Research ArticleAn Inertial Accelerated Algorithm for Solving Split FeasibilityProblem with Multiple Output Sets

Huijuan Jia12 Shufen Liu1 and Yazheng Dang 3

1College of Computer Science and Technology Jilin University 2699 Qianjin Street Chaoyang DistrictChangchun 130012 China2College of Computer Science and Technology Henan Polytechnic University 2001 Century AvenueShanyang District Jiaozuo 454003 China3Department of Management Shanghai University of Science and Technology Shanghai PRC 200093 China

Correspondence should be addressed to Yazheng Dang jgdyz163com

Received 4 June 2021 Accepted 6 October 2021 Published 5 November 2021

Academic Editor Antonio Di Crescenzo

Copyright copy 2021 Huijuan Jia et al is is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

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

1 Introduction

Let H1 and H2 be two real Hilbert spaces and C and Q benonempty closed and convex subsets of H1 and H2 re-spectively Let B H1⟶ H2 be a nonzero bounded linearoperator and BT be its adjoint

e split feasibility problem (SFP) is formulated to find apoint xlowast isin H1 satisfying

xlowast isin C such thatBx

lowast isin Q (1)

e SFP was introduced in [1] which has broad ap-plications in many fields such as image reconstructionproblem [2 3] and intensity-modulated radiation therapy(IMRT) [4ndash6] Byrne [2] introduced the CQ-algorithm forsolving (1) as follows

x0 isin H1 x

k+1 ≔ PC xk

minus τkBT

I minus PQ1113872 1113873Bxk

1113872 1113873 (2)

where PC and PQ are the metric projections onto C and Qrespectively that is PC(x) argminyisinCx minus y over all xC τk isin (0 2B2) where B2 is the spectral radius of thematrix BTB

In 2004 Yang [7] made the following assumptions

(1) e sets C and Q are denoted as

C ≔ x isin H1 c(x)le 01113864 1113865 andQ y isin H2 q(y)le 01113864 1113865

(3)

where c H1⟶ R and q H2⟶ R are convexand subdifferential functions on H1 and H2respectively

(2) e subgradients zc(x) and zq(y) of c and q re-spectively can be calculated they are

HindawiJournal of MathematicsVolume 2021 Article ID 6252984 12 pageshttpsdoiorg10115520216252984

zc(x) ≔ ξ isin H1 c(z)ge c(x) + ξ z1113864

minus x for each z isin H11113865(4)

zq(y) ≔ η isin H2 q(z)ge q(x) + η u1113864

minus y for each u isin H21113865(5)

which are bounded operators (ie bounded on boundedsets)

Define two half-spaces Ck and Qk as

Ck ≔ x isin H1 c xk

1113872 1113873le ξk x

kminus x1113966 1113967 (6)

where ξk isinzc(xk) and

Qk ≔ y isin H2 q Bxk

1113872 1113873le ηk Bx

kminus y1113966 1113967 (7)

where ηk isin zq(Bx) It is easy to see that CkIC and QkIQfor all kge 1

Yang in [7] introduced the following relaxedCQalgorithm for solving the SFP (1) in a finite-dimensionalHilbert space

x0 isin H1 x

k+1 ≔ PCkx

kminus τknablafk x

k1113872 11138731113872 1113873 (8)

where fk(xk) ≔ (12)(I minus PQk)2 nablafk(xk) ≔ BT(I minus

PQk)Bxk and τk isin (0 2B2) Since PCk

and PQkare easily

calculated this method appears to be very practical How-ever to compute the norm of B turns out to be complicatedand costly To overcome this difficulty in 2012 Lacuteopez et al[8] introduced a relaxed CQ algorithm for solving the SFP(1) with a new adaptive way of determining the stepsizesequence τk defined as follows

τk ≔ρkfk x

k1113872 1113873

nablafk xk

1113872 11138732 (9)

where ρk1113864 1113865 isin (0 4) forallkge 1 such that infn⟶infinρk(4 minus ρk)gt 0It was proved that the sequence xk1113864 1113865 generated by (8) with τk

defined by (9) converges weakly to a solution of the SFP (1)at is their algorithm has only weak convergence in theframework of infinite-dimensional Hilbert spaces

Many authors also proposed algorithms that generate asequence xk1113864 1113865 which converges strongly to a point in thesolution set of the SFP (1) see eg [9ndash12] In particularDeepho and Kumam [11] proposed a modified Halpernrsquositerative scheme for solving the SFP (1) in the setting ofinfinite-dimensional Hilbert spaces as follows

u x0 isin H1 x

k+1 ≔ βku + δkxk

+ ckPC xk

minus τkBlowast

I minus PQ1113872 1113873Bxk

1113872 1113873 forallkge 1

(10)

where 0 lt τk lt 2B2 and βk1113864 1113865 δk1113864 1113865 and ck1113864 1113865 are three se-quences in [0 1] such that βk + δk + ck 1 Under somemild and standard assumptions it was proved that the se-quence xk1113864 1113865 generated by (10) converges strongly to a so-lution of the SFP (1)

Recently Reich et al [13] considered and studied thefollowing split feasibility problem with multiple output setsin Hilbert spaces

Let H and Hi i 1 N be real Hilbert spaces and letBi H⟶ Hi i 1 N be bounded linear operatorsLet C and Qi i 1 N be nonempty closed and convexsubsets of H and Hi i 1 N respectively Given H Hiand Bi as above the split feasibility problem with multipleoutput sets (in short SFPwMOS) is to find an element xlowast

such that

xlowast isin Γ ≔ Ccap cap Ni1B

minus 1i Qi( 11138571113872 1113873ne 0 (11)

Reich et al defined the function g H⟶ R by

g(x) ≔ 121113944N

iminus 1I minus PQi

1113872 1113873Bix2 for all x isin H (12)

It is not difficult to see that an element plowast is a solution tothe SFPwMOS (11) if and only if it is a solution to theproblem

minxisinC

g(x) (13)

is is equivalent to

0 isinnablag xlowast

( 1113857 + NC xlowast

( 1113857 (14)

where NC(x) is the normal cone of C at the point x (recall letC sub H be a closed convex subset of a real Hilbert space He normal cone of C at x denoted by NC(x) is given byNC(x) z isin H z y minus xle 0forally isin C1113864 1113865) which implies

xlowast

PC xT

minus λ1113944N

i1B

Ti I minus PQi1113872 1113873Bix

T⎛⎝ ⎞⎠ (15)

where λ is an arbitrary positive real number Motivated bythese characterizations Reich et al [13] introduced thefollowing iterative method for solving the SFPwMOS (11)For any given point x0 isin H xk1113864 1113865 is generated by the fol-lowing iteration

xk+1 ≔ αkf x

k1113872 1113873 + 1 minus αk( 1113857PC x

kminus λk 1113944

N

i1B

Ti I minus PQi1113872 1113873Bix

k⎛⎝ ⎞⎠

(16)

where f C⟶ C is a strict contraction mapping of H intoitself with the contraction constant θ isin [0 1) λk sub (0 infin)

and αk1113864 1113865 sub (0 1) It was proved that if the sequences λk1113864 1113865

and αk1113864 1113865 satisfy the conditions

(I minus f)xlowast x minus px

lowast ge 0forally isin Γ (17)

An important observation here is that the method(Scheme (17)) introduced by Reich et al [13] requires tocompute the metric projections on to the sets C and QiMoreover it needs to compute the operator norm and itonly employs current iterative point to obtain the next it-eration which leads to slow convergence

Inertial effect was proposed by Polyak [14] to speed upconvergence of smooth convex minimization problem e

2 Journal of Mathematics

main idea is to make use of two previous iterates in order toupdate the next iterate Due to its good acceleration char-acter inertial type algorithms have been widely studied byauthors [15ndash19]

Motivated by the modified Halpernrsquos iterative schemeproposed by [11] for the SFP (1) the relaxed CQ algorithm isproposed by He et al [11] to solve the SFP (1) and inertialeffect is proposed by Polyak [14] In this paper we propose anew inertial self-adaptive relaxed CQ algorithm for solvingthe SFPwMOS (11) in general Hilbert spaces

In Section 2 we recall some necessary tools which areused in establishing our main results In Section 3 wepropose an inertial self-adaptive relaxed CQ algorithm forsolving the SFPwMOS (11) and we establish and analyze astrong convergence theorem for the proposed algorithm InSection 4 we present some newly derived results for solvingthe SFPwMOS (11) Finally in Section 5 we provide onenumerical experiment to illustrate the implementation ofour proposed method and compare it with some existingresults

2 Preliminaries

In this section we recall some preliminaries which areneeded in the sequel LetH be a real Hilbert space with theinner product lang rang and induced norm| middot | Let I stand for theidentity operator on H We denote the fixed point set of anoperator T H⟶ H (if T has fixed point) by Fix(T) ieFix(T) x isin H Tx x Let the symbolsldquordquoand ldquo⟶ rdquo denote the weak and strong convergence re-spectively For any sequence xk1113864 1113865 sub H ωω (xk)

x isin H exist xkj

1113966 1113967 sub xk1113864 11138651113966 1113967 such that xkjxk1113966 1113967 denotes theweak ω-limit set of xk1113864 1113865

Definition 1 (see [3]) LetH be a real Hilbert space with innerproduct lang rang and induced norm | middot | Let C be a nonemptyclosed convex subset of H Let T C⟶ H be a given op-erator =en T is called

(1) Lipschitz continuous with constant σ gt 0 on C if

Tx minus Tyle σx minus y forallx y isin C (18)

(2) Nonexpansive on C if

Tx minus Tylex minus y forallx y isin C (19)

(3) Firmly nonexpansive on C if

Tx minus Ty2 lex minus y

2minus (I minus T)x minus (I minus T)y

2forallx y isin C

(20)

which is equivalent to

Tx minus Ty2 leTx minus Ty x minus y forallx y isin C (21)

(4) Averaged if there exists a number σ isin (0 1) and anonexpansive operator F C⟶ H such that

T F +(1 minus σ)I (22)

where I is the identity operator

In this case we say that T is σ-averaged

Lemma 1 Let C sub H be a nonempty closed and convex setand PC(x) denotes the metric (orthogonal) projection of x

onto C =en the following assertions hold for any x y isin H

and z isin C

(1) x minus PC(x) z minus PC(x)le 0

(2) PC(x) minus PC(y)lex minus y

(3) PC(x) minus PC(y)2 lePC(x) minus PC(y) x minus y

(4) PC(x) minus z2 lex minus z

2minus x minus PC(x)

2

(23)

Evidently I minus PC is firmly nonexpansive and alsononexpansive

Definition 2 Let δ isin (0 1) and f H⟶ R be a properfunction en

(1) f is convex if

f(δx +(1 minus δ)y)le δf(x) +(1 minus δ)f(y) forallx y isin H

(24)

(2) f is strongly convex with constant σ where σ gt 0 if

f(δx +(1 minus δ)y) +σ2δ(1 minus δ)x minus y

2

le δf(x) +(1 minus δ)f(y) forallx y isin H

(25)

(3) A vector ω isin H is a subgradient of f at a point x if

f(y) gef(x) + ω y minus x forally isin H (26)

(4) e set of all subgradients of a convex functionf H⟶ R at x isin H denoted by zf(x) is calledthe subdifferential of f and is defined by

zf(x) ω isin H f(y)gef(x) + ω y minus x for eachy isin H1113864 1113865

(27)

(5) If zf(x)ne 0 f is said to be subdifferentiable at x

If the function f is continuously differentiable thenzf(x) nablaf(x)1113864 1113865 e convex function is subdifferentiableeverywhere

Definition 3 Let f H⟶ [minus infin +infin) be a properfunction

(1) f is lower semicontinuous (lsc) at x if xk⟶ x

implies

f(x)le liminfk⟶infin

f xk

1113872 1113873 (28)

(2) f is weakly lower semicontinuous (ω-lsc) at x ifxkx implies

f(x)le liminfk⟶infin

f xk

1113872 1113873 (29)

Journal of Mathematics 3

(3) f is lower semicontinuous on H if it is lowersemicontinuous at every point x isin H and f is weaklylower semicontinuous on H if it is weakly lowersemicontinuous at every point x isin H

Lemma 2 (see [20]) Let C and Q be closed convex subsets ofreal Hilbert spaces H1 and H2 respectively andf H1⟶ R is given byf(x) (I minus PQ)Ax2 whereA H1⟶ H2 be a bounded linear operator =en for δ gt 0and xlowast isin H1 the following statements are equivalent

(1) =e point xlowast solves the SFP (1) iexlowast isin x isin C Ax isin Q

(2) =e point xlowast is the fixed point of the mappingPC(I minus δnablaf)

(3) =e point xlowast solves the variational inequality problemwith respect to the gradient of f that is find a pointxlowast isin C such that

nablaf(x) y minus xge 0 forally isin C (30)

Lemma 3 (see [6]) Let H1 and H2 be real Hilbert spacesand f H1⟶ R is given byf(x) 12(I minus PQ)Ax2 whereQ is closed convex subset of H2 and A H1⟶ H2 is abounded linear operator =en the following assertions hold

(1) f is convex and differentiable(2) f is weakly lower semicontinuous on H1

(3) nablaf(x) AT(I minus PQ)Ax for x isin H1

(4) nablaf(x) is A2-Lipschitz ienablaf(x) minus nablaf(y)leA2x minus y forallx y isin H1

Definition 4 Let Λk1113864 1113865 be a real sequence en Λk1113864 1113865 de-creases at infinity if there exists k0 isin N such that Λk+1 leΛkfor kge k0 In other words the sequence Λk1113864 1113865 does not de-crease at infinity if there exists a subsequence Λkt

1113966 1113967tge1 of Λk1113864 1113865

such that ΛktltΛkt+1 for all tge 1

Lemma 4 Let Λk1113864 1113865 be a sequence of real numbers that doesnot decrease at infinity Also consider the sequence of integersφ(k)1113864 1113865kgek0 defined by

φ(k) max m isin N mle k Λm leΛm+11113864 1113865 (31)

en φ(k)1113864 1113865kgek0 is a nondecreasing sequence verifyinglim

k⟶infinφ(k) infin and for all kge k0 the following two esti-

mates hold

Λφ(k) leΛφ(k)+1 andΛk leΛφ(k)+1 (32)

Lemma 5 (see [21]) Let sk1113864 1113865 be a sequence of nonnegativereal numbers satisfying the following relation

sk+1 le 1 minus σk( 1113857sk + σkμk + βk kge 1 (33)

where σk1113864 1113865 μk1113864 1113865 and βk1113864 1113865 satisfy the conditions

(1) σk1113864 1113865 sub [0 1] 1113944infin

k1σk infin

(2) limsupk⟶infin

μk le 0

(3) βk ge 0 1113944infin

k1βk ltinfinThen limsup

k⟶infinsk 0

(34)

3 Main Results

In this section we consider a general case of the SFPwMOS(11) where the nonempty closed and convex sets C andQi(i 1 N) are given by level sets of convex functionsas in [7] roughout this section we assume thatc H⟶ R and qi Hi⟶ R are lower semicontinuousconvex functions and the sets C and Qi are given by

C ≔ x isin H c(x)le 0 andQi y isin Hi qi(y)le 01113864 1113865 (35)

e relaxed set (half spaces) Ck at xk is defined as

Ck ≔ x isin H c xk

1113872 1113873le ξk x

kminus x1113966 1113967 (36)

where ξk isinzc(xk)e relaxed set (half spaces) Qk

i (i 1 middot middot middot N) at xk aredefined as

Qki ≔ x isin H c x

k1113872 1113873le ξk

xk

minus x1113966 1113967

y isin Hi qi Bixk

1113872 1113873le ηki Bix

kminus y1113966 1113967

(37)

where ηki isin zqi(Bix

k) It follows that C sub Ck and Qi sub Qki

hold for every kge 0 Now we define the following (relaxed)proximity function for x isin H

gk(x) ≔12

1113944

N

i1I minus PQk

i1113874 1113875Bix

2 (38)

We note that gk() is differentiable with its gradientgiven by

nablagk(x) ≔ 1113944N

i

BTi I minus PQk

i1113874 1113875Bix (39)

where each Qki are half spaces given in (37) We note that gk

is weakly lower semicontinuous convex and differentiablefunction and nablagk is Lipschitz continuous

Now we present an inertial self-adaptive relaxed CQ

algorithmWe can see that Algorithm 1 terminates at some iterate

(say k) when nablagk(xk) 0e following lemma is important for the convergence

analysis

Lemma 6 If zk ≔ xk + θk(xk minus xkminus 1) where 0le θk lt 1 forall k isin N then for all z isin H

zk

minus z2 le x

kminus z

2+ θk x

kminus z

2minus x

kminus 1minus z

21113872 1113873 + 2θkx

kminus x

kminus 12

(40)

4 Journal of Mathematics

Proof Using the identity 2a b a2 + b2 minus a minus b2 we have

zk

minus z2

xk

minus z + θk xk

minus xkminus 1

1113872 11138732

xk

minus z2

+ 2θkxk

minus z xk

minus xkminus 1

+ θ2kxk

minus xkminus 12

xk

minus z2

+ θkxk

minus z2

+ xk

minus xkminus 12

minus xkminus 1

minus z2

+ θ2kxk

minus xkminus 12

xk

minus z2

+ θk xk

minus z2

minus xkminus 1

minus z2

1113872 1113873

+ θk 1 + θk( 1113857xk

minus xkminus 12

lexk

minus z2

+ θk xk

minus z2

minus xkminus 1

minus z2

1113872 1113873

+ 2θkxk

minus xkminus 12

(41)

Next we show the following strong convergence theo-rem for Algorithm 1

Theorem 1 Let H Hi i 1 N be real Hilbert spacesand let Bi H⟶ Hi i 1 N be bounded linear op-erators Denote Γ as the solution set of the problem (11) Let C

and Qi i 1 N be nonempty closed and convexsubsets of H and Hi i 1 N respectively Assume thatthe SFPwMOS (11) is consistent Suppose the sequences ρk1113864 1113865βk1113864 1113865 δk1113864 1113865 and εk1113864 1113865 in Algorithm 1 satisfy the followingconditions

(A1) limk⟶infin

βk 0

and 1113944infin

k1βk +infin

βk+1

βk

le l l is a constan t

(42)

(A2) 0lt liminfk⟶infin

εk le limsupfk⟶infin

εk lt 1 (43)

(A3) liminfk⟶infin

ρk 4 minus ρk( 1113857gt 0 (44)

(A4) limk⟶infin

θk

βk

xk

minus xkminus 1

0 (45)

en the sequence xk1113864 1113865 generated by Algorithm 1converges strongly to the solution p isin Γ where p PΓ(u)

Proof Wemay assume that Algorithm 1 does not terminatein a finite number of iterations us nablagk(xk)ne 0 for allk ge 0 Denote Γ as the solution set of problem (11) In the

consistent case of problem (11) Γ is nonempty closed andconvex us the metric projection PΓ is well defined

Let p isin Γ and set sk zk minus τknablagk(zk) Note that I minus PQki

for each i 1 N is firmly nonexpansive and nablagk(p)

0 Hence we have from Lemma 1 that

langnablagk zk

1113872 1113873 zk

minus prang lang1113944N

i1B

Ti I minus PQk

i1113874 1113875Biz

k z

kminus prang

1113944N

i1B

Ti I minus PQk

i1113874 1113875Biz

k z

kminus p

1113944N

i1I minus PQk

i1113874 1113875Biz

k Biz

kminus Bip

ge 1113944N

i1I minus PQk

i1113874 1113875Biz

k2 2gk z

k1113872 1113873

(46)

which implies that

sk

minus p

2

zk

minus p minus τknablagk zk

1113872 1113873

2

zk

minus p2

+ τ2knablagk zk

1113872 11138732

minus 2τknablagk zk

1113872 1113873 zk

minus p

le zk

minus p2

+ρ2kg

2k z

k1113872 1113873

nablagk zk

1113872 11138732 minus

2ρkgk zk

1113872 1113873

nablagk zk

1113872 11138732 2gk z

k1113872 11138731113872 1113873

zk

minus p2

+ρ2kg

2k z

k1113872 1113873

nablagk zk

1113872 11138732 minus

4ρkg2k z

k1113872 1113873

nablagk zk

1113872 11138732

zk

minus p2

minus ρk 4 minus ρk( 1113857g2k z

k1113872 1113873

nablagk zk

1113872 11138732

(47)

Using condition (A3) we have

sk

minus p2 le z

kminus p

2 forallkge 0 (48)

Initialization choose positive sequences 0le θk lt 1 ρk1113864 1113865 sub (0 4) βk1113864 1113865 sub (0 1) δk1113864 1113865 sub [0 1) and εk1113864 1113865 sub (0 1) such thatβk + δk + εk 1 Let u isin H be a fixed point Select arbitrary starting points x0 x1 isin H and set k 0

Step 1 given the current iterate xk isin H if nablagk(xk) 0 for some k isin N then stop Otherwise continue and calculateτk ≔ ρkgk(zk)nablagk(zk)2

Step 2 compute the next iterate asz

k x

k+ θk(x

kminus x

kminus 1)

xk+1 ≔ PCk

(βku + δkzk

+ εk(zk

minus τknablagk(zk)))

1113896

where Ck is the half space given as in (36)

ALGORITHM 1 Inertial self-adaptive relaxed CQ algorithm for SFPwMOS

Journal of Mathematics 5

Next we show zk1113864 1113865 is bounded Since p isin Γ sub Ck and theprojection operator PCk

is nonexpansive we obtain fromAlgorithm 1 and (48) that

xk+1

minus p PCkβku + δkz

k+ εk z

kminus τknablagk z

k1113872 11138731113872 11138731113872 1113873 minus p

le βku minus p + δkzk

minus p + εksk

minus p

le βku minus p + δkzk

minus p + εkzk

minus p

βku minus p + 1 minus βk( 1113857zk

minus p

βku minus p + 1 minus βk( 1113857xk

+ θk xk

minus xkminus 1

1113872 1113873 minus p

le βku minus p + 1 minus βk( 1113857xk

minus p + θk xk

minus xkminus 1

1 minus βk( 1113857 xk

minus p

+ βk u minus p +θk

βk

xk

minus xkminus 1

1113890 1113891

(49)

Condition (A4) implies the sequence θkβkxk minus xkminus 1 isbounded Let an upper bound of u minus p + θkβkxk minus xkminus 11113864 1113865 beM en we rewrite (49) as

xk+1

minus plemax xk

minus p M1113966 1113967 (50)

By induction we have that

xk+1

minus plemax x1

minus p M1113966 1113967 (51)

Hence xk minus p1113864 1113865 is bounded so is zk minus p1113864 1113865 and sk minus p1113864 1113865And zk1113864 1113865 is bounded Consequently sk1113864 1113865 and Biz

k1113864 1113865N

i1 arealso bounded The rest of the proof will be

divided into two parts

Case 1 Suppose that there exists k0 isin N such thatxk minus p21113864 1113865

infinkk0

is nonincreasing en xk minus p21113864 1113865infink1 converges

and xk minus p2 minus xk+1 minus p2⟶ 0 as k⟶infin From (47) weobtain

ρk 4 minus ρk( 1113857g2k z

k1113872 1113873

nablagk zk

1113872 11138732 le z

kminus p

2minus s

kminus p

2 (52)

Since p isin Γ sub Ck and the projection operator PCkis

nonexpansive from Algorithm 1 we obtain

xk+1

minus p2

PCkβku + δkz

k+ εksk1113872 1113873 minus p

2

le βku + δkzk

+ εksk1113872 1113873 minus p2

le βku minus p2

+ δkzk

minus p2

+ εksk minus p2

(53)

Since βk + δk + εk 1 from (53) and using the inequalityx + y2 lex2 + 2y x + y we have the following estimation

zk

minus p2

minus sk

minus p2 le

βk

εk

u minus p2

+1 minus βk

εk

zk

minus p2

minus1εk

xk+1

minus p2

βk

εk

u minus p2

minusβk

εk

zk

minus p2

+1εk

zk

minus p2

minus xk+1

minus p2

1113960 1113961

leβk

εk

u minus p2

+1εk

zk

minus p2

minus xk+1

minus p2

1113960 1113961

1εk

βku minus p2

+ zk

minus p2

minus xk+1

minus p2

1113960 11139611113960 1113961

1εk

βku minus p2

+ xk

minus p + θk xk

minus xkminus 1

1113872 11138732

minus xk+1

minus p2

1113876 11138771113876 1113877

le1εk

βku minus p2

+ xk

minus p2

+ 2θkxk

minus xkminus 1

zk

minus p minus xk+1

minus p2

1113960 11139611113960 1113961

le1εk

βku minus p2

+ xk

minus p2

+ 2θkxk

minus xkminus 1

zk

minus p minus xk+1

minus p2

1113960 11139611113960 1113961

1εk

βku minus p2

+ xk

minus p2

minus xk+1

minus p2

+ 2θkxk

minus xkminus 1

zk

minus p1113960 11139611113960 1113961

(54)

Combining (52) and (54) we obtain

6 Journal of Mathematics

ρk 4 minus ρk( 1113857g2k z

k1113872 1113873

nablagk zk

1113872 11138732 le z

kminus p

2minus s

kminus p

2

le1εk

βku minus p2

+ xk

minus p2

minus xk+1

minus p2

11139601113960

+ 2θkxk

minus xkminus 1

zk

minus p11139611113961

(55)

By condition (A4) βk isin (0 1) imply the sequencelimk⟶infinθkxk minus xkminus 1⟶ 0 and we have proved zk minus p1113864 1113865 isbounded so 2θkxk minus xkminus 1zk minus p⟶ 0 By conditions (A2)and (A3) and (55) we have as k⟶infin

0lt ρk 4 minus ρk( 1113857g2k z

k1113872 1113873

nablagk zk

1113872 11138732

le1εk

βku minus p2

+ xk

minus p2

minus xk+1

minus p2

11139601113960

+ 2θkxk

minus xkminus 1

zk

minus p11139611113961⟶ 0

(56)

which implies that

limk⟶infin

g2k zk

1113872 1113873

nablagk zk

1113872 11138732 0 (57)

We note that for each i 1 N BTi (I minus PQk

i)Bi() is

Lipschitz continuous Since the sequence zk1113864 1113865 is boundedand

BTi I minus PQk

i1113874 1113875 Biz

k B

Ti I minus PQk

i1113874 1113875Biz

k

minus BTi I minus PQk

i1113874 1113875Biple max

1leileNBi1113874 1113875z

kminus p

(58)

for all i 1 N we have the sequence

BTi (I minus PQk

i)Biz

k1113882 1113883infin

i1is bounded Hence nablagk(zk)1113864 1113865

infink1 is

bounded Consequently we have from (57) that

limk⟶infin

I minus PQki

1113874 1113875Bizk

0 (59)

For

zk

minus xk

xk

+ θk xk

minus xkminus 1

1113872 1113873 minus xk

θkxk

minus xkminus 1

(60)

which implies

zk

minus xk⟶ 0 as k⟶infin (61)

for each i 1 N since sk zk minus τknablagk(zk) we havefrom (59) and (61) that

sk

minus zk

τknablagk zk

1113872 1113873⟶ 0 as k⟶infin

sk

minus xk le s

kminus z

k+ z

kminus x

k⟶ 0 as k⟶infin(62)

at is

limk⟶infin

sk

minus xk

0 (63)

limk⟶infin

BTi I minus PQk

i1113874 1113875Biz

k 0 (64)

Let

lk

βku + δkzk

+ εksk

βku + 1 minus βk( 1113857vk (65)

where

vk

δk

1 minus βk

zk

+εk

1 minus βk

sk (66)

is gives

vk

minus zk le

εk

1 minus βk

sk

minus zk⟶ 0 k⟶infin

lk

minus vk

le βk u minus vk

⟶ 0 k⟶infin

(67)

lk

minus zk le l

kminus v

k+ v

kminus z

k⟶ 0 k⟶infin

lk

minus xk le l

kminus z

k+ z

kminus x

k⟶ 0 as k⟶infin(68)

Hence

limk⟶infin

lk

minus zk

limk⟶infin

lk

minus xk

limk⟶infin

lk

minus vk

limk⟶infin

vk

minus zk

0

(69)

Since xk1113864 1113865 is bounded there exists a subsequence xkj1113966 1113967

of xk1113864 1113865 such that xkj1113966 1113967plowast isin ωω(xk) Next we need toshow plowast isin C and Bip

lowast isin Qi for each i 1 NWe first show that xk+1 minus xk⟶ 0 For p isin Γ

12x

k+1minus p

212x

k+1minus x

k+ x

kminus p

2

12x

kminus p

2+ x

k+1minus x

k x

k+1minus p minus

12x

k+1minus x

k2

12x

kminus p

2+ x

k+1minus l

k+ l

kminus x

k x

k+1

minus p minus12x

k+1minus x

k2

12x

kminus p

2+ x

k+1minus l

k x

k+1minus p + l

k

minus xk x

k+1minus p minus

12x

k+1minus x

k2

(70)

then

12x

k+1minus x

k2 le12x

kminus p

2minus12x

k+1minus p

2+ x

k+1minus l

k x

k+1

minus p + lk

minus xk x

k+1minus p

(71)

from xk+1 minus lk xk+1 minus plt 0

Journal of Mathematics 7

12x

k+1minus x

k2 le12x

kminus p

2minus12x

k+1minus p

2+ l

kminus x

k x

k+1minus p

(72)

Since xk minus p2 minus xk+1 minus p2⟶ 0 as k⟶infinlim

k⟶infinlk minus xk 0 we obtain

xk+1

minus xk⟶ 0 as k⟶infin (73)

Since xkj+1 isin Ckj we obtain

c xkj1113872 1113873 + ξkj x

kj+1minus x

kj le 0 (74)

us

c xkj1113872 1113873le minus ξkj x

kj+1minus x

kj le ξxkj+1

minus xkj (75)

where ξ satisfies ξk le ξ for all k By virtue of the continuity offunction c and xkj+1 minus xkj⟶ 0 we get that

c plowast

( 1113857 limj⟶infin

c xkj1113872 1113873le 0 (76)

erefore plowast isin CNow we show that Bip

lowast isin Qi to do this let hki Biz

k minus

PQkiBiz

k⟶ 0 and let ηki be such that ηk

i le η for all k SinceBiz

kj minus hkj

i PQ

kj

i

(Bizkj ) isin Q

kj

i we have

qi Bixkj1113872 1113873 + ηkj

i Bizkj minus h

kj

i1113874 1113875 minus Bixkj le 0 (77)

Hence

qi Bixkj1113872 1113873le η

kj

i Bixkj minus Biz

kj + ηkj

i hkj

i le ηBixkj

minus xkjminus 1

+ ηhkj

i ⟶ 0

(78)

By the continuity of qi and Bixkj⟶ Bix

lowast we arrive atthe conclusion

qi Biplowast

( 1113857 limj⟶infin

qi Bixkj1113872 1113873le 0 (79)

namely Biplowast isin Qi for all i 1 N Hence plowast isin Γ

Moreover for p PΓ u we can see that

limsupk⟶infin

lk

minus p u minus p limk⟶infin

lkj minus p u minus plep

lowast

minus p u minus ple 0

(80)

By Lemma 1 and (65) we have

xk+1

minus p2

PC lk

1113872 1113873 minus p2 le l

kminus p

2 βku + 1 minus βk( 1113857v

kminus p l

kminus p

βku minus p lk

minus p + 1 minus βk( 1113857vk

minus p lk

minus p

le βku minus p lk

minus p + 1 minus βk( 1113857vk

minus plk

minus p

1 minus βk( 1113857 xk

+ θk xk

minus xkminus 1

1113872 1113873 minus p

2

+ βklangu minus p lk

minus prang +βk

2u minus p

2

le 1 minus βk( 1113857 xk

minus p

2

+ 2θklangxk

minus xkminus 1

zk

minus prang1113876 1113877 + βklangu minus p lk

minus prang +βk

2u minus p

2

le 1 minus βk( 1113857 xk

minus p

2

+ 2θk xk

minus xkminus 1

zk

minus p

1113876 1113877 + βklangu minus p lk

minus prang +βk

2u minus p

2

(81)

By conditions (A1) and (A4) βk isin (0 1) imply the se-quence lim

k⟶infinθkxk minus xkminus 1⟶ 0 βk2u minus p⟶ 0 and

using (80) in (81) and applying Lemma 5 we obtain

limk⟶infin

xk

minus p2

0 (82)

erefore as k⟶infin xk⟶ p PΓu

Case 2 SetΛk xk minus p2 Assume that Λk1113864 1113865 is not decreasingat infinity Let ϕ N⟶ N be a mapping for all kge k0 (forsome k0 large enough) defined by

ϕ(k) max t isin N tle kΛt leΛt+11113864 1113865 (83)

By Lemma 4 ϕ(k)1113864 1113865infinkk0is a nondecreasing sequence

such that ϕ(k)⟶infin as k⟶infin and

max Λϕ(k)Λk1113966 1113967leΛϕ(k)+1forallkge k0 (84)

After a similar conclusion from (59) it is easy to see that

limk⟶infin

I minus PQ

ϕ(k)

i

1113874 1113875Bizϕ(k)

0 (85)

By the similar argument as the above in Case 1 weconclude immediately that

8 Journal of Mathematics

limk⟶infin

Blowasti I minus P

Qϕ(k)

i

1113874 1113875Bizϕ(k)

0 (86)

limsupk⟶infin

lϕ(k)

minus p u minus ple 0 (87)

Since xϕ(k)1113864 1113865 is bounded there exists a subsequence ofxϕ(k)1113864 1113865 still denoted by xϕ(k)1113864 1113865 which converges weakly to

plowast By similar argument as above in Case 1 we concludeimmediately that plowast isin C and Bip

lowast isin QirArrplowast isin ΓFrom (57) we have that

xϕ(k)+1

minus p2 le 1 minus βϕ(k)1113872 1113873z

ϕ(k)minus p

2+ βϕ(k)u minus p l

ϕ(k)minus p

le 1 minus βϕ(k)1113872 1113873xϕ(k)

