2 Preliminaries
Let
be a metric space and
with
. A
geodesic path from
x to
y is an isometry
such that
and
. The image of a geodesic path is called a
geodesic segment. A metric space
X is a (uniquely)
geodesic space if every two points of
X are joined by only one geodesic segment. A
geodesic triangle in a geodesic space
X consists of three points
,
,
of
X and three geodesic segments joining each pair of vertices. A
comparison triangle of a geodesic triangle
is the triangle
in the Euclidean space
such that
A geodesic space
X is a
space if for each geodesic triangle
in
X and its comparison triangle
in
, the
inequality
(2.1)
is satisfied for all and .
A thorough discussion of these spaces and their important role in various branches of mathematics are given in [
11‐
15].
Let
C be a nonempty subset of a metric space
. Recall that a mapping
is said to be
nonexpansive if
(2.2)
T is said to be an
asymptotically nonexpansive nonself mapping if there exists a sequence
with
such that
(2.3)
Let be a metric space, and C be a nonempty and closed subset of X. Recall that C is said to be a retract of X if there exists a continuous map such that , . A map is said to be a retraction if . If P is a retraction, then for all y in the range of P.
Let
X and
C be the same as above. A mapping
is said to be
-
total asymptotically nonexpansive nonself mapping if there exist nonnegative sequences
,
with
,
and a strictly increasing continuous function
with
such that
(2.4)
where P is a nonexpansive retraction of X onto C.
Remark 2.2 From the definitions, it is to know that each nonexpansive mapping is an asymptotically nonexpansive nonself mapping with a sequence , and each asymptotically nonexpansive nonself mapping is a -total asymptotically nonexpansive nonself mapping with , , and , .
A nonself mapping
is said to be
uniformly L-
Lipschitzian if there exists a constant
such that
(2.5)
The following lemma plays an important role in our paper.
In this paper, we write
for the unique point
z in the geodesic segment joining from
x to
y such that
(2.6)
We also denote by the geodesic segment joining from x to y, that is, .
A subset C of a space is convex if for all .
A geodesic space X is a space if and only if the following inequality holds:
(2.7)
for all and all .
In particular,
if x,
y,
z are points in a space and ,
then (2.8)
Let
be a bounded sequence in a
space
X. For
, we set
The
asymptotic radius of
is given by
(2.9)
The
asymptotic radius of with respect to is given by
(2.10)
The
asymptotic center of
is the set
(2.11)
And the
asymptotic center of with respect to is the set
(2.12)
Recall that a bounded sequence in X is said to be regular if for every subsequence of .
Let X be a complete space,
be a bounded sequence in X and C be a closed convex subset of X.
Then (1)
there exists a unique point
such that
(2)
and both are singleton.
Let X be a space. A sequence in X is said to △-converge to if q is the unique asymptotic center of for each subsequence of . In this case we write and call q the △-limit of .
Lemma 2.7
(1)
Every bounded sequence in a complete space always has a △-
convergent subsequence [
8].
(2)
Let X be a complete space,
C be a closed convex subset of X.
If is a bounded sequence in C,
then the asymptotic center of is in C [
20].
Remark 2.8
(1)
Let
X be a
space and
C be a closed convex subset of
X. Let
be a bounded sequence in
C. In what follows, we define
(2.13)
where
.
(2)
It is easy to know that if and only if .
Nanjaras
et al. [
21] established the following relation between △-convergence and weak convergence in a
space.
Let be a bounded sequence in a space X,
and let C be a closed convex subset of X which contains .
Then (i)
implies that ;
(ii)
the converse of (i) is true if is regular.
Let C be a closed and convex subset of a complete space X, and let be a uniformly L-Lipschitzian and -total asymptotically nonexpansive nonself mapping. Let be a bounded sequence in C such that and . Then .
Let C be a closed and convex subset of a complete space X, and let be an asymptotically nonexpansive nonself mapping with a sequence , . Let be a bounded sequence in C such that and . Then .
Let X be a space,
be a given point and be a sequence in with and .
Let and be any sequences in X such that
and
for some .
Then (2.14)
Let ,
and be the sequences of nonnegative numbers such that
If and , then the limit exists. If there exists a subsequence of which converges to 0, then .
Let X be a complete space, be a bounded sequence in X with , and let be a subsequence of with and let the sequence converge, then .
