An Algorithm Based on Resolvant Operators for Solving Positively Semidefinite Variational Inequalities

Volume 2007, Article ID 76040,15pages doi:10.1155/2007/76040

Research Article

An Algorithm Based on Resolvant Operators for Solving

Positively Semidefinite Variational Inequalities

Juhe Sun, Shaowu Zhang, and Liwei Zhang

Received 16 June 2007; Accepted 19 September 2007 Recommended by Nan-Jing Huang

A new monotonicity,M-monotonicity, is introduced, and the resolvant operator of an M-monotone operator is proved to be single-valued and Lipschitz continuous. With the help of the resolvant operator, the positively semidefinite general variational inequality (VI) problem VI (Sn

+,F+G) is transformed into a fixed point problem of a nonexpan-sive mapping. And a proximal point algorithm is constructed to solve the fixed point problem, which is proved to have a global convergence under the condition thatFin the VI problem is strongly monotone and Lipschitz continuous. Furthermore, a convergent path Newton method is given for calculating-solutions to the sequence of fixed point problems, enabling the proximal point algorithm to be implementable.

Copyright © 2007 Juhe Sun et al. This is an open access article distributed under the Cre-ative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

1. Introduction

In recent years, the variational inequality has been addressed in a large variety of prob-lems arising in elasticity, structural analysis, economics, transportation equilibrium, op-timization, oceanography, and engineering sciences [1,2]. Inspired by its wide applica-tions, many researchers have studied the classical variational inequality and generalized it in various directions. Also, many computational methods for solving variational inequal-ities have been proposed (see [3–8] and the references therein). Among these methods, resolvant operator technique is an important one, which was studied in the 1990s by many researchers (such as [4,6,9]), and further studies developed recently [3,10,11].

proved to be Lipschitz-continuous. Applying the resolvant operator technique, we trans-form the positively semidefinite variational inequality (VI) problemVI(Sn+,F+G) into a fixed point problem of a nonexpansive mapping and suggest a proximal point algo-rithm to solve the fixed point problem. Under the condition thatFin theVIproblem is strongly monotone and Lipschitz-continuous, we prove that the algorithm has a global convergence. To ensure the proposed proximal point algorithm is implementable, we in-troduce a path Newton algorithm whose step size is calculated by Armijo rule.

In the next section, we recall some results and concepts that will be used in this paper. InSection 3, we introduce the definition of anM-monotone operator, and discuss prop-erties of this kind of operators, especially the Lipschitz continuity of the resolvant opera-tor of anM-monotone operator. InSection 4, we construct a proximal point algorithm, based on the results inSection 3, forVI(Sn+,F+G), and prove its global convergence. To ensure that the proposed proximal point algorithm inSection 4is implementable, we in-troduce a path Newton algorithm, inSection 5, in which the step size is calculated by Armijo rule.

2. Preliminaries

Throughout this paper, we assume thatSndenotes the space ofn×nsymmetric matrices andSn+denote the cone ofn×nsymmetric positive semidefinite matrices. ForA,B∈Sn, we define an inner productA,B =tr(AB) which induces the normA =A,A. Let 2Sn denote the family of all the nonempty subsets ofSn. We recall the following concepts, which will be used in the sequel.

Definition 2.1. LetA,B,C:Sn→Snbe single-valued operators and letM:Sn×Sn→Snbe mapping.

(i)M(A,·) is said to beα-strongly monotone with respect toAif there exists a con-stantα >0 satisfying

M(Ax,u)−M(Ay,u),x−y≥αx−y2_{, ∀x,y,u∈S}n_{; (2.1)}

(ii)M(·,B) is said to beβ-relaxed monotone with respect toBif there exists a constant β >0 satisfying

M(u,Bx)−M(u,By),x−y≥ −βx−y2_{, ∀x,y,u∈S}n_{; (2.2)}

(iii)M(·,·) is said to beαβ-symmetric monotone with respect toAandBifM(A,·) is α-strongly monotone with respect toA; andM(·,B) isβ-relaxed monotone with respect toBwithα≥βandα=βif and only ifx=y, for allx,y,u∈Sn_;

(iv)M(·,·) is said to beξ-Lipschitz-continuous with respect to the first argument if there exists a constantξ >0 satisfying

(iv)Ais said to bet-Lipschitz-continuous if there exists a constantt >0 satisfying

Ax−Ay ≤tx−y, ∀x,y∈Sn_{; (2.4)}

(vi)Bis said to bel-cocoercive if there exists a constantl >0 satisfying

Bx−By,x−y ≥lBx−By2_{, ∀x,y∈S}n_{; (2.5)}

(vii)Cis said to ber-strongly monotone with respect toM(A,B) if there exists a con-stantr >0 satisfying

Cx−Cy,M(Ax,Bx)−M(Ay,By)≥rx−y2_{, ∀x,y∈S}n_{. (2.6)}

In a similar way to (v), we can define the Lipschitz continuity of the mappingM with respect to the second argument.

