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.

**Definition 2.1:** *Let* (*X, ρ*_{b}) *be a quasi-b-metric-like space with a constant s* ≥ 1 *and let α* : *Χ* × *X* → [0, ∞*) and**φ*_{b} ∈ Φ_{b} *be two functions*.(i) *An α — φ*_{b} contractive mapping Τ : *X* → *X is of type* (*A*)* if*
$$\alpha (\xi ,\eta ){\rho}_{b}(T\xi ,T\eta )\le {\phi}_{b}(M(\xi ,\eta )),\mathit{f}\mathit{o}\mathit{r}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathit{a}\mathit{l}\mathit{l}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\xi ,\eta \in X,$$(3)*where*
$$M(\xi ,\eta )=max\{{\rho}_{b}(\xi ,\eta ),{\rho}_{b}(T\xi ,\xi ),{\rho}_{b}(T\eta ,\eta ),\frac{1}{4s}[{\rho}_{b}(T\xi ,\eta )+{\rho}_{b}(T\eta ,\xi )]\}.$$(ii) *An α — φ*_{b} contractive mapping Τ : *X* → *X is of type* (*B*)* if*
$$\alpha (\xi ,\eta ){\rho}_{b}(T\xi ,T\eta )\le {\phi}_{b}(N(\xi ,\eta )),forall\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\xi ,\eta \in X,$$(4)*where*
$$N(\xi ,\eta )=max\{{\rho}_{b}(\xi ,\eta ),\frac{1}{2s}[{\rho}_{b}(T\xi ,\xi )+{\rho}_{b}(T\eta ,\eta )],\frac{1}{4s}[{\rho}_{b}(T\xi ,\eta )+{\rho}_{b}(T\eta ,\xi )]\}.$$(iii) *An α —φ*_{b} contractive mapping Τ : X →* X is of type* (*C*)* if*
$$\alpha (\xi ,\eta ){\rho}_{b}(T\xi ,T\eta )\le {\phi}_{b}(K(\xi ,\eta )),for\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}all\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\xi ,\eta \in X,$$(5)*where*
$$K(\xi ,\eta )=max\{{\rho}_{b}(\xi ,\eta ),{\rho}_{b}(T\xi ,\xi ),{\rho}_{b}(T\eta ,\eta )\}$$

**Theorem 2.3:** *Let (X, ρ*_{b}) *be a 0-complete quasi-b-metric-like space with a constant s* ≥ 1. *Suppose that Τ* : *X* → *X is an α — φ*_{b} contractive mapping of type (*A*)* satisfying the following*:(i) *T is α-admissible;*(ii) *there exists* ξ_{0} ∈ *X such that α*(*Τ* ξ_{0},ξ_{0})≥ 1 *and α*(ξ_{0}*, Τ* ξ_{0})≥ 1;(iii) *T is continuous;**Then Τ has a fixed point*.

**Proof:** *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*.

**Theorem 2.4:** *Adding the condition*(*U*) *For every pair ξ and η of fixed points of T, α*(*ξ, η*)≥ 1 *and* α(*η*, ξ*)* ≥ 1.*to the* *Statement of Theorem 2.3 we obtain the uniqueness of the fixed point*.

