Fischer-Burmeister Complementarity Function on Euclidean Jordan Algebras

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1 Fischer-Burmeister Complementarity Function on Euclidean Jordan Algebras Lingchen Kong, Levent Tunçel, and Naihua Xiu 3 (November 6, 7; Revised December, 7) Abstract Recently, Gowda et al. [] established the Fischer-Burmeister (FB) complementarity function (C-function) on Euclidean Jordan algebras. In this paper, we prove that FB C-function as well as the derivatives of the squared norm of FB C-function are Lipschitz continuous. Keywords: Fischer-Burmeister function, Euclidean Jordan algebra, Lipschitz continuity. MSC Subject Classification: primary: 65K5, 9C33; secondary: 6B5, 5A9. Abbreviated Title: FB C-function on Euclidean Jordan algebras Introduction It is well-known that the scalar-valued FB function φ : R R R is specified by φ(a, b) : a + b a + b, a, b R, (.) which is attributed by Fischer to Burmeister (see [5, 6, 7]). It is a complementarity function for nonlinear complementarity problem (NCP)(called C-function or NCP function), that is, φ(a, b) a, b, ab. (.) FB function has been much studied in the context of NCP, because it has nice properties, such as strong semismoothness. Moreover, the squared norm of FB function has a Lipschitz continuous gradient, which can be effectively employed in the algorithmic development, see, e.g., [3, 8, 3]. Recently, FB function has been generalized to solve the semidefinite complementarity problem (SDCP) and the second-order cone complementarity problem (SOCCP). For instance, Tseng [] (also see Borwein and Lewis []) proved that FB function is a C-function for SDCP, and The work was partly supported by a Discovery Grant from NSERC, and the National Natural Science Foundation of China (67, ). Lingchen Kong, Department of Combinatorics and Optimization, Faculty of Mathematics, University of Waterloo, Waterloo, Ontario NL 3G, Canada; Department of Applied Mathematics, Beijing Jiaotong University, Beijing 44, P. R. China ( konglchen@6.com) Levent Tunçel, Department of Combinatorics and Optimization, Faculty of Mathematics, University of Waterloo, Waterloo, Ontario NL 3G, Canada ( ltuncel@math.uwaterloo.ca) 3 Naihua Xiu, Department of Applied Mathematics, Beijing Jiaotong University, Beijing 44, P. R. China ( nhxiu@center.njtu.edu.cn)

2 Fukushima, Luo and Tseng [9] showed that this is true in the setting of SOCCP. It was proved by Sim, Sun and Ralph [7] and Chen, Sun and Sun [] that the squared norm of FB function has a Lipschitz continuous gradient in the settings of SDCP and SOCCP, respectively. Gowda, Sznajder and Tao [] proposed the following (vector-valued) FB function on Euclidean Jordan algebras as Φ F B (x, y) : x + y (x + y ), (.3) (detailed description is in the next section) and showed that it is a C-function for symmetric cone complementarity problem (SCCP) which is to find a vector x J such that x K, y K, x, y, y F (x), (.4) where J is a space of n-dimensional real column vectors, (J,,, ) is a Euclidean Jordan algebra, K is the symmetric cone in V (see Section ), and F : J J is a given continuously differentiable mapping. SCCP provides a simple, natural, and unified framework for various complementarity problems, such as NCP, SOCCP and SDCP. Because of wide applications in engineering, management science and other fields, it has attracted much attention, see, e.g., [,,, 5, 6,, ]. Here, we say Φ : J J J is a C-function (for SCCP) if it satisfies Φ(x, y) x K, y K, x, y. (.5) Liu, Zhang and Wang [6] showed that the squared norm of FB function, Ψ F B : J J R, defined by Ψ F B (x, y) : Φ F B(x, y) (.6) is differentiable. Motivated by all of the cited work above, a natural question arises: Are the derivatives of the squared norm of FB function Ψ F B (given by (.6)) Lipschitz continuous? We answer the above question in the affirmative. To do so, we establish useful inequalities on the Lyapunov operator, employing the norm induced by the underlying inner product. In Section we establish the preliminaries and present some useful results about Lyapunov transformation. We show that FB function is Lipschitz continuous in Section 3. Section 4 establishes that the derivatives of squared norm of FB function are Lipschitz continuous. We conclude the paper in Section 5 and raise an open question. Preliminaries We review some results on Euclidean Jordan algebras (see for instance [4, 4]) and develop some basic inequalities on Euclidean Jordan algebras. A Euclidean Jordan algebra is a triple (J,,, )(V for short), where (J,, ) is a n- dimensional inner product space over real field R and (x, y) x y : J J J is a bilinear mapping which satisfies the following conditions:

3 (i) x y y x for all x, y J, (ii) x (x y) x (x y) for all x, y J where x : x x and (iii) x y, z x, y z for all x, y, z J. We call x y the Jordan product of x and y. In general, the Jordan product is not associative, i.e., (x y) z x (y z) for all x, y, z J. We assume that there exists an element e (called the identity element) such that x e e x x for all x J. Define the set of squares as K : {x : x J }. It is well-known that K is a symmetric cone in V, i.e., K is a closed, convex, homogeneous and self-dual cone. For x J, the degree of x denoted by deg(x) is the smallest positive integer m such that the set {e, x, x,, x m } is linearly dependent. The rank of V is defined as max{deg(x) : x J }. In this paper, r will denote the rank of the underlying Euclidean Jordan algebra. Let dim(j ) denote the dimension of J. Obviously, r dim(j ). Recall that an element c J is idempotent if c c. It is also primitive if it cannot be written as a sum of two idempotents. A complete system of orthogonal idempotents is a finite set {c, c,, c k } of idempotents with c i c j (i j) and k i c i e. A complete system of orthogonal primitive idempotents is called a Jordan frame of V. Thus, for any element x J, we have the following important spectral decomposition theorem. Theorem. (Theorem III.., [4]) Let V be a Euclidean Jordan algebra of rank r. Then for every vector x J there exist a Jordan frame {c (x), c (x),, c r (x)} and real numbers λ (x), λ (x),, λ r (x), the eigenvalues of x, such that We call (.) the spectral decomposition of x. x λ (x)c (x) + λ (x)c (x) + + λ r (x)c r (x). (.) Let x r j λ j(x)c j (x) and be the norm on J induced by the inner product, i.e., x : x, x λ j (x). We have c j (x) for j {,,, r}. as Let g : R R be a real-valued function. We define the vector-valued function G : J J G(x) : g(λ j (x))c j (x) g(λ (x))c (x) + g(λ (x))c (x) + + g(λ r (x))c r (x), j which is a Löwner operator. In particular, taking t + : max{, t}, we can define the projection of x onto K as x + : (λ j (x)) + c j (x). j Note that x K if and only if λ i (x), i {,,, r}. Letting g(t) : t for t R +, we define x : λ j (x)c j (x) for x K. j Therefore, FB function (.3) and its squared norm (.6) are well-defined. We next recall the Peirce decomposition theorem on the space J, where the Jordan frame {c, c,, c r } can be fixed beforehand. 3 j

4 Theorem. (Theorem IV.., [4]) Let {c, c,, c r } be a given Jordan frame in a Euclidean Jordan algebra V of rank r. Then J is the orthogonal direct sum of spaces J ij (i j), where the subspaces J ij for i, j {,,, r} are defined by { J ii : {x J : x c i x} and J ij : x J : x c i } x x c j, i j. Furthermore, (i) J ij J ij J ii + J jj ; (ii) J ij J jk J ik, if i k; (iii) J ij J kl {}, if {i, j} {k, l} Ø. Based on the result above and Lemma IV.. in [4], we have the following connection between x y and x y, which is useful in the subsequent analysis. Lemma.3 Let x J ij, y J kl with i < j and k < l. Then x y x y. Furthermore, if {i, j} {k, l} Ø, x y x y if i k, j l, 8 x y otherwise. Proof. Note that if {i, j} {k, l} Ø, then x y. If x J ij, y J jl with i, j, l all distinct, by Lemma IV.. in [4], x y 8 x y. We only need to prove the conclusion in the case of x, y J ij. By Theorem., x y δ c i + δ c j, for some δ, δ R. Thus, x y δ + δ. Meanwhile, by direct computation, we have x y x y, x y δ c i + δ c j, x y (δ c i + δ c j ) x, y (δ + δ )x, y (by Theorem.) δ + δ x y. Therefore, we conclude in this case that x y x y. Below we consider a very fundamental linear operator, Lyapunov transformation, and derive some inequalities on it that will be useful to us. For each x J, we define the Lyapunov transformation (operator) L(x) : J J by L(x)y x y, for all y J, which is a symmetric operator in the sense that L(x)y, z y, L(x)z for all y, z J. Given a r i λ i(a)c i (a) with λ (a) λ > λ + λ r (a), where (a) : {i : λ i (a) > }, we define a subspace J a : J(e (a), ) : {x J : x e (a) x} with e (a) : 4 i c i (a). (.)

