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DOI 10.1007/s10589-015-9781-1

Constructions of complementarity functions and meritfunctions for circular cone complementarity problem

Xin-He Miao1 · Shengjuan Guo1 · Nuo Qi1 · Jein-Shan Chen2

Received: 29 March 2015 / Published online: 20 August 2015

Abstract In this paper, we consider complementarity problem associated with circu- lar cone, which is a type of nonsymmetric cone complementarity problem. The main purpose of this paper is to show the readers how to construct complementarity func- tions for such nonsymmetric cone complementarity problem, and propose a few merit functions for solving such a complementarity problem. In addition, we study the con- ditions under which the level sets of the corresponding merit functions are bounded, and we also show that these merit functions provide an error bound for the circular cone complementarity problem. These results ensure that the sequence generated by descent methods has at least one accumulation point, and build up a theoretical basis for designing the merit function method for solving circular cone complementarity problem.

Keywords Circular cone complementarity problem· Complementarity function · Merit function· The level sets · Strong coerciveness

B

Jein-Shan Chen jschen@math.ntnu.edu.tw Xin-He Miao

xinhemiao@tju.edu.cn Shengjuan Guo gshengjuan@163.com Nuo Qi

qinuo@163.com

1 Department of Mathematics, School of Science, Tianjin University, Tianjin 300072, People’s Republic of China

2 Department of Mathematics, National Taiwan Normal University, Taipei 11677, Taiwan

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1 Motivation and introduction

The general conic complementarity problem is to find an element x ∈ IRnsuch that x∈ K, F(x) ∈ K and x, F(x) = 0, (1) where·, · denotes the Euclidean inner product, F : IRn → IRn is a continuously differentiable mapping,K represents a closed convex cone, and Kis the dual cone ofK given by

K:= {v ∈ IRn| v, x ≥ 0, ∀x ∈ K}.

WhenK is a symmetric cone, the problem (1) is called the symmetric cone comple- mentarity problem [12,14,18,20]. In particular, whenK is the so-called second-order cone which is defined as

Kn:= {(x1, x2) ∈ IR × IRn−1| x2 ≤ x1},

the problem (1) reduces to the second-order cone complementarity problem [1,3–5,10, 11]. In contrast to symmetric cone programming and symmetric cone complementarity problem, we are not familiar with their nonsymmetric counterparts. Referring the reader to [16,19] and the bibliographies therein, we observe that there is no any unified way to handle nonsymmetric cone constraints, and the study on each item for such problems usually uses certain specific features of the nonsymmetric cones under consideration.

In this paper, we pay attention to a special nonsymmetric coneK for problem (1).

In particular, we focus on the case of K being the circular cone defined as below, which enables the problem (1) reduce to the circular cone complementarity problem (CCCP for short). Indeed in IRn, the circular cone [7,23] is a pointed closed convex cone having hyper-spherical sections orthogonal to its axis of revolution about which the cone is invariant to rotation. Let its half-aperture angle beθ with θ ∈ (0,π2). Then, the circular cone denoted byLθ can be expressed as

Lθ :=

x= (x1, x2) ∈ IR × IRn−1| x cos θ ≤ x1



= 

x= (x1, x2) ∈ IR × IRn−1| x2 ≤ x1tanθ

. (2)

Whenθ = π4, the circular cone is exactly the second-order cone, which means the circular cone complementarity problem is actually the second-order cone comple- mentarity problem. Thus, the circular cone complementarity problem (CCCP) can be viewed as the generalization of the second-order cone complementarity problem.

Moreover, the CCCP includes the KKT system of the circular programming problem [13] as a special case. For real world applications of optimization problems involving circular cones, please refer to [6]. Note that in [23], Zhou and Chen characterize the relation between circular coneLθand second-order cone as follows:

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Lθ = A−1Kn and Kn= ALθ with A=

tanθ 0 0 I

 .

