R E S E A R C H
Open Access
Iterative approximation of common element
of solution sets of various nonlinear operator
problems
Atid Kangtunyakarn
**Correspondence: [email protected] Department of Mathematics, Faculty of Science, King Mongkut’s Institute of Technology Ladkrabang, Bangkok 10520, Thailand
Abstract
In this paper, we prove strong convergence theorem for finding a common element of the set of fixed point of a finite family of nonexpansive mappings and a finite family of
κ
i-strictly pseudocontractive mappings and the set of a finite family of the set ofsolution of equilibrium problems by using the new mapping generated by a finite family of nonexpansive mappings and a finite family of
κ
i-strictly pseudocontractivemappings and a sequences of positive real numbers. Furthermore, by using our main result, we obtain two interesting theorems involving variational inequality problems and variational inclusion problems. In the last section, we give numerical examples to support our main results.
Keywords: nonexpansive mapping; strictly pseudocontractive mapping; equilibrium problem; variational inequality problem; variational inclusion problem
1 Introduction
LetH be a real Hilbert space andC be a nonempty closed convex subset ofH. A self mappingf :C→C is acontractiononC if there exists a constantk∈[, ) such that
f(x) –f(y) ≤kx–y,∀x,y∈C. LetT:C→Cbe a mapping, a pointx∈Cis called a fixed point ofT if and only ifTx=x. In this paper, we useF(T) to denote the set of fixed point ofT. Recall the following definitions.
Definition . A mappingT:C→Cis called nonexpansive if and only if for allx,y∈C,
Tx–Ty ≤ x–y.
Definition . A mappingT:C→Cis calledκ-strictly pseudocontractive [] if and only if there exists a constantκ∈[, ) such that for allx,y∈C,
Tx–Ty≤ x–y+κ(I–T)x– (I–T)y. (.)
For such case,Tis also said to be aκ-strictly pseudo contraction.
Note that the class of κ-strict pseudo-contractions strictly includes the class of non-expansive mappings, that is T is nonexpansive if and only if T is -strict pseudo-contractive.
LetF:C×C→Rbe a bifunction. The equilibrium problem forFis to determine its equilibrium points,i.e., the set
EP(F) =x∈C:F(x,y)≥,∀y∈C. (.)
Given T :C→H, letF(x,y) =Tx,y–x for allx,y∈C. Thenz∈EP(F) if and only if
Tz,y–z ≥ for ally∈C, that is,zis a solution of the variational inequality.
Equilibrium problems, which were introduced in [] in , have had a great impact and influence in the development of several branches of pure and applied sciences. Numerous problems in physics, minimization problems, Nash equilibria in noncooperative games, optimization and economics reduce to find a solution ofEP(F) (see, for example, [–]). Some methods have been proposed to solve the equilibrium problem (see, for example, [–]).
In , Takahashi and Takahashi [] proved the following theorem.
Theorem . Let C be a nonempty closed convex subset of H.Let F be a bifunction from C×C toRsatisfying
(A) F(x,x) = ,∀x∈C;
(A) Fis monotone,i.e.,F(x,y) +F(y,x)≤,∀x,y∈C; (A) ∀x,y,z∈C,
lim
t→+F
tz+ ( –t)x,y≤F(x,y);
(A) ∀x∈C,y→F(x,y)is convex and lower semicontinuous;
and let S be a nonexpansive mapping of C into H such that F(S)∩EP(G)=∅.Let f be a contraction of H into itself,and let{xn}and{un}be sequences generated by x∈H and
F(un,y) +
rn
y–un,un–xn ≥, ∀y∈C,
xn+=αnf(xn) + ( –αn)Sun
for all n∈N,where{αn} ⊂[, ]and{rn} ⊂(, )satisfy(C)-(C)as follows:
(C) αn→; (C) ∞n=αn=∞;
(C) either∞n=|αn+–αn|<∞orlimn→∞ααn+n = , andlim infn→∞rn> and
∞
n=|rn+–rn|<∞.
Then{xn}and{un}converge strongly to z∈F(S)∩EP(F),where z=PF(S)∩EP(F)f(z).
In , Kangtunyakarn and Suantai [] proved the strong convergence theorem by us-ing theS-mapping generated by a finite family of strictly pseudocontractive mappings and a finite family of real number as follows.