+ θϕ(k) xϕ(k)

minus xϕ(k)minus 1

1113872 1113873 minus p2

+ βϕ(k)u minus p lϕ(k)

minus p

le 1 minus βϕ(k)1113872 1113873 xϕ(k)

minus p2

+ 2θϕ(k)xϕ(k)

minus xϕ(k)minus 1

zϕ(k)

minus p1113960 1113961 + βϕ(k)u minus p lϕ(k)

minus p

le 1 minus βϕ(k)1113872 1113873 xϕ(k)

minus p2

+ 2θϕ(k)xϕ(k)

minus xϕ(k)minus 1

zϕ(k)

minus p1113960 1113961 + βϕ(k)u minus p lϕ(k)

minus p

(88)

which implies by Lemma 5

limk⟶infin

xφ(k)

minus p

2

0 and limk⟶infin

xϕ(k)+1

minus p2

0 (89)

Moreover for kge k0 it is easy to see thatΛϕ(k) minus Λϕ(k)+1 le 0 if kneϕ(k) (that is kge ϕ(k)) because forϕ(k) + 1lemle n Λm gtΛm+1 As a consequence we obtain

0leΛk lemax Λϕ(k)Λϕ(k)+11113966 1113967 Λϕ(k)+1 forallkge k0 (90)

erefore we obtain limk⟶infinΛk 0 that is xk1113864 1113865 converges

strongly to p is completes the proof

4 Some Extensive Results

For the SFPwMOS (11) when N 1 it becomes the SFP(1) us we have the following corollary for solving the SFP(1) which is an immediate consequence of eorem 1

Corollary 1 Let H1 and H2 be two real Hilbert spaces andlet B H1⟶ H2 be bounded linear operator Let C and Q

be nonempty closed and convex subsets of H1 and H2 re-spectively Assume that Ω Ccap Bminus 1(Q)ne 0 Let u isin H1 be afixed point For any starting point x0 x1 isin H1 let xk1113864 1113865 be thesequence generated by

zk

xk

+ θk xk

minus xkminus 1

1113872 1113873

xk+1 ≔ PCk

βku + δkzk

+ εk zk

minus τknablagk zk

1113872 11138731113872 11138731113872 1113873

⎧⎪⎨

⎪⎩(91)

where θk Ck τk and nablafk are given as in (8) Suppose thesequences βk1113864 1113865 δk1113864 1113865 and εk1113864 1113865 satisfy the conditions in =e-orem 1 =en the sequence xk1113864 1113865 converges strongly to thesolution p isin Ω where p PΩ(u)

When we take u x0 in Algorithm 1 we note also thefollowing results regarding to the SFPwMOS (11)

Corollary 2 Let H Hi i 1 N be real Hilbert spacesand let Bi H⟶ Hi i 1 N be bounded linearoperators Let C and Qi i 1 N be nonempty closedand convex subsets of H and Hi i 1 N respectivelyAssume that problem (11) is consistent For any initial guessx0 x1 isin H let xk1113864 1113865 be the sequence generated by

zk

xk

+ θk xk

minus xkminus 1

1113872 1113873

xk+1 ≔ PCk

βkx0

+ δkzk

+ εk zk

minus τknablagk zk

1113872 11138731113872 11138731113872 1113873

⎧⎪⎨

⎪⎩(92)

whereθk Ck τk and nablagk are given by (36) and (39) andAlgorithm 1 respectively Suppose the sequences βk1113864 1113865 δk1113864 1113865and εk1113864 1113865 satisfy the conditions in =eorem 1 =en the se-quence xk1113864 1113865 generated by (92) strongly converges to the so-lution p PΩ(x0) isin Γ

When we take δk equiv 0 in Algorithm 1 we obtain thefollowing result regarding the SFPwMOS (11)

Corollary 3 Let H Hi i 1 N be real Hilbertspaces and let Ti H⟶ Hi i 1 N be boundedlinear operators Let C and Qi i 1 N be nonemptyclosed and convex subsets of H and Hi i 1 N re-spectively Assume that problem (11) is consistent For a fixedpoint u isin H and any initial guess x0 isin H let xk1113864 1113865 be thesequence generated by

zk

xk

+ θk xk

minus xkminus 1

1113872 1113873

xk+1 ≔ PCk

βku + 1 minus βk( 1113857 zk

minus τknablagk zk

1113872 11138731113872 11138731113872 1113873

⎧⎪⎨

⎪⎩(93)

where θk Ck τk and nablagk are given by (36) and (39) andAlgorithm 1 respectively Suppose the sequences satisfyconditions (A1) =en the sequence xk1113864 1113865 generated by (93)strongly converges to the solution p PΩ(u) isin Γ

Of course when we take u x0 we get the followingresult regarding the SFPwMOS (11)

Corollary 4 Let H Hi i 1 N be real Hilbert spacesand let Bi H⟶ Hi i 1 N be bounded linear op-erators Let C and Qi i 1 N be nonempty closed andconvex subsets of H and Hi i 1 N respectively As-sume that problem (11) is consistent For any initial guessx0 isin H let xk1113864 1113865 be the sequence generated by

zk

xk

+ θk xk

minus xkminus 1

1113872 1113873

xk+1 ≔ PCk

βkx0

+ 1 minus tarts minus with(primehskip

prime)]))hskipsubstring minus after(preceding minus sibling comment()[starts minus with(prime hskipprime)]prime hskipprime)pt gt βk) z

kminus τknablagk z

k1113872 11138731113872 1113873)11138741113874

⎧⎪⎪⎨

⎪⎪⎩

(94)

Journal of Mathematics 9

where θk Ck τk and nablagk are given by (36) and (39) andAlgorithm 1 respectively Suppose the sequence βk1113864 1113865 satisfiesconditions (A1) =en the sequence xk1113864 1113865 generated by (94)strongly converges to the solution p PΩ(u) isin Γ

Remark 1 In Corollary 4 for the particular casewhereN 1 the iterative scheme (94) reduced exactly toiterative scheme proposed by He et al in eorem 32 of[10]

5 Numerical Experiment

In this section we provide one numerical experiment toillustrate the implementation and efficiency of our proposedmethod First we study the behavior of Algorithm 1 fordifferent choices of θkand ρk Next we compare Algorithm 1(θk 08) with no inertial Algorithm 1 (θk 0) and Scheme(17) for different initial points e numerical results arecompleted on a standard TOSHIBA laptop with Intel(R)Core(TM) i5-2450M CPU25GHz 25GHz with memory4GB e code is implemented in MATLAB R2020a

Let

A1

minus 04 minus 02 minus 02

04 05 minus 01

02 minus 05 03

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

A2

03 minus 01 03

minus 03 02 minus 02

0 02 03

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

C x isin R3|x1 minus x

22 + 2x3 le 01113966 1113967

Q1 x isin R3

|x21 + x2 minus x3 le 01113966 1113967

Q2 x isin R3|x1 + x

22 minus x3 le 01113966 1113967

(95)

Find xlowast such that

xlowast isin Γ ≔ Ccap cap 2i1A

minus 1i Qi( 11138571113872 1113873ne 0 (96)

Define the error function as

Ek 13x

kminus PCk

xk

1113872 11138732

+13A1x

kminus PQ1k

A1xk

1113872 11138732

+13A2x

kminus PQ2k

A2xk

1113872 11138732

(97)

Note that if Ek 0 at the kth step then xk isin ΓFor Algorithm 1 for different choices of ρk and θk we

choose βk 19k + 1 δk k9k + 1 εk 8k9k + 1u (01 01 01)T and x0 (minus 01 minus 01 minus 01)T andx1 [03 05 minus 06]T Using Ek lt ε as stopping criteriawhere ε is a small enough positive number the results of thenumerical experiment are reported in Table 1e numericalresults can be seen from Table 1 and Figures 1 and 2 InTable 1 ldquoIterrdquo denotes the number of iterations

e behavior of the function Ek in Table 1 is described inFigures 1 and 2

From Table 1 and Figures 1 and 2 we can observe thatAlgorithm 1 for inertial effect case has less number of it-erations than Algorithm 1 for θk 0(that is no inertialeffect) case Algorithm 1 with bigger θk has less number ofiterations than Algorithm 1 for small case for the same θkbig ρk case has less number of iterations than Algorithm 1for small case

Table 1 Algorithm 1 for different choices of θk ε and ρk

ε θkρk 398 ρk 300

Iter Ek Iter Ek

10minus 3θk 08 5 70906e minus 05

6 58048e minus 05θk 03 7 69397e minus 04θk 0 9 75095e minus 04

10minus 4θk 08 5 70906e minus 05

6 58048e minus 05θk 03 11 91282e minus 05θk 0 15 96150e minus 05

10minus 5θk 08 12 44752e minus 06

14 25296e minus 06θk 03 16 91012e minus 06θk 0 24 88064e minus 06

0 5 10 15 20 25The number of Iteration (k)

0

001

002

003

004

005

006

E k

θk=08θk=03θk=00

Figure 1 Error against iterations for Algorithm 1 with differencechoices of θk under ρk 398 and ε 10minus 5

10 Journal of Mathematics

Following we compare our algorithmwith iteration (16)For the example iteration (16) can be written as

xk+1 ≔ αkf x

k1113872 1113873 + 1 minus αk( 11138571113857PC x

kminus λk B

lowast1 I minus PQ11113872 11138731113872 1113873B1x

k1113872

+ Blowast2 I minus PQ21113872 11138731113872 1113873B2x

k1113873

(98)In the process we take αk 08 and f(xk)

u (111)Tλk 000005 for Scheme (97) We choose βk

12k + 1 δk εk k2k + 1 and ρk 3for Algorithm 1From Table 2 and Figure 3 we can observe that Algo-

rithm 1 has less number of iterations than Scheme (97)Moreover our iterative method is advantageous overScheme (98) because Algorithm 1 does not require com-puting of the operator norm

Data Availability

No data were used to support this study

Conflicts of Interest

e authors declare that they have no conflicts of interest

Authorsrsquo Contributions

is entire work has been completed by the authors HuijuanJia Shufen Liu and Yazheng Dang e authors read andapproved the finial manuscript

0 5 10 15e number of Iteration (k)

ρk=398ρk=300

0

001

002

003

004

005

006

007

E k

Figure 2 Error against iterations for Algorithm 1 with difference choices of ρk under θk 08 and ε 10minus 5

Table 2 Comparsion of Algorithm 1 and Scheme (97) for different choice initial point for ε 10minus 4

Initiative point x1 Algorithm 1 with θk 08 Algorithm 1 with θk 00 (98)Iter Ek Iter Ek Iter Ek

[04 09ndash09] 11 41830e minus 06 25 96999e minus 06 51 88226e minus 06[09 03ndash05] 6 14290e minus 07 22 86399e minus 06 26 96719e minus 06[minus 07 03ndash02] 12 78838e minus 06 24 92013e minus 06 48 84198e minus 06[01 01ndash09]prime 12 15353e minus 07 26 87544e minus 06 52 99722e minus 06

0 10 20 30 5040e number of Iteration (k)

Algorithm 1 with θ=08Algorithm 1 with θ=00(52)

0

004

002

006

008

01

012

014

E k

Figure 3 Error against of iterations for the comparision of Al-gorithm 1 and Scheme (97)

Journal of Mathematics 11

Acknowledgments

is research was supported by the National Natural ScienceFoundation of China (Grant no 61872126) Henan ProvinceKey Science and Technology Project (Grant no192102210123) and Young Backbone Teachers in Univer-sities of Henan Province (Grant no 2019GGJS061)

References

[1] Y Censor and T Elfving ldquoA multiprojection algorithm usingbregman projections in a product spacerdquo Numerical Algo-rithms vol 8 no 2 pp 221ndash239 1994

[2] C Byrne ldquoIterative oblique projection onto convex sets andthe split feasibility problemrdquo Inverse Problems vol 18 no 2pp 441ndash453 2002

[3] G T Herman ldquoImage reconstruction from projections thefundamentals of computerized tomographyrdquo 1980

[4] Y Censor and A Segal ldquoIterative projection methods inbiomedical inverse problemsrdquo 2008

[5] Y Censor T Bortfeld B Martin and A Trofimov ldquoA unifiedapproach for inversion problems in intensity-modulated ra-diation therapyrdquo Physics in Medicine and Biology vol 51no 10 pp 2353ndash2365 2006

[6] C Yair and Tommy ldquoe multiple-sets split feasibilityproblem and its applications for inverse problemsrdquo InverseProblems vol 21 no 6 pp 2071ndash2084 2005

[7] Q Yang ldquoe relaxed CQ algorithm solving the split feasi-bility problemrdquo Inverse Problems vol 20 no 4pp 1261ndash1266 2004

[8] G Lopez V Martın-Marquez and F H Wang ldquoSolving thesplit feasibility problem without prior knowledge of matrixnormsrdquo Inverse Problems vol 28 no 8 pp 374ndash389 2012

[9] J Deepho and P Kumam ldquoA modified halpernrsquos iterativescheme for solving split feasibility problemsrdquo Abstract andApplied Analysis vol 2012 Article ID 876069 8 pages 2012

[10] Y Dang and Y Gao ldquoe strong convergence of a KM-CQ-like algorithm for a split feasibility problemrdquo Inverse Prob-lems vol 27 no 1 Article ID 015007 2011

[11] S N He and Z Y Zhao ldquoStrong convergence of a relaxed CQalgorithm for the split feasibility problemrdquo Journal of In-equalities and Applications vol 2013 no 1 11 pages 2013

[12] Y H Yao and Mihai Strong Convergence of a Self-AdaptiveMethod for the Split Feasibility Problem Fixed Point =eory ampApplications Springer Berlin Germany 2013

[13] S Reich M T Truong and T Mai ldquoe split feasibilityproblem with multiple output sets in Hilbert spacesrdquo Opti-mization Letters vol 36 2020

[14] S Suantai N Pholasa N Pholasa and P Cholamjiak ldquoemodified inertial relaxed CQ algorithm for solving the splitfeasibility problemsrdquo Journal of Industrial and ManagementOptimization vol 14 no 4 pp 1595ndash1615 2018

[15] W Cholamjiak P Cholamjiak and S Suantai ldquoAn inertialforward-backward splitting method for solving inclusionproblems in Hilbert spacesrdquo Journal of Fixed Point=eory andApplications vol 20 no 1 p 42 2018

[16] Q L Dong Y J Cho and L L Zhong ldquoInertial projectionand contraction algorithms for variational inequalitiesrdquoJournal of Global Optimization vol 70 no 3 pp 1ndash18 2018

[17] D R Sahu Y J Cho and Q L Dong ldquoInertial relaxed CQalgorithms for solving a split feasibility problem in Hilbertspacesrdquo Numerical Algorithms vol 3 2020

[18] Y Shehu and A Gibali ldquoInertial krasnoselskiindashmann methodin banach spacesrdquo Mathematics vol 8 no 4 2020

[19] Y Shehu P T Vuong and P Cholamjiak ldquoA self-adaptiveprojection method with an inertial technique for split feasi-bility problems in Banach spaces with applications to imagerestoration problemsrdquo Journal of Fixed Point =eory andApplications vol 21 no 2 2019

[20] H-K Xu ldquoIterative methods for the split feasibility problemin infinite-dimensional Hilbert spacesrdquo Inverse Problemsvol 26 no 10 Article ID 105018 2010

[21] R I Bot and E R Csetnek ldquoAn inertial Tsengrsquos type proximalalgorithm for nonsmooth and nonconvex optimizationproblemsrdquo Journal of Optimization =eory and Applicationsvol 171 no 2 pp 600ndash616 2016

12 Journal of Mathematics

main idea is to make use of two previous iterates in order toupdate the next iterate Due to its good acceleration char-acter inertial type algorithms have been widely studied byauthors [15ndash19]

Motivated by the modified Halpernrsquos iterative schemeproposed by [11] for the SFP (1) the relaxed CQ algorithm isproposed by He et al [11] to solve the SFP (1) and inertialeffect is proposed by Polyak [14] In this paper we propose anew inertial self-adaptive relaxed CQ algorithm for solvingthe SFPwMOS (11) in general Hilbert spaces

In Section 2 we recall some necessary tools which areused in establishing our main results In Section 3 wepropose an inertial self-adaptive relaxed CQ algorithm forsolving the SFPwMOS (11) and we establish and analyze astrong convergence theorem for the proposed algorithm InSection 4 we present some newly derived results for solvingthe SFPwMOS (11) Finally in Section 5 we provide onenumerical experiment to illustrate the implementation ofour proposed method and compare it with some existingresults

2 Preliminaries

In this section we recall some preliminaries which areneeded in the sequel LetH be a real Hilbert space with theinner product lang rang and induced norm| middot | Let I stand for theidentity operator on H We denote the fixed point set of anoperator T H⟶ H (if T has fixed point) by Fix(T) ieFix(T) x isin H Tx x Let the symbolsldquordquoand ldquo⟶ rdquo denote the weak and strong convergence re-spectively For any sequence xk1113864 1113865 sub H ωω (xk)

x isin H exist xkj

1113966 1113967 sub xk1113864 11138651113966 1113967 such that xkjxk1113966 1113967 denotes theweak ω-limit set of xk1113864 1113865

Definition 1 (see [3]) LetH be a real Hilbert space with innerproduct lang rang and induced norm | middot | Let C be a nonemptyclosed convex subset of H Let T C⟶ H be a given op-erator =en T is called

(1) Lipschitz continuous with constant σ gt 0 on C if

Tx minus Tyle σx minus y forallx y isin C (18)

(2) Nonexpansive on C if

Tx minus Tylex minus y forallx y isin C (19)

(3) Firmly nonexpansive on C if

Tx minus Ty2 lex minus y

2minus (I minus T)x minus (I minus T)y

2forallx y isin C

(20)

which is equivalent to

Tx minus Ty2 leTx minus Ty x minus y forallx y isin C (21)

(4) Averaged if there exists a number σ isin (0 1) and anonexpansive operator F C⟶ H such that

T F +(1 minus σ)I (22)

where I is the identity operator

In this case we say that T is σ-averaged

Lemma 1 Let C sub H be a nonempty closed and convex setand PC(x) denotes the metric (orthogonal) projection of x

onto C =en the following assertions hold for any x y isin H

and z isin C

(1) x minus PC(x) z minus PC(x)le 0

(2) PC(x) minus PC(y)lex minus y

(3) PC(x) minus PC(y)2 lePC(x) minus PC(y) x minus y

(4) PC(x) minus z2 lex minus z

2minus x minus PC(x)

2

(23)

Evidently I minus PC is firmly nonexpansive and alsononexpansive

Definition 2 Let δ isin (0 1) and f H⟶ R be a properfunction en

(1) f is convex if

f(δx +(1 minus δ)y)le δf(x) +(1 minus δ)f(y) forallx y isin H

(24)

(2) f is strongly convex with constant σ where σ gt 0 if

f(δx +(1 minus δ)y) +σ2δ(1 minus δ)x minus y

2

le δf(x) +(1 minus δ)f(y) forallx y isin H

(25)

(3) A vector ω isin H is a subgradient of f at a point x if

f(y) gef(x) + ω y minus x forally isin H (26)

(4) e set of all subgradients of a convex functionf H⟶ R at x isin H denoted by zf(x) is calledthe subdifferential of f and is defined by

zf(x) ω isin H f(y)gef(x) + ω y minus x for eachy isin H1113864 1113865

(27)

(5) If zf(x)ne 0 f is said to be subdifferentiable at x

If the function f is continuously differentiable thenzf(x) nablaf(x)1113864 1113865 e convex function is subdifferentiableeverywhere

Definition 3 Let f H⟶ [minus infin +infin) be a properfunction

(1) f is lower semicontinuous (lsc) at x if xk⟶ x

implies

f(x)le liminfk⟶infin

f xk

1113872 1113873 (28)

(2) f is weakly lower semicontinuous (ω-lsc) at x ifxkx implies

f(x)le liminfk⟶infin

f xk

1113872 1113873 (29)

Journal of Mathematics 3

(3) f is lower semicontinuous on H if it is lowersemicontinuous at every point x isin H and f is weaklylower semicontinuous on H if it is weakly lowersemicontinuous at every point x isin H

Lemma 2 (see [20]) Let C and Q be closed convex subsets ofreal Hilbert spaces H1 and H2 respectively andf H1⟶ R is given byf(x) (I minus PQ)Ax2 whereA H1⟶ H2 be a bounded linear operator =en for δ gt 0and xlowast isin H1 the following statements are equivalent

(1) =e point xlowast solves the SFP (1) iexlowast isin x isin C Ax isin Q

(2) =e point xlowast is the fixed point of the mappingPC(I minus δnablaf)

(3) =e point xlowast solves the variational inequality problemwith respect to the gradient of f that is find a pointxlowast isin C such that

nablaf(x) y minus xge 0 forally isin C (30)

Lemma 3 (see [6]) Let H1 and H2 be real Hilbert spacesand f H1⟶ R is given byf(x) 12(I minus PQ)Ax2 whereQ is closed convex subset of H2 and A H1⟶ H2 is abounded linear operator =en the following assertions hold

(1) f is convex and differentiable(2) f is weakly lower semicontinuous on H1

(3) nablaf(x) AT(I minus PQ)Ax for x isin H1

(4) nablaf(x) is A2-Lipschitz ienablaf(x) minus nablaf(y)leA2x minus y forallx y isin H1

Definition 4 Let Λk1113864 1113865 be a real sequence en Λk1113864 1113865 de-creases at infinity if there exists k0 isin N such that Λk+1 leΛkfor kge k0 In other words the sequence Λk1113864 1113865 does not de-crease at infinity if there exists a subsequence Λkt

1113966 1113967tge1 of Λk1113864 1113865

such that ΛktltΛkt+1 for all tge 1

Lemma 4 Let Λk1113864 1113865 be a sequence of real numbers that doesnot decrease at infinity Also consider the sequence of integersφ(k)1113864 1113865kgek0 defined by

φ(k) max m isin N mle k Λm leΛm+11113864 1113865 (31)

en φ(k)1113864 1113865kgek0 is a nondecreasing sequence verifyinglim

k⟶infinφ(k) infin and for all kge k0 the following two esti-

mates hold

Λφ(k) leΛφ(k)+1 andΛk leΛφ(k)+1 (32)

Lemma 5 (see [21]) Let sk1113864 1113865 be a sequence of nonnegativereal numbers satisfying the following relation

sk+1 le 1 minus σk( 1113857sk + σkμk + βk kge 1 (33)

where σk1113864 1113865 μk1113864 1113865 and βk1113864 1113865 satisfy the conditions

(1) σk1113864 1113865 sub [0 1] 1113944infin

k1σk infin

(2) limsupk⟶infin

μk le 0

(3) βk ge 0 1113944infin

k1βk ltinfinThen limsup

k⟶infinsk 0

(34)

3 Main Results

In this section we consider a general case of the SFPwMOS(11) where the nonempty closed and convex sets C andQi(i 1 N) are given by level sets of convex functionsas in [7] roughout this section we assume thatc H⟶ R and qi Hi⟶ R are lower semicontinuousconvex functions and the sets C and Qi are given by

C ≔ x isin H c(x)le 0 andQi y isin Hi qi(y)le 01113864 1113865 (35)

e relaxed set (half spaces) Ck at xk is defined as

Ck ≔ x isin H c xk

1113872 1113873le ξk x

kminus x1113966 1113967 (36)

where ξk isinzc(xk)e relaxed set (half spaces) Qk

i (i 1 middot middot middot N) at xk aredefined as

Qki ≔ x isin H c x

k1113872 1113873le ξk

xk

minus x1113966 1113967

y isin Hi qi Bixk

1113872 1113873le ηki Bix

kminus y1113966 1113967

(37)

where ηki isin zqi(Bix

k) It follows that C sub Ck and Qi sub Qki

hold for every kge 0 Now we define the following (relaxed)proximity function for x isin H

gk(x) ≔12

1113944

N

i1I minus PQk

i1113874 1113875Bix

2 (38)

We note that gk() is differentiable with its gradientgiven by

nablagk(x) ≔ 1113944N

i

BTi I minus PQk

i1113874 1113875Bix (39)

where each Qki are half spaces given in (37) We note that gk

is weakly lower semicontinuous convex and differentiablefunction and nablagk is Lipschitz continuous

Now we present an inertial self-adaptive relaxed CQ

algorithmWe can see that Algorithm 1 terminates at some iterate

(say k) when nablagk(xk) 0e following lemma is important for the convergence

analysis

Lemma 6 If zk ≔ xk + θk(xk minus xkminus 1) where 0le θk lt 1 forall k isin N then for all z isin H

zk

minus z2 le x

kminus z

2+ θk x

kminus z

2minus x

kminus 1minus z

21113872 1113873 + 2θkx

kminus x

kminus 12

(40)

4 Journal of Mathematics

Proof Using the identity 2a b a2 + b2 minus a minus b2 we have

zk

minus z2

xk

minus z + θk xk

minus xkminus 1

1113872 11138732

xk

minus z2

+ 2θkxk

minus z xk

minus xkminus 1

+ θ2kxk

minus xkminus 12

xk

minus z2

+ θkxk

minus z2

+ xk

minus xkminus 12

minus xkminus 1

minus z2

+ θ2kxk

minus xkminus 12

xk

minus z2

+ θk xk

minus z2

minus xkminus 1

minus z2

1113872 1113873

+ θk 1 + θk( 1113857xk

minus xkminus 12

lexk

minus z2

+ θk xk

minus z2

minus xkminus 1

minus z2

1113872 1113873

+ 2θkxk

minus xkminus 12

(41)

Next we show the following strong convergence theo-rem for Algorithm 1

Theorem 1 Let H Hi i 1 N be real Hilbert spacesand let Bi H⟶ Hi i 1 N be bounded linear op-erators Denote Γ as the solution set of the problem (11) Let C

and Qi i 1 N be nonempty closed and convexsubsets of H and Hi i 1 N respectively Assume thatthe SFPwMOS (11) is consistent Suppose the sequences ρk1113864 1113865βk1113864 1113865 δk1113864 1113865 and εk1113864 1113865 in Algorithm 1 satisfy the followingconditions

(A1) limk⟶infin

βk 0

and 1113944infin

k1βk +infin

βk+1

βk

le l l is a constan t

(42)

(A2) 0lt liminfk⟶infin

εk le limsupfk⟶infin

εk lt 1 (43)

(A3) liminfk⟶infin

ρk 4 minus ρk( 1113857gt 0 (44)

(A4) limk⟶infin

θk

βk

xk

minus xkminus 1

0 (45)

en the sequence xk1113864 1113865 generated by Algorithm 1converges strongly to the solution p isin Γ where p PΓ(u)

Proof Wemay assume that Algorithm 1 does not terminatein a finite number of iterations us nablagk(xk)ne 0 for allk ge 0 Denote Γ as the solution set of problem (11) In the

consistent case of problem (11) Γ is nonempty closed andconvex us the metric projection PΓ is well defined

Let p isin Γ and set sk zk minus τknablagk(zk) Note that I minus PQki

for each i 1 N is firmly nonexpansive and nablagk(p)

0 Hence we have from Lemma 1 that

langnablagk zk

1113872 1113873 zk

minus prang lang1113944N

i1B

Ti I minus PQk

i1113874 1113875Biz

k z

kminus prang

1113944N

i1B

Ti I minus PQk

i1113874 1113875Biz

k z

kminus p

1113944N

i1I minus PQk

i1113874 1113875Biz

k Biz

kminus Bip

ge 1113944N

i1I minus PQk

i1113874 1113875Biz

k2 2gk z

k1113872 1113873

(46)

which implies that

sk

minus p

2

zk

minus p minus τknablagk zk

1113872 1113873

2

zk

minus p2

+ τ2knablagk zk

1113872 11138732

minus 2τknablagk zk

1113872 1113873 zk

minus p

le zk

minus p2

+ρ2kg

2k z

k1113872 1113873

nablagk zk

1113872 11138732 minus

2ρkgk zk

1113872 1113873

nablagk zk

1113872 11138732 2gk z

k1113872 11138731113872 1113873

zk

minus p2

+ρ2kg

2k z

k1113872 1113873

nablagk zk

1113872 11138732 minus

4ρkg2k z

k1113872 1113873

nablagk zk

1113872 11138732

zk

minus p2

minus ρk 4 minus ρk( 1113857g2k z

k1113872 1113873

nablagk zk

1113872 11138732

(47)

Using condition (A3) we have

sk

minus p2 le z

kminus p

2 forallkge 0 (48)

Initialization choose positive sequences 0le θk lt 1 ρk1113864 1113865 sub (0 4) βk1113864 1113865 sub (0 1) δk1113864 1113865 sub [0 1) and εk1113864 1113865 sub (0 1) such thatβk + δk + εk 1 Let u isin H be a fixed point Select arbitrary starting points x0 x1 isin H and set k 0

Step 1 given the current iterate xk isin H if nablagk(xk) 0 for some k isin N then stop Otherwise continue and calculateτk ≔ ρkgk(zk)nablagk(zk)2

Step 2 compute the next iterate asz

k x

k+ θk(x

kminus x

kminus 1)

xk+1 ≔ PCk

(βku + δkzk

+ εk(zk

minus τknablagk(zk)))

1113896

where Ck is the half space given as in (36)

ALGORITHM 1 Inertial self-adaptive relaxed CQ algorithm for SFPwMOS

Journal of Mathematics 5

Next we show zk1113864 1113865 is bounded Since p isin Γ sub Ck and theprojection operator PCk

is nonexpansive we obtain fromAlgorithm 1 and (48) that

xk+1

minus p PCkβku + δkz

k+ εk z

kminus τknablagk z

k1113872 11138731113872 11138731113872 1113873 minus p

le βku minus p + δkzk

minus p + εksk

minus p

le βku minus p + δkzk

minus p + εkzk

minus p

βku minus p + 1 minus βk( 1113857zk

minus p

βku minus p + 1 minus βk( 1113857xk

+ θk xk

minus xkminus 1

1113872 1113873 minus p

le βku minus p + 1 minus βk( 1113857xk

minus p + θk xk

minus xkminus 1

1 minus βk( 1113857 xk

minus p

+ βk u minus p +θk

βk

xk

minus xkminus 1

1113890 1113891

(49)

Condition (A4) implies the sequence θkβkxk minus xkminus 1 isbounded Let an upper bound of u minus p + θkβkxk minus xkminus 11113864 1113865 beM en we rewrite (49) as

xk+1

minus plemax xk

minus p M1113966 1113967 (50)

By induction we have that

xk+1

minus plemax x1

minus p M1113966 1113967 (51)

Hence xk minus p1113864 1113865 is bounded so is zk minus p1113864 1113865 and sk minus p1113864 1113865And zk1113864 1113865 is bounded Consequently sk1113864 1113865 and Biz

k1113864 1113865N

i1 arealso bounded The rest of the proof will be

divided into two parts

Case 1 Suppose that there exists k0 isin N such thatxk minus p21113864 1113865

infinkk0

is nonincreasing en xk minus p21113864 1113865infink1 converges

and xk minus p2 minus xk+1 minus p2⟶ 0 as k⟶infin From (47) weobtain

ρk 4 minus ρk( 1113857g2k z

k1113872 1113873

nablagk zk

1113872 11138732 le z

kminus p

2minus s

kminus p

2 (52)

Since p isin Γ sub Ck and the projection operator PCkis

nonexpansive from Algorithm 1 we obtain

xk+1

minus p2

PCkβku + δkz

k+ εksk1113872 1113873 minus p

2

le βku + δkzk

+ εksk1113872 1113873 minus p2

le βku minus p2

+ δkzk

minus p2

+ εksk minus p2

(53)

Since βk + δk + εk 1 from (53) and using the inequalityx + y2 lex2 + 2y x + y we have the following estimation

zk

minus p2

minus sk

minus p2 le

βk

εk

u minus p2

+1 minus βk

εk

zk

minus p2

minus1εk

xk+1

minus p2

βk

εk

u minus p2

minusβk

εk

zk

minus p2

+1εk

zk

minus p2

minus xk+1

minus p2

1113960 1113961

leβk

εk

u minus p2

+1εk

zk

minus p2

minus xk+1

minus p2

1113960 1113961

1εk

βku minus p2

+ zk

minus p2

minus xk+1

minus p2

1113960 11139611113960 1113961

1εk

βku minus p2

+ xk

minus p + θk xk

minus xkminus 1

1113872 11138732

minus xk+1

minus p2

1113876 11138771113876 1113877

le1εk

βku minus p2

+ xk

minus p2

+ 2θkxk

minus xkminus 1

zk

minus p minus xk+1

minus p2

1113960 11139611113960 1113961

le1εk

βku minus p2

+ xk

minus p2

+ 2θkxk

minus xkminus 1

zk

minus p minus xk+1

minus p2

1113960 11139611113960 1113961

1εk

βku minus p2

+ xk

minus p2

minus xk+1

minus p2

+ 2θkxk

minus xkminus 1

zk

minus p1113960 11139611113960 1113961

(54)

Combining (52) and (54) we obtain

6 Journal of Mathematics

ρk 4 minus ρk( 1113857g2k z

k1113872 1113873

nablagk zk

1113872 11138732 le z

kminus p

2minus s

kminus p

2

le1εk

βku minus p2

+ xk

minus p2

minus xk+1

minus p2

11139601113960

+ 2θkxk

minus xkminus 1

zk

minus p11139611113961

(55)

By condition (A4) βk isin (0 1) imply the sequencelimk⟶infinθkxk minus xkminus 1⟶ 0 and we have proved zk minus p1113864 1113865 isbounded so 2θkxk minus xkminus 1zk minus p⟶ 0 By conditions (A2)and (A3) and (55) we have as k⟶infin

0lt ρk 4 minus ρk( 1113857g2k z

k1113872 1113873

nablagk zk

1113872 11138732

le1εk

βku minus p2

+ xk

minus p2

minus xk+1

minus p2

11139601113960

+ 2θkxk

minus xkminus 1

zk

minus p11139611113961⟶ 0

(56)

which implies that

limk⟶infin

g2k zk

1113872 1113873

nablagk zk

1113872 11138732 0 (57)

