Let
\(\tilde{x}^{0}_{\lambda_{0}} \in\mathcal{SP}(\tilde{X})\) be given. Put
$$\tilde{x}^{n+1}_{\lambda_{n+1}}=\bigl((f,\varphi) \bigl( \tilde{x}^{n}_{\lambda _{n}}\bigr)\bigr)=\bigl(f^{n+1}\bigl( \tilde{x}^{0}_{\lambda_{0}}\bigr)\bigr)_{\varphi^{n+1}(\lambda_{0})}, $$
for each
\(n\in\mathbb{N}\cup\{0\}\). Then by the inequality (
1), we have
$$\begin{aligned} \tilde{d}\bigl(\tilde{x}^{n}_{\lambda_{n}},\tilde{x}^{n+1}_{\lambda _{n+1}} \bigr)&=\tilde{d}\bigl((f,\varphi) \bigl(\tilde{x}^{n-1}_{\lambda _{n-1}} \bigr),(f,\varphi) \bigl(\tilde{x}^{n}_{\lambda_{n}}\bigr)\bigr) \\ &\, \tilde{< } \, \tilde {d}\bigl(\tilde{x}^{n-1}_{\lambda_{n-1}}, \tilde{x}^{n}_{\lambda_{n}}\bigr), \end{aligned}$$
for each
\(n\in\mathbb{N}\cup\{0\}\). Therefore the sequence
\(\{\tilde {d}(\tilde{x}^{n}_{\lambda_{n}},\tilde{x}^{n+1}_{\lambda_{n+1}})\}_{n}\) is decreasing, it must converge to some soft real number
\(\tilde{\eta} \, \tilde{\geq}\, \overline{0}\), that is,
\(\tilde{d}(\tilde{x}^{n}_{\lambda _{n}},\tilde{x}^{n+1}_{\lambda_{n+1}})\rightarrow\tilde{\eta}\), as
\(n\rightarrow\infty\). Notice that
\(\tilde{\xi}=\inf\{\tilde{d}(\tilde {x}^{n}_{\lambda_{n}},\tilde{x}^{n+1}_{\lambda_{n+1}}):n\in\mathbb {N}\cup\{0\}\}\). We claim that
\(\tilde{\eta}=\overline{0}\). Suppose, on the contrary, that
\(\tilde{\eta}\, \tilde{>}\, \overline{0}\). Since
f is a soft Meir-Keeler contractive mapping, corresponding to
\(\tilde{\eta }\), there exist
\(\tilde{\gamma} >\overline{0}\) and
\(k\in\mathbb{N}\) such that
$$\begin{aligned}& \tilde{\eta}\, \tilde{\leq}\, \tilde{d}\bigl(\tilde{x}^{k}_{\lambda_{k}}, \tilde {x}^{k+1}_{\lambda_{k+1}}\bigr)\, \tilde{< }\, \tilde{\eta}+\tilde{ \gamma} \\& \quad \Longrightarrow\quad \tilde{d}\bigl(\tilde{x}^{k+1}_{\lambda_{k+1}}, \tilde {x}^{k+2}_{\lambda_{k+2}}\bigr)=\tilde{d}\bigl((f,\varphi) \bigl( \tilde{x}^{k}_{\lambda _{k}}\bigr),(f,\varphi) \bigl( \tilde{x}^{k+1}_{\lambda_{k+1}}\bigr)\bigr)\, \tilde{< }\, \tilde {\eta}. \end{aligned}$$
This is a contradiction since
\(\tilde{\eta}=\inf\{\tilde{d}(\tilde {x}^{n}_{\lambda_{n}},\tilde{x}^{n+1}_{\lambda_{n+1}}):n\in\mathbb {N}\cup\{0\}\}\). Thus, we obtain
\(\tilde{d}(\tilde{x}^{n}_{\lambda _{n}}, \tilde{x}^{n+1}_{\lambda_{n+1}})\rightarrow\overline{0}\), as
\(n\rightarrow\infty\).
