Các phương pháp lặp để giải bài toán tựa cân bằng và bài toán điểm bất động viện khoa học và công nghệ tính toán

SỞ KHOA HỌC VÀ CƠNG NGHỆ TP HỒ CHÍ MINH VIỆN KHOA HỌC VÀ CƠNG NGHỆ TÍNH TỐN BÁO CÁO TỔNG KẾT Các phương pháp lặp để giải toán tựa cân toán điểm bất động Đơn vị thực hiện: PTN Cơng Nghệ Tốn Ứng dụng Chủ nhiệm đề tài: TS Nguyễn Thị Thu Vân TP HỒ CHÍ MINH, THÁNG 01/2018 SỞ KHOA HỌC VÀ CƠNG NGHỆ TP HỒ CHÍ MINH VIỆN KHOA HỌC VÀ CƠNG NGHỆ TÍNH TỐN BÁO CÁO TỔNG KẾT Các phương pháp lặp để giải toán tựa cân toán điểm bất động Viện trưởng: Đơn vị thực hiện: PTN Cơng nghệ Tốn Ứng dụng Chủ nhiệm đề tài: TS Nguyễn Thị Thu Vân Nguyễn Kỳ Phùng Nguyễn Thị Thu Vân TP HỒ CHÍ MINH, THÁNG 01/2018 Các phương pháp lặp để giải toán tựa cân toán điểm bất động MỤC LỤC Trang MỞ ĐẦU ĐƠN VỊ THỰC HIỆN KẾT QUẢ NGHIÊN CỨU I Báo cáo khoa học II Tài liệu khoa học xuất 15 III Chương trình giáo dục đào tạo 16 IV Hội nghị, hội thảo 17 V File liệu 18 TÀI LIỆU THAM KHẢO 19 CÁC PHỤ LỤC 22 PHỤ LỤC 1: “An extragradient-type method for solving nonmonotone quasi-equilibrium problems” PHỤ LỤC 2: “Strong convergence of an iterative method for solving the multiple-set split equality fixed point problem in a real Hilbert space” Viện Khoa học Cơng nghệ Tính tốn TP Hồ Chí Minh Page Các phương pháp lặp để giải toán tựa cân toán điểm bất động MỞ ĐẦU Mục đích chúng tơi nghiên cứu gồm phần: Phần 1, nghiên cứu phương pháp kiểu chiếu tăng cường để giải tốn tựa cân khơng đơn điệu, nghĩa toán tối ưu theo nghĩa Blum Oettli [1] hay bất đẳng thức Ky Fan [2], với tập ràng buộc phụ thuộc vào điểm khơng có giả thiết đơn điệu cho song hàm Phần 2, nghiên cứu phương pháp giải toán cực tiểu hàm lồi không âm phần giao vô hạn tập điểm bất động Bài toán cân nghiên cứu rộng rãi năm gần đây, ví dụ xem [3,4,5,6,7,8] tài liệu tham khảo đính kèm Các tốn cân bao gồm trường hợp đặc biệt toán tối ưu vơ hướng hay vector, tốn điểm n ngựa, bất đẳng thức biến phân, cân Nash, toán bù, toán điểm bất động, … Ta biết toán tựa bất đẳng thức biến phân với tập ràng buộc tập thay đổi khơng thuộc vào phạm vi tốn cân mà thuộc vào tốn tựa cân Có lẽ ví dụ quan trọng toán toán cân Nash tổng quát [9,10] , dạng mơ hình tốn học số lớn ứng dụng kỹ thuật, kinh tế, khoa học quản lý, lĩnh vực khác (ví dụ xem [11,12,13]) Bài tốn tựa cân mở rộng khái niệm cân cổ điển, điều dẫn đến việc ta mơ hình nghiên cứu trường hợp liệu tổng quát Nhiều phương pháp số đề nghị để giải toán cân phương pháp chiếu [14,15,16], phương pháp xấp xỉ điểm [17,18], phương pháp chiếu tăng cường có khơng có tìm kiếm tia [19,20,21] phương pháp hàm gap [22,23,24] Mỗi phương pháp nghiệm thích hợp với lớp tốn cân Để có tóm tắc chi tiết phương pháp cho tốn cân bằng, người đọc tham khảo thêm [25] tài liệu tham khảo đính kèm Trong hầu hết phương pháp này, hàm cân giả sử giả đơn điệu, trường hợp tập nghiệm toán cân trùng với tập nghiệm toán cân Minty [26,27] Điều thỏa cho trường hợp toán tựa cân hàm cân giả đơn điệu [28,29] Gần cơng trình [30], phương pháp kiểu chiếu tăng cường đề nghị để giải tốn cân hàm cân khơng giả sử giả đơn điệu (có thể xem thêm Viện Khoa học Cơng nghệ Tính tốn TP Hồ Chí Minh Page Các phương pháp lặp để giải toán tựa cân toán điểm bất động [31]) Trong trường hợp này, ngược lại với trường hợp tựa cân bằng, bước lặp, tất siêu phẳng tách tập nghiệm từ điểm lặp trước phải giữ lại để xây dựng bước Cho phép đề cập tác giả cơng trình [32] gần đề nghị phương pháp tương tự để giải toán cân trường hợp không đơn điệu Trong nghiên cứu này, phần chúng tơi giải tốn tựa cân mà khơng địi hỏi giả thiết đơn điệu hàm cân ngoại trừ tập ràng buộc thuộc vào thuộc tập điểm bất động phụ không gian Hilbert Để làm vậy, chúng tơi xét tốn cân dạng đối ngẫu Minty tập Khi hàm cân giả đơn điệu, tập nghiệm hai toán trùng Hàm cân giả sử giả đơn điệu để có tập nghiệm Minty chứa tập nghiệm toán cân tốn cân bằng, bao hàm thức ngặt Xem chi tiết [17], đặc biệt Mệnh đề Để xem xét trường hợp này, chiến thuật sử dụng xây dựng dãy bước lặp hội tụ mạnh giả thiết tập khăc rống sau chứng minh điểm tụ yếu dãy nghiệm toán tựa cân Làm hi vọng nhận dãy lặp hội tụ đến nghiệm toán tựa cân Trong phần hai xét tốn cực tiểu hàm lồi khơng cần thiết khả vi tập giao tập điểm bất động liên kết với họ vô hạn ánh xạ đa trị tựa không giãn không gian Hilbert thực Chúng tơi đề nghị thuật tốn để giải tốn ánh xạ khơng cần thiết tốn tử chiếu hay việc tính tốn phép chiếu không dễ thực Thông tin chuẩn tốn tử khơng cần điều kiện để đạt kết hội tụ mạnh thuật toán nghiên cứu Cuối xét trường hợp đặc biệt toán điểm bất động phương trình tách biến cho họ ánh xạ đa trị Viện Khoa học Cơng nghệ Tính tốn TP Hồ Chí Minh Page Các phương pháp lặp để giải toán tựa cân toán điểm bất động Lời cảm ơn đến ICST Đề tài nhận tài trợ tài Sở Khoa học Cơng nghệ Tp.Hồ Chí Minh theo mã số hợp đồng thuê khoán thực nhiệm vụ nghiên cứu khoa học công nghệ số 20/2017/HĐ-KHCNTT ngày 18 tháng năm 2017 