for the second-order cone complementarity problem

DOI 10.1007/s10589-008-9182-9

A one-parametric class of merit functions

for the second-order cone complementarity problem

Jein-Shan Chen· Shaohua Pan

Received: 11 January 2007 / Revised: 21 April 2008 / Published online: 8 May 2008

© Springer Science+Business Media, LLC 2008

Abstract We investigate a one-parametric class of merit functions for the second- order cone complementarity problem (SOCCP) which is closely related to the popular Fischer–Burmeister (FB) merit function and natural residual merit function. In fact, it will reduce to the FB merit function if the involved parameter τ equals 2, whereas as τtends 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 prop- erties 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 function corresponding to τ= 2.5 or 3 works better than the FB merit function, whereas for the dense convex SOCPs, the merit function with τ= 0.1, 0.5 or 1.0 seems to have better numerical performance.

Keywords Second-order cone· Complementarity · Merit function · Jordan product

J.-S. Chen (



Department of Mathematics, National Taiwan Normal University, 88 Section 4, Ting-Chou Road, 11677 Taipei, Taiwan

e-mail:jschen@math.ntnu.edu.tw J.-S. Chen

Mathematics Division, National Center for Theoretical Sciences, Taipei Office, Taipei, Taiwain S. Pan

School of Mathematical Sciences, South China University of Technology, Guangzhou 510640, China e-mail:shhpan@scut.edu.cn

1 Introduction

We consider the conic complementarity problem of finding a vector ζ∈ Rⁿsuch that F (ζ )∈ K, G(ζ )∈ K, F (ζ ), G(ζ ) = 0, (1) where·, · is the Euclidean inner product, F : Rⁿ→ Rⁿ and G: Rⁿ→ Rⁿ are the mappings assumed to be continuously differentiable throughout this paper, andK is the Cartesian product of second-order cones (SOCs). In other words,

K = Kⁿ¹× Kⁿ²× · · · × K^nN, (2) where N, n1, . . . , n_N≥ 1, n1+ · · · + nN= n, and

Kⁿⁱ:=

(x1, x2)∈ R × Rⁿⁱ⁻¹| x2 ≤ x1

,

with ·  denoting the Euclidean norm and K¹ denoting the set of nonnegative re- alsR+. We will refer to (1)–(2) as the second-order cone complementarity problem (SOCCP).

An important special case of the SOCCP corresponds to G(ζ )= ζ for all ζ ∈ Rⁿ. Then (1) reduces to

F (ζ )∈ K, ζ ∈ K, F (ζ ), ζ = 0, (3) which is a natural extension of the nonlinear complementarity problem (NCP) [7,8]

withK = Rⁿ₊, the nonnegative orthant cone ofRⁿ. Another important special case corresponds to the KKT optimality conditions of the convex second-order cone pro- gram (CSOCP):

minimize g(x)

subject to Ax= b, x ∈ K, (4)

where g: Rⁿ→ R is a convex twice continuously differentiable function, A ∈ R^m×n has full row rank and b∈ R^m. From [4], we know that the KKT conditions of (4), which are sufficient but not necessary for optimality, can be reformulated as (1) with

F (ζ ):= ¯x +

I− A^T(AA^T)⁻¹A

ζ, G(ζ ):= ∇g(F (ζ )) − A^T(AA^T)⁻¹Aζ, (5) where ¯x ∈ Rⁿ is any point such that A¯x = b. When g is linear, the CSOCP reduces to the linear SOCP which arises in numerous applications in engineering design, finance, robust optimization, and includes as special cases convex quadratically con- strained quadratic programs and linear programs; see [1,13] and references therein.

There have been various methods proposed for solving SOCPs and SOCCPs. They include the interior-point methods [2,3,15,16,19], the non-interior smoothing New- ton methods [6,9], and the smoothing-regularization method [11]. Recently, there was an alternative method [4] based on reformulating the SOCCP as an unconstrained minimization problem. In that approach, it aims to find a function ψ: Rⁿ×Rⁿ→ R₊ satisfying

ψ (x, y)= 0 ⇐⇒ x ∈ K, y ∈ K, x, y = 0, (6)

so that the SOCCP can be reformulated as an unconstrained minimization problem min

ζ∈Rⁿf (ζ ):= ψ(F (ζ ), G(ζ )).

