An algorithm approach to maximal monotone

Available online at www.tjnsa.com J. Nonlinear Sci. Appl. 9 (2016), 2136–2148 Research Article

An algorithm approach to maximal monotone operators and pseudo-contractions Xinhe Zhua , Zhangsong Yaob , Abdelouahed Hamdic,∗ a

Department of Mathematics, Tianjin Polytechnic University, Tianjin 300387, China.

b

School of Information Engineering, Nanjing Xiaozhuang University, Nanjing 211171, China.

c

Department of Mathematics, Statistics and Physics, College of Arts and Sciences, Qatar University, P. O. Box 2713, Doha, Qatar. Communicated by Y. Yao

Abstract The purpose of this article is to find the minimum norm solution of maximal monotone operators and strict pseudo-contractions in Hilbert spaces. A parallel algorithm is constructed. Some analysis techniques c are used to show the convergence of the presented algorithm. 2016 All rights reserved. Keywords: Maximal monotone operator, strict pseudo-contractions, zero point, fixed point, minimumnorm. 2010 MSC: 49J40, 47J20, 47H09, 65J15.

1. Introduction 1.1. Problem statement Let H be a real Hilbert space. Its inner product and norm are denoted by h·, ·i and k · k, respectively. Let N Ci (i = 1, · · · , N ) be the nonempty TN closed convex subset of H. Suppose the intersection of {Ci }i=1 denoted by D is nonempty, i.e., D = i=1 Ci 6= ∅. The prototype of the problem of image recovery can be stated as follows. The original (unknown) image x† is known a priori to belong D; given only the metric projections PCi recover x† by an iterative scheme. This problem is referred to as the convex feasibility problem (CFP), see for instance [1, 4, 5, 7, 10, 18]. One effective approach for solving (CFP) is algorithmic iteration. In this article, our purpose is to find the common fixed points or/and zero points of two nonlinear mappings by using algorithmic approach. Next, we recall some existing results in the literature. ∗

Corresponding author Email addresses: [email protected] (Xinhe Zhu), [email protected] (Zhangsong Yao), [email protected] (Abdelouahed Hamdi) Received 2015-10-10

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1.2. Existing results P α [β P + (1 − βn )I] where PCn (n = 1, · · · , N ) is the metric projection of H onto Let T = α0 I + N PNn=1 n n Cn Cn , {αn } ⊂ (0, 1), n=1 αn = 1 and {βn } ⊂ (0, 2). Iteration 1.(Picard’s iteration) Initialization x0 ∈ H and iterative step xn+1 = T xn , ∀n ∈ N.

(1.1)

Crombez [8] proved that the sequence {xn } generated by (1.1) converges weakly to an element of (D = T ) N C . i=1 i Let (H ⊃)C 6= ∅ be a closed convex set. Let S, T : C → C be two nonlinear mappings. We use F (S) and F (T ) to denote the set of fixed points of S and T , respectively. In [15], Takahashi and Tamura proved that the following Das and Debata’s iteration converges weakly to x† ∈ (D =)F (S) ∩ F (T ). Iteration 2.(Das and Debata’s iteration [9]) Initialization x1 ∈ C and iterative step xn+1 = βn xn + (1 − βn )S(αn xn + (1 − αn )T xn ), ∀n ∈ N,

(1.2)

where S and T are two nonexpansive mappings and {αn } ⊂ (0, 1) and {βn } ∈ (0, 1) are two iterative parameters. Let A : C → H be an inverse strongly monotone mapping with coefficient α > 0. Let B : H → H be a maximal monotone mapping. we denote by (A + B)−1 0 zero points of A + B, by JλB = (I + λB)−1 the resolvent of B for λ > 0. For finding x† ∈ (D =)F (S) ∩ (A + B)−1 0, Takahashi, Takahashi and Toyoda [14] constructed the following iteration. Iteration 3. Initialization x1 ∈ C and iterative step xn+1 = βn xn + (1 − βn )S(αn x + (1 − αn )JλBn (xn − λn Axn )), ∀n ∈ N,

(1.3)

where the parameters {λn } ⊂ (0, 2α), {αn } ⊂ (0, 1) and {βn } ⊂ (0, 1). Subsequently, Takahashi, Takahashi and Toyoda proved the strong convergence of the sequence {xn } to x† ∈ (D =)F (S) ∩ (A + B)−1 0 provided the domain of B is included in C and the parameters satisfy the conditions: (i) 0 < a ≤ λn ≤ b < 2α and limn→∞ (λn+1 − λn ) = 0; P (ii) limn→∞ αn = 0 and n αn = ∞; (iii) 0 < c ≤ βn ≤ d < 1. One purpose of this article is to extend the above algorithm to a general case in which S is a strict pseudo-contraction. 1.3. Minimization problem At the same time, in the practical problem, it is always needed to find minimum norm solution. A typical example is the least-squares solution to the constrained linear inverse problem ([12]) which is stated below. Example 1.1. Let H1 and H2 be two real Hilbert spaces. Let A : H1 → H2 be a bounded linear operator. For given b ∈ H2 , finding x† ∈ C such that Ax† = b. (1.4) Note that (1.4) can be reduced to solve the minimization problem of finding a point x† ∈ Γ such that x† = arg min kxk. x∈Γ