Definition 2.2. LetA,B:Sn→Sn,M:Sn×Sn→Snbe mappings.Mis said to be coercive with respect toAandBif

lim

x→∞

M(Ax,Bx),x

x =+∞. (2.7)

Definition 2.3. LetA,B:Sn→Sn_,M:Sn_×Sn_→Sn_{be mappings.Mis said to be bounded}

with respect toAandBifM(A(P),B(P)) is bounded for every bounded subsetPofSn. M is said to be semicontinuous with respect toAandBif for any fixedx,y,z∈Sn_{, the}

functiont→ M(A(x+ty),B(x+ty)),zis continuous at 0+_. Definition 2.4. T:Sn→2Snis said to be monotone if

x−y,u−v ≥0, ∀u,v∈Sn,x∈Tu, y∈Tv; (2.8)

and it is said to be maximal monotone ifT is monotone and (I+cT)(Sn)=Sn _{for all}

c >0, whereIdenotes the identity mapping onSn.

3.M-Monotone operators

In this section, we introduceM-monotonicity of operators and discuss its properties. Definition 3.1. LetA,B:Sn→Sn_{be single-valued operators,M:S}n_×Sn_→Sn_{a mapping,}

andT:Sn→2Sna multivalue operator.Tis said to beM-monotone with respect toAand BifTis monotone and (M(A,B) +cT)(Sn)=Sn_{holds for everyc >0.}

Remark 3.2. IfM(A,B)=H, then the above definition reduces toH-monotonicity, which was studied in [5]. IfM(A,B)=I, then the definition ofI-monotonicity is just the maxi-mal monotonicity.

Remark 3.3. LetTbe a monotone operator and letcbe a positive constant. IfT:Sn→2Sn is anM-monotone operator with respect toAandB, every matrixz∈Sn_{can be written}

Proposition3.4. LetM beαβ-symmetric monotone with respect toAandBand letT: Sn_→2Sn

be anM-monotone operator with respect toAandB, thenTis maximal monotone.

Proof. SinceT is monotone, it is sufficient to prove the following property; inequality

x−y,u−v ≥0 for (v,y)∈Graph(T) implies that

x∈Tu. (3.1)

Suppose, by contradiction, that there exists some (u0,x0)∈Graph(T) such that

x0−y,u0−v

≥0, ∀(v,y)∈Graph(T). (3.2)

SinceTisM-monotone with respect toAandB, (M(A,B) +cT)(Sn)=Sn_{holds for every}

c >0, there exists (u1,x1)∈Graph(T) such that

MAu1,Bu1

+cx1=M

Au0,Bu0

+cx0∈Sn. (3.3)

It follows form (3.2) and (3.3) that 0≤cx0−x1,u0−u1

= −MAu0,Bu0

−MAu1,Bu1

,u0−u1

= −MAu0,Bu0

−MAu1,Bu0

,u0−u1

−MAu1,Bu0

−MAu1,Bu1

,u0−u1

≤ −(α−β)u0−u1

≤0,

(3.4)

which yieldsu1=u0. By (3.3), we have thatx1=x0. Hence (u0,x0)∈Graph(T), which is a contradiction. Therefore (3.1) holds andT is maximal monotone. This completes the

proof.

The following example shows that a maximal monotone operator may not beM -monotone for someAandB.

Example 3.5. LetSn=S2_{,T=I, andM(Ax,Bx)=x}2_{+ 2E−xfor allx∈S}2_{, whereEis an} identity matrix. Then it is easy to see thatIis maximal monotone. For allx∈S2_{, we have} that

_{M(A,B) +I}

(x)2=x2_{+ 2E−x+x}2_=x2_{+ 2E}2_=trx2_{+ 2E}2 _{≥8, (3.5)}

which means that 0 ¯∈(M(A,B) +I)(S2_{) andIis notM-monotone with respect toAand} B.