Chúng biết ơn nguồn tài ngun tính tốn hỗ trợ Viện Khoa học Cơng nghệ Tính tốn thành phố Hồ Chí Minh Viện Khoa học Cơng nghệ Tính tốn TP Hồ Chí Minh Page Các phương pháp lặp để giải toán tựa cân tốn điểm bất động ĐƠN VỊ THỰC HIỆN Phịng thí nghiệm: Cơng nghệ Tốn Ứng dụng Chủ nhiệm đề tài: TS Nguyễn Thị Thu Vân Viện Khoa học Cơng nghệ Tính Tốn, Sở Khoa học Cơng nghệ Tp.Hồ Chí Minh Đại học Khoa học Tự nhiên, VNU-HCM Thành viên đề tài: TS Nguyễn Thị Thu Vân Viện Khoa học Cơng nghệ Tính Tốn, Sở Khoa học Cơng nghệ Tp.Hồ Chí Minh Đại học Khoa học Tự nhiên, VNU-HCM Giáo sư – Tiến sĩ Jean Jacques Strodiot Viện Khoa học Cơng nghệ Tính Tốn, Sở Khoa học Cơng nghệ Tp.Hồ Chí Minh Đại học Namur, Vương Quốc Bỉ Thạc sỹ Đinh Minh Giang Viện Khoa học Cơng nghệ Tính Tốn, Sở Khoa học Cơng nghệ Tp.Hồ Chí Minh Cơ quan phối hợp: Đại học Khoa học Tự nhiên, VNU-HCM Đại học Namur, Vương Quốc Bỉ Viện Khoa học Cơng nghệ Tính tốn TP Hồ Chí Minh Page Các phương pháp lặp để giải toán tựa cân toán điểm bất động KẾT QUẢ NGHIÊN CỨU I BÁO CÁO KHOA HỌC Cho X tập lồi, đóng, khác rỗng khơng gian , cho rỗng ánh xạ đa trị từ X vào X cho với cho song hàm và với chuẩn tập lồi đóng khác , thỏa với hàm lồi Bài toán tựa cân liên kết với tìm với tích vơ hướng ký hiệu toán thỏa Gần cơng trình [39], phương pháp kiểu chiếu tăng cường đề nghị để giải tốn cân song hàm khơng cần thỏa điều kiện giả đơn điệu (xem thêm [40]) Trong trường hợp này, ngược lại với trường hợp tựa đơn điệu, bước lặp, tất siêu phẳng tách tập nghiệm từ bước lặp trước phải lưu trữ để xây dựng bước lặp Cũng cần đề cập tác giả công trình [41] gần đề nghị phương pháp lặp tương tự cho trường hợp toán cân khơng đơn điệu Mục đích chúng tơi phần đề tài giải toán tựa cân với tập ràng buộc phụ thuộc vào biến chứa tập ràng buộc không gian mà khơng địi hỏi giả thiết đơn điệu tập liệu Để thực hiện, xét toán tựa cân Minty để đảm bảo tập nghiệm toán tựa cân gốc tốn Minty khác rỗng, chúng tơi giả thiết tồn điểm bất động ánh xạ nghiệm toán tựa cân cổ điển Minty Điều kiện xem dạng đối ngẫu điều kiện xét cơng trình [38.,42] để giải tốn tựa cân giả đơn điệu Với giả thiết này, chứng minh hội tụ thuật toán hai bước Trước hết chứng minh dãy sinh thuật toán hội tụ đến điểm bất động ánh xạ đa trị sau chúng tơi chứng minh dãy hội tụ nghiệm toán tựa cân Trong báo cáo cho phép chúng tơi giới thiệu ngắn gọn kết chính, định nghĩa, bổ đề, chứng minh người đọc xem chi tiết cơng trình liệt kê phần phụ lục Thuật toán hội tụ Sau thuật toán hội tụ đề nghị để tìm nghiệm tốn tựa cân khơng đơn điệu mà khơng địi hỏi giả thiết đơn điệu tập liệu Viện Khoa học Công nghệ Tính tốn TP Hồ Chí Minh Page Các phương pháp lặp để giải toán tựa cân toán điểm bất động Data Let Step and then stop Otherwise go to Step Find the smallest nonnegative integer such that and set Step Compute If Step Set Consider the half-space Find where denotes the convex closed set Calculate where Set and go back to Step Để thuật toán hội tụ, cần số giả thiết sau: Giả thiết A A is defined on for all A where and is convex on is jointly continuous on , A For all A The multivalued mapping continuous at each A The set is an open convex subset of for all containing , is a nonempty closed convex bounded subset of is *-nonexpansive on and lower semi is nonempty Mệnh đề sau chứng tỏ thuật tốn đề nghị hồn tồn khả thi Mệnh đề 3.1 Each subset In addition, Mệnh đề 3.2 For all is convex, closed, and contains the nonempty set is nonempty, convex and closed due to the assumption A5  and for every , we have the following inequality In particular, when for some , the vector is a solution of problem Từ Mệnh đề 3.2, phần cịn lại chúng tơi giả sử Mệnh đề 3.3 Let  với Then, for all k, the following inequalities hold: Viện Khoa học Cơng nghệ Tính tốn TP Hồ Chí Minh Page Các phương pháp lặp để giải toán tựa cân toán điểm bất động i ii Furthermore, exists and the sequences Mệnh đề 3.4 Let and are bounded  for all Then the sequences and  tends to as Mệnh đề 3.5 Every limit point of the sequence generated by the algorithm belongs to  Từ Mệnh đề 3.1 – 3.5, thu kết hội tụ sau: Định lý 3.6 Assume that Assumption A holds Then the whole sequence algorithm converges to some  belonging to Tiếp theo chứng minh điểm tụ dãy sinh thuật toán toán generated bby the nghiệm Mệnh đề 3.7 Assume that Then the two associated subsequences and are bounded  Mệnh đề 3.8 Let be a limit point of the bounded sequence that a subsequence of converges to Then Assume that and is a solution of problem  Mệnh đề 3.9 The following inequality holds for all where Furthermore, as Mệnh đề 3.10 Let as j to as  Then there exists a subsequence of Furthermore, is a solution of problem Viện Khoa học Công nghệ Tính tốn TP Hồ Chí Minh which converges  Page RACSAM DOI 10.1007/s13398-016-0338-7 ORIGINAL PAPER Strong convergence of an iterative method for solving the multiple-set split equality fixed point problem in a real