# Common Fixed Point Theorems in Modified Intuitionistic Fuzzy Metric Spaces

Authors: Saurabh Manro; Sanjay Kumar; S. S. Bhatia; Kenan Tas
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

This paper consists of main two sections. In the first section, we prove a common fixed point theorem in modified intuitionistic fuzzy metric space by combining the ideas of pointwiseR-weak commutativity and reciprocal continuity of mappings satisfying contractive conditions. In the second section, we prove common fixed point theorems in modified intuitionistic fuzzy metric space from the class of compatible continuous mappings to noncompatible and discontinuous mappings. Lastly, as an application, we prove fixed point theorems using weakly reciprocally continuous noncompatible self-mappings on modified intuitionistic fuzzy metric space satisfying some implicit relations.


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## Body

## 1. Introduction and Preliminaries

Recently, Saadati et al. [1] introduced the modified intuitionistic fuzzy metric space and proved some fixed point theorems for compatible and weakly compatible maps. Consequently, in this modified setting of intuitionistic fuzzy metric space, Jain et al. [2] discussed the notion of the compatibility of type (P); Sedghi et al. [3] proved some common fixed point theorems for weakly compatible maps using contractive conditions of integral type. The paper [1] is the inspiration of a large number of papers [4–7] that employ the use of modified intuitionistic fuzzy metric space and its applications.In this paper, we prove some new common fixed point theorems in modified intuitionistic fuzzy metric spaces. While proving our results, we utilize the idea of compatibility due to Jungck [8] together with weakly reciprocal continuity due to Pant et al. [16]. Consequently, our results improve and sharpen many known common fixed point theorems available in the existing literature of modified intuitionistic fuzzy fixed point theory.Firstly, we recall the following notions that will be used in the sequel.Lemma 1 (see [10]).
Consider the setL* and the operation ≤L* defined by
(1)L*={(x1,x2):(x1,x2)∈[0,1]2,x1+x2≤1}(x1,x2)≤L*(y1,y2)⟺x1≤y1,x2≥y2,
for every (x1,x2), (y1,y2) in L*. Then, (L*,≤L*) is a complete lattice.
One denotes its units by0L*=(0,1) and 1L*=(1,0).Definition 2 (see [11]).
A triangular norm (t-norm) on L* is a mapping ℱ:(L*)2→L* satisfying the following conditions:(1)
ℱ(x,1L*)=x for all x in L*,(2)
ℱ(x,y)=ℱ(y,x) for all x, y in L*,(3)
ℱ(x,ℱ(y,z))=ℱ(ℱ(x,y),z) for all x, y, z in L*,(4)
if for allx,x′,y,y′ in L*, x≤L*x′ and y≤L*y′ imply ℱ(x,y)≤L*ℱ(x′,y′).Definition 3 (see [10, 11]).
A continuoust-norm ℱ on L* is called continuous t-representable if and only if there exist a continuous t-norm * and a continuous t-conorm ⋄  on [0,1] such that for all x=(x1,x2),y=(y1,y2)∈L*[0,1]2,ℱ(x,y)=(x1*y1,x2⋄y2).Definition 4 (see [1]).
LetM,N are fuzzy sets from X2×(0,+∞)→[0,1] such that M(x,y,t)+N(x,y,t)≤1 for all x,y in X and t>0. The 3-tuple (X,ζM,N,ℱ) is said to be a modified intuitionistic fuzzy metric space if X is an arbitrary  nonempty set, ℱ is a continuous t-representable, and  ζM,N is a mapping X2×(0,+∞)→L* satisfying the following conditions for every x,y in X and t,s>0:(a)
ζM,N(x,y,t)>L*0L*,(b)
ζM,N(x,y,t)=1L* if and only if x=y,(c)
ζM,N(x,y,t)=ζM,N(y,x,t),(d)
ζM,N(x,y,t+s)≥L*ℱ(ζM,N(x,z,t),ζM,N(z,y,s)),(e)
ζM,N(x,y,·):(0,+∞)→L* is continuous.
In this case,ζM,N is called a modified intuitionistic fuzzy metric. Here, ζM,N(x,y,t)=(M(x,y,t),N(x,y,t)).Remark 5 (see [12]).
In a modified intuitionistic fuzzy metric space(X,ζM,N,ℱ), M(x,y,·) is nondecreasing, and N(x,y,·) is nonincreasing for all x, y in X. Hence, ζM,N(x,y,t) is nondecreasing with respect to t for all x, y in X.Definition 6 (see [1]).
A sequence{xn} in a modified intuitionistic fuzzy metric space (X,ζM,N,ℱ) is called a Cauchy sequence if for each  ε>0 and t>0, there exists n0∈ℕ such that  ζM,N(xn,xm,t)>L*(1-ε,ε) for each n,m≥n0 and for all t.Definition 7 (see [1]).
A sequence{xn} in a modified intuitionistic fuzzy metric space (X,ζM,N,ℱ) is said to be convergent to x in X, denoted by xn→x if limn→∞ζM,N(xn,x,t)=1L*   for all t.Definition 8 (see [1]).
A modified intuitionistic fuzzy metric space(X,ζM,N,ℱ) is said to be complete if and only if every Cauchy sequence is convergent to a point of it.Definition 9 (see [1, 13]).
A pair of self-mappings(f,g) of modified intuitionistic fuzzy metric space (X,ζM,N,ℱ) is said to be compatible if limn→∞ζM,N(fgxn,gfxn,t)=1L* whenever {xn} is a sequence in X such that limn→∞f(xn)=limn→∞g(xn)=z for some z in X.Definition 10 (see [13]).
Two self-mappingsf and g are called noncompatible if there exists at least one sequence {xn} such that limn→∞f(xn)=limn→∞g(xn)=z for some z in X but either limn→∞ζM,N(fgxn,gfxn,t)≠1L* or the limit does not exist for all z in X.Definition 11 (see [14, 15]).
A pair of self mappings(f,g) of a modified intuitionistic fuzzy metric space (X,ζM,N,ℱ) is said to be R-weakly commuting at a point x in X if ζM,N(fgxn,gfxn,t)≥L*ζM,N(fxn,gxn,t/R)  for some R>0.Definition 12 (see [14, 15]).
The two self-mapsf and g of a modified intuitionistic fuzzy metric space (X,ζM,N,ℱ) are called pointwise R-weakly commuting on X if given x in X, there exists R>0 such that ζM,N(fgx,gfx,t)≥L*ζM,N(fx,gx,t/R).Definition 13 (see [15]).
The two self-mapsf and g of a modified intuitionistic fuzzy metric space (X,ζM,N,ℱ) are called R-weakly commuting of type (Ag) if there exists some R>0 such that ζM,N(ffx,gfx,t)≥ζM,N(fx,gx,t/R) for all x in X. Similarly, two self-mappings f and g of a modified intuitionistic fuzzy metric space (X,ζM,N,ℱ) are called R-weakly commuting of type (Af) if there exists some R>0 such that ζM,N(fgx,ggx,t)≥L*ζM,N(fx,gx,t/R) for all x in X.
It is obvious that pointwiseR-weakly commuting maps commute at their coincidence points and pointwise R-weak commutativity is equivalent to commutativity at coincidence points. It may be noted that both compatible and noncompatible mappings can be R-weakly commuting of type (Ag) or (Af), but the converse needs not be true.Definition 14 (see [2]).
Two self-mappingsf and g of a modified intuitionistic fuzzy metric space  (X,ζM,N,ℱ) are called R-weakly commuting of type (P) if there exists some R>0 such that ζM,N(ffx,ggx,t)≥L*ζM,N(fx,gx,t/R) for all x in X.In 1999, Pant [9] introduced a new continuity condition, known as reciprocal continuity as follows.Definition 15 (see [9]).
Two self-mappingsf and g are called reciprocally continuous if limn→∞fgxn=fz and limn→∞gfxn=gz, whenever {xn} is a sequence such that limn→∞fxn=limn→∞gxn=z for some z in X.
Iff and g are both continuous, then they are obviously reciprocally continuous, but the converse is not true.Recently, Pant et al. [16] generalized the notion of reciprocal continuity to weak reciprocal continuity as follows.Definition 16 (see [16]).
Two self-mappingsf and g are called weakly reciprocally continuous if limn→∞fgxn=fz or limn→∞gfxn=gz whenever {xn} is a sequence such that limn→∞fxn=limn→∞gxn=z for some z in X.Iff and g are reciprocally continuous, then they are obviously weak reciprocally continuous, but the converse is not true. Now, with an application of weak reciprocal continuity, we prove common fixed point theorems under contractive conditions that extend the scope of the study of common fixed point theorems from the class of compatible continuous mappings to a wider class of mappings which also includes noncompatible mappings.


## 2. Lemmas

The proof of our result is based upon the following lemmas.Lemma 17 (see [2]).
Let(X,ζM,N,𝒯)  be a modified intuitionistic fuzzy metric space and for all x,y∈X, t>0 and if for a number k∈(0,1),
(2)ζM,N(x,y,kt)≥L*ζM,N(x,y,t).
Then  x=y.Lemma 18 (see [2]).
Let(X,ζM,N,𝒯) be a modified intuitionistic fuzzy metric space and {yn} a sequence in X. If there exist a number k∈(0,1) such that
(3)ζM,N(yn,yn+1,kt)≥L*ζM,N(yn-1,yn,t)∀t>0,n=1,2,3,…,
then {yn} is a Cauchy sequence in X.


