Journal of Applied Mathematics & Bioinformatics, vol.3, no.4, 2013, 285-300
ISSN: 1792-6602 (print), 1792-6939 (online)
Scienpress Ltd, 2013
A New Iterative Scheme for Approximating
Common Fixed Points of Two Nonexpansive
Mappings in Banach Space
Jinzuo Chen1 and Dingping Wu2
Abstract
In this paper, we introduce the modified iterations of Mann’s type for
nonexpansive mappings to have the strong convergence in a uniformly
convex Banach space. We study approximation of common fixed point
of nonexpansive mappings in Banach space by using a new iterative
scheme.
Mathematics Subject Classification: 47H09; 47H10; 47J25
Keywords: Strong convergence; Modified Mann iteration; Uniformly convex
Banach space; Nonexpansive mapping; Common fixed point
1
2
College of Applied Mathematics, Chengdu University of Information Technology.
E-mail: chanjanegeoger@163.com.
College of Applied Mathematics, Chengdu University of Information Technology.
E-mail: wdp68@163.com
Article Info: Received : July 22, 2013. Revised : September 11, 2013.
Published online : December 10, 2013.
286
1
A New Iterative Scheme for Approximating Common Fixed Points...
Introduction
Let E be a real Banach space, C a nonempty closed convex subset of E,
and T : C → C a mapping. Recall that T is nonexpansive mapping[1] if
kT x − T yk ≤ kx − yk for all x, y ∈ C.
A point x ∈ C is a f ixed point of T provided T x = x. Denote by F ix(T ) the
set of fixed point of T ; that is, F ix(T ) = {x ∈ C : T x = x}. It is assumed
throughout the paper that T is a nonexpansive mapping such that F ix(T ) 6= ∅.
Iterative methods are often used to solve the fixed point equation T x = x.
The most well-known method is perhaps the Picard successive iteration method
when T is a contraction. Picard’s method generates a sequence {xn } successively as xn = T xn−1 for n ≥ 1 with x0 arbitrary, and this sequence converges
in norm to the unique fixed point of T . However, if T is not a contraction
(for instance, if T is nonexpansive), then Picard’s successive iteration fail, in
general, to converge. Instead, Mann’s iteration method [10] prevails. Mann’s
method, an averaged process in nature, generates a sequence {xn } recursively
via
xn+1 = αn xn + (1 − αn )T xn , n ≥ 0,
(1.1)
where the initial guess x0 ∈ C is arbitrarily chosen.
Mann’s iteration method (1.1) has been proved to be a powerful method for
solving nonlinear operator equations involving nonexpansive mapping, asymptotically nonexpansive mapping, and other kinds of nonlinear mapping; see [2,
3, 6-8, 10, 12, 13, 15, 16] and the references therein.
It is known that Mann’s iteration method (1.1) is in general not strongly
convergent [5] for nonexpansive mappings. So to get strong convergence, one
has to modify the iteration method (1.1). In this regard, we will show in
Section 3.
Motivated and inspired by the research going on in these fields, we suggest
and analyze now new modified Mann’s iteration for finding the common fixed
point of the nonexpansive mappings in Banach space. We propose the modified Mann’s iteration and consider the strong convergence of the approximate
solutions for nonexpansive in Banach space.
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Jinzuo Chen and Dingping Wu
We suggest and analyze the following two iterative methods:


x0 ∈ C chosen arbitrarily,




 y = β x + (1 − β )T x ,
n
n n
n
n


zn = γn xn + (1 − γn )Sxn ,




xn+1 = αn yn + (1 − αn )zn , n ≥ 0,
(1.2)
and if there exists two sequences {x0n } and {x00n } generated by


x0 ∈ C chosen arbitrarily,





0
0


 yn = βn xn + (1 − βn )T xn ,
zn = γn x00n + (1 − γn )Sx00n ,




 x0n+1 = αn zn + (1 − αn )yn ,



 00
xn+1 = αn yn + (1 − αn )zn , n ≥ 0,
(1.3)
where αn , βn , γn are constants satisfying certain conditions.
We write xn → x to indicate that the sequence {xn } converges strongly to
x. We use F to denote the set of common fixed point of the mappings T and
S.
