Convergence of Iterative Scheme for Asymptotically Nonexpansive
Mapping in Hadamard Spaces
1ANANTACHAI PADCHAROEN, 2,∗PAKEETA SUKPRASERT
1Department of Mathematics, Faculty of Science and Technology,
Rambhai Barni Rajabhat University, Chanthaburi 22000, THAILAND
2,∗Department of Mathematics and Computer Science,
Faculty of Science and Technology,
Rajamangala University of Technology Thanyaburi (RMUTT),
Thanyaburi, Pathumthani 12110, THAILAND
Abstract: -In this paper, we introduce and prove the convergence of a novel iterative scheme for asymptotically
nonexpansive mapping under some suitable conditions in the context of Hadamard spaces. We also present a
numerical experiment in which the rate of convergence of the new iterative scheme is compared to that of an
existing iterative scheme.
Key-Words: Hadamard space, CAT(0) space, Asymptotically nonexpansive mappings, Weak and strong
convergence
Received: September 19, 2022. Revised: November 7, 2022. Accepted: December 4, 2022. Available online: December 22, 2022.
1 Introduction
Ageodesic triangle ∆(u1, u2, u3)in a geodesic met-
ric space (X, d)consists of three points in X(called
vertices of ∆) and a geodesic segment between each
pair of vertices (the edges of ∆). A comparison tri-
angle for geodesic triangle ∆(u1, u2, u3)in (X, d)is
a triangle ¯
∆(u1, u2, u3) := ∆(¯u1,¯u2,¯u3)in R2such
that dR2(¯ui,¯uj) = d(ui, uj)for i, j ∈ {1,2,3}.Such
a triangle always exists, [1].
Let ∆be a geodesic triangle in Xand ∆its
comparison triangle in R2.Then ∆is said to satisfy
CAT(0) inequality if for all u, v ∈∆and all com-
parison points u, v ∈∆, d(u, v)≤dR2(u, v).A
geodesic metric space X is called a CAT(0) space if all
geodesic triangles satisfy the above comparison ax-
iom (i.e. CAT(0) inequality). Some well known ex-
amples of CAT(0) spaces are complete. The complete
CAT(0) spaces are often called Hadamard spaces.
Fixed point theory in a CAT(0) space has been first
studied by Kirk (see [2]). He showed that every non-
expansive mapping defined on a bounded closed con-
vex subset of a complete CAT(0) space always has a
fixed point.
Let Kbe a nonempty closed subset of a CAT(0)
space X, and Tbe a self map defined on K.Then T
is said to be:
• nonexpansive if
d(Tu, Tv)≤d(u, v),∀u, v ∈ K,
• asymptotically nonexpansive if there exists a
sequence{ζn}in [1,∞)with limn→∞ ζn= 1
such that
d(Tnu, Tnv)≤ζnd(u, v),∀u, v ∈ K,∀n≥1,
• uniformly L-Lipschitzian if there exists a con-
stant L>0such that
d(Tnu, Tnv)≤ L(u, v),∀u, v ∈ K ∀n≥1.
Moreover, every asymptotically nonexpansive
mapping is a uniformly L-Lipschitzian mapping with
L=supn∈N{ζn}.
A mapping Tis said to have a fixed point u∗if
Tu∗=u∗and a sequence {un}is said to be asymp-
totic fixed point sequence if
lim
n→∞ d(un,Tun) = 0.
Authors create many new iterative processes to
achieve a relatively effective rate of convergence
and overcome such difficulties (see, e.g., Mann [4],
Ishikawa [5], Agarwal et al. [6], Noor [7], Abbas and
Nazir [8] and Thakur et al. [9]).
Şahin and Basarir, [10], suggested an effective two
step iterative scheme for approximating fixed points
of asymptotically quasi-nonexpansive mapping and
sequence {un}as follows:
{u1∈ K,
vn= (1 −ρn)un⊕ρnTnun,
un+1 = (1 −λn)Tnun⊕λnTnvn,∀n≥1,
(1)
where and throughout the paper {λn},{ρn}are the
sequence such that 0≤λn, ρn≤1for all n≥1.