Hilbert space Dinh Minh Giang1 · Jean Jacques Strodiot1,2 Van Hien Nguyen1,2 · Received: March 2016 / Accepted: 15 September 2016 © Springer-Verlag Italia 2016 Abstract In this paper we consider the problem of minimizing a non necessarily differentiable convex function over the intersection of fixed point sets associated with an infinite family of multivalued quasi-nonexpansive mappings in a real Hilbert space The new algorithm allows us to solve problems when the mappings are not necessarily projection operators or when the computation of projections is not an easy task The a priori knowledge of operator norms is avoided and conditions to get the strong convergence of the new algorithm are given Finally the particular case of split equality fixed point problems for family of multivalued mappings is displayed Our general algorithm can be considered as an extension of Shehu’s method to a larger class of problems Keywords Multiple-set split equality fixed point problem · Quasi-nonexpansive operator · Demiclosed operator · Strong convergence Mathematics Subject Classification 65K05 · 90C25 · 47H10 The authors would like to thank the referee and the Associate Editor for their valuable comments This research is funded by the Department of Science and Technology at Ho Chi Minh City, Vietnam Support provided by the Institute for Computational Science and Technology (ICST) at Ho Chi Minh City is gratefully acknowledged B Jean Jacques Strodiot jjstrodiot@fundp.ac.be Dinh Minh Giang giang.dm@icst.org.vn Van Hien Nguyen vhnguyen@fundp.ac.be Institute for Computational Science and Technology at Ho Chi Minh City (ICST), Ho Chi Minh City, Vietnam University of Namur, Namur, Belgium D M Giang et al Introduction Let H be a real Hilbert space In this paper we consider the following problem: ∞ (P) Find z ∗ ∈ ∩i=1 Fix(Wi ) such that f (z ∗ ) = where f : H → R is a nonnegative lower semicontinuous ( l.s.c.) convex function defined on H and each Wi (i ≥ 1) : H → H is a quasi-nonexpansive multivalued mapping such that ∞ Fix(W )  = ∅ Here, Fix(W ) denotes, for each i, the set of fixed points of the mapping ∩i=1 i i Wi An important particular case of problem (P) is the multiple-set split equality fixed point problem which can be expressed as ∞ ∞ Fix(Ri ) and y ∗ ∈ ∩i=1 Fix(Ti ) such that Ax ∗ = By ∗ (M S E F P) Find x ∗ ∈ ∩i=1 where A : H1 → H3 and B : H2 → H3 are two bounded linear operators, and Ri : H1 → C B(H1 ) and Ti : H2 → C B(H2 ) are, for each i ≥ 1, quasi-nonexpansive mappings defined on H1 and H2 , respectively Here H1 , H2 and H3 are three infinite-dimensional real Hilbert spaces and C B(H1 ) and C B(H2 ) denote the collection of all nonempty closed and bounded subsets of H1 and H2 , respectively When H = H1 × H2 and the function f and the operators Wi (i ≥ 1) are defined for all (x, y) ∈ H by f :H →R Wi : H → H f (x, y) = 21 Ax − By Wi (x, y) = (Ri (x), Ti (y)), it is easy to see that Fix(Wi ) = Fix(Ri )× Fix(Ti ) and that problem (P) coincides with problem (M S E F P) When B is the identity operator and H2 = H3 , problem (M S E F P) becomes ∞ ∞ Fix(Ri ) such that Ax ∗ ∈ ∩i=1 Fix(Ti ) (M SC F P) Find x ∗ ∈ ∩i=1 Taking H = H1 and Wi = Ri for all i and defining f : H → R by ∞ f (z) = 1 (I − Ti )Az for all z ∈ H, i=1 it is immediate that problem (P) coincides with problem (M SC F P) Furthermore, when for each i, the operators Ri and Ti are orthogonal projections onto nonempty closed convex subsets Ci and Q i , respectively, it follows that for each i Fix(Ri ) = Ci and Fix(Ti ) = Q i So, in that case, problem (M S E F P) coincides with the general split equality problem: ∞ ∞ Ci and y ∗ ∈ ∩i=1 Q i such that Ax ∗ = By ∗ , (G S E P) Find x ∗ ∈ ∩i=1 and problem (M SC F P) coincides with the general split feasibility problem: ∞ ∞ Find x ∗ ∈ ∩i=1 Ci such that Ax ∗ ∈ ∩i=1 Qi When the mappings Ri = R and Ti = T for all i, where R : H1 → H1 and T : H2 → H2 are two quasi-nonexpansive mappings, problem (M S E F P) becomes the following problem (S E F P) Find x ∗ ∈ Fix(R) and y ∗ ∈ Fix(T ) such that Ax ∗ = By ∗ Strong convergence of an iterative method for solving and, when H2 = H3 and B = I , (SC F P) Find x ∗ ∈ Fix(R) such that Ax ∗ ∈ Fix(T ) Problem (S E F P) is called the split equality fixed point problem and has been recently studied by Moudafi in [18,19] while problem (SC F P) corresponds to the split common fixed point problem introduced by Censor and Segal in [4] Problem (SC F P) is at the core of the modelling of many significant real-world inverse problems such as radiation therapy treatment planning, data compression, magnetic resonance imaging, neural networks and graph matching (more details and other examples can be found in [1,16]) On the other hand, as mentioned in [2], problem (S E F P) has also other important applications in different areas of applied mathematics, such as fully discretized models of inverse