# An Inertial Accelerated Algorithm for Solving Split Feasibility Problem with Multiple Output Sets

Authors: Huijuan Jia; Shufen Liu; Yazheng Dang
Journal: Journal of Mathematics
 (2021)

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## Abstract

The paper proposes an inertial accelerated algorithm for solving split feasibility problem with multiple output sets. To improve the feasibility, the algorithm involves computing of projections onto relaxed sets (half spaces) instead of computing onto the closed convex sets, and it does not require calculating matrix inverse. To accelerate the convergence, the algorithm adopts self-adaptive rules and incorporates inertial technique. The strong convergence is shown under some suitable conditions. In addition, some newly derived results are presented for solving the split feasibility problem and split feasibility problem with multiple output sets. Finally, numerical experiments illustrate that the algorithm converges more quickly than some existing algorithms. Our results extend and improve some methods in the literature.


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## Body

## 1. Introduction

LetH1 and H2 be two real Hilbert spaces and C and Q be nonempty, closed, and convex subsets of H1 and H2, respectively. Let B:H1⟶H2 be a nonzero bounded linear operator and BT be its adjoint.The split feasibility problem (SFP) is formulated to find a pointx∗∈H1 satisfying(1)x∗∈C such thatBx∗∈Q.The SFP was introduced in [1], which has broad applications in many fields such as image reconstruction problem [2, 3] and intensity-modulated radiation therapy (IMRT) [4–6]. Byrne [2] introduced the CQ-algorithm for solving (1) as follows:(2)x0∈H1;xk+1≔PCxk−τkBTI−PQBxk,where PC and PQ are the metric projections onto C and Q, respectively, that is, PCx=argminy∈Cx−y, over all x C. τk∈0,2/B2, where B2 is the spectral radius of the matrix BTB.In 2004, Yang [7] made the following assumptions.(1)
The setsC and Q are denoted as(3)C≔x∈H1:cx≤0andQ=y∈H2:qy≤0,wherec:H1⟶ℝ and q:H2⟶ℝ are convex and subdifferential functions on H1 and H2, respectively.(2)
The subgradients∂cx and ∂qyof c and q, respectively, can be calculated; they are(4)∂cx≔ξ∈H1:cz≥cx+ξ,z−x,for eachz∈H1,(5)∂qy≔η∈H2:qz≥qx+η,u−y,for eachu∈H2,which are bounded operators (i.e., bounded on bounded sets).Define two half-spacesCk and Qk as(6)Ck≔x∈H1:cxk≤ξk,xk−x,where ξk∈∂cxk and(7)Qk≔y∈H2:qBxk≤ηk,Bxk−y,where ηk∈∂qBx. It is easy to see that Ck⊃C and Qk⊃Q, for all k≥1.Yang, in [7], introduced the following relaxed CQalgorithm for solving the SFP (1) in a finite-dimensional Hilbert space:(8)x0∈H1;xk+1≔PCkxk−τk∇fkxk,where fkxk≔1/2I−PQk2,∇fkxk≔BTI−PQkBxk, and τk∈0,2/B2. Since PCk and PQk are easily calculated, this method appears to be very practical. However, to compute the norm of B turns out to be complicated and costly. To overcome this difficulty, in 2012, L´opez et al. [8] introduced a relaxed CQ algorithm for solving the SFP (1) with a new adaptive way of determining the stepsize sequence τk defined as follows:(9)τk≔ρkfkxk∇fkxk2,where ρk∈0,4, ∀k≥1, such that infn⟶∞ρk4−ρk>0. It was proved that the sequence xk generated by (8) with τk defined by (9) converges weakly to a solution of the SFP (1). That is, their algorithm has only weak convergence in the framework of infinite-dimensional Hilbert spaces.Many authors also proposed algorithms that generate a sequencexk which converges strongly to a point in the solution