Hindawi Publishing Corporation Journal of Function Spaces and Applications Volume 2013, Article ID 735941, 6 pages http://dx.doi.org/10.1155/2013/735941
Research Article Damped Algorithms for the Split Fixed Point and Equilibrium Problems Li-Jun Zhu1,2 and Minglun Ren1 1 2
School of Management, Hefei University of Technology, Hefei 230009, China School of Mathematics and Information Science, Beifang University of Nationalities, Yinchuan 750021, China
Correspondence should be addressed to Minglun Ren;
[email protected] Received 23 June 2013; Accepted 5 August 2013 Academic Editor: J. Liang Copyright © 2013 L.-J. Zhu and M. Ren. 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. The main purpose of this paper is to study the split fixed point and equilibrium problems which includes fixed point problems, equilibrium problems, and variational inequality problems as special cases. A damped algorithm is presented for solving this split common problem. Strong convergence analysis is shown.
1. Introduction Very recently, the split problems (e.g., the split feasibility problem, the split common fixed points problem, and the split variational inequality problem) have been studied extensively, see, for instance, [1–19]. Now we recall the related history. Let 𝐻1 and 𝐻2 be two Hilbert spaces and 𝐶 ⊂ 𝐻1 and 𝑄 ⊂ 𝐻2 two nonempty closed convex subsets. Let 𝐴 : 𝐻1 → 𝐻2 be a bounded linear operator. The split feasibility problem is to solve the inclusion: 𝑥 ∈ 𝐶 ∩ 𝐴−1 (𝑄)
Cui et al., [6] extended the damped projection algorithm to the split common fixed point problems. Let 𝜓 : 𝐶 × 𝐶 → R be a bifunction. The equilibrium problem is to find 𝑥† ∈ 𝐶 such that 𝜓 (𝑥† , 𝑥) ≥ 0,
∀𝑥 ∈ 𝐶.
(4)
We will indicate with EP(𝜓) the set of solutions of (4). In the present paper, our main purpose is to study the following split fixed point and equilibrium problem.
(1)
which arise in the field of intensity-modulated radiation therapy and was presented in [1]. The iteration 𝜌𝑛+1 = proj𝐶(𝜌𝑛 − 𝜍𝐴∗ (𝐼 − 𝑃𝑄)𝐴𝜌𝑛 ) is popular with 𝜍 ∈ (0, 2/‖𝐴‖2 ). Further, Xu [3] suggested a single step regularized method. Dang and Gao [4] developed a damped projection algorithm. If 𝐶 and 𝑄 are the fixed point sets of mappings 𝑈 and 𝑇, respectively, then (1) becomes a special case of the split common fixed point problem: Find 𝑥 ∈ Fix (𝑈) ∩ 𝐴−1 (Fix (𝑇)) .
(2)
Find a point 𝑢§ ∈ Fix (𝑊) ∩ EP (𝜓) such that 𝐴𝑢§ ∈ Fix (𝑆) ∩ EP (𝜑) ,
(5)
where Fix(𝑆) and Fix(𝑊) are the sets of fixed points of two nonlinear mappings 𝑆 and 𝑊, respectively; EP(𝜓) and EP(𝜑) are the solution sets of two equilibrium problems with bifunctions 𝜓 and 𝜑, respectively, and 𝐴 is a bounded linear mapping. Denote the solution set of (5) by Θ = {𝑥 ∈ Fix (𝑊) ∩ EP (𝜓) : 𝐴𝑥 ∈ Fix (𝑆) ∩ EP (𝜑)} . (6)
Censor and Segal [5] invented a scheme below to solve (2): 𝜌𝑛+1 = 𝑈 (𝜌𝑛 − 𝜍𝐴∗ (𝐼 − 𝑇) 𝐴𝜌𝑛 ) ,
𝑛 ∈ N.
(3)
We develop a damped algorithm to solve this split fixed point and equilibrium problem. Strong convergence of the suggested damped algorithm is demonstrated.
2
Journal of Function Spaces and Applications
2. Concepts and Lemmas 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 𝑊 : 𝐶 → 𝐶 is called nonexpansive if † 𝑊𝑥 − 𝑊𝑥‡ ≤ 𝑥† − 𝑥‡ , (7) for all 𝑥† , 𝑥‡ ∈ 𝐶. We call proj𝐶 : 𝐻 → 𝐶 the metric projection if for each 𝑥♭ ∈ 𝐻 ♭ 𝑥 − proj𝐶 (𝑥♭ ) = inf {𝑥♭ − 𝑥† : 𝑥† ∈ 𝐶} .
