A splitting algorithm for coupled system of primal dual monotone inclusions

Transcription

1 A splitting algorithm for coupled system of primal dual monotone inclusions B`ăng Công Vũ UPMC Université Paris 06 Laboratoire Jacques-Louis Lions UMR CNRS Paris, France Abstract We propose a splitting algorithm for solving a coupled system of primal-dual monotone inclusions in real Hilbert spaces. The weak convergence of the algorithm proposed is proved. Applications to minimization problems is demonstrated. Keywords: coupled system, monotone inclusion, monotone operator, operator splitting, cocoercivity, forward-backward algorithm, composite operator, duality, primal-dual algorithm Mathematics Subject Classifications H05, 49M29, 49M27, 90C25 1 Introduction Various problems in applied mathematics such as evolution inclusions [2], partial differential equations[1, 30, 32], mechanics[31], variational inequalities[10, 29], Nash equilibria[4], and optimization problems [6, 16, 22, 27, 37, 43], reduce to solving monotone inclusions. The simplest monotone inclusion is to find a zero point of a maximally monotone operator A acting on a real Hilbert space H. This problem can be solved efficiently by the proximal point algorithm when the resolvent of A is easy to implement numerically [41] see [11, 13, 14, 25, 34, 35, 38] in the context of variable metric. Thisproblemwas thenextended to theproblem of findingazero of thesumof a maximally monotone operator A and a cocoercive operator B. In this case, we can used the forward-backward splitting algorithm [2, 18, 32, 43] see [26] in the context of variable metric. When A has a structure, for examples, mixtures of composite, Lipschitzian or cocoercive, and parallel-sum type monotone operators as in [23, 26, 44, 45], existing purely primal splitting methods do not offer satisfactory options to solve the problem due to the appearance of the composite components and hence alternative primal-dual strategies must be explored. Very recently, these 1

2 frameworks are unified into a system of monotone inclusions with mixtures of composite, Lipschitzian, and parallel-sum type monotone operators in [19]. In this paper, we address to the numerical solutions of coupled system of primal-dual inclusions in real Hilbert spaces. Problem 1.1 Let m,s be strictly positive integers. For every i {1,...,m}, let H i, be a realhilbertspace, letz i H i, leta i : H i 2 H i bemaximallymonotone, letc i : H 1... H m H i be such that ν 0 ]0,+ [ x i 1 i m H 1... H m y i 1 i m H 1... H m m m x i y i C i x 1,...,x m C i y 1,...,y m ν 0 C i x 1,...,x m C i y 1,...,y m For every k {1,...,s}, let G k, be a real Hilbert space, let r k G k, let B k : G k 2 G k be maximally monotone, let S k : G 1... G s G k be such that µ 0 ]0,+ [ v k 1 k s G 1... G s w k 1 k s G 1... G s v k w k S k v 1,...,v s S k w 1,...,w s µ 0 S k v 1,...,v s S k w 1,...,w s For every i {1,...,m} and every k {1,...,s}, let L k,i : H i G k be a bounded linear operator. The problem is to solve the following system of primal-dual inclusions: find x 1 H 1,...,x m H m and v 1 G 1,...,v s G s such that m z 1 L k,1 v k A 1 x 1 +C 1 x 1,...,x m L 1,i x i r 1 B 1 v 1 +S 1 v 1,...,v s. and. m z m L k,m v k A m x m +C m x 1,...,x m, L s,i x i r s B s v s +S s v 1,...,v s. We denote by Ω the set of solutions to 1.3. In the case when every linear operators L k,i 1 k s 1 i m are zeros, we can use the algorithm in [2] to solve the inclusions in the left hand side and in the right hand side of 1.3 separately. Let us note that the non-linear coupling terms C i 1 i m and S k 1 k s are introduced in [2] and they are cocoercive operators which often play a central role; see for instance [2, 10, 17, 18, 29, 30, 31, 32, 42, 43, 46]. Let us add that the general algorithm in [19] can solve Problem 1.1 for the case when C i and S k are univariate, monotone and Lipschitzian. Furthermore, the primal-dual algorithm in [26, Section 6] can solve Problem 1.1 for the case when m = 1 and each S k are univariate, monotone and Lipschitzian. To sum up, the recent general frameworks can solve special cases of the above problem and no existing algorithm can solve it in the general case. In the present paper, we propose a primal-dual splitting algorithm for solving Problem 1.1 in Section 3. We recall some notations and background on the monotone operator theory in Section 2. In Section 4, we provide application to coupled system of monotone inclusions in duality. Section 5 is devoted to applications to minimization problems. In the last section, an application to multidictionary signal representation is presented

3 2 Notation and background, and technical results 2.1 Notation and background Throughout, H, G, and G i 1 i m are real Hilbert spaces. Their scalar products and associated normsarerespectively denoted by and. WedenotebyBH,G thespaceofboundedlinear operators from H to G. The adjoint of L BH,G is denoted by L. We set BH = BH,H. The symbols and denote respectively weak and strong convergence, and Id denotes the identity operator, we denote by l 1 +N the set of summable sequences in [0,+ [ and by l 2 K K N the set of square summable sequences, indexed by K, in R. Let M 1 and M 2 be self-adjoint operators in BH, we write M 1 M 2 if and only if x H M 1 x x M 2 x x. Let α ]0,+ [. We set P α H = { M BH M = M and M αid }. 2.1 The square root of M in P α H is denoted by M. Moreover, for every M P α H, we define respectively a scalar product and a norm by x H y H x y M = Mx y and x M = Mx x, 2.2 and, for any L BH, we define L M = sup Lx M. 2.3 x M 1 Let A: H 2 H be a set-valued operator. The domain of A is doma = { x H Ax }, and the graph of A is graa = { x,u H H u Ax }. The set of zeros of A is zera = { x H 0 Ax }, and the range of A is rana = { u H x H u Ax }. The inverse of A is A 1 : H 2 H : u { x H u Ax }, and the resolvent of A is Moreover, A is monotone if J A = Id+A x,y H H u,v Ax Ay x y u v 0, 2.5 and maximally monotone if it is monotone and there exists no monotone operator B: H 2 H such that graa grab and A B. A single-valued operator B: H H is β-cocoercive, for some β ]0,+ [, if x H y H x y Bx By β Bx By The parallel sum of A: H 2 H and B: H 2 H is A B = A 1 +B Let Γ 0 H be the class of proper lower semicontinuous convex functions from H to ],+ ]. For any U P α H and f Γ 0 H, we define J U 1 f = prox U f 1 : H H: x argmin fy+ y H 2 x y 2 U, 2.8 3

