R E S E A R C H
Open Access
Parallel algorithms for variational inclusions
and fixed points with applications
Afrah AN Abdou
1*, Badriah AS Alamri
1, Yeol Je Cho
1,2, Yonghong Yao
3and Li-Jun Zhu
4*Correspondence:
[email protected]; [email protected]
1Department of Mathematics, King
Abdulaziz University, Jeddah, 21589, Saudi Arabia
Full list of author information is available at the end of the article
Abstract
In this paper, we introduce two parallel algorithms for finding a zero of the sum of two monotone operators and a fixed point of a nonexpansive mapping in Hilbert spaces and prove some strong convergence theorems of the proposed algorithms. As special cases, we can approach the minimum-norm common element of the zero of the sum of two monotone operators and the fixed point of a nonexpansive mapping without using the metric projection. Further, we give some applications of our main results.
MSC: 49J40; 47J20; 47H09; 65J15
Keywords: monotone operator; nonexpansive mapping; zero point; fixed point; resolvent
1 Introduction
LetHbe a real Hilbert space. LetA:H→Hbe a single-valued nonlinear mapping and
B:H→Hbe a set-valued mapping.
Now, we are concerned with the following variational inclusion:
Find a zerox∈Hof the sum of two monotone operatorsAandBsuch that
∈Ax+Bx, (.)
where is the zero vector inH.
The set of solutions of the problem (.) is denoted by (A+B)–. IfH=Rm, then the
problem (.) becomes the generalized equation introduced by Robinson []. IfA= , then the problem (.) becomes the inclusion problem introduced by Rockafellar []. It is well known that the problem (.) is among the most interesting and intensively studied classes of mathematical problems and has wide applications in the fields of optimization and con-trol, economics and transportation equilibrium, engineering science, and many others. For the past years, many existence results and iterative algorithms for various variational inequality and variational inclusion problems have been extended and generalized in var-ious directions using novel and innovative techniques. A useful and important general-ization is called the general variational inclusion involving the sum of two nonlinear op-erators. Moudafi and Noor [] studied the sensitivity analysis of variational inclusions by using the technique of resolvent equations. Recently much attention has been given to de-veloping iterative algorithms for solving the variational inclusions. Donget al.[] analyzed the solution’s sensitivity for variational inequalities and variational inclusions by using a
resolvent operator technique. By using the concept and technique of resolvent operators, Agarwalet al.[] and Jeong [] introduced and studied a new system of parametric gener-alized nonlinear mixed quasi-variational inclusions in a Hilbert space. Lan [] introduced and studied a stable iteration procedure for a class of generalized mixed quasi-variational inclusion systems in Hilbert spaces. Recently, Zhanget al.[] introduced a new iterative scheme for finding a common element of the set of solutions to the problem (.) and the set of fixed points of nonexpansive mappings in Hilbert spaces. Penget al.[] introduced another iterative scheme by the viscosity approximate method for finding a common ele-ment of the set of solutions of a variational inclusion with set-valued maximal monotone mapping and inverse strongly monotone mappings, the set of solutions of an equilibrium problem, and the set of fixed points of a nonexpansive mapping. For some related work, see [–] and the references therein.
Recently, Takahashiet al.[] introduced the following iterative algorithm for finding a zero of the sum of two monotone operators and a fixed point of a nonexpansive map-ping:
xn+=βnxn+ ( –βn)S
αnx+ ( –αn)JλBn(xn–λnAxn)
(.)
for all n≥. Under some assumptions, they proved that the sequence {xn} converges
strongly to a point ofF(S)∩(A+B)–.
Remark . We note that the algorithm (.) cannot be used to find the minimum-norm element due to the facts thatx∈CandSis a self-mapping ofC. However, there exist a large number of problems for which one needs to find the minimum-norm solution (see, for ex-ample, [–]). A useful path to circumvent this problem is to use projection. Bauschke and Browein [] and Censor and Zenios [] provide reviews of the field. The main diffi-culty is in the computation. Hence it is an interesting problem to find the minimum-norm element without using the projection.
Motivated and inspired by the works in this field, we first suggest the following two algorithms without using projection:
xt= ( –κ)Sxt+κJλB
tγf(xt) + ( –t)xt–λAxt
for allt∈(, ) and
xn+= ( –κ)Sxn+κJλBn
αnγf(xn) + ( –αn)xn–λnAxn
for all n≥. Notice that these two algorithms are indeed well defined (see the next
section). We show that the suggested algorithms converge strongly to a point x˜ =
PF(S)∩(A+B)–(γf(x˜)) which solves the following variational inequality:
γf(x˜) –x˜,x˜–z≥
As special cases, we can approach the minimum-norm element in F(S)∩(A+B)– without using the metric projection and give some applications.
2 Preliminaries
LetHbe a real Hilbert space with the inner product·,·and the norm · , respectively. LetCbe a nonempty closed convex subset ofH.
() A mappingS:C→Cis said to benonexpansiveif
Sx–Sy ≤ x–y
for allx,y∈C. We denote byF(S)the set of fixed points ofS.
() A mappingA:C→His said to beα-inverse strongly monotoneif there existsα> such that
Ax–Ay,x–y ≥αAx–Ay
for allx,y∈C.
It is well known that, if A isα-inverse strongly monotone, thenAx–Ay ≤αx–y
for allx,y∈C.
LetBbe a mapping fromHinto H. The effective domain ofBis denoted bydom(B),
that is,dom(B) ={x∈H:Bx=∅}.
() A multi-valued mappingBis said to be amonotone operatoronHif
x–y,u–v ≥
for allx,y∈dom(B),u∈Bx, andv∈By.
