to appear in Computational Optimization and Applications, 2017
On merit functions for p-order cone complementarity problem
Xin-He Miao1
Department of Mathematics Tianjin University, China
Tianjin 300072, China
Yu-Lin Chang2 Department of Mathematics National Taiwan Normal University
Taipei 11677, Taiwan
Jein-Shan Chen 3 Department of Mathematics National Taiwan Normal University
Taipei 11677, Taiwan
August 17, 2016
(revised on December 12, 2016)
Abstract Merit function approach is a popular method to deal with complementarity problems, in which the complementarity problem is recast as an unconstrained mini- mization via merit function or complementarity function. In this paper, for the comple- mentarity problem associated with p-order cone, which is a type of nonsymmetric cone complementarity problem, we show the readers how to construct merit functions for solv- ing p-order cone complementarity problem. In addition, we study the conditions under which the level sets of the corresponding merit functions are bounded, and we also assert that these merit functions provide an error bound for the p-order cone complementarity problem. These results build up a theoretical basis for the merit method for solving p-order cone complementarity problem.
1E-mail: xinhemiao@tju.edu.cn. The author’s work is supported by National Natural Science Foun- dation of China (No. 11471241).
2E-mail: ylchang@math.ntnu.edu.tw
3Corresponding author. The author’s work is supported by Ministry of Science and Technology, Taiwan. E-mail: jschen@math.ntnu.edu.tw
Keywords. p-order cone complementarity problem, merit function, error bound.
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 complementarity problem [9, 10, 11, 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, 2, 3, 4, 5, 7, 8]. In contrast to symmetric cone programming and symmetric cone complementarity problem, we are not familiar with their nonsymmetric counterparts. Referring the reader to [16, 22, 14, 19] and the bibliographies therein, we observe that there is no 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 focus on a special nonsymmetric cone K for problem (1), i.e., p- order cone. Then, the problem (1) reduces to the p-order cone complementarity problem (POCCP for short). Indeed , the p-order cone [17, 22] is a generalization of the second- order cone in IRn, denoted by Kp, and can be expressed as
Kp :=
x ∈ IRn
x1 ≥
n
X
i=2
|xi|p
!1p
(p > 1).
If we write x := (x1, ¯x) ∈ IR × IRn−1, the p-order cone Kp can be equivalently expressed as
Kp =x = (x1, ¯x) ∈ IR × IRn−1| x1 ≥ k¯xkp , (p > 1).
When p = 2, it is obvious that the p-order cone is exactly the second-order cone, which means the p-order cone complementarity problem is actually the second-order cone com- plementarity problem. Thus, the p-order cone complementarity problem (POCCP) can be viewed as the generalization of the second-order cone complementarity problem. As shown in [15, 17], Kp is a convex cone and its dual cone is given by
Kp∗ =
y ∈ IRn
y1 ≥
n
X
i=2
|yi|q
!1q
or equivalently
Kp∗ =y = (y1, ¯y) ∈ IR × IRn−1| y1 ≥ k¯ykq = Kq,
where q > 1 and satisfies 1p +1q = 1. In addition, the dual cone K∗p is also a convex cone.
For more details regarding p-order cone and its involved optimization problems, please refer to [15, 17, 22].
During the past decade, there had active research and various methods for com- plementarity problems, which include the interior-point methods, the smoothing New- ton methods, the semismooth Newton methods, and the merit function methods, see [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 21] and references therein. As seen in the literature, almost all the attention was paid to symmetric cone complementarity problems, that is, nonlinear complementarity problem (NCP), positive semi-definite complementarity problem (SDCP), second-order cone complementarity problem (SOCCP). As mentioned earlier, there is no unified framework to deal with general nonsymmetric cone comple- mentarity problems. Consequently, the study about nonsymmetric cone complementarity problem is very limited. Nonetheless, we believe that that merit function approach, in which the complementarity problem is recast as an unconstrained minimization via merit function or complementarity function, may be appropriately viewed as a unified way to deal with nonsymmetric cone complementarity problem. Indeed, the main difficulty lies on how to construct complementarity functions or merit functions for nonsymmetric cone complementarity problem. For circular cone setting, several successful ways were shown in [16]. Inspired by the work [16], we employ the similar ways to construct merit func- tions for solving p-order cone complementarity problem. For completeness, the idea is roughly described again as below.
