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

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### 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.

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### 1 Introduction

We consider the conic complementarity problem of finding a vector ζ ∈ IRn such that F (ζ) ∈ K, G(ζ) ∈ K, hF (ζ), G(ζ)i = 0 (1) where h·, ·i is the Euclidean inner product, F : IRn → IRn and G : IRn → IRn are smooth mappings, and K is the Cartesian product of second-order cones (SOCs). In other words,

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

Kni :=n(x1, x2) ∈ IR × IRni−1 | kx2k ≤ x1

o, (3)

with k · k denoting the Euclidean norm and K1 denoting the set of nonnegative reals IR+. A special case of (2) is K = IRn+, the nonnegative orthant in IRn, which corresponds to N = n and n1 = · · · = nN = 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 = Kn. 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 ∈ IRn. 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 = IRn+. 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 : IRn → IR is a convex twice continuously differentiable function, A ∈ IRm×n has full row rank and b ∈ IRm. 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 : IRn× IRn→ IR+ such that ψ(x, y) = 0 ⇐⇒ x ∈ K, y ∈ K, hx, yi = 0. (6)

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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:

ψFB(x, y) = 1

2FB(x, y)k2, (7)

where φFB : IRn× IRn→ IRn is the well-known FB function [12, 13] given by

φFB(x, y) = (x2+ y2)1/2− x − y, (8) with x2 to mean x ◦ x and x + y to mean the usual componentwise addition of vectors.

More specifically, for any x = (x1, x2), y = (y1, y2) ∈ IR × IRn−1, their Jordan product x ◦ y associated with Kn is defined as

x ◦ y := (hx, yi, y1x2+ x1y2). (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 ∈ IRn. The function ψFB was studied in [6] and particularly shown to be a smooth merit function for the SOCCP. Another popular choice of ψ is

ψNR(x, y) := 1

2NR(x, y)k2, (10)

which is induced by the natural residual function φNR : IRn× IRn → IRn

φNR(x, y) := x − [x − y]+, (11)

where [ · ]+ means the projection in the Euclidean norm onto Kn. The function φNR was studied in [14, 16] which is involved in smoothing methods for the SOCCP. Compared with the FB merit function ψFB, the function ψNR has a drawback, i.e., its non-differentiability.

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

2τ(x, y)k2, (12)

where φτ : IRn× IRn→ IRn is a family of functions defined by

φτ(x, y) := h(x − y)2 + τ (x ◦ y)i1/2− (x + y) (13) with τ being a fixed parameter such that τ ∈ (0, 4). Notice that, for any x, y ∈ IRn,

(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, (14)

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and hence φτ and ψτ are well-defined. We will prove that ψτ 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∈IRminnfτ(ζ) := ψτ(F (ζ), G(ζ)). (15) Also, we will show that every stationary point of fτ solves the SOCCP if ∇F and −∇G are column monotone (see Proposition 4.2). Observe that φτ reduces to φFB when τ = 2, whereas its limit as τ → 0 becomes a multiple of φNR. 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 φτ is exactly a multiple of φNR whose squared norm is not even differentiable. This violates our expectation that τk2 is smooth and that is why we exclude τ = 0 from the interval (0, 4). This study is motivated by the work [18] 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 smooth minimization problem (15). Numerical results are also reported by solving some convex SOCPs, which indicate that the merit function ψτ can be an alterna- tive for the FB merit function if a suitable τ is selected.

Throughout this paper, IRn denotes the space of n-dimensional real column vectors, IRn1 × · · · × IRnm is identified with IRn1+···+nm, and int(Kn) denotes the interior of Kn. For any x, y in IRn, we write x ºKn y if x − y ∈ Kn; and write x ÂKn y if x − y ∈ int(Kn).

For any differentiable mapping F : IRn → IRm, ∇F (x) ∈ IRn×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 = (x1, x2) ∈ IR × IRn−1, we define its determinant and trace as follows:

det(x) := x21− kx2k2, tr(x) = 2x1.