() A monotone operatorBonHis said to bemaximalif its graph is not strictly contained in the graph of any other monotone operator onH.
LetBbe a maximal monotone operator onHandB– ={x∈H: ∈Bx}. For a maximal monotone operatorBonHandλ> , we may define a single-valued operatorJB
λ = (I+ λB)–:H→dom(B), which is called theresolventofBforλ. It is well known that the resolventJB
λ is firmly nonexpansive,i.e.,
JλBx–JλBy
≤JλBx–JλBy,x–y
for allx,y∈CandB– =F(JB
λ) for allλ> . The following resolvent identity is well known: for anyλ> andμ> , the following identity holds:
JλBx=JμB
μ λx+
–μ
λ
JλBx
(.)
for allx∈H.
We use the notation thatxnxstands for the weak convergence of (xn) toxandxn→x
stands for the strong convergence of (xn) tox, respectively.
Lemma .([]) Let C be a nonempty closed convex subset of a real Hilbert space H.Let A:C→H be anα-inverse strongly monotone mapping andλ> be a constant.Then we have
(I–λA)x– (I–λA)y≤ x–y+λ(λ– α)Ax–Ay
for all x,y∈C.In particular,if≤λ≤α,then I–λA is nonexpansive.
Lemma .([]) Let C be a closed convex subset of a Hilbert space H.Let S:C→C be a nonexpansive mapping.Then F(S)is a closed convex subset of C and the mapping I–S is demiclosed at,i.e. whenever{xn} ⊂C is such that xnx and(I–S)xn→,then
(I–S)x= .
Lemma .([]) Let C be a nonempty closed convex subset of a real Hilbert space H. As-sume that the mapping F :C→H is monotone and weakly continuous along segments,
that is,F(x+ty)→F(x)weakly as t→.Then the variational inequality
x∗∈C, Fx∗,x–x∗≥
for all x∈C is equivalent to the dual variational inequality
x∗∈C, Fx,x–x∗≥
for all x∈C.
Lemma .([]) Let{xn},{yn}be bounded sequences in a Banach space X and{βn}be a
sequence in[, ]with
<lim inf
n→∞ βn≤lim supn→∞ βn< .
Suppose that xn+= ( –βn)yn+βnxnfor all n≥and
lim sup
n→∞
yn+–yn–xn+–xn
≤.
Thenlimn→∞yn–xn= .
Lemma .([]) Assume that{an}is a sequence of nonnegative real numbers such that
an+≤( –γn)an+δnγn
for all n≥,where{γn}is a sequence in(, )and{δn}is a sequence such that
(a) ∞n=γn=∞;
3 Main results
In this section, we prove our main results.
Theorem . Let C be a nonempty closed convex subset of a real Hilbert space H.Let A be anα-inverse strongly monotone mapping from C into H.Let f :C→H be aρ-contraction andγ be a constant such that <γ <ρ.Let B be a maximal monotone operator on H such that the domain of B is included in C.Let JB
λ= (I+λB)–be the resolvent of B for anyλ>
and S be a nonexpansive mapping from C into itself such that F(S)∩(A+B)–=∅.Letλ
andκ be two constants satisfying a≤λ≤b,where[a,b]⊂(, α)andκ∈(, ).For any t∈(, – λ
α),let{xt} ⊂C be a net generated by
xt= ( –κ)Sxt+κJλB
tγf(xt) + ( –t)xt–λAxt. (.)
Then the net{xt}converges strongly as t→+to a pointx˜=PF(S)∩(A+B)–(γf(x˜)),which
solves the following variational inequality:
γf(x˜) –x˜,x˜–z≥
for all z∈F(S)∩(A+B)–.
Proof First, we show that the net{xt} is well defined. For anyt∈(, – αλ), we define a mappingW := ( –κ)S+κJB
λ(tγf + ( –t)I–λA). Note thatJλB, S, and I– λ –tA(see
Lemma .) are nonexpansive. For anyx,y∈C, we have
Wx–Wy
=( –κ)(Sx–Sy) +κ
JλB
tγf(x) + ( –t)
I– λ –tA
x
–JλB
tγf(y) + ( –t)
I– λ –tA
y
≤( –κ)Sx–Sy
+κtγf(x) –f(y)+ ( –t)
I– λ –tA
x–
I– λ –tA
y
≤( –κ)x–y+κtγf(x) –f(y)
+ ( –t)κ
I– λ –tA
x–
I– λ –tA
y
≤( –κ)x–y+tκγρx–y+ ( –t)κx–y
= – ( –γρ)κtx–y,
which implies the mappingWis a contraction onC. We usextto denote the unique fixed point ofWinC. Therefore,{xt}is well defined. Setyt=JλButandut=γf(xt) + ( –t)xt– λAxtfor allt> . Takingz∈F(S)∩(A+B)–, it is obvious thatz=Sz=JB
λ(z–λAz) for all λ> and so
z=Sz=JλB(z–λAz) =JλB
tz+ ( –t)
I– λ –tA
z
for allt∈(, –αλ). From (.), it follows that
xt–z=( –κ)(Sxt–z) +κ(yt–z)
≤( –κ)Sxt–z+κyt–z ≤( –κ)xt–z+κyt–z.
