# Damped Algorithms for the Split Fixed Point and Equilibrium Problems

Authors: Li-Jun Zhu; Minglun Ren
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

The main purpose of this paper is to study the split fixed point and equilibrium problems which includes fixed point problems, equilibrium problems, and variational inequality problems as special cases. A damped algorithm is presented for solving this split common problem. Strong convergence analysis is shown.


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## Body

## 1. Introduction

Very recently, the split problems (e.g., the split feasibility problem, the split common fixed points problem, and the split variational inequality problem) have been studied extensively, see, for instance, [1–19]. Now we recall the related history. Let H1 and H2 be two Hilbert spaces and C⊂H1 and Q⊂H2 two nonempty closed convex subsets. Let A:H1→H2 be a bounded linear operator. The split feasibility problem is to solve the inclusion:
(1)x∈C∩A-1(Q)
which arise in the field of intensity-modulated radiation therapy and was presented in [1]. The iteration ρn+1=projC(ρn-ςA*(I-PQ)Aρn) is popular with ς∈(0,2/∥A∥2). Further, Xu [3] suggested a single step regularized method. Dang and Gao [4] developed a damped projection algorithm. If C and Q are the fixed point sets of mappings U and T, respectively, then (1) becomes a special case of the split common fixed point problem:
(2)Findx∈Fix(U)∩A-1(Fix(T)).Censor and Segal [5] invented a scheme below to solve (2):
(3)ρn+1=U(ρn-ςA*(I-T)Aρn),n∈ℕ.Cui et al., [6] extended the damped projection algorithm to the split common fixed point problems. Let ψ:C×C→ℝ be a bifunction. The equilibrium problem is to find x†∈C such that
(4)ψ(x†,x)≥0,∀x∈C.We will indicate withEP(ψ) the set of solutions of (4).In the present paper, our main purpose is to study the following split fixed point and equilibrium problem.(5)Findapointu§∈Fix(W)∩EP(ψ)suchthatAu§∈Fix(S)∩EP(φ),
where Fix(S) and Fix(W) are the sets of fixed points of two nonlinear mappings S and W, respectively; EP(ψ) and EP(φ) are the solution sets of two equilibrium problems with bifunctions ψ and φ, respectively, and A is a bounded linear mapping. Denote the solution set of (5) by
(6)Θ={x∈Fix(W)∩EP(ψ):Ax∈Fix(S)∩EP(φ)}.We develop a damped algorithm to solve this split fixed point and equilibrium problem. Strong convergence of the suggested damped algorithm is demonstrated.