A vector x = (x1, x2) ∈ IR × IRn−1 is said to be invertible if det(x) 6= 0. If x is invertible, then there exists a unique y = (y1, y2) ∈ IR × IRn−1 satisfying x ◦ y = y ◦ x = e. We call this y the inverse of x and denote it by x−1. For any x = (x1, x2) ∈ IR × IRn−1, let

Lx :=

"

x1 xT2 x2 x1I

#

(16) which can be regarded a linear mapping from IRnto IRn. It is easily verified that Lxy = x◦y and Lx+y = Lx+ Ly for any x, y ∈ IRn, but generally L2x = LxLx 6= Lx2 and L−1x 6= Lx−1.

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We next recall from [14] that each x = (x1, x2) ∈ IR × IRn−1 admits a spectral factor- ization, associated with Kn, of the form

x = λ1(x)u(1)x + λ2(x)u(2)x ,

where λi(x) and u(i)x are the spectral values and the associated spectral vectors of x:

λi(x) = x1+ (−1)ikx2k, u(i)x = 1 2

³1, (−1)ix¯2

´ for i = 1, 2,

with ¯x2 = kxx22k if x2 6= 0, and otherwise ¯x2 being any vector in IRn−1 such that k¯x2k = 1. If x2 6= 0, the factorization is unique. The spectral factorization of x and the matrix Lx have various interesting properties, and we list several ones that will be used later.

Property 2.1 For any x = (x1, x2) ∈ IR × IRn−1, the following results always hold:

(a) x2 = λ21(x)u(1)x + λ22(x)u(2)x ∈ Kn and x1/2 =qλ1(x) u(1)x +qλ2(x) u(2)x ∈ Kn if x ∈ Kn; (b) x ºKn 0 ⇐⇒ λ1(x) ≥ 0 ⇐⇒ Lx º O;

(c) x ÂKn 0 ⇐⇒ λ1(x) > 0 ⇐⇒ Lx Â O, and the inverse of Lx is given by

L−1x = 1 det(x)

x1 −xT2

−x2 det(x) x1 I + 1

x1x2xT2

. (17)

### 3 Smoothness of Merit Functions

In this section we show that ψτ defined by (12) is a smooth merit function for the SOCCP.

First, we show that ψτ is a merit function, which is direct by the following proposition.

Proposition 3.1 The φτ given by (13) is a complementarity function of the SOCCP, i.e., φτ(x, y) = 0 ⇐⇒ x ∈ Kn, y ∈ Kn, 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 φτ(x, y) yields that

φτ(x, y) = (x2+ y2)1/2− (x + y) = φFB(x, y).

Applying Proposition 2.1 of [14], we readily obtain φτ(x, y) = 0.

“⇒”. Suppose that φτ(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)i1/2 = (x2+ y2)1/2,

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i.e., φFB(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 = (x1, x2), y = (y1, y2) ∈ IR × IRn−1 and τ ∈ (0, 4), let

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

z = (z1, z2) = z(x, y) := h(x − y)2+ τ (x ◦ y)i1/2. (18) Then w ∈ Kn and z ∈ Kn always hold. Moreover, by the definition of Jordan product,

w1 = w1(x, y) = kxk2+ kyk2+ (τ − 2)xTy,

w2 = w2(x, y) = 2(x1x2 + y1y2) + (τ − 2)(x1y2+ y1x2). (19) Let λ1(w) and λ2(w) be the spectral values of w. From Property 2.1 (a),

z1 = z1(x, y) =

q

λ1(w) +qλ2(w)

2 , z2 = z2(x, y) =

q

λ2(w) −qλ1(w)

2 w¯2, (20)

where ¯w2 := kww22k if w2 6= 0 and otherwise ¯w2 is any vector in IRn−1 satisfying k ¯w2k = 1.

In what follows, we concentrate on the proof of the smoothness of ψτ. 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 Kn. In fact, it may be viewed as an extension of [6, Lemma 3.2].