We note that for each i 1 N BTi (I minus PQk

i)Bi() is

Lipschitz continuous Since the sequence zk1113864 1113865 is boundedand

BTi I minus PQk

i1113874 1113875 Biz

k B

Ti I minus PQk

i1113874 1113875Biz

k

minus BTi I minus PQk

i1113874 1113875Biple max

1leileNBi1113874 1113875z

kminus p

(58)

for all i 1 N we have the sequence

BTi (I minus PQk

i)Biz

k1113882 1113883infin

i1is bounded Hence nablagk(zk)1113864 1113865

infink1 is

bounded Consequently we have from (57) that

limk⟶infin

I minus PQki

1113874 1113875Bizk

0 (59)

For

zk

minus xk

xk

+ θk xk

minus xkminus 1

1113872 1113873 minus xk

θkxk

minus xkminus 1

(60)

which implies

zk

minus xk⟶ 0 as k⟶infin (61)

for each i 1 N since sk zk minus τknablagk(zk) we havefrom (59) and (61) that

sk

minus zk

τknablagk zk

1113872 1113873⟶ 0 as k⟶infin

sk

minus xk le s

kminus z

k+ z

kminus x

k⟶ 0 as k⟶infin(62)

at is

limk⟶infin

sk

minus xk

0 (63)

limk⟶infin

BTi I minus PQk

i1113874 1113875Biz

k 0 (64)

Let

lk

βku + δkzk

+ εksk

βku + 1 minus βk( 1113857vk (65)

where

vk

δk

1 minus βk

zk

+εk

1 minus βk

sk (66)

is gives

vk

minus zk le

εk

1 minus βk

sk

minus zk⟶ 0 k⟶infin

lk

minus vk

le βk u minus vk

⟶ 0 k⟶infin

(67)

lk

minus zk le l

kminus v

k+ v

kminus z

k⟶ 0 k⟶infin

lk

minus xk le l

kminus z

k+ z

kminus x

k⟶ 0 as k⟶infin(68)

Hence

limk⟶infin

lk

minus zk

limk⟶infin

lk

minus xk

limk⟶infin

lk

minus vk

limk⟶infin

vk

minus zk

0

(69)

Since xk1113864 1113865 is bounded there exists a subsequence xkj1113966 1113967

of xk1113864 1113865 such that xkj1113966 1113967plowast isin ωω(xk) Next we need toshow plowast isin C and Bip

lowast isin Qi for each i 1 NWe first show that xk+1 minus xk⟶ 0 For p isin Γ

12x

k+1minus p

212x

k+1minus x

k+ x

kminus p

2

12x

kminus p

2+ x

k+1minus x

k x

k+1minus p minus

12x

k+1minus x

k2

12x

kminus p

2+ x

k+1minus l

k+ l

kminus x

k x

k+1

minus p minus12x

k+1minus x

k2

12x

kminus p

2+ x

k+1minus l

k x

k+1minus p + l

k

minus xk x

k+1minus p minus

12x

k+1minus x

k2

(70)

then

12x

k+1minus x

k2 le12x

kminus p

2minus12x

k+1minus p

2+ x

k+1minus l

k x

k+1

minus p + lk

minus xk x

k+1minus p

(71)

from xk+1 minus lk xk+1 minus plt 0

Journal of Mathematics 7

12x

k+1minus x

k2 le12x

kminus p

2minus12x

k+1minus p

2+ l

kminus x

k x

k+1minus p

(72)

Since xk minus p2 minus xk+1 minus p2⟶ 0 as k⟶infinlim

k⟶infinlk minus xk 0 we obtain

xk+1

minus xk⟶ 0 as k⟶infin (73)

Since xkj+1 isin Ckj we obtain

c xkj1113872 1113873 + ξkj x

kj+1minus x

kj le 0 (74)

us

c xkj1113872 1113873le minus ξkj x

kj+1minus x

kj le ξxkj+1

minus xkj (75)

where ξ satisfies ξk le ξ for all k By virtue of the continuity offunction c and xkj+1 minus xkj⟶ 0 we get that

c plowast

( 1113857 limj⟶infin

c xkj1113872 1113873le 0 (76)

erefore plowast isin CNow we show that Bip

lowast isin Qi to do this let hki Biz

k minus

PQkiBiz

k⟶ 0 and let ηki be such that ηk

i le η for all k SinceBiz

kj minus hkj

i PQ

kj

i

(Bizkj ) isin Q

kj

i we have

qi Bixkj1113872 1113873 + ηkj

i Bizkj minus h

kj

i1113874 1113875 minus Bixkj le 0 (77)

Hence

qi Bixkj1113872 1113873le η

kj

i Bixkj minus Biz

kj + ηkj

i hkj

i le ηBixkj

minus xkjminus 1

+ ηhkj

i ⟶ 0

(78)

By the continuity of qi and Bixkj⟶ Bix

lowast we arrive atthe conclusion

qi Biplowast

( 1113857 limj⟶infin

qi Bixkj1113872 1113873le 0 (79)

namely Biplowast isin Qi for all i 1 N Hence plowast isin Γ

Moreover for p PΓ u we can see that

limsupk⟶infin

lk

minus p u minus p limk⟶infin

lkj minus p u minus plep

lowast

minus p u minus ple 0

(80)

By Lemma 1 and (65) we have

xk+1

minus p2

PC lk

1113872 1113873 minus p2 le l

kminus p

2 βku + 1 minus βk( 1113857v

kminus p l

kminus p

βku minus p lk

minus p + 1 minus βk( 1113857vk

minus p lk

minus p

le βku minus p lk

minus p + 1 minus βk( 1113857vk

minus plk

minus p

1 minus βk( 1113857 xk

+ θk xk

minus xkminus 1

1113872 1113873 minus p

2

+ βklangu minus p lk

minus prang +βk

2u minus p

2

le 1 minus βk( 1113857 xk

minus p

2

+ 2θklangxk

minus xkminus 1

zk

minus prang1113876 1113877 + βklangu minus p lk

minus prang +βk

2u minus p

2

le 1 minus βk( 1113857 xk

minus p

2

+ 2θk xk

minus xkminus 1

zk

minus p

1113876 1113877 + βklangu minus p lk

minus prang +βk

2u minus p

2

(81)

By conditions (A1) and (A4) βk isin (0 1) imply the se-quence lim

k⟶infinθkxk minus xkminus 1⟶ 0 βk2u minus p⟶ 0 and

using (80) in (81) and applying Lemma 5 we obtain

limk⟶infin

xk

minus p2

0 (82)

erefore as k⟶infin xk⟶ p PΓu

Case 2 SetΛk xk minus p2 Assume that Λk1113864 1113865 is not decreasingat infinity Let ϕ N⟶ N be a mapping for all kge k0 (forsome k0 large enough) defined by

ϕ(k) max t isin N tle kΛt leΛt+11113864 1113865 (83)

By Lemma 4 ϕ(k)1113864 1113865infinkk0is a nondecreasing sequence

such that ϕ(k)⟶infin as k⟶infin and

max Λϕ(k)Λk1113966 1113967leΛϕ(k)+1forallkge k0 (84)

After a similar conclusion from (59) it is easy to see that

limk⟶infin

I minus PQ

ϕ(k)

i

1113874 1113875Bizϕ(k)

0 (85)

By the similar argument as the above in Case 1 weconclude immediately that

8 Journal of Mathematics

limk⟶infin

Blowasti I minus P

Qϕ(k)

i

1113874 1113875Bizϕ(k)

0 (86)

limsupk⟶infin

lϕ(k)

minus p u minus ple 0 (87)

Since xϕ(k)1113864 1113865 is bounded there exists a subsequence ofxϕ(k)1113864 1113865 still denoted by xϕ(k)1113864 1113865 which converges weakly to

plowast By similar argument as above in Case 1 we concludeimmediately that plowast isin C and Bip

lowast isin QirArrplowast isin ΓFrom (57) we have that

xϕ(k)+1

minus p2 le 1 minus βϕ(k)1113872 1113873z

ϕ(k)minus p

2+ βϕ(k)u minus p l

ϕ(k)minus p

le 1 minus βϕ(k)1113872 1113873xϕ(k)

+ θϕ(k) xϕ(k)

minus xϕ(k)minus 1

1113872 1113873 minus p2

+ βϕ(k)u minus p lϕ(k)

minus p

le 1 minus βϕ(k)1113872 1113873 xϕ(k)

minus p2

+ 2θϕ(k)xϕ(k)

minus xϕ(k)minus 1

zϕ(k)

minus p1113960 1113961 + βϕ(k)u minus p lϕ(k)

minus p

le 1 minus βϕ(k)1113872 1113873 xϕ(k)

minus p2

+ 2θϕ(k)xϕ(k)

minus xϕ(k)minus 1

zϕ(k)

minus p1113960 1113961 + βϕ(k)u minus p lϕ(k)

minus p

(88)

which implies by Lemma 5

limk⟶infin

xφ(k)

minus p

2

0 and limk⟶infin

xϕ(k)+1

minus p2

0 (89)

Moreover for kge k0 it is easy to see thatΛϕ(k) minus Λϕ(k)+1 le 0 if kneϕ(k) (that is kge ϕ(k)) because forϕ(k) + 1lemle n Λm gtΛm+1 As a consequence we obtain

0leΛk lemax Λϕ(k)Λϕ(k)+11113966 1113967 Λϕ(k)+1 forallkge k0 (90)

erefore we obtain limk⟶infinΛk 0 that is xk1113864 1113865 converges

strongly to p is completes the proof

4 Some Extensive Results

For the SFPwMOS (11) when N 1 it becomes the SFP(1) us we have the following corollary for solving the SFP(1) which is an immediate consequence of eorem 1

Corollary 1 Let H1 and H2 be two real Hilbert spaces andlet B H1⟶ H2 be bounded linear operator Let C and Q

be nonempty closed and convex subsets of H1 and H2 re-spectively Assume that Ω Ccap Bminus 1(Q)ne 0 Let u isin H1 be afixed point For any starting point x0 x1 isin H1 let xk1113864 1113865 be thesequence generated by

zk

xk

+ θk xk

minus xkminus 1

1113872 1113873

xk+1 ≔ PCk

βku + δkzk

+ εk zk

minus τknablagk zk

1113872 11138731113872 11138731113872 1113873

⎧⎪⎨

⎪⎩(91)

where θk Ck τk and nablafk are given as in (8) Suppose thesequences βk1113864 1113865 δk1113864 1113865 and εk1113864 1113865 satisfy the conditions in =e-orem 1 =en the sequence xk1113864 1113865 converges strongly to thesolution p isin Ω where p PΩ(u)

When we take u x0 in Algorithm 1 we note also thefollowing results regarding to the SFPwMOS (11)

Corollary 2 Let H Hi i 1 N be real Hilbert spacesand let Bi H⟶ Hi i 1 N be bounded linearoperators Let C and Qi i 1 N be nonempty closedand convex subsets of H and Hi i 1 N respectivelyAssume that problem (11) is consistent For any initial guessx0 x1 isin H let xk1113864 1113865 be the sequence generated by

zk

xk

+ θk xk

minus xkminus 1

1113872 1113873

xk+1 ≔ PCk

βkx0

+ δkzk

+ εk zk

minus τknablagk zk

1113872 11138731113872 11138731113872 1113873

⎧⎪⎨

⎪⎩(92)

whereθk Ck τk and nablagk are given by (36) and (39) andAlgorithm 1 respectively Suppose the sequences βk1113864 1113865 δk1113864 1113865and εk1113864 1113865 satisfy the conditions in =eorem 1 =en the se-quence xk1113864 1113865 generated by (92) strongly converges to the so-lution p PΩ(x0) isin Γ

When we take δk equiv 0 in Algorithm 1 we obtain thefollowing result regarding the SFPwMOS (11)

Corollary 3 Let H Hi i 1 N be real Hilbertspaces and let Ti H⟶ Hi i 1 N be boundedlinear operators Let C and Qi i 1 N be nonemptyclosed and convex subsets of H and Hi i 1 N re-spectively Assume that problem (11) is consistent For a fixedpoint u isin H and any initial guess x0 isin H let xk1113864 1113865 be thesequence generated by

zk

xk

+ θk xk

minus xkminus 1

1113872 1113873

xk+1 ≔ PCk

βku + 1 minus βk( 1113857 zk

minus τknablagk zk

1113872 11138731113872 11138731113872 1113873

⎧⎪⎨

⎪⎩(93)

where θk Ck τk and nablagk are given by (36) and (39) andAlgorithm 1 respectively Suppose the sequences satisfyconditions (A1) =en the sequence xk1113864 1113865 generated by (93)strongly converges to the solution p PΩ(u) isin Γ

Of course when we take u x0 we get the followingresult regarding the SFPwMOS (11)

Corollary 4 Let H Hi i 1 N be real Hilbert spacesand let Bi H⟶ Hi i 1 N be bounded linear op-erators Let C and Qi i 1 N be nonempty closed andconvex subsets of H and Hi i 1 N respectively As-sume that problem (11) is consistent For any initial guessx0 isin H let xk1113864 1113865 be the sequence generated by

zk

xk

+ θk xk

minus xkminus 1

1113872 1113873

xk+1 ≔ PCk

βkx0

+ 1 minus tarts minus with(primehskip

prime)]))hskipsubstring minus after(preceding minus sibling comment()[starts minus with(prime hskipprime)]prime hskipprime)pt gt βk) z

kminus τknablagk z

k1113872 11138731113872 1113873)11138741113874

⎧⎪⎪⎨

⎪⎪⎩

(94)

Journal of Mathematics 9

where θk Ck τk and nablagk are given by (36) and (39) andAlgorithm 1 respectively Suppose the sequence βk1113864 1113865 satisfiesconditions (A1) =en the sequence xk1113864 1113865 generated by (94)strongly converges to the solution p PΩ(u) isin Γ

Remark 1 In Corollary 4 for the particular casewhereN 1 the iterative scheme (94) reduced exactly toiterative scheme proposed by He et al in eorem 32 of[10]

5 Numerical Experiment

In this section we provide one numerical experiment toillustrate the implementation and efficiency of our proposedmethod First we study the behavior of Algorithm 1 fordifferent choices of θkand ρk Next we compare Algorithm 1(θk 08) with no inertial Algorithm 1 (θk 0) and Scheme(17) for different initial points e numerical results arecompleted on a standard TOSHIBA laptop with Intel(R)Core(TM) i5-2450M CPU25GHz 25GHz with memory4GB e code is implemented in MATLAB R2020a

Let

A1

minus 04 minus 02 minus 02

04 05 minus 01

02 minus 05 03

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

A2

03 minus 01 03

minus 03 02 minus 02

0 02 03

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

C x isin R3|x1 minus x

22 + 2x3 le 01113966 1113967

Q1 x isin R3

|x21 + x2 minus x3 le 01113966 1113967

Q2 x isin R3|x1 + x

22 minus x3 le 01113966 1113967

(95)

Find xlowast such that

xlowast isin Γ ≔ Ccap cap 2i1A

minus 1i Qi( 11138571113872 1113873ne 0 (96)

Define the error function as

Ek 13x

kminus PCk

xk

1113872 11138732

+13A1x

kminus PQ1k

A1xk

1113872 11138732

+13A2x

kminus PQ2k

A2xk

1113872 11138732

(97)

Note that if Ek 0 at the kth step then xk isin ΓFor Algorithm 1 for different choices of ρk and θk we

choose βk 19k + 1 δk k9k + 1 εk 8k9k + 1u (01 01 01)T and x0 (minus 01 minus 01 minus 01)T andx1 [03 05 minus 06]T Using Ek lt ε as stopping criteriawhere ε is a small enough positive number the results of thenumerical experiment are reported in Table 1e numericalresults can be seen from Table 1 and Figures 1 and 2 InTable 1 ldquoIterrdquo denotes the number of iterations

e behavior of the function Ek in Table 1 is described inFigures 1 and 2

From Table 1 and Figures 1 and 2 we can observe thatAlgorithm 1 for inertial effect case has less number of it-erations than Algorithm 1 for θk 0(that is no inertialeffect) case Algorithm 1 with bigger θk has less number ofiterations than Algorithm 1 for small case for the same θkbig ρk case has less number of iterations than Algorithm 1for small case

Table 1 Algorithm 1 for different choices of θk ε and ρk

ε θkρk 398 ρk 300

Iter Ek Iter Ek

10minus 3θk 08 5 70906e minus 05

6 58048e minus 05θk 03 7 69397e minus 04θk 0 9 75095e minus 04

10minus 4θk 08 5 70906e minus 05

6 58048e minus 05θk 03 11 91282e minus 05θk 0 15 96150e minus 05

10minus 5θk 08 12 44752e minus 06

14 25296e minus 06θk 03 16 91012e minus 06θk 0 24 88064e minus 06

0 5 10 15 20 25The number of Iteration (k)

0

001

002

003

004

005

006

E k

θk=08θk=03θk=00

Figure 1 Error against iterations for Algorithm 1 with differencechoices of θk under ρk 398 and ε 10minus 5

10 Journal of Mathematics

Following we compare our algorithmwith iteration (16)For the example iteration (16) can be written as

xk+1 ≔ αkf x

k1113872 1113873 + 1 minus αk( 11138571113857PC x

kminus λk B

lowast1 I minus PQ11113872 11138731113872 1113873B1x

k1113872

+ Blowast2 I minus PQ21113872 11138731113872 1113873B2x

k1113873

(98)In the process we take αk 08 and f(xk)

u (111)Tλk 000005 for Scheme (97) We choose βk

12k + 1 δk εk k2k + 1 and ρk 3for Algorithm 1From Table 2 and Figure 3 we can observe that Algo-

rithm 1 has less number of iterations than Scheme (97)Moreover our iterative method is advantageous overScheme (98) because Algorithm 1 does not require com-puting of the operator norm

Data Availability

No data were used to support this study

Conflicts of Interest

e authors declare that they have no conflicts of interest

Authorsrsquo Contributions

is entire work has been completed by the authors HuijuanJia Shufen Liu and Yazheng Dang e authors read andapproved the finial manuscript

0 5 10 15e number of Iteration (k)

ρk=398ρk=300

0

001

002

003

004

005

006

007

E k

Figure 2 Error against iterations for Algorithm 1 with difference choices of ρk under θk 08 and ε 10minus 5

Table 2 Comparsion of Algorithm 1 and Scheme (97) for different choice initial point for ε 10minus 4

Initiative point x1 Algorithm 1 with θk 08 Algorithm 1 with θk 00 (98)Iter Ek Iter Ek Iter Ek

[04 09ndash09] 11 41830e minus 06 25 96999e minus 06 51 88226e minus 06[09 03ndash05] 6 14290e minus 07 22 86399e minus 06 26 96719e minus 06[minus 07 03ndash02] 12 78838e minus 06 24 92013e minus 06 48 84198e minus 06[01 01ndash09]prime 12 15353e minus 07 26 87544e minus 06 52 99722e minus 06

0 10 20 30 5040e number of Iteration (k)

Algorithm 1 with θ=08Algorithm 1 with θ=00(52)

0

004

002

006

008

01

012

014

E k

Figure 3 Error against of iterations for the comparision of Al-gorithm 1 and Scheme (97)

Journal of Mathematics 11

Acknowledgments

is research was supported by the National Natural ScienceFoundation of China (Grant no 61872126) Henan ProvinceKey Science and Technology Project (Grant no192102210123) and Young Backbone Teachers in Univer-sities of Henan Province (Grant no 2019GGJS061)

References

[1] Y Censor and T Elfving ldquoA multiprojection algorithm usingbregman projections in a product spacerdquo Numerical Algo-rithms vol 8 no 2 pp 221ndash239 1994

[2] C Byrne ldquoIterative oblique projection onto convex sets andthe split feasibility problemrdquo Inverse Problems vol 18 no 2pp 441ndash453 2002

[3] G T Herman ldquoImage reconstruction from projections thefundamentals of computerized tomographyrdquo 1980

[4] Y Censor and A Segal ldquoIterative projection methods inbiomedical inverse problemsrdquo 2008

[5] Y Censor T Bortfeld B Martin and A Trofimov ldquoA unifiedapproach for inversion problems in intensity-modulated ra-diation therapyrdquo Physics in Medicine and Biology vol 51no 10 pp 2353ndash2365 2006

[6] C Yair and Tommy ldquoe multiple-sets split feasibilityproblem and its applications for inverse problemsrdquo InverseProblems vol 21 no 6 pp 2071ndash2084 2005

[7] Q Yang ldquoe relaxed CQ algorithm solving the split feasi-bility problemrdquo Inverse Problems vol 20 no 4pp 1261ndash1266 2004

[8] G Lopez V Martın-Marquez and F H Wang ldquoSolving thesplit feasibility problem without prior knowledge of matrixnormsrdquo Inverse Problems vol 28 no 8 pp 374ndash389 2012

[9] J Deepho and P Kumam ldquoA modified halpernrsquos iterativescheme for solving split feasibility problemsrdquo Abstract andApplied Analysis vol 2012 Article ID 876069 8 pages 2012

[10] Y Dang and Y Gao ldquoe strong convergence of a KM-CQ-like algorithm for a split feasibility problemrdquo Inverse Prob-lems vol 27 no 1 Article ID 015007 2011

[11] S N He and Z Y Zhao ldquoStrong convergence of a relaxed CQalgorithm for the split feasibility problemrdquo Journal of In-equalities and Applications vol 2013 no 1 11 pages 2013

[12] Y H Yao and Mihai Strong Convergence of a Self-AdaptiveMethod for the Split Feasibility Problem Fixed Point =eory ampApplications Springer Berlin Germany 2013

[13] S Reich M T Truong and T Mai ldquoe split feasibilityproblem with multiple output sets in Hilbert spacesrdquo Opti-mization Letters vol 36 2020

[14] S Suantai N Pholasa N Pholasa and P Cholamjiak ldquoemodified inertial relaxed CQ algorithm for solving the splitfeasibility problemsrdquo Journal of Industrial and ManagementOptimization vol 14 no 4 pp 1595ndash1615 2018

[15] W Cholamjiak P Cholamjiak and S Suantai ldquoAn inertialforward-backward splitting method for solving inclusionproblems in Hilbert spacesrdquo Journal of Fixed Point=eory andApplications vol 20 no 1 p 42 2018

[16] Q L Dong Y J Cho and L L Zhong ldquoInertial projectionand contraction algorithms for variational inequalitiesrdquoJournal of Global Optimization vol 70 no 3 pp 1ndash18 2018

[17] D R Sahu Y J Cho and Q L Dong ldquoInertial relaxed CQalgorithms for solving a split feasibility problem in Hilbertspacesrdquo Numerical Algorithms vol 3 2020

[18] Y Shehu and A Gibali ldquoInertial krasnoselskiindashmann methodin banach spacesrdquo Mathematics vol 8 no 4 2020

[19] Y Shehu P T Vuong and P Cholamjiak ldquoA self-adaptiveprojection method with an inertial technique for split feasi-bility problems in Banach spaces with applications to imagerestoration problemsrdquo Journal of Fixed Point =eory andApplications vol 21 no 2 2019

[20] H-K Xu ldquoIterative methods for the split feasibility problemin infinite-dimensional Hilbert spacesrdquo Inverse Problemsvol 26 no 10 Article ID 105018 2010

[21] R I Bot and E R Csetnek ldquoAn inertial Tsengrsquos type proximalalgorithm for nonsmooth and nonconvex optimizationproblemsrdquo Journal of Optimization =eory and Applicationsvol 171 no 2 pp 600ndash616 2016

12 Journal of Mathematics

(3) f is lower semicontinuous on H if it is lowersemicontinuous at every point x isin H and f is weaklylower semicontinuous on H if it is weakly lowersemicontinuous at every point x isin H

Lemma 2 (see [20]) Let C and Q be closed convex subsets ofreal Hilbert spaces H1 and H2 respectively andf H1⟶ R is given byf(x) (I minus PQ)Ax2 whereA H1⟶ H2 be a bounded linear operator =en for δ gt 0and xlowast isin H1 the following statements are equivalent

(1) =e point xlowast solves the SFP (1) iexlowast isin x isin C Ax isin Q

(2) =e point xlowast is the fixed point of the mappingPC(I minus δnablaf)

(3) =e point xlowast solves the variational inequality problemwith respect to the gradient of f that is find a pointxlowast isin C such that

nablaf(x) y minus xge 0 forally isin C (30)

Lemma 3 (see [6]) Let H1 and H2 be real Hilbert spacesand f H1⟶ R is given byf(x) 12(I minus PQ)Ax2 whereQ is closed convex subset of H2 and A H1⟶ H2 is abounded linear operator =en the following assertions hold

(1) f is convex and differentiable(2) f is weakly lower semicontinuous on H1

(3) nablaf(x) AT(I minus PQ)Ax for x isin H1

(4) nablaf(x) is A2-Lipschitz ienablaf(x) minus nablaf(y)leA2x minus y forallx y isin H1

Definition 4 Let Λk1113864 1113865 be a real sequence en Λk1113864 1113865 de-creases at infinity if there exists k0 isin N such that Λk+1 leΛkfor kge k0 In other words the sequence Λk1113864 1113865 does not de-crease at infinity if there exists a subsequence Λkt

1113966 1113967tge1 of Λk1113864 1113865

such that ΛktltΛkt+1 for all tge 1

Lemma 4 Let Λk1113864 1113865 be a sequence of real numbers that doesnot decrease at infinity Also consider the sequence of integersφ(k)1113864 1113865kgek0 defined by

φ(k) max m isin N mle k Λm leΛm+11113864 1113865 (31)

en φ(k)1113864 1113865kgek0 is a nondecreasing sequence verifyinglim

k⟶infinφ(k) infin and for all kge k0 the following two esti-

mates hold

Λφ(k) leΛφ(k)+1 andΛk leΛφ(k)+1 (32)

Lemma 5 (see [21]) Let sk1113864 1113865 be a sequence of nonnegativereal numbers satisfying the following relation

sk+1 le 1 minus σk( 1113857sk + σkμk + βk kge 1 (33)

where σk1113864 1113865 μk1113864 1113865 and βk1113864 1113865 satisfy the conditions

(1) σk1113864 1113865 sub [0 1] 1113944infin

k1σk infin

(2) limsupk⟶infin

μk le 0

(3) βk ge 0 1113944infin

k1βk ltinfinThen limsup

k⟶infinsk 0

(34)

3 Main Results

In this section we consider a general case of the SFPwMOS(11) where the nonempty closed and convex sets C andQi(i 1 N) are given by level sets of convex functionsas in [7] roughout this section we assume thatc H⟶ R and qi Hi⟶ R are lower semicontinuousconvex functions and the sets C and Qi are given by

C ≔ x isin H c(x)le 0 andQi y isin Hi qi(y)le 01113864 1113865 (35)

e relaxed set (half spaces) Ck at xk is defined as

Ck ≔ x isin H c xk

1113872 1113873le ξk x

kminus x1113966 1113967 (36)

where ξk isinzc(xk)e relaxed set (half spaces) Qk

i (i 1 middot middot middot N) at xk aredefined as

Qki ≔ x isin H c x

k1113872 1113873le ξk

xk

minus x1113966 1113967

y isin Hi qi Bixk

1113872 1113873le ηki Bix

kminus y1113966 1113967

(37)

where ηki isin zqi(Bix

k) It follows that C sub Ck and Qi sub Qki

hold for every kge 0 Now we define the following (relaxed)proximity function for x isin H

gk(x) ≔12

1113944

N

i1I minus PQk

i1113874 1113875Bix

2 (38)

We note that gk() is differentiable with its gradientgiven by

nablagk(x) ≔ 1113944N

i

BTi I minus PQk

i1113874 1113875Bix (39)

where each Qki are half spaces given in (37) We note that gk

is weakly lower semicontinuous convex and differentiablefunction and nablagk is Lipschitz continuous

Now we present an inertial self-adaptive relaxed CQ

algorithmWe can see that Algorithm 1 terminates at some iterate

(say k) when nablagk(xk) 0e following lemma is important for the convergence

analysis

Lemma 6 If zk ≔ xk + θk(xk minus xkminus 1) where 0le θk lt 1 forall k isin N then for all z isin H

zk

minus z2 le x

kminus z

2+ θk x

kminus z

2minus x

kminus 1minus z

21113872 1113873 + 2θkx

kminus x

kminus 12

(40)

4 Journal of Mathematics

Proof Using the identity 2a b a2 + b2 minus a minus b2 we have

zk

minus z2

xk

minus z + θk xk

minus xkminus 1

1113872 11138732

xk

minus z2

+ 2θkxk

minus z xk

minus xkminus 1

+ θ2kxk

minus xkminus 12

xk

minus z2

+ θkxk

minus z2

+ xk

minus xkminus 12

minus xkminus 1

minus z2

+ θ2kxk

minus xkminus 12

xk

minus z2

+ θk xk

minus z2

minus xkminus 1

minus z2

1113872 1113873

+ θk 1 + θk( 1113857xk

minus xkminus 12

lexk

minus z2

+ θk xk

minus z2

minus xkminus 1

minus z2

1113872 1113873

+ 2θkxk

minus xkminus 12

(41)

Next we show the following strong convergence theo-rem for Algorithm 1

Theorem 1 Let H Hi i 1 N be real Hilbert spacesand let Bi H⟶ Hi i 1 N be bounded linear op-erators Denote Γ as the solution set of the problem (11) Let C

and Qi i 1 N be nonempty closed and convexsubsets of H and Hi i 1 N respectively Assume thatthe SFPwMOS (11) is consistent Suppose the sequences ρk1113864 1113865βk1113864 1113865 δk1113864 1113865 and εk1113864 1113865 in Algorithm 1 satisfy the followingconditions

(A1) limk⟶infin

βk 0

and 1113944infin

k1βk +infin

βk+1

βk

le l l is a constan t

(42)

(A2) 0lt liminfk⟶infin

εk le limsupfk⟶infin

εk lt 1 (43)

(A3) liminfk⟶infin

ρk 4 minus ρk( 1113857gt 0 (44)

(A4) limk⟶infin

θk

βk

xk

minus xkminus 1

0 (45)

en the sequence xk1113864 1113865 generated by Algorithm 1converges strongly to the solution p isin Γ where p PΓ(u)

Proof Wemay assume that Algorithm 1 does not terminatein a finite number of iterations us nablagk(xk)ne 0 for allk ge 0 Denote Γ as the solution set of problem (11) In the

consistent case of problem (11) Γ is nonempty closed andconvex us the metric projection PΓ is well defined

Let p isin Γ and set sk zk minus τknablagk(zk) Note that I minus PQki

for each i 1 N is firmly nonexpansive and nablagk(p)

0 Hence we have from Lemma 1 that

langnablagk zk

1113872 1113873 zk

minus prang lang1113944N

i1B

Ti I minus PQk

i1113874 1113875Biz

k z

kminus prang

1113944N

i1B

Ti I minus PQk

i1113874 1113875Biz

k z

kminus p

1113944N

i1I minus PQk

i1113874 1113875Biz

k Biz

kminus Bip

ge 1113944N

i1I minus PQk

i1113874 1113875Biz

k2 2gk z

k1113872 1113873

(46)

which implies that

sk

minus p

2

zk

minus p minus τknablagk zk

1113872 1113873

2

zk

minus p2

+ τ2knablagk zk

1113872 11138732

minus 2τknablagk zk

1113872 1113873 zk

minus p

le zk

minus p2

+ρ2kg

2k z

k1113872 1113873

nablagk zk

1113872 11138732 minus

2ρkgk zk

1113872 1113873

nablagk zk

1113872 11138732 2gk z

k1113872 11138731113872 1113873

zk

minus p2

+ρ2kg

2k z

k1113872 1113873

nablagk zk

1113872 11138732 minus

4ρkg2k z

k1113872 1113873

nablagk zk

1113872 11138732

zk

minus p2

minus ρk 4 minus ρk( 1113857g2k z

k1113872 1113873

nablagk zk

1113872 11138732

(47)

Using condition (A3) we have

sk

minus p2 le z

kminus p

2 forallkge 0 (48)

Initialization choose positive sequences 0le θk lt 1 ρk1113864 1113865 sub (0 4) βk1113864 1113865 sub (0 1) δk1113864 1113865 sub [0 1) and εk1113864 1113865 sub (0 1) such thatβk + δk + εk 1 Let u isin H be a fixed point Select arbitrary starting points x0 x1 isin H and set k 0

Step 1 given the current iterate xk isin H if nablagk(xk) 0 for some k isin N then stop Otherwise continue and calculateτk ≔ ρkgk(zk)nablagk(zk)2

Step 2 compute the next iterate asz

k x

k+ θk(x

kminus x

kminus 1)

xk+1 ≔ PCk

(βku + δkzk

+ εk(zk

minus τknablagk(zk)))

1113896

where Ck is the half space given as in (36)

ALGORITHM 1 Inertial self-adaptive relaxed CQ algorithm for SFPwMOS

Journal of Mathematics 5

Next we show zk1113864 1113865 is bounded Since p isin Γ sub Ck and theprojection operator PCk

is nonexpansive we obtain fromAlgorithm 1 and (48) that

xk+1

minus p PCkβku + δkz

k+ εk z

kminus τknablagk z

k1113872 11138731113872 11138731113872 1113873 minus p

le βku minus p + δkzk

minus p + εksk

minus p

le βku minus p + δkzk

minus p + εkzk

minus p

βku minus p + 1 minus βk( 1113857zk

minus p

βku minus p + 1 minus βk( 1113857xk

+ θk xk

minus xkminus 1

1113872 1113873 minus p

le βku minus p + 1 minus βk( 1113857xk

minus p + θk xk

minus xkminus 1

1 minus βk( 1113857 xk

minus p

+ βk u minus p +θk

βk

xk

minus xkminus 1

1113890 1113891

(49)

Condition (A4) implies the sequence θkβkxk minus xkminus 1 isbounded Let an upper bound of u minus p + θkβkxk minus xkminus 11113864 1113865 beM en we rewrite (49) as

xk+1

minus plemax xk

minus p M1113966 1113967 (50)

