to appear in Computational Optimization and Applications, 2010
A new class of penalized NCP-functions and its properties
Jein-Shan Chen 1 Department of Mathematics National Taiwan Normal University
Taipei, Taiwan 11677 E-mail: jschen@math.ntnu.edu.tw
Zheng-Hai Huang 2 Department of Mathematics
Tianjin University Tianjin 300072, China E-mail: huangzhenghai@tju.edu.cn
Chin-Yu She
Department of Mathematics National Taiwan Normal University
Taipei, Taiwan 11677 E-mail: bbgmmshe@hotmail.com
April 17, 2009
(first revised on September 15, 2009) (second revised on December 18, 2009)
Abstract. In this paper, we consider a class of penalized NCP-functions, which includes several existing well-known NCP-functions as special cases. The merit function 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 proposed, its global
1Member of Mathematics Division, National Center for Theoretical Sciences, Taipei Office. The author’s work is partially supported by National Science Council of Taiwan.
2This author’s work is partly supported by by the National Natural Science Foundation of China (Grant No. 10871144).
convergence is proved and numerical performance compared with those based on some existing NCP-functions is reported.
Key Words. NCP-function, penalized, bounded level sets, error bounds.
1 Introduction
The nonlinear complementarity problem (NCP) is to find a point x∈ IRn such that
x≥ 0, F (x) ≥ 0, ⟨x, F (x)⟩ = 0, (1)
where ⟨·, ·⟩ is the Euclidean inner product and F = (F1, . . . , Fn)T is a map from IRn to IRn. We assume that F is continuously differentiable throughout this paper. The 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, 5, 14, 16, 17, 18, 20, 23, 26, 30, 31]. 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 nonlinear equations [24, 25, 31], or as an unconstrained minimization problem [7, 10, 11, 12, 19, 21, 27, 30]. 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 function ϕ : IR2 → IR is 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) ∈ IR2. (3) Through this NCP-function ϕFB, the NCP (1) can be reformulated as a system of nons- mooth equations:
ΦFB(x) :=
ϕFB(x1 , F1(x))
··
·
ϕFB(xn , Fn(x))
= 0. (4)
Thus, the function ΨFB : IRn → IR+ defined as below is a merit function for the NCP:
ΨFB(x) := 1
2∥ΦFB(x)∥2 =
∑n i=1
ψFB(xi, Fi(x)), (5)
where ψFB : IR2 → IR+ is the square of ϕFB, i.e., ψFB(a, b) = 1
2 √
a2+ b2− (a + b) 2. (6) Consequently, the NCP is equivalent to the unconstrained minimization problem:
xmin∈IRnΨFB(x). (7)
There are several generalizations of the FB function in the literature. For example, Kanzow and Kleinmichel [22] extend ϕFB function to
ϕθ(a, b) :=√
(a− b)2+ θab− (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, 5] 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) where p > 1, θ∈ (0, 1].
All the aforementioned functions are NCP-functions. The corresponding function ψθ, ψλ, ψp, and ψθ,p is square of ϕθ, ϕλ, ϕp, and ϕθ,p, respectively, and naturally induces a merit function Ψθ, Ψλ, Ψp, and Ψθ,p like what ψFB function does. Along this track, in this paper, we study the following merit function Ψα,θ,p : IRn→ IR+ for the NCP:
Ψα,θ,p(x) :=
∑n i=1
ψα,θ,p(xi , Fi(x)), (10)
where ψα,θ,p : IR2 → IR+ is an NCP-function defined by ψα,θ,p(a, b) := α
2(max{0, ab})2 + ψθ,p(a, b) (11)
with α≥ 0 being a real parameter. Note that ψα,θ,p includes all the above functions ψFB, ψp, ψθ, ψθ,p (and ψ7 in [28]) as special cases. Although ψα,θ,p is obtained by penalizing the function ψθ,p considered in [15], more favorable properties of ψα,θ,p are explored in this work. In particular, Ψα,θ,p has property of bounded level sets and provides a global error bound for the NCP under mild condition which were not studied 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 Section 2, we review some definitions and preliminary results to be used in the subsequent analysis. In Section 3, we show some properties of the proposed merit function. In Section 4, we propose a derivative free algorithm based on this merit function Ψα,θ,p, show its global convergence, and report some numerical results. In Section 5, we make concluding remarks.