In other words, for any x = (x1, x2) ∈ IR × IRn−1and y= (y1, y2) ∈ IR × IRn−1, there have

x∈ Lθ ⇐⇒ Ax ∈ Kn, y∈ Lθ ⇐⇒ A−1y∈ Kn. (3) Relation (3) indicates that after scaling the circular cone complementarity problem and the second-order cone complementarity problem are equivalent. However, when dealing with the circular cone complementarity problem, this approach may not be acceptable from both theoretical and numerical viewpoints. Indeed, if the appropriate scaling is not found or checked, some scaling step can cause undesirable numeri- cal performance due to round-off errors in computers, which has been confirmed by experiments. Moreover, it usually need to exploits its associated merit functions or complementarity functions, which plays an important role in tackling complementar- ity problem. To this end, we are devoted to seeking a way to construct complementarity functions and merit functions for the circular cone complementarity problem directly.

Thus, we pay our attention to the circular cone complementarity problem and the structure ofLθ mainly. There is another relationship between the circular cone and the (nonsymmetric) matrix cone introduced in [8,9], where the authors study the epi- graph of six different matrix norms, such as the Frobeninus norm, the lnorm, l1

norm, the spectral or the operator norm, the nuclear norm, the Ky Fan k-norm. If we regard a matrix as a high-dimensional vector, then the circular cone is equivalent to the matrix cone with Frobeninus norm, see [24] for more details.

While there have been much attention to the symmetric cone complementarity problem and the second-order cone complementarity problem, the study about non- symmetric cone complementarity problem is very limited. The main difficulty is that the idea for constructing complementarity functions (C-functions for short) and merit functions is not clear. Hence, The main goal of this paper is showing the readers how to construct C-functions and merit functions for such complementarity problem, and studying the properties of these merit functions. To our best knowledge, the idea is new and we believe that it will help in analyzing other types of nonsymmetric cone complementarity problems.

Recall that for solving the problem (1), a popular approach is to reformulate it as an unconstrained smooth minimization problem or a system of nonsmooth equations.

In this category of methods, it is important to adapt a merit function. Officially, a merit function for the circular cone complementarity problem is a function h: IRn[0, +∞), provided that

h(x) = 0 ⇐⇒ x solves the CCCP (1).

Hence, solving the problem (1) is equivalent to handling the unconstrained minimiza- tion problem

xmin∈IRnh(x)

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with the optimal value zero. For constructing the merit functions in finite dimensional vector space, please refer to [17]. Until now, for solving symmetric cone comple- mentarity problem, a number of merit functions have been proposed. Among them, one of the most popular merit functions is the natural residual (NR) merit function

N R : IRn→ IR, which is defined as

N R(x) := 1

2 φNR(x, F(x)) 2= 1

2 x − (x − F(x))+ 2,

where(·)+denotes the projection onto the symmetric coneK. It is well known that

N R(x) = 0 if and only if x is a solution to the symmetric cone complementarity problem. In this paper, we present two classes of complementarity functions and four types of merit functions for the circular cone complementarity problem. Moreover, we investigate the properties of these proposed merit functions, and study conditions under which these merit functions provide bounded level sets. Note that such prop- erties will guarantee that the sequence generated by descent methods has at least one accumulation point, and build up a theoretical basis for designing the merit function method for solving circular cone complementarity problem.

2 Preliminaries

In this section, we briefly review some basic concepts and background materials about the circular cone and second-order cone, which will be extensively used in subsequent analysis.

As defined in (2), the circular coneLθis a pointed closed convex cone and has a rev- olution axis which is the ray generated by the canonical vector e1:= (1, 0, . . . , 0)T ∈ IRn. Its dual cone denoted byLθis given as

Lθ := {y = (y1, y2) ∈ IR × IRn−1| y sin θ ≤ y1}.

Note that the circular coneLθ is not a self-dual cone whenθ = π4, that is,Lθ = Lθ, wheneverθ = 45. Hence,Lθ is not a symmetric cone forθ ∈

0,π2

\{π4}. It is also known from [23] that the dual cone ofLθcan be expressed as

Lθ = {y = (y1, y2) ∈ IR × IRn−1| y2 ≤ y1cotθ} = Lπ2−θ.

