R E S E A R C H
Open Access
On hybrid split problem and its nonlinear
algorithms
Zhenhua He
1and Wei-Shih Du
2**Correspondence:
2Department of Mathematics,
National Kaohsiung Normal University, Kaohsiung, 824, Taiwan Full list of author information is available at the end of the article
Abstract
In this paper, we study a hybrid split problem (HSP for short) for equilibrium problems and fixed point problems of nonlinear operators. Some strong and weak convergence theorems are established.
MSC: 47J25; 47H09; 65K10
Keywords: fixed point problem; equilibrium problem; hybrid split problem; iterative algorithm; strong (weak) convergence theorem
1 Introduction
Throughout this paper, we assume thatHis a real Hilbert space with zero vectorθ, whose inner product and norm are denoted by·,·and · , respectively. LetCbe a nonempty subset ofHandT :C→H be a mapping. Denote byF(T) the set of fixed points ofT. The symbolsNandRare used to denote the sets of positive integers and real numbers, respectively.
LetHbe a Hilbert space andCbe a closed convex subset ofH. Letf :C×C→Rbe a bi-function. The classical equilibrium problem (EP for short) is defined as follows.
Findp∈Csuch thatf(p,y)≥, ∀y∈C. (EP)
The symbolEP(f) is used to denote the set of all solutions of the problem (EP), that is,
EP(f) =u∈K:f(u,v)≥,∀v∈K.
It is known that the problem (EP) contains optimization problems, complementary prob-lems, variational inequalities probprob-lems, saddle point probprob-lems, fixed point probprob-lems, bilevel problems, semi-infinite problems and others as special cases and have many ap-plications in physics and economics problems; for detail, one can refer to [–] and refer-ences therein.
In last ten years or so, the problem (EP) has been generalized and improved to find a common element of the set of fixed points of a nonlinear operator and the set of solutions of the problem (EP). More precisely, many authors have studied the following problem (FTEP) (see, for instance, [–]):
Findp∈Csuch thatTp=pandf(p,y)≥, ∀y∈C, (FTEP)
whereCis a closed convex subset of a Hilbert spaceH,f:C×C→Ris a bi-function and T:C→Cis a nonlinear operator.
In this paper, motivated by the problems (EP) and (FTEP), we study the following math-ematical model about a hybrid split problem for equilibrium problems and fixed point problems of nonlinear operators (HSP for short).
LetE andE be two real Banach spaces. LetC be a closed convex subset ofE and
K be a closed convex subset ofE. Letf :C×C→Randg:K×K→Rbe two
bi-functions,A:E→Ebe a bounded linear operator. LetT:C→CandS:K→Kbe two
nonlinear operators withF(T) =∅andF(S) =∅. The mathematical model about a hybrid split problem for equilibrium problems and fixed point problems of nonlinear operators (HSP for short) is defined as follows:
Findp∈Csuch thatTp=p,f(p,y)≥, ∀y∈C, and
u:=ApsatisfyingSu=u∈K,g(u,v)≥, ∀v∈K.
(HSP)
In fact, (HSP) contains several important problems as special cases. We give some ex-amples to explain about it.
Example A If T is an identity operator onC, then (HSP) will reduce to the following problem (P):
(P) Findp∈Csuch thatf(p,y)≥,∀y∈C, andu:=ApsatisfyingSu=u∈K,g(u,v)≥,
∀v∈K.
Example B IfSis an identity operator onK, then (HSP) will reduce to the following prob-lem (P):
(P) Findp∈Csuch thatTp=p,f(p,y)≥,∀y∈C, andu:=Ap∈Ksatisfyingg(u,v)≥,
∀v∈K.
Example C IfT,Sare all identity operators, then (HSP) will reduce to the following split equilibrium problem (P) which has been considered in []:
(P) Findp∈Csuch thatf(p,y)≥,∀y∈C, andu:=Ap∈Ksatisfyingg(u,v)≥,∀v∈K.
Example D IfSis an identity operator andf(x,y)≡ for all (x,y)∈C×C, then (HSP) will reduce to the following problem (P) which has been studied in []:
(P) Findp∈Csuch thatTp=pandu:=Ap∈Ksatisfyingg(u,v)≥,∀v∈K.
In this paper, we introduce some new iterative algorithms for (HSP) and some strong and weak convergence theorems for (HSP) will be established.
2 Preliminaries
In what follows, the symbolsand→will symbolize weak convergence and strong con-vergence as usual, respectively. A Banach space (X, · ) is said to satisfy Opial’s condition if for each sequence{xn}inXwhich converges weakly to a pointx∈X, we have
lim inf
It is well known that any Hilbert space satisfies Opial’s condition. LetK be a nonempty subset of real Hilbert spacesH. Recall that a mappingT:K→Kis said to be nonexpansive ifTx–Ty ≤ x–yfor allx,y∈K.
LetHandHbe two Hilbert spaces. LetA:H→HandB:H→Hbe two bounded
linear operators.Bis called the adjoint operator (or adjoint) ofAif for allz∈H,w∈H,
BsatisfiesAz,w=z,Bw. It is known that the adjoint operator of a bounded linear op-erator on a Hilbert space always exists and is bounded linear and unique. Moreover, it is not hard to show that ifBis an adjoint operator ofA, thenA=B.