Throughout this paper, IRn denotes the space of n-dimensional real column vectors and T denotes transpose. For every differentiable function f : IRn → IR, ∇f(x) denotes the gradient of f at x. For every differentiable mapping F = (F1, . . . , Fn)T : IRn → IRn,
∇F (x) = (∇F1(x) . . . ∇Fn(x)) denotes the transpose Jacobian of F at x. We use ∥x∥p
to denote the p-norm of x and denote ∥x∥ the Euclidean norm of x. The level set of a function Ψ : IRn → IR is denoted by L(Ψ, c) := {x ∈ IRn | Ψ(x) ≤ c}. In addition, we will frequently mention two merit functions. One is the natural residual merit function ΨNR : IRn → IR+ defined by
ΨNR(x) := 1 2
∑n i=1
ϕ2
NR(xi , Fi(x)), (12)
where ϕNR : IR2 → IR denotes the minimum NCP-function min{a, b}. Another one is Ψθ,p : IRn→ IR+ induced by ψθ,p:
Ψθ,p(x) := 1 2
∑n i=1
ϕ2θ,p(xi , Fi(x)). (13)
Unless otherwise stated, in the sequel, we always suppose that p is a fixed real number in (1,∞).
2 Preliminaries
This section briefly recalls some concepts about the mapping F that will be used later.
A matrix is said to be P -matrix if each of its principal minors is positive, and is called P0-matrix if each of its principal minors is nonnegative. Obviously, P -matrix is a gen- eralization of positive definite matrix, while P0-matrix is a generalization of positive semidefinite matrix. Such concepts of P -matrix and P0-function can be further extended to nonlinear mapping, which we call them P -function and P0-function.
Definition 2.1 Let F = (F1, . . . , Fn)T with Fi : IRn→ IR for i = 1, . . . , n. We say that
(a) F is monotone if ⟨x − y, F (x) − F (y)⟩ ≥ 0 for all x, y ∈ IRn.
(b) F is strongly monotone if ⟨x − y, F (x) − F (y)⟩ ≥ µ∥x − y∥2 for some µ > 0 and for all x, y ∈ IRn.
(c) F is a P0-function if max
1≤i≤n xi̸=yi
(xi− yi)(Fi(x)− Fi(y))≥ 0 for all x, y ∈ IRn and x̸= y.
(d) F is a uniform P -function with modulus µ > 0 if max
1≤i≤n(xi − yi)(Fi(x)− Fi(y)) ≥ µ∥x − y∥2 for all x, y∈ IRn.
(e) F is Lipschitz continuous if there exists a constant L > 0 such that ∥F (x)−F (y)∥ ≤ L∥x − y∥ for all x, y ∈ IRn.
It is well-known that every monotone function is an P0 function and every strongly monotone function is a uniform P -function. For a continuously differentiable function F , if its (transpose) Jacobian ∇F (x) is an P -matrix then F is an P -function (the converse may not be true), whereas the (transpose) Jacobian ∇F (x) is an P0-matrix if and only if F is an P0-function. For more detailed properties of various monotone and P (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 function Ψα,θ,phas 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 For p > 1, a > 0, b > 0, we have ap+ bp ≤ (a + b)p
Proof. We present two different ways to prove this lemma.
