Definition 2.1
Let be a bmetric space. A selfmapping is called admissible interpolative contraction (\(l=1,2\)) if there exist \(\phi \in \Theta \) and such that
where , \(i=1, 2, 3, 4, 5\), are such that and
and
for any . (.)
The first main results of this paper is given in the following theorem.
Theorem 2.2
Let be a complete bmetric space and be an admissible interpolative contraction such that
 \((i)\):

is αorbital admissible;
 \((\mathit{ii})\):

there exists such that ;
 \((\mathit{iii}_{1})\):

is mcontinuous for \(m\geq 1\), or
 \((\mathit{iii}_{2})\):

is orbitally continuous.
Then possesses a fixed point and the sequence converges to this point ϖ.
Proof
Let in be an arbitrary point and the sequence \(\{ \eta _{n} \} \) be defined as , for all \(n\in \mathbb{N}\). If we can find some \(q\in \mathbb{N}\) such that , then it follows that \(\eta _{q}\) is a fixed point of and the proof is closed. For this reason, we can assume from now on that \(\eta _{n}\neq \eta _{n1}\) for any \(n\in \mathbb{N}\). Using assumption \((i)\), is αorbital admissible, we have
On the other hand, we have that
Now, taking into account the main assumption that is an admissible interpolative contraction, if we substitute with \(\eta _{n1}\) and ω with \(\eta _{n}\) in (2.1), we get
But by \((B)\), together with the monotony of the function ϕ, it follows
moreover, by \((\phi 1)\) we have
If there exists \(m_{0}\in \mathbb{N}\) such that , then the above inequality becomes
which is a contradiction since (keeping in mind that ) it is equivalent with
Therefore, for any \(n\in \mathbb{N}\),
Furthermore, returning to inequality (2.5), we have
Let \(q\in \mathbb{N}\). Then, by \((B)\), together with (2.6), we obtain
It follows that \(\{ \eta _{n} \} \) is a Cauchy sequence in a orbitally complete bmetric space. Therefore, we can find such that .
We claim that ϖ is a fixed point of the mapping under of any hypothesis, \((\mathit{iii})_{1}\) or \((\mathit{iii})_{2}\).
Indeed,
Moreover,
If is mcontinuous, then , and by (2.7) it follows that .
If is assumed to be orbitally continuous on , then
Therefore, . □
Example
Let and be the bmetric defined as for all . Let the mapping be defined by
and a function , where
Let also the comparison function \(\phi :[0,\infty )\rightarrow [0,\infty )\), \(\phi (t)=t/3\), and we choose , , . Thus, we can easily observe that assumptions (i) and (ii) are satisfied, and since is continuous, assumption (iv) is also verified.
Case \((i.)\) For , we have , so inequality (2.1) holds.
Case \((\mathit{ii}.)\) For and \(\omega =2\), we have and . Thus, (2.1) holds.
Case \((\mathit{iii}.) \) For and \(\omega =3\), we have ⇒
Case \((\mathit{iv}.) \) For and \(\omega =9\), we have ⇒
All other cases are of no interest because and (2.1) is satisfied.
Therefore, the mapping is an admissible interpolative contraction. On the other hand, since is continuous and is αorbital continuous, by Theorem 2.2 we get that there exists a fixed point of the mapping ; that is, .
Theorem 2.3
Let be a complete bmetric space and be an admissible interpolative contraction such that
 \((i)\):

is αorbital admissible;
 \((\mathit{ii})\):

there exists such that ;
 \((\mathit{iii}_{1})\):

is mcontinuous for \(m\geq 1\), or
 \((\mathit{iii}_{2})\):

is orbitally continuous.
Then possesses a fixed point .
Proof
As in the previous proof, for , we build the sequence \(\{ \eta _{n} \} \), where and for any \(n\in \mathbb{N}\). Since \(\eta _{n1}\neq \eta _{n}\) for any \(n\in \mathbb{N}\cup {0}\), taking into account that the mapping is supposed to be admissible interpolative contraction, we have
where
Therefore, since by assumption \((i)\) it follows that \(\alpha (\eta _{n1},\eta _{n})\geq 1\) for all \(n\in \mathbb{N}\), we have
(Here, we used the property \((\phi 1)\) of the function ϕ.)
Thus,
and then for any \(n\in \mathbb{N}\). Furthermore, by (2.8) and keeping in mind \((\phi 2)\), we obtain
and following the same steps as in the proof of Theorem 2.2, we can easily find that the sequence \(\{ \eta _{n} \} \) is Cauchy. Moreover, since is supposed to be orbitally complete, we can find a point such that . Assuming that is mcontinuous, we have
and assuming that is orbitally continuous, we get
that is, ϖ is a fixed point of . □
In case we replace the continuity condition of the mapping with the continuity of the bmetric , we get the following results.
Theorem 2.4
Let be a complete, αregular bmetric space, where the bmetric is continuous, and is such that
where \(\phi \in \Theta \) and , for \(l=1,2\) are given by (2.2) and (2.3). If
 \((i)\):

is αorbital admissible;
 \((\mathit{ii})\):

there exists such that .
Then possesses a fixed point , and the sequence converges to this point ϖ.
Proof
From the proof of Theorem 2.2 we know that the sequence \(\{ \eta _{n} \} \), where converges to a point , and we claim that ϖ is a fixed point of the mapping . For this purpose, we claim that
or
Indeed, supposing the contrary
we get that
This is a contradiction, and then (2.10) or (2.11) holds. Under the regularity assumption of the space , we have that \(\alpha (\eta _{n},\varpi )\geq 1\) for any \(n\in \mathbb{N}\).
Case 1. (\(l=1\))
 \((1.a)\):

If (2.10) holds, we get
 \((1.b)\):

If (2.11) holds,
We can distinguish the following two situations:

(i)
.
Letting \(n\rightarrow \infty \) in (2.12) respectively (2.13), we obtain . Thus, .

(ii)
.
In this case, when \(n\rightarrow \infty \), from (2.12), (2.13) and keeping in mind the continuity of bmetric , we get
which is a contradiction.
Consequently, , that is, ϖ is a fixed point of the mapping .
Case 2. (\(l=2\))
 \((2.a)\):

If (2.10) holds, we get
 \((2.b)\):

If (2.11) holds,
We can distinguish the following two situations:

(i)
.
Letting \(n\rightarrow \infty \) in (2.14), respectively (2.15), we obtain . Thus, .

(ii)
.
In this case, when \(n\rightarrow \infty \), from (2.14) and (2.15), we get
which is a contradiction.
Consequently, , that is, ϖ is a fixed point of the mapping . □
Example
Let and be a bmetric space (\(s=2\)), defined by
Let be a selfmapping on , with and . Taking , for all , \(\phi (t)=t/2\) and the constants for \(i\in \{ 1,2,3,4,5 \} \), we have
Thus, by Theorem 2.4, the mapping has (at least) a fixed point.