to appear in Computational Optimization and Applications, 2015
Constructions of complementarity functions and merit functions for circular cone complementarity problem
Xin-He Miao1
Department of Mathematics School of Science Tianjin University Tianjin 300072, P.R. China E-mail: xinhemiao@tju.edu.cn
Shengjuan Guo Department of Mathematics
School of Science Tianjin University Tianjin 300072, P.R. China E-mail: gshengjuan@163.com
Nuo Qi
Department of Mathematics School of Science Tianjin University Tianjin 300072, P.R. China
E-mail: qinuo@163.com
Jein-Shan Chen 2 Department of Mathematics National Taiwan Normal University
Taipei 11677, Taiwan E-mail: jschen@math.ntnu.edu.tw
March 30, 2015
1The author’s work is supported by National Young Natural Science Foundation (No. 11101302 and No. 61002027) and National Natural Science Foundation of China (No. 11471241).
2Corresponding author. The author’s work is supported by Ministry of Science and Technology, Taiwan.
(revised on August 7, 2015)
Abstract In this paper, we consider complementarity problem associated with circular 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 functions for such nonsymmetric cone complementarity problem, and propose a few merit functions for solving such a complementarity problem. In addition, we study the conditions 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.
1 Motivation and Introduction
The general conic complementarity problem is to find an element x ∈ IRn such that x ∈ K, F (x) ∈ K∗ and hx, F (x)i = 0, (1) where h·, ·i denotes the Euclidean inner product, F : IRn→ IRn is a continuously differ- entiable mapping, K represents a closed convex cone, and K∗ is the dual cone of K given by
K∗ := {v ∈ IRn| hv, xi ≥ 0, ∀x ∈ K}.
When K is a symmetric cone, the problem (1) is called the symmetric cone complemen- tarity problem [12, 14, 18, 20]. In particular, when K is the so-called second-order cone which is defined as
Kn:= {(x1, x2) ∈ IR × IRn−1| kx2k ≤ x1},
the problem (1) reduces to the second-order cone complementarity problem [1, 3, 4, 5, 10, 11]. In contrast to symmetric cone programming and symmetric cone complemen- tarity 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 prob- lems usually uses certain specific features of the nonsymmetric cones under consideration.
In this paper, we pay attention to a special nonsymmetric cone K 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 be θ with θ ∈ (0,π2). Then, the circular cone denoted by Lθ can be expressed as
Lθ := x = (x1, x2) ∈ IR × IRn−1| kxk cos θ ≤ x1
(2)
= x = (x1, x2) ∈ IR × IRn−1| kx2k ≤ x1tan θ .
When θ = π4, the circular cone is exactly the second-order cone, which means the circular cone complementarity problem is actually the second-order cone complementarity prob- lem. 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 cone Lθ and second-order cone as follows:
Lθ = A−1Kn and Kn= ALθ with A =tan θ 0
0 I
.
In other words, for any x = (x1, x2) ∈ IR × IRn−1 and y = (y1, y2) ∈ IR × IRn−1, there have
x ∈ Lθ ⇐⇒ Ax ∈ Kn, y ∈ L∗θ ⇐⇒ 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 numerical 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 complementarity 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 of Lθ mainly. There is another relationship between the circular cone and the (nonsymmetric) matrix cone introduced in [8, 9], where the authors study the epigraph of six different matrix norms, such as the Frobeninus norm, the l∞ norm, 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 prob- lem and the second-order cone complementarity problem, the study about nonsymmetric 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 be- lieve 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 minimization problem
x∈IRminnh(x)
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 complementarity 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 ΨN R : IRn → IR, which is defined as
ΨN R(x) := 1
2kφNR(x, F (x))k2 = 1
2kx − (x − F (x))+k2,
where (·)+ denotes the projection onto the symmetric cone K. It is well known that ΨN R(x) = 0 if and only if x is a solution to the symmetric cone complementarity prob- lem. 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 properties 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 cir- cular 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 cone Lθ is a pointed closed convex cone and has a revolution axis which is the ray generated by the canonical vector e1 := (1, 0, · · · , 0)T ∈ IRn. Its dual cone denoted by L∗θ is given as
L∗θ := {y = (y1, y2) ∈ IR × IRn−1| kyk sin θ ≤ y1}.
Note that the circular cone Lθ is not a self-dual cone when θ 6= π4, that is, L∗θ 6= Lθ, whenever θ 6= 45◦. Hence, Lθ is not a symmetric cone for θ ∈ 0,π2 \{π4}. It is also known from [23] that the dual cone of Lθ can be expressed as
L∗θ = {y = (y1, y2) ∈ IR × IRn−1| ky2k ≤ y1cot θ} = Lπ
2−θ.