3 Main results
Theorem 3.1 Let X be a complete space, C be a nonempty bounded closed and convex subset of X. If is a uniformly Lipschitzian and total asymptotically nonexpansive nonself mapping, then T has a fixed point in C.
Proof For any given point
, define
where P is a nonexpansive retraction of X onto C.
Since
T is a total asymptotically nonexpansive nonself mapping,
ζ is a strictly increasing continuous function, one gets
for any
. Letting
and taking superior limit on the both sides of the above inequality, we have
(3.1)
It is easy to know that the function
is lower semi-continuous, and
C is bounded closed and convex, there exists a point
such that
. Letting
in (3.1), for each
, we have
(3.2)
By using (2.7) with
, for any positive integers
, we obtain
(3.3)
Let
and take superior limit in (3.3). It follows from (3.2) that
This implies that
As
T is a total asymptotically nonexpansive nonself mapping, so
which implies that
is a Cauchy sequence in
C. Since
C is complete, let
. In view of the continuity of
TP, we have
Since , , this shows that , i.e., v is a fixed point of T in C.
The proof is completed. □
Remark 3.2 Theorem 3.1 is a generalization of Kirk [
2,
3] and Abbas
et al. [
10] from nonexpansive mappings and asymptotically nonexpansive mappings in the intermediate sense to total asymptotically nonexpansive nonself mappings.
Theorem 3.3 Let X be a complete space, C be a nonempty bounded closed and convex subset of X. If is a uniformly Lipschitzian and total asymptotically nonexpansive nonself mapping, then the fixed point set of T, , is closed and convex.
Proof As
T is continuous, so
is closed. In order to prove that
is convex, it is enough to prove that
whenever
. Setting
, by using (2.7) with
, for any
, we have
(3.4)
Since
T is a total asymptotically nonexpansive nonself mapping, using (2.8), we obtain
(3.5)
Similarly,
(3.6)
Substituting (3.5) and (3.6) into (3.4) and simplifying, we have
for any
. Hence
, in view of the continuity of
TP, we have
Since C is convex, this shows that . Therefore , which implies that , i.e., .
The proof is completed. □
Now we prove a △-convergence theorem for the following implicit iterative scheme:
(3.7)
where C is a nonempty closed and convex subset of a complete space X for each , is a uniformly -Lipschitzian and -total asymptotically nonexpansive nonself mapping defined by (2.4), and for each positive integer n, and are the solutions to the positive integer equation . It is easy to see that (as ).
Remark 3.4 Letting , , and , then is a finite family of uniformly L-Lipschitzian and -total asymptotically nonexpansive nonself mappings defined by (2.4).
Theorem 3.5 Let X be a complete space,
C be a nonempty bounded closed and convex subset of X.
If is a finite family of uniformly L-
Lipschitzian and -
total asymptotically nonexpansive nonself mappings satisfying the following conditions:
(i)
, ;
(ii)
there exists a constant such that , ;
(iii)
there exist constants with such that .
If , then the sequence defined by (3.7) △-converges to some point .
Proof Since for each
,
is a
-total asymptotically nonexpansive nonself mapping, by condition (ii), for each
and any
, we have
(3.8)
where P is a nonexpansive retraction of X onto C.
(I) We first prove that the following limits exist:
(3.9)
In fact, since
and
,
, is a total asymptotically nonexpansive nonself mapping,
is nonexpansive, it follows from Lemma 2.4 and (3.8) that
(3.10)
Simplifying it and using condition (iii), we have
Since
(as
), there exists a positive integer
such that
for all
. Therefore one has
(3.11)
and so
(3.12)
where and . By condition (i), and . Therefore it follows from Lemma 2.13 that the limits and exist for each .
(II) Next we prove that for each
,
(3.13)
For each
, from the proof of (I), we know that
exists, we may assume that
(3.14)
From (3.11) we get
which implies that
(3.15)
In addition, since
From (3.14), we have
(3.16)
It follows from (3.14)-(3.16) and Lemma 2.12 that
(3.17)
Now, by Lemma 2.4, we obtain
(3.18)
Hence, from (3.17) and (3.18), one gets
(3.19)
and for each
,
(3.20)
Since
,
, is uniformly
L-Lipschitzian and for each
and
P is a nonexpansive retraction of
X onto
C, we have
, where
,
and
. Hence it follows from (3.19) and (3.20) that
(3.21)
where
. Consequently, for any
, from (3.20) and (3.21) it follows that
This implies that the sequence
Since for each
,
is a subsequence of
, therefore we have
Conclusion (3.13) is proved.