(1.5)

It is clear that (1.5) is equivalent to x† = PΓ (0). This indicates that one can use projection technique

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to find the minimum norm solution. In this respect, there are a large number references in which the projection technique is applied to find the minimum norm solution of some nonlinear operators, see e.g., [2, 3, 19, 20, 21]. Remark 1.2. We observe that the above algorithm (1.3) can also find the minimum-norm solution x† ∈ (D = )F (S) ∩ (A + B)−1 0 provided 0 ∈ C. However, if 0 6∈ C, then this algorithm (1.3) does not work to find the minimum-norm solution. A natural problem arise in the mind if we can find the minimum norm solution without using the projection technique. This is our another purpose of this article. We will devote to find the minimum norm solution x† ∈ (D =)F (S) ∩ (A + B)−1 0 where S is a strict pseudo-contraction. We suggest the following algorithm: for initialization x0 ∈ C, let the sequence {xn } be generated by un+1 = σn un + ςn Sun + δn JµBn ((1 − ξn )un − µn Aun ), ∀n ≥ 0. We will show the above algorithm converges strongly to x† = PF (S)∩(A+B)−1 0 (0) which is the minimum-norm element in F (S) ∩ (A + B)−1 0.

2. Preliminaries 2.1. Notations Throughout this paper, we assume that H is a real Hilbert space equipped up its inner product h·, ·i and norm k · k. Let (H ⊃)C 6= ∅ be a closed convex set. • A self-mapping Ψ on C is said to be nonexpansive if kΨu − Ψu† k ≤ ku − u† k, for all u, u† ∈ C. • A mapping Ψ : C → C is said to be strictly pseudo-contractive if kΨu − Ψu† k2 ≤ ku − u† k2 + κk(I − Ψ)u − (I − Ψ)u† k2 ,

(2.1)

for all u, u† ∈ C and for some constant κ ∈ [0, 1). In this case, we always say that Ψ is a κ-strict pseudocontraction. Remark 2.1. It is obvious that (2.1) equals hΨu − Ψu† , u − u† i ≤ ku − u† k2 −

1−κ k(I − Ψ)u − (I − Ψ)u† k2 , 2

(2.2)

for all u, u† ∈ C. • A single-valued mapping G : C → H is said to be inverse strongly monotone if hGu − Gu† , u − u† i ≥ ζkGu − Gu† k2

(2.3)

for some ζ > 0 and for all u, u† ∈ C. Subsequently, we call G is ζ-inverse strongly monotone. Remark 2.2. From (2.3), we deduce that kGu − Gu† kku − u† k ≥ hGu − Gu† , u − u† i ≥ ζkGu − Gu† k2 . Thus, kGu − Gu† k ≤ 1/ζku − u† k for all u, u† ∈ C. That is, G is 1/ζ-Lipschitz continuous. Let W : H → 2H be a multi-valued mapping. We denote by dom(W ) the effective domain of W , i.e., dom(W ) = {u ∈ H : W u 6= ∅}. • W is said to be monotone if hx − x† , u − u† i ≥ 0 for all x, x† ∈ dom(W ), u ∈ W x and u† ∈ W x† .

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A monotone mapping W on H is said to be maximal iff its graph is not strictly contained in the graph of any other monotone mapping on H. We use W −1 0 to denote the set of zero points of W , that is, W −1 (0) = {u ∈ H : 0 ∈ W u}, see [6] and [11]. Let W be a maximal monotone mapping on H and let λ be a positive constant. Now we know that (I + λW )−1 , the resolvent of W is a single-valued mapping from H onto dom(W ) which is denoted by JλW . That is, JλW = (I + λW )−1 . 2.2. Lemmas Next, we collect several useful lemmas which will be cited in the next section. Lemma 2.3. Properties of the resolvent JλW are listed as follows. (i) (Firmly-nonexpansive) kJλW u − JλW u† k2 ≤ hJλW u − JλW u† , u − u† i, ∀u, u† ∈ C. (ii) F (JλW ) = W −1 0, ∀λ > 0. (iii) (Resolvent identity) ∀λ > 0 and ∀µ > 0, we have the following identity µ  µ JλW x† = JµW x† + (1 − )JλW x† , ∀x† ∈ H. λ λ

(2.4)

