In this section we present our main results. We concentrate on existence and uniqueness of fixed points for a general class of *α*-admissible contractive mappings.

As usual, we take ξ_{0} ∈ *X* such that *α* (*Τ* ξ_{0}, ξ_{0}) ≥ 1 and α (ξ_{0}, *T* ξ_{0}) ≥ 1 and construct the sequence {ξ_{n}} as
$${\xi}_{n+1}=T{\xi}_{n}\text{\hspace{0.17em}for\hspace{0.17em}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}n\in \mathbb{N}.$$

Notice that if for some *n*_{0} ≥ 0 we have ${{\xi}_{n}}_{0}={{\xi}_{n}}_{0+1}$
then the proof is done, i.e., ${{\xi}_{n}}_{0}$
is a fixed point of *Τ*. Assume that ξ_{n} ≠ξ_{n}_{+}_{1}for all *n* ≥ 0. Because of the lack of symmetry condition in quasi-b-metric-like spaces, we will show that the sequence {ξ_{n}} is both left- and right-Cauchy.

From the conditions (*i*) and (*ii*), we have
$$\alpha ({\xi}_{1},{\xi}_{0})=\alpha (T{\xi}_{0},{\xi}_{0})\ge 1\Rightarrow \alpha (T{\xi}_{1},T{\xi}_{0})=\alpha ({\xi}_{2},{\xi}_{1})\ge {1}_{},$$(6)

and
$$\alpha ({\xi}_{0},{\xi}_{1})=\alpha ({\xi}_{0},T{\xi}_{0})\ge 1\Rightarrow \alpha (T{\xi}_{0},T{\xi}_{1})=\alpha ({\xi}_{1},{\xi}_{2})\ge {1}_{},$$(7)

or, in general
$$\alpha ({\xi}_{n+1},{\xi}_{n})\ge 1\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\alpha ({\xi}_{n},{\xi}_{n+1})\ge 1\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{\forall}n\in \mathbb{N}.$$(8)

Regarding (8), the contractive condition (3) with ξ* =* ξ_{n+}_{1}and *n =* ξ_{n} becomes
$${\rho}_{b}({\xi}_{n+1},{\xi}_{n})={\rho}_{b}(T{\xi}_{n},T{\xi}_{n-1})\le \alpha ({\xi}_{n},{\xi}_{n-1}){\rho}_{b}(T{\xi}_{n},T{\xi}_{n-1})\le \phantom{\rule{thinmathspace}{0ex}}{\phi}_{b}(M({\xi}_{n},{\xi}_{n-1}){)}_{},$$(9)

where
$$\begin{array}{ll}M({\xi}_{n},{\xi}_{n-1})& =max\{{\rho}_{b}({\xi}_{n},{\xi}_{n-1}{)}_{},{\rho}_{b}(T{\xi}_{n},{\xi}_{n}{)}_{},{\rho}_{b}(T{\xi}_{n-1},{\xi}_{n-1}),\\ & \frac{1}{4s}[{\rho}_{b}(T{\xi}_{n},{\xi}_{n-1})+{\rho}_{b}(T{\xi}_{n-1},{\xi}_{n})]\}\\ & =max\{{\rho}_{b}({\xi}_{n},{\xi}_{n-1}{)}_{},{\rho}_{b}({\xi}_{n+1},{\xi}_{n}{)}_{},{\rho}_{b}({\xi}_{n},{\xi}_{n-1}{)}_{},\\ & \frac{1}{4s}[{\rho}_{b}({\xi}_{n+1},{\xi}_{n-1})+{\rho}_{b}({\xi}_{n},{\xi}_{n})]\}\end{array}$$

Observe that for the last term in *Μ*(ξ_{n}, ξ_{n-1}), by the triangle inequality we have
$$\begin{array}{l}\frac{1}{4s}[{\rho}_{b}({\xi}_{n+1},{\xi}_{n-1})+{\rho}_{b}({\xi}_{n},{\xi}_{n})]\\ \le \frac{1}{4}[{\rho}_{b}({\xi}_{n+1},{\xi}_{n})+{\rho}_{b}({\xi}_{n-1},{\xi}_{n})+{\rho}_{b}({\xi}_{n},{\xi}_{n+1}){\rho}_{b}({\xi}_{n+1},{\xi}_{n})]\\ \frac{1}{4}[{\rho}_{b}({\xi}_{n+1},{\xi}_{n})+{\rho}_{b}({\xi}_{n},{\xi}_{n-1})+{\rho}_{b}({\xi}_{n},{\xi}_{n+1})+{\rho}_{b}({\xi}_{n+1},{\xi}_{n})]\\ =\frac{1}{4}[2{\rho}_{b}({\xi}_{n+1},{\xi}_{n})+{\rho}_{b}({\xi}_{n},{\xi}_{n-1})+{\rho}_{b}({\xi}_{n},{\xi}_{n+1})]\\ \le max\{{\rho}_{b}({\xi}_{n+1},{\xi}_{n}{)}_{},{\rho}_{b}({\xi}_{n},{\xi}_{n-1}{)}_{},{\rho}_{b}({\xi}_{n},{\xi}_{n+1})\},\end{array}$$(10)

