DSpace at VNU: Parallel methods for regularizing systems of equations involving accretive operators

DSpace at VNU: Parallel methods for regularizing systems of equations involving accretive operators tài liệu, giáo án, b...

This article was downloaded by: [University of Newcastle (Australia)] On: 28 September 2014, At: 08:58 Publisher: Taylor & Francis Informa Ltd Registered in England and Wales Registered Number: 1072954 Registered office: Mortimer House, 37-41 Mortimer Street, London W1T 3JH, UK Applicable Analysis: An International Journal Publication details, including instructions for authors and subscription information: http://www.tandfonline.com/loi/gapa20 Parallel methods for regularizing systems of equations involving accretive operators a b Pham Ky Anh , Nguyen Buong & Dang Van Hieu a a Department of Mathematics, Vietnam National University, Hanoi, 334 Nguyen Trai, Thanh Xuan, Hanoi 10000, Vietnam b Vietnamese Academy of Science & Technology, Institute of Information Technology, 18 Hoang Quoc Viet, Cau Giay, Hanoi 10000, Vietnam Published online: 20 Jan 2014 To cite this article: Pham Ky Anh, Nguyen Buong & Dang Van Hieu (2014) Parallel methods for regularizing systems of equations involving accretive operators, Applicable Analysis: An International Journal, 93:10, 2136-2157, DOI: 10.1080/00036811.2013.872777 To link to this article: http://dx.doi.org/10.1080/00036811.2013.872777 PLEASE SCROLL DOWN FOR ARTICLE Taylor & Francis makes every effort to ensure the accuracy of all the information (the “Content”) contained in the publications on our platform However, Taylor & Francis, our agents, and our licensors make no representations or warranties whatsoever as to the accuracy, completeness, or suitability for any purpose of the Content Any opinions and views expressed in this publication are the opinions and views of the authors, and are not the views of or endorsed by Taylor & Francis The accuracy of the Content should not be relied upon and should be independently verified with primary sources of information Taylor and Francis shall not be liable for any losses, actions, claims, proceedings, demands, costs, expenses, damages, and other liabilities whatsoever or howsoever caused arising directly or indirectly in connection with, in relation to or arising out of the use of the Content This article may be used for research, teaching, and private study purposes Any substantial or systematic reproduction, redistribution, reselling, loan, sub-licensing, systematic supply, or distribution in any form to anyone is expressly forbidden Terms & Downloaded by [University of Newcastle (Australia)] at 08:58 28 September 2014 Conditions of access and use can be found at http://www.tandfonline.com/page/termsand-conditions Applicable Analysis, 2014 Vol 93, No 10, 2136–2157, http://dx.doi.org/10.1080/00036811.2013.872777 Parallel methods for regularizing systems of equations involving accretive operators Downloaded by [University of Newcastle (Australia)] at 08:58 28 September 2014 Pham Ky Anha∗ , Nguyen Buongb and Dang Van Hieua a Department of Mathematics, Vietnam National University, Hanoi, 334 Nguyen Trai, Thanh Xuan, Hanoi 10000, Vietnam; b Vietnamese Academy of Science & Technology, Institute of Information Technology, 18 Hoang Quoc Viet, Cau Giay, Hanoi 10000, Vietnam Communicated by Prof Boris Mordukhovich (Received 21 September 2013; accepted 22 November 2013) In this paper, two parallel methods for solving systems of accretive operator equations in Banach spaces are studied The convergence analysis of the methods in both free-noise and noisy data cases is provided Keywords: uniformly smooth and uniformly convex Banach spaces; accretive and inverse uniformly accretive operators; iterative regularization method; Newton-type method; parallel computation AMS Subject Classifications: 47J06; 47J25; 65J15; 65J20; 65Y05 Introduction, preliminaries, and notations Various problems of science and engineering, including a multi-parameter identification problem, the convex feasibility problem, a common fixed point problem, etc., lead to a system of ill-posed operator equations Ai (x) = 0, x ∈ X, (i = 1, 2, , N ), (1) where X is a real Banach space and Ai : D(Ai ) = X → X are possibly nonlinear operators on X Very recently, several sequential and parallel regularizing methods for solving system (1) have been proposed The Kaczmarz method,[1,2] the Newton-Kacmarz method,[3] the steepest-descent-Kaczmarz method,[4] parallel iterative regularization methods,[5] parallel regularized Newton-type methods,[6,7] and parallel hybrid methods,[8] to name only few However, most of the investigation of available methods was carried out in the framework of Hilbert spaces In this paper, we study parallel methods extended to system (1) involving m-accretive operators in the setting of Banach spaces In the sequel, we always assume that system (1) is consistent, i.e the solution set S of (1) is not empty It is known that if Ai (i = 1, , N ) are not strongly or uniformly accretive, then system (1) in general is ill-posed, i.e the ∗ Corresponding author Email: anhpk@vnu.edu.vn © 2013 Taylor & Francis Applicable Analysis 2137 solution set S of (1) may not depend continuously on data In that case, a process known as regularization should be applied for stable solution of (1) In what follows, for the reader’s convenience, we collect some definitions and results