Lemma 2.4 ([22]). Let H be a real Hilbert space. Let (H ⊃)C 6= ∅ be a closed convex set. Let R : C → H be a ρ-strict pseudo-contraction. Set U = γI + (1 − γ)R, ∀γ ∈ (0, 1). Then, F (U ) = F (R) and U is nonexpansive when γ ∈ [ρ, 1). Lemma 2.5 ([16]). Let H be a real Hilbert space. Let (H ⊃)C = 6 ∅ be a closed convex set. Let U : C → H be an inverse strongly monotone mapping with coefficient α > 0. Then, we have k(I − ςU )x − (I − ςU )x† k2 ≤ kx − x† k2 + ς(ς − 2α)kU x − U x† k2 , ∀x, x† ∈ C. Especially, I − ςU is nonexpansive when 0 ≤ ς ≤ 2α. Lemma 2.6 ([22]). Let H be a real Hilbert space. Let (H ⊃)C 6= ∅ be a closed convex set. Let R : C → C be a λ-strict pseudo-contraction. Then I − R is demi-closed at 0, i.e., ( xn * x ∈ C =⇒ x ∈ F (R). xn − Rxn → 0 Lemma 2.7 ([13]). Let X be a Banach space. Let {un } ⊂ X and {vn } ⊂ X be two bounded sequences satisfying un+1 = (1 − δn )vn + δn un , ∀n ≥ 0 where {δn } ⊂ ($1 , $2 ) ⊂ (0, 1) is a real sequence. Then, lim supn→∞ (kvn+1 − vn k − kun+1 − un k) ≤ 0 implies that limn→∞ kun − vn k = 0. Lemma 2.8 ([17]). Let {%n } ⊂ [0, +∞), {ϑn } ⊂ (0, 1) and {ηn } be three real number sequences. Suppose {%n }, {ϑn } and {ηn } satisfy the following three conditions (i) %n+1 ≤ (1 − ϑn )%n + ηn ϑn , P∞ (ii) n=1 ϑn = ∞, P (iii) lim supn→∞ ηn ≤ 0 or ∞ n=1 |ηn ϑn | < ∞. Then limn→∞ %n = 0. 3. Algorithm and Convergence Let H be a real Hilbert space. Let (H ⊃)C 6= ∅ be a closed convex set. Let A : C → H be an inverse strongly monotone mapping with coefficient α > 0 and let B : H → 2H be a maximal monotone mapping

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with its resolvent JµB = (I + µB)−1 . Let S : C → C be a strict pseudo-contraction with coefficient κ ∈ [0, 1). Our objective is to find x† ∈ F (S) ∩ (A + B)−1 such that its norm is minimal in F (S) ∩ (A + B)−1 . Our means is to use algorithmic approach. Now, we first introduce our algorithm. Algorithm 3.1. Initialization x0 ∈ C and iterative step un+1 = σn un + ςn Sun + δn JµBn ((1 − ξn )un − µn Aun ), ∀n ≥ 0,

(3.1)

where {µn } ⊂ (0, 2α) and {ξn }, {σn }, {ςn } and {δn } are four real number sequences in (0, 1). Theorem 3.2. Suppose F (S) ∩ (A + B)−1 0 6= ∅ and dom(B) ⊂ C. Assume that the following restrictions are satisfied. (i) (ii) (iii) (iv) (v)

σn + ςn + δn = 1 for all n ≥ 0, P limn→∞ ξn = 0 and n ξn = ∞, σn ∈ [c, d] ⊂ (κ, 1) and 0 < lim inf n→∞ ςn ≤ lim supn→∞ ςn < 1 − κ, a(1 − ξn ) ≤ µn ≤ b(1 − ξn ) where [a, b] ⊂ (0, 2α) and limn→∞ (µn+1 − µn ) = 0, δn+1 ςn δn n+1 limn→∞ ( 1−κ−σςn+1 +κδn+1 − 1−κ−σn +κδn ) = 0 and limn→∞ ( 1−κ−σn+1 +κδn+1 − 1−κ−σn +κδn ) = 0.

Then the sequence {un } generated by (3.1) converges strongly to x ˜ = PF (S)∩(A+B)−1 0 (0) which is the mini−1 mum norm element in F (S) ∩ (A + B) 0.  Proof. Let any u† ∈ F (S) ∩ (A + B)−1 0. Then, we have u† = JµBn (u† − µn Au† ) = JµBn ξn u† + (1 − ξn )(u† −  µn Au† /(1 − ξn )) for all n ≥ 0. Since JµB is nonexpansive for all µ > 0, we deduce kJµBn ((1 − ξn )un − µn Aun ) − u† k = kJµBn ((1 − ξn )(un − µn Aun /(1 − ξn ))) − JµBn (ξn u† + (1 − ξn )(u† − µn Au† /(1 − ξn )))k ≤ k((1 − ξn )(un − µn Aun /(1 − ξn ))) − (ξn u† + (1 − ξn )(u† − µn Au† /(1 − ξn )))k †

†

(3.2)