By induction we have that

xk+1

minus plemax x1

minus p M1113966 1113967 (51)

Hence xk minus p1113864 1113865 is bounded so is zk minus p1113864 1113865 and sk minus p1113864 1113865And zk1113864 1113865 is bounded Consequently sk1113864 1113865 and Biz

k1113864 1113865N

i1 arealso bounded The rest of the proof will be

divided into two parts

Case 1 Suppose that there exists k0 isin N such thatxk minus p21113864 1113865

infinkk0

is nonincreasing en xk minus p21113864 1113865infink1 converges

and xk minus p2 minus xk+1 minus p2⟶ 0 as k⟶infin From (47) weobtain

ρk 4 minus ρk( 1113857g2k z

k1113872 1113873

nablagk zk

1113872 11138732 le z

kminus p

2minus s

kminus p

2 (52)

Since p isin Γ sub Ck and the projection operator PCkis

nonexpansive from Algorithm 1 we obtain

xk+1

minus p2

PCkβku + δkz

k+ εksk1113872 1113873 minus p

2

le βku + δkzk

+ εksk1113872 1113873 minus p2

le βku minus p2

+ δkzk

minus p2

+ εksk minus p2

(53)

Since βk + δk + εk 1 from (53) and using the inequalityx + y2 lex2 + 2y x + y we have the following estimation

zk

minus p2

minus sk

minus p2 le

βk

εk

u minus p2

+1 minus βk

εk

zk

minus p2

minus1εk

xk+1

minus p2

βk

εk

u minus p2

minusβk

εk

zk

minus p2

+1εk

zk

minus p2

minus xk+1

minus p2

1113960 1113961

leβk

εk

u minus p2

+1εk

zk

minus p2

minus xk+1

minus p2

1113960 1113961

1εk

βku minus p2

+ zk

minus p2

minus xk+1

minus p2

1113960 11139611113960 1113961

1εk

βku minus p2

+ xk

minus p + θk xk

minus xkminus 1

1113872 11138732

minus xk+1

minus p2

1113876 11138771113876 1113877

le1εk

βku minus p2

+ xk

minus p2

+ 2θkxk

minus xkminus 1

zk

minus p minus xk+1

minus p2

1113960 11139611113960 1113961

le1εk

βku minus p2

+ xk

minus p2

+ 2θkxk

minus xkminus 1

zk

minus p minus xk+1

minus p2

1113960 11139611113960 1113961

1εk

βku minus p2

+ xk

minus p2

minus xk+1

minus p2

+ 2θkxk

minus xkminus 1

zk

minus p1113960 11139611113960 1113961

(54)

Combining (52) and (54) we obtain

6 Journal of Mathematics

ρk 4 minus ρk( 1113857g2k z

k1113872 1113873

nablagk zk

1113872 11138732 le z

kminus p

2minus s

kminus p

2

le1εk

βku minus p2

+ xk

minus p2

minus xk+1

minus p2

11139601113960

+ 2θkxk

minus xkminus 1

zk

minus p11139611113961

(55)

By condition (A4) βk isin (0 1) imply the sequencelimk⟶infinθkxk minus xkminus 1⟶ 0 and we have proved zk minus p1113864 1113865 isbounded so 2θkxk minus xkminus 1zk minus p⟶ 0 By conditions (A2)and (A3) and (55) we have as k⟶infin

0lt ρk 4 minus ρk( 1113857g2k z

k1113872 1113873

nablagk zk

1113872 11138732

le1εk

βku minus p2

+ xk

minus p2

minus xk+1

minus p2

11139601113960

+ 2θkxk

minus xkminus 1

zk

minus p11139611113961⟶ 0

(56)

which implies that

limk⟶infin

g2k zk

1113872 1113873

nablagk zk

1113872 11138732 0 (57)

We note that for each i 1 N BTi (I minus PQk

i)Bi() is

Lipschitz continuous Since the sequence zk1113864 1113865 is boundedand

BTi I minus PQk

i1113874 1113875 Biz

k B

Ti I minus PQk

i1113874 1113875Biz

k

minus BTi I minus PQk

i1113874 1113875Biple max

1leileNBi1113874 1113875z

kminus p

(58)

for all i 1 N we have the sequence

BTi (I minus PQk

i)Biz

k1113882 1113883infin

i1is bounded Hence nablagk(zk)1113864 1113865

infink1 is

bounded Consequently we have from (57) that

limk⟶infin

I minus PQki

1113874 1113875Bizk

0 (59)

For

zk

minus xk

xk

+ θk xk

minus xkminus 1

1113872 1113873 minus xk

θkxk

minus xkminus 1

(60)

which implies

zk

minus xk⟶ 0 as k⟶infin (61)

for each i 1 N since sk zk minus τknablagk(zk) we havefrom (59) and (61) that

sk

minus zk

τknablagk zk

1113872 1113873⟶ 0 as k⟶infin

sk

minus xk le s

kminus z

k+ z

kminus x

k⟶ 0 as k⟶infin(62)

at is

limk⟶infin

sk

minus xk

0 (63)

limk⟶infin

BTi I minus PQk

i1113874 1113875Biz

k 0 (64)

Let

lk

βku + δkzk

+ εksk

βku + 1 minus βk( 1113857vk (65)

where

vk

δk

1 minus βk

zk

+εk

1 minus βk

sk (66)

is gives

vk

minus zk le

εk

1 minus βk

sk

minus zk⟶ 0 k⟶infin

lk

minus vk

le βk u minus vk

⟶ 0 k⟶infin

(67)

lk

minus zk le l

kminus v

k+ v

kminus z

k⟶ 0 k⟶infin

lk

minus xk le l

kminus z

k+ z

kminus x

k⟶ 0 as k⟶infin(68)

Hence

limk⟶infin

lk

minus zk

limk⟶infin

lk

minus xk

limk⟶infin

lk

minus vk

limk⟶infin

vk

minus zk

0

(69)

Since xk1113864 1113865 is bounded there exists a subsequence xkj1113966 1113967

of xk1113864 1113865 such that xkj1113966 1113967plowast isin ωω(xk) Next we need toshow plowast isin C and Bip

lowast isin Qi for each i 1 NWe first show that xk+1 minus xk⟶ 0 For p isin Γ

12x

k+1minus p

212x

k+1minus x

k+ x

kminus p

2

12x

kminus p

2+ x

k+1minus x

k x

k+1minus p minus

12x

k+1minus x

k2

12x

kminus p

2+ x

k+1minus l

k+ l

kminus x

k x

k+1

minus p minus12x

k+1minus x

k2

12x

kminus p

2+ x

k+1minus l

k x

k+1minus p + l

k

minus xk x

k+1minus p minus

12x

k+1minus x

k2

(70)

then

12x

k+1minus x

k2 le12x

kminus p

2minus12x

k+1minus p

2+ x

k+1minus l

k x

k+1

minus p + lk

minus xk x

k+1minus p

(71)

from xk+1 minus lk xk+1 minus plt 0

Journal of Mathematics 7

12x

k+1minus x

k2 le12x

kminus p

2minus12x

k+1minus p

2+ l

kminus x

k x

k+1minus p

(72)

Since xk minus p2 minus xk+1 minus p2⟶ 0 as k⟶infinlim

k⟶infinlk minus xk 0 we obtain

xk+1

minus xk⟶ 0 as k⟶infin (73)

Since xkj+1 isin Ckj we obtain

c xkj1113872 1113873 + ξkj x

kj+1minus x

kj le 0 (74)

us

c xkj1113872 1113873le minus ξkj x

kj+1minus x

kj le ξxkj+1

minus xkj (75)

where ξ satisfies ξk le ξ for all k By virtue of the continuity offunction c and xkj+1 minus xkj⟶ 0 we get that

c plowast

( 1113857 limj⟶infin

c xkj1113872 1113873le 0 (76)

erefore plowast isin CNow we show that Bip

lowast isin Qi to do this let hki Biz

k minus

PQkiBiz

k⟶ 0 and let ηki be such that ηk

i le η for all k SinceBiz

kj minus hkj

i PQ

kj

i

(Bizkj ) isin Q

kj

i we have

qi Bixkj1113872 1113873 + ηkj

i Bizkj minus h

kj

i1113874 1113875 minus Bixkj le 0 (77)

Hence

qi Bixkj1113872 1113873le η

kj

i Bixkj minus Biz

kj + ηkj

i hkj

i le ηBixkj

minus xkjminus 1

+ ηhkj

i ⟶ 0

(78)

By the continuity of qi and Bixkj⟶ Bix

lowast we arrive atthe conclusion

qi Biplowast

( 1113857 limj⟶infin

qi Bixkj1113872 1113873le 0 (79)

namely Biplowast isin Qi for all i 1 N Hence plowast isin Γ

Moreover for p PΓ u we can see that

limsupk⟶infin

lk

minus p u minus p limk⟶infin

lkj minus p u minus plep

lowast

minus p u minus ple 0

(80)

By Lemma 1 and (65) we have

xk+1

minus p2

PC lk

1113872 1113873 minus p2 le l

kminus p

2 βku + 1 minus βk( 1113857v

kminus p l

kminus p

βku minus p lk

minus p + 1 minus βk( 1113857vk

minus p lk

minus p

le βku minus p lk

minus p + 1 minus βk( 1113857vk

minus plk

minus p

1 minus βk( 1113857 xk

+ θk xk

minus xkminus 1

1113872 1113873 minus p

2

+ βklangu minus p lk

minus prang +βk

2u minus p

2

le 1 minus βk( 1113857 xk

minus p

2

+ 2θklangxk

minus xkminus 1

zk

minus prang1113876 1113877 + βklangu minus p lk

minus prang +βk

2u minus p

2

le 1 minus βk( 1113857 xk

minus p

2

+ 2θk xk

minus xkminus 1

zk

minus p

1113876 1113877 + βklangu minus p lk

minus prang +βk

2u minus p

2

(81)

By conditions (A1) and (A4) βk isin (0 1) imply the se-quence lim

k⟶infinθkxk minus xkminus 1⟶ 0 βk2u minus p⟶ 0 and

using (80) in (81) and applying Lemma 5 we obtain

limk⟶infin

xk

minus p2

0 (82)

erefore as k⟶infin xk⟶ p PΓu

Case 2 SetΛk xk minus p2 Assume that Λk1113864 1113865 is not decreasingat infinity Let ϕ N⟶ N be a mapping for all kge k0 (forsome k0 large enough) defined by

ϕ(k) max t isin N tle kΛt leΛt+11113864 1113865 (83)

By Lemma 4 ϕ(k)1113864 1113865infinkk0is a nondecreasing sequence

such that ϕ(k)⟶infin as k⟶infin and

max Λϕ(k)Λk1113966 1113967leΛϕ(k)+1forallkge k0 (84)

After a similar conclusion from (59) it is easy to see that

limk⟶infin

I minus PQ

ϕ(k)

i

1113874 1113875Bizϕ(k)

0 (85)

By the similar argument as the above in Case 1 weconclude immediately that

8 Journal of Mathematics

limk⟶infin

Blowasti I minus P

Qϕ(k)

i

1113874 1113875Bizϕ(k)

0 (86)

limsupk⟶infin

lϕ(k)

minus p u minus ple 0 (87)

Since xϕ(k)1113864 1113865 is bounded there exists a subsequence ofxϕ(k)1113864 1113865 still denoted by xϕ(k)1113864 1113865 which converges weakly to

plowast By similar argument as above in Case 1 we concludeimmediately that plowast isin C and Bip

lowast isin QirArrplowast isin ΓFrom (57) we have that

xϕ(k)+1

minus p2 le 1 minus βϕ(k)1113872 1113873z

ϕ(k)minus p

2+ βϕ(k)u minus p l

ϕ(k)minus p

le 1 minus βϕ(k)1113872 1113873xϕ(k)

+ θϕ(k) xϕ(k)

minus xϕ(k)minus 1

1113872 1113873 minus p2

+ βϕ(k)u minus p lϕ(k)

minus p

le 1 minus βϕ(k)1113872 1113873 xϕ(k)

minus p2

+ 2θϕ(k)xϕ(k)

minus xϕ(k)minus 1

zϕ(k)

minus p1113960 1113961 + βϕ(k)u minus p lϕ(k)

minus p

le 1 minus βϕ(k)1113872 1113873 xϕ(k)

minus p2

+ 2θϕ(k)xϕ(k)

minus xϕ(k)minus 1

zϕ(k)

minus p1113960 1113961 + βϕ(k)u minus p lϕ(k)

minus p

(88)

which implies by Lemma 5

limk⟶infin

xφ(k)

minus p

2

0 and limk⟶infin

xϕ(k)+1

minus p2

0 (89)

Moreover for kge k0 it is easy to see thatΛϕ(k) minus Λϕ(k)+1 le 0 if kneϕ(k) (that is kge ϕ(k)) because forϕ(k) + 1lemle n Λm gtΛm+1 As a consequence we obtain

0leΛk lemax Λϕ(k)Λϕ(k)+11113966 1113967 Λϕ(k)+1 forallkge k0 (90)

erefore we obtain limk⟶infinΛk 0 that is xk1113864 1113865 converges

strongly to p is completes the proof

4 Some Extensive Results

For the SFPwMOS (11) when N 1 it becomes the SFP(1) us we have the following corollary for solving the SFP(1) which is an immediate consequence of eorem 1

Corollary 1 Let H1 and H2 be two real Hilbert spaces andlet B H1⟶ H2 be bounded linear operator Let C and Q

be nonempty closed and convex subsets of H1 and H2 re-spectively Assume that Ω Ccap Bminus 1(Q)ne 0 Let u isin H1 be afixed point For any starting point x0 x1 isin H1 let xk1113864 1113865 be thesequence generated by

zk

xk

+ θk xk

minus xkminus 1

1113872 1113873

xk+1 ≔ PCk

βku + δkzk

+ εk zk

minus τknablagk zk

1113872 11138731113872 11138731113872 1113873

⎧⎪⎨

⎪⎩(91)

where θk Ck τk and nablafk are given as in (8) Suppose thesequences βk1113864 1113865 δk1113864 1113865 and εk1113864 1113865 satisfy the conditions in =e-orem 1 =en the sequence xk1113864 1113865 converges strongly to thesolution p isin Ω where p PΩ(u)

When we take u x0 in Algorithm 1 we note also thefollowing results regarding to the SFPwMOS (11)

Corollary 2 Let H Hi i 1 N be real Hilbert spacesand let Bi H⟶ Hi i 1 N be bounded linearoperators Let C and Qi i 1 N be nonempty closedand convex subsets of H and Hi i 1 N respectivelyAssume that problem (11) is consistent For any initial guessx0 x1 isin H let xk1113864 1113865 be the sequence generated by

zk

xk

+ θk xk

minus xkminus 1

1113872 1113873

xk+1 ≔ PCk

βkx0

+ δkzk

+ εk zk

minus τknablagk zk

1113872 11138731113872 11138731113872 1113873

⎧⎪⎨

⎪⎩(92)

whereθk Ck τk and nablagk are given by (36) and (39) andAlgorithm 1 respectively Suppose the sequences βk1113864 1113865 δk1113864 1113865and εk1113864 1113865 satisfy the conditions in =eorem 1 =en the se-quence xk1113864 1113865 generated by (92) strongly converges to the so-lution p PΩ(x0) isin Γ

When we take δk equiv 0 in Algorithm 1 we obtain thefollowing result regarding the SFPwMOS (11)

Corollary 3 Let H Hi i 1 N be real Hilbertspaces and let Ti H⟶ Hi i 1 N be boundedlinear operators Let C and Qi i 1 N be nonemptyclosed and convex subsets of H and Hi i 1 N re-spectively Assume that problem (11) is consistent For a fixedpoint u isin H and any initial guess x0 isin H let xk1113864 1113865 be thesequence generated by

zk

xk

+ θk xk

minus xkminus 1

1113872 1113873

xk+1 ≔ PCk

βku + 1 minus βk( 1113857 zk

minus τknablagk zk

1113872 11138731113872 11138731113872 1113873

⎧⎪⎨

⎪⎩(93)

where θk Ck τk and nablagk are given by (36) and (39) andAlgorithm 1 respectively Suppose the sequences satisfyconditions (A1) =en the sequence xk1113864 1113865 generated by (93)strongly converges to the solution p PΩ(u) isin Γ

Of course when we take u x0 we get the followingresult regarding the SFPwMOS (11)

Corollary 4 Let H Hi i 1 N be real Hilbert spacesand let Bi H⟶ Hi i 1 N be bounded linear op-erators Let C and Qi i 1 N be nonempty closed andconvex subsets of H and Hi i 1 N respectively As-sume that problem (11) is consistent For any initial guessx0 isin H let xk1113864 1113865 be the sequence generated by

zk

xk

+ θk xk

minus xkminus 1

1113872 1113873

xk+1 ≔ PCk

βkx0

+ 1 minus tarts minus with(primehskip

prime)]))hskipsubstring minus after(preceding minus sibling comment()[starts minus with(prime hskipprime)]prime hskipprime)pt gt βk) z

kminus τknablagk z

k1113872 11138731113872 1113873)11138741113874

⎧⎪⎪⎨

⎪⎪⎩

(94)

Journal of Mathematics 9

where θk Ck τk and nablagk are given by (36) and (39) andAlgorithm 1 respectively Suppose the sequence βk1113864 1113865 satisfiesconditions (A1) =en the sequence xk1113864 1113865 generated by (94)strongly converges to the solution p PΩ(u) isin Γ

Remark 1 In Corollary 4 for the particular casewhereN 1 the iterative scheme (94) reduced exactly toiterative scheme proposed by He et al in eorem 32 of[10]

5 Numerical Experiment

In this section we provide one numerical experiment toillustrate the implementation and efficiency of our proposedmethod First we study the behavior of Algorithm 1 fordifferent choices of θkand ρk Next we compare Algorithm 1(θk 08) with no inertial Algorithm 1 (θk 0) and Scheme(17) for different initial points e numerical results arecompleted on a standard TOSHIBA laptop with Intel(R)Core(TM) i5-2450M CPU25GHz 25GHz with memory4GB e code is implemented in MATLAB R2020a

Let

A1

minus 04 minus 02 minus 02

04 05 minus 01

02 minus 05 03

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

A2

03 minus 01 03

minus 03 02 minus 02

0 02 03

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

C x isin R3|x1 minus x

22 + 2x3 le 01113966 1113967

Q1 x isin R3

|x21 + x2 minus x3 le 01113966 1113967

Q2 x isin R3|x1 + x

22 minus x3 le 01113966 1113967

(95)

Find xlowast such that

xlowast isin Γ ≔ Ccap cap 2i1A

minus 1i Qi( 11138571113872 1113873ne 0 (96)

Define the error function as

Ek 13x

kminus PCk

xk

1113872 11138732

+13A1x

kminus PQ1k

A1xk

1113872 11138732

+13A2x

kminus PQ2k

A2xk

1113872 11138732

(97)

Note that if Ek 0 at the kth step then xk isin ΓFor Algorithm 1 for different choices of ρk and θk we

choose βk 19k + 1 δk k9k + 1 εk 8k9k + 1u (01 01 01)T and x0 (minus 01 minus 01 minus 01)T andx1 [03 05 minus 06]T Using Ek lt ε as stopping criteriawhere ε is a small enough positive number the results of thenumerical experiment are reported in Table 1e numericalresults can be seen from Table 1 and Figures 1 and 2 InTable 1 ldquoIterrdquo denotes the number of iterations

e behavior of the function Ek in Table 1 is described inFigures 1 and 2

From Table 1 and Figures 1 and 2 we can observe thatAlgorithm 1 for inertial effect case has less number of it-erations than Algorithm 1 for θk 0(that is no inertialeffect) case Algorithm 1 with bigger θk has less number ofiterations than Algorithm 1 for small case for the same θkbig ρk case has less number of iterations than Algorithm 1for small case

Table 1 Algorithm 1 for different choices of θk ε and ρk

ε θkρk 398 ρk 300

Iter Ek Iter Ek

10minus 3θk 08 5 70906e minus 05

6 58048e minus 05θk 03 7 69397e minus 04θk 0 9 75095e minus 04

10minus 4θk 08 5 70906e minus 05

6 58048e minus 05θk 03 11 91282e minus 05θk 0 15 96150e minus 05

10minus 5θk 08 12 44752e minus 06

14 25296e minus 06θk 03 16 91012e minus 06θk 0 24 88064e minus 06

0 5 10 15 20 25The number of Iteration (k)

0

001

002

003

004

005

006

E k

θk=08θk=03θk=00

Figure 1 Error against iterations for Algorithm 1 with differencechoices of θk under ρk 398 and ε 10minus 5

10 Journal of Mathematics

Following we compare our algorithmwith iteration (16)For the example iteration (16) can be written as

xk+1 ≔ αkf x

k1113872 1113873 + 1 minus αk( 11138571113857PC x

kminus λk B

lowast1 I minus PQ11113872 11138731113872 1113873B1x

k1113872

+ Blowast2 I minus PQ21113872 11138731113872 1113873B2x

k1113873

(98)In the process we take αk 08 and f(xk)

u (111)Tλk 000005 for Scheme (97) We choose βk

12k + 1 δk εk k2k + 1 and ρk 3for Algorithm 1From Table 2 and Figure 3 we can observe that Algo-

rithm 1 has less number of iterations than Scheme (97)Moreover our iterative method is advantageous overScheme (98) because Algorithm 1 does not require com-puting of the operator norm

Data Availability

No data were used to support this study

Conflicts of Interest

e authors declare that they have no conflicts of interest

Authorsrsquo Contributions

is entire work has been completed by the authors HuijuanJia Shufen Liu and Yazheng Dang e authors read andapproved the finial manuscript

0 5 10 15e number of Iteration (k)

ρk=398ρk=300

0

001

002

003

004

005

006

007

E k

Figure 2 Error against iterations for Algorithm 1 with difference choices of ρk under θk 08 and ε 10minus 5

Table 2 Comparsion of Algorithm 1 and Scheme (97) for different choice initial point for ε 10minus 4

Initiative point x1 Algorithm 1 with θk 08 Algorithm 1 with θk 00 (98)Iter Ek Iter Ek Iter Ek

[04 09ndash09] 11 41830e minus 06 25 96999e minus 06 51 88226e minus 06[09 03ndash05] 6 14290e minus 07 22 86399e minus 06 26 96719e minus 06[minus 07 03ndash02] 12 78838e minus 06 24 92013e minus 06 48 84198e minus 06[01 01ndash09]prime 12 15353e minus 07 26 87544e minus 06 52 99722e minus 06

0 10 20 30 5040e number of Iteration (k)

Algorithm 1 with θ=08Algorithm 1 with θ=00(52)

0

004

002

006

008

01

012

014

E k

Figure 3 Error against of iterations for the comparision of Al-gorithm 1 and Scheme (97)

Journal of Mathematics 11

Acknowledgments

is research was supported by the National Natural ScienceFoundation of China (Grant no 61872126) Henan ProvinceKey Science and Technology Project (Grant no192102210123) and Young Backbone Teachers in Univer-sities of Henan Province (Grant no 2019GGJS061)

References

[1] Y Censor and T Elfving ldquoA multiprojection algorithm usingbregman projections in a product spacerdquo Numerical Algo-rithms vol 8 no 2 pp 221ndash239 1994

[2] C Byrne ldquoIterative oblique projection onto convex sets andthe split feasibility problemrdquo Inverse Problems vol 18 no 2pp 441ndash453 2002

[3] G T Herman ldquoImage reconstruction from projections thefundamentals of computerized tomographyrdquo 1980

[4] Y Censor and A Segal ldquoIterative projection methods inbiomedical inverse problemsrdquo 2008

[5] Y Censor T Bortfeld B Martin and A Trofimov ldquoA unifiedapproach for inversion problems in intensity-modulated ra-diation therapyrdquo Physics in Medicine and Biology vol 51no 10 pp 2353ndash2365 2006

[6] C Yair and Tommy ldquoe multiple-sets split feasibilityproblem and its applications for inverse problemsrdquo InverseProblems vol 21 no 6 pp 2071ndash2084 2005

[7] Q Yang ldquoe relaxed CQ algorithm solving the split feasi-bility problemrdquo Inverse Problems vol 20 no 4pp 1261ndash1266 2004

[8] G Lopez V Martın-Marquez and F H Wang ldquoSolving thesplit feasibility problem without prior knowledge of matrixnormsrdquo Inverse Problems vol 28 no 8 pp 374ndash389 2012

[9] J Deepho and P Kumam ldquoA modified halpernrsquos iterativescheme for solving split feasibility problemsrdquo Abstract andApplied Analysis vol 2012 Article ID 876069 8 pages 2012

[10] Y Dang and Y Gao ldquoe strong convergence of a KM-CQ-like algorithm for a split feasibility problemrdquo Inverse Prob-lems vol 27 no 1 Article ID 015007 2011

[11] S N He and Z Y Zhao ldquoStrong convergence of a relaxed CQalgorithm for the split feasibility problemrdquo Journal of In-equalities and Applications vol 2013 no 1 11 pages 2013

[12] Y H Yao and Mihai Strong Convergence of a Self-AdaptiveMethod for the Split Feasibility Problem Fixed Point =eory ampApplications Springer Berlin Germany 2013

[13] S Reich M T Truong and T Mai ldquoe split feasibilityproblem with multiple output sets in Hilbert spacesrdquo Opti-mization Letters vol 36 2020

[14] S Suantai N Pholasa N Pholasa and P Cholamjiak ldquoemodified inertial relaxed CQ algorithm for solving the splitfeasibility problemsrdquo Journal of Industrial and ManagementOptimization vol 14 no 4 pp 1595ndash1615 2018

[15] W Cholamjiak P Cholamjiak and S Suantai ldquoAn inertialforward-backward splitting method for solving inclusionproblems in Hilbert spacesrdquo Journal of Fixed Point=eory andApplications vol 20 no 1 p 42 2018

[16] Q L Dong Y J Cho and L L Zhong ldquoInertial projectionand contraction algorithms for variational inequalitiesrdquoJournal of Global Optimization vol 70 no 3 pp 1ndash18 2018

[17] D R Sahu Y J Cho and Q L Dong ldquoInertial relaxed CQalgorithms for solving a split feasibility problem in Hilbertspacesrdquo Numerical Algorithms vol 3 2020

[18] Y Shehu and A Gibali ldquoInertial krasnoselskiindashmann methodin banach spacesrdquo Mathematics vol 8 no 4 2020

[19] Y Shehu P T Vuong and P Cholamjiak ldquoA self-adaptiveprojection method with an inertial technique for split feasi-bility problems in Banach spaces with applications to imagerestoration problemsrdquo Journal of Fixed Point =eory andApplications vol 21 no 2 2019

[20] H-K Xu ldquoIterative methods for the split feasibility problemin infinite-dimensional Hilbert spacesrdquo Inverse Problemsvol 26 no 10 Article ID 105018 2010

[21] R I Bot and E R Csetnek ldquoAn inertial Tsengrsquos type proximalalgorithm for nonsmooth and nonconvex optimizationproblemsrdquo Journal of Optimization =eory and Applicationsvol 171 no 2 pp 600ndash616 2016

12 Journal of Mathematics

Proof Using the identity 2a b a2 + b2 minus a minus b2 we have

zk

minus z2

xk

minus z + θk xk

minus xkminus 1

1113872 11138732

xk

minus z2

+ 2θkxk

minus z xk

minus xkminus 1

+ θ2kxk

minus xkminus 12

xk

minus z2

+ θkxk

minus z2

+ xk

minus xkminus 12

minus xkminus 1

minus z2

+ θ2kxk

minus xkminus 12

xk

minus z2

+ θk xk

minus z2

minus xkminus 1

minus z2

1113872 1113873

+ θk 1 + θk( 1113857xk

minus xkminus 12

lexk

minus z2

+ θk xk

minus z2

minus xkminus 1

minus z2

1113872 1113873

+ 2θkxk

minus xkminus 12

(41)

Next we show the following strong convergence theo-rem for Algorithm 1

Theorem 1 Let H Hi i 1 N be real Hilbert spacesand let Bi H⟶ Hi i 1 N be bounded linear op-erators Denote Γ as the solution set of the problem (11) Let C

and Qi i 1 N be nonempty closed and convexsubsets of H and Hi i 1 N respectively Assume thatthe SFPwMOS (11) is consistent Suppose the sequences ρk1113864 1113865βk1113864 1113865 δk1113864 1113865 and εk1113864 1113865 in Algorithm 1 satisfy the followingconditions

(A1) limk⟶infin

βk 0

and 1113944infin

k1βk +infin

βk+1

βk

le l l is a constan t

(42)

(A2) 0lt liminfk⟶infin

εk le limsupfk⟶infin

εk lt 1 (43)

(A3) liminfk⟶infin

ρk 4 minus ρk( 1113857gt 0 (44)

(A4) limk⟶infin

θk

βk

xk

minus xkminus 1

0 (45)

en the sequence xk1113864 1113865 generated by Algorithm 1converges strongly to the solution p isin Γ where p PΓ(u)

Proof Wemay assume that Algorithm 1 does not terminatein a finite number of iterations us nablagk(xk)ne 0 for allk ge 0 Denote Γ as the solution set of problem (11) In the

consistent case of problem (11) Γ is nonempty closed andconvex us the metric projection PΓ is well defined

Let p isin Γ and set sk zk minus τknablagk(zk) Note that I minus PQki

for each i 1 N is firmly nonexpansive and nablagk(p)

0 Hence we have from Lemma 1 that

langnablagk zk

1113872 1113873 zk

minus prang lang1113944N

i1B

Ti I minus PQk

i1113874 1113875Biz

k z

kminus prang

1113944N

i1B

Ti I minus PQk

i1113874 1113875Biz

k z

kminus p

1113944N

i1I minus PQk

i1113874 1113875Biz

k Biz

kminus Bip

ge 1113944N

i1I minus PQk

i1113874 1113875Biz

k2 2gk z

k1113872 1113873

(46)

which implies that

sk

minus p

2

zk

minus p minus τknablagk zk

1113872 1113873

2

zk

minus p2

+ τ2knablagk zk

1113872 11138732

minus 2τknablagk zk

1113872 1113873 zk

minus p

le zk

minus p2

+ρ2kg

2k z

k1113872 1113873

nablagk zk

1113872 11138732 minus

2ρkgk zk

1113872 1113873

nablagk zk

1113872 11138732 2gk z

k1113872 11138731113872 1113873

zk

minus p2

+ρ2kg

2k z

k1113872 1113873

nablagk zk

1113872 11138732 minus

4ρkg2k z

k1113872 1113873

nablagk zk

1113872 11138732

zk

minus p2

minus ρk 4 minus ρk( 1113857g2k z

k1113872 1113873

nablagk zk

1113872 11138732

(47)

Using condition (A3) we have

sk

minus p2 le z

kminus p

2 forallkge 0 (48)

Initialization choose positive sequences 0le θk lt 1 ρk1113864 1113865 sub (0 4) βk1113864 1113865 sub (0 1) δk1113864 1113865 sub [0 1) and εk1113864 1113865 sub (0 1) such thatβk + δk + εk 1 Let u isin H be a fixed point Select arbitrary starting points x0 x1 isin H and set k 0

Step 1 given the current iterate xk isin H if nablagk(xk) 0 for some k isin N then stop Otherwise continue and calculateτk ≔ ρkgk(zk)nablagk(zk)2

Step 2 compute the next iterate asz

k x

k+ θk(x

kminus x

kminus 1)

xk+1 ≔ PCk

(βku + δkzk

+ εk(zk

minus τknablagk(zk)))

1113896

where Ck is the half space given as in (36)

ALGORITHM 1 Inertial self-adaptive relaxed CQ algorithm for SFPwMOS

Journal of Mathematics 5

Next we show zk1113864 1113865 is bounded Since p isin Γ sub Ck and theprojection operator PCk

is nonexpansive we obtain fromAlgorithm 1 and (48) that

xk+1

minus p PCkβku + δkz

k+ εk z

kminus τknablagk z

k1113872 11138731113872 11138731113872 1113873 minus p

le βku minus p + δkzk

minus p + εksk

minus p

le βku minus p + δkzk

minus p + εkzk

minus p

βku minus p + 1 minus βk( 1113857zk

minus p

βku minus p + 1 minus βk( 1113857xk

+ θk xk

minus xkminus 1

1113872 1113873 minus p

le βku minus p + 1 minus βk( 1113857xk

minus p + θk xk

minus xkminus 1

1 minus βk( 1113857 xk

minus p

+ βk u minus p +θk

βk

xk

minus xkminus 1

1113890 1113891

(49)

Condition (A4) implies the sequence θkβkxk minus xkminus 1 isbounded Let an upper bound of u minus p + θkβkxk minus xkminus 11113864 1113865 beM en we rewrite (49) as

xk+1

minus plemax xk

minus p M1113966 1113967 (50)

By induction we have that

xk+1

minus plemax x1

minus p M1113966 1113967 (51)

Hence xk minus p1113864 1113865 is bounded so is zk minus p1113864 1113865 and sk minus p1113864 1113865And zk1113864 1113865 is bounded Consequently sk1113864 1113865 and Biz

k1113864 1113865N

i1 arealso bounded The rest of the proof will be

divided into two parts

Case 1 Suppose that there exists k0 isin N such thatxk minus p21113864 1113865

infinkk0

is nonincreasing en xk minus p21113864 1113865infink1 converges

and xk minus p2 minus xk+1 minus p2⟶ 0 as k⟶infin From (47) weobtain

ρk 4 minus ρk( 1113857g2k z

k1113872 1113873

nablagk zk

1113872 11138732 le z

kminus p

2minus s

kminus p

2 (52)

Since p isin Γ sub Ck and the projection operator PCkis

nonexpansive from Algorithm 1 we obtain

xk+1

minus p2

PCkβku + δkz

k+ εksk1113872 1113873 minus p

2

le βku + δkzk

+ εksk1113872 1113873 minus p2

le βku minus p2

+ δkzk

minus p2

+ εksk minus p2

(53)

