Journal of Applied MathematicsVolume 2013, Article ID 647524, 6 pages http://dx.doi.org/10.1155/2013/647524 Research Article An Iterative Method with Norm Convergence for a Class of Gene
Trang 1Journal of Applied Mathematics
Volume 2013, Article ID 647524, 6 pages
http://dx.doi.org/10.1155/2013/647524
Research Article
An Iterative Method with Norm Convergence for a Class of
Generalized Equilibrium Problems
Haixia Zhang1and Fenghui Wang2
1 Department of Mathematics, Henan Normal University, Xinxiang 453007, China
2 Department of Mathematics, Luoyang Normal University, Luoyang 471022, China
Correspondence should be addressed to Fenghui Wang; wfenghui@163.com
Received 12 January 2013; Accepted 1 July 2013
Academic Editor: Filomena Cianciaruso
Copyright © 2013 H Zhang and F Wang 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
Recently, Takahashi and Takahashi proposed an iterative algorithm for solving a problem for finding common solutions of generalized equilibrium problems governed by inverse strongly monotone mappings and of fixed point problems for nonexpansive mappings In this paper, we provide a result that allows for the removal of one condition ensuring the strong convergence of the algorithm
1 Introduction
LetH be a real Hilbert space and 𝐶 a nonempty closed convex
subset A generalized equilibrium problem is formulated as a
problem of finding a point𝑥∗∈ 𝐶 with the property
𝐹 (𝑥∗, 𝑦) + ⟨𝐴𝑥∗, 𝑦 − 𝑥∗⟩ ≥ 0, ∀𝑦 ∈ 𝐶, (1)
where𝐹 : 𝐶 × 𝐶 → R is a bifunction and 𝐴 : 𝐶 → H is
a nonlinear mapping In particular, if𝐴 is the zero mapping,
then problem (1) is reduced to an equilibrium problem; find
a point𝑥∗ ∈ 𝐶 with the property
We will denote by EP(𝐹; 𝐴) and EP(𝐹) the solution set of
problem (1) and problem (2), respectively A fixed point
problem (FPP) is to find a point𝑥∗with the property
where𝑆 : 𝐶 → 𝐶 is a nonlinear mapping The set of fixed
points of𝑆 is denoted as Fix(𝑆)
The problem under consideration in this paper is to find a
common solution of problem (1) and of FPP (3) Namely, we
seek a point𝑥∗such that
We consider problem (4) in the case whenever 𝐴 is a ]-inverse strongly monotone mapping and𝑆 is a nonexpansive mapping To solve problem (4), Takahashi and Takahashi [1
introduced an algorithm which generates a sequence(𝑥𝑛) by the iterative procedure
𝐹 (𝑧𝑛, 𝑦) + ⟨𝐴𝑥𝑛, 𝑦 − 𝑧𝑛⟩ + 1