Example .([]) LetH=Rwith the standard norm| · |andH=R with the norm
α= (a+a) for someα= (a,a)∈R.x,y=xydenotes the inner product ofH
for somex,y∈Handα,β=
i=aibi denotes the inner product ofH for someα=
(a,a),β= (b,b)∈H. LetAα=a–aforα= (a,a)∈HandBx= (–x,x) forx∈H,
thenBis an adjoint operator ofA.
Example .([]) LetH=Rwith the normα= (a+a)
for someα= (a,a)∈R
andH=Rwith the normγ= (c+c+c)
for someγ = (c,c,c)∈R. Letα,β=
i=aibiandγ,η=
i=cididenote the inner product ofHandH, respectively, where
α= (a,a),β= (b,b)∈H,γ = (c,c,c),η= (d,d,d)∈H. LetAα= (a,a,a–a)
forα= (a,a)∈H andBγ = (c+c,c–c) forγ = (c,c,c)∈H. Obviously,Bis an
adjoint operator ofA.
LetK be a closed convex subset of a real Hilbert spaceH. For each pointx∈H, there exists a unique nearest point inK, denoted byPKx, such thatx–PKx ≤ x–y ∀y∈K. The mappingPKis called themetric projectionfromHontoK. It is well known thatPK has the following characteristics:
(i) x–y,PKx–PKy ≥ PKx–PKyfor everyx,y∈H;
(ii) forx∈Handz∈K,z=PK(x)⇔ x–z,z–y ≥,∀y∈K; (iii) forx∈Handy∈K,
y–PK(x)+x–PK(x)≤ x–y. (.)
Lemma .(see []) Let K be a nonempty closed convex subset of H and F be a bi-function of K×K intoRsatisfying the following conditions:
(A) F(x,x) = for allx∈K;
(A) Fis monotone,that is,F(x,y) +F(y,x)≤for allx,y∈K; (A) for eachx,y,z∈K,lim supt↓F(tz+ ( –t)x,y)≤F(x,y); (A) for eachx∈K,y→F(x,y)is convex and lower semi-continuous.
Let r> and x∈H.Then there exists z∈K such that F(z,y) + ry–z,z–x ≥for all y∈K.
Lemma .(see []) Let K be a nonempty closed convex subset of H and let F be a bi-function of K×K intoRsatisfying(A)-(A).For r> ,define a mapping TF
r :H→K as follows:
TrF(x) =
z∈K:F(z,y) +
ry–z,z–x ≥,∀y∈K
for all x∈H.Then the following hold:
(i) TF
r is single-valued andF(TrF) =EP(F)for∀r> andEP(F)is closed and convex;
(ii) TF
r is firmly non-expansive,that is,for anyx,y∈H, TF
rx–TrFy≤ TrFx–TrFy,x–y.
Lemma .(see,e.g., []) Let H be a real Hilbert space.Then the following hold:
(a) x+y≤ y+ x,x+y;
(b) x–y=x+y– x,yfor allx,y∈H;
(c) αx+ ( –α)y=αx+ ( –α)y–α( –α)x–yfor allx,y∈Hand
α∈[, ].
Lemma . Let FF
r be the same as in Lemma..IfF(TrF) =EP(F)=∅,then for any x∈H and x*∈F(TrF),TrFx–x≤ x–x*–TrFx–x*.
Proof By (ii) of Lemma . and (b) of Lemma .,
TrFx–x*≤TrFx–x*,x–x* = T
F rx–x*
+x–x*–TrFx–x,
which shows thatTF
rx–x≤ x–x*–TrFx–x*.
Lemma .([, ]) Let the mapping TF
r be defined as in Lemma..Then,for r,s> and x,y∈H,
TrF(x) –TsF(y)≤ x–y+|s–r| s T
F s(y) –y.
In particular,TF
r(x) –TrF(y) ≤ x–yfor any r> and x,y∈H,that is,TrFis nonexpan-sive for any r> .
Remark . In fact, Lemma . is motivated by a proof of [, Theorem .]. In order to the sake of convenience for proving, we restated the fact and gave its proof in Lemma . [, ].
Lemma .([]) Let{an}be a nonnegative real sequence satisfying the following condi-tion:
an+≤( –λn)an+λnbn, ∀n≥n,
where nis some nonnegative integer,{λn}is a sequence in(, )and{bn}is a sequence inR
such that
(i) ∞n=λn=∞;
(ii) lim supn→∞bn≤or∞n=λnbnis convergent.Thenlimn→∞an= .
Lemma .([]) Let{xn}and{yn}be bounded sequences in a Banach space E and let{βn}
3 Weak convergence iterative algorithms for (HSP)
In this section, we will introduce some weak convergence iterative algorithms for the hy-brid split problem.