They established some strong convergence results un-
der some suitable conditions such that generalizing
some results of Khan and Abbas, [11].
Niwongsa and Panyanak, [12], suggested an effec-
tive two step iterative scheme for approximating fixed
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1anantachai.p@rbru.ac.th, 2pakeeta_s@rmutt.ac.th
points of asymptotically nonexpansive mapping and
sequence {un}as follows:
u1∈ K,
yn=τnTnun⊕(1 −τn)un,
vn=ρnTnyn⊕(1 −ρn)un,
un+1 =λnTnvn⊕(1 −λn)un,∀n≥1,
(2)
where {λn},{ρn}and {τn}are real sequence in
[0,1].They proved ∆and strong convergence theo-
rems of the following Noor iteration for an asymptot-
ically nonexpansive mapping in CAT(0) spaces.
Recently, Yambangwai et al., [13], suggested an
effective three step iterative scheme for approximat-
ing fixed points of asymptotically nonexpansive map-
ping and sequence {un}as follows:
u1∈ K,
yn=τnTnun⊕(1 −τn)un,
vn=ρnTnyn⊕(1 −ρn)yn,
un+1 =λnTnvn⊕(1 −λn)Tnyn,∀n≥1,
(3)
where {λn},{ρn}and {τn}are real sequence in
[0,1].They established some convergence theorems
to approximate the fixed points of asymptotically
nonexpansive mapping in the setting CAT(0) spaces.
Motivated by the preceding work, we present a
new iterative scheme, which is defined as follows:
u1∈ K,
yn=Tn(τnTnun⊕(1 −τn)un),
vn=Tn(ρnTnyn⊕(1 −ρn)yn),
un+1 =Tn(λnTnvn⊕(1 −λn)vn),∀n≥1,
(4)
where {λn},{ρn}and {τn}are real sequence in [0,1]
and Tis an asymptotically nonexpansive mapping on
a nonempty closed bounded and convex subset of a
Hadamard space X.
2 Preliminaries
Definition 2.1. [14] A sequence {un}in Xis said to
∆-converge to u∗∈Xif u∗is the unique asymptotic
center of {wn}for every subsequence {wn}of {un}.
In this case we
write ∆-limn→∞ un=u∗and we call u∗the ∆-
limn→∞ un=u∗
Lemma 2.2. [14]
(i) Every bounded sequence in Xhas ∆-
convergence subsequence.
(ii) If Kis a closed convex subset of Xand if {un}
is a bounded sequence in K,then the asymptotic
center of {un}is in K.
The asymptotic radius r({un})of {un}is given
by
r({un}) = inf{r(u, {un}) : u∈X},
and the asymptotic center A({un})of {un}is the set
A({un}) = {u∈X:r(u, {un}) = r({un})}.
Lemma 2.3. [15] Let Xbe a complete CAT(0)
space and {un}be a bounded sequence in X. If
A({un}) = {u},{wn}is a subsequence of {un}
such that A({wn}) = {w}and d(un, w)converges,
then u=w.
Lemma 2.4. [16] Let Kbe a closed and convex
subset of a complete CAT(0) space Xand T:
K → K be an asymptotically nonexpansive map-
ping. Let {un}be a bounded sequence in Ksuch that
limn→∞ d(un,Tun) = 0 and ∆-limn→∞ un=u∗.
Then u∗=Tu∗.
Lemma 2.5. [17] Let Xbe a CAT(0) space, u, v, w ∈
Xand t∈[0,1].Then
(i) d((1 −t)u⊕tv, w)≤(1 −t)d(u, w) + td(v, w).
(ii) d2((1 −t)u⊕tv, w)≤(1 −t)d2(u, w) +
td2(v, w)−t(1 −t)d2(u, v).