## 2. Concepts and Lemmas

LetH be a real Hilbert space with inner product 〈·,·〉 and norm ∥·∥, respectively. Let C be a nonempty closed convex subset of H. A mapping W:C→C is called nonexpansive if
(7)∥Wx†-Wx‡∥≤∥x†-x‡∥,
for all x†,x‡∈C. We call projC:H→C the metric projection if for each x♭∈H(8)∥x♭-projC(x♭)∥=inf{∥x♭-x†∥:x†∈C}.
It is well known that the metric projection projC:H→C is firmly nonexpansive, that is,
(9)∥projC(x†)-projC(x‡)∥2≤〈x†-x‡,projC(x†)-projC(x‡)〉
for all x†,x‡∈H. Hence projC is also nonexpansive.Lemma 1 (see [20]).
LetC be a nonempty closed convex subset of a real Hilbert space H. Let ψ:C×C→ℝ be a bifunction which satisfies the following conditions: (H1)
ψ(x‡,x‡)=0 for all x‡∈C;(H2)
ψ is monotone, that is, ψ(x‡,x†)+ψ(x†,x‡)≤0 for all x‡,x†∈C;(H3)
for eachx†,x‡,x♮∈C, limt↓0ψ(tx♮+(1-t)x†,x‡)≤ψ(x†,x‡);(H4)
for eachx†∈C, x‡↦ψ(x†,x‡) is convex and lower semicontinuous.
Letϖ>0 and x†∈C. Then, there exists x♮∈C such that
(10)ψ(x♮,x‡)+1ϖ〈x‡-x♮,x♮-x†〉≥0,∀x‡∈C.
Further, ifUϖψ(x†)={x♮∈C:ψ(x♮,x‡)+(1/ϖ)〈x‡-x♮,x♮-x†〉≥0,forallx‡∈C}, then the following hold: (i)
Uϖψ is single-valued and Uϖψ is firmly nonexpansive, that is, for any x†,x‡∈H, ∥Uϖψx†-Uϖψx‡∥2≤〈Uϖψx†-Uϖψx‡,x†-x‡〉;(ii)
EP(ψ) is closed and convex and EP(ψ)=Fix(Uϖψ).Lemma 2 (see [21]).
LetH be a Hilbert space and C⊂H a closed convex subset. Let W:C→C be a nonexpansive mapping. Then, the mapping I-W is demiclosed. That is, if {ρn} is a sequence in C such that ρn→ν weakly and (I-W)ρn→u strongly, then (I-W)ν=u.Lemma 3 (see [22]).
Assume that{ηn} is a sequence of nonnegative real numbers such that
(11)ηn+1≤(1-κn)ηn+ςn,n∈ℕ,
where {κn} is a sequence in (0,1) and {ςn} is a sequence such that(1)
∑n=1∞‍κn=∞;(2)
limsupn→∞(ςn/κn)≤0 or ∑n=1∞‍|ςn|<∞.
Thenlimn→∞ηn=0.