Theorem . Let H and Hbe two real Hilbert spaces.Let C⊂Hand K⊂Hbe two
nonempty closed convex sets.Let T:C→C and S:K→K be non-expansive mappings and f :C×C→Rand g:K×K→Rbe bi-functions satisfying the conditions(A)-(A). Let A:H→Hbe a bounded linear operator with its adjoint B.Let x∈C,{xn}and{un}
be sequences generated by
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
un=Trfnxn,
yn= ( –α)un+αTun,
wn=TrgnAyn,
xn+=PC(yn+ξB(Swn–Ayn)), ∀n∈N,
(.)
whereα∈(, ),ξ∈(,B)and{rn} ⊂(, +∞)withlim infn→+∞rn> ,PCis a projection
operator from Hinto C.Suppose that={p∈F(T)∩EP(f) :Ap∈F(S)∩EP(g)} =∅,
then xn,unq∈and wnAq∈F(S)∩EP(g).
Proof Letp∈, the following several inequalities can be proved easily:
yn–p ≤ un–p ≤ xn–p, wn–Ap ≤ Ayn–Ap. (.)
By Lemma .,TrgnAyn–Ayn≤ Ayn–Ap–T
g
rnAyn–Ap, hence
Swn–Ap=STrgnAyn–Ap≤TrgnAyn–Ap
≤ Ayn–Ap–TrgnAyn–Ayn. (.)
By (b) of Lemma . and (.), for eachn∈N, we have
ξyn–p,B
STrgn–IAyn
= ξA(yn–p) +
STrgn–IAyn–
STrgn–IAyn,
STrgn–IAyn
= ξ
ST
g
rnAyn–Ap
+ ST
g rn–I
Ayn
–
Ayn–Ap
–STg rn–I
Ayn
≤ξ
–
T g
rnAyn–Ayn
+ ST
g rn–I
Ayn–STrgn–IAyn
= –ξSTrgn–IAyn
–ξTrgnAyn–Ayn
. (.)
On the other hand,B(STrgn–I)Ayn≤ B(ST
g
rn–I)Ayn, so from (.)-(.), we have
xn+–p =PC
yn+ξBSTrgn–IAyn–p≤yn+ξBSTrgn–IAyn–p
≤ yn–p+ξBSTrgn–IAyn–ξSTrgn–IAyn
–ξTrgn–IAyn
=yn–p–ξ –ξBSTrfn–IAyn–ξTrgn–IAyn
≤ xn–p–ξ –ξBSTrfn–IAyn–ξTrgn–IAyn. (.)
Sinceξ∈(,B),ξ( –ξB) > , by (.) and (.), we have
xn+–p ≤ yn–p ≤ un–p ≤ xn–p (.)
and
ξ –ξBSTrgn–IAyn+ξTrgn–IAyn≤ xn–p–xn+–p. (.)
The inequality (.) implieslimn→∞xn–pexists. Further, from (.) and (.), we get
lim
n→∞xn–p=nlim→∞yn–p=nlim→∞un–p,
lim
n→∞ST
g rn–I
Ayn= lim
n→∞T
g rn–I
Ayn= lim
n→∞wn–Ayn= .
(.)
The inequality (.) also implies that
lim
n→∞Swn–wn= . (.)
Using Lemma . and (.), we have
un–xn=Trfnxn–xn≤ xn–p–Trfnxn–p
=xn–p–un–p→. (.)
Notice that
yn–p = ( –α)un–p+αTun–p–α( –α)Tun–un ≤ un–p–α( –α)Tun–un,
hence,
lim
n→∞Tun–un= . (.)
From (.) and (.), we also have
yn–xn ≤ yn–un+un–xn
=αTun–un+un–xn → asn→ ∞. (.)
The existence oflimn→∞xn–pimplies that {xn}is bounded, hence {xn}has a weak convergence subsequence{xnj}. Assume thatxnjqfor someq∈C, thenynjq,Aynj
Aq∈Kandwnj=T
g
We sayq∈, in other words,q∈F(T)∩EP(f) andAq∈F(S)∩EP(g). By (.), we also obtainunjq. IfTq =q, then, by Opial’s condition and (.), we get
lim inf
j→∞ unj–q <lim inf
j→∞ unj–Tq
≤lim inf
j→∞ unj–Tunj+Tunj–Tq
≤lim inf
j→∞ unj–q,
which is a contradiction. HenceTq=qorq∈F(T). On the other hand, from Lemma ., we knowEP(f) =F(Trf) for anyr> . Hence, ifTrfq =qforr> , then by Opial’s condition and (.) and Lemma ., we have
lim inf
j→∞ xnj–q<lim inf
j→∞ xnj–T
f rq
=lim inf
j→∞ xnj–T
f rnjxnj+T
f rnjxnj–T
f rq
≤lim inf
j→∞ xnj–T
f
rnjxnj+T
f
rq–Trfnjxnj
≤lim inf
j→∞
xnj–T
f
rnjxnj+xnj–q+
|r–rnj|
rnj
Trf
njxnj–xnj
=lim inf
j→∞ xnj–q,
which is also a contradiction. So, for eachr> ,Trfq=q, namelyq∈EP(f). Thus, we have provedq∈F(T)∩EP(f). Similarly, we can also proveAq∈F(S)∩EP(g). Hence,q∈.
Finally, we prove{xn}converges weakly toq∈. Otherwise, if there exists another sub-sequence of {xn}, which is denoted by{xnl}, such thatxnl x¯∈withx¯ =q, then by Opial’s condition,
lim inf
l→∞ xnl–x¯ <lim infl→∞ xnl–q=lim infj→∞ xnj–q<lim inf
l→∞ xnl–x.¯
This is a contradiction. Hence{xn}converges weakly to an elementq∈. Together with un–xn → (see (.)), we also getunq.
