DOI 10.1007/s10589-009-9315-9
A new class of penalized NCP-functions and its properties
J.-S. Chen·Z.-H. Huang·C.-Y. She
Received: 16 April 2009 / Published online: 23 January 2010
© Springer Science+Business Media, LLC 2010
Abstract In this paper, we consider a class of penalized NCP-functions, which in- cludes several existing well-known NCP-functions as special cases. The merit func- tion induced by this class of NCP-functions is shown to have bounded level sets and provide error bounds under mild conditions. A derivative free algorithm is also pro- posed, its global convergence is proved and numerical performance compared with those based on some existing NCP-functions is reported.
Keywords NCP-function·Penalized·Bounded level sets·Error bounds
1 Introduction
The nonlinear complementarity problem (NCP) is to find a pointx∈Rnsuch that
x≥0, F (x)≥0, x, F (x) =0, (1)
where·,·is the Euclidean inner product andF =(F1, . . . , Fn)T is a map fromRn toRn. We assume thatF is continuously differentiable throughout this paper. The
J.-S. Chen member of Mathematics Division, National Center for Theoretical Sciences, Taipei Office. The author’s work is partially supported by National Science Council of Taiwan.
Z.-H. Huang’s work is partly supported by the National Natural Science Foundation of China (Grant No. 10871144).
J.-S. Chen (
)·C.-Y. SheDepartment of Mathematics, National Taiwan Normal University, Taipei 11677, Taiwan e-mail:jschen@math.ntnu.edu.tw
C.-Y. She
e-mail:bbgmmshe@hotmail.com Z.-H. Huang
Department of Mathematics, Tianjin University, Tianjin 300072, China e-mail:huangzhenghai@tju.edu.cn
NCP has attracted much attention because of its wide applications in the fields of economics, engineering, and operations research [6,14].
Many methods have been proposed to solve the NCP; see [3,4,14,16–18,20,23, 26,29,30]. For more details, please refers to the excellent monograph [9]. One of the most powerful and popular methods is to reformulate the NCP as a system of nonlin- ear equations [24,25,30], or as an unconstrained minimization problem [7,10–12, 19,21,27,29]. The objective function that can constitute an equivalent unconstrained minimization problem is called a merit function, whose global minima are coincident with the solutions of the original NCP (1). To construct a merit function, a class of functions called NCP-functions and defined below, plays a significant role.
Definition 1.1 A functionφ:R2→Ris called an NCP-function if it satisfies φ (a, b)=0 ⇐⇒ a≥0, b≥0, ab=0. (2) Many NCP-functions have been proposed in the literature. Among them, the Fischer-Burmeister (FB) function is one of the most popular NCP-functions, which is defined by
φFB(a, b)=
a2+b2−(a+b), ∀(a, b)∈R2. (3) Through this NCP-functionφFB, the NCP (1) can be reformulated as a system of nonsmooth equations:
FB(x):=
⎛
⎝
φFB(x1, F1(x)) ... φFB(xn, Fn(x))
⎞
⎠=0. (4)
Thus, the functionFB:Rn→R+defined as below is a merit function for the NCP:
FB(x):=1
2FB(x)2= n
i=1
ψFB(xi, Fi(x)), (5)
whereψFB:R2→R+is the square ofφFB, i.e., ψFB(a, b)=1
2a2+b2−(a+b)2. (6)
Consequently, the NCP is equivalent to the unconstrained minimization problem:
xmin∈RnFB(x). (7)
There are several generalizations of the FB function in the literature. For example, Kanzow and Kleinmichel [22] extendφFBfunction to
φθ(a, b):=
(a−b)2+θ ab−(a+b), θ∈(0,4).
Chen, Chen, and Kanzow [2] study a penalized FB function
φλ(a, b):=λφFB(a, b)+(1−λ)a+b+, λ∈(0,1).
Some other types of penalized FB functions are also investigated by Sun and Qi in [28]. Recently, Chen and Pan [3,4] consider the following generalization of the FB function:
φp(a, b):= (a, b)p−(a+b), (8) where p > 1 and (a, b)p denotes the p-norm of (a, b), i.e., (a, b)p =
√p
|a|p+ |b|p. Another further generalization is proposed by Hu, Huang and Chen in [15]:
φθ,p(a, b):=p
θ (|a|p+ |b|p)+(1−θ )(|a−b|p)−(a+b), (9) wherep >1,θ∈(0,1].
