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Fixed points of mappings with a contractive iterate at a point in partial metric spaces

Abstract

In 1994, Matthews introduced and studied the concept of partial metric space and obtained a Banach-type fixed point theorem on complete partial metric spaces. In this paper we study fixed point results of new mappings with a contractive iterate at a point in partial metric spaces. Our results generalize and unify some results of Sehgal, Guseman and Ćirić for mappings with a generalized contractive iterate at a point to partial metric spaces. We give some generalized versions of the fixed point theorem of Matthews. The theory is illustrated by some examples.

1 Introduction

In 1922, Banach proved the following famous fixed point theorem [1]. Let (X,d) be a complete metric space. Let T be a contractive mapping on X, that is, one for which exists q[0,1) satisfying

d(Tx,Ty)qd(x,y)
(1.1)

for all x,yX. Then there exists a unique fixed point x 0 X of T. This theorem, called the Banach contraction principle, is a forceful tool in nonlinear analysis. This principle has many applications and has been extended by a great number of authors. For the convenience of the reader, let us recall the following results [24].

In 1969, Sehgal [4] proved the following interesting generalization of the contraction mapping principle.

Theorem 1.1 ([4])

Let (X,d) be a complete metric space, q[0,1) and T:XX be a continuous mapping. If for each xX there exists a positive integer n=n(x) such that

d ( T n x , T n y ) qd(x,y)
(1.2)

for all yX, then T has a unique fixed point uX. Moreover, for any xX, u= lim m T m x.

In 1970, Guseman [3] (see also [5]) generalized the result of Sehgal to mappings which are both necessarily continuous and which have a contractive iterate at each point in a (possibly proper) subset of the space.

In 1983, Ćirić [2], among other things, proved the following interesting generalization of Sehgal’s result.

Theorem 1.2 ([2])

Let (X,d) be a complete metric space, q[0,1) and T:XX. If for each xX there exists a positive integer n=n(x) such that

d ( T n x , T n y ) qmax { d ( x , y ) , d ( x , T y ) , , d ( x , T n y ) , d ( x , T n x ) }
(1.3)

holds for all yX, then T has a unique fixed point uX. Moreover, for every xX, u= lim m T m x.

Partial metric spaces were introduced in [6] by Matthews as part of the study of denotational semantics of dataflow networks and served as a device to solve some difficulties in domain theory of computer science (see also [7, 8]), in particular the ones which arose in the modeling of a parallel computation program given in [9]. The concept has since proved extremely useful in domain theory (see, e.g., [1013]) and in constructing models in the theory of computation (see, e.g., [1417]).

On the other hand, fixed point theory of mappings defined on partial metric spaces since the first results obtained in [6] has flourished in the meantime, a fact evidenced by quite a number of papers dedicated to this subject (see, e.g., [1847]). The task seems to have been set forth of determining what known fixed point results from the usual metric setting remain valid - after adequate modifications that should as much as possible reflect the nature of the concept of partial metric - when formulated in the partial metric setting.

The potentially nonzero self-distance, built into Matthew’s definition of partial metrics, was taken into account in [29] in an essential way by a rather mild variation of the classical Banach contractive condition, and in [30] further considerations in this direction were carried out which were in turn generalized by Chi et al. in [27]. This paper represents a continuation of the previous work by the authors. Now we study fixed point results of new mappings with a contractive iterate at a point in partial metric spaces. Our results generalize and unify some results of Sehgal, Guseman and Ćirić for mappings with a generalized contractive iterate at a point to partial metric spaces. We give some generalized versions of the fixed point theorem of Matthews. The theory is illustrated by some examples.

2 Preliminaries

Throughout this paper the letters and will denote the set of real numbers and positive integers, respectively.

Let us recall [7] that a nonnegative mapping p:X×XR, where X is a nonempty set, is said to be a partial metric on X if for any x,y,zX the following four conditions hold true:

  1. (P1)

    p(x,y)=p(y,x),

  2. (P2)

    p(x,x)p(x,y),

  3. (P3)

    if p(x,x)=p(y,y)=p(x,y), then x=y,

  4. (P4)

    p(x,z)p(x,y)+p(y,z)p(y,y).

The pair (X,p) is then called a partial metric space. A sequence { x m } m = 0 of elements of X is called p-Cauchy if the limit lim m , n p( x n , x m ) exists and is finite. The partial metric space (X,p) is called complete if for each p-Cauchy sequence { x m } m = 0 there is some zX such that

p(z,z)= lim n p(z, x n )= lim n , m p( x n , x m ).
(2.1)

Observe that condition (P4) is a strengthening of the triangle inequality and that p(x,y)=0 implies x=y as in the case of an ordinary metric.

It can be shown that if (X,p) is a partial metric space, then by p s (x,y)=2p(x,y)p(x,x)p(y,y), for x,yX, a metric p s is defined on the set X such that { x n } n 1 converges to zX with respect to p s if and only if (2.1) holds. Also (X,p) is a complete partial metric space if and only if (X, p s ) is a complete metric space. For proofs of these facts, see [7, 44].

A paradigm for partial metric spaces is the pair (X,p) where X=[0,+) and p(x,y)=max{x,y} for x,y0. Below we give two more examples of partial metrics both of which are taken from [7].

Example 2.1 If X:={[a,b]a,bR,ab}, then p([a,b],[c,d])=max{b,d}min{a,b} defines a partial metric p on X.

Example 2.2 Let X:= R N 0 n 1 R { 0 , 1 , , n 1 } , where N 0 is the set of nonnegative integers.

By L(x) denote the set {0,1,,n} if x R { 0 , 1 , , n 1 } for some nN, and the set N 0 if x R N 0 . Then a partial metric is defined on X by

p(x,y)=inf { 2 i i L ( x ) L ( y )  and  j N 0 ( j < i x ( j ) = y ( j ) ) } .

