1. Introduction and preliminaries

^{Zadeh (1965)} investigated fuzzy set theory. Many authors utilized the concept of fuzzy set theory in metric space in number of ways. Banach contraction principal is the elate result of fixed point theory. Several authors have developed different contractive conditions to find fixed point in metric space (^{Dutta & Choudhury, 2008}; ^{Gupta & Mani, 2014}; ^{Gupta, Mani & Tripathi, 2012}; ^{Gupta, Saini, Mani & Tripathi, 2015}; ^{Song, 2007}; ^{Song & Xu, 2007}). ^{Kramosil and Michalek (1975)} defined fuzzy metric space using the concept of t-norm. ^{George and Veeramani (1994)} modified the notion of fuzzy metric spaces by using continuous t-norm. ^{Gregori and Sapena (2002)} also explored the Banach contraction principal to fuzzy contractive mapping on complete fuzzy metric space. Particularly, ^{Mihet (2008)} introduced the concept of fuzzy contractive mappings, which is one of the weak contractions in fuzzy metric space. In 1997, the concept of weak contraction was defined (^{Alber & Guerre-Delabriere, 1997}) for single valued maps on Hilbert spaces.

^{Rhoades (2001)} introduced weakly contractive mapping in metric space by defining a map T:X→X, which satisfy the condition d(Tx,Ty)≤d(x,y)-ϕ(d(x,y)), where x,y∈X and ϕ:[0,∞)→[0,∞) is a continuous and nondecreasing function such that ϕ(t)=0 if and only if t=0 .

^{Zhang and Song (2009)} proved a unique common fixedpoint theorem of hybrid generalized ϕ-weak contraction for two maps T,S on complete metric space X. The result is given below.

Theorem 1.1 (^{Zhang & Song, 2009}) Let (X,d) be a complete metric space and T,S:X→X be two mappings such that for all x,y∈X,

where ϕ:[0,∞)→[0,∞) is a lower semi-continuous function with ϕ(t)>0 for t∈(0,∞) and

ϕ(0)=0,

Then there exists a unique u∈X such that u=Tu=Su .

^{Doric (2009)} established a fixed point theorem which generalized the result of ^{Zhang and Song (2009)} using control functions, which is given below;

Theorem 1.2 (^{Doric, 2009}) Let (X,d) be a complete metric space and T,S:X→X be two mappings such that for all x,y∈X,

where

Then there exists a unique u∈X such that u=Tu=Su . The aim of our work is to prove above results in fuzzy metric spaces. The first theorem is the extension of the result of ^{Zhang and Song (2009)} under the different contractive conditions using control functions. Second result is analogous to the result of ^{Doric (2009)} in metric spaces.

Definition 1.1 (^{Schweizer & Sklar, 1960}) A binary operation *:[0,1]×[0,1]→[0,1] is continuous t- norm if * satisfies the following conditions

Definition 1.2 (^{George & Veeramani, 1994}) The 3-tuple (X,M,*) is said to be fuzzy metric space if X is an arbitrary set, * is continuous t -norm and M is fuzzy set on X2×[0,∞) satisfying the following conditions for all x,y,z∈X and s,t>0,

(FM-1) M(x,y,t)>0

(FM-2)M(x,y,t)=1 , ∀ t>0 iff x=y ;

(FM-3)M(x,y,t)=M(y,x,t);

(FM-4)M(x,y,t)*M(y,z,s)≤M(x,z,t+s);

(FM-5)  M(x,y,.):[0,∞)→[0,1] is continuous.

The triplet M(x,y,t) can be taken as the degree of nearness between x and y with respect to t≥0 .

Remark 1.1 (^{Shen, Qiu & Chen, 2013}) Since * is continuous, it follows from (FM-4) that the limit of the sequence in fuzzy metric space is uniquely determined.

Let (X,M,*) is a fuzzy metric space then the following condition also holds:

(FM-6) limt→∞⁡M(x,y,t)=1.

Lemma 1.1 (^{Grabiec, 1988}) In fuzzy metric space (X,M,*) , M(x,y,⋅) is non-decreasing for all x,y∈X.

Definition 1.3 (^{George & Veeramani, 1994}) Let (X,M,*) be a fuzzy metric space. Then a sequence xn ∈X is said to be

2. Main results

Theorem 2.1 Let (X,M,*) be a complete fuzzy metric space and T,S:X→X be two mappings such that for all x,y∈X ,

where ϕ:[0,1]→[0,1] is a upper semi-continuous function such that ϕ(t)>0 for t∈(0,1) and ϕ(1)=0, and

      Nx,y,t=min⁡{Mx,y,t,MTx,x,t,

then there exists a unique u∈X such that u=Tu=Su .