(8)
It is well known that the metric projection proj𝐶 : 𝐻 → 𝐶 is firmly nonexpansive, that is, 2 proj𝐶 (𝑥† ) − proj𝐶 (𝑥‡ )
(9)
≤ ⟨𝑥† − 𝑥‡ , proj𝐶 (𝑥† ) − proj𝐶 (𝑥‡ )⟩
Lemma 1 (see [20]). Let 𝐶 be a nonempty closed convex subset of a real Hilbert space 𝐻. Let 𝜓 : 𝐶 × 𝐶 → R be a bifunction which satisfies the following conditions: (H1) 𝜓(𝑥‡ , 𝑥‡ ) = 0 for all 𝑥‡ ∈ 𝐶; (H2) 𝜓 is monotone, that is, 𝜓(𝑥‡ , 𝑥† ) + 𝜓(𝑥† , 𝑥‡ ) ≤ 0 for all 𝑥‡ , 𝑥† ∈ 𝐶; (H3) for each 𝑥† , 𝑥‡ , 𝑥♮ ∈ 𝐶, lim𝑡↓0 𝜓(𝑡𝑥♮ + (1 − 𝑡)𝑥† , 𝑥‡ ) ≤ 𝜓(𝑥† , 𝑥‡ ); (H4) for each 𝑥† ∈ 𝐶, 𝑥‡ → 𝜓(𝑥† , 𝑥‡ ) is convex and lower semicontinuous. Let 𝜛 > 0 and 𝑥† ∈ 𝐶. Then, there exists 𝑥♮ ∈ 𝐶 such that 1 ‡ ⟨𝑥 − 𝑥♮ , 𝑥♮ − 𝑥† ⟩ ≥ 0, 𝜛
∀𝑥‡ ∈ 𝐶.
(10)
𝜓
Further, if 𝑈𝜛 (𝑥† ) = {𝑥♮ ∈ 𝐶 : 𝜓(𝑥♮ , 𝑥‡ ) + (1/𝜛)⟨𝑥‡ − 𝑥 , 𝑥♮ − 𝑥† ⟩ ≥ 0, 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥‡ ∈ 𝐶}, then the following hold: ♮
𝜓
𝜓
(i) 𝑈𝜛 is single-valued and 𝑈𝜛 is firmly nonexpansive, that 2 𝜓 𝜓 𝜓 is, for any 𝑥† , 𝑥‡ ∈ 𝐻, ‖𝑈𝜛 𝑥† − 𝑈𝜛 𝑥‡ ‖ ≤ ⟨𝑈𝜛 𝑥† − 𝜓 ‡ † ‡ 𝑈𝜛 𝑥 , 𝑥 − 𝑥 ⟩; 𝜓
(ii) EP(𝜓) is closed and convex and EP(𝜓) = Fix(𝑈𝜛 ). Lemma 2 (see [21]). Let 𝐻 be a Hilbert space and 𝐶 ⊂ 𝐻 a closed convex subset. Let 𝑊 : 𝐶 → 𝐶 be a nonexpansive mapping. Then, the mapping 𝐼−𝑊 is demiclosed. That is, if {𝜌𝑛 } is a sequence in 𝐶 such that 𝜌𝑛 → ] weakly and (𝐼 − 𝑊)𝜌𝑛 → 𝑢 strongly, then (𝐼 − 𝑊)] = 𝑢. Lemma 3 (see [22]). Assume that {𝜂𝑛 } is a sequence of nonnegative real numbers such that 𝜂𝑛+1 ≤ (1 − 𝜅𝑛 ) 𝜂𝑛 + 𝜍𝑛 ,
𝑛 ∈ N,
(1) ∑∞ 𝑛=1 𝜅𝑛 = ∞; (2) lim sup𝑛 → ∞ (𝜍𝑛 /𝜅𝑛 ) ≤ 0 or ∑∞ 𝑛=1 |𝜍𝑛 | < ∞. Then lim𝑛 → ∞ 𝜂𝑛 = 0.
3. Main Results Let 𝐻1 and 𝐻2 be two Hilbert spaces and 𝐶 ⊂ 𝐻1 and 𝑄 ⊂ 𝐻2 two nonempty closed convex subsets. Let 𝐴 : 𝐻1 → 𝐻2 be a bounded linear operator with its adjoint 𝐴∗ . Let 𝜓 : 𝐶 × 𝐶 → R and let 𝜑 : 𝐷 × 𝐷 → R be two bifunctions satisfying the conditions (H1)–(H4) in Lemma 1. Let 𝑆 : 𝐷 → 𝐷 and 𝑊 : 𝐶 → 𝐶 be two nonexpansive mappings. Algorithm 4. Let 𝑥0 ∈ 𝐻1 . Define a sequence {𝑥𝑛 } as follows:
for all 𝑥† , 𝑥‡ ∈ 𝐻. Hence proj𝐶 is also nonexpansive.