For applications of partial metrics to problems in theoretical computer science, the reader is referred to [8, 11, 14, 16].

In [7] Matthews proved the following extension of the Banach contraction principle to the setting of partial metric spaces.

Theorem 2.1 Let (X,p) be a complete partial metric space, α[0,1) and T:XX be a given mapping. Suppose that for each x,yX the following condition holds

p(Tx,Ty)αp(x,y).
(2.2)

Then there is a unique zX such that Tz=z. Also p(z,z)=0 and for each xX the sequence { T n x } n 1 converges with respect to the metric p s to z.

A variant of the result above concerning the so-called dualistic partial metric spaces was later given in [44]. Altun et al. [23] further generalized the result of Matthews as well as extended to partial metric spaces several other well-known results about fixed points of mappings on metric spaces.

Taking a different approach to the way in which contractive condition (2.2) can be generalized for partial metrics, we [29, 30] have obtained other extensions of Theorem 2.1. To state one of them, we will use the following notation. Given a partial metric space (X,p) set r p :=inf{p(x,y):x,yX}=inf{p(x,x):xX} and R p :={xX:p(x,x)= r p }. Notice that R p may be empty and that if p is a metric, then, clearly, r p =0 and R p =X.

Theorem 2.2 (Theorem 3.1 of [29])

Let (X,p) be a complete partial metric space, α[0,1) and T:XX be a given mapping. Suppose that for each x,yX the following condition holds

p(Tx,Ty)max { α p ( x , y ) , p ( x , x ) , p ( y , y ) } .
(2.3)

Then the set R p is nonempty. There is a unique u R p such that Tu=u. For each x R p , the sequence { T n x } n 1 converges with respect to the metric p s to u.

Remark 2.1 Although Theorem 2.2 does not imply uniqueness of the fixed point, it is easy to see that, under the assumptions made, if u and v are both fixed points satisfying p(u,u)=p(v,v), then u=v.

Remark 2.2 Completeness of a partial metric does not necessarily entail that R p is nonempty. A (class of) counterexample(s) is easily constructed as follows.

Let (X,d) be a partial metric space, a>0 and f:X[0,a) be an arbitrary mapping. If x,yX are such that xy, define p(x,y)=d(x,y)+a and p(x,x)=f(x). Then (X,p) is a partial metric space, as is easily verified.

Now if b:=supf[X]<a, then, given a sequence { x n } n 1 , we have lim sup n p( x n , x n )b<a and p( x n , x m )a whenever x n x m . Thus there are no nonstationary p-Cauchy sequences. Hence (X,p) is complete. But R p = whenever inff[X]f[X].

If condition (2.3) is replaced by the somewhat stronger condition below, then the uniqueness of the fixed point is guaranteed.

Theorem 2.3 (Theorem 3.2 of [29])

Let (X,p) be a complete partial metric space, α[0,1) and T:XX be a given mapping. Suppose that for each x,yX the following condition holds

p(Tx,Ty)max { α p ( x , y ) , p ( x , x ) + p ( y , y ) 2 } .
(2.4)

Then there is a unique zX such that Tz=z. Furthermore, z R p and for each x R p the sequence { T n x } n 1 converges with respect to the metric p s to z.

3 Auxiliary results

We now introduce the two types of contractive conditions that we shall be considering in this paper. Let us remark that if T:XX, then we write T 0 =I for the identity mapping I:XX, i.e., I(x)=x, xX.

Definition 3.1 Let (X,p) be a partial metric space, α(0,1) and T:XX.

  1. (i)

    We say that T is a C 1 -operator on X if for each xX there is some n(x)N such that for each yX there holds

    p ( T n ( x ) x , T n ( x ) y ) max { α p ( x , T j y ) , α p ( x , T n ( x ) x ) , p ( x , x ) , p ( T n ( x ) 1 y , T n ( x ) 1 y ) }
    (3.1)

    for some j{0,1,,n(x)}.

  2. (ii)

    We say that T is a C-operator on X if for each xX there is some n(x)N such that for each yX there holds

    p ( T n ( x ) x , T n ( x ) y ) αmax { p ( x , T j y ) , p ( x , T n ( x ) x ) }
    (3.2)

    for some j{0,1,,n(x)}.

For a C 1 -operator T on a partial metric space (X,p) and xX define the supporting sequence at the point x as the sequence { s k } k 0 , where s 0 =0 and s k + 1 = s k +n( T s k x). Clearly, this is a strictly increasing sequence. Also set R T (X):={xX T m x= T m + 1 x for some mN}.

Lemma 3.1 Let T be a C 1 -operator on a partial metric space (X,p), xX R T (X), { s k } k 0 be the supporting sequence at x and k1 and i s k be given integers. Then we must have

p ( T s k x , T i x ) max { α p ( T s k 1 x , T j x ) , p ( T s k 1 x , T s k 1 x ) } for some  j s k 1 .
(3.3)

Proof Case 1. Suppose i= s k +2. By (3.1) we know that if

p ( T s k x , T s k + 2 x ) > max { α p ( T s k 1 x , T j x ) , p ( T s k 1 x , T s k 1 x ) } for all  j { s k 1 , , s k + 2 } ,
(3.4)

then

p ( T s k x , T s k + 2 x ) p ( T s k + 1 x , T s k + 1 x ) .
(3.5)

Likewise, if

p ( T s k x , T s k + 1 x ) > max { α p ( T s k 1 x , T j x ) , p ( T s k 1 x , T s k 1 x ) } for all  j { s k 1 , , s k + 1 } ,
(3.6)

then

p ( T s k x , T s k + 1 x ) p ( T s k x , T s k x ) .
(3.7)

Now if (3.4) were to hold, then (3.5) and (P2) would imply that (3.6) is true as well. So (3.7) also holds and thus

p ( T s k x , T s k + 2 x ) p ( T s k + 1 x , T s k + 1 x ) p ( T s k x , T s k + 1 x ) p ( T s k x , T s k x ) p ( T s k x , T s k + 2 x )

meaning (by (P3)) that T s k x= T s k + 1 x. But this contradicts the assumption x R T (X), so (3.4) must be true.