𝜆𝑛⟨𝑦 − 𝑧𝑛, 𝑧𝑛− 𝑥𝑛⟩ ≥ 0,
∀𝑦 ∈ 𝐶,
𝑥𝑛+1= 𝛽𝑛𝑥𝑛+ (1 − 𝛽𝑛) 𝑆 [𝛼𝑛𝑢 + (1 − 𝛼𝑛) 𝑧𝑛] ,
(5)
where(𝛼𝑛) ⊆ [0, 1], (𝛽𝑛) ⊆ [0, 1], and (𝜆𝑛) ⊆ [0, 2]] are chosen so that
0 < 𝑎 ≤ 𝜆𝑛≤ 𝑏 < 2], 0 < 𝑐 ≤ 𝛽𝑛≤ 𝑑 < 1, lim
𝑛 → ∞𝛼𝑛= 0, ∑∞
𝑛=0
𝛼𝑛= ∞,
𝜆𝑛− 𝜆𝑛+1 → 0
(6)
Under these conditions, they proved that the sequence(𝑥𝑛) generated by (5) can be strongly convergent to a solution of problem (4)
It is the aim of this paper to continue the study of algorithm (5) We will show that problem (4) is in fact
Trang 2a special fixed point problem for a nonexpansive mapping
(a composition of a nonexpansive mapping and an averaged
mapping) Our approach mainly uses the properties of
aver-aged mappings, which is different from the existing methods
invented by Takahashi and Takahashi Moreover, we shall
prove that condition|𝜆𝑛− 𝜆𝑛+1| → 0 sufficient to guarantee
the convergence of algorithm (5) is superfluous
2 Preliminaries and Notations
Notation 1. → strong convergence, ⇀ weak convergence and
𝜔𝑤(𝑥𝑛) the set of the weak cluster points of (𝑥𝑛)
Denote by𝑃𝐶the projection fromH onto 𝐶; namely, for
𝑥 ∈ H, 𝑃𝐶𝑥 is the unique point in 𝐶 with the property
𝑥 − 𝑃𝐶𝑥 = min𝑦∈𝐶𝑥 − 𝑦 (7)
It is well known that𝑃𝐶𝑥 is characterized by the inequality
𝑃𝐶𝑥 ∈ 𝐶,
⟨𝑥 − 𝑃𝐶𝑥, 𝑧 − 𝑃𝐶𝑥⟩ ≤ 0, ∀𝑧 ∈ 𝐶 (8)
We will use the following notions on nonlinear mappings
𝑇 : 𝐶 → H
(i)𝑇 is nonexpansive if
𝑇𝑥 − 𝑇𝑦 ≤ 𝑥 − 𝑦, ∀𝑥,𝑦 ∈ 𝐶 (9)
(ii)𝑇 is firmly nonexpansive if
⟨𝑇𝑥 − 𝑇𝑦, 𝑥 − 𝑦⟩ ≥ 𝑇𝑥 − 𝑇𝑦2, ∀𝑥, 𝑦 ∈ 𝐶 (10)
(iii)𝑇 is 𝛼-averaged if there exist a constant 𝛼 ∈ (0, 1) and
a nonexpansive mapping𝑆 such that 𝑇 = (1−𝛼)𝐼+𝛼𝑆,
where𝐼 is the identity mapping on H
(iv)𝑇 is ]-inverse strongly monotone if there is a constant
] > 0 such that
⟨𝑇𝑥 − 𝑇𝑦, 𝑥 − 𝑦⟩ ≥ ]𝑇𝑥 − 𝑇𝑦2, ∀𝑥, 𝑦 ∈ 𝐶 (11)
The next lemma is referred to as the demiclosedness
principle for nonexpansive mappings (see [2])
Lemma 1 Let 𝐶 be a nonempty closed convex subset of H and
𝑇 : 𝐶 → H a nonexpansive mapping with Fix(𝑇) ̸= 0 If (𝑥𝑛)
is a sequence in 𝐶 such that 𝑥𝑛⇀ 𝑥 and (𝐼 − 𝑇)𝑥𝑛 → 0, then
(𝐼 − 𝑇)𝑥 = 0; that is, 𝑥 ∈ Fix(𝑇).
Averaged mappings will play important role in our
con-vergence analysis We therefore collect some useful properties
of averaged mappings (see, e.g., [3–5])
Lemma 2 The following assertions hold.
(i)𝑇 is firmly nonexpansive if and only if 𝑇 is
1/2-averaged.
(ii) If 𝑇𝑖is]𝑖-averaged, 𝑖 = 1, 2, then 𝑇1𝑇2is(]1+]2−]1]2
)-averaged.
