DOI 10.1007/s10957-008-9495-y
Growth Behavior of Two Classes of Merit Functions for Symmetric Cone Complementarity Problems
S.H. Pan· J.-S. Chen
Published online: 14 January 2009
© Springer Science+Business Media, LLC 2009
Abstract In the solution methods of the symmetric cone complementarity problem (SCCP), the squared norm of a complementarity function serves naturally as a merit function for the problem itself or the equivalent system of equations reformulation.
In this paper, we study the growth behavior of two classes of such merit functions, which are induced by the smooth EP complementarity functions and the smooth im- plicit Lagrangian complementarity function, respectively. We show that, for the linear symmetric cone complementarity problem (SCLCP), both the EP merit functions and the implicit Lagrangian merit function are coercive if the underlying linear transfor- mation has the P -property; for the general SCCP, the EP merit functions are coercive only if the underlying mapping has the uniform Jordan P -property, whereas the co- erciveness of the implicit Lagrangian merit function requires an additional condition for the mapping, for example, the Lipschitz continuity or the assumption as in (45).
Keywords Symmetric cone complementarity problem· Jordan algebra · EP merit functions· Implicit Lagrangian function · Coerciveness
Communicated by M. Fukushima.
The authors would like to thank the two anonymous referees for their helpful comments which improved the presentation of this paper greatly.
The research of J.-S. Chen was partially supported by National Science Council of Taiwan.
S.H. Pan
School of Mathematical Sciences, South China University of Technology, Guangzhou 510640, China e-mail:shhpan@scut.edu.cn
J.-S. Chen (
)Department of Mathematics, National Taiwan Normal University, Taipei 11677, Taiwan e-mail:jschen@math.ntnu.edu.tw
1 Introduction
Given a Euclidean Jordan algebraA = (V, ◦, ·, ·), where ‘◦’ denotes the Jordan product andV is a finite-dimensional vector space over the real field R equipped with the inner product·, ·, let K be a symmetric cone in V and let F : V → V be a continuous mapping. The symmetric cone complementarity problem (SCCP) is to find ζ∈ V such that
ζ∈ K, F (ζ ) ∈ K, ζ, F (ζ ) = 0. (1) The model provides a simple unified framework for various existing complementar- ity problems such as the nonlinear complementarity problem over nonnegative or- thant cone (NCP), the second-order cone complementarity problem (SOCCP) and the semidefinite complementarity problem (SDCP), and hence has extensive applica- tions in engineering, economics, management science, and other fields; see [1–4] and references therein. When F (ζ )= L(ζ ) + b, L : V → V being a linear transformation and b∈ V, the SCCP becomes the linear complementarity problem over symmetric cones (SCLCP),
ζ∈ K, L(ζ ) + b ∈ K, ζ, L(ζ ) + b = 0. (2) Recently, there is much interest in the study of merit functions or complemen- tarity functions associated with symmetric cones and the development of the merit function approach or the smoothing method for solving the SCCP. For example, Liu, Zhang and Wang [5] extended a class of merit functions proposed in [6] to the SCCP, Kong, Tuncel and Xiu [7] studied the extension of the implicit Lagrangian function proposed by Mangasarian and Solodov [8] to symmetric cones; Kong, Sun and Xiu [9] proposed a regularized smoothing method by the natural residual complementar- ity function associated with symmetric cones; and Huang and Ni [10] developed a smoothing-type algorithm with the regularized CHKS smoothing function over the symmetric cone.
A mapping φ: V × V → V is called a complementarity function associated with the symmetric coneK if the following equivalence holds:
φ (x, y)= 0 ⇐⇒ x ∈ K, y ∈ K, x, y = 0. (3) By Propositions III.4.4–4.5 and Theorem V.3.7 of [11], the Euclidean Jordan algebra V and the corresponding symmetric cone K can be written as
V = V1× V2× · · · × Vm and K = K1× K2× · · · × Km, (4) where each (Vi,◦, ·, ·) is a simple Euclidean Jordan algebra and Ki is the sym- metric cone inVi. Moreover, for any x= (x1, . . . , xm), y= (y1, . . . , ym)∈ V with xi, yi∈ Vi,
x◦ y = (x1◦ y1, . . . , xm◦ ym) and x, y = x1, y1 + · · · + xm, ym.
