Journal of Mathematical Analysis and Applications. A one-parametric class of merit functions for the symmetric cone complementarity problem

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1 J. Math. Anal. Appl Contents lists available at ScienceDirect Journal of Mathematical Analysis and Applications A one-parametric class of merit functions for the symmetric cone complementarity problem Shaohua Pan a,1, Jein-Shan Chen b,, a School of Mathematical Sciences, South China University of Technology, Guangzhou 51060, China b Department of Mathematics, National Taiwan Normal University, Taipei, Taiwan article info abstract Article history: Received 3 June 008 Available online 5 February 009 Submitted by H. Franowsa Keywords: Symmetric cone complementarity problem Merit function Jordan algebra Smoothness Lipschitz continuity Cartesian P -properties In this paper, we extend the one-parametric class of merit functions proposed by Kanzow and Kleinmichel [C. Kanzow, H. Kleinmichel, A new class of semismooth Newtontype methods for nonlinear complementarity problems, Comput. Optim. Appl ] for the nonnegative orthant complementarity problem to the general symmetric cone complementarity problem SCCP. We show that the class of merit functions is continuously differentiable everywhere and has a globally Lipschitz continuous gradient mapping. From this, we particularly obtain the smoothness of the Fischer Burmeister merit function associated with symmetric cones and the Lipschitz continuity of its gradient. In addition, we also consider a regularized formulation for the class of merit functions which is actually an extension of one of the NCP function classes studied by [C. Kanzow, Y. Yamashita, M. Fuushima, New NCP functions and their properties, J. Optim. Theory Appl ] to the SCCP. By exploiting the Cartesian P -properties for a nonlinear transformation, we show that the class of regularized merit functions provides a global error bound for the solution of the SCCP, and moreover, has bounded level sets under a rather wea condition which can be satisfied by the monotone SCCP with a strictly feasible point or the SCCP with the joint Cartesian R 0 -property. All of these results generalize some recent important wors in [J.-S. Chen, P. Tseng, An unconstrained smooth minimization reformulation of the second-order cone complementarity problem, Math. Program ; C.-K. Sim, J. Sun, D. Ralph, A note on the Lipschitz continuity of the gradient of the squared norm of the matrix-valued Fischer Burmeister function, Math. Program ; P. Tseng, Merit function for semidefinite complementarity problems, Math. Program ] under a unified framewor. 009 Elsevier Inc. All rights reserved. 1. Introduction Given a Euclidean Jordan algebra A = V,,, where V is a finite-dimensional vector space over the real field R endowed with the inner product, and denotes the Jordan product. Let K be a symmetric cone in V and G, F : V V be nonlinear transformations assumed to be continuously differentiable throughout this paper. Consider the symmetric cone complementarity problem SCCP of finding ζ V such that Gζ K, F ζ K, Gζ, F ζ = 0. 1 * Corresponding author. addresses: shhpan@scut.edu.cn S. Pan, jschen@math.ntnu.edu.tw J.-S. Chen. 1 Partially supported by the Doctoral Starting-up Foundation B13B of GuangDong Province. Member of Mathematics Division, National Center for Theoretical Sciences, Taipei Office. Partially supported by National Science Council of Taiwan. 00-7X/$ see front matter 009 Elsevier Inc. All rights reserved. doi: /j.jmaa

2 196 S. Pan, J.-S. Chen / J. Math. Anal. Appl The model provides a simple, natural and unified framewor for various existing complementarity problems such as the nonnegative orthant nonlinear complementarity problem NCP, the second-order cone complementarity problem SOCCP, and the semidefinite complementarity problem SDCP. In addition, the model itself is closely related to the KKT optimality conditions for the convex symmetric cone program CSCP: minimize gx, subject to a i, x =b i, i = 1,,...,m, x K, where a i V, b i R for i = 1,,...,m, and g : V R is a convex twice continuously differentiable function. Therefore, the SCCP has wide applications in engineering, economics, management science and other fields; see [1,11,0,9] and references therein. During the past several years, interior-point methods have been well used for solving the symmetric cone linear programming problem SCLP, i.e., the CSCP with g being a linear function see [7,8,3,]. However, in view of the wide applications of the SCCP, it is worthwhile to explore other solution methods for the more general CSCP and SCCP. Recently, motivated by the successful applications of the merit function approach in the solution of NCPs, SOCCPs and SDCPs see, e.g., [,10,,8], some researchers started with the investigation of merit functions or complementarity functions associated with symmetric cones. For example, Liu, Zhang and Wang [1] extended a class of merit functions proposed in [18] to the following special SCCP: ζ K, F ζ K, ζ, F ζ = 0; 3 Kong, Tuncel and Xiu [17] studied the extension of the implicit Lagrangian function proposed by Mangasarian and Solodov [] to symmetric cones; and Kong, Sun and Xiu [16] proposed a regularized smoothing method for the SCCP 3 based on the natural residual complementarity function associated with symmetric cones. Following this line, in this paper we will consider the extension of the one-parametric class of merit functions proposed by Kanzow and Kleinmichel [1] and a class of regularized functions based on it. We define the one-parametric class of vector-valued functions φ τ : V V V by φ τ x, y := x + y + τ x y 1/ x + y, where τ 0, is an arbitrary but fixed parameter, x = x x, x 1/ is a vector such that x 1/ = x, and x + y means the usual componentwise addition of vectors. When τ =, φ τ reduces to the vector-valued Fischer Burmeister function given by φ FB x, y := x + y 1/ x + y; 5 whereas as τ 0 it will become a multiple of the vector-valued residual function ψ NR x, y := x x y +, where + denotes the metric projection on K. In this sense, the one-parametric class of vector-valued functions covers the two popular complementarity functions associated with the symmetric cone K. In fact, from Lemma 3.1 later, it follows that the function φ τ with any τ 0, is a complementarity function associated with K, that is, φ τ x, y = 0 x K, y K, x, y =0. Consequently, its squared norm yields a merit function associated with K ψ τ x, y := 1 φτ x, y, 6 where is the norm induced by,, and the SCCP can be reformulated as min f τ ζ := ψ τ Gζ, F ζ. ζ V 7 To apply the effective unconstrained optimization methods, such as the quasi-newton method, the trust-region method and the conjugate gradient method, for solving the unconstrained minimization reformulation 7 of the SCCP, the smoothness of the merit function ψ τ and the Lipschitz continuity of its gradient will play an important role. In Sections 3 and, we show that the function ψ τ defined by 6 is continuously differentiable everywhere and has a globally Lipschitz continuous gradient with the Lipschitz constant being a positive multiple of 1 + τ 1. These results generalize some recent important wors in [,5,8] under a unified framewor, as well as improve the wor [1] greatly in which only the differentiability of the merit function ψ FB was given. In addition, we also consider a class of regularized functions for f τ defined as ˆfτ ζ := ψ 0 Gζ F ζ + ψτ Gζ, F ζ, 8