## 3. Main Results

LetH1 and H2 be two Hilbert spaces and C⊂H1 and Q⊂H2 two nonempty closed convex subsets. Let A:H1→H2 be a bounded linear operator with its adjoint A*. Let ψ:C×C→ℝ and let φ:D×D→ℝ be two bifunctions satisfying the conditions (H1)–(H4) in Lemma 1. Let S:D→D and W:C→C be two nonexpansive mappings.Algorithm 4.
Letx0∈H1. Define a sequence {xn} as follows:
(12)ρn+1=WUιψ[(1-ζn)×(ρn+ςA*(SUκφ-I)Aρn)],∀n∈ℕ,
where ι,κ, and ς are three constants satisfying ι∈(0,∞), κ∈(0,∞), ς∈(0,1/∥A∥2), and {ζn} is a real number sequence in (0,1).In the sequel, we assume that(13)Θ={x∈Fix(W)∩EP(ψ):Ax∈Fix(S)∩EP(φ)}≠∅.Theorem 5.
If{ζn} satisfies limn→∞ζn=0, ∑n=1∞‍ζn=∞ and limn→∞ζn+1/ζn=1, then {ρn} generated by algorithm (12) converges strongly to projΘ(0) which is the minimum-norm element in Θ.Proof.
Letp=projΘ(0). Then, p∈Fix(W)∩EP(ψ) and Ap∈Fix(S)∩EP(φ). Set zn=UκφAρn, yn=(1-ζn)(ρn+ςA*(SUκφ-I)Aρn) and un=Uιψ[(1-ζn)(ρn+ςA*(SUκφ-I)Aρn)] for all n∈ℕ. Then un=Uιψyn. From Lemma 1, we know that Uιψ and Uκφ are firmly nonexpansive. Thus, we have
(14)∥zn-Ap∥=∥UκφAρn-Ap∥≤∥Aρn-Ap∥,(15)∥un-p∥=∥Uιψyn-p∥≤∥yn-p∥,(16)∥SUκφAρn-Ap∥2=∥SUκφAρn-SUκφAp∥≤∥UκφAρn-UκφAp∥2≤∥Aρn-Ap∥2-∥UκφAρn-Aρn∥2.
Note that
(17)∥un+1-un∥=∥Uιψyn+1-Uιψyn∥≤∥yn+1-yn∥,(18)∥zn+1-zn∥=∥UκφAρn+1-UκφAρn∥≤∥Aρn+1-Aρn∥.
From (12) and (15), we have
(19)∥ρn+1-p∥=∥Wun-p∥≤∥un-p∥≤∥yn-p∥.
Observe that
(20)∥yn-p∥2=∥(1-ζn)×(ρn-p+ςA*(Szn-Aρn))-ζnp∥2≤(1-ζn)∥(ρn-p+ςA*(Szn-Aρn))∥2+ζn∥p∥2=(1-ζn)[∥A*(Szn-Aρn)∥2∥ρn-p∥+2ς×〈ρn-p,A*(Szn-Aρn)〉+ς2∥A*(Szn-Aρn)∥2]+ζn∥p∥2.
Since A* is the adjoint of A, we have
(21)〈ρn-p,A*(Szn-Aρn)〉=〈A(ρn-p),Szn-Aρn〉=〈Aρn-Ap+Szn-Aρn-(Szn-Aρn),Szn-Aρn〉=〈Szn-Ap,Szn-Aρn〉-∥Szn-Aρn∥2.
Using parallelogram law, we obtain
(22)〈Szn-Ap,Szn-Aρn〉=12(∥Szn-Ap∥2+∥Szn-Aρn∥2kkkkkkk-∥Aρn-Ap∥2).
From (16), (21) and (22), we have
(23)〈ρn-p,A*(Szn-Aρn)〉=12(∥Szn-Ap∥2+∥Szn-Aρn∥2kkkkkkkk-∥Aρn-Ap∥2)-∥Szn-Aρn∥2≤12(∥Aρn-Ap∥2-∥zn-Aρn∥2kkkkkkk+∥Szn-Aρn∥2-∥Aρn-Ap∥2)-∥Szn-Aρn∥2=-12∥zn-Aρn∥2-12∥Szn-Aρn∥2.
By (20) and (23), we deduce
(24)∥yn-p∥2≤(1-ζn)[(-12∥zn-Aρn∥2-12∥Szn-Aρn∥2)∥ρn-p∥2+ς2∥A∥2∥Szn-Aρn∥2+2ς(-12∥zn-Aρn∥2-12∥Szn-Aρn∥2)]+ζn∥p∥2=(1-ζn)×[∥ρn-p∥2+(ς2∥A∥2-ς)∥Szn-Aρn∥2-ς∥zn-Aρn∥2]+ζn∥p∥2≤(1-ζn)∥ρn-p∥2+ζn∥p∥2.
It follows from (19), we get
(25)∥ρn+1-p∥2≤∥yn-p∥2≤(1-ζn)∥ρn-p∥2+ζn∥p∥2≤max{∥ρn-p∥2,∥p∥2}.
