The SC 1 property of the squared norm of the SOC Fischer–Burmeister function

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www.elsevier.com/locate/orl

The SC ¹ property of the squared norm of the SOC Fischer–Burmeister function

Jein-Shan Chen

^a,^∗

, Defeng Sun

^b

, Jie Sun

^c

aDepartment of Mathematics, National Taiwan Normal University, Taipei, 11677, Taiwan bDepartment of Mathematics, National University of Singapore, Singapore 117543, Singapore cDepartment of Decision Sciences, National University of Singapore, Singapore 117592, Singapore

Received 19 May 2007; accepted 27 August 2007 Available online 23 December 2007

Abstract

We show that the gradient mapping of the squared norm of Fischer–Burmeister function is globally Lipschitz continuous and semismooth, which provides a theoretical basis for solving nonlinear second-order cone complementarity problems via the conjugate gradient method and the semismooth Newton’s method.

c

Keywords:Second-order cone; Merit function; Spectral factorization; Lipschitz continuity; Semismoothness

1. Introduction

A popular approach to solving the nonlinear complementarity problem (NCP) is to reformulate it as the global minimization via a certain merit function over Rⁿ. For this approach to be effective, the choice of the merit function is crucial. A popular choice of the merit function is the squared norm of the Fischer–Burmeister (FB) function Ψ : Rⁿ× Rⁿ→ R+defined by

Ψ(a, b) := 1 2

n

X

i =1

|φ(ai, bi)|², (1)

for all a = (a1, . . . , an)^T ∈ Rⁿ and b = (b1, . . . , bn)^T ∈ Rⁿ. The aforementioned Fischer–Burmeister function is denoted by Φ : Rⁿ× Rⁿ→ Rⁿwhose i th component function is Φi(a, b) = φ(ai, bi) with φ : R²→ R given by

φ(ai, bi) =q

a²_i +b²_i −ai−bi. (2)

It is well known that the FB function satisfies

φ(ai, bi) = 0 ⇐⇒ ai ≥0, bi ≥0, aib_i =0. (3)

It has been shown thatφ² is smooth (continuously differentiable) even thoughφ is not differentiable. This merit function and its analysis were subsequently extended by Tseng [11] to the semidefinite complementarity problem (SDCP) although only differentiability, not continuous differentiability, was established. More recently, the squared norm of the FB function for SDCP was reported in [9] to be a smooth function and its gradient is Lipschitz continuous.

∗Corresponding author.

E-mail addresses:[email protected](J.-S. Chen),[email protected](D. Sun),[email protected](J. Sun).

doi:10.1016/j.orl.2007.08.005

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The second-order cone (SOC), also called the Lorentz cone, in Rⁿis defined as

Kⁿ:= {(x1, x2) ∈ R × Rⁿ⁻¹| kx2k ≤x1}, (4)

where k·k denotes the Euclidean norm. By definition, K¹is the set of nonnegative reals R+. The second-order cone complementarity problem (SOCCP) which is to find x, y ∈ Rⁿsatisfying

x = F(ζ), y = G(ζ), hx, yi = 0, x ∈ Kⁿ, y ∈ Kⁿ, (5)

where h·, ·i is the Euclidean inner product and F, G : Rⁿ→ Rⁿare continuous (possibly nonlinear) functions. The FB function for the SOCCP is the vector-valued functionφFB: Rⁿ× Rⁿ→ Rⁿdefined by

φFB(x, y) := (x²+y²)¹^/2−(x + y), (6)

and the squared norm of the FB function for the SOCCP isψFB: Rⁿ× Rⁿ→ R⁺given by ψFB(x, y) := 1

2kφFB(x, y)k². (7)

Note that x²and y²in(6)mean x ◦x and y◦y, respectively (“◦” is introduced in Section2); and x +y means the usual componentwise addition of vectors. It is known that x²∈Kⁿfor all x ∈ Rⁿ. Moreover, if x ∈ Kⁿthen there exists a unique vector in Kⁿdenoted by x¹^/2such that(x¹^/2)² =x¹^/2◦x¹^/2 =x. Therefore, the FB function given as in(6)is well defined for all(x, y) ∈ Rⁿ× Rⁿ. Besides, it was shown in [4] that property(3)ofφ can be extended to φFB. Thus,ψFBis a merit function for the SOCCP since the SOCCP can be expressed as an unconstrained minimization problem:

ζ ∈Rminⁿ f(ζ) := ψFB(F(ζ ), G(ζ )). (8)

Like in the NCP and the SDCP cases,ψFBis shown to be smooth, and when ∇ F and −∇G are column monotone, every stationary point of(8)solves SOCCP; see [2].

The last hurdle to cross in applying(8) to solve(5)is to show that the gradient ofψFB is sufficiently smooth to warrant the convergence of appropriate computational methods. In particular, we are concerned with the conjugate gradient methods and the semismooth Newton’s methods [3]. The former methods generally require the Lipschitz continuity of the gradient ( f ∈ LC¹for short since f : Rⁿ → R is said to be an LC¹ function if it is continuously differentiable and its gradient is locally Lipschitz continuous), while the latter requires that the gradient is semismooth ( f ∈ SC¹for short since f is called an SC¹function if it is continuously differentiable and its gradient is semismooth), in addition to being Lipschitz continuous.

