# Double Controlled Partial Metric Type Spaces and Convergence Results

Authors: Haroon Ahmad; Mudasir Younis; Mehmet Emir Köksal
Journal: Journal of Mathematics
 (2021)

---

## Abstract

In this paper, we firstly propose the notion of double controlled partial metric type spaces, which is a generalization of controlled metric type spaces, partial metric spaces, and double controlled metric type spaces. Secondly, our aim is to study the existence of fixed points for Kannan type contractions in the context of double controlled partial metric type spaces. The proposed results enrich, theorize, and sharpen a multitude of pioneer results in the context of metric fixed point theory. Additionally, we provide numerical examples to illustrate the essence of our obtained theoretical results.


---

## Body

## 1. Introduction and Preliminaries

The study of fixed points of given mappings satisfying certain contractive conditions in various abstract spaces has been at the middle of vigorous research activity. Banach contraction mapping principle has attracted the eye of the many authors to generalize, extend, and improve the metric fixed point theory. For this purpose, the authors considered the extension of metric fixed point theory to different abstract spaces such as symmetric spaces, quasimetric spaces, fuzzy metric spaces, partial metric spaces, probabilistic metric spaces, and spaces with graph.The notion ofb-metric spaces was first presented by Bakhtin [1] and Czerwik [2]. Many writers have since obtained a number of fixed point solutions in b-metric spaces for single and multivalued operators. We reference Kamran et al. [3] (see also [4, 5]), who presented extended b-metric spaces by manipulating the triangle inequality rather than utilizing control functions, as one of the generalizations concerning b-metric spaces. Following that, in 2018, Abdeljawad et al. [6, 7] established the concepts of controlled metric type spaces and double controlled metric type spaces, respectively. Souayah and Mrad [8] proposed a more broad idea of controlled partial metric type spaces in 2019. It is useful to establish the extensions of the contraction principle from metric spaces to b-metric spaces, and therefore the controlled metric type of spaces is useful to prove the existence and uniqueness of theorems for many forms of integral and differential equations. Some interesting applications can be found in the recent papers [4, 9–15]. It is always interesting to find novel applications dealing with engineering science and technology using fixed point technique.On the other hand, the notion of partial metric space was given by Matthews [16, 17] in 1992, which is the generalization of the usual metric space in which d (x, x) is not zero. After that, many researchers worked on the partial metric type spaces to discover the existence of fixed point and their uniqueness. In 2019, Gu and Shatanawi [18] expounded some coupled fixed point theorems in the context of partial metric spaces for hybrid pairs of mappings satisfying a symmetric type contraction. In 2020, Nguyen and Tram [19] demonstrated various fixed point results involving involution mappings. Recently, in 2021, Javaid et al. [20] propounded fixed point results in the setting orthogonal partial metric spaces with application. Researchers can refer to [14, 21–23] for further information on fixed points in partial type metric spaces.Taking into consideration the facts mentioned above, in this article, we introduce the concept of double controlled partial metric type space, which is an extension of the controlled metric type spaces, double controlled metric type spaces, and controlled partial metric type spaces. We also look into the existence and uniqueness of fixed point results, which are Kannan contractions’ extensions.Let us begin by reviewing the definition of double controlled metric space as follows.Definition 1 (see [6]).
LetX be a nonempty set and consider the functions α,µ:X×X⟶1,∞.
Letd:X×X⟶0,∞ satisfy(1)
dx1,x2=0if and only ifx1=x2,(2)
dx1,x2=dx2,x1,(3)
dx1,x2≤αx1,x3dx1,x3+µx3,x2dx3,x2, for all x1,x2,x3∈X, then X,d is called a double controlled metric type space.


