In this section, we prove a strong convergence theorem which solves the problem for finding a common solution of the system of equilibrium problems and variational inequality problems and fixed point problems in Banach spaces.
Proof.
We will complete this proof by seven steps below.
Step 1. We will show that
C
n
is closed and convex for each
n
≥
0
.
From the definition of Cn, it is obvious that Cn is closed. Moreover, since
(33)ϕ(z,yn)≤ϕ(z,xn) is equivalent to2〈z,Jxn-Jyn〉-∥xn∥2+∥yn∥2≤0.
It follows that Cn is convex for each n≥0. Therefore, Cn is closed and convex for each n≥0.
Step 2. We will show that
Ω
⊂
C
n
for each
n
≥
0
.
From the assumption, we see that F⊂C0=C. Suppose that F⊂Ck for some k≥1. Now, for p∈Ω, Since Sk and Tk are quasi-ϕ-nonexpansive and by Lemmas 10 and 11, we compute
(34)ϕ(p,yk)=ϕ(p,J-1(αJxk+βJTkzk+γJSkuk))=∥p∥2-2〈p,αJxk+βJTkzk+γJSkuk〉+∥αJxk+βJTkzk+γJSkuk∥2≤∥p∥2-2α〈p,Jxk〉-2β〈p,JTkzk〉-2γ〈p,JSkuk〉+α∥Jxk∥2+β∥JTkzk∥2+γ∥JSkuk∥2=αϕ(p,xk)+βϕ(p,Tkzk)+γϕ(p,Skuk)≤αϕ(p,xk)+βϕ(p,zk)+γϕ(p,uk)=αϕ(p,xk)+βϕ(p,Krkxk)+γϕ(p,Trkxk)≤αϕ(p,xk)+βϕ(p,xk)+γϕ(p,xk)=(α+β+γ)ϕ(p,xk)=ϕ(p,xk).
Therefore, p∈Ck+1. By the induction, this implies that Ω⊂Cn for each n≥0.
Hence, the sequence {xn} is well defined.
Step 3. We will show that the sequence
{
x
n
}
is bounded.
Let G:=⋂n=0∞Cn. From Ω⊂Cn for each n≥0 and {xn} is well defined. From the assumption of G, we see that G is closed and convex subset of C.
Let p^=ΠG(x0), where p^ is the unique element that satisfies infx∈G ϕ(x,x0)=ϕ(p^,x0). Now, we will show that ∥xn-p^∥→0 as n→∞.
From the assumption of Cn, we know that C⊃C1⊃C2⊃C3⊃⋯ and since xn=ΠCn(x0) and xn+1=ΠCn+1(x0)∈Cn+1⊂Cn. Using Lemma 4, we get
(35)ϕ(xn,x0)≤ϕ(xn+1,x0)≤⋯≤ϕ(p^,x0),
for all p^∈Ω⊂Cn, where n≥1. Then, the sequence {ϕ(xn,x0)} is bounded. Hence, the sequence {xn} is also bounded.
Step 4. We will show that there exists
p
^
∈
C
such that
x
n
→
p
^
, as n→∞.
Since xn=ΠCn(x0) and xn+1=ΠCn+1x0∈Cn+1⊂Cn, we have
(36)ϕ(xn,x0)≤ϕ(xn+1,x0), ∀n≥0.
Therefore, the sequence {ϕ(xn,x0)} is nondecreasing. Hence limn→∞ϕ(xn,x0) exists. By the definition of Cn, one has that Cm⊂Cn and xm=ΠCmx0∈Cn for any positive integer m≥n.
It follows that
(37)ϕ(xm,xn)=ϕ(xm,ΠCnx0)≤ϕ(xm,x0)-ϕ(ΠCnx0,x0)=ϕ(xm,x0)-ϕ(xn,x0).
Since limn→∞ϕ(xn,x0) exists, by taking m,n→∞ in (37), we have ϕ(xm,xn)→0.
From Lemma 7, it follows that ∥xm-xn∥→0 as m,n→∞. Thus {xn} is a Cauchy sequence.
Without a loss of generalization, we can assume that xn⇀p0∈E. Since {xn} is bounded and E is reflexive. Since G:=⋂n=0∞Cn is closed and convex, it follows that p0∈G, ∀n≥0. Moreover, by using the weak lower semicontinuous of the norm on E and (35), we obtain
(38)ϕ(p^,x0)≤ϕ(p0,x0)≤liminfn→∞ ϕ(xn,x0)≤limsupn→∞ ϕ(xn,x0)≤infx∈G ϕ(x,x0)=ϕ(p^,x0).
This implies that
(39)limn→∞ϕ(xn,x0)=ϕ(p^,x0)=ϕ(p0,x0)=infx∈G ϕ(x,x0).
