1 Introduction

The fixed point theorem, generally known as the Banach contraction principle, appeared in an explicit form in Banach’s thesis in 1922 [1], where it was used to establish the existence of a solution to an integral equation. Since then, because of its simplicity and usefulness, it has become a very popular tool in solving existence problems in many branches of mathematical analysis. This principle states that if (X,d) is a complete metric space and T:XX is a contraction map (i.e., d(Tx,Ty)λd(x,y) for all x,yX, where λ(0,1) is a constant), then T has a unique fixed point.

The Banach contraction principle has been generalized in many ways over the years. In some generalizations, the contractive nature of the map is weakened; see [29] and others. In other generalizations, the topology is weakened; see [1023] and others. In [24], Nadler extended the Banach fixed point theorem from single-valued maps to set-valued contractive maps. Other fixed point results for set-valued maps can be found in [2530] and the references therein.

In 2000, Branciari [11] introduced the concept of generalized metric spaces, where the triangle inequality is replaced by the inequality d(x,y)d(x,u)+d(u,v)+d(v,y) for all pairwise distinct points x,y,u,vX. Various fixed point results were established on such spaces; see [10, 13, 1720, 22, 3133] and the references therein.

We recall the following definitions introduced in [11].

Definition 1.1 Let X be a non-empty set and d:X×X[0,) be a mapping such that for all x,yX and for all distinct points u,vX, each of them different from x and y, one has

  1. (i)

    d(x,y)=0x=y;

  2. (ii)

    d(x,y)=d(y,x);

  3. (iii)

    d(x,y)d(x,u)+d(u,v)+d(v,y).

Then (X,d) is called a generalized metric space (or for short g.m.s).

Definition 1.2 Let (X,d) be a g.m.s, { x n } be a sequence in X and xX. We say that { x n } is convergent to x if and only if d( x n ,x)0 as n. We denote this by x n x.

Definition 1.3 Let (X,d) be a g.m.s and { x n } be a sequence in X. We say that { x n } is Cauchy if and only if d( x n , x m )0 as n,m.

Definition 1.4 Let (X,d) be a g.m.s. We say that (X,d) is complete if and only if every Cauchy sequence in X converges to some element in X.

The following result was established in [17] (see also [34]).

Lemma 1.1 Let (X,d) be a g.m.s and { x n } be a Cauchy sequence in (X,d) such that d( x n ,x)0 as n for some xX. Then d( x n ,y)d(x,y) as n for all yX. In particular, { x n } does not converge to y if yx.

We denote by Θ the set of functions θ:(0,)(1,) satisfying the following conditions:

( Θ 1 ) θ is non-decreasing;

( Θ 2 ) for each sequence { t n }(0,), lim n θ( t n )=1 if and only if lim n t n = 0 + ;

( Θ 3 ) there exist r(0,1) and (0,] such that lim t 0 + θ ( t ) 1 t r =;

( Θ 4 ) θ is continuous.

Recently, Jleli and Samet [35] established the following generalization of the Banach fixed point theorem in the setting of Branciari metric spaces.

Theorem 1.1 (Jleli and Samet [35])

Let (X,d) be a complete g.m.s and T:XX be a given map. Suppose that there exist θΘ and k(0,1) such that

x,yX,d(Tx,Ty)0θ ( d ( T x , T y ) ) [ θ ( d ( x , y ) ) ] k .

Then T has a unique fixed point.

Note that the condition ( Θ 4 ) is not supposed in Theorem 1.1.

The aim of this paper is to extend the result given by Theorem 1.1.

2 Result and proof

Now, we are ready to state and prove our main result.

Theorem 2.1 Let (X,d) be a complete g.m.s and T:XX be a given map. Suppose that there exist θΘ and k(0,1) such that

x,yX,d(Tx,Ty)0θ ( d ( T x , T y ) ) [ θ ( M ( x , y ) ) ] k ,
(1)

where

M(x,y)=max { d ( x , y ) , d ( x , T x ) , d ( y , T y ) } .
(2)

Then T has a unique fixed point.