Proposition3.6. LetT:Sn→2Snbe a maximal monotone operator and letM:Sn×Sn_→

Sn_{be a bounded, coercive, semicontinuous, andαβ-symmetric monotone operator with}

re-spect toAandB. ThenTisM-monotone with respect toAandB.

toAandB, it follows from [9, Corollary 32.26] thatM(A,B) +cT is surjective, that is, (M(A,B) +cT)(Sn_)=Sn_{holds for everyc >0. Thus,Tis anM-monotone operator with}

respect toAandB. The proof is complete. Theorem3.7. LetMbe anαβ-symmetric monotone with respect toAandBand letT be anM-monotone operator with respect toAandB. Then the operator(M(A,B) +cT)−1_is single-valued.

Proof. For any givenu∈Sn_{, letx,y∈(M(A,B) +cT)}−1_{(u). It follows that−M(Ax,Bx) +} u∈Txand−M(Ay,By) +u∈T y. The monotonicity ofTandMimplies that

0≤−M(Ax,Bx) +u−−M(Ay,By) +u,x−y

= −M(Ay,By)−M(Ax,Bx),x−y

≤ −(α−β)u0−u1

≤0.

(3.6)

From the symmetric monotonicity ofM, we get thatx=y. Thus (M(A,B) +cT)−1 _is single-valued. This completes the proof. Definition 3.8. LetMbe anαβ-symmetric monotone with respect toAandBand letTbe anM-monotone operator with respect toAandB. The resolvant operatorJ_cTM:Sn→Sn_is

defined by

JM

cT(u)=M(A,B) +cT−1(u), ∀u∈Sn. (3.7)

Theorem3.9. LetM(A,B)beα-strongly monotone with respect toAandβ-relaxed mono-tone with respect toBwithα > β. Suppose thatT:Sn→2Sn is anM-monotone operator. Then the resolvant operatorJ_cTM :Sn→Snis Lipschitz-continuous with constant1/(α−β), that is,

_JM

cT(u)−JcTM(v)≤_α−1_βu−v, ∀u,v∈Sn. (3.8)

Since the proof ofTheorem 3.9is similar as that of [5, Theorem 2.2], we here omit it.

4. An algorithm for variational inequalities

Let F,G:Sn+→Sn be operators. Consider the general variational inequality problem VI(Sn+,F+G), defined by findingu∈Sn+such that

F(u) +G(u),v−u≥0, ∀v∈Sn+. (4.1)

We can rewrite it as the problem of findingu∈Sn

+such that

0∈G(u) +T(u), (4.2)

whereT≡F+ᏺ(·;Sn

Proposition4.1. LetF,G:Sn+→Snbe continuous and letM:Sn×Sn→Snbe a bounded, coercive, semicontinuous, andαβ-symmetric monotone operator with respect toA:Sn→Sn

andB:Sn→Sn. Then the following two properties hold for the mapT≡F+ᏺ(·;Sn+): (a)J_cTM(M(Ax,Bx)−cG(x))=Sol(Sn+,Fcx), whereFcx(y)=M(Ay,By)−M(Ax,Bx) +

c(F(y) +G(x));

(b)IfFis monotone, thenTisM-monotone with respect toAandB. Proof. We have that the inclusion

y∈JM

cTM(Ax,Bx)−cG(x)=M(A,B) +cT−1M(Ax,Bx)−cG(x) (4.3)

is equivalent to

M(Ax,Bx)∈M(A,B) +cF+cᏺ·;Sn+

(y) +cG(x), (4.4)

or in other words,

0∈M(Ay,By)−M(Ax,Bx) +cF(y) +G(x)+ᏺy;Sn+

. (4.5)

This establishes (a).

By [10, Proposition 12.3.6], we can deduce thatT is maximal monotone, it follows fromProposition 3.6, we get thatT isM-monotone with respect toAandB. This

com-pletes the proof.

Lemma4.2. LetMbe anαβ-symmetric monotone with respect toAandBand letTbe an M-monotone operator with respect toAandB. Thenu∈Sn

+is a solution of0∈G(u) +T(u) if and only if

u=JM

cTM(Au,Bu)−cG(u), (4.6)

whereJ_cTM=(M(A,B) +cT)−1_{andc >0is a constant.}

In order to obtain our results, we need the following assumption.

