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A Counterexample on a Theorem by Khojasteh, Goodarzi, and Razani


In the paper by Khojasteh et al. (2010), the authors tried to generalize Branciari's theorem, introducing the new integral type contraction mappings. In this note we give a counterexample on their main statement (Theorem 2.9). Also we give a comment explaining what the mistake in the proof is, and suggesting what conditions might be appropriate in generalizing fixed point results to cone spaces, where the cone is taken from the infinite dimensional space.

1. Introduction

In the paper [1], Branciari proved the following fixed point theorem with integral-type contraction condition.

Theorem 1.1.

Let be a complete metric space, , and is a mapping such that for all ,


where is nonnegative measurable mapping, having finite integral on each compact subset of such that for each , . Then has a unique fixed point , such that for each , .

There are many generalizations of fixed point results to the so-called cone metric spaces, introduced by several Russian authors in mid-20th. These spaces are re-introduced by Huang and Zhang [2]. In the same paper, the notion of convergent and Cauchy sequences are given.

Definition 1.2.

Let be a Banach space. By we denote the zero element of . A subset of is called a cone if

(1) is closed, nonempty, and ;

(2), , and imply ;


Given a cone , we define partial ordering on with respect to by if and only if . We will write to indicate that and , whereas will stand for (interior of ).

We say that is a solid cone if and only if .

Let be a solid cone in and let be the corresponding partial ordering on with respect to . We say that is a normal cone if and only if there exists a real number such that implies


for each . The least positive satisfying (1.2) is called the normal constant of .

Definition 1.3.

Let be a nonempty set. Suppose that a mapping satisfies:

(1) for all and if and only if ;

(2) for all ;

(3) for all .

Then, is called a cone metric on , and is called a cone metric space.

Definition 1.4.

Let be a solid cone metric space, let , and let be a sequence in . Then

(1) converges to if for every there exists a positive integer such that for all . We denote this by or ;

(2) is a cone Cauchy sequences if for every there exists a positive integer such that for all ;

(3) is a complete cone metric space if every Cauchy sequence is convergent.

In the paper [3] Khojasteh et al. tried to generalize Branciari fixed point result to the cone metric spaces. They introduce the concept of integration along the interval as a limit of Cauchy sums.

Definition 1.5 (see [3]).

Let be a normal solid cone, and let . We say that is integrable on if and only if the following sums:


converge to the same element of , where form a partition of . Clearly, stands for . This element is denoted by


We say that is subadditive if and only if for any there holds


Using this concept, they stated the following statement (Theorem in [3]).

Theorem 1.6 (see [3]).

Let be a complete cone metric space and let be a normal cone. Suppose that is a nonvanishing map which is subadditive cone integrable on each and such that for each , . If is a map such that for all ,


for some , then has a unique fixed point in .

However, the last statement is not true. This will be proved in the next section.

2. Constructing the Counterexample

Consider the Banach space


with the norm , and the cone


It is obvious that is a normal solid cone with normal constant equals to .

Consider the set


and the mapping given by


Proposition 2.1.


  1. (b)

    is a cone metric space.

  2. (c)

    A sequence is convergent (in ) to if and only if . Also is Cauchy sequence (in ) if and only if is a Cauchy sequence with respect to norm in .

  3. (d)

    is a complete cone metric space.


  1. (a)

    and (b) obvious.

  2. (c)

    Let (in ), and let . Then, the function , and we have that for all and for all there holds


Let , and let . Then there exists such that for all , . Also for all there holds


Hence, implying .

The similar argument proves the second part of the statement concerning Cauchy sequences.

  1. (d)

    The set can be represented as , where is the bounded linear functional given by . Therefore, is a closed subset of and hence complete in the norm. By part (c) of this proposition, it implies that is a complete cone metric space.

Let denote the function identically equal to . Consider the mapping given by


for and .

Proposition 2.2.

(a) is integrable on every segment and .

(b) is a nonvanishing subadditive function such that for all there holds .


  1. (a)

    The integrability of on follows immediately from its continuity. Further, let be a partition of . Then for some partition of , and we have


and similarly .

  1. (b)

    Follows from the part (a).

Proposition 2.3.

Let the space be defined by (2.1) and (2.3). Let be given by for , and , otherwise, and let .

The space together with the mappings and satisfies all assumptions of Theorem 1.6. On the other hand, has no fixed point.


We only have to check the inequality (1.6). Note that for all and all we have . Also, note that is a linear mapping, and . Therefore . Thus (1.6) becomes


Taking into account Proposition 2.2, part (a), we have for all


Putting , we obtain


which completes the proof of the first statement.

On the other hand, has no fixed point. Namely, if we suppose that is a fixed point for , it means that for all , and moreover for all , and also for all , by induction. By continuity of , it follows that implying !

3. A Comment

The mistake in the proof of Theorem 1.6 given in [3] is in the following. The authors from conclude that also, which is unjustifiable. The original Branciari's proof [1] deals with one-dimensional integral, and such conclusion is valid due to the implication


and the existence of the total ordering on . However, in infinite dimensional case, such conclusion invokes continuity of the function inverse to . Even for the linear mappings this is not always true, but only under additional assumption that initial mapping is bijective. This asserts the well known Banach open mapping theorem. In the absence of some generalization of the open mapping theorem to nonlinear case, it is necessary to include continuity of the inverse function in the assumptions, as it was done in [4].


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Correspondence to IvanD Arandelović.

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Arandelović, I., Kečkić, D. A Counterexample on a Theorem by Khojasteh, Goodarzi, and Razani. Fixed Point Theory Appl 2010, 470141 (2010).

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