## 3. Main Results

### 3.1. Section I: PointwiseR-weakly Commuting Pairs and Fixed Point

Lemma 19.
Let(X,ζM,N,𝒯) be a  modified intuitionistic fuzzy metric space, and let (A,S) and (B,T) be  pairs of self-mappings on X satisfying
(4)A(X)⊆T(X),B(X)⊆S(X),(5)thereexistsaconstantk∈(0,1)suchthatζM,N(Ax,By,kt)≥L*min{ζM,N(Ty,By,t),ζM,N(Sx,Ax,t),ζM,N(Sx,By,αt),ζM,N(Ty,Ax,(2-α)t),ζM,N(Ty,Sx,t)}
for all x,y∈X, t>0 and  α∈(0,2). Then, the continuity of one of the mappings in compatible pair (A,S) or (B,T) on (X,ζM,N,𝒯) implies their reciprocal continuity.Proof.
First, assume thatA and S are compatible and S is continuous. We show that A and S are reciprocally continuous. Let {un} be a sequence such that Aun→z and Sun→z for some z in X as n→∞. Since S is continuous, we have SAun→Sz and SSun→Sz as n→∞ and since (A,S) is compatible, we have for all t>0,
(6)limn→∞ζM,N(ASun,SAun,t)=1L*,limn→∞ζM,N(ASun,Sz,t)=1L*.
That is, ASun→Sz as n→∞. By (4), for each n, there exists vn in X such that ASun=Tvn. Thus, we have SSun→Sz, SAun→Sz, ASun→Sz, and Tvn→Sz as n→∞ whenever ASun=Tvn.
Now, we claim thatBvn→Sz as n→∞.
Suppose not, then, by (5), taking α=1,
(7)ζM,N(ASun,Bvn,kt)≥L*min{ζM,N(Tvn,Bvn,t),ζM,N(SSun,ASun,t),ζM,N(SSun,Bvn,t),ζM,N(Tvn,ASun,t),ζM,N(Tvn,SSun,t)}.
Taking n→∞, we have
(8)ζM,N(Sz,limn→∞Bvn,kt)≥L*min{ζM,N(Sz,limn→∞Bvn,t),ζM,N(Sz,Sz,t),ζM,N(Sz,limn→∞Bvn,t),ζM,N(Sz,Sz,t),ζM,N(Sz,Sz,t)limn→∞}=min{ζM,N(Sz,limn→∞Bvn,t),1L*,ζM,N(Sz,limn→∞Bvn,t),1L*,1L*}=ζM,N(Sz,limn→∞Bvn,t).
By Lemma 17, we have limn→∞Bvn=Sz.
Claim thatAz=Sz. Again, by (5), taking α=1,
(9)ζM,N(Az,Bvn,kt)≥L*min{ζM,N(Tvn,Bvn,t),ζM,N(Sz,Az,t),ζM,N(Sz,Bvn,t),ζM,N(Tvn,Az,t),ζM,N(Tvn,Sz,t)}.
Taking n→∞, we get
(10)ζM,N(Az,Sz,kt)≥L*min{ζM,N(Sz,Sz,t),ζM,N(Sz,Az,t),ζM,N(Sz,Sz,t),ζM,N(Sz,Az,t),ζM,N(Sz,Sz,t)}=min{1L*,ζM,N(Sz,Az,t),1L*,ζM,N(Sz,Az,t),1L*}=ζM,N(Sz,Az,t).
By Lemma 17, Az=Sz.
Therefore,SAun→Sz, ASun→Sz=Az as n→∞.
Hence,A and S are reciprocally continuous on X. If the pair (B,T) is assumed to be compatible and T is continuous, the proof is similar.Theorem 20.
Let(X,ζM,N,𝒯) be a complete modified intuitionistic fuzzy metric space. Further, let (A,S) and (B,T) be pointwise R-weakly commuting pairs of self-mappings of X satisfying (4), (5). If one of the mappings in compatible pair (A,S) or (B,T) is continuous,then A, B, S, and T have a unique common fixed point in X.Proof.
Letx0∈X. By (4), we define the sequences {xn} and {yn} in X such that for all n=0,1,2,…,
(11)y2n=Ax2n=Tx2n+1,y2n+1=Bx2n+1=Sx2n+2.
We show that {yn} is a Cauchy sequence in X. By (5) taking α=1-β,β∈(0,1), we have
(12)ζM,N(y2n+1,y2n+2,kt)=ζM,N(Bx2n+1,Ax2n+2,kt)=ζM,N(Ax2n+2,Bx2n+1,kt)≥L*min{ζM,N(Tx2n+1,Bx2n+1,t),ζM,N(Sx2n+2,Ax2n+2,t),ζM,N(Sx2n+2,Bx2n+1,(1-β)t),ζM,N(Tx2n+1,Ax2n+2,(1+β)t),ζM,N(Tx2n+1,Sx2n+2,t)}=min{ζM,N(y2n,y2n+1,t),ζM,N(y2n+1,y2n+2,t),ζM,N(y2n+1,y2n+1,(1-β)t),ζM,N(y2n,y2n+2,(1+β)t),ζM,N(y2n,y2n+1,t)}=min{ζM,N(y2n,y2n+1,t),ζM,N(y2n+1,y2n+2,t),1L*,ζM,N(y2n,y2n+1,t),ζM,N(y2n+1,y2n+2,βt),ζM,N(y2n,y2n+1,t)}≥L*min{ζM,N(y2n,y2n+1,t),ζM,N(y2n+1,y2n+2,t),ζM,N(y2n+1,y2n+2,βt)}.
Taking β→1, we have
(13)ζM,N(y2n+1,y2n+2,kt)≥L*min{ζM,N(y2n,y2n+1,t),ζM,N(y2n+1,y2n+2,t),ζM,N(y2n+1,y2n+2,t)},ζM,N(y2n+1,y2n+2,kt)≥L*min{ζM,N(y2n,y2n+1,t),ζM,N(y2n+1,y2n+2,t)}ζM,N(y2n+1,y2n+2,kt)≥L*ζM,N(y2n,y2n+1,t).
Similarly,
(14)ζM,N(y2n+2,y2n+3,kt)≥L*ζM,N(y2n+1,y2n+2,t).
Therefore, for any n and t, we have
(15)ζM,N(yn,yn+1,kt)≥L*ζM,N(yn-1,yn,t).
Hence, by Lemma 18, {yn} is a Cauchy sequence in X. Since X is complete, {yn} converges to z in X. Its subsequences {Ax2n}, {Tx2n+1}, {Bx2n+1}, and  {Sx2n+2} also converge to z.
Now, suppose that(A,S) is a compatible pair and S is continuous. Then, by Lemma 17, A and S are reciprocally continuous, then  SAxn→Sz, ASxn→Az as n→∞. As, (A,S), is a compatible pair. This implies that
(16)limn→∞ζM,N(ASxn,SAxn,t)=1L*,ζM,N(Az,Sz,t)=1L*.
Hence, Az=Sz.
SinceA(X)⊆T(X), there exists a point p in X such that Az=Tp=Sz.
By (5), taking α=1,
(17)ζM,N(Az,Bp,kt)≥L*min{ζM,N(Tp,Bp,t),ζM,N(Sz,Az,t),ζM,N(Sz,Bp,t),ζM,N(Tp,Az,t),ζM,N(Tp,Sz,t)},ζM,N(Az,Bp,kt)≥L*min{ζM,N(Az,Bp,t),ζM,N(Az,Az,t),ζM,N(Az,Bp,t),ζM,N(Az,Az,t),ζM,N(Az,Az,t)},ζM,N(Az,Bp,kt)≥L*ζM,N(Az,Bp,t).
Thus, by Lemma 17, we have Az=Bp.
Thus,Az=Bp=Sz=Tp.
SinceA and S are pointwise R-weakly commuting mappings, there exists R>0, such that
(18)ζM,N(ASz,SAz,t)≥L*ζM,N(Az,Sz,tR)=1L*.
Therefore, ASz=SAz and AAz=ASz=SAz=SSz.
Similarly, sinceB and T are pointwise R-weakly commuting mappings, we have BBp=BTp=TBp=TTp.
Again, by (5), taking α=1,
(19)ζM,N(AAz,Bp,kt)≥L*min{ζM,N(Tp,Bp,t),ζM,N(SAz,AAz,t),ζM,N(SAz,Bp,t),ζM,N(Tp,AAz,t),ζM,N(Tp,SAz,t)},ζM,N(AAz,Az,kt)≥L*min{ζM,N(Tp,Tp,t),ζM,N(AAz,AAz,t),ζM,N(AAz,Az,t),ζM,N(Az,AAz,t),ζM,N(Az,AAz,t)},ζM,N(AAz,Az,kt)≥L*ζM,N(AAz,Az,t).
By Lemma 17, we have AAz=Az=SAz.Hence Az is a common fixed point of A and S. Similarly, by (5), Bp=Az is a common fixed point of B and T. Hence, Az is a common fixed point of A, B, S, and T.For Uniqueness. Suppose that Ap(≠Az)  is another common fixed point of A, B, S, and T. Then, by (5), taking α=1,
(20)ζM,N(AAz,BAp,kt)≥L*min{ζM,N(TAp,BAp,t),ζM,N(SAz,AAz,t),ζM,N(SAz,BAp,t),ζM,N(TAp,AAz,t),M(TAp,SAz,t)},ζM,N(Az,Ap,kt)≥L*min{ζM,N(Ap,Ap,t),ζM,N(Az,Az,t),ζM,N(Az,Ap,t),ζM,N(Ap,Az,t),ζM,N(Ap,Az,t)},ζM,N(Az,Ap,kt)≥L*ζM,N(Az,Ap,t).
By Lemma 17,  Az=Ap.Thus, the uniqueness follows.Corollary 21.
Let(X,ζM,N,𝒯) be a complete modified intuitionistic fuzzy metric space. Further, let A and B be reciprocally continuous mappings on X satisfying
(21)ζM,N(Ax,By,kt)≥L*min{ζM,N(y,By,t),ζM,N(x,Ax,t),ζM,N(x,By,αt),ζM,N(y,Ax,(2-α)t),ζM,N(y,x,t)}
for all x,y∈X, t>0 and α∈(0,2), then pair A and B has a unique common fixed point.Example 22.
LetX=[2,20] and for each t>0,  define
(22)ζM,N(x,y,t)=(tt+|x-y|,|x-y|t+|x-y|).
Then, (X,ζM,N,𝒯) is complete modified intuitionistic fuzzy metric space. Let A,B,S, and T be self-mappings of  X defined as
(23)A(2)=2,Au=3ifu>2,B(u)=2ifu=2oru>5,Bu=6if2<u≤5,S(2)=2,S(u)=6ifu>2,T(2)=2,T(u)=12if2<u≤5,T(u)=u-3ifu>5.
Then, A,B,S, and T satisfy all the conditions of the above theorem with  k∈(0,1) and have a unique common fixed point u=2.


