### A one-parametric class of merit functions for the second-order cone complementarity problem

Jein-Shan Chen ^{1}
Department of Mathematics
National Taiwan Normal University

Taipei, Taiwan 11677 E-mail: jschen@math.ntnu.edu.tw

Shaohua Pan

School of Mathematical Sciences South China University of Technology

Guangzhou 510640, China E-mail: shhpan@scut.edu.cn

January 10, 2007

(first revised September 3, 2007) (second revised December 24, 2007)

Abstract: We investigate a one-parametric class of merit functions for the second-order
cone complementarity problem (SOCCP) which is closely related to two popular merit
functions, i.e., the Fischer-Burmeister (FB) merit function and the natural residual merit
*function. In fact, it will reduce to the FB merit function if the parameter τ is equal to 2,*
*whereas as τ tends to zero, its limit will become a multiple of the natural residual merit*
function. In this paper, we show that this class of merit functions enjoys several favorable
properties as the FB merit function holds, for example, the smoothness. These properties
play an important role in the reformulation method of an unconstrained minimization or
a nonsmooth system of equations for the SOCCP. Numerical results are reported for some
convex second-order cone programs (SOCPs) by solving the unconstrained minimization
reformulation of the KKT optimality conditions, which indicate that the FB merit function
*is not the best. For the sparse linear SOCPs, the merit functions associated with τ ∈ (2, 3]*

work as well as, even better than, the FB merit function; whereas for the dense convex
*SOCPs, the merit functions with τ ∈ [0.1, 1.5] have better numerical performance.*

Key words. Second-order cone complementarity problem, merit function, smoothness, Jordan product.

AMS subject classifications. 26B05, 26B35, 90C33, 65K05

1Member of Mathematics Division, National Center for Theoretical Sciences, Taipei Office. The author’s work is partially supported by National Science Council of Taiwan.

### 1 Introduction

*We consider the conic complementarity problem of finding a vector ζ ∈ IR** ^{n}* such that

*F (ζ) ∈ K,*

*G(ζ) ∈ K,*

*hF (ζ), G(ζ)i = 0*(1)

*where h·, ·i is the Euclidean inner product, F : IR*

^{n}*→ IR*

^{n}*and G : IR*

^{n}*→ IR*

*are smooth*

^{n}*mappings, and K is the Cartesian product of second-order cones (SOCs). In other words,*

*K = K*^{n}^{1} *× K*^{n}^{2} *× · · · × K*^{n}^{N}*,* (2)
*where N, n*_{1}*, . . . , n*_{N}*≥ 1, n*_{1}*+ · · · + n*_{N}*= n, and*

*K*^{n}* ^{i}* :=

^{n}

*(x*1

*, x*2

*) ∈ IR × IR*

^{n}

^{i}

^{−1}*| kx*2

*k ≤ x*1

o*,* (3)

*with k · k denoting the Euclidean norm and K*^{1} denoting the set of nonnegative reals IR+. A
*special case of (2) is K = IR*^{n}_{+}, the nonnegative orthant in IR^{n}*, which corresponds to N = n*
*and n*_{1} *= · · · = n*_{N}*= 1. We will refer to (1)-(2) as the second-order cone complementarity*
*problem (SOCCP). For convenience, in the sequel, we will focus on K = K** ^{n}*. All analysis

*can be carried over to the general case where K has the direct product structure as (2).*

*An important special case of the SOCCP corresponds to G(ζ) = ζ for all ζ ∈ IR** ^{n}*. Then
(1) reduces to

*hF (ζ), ζi = 0,* *F (ζ) ∈ K,* *ζ ∈ K,* (4)

which is a natural extension of the nonlinear complementarity problem (NCP) [9, 11] with
*K = IR*^{n}_{+}. Another important special case corresponds to the Karush-Kuhn-Tucker (KKT)
conditions for the convex second-order cone program (CSOCP):

minimize *g(x)*

*subject to Ax = b,* *x ∈ K,* (5)

*where g : IR*^{n}*→ IR is a convex twice continuously differentiable function, A ∈ IR** ^{m×n}* has

*full row rank and b ∈ IR*

^{m}*. When g is linear, this reduces to the linear SOCP which arises*in numerous applications in engineering design, finance, robust optimization, and includes as special cases convex quadratically constrained quadratic programs and linear programs;

see [1, 19] and references therein.

There have been various methods proposed for solving SOCPs and SOCCPs. They include interior-point methods [2, 3, 19, 21, 22, 27], non-interior smoothing Newton methods [8, 14, 15], and smoothing–regularization methods [16]. Recently, there was an alternative approach [6] based on reformulating the SOCCP as an unconstrained minimization problem.

*In that approach, it aimed to find a smooth function ψ : IR*^{n}*× IR*^{n}*→ IR*+ such that
*ψ(x, y) = 0* *⇐⇒* *x ∈ K,* *y ∈ K,* *hx, yi = 0.* (6)

*We call such a ψ a smooth merit function. Consequently, the SOCCP can be expressed as*
*an unconstrained smooth minimization problem of f (ζ) := ψ(F (ζ), G(ζ)).*

*A popular choice of ψ is the squared norm of Fischer-Burmeister (FB) function:*

*ψ*_{FB}*(x, y) =* 1

2*kφ*_{FB}*(x, y)k*^{2}*,* (7)

*where φ*_{FB} : IR^{n}*× IR*^{n}*→ IR** ^{n}* is the well-known FB function [12, 13] given by

*φ*_{FB}*(x, y) = (x*^{2}*+ y*^{2})^{1/2}*− x − y,* (8)
*with x*^{2} *to mean x ◦ x and x + y to mean the usual componentwise addition of vectors.*

*More specifically, for any x = (x*_{1}*, x*_{2}*), y = (y*_{1}*, y*_{2}*) ∈ IR × IR*^{n−1}*, their Jordan product x ◦ y*
*associated with K** ^{n}* is defined as