Hence we getxt–z ≤ yt–z. SinceJλBis nonexpansive, we have
yt–z
=JλB
tγf(xt) + ( –t)
xt– λ –tAxt
–JλB
tz+ ( –t)
z– λ –tAz
≤
tγf(xt) + ( –t)
xt– λ –tAxt
–
tz+ ( –t)
z– λ –tAz
=( –t)
xt– λ –tAxt
–
z– λ –tAz
+tγf(xt) –z
≤( –t)
I– λ –tA
xt–
I– λ –tA
z+tγf(xt) –f(z)+tγf(z) –z
≤( –t)xt–z+tγρxt–z+tγf(z) –z. (.)
Thus it follows that
xt–z ≤
–γργf(z) –z.
Therefore,{xt}is bounded. We deduce immediately that{f(xt)},{Axt},{Sxt},{ut}, and{yt}
are also bounded. By using the convexity of · and theα-inverse strong monotonicity of
A, from (.), we derive
xt–z
≤ yt–z
≤( –t)
xt– λ –tAxt
–
z– λ –tAz
+tγf(xt) –z
≤( –t)
xt– λ –tAxt
–
z– λ –tAz
+tγf(xt) –z
= ( –t)(xt–z) –
λ
–t(Axt–Az)
+tγf(xt) –z
= ( –t)
xt–z– λ
–tAxt–Az,xt–z+
λ
( –t)Axt–Az
+tγf(xt) –z
≤( –t)
xt–z–
αλ
–tAxt–Az
+ λ
( –t)Axt–Az
= ( –t)
xt–z+ λ ( –t)
λ– ( –t)αAxt–Az
+tγf(xt) –z
≤( –t)xt–z+
λ ( –t)
λ– ( –t)αAxt–Az+tγf(xt) –z
(.)
and so
λ ( –t)
( –t)α–λAxt–Az≤tγf(xt) –z
–txt–z→.
By the assumption, we have ( –t)α–λ> for allt∈(, – λ
α) and so we obtain
lim
t→+Axt–Az= . (.)
Next, we showxt–Sxt →. By using the firm nonexpansivity ofJλB, we have
yt–z=JλB
tγf(xt) + ( –t)xt–λAxt–z
=JλB
tγf(xt) + ( –t)xt–λAxt–JλB(z–λAz)
≤tγf(xt) + ( –t)xt–λAxt– (z–λAz),yt–z
=
tγf(xt) + ( –t)xt–λAxt– (z–λAz)
+yt–z
–tγf(xt) + ( –t)xt–λ(Axt–Az) –yt.
Thus it follows that
yt–z≤tγf(xt) + ( –t)xt–λAxt– (z–λAz)
–tγf(xt) + ( –t)xt–λ(Axt–Az) –yt.
By the nonexpansivity ofI– λ
–tA, we have
tγf(xt) + ( –t)xt–λAxt– (z–λAz)
=( –t)
xt– λ –tAxt
–
z– λ –tAz
+tγf(xt) –z
≤( –t)
xt– λ –tAxt
–
z– λ –tAz
+tγf(xt) –z ≤( –t)xt–z+tγf(xt) –z
and thus
xt–z≤ yt–z
≤( –t)xt–z+tγf(xt) –z
–tγf(xt) + ( –t)xt–λ(Axt–Az) –yt.
Hence it follows that
SinceAxt–Az →, we deducelimt→+xt–yt= , which implies that
lim
t→+xt–Sxt= . (.)
From (.), we have
yt–z
≤( –t)
xt– λ –tAxt
–
z– λ –tAz
+tγf(xt) –z
= ( –t)
xt– λ –tAxt
–
z– λ –tAz
+ t( –t)
γf(xt) –z,
xt–
λ –tAxt
–
z– λ –tAz
+tγf(xt) –z
≤( –t)xt–z+ t( –t)
γf(xt) –z,xt– λ
–t(Axt–Az) –z
+tγf(x
t) –z
= ( –t)xt–z+ t( –t)γ
f(xt) –f(z),xt– λ
–t(Axt–Az) –z
+ t( –t)
γf(z) –z,xt– λ
–t(Axt–Az) –z
+tγf(xt) –z.
Note thatxt–z ≤ yt–z. Then we obtain
xt–z≤( –t)xt–z+ t( –t)γf(xt) –f(z)
xt–z+
–λt(Axt–Az)
+ t( –t)
γf(z) –z,xt– λ
–t(Axt–Az) –z
+tγf(xt) –z
≤( –t)xt–z+ t( –t)γρxt–z+ tλγρxt–zAxt–Az
+ t( –t)
γf(z) –z,xt–
λ
–t(Axt–Az) –z
+tγf(xt) –z
≤ – ( –γρ)txt–z+ t
( –t)
γf(z) –z,xt– λ
–t(Axt–Az) –z
+ t
γf(xt) –z
+xt–z+λγρxt–zAxt–Az
.
Thus it follows that
xt–z≤ –γρ
γf(z) –z,xt– λ
–t(Axt–Az) –z
+ t
γf(xt) –z
+xt–z+tγf(z) –zxt– λ
–t(Axt–Az) –z
+λγρxt–zAxt–Az
≤
–γρ
whereMis some constant such that
sup
–γρ
γf(xt) –z
+xt–z
+γf(z) –zxt–
λ
–t(Axt–Az) –z
,
λγρxt–z:t∈
, – λ
α
≤M.
Next, we show that{xt}is relatively norm-compact ast→+. Assume that{tn} ⊂(, – λ
α) is such thattn→+ asn→ ∞. Putxn:=xtn. From (.), we have
xn–z≤ –γρ
γf(z) –z,xn–z+tn+Axn–AzM. (.)
Since{xn}is bounded, without loss of generality, we may assume thatxnjx˜∈C. Hence
ynjx˜because ofxn–yn →. From (.), we have
lim
n→∞xn–Sxn= . (.)