### 3.2. Section II: Weak Reciprocal Continuity and Fixed Point Theorem

Theorem 23.
Letf and g be weakly reciprocally continuous self-mappings of a complete modified intuitionistic fuzzy metric space (X,ζM,N,𝒯)  satisfying the following conditions:
(24)f(X)⊂g(X)(25)foranyx,y∈X,t>0,k∈(0,1)suchthatζM,N(fx,fy,kt)≥L*min{ζM,N(gx,gy,t),ζM,N(gx,fy,2t),ζM,N(fx,gx,t),ζM,N(fx,gy,t),ζM,N(fy,gy,t)}.
If f and g are either compatible or R-weakly commuting of type (Ag) or R-weakly commuting of type (Af) or R-weakly commuting of type (P), then f and g have a unique common fixed point.Proof.
Letx0 be any point in X. Then, as f(X)⊂g(X), there exist a sequence of points {xn} such that  f(xn)=g(xn+1).
Also, define a sequence{yn} in X as
(26)yn=f(xn)=g(xn+1).
Now, we show that {yn} is a Cauchy sequence in X. For proving this, by (25), we have
(27)ζM,N(yn,yn+1,kt)=ζM,N(fxn,fxn+1,kt)≥L*min{ζM,N(gxn,gxn+1,t),ζM,N(gxn,fxn+1,2t),ζM,N(fxn,gxn,t),ζM,N(fxn,gxn+1,t),ζM,N(fxn+1,gxn+1,t)}=min{ζM,N(yn-1,yn,t),ζM,N(yn-1,yn+1,2t),ζM,N(yn,yn-1,t),ζM,N(yn,yn,t),ζM,N(yn+1,yn,t)}≥L*min{ζM,N(yn-1,yn,t),ζM,N(yn-1,yn,t),ζM,N(yn,yn+1,t),ζM,N(yn,yn-1,t),1,ζM,N(yn+1,yn,t)}=min{ζM,N(yn-1,yn,t),ζM,N(yn,yn+1,t)}ζM,N(yn,yn+1,kt)≥L*ζM,N(yn-1,yn,t).
Then, by Lemma 18, {yn} is a Cauchy sequence in X.   As X is complete, there exists a point z in X such that limn→∞yn=z. Therefore, by (26), we have limn→∞yn=limn→∞f(xn)=limn→∞g(xn+1)=z.
Suppose thatf and g are compatible mappings. Now, by weak reciprocal continuity of f and g, it implies that limn→∞fg(xn)=fz or limn→∞gf(xn)=gz. Let limn→∞gf(xn)=gz, then the compatibility of f and g gives
(28)limn→∞ζM,N(fgxn,gfxn,t)=1L*.
This gives
(29)limn→∞ζM,N(fgxn,gz,t)=1L*.
Hence, limn→∞fg(xn)=gz. By (26), we get limn→∞fg(xn+1)=limn→∞ff(xn)=gz.
Therefore, by (25), we get
(30)ζM,N(fz,ffxn,kt)≥L*min{ζM,N(gz,gfxn,t),ζM,N(gz,ffxn,2t),ζM,N(fz,gz,t),ζM,N(fz,gfxn,t),ζM,N(ffxn,gfxn,t)}.
Taking n→∞, we get
(31)ζM,N(fz,gz,kt)≥L*min{ζM,N(gz,gz,t),ζM,N(gz,gz,2t),ζM,N(fz,gz,t),ζM,N(fz,gz,t),ζM,N(gz,gz,t)},ζM,N(fz,gz,kt)≥L*ζM,N(fz,gz,t).
Hence, by Lemma 17, we get fz=gz. Again, the compatibility of f and g implies commutativity at a coincidence point. Hence gfz=fgz=ffz=ggz. Using (25), we obtain
(32)ζM,N(fz,ffz,kt)≥L*min{ζM,N(gz,gfz,t),ζM,N(gz,ffz,2t),ζM,N(fz,gz,t),ζM,N(fz,gfz,t),ζM,N(ffz,gfz,t)}=min{ζM,N(fz,ffz,t),ζM,N(fz,ffz,2t),1L*,ζM,N(fz,ffz,t),1L*}≥min{ζM,N(fz,ffz,t),ζM,N(fz,ffz,t),1L*,ζM,N(fz,ffz,t),1L*},=ζM,N(fz,ffz,t).
That is, fz=ffz. Hence, fz=ffz=gfz and fz is a common fixed point of  f and g.
Next, suppose thatlimn→∞fg(xn)=fz. Then, f(X)⊂g(X) implies that fz=gu for some u∈X and, therefore, limn→∞fg(xn)=gu.
The compatibility off and g implies that limn→∞gf(xn)=gu. By the virtue of (26), this gives
(33)limn→∞fg(xn+1)=limn→∞ff(xn)=gu.
Using (25), we get
(34)ζM,N(fu,ffxn,kt)≥L*min{ζM,N(gu,gfxn,t),ζM,N(gu,ffxn,2t),ζM,N(fu,gu,t),ζM,N(fu,gfxn,t),ζM,N(ffxn,gfxn,t)}.
Taking n→∞, we have
(35)ζM,N(fu,gu,kt)≥L*min{ζM,N(gu,gu,t),ζM,N(gu,gu,2t),ζM,N(fu,gu,t),ζM,N(fu,gu,t),ζM,N(gu,gu,t)},ζM,N(fu,gu,kt)≥L*ζM,N(fu,gu,t).
By Lemma 17, we get fu=gu. The compatibility of f and g yields fgu=ggu=ffu=gfu. Finally, using (25), we obtain
(36)ζM,N(fu,ffu,kt)≥L*min{ζM,N(gu,gfu,t),ζM,N(gu,ffu,2t),ζM,N(fu,gu,t),ζM,N(fu,gfu,t),ζM,N(ffu,gfu,t)}=min{ζM,N(fu,ffu,t),ζM,N(fu,ffu,2t),1L*,ζM,N(fu,ffu,t),1L*}≥L*min{ζM,N(fu,ffu,t),ζM,N(fu,ffu,t),1L*,ζM,N(fu,ffu,t),1L*}=ζM,N(fu,ffu,t).
That is, fu=ffu. Hence fu=ffu=gfu and fu is a common fixed point of f and g.
Now, suppose thatf and g are R-weakly commuting of type (Ag). Now, weak reciprocal continuity of f and g implies that limn→∞fg(xn)=fz or limn→∞gf(xn)=gz. Let us first assume that limn→∞gf(xn)=gz. Then, R-weak commutativity of type (Ag) of f and g yields
(37)ζM,N(ffxn,gfxn,t)≥L*ζM,N(fxn,gxn,tR),limn→∞ζM,N(ffxn,gz,t)≥L*ζM,N(z,z,tR)=1L*.  This gives limn→∞ffxn=gz. Also, using (25), we get
(38)ζM,N(fz,ffxn,kt)≥L*min{ζM,N(gz,gfxn,t),ζM,N(gz,ffxn,2t),ζM,N(fz,gz,t),ζM,N(fz,gfxn,t),ζM,N(ffxn,gfxn,t)}.
Taking n→∞, we have
(39)ζM,N(fz,gz,kt)≥L*min{ζM,N(gz,gz,t),ζM,N(gz,gz,2t),ζM,N(fz,gz,t),ζM,N(fz,gz,t),ζM,N(gz,gz,t)},ζM,N(fz,gz,kt)≥L*ζM,N(fz,gz,t).
Hence, by Lemma 17, we get fz=gz. Again, by using  R-weak commutativity of type (Ag),(40)ζM,N(ffz,gfz,t)≥L*ζM,N(fz,gz,tR)=1L*.
This yieldsffz=gfz. Therefore, ffz=fgz=gfz=ggz. Using (25), we get
(41)ζM,N(fz,ffz,kt)≥L*min{ζM,N(gz,gfz,t),ζM,N(gz,ffz,2t),ζM,N(fz,gz,t),ζM,N(fz,gfz,t),ζM,N(ffz,gfz,t)}=min{ζM,N(fz,ffz,t),ζM,N(fz,ffz,2t),1L*,ζM,N(fz,ffz,t),1L*}≥L*min{ζM,N(fz,ffz,t),ζM,N(fz,ffz,t),1L*,ζM,N(fz,ffz,t),1L*}=ζM,N(fz,ffz,t).
That is, fz=ffz. Hence, fz=ffz=gfz and fz is a common fixed point of f and g.
Similarly, we prove iflimn→∞fg(xn)=fz.
Suppose thatf and g are R-weakly commuting of type (Af). Again, as done above, we can easily prove that  fz is a common fixed point of f and g.
Finally, suppose thatf and g are R-weakly commuting of type (P). Weak reciprocal continuity of f and g implies that limn→∞fg(xn)=fz or limn→∞gf(xn)=gz. Let us assume that limn→∞gf(xn)=gz. Then, R-weak commutativity of type (P) of f and g yields
(42)ζM,N(ffxn,ggxn,t)≥L*ζM,N(fxn,gxn,tR).Takinglimitasn⟶∞,limn→∞ζM,N(ffxn,ggxn,t)≥L*ζM,N(z,z,tR)=1L*.
This gives limn→∞ζM,N(ffxn,ggxn,t)=1L*.
Using (24) and (26), we have gfxn-1=ggxn→gz as n→∞; this gives ffxn→gz as n→∞. Also, using (25) we get
(43)ζM,N(fz,ffxn,kt)≥L*min{ζM,N(gz,gfxn,t),ζM,N(gz,ffxn,2t),ζM,N(fz,gz,t),ζM,N(fz,gfxn,t),ζM,N(ffxn,gfxn,t)}.
Taking n→∞, we have
(44)ζM,N(fz,gz,kt)≥L*min{ζM,N(gz,gz,t),ζM,N(gz,gz,2t),ζM,N(fz,gz,t),ζM,N(fz,gz,t),ζM,N(gz,gz,t)},ζM,N(fz,gz,kt)≥L*min{1L*,1L*,ζM,N(fz,gz,t),ζM,N(fz,gz,t),1L*},ζM,N(fz,gz,kt)≥L*ζM,N(fz,gz,t).
By Lemma 17, we get fz=gz. Again, by using R-weak commutativity of type (P),
(45)ζM,N(ffz,ggz,t)≥L*ζM,N(fz,gz,tR)=1L*.
This yields ffz=ggz.
Therefore,ffz=fgz=gfz=ggz. Using (25), we get
