1 Introduction and preliminaries

The evolution of fuzzy mathematics commenced with an introduction of the notion of fuzzy sets by Zadeh [1] in 1965 as a new way to represent vagueness in every day life. The idea of intuitionistic fuzzy sets (IFS) was introduced by Atanassov [2]. Saadati and Park [3, 4] introduced intuitionistic fuzzy normed spaces (IFNS). For the detailed survey on fixed point results in fuzzy metric spaces, fuzzy normed spaces and IFNS, we refer the reader to [58]. Recently coupled fixed point theorems have been proved in IFNS; for details of these we refer to Gordji [9] and Sintunavarat et al. [10]. More recently, tripled fixed point theorems have been introduced in partially ordered metric spaces by Berinde [11]. In this paper, we have proved tripled fixed point and tripled coincidence point theorems in IFNS. Now we give some definitions, examples and lemmas for our main results.

For the sake of completeness, we recall some definitions and known results in a fuzzy metric space.

Definition 1.1 ([1])

Let X be any set. A fuzzy set A in X is a function with domain X and values in [0,1].

Definition 1.2 ([12])

A binary operation :[0,1]×[0,1][0,1] is called a continuous t-norm if

  1. (1)

    ∗ is associative and commutative;

  2. (2)

    ∗ is continuous;

  3. (3)

    a1=a for all a[0,1];

  4. (4)

    abcd whenever ac and bd.

Example 1.3 Three typical examples of continuous t-norms are ab=min{a,b} (minimum t-norm), ab=ab (product t-norm), and ab=max{a+b1,0} (Lukasiewicz t-norm).

Definition 1.4 ([12])

A binary operation :[0,1]×[0,1][0,1] is called a continuous t-conorm if

  1. (1)

    ⋄ is associative and commutative;

  2. (2)

    ⋄ is continuous;

  3. (3)

    a0=a for all a[0,1];

  4. (4)

    abcd whenever ac and bd.

Example 1.5 Two typical examples of continuous t-conorms are ab=min{a+b,1} and ab=max{a,b}.

Using the continuous t-norm and continuous t-conorm, Saadati and Park [3] introduced the concept of intuitionistic fuzzy normed spaces.

Definition 1.6 ([3])

The 5-tuple (X,μ,υ,,) is called an intuitionistic fuzzy normed space (for short, IFNS) if X is a vector space, ∗ and ⋄ are continuous t-norm and continuous t-conorm respectively and μ, υ are fuzzy sets on X×(0,) satisfying the following conditions: for all x,yX and s,t>0,

(IF1) μ(x,t)+υ(x,t)1;

(IF2) μ(x,t)>0;

(IF3) μ(x,t)=1 if and only if x=0;

(IF4) μ(αx,t)=μ(x, t | α | ) for all α0;

(IF5) μ(x,t)μ(y,s)μ(x+y,t+s);

(IF6) μ(x,):(0,)[0,1] is continuous;

(IF7) μ is a non-decreasing function on R + ,

lim t μ(x,t)=1and lim t 0 μ(x,t)=0,xX,t>0;

(IF8) υ(x,t)<1;

(IF9) υ(x,t)=0 if and only if x=0;

(IF10) υ(αx,t)=υ(x, t | α | ) for all α0;

(IF11) υ(x,t)υ(y,s)υ(x+y,t+s);

(IF12) υ(x,):(0,)[0,1] is continuous;

(IF13) υ is a non-increasing function on R + ,

lim t υ(x,t)=0and lim t 0 υ(x,t)=1,xX,t>0.

In this case (μ,υ) is called an intuitionistic fuzzy norm.

Definition 1.7 ([3])

Let (X,μ,υ,,) be an IFNS. A sequence { x n } in X is said to be:

  1. (1)

    convergent to a point xX with respect to an intuitionistic fuzzy norm (μ,υ) if for any ϵ>0 and t>0, there exists kN such that

    μ( x n x,t)>1ϵandυ( x n x,t)<ϵ,nk.