3 S. Pan, J.-S. Chen / J. Math. Anal. Appl where ψ 0 : V R + is continuously differentiable and satisfies ψ 0 u = 0 u K and ψ 0 u β u+ u V 9 for some constant β>0. Using the properties of ψ 0 in 9, it is not hard to verify that ˆf τ is a merit function for the SCCP. The class of functions will reduce to the one studied in [1] if τ = and G degenerates into an identity transformation. In Section 5, we show that the class of merit functions can provide a global error bound for the solution of the SCCP under the condition that G and F have the joint uniform Cartesian P -property. In Section 6, we establish the boundedness of the level sets of ˆf τ under a weaer condition than the one used by [1], which can be satisfied by the monotone SCCP with a strictly feasible point or the SCCP with G and F having the joint Cartesian R 0 -property. Throughout this paper, I denotes an identity operator, represents the norm induced by the inner product,, and intk denotes the interior of the symmetric cone K. All vectors are column ones and write the column vector x T 1,...,xT m T as x 1,...,x m, where x i is a column vector from the subspace V i.foranyx V, wedenotex + and x by the metric projection of x onto K and K, respectively, i.e., x + := arg min y K { x y }. For any symmetric matrix A, the notation A O means that A is positive semidefinite. For a differentiable mapping F : V V, the notation F x denotes the transposed Jacobian operator of F at a point x. We write x = oα respectively, x = Oα if x / α 0respectively, uniformly bounded as α 0.. Preliminaries In this section, we recall some concepts and materials of Euclidean Jordan algebras that will be used in the subsequent sections. More detailed expositions of Euclidean Jordan algebras can be found in the monograph by Faraut and Korányi [9]. Besides, one can find excellent summaries in the articles [,1,,6]. A Euclidean Jordan algebra is a triple V,,, V, where V is a finite-dimensional inner product space over the real field R and x, y x y : V V V is a bilinear mapping satisfying the following conditions: i x y = y x for all x, y V, ii x x y = x x y for all x, y V, where x := x x, and iii x y, z V = x, y z V for all x, y, z V. We call x y the Jordan product of x and y. We also assume that there is an element e V, called the unit element, such that x e = x for all x V. Forx V, letζx bethedegreeoftheminimalpolynomialofx, which can be equivalently defined as ζx := min { : { e, x, x,...,x } are linearly dependent }. Since ζx dimv where dimv denotes the dimension of V, the ran of V is well defined by r := max{ζx: x V}. Ina Euclidean Jordan algebra A = V,,, V,wedefinethesetofsquaresasK:= {x : x V}. Then, by Theorem III..1 of [9], K is a symmetric cone. This means that K is a self-dual closed convex cone with nonempty interior intk and for any two elements x, y intk, there exists an invertible linear transformation T : V V such that T K = K and T x = y. A Euclidean Jordan algebra is said to be simple if it is not the direct sum of two Euclidean Jordan algebras. By Proposition III.. of [9], each Euclidean Jordan algebra is, in a unique way, a direct sum of simple Euclidean Jordan algebras. A common simple Euclidean Jordan algebra is S n,,, S n, where S n is the space of n n real symmetric matrices with the inner product X, Y S n := TrXY, and the Jordan product is defined by X Y := XY + YX/. Here, XY is the usual matrix multiplication of X and Y and TrX is the trace of X. The associate cone K is the set of all positive semidefinite matrices. Another one is the Lorentz algebra R n,,, R n, where R n is the Euclidean space of dimension n with the standard inner product x, y R n = x T y, and the Jordan product is defined by x y := x, y R n, x 1 y + y 1 x for any x = x 1, x, y = y 1, y R R n 1. The associate cone, called the Lorentz cone or the second-order cone, is K := { x = x 1, x R R n 1 : x x 1 }. Recall that an element c V is said to be idempotent if c = c. Twoidempotentsc and d are said to be orthogonal if c d = 0. One says that {c 1, c,...,c } isacompletesystemoforthogonalidempotentsif c j = c j, c j c i = 0 if j i, j, i = 1,,...,, and c j = 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.

4 198 S. Pan, J.-S. Chen / J. Math. Anal. Appl Theorem.1. See [9, Theorem III.1.]. Suppose that A = V,,, V is a Euclidean Jordan algebra and the ran of A is r. Then for any x V, there exist a Jordan frame {c 1, c,...,c r } and real numbers λ 1 x, λ x,...,λ r x, arranged in the decreasing order λ 1 x λ x λ r x, suchthatx= r λ jxc j. The numbers λ j x counting multiplicities, which are uniquely determined by x, are called the eigenvalue, and we write the maximum eigenvalue and the minimum eigenvalue of x as λ max x and λ min x, respectively. The trace of x, denotedas trx, isdefinedbytrx := r λ jx; whereas the determinant of x is defined by detx := r λ jx. By Proposition III.1.5 of [9], a Jordan algebra over R with a unit element e V is Euclidean if and only if the symmetric bilinear form trx y is positive definite. Hence, we may define an inner product, on V by x, y :=trx y, x, y V. Unless otherwise states, the inner product, appearing in this paper always means the one defined by 10. By the associativity of tr see [9, Proposition II..3], the inner product, is associative, i.e., x, y z = y, x z for all x, y, z V. Let Lxy := x y for every y V. Then, the linear operator Lx for each x V is symmetric with respect to the inner product, in the sense that Lxy, z = y, Lxz for any y, z V. Let be the norm on V induced by the inner product,, namely, x := x, x = λ j x 1/, x V. It is not difficult to verify that for any x, y V, there always holds that x, y 1 x + y and x + y x + y. 11 Let ϕ : R R be a scalar valued function. Then, it is natural to define a vector-valued function associated with the Euclidean Jordan algebra A = V,,, by ϕ V x := ϕ λ 1 x c 1 + ϕ λ x c + +ϕ λ r x c r, 1 where x V has the spectral decomposition x = r λ jxc j. The function ϕ V is also called the Löwner operator [6]. When ϕt is chosen as max{0, t} and min{0, t} for t R, respectively, ϕ V becomes the metric projection operator onto K and K: x + := { max 0,λ j x } c j and x := { min 0,λ j x } c j. 13 Lemma.1. See [6, Theorem 13]. For any x = r λ jxc j,letϕ V be given as in 1. Thenϕ V is continuously differentiable at xifandonlyifϕ is continuously differentiable at each λ j x, j= 1,,...,r. The derivative of ϕ V at x, for any h V,is where ϕ V xh = [ ϕ [1] λx ] jj c j, h c j + [ ϕ [1] λx ] { ϕλi x ϕλ j x := λ i x λ j x ij ϕ λ i x 1 j<l r if λ i x λ j x, if λ i x = λ j x, [ ϕ [1] λx ] jl c j c l h, i, j = 1,,...,r. In fact, the Jacobian ϕ V is a linear and symmetric operator, which can be written as ϕ V x = ϕ λ j x Qc j + i,, i j [ ϕ [1] λx ] ij Lc jlc i, 1 where Qx := L x Lx for any x V is called the quadratic representation of V. In the sequel, unless otherwise stated, we assume that A = V,,, is a simple Euclidean Jordan algebra of ran r and dimv = n. An important part in the theory of Euclidean Jordan algebras is the Peirce decomposition. Letc be a nonzero idempotent in A. Then, by [9, Proposition III.1.3], c satisfies L 3 c 3L c + Lc = 0 and the distinct eigenvalues of the symmetric 1 operator Lc are 0, and 1. Let Vc, 1, Vc, 1 and Vc, 0 be the three corresponding eigenspaces, i.e., 10