All the aforementioned functions are NCP-functions. The corresponding function ψθ,ψλ,ψp, andψθ,p is square ofφθ,φλ,φp, andφθ,p, respectively, and naturally induces a merit functionθ,λ,p, andθ,plike whatψFBfunction does. Along this track, in this paper, we study the following merit functionα,θ,p:Rn→R+for the NCP:
α,θ,p(x):=
n i=1
ψα,θ,p(xi, Fi(x)), (10)
whereψα,θ,p:R2→R+is an NCP-function defined by ψα,θ,p(a, b):=α
2(max{0, ab})2+ψθ,p(a, b) (11) withα≥0 being a real parameter. Note thatψα,θ,pincludes all the above functions ψFB,ψp,ψθ,ψθ,p(andψ7in [28]) as special cases. Althoughψα,θ,pis obtained by penalizing the functionψθ,pconsidered in [15], more favorable properties ofψα,θ,p are explored in this work. In particular,α,θ,phas property of bounded level sets and provides a global error bound for the NCP under mild condition which were not stud- ied in [15]. Thus, this paper can be viewed as a follow-up of [15]. On the other hand, as remarked in [2], penalized Fischer-Burmeister (FB) function not only possesses stronger properties than FB function but also gives extremely promising numerical performance, which is another motivation of our considering this generalization of several NCP-functions.
This paper is organized as follows. In Sect. 2, we review some definitions and preliminary results to be used in the subsequent analysis. In Sect.3, we show some properties of the proposed merit function. In Sect.4, we propose a derivative free al- gorithm based on this merit functionα,θ,p, show its global convergence, and report some numerical results. In Sect.5, we make concluding remarks.
Throughout this paper,Rn denotes the space ofn-dimensional real column vec- tors andT denotes transpose. For every differentiable functionf :Rn→R,∇f (x) denotes the gradient off atx. For every differentiable mappingF =(F1, . . . , Fn)T :
Rn→Rn,∇F (x)=(∇F1(x) . . .∇Fn(x))denotes the transpose Jacobian ofF atx.
We usexpto denote thep-norm ofxand denotexthe Euclidean norm ofx. The level set of a function:Rn→Ris denoted byL(, c):= {x∈Rn|(x)≤c}. In addition, we will frequently mention two merit functions. One is the natural residual merit functionNR:Rn→R+defined by
NR(x):=1 2
n i=1
φ2NR(xi, Fi(x)), (12) whereφNR:R2→Rdenotes the minimum NCP-function min{a, b}. Another one is θ,p:Rn→R+induced byψθ,p:
θ,p(x):=1 2
n i=1
φθ,p2 (xi, Fi(x)). (13) Unless otherwise stated, in the sequel, we always suppose thatpis a fixed real num- ber in(1,∞).
2 Preliminaries
This section briefly recalls some concepts about the mappingF that will be used later. A matrix is said to beP-matrix if each of its principal minors is positive, and is calledP0-matrix if each of its principal minors is nonnegative. Obviously,P-matrix is a generalization of positive definite matrix, whileP0-matrix is a generalization of positive semidefinite matrix. Such concepts ofP-matrix andP0-function can be fur- ther extended to nonlinear mapping, which we call themP-function andP0-function.
Definition 2.1 LetF =(F1, . . . , Fn)T withFi :Rn→Rfori=1, . . . , n. We say that
(a) F is monotone ifx−y, F (x)−F (y) ≥0 for allx, y∈Rn.
(b) F is strongly monotone ifx−y, F (x)−F (y) ≥μx−y2for someμ >0 and for allx, y∈Rn.
(c) F is aP0-function if max1≤i≤n xi=yi
(xi −yi)(Fi(x)−Fi(y))≥0 for allx, y∈Rn andx=y.
(d) F is a uniform P-function with modulusμ >0 if max1≤i≤n(xi−yi)(Fi(x)− Fi(y))≥μx−y2for allx, y∈Rn.
(e) F is Lipschitz continuous if there exists a constant L >0 such that F (x)− F (y) ≤Lx−yfor allx, y∈Rn.
It is well-known that every monotone function is anP0function and every strongly monotone function is a uniformP-function. For a continuously differentiable func- tionF, if its (transpose) Jacobian∇F (x)is anP-matrix thenF is anP-function (the converse may not be true), whereas the (transpose) Jacobian∇F (x)is anP0-matrix if and only ifF is anP0-function. For more detailed properties of various monotone andP (P0)-function, please refer to [9].
3 Properties of the new NCP-function
In this section, we study some favorable properties of the merit function ψα,θ,p, and then present some mild conditions under which the merit functionα,θ,p has bounded level sets and provides a global error bound, respectively. To this end, we present some technical lemmas which are needed for subsequent analysis.
Lemma 3.1 Forp >1,a >0,b >0, we haveap+bp≤(a+b)p. Proof We present two different ways to prove this lemma.
(1) For anyp >1, p=n+m, where n= [p](the greatest integer less than or equal top) andm=p−n, applying binomial theorem gives
(a+b)p=(a+b)n(a+b)m
≥(an+bn)(a+b)m
=an(a+b)m+bn(a+b)m
≥anam+bnbm
=ap+bp.
