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On fixed point theorems for mappings with PPF dependence

Abstract

In this paper, we consider two families Ψ 1 and Ψ 2 of mappings defined on [0,+) satisfy some certain properties. Using the mentioned properties for Ψ 1 and Ψ 2 , we prove the analogous PPF dependent fixed point theorems for mappings as in Drici et al. (Nonlinear Anal. 67:641-647, 2007) in partially ordered Banach spaces where mappings satisfy the weaker contractive conditions without assuming the topological closedness with respect to the norm topology for the Razumikhin class R c .

MSC:54H25, 55M20.

1 Introduction and preliminary results

The fixed point theorems for mappings satisfying certain contractive conditions have been continually studied for decade (see [19] and references contained therein). Bernfeld et al. [10] proved the existence of PPF (past, present and future) dependent fixed points in the Razumikhin class for mappings that have different domains and ranges. After that Dhage [11] extended the existence of PPF dependent fixed points to PPF common dependent fixed points for mappings satisfying the weaker contractive conditions. In 2007, Drici et al. [2] proved the fixed point theorems in partially ordered metric spaces for mappings with PPF dependence. In this paper, we consider two families Ψ 1 and Ψ 2 of mappings defined on [0,+) satisfy some certain properties. Moreover, the PPF dependent fixed point theorems for mappings satisfying some generalized contractive conditions in partially ordered Banach spaces are proven using the mentioned properties for Ψ 1 and Ψ 2 .

Suppose that E is a real Banach space with the norm E and I is a closed interval [a,b] in . Let E 0 =C(I,E) be the set of all continuous E-valued mappings on I equipped with the supremum norm E 0 defined by

ϕ E 0 = sup t I ϕ ( t ) E ,
(1.1)

for all ϕ E 0 . For a fixed element cI, the Razumikhin class of mappings in E 0 is defined by

R c = { ϕ E 0 : ϕ E 0 = ϕ ( c ) E } .
(1.2)

Recall that a point ϕ E 0 is said to be a PPF dependent fixed point or a fixed point with PPF dependence of T: E 0 E if Tϕ=ϕ(c) for some cI.

Example 1.1 Let T:C([0,1],R)R be defined by

Tϕ= 1 2 ( sup t [ 0 , 1 ] | ϕ ( t ) | ) for all ϕC ( [ 0 , 1 ] , R ) .

Therefore T is a contraction with a constant 1 2 . Suppose that ϕ(t)= t 2 +1 for all t[0,1]. Since Tϕ= 1 2 ( sup t [ 0 , 1 ] |ϕ(t)|)=1=ϕ(0), we find that ϕ is a fixed point with dependence of T.

Definition 1.2 Let A be a subset of E. Then

  1. (i)

    A is said to be topologically closed with respect to the norm topology if for each sequence { x n } in A with x n x as n implies xA.

  2. (ii)

    A is said to be algebraically closed with respect to the difference if xyA for all x,yA.

Recently, Dhage [11] proved the existence of PPF fixed points for mappings satisfying the condition of Cirić type generalized contraction assuming topological closedness with respect to the norm topology for a Razumikhin class.

Definition 1.3 (Dhage, [11])

A mapping T: E 0 E is said to satisfy the condition of Cirić type generalized contraction if there exists a real number λ[0,1) satisfying

T ϕ T α λ max { ϕ α E 0 , ϕ ( c ) T ϕ E , α ( c ) T α E , 1 2 [ ϕ ( c ) T α E + α ( c ) T ϕ E ] } ,
(1.3)

for all ϕ,α E 0 and for some cI.

Theorem 1.4 (Dhage, [11])

Suppose that T: E 0 E satisfies the condition of Cirić type generalized contraction. Assume that R c is topologically closed with respect to the norm topology and is algebraically closed with respect to the difference, then T has a unique PPF dependent fixed point in R c .

It is a natural question that the result of the previous theorem is still valid by omitting the topological closedness of R c . In the next result, we will answer the question.

Proposition 1.5 The Razumikhin class R c is topologically closed with respect to the norm topology.