5 S. Pan, J.-S. Chen / J. Math. Anal. Appl Vc, α := { x V: Lcx = αx }, α = 1, 1, 0. Then V is the orthogonal direct sum of Vc, 1, Vc, 1 and Vc, 0. The decomposition V = Vc, 1 V c, 1 Vc, 0 is called the Peirce decomposition of V with respect to the nonzero idempotent c. Let {c 1, c,...,c r } be a Jordan frame of A. Fori, j {1,...,r}, define the eigenspaces V ii := Vc i, 1 = Rc i, V ij := V c i, 1 V c j, 1, i j. Then, from [9, Theorem IV..1], it follows that the following conclusion holds. Theorem.. The space V is the orthogonal direct sum of subspaces V ij 1 i j r, i.e., V = i j V ij.furthermore, V ij V ij V ii + V jj, V ij V j V i, if i, V ij V l ={0}, if {i, j} {,l}=. Let x V have the spectral decomposition x = r λ jxc j.fori, j {1,,...,r}, letc ij x be the orthogonal projection operator onto V ij. Then, C ij x = C ij x, C ij x = C ijx, C ij xc l x = 0 if {i, j} {,l}, i, j,,l = 1,...,r, 15 and 1 i j r C ij x = I, 16 where C ij is the adjoint operator of C ij. In addition, by [9, Theorem IV..1], C jj x = Qc j and C ij x = Lc i Lc j = Lc j Lc i = C ji x, i, j = 1,,...,r. Note that the original notation in [9] for orthogonal projection operator is P ij. However, to avoid confusion with another orthogonal projector P i c j onto Vc, α and orthogonal matrix P which will be used later Sections 3 and, we adopt C ij instead. With the orthogonal projection operators {C ij x: i, j = 1,,...,r}, we have the following spectral decomposition theorem for Lx and Lx ; see [15, Chapters VI V]. Lemma.. Let x V have the spectral decomposition x = r λ jxc j. Then the symmetric operator Lx has the spectral decomposition Lx = λ j xc jj x + 1 j<l r 1 λ j x + λ l x C jl x with the spectrum σ Lx consisting of all distinct numbers in { 1 λ jx + λ l x: j,l = 1,,...,r}, andlx has the spectral decomposition L x = λ j xc jjx j<l r λ j x + λ l x C jl x 17 with the spectrum σ Lx consisting of all distinct numbers in { 1 λ j x + λ x: j,l = 1,,...,r}. l Proposition.1. For any x V, the operator Lx L x is positive semidefinite.

6 00 S. Pan, J.-S. Chen / J. Math. Anal. Appl Proof. By Lemma. and 15, we can verify that L x has the spectral decomposition L x = λ j xc jjx + 1 j<l r This means that the operator Lx L x has the spectral decomposition L x L x = [ 1 λ j x + λ l x 1 λ j x + λ l x ] C jl x. 1 j<l r 1 λ j x + λ l x C jl x. 18 Noting that the orthogonal projection operator is positive semidefinite on V and λ j x + λ l x λ jx + λ l x for all j,l = 1,,...,r, we readily obtain the conclusion from the spectral decomposition of Lx L x. 3. Differentiability of the function ψ τ In this section, we show that ψ τ is a merit function associated with K, and moreover, it is differentiable everywhere on V V. By the definition of Jordan product, x + y + τ x y = x + τ y + τ τ y = y + τ x + τ τ x K 19 for any x, y V, and consequently the function φ τ in is well defined. The following lemma states that φ τ and ψ τ is respectively a complementarity function and a merit function associated with K. Lemma 3.1. For any x, y V,letφ τ and ψ τ be given by and 6, respectively. Then, ψ τ x, y = 0 φ τ x, y = 0 x K, y K, x, y =0. Proof. The first equivalence is clear by the definition of ψ τ, and we only need to prove the second equivalence. Suppose that φ τ x, y = 0. Then, [ x + y + τ x y ] 1/ = x + y. 0 Squaring the two sides of 0 yields that x + y + τ x y = x + y + x y, which implies x y = 0sinceτ 0,. Substituting x y = 0 into 0, we have that x = x + y 1/ y and y = x + y 1/ x. Since x + y K, x K and y K, from [1, Proposition 8] or [19, Corollary 9] it follows that x, y K. Consequently, the necessity holds. For the other direction, suppose x, y K and x y = 0. Then, x + y = x + y. This, together with x y = 0, implies that [ x + y + τ x y ] 1/ x + y = 0. Consequently, the sufficiency follows. The proof is thus completed. In what follows, we concentrate on the differentiability of the merit function ψ τ. For this purpose, we need the following two crucial technical lemmas. Lemma 3.. For any x, y V, letux, y := x + y 1/. Then, the function ux, y is continuously differentiable at any point x, y such that x + y intk.furthermore, x ux, y = LxL 1 ux, y and y ux, y = LyL 1 ux, y. 1

7 S. Pan, J.-S. Chen / J. Math. Anal. Appl Proof. The first part is due to Lemma.1. It remains to derive the formulas in 1. From the definition of ux, y, it follows that u x, y = x + y, x, y V. By the formula 1, it is easy to verify that x x = Lx. Differentiating on both sides of with respect to x then yields that x ux, yl ux, y = Lx. This implies that x ux, y = LxL 1 ux, y since, by ux, y intk, Lux, y is positive definite on V. Similarly, we have that y ux, y = LyL 1 ux, y. To present another lemma, we first introduce some related notations. For any 0 z K and z / intk, suppose that z has the spectral decomposition z = r λ jzc j, where {c 1, c,...,c r } is a Jordan frame andλ 1 z,...,λ r z are the eigenvalues arranged in the decreasing order λ 1 z λ z λ r z = 0. Define the index j := { min j λ j z = 0, j = 1, },...,r 3 and let j 1 c J := c l. l=1 Clearly, j and c J are well defined since 0 z K and z / intk. Sincec J is an idempotent and c J 0otherwisez = 0, V can be decomposed as the orthogonal direct sum of the subspaces Vc J, 1, Vc J, 1 and Vc J, 0. Inthesequel,we write P 1 c J, P 1 c J and P 0 c J as the orthogonal projection onto Vc J, 1, Vc J, 1 and Vc J, 0, respectively. From [1], we now that Lz is positive definite on Vc J, 1 and is a one-to-one mapping from Vc J, 1 to Vc J, 1. This means that Lz an inverse L 1 z on Vc J, 1, i.e., for any u Vc J, 1, L 1 zu is the unique v Vc J, 1 such that z v = u. Lemma 3.3. For any x, y V,letz: V V V be the mapping defined as z = zx, y := [ x + y + τ x y ] 1/. 5 If x, y 0, 0 such that zx, y / intk, then the following results hold: a The vectors x, y, x + y, x + τ yandy+ τ xbelongtothesubspacevc J, 1. b For any h V such that z x, y + h K, letw= wx, y := [z x, y + h] 1/ zx, y. Then,P 1 c J w = 1 L 1 zx, y [P 1 c J h]+o h. Proof. From 19 and the definition of z, it is clear that zx, y K for all x, y V. Hence, using the similar arguments as Lemma 11 of [1] yields the desired result. Now by Lemmas 3. and 3.3, we prove the differentiability of the merit function ψ τ. Proposition 3.1. The function ψ τ defined by 6 is differentiable everywhere on V V. Furthermore, x ψ τ 0, 0 = y ψ τ 0, 0 = 0, and if x, y 0, 0, then [ x ψ τ x, y = L x + τ y L 1 zx, y I [ y ψ τ x, y = L y + τ x L 1 zx, y I where zx, y is given by 5. ] φ τ x, y, ] φ τ x, y, 6 Proof. We prove the conclusion by the following three cases. Case 1: x, y = 0, 0. For any u, v V, suppose that u + v + τ u v has the spectrum decomposition u + v + τ u v = r μ jd j, where {d 1, d,...,d r } is the corresponding Jordan frame. Then, for j = 1,,...,r, we have