Therefore, the characterization (3) of complementarity function is equivalent to φ (x, y)= 0 ⇐⇒ xi∈ Ki, yi∈ Ki, xi, yi = 0 for all i = 1, 2, . . . , m. (5)
This means that, if φ is a complementarity function associated with the coneK, then φ(x, y) for any x= (x1, . . . , xm), y= (y1, . . . , ym)∈ V with xi, yi∈ Vi can be written as
φ (x, y):=
φ(1)(x1, y1), φ(2)(x2, y2), . . . , φ(m)(xm, ym)
,
where φ(i): Vi× Vi→ Viis a complementarity function associated withKi, i.e., φ(i)(xi, yi)= 0 ⇐⇒ xi∈ Ki, yi∈ Ki, xi, yi = 0. (6) Consequently, the SCCP can be reformulated as the following system of equations:
(ζ ):= φ(ζ, F (ζ )) =
⎛
⎜⎝
φ(1)(ζ1, F1(ζ )) ... φ(m)(ζm, Fm(ζ ))
⎞
⎟⎠ = 0,
which naturally induces a merit function f: V → R+for the SCCP, defined as
f (ζ ):= (1/2) (ζ ) 2= (1/2) m i=1
φ(i)(ζi, Fi(ζ )) 2.
In the rest of this paper, corresponding to the Cartesian structure ofV, we always write F= (F1, . . . , Fm)with Fi: V → Viand ζ= (ζ1, . . . , ζm)with ζi∈ Vi.
The merit function f is often involved in the design of the merit function methods or the equation reformulation methods for the SCCP. For these methods, the coer- civeness of f plays a crucial role in establishing the global convergence results. In this paper, we will study the growth behavior of two classes of such merit functions, which respectively correspond to the EP-functions introduced by Evtushenko and Purtov [12] and the implicit Lagrangian function by Mangasarian and Solodov [8].
The EP-functions over the symmetric cone K were first introduced by Kong and Xiu [13], defined by
φα(x, y):= −x ◦ y + (1/2α)
(x+ y)− 2
, 0 < α≤ 1, (7)
φβ(x, y):= −x ◦ y + (1/2β)
(x−)2+ (y−)2
, 0 < β < 1, (8) where (·)− denotes the minimum metric projection onto−K. They showed that φα
and φβ are continuously differentiable and strongly semismooth complementarity functions associated with K. In addition, Kong, Tuncel and Xiu [7] extended the implicit Lagrangian function to the symmetric cone K and studied its continuous differentiability and strongly semismoothness. The function is defined as follows:
φMS(x, y):= x ◦ y + (1/2α)
[(x − αy)+]2− x2+ [(y − αx)+]2− y2 , (9) where α > 0 ( = 1) is a fixed constant, and (·)+denotes the minimum metric projec- tion onK. Particularly, for the implicit Lagrangian merit function of the SCCP, they presented a mild stationary point condition and proved that it can provide a global error bound under the uniform Cartesian P -property and Lipschitz continuity of F .
This paper is mainly concerned with the growth behavior of the merit functions induced by the above three types of smooth complementarity functions, that is,
fα(ζ ):= (1/2) φα(ζ, F (ζ )) 2= (1/2) m i=1
φα(i)(ζi, Fi(ζ )) 2, (10)
fβ(ζ ):= (1/2) φβ(ζ, F (ζ )) 2= (1/2) m i=1
φβ(i)(ζi, Fi(ζ )) 2, (11)
fMS(ζ ):= (1/2) φMS(ζ, F (ζ )) 2= (1/2) m
i=1
φ(i)MS(ζi, Fi(ζ )) 2, (12)
where φ(i)α , φ(i)
β , φMS(i)defined as in (7), (8), (9), respectively, are a complementarity function associated withKi. Specifically, we show that for the SCLCP (2), the EP merit functions fαand fβ and the implicit Lagrangian function fMSare coercive only if the linear transformation L has the P -property; for the general SCCP, fαand fβ are coercive if the mapping F has the uniform Jordan P -property, but the coerciveness of fMSneeds an additional condition of F , for example, the Lipschitz continuity or the assumption as in (45). WhenV = Rnand “◦” denotes the componentwise product of the vectors, the obtaining results precisely reduce to those of Theorems 2.1 and 2.3 in [14] and Theorem 4.1 in [15]. However, for the general Euclidean Jordan algebra even the Lorentz algebra, to the best of our knowledge, similar results have not been established for these merit functions.
Throughout this paper, · represents the norm induced by the inner prod- uct·, ·, int(K) denotes the interior of the symmetric cone K, and (x1, . . . , xm)∈ V1× · · · × Vmis viewed as a column vector inV = V1× · · · × Vm. For any x∈ V, (x)+and (x)−denotes the metric projection of x ontoK and −K, respectively, i.e., (x)+:= argminy∈K{ x − y }.
2 Preliminaries
This section recalls some concepts and materials of Euclidean Jordan algebras that will be used in the subsequent analysis. More detailed expositions of Euclidean Jor- dan algebras can be found in Koecher’s lecture notes [16] and the monograph by Faraut and Korányi [11]. Besides, one can find excellent summaries in [17–19].