Since βk + δk + εk 1 from (53) and using the inequalityx + y2 lex2 + 2y x + y we have the following estimation

zk

minus p2

minus sk

minus p2 le

βk

εk

u minus p2

+1 minus βk

εk

zk

minus p2

minus1εk

xk+1

minus p2

βk

εk

u minus p2

minusβk

εk

zk

minus p2

+1εk

zk

minus p2

minus xk+1

minus p2

1113960 1113961

leβk

εk

u minus p2

+1εk

zk

minus p2

minus xk+1

minus p2

1113960 1113961

1εk

βku minus p2

+ zk

minus p2

minus xk+1

minus p2

1113960 11139611113960 1113961

1εk

βku minus p2

+ xk

minus p + θk xk

minus xkminus 1

1113872 11138732

minus xk+1

minus p2

1113876 11138771113876 1113877

le1εk

βku minus p2

+ xk

minus p2

+ 2θkxk

minus xkminus 1

zk

minus p minus xk+1

minus p2

1113960 11139611113960 1113961

le1εk

βku minus p2

+ xk

minus p2

+ 2θkxk

minus xkminus 1

zk

minus p minus xk+1

minus p2

1113960 11139611113960 1113961

1εk

βku minus p2

+ xk

minus p2

minus xk+1

minus p2

+ 2θkxk

minus xkminus 1

zk

minus p1113960 11139611113960 1113961

(54)

Combining (52) and (54) we obtain

6 Journal of Mathematics

ρk 4 minus ρk( 1113857g2k z

k1113872 1113873

nablagk zk

1113872 11138732 le z

kminus p

2minus s

kminus p

2

le1εk

βku minus p2

+ xk

minus p2

minus xk+1

minus p2

11139601113960

+ 2θkxk

minus xkminus 1

zk

minus p11139611113961

(55)

By condition (A4) βk isin (0 1) imply the sequencelimk⟶infinθkxk minus xkminus 1⟶ 0 and we have proved zk minus p1113864 1113865 isbounded so 2θkxk minus xkminus 1zk minus p⟶ 0 By conditions (A2)and (A3) and (55) we have as k⟶infin

0lt ρk 4 minus ρk( 1113857g2k z

k1113872 1113873

nablagk zk

1113872 11138732

le1εk

βku minus p2

+ xk

minus p2

minus xk+1

minus p2

11139601113960

+ 2θkxk

minus xkminus 1

zk

minus p11139611113961⟶ 0

(56)

which implies that

limk⟶infin

g2k zk

1113872 1113873

nablagk zk

1113872 11138732 0 (57)

We note that for each i 1 N BTi (I minus PQk

i)Bi() is

Lipschitz continuous Since the sequence zk1113864 1113865 is boundedand

BTi I minus PQk

i1113874 1113875 Biz

k B

Ti I minus PQk

i1113874 1113875Biz

k

minus BTi I minus PQk

i1113874 1113875Biple max

1leileNBi1113874 1113875z

kminus p

(58)

for all i 1 N we have the sequence

BTi (I minus PQk

i)Biz

k1113882 1113883infin

i1is bounded Hence nablagk(zk)1113864 1113865

infink1 is

bounded Consequently we have from (57) that

limk⟶infin

I minus PQki

1113874 1113875Bizk

0 (59)

For

zk

minus xk

xk

+ θk xk

minus xkminus 1

1113872 1113873 minus xk

θkxk

minus xkminus 1

(60)

which implies

zk

minus xk⟶ 0 as k⟶infin (61)

for each i 1 N since sk zk minus τknablagk(zk) we havefrom (59) and (61) that

sk

minus zk

τknablagk zk

1113872 1113873⟶ 0 as k⟶infin

sk

minus xk le s

kminus z

k+ z

kminus x

k⟶ 0 as k⟶infin(62)

at is

limk⟶infin

sk

minus xk

0 (63)

limk⟶infin

BTi I minus PQk

i1113874 1113875Biz

k 0 (64)

Let

lk

βku + δkzk

+ εksk

βku + 1 minus βk( 1113857vk (65)

where

vk

δk

1 minus βk

zk

+εk

1 minus βk

sk (66)

is gives

vk

minus zk le

εk

1 minus βk

sk

minus zk⟶ 0 k⟶infin

lk

minus vk

le βk u minus vk

⟶ 0 k⟶infin

(67)

lk

minus zk le l

kminus v

k+ v

kminus z

k⟶ 0 k⟶infin

lk

minus xk le l

kminus z

k+ z

kminus x

k⟶ 0 as k⟶infin(68)

Hence

limk⟶infin

lk

minus zk

limk⟶infin

lk

minus xk

limk⟶infin

lk

minus vk

limk⟶infin

vk

minus zk

0

(69)

Since xk1113864 1113865 is bounded there exists a subsequence xkj1113966 1113967

of xk1113864 1113865 such that xkj1113966 1113967plowast isin ωω(xk) Next we need toshow plowast isin C and Bip

lowast isin Qi for each i 1 NWe first show that xk+1 minus xk⟶ 0 For p isin Γ

12x

k+1minus p

212x

k+1minus x

k+ x

kminus p

2

12x

kminus p

2+ x

k+1minus x

k x

k+1minus p minus

12x

k+1minus x

k2

12x

kminus p

2+ x

k+1minus l

k+ l

kminus x

k x

k+1

minus p minus12x

k+1minus x

k2

12x

kminus p

2+ x

k+1minus l

k x

k+1minus p + l

k

minus xk x

k+1minus p minus

12x

k+1minus x

k2

(70)

then

12x

k+1minus x

k2 le12x

kminus p

2minus12x

k+1minus p

2+ x

k+1minus l

k x

k+1

minus p + lk

minus xk x

k+1minus p

(71)

from xk+1 minus lk xk+1 minus plt 0

Journal of Mathematics 7

12x

k+1minus x

k2 le12x

kminus p

2minus12x

k+1minus p

2+ l

kminus x

k x

k+1minus p

(72)

Since xk minus p2 minus xk+1 minus p2⟶ 0 as k⟶infinlim

k⟶infinlk minus xk 0 we obtain

xk+1

minus xk⟶ 0 as k⟶infin (73)

Since xkj+1 isin Ckj we obtain

c xkj1113872 1113873 + ξkj x

kj+1minus x

kj le 0 (74)

us

c xkj1113872 1113873le minus ξkj x

kj+1minus x

kj le ξxkj+1

minus xkj (75)

where ξ satisfies ξk le ξ for all k By virtue of the continuity offunction c and xkj+1 minus xkj⟶ 0 we get that

c plowast

( 1113857 limj⟶infin

c xkj1113872 1113873le 0 (76)

erefore plowast isin CNow we show that Bip

lowast isin Qi to do this let hki Biz

k minus

PQkiBiz

k⟶ 0 and let ηki be such that ηk

i le η for all k SinceBiz

kj minus hkj

i PQ

kj

i

(Bizkj ) isin Q

kj

i we have

qi Bixkj1113872 1113873 + ηkj

i Bizkj minus h

kj

i1113874 1113875 minus Bixkj le 0 (77)

Hence

qi Bixkj1113872 1113873le η

kj

i Bixkj minus Biz

kj + ηkj

i hkj

i le ηBixkj

minus xkjminus 1

+ ηhkj

i ⟶ 0

(78)

By the continuity of qi and Bixkj⟶ Bix

lowast we arrive atthe conclusion

qi Biplowast

( 1113857 limj⟶infin

qi Bixkj1113872 1113873le 0 (79)

namely Biplowast isin Qi for all i 1 N Hence plowast isin Γ

Moreover for p PΓ u we can see that

limsupk⟶infin

lk

minus p u minus p limk⟶infin

lkj minus p u minus plep

lowast

minus p u minus ple 0

(80)

By Lemma 1 and (65) we have

xk+1

minus p2

PC lk

1113872 1113873 minus p2 le l

kminus p

2 βku + 1 minus βk( 1113857v

kminus p l

kminus p

βku minus p lk

minus p + 1 minus βk( 1113857vk

minus p lk

minus p

le βku minus p lk

minus p + 1 minus βk( 1113857vk

minus plk

minus p

1 minus βk( 1113857 xk

+ θk xk

minus xkminus 1

1113872 1113873 minus p

2

+ βklangu minus p lk

minus prang +βk

2u minus p

2

le 1 minus βk( 1113857 xk

minus p

2

+ 2θklangxk

minus xkminus 1

zk

minus prang1113876 1113877 + βklangu minus p lk

minus prang +βk

2u minus p

2

le 1 minus βk( 1113857 xk

minus p

2

+ 2θk xk

minus xkminus 1

zk

minus p

1113876 1113877 + βklangu minus p lk

minus prang +βk

2u minus p

2

(81)

By conditions (A1) and (A4) βk isin (0 1) imply the se-quence lim

k⟶infinθkxk minus xkminus 1⟶ 0 βk2u minus p⟶ 0 and

using (80) in (81) and applying Lemma 5 we obtain

limk⟶infin

xk

minus p2

0 (82)

erefore as k⟶infin xk⟶ p PΓu

Case 2 SetΛk xk minus p2 Assume that Λk1113864 1113865 is not decreasingat infinity Let ϕ N⟶ N be a mapping for all kge k0 (forsome k0 large enough) defined by

ϕ(k) max t isin N tle kΛt leΛt+11113864 1113865 (83)

By Lemma 4 ϕ(k)1113864 1113865infinkk0is a nondecreasing sequence

such that ϕ(k)⟶infin as k⟶infin and

max Λϕ(k)Λk1113966 1113967leΛϕ(k)+1forallkge k0 (84)

After a similar conclusion from (59) it is easy to see that

limk⟶infin

I minus PQ

ϕ(k)

i

1113874 1113875Bizϕ(k)

0 (85)

By the similar argument as the above in Case 1 weconclude immediately that

8 Journal of Mathematics

limk⟶infin

Blowasti I minus P

Qϕ(k)

i

1113874 1113875Bizϕ(k)

0 (86)

limsupk⟶infin

lϕ(k)

minus p u minus ple 0 (87)

Since xϕ(k)1113864 1113865 is bounded there exists a subsequence ofxϕ(k)1113864 1113865 still denoted by xϕ(k)1113864 1113865 which converges weakly to

plowast By similar argument as above in Case 1 we concludeimmediately that plowast isin C and Bip

lowast isin QirArrplowast isin ΓFrom (57) we have that

xϕ(k)+1

minus p2 le 1 minus βϕ(k)1113872 1113873z

ϕ(k)minus p

2+ βϕ(k)u minus p l

ϕ(k)minus p

le 1 minus βϕ(k)1113872 1113873xϕ(k)

+ θϕ(k) xϕ(k)

minus xϕ(k)minus 1

1113872 1113873 minus p2

+ βϕ(k)u minus p lϕ(k)

minus p

le 1 minus βϕ(k)1113872 1113873 xϕ(k)

minus p2

+ 2θϕ(k)xϕ(k)

minus xϕ(k)minus 1

zϕ(k)

minus p1113960 1113961 + βϕ(k)u minus p lϕ(k)

minus p

le 1 minus βϕ(k)1113872 1113873 xϕ(k)

minus p2

+ 2θϕ(k)xϕ(k)

minus xϕ(k)minus 1

zϕ(k)

minus p1113960 1113961 + βϕ(k)u minus p lϕ(k)

minus p

(88)

which implies by Lemma 5

limk⟶infin

xφ(k)

minus p

2

0 and limk⟶infin

xϕ(k)+1

minus p2

0 (89)

Moreover for kge k0 it is easy to see thatΛϕ(k) minus Λϕ(k)+1 le 0 if kneϕ(k) (that is kge ϕ(k)) because forϕ(k) + 1lemle n Λm gtΛm+1 As a consequence we obtain

0leΛk lemax Λϕ(k)Λϕ(k)+11113966 1113967 Λϕ(k)+1 forallkge k0 (90)

erefore we obtain limk⟶infinΛk 0 that is xk1113864 1113865 converges

strongly to p is completes the proof

4 Some Extensive Results

For the SFPwMOS (11) when N 1 it becomes the SFP(1) us we have the following corollary for solving the SFP(1) which is an immediate consequence of eorem 1

Corollary 1 Let H1 and H2 be two real Hilbert spaces andlet B H1⟶ H2 be bounded linear operator Let C and Q

be nonempty closed and convex subsets of H1 and H2 re-spectively Assume that Ω Ccap Bminus 1(Q)ne 0 Let u isin H1 be afixed point For any starting point x0 x1 isin H1 let xk1113864 1113865 be thesequence generated by

zk

xk

+ θk xk

minus xkminus 1

1113872 1113873

xk+1 ≔ PCk

βku + δkzk

+ εk zk

minus τknablagk zk

1113872 11138731113872 11138731113872 1113873

⎧⎪⎨

⎪⎩(91)

where θk Ck τk and nablafk are given as in (8) Suppose thesequences βk1113864 1113865 δk1113864 1113865 and εk1113864 1113865 satisfy the conditions in =e-orem 1 =en the sequence xk1113864 1113865 converges strongly to thesolution p isin Ω where p PΩ(u)

When we take u x0 in Algorithm 1 we note also thefollowing results regarding to the SFPwMOS (11)

Corollary 2 Let H Hi i 1 N be real Hilbert spacesand let Bi H⟶ Hi i 1 N be bounded linearoperators Let C and Qi i 1 N be nonempty closedand convex subsets of H and Hi i 1 N respectivelyAssume that problem (11) is consistent For any initial guessx0 x1 isin H let xk1113864 1113865 be the sequence generated by

zk

xk

+ θk xk

minus xkminus 1

1113872 1113873

xk+1 ≔ PCk

βkx0

+ δkzk

+ εk zk

minus τknablagk zk

1113872 11138731113872 11138731113872 1113873

⎧⎪⎨

⎪⎩(92)

whereθk Ck τk and nablagk are given by (36) and (39) andAlgorithm 1 respectively Suppose the sequences βk1113864 1113865 δk1113864 1113865and εk1113864 1113865 satisfy the conditions in =eorem 1 =en the se-quence xk1113864 1113865 generated by (92) strongly converges to the so-lution p PΩ(x0) isin Γ

When we take δk equiv 0 in Algorithm 1 we obtain thefollowing result regarding the SFPwMOS (11)

Corollary 3 Let H Hi i 1 N be real Hilbertspaces and let Ti H⟶ Hi i 1 N be boundedlinear operators Let C and Qi i 1 N be nonemptyclosed and convex subsets of H and Hi i 1 N re-spectively Assume that problem (11) is consistent For a fixedpoint u isin H and any initial guess x0 isin H let xk1113864 1113865 be thesequence generated by

zk

xk

+ θk xk

minus xkminus 1

1113872 1113873

xk+1 ≔ PCk

βku + 1 minus βk( 1113857 zk

minus τknablagk zk

1113872 11138731113872 11138731113872 1113873

⎧⎪⎨

⎪⎩(93)

where θk Ck τk and nablagk are given by (36) and (39) andAlgorithm 1 respectively Suppose the sequences satisfyconditions (A1) =en the sequence xk1113864 1113865 generated by (93)strongly converges to the solution p PΩ(u) isin Γ

Of course when we take u x0 we get the followingresult regarding the SFPwMOS (11)

Corollary 4 Let H Hi i 1 N be real Hilbert spacesand let Bi H⟶ Hi i 1 N be bounded linear op-erators Let C and Qi i 1 N be nonempty closed andconvex subsets of H and Hi i 1 N respectively As-sume that problem (11) is consistent For any initial guessx0 isin H let xk1113864 1113865 be the sequence generated by

zk

xk

+ θk xk

minus xkminus 1

1113872 1113873

xk+1 ≔ PCk

βkx0

+ 1 minus tarts minus with(primehskip

prime)]))hskipsubstring minus after(preceding minus sibling comment()[starts minus with(prime hskipprime)]prime hskipprime)pt gt βk) z

kminus τknablagk z

k1113872 11138731113872 1113873)11138741113874

⎧⎪⎪⎨

⎪⎪⎩

(94)

Journal of Mathematics 9

where θk Ck τk and nablagk are given by (36) and (39) andAlgorithm 1 respectively Suppose the sequence βk1113864 1113865 satisfiesconditions (A1) =en the sequence xk1113864 1113865 generated by (94)strongly converges to the solution p PΩ(u) isin Γ

Remark 1 In Corollary 4 for the particular casewhereN 1 the iterative scheme (94) reduced exactly toiterative scheme proposed by He et al in eorem 32 of[10]

5 Numerical Experiment

In this section we provide one numerical experiment toillustrate the implementation and efficiency of our proposedmethod First we study the behavior of Algorithm 1 fordifferent choices of θkand ρk Next we compare Algorithm 1(θk 08) with no inertial Algorithm 1 (θk 0) and Scheme(17) for different initial points e numerical results arecompleted on a standard TOSHIBA laptop with Intel(R)Core(TM) i5-2450M CPU25GHz 25GHz with memory4GB e code is implemented in MATLAB R2020a

Let

A1

minus 04 minus 02 minus 02

04 05 minus 01

02 minus 05 03

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

A2

03 minus 01 03

minus 03 02 minus 02

0 02 03

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

C x isin R3|x1 minus x

22 + 2x3 le 01113966 1113967

Q1 x isin R3

|x21 + x2 minus x3 le 01113966 1113967

Q2 x isin R3|x1 + x

22 minus x3 le 01113966 1113967

(95)

Find xlowast such that

xlowast isin Γ ≔ Ccap cap 2i1A

minus 1i Qi( 11138571113872 1113873ne 0 (96)

Define the error function as

Ek 13x

kminus PCk

xk

1113872 11138732

+13A1x

kminus PQ1k

A1xk

1113872 11138732

+13A2x

kminus PQ2k

A2xk

1113872 11138732

(97)

Note that if Ek 0 at the kth step then xk isin ΓFor Algorithm 1 for different choices of ρk and θk we

choose βk 19k + 1 δk k9k + 1 εk 8k9k + 1u (01 01 01)T and x0 (minus 01 minus 01 minus 01)T andx1 [03 05 minus 06]T Using Ek lt ε as stopping criteriawhere ε is a small enough positive number the results of thenumerical experiment are reported in Table 1e numericalresults can be seen from Table 1 and Figures 1 and 2 InTable 1 ldquoIterrdquo denotes the number of iterations

e behavior of the function Ek in Table 1 is described inFigures 1 and 2

From Table 1 and Figures 1 and 2 we can observe thatAlgorithm 1 for inertial effect case has less number of it-erations than Algorithm 1 for θk 0(that is no inertialeffect) case Algorithm 1 with bigger θk has less number ofiterations than Algorithm 1 for small case for the same θkbig ρk case has less number of iterations than Algorithm 1for small case

Table 1 Algorithm 1 for different choices of θk ε and ρk

ε θkρk 398 ρk 300

Iter Ek Iter Ek

10minus 3θk 08 5 70906e minus 05

6 58048e minus 05θk 03 7 69397e minus 04θk 0 9 75095e minus 04

10minus 4θk 08 5 70906e minus 05

6 58048e minus 05θk 03 11 91282e minus 05θk 0 15 96150e minus 05

10minus 5θk 08 12 44752e minus 06

14 25296e minus 06θk 03 16 91012e minus 06θk 0 24 88064e minus 06

0 5 10 15 20 25The number of Iteration (k)

0

001

002

003

004

005

006

E k

θk=08θk=03θk=00

Figure 1 Error against iterations for Algorithm 1 with differencechoices of θk under ρk 398 and ε 10minus 5

10 Journal of Mathematics

Following we compare our algorithmwith iteration (16)For the example iteration (16) can be written as

xk+1 ≔ αkf x

k1113872 1113873 + 1 minus αk( 11138571113857PC x

kminus λk B

lowast1 I minus PQ11113872 11138731113872 1113873B1x

k1113872

+ Blowast2 I minus PQ21113872 11138731113872 1113873B2x

k1113873

(98)In the process we take αk 08 and f(xk)

u (111)Tλk 000005 for Scheme (97) We choose βk

12k + 1 δk εk k2k + 1 and ρk 3for Algorithm 1From Table 2 and Figure 3 we can observe that Algo-

rithm 1 has less number of iterations than Scheme (97)Moreover our iterative method is advantageous overScheme (98) because Algorithm 1 does not require com-puting of the operator norm

Data Availability

No data were used to support this study

Conflicts of Interest

e authors declare that they have no conflicts of interest

Authorsrsquo Contributions

is entire work has been completed by the authors HuijuanJia Shufen Liu and Yazheng Dang e authors read andapproved the finial manuscript

0 5 10 15e number of Iteration (k)

ρk=398ρk=300

0

001

002

003

004

005

006

007

E k

Figure 2 Error against iterations for Algorithm 1 with difference choices of ρk under θk 08 and ε 10minus 5

Table 2 Comparsion of Algorithm 1 and Scheme (97) for different choice initial point for ε 10minus 4

Initiative point x1 Algorithm 1 with θk 08 Algorithm 1 with θk 00 (98)Iter Ek Iter Ek Iter Ek

[04 09ndash09] 11 41830e minus 06 25 96999e minus 06 51 88226e minus 06[09 03ndash05] 6 14290e minus 07 22 86399e minus 06 26 96719e minus 06[minus 07 03ndash02] 12 78838e minus 06 24 92013e minus 06 48 84198e minus 06[01 01ndash09]prime 12 15353e minus 07 26 87544e minus 06 52 99722e minus 06

0 10 20 30 5040e number of Iteration (k)

Algorithm 1 with θ=08Algorithm 1 with θ=00(52)

0

004

002

006

008

01

012

014

E k

Figure 3 Error against of iterations for the comparision of Al-gorithm 1 and Scheme (97)

Journal of Mathematics 11

Acknowledgments

is research was supported by the National Natural ScienceFoundation of China (Grant no 61872126) Henan ProvinceKey Science and Technology Project (Grant no192102210123) and Young Backbone Teachers in Univer-sities of Henan Province (Grant no 2019GGJS061)

References

[1] Y Censor and T Elfving ldquoA multiprojection algorithm usingbregman projections in a product spacerdquo Numerical Algo-rithms vol 8 no 2 pp 221ndash239 1994

[2] C Byrne ldquoIterative oblique projection onto convex sets andthe split feasibility problemrdquo Inverse Problems vol 18 no 2pp 441ndash453 2002

[3] G T Herman ldquoImage reconstruction from projections thefundamentals of computerized tomographyrdquo 1980

[4] Y Censor and A Segal ldquoIterative projection methods inbiomedical inverse problemsrdquo 2008

[5] Y Censor T Bortfeld B Martin and A Trofimov ldquoA unifiedapproach for inversion problems in intensity-modulated ra-diation therapyrdquo Physics in Medicine and Biology vol 51no 10 pp 2353ndash2365 2006

[6] C Yair and Tommy ldquoe multiple-sets split feasibilityproblem and its applications for inverse problemsrdquo InverseProblems vol 21 no 6 pp 2071ndash2084 2005

[7] Q Yang ldquoe relaxed CQ algorithm solving the split feasi-bility problemrdquo Inverse Problems vol 20 no 4pp 1261ndash1266 2004

[8] G Lopez V Martın-Marquez and F H Wang ldquoSolving thesplit feasibility problem without prior knowledge of matrixnormsrdquo Inverse Problems vol 28 no 8 pp 374ndash389 2012

[9] J Deepho and P Kumam ldquoA modified halpernrsquos iterativescheme for solving split feasibility problemsrdquo Abstract andApplied Analysis vol 2012 Article ID 876069 8 pages 2012

[10] Y Dang and Y Gao ldquoe strong convergence of a KM-CQ-like algorithm for a split feasibility problemrdquo Inverse Prob-lems vol 27 no 1 Article ID 015007 2011

[11] S N He and Z Y Zhao ldquoStrong convergence of a relaxed CQalgorithm for the split feasibility problemrdquo Journal of In-equalities and Applications vol 2013 no 1 11 pages 2013

[12] Y H Yao and Mihai Strong Convergence of a Self-AdaptiveMethod for the Split Feasibility Problem Fixed Point =eory ampApplications Springer Berlin Germany 2013

[13] S Reich M T Truong and T Mai ldquoe split feasibilityproblem with multiple output sets in Hilbert spacesrdquo Opti-mization Letters vol 36 2020

[14] S Suantai N Pholasa N Pholasa and P Cholamjiak ldquoemodified inertial relaxed CQ algorithm for solving the splitfeasibility problemsrdquo Journal of Industrial and ManagementOptimization vol 14 no 4 pp 1595ndash1615 2018

[15] W Cholamjiak P Cholamjiak and S Suantai ldquoAn inertialforward-backward splitting method for solving inclusionproblems in Hilbert spacesrdquo Journal of Fixed Point=eory andApplications vol 20 no 1 p 42 2018

[16] Q L Dong Y J Cho and L L Zhong ldquoInertial projectionand contraction algorithms for variational inequalitiesrdquoJournal of Global Optimization vol 70 no 3 pp 1ndash18 2018

[17] D R Sahu Y J Cho and Q L Dong ldquoInertial relaxed CQalgorithms for solving a split feasibility problem in Hilbertspacesrdquo Numerical Algorithms vol 3 2020

[18] Y Shehu and A Gibali ldquoInertial krasnoselskiindashmann methodin banach spacesrdquo Mathematics vol 8 no 4 2020

[19] Y Shehu P T Vuong and P Cholamjiak ldquoA self-adaptiveprojection method with an inertial technique for split feasi-bility problems in Banach spaces with applications to imagerestoration problemsrdquo Journal of Fixed Point =eory andApplications vol 21 no 2 2019

[20] H-K Xu ldquoIterative methods for the split feasibility problemin infinite-dimensional Hilbert spacesrdquo Inverse Problemsvol 26 no 10 Article ID 105018 2010

[21] R I Bot and E R Csetnek ldquoAn inertial Tsengrsquos type proximalalgorithm for nonsmooth and nonconvex optimizationproblemsrdquo Journal of Optimization =eory and Applicationsvol 171 no 2 pp 600ndash616 2016

12 Journal of Mathematics

Next we show zk1113864 1113865 is bounded Since p isin Γ sub Ck and theprojection operator PCk

is nonexpansive we obtain fromAlgorithm 1 and (48) that

xk+1

minus p PCkβku + δkz

k+ εk z

kminus τknablagk z

k1113872 11138731113872 11138731113872 1113873 minus p

le βku minus p + δkzk

minus p + εksk

minus p

le βku minus p + δkzk

minus p + εkzk

minus p

βku minus p + 1 minus βk( 1113857zk

minus p

βku minus p + 1 minus βk( 1113857xk

+ θk xk

minus xkminus 1

1113872 1113873 minus p

le βku minus p + 1 minus βk( 1113857xk

minus p + θk xk

minus xkminus 1

1 minus βk( 1113857 xk

minus p

+ βk u minus p +θk

βk

xk

minus xkminus 1

1113890 1113891

(49)

Condition (A4) implies the sequence θkβkxk minus xkminus 1 isbounded Let an upper bound of u minus p + θkβkxk minus xkminus 11113864 1113865 beM en we rewrite (49) as

xk+1

minus plemax xk

minus p M1113966 1113967 (50)

By induction we have that

xk+1

minus plemax x1

minus p M1113966 1113967 (51)

Hence xk minus p1113864 1113865 is bounded so is zk minus p1113864 1113865 and sk minus p1113864 1113865And zk1113864 1113865 is bounded Consequently sk1113864 1113865 and Biz

k1113864 1113865N

i1 arealso bounded The rest of the proof will be

divided into two parts

Case 1 Suppose that there exists k0 isin N such thatxk minus p21113864 1113865

infinkk0

is nonincreasing en xk minus p21113864 1113865infink1 converges

and xk minus p2 minus xk+1 minus p2⟶ 0 as k⟶infin From (47) weobtain

ρk 4 minus ρk( 1113857g2k z

k1113872 1113873

nablagk zk

1113872 11138732 le z

kminus p

2minus s

kminus p

2 (52)

Since p isin Γ sub Ck and the projection operator PCkis

nonexpansive from Algorithm 1 we obtain

xk+1

minus p2

PCkβku + δkz

k+ εksk1113872 1113873 minus p

2

le βku + δkzk

+ εksk1113872 1113873 minus p2

le βku minus p2

+ δkzk

minus p2

+ εksk minus p2

(53)

Since βk + δk + εk 1 from (53) and using the inequalityx + y2 lex2 + 2y x + y we have the following estimation

zk

minus p2

minus sk

minus p2 le

βk

εk

u minus p2

+1 minus βk

εk

zk

minus p2

minus1εk

xk+1

minus p2

βk

εk

u minus p2

minusβk

εk

zk

minus p2

+1εk

zk

minus p2

minus xk+1

minus p2

1113960 1113961

leβk

εk

u minus p2

+1εk

zk

minus p2

minus xk+1

minus p2

1113960 1113961

1εk

βku minus p2

+ zk

minus p2

minus xk+1

minus p2

1113960 11139611113960 1113961

1εk

βku minus p2

+ xk

minus p + θk xk

minus xkminus 1

1113872 11138732

minus xk+1

minus p2

1113876 11138771113876 1113877

le1εk

βku minus p2

+ xk

minus p2

+ 2θkxk

minus xkminus 1

zk

minus p minus xk+1

minus p2

1113960 11139611113960 1113961

le1εk

βku minus p2

+ xk

minus p2

+ 2θkxk

minus xkminus 1

zk

minus p minus xk+1

minus p2

1113960 11139611113960 1113961

1εk

βku minus p2

+ xk

minus p2

minus xk+1

minus p2

+ 2θkxk

minus xkminus 1

zk

minus p1113960 11139611113960 1113961

(54)

Combining (52) and (54) we obtain

6 Journal of Mathematics

ρk 4 minus ρk( 1113857g2k z

k1113872 1113873

nablagk zk

1113872 11138732 le z

kminus p

2minus s

kminus p

2

le1εk

βku minus p2

+ xk

minus p2

minus xk+1

minus p2

11139601113960

+ 2θkxk

minus xkminus 1

zk

minus p11139611113961

(55)

By condition (A4) βk isin (0 1) imply the sequencelimk⟶infinθkxk minus xkminus 1⟶ 0 and we have proved zk minus p1113864 1113865 isbounded so 2θkxk minus xkminus 1zk minus p⟶ 0 By conditions (A2)and (A3) and (55) we have as k⟶infin

0lt ρk 4 minus ρk( 1113857g2k z

k1113872 1113873

nablagk zk

1113872 11138732

le1εk

βku minus p2

+ xk

minus p2

minus xk+1

minus p2

11139601113960

+ 2θkxk

minus xkminus 1

zk

minus p11139611113961⟶ 0

(56)

which implies that

limk⟶infin

g2k zk

1113872 1113873

nablagk zk

1113872 11138732 0 (57)

We note that for each i 1 N BTi (I minus PQk

i)Bi() is

Lipschitz continuous Since the sequence zk1113864 1113865 is boundedand

BTi I minus PQk

i1113874 1113875 Biz

k B

Ti I minus PQk

i1113874 1113875Biz

k

minus BTi I minus PQk

i1113874 1113875Biple max

1leileNBi1113874 1113875z

kminus p

(58)

for all i 1 N we have the sequence

BTi (I minus PQk

i)Biz

k1113882 1113883infin

i1is bounded Hence nablagk(zk)1113864 1113865

infink1 is

bounded Consequently we have from (57) that

limk⟶infin

I minus PQki

1113874 1113875Bizk

0 (59)

For

zk

minus xk

xk

+ θk xk

minus xkminus 1

1113872 1113873 minus xk

θkxk

minus xkminus 1

(60)

which implies

zk

minus xk⟶ 0 as k⟶infin (61)

for each i 1 N since sk zk minus τknablagk(zk) we havefrom (59) and (61) that

sk

minus zk

τknablagk zk

1113872 1113873⟶ 0 as k⟶infin

sk

minus xk le s

kminus z

k+ z

kminus x

k⟶ 0 as k⟶infin(62)

at is

limk⟶infin

sk

minus xk

0 (63)

limk⟶infin

BTi I minus PQk

i1113874 1113875Biz

k 0 (64)

Let

lk

βku + δkzk

+ εksk

βku + 1 minus βk( 1113857vk (65)

where

vk

δk

1 minus βk

zk

+εk

1 minus βk

sk (66)

is gives

vk

minus zk le

εk

1 minus βk

sk

minus zk⟶ 0 k⟶infin

lk

minus vk

le βk u minus vk

⟶ 0 k⟶infin

(67)

lk

minus zk le l

kminus v

k+ v

kminus z

k⟶ 0 k⟶infin

lk

minus xk le l

kminus z

k+ z

kminus x

k⟶ 0 as k⟶infin(68)

Hence

limk⟶infin

lk

minus zk

limk⟶infin

lk

minus xk

limk⟶infin

lk

minus vk

limk⟶infin

vk

minus zk

0

(69)

Since xk1113864 1113865 is bounded there exists a subsequence xkj1113966 1113967

of xk1113864 1113865 such that xkj1113966 1113967plowast isin ωω(xk) Next we need toshow plowast isin C and Bip

lowast isin Qi for each i 1 NWe first show that xk+1 minus xk⟶ 0 For p isin Γ

12x

k+1minus p

212x

k+1minus x

k+ x

kminus p

2

12x

kminus p

2+ x

k+1minus x

k x

k+1minus p minus

12x

k+1minus x

k2

12x

kminus p

2+ x

k+1minus l

k+ l

kminus x

k x

k+1

minus p minus12x

k+1minus x

k2

12x

kminus p

2+ x

k+1minus l

k x

k+1minus p + l

k

minus xk x

k+1minus p minus

12x

k+1minus x

k2

(70)

then

12x

k+1minus x

k2 le12x

kminus p

2minus12x

k+1minus p

2+ x

k+1minus l

k x

k+1

minus p + lk

minus xk x

k+1minus p

(71)

from xk+1 minus lk xk+1 minus plt 0

Journal of Mathematics 7

12x

k+1minus x

k2 le12x

kminus p

2minus12x

k+1minus p

2+ l

kminus x

k x

k+1minus p

(72)