†

= k(1 − ξn )((un − µn Aun /(1 − ξn )) − (u − µn Au /(1 − ξn ))) + ξn (−u )k. Noting that the norm k · k is convex and the mapping A is α-inverse strongly monotone, we obtain k(1 − ξn )((un − µn Aun /(1 − ξn )) − (u† − µn Au† /(1 − ξn ))) + ξn (−u† )k2 ≤ (1 − ξn )k(un − µn Aun /(1 − ξn )) − (u† − µn Au† /(1 − ξn ))k2 + ξn ku† k2 = (1 − ξn )k(un − u† ) − µn (Aun − Au† )/(1 − ξn )k2 + ξn ku† k2   2µn µ2n † 2 = (1 − ξn ) kun − u† k2 − hAun − Au† , un − u† i + kAu − Au k n 1 − ξn (1 − ξn )2

(3.3)

+ ξn ku† k2   2αµn µ2n † 2 ≤ (1 − ξn ) kun − u† k2 − kAun − Au† k2 + kAu − Au k + ξn ku† k2 n 1 − ξn (1 − ξn )2   µn † 2 = (1 − ξn ) kun − u† k2 + (µ − 2(1 − ξ )α)kAu − Au k + ξn ku† k2 . n n n (1 − ξn )2 By condition (iv), we derive that µn − 2(1 − ξn )α ≤ 0 for all n ≥ 0. Thus, from (3.2) and (3.3), we obtain kJµBn ((1 − ξn )un − µn Aun ) − u† k2   µn † 2 † 2 (µn − 2(1 − ξn )α)kAun − Au k + ξn ku† k2 ≤ (1 − ξn ) kun − u k + (1 − ξn )2 ≤ (1 − ξn )kun − u† k2 + ξn ku† k2 .

(3.4)

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Applying (2.1) and (2.2), we obtain kσn (un − u† ) + ςn (Sun − u† )k2 = σn2 kun − u† k2 + ςn2 kSun − u† k2 + 2σn ςn hSun − u† , un − u† i ≤ σn2 kun − u† k2 + ςn2 [kun − u† k2 + κkun − Sun k2 ] 1−κ + 2σn ςn [kun − u† k2 − kun − Sun k2 ] 2 = (σn + ςn )2 kun − u† k2 + [ςn2 κ − (1 − κ)σn ςn ]kun − Sun k2

(3.5)

= (σn + ςn )2 kun − u† k2 + ςn [(σn + ςn )κ − σn ]kun − Sun k2 ≤ (σn + ςn )2 kun − u† k2 , which implies that kσn (un − u† ) + ςn (Sun − u† )k ≤ (σn + ςn )kun − u† k.

(3.6)

Note that I − µn A/(1 − ξn ) is nonexpansive by Lemma 2.5. From (3.2), we have kJµBn ((1 − ξn )un − µn Aun ) − u† k ≤ k(1 − ξn )((un − µn Aun /(1 − ξn )) − (u† − µn Au† /(1 − ξn ))) + ξn (−u† )k

(3.7)

≤ (1 − ξn )k(un − µn Aun /(1 − ξn )) − (u† − µn Au† /(1 − ξn ))k + ξn ku† k ≤ (1 − ξn )kun − u† k + ξn ku† k. By (3.1) and (3.7), we get kun+1 − u† k = kσn (un − u† ) + ςn (Sun − u† ) + δn (JµBn ((1 − ξn )un − µn Aun ) − u† )k ≤ kσn (un − u† ) + ςn (Sun − u† )k + δn kJµBn ((1 − ξn )un − µn Aun ) − u† k ≤ (σn + ςn )kun − u† k + δn (1 − ξn )kun − u† k + ξn ku† k = (1 − δn ξn )kun − u† k + δn ξn ku† k ≤ max{kun − u† k, ku† k}. By induction, we have kun+1 − u† k ≤ max{kx0 − u† k, ku† k}. So, {un } is bounded. We also deduce that {Aun } is bounded according to the Lipschitzian continuity of A. Set xn = (1 − ξn )un − µn Aun and yn = JµBn xn for all n ≥ 0. It is easy to see that {xn }, {JµBn un }, {Sun } and {yn } are all bounded. (3.1) can be rewritten as     κςn ςn un+1 = σn − un + κun + (1 − κ)Sun + δn yn , for all n ≥ 0. (3.8) 1−κ 1−κ Observe that (σn −