𝜓 (𝑥♮ , 𝑥‡ ) +
where {𝜅𝑛 } is a sequence in (0, 1) and {𝜍𝑛 } is a sequence such that
(11)
𝜌𝑛+1 = 𝑊𝑈𝜄𝜓 [(1 − 𝜁𝑛 ) × (𝜌𝑛 + 𝜍𝐴∗ (𝑆𝑈𝜅𝜑 − 𝐼) 𝐴𝜌𝑛 )] ,
∀𝑛 ∈ N, (12)
where 𝜄, 𝜅, and 𝜍 are three constants satisfying 𝜄 ∈ (0, ∞), 𝜅 ∈ (0, ∞), 𝜍 ∈ (0, 1/‖𝐴‖2 ), and {𝜁𝑛 } is a real number sequence in (0, 1). In the sequel, we assume that Θ = {𝑥 ∈ Fix (𝑊) ∩ EP (𝜓) : 𝐴𝑥 ∈ Fix (𝑆) ∩ EP (𝜑)} ≠ 0. (13) Theorem 5. If {𝜁𝑛 } satisfies lim𝑛 → ∞ 𝜁𝑛 = 0, ∑∞ 𝑛=1 𝜁𝑛 = ∞ and lim𝑛 → ∞ 𝜁𝑛+1 /𝜁𝑛 = 1, then {𝜌𝑛 } generated by algorithm (12) converges strongly to projΘ (0) which is the minimum-norm element in Θ. Proof. Let 𝑝 = projΘ (0). Then, 𝑝 ∈ Fix(𝑊) ∩ EP(𝜓) and 𝐴𝑝 ∈ Fix(𝑆) ∩ EP(𝜑). Set 𝑧𝑛 = 𝑈𝜅𝜑 𝐴𝜌𝑛 , 𝑦𝑛 = (1 − 𝜁𝑛 )(𝜌𝑛 + 𝜍𝐴∗ (𝑆𝑈𝜅𝜑 − 𝐼)𝐴𝜌𝑛 ) and 𝑢𝑛 = 𝑈𝜄𝜓 [(1 − 𝜁𝑛 )(𝜌𝑛 + 𝜍𝐴∗ (𝑆𝑈𝜅𝜑 − 𝐼)𝐴𝜌𝑛 )] for all 𝑛 ∈ N. Then 𝑢𝑛 = 𝑈𝜄𝜓 𝑦𝑛 . From Lemma 1, we know that 𝑈𝜄𝜓 and 𝑈𝜅𝜑 are firmly nonexpansive. Thus, we have 𝑛 (14) 𝑧 − 𝐴𝑝 = 𝑈𝜅𝜑 𝐴𝜌𝑛 − 𝐴𝑝 ≤ 𝐴𝜌𝑛 − 𝐴𝑝 , 𝑢𝑛 − 𝑝 = 𝑈𝜓 𝑦𝑛 − 𝑝 ≤ 𝑦𝑛 − 𝑝 , (15) 𝜄 𝜑 𝑛 2 𝜑 𝑛 𝜑 𝑆𝑈𝜅 𝐴𝜌 − 𝐴𝑝 = 𝑆𝑈𝜅 𝐴𝜌 − 𝑆𝑈𝜅 𝐴𝑝 2 ≤ 𝑈𝜅𝜑 𝐴𝜌𝑛 − 𝑈𝜅𝜑 𝐴𝑝 2 2 ≤ 𝐴𝜌𝑛 − 𝐴𝑝 − 𝑈𝜅𝜑 𝐴𝜌𝑛 − 𝐴𝜌𝑛 . (16)
Journal of Function Spaces and Applications
3
Note that
From (16), (21) and (22), we have 𝑛+1 𝑢 − 𝑢𝑛 = 𝑈𝜄𝜓 𝑦𝑛+1 − 𝑈𝜄𝜓 𝑦𝑛 𝑛+1 𝑛 ≤ 𝑦 − 𝑦 , 𝑛+1 𝑧 − 𝑧𝑛 = 𝑈𝜅𝜑 𝐴𝜌𝑛+1 − 𝑈𝜅𝜑 𝐴𝜌𝑛 𝑛+1 ≤ 𝐴𝜌 − 𝐴𝜌𝑛 .
(17)
⟨𝜌𝑛 − 𝑝, 𝐴∗ (𝑆𝑧𝑛 − 𝐴𝜌𝑛 )⟩
(18)
2 − 𝑆𝑧𝑛 − 𝐴𝜌𝑛 1 𝑛 2 2 (𝐴𝜌 − 𝐴𝑝 − 𝑧𝑛 − 𝐴𝜌𝑛 2 2 2 + 𝑆𝑧𝑛 − 𝐴𝜌𝑛 −𝐴𝜌𝑛 − 𝐴𝑝 )
≤
From (12) and (15), we have 𝑛+1 𝜌 − 𝑝 = 𝑊𝑢𝑛 − 𝑝 ≤ 𝑢𝑛 − 𝑝 ≤ 𝑦𝑛 − 𝑝 .