concerning the geometry of Banach spaces and accretive operators, which are used in this paper We refer the reader to [9–13] for more details Definition 1.1 A Banach space X is called Downloaded by [University of Newcastle (Australia)] at 08:58 28 September 2014 (1) (2) strictly convex if the unit sphere S1 (0) = {x ∈ X : ||x|| = 1} is strictly convex, i.e the inequality ||x + y|| < holds for all x, y ∈ S1 (0), x = y; uniformly convex if for any given > there exists δ = δ( ) > such that for all x, y ∈ X with x ≤ 1, y ≤ 1, x − y = the inequality x + y ≤ 2(1 − δ) holds The modulus of convexity of X is defined by δ X ( ) = inf − x−y : x = y = 1, x − y = The modulus of smoothness of X is defined by ρ X (τ ) = sup x+y + x−y − : x = 1, y = τ Definition 1.2 A Banach space X is called uniformly smooth if lim h X (τ ) := lim τ →0 τ →0 ρ X (τ ) = τ Observe that if X is a real uniformly convex and uniformly smooth Banach space, then the modulus of convexity δ X is a continuous and strictly increasing function on the whole segment [0, 2] (see, e.g [14]) Definition 1.3 A Banach space X possesses the approximation if there exists a directed family of finite dimensional subspaces X n ordered by inclusion and a corresponding family of projectors Pn : X → X n , such that ||Pn || = for all n ≥ and ∪n X n is dense in X Let X be a real Banach space with the dual space X ∗ Throughout this paper, we assume that the so-called normalized duality mapping J : X → X ∗ , satisfying the relation x, J (x) = x = J (x) , ∀x ∈ X, is single-valued This assumption will be fulfilled if X is smooth A Banach space X is said to have a uniformly Gateaux differetiable norm if for every is attained uniformly for x ∈ S1 (0) It is well known y ∈ S1 (0) the limit limt→0 ||x+t y||−||x|| t (see [11]) that if the norm of X is uniformly Gateaux differentiable, then the normalized duality mapping is single-valued and norm to weak star uniformly continuous on every bounded subset of X 2138 P.K Anh et al For the sake of simpicity, we will denote norms of both spaces X and X ∗ by the same symbol The dual product of f ∈ X ∗ and x ∈ X will be denoted by x, f or f, x Besides, we put R+ := (0, ∞), R+ ∗ := [0, ∞) Definition 1.4 An operator A : X → X is called (1) accretive, if A(x) − A(y), J (x − y) ≥ Downloaded by [University of Newcastle (Australia)] at 08:58 28 September 2014 (2) (3) (4) ∀x, y ∈ X ; maximal accretive, if it is accretive and its graph is not the right part of the graph of any other accretive operator; m-accretive, if it is accretive and R(A + α I ) = X for all α > 0, where I is the identity operator in X ; uniformly accretive, if there exists a strictly increasing function ψ : R+ ∗ → , ψ(0) = 0, such that R+ ∗ A(x) − A(y), J (x − y) ≥ ψ(||x − y||) (5) (6) ∀x, y ∈ X ; (2) strongly accretive, if there exists a positive constant c, such that in (2), ψ(t) = ct ; inverse strongly accretive, if there exists a positive constant c, such that A(x) − A(y), J (x − y) ≥ c||A(x) − A(y)||2 ∀x, y ∈ X If X is a Hilbert space then J is an identity operator and accretive operators are also called monotone Definition 1.5 A continuous operator A mapping a Banach space X into itself is called ϕ-inverse uniformly accretive (or simply, inverse uniformly accretive), if there exists a + function ϕ : R+ × R+ ∗ → R∗ , which is continuous and strictly increasing in the second variable and ϕ(s, t) = if and only if t = for every fixed s > 0, such that A(x) − A(y), J (x − y) ≥ ϕ (R, A(x) − A(y)) ) ∀x, y ∈ X, x , y ≤ R, ∀R > (3) Example 1.6 Any inverse strongly accretive operator is inverse uniformly accretive, hence, is accretive Indeed, let A be a c-inverse strongly accretive operator Then, A is Lipschitz continuous with the Lipschit constant c−1 and the inequality (3) holds with the function ϕ(s, t) = ct Example 1.7 Let T be a nonexpansive operator on a uniformly convex and uniformly smooth Banach space X Then A := I − T is a Lipschitz continuous operator Moreover, according to Alber [15], A(x) − A(y), J (x − y) ≥ L −1 R δ X A(x) − A(y) 4R ∀x, y ∈ X, x , y ≤ R, where L ∈ (1; 1.7) is the Figiel constant and δ X ( ) is the modulus of the convexity of X Observe that := A(x)−A(y) ≤ for any x, y ∈ X ; x , y ≤ R and inequality (3) 4R holds with the function ϕ(s, t) = L −1 s δ X 4st , s ∈ R+ ; t ∈ [0; 2s] Applicable Analysis 2139 Example 1.8 Now let X in Example 1.7 be one of the following Banach spaces L p , l p , W pm , where < p < ∞ Then X is uniformly smooth and uniformly convex, and it is well known that (see [9]) p−1 t , < p < 2; 16 p δ X (t) ≥ t , p ≥ p2 p δ X (t) ≥ Downloaded by [University of Newcastle (Australia)] at 08:58 28 September 2014 Thus, for all x, y ∈ X, x , y ≤ R, one gets p−1 A(x) − A(y) , < p < 2; 256L A(x) − A(y) p A(x) − A(y), J (x − y) ≥ , p ≥ p L p.8 R p−2 A(x) − A(y), J (x − y) ≥ So, the operator A = I − T, where T : l p → l p is a nonexpansive operator, is inverse p strongly accretive if < p < 2, and is inverse uniformly accretive with ϕ(s, t) = pL8tp s p−2 , if p ≥ Definition 1.9 An operator, B : D(B) ⊂ X → X is called Hemicontinuous at a point x0 ∈ D(B), if B(x0 + tn h) x0 as tn → for any vector h, such that x0 + tn h ∈ D(B) and ≤ tn ≤ t (x0 ) x0 implies that B(x) B(x0 ) (2) Weakly continuous at x0 ∈ D(B), if D(B) x (1) If B is hemicontinuous (weakly continuous) at every point of D(B), then B is said to be hemicontinuous (weakly continuous), respectively For regularizing accretive operator equations one needs the following fact.