Step 1. If
\(H([\theta s_{0}]_{\alpha _{\theta s_{0}}},[\theta s_{1}]_{\alpha _{ \theta s_{1}}})=0\) then
\([\theta s_{0}]_{\alpha _{\theta s_{0}}}=[\theta s_{1}]_{\alpha _{ \theta s_{1}}}\).
θ. Thus,
\(s_{1}\) is a FP of
θ. If
\(H([\theta s_{0}]_{\alpha _{\theta s_{0}}},[\theta s_{1}]_{\alpha _{ \theta s_{1}}})>0\), by Lemma
3.9 then for each
\(g_{1}>1\), there exists
\(s_{2} \in [\theta s_{1}]_{\alpha _{\theta s_{1}}}\) such that
$$ \varpi (s_{1},s_{2})< g_{1}H\bigl([\theta s_{0}]_{\alpha _{\theta s_{0}}},[ \theta s_{1}]_{\alpha _{\theta s_{1}}}\bigr). $$
Step 2. Similarly, if
\(H([\theta s_{1}]_{\alpha _{\theta s_{1}}},[\theta s_{2}]_{\alpha _{ \theta s_{2}}})=0\) then
\([\theta s_{1}]_{\alpha _{\theta s_{1}}}=[\theta s_{2}]_{\alpha _{ \theta s_{2}}}\). Thus,
\(s_{2}\) is a FP of
θ. If
\(H([\theta s_{1}]_{\alpha _{\theta s_{1}}},[\theta s_{2}]_{\alpha _{ \theta s_{2}}})>0\), by Lemma
3.9 then for each
\(g_{2}>1\), there exists
\(s_{3} \in [\theta s_{2}]_{\alpha _{\theta s_{2}}}\) such that
$$ \varpi (s_{2},s_{3})< g_{2}H\bigl([\theta s_{1}]_{\alpha _{\theta s_{1}}},[ \theta s_{2}]_{\alpha _{\theta s_{2}}}\bigr). $$
Step n. Continuing in this manner, if
\(H([\theta s_{n-1}]_{\alpha _{\theta s_{n-1}}},[\theta s_{n}]_{ \alpha _{\theta s_{n}}})=0\). Thus,
\(s_{n}\) is a FP of
θ. If
\(H([\theta s_{n-1}]_{\alpha _{\theta s_{n-1}}},[\theta s_{n}]_{ \alpha _{\theta s_{n}}})>0\), by Lemma
3.9 then for each
\(g_{n}>1\), there exists
\(s_{n+1} \in [\theta s_{n}]_{\alpha _{\theta s_{n}}}\) such that
$$ \varpi (s_{n},s_{n+1})< g_{n}H\bigl([\theta s_{n-1}]_{\alpha _{\theta s_{n-1}}},[ \theta s_{n}]_{\alpha _{\theta s_{n}}}\bigr). $$
The above process continues, if at step
k satisfying
\(H([\theta s_{k-1}]_{\alpha _{\theta s_{k-1}}},[\theta s_{k}]_{ \alpha _{\theta s_{k}}})=0\), then
\(s_{k}\) is a FP of
θ. If not, we obtain two sequences
\(\{s_{n}\}\) and
\(\{g_{n}\}\) such that
\(s_{n} \in [\theta s_{n-1}]_{\alpha _{\theta s_{n-1}}}\),
\(g_{n}>1\) and
$$ \varpi (s_{n},s_{n+1})< g_{n}H \bigl([\theta s_{n-1}]_{\alpha _{\theta s_{n-1}}},[ \theta s_{n}]_{\alpha _{\theta s_{n}}} \bigr),\quad \forall n\geq 1. $$
(3.7)
Since
\(\frac{1}{\sigma +1}d(s_{n-1},[\theta s_{n-1}]_{\alpha _{\theta s_{n-1}}}) \leq \frac{1}{\sigma +1}d(s_{n-1},s_{n})\leq \varpi (s_{n-1},s_{n})\) and by hypothesis, we obtain