We can use Lemma . to (.) to deduce x˜∈F(S). Further, we show thatx˜ is also in (A+B)–. Letv∈Bu. Note thatyn=JB
λ(tnγf(xn) + ( –tn)xn–λAxn) for alln≥. Then we have
tnγf(xn) + ( –tn)xn–λAxn∈(I+λB)yn
⇒ tnγf(xn)
λ +
–tn
λ xn–Axn–
yn
λ ∈Byn.
SinceBis monotone, we have, for all (u,v)∈B,
tnγf(xn)
λ +
–tn
λ xn–Axn–
yn
λ –v,yn–u
≥
⇒ tnγf(xn) + ( –tn)xn–λAxn–yn–λv,yn–u
≥
⇒ Axn+v,yn–u ≤
λxn–yn,yn–u–
tn
λ
xn–γf(xn),yn–u
⇒ A˜x+v,yn–u ≤
λxn–yn,yn–u–
tn
λ
xn–γf(xn),yn–u
+A˜x–Axn,yn–u
⇒ Ax˜+v,yn–u ≤
λxn–ynyn–u+
tn
λxn–γf(xn)yn–u
+A˜x–Axnyn–u.
Thus it follows that
A˜x+v,x˜–u ≤
λxnj–ynjynj–u+
tnj
λxnj–γf(xnj)ynj–u
+A˜x–Axnjynj–u+A˜x+v,x˜–ynj. (.)
Sincexnj–x˜,Axnj–A˜x ≥αAxnj–A˜x,Axnj→Az, andxnjx˜, it follows thatAxnj→
v,x˜–u ≤, that is,–A˜x–v,x˜–u ≥. SinceBis maximal monotone, we have –A˜x∈B˜x. This shows that ∈(A+B)x˜. Hence we havex˜∈F(S)∩(A+B)–. Therefore, substituting
˜
xforzin (.), we get
xn–x˜≤
–γρ
γf(x˜) –x˜,xn–x˜
+tn+Axn–Ax˜
M.
Consequently, the weak convergence of{xn}tox˜actually implies thatxn→ ˜x. This proved the relative norm-compactness of the net{xt}ast→+.
Now, we return to (.) and, taking the limit asn→ ∞, we have
˜x–z≤
–γρ
γf(z) –z,x˜–z
for allz∈F(S)∩(A+B)–. In particular,x˜solves the following variational inequality:
˜
x∈F(S)∩(A+B)–, γf(z) –z,x˜–z≥
for allz∈F(S)∩(A+B)– or the equivalent dual variational inequality (see Lemma .):
˜
x∈F(S)∩(A+B)–, γf(x˜) –x˜,x˜–z≥
for allz∈F(S)∩(A+B)–. Hence x˜=P
F(S)∩(A+B)–(γf(x˜)). Clearly, this is sufficient to
conclude that the entire net{xt}converges tox˜. This completes the proof.
Theorem . Let C be a nonempty closed convex subset of a real Hilbert space H.Let A be anα-inverse strongly monotone mapping from C into H.Let f :C→H be aρ-contraction andγ be a constant such that <γ <
ρ.Let B be a maximal monotone operator on H such
that the domain of B is included in C.Let JB
λ= (I+λB)–be the resolvent of B for anyλ>
and S be a nonexpansive mapping from C into itself such that F(S)∩(A+B)–=∅.For
any x∈C,let{xn} ⊂C be a sequence generated by
xn+= ( –κ)Sxn+κJλBn
αnγf(xn) + ( –αn)xn–λnAxn
(.)
for all n≥,whereκ∈(, ),{λn} ⊂(, α)and{αn} ⊂(, )satisfy the following condi-tions:
(a) limn→∞αn= ,limn→∞ααn+n = and nαn=∞;
(b) a( –αn)≤λn≤b( –αn),where[a,b]⊂(, α)andlimn→∞λnα+–λn+n = .
Then the sequence{xn}converges strongly to a pointx˜=PF(S)∩(A+B)–(γf(x˜)),which solves
the following variational inequality:
γf(x˜) –x˜,x˜–z≥
for all z∈F(S)∩(A+B)–.
Proof Setyn=JB
λnun,un=αnγf(xn) + ( –αn)xn–λnAxnfor alln≥. Pick upz∈F(S)∩
(A+B)–. It is obvious that
z=Sz=JλBn(z–λnAz) =JλBn
αnz+ ( –αn)
z– λn –αn
Az
for alln≥. SinceJB
λn,S, andI–
λn
–αnAare nonexpansive for allλ> andn≥, we have
yn–z
=JλBn
αnγf(xn) + ( –αn)xn–λnAxn
–z
=JλBn
αnγf(xn) + ( –αn)
xn– λn –αn
Axn
–JλBn
αnz+ ( –αn)
z– λn –αn
Az
≤
αnγf(xn) + ( –αn)
xn– λn –αn
Axn
–
αnz+ ( –αn)
z– λn –αn
Az
=( –αn)
xn– λn –αn
Axn
–
z– λn –αn
Az
+αn
γf(xn) –z
≤( –αn)xn–z+αnγf(xn) –γf(z)+αnγf(z) –z ≤ – ( –γρ)αn
xn–z+αnγf(z) –z. (.)
Hence we have
xn+–z ≤( –κ)Sxn–z+κyn–z ≤( –κ)xn–z+κ – ( –γρ)αn
xn–z+καnγf(z) –z
= – ( –γρ)καn
xn–z+καnγf(z) –z.