8 0 S. Pan, J.-S. Chen / J. Math. Anal. Appl μ j = 1 μ d j j d j, d j = u + v + τ u v, d j = + τ τ u + τ v v, e u + τ v + τ τ v, d j = u + τ u, v + v τ / u + v, 7 where the second equality is by d j =1, the first inequality is due to e = r d j and d j K for j = 1,,...,r, and the last inequality is due to 11. Therefore, ψ τ u, v ψ τ 0, 0 = 1 [ u + v + τ u v ] 1/ u + v 1 = μ j d j u + v μ j d j + u + v μ j d j + u + v 1 u τ r + + v, where the first two inequalities are due to 11, and the last one is from 7. This shows that ψ τ is differentiable at 0, 0 with x ψ τ 0, 0 = y ψ τ 0, 0 = 0. Case : zx, y intk. Sinceφ τ x, y = zx, y x+ y, we have from Lemma.1 that φ τ is continuously differentiable under this case. Notice that ψ τ x, y = 1 e,φ τ x, y, and hence the function ψ τ is continuously differentiable. Applying the chain rule yields x ψ τ x, y = x φ τ x, yl φ τ x, y e = x φ τ x, yφ τ x, y. 8 On the other hand, from 19 it follows that [ φ τ x, y = x + τ y + τ τ and therefore using the formulas in 1 gives that x φ τ x, y = L x + τ y L 1 zx, y I. y ] 1/ x + y, This, together with 8, immediately yields that [ x ψ τ x, y = L x + τ y L 1 zx, y ] I φ τ x, y. For symmetry of x and y in ψ τ x, y, we also have that [ y ψ τ x, y = L y + τ x L 1 zx, y ] I φ τ x, y. Case 3: x, y 0, 0 and zx, y / intk. Foranyu, v V, define ẑ := ˆx u + ŷ v + u + v + τ u v with ˆx = x + τ y and ŷ = y + τ x.itisnotdifficulttoverifythat z x, y + ẑ = x + u + τ y + v + τ τ y + v = z x + u, y + v K. Let wx, y := z x, y + ẑ 1/ zx, y. From the definitions of ψ τ and zx, y, it then follows that

9 S. Pan, J.-S. Chen / J. Math. Anal. Appl ψ τ x + u, y + v ψ τ x, y = 1 [[ z x, y + ] 1/ ẑ x + u + y + v zx, y x + y ] = 1 [ ] ẑ, e + u + v wx, y, x + u + y + v + x + y zx, y, u + v = wx, y, x + y + x + y zx, y, u + v + ˆx, u + ŷ, v +o u, v. 9 By Lemma 3.3a, x + y Vc J, 1. Thus, using Lemma 3.3b, we have that wx, y, x + y = P 1 c J wx, y, x + 1 y = L 1 zx, y [ ] P 1 c J ẑ + o ẑ, x + y = 1 P 1 c J ẑ, L 1 zx, y [x + y] + o ẑ = P 1 c J [ˆx u + ŷ v], L 1 zx, y [x + y] + o u, v = ˆx u + ŷ v, P 1 c J [ L 1 zx, y x + y ] + o u, v = ˆx u + ŷ v, L 1 zx, y x + y + o u, v = [ L 1 zx, y x + y ] ˆx, u + [ L 1 zx, y x + y ] ŷ, v + o u, v, 30 where the first equality is since V = Vc J, 1 Vc J, 1 Vc J, 0, thefifthoneisduetop 1 c J = P 1 c J, and the sixth is from the fact that L 1 zx, yx + y Vc J, 1. Combining 9 with 30, we obtain that ψ τ x + u, y + v ψ τ x, y = ˆx + x + y zx, y [ L 1 zx, y x + y ] ˆx, u + ŷ + x + y zx, y [ L 1 zx, y x + y ] ŷ, v + o u, v. This implies that the function ψ τ is differentiable at x, y, and furthermore, x ψ τ x, y = ˆx + x + y zx, y [ L 1 zx, y x + y ] ˆx, y ψ τ x, y = ŷ + x + y zx, y [ L 1 zx, y x + y ] ŷ. Notice that ˆx + x + y zx, y [ L 1 zx, y x + y ] ˆx = ˆx φ τ x, y [ L 1 zx, y x + y ] x + τ y = x + τ y φ τ x, y L x + τ [L 1 y zx, y x + y ] = L x + τ y L 1 zx, y [ zx, y x ] y φ τ x, y [ = L x + τ y L 1 zx, y ] I φ τ x, y, where the third equality is due to L 1 zx, yzx, y = e and the fact that x + τ y = L x + τ y e = L x + τ y L 1 zx, y zx, y. Therefore, x ψ τ x, y = [ L x + τ y L 1 zx, y I ] φ τ x, y. Similarly, we also have that [ y ψ τ x, y = L y + τ x L 1 zx, y ] I φ τ x, y. This shows that the conclusion holds under this case. The proof is thus completed. It should be pointed out that the formula 6 is well defined even if zx, y / intk since in this case φ τ x, y Vc J, 1 by Lemma 3.3a. When A is specified as the Lorentz algebra R n,,, R n, the formula reduces to the one of [3, Proposition 3.]; whereas when A is specified as S n,,, S n and τ =, the formula is same as the one in [8, Lemma 6.3b] by noting that zx, y = x + y 1/ and

10 0 S. Pan, J.-S. Chen / J. Math. Anal. Appl x ψ τ x, y = LxL 1 zx, y φ FB x, y φ FB x, y = x [ L 1 zx, y φ FB x, y ] L zx, y L 1 zx, y φ FB x, y = x [ L 1 zx, y φ FB x, y ] zx, y [ L 1 zx, y φ FB x, y ] = [ L 1 zx, y φ FB x, y ] x zx, y. Thus, the formula 6 provides a unified framewor for the SOCCP and the SDCP cases. From Proposition 3.1, we readily obtain the following properties of ψ τ, which have been given in the setting of NCP [1] and the SOCCP [3], respectively. Proposition 3.. Let ψ τ be given as in 6. Then,foranyx, y V V,wehave a x, x ψ τ x, y + y, y ψ τ x, y = φ τ x, y. b ψ τ x, y = 0 if and only if x K, y K, x, y =0. Proof. a If x, y = 0, 0, the result is clear. Otherwise, from 6 it follows that x, x ψ τ x, y + y, y ψ τ x, y = x, x + τ y [ L 1 zx, y φ τ x, y ] x,φ τ x, y + y, y + τ x [ L 1 zx, y φ τ x, y ] y,φ τ x, y = x x + τ y, L 1 zx, y φ τ x, y x,φ τ x, y + y y + τ x, L 1 zx, y φ τ x, y y,φ τ x, y = z x, y, L 1 zx, y φ τ x, y x + y,φ τ x, y = zx, y, φ τ x, y x + y,φ τ x, y = φτ x, y, where the next to last equality is by z = Lzz and the symmetry of Lz. b The proof is direct by part a, Lemma 3.1 and Proposition Lipschitz continuity of ψ τ In this section, we investigate the continuity of the gradients x ψ τ x, y and y ψ τ x, y. To this end, for any ɛ > 0, we define the mapping z ɛ : V V V by z ɛ = z ɛ x, y := x + y + τ x y + ɛe 1/. 31 From 19, clearly, z ɛ x, y intk for any x, y V, and hence the operator Lz ɛ x, y is positive definite on V. Sincethe spectral function induced by ϕt = t t 0 is continuous by Lemma.1, it follows that z ɛ x, y zx, y as ɛ 0 + for any x, y V V, where zx, y is given by 5. This means that Lz ɛ x, y Lzx, y as ɛ 0 +. In what follows, we prove that the gradients x ψ τ x, y and y ψ τ x, y are Lipschitz continuous by arguing the Lipschitz continuity of z ɛ x, y and the mapping H ɛ x, y := L x + τ y L 1 z ɛ x, y x + y. 3 To establish the Lipschitz continuity of z ɛ x, y, we need the following crucial lemma. Lemma.1. For any x, y V V and ɛ > 0, letz ɛ x, y be defined as in 31. Then the function z ɛ x, y is continuously differentiable everywhere with x z ɛ x, y = L x + τ y L 1 z ɛ x, y, y z ɛ x, y = L y + τ x L 1 z ɛ x, y. 33 Furthermore, there exists a constant C > 0, independentof x, y and ɛ, τ, such that x z ɛ x, y C and y z ɛ x, y C.