Lemma 3.1 For any x = (x1, x2), y = (y1, y2) ∈ IR × IRn−1, let w be given as in (18). If (x − y)2+ τ (x ◦ y) /∈ int(Kn), then then there hold that

x21 = kx2k2, y12 = ky2k2, x1y1 = xT2y2, x1y2 = y1x2; (21) x21+ y12+ (τ − 2)x1y1 = kx1x2+ y1y2+ (τ − 2)x1y2k

= kx2k2+ ky2k2+ (τ − 2)xT2y2. (22) If, in addition, (x, y) 6= (0, 0), then w2 = w2(x, y) 6= 0, and furthermore,

xT2 w2

kw2k = x1, x1 w2

kw2k = x2, y2T w2

kw2k = y1, y1 w2

kw2k = y2. (23) Proof. Since (x − y)2+ τ (x ◦ y) /∈ int(Kn), using [6, Lemma 3.2] and (14) yields that

µ

x1 +τ − 2 2 y1

2

=

°°

°°x2+τ − 2 2 y2

°°

°°

2

, y21 = ky2k2,

µ

x1+ τ − 2 2 y1

y2 =

µ

x2+τ − 2 2 y2

y1,

µ

x1+ τ − 2 2 y1

y1 =

µ

x2+τ − 2 2 y2

T

y2;

µ

y1+τ − 2 2 x1

2

=

°°

°°y2+ τ − 2 2 x2

°°

°°

2

, x21 = kx2k2,

µ

y1+ τ − 2 2 x1

x2 =

µ

y2+τ − 2 2 x2

x1,

µ

y1+τ − 2 2 x1

x1 =

µ

y2+τ − 2 2 x2

T

x2.

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From the above equalities, we immediately obtain the results in (21). In addition, since w /∈ int(Kn) but w ∈ Kn, we have λ1(w) = 0, which implies that

kxk2+ kyk2+ (τ − 2)xTy = k2x1x2+ 2y1y2+ (τ − 2)(x1y2+ y1x2)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 kx1x2+ y1y2+ (τ − 2)x1y2k = x21+ y21+ (τ − 2)x1y1 6= 0. To prove the equalities in (23), we only need to verify that xT2kww22k = x1 and x1kww22k = x2 in view of the symmetry of x and y in w. The verifications are straightforward due to x1y2 = y1x2 and equation (22). 2

By Lemma 3.1, when w(x, y) = (x − y)2+ τ (x ◦ y) /∈ int(Kn), the spectral values of w can be further simplified. Clearly, λ1(w) = 0 and λ2(w) can be rewritten as

λ2(w) = 2³x21+ y12+ (τ − 2)x1y1´+ 2kx1x2+ y1y2+ (τ − 2)x1y2k

= 4³x21+ y12+ (τ − 2)x1y1

´. (24)

Therefore, if (x, y) 6= (0, 0) also holds, using equations (20), (22) and (24) yields that z1(x, y) =

q

x21+ y12+ (τ − 2)x1y1, z2(x, y) = x1x2+ y1y2+ (τ − 2)x1y2

q

x21+ y21 + (τ − 2)x1y1 .

Thus, if (x, y) 6= (0, 0) and (x − y)2+ τ (x ◦ y) /∈ int(Kn), the function φτ is rewritten as

φτ(x, y) = z(x, y) − (x + y) =

qx21+ y12+ (τ − 2)x1y1− (x1+ y1) x1x2+ y1y2+ (τ − 2)x1y2

q

x21+ y12+ (τ − 2)x1y1 − (x2+ y2)

. (25)

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

Proposition 3.2 The function ψτ given by (12) is differentiable at every (x, y) ∈ IRn×IRn. Moreover, ∇xψτ(0, 0) = ∇yψτ(0, 0) = 0; and if (x − y)2+ τ (x ◦ y) ∈ int(Kn), then

xψτ(x, y) = hLx+τ −2

2 yL−1z − Iiφτ(x, y),

yψτ(x, y) = hLy+τ −2

2 xL−1z − Iiφτ(x, y). (26) If (x, y) 6= (0, 0) and (x − y)2+ τ (x ◦ y) 6∈ int(Kn), then x21+ y12+ (τ − 2)x1y1 6= 0 and

xψτ(x, y) =

x1+τ −22 y1

q

x21+ y21+ (τ − 2)x1y1 − 1

φτ(x, y),

yψτ(x, y) =

y1 +τ −22 x1

qx21+ y21+ (τ − 2)x1y1 − 1

φτ(x, y). (27)