We call such ψ a merit function associated with the coneK.

A popular choice of ψ is the Fischer–Burmeister (FB) merit function ψ_FB(x, y):=1

2φFB(x, y)², (7)

where φ_FB: Rⁿ× Rⁿ→ Rⁿis the vector-valued FB function defined by

φ_FB(x, y):= (x²+ y²)^1/2− (x + y), (8) with x²= x ◦ x denoting the Jordan product between x and itself, x^1/2being a vector such that (x^1/2)²= x, and x + y meaning the usual componentwise addition of vec- tors. The function ψ_FB was studied in [4] and particularly shown to be continuously differentiable (smooth). Another popular choice of ψ is the natural residual merit function

ψ_NR(x, y):=1

2φNR(x, y)²,

where φ_NR: Rⁿ× Rⁿ→ Rⁿis the vector-valued natural residual function given by φ_NR(x, y):= x − (x − y)₊

with (·)₊meaning the projection in the Euclidean norm onto K. The function φNR

was studied in [9,11] which is involved in smoothing methods for the SOCCP. Com- pared with the FB merit function ψ_FB, the function ψ_NR has a drawback, i.e., its non-differentiability.

In this paper, we will investigate the following one-parametric class of functions ψτ(x, y) := 1

2φτ(x, y)², (9)

where τ is a fixed parameter from (0, 4) and φτ: Rⁿ× Rⁿ→ Rⁿis defined by φ_τ(x, y):=

(x− y)²+ τ(x ◦ y)1/2

− (x + y). (10)

Specifically, we prove that ψτ is a merit function associated withK which is continu- ously differentiable everywhere with computable gradient formulas (see Propositions 3.1–3.3), and hence the SOCCP can be reformulated as an unconstrained smooth minimization

ζmin∈Rⁿfτ(ζ ):= ψτ(F (ζ ), G(ζ )). (11) Also, we show that every stationary point of fτ solves the SOCCP under the condi- tion that∇F and −∇G are column monotone (see Proposition4.1). Observe that φτ

reduces to φ_FB when τ= 2, whereas its limit as τ → 0 becomes a multiple of φNR.

Thus, this class of merit functions has a close relation to two of the most important merit functions so that a closer look and study for it is worthwhile. In addition, this study is motivated by the work [12] where φτ 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 minimization problem (11). Numerical results are also reported for some convex SOCPs, which indicate that ψτ can be an alterna- tive for ψ_FBif a suitable τ is selected.

Throughout this paper,Rⁿ denotes the space of n-dimensional real column vec- tors, andRⁿ¹× · · · × R^nm is identified withR^n1+···+nm. Thus, (x1, . . . , x_m)∈ Rⁿ¹×

· · · × R^nm is viewed as a column vector inR^n1+···+nm. The notation I denotes an identity matrix of suitable dimension, and int(Kⁿ)denotes the interior ofKⁿ. For any differentiable mapping F: Rⁿ→ R^m,∇F (x) ∈ R^n×mdenotes the transposed Jaco- bian of F at x. For a symmetric matrix A, we write A O (respectively, A  O) to mean A is positive semidefinite (respectively, positive definite). For nonnegative αand β, we write α= O(β) to mean α ≤ Cβ, with C > 0 independent of α and β.

Without loss of generality, in the rest of this paper we assume thatK = Kⁿ(n >1).

All analysis can be carried over to the general case whereK has the structure as (2).

In addition, we always assume that τ satisfies 0 < τ < 4.

2 Preliminaries

It is known thatKⁿ(n >1) is a closed convex self-dual cone with nonempty interior int(Kⁿ):=

x= (x1, x2)∈ R × Rⁿ⁻¹| x1>x2 .