κςn ςn )+ + δn = 1 1−κ 1−κ

and 0 < lim inf (σn − n→∞

Set un+1 = (σn −

κςn 1−κ )un

+ (1 − σn +

zn+1 − zn =

κςn κςn ) ≤ lim sup(σn − ) < 1. 1−κ 1−κ n→∞

κςn 1−κ )zn

κςn+1 1−κ )un+1 κςn+1 σn+1 + 1−κ

un+2 − (σn+1 − 1−

for all n ≥ 0. It follows that −

κςn un+1 − (σn − 1−κ )un κςn 1 − σn + 1−κ

X. Zhu, Z. Yao, A. Hamdi, J. Nonlinear Sci. Appl. 9 (2016), 2136–2148 =

ςn+1 1−κ (κun+1

2142

+ (1 − κ)Sun+1 ) + δn+1 yn+1 1 − σn+1 +

κςn+1 1−κ

ςn 1−κ (κun

+ (1 − κ)Sun ) + δn yn κςn 1 − σn + 1−κ   δn+1 (yn+1 − yn ) δn δn+1 = + − yn κςn n+1 n+1 1 − σn + 1−κ 1 − σn+1 + κς1−κ 1 − σn+1 + κς1−κ   ςn+1 1−κ + κ(un+1 − un ) + (1 − κ)(Sun+1 − Sun ) n+1 1 − σn+1 + κς1−κ   ςn+1 ςn 1−κ 1−κ + − (κun + (1 − κ)Sun ). κςn n+1 1 − σn + 1−κ 1 − σn+1 + κς1−κ −

Applying Lemma 2.4, we deduce that κI + (1 − κ)S is nonexpansive. Hence, we get δn+1 kyn+1 − yn k δn+1 δn kzn+1 − zn k ≤ κςn+1 + κςn+1 − κςn kyn k 1 − σn + 1−κ 1 − σn+1 + 1−κ 1 − σn+1 + 1−κ ςn+1 1−κ

kκ(un+1 − un ) + (1 − κ)(Sun+1 − Sun )k n+1 1 − σn+1 + κς1−κ ςn+1 ςn 1−κ 1−κ + κςn+1 − κςn kκun + (1 − κ)Sun k 1 − σn + 1−κ 1 − σn+1 + 1−κ δn δn+1 δn+1 kyn+1 − yn k ≤ κςn+1 + κςn+1 − κςn kyn k 1 − σn + 1−κ 1 − σn+1 + 1−κ 1 − σn+1 + 1−κ ςn+1 ςn 1−κ 1−κ + κςn+1 − κςn kκun + (1 − κ)Sun k 1 − σn + 1−κ 1 − σn+1 + 1−κ +

+

ςn+1 1−κ

1 − σn+1 +

κςn+1 1−κ

kun+1 − un k.

Notice that kyn+1 − yn k = kJµBn+1 xn+1 − JµBn xn k ≤ kJµBn+1 xn+1 − JµBn+1 xn k + kJµBn+1 xn − JµBn xn k ≤ kxn+1 − xn k + kJµBn+1 xn − JµBn xn k = k(I − µn+1 A)un+1 − (I − µn+1 A)un + (µn − µn+1 )Aun + ξn un − ξn+1 un+1 k + kJµBn+1 xn − JµBn xn k ≤ k(I − µn+1 A)un+1 − (I − µn+1 A)un k + |µn+1 − µn |kAun k + ξn kun k + ξn+1 kun+1 k + kJµBn+1 xn − JµBn xn k. By the assumption, µn+1 ∈ (0, 2α). Thus I − µn+1 A is nonexpansive. So, we have k(I − µn+1 A)un+1 − (I − µn+1 A)un k ≤ kun+1 − un k. Applying the property of the resolvent (2.4), we have   µ µn n xn + (1 − )JµBn+1 xn . JµBn+1 xn = JµBn µn+1 µn+1 It follows that

 µ  µn

n kJµBn+1 xn − JµBn xn k = JµBn xn + (1 − )JµBn+1 xn − JµBn xn µn+1 µn+1

(3.9)

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 µ  µn

n xn + (1 − )JµBn+1 xn − xn ≤ µn+1 µn+1 |µn+1 − µn | ≤ kxn − JµBn+1 xn k. µn+1 So, kyn+1 − yn k ≤kun+1 − un k + |µn+1 − µn |kAun k + ξn kun k + ξn+1 kun+1 k |µn+1 − µn | + kxn − JµBn+1 xn k. µn+1 Substituting the last inequality into (3.9) to deduce kzn+1 − zn k

ςn+1 1−κ ≤ kun+1 n+1 1 − σn+1 + κς1−κ ςn+1 1−κ − + n+1 1 − σn+1 + κς1−κ

δn+1 +

δn δn+1 − un k + κςn+1 − 1 − σn + 1 − σn+1 + 1−κ ςn 1−κ κςn kκun + (1 − κ)Sun k 1 − σn + 1−κ

+ |µn+1 − µn |kAun k + ξn kun k + ξn+1 kun+1 k + Since

δn+1 1 − σn+1 +

κςn+1 1−κ

ςn+1 1−κ

1 − σn+1 +

κςn+1 1−κ

−

−

δn 1 − σn +

1−

κςn 1−κ

ςn 1−κ

1 − σn +

ςn+1 1−κ n+1 σn+1 + κς1−κ

δn+1 +

0
0, we obtain lim sup(kzn+1 − zn k − kun+1 − un k) ≤ 0. n→∞

This together with Lemma 2.7 imply that lim kzn − un k = 0.