By using Lemma 5, we have 〈p^-p0,Jp^-Jp0〉=0 and hence, p^=p0. By the definition of ϕ, we get
(40)liminfn→∞ ϕ(xn,x0)=liminfn→∞(∥xn∥2-2〈xn,Jx0〉+∥x0∥2)≥∥p^∥2-2〈p^,Jx0〉+∥x0∥2=ϕ(p^,x0).
Therefore, ∥xn∥→∥p^∥. Since xn⇀p^, by the Kadec-Klee property of E, we obtain
(41)limn→∞xn=p^.
From J is uniformly continuous, we also have
(42)limn→∞Jxn=Jp^.Step 5. We will show that yn→p^, zn→p^ and un→p^ as n→∞.
Since xn+1∈Cn+1, we have ϕ(xn+1,yn)≤ϕ(xn+1,xn)→0, as n→∞. Thus, from (21), we obtain
(43)∥yn∥→∥p^∥, as n→∞,
and so
(44)∥Jyn∥→∥Jp^∥, as n→∞.
This implies that {Jyn} is bounded. Note that reflexivity of E implies reflexivity of E*. Thus, we assume that Jyn⇀y∈E*. Furthermore, the reflexivity of E implies there exists x∈E such that y=Jx. Then, it follows that
(45)ϕ(xn+1,yn)=∥xn+1∥2-2〈xn+1,Jyn〉+∥yn∥2=∥xn+1∥2-2〈xn+1,Jyn〉+∥Jyn∥2.
Taking the liminfn→∞ on both sides of (45) and using weak lower semicontinuous of norm to get that
(46)0≥∥p^∥2-2〈p^,y〉+∥y∥2=∥p^∥2-2〈p^,Jx〉+∥Jx∥2=∥p^∥2-2〈p^,Jx〉+∥x∥2=ϕ(p^,x).
Thus p^=x, and so y=Jx=Jp^. It follows that Jyn⇀Jp^. Now, from (44) and the Kadec-Klee property of E*, we obtain
(47)Jyn→Jp^, as n→∞.
Thus the demicontinuity of J-1 implies that yn⇀p^. Now, from (43) and the fact that E has the Kadec-Klee property, we obtain
(48)limn→∞yn=p^.
In the fact that xn→p^ and yn→p^ as n→∞, we get
(49)limn→∞ϕ(p,zn)=ϕ(p,p^).
Since zn=Krnxn, it follows from Lemma 11 that
(50)ϕ(zn,xn)=ϕ(Krnxn,xn)≤ϕ(p,xn)-ϕ(p,Krnxn)≤ϕ(p,xn)-ϕ(p,zn)→0, as n→∞.
From (21), we obtain
(51)∥zn∥→∥p^∥,
and so {zn} is bounded. Since E is reflexive, we assume that zn⇀z∈E. It follows that
(52)ϕ(zn,xn)=∥zn∥2-2〈zn,Jxn〉+∥Jxn∥2.
Taking the liminfn→∞ on both sides of (52) and using the continuity of J, we get that
(53)0≥∥z∥2-2〈z,Jp^〉+∥Jp^∥2=ϕ(z,p^).
This implies that p^=z and hence zn⇀p^.
Now, from (51) and the Kadec-Klee property of E, we obtain
(54)limn→∞zn=p^.
In the fact that xn→p^, yn→p^ and zn→p^ as n→∞, we get
(55)limn→∞ϕ(p,un)=ϕ(p,p^).
Since un=Trnxn, it follows from Lemma 10 that
(56)ϕ(un,xn)=ϕ(Trnxn,xn)≤ϕ(p,xn)-ϕ(p,Trnxn)≤ϕ(p,xn)-ϕ(p,un)→0, as n→∞.
From (21), we obtain
(57)∥un∥→∥p^∥,
and so {un} is bounded. Since E is reflexive, we assume that un⇀u∈E. It follows that
(58)ϕ(un,xn)=∥un∥2-2〈un,Jxn〉+∥Jxn∥2.
Taking the liminfn→∞ on both sides of (58) and using the continuity of J, we get that
(59)0≥∥u∥2-2〈u,Jp^〉+∥Jp^∥2=ϕ(u,p^).
This implies that p^=u and hence un⇀p^.
Now, from (57) and the Kadec-Klee property of E, we obtain
(60)limn→∞un=p^.Step 6. We will show that p^∈Ω.
Substep 1. We will show that
p
^
∈
⋂
n
=
1
N
VI
(
C
,
A
n
)
.
From the definition of Krn of algorithm (32), we have
(61)〈y-zn,Anzn〉+1rn〈y-zn,Jzn-Jxn〉≥0, ∀y∈C.