4 and J f = prox f : H H: x argmin y H fy+ 1 2 x y 2, 2.9 and the conjugate function of f is f : a sup a x fx x H Note that, or equivalently, f Γ 0 Hx Hy H y fx x f y, 2.11 f Γ 0 H f 1 = f The infimal convolution of the two functions f and g from H to ],+ ] is f g: x inf fy+gx y y H The indicator function of a nonempty closed convex set C is denoted by ι C, its dual function is the support function σ C, the distance function of C is denoted by d C. Finally, the strong relative interior of a subset C of H is the set of points x C such that the cone generated by x+c is a closed vector subspace of H. 2.2 Technical results We recall some results on monotone operators. Definition 2.1 [2, Definition 2.3] An operator A: H 2 H is demiregular at x doma if, for every sequence x n,u n in graa and every u Ax such that x n x and u n u, we have x n x. Lemma 2.2 [2, Proposition 2.4] Let A: H 2 H be monotone and suppose that x doma. Then A is demiregular at x in each of the following cases. i A is uniformly monotone at x, i.e., there exists an increasing function φ: [0,+ [ [0,+ ] that vanishes only at 0 such that u Ax y,v graa x y u v φ x y. ii A is strongly monotone, i.e., there exists α ]0,+ [ such that A αid is monotone. iii J A is compact, i.e., for every bounded set C H, the closure of J A C is compact. In particular, dom A is boundedly relatively compact, i.e., the intersection of its closure with every closed ball is compact. iv A: H H is single-valued with a single-valued continuous inverse. v A is single-valued on doma and Id A is demicompact, i.e., for every bounded sequence x n in doma such that Ax n converges strongly, x n admits a strong cluster point. 4

5 vi A = f, where f Γ 0 H is uniformly convex at x, i.e., there exists an increasing function φ: [0,+ [ [0,+ ] that vanishes only at 0 such that α ]0,1[ y domf f αx + 1 αy +α1 αφ x y αfx+1 αfy. vii A = f, where f Γ 0 H and, for every ξ R, { x H fx ξ } is boundedly compact. Lemma 2.3 [26, Lemma 3.7] Let A: H 2 H be maximally monotone, let α ]0,+ [, let U P α H, and let G be the real Hilbert space obtained by endowing H with the scalar product x,y x y U 1 = x U 1 y. Then the following hold. i UA: G 2 G is maximally monotone. ii J UA : G G is 1-cocoercive, i.e., firmly nonexpansive, hence nonexpansive. iii J UA = U 1 +A 1 U 1. Lemma 2.4 Let α and β be strictly positive reals, let B: H H be β-cocoercive, let U P α H such that U 1 < 2β and set P = Id U 1 B. Then, x Hy H Px Py 2 U x y 2 U 2β U 1 Bx By Hence, P is nonexpansive with respective to the norm U. Proof. Let x H and y H. Then using the cocoercivity of B, we have which proves Px Py 2 U = x y 2 U 2 x y Bx By + U 1 Bx By 2 U x y 2 U 2β Bx By 2 + Bx By U 1 Bx By x y 2 U 2β U 1 Bx By 2, 2.15 Theorem 2.5 [26, Theorem 4.1] Let K be a real Hilbert space with scalar product and the associated norm. Let A: K 2 K be maximally monotone, let α ]0,+ [, let β ]0,+ [, let B: K K be β-cocoercive, let η n l 1 + N, and let U n be a sequence in P α K such that µ = sup U n < + and n N 1+η n U n+1 U n Let ε ]0,min{1,2β/µ +1}], let λ n be a sequence in [ε,1], let γ n be a sequence in [ε,2β ε/µ], let x 0 K, and let a n and b n be absolutely summable sequences in K. Suppose that Z = zera+b, 2.17 and set n N Then the following hold for some x Z. i x n x. ii Bx n Bx 2 < +. yn = x n γ n U n Bx n +b n x n+1 = x n +λ n JγnU nay n +a n x n iii Suppose that at every point in Z, A or B is demiregular, then x n x. 5

6 3 Algorithm and convergence We propose the following algorithm for solving Problem 1.1. Algorithm 3.1 Let α ]0,+ [ and, for every i {1,...,m} and every k {1,...,s}, let U i,n be a sequence in P α H i and let V k,n be a sequence in P α G k. Set β = min{ν 0,µ 0 }, and let ε ]0,min{1,β}[, let λ n be a sequence in [ε,1]. Let x i,0 1 i m H 1... H m and v k,0 1 k s G 1... G s. Set For n = 0,1,... For i = 1,...,m p i,n = J Ui,n A i x i,n U s i,n L k,i v k,n +C i,n x 1,n,...,x m,n +c i,n z i +a i,n y i,n = 2p i,n x i,n x i,n+1 = x i,n +λ i,n p i,n x i,n For k = 1,...,s q k,n = J Vk,n B k v k,n +V m k,n L k,iy i,n S k,n v 1,n,...,v s,n d k,n r k +b k,n v k,n+1 = v k,n +λ m+k,n q k,n v k,n, where, for every i {1,...,m} and every k {1,...,s}, the following conditions hold 3.1 i n N U i,n+1 U i,n and V k,n+1 V k,n, and µ = sup{ U 1,n,..., U m,n, V 1,n,..., V s,n } < ii C i,n are operators from H 1... H m to H i such that a C i,n C i are Lipschitz continuous with respective constants κ i,n ]0,+ [ satisfying κ i,n < +. b There exists s H 1... H m not depending i such that n N C i,n s = C i s. iii S k,n are operators from G 1...G s to G k such that a S k,n S k are Lipschitz continuous with respective constants η k,n ]0,+ [ satisfying η k,n < +. b There exists w G 1... G s not depending k such that n N S k,n w = S k w. iv a i,n and c i,n are absolutely summable sequences in H i. v b k,n and d k,n are absolutely summable sequences in G k. vi λ i,n and λ m+k,n are in ]0,1] such that λ i,n λ n + λ m+k,n λ n < Remark 3.2 Here are some remarks 6