1 𝑛 2 2 (𝑆𝑧 − 𝐴𝑝 + 𝑆𝑧𝑛 − 𝐴𝜌𝑛 2 2 −𝐴𝜌𝑛 − 𝐴𝑝 )
=
(19)
(23)
2 − 𝑆𝑧𝑛 − 𝐴𝜌𝑛
Observe that
1 2 = − 𝑧𝑛 − 𝐴𝜌𝑛 2
𝑛 2 𝑦 − 𝑝 = (1 − 𝜁𝑛 )
1 2 − 𝑆𝑧𝑛 − 𝐴𝜌𝑛 . 2
2 × (𝜌𝑛 − 𝑝 + 𝜍𝐴∗ (𝑆𝑧𝑛 − 𝐴𝜌𝑛 )) − 𝜁𝑛 𝑝 2 ≤ (1 − 𝜁𝑛 ) (𝜌𝑛 − 𝑝 + 𝜍𝐴∗ (𝑆𝑧𝑛 − 𝐴𝜌𝑛 )) 2 + 𝜁𝑛 𝑝 = (1 − 𝜁𝑛 ) [ 𝜌𝑛 − 𝑝 + 2𝜍
By (20) and (23), we deduce (20)
2 𝑛 𝑦 − 𝑝 2 2 ≤ (1 − 𝜁𝑛 ) [𝜌𝑛 − 𝑝 + 𝜍2 ‖𝐴‖2 𝑆𝑧𝑛 − 𝐴𝜌𝑛
× ⟨𝜌𝑛 − 𝑝, 𝐴∗ (𝑆𝑧𝑛 − 𝐴𝜌𝑛 )⟩ 2 +𝜍2 𝐴∗ (𝑆𝑧𝑛 − 𝐴𝜌𝑛 ) ]
1 2 + 2𝜍 (− 𝑧𝑛 − 𝐴𝜌𝑛 2
2 + 𝜁𝑛 𝑝 .
1 2 2 − 𝑆𝑧𝑛 − 𝐴𝜌𝑛 )] + 𝜁𝑛 𝑝 (24) 2
Since 𝐴∗ is the adjoint of 𝐴, we have
= (1 − 𝜁𝑛 ) 2 2 × [𝜌𝑛 − 𝑝 + (𝜍2 ‖𝐴‖2 − 𝜍) 𝑆𝑧𝑛 − 𝐴𝜌𝑛
⟨𝜌𝑛 − 𝑝, 𝐴∗ (𝑆𝑧𝑛 − 𝐴𝜌𝑛 )⟩
2 2 −𝜍𝑧𝑛 − 𝐴𝜌𝑛 ] + 𝜁𝑛 𝑝
= ⟨𝐴 (𝜌𝑛 − 𝑝) , 𝑆𝑧𝑛 − 𝐴𝜌𝑛 ⟩ = ⟨𝐴𝜌𝑛 − 𝐴𝑝 + 𝑆𝑧𝑛 − 𝐴𝜌𝑛
(21)
2 2 ≤ (1 − 𝜁𝑛 ) 𝜌𝑛 − 𝑝 + 𝜁𝑛 𝑝 .
− (𝑆𝑧𝑛 − 𝐴𝜌𝑛 ) , 𝑆𝑧𝑛 − 𝐴𝜌𝑛 ⟩ 2 = ⟨𝑆𝑧𝑛 − 𝐴𝑝, 𝑆𝑧𝑛 − 𝐴𝜌𝑛 ⟩ − 𝑆𝑧𝑛 − 𝐴𝜌𝑛 .
It follows from (19), we get
Using parallelogram law, we obtain ⟨𝑆𝑧𝑛 − 𝐴𝑝, 𝑆𝑧𝑛 − 𝐴𝜌𝑛 ⟩ =
1 𝑛 2 2 (𝑆𝑧 − 𝐴𝑝 + 𝑆𝑧𝑛 − 𝐴𝜌𝑛 2 2 −𝐴𝜌𝑛 − 𝐴𝑝 ) .
(22)
𝑛+1 2 𝜌 − 𝑝 ≤ 𝑦𝑛 − 𝑝2 2 2 ≤ (1 − 𝜁𝑛 ) 𝜌𝑛 − 𝑝 + 𝜁𝑛 𝑝 2 2 ≤ max {𝜌𝑛 − 𝑝 , 𝑝 } . The boundedness of the sequence {𝜌𝑛 } yields.
