Volume 2011, Article ID 208434, 16 pagesdoi:10.1155/2011/208434 Research Article A Method for a Solution of Equilibrium Problem and Fixed Point Problem of a Nonexpansive Semigroup in Hil
Trang 1Volume 2011, Article ID 208434, 16 pages
doi:10.1155/2011/208434
Research Article
A Method for a Solution of Equilibrium
Problem and Fixed Point Problem of
a Nonexpansive Semigroup in Hilbert’s Spaces
1 Vietnamese Academy of Science and Technology, Institute of Information Technology,
18 Hoang Quoc Viet Road, Cau Giay, Hanoi, Vietnam
2 Department of Mathematics, Vietnam Maritime University, Hai Phong 35000, Vietnam
Correspondence should be addressed to Nguyen Buong,nbuong@ioit.ac.vn
Received 3 October 2010; Accepted 13 January 2011
Academic Editor: Ljubomir B Ciric
Copyright q 2011 N Buong and N D Duong 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
We introduce a viscosity approximation method for finding a common element of the set of solutions for an equilibrium problem involving a bifunction defined on a closed, convex subset and the set of fixed points for a nonexpansive semigroup on another one in Hilbert’s spaces
1 Introduction
Let C be a nonempty, closed, and convex subset of a real Hilbert space H Denote the metric projection from x ∈ H onto C by P C x Let T : C → C be a nonexpansive mapping on C, that
is, T : C → C and Tx − Ty ≤ x − y, for all x, y ∈ C We use FT to denote the set of fixed points of T, that is, FT {x ∈ C : x Tx}.
Let {Ts : s > 0} be a nonexpansive semigroup on a closed convex subset C, that is,
1 for each s > 0, Ts is a nonexpansive mapping on C,
2 T0x x for all x ∈ C,
3 Ts1 s2 Ts1 ◦ Ts2 for all s1, s2> 0,
4 for each x ∈ C, the mapping T·x from 0, ∞ into C is continuous.
Denote by F
s>0 FTs We know 1,2 that F is a closed, convex subset in H and
F / ∅ if C is bounded.
Trang 2The equilibrium problem is for a bifunction Gu, v defined on C × C to find u∗ ∈ C
such that
Assume that the bifunction G satisfies the following set of standard properties:
A1 Gu, u 0, for all u ∈ C,
A2 Gu, v Gv, u ≤ 0 for all u, v ∈ C × C,
A3 for every u ∈ C, Gu, · : C → −∞, ∞ is weakly lower semicontinuous and
convex,
A4 limt → 0 G1 − tu tz, v ≤ Gu, v, for all u, z, v ∈ C × C × C.
Denote the set of solutions of 1.1 by EPG We also know 3 that EPG is a closed convex subset in H.
The problem studied in this paper is formulated as follows Let C1 and C2 be closed
convex subsets in H Let Gu, v be a bifunction satisfying conditions A1–A4 with C replaced by C1and let {Ts : s > 0} be a nonexpansive semigroup on C2 Find an element
where EPG and F denote the set of solutions of an equilibrium problem involving by a bifunction Gu, v on C1× C1and the fixed point set of a nonexpansive semigroup {Ts : s > 0} on a closed convex subset C2, respectively
In the case that C1 ≡ H, Gu, v 0, C2 C, and Ts T, a nonexpansive mapping on
C, for all s > 0, 1.2 is the fixed point problem of a nonexpansive mapping In 2000, Moudafi
4 proved the following strong convergence theorem
Theorem 1.1. Let C be a nonempty, closed, convex subset of a Hilbert space H and let T be a nonexpansive mapping on C such that FT / ∅ Let f be a contraction on C and let {x k } be a sequence
1 ε k fx k
1
1 ε k Tx k , k ≥ 1, 1.3
where {ε k } ∈ 0, 1 satisfies
lim
k → ∞ ε k 0,
∞
k1
k → ∞
ε k11 − 1
ε k
0. 1.4
Then, {x k } converges strongly to p ∈ FT, where p P FT fp.
