Suppose that is a metric space and that is a map. We say that is contractive if there exists such that for all . The well-known Banach fixed point theorem states that has a fixed point if and is complete. In 1962, Rakotch [1] obtained an extension of Banach theorem replacing the constant by a function of , , provided that is nonincreasing and for all (for a recent refinement of this result see [2]). A similar generalization of the contractive condition was considered by Dugundji and Granas [3], who extended Banach theorem to the class of weakly contractive mappings (i.e., , with for all ).

Another focus of attention in Fixed Point Theory is to establish fixed point theorems for non-self mappings. In the setting of a Banach space, Gatica and Kirk [4] proved that if is contractive, with an open neighborhood of the origin, then has a fixed point if it satisfies the well-known Leray-Schauder condition:

Recently, Kirk [5] has extended this result to the abstract setting of a certain class of metric spaces: the CAT(0) spaces. In the proof, the author uses a homotopy result due to Granas [6], which is known as continuation method for contractive maps. In fact, the jump from a Banach space setting to the metric space setting was given by Granas himself in [6] (for more information on this topic see, for instance, [7–9]). After Granas, Frigon [8] gave a similar result for weakly contractive maps.

A variant of the Banach contraction principle was given by Kannan [10], who proved that a map , where is a complete metric space, has a unique fixed point if is what we call a Kannan map, that is, there exists such that, for all ,

In this note, following the pattern of Dugundji and Granas [3], we extend Kannan theorem to the class of weakly Kannan maps (i.e., , with for all ). This is done in Section 2. In Section 3 we use a local version of the previous result to obtain a continuation method for weakly Kannan maps.

In this section we follow the pattern of Dugundji and Granas [3] to introduce the concept of weakly Kannan maps.

Definition 2.1.

Let be a metric space, , and . Therefore is a weakly Kannan map if there exists , with for every such that, for all ,

Remark 2.2.

Clearly, any weakly Kannan map has at most one fixed point: if and , then

Remark 2.3.

Notice that if is a weakly Kannan map and we define on as

then is well defined, takes values in , satisfies for all (for is smaller than any associated to ), and also satisfies (2.1), with replaced by , for all . Conversely, if is defined as in (2.3) and satisfies the above set of conditions, then is a weakly Kannan map, establishing in this way an equivalent definition for Kannan maps.

Remark 2.4.

Although Kannan showed that the concept of Kannan map is independent of the concept of contractive map, Janos [11] observed that any contractive map whose Lipschitz constant defined by

is less than is a Kannan map. Next, we exhibit an example of a weakly Kannan map , with , which is not a Kannan map, thus showing that the constant in the aforementioned result by Janos is sharp.

Example 2.5.

Consider the metric space with the usual metric , and let be the function defined as . Then, and is a weakly Kannan map, but not a Kannan map.

The equality follows from the fact that for all together with

We also have that is not a Kannan map because

To check that is a weakly Kannan map, consider the function given by (2.3). This function is well defined and also takes values in since . Next, assume that and let us see that . To see this, observe that as , so there is such that for all . Observe also that , the restriction of to , is a Kannan map with constant , due to the fact that , for is continuously differentiable on and for all . We will see . To do it, suppose that with and . Then, if , use and that to obtain . Otherwise, we would have and then .

Although the way we have introduced the concept of weakly Kannan map has been by analogy with the work done by Dugundji and Granas in [3], we would like to mention that this extension may be done in some different ways. For instance, Pathak et al. [12, Theorem 3.1] have proved the following result.

Theorem 2 A.

Let be a complete metric space and suppose that is a map such that

for all , where . If, in addition, there exists a sequence in with , then has a fixed point in .

Observe that relation (2.7) can be written in the following more general form:

for all , where , , and notice that any map satisfying (2.8) also satisfies the relation (2.1) with . In fact, the arguments used by the authors in the proof of Theorem A are also valid for this class of maps. Next, we state this slightly more general result and include the proof for the sake of completeness. Then, we obtain, as a consequence, a fixed point theorem for weakly Kannan maps.

Theorem 2.6.

Let be a complete metric space and assume that is a bounded function satisfying the following condition: for any sequence in and ,

Assume also that is a map such that

for all . If there exists a sequence in with , then has a unique fixed point in , and .

Proof.

Since is bounded, there exists such that for all . Suppose that is a sequence in with and use (2.9) to obtain that, for all ,

This implies that is a Cauchy sequence. Since is complete, the sequence is convergent, say to . Then because . Thus, by (*), .

That is a consequence of the following relation and the fact that , then

Finally, is the unique fixed point of because if :

Corollary 2.7.

Let be a complete metric space and suppose that is a weakly Kannan map. Then, has a unique fixed point and, for any , the sequence of iterates converges to .

Proof.

Since is a weakly Kannan map, there exists a function with for all , satisfying (2.1) for all . Hence, the function given as is bounded and satisfies the conditions (*) and (2.9).

Consider any and define , We may assume that because otherwise we have finished. We will prove that and hence, by Theorem 2.6, will converge to a point which is the unique fixed point of .