and hence, either *Μ*(ξ_{n}, ξ_{n-1})* =* max{*ρ*_{b}(ξ_{n+1},ξ_{n})*, ρ*_{b}(ξ_{n}, ξ_{n-1})}or *Μ*(ξ_{n}, ξ_{n-1})≤* ρ*_{b}(ξ_{n}, ξ_{n+1}). Now, we will examine all three cases.

Case 1. Suppose that *Μ*(ξ_{n}, ξ_{n-1})* = ρ*_{b}(ξ_{n+1},ξ_{n})for some *n* ≥ 1. Since *ρ*_{b}(ξ_{n+1},ξ_{n})* >* 0, from (9), we have
$${\rho}_{b}({\xi}_{n+1},{\xi}_{n})\le {\phi}_{b}(M({\xi}_{n},{\xi}_{n-1}))={\phi}_{b}({\rho}_{b}({\xi}_{n+1},{\xi}_{n}))<{\rho}_{b}({\xi}_{n+1},{\xi}_{n})$$

which is a contradiction. Hence, for all *n* ≥ 1 either *M*(ξ_{n}, ξ_{n-1})* = ρ*_{b}(ξ_{n}, ξ_{n-1}) or *M*(ξ_{n}, ξ_{n-1}) ≤ *ρ*_{b}(ξ_{n}, ξ_{n+1}).

Case 2. Suppose that *Μ*(ξ_{n}, ξ_{n-}_{1})* = ρ*_{b}(ξ_{n}, ξ_{n-}_{1}) for some *n* ≥ 1. Regarding the properties of *φ*_{b} ∈ Φ_{b}, and (9), we get
$${\rho}_{b}({\xi}_{n+1},{\xi}_{n})\le {\phi}_{b}(M({\xi}_{n},{\xi}_{n-1}))\le {\phi}_{b}({\rho}_{b}({\xi}_{n},{\xi}_{n-1}))<{\rho}_{b}({\xi}_{n},{\xi}_{n-1})$$

for all n≥ 1. Inductively, we obtain
$${\rho}_{b}({\xi}_{n+1},{\xi}_{n})\le {\phi}_{b}^{n}({\rho}_{b}({\xi}_{1},{\xi}_{0}))\phantom{\rule{0.056em}{0ex}},\mathrm{f}\mathrm{o}\mathrm{r}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{0.056em}{0ex}}\mathrm{a}\mathrm{l}\mathrm{l}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{0.056em}{0ex}}\mathrm{n}\ge 1.$$(11)

Applying repeatedly triangle inequality (*QBML*_{2})and regarding (11), for all *k* ≥ 1, we get
$$\begin{array}{ll}{\rho}_{b}({{\xi}_{n}}_{+k},{\xi}_{n})& \le s[{\rho}_{b}({{\xi}_{n}}_{+k},{{\xi}_{n}}_{+1})+{\rho}_{b}({{\xi}_{n}}_{+1},{\xi}_{n})]\\ & \le {s}^{2}[{\rho}_{b}({{\xi}_{n}}_{+k},{{\xi}_{n}}_{+2})+{\rho}_{b}({{\xi}_{n}}_{+2},{{\xi}_{n}}_{+1})]+s{\rho}_{b}({{\xi}_{n}}_{+1},{\xi}_{n})\\ & \vdots \\ & \le {s}^{k}{\rho}_{b}({{\xi}_{n}}_{+k},{\xi}_{n+k-1})+{s}^{k-1}{\rho}_{b}({{\xi}_{n}}_{+k-1},{{\xi}_{n}}_{+k-2})\\ & +\dots +{s}^{2}{\rho}_{b}({\xi}_{n+2},{\xi}_{n+1})+s{\rho}_{b}({\xi}_{n+1},{\xi}_{n})\\ & \le {s}^{k}{\phi}_{b}^{n+k-1}({\rho}_{b}({\xi}_{1},{\xi}_{0}))+{s}^{k-1}{\phi}_{b}^{n+k-2}({\rho}_{b}({\xi}_{1},{\xi}_{0}))\\ & +\dots +s{\phi}_{b}^{n}({\rho}_{b}({\xi}_{1},{\xi}_{0}))\\ & =\frac{1}{{s}^{n-1}}\left[{s}^{n+k-1}{\phi}_{b}^{n+k-1}({\rho}_{b}({\xi}_{1},{\xi}_{0}))\right.\\ & \left.+{s}^{n+k-2}{\phi}_{b}^{n+k-2}({\rho}_{b}({\xi}_{1},{\xi}_{0}))+\dots +{s}^{n}{\phi}_{b}^{n}({\rho}_{b}({\xi}_{1},{\xi}_{0}))\right]\end{array}$$(12)