For any x= (x1, x₂), y= (y1, y₂)∈ R × Rⁿ⁻¹, the Jordan product of x and y is defined by

x◦ y := (x, y, y1x2+ x1y2). (12) 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∈ Rⁿ. For any x= (x1, x2)∈ R × Rⁿ⁻¹, the determinant of x is defined by det(x):= x₁²−x2². If det(x)= 0, then x is said to be invertible. If x is invertible, there exists a unique y= (y1, y₂)∈ R × Rⁿ⁻¹satisfying x◦ y = y ◦ x = e. We call this y the inverse of x and denote it by x⁻¹. For each x= (x1, x2)∈ R × Rⁿ⁻¹, let

Lx:=

x1 x₂^T x2 x1I

. (13)

It is easily verified that Lxy = x ◦ y and Lx+y= Lx+ Ly for any x, y∈ Rⁿ, but generally L²_x= LxLx= L_x² and L⁻¹_x = L_x−1. If Lxis invertible, then the inverse of L_xis given by

L⁻¹_x = 1 det(x)

⎡

⎣ x1 −x₂^T

−x2

det(x) x₁ I+ 1

x₁x2x₂^T

⎤

⎦ . (14)

We next recall from [9] that each x= (x1, x₂)∈ R × Rⁿ⁻¹admits a spectral fac- torization, associated withKⁿ, of the form

x= λ1(x)· u⁽¹⁾_x + λ2(x)· u⁽²⁾_x ,

where λ1(x), λ2(x)and u⁽¹⁾x , u⁽²⁾x are the spectral values and the associated spectral vectors of x given by

λ_i(x)= x1+ (−1)ⁱx2, u⁽ⁱ⁾_x =1

2(1, (−1)ⁱ¯x2) for i= 1, 2,

with ¯x2= _x^x2_2 if x2= 0, and otherwise ¯x2 being any vector in Rⁿ⁻¹ such that

 ¯x2 = 1. If x2= 0, the factorization is unique. The spectral factorization of x has various interesting properties; see [9]. We list three properties that will be used later.

Property 2.1

(a) x²= λ²₁(x)· u⁽¹⁾x + λ²₂(x)· u⁽²⁾x ∈ Kⁿfor any x∈ Rⁿ. (b) If x∈ Kⁿ, then x^1/2=√

λ1(x)· u⁽¹⁾x +√

λ2(x)· u⁽²⁾x ∈ Kⁿ.

(c) x∈ Kⁿ⇐⇒ λ1(x)≥ 0 ⇐⇒ Lx O, x ∈ int(Kⁿ)⇐⇒ λ1(x) >0⇐⇒ Lx O.

3 Smoothness of the function ψτ

In this section we will show that ψτ defined by (9) is a smooth merit function. First, by Properties2.1(a) and (b), φτ and ψτ are well-defined since for any x, y∈ Rⁿ, we can verify that

(x− y)²+ τ(x ◦ y) =



x+τ− 2 2 y

2

+τ (4− τ) 4 y²

=



y+τ− 2 2 x

2

+τ (4− τ)

4 x²∈ Kⁿ. (15) The following proposition shows that ψτ is indeed a merit function associated withKⁿ.

Proposition 3.1 Let ψτ and φτbe given as in (9) and (10), respectively. Then,

ψτ(x, y)= 0 ⇐⇒ φτ(x, y)= 0 ⇐⇒ x ∈ Kⁿ, y∈ Kⁿ, x, y = 0.

Proof The first equivalence is clear by the definition of ψτ. We consider the second one.

“⇐”. Since x ∈ K, y ∈ K and x, y = 0, we have x ◦ y = 0. Substituting it into the expression of φτ(x, y)then yields that φτ(x, y)= (x²+ y²)^1/2− (x + y) = φ_FB(x, y). From Proposition 2.1 of [9], we immediately obtain φτ(x, y)= 0.

“⇒”. Suppose that φτ(x, y)= 0. Then, x + y =

(x− y)²+ τ(x ◦ y)1/2

. Squar- ing both sides yields x◦ y = 0. This implies that x + y = (x²+ y²)^1/2, i.e.

φ_FB(x, y)= 0. From Proposition 2.1 of [9], it then follows that x∈ Kⁿ, y∈ Kⁿand

x, y = 0. 

In what follows, we focus on the proof of the smoothness of ψτ. We first introduce some notation that will be used in the sequel. For any x= (x1, x2), y= (y1, y2)∈ R × Rⁿ⁻¹, let

w= (w1, w₂)= w(x, y) := (x − y)²+ τ(x ◦ y),

(16) z= (z1, z2)= z(x, y) :=

(x− y)²+ τ(x ◦ y)1/2

.

Then, w∈ Kⁿand z∈ Kⁿ. Moreover, by the definition of Jordan product, w1= w1(x, y)= x²+ y²+ (τ − 2)x^Ty,

(17) w2= w2(x, y)= 2(x1x2+ y1y2)+ (τ − 2)(x1y2+ y1x2).