set of the SFP (1), see, e.g., [9–12]. In particular, Deepho and Kumam [11] proposed a modified Halpern’s iterative scheme for solving the SFP (1) in the setting of infinite-dimensional Hilbert spaces as follows:(10)u,x0∈H1;xk+1≔βku+δkxk+γkPCxk−τkB∗I−PQBxk,∀k≥1,where 0<τk<2/B2 and βk, δk, and γk are three sequences in 0,1 such that βk+δk+γk=1. Under some mild and standard assumptions, it was proved that the sequence xk generated by (10) converges strongly to a solution of the SFP (1).Recently, Reich et al. [13] considered and studied the following split feasibility problem with multiple output sets in Hilbert spaces.LetH and Hi,i=1,…,N, be real Hilbert spaces and let Bi:H⟶Hi,i=1,…,N, be bounded linear operators. Let C and Qi,i=1,…,N, be nonempty, closed, and convex subsets of H and Hi,i=1,…,N, respectively. Given H, Hi, and Bi as above, the split feasibility problem with multiple output sets (in short, SFPwMOS) is to find an element x∗ such that(11)x∗∈Γ≔C∩​∩​i=1NBi−1Qi≠0.Reich et al. defined the functiong:H⟶ℝ by(12)gx≔1/2∑i−1NI−PQiBix2,for allx∈H.It is not difficult to see that an elementp∗ is a solution to the SFPwMOS (11) if and only if it is a solution to the problem(13)minx∈Cgx,This is equivalent to(14)0∈∇gx∗+NCx∗,where NCx is the normal cone of C at the point xrecall: letC⊂H be a closed convex subset of a real Hilbert space H. The normal cone of C at x denoted by NCx is given by NCx=z∈H:z,y−x≤0，∀y∈C, which implies(15)x∗=PCxT−λ∑i=1NBiTI−PQiBixT,where λ is an arbitrary positive real number. Motivated by these characterizations, Reich et al. [13] introduced the following iterative method for solving the SFPwMOS (11). For any given point, x0∈H,xk is generated by the following iteration:(16)xk+1≔αkfxk+1−αkPCxk−λk∑i=1NBiTI−PQiBixk,where f:C⟶C is a strict contraction mapping of H into itself with the contraction constant θ∈0,1,λk⊂0,∞ andαk⊂0,1. It was proved that if the sequences λk and αk satisfy the conditions,(17)I−fx∗,x−px∗≥0∀y∈Γ.An important observation here is that the method (Scheme (17)) introduced by Reich et al. [13] requires to compute the metric projections on to the sets C and Qi. Moreover, it needs to compute the operator norm, and it only employs current iterative point to obtain the next iteration, which leads to slow convergence.Inertial effect was proposed by Polyak [14] to speed up convergence of smooth convex minimization problem. The main idea is to make use of two previous iterates in order to update the next iterate. Due to its good acceleration character, inertial type algorithms have been widely studied by authors [15–19].Motivated by the modified Halpern’s iterative scheme proposed by [11] for the SFP (1), the relaxed CQ algorithm is proposed by He et al. [11] to solve the SFP (1) and inertial effect is proposed by Polyak [14]. In this paper, we propose a new inertial self-adaptive relaxed CQ algorithm for solving the SFPwMOS (11) in general Hilbert spaces.In Section2, we recall some necessary tools which are used in establishing our main results. In Section 3, we propose an inertial self-adaptive relaxed CQ algorithm for solving the SFPwMOS (11), and we establish and analyze a strong convergence theorem for the proposed algorithm. In Section 4, we present some newly derived results for solving the SFPwMOS (11). Finally, in Section 5, we provide one numerical experiment to illustrate the implementation of our proposed method and compare it with some existing results.