n→∞

Therefore, lim kun+1 − un k = lim (1 − σn +

n→∞

n→∞

κςn )kzn − un k = 0. 1−κ

From (3.3) and (3.8), we have κςn ςn )kun − u† k2 + kκun + (1 − κ)Sun − u† k2 1−κ 1−κ + δn kJµBn xn − u† k2

kun+1 − u† k2 ≤ (σn −

≤ (σn + ςn )kun − u† k2 + δn kJµBn xn − u† k2    µn † 2 ≤ δn (1 − ξn ) (µn − 2(1 − ξn )α)kAun − Au k (1 − ξn )2  † 2 † 2 + kun − u k + ξn ku k + (σn + ςn )kun − u† k2

(3.11)

X. Zhu, Z. Yao, A. Hamdi, J. Nonlinear Sci. Appl. 9 (2016), 2136–2148 = (1 − δn ξn )kun − u† k2 +

2144

δn µn (µn − 2(1 − ξn )α)kAun − Au† k2 (1 − ξn )2

+ δn ξn ku† k2 . It follows that δn µn (2(1 − ξn )α − µn )kAun − Au† k2 (1 − ξn )2 ≤ kun − u† k2 − kun+1 − u† k2 + δn ξn ku† k2 ≤ (kun − u† k − kun+1 − u† k)kun+1 − un k + δn ξn ku† k2 . Since limn→∞ ξn = 0, limn→∞ kun+1 − un k = 0 and lim inf n→∞

δn µ n (2(1 (1−ξn )2

− ξn )α − µn ) > 0, we deduce

lim kAun − Au† k = 0.

(3.12)

n→∞

Using Lemma 2.3, we have kJµBn xn − u† k2 = kJµBn xn − JµBn (u† − µn Au† )k2 ≤ hxn − (u† − µn Au† ), JµBn xn − u† i  1 = kxn − (u† − µn Au† )k2 + kJµBn xn − u† k2 2 †

JµBn xn k2



− k(1 − ξn )un − µn (Aun − µn Au ) −  1 ≤ (1 − ξn )kun − u† k2 + ξn ku† k2 + kJµBn xn − u† k2 2  B † 2 − k(1 − ξn )un − Jµn xn − µn (Aun − µn Au )k . It follows that kJµBn xn − u† k2 ≤ (1 − ξn )kun − u† k2 + ξn ku† k2 − k(1 − ξn )un − JµBn xn − µn (Aun − µn Au† )k2 = (1 − ξn )kun − u† k2 + ξn ku† k2 − k(1 − ξn )un − JµBn xn k2 + 2µn h(1 − ξn )un − JµBn xn , Aun − Au† i − µ2n kAun − Au† k2 ≤ (1 − ξn )kun − u† k2 + ξn ku† k2 − k(1 − ξn )un − JµBn xn k2 + 2µn k(1 − ξn )un − JµBn xn kkAun − Au† k. This together with (3.11) imply that kun+1 − u† k2 ≤ (σn + ςn )kun − u† k2 + δn (1 − ξn )kun − u† k2 + δn ξn ku† k2 − δn k(1 − ξn )un − JµBn xn k2 + 2µn δn k(1 − ξn )un − JµBn xn kkAun − Au† k = (1 − δn ξn )kun − u† k2 + δn ξn ku† k2 − δn k(1 − ξn )un − JµBn xn k2 + 2µn δn k(1 − ξn )un − JµBn xn kkAun − Au† k. Hence, δn k(1 − ξn )un − JµBn xn k2 ≤ kun − u† k2 − kun+1 − u† k2 − δn ξn kun − u† k2 + δn ξn ku† k2

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+ 2µn δn k(1 − ξn )un − JµBn xn kkAun − Au† k ≤ (kun − u† k + kun+1 − u† k)kun+1 − un k + δn ξn ku† k2 + 2µn δn k(1 − ξn )un − JµBn xn kkAun − Au† k. By assumptions (i) and (iii), we get lim supn→∞ δn < 1. Since kun+1 −un k → 0, ξn → 0 and kAun −Au† k → 0, we deduce lim k(1 − ξn )un − JµBn xn k = 0. n→∞

This implies that lim kun − yn k = lim kun − JµBn ((1 − ξn )un − µn Aun )k = 0.

n→∞

n→∞

(3.13)

Observe that kun − Sun k ≤ kun+1 − un k + kun+1 − Sun k ≤ kun+1 − un k + σn kun − Sun k + δn kyn − un k. Then, 1 (kun+1 − un k + δn kyn − un k) 1 − σn → 0.

kun − Sun k ≤

(3.14)