7 i Our algorithm has basically a structure of the variable metric forward-backward splitting since the multi-valued operators are used individually in the backward steps via their resolvents, the single-valued operators are used individually in the forward steps via their values. ii The algorithm allows the metric to vary over the course of the iterations. Even when restricted to the constant metric case which is the case where U i,n 1 i m and V k,n 1 k s are identity operators, the algorithm is new. iii Condition i is used in [26, 45] while conditions ii, iii and vi are used in [2], and conditions iv and v which quantify the tolerance allowed in the inexact implementation of the resolvents and the approximations of single-valued are widely used in the literature. iv Algorithm 3.1 is an extension of [26, Corollary 6.2] where m = 1 and every n N: C 1,n = C and for every k {1,...,s}, S k,n = D 1 k are restricted to univariate and cocoercive, and B k is replaced by B 1 k, and for every j {1,...,m+s}, λ j,n = λ n. The main result of the paper can be now stated. Theorem 3.3 Suppose in Problem 1.1 that Ω and there exists L k0,i 0 0 for some i 0 {1,...,m} and k 0 {1,...,s}. For every n N, set and suppose that ζ n = δ n = m 1 2 V k,n L k,i Ui,n 1, 3.4 δ n 1+δ n max 1 i m,1 k s { U i,n, V k,n } 1 2β ε. 3.5 For every i {1,...,m} and every k {1,...,s}, let x i,n and v k,n be sequences generated by Algorithm 3.1. Then the following hold for some x 1,...,x m,v 1,...,v s Ω. i i {1,...,m} x i,n x i and k {1,...,s} v k,n v k. ii Suppose that the operator x i 1 i m C j x i 1 i m 1 j m is demiregular see Lemma 2.2 for special cases at x 1,...,x m, then i {1,...,m} x i,n x i. iii Suppose that the operator v k 1 k s S j v k 1 k s 1 j s is demiregular see Lemma 2.2 for special cases at v 1,...,v s, then k {1,...,s} v k,n v k. iv Suppose that there exists j {1,...,m} and an operator C: H j H j such that x i 1 i m H i 1 i m C j x 1,...,x m = Cx j and C is demiregular see Lemma 2.2 for special cases at x j, then x j,n x j. v Suppose that there exists j {1,...,s} and an operator D: G j G j such that v k 1 k s G k 1 k s S j v 1,...,v s = Dv j and D is demiregular see Lemma 2.2 for special cases at v j, then v j,n v j. 7

8 Proof. Let us introduce the Hilbert direct sums H = H 1... H m, G = G 1... G s, and K = H G. 3.6 We denote by x = x i 1 i m, y = y i 1 i m and v = v k 1 k s, w = w k 1 k s the generic elements in H and G, respectively. The generic elements in K will be in the form p = x,v. The scalar product and the norm of H are respectively defined by : x,y m x i y i, 3.7 and : x x x. 3.8 The scalar product and the norm of G are defined by the same fashion as those of H, : v,w v k w k, 3.9 and : v v v We next define the scalar product and the norm of K are respectively defined by : x,v,y,w m x i y i + v k w k 3.11 and : x,v x,v x,v Set A: H 2 H : x m A ix i C: H H: x C i x 1 i m L: H G: x m z = z 1,...,z m, L k,ix i 1 k s B: G 2 G : v s B kv k and D: G 2 G : v S k v 1 k s 3.13 r = r 1,...,r s, and n N C n : H H: x C i,n x 1 i m and D n : G G: v S k,n v 1 k s Then, it follows from 1.1 that x Hy H x y Cx Cy ν 0 Cx Cy 2, 3.15 from 1.2 that v Hw H v w Dv Dw µ 0 Dv Dw 2,

9 which shows that C and D are respectively ν 0 -cocoercive and µ 0 -cocoercive and hence they are maximally monotone [10, Example 20.28]. Moreover, it follows from [10, Proposition 20.23] that A and B are maximally monotone. Furthermore, L : G H: v L k,i v k i m Then, using 3.13, we can rewrite the system of monotone inclusions 1.3 as a monotone inclusion in K, Set and find x,v K such that z L v A+Cx and Lx r B +Dv n N M: K 2 K : x,v z +Ax,r +Bv S: K K: x,v L v, Lx Q: K K: x,v Cx,Dv, Q n : K K: x,v C n x,d n v Λ n : K K: x,v λ i,n x i 1 i m,λ m+k,n v k 1 k s U n : K K: x,v U i,n x i 1 i m,v k,n v k 1 k s V n : K K: x,v U 1 n x,v L v,lx Then M, S are maximally monotone operators and 3.15, 3.16 implies that Q is β-cocoercive and hence it is maximally monotone [10, Example 20.28]. Therefore, M + S + Q is maximally monotone [10, Corollary 24.4]. Furthermore, the problem 3.18 is reduced to find a zero point of M +S +Q. Note that Ω implies that Moreover, we also have Hence, where zerm +S +Q 3.21 n N Λ n V n = max 1 j m+s λ j,n 1 and Id Λ n V n = 1 min 1 j m+s λ j,n Λ n V n + Id Λ n V n = 1+ max 1 j m+s λ j,n λ n min 1 j m+s λ j,n λ n 1+τ n, 3.23 We derive from the condition vi in Algorithm 3.1 that n N τ n = 2 max 1 j m+s λ j,n λ n τ n 2 m+s λ j,n λ n < j=1 We next derive from the condition i in Algorithm 3.1 that µ = sup U n < +, and n N U n+1 U n P α K,