Finally, we prove{wn=TrgnAyn}converges weakly toAq∈F(S)∩EP(g). From (.), we
haveynq, soAynAq. Thus, from (.) we havewn=TrgnAynAq∈F(S)∩EP(g).
The proof is completed.
If T=I orS=I, whereI denotes an identity operator, then the following corollaries follow from Theorem ..
Corollary . Let Hand Hbe two real Hilbert spaces.Let C⊂H and K⊂Hbe two
nonempty closed convex sets.Let S:K→K be a non-expansive mapping and f :C×C→R and g:K×K→Rbe bi-functions satisfying the conditions(A)-(A).Let A:H→Hbe a
by
⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
un=Trfnxn,
wn=TrgnAun,
xn+=PC(un+ξB(Swn–Aun)), ∀n∈N,
whereξ∈(,B)and{rn} ⊂(, +∞)withlim infn→+∞rn> ,PCis a projection operator
from Hinto C.Suppose that={p∈EP(f) :Ap∈F(S)∩EP(g)} =∅,then xn,unq∈
and wnAq∈F(S)∩EP(g).
Corollary . Let Hand Hbe two real Hilbert spaces.Let C⊂Hand K⊂Hbe two
nonempty closed convex sets.Let T:C→C be a non-expansive mapping and f :C×C→ Rand g:K×K→Rbe bi-functions satisfying the conditions(A)-(A).Let A:H→H
be a bounded linear operator with its adjoint B.Let x∈C,{xn} and{un} be sequences
generated by
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
un=Trfnxn,
yn= ( –α)un+αTun,
wn=TrgnAyn,
xn+=PC(yn+ξB(wn–Ayn)), ∀n∈N,
whereα∈(, ),ξ ∈(,B)and{rn} ⊂(, +∞)withlim infn→+∞rn> ,PC is a
projec-tion operator from Hinto C.Suppose that={p∈F(T)∩EP(f) :Ap∈EP(g)} =∅,then
xn,unq∈and wnAq∈EP(g).
Corollary . Let C⊂Hand K⊂Hbe two nonempty closed convex sets.Let f :C×C→
Rand g:K×K→Rbe bi-functions satisfying the conditions(A)-(A).Let A:H→H
be a bounded linear operator with its adjoint B.Let x∈C,{xn} and{un} be sequences
generated by
⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
un=Trfnxn,
wn=TrgnAun,
xn+=PC(un+ξB(wn–Aun)), ∀n∈N,
whereξ∈(,B)and{rn} ⊂(, +∞)withlim infn→+∞rn> ,PCis a projection operator
from H into C.Suppose that={p∈EP(f) :Ap∈EP(g)} =∅,then xn,unq∈and wnAq∈EP(g).
4 Strong convergence iterative algorithms for (HSP)
In this section, we introduce two strong convergence algorithms for (HSP); see Theo-rem . and TheoTheo-rem ..
Theorem . Let Hand Hbe two real Hilbert spaces.Let C⊂Hand K⊂Hbe two
and f :C×C→Rand g:K×K→Rbe bi-functions satisfying the conditions(A)-(A). Let A:H→Hbe a bounded linear operator with its adjoint B.Let x∈C:=C,{xn}and
{un}be sequences generated by ⎧
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
un=Trfnxn,
yn= ( –α)un+αTun,
wn=TrgnAyn,
zn=PC(yn+ξB(Swn–Ayn)),
Cn+={v∈Cn:zn–v ≤ yn–v ≤ xn–v},
xn+=PCn+(x), n∈N,
(.)
whereα∈(, ),ξ∈(,B)and{rn} ⊂(, +∞)withlim infn→+∞rn> ,PCis a projection
operator from Hinto C.Suppose that={p∈F(T)∩EP(f) :Ap∈F(S)∩EP(g)} =∅,
then xn,un→q∈and wn→Aq∈F(S)∩EP(g).
Proof We claim thatCnis a nonempty closed convex set forn∈N. In fact, letp∈, it follows from (.) that
ξyn–p,B(Swn–Ayn) ≤–ξTrgn–I
Axn
–ξSwn–Ayn. (.)
By (.), (.) and (.), we obtain
zn–p≤yn+ξB(Swn–Ayn) –p
=yn–p+ξB(Swn–Ayn)+ ξyn–p,B(Swn–Ayn)
≤ yn–p+ξBSwn–Ayn–ξTrgn–I
Ayn
–ξSwn–Ayn
=yn–p–ξ –ξBSTrgn–IAyn–ξTrgn–IAyn
≤ un–p– ( –α)αun–Tun
–ξ –ξBSTrgn–IAyn
–ξTrgn–IAyn
≤ xn–p–ξ –ξBSTrgn–IAyn
–ξTrgn–IAyn– ( –α)αun–Tun. (.)
Noticeξ∈(,B),ξ( –ξB) > . It follows from (.) that
zn–p ≤ yn–p ≤ un–p ≤ xn–p for alln∈N,
hencep∈Cn, which yields that⊂CnandCn =∅forn∈N.