*x ◦ y := (hx, yi, y*_{1}*x*_{2}*+ x*_{1}*y*_{2}*).* (9)
*The Jordan product ◦, unlike scalar or matrix multiplication, is not associative, which is*
a main source on complication in the analysis of SOCCP. The identity element under this
*product is e := (1, 0, . . . , 0)*^{T}*∈ IR*^{n}*. The function ψ*_{FB} was studied in [6] and particularly
*shown to be a smooth merit function for the SOCCP. Another popular choice of ψ is*

*ψ*_{NR}*(x, y) :=* 1

2*kφ*_{NR}*(x, y)k*^{2}*,* (10)

*which is induced by the natural residual function φ*_{NR} : IR^{n}*× IR*^{n}*→ IR*^{n}

*φ*_{NR}*(x, y) := x − [x − y]*+*,* (11)

*where [ · ]*_{+} *means the projection in the Euclidean norm onto K*^{n}*. The function φ*_{NR} was
studied in [14, 16] which is involved in smoothing methods for the SOCCP. Compared with
*the FB merit function ψ*_{FB}*, the function ψ*_{NR} has a drawback, i.e., its non-differentiability.

In this paper, we investigate the following one-parametric class of merit functions
*ψ**τ**(x, y) :=* 1

2*kφ**τ**(x, y)k*^{2}*,* (12)

*where φ** _{τ}* : IR

^{n}*× IR*

^{n}*→ IR*

*is a family of functions defined by*

^{n}*φ*_{τ}*(x, y) :=* ^{h}*(x − y)*^{2} *+ τ (x ◦ y)*^{i}^{1/2}*− (x + y)* (13)
*with τ being a fixed parameter such that τ ∈ (0, 4). Notice that, for any x, y ∈ IR** ^{n}*,

*(x − y)*^{2}*+ τ (x ◦ y) =*

·

*x +* *τ − 2*
2 *y*

¸_{2}

+*τ (4 − τ )*
4 *y*^{2}

=

·

*y +* *τ − 2*
2 *x*

¸_{2}

+*τ (4 − τ )*

4 *x*^{2} *∈ K*^{n}*,* (14)

*and hence φ*_{τ}*and ψ*_{τ}*are well-defined. We will prove that ψ** _{τ}* is a smooth merit function for
the SOCCP with computable gradient formulas (see Propositions 3.1–3.3). In other words,
the SOCCP can be expressed as an unconstrained smooth minimization problem:

*ζ∈IR*min^{n}*f**τ**(ζ) := ψ**τ**(F (ζ), G(ζ)).* (15)
*Also, we will show that every stationary point of f*_{τ}*solves the SOCCP if ∇F and −∇G*
*are column monotone (see Proposition 4.2). Observe that φ*_{τ}*reduces to φ*_{FB} *when τ = 2,*
*whereas its limit as τ → 0 becomes a multiple of φ*_{NR}. Thus, this class of merit functions
covers two of the most important merit functions for SOCCPs under this sense so that a
*closer look and study for it is worthwhile. Indeed, if τ = 0 then φ** _{τ}* is exactly a multiple

*of φ*

_{NR}whose squared norm is not even differentiable. This violates our expectation that

*kφ*

_{τ}*k*

^{2}

*is smooth and that is why we exclude τ = 0 from the interval (0, 4). This study is*

*motivated by the work [18] where φ*

*τ*was used to develop a nonsmooth Newton method for the NCP. This paper is mainly concerned with the merit function approach based on the unconstrained smooth minimization problem (15). Numerical results are also reported by

*solving some convex SOCPs, which indicate that the merit function ψ*

*can be an alterna-*

_{τ}*tive for the FB merit function if a suitable τ is selected.*

Throughout this paper, IR^{n}*denotes the space of n-dimensional real column vectors,*
IR^{n}^{1} *× · · · × IR*^{n}* ^{m}* is identified with IR

^{n}^{1}

^{+···+n}

^{m}*, and int(K*

^{n}*) denotes the interior of K*

*. For*

^{n}*any x, y in IR*

^{n}*, we write x º*

_{Kn}*y if x − y ∈ K*

^{n}*; and write x Â*

_{Kn}*y if x − y ∈ int(K*

*).*

^{n}*For any differentiable mapping F : IR*^{n}*→ IR*^{m}*, ∇F (x) ∈ IR** ^{n×m}* denotes the transposed

*Jacobian of F at x. For a symmetric matrix A, we write A º O (respectively, A Â O) to*

*mean A is positive semidefinite (respectively, positive definite). For nonnegative α and β,*

*we write α = O(β) to mean α ≤ Cβ, with C > 0 independent of α and β.*

### 2 Prelimiaries

*For any x = (x*1*, x*2*) ∈ IR × IR*^{n−1}*, we define its determinant and trace as follows:*

*det(x) := x*^{2}_{1}*− kx*2*k*^{2}*,* *tr(x) = 2x*1*.*

*A vector x = (x*1*, x*2*) ∈ IR × IR*^{n−1}*is said to be invertible if det(x) 6= 0. If x is invertible,*
*then there exists a unique y = (y*_{1}*, y*_{2}*) ∈ IR × IR*^{n−1}*satisfying x ◦ y = y ◦ x = e. We call*
*this y the inverse of x and denote it by x*^{−1}*. For any x = (x*_{1}*, x*_{2}*) ∈ IR × IR** ^{n−1}*, let

*L**x* :=

"

*x*_{1} *x*^{T}_{2}
*x*_{2} *x*_{1}*I*

#

(16)
which can be regarded a linear mapping from IR* ^{n}*to IR

^{n}*. It is easily verified that L*

_{x}*y = x◦y*

*and L*

_{x+y}*= L*

_{x}*+ L*

_{y}*for any x, y ∈ IR*

^{n}*, but generally L*

^{2}

_{x}*= L*

_{x}*L*

_{x}*6= L*

_{x}^{2}

*and L*

^{−1}

_{x}*6= L*

_{x}*.*

^{−1}*We next recall from [14] that each x = (x*_{1}*, x*_{2}*) ∈ IR × IR** ^{n−1}* admits a spectral factor-

*ization, associated with K*

*, of the form*

^{n}*x = λ*_{1}*(x)u*^{(1)}_{x}*+ λ*_{2}*(x)u*^{(2)}_{x}*,*

*where λ*_{i}*(x) and u*^{(i)}_{x}*are the spectral values and the associated spectral vectors of x:*

*λ**i**(x) = x*1*+ (−1)*^{i}*kx*2*k, u*^{(i)}* _{x}* = 1
2

³*1, (−1)*^{i}*x*¯2

´ *for i = 1, 2,*

with ¯*x*2 = _{kx}^{x}^{2}_{2}_{k}*if x*2 *6= 0, and otherwise ¯x*2 being any vector in IR^{n−1}*such that k¯x*2*k = 1. If*
*x*_{2} *6= 0, the factorization is unique. The spectral factorization of x and the matrix L** _{x}* have
various interesting properties, and we list several ones that will be used later.