11 S. Pan, J.-S. Chen / J. Math. Anal. Appl Proof. The first part follows from Lemma 3. and the following fact that [ z ɛ x, y = x + τ y + τ τ 1/ [ y + ɛe] = y + τ x + τ τ 1/ x + ɛe]. 3 We next prove that the operator x z ɛ x, y is bounded for any x, y V and ɛ > 0. Let {u 1, u,...,u n } be an orthonormal basis of V. Foranyx, y V, letlz, Lx + τ y, Lz ɛ and Lx + τ y be the corresponding matrix representation of the operators Lz, Lx + τ y, Lz ɛ and Lx + τ y with respect to the basis {u 1, u,...,u n }. Then, by the formula 33, it suffices to prove that the matrix Lx + τ yl 1 z ɛ is bounded for any x, y V and ɛ > 0. The verifications are given as below. Suppose that z = zx, y has the spectral decomposition z = r λ jzc j, where λ 1 z λ z λ r z 0arethe eigenvalue of z and {c 1, c,...,c r } is the corresponding Jordan frame. From Lemma., Lz has the spectral decomposition Lz = λ j zc jj z + 1 j<l r 1 λ j z + λ l z C jl z 35 with the spectrum σ Lz consisting of all distinct numbers in { 1 λ jz + λ l z: spectral decomposition L z = λ j zc jjz j<l r j,l = 1,,...,r}, and Lz has the λ j z + λ l z C jl z 36 with σ Lz consisting of all distinct numbers in { 1 λ j z + λ l z: j,l = 1,,...,r}. By the definition of z ɛx, y, itis easy to verify that z ɛ = r λ j z + ɛc j, and consequently the symmetric operator Lz ɛ has the spectral decomposition Lz ɛ = λ j z + ɛc jjz j<l r with the spectrum σ Lz ɛ consisting of all distinct numbers in { 1 } λ j z + ɛ + λ l z + ɛ : j,l = 1,,...,r. λ j z + ɛ + λ l z + ɛ C jl z 37 We first prove that the matrix Lx + τ ylz + ɛ I 1/ is bounded for any x, y V and ɛ > 0. For this purpose, let P be an n n orthogonal matrix such that PL z P T = diag λ 1 L z,λ L z,...,λ n L z, 38 where λ 1 Lz λ Lz λ n Lz 0 are the eigenvalues of Lz. Then, it is not hard to verify that for any ɛ > 0, P L z + ɛ 1/ I P T 1 = diag λ1 Lz + ɛ,..., 1. λn Lz + ɛ Write Ũ := PLx + τ yp T. We can compute that L x + τ L y z + ɛ 1/ I = P T 1 Ũ diag λ1 Lz + ɛ,..., 1 P λn Lz + ɛ [ ] = P T Ũ i P. 39 λ Lz 1 i n + ɛ 1 n Since Lz = Lx + τ y + L τ τ y and Ly is positive semidefinite, we get L z L x + τ y O. In addition, by Proposition.1 L[x + τ y ] Lx + τ ylx + τ y is positive semidefinite, and hence we have that L x + τ y L x + τ y L x + τ y O. The last two equations thus imply that

12 06 S. Pan, J.-S. Chen / J. Math. Anal. Appl L z L x + τ y L x + τ y O. 0 Now, for any given {1,,...,n}, from 38 and 0 it follows that λ L z = [ PL z ] P [PL T x + τ y L x + τ ] y P T n =[ŨŨ] = Ũ i, i=1 where the inequality is by the fact that the diagonal entries of a positive semidefinite matrix are nonnegative. This immediately implies that λ L z + ɛ n Ũ i Ũ i i = 1,,...,n. i=1 Combining with Eq. 39, there exists a constant C 1 > 0 such that L x + τ L y z + ɛ 1/ I C 1 x, y V and ɛ > 0. 1 We next prove that the matrix Lz + ɛ I 1/ L 1 z ɛ is bounded for any x, y V and ɛ > 0. Let C jl z for 1 j,l r be the matrix representation of C jl z with respect to the basis {u 1, u,...,u n }. From Eqs , it then follows that L z + ɛ 1/ I = λ j z + ɛc jjz + 1 λ j z + λ l z + ɛ C jl z, 1 j<l r L 1 1 z ɛ = λ j z + ɛ C 1 jjz + C jl z. 1 j<l r λ λ j z + ɛ + l z + ɛ/ Using the last two equalities and 15, it is easy to compute that L z + ɛ 1/ I L 1 z ɛ = C jj z + λ j z + λ z + ɛ l C jl z. λ λ j z + ɛ + l z + ɛ 1 j<l r Notice that the projection matrix C jl z with 1 j,l r is bounded for any x, y V, andforanyx, y V and ɛ > 0 λ λ j z + ɛ + l z + ɛ λ j z + λ z + ɛ 1 j,l r. l Hence, from we can deduce that Lz + ɛ I 1/ L 1 z ɛ is bounded for any x, y V and ɛ > 0, i.e., there exists a positive constant C such that L z + ɛ I 1/ L 1 z ɛ C x, y V and ɛ > 0. 3 Combining 3 and 1, we have that the matrix Lx + τ yl 1 z ɛ is bounded for any x, y V and ɛ > 0, because L x + τ y L 1 z ɛ = L x + τ [L y z + ɛ 1/ I L z + ɛ 1/] I L 1 z ɛ [ = L x + τ L y z + ɛ ] 1/ [L I z + ɛ 1/ I L 1 z ɛ ] L x + τ L y z + ɛ 1/ I L z + ɛ 1/ I L 1 z ɛ C 1 C x, y V and ɛ > 0. Consequently, there exists a constant C > 0 such that x z ɛ x, y C for any x, y V and ɛ > 0. For the symmetry, y z ɛ x, y C also holds for any x, y V and ɛ > 0. From the discussions above, we see that the constant C is also independent of τ. By Lemma.1 and the Mean-Value Theorem, we can establish the global Lipschitz continuity of the mapping z ɛ x, y, which is summarized in the following proposition.