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. A merit function for the p-order cone complementarity problem is a function h : IRn→ [0, +∞), provided that
h(x) = 0 ⇐⇒ x solves the POCCP (1).
Hence, solving the problem (1) is equivalent to handling the unconstrained minimization problem
x∈IRminnh(x)
with the optimal value zero. Until now, for solving symmetric cone complementarity problem, a large number of merit functions have been proposed. Among them, one of the most popular merit functions is the natural residual (NR) merit function ΨNR : IRn→ IR, which is defined as
ΨNR(x) := 1
2kφNR(x, F (x))k2 = 1 2
x − (x − F (x))K+
2,
where (·)K+ denotes the projection onto the symmetric cone K. Then, we know that ΨNR(x) = 0 if and only if x is a solution to the symmetric cone complementarity prob- lem. As remarked in [16], this function ΨNR (or φNR) can also serve as merit function (or complementarity function) for general conic complementarity problem. Hence, it is also applicable to p-order cone complementarity problem. Under this setting, for any x ∈ IRn, we denote x+ be the projection of x onto the p-order cone Kp, and x− be the projection of −x onto the dual cone K∗p of Kp. By properties of projection onto the closed convex cone, it can be verified that x = x+− x−. Moreover, the formula of projection of x ∈ IRn onto Kp is obtained in [17]. Besides the NR merit function ΨNR, are there any other types of merit functions for POCCP? In this paper, we answer this question by presenting other types of merit functions for the p-order 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 p-order cone complementarity problem.
2 Preliminaries
In this section, we briefly review some basic concepts and background materials about the p-order cone, and define one type of product associated with p-order cone, which will be extensively used in subsequent analysis.
As mentioned, the p-order cone Kp is a pointed closed convex cone, and its dual cone denoted by Kp∗ is given as
Kp∗ =
y ∈ IRn
y1 ≥
n
X
i=2
|yi|q
!1q
or equivalently
Kp∗ =y = (y1, ¯y) ∈ IR × IRn−1| y1 ≥ k¯ykq = Kq,
where q > 1 and satisfies 1p +1q = 1. From the expression of the dual cone K∗p, it is easy to know that the dual cone K∗p is also a closed convex cone. In addition, when p 6= q, we have Kp 6= Kq = Kp∗, i.e., the p-order cone Kp is not a self-dual cone. That is to say, the p-order cone Kp is not a symmetric cone for p 6= 2.
It is well known that Jordan product plays a critical role in the study of symmetric cone programming or symmetric cone complementarity problems. However, there is no
Jordan product for the setting of the p-order cone so far. Hence, we need to find one type of special product for the setting of the p-order cone, which is similar to the one for the setting of symmetric cone. To this end, for any x = (x1, · · · , xn)T ∈ IRn and y = (y1, · · · , yn)T ∈ IRn, we define one type of product of x and y associated with p-order cone Kp as follows:
x • y = hx, yi w
where w := (w2, · · · , wn)T with wi = |x1|pq|yi| − |y1||xi|pq. (2) From the above definition (2) of product, when p = q = 2, it is not hard to see that the product x•y is exactly the Jordan product in the setting of second-order cone. According to the product “•” defined as in (2), we have the following equivalence.
Proposition 2.1. For any x = (x1, ¯x) ∈ IR × IRn−1 with ¯x = (x2, · · · , xn)T ∈ IRn−1 and y = (y1, ¯y) ∈ IR × IRn−1 with ¯y = (y2, · · · , yn)T ∈ IRn−1, the following statements are equivalent:
(a) x ∈ Kp, y ∈ Kp∗ = Kq and hx, yi = 0.
(b) x ∈ Kp, y ∈ K∗p = Kq and x • y = 0.
In each case, x and y satisfy the condition that there is c ≥ 0 such that |xi|p = c|yi|q or
|yi|q= c|xi|p for any i = 2, · · · , n.
Proof. (b) ⇒ (a) From the definition of product x • y of x and y associated with Kp, the implication is obvious.
(a) ⇒ (b) When ¯x = 0 or ¯y = 0, from (a), we know x ∈ Kp, y ∈ Kq and hx, yi = 0.