Proof Let { ϕ n } be a sequence in R c converging to ϕ. This implies that

lim n ϕ n ϕ E 0 =0where  ϕ n ϕ E 0 = sup t I ϕ n ( t ) ϕ ( t ) E .

Therefore

lim n ϕ n E 0 = ϕ E 0 and lim n ϕ n ( t ) E = ϕ ( t ) E for all tI.

Since ϕ n R c for all nN, we obtain ϕ n E 0 = ϕ n ( c ) E . Therefore

lim n ϕ n ( c ) E = ϕ E 0 .

By the uniqueness of the limit, we have ϕ E 0 = ϕ ( c ) E . Hence ϕ R c and thus R c is topologically closed with respect to the norm topology. □

Hence, using Proposition 1.5, we can drop the topological closedness with respect to the norm topology for R c in Theorem 1.4.

The following example shows that the algebraic closedness with respect to the difference of Razumikhin class R c may fail.

Example 1.6 Let E 0 =C([0,1],R) and c=1. If we take ϕ(t)= t 2 and α(t)=t for all t[0,1], then ϕ,α R c while ϕα R c .

Proposition 1.7 If the Razumikhin class R c is algebraically closed with respect to the difference, then R c is a convex set.

Proof Since R c is algebraically closed with respect to the difference, we have R c R c R c . Using the fact that R c = R c , we obtain R c + R c R c . Since λ R c R c for all λ[0,1], we get λ R c +(1λ) R c R c . Hence R c is a convex set. □

One can verify that Razumikhin class R c is a cone (i.e., λϕ R c , for each ϕ R c and λ0). Then by applying the previous theorem Razumikhin class R c is a convex cone (also closed).

In 2007, Drici et al. [2] proved the following the fixed point theorems in partially ordered complete metric spaces for mappings with PPF dependence.

Theorem 1.8 ([2])

Let (E,d,) be a partially ordered complete metric space and T: E 0 E where E 0 =C(I,E) and I=[a,b]. Assume that

  1. (i)

    T is a nondecreasing mapping;

  2. (ii)

    for all ϕ,α E 0 with ϕα, d(Tϕ,Tα)k d 0 (ϕ,α) where d 0 (ϕ,α)= max s I d(ϕ(s),α(s)) and k[0,1);

  3. (iii)

    there exists a lower solution ϕ 0 such that ϕ 0 (c)T ϕ 0 ;

  4. (iv)

    T is a continuous mapping or if { ϕ n } is a nondecreasing sequence in E 0 converging to ϕ E 0 , then ϕ n ϕ for all nN.

Then T has a PPF dependent fixed point in E 0 .

It is a natural question if one can obtain the result of the aforementioned theorem for a generalized contraction (that is, condition (ii)). One of the aims of this paper is to answer the question by considering a general case of Cirić type generalized contractions.

In this paper, we consider two families Ψ 1 and Ψ 2 of mappings defined on [0,+) satisfying some certain properties. Using the mentioned properties for Ψ 1 and Ψ 2 , we prove the analogous PPF dependent fixed point theorems for mappings as in [2] in partially ordered real Banach spaces where mappings satisfy the weaker contractive conditions.

2 PPF dependent fixed points in partially ordered Banach spaces

We begin this part with the consideration of the example of a partially ordered real Banach space. Recall that the set B(X,R) of all bounded linear operators from a normed space X into is a real Banach space with a norm defined by

f= sup x X , x = 1 | f ( x ) | for all fB(X,R).

We know that B(X,R) is a partially ordered real Banach space with a partial order defined as follows:

fgif and only iff(x)g(x)for all xX.

From now on, let (E,) be a partially ordered real Banach space. In this paper, we use the following notations:

Ψ 1 = { ψ : ψ : [ 0 , + ) [ 0 , + )  is nondecreasing with n = 1 ψ n ( t ) <  for all  t ( 0 , + ) } and Ψ 2 = { ψ : ψ : [ 0 , + ) [ 0 , + )  is continuous nondecreasing with ψ ( t ) < t  for all  t ( 0 , + ) } .

Lemma 2.1 ([12])

Suppose that ψ:[0,+)[0,+). If ψ is nondecreasing, then for each t(0,+), lim n ψ n (t)=0 implies ψ(t)<t.