13 S. Pan, J.-S. Chen / J. Math. Anal. Appl Proposition.1. For any x, y V and ɛ > 0, letz ɛ x, y be defined as in 31. Then the function z ɛ x, y is globally Lipschitz continuous. Proof. For any x, y, a, b V V, by the Mean-Value Theorem we have that zɛ x, y z ɛ a, b = [ zɛ x, y z ɛ a, y ] + [ z ɛ a, y z ɛ a, b ] = x z ɛ a + tx a, y x a dt y z ɛ a, b + ty b y b dt 1 x z ɛ a + tx a, y x a dt + y z ɛ a, b + ty b y b dt C x a + y b C x, y a, b, where the last two inequalities are respectively by Lemma.1 and 11. This shows that the function z ɛ x, y is globally Lipschitz continuous. 0 Next, we focus on the Lipschitz continuity of the mapping H ɛ given by 3. We achieve the goal by proving that x H ɛ x, y and y H ɛ x, y are bounded for any x, y V and ɛ > 0. To compute x H ɛ x, y and y H ɛ x, y, the following lemma is needed. Lemma.. For any x, y V and ɛ > 0, leth V be such that z ɛ x, y + h K and write w := [z ɛ x, y + h]1/ z ɛ x, y. Then, w = 1 L 1 z ɛ x, yh + o h. Proof. From the definition of w, it immediately follows that [ w + zɛ x, y ] = z ɛ x, y + h, which is equivalent to saying that w + w z ɛ x, y = h or h = L z ɛ x, y w + w. 5 We claim that, as h 0, we must have w 0. Indeed, let h 0, then we obtain from that w + w z ɛ x, y = 0. Adding zɛ x, y to both sides gives w + zɛ x, y = z ɛ x, y. This, by the fact that w + z ɛ x, y K and z ɛ x, y intk, implies that w + z ɛ x, y = z ɛ x, y, and hence w = 0. Since Lz ɛ x, y is invertible on V and w 0as h 0, using the implicit function theorem and Eq. 5 yields that w ɛ = 1 L 1 z ɛ x, y h + o h. Consequently, the proof is completed. Lemma.3. For any x, y V and ɛ > 0, leth ɛ x, y be given as in 3. Then,H ɛ x, y is differentiable everywhere. Moreover, for any given u, v V, x H ɛ x, yu = [ L 1 z ɛ x, y x + y ] u + L x + τ y [ u L 1 z ɛ x, y x + y L 1 z ɛ x, y L L 1 z ɛ x, y x + τ y ] u,

14 08 S. Pan, J.-S. Chen / J. Math. Anal. Appl y H ɛ x, yv = τ [ L 1 z ɛ x, y x + y ] v + L x + τ y L 1 z ɛ x, y [ v L 1 z ɛ x, y x + y L 1 z ɛ x, y L y + τ ] x v. 6 Proof. For any x, y V and any given u, v V, letx = x + τ y, y = y + τ x and h := x u + y v + u + v + τ u v. It is easy to compute that Let Then, z ɛ x, y + h = x + u + y + v + τ [ x + u y + v ] + ɛe = zɛ x + u, y + v intk. w := [ z ɛ x, y + h] 1/ zɛ x, y. w + z ɛ x, y = [ z ɛ x, y + h] 1/ = zɛ x + u, y + v intk. Applying Lemma. then yields that w = 1 L 1 z ɛ x, y h + o u, v, 7 which implies that w 0asu 0, v 0 and w = O u, v. Write g := L 1 z ɛ x, y x + y and g + s := L 1 z ɛ x, y + w x + u + y + v. 8 We next express s in terms of g, w, u, v and z ɛ x, y. By 8, clearly, L z ɛ x, y g = x + y and L z ɛ x, y + w g + s = x + u + y + v, which in turn implies that and L z ɛ x, y s = u + v w g w s, s = L 1 z ɛ x, y u + v w g w s. 9 Using 7 and 9, we have that s 0 as u, v 0. This, together with 7, means that w s = o w = o u, v. Therefore, and L 1 z ɛ x, y w s = o u, v s = L 1 z ɛ x, y u + v w g + o u, v. Now, from the above discussions and the definition of H ɛ, it follows that H ɛ x + u, y + v H ɛ x, y = L x + u + τ y + v L 1 z ɛ x, y + w x + u + y + v L x + τ y L 1 z ɛ x, y x + y = L x + u + τ y + v g + s L x + τ y g = L x + τ y s + L u + τ v g + s = L x + τ [L y 1 z ɛ x, y u + v g w ] + L u + τ v g + o u, v

15 S. Pan, J.-S. Chen / J. Math. Anal. Appl = L x + τ y L 1 z ɛ x, y u + v L x + τ y L 1 z ɛ x, y [ L 1 z ɛ x, y x + y L 1 z ɛ x, y Lx u + L 1 z ɛ x, y Ly ] v + L u + τ v L 1 z ɛ x, y x + y + o u, v. This means that H ɛ is differentiable at the point x, y. Also, the formulas of x H ɛ x, yu and x H ɛ x, yv are exactly given by 6. The proof is then completed. Lemma.. For any x, y V and ɛ > 0,letH ɛ x, y be defined as in 3.Then,foranygivenu, v V, there exists a constant C > 0 independent of x, yandɛ, τ such that x H ɛ x, yu Cτ 1 u and y H ɛ x, yv Cτ 1 v. Proof. By Lemma.1, there exists a constant C > 0 independent of x, y, ɛ, τ such that L x + τ y L 1 z ɛ x, y C and L y + τ x L 1 z ɛ x, y C. Hence, their adjoint operators L 1 z ɛ x, ylx + τ y and L 1 z ɛ x, yly + τ x are also bounded for any x, y V and ɛ > 0, i.e., L 1 z ɛ x, y L x + τ y C and L 1 z ɛ x, y L y + τ x C. Noting that L 1 z ɛ x, y x + y = τ L 1 z ɛ x, y [ L x + τ y e + L we also have L 1 z ɛ x, y x + y Cτ 1. y + τ ] x e, Thus, by the formulas of x H ɛ x, yu and y H ɛ x, yv, wegetthedesiredresult. Using Lemmas.3 and. and the same arguments as for Proposition.1, we obtain the global Lipschitz continuity of H ɛ x, y, which is summarized in the following proposition. 50 Proposition.. For any x, y V and ɛ > 0, leth ɛ x, y be defined as in 3. Then the function H ɛ x, y is globally Lipschitz continuous with the Lipschitz constant being Cτ 1,whereC> 0 is independent of x, yandɛ, τ. Now we are in a position to establish the Lipschitz continuity of x ψ τ and y ψ τ. Proposition.3. The function ψ τ has a Lipschitz continuous gradient with the Lipschitz constant being positive multiple of 1 + τ 1, i.e., there exists a constant C > 0 such that x ψ τ x, y x ψ τ a, b C 1 + τ 1 x, y a, b, y ψ τ x, y y ψ τ a, b C 1 + τ 1 x, y a, b for any x, y, a, b V V, where C is independent of x, y, a, b and ɛ, τ. Proof. For the symmetry, we only need to prove the first inequality. By 6, x ψ τ x, y = x + τ y zx, y L x + τ y L 1 zx, y x + y. For any ɛ > 0, let G ɛ : V V V be the mapping defined by G ɛ x, y := x + τ y z ɛx, y H ɛ x, y. Then, from Propositions.1 and., it follows that G ɛ x, y is globally Lipschitz continuous with the Lipschitz constant being C1 + τ 1, i.e., for all x, y, a, b V V,