## 2. Preliminaries

In this section, we recall some preliminaries which are needed in the sequel. LetH be a real Hilbert space with the inner product .,. and induced norm⋅. Let I stand for the identity operator on H. We denote the fixed point set of an operator T:H⟶H (if T has fixed point) by FixT, i.e., FixT=x∈H:Tx=x. Let the symbols“⇀” and“⟶” denote the weak and strong convergence, respectively. For any sequence, xk⊂H,ωωxk=x∈H:∃xkj⊂xk such that xkj⇀xk denotes the weak ω-limit set of xk.Definition 1 (see [3]).
LetH be a real Hilbert space with inner product , and induced norm ⋅. Let C be a nonempty closed convex subset of H. Let T:C⟶H be a given operator. Then, T is called(1)
Lipschitz continuous with constantσ>0 on C if(18)Tx−Ty≤σx−y,∀x,y∈C.(2)
Nonexpansive onC if(19)Tx−Ty≤x−y,∀x,y∈C.(3)
Firmly nonexpansive onC if(20)Tx−Ty2≤x−y2−I−Tx−I−Ty2,∀x,y∈C,which is equivalent to(21)Tx−Ty2≤Tx−Ty,x−y,∀x,y∈C.(4)
Averaged if there exists a numberσ∈0,1 and a nonexpansive operator F:C⟶H such that(22)T=F+1−σI,whereI is the identity operator.
In this case, we say thatT is σ-averaged.Lemma 1.
LetC⊂H be a nonempty, closed, and convex set, and PCx denotes the metric (orthogonal) projection of x onto C. Then, the following assertions hold for any x,y∈H and z∈C:(23)1x−PCx,z−PCx≤0,2PCx−PCy≤x−y,3PCx−PCy2≤PCx−PCy,x−y,4PCx−z2≤x−z2−x−PCx2.
Evidently,I−PC is firmly nonexpansive and also nonexpansive.Definition 2.
Letδ∈0,1 and f:H⟶R be a proper function. Then,(1)
f is convex if(24)fδx+1−δy≤δfx+1−δfy,∀x,y∈H.(2)
f is strongly convex with constant σ, where σ>0, if(25)fδx+1−δy+σ2δ1−δx−y2≤δfx+1−δfy,∀x,y∈H.(3)
A vectorω∈H is a subgradient of f at a point x if(26)fy≥fx+ω,y−x,∀y∈H.(4)
The set of all subgradients of a convex functionf:H⟶R at x∈H, denoted by ∂fx, is called the subdifferential of f and is defined by(27)∂fx=ω∈H:fy≥fx+ω,y−x,for eachy∈H.(5)
If∂fx≠0, f is said to be subdifferentiable at x.
If the functionf is continuously differentiable, then ∂fx=∇fx. The convex function is subdifferentiable everywhere.Definition 3.
Letf:H⟶−∞,+∞ be a proper function.(1)
f is lower semicontinuous (lsc) at x if xk⟶x implies(28)fx≤liminfk⟶∞fxk.(2)
f is weakly lower semicontinuous (ω-lsc) at x if xk⇀x implies(29)fx≤liminfk⟶∞fxk.(3)
f is lower semicontinuous on H if it is lower semicontinuous at every point x∈H and f is weakly lower semicontinuous on H if it is weakly lower semicontinuous at every point x∈H.Lemma 2 (see [20]).
LetC and Q be closed convex subsets of real Hilbert spaces H1 and H2, respectively, and f:H1⟶R is given byfx=I−PQAx2, where A:H1⟶H2 be a bounded linear operator. Then, for δ>0 and x∗∈H1, the following statements are equivalent.(1)
The pointx∗ solves the SFP (1), i.e., x∗∈x∈C:Ax∈Q.(2)
The pointx∗ is the fixed point of the mapping PCI−δ∇f.(3)
The pointx∗ solves the variational inequality problem with respect to the gradient of f, that is, find a point x∗∈C such that(30)∇fx,y−x≥0,∀y∈C.Lemma 3 (see [6]).
LetH1 and H2 be real Hilbert spaces, and f:H1⟶R is given by fx=1/2I−PQAx2, where Q is closed convex subset of H2 and A:H1⟶H2 is a bounded linear operator. Then, the following assertions hold:(1)
f is convex and differentiable(2)
f is weakly lower semicontinuous on H1(3)
∇fx=ATI−PQAx, for x∈H1(4)
∇fx is A2-Lipschitz, i.e., ∇fx−∇fy≤A2x−y,∀x,y∈H1Definition 4.
LetΛk be a real sequence. Then, Λk decreases at infinity if there exists k0∈ℕ such that Λk+1≤Λk, for k≥k0. In other words, the sequence Λk does not decrease at infinity if there exists a subsequence Λktt≥1 of Λk such that Λkt<Λkt+1, for all t≥1.Lemma 4.
LetΛk be a sequence of real numbers that does not decrease at infinity. Also, consider the sequence of integers φkk≥k0 defined by(31)φk=maxm∈ℕ:m≤k,Λm≤Λm+1.Then,φkk≥k0 is a nondecreasing sequence verifying limk⟶∞φk=∞, and for all k≥k0, the following two estimates hold:(32)Λφk≤Λφk+1andΛk≤Λφk+1.Lemma 5 (see [21]).
Letsk be a sequence of nonnegative real numbers satisfying the following relation:(33)sk+1≤1−σksk+σkμk+βk,k≥1,where σk, μk, and βk satisfy the conditions:(34)1σk⊂0,1,∑k=1∞σk=∞,2limsupk⟶∞μk≤0,3βk≥0∑k=1∞βk<∞Then,limsupk⟶∞sk=0.


