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 parameterτ 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 toτ =2.5 or 3 works better than the FB merit function, whereas for the dense convex SOCPs, the merit function withτ =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:[email protected] 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:[email protected]

1 Introduction

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

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

Kⁿⁱ:=

(x1, x2)∈R×Rⁿⁱ⁻¹| x2 ≤x1

,

with · denoting the Euclidean norm andK¹ 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 toG(ζ )=ζ for allζ ∈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):

minimizeg(x)

subject to Ax=b, x∈K, (4)

whereg:Rⁿ→Ris a convex twice continuously differentiable function,A∈R^m×n has full row rank andb∈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) wherex¯∈Rⁿ is any point such thatAx¯=b. Whengis 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 functionψ: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 suchψa merit function associated with the coneK.

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

2φ_FB(x, y)², (7)

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

φ_FB(x, y):=(x²+y²)^1/2−(x+y), (8) withx²=x◦xdenoting the Jordan product betweenxand itself,x^1/2being a vector such that(x^1/2)²=x, andx+ymeaning the usual componentwise addition of vec- tors. The functionψ_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

2φ_NR(x, y)²,

whereφ_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 functionφ_NR 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)

whereτ is a fixed parameter from(0,4)andφ_τ:Rⁿ×Rⁿ→Rⁿis defined by φ_τ(x, y):=

(x−y)²+τ (x◦y)1/2

−(x+y). (10) Specifically, we prove thatψ_τ is a merit function associated withKwhich 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 off_τ solves the SOCCP under the condi- tion that∇F and−∇Gare column monotone (see Proposition4.1). Observe thatφ_τ reduces toφ_FB whenτ =2, whereas its limit as τ→0 becomes a multiple ofφ_NR.

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 thatψ_τ can be an alterna- tive forψ_FBif a suitableτ is selected.

Throughout this paper,Rⁿ denotes the space ofn-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 notationI denotes an identity matrix of suitable dimension, and int(Kⁿ)denotes the interior ofKⁿ. For any differentiable mappingF :Rⁿ→R^m,∇F (x)∈R^n×mdenotes the transposed Jaco- bian ofF atx. For a symmetric matrixA, we writeAO (respectively,AO) to meanAis positive semidefinite (respectively, positive definite). For nonnegative αandβ, we writeα=O(β)to meanα≤Cβ, withC >0 independent ofαandβ. 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 whereKhas the structure as (2).

In addition, we always assume thatτ 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 anyx =(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 ise:=(1,0, . . . ,0)^T∈Rⁿ. For anyx=(x1, x2)∈R×Rⁿ⁻¹, the determinant ofxis defined by det(x):=x₁²−x2². If det(x)=0, thenxis said to be invertible. Ifxis invertible, there exists a uniquey=(y₁, y₂)∈R×Rⁿ⁻¹satisfying x◦y=y◦x =e. We call this y the inverse ofx 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 thatLxy =x◦y andLx+y=Lx+Ly for anyx, y∈Rⁿ, but generallyL²_x=LxLx=L_x2 andL⁻_x¹=L_x−1. IfLxis invertible, then the inverse of L_xis given by

L⁻_x¹= 1 det(x)

⎡

⎣ x1 −x₂^T

−x2

det(x) x₁ I+ 1

x₁x2x₂^T

⎤

⎦. (14)

We next recall from [9] that eachx=(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 ,

whereλ1(x), λ2(x)andu⁽¹⁾x , u⁽²⁾x are the spectral values and the associated spectral vectors ofxgiven by

λ_i(x)=x1+(−1)ⁱx2, u⁽ⁱ⁾_x =1

2(1, (−1)ⁱx¯2) fori=1,2,

with x¯2= ^x_x²₂ if x2=0, and otherwise x¯2 being any vector in Rⁿ⁻¹ such that ¯x₂ =1. Ifx₂=0, the factorization is unique. The spectral factorization ofx 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 anyx∈Rⁿ. (b) Ifx∈Kⁿ, thenx^1/2=√

λ1(x)·u⁽¹⁾_x +√

λ2(x)·u⁽²⁾_x ∈Kⁿ.