## 2. Double Controlled Partial Metric Type Spaces

The following is the formal definition of the double controlled partial metric type space which generalizes the notation of controlled metric type spaces, double controlled metric type spaces, and partial metric spaces.Definition 2.
LetX be a nonempty set consider α,µ:X×X⟶1,∞ be a function.
Letd:X×X⟶0,∞ satisfy(1)
dx1,x2=0if and only ifx1=x2,(2)
dx1,x2=dx2,x1,(3)
dx1,x1≤dx1,x2,(4)
dx1,x2≤αx1,x3dx1,x3+µx3,x2dx3,x2, for all x1, x2, x3 ∈ X, then (X, d) is called a double controlled partial metric type space.
Note that double controlled partial metric type space is more extensive than the double controlled metric type space.Example 1.
A double controlled partial metric type space is not necessarily a double controlled metric type space.
LetX=0,1,2,3,4 and take d:X×X⟶0,∞. Consider α,µ:X×X⟶1,∞, where(1)αx,y=dx,y+5,µx,y=dx,y+7.
Let the metricd be defined by the following (Table 1).
It is easy to verify thatp1 and p2 are true.
We prove condition (3) with different cases, that is,dx1,x1≤dx1,x2, for all x1,x2∈X and x1≠x2.Case (i): letdx1,x1=d0,0=1/27, d0,0≤dx1,x2, satisfied for all x1,x2∈X and x1≠x2.Case (ii): letdx1,x1=d1,1=1/28, d1,1≤dx1,x2, satisfied for all x1,x2∈X and x1≠x2Case (iii): letdx1,x1=d2,2=1/29, d2,2≤dx1,x2, satisfied for all lx1,x2∈X and x1≠x2.Case (iv): letdx1,x1=d3,3=1/28, d3,3≤dx1,x2, satisfied for all x1,x2∈X and x1≠x2.Case (v): letdx1,x1=d4,4=1/27, d4,4≤dx1,x2, satisfied for all x1,x2∈X and x1≠x2.
Now, we will prove the propertyp4.Case (i): to satisfyd0,0, we have(2)d0,0≤α0,0d0,0+μ0,0d0,00.0370≤0.4513,d0,0≤α0,1d0,1+μ1,0d1,00.0370≤3.1250,d0,0≤α0,2d0,2+μ2,0d2,00.0370≤2.48,d0,0≤α0,3d0,3+μ3,0d3,00.0370≤2.0555,d0,0≤α0,4d0,4+μ4,0d4,00.0370≤1.7551.Case (ii): now, we have to satisfyd0,1=d1,0:(3)d0,1≤α0,0d0,0+μ0,1d0,10.25≤1.9990,d0,1≤α0,1d0,1+μ1,1d1,10.25≤1.5637,d0,1≤α0,2d0,2+μ2,1d2,10.25≤3.1216,d0,1≤α0,3d0,3+μ3,1d3,10.25≤2.4660,d0,1≤α0,4d0,4+μ4,1d4,10.25≤2.0404.Case (iii): to proved0,2=d2,0, we have(4)d0,2≤α0,0d0,0+μ0,2d0,20.2≤1.6265,d0,2≤α0,1d0,1+μ1,2d1,20.2≤3.3941,d0,2≤α0,2d0,2+μ2,2d2,20.2≤1.2825,d0,2≤α0,3d0,3+μ3,2d3,20.2≤2.8445,d0,2≤α0,4d0,4+μ4,2d4,20.2≤2.4033.Case (iv): in order to showd0,3=d3,0, we proceed as follows:(5)d0,3≤α0,0d0,0+μ0,3d0,30.1666≤1.3810,d0,3≤α0,1d0,1+μ1,3d1,30.1666≤2.9174,d0,3≤α0,2d0,2+μ2,3d2,30.1666≤3.0234,d0,3≤α0,3d0,3+μ3,3d3,30.1666≤1.1123,d0,3≤α0,4d0,4+μ4,3d4,30.1666≤2.9832.Case (v): now, we have to satisfyd0,4=d4,0:(6)d0,4≤α0,0d0,0+μ0,4d0,40.1428≤1.2069,d0,4≤α0,1d0,1+μ1,4d1,40.1428≤2.6182,d0,4≤α0,2d0,2+μ2,4d2,40.1428≤2.7086,d0,4≤α0,3d0,3+μ3,4d3,40.1428≤3.1096,d0,4≤α0,4d0,4+μ4,4d4,40.1428≤0.9953.Case (vi): for the cased1,1, we have(7)d1,1≤α1,0d1,0+μ0,1d0,10.03571≤3.125,d1,1≤α1,1d1,1+μ1,1d1,10.03571≤0.4311,d1,1≤α1,2d1,2+μ2,1d2,10.03571≤3.5918,d1,1≤α1,3d1,3+μ3,1d3,10.03571≤2.7654,d1,1≤α1,4d1,4+μ1,4d4,10.03571≤2.2479.Case (vii): to satisfyd1,2=d2,1, we have(8)d1,2≤α1,0d1,0+μ0,2d0,20.2857≤2.7525,d1,2≤α1,1d1,1+μ1,2d1,20.2857≤2.2614,d1,2≤α1,2d1,2+μ2,2d2,20.2857≤1.7527,d1,2≤α1,3d1,3+μ3,2d3,20.2857≤3.1439,d1,2≤α1,4d1,4+μ4,2d4,20.2857≤2.6107.Case (viii): now, we have