Proof.
We first show that Cn, for all n≥0 is closed and convex. Clearly C0=C is closed and convex. Suppose that Ck is closed and convex for some k>1. For each z∈Ck, we see that ϕ(z,uk(i))≤ϕ(z,xk) is equivalent to
(3.3)2(〈z,xk〉-〈z,uk(i)〉)≤‖xk‖2-‖uk(i)‖2.
By the set of Ck+1, we have
(3.4)Cn+1={z∈Cn:maxi=1,2,…,Nϕ(z,un(i))≤ϕ(z,xn)+θn}=⋂i=1N{z∈C:ϕ(z,un(i))≤ϕ(z,xn)+θn}.
Hence, Cn+1 is also closed and convex.
By taking Θnj=Krj,fiKrj-1,fj-1⋯Kr1,f1 for any j∈{1,2,…,i} and Θn0=I for all n≥1. We note that un(i)=Θnizn.
Next, we show that ℱ⊂Cn,for all n≥1. For n≥1, we have ℱ⊂C=C1. For any given p∈ℱ:=(∩i=1NΩi)∩(∩i=1∞F(Ti))∩(∩i=1∞F(Si)). By (3.1) and Lemma 2.4, we have
(3.5)ϕ(p,yn)=ϕ(p,J-1(∑i=0∞βn,iJTinxn))=‖p‖2-∑i=0∞βn,i2〈p,JTinxn〉+‖∑i=0∞βn,iJTinxn‖2≤‖p‖2-∑i=0∞βn,i2〈p,JTinxn〉+∑i=0∞βn,i‖JTinxn‖2-βn,0βn,ig(‖JT0nxn-JTinxn‖)=‖p‖2-∑i=0∞βn,i2〈p,JTinxn〉+∑i=0∞βn,i‖Tinxn‖2-βn,0βn,ig(‖Jxn-JTinxn‖)=∑i=0∞βn,iϕ(p,Tinxn)-βn,0βn,ig(‖Jxn-JTinxn‖)≤knϕ(p,xn)-βn,0βn,ig(‖Jxn-JTinxn‖)≤ϕ(p,xn)+supp∈F(kn-1)ϕ(p,xn)-βn,0βn,ig(‖Jxn-JTinxn‖)≤ϕ(p,xn)+ξn-βn,0βn,ig(‖Jxn-JTinxn‖)≤ϕ(p,xn)+ξn,
where ξn=sup p∈ℱ(kn-1)ϕ(p,xn).
By (3.1) and (3.5), we note that
(3.6)ϕ(p,un(i))=ϕ(p,Θnizn)≤ϕ(p,zn)≤ϕ(p,J-1(αn,0Jxn+∑i=1∞JSinyn))=‖p‖2-2〈p,αn,0Jxn+∑i=1∞JSinyn〉+‖αn,0Jxn+∑i=1∞JSinyn‖2≤‖p‖2-2αn,0〈p,Jxn〉-2∑i=1∞αn,i〈p,JSinyn〉+αn,0‖xn‖2+∑i=1∞‖Sinyn‖2 -αn,0αn,ig‖Jxn-JSinyn‖≤αn,0ϕ(p,xn)+∑i=1∞αn,iϕ(p,Sinyn)-αn,0αn,ig‖Jxn-JSinyn‖≤αn,0ϕ(p,xn)+ζn∑i=1∞αn,iϕ(p,yn)-αn,0αn,ig‖Jxn-JSinyn‖≤αn,0ϕ(p,xn)+ζn∑i=1∞αn,i(ϕ(p,xn)+ξn)-αn,0αn,ig‖Jxn-JSinyn‖≤αn,0ϕ(p,xn)+ζn∑i=1∞αn,iϕ(p,xn)+ξnζn∑i=1∞αn,i-αn,0αn,ig‖Jxn-JSinyn‖≤ζnϕ(p,xn)+ξnζn∑i=1∞αn,i-αn,0αn,ig‖Jxn-JSinyn‖≤ϕ(p,xn)+supp∈F(ζn-1)ϕ(p,xn)+ξnζn∑i=1∞αn,i-αn,0αn,ig‖Jxn-JSinyn‖≤ϕ(p,xn)+δn+ξnζn-αn,0αn,ig‖Jxn-JSinyn‖≤ϕ(p,xn)+θn,
where δn=sup p∈ℱ(ζn-1)ϕ(p,xn), θn=δn+ξnζn. By assumptions on {kn} and {ζn}, we have
(3.7)ξn=supp∈F(kn-1)ϕ(p,xn)≤supp∈F(kn-1)(‖p‖+M)2⟶0 as n⟶∞,(3.8)δn=sup p∈ℱ(ζn-1)ϕ(p,xn)≤sup p∈ℱ(ζn-1)(∥p∥+M)2⟶0 as n⟶∞,
where M=sup n≥0∥xn∥.
