1. Introduction
The theory of fixed points plays a vital role in solving the nonlinear problems that arise in pure and applied sciences. Banach [
1] studied the existence of fixed points for contraction mappings in 1922. He proved that a contraction mapping
T defined on a closed nonempty subset
N of a Banach space
has a unique fixed point. To approximate the fixed point, he used the Picard iterative method (or successive approximation method), i.e.,
. But in the case of nonexpansive mappings, the Picard iterative method fails to converge to the fixed point. So, for the case of nonexpansive mappings, Browder [
2] and Gohde [
3] independently proved the same results in the framework of a uniformly convex Banach space. They stated that a nonexpansive mapping
T on a nonempty bounded closed and convex subset
N of a uniformly convex Banach space
admits a fixed point. Similarly, for the fixed point approximation of different mappings, many iterative methods have been developed. Some of these iteration processes are in Maan [
4], Ishikawa [
5], Agarwal et al. [
6], Thakur [
7], Noor [
8] and Abbas [
9]. For more details on iteration processes, see [
10,
11,
12,
13,
14,
15,
16,
17] and references therein.
In 2008, Suzuki [
18] studied the non-expansive mappings and introduced a new notion of Suzuki generalized non-expansive mapping (or mappings satisfying condition-
). Many results have been obtained for this type of mapping in
spaces (see, [
9,
19,
20]).
In 2012, Karapinar [
21] studied the condition
and introduced a new condition called
condition (Riech Chatterjee Suzuki
) as a generalization of condition (C).
In 2018, Ullah and Arshad [
22] introduced a new three-step iteration process to approximate the fixed point of mapping satisfying condition (C), known as the M-iteration process, which is defined as:
Let
(Banach space) and
be a selfmap. Then, for any real sequence
for
, we have
They proved the speedier convergence behavior of the M-iteration process and compared it with Picard- and -iterative schemes with the help of a numerical example. They also proved the weak and strong convergence results in Banach spaces.
Let and be a metric space. Then a mapping is a geodesic joining the two points such that , and for all . The mapping is isometry and . The image of the geodesic is called a geodesic segment joining the points denoted by whenever it exists uniquely. The space is called a geodesic space whenever for every two points there exists a geodesic joining these points. If the space has only one geodesic joining each two points , then the space is called uniquely geodesic. A subset N is convex in if each geodesic segment (joining ) contained in N.
Let
is a geodesic space, then a triangle
in
is called a geodesic triangle having three elements of
, which denote the vertices of the triangle
, i.e.,
and three geodesic segments, which denote the edges of
, i.e.,
,
. For any geodesic triangle
, a triangle
is called a comparison triangle if it satisfies
If , then an element is called a comparison element, if .
A geodesic space
is called a
space if there exists a comparison triangle
for a geodesic triangle
such that
where
and
.
This research article aims to prove some strong and
-convergence theorems for mapping with
condition using the M-iteration process in the
spaces. The M-iteration in
space is defined as:
where
and
can be any real sequence in (0,1).
2. Preliminaries
Let where is a metric space and is a self mapping, then a point is a fixed point of the mapping T if . The set , i.e., = denotes the set of all fixed points of mapping T.
Definition 1. The mapping T is called non-expansive mapping if for , we have Definition 2. A self-mapping is said to be quasi if the set of fixed point and for all and , we have Definition 3. Let where is a metric space. Then satisfies condition (C) if for each , we have Definition 4 ([
21]).
Let (metric space). Then satisfies condition if for the mapping T satisfy Example 1. Suppose that and defined by It is easy to show that T satisfies the (RCSC) condition but does not satisfy condition (C) at and .
Remark 1. The above example shows that condition is the extension of condition (C).
Lemma 1 ([
23]).
Let be a space. Then for and , there is a unique line segment i.e., such that Lemma 2 ([
23]).
For any and in Lemma 1, we have Lemma 3 ([
21]).
Let where is a space and be a self mapping with condition. Then for any and , T satisfies that Lemma 4 ([
21]).
Let such that N is closed and convex in space . Then for any , the self mapping with condition holds that Lemma 5 ([
9]).
Let such that N is closed in space . Let be a mapping with condition and a sequence bounded in N with and , then . Lemma 6 ([
24]).
Suppose that be a sequence in the interval for some . Let are sequences in the space such that, for some then Let is a space and is closed in . If a sequence is bounded in N, then for any , we define
(i) Asymptotic radius of the sequence
at the point
(ii) Asymptotic radius of the sequence
relative to the set
N(iii) Asymptotic center of the sequence
relative to the set
NWe know that
has only one element (see [
25]).
In 2008, Kirk and Panyanak [
26] proved a weak convergence result in the framework of Banach spaces. They studied Lim’s [
27] concept of
-convergence and proved that a nonexpansive mapping
admits a fixed point, where
N is closed, convex, and bounded in
space
.
Definition 5 ([
26]).
Let be a space. Then a sequence Δ-
converges to an element if r is the unique asymptotic center of every sub sequence of . Here r is the called of and written as . From the definition of
-convergence we know that each
space
satisfies the Opial’s condition [
28].
Lemma 7 ([
26]).