to satisfyd1,3=d3,1:(9)d1,3≤α1,0d1,0+μ0,3d0,30.2222≤2.5069,d1,3≤α1,1d1,1+μ1,3d1,30.2222≤1.7847,d1,3≤α1,2d1,2+μ2,3d2,30.2222≤3.4936,d1,3≤α1,3d1,3+μ3,3d3,30.2222≤1.4117,d1,3≤α1,4d1,4+μ4,3d4,30.2222≤3.19066.Case (ix): for the cased1,4=d4,1, consider the following:(10)d1,4≤α1,0d1,0+μ0,4d0,40.1818≤2.3329,d1,4≤α1,1d1,1+μ1,4d1,40.1818≤1.4856,d1,4≤α1,2d1,2+μ2,4d2,40.1818≤3.1788,d1,4≤α1,3d1,3+μ3,4d3,40.1818≤3.4090,d1,4≤α1,4d1,4+μ4,4d4,40.1818≤1.2027.Case (x): for the cased2,2, we have(11)d2,2≤α2,0d2,0+μ0,2d0,20.0344≤2.48,d2,2≤α2,1d2,1+μ1,2d1,20.0344≤3.5918,d2,2≤α2,2d2,2+μ2,2d2,20.0344≤0.4161,d2,2≤α2,3d2,3+μ2,3d2,30.0344≤3.4214,d2,2≤α2,4d2,4+μ4,2d4,20.0344≤2.8757.Case (xi): to satisfyd2,3=d3,2, we proceed as follows:(12)d2,3≤α2,0d2,0+μ0,3d0,30.2727≤2.2344,d2,3≤α2,1d2,1+μ1,3d1,30.2727≤3.1151,d2,3≤α2,2d2,2+μ2,3d2,30.2727≤2.1570,d2,3≤α2,3d2,3+μ3,3d3,30.2727≤1.6892,d2,3≤α2,4d2,4+μ4,3d4,30.2727≤3.4556.Case (xii): next, we have to satisfyd2,4=d4,2:(13)d2,4≤α2,0d2,0+μ0,4d0,40.2307≤2.0604,d2,4≤α2,1d2,1+μ1,4d1,40.2307≤2.8159,d2,4≤α2,2d2,2+μ2,4d2,40.2307≤1.8422,d2,4≤α2,3d2,3+μ3,4d3,40.2307≤3.6865,d2,4≤α2,4d2,4+μ4,4d4,40.2307≤1.4677.Case(xiii): now, for the cased3,3, we consider(14)d3,3≤α3,0d3,0+μ0,3d0,30.03571≤2.0555,d3,3≤α3,1d3,1+μ1,3d1,30.03571≤2.7654,d3,3≤α3,2d3,2+μ2,3d2,30.03571≤3.4214,d3,3≤α3,3d3,3+μ3,3d3,30.03571≤0.4311,d3,3≤α3,4d3,4+μ4,3d4,30.03571≤3.8816.Case (xiv): now, we have to satisfyd3,4=d4,3:(15)d3,4≤α3,0d3,0+μ0,4d0,40.3076≤1.8855,d3,4≤α3,1d3,1+μ1,4d1,40.3076≤2.4662,d3,4≤α3,2d3,2+μ2,4d2,40.3076≤3.1066,d3,4≤α3,3d3,3+μ3,4d3,40.3076≤2.4283,d3,4≤α3,4d3,4+μ4,4d4,40.3076≤1.8937.Case (xv): lastly, for the cased4,4, we have(16)d4,4≤α4,0d4,0+μ0,4d0,40.0370≤1.7551,d4,4≤α4,1d4,1+μ1,4d1,40.0370≤2.2479,d4,4≤α4,2d4,2+μ2,4d2,40.0370≤2.8757,d4,4≤α4,3d4,3+μ3,4d3,40.0370≤3.8816,d4,4≤α4,4d4,4+μ4,4d4,40.0370≤0.4471.
Therefore,X,d is a double controlled partial metric type space but is not a double controlled metric type space since dx,x is not equal to zero all the time.
We define Cauchy and convergent sequence in double controlled partial metric type spaces as follows.Table 1
Metricd defined in Example 1.
d0123401/271/41/51/61/711/41/282/72/92/1121/52/71/293/113/1331/62/93/111/284/1341/72/113/134/131/27Definition 3.
LetX,d be a double controlled partial metric type space; the sequence xnn≥0 converges to some x in X, if limn,m⟶∞dxn,x=dx,x; in this case, we write limn⟶∞xn=x.Definition 4.
The sequencexn in a double controlled partial metric type space X,d is said to be Cauchy sequence, if limn,m⟶∞dxn,xm exists and is finite.Definition 5.
A double controlled partial metric type spaceX,d is said to be complete if every Cauchy sequence x in X converges to a point x∈X, that is, dx,x=limn,m⟶∞dxn,xm.Definition 6.
LetX,d be a double controlled partial metric type space. Let x∈X and ε>0.(i)
The open ballBpx,ε is(17)Bpx,ε=y∈X,dx,y<dx,x+ε.(ii)
The mappingT:X⟶X is said to be continuous at x∈X if for all ε>0, there exists δ>0 such that(18)TBpx,δ⊆BpTx,ε.
Therefore, ifT is continuous at x in the double controlled partial metric type space X,d, then xn⟶x implies that Txn⟶Tx as n⟶∞