Case 2. Suppose i= s k . Since p( T s k x, T s k x)p( T s k x, T s k + 2 x), the assertion here follows from the previous case.

Case 3. Suppose now i= s k +1. Assume that

p ( T s k x , T s k + 1 x ) > max { α p ( T s k 1 x , T j x ) , p ( T s k 1 x , T s k 1 x ) } for all  j { s k 1 , , s k + 1 }

since otherwise there is nothing to prove. Then by (3.1) we must have p( T s k x, T s k + 1 x)p( T s k x, T s k x). But then, by the previous case, there is some j s k 1 such that

p ( T s k x , T s k + 1 x ) p ( T s k x , T s k x ) max { α p ( T s k 1 x , T j x ) , p ( T s k 1 x , T s k 1 x ) }

and we are done.

For i> s k +2, the argument carries on by induction. Suppose that (3.3) holds for some i s k . If p( T s k x, T i + 1 x)>{αp( T s k 1 x, T j x),p( T s k 1 x, T s k 1 x)} for all j{ s k 1 ,,i+1}, then we must have p( T s k x, T i + 1 x)p( T i x, T i x). By the induction hypothesis, there is some j s k 1 such that

p ( T s k x , T i x ) { α p ( T s k 1 x , T j x ) , p ( T s k 1 x , T s k 1 x ) } .

But since p( T s k x, T i + 1 x)p( T i x, T i x)p( T s k x, T i x), the last inequality clashes with our assumption. □

To shorten the foregoing considerations, we introduce some auxiliary notions as follows. Fix xX R T (X). For integers k1 and i s k , use Lemma 3.1 repeatedly to fix integers l j s j , 0j<k and t 1 ,, t k {0,1} such that, putting l k :=i, there holds

p ( T s j x , T l j x ) α t j p ( T s j 1 x , T l j 1 x )

for all 1jk, where

t j = { 1 if  s j 1 < l j 1 , 0 if  s j 1 = l j 1 .

We shall refer to ( l 0 ,, l k 1 ) and ( t 1 ,, t k ) as the (k,i)-descent and the (k,i)-signature at x, respectively. Set S k , i x :={j{1,,k} t j =1}. We shall say that x is of type 1 if there are sequences of positive integers { k m } m 0 and { i m } m 0 , the first one strictly increasing, such that for all m0 we have i m s k m and card( S k m , i m x )<card( S k m + 1 , i m + 1 x ); here and henceforth, for a finite set P, we denote by card(P) the number of its elements. We shall say that x is of type 2 if x is not of type 1, i.e., if there are k 0 ,DN such that for all k k 0 and all i s k there holds card( S k , i x )<D.

To make the proof of our main result more transparent, we have extracted from it several parts and presented them first in form of the next five lemmas.

Lemma 3.2 Let T be a C 1 -operator on a partial metric space (X,p), x R T (X), and let { s k } k 0 be the supporting sequence at x. Then:

  1. (a)

    If ( l 0 ,, l k 1 ) is the (k, i 0 )-descent at x, then

    p ( T s k x , T i 0 x ) α card ( S k , i 0 x ) p ( x , T l 0 x ) and p ( T s k x , T i 0 x ) p ( T s j x , T l j x ) for all  0 j k ,

    where we set l k := i 0 .

  2. (b)

    If P{0,,k1} is such that card( S k , i 0 x )<card(P), then for some j 0 P there must hold

    p ( T s k x , T i 0 x ) p ( T s j 0 x , T s j 0 x ) .

Proof Regarding Lemma 3.1, we get p( T s k x, T i 0 x) α i = j + 1 k t i p( T s j x, T l j x) for all j such that k>j0, recursively. Now (a) follows directly.

To prove (b) simply observe that card( S k , i 0 x )<card(P) implies that the set {j+1jP} is a subset of {1,,k} with card(P)>card( S k , i 0 x ) elements so that there must be some j 0 P with t j 0 + 1 =0. Hence

p ( T s k x , T i 0 x ) p ( T s j 0 + 1 x , T l j 0 + 1 x ) α t j 0 + 1 p ( T s j 0 x , T l j 0 x ) =p ( T s j 0 x , T s j 0 x ) ,

where we used (a) and the fact that l j 0 = s j 0 . □

Lemma 3.3 If T is a C 1 -operator on a partial metric space (X,p) and xX, then there is some M x >0 such that for all i0 we must have p(x, T i x) M x .

Proof If x R T (X), then this is obvious. Thus suppose x R T (X) and set . If k=n(x), then it is certainly true that p(x, T i x) M x for all 0ik. Now suppose that the same is valid for some kn(x).

If p(x, T k + 1 x)p(x, T i x) for some 0ik, then by the induction hypothesis there holds p(x, T k + 1 x)p(x, T i x) M x . Otherwise, we must have

p ( x , T k + 1 x ) >max { p ( x , T i x ) 0 i k } .
(3.8)

Now using (3.1)

p ( x , T k + 1 x ) p ( x , T n ( x ) x ) + p ( T n ( x ) x , T k + 1 x ) p ( x , T n ( x ) x ) + max { α p ( x , T j x ) , p ( x , x ) , p ( T k x , T k x ) , α p ( x , T n ( x ) x ) }

for some k+1n(x)jk+1. Hence we either have

p ( x , T k + 1 x ) p ( x , T n ( x ) x ) +p(x,x) M x

or, using (3.8) and the fact that p( T k x, T k x)p(x, T k x), it must be that

p ( x , T k + 1 x ) p ( x , T n ( x ) x ) + α p ( x , T k + 1 x ) , i.e. , p ( x , T k + 1 x ) 1 1 α p ( x , T n ( x ) x ) M x .

