Characterizations of Boundary Conditions on Some Non-Symmetric Cones

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Characterizations of Boundary Conditions on Some Non-Symmetric Cones

Yu-Lin Chang, Chu-Chin Hu, Ching-Yu Yang, and Jein-Shan Chen Department of Mathematics, National Taiwan Normal University, Taipei, Taiwan

ABSTRACT

In contrast to symmetric cone optimization, there has no unified framework for non-symmetric cone optimization. One main reason is that the structure of various non-symmetric cone differs case by case. Especially, their boundary conditions are usually mysterious. In this paper, we provide characterizations of boundary conditions on some non-symmetric cones, includingp-order cone, ellipsoidal cone, power cone and general closed convex cone. These results will be key bricks for further investigations on non-symmetric cone optimization accordingly.

ARTICLE HISTORY Received 25 June 2021 Revised 10 November 2021 Accepted 10 November 2021 KEYWORDS

Power cone;p-order cone;

ellipsoidal cone; non- symmetric cone

1. Introduction

In contrast to symmetric cone optimization, there has no unified framework for non-symmetric cone optimization. One main reason is that the structure of various non-symmetric cone differs case by case. In the literature, the study regarding non-symmetric cones, focuses on homogeneous cones [1–3], matrix norm cones [4], p-order cones [5–8], hyperbolicity cones [9–11], circular cones [12, 13] and copositive cones [14], etc. In particular, there seems no systematic study due to the various features and very few algorithms are proposed to solve optimization problems with these non-symmetric cones constraints, except for some interior-point type methods [1, 8,15,16]. For instance, Xue and Ye [8] study an optimization problem of minimizing a sum of p-norms, in which two new barrier functions are introduced for p-order cones and a primal-dual potential reduction algorithm is presented. Chua [1] combines the T-algebra with the primal- dual interior-point algorithm to solve the homogeneous conic programming problems. In light of the concept of self-concordant barriers and the efficient computational experience of the long path-following steps, Nesterov [15] proposes a new predictor-corrector path-following method.

CONTACT Jein-Shan Chen jschen@math.ntnu.edu.tw Department of Mathematics, National Taiwan Normal University, Taipei 11677, Taiwan

ß 2021 Taylor & Francis Group, LLC

https://doi.org/10.1080/01630563.2021.2006693

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Skajaa and Ye [16] investigate a homogeneous interior-point algorithm for non-symmetric convex conic optimization.

Besides the aforementioned interior-point type methods, are there any other possible algorithms that we can explore? To answer this question, we recall that one category of algorithms to deal with optimization problems rely on so-called complementarity functions, which play an important role in recasting the corresponding KKT conditions as a system of nonsmooth equations or an unconstrained minimization problem. Hence, looking for appropriate complementarity functions is an important issue from a computational viewpoint. In [17], the authors establish novel constructions of complementarity functions associated with symmetric cones, in which the decomposition associated with symmetric cones and the boundary conditions on symmetric cones are crucial in the analysis. Can these ideas be employed in non-symmetric cone setting? Indeed, for non-symmetric cones, their corresponding decompositions are figured out only in a few cases, see [13, 18]. The big hurdle is still the miscellaneous structures of non-symmetric cones. In view of this, we pay attention to the boundary conditions on some non-symmetric cones, which may help building up more useful links to explore possible algorithms for solving non-symmetric cone optimization.

It is well-known that the KKT conditions of an optimization problem is closely related to complementarity problem. More specifically, for a nonlinear programming, its KKT conditions can be rewritten as a nonlinear complementarity problem (NCP) as below: find a solution x 2Rⁿ to the system

x 0, FðxÞ 0, hx, FðxÞi ¼ 0,

where h, i is the Euclidean inner product and F is a map from Rⁿ to Rⁿ: There are essentially three popular ways to solve the NCP: (i) smoothing approach, (ii) merit functions approach, and (iii) projection-type approach.

