Volume 2011, Article ID 368137, 15 pagesdoi:10.1155/2011/368137 Research Article Convergence of Iterative Sequences for Fixed Point and Variational Inclusion Problems Li Yu and Ma Liang
Trang 1Volume 2011, Article ID 368137, 15 pages
doi:10.1155/2011/368137
Research Article
Convergence of Iterative Sequences for Fixed Point and Variational Inclusion Problems
Li Yu and Ma Liang
School of Management, University of Shanghai for Science and Technology, Shanghai 200093, China
Correspondence should be addressed to Li Yu,brucemath@139.com
Received 14 November 2010; Accepted 8 February 2011
Academic Editor: Yeol J Cho
Copyrightq 2011 L Yu and M Liang This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited
An iterative process is considered for finding a common element in the fixed point set of a strict pseudocontraction and in the zero set of a nonlinear mapping which is the sum of a maximal monotone operator and an inverse strongly monotone mapping Strong convergence theorems of common elements are established in real Hilbert spaces
1 Introduction and Preliminaries
Throughout this paper, we always assume that H is a real Hilbert space with the inner
product·, · and the norm · .
Let C be a nonempty closed convex subset of H and S : C → C a nonlinear mapping.
In this paper, we use FS to denote the fixed point set of S Recall that the mapping S is said
to be nonexpansive if
S is said to be κ-strictly pseudocontractive if there exists a constant κ ∈ 0, 1 such that
Sx − Sy2≤x − y2 κx − Sx − y − Sy2
The class of strictly pseudocontractive mappings was introduced by Browder and Petryshyn
1 in 1967 It is easy to see that every nonexpansive mapping is a 0-strictly pseudocontractive mapping
Trang 2Let A : C → H be a mapping Recall that A is said to be monotone if
A is said to be inverse strongly monotone if there exists a constant α > 0 such that
Ax − Ay, x − y ≥ αAx − Ay2
For such a case, A is also said to be α-inverse strongly monotone.
Let M : H → 2H be a set-valued mapping The set DM defined by DM {x ∈ H :
Mx / ∅} is said to be the domain of M The set RM defined by RM x ∈H Mx is said to
be the range of M The set GM defined by GM {x, y ∈ H ×H : x ∈ DM, y ∈ RM}
is said to be the graph of M.
Recall that M is said to be monotone if
x − y, f − g> 0, ∀x, f
,
y, g
M is said to be maximal monotone if it is not properly contained in any other monotone
operator Equivalently, M is maximal monotone if RI rM H for all r > 0 For a maximal monotone operator M on H and r > 0, we may define the single-valued resolvent J r
I rM−1: H → DM It is known that J r is firmly nonexpansive and M−10 FJ r
Recall that the classical variational inequality problem is to find x ∈ C such that
Ax, y − x≥ 0, ∀y ∈ C. 1.6
Denote by VIC, A of the solution set of 1.6 It is known that x ∈ C is a solution to 1.6 if
and only if x is a fixed point of the mapping P C I − λA, where λ > 0 is a constant and I is the
identity mapping
Recently, many authors considered the convergence of iterative sequences for the variational inequality1.6 and fixed point problems of nonlinear mappings see, for example,
1 32
In 2005, Iiduka and Takahashi7 proved the following theorem
Theorem IT Let C be a closed convex subset of a real Hilbert space H Let A be an
α-inverse-strongly monotone mapping of C into H, and let S be a nonexpansive mapping of C into itself such that F S ∩ VIC, A / ∅ Suppose that x1 x ∈ C and {x n } is given by
x n1 α n x 1 − α n SP C x n − λ n Ax n , 1.7
for every n 1, 2, , where {α n } is a sequence in 0, 1 and {λ n } is a sequence in a, b If {α n } and {λ n } are chosen so that {λ n } ∈ a, b for some a, b with 0 < a < b < 2α,
lim
n→ ∞α n 0, ∞
n1
α n ∞, ∞
n1
|α n1− α n | < ∞, ∞
n1
|λ n1− λ n | < ∞, 1.8
then {x n } converges strongly to P F S∩VIC,A x.
Trang 3In 2007, Y Yao and J.-C Yao31 further obtained the following theorem.