## 3. Some Novel Results

This section is devoted to discuss some fixed point results in double controlled partial metric type spaceX,d. The main result of this article is given by the following theorem.Theorem 1.
LetX,d be a complete double controlled partial metric type space by the functions α,μ:X×X⟶1,∞. Suppose that f:X⟶X satisfies(19)dfx,fy≤βdx,fx+y,fy,for all x,y∈X, where β∈0,1/2. For x0∈X, take xn=fnx0, assuming that(20)supm≥1limi⟶∞αxi+1,xi+2αxi,xi+1μxi,xm<1k,wherek∈0,1.Furthermore, assume that for everyx∈X, limn⟶∞αx,xn, limn⟶∞αxn,x, limn⟶∞μx,xn, and limn⟶∞μxn,x exist and are finite. Then, the sequence xn converges to some u∈X; moreover, if α and μ satisfy the following assumptions,(21)limn⟶∞αu,xn+11−βμxn+1,fu≤0,then f has a unique fixed point.Proof.
Considerxn=fnx0, let x1∈X be arbitrary, and let x2=fx1 and let x3=fx2 be chosen.
By using (19), we get(22)dx2,x3=dfx1,fx2≤βdx1,fx1+dx2,fx2=βdx1,x2+dx2,x3.
Then,(23)dx2,x3≤β1−βdx1,x2,whereβ1−β=η∈0,1.
By repeating the same procedure in inequality (23), we obtain(24)dxn,xn+1≤ηn−1dx1,x2.
Now, we have to show thatxn is Cauchy sequence. Since X,d is a double controlled partial metric type space, for all natural numbers n,m∈N with n<m, we acquire(25)dxn,xm≤αxn,xn+1dxn,xn+1+μxn+1,xmdxn+1,xm≤αxn,xn+1dxn,xn+1+αxn+1,xn+2μxn+1,xmdxn+1,xn+2+μxn+1,xmμxn+2,xmdxn+2,xm≤αxn,xn+1dxn,xn+1+αxn+1,xn+2μxn+1,xmdxn+1,xn+2+αxn+2,xn+3μxn+1,xmμxn+2,xmdxn+2,xn+3+μxn+1,xmμxn+2,xmμxn+3,xmdxn+3,xm≤αxn,xn+1dxn,xn+1+∑i=n+1m−2∏j=n+1iμxj,xmαxi,xi+1dxi,xi+1+∏k=n+1m−1μxk,xmdxm−1,xm≤αxn,xn+1ηndx0,x1+∑i=n+1m−2∏j=n+1iμxj,xmαxi,xi+1ηidx0,x1+∏k=n+1m−1μxk,xmηm−1dx0,x1+∏k=n+1m−1μxk,xmαxm−1,xmηm−1dx0,x1=αxn,xn+1ηndx0,x1+∑i=n+1m−1∏j=n+1iμxj,xmαxi,xi+1ηidx0,x1≤αxn,xn+1ηndx0,x1+∑i=n+1m−1∏j=n+1iμxj,xmαxi,xi+1ηidx0,x1≤αxn,xn+1ηndx0,x1+∑i=n+1m−1∏j=n+1iμxj,xmαxi,xi+1ηidx0,x1.
Assume that(26)Sp=∑i=n+1m−1∏j=0iμxj,xmαxi,xi+1ηidx0,x1.
Then, we obtain(27)dxn,xm≤dx0,x1ηnαxn,xn+1+Sm−1−Sn.
Using ratio test, we have(28)ai=∏j=0iμxj,xmαxi,xi+1ηidx0,x1,whereai+1ai<1η.
Takinglimn,m⟶∞, (27) becomes(29)limn,m⟶∞dxn,xm=0.
This implies thatxn is a Cauchy sequence in a complete double controlled metric type space X,d, so xn converges to some u∈X. Now, we have to prove that u is a fixed point of T, so we need to verify that(30)du,fu=du,u=dfu,fu.
From thep3, we have(31)du,u≤du,fu,dfu,fu≤du,fu.