Lemma 2.6. [18] Let {δn}and {σn}be sequences of
nonnegative real numbers satisfying the inequality
δn+1 ≤(1 + σn)δn, n ≥1.
If ∑∞
n=1 σn<∞,then limn→∞ δnexists.
Lemma 2.7. [19] Let Xbe a complete CAT(0) space
and let u∗∈X. Suppose {αn}is a sequence in
[e, f]for some e, f ∈(0,1) and {un},{wn}are
sequences in Xsuch that lim supn→∞ d(un, u∗)≤
l, lim supn→∞ d(wn, u∗)≤land limn→∞ d((1 −
αn)un⊕αnwn, u∗) = lfor some l≥0.Then
d(un, wn) = 0.
Theorem 2.8. [20] Let Kbe a nonempty bounded
closed and convex subset of a complete CAT(0) space
Xand T:K → K be asymptotically nonexpansive.
Then Thas a fixed point.
.
3 Main results
Theorem 3.1. Let Kbe a closed bounded and con-
vex subset of a Hadamard space Xand a self map T
defined on Kbe an asymptotically nonexpansive map-
ping with {ζn}.Assume that the following conditions
hold:
(i) {ζn} ≥ 1and ∑∞
n=1(ζn−1) <∞,
(ii) there exist constants c1, c2with 0< c1≤τn≤
c2<1for each n∈N,
(iii) there exist constants b1, b2with 0< b1≤ρn≤
b2<1for each n∈N,
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(iv) there exist constants a1, a2with 0< a1≤λn≤
a2<1for each n∈N.
For the sequence {un}given by (4). Then
limn→∞ d(un, u∗)exists for all u∗∈ F(T).
Proof. Using Theorem 2.8, we note that F(T)=∅.
Putting ζn= 1 + κnfor all n≥1. Using ∑∞
n=1(ζn−
1) <∞, we have ∑∞
n=1 κn<∞. For each u∗∈
F(T), we obtain that
d(yn, u∗)(5)
=d(Tn(τnTnun⊕(1 −τn)un), u∗)
≤(1 + κn)d(τnTnun⊕(1 −τn)un, u∗)
≤(1 + κn)[τnd(Tnun, u∗)
+ (1 −τn)d(un, u∗)]
≤(1 + κn)[τn(1 + κn)d(un, u∗)
+ (1 −τn)d(un, u∗)]
= (1 + κn)(1 + τnκn)d(un, u∗)
≤(1 + κn)2d(un, u∗)(6)
and
d(vn, u∗)(7)
=d(Tn(ρnTnyn⊕(1 −ρn)yn), u∗)
≤(1 + κn)d(ρnTnyn⊕(1 −ρn)yn, u∗)
≤(1 + κn)[ρnd(Tnyn, u∗)
+ (1 −ρn)d(yn, u∗)]
≤(1 + κn)[ρn(1 + κn)d(yn, u∗)
+ (1 −ρn)d(yn, u∗)]
= (1 + κn)(1 + ρnκn)d(yn, u∗)
≤(1 + κn)2d(yn, u∗).(8)
Using (5) and (7), we have
d(un+1, u∗)(9)
=d(Tn(λnTnvn⊕(1 −λn)vn), u∗)
≤(1 + κn)d(λnTnvn⊕(1 −λn)vn, u∗)
≤(1 + κn)[λnd(Tnvn, u∗)
+ (1 −λn)d(Tnvn, u∗)]
≤(1 + κn)[λn(1 + κn)d(vn, u∗)
+ (1 −λn)d(vn, u∗)]
= (1 + κn)(1 + λnκn)d(vn, u∗)
≤(1 + κn)2d(vn, u∗)
≤(1 + κn)4d(yn, u∗)
≤(1 + κn)6d(un, u∗).(10)
Because ∑∞
n=1 κn<∞and using Lemma 2.6, we
have limn→∞ d(un, u∗)exists.
