A viscosity method for hierarchical fixed point problems is presented to solve variational inequalities, where the involved mappings are nonexpansive nonself-mappings. Solutions are sought in the set of the common fixed points of an infinite family of nonexpansive nonself-mappings. The results generalize and improve the recent results announced by many other authors.
1. Introduction and Preliminaries
Let X a real Banach space and J be the normalized duality mapping from X into 2X* given by
(1)J(x)={x*∈X*:〈x,x*〉=∥x∥∥x*∥,∥x∥=∥x*∥}
for all x∈X, where X* denotes the dual space of X and 〈·,·〉 the generalized duality pairing between X and X*. If X=H is a Hilbert space, then J becomes the identity mapping on H. A point x∈C is a fixed point of T:C⊂X→X provided Tx=x. Denote by F(T) the set of fixed points of T; that is, F(T)={x∈C:Tx=x}.
Let X be a normed linear space with dimX≥2. The modulus of smoothness of X is the function ρX:[0,+∞)→[0,+∞) defined by
(2)ρX(τ):=sup{∥x+y∥+∥x-y∥2-1:∥x∥=1,∥y∥=τ}.
The space X is said to be smooth if ρX(τ)>0,forallτ>0. It is well known that if X is smooth then J is single valued. A Banach space X is said to be strictly convex if ∥x∥=∥y∥=1, x≠y, implies ∥x+y∥/2<1.
Let C be a nonempty closed convex subset of a real Banach space X. Recall the following concepts.
Definition 1.
(i) A mapping f:C→C is a ρ-contraction if ρ∈[0,1) and if the following property is satisfied
(3)∥f(x)-f(y)∥≤ρ∥x-y∥,∀x,y∈C.
A mapping T:C→E is nonexpansive provided
(4)∥Tx-Ty∥≤∥x-y∥,∀x,y∈C.
A mapping S:C→X is
accretive if for any x,y∈C there exists j(x-y)∈J(x-y) such that
(5)〈Sx-Sy,j(x-y)〉≥0;
β-strongly accretive if for any x,y∈C there exists j(x-y)∈J(x-y) such that
(6)〈Sx-Sy,j(x-y)〉≥β∥x-y∥2,
for some real constant β>0.
Noting that if S:C→X is nonexpansive, then I-S is accretive; if f:C→C is a ρ-contraction, then I-f is (1-ρ)-strongly accretive. particulary, if X=H is a Hilbert space, then (strongly) accretive mappings become (strongly) monotone mappings.
Definition 2.
Let C and D be nonempty subsets of a Banach space X such that C is nonempty closed convex and D⊂C.
A mapping Q:C→D is called sunny, if Q(Qx+t(x-Qx))=Qx for each x∈C and t≥0 with Q(Qx+t(x-Qx))∈C.
A mapping Q:C→D is called a retraction from C to D if Q is continuous and F(Q)=D.
A subset D of C⊂E is said to be a sunny nonexpansive retract of C if there exists a sunny nonexpansive retraction Q of C onto D. For details, see [1–3].
Note that if X=H is a Hilbert space, Q becomes the projection on C, denoted by PC.
Let P:C→C a nonexpansive self-mapping on C and {Tn} be a countable family of nonexpansive nonself-mappings of C into X such that ℱ=⋂n=1∞F(Tn)≠∅. Then we consider the following problem: find hierarchically a common fixed point of the infinite family {Tn} with respect to a nonexpansive mapping P; namely, find x*∈ℱ, such that
(7)〈x*-Px*,J(x-x*)〉≥0,∀x∈ℱ.
Particularly,if {Tn} is a finite family of nonexpansive nonself-mappings, problem (7) has been studied by Ceng and Petruşel [4]. If X=H and {Tn} is an infinite family of nonexpansive self-mappings, Problem (7) reduces to the following problem: find hierarchically a common fixed point of {Tn} with respect to a nonexpansive mapping P, namely, find x*∈ℱ, such that
(8)〈x*-Px*,x-x*〉≥0,∀x∈ℱ,
which was studied by Zhang et al. [5]. If X=H is a Hilbert space and Tn=T,foralln≥1, where T is a nonexpansive mapping on C, then problem (7) reduces to the following problem: finding hierarchically a fixed point of T with respect to another nonexpansive mapping P; namely, find x*∈F(T) such that
(9)〈x*-Px*,x-x*〉≥0,∀x∈F(T).