Then, it is easy to see that x • y = 0. When ¯x 6= 0 and ¯y 6= 0, by x ∈ Kp and y ∈ Kq, we have x1 ≥ k¯xkp and y1 ≥ k¯ykq. Hence, it follows from hx, yi = 0 that
0 = hx, yi
= x1y1+ h¯x, ¯yi
≥ k¯xkpk¯ykq− k¯xkpkk¯ykq
= 0.
This implies that x1 = k¯xkp, y1 = k¯ykq and |xi|p = c|yi|q or |yi|q = c|xi|p with some c ≥ 0 for any i = 2, · · · , n. Next, we only consider the case |xi|p = c|yi|q, and the same arguments apply for the case |yi|q = c|xi|p. Because |xi|p = c|yi|q, we have |xi|pq = c1q|yi|.
This yields that, for any i = 2, · · · , n,
|x1|pq|yi| − |y1||xi|pq = |x1|pq|yi| − |y1|c1q|yi|
=
n
X
k=2
|xk|p
!1q
|yi| −
n
X
k=2
|yk|q
!1q c1q|yi|
= c1q
n
X
k=2
|yk|q
!1q
|yi| − c1q
n
X
k=2
|yk|q
!1q
|yi|
= 0,
where the second equality holds due to x1 = k¯xkp, y1 = k¯ykq. Then, it follows that x • y = 0, and the proof is complete. 2
To close this section, we introduce some other concepts that will be needed in subse- quent analysis. 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 result is crucial for achieving the property of bounded level sets. Although the analysis technique is similar to [16, Lemma 4.1], we present the details for completeness.
Proposition 2.2. Suppose that the POCCP has a strictly feasible point z, i.e., z ∈ int(Kp) and F (z) ∈ int(K∗p) 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) → ∞.
Proof. Since F is monotone, for any xk ∈ IRn, we have xk− z, F (xk) − F (z) ≥ 0, which leads to
xk, F (xk) + hz, F (z)i ≥ xk, F (z) + z, F (xk) . (3)
From properties of projection, we write xk = xk+ − xk− and F (xk) = (−F (xk))− − (−F (xk))+. Then, it follows from (3) that
xk, F (xk) + hz, F (z)i
≥ xk+, F (z) − xk−, F (z) + z, (−F (xk))− − z, (−F (xk))+ . (4) Now, we denote xk+ :=
[xk+]1, xk+TT
and F (z) :=
[f (z)]1, f (z)TT
. With these nota- tions, we look into the first term on the right-hand side of (4):
xk+, F (z)
= xk+
1[f (z)]1+D
xk+, f (z)E
≥ xk+
1[f (z)]1− xk+
p·
f (z)
q
≥ xk+
1[f (z)]1−xk+
1
f (z)
q
= xk+
1
[f (z)]1− f (z)
q
(5)
≥ 0.
Note that xk = xk+−xk−, which gives kxk+k ≥ kxkk−kxk−k. Using the assumptions on {xk}, i.e., kxkk → ∞, and lim supk→∞kxk−k < ∞, we see that kxk+k → ∞, and hence [xk+]1 →
∞. Because the POCCP has a strictly feasible point z, we know [f (z)]1 − kf (z)kq > 0, which together with (5) implies that
hxk+, F (z)i → ∞ as k → ∞. (6)
Moreover, we also observe that lim sup
k→∞
hxk−, F (z)i ≤ lim sup
k→∞
kxk−kkF (z)k < ∞, lim sup
k→∞
hz, (−F (xk))+i ≤ lim sup
k→∞
kzkk(−F (xk))+k < ∞ and hz, (−F (xk))−i ≥ 0. All of these together with (4) and (6) yield
xk, F (xk) → ∞.
Then, the proof is complete. 2
3 Merit functions for POCCP
In this section, based on the product (2) of x and y associated with p-order cone in IRn and employing the same idea in [16], we propose several classes of merit functions for the p-order cone complementarity problem and investigate their favorable properties, respectively.
