Lipschitz continuity of the gradient of a one-parametric class of SOC merit functions

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Vol. 59, No. 5, July 2010, 661–676

Lipschitz continuity of the gradient of a one-parametric class of SOC merit functions

Jein-Shan Chen^a* and Shaohua Pan^b

aDepartment of Mathematics, National Taiwan Normal University, Taipei, Taiwan;^bSchool of Mathematical Sciences, South China

University of Technology, Guangzhou, China

(Received 27 March 2007; final version received 11 March 2008)

In this article, we show that a one-parametric class of SOC merit functions has a Lipschitz continuous gradient; and moreover, the Lipschitz constant is related to the parameter in this class of SOC merit functions. This fact will lay a building block when the merit function approach as well as the Newton-type method are employed for solving the second-order cone complementarity problem with this class of merit functions.

Keywords: second-order cone; merit function; spectral factorization; Lipschitz continuity

AMS Subject Classifications: 26B05; 26B35; 90C33; 65K05

1. Introduction

A well-known approach to solving the non-linear complementarity problem (NCP) is to reformulate it as the global minimization via a certain merit function over IRⁿ. For the approach to be effective, the choice of the merit function is crucial. A popular choice is the squared norm of the Fischer–Burmeister (FB) function FB: IRⁿIRⁿ!IRþdefined by

FBða, bÞ :¼1 2

Xⁿ

i¼1

FBðai, biÞ

½ ² ð1Þ

for all a ¼ (a1, . . . , an)^T2IRⁿ and b ¼ (b1, . . . , bn)^T2IRⁿ, where FB: IR IR ! IR is the Fischer–Burmeister NCP function given as

_FBða_i, b_iÞ ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a²_i þb²_i q

a_ib_i: ð2Þ

It has been shown that FB enjoys many desirable properties [8,9], for example, smoothness (continuous differentiability). This merit function and its analysis were subsequently extended by Tseng [18] to the semidefinite complementarity problem (SDCP)

*Corresponding author. Email: jschen@math.ntnu.edu.tw

ISSN 0233–1934 print/ISSN 1029–4945 online

2010 Taylor & Francis DOI: 10.1080/02331930802180319 http://www.informaworld.com

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although only differentiability, not continuous differentiability, was established. In fact, the FB function for the SDCP is the matrix-valued function FB: Sⁿ Sⁿ! Sⁿdefined by

_FBðX, YÞ :¼ ðX²þY²Þ¹⁼² ðX þ YÞ,

while the squared norm of the FB function for the SDCP is the function FB: Sⁿ Sⁿ! IRþgiven by

_FBðX, YÞ :¼1

2kðX, YÞk²,

where Sⁿ denotes the set of real n n symmetric matrices. The function FB has been proved to be strongly semismooth [17]. More recently, the squared norm of the matrix- valued FB function FB was reported in [16] to be smooth and its gradient is Lipschitz continuous.

The second-order cone (SOC), also called the Lorentz cone, in IRⁿis defined as Kⁿ:¼ ðx₁, x₂Þ 2 R Rⁿ¹j kx2k x1

, ð3Þ

where k k denotes the Euclidean norm. By definition, K¹ is the set of non-negative reals IRþ. The second-order cone complementarity problem (SOCCP) is to find x, y 2 IRⁿ satisfying

x ¼ FðÞ, y ¼ GðÞ, hx, yi ¼ 0, x 2 Kⁿ, y 2 Kⁿ, ð4Þ where h, i is the Euclidean inner product and F, G : IRⁿ!IRⁿare continuous (possibly non-linear) mappings. The merit function approach based on reformulating the NCP as an equivalent unconstrained minimization can be extended to the SOCCP case [6].

This approach aims to find a smooth function : IRⁿIRⁿ!IRþsuch that

ðx, yÞ ¼ 0 () x 2 Kⁿ, y 2 Kⁿ, hx, yi ¼ 0: ð5Þ We call such a SOC merit function. Then the SOCCP can be expressed as an unconstrained smooth (global) minimization problem:

2IRminⁿ f ðÞ:¼ ðFðÞ, GðÞÞ: ð6Þ Analogously, the squared norm of FB function can be considered in the SOCCP setting.

We define FB: IRⁿIRⁿ!IRþassociated with the second-order cone Kⁿas

FBðx, yÞ :¼ 1

2kFBðx, yÞk², ð7Þ

where FB: IRⁿIRⁿ!IRⁿis the FB function defined by

_FBðx, yÞ :¼ ðx²þy²Þ¹⁼²x y: ð8Þ More specifically, for any x ¼ (x1, x2), y ¼ ( y1, y2) 2 IR IRⁿ¹, we define their Jordan productassociated with Kⁿas

x y:¼ ðhx, yi, y1x2þx1y2Þ: ð9Þ The Jordan product , unlike scalar or matrix multiplication, is not associative, which is a main source of complication in the analysis of SOCCP. The identity element under this

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product is e :¼ (1, 0, . . . , 0)^T2IRⁿ. We write x²to mean x x and write x þ y to mean the usual componentwise addition of vectors. It is known that x²2 Kⁿ for all x 2 IRⁿ. Moreover, if x 2 Kⁿ, then there exists a unique vector in Kⁿ, denoted by x^1/2, such that (x^1/2)²¼x^1/2x^1/2¼x. Thus, FBdefined as in (8) is well-defined for all (x, y) 2 IRⁿIRⁿ and maps IRⁿIRⁿto IRⁿ. It was shown in [10] that FBsatisfies the relation (5). Hence,

FBas defined in (7) is a merit function for the SOCCP. In the recent manuscript [5], this SOC merit function was shown to be an LC¹function (a smooth function with its gradient being locally Lipschitz continuous).