*Property 2.1 For any x = (x*_{1}*, x*_{2}*) ∈ IR × IR*^{n−1}*, the following results always hold:*

*(a) x*^{2} *= λ*^{2}_{1}*(x)u*^{(1)}_{x}*+ λ*^{2}_{2}*(x)u*^{(2)}_{x}*∈ K*^{n}*and x** ^{1/2}* =

^{q}

*λ*

_{1}

*(x) u*

^{(1)}

*+*

_{x}^{q}

*λ*

_{2}

*(x) u*

^{(2)}

_{x}*∈ K*

^{n}*if x ∈ K*

^{n}*;*

*(b) x º*

_{Kn}*0 ⇐⇒ λ*

_{1}

*(x) ≥ 0 ⇐⇒ L*

_{x}*º O;*

*(c) x Â*_{Kn}*0 ⇐⇒ λ*_{1}*(x) > 0 ⇐⇒ L*_{x}*Â O, and the inverse of L*_{x}*is given by*

*L*^{−1}* _{x}* = 1

*det(x)*

*x*_{1} *−x*^{T}_{2}

*−x*_{2} *det(x)*
*x*_{1} *I +* 1

*x*_{1}*x*_{2}*x*^{T}_{2}

*.* (17)

### 3 Smoothness of Merit Functions

*In this section we show that ψ** _{τ}* defined by (12) is a smooth merit function for the SOCCP.

*First, we show that ψ** _{τ}* is a merit function, which is direct by the following proposition.

*Proposition 3.1 The φ*_{τ}*given by (13) is a complementarity function of the SOCCP, i.e.,*
*φ*_{τ}*(x, y) = 0 ⇐⇒ x ∈ K*^{n}*,* *y ∈ K*^{n}*,* *hx, yi = 0.*

*Proof. “⇐”. Since the condition that x ∈ K, y ∈ K and hx, yi = 0 implies x ◦ y = 0,*
*substituting it into the expression of φ*_{τ}*(x, y) yields that*

*φ**τ**(x, y) = (x*^{2}*+ y*^{2})^{1/2}*− (x + y) = φ*_{FB}*(x, y).*

*Applying Proposition 2.1 of [14], we readily obtain φ*_{τ}*(x, y) = 0.*

*“⇒”. Suppose that φ*_{τ}*(x, y) = 0. Then, x + y = [(x − y)*^{2}*+ τ (x ◦ y)]** ^{1/2}*. Squaring both

*sides yields (τ − 4)(x ◦ y) = 0. Since τ − 4 6= 0, we have x ◦ y = 0. This means that*

*x + y =*^{h}*(x − y)*^{2}*+ τ (x ◦ y)*^{i}^{1/2}*= (x*^{2}*+ y*^{2})^{1/2}*,*

*i.e., φ*_{FB}*(x, y) = 0. Consequently, the desired result is from Proposition 2.1 of [14].* *2*
Now we introduce some notation that will be used in the rest of this paper. For any
*x = (x*_{1}*, x*_{2}*), y = (y*_{1}*, y*_{2}*) ∈ IR × IR*^{n−1}*and τ ∈ (0, 4), let*

*w = (w*_{1}*, w*_{2}*) = w(x, y) := (x − y)*^{2}*+ τ (x ◦ y),*

*z = (z*_{1}*, z*_{2}*) = z(x, y) :=* ^{h}*(x − y)*^{2}*+ τ (x ◦ y)*^{i}^{1/2}*.* (18)
*Then w ∈ K*^{n}*and z ∈ K** ^{n}* always hold. Moreover, by the definition of Jordan product,

*w*_{1} *= w*_{1}*(x, y) = kxk*^{2}*+ kyk*^{2}*+ (τ − 2)x*^{T}*y,*

*w*_{2} *= w*_{2}*(x, y) = 2(x*_{1}*x*_{2} *+ y*_{1}*y*_{2}*) + (τ − 2)(x*_{1}*y*_{2}*+ y*_{1}*x*_{2}*).* (19)
*Let λ*_{1}*(w) and λ*_{2}*(w) be the spectral values of w. From Property 2.1 (a),*

*z*1 *= z*1*(x, y) =*

q

*λ*1*(w) +*^{q}*λ*2*(w)*

2 *, z*2 *= z*2*(x, y) =*

q

*λ*2*(w) −*^{q}*λ*1*(w)*

2 *w*¯2*,* (20)

where ¯*w*_{2} := _{kw}^{w}^{2}_{2}_{k}*if w*_{2} *6= 0 and otherwise ¯w*_{2} is any vector in IR^{n−1}*satisfying k ¯w*_{2}*k = 1.*

*In what follows, we concentrate on the proof of the smoothness of ψ** _{τ}*. First, we state

*an important lemma which describes the behavior of x, y when (x − y)*

^{2}

*+ τ (x ◦ y) is on the*

*boundary of K*

*. In fact, it may be viewed as an extension of [6, Lemma 3.2].*

^{n}*Lemma 3.1 For any x = (x*_{1}*, x*_{2}*), y = (y*_{1}*, y*_{2}*) ∈ IR × IR*^{n−1}*, let w be given as in (18). If*
*(x − y)*^{2}*+ τ (x ◦ y) /∈ int(K*^{n}*), then then there hold that*