Theorem . Let H be a Hilbert space,let f be anα-contraction on H,and let A be a strongly positive linear bounded self-adjoint operator with coefficientγ > .Assume that <γ < γα.Let{Ti}Ni= be a finite family ofκi-strictly pseudo contraction of H into itself
for someκi∈[, )andκ =max{κi:i= , , . . . ,N} with
N
i=F(Ti)=∅.Let Sn be the
S-mappings generated by T,T, . . . ,TN andα(n),α (n) , . . . ,α
(n)
N ,whereα
(n)
j = (α n,j
,α
n,j
,α
n,j
I×I×I,I= [, ],αn,j+αn,j+αn,j= andκ<a≤αn,j,αn,j≤b< for all j= , , . . . ,N– ,
κ<c≤αn,N ≤,κ≤αn,N ≤d< ,κ≤αn,j≤e< for all j= , , . . . ,N.For a point u∈H and x∈H,let{xn}and{yn}be the sequences defined iteratively by
⎧ ⎨ ⎩
yn=βnxn+ ( –βn)Snxn,
xn+=αnγ(anu+ ( –an)f(xn)) + (I–αnA)yn, n≥,
(.)
where{βn},{αn}and{an}are sequences in[, ].Assume that the following conditions hold:
(i) limn→∞αn= ,∞n=αn=∞andlimn→∞an= ;
(ii) ∞n=|αn+,j–αn,j|<∞,n∞=|αn+,j–αn,j|<∞for allj∈ {, , , . . . ,N}and
∞
n=|αn+–αn|<∞,
∞
n=|βn+–βn|<∞and
∞
n=|an+–an|<∞;
(iii) ≤κ≤βn<θ< for alln≥for someθ∈(, ).
Then both{xn}and{yn}strongly converge to q∈Ni=F(Ti),which solves the following
vari-ational inequality
γf(q) –Aq,p–q≤, ∀p∈
N
i= F(Ti).
Question Can we prove a strong convergence theorem for finding a common solution of the set of fixed point of a finite family of nonexpansive mappings and a finite family of strictly pseudocontractive mappings and a finite family of the set of solution of equilibrium problems?
LetC be a nonempty closed convex subset of Hilbert spaceH. Let{Ti}Ni= be a finite family of κi-strict pseudo-contractions ofC into itself, and let{Si}Ni= be a finite family of nonexpansive mappings ofC into itself. For eachn∈Nandj= , , . . . ,N, letαj(n)= (αn,j,αn,j,αn,j)∈I×I×I, whereI= [, ],αn,j+αn,j+αn,j= . We define the mapping SAn:C→Cas follows:
Un,=I,
Un,=S
αn,TUn,+αn,Un,+αn,I
,
Un,=S
αn,TUn,+αn,Un,+αn,I
,
Un,=S
αn,TUn,+αn,Un,+αn,I
,
.. .
Un,N–=SN–
αn,N–TN–Un,N–+αn,N–Un,N–+αn,N–I
,
SAn=Un,N=SN
αn,NTNUn,N–+αn,NUn,N–+αn,NI
. (.)
In Lemma ., under suitable conditions of the real sequences{αn,j},{αn,j}and{αn,j}for everyj= , , . . . ,N, we show thatF(SA
n) =
N
i=F(Si)∩
N
i=F(Ti) andSAnis a nonexpansive
mapping.
mappings and a finite family of the set of solution of equilibrium problems by using the mapping defined by (.). Furthermore, in the last section, we prove two interesting theo-rems involving a finite family of the set of solutions of variational inequality problem and variational inclusion problem. In the last section, we give numerical examples to support our main results.
2 Preliminaries
In this section, we need the following lemmas to prove our main result. LetCbe a closed convex subset of a real Hilbert spaceH, letPCbe the metric projection ofHontoC,i.e.,
forx∈H,PCxsatisfies the property
x–PCx=min y∈Cx–y.
The following characterizes the projectionPC.
Lemma .(See []) Given x∈H and y∈C.Then PCx=y if and only if the following
inequality holds
x–y,y–z ≥, ∀z∈C.
Lemma .(See []) Let{sn}be a sequence of nonnegative real numbers satisfying
sn+= ( –αn)sn+αnβn, ∀n≥,
where{αn},{βn}satisfy the conditions
() {αn} ⊂[, ],
∞
n=
αn=∞;
() lim sup
n→∞
βn≤ or
∞
n=
|αnβn|<∞.
Thenlimn→∞sn= .
Lemma .(See []) Let{sn}be a sequence of nonnegative real numbers satisfying
sn+= ( –αn)sn+δn, ∀n≥,
where{αn}is a sequence in(, )and{δn}is a sequence such that
()
∞
n=
αn=∞;
() lim sup
n→∞
δn αn≤
or
∞
n=
|δn|<∞.
Lemma .(See []) Let C be a nonempty closed convex subset of a real Hilbert space H, and let S:C→C be a self-mapping of C.If S is aκ-strict pseudo-contraction mapping, then S satisfies the Lipschitz condition
Sx–Sy ≤ +κ
–κx–y, ∀x,y∈C.
For solving the equilibrium problem for a bifunctionF:C×C→R, let us assume that Fsatisfies the following conditions:
(A) F(x,x) = ,∀x∈C;
(A) Fis monotone,i.e.,F(x,y) +F(y,x)≤,∀x,y∈C; (A) ∀x,y,z∈C,
lim
t→+F
tz+ ( –t)x,y≤F(x,y);
(A) ∀x∈C,y→F(x,y)is convex and lower semicontinuous.
The following lemma appears implicitly in [].