Let λ1(w)and λ2(w)be the spectral values of w. By Property2.1(b), we have that z1= z1(x, y)=

√λ₂(w)+√ λ₁(w)

2 ,

(18) z₂= z2(x, y)=

√λ2(w)−√ λ1(w)

2 ¯w2,

where ¯w2:= _w^w2_2 if w2= 0 and otherwise ¯w2 is any vector in Rⁿ⁻¹ satisfying

 ¯w2 = 1.

The following technical lemma describes the behavior of x, y when w= (x − y)²+ τ(x ◦ y) is on the boundary of Kⁿ. In fact, it may be viewed as an extension of [4, Lemma 3.2].

Lemma 3.1 For any x= (x1, x2), y= (y1, y2)∈ R × Rⁿ⁻¹, if w /∈ int(Kⁿ), then x₁²= x2², y₁²= y2², x₁y₁= x₂^Ty₂, x₁y₂= y1x₂; (19) x₁²+ y1²+ (τ − 2)x1y1= x1x2+ y1y2+ (τ − 2)x1y2

= x2²+ y2²+ (τ − 2)x₂^Ty₂. (20) If, in addition, (x, y)= (0, 0), then w2= 0, and furthermore,

x^T₂ w2

w2= x1, x₁ w2

w2 = x2, y₂^T w2

w2= y1, y₁ w2

w2 = y2. (21) Proof Since w= (x −y)²+τ(x ◦y) /∈ int(Kⁿ), using (15) and [4, Lemma 3.2] yields



x₁+τ− 2 2 y₁

2

=

x²+τ− 2 2 y₂

², y₁²= y2²,



x₁+τ− 2 2 y₁

 y₂=



x₂+τ− 2 2 y₂

 y₁,



x1+τ− 2 2 y1

 y1=



x2+τ− 2 2 y2

T

y2;



y1+τ− 2 2 x1

2

=

y²+τ− 2 2 x2

², x₁²= x2²,



y1+τ− 2 2 x1

 x2=



y2+τ− 2 2 x2

 x1,



y₁+τ− 2 2 x₁

 x₁=



y₂+τ− 2 2 x₂

T

x₂.

From these equalities, we readily get the results in (19). Since w ∈ Kⁿ but w /∈ int(Kⁿ), we havex²+ y²+ (τ − 2)x^Ty= 2x1x2+ 2y1y2+ (τ − 2)(x1y2+ y1x2) by λ1(w)= 0. Applying the relations in (19) then gives the equalities in (20).

If, in addition, (x, y)= (0, 0), then it is clear that x1x₂+ y1y₂+ (τ − 2)x1y₂ = x₁²+ y₁²+ (τ − 2)x1y1= 0. To prove the equalities in (21), it suffices to verify that x₂^T_w^w2

2= x1and x1 w₂

w2= x2 by the symmetry of x and y in w. The verifications

are straightforward by (20) and x1y2= y1x2. 

By Lemma3.1, when w /∈ int(Kⁿ), the spectral values of w are calculated as fol- lows:

λ₁(w)= 0, λ₂(w)= 4

x₁²+ y₁²+ (τ − 2)x1y₁

. (22)

If (x, y)= (0, 0) also holds, then using (18), (20) and (22) yields that

z₁(x, y)=

x₁²+ y₁²+ (τ − 2)x1y₁, z₂(x, y)=x1x2+ y1y2+ (τ − 2)x1y2



x₁²+ y₁²+ (τ − 2)x1y1

.

Thus, if (x, y)= (0, 0) and (x − y)²+ τ(x ◦ y) /∈ int(Kⁿ), φτ(x, y)can be rewritten as

φ_τ(x, y)= z(x, y) − (x + y) =

⎛

⎜⎝



x₁²+ y₁²+ (τ − 2)x1y₁− (x1+ y1)

x1x2+y1y2+(τ−2)x1y2

x₁²+y1²+(τ−2)x1y1

− (x2+ y2)

⎞

⎟⎠ . (23)

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

Proposition 3.2 The function ψτgiven by (9) is differentiable at every (x, y)∈ Rⁿ× Rⁿ. Moreover,∇xψτ(0, 0)= ∇yψτ(0, 0)= 0; if (x − y)²+ τ(x ◦ y) ∈ int(Kⁿ), then