Theorem YY Let C be a closed convex subset of a real Hilbert space H Let A be an
α-inverse-strongly monotone mapping of C into H, and let S be a nonexpansive mapping of C into itself such that F S ∩ Ω / ∅, where Ω denotes the set of solutions of a variational inequality for the
α-inverse-strongly monotone mapping Suppose that x1 u ∈ C and {x n }, {y n } are given by
x1 u ∈ C,
y n P C x n − λ n Ax n ,
x n1 α n u β n x n γ n SP C I − λ n A y n , n ≥ 1,
1.9
where {α n }, {β n }, and {γ n } are three sequences in 0, 1 and {λ n } is a sequence in 0, 2a If {α n }, {β n }, {γ n }, and {λ n } are chosen so that λ n ∈ a, b for some a, b with 0 < a < b < 2a and
a α n β n γ n 1, for all n ≥ 1,
b limn→ ∞α n 0, ∞n1α n ∞,
c 0 < lim inf n→ ∞β n≤ lim supn→ ∞β n < 1,
d limn→ ∞λ n1− λ n 0,
then {x n } converges strongly to P F S∩Ω u.
In this work, motivated by the above results, we consider the problem of finding a common element in the fixed point set of a strict pseudocontraction and in the zero set of a nonlinear mapping which is the sum of a maximal monotone operator and a inverse strongly monotone mapping Strong convergence theorems of common elements are established in real Hilbert spaces The results presented in this paper improve and extend the corresponding results announced by Iiduka and Takahashi7 and Y Yao and J.-C Yao 31
In order to prove our main results, we also need the following lemmas
Lemma 1.1 see 22 Let C be a nonempty closed convex subset of a Hilbert space H, A : C → H
a mapping, and M : H → 2H a maximal monotone mapping Then,
Lemma 1.2 see 1 Let C be a nonempty closed convex subset of a real Hilbert space H and S :
C → C a κ-strict pseudocontraction with a fixed point Define S : C → C by S a x ax 1 − aSx
for each x ∈ C If a ∈ κ, 1, then S a is nonexpansive with F S a FS.
Lemma 1.3 see 25 Let C be a nonempty closed convex subset of a Hilbert space H and S : C →
C a κ-strict pseudocontraction Then,
a S is 1 κ/1 − κ-Lipschitz,
b I − S is demi-closed, this is, if {x n } is a sequence in C with x n x and x n − Sx n → 0,
then x ∈ FS.
Trang 4Lemma 1.4 see 28 Let {x n } and {y n } be bounded sequences in a Hilbert space H, and let {β n}
be a sequence in 0, 1 with
0 < lim inf
n→ ∞ β n≤ lim sup
n→ ∞ β n < 1. 1.11
Suppose that x n1 1 − β n y n β n x n for all integers n ≥ 1 and
lim sup
n→ ∞
y n1− y n − x n1− x n≤ 0. 1.12
Then, lim n→ ∞y n − x n 0.
Lemma 1.5 see 29 Assume that {α n } is a sequence of nonnegative real numbers such that
α n1≤1− γ n
where {γ n } is a sequence in 0, 1 and {δ n } is a sequence such that
a ∞
n1γ n ∞,
b lim supn→ ∞δ n /γ n ≤ 0 or ∞n1|δ n | < ∞.
Then, lim n→ ∞α n 0.
Lemma 1.6 see 24 Let H be a Hilbert space and M a maximal monotone operator on H Then,
the following holds:
J r x − J s x2 ≤ r − x
r J r x − J s x, J r x − x, ∀s, t > 0, x ∈ H, 1.14
where J r I rM−1and J s I sM−1.
2 Main Results
Theorem 2.1 Let H be a real Hilbert space H and C a nonempty close and convex subset of H Let
M : H → 2H and W : H → 2H two maximal monotone operators such that D M ⊂ C and
D W ⊂ C, respectively Let S : C → C be a κ-strict pseudocontraction, A : C → H an α-inverse
strongly monotone mapping, and B : C → H a β-inverse strongly monotone mapping Assume that
F : FS ∩ A M−10 ∩ B W−10 / ∅ Let {x n } be a sequence generated in the following
manner:
x1∈ C,
y n J s n x n − s n Bx n ,
x n1 α n u β n x n γ n
δ n J r n
y n − r n Ay n
1 − δ n SJ r n
y n − r n Ay n
, ∀n ≥ 1,
2.1
where u ∈ C is a fixed element, J r n I r n M−1 and J s n I s n W−1, {r n } is a sequence
in 0, 2α, {s n } is a sequence in 0, 2β and {α n }, {β n }, {γ n }, and {δ n } are sequences in 0, 1.