Hence the difference between an element of Ψ 1 and an element of Ψ 2 is continuity.

Remark 2.2

  1. (i)

    It is easily seen that if ψ:[0,+)[0,+) is nondecreasing and ψ(t)<t for all t(0,+), then ψ(0)=0.

  2. (ii)

    We can see that if ψ:[0,+)[0,+), ψ(t)<t for all t(0,+) and ψ(0)=0, then ψ is continuous at 0.

We now prove the PPF dependent fixed point theorems for mappings satisfying the generalized contractive conditions concerning with ψ Ψ 1 without assuming the topological closedness with respect to the norm topology for the Razumikhin class R c .

Theorem 2.3 Suppose that ψ Ψ 1 , cI and T: E 0 E satisfies the following conditions:

  1. (i)

    T is a nondecreasing mapping;

  2. (ii)

    for all ϕ,α E 0 with ϕα, we have

    T ϕ T α E ψ ( max { ϕ α E 0 , ϕ ( c ) T ϕ E , α ( c ) T α E , 1 2 [ ϕ ( c ) T α E + α ( c ) T ϕ E ] } ) ;
  3. (iii)

    there exists a lower solution ϕ 0 R c such that ϕ 0 (c)T ϕ 0 ;

  4. (iv)

    T is a continuous mapping.

Assume that R c is algebraically closed with respect to the difference. Then T has a PPF dependent fixed point in R c .

Proof Since ϕ 0 R c and T ϕ 0 E, there exists x 1 E such that T ϕ 0 = x 1 . Choose ϕ 1 R c such that x 1 = ϕ 1 (c). Since ϕ 0 is a lower solution in R c such that ϕ 0 (c)T ϕ 0 , it follows that ϕ 0 ϕ 1 . Using the algebraic closedness with respect to the difference of R c , this yields

ϕ 0 ϕ 1 E 0 = ϕ 0 ( c ) ϕ 1 ( c ) E .

By the fact that T is nondecreasing, we obtain

ϕ 1 (c)=T ϕ 0 T ϕ 1 .

By induction, we can construct the sequence { ϕ n } such that

T ϕ 0 = ϕ 1 ( c ) T ϕ 1 = ϕ 2 ( c ) T ϕ n = ϕ n + 1 ( c ) T ϕ n + 1 , ϕ n ϕ n + 1 and ϕ n ϕ n + 1 E 0 = ϕ n ( c ) ϕ n + 1 ( c ) E ,

for all nN{0}. Assume that ϕ n 1 = ϕ n for some nN. It follows that ϕ n 1 (c)= ϕ n (c)=T ϕ n 1 . Therefore T has a fixed point in R c . Suppose that ϕ n 1 ϕ n for all nN. Therefore, for each nN, we obtain

ϕ n ϕ n + 1 E 0 = ϕ n ( c ) ϕ n + 1 ( c ) E = T ϕ n T ϕ n 1 E ψ ( max { ϕ n ϕ n 1 E 0 , ϕ n ( c ) T ϕ n E , ϕ n 1 ( c ) T ϕ n 1 E , 1 2 [ ϕ n ( c ) T ϕ n 1 E + ϕ n 1 ( c ) T ϕ n E ] } ) = ψ ( max { ϕ n ϕ n 1 E 0 , ϕ n ( c ) ϕ n + 1 ( c ) E , ϕ n 1 ( c ) ϕ n ( c ) E , 1 2 [ ϕ n ( c ) ϕ n ( c ) E + ϕ n 1 ( c ) ϕ n + 1 ( c ) E ] } ) ψ ( max { ϕ n ϕ n 1 E 0 , ϕ n ϕ n + 1 E 0 , 1 2 ϕ n 1 ϕ n + 1 E 0 } ) ψ ( max { ϕ n ϕ n 1 E 0 , ϕ n ϕ n + 1 E 0 , 1 2 ϕ n 1 ϕ n E 0 + ϕ n ϕ n + 1 E 0 } ) ψ ( max { ϕ n ϕ n 1 E 0 , ϕ n ϕ n + 1 E 0 } ) .