Another popular SOC merit function is the natural residual merit function

NRðx, yÞ :¼ k_NRðx, yÞk², ð10Þ

which is induced by the natural residual function NR: IRⁿIRⁿ!IRⁿ

NRðx, yÞ :¼ x ½x y_þ, ð11Þ

where [ ]þmeans the projection onto Kⁿ. The natural residual function NRwas studied in [10,11] which is involved in smoothing methods for the SOCCP. A drawback of NRis its non-differentiability compared to FB. Some other classes of SOC merit functions for the SOCCP are also recently studied in [1,2].

In this article, we consider the following one-parametric class of SOC merit functions which was originally proposed in [12] for the NCP case:

ðx, yÞ :¼1

2kðx, yÞk² ð12Þ

where : IRⁿIRⁿ!IRⁿis a family of functions associated with the SOC, defined by

ðx, yÞ :¼ ðx yÞ ²þðx yÞ1=2

ðx þ yÞ, ð13Þ

and is a fixed parameter such that 2 (0, 4). It can be verified that for any x, y 2 IRⁿ ðx yÞ²þðx yÞ ¼ x þ 2

2 y

2

þð4 Þ 4 y²

¼ y þ 2 2 x

2

þð4 Þ 4 x²

_Kn 0, ð14Þ

where the inequality holds because 2 (0, 4). Therefore, in (13) is well-defined. Notice that reduces to the FB function FBwhen ¼ 2, whereas it becomes a multiple of the natural residual function NR when ! 0. Thus, this class of SOC complementarity functions covers the current two most important SOC complementarity functions so that a closer look and study for this new class of functions is worthwhile.

In fact, as mentioned in [5], when solving the equivalent unconstrained minimization via SOC merit functions, it is very important to show that the gradient of the employed SOC merit function is sufficiently smooth so as to warrant the convergence of appropriate computational methods. Here, we are particularly concerned with the conjugate gradient method. The method generally requires the Lipschitz continuity of the gradient ( f 2 LC¹by our notation). The main purpose of this article is to show that the function as defined in (12) has a globally Lipschitz continuous gradient. Thus, this article can be regarded as

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a follow-up of [5] since it extends the LC¹property of FBto the class of SOC merit functions

. Nonetheless, the technique used here is a bit different and the analysis is more tedious and subtle. In particular, from the extension work, we see that the Lipschitz continuity of the gradient of becomes worse when ! 0. This fact will provide an instructional help for the design of algorithms with this class of SOC merit functions.

Throughout this article, IRⁿdenotes the space of n-dimensional real column vectors and the supscript ‘T’ represents the transpose. For any differentiable function f : IRⁿ!IR, rf(x) denotes the gradient of f at x. For any differentiable mapping F ¼(F1, . . . , Fm)^T: IRⁿ!IR^m, rF(x) ¼ [rF1(x) rFm(x)] is a n m matrix denoting the transposed Jacobian of F at x. For non-negative scalars and , we write ¼ O() to mean C, with C independent of and .

2. Preliminaries

It is known that Kⁿis a closed convex self-dual cone with non-empty interior given as Kⁿ:¼ ðx1, x2Þ 2 R Rⁿ¹j kx2k x1

:

For any x ¼ (x1, x2) 2 IR IRⁿ¹, we define the determinant and the trace of x as follows:

detðxÞ :¼ x²₁ kx2k², trðxÞ ¼ 2x1:

In general, det(x y) 6¼ det(x)det( y) unless x and y are collinear, i.e. x ¼ y for some 2 IR.

A vector x ¼ (x1, x2) 2 IR IRⁿ¹is said to be invertible if det(x) 6¼ 0. If x is invertible, then there exists a unique y ¼ ( y1, y2) 2 IR IRⁿ¹satisfying x y ¼ y x ¼ e. We call this y the inverse of x and denote it by x¹. In fact, we have

x¹¼ 1

x²₁ kx₂k²ðx1, x2Þ ¼ 1 detðxÞ

trðxÞe x

:

It is not difficult to see that x 2 int(Kⁿ) if and only if x¹2int(Kn). For any x ¼(x1, x2) 2 IR IRⁿ¹, we define the matrix Lxby

L_x:¼ x1 x^T₂ x₂ x₁I

:

It is easily verified that Lxþy¼LxþLyand x y ¼ Lxyfor any x, y 2 IRⁿ, and Lxis positive definite (and hence invertible) if and only if x 2 int(Kⁿ). However, L¹_x y 6¼ x¹y, for some x 2int(Kⁿ) and y 2 IRⁿ, i.e. L¹_x 6¼L_x¹.

We next recall from [10] that each x ¼ (x1, x2) 2 IR IRⁿ¹ admits a spectral factorization, associated with Kⁿ, of the form

x ¼ 1ðxÞu^ð1Þ_x þ2ðxÞu^ð2Þ_x , ð15Þ where 1(x), 2(x) and u^ð1Þ_x , u^ð2Þ_x are the spectral values and the associated spectral vectors of x, with respect to Kⁿ, given by

_iðxÞ ¼ x₁þ ð1Þⁱkx₂k, ð16Þ

u^ðiÞ_x ¼ 1 2

1, ð1Þⁱ x2

kx2k

, if x26¼0, 1

2 1, ð1Þⁱw2

, if x2¼0, 8>

><

>>

:

ð17Þ

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for i ¼ 1, 2, with w2 being any vector in IRⁿ¹ satisfying k w2k ¼1. The spectral factorization of x, x²and x^1/2as well as the matrix Lxhave various interesting properties (cf. [10]). We list some properties that we will use later.