It is not hard to verify thatCnis closed forn∈N, so it suffices to verifyCnis convex for n∈N. Indeed, letw,w∈Cn+andγ∈[, ], we have
zn–
γw+ ( –γ)w
=γzn–w+ ( –γ)zn–w–γ( –γ)w–w
≤γyn–w+ ( –γ)yn–w–γ( –γ)w–w
=yn–γw+ ( –γ)w
,
namelyzn– (γw+ ( –γ)w) ≤ yn– (γw+ ( –γ)w). Similarly,yn– (γw+ ( –
γ)w) ≤ xn– (γw+ ( –γ)w), which impliesγw+ ( –γ)w∈Cn+ andCn+ is a
convex set,n∈N.
Notice thatCn+⊂Cnandxn+=PCn+(x)⊂Cn, thenxn+–x ≤ xn–xforn> .
It follows thatlimn→∞xn–xexists. Hence{xn}is bounded, which yields that{zn}and
{yn}are bounded. For somek,n∈Nwithk>n> , fromxk=PCk(x)⊂Cnand (.), we
have
xn–xk+x–xk=xn–PCk(x)
+x–PCk(x)
≤ xn–x. (.)
Bylimn→∞xn–xexists and (.), we havelimn→∞xn–xk= , so{xn}is a Cauchy
sequence.
Letxn→q, thenq∈. Firstly, byxn+=PCn+(x)∈Cn+⊂Cn, from (.) we have
zn–xn ≤ zn–xn++xn+–xn ≤xn+–xn →,
yn–xn ≤ yn–xn++xn+–xn ≤xn+–xn →.
(.)
Settingρ=ξ( –ξB), by (.) again, we have
ρSTrgn–IAyn+ξTrgn–IAyn+ ( –α)αun–Tun
≤ xn–p–zn–p≤ xn–znxn–p+zn–p→. (.)
So,
lim
n→∞Tun–un= , nlim→∞wn–Ayn=nlim→∞T g rn–I
Ayn= ,
lim
n→∞Swn–Ayn=nlim→∞ST g rn–I
Ayn= , lim
n→∞Swn–wn= .
(.)
Notice thatlimn→∞Tun–un= andyn–un=αTun–un, so
lim
n→∞yn–un= . (.)
Further, from (.) and (.),
lim
n→∞xn–un= . (.)
Sincexn→q, we haveun→qby (.). Thus
namelyTq=qandq∈F(T). On the other hand, forr> , by Lemma ., we have
Tf
rq–q≤Trfq–Trfnxn+T
f
rnxn–xn+xn–q
≤ xn–q+|rn–r| rn T
f
rnxn–xn+T
f
rnxn–xn+xn–q →,
which yieldsq∈F(Trf) =EP(f). We have verifiedq∈F(T)∩EP(f).
Next, we proveAq∈F(S)∩EP(g). Sincexn→qandxn–yn→ by (.) and (.) and wn–Ayn→ by (.), we haveyn→qandAyn→Aqandwn→Aq. So,
SAq–Aq ≤ SAq–Swn+Swn–wn+wn–Aq →,
namelySAq=AqandAq∈F(S). On the other hand, forr> , by Lemma . again, we have
TrgAq–Aq≤TrgAq–TrgnAyn+TrgnAyn–Ayn+Ayn–Aq
≤ Ayn–Aq+|rn–r| rn T
g
rnAyn–Ayn
+TrgnAyn–Ayn+Ayn–Aq →,
which implies thatAq∈F(Trg) =EP(g). We have verifiedAq∈F(S)∩EP(g).
So, we have obtainedq∈andxn,un→qandwn→Aq, the proof is completed.
If T=I orS=I, whereI denotes an identity operator, then the following corollaries follow from Theorem ..
Corollary . Let Hand Hbe two real Hilbert spaces.Let C⊂Hand K⊂Hbe two
nonempty closed convex sets.Let f :C×C→Rand g:K×K→Rbe bi-functions satisfy-ing the conditions(A)-(A)and S:K→K be a non-expansive mapping.Let A:H→H
be a bounded linear operator with its adjoint B.Let x∈C:=C,{xn}and{un}be sequences
generated by
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
un=Trfnxn,
wn=TrgnAun,
zn=PC(un+ξB(Swn–Aun)),
Cn+={v∈Cn:zn–v ≤ un–v ≤ xn–v},
xn+=PCn+(x), n∈N,
whereξ∈(,
B)and{rn} ⊂(, +∞)withlim infn→+∞rn> ,PCis a projection operator from Hinto C.Suppose that={p∈EP(f) :Ap∈F(S)∩EP(g)} =∅,then xn,un→q∈
and wn→Aq∈F(S)∩EP(g).
Corollary . Let Hand Hbe two real Hilbert spaces.Let C⊂Hand K⊂Hbe two
be a bounded linear operator with its adjoint B.Let x∈C:=C,{xn}and{un}be sequences generated by ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
un=Trfnxn,
yn= ( –α)un+αTun,
wn=TrgnAyn,
zn=PC(yn+ξB(wn–Ayn)),
Cn+={v∈Cn:zn–v ≤ yn–v ≤ xn–v},
xn+=PCn+(x), n∈N,
where,α∈(, ),ξ ∈(,B)and{rn} ⊂(, +∞)withlim infn→+∞rn> ,PCis a
projec-tion operator from Hinto C.Suppose that={p∈F(T)∩EP(f) :Ap∈EP(g)} =∅,then
xn,un→q∈and wn→Aq∈EP(g).