Proof Let xX be an arbitrary point in X. If for some pN we have T p x= T p + 1 x, then T p x will be a fixed point of T. So, without restriction of the generality, we can suppose that d( T n x, T n + 1 x)>0 for all nN. Now, from (1), for all nN, we have

θ ( d ( T n x , T n + 1 x ) ) [ θ ( M ( T n 1 x , T n x ) ) ] k ,
(3)

where from (2)

M ( T n 1 x , T n x ) = max { d ( T n 1 x , T n x ) , d ( T n 1 x , T T n 1 x ) , d ( T n x , T T n x ) } = max { d ( T n 1 x , T n x ) , d ( T n 1 x , T n x ) , d ( T n x , T n + 1 x ) } = max { d ( T n 1 x , T n x ) , d ( T n x , T n + 1 x ) } .
(4)

If M( T n 1 x, T n x)=d( T n x, T n + 1 x), then inequality (3) turns into

θ ( d ( T n x , T n + 1 x ) ) [ θ ( d ( T n x , T n + 1 x ) ) ] k ,

which implies that

ln [ θ ( d ( T n x , T n + 1 x ) ) ] kln [ θ ( d ( T n x , T n + 1 x ) ) ] ,

that is a contradiction with k(0,1). Hence, from (4) we have M( T n 1 x, T n x)=d( T n 1 x, T n x), and inequality (3) yields

θ ( d ( T n x , T n + 1 x ) ) [ θ ( d ( T n 1 x , T n x ) ) ] k [ θ ( d ( T n 2 x , T n 1 x ) ) ] k 2 [ θ ( d ( x , T x ) ) ] k n .

Thus we have

1θ ( d ( T n x , T n + 1 x ) ) [ θ ( d ( x , T x ) ) ] k n for all nN.
(5)

Letting n in (5), we obtain

θ ( d ( T n x , T n + 1 x ) ) 1as n,
(6)

which implies from ( Θ 2 ) that

lim n d ( T n x , T n + 1 x ) =0.

From condition ( Θ 3 ), there exist r(0,1) and (0,] such that

lim n θ ( d ( T n x , T n + 1 x ) ) 1 [ d ( T n x , T n + 1 x ) ] r =.

Suppose that <. In this case, let B=/2>0. From the definition of the limit, there exists n 0 N such that

| θ ( d ( T n x , T n + 1 x ) ) 1 [ d ( T n x , T n + 1 x ) ] r |Bfor all n n 0 .

This implies that

θ ( d ( T n x , T n + 1 x ) ) 1 [ d ( T n x , T n + 1 x ) ] r B=Bfor all n n 0 .

Then

n [ d ( T n x , T n + 1 x ) ] r An [ θ ( d ( T n x , T n + 1 x ) ) 1 ] for all n n 0 ,

where A=1/B.

Suppose now that =. Let B>0 be an arbitrary positive number. From the definition of the limit, there exists n 0 N such that

θ ( d ( T n x , T n + 1 x ) ) 1 [ d ( T n x , T n + 1 x ) ] r Bfor all n n 0 .

This implies that

n [ d ( T n x , T n + 1 x ) ] r An [ θ ( d ( T n x , T n + 1 x ) ) 1 ] for all n n 0 ,

where A=1/B.

Thus, in all cases, there exist A>0 and n 0 N such that

n [ d ( T n x , T n + 1 x ) ] r An [ θ ( d ( T n x , T n + 1 x ) ) 1 ] for all n n 0 .

Using (5), we obtain

n [ d ( T n x , T n + 1 x ) ] r An ( [ θ ( d ( x , T x ) ) ] k n 1 ) for all n n 0 .

Letting n in the above inequality, we obtain

lim n n [ d ( T n x , T n + 1 x ) ] r =0.