We next show that
\(\{{\tilde{x}^{n}_{\lambda_{n}}}\}_{n}\) is a Cauchy sequence in
\((\tilde{X},\tilde{d},E)\). To prove this, on the contrary, assume that a soft real number
\(\tilde{\epsilon}\, \tilde{>}\, \overline{0}\) such that for any
\(k\in\mathbb{N}\), there are
\(m_{k}, n_{k}\in\mathbb{N}\) with
\(n_{k} > m_{k} \geq k\) satisfying
$$\tilde{d}\bigl(\tilde{x}^{m_{k}}_{\lambda_{m_{k}}},\tilde{x}^{n_{k}}_{\lambda _{n_{k}}} \bigr)\, \tilde{\geq} \, \tilde{\epsilon}. $$
Further, corresponding to
\(m_{k}\geq k\), we can choose
\(n_{k}\) in such a way that it is the smallest integer with
\(n_{k}>m_{k}\geq k\) and
\(\tilde {d}(\tilde{x}^{m_{k}}_{\lambda_{m_{k}}},\tilde{x}^{n_{k}}_{\lambda _{n_{k}}})\, \tilde{\geq} \, \tilde{\epsilon}\). Therefore,
$$\tilde{d}\bigl(\tilde{x}^{m_{k}}_{\lambda_{m_{k}}},\tilde{x}^{n_{k}-2}_{\lambda _{n_{k}-2}} \bigr)\, \tilde{< }\, \tilde{\epsilon}. $$
Now we have, for all
\(k\in\mathbb{N}\),
$$\begin{aligned} \tilde{\epsilon}&\, \tilde{\leq}\, \tilde{d}\bigl(\tilde{x}^{m_{k}}_{\lambda _{m_{k}}}, \tilde{x}^{n_{k}}_{\lambda_{n_{k}}}\bigr) \\ &\, \tilde{\leq}\, \tilde{d}\bigl(\tilde {x}^{m_{k}}_{\lambda_{m_{k}}}, \tilde{x}^{n_{k}-2}_{\lambda_{n_{k}-2}}\bigr)+\tilde {d}\bigl( \tilde{x}^{n_{k}-2}_{\lambda_{n_{k}-2}},\tilde{x}^{n_{k}-1}_{\lambda _{n_{k}-1}} \bigr)+\tilde{d}\bigl(\tilde{x}^{n_{k}-1}_{\lambda_{n_{k}-1}},\tilde {x}^{n_{k}}_{\lambda_{n_{k}}}\bigr) \\ &\, \tilde{< }\, \tilde{\epsilon}+\tilde{d}\bigl(\tilde {x}^{n_{k}-2}_{\lambda_{n_{k}-2}}, \tilde{x}^{n_{k}-1}_{\lambda _{n_{k}-1}}\bigr)+\tilde{d}\bigl(\tilde{x}^{n_{k}-1}_{\lambda_{n_{k}-1}}, \tilde {x}^{n_{k}}_{\lambda_{n_{k}}}\bigr). \end{aligned}$$
Letting
\(k\rightarrow\infty\) in the above inequality, we get
$$\tilde{d}\bigl(\tilde{x}^{m_{k}}_{\lambda_{m_{k}}},\tilde{x}^{n_{k}}_{\lambda _{n_{k}}} \bigr)\rightarrow\tilde{\epsilon} \quad \mbox{as } n\rightarrow\infty. $$
On the other hand, we have
$$\begin{aligned} \begin{aligned} \tilde{\epsilon}\, \tilde{\leq}\,{}& \tilde{d}\bigl(\tilde{x}^{m_{k}}_{\lambda _{m_{k}}}, \tilde{x}^{n_{k}}_{\lambda_{n_{k}}}\bigr) \\ \, \tilde{\leq}\,{}& \tilde{d}\bigl(\tilde{x}^{m_{k}}_{\lambda_{m_{k}}}, \tilde {x}^{m_{k}+1}_{\lambda_{m_{k}+1}}\bigr)+\tilde{d}\bigl( \tilde{x}^{m_{k}+1}_{\lambda _{m_{k}+1}},\tilde{x}^{n_{k}+1}_{\lambda_{n_{k}+1}} \bigr)+\tilde{d}\bigl(\tilde {x}^{n_{k}+1}_{\lambda_{n_{k}+1}}, \tilde{x}^{n_{k}}_{\lambda_{n_{k}}}\bigr) \\ \, \tilde{\leq}\,{}& \tilde{d}\bigl(\tilde{x}^{m_{k}}_{\lambda_{m_{k}}}, \tilde {x}^{m_{k}+1}_{\lambda_{m_{k}+1}}\bigr)+\tilde{d}\bigl( \tilde{x}^{m_{k}+1}_{\lambda _{m_{k}+1}},\tilde{x}^{m_{k}}_{\lambda_{m_{k}}} \bigr)+\tilde{d}\bigl(\tilde {x}^{m_{k}}_{\lambda_{m_{k}}}, \tilde{x}^{n_{k}}_{\lambda_{n_{k}}}\bigr) \\ &{} +\tilde {d}\bigl(\tilde{x}^{n_{k}}_{\lambda_{n_{k}}}, \tilde{x}^{n_{k}+1}_{\lambda _{n_{k}+1}}\bigr)+\tilde{d}\bigl(\tilde{x}^{n_{k}+1}_{\lambda_{n_{k}+1}}, \tilde {x}^{n_{k}}_{\lambda_{n_{k}}}\bigr). \end{aligned} \end{aligned}$$