10 and it follows from 3.12 and [25, Lemma 2.1ii] that n N p = x,v K p 2 U 1 n = m x i 2 + U 1 i,n v k 2 V 1 k,n m x i 2 Ui,n 1 + v k 2 V 1 k,n p 2 min { U i,n 1, V k,n 1 } i m,1 k s Note that V n are self-adjoint, let us check that V n are strongly monotone. To this end, let us introduce m T n : H G: x Vk,n L k,i x i n N 1 k s V1,n 1 R n : G G: v v1,..., V s,n vs. Then, by using Cauchy-Schwartz s inequality, we have where we set n Nx H T n x 2 = m 1 2 Vk,n L k,i Ui,n Ui,n xi m 2 1 V k,n L k,i Ui,n Ui,n xi m 2 m V k,n L k,i Ui,n 1 2 U i,n xi m = x i 2 U 1 i,n = β n m n N β n = which together with 3.4 imply that Moreover, m 2 V k,n L k,i Ui,n x i 2, 3.29 U 1 i,n m 2 V k,n L k,i Ui,n, 3.30 n N 1+δ n β n = 1 1+δ n n Nv G R n v 2 = = 1 V k,n vk 2 v k V 1 k,n 10

11 Therefore, for every p = x,v K, and every n N, it follows from 3.20, 3.28, 3.31, 3.32 and 3.27, 3.5 that p V n p = p 2 U 1 n = p 2 U 1 n = p 2 U 1 n p 2 U 1 n p 2 U 1 n 2 Lx v m 2 Vk,n L k,i x i 1 V k,n vk δn β n Tn x 1+δ n β n R n v T n x 2 +1+δ n β n R n v 2 1+δ n β n m x i 2 U 1 i,n +1+δ n β n v k 2 1+δ n V 1 k,n = δ m n x i δ U 1 n i,n v k 2 V 1 k,n ζ n p In turn, V n are invertible, by [25, Lemma 2.1iii] and 3.5, n N V 1 n 1 ζ n 2β ε, 3.34 and by [25, Lemma 2.1i], 3.26, n N U n+1 U n U 1 n U 1 1 V n. Furthermore, we derive from [25, Lemma 2.1ii] that V 1 n+1 Altogether, n+1 V n V n+1 p K V 1 n p p V n 1 p 2 1 ρ p 2, where ρ = α 1 + S sup V 1 n 2β ε and n N V 1 1 n+1 V n P 1/ρ K Moreover, using [25, Lemma 2.1iii] and 3.36, we obtain zn K N z n < + z n V 1 n < and and zn K N z n < + z n V n < +, 3.38 p K sup p V n <

12 Now we can reformulate the algorithm 3.1 as iterations in the space K. We first observe that 3.1 is equivalent to For n = 0,1... For i = 1,...,m For k = 1,...,s U 1 i,n x i,n p i,n s L k,i v k,n C i,n x 1,n,...,x m,n z i +A i p i,n a i,n +c i,n U 1 i,n a i,n x i,n+1 = x i,n +λ i,n p i,n x i,n V 1 k,n v k,n q k,n m L k,ix i,n p i,n S k,n v 1,n,...,v s,n r k +B k q k,n b k,n m L k,ip i,n +d k,n V 1 k,n b k,n v k,n+1 = v k,n +λ m+k,n q k,n v k,n Set p n = x 1,n,...x m,n,v 1,n,...,v s,n y n = p 1,n,...,p m,n,q 1,n,...,q s,n a n = a 1,n,...,a m,n,b 1,n,...,b s,n c n = c 1,n,...,c m,n,d 1,n,...,d s,n d n = U1,n 1 a 1,n,...,Um,n 1 a m,n,v1,n 1 b 1,n,...,V 1 b n = S +V n a n +c n d n. s,n b s,n Then, using the same arguments as in [44, Eqs ], using 3.19, 3.20, 3.40 yields We have n N V n p n y n Q n p n M +Sy n a n +Sa n +c n d n p n+1 = p n +Λ n y n p n n N V n p n y n Q n p n M +Sy n a n +Sa n +c n d n n N V n Q n p n M +S +V n y n a n +S +V n a n +c n d n n N y n = 1 M +S +V n V n Q n p n S +V n a n c n +d n +a n n N y n = Id+V 1 n N y n = J V 1 n M+S Therefore, 3.41 becomes +a n 1 Id V n M +S 1 n Q n pn V 1 n b n Id V 1 n Q n pn V 1 n b n +a n n N p n+1 = p n +Λ n J V 1 n M+S p n V 1 n Q np n +b n +a n p n By setting n N M: K 2 K : x,v Mx,v+Sx,v, P n = Id V 1 n Q n and P n = Id V 1 n Q, E n = Q n Q and Qn = V 1 n E n, e 1,n = Q n p n +V 1 n b n, e n = a n + 1 λ λn n Id Λ n p n J V 1 Pn p n e 1,n an, n M

13 we have 3.43 n N p n+1 = p n +Λ n J V 1 Pn M p n n V 1 n b n +an p n = Id Λ n p n +Λ n J V 1 Pn M p n n V 1 n b n +Λn a n = Id Λ n p n +Λ n J V 1 n M Pn p n e 1,n +Λn a n 3.45 = 1 λ n p n +λ n J V 1 M Pn p n n e 1,n +en Algorithm 3.46 is a special instance of the variable metric forward-backward splitting 2.18 with [ ] n N γ n = 1 ε,2β ε/sup V 1 n see Moreover, since M is maximally monotone, Q is β-cocoercive, and n N λ n [ε,1], since 3.36 and 3.21 respectively show that 2.16 and 2.17 are satisfied. In view of Theorem 2.5, it is sufficient to prove that e 1,n and e n are absolutely summable in K, i.e, we prove that and e 1,n < +, 3.48 e n < For every i {1,...,m} and every k {1,...,s}, since a i,n, c i,n and b k,n and d k,n are absolutely summable, we have and a n m a i,n + c n m c i,n + b k,n < +, 3.50 d k,n < Moreover, for every n N, U n P α K, it follows from [25, Lemma 2.1iii] that U 1 n α 1. Hence, d n α 1 a n < and b n ρ a n + cn + d n < Therefore, a n, b n, c n and d n are absolutely summable in K. Next it follows fromtheconditionsii, iiiinalgorithm 3.1 and3.36, 3.33, 3.5that, forevery p = x,v K 13