By induction, we have
xn+–z ≤max
x–z,
–γργf(z) –z
.
Therefore,{xn}is bounded. SinceAisα-inverse strongly monotone, it is α-Lipschitz con-tinuous. We deduce immediately that{f(xn)},{Sxn},{Axn},{un}, and{yn}are also bounded.
By using the convexity of · and theα-inverse strong monotonicity ofA, it follows from (.) that
( –αn)
xn–
λn
–αn Axn
–
z– λn –αn
Az
+αn
γf(xn) –z
≤( –αn)
xn– λn –αn
Axn
–
z– λn –αn
Az
+αnγf(xn) –z
= ( –αn)
(xn–z) – λn –αn
(Axn–Az)
+αnγf(xn) –z
= ( –αn)
xn–z–
λn
–αn
Axn–Az,xn–z+
λ
n
( –αn)
Axn–Az
+αnγf(xn) –z
≤( –αn)
xn–z– αλn –αn
Axn–Az+ λ
n
( –αn)
Axn–Az
+αnγf(xn) –z
= ( –αn)
xn–z+ λn ( –αn)
λn– ( –αn)α
Axn–Az
+αnγf(xn) –z. (.)
By the condition (c), we getλn– ( –αn)α≤ for alln≥. Then, from (.) and (.),
we obtain
JλBnun–z
≤( –αn)
xn–z+ λn ( –αn)
λn– ( –αn)α
Axn–Az
+αnγf(xn) –z. (.)
From (.), it follows that
xn+–z=( –κ)(Sxn–z) +κ
JλBnun–z
≤( –κ)xn–z+κJλBnun–z
. (.)
Next, we estimatexn+–xn. In fact, we have
xn+–xn+=( –κ)(Sxn+–Sxn) +κ(yn+–yn)
≤( –κ)xn+–xn+κyn+–yn
and
yn+–yn=JλBn+un+–J
B
λnun
≤JλBn+un+–J
B
λn+un+J
B
λn+un–J
B
λnun
≤αn+γf(xn+) + ( –αn+)xn+–λn+Axn+
–αnγf(xn) + ( –αn)xn–λnAxn+JλBn+un–J
B
λnun
=αn+γ
f(xn+) –f(xn)+ (αn+–αn)γf(xn)
+ ( –αn+)
I– λn+ –αn+
A
xn+–
I– λn+ –αn+
A
xn
+ (αn–αn+)xn+ (λn–λn+)Axn
+JλBn+un–J
B
λnun
≤αn+γρxn+–xn+|αn+–αn|γf(xn)+xn
+ ( –αn+)xn+–xn+|λn–λn+|Axn+JλBn+un–J
B
λnun
= – ( –γρ)αn+
xn+–xn+|αn+–αn|γf(xn)+xn
+|λn–λn+|Axn+JλBn+un–J
B
By the resolvent identity (.), we have
JλBn+un=J
B
λn
λn
λn+
un+
– λn λn+
JλBn+un
.
Thus it follows that
JλBn+un–JλBnun=
JλBn
λn
λn+
un+
– λn
λn+
JλBn+un
–JλBnun
≤
λn
λn+
un+
– λn λn+
JλBn+un
–un
≤|λn+–λn| λn+
un–JλBn+un
and so
xn+–xn+
≤( –κ)xn+–xn+κyn+–yn
≤( –κ)xn+–xn+κ
– ( –γρ)αn+
xn+–xn
+κ|αn+–αn|γf(xn)+xn
+κ|λn–λn+|Axn
+κ|λn+–λn| λn+
un–JλBn+un
≤ – ( –γρ)καn+
xn+–xn+ ( –γρ)καn+ |
αn+–αn| αn+
γf(xn)+xn –γρ
+|λn–λn+| αn+
Axn
–γρ +
|λn+–λn| αn+λn+
un–JB
λn+un –γρ
.
By the assumptions, we know that |αn+–αn|
αn+ → and
|λn+–λn|
αn+ →. Then, from Lemma .,
we get
lim
n→∞xn+–xn= . (.)
Thus, from (.) and (.), it follows that
xn+–z
≤( –κ)xn–z+κJλBnun–z
≤κ
( –αn)
xn–z+ λn ( –αn)
λn– ( –αn)α
Axn–Az
+αnγf(xn) –z
+ ( –κ)xn–z
= [ –καn]xn–z+
κλn
–αn
λn– ( –αn)α
Axn–Az+καnγf(xn) –z
≤ xn–z+ κλn –αn
λn– ( –αn)α
and so
κλn
( –αn)
( –αn)α–λn
Axn–Az
≤ xn–z–xn+–z+καnγf(xn) –z
≤xn–z–xn+–z
xn+–xn+καnγf(xn) –z
.
Sincelimn→∞αn= ,limn→∞xn+–xn= , andlim infn→∞(–ακλnn)(( –αn)α–λn) > ,
we have
lim
n→∞Axn–Az= . (.)
Next, we showxn–Sxn →. By using the firm nonexpansivity ofJλBn, we have
JλBnun–z
=JλBn
αnγf(xn) + ( –αn)xn–λnAxn
–JλBn(z–λnAz)
≤αnγf(xn) + ( –αn)xn–λnAxn– (z–λnAz),JλBnun–z
=
αnγf(xn) + ( –αn)xn–λnAxn– (z–λnAz)
+JλBnun–z
–αnγf(xn) + ( –αn)xn–λn(Axn–Az) –JλBnun
.
