A modified iterative algorithm for nonexpansive mappings
Youli Yu ∗, Ching-Feng Wen †,
Abstract
A modified iterative algorithm is presented based on the semi-implicit midpoint rule. Strong
convergence analysis is demonstrated. Our method gives a unified framework related to the implicit midpoint rule. Our results improve and extend the corresponding results in the literature.
Keywords: nonexpansive mapping; implicit midpoint rule; fixed point.
2010 Mathematics Subject Classification: 47J25; 47N20; 34G20; 65J15.
1
Introduction
The purpose of the paper is to construct iterative methods for finding the fixed points of nonexpansive
mappings. Fixed point methods for nonexpansive mappings have been studied extensively by many
researchers due to its applications in engineering and natural science. Especially, the following fixed
point methods have attracted so much attention.
Browder’s method ([1]): for fixed u ∈ C,
xt = tu + (1 − t)T xt ,
where t ∈ (0, 1). Mann’s method ([2]):
xn+1 = αn xn + (1 − αn )T xn ,
n ≥ 0,
where αn ∈ (0, 1). Ishikawa’s methd ([3]):
yn = αn xn + (1 − αn )T xn ,
xn+1 = βn xn + (1 − βn )T yn , n ≥ 0,
where αn and βn are in (0, 1). Halpern’s method ([4]):
xn+1 = αn u + (1 − αn )T xn ,
n ≥ 0,
∗ School of Mathematics and Information Engineering, Taizhou University, Linhai 317000, China. e-mail: yuyouli@tzc.edu.cn
† Corresponding Author, Center for Fundamental Science, and Research Center for Nonlinear Analysis and Optimizatio, Kaohsiung Medical University, Kaohsiung, 807, Taiwan. e-mail: cfwen@kmu.edu.tw
1
where u ∈ C is a fixed point and αn ∈ (0, 1). Moudafi’s viscosity method ([5])
xn+1 = αn Q(xn ) + (1 − αn )T xn ,
n ≥ 0,
where Q : C → C is a contraction and αn ∈ (0, 1). Modified Mann’s method ([6]):
yn = αn xn + (1 − αn )T xn ,
xn+1 = βn u + (1 − βn )yn , n ≥ 0.
Many researchers demonstrated the convergence results of the above methods and their variant
forms. Related references, please refer to [7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22,
23, 24, 25, 26, 33]. Very recently, in [27] and [28], the authors presented the following semi-implicit
midpoint rule for nonexpansive mappings:
(
xn+1 = (1 − αn )xn + αn T
)
xn + xn+1
, n ≥ 0,
2
(1.1)
and
(
xn+1 = αn Q(xn ) + (1 − αn )T
)
xn + xn+1
, n ≥ 0,
2
(1.2)
where αn ∈ (0, 1) and Q is a contraction. Further, Yao, Shahzad and Liou [29] introduced the
following semi-implicit midpoint method:
(
xn+1
)
xn + xn+1
= αn Q(xn ) + βn xn + γn T
, n ≥ 0.
2
(1.3)
Motivated and inspired by the above works, the purpose of the paper is to construct the following
unified iterative algorithm for finding the fixed points of nonexpansive mappings
xn+1 = αn Q(xn ) + βn xn + γn T (δn xn + (1 − δn )xn+1 ), n ≥ 0.
We prove that the above algorithm converges strongly to a fixed point of nonexpansive mappings T .
2
Preliminaries
Let H be a real Hilbert space with inner product ⟨·, ·⟩ and norm ∥ · ∥, respectively. Let C be a
nonempty closed convex subset of H. A mapping T : C → C is said to be nonexpansive if
∥T x − T y∥ ≤ ∥x − y∥
for all x, y ∈ C. We use F ix(T ) to denote the set of fixed points of T .
A mapping Q : C → C is said to be contractive if there exists a constant α ∈ (0, 1) such that
∥Q(x) − Q(y)∥ ≤ α∥x − y∥
for all x, y ∈ C. In this case, Q is called α-contraction.