[9] Lem m a 1.10 Suppose that the Banach space X possesses the approximation, A : X → X is a hemicontinuous accretive operator with D(A) = X, and the normalized duality mapping J : X → X ∗ is sequentially weakly continuous and continuous Then the problem A(x) + αx = y, (4) where α is a fixed positive parameter and y ∈ X , is well-posed The unique solvability of (4) is established in [9] The continuous dependence of the solution xα of (4) on the right-hand side y follows from the inequality ||x α,1 − xα,2 || ≤ ||y1 −y2 || , where xα,i are the unique solution of (4) with respect to the right-hand side α y = yi , i = 1, The next five lemmas will be used in Section for establishing the convergence of implicit and explicit parallel iterative regularization methods Lem m a 1.11 [16] Let X be a real uniformly smooth Banach space Then for any x, y ∈ X such that x ≤ R, y ≤ R, the following inequality holds: J (x) − J (y) ≤ 8Rh X 16L x − y R , 2140 P.K Anh et al where L is Figiel constant, (1 < L < 1.7) Lem m a 1.12 [9] x If X is a real uniformly smooth Banach space, then the inequality ≤ y + x − y, J (x) ≤ y + x − y, J (y) + x − y, J (x) − J (y) holds for every x, y ∈ X Let X be a uniformly smooth Banach space Then for x, y ∈ X Downloaded by [University of Newcastle (Australia)] at 08:58 28 September 2014 Lem m a 1.13 [9] x − y, J (x) − J (y) ≤ x − y + C ( x , y ) ρ X ( x − y ), where C ( x , y ) ≤ max {2L , x + y } In a uniformly smooth Banach space X , for x, y ∈ X, Lem m a 1.14 [9] x − y, J (x) − J (y) ≤ R ( x , y )ρ X where R( x , y ) = 2−1 ( x If x ≤ R, y ≤ R, then x−y , R( x , y ) + y ) x − y, J (x) − J (y) ≤ 2L R ρ X x−y R Lem m a 1.15 [9,17] Let {λn } and { pn } be sequences of nonnegative numbers, {bn } be a sequence of positive numbers, satisfying the inequalities λn+1 ≤ (1 − pn ) λn + bn , where pn ∈ (0; 1) , bn pn → +∞ (n → +∞) and ∞ n=1 ∀n ≥ 0, pn = +∞ Then λn → (n → +∞) In Section 3, when dealing with a parallel Newton-type regularization method, we need some more results Lem m a 1.16 [18] Suppose A : D(A) = X → X is a continuously Frechet differentiable accretive operator and let L := A (h), h ∈ X , and α be a real positive number Then (α I + L)−1 ≤ Lem m a 1.17 [7,19] relations , α (α I + L)−1 L ≤ Let {ωn } be a sequence of nonnegative numbers satisfying the ωn+1 ≤ a + bωn + cωn2 , n ≥ 0, for some a, b, c > Let M+ := − b + (1 − b)2 − 4ac /2c, M− := (1 − b − √ (1 − b)2 − 4ac)/2c If b + ac < and ω0 ≤ M+ , then ωn ≤ l := max {ω0 , M− } for all n ≥ Applicable Analysis 2141 Downloaded by [University of Newcastle (Australia)] at 08:58 28 September 2014 An outline of the remainder of the paper is as follows: in Section 2, we propose two parallel iterative regularizations methods (PIRMs) for system (1), namely implicit PIRM and explicit PIRM The convergence of these PIRMs is established for both exact and noisy data cases Section studies a parallel regularizing Newton-type method for system (1) The convergence analysis of the proposed method in exact and noisy data cases is also studied Equations with inverse uniformly accretive operators In this section, we consider system (1) with inverse uniformly accretive operators Clearly, if each operator Ai is ϕi -inverse uniformly accretive, then it is ϕ-inverse uniformly accretive with ϕ(s, t) := mini=1, ,N ϕi (s, t) Thus, without loss of generality we can assume that all the operators Ai , i = 1, , N are ϕ-inverse uniformly accretive with the same funtion ϕ We begin with the following simple fact (cf [6]) Lem m a 2.1 Suppose Ai , i = 1, 2, , N , are inverse uniformly accretive operators If system (1) is consistent, then it is equivalent to the operator equation N Ai (x) = A(x) := (5) i=1 Proof Let the opeartors Ai , i = 1, , N , be ϕ-inverse uniformly accretive with the same funtion ϕ Obviously, any solution of (1) is a solution of (5) Conversely, let y be a solution of (5), i.e N Ai (y) = i=1 Let z be a solution of system (1), i.e Ai (z) = 0, i = 1, N Then, Ai (z)) = 0, and since Ai are inverse uniformly accretive, one gets N − N ϕ (R, Ai (y) − Ai (z) ) ≤ i=1 N i=1 (Ai (y) Ai (y) − Ai (z), J (y − z) = 0, i=1 where R = max { y , z } Thus, ϕ (R, Ai (y) − Ai (z) ) = 0, and hence Ai (y) = Ai (z) = 0, i = 1, 2, N Therefore, y is a solution of system (1) In this section, we need the following lemma, where the sequential weak continuity of the normalized duality mapping is not required (cf [9, Theorem 2.7.1]) Lem m a 2.2 [20, Theorem 2.1] Let X be a real, reflexive and strictly convex Banach space with a uniformly Gateaux differentiable norm and let A be an m-accretive mapping on X Then for each α > and a fixed y ∈ X , Equation (4) possesses a unique solution xα , and in addition, if the solution set S A of the equation A(x) = y is nonempty, then the net {xα } converges strongly to the unique element x ∗ solving the following variational inequality x ∗ , J (x ∗ − x ∗ ) ≤ 0, ∀x ∗ ∈ S A Moreover we have ||xαδ − xα || ≤ δ/α, where xαδ is the unique solution of the equation A(x) + αx = yδ , for any α > and yδ ∈ X satisfying ||yδ − y|| ≤ δ 2142 P.K Anh et al In the remainder of Section 2, we impose two sets of conditions on the space X , the duality mapping J and the operators Ai , i = 1, 2, , N Conditions (AJX) Downloaded by [University of Newcastle (Australia)] at 08:58 28 September 2014 A1 Ai , i = 1, 2, , N , are ϕ-inverse uniformly accretive