2
Preliminaries
This section collects some lemmas which will be used in the proofs for the
main results in the next section.
Lemma 2.1. [12] Let {an }, {bn } and {δn } be sequences of nonnegative real
numbers satisfying the inequality
an+1 ≤ (1 + δn )an + bn , n ≥ 1.
If
∞
P
δn < ∞ and
n=1
(1) lim an exists;
∞
P
bn < ∞, then
n=1
n→∞
(2) lim an = 0 whenever lim inf an = 0.
n→∞
n→∞
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A New Iterative Scheme for Approximating Common Fixed Points...
Lemma 2.2. [14] Suppose that E is a uniformly convex Banach space and
0 < tn < 1 for all n ∈ N . Let {xn } and {yn } be two sequences of E such that
lim sup kxn k ≤ r, lim sup kyn k ≤ r and lim ktn xn + (1 − tn )yn k = r hold for
n→∞
n→∞
n→∞
some r ≥ 0, then lim kxn − yn k = 0.
n→∞
Lemma 2.3. [15] A mapping T :C→C with nonempty fixed point set F in C
will be said to satisfy Condition (I):
If there is a nondecreasing function f : [0, ∞) → [0, ∞) with f (0) =
0, f (r) > 0 for all r ∈ (0, ∞) such that kx − T xk ≥ f (d(x, F )) f or all x ∈
C, where d(x, F ) = inf{kx − pk : p ∈ F }.
Lemma 2.4. [18] Given a number r > 0. A real Banach space E is uniformly
convex if and only if there exists a continuous strictly increasing function ϕ :
[0, ∞] → [0, ∞], ϕ(0) = 0, such that
kλx + (1 − λ)yk2 ≤ λkxk2 + (1 − λ)kyk2 − λ(1 − λ)ϕ(kx − yk)
for all λ ∈ [0, 1] and x, y ∈ E such that kxk ≤ r and kyk ≤ r.
Note that the inequality in Lemma 2.4 is known as Xu0 s inequality. It
follows from Lemma 2.4 that we get the following lemma can be found in [4].
Lemma 2.5. Given a number r > 0. Let E is a uniformly convex Banach
space then there exists a continuous strictly increasing function ϕ : [0, ∞] →
[0, ∞] with ϕ(0) = 0, such that
kλx + µy + γzk2 ≤ λkxk2 + µkyk2 + γkzk2 − λµϕ(kx − yk)
for all λ, µ, γ ∈ [0, 1] and x, y, z ∈ E such that kxk ≤ r, kyk ≤ r and kzk ≤ r.
Lemma 2.6. [9] Let {αn }, {βn } be sequences of nonnegative real numbers
∞
∞
P
P
such that
αn = ∞. If
αn βn < ∞, then lim inf βn = 0.
n=1
n=1
n→∞
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Jinzuo Chen and Dingping Wu
3
Convergence to a common fixed point of
nonexpansive mappings
3.1
There exists one sequence {xn }
In this part, we prove our main theorem for finding a common fixed point
of nonexpansive mappings in Banach space in the case of one sequence.
Theorem 3.1. Let C be a nonempty closed convex subset of a uniformly convex
Banach space E and let T and S be two nonexpansive self mappings of C satisfy
Condition (I) and F 6= ∅. Given {αn }, {βn } and {γn } are sequences in (0,1),
P
P
P
such that
αn < ∞,
γn βn = ∞, (1 − γn ) < ∞ for all n ≥ 1.
∞
Define a sequence {xn }∞
n=0 in C by the algorithm (1.2), then {xn }n=0
strongly converges to a common fixed point of T and S.
Proof. First, we observe that {xn } is bounded, if we take an arbitrary fixed
point q of F , noting that kyn − qk ≤ kxn − qk and kzn − qk ≤ kxn − qk, we
have
kxn+1 − qk = kαn yn + (1 − αn )zn − qk
≤ αn kyn − zn k + kzn − qk
≤ αn kyn − qk + αn kzn − qk + kzn − qk
By Lemma 2.1 and
P
(3.1)
≤ (1 + 2αn )kxn − qk.
αn < ∞, thus lim kxn − qk exists. Denote
n→∞
lim kxn − qk = c.
n→∞
Hence, {xn } is bounded, so are {yn } and {zn }. Now
kxn+1 − qk = kαn yn + (1 − αn )zn − qk
= kαn (yn − zn ) + (zn − q)k
≤ αn kyn − zn k + kzn − qk.