*x*^{2}_{1} *= kx*_{2}*k*^{2}*, y*_{1}^{2} *= ky*_{2}*k*^{2}*, x*_{1}*y*_{1} *= x*^{T}_{2}*y*_{2}*, x*_{1}*y*_{2} *= y*_{1}*x*_{2}; (21)
*x*^{2}_{1}*+ y*_{1}^{2}*+ (τ − 2)x*_{1}*y*_{1} *= kx*_{1}*x*_{2}*+ y*_{1}*y*_{2}*+ (τ − 2)x*_{1}*y*_{2}*k*

*= kx*_{2}*k*^{2}*+ ky*_{2}*k*^{2}*+ (τ − 2)x*^{T}_{2}*y*_{2}*.* (22)
*If, in addition, (x, y) 6= (0, 0), then w*_{2} *= w*_{2}*(x, y) 6= 0, and furthermore,*

*x*^{T}_{2} *w*_{2}

*kw*_{2}*k* *= x*_{1}*,* *x*_{1} *w*_{2}

*kw*_{2}*k* *= x*_{2}*,* *y*_{2}^{T}*w*_{2}

*kw*_{2}*k* *= y*_{1}*,* *y*_{1} *w*_{2}

*kw*_{2}*k* *= y*_{2}*.* (23)
*Proof. Since (x − y)*^{2}*+ τ (x ◦ y) /∈ int(K** ^{n}*), using [6, Lemma 3.2] and (14) yields that

µ

*x*_{1} +*τ − 2*
2 *y*_{1}

¶_{2}

=

°°

°°*x*_{2}+*τ − 2*
2 *y*_{2}

°°

°°

2

*, y*^{2}_{1} *= ky*_{2}*k*^{2}*,*

µ

*x*_{1}+ *τ − 2*
2 *y*_{1}

¶

*y*_{2} =

µ

*x*_{2}+*τ − 2*
2 *y*_{2}

¶

*y*_{1}*,*

µ

*x*_{1}+ *τ − 2*
2 *y*_{1}

¶

*y*_{1} =

µ

*x*_{2}+*τ − 2*
2 *y*_{2}

¶_{T}

*y*_{2};

µ

*y*_{1}+*τ − 2*
2 *x*_{1}

¶_{2}

=

°°

°°*y*_{2}+ *τ − 2*
2 *x*_{2}

°°

°°

2

*, x*^{2}_{1} *= kx*_{2}*k*^{2}*,*

µ

*y*_{1}+ *τ − 2*
2 *x*_{1}

¶

*x*_{2} =

µ

*y*_{2}+*τ − 2*
2 *x*_{2}

¶

*x*_{1}*,*

µ

*y*_{1}+*τ − 2*
2 *x*_{1}

¶

*x*_{1} =

µ

*y*_{2}+*τ − 2*
2 *x*_{2}

¶_{T}

*x*_{2}*.*

From the above equalities, we immediately obtain the results in (21). In addition, since
*w /∈ int(K*^{n}*) but w ∈ K*^{n}*, we have λ*_{1}*(w) = 0, which implies that*

*kxk*^{2}*+ kyk*^{2}*+ (τ − 2)x*^{T}*y = k2x*_{1}*x*_{2}*+ 2y*_{1}*y*_{2}*+ (τ − 2)(x*_{1}*y*_{2}*+ y*_{1}*x*_{2}*)k.*

*Applying the relations in (21) then gives the equalities in (22). If, in addition, (x, y) 6= (0, 0),*
*then it is clear that kx*_{1}*x*_{2}*+ y*_{1}*y*_{2}*+ (τ − 2)x*_{1}*y*_{2}*k = x*^{2}_{1}*+ y*^{2}_{1}*+ (τ − 2)x*_{1}*y*_{1} *6= 0. To prove the*
*equalities in (23), we only need to verify that x*^{T}_{2}_{kw}^{w}^{2}_{2}_{k}*= x*_{1} *and x*_{1}_{kw}^{w}^{2}_{2}_{k}*= x*_{2} in view of the
*symmetry of x and y in w. The verifications are straightforward due to x*1*y*2 *= y*1*x*2 and
equation (22). *2*

*By Lemma 3.1, when w(x, y) = (x − y)*^{2}*+ τ (x ◦ y) /∈ int(K*^{n}*), the spectral values of w*
*can be further simplified. Clearly, λ*_{1}*(w) = 0 and λ*_{2}*(w) can be rewritten as*

*λ*_{2}*(w) = 2*^{³}*x*^{2}_{1}*+ y*_{1}^{2}*+ (τ − 2)x*_{1}*y*_{1}^{´}*+ 2kx*_{1}*x*_{2}*+ y*_{1}*y*_{2}*+ (τ − 2)x*_{1}*y*_{2}*k*

= 4^{³}*x*^{2}_{1}*+ y*_{1}^{2}*+ (τ − 2)x*1*y*1

´*.* (24)

*Therefore, if (x, y) 6= (0, 0) also holds, using equations (20), (22) and (24) yields that*
*z*_{1}*(x, y) =*

q

*x*^{2}_{1}*+ y*_{1}^{2}*+ (τ − 2)x*_{1}*y*_{1}*, z*_{2}*(x, y) =* *x*_{1}*x*_{2}*+ y*_{1}*y*_{2}*+ (τ − 2)x*_{1}*y*_{2}

q

*x*^{2}_{1}*+ y*^{2}_{1} *+ (τ − 2)x*_{1}*y*_{1} *.*

*Thus, if (x, y) 6= (0, 0) and (x − y)*^{2}*+ τ (x ◦ y) /∈ int(K*^{n}*), the function φ** _{τ}* is rewritten as

*φ*_{τ}*(x, y) = z(x, y) − (x + y) =*

q*x*^{2}_{1}*+ y*_{1}^{2}*+ (τ − 2)x*_{1}*y*_{1}*− (x*_{1}*+ y*_{1})
*x*_{1}*x*_{2}*+ y*_{1}*y*_{2}*+ (τ − 2)x*_{1}*y*_{2}

q

*x*^{2}_{1}*+ y*_{1}^{2}*+ (τ − 2)x*_{1}*y*_{1} *− (x*_{2}*+ y*_{2})

*.* (25)

This specific expression will be employed in the proof of the following main result.