Since F (S) ∩ (A + B)−1 0 is convex, PF (S)∩(A+B)−1 0 (0) exists and is unique which is denoted by x ˜, i.e., x ˜ = PF (S)∩(A+B)−1 0 (0). µn Set vn = un − 1−ξ (Aun − A˜ x) for all n ≥ 0. In (3.12), we choose u† = x ˜. From (3.12), we get n kAun − A˜ xk → 0. Next, we first prove lim supn→∞ h˜ x, vn − x ˜i ≥ 0. Let {vni } be a subsequence of {vn } such that lim suph˜ x, vn − x ˜i = lim h˜ x, vni − x ˜i. i→∞

n→∞

Since {un } is bounded and kAun − A˜ xk → 0, we deduce that {vn } is bounded. Thus, there exists a subsequence {vnij } of {vni } such that vnij * w ∈ C. It is easy to check that {unij } and {ynij } also converge weakly to w. From (3.14), we have lim kunij − Sunij k = 0. (3.15) j→∞

Applying Lemma 2.6 to (3.15), we deduce w ∈ F (S). Let v ∈ Bu. Then, we have (1 − ξn )un − µn Aun ∈ (I + µn B)yn ⇒

1 − ξn yn un − Aun − ∈ Byn . µn µn

Since B is monotone, we have, for (u, v) ∈ B, D1 − ξ

E yn − v, yn − u ≥ 0 µn µn ⇒h(1 − ξn )un − µn Aun − yn − µn v, yn − ui ≥ 0 1 ξn hun − yn , yn − ui − hun , yn − ui ⇒hAun + v, yn − ui ≤ µn µn 1 ξn ⇒hAw + v, yn − ui ≤ hun − yn , yn − ui − hun , yn − ui + hAw − Aun , yn − ui µn µn ξn 1 ⇒hAw + v, yn − ui ≤ kun − yn kkyn − uk + kun kkyn − uk + kAw − Aun kkyn − uk. µn µn n

un − Aun −

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It follows that hAw + v, w − ui ≤

1 ξn kun − yn kkyn − uk + kun kkyn − uk µn µn + kAw − Aun kkyn − uk + hAw + v, w − yn i.

(3.16)

Since Aun is strongly convergent and un is weakly convergent to w, we have limn→∞ hun − w, Aun − Awi = 0. By the inverse strong monotonicity of A, we have αkAun − Awk2 ≤ hun − w, Aun − Awi → 0. So, Aun → Aw. Hence, from (3.16), we derive hAw + v, w − ui ≤ 0. By the maximal monotonicity of B, we obtain immediately that −Aw ∈ Bw. Therefore, 0 ∈ (A + B)w. Hence, we have w ∈ F (S) ∩ (A + B)−1 0. Noting that x ˜ = PF (S)∩(A+B)−1 0 (0), we get h˜ x, w − x ˜i ≥ 0. Therefore, lim suph˜ x, vn − x ˜i = lim h˜ x, vnij − x ˜i = h˜ x, w − x ˜i ≥ 0. j→∞

n→∞

From (3.11), we have kun+1 − x ˜k2 ≤ (σn + ςn )kun − x ˜k2 + δn kJµBn xn − x ˜k2

2  

µn µn

≤ δn (1 − ξn ) (un − ˜ Aun ) − (˜ x− A˜ x ) − ξn x

1 − ξn 1 − ξn + (σn + ςn )kun − x ˜ k2

2 

µn µn 2 Aun ) − (˜ x− A˜ x) = δn (1 − ξn ) (un −

1 − ξn 1 − ξn    µn µn 2 2 − 2ξn (1 − ξn ) x ˜, (un − Aun ) − (˜ x− A˜ x) + ξn k˜ xk 1 − ξn 1 − ξn

(3.17)

+ (σn + ςn )kun − x ˜ k2  ≤ (σn + ςn )kun − x ˜k + δn (1 − ξn )2 kun − x ˜ k2    µn 2 2 − 2ξn (1 − ξn ) x ˜, un − (Aun − A˜ x) − x ˜ + ξn k˜ xk 1 − ξn 2

≤ (1 − δn ξn )kun − x ˜k2 + δn ξn [−2(1 − ξn )h˜ x, vn − x ˜i + ξn k˜ xk2 ]. P It is easy to check that n (1 − σn )ξn = ∞ and lim supn→∞ (−2(1 − ξn )h˜ x, vn − x ˜i + ξn k˜ xk2 ) ≤ 0. Applying Lemma 2.8 to (3.17), we conclude that un → x ˜. This completes the proof. Corollary 3.3. Let H be a real Hilbert space. Let (H ⊃)C = 6 ∅ be a closed convex set. Let A : C → H be an inverse strongly monotone mapping with coefficient α > 0 and let B : H → 2H be a maximal monotone mapping such that dom(B) ⊂ C. Let S : C → C be a nonexpansive mapping. Suppose F (S)∩(A+B)−1 0 6= ∅. For given x0 ∈ C, let the sequence {un } ⊂ C be generated by un+1 = σn un + ςn Sun + δn JµBn ((1 − ξn )un − µn Aun ), ∀n ≥ 0,