Trang 3Such a method for approximation of fixed points is called the viscosity approximation method It has been developed by Chen and Song 5 to find p ∈ F, the set of fixed points for
a semigroup {Ts : s > 0} on C They proposed the following algorithm: x1∈ C and
x k1 μ k fx k
1 − μ k 1
s k
s k
0
where f : C → C, is a contraction, {μ k } ⊂ 0, 1 and {s k} are sequences of positive real
numbers satisfying the conditions: μ k → 0,∞
k1 μ k ∞, and s k → ∞ as k → ∞.
Recently, Yao and Noor 6 proposed a new viscosity approximation method
where {μ k }, {β k }, and {γ k } are in 0, 1, s k → ∞, for finding p ∈ F, when {Ts : s > 0}
satisfies the uniformly asymptotically regularity condition
lim
s → ∞sup
x∈ C
TtTsx − Tsx 0, 1.7
uniformly in t, and C is any bounded subset of C Further, Plubtieng and Pupaeng in 7 studied the following algorithm:
x k1 μ k fx k β k x k
1 − β k − μ ks k
0
where {μ k } and {β k } are in 0, 1 satisfying the following conditions: μ k β k < 1, lim k → ∞ μ k limk → ∞ β k 0,
k≥1 μ k ∞, and {s k} is a positive divergent real sequence
There were some methods proposed to solve equilibrium problem 1.1; see for instance 8 12 In particular, Combettes and Histoaga 3 proposed several methods for solving the equilibrium problem
In 2007, S Takahashi and W Takahashi 13 combinated the Moudafi’s method with the Combettes and Histoaga’s result in 3 to find an element p ∈ EPG ∩ FT They proved
the following strong convergence theorem
Theorem 1.2. Let C be a nonempty, closed, convex subset of a Hilbert space H, let T be a nonexpansive mapping on C and let G be a bifunction from C × C to −∞, ∞ satisfying (A1)– (A4) such that EPG ∩ FT / ∅ Let f be a contraction on C and let {x k } and {u k } be sequences
G
u k , y
1
r k u k − x k , y − u k
≥ 0, ∀y ∈ C,
x k1 μ k fx k
1 − μ k
1.9
Trang 4where {μ k } ∈ 0, 1 and {r k } ⊂ 0, ∞ satisfy
lim
k → ∞ μ k 0,
∞
k1
μ k ∞, lim inf
k → ∞ r k > 0,
∞
k1
μ k1 − μ k
< ∞, ∞
k1
|r k1 − r k | < ∞.
1.10
Then, {x k } and {u k } converge strongly to p ∈ EPG ∩ FT, where p P EPG∩FT fp.
Very recently, Ceng and Wong in 14 combined algorithm 1.6 with the result in 3
to propose the following procudure:
G
u k , y
1
r k u k − x k , y − u k
≥ 0, ∀y ∈ C,
x k1 μ k fx k β k x k γ k Ts ku k , k ≥ 1,
1.11
for finding an element p ∈ EPG ∩ F in the case that C1 C2 C under the uniformly asymptotic regularity condition on the nonexpansive semigroup {Ts : s > 0} on C.
In this paper, motivated by the above results, to solve 1.2, we introduce the following algorithm:
x1∈ H, any element,
u k ∈ C1: G
u k , y
1
r k u k − x k , y − u k
≥ 0, ∀y ∈ C1,
x k1 μ k fu k β k x k γ k T k P C2u k , k ≥ 1,
1.12
where f is a contraction on H, that is, f : H → H and fx − fy ≤ ax − y, for all
x, y ∈ H, 0 ≤ a < 1,
T k x 1
s k
s k
0
for all x ∈ C2, {μ k }, {β k }, and {γ k } be the sequences in 0,1, and {r k }, {s k} are the sequences
in 0, ∞ satisfy the following conditions:
i μ k β k γ k 1,
ii limk → ∞ μ k 0,
k≥1 μ k ∞,
iii 0 < lim inf k → ∞ β k≤ lim supk → ∞ β k < 1,
iv limk → ∞ s k ∞ with bounded supk≥1 |s k − s k1|,
v lim infk → ∞ r k > 0 and lim k → ∞ |r k − r k1| 0
The strong convergence of 1.12-1.13 and its corollaries are showed in the next section
Trang 52 Main Results
We formulate the following facts needed in the proof of our results
Lemma 2.1. Let H be a real Hilbert space H There holds the following identity:
x y 2≤ x2 2 y, x y
, ∀x, y ∈ H. 2.1
Lemma 2.2 see 15 Let C be a nonempty, closed, convex subset of a real Hilbert space H For any
x ∈ H, there exists a unique z ∈ C such that z − x ≤ y − x, for all y ∈ C, and z ∈ P C x if and only if z − x, y − z ≥ 0 for all y ∈ C.