(46)ζM,N(fz,ffz,kt)≥L*min{ζM,N(gz,gfz,t),ζM,N(gz,ffz,2t),ζM,N(fz,gz,t),ζM,N(fz,gfz,t),ζM,N(ffz,gfz,t)}=min{ζM,N(fz,ffz,t),ζM,N(fz,ffz,2t),ζM,N(fz,fz,t),ζM,N(fz,ffz,t),ζM,N(ffz,ffz,t)}≥L*min{ζM,N(fz,ffz,t),ζM,N(fz,ffz,t),ζM,N(fz,fz,t),ζM,N(fz,ffz,t),ζM,N(ffz,ffz,t)}=ζM,N(fz,ffz,t).
By Lemma 17, we get  fz=ffz. Hence, fz=ffz=gfz and fz is a common fixed point of f and g.
Similarly, we prove that iflimn→∞fg(xn)=fz.Uniqueness of the common fixed point theorem follows easily in each of the three cases by using (5).We now give an example to illustrate Theorem23.Example 24 (see [16]).
Let(X,ζM,N,𝒯)  be modified intuitionistic fuzzy metric space, where X=[2,20], as defined in  Example 22.
Definef,g:X→X by
(47)fx=2ifx=2orx>5,fx=6if2<x≤5,g2=2,gx=11if2<x≤5,gx=(x+1)3ifx>5.
Let {xn} be a sequence in X such that either  xn=2 or xn=5+1/n for each n.
Then, clearly,f and g satisfy all the conditions of Theorem 23 and have a unique common fixed point at x=2.Theorem 25.
Letf and g be weakly reciprocally continuous noncompatible self-mappings of a modified intuitionistic fuzzy metric space (X,ζM,N,𝒯) satisfying (24) and
(48)ζM,N(fx,fy,kt)≥L*ζM,N(gx,gy,t)
for all k≥0(49)ζM,N(fx,f2x,t)>L*max{(gx,f2x,t)ζM,N(gx,gfx,t),ζM,N(fx,gx,t),ζM,N(f2x,gfx,t),ζM,N(fx,gfx,t),ζM,N(gx,f2x,t)},
whenever fx≠f2x for all x,y∈X and t>0.
Iff and g are R-weakly commuting of type (Ag) or R-weakly commuting of type (Af) or R-weakly commuting of type (P), then f and g have common fixed point.Proof.
Sincef and g are noncompatible maps, there exists a sequence {xn} in X such that fxn→z and gxn→z for some z in X as n→∞ but either limn→∞ζM,N(fgxn,gfxn,t)≠1L* or the limit does not exist. Since f(X)⊂g(X), for each {xn} there exists {yn}, in X such that fxn=gyn. Thus, fxn→z,gxn→z and gyn→z as n→∞. By the virtue of this and using (48), we obtain
(50)ζM,N(fxn,fyn,kt)≥L*ζM,N(gxn,gyn,t),limn→∞ζM,N(z,fyn,kt)≥L*ζM,N(z,z,t)=1L*.
This gives fyn→z  as n→∞. Therefore, we have fxn→z,gxn→z,gyn→z,fyn→z.
Suppose thatf and g are R-weakly commuting of type (Ag). Then, weak reciprocal continuity of f and g implies that fgxn→fz or gfxn→gz. Similarly, fgyn→fz or gfyn→gz. Let us first assume that gfyn→gz. Then, R-weak commutativity of type (Ag) of f and g yields
(51)ζM,N(ffyn,gfyn,t)≥L*ζM,N(fyn,gyn,tR),limn→∞ζM,N(ffyn,gz,t)≥L*ζM,N(z,z,tR)=1L*.
This gives ffyn→gz. Using (48), we get
(52)ζM,N(ffyn,fz,kt)≥L*ζM,N(gfyn,gz,t)n⟶∞ζM,N(gz,fz,kt)≥L*ζM,N(gz,gz,t)=1L*.
This implies that fz=gz.  Again, by the virtue of R-weak commutativity of type (Ag),
(53)ζM,N(ffz,gfz,t)≥L*ζM,N(fz,gz,tR)=1L*.
This yields ffz=gfz and ffz=fgz=gfz=ggz. If fz≠ffz, then by using (49), we get
(54)ζM,N(fz,f2z,t)>L*max{(gz,f2z,t)ζM,N(gz,gfz,t),ζM,N(fz,gz,t),ζM,N(f2z,gfz,t),ζM,N(fz,gfz,t),ζM,N(gz,f2z,t)}=max{ζM,N(fz,ffz,t),ζM,N(fz,gz,t),ζM,N(ffz,gfz,t),ζM,N(fz,ffz,t),ζM,N(fz,ffz,t)},ζM,N(fz,f2z,t)>L*1L*,
which  is a contradiction. Hence, fz=ffz=gfz and fz is a common fixed point of f and g.
Similarly, we can prove that iffgyn→fz, then again fz is a common fixed point of f and g. Proof is similar if  f and g are R-weakly commuting of type (Af) or (P).We now give an example to illustrate Theorem25.Example 26 (see [16]).
Let(X,ζM,N,𝒯) be modified intuitionistic fuzzy metric space as defined in Example 22 and let {xn} be a sequence in X as in  Example 22.
Definef,g:X→X by
(55)fx=2ifx=2orx>5,fx=6if2<x≤5,g2=2,gx=11if2<x≤5,gx=(x+1)3ifx>5.
Let {xn} be a sequence in X such that either xn=2 or xn=5+1/n for each n. Then, clearly, f and g satisfy all the conditions of Theorem 25 and have a  common fixed point at x=2.Theorem 27.
Letf and g be weakly reciprocally continuous noncompatible self-mappings of a modified intuitionistic fuzzy metric space (X,ζM,N,𝒯) satisfying the conditions (24) and
(56)ζM,N(fx,fy,kt)≥L*ζM,N(gx,gy,t)∀k≥0,(57)ζM,N(fx,f2x,t)>L*ζM,N(gx,g2x,t),
whenever fx≠f2x for all x,y∈X and t>0.
Iff and g are R-weakly commuting of type (Ag) or R-weakly commuting of type (Af) or R-weakly commuting of type (P), then f and g have common fixed point.Proof.
Sincef and g are noncompatible maps, there exists a sequence {xn} in X such that fxn→z and gxn→z for some z in X  as n→∞ but either
(58)limn→∞ζM,N(fgxn,gfxn,t)≠1L*
or the limit does not exist.
Sincef(X)⊂g(X), for each {xn}, there exists {yn} in X such that fxn=gyn. Thus, fxn→z,gxn→z and gyn→z as n→∞. By the virtue of this and (56), we obtain fyn→z. Therefore, we have fxn→z,gxn→z,gyn→z,fyn→z.
Suppose thatf and g are R-weakly commuting of type (Ag). Then, weak reciprocal continuity of f and g implies that fgxn→fz or gfxn→gz. Similarly, fgyn→fz or gfyn→gz. Let us first assume that gfyn→gz. Then, R-weak commutativity of type (Ag) of f and g yields
(59)ζM,N(ffyn,gfyn,t)≥L*ζM,N(fyn,gyn,tR),limn→∞ζM,N(ffyn,gz,t)≥L*ζM,N(z,z,tR)=1L*.
This gives ffyn→gz. Using (56), we get
(60)ζM,N(ffyn,fz,kt)≥L*ζM,N(gfyn,gz,t).
Taking n→∞, we have
(61)ζM,N(gz,fz,kt)≥L*ζM,N(gz,gz,t)=1L*.
This implies that fz=gz.  Again, by the virtue of R-weak commutativity of type (Ag),(62)ζM,N(ffz,gfz,t)≥L*ζM,N(fz,gz,tR)=1L*.
This yields ffz=gfz and ffz=fgz=gfz=ggz. If fz≠ffz, then by using (57), we get
(63)ζM,N(fz,f2z,t)>L*ζM,N(gz,g2z,t)=ζM,N(fz,f2z,t),
which is a contradiction. Hence fz=ffz=gfz and fz is a common fixed point of f and g.
Similarly, we can prove that iffgyn→fz, then, again fz is a common fixed point of f and g. Proof is similar if f and g are R-weakly commuting of type (Af) or R-weakly commuting of type (P).We now give an example to illustrate Theorem27.Example 28 (see [16]).
Let(X,ζM,N,𝒯)  be modified intuitionistic fuzzy metric space as defined in Example 22 and let {xn} be a sequence in X as in Example 22.
Definef,g:X→X as follows:
(64)fx=2ifx=2orx>5,fx=4if2<x≤5,g2=2,gx=4if2<x≤5,gx=(x+1)3ifx>5.
Then, f and g satisfy all the conditions of Theorem 27 and have two common fixed points at x=2 and x=4.Remark 29.
Theorems23, 25, and 27 generalize the result of Pant et al. [16] for a pair of mappings in modified intuitionistic fuzzy metric space.