By induction the desired conclusion follows. □

Lemma 3.4 Let T be a C 1 -operator on a partial metric space (X,p) and xX R T (X). If x is of type 1, then lim i , j p( T i x, T j x)=0.

Proof Fix m0. If ( l 0 ,, l k m 1 ) is the ( s k m , i m )-descent, then by (a) of Lemma 3.2 we have

p ( T s k m x , T i m x ) α card ( S k m , i m x ) p ( x , T l 0 x ) α card ( S k m , i m x ) M x .

Since lim m card( S k m , i m x )=, this implies lim m p( T s k m x, T i m x)= lim m p( T s k m x, T s k m x)=0.

Now given ε>0 choose m 0 1 such that α m 0 M x <ε and such that for all m m 0 it holds p( T s k m x, T s k m x)<ε. Let i s k 2 m 0 be arbitrary.

Suppose first that card( S k 2 m 0 , i x ) m 0 . Then

p ( T s k 2 m 0 x , T i x ) α card ( S k 2 m 0 , i x ) M x α m 0 M x <ε.

Suppose now that card( S k 2 m 0 , i x )< m 0 . For P:={ k m 0 ,, k 2 m 0 1 }{0,1,, k 2 m 0 1}, we have card(P)>card( S k 2 m 0 , i x ), so by (b) of Lemma 3.2 there must be some m 0 j2 m 0 1 such that p( T s k 2 m 0 x, T i x)p( T s k j x, T s k j x)<ε.

We have thus shown that p( T s k 2 m 0 x, T i x)<ε must hold for all i s k 2 m 0 . Therefore if i,j s k 2 m 0 , then

p ( T i x , T j x ) p ( T s k 2 m 0 x , T i x ) +p ( T s k 2 m 0 x , T j x ) <2ε.

The previous analysis proves lim i , j p( T i x, T j x)=0. □

Lemma 3.5 Let T be a C 1 -operator on a partial metric space (X,p) and xX R T (X). If x is of type 2, then the sequence { T i x } i 0 is p-Cauchy.

Proof Let { s k } k 0 be the supporting sequence at x.

We first show lim inf m p( T s m x, T s m x)= lim sup m p( T s m x, T s m x). Suppose that this is not true and pick a real θ with lim inf m p( T s m x, T s m x)<θ< lim sup m p( T s m x, T s m x). Let k 1 < k 2 << k D < k D + 1 and i> k D + 1 be positive integers, where k D + 1 k 0 , such that

p ( T s k j x , T s k j x ) <θfor all 1jDandp ( T s i x , T s i x ) >θ.

The fact that s i > s k D + 1 implies that S k D + 1 , s i x is defined and since k D + 1 k 0 , we have card( S k D + 1 , s i x )<D. For P:={ k 1 ,, k D }{0,1,, k D + 1 1}, we have card(P)=D>card( S k D + 1 , s i x )| so, by (b) of Lemma 3.2, there is some j{1,,D} such that

θ<p ( T s i x , T s i x ) p ( T s k D + 1 x , T s i x ) p ( T s k j x , T s k j x ) <θ,

a contradiction.

By the preceding part and since 0p( T s m x, T s m x)2 M x , with M x as in Lemma 3.3, we have r x := lim m p( T s m x, T s m x)R.

Let us prove

ε>0 m 0 m m 0 i s m p ( T s m x , T i x ) ( r x ε, r x +ε).
(3.9)

Given ε>0, take m 1 k 0 such that p( T s m x, T s m x)( r x ε, r x +ε) for all m m 1 . Let m m 1 +D and i s m be arbitrary. For P:={ m 1 ,, m 1 +D1}{0,1,,m1}, we have card(P)=D>card( S m , i x ), so for some m 1 j m 1 +D1 it must be

r x ε<p ( T s m x , T s m x ) p ( T s m x , T i x ) p ( T s j x , T s j x ) < r x +ε

and we are done.

From (3.9) it now immediately follows that

ε>0ki,jkp ( T i x , T j x ) < r x +ε.
(3.10)

Indeed, given ε>0, consider m 0 as in (3.9) and let i,j s m 0 be arbitrary. Then

p ( T i x , T j x ) p ( T s m 0 x , T i x ) + [ p ( T s m 0 x , T j x ) p ( T s m 0 x , T s m 0 x ) ] < r x + ε + 2 ε = r x + 3 ε .

To prove

lim i , j p ( T i x , T j x ) = r x ,
(3.11)

we now only need to show that

ε>0ki,jk r x ε<p ( T i x , T j x ) .
(3.12)

Let ε(0, r x ( 1 α ) 1 + α ) be arbitrary and let kN be as in (3.10). We claim that r x ε<p( T i x, T i x) holds for all ik. This would prove (3.12) since p( T i x, T i x)p( T i x, T j x).