Now, we talk about the projection ontoLθandLθ. To this end, we let x+denote the projection of x onto the circular coneLθ, and xbe the projection of−x onto the dual coneLθ. With these notations, for any x ∈ IRn, it can be verified that x= x+− x. Moreover, due to the special structure of the circular coneLθ, the explicit formula of projection of x∈ IRnontoLθ is obtained in [23] as below:

x+=

⎧⎨

x if x∈ Lθ, 0 if x ∈ −Lθ,

u otherwise, (4)

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where

u=

⎢⎣

x1+ x2 tan θ 1+ tan2θ

x1+ x2 tan θ 1+ tan2θ tanθ

 x2

x2

⎦ .

Similarly, we can obtain the expression of xas below:

x=

⎧⎨

0 if x ∈ Lθ,

−x if x ∈ −Lθ,

w otherwise, (5)

where

w =

⎢⎣−x1− x2 cot θ 1+ cot2θ

x1− x2 cot θ 1+ cot2θ cotθ

 x2

x2

⎦ .

From the expressions (4)–(5) for x+and x, it is easy to verity thatx+, x = 0 for any x ∈ IRn.

Next, we introduce the Jordan product associated with second-order cone. As men- tioned earlier, the SOC in IRn(also called Lorentz cone or ice-cream cone) is defined by

Kn := {x = (x1, x2) ∈ IR × IRn−1| x2 ≤ x1}.

It is well known that the dual cone ofKnis itself, and the second-order coneKnbelongs to a class of symmetric cones. In addition,Knis a special case ofLθ corresponding toθ = π4. In fact, there is a relationship betweenLθ andKn, which is described in (3). In the SOC setting, there is so-called Jordan algebra associated with SOC. More specifically, for any x= (x1, x2) ∈ IR × IRn−1and y= (y1, y2) ∈ IR × IRn−1, in the setting of the SOC, the Jordan product of x and y is defined as

x◦ y :=

x, y

y1x2+ x1y2

 .

The Jordan product “◦”, unlike scalar or matrix multiplication, is not associative. The identity element under Jordan product is e= (1, 0, . . . , 0)T ∈ IRn. In this paper, we write x2to mean x◦ x. It is known that x2∈ Knfor any x ∈ IRn, and if x ∈ Kn, there exists a unique vector denoted by x12 inKnsuch that(x12)2= x21 ◦ x12 = x. For any x ∈ IRn, we denote|x| :=

x2and x+socmeans the orthogonal projection of x onto the second-order coneKn. Then, it follows that x+soc = x+ |x|

2 . For further details regarding the SOC and Jordan product, please refer to [1,3,5,10].

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Lemma 2.1 ([10, Proposition 2.1]) For any x, y ∈ IRn, the following holds:

x∈ Kn, y ∈ Kn, and x, y = 0 ⇐⇒ x ∈ Kn, y ∈ Kn, and x ◦ y = 0.

With the help of (3) and Lemma2.1, we obtain the following theorem which explains the relationship between SOCCP and CCCP.

Theorem 2.1 Let A =

tanθ 0 0 I



. For any x = (x1, x2) ∈ IR × IRn−1 and y = (y1, y2) ∈ IR × IRn−1, the following are equivalent:

(a) x ∈ Lθ, y ∈ Lθandx, y = 0.

(b) Ax∈ Kn, A−1y∈ KnandAx, A−1y = 0.

(c) Ax∈ Kn, A−1y∈ Knand Ax◦ A−1y= 0.

(d) x ∈ Lθ, y ∈ Lθand Ax◦ A−1y= 0.

In each case, elements x and y satisfy the condition that either y2is a multiple of x2

or x2is a multiple of y2.

Proof From the relation betweenKnandLθgiven as in (3), we know that x∈ Lθ ⇐⇒ Ax ∈ Kn and y∈ Lθ ⇐⇒ A−1y∈ Kn. Moreover, under condition (a), there holds

Ax, A−1y = A−1Ax, y = x, y = 0.

Hence, it follows that (a) and (b) are equivalent. The equivalence of (b) and (c) has been shown in Lemma2.1. In addition, based on the relation betweenKn andLθ again, the equivalence of (c) and (d) is obvious.

Now, under condition (a), we prove that either y2is a multiple of x2or x2is a multiple of y2. To see this, note that x ∈ Lθand y∈ Lθ which gives

x2 ≤ x1tanθ and y2 ≤ y1cotθ.