$$\begin{aligned} H\bigl([\theta s_{n-1}]_{\alpha _{\theta s_{n-1}}},[\theta s_{n}]_{\alpha _{ \theta s_{n}}}\bigr)&\leq x\bigl\{ d\bigl(s_{n-1},[ \theta s_{n-1}]_{\alpha _{\theta s_{n-1}}}\bigr)+d\bigl(s_{n},[ \theta s_{n}]_{\alpha _{\theta s_{n}}}\bigr)\bigr\} \\ &\leq x\bigl\{ d(s_{n-1},s_{n})+d(s_{n},s_{n+1}) \bigr\} . \end{aligned}$$
(3.8)
From (
3.7) and (
3.8), we have
$$ \varpi (s_{n},s_{n+1})< g_{n}x\bigl\{ \varpi (s_{n-1},s_{n})+ \varpi (s_{n},s_{n+1}) \bigr\} . $$
We can choose
\(g_{n}=\frac{k}{x}>1\) with
\(x \in (0,k)\) and
\(0< k<\frac{1}{2}\). Then, we obtain
\(\varpi _{n}<\frac{k}{1-k} \varpi _{n-1}\), where
\(\frac{k}{1-k}<1\) and
\(\varpi _{n}= \varpi (s_{n},s_{n+1})\). Thus,
\(\varpi _{n}<(\frac{k}{1-k})^{n} \varpi _{0}\) for all
\(n\geq 1\). Hence,
$$ \sum ^{\infty}_{n=1} \varpi _{n}\leq \varpi _{0}\sum ^{\infty}_{n=1}\biggl( \frac{k}{1-k} \biggr)^{n}< +\infty . $$
By Proposition
2.5,
\(\{s_{n}\}\) is a CS in
S. Since
S is complete, ∃
\(r \in S\) such that
\(\lim_{n\rightarrow \infty}s_{n}=r\). Now, we show that for any
\(n\geq 0\), either
$$ \frac{1}{\sigma +1} \varpi \bigl(s_{n},[\theta s_{n}]_{\alpha _{\theta s_{n}}}\bigr) \leq \varpi (s_{n},r)\quad \text{or} \quad \frac{1}{\sigma +1} \varpi \bigl(s_{n+1},[ \theta s_{n+1}]_{\alpha _{\theta s_{n+1}}}\bigr)\leq \varpi (s_{n+1},r). $$
(3.9)
Arguing by contradiction, we suppose that for some
\(n\geq 0\),
$$ \varpi (s_{n},r)< \frac{1}{\sigma +1}d\bigl(s_{n},[\theta s_{n}]_{\alpha _{ \theta s_{n}}}\bigr)\quad \text{or} \quad \varpi (s_{n+1},r)< \frac{1}{\sigma +1}d\bigl(s_{n+1},[\theta s_{n+1}]_{\alpha _{\theta s_{n+1}}}\bigr). $$
Then, by the triangular inequality, we obtain
$$\begin{aligned} \varpi _{n}&= \varpi (s_{n},s_{n+1})\leq \varpi (s_{n},r)+ \sigma \varpi (s_{n+1},r) \\ & < \frac{1}{\sigma +1}d\bigl(s_{n},[\theta s_{n}]_{\alpha _{\theta s_{n}}} \bigr)+ \frac{\sigma }{\sigma +1}d\bigl(s_{n+1},[\theta s_{n+1}]_{\alpha _{\theta s_{n+1}}} \bigr) \\ &\leq \frac{1}{\sigma +1} \varpi (s_{n},s_{n+1})+ \frac{\sigma }{\sigma +1} \varpi (s_{n+1},s_{n+2}) \\ &\leq \varpi _{n}. \end{aligned}$$
This is a contradiction. Hence, by hypothesis for each
\(n\geq 0\) and from (
3.9), either
$$ H\bigl([\theta s_{n}]_{\alpha _{\theta s_{n}}},[\theta r]_{\alpha _{ \theta r}}\bigr)\leq x\bigl\{ d\bigl(s_{n},[\theta s_{n}]_{\alpha _{\theta s_{n}}}\bigr)+d\bigl(r,[ \theta r]_{\alpha _{\theta r}}\bigr) \bigr\} , $$