5 It is well-known that L a : L(a) is a one-to-one mapping from J a to J a and therefore it has an inverse L a on J a, i.e., for any x J a, L a (x) is the unique d J a such that a d x. Using Lemma in [], any x J a can be expressed as x i x i c i (a) + i<j x ij, (.3) where x i R and x ij J ij with the given Jordan frame {c (a), c (a),, c r (a)}. The following proposition gives a formula for L a (x). Proposition.4 Let a r i λ i(a)c i (a) with λ (a) λ > λ + λ r (a). Let J a and e (a) be given by (.). Then every x J a can be written as in (.3) and L a (x) i x i λ i (a) c i(a) + j<l λ j (a) + λ l (a) x jl. (.4) In particular, L a (a) e (a) is the identity element in J a, and L a (a k ) a k for k >. Proof. As we noted before, the fact that every x J a can be written in the form (.3) is given by Lemma in []. Let d : L a (x). Then d i d i c i (a) + j<l for some d i R and d jl J jl. By Theorem., direct calculation yields that a d λ i (a)d i c i (a) + λ i (a)c i (a) i i i λ i (a)d i c i (a) + λ i (a)d i c i (a) + i j<l j<l i d jl, j<l λ i (a)c i (a) d jl λ j (a) + λ l (a) d jl. d jl This together with a d x establishes (.4). Likewise, for the above a and (a), we define subspaces J a : J(e (a), ) : {x J : x e (a) }, J a : J(e (a), ) : {x J : x e (a) x}. It is easy to see that Ja J(e e (a), ). Similarly, applying Lemma in [], any x Ja can be expressed as x x i c i (a) + x ij, i+ +i<jr 5

6 where x i R and x ij J ij. It is known that J is the orthogonal direct sum of spaces J a, J a and J a (see Page 6 of [4]). From Theorem., we obtain J a jl J jl, J a j,+lr J jl, J a +jlr J jl. (.5) Thus any x J can be expressed as x x () + x ( ) + x () where x () J a, x ( ) J a x () J a. Observe that a x ( ) J a. Moreover, L a is a one-to-one mapping from J a to J a, which is shown by the following. Proposition.5 Let a r i λ i(a)c i (a) with λ (a) λ > λ + λ r (a). Then every y J a can be written as y j,+lr and y jl (.6) and where y jl J jl. L a (y) j,+lr λ j (a) y jl, (.7) Proof. By Theorem. and (.5), every y J a can be written in the form (.6). Let d : L a (y). Then d d jl, j,+lr for d jl J jl. As in the proof of Proposition.4, we have L a (y) j,+lr λ j (a) + λ l (a) y jl. The desired conclusion follows from λ l (a) for + l r. Next, we consider some continuity property of L a ε where a ε : (a + ε e). Proposition.6 Let a r i λ i(a)c i (a) with λ (a) λ > λ + λ r (a). Then for any x J a and y J a, L (x + y) is well-defined and Let a ε : (a + ε e). Then L a a (x + y) L (x) + L (y). a a lim ε L a ε (x + y) L a (x + y). (.8) Proof. The first part of the theorem is obvious by Propositions.4 and.5. For the second part, since x J a and y J a, we can take x and y as in the forms (.3) and (.6), respectively. 6