14 and q = y,w K, n N Qn p Q n q 2 V n = Q n p Q n q V n Qn p Q n q = E n p E n q V 1 n E np V 1 n E nq V 1 n E n p E n q 2 2β ε C n Cx C n Cy 2 + D n Dv D n Dw 2 m = 2β ε C i,n C i x C i,n C i y 2 + S k,n S k v S k,n S k w 2 m 2β ε κ 2 i,n x y 2 + m 2β ε κ 2 i,n + 2β εζ 1 n η 2 k,n m κ 2 i,n + m 2β ε 2 κ 2 i,n + η 2 k,n ηk,n 2 v w 2 p q 2 η 2 k,n p q 2 V n p q 2 V n, 3.54 which implies that Q n is Lipschitz continuous in the norm V n with respectively constant κ n = 2β ε m κ 2 i,n + ηk,n 2, 3.55 that satisfies κ n < Let p = x,v zerm +S +Q and noting that n N Q n s,w = 0, n N e 1,n V n Q n p n V n + V 1 n b n V n Since p zerm +S +Q, we have Q n p n Q n p V n + Q n p Q n s,w V n + V 1 n b n V n κ n p n p V n +κ n p s,w V n + V 1 n b n V n = κ n p n p V n +κ n p s,w V n + b n V 1 n n N p = J V 1 n M P np

15 Hence, since J V 1 n M and P n are nonexpansive with respect to the norm V n by Lemma 2.3ii and Lemma 2.4, on one hand, we have J V 1 n M Pn p n e 1,n p V n = J V 1 n M Pn p n e 1,n JV 1 n M P np V n p n p V n + e 1,n V n, 3.59 which and 3.45, 3.22, 3.23 imply that n N p n+1 p V n Id Λn p n p V n where + Λ M n JV 1 Pn p n n e 1,n p V n + Λ n a n V n Id Λ n + V Λ n n V p n n p + V Λ n n V e 1,n n V n + Λ n a n V n 1+τ n +κ n p n p V n +α n, 3.60 n N α n = κ n p s,w V n + bn V 1 n + an V n Noting that, by 3.25, 3.56, 3.39, 3.38, 3.37 and 3.53, 3.50, we have α n < + and τ n +κ n < Therefore, we derive from 3.60 and n N V n V n+1 that n N pn+1 p V n+1 1+τ n +κ n p n p V n +α n, 3.63 and hence, by [36, Lemma 2.2.2], sup p n p V n < +, 3.64 which and 3.57,3.56,3.62, 3.53, 3.37, 3.38, 3.39 imply that e 1,n V n < On the other hand, p n J V 1 M Pn p n n e 1,n an V n 2 p n p V n + e 1,n V n + a n V n, 3.66 which and 3.64, 3.65 imply that ν = sup p n J V 1 n M Pn p n e 1,n an V n < Now using the condition vi, 3.67 and the definition of e 1,n in 3.44, we obtain e n V n a n V n +νε m 1 λ i,n λ n + λ m+k,n λ n <

16 By using 3.38, we derive from 3.68 and 3.65 that e n < + and e 1,n < +, 3.69 which prove 3.48 and i: By Theorem 2.5i, p n p zerm +S +Q. iiiii: By Theorem 2.5ii and iii, Qp n Qp 0, 3.70 which implies that, for every i {1,...,m} and every k {1,...,s}, C i x 1,n,...,x m,n C i x 1,...,x m 0 and S k v 1,n,...,v s,n S k v 1,...,v s Moreover, by i, i {1,...,m} x i,n x i and k {1,...,s} v k,n v k. Therefore, the conclusions follow from the definition of the demiregular operators. ivv: The conclusions follow from our assumptions the definition of the demiregular operators. 4 Application to coupled system of monotone inclusions in duality We provide an application to coupled system of monotone inclusions. Problem 4.1 Let m,s be strictly positive integers. For every i {1,...,m}, let H i, be a realhilbertspace, letz i H i, leta i : H i 2 H i bemaximallymonotone, letc i : H 1... H m H i be such that ν 0 ]0,+ [ x i 1 i m H 1... H m y i 1 i m H 1... H m m m x i y i C i x 1,...,x m C i y 1,...,y m ν 0 C i x 1,...,x m C i y 1,...,y m For every k {1,...,s}, let G k, be a real Hilbert space, let r k G k, let D k : G k 2 G k be maximally monotone and ν k -strongly monotone for some ν k ]0,+ [, let B k : G k 2 G k be maximally monotone. For every i {1,...,m} and every k {1,...,s}, let L k,i : H i G k be a bounded linear operator. The primal problem is to solve the primal inclusion: find x 1 H 1,...,x m H m such that m z 1 A 1 x 1 + L k,1 D k B k L k,i x i r k +C 1 x 1,...,x m. z m A m x m + L k,m m D k B k L k,i x i r k +C m x 1,...,x m

17 We denote by P the set of solutions to 4.2. The dual problem is to solve the dual inclusion: find v 1 G 1,...,v s G s such that x i 1 i m H i 1 i m m z 1 L k,1 v k A 1 x 1 +C 1 x 1,...,x m L 1,i x i r 1 B1 1 v 1 +D1 1 v 1. and. m z m L k,m v k A m x m +C m x 1,...,x m, L s,i x i r s Bs 1 v s +Ds 1 v s. The set of solutions to 4.3 is denoted by D. 4.3 Problem 4.1 covers not only a wide class of monotone inclusions and duality frameworks in the literature [3, 5, 6, 12, 17, 18, 26, 28, 33, 37, 39, 40, 42, 43, 44] and coupled system of monotone inclusions unified in [2] and the references therein, but also a wide class of minimization formulations, in particular, in the multi-component signal decomposition and recovery [2, 5, 7] and the references therein. Algorithm 4.2 Let α ]0,+ [ and, for every i {1,...,m} and every k {1,...,s}, let U i,n be a sequence in P α H i and let V k,n be a sequence in P α G k. Set β = min{ν 0,ν 1,...,ν s }, and let ε ]0,min{1,β}[, let λ n be a sequence in [ε,1]. Let x i,0 1 i m H 1... H m and v k,0 1 k s G 1... G s. Set For n = 0,1,... For i = 1,...,m y i,n = 2p i,n x i,n x i,n+1 = x i,n +λ i,n p i,n x i,n For k = 1,...,s q k,n = J Vk,n B 1 k v k,n+1 = v k,n +λ m+k,n q k,n v k,n, p i,n = J Ui,n A i x i,n U i,n s L k,i v k,n +C i x 1,n,...,x m,n +c i,n z i +a i,n v k,n +V k,n m L k,iy i,n D 1 k v k,n d k,n r k +b k,n 4.4 where, for every i {1,...,m} and every k {1,...,s}, the following conditions hold i n N U i,n+1 U i,n and V k,n+1 V k,n, and µ = sup{ U 1,n,..., U m,n, V 1,n,..., V s,n } < ii a i,n and c i,n are absolutely summable sequences in H i. iii b k,n and d k,n are absolutely summable sequences in G k. iv λ i,n and λ m+k,n are in ]0,1] such that λ i,n λ n + λ m+k,n λ n <