(III) Now we show that △-converges to a point in ℱ.
Let . We first prove that .
In fact, let , then there exists a subsequence of such that . By Lemma 2.7, there exists a subsequence of such that . In view of (3.13), . It follows from Lemma 2.10 that ; so, by (3.9), the limit exists. By Lemma 2.14, . This implies that .
Next let be a subsequence of with , and let . Since , from (3.9) the limit exists. In view of Lemma 2.14, . This implies that consists of exactly one point. We know that △-converges to some point .
The conclusion of Theorem 3.5 is proved. □
Now we prove the strong convergence results for the following iterative scheme:
(3.22)
where
C is a nonempty closed and convex subset of a complete
space
X for each
,
is a uniformly
-Lipschitzian and
-total asymptotically nonexpansive nonself mapping defined by (2.4), and for each positive integer
,
and
are the unique solutions to the following positive integer equation:
(3.23)
Lemma 3.6 [
23]
(1)
The unique solutions to the positive integer equation (3.23)
are
where denotes the maximal integer that is not larger than x.
(2)
For each ,
denote
then , .
Theorem 3.7 Let X be a complete space,
C be a nonempty bounded closed and convex subset of X,
and for each ,
let be a uniformly -
Lipschitzian and -
total asymptotically nonexpansive nonself mapping defined by (2.4),
satisfying the following conditions:
(i)
, ;
(ii)
there exists a constant such that , , ;
(iii)
there exist constants with such that .
If and there exist a mapping and a nondecreasing function with and ,
,
such that (3.24)
then the sequence defined by (3.22) converges strongly (i.e., in metric topology) to some point .
Proof We observe that for each
,
is a
-total asymptotically nonexpansive nonself mapping. By condition (ii), for each
and any
, we have
(3.25)
(I) We first prove that the following limits exist:
(3.26)
In fact, since
and
is nonexpansive, it follows from Lemma 2.4, (3.25) that
(3.27)
and
(3.28)
Substituting (3.27) into (3.28) and simplifying it, we have
(3.29)
and so
(3.30)
where , . By condition (i), and . By Lemma 2.13, the limits and exist for each .
(II) Next we prove that for each
, there exists a subsequence
such that
(3.31)
In fact, for each given
, from the proof of (I), we know that
exists. Without loss of generality, we may assume that
(3.32)
From (3.27) one gets
(3.33)
Since
we have
(3.34)
In addition, it follows from (3.29) that
which implies that
(3.35)
From (3.32)-(3.35) and Lemma 2.12, one has
(3.36)
Since
Taking lim inf as
on both sides in the inequality above, from (3.36) we have
which combined with (3.33) implies that
(3.37)
Using (3.27) we have
(3.38)
This implies that
(3.39)
Similarly, we have that
This together with (3.32) and (3.39) and Lemma 2.12 yields that
(3.40)
Therefore we obtain
(3.41)
Furthermore, it follows from (3.36) that
(3.42)
Now combined with (3.41) this shows that
(3.43)
From Lemma 3.6, (3.37), (3.40), (3.42) and (3.43), we have that for each given positive integer
, there exist subsequences
,
and
such that
Conclusion (3.31) is proved.
(III) Now we prove that converges strongly (i.e., in metric topology) to some point .
In fact, it follows from (3.31) and (3.24) that for a given mapping
, there exists a subsequence
of
such that
and
Taking lim sup on both sides of the above inequality, one has
By the property of
f, this implies that
(3.44)
Next we prove that is a Cauchy sequence in C.
In fact, it follows from (3.29) that for any
,
where
and
. Hence, for any positive integers
m,
k, we have
Since for each
,
, one gets
and
where
. By (3.44) we have
This shows that the subsequence is a Cauchy sequence in C. Since C is a closed subset in a space X, it is complete. Without loss of generality, we can assume that the subsequence converges strongly (i.e., in metric topology in X) to some point . By Theorem 3.3, we know that ℱ is a closed subset in C. Since , . By using (3.26), it yields that the whole sequence converges in the metric topology to some point .
The proof is completed. □