## 3. Main Results

In this section, we consider a general case of the SFPwMOS (11), where the nonempty, closed, and convex sets C and Qii=1,…,N are given by level sets of convex functions as in [7]. Throughout this section, we assume that c:H⟶ℝ and qi:Hi⟶ℝare lower semicontinuous convex functions and the sets C and Qi are given by(35)C≔x∈H:cx≤0andQi:y∈Hi:qiy≤0.The relaxed set (half spaces)Ck at xk is defined as(36)Ck≔x∈H:cxk≤ξk,xk−x,where ξk∈∂cxk.The relaxed set (half spaces)Qiki=1,⋯,N at xk are defined as(37)Qik≔x∈H:cxk≤ξk,xk−x,y∈Hi:qiBixk≤ηik,Bixk−y,where ηik∈∂qiBixk. It follows that C⊂Ck and Qi⊂Qik hold, for every k≥0. Now, we define the following (relaxed) proximity function; for x∈H,(38)gkx≔12∑i=1NI−PQikBix2.We note thatgk. is differentiable with its gradient given by(39)∇gkx≔∑iNBiTI−PQikBix,where each Qik are half spaces given in (37). We note that gk is weakly lower semicontinuous, convex, and differentiable function and ∇gk is Lipschitz continuous.Now, we present an inertial self-adaptive relaxedCQ algorithm.We can see that Algorithm1 terminates at some iterate (say k) when ∇gkxk=0.Algorithm 1: Inertial self-adaptive relaxed CQ algorithm for SFPwMOS.
Initialization: choose positive sequences0≤θk<1,ρk⊂0,4,βk⊂0,1,δk⊂0,1 and εk⊂0,1 such that βk+δk+εk=1. Let u∈H be a fixed point. Select arbitrary starting points x0,x1∈H, and set k=0Step 1: given the current iteratexk∈H, if ∇gkxk=0, for some k∈ℕ, then stop. Otherwise, continue and calculateτk≔ρkgkzk/∇gkzk2.Step 2: compute the next iterate aszk=xk+θkxk−xk−1xk+1≔PCkβku+δkzk+εkzk−τk∇gkzkwhereCk is the half space given as in (36)The following lemma is important for the convergence analysis.Lemma 6.
Ifzk≔xk+θkxk−xk−1, where 0≤θk<1, for all k∈ℕ, then, for all z∈H,(40)zk−z2≤xk−z2+θkxk−z2−xk−1−z2+2θkxk−xk−12.Proof.
Using the identity2a,b=a2+b2−a−b2, we have(41)zk−z2=xk−z+θkxk−xk−12=xk−z2+2θkxk−z,xk−xk−1+θk2xk−xk−12,=xk−z2+θkxk−z2+xk−xk−12−xk−1−z2+θk2xk−xk−12,=xk−z2+θkxk−z2−xk−1−z2+θk1+θkxk−xk−12,≤xk−z2+θkxk−z2−xk−1−z2+2θkxk−xk−12.
‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬Next, we show the following strong convergence theorem for Algorithm1.Theorem 1.
LetH,Hi,i=1,…,N, be real Hilbert spaces and let Bi:H⟶Hi,i=1,…,N, be bounded linear operators. Denote Γ as the solution set of the problem (11). Let C and Qi,i=1,…,N, be nonempty, closed, and convex subsets of H and Hi,i=1,…,N, respectively. Assume that the SFPwMOS (11) is consistent. Suppose the sequences ρk, βk, δk, and εk in Algorithm 1 satisfy the following conditions:(42)A1limk⟶∞βk=0and∑k=1∞βk=+∞βk+1βk≤l,lis a constant,(43)A20<liminfk⟶∞εk≤limsupfk⟶∞εk<1,(44)A3liminfk⟶∞ρk4−ρk>0,(45)A4limk⟶∞θkβkxk−xk−1=0.Then, the sequencexk generated by Algorithm 1 converges strongly to the solution p∈Γ, where p=PΓu.Proof.
We may assume that Algorithm1 does not terminate in a finite number of iterations. Thus, ∇gkxk≠0, for all k≥0. Denote Γ as the solution set of problem (11). In the consistent case of problem (11), Γ is nonempty, closed, and convex. Thus, the metric projection PΓ is well defined.