Problem (7) includes many problems as special cases, so it is very important in the area of optimization and related fields, such as signal processing and image reconstruction (see [6–9]).
In 2007, Moudafi [10] introduced the following Krasnoselski-Mann's algorithm in Hilbert spaces:
(10)xn+1=(1-αn)xn+αn(σnPxn+(1-σn)Txn),∀n≥0,
where {αn} and {σn} are two real sequences in (0,1) and T and P are two nonexpansive mappings of C into itself. Furthermore, he established a weak convergence result for Algorithm (10) for solving problem (9).
Subsequently, Yao and Liou [11] derived a weak convergence result of algorithm (10) under the restrictions on parameters weaker than those in [10, Theorem 2.1].
Recently, Marino and Xu [12] introduced the following explicit hierarchical fixed point algorithm in Hilbert spaces:
(11)xn+1=λnf(xn)+(1-λn)(αnVxn+(1-αn)Txn),∀n≥0,
where f is a contraction on C and V,T are two nonexpansive mappings of C into itself and proved that the sequence {xn} generated by (11) converges strongly to a solution of problem (9).
Very recently, Zhang et al. [5] introduced the following iterative algorithm in order to find hierarchically a fixed point of Problem (8):
(12)x0∈C,xn+1=αnf(xn)+(1-αn)yn,yn=βnP(xn)+(1-βn)Txn,
where f:C→C is a contraction, P:C→C is a nonexpansive mapping, {Tn}:C→C is a countable family of nonexpansive mappings, and T:C→C is a mapping defined by
(13)T=∑n=1∞λnTn,λn≥0(n=1,2,…)with∑n=1∞λn=1.
Under suitable conditions on parameters {αn} and {βn}, they established some strong and weak convergence theorems. Note that, in [5], {Tn} is an infinite family of self-mappings and P is also a self-mapping. And they obtained the results in the setting of Hilbert spaces.
Motivated and inspired by the above researches, in a reflexive Banach space which admits a weakly sequentially continuous duality mapping J, we propose and analyze an iteration process for a countable family of nonexpansive nonself-mappings {Tn}:C→X and S:C→X is a nonexpansive nonself-mapping as follows:
(14)x0∈C,xn+1=Q(αnf(xn)+(1-αn)yn),yn=βnSxn+(1-βn)Txn,n≥0,
where Q is a sunny nonexpansive retraction of X onto C and establishes a convergence theorem. particularly, if X=H is a Hilbert space, we obtain some convergence results.
To prove the main results, we need the following lemmas.
Lemma 3 (see [1]).
Let C be a nonempty and convex subset of a smooth Banach space X, D⊂C, J:X→X* the normalized duality mapping of X, and Q:C→D a retraction. Then the following conditions are equivalent:
〈x-Qx,J(y-Qx)〉≤0, for all x∈C and y∈D;
Q is both sunny and nonexpansive.
Lemma 4 (see [13, Lemma 3.1, 3.3]).
Let X be a real smooth and strictly convex Banach space and C a nonempty closed and convex subset of X which is also a sunny nonexpansive retract of X. Assuming that T:C→X is a nonexpansive mapping and Q is a sunny nonexpansive retraction of X onto C, then F(T)=F(QT).
Lemma 5 (see [1]).
Let X be a real Banach space and J:X→2X* the normalized duality mapping. Then for any x,y∈X, the following hold:
∥x+y∥2≤∥x∥2+2〈y,j(x+y)〉, for allj(x+y)∈J(x+y);
∥x∥2+2〈y,j(x)〉≤∥x+y∥2, for all j(x)∈J(x).
Lemma 6 (see [14]).
Let {an} and {bn} be two sequences of nonnegative real numbers satisfying
(15)∑n=0∞bn<∞,an+1≤an+bn,n=0,1,2,….
Then limn→∞an exists.
Lemma 7 (see [15]).
Let {an} be a sequence of nonnegative real numbers satisfying
(16)an+1≤(1-λn)an+λnbn+cn,∀n≥0,
where {λn},{bn} and {cn} satisfy the following conditions:
If Banach space X admits sequentially continuous duality mapping J from weak topology to weak * topology, then by [16, Lemma 1] we get that duality mapping J is single-valued. In this case, duality mapping J is also said to be weakly sequentially continuous, that is, for each {xn}⊂X with xn⇀x, then J(xn)⇁Jx [16, 17].