18 Corollary 4.3 Suppose that P and there exists L k0,i 0 0 for some i 0 {1,...,m} and k 0 {1,...,s}, and 3.5 is satisfied. For every i {1,...,m} and every k {1,...,s}, let x i,n and v k,n be sequences generated by Algorithm 4.2. Then the following hold for some x 1,...,x m P and v 1,...,v s D. i i {1,...,m} x i,n x i and k {1,...,s} v k,n v k. ii Suppose that the operator x i 1 i m C j x i 1 i m 1 j m is demiregular see Lemma 2.2 for special cases at x 1,...,x m, then i {1,...,m} x i,n x i. iii Suppose that D 1 j is demiregular at v j, for some j {1,...,s}, then v j,n v j. iv Suppose that there exists j {1,...,m} and operator C: H j H j such that x i 1 i m H i 1 i m C j x 1,...,x m = Cx j and C is demiregular see Lemma 2.2 for special cases at x j, then x j,n x j. Proof. Set µ 0 = min{ν 1,...,ν s } and define k {1,...,s} S k : G 1...G s G k : v 1,...,v s D 1 k v k. 4.7 Then, for every v k 1 k s G 1... G s and every w k 1 k s G 1... G s, we obtain v k w k S k v 1,...,v s S k w 1,...,w s = µ 0 vk w k D 1 k v k D 1 k w k ν k D 1 k v k D 1 k w k 2 D 1 k v k D 1 k w k 2 which shows that 1.2 is satisfied. Moreover, upon setting { i {1,...,m} C i,n = C i n N k {1,...,s} S k,n = S k, = µ 0 S k v 1,...,v s S k w 1,...,w s 2, 4.8 the conditions ii and iii in Algorithm 3.1 are satisfied. Note that the conditions i, ii, iii, and iv in Algorithm 4.2 are the same as in Algorithm 3.1. Moreover, the algorithm 3.1 reduces to 4.4 where B k is replaced by B 1 k. Next, since P, we derive from 4.2 that, for every k {1,...,s}, there exists v k G k such that m m v k D k B k L k,i x i r k L k,i x i r k B 1 k v k +D 1 k v k, 4.10 and i {1,...,m} z i 4.9 L k,i v k A i x i +C i x 1,...,x m,

19 which shows that Ω and D. Inversely, if x 1,...,x m,v 1,...,v s Ω, then the inclusions 4.10 and 4.11 are satisfied. Hence v 1,...,v s D and x 1,...,x m P. Therefore, the conclusions follow from Theorem Application to minimization problems We provide applications to minimization problems involving infimal convolutions, composite functions and coupling. Problem 5.1 Let m,s be strictly positive integers. For every i {1,...,m}, let H i be a real Hilbert space, let z i H i, let f i Γ 0 H i. For every k {1,...,s}, let G k be a real Hilbert space, let r k G k, let l k Γ 0 G k be ν k -strongly convex function, for some ν k ]0,+ [, let g k Γ 0 G k. For every i {1,...,m}, and every k {1,...,s}, let L k,i : H i G k be a bounded linear operator. Let ϕ: H 1... H m R beconvex differentiable function with ν0 1 -Lipschitz continuous gradient. The primal problems is to minimize x 1 H 1,...,x m H m m fi x i x i z i + under the the assumption that, i {1,...,m} z i ran f i + m lk g k L k,i x i r k +ϕx 1,...,x m, 5.1 L k,i m l k g k L k,j r k + i ϕ, 5.2 where i ϕ is the ith component of the gradient ϕ, and the dual problem is to m minimize ϕ fi z i v 1 G 1,...,v s G s L k,i v k 1 i m + j=1 l k v k+gk v k+ v k r k. 5.3 In the case when the infimal convolutions are absent, Problem 5.1 often appears in the multicomponents signal decomposition and recovery problems [2, 5, 4] and the references therein. Example 5.2 Some special cases of this problem are listed in the following: i In the case when ϕ: x 1,...,x m m h ix i, where for every i {1,...,m}, h i : H i R is a convex differential function with τi 1 -Lipschitz continuous gradient, for some τi 1 ]0, + [, Problem 5.1 reduces to the general minimization problem [19, Problem 5.1] which covers a wide class of the convex minimization problems in the literature. ii In the case when ϕ: x 1,...,x m 0 and, for every k {1,...,s}, l k = ι {0} and g k is differentiable with τ 1 k -Lipschitz continuous gradient, for some τ k ]0,+ [, Problem 5.1 reduces to [5, Problem 1.1]. iii In the case when m = 1, Problem 5.1 reduces to [23, Problem 4.1] which is also studied in [26, 44]. 19