Thus, there exists n 1 N such that

d ( T n x , T n + 1 x ) 1 n 1 / r for all n n 1 .
(7)

Now, we shall prove that T has a periodic point. Suppose that it is not the case, then T n x T m x for every n,mN such that nm. Using (1), we obtain

θ ( d ( T n x , T n + 2 x ) ) [ θ ( M ( T n 1 x , T n + 1 x ) ) ] k ,
(8)

where from (2)

M ( T n 1 x , T n + 1 x ) =max { d ( T n 1 x , T n + 1 x ) , d ( T n 1 x , T n x ) , d ( T n + 1 x , T n + 2 x ) } .
(9)

Since θ is non-decreasing, we obtain from (8) and (9)

θ ( d ( T n x , T n + 2 x ) ) [ max { θ ( d ( T n 1 x , T n + 1 x ) ) , θ ( d ( T n 1 x , T n x ) ) , θ ( d ( T n + 1 x , T n + 2 x ) ) } ] k .
(10)

Let I be the set of nN such that

u n : = max { θ ( d ( T n 1 x , T n + 1 x ) ) , θ ( d ( T n 1 x , T n x ) ) , θ ( d ( T n + 1 x , T n + 2 x ) ) } = θ ( d ( T n 1 x , T n + 1 x ) ) .

If |I|<, then there exists NN such that for every nN,

max { θ ( d ( T n 1 x , T n + 1 x ) ) , θ ( d ( T n 1 x , T n x ) ) , θ ( d ( T n + 1 x , T n + 2 x ) ) } = max { θ ( d ( T n 1 x , T n x ) ) , θ ( d ( T n + 1 x , T n + 2 x ) ) } .

In this case, we obtain from (10)

1θ ( d ( T n x , T n + 2 x ) ) [ max { θ ( d ( T n 1 x , T n x ) ) , θ ( d ( T n + 1 x , T n + 2 x ) ) } ] k

for all nN. Letting n in the above inequality and using (6), we get

θ ( d ( T n x , T n + 2 x ) ) 1as n.

If |I|=, we can find a subsequence of { u n }, that we denote also by { u n }, such that

u n =θ ( d ( T n 1 x , T n + 1 x ) ) for n large enough.

In this case, we obtain from (10)

1 θ ( d ( T n x , T n + 2 x ) ) [ θ ( d ( T n 1 x , T n + 1 x ) ) ] k [ θ ( d ( T n 2 x , T n x ) ) ] k 2 [ θ ( d ( x , T 2 x ) ) ] k n

for n large enough. Letting n in the above inequality, we obtain

θ ( d ( T n x , T n + 2 x ) ) 1as n.
(11)

Then in all cases, (11) holds. Using (11) and the property ( Θ 2 ), we obtain

lim n d ( T n x , T n + 2 x ) =0.

Similarly, from condition ( Θ 3 ), there exists n 2 N such that

d ( T n x , T n + 2 x ) 1 n 1 / r for all n n 2 .
(12)

Let N=max{ n 0 , n 1 }. We consider two cases.

Case 1. If m>2 is odd, then writing m=2L+1, L1, using (7), for all nN, we obtain

d ( T n x , T n + m x ) d ( T n x , T n + 1 x ) + d ( T n + 1 x , T n + 2 x ) + + d ( T n + 2 L x , T n + 2 L + 1 x ) 1 n 1 / r + 1 ( n + 1 ) 1 / r + + 1 ( n + 2 L ) 1 / r i = n 1 i 1 / r .

Case 2. If m>2 is even, then writing m=2L, L2, using (7) and (12), for all nN, we obtain

d ( T n x , T n + m x ) d ( T n x , T n + 2 x ) + d ( T n + 2 x , T n + 3 x ) + + d ( T n + 2 L 1 x , T n + 2 L x ) 1 n 1 / r + 1 ( n + 2 ) 1 / r + + 1 ( n + 2 L 1 ) 1 / r i = n 1 i 1 / r .

Thus, combining all the cases, we have

d ( T n x , T n + m x ) i = n 1 i 1 / r for all nN,mN.

From the convergence of the series i 1 i 1 / r (since 1/r>1), we deduce that { T n x} is a Cauchy sequence. Since (X,d) is complete, there is zX such that T n xz as n. Without restriction of the generality, we can suppose that T n xz for all n (or for n large enough). Suppose that d(z,Tz)>0, using (1), we get

θ ( d ( T n + 1 x , T z ) ) [ θ ( M ( T n x , z ) ) ] k for all nN,

where

M ( T n x , z ) =max { d ( T n x , z ) , d ( T n x , T n + 1 x ) , d ( z , T z ) } .