Define
$${S}_{n}=\sum _{p=0}^{n}{s}^{p}{\phi}_{b}^{p}({\rho}_{b}({\xi}_{1},{\xi}_{0}))\phantom{\rule{thinmathspace}{0ex}}\text{for}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}n\ge 1.$$(13)

We obtain
$${\rho}_{b}({\xi}_{n+k},{\xi}_{n})\le \frac{1}{{s}^{n-1}}[{S}_{n+k-1}-{S}_{n-1}{]}_{},n\ge {1}_{},k\ge 1.$$(14)

Due to the assumption ξ_{n} ≠ξ_{n}_{+1} for all *n* ∈ ℕ and Lemma 1.13, we conclude that the series ${\sum}_{p=0}^{\mathrm{\infty}}{s}^{p}{\phi}_{b}^{p}({\rho}_{b}({\xi}_{1},{\xi}_{0}))$ is convergent to some *S* ≥ 0. Thus, $\underset{n\to \mathrm{\infty}}{lim}{\rho}_{b}({\xi}_{n+k},{\xi}_{n})=0$ or, in other words, for *m > n*
$$\underset{m,n\to \mathrm{\infty}}{lim}{\rho}_{b}({\xi}_{m},{\xi}_{n})=0.$$(15)

Case 3. Suppose that *Μ*(ξ_{n}, ξ_{n}_{-1})≤ *ρ*_{b}(ξ_{n}, ξ_{n}_{+1})for some *n* ≥ 1. Using the fact that *φ*_{b} ∈Φ_{b} and the inequality (9), we get
$${\rho}_{b}({\xi}_{n+1},{\xi}_{n})\le {\phi}_{b}(M({\xi}_{n},{\xi}_{n-1}))\le {\phi}_{b}({\rho}_{b}({\xi}_{n},{\xi}_{n+1}))<{\rho}_{b}({\xi}_{n},{\xi}_{n+1})$$(16)

for all *n* ≥ 1. On the other hand, if we put ξ* =* ξ_{n} and *η =* ξ_{n}_{+1} in (3), taking into account (8), we find
$$\begin{array}{ll}{\rho}_{b}({\xi}_{n},{\xi}_{n+1})& ={\rho}_{b}(T{\xi}_{n-1},T{\xi}_{n})\\ & \le \alpha ({\xi}_{n-1},{\xi}_{n}){\rho}_{b}(T{\xi}_{n-1},T{\xi}_{n})\le {\phi}_{b}(M({\xi}_{n-1},{\xi}_{n})),\end{array}$$(17)

where
$$\begin{array}{ll}M({\xi}_{n-1},{\xi}_{n})& =max\left\{{\rho}_{b}({\xi}_{n-1},{\xi}_{n}),{\rho}_{b}(T{\xi}_{n-1},{\xi}_{n-1}),{\rho}_{b}(T{\xi}_{n},{\xi}_{n}),\right.\\ & \left.\frac{1}{4s}[{\rho}_{b}(T{\xi}_{n-1},{\xi}_{n})+{\rho}_{b}(T{\xi}_{n},{\xi}_{n-1})]\right\}\\ & =max\left\{{\rho}_{b}({\xi}_{n-1},{\xi}_{n}),{\rho}_{b}({\xi}_{n},{\xi}_{n-1}),{\rho}_{b}({\xi}_{n+1},{\xi}_{n}),\right.\\ & \left.\frac{1}{4s}[{\rho}_{b}({\xi}_{n},{\xi}_{n})+{\rho}_{b}({\xi}_{n+1},{\xi}_{n-1})]\right\}\end{array}$$