Lemma 2.3 see 16 Let {a k } be a sequence of nonnegative real numbers satisfying the following
condition:
a k1 ≤ 1 − b ka k b k c k , 2.2
k1 b k ∞, and
lim supk → ∞ c k ≤ 0 Then, lim k → ∞ a k 0.
Lemma 2.4 see 9 Let C be a nonempty, closed, convex subset of H and G be a bifunction of C ×C
into −∞, ∞ satisfying the conditions (A1)–(A4) Let r > 0 and x ∈ H Then, there exists z ∈ C such that
r z − x, v − z ≥ 0, ∀v ∈ C. 2.3
Lemma 2.5 see 9 Assume that G : C × C → −∞, ∞ satisfies the conditions (A1)–(A4) For
r > 0 and x ∈ H, define a mapping T r : H → C as follows:
T rx z ∈ C : Gz, v 1
r z − x, v − z ≥ 0 , ∀v ∈ C
. 2.4
Then, the following statements hold:
i T r is single-valued,
ii T r is firmly nonexpansive, that is, for any x, y ∈ H,
T r x − T r y 2
≤ T rx − T r
y
, x − y
iii FT r EPG,
iv EPG is closed and convex.
Trang 6Lemma 2.6 see 17 Let C be a nonempty bounded closed convex subset in a real Hilbert space H
and let {Ts : s > 0} be a nonexpansive semigroup on C Then, for any h > 0,
lim sup
t → ∞
sup
y∈C
Th
1
t
t
0
Tsyds
−1
t
t
0
Tsyds
0. 2.6
Lemma 2.7 Demiclosedness Principle 18 If C is a closed convex subset of H, T is a
nonexpansive mapping on C, {x k } is a sequence in C such that x k ⇀ x ∈ C and x k − Tx k → 0, then
x − Tx 0.
Lemma 2.8 see 19 Let {x k } and {z k } be bounded sequences in a Banach space E and {β k } be a
sequence in 0, 1 with 0 < lim inf k → ∞ β k≤ lim supk → ∞ β k < 1 Suppose x k1 β k x k 1 − β k z k
for all k ≥ 1 and lim sup k → ∞ z k1 − z k − x k1 − x k ≤ 0 Then, lim k → ∞ z k − x k 0.
Now, we are in a position to prove the following result
Theorem 2.9. Let C1 and C2 be two nonempty, closed, convex subsets in a real Hilbert space H Let G
be a bifunction from C1 × C1to −∞, ∞ satisfying conditions (A1)–(A4) with C replaced by C1, let
{Ts : s > 0} be a nonexpansive semigroup on C2such that EPG ∩ F / ∅ and let f be a contraction
of H into itself Then, {x k } and {u k } generated by 1.12-1.13 converge strongly to p ∈ EPG ∩ F,
where p PEPG∩Ffp.