Letp∈Γ and set sk=zk−τk∇gkzk. Note that I−PQik, for each i=1,…,N, is firmly nonexpansive and ∇gkp=0. Hence, we have, from Lemma 1, that(46)∇gkzk,zk−p=∑i=1NBiTI−PQikBizk,zk−p,=∑i=1NBiTI−PQikBizk,zk−p,=∑i=1NI−PQikBizk,Bizk−Bip,≥∑i=1NI−PQikBizk2=2gkzk,which implies that(47)sk−p2=zk−p−τk∇gkzk2,=zk−p2+τk2∇gkzk2−2τk∇gkzk,zk−p≤zk−p2+ρk2gk2zk∇gkzk2−2ρkgkzk∇gkzk22gkzk=zk−p2+ρk2gk2zk∇gkzk2−4ρkgk2zk∇gkzk2=zk−p2−ρk4−ρkgk2zk∇gkzk2.
Using condition (A3), we have(48)sk−p2≤zk−p2,∀k≥0.
Next, we showzk is bounded. Since p∈Γ⊂Ck and the projection operator PCk is nonexpansive, we obtain from Algorithm 1 and (48) that(49)xk+1−p=PCkβku+δkzk+εkzk−τk∇gkzk−p,≤βku−p+δkzk−p+εksk−p,≤βku−p+δkzk−p+εkzk−p,=βku−p+1−βkzk−p,=βku−p+1−βkxk+θkxk−xk−1−p,≤βku−p+1−βkxk−p+θkxk−xk−1,=1−βkxk−p+βku−p+θkβkxk−xk−1.
Condition (A4) implies the sequenceθk/βkxk−xk−1 is bounded. Let an upper bound of u−p+θk/βkxk−xk−1 be M. Then, we rewrite (49) as(50)xk+1−p≤maxxk−p,M,
By induction, we have that(51)xk+1−p≤maxx1−p,M.
Hence,xk−p is bounded, so is zk−p and sk−p. And, zk is bounded. Consequently, sk and Bizki=1N are also bounded. The rest of the proof will be divided into two parts.Case 1.
Suppose that there existsk0∈ℕ such that xk−p2k=k0∞is nonincreasing. Then, xk−p2k=1∞ converges and xk−p2−xk+1−p2⟶0 as k⟶∞. From (47), we obtain(52)ρk4−ρkgk2zk∇gkzk2≤zk−p2−sk−p2.
Sincep∈Γ⊂Ck and the projection operator PCk is nonexpansive, from Algorithm 1, we obtain(53)xk+1−p2=PCkβku+δkzk+εksk−p2,≤βku+δkzk+εksk−p2,≤βku−p2+δkzk−p2+εksk−p2.
Sinceβk+δk+εk=1, from (53) and using the inequality x+y2≤x2+2y,x+y, we have the following estimation:(54)zk−p2−sk−p2≤βkεku−p2+1−βkεkzk−p2−1εkxk+1−p2,=βkεku−p2−βkεkzk−p2+1εkzk−p2−xk+1−p2,≤βkεku−p2+1εkzk−p2−xk+1−p2,=1εkβku−p2+zk−p2−xk+1−p2,=1εkβku−p2+xk−p+θkxk−xk−12−xk+1−p2,≤1εkβku−p2+xk−p2+2θkxk−xk−1,zk−p−xk+1−p2,≤1εkβku−p2+xk−p2+2θkxk−xk−1zk−p−xk+1−p2,=1εkβku−p2+xk−p2−xk+1−p2+2θkxk−xk−1zk−p.
Combining (52) and (54), we obtain(55)ρk4−ρkgk2zk∇gkzk2≤zk−p2−sk−p2,≤1εkβku−p2+xk−p2−xk+1−p2+2θkxk−xk−1zk−p.
By condition (A4),βk∈0,1 imply the sequence limk⟶∞θkxk−xk−1⟶0, and we have proved zk−p is bounded, so 2θkxk−xk−1zk−p⟶0. By conditions (A2) and (A3) and (55), we have, as k⟶∞,(56)0<ρk4−ρkgk2zk∇gkzk2,≤1εkβku−p2+xk−p2−xk+1−p2+2θkxk−xk−1zk−p⟶0,which implies that(57)limk⟶∞gk2zk∇gkzk2=0.
We note that, for eachi=1,...,N, BiTI−PQikBi. is Lipschitz continuous. Since the sequence zk is bounded and(58)BiTI−PQikBizk=BiTI−PQikBizk−BiTI−PQikBip≤max1≤i≤NBizk−p,for all i=1,…,N, we have the sequence BiTI−PQikBizki=1∞ is bounded. Hence, ∇gkzkk=1∞ is bounded. Consequently, we have from (57) that(59)limk⟶∞I−PQikBizk=0
For(60)zk−xk=xk+θkxk−xk−1−xk=θkxk−xk−1,which implies(61)zk−xk⟶0,ask⟶∞,for each i=1,…,N, since sk=zk−τk∇gkzk, we have from (59) and (61) that(62)sk−zk=τk∇gkzk⟶0,ask⟶∞,sk−xk≤sk−zk+zk−xk⟶0,ask⟶∞.
That is,(63)limk⟶∞sk−xk=0,(64)limk⟶∞BiTI−PQikBizk=0
Let(65)lk=βku+δkzk+εksk=βku+1−βkvk,where(66)vk=δk1−βkzk+εk1−βksk.