(2) Letf (t )=(t+1)p−(tp+1). It is easy to verify that f is increasing on [0,∞)whenp >1. Hence,f (a/b)≥f (0)=0 which yields(a+b)p≥ap+bp. Lemma 3.2 The functionψα,θ,pdefined by (11) has the following favorable proper- ties:
(a) ψα,θ,pis an NCP-function andψα,θ,p≥0 for all(a, b)∈R2.
(b) ψα,θ,pis continuously differentiable everywhere. Moreover, if(a, b)=(0,0),
∇aψα,θ,p(a, b)
=αb(ab)++ θsgn(a)· |a|p−1+(1−θ )sgn(a−b)|a−b|p−1 [θ (|a|p+ |b|p)+(1−θ )|a−b|p)](p−1)/p −1
φθ,p(a, b),
∇bψα,θ,p(a, b)
=αa(ab)++ θsgn(b)· |b|p−1−(1−θ )sgn(a−b)|a−b|p−1 [θ (|a|p+ |b|p)+(1−θ )|a−b|p)](p−1)/p −1
φθ,p(a, b), (14) and otherwise,∇aψα,θ,p(0,0)= ∇bψα,θ,p(0,0)=0.
(c) For p≥2, the gradient of ψα,θ,p is Lipschitz continuous on any nonempty bounded setS, i.e., there existsL >0 such that for any(a, b), (c, d)∈S,
∇ψα,θ,p(a, b)− ∇ψα,θ,p(c, d) ≤L(a, b)−(c, d).
(d) ∇aψα,θ,p(a, b)· ∇bψα,θ,p(a, b)≥0 for any(a, b)∈R2, and the equality holds if and only ifψα,θ,p(a, b)=0.
(e) ∇aψα,θ,p(a, b)=0⇐⇒ ∇bψα,θ,p(a, b)=0⇐⇒ψα,θ,p(a, b)=0.
(f) Suppose thatα >0. Ifa→ −∞orb→ −∞orab→ ∞, thenψα,θ,p(a, b)→
∞.
Proof (a) It is clear thatψα,θ,p(a, b)≥0 for all(a, b)∈R2 from the definition of ψα,θ,p. Then by [15, Proposition 2.1], we have
ψα,θ,p(a, b)=0 ⇐⇒ α 2
max{0, ab}2
=0 and ψθ,p(a, b)=0 ⇐⇒ a≥0, b≥0, ab=0.
Hence,ψα,θ,pis an NCP-function.
(b) First, direct calculations give the partial derivatives of ψα,θ,p. Then, using αb(ab)+→(0,0)andαa(ab)+→(0,0)as(a, b)→(0,0), we haveα2(max{0, ab})2 is continuously differentiable everywhere. By [15, Proposition 2.5], it is known that ψθ,p is continuously differentiable everywhere. In view of the expression of
∇ψα,θ,p(a, b),ψα,θ,pis also continuously differentiable everywhere.
(c) First, we claim thata(ab)+ for anya, b∈Ris Lipschitz continuous on any nonempty bounded setS. For any(a, b)∈Sand(c, d)∈S, without loss of generality, we may assume thata2+b2≤kandc2+d2≤kwhich imply|a| ≤k+1,|b| ≤k+1,
|c| ≤k+1 and|d| ≤k+1. Then, a(ab)+−c(cd)+
=1
2a2b+a|ab| −c2d−c|cd|
=1
2a2b−a2d+a2d−c2d+a|ab| −c|ab| +c|ab| −c|cd|
≤1 2
|a2b−a2d| + |a2d−c2d| +a|ab| −c|ab|+c|ab| −c|cd|
=1 2
a2|b−d| + |a+c||d||a−c| + |ab||a−c| + |c||ab−cd|
≤1 2
k|b−d| +(|a| + |c|)|d||a−c| +k|a−c| +(k+1)|ab−ad+ad−cd|
≤1 2
k|b−d| +2(k+1)2|a−c| +k|a−c| +(k+1)2(|b−d| + |a−c|)
=1 2
2(k+1)2+k+(k+1)2
|a−c| +
k+(k+1)2
|b−d|
≤l
|a−c| + |b−d|
≤√
2l(a, b)−(c, d),
wherel=2(k+1)2+k+(k+1)2. Hence, the mappinga(ab)+is Lipschitz con- tinuous on any nonempty bounded setS and so isαa(ab)+. Similarly,αb(ab)+ is Lipschitz continuous on any nonempty bounded setS. All of these imply the gradient
function of the function α2(max{0, ab})2is Lipschitz continuous on any bounded set S. On the other hand, by [15, Theorem 2.1], the gradient function of the function ψθ,p withp≥2, θ∈(0,1]is Lipschitz continuous. Thus, the gradient ofψα,θ,p is Lipschitz continuous on any nonempty bounded setS.