The boundedness of the sequence {ρn} yields.
Setvn=ρn+ςA*(SUκφ-I)Aρn. Then, we have
(26)∥vn+1-vn∥2=∥ρn+1-ρn+ς[A*(Szn+1-Aρn+1)-A*(Szn-Aρn)]∥2=∥ρn+1-ρn∥2+2ς〈ρn+1-ρn,A*[(Szn+1-Aρn+1)-(Szn-Aρn)]〉+ς2∥A*(Szn+1-Aρn+1)-A*(Szn-Aρn)A*(Szn+1-Aρn+1)∥2≤∥ρn+1-ρn∥2+2ς〈Aρn+1-Aρn,Szn+1-Szn-(Aρn+1-Aρn)〉+ς2∥A∥2∥Szn+1-Szn-(Aρn+1-Aρn)∥2=∥ρn+1-ρn∥2+ς2∥A∥2∥Szn+1-Szn-(Aρn+1-Aρn)∥2+2ς〈Szn+1-Szn,Szn+1-Szn-(Aρn+1-Aρn)〉-2ς∥Szn+1-Szn-(Aρn+1-Aρn)∥2=∥ρn+1-ρn∥2+ς2∥A∥2∥Szn+1-Szn-(Aρn+1-Aρn)∥2+ς(∥Szn+1-Szn∥2+∥Szn+1-Szn-(Aρn+1-Aρn)∥2-∥Aρn+1-Aρn∥2)-2ς∥Szn+1-Szn-(Aρn+1-Aρn)∥2=∥ρn+1-ρn∥2+(ς2∥A∥2-ς)×∥Szn+1-Szn-(Aρn+1-Aρn)∥2+ς(∥Szn+1-Szn∥2-∥(Aρn+1-Aρn)∥2)≤∥ρn+1-ρn∥2+(ς2∥A∥2-ς)×∥Szn+1-Szn-(Aρn+1-Aρn)∥2+ς(∥zn+1-zn∥2-∥Aρn+1-Aρn∥2).
Since ς∈(0,1/∥A∥2), we derive by virtue of (18) and (26) that
(27)∥vn+1-vn∥≤∥ρn+1-ρn∥.
According to (17) and (27), we have
(28)∥ρn+1-ρn∥=∥Wun+1-Wun∥≤∥un+1-un∥≤∥yn+1-yn∥=∥(1-ζn+1)vn+1-(1-ζn)vn∥=∥(1-ζn+1)(vn+1-vn)+(ζn-ζn+1)vn∥≤(1-ζn+1)∥vn+1-vn∥+|ζn+1-ζn|∥vn∥≤(1-ζn+1)∥ρn+1-ρn∥+|ζn+1-ζn|∥vn∥.
It follows that
(29)∥ρn+1-ρn∥≤|ζn+1-ζn|ζn+1∥vn∥.
Since {ρn} is bounded, we can deduce {vn} is also bounded. From (29), we have
(30)limn→∞∥ρn+1-ρn∥=0.
Hence,
(31)limn→∞∥ρn-Wun∥=0.
Using the firmly-nonexpansivenessity of Uιψ, we have
(32)∥un-p∥2=∥Uιψyn-p∥2≤∥yn-p∥2-∥Uιψyn-yn∥2=∥yn-p∥2-∥un-yn∥2.
Thus, we get
(33)∥ρn+1-p∥2≤∥un-p∥2≤∥yn-p∥2-∥un-yn∥2≤(1-ζn)∥ρn-p∥2+ζn∥p∥2-∥un-yn∥2.
It follows that
(34)∥un-yn∥2≤∥ρn-p∥2-∥ρn+1-p∥2+ζn∥p∥2≤(∥ρn-p∥+∥ρn+1-p∥2)∥ρn+1-p∥2+ζn∥p∥2.
This together with (30) and (C1) implies that
(35)limn→∞∥un-yn∥=0.
Note that
(36)∥ρn+1-p∥2≤∥yn-p∥2≤(1-ζn)∥ρn-p∥2+(1-ζn)(ς2∥A∥2-ς)∥Szn-Aρn∥2-(1-ζn)ς∥zn-Aρn∥2+ζn∥p∥2.
Hence,
(37)(1-ζn)(ς-ς2∥A∥2)∥Szn-Aρn∥2+(1-ζn)ς∥zn-Aρn∥2≤∥ρn-p∥2-∥ρn+1-p∥2+ζn∥p∥2≤(∥ρn-p∥+∥ρn+1-p∥)∥ρn+1-ρn∥+ζn∥p∥2,
which implies that
(38)limn→∞∥Szn-Aρn∥=limn→∞∥zn-Aρn∥=0.
So, we get
(39)limn→∞∥Szn-zn∥=0.
Since
(40)∥yn-pn∥=∥ςA*(SUκφ-I)Aρn+ζnvn∥≤ς∥A∥∥Szn-Aρn∥+ζn∥vn∥,
we get
(41)limn→∞∥ρn-yn∥=0.
From (31), (35), and (41), we get
(42)limn→∞∥ρn-Wρn∥=0.
Now, we show that
(43)limsupn→∞〈p,yn-p〉≥0.
Choose a subsequence {yni} of {yn} such that
(44)limsupn→∞〈p,yn-p〉=limi→∞〈p,yni-p〉.
Notice that {yni} is bounded, we can choose {ynij} of {yni} such that ynij⇀z. Without loss of generality, we assume that yni⇀z. From the above conclusions, we derive that
(45)ρni⇀z,uni⇀z,Aρni⇀Az,zni⇀Az.