Now, we talk about the projection onto Lθ and L∗θ. To this end, we let x+ denote the projection of x onto the circular cone Lθ, and x− be the projection of −x onto the dual cone L∗θ. With these notations, for any x ∈ IRn, it can be verified that x = x+− x−. Moreover, due to the special structure of the circular cone Lθ, the explicit formula of projection of x ∈ IRn onto Lθ is obtained in [23] as below:
x+=
x if x ∈ Lθ, 0 if x ∈ −L∗θ, u otherwise,
(4)
where
u =
x1 + kx2k tan θ 1 + tan2θ
x1+ kx2k tan θ 1 + tan2θ tan θ
x2 kx2k
. Similarly, we can obtain the expression of x− as below:
x−=
0 if x ∈ Lθ,
−x if x ∈ −L∗θ, w otherwise,
(5)
where
w =
−x1− kx2k cot θ 1 + cot2θ
x1− kx2k cot θ 1 + cot2θ cot θ
x2
kx2k
.
From the expressions (4)-(5) for x+ and x−, it is easy to verity that hx+, x−i = 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| kx2k ≤ x1}.
It is well known that the dual cone of Kn is itself, and the second-order cone Kn belongs to a class of symmetric cones. In addition, Kn is a special case of Lθ corresponding to θ = π4. In fact, there is a relationship between Lθ and Kn, 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−1 and y = (y1, y2) ∈ IR × IRn−1, in the setting of the SOC, the Jordan product of x and y is defined as
x ◦ y :=
hx, yi 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 x2 to mean x ◦ x. It is known that x2 ∈ Kn for any x ∈ IRn, and if x ∈ Kn, there exists a unique vector denoted by x12 in Kn such that (x12)2 = x12 ◦ x12 = x. For any x ∈ IRn, we denote |x| := √
x2 and xsoc+ means the orthogonal projection of x onto the second-order cone Kn. Then, it follows that xsoc+ = x + |x|
2 . For further details regarding the SOC and Jordan product, please refer to [1, 3, 5, 10].
Lemma 2.1. ([10, Proposition 2.1]) For any x, y ∈ IRn, the following holds:
x ∈ Kn, y ∈ Kn, and hx, yi = 0 ⇐⇒ x ∈ Kn, y ∈ Kn, and x ◦ y = 0.
With the help of (3) and Lemma 2.1, we obtain the following theorem which explains the relationship between SOCCP and CCCP.
Theorem 2.1. Let A =tan θ 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 ∈ L∗θ and hx, yi = 0.
(b) Ax ∈ Kn, A−1y ∈ Kn and hAx, A−1yi = 0.
(c) Ax ∈ Kn, A−1y ∈ Kn and Ax ◦ A−1y = 0.
(d) x ∈ Lθ, y ∈ L∗θ and Ax ◦ A−1y = 0.
In each case, elements x and y satisfy the condition that either y2 is a multiple of x2 or x2 is a multiple of y2.
Proof. From the relation between Kn and Lθ given as in (3), we know that x ∈ Lθ ⇐⇒ Ax ∈ Kn and y ∈ L∗θ ⇐⇒ A−1y ∈ Kn. Moreover, under condition (a), there holds
hAx, A−1yi = hA−1Ax, yi = hx, yi = 0.
Hence, it follows that (a) and (b) are equivalent. The equivalence of (b) and (c) has been shown in Lemma 2.1. In addition, based on the relation between Kn and Lθ again, the equivalence of (c) and (d) is obvious.
Now, under condition (a), we prove that either y2 is a multiple of x2 or x2 is a multiple of y2. To see this, note that x ∈ Lθ and y ∈ L∗θ which gives
kx2k ≤ x1tan θ and ky2k ≤ y1cot θ.
This together with hx, yi = 0 yields
0 = hx, yi
= x1y1+ hx2, y2i
≥ x1y1− kx2kky2k
≥ x1y1− x1y1
= 0
which implies hx2, y2i = kx2kky2k. This says that either y2 is a multiple of x2 or x2 is a multiple of y2. Thus, the proof is complete. 2
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 construct some C- functions which play an important role in solving the circular cone complementarity problems by merit function methods.
Definition 3.1. Given a mapping φ : IRn × IRn → IRn, we call φ an C-function for CCCP if, for any (x, y) ∈ IRn× IRn, it satisfies
φ(x, y) = 0 ⇐⇒ x ∈ Lθ, y ∈ L∗θ, hx, yi = 0.