*Proposition 3.2 The function ψ*_{τ}*given by (12) is differentiable at every (x, y) ∈ IR*^{n}*×IR*^{n}*.*
*Moreover, ∇*_{x}*ψ*_{τ}*(0, 0) = ∇*_{y}*ψ*_{τ}*(0, 0) = 0; and if (x − y)*^{2}*+ τ (x ◦ y) ∈ int(K*^{n}*), then*

*∇*_{x}*ψ*_{τ}*(x, y) =* ^{h}*L*_{x+}^{τ −2}

2 *y**L*^{−1}_{z}*− I*^{i}*φ*_{τ}*(x, y),*

*∇*_{y}*ψ*_{τ}*(x, y) =* ^{h}*L*_{y+}^{τ −2}

2 *x**L*^{−1}_{z}*− I*^{i}*φ*_{τ}*(x, y).* (26)
*If (x, y) 6= (0, 0) and (x − y)*^{2}*+ τ (x ◦ y) 6∈ int(K*^{n}*), then x*^{2}_{1}*+ y*_{1}^{2}*+ (τ − 2)x*_{1}*y*_{1} *6= 0 and*

*∇*_{x}*ψ*_{τ}*(x, y) =*

*x*_{1}+^{τ −2}_{2} *y*_{1}

q

*x*^{2}_{1}*+ y*^{2}_{1}*+ (τ − 2)x*_{1}*y*_{1} *− 1*

*φ*_{τ}*(x, y),*

*∇*_{y}*ψ*_{τ}*(x, y) =*

*y*1 +^{τ −2}_{2} *x*1

q*x*^{2}_{1}*+ y*^{2}_{1}*+ (τ − 2)x*_{1}*y*_{1} *− 1*

*φ*_{τ}*(x, y).* (27)

*Proof. Case (1): (x, y) = (0, 0). For any h, k ∈ IR*^{n}*, let µ*_{1} *≤ µ*_{2} be the spectral values of
*(h − k)*^{2}*+ τ (h ◦ k) and v*^{(1)}*, v*^{(2)} be the corresponding spectral vectors. Then,

*ψ**τ**(h, k) − ψ**τ**(0, 0) =* 1
2

°°

°*[h*^{2}*+ k*^{2} *+ (τ − 2)(h ◦ k)]*^{1/2}*− h − k*^{°}^{°}°^{2}

= 1 2

°°

°*√*

*µ*1 *v*^{(1)}+*√*

*µ*2 *v*^{(2)}*− h − k*^{°}^{°}°^{2}

*≤* 1
2

·q

*2µ*_{2}*+ khk + kkk*

¸_{2}

*.*

*In addition, by the definition of spectral value µ*_{2}, it is easy to verify that
*µ*_{2} *≤ 2khk*^{2}*+ 2kkk*^{2}*+ 3|τ − 2|khkkkk ≤ 5(khk*^{2}*+ kkk*^{2}*).*

*Combining the last two equations then yields ψ*_{τ}*(h, k) − ψ*_{τ}*(0, 0) = O(khk*^{2}*+ kkk*^{2}). This
*shows that ψ*_{τ}*is differentiable at (0, 0) with ∇*_{x}*ψ*_{τ}*(0, 0) = ∇*_{y}*ψ*_{τ}*(0, 0) = 0.*

*Case (2): (x−y)*^{2}*+τ (x◦y) ∈ int(K** ^{n}*). By [7, Proposition 5] or [14, Proposition 5.2], we know

*that z(x, y) defined by (20) is continuously differentiable at such (x, y), and consequently,*

*φ*

*τ*

*(x, y) = z(x, y) − (x + y) is also continuously differentiable at such (x, y). Notice that*

*z*^{2}*(x, y) =*

µ

*x +τ − 2*
2 *y*

¶_{2}

+*τ (4 − τ )*
4 *y*^{2}*.*

*Differentiating on both sides about x, it then follows that ∇*_{x}*z(x, y)L*_{z}*= L*_{x+}^{τ −2}

2 *y*. Using
*Property 2.1 (c) and noting that ∇*_{x}*φ*_{τ}*(x, y) = ∇*_{x}*z(x, y) − I, we have*

*∇*_{x}*φ*_{τ}*(x, y) = L*_{x+}^{τ −2}

2 *y**L*^{−1}_{z}*− I,*

*which, together with ∇**x**ψ**τ**(x, y) = ∇**x**φ**τ**(x, y)φ**τ**(x, y), yields the first formula in (26). For*
*the symmetry of x and y in ψ** _{τ}*, the second formula in (26) also holds.

*Case (3): (x, y) 6= (0, 0) and (x − y)*^{2}*+ τ (x ◦ y) /∈ int(K*^{n}*). For any (x*^{0}*, y*^{0}*) ∈ IR*^{n}*× IR** ^{n}*,

*2ψ*

_{τ}*(x*

^{0}*, y*

*) =*

^{0}°°

°°

h*x*^{02}*+ y*^{02}*+ (τ − 2)(x*^{0}*◦ y** ^{0}*)

^{i}

^{1/2}°°

°°

2*+ kx*^{0}*+ y*^{0}*k*^{2}

*−2*^{¿h}*x*^{02}*+ y*^{02}*+ (τ − 2)(x*^{0}*◦ y** ^{0}*)

^{i}

^{1/2}*, x*

^{0}*+ y*

^{0}À

*= kx*^{0}*k*^{2}*+ ky*^{0}*k*^{2}*+ (τ − 2)hx*^{0}*, y*^{0}*i + kx*^{0}*+ y*^{0}*k*^{2}

*−2*^{D}*[x*^{02}*+ y*^{02}*+ (τ − 2)(x*^{0}*◦ y** ^{0}*)]