All of these approaches rely on so-called NCP-functions and their corresponding merit functions. A function / : R²! R is called a NCP-function if

/ða, bÞ ¼ 0 () a, b 0 and ab ¼ 0:

From nonlinear programming problems to symmetric cone optimization, the above ideas can be employed if we extend the concepts of NCP-function and complementarity problem to the setting of symmetric cones.

Accordingly, as a natural extension of the NCP, the symmetric cone complementarity problem (SCCP) is to find a point v 2E such that

v 2 K, FðvÞ 2 K and hv, FðvÞi ¼ 0,

where F is an operator on Rⁿ; E is a Euclidean Jordan algebra and K is the

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corresponding symmetric cone in E: In particular, for E ¼ Rⁿ and K ¼ Kⁿ, i.e., the second-order cone (SOC, also called Lorentz cone, see Figure 1), defined by

Kⁿ :¼ fx ¼ ðx, xnÞ 2 Rⁿ¹ R j xn jjxjjg,

then the SCCP reduces to the second-order cone complementarity problem (SOCCP), which is to find a point z 2 Rⁿ satisfying

z 2 Kⁿ, FðzÞ 2 Kⁿ and hz, FðzÞi ¼ 0:

There exists a special Jordan product associated with SOC of x ¼ ðx, xnÞ 2 Rⁿ¹ R and y ¼ ðy, ynÞ 2 Rⁿ¹ R, which is defined by

x y:¼ y nx þ xny, hx, yi : Then, by [19, Proposition 2.1], there holds

x 2 Kⁿ, y 2 Kⁿ and hx, yi ¼ 0 () x 2 Kⁿ, y 2 Kⁿ and x y ¼ 0: (1) As a result, a so-called C-function (parallel to NCP-function) associated with SOC is a function u : Rⁿ Rⁿ ! Rⁿ satisfying

uðx, yÞ ¼ 0 () x 2 Kⁿ, y 2 Kⁿ and x y ¼ 0:

In the setting of non-symmetric cone, the corresponding complementarity problem becomes finding an element z such that

z 2 K, FðzÞ 2 K and hz, FðzÞi ¼ 0,

where K means the dual cone of K: To tackle this non-symmetric cone complementarity problem, we observe that the following parts are key com- ponents to do the analysis.

i. z ¼ 0 and Fð0Þ 2 K:

Figure 1. The graph of a 3-dimensional second-order cone.

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ii. z 2 K and F(z) ¼ 0.

iii. z 2 Knf0g, FðzÞ 2 Knf0g:

For part (iii), it leads to investigate z being on the boundary of the cone

@K and F(z) being on the boundary of the dual cone @K (this will be ela- borated more in Section 4). This is another reason why we look into the boundary conditions of non-symmetric cones. Apparently, our results on the boundary conditions of non-symmetric cones are connected to non- symmetric cone complementarity problem, and hence possible algorithms can be adopted in light of the complementarity problem. In other words, the results established in this paper will be key bricks for further investigations on non-symmetric cone optimization accordingly, which has important contribution to the development of non-symmetric cone optimization.

2. Boundary conditions on ellipsoidal cone

In this section, we provide characterizations of boundary conditions on ellipsoidal cone [20, 21]. Before showing out the characterizations, we pre- sent the boundary conditions on second-order cone in the below proposition, which are already studied in [17, Lemmas 2.3, 2.4 and 2.5]. However, we hereby offer an alternative proof without using the spectral decomposition of vectors. From which, the similar idea and technique will be applied to our subsequent analysis.

Proposition 2.1. Let x ¼ ðx, xnÞ, y ¼ ðy, ynÞ 2 Rⁿ¹ R with xn 6¼ 0 and yn 6¼ 0. Then, x 2 Kⁿ, y 2 Kⁿ and x y ¼ 0 if and only if x, y are both on the boundary of Kⁿ (that is, xn ¼ jjxjj, yn ¼ jjyjj) and ynx þ xny ¼ 0.

