In this section, we firstly introduce a hybrid iterative algorithm for finding the common element of the fixed point problem and the variational inequality problem.
Proof.
The outline of our proof is as follows.
Step
1.
Λ
⊂
C
n
for all n∈N;
Step
2.
C
n
is closed and convex for all n∈N;
Step
3.
lim
n
→
∞
∥
x
n

ν
∥
=
0
where ν=proj⋂n=1∞Cnψ(ν);
Step
4.
ν
∈
Fix
(
T
)
;
Step
5.
ν
∈
VI
(
C
,
A
)
;
Step
6.
ν
=
proj
Λ
ψ
(
ν
)
=
x
†
.
Proof of Step
1. We prove this step by induction. (i) Λ⊂C0 is obvious. (ii) Suppose that Λ⊂Ck for some k∈N. Pick up x*∈Λ⊂Ck. Then, we have
(15)∥unx*∥=∥projC[xnαAxn]projC[x*αAx*]∥≤∥(xnαAxn)(x*αAx*)∥≤∥xnx*∥.
By (2), we have
(16)∥Tunx*∥2≤ ∥unx*∥2+∥Tunun∥2,(17)∥Tvnx*∥2= ∥T((1ϱn)un+ϱnTun)x*∥2≤ ∥(1ϱn)(unx*)+ϱn(Tunx*)∥2+∥(1ϱn)un+ϱnTunTvn∥2.
From (10), we obtain
(18)∥(1ϱn)un+ϱnTunTvn∥2 =∥(1ϱn)(unTvn)+ϱn(TunTvn)∥2 =(1ϱn)∥unTvn∥2+ϱn∥TunTvn∥2 ϱn(1ϱn)∥unTun∥2.
Since T is κLipschitzian and unvn=ϱn(unTun), by (18), we get
(19)∥(1ϱn)un+ϱnTunTvn∥2 ≤(1ϱn)∥unTvn∥2+ϱn3κ2∥unTun∥2 ϱn(1ϱn)∥unTun∥2 =(1ϱn)∥unTvn∥2 +(ϱn3κ2+ϱn2ϱn)∥unTun∥2.
By (10) and (16), we have
(20)∥(1ϱn)(unx*)+ϱn(Tunx*)∥2 =∥(1ϱn)(unx*)+ϱn(Tunx*)∥2 =(1ϱn)∥unx*∥2+ϱn∥Tunx*∥2 ϱn(1ϱn)∥unTun∥2 ≤(1ϱn)∥unx*∥2+ϱn(∥unx*∥2+∥unTun∥2) ϱn(1ϱn)∥unTun∥2 =∥unx*∥2+ϱn2∥unTun∥2.
From (17), (19), and (20), we deduce
(21)∥Tvnx*∥2≤ ∥unx*∥2+(1ϱn)∥unTvn∥2ϱn(12ϱnϱn2κ2)∥unTun∥2.
Since ϱn<c2<1/(1+κ2+1), we have
(22)12ϱnϱn2κ2>0
for all n∈N. This together with (21) implies that
(23)∥Tvnx*∥2≤∥unx*∥2+(1ϱn)∥unTvn∥2.
By (10), (15), and (23) and noting that ϖn≤ϱn, we have
(24)∥wnx*∥2= ∥(1ϖn)un+ϖnTvnx*∥2= (1ϖn)∥unx*∥2+ϖn∥Tvnx*∥2ϖn(1ϖn)∥unTvn∥2≤ ∥unx*∥2ϖn(ϱnϖn)∥Tvnx*∥2≤ ∥unx*∥2≤ ∥xnx*∥2
and hence x*∈Ck+1. This indicates that Λ⊂Cn for all n∈N.
Proof of Step
2. In fact, it is obvious from the assumption that C0=C is closed convex. Suppose that Ck is closed and convex for some k∈N. For any μ∈Ck, we know that ∥ykμ∥≤∥xkμ∥ is equivalent to
(25)∥ykxk∥2+2〈ykxk,xkμ〉≤0.
So Ck+1 is closed and convex. By induction, we deduce that Cn is closed and convex for all n∈N.