Suppose to the contrary that there is some i 0 k with p( T i 0 x, T i 0 x) r x ε. Put z:= T i 0 x. x R T (X) implies z R T (X). If z is of type 1, then by Lemma 3.4 we have 0= lim i , j p( T i z, T j z)= lim i , j p( T i x, T j x), so { T i x } i 0 is p-Cauchy and we are finished. Suppose now that z is of type 2, so that by what we have proved thus far we know that r z = lim m p( T s m z, T s m z)R and also that (3.10) holds with z taken instead of x. It cannot be p( T n ( z ) z, T n ( z ) z)> r x ε because this would mean that p(z,z)<p( T n ( z ) z, T n ( z ) z), so using Lemma 3.1, it would follow r x ε<p( T n ( z ) z, T n ( z ) z)αp(z, T j z) for some j N 0 , i.e., r x ε<αp( T i 0 x, T i 0 + j x)α( r x +ε), giving r x ( 1 α ) 1 + α <ε, a contradiction. So p( T n ( z ) z, T n ( z ) z) r x ε. The argument actually shows that p( T q m z, T q m z) r x ε holds for every m0, where { q m } m 0 is the supporting sequence at the point z. So r z = lim m p( T q m z, T q m z) r x ε. Now use the fact that r z < r z + r x 2 and (3.10) (with z taken instead of x and r x r z 2 instead of ε of course) to find j 0 N such that

p ( T j z , T j z ) < r z + r x 2 for all j j 0 .
(3.13)

As lim m p( T s m x, T s m x)= r x and r z + r x 2 < r x , there is some m i 0 + j 0 with p( T s m x, T s m x)> r z + r x 2 . Now, using s m i 0 m i 0 j 0 , we obtain

r z + r x 2 <p ( T s m x , T s m x ) =p ( T s m i 0 z , T s m i 0 z ) < r z + r x 2 ,

which is not possible. □

Lemma 3.6 Let T:XX and let p:X×XR be any mapping satisfying (P3). Suppose that xX is such that T k x=x holds for some positive integer k, and that there exists yX such that

p(y,y)= lim i p ( y , T i x ) = lim i , j p ( T i x , T j x ) .
(3.14)

Then Tx=x.

Proof From T k i x=x, i0, we have

p(y,y)= lim i p ( y , T k i x ) =p(y,x)andp(y,y)= lim i p ( T k i x , T k i x ) =p(x,x),

hence y=x. But (3.14) now gives

p(x,x)= lim i p ( x , T k i + 1 x ) =p(x,Tx)andp(x,x)= lim i p ( T k i + 1 x , T k i + 1 x ) =p(Tx,Tx)

so Tx=x. □

4 Main results

Having made the necessary preparations, we are now able to prove fixed point results for C 1 -operators on complete partial metric spaces. But first we prove a proposition giving some insight into the structure of this type of mappings.

Proposition 4.1 If T is a C 1 -operator on a complete partial metric space (X,p), then

  1. (1)

    for each xX, the sequence { T i x } i 0 p s -converges to some v x X;

  2. (2)

    for all x,yX, there holds p( v x , v y )=max{p( v x , v x ),p( v y , v y )}.

Proof The existence of such points v x is assured by Lemmas 3.4 and 3.5 and completeness if xX R T (X), and is self-evident if x R T (X).

To prove (2), let x,yX be arbitrary and suppose that p( v x , v x )p( v y , v y ). If p( v x , v y )=0, then v x = v y (by (P2) and (P3)) and we are done. Thus assume that p( v x , v y )>0 and let ε>0 be arbitrary such that 2 ε ( 2 + α ) 1 α <p( v x , v y ). There is some m 0 N such that for all i,j m 0 there holds

max { | p ( T i y , T j y ) p ( v y , v y ) | , | p ( v y , T j y ) p ( v y , v y ) | } < ε , max { | p ( T i x , T j x ) p ( v x , v x ) | , | p ( v x , T j x ) p ( v x , v x ) | } < ε .

For i,j m 0 we have

p ( T i y , T j x ) p ( T i y , v y ) p( v y , v y )+p( v y , v x )p( v x , v x )+p ( v x , T j x ) <2ε+p( v y , v x )

and, similarly,

p ( v y , v x ) p ( v y , T i y ) p ( T i y , T i y ) + p ( T i y , T j x ) p ( T j x , T j x ) + p ( T j x , v x ) < 4 ε + p ( T i y , T j x ) .

Fix any i 1 m 0 and set i 1 :=n( T i 0 x). By (3.1) there is some j 0 { i 0 ,, i 0 + i 1 } such that

p ( v y , v x ) 4 ε < p ( T i 0 + i 1 x , T i 0 + i 1 y ) max { α p ( T i 0 x , T j 0 y ) , α p ( T i 0 x , T i 0 + i 1 x ) , p ( T i 0 x , T i 0 x ) , p ( T i 0 + i 1 1 y , T i 0 + i 1 1 y ) } max { α [ 2 ε + p ( v y , v x ) ] , α [ p ( v x , v x ) + ε ] , p ( v x , v x ) + ε , p ( v y , v y ) + ε } = max { α [ 2 ε + p ( v y , v x ) ] , p ( v x , v x ) + ε } .

Now p( v y , v x )4ε<α[2ε+p( v y , v x )] is just 2 ε ( 2 + α ) 1 α >p( v x , v y ), which is false by our choice of ε. This leaves us with the only other possibility: p( v y , v x )4ε<p( v x , v x )+ε, i.e., p( v y , v x )<p( v x , v x )+5ε.

From the preceding analysis it follows that p( v y , v x )p( v x , v x ), which by (P2) actually means that p( v y , v x )=p( v x , v x )=max{p( v x , v x ),p( v y , v y )}. □

Theorem 4.1 If T is a C 1 -operator on a complete partial metric space (X,p), then there is a fixed point zX of T such that p(z,z)= inf x X p( v x , v x ), where v x are as in Proposition  4.1.

Proof For xX put r x :=p( v x , v x ) (this is consistent with the notation of Lemma 3.5). Set I:= inf x X r x . For m1 pick x m X such that for all i,j0 it holds

p ( T i x m , T j x m ) (I1/m,I+1/m).
(4.1)

(You can first pick x m X such that lim i , j p( T i x m , T j x m )= v x m [I,I+ 1 m ), then choose k(m)N such that for all i,jk(m) there holds I 1 m <p( T i x m , T j x m )<I+ 1 m and finally put x m := T k ( m ) x m .)