for all *n* ≥ 1. Observe that, applying triangle inequality (*QBML*_{2})to the last term in *Μ*(ξ_{n}_{-1}, ξ_{n})we have
$$\begin{array}{l}\frac{1}{4s}[{\rho}_{b}({\xi}_{n},{\xi}_{n})+{\rho}_{b}({\xi}_{n+1},{\xi}_{n-1})]\\ \le \frac{1}{4s}[{\rho}_{b}({\xi}_{n},{\xi}_{n+1})+{\rho}_{b}({\xi}_{n+1},{\xi}_{n})+{\rho}_{b}({\xi}_{n+1},{\xi}_{n})+{\rho}_{b}({\xi}_{n},{\xi}_{n-1})]\\ =\frac{1}{4}[{\rho}_{b}({\xi}_{n},{\xi}_{n+1})+2{\rho}_{b}({\xi}_{n+1},{\xi}_{n})+{\rho}_{b}({\xi}_{n},{\xi}_{n-1})]\\ \le max\{{\rho}_{b}({\xi}_{n},{\xi}_{n+1}),{\rho}_{b}({\xi}_{n+1},{\xi}_{n}),{\rho}_{b}({\xi}_{n},{\xi}_{n-1})\}={\rho}_{b}({\xi}_{n},{\xi}_{n+1}),\end{array}$$(18)

since we are in Case 3. On the other hand, from (16) we see that
$${\rho}_{b}({\xi}_{n},{\xi}_{n-1})\le {\rho}_{b}({\xi}_{n-1},{\xi}_{n}{)}_{},\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}{\rho}_{b}({\xi}_{n+1},{\xi}_{n})\le {\rho}_{b}({\xi}_{n},{\xi}_{n+1}).$$

Therefore, either *M*(ξ_{n}_{-1}, ξ_{n}) ≤ *ρ*_{b}(ξ_{n}, ξ_{n+1})or *M*(ξ_{n}_{-1}, ξ_{n}) = *ρ*_{b}(ξ_{n}_{-1}, ξ_{n}).

If *Μ*(ξ_{n}_{-1}, ξ_{n}) ≤* ρ*_{b}(ξ_{n}, ξ_{n+1})for some *n* ∈ ℕ, *since ρ*_{b}(ξ_{n}_{+1}, ξ_{n})> 0, the inequality (17) implies
$${\rho}_{b}({\xi}_{n},{\xi}_{n+1})\le {\phi}_{b}(M({\xi}_{n-1},{\xi}_{n}))\le {\phi}_{b}({\rho}_{b}({\xi}_{n},{\xi}_{n+1}))<{\rho}_{b}({\xi}_{n},{\xi}_{n+1}{)}_{},$$

which is a contradiction.

Thus, we should have *M*(ξ_{n-1}, ξ_{n})* = ρ*_{b} (ξ_{n-1}, ξ_{n}) for all *n* ≥ 1. The inequality (17) becomes
$${\rho}_{b}({\xi}_{n},{\xi}_{n+1})\le {\phi}_{b}(M({\xi}_{n-1},{\xi}_{n}))\le {\phi}_{b}({\rho}_{b}({\xi}_{n-1},{\xi}_{n}))<{\rho}_{b}({\xi}_{n-1},{\xi}_{n})$$

for all *n* ≥ 1, using the fact that *φ*_{b} ∈Φ_{b}.

By induction, we get
$${\rho}_{b}({\xi}_{n},{\xi}_{n+1})\le {\phi}_{b}^{n}({\rho}_{b}({\xi}_{0},{\xi}_{1})),\phantom{\rule{thinmathspace}{0ex}}\text{for all}\phantom{\rule{thinmathspace}{0ex}}n\ge 1.$$(19)

Hence, combining (16) and (19) we deduce
$${\rho}_{b}({\xi}_{n+1},{\xi}_{n})\le {\phi}_{b}^{n}({\rho}_{b}({\xi}_{0},{\xi}_{1})),\phantom{\rule{thinmathspace}{0ex}}\text{for all}\phantom{\rule{thinmathspace}{0ex}}n\ge 1.$$(20)

Due to *(QBML*_{2}), together with (20), for all *k* ≤ 1, we get
$$\begin{array}{ll}{\rho}_{b}({{\xi}_{n}}_{+k},{\xi}_{n})& \le s[{\rho}_{b}({{\xi}_{n}}_{+k},{\xi}_{n+1})+{\rho}_{b}({\xi}_{n+1},{\xi}_{n})]\\ & \le {s}^{2}[{\rho}_{b}({{\xi}_{n}}_{+k},{\xi}_{n+2})+{\rho}_{b}({\xi}_{n+2},{\xi}_{n+1})]+s{\rho}_{b}({\xi}_{n+1},{\xi}_{n})\\ & \vdots \\ & \le {s}^{k}{\rho}_{b}({{\xi}_{n}}_{+k},{\xi}_{n+k-1})+{s}^{k-1}{\rho}_{b}({{\xi}_{n}}_{+k-1},{{\xi}_{n}}_{+k-2})\\ & +\dots +{s}^{2}{\rho}_{b}({\xi}_{n+2},{\xi}_{n+1})+s{\rho}_{b}({\xi}_{n+1},{\xi}_{n})\\ & \le {s}^{k}{\phi}_{b}^{n+k-1}({\rho}_{b}({\xi}_{0},{\xi}_{1}))+{s}^{k-1}{\phi}_{b}^{n+k-2}({\rho}_{b}({\xi}_{0},{\xi}_{1}))\\ & +\dots +s{\phi}_{b}^{n}({\rho}_{b}({\xi}_{0},{\xi}_{1}))\\ & =\frac{1}{{s}^{n-1}}\left[{s}^{n+k-1}{\phi}_{b}^{n+k-1}({\rho}_{b}({\xi}_{0},{\xi}_{1}))\right.\\ & \left.+{s}^{n+k-2}{\phi}_{b}^{n+k-2}({\rho}_{b}({\xi}_{0},{\xi}_{1}))+\dots +{s}^{n}{\phi}_{b}^{n}({\rho}_{b}({\xi}_{0},{\xi}_{1}))\right]\end{array}$$