Property 2.1 For any x ¼(x1, x2) 2 IR IRⁿ¹with spectral values 1(x), 2(x) and spectral vectors u^ð1Þ_x , u^ð2Þ_x , the following results hold.

(a) x²¼²₁ðxÞu^ð1Þ_x þ²₂ðxÞu^ð2Þ_x 2 Kⁿ.

(b) If x 2 Kⁿ, then 0 1(x) 2(x) and x¹⁼² ¼ ffiffiffiffiffiffiffiffiffiffiffi

1ðxÞ

p u^ð1Þ_x þ ffiffiffiffiffiffiffiffiffiffiffi

2ðxÞ p u^ð2Þ_x . (c) If x 2 int(Kⁿ, then 0 5 ₁(x) ₂(x), and L_xis invertible with

L¹_x ¼ 1 detðxÞ

x1 x^T₂

x2

detðxÞ x1

I þ 1 x1

x2x^T₂ 2

4

3 5:

(d) The determinant, the trace and the Euclidean norm of x can be denoted by 1(x),

2(x):

detðxÞ ¼ ₁ðxÞ₂ðxÞ, trðxÞ ¼ ₁ðxÞ þ ₂ðxÞ, kxk²¼²₁ðxÞ þ ²₂ðxÞ

2 :

Before giving out several technical lemmas that will be applied in the next section, we introduce some notations that will be frequently used in the subsequent analysis.

Unless otherwise stated, in this article, we always write

w ¼ wðx, yÞ :¼ ðx yÞ²þðx yÞ and z ¼ zðx, yÞ :¼ ½ðx yÞ²þðx yÞ¹⁼²: ð18Þ Since (x y)²þ(x y) ¼ x²þy²þ( 2) (x y) 2 Kⁿfor any x, y 2 IRⁿ, we have

w:¼ w1

w2

¼ kxk²þ kyk²þ ð 2Þx^Ty 2ðx1x2þy1y2Þ þ ð 2Þðx1y2þy1x2Þ

!

2 Kⁿ: ð19Þ

From this, it follows that the spectral values of w are given by

1ðwÞ:¼ kxk²þ kyk²þ ð 2Þx^Ty k2ðx1x2þy1y2Þ þ ð 2Þðx1y2þy1x2Þk,

₂ðwÞ:¼ kxk²þ kyk²þ ð 2Þx^Ty þ k2ðx₁x₂þy₁y₂Þ þ ð 2Þðx₁y₂þy₁x₂Þk: ð20Þ By Property 2.1(b), the vector z has the spectral values ffiffiffiffiffiffiffiffiffiffiffi

1ðwÞ

p , ffiffiffiffiffiffiffiffiffiffiffi

2ðwÞ

p and

z:¼ ðz1, z2Þ ¼

ffiffiffiffiffiffiffiffiffiffiffi

₁ðwÞ

p þ ffiffiffiffiffiffiffiffiffiffiffi

₂ðwÞ p

2 ,

₂ðwÞ

p ffiffiffiffiffiffiffiffiffiffiffi

₁ðwÞ p

2 w2

!

, ð21Þ

where w₂ :¼ ðw₂=kw₂kÞ if w26¼0 and otherwise w₂ is any vector in IRⁿ¹ satisfying kw2k ¼1.

The following four technical lemmas are crucial in proving our main results.

Lemma 2.1 measures how close w comes to the boundary of Kⁿ, and Lemma 2.2 describes the behaviour of (x, y) when w lies on the boundary of Kⁿ. Lemma 2.3 talks about the differential rule for the Jordan product function. Lemma 2.4 gives the gradient of the function z(x, y).

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LEMMA 2.1 [4, Lemma 3.4] For any x ¼(x1, x2), y ¼ ( y1, y2) 2 IR IRⁿ¹and 2(0, 4), if w2¼2(x1x2þy1y2) þ ( 2)(x1y2þy1x2) 6¼ 0, then we have

x₁þ 2 2 y₁

þ ð1Þⁱ

x₂þ 2

2 y₂T w2

kw₂k

2

x2þ 2 2 y2

þ ð1Þⁱ

x1þ 2 2 y1

w₂ kw2k

2

kxk²þ kyk²þ ð 2Þhx, yi þ ð1Þⁱkw2k

_iðwÞ

for i ¼1, 2, and furthermore these relations also hold when interchanging x and y.

LEMMA 2.2 [5, Lemma 3.2] For any x ¼(x1, x2), y ¼ ( y1, y2) 2 IR IRⁿ¹and 2(0, 4), if w ¼(x y)²þ(x y) =2int(Kⁿ), then there always holds that

x²₁¼ kx2k², y²₁¼ ky2k², x1y1¼x^T₂y2, x1y2 ¼y1x2; x²₁þy²₁þ ð 2Þx₁y₁¼ kx₁x₂þy₁y₂þ ð 2Þx₁y₂k

¼ kx2k²þ ky2k²þ ð 2Þx^T₂y2:

If, in addition, (x, y) 6¼ (0, 0), then w2¼2(x1x2þy1y2þ( 2)x1y2) 6¼ 0, and furthermore, x^T₂ w2

kw2k¼x1, x1

w2

kw2k¼x2, y^T₂ w2

kw2k¼y1, y1

w2

kw2k¼y2:

LEMMA 2.3 [6, Lemma 3.1] Let !: IRⁿIRⁿ!IR^mbe given by !(x, y) :¼ u(x, y) v(x, y), where u, v : IRⁿIRⁿ!IR^mare differentiable mappings. Then, ! is differentiable and

r_x!ðx, yÞ ¼ r_xuðx, yÞLvðx,yÞþ r_xvðx, yÞLuðx,yÞ, r_y!ðx, yÞ ¼ r_yuðx, yÞL_vðx,yÞþ r_yvðx, yÞL_uðx,yÞ: In particular, when !(x, y) ¼ x y, there holds

r_x!ðx, yÞ ¼ L_y, r_y!ðx, yÞ ¼ L_x; and when !(x, y) ¼ x²y², there holds

r_x!ðx, yÞ ¼ 2LxL_y², r_y!ðx, yÞ ¼ 2LyL_x²:

LEMMA 2.4 For any x, y 2 IRⁿ and 2(0, 4), let z(x, y) be defined as in (18). Then the function z(x, y) is continuously differentiable at a point (x, y) satisfying (x y)²þ(x y) 2 int(Kⁿ). Moreover, we have that

r_xzðx, yÞ ¼ Lxþ 2 2 Ly

L¹_zðx,yÞ, r_yzðx, yÞ ¼ L_yþ 2

2 L_x

L¹_zðx,yÞ:

Proof The differentiability of z(x, y) is an immediate consequence of [13], see also [3, Prop. 4]. Since z²(x, y) ¼ (x y)²þ(x y), applying Lemma 2.3 yields

2rxzðx, yÞLzðx,yÞ¼2LxyþLy ¼2Lxþ ð 2ÞLy:

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Hence, rxzðx, yÞ ¼ ðLxþ ð 2=2ÞLyÞL¹_zðx,yÞ. In view of the symmetry of x and y in the z(x, y), there also holds that r_yzðx, yÞ ¼ ðL_yþ ð 2=2ÞL_xÞL¹_zðx,yÞ. œ

3. Main results

In this section, we present the proof showing that the gradient function of is Lipschitz continuous. By the notation in Section 2, the function can be rewritten as

ðx, yÞ ¼1

2½ðx yÞ²þðx yÞ¹⁼² ðx þ yÞ²

¼1

2zðx, yÞ²ðx þ yÞ^T½ðx yÞ²þðx yÞ¹⁼²þ1

2kx þ yk²

¼1 2

2ðwÞ þ 1ðwÞ

2 þ kx þ yk²

ðx þ yÞ^T½ðx yÞ²þðx yÞ¹⁼²

¼1

22kxk²þ2kyk²þðx yÞ

ðx þ yÞ^T½ðx yÞ²þðx yÞ¹⁼²

¼ kxk²þ kyk²þ

2ðx yÞ ðx þ yÞ^T½ðx yÞ²þðx yÞ¹⁼², ð22Þ where the third equality is due to Property 2.1(d). Clearly, the gradient of the function kxk²þ kyk²þ ð=2Þðx yÞ is globally Lipschitz continuous. Therefore, to show that the gradient of is globally Lipschitz continuous, we only need to show that following function

Fðx, yÞ :¼ ðx þ yÞ^T½ðx yÞ²þðx yÞ¹⁼² ð23Þ has a Lipschitz continuous gradient. The following lemma states the gradient of F(x, y).

LEMMA 3.1 For any x, y 2 IRⁿand 2(0, 4), let F : IRⁿIRⁿ!IR be defined as in (23).

Then, the function F(x, y) is continuously differentiable everywhere. Moreover, r_xF(0, 0) ¼ r_yF(0, 0) ¼ 0. If (x, y) 6¼ (0, 0) and (x y)²þ(x y) 2 int(Kⁿ), then

r_xFðx, yÞ ¼ zðx, yÞ þ Lxþ 2 2 Ly

L¹_zðx,yÞðx þ yÞ r_yFðx, yÞ ¼ zðx, yÞ þ L_yþ 2

2 L_x

L¹_zðx,yÞðx þ yÞ: ð24Þ

If(x, y) 6¼ (0, 0) and (x y)²þ(x y) =2int(Kⁿ), then r_xFðx, yÞ ¼ zðx, yÞ þ x1þ²₂ y1

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x²₁þy²₁þ ð 2Þx₁y₁

q ðx þ yÞ

r_yFðx, yÞ ¼ zðx, yÞ þ y1þ²₂ x1

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x²₁þy²₁þ ð 2Þx1y1

q ðx þ yÞ: ð25Þ

Proof Case 1 x ¼ y ¼0. For any h, k 2 IRⁿ, let 12be the spectral values and v⁽¹⁾and v⁽²⁾be the corresponding vectors of (h k)²þ(h k). Then, by Property 2.1(b),

k½ðh kÞ²þðh kÞ¹⁼²k ¼ k ffiffiffiffiffiffi

1

p v^ð1Þþ ffiffiffiffiffiffi

2

p v^ð2Þk ð ffiffiffiffiffiffi

1

p þ ffiffiffiffiffiffi

2

p Þ= ffiffiffi 2 p

ffiffiffiffiffiffiffiffi 22

p :

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Therefore,

Fðh, kÞ Fð0, 0Þ ¼ ðh þ kÞ^T½ðh kÞ²þðh kÞ¹⁼²

k½ðh kÞ²þðh kÞ¹⁼²kkh þ kk

ffiffiffiffiffiffiffiffi 22

p ðkhk þ kkkÞ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðkhk²þ kkk²þ ð 2Þh^TkÞ q

ðkhk þ kkkÞ

¼Oðkhk²þ kkk²Þ:

This shows that F(x, y) is differentiable at (0, 0) with rxF(0, 0) ¼ ryF(0, 0) ¼ 0.