A Euclidean Jordan algebra is a triple (V, ◦, ·, ·V), where (V, ·, ·V)is a finite- dimensional inner product space over the real fieldR and (x, y) → x ◦y : V×V → V is a bilinear mapping satisfying the following three conditions:
(i) x◦ y = y ◦ x for all x, y ∈ V;
(ii) x◦ (x2◦ y) = x2◦ (x ◦ y) for all x, y ∈ V, where x2:= x ◦ x;
(iii) x ◦ y, zV= y, x ◦ zVfor all x, y, z∈ V.
We assume that there is an element e∈ V such that x ◦ e = x for all x ∈ V and call e the unit element. Let ζ (x) be the degree of the minimal polynomial of x∈ V,
which can be equivalently defined as ζ (x):= min{k : {e, x, x2, . . . , xk} are linearly dependent}. Since ζ (x)≤ dim(V) where dim(V) denotes the dimension of V, the rank of (V, ◦) is well defined by q := max{ζ(x) : x ∈ V}. In a Euclidean Jordan algebra (V, ◦, ·, ·V), we denoteK := {x2: x ∈ V} by the set of squares. From The- orem III.2.1 of [11],K is a symmetric cone. This means that K is a self-dual closed convex cone with nonempty interior int(K), and for any x, y ∈ int(K), there exists an invertible linear transformationT : V → V such that T (K) = K.
A Euclidean Jordan algebra is said to be simple if it is not the direct sum of two Euclidean Jordan algebras. By Propositions III.4.4–III.4.5 and Theorem V.3.7 of [11], any Euclidean Jordan algebra is, in a unique way, a direct sum of simple Euclidean Jordan algebras. Moreover, the symmetric cone in a given Euclidean Jordan alge- bra is, in a unique way, a direct sum of symmetric cones in the constituent simple Euclidean Jordan algebras. Here are two popular examples of simple Euclidean Jor- dan algebras. One is the algebraSn of n× n real symmetric matrices with the inner productX, Y Sn:= Tr(XY ) and the Jordan product X ◦ Y := (XY + Y X)/2, where Tr(X) is the trace of X and XY is the usual matrix multiplication of X and Y . In this case, the unit element is the identity matrix inSn and the coneK is the set of all positive semidefinite matrices. The other is the Lorentz algebra Ln, also called the quadratic forms algebra, withV = Rn,·, ·Vbeing the usual inner product inRnand the Jordan product defined by
x◦ y := (x, yRn, x1y2+ y1x2) , (13) for any x= (x1, x2), y= (y1, y2)∈ R × Rn−1. Under this case, the unit element e= (1, 0, . . . , 0) ∈ Rn, and the associate cone, called the Lorentz cone (or the second- order cone), is given byK := {x = (x1, x2)∈ R × Rn−1: x2 ≤ x1}.
Recall that an element c∈ V is said to be idempotent if c2= c. Two idempotents c and d are said to be orthogonal if c◦ d = 0. We say that {c1, c2, . . . , ck} is a complete system of orthogonal idempotents if
c2j= cj, cj◦ ci= 0 if j = i, j, i = 1, 2, . . . , k, and k j=1
cj= e.
A nonzero idempotent is said to be primitive if it cannot be written as the sum of two other nonzero idempotents. We call a complete system of orthogonal primitive idempotents a Jordan frame. Then, we have the following spectral decomposition theorem (see Theorem III.1.2 in [11]).
Theorem 2.1 Suppose thatA = (V, ◦, ·, ·V)is a Euclidean Jordan algebra with rank q. Then, for each x∈ V, there exist a Jordan frame {c1, c2, . . . , cq} and real numbers λ1(x), λ2(x), . . . , λq(x)such that x=q
j=1λj(x)cj. The numbers λj(x) (counting multiplicities), which are uniquely determined by x, are called the eigen- values of x.
In the sequel, we denote by λmax(x)and λmin(x)the maximum eigenvalue and the minimum eigenvalue of x respectively and by tr(x):=q
j=1λj(x)the trace of x.
By Proposition III.1.5 of [11], a Jordan algebraA = (V, ◦) over R with a unit element e∈ V is Euclidean if and only if the symmetric bilinear form tr(x ◦ y) is positive definite. Therefore, we may define an inner product·, · on V by
x, y := tr(x ◦ y), ∀x, y ∈ V.
Let · be the norm on V induced by the inner product ·, ·, namely,
x :=
x, x =
q
j=1
λ2j(x)
1/2
, ∀x ∈ V.
Then, by the Schwartz inequality, it is easy to verify that
x ◦ y ≤ x · y , ∀x, y ∈ V. (14) For a given x∈ V, we define the linear operator L : V → V by
L(x)y := x ◦ y, for every y ∈ V.