From the condition (c) and theα-inverse strongly monotonicity ofA, we know thatI– λnA/( –αn) is nonexpansive. Hence it follows that
αnγf(xn) + ( –αn)xn–λnAxn– (z–λnAz)
=( –αn)
xn–
λn
–αn Axn
–
z– λn –αn
Az
+αn
γf(xn) –z
≤( –αn)
xn– λn –αn
Axn
–
z– λn –αn
Az
+αnγf(xn) –z
≤( –αn)xn–z+αnγf(xn) –z
and thus
JλBnun–z
≤
( –αn)xn–z+αnγf(xn) –z
+JλBnun–z
–αnγf(xn) + ( –αn)xn–JλBnun–λn(Axn–Az)
,
that is,
JλBnun–z
≤( –αn)xn–z+αnγf(xn) –z
–αnγf(xn) + ( –αn)xn–JλBnun–λn(Axn–Az)
= ( –αn)xn–z+αnγf(xn) –z
–αnγf(xn) + ( –αn)xn–JλBnun
+ λn
αnγf(xn) + ( –αn)xn–JλBnun,Axn–Az
–λnAxn–Az
≤( –αn)xn–z+αnγf(xn) –z
–αnγf(xn) + ( –αn)xn–JλBnun
+ λnαnγf(xn) + ( –αn)xn–JλBnunAxn–Az.
This together with (.) implies that
xn+–z≤( –κ)xn–z+κ( –αn)xn–z+καnγf(xn) –z
–καnγf(xn) + ( –αn)xn–JλBnun
+ λnκαnγf(xn) + ( –αn)xn–JλBnunAxn–Az
= [ –καn]xn–z+καnγf(xn) –z
–καnγf(xn) + ( –αn)xn–JλBnun
+ λnκαnγf(xn) + ( –αn)xn–JλBnunAxn–Az
and hence
καnγf(xn) + ( –αn)xn–JλBnun
≤ xn–z–xn+–z–καnxn–z+καnγf(xn) –z
+ λnκαnγf(xn) + ( –αn)xn–JλBnunAxn–Az
≤xn–z+xn+–z
xn+–xn+καnγf(xn) –z
+ λnκαnγf(xn) + ( –αn)xn–JλBnunAxn–Az.
Sincexn+–xn →,αn→, andAxn–Az → (by (.)), we deduce
lim
n→∞αnγf(xn) + ( –αn)xn–J B
λnun= .
This implies that
lim
n→∞xn–J B
λnun= . (.)
Combining (.), (.), and (.), we get
lim
n→∞xn–Sxn= . (.)
Putx˜=limt→+xt=PF(S)∩(A+B)–(γf(x˜)), where{xt}is the net defined by (.).
Finally, we show thatxn→ ˜x. Takingz=x˜in (.), we getAxn–Ax˜ →. First, we
provelim supn→∞γf(x˜) –x˜,xn–x ≤˜ . We take a subsequence{xni}of{xn}such that
lim sup
n→∞
γf(x˜) –x˜,xn–x˜
=lim
i→∞
γf(x˜) –x˜,xni–x˜
.
There exists a subsequence{xnij}of{xni}which converges weakly to a pointw∈C. Hence
principle of the nonexpansive mapping (see Lemma .) and (.), we deducew∈F(S). Furthermore, by a similar argument to that of Theorem ., we can show thatwis also in (A+B)–. Hence we havew∈F(S)∩(A+B)–. This implies that
lim sup
n→∞
γf(x˜) –x˜,xn–x˜=lim
j→∞
γf(x˜) –x˜,xnij–x˜=γf(x˜) –x˜,w–x˜.
Note thatx˜=PF(S)∩(A+B)–(γf(x˜)). Then we have
γf(x˜) –x˜,w–x˜≤
for allw∈F(S)∩(A+B)–. Therefore, it follows that
lim sup
n→∞
γf(x˜) –x˜,xn–x˜≤.
From (.), we have
xn+–x˜
≤( –κ)xn–x˜ +κJλBnun–x˜
= ( –κ)xn–x˜ +κJλBnun–J
B
λn(x˜–λnA˜x)
≤( –κ)xn–x˜ +κun– (x˜–λnA˜x)
= ( –κ)xn–x˜ +καnγf(xn) + ( –αn)xn–λnAxn– (x˜–λnA˜x)
=κ( –αn)
xn– λn –αn
Axn
–
˜ x– λn
–αn Ax˜
+αn
γf(xn) –x˜
+ ( –κ)xn–x˜
= ( –κ)xn–x˜ +κ
( –αn)
xn– λn –αn
Axn
–
˜ x– λn
–αn A˜x
+ αn( –αn)
γf(xn) –x˜,
xn– λn –αn
Axn
–
˜ x– λn
–αnA˜ x
+αnγf(xn) –x˜
≤( –κ)xn–x˜ +κ( –αn)xn–x˜ + αnλn
γf(xn) –x˜,Axn–A˜x
+ αn( –αn)γ
f(xn) –f(x˜),xn–x˜+ αn( –αn)
γf(x˜) –x˜,xn–x˜
+αnγf(xn) –x˜
≤( –κ)xn–x˜+κ
( –αn)xn–x˜+ αnλnγf(xn) –x˜Axn–Ax˜
+ αn( –αn)γρxn–x˜ + αn( –αn)
γf(x˜) –x˜,xn–x˜+αnγf(xn) –x˜
≤ – κ( –γρ)αn
xn–x˜ + αnκλnγf(xn) –x˜Axn–Ax˜
+ αnκ( –αn)
γf(x˜) –x˜,xn–x˜
+καnγf(xn) –x˜
+xn–x˜
= – κ( –γρ)αn
+ κ( –γρ)αn
λn
–γργf(xn) –x˜Axn–A˜x
+ –αn –γρ
γf(x˜) –x˜,xn–x˜+ αn
( –γρ)γf(xn) –x˜
+xn–x˜ .