So, we have p∈Cn+1. This implies that ℱ∈Cn,for all n≥0 and also {xn} is well defined.
From Lemma 2.2 and xn=ΠCnx0, we have
(3.9)〈xn-z,Jx0-Jxn〉≥0, ∀z∈Cn,〈xn-p,Jx0-Jxn〉≥0, ∀p∈Cn.
From Lemma 2.3, one has
(3.10)ϕ(xn,x0)=ϕ(ΠCnx0,x0)≤ϕ(p,x0)-ϕ(p,xn)≤ϕ(p,x0)
for all p∈ℱ⊂Cn and n≥1. Then, the sequence {ϕ(xn,x0)} is also bounded. Thus {xn} is bounded. Since xn=ΠCnx0 and xn+1=ΠCn+1x0∈Cn+1⊂Cn, we have
(3.11)ϕ(xn,x0)≤ϕ(xn+1,x0), ∀n∈N.
Therefore, {ϕ(xn,x0)} is nondecreasing. Hence, the limit of {ϕ(xn,x0)} exists. By the construction of Cn, one has that Cm⊂Cn and xm=ΠCmx0∈Cn for any positive integer m≥n. It follows that
(3.12)ϕ(xm,xn)=ϕ(xm,ΠCnx0)≤ϕ(xm,x0)-ϕ(ΠCnx0,x0)=ϕ(xm,x0)-ϕ(xn,x0).
Letting m,n→0 in (3.12), we get ϕ(xm,xn)→0. It follows from Lemma 2.1, that ∥xm-xn∥→0 as m,n→∞. That is, {xn} is a Cauchy sequence.
Since {xn} is bounded and E is reflexive, there exists a subsequence {xni}⊂{xn} such that xni⇀u. Since Cn is closed and convex and Cn+1⊂Cn, this implies that Cn is weakly closed and u∈Cn for each n≥0. since xn=ΠCnx0, we have
(3.13)ϕ(xni,x0)≤ϕ(u,x0), ∀ni≥0.
Since
(3.14)liminfni→∞ ϕ(xni,x0)=liminfni→∞{‖xni‖2-2〈xni,Jx0〉+‖x0‖2}≤‖u‖2-2〈u,Jx0〉+‖x0‖2=ϕ(u,x0).
We have
(3.15)ϕ(u,x0)≤liminfni→∞ ϕ(xni,x0)≤limsupni→∞ ϕ(xni,x0)≤ϕ(u,x0).
This implies that lim ni→∞ϕ(xni,x0)=ϕ(u,x0). That is, ∥xni∥→∥u∥. Since xni⇀u, by the Kadec-klee property of E, we obtain that
(3.16)limn→∞xni=u.
If there exists some subsequence {xnj}⊂{xn} such that xnj→q, then we have
(3.17)ϕ(u,q)=limni→∞,nj→∞ϕ(xni,xnj)≤limni→∞,nj→∞(ϕ(xni,x0)-ϕ(ΠCnjx0,x0))=limni→∞,nj→∞(ϕ(xni,x0)-ϕ(xnjx0,x0))=0.
Therefore, we have u=q. This implies that
(3.18)limn→∞xn=u.
Since
(3.19)ϕ(xn+1,xn)=ϕ(xn+1,ΠCnx0)≤ϕ(xn+1,x0)-ϕ(ΠCnx0,x0)=ϕ(xn+1,x0)-ϕ(xn,x0)
for all n∈ℕ, we also have
(3.20)limn→∞ϕ(xn+1,xn)=0.