Since xk minus p2 minus xk+1 minus p2⟶ 0 as k⟶infinlim

k⟶infinlk minus xk 0 we obtain

xk+1

minus xk⟶ 0 as k⟶infin (73)

Since xkj+1 isin Ckj we obtain

c xkj1113872 1113873 + ξkj x

kj+1minus x

kj le 0 (74)

us

c xkj1113872 1113873le minus ξkj x

kj+1minus x

kj le ξxkj+1

minus xkj (75)

where ξ satisfies ξk le ξ for all k By virtue of the continuity offunction c and xkj+1 minus xkj⟶ 0 we get that

c plowast

( 1113857 limj⟶infin

c xkj1113872 1113873le 0 (76)

erefore plowast isin CNow we show that Bip

lowast isin Qi to do this let hki Biz

k minus

PQkiBiz

k⟶ 0 and let ηki be such that ηk

i le η for all k SinceBiz

kj minus hkj

i PQ

kj

i

(Bizkj ) isin Q

kj

i we have

qi Bixkj1113872 1113873 + ηkj

i Bizkj minus h

kj

i1113874 1113875 minus Bixkj le 0 (77)

Hence

qi Bixkj1113872 1113873le η

kj

i Bixkj minus Biz

kj + ηkj

i hkj

i le ηBixkj

minus xkjminus 1

+ ηhkj

i ⟶ 0

(78)

By the continuity of qi and Bixkj⟶ Bix

lowast we arrive atthe conclusion

qi Biplowast

( 1113857 limj⟶infin

qi Bixkj1113872 1113873le 0 (79)

namely Biplowast isin Qi for all i 1 N Hence plowast isin Γ

Moreover for p PΓ u we can see that

limsupk⟶infin

lk

minus p u minus p limk⟶infin

lkj minus p u minus plep

lowast

minus p u minus ple 0

(80)

By Lemma 1 and (65) we have

xk+1

minus p2

PC lk

1113872 1113873 minus p2 le l

kminus p

2 βku + 1 minus βk( 1113857v

kminus p l

kminus p

βku minus p lk

minus p + 1 minus βk( 1113857vk

minus p lk

minus p

le βku minus p lk

minus p + 1 minus βk( 1113857vk

minus plk

minus p

1 minus βk( 1113857 xk

+ θk xk

minus xkminus 1

1113872 1113873 minus p

2

+ βklangu minus p lk

minus prang +βk

2u minus p

2

le 1 minus βk( 1113857 xk

minus p

2

+ 2θklangxk

minus xkminus 1

zk

minus prang1113876 1113877 + βklangu minus p lk

minus prang +βk

2u minus p

2

le 1 minus βk( 1113857 xk

minus p

2

+ 2θk xk

minus xkminus 1

zk

minus p

1113876 1113877 + βklangu minus p lk

minus prang +βk

2u minus p

2

(81)

By conditions (A1) and (A4) βk isin (0 1) imply the se-quence lim

k⟶infinθkxk minus xkminus 1⟶ 0 βk2u minus p⟶ 0 and

using (80) in (81) and applying Lemma 5 we obtain

limk⟶infin

xk

minus p2

0 (82)

erefore as k⟶infin xk⟶ p PΓu

Case 2 SetΛk xk minus p2 Assume that Λk1113864 1113865 is not decreasingat infinity Let ϕ N⟶ N be a mapping for all kge k0 (forsome k0 large enough) defined by

ϕ(k) max t isin N tle kΛt leΛt+11113864 1113865 (83)

By Lemma 4 ϕ(k)1113864 1113865infinkk0is a nondecreasing sequence

such that ϕ(k)⟶infin as k⟶infin and

max Λϕ(k)Λk1113966 1113967leΛϕ(k)+1forallkge k0 (84)

After a similar conclusion from (59) it is easy to see that

limk⟶infin

I minus PQ

ϕ(k)

i

1113874 1113875Bizϕ(k)

0 (85)

By the similar argument as the above in Case 1 weconclude immediately that

8 Journal of Mathematics

limk⟶infin

Blowasti I minus P

Qϕ(k)

i

1113874 1113875Bizϕ(k)

0 (86)

limsupk⟶infin

lϕ(k)

minus p u minus ple 0 (87)

Since xϕ(k)1113864 1113865 is bounded there exists a subsequence ofxϕ(k)1113864 1113865 still denoted by xϕ(k)1113864 1113865 which converges weakly to

plowast By similar argument as above in Case 1 we concludeimmediately that plowast isin C and Bip

lowast isin QirArrplowast isin ΓFrom (57) we have that

xϕ(k)+1

minus p2 le 1 minus βϕ(k)1113872 1113873z

ϕ(k)minus p

2+ βϕ(k)u minus p l

ϕ(k)minus p

le 1 minus βϕ(k)1113872 1113873xϕ(k)

+ θϕ(k) xϕ(k)

minus xϕ(k)minus 1

1113872 1113873 minus p2

+ βϕ(k)u minus p lϕ(k)

minus p

le 1 minus βϕ(k)1113872 1113873 xϕ(k)

minus p2

+ 2θϕ(k)xϕ(k)

minus xϕ(k)minus 1

zϕ(k)

minus p1113960 1113961 + βϕ(k)u minus p lϕ(k)

minus p

le 1 minus βϕ(k)1113872 1113873 xϕ(k)

minus p2

+ 2θϕ(k)xϕ(k)

minus xϕ(k)minus 1

zϕ(k)

minus p1113960 1113961 + βϕ(k)u minus p lϕ(k)

minus p

(88)

which implies by Lemma 5

limk⟶infin

xφ(k)

minus p

2

0 and limk⟶infin

xϕ(k)+1

minus p2

0 (89)

Moreover for kge k0 it is easy to see thatΛϕ(k) minus Λϕ(k)+1 le 0 if kneϕ(k) (that is kge ϕ(k)) because forϕ(k) + 1lemle n Λm gtΛm+1 As a consequence we obtain

0leΛk lemax Λϕ(k)Λϕ(k)+11113966 1113967 Λϕ(k)+1 forallkge k0 (90)

erefore we obtain limk⟶infinΛk 0 that is xk1113864 1113865 converges

strongly to p is completes the proof

4 Some Extensive Results

For the SFPwMOS (11) when N 1 it becomes the SFP(1) us we have the following corollary for solving the SFP(1) which is an immediate consequence of eorem 1

Corollary 1 Let H1 and H2 be two real Hilbert spaces andlet B H1⟶ H2 be bounded linear operator Let C and Q

be nonempty closed and convex subsets of H1 and H2 re-spectively Assume that Ω Ccap Bminus 1(Q)ne 0 Let u isin H1 be afixed point For any starting point x0 x1 isin H1 let xk1113864 1113865 be thesequence generated by

zk

xk

+ θk xk

minus xkminus 1

1113872 1113873

xk+1 ≔ PCk

βku + δkzk

+ εk zk

minus τknablagk zk

1113872 11138731113872 11138731113872 1113873

⎧⎪⎨

⎪⎩(91)

where θk Ck τk and nablafk are given as in (8) Suppose thesequences βk1113864 1113865 δk1113864 1113865 and εk1113864 1113865 satisfy the conditions in =e-orem 1 =en the sequence xk1113864 1113865 converges strongly to thesolution p isin Ω where p PΩ(u)

When we take u x0 in Algorithm 1 we note also thefollowing results regarding to the SFPwMOS (11)

Corollary 2 Let H Hi i 1 N be real Hilbert spacesand let Bi H⟶ Hi i 1 N be bounded linearoperators Let C and Qi i 1 N be nonempty closedand convex subsets of H and Hi i 1 N respectivelyAssume that problem (11) is consistent For any initial guessx0 x1 isin H let xk1113864 1113865 be the sequence generated by

zk

xk

+ θk xk

minus xkminus 1

1113872 1113873

xk+1 ≔ PCk

βkx0

+ δkzk

+ εk zk

minus τknablagk zk

1113872 11138731113872 11138731113872 1113873

⎧⎪⎨

⎪⎩(92)

whereθk Ck τk and nablagk are given by (36) and (39) andAlgorithm 1 respectively Suppose the sequences βk1113864 1113865 δk1113864 1113865and εk1113864 1113865 satisfy the conditions in =eorem 1 =en the se-quence xk1113864 1113865 generated by (92) strongly converges to the so-lution p PΩ(x0) isin Γ

When we take δk equiv 0 in Algorithm 1 we obtain thefollowing result regarding the SFPwMOS (11)

Corollary 3 Let H Hi i 1 N be real Hilbertspaces and let Ti H⟶ Hi i 1 N be boundedlinear operators Let C and Qi i 1 N be nonemptyclosed and convex subsets of H and Hi i 1 N re-spectively Assume that problem (11) is consistent For a fixedpoint u isin H and any initial guess x0 isin H let xk1113864 1113865 be thesequence generated by

zk

xk

+ θk xk

minus xkminus 1

1113872 1113873

xk+1 ≔ PCk

βku + 1 minus βk( 1113857 zk

minus τknablagk zk

1113872 11138731113872 11138731113872 1113873

⎧⎪⎨

⎪⎩(93)

where θk Ck τk and nablagk are given by (36) and (39) andAlgorithm 1 respectively Suppose the sequences satisfyconditions (A1) =en the sequence xk1113864 1113865 generated by (93)strongly converges to the solution p PΩ(u) isin Γ

Of course when we take u x0 we get the followingresult regarding the SFPwMOS (11)

Corollary 4 Let H Hi i 1 N be real Hilbert spacesand let Bi H⟶ Hi i 1 N be bounded linear op-erators Let C and Qi i 1 N be nonempty closed andconvex subsets of H and Hi i 1 N respectively As-sume that problem (11) is consistent For any initial guessx0 isin H let xk1113864 1113865 be the sequence generated by

zk

xk

+ θk xk

minus xkminus 1

1113872 1113873

xk+1 ≔ PCk

βkx0

+ 1 minus tarts minus with(primehskip

prime)]))hskipsubstring minus after(preceding minus sibling comment()[starts minus with(prime hskipprime)]prime hskipprime)pt gt βk) z

kminus τknablagk z

k1113872 11138731113872 1113873)11138741113874

⎧⎪⎪⎨

⎪⎪⎩

(94)

Journal of Mathematics 9

where θk Ck τk and nablagk are given by (36) and (39) andAlgorithm 1 respectively Suppose the sequence βk1113864 1113865 satisfiesconditions (A1) =en the sequence xk1113864 1113865 generated by (94)strongly converges to the solution p PΩ(u) isin Γ

Remark 1 In Corollary 4 for the particular casewhereN 1 the iterative scheme (94) reduced exactly toiterative scheme proposed by He et al in eorem 32 of[10]

5 Numerical Experiment

In this section we provide one numerical experiment toillustrate the implementation and efficiency of our proposedmethod First we study the behavior of Algorithm 1 fordifferent choices of θkand ρk Next we compare Algorithm 1(θk 08) with no inertial Algorithm 1 (θk 0) and Scheme(17) for different initial points e numerical results arecompleted on a standard TOSHIBA laptop with Intel(R)Core(TM) i5-2450M CPU25GHz 25GHz with memory4GB e code is implemented in MATLAB R2020a

Let

A1

minus 04 minus 02 minus 02

04 05 minus 01

02 minus 05 03

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

A2

03 minus 01 03

minus 03 02 minus 02

0 02 03

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

C x isin R3|x1 minus x

22 + 2x3 le 01113966 1113967

Q1 x isin R3

|x21 + x2 minus x3 le 01113966 1113967

Q2 x isin R3|x1 + x

22 minus x3 le 01113966 1113967

(95)

Find xlowast such that

xlowast isin Γ ≔ Ccap cap 2i1A

minus 1i Qi( 11138571113872 1113873ne 0 (96)

Define the error function as

Ek 13x

kminus PCk

xk

1113872 11138732

+13A1x

kminus PQ1k

A1xk

1113872 11138732

+13A2x

kminus PQ2k

A2xk

1113872 11138732

(97)

Note that if Ek 0 at the kth step then xk isin ΓFor Algorithm 1 for different choices of ρk and θk we

choose βk 19k + 1 δk k9k + 1 εk 8k9k + 1u (01 01 01)T and x0 (minus 01 minus 01 minus 01)T andx1 [03 05 minus 06]T Using Ek lt ε as stopping criteriawhere ε is a small enough positive number the results of thenumerical experiment are reported in Table 1e numericalresults can be seen from Table 1 and Figures 1 and 2 InTable 1 ldquoIterrdquo denotes the number of iterations

e behavior of the function Ek in Table 1 is described inFigures 1 and 2

From Table 1 and Figures 1 and 2 we can observe thatAlgorithm 1 for inertial effect case has less number of it-erations than Algorithm 1 for θk 0(that is no inertialeffect) case Algorithm 1 with bigger θk has less number ofiterations than Algorithm 1 for small case for the same θkbig ρk case has less number of iterations than Algorithm 1for small case

Table 1 Algorithm 1 for different choices of θk ε and ρk

ε θkρk 398 ρk 300

Iter Ek Iter Ek

10minus 3θk 08 5 70906e minus 05

6 58048e minus 05θk 03 7 69397e minus 04θk 0 9 75095e minus 04

10minus 4θk 08 5 70906e minus 05

6 58048e minus 05θk 03 11 91282e minus 05θk 0 15 96150e minus 05

10minus 5θk 08 12 44752e minus 06

14 25296e minus 06θk 03 16 91012e minus 06θk 0 24 88064e minus 06

0 5 10 15 20 25The number of Iteration (k)

0

001

002

003

004

005

006

E k

θk=08θk=03θk=00

Figure 1 Error against iterations for Algorithm 1 with differencechoices of θk under ρk 398 and ε 10minus 5

10 Journal of Mathematics

Following we compare our algorithmwith iteration (16)For the example iteration (16) can be written as

xk+1 ≔ αkf x

k1113872 1113873 + 1 minus αk( 11138571113857PC x

kminus λk B

lowast1 I minus PQ11113872 11138731113872 1113873B1x

k1113872

+ Blowast2 I minus PQ21113872 11138731113872 1113873B2x

k1113873

(98)In the process we take αk 08 and f(xk)

u (111)Tλk 000005 for Scheme (97) We choose βk

12k + 1 δk εk k2k + 1 and ρk 3for Algorithm 1From Table 2 and Figure 3 we can observe that Algo-

rithm 1 has less number of iterations than Scheme (97)Moreover our iterative method is advantageous overScheme (98) because Algorithm 1 does not require com-puting of the operator norm

Data Availability

No data were used to support this study

Conflicts of Interest

e authors declare that they have no conflicts of interest

Authorsrsquo Contributions

is entire work has been completed by the authors HuijuanJia Shufen Liu and Yazheng Dang e authors read andapproved the finial manuscript

0 5 10 15e number of Iteration (k)

ρk=398ρk=300

0

001

002

003

004

005

006

007

E k

Figure 2 Error against iterations for Algorithm 1 with difference choices of ρk under θk 08 and ε 10minus 5

Table 2 Comparsion of Algorithm 1 and Scheme (97) for different choice initial point for ε 10minus 4

Initiative point x1 Algorithm 1 with θk 08 Algorithm 1 with θk 00 (98)Iter Ek Iter Ek Iter Ek

[04 09ndash09] 11 41830e minus 06 25 96999e minus 06 51 88226e minus 06[09 03ndash05] 6 14290e minus 07 22 86399e minus 06 26 96719e minus 06[minus 07 03ndash02] 12 78838e minus 06 24 92013e minus 06 48 84198e minus 06[01 01ndash09]prime 12 15353e minus 07 26 87544e minus 06 52 99722e minus 06

0 10 20 30 5040e number of Iteration (k)

Algorithm 1 with θ=08Algorithm 1 with θ=00(52)

0

004

002

006

008

01

012

014

E k

Figure 3 Error against of iterations for the comparision of Al-gorithm 1 and Scheme (97)

Journal of Mathematics 11

Acknowledgments

is research was supported by the National Natural ScienceFoundation of China (Grant no 61872126) Henan ProvinceKey Science and Technology Project (Grant no192102210123) and Young Backbone Teachers in Univer-sities of Henan Province (Grant no 2019GGJS061)

References

[1] Y Censor and T Elfving ldquoA multiprojection algorithm usingbregman projections in a product spacerdquo Numerical Algo-rithms vol 8 no 2 pp 221ndash239 1994

[2] C Byrne ldquoIterative oblique projection onto convex sets andthe split feasibility problemrdquo Inverse Problems vol 18 no 2pp 441ndash453 2002

[3] G T Herman ldquoImage reconstruction from projections thefundamentals of computerized tomographyrdquo 1980

[4] Y Censor and A Segal ldquoIterative projection methods inbiomedical inverse problemsrdquo 2008

[5] Y Censor T Bortfeld B Martin and A Trofimov ldquoA unifiedapproach for inversion problems in intensity-modulated ra-diation therapyrdquo Physics in Medicine and Biology vol 51no 10 pp 2353ndash2365 2006

[6] C Yair and Tommy ldquoe multiple-sets split feasibilityproblem and its applications for inverse problemsrdquo InverseProblems vol 21 no 6 pp 2071ndash2084 2005

[7] Q Yang ldquoe relaxed CQ algorithm solving the split feasi-bility problemrdquo Inverse Problems vol 20 no 4pp 1261ndash1266 2004

[8] G Lopez V Martın-Marquez and F H Wang ldquoSolving thesplit feasibility problem without prior knowledge of matrixnormsrdquo Inverse Problems vol 28 no 8 pp 374ndash389 2012

[9] J Deepho and P Kumam ldquoA modified halpernrsquos iterativescheme for solving split feasibility problemsrdquo Abstract andApplied Analysis vol 2012 Article ID 876069 8 pages 2012

[10] Y Dang and Y Gao ldquoe strong convergence of a KM-CQ-like algorithm for a split feasibility problemrdquo Inverse Prob-lems vol 27 no 1 Article ID 015007 2011

[11] S N He and Z Y Zhao ldquoStrong convergence of a relaxed CQalgorithm for the split feasibility problemrdquo Journal of In-equalities and Applications vol 2013 no 1 11 pages 2013

[12] Y H Yao and Mihai Strong Convergence of a Self-AdaptiveMethod for the Split Feasibility Problem Fixed Point =eory ampApplications Springer Berlin Germany 2013

[13] S Reich M T Truong and T Mai ldquoe split feasibilityproblem with multiple output sets in Hilbert spacesrdquo Opti-mization Letters vol 36 2020

[14] S Suantai N Pholasa N Pholasa and P Cholamjiak ldquoemodified inertial relaxed CQ algorithm for solving the splitfeasibility problemsrdquo Journal of Industrial and ManagementOptimization vol 14 no 4 pp 1595ndash1615 2018

[15] W Cholamjiak P Cholamjiak and S Suantai ldquoAn inertialforward-backward splitting method for solving inclusionproblems in Hilbert spacesrdquo Journal of Fixed Point=eory andApplications vol 20 no 1 p 42 2018

[16] Q L Dong Y J Cho and L L Zhong ldquoInertial projectionand contraction algorithms for variational inequalitiesrdquoJournal of Global Optimization vol 70 no 3 pp 1ndash18 2018

[17] D R Sahu Y J Cho and Q L Dong ldquoInertial relaxed CQalgorithms for solving a split feasibility problem in Hilbertspacesrdquo Numerical Algorithms vol 3 2020

[18] Y Shehu and A Gibali ldquoInertial krasnoselskiindashmann methodin banach spacesrdquo Mathematics vol 8 no 4 2020

[19] Y Shehu P T Vuong and P Cholamjiak ldquoA self-adaptiveprojection method with an inertial technique for split feasi-bility problems in Banach spaces with applications to imagerestoration problemsrdquo Journal of Fixed Point =eory andApplications vol 21 no 2 2019

[20] H-K Xu ldquoIterative methods for the split feasibility problemin infinite-dimensional Hilbert spacesrdquo Inverse Problemsvol 26 no 10 Article ID 105018 2010

[21] R I Bot and E R Csetnek ldquoAn inertial Tsengrsquos type proximalalgorithm for nonsmooth and nonconvex optimizationproblemsrdquo Journal of Optimization =eory and Applicationsvol 171 no 2 pp 600ndash616 2016

12 Journal of Mathematics

ρk 4 minus ρk( 1113857g2k z

k1113872 1113873

nablagk zk

1113872 11138732 le z

kminus p

2minus s

kminus p

2

le1εk

βku minus p2

+ xk

minus p2

minus xk+1

minus p2

11139601113960

+ 2θkxk

minus xkminus 1

zk

minus p11139611113961

(55)

By condition (A4) βk isin (0 1) imply the sequencelimk⟶infinθkxk minus xkminus 1⟶ 0 and we have proved zk minus p1113864 1113865 isbounded so 2θkxk minus xkminus 1zk minus p⟶ 0 By conditions (A2)and (A3) and (55) we have as k⟶infin

0lt ρk 4 minus ρk( 1113857g2k z

k1113872 1113873

nablagk zk

1113872 11138732

le1εk

βku minus p2

+ xk

minus p2

minus xk+1

minus p2

11139601113960

+ 2θkxk

minus xkminus 1

zk

minus p11139611113961⟶ 0

(56)

which implies that

limk⟶infin

g2k zk

1113872 1113873

nablagk zk

1113872 11138732 0 (57)

We note that for each i 1 N BTi (I minus PQk

i)Bi() is

Lipschitz continuous Since the sequence zk1113864 1113865 is boundedand

BTi I minus PQk

i1113874 1113875 Biz

k B

Ti I minus PQk

i1113874 1113875Biz

k

minus BTi I minus PQk

i1113874 1113875Biple max

1leileNBi1113874 1113875z

kminus p

(58)

for all i 1 N we have the sequence

BTi (I minus PQk

i)Biz

k1113882 1113883infin

i1is bounded Hence nablagk(zk)1113864 1113865

infink1 is

bounded Consequently we have from (57) that

limk⟶infin

I minus PQki

1113874 1113875Bizk

0 (59)

For

zk

minus xk

xk

+ θk xk

minus xkminus 1

1113872 1113873 minus xk

θkxk

minus xkminus 1

(60)

which implies

zk

minus xk⟶ 0 as k⟶infin (61)

for each i 1 N since sk zk minus τknablagk(zk) we havefrom (59) and (61) that

sk

minus zk

τknablagk zk

1113872 1113873⟶ 0 as k⟶infin

sk

minus xk le s

kminus z

k+ z

kminus x

k⟶ 0 as k⟶infin(62)

at is

limk⟶infin

sk

minus xk

0 (63)

limk⟶infin

BTi I minus PQk

i1113874 1113875Biz

k 0 (64)

Let

lk

βku + δkzk

+ εksk

βku + 1 minus βk( 1113857vk (65)

where

vk

δk

1 minus βk

zk

+εk

1 minus βk

sk (66)

is gives

vk

minus zk le

εk

1 minus βk

sk

minus zk⟶ 0 k⟶infin

lk

minus vk

le βk u minus vk

⟶ 0 k⟶infin

(67)

lk

minus zk le l

kminus v

k+ v

kminus z

k⟶ 0 k⟶infin

lk

minus xk le l

kminus z

k+ z

kminus x

k⟶ 0 as k⟶infin(68)

Hence

limk⟶infin

lk

minus zk

limk⟶infin

lk

minus xk

limk⟶infin

lk

minus vk

limk⟶infin

vk

minus zk

0

(69)

Since xk1113864 1113865 is bounded there exists a subsequence xkj1113966 1113967

of xk1113864 1113865 such that xkj1113966 1113967plowast isin ωω(xk) Next we need toshow plowast isin C and Bip

lowast isin Qi for each i 1 NWe first show that xk+1 minus xk⟶ 0 For p isin Γ

12x

k+1minus p

212x

k+1minus x

k+ x

kminus p

2

12x

kminus p

2+ x

k+1minus x

k x

k+1minus p minus

12x

k+1minus x

k2

12x

kminus p

2+ x

k+1minus l

k+ l

kminus x

k x

k+1

minus p minus12x

k+1minus x

k2

12x

kminus p

2+ x

k+1minus l

k x

k+1minus p + l

k

minus xk x

k+1minus p minus

12x

k+1minus x

k2

(70)

then

12x

k+1minus x

k2 le12x

kminus p

2minus12x

k+1minus p

2+ x

k+1minus l

k x

k+1

minus p + lk

minus xk x

k+1minus p

(71)

from xk+1 minus lk xk+1 minus plt 0

Journal of Mathematics 7

12x

k+1minus x

k2 le12x

kminus p

2minus12x

k+1minus p

2+ l

kminus x

k x

k+1minus p

(72)

Since xk minus p2 minus xk+1 minus p2⟶ 0 as k⟶infinlim

k⟶infinlk minus xk 0 we obtain

xk+1

minus xk⟶ 0 as k⟶infin (73)

Since xkj+1 isin Ckj we obtain

c xkj1113872 1113873 + ξkj x

kj+1minus x

kj le 0 (74)

us

c xkj1113872 1113873le minus ξkj x

kj+1minus x

kj le ξxkj+1

minus xkj (75)

where ξ satisfies ξk le ξ for all k By virtue of the continuity offunction c and xkj+1 minus xkj⟶ 0 we get that

c plowast

( 1113857 limj⟶infin

c xkj1113872 1113873le 0 (76)

erefore plowast isin CNow we show that Bip

lowast isin Qi to do this let hki Biz

k minus

PQkiBiz

k⟶ 0 and let ηki be such that ηk

i le η for all k SinceBiz

kj minus hkj

i PQ

kj

i

(Bizkj ) isin Q

kj

i we have

qi Bixkj1113872 1113873 + ηkj

i Bizkj minus h

kj

i1113874 1113875 minus Bixkj le 0 (77)

Hence

qi Bixkj1113872 1113873le η

kj

i Bixkj minus Biz

kj + ηkj

i hkj

i le ηBixkj

minus xkjminus 1

+ ηhkj

i ⟶ 0

(78)

By the continuity of qi and Bixkj⟶ Bix

lowast we arrive atthe conclusion

qi Biplowast

( 1113857 limj⟶infin

qi Bixkj1113872 1113873le 0 (79)

namely Biplowast isin Qi for all i 1 N Hence plowast isin Γ

Moreover for p PΓ u we can see that

limsupk⟶infin

lk

minus p u minus p limk⟶infin

lkj minus p u minus plep

lowast

minus p u minus ple 0

(80)

By Lemma 1 and (65) we have

xk+1

minus p2

PC lk

1113872 1113873 minus p2 le l

kminus p

2 βku + 1 minus βk( 1113857v

kminus p l

kminus p

βku minus p lk

minus p + 1 minus βk( 1113857vk

minus p lk

minus p

le βku minus p lk

minus p + 1 minus βk( 1113857vk

minus plk

minus p

1 minus βk( 1113857 xk

+ θk xk

minus xkminus 1

1113872 1113873 minus p

2

+ βklangu minus p lk

minus prang +βk

2u minus p

2

le 1 minus βk( 1113857 xk

minus p

2

+ 2θklangxk

minus xkminus 1

zk

minus prang1113876 1113877 + βklangu minus p lk

minus prang +βk

2u minus p

2

le 1 minus βk( 1113857 xk

minus p

2

+ 2θk xk

minus xkminus 1

zk

minus p

1113876 1113877 + βklangu minus p lk

minus prang +βk

2u minus p

2

(81)

By conditions (A1) and (A4) βk isin (0 1) imply the se-quence lim

k⟶infinθkxk minus xkminus 1⟶ 0 βk2u minus p⟶ 0 and

using (80) in (81) and applying Lemma 5 we obtain

limk⟶infin

xk

minus p2

0 (82)

erefore as k⟶infin xk⟶ p PΓu

Case 2 SetΛk xk minus p2 Assume that Λk1113864 1113865 is not decreasingat infinity Let ϕ N⟶ N be a mapping for all kge k0 (forsome k0 large enough) defined by

ϕ(k) max t isin N tle kΛt leΛt+11113864 1113865 (83)

By Lemma 4 ϕ(k)1113864 1113865infinkk0is a nondecreasing sequence

such that ϕ(k)⟶infin as k⟶infin and

max Λϕ(k)Λk1113966 1113967leΛϕ(k)+1forallkge k0 (84)

After a similar conclusion from (59) it is easy to see that

limk⟶infin

I minus PQ

ϕ(k)

i

1113874 1113875Bizϕ(k)

0 (85)

By the similar argument as the above in Case 1 weconclude immediately that

8 Journal of Mathematics

limk⟶infin

Blowasti I minus P

Qϕ(k)

i

1113874 1113875Bizϕ(k)

0 (86)

limsupk⟶infin

lϕ(k)

minus p u minus ple 0 (87)

Since xϕ(k)1113864 1113865 is bounded there exists a subsequence ofxϕ(k)1113864 1113865 still denoted by xϕ(k)1113864 1113865 which converges weakly to

plowast By similar argument as above in Case 1 we concludeimmediately that plowast isin C and Bip

lowast isin QirArrplowast isin ΓFrom (57) we have that

xϕ(k)+1

minus p2 le 1 minus βϕ(k)1113872 1113873z

ϕ(k)minus p

2+ βϕ(k)u minus p l

ϕ(k)minus p

le 1 minus βϕ(k)1113872 1113873xϕ(k)

+ θϕ(k) xϕ(k)

minus xϕ(k)minus 1

1113872 1113873 minus p2

+ βϕ(k)u minus p lϕ(k)

minus p

le 1 minus βϕ(k)1113872 1113873 xϕ(k)

minus p2

+ 2θϕ(k)xϕ(k)

minus xϕ(k)minus 1

zϕ(k)

minus p1113960 1113961 + βϕ(k)u minus p lϕ(k)

minus p

le 1 minus βϕ(k)1113872 1113873 xϕ(k)

minus p2

+ 2θϕ(k)xϕ(k)

minus xϕ(k)minus 1

zϕ(k)

minus p1113960 1113961 + βϕ(k)u minus p lϕ(k)

minus p

(88)

which implies by Lemma 5

limk⟶infin

xφ(k)

minus p

2

0 and limk⟶infin

xϕ(k)+1

minus p2

0 (89)

Moreover for kge k0 it is easy to see thatΛϕ(k) minus Λϕ(k)+1 le 0 if kneϕ(k) (that is kge ϕ(k)) because forϕ(k) + 1lemle n Λm gtΛm+1 As a consequence we obtain

0leΛk lemax Λϕ(k)Λϕ(k)+11113966 1113967 Λϕ(k)+1 forallkge k0 (90)

erefore we obtain limk⟶infinΛk 0 that is xk1113864 1113865 converges

strongly to p is completes the proof

4 Some Extensive Results

For the SFPwMOS (11) when N 1 it becomes the SFP(1) us we have the following corollary for solving the SFP(1) which is an immediate consequence of eorem 1

Corollary 1 Let H1 and H2 be two real Hilbert spaces andlet B H1⟶ H2 be bounded linear operator Let C and Q

be nonempty closed and convex subsets of H1 and H2 re-spectively Assume that Ω Ccap Bminus 1(Q)ne 0 Let u isin H1 be afixed point For any starting point x0 x1 isin H1 let xk1113864 1113865 be thesequence generated by

zk

xk

+ θk xk

minus xkminus 1

1113872 1113873

xk+1 ≔ PCk

βku + δkzk

+ εk zk

minus τknablagk zk

1113872 11138731113872 11138731113872 1113873

⎧⎪⎨

⎪⎩(91)

where θk Ck τk and nablafk are given as in (8) Suppose thesequences βk1113864 1113865 δk1113864 1113865 and εk1113864 1113865 satisfy the conditions in =e-orem 1 =en the sequence xk1113864 1113865 converges strongly to thesolution p isin Ω where p PΩ(u)

When we take u x0 in Algorithm 1 we note also thefollowing results regarding to the SFPwMOS (11)

Corollary 2 Let H Hi i 1 N be real Hilbert spacesand let Bi H⟶ Hi i 1 N be bounded linearoperators Let C and Qi i 1 N be nonempty closedand convex subsets of H and Hi i 1 N respectivelyAssume that problem (11) is consistent For any initial guessx0 x1 isin H let xk1113864 1113865 be the sequence generated by

zk

xk

+ θk xk

minus xkminus 1

1113872 1113873

xk+1 ≔ PCk

βkx0

+ δkzk

+ εk zk

minus τknablagk zk

1113872 11138731113872 11138731113872 1113873

⎧⎪⎨

⎪⎩(92)

whereθk Ck τk and nablagk are given by (36) and (39) andAlgorithm 1 respectively Suppose the sequences βk1113864 1113865 δk1113864 1113865and εk1113864 1113865 satisfy the conditions in =eorem 1 =en the se-quence xk1113864 1113865 generated by (92) strongly converges to the so-lution p PΩ(x0) isin Γ

When we take δk equiv 0 in Algorithm 1 we obtain thefollowing result regarding the SFPwMOS (11)

Corollary 3 Let H Hi i 1 N be real Hilbertspaces and let Ti H⟶ Hi i 1 N be boundedlinear operators Let C and Qi i 1 N be nonemptyclosed and convex subsets of H and Hi i 1 N re-spectively Assume that problem (11) is consistent For a fixedpoint u isin H and any initial guess x0 isin H let xk1113864 1113865 be thesequence generated by

zk

xk

+ θk xk

minus xkminus 1

1113872 1113873

xk+1 ≔ PCk

βku + 1 minus βk( 1113857 zk

minus τknablagk zk

1113872 11138731113872 11138731113872 1113873

⎧⎪⎨

⎪⎩(93)

where θk Ck τk and nablagk are given by (36) and (39) andAlgorithm 1 respectively Suppose the sequences satisfyconditions (A1) =en the sequence xk1113864 1113865 generated by (93)strongly converges to the solution p PΩ(u) isin Γ

Of course when we take u x0 we get the followingresult regarding the SFPwMOS (11)

