Let ℱ denotes the class of all functions satisfying the following condition:
(2.1)
Definition 7 Let be a nonself-mapping and be a function. We say T is a rational Geraghty contraction of type I if there exist and such that
for all , where
Theorem 2 Let and be two mappings satisfying the following assertions:
-
(a)
there exists such that is topologically closed and algebraically closed with respect to difference,
-
(b)
T is an -admissible,
-
(c)
T is a rational Geraghty contractive mapping of type I,
-
(d)
if is a sequence in such that as and , then for all ,
-
(e)
there exists such that .
Then T has a unique PPF dependent fixed point . Moreover, for a fixed , if the sequence of iterates of T is defined by for all , then converges to .
Proof Let is a point in such that . Since , there exists such that . Choose such that . Since and, by hypothesis, we get . This implies that there exists such that . Thus, we can choose such that . Continuing this process, by induction, we can build the sequence in such that for all . It follows from the fact that is algebraically closed with respect to difference
Since T is -admissible and , we deduce that
By continuing this process, we get for all . Since T is a rational Geraghty contraction of type I, we have
(2.2)
On the other hand,
If
from (2.2) we have
(2.3)
which is a contradiction. So,
By (2.2) we conclude
(2.4)
for all . This implies that the sequence is decreasing in . So, it is convergent. Suppose that there exists such that . Assume that . Taking the limit as from (2.4) we conclude
which implies . So,
and since , , which is a contradiction. Hence, . This means
(2.5)
We prove that the sequence is a Cauchy sequence in . Assume that is not a Cauchy sequence, then
(2.6)
Since T is a rational Geraghty contraction of type I, we have
Taking the limit when in the above inequality and applying (2.5) we deduce
(2.7)
where
(2.8)
Letting in the above inequality and applying (2.5), we get
(2.9)
So, by (2.7) and (2.9), we have
and hence from (2.6) we get . This means
and since , we conclude
which is a contradiction. Consequently,
and hence is a Cauchy sequence in . By Completeness of , we find that converges to a point , this means , as . Since is topologically closed, we deduce, . By condition b, we have . Now, since T is a rational Geraghty contraction of type I, we have
Taking the limit as in the above inequality, we get
(2.10)
But
(2.11)
Therefore, from (2.10) and (2.11), we deduce
that is,
which implies that is a PPF dependent fixed point of T in . Now, we show that T has a unique PPF dependent fixed point in . Suppose on the contrary that and are two PPF dependent fixed points of T in such that . Then
where
Therefore,
which is a contradiction. Hence, . Then T has a unique PPF dependent fixed point in . □
Definition 8 Let and . We say that T is a rational Geraghty contraction of type II if there exist and such that
for all , where
Theorem 3 Let and be two mappings satisfying the following assertions:
-
(a)
there exists such that is topologically closed and algebraically closed with respect to difference,
-
(b)
T is an -admissible,
-
(c)
T is a rational Geraghty contractive mapping of type II,
-
(d)
if is a sequence in such that as and , then for all ,
-
(e)
there exists such that .
Then T has a unique PPF dependent fixed point . Moreover, for a fixed , if the sequence of iterates of T is defined by for all , then converges to .
Proof Suppose that is a point in such that . Since , there exists such that . Choose such that . Since and, by hypothesis, we get . This implies that there exists such that . Thus, we can choose such that . Continuing this process, by induction, we can build the sequence in such that for all . It follows from the fact that is algebraically closed with respect to difference
Since T is -admissible and , we deduce that
Continuing this process, we get for all . Since T is a rational Geraghty contraction of type II, we have
(2.12)
On the other hand,
From (2.12) we conclude
(2.13)
for all . So, the sequence is decreasing in and there exists such that . Reviewing the proof of Theorem 2, we can show that , i.e.,
(2.14)
Now, we prove that the sequence is Cauchy in . If not, then
(2.15)
From the fact that T is a rational Geraghty contraction of type II, we have
Letting in the above inequality and applying (2.14) we deduce
(2.16)
where
(2.17)
Letting in the above inequality and applying (2.14), we get
(2.18)
So, from (2.16) and (2.18), we obtain
and so by (2.15) we get, . That is,
and since , we deduce
which is a contradiction. Consequently,
and hence is a Cauchy sequence in . By completeness of , we find that converges to a point , this means that , as . Since is topologically closed, we deduce that . Now, since T is a rational Geraghty contraction of type II, we have
Taking the limit as in the above inequality, we get
(2.19)
But
(2.20)
So,
and by (2.19) and (2.20), we conclude
that is,
which implies that is a PPF dependent fixed point of T in . Finally, we prove the uniqueness of the PPF dependent fixed point of T in . Let and be two PPF dependent fixed points of T in such that . So, we obtain
where
Therefore,
which is a contradiction. Hence, . Therefore, T has a unique PPF dependent fixed point in . This completes the proof. □
Definition 9 Let and . We say that T is a rational Geraghty contraction of type III if there exist and such that
for all , where
Theorem 4 Let and be two mappings satisfying the following assertions:
-
(a)
there exists such that is topologically closed and algebraically closed with respect to difference,
-
(b)
T is an -admissible,
-
(c)
T is a rational Geraghty contractive mapping of type III,
-
(d)
if is a sequence in such that as and , then for all ,
-
(e)
there exists such that .
Then T has a unique PPF dependent fixed point . Moreover, for a fixed , if the sequence of iterates of T defined by for all , then converges to the PPF dependent fixed point of T in .
Proof Suppose that be a point in such that . Since , there exists such that . Choose such that . Since and, by hypothesis, we get . This implies that there exists such that . Thus, we can choose such that . Repeating this process, by induction, we can construct the sequence in such that for all . From the fact that is algebraically closed with respect to difference it follows that
Since T is -admissible and , we deduce
Continuing this process, we get for all . By the fact that T is a rational Geraghty contraction of type III, we have
(2.21)
On the other hand,
From (2.21) we conclude
(2.22)
for all . This implies that the sequence is decreasing in . Then there exists such that . Repeating the proof of Theorem 2, we conclude that . That is,
(2.23)
Now, we prove that the sequence is Cauchy in . If not, then
(2.24)
Since T is a rational Geraghty contraction of type III, we have
Making in the above inequality and applying (2.23) we have
(2.25)
Also,
Letting in the above inequality and applying (2.23), we get
(2.26)
Hence, from (2.25) and (2.26), we obtain
and so by (2.24) we get . That is,
and since , we deduce
which is a contradiction. Consequently,
and hence is a Cauchy sequence in . Completeness of shows that converges to a point , this means that , as . Since is topologically closed, we deduce that . Now, since T is a rational Geraghty contraction of type III, we have
Taking the limit as in the above inequality, we get
(2.27)
But
(2.28)
Therefore, from (2.27) and (2.28), we deduce that
that is,
which implies that is a PPF dependent fixed point of T in . Suppose that and are two PPF dependent fixed points of T in such that . So,
where
Therefore,
which is a contradiction. Hence, . Then T has a unique PPF dependent fixed point in . □
Corollary 1 Let and be two mappings satisfying the following assertions:
-
(a)
there exists such that is topologically closed and algebraically closed with respect to difference,
-
(b)
T is an -admissible,
-
(c)
assume that
for all , where and
or