# Fixed Point Results viaG-Function over the Complete Partial b-Metric Space

Authors: Sara Salem Alzaid; Andreea Fulga; Badr Alqahtani
Journal: Journal of Function Spaces
 (2020)

---

## Abstract

In this paper, we consider an auxiliary functionG to combine and unify several existing fixed point theorems in the setting of the complete partial b-metric space. We consider also some examples to support the observed main results.


---

## Body

## 1. Introduction and Preliminaries

The notion of the distance has been investigated and improved from the beginning of the mathematics sciences. The first formal definition was given by Hausdorff and Frechet under the name of metric spaces. The formal definition was extended, improved, and generalized in several ways. In this paper, we shall consider the combination of notions of partial metric space andb-metric space. Partial metric space, defined by Matthews [1, 2] is the most economical way to calculate the distance in computer science. So, it is important in the setting of theoretical computer science. On the other hand, b-metric is the most interesting and real generalization of metric spaces; in this case, the triangle inequality is replaced by a modified version of triangle inequality.For more details on the advances of fixed point theory in the setting of b-metric spaces, see e.g. [13]-[27].In this paper, we shall propose a fixed point theorem by using an auxiliary functionG to combine, generalize, and unify several fixed point results in the setting of the complete partial b-metric spaces.In [3], the authors proposed a new fixed point theorem in the setting of metric spaces.We consider the follow sets of functions:(1)
G be the set of the functions G:0,∞3⟶0,∞ that satisfy the following conditions:(f1) G is continuous,(f2) G0,0,0=0,(f3) maxτ,υ≤Gτ,υ,ω, for all τ,υ,ω∈0,∞.In [3], some examples of such a function were given.
(i)
Gτ,υ,ω=τ+υ+ω(ii)
Gτ,υ,ω=maxτ,υ,ω(iii)
Gτ,υ,ω=τ+υ1+ω(2)
Φ be the set of functions ψ:0,∞⟶0,∞ that satisfy the following conditions:(b1) ψ is nondecreasing,(b2) ∑i≥1ψiu<∞ for each u>0. (Here, by ψi, we denote the ith iterate ohψ.)We mention that the functionsψ∈Φ are called c-comparison functions. Moreover, it is not difficult to check that ϕu<u for every u>0.(3)
Γ=γ:X⟶0,∞∣γislowersemicontinuousTheorem 1 (see [3]).
LetX,d be a complete metric space, a lower semicontinuous function γ:X⟶0,∞, and a self-mapping T:X⟶X. If there exist ψ∈Φ and G∈G such that
(1)GdTx,Ty,γTx,γTy≤ψmaxGdx,y,γx,γy,Gdx,Tx,γTx,γx+Gdy,Ty,γTy,γy2,for every x,y∈X, then T has a unique fixed point.LetX be a nonempty set.
(i)
A functionb:X×X⟶0,∞ is a b-metric on X if for a given real number s≥1 and for all x,y,z∈X the following conditions hold:
(2)b1bx,y=0⇔x=y,b2bx,y=by,x,b3bx,y≤sbx,z+bz,y.The tripletX,b,s≥1 is called a b-metric space.
(ii)
A functionρ:X×X⟶0,∞ is a partial metric on X if for all x,y,z∈X the following conditions hold:
(3)ρ1x=y⇔ρx,x=ρy,y=ρx,y⇔x=y,ρ2ρx,x≤ρx,y,ρ3ρx,y=ρy,xρ4ρx,y≤ρx,z+ρz,y−ρz,z.The pairX,ρ is said to be a partial metric space.Combining these two concepts, Shukla [4] introduced the notion of partial b-metric space as follows.
(iii)
A functionρb:X×X⟶0,∞ is a partial b-metric on X if for all x,y,z∈X the following conditions hold:
(4)ρb1x=y⇔ρbx,x=ρby,y=ρbx,y,ρb2ρbx,x≤ρbx,y,ρb3ρbx,y=ρby,x,ρb4ρbx,y≤sρbx,z+ρbz,y−ρbz,z.The tripletX,ρb,s≥1 is said to be a partial b-metric space.On a partialb-metric space X,ρb,s≥1 a sequence xn is said to be
(i)
convergent tox∈X if limn→∞ρbxn,x=bx,x (the limit of a convergent sequence is not necessarily unique)(ii)
Cauchy iflimn,m→∞ρbxn,xp exists and its finiteMoreover, the partialb-metric space is complete if for every Cauchy sequence axn there exists x∈X such that
(5)limn,p→∞ρbxn,xp=limn→∞ρbxn,x=ρbx,x.LetX,ρb,s≥1 be a partial b-metric space. We say that a self-mapping T on X is continuous if for every sequence xn in X which converges to a point x∈X we have
(6)limn→∞ρbTxn,Tx=limn→∞ρbTxn,Txn+j=ρbTx,Tx.In [5], the authors introduced the following new notions.
(i)
On a partialb-metric space, a sequence xn is a 0-Cauchy sequence if limn→∞ρbxn,xp=0(ii)
The spaceX,ρb,s≥1 is said to be 0-complete if for each 0-Cauchy sequence xn in X, there exists a point x∈X such that
(7)limn,p→∞ρbxn,xp=limn→∞ρbxn,x=ρbx,x=0.Moreover, they proved that if the partialb-metric space X,ρb,s≥1 is complete, then it is 0-complete.For a better understanding of the connections between these spaces (partial metric space,b-metric space, and partial b-metric space), we mention some papers that can be consulted [6–12].LetΦb be the set of functions ϕ:0,∞⟶0,∞ that satisfy the following conditions:ϕ1ϕ is nondecreasing,ϕ2∑i≥1siϕiu<∞ for each u>0. (Here, by ϕi, we denote the ith iterate oh ϕ.)