Let {nk}k∈ℕ⊂ℕ be such that Ank=A1, for all k∈ℕ. Then from (61), we obtain
(62)〈y-znk,A1znk〉+1rnk〈y-znk,Jznk-Jxnk〉≥0, ∀y∈C,
and that is
(63)〈y-znk,A1znk〉+〈y-znk,Jznk-Jxnkrnk〉≥0, ∀y∈C.
Now, we set vt=tv+(1-t)p^, for all t∈(0,1] and v∈C.
Therefore, we get vt∈C. From (63), it follows that
(64)〈vt-znk,A1vt〉≥〈vt-znk,A1vt〉-〈vt-znk,A1znk〉-〈vt-znk,Jznk-Jxnkrnk〉=〈vt-znk,A1vt-A1znk〉-〈vt-znk,Jznk-Jxnkrnk〉.
From the continuity of J and (41) and (54), we have xnk→p^, znk→p^, as k→∞, we obtain
(65)Jznk-Jxnkrnk→0, as k→∞.
Since A1 is a monotone mapping, we also have 〈vt-znk,A1vt-A1znk〉≥0.
Thus, it follows that
(66)0≤limn→∞〈vt-znk,A1vt〉=〈vt-p^,A1vt〉,
and hence
(67)〈y-p^,A1vt〉≥0, ∀y∈C.
If t→0, we obtain
(68)〈y-p^,A1p^〉≥0, ∀y∈C.
This implies that p^∈VI(C,A1).
Similarly, let {nk}k∈ℕ⊂ℕ be such that Ank=A2, for all k∈ℕ.
Then, we have again that p^∈VI(C,A2).
Continuing in the same way, we obtain that p^∈VI(C,An), where n=3,4,5,…,N.
Hence, p^∈⋂n=1NVI(C,An).
Substep 2. We will show that
p
^
∈
⋂
k
=
1
M
SEP
(
F
k
)
.
From the definition of Trn of algorithm (32) and (A2), we have
(69)1rn〈y-un,Jun-Jxn〉≥-Fn(un,y)≥F(y,un), ∀y∈C.
Let {nk}k∈ℕ⊂ℕ be such that Fnk=F1, for all k∈ℕ. Then from (69), we obtain
(70)1rnk〈y-unk,Junk-Jxnk〉=〈y-unk,Junk-Jxnkrnk〉 ≥F1(y,unk), ∀y∈C.
From the continuity of J and (41) and (60), we have xnk→p^, unk→p^, as k→∞ and we obtain
(71)Junk-Jxnkrnk→0, as k→∞.
Therefore, F1(y,unk)≤0, ∀y∈C.
Now, we set wt=tw+(1-t)p^, for all t∈(0,1] and w∈C.
Consequently, we get wt∈C. And so F1(wt,p^)≤0.
Therefore, from (A1), we obtain
(72)0=F1(wt,wt)≤tF1(wt,y)+(1-t)F1(wt,p^)≤tF1(wt,y).
Thus, F1(wt,y)≥0, ∀y∈C. From (A3), if t→0, then we get F1(p^,y)≥0, ∀y∈C. This implies that p^∈EP(F1).
Similarly, let {nk}k∈ℕ⊂ℕ be such that Fnk=F2, for all k∈ℕ. Then, we have again that p^∈EP(F2).
Continuing in the same way, we obtain that p^∈EP(Fk), where k=3,4,5,…,M. Hence, p^∈⋂k=1MSEP(Fk).
Substep 3. We will show that
p
^
∈
⋂
i
=
1
D
F
(
S
i
)
.
From algorithm (32) and Lemma 12, we compute
(73)ϕ(p,yn)=ϕ(p,J-1(αJxn+βJTnzn+γJSnun))=∥p∥2-2〈p,αJxn+βJTnzn+γJSnun〉+∥αJxn+βJTnzn+γJSnun∥2≤∥p∥2-2α〈p,Jxn〉-2β〈p,JTnzn〉-2γ〈p,JSnun〉+α∥xn∥2+β∥Tnzn∥2+γ∥Snun∥2-αβg(∥Jxn-JTnzn∥)-αγg(∥Jxn-JSnun∥)=αϕ(p,xn)+βϕ(p,Tnzn)+γϕ(p,Snun)-αβg(∥Jxn-JTnzn∥)-αγg(∥Jxn-JSnun∥)≤αϕ(p,xn)+βϕ(p,zn)+γϕ(p,un)-αβg(∥Jxn-JTnzn∥)-αγg(∥Jxn-JSnun∥)=αϕ(p,xn)+βϕ(p,Krnxn)+γϕ(p,Trnxn)-αβg(∥Jxn-JTnzn∥)-αγg(∥Jxn-JSnun∥)≤αϕ(p,xn)+βϕ(p,xn)+γϕ(p,xn)-αβg(∥Jxn-JTnzn∥)-αγg(∥Jxn-JSnun∥)=(α+β+γ)ϕ(p,xn)-αβg(∥Jxn-JTnzn∥)-αγg(∥Jxn-JSnun∥)=ϕ(p,xn)-αβg(∥Jxn-JTnzn∥)-αγg(∥Jxn-JSnun∥)≤ϕ(p,xn).