(1) For any p > 1, p = n + m, where n = [p] (the greatest integer less than or equal to p) and m = 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) Let f (t) = (t + 1)p− (tp+ 1). It is easy to verify that f is increasing on [0,∞) when p > 1. Hence, f (a/b)≥ f(0) = 0 which yields (a + b)p ≥ ap+ bp. 2
Lemma 3.2 The function ψα,θ,p defined by (11) has the following favorable properties:
(a) ψα,θ,p is an NCP-function and ψα,θ,p≥ 0 for all (a, b) ∈ IR2.
(b) ψα,θ,p is 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 set S, i.e., there exists L > 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) ∈ IR2, and the equality holds if and only if ψα,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 that α > 0. If a→ −∞ or b → −∞ or ab → ∞, then ψα,θ,p(a, b)→ ∞.
Proof. (a) It is clear that ψα,θ,p(a, b) ≥ 0 for all (a, b) ∈ IR2 from the definition of ψα,θ,p. Then by [15, Prop. 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,p is an NCP-function.
(b) First, direct calculations give the partial derivatives of ψα,θ,p. Then, using αb(ab)+ → (0, 0) and αa(ab)+ → (0, 0) as (a, b) → (0, 0), we have α2(max{0, ab})2 is continuously differentiable everywhere. By [15, Prop. 2.5], it is known that ψθ,p is continuously differ- entiable everywhere. In view of the expression of ∇ψα,θ,p(a, b), ψα,θ,p is also continuously differentiable everywhere.
(c) First, we claim that a(ab)+ for any a, b∈ IR is Lipschitz continuous on any nonempty bounded set S. For any (a, b) ∈ S and (c, d) ∈ S, without loss of generality, we may assume that a2+ b2 ≤ k and c2+ d2 ≤ k which imply |a| ≤ k + 1, |b| ≤ k + 1, |c| ≤ k + 1 and |d| ≤ k + 1. Then,
a(ab)+− c(cd)+
= 1 2
a2b + a|ab| − c2d− c|cd|
= 1 2
a2b− 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)∥,
where l = 2(k + 1)2 + k + (k + 1)2. Hence, the mapping a(ab)+ is Lipschitz continuous on any nonempty bounded set S and so is αa(ab)+. Similarly, αb(ab)+ is Lipschitz continuous on any nonempty bounded set S. All of these imply the gradient function of the function α2(max{0, ab})2 is Lipschitz continuous on any bounded set S. On the
other hand, by [15, Theorem 2.1], the gradient function of the function ψθ,p with p≥ 2, θ ∈ (0, 1] is Lipschitz continuous. Thus, the gradient of ψα,θ,p is Lipschitz continuous on any nonempty bounded set S.
(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) (15)
= cdϕ2θ,p(a, b) + α2ab(ab)+2+ αa(ab)+cϕθ,p(a, b) + αb(ab)+dϕθ,p(a, b), 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, Prop. 2.5 ], we know ab(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, (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, (16)
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). If a ≤ 0 and b ≤ 0, then ϕθ,p(a, b) ≥ 0, which together with the second inequality in (16) implies that (17) holds. If a ≤ 0 and b ≥ 0, then (ab)+ = 0, which says that (17) holds. If a > 0 and b > 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). If a > 0 and b≤ 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, Prop. 2.5], we know c = 0 if and only if b = 0 and a > 0; d = 0 if and only if a = 0 and b > 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 and α2ab(ab)+2 = 0} or {c = 0} or {d = 0} if and only if {ψθ,p(a, b) = 0 and ab ≤ 0} or {c = 0} or {d = 0} if and only if ψθ,p(a, b) = 0 and
α
2(max{0, ab})2 = 0 if and only if ψα,θ,p(a, b) = 0.
(e) If ψα,θ,p(a, b) = 0, then α2(max{0, ab})2 = 0 and ψθ,p(a, b) = 0, which imply ab ≤ 0 and ϕθ,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 implying ψα,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 implies b = 0)
which are helpful for the following discussions.
Case 1: b = 0 and a < 0. Then, c = 0 but d̸= 0 which violates (18).
Case 2: b < 0 and a ≥ 0. Then, we have e < 0, and hence c = 0 but d ̸= 0, which violates (18).