This together withx, y = 0 yields 0= x, y

= x1y1+ x2, y2

≥ x1y1− x2 y2

≥ x1y1− x1y1

= 0

which impliesx2, y2 = x2 y2 . This says that either y2is a multiple of x2or x2

is a multiple of y2. Thus, the proof is complete. 

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3 C-functions for CCCP

In this section, we define C-functions for CCCP and the product of elements in the setting of the circular cone. Moreover, based on the product of elements, we con- struct some C-functions which play an important role in solving the circular cone complementarity problems by merit function methods.

Definition 3.1 Given a mappingφ : IRn× IRn → IRn, we callφ an C-function for CCCP if, for any(x, y) ∈ IRn× IRn, it satisfies

φ(x, y) = 0 ⇐⇒ x ∈ Lθ, y ∈ Lθ, x, y = 0.

Whenθ =π4, an C-function for CCCP reduces to an C-function for SOCCP, i.e., φ(x, y) = 0 ⇐⇒ x ∈ Kn, y ∈ Kn, x, y = 0.

Two popular and well-known C-functions for SOCCP are Fischer-Burmeister (FB) function and natural residual (NR) function:

φFB(x, y) =

x2+ y21/2

− (x + y), φNR(x, y) = x − (x − y)soc+ .

We may ask whether we can modify the above two C-functions for SOCCP to form C-functions for CCCP. The answer is affirmative. In fact, we consider

φFB(x, y) :=

(Ax)2+ (A−1y)21

2 − (Ax + A−1y), φNR(x, y) := Ax − [Ax − A−1y]soc+ .

Then, these two functions are C-functions for CCCP.

Proposition 3.1 Let φFBandφNRbe defined as above where(Ax)2equals(Ax)◦(Ax) under Jordan product. Then, φFBandφNR are both C-functions for CCCP.

Proof In view of Theorem2.1and Definition3.1, it is not hard to verify that

φFB(x, y) = 0 ⇐⇒ x ∈ Lθ, y ∈ Lθ, x, y = 0, φNR(x, y) = 0 ⇐⇒ x ∈ Lθ, y ∈ Lθ, x, y = 0,

which says that these two functions are C-functions for CCCP. 

We point out that if we consider directly the FB functionφFB(x, y) for CCCP, unfor- tunately, it cannot be C-function for CCCP because x2is not well-defined associated

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with the circular coneLθ for any x ∈ IRn. More specifically, because x2is defined under the Jordan product in the setting of SOC, i.e.,

x2:= x ◦ x =

x, y

x1y2+ y1x2

 ,

it follows that x2∈ Kn, which implies x2may not belong toLθ orLθ. Furthermore, whenφFB(x, y) = 0, we have x + y =

x2+ y21

2 ∈ Kn, which yields that x, y ∈ Kn. This says that either x /∈ Lθor y /∈ Lθ. All the above explains that the FB functionφFB

cannot be an C-function for CCCP. Nonetheless, the NR functionφNR : IRn× IRn→ IRngiven by

φNR(x, y) := x − (x − y)+ (6)

is always an C-function for CCCP. Moreover, it is also an C-function for general cone complementarity problem, see [11, Proposition 1.5.8].

Are there any other types of C-functions for CCCP and how to construct an C- function for CCCP? As mentioned earlier, The FB functionφFB cannot serve as C- functions for CCCP because “x2” is not well-defined in the setting of circular cone.

This inspires us to define a special product associated with circular cone, and find other C-functions for CCCP.

For any x = (x1, x2) ∈ IR × IRn−1and y= (y1, y2) ∈ IR × IRn−1, we define one type of product of x and y as follows:

x• y =

x1

x2



y1

y2



=

x, y

max{tan2θ, 1} x1y2+ max{cot2θ, 1} y1x2

 . (7)

From the above product and direct calculation, it is easy to verify that

x • y, z = x, z • y, ∀z ∈ IRnwithθ ∈ 0,π

4



(8) and

x • y, z = y, x • z, ∀z ∈ IRnwithθ ∈π 4

2

. (9)

Moreover, we also obtain the following inequalities which are crucial to establishing our main results.

Lemma 3.1 For any x, y ∈ IRn,

(a) ifθ ∈ (0,π4], we have x, x+• (−y) ≤ 0;

(b) ifθ ∈ [π4,π2), we have (−y)+, x+• (−y) ≤ 0.