Recall that a Banach space X is said to be satisfying Opial’s condition if for any sequence {xn} in E, xn⇀x(n→∞) implies that
(17)limsupn→∞∥xn-x∥<limsupn→∞∥xn-y∥,∀y∈E,withy≠x.
By [16, Lemma 1], we know that if X admits a weakly sequentially continuous duality mapping, then X satisfies Opial’s condition.
In the sequel, we also need the following lemmas.
Lemma 8 (see [17]).
Let C be a nonempty, closed and convex subset of a reflexive Banach space X which satisfies Opial’s condition and T:C→X a nonexpansive mapping. Then the mapping I-T is demiclosed at zero, that is,
(18)xn⇀xxn-Txn⟶0imply x=Tx.
Let C be a nonempty and convex subset of a Banach space X. Then for x∈C, one defines the inward set IC(x) as follows [2, 3]:
(19)IC(x)={y∈X:y=x+λ(z-x),z∈C,λ≥0}.
A mapping T:C→X is said to satisfy the inward condition if Tx∈IC(x) for all x∈C. T is also said to satisfy the weakly inward condition if for each x∈C, Tx∈IC(x)¯(IC(x)¯ is the closure of IC(x)). Clearly C⊂IC(x) and it is not hard to show that IC(x) is a convex set if C does.
Lemma 9 (see [18, Theorem 2.4]).
Let X be a reflexive Banach space which admits a weakly sequentially continuous duality mapping J from X to X*. Suppose C is a nonempty closed convex subset of X which is also a sunny nonexpansive retract of X, and T:C→X is a nonexpansive mapping satisfying the weakly inward condition and F(T)≠∅. Let {un} be defined by
(20)u0∈C,un+1=Q(αnf(un)+(1-αn)Tun),
where Q is a sunny nonexpansive retract of X onto C and αn∈(0,1) satisfy the following conditions:
αn→0,asn→∞;
∑n=0∞αn=∞;
either ∑n=0∞|αn+1-αn|<∞ or limn→∞(αn/αn+1)=1.
Then {xn} converges strongly to a fixed point p of T such that p is the unique solution in F(T) to the following variational inequality:
(21)〈(I-f)p,j(p-u)〉≤0,∀u∈F(T).
Remark 10.
If a Banach space X admits a sequentially continuous duality mapping J from weak topology to weak star topology, from Lemma 1 of [16] it follows that X is smooth. So for Lemma 9, if X is a reflexive and strictly convex Banach space which admits a weakly sequentially continuous duality mapping J, by Lemma 4, the weakly inward condition of T can be removed.
2. Main Results Theorem 11.
Let X be a reflexive and strictly convex Banach space which admits a weakly sequentially continuous duality mapping J:X→X* and C a nonempty, closed and convex subset of X which is also a sunny nonexpansive retract of X. Let S:C→X be a nonexpansive nonself-mapping, f:C→C a contractive mapping with a contractive constant ρ∈(0,1) and Ti:C→X(i={1,2,…}) an infinite family of nonexpansive nonself-mappings such that ⋂i=1∞F(Ti)≠∅. Let T:C→X be defined by (13) and Q a sunny nonexpansive retraction of X onto C. Let {xn} be the sequence generated by (14), and {αn} and {βn} the sequences in(0,1)satisfying the following conditions:
αn→0(n→∞),∑n=0∞αn=∞;
limn→∞(βn/αn)=0;
∑n=0∞|αn+1-αn|<∞.
Then {xn} converges strongly to some point x*∈F(T)=⋂i=1∞F(Ti), which is the unique solution to the following variational inequality:
(22)〈(I-f)x*,J(x-x*)〉≥0,∀x∈F(T).
Proof.
From condition (ii), without loss of generality, we can assume that βn≤αn,foralln≥0.
First we prove that the sequence {xn} is bounded.