Let is a space. Then every sequence bounded in has a sub-sequence , which is Δ-
convergent. Lemma 8 ([
23]).
Let N be a closed convex and non-empty subset of . Then the asymptotic center of a bounded sequence is in the set N. 3. Convergence Results for Mapping with (RCSC) Condition
Some results of strong and -convergence are proved for the mapping satisfying condition via M iteration process in the framework of space . We start from a key lemma.
Lemma 9. Let be a complete space and , closed and convex in . Assume that is a self mapping with (RCSC) condition and . If is a sequence defined by (2), then exists for each . Proof. Assume that
and
. Since
T is a self-mapping with
condition by Lemmas 2 and 3, we obtain
Using the inequality (
3) we have
Similarly using the inequality (
4) we have,
Hence the sequence is a non-increasing and bounded, which implies that, exists for each . □
Theorem 1. Let be a complete space and , closed and convex in . Assume that is a self mapping with (RCSC) condition. Then a sequence defined by (2), is bounded if and only if and . Proof. Suppose that a sequence
is bounded and
. We need to prove that
. For this we choose any
. We are going to show that
(i.e.,
). Since
T satisfies
condition, so Lemma 4 suggests that
Hence . Since the space is uniformly convex, so contains only one element and thus we have . Hence it is proved that is the element of , i.e., .
Conversely, suppose that
. We need to prove that
and the sequence
is bounded. For this we assume that
be fix, then in Lemma 9, we have already proved that the sequence
is bounded and
exists. Put
Using inequality (
3) and Equation (
6), we have
Taking limsup on both sides, we have
By using Lemma 3, we have
Also by Lemma 9, we know that
Using inequalities (
7) and (
9), we have
From Equation (
10), we have
This completes the proof. □
Next, we prove a strong convergence result for mapping with condition in a compact domain.
Theorem 2. Let be a complete space, is a closed convex and compact subset of , a self mapping satisfying (RCSC) condition and . Then, is a sequence defined by iteration (2) that will strongly converge to the fixed point of T. Proof. Since , so by Theorem 1, we know that is bounded and . Since N is a compact in , so has a sub-sequence such that , for some i.e., converges strongly to .
Taking
on both sides we have
In the view of Theorem 1,
. Now using
and
in the inequality (
12), we obtain
This means that the sequence converges to , so i.e., . Lemma 9, suggests that exists. Therefore the strong limit of is . □
Now, we will prove the strong convergence results based on condition
which was defined by Sentor and Dotson in [
29].
Definition 6 ([
29]).
Let subset of . Then a self mapping satisfies condition , if there exists a non-decreasing mapping with , for each and for all , where = . Theorem 3. Let is a complete space and , closed and convex in . Since be a self mapping satisfy (RCSC) condition with and a sequence defined by iteration (2). If the mapping T satisfies condition , then the sequence strongly converges to the fixed point of T. Proof. By Lemma 9, we know that
exists for all
, so
exists. Let
for some
. If
then the equation is true. Let
. Now,
for all
gives that
yields the inequality
This shows that
is bounded below and non-increasing, so
exists. Also, Theorem 1 suggests that
. It follows from the condition
that
Also by the proof of Theorem 1, we know that
. So inequality (
14) becomes
Since the mapping
is nondecreasing with
and
for each
. Now from (
15), we have
Next, we need to prove that
is a Cauchy sequence in
N. Let
be chosen arbitrarily. Since
, so there exists a natural
such that for each
So there must be
such that
Now, for any
, we have
Thus we see that the sequence is a Cauchy sequence in N. Since N is closed in , so N is complete and therefore must be converged to the point . As , this gives that . Since N is closed and T satisfies condition, so is closed and hence r is the fixed point of T. Thus, the sequence strongly converges to fixed point r of T. □
Finally, we suggest the -convergence results for mapping having -condition via the M-iteration process.
Theorem 4. Let is complete space and , closed and convex in . Assume that is a self mapping satisfying (RCSC) condition with . Then a sequence defined by iteration (2) Δ-
converges to a fixed point of T. Proof. By Theorem 1, we know that the sequence is bounded and . Now we assume that where is the subsequence of . In order to prove that -converges to the fixed point of T, we follow the two steps.
- (I)
We prove that .
- (II)
contains exactly one point.
Step-I To show
. Let
, then there is a sub-sequence
of
such that
. Using Lemmas 7 and 8, the sequence
has a sub-sequence
with
. Since
is a sub-sequence of
and
so
. Also, the mapping
T satisfies the condition
, using Lemma 4, we have
Taking limsup on both sides of the above inequality, we have
As
, using Opial condition, we obtain
Hence, , i.e., .
Now, from Lemma 9, we have
exists. So we need to prove that
. On contrary, assume that
, then by uniqueness property of asymptotic centers, we have
Consequently, . This is a contradiction and so . Hence, .
Step-II: To prove the
-convergent of the sequence
in the set
we will prove that
has only one element. If
is a sub-sequence of
, then Lemmas 7 and 8 suggest that the sequence
has a subsequence
such that
. Let
and
. We have already proved that
and
. We contrarily suppose that
, then
exists and the asymptotic centers are unique, so we have
which is a contradiction, so
. Therefore,
. This completes the proof. □