3.1 The first class of merit functions
In this subsection, we focus on the natural residual (NR) function φNR : IRn× IRn→ IRn, which is given by:
φNR(x, y) := x − (x − y)+,
where (·)+denotes the projection function. We know that the NR function φNR is always an complementarity function for general conic complementarity problem, see [16] or [8, Proposition 1.5.8]. In light of this, it is clear that the function ΨNR(x) = 12kφNR(x, F (x))k2 serves a merit function for the POCCP.
Lemma 3.1. Let x, y ∈ IRn and φNR(x, y) = x − (x − y)+. For any closed convex cone K, we have
kφNR(x, y)k ≥ maxkxK−∗k, k(−y)K+k ,
where z+K denotes the projection of z onto the closed convex cone K, and z−K∗ means the projection of −z onto its dual cone K∗.
Proof. The proof is similar to [16, Lemma 3.2] because the cone therein can be replaced by any closed convex cone. Hence, we omit it here. 2
In fact, Lu and Huang considered a more general NR merit function in [12], whose format is as bellow:
Ψα(x) = 1
2kx − (x − αF (x))+k2, (α > 0).
They also showed the property of error bound under the strong monotonicity and the global Lipschitz continuity of F .
Theorem 3.1. [12, Theorem 3.3] Suppose that F is strongly monotone with modulus ρ > 0 and is Lipschitz continuous with constant L > 0, Then for any fixed α > 0, the following inequality holds
1 2 + αL
pΨα(x) ≤ kx − x∗k ≤ 1 + αL αρ
pΨα(x),
where x∗ is the unique solution of the generally closed convex cone complementarity prob- lems.
From Theorem 3.1, we know that the NR merit function ΨNR (i.e., α = 1) provides an error bound for the POCCP. Unfortunately, when considering the boundedness of the level set for the NR function φNR, if under the same conditions used in Proposition 2.2, we cannot guarantee the boundedness of the level set for the function φNR. For example,
as mentioned in [16], taking F (x) = 1 − 1
x and x > 0, it is easy to verify that the level set
LNR(2) = {x ∈ IRn| kφNR(x, F (x))k ≤ 2}
is unbounded. Thus, a different condition is needed. In fact, in order to establish the boundedness of the level set for the natural residual function φNR or the merit function Ψα, we need the following concept.
Definition 3.1. 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 3.2. Suppose that F is strongly coercive. Then, the level set LNR(γ) = {x ∈ IRn| kφNR(x, F (x))k ≤ γ}
or
LΨα(γ) = {x ∈ IRn| Ψα(x) ≤ γ}
is bounded for all γ ≥ 0.
Proof. The proof is similar to [16, Theorem 4.2]. Hence, we omit it. 2
3.2 The second class of merit functions
For any x ∈ IRn, we denote fLT the LT (standing for Luo-Tseng) merit function associated with the p-order cone complementarity problem, whose mathematical formula is given as follows:
fLT(x) := ϕ(hx, F (x)i) + 1
2k(x)−k2+ k(−F (x))+k2, (7) where ϕ : IR → IR+ is an arbitrary smooth function satisfying
ϕ(t) = 0, ∀t ≤ 0 and ϕ0(t) > 0, ∀t > 0.
It is easy to see that ϕ(t) ≥ 0 for all t ∈ IR from the above condition. This class of functions has been considered by Tseng for the positive semidefinite complementarity problem in [21], for the second-order cone complementarity problem by Chen in [2], and for the general SCCP case by Pan and Chen in [18], respectively. For the setting of general closed convex cone complementarity problems, the LT merit function has also been studied by Lu and Huang in [12], with some favorable properties shown as below.
Property 3.1. ([12, Lemma 3.1 and Theorem 3.4]) Let fLT : IRn → IR be given as in (7). 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 p-order 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.
Property 3.2. ([12, Theorem 3.6]) Let fLT be given as in (7). Suppose that F : IRn → IRn is a strongly monotone mapping and that the p-order 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.
Although the above properties were established in [12] for general closed convex cone setting, there is no study about bounded level set for fLT therein. Hence, we hereby present the condition which ensures the boundedness of level sets for the LT merit func- tion fLT to solve the p-order cone complementarity problem.
Theorem 3.3. Suppose that the p-order cone complementarity problem has a strictly feasible point and that F is monotone. Then, the level set
LfLT(γ) := {x ∈ IRn| fLT(x) ≤ γ}
is bounded for all γ ≥ 0.