This gives(67)vk−zk≤εk1−βksk−zk⟶0,k⟶∞,lk−vk≤βku−vk⟶0,k⟶∞,(68)lk−zk≤lk−vk+vk−zk⟶0,k⟶∞,lk−xk≤lk−zk+zk−xk⟶0,ask⟶∞.
Hence,(69)limk⟶∞lk−zk=limk⟶∞lk−xk=limk⟶∞lk−vk=limk⟶∞vk−zk=0.
Sincexk is bounded, there exists a subsequence xkj of xk such that xkj⇀p∗∈ωωxk. Next, we need to show p∗∈C and Bip∗∈Qi, for each i=1,…,N.
We first show thatxk+1−xk⟶0. For p∈Γ,(70)12xk+1−p2=12xk+1−xk+xk−p2,=12xk−p2+xk+1−xk,xk+1−p−12xk+1−xk2,=12xk−p2+xk+1−lk+lk−xk,xk+1−p−12xk+1−xk2,=12xk−p2+xk+1−lk,xk+1−p+lk−xk,xk+1−p−12xk+1−xk2,then(71)12xk+1−xk2≤12xk−p2−12xk+1−p2+xk+1−lk,xk+1−p+lk−xk,xk+1−p,from xk+1−lk,xk+1−p<0,(72)12xk+1−xk2≤12xk−p2−12xk+1−p2+lk−xk,xk+1−p.
Sincexk−p2−xk+1−p2⟶0 as k⟶∞, limk⟶∞lk−xk=0, we obtain(73)xk+1−xk⟶0,ask⟶∞.
Sincexkj+1∈Ckj, we obtain(74)cxkj+ξkj,xkj+1−xkj≤0.
Thus,(75)cxkj≤−ξkj,xkj+1−xkj≤ξxkj+1−xkj,where ξ satisfies ξk≤ξ, for all k. By virtue of the continuity of function c and xkj+1−xkj⟶0, we get that(76)cp∗=limj⟶∞cxkj≤0.
Therefore,p∗∈C.
Now, we show thatBip∗∈Qi; to do this, let hik=Bizk−PQikBizk⟶0 and let ηik be such that ηik≤η, for all k. Since Bizkj−hikj=PQikjBizkj∈Qikj, we have(77)qiBixkj+ηikj,Bizkj−hikj−Bixkj≤0.
Hence,(78)qiBixkj≤ηikj,Bixkj−Bizkj+ηikj,hikj≤ηBixkj−xkj−1+ηhikj⟶0.
By the continuity ofqi and Bixkj⟶Bix∗, we arrive at the conclusion(79)qiBip∗=limj⟶∞qiBixkj≤0,namely, Bip∗∈Qi, for all i=1,…,N. Hence, p∗∈Γ.
Moreover, forp=PΓu, we can see that(80)limsupk⟶∞lk−p,u−p=limk⟶∞lkj−p,u−p≤p∗−p,u−p≤0.
By Lemma1 and (65), we have(81)xk+1−p2=PClk−p2≤lk−p2=βku+1−βkvk−p,lk−p=βku−p,lk−p+1−βkvk−p,lk−p≤βku−p,lk−p+1−βkvk−plk−p=1−βkxk+θkxk−xk−1−p2+βku−p,lk−p+βk2u−p2,≤1−βkxk−p2+2θkxk−xk−1,zk−p+βku−p,lk−p+βk2u−p2,≤1−βkxk−p2+2θkxk−xk−1zk−p+βku−p,lk−p+βk2u−p2.
By conditions (A1) and (A4),βk∈0,1 imply the sequence limk⟶∞θkxk−xk−1⟶0,βk/2u−p⟶0, and using (80) in (81) and applying Lemma 5, we obtain(82)limk⟶∞xk−p2=0.
Therefore, ask⟶∞,xk⟶p=PΓuCase 2.
SetΛk=xk−p2. Assume that Λk is not decreasing at infinity. Let ϕ:ℕ⟶ℕ be a mapping for all k≥k0 (for some k0 large enough) defined by(83)ϕk=maxt∈ℕ:t≤k,Λt≤Λt+1.
By Lemma4, ϕkk=k0∞ is a nondecreasing sequence such that ϕk⟶∞ as k⟶∞ and(84)maxΛϕk,Λk≤Λϕk+1,∀k≥k0.
After a similar conclusion from (59), it is easy to see that(85)limk⟶∞I−PQiϕkBizϕk=0.
By the similar argument as the above in Case1, we conclude immediately that(86)limk⟶∞Bi∗I−PQiϕkBizϕk=0,(87)limsupk⟶∞lϕk−p,u−p≤0.
Sincexϕk is bounded, there exists a subsequence of xϕk, still denoted by xϕk which converges weakly to p∗. By similar argument as above in Case 1, we conclude immediately that p∗∈C and Bip∗∈Qi⇒p∗∈Γ.
From (57), we have that(88)xϕk+1−p2≤1−βϕkzϕk−p2+βϕku−p,lϕk−p≤1−βϕkxϕk+θϕkxϕk−xϕk−1−p2+βϕku−p,lϕk−p,≤1−βϕkxϕk−p2+2θϕkxϕk−xϕk−1,zϕk−p+βϕku−p,lϕk−p≤1−βϕkxϕk−p2+2θϕkxϕk−xϕk−1zϕk−p+βϕku−p,lϕk−p,which implies by Lemma 5(89)limk⟶∞xφk−p2=0andlimk⟶∞xϕk+1−p2=0.
Moreover, fork≥k0, it is easy to see that Λϕk−Λϕk+1≤0 if k≠ϕk (that is, k≥ϕk) because, for ϕk+1≤m≤n, Λm>Λm+1. As a consequence, we obtain(90)0≤Λk≤maxΛϕk,Λϕk+1=Λϕk+1,∀k≥k0.
Therefore, we obtainlimk⟶∞Λk=0, that is, xk converges strongly to p. This completes the proof.