Since the inner product·, · is associative by the associativity of tr(·) (see Propo- sition II.4.3 of [11]), i.e., for all x, y, z∈ V, it holds that x, y ◦ z = y, x ◦ z, the linear operatorL(x) for each x ∈ V is symmetric with respect to ·, · in the sense that
L(x)y, z = y, L(x)z, ∀y, z ∈ V.
We say that elements x and y operator commute ifL(x) and L(y) commute, i.e., L(x)L(y) = L(y)L(x).
Let ϕ: R → R be a real-valued function. Then, it is natural to define a vector- valued function associated with the Euclidean Jordan algebraA = (V, ◦, ·, ·) by
ϕV(x):= ϕ(λ1(x))c1+ ϕ(λ2(x))c2+ · · · + ϕ(λq(x))cq, (15) where x∈ V has the spectral decomposition x =q
j=1λj(x)cj. The function ϕV is also called the Löwner operator in [19] and shown to inherit many properties from ϕ. Especially, when ϕ(t) is chosen as max{0, t} and min{0, t} for t ∈ R, respectively, ϕV becomes the metric projection operator ontoK and −K,
(x)+:=
q j=1
max
0, λj(x)
cj, (x)−:=
q j=1
min
0, λj(x)
cj. (16)
It is easy to verify that x= (x)++ (x)−,|x| = (x)+− (x)−and x 2= (x)+ 2+ (x)− 2.
An important part in the theory of Euclidean Jordan algebras is the Peirce decom- position theorem which is stated as follows (see Theorem IV.2.1 of [11]).
Theorem 2.2 LetA = (V, ◦, ·, ·) be a Euclidean Jordan algebra with rank q and let{c1, c2, . . . , cq} be a Jordan frame in V. For i, j ∈ {1, 2, . . . , q}, define
Vii:=
x∈ V : x ◦ ci= x
, Vij:=
x∈ V : x ◦ ci= (1/2)x = x ◦ cj
, i = j.
Then, the spaceV is the orthogonal direct sum of subspaces Vij (i≤ j). Further- more,
(a) Vij◦ Vij⊆ Vii+ Vjj; (b) Vij◦ Vj k⊆ Vikif i = k;
(c) Vij◦ Vkl= {0} if {i, j} ∩ {k, l} = ∅.
To close this section, we recall the concepts of the P -property and the uniform Jordan P -property for a linear transformation and a nonlinear mapping.
Definition 2.1 A linear transformation L: V → V is said to have the P -property if ζ and L(ζ ) operator commute
ζ◦ L(ζ ) ∈ −K ⇒ x = 0.
Definition 2.2 A mapping F= (F1, . . . , Fm)with Fi: V → Vi is said to have:
(i) the uniform Cartesian P -property if there is a positive scalar ρ such that, for any ζ, ξ∈ V, there is an index ν ∈ {1, 2, . . . , m} such that
ζν− ξν, Fν(ζ )− Fν(ξ ) ≥ ρ ζ − ξ 2,
(ii) the uniform Jordan P -property if there is a positive scalar ρ such that, for any ζ, ξ∈ V, there is an index ν ∈ {1, 2, . . . , m} such that
λmax[(ζν− ξν)◦ (Fν(ζ )− Fν(ξ ))]≥ ρ ζ − ξ 2.
Unless otherwise stated, in the subsequent analysis, we assume that A = (V, ◦, ·, ·) is a simple Euclidean Jordan algebra of rank q and dim(V) = n.
3 Coerciveness of fα and fβ
In this section, we study the conditions under which the EP merit functions fα and fβ are coercive. For this purpose, we need Lemma 3.1 of [13], which is stated as follows.
Lemma 3.1 For a given Jordan frame{c1, c2, . . . , cq}, if z ∈ V can be written as
z= q i=1
zici+
1≤i<j≤q
zij,
with zi∈ R for i = 1, 2, . . . , q and zij∈ Vij for 1≤ i < j ≤ q, then
z+= q i=1
sici+
1≤i<j≤q
sij, z−= q i=1
wici+
1≤i<j≤q
wij,
where si ≥ (zi)+≥ 0, 0 ≥ (zi)−≥ wi with si + wi = zi for i = 1, . . . , q, and sij, wij∈ Vij with sij+ wij= zij for 1≤ i < j ≤ q.
The following lemma summarizes some important inequalities involved in the maximum eigenvalue and the minimum eigenvalue for any x∈ V. Since their proofs can be found in Lemma 14 of [17] and Proposition 2.1 of [20], we here omit them.
Lemma 3.2 For any x, y∈ V, the following inequalities always hold:
(a) λmin(x) c 2≤ x, c ≤ λmax(x) c 2for any nonzero idempotent c;
(b) |λmax(x+ y) − λmax(x)| ≤ y and |λmin(x+ y) − λmin(x)| ≤ y ; (c) λmax(x+ y) ≤ λmax(x)+ λmax(y)and λmin(x+ y) ≥ λmin(x)+ λmin(y).