When θ = π4, an C-function for CCCP reduces to an C-function for SOCCP, i.e., φ(x, y) = 0 ⇐⇒ x ∈ Kn, y ∈ Kn, hx, yi = 0.
Two popular and well-known C-functions for SOCCP are Fischer-Burmeister (FB) func- tion 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
φgFB(x, y) := (Ax)2+ (A−1y)212
− (Ax + A−1y), φgNR(x, y) := Ax − [Ax − A−1y]soc+ .
Then, these two functions are C-functions for CCCP.
Proposition 3.1. Let gφFB and gφNR be defined as above where (Ax)2 equals (Ax) ◦ (Ax) under Jordan product. Then, gφFB and gφNR are both C-functions for CCCP.
Proof. In view of Theorem 2.1 and Definition 3.1, it is not hard to verify that gφFB(x, y) = 0 ⇐⇒ x ∈ Lθ, y ∈ L∗θ, hx, yi = 0,
φgNR(x, y) = 0 ⇐⇒ x ∈ Lθ, y ∈ L∗θ, hx, yi = 0, which says that these two functions are C-functions for CCCP. 2
We point out that if we consider directly the FB function φFB(x, y) for CCCP, un- fortunately, it cannot be C-function for CCCP because x2 is not well-defined associated with the circular cone Lθ for any x ∈ IRn. More specifically, because x2 is defined under the Jordan product in the setting of SOC, i.e.,
x2 := x ◦ x =
hx, yi x1y2+ y1x2
,
it follows that x2 ∈ Kn, which implies x2 may not belong to Lθ or L∗θ. Furthermore, when φFB(x, y) = 0, we have x + y = (x2+ y2)12 ∈ Kn, which yields that x, y ∈ Kn. This says that either x /∈ Lθ or y /∈ L∗θ. All the above explains that the FB function φFB cannot be an C-function for CCCP. Nonetheless, the NR function φNR : IRn× IRn → IRn given 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 function φFB 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−1 and y = (y1, y2) ∈ IR × IRn−1, we define one type of product of x and y as follows:
x • y = x1 x2
• y1 y2
=
hx, yi
max{tan2θ, 1} x1y2+ max{cot2θ, 1} y1x2
. (7) From the above product and direct calculation, it is easy to verify that
hx • y, zi = hx, z • yi, ∀z ∈ IRn with θ ∈ 0,π
4 i
(8) and
hx • y, zi = hy, x • zi, ∀z ∈ IRn with θ ∈hπ 4,π
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) if θ ∈ (0,π4], we have hx−, x+• (−y)−i ≤ 0;
(b) if θ ∈ [π4,π2), we have h(−y)+, x+• (−y)−i ≤ 0.
Proof. (a) When θ ∈ (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+, x− and (−y)−, if there exist at least one in them is zero, it is easy to obtain
hx−, x+• (−y)−i = 0.
If all the three elements are not equal to zero, from the definition of x+, x−, and (−y)−, we have k cot θ ≥ kwk, s tan θ = kuk, t cot θ = kvk 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
hx−, x+• (−y)−i
= stk + thu, wi + shv, wi + khu, vi cot2θ
= kukkvkk − kkukkvk cot2θ − kukhv, wi tan θ + kukhv, wi cot θ
= (1 − cot2θ)kkukkvk − (1 − cot2θ)(kukhv, wi tan θ)
= (1 − cot2θ)[kkukkvk − kukhv, wi tan θ]
≤ (1 − cot2θ)[kkukkvk − kukkvkkwk tan θ]
= (1 − cot2θ)kukkvk[k − kwk tan θ]
≤ 0.
Here the second equality is true due to αt = αkvk tan θ = −kuk tan θ. The last inequality holds due to k cot θ ≥ kwk and θ ∈ (0,π4]. Hence, the desired result follows.
(b) When θ ∈ [π4,π2), with the same skills, we also conclude that h(−y)+, x+• (−y)−i ≤ 0.
Then, the desired result follows. 2
Besides the inequalities in Lemma 3.1, “•” 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−1 and y = (y1, y2) ∈ IR × IRn−1, the following statements are equivalent:
(a) x ∈ Lθ, y ∈ L∗θ and hx, yi = 0.
(b) x ∈ Lθ, y ∈ L∗θ and x • y = 0.
In each case, x and y satisfy the condition that either y2 is a multiple of x2 or x2 is a multiple of y2.
Proof. In view of Theorem 2.1, we know that part (a) is equivalent to x ∈ Lθ, y ∈ L∗θ and Ax ◦ A−1y = 0.