First of all, observe that the inequality

holds for all . In fact, it is a consequence of the following one, which is true by (2.1):

From (2.13) we obtain that the sequence is nonincreasing, for , and then it is convergent to the real number

To prove that , suppose that and arrive to a contradiction as follows: use

and the definition of to obtain for all This, together with (2.13), gives that

for all , which is impossible since and .

Remark 2.8.

We do not know whether Theorem A is, or not, a particular case of Theorem 2.6, although that is the case if the functions satisfy the additional assumption . To see this, suppose that the map is in the conditions of Theorem A, that is, satisfies relation (2.7) for some given functions , , and suppose also that the functions satisfy in addition . Define as , where is given by

Let us see that, with this function , satisfies the hypotheses of Theorem 2.6. Indeed, is clearly bounded and also satisfies (*); if is a sequence in and , with , then

Since we also have that , we obtain that .

Finally, to see that satisfies relation (2.9), use relation (2.7) with , together with the same relation interchanging the roles of and , and the fact that , to obtain that

from which the result follows.

To prove the homotopy result of the next section, we will need the following local version of Corollary 2.7.

Corollary 2.9.

Assume that is a complete metric space, , and is a weakly Kannan map with associated function satisfying (2.1). If is defined as usual, and

then has a fixed point.

Proof.

In view of Corollary 2.7, it suffices to show that the closed ball is invariant under . To prove it, consider any and obtain the relation

from which, having in mind that ,

To end the proof, obtain that through the above inequality by considering two cases: if , then because . Otherwise, we would have , and consequently , from which

3. A Homotopy Result

In 1974 Ćirić [13] introduced the concept of quasicontractions and proved the following fixed point theorem: suppose that is a complete metric space and that is a quasicontraction, that is, there exists such that, for all ,

Then, has a fixed point in .

Observe that any contractive map, as well as any Kannan map, is a quasicontraction; thus, the theorem by Ćirić generalizes the well known fixed point theorems by Banach and Kannan.

On the other hand, Agarwal and O'Regan [14] considered a certain class of quasicontractions: those maps , where is a metric space, for which there exists such that, for all ,

and gave the following homotopy result.

Theorem 3 B.

Let be a complete metric space, an open subset of , and satisfying the following properties:

(i) for all and all ,

(ii)there exists such that for all and we have

(iii) is continuous in , uniformly for .

If has a fixed point in , then also has a fixed point in for all .

The above homotopy result includes the corresponding one for the class of Kannan maps, and in the following theorem we show that an analogous result is true for the wider class of weakly Kannan maps.

Theorem 3.1.

Let be a complete metric space, an open subset of , and satisfying the following properties:

(P1) for all and all ,

(P2)there exists such that for all and one has

and for all ,

(P3)there exists a continuous function such that, for every and , .

If has a fixed point in , then also has a fixed point in for all .

Proof.

Consider the nonempty set

We will prove that , and for this it suffices to show that is both closed and open in .

We start showing that is closed in : suppose that is a sequence in converging to and let us show that . By definition of , there exists a sequence in with . We will prove that converges to a point with , thus showing that .

That is a Cauchy sequence is a consequence of the following relation, where we have used (P2), (P3), and the fact that :

Write and let us see that and also that . That is a consequence of the following relation:

and that is straightforward from (P1).

Next we prove that is open in : suppose that and let us show that , for some . Since , there exists with . Consider with and use the continuity of to obtain such that

for all .

To show now that any is also in , it suffices to prove that the map has a fixed point. And this is true by Corollary 2.9, since

Remark 3.2.

A careful reading of the proof shows that hypothesis (P3) in Theorem 3.1 can be easily replaced by the weaker hypothesis (iii) in Theorem B.

Remark 3.3.

The counterpart to Theorem 3.1 for weakly contractive maps was proved by Frigon [8]. In that result, it was assumed, in place of our (3.3), an equivalent formulation of the following condition (H'):

Observe that condition (H') means that all the maps , are weakly contractive, and with the same function . Our condition (3.3) is no surprise then. It also means that all the maps are of weakly Kannan type, and with the same function .

We end the section with an example of a homotopy satisfying (P1), (P2), and (P3) but not the hypotheses of Theorem B. In fact, the function will be of weakly Kannan type, but will not satisfy the quasicontractivity condition (Q) (hence, it will not be of Kannan type since any Kannan map satisfies (Q)). Moreover, will not be of weakly contractive type.

Example 3.4.

Consider the metric space , where and , and let be the map given as

First of all, we will see that the map does not satisfy condition (Q). Define, for ,

Then, for , we have that , since . Hence,

showing that no can be found to satisfy (Q).

Secondly, observe that is not weakly contractive, since any weakly contractive map is continuous.

Next, let us check that is a weakly Kannan map. Since has as unique fixed point then, the function given by if , , is well defined. We have to check that only takes values in and that for all . In fact, all this will follow if we just show that, for ,

Thus, take and assume that , with . If any of the points equals , for example , then use and to obtain that

Otherwise, we would have that . In this case, since , then we may assume additionally that , and we claim that

To be convinced of this, check the following chain of inequalities having in mind that for all , that , and also that :

Next, define by and let us see that satisfies (P1), (P2), and (P3).

It is obvious that satisfies (P1). To check (P2), observe that

for all and all , and hence, if is the function previously defined, we have that, for all and all ,

Finally, (P3) is trivially satisfied with .