By
P
αn < ∞ and the boundedness of {zn } and {yn }, we obtain
lim kxn − qk ≤ lim inf kzn − qk.
n→∞
n→∞
(3.2)
Since kzn − qk ≤ kxn − qk, which implies that
lim sup kzn − qk ≤ lim kxn − qk,
n→∞
n→∞
(3.3)
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A New Iterative Scheme for Approximating Common Fixed Points...
so that (3.2) and (3.3) give
lim kzn − qk = lim kxn − qk = c.
n→∞
n→∞
Moreover, kSxn − qk ≤ kxn − qk implies that
lim sup kSxn − qk ≤ c.
n→∞
Thus
c = lim kzn − qk = lim kγn xn + (1 − γn )Sxn − qk
n→∞
n→∞
= lim kγn (xn − q) + (1 − γn )(Sxn − q)k,
n→∞
given by Lemma 2.2 that
lim kSxn − xn k = 0.
n→∞
By (3.1) and
P
(3.4)
αn < ∞, then we have
kxn+m − qk ≤ (1 + 2αn+m−1 )kxn+m−1 − qk
≤ e2αn+m−1 kxn+m−1 − qk
≤ e2αn+m−1 e2αn+m−2 kxn+m−2 − qk
≤ ...
2
n+m−1
P
≤e
i=n
αi
kxn − qk.
That is
kxn+m − qk ≤ M kxn − qk,
2
n+m−1
P
(3.5)
αi
where M = e i=n , for all m, n ≥ 1, for all q ∈ F and for M > 0.
Next we prove that {xn }∞
n=0 is Cauchy sequence.
Since q ∈ F arbitrarily and lim kxn − qk exists, consequently d(xn , F )
n→∞
exists by Lemma 2.3. From the Lemma 2.3 and (3.4), we get
lim f (d(xn , F )) ≤ lim kxn − Sxn k = 0
n→∞
n→∞
Since f : [0, ∞) → [0, ∞) is a nondecreasing function satisfy f (0) = 0, f (r) > 0
for all r ∈ (0, ∞), therefore we have
lim d(xn , F ) = 0.
n→∞
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Jinzuo Chen and Dingping Wu
Let ε > 0, since lim d(xn , F ) = 0 and therefore exists a constant n0 such
n→∞
that for all n ≥ n0 , we have
d(xn , F ) ≤
ε
,
2M
d(xn0 , F ) ≤
ε
.
2M
in particular,
There must exist p1 ∈ F , such that
d(xn0 , p1 ) ≤
ε
.
2M
From (3.5), it can be obtained that when n ≥ n0 ,
kxn+m − xn k ≤ kxn+m − p1 k + kxn − p1 k
≤ 2M kxn0 − p1 k
ε
= ε.
≤ 2M ·
2M
This implies {xn }∞
n=0 is a Cauchy sequence in a closed convex subset C of a
Banach space E. Thus, it must converges to a point in C, let lim xn = p.
n→∞
For all ² > 0, as lim xn = p, thus there exists a number n1 such that when
n→∞
n2 ≥ n1 ,
²
kxn2 − pk ≤ .
(3.6)
4
In fact, lim d(xn , F ) = 0 implies that using the n2 above, when n ≥ n2 ,
n→∞
we get
²
d(xn , F ) ≤ .
8
In particular, d(xn2 , F ) ≤ 8² . Thus, there must exist p̄ ∈ F , such that
²
kxn2 − p̄k = d(xn2 , p̄) = .
8
From (3.6) and (3.7), we get
kSp − pk = kSp − p̄ + Sxn2 − p̄ + p̄ − xn2 + xn2 − p + p̄ − Sxn2 k
≤ kSp − p̄k + kxn2 − p̄k + kxn2 − pk + 2kSxn2 − p̄k
≤ kp − p̄k + 3kxn2 − p̄k + kxn2 − pk
≤ kxn2 − pk + kxn2 − p̄k + 3kxn2 − p̄k + kxn2 − pk
= 4kxn2 − p̄k + 2kxn2 − pk
4² 2²
+
= ².
≤
8
4
(3.7)
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A New Iterative Scheme for Approximating Common Fixed Points...