Lemma .(See []) Let C be a nonempty closed convex subset of H,and let F be a bi-function of C×C intoRsatisfying(A)-(A).Let r> and x∈H.Then there exists z∈C such that
F(z,y) +
ry–z,z–x (.)
for all y∈C.
Lemma .(See []) Assume that F:C×C→Rsatisfies(A)-(A).For r> and x∈H, define a mapping Tr:H→C as follows:
Tr(x) =
z∈C:F(z,y) +
ry–z,z–x ≥,∀y∈C
for all z∈H.Then the following hold:
() Tris single-valued;
() Tris firmly nonexpansive,i.e., Tr(x) –Tr(y)
≤Tr(x) –Tr(y),x–y
, ∀x,y∈H;
() F(Tr) =EP(F);
() EP(F)is closed and convex.
Lemma .(See []) Let E be a uniformly convex Banach space,C be a nonempty closed convex subset of E,and S:C→C be a nonexpansive mapping.Then I–S is demi-closed at zero.
Definition . LetCbe a nonempty convex subset of real Hilbert space. Let{Ti}N i=be a finite family ofκi-strict pseudo-contractions ofCinto itself, and let{Si}Ni=be a finite family of nonexpansive mappings ofC into itself. For eachj= , , . . . ,N, letαj= (αj,α
j
,α
j
I×I×I, whereI∈[, ] andαj+αj+αj= . We define the mappingSA:C→Cas
follows:
U=I,
U=S
αTU+αU+αI
,
U=S
αTU+αU+αI
,
U=S
αTU+αU+αI
,
.. .
UN–=SN–
αN–TN–UN–+αN–UN–+αN–I
,
SA=UN=SN
αNTNUN–+αNUN–+αNI
.
This mapping is called the SA-mapping generated byS
,S, . . . ,SN,T,T, . . . ,TN and α,α, . . . ,αN.
Lemma . Let C be a nonempty closed convex subset of a real Hilbert space.Let{Ti}N i= be a finite family ofκi-strict pseudo-contractions of C into itself,and let{Si}Ni=be a finite family of nonexpansive mappings of C into itself withNi=F(Si)∩
N
i=F(Ti)=∅andκ=
max{κi:i= , , . . . ,N},and letαj= (αj,α
j
,α
j
)∈I×I×I,j= , , , . . . ,N,where I= [, ], αj+αj+αj= ,αj,αj∈(κ, )for all j= , , . . . ,N– andαN∈(κ, ],αN∈[κ, ),αj∈(κ, ) for all j= , , . . . ,N.Let SAbe the SA-mapping generated by S
,S, . . . ,SN,T,T, . . . ,TNand α,α, . . . ,αN.Then F(SA) =
N
i=F(Si)∩ N
i=F(Ti),and SAis a nonexpansive mapping.
Proof It is easy to see that Ni=F(Si)∩Ni=F(Ti)⊆ F(SA). Let x∈F(SA) and x∗ ∈
N
i=F(Si)∩
N
i=F(Ti). Then we have SAx
–x∗
=SN
αNTNUN–x+αNUN–x+αNx
–x∗
≤αNTNUN–x+αNUN–x+αNx–x∗
=αNTNUN–x–x∗
+αNUN–x–x∗
+αNx–x∗
=αNTNUN–x–x∗
+αNUN–x–x∗
+αNx–x∗
–αNαNTNUN–x–UN–x–αNαNTNUN–x–x
–αNαNUN–x–x
≤αNUN–x–x∗+κ(I–TN)UN–x– (I–TN)x∗
+αNUN–x–x∗
+αNx–x∗
–αNαNTNUN–x–UN–x
–αNαNTNUN–x–x–αNαNUN–x–x
= –αNUN–x–x∗+αN
κ–αN(I–TN)UN–x
+ – –αNx–x∗–αN αNTNUN–x–x
–αNαNUN–x–x
≤ –αNUN–x–x∗
+ – –αNx–x∗
–αNαNUN–x–x
≤ N
j=N–
–αjUN–x–x∗
+ –αNαN–κ–αN–(I–TN–)UN–x
– –αNαN–αN–UN–x–x
+
–
N
j=N–
–αj
x–x∗
≤ N
j=N–
–αjUN–x–x∗ + – N
j=N–
–αj
x–x∗
≤ N
j=N–
–αjUN–x–x∗
+
N
j=N–
–αjαN–κ–αN–(I–TN–)UN–x
–
N
j=N–
–αjαN–αN–UN–x–x
+
–
N
j=N–
–αj
x–x∗
≤ N
j=N–
–αjUN–x–x∗ + – N
j=N–
–αj
x–x∗ .. . ≤ N j=
–αjUx–x∗
+
N
j=
–αjακ–α(I–T)Ux – N j=
–αjααUx–x
+ – N j=
–αj
x–x∗ (.)