Using Lemmas 3.1–3.2, we may establish a lower bound for φα(x, y) and φβ(x, y) .
Lemma 3.3 Let φα and φβ be given by (7) and (8), respectively. Then, for any x, y∈ V,
φα(x, y) ≥
(2α− α2)/(2α)
max
[(λmin(x))−]2, [(λmin(y))−]2 , (17) φβ(x, y) ≥
(1− β2)/(2β)
max
[(λmin(x))−]2,[(λmin(y))−]2
. (18) Proof Suppose that x has the spectral decomposition x=q
i=1xici with xi∈ R and {c1, c2, . . . , cq} being a Jordan frame. From Theorem2.2, y∈ V can be expressed by
y= q i=1
yici+
1≤i<j≤q
yij, (19)
where yi∈ R for i = 1, 2, . . . , q and yij∈ Vij. Therefore, for any l∈ {1, 2, . . . , q},
cl, x◦ y = cl◦ x, y =
xlcl,
q i=1
yici+
1≤i<j≤q
yij
= xl
cl,
q i=1
yici
+ xl
cl,
1≤i<j≤q
yij
= xlyl, (20)
where the last equality is due to the fact thatcl,
1≤i<j≤qyij = 0 by the orthogo- nality of Vij (i≤ j).
We next prove the inequality (17). From (19) and the spectral decomposition of x,
x+ y = q i=1
(xi+ yi)ci+
1≤i<j≤q
yij,
which together with Lemma3.1implies that
(x+ y)−= q i=1
uici+
1≤i<j≤q
uij,
where ui≤ (xi+ yi)−≤ 0 for i = 1, 2, . . . , q and uij∈ Vij. By this, we can compute
cl,[(x + y)−]2
=
cl◦
q
i=1
uici+
1≤i<j≤q
uij
, (x+ y)−
=
ulcl+
cl◦
1≤i<j≤q
uij
,
q i=1
uici+
1≤i<j≤q
uij
= u2l + ul
cl,
1≤i<j≤q
uij
+
1≤i<j≤q
uij, cl◦ q
i=1
uici
+
cl◦
1≤i<j≤q
uij,
1≤i<j≤q
uij
= u2l +
cl,
1≤i<j≤q
uij
2
, ∀l = 1, 2, . . . , q, (21)
where the last equality is due to the fact thatcl,
1≤i<j≤quij = 0 by the orthogo- nality of Vij (i≤ j). Now, using (20)–(21), we obtain that
cl,−φα(x, y) =
cl, x◦ y − (1/2α)
(x+ y)− 2
= xlyl− (1/2α)
u2l +
cl,
1≤i<j≤q
uij
2
≤ xlyl− (1/2α)
(xl+ yl)− 2
, ∀l = 1, 2, . . . , q, (22) where the inequality is due to the following facts
ul≤ (xl+ yl)−≤ 0 and
cl,
1≤i<j≤q
uij
2
≥ 0.
On the other hand, from Lemma3.2(a) we have that
cl,−φα(x, y) ≥ λmin(−φα(x, y)) cl 2= λmin(−φα(x, y)), ∀l = 1, 2, . . . , q.
(23) Thus, combining (22) with (23), it follows that
2αλmin(−φα(x, y))≤ 2αxlyl− [(xl+ yl)−]2, ∀l = 1, 2, . . . , q.
Let λmin(x)= xν with ν∈ {1, 2, . . . , q}. Then, we have particularly that
2αλmin(−φα(x, y))≤ 2αλmin(x)yν− [(λmin(x)+ yν)−]2. (24) We next proceed to the proof for two cases: λmin(x)≤ 0 and λmin(x) >0.
Case (i): λmin(x)≤ 0. Under this case, we prove the following inequality:
2αλmin(x)yν− [(λmin(x)+ yν)−]2≤ −(2α − α2)[(λmin(x))−]2, (25) which, together with (24), implies immediately
φα(x, y) ≥!!λmin(−φα(x, y))!! ≥
(2α− α2)/(2α)
[(λmin(x))−]2. (26)
In fact, if λmin(x)+ yν≥ 0, then we can deduce that
2αλmin(x)yν− [(λmin(x)+ yν)−]2= 2α(λmin(x))−(yν)+
≤ −(2α − α2)[(λmin(x))−]2; otherwise, we have that
2αλmin(x)yν− [(λmin(x)+ yν)−]2= 2αλmin(x)yν− [(λmin(x)+ yν)]2
≤ −(2α − α2)[λmin(x)]2
= −(2α − α2)[(λmin(x))−]2. Case (ii): λmin(x) >0. Under this case, the inequality (26) clearly holds.