Notice that if i(m) and j(m) are nonnegative integers for mN, then we have

lim m p ( T i ( m ) x m , T j ( m ) x m ) =0.

First we prove that lim m , k p( x m , x k )=I.

For m,k2 let C m , k >0 be such that p( T i x m , T j x k )< C m , k holds for all i,j0.

Fix m,k2 and let { s q } q N be the supporting sequence at x m . Let l1 be any integer such that α l C k < 1 k + m . We have

p ( x m , x k ) p ( x m , T s l x m ) p ( T s l x m , T s l x m ) + p ( T s l x m , T s l x k ) + p ( x k , T s l x k ) p ( T s l x k , T s l x k ) .

Now

δ m , k :=p ( x m , T s l x m ) p ( T s l x m , T s l x m ) <2/m

and

μ m , k :=p ( x k , T s l x k ) p ( T s l x k , T s l x k ) <2/k.

First suppose that p( T s l x m , T s l x k )>p( T i x k , T i x k ) for all i{0,, s l }, and p( T s l x m , T s l x k )>p( T i x m , T j x m ) for all i,j{0,, s l }. Then, by repeated use of (3.1), we obtain

p ( T s l x m , T s l x k ) α p ( T s l 1 x m , T i 1 x k ) for some  i 1 s l 1 , p ( T s l x m , T s l x k ) α 2 p ( T s l 2 x m , T i 2 x k ) for some  i 2 s l 2 ,

and continuing in this manner finally

p ( T s l x m , T s l x k ) α l p ( x m , T i l x k ) for some  i l 0.

Thus p( T s l x m , T s l x k ) α l C m , k < 1 k + m .

On the other hand, if p( T s l x m , T s l x k )p( T i x k , T i x k ) for some i{0,, s l }, or p( T s l x m , T s l x k )p( T i x m , T j x m ) for some i,j{0,, s l }, then by (4.1) we must have p( T s l x m , T s l x k )<I+max{ 1 m , 1 k }.

Therefore

p ( x m , x k ) δ m , k + μ m , k + p ( T s l x m , T s l x k ) < 2 ( 1 m + 1 k ) + I + max { 1 m , 1 k } .

From the above considerations and from I1/m<p( x m , x m )p( x m , x k ), it is now clear that lim m , k p( x m , x k )=I.

So by completeness there is some uX such that

I= lim m , k p( x m , x k )= lim k p(u, x k )=p(u,u).
(4.2)

Let { s m } m 0 be the supporting sequence at u.

Let us show by induction on k that if f:N N 0 is such that f(m) s k , for all mN, then

I= lim m p ( T s k u , T f ( m ) x m ) =p(u,u).
(4.3)

Suppose first that k=0.

p ( u , u ) p ( u , T f ( m ) x m ) p ( u , x m ) + p ( x m , T f ( m ) x m ) p ( x m , x m ) < p ( u , x m ) + 2 m ,

so the desired conclusion immediately follows. Now suppose that the assertion is true for some k0, take any f:N N 0 such that f(m) s k + 1 , m1, and proceed as follows.

We have

p ( T s k u , T s k u ) 2p ( T s k u , T f ( m ) x m ) p ( T f ( m ) x m , T f ( m ) x m ) 2p ( T s k u , T f ( m ) x m ) I+ 1 m

for all mN, so that taking the limit above as m approaches infinity and using (4.3) (which is justified since f(m) s k + 1 > s k ) it follows that p( T s k u, T s k u)I.

Now for each mN, since f(m)( s k + 1 s k ) s k , there must be some h(m){ s k ,,f(m)} such that

p ( T s k + 1 u , T f ( m ) x m ) max { α p ( T s k u , T h ( m ) x m ) , p ( T s k u , T s k u ) , p ( T f ( m ) 1 x m , T f ( m ) 1 x m ) , α p ( T s k u , T s k + 1 u ) } .

Using p( T s k u, T s k + 1 u)p( T s k u, T f ( m ) x m )p( T f ( m ) x m , T f ( m ) x m )+p( T s k + 1 u, T f ( m ) x m ), we proceed to obtain

p ( T f ( m ) x m , T f ( m ) x m ) p ( T s k + 1 u , T f ( m ) x m ) max { α p ( T s k u , T h ( m ) x m ) , I , I + 1 m , α 1 α [ p ( T s k u , T f ( m ) x m ) p ( T f ( m ) x m , T f ( m ) x m ) ] } .

Now we have I1/m<p( T f ( m ) x m , T f ( m ) x m )<I+1/m and also h(m) s k and f(m) s k + 1 > s k for all m1. Hence, in view of the induction hypothesis, we finally arrive at lim m p( T s k + 1 u, T f ( m ) x m )=I.

Using (4.3) it is straightforward to see that for all k 1 , k 2 0 there holds p( T s k 1 u, T s k 2 u)I=p(u,u): indeed this follows by letting mN tend to infinity in

p ( T s k 1 u , T s k 2 u ) p ( T s k 1 u , T s k 2 x m ) +p ( T s k 2 u , T s k 2 x m ) p ( T s k 2 x m , T s k 2 x m ) .

Thus r u I. But by definition of I we must actually have I= r u .

We now claim that there are positive integers k 1 < k 2 such that

p ( T s k 1 u , T s k 1 u ) =p ( T s k 2 u , T s k 2 u ) =I.

Assume this is not the case. Then p( T s k u, T s k u)=I can hold for at most one kN. As we have 0p( T s k u, T s k u)I for all kN, our assumption implies in particular that r u =I>0. Thus we can take some ε>0 such that r u ε>α( r u +ε). The assumption also allows us to find some m 0 N such that for all k with s k m 0 we have p( T s k u, T s k u)<I, and such that for all i,j m 0 it holds p( T i u, T j u)( r u ε, r u +ε) (remember that r u =p( v u , v u )= lim i , j p( T i u, T j u)).