In fact, for any u∈F(T), we have
(23)∥xn+1-u∥=∥Q(αnf(xn)+(1-αn)yn)-Qu∥≤αn∥f(xn)-u∥+(1-αn)∥βnSxn+(1-βn)Txn-u∥≤αn(ρ∥xn-u∥+∥f(u)-u∥)+(1-αn)(βn∥Sxn-u∥+(1-βn)∥Txn-u∥)≤(1-αn(1-ρ))∥xn-u∥+αn∥f(u)-u∥+(1-αn)βn∥Su-u∥≤(1-αn(1-ρ))∥xn-u∥+αn(∥f(u)-u∥+∥Su-u∥)≤max{∥xn-u∥,∥f(u)-u∥+∥Su-u∥1-ρ}.
By induction,
(24)∥xn+1-u∥≤max{∥x0-u∥,∥f(u)-u∥+∥Su-u∥1-ρ}.
Thus {xn} is bounded, so {Sxn} and {Txn} are also bounded.
Next we prove that ∥xn-un∥→0,asn→∞, where the sequence {un} is defined by
(25)u0=x0∈C,un+1=Q(αnf(un)+(1-αn)Tun).
By Lemma 9 and Remark 10, {un} converges strongly to some point x*∈F(T), which is the unique solution to the following variational inequality:
(26)〈(I-f)x*,j(x*-x)〉≤0,∀x∈F(T).
Furthermore, we obtain
(27)∥xn+1-un+1∥≤∥Q(αnf(xn)+(1-αn)yn)-Q(αnf(un)+(1-αn)Tun)∥≤∥αn(f(xn)-f(un))+(1-αn)(yn-Tun)∥≤αnρ∥xn-un∥+(1-αn)×(βn∥Sxn-Tun∥+(1-βn)∥Txn-Tun∥)≤(1-αn(1-ρ))∥xn-un∥+(1-αn)βnM≤(1-αn(1-ρ))∥xn-un∥+βnM,
where M=supn≥0∥Sxn-Tun∥. It follows from conditions (i)-(ii) and Lemma 7 we have limn→∞∥xn-un∥=0. Since as n→∞, un→x*∈F(T), we get xn→x*(n→∞), which is the unique solution to the variational inequality (22).
Remark 12.
Theorem 11 extends Theorem 2.1 in [5] from the following aspects: (i) from Hilbert spaces to reflexive and strictly convex Banach spaces which admits a weakly sequentially continuous duality mapping; (ii) for the infinite family of mappings {Ti} from self-mappings to nonself-mappings. In addition, the existence of the sunny nonexpansive retraction has been proved in [19, Theorem 3.10].
Remark 13.
If we take
(28)αn=1(1+n)α,βn=1(1+n)β,0<α<β<1,
then since |αn+1-αn|≈1/nα+1 and |βn+1-βn|≈1/nβ+1(as n→∞), it is not hard to find that the conditions (i)–(iii) are satisfied. For details, see [12, Remark 3.2].
In the sequel, we consider the result in the setting of Hilbert spaces.
Theorem 14.
Let H be a Hilbert space and C a nonempty, closed and convex subset of H. Let S:C→H be a nonexpansive nonself-mapping, f:C→C a contractive mapping with a contractive constant ρ∈(0,1), and Ti:C→H(i={1,2,…}) an infinite family of nonexpansive nonself-mappings such that F(T)=⋂i=1∞F(Ti)≠∅. Let {xn} be the sequence generated by (14) and {αn} and {βn} the sequences in(0,1)satisfying the following conditions:
αn→0,∑n=0∞αn=∞;
limn→∞(βn/αn)=τ∈(0,+∞);
limn→∞((|βn-βn-1|+|αn-αn-1|)/αnβn)=0;
there exists a constant K>0 such that 1/αn|(1/βn)-(1/βn-1)|≤K for all n>0.
Then {xn} converges strongly to some point x*∈F(T), which is the unique solution to the following variational inequality:
(29)〈1τ(I-f)x*+(I-S)x*,x-x*〉≥0,∀x∈F(T).
Proof.
By condition (ii), without loss of generality, we can assume that βn≤(τ+1)αn,foralln≥0. Similar to the proof of (24), for any u∈F(T), we have
(30)∥xn+1-u∥≤max{∥x0-u∥,(τ+1)(∥f(u)-u∥+∥Su-u∥)1-ρ}.
Thus {xn} is bounded. Furthermore, {f(xn)},{Txn},{yn},{Sxn} are all bounded. Put un=αnf(xn)+(1-αn)yn and M=supn≥0{∥f(xn)∥+∥yn∥,∥Txn∥+∥Sxn∥}. So {un} and {PC(un)} are also bounded.