(3.18)

where {µn } ⊂ (0, 2α), {ξn } ⊂ (0, 1), {σn } ⊂ (0, 1), {ςn } ⊂ (0, 1) and {δn } ⊂ (0, 1) satisfy the following conditions (i) σn + ςn + δn = 1 for all n ≥ 0;

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P limn→∞ ξn = 0 and n ξn = ∞; σn ∈ [c, d] ⊂ (0, 1) and 0 < lim inf n→∞ ςn ≤ lim supn→∞ ςn < 1; a(1 − ξn ) ≤ µn ≤ b(1 − ξn ) where [a, b] ⊂ (0, 2α) and limn→∞ (µn+1 − µn ) = 0; ςn+1 δn+1 ςn δn limn→∞ ( 1−σ − 1−σ ) = 0 and limn→∞ ( 1−σ − 1−σ ) = 0. n n n+1 n+1

Then the sequence {un } generated by (3.18) converges strongly to x ˜ = PF (S)∩(A+B)−1 0 (0). Corollary 3.4. Let H be a real Hilbert space. Let (H ⊃)C = 6 ∅ be a closed convex set. Let A : C → H be an inverse strongly monotone mapping with coefficient α > 0 and let B : H → 2H be a maximal monotone mapping such that dom(B) ⊂ C. Let JµB = (I + µB)−1 be the resolvent of B for µ. Suppose (A + B)−1 0 6= ∅. For given x0 ∈ C, let the sequence {un } ⊂ C be generated by un+1 = σn un + (1 − σn )JµBn ((1 − ξn )un − µn Aun ), ∀n ≥ 0,

(3.19)

where {µn } ⊂ (0, 2α), {ξn } ⊂ (0, 1) and {σn } ⊂ (0, 1) satisfy the following conditions P (i) limn→∞ ξn = 0 and n ξn = ∞; (ii) σn ∈ [c, d] ⊂ (0, 1); (iii) a(1 − ξn ) ≤ µn ≤ b(1 − ξn ) where [a, b] ⊂ (0, 2α) and limn→∞ (µn+1 − µn ) = 0. Then {un } generated by (3.19) converges strongly to x ˜ = P(A+B)−1 0 (0).

Acknowledgments Abdelouahed Hamdi would like to thank Qatar University for providing excellent research facilities under the Grant: QUUG-CAS-DMSP-14/15-4. Zhangsong Yao was supported by the Scientific Research Project of Nanjing Xiaozhuang University (2015NXY46). References [1] H. H. Bauschke, J. M. Borwein, On projection algorithms for solving convex feasibility problems, SIAM Rev., 38 (1996), 367–426. 1.1 [2] L. C. Ceng, Q. H. Ansari, J. C. Yao, Relaxed extragradient methods for finding minimum-norm solutions of the split feasibility problem, Nonlinear Anal., 75 (2012), 2116–2125. 1.3 [3] L. C. Ceng, S. M. Guu, J. C. Yao, Hybrid methods with regularization for minimization problems and asymptotically strict pseudocontractive mappings in the intermediate sense, J. Global Optim., 60 (2014), 617–634. 1.3 [4] Y. Censor, Parallel application of block-iterative methods in medical imaging and radiation therapy, Math. Programming, 42 (1988), 307–325. 1.1 [5] S. S. Chang, J. Kim, Y. J. Cho, J. Sim, Weak and strong convergence theorems of solutions to split feasibility problem for nonspreading type mapping in Hilbert spaces, Fixed Point Theory Appl., 2014 (2014), 12 pages. 1.1 [6] Y. J. Cho, Narin Petrot, Regularization method for Noor’s variational inequality problem induced by a hemicontinuous monotone operator, Fixed Point Theory Appl., 2012 (2012), 13 pages. 2.1 [7] P. L. Combettes, Hilbertian convex feasibility problem: Convergence of projection methods, Appl. Math. Optim., 35 (1997), 311–330. 1.1 [8] G. Crombez, Image recovery by convex combinations of projections, J. Math. Anal. Appl., 155 (1991), 413–419. 1.2 [9] G. Das, J. P. Debata, Fixed points of quasi-nonexpansive mappings, Indian J. Pure Appl. Math., 17 (1986), 1263–1269. 1.2 [10] A. Hamdi, Y. C. Liou, Y. Yao, C. Luo, The common solutions of the split feasibility problems and fixed point problems, J. Inequal. Appl., 2015 (2015), 17 pages. 1.1 [11] H. Pathak, Y. J. Cho, Strong convergence of a proximal-type algorithm for an occasionally pseudomonotone operator in Banach spaces, Fixed Point Theory Appl., 2012 (2012), 13 pages. 2.1 [12] A. Sabharwal, L. C. Potter, Convexly constrained linear inverse problems: iterative least-squares and regularization, IEEE Trans. Signal Process., 46 (1998), 2345–2352. 1.3 [13] T. Suzuki, Strong convergence theorems for infinite families of nonexpansive mappings in general Banach spaces, Fixed Point Theory Appl., 2005 (2005), 103–123. 2.7

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