In this section we follow the pattern of Dugundji and Granas  to introduce the concept of weakly Kannan maps.
Let be a metric space, , and . Therefore is a weakly Kannan map if there exists , with for every such that, for all ,
Clearly, any weakly Kannan map has at most one fixed point: if and , then
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.
Although Kannan showed that the concept of Kannan map is independent of the concept of contractive map, Janos  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.
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 , 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.
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 .
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 :
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 .
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 .
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.
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.
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