Proof. We prove this result by contradiction. Suppose there exists an unbounded sequence {xk} ⊆LfLT(γ) for some γ ≥ 0. Then, the sequences {xk−} and {(−F (xk))+} must be bounded. If not, from the expression (7) of fLT and the property ϕ(t) ≥ 0 for all t ∈ IR, it follows that
fLT(xk) ≥ 1
2kxk−k2+ k(−F (xk))+k2 → ∞, which contradicts {xk} ⊆LfLT(γ). Hence, we have
lim sup
k→∞
kxk−k < ∞ and lim sup
k→∞
k(−F (xk))+k < ∞.
Then, applying Proposition 2.2 yields
hxk, F (xk)i → ∞.
Using the properties of the function ϕ again, we have ϕ(hxk, F (xk)i) → ∞, which leads to fLT(xk) → ∞. It contradicts {xk} ⊆LfLT(γ). Thus, the level setLfLT(γ) is bounded for all γ ≥ 0 and the proof is complete. 2
3.3 The third class of merit functions
Motivated by the construction way of the merit function fLT, we make a slight modifi- cation on the LT merit function fLT associated with the p-order cone complementarity problem, which leads to the third class of merit functions. More specifically, we first look into the set Ω := Kp∩ K∗p. Indeed, the set Ω is characterized as follows:
Ω := Kp∩ K∗p = Kp for 1 ≤ p ≤ 2, Kp∗ = Kq for p ≥ 2,
where q satisfies the condition q ≥ 1 and 1p + 1q = 1. Moreover, it is easy to check that Ω is also a closed convex cone. In light of this closed convex cone Ω, another function is considered:
fdLT(x) := 1
2k(x • F (x))Ω+k2+ 1
2kx−k2+ k(−F (x))+k2 , (8) where (x • y)Ω+ denotes the projection of x • y onto Ω. As shown in the following theo- rem, we see that the function dfLT is also a type of merit functions for the p-order cone complementarity problem.
Theorem 3.4. Let the function dfLT be given as in (8). Then, for all x ∈ IRn, we have fdLT(x) = 0 ⇐⇒ x ∈ Kp, F (x) ∈ Kp∗ and hx, F (x)i = 0,
where K∗p denotes the dual cone of Kp, i.e., Kp∗ = Kq with p, q ≥ 1 and 1p + 1q = 1.
Proof. From the definition of the function dfLT given in (8), 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) ∈ −Kp or x • F (x) ∈ −K∗p , x ∈ Kp and F (x) ∈ K∗p,
⇐⇒ −x • F (x) ∈ Kp or − x • F (x) ∈ K∗p , x ∈ Kp and F (x) ∈ Kp∗,
⇐⇒ x ∈ Kp, F (x) ∈ K∗p and hx, F (x)i = 0,
where the last equivalence holds due to the properties of Kp and K∗p. Thus, the proof is complete. 2
In the following, we investigate the error bound property and the boundedness prop- erty of level sets of the merit function dfLT for the p-order cone complementarity problem.
In order to achieve these results, we need a novel lemma as below.
Lemma 3.2. For any x, y ∈ IRn, we have hx, yi ≤
(x • y)Ω+ , where the product x • y is defined as in (2).
Proof. Given any x, y ∈ IRn, let x = (x1, · · · , xn)T and y = (y1, · · · , yn)T. Recall from the product (2), we know
x • y = hx, yi w
where w :=
|x1|pq|yi| − |y1||xi|pqn i=2. To proceed the arguments, we consider the following three cases.
Case 1 When x • y ∈ Ω, we have (x • y)Ω+ = x • y. Then, it is easy to verify that (x • y)Ω+
≥ hx, yi.
Case 2 When x • y ∈ −Ω∗, where Ω∗ denotes the dual cone of Ω, we have (x • y)Ω+= 0 and hx, yi ≤ 0. This clearly implies that
(x • y)Ω+
≥ hx, yi.
Case 3 When x • y /∈ Ω ∪ (−Ω∗), let (x • y)Ω+ := (v1, ¯vT)T. If hx, yi ≤ 0, then the result is obvious. Thus, we only need to look into the case of hx, yi > 0. If Ω = Kp, by the property of projection onto the p-order cone, we have
(x • y)Ω+− x • y = v1− hx, yi
¯ v − w
∈ Ω∗ = K∗p = Kq.