≤ N
j=
–αjUx–x∗+
– N j=
–αj
x–x∗
≤ N
j=
–αjUx–x∗ + N j=
–αjακ–α(I–T)Ux – N j=
+ – N j=
–αj
x–x∗ (.) ≤ N j=
–αjUx–x∗ + – N j=
–αj
x–x∗ ≤ N j=
–αjUx–x∗ + N j=
–αjακ–α(I–T)Ux – N j=
–αjααUx–x+
– N j=
–αj
x–x∗
≤x–x∗+
N
j=
–αjακ–α(I–T)x. (.)
By (.), we have
N
j=
–αjαα–κ(I–T)x
≤x–x∗
–x–x∗
= ,
which implies thatTx=x, that is,x∈F(T). It implies that
Ux=S
αTUx+αUx+αx
=Sx. (.)
By (.) and (.), we have
SAx
–x∗ ≤ N j=
–αjSx–x∗ + N j=
–αjακ–α(I–T)Ux – N j=
–αjααUx–x
+ – N j=
–αj
x–x∗
≤x–x∗–
N
j=
–αjααUx–x. (.)
By (.), we have
N
j=
It implies that
x=Ux. (.)
By (.) and (.), we havex∈F(S). Hence, we have
x∈F(S)∩F(T). (.)
Sincex=Uxand (.), we have
SAx–x∗
≤ N
j=
–αjUx–x∗
+
N
j=
–αjακ–α(I–T)Ux
–
N
j=
–αjααUx–x
+
–
N
j=
–αj
x–x∗
=
N
j=
–αjx–x∗
+
N
j=
–αjακ–α(I–T)x
+
–
N
j=
–αj
x–x∗.
It follows that
N
j=
–αjαα–κ(I–T)x
≤,
which implies thatx=Tx, that is,x∈F(T). Sincex=Ux=Tx, we have
Ux=S
αTUx+αUx+αx
=Sx. (.)
By (.), we have
SAx–x∗
≤ N
j=
–αjUx–x∗
+
N
j=
–
N
j=
–αjααUx–x
+
–
N
j=
–αj
x–x∗
≤ N
j=
–αjSx–x∗
–
N
j=
–αjααUx–x
+
–
N
j=
–αj
x–x∗
≤x–x∗–
N
j=
–αjααUx–x.
It follows that
N
j=
–αjααUx–x≤.
It implies that
x=Ux. (.)
By (.) and (.), we havex∈F(S). Hence, we have
x∈F(S)∩F(T). (.)
By continuing in this way, we can show thatx∈F(Si)∩F(Ti) andx=Uixfor alli= , , . . . ,N– . Finally, we shall show thatx∈F(SN)∩F(TN). Since
SAx–x∗≤
–αNUN–x–x∗
+αNκ–αN(I–TN)UN–x
+ – –αNx–x∗
= –αNSN–x–x∗
+αN κ–αN(I–TN)x
+ – –αNx–x∗
≤x–x∗
+αNκ–αN(I–TN)x
.
It implies that
αNαN –κ(I–TN)x
which implies thatx=TNx, that is,x∈F(TN). It implies that
x=SAx=SN
αNTNUN–x+αNUN–x+αNx
=SNx. (.)
Then we havex∈F(SN)∩F(TN). HenceF(SA)⊆ N
i=F(Si)∩ N
i=F(Ti).
Applying (.), we have that the mappingSAis a nonexpansive.
Lemma . Let C be a nonempty closed convex subset of a real Hilbert space.Let{Ti}N i= be a finite family ofκi-strict pseudo-contractions of C into itself,and let{Si}Ni=be a finite family of nonexpansive mappings of C into itself withκ=max{κi:i= , , . . . ,N},and let α(jn)= (αn,j,αn,j,αn,j),αj= (αj,α
j
,α
j
)∈I×I×I,where I= [, ],α
n,j
+α
n,j
+α
n,j
= andα
j
+ αj+αj= such thatαin,j→αji∈[, ]as n→ ∞for i= , and j= , , , . . . ,N.Moreover, for every n∈N,let SAand SA
n be the SA-mapping generated by S,S, . . . ,SN,T,T, . . . ,TN
andα,α, . . . ,αNand S,S, . . . ,SN,T,T, . . . ,TN andα(n),α (n) , . . . ,α
(n)
N ,respectively.Then
limn→∞SA
nxn–SAxn= for every bounded sequence{xn}in C.
Proof Let{xn} be a bounded sequence inC,Uk andUn,k be generated byS,S, . . . ,SN,
T,T, . . . ,TN andα,α, . . . ,αN andS,S, . . . ,SN,T,T, . . . ,TN andα(n),α (n) , . . . ,α
(n)
N ,
re-spectively. For eachn∈N, we have
Un,xn–Uxn=S
αn,Txn+
–αn,xn
–S
αTxn+
–αxn
≤αn,Txn+
–αn,xn–αTxn–
–αxn
=αn,–αTxn–xn, (.)