operators with D(Ai ) = X ; A2 the normalized duality mapping J is sequentially weakly continuous and continuous; A3 X is a smooth and reflexive Banach space, possessing the approximation Conditions (AX) B1 Ai , i = 1, 2, , N , are m-accretive and ϕ-inverse uniformly accretive operators with D(Ai ) = X ; B2 X is a uniformly smooth and uniformly convex Banach space Together with Equation (5), we consider the following regularized one N Ai (x) + αn x = 0, (6) i=1 where the regularization parameter αn → as n → ∞ Lem m a 2.3 hold: (i) (ii) (iii) (iv) (v) Let conditions A1–A3 or B1–B2 be fulfilled Then the following statements For every αn > 0, Equation (6) has a unique solution xn∗ xn∗ ≤ x , where x is an arbitrary element of S ∗ xn → x ∗ as n → +∞, where x ∗ is an unique solution of the inequality x ∗ , J (x ∗ − x ∗ ) ≤ 0, ∀x ∗ ∈ S |αn+1 −αn | ∗ ≤ x∗ xn∗ − xn+1 αn 6αn x ∗ Ai (xn∗ ) ≤ ϕ −1 R i = 1, 2, , N , where R > is a fixed number satisfying an a-priori estimate R ≥ x ∗ and ϕs−1 denotes the inverse function of ϕ (s, t) with respect to the second variable t for fixed s > Proof (1) Suppose that conditions A1–A3 hold We perform the regularization process N Ai For the proofs of statements (6) for Equation (5) with the accretive operator A = i=1 (i)–(iv) we refer the reader to [9] Concerning the last part (v) we observe that N (Ai (xn∗ ) − Ai (x ∗ )) + αn xn∗ = 0, i=1 hence N i=1 Ai (xn∗ ) − Ai (x ∗ ), J (xn∗ − x ∗ ) + αn xn∗ , J (xn∗ − x ∗ ) = Applicable Analysis 2143 Using the inverse uniform accretiveness of Ai , from the last inequality we have N ϕ R, Ai (xn∗ ) − Ai (x ∗ ) ≤ −αn xn∗ , J (xn∗ − x ∗ ) ≤ αn ||xn∗ ||||xn∗ − x ∗ ||, (7) i=1 Downloaded by [University of Newcastle (Australia)] at 08:58 28 September 2014 where R ≥ x ∗ By part (ii) xn∗ ≤ x ∗ , hence xn∗ − x ∗ ≤ x ∗ Combining the last inequalities with (7), we get ϕ R, Ai (xn∗ ) ≤ 6αn x ∗ Thus, Ai (xn∗ ) ≤ 6αn x ∗ ϕ −1 R (2) Now suppose that conditions B1–B2 hold Since all Ai are inverse uniformly accretive, they are continuous, and hence locally bounded Besides, Ai , i = 1, N , are m-accretive, D(Ai ) = X, and the spaces X and X ∗ are uniformly convex; then by [9, Theorem 1.15.22], N Ai is also m-accretive Lemma 2.2 applied to Equation (5) ensures the operator A = i=1 the convergence of regularized solutions x n∗ –x ∗ The remaining statements can be argued similarly as in part Following [5], we consider an implicit PIRM consisting of solving simultaneously N regularized equations αn + γn xni = γn xn , i = 1, 2, , N , (8) Ai xni + N where αn > and γn > are regularization and parallel splitting up parameters, respectively, and defining the next approximation as an average of the regularized solutions x ni , xn+1 = N N xni , x0 ∈ X (9) i=1 According to Lemmas 2.2 and 2.3, all the problems (8) are well posed and independent from each other; hence the regularized solutions x ni can be found stably and simultaneously by parallel processors We first prove the boundedness of the sequence {xn } defined by the implicit PIRM (8) and (9) Lem m a 2.4 Under conditions A1–A3 or B1–B2, the sequence {xn } generated by (8) and (9) is bounded Proof If conditions A1–A3 (B1–B2) hold, then by Lemma 1.10 (Lemma 2.2), respectively, the regularized Equation (8) has a unique solution denoted by xni Let Br (x ∗ ) be the closed ball with center x ∗ and radius r Choose r > sufficiently large such that r ≥ x ∗ and x0 ∈ Br (x ∗ ) Supposing for some n > 0, xn ∈ Br (x ∗ ), we will show that xn+1 ∈ Br (x ∗ ) Indeed, from (8) and Ai (x ∗ ) = 0, we get αn αn ∗ + γn xni − x ∗ = γn (xn − x ∗ ) − x Ai (xni ) − Ai (x ∗ ) + N N Thus, αn + γn xni − x ∗ , J (xni − x ∗ ) N αn ∗ xn − x ∗ , J (xni − x ∗ ) − x , J (xni − x ∗ ) N Ai (xni ) − Ai (x ∗ ), J (xni − x ∗ ) + = γn 2144 P.K Anh et al By the accretiveness of Ai , we get αn + γn N xni − x ∗ ≤ γn xn − x ∗ xni − x ∗ + αn ∗ x N xni − x ∗ , hence Downloaded by [University of Newcastle (Australia)] at 08:58 28 September 2014 αn αn ∗ + γn xni − x ∗ ≤ γn xn − x ∗ + x N N Using the inequalities xn − x ∗ ≤ r and r ≥ x ∗ , we have αn αn αn + γn xni − x ∗ ≤ γn r + r ≤ + γn r, N N N which gives xni − x ∗ ≤ r By (9), one gets xn+1 − x ∗ ≤ Therefore, xn+1 ∈ Br (x ∗ ) N N xni − x ∗ ≤ r i=1 Thus, {xn } is bounded Theorem 2.5 Suppose conditions A1–A3 or B1–B2 are fulfilled Let {αn } and {γn } be real sequences, such that (i) (ii) αn → 0, γn → +∞ as n → +∞, γn |αn+1 −αn | → as n → +∞, ∞ n=1 α2 n (iii) h X (τn )ϕ −1 R (R1 αn ) αn αn γn = +∞, → as n → +∞, where R ≥ 2||x ∗ ||, R1 := 3R 2 and τn = γn−1 If in addition, the function ϕ(s,t) is coercive in t for any fixed s > 0, i.e ϕ(s,t) → +∞ as t t t → +∞, then starting from arbitrary x ∈ X , the sequence {xn } defined by (8) and (9) converges strongly to x ∗ Proof Let xn∗ be the unique solution of (6) Setting eni = xni − xn∗ ; en = xn − xn∗ ; we can rewrite (8) as xni + τn Ai xni + n xni = xn , n = αn τn N , or eni − en + τn Ai xni − Ai xn∗ + = −τn Ai xn∗ − i n en ∗ n xn , i = 1, 2, N From the last relation, using the accretiveness of Ai we find eni − en , J eni +2 n eni ≤ −2 τn Ai (xn∗ ) + ∗ n xn , J (eni ) From Lemma 1.12, we get eni − en , J eni ≥ eni − en Combining this inequality with (10), we obtain (1 + n ) eni − en ≤ −2τn Ai (xn∗ ) + αn ∗ x , J eni N n , (10) Applicable Analysis 2145 hence N (1 + n ) eni N − N en Ai (xn∗ ) + ≤ −2τn i=1 i=1 αn ∗ x , J eni N n (11) Observing that xn∗ is the solution of (6) and using Lemma 2.3, we can