Hence, for provingfu=u, it is sufficient to prove that du,u≥du,fu and dfu,fu≥du,fu. The triangular inequality yields that(32)du,fu≤αu,xn+1du,xn+1+μxn+1,fudxn+1,fu≤αu,xn+1du,xn+1+μxn+1,fudfxn,fu≤αu,xn+1du,xn+1+βμxn+1,fudxn,fxn+βμxn+1,fudu,fu.
Taking limit asn⟶∞, we obtain(33)du,fu≤limn⟶∞αu,xn+11−βμxn+1,fudu,fudu,u.
Utilizing condition (21), we get(34)du,fu≤du,u.
On the other hand,(35)du,fu≤αu,fudu,fu+μfu,fudfu,fu≤αu,fudu,fu+μfu,fuβdu,fu+du,fu≤αu,fudu,fu+βμfu,fudu,fu+βμfu,fudu,fu≤αu,fu1−βμfu,fudfu,fu.
Hence, we get(36)du,fu≤du,u.
From (31)–(36), we obtain(37)u=fu.
Uniqueness: assume that there are two fixed pointsu and v of T, then(38)du,v=dfu,fv≤βdu,fu+dv,fv=βdu,u+dv,v.
Furthermore, we have(39)du,u=dfu,fu≤2βdu,fu=2βdu,u,whereβ>1,thendu,u=0,similarly(40)dv,v=dfv,fv≤2βdv,fv=2βdv,v.
Then,dv,v=0. Since du,u=dv,v=0, then du,v=0. Therefore, du,u=dv,v=du,v, which gives u=v and T has a unique fixed point.Definition 7.
LetX,d be complete double controlled partial type metric space; a mapping T:X⟶X is sequentially convergent. For every sequence xn, if fxn is convergent, then xn also converges. Also, f is said to be subsequentially convergent. For every sequence xn, if fxn is convergent, then xn has a convergent subsequence.Theorem 2.
LetX,d be a complete double controlled partial metric type space and f,g:X⟶X be mapping such that f is continuous, one-to-one, and subsequentially convergent(41)dfgx,fgy≤βdfx,fgx+fy,fgy.For allx,y∈X, where β∈0,1/2. For x0∈X, take xn=gnx0, assuming that(42)supm≥1limi⟶∞αfxi+1,fxi+2αfxi,fxi+1μfxi,fxm<1k,wherek∈0,1.Furthermore, assume that for everyx∈X, limn⟶∞αx,xn, limn⟶∞αxn,x, limn⟶∞μx,xn, and limn⟶∞μxn,x exist and are finite. Then, g has a unique fixed point.Proof.
Letx0 be an arbitrary point in X and consider the sequence xn defined in the hypothesis of the theorem. From (41), we obtain(43)dfxn,fxn+1=dfgxn−1,fxn≤βdfxn−1,fgxn−1+dfxn,fgxn=βdfxn−1,fgxn−1+dfxn,fxn+1=β1−βdfxn−1,fgxn−1.
By induction, we get(44)dfxn,fxn+1≤β1−βndfgxn−1,fxn=ηndfgx0,fx1,where(45)β1−β=η∈0,1.
Now, we have to show thatfxn is a Cauchy sequence. Since X,d is double controlled partial metric type space for all natural numbers n,m∈N with n<m, we get(46)dfxn,fxm≤αfxn,fxn+1dfxn,fxn+1+μfxn+1,fxmdfxn+1,fxm≤αfxn,fxn+1dfxn,fxn+1+αfxn+1,fxn+2μfxn+1,fxmdfxn+1,fxn+2+μfxn+1,fxmμfxn+2,fxmdfxn+2,fxm≤αfxn,fxn+1dfxn,fxn+1+αfxn+1,fxn+2μfxn+1,fxmdfxn+1,fxn+2+αfxn+2,fxn+3μfxn+1,fxmμfxn+2,fxmdfxn+2,fxn+3+μfxn+1,fxmμfxn+2,fxmμfxn+3,fxmdfxn+3,fxm≤αfxn,fxn+1dfxn,fxn+1+∑i=n+1m−2∏j=n+1iμfxj,fxmαfxi,fxi+1dfxi,fxi+1+∏k=n+1m−1μfxk,fxmdfxm−1,fxm≤αfxn,fxn+1ηndfx0,fx1+∑i=n+1m−2∏j=n+1iμfxj,fxmαfxi,fxi+1ηidfx0,fx1+∏k=n+1m−1μfxk,fxmηm−1dfx0,fx1≤αfxn,fxn+1ηndfx0,fx1+∑i=n+1m−2∏j=n+1iμfxj,fxmαfxi,fxi+1ηidfx0,fx1+∏k=n+1m−1μfxk,fxmαfxm−1,fxmηm−1dfx0,fx1=αfxn,fxn+1ηndfx0,fx1+∑i=n+1m−1∏j=n+1iμfxj,fxmαfxi,fxi+1ηidfx0,fx1≤αfxn,fxn+1ηndfx0,fx1+∑i=n+1m−1∏j=n+1iμfxj,fxmαfxi,fxi+1ηidfx0,fx1≤αfxn,fxn+1ηndfx0,fx1+∑i=n+1m−1∏j=n+1iμfxj,fxmαfxi,fxi+1ηidfx0,fx1.
Assume that(47)Sp=∑i=n+1m−1∏j=0iμfxj,fxmαfxi,fxi+1ηidfx0,fx1.
Then, we obtain(48)dfxn,fxm≤dfx0,fx1ηnαfxn,fxn+1+Sm−1−Sn.
Using ratio test, we have(49)ai=∏j=0iμfxj,fxmαfxi,fxi+1ηidfx0,fx1,whereai+1ai<1η.
Takinglimn,m⟶∞ inequality, (48) reduces to(50)limn,m⟶∞dfxn,fxm=0.
This amounts to say thatfxn is a Cauchy sequence in a complete double controlled partial metric type space X,d, hence there exists v∈X such that(51)limn⟶∞fxn=v.
Sincef is convergent, the sequence xn has a convergent subsequence denoted by xnkk=1∞ such that(52)limk⟶∞xnk=u.
Using the continuity off, we obtain(53)limk⟶∞fxnk=fu.
From (51) and (53), we conclude that fu=v. Making use of triangular inequality, we get(54)dfgu,fu≤αfgu,fgnkx0dfgu,fgnkx0+μfgnkx0,fudfgnkx0,fu≤αfgu,fgnkx0βdfu,fgu+dfgnk−1x0,fgnkx0+μfgnkx0,fudfgnkx0,fu≤βαfgu,fgnkx0dfu,fgu+βαfgu,fgnkx0dfgnk−1x0,fgnkx0+μfgnkx0,fudfgnkx0,fu≤βαfgu,fxnk1−βαfgu,fxnkdfxnk−1,fxnk+βμfu,fxnk1−βαfgu,fxnkdfxnk,fu≤βαfgu,fxnk1−βαfgu,fxnkβ1−βnk−1dfx0,fx1+βμfu,fxnk1−βαfgu,fxnkdfxnk,fu.
Proceeding thelimk⟶∞, we obtain(55)dfgu,fu≤constant×dfu,fu,which proves that dfu,fu=0. From the triangular inequality, we have(56)dfu,fu≤αfu,udfu,u+μu,fudu,fu.
Suppose thatαfu,u≤μu,fu, then(57)dfu,fu≤2αfu,udfu,u.
On the other hand,(58)du,fu≤αu,udu,u+μu,fudu,fu≤αu,u1−μu,fudu,u.
Note that ifμ:X×X⟶1,∞, then 1−μu,fu≤0 and we get du,fu=0. Thus, from (57), we obtain(59)dfu,fu=0.
From (55) and (57), we deduce that dfgu,fu=0. To check the property p1, i.e.,(60)dfgu,fu=dfu,fu=dfgu,fgu=0.
It is easy to see that(61)dfgu,fgu≤βdfu,fgu+dfu,fgu=2βdfu,fgu=0.
Thus,fgu=fu, since f is one-to-one, gu=u. Therefore, u is a fixed point of g.
Uniqueness: letu,v be two fixed points of g, then gu=u and gv=v. From the condition p3, we have(62)dfv,fv≤dfu,fv,(63)dfu,fu≤dfu,fv.
On the other hand, using the triangular inequality, we get(64)dfu,fv=du,v≤au,udu,u+μu,vdu,v≤au,u1−μu,vdu,u.
Sinceμ:X×X⟶1,∞, then 1−μu,v≤0 and we get dfu,fv=0. Therefore, from (62) and (63), we obtain that(65)dfu,fv=dfu,fu=dfv,fv=0.