∇xψ_τ(x, y)= L_x+τ−2

2 yL⁻¹_z − I

φ_τ(x, y),

∇yψ_τ(x, y)= (24) L_y+τ−2

2 xL⁻¹_z − I

φ_τ(x, y);

if (x, y)= (0, 0) and (x − y)²+ τ(x ◦ y) ∈ int(Kⁿ), then x₁²+ y₁²+ (τ − 2)x1y₁= 0 and

∇xψτ(x, y)=

 x1+^τ−2₂ y1



x₁²+ y₁²+ (τ − 2)x1y1

− 1



φτ(x, y),

(25)

∇yψ_τ(x, y)=

 y1+^τ−2₂ x1



x₁²+ y₁²+ (τ − 2)x1y1

− 1



φ_τ(x, y).

Proof Case (1): (x, y)= (0, 0). For any u = (u1, u₂), v= (v1, v₂)∈ R × Rⁿ⁻¹, let μ₁, μ₂ be the spectral values of (u− v)²+ τ(u ◦ v) and ξ⁽¹⁾, ξ⁽²⁾ be the spectral vectors. Then,

2

ψ_τ(u, v)− ψτ(0, 0)

=u²+ v²+ (τ − 2)(u ◦ v)1/2

− u − v²

=√μ₁ξ⁽¹⁾+√

μ₂ξ⁽²⁾− u − v²

≤

2μ2+ u + v2

.

In addition, from the definition of spectral value, it follows that

μ2= u²+ v²+ (τ − 2)u^Tv+ 2(u1u2+ v1v2)+ (τ − 2)(u1v2+ v1u2)

≤ 2u²+ 2v²+ 3|τ − 2|uv ≤ 5(u²+ v²).

Now combining the last two equations, we have ψτ(u, v)− ψτ(0, 0)= O(u²+

v²). This shows that ψτ is differentiable at (0, 0) with ∇xψτ(0, 0) =

∇yψ_τ(0, 0)= 0.

Case (2): (x− y)²+ τ(x ◦ y) ∈ int(Kⁿ). By [5, Proposition 5], z(x, y) defined by (18) is continuously differentiable at such (x, y), and consequently φτ(x, y)is also continuously differentiable at such (x, y) since φτ(x, y)= z(x, y) − (x + y). Notice that

z²(x, y)=



x+τ − 2 2 y

2

+τ (4− τ) 4 y², which leads to ∇xz(x, y)L_z = L_x+τ−2

2 y by taking differentiation on both sides about x. Since Lz O by Property2.1(c), it follows that∇xz(x, y)= L_x+^τ−2

2 yL⁻¹_z . Consequently,

∇xφ_τ(x, y)= ∇xz(x, y)− I = L_x+τ−2

2 yL⁻¹_z − I.

This together with∇xψτ(x, y)= ∇xφτ(x, y)φτ(x, y)proves the first formula of (24).

For the symmetry of x and y in ψτ, the second formula also holds.

Case (3): (x, y)= (0, 0) and (x − y)²+ τ(x ◦ y) /∈ int(Kⁿ). For any x^ = (x₁^, x₂^), y^= (y^₁, y₂^)∈ R × Rⁿ⁻¹, it is easy to verify that

2ψτ(x^, y^)=x^2+ y^2+ (τ − 2)(x^◦ y^)1/2²+ x^+ y^²

− 2

x^2+ y^2+ (τ − 2)(x^◦ y^)1/2

, x^+ y^

= x^²+ y^²+ (τ − 2)x^, y^ + x^+ y^²

− 2

x^2+ y^2+ (τ − 2)(x^◦ y^)1/2

, x^+ y^ ,

where the second equality uses the fact that z²= z², e for any z ∈ Rⁿ. Since

x^²+ y^²+ (τ − 2)x^, y^ + x^ + y^² is clearly differentiable in (x^, y^), it suffices to show that [x^2+ y^2 + (τ − 2)(x^ ◦ y^)]^1/2, x^ + y^ is differen- tiable at (x^, y^)= (x, y). By Lemma3.1, w2= w2(x, y)= 0, which implies w₂^ = w2(x^, y^)= 2x^₁x₂^+2y₁^y₂^+(τ −2)(x₁^y₂^+y₁^x₂^)= 0 for all (x^, y^)∈ Rⁿ×Rⁿsuf- ficiently near to (x, y). Let μ1, μ₂be the spectral values of x^2+y^2+(τ −2)(x^◦y^).