From the definition of dual order cone K∗p again, it follows that v1− hx, yi ≥ 0, i.e., v1 ≥ hx, yi. Hence, this yields that
(x • y)Ω+
≥ |v1| ≥ v1 ≥ hx, yi.
With similar arguments, for the case of Ω = Kp∗ = Kq, we also obtain that hx, yi ≤
(x • y)Ω+ . From all the above cases, we have shown that hx, yi ≤
(x • y)Ω+
. Thus, the proof is complete. 2
Theorem 3.5. Let the function dfLT be given as in (8). Suppose that F : IRn → IRn is strongly monotone mapping and that x∗ is a solution to the p-order cone complementarity problem. Then, there exists a scalar τ > 0 such that
τ kx − x∗k2 ≤ 2 +√
2 h
fdLT(x)i12 .
Proof. Since the function F is strongly monotone and x∗ is a solution to the p-order cone complementarity problem, there exists a scalar ρ > 0 such that, for any x ∈ IRn,
ρ kx − x∗k2 ≤ hF (x) − F (x∗), x − x∗i
= hF (x), xi + hF (x∗), −xi + h−F (x), x∗i
= hF (x), xi + hF (x∗), x−− x+i + h(−F (x))+− (−F (x))−, x∗i
≤ hF (x), xi + hF (x∗), x−i + h(−F (x))+, x∗i
≤
(x • F (x))Ω+
+ kx−kkF (x∗)k + kx∗k k(−F (x))+k
≤ max {1, kF (x∗)k, kx∗k}
(x • F (x))Ω+
+ kx−k + k(−F (x))+k , where the second inequality holds due to the properties of Kp and K∗p, and the third inequality follows from Lemma 3.2. Then, setting τ := ρ
max {1, kF (x∗)k, kx∗k} yields τ kx − x∗k2 ≤
(x • F (x))Ω+
+ kx−k + k(−F (x))+k.
Moreover, we observe that (x • F (x))Ω+
=√ 2 1
2k(x • F (x))Ω+k2
12
≤√ 2h
fdLT(x)i12 , and
kx−k + k(−F (x))+k ≤√
2 kx−k2+ k(−F (x))+k212
≤ 2h
fdLT(x) i12
. Putting all the above together gives
τ kx − x∗k2 ≤ (2 +√ 2)h
fdLT(x)i12 , which is the desired result. 2
Next, we study the boundedness of level sets of merit function dfLT.
Theorem 3.6. Let the merit function dfLT be given as in (8). Suppose that the p-order cone complementarity problem has a strictly feasible point and that F is monotone. Then, the level set
LfdLT(γ) =
x ∈ IRn
fdLT(x) ≤ γ
is bounded for all γ ≥ 0.
Proof. Like the proof of Theorem 3.3, we prove this result by contradiction. Suppose there exists an unbounded sequence {xk} ⊆LfdLT(γ) for some γ ≥ 0. We claim that the sequences {xk−} and {(−F (xk))+} are bounded. If not, by the expression (8) of dfLT, we obtain
fdLT(xk) ≥ 1 2
kxk−k2+ k(−F (xk))+k2
→ ∞, which contradicts {xk} ⊆LfdLT(γ). Therefore, it follows that
lim sup
k→∞
kxk−k < ∞ and lim sup
k→∞
k(−F (xk))+k < ∞.
Then, applying Proposition 2.2 yields hxk, F (xk)i → ∞. This together with Lemma 3.2 implies
(xk• F (xk))Ω+
≥xk, F (xk) → ∞,
which leads to dfLT(xk) → ∞. This clearly contradicts {xk} ⊆LfdLT(γ). Hence, the level setLfdLT(γ) is bounded and the proof is complete. 2
Remark 3.1. In fact, if the term xk• F (xk)Ω
+ in the expression of dfLT is replaced by xk• F (xk), all Theorem 3.4, Lemma 3.2, Theorem 3.5, and Theorem 3.6 still hold.