(iii) If 𝑇 : 𝐶 → H is ]-averaged, then for any 𝑧 ∈ Fix(𝑇) and for all 𝑥 ∈ 𝐶,
‖𝑇𝑥 − 𝑧‖2≤ ‖𝑥 − 𝑧‖2−1 − ]] ‖𝑇𝑥 − 𝑥‖2 (12) From now on, we assume that 𝐹 : 𝐶 × 𝐶 → R is a bifunction so that
(A1)𝐹(𝑥, 𝑥) = 0, for all 𝑥 ∈ 𝐶;
(A2)𝐹 is monotone; that is, 𝐹(𝑥, 𝑦) + 𝐹(𝑦, 𝑥) ≤
0, for all 𝑥, 𝑦 ∈ 𝐶;
(A3) lim𝑡↓0𝐹(𝑡𝑧 + (1 − 𝑡)𝑥, 𝑦) ≤ 𝐹(𝑥, 𝑦), for all 𝑥, 𝑦 ∈ 𝐶; (A4) for each 𝑥 ∈ 𝐶, 𝑦 → 𝐹(𝑥, 𝑦) is convex and lower semicontinuous
Under these assumptions, the following results hold (see [6,
7])
Lemma 3 Let 𝐹 : 𝐶×𝐶 → R satisfy (A1)–(A4) Then for any
𝜆 > 0 and 𝑥 ∈ H, there exists 𝑧 ∈ 𝐶 so that
𝐹 (𝑧, 𝑦) +1
𝜆⟨𝑦 − 𝑧, 𝑧 − 𝑥⟩ ≥ 0, ∀𝑦 ∈ 𝐶. (13)
Moreover if𝑆𝜆𝑥 = {𝑧 ∈ 𝐶 : 𝐹(𝑧, 𝑦) + 1/𝜆⟨𝑦 − 𝑧, 𝑧 − 𝑥⟩ ≥ 0,
for all 𝑦 ∈ 𝐶}, then
(i)𝑆𝜆is single valued and Fix(𝑆𝜆) = EP(𝐹);
(ii)𝑆𝜆is firmly nonexpansive;
(iii) EP(𝐹) is closed and convex
We end this section by a useful lemma (see Xu [8])
Lemma 4 Let (𝑎𝑛) be a nonnegative real sequence satisfying
𝑎𝑛+1≤ (1 − 𝛼𝑛) 𝑎𝑛+ 𝛼𝑛𝑏𝑛, (14)
where(𝛼𝑛) ⊂ (0, 1) and (𝑏𝑛) are real sequences Then 𝑎𝑛 → 0
provided that
(i)∑𝑛𝛼𝑛= ∞, lim𝑛𝛼𝑛 = 0;
(ii) lim sup𝑛𝑏𝑛≤ 0 or ∑ 𝛼𝑛|𝑏𝑛| < ∞.
3 Algorithm and Its Convergence
We begin with the following lemma
Lemma 5 Assume that 𝐴 : 𝐶 → H is ]-inverse strongly
monotone mapping for some ] > 0 Given a real number 𝜆 such that 0 < 𝜆 < 2], set 𝑇𝜆 = 𝑆𝜆(𝐼 − 𝜆𝐴) with 𝑆𝜆defined as
in Lemma 3 Then the following assertions hold:
(a)𝑇𝜆is single valued and Fix(𝑇𝜆) = EP(𝐹; 𝐴);
(b)𝑇𝜆is (2] + 𝜆)/4]-averaged;
(c) given 𝑧 ∈ EP(𝐹; 𝐴), it follows that
𝑇𝜆𝑥 − 𝑧2≤ ‖𝑥 − 𝑧‖2−2] − 𝜆2] + 𝜆𝑇𝜆𝑥 − 𝑥2; (15)
Trang 3(d) if0 < 𝜆 ≤ 𝜆< 2], then for all 𝑥 ∈ 𝐶
𝑇𝜆𝑥 − 𝑥 ≤ 2𝑇𝜆 𝑥 − 𝑥 (16)
Proof (a) It is readily seen that𝑇𝜆is single valued because𝑆𝜆
is single valued The equality follows from the definition of