16 10 S. Pan, J.-S. Chen / J. Math. Anal. Appl Gɛ x, y G ɛ a, b C 1 + τ 1 x, y a, b. 51 We next show that for any x, y V V, G ɛ x, y x ψ τ x, y as ɛ 0 +.Indeed,ifx, y = 0, 0, thenh ɛ 0, 0 0 by 3 and 50, and so G ɛ 0, 0 0 = x ψ τ 0, 0. Ifx, y 0, 0, noting that z ɛ x, y zx, y as ɛ 0 +,itsufficesto prove that L 1 z ɛ x, y x + y L 1 zx, y x + y. 5 If x, y 0, 0 such that z intk, thenlz is positive definite on V. Ifx, y 0, 0 such that z / intk, thenfrom Section 3 it follows that Lz is positive definite on the subspace Vc J, 1. By the proof of [13, Lemma.1ii], L 1 z is then continuous on V or Vc J, 1, which implies the result of 5. Thus, we show that G ɛ x, y x ψ τ x, y as ɛ 0 + for any x, y V V. Nowtaingɛ 0 + in 51 and applying the relation between G ɛ x, y and x ψ τ x, y shown as above, we get the desired result. Propositions 3.1 and.3 show that the one-parametric class of merit functions ψ τ is smooth everywhere on V V and has a Lipschitz continuous gradient with the Lipschitz constant related to τ 1. From this, we particularly obtain that the squarednormofthefbfunctionφ FB in 5 is also a smooth merit function with a global Lipschitz continuous gradient, generalizing the main results of [,5]. 5. Global error bound From this section, we start with the case that A = V,,, is a general Euclidean Jordan algebra. By Propositions III.. and III..5 and Theorem V.3.7 of [9], we now that for a given Euclidean Jordan algebra A and the corresponding symmetric cone K, V = V 1 V V m and K = K 1 K K m, with each A i = V i,,, being a simple Euclidean Jordan algebra where dimv i = n i and the corresponding symmetric cone is K i.moreover,foranyx = x 1,...,x m, y = y 1,...,y m V with x i, y i V i, one also has x y = x 1 y 1,...,x m y m and x, y = x 1, y x m, y m. Therefore, the complementarity condition involved in the SCCP is equivalent to x i K i, y i K i, x i, y i =0 for all i = 1,,...,m, and the functions φ τ x, y and ψ τ x, y can be respectively rewritten as φ τ x, y = φ τ x 1, y 1,...,φ τ x m, y m and ψ τ x, y = The gradients of ψ τ with respect to x and y then become respectively x ψ τ x, y = x1 ψ τ x 1, y 1,..., xm ψ τ x m, y m, y ψ τ x, y = y1 ψ τ x 1, y 1,..., ym ψ τ x m, y m. m ψ τ x i, y i. By this, it is not hard to verify that all the results in Sections 3 and are valid in the general Euclidean Jordan algebra A since they hold in each simple Euclidean Jordan algebra A i. In the sequel, corresponding to the structure of V, wewritethe mappings G and F as G = G 1,...,G m, F = F 1,...,F m with G i, F i : V V i. From the last two sections, we see that the SCCP is equivalent to the global minimization of the smooth functions f τ or ˆf τ. In this section, we show that the regularized merit function ˆf τ can provide a global error bound for the solution of the SCCP under a different condition from [1]. To the end, we need the concepts of Cartesian P -properties introduced in [17] for a nonlinear transformation, which are natural extensions of the P -properties on Cartesian products in R n established by Facchinei and Pang [11] and the Cartesian P -properties in the setting of S n developed by Chen and Qi [6]. Definition 5.1. The mappings G = G 1,...,G m and F = F 1,...,F m are said to have a the joint uniform Cartesian P -property if there exists a constant ρ > 0suchthat,foreveryζ,ξ V, there is an index i {1,,...,m} such that Gi ζ G i ξ, F i ζ F i ξ ρ ζ ξ ; b the joint Cartesian P -property if for every ζ,ξ V with ζ ξ, there is an index i {1,,...,m} such that ζ i ξ i and G i ζ G i ξ, F i ζ F i ξ > 0. i=1

17 S. Pan, J.-S. Chen / J. Math. Anal. Appl To establish the main result of this section, we also need the following two lemmas. Since the proof of the first lemma is similar to that of Lemma 7 in [], we here omit it. Lemma 5.1. For any x V, letx + and x be the metric projection of x onto K and K, respectively. Then, the following results hold: a x = x + + x, x +,x =0,and x = x + + x. b For any x K and y V with x y K,wehavex y K. c For any x V and y K,wehavethat x, y x +, y and x + y + x +. Lemma 5.. For any x, y V,letψ τ be defined as in 6. Then, ψ τ x, y [ φ τ x, y ] τ [ x y + ]. Proof. The first inequality is due to Lemma 5.1a and the definition of ψ τ. We next prove the second inequality. From 19 and Lemma 5.1b, it follows that [ x + y + τ x y ] 1/ x + τ y K, [ x + y + τ x y ] 1/ y + τ x K. Combining with Lemma 5.1c, we then obtain that [ x + y + τ x [ 1/ y x y ]+ x = + y + τ x y 1/ x + τ y τ ] y + [ x + + y + τ x y 1/ y + τ x τ ] x + τ y + τ + x +. Thus, we complete the proof of the second inequality. Now we prove that the regularized merit function ˆf τ provides a global error bound for the solution of the SCCP if G and F have the joint uniform Cartesian P -property. Proposition 5.1. Let ˆf τ be defined as in 8 9. Suppose that G and F have the joint uniform Cartesian P -property and the SCCP has a solution, denoted by ζ. Then, there exists a constant κ > 0 such that for any ζ V, κ ζ ζ β 1 ˆfτ ζ + τ ˆfτ ζ 1/. 53 Proof. Since G and F have the joint uniform Cartesian P -property, there exists a constant ρ > 0 such that, for any ζ V, there is an index i {1,,...,m} such that ρ ζ ζ G i ζ G i ζ, F i ζ F i ζ = G i ζ, F i ζ + F i ζ, G i ζ + F i ζ, G i ζ G i ζ, F i ζ + [ G i ζ ] +, F iζ + [ F i ζ ] +, G iζ [ λ max Gi ζ F i ζ ] + Fi ζ [ G i ζ ] + Gi ζ [ F + i ζ ] + { max 1, Gi ζ, Fi ζ }[ [ λ max Gi ζ F i ζ [ +] + Gi ζ ] [ + + Fi ζ ] ] + { max 1, Gζ, F ζ }[ [ G i ζ F i ζ ] [ + + Gi ζ ] [ + + Fi ζ ] ], + where the equality is since G i ζ, F i ζ =0, the second inequality is due to Lemma 5.1b, and the third one follows ρ from Proposition.1 of [7] and the Cauchy Schwartz inequality. Setting κ := max{1, Gζ, F ζ }, we immediately obtain κ ζ ζ [ G i ζ F i ζ ] [ + + Gi ζ ] [ + + Fi ζ ] +. From the conditions given by 9, clearly, for any i {1,,...,m} we have [ G i ζ F i ζ ] + β 1 ψ 0 Gi ζ F i ζ β 1 ˆfτ ζ.