(d) If(a, b)=(0,0), part (d) clearly holds. Now we assume that(a, b)=(0,0).
Then,
∇aψα,θ,p(a, b)· ∇bψα,θ,p(a, b)
=cdφθ,p2 (a, b)+α2ab(ab)+2+αa(ab)+cφθ,p(a, b)+αb(ab)+dφθ,p(a, b), (15) where
c= θsgn(a)· |a|p−1+(1−θ )sgn(a−b)|a−b|p−1 [θ (|a|p+ |b|p)+(1−θ )|a−b|p)](p−1)/p −1
,
d= θsgn(b)· |b|p−1−(1−θ )sgn(a−b)|a−b|p−1 [θ (|a|p+ |b|p)+(1−θ )|a−b|p)](p−1)/p −1
. From the proof of [15, Proposition 2.5 ], we knowab(ab)2+≥0 and
θsgn(a)· |a|p−1+(1−θ )sgn(a−b)|a−b|p−1 [θ (|a|p+ |b|p)+(1−θ )|a−b|p)](p−1)/p −1
≤0,
(16) θsgn(b)· |b|p−1−(1−θ )sgn(a−b)|a−b|p−1
[θ (|a|p+ |b|p)+(1−θ )|a−b|p)](p−1)/p −1
≤0,
it suffices to show that the last two terms of (15) are nonnegative. For this purpose, we claim that
αa(ab)+ θsgn(a)· |a|p−1+(1−θ )sgn(a−b)|a−b|p−1 [θ (|a|p+ |b|p)+(1−θ )|a−b|p)](p−1)/p −1
φθ,p(a, b)≥0 (17) for all(a, b)=(0,0). Ifa≤0 andb≤0, thenφθ,p(a, b)≥0, which together with the second inequality in (16) implies that (17) holds. Ifa≤0 andb≥0, then(ab)+=0, which says that (17) holds. Ifa >0 andb >0, then|a|p+ |b|p≥ |a−b|p. Thus, φθ,p(a, b)≤φp(a, b)≤0, which together with the second inequality in (16) yields (17). Ifa >0 andb≤0, then(ab)+=0, and hence (17) holds. Similarly, we also have
αb(ab)+ θsgn(b)· |b|p−1−(1−θ )sgn(a−b)|a−b|p−1 [θ (|a|p+ |b|p)+(1−θ )|a−b|p)](p−1)/p −1
φθ,p(a, b)≥0 for all(a, b)=(0,0). Consequently,∇aψα,θ,p(a, b)· ∇bψα,θ,p(a, b)≥0. Besides, by the proof of [15, Proposition 2.5], we knowc=0 if and only ifb=0 anda >0;
d=0 if and only ifa=0 andb >0. This together with (15) says∇aψα,θ,p(a, b)·
∇bψα,θ,p(a, b)=0 if and only if{ψθ,p(a, b)=0 andα2ab(ab)+2=0}or{c=0}or {d=0}if and only if{ψθ,p(a, b)=0 andab≤0}or{c=0}or{d=0}if and only ifψθ,p(a, b)=0 and α2(max{0, ab})2=0 if and only ifψα,θ,p(a, b)=0.
(e) Ifψα,θ,p(a, b)=0, thenα2(max{0, ab})2=0 andψθ,p(a, b)=0, which imply ab≤0 andφθ,p(a, b)=0. Hence,∇aψα,θ,p(a, b)=0 and∇bψα,θ,p(a, b)=0. Now, it remains to show that∇aψα,θ,p(a, b)=0 implyingψα,θ,p(a, b)=0. Suppose that
∇aψα,θ,p(a, b)=0. Then,
αb(ab)+= − θsgn(a)· |a|p−1+(1−θ )sgn(a−b)|a−b|p−1 [θ (|a|p+ |b|p)+(1−θ )|a−b|p)](p−1)/p −1
φθ,p(a, b).
(18) We will argue that the equality (18) implies(b=0, a≥0)or(b >0, a=0). To see this, we let
c=αb(ab)+,
d= − θsgn(a)· |a|p−1+(1−θ )sgn(a−b)|a−b|p−1 [θ (|a|p+ |b|p)+(1−θ )|a−b|p)](p−1)/p −1
φθ,p(a, b),
e= θsgn(a)· |a|p−1+(1−θ )sgn(a−b)|a−b|p−1 [θ (|a|p+ |b|p)+(1−θ )|a−b|p)](p−1)/p −1
.
It is not hard to observe that(e≤0)and (e=0 impliesb=0) which are helpful for the following discussions.
Case 1: b=0 anda <0. Then,c=0 butd=0 which violates (18).
Case 2: b <0 anda≥0. Then, we havee <0, and hencec=0 butd=0, which violates (18).