and fork∈ {, , . . . ,N}, by using Lemma ., we obtain
Un,kxn–Ukxn=Sk
αn,kTkUn,k–xn+αn,kUn,k–xn+αn,kxn
–Sk
αkTkUk–xn+αkUk–xn+αkxn
≤αn,kTkUn,k–xn+αn,kUn,k–xn+αn,kxn
–αkTkUk–xn–αkUk–xn–αkxn
=αn,k(TkUn,k–xn–TkUk–xn) +
αn,k–αkTkUk–xn
+αn,k–αkxn+αn,k(Un,k–xn–Uk–xn)
+αn,k–αkUk–xn
≤αn,kTkUn,k–xn–TkUk–xn+αn,k–αkTkUk–xn
+αn,k–αkxn+αn,kUn,k–xn–Uk–xn
+αn,k–αkUk–xn
=αn,kTkUn,k–xn–TkUk–xn+αn,k–αkTkUk–xn
+αn,kUn,k–xn–Uk–xn+ –αn,k–αn,k–
+αk+αkUk–xn+αn,k–αkxn
≤αn,k +κ
+αn,k–αkTkUk–xn+αn,kUn,k–xn–Uk–xn
+αk–αn,k+αn,k–αkUk–xn+αn,k–αkxn
≤ +κ
–κUn,k–xn–Uk–xn+α n,k
–αkTkUk–xn
+ –κ
–κUn,k–xn–Uk–xn+α k
–α
n,k
+αn,k–αkUk–xn+αn,k–αkxn
≤
–κUn,k–xn–Uk–xn+α n,k
–αkTkUk–xn+Uk–xn
+αn,k–αkUk–xn+xn
. (.)
By (.) and (.), we have
SnAxn–SAxn=Un,Nxn–UNxn
≤
–κUn,N–xn–UN–xn+α n,N
–αNTNUN–xn
+UN–xn
+αn,N–αNUN–xn+xn
≤
–κ
–κUn,N–xn–UN–xn
+αn,N––αN–TN–UN–xn+UN–xn
+αn,N––αN–UN–xn+xn
+αn,N–αNTNUN–xn+UN–xn
+αn,N–αNUN–xn+xn
=
–κ
Un,N–xn–UN–xn
+
N
j=N–
–κ
N–j
αn,j–αjTjUj–xn+Uj–xn
+
N
j=N–
–κ
N–j
αn,j–αjUj–xn+xn
≤ · · ·
≤
–κ
N–
Un,xn–Uxn
+
N
j=
–κ
N–j
αn,j–αjTjUj–xn+Uj–xn
+
N
j=
–κ
N–j
αn,j–αjUj–xn+xn
=
–κ
N–
+
N
j=
–κ
N–j
αn,j–αjTjUj–xn+Uj–xn
+
N
j=
–κ
N–j
αn,j–αjUj–xn+xn
. (.)
This together with the assumption αin,j→αijasn→ ∞(i= , ,j= , , . . . ,N), we can conclude that
lim
n→∞S A
nxn–SAxn= .
Lemma . Let C be a nonempty closed convex subset of a real Hilbert space.Let{Ti}Ni= be a finite family ofκi-strict pseudo-contractions of C into itself,and let{Si}Ni=be a finite family of nonexpansive mappings of C into itself withκ=max{κi:i= , , . . . ,N},and let α(jn)= (αn,j,αn,j,αn,j),αj= (αj,α
j
,α
j
)∈I×I×I,where I= [, ],α
n,j
+α
n,j
+α
n,j
= andα
j
+ αj+αj= such that∞n=|αn+,j–αn,j|<∞,∞n=|αn+,j–αn,j|<∞for all j∈ {, , , . . . ,N}. For every n∈N,let SAn be the SA-mapping generated by S,S, . . . ,SN,T,T, . . . ,TN and α(n),α(n), . . . ,αN(n).Then∞n=SA
n+zn–SnAzn<∞for every bounded sequence{zn}in C.
Proof Let{zn}be a bounded sequence inC. For eachn∈Nand the definition ofSA, we
have
Un+,zn–Un,zn =S
αn+,Tzn+
–αn+,zn
–S
αn,Tzn+
–αn,zn
≤αn+,Tzn+
–αn+,zn–αn,Tzn–
–αn,zn
=αn+,–αn,Tzn–zn. (.)
Fork∈ {, , . . . ,N}, and using the same method as (.) in Lemma ., we have
Un+,kzn–Un,kzn ≤
–κUn+,k–zn–Un,k–zn+α n+,k
–αn,kTkUn,k–zn
+Un,k–zn
+αn+,k–αn,kUn,k–zn+zn
. (.)
From (.), (.), and using the same method as (.) in Lemma ., we have
SnA+zn–SAnzn≤
–κ
N–
αn+,–αn,Tzn–zn
+
N
j=
–κ
N–j
αn+,j–αn,jTjUn,j–zn+Un,j–zn
+
N
j=
–κ
N–j
αn+,j–αn,jUn,j–zn+zn
.