Then we can compute that 2

x^2+ y^2+ (τ − 2)(x^◦ y^)1/2

, x^+ y^

=√ μ₂



x₁^+ y₁^ +[2(x₁^x₂^ + y₁^y₂^)+ (τ − 2)(x₁^y₂^ + y₁^x₂^)]^T(x₂^ + y₂^)

2(x^₁x₂^ + y₁^y₂^)+ (τ − 2)(x^₁y₂^ + y₁^x₂^)



+√ μ1



x^₁+ y1^−[2(x₁^x₂^+ y₁^y₂^)+ (τ − 2)(x₁^y₂^+ y₁^x₂^)]^T(x₂^+ y₂^)

2(x₁^x₂^+ y₁^y₂^)+ (τ − 2)(x₁^y₂^+ y₁^x₂^)

 .

(26) Since λ2(w) >0 and w2(x, y)= 0, the first term on the right-hand side of (26) is differentiable at (x^, y^)= (x, y). Now, we claim that the second term is o(x^− x + y^− y), i.e., it is differentiable at (x, y) with zero gradient. To see this, no- tice that w2(x, y)= 0, and hence μ1= x^²+ y^²+ (τ − 2)x^, y^ − 2(x₁^x₂^+ y₁^y₂^)+ (τ − 2)(x^₁y^₂+ y₁^x^₂), viewed as a function of (x^, y^), is differentiable at (x^, y^)= (x, y). Moreover, μ1= λ1(w)= 0 when (x^, y^)= (x, y). Thus, the first- order Taylor’s expansion of μ1at (x, y) yields

μ1 = O(x^− x + y^− y).

Also, since w2(x, y)= 0, by the product and quotient rules for differentiation, the function

x₁^+ y1^−[2(x₁^x₂^+ y^₁y₂^)+ (τ − 2)(x₁^y₂^+ y^₁x₂^)]^T(x₂^+ y^₂)

2(x₁^x₂^ + y^₁y₂^)+ (τ − 2)(x₁^y₂^ + y^₁x₂^) (27) is differentiable at (x^, y^)= (x, y), and it has value 0 at (x^, y^)= (x, y) due to

x1+ y1−[x1x₂+ y1y₂+ (τ − 2)x1y₂]^T(x₂+ y2)

x1x2+ y1y2+ (τ − 2)x1y2

= x1− x₂^T w2

w2+ y1− y₂^T w2

w2= 0.

Hence, the function in (27) is O(x^− x + y^− y) in magnitude, which together with μ1= O(x^− x + y^− y) shows that the second term on the right-hand side

of (26) is O

(x^− x + y^− y)^3/2

= o

x^− x + y^− y .

Thus, we have shown that ψτ is differentiable at (x, y). Moreover, we see that 2∇ψτ(x, y)is the sum of the gradient ofx^²+ y^²+ (τ − 2)x^, y^ + x^+ y^² and the gradient of the first term on the right-hand side of (26), evaluated at (x^, y^)= (x, y).

The gradient ofx^²+ y^²+ (τ − 2)x^, y^ + x^+ y^²with respect to x^, evaluated at (x^, y^)= (x, y), is 2x + (τ − 2)y + 2(x + y). The derivative of the first term on the right-hand side of (26) with respect to x^₁, evaluated at (x^, y^)= (x, y), works out to be

√ 1 λ2(w)



x1+τ− 2 2 y1

 +



x2+τ− 2 2 y2

T

w₂

w2



×



x₁+ y1+ (x2+ y2)^T w2

w2



+ λ₂(w)



1+ (x2+^τ−2₂ y2)^T(x2+ y2)

x1x2+ y1y2+ (τ − 2)x1y2

− w^T₂(x2+ y2)· w₂^T(x2+^τ−2₂ y2)

x1x₂+ y1y₂+ (τ − 2)x1y₂ · w2²



= 2(x1+^τ−2₂ y₁)(x₁+ y1)



x²₁+ y₁²+ (τ − 2)x1y₁ + 2

x₁²+ y₁²+ (τ − 2)x1y1,

where the equality follows from Lemma3.1. Similarly, the gradient of the first term on the right of (26) with respect to x^₂, evaluated at (x^, y^)= (x, y), works out to be