To proceed the proof, we discuss the following two cases.
Case 1: For θ ∈ (0,π4], from the definition of the product of x and y, we have x • y =
hx, yi
x1y2+ cot2θ y1x2
which implies
Ax ◦ A−1y =
hx, yi
x1tan θ y2+ cot θ y1x2
= 1 0
0 (tan θ)I
(x • y).
This together with Theorem 2.1 yields the conclusion.
Case 2: For θ ∈ [π4,π2), from the definition of the product of x and y again, we have x • y =
hx, yi
tan2θ x1y2+ y1x2
which says
Ax ◦ A−1y = 1 0 0 (cot θ)I
(x • y).
Then, applying Theorem 2.1 again, the desired result follows. 2
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 function φp is an C-function for CCCP. To achieve the conclusion, a technical lemma is needed.
Lemma 3.2. Let φp : IRn× IRn → IRn be defined as in (10). Then, for any x, y ∈ IRn, we have
kφp(x, y)k ≥ max {kx−k, k(−y)+k} .
Proof. First, we prove that kφp(x, y)k ≥ kx−k. To see this, we observe that kφp(x, y)k2
= hx − (x − y)++ p x+• (−y)−, x − (x − y)++ p x+• (−y)−i
= hx+− x−− (x − y)++ p x+• (−y)−, x+− x−− (x − y)++ p x+• (−y)−i
= kx−k2+ kx+− (x − y)++ p x+• (−y)−k2− 2 hx−, x+− (x − y)++ p x+• (−y)−i
≥ kx−k2− 2hx−, x+i + 2 hx−, (x − y)+i − 2 hx−, p x+• (−y)−i
≥ kx−k2− 2p hx−, x+• (−y)−i .
Here, the last inequality is true due to x+, (x − y)+∈ Lθ, x− ∈ L∗θ, hx+, x−i = 0 and the relation between Lθ and L∗θ. When θ ∈ (0,π4], by Lemma 3.1(a), we have
hx−, x+• (−y)−i ≤ 0.
When θ ∈ [π4,π2), from equation (9), we have
hx−, x+• (−y)−i = h(−y)−, x+• x−i = 0
where the second equality holds due to x+• x−= 0. In summary, from all the above, we prove that
kφp(x, y)k2 ≥ kx−k2.
With similar arguments, we also obtain kφp(x, y)k2
= hx − (x − y)++ p x+• (−y)−, x − (x − y)++ p x+• (−y)−i
= hy − (x − y)−+ p x+• (−y)−, y − (x − y)−+ p x+• (−y)−i
= h(−y)−− (−y)+− (x − y)−+ p x+• (−y)−, (−y)−− (−y)+− (x − y)−
+px+ • (−y)−i
= k(−y)+k2+ k(−y)−− (x − y)−+ p x+• (−y)−k2− 2h(−y)+, (−y)−− (x − y)−
+px+ • (−y)−i
≥ k(−y)+k2− 2h(−y)+, (−y)−i + 2h(−y)+, (x − y)−i − 2h(−y)+, p x+• (−y)−i
≥ k(−y)+k2− 2p h(−y)+, x+• (−y)−i
≥ k(−y)+k2,
where the second inequality holds due to due to (−y)+ ∈ Lθ, (−y)−, (x − y)− ∈ L∗θ, h(−y)+, (−y)−i = 0 and the relation between Lθ and L∗θ. The last inequality holds due to equation (8) and Lemma 3.1(b). Therefore, we prove that kφp(x, y)k ≥ k(−y)+k.
Then, the proof is complete. 2
Remark 3.1. From the proof of Lemma 3.2, it also can be seen that kφNR(x, y)k ≥ max{kx−k, k(−y)+k}.
Theorem 3.2. Let φp : 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 ∈ L∗θ and hx, yi = 0.
Proof. “=⇒” Suppose that φp(x, y) = 0. If either x /∈ Lθ or y /∈ L∗θ, applying Lemma 3.2 yields
kφp(x, y)k ≥ max{kx−k, k(−y)+k} > 0.
This contradicts with φp(x, y) = 0. Hence, there must have x ∈ Lθ and y ∈ L∗θ. Next, we argue that hx, yi = 0. To see this, we consider the first component of φp(x, y), which is denoted by [φp(x, y)]1. In other words,
[φp(x, y)]1 = [x − (x − y)++ p x • y]1
=
y1+ p hx, yi if x − y ∈ Lθ, x1+ p hx, yi if x − y ∈ −L∗θ, w + p hx, yi otherwise,
where
w = x1− x1− y1+ kx2− y2k tan θ
1 + tan2θ = x1tan2θ + y1− kx2− y2k tan θ 1 + tan2θ . Since x ∈ Lθ and y ∈ L∗θ, it follows that x1, y1 ≥ 0, hx, yi ≥ 0 and
x1tan2θ + y1− kx2− y2k tan θ
1 + tan2θ ≥ tan θ(x1tan θ − kx2k + y1cot θ − ky2k)
1 + tan2θ ≥ 0.