Take any k with s k m 0 . Then L:=max{p( T s k u, T s k u),p( T s k + 1 u, T s k + 1 u)}<I= r y . There is some positive ε 1 <ε such that L< r u ε 1 . Let i be the smallest integer with i> s k + 1 such that p( T i u, T i u)> r u ε 1 , and let mN be the greatest integer such that s m i. So mk+1. By (3.1) there is some j s m 1 such that

p ( T i u , T i u ) p ( T s m u , T i u ) max { α p ( T s m 1 u , T j u ) , p ( T s m 1 u , T s m 1 u ) , p ( T i 1 u , T i 1 u ) } .

Clearly, we have s m 1 s k m 0 and i1 s k + 1 m 0 . The minimality of i and m and the fact that L< r u ε 1 can now easily be used to deduce that p( T i u, T i u)>max{p( T s m 1 u, T s m 1 u),p( T i 1 u, T i 1 u)}. Therefore r u ε< r u ε 1 <p( T i u, T i u)αp( T s m 1 u, T j u)α( r u +ε) and this cannot be true by the choice of ε.

So we have proved that there are positive integers k 1 < k 2 such that p( T s k 1 u, T s k 1 u)=p( T s k 2 u, T s k 2 u)=I. Since p( T s k 1 u, T s k 2 u)I, we must have T s k 1 u= T s k 2 u (by (P2) and (P3)), i.e., T s k 2 s k 1 z=z for z:= T s k 1 u. As p( v z , v z )= lim i p( v z , T i z)= lim i , j p( T i z, T j z) and s k 2 s k 1 N, by Lemma 3.6 it follows Tz=z. Of course lim i , j p( T i z, T j z)= lim i , j p( T i u, T j u)= r u =I= inf x X p( v x , v x ). □

Remark 4.1 To ensure uniqueness of the fixed point, we can strengthen condition (3.1) as follows. Given a partial metric space (X,p), call T:XX a C 2 -operator if for each xX there is some n(x)N such that for each yX there holds

p ( T n ( x ) x , T n ( x ) y ) max { α p ( x , T j y ) , α p ( x , T n ( x ) x ) , p ( x , x ) + p ( T n ( x ) 1 y , T n ( x ) 1 y ) 2 }
(4.4)

for some j{0,1,,n(x)}. Evidently, each C 2 -operator is a C 1 -operator as well so that if (X,p) is complete, the conclusion of Theorem 4.1 holds. But now, in addition, if Ta=a and Tb=b, then

p(a,b)=p ( T n ( a ) a , T n ( b ) b ) max { α p ( a , b ) , α p ( a , a ) , p ( a , a ) + p ( b , b ) 2 }

so that either (1α)p(a,b)0 or p s (a,b)=2p(a,b)p(a,a)p(b,b)=0, meaning that in any case we must have a=b.

Recall that a sequence x n in a partial metric space (X,p) is called 0-Cauchy with respect to p (see, e.g., [29]) if lim m , n p( x n , x m )=0. We say that (X,p) is 0-complete if every 0-Cauchy sequence in X p s -converges to some xX (for which we then necessarily must have p(x,x)=0). Note that every 0-Cauchy sequence in (X,p) is Cauchy in (X, p s ), and that every complete partial metric space is 0-complete.

Remark 4.2 Recently a very interesting paper by Haghi, Rezapour and Shahzad [45] showed up in which the authors associated to each partial metric space (X,p) a metric space (X,d) by setting d(x,x)=0 and d(x,y)=p(x,y) if xy, and proved that (X,p) is 0-complete if and only if (X,d) is complete. They then proceeded to demonstrate how using the associated metric d some of the fixed point results in partial metric spaces can easily be deduced from the corresponding known results in metric spaces.

Let us point out that these considerations cannot apply to C 1 -operators since the terms p(x,x) and p( T n ( x ) 1 y, T n ( x ) 1 y) on the right-hand side of (3.1) are not multiplied by α. Thus our Theorem 4.1 cannot follow from the result of Ćirić it generalizes.

If we completely neglect the role of self-distances in (3.1), we can easily verify that the statement of Theorem 1.2 remains valid upon substituting the words ‘partial metric’ for ‘metric’ and ‘0-complete’ for ‘complete’. We will prove this using the approach of Haghi, Rezapour and Shahzad [45] that will allow us to deduce Theorem 4.2 directly from Ćirić’s result (Theorem 1.2).

Theorem 4.2 If T is a C-operator on a 0-complete partial metric space (X,p), then there is a unique fixed point z of T. Furthermore, we have p(z,z)=0 and for each xX the sequence { T i x } i 0 p s -converges to z.

Proof Let d be defined as in Remark 4.2. So (X,d) is a complete metric space (see Proposition 2.1 of [45]). Observe that we have d(x,y)p(x,y) for all x,yX. For xX let n(x)N be as in (3.2), and for x,yX set

S(x,y)= { y , T y , T 2 y , , T n ( x ) y , T n ( x ) x }

and M p =max{p(x,z)zS(x,y)}, M d =max{d(x,z)zS(x,y)}. We thus have that

p ( T n ( x ) x , T n ( x ) y ) α M p (x,y)

for all x,yX. We check that for all x,yX it holds that d( T n ( x ) x, T n ( x ) y)α M d (x,y), so that Theorem 1.2 can immediately be applied.

Case 1. There is some zS(x,y) such that xz and M p (x,y)=p(x,z). Here we have

d ( T n ( x ) x , T n ( x ) y ) p ( T n ( x ) x , T n ( x ) y ) α M p (x,y)=αp(x,z)=αd(x,z)α M d (x,y).