7 Noting that a ε r i λ i(a ε )c i (a) with λ i (a ε ) λ i (a) + ε, and employing an argument similar to the one in the proof of Proposition.4, we have L a ε (x + y) i + x i λ i (a ε ) c i(a) + j,+lr j<l λ j (a ε ) + λ l (a ε ) x jl λ j (a ε ) + λ l (a ε ) y jl. This together with the facts λ l (a ε ) ε for + l r, (.4) and (.7) yields (.8). We end this section by presenting various useful inequalities on L. Lemma.7 For x, y J, let a ε (x, y) : (x + y + ε e) with ε. Then for every u, v J, we have L L a (x + y) u ε(x,y) β u, L a (v) x ε(x,y) β v and a (x u) ε(x,y) γ u, where β and γ are positive constants only dependent on the rank of J, which can be taken as β r 4 and γ 3r. Proof. For x, y J, define a : (x +y ) and let a r i λ i(a)c i. For u, v J, by Theorem., we have x x i c i + x jl, y y i c i + y jl, u i u i c i + i j<lr j<lr u jl, v i v i c i + i j<lr j<lr where x i, y i, u i, v i R and x jl, y jl, u jl, v jl J jl. Note that a ε (x, y) (x + y + ε e) λ i (a) + ε c i and a ε(x, y) x + y + ε e. Thus, r i λ i (a) + ε c i, a ε(x, y) c i, x + y ) + ε v jl, x c i, x + y c i, y + ε x i + yi + ( x jl + y jl ) + ε, j<lr,i {j,l} where the second equality holds by c i, e, the fourth follows from the facts that x c i x i c i + j<lr,i {j,l} x jl by Theorem. and x i c i + x jl, x x i + x jl. j<lr,i {j,l} j<lr,i {j,l} This implies that λ i (a) + ε max { x i, y i, x jl, y jl, i {j, l}}, (.9) 7

8 and for j l, j, l {,,, r} we obtain λ j (a) + ε + λ l (a) + ε max { x jl, y jl }. (.) From Proposition.4, we have L a ε (x,y) (x) i x i λ i (a) + ε c i + i j<lr x jl. λ j (a) + ε + λ l (a) + ε Thus, by Theorem., direct calculation yields L a (x) u x i u i ε(x,y) c i + x i c i λ i λ (a) + ε i (a) + ε ( ) + u i c i + i + i j<lr j<lr i λ j (a) + ε + x i u i λ i (a) + ε c i + i j<lr,i {j,l} + j<lr i<kr j<lr u jl x jl λ j (a) + ε + λ l (a) + ε x jl λ l (a) + ε i j<lr,i {j,l} u i λ j (a) + ε + λ j (a) + ε + x i j<lr u jl λ i (a) + ε λ l (a) + ε x jl x jl u ik, λ l (a) + ε where the second equality follows from c i u jl u jl with i {j, l}. Therefore, we have L a ε (x,y) (x) u x i u i c i i λ i (a) + ε + x i u jl i j<lr,i {j,l} λ i (a) + ε u i + x jl i j<lr,i {j,l} λ j (a) + ε + λ l (a) + ε + x jl u ik j<lr i<kr λ j (a) + ε + λ l (a) + ε u i + u jl + u i i + i j<lr,i {j,l} j<lr i<kr λ j (a) + ε + i j<lr,i {j,l} u jl x jl u λ l (a) + ik ε 8

9 r u + r(r ) u + r(r ) u + 4 [r(r )] u, where the second inequality holds by Lemma.3, the third by the fact u max{ u i, u jl }, (.9) and (.). Let β r + 3 r(r ) + 4 [r(r )]. Then L a ε (x,y) (x) u β u. Likewise, we have Similarly, noting that we obtain L (y) u β u. Hence, L a ε (x,y) (x + y) u L a ε (x,y) (x) u + L a ε (x,y) (y) u β u. L a ε (x,y) (v) We next show x u L i v i λ i (a) + ε c i + j<lr L a ε (x,y) (v) x β v. v jl, λ j (a) + ε + λ l (a) + ε a (x u) ε(x,y) γ u with γ only dependent on r. Note that ( ) x i u i c i + x i c i ( ) + u i c i i + j<lr x i u i c i + i + x jl i j<lr j<lr x j + x l i<jr k<lr,{i,j} {k,l} x i u i c i + i + i<j<kr j<lr ( xj + x l j<lr u jl u jl + u jl j<lr x ij u kl + i u j + u l x jl j<lr u jl + u ) j + u l x jl (x ij u jk + x jk u ij ) + j<lr x jl u jl x jl u jl. j<lr By Theorem., we can write x jl u jl c j + c l with, R. Thus, by Proposition.4, L x jl u jl L ( c j + c l) j<lr j<lr j<lr x jl c j + c l. λ j λ (a) + ε l (a) + ε 9