2
Lemma 2.1. ([30]) Let C be a nonempty closed convex subset of a Hilbert space H, and let T :
C → C be a nonexpansive mapping with F ix(T ) ̸= ∅. Assume that {yn } is a sequence in C such
that yn ⇀ x† and (I − T )yn → 0. Then x† ∈ F ix(T ).
Lemma 2.2. ([31]) Let {xn } and {yn } be bounded sequences in a Banach space E and {βn } be a
sequence in [0, 1] with 0 < lim inf n→∞ βn ≤ lim supn→∞ βn < 1. Suppose that xn+1 = (1 − βn )xn +
βn zn for all n ≥ 0 and lim supn→∞ (∥zn+1 − zn ∥ − ∥xn+1 − xn ∥) ≤ 0. Then limn→∞ ∥zn − xn ∥ = 0.
Lemma 2.3. ([32]) Let {an }n∈N be a sequence of non-negative real numbers satisfying the following
relation:
an+1 ≤ (1 − αn )an + αn σn + δn ,
where
(i) {αn }n∈N ⊂ [0, 1] and
∑∞
n=1
n ≥ 0,
αn = ∞;
(ii) lim supn→∞ σn ≤ 0;
∑∞
(iii) n=1 δn < ∞.
Then limn→∞ an = 0.
3
Main results
Throughout, we assume that H is a real Hilbert space and C ⊂ H is a nonempty closed convex
set. Let T : C → C be a nonexpansive mapping with its fixed points set being nonempty, that is,
F ix(T ) ̸= ∅. Let Q : C → C be an α-contraction.
Now, we firstly present the following unified iterative algorithm.
Algorithm 3.1. For given x0 ∈ C arbitrarily, let the sequence {xn } be generated iteratively by the
manner
xn+1 = αn Q(xn ) + βn xn + γn T (δn xn + (1 − δn )xn+1 ), n ≥ 0,
(3.1)
where {αn } ⊂ (0, 1), {βn } ⊂ [0, 1), {γn } ⊂ (0, 1) and {δn } ⊂ [κ1 , κ2 ] ⊂ (0, 1) are four sequences
satisfying αn + βn + γn = 1 for all n ≥ 0.
Remark 3.2. (3.1) is well-defined. As a matter of fact, for fixed u ∈ C, we can define a mapping
x 7→ Tu x := αQ(u) + βu + γT (δu + (1 − δ)x), ∀x ∈ C.
Then, we have
∥Tu x − Tu y∥ = γ∥T (δu + (1 − δ)x) − T (δu + (1 − δ)y)∥
≤ (1 − δ)γ∥x − y∥.
This means Tu is a contraction with coefficient (1−δ)γ ∈ (0, 1). Hence, Algorithm 3.1 is well-defined.
3
Next, we show the boundedness of the sequence {xn }.
Proposition 3.3. The sequence {xn } generated by (3.1) is bounded.
Proof. Let z ♯ ∈ F ix(T ). From (3.1), we get
∥xn+1 − z ♯ ∥ = ∥αn (Q(xn ) − Q(z ♯ )) + αn (Q(z ♯ ) − z ♯ ) + βn (xn − z ♯ )
+ γn (T (δn xn + (1 − δn )xn+1 ) − z ♯ )∥
≤ αn ∥Q(xn ) − Q(z ♯ )∥ + αn ∥Q(z ♯ ) − z ♯ ∥ + βn ∥xn − z ♯ ∥
+ γn ∥T (δn xn + (1 − δn )xn+1 ) − z ♯ ∥
≤ αn α∥xn − z ♯ ∥ + αn ∥Q(z ♯ ) − z ♯ ∥ + βn ∥xn − z ♯ ∥
+ γn δn ∥xn − z ♯ ∥ + γn (1 − δn )∥xn+1 − z ♯ ∥.
From the last inequality, we obtain
αn α + βn + γn δn
αn
∥xn − z ♯ ∥ +
∥Q(z ♯ ) − z ♯ ∥
1 − γn (1 − δn )
1 − γn (1 − δn )
[
]
(1 − α)αn
(1 − α)αn
1
= 1−
∥xn − z ♯ ∥ +
∥Q(z ♯ ) − z ♯ ∥
1 − γn (1 − δn )
1 − γn (1 − δn ) 1 − α
1
≤ max{∥xn − z ♯ ∥,
∥Q(z ♯ ) − z ♯ ∥}.