Theorem 3.2. Let Kbe a closed bounded and con-
vex subset of a Hadamard space Xand a self map T
defined on Kbe an asymptotically nonexpansive map-
ping with {ζn}.Assume that the following conditions
hold:
(i) {ζn} ≥ 1and ∑∞
n=1(ζn−1) <∞,
(ii) there exist constants c1, c2with 0< c1≤τn≤
c2<1for each n∈N,
(iii) there exist constants b1, b2with 0< b1≤ρn≤
b2<1for each n∈N,
(iv) there exist constants a1, a2with 0< a1≤λn≤
a2<1for each n∈N.
For the sequence {un}given by (4). Then
limn→∞ d(un,Tun) = 0.
Proof. For each u∗∈ F(T). Putting ζn= 1 + κn
for all n≥1. Using ∑∞
n=1(ζn−1) <∞, we have
∑∞
n=1 κn<∞. From Theorem 3.1, we support that
lim
n→∞ d(un, u∗) = l≥0.(11)
From (5), we have
lim sup
n→∞
d(yn, u∗)≤l. (12)
Because Tbe an asymptotically nonexpansive
d(Tyn, u∗)≤(1 + κn)d(yn, u∗)(13)
Using (12) and (13), we have
lim sup
n→∞
d(Tnyn, u∗)≤l. (14)
Similar to that,
lim sup
n→∞
d(Tnun, u∗)≤l(15)
and
lim sup
n→∞
d(Tnvn, u∗)≤l. (16)
Because
d(un+1, u∗)≤(1 + κn)4d(yn, u∗).(17)
Taking limit infimum both sides, we obtain,
l≤lim inf
n→∞ d(yn, u∗).(18)
Using (12) and (18), we obtain that
l=lim
n→∞ d(yn, u∗)
=lim
n→∞ d(Tn(τnTnun⊕(1 −τn)un), u∗).(19)
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Also,
d(Tn(τnTnun⊕(1 −τn)un), u∗)
≤(1 + κn)d(τnTnun⊕(1 −τn)un, u∗)
and
l≤lim inf
n→∞ d(Tn(τnTnun⊕(1 −τn)un), u∗)
≤lim inf
n→∞ d(τnTnun⊕(1 −τn)un, u∗).(20)
Using (11) and (15), we have
d(τnTnun⊕(1 −τn)un, u∗)
≤(1 + κn)[τnd(Tnun, u∗) + (1 −τn)d(un, u∗)]
and
lim sup
n→∞
d(τnTnun⊕(1 −τn)un, u∗)≤l. (21)
Using (20) and (21), we have
lim
n→∞ d(τnTnun⊕(1 −τn)un, u∗) = l. (22)
Using (11), (15), (22) and Lemma 2.7, we have
lim
n→∞ d(un,Tnun) = 0.(23)
From (9), we have
d(un+1, u∗)≤(1 + κn)2d(vn, u∗)
and
l≤lim inf
n→∞ d(un+1, u∗)≤lim inf
n→∞ d(vn, u∗)(24)
From (7), we have
d(vn, u∗)≤(1 + κn)2d(yn, u∗)
and
lim sup
n→∞
d(vn, u∗)≤lim sup
n→∞
d(yn, u∗)≤l. (25)
Using (24) and (25), we have
lim
n→∞ d(vn, u∗) = l(26)
and
l=lim
n→∞ d(vn, u∗)
=lim
n→∞ d(Tn(ρnTnyn⊕(1 −ρn)yn), u∗)
which
d(Tn(ρnTnyn⊕(1 −ρn)yn), u∗)
≤(1 + κn)d(ρnTnyn⊕(1 −ρn)yn, u∗).