Since xn+1=ΠCn+1x0∈Cn+1 and by the definition of Cn+1, for i=1,2,…,N, we have
(3.21)ϕ(xn+1,uni)≤ϕ(xn+1,xn)+θn.
Noticing that lim n→∞ϕ(xn+1,xn)=0, we obtain
(3.22)limn→∞ϕ(xn+1,uni)=0, for i=1,2,…,N.
It then yields that lim n→∞(∥xn+1∥-∥uni∥)=0,for all i=1,2,…,N. Since lim n→∞∥xn+1∥=∥u∥, we have
(3.23)limn→∞‖uni‖=‖u‖, ∀i=1,2,…,N.
Hence,
(3.24)limn→∞‖Juni‖=‖Ju‖, ∀i=1,2,…,N.
From Lemma 2.1 and (3.22), we have
(3.25)limn→∞‖xn+1-xn‖=limn→∞‖xn+1-uni‖=0, ∀i=1,2,…,N.
By the triangle inequality, we get
(3.26)limn→∞‖xn-uni‖=0, ∀i=1,2,…,N.
Since J is uniformly norm-to-norm continuous on bounded sets, we note that
(3.27)limn→∞‖Jxn-Juni‖=limn→∞‖Jxn+1-Juni‖=0, ∀i=1,2,…,N.
Now, we prove that u∈(∩i=1∞F(Ti))∩(∩i=1∞F(Si)). From the construction of Cn, we obtain that
(3.28)ϕ(xn+1,yn)≤ϕ(xn+1,xn)+ξn.
From (3.7) and (3.20), we have
(3.29)limn→∞ϕ(xn+1,yn)=0.
By Lemma 2.1, we also have
(3.30)limn→∞‖xn+1-yn‖=0.
Since J is uniformly norm-to-norm continuous on bounded sets, we note that
(3.31)limn→∞‖Jxn+1-Jyn‖=0.
From (2.4) and (3.29), we have (∥xn+1∥-∥yn∥)2→0. Since ∥xn+1∥→∥u∥, it yields that
(3.32)‖yn‖⟶‖u‖ as n⟶∞.
Since J is uniformly norm-to-norm continuous on bounded sets, it follows that
(3.33)‖Jyn‖⟶‖Ju‖ as n⟶∞.
This implies that {Jyn} is bounded in E*. Since E is reflexive, there exists a subsequence {Jyni}⊂{Jyn} such that Jyni⇀r∈E*. Since E is reflexive, we see that J(E)=E*. Hence, there exists x∈E such that Jx=r. We note that
(3.34)ϕ(xni+1,yni)=‖xni+1‖2-2〈xni+1,Jyni〉+‖yni‖2=‖xni+1‖2-2〈xni+1,Jyni〉+‖Jyni‖2.
Taking the limit interior of both side and in view of weak lower semicontinuity of norm ∥·∥, we have
(3.35)0≥‖u‖2-2〈u,r〉+‖r‖2=‖u‖2-2〈u,Jx〉+‖Jx‖2=‖u‖2-2〈u,Jx〉+‖x‖2=ϕ(u,x),
that is, u=x. This implies that r=Ju and so Jyn⇀Jp. It follows from lim n→∞∥Jyn∥=∥Ju∥, as n→∞ and the Kadec-Klee property of E* that Jyni→Ju as n→∞. Note that J-1:E*→E is hemicontinuous, it yields that yni⇀u. It follows from lim n→∞∥un∥=∥u∥, as n→∞ and the Kadec-Klee property of E that lim ni→∞yni=u.
By similar, we can prove that
(3.36)limn→∞yn=u.
By (3.20) and (3.30), we obtain
(3.37)limn→∞‖xn-yn‖=0.
Since J is uniformly norm-to-norm continuous on bounded sets, we note that
(3.38)limn→∞‖Jxn-Jyn‖=0.
So, from (3.27) and (3.31), by the triangle inequality, we get
(3.39)limn→∞‖Jyn-Juni‖=0, for i=1,2,…,N.