^{1/2}*, x*

^{0}*+ y*

^{0}^{E}

*,*

*where the second equality uses the observation that kzk*^{2} *= hz*^{2}*, ei for any z ∈ IR** ^{n}*. Since

*kx*

^{0}*k*

^{2}

*+ ky*

^{0}*k*

^{2}

*+ (τ − 2)hx*

^{0}*, y*

^{0}*i + kx*

^{0}*+ y*

^{0}*k*

^{2}

*is clearly differentiable in (x*

^{0}*, y*

*), it suffices to*

^{0}*show that h[x*

^{02}*+ y*

^{02}*+ (τ − 2)(x*

^{0}*◦ y*

*)]*

^{0}

^{1/2}*, x*

^{0}*+ y*

^{0}*i is differentiable at (x*

^{0}*, y*

^{0}*) = (x, y). By*

*Lemma 3.1, w*

_{2}

*(x, y) = 2(x*

_{1}

*x*

_{2}

*+ y*

_{1}

*y*

_{2}

*) + 2(τ − 2)x*

_{1}

*y*

_{2}

*6= 0, which implies*

*w*2*(x*^{0}*, y*^{0}*) = 2x*^{0}_{1}*x*^{0}_{2}*+ 2y*_{1}^{0}*y*^{0}_{2}*+ (τ − 2)(x*^{0}_{1}*y*^{0}_{2}*+ y*_{1}^{0}*x*^{0}_{2}*) 6= 0*

*for all (x*^{0}*, y*^{0}*) ∈ IR*^{n}*× IR*^{n}*sufficiently near to (x, y). Let µ*_{1}*, µ*_{2} be the spectral values of
*x*^{02}*+ y*^{02}*+ (τ − 2)(x*^{0}*◦ y** ^{0}*). Then we can compute that

2^{Dh}*x*^{02}*+ y*^{02}*+ (τ − 2)(x*^{0}*◦ y** ^{0}*)

^{i}

^{1/2}*, x*

^{0}*+ y*

^{0}^{E}

= *√*

*µ*_{2}

*x*^{0}_{1} *+ y*^{0}_{1}+

h*2(x*^{0}_{1}*x*^{0}_{2}*+ y*_{1}^{0}*y*_{2}^{0}*) + (τ − 2)(x*^{0}_{1}*y*_{2}^{0}*+ y*_{1}^{0}*x*^{0}_{2})^{i}^{T}*(x*^{0}_{2}*+ y*_{2}* ^{0}*)

*k2(x*

^{0}_{1}

*x*

^{0}_{2}

*+ y*

_{1}

^{0}*y*

_{2}

^{0}*) + (τ − 2)(x*

^{0}_{1}

*y*

_{2}

^{0}*+ y*

_{1}

^{0}*x*

^{0}_{2}

*)k*

+*√*
*µ*_{1}

*x*^{0}_{1}*+ y*_{1}^{0}*−*

h*2(x*^{0}_{1}*x*^{0}_{2}*+ y*_{1}^{0}*y*_{2}^{0}*) + (τ − 2)(x*^{0}_{1}*y*_{2}^{0}*+ y*_{1}^{0}*x*^{0}_{2})^{i}^{T}*(x*^{0}_{2}*+ y*_{2}* ^{0}*)

*k2(x*

^{0}_{1}

*x*

^{0}_{2}

*+ y*

_{1}

^{0}*y*

_{2}

^{0}*) + (τ − 2)(x*

^{0}_{1}

*y*

_{2}

^{0}*+ y*

_{1}

^{0}*x*

^{0}_{2}

*)k*

*.* (28)

*Since λ*_{2}*(w) > 0 and w*_{2}*(x, y) 6= 0, the first term on the right-hand side of (28) is differen-*
*tiable at (x*^{0}*, y*^{0}*) = (x, y). We claim that the second term is o(khk+kkk) with h := x*^{0}*−x, k :=*

*y*^{0}*−y, i.e., it is differentiable at (x, y) with zero gradient. To see this, note that w*_{2}*(x, y) 6= 0,*
*and hence µ*1 *= kx*^{0}*k*^{2}*+ky*^{0}*k*^{2}*+(τ −2)hx*^{0}*, y*^{0}*i−k2(x*^{0}_{1}*x*^{0}_{2}*+y*^{0}_{1}*y*_{2}^{0}*)+(τ −2)(x*^{0}_{1}*y*^{0}_{2}*+y*^{0}_{1}*x*^{0}_{2}*)k, viewed*
*as a function of (x*^{0}*, y*^{0}*), is differentiable at (x*^{0}*, y*^{0}*) = (x, y). Moreover, µ*_{1} *= λ*_{1}*(w) = 0 when*
*(x*^{0}*, y*^{0}*) = (x, y). Thus, the first-order Taylor’s expansion of µ*_{1} *at (x, y) yields*

*µ*_{1} *= O(kx*^{0}*− xk + ky*^{0}*− yk) = O(khk + kkk).*

*Also, since w*2*(x, y) 6= 0, by the product and quotient rules for differentiation, the function*

*x*^{0}_{1}*+ y*_{1}^{0}*−*

h*2(x*^{0}_{1}*x*^{0}_{2} *+ y*^{0}_{1}*y*_{2}^{0}*) + (τ − 2)(x*^{0}_{1}*y*_{2}^{0}*+ y*^{0}_{1}*x*^{0}_{2})^{i}^{T}*(x*^{0}_{2}*+ y*^{0}_{2})

*k2(x*^{0}_{1}*x*^{0}_{2} *+ y*^{0}_{1}*y*_{2}^{0}*) + (τ − 2)(x*^{0}_{1}*y*_{2}^{0}*+ y*^{0}_{1}*x*^{0}_{2}*)k* (29)
*is also differentiable at (x*^{0}*, y*^{0}*) = (x, y), and it has value 0 at (x*^{0}*, y*^{0}*) = (x, y) because*

*x*_{1}*+ y*_{1}*−*

h*x*_{1}*x*_{2}*+ y*_{1}*y*_{2}*+ (τ − 2)x*_{1}*y*_{2}^{i}^{T}*(x*_{2}*+ y*_{2})