## 3.1. Section I: PointwiseR-weakly Commuting Pairs and Fixed Point

Lemma 19.
Let(X,ζM,N,𝒯) be a  modified intuitionistic fuzzy metric space, and let (A,S) and (B,T) be  pairs of self-mappings on X satisfying
(4)A(X)⊆T(X),B(X)⊆S(X),(5)thereexistsaconstantk∈(0,1)suchthatζM,N(Ax,By,kt)≥L*min{ζM,N(Ty,By,t),ζM,N(Sx,Ax,t),ζM,N(Sx,By,αt),ζM,N(Ty,Ax,(2-α)t),ζM,N(Ty,Sx,t)}
for all x,y∈X, t>0 and  α∈(0,2). Then, the continuity of one of the mappings in compatible pair (A,S) or (B,T) on (X,ζM,N,𝒯) implies their reciprocal continuity.Proof.
First, assume thatA and S are compatible and S is continuous. We show that A and S are reciprocally continuous. Let {un} be a sequence such that Aun→z and Sun→z for some z in X as n→∞. Since S is continuous, we have SAun→Sz and SSun→Sz as n→∞ and since (A,S) is compatible, we have for all t>0,
(6)limn→∞ζM,N(ASun,SAun,t)=1L*,limn→∞ζM,N(ASun,Sz,t)=1L*.
That is, ASun→Sz as n→∞. By (4), for each n, there exists vn in X such that ASun=Tvn. Thus, we have SSun→Sz, SAun→Sz, ASun→Sz, and Tvn→Sz as n→∞ whenever ASun=Tvn.
Now, we claim thatBvn→Sz as n→∞.
Suppose not, then, by (5), taking α=1,
(7)ζM,N(ASun,Bvn,kt)≥L*min{ζM,N(Tvn,Bvn,t),ζM,N(SSun,ASun,t),ζM,N(SSun,Bvn,t),ζM,N(Tvn,ASun,t),ζM,N(Tvn,SSun,t)}.
Taking n→∞, we have
(8)ζM,N(Sz,limn→∞Bvn,kt)≥L*min{ζM,N(Sz,limn→∞Bvn,t),ζM,N(Sz,Sz,t),ζM,N(Sz,limn→∞Bvn,t),ζM,N(Sz,Sz,t),ζM,N(Sz,Sz,t)limn→∞}=min{ζM,N(Sz,limn→∞Bvn,t),1L*,ζM,N(Sz,limn→∞Bvn,t),1L*,1L*}=ζM,N(Sz,limn→∞Bvn,t).
By Lemma 17, we have limn→∞Bvn=Sz.
Claim thatAz=Sz. Again, by (5), taking α=1,
(9)ζM,N(Az,Bvn,kt)≥L*min{ζM,N(Tvn,Bvn,t),ζM,N(Sz,Az,t),ζM,N(Sz,Bvn,t),ζM,N(Tvn,Az,t),ζM,N(Tvn,Sz,t)}.
Taking n→∞, we get
(10)ζM,N(Az,Sz,kt)≥L*min{ζM,N(Sz,Sz,t),ζM,N(Sz,Az,t),ζM,N(Sz,Sz,t),ζM,N(Sz,Az,t),ζM,N(Sz,Sz,t)}=min{1L*,ζM,N(Sz,Az,t),1L*,ζM,N(Sz,Az,t),1L*}=ζM,N(Sz,Az,t).
By Lemma 17, Az=Sz.
Therefore,SAun→Sz, ASun→Sz=Az as n→∞.
Hence,A and S are reciprocally continuous on X. If the pair (B,T) is assumed to be compatible and T is continuous, the proof is similar.Theorem 20.
Let(X,ζM,N,𝒯) be a complete modified intuitionistic fuzzy metric space. Further, let (A,S) and (B,T) be pointwise R-weakly commuting pairs of self-mappings of X satisfying (4), (5). If one of the mappings in compatible pair (A,S) or (B,T) is continuous,then A, B, S, and T have a unique common fixed point in X.Proof.
Letx0∈X. By (4), we define the sequences {xn} and {yn} in X such that for all n=0,1,2,…,
(11)y2n=Ax2n=Tx2n+1,y2n+1=Bx2n+1=Sx2n+2.
We show that {yn} is a Cauchy sequence in X. By (5) taking α=1-β,β∈(0,1), we have
(12)ζM,N(y2n+1,y2n+2,kt)=ζM,N(Bx2n+1,Ax2n+2,kt)=ζM,N(Ax2n+2,Bx2n+1,kt)≥L*min{ζM,N(Tx2n+1,Bx2n+1,t),ζM,N(Sx2n+2,Ax2n+2,t),ζM,N(Sx2n+2,Bx2n+1,(1-β)t),ζM,N(Tx2n+1,Ax2n+2,(1+β)t),ζM,N(Tx2n+1,Sx2n+2,t)}=min{ζM,N(y2n,y2n+1,t),ζM,N(y2n+1,y2n+2,t),ζM,N(y2n+1,y2n+1,(1-β)t),ζM,N(y2n,y2n+2,(1+β)t),ζM,N(y2n,y2n+1,t)}=min{ζM,N(y2n,y2n+1,t),ζM,N(y2n+1,y2n+2,t),1L*,ζM,N(y2n,y2n+1,t),ζM,N(y2n+1,y2n+2,βt),ζM,N(y2n,y2n+1,t)}≥L*min{ζM,N(y2n,y2n+1,t),ζM,N(y2n+1,y2n+2,t),ζM,N(y2n+1,y2n+2,βt)}.
Taking β→1, we have
(13)ζM,N(y2n+1,y2n+2,kt)≥L*min{ζM,N(y2n,y2n+1,t),ζM,N(y2n+1,y2n+2,t),ζM,N(y2n+1,y2n+2,t)},ζM,N(y2n+1,y2n+2,kt)≥L*min{ζM,N(y2n,y2n+1,t),ζM,N(y2n+1,y2n+2,t)}ζM,N(y2n+1,y2n+2,kt)≥L*ζM,N(y2n,y2n+1,t).
Similarly,
(14)ζM,N(y2n+2,y2n+3,kt)≥L*ζM,N(y2n+1,y2n+2,t).
Therefore, for any n and t, we have
(15)ζM,N(yn,yn+1,kt)≥L*ζM,N(yn-1,yn,t).
Hence, by Lemma 18, {yn} is a Cauchy sequence in X. Since X is complete, {yn} converges to z in X. Its subsequences {Ax2n}, {Tx2n+1}, {Bx2n+1}, and  {Sx2n+2} also converge to z.
Now, suppose that(A,S) is a compatible pair and S is continuous. Then, by Lemma 17, A and S are reciprocally continuous, then  SAxn→Sz, ASxn→Az as n→∞. As, (A,S), is a compatible pair. This implies that
(16)limn→∞ζM,N(ASxn,SAxn,t)=1L*,ζM,N(Az,Sz,t)=1L*.
Hence, Az=Sz.
SinceA(X)⊆T(X), there exists a point p in X such that Az=Tp=Sz.
By (5), taking α=1,
(17)ζM,N(Az,Bp,kt)≥L*min{ζM,N(Tp,Bp,t),ζM,N(Sz,Az,t),ζM,N(Sz,Bp,t),ζM,N(Tp,Az,t),ζM,N(Tp,Sz,t)},ζM,N(Az,Bp,kt)≥L*min{ζM,N(Az,Bp,t),ζM,N(Az,Az,t),ζM,N(Az,Bp,t),ζM,N(Az,Az,t),ζM,N(Az,Az,t)},ζM,N(Az,Bp,kt)≥L*ζM,N(Az,Bp,t).
Thus, by Lemma 17, we have Az=Bp.
Thus,Az=Bp=Sz=Tp.
SinceA and S are pointwise R-weakly commuting mappings, there exists R>0, such that
(18)ζM,N(ASz,SAz,t)≥L*ζM,N(Az,Sz,tR)=1L*.
Therefore, ASz=SAz and AAz=ASz=SAz=SSz.
Similarly, sinceB and T are pointwise R-weakly commuting mappings, we have BBp=BTp=TBp=TTp.
Again, by (5), taking α=1,
(19)ζM,N(AAz,Bp,kt)≥L*min{ζM,N(Tp,Bp,t),ζM,N(SAz,AAz,t),ζM,N(SAz,Bp,t),ζM,N(Tp,AAz,t),ζM,N(Tp,SAz,t)},ζM,N(AAz,Az,kt)≥L*min{ζM,N(Tp,Tp,t),ζM,N(AAz,AAz,t),ζM,N(AAz,Az,t),ζM,N(Az,AAz,t),ζM,N(Az,AAz,t)},ζM,N(AAz,Az,kt)≥L*ζM,N(AAz,Az,t).
By Lemma 17, we have AAz=Az=SAz.Hence Az is a common fixed point of A and S. Similarly, by (5), Bp=Az is a common fixed point of B and T. Hence, Az is a common fixed point of A, B, S, and T.For Uniqueness. Suppose that Ap(≠Az)  is another common fixed point of A, B, S, and T. Then, by (5), taking α=1,
(20)ζM,N(AAz,BAp,kt)≥L*min{ζM,N(TAp,BAp,t),ζM,N(SAz,AAz,t),ζM,N(SAz,BAp,t),ζM,N(TAp,AAz,t),M(TAp,SAz,t)},ζM,N(Az,Ap,kt)≥L*min{ζM,N(Ap,Ap,t),ζM,N(Az,Az,t),ζM,N(Az,Ap,t),ζM,N(Ap,Az,t),ζM,N(Ap,Az,t)},ζM,N(Az,Ap,kt)≥L*ζM,N(Az,Ap,t).
By Lemma 17,  Az=Ap.Thus, the uniqueness follows.Corollary 21.
Let(X,ζM,N,𝒯) be a complete modified intuitionistic fuzzy metric space. Further, let A and B be reciprocally continuous mappings on X satisfying
(21)ζM,N(Ax,By,kt)≥L*min{ζM,N(y,By,t),ζM,N(x,Ax,t),ζM,N(x,By,αt),ζM,N(y,Ax,(2-α)t),ζM,N(y,x,t)}
for all x,y∈X, t>0 and α∈(0,2), then pair A and B has a unique common fixed point.Example 22.
LetX=[2,20] and for each t>0,  define
(22)ζM,N(x,y,t)=(tt+|x-y|,|x-y|t+|x-y|).
Then, (X,ζM,N,𝒯) is complete modified intuitionistic fuzzy metric space. Let A,B,S, and T be self-mappings of  X defined as
(23)A(2)=2,Au=3ifu>2,B(u)=2ifu=2oru>5,Bu=6if2<u≤5,S(2)=2,S(u)=6ifu>2,T(2)=2,T(u)=12if2<u≤5,T(u)=u-3ifu>5.
Then, A,B,S, and T satisfy all the conditions of the above theorem with  k∈(0,1) and have a unique common fixed point u=2.