## 2. Main Results

The following is the main result of the paper.Theorem 2.
LetX,ρb,s≥1be a 0-complete partialb-metric space, a functionγ∈Γ, G∈G, and a self-mappingT:X⟶X. If there existsϕ∈Φbsuch that(8)GρbTx,Ty,γTx,γTy≤ϕmaxGρbx,y,γx,γy,Gρbx,Tx,γTx,γx+Gdy,Ty,γTy,γy2s,for everyx,y∈X. IfTis continuous orρbis continuous, thenThas a unique fixed point.Proof.
Starting with a pointx0∈X, we consider the sequence x defined by xn=Txn−1, n∈ℕ. Without losing the generality, we can assume that for any n∈ℕ, we have bxn,xn+1>0. Indeed, on the contrary, if there exists a positive integer j0 such that xj0=xj0+1, we get that xj0 is a fixed point of T, because due to the way the sequence was x defined, it follows that xj0=Txj0. Moreover, using this remark, we can easily see that
(9)Gρbxn,xn+1,γxn,γxn+1>0,foreveryn∈ℕ.
Again supposing thatGρbxj0,xj0+1,γxj0,γxj0+1=0 for some j0 from f3, we have
(10)0<ρbxj0,xj0+1≤maxρbxj0,xj0+1,γxj0≤Gρbxj0,xj0+1,γxj0,γxj0+1,which is a contradiction. Taking x=xn and y=xn+1 in (8) we get
(11)Gρbxn+1,xn+2,γxn+1,γxn+2≤GρbTxn,Txn+1,γTxn,γTxn+1≤ϕmaxGρbxn,xn+1,γxn,γxn+1,Gρbxn,Txn,γxn,γTxn+Gρbxn+1,Txn+1,γxn+1,γTxn+12s≤ϕmaxGρbxn,xn+1,γxn,γxn+1,Gρbxn,xn+1,γxn,γxn+1+Gρbxn+1,xn+2,γxn+1,γxn+22s≤ϕmaxGρbxn,xn+1,γxn,γxn+1,Gρbxn+1,xn+2,γxn+1,γxn+2.
There are two possibilities, namely,(12)maxGρbxn,xn+1,γxn,γxn+1,Gρbxn+1,xn+2,γxn+1,γxn+2=Gdxn+1,xn+2,γxn+1,γxn+2,which leads us (since ϕu<u for any u>0) to
(13)Gdxn+1,xn+2,γxn+1,γxn+2≤ϕGρbxn+1,xn+2,γxn+1,γxn+2<Gρbxn+1,xn+2,γxn+1,γxn+2.
But, this is a contradiction, and then(14)maxGdxn,xn+1,γxn,γxn+1,Gdxn+1,xn+2,γxn+1,γxn+2=Gdxn,xn+1,γxn,γxn+1.
Therefore, by (11) and taking into account f3, we have
(15)ρbxn+1,xn+2≤maxρbxn+1,xn+2,γxn+1≤Gρbxn+1,xn+2,γxn+1,γxn+2≤ϕGρbxn,xn+1,γxn,γxn+1,foreveryn∈ℕ∪0.
Consequently, for everyn∈ℕ, we obtain
(16)ρbxn,xn+1≤ϕGρbxn−1,xn,γxn−1,γxn≤ϕnGρbx0,x1,γx0,γx1.
Letp,m∈ℕ such that p<m. By applying the (triangle-type inequality) ρb4, we have
(17)ρbxp,xm≤sρbxp,xp+1+ρbxp+1,xm−ρbxp+1,xp+1≤sρbxp,xp+1+ρbxp+1,xm≤sρbxp,xp+1+s2ρbxp+1,xp+2+sρbxp+2,xm−ρbxp+2,xp+2≤sρbxp,xp+1+s2ρbxp+1,xp+2+sρbxp+2,xm⋯≤sρbxp,xp+1+s2ρbxp+1,xp+2+⋯+sm−p−1ρbxm−1,xm,and (17) leads us to
(18)ρbxp,xm≤1sp−1∑i=pm−1siϕiGρbx0,x1,γx0,γx1<1sp−1∑i=pm−1siϕiGρbx0,x1,γx0,γx1=1sp−1Sm−1−Sp−1,where Sn=∑i=0nsiϕiGρbx0,x1,γx0,γx1. Keeping in mind ϕ2, we deduce that there exists Sn⟶S as n⟶∞, and from (18), we get
(19)limp,m→∞ρbxp,xm=0.