Proof. To prove our result, we follow the following steps.

Step- I We show that N(x,y,t)=1 if and only if x=y , is a common fixed point of T and S .

Infact, if x=y=Tx=Ty=Sx=Sy, then N(x,y,t)=1. Let N(x,y,t)=1 , then

Mx,y,t≥Nx,y,t,MTx,x,t≥Nx,y,t,M(Sy,y,t)≥N(x,y,t)

we have x=y=Tx=Ty=Sx=Sy.

Step- II Let x0 be a given point in X . We construct the sequence xn for n≥0 inductively as follows.

Choose a sequence xn∈X so that x2n+1=Sx2n , x2n+2=Tx2n+1 and prove that M(xn+1,xn,t)→1 as n→∞ .

Suppose that n is an odd number. Substituting x=xn and y=xn-1 in (5), we obtain

M(xn+1,xn,t)=M(Txn,Sxn-1,t)≥N(xn,xn-1,t)+ϕ(N(xn,xn-1,t))≥min⁡{M(xn,xn-1,t),M(Txn,xn,t),M(Sxn-1,xn-1,t),M(xn-1,Txn,t)*M(xn,Txn-1,t)}min⁡{M(xn,xn-1,t),M(xn+1,xn,t),M(xn,xn-1,t),M(xn-1,xn,t)*M(xn,xn-1,t)},

which implies that M(x_n+1,x_n,t)≥M(x_n,x_n-1,t),

If we take M(xn+1,xn,t)<M(xn,xn-1,t),

Then N(x_n,x_n-1,t)=M(x_n+1,x_n,t) and moreover

Mxn+1,xn,t≥Mxn+1,xn,t+∅(Mxn+1,xn,t),

this is a contradiction. Therefore we have

Similarly, we can also obtain inequalities (7) in case when n is an even number.

So, the sequence {Mxn+1,xn,t} is non-decreasing sequence and bounded above. So there exists r>0, such that limn→∞M(xn+1,xn,t)=limn→∞M(xn,xn-1,t)=r, then upper continuity of ϕ implies that ϕr≤limn→∞supϕMxn,xn-1,t.

We claim that r=1 In fact, taking upper limit as n→∞ on either side of the following inequality, we have

Mxn+1,xn,t≥Nxn,xn-1,t+ϕ(Nxn,xn-1,t)

r≥r+limn→∞sup⁡ϕMxn,xn-1,t,r≥r+ϕ(r).

this is a contradiction unless ϕ(r)=0 at r=1.

Hence

Step -III Next we prove that xn is a Cauchy sequence. To prove this, it is sufficient to prove that the sub-sequence x2n of xn is a Cauchy sequence. Suppose opposite that the sequence xn is not a Cauchy sequence. Then there exists ϵ>0 such that n(k) is the smallest index for which n(k)>m(k)>k , we have M(x2m(k),x2n(k),t)≤ϵ.

Therefore,

M(x2m(k),x2n(k),t)=M(Tx2m-1,Sx2n-1,t)≥N(x2m-1,x2n-1,t)+ϕ(N(x2m-1,x2n-1,t),

this gives ϵ≥ϵ+ϕ(ϵ),

this gives a contradiction with ∈>0.

Thus x2n is a Cauchy equence and hence xn is a Cauchy sequence.

In complete fuzzy metric space (X,M,*) , there exists u∈X such that sequence xn→u as n→∞ .

Step - IV Now we prove that u is a fixed point of T and S . For this suppose that u≠Tu , then for M(u,Tu,t)<1 , there exist N1∈N such that for any n≥N1

M(x2n+1,u,t)>M(u,Tu,t),M(x2n,u,t)>M(u,Tu,t),M(x2n,x2n+1,t)>M(u,Tu,t).

Under this consideration, we have

M(u,Tu,t)≥N(u,xn,t)=min⁡{M(u,x2n,t),M(Tu,u,t),M(Sx2n,x2n,t),   M(x2n,Tu,t)*M(u,Sx2n,t)}=min⁡{M(u,x2n,t),M(Tu,u,t),M(x2n+1,x2n,t), M(x2n,Tu,t)*M(u,x2n+1,t)}.