Trang 5Assume that the following restrictions are satisfied:
a 0 < a ≤ r n ≤ b < 2α, lim n→ ∞r n − r n1 0,
b 0 < c ≤ s n ≤ d < 2β n , lim n→ ∞s n − s n1 0,
c 0 ≤ κ ≤ δ n < e < 1, lim n→ ∞δ n − δ n1 0,
d limn→ ∞α n 0, ∞n1α n ∞,
e 0 < lim inf n→ ∞β n≤ lim infn→ ∞β n < 1.
Then, the sequence {x n } converges strongly to q PFu.
Proof The proof is split into five steps.
Step 1 Show that {x n} is bounded
Note thatI − r n A and I − s n B are nonexpansive for each fixed n ≥ 1 Indeed, we see
from the restrictiona that
I − r n A x − I − r n A y2x − y2− 2r n
x − y, Ax − Ay r2
nAx − Ay2
≤x − y2− r n 2α − r nAx − Ay2
≤x − y2
, ∀x, y ∈ C.
2.2
This shows thatI − r n A is nonexpansive for each fixed n ≥ 1, so is I − s n B Put
In view of the restrictionc, we obtain fromLemma 1.2that S nis a nonexpansive mapping
with FS n FS for each fixed n ≥ 1 Fixing p ∈ F and since J r n and I − r n A are
nonexpansive, we see that
x n1− p ≤ α n u − p β n x n − p γ n S n J r n
y n − r n Ay n
− p
≤ α n u − p β n x n − p γ n J r n
y n − r n Ay n
− p
≤ α n u − p β n x n − p γ n y n − p
≤ α n u − p 1 − α n x n − p.
2.4
By mathematical inductions, we see that {x n } is bounded and so is {y n} This completes
Step 1
Trang 6Step 2 Show that x n1− x n → 0 as n → ∞.
Notice fromLemma 1.6that
y n1− y n ≤ x n1− s n1Bx n1 − x n − s n Bx n
J s n1x n − s n Bx n − J s n x n − s n Bx n
≤ x n1− x n |s n1− s n |Bx n
|s n1− s n|
s n1 J s n1x n − s n Bx n − x n − s n Bx n
≤ x n1− x n 2M1|s n1− s n |,
2.5
where M1is an appropriate constant such that
M1 max sup
n≥1{Bx n }, sup
n≥1
J s n1x n − s n Bx n − x n − s n Bx n
Put
z n J r n
y n − r n Ay n
In a similar way, we can obtain fromLemma 1.6that
z n1− z n ≤ y n1− r n1Ay n1
−y n − r n Ay n
J r n1
y n − r n Ay n
− J r n
y n − r n Ay n
≤y n1− −y n |r n1− r n|Ay n
|r n1− r n|
r n1 J r n1
y n − r n Ay n
−y n − r n Ay n
≤ y n1− y n 2M2|r n1− r n |,
2.8
where M2is an appropriate constant such that
M2 max sup
n≥1
Ay n, sup
n≥1
J r n1
y n − r n Ax n
−y n − r n Ay n
Substituting2.5 into 2.8 yields that
z n1− z n ≤ x n1− x n M3|s n1− s n | |r n1− r n |, 2.10
Trang 7where M3is an appropriate constant such that
It follows from2.10 that
S n1z n1− S n z n ≤ z n1− z n z n − Sz n |δ n − δ n1|
≤ x n1− x n M4|s n1− s n | |r n1− r n | |δ n − δ n1|, 2.12 where M4is an appropriate constant such that
M4 max sup
Put
l n x n1− β n x n
1− β n
Note that
l n1− l n α n1u γ n1S n1z n1
1− β n1 −α n u γ n S n z n
1− β n
α n1
1− β n1 − α n
1− β n
u γ n1
1− β n1S n1z n1− S n z n
n1
1− β n1− γ n
1− β n
S n z n
α n1
1− β n1 − α n
1− β n
u − S n z n γ n1
1− β n1S n1z n1− S n z n .
2.15
It follows from2.12 that
l n1− l n ≤
α n1
1− β n1− α n
1− β n
u − S n z n γ n1
1− β n1S n1z n1− S n z n
≤
α n1
1− β n1− α n
1− β n
u − S n z n x n1− x n
M4|s n1− s n | |r n1− r n | |δ n − δ n1|.
2.16
This in turn implies from the restrictionsa–e that
lim sup
n→ ∞ l n1− l n − x n1− x n ≤ 0. 2.17
Trang 8FromLemma 1.4, we obtain that
lim
Notice that
x n1− x n1− β n
It follows that
lim
This completesStep 2
Step 3 Show that x n − Sx n → 0 as n → ∞.