It is clear that nκ( –γρ)αn=∞and
lim sup
n→∞
λn
–γργf(xn) –x˜Axn–Ax˜+ –αn
–γρ
γf(x˜) –x˜,xn–x˜
+ αn
( –γρ)γf(xn) –x˜
+xn–x˜ ≤.
Therefore, we can apply Lemma . to conclude thatxn→ ˜x. This completes the proof.
Remark . One quite often seeks a particular solution of a given nonlinear problem, in particular, the minimum-norm element. For instance, given a closed convex subsetCof a Hilbert spaceHand a bounded linear operatorW:H→H, whereHis another Hilbert space. TheC-constrained pseudoinverse ofW,WC†, is then defined as the minimum-norm solution of the constrained minimization problem
WC†(b) :=arg min
x∈CWx–b,
which is equivalent to the fixed point problem
u=projCu–μW∗(Wu–b),
whereW∗is the adjoint ofW andμ> is a constant, andb∈His such thatPW(C)(b)∈
W(C). From Theorems . and ., we get the following corollaries which can find the minimum-norm element inF(S)∩(A+B)–; that is, findx˜∈F(S)∩(A+B)– such that
˜
x=arg min
x∈F(S)∩(A+B)–x.
Corollary . Let C be a nonempty closed convex subset of a real Hilbert space H.Let A be anα-inverse strongly monotone mapping from C into H.Let B be a maximal monotone operator on H such that the domain of B is included in C.Let JλB= (I+λB)–be the resolvent
of B for anyλ> and S be a nonexpansive mapping from C into itself such that F(S)∩(A+
B)–=∅.Letλandκ be two constants satisfying a≤λ≤b,where[a,b]⊂(, α)and κ∈(, ).For any t∈(, – λ
α),let{xt} ⊂C be a net generated by
xt= ( –κ)Sxt+κJλB
( –t)xt–λAxt.
Then the net{xt}converges strongly as t→+to a pointx˜=PF(S)∩(A+B)–()which is the
minimum-norm element in F(S)∩(A+B)–.
of B for anyλ> such that(A+B)–= ∅.Letλbe a constant satisfying a≤λ≤b,where [a,b]⊂(, α).For any t∈(, –αλ),let{xt} ⊂C be a net generated by
xt=JB
λ
( –t)xt–λAxt
.
Then the net {xt} converges strongly as t→+to a pointx˜=P(A+B)–(), which is the
minimum-norm element in(A+B)–.
Corollary . Let C be a nonempty closed convex subset of a real Hilbert space H.Let A be anα-inverse strongly monotone mapping from C into H.Let B be a maximal monotone operator on H such that the domain of B is included in C.Let JB
λ= (I+λB)–be the resolvent
of B for anyλ> and let S be a nonexpansive mapping from C into itself such that F(S)∩ (A+B)–=∅.For any x
∈C,let{xn} ⊂C be a sequence generated by
xn+= ( –κ)Sxn+κJλBn
( –αn)xn–λnAxn
for all n≥,whereκ∈(, ),{λn} ⊂(, α),and{αn} ⊂(, )satisfy the following condi-tions:
(a) limn→∞αn= ,limn→∞ααn+n = ,and nαn=∞;
(b) a( –αn)≤λn≤b( –αn),where[a,b]⊂(, α)andlimn→∞λn+–λαn n = .
Then the sequence {xn}converges strongly to a point x˜=PF(S)∩(A+B)–(),which is the
minimum-norm element in F(S)∩(A+B)–.
Corollary . Let C be a closed convex subset of a real Hilbert space H.Let A be anα -inverse strongly monotone mapping from C into H and let B be a maximal monotone oper-ator on H such that the domain of B is included in C.Let JλB= (I+λB)–be the resolvent of B
for anyλ> such that(A+B)–=∅.For any x
∈C,let{xn} ⊂C be a sequence generated
by
xn+= ( –κ)xn+κJλBn
( –αn)xn–λnAxn
for all n≥,whereκ∈(, ),{λn} ⊂(, α),and{αn} ⊂(, )satisfy the following condi-tions:
(a) limn→∞αn= ,limn→∞ααn+n = ,and nαn=∞;
(b) a( –αn)≤λn≤b( –αn),where[a,b]⊂(, α)andlimn→∞λn+–λαn n = .
Then the sequence {xn} converges strongly to a point x˜ =P(A+B)–(), which is the
minimum-norm element in(A+B)–.
Remark . The present paper provides some methods which do not use projection for finding the minimum-norm solution problem.
4 Applications
Next, we consider the problem for finding the minimum-norm solution of a mathematical model related to equilibrium problems. LetCbe a nonempty closed convex subset of a Hilbert space andG:C×C→Rbe a bifunction satisfying the following conditions:
(E) G(x,x) = for allx∈C;
(E) for allx,y,z∈C,lim supt↓G(tz+ ( –t)x,y)≤G(x,y); (E) for allx∈C,G(x,·)is convex and lower semicontinuous.
Then the mathematical model related to the equilibrium problem (with respect toC) is as follows:
Findx˜∈Csuch that
G(x˜,y)≥ (.)
for ally∈C. The set of such solutionsx˜is denoted byEP(G). The following lemma appears implicitly in Blum and Oettli [].