Proof of Step
3. Firstly, from Step 2, we note that {xn} is well defined. Since ⋂n=1∞Cn is closed convex, we also have that proj⋂n=1∞Cn is well defined and so proj⋂n=1∞Cnψ is a MeirKeeler contraction on C. By Lemma 7, there exists a unique fixed point ν∈⋂n=1∞Cn of proj⋂n=1∞Cnψ. Since Cn is a nonincreasing sequence of nonempty closed convex subsets of H with respect to inclusion, it follows that
(26)∅≠Λ⊂⋂n=1∞Cn=Mlimn→∞Cn.
Setting sn∶=projCnψ(ν) and applying Lemma 6, we can conclude that
(27)limn→∞sn=proj⋂n=1∞Cnψ(ν)=ν.
Now, we show that limn→∞∥xnν∥=0. Assume that M=lim¯n→∞∥xnν∥>0. Then, for any ϵ with 0<ϵ<M, we can choose δ1>0 such that
(28)limn→∞¯∥xnν∥>ϵ+δ1.
Since ψ is a MeirKeeler contraction, for the positive ϵ, there exists another δ2>0 such that
(29)∥xy∥<ϵ+δ2⟹∥ψ(x)ψ(y)∥<ϵ
for all x,y∈C.
In fact, we can choose a common δ>0 such that (28) and (29) hold. If δ1>δ2, then
(30)limn→∞¯∥xnν∥>ϵ+δ1>ϵ+δ2.
If δ1≤δ2, then, from (29), it follows that
(31)∥xy∥<ϵ+δ1⟹∥ψ(x)ψ(y)∥<ϵ
for all x,y∈C. Thus, we have
(32)limn→∞¯∥xnν∥>ϵ+δ,(33)∥xy∥<ϵ+δ⟹∥ψ(x)ψ(y)∥<ϵ
for all x,y∈C. Since sn→ν, there exists n0∈N such that
(34)∥snν∥<δ
for all n≥n0.
Now, we consider two possible cases.
Case
1. There exists n1≥n0 such that
(35)∥xn1ν∥≤ϵ+δ.
By (33) and (34), we get
(36)∥xn1+1ν∥≤∥xn1+1sn1+1∥+∥sn1+1ν∥=∥projCn1+1ψ(xn1)projCn1+1ψ(ν)∥+∥sn1+1ν∥≤∥ψ(xn1)ψ(ν)∥+∥sn1+1ν∥≤ ϵ+δ.
By induction, we can obtain that
(37)∥xn1+mν∥≤ϵ+δ
for all m≥1, which implies that
(38)limn→∞¯∥xnν∥≤ϵ+δ,
which contradicts (32). Therefore, we conclude that ∥xnν∥→0 as n→∞.
Case
2 (∥xnν∥>ϵ+δ for all n≥n0). Now, we prove that Case 2 is impossible. Suppose that Case 2 is true. By Lemma 8, there exists σ∈(0,1) such that
(39)∥ψ(xn)ψ(ν)∥≤σ∥xnν∥
for all n≥n0. Thus we have
(40)∥xn+1sn+1∥=∥projCn+1ψ(xn)projCn+1ψ(ν)∥≤∥ψ(xn)ψ(ν)∥≤σ∥xnν∥
for all n≥n0. It follows that
(41)limn→∞¯∥xn+1ν∥=limn→∞¯∥xn+1sn+1∥≤σlimn→∞¯∥xnν∥<limn→∞¯∥xnν∥,
which gives a contradiction. Hence we obtain
(42)limn→∞∥xnν∥=0.
Proof of Step
4. By Step 3, we deduce immediately that {xn} is bounded. Observe that
(43)∥xn+1xn∥≤∥xnν∥+∥νsn+1∥+∥sn+1xn+1∥=∥xnν∥+∥νsn+1∥+∥projCn+1ψ(xn)projCn+1ψ(ν)∥≤∥xnν∥+∥νsn+1∥+∥ψ(xn)ψ(ν)∥.
Therefore, we have
(44)limn→∞∥xn+1xn∥=0.
Since xn+1∈Cn+1, we have
(45)∥wnxn+1∥≤∥xnxn+1∥.
This together with (44) implies that
(46)limn→∞∥wnxn+1∥=limn→∞∥wnxn∥=0.