## 4. Some Extensive Results

For the SFPwMOS (11), when N=1, it becomes the SFP (1). Thus, we have the following corollary for solving the SFP (1), which is an immediate consequence of Theorem 1.Corollary 1.
LetH1 and H2 be two real Hilbert spaces, and let B:H1⟶H2 be bounded linear operator. Let C and Q be nonempty, closed, and convex subsets of H1 and H2, respectively. Assume that Ω=C∩​B−1Q≠0. Let u∈H1 be a fixed point. For any starting point x0,x1∈H1, let xk be the sequence generated by(91)zk=xk+θkxk−xk−1,xk+1≔PCkβku+δkzk+εkzk−τk∇gkzk,where θk, Ck, τk, and ∇fk are given as in (8). Suppose the sequences βk,δk, and εk satisfy the conditions in Theorem 1. Then, the sequence xk converges strongly to the solution p∈Ω, where p=PΩu.When we takeu=x0 in Algorithm 1, we note also the following results regarding to the SFPwMOS (11).Corollary 2.
LetH,Hi,i=1,…,N, be real Hilbert spaces and let Bi:H⟶Hi,i=1,…,N, be bounded linear operators. Let C and Qi,i=1,…,N, be nonempty, closed, and convex subsets of H and Hi,i=1,…,N, respectively. Assume that problem (11) is consistent. For any initial guess x0,x1∈H, let xk be the sequence generated by(92)zk=xk+θkxk−xk−1,xk+1≔PCkβkx0+δkzk+εkzk−τk∇gkzk,whereθk, Ck, τk, and ∇gk are given by (36) and (39) and Algorithm 1, respectively. Suppose the sequences βk,δk, and εk satisfy the conditions in Theorem 1. Then, the sequence xk generated by (92) strongly converges to the solution p=PΩx0∈Γ.When we takeδk≡0 in Algorithm 1, we obtain the following result regarding the SFPwMOS (11).Corollary 3.
LetH,Hi,i=1,…,N, be real Hilbert spaces, and let Ti:H⟶Hi,i=1,…,N, be bounded linear operators. Let C and Qi,i=1,…,N, be nonempty, closed, and convex subsets of H and Hi,i=1,…,N, respectively. Assume that problem (11) is consistent. For a fixed point u∈H and any initial guess x0∈H, let xk be the sequence generated by(93)zk=xk+θkxk−xk−1,xk+1≔PCkβku+1−βkzk−τk∇gkzk,whereθk, Ck, τk, and ∇gk are given by (36) and (39) and Algorithm 1, respectively. Suppose the sequences satisfy conditions (A1). Then, the sequence xk generated by (93) strongly converges to the solution p=PΩu∈Γ.Of course, when we takeu=x0, we get the following result regarding the SFPwMOS (11).Corollary 4.
LetH,Hi,i=1,…,N, be real Hilbert spaces and let Bi:H⟶Hi,i=1,…,N, be bounded linear operators. Let C and Qi,i=1,…,N, be nonempty, closed, and convex subsets of H and Hi,i=1,…,N, respectively. Assume that problem (11) is consistent. For any initial guess x0∈H, let xk be the sequence generated by(94)zk=xk+θkxk−xk−1,xk+1≔PCkβkx0+1−βkzk−τk∇gkzk,where θk,Ck, τk, and ∇gk are given by (36) and (39) and Algorithm 1, respectively. Suppose the sequence βk satisfies conditions (A1). Then, the sequence xk generated by (94) strongly converges to the solution p=PΩu∈Γ.Remark 1.
In Corollary4, for the particular case, whereN=1, the iterative scheme (94) reduced exactly to iterative scheme proposed by He et al. in Theorem 3.2 of [10].