**Proof:** *By Theorem 2.3 we know that the mapping **Τ* has at least one fixed point. To show the uniqueness, we assume that *η* is another fixed point of *T*, such that *ξ* ≠* η*. From the condition (*U*) we have, *α*(*ξ, η*)≥ 1, hence, the contractive condition for the fixed points *ξ* and *η*, that is,
$${\rho}_{b}({\xi}_{},\eta )=\rho {}_{(}T{\xi}_{},T\eta )\le \alpha ({\xi}_{},\eta ){\rho}_{b}(T{\xi}_{},T\eta )\le {\phi}_{b}(M({\xi}_{},\eta )),$$where,
$$M({\xi}_{},\eta )=max\{{\rho}_{b}({\xi}_{},\eta {)}_{},{\rho}_{b}(T{\xi}_{},\xi {)}_{},{\rho}_{b}(T{\eta}_{},\eta {)}_{},\frac{1}{4s}[{\rho}_{b}(T{\xi}_{},\eta )+{\rho}_{b}(T{\eta}_{},\xi )]\}={\rho}_{b}({\xi}_{},\eta ).$$This yields
$${\rho}_{b}({\xi}_{},\eta )\le {\phi}_{b}({\rho}_{b}({\xi}_{},\eta ))<{\rho}_{b}({\xi}_{},\eta {)}_{},$$which implies *ρ*_{b}(ξ_{n}, η) * =* 0. In a similar way, by changing the roles of *ξ* and *η* in the contractive condition we get
$${\rho}_{b}({\eta}_{},\xi )=\rho {}_{(}T{\eta}_{},T\xi )\le \alpha ({\eta}_{},\xi ){\rho}_{b}(T{\eta}_{},T\xi )\le {\phi}_{b}(M({\eta}_{},\xi )),$$where,
$$M({\eta}_{},\xi )=max\{{\rho}_{b}({\eta}_{},\xi {)}_{},{\rho}_{b}(T{\eta}_{},\eta {)}_{},{\rho}_{b}(T{\xi}_{},\xi {)}_{},\frac{1}{4s}[{\rho}_{b}(T{\eta}_{},\xi )+{\rho}_{b}(T{\xi}_{},\eta )]\}={\rho}_{b}({\eta}_{},\xi {)}_{},$$and thus,
$${\rho}_{b}({\eta}_{},\xi )\le {\phi}_{b}({\rho}_{b}({\eta}_{},\xi ))<{\rho}_{b}({\eta}_{},\xi ).$$Therefore, we deduce*ρ*_{b}**(***η, ξ*)* =* 0 and hence, *ξ = η*. This completes the uniqueness proof of the Theorem.Regarding Remark 2.2, we can state the following result as an immediate consequence.

**Corollary 2.5:** *Let(X, ρ*_{b}) be a 0-complete quasi-b-metric-like space with a constant s ≥ 1. *Suppose that T* : *X* → *X is an α — φ*_{b} contractive mapping of type (B) or type (C) satisfying the following:(i) *Τ is α-admissible;*(ii) *there exists ξ*_{0} ∈ *X such that α*(*Τ ξ*_{0}, ξ_{0})≥ 1 *and α*(*ξ*_{0}*, Τξ*_{0})≥ 1;(iii) *T is continuous;*(iv) *For every pair ξ and η of fixed points of T, α*(*ξ, η*)≥ 1 *and α*(*η, ξ*)≥ 1. *Then Τ has a unique fixed point*.We also state another consequence of the main result obtained by taking the function *α(x, y*) = 1.

**Corollary 2.6:** *Let* (*X, ρ*_{b})* be a 0-complete quasi-b-metric-like space with a constant s* ≥ 1. *Suppose that Τ* : *X* → *X is a continuous mapping satisfying*
$${\rho}_{b}(T{\xi}_{},T\eta )\le {\phi}_{b}(M({\xi}_{},\eta )),\phantom{\rule{thinmathspace}{0ex}}for\phantom{\rule{thinmathspace}{0ex}}all\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\xi}_{},\eta \in {X}_{},$$(35)*where*
$$M(\xi ,\eta )=max\{{\rho}_{b}(\xi ,\eta ),{\rho}_{b}(T\xi ,\xi ),{\rho}_{b}(T\eta ,\eta ),\frac{1}{4s}[{\rho}_{b}(T\xi ,\eta )+{\rho}_{b}(T\eta ,\xi )]\}.$$*Then Τ has a unique fixed point*.

**Corollary 2.7:** *Let* (*X, ρ*_{b})* be a 0-complete quasi-b-metric-like space with a constant s* ≥ 1. *Suppose that T* : *X* → *X is a continuous mapping satisfying*
$${\rho}_{b}(T{\xi}_{},T\eta )\le {\phi}_{b}(N({\xi}_{},\eta ){)}_{},\phantom{\rule{thinmathspace}{0ex}}for\phantom{\rule{thinmathspace}{0ex}}all\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\xi}_{},\eta \in X$$(36)*where*
$$N({\xi}_{},\eta )=max\{{\rho}_{b}({\xi}_{},\eta {)}_{},\frac{1}{2s}[{\rho}_{b}(T{\xi}_{},\xi )+{\rho}_{b}(T{\eta}_{},\eta )],\frac{1}{4s}[{\rho}_{b}(T{\xi}_{},\eta )+{\rho}_{b}(T{\eta}_{},\xi )]\}.$$*or*
$${\rho}_{b}(T{\xi}_{},T\eta )\le {\phi}_{b}(K({\xi}_{},\eta )),\phantom{\rule{thinmathspace}{0ex}}for\phantom{\rule{thinmathspace}{0ex}}all\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\xi}_{},\eta \in {X}_{},$$(37)*where*
$$K(\xi ,\eta )=max\{{\rho}_{b}(\xi ,\eta ),{\rho}_{b}(T\xi ,\xi ),{\rho}_{b}(T\eta ,\eta )\}.$$*Then Τ has a unique fixed point*.We also point out that the main result of Gülyaz [8], that is the Theorem 1.16, follows from our main Theorem 2.3. In addition, by taking *α*(*x, y*)* =* 1, we deduce another corollary.

