We discuss the newly introduced concept of cone metric spaces. We also discuss the fixed point existence results of contractive mappings defined on such metric spaces. In particular, we show that most of the new results are merely copies of the classical ones.
Cone metric spaces were introduced in . A similar notion was also considered by Rzepecki in . After carefully defining convergence and completeness in cone metric spaces, the authors proved some fixed point theorems of contractive mappings. Recently, more fixed point results in cone metric spaces appeared in [3–8]. Topological questions in cone metric spaces were studied in  where it was proved that every cone metric space is first countable topological space. Hence, continuity is equivalent to sequential continuity and compactness is equivalent to sequential compactness. It is worth mentioning the pioneering work of Quilliot  who introduced the concept of generalized metric spaces. His approach was very successful and used by many (see references in ). It is our belief that cone metric spaces are a special case of generalized metric spaces. In this work, we introduce a metric type structure in cone metric spaces and show that classical proofs do carry almost identically in these metric spaces. This approach suggest that any extension of known fixed point result to cone metric spaces is redundant. Moreover the underlying Banach space and the associated cone subset are not necessary.
For more on metric fixed point theory, the reader may consult the book .
2. Basic Definitions and Results
First let us start by making some basic definitions.
Let be a real Banach space with norm and a subset of . Then is called a cone if and only if
(1) is closed, nonempty, and , where is the zero vector in ;
(2)if , and , then ;
(3)if and , then .
Given a cone in a Banach space , we define a partial ordering with respect to by
We also write whenever and , while will stand for (where Int() designate the interior of ). The cone is called normal if there is a number , such that for all , we have
The least positive number satisfying this inequality is called the normal constant of . The cone is called regular if every increasing sequence which is bounded from above is convergent. Equivalently the cone is called regular if every decreasing sequence which is bounded from below is convergent. Regular cones are normal and there exist normal cones which are not regular.
Throughout the Banach space and the cone will be omitted.
A cone metric space is an ordered pair , where is any set and is a mapping satisfying
(1), that is, , for all , and if and only if ;
(2) for all ;
(3), for all .
Convergence is defined as follows.
Let be a cone metric space, let be a sequence in and . If for any with , there is such that for all , , then is said to be convergent. We will say converges to and write .
It is easy to show that the limit of a convergent sequence is unique. Cauchy sequences and completeness are defined by
Let be a cone metric space, be a sequence in . If for any with , there is such that for all , , then is called Cauchy sequence. If every Cauchy sequence is convergent in , then is called a complete cone metric space.
The basic properties of convergent and Cauchy sequences may be found at . In fact the properties and their proofs are identical to the classical metric ones. Since this work concerns the fixed point property of mappings, we will need the following property.
Let be a cone metric space. A mapping is called Lipschitzian if there exists such that
for all . The smallest constant which satisfies the above inequality is called the Lipschitz constant of , denoted . In particular is a contraction if .
As we mentioned earlier cone metric spaces have a metric type structure. Indeed we have the following result.
Let be a metric cone over the Banach space with the cone which is normal with the normal constant . The mapping defined by satisfies the following properties:
(1) if and only if ;
(2), for any ;
(3), for any points , .
The proofs of (1) and (2) are easy and left to the reader. In order to prove (3), let be any points in . Using the triangle inequality satisfied by , we get
Since is normal with constant we get
This completes the proof of the theorem.
Note that the property (3) is discouraging since it does not give the classical triangle inequality satisfied by a distance. But there are many examples where the triangle inequality fails (see, e.g., ).
The above result suggest the following definition.
Let be a set. Let be a function which satisfies
(1) if and only if ;
(2), for any ;
(3), for any points , , for some constant .
The pair is called a metric type space.
Similarly we define convergence and completeness in metric type spaces.
Let be a metric type space.
(1)The sequence converges to if and only if .
(2)The sequence is Cauchy if and only if .
is complete if and only if any Cauchy sequence in is convergent.
3. Some Fixed Point Results
Let be a map. is called Lipschitzian if there exists a constant such that
for any . The smallest constant will be denoted .
Let be a complete metric type space. Let be a map such is Lipschitzian for all and that . Then has a unique fixed point . Moreover for any , the orbit converges to .
Let . For any , we have
Since is convergent, then . This forces to be a Cauchy sequence. Since is complete, then converges to some point . First let us show that is a fixed point of . Since
If we let , we get , or . Next we show that has at most one fixed point. Indeed let and be two fixed points of . Then we have
for any . Since , we get , or . Therefore we have for any , which completes the proof of the theorem.
The condition is needed because of the condition (3) satisfied by . In fact a more natural condition should be
(), for any points , for some constant .
An example of such satisfying is given below.
Let be the set of Lebesgue measurable functions on such that
Then satisfies the following properties:
if and only if ;
, for any ;
, for any points .
In the next result we consider the case of metric type spaces when satisfies . Recall that a subset of is said to be bounded whenever .
Let be a complete metric type space, where satisfies instead of (3). Let be a map such that is Lipschitzian for any and . Then has a unique fixed point if and only if there exists a bounded orbit. Moreover if has a fixed point , then for any , the orbit converges to .
Clearly if has a fixed point, then its orbit is bounded. Conversely let such that is bounded, that is, there exists such that , for any . Let , we have
Since , then is a Cauchy sequence. Hence converges to some point since is complete. The remaining part of the proof follows the same as in the previous theorem.
The connection between the above results and the main theorems of  are given in the following corollary.
Let be a metric cone over the Banach space with the cone which is normal with the normal constant . Consider defined by . Let be a contraction with constant . Then
for any and . Hence , for any . Therefore is convergent, which implies has a unique fixed point , and any orbit converges to .
From the definition of in the above Corollary, we easily see that -convergence and -convergence are identical.
In  the authors gave an example of a map which is contraction for but not for the euclidian distance. From the above corollary, we see that . Since may not be less than 1, then may not be a contraction for . This is why the above theorems were stated in terms of .
Using the ideas described above one can prove fixed point results for mappings which contracts orbits and obtain similar results as Theorem for example in .
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