As ² is an arbitrary positive number, thus, Sp = p.
Let
un+1 = γn xn + βn T xn + (1 − βn − γn )T xn .
Then we have
kun+1 − qk2
= kγn xn + βn T xn + (1 − βn − γn )T xn − qk2
= kγn (xn − q) + βn (T xn − q) + (1 − βn − γn )(T xn − q)k2
≤ γn kxn − qk2 + βn kT xn − qk2 + (1 − βn − γn )kT xn − qk2 − γn βn ϕkxn − T xn k
≤ kxn − qk2 − γn βn ϕkxn − T xn k,
and hence
γn βn ϕkxn − T xn k ≤ kxn − qk2 − kun+1 − qk2
for q ∈ F . Summing from n=1 to ∞, we have
∞
X
γn βn ϕkxn − T xn k =
∞
X
(kxn − qk2 − kun+1 − qk2 )
n=1
n=1
=
∞
X
(kxn − qk + kun+1 − qk)(kxn − qk − kun+1 − qk)
n=1
≤ 2K
≤ 2K
≤ 4K
∞
X
kun+1 − xn k
n=1
∞
X
(1 − γn )kxn − T xn k
n=1
∞
X
2
(1 − γn ),
n=1
where K = sup{kxn − qk}, from
n∈
N
∞
X
∞
P
n=1
(1 − γn ) < ∞, we get
n=1
∞
X
γn βn ϕkxn − T xn k ≤ 4K
(1 − γn ) < ∞.
2
n=1
Since
∞
P
n=1
γn βn = ∞, from Lemma 2.6, we get lim inf ϕkT xn − xn k = 0. Hence
n→∞
lim inf kT xn − xn k = 0.
n→∞
(3.8)
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Jinzuo Chen and Dingping Wu
Since T is a nonexpansive mapping, we have
kT xn+1 − xn+1 k
= kT xn+1 − αn yn − (1 − αn )zn k
= kT xn+1 − T xn + T xn + αn T xn − αn T xn − αn yn − (1 − αn )zn k
≤ kT xn+1 − T xn k + (1 − αn )kT xn − zn k + αn kT xn − yn k
≤ kxn+1 − xn k + (1 − αn )kT xn − zn k + αn kT xn − yn k
= kαn yn + (1 − αn )zn − xn k + (1 − αn )kT xn − zn k + αn kT xn − yn k
≤ αn kyn − xn k + (1 − αn )kzn − xn k + (1 − αn )kT xn − zn k + αn kT xn − yn k
= αn kβn xn + (1 − βn )T xn − xn k + (1 − αn )kγn xn + (1 − γn )Sxn − xn k
+ (1 − αn )kT xn − γn xn − (1 − γn )Sxn k + αn kT xn − βn xn − (1 − βn )T xn k
≤ αn (1 − βn )kT xn − xn k + (1 − αn )(1 − γn )kSxn − xn k + αn βn kT xn − xn k
+ γn (1 − αn )kT xn − xn k + (1 − αn )(1 − γn )kT xn − Sxn k
≤ αn (1 − βn )kT xn − xn k + (1 − γn )kSxn − xn k + αn βn kT xn − xn k
+ (1 − αn )kT xn − xn k + (1 − γn )kT xn − Sxn k
≤ kT xn − xn k + (1 − γn )(kSxn − xn k + kT xn − Sxn k).
Since
∞
P
(1 − γn ) < ∞, it follows from Lemma 2.1 that lim kT xn − xn k exists.
n→∞
n=1
Therefore, from (3.8), we get
lim kT xn − xn k = 0.
n→∞
Then using the same argument we can show that {xn }∞
n=0 converges strongly
to a common fixed point of T and S.
Remark 3.1. It is to be noted that (1.2) reduces to (1.1) when T or S equal
to I.
3.2
There exists two sequences {x0n } and {x00n }
In this part, we prove our main theorem for finding a common fixed point
of nonexpansive mappings in Banach space in the case of two sequences.
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A New Iterative Scheme for Approximating Common Fixed Points...
Theorem 3.2. Let C be a nonempty bounded closed convex subset of a uniformly convex Banach space E and let T and S be two nonexpansive self mappings of C satisfy Condition (I) and F 6= ∅. Given {αn }, {βn } and {γn } are
P
P
sequences in (0,1), such that
αn < ∞,
βn < ∞, βn > γn for all n ≥ 1.