Proof Let Q P EPG∩F Then, Qf is a contraction of H into itself In fact, from fx − fy ≤
H, it implies that
Qfx − Qf
y
≤ fx − f
y
≤ ax − y. 2.7
Hence, Qf is a contraction of H into itself Since H is complete, there exists a unique element
ByLemma 2.4, {u k } and {x k } are well defined For each u ∈ EPG ∩ F, by putting
u k T r k x kand using ii and iii inLemma 2.5, we have that
u k − u T r k x k − T r k u ≤ x k − u. 2.8
Put M u max{x1−u, 1/1−afu−u} Clearly, x1−u ≤ M u Suppose that x k −u ≤
M u Then, we have, from the nonexpansive property of T k P C2, condition i and 2.8, that
x k1 − u μ k
fu k −u
β kx k − u γ kT k P C2u k − u
≤ μ k fu k −u β k x k − u γ k T k P C2u k − T k P C2u
≤ μ k
fu k −fu fu − u
β k x k − u γ k u k − u
≤ μ k
au k − u fu − u
1 − μ k
x k − u
Trang 71 − μ k1 − a
x k − u μ k1 − a 1
1 − a fu − u
≤
1 − μ k1 − a
M u μ k1 − aMu M u
2.9
So, x k − u ≤ M u for all k ≥ 1 and hence {x k } is bounded Therefore, {u k }, {T k P C2u k}, and
{fu k} are also bounded
Next, we show that x k1 − x k → 0 as k → ∞ For this purpose, we define a sequence {x k} by
x k1 β k x k
1 − β k
Then, we observe that
z k1 − z k μ k1 fu k1 γ k1 T k1 P C2u k1
1 − β k1
−μ k fu k γ k T k P C2u k
1 − β k
μ k1
1 − β k1 fu k1 − μ k
1 − β k fu k
γ k1
1 − β k1
T k1 P C2u k1 − T k1 P C2u k
γ k1
1 − β k1
T k1 P C2u k− γ k
1 − β k
T k P C2u k
μ k1
1 − β k1 fu k1 −
μ k
1 − β k fu k
γ k1
1 − β k1T k1 P C2u k1 − T k1 P C2u k T k1 P C2u k
− μ k1
1 − β k1 T k1 P C2u k − T k P C2u k μ k
1 − β k T k P C2u k ,
2.11
and, hence,
z k1 − z k − x k1 − x k ≤ μ k1
1 − β k1
fu k1 T k1 P C2u k
μ k
1 − β k
fu k T k P C2u k γ k1
1 − β k1 u k1 − u k
T k1 P C u k − T k P C u k − x k1 − x k .
2.12
Trang 8Now, we estimate the value u k1 − u k by using u k T r k x k and u k1 T r k1 x k1 We have from 2.4 that
G
u k , y
1
r k u k − x k , y − u k
≥ 0, ∀y ∈ C1, 2.13
G
u k1 , y
1
r k1
u k1 − x k1 , y − u k1
≥ 0, ∀y ∈ C1. 2.14
Putting y u k1in 2.13 and y u k in 2.14, adding the one to the other obtained result and using A2, we obtain that
u
k − x k
r k
−u k1 − x k1
r k1 , u k1 − u k
and, hence,
u k − u k1 u k1 − x k− r k
r k1
u k1 − x k1, u k1 − u k
≥ 0. 2.16
Without loss of generality, let us assume that there exists a real number b such that r k > b > 0
for all k ≥ 1 Then, we have
u k1 − u k2≤
x k1 − x k
1 − r k
r k1
u k1 − x k1, u k1 − u k
≤
x k1 − x k
1 − r r k1 k
u k1 − x k1
u k1 − u k
2.17
and, hence,
u k1 − u k ≤ x k1 − x k 1
r k1
|r k1 − r k |u k1 − x k1
≤ x k1 − x k 2M u
b |r k1 − r k |.
