}$$
implies that
$$\begin{aligned} \mathfrak{F}^{2}a=e^{-\kappa}\biggl(\tau \biggl( \frac{2}{3}a_{3}\biggr),a_{4},\ldots ,a_{i},0,0, \ldots \biggr). \end{aligned}$$
Then, for all \(m>i\), we have \(\mathfrak{F}^{m}a=(0,0,\ldots ,0)\). That is, \(\{\mathfrak{F}^{m}a\}_{m=1}^{+\infty}=(0,0,\ldots )\) is a unique fixed point.