Corollary . Let Hand Hbe two real Hilbert spaces.Let C⊂Hand K⊂Hbe two
nonempty closed convex sets.Let f :C×C→Rand g:K×K→Rbe bi-functions sat-isfying the conditions (A)-(A).Let A:H→H be a bounded linear operator with its
adjoint B.Let x∈C:=C,{xn}and{un}be sequences generated by
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
un=Trfnxn, wn=T
g rnAun,
zn=PC(yn+ξB(wn–Aun)),
Cn+={v∈Cn:zn–v ≤ un–v ≤ xn–v},
xn+=PCn+(x), n∈N,
whereξ∈(,B)and{rn} ⊂(, +∞)withlim infn→+∞rn> ,PCis a projection operator from H into C.Suppose that={p∈EP(f) :Ap∈EP(g)} =∅,then xn,un→q∈and
wn→Aq∈EP(g).
It is well known that the viscosity iterative method is always applied to study the iterative solution for the fixed point problem of nonlinear operators, for example, [, , , , ]. Similarly, the viscosity iterative method can also be used to study the hybrid split prob-lem (HSP). So, at the end of this paper, we introduce a viscosity iterative algorithm which can converge strongly to a solution of (HSP).
Theorem . Let Hand Hbe two real Hilbert spaces.Let C⊂Hand K⊂Hbe two
nonempty closed convex sets.Let h:C→C be aα-contraction mapping,T:C→C and S:K→K be non-expansive mappings and f :C×C→Rand g:K×K→Rbe bi-functions satisfying the conditions(A)-(A).Let A:H→Hbe a bounded linear operator
with its adjoint B.Let x∈C,{xn}and{un}be sequences generated by
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
un=Trfnxn,
wn=TrgnAun,
yn=PC(un+ξB(Swn–Aun)),
zn= ( –r)xn+rTyn,
xn+=αnh(xn) + ( –αn)zn, n∈N,
where r∈(, ),ξ∈(,B)and{rn} ⊂(, +∞),PCis a projection operator from Hinto C,
and the coefficients{αn}and{rn}satisfy the following conditions:
() {αn} ⊂(, ),limn→∞αn= , ∞
n=αn=∞;
() lim infn→+∞rn> ,limn→∞|rn+–rn|= .
Suppose that={p∈F(T)∩EP(f) :Ap∈F(S)∩EP(g)} =∅,then xn,un→q∈and wn→Aq∈F(S)∩EP(g),where q=Ph(q).
Proof Letp∈. The following inequalities are easily verified:
un–p ≤ xn–p, wn–Ap ≤ Aun–Ap. (.)
By Lemma .,
un–p≤ xn–p–Trgnxn–xn=xn–p–un–xn;
Swn–Ap=STrgnAun–Ap≤TrgnAun–Ap (.) ≤ Aun–Ap–TrgnAun–Aun.
From (.) and (.), we have
ξun–p,B(Swn–Aun)
= ξA(un–p) +Swn–Aun– (Swn–Aun),Swn–Aun
= ξ
Swn–Ap
+
Swn–Aun
–
Aun–Ap
–Swn–Aun
≤ξ
–
T g
rnAun–Aun
–
Swn–Aun
= –ξSwn–Aun–ξTrgnAun–Aun
= –ξSwn–Aun–ξwn–Aun (.)
and
yn–p =PC(un+ξB(Swn–Aun) –PCp
≤un–p+ξB(Swn–Aun)
=un–p+ξB(Swn–Aun)
+ ξun–p,B(Swn–Aun)
≤ un–p–ξ –ξBSwn–Aun–ξTrgnAun–Aun ≤ xn–p–ξ –ξBSwn–Aun–ξTrgnAun–Aun
=xn–p–ξ –ξBSwn–Aun–ξwn–Aun. (.)
So, from (.)-(.) and (.), we have
We say{xn}is bounded. In fact, from (.) and (.), we have
xn+–p=αn
f(xn) –p+ ( –αn)(zn–p)≤( –αn)zn–p+αnf(xn) –p
≤( –αn)xn–p+αnαxn–p+αnf(p) –p
= –αn( –α)xn–p+αn( –α)f(p) –p –α ,
which implies that
xn–p ≤max
x–p,f
(p) –p –α
, ∀n∈N, (.)
so{xn}is bounded. Further,{un},{wn}and{yn}are also bounded by (.). By Lemma ., from (.) we have
un+–un=Trfn+xn+–T
f rnxn
≤
xn+–xn+|
rn–rn+|
rn T f
rnxn–xn
≤ xn+–xn+|rn
–rn+|
rn M,
wn+–wn=Trgn+Aun+–T
g rnAun
≤
Aun+–Aun+|rn
–rn+|
rn T g
rnAun–Aun
≤ Aun+–Aun+|rn
–rn+|
rn M
(.)