Summing up the above discussions, the inequality (26) holds for any x, y∈ V. In view of the symmetry of x and y in φα(x, y), we also have that
φα(x, y) ≥
(2α− α2)/(2α)
[(λmin(y))−]2,
for any x, y∈ V. Thus, the proof of the inequality (17) is completed.
We next prove the inequality (18). By the spectral decomposition of x, we have that (x−)2=q
i=1[(xi)−]2ci, which in turn implies
cl, (x−)2 = [(xl)−]2, ∀l = 1, 2, . . . , q. (27) In addition, from Lemma3.1and the expression of y given by (19), it follows that
y−= q i=1
vici+
1≤i<j≤q
vij,
where vi ≤ (yi)−≤ 0 for i = 1, 2, . . . , q and vij ∈ Vij. By the same arguments as (21),
cl, (y−)2 = vl2+
cl,
1≤i<j≤q
vij
2
, ∀l = 1, 2, . . . , q. (28)
Now, from (20), (27) and (28), it follows that
cl,−φβ(x, y) =
cl, x◦ y − (1/2β)
(x−)2+ (y−)2
= xlyl− (1/2β)
((xl)−)2+ vl2+
cl,
1≤i<j≤q
vij
2
≤ xlyl− (1/2β)
((xl)−)2+ (vl)2
≤ xlyl− (1/2β)
((xl)−)2+ ((yl)−)2
, ∀l = 1, 2, . . . , q, where the first inequality is due to the nonnegativity ofcl, (
1≤i<j≤qvij)2 and the second one is due to the fact that vl≤ (yl)−≤ 0. On the other hand, by Lemma3.2(a),
cl,−φβ(x, y) ≥ λmin(−φβ(x, y)) cl 2= λmin(−φβ(x, y)), ∀l = 1, 2, . . . , q.
Combining the last two inequalities leads immediately to λmin(−φβ(x, y))≤ xlyl− (1/2β)
((xl)−)2+ ((yl)−)2
, ∀l = 1, 2, . . . , q.
Let λmin(x)= xν with ν∈ {1, 2, . . . , q}, and suppose that λmin(x)≤ 0. Then, λmin(−φβ(x, y))≤ λmin(x)yν− (1/2β)
((λmin(x))−)2+ ((yν)−)2
≤ [(λmin(x))−][(yν)−] − (1/2β)
((λmin(x))−)2+ ((yν)−)2
= −(1/2β)
β(λmin(x))−− (yν)− 2
+ (1 − β2)[(λmin(x))−]2
≤ −
(1− β2)/(2β)
[(λmin(x))−]2,
which in turn implies
φβ(x, y) ≥!!λmin(−φβ(x, y))!! ≥
(1− β2)/(2β)
[(λmin(x))−]2. (29)
If λmin(x)= xν>0, then the inequality (29) is obvious. Thus, (29) holds for any x, y∈ V. In view of the symmetry of x and y in φβ(x, y), we also have
φβ(x, y) ≥ |λmin(−φβ(x, y))| ≥
(1− β2)/(2β)
[(λmin(y))−]2
for any x, y∈ V. Consequently, the desired result follows.
The following proposition characterizes an important property for the smooth EP complementarity functions φα and φβ under a unified framework.
Proposition 3.1 Let φα and φβ be given as in (7) and (8), respectively. Let{xk} ⊂ V and{yk} ⊂ V be sequences satisfying one of the following conditions:
(i) either λmin(xk)→ −∞ or λmin(yk)→ −∞;
(ii) λmin(xk), λmin(yk) >−∞, λmax(xk), λmax(yk)→ +∞ and xk◦ yk → +∞.
Then, φα(xk, yk) → +∞ and φβ(xk, yk) → +∞.
Proof Under Case (i) the assertion is direct by Lemma3.3. In what follows, we will prove the assertion under Case (ii). Notice that, in this case, the sequences {xk}, {yk} and {xk+ yk} are all bounded below since λmin(xk), λmin(yk) >−∞
and λmin(xk+ yk)≥ λmin(xk)+ λmin(yk) >−∞. Therefore, the sequences {[(xk+ yk)−]2}, {((xk)−)2} and {((yk)−)2} are bounded. In addition, we also have λmin(xk◦ yk)→ −∞ or λmax(xk◦ yk)→ +∞, since xk◦ yk → +∞.
If λmin(xk◦ yk)→ −∞ as k → ∞, then by Lemma3.2(c) there holds that λmin(−φα(x, y))= λmin
(xk◦ yk)− (1/2α)((xk+ yk)−)2
≤ λmin(xk◦ yk)+ (1/2α)"""((xk+ yk)−)2""" , λmin(−φβ(x, y))= λmin
(xk◦ yk)− (1/2β)
((xk)−)2+ ((yk)−)2
≤ λmin(xk◦ yk)+ (1/2β)"""((xk)−)2+ ((yk)−)2""" ,
which, together with the boundedness of ((xk+yk)−)2 and ((xk)−)2+((yk)−)2 , implies that λmin(−φα(xk, yk))→ −∞ and λmin(−φβ(xk, yk))→ −∞. Since
φα(xk, yk) ≥ |λmin(−φα(x, y))| and φβ(xk, yk) ≥ |λmin(−φβ(x, y))|, we obtain immediately that φα(xk, yk) → +∞ and φβ(xk, yk) → +∞.