𝑆𝜆
(b) It follows that
(𝐼 − 2]𝐴)𝑥 − (𝐼 − 2]𝐴)𝑦2
= (𝑥 − 𝑦) − 2](𝐴𝑥 − 𝐴𝑦)2
= 𝑥 − 𝑦2+ 4]2𝐴𝑥 − 𝐴𝑦2
− 4] ⟨𝑥 − 𝑦, 𝐴𝑥 − 𝐴𝑦⟩
(17)
Since𝐴 is ]-inverse strongly monotone, 𝐼−2]𝐴 is
nonexpan-sive Observe that
𝐼 − 𝜆𝐴 = (1 − 𝜆
2]) 𝐼 +
𝜆 2](𝐼 − 2]𝐴) , (18) which implies that𝐼−𝜆𝐴 is 𝜆/2]-averaged Consequently (b)
follows from part (ii) ofLemma 2and (c) follows from part
(iii) ofLemma 2
(d) Let𝑧1= 𝑇𝜆𝑥 and 𝑧2= 𝑇𝜆𝑥 By definition of 𝑆𝜆,
𝐹 (𝑧1, 𝑦) + ⟨𝐴𝑥, 𝑦 − 𝑧1⟩ + 1
𝜆⟨𝑦 − 𝑧1, 𝑧1− 𝑥⟩ ≥ 0,
∀𝑦 ∈ 𝐶
(19)
Letting𝑦 = 𝑧2in (19) yields
𝐹 (𝑧1, 𝑧2) + ⟨𝐴𝑥, 𝑧2− 𝑧1⟩ + 1𝜆⟨𝑧2− 𝑧1, 𝑧1− 𝑥⟩ ≥ 0 (20)
Similarly,
𝐹 (𝑧2, 𝑧1) + ⟨𝐴𝑥, 𝑧1− 𝑧2⟩ +𝜆1⟨𝑧1− 𝑧2, 𝑧2− 𝑥⟩ ≥ 0 (21)
Adding up these inequalities and using the monotonicity of
𝐹,
1
𝜆⟨𝑧2− 𝑧1, 𝑧1− 𝑥⟩ +
1
𝜆⟨𝑧1− 𝑧2, 𝑧2− 𝑥⟩ ≥ 0, (22)
or equivalently,
𝑧2− 𝑧12≤ (1 −𝜆𝜆) ⟨𝑧2− 𝑧1, 𝑧2− 𝑥⟩ (23)
Hence,‖𝑧2− 𝑧1‖ ≤ ‖𝑧2− 𝑥‖ By the triangle inequality,
𝑧1− 𝑥 ≤𝑧1− 𝑧2 + 𝑧2− 𝑥 ≤ 2𝑧2− 𝑥 , (24)
which is the result as desired
For every𝑛 ≥ 0, if we define 𝑇𝑛 = 𝑆𝜆𝑛(𝐼 − 𝜆𝑛𝐴), where 𝑆𝜆𝑛
is defined as inLemma 3, then we can rewrite algorithm (5)
as
𝑦𝑛 = 𝛼𝑛𝑢 + (1 − 𝛼𝑛) 𝑇𝑛𝑥𝑛,
𝑥𝑛+1= 𝛽𝑛𝑥𝑛+ (1 − 𝛽𝑛) 𝑆𝑦𝑛 (25)
Theorem 6 Let 𝐹 : 𝐶 × 𝐶 → R be a bifunction satisfying
(A1)–(A4), 𝐴 : 𝐶 → H a ]-inverse strongly monotone mapping for some ] > 0, and 𝑆 : 𝐶 → 𝐶 a nonexpansive mapping so that the solution set Ω := Fix(𝑆) ∩ EP(𝐹; 𝐴) is nonempty If the following conditions hold:
0 < 𝑎 ≤ 𝜆𝑛≤ 𝑏 < 2], 0 < 𝑐 ≤ 𝛽𝑛≤ 𝑑 < 1, lim
𝑛 → ∞𝛼𝑛= 0, ∑∞
𝑛=0
then the sequence(𝑥𝑛) generated by (25) converges strongly to
𝑥∗ = 𝑃Ω𝑢.
Before proving the theorem, we need some lemmas
Lemma 7 Let the conditions in Theorem 6 be satisfied If(𝑥𝑛)
and(𝑦𝑛) are the sequences generated by (25), then both(𝑥𝑛)
and(𝑦𝑛) are bounded.