and
yn+–yn≤un++ξB(Swn+–Aun+) –un–ξB(Swn–Aun)
=un+–un+ξB
Swn+–Aun+– (Swn–Aun)
=un+–un+ξB
Swn+–Aun+– (Swn–Aun)
+ ξun+–un,B
Swn+–Aun+– (Swn–Aun)
≤ un+–un+ξBSwn+–Aun+– (Swn–Aun)
+ ξA(un+–un),Swn+–Aun+– (Swn–Aun)
=un+–un+ξBSwn+–Aun+– (Swn–Aun)
+ ξA(un+–un) +Swn+–Aun+– (Swn–Aun),Swn+
–Aun+– (Swn–Aun)
– ξSwn+–Aun+– (Swn–Aun),Swn+–Aun+– (Swn–Aun)
=un+–un+ξBSwn+–Aun+– (Swn–Aun)
+ ξSwn+–Swn,Swn+–Aun+– (Swn–Aun)
=un+–un+ξBSwn+–Aun+– (Swn–Aun)
+ ξ
Swn+–Swn+Swn+–Aun+– (Swn–Aun)
–Aun+–Aun
– ξSwn+–Aun+– (Swn–Aun)
=un+–un+ξBSwn+–Aun+– (Swn–Aun)
+ξSwn+–Swn–Aun+–Aun
–ξSwn+–Aun+– (Swn–Aun)
≤ un+–un–ξ
–ξBSwn+–Aun+– (Swn–Aun)
+ξwn+–wn–Aun+–Aun
≤ un+–un–ξ
–ξBSwn+–Aun+– (Swn–Aun)
+ξ
Aun+–Aun+|rn
–rn+|
rn M–Aun+–Aun
=un+–un–ξ
–ξBSwn+–Aun+– (Swn–Aun)
+ξ|rn–rn+|
rn M
≤ xn+–xn–ξ
–ξBSwn+–Aun+– (Swn–Aun)
+|rn–rn+|
rn (ξM+M), (.)
whereMis a constant satisfying
sup
n∈N
xn+–xnTrfnxn–xn+
|rn–rn+|
rn T f
rnxn–xn
,
Aun+–AunTrgnAun–Aun+
|rn–rn+|
rn T g
rnAun–Aun
≤M.
Provingxn+–xn → asn→ ∞. Settingβn= – ( –αn)( –r) andvn=xn+–xβnn+βnxn,
namely vn = αnf(xn)+(–αn)rTyn
βn . Let M be a constant satisfying supn∈N{fβ(xnn++),
f(xn) βn ,
Tyn} ≤Mfor alln∈N. Then
vn+–vn =
αn+f(xn+) + ( –αn+)rTyn+
βn+
–αnf(xn) + ( –αn)rTyn
βn
≤αn+
f(xn+)
βn+
+αn
f(xn)
βn
+r( –αn+)Tyn+
βn+
–( –αn)Tyn
βn
≤(αn++αn)M+r
( –αn+)(Tyn+–Tyn)
βn+
+( –αn+)Tyn
βn+
–( –αn)Tyn
βn
≤(αn++αn)M+r
( –αn+)yn+–yn
βn+
+( –αn+)
βn+
–( –αn)
= (αn++αn)M+r
( –αn+)yn+–yn
βn+
+( –r)(αn–αn+) +βn+αn–βnαn+
βnβn+
M
≤(αn++αn)M+r
( –αn+)yn+–yn
βn+
+ αn+αn+
βnβn+
M
:=ρn+r
( –αn+)yn+–yn
βn+
. (.)
From (.) and (.), we have
vn+–vn≤
ρn+r
( –αn+)yn+–yn
βn+
=ρn+ ρnr( –αn+)yn+–yn
βn+
+r( –αn+)
yn
+–yn
βn+ ,
≤ρn+ ρnr
( –αn+)yn+–yn
βn+
+r( –αn+)
βn+ xn+–xn
+r( –αn+)
βn+
|rn–rn+|
rn ( +ξ)M. (.)
By the conditions () and () and (.), we obtain
lim sup
n→∞
vn+–vn–xn+–xn
≤. (.)
Noticevn+–vn–xn+–xn= (vn+–vn–xn+–xn)(vn+–vn+xn+–xn),
hence from (.) we have
lim sup
n→∞
vn+–vn–xn+–xn
≤. (.)
By Lemma . and (.), we havelimn→∞vn–xn= , which implies that
lim
n→∞xn+–xn= (.)
by the definition ofvn. Sincexn+–zn →, together with (.), we have
lim
n→∞xn–zn= . (.)
Using (.), (.) and (.),
xn+–p =αn
f(xn) –p+ ( –αn)(zn–p)
≤( –αn)zn–p+αnf(xn) –p
≤( –r)xn–p+run–p+αnf(xn) –p
≤ xn–p–run–xn+αnf(xn) –p
which yields
run–xn≤ xn–p–xn+–p+αnf(xn) –p
=xn–p+xn+–p
xn–p–xn+–p
+αnf(xn) –p
≤xn–p+xn+–p
xn–xn++αnf(xn) –p
. (.)
From (.) we have
lim
n→∞T
f
rnxn–xn=nlim→∞un–xn= . (.)
Again, applying (.), (.) and (.), we have
xn+–p≤( –αn)zn–p+αnf(xn) –p
≤( –r)xn–p+ryn–p+αnf(xn) –p
≤ xn–p–rξ –ξBSwn–Aun
–rξwn–Aun+αnf(xn) –p, (.)
which implies that
rξ –ξBSwn–Aun+rξwn–Aun
≤xn–p+xn+–p
xn–xn++αnf(xn) –p
. (.)