If max{ ϕ n ϕ n 1 E 0 , ϕ n ϕ n + 1 E 0 }= ϕ n ϕ n + 1 E 0 , then

ϕ n ϕ n + 1 E 0 ψ ( ϕ n ϕ n + 1 E 0 ) < ϕ n ϕ n + 1 E 0 ,

which leads to a contradiction. Therefore, for each nN, we have

ϕ n ϕ n + 1 E 0 ψ ( ϕ n ϕ n 1 E 0 ) .

By induction, we obtain

ϕ n ϕ n + 1 E 0 ψ n ( ϕ 0 ϕ 1 E 0 ) .

Fix ε>0. This implies that there exists NN such that

n N ψ n ( ϕ 0 ϕ 1 E 0 ) <ε.

For each m,nN with m>n>N, we obtain

ϕ n ϕ m E 0 k = n m 1 ϕ k ϕ k + 1 E 0 k = n m 1 ψ k ( ϕ 0 ϕ 1 E 0 ) n N ψ n ( ϕ 0 ϕ 1 E 0 ) <ε.

This implies that { ϕ n } is a Cauchy sequence. By the completeness of E 0 , we have lim n ϕ n =ϕ for some ϕ E 0 and

lim n T ϕ n = lim n ϕ n + 1 (c)=ϕ(c).

Since R c is algebraically closed with respect to the norm topology, we have ϕ R c . We next prove that ϕ is a PPF dependent fixed point of T. Using the continuity of T, we obtain lim n T ϕ n =Tϕ. By the uniqueness of the limit, we have Tϕ=ϕ(c). □

Remark 2.4

  1. (i)

    From the proof of Theorem 2.3, we assume that the Razumikhin class R c is algebraically closed with respect to difference, that is, ϕα R c for all ϕ,α R c , in order to construct the sequence { ϕ n } satisfying

    ϕ n ϕ n + 1 E 0 = ϕ n ( c ) ϕ n + 1 ( c ) E for all nN{0}.
  2. (ii)

    In the proof of Theorem 2.3, if we choose ϕ n R c to be a constant mapping for each nN{0}, then

    ϕ n ϕ n + 1 E 0 = ϕ n ( c ) ϕ n + 1 ( c ) E for all nN{0}.

    Therefore the algebraic closedness with respect to the difference of R c can be dropped.

Example 2.5 Let E= R 2 with respect to the norm (x,y)=|x|+|y| and E 0 =C([0,1], R 2 ). Define a mapping ψ:[0,)[0,) by

ψ(t)= t 3 for all t[0,).

We see that ψ is a nondecreasing mapping with n = 1 ψ n (t)< for all t(0,+). Let ϕ E 0 . Therefore, for each t[0,1], we obtain ϕ(t) R 2 . Thus we can define mappings f ϕ , g ϕ :[0,1]R such that

ϕ(t)= ( f ϕ ( t ) , g ϕ ( t ) ) for all t[0,1].

Define a mapping T: E 0 E by

Tϕ= ( 1 4 f ϕ , 1 4 g ϕ ) for all ϕ E 0 ,

where f ϕ = sup t [ 0 , 1 ] | f ϕ (t)| and g ϕ = sup t [ 0 , 1 ] | g ϕ (t)|. Suppose that ϕ,α E 0 with ϕα. For each c[0,1], we obtain

T ϕ T α E = ( 1 4 f ϕ , 1 4 g ϕ ) ( 1 4 f α , 1 4 g α ) E = | 1 4 f ϕ 1 4 f α | + | 1 4 g ϕ 1 4 g α | 1 4 ( f ϕ f α + g ϕ g α ) = 1 4 ( sup t [ 0 , 1 ] | f ϕ ( t ) f α ( t ) | + sup t [ 0 , 1 ] | g ϕ ( t ) g α ( t ) | ) = 1 4 ( sup t [ 0 , 1 ] ( | f ϕ ( t ) f α ( t ) | + | g ϕ ( t ) g α ( t ) | ) ) = 1 4 ( sup t [ 0 , 1 ] ( f ϕ ( t ) f α ( t ) , g ϕ ( t ) g α ( t ) ) E ) = 1 4 ( sup t [ 0 , 1 ] ( f ϕ ( t ) , g ϕ ( t ) ) ( f α ( t ) , g α ( t ) ) E ) = 1 4 ( sup t [ 0 , 1 ] ϕ ( t ) α ( t ) E ) = 1 4 ( ϕ α E 0 ) ψ ( max { ϕ α E 0 , ϕ ( c ) T ϕ E , α ( c ) T α E , 1 2 [ ϕ ( c ) T α E + α ( c ) T ϕ E ] } ) .