Proof (a) Whenθ ∈ (0,π4], let x+:= (s, u) ∈ IR × IRn−1, x:= (t, v) ∈ IR × IRn−1 and(−y):= (k, w) ∈ IR × IRn−1. For the elements x+, xand(−y), if there exist at least one in them is zero, it is easy to obtain

x, x+• (−y) = 0.

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If all the three elements are not equal to zero, from the definition of x+, x, and(−y), we have k cotθ ≥ w , s tan θ = u , t cot θ = v and

u= αv or v = αu with α < 0.

Without loss of generality, we consider the case u= αv with α < 0 for the following analysis. In fact, using this, we know that

x, x+• (−y)

= stk + tu, w + sv, w + ku, v cot2θ

= u v k − k u v cot2θ − u v, w tan θ + u v, w cot θ

= (1 − cot2θ)k u v − (1 − cot2θ)( u v, w tan θ)

= (1 − cot2θ)[k u v − u v, w tan θ]

≤ (1 − cot2θ)[k u v − u v w tan θ]

= (1 − cot2θ) u v [k − w tan θ]

≤ 0.

Here the second equality is true due to αt = α v tan θ = − u tan θ. The last inequality holds due to k cotθ ≥ w and θ ∈ (0,π4]. Hence, the desired result follows.

(b) Whenθ ∈ [π4,π2), with the same skills, we also conclude that

(−y)+, x+• (−y) ≤ 0.

Then, the desired result follows. 

Besides the inequalities in Lemma3.1, “•” defined as in (7) plays the similar role like what “◦” does in the setting of second-order cone. This is shown as below.

Theorem 3.1 For any x = (x1, x2) ∈ IR × IRn−1and y= (y1, y2) ∈ IR × IRn−1, the following statements are equivalent:

(a) x ∈ Lθ, y ∈ Lθ andx, y = 0.

(b) x ∈ Lθ, y ∈ Lθ and x• y = 0.

In each case, x and y satisfy the condition that either y2is a multiple of x2or x2is a multiple of y2.

Proof In view of Theorem2.1, we know that part (a) is equivalent to x∈ Lθ, y ∈ Lθ and Ax◦ A−1y= 0.

To proceed the proof, we discuss the following two cases.

Case 1 Forθ ∈ (0,π4], from the definition of the product of x and y, we have

x• y =

x, y

x1y2+ cot2θ y1x2



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which implies

Ax ◦ A−1y=

x, y

x1tanθ y2+ cot θ y1x2



=

1 0 0(tan θ)I

 (x • y).

This together with Theorem2.1yields the conclusion.

Case 2 Forθ ∈ [π4,π2), from the definition of the product of x and y again, we have

x• y =

x, y

tan2θ x1y2+ y1x2



which says

Ax◦ A−1y=

1 0 0(cot θ)I

 (x • y).

Then, applying Theorem2.1again, the desired result follows. 

Based on the product x • y of x and y. we now introduce a class of functions φp : IRn× IRn → IRn, which is called the penalized natural residual function and defined as

φp(x, y) = x − (x − y)++ p (x+• (−y)) , p > 0. (10) Note that when p= 0, φp(x, y) reduces to φNR(x, y). In the following, we show that the functionφp is an C-function for CCCP. To achieve the conclusion, a technical lemma is needed.

Lemma 3.2 Letφp: IRn×IRn → IRnbe defined as in (10). Then, for any x, y ∈ IRn, we have

φp(x, y) ≥ max { x , (−y)+ } .

Proof First, we prove that φp(x, y) ≥ x . To see this, we observe that

φp(x, y) 2

= x − (x − y)++ p x+• (−y), x − (x − y)++ p x+• (−y)

= x+− x− (x − y)++ p x+• (−y), x+− x− (x − y)++ p x+• (−y)

= x 2+ x+− (x − y)++ p x+• (−y) 2− 2 x, x+− (x − y)++ p x+• (−y)

≥ x 2− 2x, x+ + 2 x, (x − y)+ − 2 x, p x+• (−y)

≥ x 2− 2p x, x+• (−y) .