Since J r n and J s n are nonexpansive, we see that
z n − p2≤x n − p2− r n 2α − r nAy n − Ap2
y n − p2≤x n − p2− s n
2β − s nBx n − Bp2
It follows from2.21 that
x n1− p2≤ α nu − p2 β nx n − p2 γ nS n z n − p2
≤ α nu − p2 β nx n − p2 γ nz n − p2
≤ α nu − p2x n − p2− γ n r n 2α − r nAy n − Ap2
.
2.23
This in turn implies that
γ n r n 2α − r nAy n − Ap2≤ α nu − p2 x n − x n1x n − p x n1− p.
2.24
In view of2.20, we see from the restrictions a, d, and e that
lim
Trang 9It follows from2.22 that
x n1− p2≤ α nu − p2 β nx n − p2 γ nS n z n − p2
≤ α nu − p2 β nx n − p2 γ nJ r n y n − r n Ay n − p2
≤ α nu − p2 β nx n − p2 γ ny n − p2
≤ α nu − p2x n − p2− γ n s n
2β − s nBx n − Bp2
.
2.26
This in turn implies that
γ n s n
2β − s nBx n − Bp2≤ α nu − p2 x n − x n1x n − p x n1− p.
2.27
In view of2.20, we see from the restrictions a, d, and e that
lim
Since J r nis firmly nonexpansive, we obtain that
z n − p2J r
n y n − r n Ay n − J r n p − r n Ap2
≤z n − p,y n − r n Ay n
−p − r n Ap
1
2
z
n − p2y n − r n Ay n − p − r n Ap2
−z n − p
−y n − r n Ay n
−p − r n Ap2
≤ 1
2
z
n − p2y n − p2−z n − y n r n
Ay n − Ap2
1
2
z
n − p2y n − p2−z n − y n2− r2
nAy n − Ap2
−2r n
z n − y n , Ay n − Ap
≤ 1
2
z
n − p2y n − p2−z n − y n2 2r nz n − y nAy n − Ap
≤ 1
2
z
n − p2x n − p2−z n − y n2 2r n z n − y n Ay n − Ap.
2.29
This in turn implies that
z n − p2≤x n − p2−z n − y n2 2r n z n − y n Ay n − Ap. 2.30
Trang 10In a similar way, we can obtain that
y n − p2≤x n − p2−y n − x n2 2s n y n − x n Bx n − Bp. 2.31
In view of2.30, we see that
x n1− p2≤ α nu − p2 β nx n − p2 γ nS n z n − p2
≤ α nu − p2 β nx n − p2 γ nz n − p2
≤ α nu − p2x n − p2− γ nz n − y n2 2r n z n − y n Ay n − Ap.
2.32
It follows that
γ nz n − y n2≤ α nu − p2 x n − x n1x n − p x n1− p
In view of2.25, we obtain from the restrictions d and e that
lim
Notice from2.31, we see that
x n1− p2≤ α nu − p2 β nx n − p2 γ nS n z n − p2
≤ α nu − p2 β nx n − p2 γ nz n − p2
≤ α nu − p2 β nx n − p2 γ ny n − p2
≤ α nu − p2x n − p2− γ ny n − x n2 2s n y n − x n Bx n − Bp.
2.35
It follows that
γ ny n − x n2≤ α nu − p2 x n − x n1x n − p x n1− p
In view of2.28, we obtain from the restrictions d and e that
lim
Trang 11Combining2.34 with 2.37 yields that
lim
Note that
x n1− x n α n u − x n γ n S n z n − x n . 2.39
In view of2.20, we see from the restriction d that
lim
Note that
Sz n − x n S n z n − x n
1− δ n δ n x n − z n
1− δ n
From2.38 and 2.40, we get from the restriction c that
lim
Notice that
Sx n − x n ≤ Sx n − Sz n Sz n − x n . 2.43
In view of2.38 and 2.42, we see fromLemma 1.3that
lim
This completesStep 3
Step 4 Show that lim sup n→ ∞u − q, x n − q ≤ 0, where q PFu.
To show it, we may choose a subsequence{x n i } of {x n} such that
lim sup
n→ ∞
u − q, x n − q lim sup
i→ ∞
Since{x n i } is bounded, we can choose a subsequence {x n ij } of {x n i} converging weakly to
x We may, without loss of generality, assume that x n i x, where denotes the weak
convergence Next, we prove thatx ∈ F In view of 2.44, we can conclude fromLemma 1.3
thatx ∈ FS easily Notice that