Corollary 4 Let H Hi i 1 N be real Hilbert spacesand let Bi H⟶ Hi i 1 N be bounded linear op-erators Let C and Qi i 1 N be nonempty closed andconvex subsets of H and Hi i 1 N respectively As-sume that problem (11) is consistent For any initial guessx0 isin H let xk1113864 1113865 be the sequence generated by

zk

xk

+ θk xk

minus xkminus 1

1113872 1113873

xk+1 ≔ PCk

βkx0

+ 1 minus tarts minus with(primehskip

prime)]))hskipsubstring minus after(preceding minus sibling comment()[starts minus with(prime hskipprime)]prime hskipprime)pt gt βk) z

kminus τknablagk z

k1113872 11138731113872 1113873)11138741113874

⎧⎪⎪⎨

⎪⎪⎩

(94)

Journal of Mathematics 9

where θk Ck τk and nablagk are given by (36) and (39) andAlgorithm 1 respectively Suppose the sequence βk1113864 1113865 satisfiesconditions (A1) =en the sequence xk1113864 1113865 generated by (94)strongly converges to the solution p PΩ(u) isin Γ

Remark 1 In Corollary 4 for the particular casewhereN 1 the iterative scheme (94) reduced exactly toiterative scheme proposed by He et al in eorem 32 of[10]

5 Numerical Experiment

In this section we provide one numerical experiment toillustrate the implementation and efficiency of our proposedmethod First we study the behavior of Algorithm 1 fordifferent choices of θkand ρk Next we compare Algorithm 1(θk 08) with no inertial Algorithm 1 (θk 0) and Scheme(17) for different initial points e numerical results arecompleted on a standard TOSHIBA laptop with Intel(R)Core(TM) i5-2450M CPU25GHz 25GHz with memory4GB e code is implemented in MATLAB R2020a

Let

A1

minus 04 minus 02 minus 02

04 05 minus 01

02 minus 05 03

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

A2

03 minus 01 03

minus 03 02 minus 02

0 02 03

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

C x isin R3|x1 minus x

22 + 2x3 le 01113966 1113967

Q1 x isin R3

|x21 + x2 minus x3 le 01113966 1113967

Q2 x isin R3|x1 + x

22 minus x3 le 01113966 1113967

(95)

Find xlowast such that

xlowast isin Γ ≔ Ccap cap 2i1A

minus 1i Qi( 11138571113872 1113873ne 0 (96)

Define the error function as

Ek 13x

kminus PCk

xk

1113872 11138732

+13A1x

kminus PQ1k

A1xk

1113872 11138732

+13A2x

kminus PQ2k

A2xk

1113872 11138732

(97)

Note that if Ek 0 at the kth step then xk isin ΓFor Algorithm 1 for different choices of ρk and θk we

choose βk 19k + 1 δk k9k + 1 εk 8k9k + 1u (01 01 01)T and x0 (minus 01 minus 01 minus 01)T andx1 [03 05 minus 06]T Using Ek lt ε as stopping criteriawhere ε is a small enough positive number the results of thenumerical experiment are reported in Table 1e numericalresults can be seen from Table 1 and Figures 1 and 2 InTable 1 ldquoIterrdquo denotes the number of iterations

e behavior of the function Ek in Table 1 is described inFigures 1 and 2

From Table 1 and Figures 1 and 2 we can observe thatAlgorithm 1 for inertial effect case has less number of it-erations than Algorithm 1 for θk 0(that is no inertialeffect) case Algorithm 1 with bigger θk has less number ofiterations than Algorithm 1 for small case for the same θkbig ρk case has less number of iterations than Algorithm 1for small case

Table 1 Algorithm 1 for different choices of θk ε and ρk

ε θkρk 398 ρk 300

Iter Ek Iter Ek

10minus 3θk 08 5 70906e minus 05

6 58048e minus 05θk 03 7 69397e minus 04θk 0 9 75095e minus 04

10minus 4θk 08 5 70906e minus 05

6 58048e minus 05θk 03 11 91282e minus 05θk 0 15 96150e minus 05

10minus 5θk 08 12 44752e minus 06

14 25296e minus 06θk 03 16 91012e minus 06θk 0 24 88064e minus 06

0 5 10 15 20 25The number of Iteration (k)

0

001

002

003

004

005

006

E k

θk=08θk=03θk=00

Figure 1 Error against iterations for Algorithm 1 with differencechoices of θk under ρk 398 and ε 10minus 5

10 Journal of Mathematics

Following we compare our algorithmwith iteration (16)For the example iteration (16) can be written as

xk+1 ≔ αkf x

k1113872 1113873 + 1 minus αk( 11138571113857PC x

kminus λk B

lowast1 I minus PQ11113872 11138731113872 1113873B1x

k1113872

+ Blowast2 I minus PQ21113872 11138731113872 1113873B2x

k1113873

(98)In the process we take αk 08 and f(xk)

u (111)Tλk 000005 for Scheme (97) We choose βk

12k + 1 δk εk k2k + 1 and ρk 3for Algorithm 1From Table 2 and Figure 3 we can observe that Algo-

rithm 1 has less number of iterations than Scheme (97)Moreover our iterative method is advantageous overScheme (98) because Algorithm 1 does not require com-puting of the operator norm

Data Availability

No data were used to support this study

Conflicts of Interest

e authors declare that they have no conflicts of interest

Authorsrsquo Contributions

is entire work has been completed by the authors HuijuanJia Shufen Liu and Yazheng Dang e authors read andapproved the finial manuscript

0 5 10 15e number of Iteration (k)

ρk=398ρk=300

0

001

002

003

004

005

006

007

E k

Figure 2 Error against iterations for Algorithm 1 with difference choices of ρk under θk 08 and ε 10minus 5

Table 2 Comparsion of Algorithm 1 and Scheme (97) for different choice initial point for ε 10minus 4

Initiative point x1 Algorithm 1 with θk 08 Algorithm 1 with θk 00 (98)Iter Ek Iter Ek Iter Ek

[04 09ndash09] 11 41830e minus 06 25 96999e minus 06 51 88226e minus 06[09 03ndash05] 6 14290e minus 07 22 86399e minus 06 26 96719e minus 06[minus 07 03ndash02] 12 78838e minus 06 24 92013e minus 06 48 84198e minus 06[01 01ndash09]prime 12 15353e minus 07 26 87544e minus 06 52 99722e minus 06

0 10 20 30 5040e number of Iteration (k)

Algorithm 1 with θ=08Algorithm 1 with θ=00(52)

0

004

002

006

008

01

012

014

E k

Figure 3 Error against of iterations for the comparision of Al-gorithm 1 and Scheme (97)

Journal of Mathematics 11

Acknowledgments

is research was supported by the National Natural ScienceFoundation of China (Grant no 61872126) Henan ProvinceKey Science and Technology Project (Grant no192102210123) and Young Backbone Teachers in Univer-sities of Henan Province (Grant no 2019GGJS061)

References

[1] Y Censor and T Elfving ldquoA multiprojection algorithm usingbregman projections in a product spacerdquo Numerical Algo-rithms vol 8 no 2 pp 221ndash239 1994

[2] C Byrne ldquoIterative oblique projection onto convex sets andthe split feasibility problemrdquo Inverse Problems vol 18 no 2pp 441ndash453 2002

[3] G T Herman ldquoImage reconstruction from projections thefundamentals of computerized tomographyrdquo 1980

[4] Y Censor and A Segal ldquoIterative projection methods inbiomedical inverse problemsrdquo 2008

[5] Y Censor T Bortfeld B Martin and A Trofimov ldquoA unifiedapproach for inversion problems in intensity-modulated ra-diation therapyrdquo Physics in Medicine and Biology vol 51no 10 pp 2353ndash2365 2006

[6] C Yair and Tommy ldquoe multiple-sets split feasibilityproblem and its applications for inverse problemsrdquo InverseProblems vol 21 no 6 pp 2071ndash2084 2005

[7] Q Yang ldquoe relaxed CQ algorithm solving the split feasi-bility problemrdquo Inverse Problems vol 20 no 4pp 1261ndash1266 2004

[8] G Lopez V Martın-Marquez and F H Wang ldquoSolving thesplit feasibility problem without prior knowledge of matrixnormsrdquo Inverse Problems vol 28 no 8 pp 374ndash389 2012

[9] J Deepho and P Kumam ldquoA modified halpernrsquos iterativescheme for solving split feasibility problemsrdquo Abstract andApplied Analysis vol 2012 Article ID 876069 8 pages 2012

[10] Y Dang and Y Gao ldquoe strong convergence of a KM-CQ-like algorithm for a split feasibility problemrdquo Inverse Prob-lems vol 27 no 1 Article ID 015007 2011

[11] S N He and Z Y Zhao ldquoStrong convergence of a relaxed CQalgorithm for the split feasibility problemrdquo Journal of In-equalities and Applications vol 2013 no 1 11 pages 2013

[12] Y H Yao and Mihai Strong Convergence of a Self-AdaptiveMethod for the Split Feasibility Problem Fixed Point =eory ampApplications Springer Berlin Germany 2013

[13] S Reich M T Truong and T Mai ldquoe split feasibilityproblem with multiple output sets in Hilbert spacesrdquo Opti-mization Letters vol 36 2020

[14] S Suantai N Pholasa N Pholasa and P Cholamjiak ldquoemodified inertial relaxed CQ algorithm for solving the splitfeasibility problemsrdquo Journal of Industrial and ManagementOptimization vol 14 no 4 pp 1595ndash1615 2018

[15] W Cholamjiak P Cholamjiak and S Suantai ldquoAn inertialforward-backward splitting method for solving inclusionproblems in Hilbert spacesrdquo Journal of Fixed Point=eory andApplications vol 20 no 1 p 42 2018

[16] Q L Dong Y J Cho and L L Zhong ldquoInertial projectionand contraction algorithms for variational inequalitiesrdquoJournal of Global Optimization vol 70 no 3 pp 1ndash18 2018

[17] D R Sahu Y J Cho and Q L Dong ldquoInertial relaxed CQalgorithms for solving a split feasibility problem in Hilbertspacesrdquo Numerical Algorithms vol 3 2020

[18] Y Shehu and A Gibali ldquoInertial krasnoselskiindashmann methodin banach spacesrdquo Mathematics vol 8 no 4 2020

[19] Y Shehu P T Vuong and P Cholamjiak ldquoA self-adaptiveprojection method with an inertial technique for split feasi-bility problems in Banach spaces with applications to imagerestoration problemsrdquo Journal of Fixed Point =eory andApplications vol 21 no 2 2019

[20] H-K Xu ldquoIterative methods for the split feasibility problemin infinite-dimensional Hilbert spacesrdquo Inverse Problemsvol 26 no 10 Article ID 105018 2010

[21] R I Bot and E R Csetnek ldquoAn inertial Tsengrsquos type proximalalgorithm for nonsmooth and nonconvex optimizationproblemsrdquo Journal of Optimization =eory and Applicationsvol 171 no 2 pp 600ndash616 2016

12 Journal of Mathematics

12x

k+1minus x

k2 le12x

kminus p

2minus12x

k+1minus p

2+ l

kminus x

k x

k+1minus p

(72)

Since xk minus p2 minus xk+1 minus p2⟶ 0 as k⟶infinlim

k⟶infinlk minus xk 0 we obtain

xk+1

minus xk⟶ 0 as k⟶infin (73)

Since xkj+1 isin Ckj we obtain

c xkj1113872 1113873 + ξkj x

kj+1minus x

kj le 0 (74)

us

c xkj1113872 1113873le minus ξkj x

kj+1minus x

kj le ξxkj+1

minus xkj (75)

where ξ satisfies ξk le ξ for all k By virtue of the continuity offunction c and xkj+1 minus xkj⟶ 0 we get that

c plowast

( 1113857 limj⟶infin

c xkj1113872 1113873le 0 (76)

erefore plowast isin CNow we show that Bip

lowast isin Qi to do this let hki Biz

k minus

PQkiBiz

k⟶ 0 and let ηki be such that ηk

i le η for all k SinceBiz

kj minus hkj

i PQ

kj

i

(Bizkj ) isin Q

kj

i we have

qi Bixkj1113872 1113873 + ηkj

i Bizkj minus h

kj

i1113874 1113875 minus Bixkj le 0 (77)

Hence

qi Bixkj1113872 1113873le η

kj

i Bixkj minus Biz

kj + ηkj

i hkj

i le ηBixkj

minus xkjminus 1

+ ηhkj

i ⟶ 0

(78)

By the continuity of qi and Bixkj⟶ Bix

lowast we arrive atthe conclusion

qi Biplowast

( 1113857 limj⟶infin

qi Bixkj1113872 1113873le 0 (79)

namely Biplowast isin Qi for all i 1 N Hence plowast isin Γ

Moreover for p PΓ u we can see that

limsupk⟶infin

lk

minus p u minus p limk⟶infin

lkj minus p u minus plep

lowast

minus p u minus ple 0

(80)

By Lemma 1 and (65) we have

xk+1

minus p2

PC lk

1113872 1113873 minus p2 le l

kminus p

2 βku + 1 minus βk( 1113857v

kminus p l

kminus p

βku minus p lk

minus p + 1 minus βk( 1113857vk

minus p lk

minus p

le βku minus p lk

minus p + 1 minus βk( 1113857vk

minus plk

minus p

1 minus βk( 1113857 xk

+ θk xk

minus xkminus 1

1113872 1113873 minus p

2

+ βklangu minus p lk

minus prang +βk

2u minus p

2

le 1 minus βk( 1113857 xk

minus p

2

+ 2θklangxk

minus xkminus 1

zk

minus prang1113876 1113877 + βklangu minus p lk

minus prang +βk

2u minus p

2

le 1 minus βk( 1113857 xk

minus p

2

+ 2θk xk

minus xkminus 1

zk

minus p

1113876 1113877 + βklangu minus p lk

minus prang +βk

2u minus p

2

(81)

By conditions (A1) and (A4) βk isin (0 1) imply the se-quence lim

k⟶infinθkxk minus xkminus 1⟶ 0 βk2u minus p⟶ 0 and

using (80) in (81) and applying Lemma 5 we obtain

limk⟶infin

xk

minus p2

0 (82)

erefore as k⟶infin xk⟶ p PΓu

Case 2 SetΛk xk minus p2 Assume that Λk1113864 1113865 is not decreasingat infinity Let ϕ N⟶ N be a mapping for all kge k0 (forsome k0 large enough) defined by

ϕ(k) max t isin N tle kΛt leΛt+11113864 1113865 (83)

By Lemma 4 ϕ(k)1113864 1113865infinkk0is a nondecreasing sequence

such that ϕ(k)⟶infin as k⟶infin and

max Λϕ(k)Λk1113966 1113967leΛϕ(k)+1forallkge k0 (84)

After a similar conclusion from (59) it is easy to see that

limk⟶infin

I minus PQ

ϕ(k)

i

1113874 1113875Bizϕ(k)

0 (85)

By the similar argument as the above in Case 1 weconclude immediately that

8 Journal of Mathematics

limk⟶infin

Blowasti I minus P

Qϕ(k)

i

1113874 1113875Bizϕ(k)

0 (86)

limsupk⟶infin

lϕ(k)

minus p u minus ple 0 (87)

Since xϕ(k)1113864 1113865 is bounded there exists a subsequence ofxϕ(k)1113864 1113865 still denoted by xϕ(k)1113864 1113865 which converges weakly to

plowast By similar argument as above in Case 1 we concludeimmediately that plowast isin C and Bip

lowast isin QirArrplowast isin ΓFrom (57) we have that

xϕ(k)+1

minus p2 le 1 minus βϕ(k)1113872 1113873z

ϕ(k)minus p

2+ βϕ(k)u minus p l

ϕ(k)minus p

le 1 minus βϕ(k)1113872 1113873xϕ(k)

+ θϕ(k) xϕ(k)

minus xϕ(k)minus 1

1113872 1113873 minus p2

+ βϕ(k)u minus p lϕ(k)

minus p

le 1 minus βϕ(k)1113872 1113873 xϕ(k)

minus p2

+ 2θϕ(k)xϕ(k)

minus xϕ(k)minus 1

zϕ(k)

minus p1113960 1113961 + βϕ(k)u minus p lϕ(k)

minus p

le 1 minus βϕ(k)1113872 1113873 xϕ(k)

minus p2

+ 2θϕ(k)xϕ(k)

minus xϕ(k)minus 1

zϕ(k)

minus p1113960 1113961 + βϕ(k)u minus p lϕ(k)

minus p

(88)

which implies by Lemma 5

limk⟶infin

xφ(k)

minus p

2

0 and limk⟶infin

xϕ(k)+1

minus p2

0 (89)

Moreover for kge k0 it is easy to see thatΛϕ(k) minus Λϕ(k)+1 le 0 if kneϕ(k) (that is kge ϕ(k)) because forϕ(k) + 1lemle n Λm gtΛm+1 As a consequence we obtain

0leΛk lemax Λϕ(k)Λϕ(k)+11113966 1113967 Λϕ(k)+1 forallkge k0 (90)

erefore we obtain limk⟶infinΛk 0 that is xk1113864 1113865 converges

strongly to p is completes the proof

4 Some Extensive Results

For the SFPwMOS (11) when N 1 it becomes the SFP(1) us we have the following corollary for solving the SFP(1) which is an immediate consequence of eorem 1

Corollary 1 Let H1 and H2 be two real Hilbert spaces andlet B H1⟶ H2 be bounded linear operator Let C and Q

be nonempty closed and convex subsets of H1 and H2 re-spectively Assume that Ω Ccap Bminus 1(Q)ne 0 Let u isin H1 be afixed point For any starting point x0 x1 isin H1 let xk1113864 1113865 be thesequence generated by

zk

xk

+ θk xk

minus xkminus 1

1113872 1113873

xk+1 ≔ PCk

βku + δkzk

+ εk zk

minus τknablagk zk

1113872 11138731113872 11138731113872 1113873

⎧⎪⎨

⎪⎩(91)

where θk Ck τk and nablafk are given as in (8) Suppose thesequences βk1113864 1113865 δk1113864 1113865 and εk1113864 1113865 satisfy the conditions in =e-orem 1 =en the sequence xk1113864 1113865 converges strongly to thesolution p isin Ω where p PΩ(u)

When we take u x0 in Algorithm 1 we note also thefollowing results regarding to the SFPwMOS (11)

Corollary 2 Let H Hi i 1 N be real Hilbert spacesand let Bi H⟶ Hi i 1 N be bounded linearoperators Let C and Qi i 1 N be nonempty closedand convex subsets of H and Hi i 1 N respectivelyAssume that problem (11) is consistent For any initial guessx0 x1 isin H let xk1113864 1113865 be the sequence generated by

zk

xk

+ θk xk

minus xkminus 1

1113872 1113873

xk+1 ≔ PCk

βkx0

+ δkzk

+ εk zk

minus τknablagk zk

1113872 11138731113872 11138731113872 1113873

⎧⎪⎨

⎪⎩(92)

whereθk Ck τk and nablagk are given by (36) and (39) andAlgorithm 1 respectively Suppose the sequences βk1113864 1113865 δk1113864 1113865and εk1113864 1113865 satisfy the conditions in =eorem 1 =en the se-quence xk1113864 1113865 generated by (92) strongly converges to the so-lution p PΩ(x0) isin Γ

When we take δk equiv 0 in Algorithm 1 we obtain thefollowing result regarding the SFPwMOS (11)

Corollary 3 Let H Hi i 1 N be real Hilbertspaces and let Ti H⟶ Hi i 1 N be boundedlinear operators Let C and Qi i 1 N be nonemptyclosed and convex subsets of H and Hi i 1 N re-spectively Assume that problem (11) is consistent For a fixedpoint u isin H and any initial guess x0 isin H let xk1113864 1113865 be thesequence generated by

zk

xk

+ θk xk

minus xkminus 1

1113872 1113873

xk+1 ≔ PCk

βku + 1 minus βk( 1113857 zk

minus τknablagk zk

1113872 11138731113872 11138731113872 1113873

⎧⎪⎨

⎪⎩(93)

where θk Ck τk and nablagk are given by (36) and (39) andAlgorithm 1 respectively Suppose the sequences satisfyconditions (A1) =en the sequence xk1113864 1113865 generated by (93)strongly converges to the solution p PΩ(u) isin Γ

Of course when we take u x0 we get the followingresult regarding the SFPwMOS (11)

Corollary 4 Let H Hi i 1 N be real Hilbert spacesand let Bi H⟶ Hi i 1 N be bounded linear op-erators Let C and Qi i 1 N be nonempty closed andconvex subsets of H and Hi i 1 N respectively As-sume that problem (11) is consistent For any initial guessx0 isin H let xk1113864 1113865 be the sequence generated by

zk

xk

+ θk xk

minus xkminus 1

1113872 1113873

xk+1 ≔ PCk

βkx0

+ 1 minus tarts minus with(primehskip

prime)]))hskipsubstring minus after(preceding minus sibling comment()[starts minus with(prime hskipprime)]prime hskipprime)pt gt βk) z

kminus τknablagk z

k1113872 11138731113872 1113873)11138741113874

⎧⎪⎪⎨

⎪⎪⎩

(94)

Journal of Mathematics 9

where θk Ck τk and nablagk are given by (36) and (39) andAlgorithm 1 respectively Suppose the sequence βk1113864 1113865 satisfiesconditions (A1) =en the sequence xk1113864 1113865 generated by (94)strongly converges to the solution p PΩ(u) isin Γ

Remark 1 In Corollary 4 for the particular casewhereN 1 the iterative scheme (94) reduced exactly toiterative scheme proposed by He et al in eorem 32 of[10]

5 Numerical Experiment

In this section we provide one numerical experiment toillustrate the implementation and efficiency of our proposedmethod First we study the behavior of Algorithm 1 fordifferent choices of θkand ρk Next we compare Algorithm 1(θk 08) with no inertial Algorithm 1 (θk 0) and Scheme(17) for different initial points e numerical results arecompleted on a standard TOSHIBA laptop with Intel(R)Core(TM) i5-2450M CPU25GHz 25GHz with memory4GB e code is implemented in MATLAB R2020a

Let

A1

minus 04 minus 02 minus 02

04 05 minus 01

02 minus 05 03

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

A2

03 minus 01 03

minus 03 02 minus 02

0 02 03

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

C x isin R3|x1 minus x

22 + 2x3 le 01113966 1113967

Q1 x isin R3

|x21 + x2 minus x3 le 01113966 1113967

Q2 x isin R3|x1 + x

22 minus x3 le 01113966 1113967

(95)

Find xlowast such that

xlowast isin Γ ≔ Ccap cap 2i1A

minus 1i Qi( 11138571113872 1113873ne 0 (96)

Define the error function as

Ek 13x

kminus PCk

xk

1113872 11138732

+13A1x

kminus PQ1k

A1xk

1113872 11138732

+13A2x

kminus PQ2k

A2xk

1113872 11138732

(97)

Note that if Ek 0 at the kth step then xk isin ΓFor Algorithm 1 for different choices of ρk and θk we

choose βk 19k + 1 δk k9k + 1 εk 8k9k + 1u (01 01 01)T and x0 (minus 01 minus 01 minus 01)T andx1 [03 05 minus 06]T Using Ek lt ε as stopping criteriawhere ε is a small enough positive number the results of thenumerical experiment are reported in Table 1e numericalresults can be seen from Table 1 and Figures 1 and 2 InTable 1 ldquoIterrdquo denotes the number of iterations

e behavior of the function Ek in Table 1 is described inFigures 1 and 2

From Table 1 and Figures 1 and 2 we can observe thatAlgorithm 1 for inertial effect case has less number of it-erations than Algorithm 1 for θk 0(that is no inertialeffect) case Algorithm 1 with bigger θk has less number ofiterations than Algorithm 1 for small case for the same θkbig ρk case has less number of iterations than Algorithm 1for small case

Table 1 Algorithm 1 for different choices of θk ε and ρk

ε θkρk 398 ρk 300

Iter Ek Iter Ek

10minus 3θk 08 5 70906e minus 05

6 58048e minus 05θk 03 7 69397e minus 04θk 0 9 75095e minus 04

10minus 4θk 08 5 70906e minus 05

6 58048e minus 05θk 03 11 91282e minus 05θk 0 15 96150e minus 05

10minus 5θk 08 12 44752e minus 06

14 25296e minus 06θk 03 16 91012e minus 06θk 0 24 88064e minus 06

0 5 10 15 20 25The number of Iteration (k)

0

001

002

003

004

005

006

E k

θk=08θk=03θk=00

Figure 1 Error against iterations for Algorithm 1 with differencechoices of θk under ρk 398 and ε 10minus 5

10 Journal of Mathematics

Following we compare our algorithmwith iteration (16)For the example iteration (16) can be written as

xk+1 ≔ αkf x

k1113872 1113873 + 1 minus αk( 11138571113857PC x

kminus λk B

lowast1 I minus PQ11113872 11138731113872 1113873B1x

k1113872

+ Blowast2 I minus PQ21113872 11138731113872 1113873B2x

k1113873

(98)In the process we take αk 08 and f(xk)

u (111)Tλk 000005 for Scheme (97) We choose βk

12k + 1 δk εk k2k + 1 and ρk 3for Algorithm 1From Table 2 and Figure 3 we can observe that Algo-

rithm 1 has less number of iterations than Scheme (97)Moreover our iterative method is advantageous overScheme (98) because Algorithm 1 does not require com-puting of the operator norm

Data Availability

No data were used to support this study

Conflicts of Interest

e authors declare that they have no conflicts of interest

Authorsrsquo Contributions

is entire work has been completed by the authors HuijuanJia Shufen Liu and Yazheng Dang e authors read andapproved the finial manuscript

0 5 10 15e number of Iteration (k)

ρk=398ρk=300

0

001

002

003

004

005

006

007

E k

Figure 2 Error against iterations for Algorithm 1 with difference choices of ρk under θk 08 and ε 10minus 5

Table 2 Comparsion of Algorithm 1 and Scheme (97) for different choice initial point for ε 10minus 4

Initiative point x1 Algorithm 1 with θk 08 Algorithm 1 with θk 00 (98)Iter Ek Iter Ek Iter Ek

[04 09ndash09] 11 41830e minus 06 25 96999e minus 06 51 88226e minus 06[09 03ndash05] 6 14290e minus 07 22 86399e minus 06 26 96719e minus 06[minus 07 03ndash02] 12 78838e minus 06 24 92013e minus 06 48 84198e minus 06[01 01ndash09]prime 12 15353e minus 07 26 87544e minus 06 52 99722e minus 06

0 10 20 30 5040e number of Iteration (k)

Algorithm 1 with θ=08Algorithm 1 with θ=00(52)

0

004

002

006

008

01

012

014

E k

Figure 3 Error against of iterations for the comparision of Al-gorithm 1 and Scheme (97)

Journal of Mathematics 11

Acknowledgments

is research was supported by the National Natural ScienceFoundation of China (Grant no 61872126) Henan ProvinceKey Science and Technology Project (Grant no192102210123) and Young Backbone Teachers in Univer-sities of Henan Province (Grant no 2019GGJS061)

References

[1] Y Censor and T Elfving ldquoA multiprojection algorithm usingbregman projections in a product spacerdquo Numerical Algo-rithms vol 8 no 2 pp 221ndash239 1994

[2] C Byrne ldquoIterative oblique projection onto convex sets andthe split feasibility problemrdquo Inverse Problems vol 18 no 2pp 441ndash453 2002

[3] G T Herman ldquoImage reconstruction from projections thefundamentals of computerized tomographyrdquo 1980

[4] Y Censor and A Segal ldquoIterative projection methods inbiomedical inverse problemsrdquo 2008

[5] Y Censor T Bortfeld B Martin and A Trofimov ldquoA unifiedapproach for inversion problems in intensity-modulated ra-diation therapyrdquo Physics in Medicine and Biology vol 51no 10 pp 2353ndash2365 2006

[6] C Yair and Tommy ldquoe multiple-sets split feasibilityproblem and its applications for inverse problemsrdquo InverseProblems vol 21 no 6 pp 2071ndash2084 2005

[7] Q Yang ldquoe relaxed CQ algorithm solving the split feasi-bility problemrdquo Inverse Problems vol 20 no 4pp 1261ndash1266 2004

[8] G Lopez V Martın-Marquez and F H Wang ldquoSolving thesplit feasibility problem without prior knowledge of matrixnormsrdquo Inverse Problems vol 28 no 8 pp 374ndash389 2012

[9] J Deepho and P Kumam ldquoA modified halpernrsquos iterativescheme for solving split feasibility problemsrdquo Abstract andApplied Analysis vol 2012 Article ID 876069 8 pages 2012

[10] Y Dang and Y Gao ldquoe strong convergence of a KM-CQ-like algorithm for a split feasibility problemrdquo Inverse Prob-lems vol 27 no 1 Article ID 015007 2011

[11] S N He and Z Y Zhao ldquoStrong convergence of a relaxed CQalgorithm for the split feasibility problemrdquo Journal of In-equalities and Applications vol 2013 no 1 11 pages 2013

[12] Y H Yao and Mihai Strong Convergence of a Self-AdaptiveMethod for the Split Feasibility Problem Fixed Point =eory ampApplications Springer Berlin Germany 2013

[13] S Reich M T Truong and T Mai ldquoe split feasibilityproblem with multiple output sets in Hilbert spacesrdquo Opti-mization Letters vol 36 2020

[14] S Suantai N Pholasa N Pholasa and P Cholamjiak ldquoemodified inertial relaxed CQ algorithm for solving the splitfeasibility problemsrdquo Journal of Industrial and ManagementOptimization vol 14 no 4 pp 1595ndash1615 2018

[15] W Cholamjiak P Cholamjiak and S Suantai ldquoAn inertialforward-backward splitting method for solving inclusionproblems in Hilbert spacesrdquo Journal of Fixed Point=eory andApplications vol 20 no 1 p 42 2018

[16] Q L Dong Y J Cho and L L Zhong ldquoInertial projectionand contraction algorithms for variational inequalitiesrdquoJournal of Global Optimization vol 70 no 3 pp 1ndash18 2018

[17] D R Sahu Y J Cho and Q L Dong ldquoInertial relaxed CQalgorithms for solving a split feasibility problem in Hilbertspacesrdquo Numerical Algorithms vol 3 2020

[18] Y Shehu and A Gibali ldquoInertial krasnoselskiindashmann methodin banach spacesrdquo Mathematics vol 8 no 4 2020

[19] Y Shehu P T Vuong and P Cholamjiak ldquoA self-adaptiveprojection method with an inertial technique for split feasi-bility problems in Banach spaces with applications to imagerestoration problemsrdquo Journal of Fixed Point =eory andApplications vol 21 no 2 2019

[20] H-K Xu ldquoIterative methods for the split feasibility problemin infinite-dimensional Hilbert spacesrdquo Inverse Problemsvol 26 no 10 Article ID 105018 2010

[21] R I Bot and E R Csetnek ldquoAn inertial Tsengrsquos type proximalalgorithm for nonsmooth and nonconvex optimizationproblemsrdquo Journal of Optimization =eory and Applicationsvol 171 no 2 pp 600ndash616 2016

12 Journal of Mathematics

limk⟶infin

Blowasti I minus P

Qϕ(k)

i

1113874 1113875Bizϕ(k)

0 (86)

limsupk⟶infin

lϕ(k)

minus p u minus ple 0 (87)

Since xϕ(k)1113864 1113865 is bounded there exists a subsequence ofxϕ(k)1113864 1113865 still denoted by xϕ(k)1113864 1113865 which converges weakly to

plowast By similar argument as above in Case 1 we concludeimmediately that plowast isin C and Bip

lowast isin QirArrplowast isin ΓFrom (57) we have that

xϕ(k)+1

minus p2 le 1 minus βϕ(k)1113872 1113873z

ϕ(k)minus p

2+ βϕ(k)u minus p l

ϕ(k)minus p

le 1 minus βϕ(k)1113872 1113873xϕ(k)

+ θϕ(k) xϕ(k)

minus xϕ(k)minus 1

1113872 1113873 minus p2

+ βϕ(k)u minus p lϕ(k)

minus p

le 1 minus βϕ(k)1113872 1113873 xϕ(k)

minus p2

+ 2θϕ(k)xϕ(k)

minus xϕ(k)minus 1

zϕ(k)

minus p1113960 1113961 + βϕ(k)u minus p lϕ(k)

minus p

le 1 minus βϕ(k)1113872 1113873 xϕ(k)

minus p2

+ 2θϕ(k)xϕ(k)

minus xϕ(k)minus 1

zϕ(k)

minus p1113960 1113961 + βϕ(k)u minus p lϕ(k)

minus p

(88)

which implies by Lemma 5

limk⟶infin

xφ(k)

minus p

2

0 and limk⟶infin

xϕ(k)+1

minus p2

0 (89)

Moreover for kge k0 it is easy to see thatΛϕ(k) minus Λϕ(k)+1 le 0 if kneϕ(k) (that is kge ϕ(k)) because forϕ(k) + 1lemle n Λm gtΛm+1 As a consequence we obtain

0leΛk lemax Λϕ(k)Λϕ(k)+11113966 1113967 Λϕ(k)+1 forallkge k0 (90)

erefore we obtain limk⟶infinΛk 0 that is xk1113864 1113865 converges

strongly to p is completes the proof

4 Some Extensive Results

For the SFPwMOS (11) when N 1 it becomes the SFP(1) us we have the following corollary for solving the SFP(1) which is an immediate consequence of eorem 1

Corollary 1 Let H1 and H2 be two real Hilbert spaces andlet B H1⟶ H2 be bounded linear operator Let C and Q

be nonempty closed and convex subsets of H1 and H2 re-spectively Assume that Ω Ccap Bminus 1(Q)ne 0 Let u isin H1 be afixed point For any starting point x0 x1 isin H1 let xk1113864 1113865 be thesequence generated by

zk

xk

+ θk xk

minus xkminus 1

1113872 1113873

xk+1 ≔ PCk

βku + δkzk

+ εk zk

minus τknablagk zk

1113872 11138731113872 11138731113872 1113873

⎧⎪⎨

⎪⎩(91)

where θk Ck τk and nablafk are given as in (8) Suppose thesequences βk1113864 1113865 δk1113864 1113865 and εk1113864 1113865 satisfy the conditions in =e-orem 1 =en the sequence xk1113864 1113865 converges strongly to thesolution p isin Ω where p PΩ(u)