Define
$${\mathcal{Q}}_{n}=\sum _{p=0}^{n}{s}^{p}{\phi}_{b}^{p}({\rho}_{b}({\xi}_{0},{\xi}_{1}))\phantom{\rule{thinmathspace}{0ex}}\text{for}\phantom{\rule{thinmathspace}{0ex}}n\ge 1.$$(21)

Then, employing (21), we get
$${\rho}_{b}({\xi}_{n+k},{\xi}_{n})\le \frac{1}{{s}^{n-1}}[{\mathcal{Q}}_{n+k-1}-{\mathcal{Q}}_{n-1}{]}_{},n\ge {1}_{},k\ge 1.$$

Using the fact that the series $\sum _{p=0}^{\mathrm{\infty}}{s}^{p}{\phi}_{b}^{p}({\rho}_{b}({\xi}_{0},{\xi}_{1}))$ converges to some 𝒬 ≤ 0, we deduce, lim_{n}_{→∞} *ρ*_{b}(ξ_{n+k}, ξ_{n})* =* 0 which means that for *m > n*,
$$\underset{m,n\to \mathrm{\infty}}{lim}{\rho}_{b}({\xi}_{m},{\xi}_{n})=0.$$(22)

Therefore, in all possible cases, we conclude that the sequence {ξ_{n}}is a left Cauchy sequence.

To show that it is also right Cauchy, we proceed as follows.

Let ξ = ξ_{n} and *η =* ξ_{n}_{+1} in (3). Using (8), we get
$$\begin{array}{ll}{\rho}_{b}({\xi}_{n},{\xi}_{n+1})& ={\rho}_{b}(T{\xi}_{n-1},T{\xi}_{n})\\ & \le \alpha ({\xi}_{n-1},{\xi}_{n}){\rho}_{b}(T{\xi}_{n-1},T{\xi}_{n})\le {\phi}_{b}(M({\xi}_{n-1},{\xi}_{n}))\end{array}$$(23)

where
$$\begin{array}{ll}M({\xi}_{n-1},{\xi}_{n})& =max\{{\rho}_{b}({\xi}_{n-1},{\xi}_{n}),{\rho}_{b}(T{\xi}_{n-1},{\xi}_{n-1}),{\rho}_{b}(T{\xi}_{n},{\xi}_{n}),\\ & \left.\frac{1}{4s}[{\rho}_{b}(T{\xi}_{n-1},{\xi}_{n})+{\rho}_{b}(T{\xi}_{n},{\xi}_{n-1})]\right\}\\ & \le max\left\{{\rho}_{b}({\xi}_{n-1},{\xi}_{n}),{\rho}_{b}({\xi}_{n},{\xi}_{n-1}),{\rho}_{b}({\xi}_{n+1},{\xi}_{n}),\right.\\ & \left.\frac{1}{4s}[{\rho}_{b}({\xi}_{n},{\xi}_{n})+{\rho}_{b}({\xi}_{n+1},{\xi}_{n-1})]\right\}.\end{array}$$

Regarding the inequality (18), we have
$$\frac{1}{4s}[{\rho}_{b}({\xi}_{n},{\xi}_{n})+{\rho}_{b}({\xi}_{n+1},{\xi}_{n-1})]\le max\{{\rho}_{b}({\xi}_{n},{\xi}_{n+1}{)}_{},{\rho}_{b}({\xi}_{n+1},{\xi}_{n}{)}_{},{\rho}_{b}({\xi}_{n},{\xi}_{n-1}){\}}_{},$$

and hence, either
$$M({\displaystyle {\xi}_{n-1},{\xi}_{n})=max\{{\rho}_{b}({\xi}_{n-1},{\xi}_{n}{)}_{},{\rho}_{b}({\xi}_{n},{\xi}_{n-1}{)}_{},{\rho}_{b}({\xi}_{n+1},{\xi}_{n}){\}}_{},}$$(24)

or *Μ*(ξ_{n}_{-1}, ξ_{n})≤ *ρ*_{b}(ξ_{n}, ξ_{n}_{+1})for all *n* ≥1. We will discuss separately these four cases.