By Lemma 2, (39), and (41), we deduce z∈Fix(W) and Az∈Fix(S).
Next, we show thatz∈EP(ψ). Since un=Uιψyn, we have
(46)ψ(un,x†)+1ι〈x†-un,un-yn〉≥0,∀x†∈C.
By the monotonicity of ψ, we have
(47)1ι〈x†-un,un-yn〉≥ψ(x†,un),
and so
(48)〈x†-uni,uni-yniι〉≥ψ(x†,uni).
Since ∥un-yn∥→0, uni⇀z, we obtain (uni-yni)/ι→0. Thus, 0≥ψ(x†,z). For t with 0<t≤1 and x†∈C, let yt=tx†+(1-t)z∈C. We obtain ψ(yt,z)≤0. Hence,
(49)0=ψ(yt,yt)≤tψ(yt,x†)+(1-t)ψ(yt,z)≤tψ(yt,x†).
So, 0≤ψ(yt,x†). And, thus, 0≤ψ(z,x†). This implies that z∈EP(ψ). Similarity, we can prove that Az∈EP(φ). To this end, we deduce z∈Fix(W)∩EP(ψ) and Az∈Fix(S)∩EP(φ). That is to say, z∈Θ. Therefore,
(50)limsupn→∞〈p,yn-p〉=limi→∞〈p,yni-p〉=limi→∞〈p,z-p〉≥0.
Finally, we prove ρn→p. From (12), we have
(51)∥ρn+1-p∥2≤∥yn-p∥2=∥(1-ζn)(vn-p)-ζnp∥2≤(1-ζn)∥vn-p∥2-2ζn〈p,yn-p〉≤(1-ζn)∥ρn-p∥2-2ζn〈p,yn-p〉.
Applying Lemma 3 and (50) to (51), we deduce ρn→p. The proof is completed.Algorithm 6.
Letρ0∈H1 arbitrarily define a sequence {ρn} by the following:
(52)ρn+1=W((1-ζn)(ρn+ςA*(S-I)Aρn)),
for all n∈ℕ, where ς∈(0,1/∥A∥2) and {ζn} is a real number sequence in (0,1).Corollary 7.
SupposeΘ1={x∈Fix(W):Ax∈Fix(S)}≠∅. If {ζn} satisfies limn→∞ζn=0, ∑n=1∞‍ζn=∞, and limn→∞ζn+1/ζn=1, then the sequence {ρn} generated by algorithm (52) converges strongly to p=projΘ1(0) which is the mum-norm element in Θ1.Algorithm 8.
Letρ0∈H1 arbitrarily define a sequence {ρn} by the following:
(53)ρn+1=Uιψ((1-ζn)(ρn+ςA*(Uκφ-I)Aρn)),
for all n∈ℕ, where ι,κ, and ς are three constants satisfying ι∈(0,∞),κ∈(0,∞), ς∈(0,1/∥A∥2), and {ζn} is a real number sequence in (0,1).Corollary 9.
SupposeΘ2={x∈EP(ψ):Ax∈EP(φ)}≠∅. If {ζn} satisfies limn→∞ζn=0, ∑n=1∞‍ζn=∞, and limn→∞ζn+1/ζn=1, then the sequence {ρn} generated by algorithm (53) converges strongly to p=projΘ2(0) which is the mum-norm element in Θ2.Algorithm 10.
Letρ0∈H1 arbitrarily define a sequence {ρn} by the following:
(54)ρn+1=projC((1-ζn)(ρn+ςA*(projQ-I)Aρn)),
for all n∈ℕ, where ς∈(0,1/∥A∥2) and {ζn} is a real number sequence in (0,1).Corollary 11.
SupposeΘ3={x∈C:Ax∈Q}≠∅. If {ζn} satisfies limn→∞ζn=0, ∑n=1∞‍ζn=∞, and limn→∞ζn+1/ζn=1, then the sequence {ρn} generated by algorithm (54) converges strongly to p=projΘ3(0) which is the mum-norm element in Θ3.



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