Letting
\(k\rightarrow\infty\) in the above inequality, we get
$$\tilde{d}\bigl(\tilde{x}^{m_{k}+1}_{\lambda_{m_{k}+1}},\tilde{x}^{n_{k}+1}_{\lambda _{n_{k}+1}} \bigr)\rightarrow\tilde{\epsilon} \quad \mbox{as } n\rightarrow\infty. $$
Since
f is soft Meir-Keeler contractive mapping, we have
$$\begin{aligned} \tilde{d}\bigl(\tilde{x}^{m_{k}+1}_{\lambda_{m_{k}+1}},\tilde{x}^{n_{k}+1}_{\lambda _{n_{k}+1}} \bigr)&=\tilde{d}\bigl((f,\varphi) \bigl(\tilde{x}^{m_{k}}_{\lambda _{m_{k}}} \bigr),(f,\varphi) \bigl(\tilde{x}^{n_{k}}_{\lambda_{n_{k}}}\bigr)\bigr) \\ & \, \tilde {< }\, \tilde{d}\bigl(\tilde{x}^{m_{k}}_{\lambda_{m_{k}}}, \tilde{x}^{n_{k}}_{\lambda_{n_{k}}}\bigr). \end{aligned}$$
Letting
\(k\rightarrow\infty\) in the above inequality, we get
\(\tilde {\epsilon}\, \tilde{<}\, \tilde{\epsilon}\), which implies a contradiction. Thus,
\(\{{\tilde{x}^{n}_{\lambda_{n}}}\}\) is a Cauchy sequence in
\((\tilde{X},\tilde{d},E)\).
Since
\((\tilde{X},\tilde{d},E)\) is complete, there exists
\(\tilde {x}^{*}_{\lambda}\in\tilde{X}\) such that
$$\tilde{x}^{n}_{\lambda_{n}}\rightarrow\tilde{x}^{*}_{\lambda}\quad \mbox{as } n\rightarrow\infty, $$
that is,
$$\tilde{d}\bigl(\tilde{x}^{n}_{\lambda_{n}},\tilde{x}^{*}_{\lambda} \bigr)\rightarrow \overline{0}\quad \mbox{as } n\rightarrow\infty. $$
We also have
$$\begin{aligned} \tilde{d}\bigl((f,\varphi) \bigl(\tilde{x}^{*}_{\lambda}\bigr), \tilde{x}^{*}_{\lambda }\bigr)&\, \tilde{\leq}\, \tilde{d}\bigl((f,\varphi) \bigl(\tilde{x}^{n}_{\lambda_{n}}\bigr),(f,\varphi ) \bigl( \tilde{x}^{*}_{\lambda}\bigr)\bigr)+\tilde{d}\bigl((f,\varphi) \bigl( \tilde{x}^{n}_{\lambda _{n}}\bigr),\tilde{x}^{*}_{\lambda}\bigr) \\ &\, \tilde{< }\, \tilde{d}\bigl(\tilde{x}^{n}_{\lambda _{n}}, \tilde{x}^{*}_{\lambda}\bigr)+\tilde{d}\bigl(\tilde{x}^{n+1}_{\lambda _{n+1}}, \tilde{x}^{*}_{\lambda}\bigr). \end{aligned}$$
Letting
\(k\rightarrow\infty\) in the above inequality, we get
$$\tilde{d}\bigl((f,\varphi) \bigl(\tilde{x}^{*}_{\lambda}\bigr), \tilde{x}^{*}_{\lambda}\bigr)= \overline{0}. $$
This implies that
\((f,\varphi)(\tilde{x}^{*}_{\lambda})=\tilde {x}^{*}_{\lambda}\), and hence
\(\tilde{x}^{*}_{\lambda}\) is a fixed point of the mapping
\((f,\varphi)\).
Finally, let
\(\tilde{y}^{*}_{\lambda}\) be another fixed point of the mapping
\((f,\varphi)\). Then
$$\tilde{d}\bigl(\tilde{y}^{*}_{\lambda},\tilde{x}^{*}_{\lambda} \bigr)=\tilde {d}\bigl((f,\varphi) \bigl(\tilde{y}^{*}_{\lambda} \bigr),(f,\varphi) \bigl(\tilde {x}^{*}_{\lambda}\bigr)\bigr)\, \tilde{< }\, \tilde{d}\bigl(\tilde{y}^{*}_{\lambda},\tilde {x}^{*}_{\lambda}\bigr), $$
which implies a contradiction. Thus,
\(\tilde{x}^{*}_{\lambda}=\tilde {y}^{*}_{\lambda}\), that is, the fixed point of the mapping
\((f,\varphi )\) is unique. □