Consequently,xn is a 0-Cauchy sequence in a 0-complete partial b-metric space, and then there exists ς∈X such that
(20)limp,m→∞ρbxp,xm=limp→∞ρbxp,ς=ρbς,ς=0.
Moreover, byf3 together with (16), we have
(21)γxn≤maxρbxn,xn+1,γxn≤Gρbxn,xn+1,γxn,γxn+1≤ϕnGρbx0,x1,γx0,γx1and using ϕ2(22)limn→∞γxn=0.
Plus, byϕ1 and (20),
(23)γς=0.
We claim that this pointς is in fact a fixed point of the mapping T. If the mapping T is continuous, then by (6), we have
(24)ρbς,ς=limp→∞ρbTxp,Tς=limp→∞ρbTxp,Txp+i=0.
Thus, applying the triangle inequalityρ4,
(25)ρbς,Tς≤sρbς,xn+1+ρbxn+1,Tς−ρbxn+1,xn+1,and together with (20) and (24), letting n⟶∞, we get ρbς,Tς=0, that is, ς is a fixed point of T.
Let assume now thatρb is continuous, that is, limn→∞ρbxn,Tς=ρbς,Tς Replacing x by xn and y by ς in (8), we have (for every n∈ℕ)
(26)Gρbxn+1,Tς,γxn+1,γTς=GρbTxn,Tς,γTxn,γTς≤ϕmaxGρbxn,ς,γxn,γς,Gρbxn,Txn,γxn,γTxn+Gρbς,Tς,γς,γTς2s≤ϕmaxGρbxn,ς,γxn,γς,Gρbxn,xn+1,γxn,γxn+1+Gρbς,Tς,γς,γTς2s<maxGρbxn,ς,γxn,γς,Gρbxn,xn+1,γxn,γxn+1+Gρbς,Tς,γς,γTς2s.
Lettingn⟶∞ and taking into account f1, we have
(27)Gρbς,Tς,0,γTς=limn→∞Gρbxn+1,Tς,γxn+1,γTς<limn→∞maxGρbxn,ς,γxn,γς,Gρbxn,xn+1,γxn,γxn+1+Gρbς,Tς,γς,γTς2s=maxG0,0,0,G0,0,0+Gρbς,Tς,0,γTς2=Gρbς,Tς,0,γTς2s.
Consequently,Gρbς,Tς,0,γTς=0. But, taking f3 into account, we get
(28)0≤maxρbς,Tς,0≤Gρbς,Tς,0,γTς=0,which means ρbς,Tς=0. Thus, Tς=ς.
As a last step, we claim thatς is the unique fixed point of T. Supposing on the contrary, that there exists another point υ∈X such that Tς=ς≠υ=Tυ. First of all, applying (8) with x=υ=y, we have
(29)G0,γυ,γυ≤ϕG0,γυ,γυ<G0,γυ,γυ,which implies that γυ=γTυ=0. Let now x=ς and y=υ in (8). We have
(30)Gρbς,υ,0,0=GρbTς,Tυ,γTς,γTυ≤ϕmaxGρbς,υ,γς,γυ,Gρbς,Tς,γς,γTς+Gρbυ,Tυ,γυ,γTυ2s=ϕmaxGρbς,υ,γς,γυ,G0,γς,γTς+G0,γυ,γTυ2s≤ϕGρbς,υ,0,0<Gρbς,υ,0,0.
This is a contradiction. Therefore,ρbς,υ=0, that is, T admits a unique fixed point.In particular, lettingGτ,υ,ω=τ+υ+ω, for τ,υ,ω∈0,∞, we can omit the continuity conditions of the mapping T or the partial b-metric ρb.Theorem 3.
LetX,ρb,s≥1be a 0-complete partialb-metric space, a functionγ∈Γ, G∈F, and a self-mappingT:X⟶X. If there existsϕ∈Φbsuch that(31)ρbTx,Ty+γTx+γTy≤ϕmaxρbx,y+γx+γy,ρbx,Tx+γTx+γx+ρby,Ty+γTy+γy2s,for everyx,y∈X, thenThas a unique fixed point.Proof.
Of course, since the functionGτ,υ,ω=τ+υ+ω∈F, by Theorem 2, we have that the sequence xn defined as xn=Txn−1 is convergent to a point ς∈X, and moreover, (22) and (23) hold. We claim that this point ς is a fixed point of T. For this purpose, by (31), for x=ς and y=ς, we get