estimate the right-hand side of (11) as follows: N − Downloaded by [University of Newcastle (Australia)] at 08:58 28 September 2014 i=1 Ai (xn∗ ) + N αn ∗ x , J (eni ) = − N n Ai (xn∗ ) + αn xn∗ , J (en ) i=1 N − Ai (xn∗ ) + i=1 N Ai (xn∗ ) + ≤ i=1 N ≤ αn ∗ x , J (eni ) − J (en ) N n αn ∗ xn N ϕ −1 6αn x ∗ R i=1 + J (eni ) − J (en ) 2αn ∗ x N J (eni ) − J (en ) (12) By Lemmas 2.3 and 2.4, the sequences xn∗ , {xn }, and xni are bounded; hence, the sequences {en } and eni are also bounded, i.e there exists a positive constant C > such that en ≤ C; eni ≤ C From Lemma 1.11, we get J (eni ) − J (en ) ≤ 8Ch X 16L xni − xn C , (13) where L ∈ (1, 1.7) is Figiel constant We show that ||Ai (z)|| ≤ Ci < +∞ for all ||z|| ≤ R0 := C + R; R ≥ 2||x ∗ || and i = 1, , N Indeed, suppose in contrary, that there exists a sequence {z n }, such that ||z n || ≤ R0 , and ||Ai (z n )|| → ∞ as n → ∞ Then tn := ||Ai (z n ) − Ai (0)|| ≥ ||Ai (z n )|| − ||Ai (0)|| → ∞ as n → ∞ Since ϕ(R0 , tn ) = ϕ(R0 , ||Ai (z n ) − Ai (0)||) ≤ Ai (z n ) − Ai (0), J (z n − 0) ≤ ||Ai (z n ) − Ai (0)||||z n || ≤ R0 tn , we get R0 ≥ ϕ(Rt0n ,tn ) , which contradicts the coerciveness of ϕ(Rt0 ,t) Thus, we can put αn x : x ≤ R0 , n = 1, 2, , i = 1, 2, , N M = sup Ai (x) + N Relation (8) yields Ai xni + αn i x = γn xn − xni , N n which gives γn xn − xni ≤ Ai (xni ) + the last inequality with (13), we obtain αn i N xn i = 1, 2, , N , ≤ M, hence xn − xni ≤ J (eni ) − J (en ) ≤ c2 h X (k0 τn ), M γn Combining (14) 2146 P.K Anh et al where c2 = 8C, k0 = N − 16L M C By (12) and (14), we have αn ∗ x , J eni N n Ai xn∗ + i=1 ≤ N c2 ϕ −1 6αn x ∗ R + 2αn ∗ x N h X (k0 τn ) (15) From (10) and (15), we get N (1 + n ) eni ≤ N en 6αn x ∗ + 2N c2 τn ϕ −1 R + Downloaded by [University of Newcastle (Australia)] at 08:58 28 September 2014 i=1 2αn ∗ x N h X (k0 τn ) (16) Taking into account relation (9), Lemma 2.3, and the inequality (a+b)2 ≤ (1+ n )(a + bn ), we find en+1 ∗ = xn+1 − xn+1 N ≤ ⎛ xn+1 − xn∗ + x ∗ ≤ N eni +2 x |αn+1 − αn | αn ∗ i=1 ≤ ⎝√ N ≤ (1 + 1/2 N eni + x∗ i=1 N n) N eni |αn+1 − αn | αn 2 ⎞2 |αn+1 − αn | ⎠ αn + x∗ αn+1 − αn n αn i=1 |αn+1 − αn |2 αn2 n Thus, N (1 + n) en+1 − 4N n x∗ N ≤ eni (17) i=1 From (17) and (16), one gets en+1 ≤ Setting λn = en αn+1 − αn ∗ 1+ n x en + 4(1 + n ) n 1+2 n n αn 2αn ∗ 2c2 (1 + n ) τn ϕ −1 x 6αn x ∗ + + R 1+2 n N ; pn = b3n n h X (k0 τn ) (18) and bn = b1n + b2n + b3n , where αn+1 − αn ∗ x ; αn n 2c2 (1 + n ) τn −1 h X (k0 τn ); = ϕ R 6αn x ∗ 1+2 n 4c2 (1 + n ) τn αn ∗ x h X (k0 τn ) = N (1 + n ) b1n = b2n n 1+2 4(1 + n) We can rewrite (18) as λn+1 ≤ (1 − pn ) λn + bn Clearly, λn , bn ≥ 0; pn ∈ (0; 1) and pn → as n → +∞ Applicable Analysis 2147 Since pn = 1+2n n and n → as n → +∞, the series ∞ n=1 pn = +∞ if and only if αn = +∞ The last fact is equivalent to the assumption ∞ n=1 γn = +∞ ∞ n=1 n n By the assumption (ii), bp1nn = (1 + n ) (1 + n ) αn+1n α−α n n → +∞ Further, using assumption (iii), we will show that the expression Downloaded by [University of Newcastle (Australia)] at 08:58 28 September 2014 b2n = 2c2 N (1 + pn n) ϕ −1 6αn x ∗ R 2 x∗ → as h X (k0 τn ) (19) αn will tend to zero as n → +∞ We first prove that there exist positive integers m and n , m such that for all n ≥ n , h X (k0 τn ) ≤ 5k0 h X (τn ) Indeed, according to [10,p.65], we have ≤ lim sup τ →0+ ρ X (2τ ) ≤ ρ X (τ ) ) Hence, there exists τ0 > 0, such that ρρXX(2τ (τ ) ≤ for all τ ≤ τ0 Since τn → as n → +∞, we can find a number n such that k0 τn ≤ τ0 for all n ≥ n Let m be a sufficiently large positive integer, such that 2m ≥ k0 Then for all n ≥ n we have ρ X (k0 τn ) = ρ X (2 k20 τ1n ) ≤ 5ρ X ( k20 τ1n ) = 5ρ X (2 k20 τ2n ) ≤ 52 ρ X ( k20 τ2n ) ≤ ≤ 5m ρ X ( k20mτn ) Because of the convexity of ρ X and 2km0 ≤ 1, we get ρ X (k0 τn ) ≤ 5m ρ X ( k20mτn ) ≤ 5m ρ X (τn ) m m Thus, we come to the relation h X (k0 τn ) = ρ Xk(k0 τ0nτn ) ≤ 5k0 ρ Xτ(τn n ) = 5k0 h X (τn ) Now using the last inequality and taking into account the fact that ϕ −1 R (t) is an increasing function and R1 := 32 R ≥ 6||x ∗ ||2 , we can estimate the expression (19) as bp2nn ≤ −1 2c2 N 5m (1+ n ) ϕ R (R1 αn )h X (τn ) k0 αn for all n ≥ n The assumption (iii) implies that b2n pn → as n → +∞ Finally, the uniform smoothness of X gives bp3nn = 4c2 (1 + n ) x ∗ h X (k0 τn ) → as n → +∞ Thus, bpnn → (n → +∞) Lemma 1.15 ensures that λn = en = xn − xn∗ → (n → +∞) Besides, by Lemma 2.3, xn∗ → x ∗ (n → +∞); hence, xn − x ∗ ≤ xn − xn∗ + xn∗ − x ∗ → (n → +∞) The proof of Theorem 2.5 is complete Example 2.6 Let Ai , i = 1, , N be inverse strongly monotone operators on a real Hilbert space X Then all the conditions A1–A3 and B1–B2 are satisfied Further, since = ct is coercive Conditions (i) and (ii) on the parameters ϕ(s, t) = ct , the function ϕ(s,t) t been already stated in [5, Theorem 2.1] On the other hand, for a Hilbert space, αn , γn have √ ρ X (t) = + t −1 ≤ t2 ; hence, the assumption (iii) of Theorem 2.5 leads to the additional 1/2 constraint γn αn → +∞ (n → +∞) An example of such a pair of parameters could be αn = (n + 1)− p , where < p < 1/2 and γn = (n + 1)1/2 In the next two examples, we suppose that X = l p , ≤ p < +∞ and Ai = I − Ti , where Ti : X → X, i = 1, , N , are nonexpansive operators In this case, both sets of conditions A1–A3 and B1–B2 are fulfilled Observe that for proving the m- accretiveness of Ai one should use the identity Ai + α I = (1 + α){I − (1 + α)−1 Ti } and the fact that (1 + α)−1 Ti is a