Moreover, we have x 6¼ 0, y 6¼ 0 and y ¼ mx, where m :¼^jj_jj^yjj_xjj:

Proof. “)” Suppose that x and y are nonzero vectors in Kⁿ with x y ¼ 0:

Then, we have

ynx þ xny ¼ 0 and 0 ¼ hx, yi ¼ xnynþ hx, yi,

which implies xnyn ¼ hx, yi: Since x, y 2 Kⁿ and xn 6¼ 0, yn 6¼ 0, we have x_n>0, yn>0, which yields

0<xnyn¼ jhx, yij jjxjj jjyjj xnyn:

This means xnyn ¼ jjxjj jjyjj and xn ¼ jjxjj, yn ¼ jjyjj; and by Cauchy- Schwartz inequality, one ofx and y is a multiple of the other.

“(” It suffices to show that if ynx þ xny ¼ 0 then hx, yi ¼ 0: Suppose ynx þ xny ¼ 0, then ynhx, yi þ xnjjyjj²¼ 0: Since yn ¼ jjyjj, ynhx, yi þ xny²_n ¼ 0 and hence hx, yi ¼ xnynþ hx, yi ¼ 0:

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Moreover, we have x 6¼ 0, y 6¼ 0, and xny ¼ ynx ) y ¼ y_n

xnx ¼ jjyjj jjxjjx:

Thus, the proof is complete. w

Remark 2.1. Note that inProposition 2.1a vector x ¼ ðx, xnÞ in the cone Kⁿ with xn 6¼ 0 is equivalent to say x 2 Kⁿ with x 6¼ 0. We use this just for convenience.

Proposition 2.1 characterizes the boundary conditions on second-order cone, which is a symmetric cone, namely it is self-dual. Here, we offer another way to verify it without using the spectral decomposition of vectors. In fact, this kind of techniques will be employed to derive analogous conditions for nonsymmetric cones, including ellipsoidal cone, p-order cone, power cone, and also for general closed convex cones.

The ellipsoidal cone (see Figure 2) is the form of K_e :¼ fx 2 Rⁿ j x^TQx 0, u^T_nx 0g,

where Q 2Rⁿⁿ is a nonsingular symmetric matrix with a single negative eigenvalue kn corresponding to the unit eigenvector un. The dual cone K_e is given as

K_e :¼ fy 2 Rⁿj y^TQ¹y 0, u^T_ny 0g:

Both K_e and K_e are closed convex cone and it is obvious that K_e is not a symmetric cone. The arising ellipsoidal cone complementarity problem (ECCP) is to find a point x 2 Rⁿ such that

Figure 2. The graph of a 3-dimensional ellipsoidal cone.

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x 2 K_e, FðxÞ 2 K_e and hx, FðxÞi ¼ 0,

where h, i is the Euclidean inner product and F: Rⁿ ! Rⁿ is a continu- ously differentiable mapping.

As an important prototype, ellipsoidal cone is a natural generalization of second-order cone, circular cone and elliptic cone. More precisely, let

Q ¼ In1 0

0 1

or In1 0 0 tan²h

or M^TM 0

0 1

,

and un ¼ ð0^,:::^,0^,1Þ^T, where In1 is the identity matrix of order n 1, M is any nonsingular matrix of order n 1, the ellipsoidal cone respectively reduces to the second-order cone:

Kⁿ :¼ x ¼ ðx, x nÞ 2 Rⁿ¹ R j xn jjxjj , the circular cone:

L_h :¼ x ¼ ðx, x nÞ 2 Rⁿ¹ R j xntanh jjxjj , and the elliptic cone:

Kⁿ_M:¼ x ¼ ðx, x nÞ 2 Rⁿ¹ R j xn jjMxjj :

Therefore, the ellipsoidal cone complementarity problem (ECCP) covers a range of nonsymmetric cone complementarity problems.