Assumption4.3. The mappingsF,G,M,A,Bsatisfy the following conditions. (1)FisL-Lipschitz-continuous andm-strongly monotone.

(2)M(A,·)isα-strongly monotone with respect toA; andM(·,B)isβ-relaxed monotone with respect toBwithα > β.

(3)M(·,·)isξ-Lipschitz-continuous with respect to the first argument andζ -Lipschitz-continuous with respect to the second argument.

(4)Aisτ-Lipschitz-continuous andBist-Lipschitz-continuous.

(5)Gisγ-Lipschitz-continuous ands-strongly monotone with respect toM(A,B). Remark 4.4. LetAssumption 4.3hold and

c−_γs₂≤

s2₋γ2₍ξτ₊ζt₎2₋₍α₋β₎2

γ2 , s

We can deduce that

_JM

cTM(Ax,Bx)−cG(x)−JcTMM(Ay,By)−cG(y) ≤_α1

−βM(Ax,Bx)−M(Ay,By)−c

G(x)−G(y)

≤

(ξτ+ζt)2₋₂cs₊c2γ2

α−β x−y

≤ x−y,

(4.8)

which implies thatJ_cTM(M(A,B)−cG) is nonexpansive. Then, it is natural to consider the recursion

xk+1_≡JM

cTMAxk,Bxk−cGxk, (4.9)

which is desired to converge to a zero ofG+T. Actually, this can be proved to be true. However, based onLemma 4.2, we construct the following proximal point algorithm for VI(Sn+,F+G).

Algorithm4.5 Data. x0_∈Sn_,c

0>0,ε0≥0, andρ0>0. Step 1. Setk=0.

Step 2. Ifxk∈Sol(Sn+,F+G), stop. Step 3. Findwksuch thatwk_−JM

ckT(M(Axk,Bxk)−ckG(xk)) ≤εk. Step 4. Setxk+1_≡(1−ρ

k)xk+ρkwkand selectck+1,εk+1andρk+1. Setk←k+ 1and go to Step 1.

The following theorem fully describes the convergence ofAlgorithm 4.5for finding a solution toVI(Sn+,F+G).

Theorem 4.6. Suppose thatAlgorithm 4.5 holds. LetM be bounded, coercive, semicon-tinuous, and αβ-symmetric monotone with respect to A and B; and let F be monotone and Lipschitz-continuous. Letx0_∈Sn_{be given, let{εk} ⊂[0,∞)satisfyE≡}∞

k=1εk<∞,

{ck} ⊂(cm,∞), wherecm>0and

ck−_γs2

<

s2₋γ2(ξτ₊ζt)2₋(α₋β)2

γ2 , s

2_{> γ}2_(ξτ+ζt)2_−(α−β)2 _{, (4.10)}

which implies that

L=

α−β−(ξτ+ζt)2₋2c

ks+ck2γ2

α−β+ 3(ξτ+ζt)2₋2c

ks+ck2γ2

2 >0. (4.11)

If{ρk} ⊆[Rm,RM], where0< Rm≤RM≤pL, for all p∈[2, +∞), then the sequence{xk}

Proof. We introduce a new map

Qk_≡I−JM ckT

M(A,B)−ckG. (4.12)

Clearly, any zero ofG+F+ᏺ(·;Sn+), being a fixed point ofJcMkT(M(A,B)−ckG), is also a zero ofQk. Now, let us prove thatQkisL-cocoercive.

Forx,y∈Sn_{we know that}

Qk_(x)−Qk_(y),x−y =x−y−JM

ckT

M(Ax,Bx)−ckG(x)−JcMkT

M(Ay,By)−ckG(y),x−y = x−y2_−JM

ckT

M(Ax,Bx)−JM ckT

M(Ay,By)−ckG(x)−G(y),x−y

≥ x−y2₋ 1

α−βM(Ax,Bx)−M(Ay,By)−ck

G(x)−G(y)x−y

≥ x−y2₋ 1 α−β

(ξτ+ζt)2₋2c

ks+ck2γ2x−y2

=

1−

(ξτ+ζt)2₋2c_ks₊c2

kγ2

α−β

x−y2_,

(4.13)