In this case, we write (μ,υ) lim n x n =x.

  1. (2)

    Cauchy sequence with respect to an intuitionistic fuzzy norm (μ,υ) if for any ϵ>0 and t>0, there exists kN such that

    μ( x n x m ,t)>1ϵandυ( x n x m ,t)<ϵ,n,mk.

Definition 1.8 ([3])

An IFNS (X,μ,υ,,) is said to be complete if every Cauchy sequence in (X,μ,υ,,) is convergent.

Definition 1.9 ([13, 14])

Let X and Y be two IFNS. A function g:XY is said to be continuous at a point x 0 X if for any sequence { x n } in X converging to a point x 0 X, the sequence {g( x n )} in Y converges to g( x 0 )Y. If g is continuous at each xX, then g:XY is said to be continuous on X.

Examples 1.10 Let (X,) be an ordinary normed space and ϕ be an increasing and continuous function from R + into (0,1) such that lim t ϕ(t)=1. Four typical examples of these functions are as follows:

ϕ(t)= t t + 1 ,ϕ(t)=sin ( π t 2 t + 1 ) ,ϕ(t)=1 e t ,ϕ(t)= e 1 t .

Let ∗ and ⋄ be a continuous t-norm and a continuous t-conorm such that

abababfor all a,b[0,1].

For any t(0,), we define

μ(x,t)= [ ϕ ( t ) ] x ,υ(x,t)=1 [ ϕ ( t ) ] x ,xX,

then (X,μ,υ,,) is an IFNS.

For further details regarding IFNS, we refer to [3].

Definition 1.11 ([9])

Let (X,μ,υ,,) be an IFNS. (μ,υ) is said to satisfy the n-property on X×(0,) if

lim n [ μ ( x , k n t ) ] n p =1, lim n [ υ ( x , k n t ) ] n p =0,

where xX, p>0, and k>1.

Throughout this paper, we assume that (μ,υ) satisfies the n-property on X×(0,).

Definition 1.12 ([11])

Let X be a non-empty set. An element (x,y,z)X×X×X is called a tripled fixed point of F:X×X×XX if

x=F(x,y,z),y=F(y,x,y)andz=F(z,y,x).

Definition 1.13 Let X be a non-empty set. An element (x,y,z)X×X×X is called a tripled coincidence point of mappings F:X×X×XX and g:XX if

g(x)=F(x,y,z),g(y)=F(y,x,y),andg(z)=F(z,y,x).

Definition 1.14 ([11])

Let (X,) be a partially ordered set. A mapping F:X×X×XX is said to have the mixed monotone property if F is monotone non-decreasing in its first and third argument and is monotone non-increasing in its second argument; that is, for any x,y,zX

and

z 1 , z 2 X, z 1 z 2 F(x,y, z 1 )F(x,y, z 2 ).

Definition 1.15 Let (X,) be a partially ordered set, and g:XX. A mapping F:X×X×XX is said to have the mixed g-monotone property if F is monotone g-non-decreasing in its first and third argument and is monotone g-non-increasing in its second argument; that is, for any x,y,zX,

and

z 1 , z 2 X,g( z 1 )g( z 2 )F(x,y, z 1 )F(x,y, z 2 ).

Lemma 1.16 ([15])

Let X be a non-empty set and g:XX be a mapping. Then there exists a subset EX such that g(E)=g(X) and g:EX is one-to-one.