3.4 The fourth class of merit function
In this subsection, in light of the product x • y and the NR merit function ΨNR, we consider another merit function as below:
fr(x) := 1
2kφNR(x, F (x))k2+1 2
(x • F (x))Ω+
2, (9)
where (x • y)Ω+denotes the projection of x • y onto Ω. As seen below, we verify that fr(x) is also a merit function for the p-order cone complementarity problem.
Theorem 3.7. Let the function fr be given as in (9). Then, for all x ∈ IRn, we have fr(x) = 0 ⇐⇒ x ∈ Kp, F (x) ∈ Kp∗ and hx, F (x)i = 0,
where K∗p denotes the dual cone of Kp, i.e., Kp∗ = Kq.
Proof. In view of the definition of fr given as in (9), we have fr(x) = 0 ⇐⇒
(x • F (x))Ω+
2 = 0 and ΨNR(x) = 1
2kφNR(x, F (x))k2 = 0,
⇐⇒ x ∈ Kp, F (x) ∈ Kp∗ and hx, F (x)i = 0,
where the second equivalence holds because ΨNR is a merit function for p-order cone complementarity problem. Thus, the proof is complete. 2
From Theorem 3.7, we see that if the squared exponent in the expression of fr is deleted, i.e.,
fer(x) := kφNR(x, F (x))k +
(x • F (x))Ω+
, (10)
then efr is also a merit function for the POCCP. In fact, for these two merit functions fr and efr, there has no big differences between them in addition to the nature that fr is better than efr. Next, we will establish the error bound properties for fr and efr.
Theorem 3.8. Let fr and efr be given as in (9) and (10), respectively. Suppose that F : IRn → IRn is strongly monotone mapping and that x∗ is a solution to the p-order cone complementarity problem. Then, there exists a scalar τ > 0 such that
τ kx − x∗k2 ≤ 3√
2fr(x)12
and τ kx − x∗k2 ≤ 2 efr(x).
Proof. From Lemma 3.1, we know
kφNR(x, F (x))k ≥ max{kx−k, k(−F (x))+k}.
Then, following similar arguments as in Theorem 3.5, we have τ kx − x∗k2 ≤
(x • F (x))Ω+
+ kx−k + k(−F (x))+k
≤ √
2(fr(x))12 + 2kφNR(x, F (x))k
= √
2(fr(x))12 + 2√ 2 1
2kφNR(x, F (x))k2
12
≤ √
2(fr(x))12 + 2√
2(fr(x))12
= 3√ 2
fr(x)
12
and
τ kx − x∗k2 ≤
(x • F (x))Ω+
+ kx−k + k(−F (x))+k
≤
(x • F (x))Ω+
+ 2 kφNR(x, F (x))k
≤ 2 efr(x),
where τ := ρ
max {1, kF (x∗)k, kx∗k}. Thus, the proof is complete. 2
The following theorem presents the boundedness of the level sets of the functions efr and fr.
Theorem 3.9. Let fr and efr be given as in (9) and (10), respectively. Suppose that that the p-order cone complementarity problem has a strictly feasible point and that F is monotone. Then, the level sets
Lfr(γ) = {x ∈ IRn| fr(x) ≤ γ}
and
Lfer(γ) = n
x ∈ IRn
efr(x) ≤ γ o are both bounded for all γ ≥ 0.
Proof. Here we only show the boundedness of the level sets of the function efr for all γ ≥ 0 because the same arguments can be easily applied to the case of fr.
As the proof in Theorem 3.3 and 3.6, we prove this result by contradiction. Suppose there exists an unbounded sequence {xk} ⊂ Lfer(γ) for some γ ≥ 0. If kxk−k → ∞ or (−F (xk))+
→ ∞, by Lemma 3.1, we know fer(xk) ≥
φNR(xk, F (xk))
→ ∞, which contradicts xk∈Lfer(γ). Hence, we have
lim sup
k→∞
xk−
< ∞ and lim sup
k→∞
(−F (xk))+
< ∞.
Then, applying Proposition 2.2 yields hxk, F (xk)i → ∞. This together with Lemma 3.2 gives
xk, F (xk) ≤
(xk• F (xk))Ω+
→ ∞,
which leads to efr(xk) → ∞. This is a contradiction because efr(xk) ≤ γ. Thus, the proof is complete. 2
Remark 3.2. As Remark 3.1, if the term xk• F (xk)Ω
+ in the expressions of fr and efr is replaced by xk• F (xk), all Theorem 3.7, Theorem 3.8, and Theorem 3.9 still hold.