Since the ellipsoidal cone is described by a symmetric matrix Q, we can change the x-coordinate to the a-coordinate by an orthogonal matrix U^T, where columns of U are eigenvectors and the corresponding eigenvalues can be chose to satisfy

k1 k2 >0>kn:

Let U :¼ ½u1u2 un and a :¼ ½a1,a2,:::^,an ^T ¼ U^Tx, i.e., ai ¼ u^T_ix for i ¼ 1, 2,:::, n, it follows that

x^TQx ¼Xⁿ

i¼1

kia²_i and u^T_nx ¼an,

and the ellipsoidal cone K_e can be expressed as K_e ¼ Ua 2 Rⁿ

Xⁿ

i¼1

kia²_i 0 and an 0

( )

: (2)

Similarly, the dual cone K_e can be expressed as K_e ¼ Ub 2 Rⁿ

Xⁿ

i¼1

k¹_i b²_i 0 and b_n 0

( )

, (3)

where b :¼ ½b_1,b_2,:::^,b_n ^T ¼ U^Ty:

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The product of x and y associated with the ellipsoidal cone K_e is defined by

x y ¼ w hx, yi

where w:¼ ðw1,:::^,wn1Þ^T with wi ¼ b_nk¹²_iaiknank_i ¹²b_i:

Let D:¼ diag k₁¹², k₂¹², , ðknÞ¹²

h i

, the diagonal entries are singular values of Q, then D is nonsingular. To establish the boundary conditions on ellipsoidal cone, we need following lemma, which provide equivalent conditions involved in the Euclidean inner product h, i, the Jordan product “” and the product “”. Indeed, this lemma is a direct consequence of [21, Theorem 2.3, 2.5]. Here, we give a new proof, which is neat and differs form the one in the literature.

Lemma 2.1. For any x, y 2Rⁿ, the following are equivalent:

a. x 2 K_e, y 2 K_e and hx, yi ¼ 0:

b. D¹U^Tx 2 Kⁿ, DU^Ty 2 Kⁿ and ðD¹U^TxÞ ðDU^TyÞ ¼ 0:

c. x 2 K_e, y 2 K_e and x y ¼ 0:

Proof. “(a) () (b)” Note that

ðD¹U^TxÞ ðDU^TyÞ ¼

k¹²₁a1

...

k¹²_n1an1

ðknÞ¹²an

0 BB BB BB

@

1 CC CC CC A

k₁¹²b1

...

ðkn1Þ¹²b_n1 ðknÞ¹²b_n 0

BB BB BB

@

1 CC CC CC A

¼ ...

...

ha, bi 0 BB BB BB

@

1 CC CC CC A

¼ ...

...

hx, yi 0 BB BB BB

@

1 CC CC CC A

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where the equalities are due to x ¼ Ua, y ¼ Ub and U being orthogonal.

In addition, from (2) and (3), it follows that x 2 K_e ()Xⁿ

i¼1

kia²_i 0 and an 0

() k1a²₁þ þ kn1a²_n1 kna²_n and an 0 () D¹U^Tx 2 Kⁿ

Similarly, it can be verified that y 2 K_e if and only if DU^Ty 2 Kⁿ: Hence, conditions (a) and (b) are equivalent.

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“(a) () (c)” For x 2 Ke, y 2 K_e, it is sufficient to show that hx, yi ¼ 0 if and only if x y ¼ 0: From the above equality (4), we see that

hx, yi ¼ 0

() ðD¹U^TxÞ ðDU^TyÞ ¼ 0, () hx, yi ¼ 0 and ðknÞ¹²bn

k¹²1a1

...

k¹²_n1an1

0 BB B@

1 CC

CAþ ðknÞ¹²an

k1¹²b1

...

ðkn1Þ¹²bn1

0 BB B@

1 CC CA¼ 0,

() hx, yi ¼ 0 and ðknÞ¹²bnk¹²iaiþ ðknÞ¹²anki¹²bi¼ 0 for all i ¼ 1, :::, n1, () hx, yi ¼ 0 and wi¼ 0 for all i ¼ 1, :::, n1,

() x y ¼ 0:

Then, the proof is complete. ^w

Now, in light of Lemma 2.1, we characterize the boundary conditions on ellipsoidal cone as below.