*kx*_{1}*x*_{2}*+ y*_{1}*y*_{2}*+ (τ − 2)x*_{1}*y*_{2}*k* *= x*_{1}*− x*^{T}_{2} *w*_{2}

*kw*_{2}*k* *+ y*_{1}*− y*^{T}_{2} *w*_{2}
*kw*_{2}*k* = 0
*by Lemma 3.1. Thus, the function (29) is O(khk + kkk) in magnitude, which together with*
*µ*_{1} *= O(khk + kkk) shows that the second term on the right-hand side of (28) is*

*O((khk + kkk)*^{3/2}*) = o(khk + kkk).*

*Then, we have shown that ψ*_{τ}*is differentiable at (x, y). Moreover, we see that 2∇ψ*_{τ}*(x, y)*
*is the sum of the gradient of kx*^{0}*k*^{2}*+ ky*^{0}*k*^{2}*+ (τ − 2)hx*^{0}*, y*^{0}*i + kx*^{0}*+ y*^{0}*k*^{2} and the gradient of
*the first term on the right-hand side of (28), evaluated at (x*^{0}*, y*^{0}*) = (x, y).*

*The gradient of kx*^{0}*k*^{2}*+ ky*^{0}*k*^{2}*+ (τ − 2)hx*^{0}*, y*^{0}*i + kx*^{0}*+ y*^{0}*k*^{2} *with respect to x** ^{0}*, evaluated

*at (x*

^{0}*, y*

^{0}*) = (x, y), is 2x + (τ − 2)y + 2(x + y). The derivative of the first term on the*

*right-hand side of (28) with respect to x*

^{0}_{1}

*, evaluated at (x*

^{0}*, y*

^{0}*) = (x, y), works out to be*

q 1
*λ*_{2}*(w)*

"µ

*x*_{1}+*τ − 2*
2 *y*_{1}

¶

+

µ

*x*_{2}+*τ − 2*
2 *y*_{2}

¶_{T}*w*_{2}
*kw*2*k*

# Ã

*x*_{1} *+ y*_{1}*+ (x*_{2}*+ y*_{2})^{T}*w*_{2}
*kw*2*k*

!

+

q

*λ*2*(w)*

"

1 + *(x*_{2}+ ^{τ −2}_{2} *y*_{2})^{T}*(x*_{2}*+ y*_{2})

*kx*_{1}*x*_{2}*+ y*_{1}*y*_{2}*+ (τ − 2)x*_{1}*y*_{2}*k* *−* *w*_{2}^{T}*(x*_{2}*+ y*_{2}*) · w*^{T}_{2}*(x*_{2}+ ^{τ −2}_{2} *y*_{2})
*kx*_{1}*x*_{2}*+ y*_{1}*y*_{2}*+ (τ − 2)x*_{1}*y*_{2}*k · kw*_{2}*k*^{2}

#

= *2(x*_{1}+ ^{τ −2}_{2} *y*_{1}*)(x*_{1}*+ y*_{1})

q

*x*^{2}_{1}*+ y*_{1}^{2}*+ (τ − 2)x*_{1}*y*_{1} + 2^{q}*x*^{2}_{1}*+ y*_{1}^{2}*+ (τ − 2)x*_{1}*y*_{1}*,*

where the equality follows from Lemma 3.1. Similarly, the gradient of the first term on the
*right of (28) with respect to x*^{0}_{2}*, evaluated at (x*^{0}*, y*^{0}*) = (x, y), works out to be*

q 1
*λ*_{2}*(w)*

"µ

*x*2+*τ − 2*
2 *y*2

¶

+

µ

*x*1+*τ − 2*
2 *y*1

¶ *w*2

*kw*_{2}*k*

# Ã

*x*1*+ y*1*+ (x*2 *+ y*2)^{T}*w*2

*kw*_{2}*k*

!

+^{q}*λ*_{2}*(w)*

"

*(2x*_{1}*+ (τ − 2)y*_{1}*)x*_{2} +^{τ}_{2}*(x*_{1}*+ y*_{1}*)y*_{2}

*kx*1*x*2*+ y*1*y*2*+ (τ − 2)x*1*y*2*k* *−* *w*^{T}_{2}*(x*_{2}*+ y*_{2}*) · (x*_{1}+^{τ −2}_{2} *y*_{1}*)w*_{2}
*kx*1*x*2*+ y*1*y*2*+ (τ − 2)x*1*y*2*k · kw*2*k*^{2}

#

= 2*(2x*1*+ (τ − 2)y*1*)x*2+^{τ}_{2}*(x*1*+ y*1*)y*2

q*x*^{2}_{1}*+ y*_{1}^{2}*+ (τ − 2)x*_{1}*y*_{1} *.*

*Since λ*_{2}*(w) = 4(x*^{2}_{1}*+ y*_{1}^{2}*+ (τ − 2)x*_{1}*y*_{1}, combining the above gradient expressions yields
*2∇*_{x}*ψ*_{τ}*(x, y) = 2x + (τ − 2)y + 2(x + y) −*

"

2^{q}*x*^{2}_{1}*+ y*_{1}^{2}*+ (τ − 2)x*_{1}*y*_{1}
0

#

*−* 2

q

*x*^{2}_{1}*+ y*_{1}^{2}*+ (τ − 2)x*_{1}*y*_{1}

"

*(x*_{1}+^{τ −2}_{2} *y*_{1}*)(x*_{1}*+ y*_{1})
*(2x*_{1}*+ (τ − 2)y*_{1}*)x*_{2} +^{τ}_{2}*(x*_{1}*+ y*_{1}*)y*_{2}

#

*.*

*Using the fact x*1*y*2 *= y*1*x*2 *and noting that φ**τ* can be simplified as the one given by (25)
*under this case, we readily rewrite the above expression for ∇*_{x}*ψ*_{τ}*(x, y) in the form of (27).*

*By symmetry, ∇*_{y}*ψ*_{τ}*(x, y) also holds as form of (27).* *2*

*Proposition 3.2 gives a formula for computing ∇ψ** _{τ}*. It is a natural question whether