10 { } Observe that since c j c l, with θ jl : min λ j (a) + ε, λ l (a) + ε, we deduce That is, c j + c l λ j λ (a) + ε l (a) + ε c j λ j (a) + + c l ε λ l (a) + ε c j + c θ jl l θ jl ( c j + c l ) θ jl θjl c j + c l (by c j c l ) x jl u jl θ jl x jl u jl θ jl x jl θ jl u jl (by Lemma.3) u jl (θ jl x jl by (.9)) u. c j + c l λ j λ (a) + ε l (a) + ε u. Therefore, we have L j<lr x jl u jl j<lr j<lr j<lr c j + c l λ j λ (a) + ε l (a) + ε c j + c l λ j λ (a) + ε l (a) + ε u r(r ) u. (.) Set ξ : r i x iu i c i + ( ) xj +x l j<lr u jl + u j+u l x jl + i<j<kr (x ij u jk + x jk u ij ). Note that by Theorem., (x ij u jk + x jk u ij ) J ik. Similarly, by Proposition.4, we have L a ε (x,y) (ξ) x i u i c i + λ i (a) + ε i + i<j<kr j<lr λ j (a) + ε + ( xj + x l λ l (a) + ε (x ij u jk + x jk u ij ). λ i (a) + ε + λ k (a) + ε u jl + u ) j + u l x jl

11 Thus, we obtain from Lemma.3 and inequalities (.9) and (.) that L a ε (x,y) (ξ) u i + i j<lr ( u jl + u j + u l ) + i<j<kr ( u jk + u ij ) r u + 3 r(r ) u + r(r ) u. (.) So, combining the above inequalities (.) and (.), we have L a ε (x,y) (x u) L a ε (x,y) ξ + x jl u jl j<lr [r u + 3 ] r(r ) u + r(r ) u + r(r ) u (3r r) u. Letting γ 3r r, we obtain the desired inequality. 3 Lipschitz continuity of FB function In this section, we establish the Lipschitz continuity of FB C-function. For this purpose, we need the following result about the derivative of x in int(k), the interior of K. Lemma 3. The function x is smooth at every x int(k). Moreover, it holds (x (L(x )) ) for every x int(k). (3.) Proof. Clearly, x is smooth at x int(k). For the second part of the lemma, suppose that (x + h) x Sh + o( h ) for some linear operator S. Multiplying both sides of this equation by (x + h) + x, we have ((x + h) + x ) ((x + h) x ) ((x + h) + x ) (Sh + o( h )). Direct computation yields h ((x + h) + x ) Sh + o( h ) or h x (Sh) + o( h ), using (x + h) x + Sh + o( h ) and Sh Sh o( h ). That is, h L(x )(Sh) + o( h ). Hence, S (L(x )) (L(x )) by the linearity of Lyapunov transformation. We now prove the Lipschitz property of (x + y ). Lemma 3. The function (x + y ) is globally Lipschitz continuous everywhere in J J. Proof. Fix x, y J, let a(x, y) : (x + y ) and a ε (x, y) : (x + y + ε e) for ε. Note that for any u, v J, a ε (x + u, y + v) a ε (x, y) a ε (x + u, y + v) a ε (x, y + v) + a ε (x, y + v) a ε (x, y)