Lemma . Let C be a nonempty closed convex subset of a Hilbert space H.Let G be a
bifunction from C×C into R satisfying the conditions(E)-(E).Then,for any r> and x∈H,there exists z∈C such that
G(z,y) +
ry–z,z–x ≥
for all y∈C.
The following lemma was given in Combettes and Hirstoaga [].
Lemma . Assume that G is a bifunction from C×C into R satisfying the conditions
(E)-(E).For any r> and x∈H,define a mapping Tr:H→C as follows:
Tr(x) =
z∈C:G(z,y) +
ry–z,z–x ≥,∀y∈C
for all x∈H.Then the following hold: (a) Tris single-valued;
(b) Tris a firmly nonexpansive mapping,i.e.,for allx,y∈H,
Trx–Try≤ Trx–Try,x–y;
(c) F(Tr) =EP(G);
(d) EP(G)is closed and convex.
We call such aTrthe resolvent ofGfor anyr> . Using Lemmas . and ., we have the following lemma (see [] for a more general result).
Lemma . Let C be a nonempty closed convex subset of a Hilbert space H.Let G be a bifunction from C×C into R satisfying the conditions(E)-(E).Let AGbe a multi-valued mapping from H into itself defined by
AGx= ⎧ ⎨ ⎩
{z∈H:G(x,y)≥ y–x,z,∀y∈C}, x∈C,
Then EP(G) =A–
G()and AGis a maximal monotone operator withdom(AG)⊂C.Further, for any x∈H and r> ,the resolvent Trof G coincides with the resolvent of AG,i.e.,
Trx= (I+rAG)–x.
Form Lemma . and Theorems . and ., we have the following results.
Theorem . Let C be a nonempty closed convex subset of a real Hilbert space H.Let G be a bifunction from C×C into R satisfying the conditions(E)-(E)and Tr be the re-solvent of G for any r> .Let S be a nonexpansive mapping from C into itself such that F(S)∩EP(G)= ∅.For any t∈(, ),let{xt} ⊂C be a net generated by
xt= ( –κ)Sxt+κTr( –t)xt.
Then the net{xt}converges strongly as t→+to a pointx˜=PF(S)∩EP(G)(),which is the minimum-norm element in F(S)∩EP(G).
Proof From Lemma ., we knowAG is maximal monotone. Thus, in Theorem ., we can setJB
λ=Tr. At the same time, in Theorem ., we can choosef = andA= , and (.) reduces to
xt= ( –κ)Sxt+κTr( –t)xt.
Consequently, from Theorem ., we get the desired result. This completes the proof.
Corollary . Let C be a nonempty closed convex subset of a real Hilbert space H.Let G be a bifunction from C×C into R satisfying the conditions(E)-(E)and Trbe the resolvent of G for any r> .Suppose that EP(G)=∅.For any t∈(, ),let{xt} ⊂C be a net generated by
xt=Tr( –t)xt.
Then the net {xt} converges strongly as t→+ to a point x˜ =PEP(G)(), which is the minimum-norm element in EP(G).
Theorem . Let C be a nonempty closed and convex subset of a real Hilbert space H.
Let G be a bifunction from C×C into R satisfying the conditions(E)-(E)and Tλbe the
resolvent of G for anyλ> .Let S be a nonexpansive mapping from C into itself such that F(S)∩EP(G)=∅.For any x∈C,let{xn} ⊂C be a sequence generated by
xn+= ( –κ)Sxn+κTλn
( –αn)xn
for all n≥,whereκ∈(, ),{λn} ⊂(,∞),and{αn} ⊂(, )satisfy the conditions: (a) limn→∞αn= ,limn→∞ααn+n = ,and nαn=∞;
(b) a≤λn≤b,where[a,b]⊂(,∞)andlimn→∞λn+–λαn n= .
Proof From Lemma ., we knowAG is maximal monotone. Thus, in Theorem ., we can setJB
λn=Tλn. At the same time, in Theorem ., we can choosef = andA= , and
(.) reduces to
xn+= ( –κ)Sxn+κTλn
( –αn)xn
for alln≥. Consequently, from Theorem ., we get the desired result. This completes
the proof.
Corollary . Let C be a nonempty closed convex subset of a real Hilbert space H.Let G be a bifunction from C×C into R satisfying the conditions(E)-(E)and Tλbe the resolvent
of G for anyλ> .Suppose EP(G)=∅.For any x∈C,let{xn} ⊂C be a sequence generated
by
xn+= ( –κ)xn+ ( –βn)Tλn
( –αn)xn
for all n≥,whereκ∈(, ),{λn} ⊂(,∞),and{αn} ⊂(, )satisfy the following condi-tions:
(a) limn→∞αn= ,limn→∞ααn+n = ,and nαn=∞;
(b) a≤λn≤b,where[a,b]⊂(,∞)andlimn→∞λn+–λαn n= .
Then the sequence{xn}converges strongly to a pointx˜=PEP(G)(),which is the minimum-norm element in EP(G).
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.
Author details
1Department of Mathematics, King Abdulaziz University, Jeddah, 21589, Saudi Arabia.2Department of Mathematics
Education and the RINS, Gyeongsang National University, Chinju, 660-701, Korea.3Department of Mathematics, Tianjin Polytechnic University, Tianjin, 300387, China.4School of Mathematics and Information Science, Beifang University of Nationalities, Yinchuan, 750021, China.
Acknowledgements
This project was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, under Grant No. (31-130-35-HiCi) and the fifth author was supported in part by NNSF of China (61362033) and NGY2012097.
Received: 19 March 2014 Accepted: 27 June 2014 Published:18 Aug 2014
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