Case I. Suppose that *M*(ξ_{n}_{-1}, ξ_{n})≤ *ρ*_{b}(ξ_{n}, ξ_{n}_{+1}) for some *n* ≥ 1. Then the inequality (23) implies
$${\rho}_{b}({\xi}_{n},{\xi}_{n+1})\le {\phi}_{b}(M({\xi}_{n-1},{\xi}_{n}))\le {\phi}_{b}({\rho}_{b}({\xi}_{n+1},{\xi}_{n}))<{\rho}_{b}({\xi}_{n},{\xi}_{n+1})$$(25)

which is a contradiction because *ρ*_{b}(ξ_{n}, ξ_{n}_{+1}) > 0.

Case II. If for some *n* ≥ 1 we have *M*(ξ_{n}_{-1}, ξ_{n})* = ρ*_{b}(ξ_{n}_{+1}, ξ_{n}), then the inequality (23) becomes
$${\rho}_{b}({\xi}_{n},{\xi}_{n+1})\le {\phi}_{b}(M({\xi}_{n-1},{\xi}_{n}))\le {\phi}_{b}({\rho}_{b}({\xi}_{n+1},{\xi}_{n}))<{\rho}_{b}({\xi}_{n+1},{\xi}_{n}).$$(26)

Recalling (9) and (10), that is,
$${\rho}_{b}({\xi}_{n+1},{\xi}_{n})\le {\phi}_{b}(M({\xi}_{n},{\xi}_{n-1}))$$(27)

where
$$M({\xi}_{n},{\xi}_{n-1})\le max\{{\rho}_{b}({\xi}_{n},{\xi}_{n-1}{)}_{},{\rho}_{b}({\xi}_{n+1},{\xi}_{n}{)}_{},{\rho}_{b}({\xi}_{n},{\xi}_{n+1}){\}}_{},$$

and since by the assumption
$$max\{{\rho}_{b}({\xi}_{n-1},{\xi}_{n}{)}_{},{\rho}_{b}({\xi}_{n},{\xi}_{n-1}{)}_{},{\rho}_{b}({\xi}_{n+1},{\xi}_{n})\}={\rho}_{b}({\xi}_{n+1},{\xi}_{n})\ge {\rho}_{b}({\xi}_{n},{\xi}_{n+1}{)}_{},$$

then, inequalities (26) and (27) yield
$${\rho}_{b}({\xi}_{n},{\xi}_{n+1})\le {\rho}_{b}({\xi}_{n+1},{\xi}_{n})\le {\phi}_{b}(M({\xi}_{n},{\xi}_{n-1}))\le {\phi}_{b}({\rho}_{b}({\xi}_{n+1},{\xi}_{n}))<{\rho}_{b}({\xi}_{n+1},{\xi}_{n})$$

due to the properties of *φ*_{b}. This is clearly impossible since *ρ*_{b}(ξ_{n}_{+1}, ξ_{n}) * >* 0 and we end up with a contradiction.

Case III. Assume that *Μ*(ξ_{n-}_{1}, ξ_{n})* = ρ*_{b}(ξ_{n-}_{1}, ξ_{n})for some n ≥ 1. Since *ρ*_{b}(ξ_{n-}_{1}, ξ_{n})* >* 0, from (9), we get
$${\rho}_{b}({\xi}_{n},{\xi}_{n+1})\le {\phi}_{b}(M({\xi}_{n-1},{\xi}_{n}))\le {\phi}_{b}({\rho}_{b}({\xi}_{n-1},{\xi}_{n}))<{\rho}_{b}({\xi}_{n-1},{\xi}_{n})$$

for all *n* ≥ 1. Recursively, we derive
$${\rho}_{b}({\xi}_{n},{\xi}_{n+1})\le {\phi}_{b}^{n}({\rho}_{b}({\xi}_{0},{\xi}_{1})),\phantom{\rule{thinmathspace}{0ex}}\mathrm{\forall}n\ge 1.$$(28)