(25)
4
Journal of Function Spaces and Applications 2 ≤ 𝜌𝑛+1 − 𝜌𝑛
Set V𝑛 = 𝜌𝑛 + 𝜍𝐴∗ (𝑆𝑈𝜅𝜑 − 𝐼)𝐴𝜌𝑛 . Then, we have
+ (𝜍2 ‖𝐴‖2 − 𝜍) 2 × 𝑆𝑧𝑛+1 − 𝑆𝑧𝑛 − (𝐴𝜌𝑛+1 − 𝐴𝜌𝑛 ) 2 2 + 𝜍 (𝑧𝑛+1 − 𝑧𝑛 − 𝐴𝜌𝑛+1 − 𝐴𝜌𝑛 ) .
𝑛+1 2 V − V𝑛 = 𝜌𝑛+1 − 𝜌𝑛 ∗
(26) 𝑛+1
+𝜍 [𝐴 (𝑆𝑧
−𝐴𝜌
𝑛+1
2 )−𝐴 (𝑆𝑧 −𝐴𝜌 )] ∗
𝑛
𝑛
2 = 𝜌𝑛+1 − 𝜌𝑛 + 2𝜍 ⟨𝜌𝑛+1 − 𝜌𝑛 , 𝐴∗ [(𝑆𝑧𝑛+1 − 𝐴𝜌𝑛+1 ) − (𝑆𝑧𝑛 − 𝐴𝜌𝑛 )]⟩ 2 + 𝜍2 𝐴∗ (𝑆𝑧𝑛+1 − 𝐴𝜌𝑛+1 )−𝐴∗ (𝑆𝑧𝑛 − 𝐴𝜌𝑛 ) 2 ≤ 𝜌𝑛+1 − 𝜌𝑛 + 2𝜍 ⟨𝐴𝜌𝑛+1 − 𝐴𝜌𝑛 , 𝑆𝑧𝑛+1 − 𝑆𝑧𝑛 − (𝐴𝜌𝑛+1 − 𝐴𝜌𝑛 )⟩ 2 + 𝜍2 ‖𝐴‖2 𝑆𝑧𝑛+1 − 𝑆𝑧𝑛 − (𝐴𝜌𝑛+1 − 𝐴𝜌𝑛 ) 2 = 𝜌𝑛+1 − 𝜌𝑛 2 + 𝜍2 ‖𝐴‖2 𝑆𝑧𝑛+1 − 𝑆𝑧𝑛 − (𝐴𝜌𝑛+1 − 𝐴𝜌𝑛 ) + 2𝜍 ⟨𝑆𝑧𝑛+1 − 𝑆𝑧𝑛 , 𝑆𝑧𝑛+1 − 𝑆𝑧𝑛 − (𝐴𝜌𝑛+1 − 𝐴𝜌𝑛 )⟩ 2 − 2𝜍𝑆𝑧𝑛+1 − 𝑆𝑧𝑛 − (𝐴𝜌𝑛+1 − 𝐴𝜌𝑛 ) 𝑛 2
= 𝜌𝑛+1 − 𝜌 2 + 𝜍2 ‖𝐴‖2 𝑆𝑧𝑛+1 − 𝑆𝑧𝑛 − (𝐴𝜌𝑛+1 − 𝐴𝜌𝑛 ) 2 + 𝜍 (𝑆𝑧𝑛+1 − 𝑆𝑧𝑛 2 + 𝑆𝑧𝑛+1 − 𝑆𝑧𝑛 − (𝐴𝜌𝑛+1 − 𝐴𝜌𝑛 ) 2 −𝐴𝜌𝑛+1 − 𝐴𝜌𝑛 ) 2 − 2𝜍𝑆𝑧𝑛+1 − 𝑆𝑧𝑛 − (𝐴𝜌𝑛+1 − 𝐴𝜌𝑛 ) 2 = 𝜌𝑛+1 − 𝜌𝑛 + (𝜍2 ‖𝐴‖2 − 𝜍) 2 × 𝑆𝑧𝑛+1 − 𝑆𝑧𝑛 − (𝐴𝜌𝑛+1 − 𝐴𝜌𝑛 ) 𝑛 2
𝑛 2
+ 𝜍 (𝑆𝑧𝑛+1 − 𝑆𝑧 − (𝐴𝜌𝑛+1 − 𝐴𝜌 ) )
2
Since 𝜍 ∈ (0, 1/‖𝐴‖ ), we derive by virtue of (18) and (26) that 𝑛+1 𝑛 𝑛 𝑛+1 (27) V − V ≤ 𝜌 − 𝜌 . According to (17) and (27), we have 𝑛+1 𝜌 − 𝜌𝑛 = 𝑊𝑢𝑛+1 − 𝑊𝑢𝑛 ≤ 𝑢𝑛+1 − 𝑢𝑛 ≤ 𝑦𝑛+1 − 𝑦𝑛 = (1 − 𝜁𝑛+1 ) V𝑛+1 − (1 − 𝜁𝑛 ) V𝑛 (28) = (1 − 𝜁𝑛+1 ) (V𝑛+1 − V𝑛 ) + (𝜁𝑛 − 𝜁𝑛+1 ) V𝑛 ≤ (1 − 𝜁𝑛+1 ) V𝑛+1 − V𝑛 + 𝜁𝑛+1 − 𝜁𝑛 V𝑛 ≤ (1 − 𝜁𝑛+1 ) 𝜌𝑛+1 − 𝜌𝑛 + 𝜁𝑛+1 − 𝜁𝑛 V𝑛 . It follows that 𝜌𝑛+1 − 𝜌𝑛 ≤ 𝜁𝑛+1 − 𝜁𝑛 V𝑛 . (29) 𝜁𝑛+1 Since {𝜌𝑛 } is bounded, we can deduce {V𝑛 } is also bounded. From (29), we have lim 𝜌𝑛+1 − 𝜌𝑛 = 0. (30) 𝑛→∞ Hence, lim 𝜌𝑛 − 𝑊𝑢𝑛 = 0. (31) 𝑛→∞
Using the firmly-nonexpansivenessity of 𝑈𝜄𝜓 , we have 𝑛 2 𝜓 𝑛 2 𝑢 − 𝑝 = 𝑈𝜄 𝑦 − 𝑝 2 2 ≤ 𝑦𝑛 − 𝑝 − 𝑈𝜄𝜓 𝑦𝑛 − 𝑦𝑛 2 2 = 𝑦𝑛 − 𝑝 − 𝑢𝑛 − 𝑦𝑛 . Thus, we get 𝑛+1 2 𝑛 2 𝜌 − 𝑝 ≤ 𝑢 − 𝑝
(32)
2 2 ≤ 𝑦𝑛 − 𝑝 − 𝑢𝑛 − 𝑦𝑛 2 2 2 ≤ (1 − 𝜁𝑛 ) 𝜌𝑛 − 𝑝 + 𝜁𝑛 𝑝 − 𝑢𝑛 − 𝑦𝑛 . (33) It follows that 𝑢𝑛 − 𝑦𝑛 2 ≤ 𝜌𝑛 − 𝑝2 − 𝜌𝑛+1 − 𝑝2 + 𝜁𝑛 𝑝2 2 2 2 ≤ (𝜌𝑛 − 𝑝 + 𝜌𝑛+1 − 𝑝 ) 𝜌𝑛+1 − 𝑝 + 𝜁𝑛 𝑝 . (34)
Journal of Function Spaces and Applications
5
This together with (30) and (C1) implies that lim 𝑢𝑛 − 𝑦𝑛 = 0.
By Lemma 2, (39), and (41), we deduce 𝑧 ∈ Fix(𝑊) and 𝐴𝑧 ∈ Fix(𝑆). Next, we show that 𝑧 ∈ EP(𝜓). Since 𝑢𝑛 = 𝑈𝜄𝜓 𝑦𝑛 , we have
(35)
𝑛→∞
Note that 𝑛+1 2 𝜌 − 𝑝 ≤ 𝑦𝑛 − 𝑝2 2 ≤ (1 − 𝜁𝑛 ) 𝜌𝑛 − 𝑝
𝜓 (𝑢𝑛 , 𝑥† ) +
+ (1 − 𝜁𝑛 ) (𝜍2 ‖𝐴‖2 − 𝜍) 𝑆𝑧𝑛 − 𝐴𝜌
⟨𝑥† − 𝑢𝑛𝑖 ,
2 (1 − 𝜁𝑛 ) (𝜍 − 𝜍2 ‖𝐴‖2 )𝑆𝑧𝑛 − 𝐴𝜌𝑛
(47)
𝑢𝑛𝑖 − 𝑦𝑛𝑖 𝜄
⟩ ≥ 𝜓 (𝑥† , 𝑢𝑛𝑖 ) .