Case 3: b < 0 and a < 0. Then, we have e < 0 and ϕθ,p(a, b) > 0, which lead to c ≤ 0 but d > 0. This contradicts to (18) too.
Case 4: b > 0 and a > 0. Then, we have e < 0 and ϕθ,p(a, b) < 0, which imply c≥ 0 but d < 0. This contradicts to (18) too.
Case 5: b > 0 and a < 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 implies ψα,θ,p(a, b) = 0.
(f) If a→ −∞ or b → −∞, from [15, Prop. 2.4], we know |ϕθ,p(a, b)| → ∞. In addition, the fact α2(max{0, ab})2 ≥ 0 gives ψα,θ,p(a, b) → ∞. If ab → ∞, since α > 0, we have
α
2(max{0, ab})2 → ∞. This together with ψθ,p(a, b)≥ 0 says ψα,θ,p(a, b)→ ∞. 2
By Lemma 3.2(a), we immediately have the following theorem.
Theorem 3.1 Let Ψα,θ,p be defined as in (10). Then Ψα,θ,p(x)≥ 0 for all x ∈ IRn and Ψα,θ,p(x) = 0 if and only if x solves the NCP. Moreover, if the NCP has at least one solution, then x is a global minimizer of Ψα,θ,p if and only if x solves the NCP.
Proof. Since ψθ,p is an NCP-function, from [15, Prop. 2.5], we have that x 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} ⇐⇒ Ψα,θ,p(x) = 0. Besides, Ψα,θ,p(x) is nonnegative. Thus, if x solves the NCP, then x is a global minimizer of Ψα,θ,p. Next, we claim that if the NCP has at least one solution, then x is a global minimizer of Ψα,θ,p =⇒ x solves the NCP.
Suppose x does not solve the NCP. From x solves the NCP ⇐⇒ Ψα,θ,p(x) = 0 and Ψα,θ,p(x) is nonnegative, it is clear Ψα,θ,p(x) > 0. However, by assumption, the NCP has a solution, say y, which makes that Ψα,θ,p(y) = 0. Then, we get a contradiction that Ψα,θ,p(x) > 0 = Ψα,θ,p(y) and x is a global minimizer of Ψα,θ,p. Thus, we complete the proof. 2
Theorem 3.1 indicates that the NCP can be recast as the unconstrained minimization:
xmin∈IRnΨα,θ,p(x). (19)
In general, it is hard to find a global minimum of Ψα,θ,p. Therefore, it is important to know under what conditions a stationary point of Ψα,θ,p is a global minimum. Using Lemma 3.2(d) and the same proof techniques as in [21, Theorem 3.5], we can establish that each stationary point of Ψα,θ,p is a global minimum only if F is a P0-function.
Theorem 3.2 Let F be a P0-function. Then x∗ ∈ IRn is a global minimum of the unconstrained optimization problem (19) if and only if x∗ is a stationary point of Ψα,θ,p.