Taking limit infimum both sides, we obtain,
l≤lim inf
n→∞ d(ρnTnyn⊕(1 −ρn)yn, u∗).(27)
Also,
d(ρnTnyn⊕(1 −ρn)yn, u∗)
≤ρnd(Tnyn, u∗) + (1 −ρn)d(yn, u∗)
Using (12) and (14), we have
lim sup
n→∞
d(ρnTnyn⊕(1 −ρn)yn, u∗)≤l. (28)
Using (27) and (28), we have
lim sup
n→∞
d(ρnTnyn⊕(1 −ρn)yn, u∗) = l. (29)
Using (12), (14), (29) and Lemma 2.7, we have
lim
n→∞ d(yn,Tnyn) = 0.(30)
From (9), we have
l=lim
n→∞ d(un+1, u∗)
=lim
n→∞ d(Tn(λnTnvn⊕(1 −λn)vn), u∗)
and
d(Tn(λnTnvn⊕(1 −λn)vn), u∗)
≤(1 + κn)d(λnTnvn⊕(1 −λn)vn, u∗).
Taking limit infimum both sides, we obtain,
l≤lim inf
n→∞ d(λnTnvn⊕(1 −λn)vn, u∗).(31)
Also,
d(λnTnvn⊕(1 −λn)vn, u∗)
≤λnd(Tnvn, u∗) + (1 −λn)d(vn, u∗).
Using (16) and (25), we have
lim sup
n→∞
d(λnTnvn⊕(1 −λn)vn, u∗)≤l. (32)
Using (31) and (32), we have
lim
n→∞ d(λnTnvn⊕(1 −λn)vn, u∗) = l. (33)
Using (16), (25), (33) and Lemma 2.7, we have
lim
n→∞ d(vn,Tnvn) = 0.(34)
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In addition, using (23), we have
d(yn, un)
=d(Tn(τnTnun⊕(1 −τn)un), un)
≤d(Tn(τnTnun⊕(1 −τn)un),Tnun)
+ (1 + κn)d(Tnun, un)
≤(1 + κn)[d(τnTnun⊕(1 −τn)un, un)]
+ (1 + κn)d(Tnun, un)
≤(1 + κn)[τnd(Tnun, un) + (1 −τn)d(un, un)]
+ (1 + κn)d(Tnun, un)
= (1 −κn)(1 −τn)d(Tnun, un)
→0as n→ ∞.(35)
Also,
d(yn,Tnun)(36)
=d(Tn(τnTnun⊕(1 −τn)un),Tnun)
≤(1 + κn)d(τnTnun⊕(1 −τn)un, un)
≤(1 + κn)[τnd(Tnun, un) + (1 −τn)d(un, un)]
≤(1 + κn)τnd(Tnun, un)
→0as n→ ∞.(37)
Using (23), (36) and (30), we have
d(un,Tnyn)
≤d(un,Tnun) + d(Tnun, yn) + d(yn,Tnyn)
→0as n→ ∞.(38)
Using (38) and (35), we have
d(vn,Tnun)
=d(Tn(ρnTnyn⊕(1 −ρn)yn),Tnun)
≤(1 + κn)d(ρnTnyn⊕(1 −ρn)yn, un)
≤(1 + κn)[ρnd(Tnyn, un) + (1 −ρn)d(yn, un)]
→0as n→ ∞.(39)
Using (23), (39) and (34), we have
d(un,Tnvn)
≤d(un,Tnun) + d(Tnun, vn) + d(vn,Tnvn)
→0as n→ ∞.(40)
Using (38), (35) and (36), we have
d(vn, un)
=d(Tn(ρnTnyn⊕(1 −ρn)yn), un)
≤d(Tn(ρnTnyn⊕(1 −ρn)yn),Tnun)
+ (1 + κn)d(Tnun, yn)
≤(1 + κn)d(ρnTnyn⊕(1 −ρn)yn), un)
+ (1 + κn)d(Tnun, yn)
≤(1 + κn)[ρnd(Tnyn, un) + (1 −ρn)d(yn, un)]
+ (1 + κn)d(Tnun, yn)
→0as n→ ∞.(41)
Using (40), (41) and (39), we have
d(un+1, un)
=d(Tn(λnTnvn⊕(1 −λn)vn), un)
≤d(Tn(λnTnvn⊕(1 −λn)vn),Tnun)
+ (1 + κn)d(Tnun, vn)
≤(1 + κn)d(λnTnvn⊕(1 −λn)vn, un)
+ (1 + κn)d(Tnun, vn)
≤1 + κn)[λnd(Tnvn, un) + (1 −λn)d(vn, un)]
+ (1 + κn)d(Tnun, vn)
→0as n→ ∞.(42)
Using (42) and (23), we have
d(un+1,Tnun+1)
≤d(un+1, un) + d(un,Tnun) + d(Tnun,Tnun+1)
≤d(un+1, un)+(1+κn)d(un+1, un) + d(un,Tnun)
≤(2 + κn)d(un+1, un) + d(un,Tnun).(43)
Using (23) and (43), we have
d(un+1,Tun+1)
≤d(un+1,Tn+1un+1) + d(Tn+1un+1,Tun+1)
≤d(un+1,Tn+1un+1)+(1+κ1)d(Tnun+1, un+1)
→0as n→ ∞,(44)
which implies limn→∞ d(xn, T xn) = 0.