Proposition 2.2. Let x, y 2Rⁿ. If x, y are nonzero vectors with x 2 K_e, y 2 K_e and hx, yi ¼ 0, then x is on the boundary of K_e and y is on the boundary of K_e:

Proof. Suppose that x, y are nonzero vectors with x 2 K_ey 2 K_e and hx, yi ¼ 0: By Lemma 2.1(b), it gives D¹U^Tx, DU^Tyare nonzero vectors in Kⁿ and ðD¹U^TxÞ ðDU^TyÞ ¼ 0: Then, applying (1) and Proposition 2.1, both D¹U^Tx, DU^Tyare on the boundary of Kⁿ: Hence, we have

Figure 3. The graph of a 3-dimensionalp-order cone with p ¼ 8.

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ðknÞ¹²an

₂

¼ k¹²₁a1

₂

þ þ k¹²_n1an1

₂

) k1a²₁þ þ kn1a²_n1þ kna²_n ¼ 0,

which implies Ua ¼ x is on the boundary of Ke: Similarly, it can be verified that Ub ¼ y is on the boundary of K_e: ^w

3. Boundary conditions on p-order cone and power cone

In this section, we establish the boundary conditions on p-order cone (see Figure 3) and power cone (see Figure 4), which are two popular cones in reality.

First, we quickly review the definition of p-order cone [22]. The p-order cone, denoted by Kp, is defined by

Kp:¼ x ¼ ðx, xn nÞ 2 Rⁿ¹ R j xn jjxjj_po , and its dual cone is given by

K_p :¼ y ¼ ðy, yn nÞ 2 Rⁿ¹ R j yn jjyjj_qo

¼ Kq,

where q 1 and ¹_pþ¹_q¼ 1: Both Kp and K_p are closed convex cone.

Indeed, the p-order cone is also a generalization of the second-order cone, but there is no Jordan product for the setting of the p-order cone yet.

There is a product for p-order cone defined in [22] as below x䉫y ¼ w

hx, yi

where w:¼ ðw1,:::^,wn1Þ^T with wi ¼ jxnj^p^qjyijjynjjxij^p^q and x ¼ ðx1,:::^,xn1Þ^T,y ¼ ðy1,:::^,yn1Þ^T: When x, y satisfy POCCP (see [22]) and

Figure 4. The graph of a 3-dimensional power cone witha ¼ 0:2:

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p ¼ q ¼ 2, it is the Jordan product in the setting of second-order cone, but it is inappropriate to call it a “Jordan product”, because it is not symmetric with respect to the inner product, that is, the condition hx䉫y, zi ¼ hy, x䉫zi is not satisfied. For example, for p ¼ 3 and q ¼³₂, let x ¼ ð1, 2Þ, y ¼ ð1, 3Þ and z ¼ ð1, 4Þ then hx䉫y, zi ¼ 29 but hy, x䉫zi ¼ 27: Moreover, the 䉫 product is even not commutative due to ð1, 2Þ䉫ð1, 3Þ 6¼ ð1, 3Þ䉫ð1, 2Þ: The following proposition describes the boundary conditions on p-order cone, and the proof is similar to that ofProposition 2.1.