0 ∞
00 ∞
Define two sequences {xn }n=0 and {xn }n=0 in C by the algorithm (1.3),
00 ∞
then {x0n }∞
n=0 and {xn }n=0 strongly converge to a common fixed point of T and
S.
Proof. By the boundedness of C, we obverse that both {x0n } and {x00n } are
bounded, if we take an arbitrary fixed point q of F , noting that kyn − qk ≤
kx0n − qk and kzn − qk ≤ kx00n − qk, we have
kx0n+1 − qk = kαn zn + (1 − αn )yn − qk
≤ αn kzn − yn k + kyn − qk
≤ αn kyn − qk + αn kzn − qk + kyn − qk
(3.9)
≤ (1 + αn )kx0n − qk + αn kx00n − qk.
By Lemma 2.1 and
P
αn < ∞, thus lim kx0n − qk exists. Denote
n→∞
lim kx0n − qk = c0 .
n→∞
Similarly, we have
lim kx00n − qk = c00 .
n→∞
Since both {x0n } and {x00n } are bounded, we get {yn } and {zn } are bounded.
Now
kx0n+1 − qk = kαn zn + (1 − αn )yn − qk
= kαn (zn − yn ) + (yn − q)k
≤ αn kzn − yn k + kyn − qk.
By
P
αn < ∞, we obtain
lim kx0n − qk ≤ lim inf kyn − qk.
n→∞
n→∞
(3.10)
Since kyn − qk ≤ kx0n − qk, which implies that
lim sup kyn − qk ≤ lim kx0n − qk,
n→∞
n→∞
(3.11)
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Jinzuo Chen and Dingping Wu
so that (3.10) and (3.11) give
lim kyn − qk = lim kx0n − qk = c0 .
n→∞
n→∞
Moreover, kT x0n − qk ≤ kx0n − qk implies that
lim sup kT x0n − qk ≤ c0 .
n→∞
Thus
c0 = lim kyn − qk = lim kβn x0n + (1 − βn )T x0n − qk
n→∞
n→∞
= lim kβn (x0n − q) + (1 − βn )(T x0n − q)k,
n→∞
given by Lemma 2.2 that
lim kT x0n − x0n k = 0.
(3.12)
n→∞
By (3.9) and
P
αn < ∞, then we have
kx0n+m − qk ≤ (1 + αn+m−1 )kx0n+m−1 − qk + αn+m−1 kx00n+m−1 − qk
≤ eαn+m−1 kx0n+m−1 − qk + αn+m−1 kx00n+m−1 − qk
≤ eαn+m−1 eαn+m−2 kxn+m−2 − qk
+ eαn+m−1 (αn+m−2 kx00n+m−2 − qk + αn+m−1 kx00n+m−1 − qk)
≤ ...
n+m−1
P
≤e
i=n
αi
n+m−1
P
kx0n
− qk + e
i=n
αi
·
n+m−1
X
αi kx00i − qk.
i=n
That is
kx0n+m − qk ≤ L(kx0n − qk +
n+m−1
X
si ),
(3.13)
i=n
n+m−1
P
αi
where L = e i=n , si = αi kx00i − qk. for all m, n ≥ 1, for all q ∈ F .
Next we prove that {x0n }∞
n=0 is Cauchy sequence.
Since q ∈ F arbitrarily and lim kx0n − qk exists, consequently d(x0n , F )
n→∞
exists by Lemma 2.3. From the Lemma 2.3 and (3.12), we get
lim f (d(x0n , F )) ≤ lim kx0n − T x0n k = 0
n→∞
n→∞
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A New Iterative Scheme for Approximating Common Fixed Points...
Since f : [0, ∞) → [0, ∞) is a nondecreasing function satisfy f (0) = 0, f (r) > 0
for all r ∈ (0, ∞), therefore we have
lim d(x0n , F ) = 0.
n→∞
Let ε > 0, since lim d(x0n , F ) = 0 and
n→∞
∞
P
si < ∞, therefore it exists a
i=0
constant n0 such that for all n ≥ n0 , we have
d(x0n , F ) ≤
ε
,
2L
d(x0n0 , F ) ≤
ε
.
2L
in particular,
There must exist p1 ∈ F , such that
d(x0n0 , p1 ) ≤
ε
.