problems which arise from phase retrieval and in medical image reconstruction (see, for example, [3,4,6,8]) Many methods have been proposed for solving problems (M S E F P) and (M SC F P) or their particular cases Let us mention here the papers [1,5,6,9,10,12,15] and [19,20,22] However, most of them use iterative methods depending on the operator norms A and B or on the largest eigenvalue of A∗ A and B ∗ B, which is not, in general, an easy task To overcome this numerical difficulty, López et al [16] and more recently Zhao [28], Zhao and Wang [29] and Zhao and Zhang [31] have proposed iterative methods for solving problems (S E F P) and (SC F P) without requiring any prior knowledge of operator norms In [28,29] a weak convergent algorithm has been derived while in [31] a strong convergence result has been obtained but under the assumption that the operator W is firmly quasi-nonexpansive (see also [22] for problem (M S E F P)) Most recent papers on the split equality fixed point problem have been written for singlevalued mappings However, fixed point theory for multivalued mappings has many useful applications in various fields as, for example, in game theory and in mathematical economics (see, for example, [6,11,21,25–27,30]) So it is natural to extend the known results on split fixed point problems for single-valued mappings to the setting of multivalued mappings This has been done in Shehu [22] for the split equality fixed point problem Our aim is to develop such a theory but for solving the more general problem (P) where f : H → R is a nonnegative l.s.c convex function and for each i, Wi : H → H is a quasi-nonexpansive (multivalued) mapping having a nonempty fixed point set Fix(Wi ) In this paper, the solution set  of this problem is assumed to be nonempty and, contrary to [31], the multivalued mappings Wi are not supposed to be firmly quasi-nonexpansive Furthermore, projections onto convex closed sets are avoided because they are not easy to compute, except in simple cases Finally, let us mention that it can be easily proved that the set of fixed points of a multivalued quasi-nonexpansive mapping T is closed and convex under the assumption that T ( p) = { p} for all fixed points p of T The paper is organized as follows In the next section, we recall some lemmas useful for proving our main results In Sect we propose and analyze an algorithm for solving problem (P) and we give conditions to obtain its strong convergence Finally we study in Sect the particular case of the multiple-set split equality fixed point problem Preliminaries Let C B(H ) be the collection of all nonempty, closed and bounded subsets of H The distance d(z, K ) from z ∈ H to K ∈ C B(H ) is defined by D M Giang et al d(z, K ) = inf { z − u | u ∈ K } and the Hausdorff metric H˜ on C B(H ) is defined by H˜ (X, Y ) = max{sup d(x, Y ) , sup d(y, X )} ∀ X, Y ∈ C B(H ) x∈X y∈Y The following definitions are well-known Definition 2.1 Let W : H → C B(H ) be a multivalued mapping An element z ∗ ∈ H is said to be a fixed point of W if z ∗ ∈ W (z ∗ ) The set of fixed points of W is denoted by Fix(W ) The mapping W is said to be (i) nonexpansive if H˜ (W (x), W (y)) ≤ x − y ∀ x, y ∈ H ; (ii) quasi-nonexpansive if Fix(W )  = ∅ and H˜ (W (x), W (z ∗ )) ≤ x − z ∗ ∀ x ∈ H, z ∗ ∈ Fix(W ); (iii) demi-closed at the origin if for any sequence {z k } ⊂ H with z k  z ∗ (weak convergence) and d(z k , W (z k )) → 0, we have z ∗ ∈ W (z ∗ ) Lemma 2.1 [7] Let E be a uniformly convex Banach space and let Br (0) = {x ∈ E | x ≤ r } be a closed ball with radius r > Then there exists a continuous, strictly increasing and convex function g : [0, ∞) → [0, ∞) with g(0) = such that ∞ 2 ∞      λk xk  ≤ λk xk − λi λ j g( xi − x j )    k=1 k=1 for any i, j ∈ N, i < j, where {xk }k≥1 ⊂ Br (0) and λk ≥ 0, ∞ k=1 λk = Lemma 2.2 Let H be a real Hilbert space The following equalities are satisfied for any x, y ∈ H : 2x, y = x + y − x − y = x + y − x − y x + y ≤ x + 2y, x + y The following lemma is due to He and Yang [13] and is a generalization of a very wellknown result due to Xu [24] Lemma 2.3 [13, Lemma 8] Assume {u k } is a sequence of nonnegative real numbers such that, for all k, u k+1 ≤ (1 − ηk )u k + ηk δk u k+1 ≤ u k − θk + μk where {ηk } ⊂ (0, 1), {θk } ⊂ R+ , and {δk } and {μk } are two sequences in R such that ∞ (i) k=1 ηk = ∞ (ii) limk→∞ μk = (iii) liml→∞ θkl = ⇒ lim supl→∞ δkl ≤ for any subsequence {kl } ⊂ {k} Then the sequence {u k } tends to as k → ∞ Strong convergence of an iterative method for solving Lemma 2.4 [17, Lemma 1.3] Let {k } be a sequence of real numbers that does not decrease at infinity, in the sense that there exists a subsequence {k j } of {k } which satisfies k j < k j +1 for all j ≥ Also consider the sequence of integers {τ (k)}k≥k0 defined, for each k ≥ k0 , by τ (k) = max{l ≤ k | l < l+1 } Then, {τ (k)}k≥k0 is a nondecreasing sequence