20 Algorithm 5.3 Let α ]0,+ [ and, for every i {1,...,m} and every k {1,...,s}, let U i,n be a sequence in P α H i and let V k,n be a sequence in P α G k. Set β = min{ν 0,ν 1,...,ν s }, and let ε ]0,min{1,β}[, let λ n be a sequence in [ε,1]. Let x i,0 1 i m H 1... H m and v k,0 1 k s G 1... G s. Set For n = 0,1,... For i = 1,...,m p i,n = prox U 1 i,n s f i x i,n U i,n L k,i v k,n + i ϕx 1,n,...,x m,n +c i,n z i +a i,n y i,n = 2p i,n x i,n x i,n+1 = x i,n +λ i,n p i,n x i,n For k = 1,...,s q k,n = prox V 1 k,n m g v k k,n +V k,n L k,iy i,n l k v k,n d k,n r k +b k,n v k,n+1 = v k,n +λ m+k,n q k,n v k,n, where, for every i {1,...,m} and every k {1,...,s}, the following conditions hold 5.4 i n N U i,n+1 U i,n and V k,n+1 V k,n, and µ = sup{ U 1,n,..., U m,n, V 1,n,..., V s,n } < ii a i,n and c i,n are absolutely summable sequences in H i. iii b k,n and d k,n are absolutely summable sequences in G k. iv λ i,n and λ m+k,n are in ]0,1] such that λ i,n λ n + λ m+k,n λ n < Corollary 5.4 Suppose that there exists L k0,i 0 0 for some i 0 {1,...,m} and k 0 {1,...,s}, and 3.5 is satisfied. For every i {1,...,m} and every k {1,...,s}, let x i,n and v k,n be sequences generated by Algorithm 5.3. Then the following hold for some solution x 1,...,x m to 5.1 and v 1,...,v s to 5.3. i i {1,...,m} x i,n x i and k {1,...,s} v k,n v k. ii Suppose that ϕ is defined as in Example 5.2i and h j is uniformly convex at x j, for some j {1,...,m}, then x j,n x j. iii Suppose that l j is uniformly convex at v j, for some j {1,...,s}, then v j,n v j. Proof. Set { i {1,...,m} A i = f i and C i = i ϕ, k {1,...,s} B k = g k and D k = l k. Then it follows from [10, Theorem 20.40] that A i 1 i m, B k 1 k s, and D k 1 k s are maximally monotone. Moreover, C 1,...,C m = ϕ is ν 0 -cocoercive [8, 9]. Moreover since, for every k

21 {1,...,s}, l k is ν k -strongly convex, l k is ν k -strongly monotone. Therefore, every conditions on the operators in Problem 1.1 are satisfied. Since, for every k {1,...,s}, doml k = G k, we next derive from [10, Proposition 20.47] that k {1,...,s} l k g k = g k l k = B k D k. 5.8 Let H and G be defined as in the proof of Theorem 3.3, and let L,z and r be defined as in 3.13, and define f: H ],+ [ : x m f ix i g: G ],+ [ : v s g kv k l: G ],+ [ : v 5.9 s l kv k. Observe that [10, Proposition 13.27], We also have f : y m fi y i, g : v Then the primal problem becomes minimize x H and the dual problem becomes minimize v G l g: v gk v k, and l : v l k v k l k g k v k fx x z +l glx r+ϕx, 5.12 ϕ f z L v+l v+g v+ v r Then, let x = x 1,...,x m be a solution to 4.2, i.e., for every i {1,...,m}, m z i f i x i + l k g k L k,j x j r k + i ϕx 1,...,x m L k,i j=1 Then, using 5.7, 5.8, [10, Corollary 16.38iii], [10, Proposition 16.8], 0 f + z x+l l glx r + ϕx Therefore, by [10, Proposition 16.5ii], we derive from 5.15 that 0 f + z +l gl r+ϕ x Hence, by Fermat s rule [10, Theorem 16.2] that x is a solution to 5.12, i.e, x is a solution to 5.1. We next let v be a solution to 4.3. Then using [10, Theorem 15.3] and 2.12, r L f + ϕ 1 z L v + g 1 v + l 1 v = L f +ϕ z L v + g v+ l v = L f ϕ z L v + g v+ l v

22 Therefore, by [10, Proposition 16.5ii], we derive from 5.16 that 0 ϕ f z L +l +g + r v Hence, by Fermat s rule [10, Theorem 16.2] that v is a solution to 5.13, i.e, v is a solution to 4.3. Now, in view of 2.8, algorithm 5.4 is a special case of the algorithm 4.4. Moreover, every specific conditions in Corollary 4.3 are satisfied. i It follows from Corollary 4.3i that x 1,n,...,x m,n x 1,...,x m which solves the primal problem 5.1, and v 1,n,...,v s,n v 1,...,v s which solves the dual problem 5.3. iiiii The conclusions follow from Corollary 4.3iiiiv and Lemma 2.2vi. Remark 5.5 Here are some remarks i Sufficient conditions which ensure that the condition 5.2 is satisfied are provided in [19, Proposition 5.3]. For instance, if 5.1 has at least one solution and r 1,...,r s belongs to the strong relative interior of { m E = L k,i x i v k 1 k s { i {1,...,m} x i domf i k {1,...,s} v k domg k +doml k } ii In the case when m = 1 and n Ni {1,...,m+s} λ i,n = λ n, the algorithm 5.4 is in [26, Eq.5.26] where the connections to existing work are available. 6 Multi-dictionary signal representation Dictionary has been used in minimization problems in signal processing in [24, Section 4.3]. Let us recall that a sequence of unit norm vectors o k k K K N in H is a dictionary with dictionary constant µ in ]0,+ [ if x H x o k 2 µ x Then the dictionary operator is defined by k K F: H l 2 K: x x o k k K 6.2 and its adjoint is F : l 2 K H: ω k k K k Kω k o k. 6.3 Dictionary extends the notion of orthonormal bases and frames which plays an important role in the theory of signal processing due to their ability to efficiently capture a wide range signal features [2, 15, 20, 21] and the references therein. The focus of this section is to explore the 22