Proposition 3.1. Let x ¼ ðx,xnÞ 2 Rⁿ¹ R, y ¼ ðy,ynÞ 2 Rⁿ¹ R with xn6¼ 0 and yn6¼ 0. If x 2 Kp, y 2 K_p and hx, yi ¼ 0, then x is on the boundary of Kpand y is on the boundary of K_p, that is, xn¼ jjxjj_pand yn¼ jjyjj_q:

Proof. Suppose that x and y are nonzero vectors with x 2 Kp, y 2 K_p and hx, yi ¼ 0, then xnynþ hx, yi ¼ 0: Applying Holder’s inequality yields

0<xny_n ¼ jhx, yij jjxjj_pjjyjj_q x_ny_n,

which says xnyn ¼ jjxjj_pjjyjj_q and xn ¼ jjxjj_p, yn ¼ jjyjj_q: Then, the proof is

complete. w

Another cone that has real applications is power cone, see [18] for more details. We describe its definition as below. Let a1,:::, am2 R be positive with a1þ þ am¼ 1: The power cone K_a is defined by

K_a:¼ ðx, yÞ 2 R ^m_þ Rⁿ j xâ₁¹xâ₂² xâ_m^m jjyjj , and its dual cone is given by

K_a :¼ ðu, vÞ 2 R^mþ Rⁿ

u1

a1

_a₁ u2

a2

_a₂

um

am

_a_m

jjvjj

( )

:

Following the similar techniques, the boundary conditions on power cone are established as below.

Proposition 3.2. Let ðx, yÞ 2 R^m_þ Rⁿ and ðu, vÞ 2 R^m_þ Rⁿ be nonzero vectors with hx, ui 6¼ 0. If ðx, yÞ 2 K_a, ðu, vÞ 2 K_a and hðx, yÞ, ðu, vÞi ¼ 0, then (x, y) is on the boundary of K_a and (u, v) is on the boundary of K_a:

Proof. Suppose that x1u1þ þ xmumþ y1v1þ þ ynvn ¼ 0, using Cauchy-Schwartz inequality, it leads to

0<x1u1þ þ xmum¼ ðy1v1þ þ ynvnÞ jjyjj jjvjj:

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Then, we have

jjyjj jjvjj x1u1þ þ xmum

¼ a1 x1

u₁ a1

þ þ am xm

u_m am

ðx1

u1

a1

Þ^a¹ ðxm

um

am

Þ^a^m

¼ x^a₁¹ x^a_m^mðu1

a1

Þ^a¹ ðum

am

Þ^a^m

jjyjj jjvjj

>0:

This indicates xâ₁¹xâ₂² x_mâ^m ¼ jjyjj and ðû_a¹

1Þ^a¹ð^u_a²

2Þ^a² ð^u_a^m

mÞ^a^m ¼ jjvjj, In other words, (x, y) is on the boundary of K_a and (u, v) is on the boundary

of K_a: ^w

4. Boundary conditions on general closed convex cones

It is a natural question whether the aforementioned analysis can be extended to general closed convex cones. In general, suppose that ðV, h, iÞ is an inner product space and K is a closed convex cone in V: Let K denote the dual cone of K, then the boundary conditions are established as follows.

Proposition 4.1. Suppose that ðV, h, iÞ is an inner product space, K is a closed convex cone in V, and both K and its dual K are solid (i.e., their interiors are nonempty). Let x, y 2V, and suppose x 2 Knf0g, y 2 Knf0g and hx, yi ¼ 0, then x is on the boundary of K and y is on the boundary of K:

Proof. Suppose on the contrary that x is an interior point of K, then there exists a radius r> 0 such that BrðxÞ ¼ x þ Brð0Þ K: Consider the point z ¼ ^r₂_jjyjj^y , then jjzjj<r and hence x þ z 2 BrðxÞ K: Thus, there holds

0 hx þ z, yi ¼ hx, yi þ hz, yi ¼ hz, yi ¼ r 2jjyjj,

which is indeed a contradiction. Therefore, x is on the boundary of K:

On the other hand, since K is a closed convex cone, we have ðKÞ ¼ K:

Consider y 2K and x 2K ¼ ðKÞ with hx, yi ¼ 0, as shown above, y

must be on the boundary of K: ^w

Remark 4.1. The condition of K being solid is essential in the proof of Proposition 4.1. Here is a counterexample, which tells that Proposition 4.1 does not hold in the relative sense. The relative interior ofK is given by

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relintðKÞ :