1−α
∥xn+1 − z ♯ ∥ ≤
By induction, we deduce
∥xn − z ♯ ∥ ≤ max{∥x0 − z ♯ ∥,
1
∥Q(z ♯ ) − z ♯ ∥}.
1−α
This indicates that {xn } is bounded. This completes the proof.
Next, we state the following theorem
Theorem 3.4. Assume {αn }, {βn } and {γn } satisfy the conditions
(C1) : limn→∞ αn = 0;
(C2) :
∑∞
n=0
αn = ∞;
(C3) : limn→∞
|αn −αn−1 |
α2n
= 0;
(C4) : limn→∞
|βn −βn−1 |
α2n
= 0;
(C5) : lim inf n→∞ γn > 0.
Then the sequence {xn } generated by (3.1) converges strongly to q = PF ix(T ) Q(q).
4
Proof. Set zn =
αn
1−βn Q(zn )
+ (1 −
αn
1−βn )T zn
for all n. By [8], we have that the sequence {zn }
converges strongly to q = PF ix(T ) Q(q) provided limn→∞ αn = 0. Note that the sequences {xn } and
{zn } are all bounded. We can rewrite yn as
zn = αn Q(zn ) + βn zn + γn T zn , n ≥ 0.
First, we note that
∥xn+1 − zn ∥ = ∥αn (Q(xn ) − Q(zn )) + βn (xn − zn ) + γn (T (δn xn + (1 − δ)xn+1 ) − T zn )∥
≤ αn α∥xn − zn ∥ + βn ∥xn − zn ∥ + γn δn ∥xn − zn ∥ + γn (1 − δn )∥xn+1 − zn ∥.
It follows that
[
]
(1 − α)αn
∥xn+1 − zn ∥ ≤ 1 −
∥xn − zn ∥
1 − γn (1 − δn )
[
]
(1 − α)αn
≤ 1−
∥xn − zn−1 ∥ + ∥zn − zn−1 ∥.
1 − γn (1 − δn )
It is easily seen that if
∑∞
n=0
αn = ∞ and limn→∞
∥zn −zn−1 ∥
αn
= 0, then we get limn→∞ ∥xn+1 −zn ∥ =
0 by Lemma 2.3. Consequently, xn → q = PF ix(T ) Q(q) provided limn→∞ αn = 0.
Next, we estimate
∥zn −zn−1 ∥
.
αn
As a matter of fact, we have
∥zn − zn−1 ∥ = ∥αn (Q(zn ) − Q(zn−1 )) + (αn − αn−1 )Q(zn−1 ) + βn (zn − zn−1 )
+ (βn − βn−1 )zn−1 + γn (T zn − T zn−1 ) + (γn − γn−1 )T zn−1 ∥
≤ (ααn + βn + γn )∥zn − zn−1 ∥ + |αn − αn−1 |∥Q(zn−1 )∥
+ |βn − βn−1 |∥zn−1 ∥ + |γn − γn−1 |∥T zn−1 ∥.
Hence,
|βn − βn−1 |
∥zn − zn−1 ∥
|αn − αn−1 |
(∥Q(zn−1 )∥ + ∥T zn−1 ∥) +
(∥zn−1 ∥ + ∥T zn−1 ∥).
≤
2
αn
(1 − α)αn
(1 − α)αn2
Since limn→∞
|αn −αn−1 |
α2n
= limn→∞
|βn −βn−1 |
α2n
= 0, we derive that limn→∞
∥zn −zn−1 ∥
αn
= 0. Therefore,
limn→∞ ∥xn+1 − zn ∥ = 0 and thus xn → q = PF ix(T ) Q(q). This completes the proof.
Theorem 3.5. Assume {αn } satisfies (C1) and (C2), {βn } and {δn } satisfies
(C1) : limn→∞ αn = 0;
(C2) :
∑∞
n=0
αn = ∞;
(C6) : 0 < lim inf n→∞ βn ≤ lim supn→∞ βn < 1;
(C7) : limn→∞ (βn+1 − βn ) = 0;
(C8) : limn→∞ (δn+1 − δn ) = 0.