Case 3: b <0 anda <0. Then, we have e <0 and φθ,p(a, b) >0, which lead to c≤0 butd >0. This contradicts to (18) too.
Case 4: b >0 anda >0. Then, we havee <0 andφθ,p(a, b) <0, which implyc≥0 butd <0. This contradicts to (18) too.
Case 5: b >0 anda <0. Similar arguments as in Case 2 cause a contradiction.
Thus, (18) implies(b=0, a≥0)or(b >0, a=0), and each of which always yields ψα,θ,p(a, b)=0. By symmetry,∇bψα,θ,p(a, b)=0 also impliesψα,θ,p(a, b)=0.
(f) Ifa→ −∞orb→ −∞, from [15, Proposition 2.4], we know|φθ,p(a, b)| →
∞. In addition, the fact α2(max{0, ab})2≥0 gives ψα,θ,p(a, b)→ ∞. Ifab→ ∞, sinceα >0, we have α2(max{0, ab})2→ ∞. This together withψθ,p(a, b)≥0 says
ψα,θ,p(a, b)→ ∞.
By Lemma3.2(a), we immediately have the following theorem.
Theorem 3.1 Letα,θ,p be defined as in (10). Thenα,θ,p(x)≥0 for allx∈Rn andα,θ,p(x)=0 if and only ifx solves the NCP. Moreover, if the NCP has at least one solution, thenx is a global minimizer ofα,θ,pif and only ifxsolves the NCP.
Proof Sinceψθ,p is an NCP-function, from [15, Proposition 2.5], we have thatx solving the NCP ⇐⇒ x ≥ 0, F (x) ≥ 0,x, F (x) = 0 ⇐⇒ x ≥ 0, F (x) ≥ 0, xiFi(x) =0 for all i ∈ {1,2, . . . , n} ⇐⇒ α,θ,p(x) =0. Besides, α,θ,p(x)is nonnegative. Thus, ifx solves the NCP, thenx is a global minimizer
ofα,θ,p. Next, we claim that if the NCP has at least one solution, thenxis a global minimizer of α,θ,p ⇒ x solves the NCP. Suppose x does not solve the NCP.
Fromx solves the NCP⇐⇒α,θ,p(x)=0 andα,θ,p(x)is nonnegative, it is clear α,θ,p(x) >0. However, by assumption, the NCP has a solution, sayy, which makes thatα,θ,p(y)=0. Then, we get a contradiction thatα,θ,p(x) >0=α,θ,p(y)and xis a global minimizer ofα,θ,p. Thus, we complete the proof.
Theorem3.1indicates that the NCP can be recast as the unconstrained minimiza- tion:
xmin∈Rnα,θ,p(x). (19)
In general, it is hard to find a global minimum ofα,θ,p. Therefore, it is important to know under what conditions a stationary point ofα,θ,p is a global minimum.
Using Lemma3.2(d) and the same proof techniques as in [21, Theorem 3.5], we can establish that each stationary point ofα,θ,p is a global minimum only if F is a P0-function.
Theorem 3.2 LetF be a P0-function. Then x∗∈Rn is a global minimum of the unconstrained optimization problem (19) if and only if x∗ is a stationary point of α,θ,p.
Theorem 3.3 The functionα,θ,phas bounded level setsL(α,θ,p, c)for allc∈R, ifF is monotone and the NCP is strictly feasible (i.e., there existsx >ˆ 0 such that F (x) >ˆ 0) whenα >0, orF is a uniformP-function whenα≥0.
Proof From [2], if F is a monotone function with a strictly feasible point, then the following condition holds: for every sequence {xk} such that xk → ∞, (−xk)+<∞, and(−F (xk))+<∞, we have max1≤i≤n{(xik)+Fi(xk)+} → ∞. Sup- pose that there exists an unbounded sequence xk⊆L(α,θ,p, c) for some c∈R.
Since α,θ,p(xk)≤c, there is no index i such that xik → −∞or Fi(xk)→ −∞
by Lemma3.2(f). Hence, max1≤i≤n{(xik)+Fi(xk)+} → ∞. Also, there is an indexj, and at least a subsequencexjk such that{(xjk)+Fj(xk)+} → ∞. However, this im- plies thatα,θ,p(xk)is unbounded by Lemma3.2(f), contracting to the assumption on level sets. Another part of the proof is similar to the proof of [4, Proposition 3.5].
In what follows, we show that the merit functionsθ,p,NRandα,θ,phave the same order of growth behavior on every bounded set. For this purpose, we need the following crucial technical lemma.
Lemma 3.3 Letφθ,p:R2→Rbe defined as in (9). Then for anyp >1 and all θ∈(0,1]we have
(2−2p1)|min{a, b}| ≤ |φθ,p(a, b)| ≤(2+2p1)|min{a, b}|. (20) Proof Without loss of generality, we assumea≥b. We will prove the desired results by considering the following two cases: (1)a+b≤0 and (2)a+b >0.