Case 2 (x, y) 6¼ (0, 0) and (x y)²þ(x y) 2 int(Kⁿ). Using Lemma 2.4, we readily obtain the formula in (24). Clearly, in this case, rxF(x, y) and ryF(x, y) are continuous, and consequently, F(x, y) is continuously differentiable at such points.

Case 3 (x, y) 6¼ (0, 0) and (x y)²þ(x y) =2int(Kⁿ). Since 1(w) ¼ 0, now it follows from (18) and (21) that

zðx, yÞ ¼ ½ðx yÞ²þðx yÞ¹⁼²¼1 2

₂ðwÞ

₂ðwÞ p w₂

T

, Moreover, by Lemma 2.2, we can compute that

w2¼2ðx₁x2þy1y2þ ð 2Þx₁y2Þ, 2ðwÞ ¼4ðx²₁þy²₁þ ð 2Þx₁y1Þ ¼2kw₂k:

Therefore, under this case, we have that

zðx, yÞ ¼ ½ðx yÞ²þðx yÞ¹⁼² ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x²₁þy²₁þ ð 2Þx₁y₁ q

x1x2þy1y2þ ð 2Þx1y2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x²₁þy²₁þ ð 2Þx1y1

q 2 66 64

3 77 75:

By this, it is easy to verify that the formula (25) holds.

Note that z(x, y) is a continuous function. This together with the proof of Case (i) and Case (iii) of [4, Proposition 3.3] means that the gradient functions rxF(x, y) and ryF(x, y) are continuous at every (x, y) 2 IRⁿIRⁿ. Hence, F(x, y) is continuously differentiable

everywhere. Thus, we complete the proof. œ

For the symmetry of x and y in rxF(x, y) and ryF(x, y), in the rest of this section, we concentrate on the proof of globally Lipschitz continuity of rxF(x, y). We first define a smooth approximation of rxF(x, y). For any 4 0, we let

^

w ¼ ^wðx, y, Þ :¼ ðx yÞ²þðx yÞ þ e,

^z ¼ ^zðx, y, Þ :¼ ½ðx yÞ²þðx yÞ þ e¹⁼², ð26Þ where e is the identity element under the Jordan product. It is not hard to see that

^

w₁¼w₁þ , w^₂¼w₂, ₁ðwÞ ¼ ^ ₁ðwÞ þ , ₂ðwÞ ¼ ^ ₂ðwÞ þ , ð27Þ and furthermore,

^z ¼ ð ^z₁, ^z2Þ ¼1 2

2ðwÞ^

1ðwÞ^

2ðwÞ^

1ðwÞ^

p

w2

ð28Þ

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where w2 :¼ ðw2=kw2kÞ if w26¼0 and otherwise w2 is any vector in IRⁿ¹ satisfying kw₂k ¼1. We define the mapping G(, , ) : IRⁿIRⁿ!IRⁿby

Gðx, y, Þ ¼ ^zðx, y, Þ þ L_xþ 2 2 L_y

L¹_^zðx,y, Þðx þ yÞ: ð29Þ

By Lemmas 3.2 and 3.4 below, G(x, y, ) is actually a smooth approximation of rxF(x, y).

Based on the relation, in the sequel, we will prove the Lipschitz continuity of rxF(x, y) through arguing that G(x, y, ) is globally Lipschitz continuous.

LEMMA 3.2 For any x, y 2 IRⁿand 4 0, let G(x, y, ) be defined as in (29). Then,

!0lim^þGðx, y, Þ ¼ r_xFðx, yÞ:

Proof If (x, y) ¼ (0, 0), then G(x, y, ) ¼ ( e)^1/2for any 4 0. Therefore,

!0lim^þGð0, 0, Þ ¼ rxFð0, 0Þ ¼ 0:

If (x, y) 6¼ (0, 0) and (x y)²þ(x y) 2 int(Kⁿ), then by (27), (28) and Property 2.1(c), it is easy to verify that lim !0^þL¹_^zðx,y, Þ¼L¹_zðx,yÞ. This together with lim !0^þL_^zðx,y, Þ¼Lzðx,yÞ

implies that the conclusion holds.

Next, we consider (x, y) 6¼ (0, 0) and (x y)²þ(x y) =2int(Kⁿ). For convenience, let

g:¼ x þ 2

2 y, h :¼ y þ 2

2 x, uðx, y, Þ :¼ L¹_^zðx,y, Þg, vðx, y, Þ :¼ L¹_^zðx,y, Þh: ð30Þ By Property 2.1(c), it is easy to compute that

u ¼ uðx, y, Þ ¼ 1 detð ^zÞ

^z1 ^z₂

^z2

detð ^zÞ

^z₁ I þ1

^z₁^z2^z₂ 2

4

3 5 g½ 1g2

¼ 1

detð ^zÞ

g1^z1g₂^z2

g1^z2þdetð ^zÞ

^z₁ g2þg₂^z2

^z₁ ^z2

2 4

3 5

:¼ u1

u2

, ð31Þ

v ¼ vðx, y, Þ ¼ 1 detð ^zÞ

^z1 ^z^T₂

^z₂ detð ^zÞ

^z₁ I þ1

^z₁ ^z₂^z^T₂ 2

4

3 5 h₁

h2

¼ 1

detð ^zÞ

h₁^z₁h^T₂^z₂

h1^z2þdetð ^zÞ

^z₁ h2þh^T₂^z2

^z₁ ^z2

2 4

3 5

:¼ v1

v2

: ð32Þ

(10)