2 Main results

Theorem 2.1 Let (X,μ,υ,,) be a complete IFNS, ⪯ be a partial order on X and suppose that

ababandaa=a
(2.1)

for all a,b[0,1]. Suppose that F:X×X×XX has the mixed monotone property and

μ ( F ( x , y , z ) F ( u , v , w ) , k t ) μ ( x u , t ) μ ( y v , t ) μ ( z w , t ) , υ ( F ( x , y , z ) F ( u , v , w ) , k t ) υ ( x u , t ) υ ( y v , t ) υ ( z w , t )
(2.2)

for all those x, y, z, u, v, w in X for which xu, yv, zw, where 0<k<1. If either

  1. (a)

    F is continuous or

  2. (b)

    X has the following property:

(bi) if { x n } is a non-decreasing sequence and (μ,υ) lim n x n =x, then x n x for all nN,

(bii) if { y n } is a non-decreasing sequence and (μ,υ) lim n y n =y, then y n y for all nN,

(biii) if { z n } is a non-decreasing sequence and (μ,υ) lim n z n =y, then z n z for all nN,

then F has a tripled fixed point provided that there exist x 0 , y 0 , z 0 X such that

x 0 F( x 0 , y 0 , z 0 ), y 0 F( y 0 , x 0 , y 0 ), z 0 F( z 0 , y 0 , x 0 ).

Proof Let x 0 , y 0 , z 0 X be such that

x 0 F( x 0 , y 0 , z 0 ), y 0 F( y 0 , x 0 , y 0 ), z 0 F( z 0 , y 0 , x 0 ).

As F(X×X×X)X, so we can construct sequences { x n }, { y n } and { z n } in X such that

x n + 1 = F ( x n , y n , z n ) , y n + 1 = F ( y n , x n , y n ) , z n + 1 = F ( z n , y n , x n ) , n 0 .
(2.3)

Now we show that

x n x n + 1 , y n y n + 1 , z n z n + 1 ,n0.
(2.4)

Since

x 0 F( x 0 , y 0 , z 0 ), y 0 F( y 0 , x 0 , y 0 ), z 0 F( z 0 , y 0 , x 0 ),

(2.4) holds for n=0. Suppose that (2.4) holds for any n0. That is,

x n x n + 1 , y n y n + 1 , z n z n + 1 .
(2.5)

As F has the mixed monotone property so by (2.5) we obtain

{ F ( x n , y , z ) F ( x n + 1 , y , z ) , (i) F ( x , y n , z ) F ( x , y n + 1 , z ) , (ii) F ( x , y , z n ) F ( x , y , z n + 1 ) , (iii) |

which on replacing y by y n and z by z n in (i) implies that F( x n , y n , z n )F( x n + 1 , y n , z n ); replacing x by x n + 1 and z by z n in (ii), we obtain F( x n + 1 , y n , z n )F( x n + 1 , y n + 1 , z n ); replacing y by y n + 1 and x by x n + 1 in (iii), we get F( x n + 1 , y n + 1 , z n )F( x n + 1 , y n + 1 , z n + 1 ). Thus, we have F( x n , y n , z n )F( x n + 1 , y n + 1 , z n + 1 ), that is, x n + 1 x n + 2 . Similarly, we have

{ F ( y , x , y n + 1 ) F ( y , x , y n ) , (iv) F ( y n + 1 , x , y ) F ( y n , x , y ) , (v) F ( y , x n + 1 , y ) F ( y , x n , y ) , (vi) |

which on replacing y by y n + 1 and x by x n + 1 in (iv) implies that F( y n + 1 , x n + 1 , y n + 1 )F( y n + 1 , x n + 1 , y n ); replacing x by x n + 1 and y by y n + 1 in (v), we obtain F( y n + 1 , x n + 1 , y n )F( y n , x n + 1 , y n ); replacing y by y n in (vi), we get F( y n , x n + 1 , y n )F( y n , x n , y n ). Thus, we have F( y n + 1 , x n + 1 , y n + 1 )F( y n , x n , y n ), that is, y n + 2 y n + 1 . Similarly, we have