3.5 The fifth class of merit functions
In this subsection, we introduce the implicit Lagrangian merit associated with the POC- COP. For any x ∈ IRn and α > 0, the implicit Lagrangian merit function is given by
Mα(x) := hx, F (x)i + 1
2αk(x − αF (x))+k2− kxk2+ k(αx − F (x))−k2− kF (x)k2 . (11)
This class of functions was first introduced by Mangasarian and Solodov [13] for solving nonlinear complementarity problems, and was extended by Kong, Tuncel and Xiu [11]
to the setting of symmetric cone complementarity problems. Moreover, for the setting of general closed convex cone complementarity problems in Hilbert space, Lu and Huang [12] further investigated this merit function. Accordingly, the corresponding results in [12] can be applied to the the setting of POCCP. For completeness, as below, the error bound property of the merit function Mα for the POCCP is also presented.
Property 3.3. ([12, Theorem 3.9]) Let Mα be given as in (11). Suppose that F : IRn→ IRn is a strongly monotone mapping with modulus ρ > 0 and is Lipschitz continuous with L > 0. Assume that the p-order cone complementarity problem has a solution x∗. Then, for any fixed α > 0, the following inequality holds
1
(α − 1)(2 + L)2 Mα(x) ≤ kx − x∗k ≤ α(1 + L)2
(α − 1)ρ2 Mα(x).
In the following theorem, we present the boundedness property of the level sets on the merit function Mα for solving the p-order cone complementarity problem.
Theorem 3.10. Suppose that the p-order cone complementarity problem has a strictly feasible point and that F is monotone. Then, the level set
LMα(γ) := {x ∈ IRn| Mα(x) ≤ γ}
is bounded for all γ ≥ 0.
Proof. First, we note that Mα(x) = hx, F (x)i + 1
2αk(x − αF (x))+k2− kxk2+ k(αx − F (x))−k2− kF (x)k2
= hx, F (x)i + 1
αhx − αF (x), (x − αF (x))+i − 1
2αk(x − αF (x))+k2+ kxk2 +
F (x), 1
α[αx − F (x) + (αx − F (x))−] − x
+ 1
2αkαx − F (x) + (αx − F (x))−− αxk2
= − hF (x), (x − αF (x))+− xi − 1
2αk(x − αF (x))+− xk2 +
F (x), 1
α(αx − F (x))+− x
+ α
2 1
α(αx − F (x))+− x
2
≥ 1 α
−
αF (x),1
α(αx − F (x))+− x
− 1 2k1
α(αx − F (x))+− xk2
+α 1
αF (x), 1
α(αx − F (x))+− x
+ α
2k1
α(αx − F (x))+− xk2
= α2− 1 2α
1
α(αx − F (x))+− x
2
= α2− 1 2α
(x − 1
αF (x))+− x
2
.
From Theorem 3.2, we know that the level set LΨα(γ) of the general NR merit function Ψα is bounded for all γ ≥ 0. With this, it is easy to see that the level set LMα(γ) is bounded for all γ ≥ 0. Then, the proof is complete. 2
4 Conclusion and future direction
Although the p-order cone complementarity problem belongs to nonsymmetric cone com- plementarity problem, for which there is no unified framework, we believe that the merit function approach may be an appropriate method that can be extended from symmetric cone complementarity problem to nonsymmetric cone complementarity problem. The key point to do such extension is constructing merit functions. Hence, in this paper, we present how to construct merit functions (by defining a new product) for the p-order cone complementarity problem. In addition, we have also shown under what conditions these merit functions have properties of error bounds and bounded level sets. These results provide a theoretical basis for designing the merit function method for solving the special nonsymmetric cone complementarity problem, i.e., p-order cone complementarity prob- lem. We leave this topic as our future direction. At last, we point out that the main idea is employed from [16] (which is for circular cone setting) and most analysis techniques look similar to those used in [16]. Nonetheless, the product is novel which contributes to the literature by providing a new way to deal with such complementarity problem.
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