From (73), we have
(74)ϕ(p,yn)≤ϕ(p,xn)-αβg(∥Jxn-JTnzn∥)-αγg(∥Jxn-JSnun∥)≤ϕ(p,xn)-αγg(∥Jxn-JSnun∥).
From xn→p^, yn→p^, and α,γ>0, we obtain
(75)αγg(∥Jxn-JSnun∥) ≤ϕ(p,xn)-ϕ(p,yn)→0, as n→∞.
Therefore,
(76)g(∥Jxn-JSnun∥)→0, as n→∞.
It follows from the property of g that
(77)∥Jxn-JSnun∥→0, as n→∞.
From (42), we have Jxn→Jp^, as n→∞. Then,
(78)JSnun→Jp^, as n→∞,
and so
(79)∥Snun∥→∥p^∥, as n→∞.
Moreover, the demicontinuity of J-1 implies that Snun⇀p^ as n→∞. Thus, the Kadec-Klee property of E, we obtain
(80)Snun→p^, as n→∞.
Let {nk}k∈ℕ⊂ℕ be such that Snk=S1, for all k∈ℕ.
Then, from (60), we have unk→p^, as n→∞. It follows from (80) and the closedness of S1 that
(81)p^=limk→∞Snkunk=limk→∞S1unk=S1p^.
This implies that p^∈F(S1).
Similarly, let {nk}k∈ℕ⊂ℕ be such that Snk=S2, for all k∈ℕ. Then, we have again that p^∈F(S2).
Continuing in the same way, we obtain that p^∈F(Si), where i=3,4,5,…,D.
Hence, p^∈⋂i=1DF(Si).
Substep 4. We will show that
p
^
∈
⋂
i
=
1
D
F
(
T
i
)
.
From (73), we have
(82)ϕ(p,yn)≤ϕ(p,xn)-αβg(∥Jxn-JTnzn∥)-αγg(∥Jxn-JSnun∥)≤ϕ(p,xn)-αβg(∥Jxn-JTnzn∥).
From xn→p^, yn→p^, and α,β>0, we obtain
(83)αβg(∥Jxn-JTnzn∥) ≤ϕ(p,xn)-ϕ(p,yn)→0, as n→∞.
Therefore,
(84)g(∥Jxn-JTnzn∥)→0, as n→∞.
It follows from the property of g that
(85)∥Jxn-JTnzn∥→0, as n→∞.
From (42), we have Jxn→Jp^, as n→∞. Then,
(86)JTnzn→Jp^, as n→∞,
and so
(87)∥Tnzn∥→∥p^∥, as n→∞.
Moreover, the demicontinuity of J-1 implies that Tnzn⇀p^ as n→∞. Thus, the Kadec-Klee property of E, we obtain
(88)Tnzn→p^, as n→∞.
Let {nk}k∈ℕ⊂ℕ be such that Tnk=T1, for all k∈ℕ.
Then, from (54), we have znk→p^, as n→∞. It follows from (88) and the closedness of T1 that
(89)p^=limk→∞Tnkznk=limk→∞T1znk=T1p^.
This implies that p^∈F(T1).
Similarly, let {nk}k∈ℕ⊂ℕ be such that Tnk=T2, for all k∈ℕ. Then, we have again that p^∈F(T2).
Continuing in the same way, we obtain that p^∈F(Ti), where i=3,4,5,…,D.
Hence, p^∈⋂i=1DF(Ti).
From Substeps (6.1)–(6.4), we can conclude that
(90)p^∈Ω:=(⋂i=1DF(Ti))⋂(⋂i=1DF(Si))⋂ (⋂k=1MSEP(Fk))⋂(⋂n=1NVI(C,An)).
Step 7. Finally, we will show that
p
^
=
Π
Ω
(
x
0
)
.
From xn=ΠCn(x0), we have
(91)〈Jx0-Jxn,xn-p〉≥0, ∀p∈Ω.
Taking n→∞ in (91), one has
(92)〈Jx0-Jp^,p^-p〉≥0, ∀p∈Ω.
Now, we have p^∈Ω and by Lemma 5, we get
(93)p^=ΠΩ(x0).
This completes the proof of Theorem 13.