Since J-1 is uniformly norm-to-norm continuous on bounded sets, we note that
(3.40)limn→∞‖yn-uni‖=0, for i=1,2,…,N.
Since
(3.41)ϕ(p,xn)-ϕ(p,yn)=‖xn‖2-‖yn‖2-2〈p,Jxn-Jyn〉≤‖xn‖2-‖yn‖2+2‖p‖‖Jxn-Jyn‖≤‖xn-yn‖(‖xn‖+‖yn‖)+2‖p‖‖Jxn-Jyn‖.
From (3.37) and (3.38), we obtain
(3.42)ϕ(p,xn)-ϕ(p,yn)⟶0, n⟶∞.
On the other hand, we observe that, for i=1,2,…,N. (3.43)ϕ(p,xn)-ϕ(p,uni)=‖xn‖2-‖uni‖2-2〈p,Jxn-Juni〉≤‖xn‖2-‖uni‖2+2‖p‖‖Jxn-Juni‖≤‖xn-uni‖(‖xn‖+‖uni‖)+2‖p‖‖Jxn-Juni‖.
From (3.22) and (3.27), we have
(3.44)ϕ(p,xn)-ϕ(p,uni)⟶0, n⟶∞, ∀i=1,2,…,N.
For any p∈∩i=1NΩi∩(∩i=1∞F(Ti))∩(∩i=1∞F(Si)), it follows from (3.5) that
(3.45)βn,0βn,ig(‖Jxn-JTinxn‖)≤ϕ(p,xn)+ξn-ϕ(p,yn).From condition, liminf n→∞βn,0βn,i>0, property of g, (3.7), and (3.42), we have that
(3.46)‖Jxn-JTinxn‖⟶0, n⟶∞, ∀i=1,2,…,N.
Since xn→u and J is uniformly norm-to-norm continuous. It yields Jxn→Jp. Hence from (3.46), we have
(3.47)‖xn-Tinxn‖⟶0, n⟶∞, ∀i=1,2,…,N.
Since xn→u, this implies that lim n→∞JTinxn→Ju as n→∞. Since J-1:E*→E is hemicontinuous, it follows that
(3.48)Tinxn⇀u, for each i≥1.
On the other hand, for each i≥1, we have
(3.49)‖Tinxn‖-‖u‖=|‖Tinxn‖-‖u‖|≤‖Tinxn-u‖⟶0, n⟶∞.
from this, together with (3.48) and the Kadec-Klee property of E, we obtain
(3.50)Tinxn⟶u, for each i≥1.
On the other hand, by the assumption that Ti is uniformly Li-Lipschitz continuous, we have
(3.51)‖Tin+1xn-Tinxn‖≤‖Tin+1xn-Tin+1xn+1‖+‖Tin+1xn+1-xn+1‖ +‖xn+1-xn‖+‖xn-Tinxn‖≤(Li+1)‖xn+1-xn‖+‖Tin+1xn+1-xn+1‖ +‖xn-Tinxn‖.
By (3.18) and (3.50), we obtain
(3.52)limn→∞‖Tin+1xn-Tinxn‖=0, ∀i≥1,
and lim n→∞Tin+1xn=u, that is, TiTnxn→u, for all i≥1. By the closeness of Ti, we have Tiu=u, for all i≥1. This implies that u∈∩i=1∞F(Ti).
By the similar way, we can prove that for each i≥1(3.53)‖Jxn-JSinyn‖⟶0, n⟶∞.
Since xn→u and J is uniformly norm-to-norm continuous. it yields Jxn→Jp. Hence from (3.53), we have
(3.54)‖xn-Sinyn‖⟶0, n⟶∞.
Since xn→u, this implies that lim n→∞JSinyn→Ju as n→∞. Since J-1:E*→E is hemicontinuous, it follows that
(3.55)Sinyn⇀u, for each i≥1.
On the other hand, for each i≥1, we have
(3.56)‖Sinyn‖-‖u‖=|‖Sinyn‖-‖u‖|≤‖Sinyn-u‖⟶0, n⟶∞.
From this, together with (3.54) and the Kadec-Klee property of E, we obtain
(3.57)Sinyn⟶u, for each i≥1.