12 L L L a ε (x+tu,y+v) L(x + tu)udt + a ε (x+tu,y+v) ((x + tu) u)dt + a ε (x+tu,y+v) ((x + tu) u) dt + γ u dt + γ( u + v ) γ (u, v), γ v dt L a ε (x,y+tv) L(y + tv)vdt L a ε (x,y+tv) ((y + tv) v)dt L a ε (x,y+tv) ((y + tv) v) dt where the second equality holds by the Mean Value Theorem and Lemma 3., the first inequality holds by Lemma.7 and γ is only dependent on r, and the last inequality follows from the fact u + v u + v (u, v). Thus, we deduce a ε (x + u, y + v) a ε (x, y) γ (u, v). The desired conclusion follows by taking ε in the inequality above. As a consequence of the lemma above, we immediately obtain the Lipschitz continuity of FB function. Theorem 3.3 The FB function Φ F B (given by (.3)) is globally Lipschitz continuous everywhere in J J. 4 Lipschitz continuity of the derivatives of Ψ F B This section deals with Lipschitz continuity of the derivatives of the squared norm of FB C- function. The main result is stated below. Theorem 4. The derivatives of the squared norm of the Fischer-Burmeister function Ψ F B (given by (.6)) are Lipschitz continuous everywhere in J J. Our proof relies on four lemmas. First, we focus on L a (x + y) x where a (x + y ) in the subsequent analysis. Observe that a may have eigenvalues that are zero. For the sake of simplicity, we look at a smoothed counterpart a ε (x, y). Let S ε (x, y) : L (x + y) x, we have the following. Lemma 4. Let u, v J be given. Then for every x, y J and ε we have [ x S ε (x, y)] u L a ε (x,y)(x + y) u + L a ε (x,y) [ ] u L a ε (x,y)(x + y) L a ε (x,y) (x u) x, [ ] [ y S ε (x, y)] v L (x + y) v + L v L (x + y) L a (y v) ε(x,y) y. Proof. Fix u J. Set z : a ε (x + u, y) a ε (x, y) and w : x u + u. Noting that a ε (x + u, y) [(x + u) + y + ε e] [(x + y + ε e) + x u + u ],

13 we have z [a ε(x, y) + w] a ε (x, y). Note that J aε(x,y) J from (.). From Lemma 6.6() in [6], it follows that z L a ε (x,y) (x u + u ) + o( u ) L a ε (x,y)(x u) + o( u ). Thus z as u and z O( u ). Let η : L a ε (x,y)(x + y) and η + h : L a ε (x,y)+z(x + u + y). It is easy to see that a ε (x, y) η x + y and [a ε (x, y) + z] (η + h) x + u + y. So, a ε (x, y) h u z η z h, or h L (u z η) L (z h). Since z as u and z O( u ), h as u and h z o( z ) o( u ). We deduce Thus, direct calculation yields L a ε (x,y)(z h) o( u ) and h L a ε (x,y)(u z η) + o( u ). S ε (x + u, y) S ε (x, y) L a ε (x+u,y)(x + u + y) (x + u) L a ε (x,y)(x + y) x (η + h) (x + u) η x η u + h (x + u) L a ε (x,y)(x + y) u + L L a ε (x,y)(x + y) u + L a ε (x,y) a ε (x,y) L (x + y) u + L (u z η) (x + u) + o( u ) (x + u) [ ] u L a ε (x,y)(x + y) L a ε (x,y) (x u) (x + u) + o( u ) [ ] u L (x + y) L a (x u) ε(x,y) x + o( u ). This establishes the formula for [ x S ε (x, y)]u. The formula for [ y S ε (x, y)]v follows from the symmetry of x and y. Lemma 4.3 For every x, y, u and v J, [ x S ε (x, y)]u ρ u, [ y S ε (x, y)]v ρ v where ρ is only dependent on the rank of J, which can be taken as ρ r + 3r 4. Proof. Using Lemmas.7 and 4., we deduce [ ] [ x S ε (x, y)]u L a ε (x,y)(x + y) u + L a ε (x,y) u L a ε (x,y)(x + y) L a ε (x,y) (x u) x [ ] L L a ε (x,y) (x + y) u + a ε (x,y) u L a ε (x,y)(x + y) L a ε (x,y) (x u) x β u + β u L a ε (x,y)(x + y) L a ε (x,y) (x u) ( ) β u + β u + L a ε (x,y)(x + y) L a ε (x,y) (x u) ( ) β u + β u + 4β L a ε (x,y) (x u) β u + β( u + 4βγ u ), where β, γ are given as in Lemma.7. Therefore, by taking ρ : 3β + 4β γ, we obtain that [ x S ε (x, y)]u ρ u with ρ only dependent on r and ρ r + 3r 4 if β r 4 and γ 3r. As in the proof of the last lemma, [ y S ε (x, y)]v ρ v follows from symmetry. 3