Case (1):a+b≤0. In this case, we need to discuss two subcases:
(i)|a|p+ |b|p≥ |a−b|p. In this subcase, we have
|φθ,p(a, b)| ≥ |p
θ (|a−b|p)+(1−θ )(|a−b|p)−(a+b)|
= |p
(|a−b|p)−(a+b)|
= |(|a−b| −(a+b)|
= |a−b−(a+b)|
= |2b|
=2|min{a, b}|
≥(2−21p)|min{a, b}|. (21) On the other hand, since|a|p+ |b|p≥ |a−b|pand by [5, Lemma 3.2], we have
|φθ,p(a, b)| ≤ |φp(a, b)| ≤(2+2p1)|min{a, b}|. (22) (ii)|a|p+ |b|p<|a−b|p. Since|a|p+ |b|p<|a−b|pand by [5, Lemma 3.2], we have
|φθ,p(a, b)|>|φp(a, b)| ≥(2−2p1)|min{a, b}|. (23) On the other hand, by the discussion of Case (1),
|φθ,p(a, b)|<2|b| ≤(2+2p1)|min{a, b}|. (24) Case (2):a+b >0. Ifab=0, then (20) clearly holds. Thus, we proceed the argu- ments by discussing two subcases:
(i)ab <0. In this subcases, we havea >0, b <0,|a|>|b|. By Lemma3.1,|a|p+
|b|p≤ |a−b|p. Then,
φθ,p(a, b)≥φp(a, b)≥ |a| −a−b≥ −b= |min{a, b}| ≥(2−2p1)|min{a, b}|.
(25) On the other hand,
φθ,p(a, b)≤ |a−b| −(a+b)= −2b=2|min{a, b}| ≤(2+2p1)|min{a, b}|. (26) (ii) ab >0. In this subcases, we have a ≥b >0,|a|p + |b|p ≥ |a−b|p. By Lemma3.1,φθ,p(a, b)≤φp(a, b)≤0. Notice thatφθ,p(a, b)≥ |a−b| −(a+b)=
−2b= −2 min{a, b}, and hence we obtain that
|φθ,p(a, b)| ≤2|min{a, b}| ≤(2+2p1)|min{a, b}|. (27) On the other hand, sinceφθ,p(a, b)≤φp(a, b)≤0 , and by [5, Lemma 3.2], and hence we obtain that
|φθ,p(a, b)| ≥ |φp(a, b)| ≥(2−2p1)|min{a, b}|. (28) All the aforementioned inequalities (21)–(28) imply that (20) holds.
Proposition 3.1 Letθ,p, NRandα,θ,p be defined as in (13), (12) and (10), re- spectively. Let S be an arbitrary bounded set. Then, for anyp >1, we have
(2−2p1)2NR(x)≤θ,p(x)≤(2+2p1)2NR(x) for allx∈Rn (29) and
(2−2p1)2NR(x)≤α,θ,p(x)≤(αB2+(2+21p)2)NR(x)
for allx∈S, (30)
whereBis a constant defined by B= max
1≤i≤n
sup
x∈S
max
|xi|,|Fi(x)|
<∞.
Proof The inequality in (29) is direct by Lemma3.3and the definitions ofθ,pand NR. In addition, from Lemma3.3and the definition ofα,θ,p, it follows that
α,θ,p(x)≥
2−2p12
NR(x) for allx∈Rn.
It remains to prove the inequality on the right hand side of (30). From the proof of [5, Proposition 3.1], we know for eachi,
(xiFi(x))+≤B|min{xi, Fi(x)}| for allx∈S. (31) By Lemma3.3and (31), for alli=1, . . . , nandx∈S,
ψα,θ,p(xi, Fi(x))≤1 2
αB2+(2+2p1)2
min{xi, Fi(x)}2
holds for anyp >1. The proof is then complete by the definitions ofα,θ,p and
NR.
From Proposition3.1, we immediately obtain the following result.
Corollary 3.1 Letθ,p and α,θ,p be defined by (13) and (10), respectively; and S be any bounded set. Then, for any p >1 and allx ∈S, we have the following inequalities:
(2−2p1)2 (αB2+(2+2p1)2)
α,θ,p(x)≤θ,p(x)≤ (2+2p1)2 (2−2p1)2
α,θ,p(x)
whereBis the constant defined as in Proposition3.1.
Sinceθ,p, NRandα,θ,phave the same order on a bounded set, one will pro- vide a global error bound for the NCP as long as the other one does. As below, we show thatα,θ,pprovides a global error bound without the Lipschitz continuity ofF whenα >0.