Letting n in the above inequality, using ( Θ 4 ) and Lemma 1.1, we obtain

θ ( d ( z , T z ) ) [ θ ( d ( z , T z ) ) ] k <θ ( d ( z , T z ) ) ,

which is a contradiction. Thus we have z=Tz, which is also a contradiction with the assumption: T does not have a periodic point. Thus T has a periodic point, say z, of period q. Suppose that the set of fixed points of T is empty. Then we have

q>1andd(z,Tz)>0.

Using (1), we obtain

θ ( d ( z , T z ) ) =θ ( d ( T q z , T q + 1 z ) ) [ θ ( z , T z ) ] k q <θ ( d ( z , T z ) ) ,

which is a contradiction. Thus, the set of fixed points of T is non-empty, that is, T has at least one fixed point. Now, suppose that z,uX are two fixed points of T such that d(z,u)=d(Tz,Tu)>0. Using (1), we obtain

θ ( d ( z , u ) ) =θ ( d ( T z , T u ) ) [ θ ( d ( z , u ) ) ] k <θ ( d ( z , u ) ) ,

which is a contradiction. Then we have one and only one fixed point. □

3 Some consequences

We start by deducing the following fixed point result.

Corollary 3.1 Let (X,d) be a complete g.m.s and T:XX be a given map. Suppose that there exists λ(0,1) such that

d(Tx,Ty)λmax { d ( x , y ) , d ( x , T x ) , d ( y , T y ) } for all x,yX.
(13)

Then T has a unique fixed point.

Proof From (13), we have

e d ( T x , T y ) [ e max { d ( x , y ) , d ( x , T x ) , d ( y , T y ) } ] λ for all x,yX.

Clearly the function θ:(0,)(1,) defined by θ(t):= e t belongs to Θ. So, the existence and uniqueness of the fixed point follows from Theorem 2.1. □

The following fixed point result established in [11] is an immediate consequence of Corollary 3.1.

Corollary 3.2 Let (X,d) be a complete g.m.s and T:XX be a given map. Suppose that there exists λ(0,1) such that

d(Tx,Ty)λd(x,y)for all x,yX.

Then T has a unique fixed point.

The following fixed point result established in [34] is an immediate consequence of Corollary 3.1.

Corollary 3.3 Let (X,d) be a complete g.m.s and T:XX be a given map. Suppose that there exist λ,μ0 with λ+μ<1 such that

d(Tx,Ty)λd(x,Tx)+μd(y,Ty)for all x,yX.

Then T has a unique fixed point.

The following fixed point result is also an immediate consequence of Corollary 3.1.

Corollary 3.4 Let (X,d) be a complete g.m.s and T:XX be a given map. Suppose that there exist λ,μ,ν0 with λ+μ+ν<1 such that

d(Tx,Ty)λd(x,y)+μd(x,Tx)+νd(y,Ty)for all x,yX.

Then T has a unique fixed point.

We note that Θ contains a large class of functions. For example, for

θ(t):=2 2 π arctan ( 1 t α ) ,0<α<1,t>0,

we obtain from Theorem 2.1 the following result.

Corollary 3.5 Let (X,d) be a complete g.m.s and T:XX be a given map. Suppose that there exist α,k(0,1) such that

2 2 π arctan ( 1 [ d ( T x , T y ) ] α ) [ 2 2 π arctan ( 1 [ M ( x , y ) ] α ) ] k for all x,yX,TxTy,

where M(x,y) is given by (2). Then T has a unique fixed point.

Finally, since a metric space is a g.m.s, from Theorem 2.1 we deduce immediately the following result.

Corollary 3.6 Let (X,d) be a complete metric space and T:XX be a given map. Suppose that there exist θΘ and k(0,1) such that

x,yX,d(Tx,Ty)0θ ( d ( T x , T y ) ) [ θ ( M ( x , y ) ) ] k ,

where M(x,y) is given by (2). Then T has a unique fixed point.