Theorem 3.3. Let Kbe a closed bounded and con-
vex subset of a Hadamard space Xand a self map T
defined on Kbe an asymptotically nonexpansive map-
ping with {ζn}.Assume that the following conditions
hold:
(i) {ζn} ≥ 1and ∑∞
n=1(ζn−1) <∞,
(ii) there exist constants c1, c2with 0< c1≤τn≤
c2<1for each n∈N,
(iii) there exist constants b1, b2with 0< b1≤ρn≤
b2<1for each n∈N,
(iv) there exist constants a1, a2with 0< a1≤λn≤
a2<1for each n∈N.
For the sequence {un}given by (4). Then {un}∆-
converges to a fixed point of T.
Proof. Let u∈ω∆(un) = ∪A({wn}).So, there ex-
ists subsequence {wn}of {un}such that A({wn}) =
{u}.By Lemmas 2.2 (i), (ii) and Theorem 2.8 there
exists a subsequence {sn}of {wn}such that ∆-
limn→∞ sn=s∈ K.From Lemma 2.4, we have
s∈ F(T).Because {d(wn, s)}converges and us-
ing Lemma 2.3, we have u=s. This implies that
ω∆(un)⊆ F(T).
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Next, we show that ω∆(un)consists of exactly
one point. Let {wn}be a subsequence of {un}with
A({wn}) = {u}and A({un}) = {p}.We have that
u=sand s∈ F(T).Finally, because {d(un, s)}
converges and using Lemma 2.3, we have p=s∈
F(T).This shows that ω∆(un) = p.
4 Numerical example
Let X=Rwith usual metric and K= [1,11].Let a
self map Ton Kas follows:
Tu=4
√u3+ 8,∀u∈ K.
It is undeniable that F(T) = {2}.Next, We demon-
strate that Tis asymptotically nonexpansive mapping
on [1,11].
We can see that the function f(u) = 4
√u3+ 8 −
u, ∀u∈[1,11] has the derivative
f′(u) = 3u2
4(u3+ 8)3/4 −1,∀u∈[1,11].
Because 1≤u, we have f′(u) = f′(u) =
3u2
4(u3+8)3/4 ≤1and so
f′(u)≤0,∀u∈[1,11],
which shows that the above function is decreasing on
[1,5].Let u, v ∈[1,5] with u≤vshows that
f(v)≤f(u),
we obtain that
4
√v3+ 8 −v≤4
√u3+ 8 −u
and change it as
4
√v3+ 8 −4
√u3+ 8 ≤v−u
4
√v3+ 8 −4
√u3+ 8≤ |v−u|
4
√u3+ 8 −4
√v3+ 8≤ |u−v|.
Therefore, we obtain that
∥T u− T v∥ ≤ ∥u−v∥.
Thus, Tsatisfies asymptotically nonexpansive
mapping because it is a nonexpansive mapping.