## 3.2. Section II: Weak Reciprocal Continuity and Fixed Point Theorem

Theorem 23.
Letf and g be weakly reciprocally continuous self-mappings of a complete modified intuitionistic fuzzy metric space (X,ζM,N,𝒯)  satisfying the following conditions:
(24)f(X)⊂g(X)(25)foranyx,y∈X,t>0,k∈(0,1)suchthatζM,N(fx,fy,kt)≥L*min{ζM,N(gx,gy,t),ζM,N(gx,fy,2t),ζM,N(fx,gx,t),ζM,N(fx,gy,t),ζM,N(fy,gy,t)}.
If f and g are either compatible or R-weakly commuting of type (Ag) or R-weakly commuting of type (Af) or R-weakly commuting of type (P), then f and g have a unique common fixed point.Proof.
Letx0 be any point in X. Then, as f(X)⊂g(X), there exist a sequence of points {xn} such that  f(xn)=g(xn+1).
Also, define a sequence{yn} in X as
(26)yn=f(xn)=g(xn+1).
Now, we show that {yn} is a Cauchy sequence in X. For proving this, by (25), we have
(27)ζM,N(yn,yn+1,kt)=ζM,N(fxn,fxn+1,kt)≥L*min{ζM,N(gxn,gxn+1,t),ζM,N(gxn,fxn+1,2t),ζM,N(fxn,gxn,t),ζM,N(fxn,gxn+1,t),ζM,N(fxn+1,gxn+1,t)}=min{ζM,N(yn-1,yn,t),ζM,N(yn-1,yn+1,2t),ζM,N(yn,yn-1,t),ζM,N(yn,yn,t),ζM,N(yn+1,yn,t)}≥L*min{ζM,N(yn-1,yn,t),ζM,N(yn-1,yn,t),ζM,N(yn,yn+1,t),ζM,N(yn,yn-1,t),1,ζM,N(yn+1,yn,t)}=min{ζM,N(yn-1,yn,t),ζM,N(yn,yn+1,t)}ζM,N(yn,yn+1,kt)≥L*ζM,N(yn-1,yn,t).
Then, by Lemma 18, {yn} is a Cauchy sequence in X.   As X is complete, there exists a point z in X such that limn→∞yn=z. Therefore, by (26), we have limn→∞yn=limn→∞f(xn)=limn→∞g(xn+1)=z.
Suppose thatf and g are compatible mappings. Now, by weak reciprocal continuity of f and g, it implies that limn→∞fg(xn)=fz or limn→∞gf(xn)=gz. Let limn→∞gf(xn)=gz, then the compatibility of f and g gives
(28)limn→∞ζM,N(fgxn,gfxn,t)=1L*.
This gives
(29)limn→∞ζM,N(fgxn,gz,t)=1L*.
Hence, limn→∞fg(xn)=gz. By (26), we get limn→∞fg(xn+1)=limn→∞ff(xn)=gz.
Therefore, by (25), we get
(30)ζM,N(fz,ffxn,kt)≥L*min{ζM,N(gz,gfxn,t),ζM,N(gz,ffxn,2t),ζM,N(fz,gz,t),ζM,N(fz,gfxn,t),ζM,N(ffxn,gfxn,t)}.
Taking n→∞, we get
(31)ζM,N(fz,gz,kt)≥L*min{ζM,N(gz,gz,t),ζM,N(gz,gz,2t),ζM,N(fz,gz,t),ζM,N(fz,gz,t),ζM,N(gz,gz,t)},ζM,N(fz,gz,kt)≥L*ζM,N(fz,gz,t).
Hence, by Lemma 17, we get fz=gz. Again, the compatibility of f and g implies commutativity at a coincidence point. Hence gfz=fgz=ffz=ggz. Using (25), we obtain
(32)ζM,N(fz,ffz,kt)≥L*min{ζM,N(gz,gfz,t),ζM,N(gz,ffz,2t),ζM,N(fz,gz,t),ζM,N(fz,gfz,t),ζM,N(ffz,gfz,t)}=min{ζM,N(fz,ffz,t),ζM,N(fz,ffz,2t),1L*,ζM,N(fz,ffz,t),1L*}≥min{ζM,N(fz,ffz,t),ζM,N(fz,ffz,t),1L*,ζM,N(fz,ffz,t),1L*},=ζM,N(fz,ffz,t).
That is, fz=ffz. Hence, fz=ffz=gfz and fz is a common fixed point of  f and g.
Next, suppose thatlimn→∞fg(xn)=fz. Then, f(X)⊂g(X) implies that fz=gu for some u∈X and, therefore, limn→∞fg(xn)=gu.
The compatibility off and g implies that limn→∞gf(xn)=gu. By the virtue of (26), this gives
(33)limn→∞fg(xn+1)=limn→∞ff(xn)=gu.
Using (25), we get
(34)ζM,N(fu,ffxn,kt)≥L*min{ζM,N(gu,gfxn,t),ζM,N(gu,ffxn,2t),ζM,N(fu,gu,t),ζM,N(fu,gfxn,t),ζM,N(ffxn,gfxn,t)}.
Taking n→∞, we have
(35)ζM,N(fu,gu,kt)≥L*min{ζM,N(gu,gu,t),ζM,N(gu,gu,2t),ζM,N(fu,gu,t),ζM,N(fu,gu,t),ζM,N(gu,gu,t)},ζM,N(fu,gu,kt)≥L*ζM,N(fu,gu,t).
By Lemma 17, we get fu=gu. The compatibility of f and g yields fgu=ggu=ffu=gfu. Finally, using (25), we obtain
(36)ζM,N(fu,ffu,kt)≥L*min{ζM,N(gu,gfu,t),ζM,N(gu,ffu,2t),ζM,N(fu,gu,t),ζM,N(fu,gfu,t),ζM,N(ffu,gfu,t)}=min{ζM,N(fu,ffu,t),ζM,N(fu,ffu,2t),1L*,ζM,N(fu,ffu,t),1L*}≥L*min{ζM,N(fu,ffu,t),ζM,N(fu,ffu,t),1L*,ζM,N(fu,ffu,t),1L*}=ζM,N(fu,ffu,t).
That is, fu=ffu. Hence fu=ffu=gfu and fu is a common fixed point of f and g.
Now, suppose thatf and g are R-weakly commuting of type (Ag). Now, weak reciprocal continuity of f and g implies that limn→∞fg(xn)=fz or limn→∞gf(xn)=gz. Let us first assume that limn→∞gf(xn)=gz. Then, R-weak commutativity of type (Ag) of f and g yields
(37)ζM,N(ffxn,gfxn,t)≥L*ζM,N(fxn,gxn,tR),limn→∞ζM,N(ffxn,gz,t)≥L*ζM,N(z,z,tR)=1L*.  This gives limn→∞ffxn=gz. Also, using (25), we get
(38)ζM,N(fz,ffxn,kt)≥L*min{ζM,N(gz,gfxn,t),ζM,N(gz,ffxn,2t),ζM,N(fz,gz,t),ζM,N(fz,gfxn,t),ζM,N(ffxn,gfxn,t)}.
Taking n→∞, we have
(39)ζM,N(fz,gz,kt)≥L*min{ζM,N(gz,gz,t),ζM,N(gz,gz,2t),ζM,N(fz,gz,t),ζM,N(fz,gz,t),ζM,N(gz,gz,t)},ζM,N(fz,gz,kt)≥L*ζM,N(fz,gz,t).
Hence, by Lemma 17, we get fz=gz. Again, by using  R-weak commutativity of type (Ag),(40)ζM,N(ffz,gfz,t)≥L*ζM,N(fz,gz,tR)=1L*.
This yieldsffz=gfz. Therefore, ffz=fgz=gfz=ggz. Using (25), we get
(41)ζM,N(fz,ffz,kt)≥L*min{ζM,N(gz,gfz,t),ζM,N(gz,ffz,2t),ζM,N(fz,gz,t),ζM,N(fz,gfz,t),ζM,N(ffz,gfz,t)}=min{ζM,N(fz,ffz,t),ζM,N(fz,ffz,2t),1L*,ζM,N(fz,ffz,t),1L*}≥L*min{ζM,N(fz,ffz,t),ζM,N(fz,ffz,t),1L*,ζM,N(fz,ffz,t),1L*}=ζM,N(fz,ffz,t).
That is, fz=ffz. Hence, fz=ffz=gfz and fz is a common fixed point of f and g.
Similarly, we prove iflimn→∞fg(xn)=fz.
Suppose thatf and g are R-weakly commuting of type (Af). Again, as done above, we can easily prove that  fz is a common fixed point of f and g.
Finally, suppose thatf and g are R-weakly commuting of type (P). Weak reciprocal continuity of f and g implies that limn→∞fg(xn)=fz or limn→∞gf(xn)=gz. Let us assume that limn→∞gf(xn)=gz. Then, R-weak commutativity of type (P) of f and g yields
(42)ζM,N(ffxn,ggxn,t)≥L*ζM,N(fxn,gxn,tR).Takinglimitasn⟶∞,limn→∞ζM,N(ffxn,ggxn,t)≥L*ζM,N(z,z,tR)=1L*.
This gives limn→∞ζM,N(ffxn,ggxn,t)=1L*.
Using (24) and (26), we have gfxn-1=ggxn→gz as n→∞; this gives ffxn→gz as n→∞. Also, using (25) we get
(43)ζM,N(fz,ffxn,kt)≥L*min{ζM,N(gz,gfxn,t),ζM,N(gz,ffxn,2t),ζM,N(fz,gz,t),ζM,N(fz,gfxn,t),ζM,N(ffxn,gfxn,t)}.
Taking n→∞, we have
(44)ζM,N(fz,gz,kt)≥L*min{ζM,N(gz,gz,t),ζM,N(gz,gz,2t),ζM,N(fz,gz,t),ζM,N(fz,gz,t),ζM,N(gz,gz,t)},ζM,N(fz,gz,kt)≥L*min{1L*,1L*,ζM,N(fz,gz,t),ζM,N(fz,gz,t),1L*},ζM,N(fz,gz,kt)≥L*ζM,N(fz,gz,t).
By Lemma 17, we get fz=gz. Again, by using R-weak commutativity of type (P),
(45)ζM,N(ffz,ggz,t)≥L*ζM,N(fz,gz,tR)=1L*.
This yields ffz=ggz.
Therefore,ffz=fgz=gfz=ggz. Using (25), we get