If λmax(xk◦ yk)→ +∞ as k → ∞, from Lemma3.2(c) it then follows that λmax(−φα(x, y))= λmax
(xk◦ yk)− (1/2α)((xk+ yk)−)2
≥ λmax(xk◦ yk)− (1/2α)"""((xk+ yk)−)2""" , λmax(−φβ(x, y))= λmax
(xk◦ yk)− (1/2β)
((xk)−)2+ ((yk)−)2
≥ λmax(xk◦ yk)− (1/2β)"""((xk)−)2+ ((yk)−)2""" ,
which, by the boundedness of ((xk+ yk)−)2 and ((xk)−)2+ ((yk)−)2 , implies that λmax(−φα(xk, yk))→ +∞ and λmax(−φβ(xk, yk))→ +∞. Noting that
φα(xk, yk) ≥ |λmax(−φα(xk, yk))| and φβ(xk, yk) ≥ |λmax(−φβ(xk, yk))|,
we obtain readily that φα(xk, yk) → +∞ and φβ(xk, yk) → +∞. WhenV = Rn, ‘◦’ being the componentwise product of the vectors, xk◦ yk → +∞ automatically holds if λmax(xk), λmax(yk)→ +∞, and Proposition3.1reduces to the result of Lemma 2.5 in [21] for the NCPs. However, for the general Euclidean Jordan algebra, this condition is necessary as illustrated by the following example.
Example 3.1 Consider the Lorentz algebra Ln = (Rn,◦, ·, ·Rn) introduced in Sect.2. Assume that n= 3 and take the sequences {xk} and {yk} as follows:
xk=
⎛
⎝k k 0
⎞
⎠ , yk=
⎛
⎝ k
−k 0
⎞
⎠ , for each k.
It is easy to verify that λmin(xk)= 0, λmin(yk)= 0, λmax(xk), λmax(yk)→ +∞, but xk ◦ yk +∞. For such {xk} and {yk}, by computation we have that φα(xk, yk) = 0 and φβ(xk, yk) = 0, i.e. the conclusion of Proposition3.1does not hold.
In the subsequent analysis, we use often the continuity of the Jordan product stated by the following lemma. Since the proof can be found in [10], we omit it.
Lemma 3.4 Let{xk} and {yk} be the sequences such that xk→ ¯x and yk→ ¯y when k→ ∞. Then, we have that xk◦ yk→ ¯x ◦ ¯y.
Now, we are in a position to establish the coerciveness of fα and fβ. Assume that A = (V, ◦, ·, ·) is a general Euclidean Jordan algebra. We consider first the SCLCP case.
Theorem 3.1 Let fα and fβ be given by (10) and (12), respectively. If F (ζ )= L(ζ )+ b, with the linear transformation L having the P -property, then fα and fβ are coercive.
Proof Let{ζk} be a sequence such that ζk → +∞. We need only to prove that fα(ζk)→ +∞, fβ(ζk)→ +∞. (30) By passing to a subsequence if necessary, we assume that ζk/ ζk → ¯ζ, and conse- quently (L(ζk)+ b)/ ζk → L(¯ζ). If λmin(ζk)→ −∞, then from Proposition3.1 it follows that φα(ζk, L(ζk)+ b) , φβ(ζk, L(ζk)+ b) → +∞, which in turn im- plies (30).
Now, assume that{ζk} is bounded below. We argue that the sequence {L(ζk)+ b}
is unbounded by contradiction. Suppose that{L(ζk)+ b} is bounded. Then, L( ¯ζ )= lim
k→∞
(L(ζk)+ b)/ ζk
= 0 ∈ K.
Since {ζk} is bounded below and λmax(ζk)→ +∞ by ζk → +∞, there is an element ¯d∈ V such that (ζk− ¯d)/ ζk− ¯d ∈ K for each k. Noting that K is closed, we have
klim→∞(ζk− ¯d)/ ζk− ¯d = ¯ζ/ ¯ζ = ¯ζ ∈ K.
Thus, ¯ζ ∈ K, L(¯ζ) ∈ K and ¯ζ ◦ L(¯ζ) = 0. From Proposition 6 of [18], it follows that ¯ζ and L( ¯ζ ) operator commute. This, together with ¯ζ◦ L(¯ζ) = 0 ∈ −K and the P-property of L, implies that ¯ζ= 0, yielding a contradiction to ¯ζ = 1. Hence, the sequence{L(ζk)+b} is unbounded. Without loss of generality, assume that L(ζk)+
b → +∞.