Suppose that ϕ 0 =0. We see that all assumptions in Theorem 2.3 are now satisfied and 0 is the PPF dependent fixed point of T in R c .

By applying Theorem 2.3, we obtain the following corollary.

Corollary 2.6 Suppose that cI and T: E 0 E satisfies the following conditions:

  1. (i)

    T is a nondecreasing mapping;

  2. (ii)

    for all ϕ,α E 0 with ϕα, we have

    T ϕ T α E k ( max { ϕ α E 0 , ϕ ( c ) T ϕ E , α ( c ) T α E , 1 2 [ ϕ ( c ) T α E + α ( c ) T ϕ E ] } ) , where  k [ 0 , 1 ) ;
  3. (iii)

    there exists a lower solution ϕ 0 R c such that ϕ 0 (c)T ϕ 0 ;

  4. (iv)

    T is a continuous mapping.

Assume that R c is algebraically closed with respect to the difference. Then T has a PPF dependent fixed point in R c .

Proof Define a function ψ:[0,+)[0,+) by ψ(t)=kt for all t[0,+). Therefore ψ is a nondecreasing mapping and

n = 1 ψ n (t)<for all t(0,+).

This implies that all assumptions in Theorem 2.3 are satisfied. Hence we obtain the desired result. □

Theorem 2.7 Suppose that ψ Ψ 1 , cI and T: E 0 E satisfies the following conditions:

  1. (i)

    T is a nondecreasing mapping;

  2. (ii)

    for all ϕ,α E 0 with ϕα, T ϕ T α E ψ( ϕ α E 0 );

  3. (iii)

    there exists a lower solution ϕ 0 R c such that ϕ 0 (c)T ϕ 0 ;

  4. (iv)

    if { ϕ n } is a nondecreasing sequence in E 0 converging to ϕ E 0 , then ϕ n ϕ for all nN.

Assume that R c is algebraically closed with respect to the difference. Then T has a PPF dependent fixed point in R c .

Proof By the analogous proof as in Theorem 2.3, we can construct a nondecreasing sequence { ϕ n } in R c converging to ϕ R c . This implies that ϕ n ϕ for all nN. Therefore, for each nN, we have

T ϕ ϕ ( c ) E T ϕ ϕ n + 1 ( c ) E + ϕ n + 1 ( c ) ϕ ( c ) E T ϕ T ϕ n E + ϕ n + 1 ϕ E 0 ψ ( ϕ ϕ n E 0 ) + ϕ n + 1 ϕ E 0 .

Since ψ is continuous at 0, we get lim n ψ( ϕ ϕ n E 0 )=ψ(0)=0. Taking the limit of the above inequality, this yields T ϕ ϕ ( c ) E =0 and so ϕ is a PPF dependent fixed point of T in R c . □

We next ensure the result on PPF dependent fixed points for mappings concerning with ψ Ψ 2 .

Theorem 2.8 Suppose that ψ Ψ 2 , cI and T: E 0 E satisfies the following conditions:

  1. (i)

    T is a nondecreasing mapping;

  2. (ii)

    for all ϕ,α E 0 with ϕα, we have

    T ϕ T α E ψ ( max { ϕ α E 0 , ϕ ( c ) T ϕ E , α ( c ) T α E , 1 2 [ ϕ ( c ) T α E + α ( c ) T ϕ E ] } ) ;
  3. (iii)

    there exists a lower solution ϕ 0 R c such that ϕ 0 (c)T ϕ 0 ;

  4. (iv)

    T is a continuous mapping.

Assume that R c is algebraically closed with respect to the difference. Then T has a PPF dependent fixed point in R c .