Utilizing the propertyp1 of the double controlled partial metric type space, we obtain fu=fv. Hence, f is one-to-one so that u=v. Finally, by replacing nk with n, we conclude that xn converges to u as n⟶∞. Thus, the sequence xn converges to the unique fixed point g.Corollary 1 (Banach contraction)..
LetX,d be a complete double controlled partial metric type space by the functions α,μ:X×X⟶1,∞. Suppose that f:X⟶X satisfies(66)dfx,fy≤βdx,y,for all x,y∈X, where β∈0,1/2. For x0∈X, take xn=fnx0, assuming that(67)supm≥1limi⟶∞αxi+1,xi+2αxi,xi+1μxi,xm<1k,wherek∈0,1.Furthermore, assume that for everyx∈X, limn⟶∞αx,xn, limn⟶∞αxn,x, limn⟶∞μx,xn, and limn⟶∞μxn,x exist and are finite. Then, the sequence xn converges to some u∈X; moreover, if α and μ satisfy the following assumptions,(68)limn⟶∞αu,xn+11−βμxn+1,fu≤0,then f has a unique fixed point.Remark 1.
Results presented in this manuscript generalize, enrich, and theorize the prominent results due to Kannan [24] and Bojor [25] in the framework of double controlled partial metric type spaces.Example 2.
LetX=0,1,2; consider the function d given as follows: (Table 2)
Givenα,µ:X×X⟶1,∞ is defined as(69)αx,y=dx,y+5,μx,y=dx,y+7.
It is easy to verify that givend equipped with X is double controlled partial metric type space but not double controlled metric type space because dx,x≠0 for all x∈X. Now, we define a mapping f:X⟶X by the following:(70)fx=1,whenx=1,2,2,whenx=0.
Choosef0=2 and f2=1, then by using (19), we acquire(71)df0,f2≤βd0,f0+d2,f2d2,1≤βd0,2+d2,115≤β27+1515≤β1735.
Sinceβ∈0,1/2, we choose β=8/17; taking x0=0 and k=1/8, it is clear that condition (20) is satisfied as follows:(72)supm≥1limi⟶∞αxi+1,xi+2αxi,xi+1μxi,xm=1768245<8=1k.
Since inequality (20) is satisfied for every x∈X, additionally, for each x∈X, we have(73)limn⟶∞αx,xn=max0,x<∞,limn⟶∞αxn,x=maxx,0<∞,limn⟶∞μx,xn=max0,x<∞,limn⟶∞μxn,x=maxx,0<∞.
Therefore, all the hypotheses of Theorem1 are contended and 1 is the unique fixed point of f.Table 2
Metricd defined in Example 2.
d01201/271/42/711/41/281/522/71/51/29


## 4. Conclusions

We launched a new concept of double controlled partial metric type spaces which expands the ideas of certain variants of metric spaces, viz., controlled metric type spaces, double controlled metric type spaces, and partial metric spaces. The introduced results sum up and broaden some previous writing, and some illustrative examples are investigated to show the potency of our work.



---

*Source: 7008737-2021-12-14.xml*