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Proof. Case (1): (x, y) = (0, 0). For any h, k ∈ IRn, let µ1 ≤ µ2 be the spectral values of (h − k)2+ τ (h ◦ k) and v(1), v(2) be the corresponding spectral vectors. Then,

ψτ(h, k) − ψτ(0, 0) = 1 2

°°

°[h2+ k2 + (τ − 2)(h ◦ k)]1/2− h − k°°°2

= 1 2

°°

°

µ1 v(1)+

µ2 v(2)− h − k°°°2

1 2

·q

2+ khk + kkk

¸2

.

In addition, by the definition of spectral value µ2, it is easy to verify that µ2 ≤ 2khk2+ 2kkk2+ 3|τ − 2|khkkkk ≤ 5(khk2+ kkk2).

Combining the last two equations then yields ψτ(h, k) − ψτ(0, 0) = O(khk2+ kkk2). This shows that ψτ is differentiable at (0, 0) with ∇xψτ(0, 0) = ∇yψτ(0, 0) = 0.

Case (2): (x−y)2+τ (x◦y) ∈ int(Kn). 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, φτ(x, y) = z(x, y) − (x + y) is also continuously differentiable at such (x, y). Notice that

z2(x, y) =

µ

x +τ − 2 2 y

2

+τ (4 − τ ) 4 y2.

Differentiating on both sides about x, it then follows that ∇xz(x, y)Lz = Lx+τ −2

2 y. Using Property 2.1 (c) and noting that ∇xφτ(x, y) = ∇xz(x, y) − I, we have

xφτ(x, y) = Lx+τ −2

2 yL−1z − I,

which, together with ∇xψτ(x, y) = ∇xφτ(x, y)φτ(x, y), yields the first formula in (26). For the symmetry of x and y in ψτ, the second formula in (26) also holds.

Case (3): (x, y) 6= (0, 0) and (x − y)2+ τ (x ◦ y) /∈ int(Kn). For any (x0, y0) ∈ IRn× IRn, τ(x0, y0) =

°°

°°

hx02+ y02+ (τ − 2)(x0◦ y0)i1/2

°°

°°

2+ kx0+ y0k2

−2¿hx02+ y02+ (τ − 2)(x0 ◦ y0)i1/2, x0+ y0

À

= kx0k2+ ky0k2+ (τ − 2)hx0, y0i + kx0 + y0k2

−2D[x02+ y02+ (τ − 2)(x0◦ y0)]1/2, x0+ y0E,

where the second equality uses the observation that kzk2 = hz2, ei for any z ∈ IRn. Since kx0k2 + ky0k2 + (τ − 2)hx0, y0i + kx0+ y0k2 is clearly differentiable in (x0, y0), it suffices to show that h[x02+ y02+ (τ − 2)(x0◦ y0)]1/2, x0 + y0i is differentiable at (x0, y0) = (x, y). By Lemma 3.1, w2(x, y) = 2(x1x2 + y1y2) + 2(τ − 2)x1y2 6= 0, which implies

w2(x0, y0) = 2x01x02+ 2y10y02+ (τ − 2)(x01y02+ y10x02) 6= 0

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for all (x0, y0) ∈ IRn× IRn sufficiently near to (x, y). Let µ1, µ2 be the spectral values of x02+ y02+ (τ − 2)(x0◦ y0). Then we can compute that

2Dhx02+ y02+ (τ − 2)(x0◦ y0)i1/2, x0+ y0E

=

µ2

x01 + y01+

h2(x01x02+ y10y20) + (τ − 2)(x01y20 + y10x02)iT(x02+ y20) k2(x01x02+ y10y20) + (τ − 2)(x01y20 + y10x02)k