Theorem 3.3 The function Ψα,θ,p has bounded level sets L(Ψα,θ,p, c) for all c∈ IR, if F is monotone and the NCP is strictly feasible (i.e., there exists ˆx > 0 such that F (ˆx) > 0) when α > 0, or F is a uniform P -function when α≥ 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 max
1≤i≤n
{(xki)+Fi(xk)+}
→ ∞. Suppose that there exists an unbounded sequence xk ⊆ L(Ψα,θ,p, c) for some c ∈ IR. Since Ψα,θ,p(xk) ≤ c, there is no index i such that xki → −∞ or Fi(xk) → −∞ by Lemma 3.2(f). Hence,
1≤i≤nmax
{(xki)+Fi(xk)+}
→ ∞. Also, there is an index j, and at least a subsequence xkj
such that {
(xkj)+Fj(xk)+}
→ ∞. However, this implies that Ψα,θ,p(xk) is unbounded by Lemma 3.2(f), contracting to the assumption on level sets. Another part of the proof is similar to the proof of [5, Prop. 3.5]. 2
In what follows, we show that the merit functions Ψθ,p, ΨNR and Ψα,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 Let ϕθ,p : IR2 → IR be defined as in (9). Then for any p > 1 and all θ ∈ (0, 1] we have
(2− 21p)| min{a, b}| ≤ |ϕθ,p(a, b)| ≤ (2 + 21p)| min{a, b}|. (20)
Proof. Without loss of generality, we assume a≥ 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|p and by [4, Lemma 3.2], we have
|ϕθ,p(a, b)| ≤ |ϕp(a, b)| ≤ (2 + 21p)| min{a, b}|. (22)
(ii) |a|p +|b|p <|a − b|p. Since |a|p+|b|p <|a − b|p and by [4, Lemma 3.2], we have
|ϕθ,p(a, b)| > |ϕp(a, b)| ≥ (2 − 21p)| min{a, b}|. (23) On the other hand, by the discussion of Case(1),
|ϕθ,p(a, b)| < 2|b| ≤ (2 + 21p)| min{a, b}|. (24) Case(2): a + b > 0. If ab=0, then (20) clearly holds. Thus, we proceed the arguments by discussing two subcases:
(i) ab < 0. In this subcases, we have a > 0, b < 0,|a| > |b|. By Lemma 3.1, |a|p+|b|p ≤
|a − b|p. Then,
ϕθ,p(a, b)≥ ϕp(a, b)≥ |a| − a − b ≥ −b = | min{a, b}| ≥ (2 − 21p)| min{a, b}|. (25)
On the other hand,
ϕθ,p(a, b)≤ |a − b| − (a + b) = −2b = 2| min{a, b}| ≤ (2 + 21p)| min{a, b}|. (26)
(ii) ab > 0. In this subcases, we have a ≥ b > 0, |a|p+|b|p ≥ |a − b|p. By Lemma 3.1, ϕθ,p(a, b) ≤ ϕp(a, b)≤ 0 . Notice that ϕθ,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, since ϕθ,p(a, b) ≤ ϕp(a, b)≤ 0 , and by [4, Lemma 3.2], and hence we obtain that
|ϕθ,p(a, b)| ≥ |ϕp(a, b)| ≥ (2 − 21p)| min{a, b}|. (28) All the aforementioned inequalities (21)-(28) imply that (20) holds. 2
Proposition 3.1 Let Ψθ,p, ΨNR and Ψα,θ,p be defined as in (13), (12) and (10), respec- tively. Let S be an arbitrary bounded set. Then, for any p > 1, we have
(2− 21p)2ΨNR(x)≤ Ψθ,p(x)≤ (2 + 21p)2ΨNR(x) for all x∈ IRn (29) and
(2− 21p)2ΨNR(x)≤ Ψα,θ,p(x)≤ (αB2+ (2 + 21p)2)ΨNR(x) for all x∈ S, (30) where B is a constant defined by B = max
1≤i≤n
{ sup
x∈S{max {|xi|, |Fi(x)|}}
}
<∞.
Proof. The inequality in (29) is direct by Lemma 3.3 and the definitions of Ψθ,p and ΨNR. In addition, from Lemma 3.3 and the definition of Ψα,θ,p, it follows that
Ψα,θ,p(x)≥(
2− 21p)2
ΨNR(x) for all x∈ IRn.
It remains to prove the inequality on the right hand side of (30). From the proof of [4, Prop. 3.1], we know for each i,
(xiFi(x))+≤ B| min{xi, Fi(x)}| for all x ∈ S. (31) By Lemma 3.3 and (31), for all i = 1, . . . , n and x∈ S,
ψα,θ,p(xi, Fi(x))≤ 1 2
{
αB2+ (2 + 21p)2 }
min{xi, Fi(x)}2
holds for any p > 1. The proof is then complete by the definitions of Ψα,θ,p and ΨNR. 2
From Proposition 3.1, we immediately obtain the following result.