√ 1 λ2(w)



x2+τ− 2 2 y2

 +



x1+τ− 2 2 y1

 w2

w2

×



x1+ y1+ (x2+ y2)^T w₂

w2



+ λ₂(w)

(2x1+ (τ − 2)y1)x2+^τ₂(x1+ y1)y2

x1x2+ y1y2+ (τ − 2)x1y2

.− w^T₂(x2+ y2)· (x1+^τ−2₂ y1)w2

x1x2+ y1y2+ (τ − 2)x1y2 · w2²

= 2(2x1+ (τ − 2)y1)x2+^τ₂(x1+ y1)y2



x₁²+ y₁²+ (τ − 2)x1y1

.

Then, combining the last two gradient expressions yields that

2∇xψ_τ(x, y)

= 2x + (τ − 2)y + 2(x + y) −

2



x₁²+ y₁²+ (τ − 2)x1y1

0

− 2



x₁²+ y₁²+ (τ − 2)x1y₁

 (x1+^τ−2₂ y1)(x1+ y1) (2x1+ (τ − 2)y1)x2+^τ₂(x1+ y1)y2

.

Using the fact that x1y2= y1x2and noting that φτ can be simplified as the one in (23) under this case, we readily rewrite the above expression for∇xψ_τ(x, y)in the form of (25). By symmetry,∇yψ_τ(x, y)also holds as the form of (25).  Proposition3.2shows that ψτ is differentiable with a computable gradient. To establish the continuity of the gradient of ψτ or the smoothness of ψτ, we need the following two crucial technical lemmas whose proofs are provided in theAppendix.

Lemma 3.2 For any x= (x1, x2), y= (y1, y2)∈ R × Rⁿ⁻¹, if w2= 0, then



x₁+τ − 2 2 y₁



+ (−1)ⁱ



x₂+τ − 2 2 y₂

T

w2

w2

2

≤



x₂+τ− 2 2 y₂



+ (−1)ⁱ



x₁+τ − 2 2 y₁

 w2

w2

² ≤ λi(w)

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

Lemma 3.3 For all (x, y) satisfying (x− y)²+ τ(x ◦ y) ∈ int(Kⁿ), we have that

Lx+^τ−22 yL⁻¹_z 

F ≤ C, L

y+^τ−22 xL⁻¹_z 

F≤ C, (28)

where C > 0 is a constant independent of x, y and τ , and·F denotes the Frobenius norm.

Proposition 3.3 The function ψτ defined by (9) is smooth everywhere onRⁿ× Rⁿ. Proof By Proposition3.2and the symmetry of x and y in∇ψτ, it suffices to show that∇xψ_τ is continuous at every (a, b)∈ Rⁿ× Rⁿ. If (a− b)²+ τ(a ◦ b) ∈ int(Kⁿ), the conclusion has been shown in Proposition3.2. We next consider the other two cases.

Case (1): (a, b)= (0, 0). By Proposition3.2, we need to show that∇xψ_τ(x, y)

→ 0 as (x, y) → (0, 0). If (x − y)²+ τ(x ◦ y) ∈ int(Kⁿ), then∇xψ_τ(x, y)is given by (24), whereas if (x, y)= (0, 0) and (x −y)²+τ(x ◦y) /∈ int(Kⁿ), then∇xψ_τ(x, y) is given by (25). Notice that

L_x+τ−2

2 yL⁻¹_z and x₁+^τ−2₂ y₁



x₁²+ y₁²+ (τ − 2)x1y₁

are bounded with bound independent of x, y and τ , using the continuity of φτ(x, y) immediately yields the desired result.

Case (2): (a, b)= (0, 0) and (a − b)²+ τ(a ◦ b) /∈ int(Kⁿ). We will show that

∇xψτ(x, y)→ ∇xψτ(a, b)by the two subcases: (2a) (x, y)= (0, 0) and (x − y)²+ τ (x◦y) /∈ int(Kⁿ)and (2b) (x−y)²+τ(x ◦y) ∈ int(Kⁿ). In subcase (2a),∇xψ_τ(x, y) is given by (25). Noting that the right-hand side of (25) is continuous at (a, b), the desired result follows.