Here, the last inequality is true due to x+, (x − y)+∈ Lθ, x ∈ Lθ, x+, x = 0 and the relation betweenLθandLθ. Whenθ ∈ (0,π4], by Lemma3.1(a), we have

x, x+• (−y) ≤ 0.

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Whenθ ∈ [π4,π2), from Eq. (9), we have

x, x+• (−y) = (−y), x+• x = 0

where the second equality holds due to x+• x= 0. In summary, from all the above, we prove that

φp(x, y) 2≥ x 2.

With similar arguments, we also obtain φp(x, y) 2

= x − (x − y)++ p x+• (−y), x − (x − y)++ p x+• (−y)

= y − (x − y)+ p x+• (−y), y − (x − y)+ p x+• (−y)

= (−y)− (−y)+− (x − y)+ p x+• (−y), (−y)− (−y)+− (x − y) +px+ • (−y)

= (−y)+ 2+ (−y)− (x − y)+ p x+• (−y) 2− 2(−y)+, (−y)

−(x − y)+ px+ • (−y)

≥ (−y)+ 2−2(−y)+, (−y)+2(−y)+, (x −y)−2(−y)+, p x+• (−y)

≥ (−y)+ 2− 2p (−y)+, x+• (−y)

≥ (−y)+ 2,

where the second inequality holds due to due to(−y)+ ∈ Lθ, (−y), (x − y)Lθ, (−y)+, (−y) = 0 and the relation between Lθ andLθ. The last inequality holds due to equation (8) and Lemma3.1(b). Therefore, we prove that φp(x, y) ≥

(−y)+ . Then, the proof is complete. 

Remark 3.1 From the proof of Lemma3.2, it also can be seen that φNR(x, y) ≥ max{ x , (−y)+ }.

Theorem 3.2 Letφp : IRn× IRn → IRn be defined as in (10). Then,φp is an C- function for CCCP, i.e., for any x, y ∈ IRn,

φp(x, y) = 0 ⇐⇒ x ∈ Lθ, y ∈ Lθ andx, y = 0.

Proof “⇒” Suppose that φp(x, y) = 0. If either x /∈ Lθor y /∈ Lθ, applying Lemma 3.2yields

φp(x, y) ≥ max{ x , (−y)+ } > 0.

This contradicts withφp(x, y) = 0. Hence, there must have x ∈ Lθand y∈ Lθ. Next, we argue thatx, y = 0. To see this, we consider the first component of φp(x, y),

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which is denoted by

φp(x, y)

1. In other words,

φp(x, y)

1=

x− (x − y)++ p x • y

1

=

⎧⎨

y1+ p x, y if x − y ∈ Lθ, x1+ p x, y if x − y ∈ −Lθ, w + p x, y otherwise, where

w = x1x1− y1+ x2− y2 tan θ

1+ tan2θ = x1tan2θ + y1− x2− y2 tan θ

1+ tan2θ .

Since x ∈ Lθ and y∈ Lθ, it follows that x1, y1≥ 0, x, y ≥ 0 and x1tan2θ + y1− x2− y2 tan θ

1+ tan2θ ≥ tanθ(x1tanθ − x2 + y1cotθ − y2 )

1+ tan2θ ≥ 0.

This together withφp(x, y) = 0 gives px, y = 0. Thus, we conclude that x, y = 0 because p> 0.

“⇐” Suppose that x ∈ Lθ, y ∈ Lθ andx, y = 0. Since φNR is always an C- function for CCCP, we have x − (x − y)+ = 0. Using Theorem 3.1again yields x+• (−y)= x • y = 0, which says φp(x, y) = 0. 

Remark 3.2 In fact, for any x = (x1, x2) ∈ IR ×IRn−1and y= (y1, y2) ∈ IR ×IRn−1, we define another type of product of x and y as follows:

x• y =

x1

x2



y1

y2



=

x, y

min{tan2θ, 1} x1y2+ min{cot2θ, 1} y1x2

 .

With the same skills, we may obtain the same results.