Proof Let𝑧 ∈ Ω be fixed We have
𝑥𝑛+1− 𝑧 ≤(1 − 𝛽𝑛) (𝑦𝑛− 𝑧) + 𝛽𝑛(𝑥𝑛− 𝑧)
≤ (1 − 𝛽𝑛) 𝑦𝑛− 𝑧 + 𝛽𝑛𝑥𝑛− 𝑧 ; (27)
on the other hand,
𝑦𝑛− 𝑧 =𝛼𝑛(𝑢 − 𝑧) + (1 − 𝛼𝑛) (𝑇𝑛𝑥𝑛− 𝑧)
≤ (1 − 𝛼𝑛) 𝑥𝑛− 𝑧 + 𝛼𝑛‖𝑢 − 𝑧‖ (28) Altogether
𝑥𝑛+1− 𝑧 ≤ [1 − 𝛼𝑛(1 − 𝛽𝑛)] 𝑥𝑛− 𝑧
+ 𝛼𝑛(1 − 𝛽𝑛) ‖𝑢 − 𝑧‖ (29)
By induction,(𝑥𝑛) is bounded and so is (𝑦𝑛)
Lemma 8 Let the conditions in Theorem 6 be satisfied If‖𝑥𝑛−
𝑇𝑛𝑥𝑛‖ → 0 and ‖𝑥𝑛− 𝑆𝑦𝑛‖ → 0, then ‖𝑥𝑛− 𝑦𝑛‖ → 0 and
𝜔𝑤(𝑥𝑛) ⊆ Ω.
Proof Let𝑇𝑎 = 𝑆𝑎(𝐼 − 𝑎𝐴) By part (d) ofLemma 5,
𝑥𝑛− 𝑇𝑎𝑥𝑛 ≤ 2𝑥𝑛− 𝑇𝑛𝑥𝑛 → 0 (30) Since𝑇𝑎is nonexpansive, applying the demiclosedness prin-ciple yields
𝜔𝑤(𝑥𝑛) ⊆ Fix (𝑇𝑎) = EP (𝐹; 𝐴) (31)
On the other hand, we see that
𝑥𝑛− 𝑦𝑛 = 𝛼𝑛(𝑢 − 𝑥𝑛) + (1 − 𝛼𝑛) (𝑇𝑛𝑥𝑛− 𝑥𝑛)
≤ 𝛼𝑛𝑢 − 𝑥𝑛 + 𝑇𝑛𝑥𝑛− 𝑥𝑛 → 0, (32) which implies that
𝑥𝑛− 𝑆𝑥𝑛 ≤ 𝑥𝑛− 𝑆𝑦𝑛 + 𝑆𝑦𝑛− 𝑆𝑥𝑛
≤ 𝑥𝑛− 𝑆𝑦𝑛 + 𝑦𝑛− 𝑥𝑛 → 0 (33) Using again the demiclosedness principle gets the desired result
Trang 4Proof of Theorem 6 Let𝑥∗ = 𝑃Ω𝑢 UsingLemma 5(c), we
have
𝑇𝑛𝑥𝑛− 𝑥∗2
≤ 𝑥𝑛− 𝑥∗2−2] − 𝜆𝑛
2] + 𝜆𝑛𝑇𝑛𝑥𝑛− 𝑥𝑛2 (34)
By the subdifferential inequality,
𝑦𝑛− 𝑥∗2
= 𝛼𝑛(𝑢 − 𝑥∗) + (1 − 𝛼𝑛) (𝑇𝑛𝑥𝑛− 𝑥∗)2
≤ (1 − 𝛼𝑛) 𝑇𝑛𝑥𝑛− 𝑥∗2
+ 2𝛼𝑛⟨𝑢 − 𝑥∗, 𝑦𝑛− 𝑥∗⟩
≤ (1 − 𝛼𝑛) 𝑥𝑛− 𝑥∗2
+ 2𝛼𝑛⟨𝑢 − 𝑥∗, 𝑦𝑛− 𝑥∗⟩
−(1 − 𝛼2] + 𝜆𝑛) (2] − 𝜆𝑛)
𝑛 𝑇𝑛𝑥𝑛− 𝑥𝑛2,
(35)
which implies that
𝑥𝑛+1− 𝑥∗2= 𝛽𝑛𝑥𝑛− 𝑥∗2+ (1 − 𝛽𝑛) 𝑆𝑦𝑛− 𝑥∗2
− 𝛽𝑛(1 − 𝛽𝑛) 𝑆𝑦𝑛− 𝑥𝑛2
≤ 𝛽𝑛𝑥𝑛− 𝑥∗2+ (1 − 𝛽𝑛) 𝑦𝑛− 𝑥∗2
− 𝛽𝑛(1 − 𝛽𝑛) 𝑆𝑦𝑛− 𝑥𝑛2
≤ 𝛽𝑛𝑥𝑛− 𝑥∗2+ (1 − 𝛽𝑛)
× (1 − 𝛼𝑛) 𝑥𝑛− 𝑥∗2
−(1 − 𝛼𝑛) (1 − 𝛽𝑛) (2] − 𝜆𝑛)
2] + 𝜆𝑛
× 𝑇𝑛𝑥𝑛− 𝑥𝑛2+ 2𝛼𝑛(1 − 𝛽𝑛)