contraction for i = 1, , N 2148 P.K Anh et al Example 2.7 Let X = l p with p ≥ 2, then ρ X (t) ≤ ( p − 1)t and h X (t) ≤ ( p − 1)t (see [9,p.48]) According to Example 1.8, all the operators Ai := I − Ti , i = 1, 2, , N , −1 are inverse uniformly accretive with ϕ(s, t) = L p8 p p−2 For any fixed s > 0, ϕs (t) = s c(s)t p , where c(s) is a positive number independent of t and the function p−1 p in t The assumption (iii) of Theorem 2.5 becomes γn αn → ∞ can choose αn = (n + 1)−k , γn = (n + 1)1/2 with < k < 12 ϕ(s,t) t is coercive (n → +∞) and we Downloaded by [University of Newcastle (Australia)] at 08:58 28 September 2014 Example 2.8 Suppose X = l p , < p < 2, then we have (see [9, p.48]) ρ X (t) ≤ t p−1 p , h X (t) ≤ p Example 1.8 shows that Ai , i = 1, , N are c− inverse strongly accretive operators with ϕ(s, t) = ct , ϕs−1 (t) = 1/2 p−1 Theorem 2.5 becomes αn γn →∞ (n + 1)1/2 with < k < min{ 12 , p − 1} √ √t , c c = p−1 256L and the assumption (iii) of (n → +∞) We can chose αn = (n + 1)−k , γn = Next we turn to the noisy data case, assuming that Ai (x) := Fi (x) − f i and the exact operators Fi (x), i = 1, , N , are inverse uniformly accretive Suppose that instead of exact data (Fi , f i ), we have only noisy ones (Fn,i , f n,i ), where the perturbed operators Fn,i : D(Fn,i ) = X → X are just accretive for all n ≥ and i = 1, , N Moreover, let Fn,i (x) − Fi (x) ≤ h n g( x ), (20) f n,i − f i ≤ δn , (21) i = 1, , N , where, g(t) is a nonnegative continuous nondecreasing function, h n > 0, δn > for all n > Starting from arbitrary z ∈ X , we perform the following implicit PIRM: αn + γn z ni = γn z n , N N z ni z n+1 = N An,i z ni + i = 1, 2, , N , (22) (23) i=1 where An,i (x) := Fn,i (x) − f n,i , i = 1, 2, , N Theorem 2.9 Assume that all the conditions of Theorem 2.5 and relations (20), (21) are n → as n → +∞, then the sequence {z n } generated by (22), fulfilled If in addition h nα+δ n (23) converges strongly to x ∗ as n → +∞ Proof Let {xn } be the sequence of approximations defined by (8) and (9) From (8) and (22), we have An,i z ni − Ai xni + αn + γn N z ni − xni = γn (z n − xn ), or Fn,i z ni − Fn,i xni = γn (z n − xn ) + Fn,i xni − Fi xni + f n,i − f i + αn + γn N z ni − xni Applicable Analysis 2149 Therefore Fn,i z ni − Fn,i xni , J z ni − xni + f n,i − f i , J z ni − xni + Fn,i xni − Fi xni , J z ni − xni αn + γn N + z ni − xni = γn z n − xn , J z ni − xni (24) Downloaded by [University of Newcastle (Australia)] at 08:58 28 September 2014 By relations (20), (21), and (24), we get αn + γn N z ni − xni ≤ hn g z ni − xni +δn z ni − xni +γn z n − xn xni Lemma 2.4 ensures the boundedness of the sequence xni Thus, xni R > Setting λn = z n − xn , from the last inequality we find z ni − xni ≤ N γn N g(R)h n N δn λn + + αn + N γn αn + N γn αn + N γn z ni − xni ≤ R for some (25) On account of (9), (23), and (25), we obtain N N γn N g(R)h n N δn λn + + αn + N γn αn + N γn αn + N γn i=1 (26) N g(R)h n αn N δn Putting pn = αn +N , b = + , from (26) we get λ = (1− p )λ +b n n+1 n n n By γn αn +N γn αn +N γn ∞ bn virtue of the hypotheses of Theorem 2.9, we have n=1 pn = +∞, pn → as n → +∞ Lemma 1.15 implies that λn = z n − xn → as n → +∞ Finally, by Theorem 2.5, xn → x ∗ (n → +∞); hence, z n − x ∗ ≤ z n − xn + xn − x ∗ → (n → +∞) The proof of Theorem 2.9 is complete λn+1 = z n+1 − xn+1 ≤ N z ni − xni ≤ We now consider an explicit PIRM for solving system (1), consisting of synchronous computations of intermediate approximations z ni z ni = z n − γn Ai (z n ) + αn αn z n = z n − τn Ai (z n ) + zn , N N i = 1, 2, , N , (27) and defining the next approximation z n+1 as an average of intermediate approximations z ni z n+1 = N N z ni , n = 1, 2, (28) i=1 Lem m a 2.10 Suppose conditions B1–B2 are satisfied Assume in addition the function ϕ(s,t) is coercive in t for every fixed s > Let {αn } and {γn } be positive sequences such t that for all n ≥ 0, αn ≤ 1, γn ≥ 1, and τn ≤ d, ρ X (τn ) ≤ d 2, τn αn (29) where τn := 1/γn and d ∈ (0, 1) is a fixed number Then starting from arbitrary z ∈ X, the sequence {z n } determined by (27), (28) is bounded 2150 P.K Anh et al Proof A simple vertification shows that the sequence {z n } defined by (27) and (28) satisfies the relation {A(z n ) + αn z n }, (30) z n+1 = z n − N γn Downloaded by [University of Newcastle (Australia)] at 08:58 28 September 2014 N where A(z) := i=1 Ai (z) By our assumptions, all the operators Ai , i = 1, , N , are continuous, m-accretive, and ϕ-inverse uniformly accretive Moreover, as it was shown in the proof of Theorem 2.5, Ai is bounded for every i; hence, the operator A : D(A) = X → X is bounded, continuous, and m-accretive According to Lemma 2.1, equation A(z) = is equivalent to the consistent system (1) By [21, Theorem 5.1] the sequence {z n } defined by (30) is bounded Theorem 2.11 Assume that all the conditions of Lemma 2.10 are fulfilled In addition, let αn → 0, τn := 1/γn → as n → +∞, such that ∞ |αn − αn+1 | →0 τn αn2 τn →0 αn αn τn = +∞, i=1 ρ X (τn ) → τn αn (31) Then the sequence {z n } generated by (27) and (28) converges to x ∗ as n → ∞ Proof Let xn∗ be the unique solution of the regularized Equation (6) It follows from Lemma 2.3 that {xn∗ } is bounded; hence, there exists a constant d˜ > such that ∗ ˜ By Lemma 1.12, we have xn∗ − xn+1 ≤ d ∗ z ni − xn+1 ≤ z ni − xn∗ ∗ + xn+1 − xn∗ , J xn∗ − z ni ∗ ∗ − xn∗ , J (xn+1 − z ni ) − J (xn∗ − z ni ) + xn+1 (32) Further, by Lemma 