_Qk_(x)−Qk_(y)2

=x−y−JM ckT

M(Ax,Bx)−ckG(x)−JcMkT

M(Ay,By)−ckG(y)2 = x−y2_−2x−y,JM

ckT

M(Ax,Bx)−ckG(x)−JcMkT

M(Ay,By)−ckG(y)

+J_cM_kTM(Ax,Bx)−ckG(x)−JcMkT

M(Ay,By)−ckG(y)2

≤ x−y2_{+ 2}

(ξτ+ζt)2₋₂c_ks₊c2

kγ2

α−β x−y2

+

(ξτ+ζt)2₋2c_ks₊c2

kγ2

α−β x−y2

=

⎛ ⎝_{1 + 3}

(ξτ+ζt)2₋2c

ks+ck2γ2

α−β

⎞

⎠_x−y2_.

(4.14)

Inequalities (4.13) and (4.14) imply that

Qk_(x)−Qk_(y),x−y

≥

⎡ ⎣₁₋

(ξτ+ζt)2₋2c

ks+c2kγ2

α−β

⎤ ⎦ ⎡ ⎣₁₊₃

(ξτ+ζt)2₋2c

ks+c2kγ2

α−β

⎤ ⎦

−1

_Qk_(x)−Qk_(y)2

=LQk_(x)−Qk_(y)2 .

For allk, we denote byxkthe point computed exactly by the resolvent. That is,

xk+1_≡1−ρ

kxk+ρkJcMkT

MAxk_,Bxk_−c

kGxk. (4.16)

For every zerox∗ofT, we obtain

_xk+1_−x∗2

=xk_−ρ

kQkxk−x∗2 =xk_−x∗2

−2ρkQkxk−Qkx∗,xk−x∗+ρk2Qkxk2 ≤xk_−x∗2

−2ρkLQkxk2+ρ2kQkxk2 ≤xk_−x∗2

−ρk2L−ρkQkxk2 ≤xk_−x∗2

−Rm2L−RMQkxk2 ≤xk_−x∗2

.

(4.17)

Sincexk−xk ≤ρkεk, we get that

_xk+1_−x∗_≤xk+1_−x∗_+xk+1_−xk+1

≤xk_−x∗_+ρ

kεk

≤x0_−x∗₊k

i=0 ρiεi

≤x0_−x∗_+pLE.

(4.18)

Therefore, the sequence{xk_{}is bounded. On the other hand, we have that}

_xk+1_−x∗2

=xk+1_−x∗_+xk+1_−xk+12

=xk+1_−x∗2

+ 2xk+1−x∗_,xk+1_−xk+1_+xk+1_−xk+12

≤xk+1_−x∗2

+ 2xk+1−x∗xk+1−xk+1+xk+1−xk+12

≤xk_−x∗2

+ 2ρkεkx0−x∗+pLE+ρ2kε2k −Rm2L−RMQkxk2.

(4.19)

LettingE2=

_∞

i=0εk2<∞, we have for everyk,

_xk+1_−x∗2

≤x0_−x∗2

+ 2p_LEx0_−x∗_+pLE

+p2L2E2−Rm2L−RM k

i=0

_Qk_xk2

. (4.20)

Passing to the limitk→ ∞, one has that∞_i=0Qk(xk)2<∞, implying that

lim

k→∞Q

According to Remark 3.3, for every k, there exists a unique pair (yk,vk) in gphT such thatzk=M(Axk_,Bxk_)−c

kG(xk)=M(Ayk,Byk) +ckvk. ThenJcMkT(M(Axk,Bxk)− ckG(xk))=yk. So thatQk(xk)→0 implies that (xk−yk)→0,vk→0.

Sinceckis bounded away from zero, it follows thatck−1Qk(xk)→0. Sincexkis bounded,

it has at least a limit point. Letx∞ be such a limit point and assume that the subse-quence{xki_:k_i∈k_{}converges to}x∞_{. It follows that{}yki_:k_i∈k_{}also converges to}x∞_. For every (y,v) in gphTby the monotonicity ofT, we have thaty−yk_,v−vk _≥0.

Let-tingki(∈k)→ ∞, we get thaty−y∞,v−vk ≥0. We see thatTisM-monotone due to Proposition 4.1, this implies that (x∞,−G(x∞_{))∈gphT, that is,−G(x}∞_)∈T(x∞_{). This}

completes the proof.