Proof Since ϕ 0 R c and T ϕ 0 E, there exists x 1 E such that T ϕ 0 = x 1 . As in the proof of Theorem 2.3, we can construct the sequence { ϕ n } such that

T ϕ 0 = ϕ 1 ( c ) T ϕ 1 = ϕ 2 ( c ) T ϕ n = ϕ n + 1 ( c ) T ϕ n + 1 , ϕ n ϕ n + 1 and ϕ n ϕ n + 1 E 0 = ϕ n ( c ) ϕ n + 1 ( c ) E ,

for all nN{0}. Assume that ϕ n 1 = ϕ n for some nN. It follows that ϕ n 1 (c)= ϕ n (c)=T ϕ n 1 . Therefore T has a fixed point in R c . Suppose that ϕ n 1 ϕ n for all nN. For each nN, we obtain

ϕ n ϕ n + 1 E 0 = ϕ n ( c ) ϕ n + 1 ( c ) E = T ϕ n T ϕ n 1 E ψ ( max { ϕ n ϕ n 1 E 0 , ϕ n ( c ) T ϕ n E , ϕ n 1 ( c ) T ϕ n 1 E , 1 2 [ ϕ n ( c ) T ϕ n 1 E + ϕ n 1 ( c ) T ϕ n E ] } ) = ψ ( max { ϕ n ϕ n 1 E 0 , ϕ n ( c ) ϕ n + 1 ( c ) E , ϕ n 1 ( c ) ϕ n ( c ) E , 1 2 [ ϕ n ( c ) ϕ n ( c ) E + ϕ n 1 ( c ) ϕ n + 1 ( c ) E ] } ) ψ ( max { ϕ n ϕ n 1 E 0 , ϕ n ϕ n + 1 E 0 , 1 2 ϕ n 1 ϕ n + 1 E 0 } ) ψ ( max { ϕ n ϕ n 1 E 0 , ϕ n ϕ n + 1 E 0 , 1 2 ϕ n 1 ϕ n E 0 + ϕ n ϕ n + 1 E 0 } ) ψ ( max { ϕ n ϕ n 1 E 0 , ϕ n ϕ n + 1 E 0 } ) .

If max{ ϕ n ϕ n 1 E 0 , ϕ n ϕ n + 1 E 0 }= ϕ n ϕ n + 1 E 0 , then

ϕ n ϕ n + 1 E 0 ψ ( ϕ n ϕ n + 1 E 0 ) < ϕ n ϕ n + 1 E 0 .

This leads to a contradiction. Therefore

ϕ n ϕ n + 1 E 0 ψ ( ϕ n ϕ n 1 E 0 ) < ϕ n ϕ n 1 E 0 .

It follows that ϕ n ϕ n + 1 E 0 ϕ n 1 ϕ n E 0 for all nN. Since the sequence { ϕ n ϕ n + 1 E 0 } is a nonincreasing sequence of nonnegative real numbers, we see that it is a convergent sequence. Suppose that

lim n ϕ n ϕ n + 1 E 0 =α,

for some nonnegative real number α. We will prove that α=0. Suppose that α>0. Since

ϕ n ϕ n + 1 E 0 ψ ( ϕ n ϕ n 1 E 0 ) ,

for all nN and the continuity of ψ, we have αψ(α)<α which leads to a contradiction. This implies that α=0. We next prove that the sequence { ϕ n } is a Cauchy sequence in  E 0 . Assume that { ϕ n } is not a Cauchy sequence. It follows that there exist ε>0 and two sequences of positive integers { m k } and { n k } satisfying m k > n k >k for each kN and

ϕ m k ϕ n k E 0 ε.
(2.1)

Let { m k } be the sequence of the least positive integers exceeding { n k } which satisfies (2.1) and

ϕ m k 1 ϕ n k E 0 <ε.
(2.2)

We will prove that lim k ϕ m k ϕ n k E 0 =ε. Since ϕ m k ϕ n k E 0 ε for all kN, we have

lim k ϕ m k ϕ n k E 0 ε.

For each kN, we obtain

ϕ m k ϕ n k E 0 ϕ m k ϕ m k 1 E 0 + ϕ m k 1 ϕ n k E 0 ϕ m k ϕ m k 1 E 0 + ε .