Corollary 3.1 Let Ψθ,p and Ψα,θ,p be defined by (13) and (10), respectively; and S be any bounded set. Then, for any p > 1 and all x∈ S, we have the following inequalities:
(2− 21p)2 (
αB2+ (2 + 21p)2
)Ψα,θ,p(x)≤ Ψθ,p(x)≤ (2 + 21p)2 (2− 21p)2
Ψα,θ,p(x)
where B is the constant defined as in Proposition 3.1.
Since Ψθ,p, ΨNR and Ψα,θ,p have the same order on a bounded set, one will provide a global error bound for the NCP as long as the other one does. As below, we show that Ψα,θ,p provides a global error bound without the Lipschitz continuity of F when α > 0.
Theorem 3.4 Let Ψα,θ,p be defined as in (10). Suppose that F is a uniform P -function with modulus µ > 0. If α > 0, then there exists a constant κ1 > 0 such that
∥x − x∗∥ ≤ κ1Ψα,θ,p(x)14 for all x∈ IRn;
if α = 0 and S is any bounded set, there exists a constant κ2 > 0 such that
∥x − x∗∥ ≤ κ2
( max
{
Ψα,θ,p(x),
√
Ψα,θ,p(x) })1
2
for all x∈ S;
where x∗ = (x∗1,· · · , x∗n) is the unique solution for the NCP.
Proof. By the proof of [4, Theorem 3.4], we have µ∥x − x∗∥2 ≤ max
1≤i≤nτi{(xiFi(x))++ (−Fi(x))++ (−xi)+}, (32) where τi := max{1, x∗i, Fi(x∗)}. We next prove that for all (a, b) ∈ IR2,
(−a)+2
+ (−b)+2 ≤ [ϕθ,p(a, b)]2. (33) To see this, without loss of generality, we assume a≥ b and discuss three cases:
(i) If a≥ 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 that α > 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)
where τ := max {6
α, 6 }
> 0. Combining (34) with (32) and letting ˆτ = max
1≤i≤nτi, we get µ∥x − x∗∥2 ≤ max
1≤i≤nτi{τψα,θ,p(xi, Fi(x))}1/2
≤ ˆττ1/2max
1≤i≤nψα,θ,p(xi, Fi(x))1/2
≤ ˆττ1/2 { n
∑
i=1
{ψα,θ,p(xi, Fi(x)) }1/2
= τ τˆ 1/2Ψα,θ,p(x)1/2.
From this, the first desired result follows immediately by setting κ1 :=[ ˆ
τ τ1/2/µ]1/2
. Suppose that α = 0. From the proof of Proposition 3.1, the inequality (31) holds.
Combining with equations (32)–(33), it then follows that for all x∈ S, µ∥x − x∗∥2 ≤ max
1≤i≤nτi[
B| min{xi, Fi(x)}| + 2(ψθ,p(xi, Fi(x)))1/2]
≤ ˆτ max
1≤i≤n
[√2 ˆB(ψθ,p(xi, Fi(x)))1/2+ 2(ψθ,p(xi, Fi(x)))1/2 ]
≤ (√
2 ˆB + 2)ˆτ (Ψθ,p(x))1/2
= (√
2 ˆB + 2)ˆτ (Ψα,θ,p(x))1/2
≤ (√
2 ˆB + 2)ˆτ (max {
Ψα,θ,p(x),
√
Ψα,θ,p(x) }
) where ˆB = B/(2− 21p), ˆτ = max
1≤i≤nτiand the second inequality is from Lemma 3.3. Letting κ2 :=
[ (√
2 ˆB + 2)ˆτ /µ ]1/2
, we obtain the desired result from the above inequality. 2
The following lemma is needed for the proof of Proposition 3.2, which we suspect is useful in analysis of convergence rate.
Lemma 3.4 For all (a, b)̸= (0, 0) and p > 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− 2p1)2
∀θ ∈ (0, 1].