5
Then the sequence {xn } generated by (3.1) converges strongly to q = PF ix(T ) Q(q).
Proof. From Proposition 3.3, we have can choose a constant M such that
)(
{(
)}
3
3
sup
+
∥Q(xn )∥ + ∥T (δn xn + (1 − δn )xn+1 )∥ + ∥xn ∥
≤ M.
1 − βn
1 − γn
n
Set yn =
xn+1 −βn xn
1−βn
for all n ≥ 0. Thus, we have
xn+2 − βn+1 xn+1
xn+1 − βn xn
−
1 − βn+1
1 − βn
αn+1 Q(xn+1 ) + (1 − αn+1 − βn+1 )T (δn+1 xn+1 + (1 − δn+1 )xn+2 )
=
1 − βn+1
αn Q(xn ) + (1 − αn − βn )T (δn xn + (1 − δn )xn+1 )
−
1 − βn
αn+1
(Q(xn+1 ) − Q(xn ))
=
1 − βn+1
1 − αn+1 − βn+1
+
(T (δn+1 xn+1 + (1 − δn+1 )xn+2 ) − T (δn xn + (1 − δn )xn+1 ))
1 − βn+1
(
)
αn+1
αn
+
−
(Q(xn ) − T (δn xn + (1 − δn )xn+1 )).
1 − βn+1
1 − βn
yn+1 − yn =
It follows that
(
)
αn+1
αn
∥yn+1 − yn ∥ ≤
+
∥Q(xn ) − T (δn xn + (1 − δn )xn+1 )∥
1 − βn+1
1 − βn
1 − αn+1 − βn+1
(δn+1 ∥xn+1 − xn ∥ + (1 − δn+1 )∥xn+2 − xn+1 ∥)
+
1 − βn+1
1 − αn+1 − βn+1
+
|δn+1 − δn |(∥xn ∥ + ∥xn+1 ∥)
1 − βn+1
ααn+1
+
∥xn+1 − xn ∥.
1 − βn+1
From (3.1), we have
∥xn+2 − xn+1 ∥ = ∥αn+1 (Q(xn+1 ) − Q(xn )) + (αn+1 − αn )Q(xn )
+ βn+1 (xn+1 − xn ) + (βn+1 − βn )(xn − T (δn xn + (1 − δn )xn+1 ))
+ γn+1 (T (δn+1 xn+1 + (1 − δn+1 )xn+2 ) − T (δn xn + (1 − δn )xn+1 ))
+ (αn − αn+1 )T (δn xn + (1 − δn )xn+1 )∥
≤ ααn+1 ∥xn+1 − xn ∥ + (αn+1 + αn )∥Q(xn )∥ + βn+1 ∥xn+1 − xn ∥
+ γn+1 (δn+1 ∥xn+1 − xn ∥ + (1 − δn+1 )∥xn+2 − xn+1 ∥)
+ γn+1 |δn+1 − δn |(∥xn ∥ + ∥xn+1 ∥) + (αn + αn+1 )∥T (δn xn + (1 − δn )xn+1 )∥
+ |βn+1 − βn |∥xn − T (δn xn + (1 − δn )xn+1 )∥.
6
(3.2)
It follows that
[
∥xn+2 − xn+1 ∥ ≤ 1 −
]
(1 − α)αn+1
∥xn+1 − xn ∥
1 − γn+1 (1 − δn+1 )
αn+1 + αn
+
(∥Q(xn )∥ + ∥T (δn xn + (1 − δn )xn+1 )∥)
1 − γn+1 (1 − δn+1 )
(3.3)
|βn+1 − βn | + |δn+1 − δn |
+
(∥xn − T (δn xn + (1 − δn )xn+1 )∥ + ∥xn ∥ + ∥xn+1 ∥)
1 − γn+1 (1 − δn+1 )
≤ ∥xn+1 − xn ∥ + M (αn + αn+1 + |βn+1 − βn | + |δn+1 − δn |).