+ µ1

x01+ y10

h2(x01x02+ y10y20) + (τ − 2)(x01y20 + y10x02)iT(x02+ y20) k2(x01x02+ y10y20) + (τ − 2)(x01y20 + y10x02)k

. (28)

Since λ2(w) > 0 and w2(x, y) 6= 0, the first term on the right-hand side of (28) is differen- tiable at (x0, y0) = (x, y). We claim that the second term is o(khk+kkk) with h := x0−x, k :=

y0−y, i.e., it is differentiable at (x, y) with zero gradient. To see this, note that w2(x, y) 6= 0, and hence µ1 = kx0k2+ky0k2+(τ −2)hx0, y0i−k2(x01x02+y01y20)+(τ −2)(x01y02+y01x02)k, viewed as a function of (x0, y0), is differentiable at (x0, y0) = (x, y). Moreover, µ1 = λ1(w) = 0 when (x0, y0) = (x, y). Thus, the first-order Taylor’s expansion of µ1 at (x, y) yields

µ1 = O(kx0 − xk + ky0− yk) = O(khk + kkk).

Also, since w2(x, y) 6= 0, by the product and quotient rules for differentiation, the function

x01+ y10

h2(x01x02 + y01y20) + (τ − 2)(x01y20 + y01x02)iT(x02+ y02)

k2(x01x02 + y01y20) + (τ − 2)(x01y20 + y01x02)k (29) is also differentiable at (x0, y0) = (x, y), and it has value 0 at (x0, y0) = (x, y) because

x1+ y1

hx1x2+ y1y2+ (τ − 2)x1y2iT(x2+ y2)

kx1x2+ y1y2+ (τ − 2)x1y2k = x1− xT2 w2

kw2k + y1− yT2 w2 kw2k = 0 by Lemma 3.1. Thus, the function (29) is O(khk + kkk) in magnitude, which together with µ1 = 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 ψτ is differentiable at (x, y). Moreover, we see that 2∇ψτ(x, y) is the sum of the gradient of kx0k2+ ky0k2+ (τ − 2)hx0, y0i + kx0+ y0k2 and the gradient of the first term on the right-hand side of (28), evaluated at (x0, y0) = (x, y).

The gradient of kx0k2+ ky0k2+ (τ − 2)hx0, y0i + kx0+ y0k2 with respect to x0, evaluated at (x0, y0) = (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 x01, evaluated at (x0, y0) = (x, y), works out to be

q 1 λ2(w)

x1+τ − 2 2 y1

+

µ

x2+τ − 2 2 y2

T w2 kw2k

# Ã

x1 + y1+ (x2+ y2)T w2 kw2k

!

(10)

+

q

λ2(w)

"

1 + (x2+ τ −22 y2)T(x2+ y2)

kx1x2+ y1y2+ (τ − 2)x1y2k w2T(x2+ y2) · wT2(x2+ τ −22 y2) kx1x2+ y1y2+ (τ − 2)x1y2k · kw2k2

#

= 2(x1+ τ −22 y1)(x1+ y1)

q

x21+ y12+ (τ − 2)x1y1 + 2qx21+ y12+ (τ − 2)x1y1,

where the equality follows from Lemma 3.1. Similarly, the gradient of the first term on the right of (28) with respect to x02, evaluated at (x0, y0) = (x, y), works out to be

q 1 λ2(w)

x2+τ − 2 2 y2

+

µ

x1+τ − 2 2 y1

w2

kw2k

# Ã

x1+ y1+ (x2 + y2)T w2

kw2k

!

+qλ2(w)

"

(2x1+ (τ − 2)y1)x2 +τ2(x1+ y1)y2

kx1x2+ y1y2+ (τ − 2)x1y2k wT2(x2+ y2) · (x1+τ −22 y1)w2 kx1x2+ y1y2+ (τ − 2)x1y2k · kw2k2

#

= 2(2x1+ (τ − 2)y1)x2+τ2(x1+ y1)y2

qx21+ y12+ (τ − 2)x1y1 .