From (.) we have
lim
n→∞T g
rnAun–Aun=nlim→∞wn–Aun= , nlim→∞Swn–Aun= (.)
and
lim
n→∞Swn–wn= . (.)
Noticeyn=PC(un+ξB(Swn–Aun)) andun∈Cfor alln∈N, so
yn–un=PCun+ξB(Swn–Aun)–PCun≤ξB(Swn–Aun) ≤ξBSwn–Aun,
so
lim
n→∞yn–un= . (.)
Further, from (.), (.) and (.), we have
lim
and
lim
n→∞yn–Tyn= by (.), (.) and (.). (.)
Letq=Pf(q). Choose a subsequence{xnk}such that
lim sup
n→∞
f(q) –q,xn–q = lim
k→∞
f(q) –q,xnk–q. (.)
Since{xn}is bounded,{f(q) –q,xn–q}is bounded. Hencelim supn→∞f(q) –q,xn–q is a constant, namelylimn→∞f(q) –q,xnk–qexists, which implies (.) is well defined.
Because{xn}is bounded,{xnk}has a weak convergence subsequence which is still denoted
by{xnk}. Supposexnkx
*, we sayx*∈. Whenx
nkx
*, from (.), (.) and (.),
we have
unkx*, ynkx*, znkx*, AunkAx*, wnkAx*. (.)
IfTx* =x*, then by (.) and (.) and Opial’s condition, we have
lim inf
k→∞
ynk–x*<lim inf
k→∞
ynk–Tx*
≤lim inf
k→∞
ynk–Tynk+Tynk–Tx *
≤lim inf
k→∞
ynk–Tynk+ynk–x
*=lim inf
k→∞ynk–x
*, (.)
which is a contradiction, soTx*=x*andx*∈F(T). Since for eachr> ,EP(f) =F(Tf
r) by Lemma ., we havex*∈F(Trf). Otherwise, if there existsr> such thatTrfx* =x*, then by (.) and Lemma . and Opial’s condition, we have
lim inf
k→∞xnk–x
*<lim inf
k→∞ xnk–T
f rx*
≤lim inf
k→∞ xnk–T
f
nkxnk+T
f nkxnk–T
f rx*
=lim inf
k→∞ T
f nkxnk–T
f rx*
≤lim inf
k→∞
xnk–x*+|rnk–r|
rnk
Tnfkxnk–xnk
=lim inf
k→∞
xnk–x*, (.)
which is also a contradiction, soTrfx*=x*andx*∈F(Trf) =EP(f). Up to now, we have provedx*∈F(T)∩EP(f). Similarly, we can also proveAx*∈F(S)∩EP(g). Hencex*∈,
because of this, we can also obtain
lim sup
n→∞
f(q) –q,xn–q = lim
k→∞
f(q) –q,xnk–q
Finally, we prove the conclusion of this theorem is right. Forq=Pf(q), from (.) we
have
xn+–q=αn
h(xn) –q+ ( –αn)(zn–q)
≤( –αn)zn–q+ αn
h(xn) –q,xn+–q
≤( –αn)xn–q+ αn
h(xn) –h(q) +h(q) –q,xn+–q
≤( –αn)xn–q+ αnαxn–qxn+–q+ αn
h(q) –q,xn+–q
≤( –αn)xn–q+αnαxn–q+αnαxn+–q
+ αn
h(q) –q,xn+–q
= ( – αn)xn–q+αnxn–q+αnαxn–q+αnαxn+–q
+ αn
h(q) –q,xn+–q. (.)
From (.) we have
xn+–q≤
–αn
– α
–αnα
xn–q+ α
n –αnαxn
–q
+ αn –αnα
h(q) –q,xn+–q, (.)
by (.) and Lemma ., we havexn→q∈. Again, from (.) and (.), we have un→q∈andwn→Aq∈F(S)∩EP(f), respectively. The proof is completed.
Remark
() In this paper, the iterative coefficientαorrcan be replaced with the sequence{ζn}if
{ζn}satisfies{ζn} ⊂[,ϑ], where,ϑ∈(, );
() Obviously, ifH=Hin this paper, these weak and strong convergence theorems are
also true;
() In this paper, ifT is a nonexpansive mapping fromHintoHandf(x,y)is a
bi-function fromH×HintoRwith the conditions (A)-(A),Sis a nonexpansive
mapping fromHintoHandg(u,v)is a bi-function fromH×HintoRwith the
conditions (A)-(A), then we may obtain a series of similar algorithms.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
Both authors contributed equally and significantly in writing this paper. Both authors read and approved the final manuscript.
Author details
1Department of Mathematics, Honghe University, Yunnan, 661100, China.2Department of Mathematics, National
Kaohsiung Normal University, Kaohsiung, 824, Taiwan.
Acknowledgements
The first author was supported by the Natural Science Foundation of Yunnan Province (2010ZC152). The second author was supported partially by grant No. NSC 101-2115-M-017-001 of the National Science Council of the Republic of China.
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doi:10.1186/1687-1812-2013-47