Theorem 3.4 Let α,θ,p be defined as in (10). Suppose that F is a uniform P-function with modulusμ >0. Ifα >0, then there exists a constantκ1>0 such that
x−x∗ ≤κ1α,θ,p(x)14 for allx∈Rn;
ifα=0 andSis any bounded set, there exists a constantκ2>0 such that x−x∗ ≤κ2
max
α,θ,p(x),
α,θ,p(x)12
for allx∈S; wherex∗=(x∗1, . . . , xn∗)is the unique solution for the NCP.
Proof By the proof of [5, Theorem 3.4], we have μx−x∗2≤ max
1≤i≤nτi{(xiFi(x))++(−Fi(x))++(−xi)+}, (32) whereτi:=max{1, xi∗, Fi(x∗)}. We next prove that for all(a, b)∈R2,
(−a)+2+(−b)+2≤ [φθ,p(a, b)]2. (33) To see this, without loss of generality, we assumea≥band discuss three cases:
(i) Ifa≥b≥0, then (33) holds obviously.
(ii) If a ≥ 0≥ b, then |a|p + |b|p ≤ |a −b|p by Lemma 3.1, which implies φθ,p(a, b)≥ (a, b)p−(a+b)≥ −b≥0. Hence,(−a)+2+(−b)+2=b2≤ [φθ,p(a, b)]2.
(iii) If 0≥a≥b, then(−a)+2+(−b)+2=a2+b2≤ [φθ,p(a, b)]2. Hence, (33) follows.
Suppose thatα >0. Using the inequality (33), we then obtain that (ab)++(−a)++(−b)+2
=(ab)2++(−b)2++(−a)2++2(ab)+(−a)+ +2(−a)+(−b)++2(ab)+(−b)+
≤(ab)2++(−b)2++(−a)2++(ab)2++(−a)2+ +(−a)2++(−b)2++(ab)2++(−b)2+
≤3
(ab)2++ [φθ,p(a, b)]2
≤τ α
2(ab)2++1 2
φθ,p(a, b)2
=τ ψα,θ,p(a, b), (34)
whereτ :=max{α6,6}>0. Combining (34) with (32) and lettingτˆ=max1≤i≤nτi, we get
μx−x∗2≤ max
1≤i≤nτi
τ ψα,θ,p
xi, Fi(x)1/2
≤ ˆτ τ1/2max
1≤i≤nψα,θ,p(xi, Fi(x))1/2
≤ ˆτ τ1/2 n
i=1
ψα,θ,p(xi, Fi(x))1/2
= ˆτ τ1/2α,θ,p(x)1/2.
From this, the first desired result follows immediately by settingκ1:= [ ˆτ τ1/2/μ]1/2. Suppose thatα=0. From the proof of Proposition3.1, the inequality (31) holds.
Combining with (32)–(33), it then follows that for allx∈S, μx−x∗2≤ max
1≤i≤nτi
B|min{xi, Fi(x)}| +2(ψθ,p(xi, Fi(x)))1/2
≤ ˆτ max
1≤i≤n
√2B(ψˆ θ,p(xi, Fi(x)))1/2+2(ψθ,p(xi, Fi(x)))1/2
≤(√
2Bˆ+2)τ (ˆ θ,p(x))1/2
=(√
2Bˆ+2)τ (ˆ α,θ,p(x))1/2
≤(√
2Bˆ+2)τˆ max
α,θ,p(x),
α,θ,p(x)
where Bˆ =B/(2−2p1), τˆ =max1≤i≤nτi and the second inequality is from Lemma3.3. Lettingκ2:= [(√
2Bˆ +2)τ /μˆ ]1/2, we obtain the desired result from
the above inequality.
The following lemma is needed for the proof of Proposition3.2, which we suspect is useful in analysis of convergence rate.
Lemma 3.4 For all(a, b)=(0,0)andp >1, we have the following inequality:
θ[sgn(a)· |a|p−1+sgn(b)· |b|p−1] [θ (|a|p+ |b|p)+(1−θ )|a−b|p)](p−1)/p −2
2
≥
2−2p12
∀θ∈(0,1]. Proof Ifa=0 orb=0, the inequality holds obviously. Then we complete the proof by considering three cases: (i)a >0 andb >0, (ii)a <0 andb <0, and (iii)ab <0.
Case (i): Sinceθ∈(0,1]andp >1, it follows thatθ1/p≤1. Now, by the proof of [5, Lemma 3.3], we have
θ[sgn(a)· |a|p−1+sgn(b)· |b|p−1] [θ (|a|p+ |b|p)+(1−θ )|a−b|p)](p−1)/p
= θ[|a|p−1+ |b|p−1]
[θ (|a|p+ |b|p)+(1−θ )|a−b|p)](p−1)/p
≤ θ[|a|p−1+ |b|p−1] [θ (|a|p+ |b|p)](p−1)/p
=θ1/p[|a|p−1+ |b|p−1] [(|a|p+ |b|p)](p−1)/p
≤21/p forp >1.