Since λ2(w) = 4(x21+ y12+ (τ − 2)x1y1, combining the above gradient expressions yields 2∇xψτ(x, y) = 2x + (τ − 2)y + 2(x + y) −

"

2qx21+ y12+ (τ − 2)x1y1 0

#

2

q

x21+ y12+ (τ − 2)x1y1

"

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

#

.

Using the fact x1y2 = y1x2 and noting that φτ can be simplified as the one given by (25) under this case, we readily rewrite the above expression for ∇xψτ(x, y) in the form of (27).

By symmetry, ∇yψτ(x, y) also holds as form of (27). 2

Proposition 3.2 gives a formula for computing ∇ψτ. It is a natural question whether ψτ 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.

Lemma 3.2 For any x = (x1, x2), y = (y1, y2), let w be given as in (18). If w2 6= 0, then

x1 +τ − 2 2 y1

+ (−1)i

µ

x2+τ − 2 2 y2

T w2

kw2k

#2

°°

°°

° µ

x2+ τ − 2 2 y2

+ (−1)i

µ

x1 +τ − 2 2 y1

w2 kw2k

°°

°°

°

2

≤ λi(w)

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

(11)

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(Kn), there exists a scalar constant C > 0 such that

°°

°Lx+τ −2

2 yL−1z °°°

F ≤ C, °°°Ly+τ −2

2 xL−1z °°°

F ≤ C, (30)

where kAkF denotes the Frobenius norm of the n × n matrix A.

Proposition 3.3 The function ψτ defined by (12) is smooth everywhere on IRn× IRn. Proof. By Proposition 3.2 and the symmetry of x and y in ∇ψτ, it suffices to show that

xψτ is continuous at every (a, b) ∈ IRn×IRn. If (a−b)2+τ (a◦b) ∈ int(Kn), 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 ∇xψτ(x, y) → 0 as (x, y) → (0, 0). If (x − y)2+ τ (x ◦ y) ∈ int(Kn), then ∇xψτ(x, y) is given by (26), whereas if (x, y) 6= (0, 0) and (x − y)2+ τ (x ◦ y) /∈ int(Kn), then ∇xψτ(x, y) is given by (27). Notice that Lx+τ −2

2 yL−1z and x1+τ −22 y1

x21+y12+(τ −2)x1y1 are uniformly bounded with bound independent of (x, y). Using the continuity of φτ(x, y) then leads to the desired result.

Case (2): (a, b) 6= (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: (i) (x, y) 6= (0, 0) and (x − y)2+ τ (x ◦ y) /∈ int(Kn) and (ii) (x − y)2+ τ (x ◦ y) ∈ int(Kn). In subcase (i), ∇xψτ(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 ∇xψτ(x, y) → ∇xψτ(a, b) in subcase (ii). From (26),

xψτ(x, y) = Lx+τ −2

2 yL−1z ³Lze − (x + y)´− φτ(x, y)

=

µ

x +τ − 2 2 y

− Lx+τ −2

2 yL−1z (x + y) − φτ(x, y). (31) On the other hand, since (a, b) 6= (0, 0) and (a − b)2+ τ (a ◦ b) /∈ int(Kn), we have

kak2+ kbk2+ (τ − 2)aTb = k2(a1a2+ b1b2) + (τ − 2)(a1b2+ b1a2)k 6= 0, (32) and moreover from (22) in Lemma 3.1, it follows that

kak2+ kbk2+ (τ − 2)aTb = 2(a21+ b21+ (τ − 2)a1b1)

= 2(ka2k2+ kb2k2+ (τ − 2)aT2b2)

= 2k(a1a2+ b1b2) + (τ − 2)a1b2k. (33) Using the equalities in (33), it is not hard to verify that

a1+ τ −22 b1

qa21+ b21+ (τ − 2)a1b1

³(a − b)2 + τ (a ◦ b)´1/2 = a + τ − 2 2 b.

Although we have obtained the global and superlinear convergence properties of Algorithm 3.1 under mild conditions, this does not mean that Algorithm 3.1 is practi- cally efficient,

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