Using the initial value u1= 9 and the specified stop-
ping criteria ∥un−2∥ ≤ 10−15. For two choices,
calculate the values of iterative scheme (3) and itera-
tive scheme (4).
Choice 1: τn=9n
√100n2+4 , ρn=4n
5n+4 and λn=
n
2n+4 .
Choice 2: τn= 1 −n
5n+4 , ρn=3n
4n+4 and λn=
7n
10n+4 .
The results of choice 1 are shown in Table 1 and Fig-
ure 1, as are the results of choice 2 in Table 2 and
Figure 2,.
Table 1: Sequences of comparison for Choice 1.
Number of
Iterations
Iterative scheme (3) Iterative scheme (4)
CPU Time (0.08 Sec.) CPU Time (0.03 Sec.)
1 9.000000000000000 9.000000000000000
23.932991495161055 2.211043635365002
3 2.436259801588707 2.002827615845894
4 2.077070700749241 2.000032201213855
5 2.012069935390678 2.000000343639970
6 2.001794224194666 2.000000003507589
7 2.000258150269182 2.000000000034648
8 2.000036209681632 2.000000000000334
9 2.000004972343876 2.000000000000003
10 2.000000670525931 2.000000000000000
11 2.000000089009141 2.000000000000000
12 2.000000011653699 2.000000000000000
13 2.000000001507295 2.000000000000000
14 2.000000000192849 2.000000000000000
15 2.000000000024434 2.000000000000000
16 2.000000000003069 2.000000000000000
17 2.000000000000382 2.000000000000000
18 2.000000000000047 2.000000000000000
19 2.000000000000006 2.000000000000000
20 2.000000000000000 2.000000000000000
Figure 1: Sequences of comparison for Choice 1.
5 Conclusions
We introduced a novel iterative scheme (4) for asymp-
totically nonexpansive mapping in Hadamard spaces
under certain conditions. The demonstrated that our
new type of iteration is more efficient than iterative
scheme (3). In addition, We have presented a numer-
WSEAS TRANSACTIONS on MATHEMATICS
DOI: 10.37394/23206.2023.22.6
Anantachai Padcharoen, Pakeeta Sukprasert
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Volume 22, 2023
Table 2: Sequences of comparison for Choice 2.
Number of
Iterations
Iterative scheme (3) Iterative scheme (4)
CPU Time (0.10 Sec.) CPU Time (0.04 Sec.)
19.000000000000000 9.000000000000000
23.389076649070681 2.170966511771209
3 2.225543383619792 2.001921267657179
4 2.029729550343821 2.000019117614857
5 2.003699439414234 2.000000181840148
6 2.000453631686722 2.000000001678215
7 2.000055244686556 .000000000015154
8 2.000006696261599 2.000000000000135
9 2.000000808714998 2.000000000000000
10 2.000000097384327 2.000000000000000
11 2.000000011698706 2.000000000000000
12 2.000000001402528 2.000000000000000
13 2.000000000167858 2.000000000000000
14 2.000000000020060 2.000000000000000
15 2.000000000002394 2.000000000000000
16 2.000000000000285 2.000000000000000
17 2.000000000000034 2.000000000000000
18 2.000000000000004 2.000000000000000
19 2.000000000000000 2.000000000000000
Figure 2: Sequences of comparison for Choice 2.
ical experiment to the reader to support our claim.
Acknowledgment
Anantachai Padcharoen (anantachai.p@rbru.ac.th)
was financially supported by the Research and De-
velopment Institute of Rambhaibarni Rajabhat
University. Finally, Pakeeta Sukprasert (pa-
keeta_s@rmutt.ac.th) was financially supported by
Rajamangala University of Technology Thanyaburi
(RMUTT).
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E-ISSN: 2224-2880
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Volume 22, 2023
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_US
Anantachai Padcharoen (anantachai.p@rbru.ac.th)
was financially supported by the Research and De-
velopment Institute of Rambhaibarni Rajabhat
University. Finally, Pakeeta Sukprasert (pa-