23 information of the original signals x i 1 i m which are assumed to be available on the coefficients of dictionaries x o i,j 1 i m j K and close to soft constraints nonempty closed convex subsets C i 1 i m modeling its prior information. The rest of the information available will be modeled by potential functions f i 1 i m hard constraints. Furthermore, the data-fitting terms are measured by non-smooth functions. Problem 6.1 Let H be a real Hilbert space, let m,s be strictly positive integers such that s > m, let γ ]0,+ [, and let K be a nonempty subset of N. For every i {1,...,m}, let G i = l 2 K, let f i Γ 0 H, let o i,j j K be a dictionary in H with associated dictionary operator F i and dictionary constant µ i, let φ i,j j K be a sequence in Γ 0 R such that j K φ i,j φ i,j 0 = 0, let C i be a nonempty closed convex subset of H. For every k {m + 1,...,s}, let Y k be a real Hilbert space, let r k Y k, let β k be in ]0,+ [. For every i {1,...,m} and every k {m+1,...,s}, let R k,i : H i Y k be a bounded linear operator. Set C = m C i. The primal problems is to minimize x 1 H,...,x m H m f i x i + m φ i,j x i o i,j j K + k=m+1 β k rk m +γdc R k,i x i x 1,...,x m 2 /2 6.4 and the dual problem is to minimize ξ 1 l 2 K,...,ξ m l 2 K,v m+1 G m+1,...v s G s v m+1 β m+1,..., v s β s m σ Ci + 1 2γ 2 fi Fi ξ i + m φ i,jξ i,j + j K k=m+1 k=m+1 R k,i v k r k v k. 6.5 Lemma 6.2 Problem 6.1 is a special case of Problem 5.1 with i {1,...,m} z i = 0 and ϕ = γd 2 C /2, ν 0 = γ, k {1,...,m}i {1,...,m} l k = ι {0} and L i,i = F i and L k,i = 0 otherwise, k {1,...,m} r k = 0,G k = l 2 K and g k : l 2 K ],+ ] : ξ k j K φ k,jξ k,j, k {m+1,...,s} G k = Y k and g k = β k, and i {1,...,m} L k,i = R k,i. 6.6 Proof. Let us note that, by [10, Corollary 12.30], ϕ is a convex differentiable function with x = x i 1 i m H i 1 i m ϕx = γx P C x = γx i P Ci x i 1 i m. 6.7 Since Id P C is firmly nonexpansive [10, Proposition 4.8], ϕ is γ-cocoercive. Next for every k {1,...,s}, G k is a real Hilbert space and l k Γ 0 G k and by [27, Example 2.19], g k Γ 0 G k. Hence the conditions imposed on the functions in Problem 5.1 are satisfied. Now we have v G k l k g k v = inf w G k lk w+g k v w = g k v

24 Therefore, in view of 6.2 and 6.6, we have i {1,...,m} x i H m m lk g k L k,i x i r k = = m g i F i x i m φ i,j x i o i,j. 6.9 j K We derive from 6.9, 6.6 and 6.8 that 5.1 reduces to 6.4. For every k {m+1,...,s}, let B k 0;β k be the closed ball of Y k, center at 0 with radius β k. Using [10, Example 13.3v], [10, Proposition 13.27] and [10, Example 13.23], we obtain g k = β k = ι Bk 0;β k and i {1,...,m} g i : ξ i,j j K j Kφ i ξ i,j, 6.10 and Moreover, ϕ = σ C +γ 2 /2 = σ C + 2 /2γ = ϕ m f i = m m σci + 2 /2γ σci + 2 /2γ f i We derive from 6.9, 6.6, 6.10, 6.11 and 6.12 that 5.3 reduces to 6.5. Lemma 6.2 allows to solve Problem 6.1 by Algorithm 5.3. More precisely, Algorithm 6.3 Let ε ]0,min{1,γ}[, let λ n be a sequence in [ε,1], let γ i 1 i s+m be a finite sequence in [ε,+ [ such that m m 2γ ε 1 γ i µ i γ m+i + γ i γ m+k R k,i 2 max {γ i,γ m+k } i m,1 k s k=m+1 For every i {1,...,m}, let α i,n,j j K be sequences in R such that α i,n,j 2 < +, 6.14 j K let a i,n be a absolutely summable sequence in H, let λ i,n be sequence in ]0,1[, and for every k {1,...,s}, let λ m+k,n be sequence in ]0,1[ such that λ i,n λ n + λ m+k,n λ n < Let x i,0 1 i m H 1... H m, and for every i {1,...,m}, let ξ i,0,j j K l 2 K and 24

25 v k,0 m+1 k s G m+1... G s. Set For n = 0,1... For i = 1,...,m p i,n = prox γi f i x i,n γ i j K ξ i,n,jo i,j + s y i,n = 2p i,n x i,n x i,n+1 = x i,n +λ i,n p i,n x i,n For k = 1,...,m For k = m+1,...,s v k,n +γ m+k v k,n+1 = v k,n +λ m+k,n k=m+1 R k,i v k,n +γx i,n P Ci x i,n For every j K ξ k,n+1,j = ξ k,n,j +λ m+k,n prox γm+k φ ξk,n,j +γ j m+k y k,n o k,j +α k,n,j ξ k,n,j β k { max m R k,iy i,n r k m } v k,n β k, v k,n +γ m+k R k,iy i,n r k +a i,n Corollary 6.4 Suppose that 6.4 has at least one solution and 0,...,0,r m+1,...,r s belongs to the strong relative interior of { m i {1,...,m} x i domf i E = L k,i x i v k k {1,...,m} v k l 1 k s 2 K, } j K φ jv k,j < +, k {m+1,...,s} v k Y k 6.17 where L k,i is defined as in 6.6. Let x 1,n,...,x m,n and ξ 1,n,...,ξ m,n,v m+1,n,...,v s,n be sequence generated by Algorithm 6.3. Then x 1,n,...,x m,n x 1,...,x m a solution to 6.4, and ξ 1,n,...,ξ m,n, v m+1,n,...,v s,n ξ 1,...,ξ m,v m+1,...,v s a solution to 6.5. Furthermore, if C j = {0}, for some j {1,...,m}, then x j,n x j. Proof. For every i {1,...,m} and every j K, we have φ i,j φ i,j 0 = 0. Therefore, we derive from 6.10 and [10, Proposition 23.31] that ξ = ξ j j K l 2 K prox g i ξ = prox φ i,j ξ j j K 6.18 Next, for every k {m+1,...,s}, using 6.10 again, we have v G k prox g k v = P B0;βk v = β k v/max{β k, v } In view of 6.18, 6.19, 6.7 and the definition of L k,i 1 k s 1 i m in 6.6, the algorithm 6.16 is a special case of 5.4 with U i,n = γ i Id and V k,n = γ m+k Id, n N i {1,...,m} k {1,...,s} c i,n = 0 and d k,n = 0, 6.20 b i,n = α i,n,j j K. Moreover, we derive from 6.14 that the sequences b i,n 1 i m are absolutely summable, and from 6.13 that 3.5 holds. Finally, since 5.1 has at least one solution and 0,...,0,r m+1,...,r s 25

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