**Corollary 2.8:** *Let* (*X, ρ*_{b}) *be a 0-complete quasi-b-metric-like space with a constant s* ≥ 1. *Suppose that T* : *X* → *X is a continuous mapping satisfying*
$${\rho}_{b}(T\xi ,T\eta )\le {\phi}_{b}({\rho}_{b}(\xi ,\eta )),\phantom{\rule{thinmathspace}{0ex}}for\phantom{\rule{thinmathspace}{0ex}}all\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\xi ,\eta \in X,$$(38)*where φ*_{b} ∈Φ_{b}. *Then Τ has a unique fixed point*.It has been shown in some recent studies that by a suitable choice of the function *α*, the fixed point results on partially ordered spaces and fixed point results of cyclic contraction mappings can be concluded from the fixed point results of *α*-admissible mappings (see [19] for details). Using the idea of these studies we present some theorems and their consequences below.We recall the definition of cyclic contraction mappings on quasi *b*-metric-like spaces introduced originally by Kirket.al. [20].

**Definition 2.9:** *Let X be nonempty set and let A and Β be nonempty subsets of X. A mapping T : A*U*B* → *A*U*B is cyclic if Τ*(*A*)C *Β and T*(*B*)⊂ *A*.We define generalized cyclic contraction of type as follows.

**Definition 2.10:** *Let* (*X, ρ*_{b}) *be a quasi-b-metric-like space with a constant s* ≥ 1 *and let φ*_{b} ∈ Φ_{b} be a given function. A generalized cyclic contractive mapping Τ : A U* Β → A* U* Β is a cyclic mapping satisfying*
$${\rho}_{b}(T\xi ,T\eta )\le {\phi}_{b}(K(\xi ,\eta )),\phantom{\rule{thinmathspace}{0ex}}for\phantom{\rule{thinmathspace}{0ex}}all\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\xi \in A,\eta \in B,$$(39)*where*
$$K(\xi ,\eta )=max\{{\rho}_{b}(\xi ,\eta ),{\rho}_{b}(T\xi ,\xi ),{\rho}_{b}(T\eta ,\eta )\}.$$A fixed point theorem for the generalized cyclic contractions defined above is given next.

**Theorem 2.11:** *Let* (*X, ρ*_{b}) *be a 0-complete quasi-b-metric-like space with a constant s* ≥ 1. *Suppose that Τ* : *A* U *B *→ *A* U *B is a generalized cyclic contractive mapping. Then Τ has a unique fixed point in Α* ⋂* Β*.

**Proof:** *Define **α : X* →* X* as follows.
$$\alpha (\xi ,\eta )=\left\{\begin{array}{l}1\phantom{\rule{0.056em}{0ex}}\phantom{\rule{0.056em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{i}\mathrm{f}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{0.056em}{0ex}}\xi \in A,\phantom{\rule{thinmathspace}{0ex}}\eta \in B\\ 0\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{0.056em}{0ex}}\mathrm{o}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{w}\mathrm{i}\mathrm{s}\mathrm{e}\end{array}.\right.$$Then all conditions of Corollary 2.5 are satisfied and *Τ* has a unique fixed point.Finally, we observe that the main result (Theorem 2.9) in [12] is a consequence of Theorem 2.11. Indeed, by choosing ${\phi}_{b}(t)=\frac{k}{s}t$ and noticing that *ρ*_{b}(ξ*, η*)≤ *Μ*(ξ*, η*), the main result of Klin-eam and Suanoom [12] follows immediately from Theorem 2.11.We end this section with illustrative examples.