This together with φp(x, y) = 0 gives phx, yi = 0. Thus, we conclude that hx, yi = 0 because p > 0.
“⇐=” Suppose that x ∈ Lθ, y ∈ L∗θ and hx, yi = 0. Since φNR is always an C-function for CCCP, we have x − (x − y)+= 0. Using Theorem 3.1 again yields x+• (−y)− = x • y = 0, which says φp(x, y) = 0. 2
Remark 3.2. In fact, for any x = (x1, x2) ∈ IR × IRn−1 and y = (y1, y2) ∈ IR × IRn−1, we define another type of product of x and y as follows:
x • y = x1 x2
• y1 y2
=
hx, yi
min{tan2θ, 1} x1y2+ min{cot2θ, 1} y1x2
. With the same skills, we may obtain the same results.
Motivated by the construction of φp given as in (10), we consider another function φr : IRn× IRn→ IRn defined by
φr(x, y) = x − (x − y)++ r (x • y)Ω+ r > 0, (11)
where Ω := Lθ∩ L∗θ = Lθ if θ ∈ (0,π4],
L∗θ if θ ∈ [π4,π2). We point out that the function φr defined as in (11) is not an C-function for CCCP. The reason come from that if φr(x, y) = 0, we have φNR(x, y) = x − (x − y)+ = −r (x • y)Ω+. Combining with the expression of φp, this implies that
−r (x • y)Ω++ p (x+• (−y)−) 6= 0
due to (x • y)Ω+ ∈ Ω = Lθ ∩ L∗θ and x+ • (−y)− ∈ K/ n ⊇ Lθ (or L∗θ) when θ ∈ (0,π4] (or θ ∈ [π4,π2)). This explains that φp(x, y) 6= 0, which contradicts φp(x, y) being an C-function for CCCP.
However, there is a merit function related to φr which 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 im- portant properties, respectively.
First, we recall that a function F : IRn → IRn is said to be monotone if, for any x, y ∈ IRn, there holds
hx − y, F (x) − F (y)i ≥ 0;
and strictly monotone if, for any x 6= y, the above inequality holds strictly; and strongly monotone with modulus ρ > 0 if, for any x, y ∈ IRn, the following inequality holds
hx − y, F (x) − F (y)i ≥ ρkx − yk2.
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(L∗θ) and that F is a monotone function. Then, for any sequence {xk} satis- fying
xk
→ ∞, lim sup
k→∞
xk−
< ∞ and lim sup
k→∞
(−F (xk))+ < ∞, we have
xk, F (xk) → ∞ and xk+, (−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) + h¯x, F (¯x)i ≥xk, F (¯x) + ¯x, F (xk) . (12) Using xk = xk+− xk− and F (xk) = (−F (xk))−− (−F (xk))+, it follows from (12) that
xk, F (xk) + h¯x, F (¯x)i
≥ xk+, 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).
xk+, F (¯x)
= xk+
1[f (¯x)]1+xk+
2, [f (¯x)]2
≥ xk+
1[f (¯x)]1− xk+
2
· k[f (¯x)]2k
≥ xk+
1[f (¯x)]1−xk+
1tan θ k[f (¯x)]2k
= xk+
1{[f (¯x)]1− tan θ k[f (¯x)]2k} . (14)
Note that xk= xk+−xk−, it gives kxk+k ≥ kxkk−kxk−k. From the assumptions on {xk}, i.e., kxkk → ∞, and lim supk→∞kxk−k < ∞, we see that kxk+k → ∞, and hence [xk+]1 → ∞.
Because CCCP has a strictly feasible point ¯x, we have [f (¯x)]1− tan θk[f (¯x)]2k > 0, which together with (14) implies that
hxk+, F (¯x)i → ∞ (k → ∞). (15)
On the other hand, we observe that lim sup
k→∞
hxk−, F (¯x)i ≤ lim sup
k→∞
kxk−kkF (¯x)k < ∞ lim sup
k→∞
h¯x, (−F (xk))+i ≤ lim sup
k→∞
k¯xkk(−F (xk))+k < ∞ and h¯x, (−F (xk))−i ≥ 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 thatxk+, (−F (xk))− → ∞. Suppose not, that is, limk→∞xk+, (−F (xk))− <
∞. Then, we obtain
xk+, (−F (xk))− kxk+k =
xk+
kxk+k, (−F (xk))−
→ 0.