verifying limk→∞ τ (k) = ∞ and, for all k ≥ k0 , it holds that τ (k) ≤ τ (k)+1 and k ≤ τ (k)+1 Main results In this section, our aim is to present a strongly convergent algorithm for finding a solution of problem (P) For this purpose, we introduce a contraction mapping in our algorithm Let us recall that c : H → H is a contraction with constant ρ ∈ [0, 1) when c(x) − c(y) ≤ ρ x − y for all x, y ∈ H A well-known example of contraction is given by c(x) = s for all x ∈ H where s is any fixed vector in H The algorithm we consider in this section for solving problem (P) has the following form: Algorithm 3.1 Given a contraction c : H → H , and a starting point z ∈ H , the sequences {sk } and {z k } are iteratively generated by  sk = z k − λk dk ∞ z k+1 = tk c(z k ) + (α0,k − tk )sk + i=1 αi,k ri,k where dk is a search direction, tk ∈ (0, 1), and for all i, αi,k ∈ (0, 1), ri,k ∈ Wi (sk ) Furthermore, the step λk is chosen in such a way that  ρk f (z k ) if dk  = dk λk = otherwise where ρk ∈ (0, 2) Let us recall here that f : H → R is a nonnegative l.s.c convex function and that for each ∞ Fix(W )  = ∅ i, Wi : H → H is a quasi-nonexpansive mapping such that ∩i=1 i When the direction dk coincides with the gradient ∇ f (z k ), the step tk = and each operator Wi = W where W is chosen as the projection onto a nonempty closed convex subset S of H , Algorithm 3.1 becomes the Gradient Projection Algorithm studied in [23] In this case, z k+1 is computed from z k as follows:  sk = z k − λk ∇ f (z k ) z k+1 = αk sk + (1 − αk )PS (sk ) Algorithm 3.1 is particularly interesting when the operator W is not a projection operator or when the projection is not easy to compute as in the following example given in [14, Problem 4.1] Let S = S1 ∩ S2 be the intersection of the two closed balls: S1 = {z ∈ R64 | z ≤ 1} and S2 = {z ∈ R64 | z − (1, 1, 0, , 0) ≤ 1} D M Giang et al The set S is nonempty, and the computations of PS1 and PS2 are easy However the calculation of PS is not easy In this case it is more judicious to consider the operator W defined by W (z) = 1 PS (z) + PS2 (z) for all z ∈ R64 2 and to use Algorithm 3.1 rather than the Gradient Projection Algorithm It can be proven that the operator W is quasi-nonexpansive, and that Fix(W ) = S In order to prove the convergence of the sequence {z k } generated by Algorithm 3.1, we denote by K the index set {k ∈ N | dk  = 0}, and towards this aim we introduce the following assumptions on the sequences {z k }, {dk } and {λk } generated by Algorithm 3.1: (A1) dk , z k − z ∗  ≥ f (z k ) for all k ∈ N and for all z ∗ ∈ ; (A2) < λ ≤ λk ≤ λ for all k ∈ K; (A3) inf k∈K [ρk (2 − ρk )] > Any vector dk ∈ ∂ f (z k ) is an example of direction satisfying assumption (A1) Indeed, since f (z ∗ ) = 0, we have by definition of the subdifferential of a convex function that ≥ f (z k ) + dk , z ∗ − z k  and thus Assumption (A1) is satisfied On the other hand, from the definition of λk and Assumption (A1), we easily observe that if k ∈ / K, then dk = 0, f (z k ) = 0, λk = and sk = z k In this case, z k is a solution of problem (P) So, in the sequel, we will assume that K = N0 = N\{0} Finally, we impose the following conditions on the sequences of parameters: ∞ (C1) i=0 αi,k = for each  k ≥ 1; (C2) limk→∞ tk = and ∞ k=0 tk = ∞; (C3) for each i ≥ 1, lim inf k→∞ α0,k αi,k > 0; (C4) tk < α0,k , for each k ≥ 1; ∞ Fix(W ) we have W (z ∗ ) = {z ∗ }; (C5) for each i ≥ and for each z ∗ ∈ ∩i=1 i i √ (C6) ρ ∈ [0, 1/ 2) As mentioned in Shehu [22], a possible choice for the sequences {αi,k }k and {tk }k is given by   1− αi,k = 2i α0,k = k 2k+1 , k 2k+1  tk = ∀ i ≥ 1, k ≥ 1; 2k+1 ∀ k ≥ Thanks to these conditions, we are now ready to prove the strong convergence of the sequence {z k } generated by Algorithm 3.1 For this purpose, we need to obtain the following intermediate results Proposition 3.1 Let z ∗ be a solution of problem (P) and let {z k }, {dk }, {sk } and {λk } be the sequences generated by Algorithm 3.1 Then, for all k ≥ 1, one has the following inequalities sk − z ∗ ≤ z k − z ∗ − ρk (2 − ρk ) [ f (z k )]2 ≤ z k − z ∗ dk (1) Strong convergence of an iterative method for solving and z k+1 − z ∗ ≤ (1 − tk (1 − 2ρ )) z k − z ∗ + 2tk c(z ∗ ) − z ∗ − (1 − tk )ρk (2 − ρk ) f (z k )2 dk (2) Proof Let k ≥ We obtain (1) by using successively assumption (A1), the definition of λk and assumption (A3) Indeed, sk − z ∗ = (z k − z ∗ ) − λk dk = z k − z ∗ − 2λk z k − z ∗ , dk  + λ2k dk ≤ z k − z ∗ − 2λk f (z k ) + λ2k dk = z k − z ∗ − 2ρk [ f (z k )]2 dk ρk2 [ f (z k ]2 dk [ f (z k )]2 dk + = z k − z ∗ − ρk (2 − ρk ) ≤ z k − z ∗ On the other hand, using (C5), the definition of H˜ and the nonexpansiveness of Wi , we can write z k+1 − z ∗ = c(z k ) + (α0,k − tk )sk + = (α0,k − tk )(sk − z ∗ ) + ≤ (α0,k − tk ) sk − z ∗ + ∞  αi,k ri,k − z ∗ i=1 ∞  αi,k (ri,k − z ∗ ) + tk (c(z k ) − z ∗ ) i=1 ∞  ri,k − z ∗ + tk s − z ∗ i=1 = (α0,k − tk ) sk − z ∗ + ∞  αi,k (d(ri,k , Wi (z ∗ ))2 + tk c(z k ) − z ∗ i=1 ≤ (α0,k − tk ) sk − z ∗ + ∞  αi,k ( H˜ (Wi (sk ), Wi (z ∗ ))2 + tk c(z k ) − z ∗ i=1 ≤ (α0,k − tk ) sk − z ∗ + ∞  αi,k sk − z ∗ + tk c(z k ) − z ∗ i=1 = (1 − tk ) sk − z ∗ + tk c(z k ) − z ∗ (3) Finally, since c is a contraction with constant ρ ∈ [0, 1), we can develop c(z k ) − z ∗ as follows: c(z k ) − z ∗ ≤ c(z k ) − c(z ∗ ) + c(z ∗ ) − z ∗ ≤ 2ρ z k − z ∗ + c(z ∗ ) − z ∗ Hence, inequality (2) can be immediately obtained after inserting (1) and (4) into (3) Let z ∗ (4)   Corollary 3.1 be a solution of problem (P) Then the sequences {z k }, {sk } and {c(z k )} generated by Algorithm 3.1 are bounded Furthermore, for all k, the following inequality holds z k+1 − z ∗ ≤ z k − z ∗ − θk + μk D M Giang et al where f (z k ) θk = (1 − tk )ρk (2 − ρk ) d , k ∗ ∗ μk = tk c(z ) − z → as k → ∞ √ Proof Since ≤ ρ < 1/ and tk ∈ (0, 1) for all k, we can deduce from the second part of Proposition 3.1 that, for all k, ∗ ∗ z k+1 − z ∗ ≤ (1 − tk (1 − 2ρ )) z k − z ∗ + tk (1 − 2ρ ) 1−2ρ c(z ) − z ≤ max{ z k − z ∗ , 1−2ρ c(z ∗ ) − z ∗ } and thus, by induction, that z k+1 − z ∗ ≤ max{ z − z ∗ , c(z ∗ ) − z ∗ } − 2ρ Consequently, the sequences {z k }, {sk } and {c(z k )} are bounded, and since for all i and k, ri,k − z ∗ ≤ sk − z ∗ , the sequences {ri,k }k are also bounded for each i Finally, since − tk (1 − 2ρ ≤ 1, the required inequality is a direct consequence of the second part of Proposition 3.1 Furthermore, μk → because tk →   In the next proposition we give another upper bound on the sequence { z k − z ∗ } Proposition 3.2 Let z ∗ be a solution of problem (P) and let {z k } be the sequence generated by Algorithm 3.1 Then there exists M > such that the following inequality holds for all k: z k+1 − z ∗ ≤ (1 − ηk ) z k − z ∗ + ηk δk (5) where 2(1 − ρ)tk tk M + z k+1 − z ∗ , c(z k ) − z ∗  and δk = − ρtk 2(1 − ρ) − ρ ∞ Furthermore, ηk → as k → ∞ and k=1 ηk = ∞ ηk = Proof Using the definition of z k+1 and the second inequality of Lemma 2.2, we obtain successively |z k+1 − z ∗ = tk c(z k ) + (α0,k − tk )sk + ∞  αi,k ri,k − z ∗ i=1 = (α0,k − tk )(sk − z ∗ ) + ∞  αi,k (ri,k − z ∗ ) + tk (c(z k ) − z ∗ ) i=1 ≤ (α0,k − tk )(sk − z ∗ ) + ≤ [(α0,k − tk ) sk − z ∗ + ∞  i=1 ∞  αi,k (ri,k − z ∗ ) + 2tk c(z k ) − z ∗ , z k+1 − z ∗  αi,k ri,k − z ∗ ]2 i=1 + 2tk c(z k ) − c(z ∗ ), z k+1 − z ∗  + 2tk c(z ∗ ) − z ∗ , z k+1 − z ∗  Strong convergence of an iterative method for solving Since, by (C1), we have ∞ i=0 αi,k = for each k and since ri,k − z ∗ ≤ sk − z ∗ ≤ z k − z ∗ , |z k+1 − z ∗ ≤ (1 − tk )2 sk − z ∗ + 2tk ρ z k − z ∗ z k+1 − z ∗ + 2tk c(z ∗ ) − z ∗ , z k+1 − z ∗  ≤ (1 − tk )2 z k − z ∗ + tk ρ[ z k − z ∗ + z k+1 − z ∗ ] + 2tk c(z ∗ ) − z ∗ , z k+1 − z ∗  Consequently, (1 − tk )2 + ρtk 2tk z k − z ∗ + c(z ∗ ) − z ∗ , z k+1 − z ∗  − ρtk − ρtk tk2 − 2tk + ρtk = z k − z ∗ + z k − z ∗ − ρtk − ρtk 2tk + c(z ∗ ) − z ∗ , z k+1 − z ∗  − ρtk 2(1 − ρ)tk tk z k − z ∗ 2(1 − ρ)tk z k − z ∗ + = 1− − ρtk − ρtk 2(1 − ρ) 2(1 − ρ)tk + c(z ∗ ) − z ∗ , z k+1 − z ∗  − ρtk − ρ z k+1 − z ∗ ≤ Since the sequence {z k } is bounded, there exists M > such that z k − z ∗ ≤ M for all k Hence, setting for all k ηk = 2(1 − ρ)tk − ρtk and δk = tk M + z k+1 − z ∗ , c(z k ) − z ∗ , 2(1 − ρ) − ρ we obtain that z k+1 − z ∗ ≤ (1 − ηk ) z k − z ∗ + ηk δk  ∞ Finally, from tk → and ∞ k=1 tk = ∞, we deduce that ηk → and k=1 ηk = ∞   Note that the mapping c is a contraction with constant ρ ∈ [0, 1), so that the mapping I − c, where I denotes the identity mapping, is strongly monotone with modulus − ρ > Consequently the variational inequality problem: Find z¯ , solution of (P), such that ¯z − c(¯z ), z − z¯  ≥ for each solution z of problem (P) has exactly one solution denoted z ∗ Proposition 3.3 below gives some important properties that will be used to prove our strong convergence theorem Proposition 3.3 Let z ∗ be the solution of problem (P) that satisfies the property z ∗ − c(z ∗ ), z − z ∗  ≥ for each solution z of problem (P) Let {kl } be a subsequence of {k} such that for each i θkl → and skl − ri,kl → as l → ∞ where ri,kl ∈ Wi (skl ) for each i and l D M Giang et al Then (i) f (z kl ) → and skl − z kl → 0, (ii) any weak limit point of {z kl } is a solution of (P), (iii) lim supl→∞ δkl ≤ Proof (i) Since θkl → when l → ∞ and < λ ≤ λkl = using (C2) and (A3), that ρkl f (z kl ) dkl for all l, we have, f (z kl ) → and λkl dkl → dkl So dkl → because λkl ≥ λ > 0, and consequently f (z kl ) = f (z kl ) dkl → dkl Finally, by definition of sk , we have for all l that skl − z kl = λkl dkl Since dkl → and {λkl } is bounded by assumption, we can immediately deduce that skl − z kl → as l → ∞ (ii) Let z¯ be a weak limit point of the sequence {z kl } Then z kl  z¯ (in fact a subsequence of {z kl }) Since skl − z kl → 0, we have that skl  z¯ Hence, using the assumption ≤ d(skl , Wi (skl )) ≤ skl − ri,kl → for each i, it follows from the demiclosedness of the operators Wi − I that z¯ ∈ Fix(Wi ) for each ∞ Fix(W ) i, and thus that z¯ ∈ ∩i=1 i On the other hand, the function f being weakly l.s.c and z kl  z¯ , we can write ≤ f (¯z ) ≤ lim inf f (z kl ) = l→∞ where we have used the fact that f (z kl ) → Consequently, f (¯z ) = and z¯ is a solution of problem (P) tk M (iii) By definition of δk and since ρ < and 2(1−ρ) → as k → ∞, we have only to prove that lim sup z kl +1 − z ∗ , c(z kl ) − z ∗  ≤ l→∞ ∞ where z kl +1 = tkl cz kl + (α0,kl − tkl )skl + i=1 αi,kl ri,kl and ri,kl ∈ Wi (skl ) for each i For this purpose, using k instead of kl , we can write that z k+1 − z ∗ , c(z k ) − z ∗  = tk c(z k ) − z ∗ , c(z ∗ ) − z ∗  ∞  + (α0,k − tk )sk − z ∗ , c(z ∗ ) − z ∗  + αi,k ri,k − z ∗ , c(z ∗ ) − z ∗  i=1 where ri,k ∈ Wi (sk ) for each i Since tk → and the sequence {c(z k )} is bounded, we have immediately that tk c(z k ) − z ∗ , c(z ∗ ) − z ∗  → as k → ∞ Strong convergence of an iterative method for solving On the other hand, we have that sk − z k → and sk − ri,k → when k → ∞ and that the following equalities hold for all k and i sk − z ∗ = sk − z k + z k − z ∗ , ∗ ri,k − z = ri,k − sk + sk − z k + z k − z ∗ So, to get (iii), it remains to prove that lim sup z kl − z ∗ , c(z ∗ ) − z ∗  ≤ l→∞ i.e., lim inf z kl − z ∗ , z ∗ − c(z ∗ ) ≥ l→∞ (6) The sequence {z k } being bounded, there exists a subsequence {z kl } of {z k } that weakly converges to some z¯ and that satisfies the equalities lim inf z kl −z ∗ , z ∗ −c(z ∗ ) = lim z kl j −z ∗ , z ∗ −c(z ∗ ) = ¯z −z ∗ , z ∗ −c(z ∗ ) (7) l→∞ j→∞ By assumption, we have that z ∗ − c(z ∗ ), z − z ∗  ≥ for every solution z of problem (P) Consequently, z¯ being a solution of (P), we obtain that z ∗ − c(z ∗ ), z¯ − z ∗  ≥ 0, and thanks to (7), that (6) is true So lim supk→∞ δk ≤   Theorem 3.1 Suppose that Assumptions (A1)–(A4) are satisfied and that each operator Wi is demi-closed at the origin Then the sequence {z k } generated by Algorithm 3.1 is strongly convergent to the solution z ∗ of (P) that satisfies the following variational inequality problem z ∗ − c(z ∗ ), z − z ∗  ≥ for each solution z of (P) (8) Proof Let z ∗ be the unique solution of the variational inequality problem (8) In order to prove the strong convergence of the sequence {z k } to z ∗ , first we observe that, for each k, we have z k+1 − z ∗ = α0,k (sk − z ∗ ) + ∞  αi,k (ri,k − z ∗ ) + tk (c(z k ) − sk ) i=1 = α0,k (sk − z ∗ ) + ∞  αi,k (ri,k − z ∗ ) + tk2 c(z k ) − sk i=1 + 2tk α0,k (sk − z ∗ ) + ∞  αi,k (ri,k − z ∗ ), c(z k ) − sk  i=1 Then, using successively Lemma 2.1 with < i, the quasi-nonexpansiveness of each Wi (namely, ri,k − z ∗ ≤ sk − z ∗ ), Condition (C1) and inequality (1), we obtain the existence of a continuous strictly increasing and convex function g : [0, ∞) → [0, ∞) with g(0) = such that D M Giang et al z k+1 − z ∗ ≤ α0,k sk − z ∗ + ∞  αi,k ri,k − z ∗ − α0,k αi,k g( sk − ri,k ) i=1 + 2tk α0,k (sk − z ∗ ) + ∞  αi,k (ri,k − z ∗ ), c(z k ) − sk  i=1 + tk2 c(z k ) − sk ≤ z k − z ∗ − α0,k αi,k g( sk − ri,k ) + tk2 c(z k ) − sk ∞  + 2tk α0,k (sk − z ∗ ) + αi,k (ri,k − z ∗ ), c(z k ) − sk  (9) i=1 Since, by Corollary 3.1, the sequences {sk }, {c(z k )}, and {ri,k } are bounded for all i, there exists M > such that for all k tk c(z k ) − sk + 2α0,k (sk − z ∗ ) + ∞  αi,k (ri,k − z ∗ ), c(z k ) − sk  ≤ M i=1 So, for all k, we deduce from (9) that z k+1 − z ∗ ≤ z k − z ∗ − α0,k αi,k g( sk − ri,k ) + tk M (10) On the other hand, from Corollary 3.1, we also have for all k that ≤ θk ≤ z k − z ∗ − z k+1 − z ∗ + μk (11) with μk → as k → ∞ Now we consider two cases: Case 1: There exists k0 such that the sequence { z k − z ∗ }k≥k0 is nonincreasing In this case, the sequence { z k − z ∗ } is convergent, and lim [ z k+1 − z ∗ − z k − z ∗ ] = k→∞ (12) Next, we prove that θk → as k → ∞ and that lim sk − ri,k = for all ri,k ∈ Wi (sk ) k→∞ Since μk → 0, we directly deduce from (11) that θk → as k → ∞ On the other hand, since tk → 0, we can write from (10) that ≤ α0,k αi,k g( sk − ri,k ) → as k → ∞ Hence, using Condition (C3), we get for all i that g( sk − ri,k ) → as k → ∞ Since g is continuous and strictly increasing with g(0) = 0, we have that lim sk − ri,k = for all i and ri,k ∈ Wi (sk ) k→∞ Consequently, thanks to Proposition 3.3, we can write that lim sup δk ≤ k→∞ Setting u k = z k − z ∗ for all k, it is now easy to see, using Proposition 3.2, that all the assumptions of Lemma 2.3 are satisfied, and thus the sequence { z k − z ∗ } → as k → ∞ Strong convergence of an iterative method for solving Case 2: There exists a subsequence { z k j − z ∗ } of { z k − z ∗ } such that z k j − z ∗ < z k j +1 − z ∗ for all j (i) First we show that θτ (k) → and sτ (k) − ri,τ (k) → as k → ∞ where τ (k) = max{l ∈ N | l ≤ k, zl − z ∗ < zl+1 − z ∗ } For this purpose, using Lemma 2.4 with k = z k − z ∗ , we