1.13, we get ∗ z ni − xn+1 ≤ z ni − xn∗ + z ni − xn∗ ∗ + 16 xn+1 − xn∗ ∗ xn+1 − xn∗ + c1 (n) ρ X ∗ xn+1 − xn∗ , (33) ∗ where c1 (n) = max 2L , z ni − xn+1 + z ni − xn∗ Taking into account Lemma 2.10 and the boundedness of the operators Ai we conclude that the sequence {z ni } is also bounded; therefore, there exist positive numbers c1 , k0 such that c1 (n) ≤ c1 and z ni − xn∗ ≤ k0 for all n ≥√0 Note that, if H is a Hilbert space, then for all < τ < τ we have ρ X (τ ) ≥ ρ H (τ ) = + τ − 1, hence ρ X (τ ) ≥ cτ , (34) √ −1 where c = + τ2 + Now, summing up both sides of (33) for i = 1, , N , and using Lemma 1.14, as ∗ − xn∗ , we obtain well as inequality (34) with τ := d˜ ≥ τ := xn+1 N i=1 ∗ z ni − xn+1 N ≤ i=1 z ni − xn∗ + 4N k0 |αn+1 − αn | ∗ x αn + N 16c−1 + c1 ρ X |αn+1 − αn | ∗ x αn Applicable Analysis 2151 From the last inequality and the fact that ρ X (τ ) ≤ τ , one gets N ∗ z ni − xn+1 N z ni − xn∗ ≤ i=1 Downloaded by [University of Newcastle (Australia)] at 08:58 28 September 2014 + c3 i=1 where c3 = 2N x ∗ 2k0 + 16c −1 + c1 Next, we shall estimate the expression we have z ni − xn∗ 2 |αn+1 − αn | , αn z ni − xn∗ N i=1 (35) By Lemma 1.12 and (27), αn zn N αn z n , J (z n − xn∗ ) ≤ z n − xn∗ − 2τn Ai (z n ) + N + z ni − z n , J (z ni − xn∗ ) − J (z n − xn∗ ) = z n − xn∗ − τn Ai (z n ) + (36) Besides, z ni − z n = τn Ai (z n ) + where M = sup Ai (z n ) + Lemma 1.14, one obtains αn N zn αn z n ≤ Mτn , N (37) : i = 1, 2, , N ; n = 1, 2, Using (37) and z ni − z n , J (z ni − xn∗ ) − J (z n − xn∗ ) ≤ 8M τn2 + c2 (n)ρ X (Mτn ) , where c2 (n) = max 2L , z ni − xn∗ + z n − xn∗ ≤ c2 because of the boundedness of the sequences {z ni } and xn∗ Here, L is the Figiel constant Therefore, z ni − z n , J (z ni − xn∗ ) − J (z n − xn∗ ) ≤ 8M τn2 + c2 ρ X (Mτn ) (38) On the other hand, since the operators Ai are accretive and xn∗ is the solution of (6), we have N i=1 αn z n , J (z n − xn∗ ) = Ai (z n ) + N N Ai (z n ) − Ai xn∗ , J (z n − xn∗ ) i=1 N + Ai xn∗ + αn xn∗ , J (z n − xn∗ ) + αn z n − xn∗ i=1 ≥ αn z n − xn∗ (39) Now, summing the both sides of (36) for i = 1, 2, , N and taking account relations (38) and (39), we get N z ni − xn∗ ≤ N z n − xn∗ − 2τn αn z n − xn∗ + 16N M τn2 + 2N c2 ρ X (Mτn ) i=1 (40) Note that z n+1 − ∗ xn+1 ≤ N N z ni − i=1 ∗ xn+1 ≤ N N i=1 ∗ z ni − xn+1 (41) 2152 P.K Anh et al From (35), (40), and (41), we get ∗ z n+1 − xn+1 ≤ z n − xn∗ − 2τn αn z n − xn∗ N + 16M τn2 + 2c2 ρ X (Mτn ) + c3 Setting λn = z n − xn∗ we can rewrite (42) as , pn = 2τn αn N , |αn+1 − αn | αn bn = 16M τn2 + 2c2 ρ X (Mτn ) + c3 (42) |αn+1 −αn | αn , Downloaded by [University of Newcastle (Australia)] at 08:58 28 September 2014 λn+1 ≤ (1 − pn )λn + bn In the same manner as in the proof of Theorem 2.5, we can find positive integers n and m, such that for all n ≥ n , ρ X (Mτn ) ≤ 5m ρ X (τn ) By Lemma 1.15 and the hypothesis (31), we conclude that λn = z n − xn∗ → as n → +∞ Finally, by Lemma 2.3, z n − x ∗ ≤ z n − xn∗ + xn∗ − x ∗ → 0, which implies that {z n } converges to x ∗ The proof of Theorem 2.11 is complete Equations with smooth accretive operators In this section, for solving system (1) with smooth accretive operators Ai , we consider a parallel regularized Newton-type method (cf [6,7]) αn αn xn − xi0 + Ai (xn ) + I xni − xn = 0, i = 1, 2, , N , (43) Ai (xn ) + N N N xn+1 = xni , n = 0, 1, (44) N i=1 The following assumptions will be needed throughout Section C1 System (1) possesses an exact solution x ∗ The operators Ai (i = 1, , N ) are accretive on a real Banach space X and Fr´echet differentiable in a closed ball Br (x ∗ ) ⊂ X with center x ∗ and radius r > Moreover, Ai (x) − Ai (y) ≤ K x − y ∀x, y ∈ Br (x ∗ ), i = 1, , N C2 The following componentwise source condition (see [3,7]) holds xi0 − x ∗ = Ai (x ∗ )vi , where xi0 ∈ Br (x ∗ ), vi ∈ X, ≤ i ≤ N C3 The parameters αn are chosen such that αn > 0, αn → 0, ≤ αn ≤ ρ, αn+1 where ρ > is a constant Theorem 3.1 Let all the assumptions C1–C3 be satisfied If small and x0 is close enough to x ∗ , then there holds the estimate xn − x ∗ = O(αn ) N i=1 vi is sufficiently (45) Applicable Analysis 2153 Proof We suppose by induction that x n ∈ Br (x ∗ ) for some n ≥ Setting eni = xni − x ∗ and en = xn − x ∗ , from Equation (43) and assumption C2, we get αn −1 αn I xn − xi0 Ai (xn ) + N N αn −1 αn αn I I en − Ai (xn ) + xn − xi0 Ai (xn ) + = Ai (xn ) + N N N αn −1 αn I xi − x ∗ + Ai (xn )en − Ai (xn ) = Ai (xn ) + N N αn −1 αn −1 αn Ai (xn ) + I I Ai (xn )en − Ai (xn ) Ai (x ∗ )vi + Ai (xn ) + = N N N Downloaded by [University of Newcastle (Australia)] at 08:58 28 September 2014 eni = en + xni − xn = en − Ai (xn ) + Using Lemma 1.16, from the last inequality we obtain eni ≤ αn N Ai (xn ) + αn I N −1 Ai (x ∗ ) vi + N αn Ai (xn )en − Ai (xn ) Obviously, if xt := xn + t (x ∗ − xn ), where ≤ t ≤ 1, then (1 − t) xn − x ∗ ) ≤ r ; hence xt ∈ Br (x ∗ ) The second term of the right-hand side of (46) can be estimated as N αn Ai (xn )en − Ai (xn ) = N αn = N αn ≤ N αn xt − x ∗ (46) = Ai (x ∗ ) − Ai (xn ) + Ai (xn )(en ) Ai (xn ) − Ai (xt ) en dt K t en dt = KN en 2αn (47) On the other hand, we have Ai (xn ) + αn I N −1 = Ai (xn ) + − Ai (x ∗ ) + αn I N −1 αn I N −1 Ai (x ∗ ) − Ai (xn ) Ai (x ∗ ) + αn I N −1 Therefore, Ai (xn ) + αn I N + −1 Ai (x ∗ ) ≤ Ai (xn ) + αn I N Ai (x ∗ ) + −1 αn I N −1 Ai (x ∗ ) Ai (x ∗ ) − Ai (xn ) Ai (x ∗ ) + αn I N −1 Ai (x ∗ ) From the last inequality and using Lemma 1.16 as well as