¼ fv 2 K j for all u

2 K, there exists some k>1 such that kv þ ð1kÞu 2 Kg:

Consider K ¼ fð0, 0, zÞ j z 0g the non-negative z-axis in R³, then the dual cone is K ¼ fðx, y, zÞ j x, y 2 R, z 0g. Let u ¼ ð0, 0, 1Þ and v ¼ ð1, 1, 0Þ, we have u 2 relintðKÞ and v 2 @K, but hu, vi ¼ 0:

In fact, the condition of a closed convex cone in Proposition 4.1 can be relaxed to a closed set in an inner product space ðV, h, iÞ: A dual cone for any subset C in V can be defined as

C¼ fy 2 V j hy, xi 0, 8x 2 Cg:

Notice that C is always a convex cone and the result of Proposition 4.1 still holds for the closed set C case.

Proposition 4.2. Suppose that ðV, h, iÞ is an inner product space, C is a closed set in V. Let x, y 2 V, and suppose x 2 Cnf0g, y 2 Cnf0g and hx, yi ¼ 0, then x is on the boundary of C and y is on the boundary of C: Proof. Note that if C is a closed set with the empty interior, then x 2 C if and only if x belongs to boundary of C. Therefore, we only consider cases:

(i) intðCÞ is nonempty and x 2 intðCÞ or (ii) intðCÞ is nonempty and y 2 intðCÞ: In each case, with almost the same arguments as in the proof of Proposition 4.1, by replacing K, K with C, C, respectively, we can verify that x is on the boundary of C, and y is on the boundary of C, without

using the convex cone property of K: ^w

Based on the observation of Proposition 4.1, we would like to propose an approach to solve the complementarity problem on general closed convex cones. Let F: V ! V be a function defined on V: Before to solve z 2 K and FðzÞ 2 K, we divide it into the following three possibilities:

I. z ¼ 0 and Fð0Þ 2K: II. z 2K and F(z) ¼ 0.

III. z 2Knf0g, FðzÞ 2 Knf0g:

For solutions of types I and II, we do not need to check hz, FðzÞi ¼ 0 since it is automatically satisfied. In fact, for solution of type I, we only have to check Fð0Þ 2K, and this is simple. On the other hand, for solution of type II, this is an inclusion problem to find a z 2 F¹ð0Þ \ K:

For solution of type III, combined with hz, FðzÞi ¼ 0, then we observe that z is on the boundary of the cone @K and F(z) is on the boundary of

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the dual cone @K by Proposition 4.1. We can take second-order cone Kⁿ as a concrete example. Thus, to find the solution is equivalent to solve the equations:

z_n ¼ jjzjj, FnðzÞ ¼ jjFðzÞjj and hz, FðzÞi ¼ 0:

A more fancy way to express the result is the following. Let C ¼ K K: It is easy to see that C is a closed convex cone inV V: Define

D ¼ fðu, vÞ 2 @K @K j u 6¼ 0, v 6¼ 0, hu, vi ¼ 0g:

Now, we denote @C :¼ ðf0g KÞ [ ðK f0gÞ [ D: Suppose z is a solution to the below complementarity problem:

z 2K, FðzÞ 2 K, hz, FðzÞi ¼ 0:

Then, we have ðz, FðzÞÞ 2 @C, and vice versa.

In summary, if we understand the boundary behavior of the cone and its dual more, then we understand the complementarity problem more. In most of the time, taking ellipsoidal cone, p-order cone and power cone for instances, the boundary conditions on these cones are always defined by algebraic equations. These are good and helpful for subsequent analysis and investigation.

Another research direction is studying on the property of the function F on the boundary behavior of the cone and its dual, which are also essential to the complementarity problem. Maybe knowledge from algebraic geom- etry would help. We leave it for future work.

Funding

The author’s work is supported by Ministry of Science and Technology, Taiwan.

ORCID

Jein-Shan Chen http://orcid.org/0000-0002-4596-9419

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