It implies that
∞
n=
3 Main result
Theorem . Let C be a nonempty closed convex subset of Hilbert spaces H,and let f be anα-contraction on H.Let Fibe a bifunction from C×C intoR,for every i= , , . . . ,N
satisfying(A)-(A).Let{Ti}Ni=be a finite family ofκi-strict pseudo-contractions of C into
itself,and let{Si}N
i=be a finite family of nonexpansive mappings of C into itself withF≡
N
i=F(Si)∩ N
i=F(Ti)∩ N
i=EP(Fi)=∅andκ =max{κi:i= , , . . . ,N},and letαj(n)=
(αn,j,αn,j,αn,j)∈I×I×I,j= , , , . . . ,N,where I= [, ],αn,j+αn,j+αn,j= ,αn,j,αn,j,αn,j∈
[a,b]⊂(κ, ) for all j= , , . . . ,N.Let SA
n be the SA-mapping generated by S,S, . . . ,SN,
T,T, . . . ,TNandα(n),α (n) , . . . ,α
(n)
N .Let{xn}and{zn}be the sequences generated by x∈C and
⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩
Fi(uin,y) +ri ny–u
i
n,uin–xn ≥, ∀y∈C and i= , , . . . ,N,
zn= N
i=δniuin,
xn+=αnf(zn) + ( –αn)SAnzn, ∀n≥,
(.)
where{αn}is a sequence in[, ].Assume that the following conditions hold:
(i) limn→∞αn= ,
∞
n=αn=∞;
(ii) ∞n=|αn+,j–αn,j|<∞,∞n=|αn+,j–αn,j|<∞,for allj∈ {, , , . . . ,N}and
∞
n=|αn+–αn|<∞;
(iii) Ni=δi n= ,
∞
n=|δni+–δin|<∞andlimn→∞δin=δi∈(κ, ),for everyi= , , . . . ,N; (iv) κ<θ≤ri
n≤η,for everyi= , , . . . ,Nand
∞
n=|rin+–rni|<∞.
Then the sequence{xn}converges strongly to x∗=PFf(x∗).
Proof Letp∈F, we havep∈Ni=EP(Fi) from Lemma ., we obtainp∈
N
i=F(Trin). Since
Fi
uin,y+ ri n
y–uin,uin–xn
≥, ∀y∈Candi= , , . . . ,N. (.)
Again from Lemma ., we haveui
n=Trinxnfor everyi= , , . . . ,N. By definition ofxn, we have
xn+–p ≤αnf(zn) –p+ ( –αn)SAnzn–p
≤αnf(zn) –f(p)+αnf(p) –p+ ( –αn)SAnzn–p ≤αnαzn–p+αnf(p) –p+ ( –αn)zn–p
=αnf(p) –p+
–αn( –α)
zn–p
=αnf(p) –p+
–αn( –α)
N
i=
δniuin–p
≤αnf(p) –p+
–αn( –α)
N
i=
δinuin–p
≤αnf(p) –p+
–αn( –α)
xn–p. (.)
Put K=max{x–p,f(–p)–αp}. By (.), we can show by induction thatxn–p ≤K,
∀n∈N. This implies that{xn}is bounded, and so are{ui
Next, we will show that
lim
n→∞xn+–xn= . (.)
By nonexpansiveness ofxn, we have
xn+–xn=αnf(zn) + ( –αn)SAnzn–αn–f(zn–) – ( –αn–)SAn–zn–
=αn
f(zn) –f(zn–)
+ (αn–αn–)f(zn–) + ( –αn)
SAnzn–SAn–zn–
+ (αn––αn)SAn–zn–
≤αnf(zn) –f(zn–)+|αn–αn–|f(zn–)+ ( –αn)SnAzn–SAn–zn–
+|αn––αn|SnA–zn–
≤αnαzn–zn–+|αn–αn–|f(zn–)
+ ( –αn)SnAzn–SAnzn–+SAnzn––SAn–zn–
+|αn––αn|SnA–zn–
≤ –αn( –α)
zn–zn–+|αn–αn–|f(zn–)
+ ( –αn)SnAzn––SnA–zn–+|αn––αn|SnA–zn–
= –αn( –α)
N
i= δniuin–
N
i=
δni–uin–
+|αn–αn–|f(zn–)
+ ( –αn)SnAzn––SnA–zn–+|αn––αn|SnA–zn–
= –αn( –α)
N
i=
δniuin–uin–+
N
i=
δni–δni–uin–
+|αn–αn–|f(zn–)+ ( –αn)SAnzn––SnA–zn–
+|αn––αn|SnA–zn–
≤ –αn( –α)
N
i=
δinuin–uin–+
N
i=
δin–δni–uin–
+|αn–αn–|f(zn–)+ ( –αn)SAnzn––SnA–zn–
+|αn––αn|SnA–zn–. (.)
From Lemma ., we have
∞
n=
SAn+zn–SnAzn<∞. (.)