× ⟨𝑢 − 𝑥∗, 𝑦𝑛− 𝑥∗⟩ − 𝛽𝑛(1 − 𝛽𝑛)
× 𝑆𝑦𝑛− 𝑥𝑛2
(36)
By our assumption, there exists𝜀 > 0 so that for all 𝑛 ≥ 0,
(1 − 𝛼𝑛) (1 − 𝛽𝑛) (2] − 𝜆𝑛)
and1 − 𝛽𝑛≥ 𝛽𝑛(1 − 𝛽𝑛) ≥ 𝜀 Consequently,
𝑥𝑛+1− 𝑥∗2≤ (1 − 𝜀𝛼𝑛) 𝑥𝑛− 𝑥∗2
− 𝜀 (𝑇𝑛𝑥𝑛− 𝑥𝑛2
+ 𝑆𝑦𝑛− 𝑥𝑛2) + 2𝛼𝑛(1 − 𝛽𝑛) ⟨𝑢 − 𝑥∗, 𝑦𝑛− 𝑥∗⟩
(38)
Set𝑠𝑛= ‖𝑥𝑛+1− 𝑥∗‖2, and let(𝑠𝑛𝑘) be a subsequence so that it
includes all elements in{𝑠𝑛} with the property; each of them
is less than or equal to the term after it Following an idea
developed by Maing´e [9], we next consider two possible cases
on(𝑠𝑛𝑘)
Case 1 Assume that{𝑠𝑛𝑘} is finite Then there exists 𝑁 ∈ N
so that𝑠𝑛 > 𝑠𝑛+1for all𝑛 ≥ 𝑁, and therefore {𝑠𝑛} must be convergent It follows from (38) that
𝜀 (𝑇𝑛𝑥𝑛− 𝑥𝑛2
+ 𝑆𝑦𝑛− 𝑥𝑛2) ≤ 𝑀𝛼𝑛+ (𝑠𝑛− 𝑠𝑛+1) , (39) where 𝑀 > 0 is a sufficiently large real number Conse-quently, both‖𝑇𝑛𝑥𝑛− 𝑥𝑛‖ and ‖𝑆𝑦𝑛 − 𝑥𝑛‖ converge to zero, and by Lemma 8 we conclude that ‖𝑦𝑛 − 𝑥𝑛‖ → 0 and
𝜔𝑤(𝑥𝑛) ⊆ Ω Hence, lim sup
𝑛 → ∞ ⟨𝑢 − 𝑥∗, 𝑦𝑛− 𝑥∗⟩ = lim sup
𝑛 → ∞ ⟨𝑢 − 𝑥∗, 𝑥𝑛− 𝑥∗⟩
= max
𝑤∈𝜔 𝑤( 𝑥 𝑛)⟨𝑢 − 𝑥
∗, 𝑤 − 𝑥∗⟩ ≤ 0,
(40) where the inequality uses (8) It then follows from (38) that
𝑠𝑛+1≤ (1 − 𝜀𝛼𝑛) 𝑠𝑛+ 2𝛼𝑛(1 − 𝛽𝑛) ⟨𝑢 − 𝑥∗, 𝑦𝑛− 𝑥∗⟩ (41)
We therefore applyLemma 4to conclude that𝑠𝑛 → 0
Case 2 Assume now that{𝑠𝑛𝑘} is infinite Let 𝑛 ∈ N be fixed Then there exists 𝑘 ∈ N so that 𝑛𝑘 ≤ 𝑛 ≤ 𝑛𝑘+1 By the choice of {𝑠𝑛𝑘}, we see that 𝑠𝑛𝑘+1 is the largest one among {𝑠𝑛𝑘, 𝑠𝑛𝑘+1, , 𝑠𝑛𝑘+1}; in particular