Substitute (3.3) into (3.2) to get
[
]
(1 − α)αn+1
∥yn+1 − yn ∥ ≤ 1 −
∥xn+1 − xn ∥ + 3M (αn+1 + αn + |βn+1 − βn | + |δn+1 − δn |).
1 − βn+1
Hence,
lim sup(∥yn+1 − yn ∥ − ∥xn+1 − xn ∥) ≤ 0.
n→∞
This together with Lemma 2.2 implies that
lim ∥yn − xn ∥ = 0.
n→∞
Note that
yn − xn =
xn+1 − xn
.
1 − βn
So,
lim ∥xn+1 − xn ∥ = 0.
n→∞
(3.4)
Again, from (3.1), we have
∥xn − T xn ∥ ≤ ∥xn − xn+1 ∥ + ∥xn+1 − T xn ∥
≤ ∥xn − xn+1 ∥ + αn ∥Q(xn ) − T xn ∥ + βn ∥xn − T xn ∥
+ γn (1 − δn )∥xn − xn+1 ∥.
It follows that
∥xn − T xn ∥ ≤
αn
1 + γn (1 − δn )
∥Q(xn ) − T xn ∥ +
∥xn+1 − xn ∥.
1 − βn
1 − βn
This together with (C1) and (3.4) imply that
lim ∥xn − T xn ∥ = 0
(3.5)
lim sup⟨q − Q(q), q − xn ⟩ ≤ 0,
(3.6)
n→∞
Next, we prove that
n→∞
7
where q ∈ F ix(T ) is the unique fixed point of the contraction PF ix(T ) Q, that is, q = PF ix(T ) Q(q).
Since {xn } is bounded, there exists a subsequence {xni } of {xn } such that {xni } converges
weakly to a point x̆ and
lim sup⟨PF ix(T ) Q(q) − Q(q), PF ix(T ) Q(q) − xn ⟩
n→∞
= lim ⟨PF ix(T ) Q(q) − Q(q), PF ix(T ) Q(q) − xni ⟩.
(3.7)
i→∞
By Lemma 2.1 and (3.5), we deduce x̆ ∈ F ix(T ). Therefore,
lim sup⟨PF ix(T ) Q(q) − Q(q), PF ix(T ) Q(q) − xn ⟩
n→∞
= lim ⟨PF ix(T ) Q(q) − Q(q), PF ix(T ) Q(q) − xni ⟩
i→∞
= ⟨PF ix(T ) Q(q) − Q(q), PF ix(T ) Q(q) − x̆⟩
≤ 0.
Finally, we prove that xn → q. From (3.1), we have
∥xn+1 − q∥2 = αn ⟨Q(xn ) − Q(q), xn+1 − q⟩ + αn ⟨Q(q) − q, xn+1 − q⟩
+ γn ⟨T (δn xn + (1 − δn )xn+1 ) − q, xn+1 − q⟩
+ βn ⟨xn − q, xn+1 − q⟩
≤ αn α∥xn − q∥∥xn+1 − q∥ + αn ⟨Q(q) − q, xn+1 − q⟩
+ γn (δn ∥xn − q∥ + (1 − δn )∥xn+1 − q∥)∥xn+1 − q∥
+ βn ∥xn − q∥∥xn+1 − q∥
1 + γn (1 − δn ) − (1 − α)αn
1 − γn (1 − δn ) − (1 − α)αn
∥xn − q∥2 +
∥xn+1 − q∥2
≤
2
2
+ αn ⟨Q(q) − q, xn+1 − q⟩.
It follows that
[
]
2(1 − α)αn
∥xn+1 − q∥ ≤ 1 −
∥xn − q∥2
1 − γn (1 − δn ) + (1 − α)αn
2αn
+
⟨Q(q) − q, xn+1 − q⟩.
1 − γn (1 − δn ) + (1 − α)αn
2
(3.8)
Applying Lemma 2.3 and (3.6) to (3.8) to deduce that xn → q. This completes the proof.
Acknowledgments
The research of the send author was partially supported by the grant MOST 104-2115-M-037-001
and the grant from Research Center for Nonlinear Analysis and Optimization, Kaohsiung Medical
University, Taiwan.
8
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