2.18
On the other hand,
T k P C2u k − T k1 P C2u k
1
s k
s k
0
TsP C2u k ds − 1
s k1
s k1
0
TsP C2u k ds
1
s k
s k
0
TsP C2u k − TsP C2uds − 1
s k1
s k1
0
TsP C2u k − TsP C2uds
Trang 9
1
s k −
1
s k1
s k1
0
TsP C2u k − TsP C2uds 1
s k
s k
s k1
TsP C2u k − TsP C2uds
≤1
s k −
1
s k1
s k1 M u|s k − s k1|
≤ supk≥1 |s k1 − s k|
s k
2M u
2.19
So, we get from 2.10, 2.12, 2.18, 2.19, and the nonexpansive property of T k1 P C2that
z k1 − z k − x k1 − x k ≤ μ k1
1 − β k1
fu k1 T k1 P C2u k
μ k
1 − β k
fu k T k P C2u k
γ k1 2M u
1 − β k1
b |r k1 − r k|
supk≥1 |s k1 − s k|
2.20
So,
lim sup
k → ∞
z k1 − z k − x k1 − x k ≤ 0, 2.21
and byLemma 2.8, we have
lim
k → ∞ z k − x k 0. 2.22
Consequently, it follows from 2.10 and condition iii that
lim
k → ∞ x k1 − x k lim
k → ∞
1 − β k
z k − x k 0. 2.23
By 2.18, 2.23, and
lim
k → ∞ |r k − r k1 | 0, 2.24
we also obtain
lim
k → ∞ u k1 − u k 0. 2.25
Trang 10We have, for every u ∈ EPG ∩ F, from iii inLemma 2.5, that
u k − u2 T r k x k − T r k u2
≤ T r k x k − T r k u, x k − u
u k − u, x k − u
1 2
u k − u2 x k − u2− u k − x k2
2.26
and, hence,
u k − u2≤ x k − u2− u k − x k2. 2.27
Therefore, from the convexity of · 2and condition i, we have
x k1 − u2≤ μ k
fu k −u 2
β k x k − u2 γ k T k P C2u k − u2
≤ μ k fu k − u 2
β k x k − u2 γ k u k − u2
≤ μ k
fu k − u 2
β k x k − u2 γ k
x k − u2− u k − x k2
≤ μ k
fu k − u 2
1 − μ k
x k − u2− γ k u k − x k2
≤ μ k
fu k −u
x k − u2− γ k u k − x k2
2.28
and, hence,
γ k u k − x k2≤ μ k fu k −u x k − u2− x k1 − u2
≤ μ k fu k −u 2M u x k − x k1 .
2.29
Without loss of generality, we assume that 0 < β∗ ≤ β k ≤ β < 1 for all k ≥ 1 Then, for
sufficiently large k,
0 ≤
1 − β − μ k
So, we have
lim
k → ∞ u k − x k 0. 2.31
Trang 11Further, since x k1 μ k fu k β k x k γ k T k P C2u k, by condition i, 2.19 and
x k1 − T k1 P C2u k1 μ k fu k β k x k γ k T k P C2u k
−
μ k β k γ k
T k P C2u k T k P C2u k − T k1 P C2u k1
μ k
fu k −T k P C2u k
β kx k − T k P C2u k
T k P C2u k − T k1 P C2u k1 ,
2.32
we obtain that
x k1 − T k1 P C2u k1 ≤ μ k fu k −T k P C2u k β k x k − T k P C2u k
u k1 − u k supk≥1 |s k1 − s k|
2.33
Then, from 2.25, 2.33 and the conditions on {μ k } and {s k}, it implies that
1 − β lim sup
k → ∞
x k − T k P C2u k ≤ 0, 2.34
and so
lim sup
k → ∞
x k − T k P C2u k ≤ 0. 2.35
Since
T k P C2u k − u k ≤ T k P C2u k − x k x k − u k , 2.36
we obtain from 2.31 that
lim
k → ∞ T k P C2u k − u k 0. 2.37
Next, we show that
lim sup
k → ∞
f
p
− p, x k − p
We choose a subsequence {u k i } of the sequence {u k} such that
lim sup
k → ∞
f
p
− p, x k − p
lim
i → ∞ f
p
− p, x k i − p
. 2.39
As {u k } is bounded, there exists a subsequence {u k j } of the sequence {u k i} which converges
weakly to z From 2.37, we also have that {T k j P C2u k j } converges weakly to z Since {u k } ⊂ C1
and {T k P C u k } ⊂ C2and C1, C2are two closed convex subsets in H, we have that z ∈ C1∩ C2
... 2.14Putting y u k1in 2.13 and y u k in 2.14, adding the one to the other obtained result and using A2, we obtain that
u... F,
where p PEPG∩Ffp.
Proof Let Q P EPG∩F Then, Qf is a contraction of H into itself In fact, from fx − fy ≤
H, it implies that... Qf is a contraction of H into itself Since H is complete, there exists a unique element
ByLemma 2.4, {u k } and {x k } are well defined For each u ∈ EPG