From (15) and (24), we have
(47)∥wnx*∥2≤ ∥unx*∥2≤ ∥(xnαAxn)(x*αAx*)∥2=∥xnx*∥2+α2∥AxnAx*∥2 2〈AxnAx*,xnx*〉≤ ∥xnx*∥2+α(α2λ)∥AxnAx*∥2.
Then we have
(48)(2λα)α∥AxnAx*∥2 ≤∥xnx*∥2∥wnx*∥2 ≤∥xnwn∥(∥xnx*∥+∥wnx*∥).
By (46) and (48), we obtain
(49)limn→∞∥AxnAx*∥=0.
Since projC is firmly nonexpansive, we have
(50)∥unx*∥2 =∥projC[xnαAxn]projC[x*αAx*]∥2 ≤〈(xnαAxn)(x*αAx*),unx*〉 =12(∥(xnαAxn)(x*αAx*)∥2+∥unx*∥2iiiiiiiiiiiii∥(xnαAxn)(x*αAx*)+x*un∥2) ≤12(∥xnx*∥2+∥unx*∥2iiiiiiiiiiiii∥(xnun)α(AxnAx*)∥2) =12(∥xnx*∥2+∥unx*∥2∥xnun∥2iiiiiiiiiiiii+2α〈xnun,AxnAx*〉α2∥AxnAx*∥2).
It follows that
(51)∥unx*∥2≤ ∥xnx*∥2∥xnun∥2+2α〈xnun,AxnAx*〉α2∥AxnAx*∥2.
From (24) and (51), we get
(52)∥wnx*∥2 ≤∥unx*∥2 ≤∥xnx*∥2∥xnun∥2 +2α〈xnun,AxnAx*〉α2∥AxnAx*∥2 ≤∥xnx*∥2∥xnun∥2 +2α∥xnun∥∥AxnAx*∥
and so
(53)∥xnun∥2≤ ∥xnx*∥2∥wnx*∥2+2α∥xnun∥∥AxnAx*∥≤∥xnwn∥(∥xnx*∥+∥wnx*∥)+2α∥xnun∥∥AxnAx*∥.
This together with (46) and (49) implies that
(54)limn→∞∥xnun∥=0.
Note that
(55)∥unTun∥≤∥unwn∥+∥wnTun∥≤∥unwn∥+(1ϖn)∥unTun∥+ϖn∥TvnTun∥≤∥unwn∥+(1ϖn)∥unTun∥+ϖnκ∥vnun∥≤∥unwn∥+(1ϖn)∥unTun∥+ϖnϱnκ∥unTun∥.
It follows that
(56)∥unTun∥≤1ϖn(1ϱnκ)∥unwn∥≤1c1(1c2κ)∥unwn∥⟶0.
Since xn→ν, we have un→ν by (54). So, from (56) and Lemma 5, we deduce that ν∈Fix(T).
Proof of Step
5. Define a mapping ϕ by
(57)ϕ(v)={Av+NCv,v∈C,∅,v∉C.
Then ϕ is maximal monotone (see [15]). Let (v,w)∈G(ϕ). Since wAv∈NCv and un∈C, we have 〈vun,wAv〉≥0. On the other hand, from un=projC[xnαAxn], we have
(58)〈vun,un(xnαAxn)〉≥0,
that is,
(59)〈vun,unxnα+Axn〉≥0.
Therefore, we have
(60)〈vun,w〉 ≥〈vun,Av〉 ≥〈vun,Av〉〈vun,unxnα+Axn〉 =〈vun,AvAxnunxnα〉 =〈vun,AvAun〉+〈vun,AunAxn〉 〈vun,unxnα〉 ≥〈vun,AunAxn〉〈vun,unxnα〉.
Noting that ∥unxn∥→0 and A is Lipschitz continuous, we obtain 〈vν,w〉≥0. Since ϕ is maximal monotone, we have ν∈ϕ1(0) and hence ν∈VI(C,A).
Proof of Step
6. Since xn+1=projCn+1ψ(xn), we have
(61)〈ψ(xn)xn+1,xn+1y〉≥0
for all y∈Cn+1. Since Λ⊂Cn+1, we get
(62)〈ψ(xn)xn+1,xn+1y〉≥0
for all y∈Λ. Noting that xn→ν∈Λ, we deduce
(63)〈ψ(ν)ν,νy〉≥0
for all y∈Λ. Thus ν=projΛψ(ν)=x†. This completes the proof.