Step 1. We prove that ∥xn+1-xn∥→0(n→∞).
From (14), we obtain
(31)∥xn+1-xn∥=∥PC(un)-PC(un-1)∥≤∥un-un-1∥≤αn∥f(xn)-f(xn-1)∥+(1-αn)∥yn-yn-1∥+|αn-αn-1|(∥f(xn-1)∥+∥yn-1∥)≤αnρ∥xn-xn-1∥+(1-αn)×∥yn-yn-1∥+|αn-αn-1|M,(32)∥yn-yn-1∥≤βn∥Sxn-Sxn-1∥+(1-βn)∥Txn-Txn-1∥+|βn-βn-1|(∥Sxn-1∥+∥Txn-1∥)≤∥xn-xn-1∥+|βn-βn-1|M.
Substituting (32) into (31), we have
(33)∥xn+1-xn∥≤(1-αn(1-ρ))∥xn-xn-1∥+αn(|αn-αn-1|+|βn-βn-1|)Mαn.
By conditions (i), (iii), and Lemma 7, we have ∥xn+1-xn∥→0(n→∞).
Step 2. We prove that ωw(xn)⊂F(T), where ωw(xn) is the ω-limit point set of {xn} in the weak topology:
(34)∥xn+1-QTxn∥≤αn∥f(xn)∥+βn∥Sxn∥+(αn+βn+αnβn)∥Txn∥.
Noting that αn→0 and βn→0, we have ∥xn+1-QTxn∥→0(n→∞). Then from Step 1 we have ∥xn-QTxn∥→0(n→∞). Furthermore, it follows from Lemmas 4 and 8 that ωw(xn)⊂F(QT)=F(T), where Q=PC.
Step 3. We show that ∥xn+1-xn∥/βn→0(n→∞).
It follows from (31) and (33) that
(35)∥xn+1-xn∥βn≤∥un-un-1∥βn≤(1-αn(1-ρ))∥xn-xn-1∥βn-1+(1-αn(1-ρ))∥xn-xn-1∥|1βn-1βn-1|+(|αn-αn-1|+|βn-βn-1|)Mβn≤(1-αn(1-ρ))∥xn-xn-1∥βn-1+αn∥xn-xn-1∥K+(|αn-αn-1|+|βn-βn-1|)Mαnβnαn.
By conditions (i) and (iii), ∥xn-xn-1∥→0(n→∞), and Lemma 7, we have
(36)∥xn+1-xn∥βn⟶0(n⟶∞).
Thus from (35), we get
(37)∥un-un-1∥βn⟶0(n⟶∞).
Step 4. We show that {xn} converges strongly to some point x′∈F(T), which is the unique solution of (29).
Setting Wn=βnS+(1-βn)T, we have
(38)xn+1=PC(un)-un+αnf(xn)+(1-αn)Wnxn.
Then
(39)xn-xn+1=un-PC(un)+αn(I-f)xn+(1-αn)(I-Wn)xn.
Letting vn=(xn-xn+1)/(1-αn)βn, from condition (i) and (36), we have vn→0(n→∞). Noting that I-Wn is monotone and I-f is (1-ρ)-strongly monotone, for any x*∈F(T), from Lemma 3 we obtain
(40)〈vn,xn-x*〉=1(1-αn)βn〈un-PC(un),xn-x*〉+αn(1-αn)βn〈(I-f)xn,xn-x*〉+1βn〈(I-Wn)xn,xn-x*〉=1(1-αn)βn〈un-PC(un),xn-x*〉+αn(1-αn)βn〈(I-f)xn,xn-x*〉+1βn〈(I-Wn)xn-(I-Wn)x*,xn-x*〉+1βn〈(I-Wn)x*,xn-x*〉=1(1-αn)βn〈un-PC(un),PC(un)-x*〉+1(1-αn)βn×〈un-PC(un),-(PC(un)-x*)+(PC(un-1)-x*)〉+αn(1-αn)βn〈(I-f)xn-(I-f)x*,xn-x*〉+αn(1-αn)βn〈(I-f)x*,xn-x*〉+1βn〈(I-Wn)xn-(I-Wn)x*,xn-x*〉+1βn〈(I-Wn)x*,xn-x*〉≥1(1-αn)βn〈un-PC(un),PC(un-1)-PC(un)〉+αn(1-ρ)(1-αn)βn∥xn-x*∥2+αn(1-αn)βn〈(I-f)x*,xn-x*〉+〈(I-S)x*,xn-x*〉.