*ψ*

*is smooth or not. In what follows, we concentrate on this issue for which two crucial technical lemmas are needed. Their proofs are provided in appendix.*

_{τ}*Lemma 3.2 For any x = (x*_{1}*, x*_{2}*), y = (y*_{1}*, y*_{2}*), let w be given as in (18). If w*_{2} *6= 0, then*

"µ

*x*_{1} +*τ − 2*
2 *y*_{1}

¶

*+ (−1)*^{i}

µ

*x*_{2}+*τ − 2*
2 *y*_{2}

¶_{T}*w*2

*kw*_{2}*k*

#_{2}

*≤*

°°

°°

° µ

*x*2+ *τ − 2*
2 *y*2

¶

*+ (−1)*^{i}

µ

*x*1 +*τ − 2*
2 *y*1

¶ *w*_{2}
*kw*_{2}*k*

°°

°°

°

2

*≤ λ**i**(w)*

*for i = 1, 2, and furthermore, these relations also hold when interchanging x and y.*

*Lemma 3.3 Let z = z(x, y) be given as in (18). Then, for any (x, y) satisfying (x − y)*^{2}+
*τ (x ◦ y) ∈ int(K*^{n}*), there exists a scalar constant C > 0 such that*

°°

°*L*_{x+}^{τ −2}

2 *y**L*^{−1}_{z}^{°}^{°}°

*F* *≤ C,* ^{°}^{°}°*L*_{y+}^{τ −2}

2 *x**L*^{−1}_{z}^{°}^{°}°

*F* *≤ C,* (30)

*where kAk*_{F}*denotes the Frobenius norm of the n × n matrix A.*

*Proposition 3.3 The function ψ*_{τ}*defined by (12) is smooth everywhere on IR*^{n}*× IR*^{n}*.*
*Proof. By Proposition 3.2 and the symmetry of x and y in ∇ψ** _{τ}*, it suffices to show that

*∇*_{x}*ψ*_{τ}*is continuous at every (a, b) ∈ IR*^{n}*×IR*^{n}*. If (a−b)*^{2}*+τ (a◦b) ∈ int(K** ^{n}*), the conclusion
has been shown in Proposition 3.2. We next consider the other two cases.

*Case (1): (a, b) = (0, 0). By Proposition 3.2, we need to show that ∇*_{x}*ψ*_{τ}*(x, y) → 0 as*
*(x, y) → (0, 0). If (x − y)*^{2}*+ τ (x ◦ y) ∈ int(K*^{n}*), then ∇*_{x}*ψ*_{τ}*(x, y) is given by (26), whereas*
*if (x, y) 6= (0, 0) and (x − y)*^{2}*+ τ (x ◦ y) /∈ int(K*^{n}*), then ∇**x**ψ**τ**(x, y) is given by (27). Notice*
*that L*_{x+}^{τ −2}

2 *y**L*^{−1}* _{z}* and

*√*

^{x}^{1}

^{+}

^{τ −2}^{2}

^{y}^{1}

*x*^{2}_{1}*+y*_{1}^{2}*+(τ −2)x*1*y*1 are uniformly bounded with bound independent of
*(x, y). Using the continuity of φ**τ**(x, y) then leads to the desired result.*

*Case (2): (a, b) 6= (0, 0) and (a − b)*^{2}*+ τ (a ◦ b) /∈ int(K*^{n}*). We will show that ∇*_{x}*ψ*_{τ}*(x, y) →*

*∇*_{x}*ψ*_{τ}*(a, b) by the two subcases: (i) (x, y) 6= (0, 0) and (x − y)*^{2}*+ τ (x ◦ y) /∈ int(K** ^{n}*) and

*(ii) (x − y)*

^{2}

*+ τ (x ◦ y) ∈ int(K*

^{n}*). In subcase (i), ∇*

*x*

*ψ*

*τ*

*(x, y) is given by (27). Noting that*

*the right hand side of (27) is continuous at (a, b), the desired result follows.*

*Next, we prove that ∇*_{x}*ψ*_{τ}*(x, y) → ∇*_{x}*ψ*_{τ}*(a, b) in subcase (ii). From (26),*

*∇*_{x}*ψ*_{τ}*(x, y) = L*_{x+}^{τ −2}

2 *y**L*^{−1}_{z}^{³}*L*_{z}*e − (x + y)*^{´}*− φ*_{τ}*(x, y)*

=

µ

*x +τ − 2*
2 *y*

¶

*− L*_{x+}^{τ −2}

2 *y**L*^{−1}_{z}*(x + y) − φ*_{τ}*(x, y).* (31)
*On the other hand, since (a, b) 6= (0, 0) and (a − b)*^{2}*+ τ (a ◦ b) /∈ int(K** ^{n}*), we have

*kak*^{2}*+ kbk*^{2}*+ (τ − 2)a*^{T}*b = k2(a*_{1}*a*_{2}*+ b*_{1}*b*_{2}*) + (τ − 2)(a*_{1}*b*_{2}*+ b*_{1}*a*_{2}*)k 6= 0,* (32)
and moreover from (22) in Lemma 3.1, it follows that

*kak*^{2}*+ kbk*^{2}*+ (τ − 2)a*^{T}*b = 2(a*^{2}_{1}*+ b*^{2}_{1}*+ (τ − 2)a*_{1}*b*_{1})

*= 2(ka*_{2}*k*^{2}*+ kb*_{2}*k*^{2}*+ (τ − 2)a*^{T}_{2}*b*_{2})

*= 2k(a*1*a*2*+ b*1*b*2*) + (τ − 2)a*1*b*2*k.* (33)
Using the equalities in (33), it is not hard to verify that

*a*1+ ^{τ −2}_{2} *b*1

q*a*^{2}_{1}*+ b*^{2}_{1}*+ (τ − 2)a*_{1}*b*_{1}

³*(a − b)*^{2} *+ τ (a ◦ b)*^{´}^{1/2}*= a +* *τ − 2*
2 *b.*