When we take u x0 in Algorithm 1 we note also thefollowing results regarding to the SFPwMOS (11)

Corollary 2 Let H Hi i 1 N be real Hilbert spacesand let Bi H⟶ Hi i 1 N be bounded linearoperators Let C and Qi i 1 N be nonempty closedand convex subsets of H and Hi i 1 N respectivelyAssume that problem (11) is consistent For any initial guessx0 x1 isin H let xk1113864 1113865 be the sequence generated by

zk

xk

+ θk xk

minus xkminus 1

1113872 1113873

xk+1 ≔ PCk

βkx0

+ δkzk

+ εk zk

minus τknablagk zk

1113872 11138731113872 11138731113872 1113873

⎧⎪⎨

⎪⎩(92)

whereθk Ck τk and nablagk are given by (36) and (39) andAlgorithm 1 respectively Suppose the sequences βk1113864 1113865 δk1113864 1113865and εk1113864 1113865 satisfy the conditions in =eorem 1 =en the se-quence xk1113864 1113865 generated by (92) strongly converges to the so-lution p PΩ(x0) isin Γ

When we take δk equiv 0 in Algorithm 1 we obtain thefollowing result regarding the SFPwMOS (11)

Corollary 3 Let H Hi i 1 N be real Hilbertspaces and let Ti H⟶ Hi i 1 N be boundedlinear operators Let C and Qi i 1 N be nonemptyclosed and convex subsets of H and Hi i 1 N re-spectively Assume that problem (11) is consistent For a fixedpoint u isin H and any initial guess x0 isin H let xk1113864 1113865 be thesequence generated by

zk

xk

+ θk xk

minus xkminus 1

1113872 1113873

xk+1 ≔ PCk

βku + 1 minus βk( 1113857 zk

minus τknablagk zk

1113872 11138731113872 11138731113872 1113873

⎧⎪⎨

⎪⎩(93)

where θk Ck τk and nablagk are given by (36) and (39) andAlgorithm 1 respectively Suppose the sequences satisfyconditions (A1) =en the sequence xk1113864 1113865 generated by (93)strongly converges to the solution p PΩ(u) isin Γ

Of course when we take u x0 we get the followingresult regarding the SFPwMOS (11)

Corollary 4 Let H Hi i 1 N be real Hilbert spacesand let Bi H⟶ Hi i 1 N be bounded linear op-erators Let C and Qi i 1 N be nonempty closed andconvex subsets of H and Hi i 1 N respectively As-sume that problem (11) is consistent For any initial guessx0 isin H let xk1113864 1113865 be the sequence generated by

zk

xk

+ θk xk

minus xkminus 1

1113872 1113873

xk+1 ≔ PCk

βkx0

+ 1 minus tarts minus with(primehskip

prime)]))hskipsubstring minus after(preceding minus sibling comment()[starts minus with(prime hskipprime)]prime hskipprime)pt gt βk) z

kminus τknablagk z

k1113872 11138731113872 1113873)11138741113874

⎧⎪⎪⎨

⎪⎪⎩

(94)

Journal of Mathematics 9

where θk Ck τk and nablagk are given by (36) and (39) andAlgorithm 1 respectively Suppose the sequence βk1113864 1113865 satisfiesconditions (A1) =en the sequence xk1113864 1113865 generated by (94)strongly converges to the solution p PΩ(u) isin Γ

Remark 1 In Corollary 4 for the particular casewhereN 1 the iterative scheme (94) reduced exactly toiterative scheme proposed by He et al in eorem 32 of[10]

5 Numerical Experiment

In this section we provide one numerical experiment toillustrate the implementation and efficiency of our proposedmethod First we study the behavior of Algorithm 1 fordifferent choices of θkand ρk Next we compare Algorithm 1(θk 08) with no inertial Algorithm 1 (θk 0) and Scheme(17) for different initial points e numerical results arecompleted on a standard TOSHIBA laptop with Intel(R)Core(TM) i5-2450M CPU25GHz 25GHz with memory4GB e code is implemented in MATLAB R2020a

Let

A1

minus 04 minus 02 minus 02

04 05 minus 01

02 minus 05 03

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

A2

03 minus 01 03

minus 03 02 minus 02

0 02 03

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

C x isin R3|x1 minus x

22 + 2x3 le 01113966 1113967

Q1 x isin R3

|x21 + x2 minus x3 le 01113966 1113967

Q2 x isin R3|x1 + x

22 minus x3 le 01113966 1113967

(95)

Find xlowast such that

xlowast isin Γ ≔ Ccap cap 2i1A

minus 1i Qi( 11138571113872 1113873ne 0 (96)

Define the error function as

Ek 13x

kminus PCk

xk

1113872 11138732

+13A1x

kminus PQ1k

A1xk

1113872 11138732

+13A2x

kminus PQ2k

A2xk

1113872 11138732

(97)

Note that if Ek 0 at the kth step then xk isin ΓFor Algorithm 1 for different choices of ρk and θk we

choose βk 19k + 1 δk k9k + 1 εk 8k9k + 1u (01 01 01)T and x0 (minus 01 minus 01 minus 01)T andx1 [03 05 minus 06]T Using Ek lt ε as stopping criteriawhere ε is a small enough positive number the results of thenumerical experiment are reported in Table 1e numericalresults can be seen from Table 1 and Figures 1 and 2 InTable 1 ldquoIterrdquo denotes the number of iterations

e behavior of the function Ek in Table 1 is described inFigures 1 and 2

From Table 1 and Figures 1 and 2 we can observe thatAlgorithm 1 for inertial effect case has less number of it-erations than Algorithm 1 for θk 0(that is no inertialeffect) case Algorithm 1 with bigger θk has less number ofiterations than Algorithm 1 for small case for the same θkbig ρk case has less number of iterations than Algorithm 1for small case

Table 1 Algorithm 1 for different choices of θk ε and ρk

ε θkρk 398 ρk 300

Iter Ek Iter Ek

10minus 3θk 08 5 70906e minus 05

6 58048e minus 05θk 03 7 69397e minus 04θk 0 9 75095e minus 04

10minus 4θk 08 5 70906e minus 05

6 58048e minus 05θk 03 11 91282e minus 05θk 0 15 96150e minus 05

10minus 5θk 08 12 44752e minus 06

14 25296e minus 06θk 03 16 91012e minus 06θk 0 24 88064e minus 06

0 5 10 15 20 25The number of Iteration (k)

0

001

002

003

004

005

006

E k

θk=08θk=03θk=00

Figure 1 Error against iterations for Algorithm 1 with differencechoices of θk under ρk 398 and ε 10minus 5

10 Journal of Mathematics

Following we compare our algorithmwith iteration (16)For the example iteration (16) can be written as

xk+1 ≔ αkf x

k1113872 1113873 + 1 minus αk( 11138571113857PC x

kminus λk B

lowast1 I minus PQ11113872 11138731113872 1113873B1x

k1113872

+ Blowast2 I minus PQ21113872 11138731113872 1113873B2x

k1113873

(98)In the process we take αk 08 and f(xk)

u (111)Tλk 000005 for Scheme (97) We choose βk

12k + 1 δk εk k2k + 1 and ρk 3for Algorithm 1From Table 2 and Figure 3 we can observe that Algo-

rithm 1 has less number of iterations than Scheme (97)Moreover our iterative method is advantageous overScheme (98) because Algorithm 1 does not require com-puting of the operator norm

Data Availability

No data were used to support this study

Conflicts of Interest

e authors declare that they have no conflicts of interest

Authorsrsquo Contributions

is entire work has been completed by the authors HuijuanJia Shufen Liu and Yazheng Dang e authors read andapproved the finial manuscript

0 5 10 15e number of Iteration (k)

ρk=398ρk=300

0

001

002

003

004

005

006

007

E k

Figure 2 Error against iterations for Algorithm 1 with difference choices of ρk under θk 08 and ε 10minus 5

Table 2 Comparsion of Algorithm 1 and Scheme (97) for different choice initial point for ε 10minus 4

Initiative point x1 Algorithm 1 with θk 08 Algorithm 1 with θk 00 (98)Iter Ek Iter Ek Iter Ek

[04 09ndash09] 11 41830e minus 06 25 96999e minus 06 51 88226e minus 06[09 03ndash05] 6 14290e minus 07 22 86399e minus 06 26 96719e minus 06[minus 07 03ndash02] 12 78838e minus 06 24 92013e minus 06 48 84198e minus 06[01 01ndash09]prime 12 15353e minus 07 26 87544e minus 06 52 99722e minus 06

0 10 20 30 5040e number of Iteration (k)

Algorithm 1 with θ=08Algorithm 1 with θ=00(52)

0

004

002

006

008

01

012

014

E k

Figure 3 Error against of iterations for the comparision of Al-gorithm 1 and Scheme (97)

Journal of Mathematics 11

Acknowledgments

is research was supported by the National Natural ScienceFoundation of China (Grant no 61872126) Henan ProvinceKey Science and Technology Project (Grant no192102210123) and Young Backbone Teachers in Univer-sities of Henan Province (Grant no 2019GGJS061)

References

[1] Y Censor and T Elfving ldquoA multiprojection algorithm usingbregman projections in a product spacerdquo Numerical Algo-rithms vol 8 no 2 pp 221ndash239 1994

[2] C Byrne ldquoIterative oblique projection onto convex sets andthe split feasibility problemrdquo Inverse Problems vol 18 no 2pp 441ndash453 2002

[3] G T Herman ldquoImage reconstruction from projections thefundamentals of computerized tomographyrdquo 1980

[4] Y Censor and A Segal ldquoIterative projection methods inbiomedical inverse problemsrdquo 2008

[5] Y Censor T Bortfeld B Martin and A Trofimov ldquoA unifiedapproach for inversion problems in intensity-modulated ra-diation therapyrdquo Physics in Medicine and Biology vol 51no 10 pp 2353ndash2365 2006

[6] C Yair and Tommy ldquoe multiple-sets split feasibilityproblem and its applications for inverse problemsrdquo InverseProblems vol 21 no 6 pp 2071ndash2084 2005

[7] Q Yang ldquoe relaxed CQ algorithm solving the split feasi-bility problemrdquo Inverse Problems vol 20 no 4pp 1261ndash1266 2004

[8] G Lopez V Martın-Marquez and F H Wang ldquoSolving thesplit feasibility problem without prior knowledge of matrixnormsrdquo Inverse Problems vol 28 no 8 pp 374ndash389 2012

[9] J Deepho and P Kumam ldquoA modified halpernrsquos iterativescheme for solving split feasibility problemsrdquo Abstract andApplied Analysis vol 2012 Article ID 876069 8 pages 2012

[10] Y Dang and Y Gao ldquoe strong convergence of a KM-CQ-like algorithm for a split feasibility problemrdquo Inverse Prob-lems vol 27 no 1 Article ID 015007 2011

[11] S N He and Z Y Zhao ldquoStrong convergence of a relaxed CQalgorithm for the split feasibility problemrdquo Journal of In-equalities and Applications vol 2013 no 1 11 pages 2013

[12] Y H Yao and Mihai Strong Convergence of a Self-AdaptiveMethod for the Split Feasibility Problem Fixed Point =eory ampApplications Springer Berlin Germany 2013

[13] S Reich M T Truong and T Mai ldquoe split feasibilityproblem with multiple output sets in Hilbert spacesrdquo Opti-mization Letters vol 36 2020

[14] S Suantai N Pholasa N Pholasa and P Cholamjiak ldquoemodified inertial relaxed CQ algorithm for solving the splitfeasibility problemsrdquo Journal of Industrial and ManagementOptimization vol 14 no 4 pp 1595ndash1615 2018

[15] W Cholamjiak P Cholamjiak and S Suantai ldquoAn inertialforward-backward splitting method for solving inclusionproblems in Hilbert spacesrdquo Journal of Fixed Point=eory andApplications vol 20 no 1 p 42 2018

[16] Q L Dong Y J Cho and L L Zhong ldquoInertial projectionand contraction algorithms for variational inequalitiesrdquoJournal of Global Optimization vol 70 no 3 pp 1ndash18 2018

[17] D R Sahu Y J Cho and Q L Dong ldquoInertial relaxed CQalgorithms for solving a split feasibility problem in Hilbertspacesrdquo Numerical Algorithms vol 3 2020

[18] Y Shehu and A Gibali ldquoInertial krasnoselskiindashmann methodin banach spacesrdquo Mathematics vol 8 no 4 2020

[19] Y Shehu P T Vuong and P Cholamjiak ldquoA self-adaptiveprojection method with an inertial technique for split feasi-bility problems in Banach spaces with applications to imagerestoration problemsrdquo Journal of Fixed Point =eory andApplications vol 21 no 2 2019

[20] H-K Xu ldquoIterative methods for the split feasibility problemin infinite-dimensional Hilbert spacesrdquo Inverse Problemsvol 26 no 10 Article ID 105018 2010

[21] R I Bot and E R Csetnek ldquoAn inertial Tsengrsquos type proximalalgorithm for nonsmooth and nonconvex optimizationproblemsrdquo Journal of Optimization =eory and Applicationsvol 171 no 2 pp 600ndash616 2016

12 Journal of Mathematics

where θk Ck τk and nablagk are given by (36) and (39) andAlgorithm 1 respectively Suppose the sequence βk1113864 1113865 satisfiesconditions (A1) =en the sequence xk1113864 1113865 generated by (94)strongly converges to the solution p PΩ(u) isin Γ

Remark 1 In Corollary 4 for the particular casewhereN 1 the iterative scheme (94) reduced exactly toiterative scheme proposed by He et al in eorem 32 of[10]

5 Numerical Experiment

In this section we provide one numerical experiment toillustrate the implementation and efficiency of our proposedmethod First we study the behavior of Algorithm 1 fordifferent choices of θkand ρk Next we compare Algorithm 1(θk 08) with no inertial Algorithm 1 (θk 0) and Scheme(17) for different initial points e numerical results arecompleted on a standard TOSHIBA laptop with Intel(R)Core(TM) i5-2450M CPU25GHz 25GHz with memory4GB e code is implemented in MATLAB R2020a

Let

A1

minus 04 minus 02 minus 02

04 05 minus 01

02 minus 05 03

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

A2

03 minus 01 03

minus 03 02 minus 02

0 02 03

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

C x isin R3|x1 minus x

22 + 2x3 le 01113966 1113967

Q1 x isin R3

|x21 + x2 minus x3 le 01113966 1113967

Q2 x isin R3|x1 + x

22 minus x3 le 01113966 1113967

(95)

Find xlowast such that

xlowast isin Γ ≔ Ccap cap 2i1A

minus 1i Qi( 11138571113872 1113873ne 0 (96)

Define the error function as

Ek 13x

kminus PCk

xk

1113872 11138732

+13A1x

kminus PQ1k

A1xk

1113872 11138732

+13A2x

kminus PQ2k

A2xk

1113872 11138732

(97)

Note that if Ek 0 at the kth step then xk isin ΓFor Algorithm 1 for different choices of ρk and θk we

choose βk 19k + 1 δk k9k + 1 εk 8k9k + 1u (01 01 01)T and x0 (minus 01 minus 01 minus 01)T andx1 [03 05 minus 06]T Using Ek lt ε as stopping criteriawhere ε is a small enough positive number the results of thenumerical experiment are reported in Table 1e numericalresults can be seen from Table 1 and Figures 1 and 2 InTable 1 ldquoIterrdquo denotes the number of iterations

e behavior of the function Ek in Table 1 is described inFigures 1 and 2

From Table 1 and Figures 1 and 2 we can observe thatAlgorithm 1 for inertial effect case has less number of it-erations than Algorithm 1 for θk 0(that is no inertialeffect) case Algorithm 1 with bigger θk has less number ofiterations than Algorithm 1 for small case for the same θkbig ρk case has less number of iterations than Algorithm 1for small case

Table 1 Algorithm 1 for different choices of θk ε and ρk

ε θkρk 398 ρk 300

Iter Ek Iter Ek

10minus 3θk 08 5 70906e minus 05

6 58048e minus 05θk 03 7 69397e minus 04θk 0 9 75095e minus 04

10minus 4θk 08 5 70906e minus 05

6 58048e minus 05θk 03 11 91282e minus 05θk 0 15 96150e minus 05

10minus 5θk 08 12 44752e minus 06

14 25296e minus 06θk 03 16 91012e minus 06θk 0 24 88064e minus 06

0 5 10 15 20 25The number of Iteration (k)

0

001

002

003

004

005

006

E k

θk=08θk=03θk=00

Figure 1 Error against iterations for Algorithm 1 with differencechoices of θk under ρk 398 and ε 10minus 5

10 Journal of Mathematics

Following we compare our algorithmwith iteration (16)For the example iteration (16) can be written as

xk+1 ≔ αkf x

k1113872 1113873 + 1 minus αk( 11138571113857PC x

kminus λk B

lowast1 I minus PQ11113872 11138731113872 1113873B1x

k1113872

+ Blowast2 I minus PQ21113872 11138731113872 1113873B2x

k1113873

(98)In the process we take αk 08 and f(xk)

u (111)Tλk 000005 for Scheme (97) We choose βk

12k + 1 δk εk k2k + 1 and ρk 3for Algorithm 1From Table 2 and Figure 3 we can observe that Algo-

rithm 1 has less number of iterations than Scheme (97)Moreover our iterative method is advantageous overScheme (98) because Algorithm 1 does not require com-puting of the operator norm

Data Availability

No data were used to support this study

Conflicts of Interest

e authors declare that they have no conflicts of interest

Authorsrsquo Contributions

is entire work has been completed by the authors HuijuanJia Shufen Liu and Yazheng Dang e authors read andapproved the finial manuscript

0 5 10 15e number of Iteration (k)

ρk=398ρk=300

0

001

002

003

004

005

006

007

E k

Figure 2 Error against iterations for Algorithm 1 with difference choices of ρk under θk 08 and ε 10minus 5

Table 2 Comparsion of Algorithm 1 and Scheme (97) for different choice initial point for ε 10minus 4

Initiative point x1 Algorithm 1 with θk 08 Algorithm 1 with θk 00 (98)Iter Ek Iter Ek Iter Ek

[04 09ndash09] 11 41830e minus 06 25 96999e minus 06 51 88226e minus 06[09 03ndash05] 6 14290e minus 07 22 86399e minus 06 26 96719e minus 06[minus 07 03ndash02] 12 78838e minus 06 24 92013e minus 06 48 84198e minus 06[01 01ndash09]prime 12 15353e minus 07 26 87544e minus 06 52 99722e minus 06

0 10 20 30 5040e number of Iteration (k)

Algorithm 1 with θ=08Algorithm 1 with θ=00(52)

0

004

002

006

008

01

012

014

E k

Figure 3 Error against of iterations for the comparision of Al-gorithm 1 and Scheme (97)

Journal of Mathematics 11

Acknowledgments

is research was supported by the National Natural ScienceFoundation of China (Grant no 61872126) Henan ProvinceKey Science and Technology Project (Grant no192102210123) and Young Backbone Teachers in Univer-sities of Henan Province (Grant no 2019GGJS061)

References

[1] Y Censor and T Elfving ldquoA multiprojection algorithm usingbregman projections in a product spacerdquo Numerical Algo-rithms vol 8 no 2 pp 221ndash239 1994

[2] C Byrne ldquoIterative oblique projection onto convex sets andthe split feasibility problemrdquo Inverse Problems vol 18 no 2pp 441ndash453 2002

[3] G T Herman ldquoImage reconstruction from projections thefundamentals of computerized tomographyrdquo 1980

[4] Y Censor and A Segal ldquoIterative projection methods inbiomedical inverse problemsrdquo 2008

[5] Y Censor T Bortfeld B Martin and A Trofimov ldquoA unifiedapproach for inversion problems in intensity-modulated ra-diation therapyrdquo Physics in Medicine and Biology vol 51no 10 pp 2353ndash2365 2006

[6] C Yair and Tommy ldquoe multiple-sets split feasibilityproblem and its applications for inverse problemsrdquo InverseProblems vol 21 no 6 pp 2071ndash2084 2005

[7] Q Yang ldquoe relaxed CQ algorithm solving the split feasi-bility problemrdquo Inverse Problems vol 20 no 4pp 1261ndash1266 2004

[8] G Lopez V Martın-Marquez and F H Wang ldquoSolving thesplit feasibility problem without prior knowledge of matrixnormsrdquo Inverse Problems vol 28 no 8 pp 374ndash389 2012

[9] J Deepho and P Kumam ldquoA modified halpernrsquos iterativescheme for solving split feasibility problemsrdquo Abstract andApplied Analysis vol 2012 Article ID 876069 8 pages 2012

[10] Y Dang and Y Gao ldquoe strong convergence of a KM-CQ-like algorithm for a split feasibility problemrdquo Inverse Prob-lems vol 27 no 1 Article ID 015007 2011

[11] S N He and Z Y Zhao ldquoStrong convergence of a relaxed CQalgorithm for the split feasibility problemrdquo Journal of In-equalities and Applications vol 2013 no 1 11 pages 2013

[12] Y H Yao and Mihai Strong Convergence of a Self-AdaptiveMethod for the Split Feasibility Problem Fixed Point =eory ampApplications Springer Berlin Germany 2013

[13] S Reich M T Truong and T Mai ldquoe split feasibilityproblem with multiple output sets in Hilbert spacesrdquo Opti-mization Letters vol 36 2020

[14] S Suantai N Pholasa N Pholasa and P Cholamjiak ldquoemodified inertial relaxed CQ algorithm for solving the splitfeasibility problemsrdquo Journal of Industrial and ManagementOptimization vol 14 no 4 pp 1595ndash1615 2018

[15] W Cholamjiak P Cholamjiak and S Suantai ldquoAn inertialforward-backward splitting method for solving inclusionproblems in Hilbert spacesrdquo Journal of Fixed Point=eory andApplications vol 20 no 1 p 42 2018

[16] Q L Dong Y J Cho and L L Zhong ldquoInertial projectionand contraction algorithms for variational inequalitiesrdquoJournal of Global Optimization vol 70 no 3 pp 1ndash18 2018

[17] D R Sahu Y J Cho and Q L Dong ldquoInertial relaxed CQalgorithms for solving a split feasibility problem in Hilbertspacesrdquo Numerical Algorithms vol 3 2020

[18] Y Shehu and A Gibali ldquoInertial krasnoselskiindashmann methodin banach spacesrdquo Mathematics vol 8 no 4 2020

[19] Y Shehu P T Vuong and P Cholamjiak ldquoA self-adaptiveprojection method with an inertial technique for split feasi-bility problems in Banach spaces with applications to imagerestoration problemsrdquo Journal of Fixed Point =eory andApplications vol 21 no 2 2019

[20] H-K Xu ldquoIterative methods for the split feasibility problemin infinite-dimensional Hilbert spacesrdquo Inverse Problemsvol 26 no 10 Article ID 105018 2010

[21] R I Bot and E R Csetnek ldquoAn inertial Tsengrsquos type proximalalgorithm for nonsmooth and nonconvex optimizationproblemsrdquo Journal of Optimization =eory and Applicationsvol 171 no 2 pp 600ndash616 2016

12 Journal of Mathematics

Following we compare our algorithmwith iteration (16)For the example iteration (16) can be written as

xk+1 ≔ αkf x

k1113872 1113873 + 1 minus αk( 11138571113857PC x

kminus λk B

lowast1 I minus PQ11113872 11138731113872 1113873B1x

k1113872

+ Blowast2 I minus PQ21113872 11138731113872 1113873B2x

k1113873

(98)In the process we take αk 08 and f(xk)

u (111)Tλk 000005 for Scheme (97) We choose βk

12k + 1 δk εk k2k + 1 and ρk 3for Algorithm 1From Table 2 and Figure 3 we can observe that Algo-

rithm 1 has less number of iterations than Scheme (97)Moreover our iterative method is advantageous overScheme (98) because Algorithm 1 does not require com-puting of the operator norm

Data Availability

No data were used to support this study

Conflicts of Interest

e authors declare that they have no conflicts of interest

Authorsrsquo Contributions

is entire work has been completed by the authors HuijuanJia Shufen Liu and Yazheng Dang e authors read andapproved the finial manuscript

0 5 10 15e number of Iteration (k)

ρk=398ρk=300

0

001

002

003

004

005

006

007

E k

Figure 2 Error against iterations for Algorithm 1 with difference choices of ρk under θk 08 and ε 10minus 5

Table 2 Comparsion of Algorithm 1 and Scheme (97) for different choice initial point for ε 10minus 4

Initiative point x1 Algorithm 1 with θk 08 Algorithm 1 with θk 00 (98)Iter Ek Iter Ek Iter Ek

[04 09ndash09] 11 41830e minus 06 25 96999e minus 06 51 88226e minus 06[09 03ndash05] 6 14290e minus 07 22 86399e minus 06 26 96719e minus 06[minus 07 03ndash02] 12 78838e minus 06 24 92013e minus 06 48 84198e minus 06[01 01ndash09]prime 12 15353e minus 07 26 87544e minus 06 52 99722e minus 06

0 10 20 30 5040e number of Iteration (k)

Algorithm 1 with θ=08Algorithm 1 with θ=00(52)

0

004

002

006

008

01

012

014

E k

Figure 3 Error against of iterations for the comparision of Al-gorithm 1 and Scheme (97)

Journal of Mathematics 11

Acknowledgments

is research was supported by the National Natural ScienceFoundation of China (Grant no 61872126) Henan ProvinceKey Science and Technology Project (Grant no192102210123) and Young Backbone Teachers in Univer-sities of Henan Province (Grant no 2019GGJS061)

References

[1] Y Censor and T Elfving ldquoA multiprojection algorithm usingbregman projections in a product spacerdquo Numerical Algo-rithms vol 8 no 2 pp 221ndash239 1994

[2] C Byrne ldquoIterative oblique projection onto convex sets andthe split feasibility problemrdquo Inverse Problems vol 18 no 2pp 441ndash453 2002

[3] G T Herman ldquoImage reconstruction from projections thefundamentals of computerized tomographyrdquo 1980

[4] Y Censor and A Segal ldquoIterative projection methods inbiomedical inverse problemsrdquo 2008

[5] Y Censor T Bortfeld B Martin and A Trofimov ldquoA unifiedapproach for inversion problems in intensity-modulated ra-diation therapyrdquo Physics in Medicine and Biology vol 51no 10 pp 2353ndash2365 2006

[6] C Yair and Tommy ldquoe multiple-sets split feasibilityproblem and its applications for inverse problemsrdquo InverseProblems vol 21 no 6 pp 2071ndash2084 2005

[7] Q Yang ldquoe relaxed CQ algorithm solving the split feasi-bility problemrdquo Inverse Problems vol 20 no 4pp 1261ndash1266 2004

[8] G Lopez V Martın-Marquez and F H Wang ldquoSolving thesplit feasibility problem without prior knowledge of matrixnormsrdquo Inverse Problems vol 28 no 8 pp 374ndash389 2012

[9] J Deepho and P Kumam ldquoA modified halpernrsquos iterativescheme for solving split feasibility problemsrdquo Abstract andApplied Analysis vol 2012 Article ID 876069 8 pages 2012

[10] Y Dang and Y Gao ldquoe strong convergence of a KM-CQ-like algorithm for a split feasibility problemrdquo Inverse Prob-lems vol 27 no 1 Article ID 015007 2011

[11] S N He and Z Y Zhao ldquoStrong convergence of a relaxed CQalgorithm for the split feasibility problemrdquo Journal of In-equalities and Applications vol 2013 no 1 11 pages 2013

[12] Y H Yao and Mihai Strong Convergence of a Self-AdaptiveMethod for the Split Feasibility Problem Fixed Point =eory ampApplications Springer Berlin Germany 2013

[13] S Reich M T Truong and T Mai ldquoe split feasibilityproblem with multiple output sets in Hilbert spacesrdquo Opti-mization Letters vol 36 2020

[14] S Suantai N Pholasa N Pholasa and P Cholamjiak ldquoemodified inertial relaxed CQ algorithm for solving the splitfeasibility problemsrdquo Journal of Industrial and ManagementOptimization vol 14 no 4 pp 1595ndash1615 2018

[15] W Cholamjiak P Cholamjiak and S Suantai ldquoAn inertialforward-backward splitting method for solving inclusionproblems in Hilbert spacesrdquo Journal of Fixed Point=eory andApplications vol 20 no 1 p 42 2018

[16] Q L Dong Y J Cho and L L Zhong ldquoInertial projectionand contraction algorithms for variational inequalitiesrdquoJournal of Global Optimization vol 70 no 3 pp 1ndash18 2018

[17] D R Sahu Y J Cho and Q L Dong ldquoInertial relaxed CQalgorithms for solving a split feasibility problem in Hilbertspacesrdquo Numerical Algorithms vol 3 2020

[18] Y Shehu and A Gibali ldquoInertial krasnoselskiindashmann methodin banach spacesrdquo Mathematics vol 8 no 4 2020

[19] Y Shehu P T Vuong and P Cholamjiak ldquoA self-adaptiveprojection method with an inertial technique for split feasi-bility problems in Banach spaces with applications to imagerestoration problemsrdquo Journal of Fixed Point =eory andApplications vol 21 no 2 2019

[20] H-K Xu ldquoIterative methods for the split feasibility problemin infinite-dimensional Hilbert spacesrdquo Inverse Problemsvol 26 no 10 Article ID 105018 2010

[21] R I Bot and E R Csetnek ldquoAn inertial Tsengrsquos type proximalalgorithm for nonsmooth and nonconvex optimizationproblemsrdquo Journal of Optimization =eory and Applicationsvol 171 no 2 pp 600ndash616 2016

12 Journal of Mathematics

Acknowledgments

is research was supported by the National Natural ScienceFoundation of China (Grant no 61872126) Henan ProvinceKey Science and Technology Project (Grant no192102210123) and Young Backbone Teachers in Univer-sities of Henan Province (Grant no 2019GGJS061)

References

[1] Y Censor and T Elfving ldquoA multiprojection algorithm usingbregman projections in a product spacerdquo Numerical Algo-rithms vol 8 no 2 pp 221ndash239 1994

[2] C Byrne ldquoIterative oblique projection onto convex sets andthe split feasibility problemrdquo Inverse Problems vol 18 no 2pp 441ndash453 2002

[3] G T Herman ldquoImage reconstruction from projections thefundamentals of computerized tomographyrdquo 1980

[4] Y Censor and A Segal ldquoIterative projection methods inbiomedical inverse problemsrdquo 2008

[5] Y Censor T Bortfeld B Martin and A Trofimov ldquoA unifiedapproach for inversion problems in intensity-modulated ra-diation therapyrdquo Physics in Medicine and Biology vol 51no 10 pp 2353ndash2365 2006

[6] C Yair and Tommy ldquoe multiple-sets split feasibilityproblem and its applications for inverse problemsrdquo InverseProblems vol 21 no 6 pp 2071ndash2084 2005

[7] Q Yang ldquoe relaxed CQ algorithm solving the split feasi-bility problemrdquo Inverse Problems vol 20 no 4pp 1261ndash1266 2004

[8] G Lopez V Martın-Marquez and F H Wang ldquoSolving thesplit feasibility problem without prior knowledge of matrixnormsrdquo Inverse Problems vol 28 no 8 pp 374ndash389 2012

[9] J Deepho and P Kumam ldquoA modified halpernrsquos iterativescheme for solving split feasibility problemsrdquo Abstract andApplied Analysis vol 2012 Article ID 876069 8 pages 2012

[10] Y Dang and Y Gao ldquoe strong convergence of a KM-CQ-like algorithm for a split feasibility problemrdquo Inverse Prob-lems vol 27 no 1 Article ID 015007 2011

[11] S N He and Z Y Zhao ldquoStrong convergence of a relaxed CQalgorithm for the split feasibility problemrdquo Journal of In-equalities and Applications vol 2013 no 1 11 pages 2013

[12] Y H Yao and Mihai Strong Convergence of a Self-AdaptiveMethod for the Split Feasibility Problem Fixed Point =eory ampApplications Springer Berlin Germany 2013

[13] S Reich M T Truong and T Mai ldquoe split feasibilityproblem with multiple output sets in Hilbert spacesrdquo Opti-mization Letters vol 36 2020

[14] S Suantai N Pholasa N Pholasa and P Cholamjiak ldquoemodified inertial relaxed CQ algorithm for solving the splitfeasibility problemsrdquo Journal of Industrial and ManagementOptimization vol 14 no 4 pp 1595ndash1615 2018

[15] W Cholamjiak P Cholamjiak and S Suantai ldquoAn inertialforward-backward splitting method for solving inclusionproblems in Hilbert spacesrdquo Journal of Fixed Point=eory andApplications vol 20 no 1 p 42 2018

[16] Q L Dong Y J Cho and L L Zhong ldquoInertial projectionand contraction algorithms for variational inequalitiesrdquoJournal of Global Optimization vol 70 no 3 pp 1ndash18 2018

[17] D R Sahu Y J Cho and Q L Dong ldquoInertial relaxed CQalgorithms for solving a split feasibility problem in Hilbertspacesrdquo Numerical Algorithms vol 3 2020

[18] Y Shehu and A Gibali ldquoInertial krasnoselskiindashmann methodin banach spacesrdquo Mathematics vol 8 no 4 2020

[19] Y Shehu P T Vuong and P Cholamjiak ldquoA self-adaptiveprojection method with an inertial technique for split feasi-bility problems in Banach spaces with applications to imagerestoration problemsrdquo Journal of Fixed Point =eory andApplications vol 21 no 2 2019

[20] H-K Xu ldquoIterative methods for the split feasibility problemin infinite-dimensional Hilbert spacesrdquo Inverse Problemsvol 26 no 10 Article ID 105018 2010

[21] R I Bot and E R Csetnek ldquoAn inertial Tsengrsquos type proximalalgorithm for nonsmooth and nonconvex optimizationproblemsrdquo Journal of Optimization =eory and Applicationsvol 171 no 2 pp 600ndash616 2016

12 Journal of Mathematics