obtain that τ (k) → ∞ and that for all k ≥ k0 z τ (k) − z ∗ < z τ (k)+1 − z ∗ and z k − z ∗ < z τ (k)+1 − z ∗ From Corollary 3.1, we have that μk → when k → ∞, and that the following inequalities hold true for all k: ≤ θτ (k) ≤ z τ (k) − z ∗ − z τ (k)+1 − z ∗ + μτ (k) ≤ μτ (k) Hence θτ (k) → 0, and using similar arguments as the ones considered in Case 1, we can easily show that for all i g( sτ (k) − ri,τ (k) ) ) → as k → ∞, and thus lim sτ (k) − ri,τ (k) = k→∞ for all i and ri,τ (k) ∈ Wi (sτ (k) ) (ii) Since θτ (k) → and sτ (k) − ri,τ (k) → as k → ∞, it follows from Proposition 3.3 that the following statements hold true: sτ (k) − z τ (k) → as k → ∞; f (z τ (k) ) → as k → ∞; any weak limit point of {z τ (k) } is a solution of (P); lim supk→∞ δτ (k) ≤ (iii) Finally, we show that the sequence {z k } strongly converges to z ∗ ∈  First, from Proposition 3.2, we have that for all k < z τ (k)+1 − z ∗ − z τ (k) − z ∗ ≤ ητ (k) (δτ (k) − z τ (k) − z ∗ ) (13) Hence ητ (k) z τ (k) − z ∗ < ητ (k) δτ (k) for all k So ητ (k) > for all k, and since lim supk→∞ δτ (k) ≤ 0, we obtain immediately that z τ (k) − z ∗ → as k → ∞, and by (13), that z τ (k)+1 − z ∗ → as k → ∞ To conclude, using Lemma 2.4 with k = z k − z ∗ for all k, we immediately deduce that ≤ z k − z ∗ ≤ z τ (k)+1 − z ∗ for all k, and thus z k − z ∗ → as k → ∞ But this means that the sequence {z k } strongly converges to z ∗ ∈    Remark 3.1 When the contraction mapping c is defined by c(z) = w for all z ∈ H where w is a fixed vector in H , the sequence {z k } generated by Algorithm 3.1 strongly converges to the projection of w onto the solution set of (P) Indeed, in that case, (8) becomes D M Giang et al z ∗ − w, z − z ∗  ≥ for each solution z of (P), which is equivalent to the fact that z ∗ is the projection of w onto the solution set of (P) A particular case In this section we examine the particular case of the multiple-set split equality fixed point problem considered by Shehu in [22] In that setting, for every (x, y) ∈ H1 × H2 , the function f and the operators Wi , i ≥ 1, are defined by f (x, y) = Ax − By and Wi (x, y) = (Ri (x), Ti (y)) for i ≥ Here the operators A : H1 → H3 and B : H2 → H3 are linear and bounded, and for each positive integer i, the operators Ri : H1 → C B(H1 ) and Ti : H2 → C B(H2 ) are two infinite families of multivalued quasi-nonexpansive mappings such that ∞ Fix(Ri )  = ∅ and ∩i=1 ∞ ∩i=1 Fix(Ti )  = ∅ With this choice of data, problem (P) becomes ∞ ∞ (M S E F P) Find x ∈ ∩i=1 Fix(Ri ) and y ∈ ∩i=1 Fix(Ti ) such that Ax = By Here the function f is convex and of class C , and for every (x, y) ∈ H1 × H2 , we have ∇ f (x, y) = (A∗ (Ax − By), −B ∗ (Ax − By)) Furthermore, we consider a point s = (u, v) in H1 × H2 and we define a contraction c on H1 × H2 by c(x, y) = (u, v) for all (x, y) ∈ H1 × H2 Then, setting z k = (xk , yk ) and sk = (u k , vk ) and choosing dk = ∇ f (xk , yk ) in Algorithm 3.1, we obtain the following algorithm for solving problem (M S E F P): Algorithm 4.1 Given s = (u, v) ∈ H1 × H2 and a starting point z = (x1 , y1 ) ∈ H1 × H2 , the sequences (u k , vk ) and (xk , yk ) are iteratively generated by ⎧ u k = xk − λk A∗ (Axk − Byk ) ⎪ ⎪ ⎨ ∞ xk+1 = tk u + (α0,k − tk )u k + i=1 αi,k ri,k ∗ v = y + λ B (Ax − By ) ⎪ k k k k k ⎪ ∞ ⎩ yk+1 = tk v + (α0,k − tk )vk + i=1 αi,k si,k where tk ∈ (0, 1) and for all i, αi,k ∈ (0, 1), ri,k ∈ Ri (u k ) and si,k ∈ Ti (vk ) Furthermore, the step λk is chosen in such a way that  ρk f (xk ,yk ) if ∇ f (xk , yk )  = λk = ∇ f (xk ,yk ) otherwise where ρk ∈ (0, 2) It is easy to see (from Proposition 3.2 in [23]) that in this particular situation, assumption (A1) is satisfied for all k and that < λ ≤ λk is true for all k whenever assumption (A3) holds Furthermore, the contraction c being constant, its contraction constant ρ can be taken equal to zero and condition (C5) becomes the following condition used by Shehu in [22, Theorem 3.1]: Strong convergence of an iterative method for solving ∞ Fix(R ) and each q ∈ ∩∞ Fix(T ), one has: (C5a) for each p ∈ ∩i=1 i i i=1 Ri ( p) = { p} and Ti (q) = {q} for each i ≥ So the following strong convergence theorem due to Shehu [22] can be easily deduced from our Theorem 3.1 Theorem 4.1 [22, Theorem 3.1] Suppose that assumptions (A3), (C1)–(C4) and (C5a) are satisfied and that each operator Ri and Ti is demi-closed at the origin Suppose also that λk ≤ λ for all k Let s = (u, v) be a fixed element of H1 × H2 Then the sequence {(xk , yk )} generated by Algorithm 4.1 is strongly convergent to the solution (x ∗ , y ∗ ) of problem (M S E F P) satisfying the condition (x ∗ , y ∗ ) = P (u, v) where  denotes the solution set of problem (M S E F P) Conclusions In this paper the minimization of a convex nonnegative function over the intersection of an infinite number of fixed point sets has been studied The 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