𝑠𝑛𝑘≤ 𝑠𝑛𝑘+1, 𝑠𝑛≤ 𝑠𝑛𝑘+1 (42) Then we deduce from (38) that
𝜀 (𝑇𝑛 𝑘𝑥𝑛𝑘− 𝑥𝑛𝑘2
+ 𝑆𝑦𝑛 𝑘− 𝑥𝑛𝑘2
) ≤ 𝑀𝛼𝑛𝑘→ 0 (43) ApplyingLemma 8yields‖𝑦𝑛𝑘− 𝑥𝑛𝑘‖ → 0 and 𝜔𝑤(𝑥𝑛𝑘) ⊆ Ω Similarly
lim sup
𝑛 → ∞ ⟨𝑢 − 𝑥∗, 𝑦𝑛𝑘− 𝑥∗⟩ ≤ 0 (44)
It follows again from (38) and inequality (42) that
𝑠𝑛𝑘≤ 2 (1 − 𝛽𝑛𝑘) ⟨𝑢 − 𝑥∗, 𝑦𝑛𝑘− 𝑥∗⟩ (45) Taking lim sup in (44) yields
lim sup
𝑘 → ∞ 𝑠𝑛𝑘≤ 0 ⇒ 𝑠𝑛𝑘 → 0 (46) Moreover, we deduce from algorithm (25) that
√𝑠𝑛 𝑘 +1= (𝑥𝑛 𝑘− 𝑥∗) − (𝑥𝑛𝑘− 𝑥𝑛𝑘+1)
≤ √𝑠𝑛 𝑘+ 𝑥𝑛 𝑘− 𝑥𝑛𝑘+1 ≤ √𝑠𝑛 𝑘+ 𝑥𝑛 𝑘− 𝑆𝑦𝑛𝑘 , (47) which together with (43) implies that𝑠𝑛𝑘+1 → 0 Conse-quently𝑠𝑛 → 0 immediately follows from (42)
Trang 54 Applications
In this section we present several applications First we
con-sider a problem for finding a common solution of equilibrium
problem (2) and fixed point problem (3); namely, find𝑥∗∈ 𝐶
so that
Taking𝐴 = 0 inTheorem 6and noting that zero mapping is
]-inverse strongly monotone for any positive number ], one
can easily get the following
Corollary 9 Let 𝐹 : 𝐶 × 𝐶 → R be a bifunction satisfying
(A1)–(A4) and 𝑆 : 𝐶 → 𝐶 a nonexpansive mapping so that
the solution set of problem (48) is nonempty Given 𝑢 ∈ 𝐶, let
(𝑥𝑛) generated by the iterative algorithm:
𝐹 (𝑧𝑛, 𝑦) + 1
𝜆𝑛⟨𝑦 − 𝑧𝑛, 𝑧𝑛− 𝑥𝑛⟩ ≥ 0, ∀𝑦 ∈ 𝐶,
𝑥𝑛+1= 𝛽𝑛𝑥𝑛+ (1 − 𝛽𝑛) 𝑆 [(1 − 𝛼𝑛) 𝑢 + 𝛼𝑛𝑧𝑛]
(49)
If the following conditions hold:
0 < 𝑎 ≤ 𝜆𝑛≤ 𝑏 < ∞, 0 < 𝑐 ≤ 𝛽𝑛≤ 𝑑 < 1,
lim
𝑛 → ∞𝛼𝑛= 0, ∑∞
𝑛=0
then the sequence (𝑥𝑛) converges strongly to a solution of
problem (48).