(46)ζM,N(fz,ffz,kt)≥L*min{ζM,N(gz,gfz,t),ζM,N(gz,ffz,2t),ζM,N(fz,gz,t),ζM,N(fz,gfz,t),ζM,N(ffz,gfz,t)}=min{ζM,N(fz,ffz,t),ζM,N(fz,ffz,2t),ζM,N(fz,fz,t),ζM,N(fz,ffz,t),ζM,N(ffz,ffz,t)}≥L*min{ζM,N(fz,ffz,t),ζM,N(fz,ffz,t),ζM,N(fz,fz,t),ζM,N(fz,ffz,t),ζM,N(ffz,ffz,t)}=ζM,N(fz,ffz,t).
By Lemma 17, we get  fz=ffz. Hence, fz=ffz=gfz and fz is a common fixed point of f and g.
Similarly, we prove that iflimn→∞fg(xn)=fz.Uniqueness of the common fixed point theorem follows easily in each of the three cases by using (5).We now give an example to illustrate Theorem23.Example 24 (see [16]).
Let(X,ζM,N,𝒯)  be modified intuitionistic fuzzy metric space, where X=[2,20], as defined in  Example 22.
Definef,g:X→X by
(47)fx=2ifx=2orx>5,fx=6if2<x≤5,g2=2,gx=11if2<x≤5,gx=(x+1)3ifx>5.
Let {xn} be a sequence in X such that either  xn=2 or xn=5+1/n for each n.
Then, clearly,f and g satisfy all the conditions of Theorem 23 and have a unique common fixed point at x=2.Theorem 25.
Letf and g be weakly reciprocally continuous noncompatible self-mappings of a modified intuitionistic fuzzy metric space (X,ζM,N,𝒯) satisfying (24) and
(48)ζM,N(fx,fy,kt)≥L*ζM,N(gx,gy,t)
for all k≥0(49)ζM,N(fx,f2x,t)>L*max{(gx,f2x,t)ζM,N(gx,gfx,t),ζM,N(fx,gx,t),ζM,N(f2x,gfx,t),ζM,N(fx,gfx,t),ζM,N(gx,f2x,t)},
whenever fx≠f2x for all x,y∈X and t>0.
Iff and g are R-weakly commuting of type (Ag) or R-weakly commuting of type (Af) or R-weakly commuting of type (P), then f and g have common fixed point.Proof.
Sincef and g are noncompatible maps, there exists a sequence {xn} in X such that fxn→z and gxn→z for some z in X as n→∞ but either limn→∞ζM,N(fgxn,gfxn,t)≠1L* or the limit does not exist. Since f(X)⊂g(X), for each {xn} there exists {yn}, in X such that fxn=gyn. Thus, fxn→z,gxn→z and gyn→z as n→∞. By the virtue of this and using (48), we obtain
(50)ζM,N(fxn,fyn,kt)≥L*ζM,N(gxn,gyn,t),limn→∞ζM,N(z,fyn,kt)≥L*ζM,N(z,z,t)=1L*.
This gives fyn→z  as n→∞. Therefore, we have fxn→z,gxn→z,gyn→z,fyn→z.
Suppose thatf and g are R-weakly commuting of type (Ag). Then, weak reciprocal continuity of f and g implies that fgxn→fz or gfxn→gz. Similarly, fgyn→fz or gfyn→gz. Let us first assume that gfyn→gz. Then, R-weak commutativity of type (Ag) of f and g yields
(51)ζM,N(ffyn,gfyn,t)≥L*ζM,N(fyn,gyn,tR),limn→∞ζM,N(ffyn,gz,t)≥L*ζM,N(z,z,tR)=1L*.
This gives ffyn→gz. Using (48), we get
(52)ζM,N(ffyn,fz,kt)≥L*ζM,N(gfyn,gz,t)n⟶∞ζM,N(gz,fz,kt)≥L*ζM,N(gz,gz,t)=1L*.
This implies that fz=gz.  Again, by the virtue of R-weak commutativity of type (Ag),
(53)ζM,N(ffz,gfz,t)≥L*ζM,N(fz,gz,tR)=1L*.
This yields ffz=gfz and ffz=fgz=gfz=ggz. If fz≠ffz, then by using (49), we get
(54)ζM,N(fz,f2z,t)>L*max{(gz,f2z,t)ζM,N(gz,gfz,t),ζM,N(fz,gz,t),ζM,N(f2z,gfz,t),ζM,N(fz,gfz,t),ζM,N(gz,f2z,t)}=max{ζM,N(fz,ffz,t),ζM,N(fz,gz,t),ζM,N(ffz,gfz,t),ζM,N(fz,ffz,t),ζM,N(fz,ffz,t)},ζM,N(fz,f2z,t)>L*1L*,
which  is a contradiction. Hence, fz=ffz=gfz and fz is a common fixed point of f and g.
Similarly, we can prove that iffgyn→fz, then again fz is a common fixed point of f and g. Proof is similar if  f and g are R-weakly commuting of type (Af) or (P).We now give an example to illustrate Theorem25.Example 26 (see [16]).
Let(X,ζM,N,𝒯) be modified intuitionistic fuzzy metric space as defined in Example 22 and let {xn} be a sequence in X as in  Example 22.
Definef,g:X→X by
(55)fx=2ifx=2orx>5,fx=6if2<x≤5,g2=2,gx=11if2<x≤5,gx=(x+1)3ifx>5.
Let {xn} be a sequence in X such that either xn=2 or xn=5+1/n for each n. Then, clearly, f and g satisfy all the conditions of Theorem 25 and have a  common fixed point at x=2.Theorem 27.
Letf and g be weakly reciprocally continuous noncompatible self-mappings of a modified intuitionistic fuzzy metric space (X,ζM,N,𝒯) satisfying the conditions (24) and
(56)ζM,N(fx,fy,kt)≥L*ζM,N(gx,gy,t)∀k≥0,(57)ζM,N(fx,f2x,t)>L*ζM,N(gx,g2x,t),
whenever fx≠f2x for all x,y∈X and t>0.
Iff and g are R-weakly commuting of type (Ag) or R-weakly commuting of type (Af) or R-weakly commuting of type (P), then f and g have common fixed point.Proof.
Sincef and g are noncompatible maps, there exists a sequence {xn} in X such that fxn→z and gxn→z for some z in X  as n→∞ but either
(58)limn→∞ζM,N(fgxn,gfxn,t)≠1L*
or the limit does not exist.
Sincef(X)⊂g(X), for each {xn}, there exists {yn} in X such that fxn=gyn. Thus, fxn→z,gxn→z and gyn→z as n→∞. By the virtue of this and (56), we obtain fyn→z. Therefore, we have fxn→z,gxn→z,gyn→z,fyn→z.
Suppose thatf and g are R-weakly commuting of type (Ag). Then, weak reciprocal continuity of f and g implies that fgxn→fz or gfxn→gz. Similarly, fgyn→fz or gfyn→gz. Let us first assume that gfyn→gz. Then, R-weak commutativity of type (Ag) of f and g yields
(59)ζM,N(ffyn,gfyn,t)≥L*ζM,N(fyn,gyn,tR),limn→∞ζM,N(ffyn,gz,t)≥L*ζM,N(z,z,tR)=1L*.
This gives ffyn→gz. Using (56), we get
(60)ζM,N(ffyn,fz,kt)≥L*ζM,N(gfyn,gz,t).
Taking n→∞, we have
(61)ζM,N(gz,fz,kt)≥L*ζM,N(gz,gz,t)=1L*.
This implies that fz=gz.  Again, by the virtue of R-weak commutativity of type (Ag),(62)ζM,N(ffz,gfz,t)≥L*ζM,N(fz,gz,tR)=1L*.
This yields ffz=gfz and ffz=fgz=gfz=ggz. If fz≠ffz, then by using (57), we get
(63)ζM,N(fz,f2z,t)>L*ζM,N(gz,g2z,t)=ζM,N(fz,f2z,t),
which is a contradiction. Hence fz=ffz=gfz and fz is a common fixed point of f and g.
Similarly, we can prove that iffgyn→fz, then, again fz is a common fixed point of f and g. Proof is similar if f and g are R-weakly commuting of type (Af) or R-weakly commuting of type (P).We now give an example to illustrate Theorem27.Example 28 (see [16]).
Let(X,ζM,N,𝒯)  be modified intuitionistic fuzzy metric space as defined in Example 22 and let {xn} be a sequence in X as in Example 22.
Definef,g:X→X as follows:
(64)fx=2ifx=2orx>5,fx=4if2<x≤5,g2=2,gx=4if2<x≤5,gx=(x+1)3ifx>5.
Then, f and g satisfy all the conditions of Theorem 27 and have two common fixed points at x=2 and x=4.Remark 29.
Theorems23, 25, and 27 generalize the result of Pant et al. [16] for a pair of mappings in modified intuitionistic fuzzy metric space.