Therefore,
2− θ[|a|p−1+ |b|p−1]
[θ (|a|p+ |b|p)+(1−θ )|a−b|p)](p−1)/p ≥2−21p forp >1. Squaring both sides then leads to the desired inequality.
Case (ii): By similar arguments as in case (i), we obtain 2−2p1 ≤2− θ[|a|p−1+ |b|p−1]
[θ (|a|p+ |b|p)+(1−θ )|a−b|p)](p−1)/p
≤2+ θ[|a|p−1+ |b|p−1]
[θ (|a|p+ |b|p)+(1−θ )|a−b|p)](p−1)/p forp >1, from which the result follows immediately.
Case (iii): Again, we suppose|a| ≥ |b|and therefore have 21p ≥ θ[|a|p−1+ |b|p−1]
[θ (|a|p+ |b|p)+(1−θ )|a−b|p)](p−1)/p
≥ θ[|a|p−1− |b|p−1]
[θ (|a|p+ |b|p)+(1−θ )|a−b|p)](p−1)/p forp >1.
Thus,
2−21p≤2− θ[|a|p−1− |b|p−1]
[θ (|a|p+ |b|p)+(1−θ )|a−b|p](p−1)/p
forp >1 and the desired result is also satisfied.
Proposition 3.2 Letψα,θ,pbe given as in (11). Then, for allx∈Rnandp >1, ∇aψα,θ,p(x, F (x))+ ∇bψα,θ,p(x, F (x))2≥2
2−21p2
θ,p(x) ∀θ∈(0,1]. In particular, for allxbelonging to any bounded setSandp >1,
∇aψα,θ,p(x, F (x))+ ∇bψα,θ,p(x, F (x))2≥ 2(2−2p1)4 (αB2+(2+2p1)2)
α,θ,p(x)
∀θ∈(0,1],
whereBis defined as in Proposition3.1and
∇aψα,θ,p(x, F (x)):=
∇aψα,θ,p(x1, F1(x)), . . . ,∇aψα,θ,p(xn, Fn(x))T
,
∇bψα,θ,p(x, F (x)):= (35)
∇bψα,θ,p(x1, F1(x)), . . . ,∇bψα,θ,p(xn, Fn(x))T
.
Proof The second part of the conclusions is direct by Corollary3.1and the first part.
Thus, it remains to show the first part. From the definitions of∇aψα,θ,p(x, F (x)),
∇bψα,θ,p(x, F (x))andθ,p(x), showing the first part is equivalent to proving that the following inequality
∇aψα,θ,p(a, b)+ ∇bψα,θ,p(a, b)2
≥2
2−2p12
ψθ,p(a, b) (36) holds for all(a, b)∈R2. When(a, b)=(0,0), the inequality (36) clearly holds. Sup- pose(a, b)=(0,0). Then, it follows from (14) that
∇aψα,θ,p(a, b)+ ∇bψα,θ,p(a, b)2
=
α(a+b)(ab)+
+(φθ,p(a, b)) θ[sgn(a)· |a|p−1+sgn(b)· |b|p−1] [θ (|a|p+ |b|p)+(1−θ )|a−b|p)](p−1)/p−2
2
=α2(a+b)2(ab)2+
+(φθ,p(a, b))2 θ[sgn(a)· |a|p−1+sgn(b)· |b|p−1] [θ (|a|p+ |b|p)+(1−θ )|a−b|p)](p−1)/p −2
2
+2α(a+b)(ab)+(φθ,p(a, b))
× θ[sgn(a)· |a|p−1+sgn(b)· |b|p−1] [θ (|a|p+ |b|p)+(1−θ )|a−b|p)](p−1)/p −2
. (37)
Now, we claim that for all(a, b)=(0,0)∈R2,
2α(a+b)(ab)+(φθ,p(a, b)) θ[sgn(a)· |a|p−1+sgn(b)· |b|p−1] [θ (|a|p+ |b|p)+(1−θ )|a−b|p)](p−1)/p−2
≥0.
(38) Ifab≤0, then(ab)+=0 and the inequality (38) is clear. Ifa, b >0, then by the proof of Lemma3.4, we have
θ[sgn(a)· |a|p−1+sgn(b)· |b|p−1] [θ (|a|p+ |b|p)+(1−θ )|a−b|p)](p−1)/p−2
≤0, ∀(a, b)=(0,0)∈R2 (39) andφθ,p(a, b)≤0, which imply the inequality (38) also holds. If a, b <0, then φθ,p(a, b)≥0, which together with (39) yields the inequality (38). Thus, we obtain that the inequality (38) holds for all(a, b)=(0,0). Now using Lemma3.4and (37)–
(38), we readily obtain the inequality (36) holds for all(a, b)=(0,0). The proof is
thus complete.