**Example 2.12:** *Let X =* [–1, 1] *and ρ*_{b}(ξ*, η*)* = |*ξ* – η|*^{2} + |2ξ + *η|*^{2}. Then ρ_{b} is a quasi-b-metric like on X with a constant s = 2. Define $T{\displaystyle \xi =\frac{\xi}{4}.}$. Then we see that
$$\begin{array}{ll}{\rho}_{b}(T\xi ,T\eta )& ={\left|\frac{\xi}{41}-\frac{\eta}{4}\right|}^{2}+{\left|2\frac{\xi}{4}+\frac{\eta}{4}\right|}^{2}\\ & =\frac{1}{16}{\left|\xi -\eta \right|}^{2}+\frac{1}{16}{\left|\xi +\eta \right|}^{2}\\ & =\frac{1}{16}({\rho}_{b}(\xi ,\eta )\le {\phi}_{b}({p}_{b}(\xi ,\eta )),\end{array}$$*where φ*_{b}(*t*)* = kt with $\frac{1}{16}\le k<\frac{1}{2}$ is clearly a (b)-comparison function. By the Corollary 2.8 Τ has a unique**fixed point which is ξ =* 0.

**Example 2.13:** *Let X =* [0, 1] *and ρ*_{b}(ξ*, η*)* = |*ξ* – η|*^{2} + 2|ξ|^{2} + |*η|*^{2}. Then ρ_{b} *is a quasi-b-metric like on X with a constant s = 2. Define* $T{\displaystyle \xi =\frac{1}{8}{\xi}^{2}{e}^{-{\xi}^{2}}.}$ *Without loss of generality assume that ξ* ≥* η. Then we have* ${e}^{-{\xi}^{2}}\le {e}^{-{\eta}^{2}}\le 1\phantom{\rule{thinmathspace}{0ex}}in\phantom{\rule{thinmathspace}{0ex}}[0,1].$
$$\begin{array}{ll}{\rho}_{b}(T\xi ,T\eta )& ={\left|\frac{1}{8}{\xi}^{2}{e}^{-{\xi}^{2}}-\frac{1}{8}{\eta}^{2}{e}^{-{\eta}^{2}}\right|}^{2}+2{\left|\frac{1}{8}{\xi}^{2}{e}^{-{\xi}^{2}}\right|}^{2}+{\left|\frac{1}{8}{\eta}^{2}{e}^{-{\eta}^{2}}\right|}^{2}\\ & \le \frac{1}{64}|{\xi}^{2}{e}^{-{\eta}^{2}}-{\eta}^{2}{e}^{-{\eta}^{2}}{|}^{2}+\frac{1}{32}|{\xi}^{2}{e}^{-{\xi}^{2}}{|}^{2}+\frac{1}{64}|{\eta}^{2}{e}^{-{\eta}^{2}}{|}^{2}\\ & =\frac{1}{64}|{\xi}^{2}-{\eta}^{2}{|}^{2}|{e}^{-{\eta}^{2}}{|}^{2}+\frac{1}{32}|{\xi}^{2}{|}^{2}|{e}^{-{\xi}^{2}}{|}^{2}+\frac{1}{64}|\eta {|}^{2}{|}^{2}|{e}^{-{\eta}^{2}}{|}^{2}\\ & \le \frac{1}{64}|\xi -\eta {|}^{2}|\xi +\eta {|}^{2}+\frac{1}{32}|\xi {|}^{2}|\xi {|}^{2}+\frac{1}{64}|\eta {|}^{2}|\eta {|}^{2}\\ & \le \frac{4}{64}|\xi -\eta {|}^{2}+\frac{1}{32}|\xi {|}^{2}+\frac{1}{64}|\eta {|}^{2}\\ & \le \frac{1}{16}(|\xi -\eta {|}^{2}+2|\xi {|}^{2}+|\eta {|}^{2})=\frac{1}{16}{\rho}_{b}(\xi ,\eta ).\end{array}$$*Choosing a (b)-comparison function φ*_{b} (*t*)* = kt where $\frac{1}{16}\le k<\frac{1}{2}$ we see that the mapping Τ satisfies the* *conditions of the Corollary 2.8. Hence, Τ has a unique fixed point which is ξ =* 0.

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