Applying repeatedly triangle inequality (*QBML*_{2})and regarding (28), we get for all *k >* 0,
$$\begin{array}{ll}{\rho}_{b}({\xi}_{n},{\xi}_{n}+k)& \le s[{\rho}_{b}({\xi}_{n},{\xi}_{n}+1)+{\rho}_{b}({{\xi}_{n}}_{+1},{\xi}_{n+k})]\\ & \vdots \\ & \le s{\phi}_{b}^{n}(\rho b({\xi}_{0},{\xi}_{1}))+{s}^{2}\phantom{\rule{thickmathspace}{0ex}}{\phi}_{b}^{n+1}({\rho}_{b}({\xi}_{0},{\xi}_{1}))+\\ & +\dots +{s}^{k-1}{\phi}_{b}^{n+k-2}({\rho}_{b}({\xi}_{0},{\xi}_{1}))+{s}^{k}{\phi}_{b}^{n+k-1}({\rho}_{b}({\xi}_{0},{\xi}_{1}))\\ & =\frac{1}{{s}^{n-1}}[{s}^{n}{\phi}_{b}^{n}({\rho}_{b}({\xi}_{0},{\xi}_{1}))+\dots +\\ & \left.+{s}^{n+k-2}{\phi}_{b}^{n+k-2}({\rho}_{b}({\xi}_{0},{\xi}_{1}))+{s}^{n+k-1}{\phi}_{b}^{n+k-1}({\rho}_{b}({\xi}_{0},{\xi}_{1}))\right]\end{array}$$(29)

Recalling (21), we see that
$${\rho}_{b}({\xi}_{n},{\xi}_{n+k})\le \frac{1}{{s}^{n-1}}[{\mathcal{Q}}_{n+k-1}-{\mathcal{Q}}_{n-1}{]}_{},n\ge {1}_{},k\ge 1.$$

As a result, we have, lim_{n}_{→∞} *ρ*_{b}(ξ_{n}, ξ_{n+k})= 0 or, in other words for *m > n*,
$$\underset{m,n\to \mathrm{\infty}}{lim}{\rho}_{b}({\xi}_{n},{\xi}_{m})=0.$$

Case IV. Suppose that *M*(ξ_{n-}_{1}, ξ_{n})* = ρ*_{b}(ξ_{n-} ξ_{n}_{-1})for some *n* ≥ 1. Then, (23) gives
$${\rho}_{b}({\xi}_{n},{\xi}_{n+1})\le {\phi}_{b}(M({\xi}_{n-1},{\xi}_{n}))\le {\phi}_{b}({\rho}_{b}({\xi}_{n},{\xi}_{n-1}))<{\rho}_{b}({\xi}_{n},{\xi}_{n-1})$$(30)

for all *n* ≥ 1. Now, rewriting (9) and (10) for *n —* 1 we have
$$\begin{array}{ll}{\rho}_{b}({\xi}_{n},{\xi}_{n-1})& ={\rho}_{b}(T{\xi}_{n-1},T{\xi}_{n-2})\\ & \le \alpha ({\xi}_{n-1},{\xi}_{n-2}){\rho}_{b}(T{\xi}_{n-1},T{\xi}_{n-2})\le {\phi}_{b}(M({\xi}_{n-1},{\xi}_{n-2}))\end{array}$$(31)

where
$$M({\xi}_{n-1},{\xi}_{n-2})\le max\{{\rho}_{b}({\xi}_{n},{\xi}_{n-1}{)}_{},{\rho}_{b}({\xi}_{n-1},{\xi}_{n-2}{)}_{},{\rho}_{b}({\xi}_{n-1},{\xi}_{n}){\}}_{},$$

where obviously, the maximum can be either *ρ*_{b}(ξ_{n}, ξ_{n}_{−1})or *ρ*_{b}(ξ_{n}_{−1}, ξ_{n}_{−2}).

The first possibility, that is, if *M*(ξ_{n}_{−1}, ξ_{n}_{−2})≤ *ρ*_{b}(ξ_{n}, ξ_{n}_{−1})for some *n* ≥ 1, results in
$${\rho}_{b}({\xi}_{n},{\xi}_{n-1})\le {\phi}_{b}(M({\xi}_{n-1},{\xi}_{n-2})\le {\phi}_{b}({\rho}_{b}({\xi}_{n},{\xi}_{n-1}))<{\rho}_{b}({\xi}_{n},{\xi}_{n-1}{)}_{},$$

which is a contradiction. Therefore, we should have *Μ*(ξ_{n}_{−1}, ξ_{n}_{−2})≤ *ρ*_{b}(ξ_{n}_{−1}, ξ_{n}_{−2}) for all *n* ≥ 1. Then the inequality (31) yields
$${\rho}_{b}({\xi}_{n},{\xi}_{n-1})\le {\phi}_{b}(M({\xi}_{n-1},{\xi}_{n-2})\le {\phi}_{b}({\rho}_{b}({\xi}_{n-1},{\xi}_{n-2}))<{\rho}_{b}({\xi}_{n-1},{\xi}_{n-2}),$$(32)

for all *n* ≥ 1. Thus, we deduce
$${\rho}_{b}({\xi}_{n},{\xi}_{n-1})\le {\phi}_{b}^{n-1}({\rho}_{b}({\xi}_{1},{\xi}_{0}))\phantom{\rule{thinmathspace}{0ex}}\text{for all}\phantom{\rule{thinmathspace}{0ex}}n\ge 1.$$(33)