Sinceuin=Tri
nxnfor everyi= , , . . . ,N. By definition ofTrin, we have
F(Tri nxn,y) +
ri n
y–Tri
nxn,Trinxn–xn ≥, ∀y∈C, (.)
similarly,
F(Tri
n+xn+,y) +
ri
n+
y–Tri
From (.) and (.), we obtain
F(Tri
nxn,Trin+xn+) +
ri n
Tri
n+xn+–Trinxn,Trinxn–xn ≥ (.)
and
F(Tri
n+xn+,Trinxn) + ri
n+
Tri
nxn–Trin+xn+,Trin+xn+–xn+ ≥. (.)
By (.) and (.), we have
ri n
Tri
n+xn+–Trnixn,Trinxn–xn +
rni+Trnixn–Trni+xn+,Trin+xn+–xn+ ≥.
It follows that
Tri
nxn–Trni+xn+,
Tri
n+xn+–xn+
rni+
–Trinxn–xn ri
n
≥.
This implies that
≤
Tri
n+xn+–Trinxn,Trnixn–Trin+xn++Trin+xn+–xn–
ri n
rin+(Trin+xn+–xn+)
.
It follows that
Tri
n+xn+–Trinxn
≤
Tri
n+xn+–Trnixn,Trin+xn+–xn–
ri n
ri n+
(Tri
n+xn+–xn+)
=
Tri
n+xn+–Trinxn,xn+–xn+
– r
i n
ri n+
(Tri
n+xn+–xn+)
≤ Tri
n+xn+–Trnixn
xn+–xn+
– r
i n
ri n+
(Tri
n+xn+–xn+)
≤ Tri
n+xn+–Trnixn
xn+–xn+
– r
i n
rin+
Tri
n+xn+–xn+
=Tri
n+xn+–Trinxn
xn+–xn+ rin+r
i
n+–rinTrin+xn+–xn+
≤ Tri
n+xn+–Trnixn
xn+–xn+ ar
i
n+–rinTrni+xn+–xn+
.
It follows that
uin+–uin≤ xn+–xn+ ar
i
n+–rniuin+–xn+ (.)
Substitute (.) into (.), we have
xn+–xn ≤
–αn( –α)
N
i=
δinuin–uin–+
N
i=
δin–δni–uin–
+|αn–αn–|f(zn–)+ ( –αn)SAnzn––SnA–zn–
+|αn––αn|SnA–zn–
≤ –αn( –α)
N
i= δin
xn+–xn+ ar
i
n+–rniuin+–xn+
+
N
i=
δni–δni–uin–
+|αn–αn–|f(zn–)+ ( –αn)SAnzn––SnA–zn–
+|αn––αn|SnA–zn–
= –αn( –α)
xn+–xn+
N
i= δni
ar
i
n+–rniuin+–xn+
+
N
i=
δni–δni–uin–
+|αn–αn–|f(zn–)+ ( –αn)SAnzn––SnA–zn–
+|αn––αn|SnA–zn–
≤ –αn( –α)
xn+–xn+
N
i= δni
ar
i
n+–rniuin+–xn+
+
N
i=
δni–δni–uin–+|αn–αn–|f(zn–)
+SnAzn––SnA–zn–+|αn––αn|SnA–zn–. (.)
By (.), (.), conditions (iii), (iv) and Lemma ., we have
lim
n→∞xn+–xn= . (.)
From (.), (.) and condition (iv), we have
lim
n→∞u
i
n+–uin= , ∀i= , , . . . ,N. (.)
Letp∈F. Fromui n=Tri
nxnfor everyi= , , . . . ,N, we have
ui n–p
=Tri
nxn–Trinp
≤ Tri
nxn–Trinp,xn–p
= u
i n–p
+xn–p–uin–xn
It implies that
ui n–p
≤ xn–p–uin–xn
. (.)
By definition of{xn}and (.), we have
xn+–p≤αnf(zn) –p
+ ( –αn)SAnzn–p
≤αnf(zn) –p
+ ( –αn)zn–p
=αnf(zn) –p
+ ( –αn)
N
i=
δinuin–p
≤αnf(zn) –p+ ( –αn) N
i=
δniuin–p
≤αnf(zn) –p
+ ( –αn) N
i=
δnixn–p–uin–xn
≤αnf(zn) –p
+xn–p– ( –αn) N
i=
δinuin–xn
.
It implies that
( –αn) N
i=
δniuin–xn≤αnf(zn) –p+xn–p–xn+–p
≤αnf(zn) –p
+xn–p
+xn+–p
xn+–xn. (.)
From conditions (i), (iii) and (.), we have
lim
n→∞u i
n–xn= , ∀i= , , . . . ,N. (.)
Since
xn+–SAnzn=αn
f(zn) –SAnzn
,
from condition (i), we have
lim
n→∞xn+–S A
nzn= . (.)
From the definition ofzn, we have
zn–xn=
N
i= δniui
n–xn
≤ N
i=
From condition (iii) and (.), we have
lim
n→∞zn–xn= . (.)
Since
zn–SAnzn≤ zn–xn+xn–xn++xn+–SAnzn,
by (.), (.) and (.), we have
lim
n→∞zn–S A
nzn= . (.)