If λmin(L(ζk)+ b) → −∞, then using Proposition 3.1 yields the desired re- sult (30). We next assume that the sequence{L(ζk)+ b} is bounded below. We prove that
(ζk/ ζk ) ◦
(L(ζk)+ b)/ ζk
0. (31)
Suppose that (31) does not hold; then, from Lemma3.4, it follows that
¯ζ ◦ L(¯ζ) = lim
k→+∞
(ζk− d)/ ζk
◦
(L(ζk)+ b − d)/ ζk
= 0 ∀d ∈ V. (32)
Since{ζk} and {L(ζk)+ b} are bounded below and λmax(ζk), λmax(L(ζk)+ b) → +∞, there is an element ˜d such that ζk− ˜d ∈ K and L(ζk)+ b − ˜d ∈ K for each k.
Therefore,
(ζk− ˜d)/ ζk
∈ K,
(L(ζk)+ b − ˜d)/ ζk
∈ K, ∀k.
Noting thatK is closed, ¯ζ = limk→∞(ζk− ˜d)/ ζk and L(¯ζ)=limk→∞[(L(ζk)+ b− ˜d)/ ζk ], we have
¯ζ ∈ K, L( ¯ζ )∈ K. (33)
From (32) and (33) and Proposition 6 of [18], it follows that ¯ζ and L( ¯ζ ) operator com- mute. Using the P -property of L and noting that ¯ζ◦ L(¯ζ) = 0 ∈ −K, we then obtain
¯ζ = 0, which clearly contradicts ¯ζ = 1. Therefore, (31) holds. Since ζk → +∞, we have ζk◦ (L(ζk)+ b) → +∞. Combining with λmin(ζk), λmin(L(ζk)+ b) >
−∞ and ζk , L(ζk)+ b → +∞, it follows that the sequences {ζk} and {L(ζk)+ b} satisfy condition (ii) of Proposition3.1. This means that the result (30) holds. Theorem 3.2 Let fα and fβ be defined as in (10) and (12), respectively. If the map- ping F has the uniform Jordan P -property, then fα and fβ are coercive.
Proof The proof technique is similar to that of Theorem 4.1 in [15]. For complete- ness, we include it. Let{ζk} be a sequence such that ζk → +∞. Corresponding to the Cartesian structure ofV, let ζk= (ζ1k, . . . , ζmk)with ζik∈ Vi for each k. Define
J:=
i∈ {1, 2, . . . , m} | {ζik} is unbounded .
Clearly, the set J = ∅, since {ζk} is unbounded. Let {ξk} be a bounded sequence with ξk= (ξ1k, . . . , ξmk)and ξik∈ Vi for i= 1, 2, . . . , m, where ξik for each k is defined as follows:
ξik=
#0, if i∈ J, ζik, otherwise,
with i= 1, 2, . . . , m. Since F has the uniform Jordan P -property, there is a constant ρ >0 such that
ρ ζk− ξk 2≤ max
i=1,...,mλmax
(ζik− ξik)◦ (Fi(ζk)− Fi(ξk))
= λmax
ζνk◦ (Fν(ζk)− Fν(ξk))
≤ ζνk◦ (Fν(ζk)− Fν(ξk))
≤ ζνk Fν(ζk)− Fν(ξk) , (34) where ν is an index from{1, 2, . . . , m} for which the maximum is attained and the last inequality is due to (14). Clearly, ν∈ J by the definition of {ξk}; consequently, {ζνk} is unbounded. Without loss of generality, we assume that
ζνk → +∞. (35)
Since
ζk− ξk 2≥ ζνk− ξνk 2= ζνk 2, for each k, (36) dividing both sides of (34) by ζνk then yields that
ρ ζνk ≤ Fν(ζk)− Fν(ξk) ≤ Fν(ζk) + Fν(ξk) .
Notice that {F (ξk)} is bounded, since the mapping F is continuous and {ξk} is bounded. Hence, the last inequality implies immediately
Fν(ζk) → +∞. (37)
In addition, we can verify by contradiction that
ζνk◦ Fν(ζk) → +∞. (38)
In fact, if{ ζνk◦ Fν(ζk) } is bounded, then on the one hand we have
klim→∞ ζνk◦ (Fν(ζk)− Fν(ξk)) / ζνk 2= 0, but on the other hand, the inequality (36) implies
k→+∞lim ρ ζk− ξk 2/ ζνk 2≥ ρ > 0,
which clearly contradicts the third inequality in (34). Thus, from (35), (37), (38), the sequences{ζνk} and {Fν(ζk)} satisfy the conditions of Proposition3.1. Therefore,