Letting n→∞ , we have

M(u,Tu,t)>min{M(u,u,t),M(Tu,u,t),

M(u,u,t), M(u,Tu,t)*1}=M(u,Tu,t),

That is

Since

M(Tu,x2n+1,t)=M(Tu,Sx2n,t)≥N(u,x2n,t)+ϕ(N(u,x2n,t)),

then letting n→∞ , we have

M(Tu,u,t)≥M(Tu,u,t)+ϕ(M(Tu,u,t)),

we obtain a contradiction. Hence Tu=u .

Also, Mu,Su,t=MTu,Su,t≥Nu,u,t +ϕ(N(u,u,t))=M(u,u,t)+ϕ(M(u,u,t))

implies Su=u

Suppose there exists another fixed point v∈X such that v=Tv=Sv , then using an argument similar to the above, we get

M(u,v,t)=M(Tu,Sv,t)≥N(u,v,t)+ϕ(N(u,v,t))≥M(u,v,t)+ϕ(M(u,v,t)).

Hence u=v . The proof is completed.

Corollary 2.1 Let (X,M,*) be a complete fuzzy metric space and T:X→X be mapping such that for all x,y∈X ,

M(Tx,Ty,t)≥N(x,y,t)+ϕ(N(x,y,t)),

where ϕ:[0,1]→[0,1] is a upper semi-continuous function such that ϕ(t)>0 for t∈(0,1) and ϕ(1)=0 , and

then there exists a unique u∈X such that u=Tu .

Example 2.1 Let (X,M,*) be a complete fuzzy metric space with metric d(x,y)=x-y and X=[0,1] . Let

Tx=x2 and Sx=x16

for each x∈[0,1] . Then

N(x,y,t)=min⁡tt+x-y,tt+x2,tt+y,tt+x-y*tt+x,=tt+x-y  x2≤y≤xtt+y   y≥x.=min⁡tt+x-y,tt+x2-x,tt+y16-y,tt+y-x2*tt+x-y16,=tt+y-x2*tt+x-y16 0≤y≤x.

For ϕ(t)=t2-1 , it is easy to show that

M(Tx,Sy,t)≥N(x,y,t)+ϕ(N(x,y,t))

for all x,y∈X . One can show that all the condition of Theorem 2.1 fulfill and T,S satisfy the Theorem 2.1.

Theorem 2.2 Let (X,M,*) be a complete fuzzy metric space and T,S:X→X be two mappings such that for all x,y∈X ,

where,

1. ψ:[0,1]→[0,1] is a continuous monotone non-decreasing function with ψ(t)=1 iff t=1 ,

2. ϕ:[0,1]→[0,1] is a upper semi-continuous function ϕ(t)>0 for t∈(0,1) and ϕ(1)=0 , and

N(x,y,t)=min⁡Mx,y,t,MTx,x,t,MSy,y,t,M(y,Tx,t)*M(x,Sy,t),

then there exists a unique u∈X such that u=Tu=Su .

Proof. For any x0∈X , we construct a sequence xn for n≥0 as x2n+1=Sx2n, x2n=Tx2n+1 and will prove that M(xn,xn-1,t)→1 as n→∞. Suppose that n is an odd number, substituting x=xn and y=xn-1 in Equation (11), we obtain

ψ(M(xn+1,xn,t))=ψ(M(Txn,Sxn-1,t))

                    ≥ψ(N(xn,xn-1,t))+ϕ(N(xn,xn-1,t)).

ψ(M(xn+1,xn,t))≥ψ(N(xn,xn-1,t)),

this implies that

From triangle inequality, we have

   Nxn,xn-1,t=min⁡{Mxn,xn-1,t,Mxn+1,xn,t,Mxn,xn-1,t,M(xn-1,xn+1,t)*M(xn,xn,t)}

= Mxn,xn-1,t,Mxn+1,xn,t,M(xn-1,xn+1,t) ≥min⁡{Mxn,xn-1,t,Mxn+1,xn,t, M(xn-1,xn,t)*M(xn,xn+1,t)}.

If we consider

Nxn,xn-1,t≥Mxn+1,xn,t≥N(xn,xn-1,t)≥M(xn+1,xn,t),

it further implies that

ψMxn+1,xn,t≥ψMxn+1,xn,t+ϕ(M(xn+1,xn,t)),

this is a contradiction. So we have

Mxn+1,xn,t≥Nxn,xn-1,t≥Mxn,xn-1,t                                          

this implies

Similarly we can prove that Equation (13) is true when n is even number.