{ F ( z n , y , x ) F ( z n + 1 , y , x ) , (vii) F ( z , y n , x ) F ( z , y n + 1 , x ) , (viii) F ( z , y , x n ) F ( z , y , x n + 1 ) , (xi) |

which on replacing y by y n and x by x n in (vii) implies that F( z n , y n , x n )F( z n + 1 , y n , x n ); replacing x by x n and z by z n + 1 in (viii), we obtain F( z n + 1 , y n , x n )F( z n + 1 , y n + 1 , x n ); replacing y by y n + 1 and z by z n + 1 in (xi), we get F( z n + 1 , y n + 1 , x n )F( z n + 1 , y n + 1 , x n + 1 ). Thus, we have F( z n , y n , x n )F( z n + 1 , y n + 1 , x n + 1 ), that is, z n + 1 z n + 2 . So, by induction, we conclude that (2.5) holds for all n0, that is,

(2.6)
(2.7)
(2.8)

Define

α n (t)=μ( x n x n + 1 ,t)μ( y n y n + 1 ,t)μ( z n z n + 1 ,t).
(2.9)

Consider

μ ( x n x n + 1 , k t ) = μ ( F ( x n 1 , y n 1 , z n 1 ) F ( x n , y n , z n ) , k t ) μ ( x n 1 x n , t ) μ ( y n 1 y n , t ) μ ( z n 1 z n , t ) = α n 1 ( t ) .
(2.10)

Also,

μ ( z n z n + 1 , k t ) = μ ( F ( z n 1 , y n 1 , x n 1 ) F ( z n , y n , x n ) , k t ) μ ( z n 1 z n , t ) μ ( y n 1 y n , t ) μ ( x n 1 x n , t ) = μ ( x n 1 x n , t ) μ ( y n 1 y n , t ) μ ( z n 1 z n , t ) = α n 1 ( t ) .
(2.11)

Now,

μ ( y n y n + 1 , k t ) = μ ( F ( y n 1 , x n 1 , y n 1 ) F ( y n , x n , y n ) , k t ) μ ( y n 1 y n , t ) μ ( x n 1 x n , t ) μ ( y n 1 y n , t ) = μ ( y n 1 y n , t ) μ ( x n 1 x n , t ) μ ( y n 1 y n , t ) 1 1 1 μ ( y n 1 y n , t ) μ ( x n 1 x n , t ) μ ( y n 1 y n , t ) μ ( z n 1 z n , t ) μ ( z n 1 z n , t ) μ ( x n 1 x n , t ) α n 1 ( t ) α n 1 ( t ) .
(2.12)

Using the properties of a t-norm, (2.9)-(2.12) and (2.1), we obtain

α n ( k t ) = μ ( x n x n + 1 , k t ) μ ( y n y n + 1 , k t ) μ ( z n z n + 1 , k t ) α n 1 ( t ) α n 1 ( t ) α n 1 ( t ) α n 1 ( t ) ( α n 1 ( t ) ) 4 n 1 ,

which implies that

α n (t) ( α n 1 ( t k ) ) 4 n1.

Now, repetition of the above process gives

α n (t) ( α n 1 ( t k ) ) 4 ( α 0 ( t k n ) ) 4 n n1.

Hence,

(2.13)

It is obvious to note that

t(1k) ( 1 + k + + k m n 1 ) <tm>n,0<k<1.

Consider

where p>0 such that m< n p . Since (μ,υ) has the n-property on X×(0,), therefore

Hence,

lim n μ( x n x m ,t)μ( y n y m ,t)μ( z n z m ,t)=1.
(2.14)

Next, we show that

lim n υ( x n x m ,t)υ( y n y m ,t)υ( z n z m ,t)=0.

Define

β n (t)=υ( x n x n + 1 ,t)υ( y n y n + 1 ,t)υ( z n z n + 1 ,t).
(2.15)

Note that

(2.16)
(2.17)

and

υ ( y n y n + 1 , k t ) = υ ( F ( y n 1 , x n 1 , y n 1 ) F ( y n , x n , y n ) , k t ) υ ( y n 1 y n , t ) μ ( x n 1 x n , t ) μ ( y n 1 y n , t ) = υ ( y n 1 y n , t ) υ ( x n 1 x n , t ) υ ( y n 1 y n , t ) 0 0 0 υ ( y n 1 y n , t ) υ ( x n 1 x n , t ) υ ( y n 1 y n , t ) υ ( z n 1 z n , t ) υ ( z n 1 z n , t ) υ ( x n 1 x n , t ) β n 1 ( t ) β n 1 ( t ) .
(2.18)