## 5. Numerical Experiment

In this section, we provide one numerical experiment to illustrate the implementation and efficiency of our proposed method. First, we study the behavior of Algorithm1 for different choices of θkand ρk. Next, we compare Algorithm 1 (θk=0.8) with no inertial Algorithm 1 (θk=0) and Scheme (17) for different initial points. The numerical results are completed on a standard TOSHIBA laptop with Intel(R) Core(TM) i5-2450M CPU@2.5GHz 2.5 GHz with memory 4 GB. The code is implemented in MATLAB R2020a.Let(95)A1=−0.4−0.2−0.20.40.5−0.10.2−0.50.3,A2=0.3−0.10.3−0.30.2−0.200.20.3,C=x∈R3|x1−x22+2x3≤0,Q1=x∈R3|x12+x2−x3≤0,Q2=x∈R3|x1+x22−x3≤0.Findx∗ such that(96)x∗∈Γ≔C∩​∩​i=12Ai−1Qi≠0.Define the error function as(97)Ek=13xk−PCkxk2+13A1xk−PQ1,kA1xk2+13A2xk−PQ2,kA2xk2.Note that ifEk=0 at the kth step, then xk∈Γ.For Algorithm1, for different choices of ρk and θk, we choose βk=1/9k+1, δk=k/9k+1, εk=8k/9k+1, u=0.10.10.1T, and x0=−0.1−0.1−0.1T and x1=0.30.5−0.6T. Using Ek<ε as stopping criteria, where ε is a small enough positive number, the results of the numerical experiment are reported in Table 1. The numerical results can be seen from Table 1 and Figures 1 and 2. In Table 1, “Iter” denotes the number of iterations.Table 1
Algorithm1 for different choices of θk, ε, and ρk.
ɛθkρk = 3.98ρk = 3.00IterEkIterEk10−3θk = 0.857.0906e − 0565.8048e − 05θk = 0.376.9397e − 04θk = 097.5095e − 0410−4θk = 0.857.0906e − 0565.8048e − 05θk = 0.3119.1282e − 05θk = 0159.6150e − 0510−5θk = 0.8124.4752e − 06142.5296e − 06θk = 0.3169.1012e − 06θk = 0248.8064e − 06Figure 1
Error against iterations for Algorithm1 with difference choices of θk, under ρk=3.98 and ε=10−5.Figure 2
Error against iterations for Algorithm1 with difference choices of ρk, under θk=0.8 and ε=10−5.The behavior of the functionEk in Table 1 is described in Figures 1 and 2.From Table1 and Figures 1 and 2, we can observe that Algorithm 1 for inertial effect case has less number of iterations than Algorithm 1 for θk=0(that is, no inertial effect) case; Algorithm 1 with bigger θk has less number of iterations than Algorithm 1 for small case; for the same θk, big ρk case has less number of iterations than Algorithm 1 for small case.Following, we compare our algorithm with iteration (16). For the example, iteration (16) can be written as(98)xk+1≔αkfxk+1−αkPCxk−λkB1∗I−PQ1B1xk+B2∗I−PQ2B2xk.In the process, we takeαk=0.8 and fxk=u=111Tλk=0.00005for Scheme (97). We choose βk=1/2k+1, δk=εk=k/2k+1, and ρk=3for Algorithm 1.From Table2 and Figure 3, we can observe that Algorithm 1 has less number of iterations than Scheme (97). Moreover, our iterative method is advantageous over Scheme (98) because Algorithm 1 does not require computing of the operator norm.Table 2
Comparsion of Algorithm1 and Scheme (97) for different choice initial point for ε=10−4
Initiative pointx1Algorithm1 with θk=0.8Algorithm1 with θk=0.0(98)IterEkIterEkIterEk[0.4 0.9–0.9]114.1830e − 06259.6999e − 06518.8226e − 06[0.9 0.3–0.5]61.4290e − 07228.6399e − 06269.6719e − 06[−0.7 0.3–0.2]127.8838e − 06249.2013e − 06488.4198e − 06[0.1 0.1–0.9]′121.5353e − 07268.7544e − 06529.9722e − 06Figure 3
Error against of iterations for the comparision of Algorithm1 and Scheme (97).



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*Source: 6252984-2021-11-05.xml*