Therefore the sequence {Mxn+1,xn,t} is monotonically non decreasing and bounded sequence. Therefore we can write

limn→∞M(xn+1,xn,t)=limn→∞M(xn,xn-1,t)=r,

Letting n→∞ in the given inequality (11), we get

ѱMxn+1,xn,t≥ѱMxn,xn-1,t+∅Mxn,xn-1,t,ѱr≥ѱr+∅r,

which is contradiction, unless r=1 and at Ǿ(r) =0 Hence

Next we prove that {x _n } is a Cauchy sequence. To prove this it is sufficient to prove that sub-sequence {x _2n } of {x _n } is a Cauchy sequence. Suppose {x _2n } is not a Cauchy sequence, then there existsor which we can find sub-sequence {x _2m(k) } and {x _2n(k) } such that n(k) is the smallest index for which, n(k)>m(k)>k we have

This means that Mx2mk,x2nk-2,t>ò . From Equation (15) and the triangle inequality, we have ò≥Mx2mk,x2nk,t

≥Mx2mk,x2nk-2,t *Mx2n(k)-2),x2nk-1,t*Mx2nk-1,x2nk,t,>ò*Mx2nk-2,x2nk-1,t*Mx2nk-1,x2nk,t,

Moreover

Mx2mk,x2nk+1,t≥Mx2nk,x2nk+1,t*Mx2mk,x2nk,t,Mx2mk-1,x2nk,t≥Mx2mk,x2nk,t*Mx2mk,x2mk-1,t.                         

Using (13) and (14), we have

Also,

Using (13) and (16), we have

Again, using (14)-(18), we have limn→∞M(x2m(k)-1,x2nk,t)=ò

Putting x= _x2m(k)-1 , y=x _2n(k) in (11), one can get

ѱMx2mk,x2nk+1,t=ѱMTx2mk-1,Sx2nk,t≥ѱMx2mk-1,x2nk,t+ϕMx2mk-1,x2nk,t,

letting k→ ∞ and using (16) and (18), we get

ѱò≥ѱò+ϕò,

this is contradiction with ò > 0. Thus {x _2n } is a Cauchy sequence and hence {x _n } is a Cauchy sequence. In complete metric space x, there exists z such that x_n→z as n→∞.

Let us now prove that z is a fixed point for T and S.

ѱMTz,x2nk+1,t=ѱMTz,Sx2nk,t≥ ѱNz,x2nk,t+ϕNz,x2nk,t,

using the same argument as in Theorem 2.1 (eq-9)and letting n→ ∞, we obtain

ѱMTz,z,t≥ ѱMTz,z,t+ϕMTz,z,t,

this implies ѱ(M(Tz,z,t))=1. Hence M(Tz,z,t)=1 gives that z is a fixed point of T.

Thus, we have

ѱMz,Sz,t= ѱ(MTz,Sz,t)≥ѱNz,z,t+ϕ(Nz,z,t)=ѱMz,Sz,t+ϕMz,Sz,t,

this implies ѱ(M(Sz,z,t))=1. Hence M(Sz,z,t)=1 or Sz = z.

To prove uniqueness, we consider another fixed point w ∊ X then

ѱMz,w,t=ѱ(MTz,Sw,t)≥ѱMz,w,t+ϕ(Nz,w,t)=ѱMz,w,t+ϕ(Mz,w,t)

Thus ѱ(M(z,w,t))=1

Hence M(z,w,t)=1 or z=w This completes the proof.

Corollary 2.2 Let (X,M,*) be a complete fuzzy metric space and T:X→X be a mapping such that for all x,y, ε X,

where

N(x, y, t) = min {M (x, y, t), M(Tx, x, t), M(Ty, y, t),

M(y, Tx, t) ∗ M(x, Ty, t)},

then there exists a unique u ∊ X such that u = Tu.

Example 2.2 Let (X,M,*) be a complete fuzzy metric space with metric dx,y=|x-y| and x=[0,1]. Let Tx=x2 and Sx=0 for each x ∊ [0,1] Then

Nx,y,t=min⁡tt+x-y,tt+x2,tt+y,tt+x-y*tt+x,={tt+|x-y| x2≤y≤xtt+y y≥x.

For ϕ(t)=t and ѱ(t)=3t, it is easy to show that

ѱMTx,Sy,t≥ѱNx,y,t+ϕNx,y,t,

All the conditions of Theorem 2.2 are satisfied.

3. Conclusions

In this paper, the use of control functions has renewed the possibility of establishing new results in fuzzy metric fixed point theory. We extend the existing result in metric space towards fuzzy metric space with a new approach of using control functions.