5. Solving an approximate fixed point toJ_cM_kT

How to calculatewkat Step 3 is the key inAlgorithm 4.5. Ifεk=0, this amounts to the

exact solution ofVI(Sn+,Fk), where

Fk(x)=M(Ax,Bx)−MAxk,Bxk+ckF(x) +G(xk. (5.1)

Now, we consider the case ofεk>0. We can prove thatJ_cM_kT(M(Axk,Bxk)−ckG(xk)) is the unique solution of theVI(Sn+,Fk). Hence,wk is an inexact solution of theVI(Sn+,Fk) satisfying dist(wk_{, Sol(S}n

+,Fk))≤εk.

Lemma5.1. LetF,G,M,A,Bsatisfy all the conditions ofAssumption 4.3. Then a constant c(k)>0exists such that

distwk, SolSn+,Fk

≤c(k)FknatSn +

wk_{. (5.2)}

Proof. ByAssumption 4.3, we can easily get that Fk isL(k)-Lipschitz-continuous and η(k)-strongly monotone, whereL(k)=ξτ+ζt+ckLandη(k)=α−β+ckm, that is,

_{Fk(x)−Fk(y)≤L(k)x−y,}

Fk(x)−Fk(y),x−y≥η(k)x−y2, ∀x,y∈Sn+.

(5.3)

Letr=(Fk)natSn₊(wk), where (F_k)nat_Sn₊ is the natural map associated with theVI(Sn₊,F_k). We have thatwk−r=ΠSn₊₍wk−F_k(wk_{)), that is,}

y−wk_+r,F

kwk−r≥0, ∀y∈Sn+. (5.4) For allx∗∈Sol(Sn

+,Fk) andwk−r∈Sn+, we also have that

wk_−r−x∗_,F

kx∗≥0. (5.5)

From (5.4) and (5.5), we get that

x∗_−wk_,F

kwk−r+r,Fkwk−r

=x∗_−wk_,Fkwk_−x∗_−wk_,r+r,Fkwk_+r2

≥0, (5.6)

Adding (5.6) and (5.7), we deduce that

η(k)wk_−x∗2

≤x∗_−wk_,F

kwk−Fkx∗ ≤ −wk_−x∗_{,r− r}2_+r,F

kx∗−Fkwk ≤ rwk_−x∗_+rL(k)wk_−x∗

=1 +L(k)wk−x∗r.

(5.8)

Hence,wk_−x∗ _≤η(k)−1_{(1 +L(k))r. This implies that}

distwk, SolSn+,Fk

≤c(k)FknatSn₊wk, (5.9)

wherec(k)=η(k)−1_{(1 +L(k)). This completes the proof.}

Consequently, the computation ofwk can be accomplished by obtaining an inexact solution ofVI(Sn+,Fk) satisfying the residual condition

_Fknat

Sn +

wk_≤ 1

c(k)εk. (5.10)

We note that the operatorSn

+(·) is directionally differentiable and strongly semismooth everywhere (see, e.g., [12]). IfFk(·) is continuously differentiable, then we get that

FknatSn₊(w)=w−

Sn₊

w−Fk(w) (5.11)

is directionally differentiable.

In what follows, we present the following path Newton method for solving the equa-tion (Fk)natSn

+(w

k_)=0.

Algorithm5.2

Data. w0_∈Sn_{,γ∈(0, 1), andρ∈(0, 1).}

Step 1. Setj=0. Step 2. If(Fk)natSn

+(w

j_{)=0, stop.}

Step 3. Select an elementVj∈∂[(Fk)natS₊n(wj)]and consider the corresponding pathpj(·)= wj_−(·)V−1

j (Fk)natSn₊(wj)with domainI_j=[0, ¯τ_j)for someτ¯_j∈(0, 1]. Find the smallest non-negative integerijsuch that withi=ij,ρiτ¯j∈Ijand

_Fknat

Sn +

pj_ρi_τ¯

j≤1−γρiτ¯jFknatSn +

wj_{. (5.12)}

Step 4. Setτj=ρijτ¯j,wj+1=pj(τj), and j←j+ 1; go to Step 2.