This means that there exists ¯x ∈ IRn such that xk+
kxk+k → x¯+ k¯x+k and
x¯+
k¯x+k, (−F (¯x))−
= 0. (16)
Denote z := x¯+
k¯x+k and apply Theorem 3.1, there exists α ∈ IR such that [(−F (¯x))−]2 = αz2 or αz2 = [(−F (¯x))−]2.
It is obvious that z ∈ Lθ and (−F (¯x))− ∈ L∗θ. Hence, equation (16) implies that α < 0, which says that z2 and [(−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))+]2 and [(−F (¯x))−]2 are in the opposite direction, to each other. These conclude that z2 and [(−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 h¯x+, (−F (¯x))+i ≥ 0.
Similarly, by the the relation between ¯x+ and ¯x−, we know [¯x−]2 and [(−F (¯x))−]2 are in the same direction. Then, combining with ¯x−, (−F (¯x))−∈ L∗θ, it leads to
h¯x−, (−F (¯x))−i ≥ 0.
Moreover, writing out the expression for h¯x, F (¯x)i, we see that
h¯x, F (¯x)i = h¯x+, (−F (¯x))−i − h¯x+, (−F (¯x))+i − h¯x−, (−F (¯x))−i + h¯x−, (−F (¯x))+i.
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→∞xk+, (−F (xk))− < ∞, we conclude that h¯x, F (¯x)i < ∞, which contradict
h¯x, F (¯x)i = lim
k→∞xk, F (xk) = ∞.
Thus, we prove that xk+, (−F (xk))− → ∞. 2
4.1 The first class of merit functions
For any x ∈ IRn, from the analysis of the section 3, we know that the function φp and φNR are complementarity function for CCCP. In this subsection, we focus on the property of bounded level sets of merit functions based on φNR and φp with the product of elements, which is a property to guarantee that the existence of accumulation points of sequence generated by some descent algorithms.
Theorem 4.1. Let φp be 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| kφp(x, F (x))k ≤ α}
is bounded for all α ≥ 0.
Proof. We prove this result by contradiction. Suppose there exists an unbounded sequence {xk} ⊂ Lp(α) for some α ≥ 0. If kxk−k → ∞ or k(−F (xk))+k → ∞, by Lemma 3.2, we have kφp(xk, F (xk))k → ∞, which contradicts kφp(xk, F (xk))k ≤ α. On the other hand, if
lim sup
k→∞
kxk−k < ∞ and lim sup
k→∞
(−F (xk))+
< ∞,
it follows from Lemma 4.1 that xk+, (−F (xk))− → ∞. From the proof of Lemma 4.1, there exists a constant κ0 such that
φNR(xk, f (xk))
1
≥
[xk+]1− κ0 if xk− F (xk) ∈ −L∗θ,
(−F (xk))−
1− κ0 if xk− F (xk) ∈ Lθ,
[xk+]1tan2θ+[(−F (xk))−]1−k[xk+]2k tan θ−k[(−F (xk))−]2k tan θ 1+tan2θ
−2κ1+tan0(1+tan θ)2θ , if xk− F (xk) /∈ Lθ∪ −L∗θ,
which means lim infφNR(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+xk+, (−F (xk))−
→ ∞, where the limit comes from
xk+, (−F (xk))− → ∞ and lim inf φNR(xk, f (xk))
1 > −∞.
Thus, we obtain that kφp(xk, F (xk))k → ∞ which contradicts kφp(xk, F (xk))k ≤ α.
Then, the proof is complete. 2
Note that, under the conditions of Lemma 4.1 or Theorem 4.1, we cannot guarantee the boundedness of the level set on the NR function φNR. For example, let F (x) = 1 −1 and x > 0, it is easy to verify that the level set x
LNR(2) = {x ∈ IRn| kφNR(x, F (x))k ≤ 2}
is unbounded. In fact, In order to establish the boundedness of the level set on the natural residual function φNR, we need the following concept.
Definition 4.1. A mapping F : IRn → IRn is said to be strongly coercive if
kxk→∞lim
hF (x), x − yi kx − yk = ∞.
holds for all y ∈ IRn.
Theorem 4.2. Suppose that F is strongly coercive. Then, the level set LNR(α) = {x ∈ IRn| kφNR(x, F (x))k ≤ α}
is bounded for all α ≥ 0.