(48)
Since ‖𝑢𝑛 − 𝑦𝑛 ‖ → 0, 𝑢𝑛𝑖 ⇀ 𝑧, we obtain (𝑢𝑛𝑖 − 𝑦𝑛𝑖 )/𝜄 → 0. Thus, 0 ≥ 𝜓(𝑥† , 𝑧). For 𝑡 with 0 < 𝑡 ≤ 1 and 𝑥† ∈ 𝐶, let 𝑦𝑡 = 𝑡𝑥† + (1 − 𝑡)𝑧 ∈ 𝐶. We obtain 𝜓(𝑦𝑡 , 𝑧) ≤ 0. Hence,
2 + (1 − 𝜁𝑛 ) 𝜍𝑧𝑛 − 𝐴𝜌𝑛 2 2 2 ≤ 𝜌𝑛 − 𝑝 − 𝜌𝑛+1 − 𝑝 + 𝜁𝑛 𝑝
(37)
≤ (𝜌𝑛 − 𝑝 + 𝜌𝑛+1 − 𝑝) 𝜌𝑛+1 − 𝜌𝑛
which implies that lim 𝑆𝑧𝑛 − 𝐴𝜌𝑛 = lim 𝑧𝑛 − 𝐴𝜌𝑛 = 0. 𝑛→∞ 𝑛→∞
(38)
lim 𝑆𝑧𝑛 − 𝑧𝑛 = 0.
lim sup ⟨𝑝, 𝑦𝑛 − 𝑝⟩ = lim ⟨𝑝, 𝑦𝑛𝑖 − 𝑝⟩ 𝑛→∞
So, we get 𝑛→∞
0 = 𝜓 (𝑦𝑡 , 𝑦𝑡 ) ≤ 𝑡𝜓 (𝑦𝑡 , 𝑥† ) + (1 − 𝑡) 𝜓 (𝑦𝑡 , 𝑧) ≤ 𝑡𝜓 (𝑦𝑡 , 𝑥† ) . (49) So, 0 ≤ 𝜓(𝑦𝑡 , 𝑥† ). And, thus, 0 ≤ 𝜓(𝑧, 𝑥† ). This implies that 𝑧 ∈ EP(𝜓). Similarity, we can prove that 𝐴𝑧 ∈ EP(𝜑). To this end, we deduce 𝑧 ∈ Fix(𝑊) ∩ EP(𝜓) and 𝐴𝑧 ∈ Fix(𝑆) ∩ EP(𝜑). That is to say, 𝑧 ∈ Θ. Therefore,
2 + 𝜁𝑛 𝑝 ,
(39)
𝑖→∞
= lim ⟨𝑝, 𝑧 − 𝑝⟩ 𝑖→∞
(50)
≥ 0.
Since 𝑛 ∗ 𝑛 𝜑 𝑛 𝑛 𝑦 − 𝑝 = 𝜍𝐴 (𝑆𝑈𝜅 − 𝐼) 𝐴𝜌 + 𝜁𝑛 V
≤ 𝜍 ‖𝐴‖ 𝑆𝑧𝑛 − 𝐴𝜌𝑛 + 𝜁𝑛 V𝑛 ,
Finally, we prove 𝜌𝑛 → 𝑝. From (12), we have (40)
we get lim 𝜌𝑛 − 𝑦𝑛 = 0.
𝑛→∞
(41)
From (31), (35), and (41), we get lim 𝜌𝑛 − 𝑊𝜌𝑛 = 0. 𝑛→∞
(42)
Now, we show that lim sup ⟨𝑝, 𝑦𝑛 − 𝑝⟩ ≥ 0.
(43)
𝑛→∞
Choose a subsequence {𝑦𝑛𝑖 } of {𝑦𝑛 } such that 𝑛𝑖
lim sup ⟨𝑝, 𝑦 − 𝑝⟩ = lim ⟨𝑝, 𝑦 − 𝑝⟩ . 𝑖→∞
(44)
𝑛
𝑢𝑛𝑖 ⇀ 𝑧, 𝑧𝑛𝑖 ⇀ 𝐴𝑧.
2 𝑛+1 𝜌 − 𝑝 ≤ 𝑦𝑛 − 𝑝2 2 = (1 − 𝜁𝑛 ) (V𝑛 − 𝑝) − 𝜁𝑛 𝑝 2 ≤ (1 − 𝜁𝑛 ) V𝑛 − 𝑝 − 2𝜁𝑛 ⟨𝑝, 𝑦𝑛 − 𝑝⟩
(45)
(51)
2 ≤ (1 − 𝜁𝑛 ) 𝜌𝑛 − 𝑝 − 2𝜁𝑛 ⟨𝑝, 𝑦𝑛 − 𝑝⟩ . Applying Lemma 3 and (50) to (51), we deduce 𝜌𝑛 → 𝑝. The proof is completed. Algorithm 6. Let 𝜌0 ∈ 𝐻1 arbitrarily define a sequence {𝜌𝑛 } by the following: 𝜌𝑛+1 = 𝑊 ((1 − 𝜁𝑛 ) (𝜌𝑛 + 𝜍𝐴∗ (𝑆 − 𝐼) 𝐴𝜌𝑛 )) ,
Notice that {𝑦𝑛𝑖 } is bounded, we can choose {𝑦 𝑖𝑗 } of {𝑦𝑛𝑖 } 𝑛 such that 𝑦 𝑖𝑗 ⇀ 𝑧. Without loss of generality, we assume that 𝑦𝑛𝑖 ⇀ 𝑧. From the above conclusions, we derive that
𝐴𝜌𝑛𝑖 ⇀ 𝐴𝑧,
(46)
and so
Hence,
𝜌𝑛𝑖 ⇀ 𝑧,
1 † ⟨𝑥 − 𝑢𝑛 , 𝑢𝑛 − 𝑦𝑛 ⟩ ≥ 𝜓 (𝑥† , 𝑢𝑛 ) , 𝜄
(36)
2 2 − (1 − 𝜁𝑛 ) 𝜍𝑧𝑛 − 𝐴𝜌𝑛 + 𝜁𝑛 𝑝 .