Theorem5.3. Let F,G,M,A, andBsatisfy all the conditions of Assumption 4.3. If for allw∈Sn

+every matrix in∂[(Fk)natS₊n(w)]is nonsingular, then the sequence{wj}generated byAlgorithm 5.2has at least one accumulation point and every accumulation point of the sequence{wj_{}is the zero point of(F}

Proof. The nonnegative sequence{(Fk)natSn₊(wj)}is monotonically decreasing; thus it is bounded. It follows fromLemma 5.1that the sequence{wj_{}is bounded. That implies}

that{wj_{}has at least one accumulation point.}

Assume that there exists a subsequence{wjm}_of{wj}_{converging to}w∗_{such that}

Fknat_Sn +

w∗_{=0, (5.13)}

that is, there exist positive constantsδandηsatisfying

_Fknat

Sn₊wjm≥η, ∀wjm∈Bw∗,δ. (5.14) By the strong semismoothness of (Fk)natSn₊, we have that

FknatSn₊(w+h)−F_knat_Sn₊(w)−V(h)=Oh2, ∀w∈Bw∗,δ,h∈Sn₊, (5.15)

whereV∈∂(Fk)natSn

+(w) is nonsingular and there is a positive constantcsuch that sup

w∈B(w∗_,δ) V∈∂(Fk)nat

Sn+(w)

maxV,V−1≤c. (5.16)

Lettingτ∈(0, ¯τjm),h=pjm(τ)−wjm, andVjm∈∂[(Fk)natSn₊(wjm)], we have that Vjm

pjm(τ)₋wjm₌F

knatSn +

pjm(τ)₋F

knatSn +

wjm₋Opjm(τ)₋wjm2. _(5.17)

It follows that

_Fknat

Sn₊pjm(τ)−V_jmpjm(τ)−wjm−F_knat_Sn₊wjm

_pjm(τ)−wjm =

Opjm(τ)₋wjm2

_pjm(τ)−wjm . (5.18) So we can choose a positivet∗small enough so that

o(t)

t ≤

1−γ

c , ∀t∈

0,t∗ . (5.19)

From the definition ofpjm_{, we know that there exists a constant}τ∗_{∈(0, ¯}τ_j

m] small enough so thatpjm(τ)₋wjm _≤t_{∗, for all}τ_∈(0,τ∗_{], which implies that}

opjm₍τ₎₋wjm

_pjm₍τ₎−wjm ≤ 1−γ

c , ∀τ∈

0,τ∗ . (5.20)

It follows from (5.16), (5.18), (5.19), and (5.20) that

_Fknat

Sn +

pjm(τ)_≤V_j m

pjm(τ)₋wjm₊F_knat

Sn +

wjm

+pjm(τ)₋wjmopjm(τ)−wjm pjm(τ)−wjm

≤(1−τ)FknatSn +

wjm₊τV−1

jmFk

nat

Sn +

wjm1−γ

c

≤(1−τγ)FknatSn +

wjm_{, ∀}τ_∈(0,τ∗_].

Consequently,

_Fknat

Sn +

pjm(τ)_≤(1−τγ)F_knat

Sn +

wjm_{, ∀}τ_∈(0,τ∗]. _(5.22)

By the definition of the step-sizeτjm, it follows that there existsξ∈(0,τ∗) such thatτjm≥ ξfor alljm. Indeed, if no suchξexists, then{τjm}converges to zero. This implies that the sequence of integers{ijm}is unbounded. Consequently, by the definition ofijm, we have, for alljmsufficiently large,

_Fknat

Sn₊pkρijm−1τ¯_k>1−γρijm−1τ¯_kF_knat_Sn₊wk; (5.23) but this contradicts (5.22) withτ≡ρijm−1_τ¯

jm. Consequently, the desiredξexists. The in-equality (5.22) implies that

_Fknat

Sn +

wjm+1_≤1−τ_j mγFk

nat

Sn +

wjm. _(5.24)

Passing to the limitm→ ∞, we deduce a contradiction because limm→∞(Fk)natSn₊(wjm) ≥ η >0 and the sequence{τjm}is bounded away from zero. This yields that (Fk)natSn₊(w∗)=0.

This completes the proof.

Remark 5.4. As stated above,Algorithm 5.2generates a sequence converging to the zero point of (Fk)nat_Sn

+, Step 3 inAlgorithm 4.5is implementable. Obviously,Algorithm 5.2stops within a finite number of iterations at awksuch that (5.10) holds.