Proof. Again, we prove this result by contradiction. Suppose there exists an unbounded sequence {xk} ⊂ LNR(α) for some α ≥ 0, i.e.,
xk
→ ∞. Note that the sequence
φNR(xk, F (xk)) = xk− (xk− F (xk))+ is bounded. It follows from the unboundedness of the sequence {xk} that the sequence {(xk− F (xk))+} is also unbounded. Then, for any y ∈ Lθ, there exist N ∈ N and β > 0 such that
(xk− F (xk))+− y
> β, ∀k > N.
From the property of projection mapping, we have
xk− F (xk) − (xk− F (xk))+, y − (xk− F (xk))+ ≤ 0 (17) for each k > N . On the other hand,
xk− F (xk) − (xk− F (xk))+, y − (xk− F (xk))+
= xk− (xk− F (xk))+, y − (xk− F (xk))+ + F (xk), (xk− F (xk))+− y
≥ −
xk− (xk− F (xk))+ ·
y − (xk− F (xk))+
+F (xk), (xk− F (xk))+− y
≥
y − (xk− F (xk))+
hF (xk), (xk− F (xk))+− yi ky − (xk− F (xk))+k − α
. Plugging in yk := xk− (xk− F (xk))+− y, we obtain
lim
k→∞
F (xk), (xk− F (xk))+− y
ky − (xk− F (xk))+k = lim
k→∞
F (xk), xk− yk kxk− ykk = ∞,
where the last equality holds due to the strong coercivity of F and [22, Theorem 2.1].
This implies that
k→∞lim xk− F (xk) − (xk− F (xk))+, y − (xk− F (xk))+ = ∞, which contradicts (17). Therefore, the level set
LNR(α) = {x ∈ IRn| kφNR(x, F (x))k ≤ α}
is bounded for all α ≥ 0. 2
4.2 The second class of merit functions
For any x ∈ IRn, LT (standing for Luo-Tseng) merit function for the circular cone complementarity problem is given as follows:
fLT(x) := ϕ(hx, F (x)i) + 1
2k(x)−k2+ k(−F (x))+k2 , (18) where ϕ : IR → IR+ is an arbitrary smooth function satisfying
ϕ(t) = 0, ∀t ≤ 0 and ϕ0(t) > 0, ∀t > 0.
Notice that we have ϕ(t) ≥ 0 for all t ∈ IR from the above condition. Indeed, this class of functions has been considered for the SDCP case (positive semidefinite complementar- ity problem) by Tseng in [21], for the SOCCP case (second-order cone complementarity problem) by Chen in [2] and for the general SCCP case by Pan and Chen in [18], respec- tively. For the case of generally closed convex cone complementarity problems, the LT merit function has been studied by Lu and Huang in [15]. In view of the results in [15], it is easy to obtain the following results directly for the circular cone complementarity problem.
Proposition 4.1. Let fLT : IRn → IR be given as in (18). Then, the following results hold.
(a) For all x ∈ IRn, we have fLT(x) ≥ 0; and fLT(x) = 0 if and only if x solves the circular cone complementarity problem.
(b) If F (·) is differentiable, then so is fLT(·). Moreover,
∇fLT(x) = ∇ϕ(hx, F (x)i)[F (x) + x∇F (x)] − x−− ∇F (x)(−F (x))+ for all x ∈ IRn.
Proof. See Lemma 3.1 and Theorem 3.4 in [15]. 2
Proposition 4.2. Let fLT be given as in (18). Suppose that F : IRn→ IRn is a strongly monotone mapping and that the circular cone complementarity problem has a solution x∗. Then, there exists a constant τ > 0 such that
τ kx − x∗k2 ≤ max{0, hx, F (x)i} + kx−k + k(−F (x))+k, ∀x ∈ IRn. Moreover,
τ kx − x∗k2 ≤ ϕ−1(fLT(x)) + 2[fLT(x)]12, ∀x ∈ IRn. Proof. See Theorem 3.6 in [15]. 2
In the following theorem, we present the condition which ensures the boundedness of the level sets for LT merit function fLT to solve the circular cone complementarity problem.
Theorem 4.3. Suppose that the circular cone complementarity problem has a strictly feasible point and that F is monotone. Then, the level set
LfLT(α) := {x ∈ IRn| fLT(x) ≤ α}
is bounded for all α ≥ 0.