Using (28) and Lemma 2.2, we have that

u₁¼ 1

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

₁ðwÞ^ ₂ðwÞ^ p

₁ðwÞ^

₂ðwÞ^ p

2 g₁þ

₁ðwÞ^

₂ðwÞ^ p

2 g^T₂w₂

" #

¼ 1

2 ffiffiffiffiffiffiffiffiffiffiffi

₂ðwÞ^

p

g₁þg^T₂w₂

þ 1

₁ðwÞ^

p

g₁g^T₂w₂

¼ g₁

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2ðwÞ þ

p , ðsince g^T₂w₂ ¼g₁Þ ð33Þ

u2¼ 1

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1ðwÞ^ 2ðwÞ^ p

1ðwÞ^

2ðwÞ^ p

2 g1w2þ 2 ffiffiffiffiffiffiffiffiffiffiffi

1ðwÞ^ p ffiffiffiffiffiffiffiffiffiffiffi

2ðwÞ^ p ffiffiffiffiffiffiffiffiffiffiffi

1ðwÞ^

2ðwÞ^

p g2

2 64

þ

1ðwÞ^

2ðwÞ^

p 2

1ðwÞ^

2ðwÞ^

p g^T₂w₂w₂ 3 75

¼ðg₁þg^T₂w₂Þw₂ 2 ffiffiffiffiffiffiffiffiffiffiffi

₂ðwÞ^

p ðg₁g^T₂w₂Þw₂ 2 ffiffiffiffiffiffiffiffiffiffiffi

₁ðwÞ^

p þ 2g₂2g^T₂w₂w₂ ffiffiffiffiffiffiffiffiffiffiffi

₁ðwÞ^

₂ðwÞ^ p

¼ g₂

₂ðwÞ þ

p : ðsince g₁w2¼g2, g^T₂w2¼g1Þ ð34Þ

Similarly, we can also obtain that v₁¼ h₁

2ðwÞ þ

p , v₂¼ h₂

2ðwÞ þ

p :

From the above expressions of u and v, it then follows that

!0lim^þL¹_^zðx,y, Þðx þ yÞ ¼2

lim

!0^þL¹_^zðx,y, Þ

x þ 2

2 y þ y þ 2 2 x

¼2

lim

!0^þ

uðx, y, Þ þ vðx, y, Þ

¼ 1

₂ðwÞ

p

x1þy1, x2þy2

T

:

This together with Lemma 2.2 yields that

!0lim^þGðx, y, Þ ¼ lim

!0^þ^zðx, y, Þ þ

L_xþ 2 2 L_y

!0lim^þL¹_^zðx,y, Þðx þ yÞ

¼zðx, yÞ þ

x₁þ 2

2 y₁ x^T₂ þ 2 2 y^T₂ x2þ 2

2 y2

x1þ 2 2 y1

I 2

64

3 75

x₁þy₁ ffiffiffiffiffiffiffiffiffiffiffi

2ðwÞ p x₂þy₂

₂ðwÞ p 0 BB B@

1 CC CA

¼zðx, yÞ þ 1 ffiffiffiffiffiffiffiffiffiffiffi

2ðwÞ p

2

x1þ 2 2 y1

ðx1þy1Þ 2

x2þ 2 2 y2

ðx1þy1Þ 0

B@

1 CA

(11)

¼zðx, yÞ þ 1 ffiffiffiffiffiffiffiffiffiffiffi

₂ðwÞ p

2

x1þ 2 2 y1

ðx1þy1Þ 2

x1þ 2 2 y₁

ðx₂þy₂Þ 0

B@

1 CA

¼zðx, yÞ þ x₁þ²₂ y₁ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x²₁þy²₁þðx1y1Þ

q ðx þ yÞ:

Thus, we complete the proof. œ

In what follows, we argue that the gradient function of G(x, y, ) is uniformly bounded, and then, by applying the mean-value theorem for vector-valued functions, we conclude that G(x, y, ) is globally Lipschitz continuous. The following two lemmas are crucial in proving our main result.

LEMMA 3.3 For any x, y 2 IRⁿand 4 0, let ^z : Rⁿ Rⁿ! Rⁿbe given as in (26). Then, the function ^zðx, y, Þ is continuously differentiable everywhere. Moreover, there exists a constant C independent of x, y and , such that

kr_x^zðx, y, Þk ¼

Lxþ 2 2 Ly

L¹_^zðx,y, Þ

C, kr_y^zðx, y, Þk ¼

Lyþ 2 2 Lx

L¹_^zðx,y, Þ

C:

Proof The first part of the conclusion follows directly from [10, Proposition 5.2].

The second part is implied by the proof of [15, Proposition 3.1]. œ LEMMA 3.4 For any x, y 2 IRⁿ and 4 0, let G : IRⁿIRⁿ!IRⁿ be defined as in (29).

Then, the function G(x, y, ) is continuously differentiable everywhere. Moreover, there exists a constant C such that krxG(x, y, )k C(1 þ 1) and kryG(x, y, )k C(1 þ ¹).