A variational inequality problem (VIP) is formulated as a
problem of finding a point𝑥∗with the property
𝑥∗∈ 𝐶, ⟨𝐴𝑥∗, 𝑧 − 𝑥∗⟩ ≥ 0, ∀𝑧 ∈ 𝐶 (51)
We will denote the solution set of VIP (51) by VI(𝐴; 𝐶) Next
we consider a problem for finding a common solution of
variational inequality problem (51) and of fixed point problem
(3), namely; find𝑥∗ ∈ 𝐶 so that
𝑥∗∈ VI (𝐴; 𝐶) ∩ Fix (𝑆) (52) Taking𝐹 = 0 in (1), we note that the generalized equilibrium
problem is reduced to the variational problem (51) Thus
applyingTheorem 6gets the following
Corollary 10 Let 𝐴 : 𝐶 → H be ]-inverse strongly
monotone mapping and 𝑆 : 𝐶 → 𝐶 a nonexpansive mapping
so that the solution set of problem (52) is nonempty Given
𝑢 ∈ 𝐶, let (𝑥𝑛) generated by the iterative algorithm:
𝑧𝑛= 𝑃𝐶(𝑥𝑛− 𝜆𝑛𝐴𝑥𝑛) ,
𝑥𝑛+1= 𝛽𝑛𝑥𝑛+ (1 − 𝛽𝑛) 𝑆 [(1 − 𝛼𝑛) 𝑢 + 𝛼𝑛𝑧𝑛] (53)
If the following conditions hold:
0 < 𝑎 ≤ 𝜆𝑛≤ 𝑏 < 2], 0 < 𝑐 ≤ 𝛽𝑛 ≤ 𝑑 < 1,
lim
𝑛 → ∞𝛼𝑛= 0, ∑∞
𝑛=0
then the sequence (𝑥𝑛) converges strongly to a solution of problem (52).
Consider the optimization problem of finding a point
𝑥∗ ∈ 𝐶 with the property
𝑓 (𝑥∗) = min
where𝑓 : H → R is a convex and differentiable function
We say that the differential∇𝑓 is 1/]-Lipschitz continuous, if
∇𝑓(𝑥) − ∇𝑓(𝑦) ≤ 1]𝑥 − 𝑦, ∀𝑥,𝑦 ∈ H (56) Denote by Argmin(𝐶; 𝑓) the solution set of problem (55) Finally we consider a problem for finding a common solution
of optimization problem (55) and of fixed point problem (3), namely; find𝑥∗ ∈ 𝐶 so that
𝑥∗ ∈ Argmin (𝐶; 𝑓) ∩ Fix (𝑆) (57)
By [10, Lemma 5.13], problem (55) is equivalent to the variational inequality problem
⟨∇𝑓 (𝑥∗) , 𝑥∗− 𝑧⟩ ≥ 0, ∀𝑧 ∈ 𝐶 (58) Taking𝐴 = ∇𝑓 inCorollary 10, we have the following result
Corollary 11 Let 𝑓 : H → R be a convex and differentiable
function so that ∇𝑓 is 1/]-Lipschitz continuous Let 𝑆 : 𝐶 → 𝐶
be a nonexpansive mapping so that the solution set of problem
(57) is nonempty Given 𝑢 ∈ 𝐶, let (𝑥𝑛) generated by
𝑧𝑛 = 𝑃𝐶(𝑥𝑛− 𝜆𝑛∇𝑓 (𝑥𝑛)) ,
𝑥𝑛+1= 𝛽𝑛𝑥𝑛+ (1 − 𝛽𝑛) 𝑆 [(1 − 𝛼𝑛) 𝑢 + 𝛼𝑛𝑧𝑛] (59)
If the following conditions hold:
0 < 𝑎 ≤ 𝜆𝑛≤ 𝑏 < 2], 0 < 𝑐 ≤ 𝛽𝑛≤ 𝑑 < 1, lim
𝑛 → ∞𝛼𝑛= 0, ∑∞
𝑛=0
then the sequence (𝑥𝑛) converges strongly to a solution of problem (57).
Proof It suffices to note that if∇𝑓 is 1/]-Lipschitz contin-uous, then it is]-inverse strongly monotone mapping (see [11, Corollary 10]) ConsequentlyCorollary 10applies and the result immediately follows
Remark 12 We can further apply the previous method to
find a common solution for fixed point and split feasibility problems, as well as for fixed point and convex constrained linear inverse problems (see [12])
Acknowledgment
This work is supported by the National Natural Science Foundation of China, Tianyuan Foundation (11226227)
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