Next, we show that
lim sup
n→∞
f(z) –z,xn–z
≤, (.)
wherez=PFf(z). To show this inequality, take a subsequence{xnk}of{xn}such that
lim sup
n→∞
f(z) –z,xn–z
= lim
k→∞
f(z) –z,xnk–z
. (.)
Without loss of generality, we may assume that a subsequence{xnk}of {xn}converges weakly to someq∈H. From (.), we have that{znk}converges weakly toq.
Sinceκ<a≤αn,j,αn,j,αn,j≤b< for allj= , , . . . ,N. Without loss of generality, we may assume that
αnk,j
→α
j
∈(κ, ), α
nk,j
→α
j
∈(κ, ) and α
nk,j
→α
j
∈(κ, ) ask→ ∞,
∀j= , , . . . ,N.
Let SA be the SA-mapping generated by S
,S, . . . ,SN, T,T, . . . ,TN and β,β, . . . ,βN,
whereβj= (αj,α
j
,α
j
),∀j= , , . . . ,N. By Lemma .,SAis a nonexpansive mapping, and F(SA) =N
i=F(Si)∩
N
i=F(Ti).
By Lemma ., we have
lim
k→∞
SnAkznk–S
Az
nk= . (.)
Since
znk–S
Az
nk≤znk–S
A
nkznk+S
A nkznk–S
Az nk,
by (.), (.), we have
lim
k→∞ znk–S
Az
nk= . (.)
Since{znk}converges weakly toqask→ ∞(.) and Lemma ., we have
q∈FSA=
N
i= F(Si)∩
N
i=
Next, we show thatq∈Ni=EP(Fi). To show this, we may assume that
lim
k→∞r i nk=r
i∈[θ,η], ∀i= , , . . . ,N.
By Lemmas . and ., for everyi= , , . . . ,N, we defineTri:H→Cby
Tri(x) =
z∈C:Fi(z,y) +
riy–z,z–x ≥,∀y∈C
, ∀x∈Handi= , , . . . ,N.
Then we have
Fi(Trixn,y) +
riy–Trixn,Trixn–xn ≥, ∀y∈Candi= , , . . . ,N.
From (.) andui
n=Trinxn, we have
Fi(Trinxn,y) + ri n
y–Tri
nxn,Trinxn–xn ≥, ∀y∈Candi= , , . . . ,N.
It implies that
Fi(Trixnk,Trnki xnk) +
riTrinkxnk–Trixnk,Trixnk–xnk ≥, ∀i= , , . . . ,N
and
Fi(Tri
nkxnk,Trixnk) + ri nk
Trixnk–Trinkxnk,Trinkxnk–xnk ≥, ∀i= , , . . . ,N.
By (A), we have
riTrnki xnk–Trixnk,Trixnk–xnk + ri nk
Trixnk–Trnki xnk,Trinkxnk–xnk ≥.
It implies that
Tri
nkxnk–Trixnk,
Trixnk–xnk
ri –
Tri
nkxnk–xnk ri
nk
≥.
It follows that
Tri
nkxnk–Trixnk,Trixnk–xnk– ri
ri nk
(Tri
nkxnk–xnk)
≥.
Then
≤
Tri
nkxnk–Trixnk,Trixnk–Tri
nkxnk+Trnki xnk–xnk– ri
ri nk
(Tri
nkxnk–xnk)
=
Tri
nkxnk–Trixnk,Trixnk–Tri nkxnk+
– r
i
ri nk
(Tri
nkxnk–xnk)
It follows that
Tri
nkxnk–Trixnk ≤
Tri
nkxnk–Trixnk,
– r
i
ri nk
(Tri
nkxnk–xnk)
≤ Tri
nkxnk–Trixnk
– r
i
ri nk
Tri
nkxnk–xnk.
It implies that
Tri
nkxnk–Trixnk ≤ ar
i nk–r
iT ri
nkxnk–xnk.
Fromlimk→∞rink=r
iand (.), we have
lim
k→∞Trinkxnk–Trixnk= , ∀i= , , . . . ,N. (.)
For everyi= , , . . . ,N, we have
xnk–Trixnk ≤ xnk–Trnki xnk+Trinkxnk–Trixnk
=xnk–u
i
nk+Trinkxnk–Trixnk,
by (.) and (.), we have
lim
k→∞xnk–Trixnk= , ∀i= , , . . . ,N. (.)
Since a subsequence{xnk}of{xn}converges weakly toqask→ ∞, from (.) and Lem-ma ., we have
q∈F(Tri), ∀i= , , . . . ,N.
Then
q∈
N
i=
F(Tri). (.)
From Lemma ., we haveEP(Fi) =F(Tri),∀i= , , . . . ,N. From (.), we have
q∈
N
i=
F(Tri) =
N
i=
EP(Fi). (.)
By (.) and (.), we have
q∈F. (.)
Sincexnk qask→ ∞andq∈Fand (.), we have
lim sup
n→∞
f(z) –z,xn–z
= lim
k→∞
f(z) –z,xnk–z