Example 5.5. Assume that there exists a positive constant ¯csuch that

sup

w∈Sn +

sup

V∈∂Sn+(xk−ck(F(w)+G(xk)))

_{VJF(w)≤c.¯ (5.25)}

Letck∈(0, 1/c¯). Suppose thatM(Aw,Bw)=w, for allw∈Sn+andFis Lipschitz-continu-ous and strongly monotone. We have (Fk)natSn

+(w)=w−

Sn

+(xk−ck(F(w) +G(xk))), for all w∈Sn

+. Then ∂[(Fk)natSn

+(w)]⊂ {I−ckVJF(w)|V∈∂

Sn +(x

k_{−ck(F(w) +G(x}k_)))},

for all w∈Sn

+. We easily get that every matrix in ∂[(Fk)natSn

+(w)] is nonsingular for all w∈Sn

+. It follows fromTheorem 5.3that every accumulation point of{wk}generated byAlgorithm 5.2is the zero point of (Fk)natSn₊.

At first sight, the M-monotonicity of T=F+ᏺ(·,Sn

+) seems having little use be-cause the algorithm based on maximal monotonicity can also solve the VI(Sn+,F+G) directly. However, we will see that in some practical cases the variational inequality us-ingAlgorithm 4.5, which is based onM-monotone operator, is actually much simpler to solve and easier to analyze than using algorithm based on maximal monotone map. We illustrate this by the following example.

Example 5.6. LetF:Sn+→Snbe defined by

F(x)=S(x) + 1

16x, G(x)= 1

8x ∀x∈S

n

We haveF is (s+ (1/16))-Lipschitz-continuous, (1/16)-strongly monotone, andG is (1/8)-Lipschitz-continuous.

Now, we takeM(Ax,Bx)=Ax+Bx, whereAx=(1 + (ck/16))x andBx= −ckS(x)−

(ck/8)xfor allx∈Snand 0< ck<16. Then, we can easily prove thatM(·,·) is

Lipschitz-continuous with first and second arguments,M(A,B) is bounded and semicontinuous; andAandBare both Lipschitz-continuous. It is also easy to see that

lim

x→∞

M(Ax,Bx),x

x =xlim→∞

−ckS(x) +1−ck/16x,x

x =+∞, (5.27)

which implies thatM(A,B) is coercive. Also, we can deduce thatM(A,B) is (1 + (ck

/16))-strongly monotone with respect toAandck(s+ (1/8))-relaxed monotone with respect to

Band (1 + (ck/16))> ck(s+ (1/8)), if we lets <(1/ck)−(1/16). Also, we can prove thatG

is strongly monotone with respect toM(A,B). We choosex0_∈Sn

+,{εk},{ck}, and{ρk}satisfyingTheorem 4.6and compute{wk}by the residual rule

_Fknat

S+

wk_=wk₋ Sn

+

wk_−MAwk_,Bwk_+MAx

k,Bxk−ckFwk+Gxk

=wk₋ Sn₊ −ckS

xk+ 1−3₁₆ck

!

xk!

≤ η(k)

1 +L(k)εk,

(5.28)

that is,wkcan be computed as follows:

wk₋

Sn₊ −ckS(xk

+ 1−3ck 16

!

xk!≤_{1 +}η(k)_L(k)εk. (5.29)

It follows fromTheorem 4.6that the sequence{xk}generated byAlgorithm 4.5converges to a solution of

"

S(x) + 1 16x+

1 8x,y−x

#

≥0, ∀y∈Sn

+. (5.30)

Note that the core of proximal point algorithm is the calculation ofwk. As we have seen, if we use [10, Algorithm 12.3.8], which is based on the maximal monotonicity ofT=

F+ᏺ(·,Sn+),wkwill be computed as

wk₋

Sn +

xk−ckSwk−c₈kxk!≤_{1 +}η(_Lk)_(k)εk, (5.31)

Acknowledgments

This work was supported by the National Natural Science Foundation of China under project Grant no. 10771026 and by the Scientific Research Foundation for the Returned Overseas Chinese Scholars, State Education Ministry.

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Juhe Sun: Department of Applied Mathematics, Dalian University of Technology, Dalian, Liaoning 116024, China

Email address:[email protected]

Shaowu Zhang: Department of Applied Mathematics, Dalian University of Technology, Dalian, Liaoning 116024, China

Email address:[email protected]

Liwei Zhang: Department of Applied Mathematics, Dalian University of Technology, Dalian, Liaoning 116024, China