Using the properties of a t-conorm, (2.15)-(2.18) and (2.1), we obtain

β n ( k t ) = υ ( x n x n + 1 , t ) υ ( y n y n + 1 , t ) υ ( z n z n + 1 , t ) β n 1 ( t ) β n 1 ( t ) β n 1 ( t ) β n 1 ( t ) = β n 1 ( t ) n 1 ,

that is,

β n (t) β n 1 ( t k ) n1.

Now, repetition of the above process gives

β n (t) β n 1 ( t k ) ( β 0 ( t k n ) ) n n1,

which further implies that

(2.19)

Using the properties of a t-conorm, we get

where p>0 such that m> n p . Since (μ,υ) has the n-property on X×(0,), we have

So,

lim n υ( x n x m ,t)υ( y n y m ,t)υ( z n z m ,t)=0.
(2.20)

Now, (2.14) and (2.20) imply that { x n }, { y n } and { z n } are Cauchy sequences in X. Since X is complete, there exist x, y and z such that lim n x n =x, lim n y n =y and lim n z n =z. If the assumption (a) does hold, then we have

and

z = lim n z n + 1 = lim n F ( z n , y n , x n ) = F ( lim n z n , lim n y n , lim n x n ) = F ( z , y , x ) .

Suppose that the assumption (b) holds then

which, on taking limit as n, gives μ(xF(x,y,z),kt)=1, x=F(x,y,z). Also,

which, on taking limit as n, implies μ(yF(y,x,y),kt)=1, y=F(y,x,y). Finally, we have

which, on taking limit as n, gives μ(zF(z,y,x),kt)=1, z=F(z,y,x). □

Theorem 2.2 Let (X,μ,υ,,) be an IFNS, ⪯ be a partial order on X, and suppose that

ababandaa=a
(2.21)

for all a,b[0,1]. Let F:X×X×XX and g:XX be mappings such that F has the mixed g-monotone property and

μ ( F ( x , y , z ) F ( u , v , w ) , k t ) μ ( g x g u , t ) μ ( g y g v , t ) μ ( g z g w , t ) and υ ( F ( x , y , z ) F ( u , v , w ) , k t ) υ ( g x g u , t ) υ ( g y g v , t ) υ ( g z g w , t )
(2.22)

for all those x, y, z, and u, v, w for which gxgu, gygv, gzgw, where 0<k<1. Assume that g(X) is complete, F(X×X×X)g(X) and g is continuous. If either

  1. (a)

    F is continuous or

  2. (b)

    X has the following property:

(bi) if { x n } is a non-decreasing sequence and (μ,υ) lim n x n =x, then x n x for all nN,

(bii) if { y n } is a non-decreasing sequence and (μ,υ) lim n y n =y, then y n y for all nN, and

(biii) if { z n } is a non-decreasing sequence and (μ,υ) lim n z n =y, then z n z for all nN.

Then F has a tripled coincidence point provided that there exist x 0 , y 0 , z 0 X such that

g( x 0 )F( x 0 , y 0 , z 0 ),g( y 0 )F( y 0 , x 0 , y 0 ),g( z 0 )F( z 0 , y 0 , x 0 ).