14 Now we are in a position to prove the Lipschitz continuity of L a (x + y) x. Lemma 4.4 L a (x + y) x is globally Lipschitz continuous for all x, y J. Proof. We first show that S ε (x, y) L x, y J. For every u, v J, (x + y) x is globally Lipschitz continuous for all S ε (x + u, y + v) S ε (x, y) S ε (x + u, y + v) S ε (x, y + v) + S ε (x, y + v) S ε (x, y). By Lemma 4., x S ε (x, y) is continuous in x. Thus, it follows by the Mean Value Theorem that S ε (x + u, y + v) S ε (x, y + v) [ x S ε (x + tu, y + v)]udt. By Lemma 4.3, it follows that [ x S ε (x + tu, y + v)]u ρ u. Hence, S ε (x + u, y + v) S ε (x, y + v) [ x S ε (x + tu, y + v)]udt [ x S ε (x + tu, y + v)]u dt ρ u dt ρ u. That is, S ε (x + u, y + v) S ε (x, y + v) ρ u with ρ only dependent on r. Likewise, we have S ε (x, y + v) S ε (x, y) ρ v. We therefore obtain that S ε (x + u, y + v) S ε (x, y) ρ( u + v ), or L a ε (x+u,y+v)(x + u + y + v) (x + u) L a ε (x,y) (x + y) x ρ( u + v ). Note that L a ε (x,y)(x + y) x L a (x + y) x as ε by Proposition.6. Letting ε in the inequality above, we obtain the desired result. Before proving Theorem 4., we need to recall a lemma by Liu, Zhang and Wang [6]. Lemma 4.5 (Lemma 6.7, [6]) Ψ F B (x, y) is differentiable at every (x, y) J J, and if (x, y) (, ), then x Ψ F B (, ) y Ψ F B (, ) ; if (x, y) (, ), then where a (x + y ). Proof of Theorem 4. It is easy to see that x Ψ F B (x, y) L a (a x y) (x a), y Ψ F B (x, y) L a (a x y) (y a), Ψ F B (x, y) x + y, e + (x + y), e (x + y ), x + y. Let R(x, y) : (x + y ), x + y. Then R(x, y) x + y, e + (x + y), e Ψ F B (x, y). 4

15 Set a : (x + y ). By Lemma 4.5, it is straightforward to derive that x R(x, y) x + (x + y) x Ψ F B (x, y) x + y L a (a x y) x + L a (a x y) a x + y [ L a (a) L a (x + y) ] x + (a x y) L a (x + y) x + (x + y ), where the third equality holds by L a (a x y) a a x y, and the fourth by L a (a) x x. Combining Lemmas 3. and 4.4, we conclude that x R(x, y) is globally Lipschitz. By symmetry in x and y, y R(x, y) is also globally Lipschitz continuous. 5 Final Remarks In this article, we studied some properties of the Lyapunov operator, and using these properties we established Lipschitz continuity of FB function and the derivatives of squared norm of FB function. Sun and Sun [8] showed that the FB function is strongly semismooth everywhere in the cases of SDCP and SOCCP. However, it is not clear whether FB function Φ F B (given by (.3)) is strongly semismooth? We leave this question as a future research topic. Acknowledgments The authors thank Defeng Sun of National University of Singapore for very helpful discussions, which helped us eliminate a very important error in the first version of this paper. References [] J.M. Borwein and A.S. Lewis, Convex Analysis and Nonlinear Optimization: Theory and Examples, Springer-Verlag, NewYork,. [] J.-S. Chen, D. Sun and J. Sun, The SC property of the squared norm of the SOC Fischer-Burmeister function, Operations Research Letters to appear (7). [3] F. Facchinei and J.-S. Pang, Finite-Dimensional Variational Inequalities and Complementarity Problems, Volume I and II, Springer-Verlag, New York, 3. [4] J. Faraut and A. Korányi, Analysis on Symmetric Cones, Oxford University Press, New York, 994. [5] A. Fischer, A special Newton-type optimization method, Optimization 4 (99) [6] A. Fischer, A Newton-type method for positive semidefinite linear complementarity problems, Journal of Optimization Theory and Applications 86 (995) [7] A. Fischer, On the local superlinear convergence of a Newton-type method for LCP under weak conditions, Optimization Methods and Software 6 (995) [8] A. Fischer and H. Jiang, Merit Functions for Complementarity and Related Problems: A Survey, Computational Optimization and Applications 7 () [9] M. Fukushima, Z.Q. Luo and P. Tseng, Smoothing functions for second-order cone complementarity problems, SIAM J. Optim. ()

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