This implies that lim k ϕ m k ϕ n k E 0 ε. Therefore

lim k ϕ m k ϕ n k E 0 =ε.

Similarly, we can prove that

lim k ϕ m k + 1 ϕ n k E 0 =ε, lim k ϕ m k ϕ n k 1 E 0 =ε,

and

lim k ϕ m k + 1 ϕ n k 1 E 0 =ε.

Since R c is algebraically closed with respect to the difference, for each kN, we obtain

ϕ n k ϕ m k + 1 E 0 = ϕ n k ( c ) ϕ m k + 1 ( c ) E = T ϕ m k T ϕ n k 1 E ψ ( max { ϕ m k ϕ n k 1 E 0 , ϕ m k ( c ) T ϕ m k E , ϕ n k 1 ( c ) T ϕ n k 1 E , 1 2 [ ϕ m k ( c ) T ϕ n k 1 E + ϕ n k 1 ( c ) T ϕ n k E ] } ) = ψ ( max { ϕ m k ϕ n k 1 E 0 , ϕ m k ( c ) ϕ m k + 1 ( c ) E , ϕ n k 1 ( c ) ϕ n k ( c ) E , 1 2 [ ϕ m k ( c ) ϕ n k ( c ) E + ϕ n k 1 ( c ) ϕ n k + 1 ( c ) E ] } ) ψ ( max { ϕ m k ϕ n k 1 E 0 , ϕ m k ϕ m k + 1 E 0 , ϕ n k 1 ϕ n k E 0 , 1 2 [ ϕ m k ϕ n k E 0 + ϕ n k 1 ϕ n k + 1 E 0 ] } ) ψ ( max { ϕ m k ϕ n k 1 E 0 , ϕ m k ϕ m k + 1 E 0 , ϕ n k 1 ϕ n k E 0 , 1 2 [ ϕ m k ϕ n k E 0 + ϕ n k 1 ϕ n k E 0 + ϕ n k ϕ n k + 1 E 0 ] } ) .

By taking the limit of both sides, we have

εψ(ε)<ε.

This leads to a contradiction. It follows that the sequence { ϕ n } is a Cauchy sequence. By the completeness of E 0 , we have lim n ϕ n =ϕ for some ϕ E 0 and

lim n T ϕ n = lim n ϕ n + 1 (c)=ϕ(c).

Since R c is algebraically closed with respect to the norm topology, we have ϕ R c . We will prove that ϕ is a PPF dependent fixed point of T. Using the continuity of T, we obtain lim n T ϕ n =Tϕ. By the uniqueness of the limit, we can conclude that Tϕ=ϕ(c). □

Example 2.9 Assume that E=R and E 0 =C([0,1],R). Define a mapping ψ:[0,)[0,) by

ψ(t)= t 2 for all t[0,).

We see that ψ is a continuous nondecreasing mapping with ψ(t)<t. Define a mapping T: E 0 E by

Tϕ= 1 3 ϕ ( 1 4 ) for all ϕ E 0 .

Suppose that ϕ,α E 0 with ψα and c[0,1]. Therefore

T ϕ T α E = 1 3 | ϕ ( 1 4 ) α ( 1 4 ) | 1 3 ϕ α E 0 ψ ( max { ϕ α E 0 , ϕ ( c ) T ϕ E , α ( c ) T α E , 1 2 [ ϕ ( c ) T α E + α ( c ) T ϕ E ] } ) .

Suppose that ϕ 0 =0. We find that all assumptions in Theorem 2.8 are now satisfied and 0 is the PPF dependent fixed point of T in R c .

For the next result, we drop the continuity of T.

Theorem 2.10 Suppose that ψ Ψ 2 , cI, and that T: E 0 E satisfies the following conditions:

  1. (i)

    T is a nondecreasing mapping;

  2. (ii)

    for all ϕ,α E 0 with ϕα, we have

    T ϕ T α E ψ ( max { ϕ α E 0 , ϕ ( c ) T ϕ E , α ( c ) T α E , 1 2 [ ϕ ( c ) T α E + α ( c ) T ϕ E ] } ) ;
  3. (iii)

    there exists a lower solution ϕ 0 R c such that ϕ 0 (c)T ϕ 0 ;

  4. (iv)

    if { ϕ n } is a nondecreasing sequence in E 0 converging to ϕ E 0 , then ϕ n ϕ for all nN.