Thus we have
(41)∥xn-x*∥2≤(1-αn)βnαn(1-ρ)〈vn,xn-x*〉-(1-αn)βnαn(1-ρ)〈(I-S)x*,xn-x*〉-1αn(1-ρ)〈un-PC(un),PC(un-1)-PC(un)〉-1(1-ρ)〈(I-f)x*,xn-x*〉≤(1-αn)βnαn(1-ρ)∥vn∥∥xn-x*∥-(1-αn)βnαn(1-ρ)〈(I-S)x*,xn-x*〉+1(1-ρ)∥un-PC(un)∥∥un-1-unαn∥-1(1-ρ)〈(I-f)x*,xn-x*〉.
Since βn≤(τ+1)αn, by (37) we have
(42)∥un-un-1∥αn⟶0(n⟶∞).
Combining condition (ii), vn→0(n→∞), (41), and (42), every weak cluster point of {xn} is also a strong cluster point. From (40), we obtain
(43)〈(I-f)xn,xn-x*〉=(1-αn)βnαn〈vn,xn-x*〉-1αn〈un-PC(un),xn-x*〉-(1-αn)αn〈(I-Wn)xn,xn-x*〉=(1-αn)βnαn〈vn,xn-x*〉-1αn〈un-PC(un),PC(un)-x*〉-1αn〈un-PC(un),PC(un-1)-PC(un)〉-(1-αn)αn×〈(I-Wn)xn-(I-Wn)x*,xn-x*〉-(1-αn)αn〈(I-Wn)x*,xn-x*〉≤(1-αn)βnαn∥vn∥∥xn-x*∥+1αn∥un-PC(un)∥PC(un-1)-PC(un)∥-(1-αn)βnαn〈(I-S)x*,xn-x*〉≤(1-αn)βnαn∥vn∥∥xn-x*∥+∥un-1-un∥αn∥un-PC(un)∥-(1-αn)βnαn〈(I-S)x*,xn-x*〉.
Note that the sequence {xn} is bounded; thus there exists a subsequence {xnj} converging to a point x′∈H. From Step 2, we have x′∈F(T). Then it follows from the above inequality, (42), and vn→0(n→∞) that
(44)〈(I-f)x′,x′-x*〉≤-τ〈(I-S)x*,x′-x*〉,∀x*∈F(T).
Replacing x* with x′+μ(x*-x′), where μ∈(0,1) and x*∈F(T), we have
(45)〈(I-f)x′,x′-x*〉≤-τ〈(I-S)(x′+μ(x*-x′)),x′-x*〉,∀x*∈F(T).
Letting μ→0, we have
(46)〈(I-f)x′,x′-x*〉≤-τ〈(I-S)x′,x′-x*〉,∀x*∈F(T).
If there exists another subsequence {xnj′} of {xn} converging to a point x′′∈H. From Step 2, we also have x′′∈F(T). Then from (46) we obtain
(47)〈(I-f)x′,x′-x′′〉≤-τ〈(I-S)x′,x′-x′′〉
and, via interchanging x′ and x′′,
(48)〈(I-f)x′′,x′′-x′〉≤-τ〈(I-S)x′′,x′′-x′〉.
Adding up these two inequalities yields
(49)(1-ρ)∥x′-x′′∥2≤〈(I-f)x′-(I-f)x′′,x′-x′′〉≤0,
which implies x′=x′′. Then {xn} converges strongly to x′∈F(T), which is the solution to the following variational inequality:
(50)〈1τ(I-f)x′+(I-S)x′,x-x′〉≥0,∀x∈F(T).
Since I-f is (1-ρ)-strongly monotone and I-S is monotone, it is easy to see that the above variational inequality has a unique solution.
Remark 15.
Theorem 14 extends Theorem 3.2 in [12] from the following aspects: (i) from a nonexpansive mapping T to an infinite family of nonexpansive mappings {Ti}; (ii) from self-mappings to nonself-mappings.
Acknowledgments
The author is extremely grateful to the referees for their useful suggestions that improved the content of the paper. Supported by the China Postdoctoral Science Foundation Funded Project (no. 2012M511928).
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