Proof By Lemma 1.16, there exists EX such that g:EX is one-to-one and g(E)=g(X). Now, define a mapping A:g(E)×g(E)×g(E)X by

A(gx,gy,gz)=F(x,y,z)x,y,zX.
(2.23)

Since g is one-to-one, so A is well defined. Now, (2.22) and (2.23) imply that

μ ( A ( g x , g y , g z ) A ( g u , g v , g w ) , k t ) μ ( g x g u , t ) μ ( g y g v , t ) μ ( g z g w , t ) , υ ( A ( g x , g y , g z ) A ( g u , g v , g w ) , k t ) υ ( g x g u , t ) υ ( g y g v , t ) υ ( g z g w , t )
(2.24)

for all x,y,z,u,v,wE for which gxgu, gygv, gzgw. Since F has the mixed g-monotone property for all x,y,zX, so we have

(2.25)

Now, from (2.23) and (2.25), we have

(2.26)

Hence, A has the mixed monotone property. Suppose that the assumption (a) holds. Since F is continuous, A is also continuous. By using Theorem 2.1, A has a tripled fixed point (u,v,w)g(E)×g(E)×g(E). If the assumption (b) holds, then using the definition of A, following similar arguments to those given in Theorem 2.1, A has a tripled fixed point (u,v,w)g(E)×g(E)×g(E). Finally, we show that F and g have a tripled coincidence point. Since A has a tripled fixed point (u,v,w)g(E)×g(E)×g(E), we get

u=A(u,v,w),v=A(v,u,v),w=A(w,u,v).
(2.27)

Hence, there exist u 1 , v 1 , w 1 X×X×X such that g u 1 =u, g v 1 =v, and g w 1 =w. Now, it follows from (2.27) that

Thus, ( u 1 , v 1 , w 1 )X×X×X is a tripled coincidence point of F and g. □

Example 2.3 Let X=R be a usual normed, :[0,1]×[0,1][0,1] and :[0,1]×[0,1][0,1] be defined by

ab=abandab=max{a,b}.

It is easy to see that ∗ is a continuous t-norm and ⋄ is a continuous t-conorm satisfy

abababfor all a,b[0,1].

Let ϕ: R + (0,1) be defined by ϕ(t)= e 1 t for all t R + . Now we have (X,μ,υ,,) is an IFNS, where

μ(x,t)= [ ϕ ( t ) ] | x | ,υ(x,t)=1 [ ϕ ( t ) ] | x | ,xX,

such that (μ,ν) satisfies the n-property on X×(0,).

If X is endowed with usual order as xyxy0, then (X,) is a partially ordered set. Define mappings F:X×X×XX and g:XX by

F(x,y,z)=2x2y+2z+1andg(x)=7x1.

Obviously, F and g both are onto maps so F(X×X×X)g(X). Also, F and g are continuous and F has the mixed g-monotone property. Indeed,

x 1 , x 2 X , g x 1 g x 2 2 x 1 2 y + 2 z + 1 2 x 2 2 y + 2 z + 1 F ( x 1 , y , z ) F ( x 2 , y , z ) .

Similarly, we can prove that

y 1 , y 2 X,g( y 1 )g( y 2 )F(x, y 2 ,z)F(x, y 1 ,z)

and

z 1 , z 2 X,g( z 1 )g( z 2 )F(x,y, z 1 )F(x,y, z 2 ).

If x 0 =0, y 0 = 2 3 , z 0 =0, then

So, there exist x 0 , y 0 , z 0 X such that

g( x 0 )F( x 0 , y 0 , z 0 ),g( y 0 )F( y 0 , x 0 , y 0 ),g( z 0 )F( z 0 , y 0 , x 0 ).

Now, for all x,y,z,u,v,wX, for which gxgu, gygv, gzgw, we have

for k= 3 3.5 <1. Hence, there exists k= 3 3.5 <1 such that

for all x,y,z,u,v,wX, for which gxgu, gygv, gzgw.

Now, for all x,y,z,u,v,wX, for which gxgu, gygv, gzgw, we have

for k= 3 3.5 <1. Hence, there exists k= 3 3.5 <1 such that

for all x,y,z,u,v,wX, for which gxgu, gygv, gzgw.

Therefore, all the conditions of Theorem 2.2 are satisfied. So, F and g have a tripled coincidence point and here ( 2 5 , 2 5 , 2 5 ) is a tripled coincidence point of F and g.