Assume that R c is algebraically closed with respect to the difference. Then T has a PPF dependent fixed point in R c .

Proof As in the proof of Theorem 2.8, we can construct a nondecreasing sequence { ϕ n } converging to ϕ R c and this yields

lim n T ϕ n = lim n ϕ n + 1 (c)=ϕ(c).

Using (iv), we have ϕ n ϕ for all nN. Therefore, for each nN, we obtain

T ϕ ϕ ( c ) E T ϕ ϕ n + 1 ( c ) E + ϕ n + 1 ( c ) ϕ ( c ) E T ϕ T ϕ n E + ϕ n + 1 ϕ E 0 ψ ( max { ϕ ϕ n E 0 , ϕ ( c ) T ϕ E , ϕ n ( c ) T ϕ n E , 1 2 [ ϕ ( c ) T ϕ n E + ϕ n ( c ) T ϕ E ] } ) + ϕ n + 1 ϕ E 0 = ψ ( max { ϕ ϕ n E 0 , ϕ ( c ) T ϕ E , ϕ n ( c ) ϕ n + 1 ( c ) E , 1 2 [ ϕ ( c ) ϕ n + 1 ( c ) E + ϕ n ( c ) T ϕ E ] } ) + ϕ n + 1 ϕ E 0 ψ ( max { ϕ ϕ n E 0 , ϕ ( c ) T ϕ E , ϕ n ϕ n + 1 E 0 1 2 [ ϕ ϕ n + 1 E 0 + ϕ n ( c ) T ϕ E ] } ) + ϕ n + 1 ϕ E 0 .

Letting n, we obtain T ϕ ϕ ( c ) E ψ( ϕ ( c ) T ϕ E ). If ϕ(c)Tϕ, then

T ϕ ϕ ( c ) E ψ ( ϕ ( c ) T ϕ E ) < ϕ ( c ) T ϕ E .

This leads to a contradiction. Therefore Tϕ=ϕ(c). This implies that ϕ is a PPF dependent fixed point of T. □

By applying Theorem 2.8 and Theorem 2.10, we obtain the following corollary.

Corollary 2.11 Suppose that cI and that T: E 0 E satisfies the following conditions:

  1. (i)

    T is a nondecreasing mapping;

  2. (ii)

    for all ϕ,α E 0 with ϕα, we have

    T ϕ T α E k ( max { ϕ α E 0 , ϕ ( c ) T ϕ E , α ( c ) T α E , 1 2 [ ϕ ( c ) T α E + α ( c ) T ϕ E ] } ) , where  k [ 0 , 1 ) ;
  3. (iii)

    there exists a lower solution ϕ 0 R c such that ϕ 0 (c)T ϕ 0 ;

  4. (iv)

    T is a continuous mapping or if { ϕ n } is a nondecreasing sequence in E 0 converging to ϕ E 0 , then ϕ n ϕ for all nN.

Assume that R c is algebraically closed with respect to the difference. Then T has a PPF dependent fixed point in R c .

Proof Define a function ψ:[0,+)[0,+) by ψ(t)=kt for all t[0,+). Therefore ψ is a continuous nondecreasing mapping and

ψ(t)<tfor all t(0,+).

This implies that all assumptions in Theorem 2.8 or Theorem 2.10 are satisfied. Hence the proof is complete. □

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Acknowledgements

The second author would like to express her deep thanks to the Centre of Excellence in Mathematics, the Commission of Higher Education and Naresuan University, Thailand for the support.

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Farajzadeh, A., Kaewcharoen, A. On fixed point theorems for mappings with PPF dependence. J Inequal Appl 2014, 372 (2014). https://doi.org/10.1186/1029-242X-2014-372

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Keywords

  • PPF dependent fixed points
  • Razumikhin classes
  • partially ordered sets
  • algebraic closedness with respect to difference