2L
From (3.13), it can be obtained that when n ≥ n0 ,
kx0n+m − x0n k ≤ kx0n+m − p1 k + kx0n − p1 k
≤ 2Lkx0n0 − p1 k
ε
= ε.
≤ 2L ·
2L
This implies {x0n }∞
n=0 is a Cauchy sequence in a closed convex subset C of a
Banach space E. Thus, it must converges to a point in C, let lim x0n = x0 .
n→∞
For all ² > 0, as lim x0n = x0 , thus there exists a number n1 such that when
n→∞
n2 ≥ n1 ,
²
(3.14)
kx0n2 − x0 k ≤ .
4
In fact, lim d(x0n , F ) = 0 implies that using the n2 above, when n ≥ n2 ,
n→∞
we get
²
d(x0n , F ) ≤ .
8
In particular, d(x0n2 , F ) ≤ 8² . Thus, there must exist x̄0 ∈ F , such that
²
kx0n2 − x̄0 k = d(x0n2 , x̄0 ) = .
8
(3.15)
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Jinzuo Chen and Dingping Wu
From (3.14) and (3.15), we get
kT x0 − x0 k = kT x0 − x̄0 + T x0n2 − x̄0 + x̄0 − x0n2 + x0n2 − x0 + x̄0 − T x0n2 k
≤ kT x0 − x̄0 k + kx0n2 − x̄0 k + kx0n2 − x0 k + 2kT x0n2 − x̄0 k
≤ kx0 − x̄0 k + 3kx0n2 − x̄0 k + kx0n2 − x0 k
≤ kx0n2 − x0 k + kx0n2 − x̄0 k + 3kx0n2 − x̄0 k + kx0n2 − x0 k
= 4kx0n2 − x̄0 k + 2kx0n2 − x0 k
4² 2²
≤
+
= ².
8
4
As ² is an arbitrary positive number, thus, T x0 = x0 .
Similarly, we have lim kSx00n −x00n k = 0, and then Sx00 = x00 (x00n → x00 as n →
n→∞
∞)
Finally, we prove x0 = x00 .
Let wn+1 = αn T x0n + (1 − αn )Sx00n and kx00n − qk ≥ kx0n − qk for all n ≥ 1.
Then
kwn+1 − qk = αn kx0n − qk + (1 − αn )kx00n − qk
≤ max{kx0n − qk, kx00n − qk}
≤ kx00n − qk.
Now,
kx00n+1 − qk = kαn yn + (1 − αn )zn − qk
= kαn βn x0n + αn (1 − βn )T x0n + (1 − αn )γn x00n + (1 − αn )(1 − γn )Sx00n − qk
≤ βn kαn x0n + (1 − αn )x00n k + kαn T x0n + (1 − αn )Sx00n − qk
= βn kαn x0n + (1 − αn )x00n k + kwn+1 − qk,
since
P
βn < ∞ and the boundedness of {x0n } and {x00n }, we get
c00 ≤ lim kwn+1 − qk.
n→∞
Then we have
lim kwn+1 − qk = c00 .
n→∞
Moreover, kT x0n − qk ≤ kx0n − qk and kSx00n − qk ≤ kx00n − qk, imply that
lim sup kT x0n − qk ≤ c00 and lim sup kSx00n − qk ≤ c00 .
n→∞
n→∞
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Thus,
c00 = lim kwn+1 − qk = kαn T x0n + (1 − αn )Sx00n − qk
n→∞
= kαn (T x0n − q) + (1 − αn )(Sx00n − q)k,
given by Lemma 2.2 that
lim kSx00n − T x0n k = 0.
n→∞
(3.16)
So
lim kx0n+1 − x00n+1 k
n→∞
= lim (2αn − 1)kzn − yn k
n→∞
= lim (2αn − 1)kγn (x00n − Sx00n ) − βn (x0n − T x0n ) + (Sx00n − T x0n )k,
n→∞
so we obtain lim kx0n+1 − x00n+1 k = 0 for (3.12) and (3.16), it means x0 = x00 .
n→∞
This completes the proof.
Acknowledgements
This work was supported by the National Natural Science Foundation of
China(Grant No. 11171046) and supported by the Scientific Research Foundation of CUIT(No. KYTZ201004).
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