(3.10)
or
$$ H\bigl([\theta s_{n+1}]_{\alpha _{\theta s_{n+1}}},[\theta r]_{\alpha _{ \theta r}}\bigr)\leq x\bigl\{ d\bigl(s_{n+1},[\theta s_{n+1}]_{\alpha _{\theta s_{n+1}}}\bigr)+d\bigl(r,[ \theta r]_{\alpha _{\theta r}}\bigr) \bigr\} . $$
(3.11)
Then, either (
3.10) holds for infinity natural numbers
n or (
3.11) holds for infinity natural numbers
n. Suppose (
3.10) holds for infinity natural numbers
n. We can choose that in that infinity set the sequence
\(\{n_{k}\}\) is a monotone strictly increasing sequence of natural numbers. Therefore, sequence
\(\{s_{n_{k}}\}\) is a subsequence of
\(\{s_{n}\}\) and
$$\begin{aligned} d\bigl(r,[\theta r]_{\alpha _{\theta r}}\bigr)&\leq d\bigl([\theta s_{n_{k}}]_{ \alpha _{\theta s_{n_{k}}}},r\bigr)+\sigma H\bigl([\theta s_{n_{k}}]_{\alpha _{ \theta s_{n_{k}}}},[\theta r]_{\alpha _{\theta r}}\bigr) \\ &\leq \varpi (s_{n_{k}+1},r)+Kx\bigl\{ d\bigl(s_{n_{k}+1},[\theta s_{n_{k}+1}]_{ \alpha _{\theta s_{n_{k}+1}}}\bigr)+d\bigl(r,[\theta r]_{\alpha _{\theta r}}\bigr) \bigr\} , \end{aligned}$$
which is equivalent to
$$ d\bigl(r,[\theta r]_{\alpha _{\theta r}}\bigr)\leq \frac{1+\sigma x}{1-\sigma x} \varpi (s_{n_{k}+1},r)+ \frac{\sigma ^{2}x}{1-\sigma x} \varpi (s_{n_{k}+2},r). $$
By taking limits on both sides of the above inequality, we obtain
\(d(r,[\theta r]_{\alpha _{\theta r}})=0\). This means that
\(r \in [\theta r]_{\alpha _{\theta r}}\). If (
3.11) holds for infinity natural numbers
n, by using an argument similar to that of above we have
r is a FP of
θ. Suppose
r̄ is another FP of
θ, then
\(0=\frac{1}{\sigma +1}d(r,[\theta r]_{\alpha _{\theta r}})\leq \varpi (r,\bar{r})\) and by hypothesis,
$$\begin{aligned} H\bigl([\theta r]_{\alpha _{\theta r}},[\theta \bar{r}]_{\alpha _{\theta \bar{r}}}\bigr)&\leq x \bigl\{ d\bigl(r,[\theta r]_{\alpha _{\theta r}}\bigr)+d\bigl(\bar{r},[ \theta \bar{r}]_{\alpha _{\theta \bar{r}}}\bigr)\bigr\} \\ &\leq x\bigl\{ d(r,r)+d(\bar{r},\bar{r}\bigr\} =0 \end{aligned}$$
and so
\(H([\theta r]_{\alpha _{\theta r}},[\theta \bar{r}]_{\alpha _{\theta \bar{r}}})=0\) implies
\([\theta r]_{\alpha _{\theta r}}=[\theta \bar{r}]_{\alpha _{\theta \bar{r}}}\) means
\(r=\bar{r}\). Hence,
θ has a unique FP
\(r \in S\). □