If we combine the inequalities (30) with (33), we derive
$${\rho}_{b}({\xi}_{n},{\xi}_{n+1})<{\rho}_{b}({\xi}_{n},{\xi}_{n-1})\le {\phi}_{b}^{n-1}({\rho}_{b}({\xi}_{1},{\xi}_{0})),$$(34)

for all *n* ≥ 1. As done in Case 2, when applying triangle inequality repeatedly for every *k* ≥ 1, we get
$$\begin{array}{ll}\rho b({\xi}_{n},{\xi}_{n+k})& \le s[{\rho}_{b}({\xi}_{n},{\xi}_{n+1})+{\rho}_{b}({{\xi}_{n}}_{+1},{{\xi}_{n}}_{+k})]\\ & \le s{\rho}_{b}({\xi}_{n},{\xi}_{n+1})+{s}^{2}[\rho b({{\xi}_{n}}_{+1},{{\xi}_{n}}_{+2})+{\rho}_{b}({{\xi}_{n}}_{+2},{{\xi}_{n}}_{+k})]\\ & \vdots \\ & \le s{\rho}_{b}({\xi}_{n},{\xi}_{n+1})+{s}^{2}{\rho}_{b}({\xi}_{n+1},{\xi}_{n+2})+\\ & +\dots +{s}^{k-1}{\rho}_{b}({\xi}_{n+k-2},{\xi}_{n+k-1})+{s}^{k}{\rho}_{b}({\xi}_{n+k-1},{\xi}_{n+k})\\ & \le s{\phi}_{b}^{n-1}({\rho}_{b}({\xi}_{1},{\xi}_{0}))+{s}^{2}{\phi}_{b}^{n}({\rho}_{b}({\xi}_{1},{\xi}_{0}))+\\ & +\dots +{s}^{k-1}{\phi}_{b}^{n+k-3}({\rho}_{b}({\xi}_{1},{\xi}_{0}))+{s}^{k}{\phi}_{b}^{n+k-2}({\rho}_{b}({\xi}_{1},{\xi}_{0}))\\ & \le \frac{1}{{s}^{n-2}}\left[{s}^{n-1}{\phi}_{b}^{n-1}({\rho}_{b}({\xi}_{1},{\xi}_{0}))+\dots \right.\\ & \left.+{s}^{n+k-3}{\phi}_{b}^{n+k-3}({\rho}_{b}({\xi}_{1},{\xi}_{0}))+{s}^{n+k-2}{\phi}_{b}^{n+k-2}({\rho}_{b}({\xi}_{1},{\xi}_{0}))\right]\\ & \le \phantom{\rule{thinmathspace}{0ex}}\le \frac{1}{{s}^{n-2}}[{\mathcal{S}}_{n+k-2}-{\mathcal{S}}_{n-2}],n\ge 1,k\ge 1,\end{array}$$

where *S*_{n} is defined in (13). Therefore, we end up with lim_{n}_{→∞} *ρ*_{b}(ξ_{n}, ξ_{n+k})* =* 0, or, equivalently, for *m > n*,
$$\underset{m,n\to \mathrm{\infty}}{lim}{\rho}_{b}({\xi}_{n},{\xi}_{m})=0.$$

We conclude that {ξ_{n}} is a right-Cauchy sequence, and hence, a Cauchy sequence. Moreover, it is a 0-Cauchy sequence in 0-complete quasi *b*-metric like space (*X, ρ*_{b}). Thus, there exists *ρ*_{b} ∈ *X* such that for *m > n* we have
$$\begin{array}{ll}{lim}_{n\to \mathrm{\infty}}{\rho}_{b}({\xi}_{n},\xi )& ={lim}_{n\to \mathrm{\infty}}{\rho}_{b}({\xi}_{},{\xi}_{n})={\rho}_{b}({\xi}_{},\xi )=0\\ & ={lim}_{m,n\to \mathrm{\infty}}{\rho}_{b}({\xi}_{m},{\xi}_{n})={lim}_{m,n\to \mathrm{\infty}}{\rho}_{b}({\xi}_{n},{\xi}_{m}).\end{array}$$

By the continuity of *T*, we obtain
$$\xi =\underset{n\to \mathrm{\infty}}{lim}{\xi}_{n+1}=\underset{n\to \mathrm{\infty}}{lim}T{\xi}_{n}=T\underset{n\to \mathrm{\infty}}{lim}{\xi}_{n}=T{\xi}_{},$$

that is, ξis a fixed point of *Τ*.

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