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

3 Smoothness of the functionψ_τ

In this section we will show thatψτ defined by (9) is a smooth merit function. First, by Properties2.1(a) and (b),φ_τ andψ_τ are well-defined since for anyx, 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 ψ_τ is indeed a merit function associated withKⁿ.

Proposition 3.1 Letψ_τ andφ_τbe 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 ofψ_τ. We consider the second one.

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

“⇒”. Suppose thatφ_τ(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 thatx∈Kⁿ, y∈Kⁿand

x, y =0.

In what follows, we focus on the proof of the smoothness ofψ_τ. We first introduce some notation that will be used in the sequel. For anyx=(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∈Kⁿandz∈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).

Letλ₁(w)andλ₂(w)be the spectral values ofw. 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 w¯₂,

where w¯₂:= ^w_w²₂ if w₂=0 and otherwise w¯₂ is any vector in Rⁿ⁻¹ satisfying ¯w₂ =1.

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

Lemma 3.1 For anyx=(x1, x2), y=(y1, y2)∈R×Rⁿ⁻¹, ifw /∈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), thenw2=0, and furthermore,

x^T₂ w2

w2=x₁, x₁ w2

w2 =x₂, y₂^T w2

w2=y₁, y₁ w2

w2 =y₂. (21) Proof Sincew=(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)byλ1(w)=0. Applying the relations in (19) then gives the equalities in (20).

If, in addition,(x, y)=(0,0), then it is clear thatx₁x₂+y₁y₂+(τ−2)x1y₂ = x₁²+y₁²+(τ−2)x1y1=0. To prove the equalities in (21), it suffices to verify that x₂^Tw_w²

2=x₁andx₁^w_w²

2=x₂ by the symmetry ofx andy inw. The verifications

are straightforward by (20) andx1y2=y1x2.

By Lemma3.1, whenw /∈int(Kⁿ), the spectral values ofware 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 functionψ_τgiven 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⁻_z¹−I

φ_τ(x, y),

∇yψ_τ(x, y)= (24) L_y+τ−2

2 xL⁻_z¹−I

φ_τ(x, y);

if(x, y)=(0,0)and(x−y)²+τ (x◦y)∈int(Kⁿ), thenx₁²+y₁²+(τ−2)x1y₁=0 and

∇xψτ(x, y)=

x1+^τ−₂²y1

x₁²+y₁²+(τ−2)x1y1

−1

φτ(x, y),

(25)

∇yψ_τ(x, y)=

y1+^τ−₂²x1

x₁²+y₁²+(τ−2)x1y1

−1

φ_τ(x, y).

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

2

ψ_τ(u, v)−ψ_τ(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 haveψ_τ(u, v)−ψ_τ(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 consequentlyφ_τ(x, y)is also continuously differentiable at such(x, y)sinceφτ(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 aboutx. SinceLzOby Property2.1(c), it follows that∇xz(x, y)=L_x+τ−2

2 yL⁻_z¹. Consequently,

∇xφ_τ(x, y)= ∇xz(x, y)−I=L_x+τ−2

2 yL⁻_z¹−I.

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

For the symmetry ofxandyinψτ, 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=(y₁, y₂)∈R×Rⁿ⁻¹, it is easy to verify that

2ψτ(x, y)=x²+y²+(τ−2)(x◦y)1/2²+ x+y²

−2

x²+y²+(τ−2)(x◦y)1/2

, x+y

= x²+ y²+(τ−2)x, y + x+y²

−2

x²+y²+(τ−2)(x◦y)1/2

, x+y ,

where the second equality uses the fact that z²= z², e for any z∈Rⁿ. Since x²+ y²+(τ −2)x, y + x +y² is clearly differentiable in (x, y), it suffices to show that [x²+y² +(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 impliesw₂ = w2(x, y)=2x₁x₂+2y₁y₂+(τ−2)(x₁y₂+y₁x₂)=0 for all(x, y)∈Rⁿ×Rⁿsuf- ficiently near to(x, y). Letμ₁, μ₂be the spectral values ofx²+y²+(τ−2)(x◦y).