Proof The first part of the conclusion is due to [10, Proposition 5.2]. For the second part, by Lemma 3.3, it suffices to prove that the gradient of the following function:

Hðx, y, Þ :¼ Lxþ 2 2 Ly

L¹_^zðx,y, Þðx þ yÞ is uniformly bounded. From the definition of H(x, y, ), we notice that

Hðx, y, Þ ¼2

L_xþ 2 2 L_y

L¹_^zðx,y, Þ x þ 2

2 y þ y þ 2 2 x

¼2

x ðu þ vÞ þ 2

2 y ðu þ vÞ

where u and v are defined as in (30). Therefore, applying Lemma 2.3 yields that r_xHðx, y, Þ ¼2

Luþvþ

r_xuðx, y, Þ þ r_xvðx, y, Þ

Lxþ 2 2 Ly

,

r_yHðx, y, Þ ¼2

2 2 L_uþvþ

r_yuðx, y, Þ þ r_yvðx, y, Þ

L_xþ 2 2 L_y

:

ð35Þ

(12)

To show that krxH(x, y, )k is uniformly bounded, we shall verify that both kLuþvk and kðr_xuðx, y, Þ þ r_xvðx, y, ÞÞðL_xþ ð 2=2ÞL_yÞk are uniformly bounded.

(i) To prove that kLuþvkis uniformly bounded, it is sufficient to argue that ju1j, ku2k and jv1j, ku2k are both uniformly bounded. First, we argue that ju1jand jv1jare uniformly bounded. From (33), we have that

u1¼ 1 2 ffiffiffiffiffiffiffiffiffiffiffi

2ðwÞ^

p

g1þg^T₂w2

þ 1

1ðwÞ^

p

g1g^T₂w2

,

v₁¼ 1 2 ffiffiffiffiffiffiffiffiffiffiffi

₂ðwÞ^

p

h₁þh^T₂w₂

þ 1

₁ðwÞ^

p

h₁h^T₂w₂ :

Note that g1¼x1þ ð 2=2Þy1, g2¼x2þ ð 2=2Þy2and h1¼y1þ ð 2=2Þx1, h2¼y2þ ð 2=2Þx2, and w2¼ ðw2=kw2kÞ. Therefore, applying Lemma 2.1 yields that

g₁g^T₂w₂ ffiffiffiffiffiffiffiffiffiffiffi

₁ðwÞ

₁ðwÞ^

p , jg₁þg^T₂w₂j ffiffiffiffiffiffiffiffiffiffiffi

₂ðwÞ

₁ðwÞ^

p ð36Þ

and

h₁h^T₂w₂

₁ðwÞ

₁ðwÞ^

p , jh₁þh^T₂w₂j ffiffiffiffiffiffiffiffiffiffiffi

₂ðwÞ

₁ðwÞ^

p : ð37Þ

Combing with the expressions of u₁and v₁ given as above, we get ju₁j 1 and jv1j 1.

Second, we argue that ku2kand kv2k are also uniformly bounded. By (34),

u₂¼

g₁þg^T₂w₂

w₂ 2 ffiffiffiffiffiffiffiffiffiffiffi

₂ðwÞ^

p

g₁g^T₂w₂

₁ðwÞ^

p þ 2g₂2g^T₂w₂w₂ ffiffiffiffiffiffiffiffiffiffiffi

₁ðwÞ^

₂ðwÞ^

p ,

v₂¼

h₁þh^T₂w₂

₂ðwÞ^

p

h₁h^T₂w₂

₁ðwÞ^

p þ 2h22h^T₂w2w2

₁ðwÞ^

₂ðwÞ^

p :

Using (36) and (37) and the fact that k w₂k ¼1, we obtain that g₁þg^T₂w₂

₂ðwÞ^

p

g₁g^T₂w₂

₁ðwÞ^

p

1

2kw₂k þ1

2kw₂k ¼1, h1þh^T₂w2

w2

2ðwÞ^

p

h1h^T₂w2

w2

1ðwÞ^

p

1

2kw2k þ1

2kw2k ¼1, and

2g₂2g^T₂w₂w^T₂ ffiffiffiffiffiffiffiffiffiffiffi

₁ðwÞ^

₂ðwÞ^ p

4kg₂k ffiffiffiffiffiffiffiffiffiffiffi

₂ðwÞ^

p 4kg₂k

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi kxk²þ kyk²þ ð 2Þx^Ty þ

p 4,

2h22h^T₂w2w^T₂ ffiffiffiffiffiffiffiffiffiffiffi

₁ðwÞ^

₂ðwÞ^

p

4kh2k ffiffiffiffiffiffiffiffiffiffiffi

₂ðwÞ^

p 4kh2k

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi kxk²þ kyk²þ ð 2Þx^Ty þ

p 4:

(13)

The above inequalities imply that ku2k and kv2k are uniformly bounded. This together with the uniform boundedness of ju1jand jv1j implies that

kL_uþvk ¼ u₁þv₁ ðu₂þv₂Þ^T u₂þv₂ ðu₁þv₁ÞI

is also uniformly bounded.