Motivated by the construction ofφpgiven as in (10), we consider another function φr : IRn× IRn → IRndefined by

φr(x, y) = x − (x − y)++ r (x • y)+ r > 0, (11)

where := Lθ∩Lθ =

Lθ ifθ ∈ (0,π4],

Lθ ifθ ∈ [π4,π2). We point out that the functionφrdefined as in (11) is not an C-function for CCCP. The reason come from that ifφr(x, y) = 0, we haveφNR(x, y) = x − (x − y)+= −r (x • y)+. Combining with the expression ofφp, this implies that

−r (x • y)++ p (x+• (−y)) = 0

due to(x • y)+∈  = Lθ∩ Lθand x+• (−y) /∈ Kn⊇ Lθ(orLθ) whenθ ∈ (0,π4] (orθ ∈ [π4,π2)). This explains that φp(x, y) = 0, which contradicts φp(x, y) being an C-function for CCCP.

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However, there is a merit function related toφrwhich possesses property of bounded level sets. We will explore it in next section.

4 Merit functions for circular cone complementarity problem

In this section, based on the product (7) of x and y in IRn, we propose four classes of merit functions for the circular cone complementarity problem and investigate their important properties, respectively.

First, we recall that a function F : IRn → IRn is said to be monotone if, for any x, y ∈ IRn, there holds

x − y, F(x) − F(y) ≥ 0;

and strictly monotone if, for any x= y, the above inequality holds strictly; and strongly monotone with modulusρ > 0 if, for any x, y ∈ IRn, the following inequality holds

x − y, F(x) − F(y) ≥ ρ x − y 2.

The following technical lemma is crucial for achieving the property of bounded level sets.

Lemma 4.1 Suppose that CCCP has a strictly feasible point ¯x, i.e., ¯x ∈ int(Lθ) and F( ¯x) ∈ int(Lθ) and that F is a monotone function. Then, for any sequence {xk} satisfying

xk → ∞, lim sup

k→∞

xk < ∞ and lim sup

k→∞



−F xk



+

 < ∞,

we have

xk, F

xk

→ ∞ and

x+k,

−F xk



→ ∞.

Proof Since F is monotone, for all xk ∈ IRn, we know

xk− ¯x, F xk

− F( ¯x)

≥ 0,

which says

xk, F

xk

+  ¯x, F( ¯x) ≥

xk, F( ¯x) +

¯x, F xk

. (12)

(14)

Using xk= x+k − xk and F xk

=

−F xk

−

−F xk

+, it follows from (12) that

xk, F

xk

+  ¯x, F( ¯x)

≥

x+k, F( ¯x)

−

xk, F( ¯x) +

¯x,

−F xk



¯x,

−F xk



+

. (13)

We look into the first term in the right-hand side of (13).

x+k, F( ¯x)

= x+k



1[ f( ¯x)]1+

x+k



2, [ f ( ¯x)]2

≥ x+k



1[ f( ¯x)]1− x+k



2

 ·[ f( ¯x)]2

≥ x+k



1[ f( ¯x)]1− x+k



1tanθ[ f( ¯x)]2

= x+k



1 [ f( ¯x)]1− tan θ[ f( ¯x)]2!. (14) Note that xk = x+k − xk, it gives x+k ≥ xk − xk . From the assumptions on{xk}, i.e., xk → ∞, and lim supk→∞ xk < ∞, we see that x+k → ∞, and hence [x+k]1 → ∞. Because CCCP has a strictly feasible point ¯x, we have [ f( ¯x)]1− tan θ [ f ( ¯x)]2 > 0, which together with (14) implies that

x+k, F( ¯x)

→ ∞ (k → ∞). (15)

On the other hand, we observe that lim sup

k→∞ xk, F( ¯x) ≤ lim sup

k→∞ xk F( ¯x) < ∞ lim sup

k→∞  ¯x,

−F xk



+ ≤ lim sup

k→∞ ¯x 

−F xk



+ < ∞ and ¯x,

−F xk

 ≥ 0. All of these together with (13) and (15) yield

xk, F

xk

→ ∞,

which is the first part of the desired result.

Next, we prove that

x+k,

−F xk

→ ∞. Suppose not, that is, limk→∞"

x+k,

−F xk

< ∞. Then, we obtain

x+k,

−F xk

x+k =

# xk+ xk+ ,

−F xk



\$

→ 0.