Proof. If a = 0 or b = 0, the inequality holds obviously. Then we complete the proof by considering three cases: (i) a > 0 and b > 0, (ii) a < 0 and b < 0, and (iii) ab < 0.
Case (i): Since θ ∈ (0, 1] and p > 1, it follows that θ1/p ≤ 1. Now, by the proof of [4, 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 for p > 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 for p > 1. Squaring both sides then leads to the desired inequality.
Case (ii): By similar arguments as in case (i), we obtain 2− 21p ≤ 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 for p > 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 for p > 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 for p > 1 and the desired result is also satisfied. 2
Proposition 3.2 Let ψα,θ,p be given as in (11). Then, for all x∈ IRn and p > 1, ∇aψα,θ,p(x, F (x)) +∇bψα,θ,p(x, F (x)) 2 ≥ 2(
2− 21p)2
Ψθ,p(x) ∀θ ∈ (0, 1].
In particular, for all x belonging to any bounded set S and p > 1, ∇aψα,θ,p(x, F (x)) +∇bψα,θ,p(x, F (x)) 2 ≥ 2(2− 21p)4
(
αB2+ (2 + 21p)2
)Ψα,θ,p(x) ∀θ ∈ (0, 1],
where B is defined as in Proposition 3.1 and
∇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)) :=
(
∇bψα,θ,p(x1, F1(x)), · · · , ∇bψα,θ,p(xn, Fn(x)) )T
. (35)
Proof. The second part of the conclusions is direct by Corollary 3.1 and 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)) and Ψθ,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− 21p)2
ψθ,p(a, b) (36) holds for all (a, b)∈ IR2. When (a, b) = (0, 0), the inequality (36) clearly holds. Suppose (a, b)̸= (0, 0). Then, it follows from equation (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) ∈ IR2,
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) If ab≤ 0, then (ab)+ = 0 and the inequality (38) is clear. If a, b > 0, then by the proof of Lemma 3.4, 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) ∈ IR2 (39)
and ϕθ,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 Lemma 3.4 and equations (37)–(38), we readily obtain the inequality (36) holds for all (a, b)̸= (0, 0). The proof is thus complete. 2
4 Algorithm and Numerical Experiments
In this section, we investigate a derivative free algorithm for complementarity problems based on the new family of NCP-function and its related merit function. In addition, we prove the global convergence of the algorithm.
Algorithm 4.1 (A Derivative Free Algorithm)
Step 0 Given real numbers α > 0, p > 1, θ ∈ (0, 1] and x0 ∈ IRn. Choose σ∈ (0, 1) and ρ, γ ∈ (0, 1). Set k := 0.
Step 1 If Ψα,θ,p(xk) = 0, stop, otherwise go to Step 2.
Step 2 Find the smallest nonnegative integer mk such that
Ψα,θ,p(xk+ ρmkdk(γmk))≤ (1 − σρ2mk)Ψα,θ,p(xk), (40) where dk(γmk) :=−∂Ψα,θ,p(xk, F (xk))
∂b − γmk∂Ψα,θ,p(xk, F (xk))
∂a .
Step 3 Set xk+1 := xk+ ρmkdk(γmk), k := k + 1 and go to Step 1.
Proposition 4.1 Let xk∈ IRnand F be a monotone function. Then the search direction defined in Algorithm 4.1 satisfies the descent condition ∇Ψα,θ,p(xk)Tdk < 0 as long as xk is not a solution of the NCP. Moreover, if F is strongly monotone with modulus µ > 0, then ∇Ψα,θ,p(xk)Tdk<−µ∥dk∥2.
Proof. The proof is similar to the one given in [5, Lemma 4.1]. 2
Proposition 4.2 Suppose that F is strongly monotone. Then the sequence {xk} gener- ated by Algorithm 4.1 has at least one accumulation point and any accumulation point is a solution of the NCP.