Proof. We prove this result by contradiction. Suppose there exists an unbounded sequence {xk} ⊆ LfLT(α) for some α ≥ 0. We may assert that the sequences {xk−} and {(−F (xk))+} are bounded. If not, from the expression (18) of LT merit function fLT and the property ϕ(t) ≥ 0 for all t ∈ IR, it follows that
fLT(xk) ≥ 1
2[kxk−k2+ k(−F (xk))+k2] → ∞,
which contradicts {xk} ⊆ LfLT(α), i.e., fLT(xk) ≤ α. Therefore, we have lim sup
k→∞
kxk−k < ∞ and lim sup
k→∞
k(−F (xk))+k < ∞.
Then, by Lemma 4.1, we get that
hxk, F (xk)i → ∞.
By the properties of the function ϕ again, we obtain that ϕ(hxk, F (xk)i) → ∞, which implies fLT(xk) → ∞. This contradicts {xk} ⊆ LfLT(α). Hence, the level set LfLT(α) is bounded for all α ≥ 0. 2
4.3 The third class of merit functions
To achieve the third class of merit functions, we make a slight modification of LT merit function fLT for the circular cone complementarity problem. More specifically, we con- sider the set Ω as follows:
Ω := Lθ∩ L∗θ = Lθ for 0 < θ ≤ π4, L∗θ for π4 < θ < π2.
Indeed, Ω is also a closed convex cone. In light of this Ω, another function is considered:
fdLT(x) := 1
2k(x • F (x))Ω+k2+ 1
2kx−k2+ k(−F (x))+k2 , (19) where (x • y)Ω+denotes the projection of x • y onto Ω. Then, together with the expressions (7) of x • y, we can verify that the function dfLT is also a type of merit function for the circular cone complementarity problem, which will be shown in following theorem.
Theorem 4.4. Let the function dfLT be given by (19). Then, for all x ∈ IRn, we have fdLT(x) = 0 ⇐⇒ x ∈ Lθ, F (x) ∈ L∗θ and hx, F (x)i = 0,
where L∗θ denotes the dual cone of Lθ, i.e., L∗θ = Lπ
2−θ.
Proof. By the definition of the function dfLT given by (19), we have fdLT(x) = 0 ⇔ k(x • F (x))Ω+k = 0, kx−k = 0 and k(−F (x))+k = 0,
⇔ (x • F (x))Ω+ = 0, x−= 0 and (−F (x))+ = 0,
⇔ x • F (x) ∈ −Lθ or x • F (x) ∈ −L∗θ, x ∈ Lθ, and F (x) ∈ L∗θ,
⇔ x ∈ Lθ, F (x) ∈ L∗θ and hx, F (x)i = 0,
where the last equivalence holds due to the properties of the cone −Lθ or −L∗θ. Thus, the proof is complete. 2
From Theorem 4.4, we know that the function dfLT is a merit function for the circular cone complementarity problem. As below, according to the type of dot product (7), we establish the differentiability of dfLT.
Theorem 4.5. Let dfLT : IRn → IR be given by (19). Suppose that the type of dot product (7) is employed. If F (·) is differentiable, then so is dfLT(·). Moreover, for all x ∈ IRn, we have
∇dfLT(x) = (Ly+ ∇F (x)Lx) · (x • F (x))Ω+− x−− ∇F (x)(−F (x))+, where
Lx =
y1 y2T
max{tan2θ, 1}y2 max{cot2θ, 1}y1I
and
Ly =
x1 xT2
max{tan2θ, 1}x2 max{cot2θ, 1}x1I
with I being the identity matrix.
Proof. From the proof of Lemma 3.1(b) in [15], we have
∇ 1
2k(z)Ω+k2
= (z)Ω+, ∀z ∈ IRn. Then, by the chain rule again, it follows that
∇ 1
2k(x • F (x))Ω+k2
= ∇x(x • F (x)) · (x • F (x))Ω+
= [Ly+ ∇F (x)Lx] · (x • F (x))Ω+, where
Lx =
y1 y2T
max{tan2θ, 1}y2 max{cot2θ, 1}y1I
and
Ly =
x1 xT2
max{tan2θ, 1}x2 max{cot2θ, 1}x1I
with I being the identity matrix. Thus, we obtain that
∇dfLT(x) = (Ly+ ∇F (x)Lx) · (x • F (x))Ω+− (x)−− ∇F (x)(−F (x))+ for all x ∈ IRn. 2
In order to establish error bound property of the merit function dfLT for the circular cone complementarity problem, we need a technical lemma as below.
Lemma 4.2. Let x = (x1, x2) ∈ IR × IRn−1 and y = (y1, y2) ∈ IR × IRn−1. Then, we have hx, yi ≤ max 1 + tan2θ
√2 ,1 + cot2θ
√2
(x • y)Ω+ where • is defined as in (7).