Then we can compute that 2

x²+y²+(τ−2)(x◦y)1/2

, x+y

=√ μ₂

x₁+y₁ +[2(x₁x₂ +y₁y₂)+(τ−2)(x₁y₂ +y₁x₂)]^T(x₂ +y₂) 2(x₁x₂ +y₁y₂)+(τ−2)(x₁y₂ +y₁x₂)

+√ μ1

x₁+y₁−[2(x₁x₂+y₁y₂)+(τ−2)(x₁y₂+y₁x₂)]^T(x₂+y₂) 2(x₁x₂+y₁y₂)+(τ−2)(x₁y₂+y₁x₂)

.

(26) Sinceλ2(w) >0 andw2(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 iso(x− x + y−y), i.e., it is differentiable at(x, y)with zero gradient. To see this, no- tice thatw2(x, y)=0, and henceμ1= x²+ y²+(τ−2)x, y − 2(x₁x₂+ y₁y₂)+(τ −2)(x₁y₂+y₁x₂), 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 ofμ₁at(x, y)yields

μ1 = O(x−x + y−y).

Also, sincew2(x, y)=0, by the product and quotient rules for differentiation, the function

x₁+y₁−[2(x₁x₂+y₁y₂)+(τ−2)(x₁y₂+y₁x₂)]^T(x₂+y₂)

2(x₁x₂ +y₁y₂)+(τ−2)(x₁y₂ +y₁x₂) (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) isO(x−x + y−y)in magnitude, which together withμ₁=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 ψ_τ is differentiable at (x, y). Moreover, we see that 2∇ψ_τ(x, y)is the sum of the gradient ofx²+ y²+(τ−2)x, y + x+y² and the gradient of the first term on the right-hand side of (26), evaluated at(x, y)= (x, y).

The gradient ofx²+ y²+(τ−2)x, y + x+y²with respect tox, 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 tox₁, 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−₂²y2)^T(x2+y2) x1x2+y1y2+(τ−2)x1y2

− w^T₂(x2+y2)·w₂^T(x2+^τ−₂²y2) x₁x₂+y₁y₂+(τ−2)x1y₂ · w₂²

= 2(x1+^τ−₂²y₁)(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 tox₂, 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−₂²y1)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−₂²y1)(x1+y1)

(2x1+(τ−2)y1)x2+^τ₂(x1+y1)y2 . Using the fact thatx1y2=y1x2and noting that φ_τ 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 ψ_τ is differentiable with a computable gradient. To establish the continuity of the gradient ofψ_τ or the smoothness ofψ_τ, we need the following two crucial technical lemmas whose proofs are provided in theAppendix.

Lemma 3.2 For anyx=(x1, x2),y=(y1, y2)∈R×Rⁿ⁻¹, ifw2=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 ≤λ_i(w)

fori=1,2. Furthermore, these relations also hold when interchangingxandy.

Lemma 3.3 For all(x, y)satisfying(x−y)²+τ (x◦y)∈int(Kⁿ), we have that Lx+^τ−22 yL⁻_z¹

F ≤C, L

y+^τ−22 xL⁻_z¹

F≤C, (28)

whereC >0 is a constant independent ofx, yandτ, and·F denotes the Frobenius norm.

Proposition 3.3 The functionψ_τ defined by (9) is smooth everywhere onRⁿ×Rⁿ. Proof By Proposition3.2and the symmetry ofx andy in∇ψ_τ, 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⁻_z¹ and x₁+^τ−₂²y₁

x₁²+y₁²+(τ−2)x1y₁

are bounded with bound independent ofx, yandτ, using the continuity ofφ_τ(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⁻_z¹(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−₂²b₁

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−₂²b₁

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⁻_z¹(x+y)→ a₁+^τ−₂²b₁

a²₁+b²₁+(τ−2)a1b₁

(a+b), (33)

which is also equivalent to proving the following three relations L_x+τ−2

2 yL⁻_z¹

x+τ−2 2 y

→ a₁+^τ−₂²b₁

a₁²+b²₁+(τ−2)a1b₁

a+τ−2 2 b

, (34)