Case 2. For all zS(x,y) we have that M p (x,y)=p(x,z)x=z. So it must be M p (x,y)=p(x,x), in particular, and hence p(x, T n ( x ) x) M p (x,y)=p(x,x)p(x, T n ( x ) x), i.e., M p (x,y)=p(x, T n ( x ) x). But by our assumption it now follows that x= T n ( x ) x. Similarly, from p(x, T n ( x ) y) M p (x,y)=p(x,x)p(x, T n ( x ) y), we obtain M p (x,y)=p(x, T n ( x ) y) and consequently x= T n ( x ) y. Now d( T n ( x ) x, T n ( x ) y)=d(x,x)=0α M d (x,y). □

Remark 4.3 It should be pointed out, however, that even though the results of Haghi et al. can deduce the same fixed point as the corresponding partial metric fixed point result, using the partial metric version computers evaluate faster since many nonsense terms are omitted. This is very important from the aspect of computer science due to its cost and explains the vast body of partial metric fixed point results found in literature.

Given a C 1 -operator and a point x, one may ask what the minimal value of n(x) is for which inequality (3.1) holds true. In the following example, for an arbitrary positive integer m, we construct a C 1 -operator on a complete partial metric space (X,p) such that for some xX it must be n(x)>m.

Example 4.1 Denote by X the set of all sequences x:NN and for nN by X n the set of all n-tuples x:{1,,n}N of positive integers. Put X:= X n N X n . For x,yX set

I(x,y)= { i N { 0 } [ j dom ( x ) dom ( y ) j i ] x ( j ) = y ( j ) }

and define p(x,y):=inf{ 1 2 i iI(x,y)} (thus if x(1)y(1), then I(x,y)={0} and p(x,y)=1). Here ‘dom(x)’ stands for the domain of the function x. Then (X,p) is a partial metric space (see [7]) and a complete one as can easily be verified.

Fix l 0 N and define T:XX as follows. For xX let I x ={iNx(i)i}.

If I x =, then set Tx=x. If I x , then define Tx=y by x X n y X n , x X y X and the following two conditions:

  • if I x is finite and has at most l 0 elements, then y(i)=i if i=max I x , and y(i)=x(i) else;

  • if I x is either infinite or finite with more than l 0 elements, then y(i)=i if i=min I x , and y(i)=x(i) else.

Let us show that T is a C 1 -operator with n(x)= l 0 for all xN. Before we proceed, observe that if k is a nonnegative integer such that k+1dom(x) and x(i)=i for 1ik, then for y= T l 0 x we must have y(i)=i for all 1ik+1.

Case 1. There is a nonnegative integer i with i+1dom(x)dom(y) such that x(i+1)i+1y(i+1)i+1. Denote by k the least such nonnegative integer.

If x(k+1)y(k+1), then p( T l 0 x, T l 0 y) 1 2 k + 1 = 1 2 1 2 k = 1 2 p(x,y).

If x(k+1)=y(k+1), then since x(k+1)k+1y(i+1)i+1 we must actually have x(k+1)=y(k+1)k+1 and thus p(x, T l 0 x)= 1 2 k . Hence p( T l 0 x, T l 0 y) 1 2 k + 1 = 1 2 p(x, T l 0 x).

Case 2. x=(1,2,,k) for some kN and xy. Here p( T l 0 x, T l 0 y)= 1 2 k =p(x,x).

Case 3. y=(1,2,,k) for some kN and yx. Here p( T l 0 x, T l 0 y)= 1 2 k =p(y,y)=p( T l 0 1 y, T l 0 1 y).

Condition (4.4) fails because the fixed point is not unique. So T is not a C 2 -operator, hence not a C-operator either.

Now suppose that l 1 < l 0 is an arbitrary positive integer and take x,y N l 1 + 1 such that x(i)=1 for all 1i l 1 +1, and y(1)=2, y(i)=1 for all 2i l 1 +1.

We have p( T l 1 x, T l 1 y)=1=p(x, T j y), for 0j l 1 , p(x, T l 1 x)= 1 2 and p( T l 1 1 y, T l 1 1 y)=p(x,x)= 1 2 l 1 + 1 . So we see that for this particular choice of x and y, substituting l 1 for n(x) in (3.1) makes the inequality false.

Let us use this very example to illustrate Proposition 4.1. Let t X be defined by t(i)=i for all iN. For nN let t n X n be defined by t n (i)=i for i= 1 , n ¯ .

If x X , we clearly have v x =t. Similarly, if x X n , then v x = t n . So p(t, t n )= 1 2 n =max{p( t n , t n ),p(t,t)} because p( t n , t n )= 1 2 n and p(t,t)=0. Also

p( t m , t n )= { 1 2 n = p ( t n , t n ) if  n m , 1 2 m = p ( t m , t m ) if  m n ,

thus p( t m , t n )=max{p( t n , t n ),p( t m , t m )}.

Remark 4.4 It should be noted that if in Theorem 4.1 we require n(x) to be equal to 1 for all xX, then Theorem 2.2 is obtained as a corollary. On the other hand, as already pointed out, if in Theorem 4.2 p is a complete (ordinary) metric on X, then the result of Ćirić (Theorem 1.2) is recovered.

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Acknowledgements

This work was supported by Grant No. 174025 of the Ministry of Science, Technology and Development, Republic of Serbia.

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Correspondence to Vladimir Pavlović.

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Ilić, D., Pavlović, V. & Rakočević, V. Fixed points of mappings with a contractive iterate at a point in partial metric spaces. Fixed Point Theory Appl 2013, 335 (2013). https://doi.org/10.1186/1687-1812-2013-335

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Keywords

  • fixed point
  • partial metric space
  • contractive iterate at a point