(15)

This means that there exists ¯x ∈ IRnsuch that x+k

x+k¯x+

¯x+ and

¯x+

¯x+ , (−F( ¯x))

= 0. (16)

Denote z:= ¯x+

¯x+ and apply Theorem3.1, there existsα ∈ IR such that

(−F( ¯x))

2= αz2 or αz2=

(−F( ¯x))

2.

It is obvious that z ∈ Lθ and(−F( ¯x)) ∈ Lθ. Hence, Eq. (16) implies thatα < 0, which says that z2and

(−F( ¯x))

2 are in opposite direction to each other. From the expression of(−F( ¯x))+and(−F( ¯x)) again, it follows that

(−F( ¯x))+

2and

(−F( ¯x))

2are in the opposite direction, to each other. These conclude that z2and

(−F( ¯x))+

2 are in the same direction, which means [ ¯x+]2 and

(−F( ¯x))+

2 are also in the same direction. Now, combining with the fact that ¯x+, (−F( ¯x))+ ∈ Lθ, we have

 ¯x+, (−F( ¯x))+ ≥ 0.

Similarly, by the the relation between ¯x+and ¯x, we know[ ¯x]2and[(−F( ¯x))]2

are in the same direction. Then, combining with ¯x, (−F( ¯x))∈ Lθ, it leads to

 ¯x, (−F( ¯x)) ≥ 0.

Moreover, writing out the expression for ¯x, F( ¯x), we see that

 ¯x, F( ¯x)= ¯x+, (−F( ¯x))− ¯x+, (−F( ¯x))+− ¯x, (−F( ¯x))+ ¯x, (−F( ¯x))+.

Note that the second and third terms of the right-hand side are nonpositive and the fourth is bounded from above. Hence, from the assumptions limk→∞

x+k,

−F xk

< ∞, we conclude that  ¯x, F( ¯x) < ∞, which contradict

 ¯x, F( ¯x) = lim

k→∞

xk, F

xk

= ∞.

Thus, we prove that

x+k,

−F xk

→ ∞. 

4.1 The first class of merit functions

For any x ∈ IRn, from the analysis of the Sect.3, we know that the functionφpandφNR

are complementarity function for CCCP. In this subsection, we focus on the property of bounded level sets of merit functions based on φNR andφp with the product of

(16)

elements, which is a property to guarantee that the existence of accumulation points of sequence generated by some descent algorithms.

Theorem 4.1 Letφpbe defined as in (10). Suppose that CCCP has a strictly feasible point and that F is monotone. Then, the level set

Lp(α) = {x ∈ IRn|φp(x, F(x)) ≤ α}

is bounded for allα ≥ 0.

Proof We prove this result by contradiction. Suppose there exists an unbounded sequence{xk} ⊂ Lp(α) for some α ≥ 0. If xk → ∞ or 

−F xk

+ → ∞, by Lemma3.2, we have φp(xk, F

xk

) → ∞, which contradicts φp(xk, F xk α. On the other hand, if ) ≤

lim sup

k→∞ xk < ∞ and lim sup

k→∞



−F xk



+

 < ∞, it follows from Lemma4.1that

x+k,

−F xk

→ ∞. From the proof of Lemma 4.1, there exists a constantκ0such that

φNR

 xk, f

xk

1

[xk+]1− κ0 if xk− F

xk

∈ −Lθ,

−F xk



1− κ0 if xk− F

xk

∈ Lθ, [x+k]1tan2θ+

−F xk



 1

− [x+k]2 tan θ− [

−F xk



]2 tan θ 1+tan2θ

2κ1+tan0(1+tan θ)2θ , if xk− F xk

/∈ Lθ∪ −Lθ,

which means lim inf

φNR(xk, f  xk

)

1> −∞. Hence, it follows that

φp

 xk, f

xk



1 =  φNR

 xk, f 

xk



1+



xk+•

−F xk





1

=  φNR

 xk, f 

xk



1+

x+k,

−F xk



→ ∞, where the limit comes from

x+k,

−F xk



→ ∞ and lim inf φNR

 xk, f 

xk



1> −∞.

Thus, we obtain that φp(xk, F xk

) → ∞ which contradicts φp(xk, F xk

α. Then, the proof is complete. ) ≤

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