On the other hand, by the assumption that Si is uniformly μi-Lipschitz continuous, we have
(3.58)‖Sin+1yn-Sinyn‖≤‖Sin+1yn-Sin+1yn+1‖+‖Sin+1yn+1-yn+1‖ +‖yn+1-yn‖+‖yn-Sinyn‖≤(μi+1)‖yn+1-yn‖+‖Sin+1yn+1-yn+1‖ +‖yn-Sinyn‖.
By (3.36) and (3.57), we obtain
(3.59)limn→∞‖Sin+1yn-Sinyn‖=0
and lim n→∞Sin+1yn=u, that is, SiTnyn→u. By the closeness of Si, we have Siu=u, for all i≥1. This implies that u∈∩i=1∞F(Si). Hence u∈(∩i=1∞F(Ti))∩(∩i=1∞F(Si)).
Next, we prove that u∈∩i=1NΩi. For any p∈ℱ, for each i=1,2,…,N, we have
(3.60)ϕ(uni,zn)=ϕ(Θnizn,zn)≤ϕ(p,zn)-ϕ(p,Θnizn)=ϕ(p,zn)-ϕ(p,uni)≤ϕ(p,xn)+θn-ϕ(p,uni)→0, as n→∞.
It then yields that lim n→∞(∥uni∥-∥zn∥)=0. Since lim n→∞∥uni∥=∥u∥, for all i≥1, we have
(3.61)limn→∞‖zn‖=‖u‖.
Hence,
(3.62)limn→∞‖Jzn‖=‖Ju‖.
This together with lim n→∞∥uni∥=∥u∥ show that for each i=1,2,…,N,
(3.63)limn→∞‖uni-uni-1‖=limn→∞‖Juni-Juni-1‖=0,
where un0=zn. On the other hand, we have
(3.64)uni=Kfi,riuni-1, for each i=2,3,…,N,
and uni is a solution of the following variational equation
(3.65)fi(uni,y)+〈Aiuni,y-uni〉+ψi(y)-ψi(uni)+1ri〈y-uni,Juni-Juni-1〉≥0, ∀y∈C.
By condition (A2), we note that
(3.66)〈Aiuni,y-uni〉+ψi(y)-ψi(uni)+1ri〈y-uni,Juni-Juni-1〉 ≥-fi(uni,y)≥fi(y,uni), ∀y∈C.
By (A4), (3.63), and uni→u for each i=2,3,…,N, we have
(3.67)〈Aiu,y-u〉+ψi(y)-ψi(u)≥fi(y,u), ∀y∈C.
For 0<t<1 and y∈C, define yt=ty+(1-t)u. Noticing that y,u∈C, we obtain yt∈C, which yields that
(3.68)〈Aiu,yt-u〉+ψi(yt)-ψi(u)≥fi(yt,u).
In view of the convexity of ϕ it yields
(3.69)t〈Aiu,y-u〉+t(ψi(y)-ψi(u))≥fi(yt,u).
It follows from (A1) and (A4) that
(3.70)0=fi(yt,yt)≤tfi(yt,y)+(1-t)fi(yt,u)≤tfi(yt,y)+(1-t)t[〈Aiu,y-u〉+(ψi(y)-ψi(u))].
Let t→0, from (A3), we obtain the following:
(3.71)fi(u,y)+〈Aiu,y-u〉+ψi(y)-ψi(u)≥0, ∀y∈C, i=1,2,…,N.
This implies that u is a solution of the system of generalized mixed equilibrium problem (3.2), that is, u∈∩i=1NΩi. Hence, u∈ℱ:=(∩i=1NΩi)∩(∩i=1∞F(Ti))∩(∩i=1∞F(Si)).
Finally, we show that xn→u=ΠFx0. Indeed from w∈F⊂Cn and xn=ΠCnx0, we have the following:
(3.72)ϕ(xn,x0)≤ϕ(w,x0), ∀n≥0.
This implies that
(3.73)ϕ(u,x0)=limn→∞ϕ(xn,x0)≤ϕ(w,x0).
From the definition of ΠFx0 and (3.73), we see that u=w. This completes the proof.