## 4. Fixed Point and Implicit Relation

LetΘ denote the class of those functions θ:(L*)5→L*  such that θ is continuous  and  θ(x,1L*,1L*,x,x)=x.There are examples ofθ∈Θ:(1)
θ1(x1,x2,x3,x4,x5)=min{x1,x2,x3,x4,x5},(2)
θ2(x1,x2,x3,x4,x5)=x1(x1+x2+x3+x4+x5)/(x1+x4+x5+2L*),(3)
θ3(x1,x2,x3,x4,x5)=x1x4x53.Now, we prove our results using this implicit relation.Theorem 30.
Letf and g be weakly reciprocally continuous noncompatible self-mappings of a modified intuitionistic fuzzy metric space (X,ζM,N,𝒯)  satisfying the conditions:
(65)f(X)⊂g(X),(66)∫0ζM,N(fx,fy,t)ψ(s)ds≥L*∫0ζM,N(gx,gy,t)ψ(s)ds,(67)∫0ζM,N(fx,f2x,t)ψ(s)ds>∫0θ{ζM,N(gx,gfx,t),ζM,N(fx,gx,t),ζM,N(f2x,gfx,t),ζM,N(fx,gfx,t),ζM,N(gx,f2x,t)}ψ(s)ds,
whenever fx≠f2x for all x,y∈X, t>0 and for some θ∈Θ, where ψ:ℝ+→ℝ is a Lebesgue integrable mapping which is summable nonnegative and such that ∫0εψ(s)ds>0 for each ε>0.
Iff and g are R-weakly commuting of type (Ag) or R-weakly commuting of type (Af) or R-weakly commuting of type (P),  then f and g have common fixed point.Proof.
Sincef and g are noncompatible maps, there exists a sequence {xn} in X such that fxn→z and gxn→z for some z in X as n→∞ but either limn→∞ζM,N(fgxn,gfxn,t)≠1L* or the limit does not exist. Since f(X)⊂g(X), for each {xn}, there exists {yn} in X such that fxn=gyn. Thus, fxn→z,gxn→z and gyn→z as n→∞. By the virtue of this and using (66), we obtain
(68)∫0ζM,N(fxn,fyn,t)ψ(s)ds≥L*∫0ζM,N(gxn,gyn,t)ψ(s)ds.
Taking n→∞, we have
(69)∫0ζM,N(z,limn→∞fyn,t)ψ(s)ds≥L*∫0ζM,N(z,z,t)ψ(s)ds,
which implies that fyn→z  as n→∞. Therefore, we have fxn→z,gxn→z,gyn→z,fyn→z.
Suppose thatf and g are R-weakly commuting of type (Ag). Then, weak reciprocal continuity of f and g implies that fgxn→fz or gfxn→gz. Similarly, fgyn→fz or gfyn→gz. Let us first assume that gfyn→gz. Then R-weak commutativity of type (Ag) of f and g yields
(70)ζM,N(ffyn,gfyn,t)≥L*ζM,N(fyn,gyn,tR),limn→∞ζM,N(ffyn,gz,t)≥L*ζM,N(z,z,tR)=1L*.
This gives ffyn→gz. Using (66), we get
(71)∫0ζM,N(ffyn,fz,kt)ψ(s)ds≥L*∫0ζM,N(gfyn,gz,t)ψ(s)ds.
Taking n→∞, we have
(72)∫0ζM,N(gz,fz,kt)ψ(s)ds≥L*∫0ζM,N(gz,gz,t)ψ(s)ds.
This implies that fz=gz.  Again, by the virtue of R-weak commutativity of type (Ag),(73)ζM,N(ffz,gfz,t)≥L*ζM,N(fz,gz,tR)=1L*.
This yields ffz=gfz and ffz=fgz=gfz=ggz. If fz≠ffz, then by using (67), we get
(74)∫0ζM,N(fz,f2z,t)ψ(s)ds>L*∫0θ{ζM,N(gz,gfz,t),ζM,N(fz,gz,t),ζM,N(f2z,gfz,t),ζM,N(fz,gfz,t),ζM,N(gz,f2z,t)}ψ(s)ds=∫0θ{ζM,N(fz,ffz,t),ζM,N(fz,fz,t),ζM,N(ffz,ffz,t),ζM,N(fz,ffz,t),ζM,N(fz,ffz,t)}ψ(s)ds=∫0θ{ζM,N(fz,f2z,t),1L*,1L*,ζM,N(fz,f2z,t),ζM,N(fz,f2z,t)}ψ(s)ds=∫0ζM,N(fz,f2z,t)ψ(s)ds
which is a contradiction. Hence, fz=ffz=gfz and fz is a common fixed point of f and g.
Similarly, we can prove iffgyn→fz, then, again, fz is a common fixed point of f and g. Proof is similar if f and g are R-weakly commuting of type (Af) or (P).If we takeθ as θ1,θ2,θ3, then we get the following corollaries.Corollary 31.
Letf and g be weakly reciprocally continuous noncompatible self-mappings of a modified intuitionistic fuzzy metric space (X,ζM,N,𝒯) satisfying the conditions (65), (66) and
(75)∫0ζM,N(fx,f2x,t)ψ(s)ds>L*∫0min{ζM,N(gx,gfx,t),ζM,N(fx,gx,t),ζM,N(f2x,gfx,t),ζM,N(fx,gfx,t),ζM,N(gx,f2x,t)}ψ(s)ds,
whenever fx≠f2x for all x,y∈X, t>0, where ψ:ℝ+→ℝ is a Lebesgue integrable mapping which is summable nonnegative and such that ∫0εψ(s)ds>0 for each ε>0.
Iff and g are R-weakly commuting of type (Ag) or R-weakly commuting of type (Af) or R-weakly commuting of type (P),  then f and g have common fixed point.Corollary 32.
Letf and g be weakly reciprocally continuous noncompatible self-mappings of a modified intuitionistic fuzzy metric space (X,ζM,N,𝒯)  satisfying the conditions (65), (66) and (76)∫0ζM,N(fx,f2x,t)ψ(s)ds>L*∫0ζM,N(gx,gfx,t)(ζM,N(gx,gfx,t)+ζM,N(fx,gx,t)+ζM,N(f2x,gfx,t)+ζM,N(fx,gfx,t)+ζM,N(gx,f2x,t))/(ζM,N(gx,gfx,t)+ζM,N(fx,gfx,t)+ζM,N(gx,f2x,t)+2L*)ψ(s)ds,whenever fx≠f2x for all x,y∈X, t>0, where ψ:ℝ+→ℝ is a Lebesgue integrable mapping which is summable nonnegative and such that ∫0εψ(s)ds>0 for each ε>0.
Iff and g are R-weakly commuting of type (Ag) or R-weakly commuting of type (Af) or R-weakly commuting of type (P), then f and g have common fixed point.Corollary 33.
Letf and g be weakly reciprocally continuous noncompatible self-mappings of a modified intuitionistic fuzzy metric space (X,ζM,N,𝒯)  satisfying the conditions (65), (66), and
(77)∫0ζM,N(fx,f2x,t)ψ(s)ds>∫0ζM,N(gx,gfx,t)·ζM,N(fx,gfx,t)·ζM,N(gx,f2x,t)3ψ(s)ds,
whenever fx≠f2x for all x,y∈X, t>0, where ψ:ℝ+→ℝ is a Lebesgue integrable mapping which is summable nonnegative and such that ∫0εψ(s)ds>0 for each ε>0.
Iff and g are R-weakly commuting of type (Ag) or R-weakly commuting of type (Af) or R-weakly commuting of type (P), then f and g have common fixed point.
LetΔ denote the class of those functions δ:(L*)4→L* such that δ is continuous and δ(x,1L*,x,1L*)=x.
There are examples ofδ∈Δ:(1)
δ1(x1,x2,x3,x4)=min{x1,x2,x3,x4},(2)
δ2(x1,x2,x3,x4)=x1x3.Now, we prove our main results.Theorem 34.
Letf and g be weakly reciprocally continuous noncompatible self-mappings of a modified intuitionistic fuzzy metric space (X,ζM,N,𝒯)  satisfying the conditions (65), (66), and
(78)∫0ζM,N(fx,f2x,t)ψ(s)ds>L*∫0δ(ζM,N(gx,gfx,t),ζM,N(fx,gx,t),ζM,N(fx,gfx,t),ζM,N(f2x,gfx,t))ψ(s)ds,
whenever fx≠f2x for all x,y∈X, t>0 and for some δ∈Δ where ψ:ℝ+→ℝ is a Lebesgue integrable mapping which is summable nonnegative and such that ∫0εψ(s)ds>0 for each ε>0.
Iff and g are R-weakly commuting of type (Ag) or R-weakly commuting of type (Af) or R-weakly commuting of type (P), then f and g have common fixed point.Proof.
Sincef and g are noncompatible maps, there exists a sequence {xn} in X such that fxn→z and gxn→z for some z in X as n→∞ but either limn→∞ζM,N(fgxn,gfxn,t)≠1L* or the limit does not exist. Since f(X)⊂g(X), for each {xn}, there exists {yn} in X such that fxn=gyn. Thus fxn→z,gxn→z and gyn→z as n→∞. By the virtue of this and using (78), we obtain
(79)∫0ζM,N(fxn,fyn,t)ψ(s)ds≥L*∫0ζM,N(gxn,gyn,t)ψ(s)ds.
Taking n→∞, we have
(80)∫0ζM,N(z,limn→∞fyn,t)ψ(s)ds≥L*∫0ζM,N(z,z,t)ψ(s)ds
which implies that, fyn→z  as n→∞. Therefore, we have fxn→z,gxn→z,gyn→z,fyn→z.
Suppose thatf and g are R-weakly commuting of type (Ag). Then, weak reciprocal continuity of f and g implies that fgxn→fz or gfxn→gz. Similarly, fgyn→fz or gfyn→gz. Let us first assume that gfyn→gz. Then R-weak commutativity of type (Ag) of f and g yields
(81)ζM,N(ffyn,gfyn,t)≥L*ζM,N(fyn,gyn,tR),limn→∞ζM,N(ffyn,gz,t)≥L*ζM,N(z,z,tR)=1L*.
This gives ffyn→gz. Using (78), we get
(82)∫0ζM,N(ffyn,fz,kt)ψ(s)ds≥L*∫0ζM,N(gfyn,gz,t)ψ(s)ds.
Taking n→∞, we have
(83)∫0ζM,N(gz,fz,kt)ψ(s)ds≥L*∫0ζM,N(gz,gz,t)ψ(s)ds.
This implies that fz=gz.  Again, by the virtue of R-weak commutativity of type (Ag), ζM,N(ffz,gfz,t)≥L*ζM,N(fz,gz,t/R)=1L*.  This yields  ffz=gfz and ffz=fgz=gfz=ggz. If fz≠ffz, then by using (78), we get
(84)∫0ζM,N(fz,f2z,t)ψ(s)ds>L*∫0δ(ζM,N(gz,gfz,t),ζM,N(fz,gz,t),ζM,N(fz,gfz,t),ζM,N(f2z,gfz,t))ψ(s)ds=∫0δ(ζM,N(fz,f2z,t),ζM,N(fz,fz,t),ζM,N(fz,f2z,t),ζM,N(f2z,f2z,t))ψ(s)ds=∫0δ(ζM,N(fz,f2z,t),1L*,ζM,N(fz,f2z,t),1L*)ψ(s)ds=∫0ζM,N(fz,f2z,t)ψ(s)ds,
which is a contradiction. Hence, fz=ffz=gfz and fz is a common fixed point of f and g.
Similarly, we can prove that iffgyn→fz, then, again, fz is a common fixed point of f and g. Proof is similar if f and g are R-weakly commuting of type (Af) or (P).If we takeδ as δ1,δ2, then we get the following corollaries.Corollary 35.
Letf and g be weakly reciprocally continuous noncompatible self-mappings of a modified intuitionistic fuzzy metric space (X,ζM,N,𝒯) satisfying the conditions (65), (66), and
(85)∫0ζM,N(fx,f2x,t)ψ(s)ds>L*∫0min{(f2x,gfx,t)ζM,N(gx,gfx,t),ζM,N(fx,gx,t),ζM,N(fx,gfx,t),ζM,N(f2x,gfx,t)}ψ(s)ds,
whenever fx≠f2x for all x,y∈X, t>0, where ψ:ℝ+→ℝ is a Lebesgue integrable mapping which is summable nonnegative and such that ∫0εψ(s)ds>0 for each ε>0.
Iff and g are R-weakly commuting of type (Ag) or R-weakly commuting of type (Af) or R-weakly commuting of type (P), then f and g have common fixed point.Corollary 36.
Letf and g be weakly reciprocally continuous noncompatible self-mappings of a modified intuitionistic fuzzy metric space (X,ζM,N,𝒯)  satisfying the conditions (65), (66) and
(86)∫0ζM,N(fx,f2x,t)ψ(s)ds>L*∫0ζM,N(gx,gfx,t)·ζM,N(fx,gfx,t)ψ(s)ds,
whenever fx≠f2x for all x,y∈X, t>0 and for some δ∈Δ, where ψ:ℝ+→ℝ is a Lebesgue integrable mapping which is summable nonnegative and such that ∫0εψ(s)ds>0 for each ε>0.
Iff and g are R-weakly commuting of type (Ag) or R-weakly commuting of type (Af) or R-weakly commuting of type (P),  then f and g have common fixed point.



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*Source: 189321-2013-05-07.xml*