(32)ρbς,Tς+γxn+1+γTς≤sρbς,Txn+ρbTxn,Tς−ρbTxn,Txn+γxn+1+γTς≤sρbς,xn+1+sρbTxn,Tς+γTxn+γTς≤sρbς,xn+1+sϕmaxρbxn,ς+γxn+γς,ρbxn,xn+1+γxn+γxn+1+ρbς,Tς+γς+γTς2s<sρbς,xn+1+smaxρbxn,ς+γxn+γς,ρbxn,xn+1+γxn+γxn+1+ρbς,Tς+γς+γTς2s.
Lettingn⟶∞, in the above inequality and keeping in mind (19), (22), and (23), we get
(33)ρbς,Tς+γTς≤sϕρbς,Tς+γTς2s<ρbς,Tς+γTς2,which is a contradiction. Therefore, ρbς,Tς=0, that is, Tς=ς.
As in the previous theorem, supposing that there existsυ, another fixed point of T, by (31), we have
(34)2γυ=ρbTυ,Tυ+γυ+γTυ≤ϕ2γυ<2γυ,which is a contradiction. Thus, γυ=0 and taking x=ς and y=υ in (31), we have
(35)ρbς,υ=ρbς,υ+γς+γυ=ρbς,υ+γς+γυ≤ϕmaxρbς,υ+γς+γυ,ρbς,Tς+γς+γTς+ρbυ,υ+γυ+γTυ2s=ϕρbς,υ<ρbς,υ.
But, this is a contraction, soρbς,υ=0 which proves the uniqueness of the fixed point.Example 4.
Let the setX=0,1 and the function ρb:X×X⟶0,∞ be defined by ρbx,1/2=ρb1/2,x=1 for any x∈X and ρbx,y=maxx,y2, otherwise. It easy to see that ρb is a partial b-metric space, with s=2. Moreover, since limn,m→∞ρbxn,xm=limn,m→∞maxxn,xm2=0implies limn,m→∞xn=0, we have
(36)limn→∞ρbxn,0=ρb0,0=0,which shows that X,ρb,s is 0-complete. On the other hand, taking, for example, the sequence xn in X, where xn=n/2n+1, we have limn,m→∞ρbxn,xm=1/4, but limn→∞ρbxn,1/2=1=ρb1/2,1/2=1. Thus, the space X,ρb,s is not complete.
Let the mappingT:X⟶X be defined as
(37)Tx=x4,ifx∈0,1,12,ifx=1.
Choosingϕu=u/2 and γu=u, we have
(i)
Ifx=y=1, then
(38)ρbT1,T1+γT1+γT1=ρb12,12+2γ12=54<32=ϕmaxρb1,1+γ1+γ1,ρb1,T1+γT1+γ12(ii)
Ifx=1, y∈0,1, then
(39)ρbTx,T1+γTx+γT1=ρbx4,12+γx4+γ12=3+x4<2+x2≤ϕmaxρbx,1+γx+γ1,ρbx,Tx+γTx+γx+ρb1,T1+γT1+γ14(iii)
Ifx,y∈0,1, then
(40)ρbTx,Ty+γTx+γTy=ρbx4,y4+γx4+γy4≤x2+4x+4y16≤x2+x+y2≤ϕmaxρbx,y+γx+γy,ρbx,Tx+γTx+γx+ρbTy,y+γTy+γy4
(We considered heremaxx,y=x. The case maxx,y=y is similar.)
Consequently, by Theorem3, the mapping T admits a unique fixed point.


## 3. Conclusion

In this paper, we investigate the uniqueness and the existence of a fixed point for certain contraction via theG-function in one of the most general frames and complete the partial b-metric space. Regarding that the metric fixed point theory has a key role in the solution of not only differential equations and fractional differential equations but also integral equations, our results can be applied in these problems.



---

*Source: 6666229-2020-12-12.xml*