18 1 S. Pan, J.-S. Chen / J. Math. Anal. Appl In addition, applying Lemma 5., we obtain that [ G i ζ ] [ + + Fi ζ ] [ ] + Gζ + [ F ζ ] 1/ 1/ + + ψ τ Gζ, F ζ τ τ ˆfτ ζ 1/. Combining the last three inequalities immediately yields the desired result. For the NCP, Kanzow and Kleinmichel [1] showed that the merit function f τ can provide a global error bound if F is a Lipschitz continuous uniform P -function. But, for the regularized merit function ˆf τ, Proposition 5.1 indicates that the Lipschitz continuity of F is not needed. The main reason is that the former is based on the global error bound result of ψ NR and the same growth behavior of ψ τ and ψ NR, whereas the regularized merit function ˆf τ attains this goal via the regularizing term ψ 0 involved in it. For the SCCP 3, by Proposition 3. and Corollary 3.1 of [7], the assumption for the existence of the solution x can be removed from Proposition 5.1, since the uniform Cartesian P -property implies the uniform Jordan P -property. In addition, we note that for the SCCP 3, a similar result was given in [1] under the assumption that F having the uniform P - property. This assumption is stronger under some cases than F having the uniform Cartesian P -property; for example, when F is a affine function Mζ + q and A is the Lorentz algebra with dimv 5. In this case, Professor Defeng Sun showed that F has the uniform P -property if and only if M is positive definite. Clearly, the positive definiteness of M implies F having Cartesian P -property, but F having Cartesian P -property does not imply M being positive definite. For example, for V = V 1 V V 3 with dimv 1 = dimv = and dimv 3 = 1, let M be a bloc diagonal matrix composed of, and 1. It is easy to verify that F ζ = Mζ + q for any q V has the Cartesian P -property, but M is not positive definite. 6. Bounded level sets In this section, we provide a condition to guarantee the boundedness of the level sets L ˆfτ γ := { ζ V ˆfτ ζ γ }, for any γ 0. Specifically, we will prove that the following condition is sufficient. Condition 6.1. For any sequence {ζ } V such that [ ζ ] [ +, G ζ ] + < +, F ζ + < +, 5 there holds that [ max λ max Gi ζ ] F i ζ +. 1 i m 55 Proposition 6.1. If the mappings G and F satisfy Condition 6.1, then the level sets L ˆfτ γ of ˆf τ for all γ 0 are bounded. Proof. Assume on the contrary that there is an unbounded sequence {ζ } L γ ˆfτ for some γ 0. Then, f τ ζ ˆf τ ζ γ for all. By Lemma 5., G and F satisfy 5. Hence, there is ν {1,...,m} such that λ max [G ν ζ F ν ζ ] +.Noting that [ λ max Gν ζ ] [ F ν ζ λ max Gν ζ [ F ν ζ +] Gν ζ ] F ν ζ + β 1 ˆfτ ζ, we have ˆf τ ζ +. This contradicts the fact that {ζ } L ˆfτ γ. Condition 6.1 is rather wea to guarantee the boundedness of the level sets of ˆfτ. Indeed, the SCCP with the jointly monotone mappings and a strictly feasible point, or the SCCP with joint Cartesian R 0 -property, all satisfy this condition. To demonstrate the fact, we need the following technical lemma which can be regarded as an extension of Lemma 9b of [] to the setting of symmetric cones. Lemma 6.1. Let {x } V be any sequence satisfying x +.Ifthesequence{λ min x } is bounded below, then x +, ˆx + for any ˆx intk.

19 S. Pan, J.-S. Chen / J. Math. Anal. Appl Proof. For every, letx have the spectral decomposition x = r λ jx q with {q j 1,...,q r } being the corresponding Jordan frame. Let ˆx have the spectral decomposition x = r λ jˆxc j with {c 1,...,c r } being the corresponding Jordan frame. Without loss of generality, suppose that λ l x = λ max x, where 1 l r. Then, for every, x +, ˆx = λ j x + q j, λ j ˆxc j λ max x + λmin ˆx q l, c j = λ max x + λmin ˆx q l, e, where the inequality holds since q j, c j K and λ j x +, λ j ˆx 0forall j = 1,,...,r. Notice that x < + as since {λ min x } is bounded below. Using the fact that x = x + + x and x +,wethenhave that x + +. This, together with x + K, immediately implies that 56 λ max x Since {q } is bounded, we assume subsequencing if necessary that lim l + q = q. By the closedness of l K and q =1 l for each, wehaveq K \{0}. From Proposition I.1. of [9], it then follows that q, e > 0sincee intk. Thus, taing the limit on the both sides of 56 and using Eq. 57, we readily obtain that x +, ˆx +. Definition 6.1. The mappings G = G 1,...,G m and F = F 1,...,F m are said to have the joint Cartesian R 0 -property if for any sequence {ζ } satisfying the condition: ζ +, [ Gζ ] + ζ 0, there exists an index ν {1,,...,m} such that λ max [G ν ζ F ν ζ ] lim inf > 0. + ζ [ F ζ ] + ζ Proposition 6.. Condition 6.1 is satisfied if one of the following assumptions holds: 0, 58 a G and F are jointly monotone mappings with Gζ + F ζ + as ζ + and there exists a point ˆζ V such that Gˆζ,F ˆζ intk; b G and F have the joint Cartesian R 0 -property. Proof. a Let {ζ } be a sequence satisfying 5. Since G and F are jointly monotone, G ζ Gˆζ,F ζ F ˆζ 0, which by Lemma 5.1a is equivalent to G ζ, F ζ + Gˆζ,F ˆζ [ G ζ ] +, F ˆζ + [ G ζ ], F ˆζ + [ F ζ ] +, Gˆζ + [ F ζ ], Gˆζ. Notice that the sequences {λ min Gζ } and {λ min F ζ } are bounded below by 5, Gζ + F ζ + and Gˆζ,F ˆζ intk. Using Lemma 6.1 then yields that [ G ζ ] +, F ˆζ + [ F ζ ] +, Gˆζ +. In addition, by 5 it is easy to verify that [ G ζ ], F ˆζ > and [ F ζ ], Gˆζ >. Therefore, from the last three equations it follows that m Gi ζ, F i ζ = G ζ, F ζ +, i=1 which in turn implies that there exists an index ν such that G ν ζ, F ν ζ +. By Proposition.1ii of [7], we have λ max [G ν ζ F ν ζ ] +, which implies 55. b The proof is direct by Definition 6.1. When Gζ = ζ for any ζ V, Liu, Zhang and Wang [1] established the boundness of the level sets of L ˆfτ γ for τ = under the condition that F is an R 0 -function. The condition, as shown by Lemma 6. below, is stronger than the one of Proposition 6.b. Thus, Proposition 6.b generalizes the result of [1, Theorem 7].

20 1 S. Pan, J.-S. Chen / J. Math. Anal. Appl Lemma 6.. Assume that Gζ ζ for any ζ V and F is an R 0 -function. Then, G and F have the joint Cartesian R 0 -property. Proof. Suppose that F is an R 0 -function. From Definition 3 of [1], for any sequence {ζ } satisfying the condition 58, there holds that λ max [ζ F ζ ] lim inf > 0. + ζ 59 For each, letz = ζ F ζ and suppose that it has the spectral decomposition z = r λ jz c j, where {c 1,...,c r } V is a Jordan frame. For convenience, we also denote z = z 1,...,z m with z V i i. By the spectral decomposition of z, clearly, z i = λ j z c j, i i = 1,,...,m, with c j = c j 1,...,c j m for every j {1,,...,r}. Now, without loss of generality we assume that λ l z = λ max z with 1 l r. Then, λ max z = λ j z c j, m c l = λ j z c j i, c l i. i=1 Combining with 59 and 60, there exists an index ν {1,,...,m} such that r λ jz c j ν,c l ν 0 < lim inf + ζ z ν =,c l lim inf ν. + ζ 61 Suppose that z ν as an element in the simple Euclidean Jordan algebra A ν = V ν,,, has the following spectral decomposition z ν = λ j z ν q ν j, where {q ν1,...,q ν r } V ν be the corresponding Jordan frame. Then, z ν, c l ν λmax z ν q ν j, c l ν = λ max z ν eν, ν c l, 6 where e ν is the identity element in V ν and the inequality is since c l ν K ν and q ν j Kν for every j = 1,,..., r. From 61 and 6, it then follows that z ν 0 <,c l lim inf ν + ζ λ max z ν lim inf + ζ e ν,c l ν. ζ Noting that e ν,c l ν is bounded for each, wehavethat e ν,c l lim inf ν = 0. + ζ From the last two inequalities, it readily follows that 60 λ max z ν lim inf > 0. + ζ By Definition 6.1, the mappings G and F have the joint Cartesian R 0 -property. 7. Conclusions and final remars We have extended the one-parametric class of merit functions proposed in [1] for the NCP to the SCCP setting, which provides a unified framewor for several existing merit functions for the SDCP and the SOCCP, for example, the squared norm of the matrix-valued FB function [8] and the vector-valued FB function []. We have established the smoothness and the global Lipschitz continuity of the class of merit functions, thereby generalizing some recent important results of [,1,5,8]. In addition, we also considered a class of regularized merit function ˆf τ based on ψ τ. The class of regularized functions was