Next, we prove that∇xψτ(x, y)→ ∇xψτ(a, b) in subcase (2b). From (24), we have that

∇xψτ(x, y)=



x+τ− 2 2 y



− L_x+^τ−2

2 yL⁻¹_z (x+ y) − φτ(x, y). (29) On the other hand, since (a, b)= (0, 0) and (a − b)²+ τ(a ◦ b) /∈ int(Kⁿ),

a²+ b²+ (τ − 2)a^Tb= 2(a1a2+ b1b2)+ (τ − 2)(a1b2+ b1a2) = 0, (30) and moreover from (20) it follows that

a²+ b²+ (τ − 2)a^Tb= 2

a²₁+ b²1+ (τ − 2)a1b1



= 2

a2²+ b2²+ (τ − 2)a₂^Tb₂

= 2(a₁a₂+ b1b₂)+ (τ − 2)a1b₂. (31)

Using the equalities in (31), it is not hard to verify that a₁+^τ−2₂ b₁



a₁²+ b²₁+ (τ − 2)a1b₁

(a− b)²+ τ(a ◦ b)1/2

= a +τ− 2 2 b.

This together with the expression of∇xψ_τ(a, b)given by (25) yields

∇xψτ(a, b)=



a+τ− 2 2 b



− a₁+^τ−2₂ b₁



a₁²+ b²₁+ (τ − 2)a1b₁

(a+ b) − φτ(a, b). (32)

Comparing (29) with (32), we see that if we wish to prove∇xψ_τ(x, y)→ ∇xψ_τ(a, b) as (x, y)→ (a, b), it suffices to show that

L_x+τ−2

2 yL⁻¹_z (x+ y) → a₁+^τ−2₂ b₁



a²₁+ b²₁+ (τ − 2)a1b₁

(a+ b), (33)

which is also equivalent to proving the following three relations

L_x+τ−2 2 yL⁻¹_z



x+τ− 2 2 y



→ a₁+^τ−2₂ b₁



a₁²+ b²₁+ (τ − 2)a1b₁



a+τ− 2 2 b

 , (34)

L_y+τ−2 2 xL⁻¹_z



y+τ− 2 2 x



→ b1+^τ−2₂ a1



a₁²+ b²₁+ (τ − 2)a1b1



b+τ− 2 2 a

 , (35)

4− τ

2 L_x−yL⁻¹_z



y+τ− 2 2 x



→

4−τ

2 (a1− b1)



a₁²+ b²₁+ (τ − 2)a1b1



b+τ− 2 2 a

 . (36)

By the symmetry of x and y in (34) and (35), we only prove (34) and (36). Let

(ζ₁, ζ₂):= L_x+^τ−2 2 yL⁻¹_z



x+τ− 2 2 y

 ,

(37) (ξ1, ξ2):= Lx−yL⁻¹_z



y+τ − 2 2 x

 .

Then showing (34) and (36) reduces to proving the following relations hold as (x, y)→ (a, b):

ζ₁→ (a1+^τ−2₂ b1)²



a²₁+ b²₁+ (τ − 2)a1b1

,

(38) ζ₂→ a1+^τ−2₂ b1



a²₁+ b²₁+ (τ − 2)a1b1



a₂+τ− 2 2 b₂

 ,

ξ1→ (a1− b1)(b1+^τ−2₂ a1)



a²₁+ b²₁+ (τ − 2)a1b1

,

(39) ξ₂→ (a1− b1)



a²₁+ b²₁+ (τ − 2)a1b1



b₂+τ− 2 2 a₂

 .

To verify (38), we take (x, y) sufficiently near to (a, b). By (30), we may assume that w2= w2(x, y)= 0. Let s = (s1, s₂)= x +^τ−2₂ y. Using (14) and (37), we can calculate that

ζ1= 1 det(z)



s₁²z1− 2s1s₂^Tz2+det(z)

z₁ s2²+(z₂^Ts2)² z₁

 ,

=s2²

z1 +(s₁z₁− s₂^Tz₂)²

z1det(z) , (40)

ζ2= 1 det(z)



s1z1s2− z^T2s2s2− s²1z2+s₁det(z) z₁ s2+s₁

z₁z^T₂s2z2



= s1

z1

s₂+(s₁z₁− s₂^Tz₂) det(z)

 s₂−s1

z1

z₂



. (41)