𝑛→∞
∀𝑥† ∈ 𝐶.
By the monotonicity of 𝜓, we have 𝑛 2
𝑛
1 † ⟨𝑥 − 𝑢𝑛 , 𝑢𝑛 − 𝑦𝑛 ⟩ ≥ 0, 𝜄
(52)
for all 𝑛 ∈ N, where 𝜍 ∈ (0, 1/‖𝐴‖2 ) and {𝜁𝑛 } is a real number sequence in (0, 1). Corollary 7. Suppose Θ1 = {𝑥 ∈ Fix(𝑊) : 𝐴𝑥 ∈ Fix(𝑆)} ≠ 0. = ∞, and If {𝜁𝑛 } satisfies lim𝑛 → ∞ 𝜁𝑛 = 0, ∑∞ 𝑛=1 𝜁𝑛 lim𝑛 → ∞ 𝜁𝑛+1 /𝜁𝑛 = 1, then the sequence {𝜌𝑛 } generated by algorithm (52) converges strongly to 𝑝 = projΘ1 (0) which is the mum-norm element in Θ1 .
6
Journal of Function Spaces and Applications
Algorithm 8. Let 𝜌0 ∈ 𝐻1 arbitrarily define a sequence {𝜌𝑛 } by the following: 𝜌𝑛+1 = 𝑈𝜄𝜓 ((1 − 𝜁𝑛 ) (𝜌𝑛 + 𝜍𝐴∗ (𝑈𝜅𝜑 − 𝐼) 𝐴𝜌𝑛 )) ,
(53)
for all 𝑛 ∈ N, where 𝜄, 𝜅, and 𝜍 are three constants satisfying 𝜄 ∈ (0, ∞), 𝜅 ∈ (0, ∞), 𝜍 ∈ (0, 1/‖𝐴‖2 ), and {𝜁𝑛 } is a real number sequence in (0, 1). Corollary 9. Suppose Θ2 = {𝑥 ∈ EP(𝜓) : 𝐴𝑥 ∈ EP(𝜑)} ≠ 0. = ∞, and If {𝜁𝑛 } satisfies lim𝑛 → ∞ 𝜁𝑛 = 0, ∑∞ 𝑛=1 𝜁𝑛 lim𝑛 → ∞ 𝜁𝑛+1 /𝜁𝑛 = 1, then the sequence {𝜌𝑛 } generated by algorithm (53) converges strongly to 𝑝 = projΘ2(0) which is the mum-norm element in Θ2 . Algorithm 10. Let 𝜌0 ∈ 𝐻1 arbitrarily define a sequence {𝜌𝑛 } by the following: 𝜌𝑛+1 = proj𝐶 ((1 − 𝜁𝑛 ) (𝜌𝑛 + 𝜍𝐴∗ (proj𝑄 − 𝐼) 𝐴𝜌𝑛 )) , (54) for all 𝑛 ∈ N, where 𝜍 ∈ (0, 1/‖𝐴‖2 ) and {𝜁𝑛 } is a real number sequence in (0, 1). Corollary 11. Suppose Θ3 = {𝑥 ∈ 𝐶 : 𝐴𝑥 ∈ 𝑄} ≠ 0. If {𝜁𝑛 } satisfies lim𝑛 → ∞ 𝜁𝑛 = 0, ∑∞ 𝑛=1 𝜁𝑛 = ∞, and lim𝑛 → ∞ 𝜁𝑛+1 /𝜁𝑛 = 1, then the sequence {𝜌𝑛 } generated by algorithm (54) converges strongly to 𝑝 = projΘ3 (0) which is the mum-norm element in Θ3 .
Acknowledgment Li-Jun Zhu was supported in part by NNSF of China (61362033), NZ13087, NGY2012097 and 2013xyz023.
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