assumption C1, we get Ai (xn ) + αn I N −1 Ai (x ∗ ) ≤ + 2K N en αn (48) NK en 2αn (49) Combining (46),(47), and (48), one has eni ≤ 2αn vi + 2K vi N en + 2154 P.K Anh et al By (44) and (49), we obtain en+1 = xn+1 − x ∗ ≤ N N i=1 + Downloaded by [University of Newcastle (Australia)] at 08:58 28 September 2014 Setting ωn = ωn+1 ≤ ≤ N en αn N i=1 vi N i=1 vi αn αn+1 where If N i=1 N i=1 2K vi en + N + 2ρ N i=1 2K + vi αn αn+1 N N i=1 2ρ K N = a + bωn + cωn2 , a= vi NK en 2αn (50) and using assumption C3, from (50) we find N 2ρ N i=1 N2 2αn eni ≤ vi N ωn + K ωn + αn αn+1 ωn2 Kρ ω n (51) N i=1 vi N , b= 2ρ K N i=1 vi N , c= Kρ vi is small enough, then a, b will be small, hence √ b + ac < 1, 2aα0 ≤r 1−b+ N (1 − b)2 − 4ac (52) Now if x0 is sufficiently close to x ∗ then ω0 = N e0 N x0 − x ∗ (1 − b + = ≤ M+ := α0 α0 (1 − b)2 − 4ac) 2c Lemma 1.17 applied to (51) ensures that ωn := where M− = (1−b− √ N en ≤ l := max {ω0 , M− } , αn (1−b)2 −4ac) 2c = In particular, (1−b+ √ 2a (1−b)2 −4ac)/2c xn+1 − x ∗ = en+1 = Observing that ω0 α0 N ∀n ≥ 0, lα0 ωn+1 αn+1 ≤ N N = x0 − x ∗ ≤ r , from (52) we have 2aα0 M− α0 = ≤ r N N − b + (1 − b)2 − 4ac Therefore, lα0 N ≤ r ; hence, xn+1 ∈ Br (x ∗ ) Thus, the estimate ωn ≤ l yields en = lαn ωn αn ≤ = O (αn ) N N The proof of Theorem 3.1 is complete (53) Applicable Analysis 2155 Now, assume that Ai (x) = Fi (x) − f i and instead of (Fi , f i ) we only have their approximations Fih , f iδ , such that f iδ − f i ≤ δ, (54) and the operators Fih , (i = 1, , N ), are accretive on a real Banach space X and Fr´echet differentiable in a closed ball Br (x ∗ ) Moreover, suppose that Fih (x) − Fih (y) ≤ K x − y ∀x, y ∈ Br (x ∗ ), i = 1, 2, , N (55) Downloaded by [University of Newcastle (Australia)] at 08:58 28 September 2014 Also, assume that Fih (x ∗ ) − Fi (x ∗ ) ≤ h, (56) Fih (x ∗ ) − Fi (x ∗ ) ≤ h (57) Given a starting point x0 ∈ Br (x ∗ ), we define a sequence {xn }: αn αn xn − xi0 + A˜ i (xn ) + I xni − xn = 0, A˜ i (xn ) + N N N xn+1 = xni , n = 1, 2, , N i = 1, 2, , N (58) (59) i=1 where A˜ i (x) = Fih (x) − f iδ We define the stopping index N (δ, h) as N (δ, h) = max n : αn2 ≥ δ+h , η (60) where η is a positive constant By the same argument as in the proof of Theorem 3.1, we come to the following estimate for iteration (58) and (59): eni ≤ 2αn vi +2 vi N Fi (x ∗ ) − Fih (xn ) + N Fih (xn )en − Fih (xn ) + f δ αn (61) From conditions (55) and (57), we have Fi (x ∗ ) − Fih (xn ) ≤ Fi (x ∗ ) − Fih (x ∗ ) + Fih (x ∗ ) − Fih (xn ) ≤ h + K en (62) Using conditions (54) and (56), we have Fih (xn )en − Fih (xn ) + f δ ≤ Fih (xn )en − Fih (xn ) + Fih (x ∗ ) + Fih (x ∗ ) − Fi (x ∗ ) + f − f δ ≤ K en 2 + h + δ (63) Combining (61), (62), and (63) and noting that en+1 ≤ en+1 ≤ N i=1 N vi (h + αn ) + h+δ αn + 2L N i=1 N N i=1 N vi eni , we obtain en + K en 2αn (64) 2156 P.K Anh et al From (60), we get ωn+1 ≤ h αn αn αn+1 N i=1 vi N N i=1 ≤ρ =a ≤ αn η ≤ α0 η Setting ωn = vi N N en αn we can rewrite (64) as N i=1 2L h h+δ +1 + + αn αn (α0 η + 1) + η + 2K vi N N i=1 vi N ωn + ωn + K ω n K ω n + bωn + cωn2 , Downloaded by [University of Newcastle (Australia)] at 08:58 28 September 2014 where a=ρ N i=1 vi N (α0 η + 1) + η , b= 2ρ K N i=1 vi N , c= Kρ N Again, if i=1 vi and η are small enough and x0 is sufficiently close to x ∗ , then arguing similarly as in the proof of Theorem 3.1, we obtain xn+1 ∈ Br (x ∗ ) and xn − x ∗ = O (αn ) for n = 1, 2, , N (δ, h) Thus, we come to the following convergence result Theorem 3.2 Assume that the assumptions C1–C3 hold for the exact operators Fi , N i = 1, , N , and conditions (54), (55), (56) and (57) are satisfied If i=1 vi is sufficiently small and x0 is close enough to x ∗ , then there holds the following estimate xn − x ∗ = O (αn ) , n = 1, 2, , N (δ, h) (65) Finally, taking into account the stopping rule (60), from Theorem 3.2 we obtain the convergence rate for the parallel regularized Newton-type method (43) and (44) in noisy data cases Corollary 3.3 Assume that all conditions of Theorem 3.2 are fulfilled Then x∗ − x ∗ = O δ 1/2 + h 1/2 , (66) where n ∗ = N (δ, h) + Conclusion Most of the existing paralell methods for systems of ill-posed operator equations deal with Hilbert spaces In this paper, we investigate two parallel iterative regularization methods and a parallel regularized Newton-type method for solving systems of equations involving m-accretive operators in Banach spaces The convergence analysis of the proposed methods in both free-noise and noisy data cases is provided Acknowledgements The authors would like to thank the referees for their careful reading and comments which improved the presentation of the paper The first author of this manuscript wishes to express his gratitude to Vietnam Institute for Advanced Study in Mathematics for financial support Applicable Analysis 2157 Downloaded by [University of Newcastle (Australia)] at 08:58 28 September 2014 References [1] Haltmeier M, Kowar R, Leitao A, Scherzer O Kaczmarz methods for regularizing nonlinear ill-posed equations, I Convergence analysis Inverse Prob Imaging 2007;1:289–298 [2] Haltmeier M, Kowar R, Leitao A, Scherzer O 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Newcastle (Australia)] at 08:58 28 September 2014... existing paralell methods for systems of ill-posed operator equations deal with Hilbert spaces In this paper, we investigate two parallel iterative regularization methods and a parallel regularized