(ii) Now, it comes to show that kðrxuðx, y, Þ þ rxvðx, y, ÞÞðLxþ ð 2=2ÞLyÞk is uniformly bounded. From the definition of u(x, y, ) and v(x, y, ) as given in (30), we have

^zðx, y, Þ ðuðx, y, Þ þ vðx, y, ÞÞ ¼ðx þ yÞ

2 :

Applying Lemma 2.3 then gives that

r_x^zðx, y, ÞL_uþvþ ðr_xuðx, y, Þ þ r_xvðx, y, ÞÞL_^zðx,y, Þ¼ ð=2ÞI:

This is equivalent to saying that

ðr_xuðx, y, Þ þ r_xvðx, yÞÞL_^zðx,y, Þ ¼ ð=2ÞI r_x^zðx, y, ÞL_uþv

¼

2I L_xþ 2 2 L_y

L¹_^zðx,y, ÞL_uþv, where the second equality is due to Lemma 2.4. Therefore,

r_xuðx, y, Þ þ rxvðx, y, Þ

Lxþ 2 2 Ly

¼

2I L_xþ 2 2 L_y

L¹_^zðx,y, ÞL_uþv

L¹_^zðx,y, Þ L_xþ 2 2 L_y

¼

2L¹_^zðx,y, Þ Lxþ 2 2 Ly

Lxþ 2 2 Ly

L¹_^zðx,y, ÞLuþvL¹_^zðx,y, Þ Lxþ 2 2 Ly

¼

2 Lxþ 2 2 Ly

L¹_^zðx,y, Þ

T

L_xþ 2 2 L_y

L¹_^zðx,y, Þ

L_uþv L_xþ 2 2 L_y

L¹_^zðx,y, Þ

T

: Now we have that

rxuðx, y, Þ þ rxvðx, y, Þ

Lxþ 2 2 Ly

2 Lxþ 2 2 Ly

L¹_^zðx,y, Þ

T

þ L_xþ 2

2 L_y

L¹_^zðx,y, Þ

kL^uþvk L_xþ 2 2 L_y

L¹_^zðx,y, Þ

T

:

From Lemma 3.3, k½ðL_xþ ð 2=2ÞL_yÞL¹_^zðx,y, Þ^Tk is uniformly bounded.

This together with the uniform boundedness of kLuþvk yields that kðr_xuðx, y, Þ þ rxvðx, y, ÞÞðLxþ ð 2=2ÞLyÞkis uniformly bounded.

(14)

From (i), (ii) and (35), we conclude that krxH(x, y, )k is uniformly bounded with the bound related to ¹. Using similar arguments, we can prove that kryH(x, y, )k is uniformly bounded with the bound related to ¹. Combining with Lemma 3.3, we then show that there exists a constant C such that krxG(x, y, )k C(1 þ ¹) and

kryG(x, y, )k C(1 þ ¹). œ

THEOREM 3.1 For any x, y 2 IRⁿand 2(0, 4), let F(x, y) be given as in (23). Then, their gradient functions rxF(x, y) and ryF(x, y) are globally Lipschitz continuous, i.e. there exists a constant C such that

kr_xFðx, yÞ r_xFða, bÞk Cð1 þ ¹Þkðx, yÞ ða, bÞk,

kr_yFðx, yÞ ryFða, bÞk Cð1 þ ¹Þkðx, yÞ ða, bÞk ð38Þ for all(x, y), (a, b) 2 IRⁿIRⁿ.

Proof We first prove that the function G(x, y, ) defined by (29) is globally Lipschitz continuous for any 4 0. For any (x, y) 2 IRⁿIRⁿand (a, b) 2 IRⁿIRⁿ, we have that

Gðx, y, Þ Gða, b, Þ ¼ Gðx, y, Þ Gða, y, Þ þ Gða, y, Þ Gða, b, Þ:

From Lemma 3.4, we know that G(x, y, ) is continuously differentiable everywhere.

Hence, from the mean-value theorem, it follows that Gðx, y, Þ Gða, y, Þ ¼

Z1 0

r_xGða þ tðx aÞ, y, Þðx aÞdt, Gða, y, Þ Gða, b, Þ ¼

Z1 0

r_yGða, b þ tð y bÞ, Þð y bÞdt Combining the last two equations and using Lemma 3.4, we then obtain that

kGðx, y, Þ Gða, b, Þk Z1

0

r_xGða þ tðx aÞ, y, Þðx aÞdt

þ Z1

0

r_yGða, b þ tð y bÞ, Þð y bÞdt

Z1

0

kr_xGða þ tðx aÞ, y, Þkkx akdt

þ Z1

0

kr_xGða þ tðx aÞ, y, Þkkðx aÞkdt

Cð1 þ ¹Þkðx, yÞ ða, bÞk, ð39Þ where C is a constant independent of x, y and , . From Lemma 3.2, we know that

!0lim^þGðx, y, Þ ¼ rxFðx, yÞ

for any x, y 2 IRⁿIRⁿ. This together with (39) immediately yields that kr_xFðx, yÞ rxFða, bÞk ¼ k lim

!0^þGðx, y, Þ lim

!0^þGða, b, Þk

¼ lim

!0^þkGðx, y, Þ Gða, b, Þk

Cð1 þ ¹Þkðx, yÞ ða, bÞk:

(15)

Thus, we prove that rxF(x, y) is globally Lipschitz continuous. Similarly, we may prove

that ryF(x, y) is also globally Lipchitz continuous. œ

From the above theorem, we immediately obtain the following corollary.

COROLLARY 3.1 Let with 2(0, 4) be defined as in (12). Then is an LC¹function, i.e.

the gradient functions rx (x, y) and ry (x, y) are globally Liptchitz continuous with the Lipschitz constant being O(1 þ ¹).

Note

1. Member of Mathematics Division, National Center for Theoretical Sciences, Taipei Office.

The author’s work is partially supported by National Science Council of Taiwan. E-mail:

jschen@math.ntnu.edu.tw.

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