The main purpose of this paper is to show that the gradient function ofψFBdefined as in(7)is globally Lipschitz continuous and semismooth, which is an important property for superlinear convergence of semismooth Newton methods [8]. It should be noted that this result is not a direct implication from a similar result on function Ψ(X, Y ) recently published in [9]. Different analysis is necessary for the proof of Lipschitz continuity.

2. Preliminaries

For any x =(x1, x2), y = (y1, y2) ∈ R × Rⁿ⁻¹, we define their Jordan product associated with Kⁿas

x ◦ y := (hx, yi, y1x₂+x₁y₂). (9)

The identity element under this product is e :=(1, 0, . . . , 0)^T∈ Rⁿ. We write x²to mean x ◦ x and write x + y to mean the usual componentwise addition of vectors.

For any x =(x1, x2) ∈ R × Rⁿ⁻¹, we define the linear mapping L_x from Rⁿto Rⁿas L_xy :=x1 x₂^T

x2 x1I

y.

It can be easily verified that x ◦ y = L_xy, ∀y ∈ Rⁿ, and L_x is positive definite (and hence invertible) if and only if x ∈ int(Kⁿ).

However, L⁻¹_x y 6= x⁻¹◦y, for some x ∈ int(Kⁿ) and y ∈ Rⁿ, i.e., L⁻¹_x 6=L_x−1. In addition, any x =(x1, x2) ∈ R × Rⁿ⁻¹can be decomposed as

x =λ1u⁽¹⁾+λ2u⁽²⁾, (10)

whereλ1,λ2and u⁽¹⁾, u⁽²⁾are the spectral values and the associated spectral vectors of x, with respect to Kⁿ, given by

λi =x1+(−1)ⁱkx2k, (11)

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u⁽ⁱ⁾=





 1 2

1, (−1)ⁱ x2

kx2k

, if x26=0, 1

2

1, (−1)ⁱw , if x₂=0,

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for i = 1, 2, with w being any vector in Rⁿ⁻¹satisfying kwk = 1.

The above spectral factorization of x, as well as x²and x¹^/2and the matrix L_x, have various interesting properties (cf. [4]). In particular, we have the following property about the invertibility of L_x.

Property 2.1. If x ∈ int(Kⁿ), then Lx is invertible with

L⁻¹_x = 1 det(x)





x1 −x₂^T

−x2

det(x) x₁ I + 1

x₁x2x₂^T



.

For any function f : R → R, the following vector-valued function associated with Kⁿ(n ≥ 1) was considered in [5,6]

f^soc(x) = f (λ1)u⁽¹⁾+ f(λ2)u⁽²⁾, x = (x1, x2) ∈ R × Rⁿ⁻¹. (13) For a recent treatment, see [1,4].

Since we aim to prove that the merit functionψFB defined as in(7)has a Lipschitz continuous gradient, we now write down the gradient function ofψFB as below. LetφFB, ψFBbe given by(6)and(7), respectively. Then, from [2, Prop. 1], we know that

∇_xψFB(0, 0) = ∇yψFB(0, 0) = 0. If (x, y) 6= (0, 0) and x²+y²∈int(Kⁿ), then

∇_xψFB(x, y) = L_xL⁻¹_(x₂

+y²)¹^/2−I φFB(x, y)

∇_yψFB(x, y) = LyL⁻¹_(x₂

+y²)¹^/2−I φFB(x, y). (14)

If(x, y) 6= (0, 0) and x²+y²6∈int(Kⁿ), then x₁²+y₁²6=0 and

∇_xψFB(x, y) =



 x₁ q

x₁²+y₁²

−1



φFB(x, y),

∇_yψFB(x, y) =



 y1

q x₁²+y₁²

−1



φFB(x, y).

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3. Main results

In this section, we will present the proof that the gradient function ofψFBis Lipschitz continuous.

Lemma 3.1. Letω : Rⁿ× Rⁿ → R^mbe defined byω(x, y) := u(x, y) ◦ v(x, y), where u, v : Rⁿ× Rⁿ→ R^m are differentiable mappings. Then,ω is differentiable and

∇_xω(x, y) = ∇xu(x, y)L_v(x,y)+ ∇_xv(x, y)Lu(x,y),

∇_yω(x, y) = ∇yu(x, y)Lv(x,y)+ ∇_yv(x, y)Lu(x,y). (16)

Proof. This is the product rule associated with Jordan product. Its proof is straightforward, so we omit it.

Lemma 3.2. For any x, y ∈ Rⁿ, let z(x, y) := (x²+y²)¹^/2, F(x, y) := LxL⁻¹_z_(x,y)(x + y), and G(x, y) := LyL⁻¹_z_(x,y)(x + y). Then, we have

(a) z is differentiable at(x, y) 6= (0, 0) ∈ Rⁿ× Rⁿwith x²+y²∈int(Kⁿ). Moreover

∇_xz(x, y) = LxL⁻¹_z_(x,y), ∇_yz(x, y) = LyL⁻¹_z_(x,y).

(b) F, G are differentiable at (x, y) 6= (0, 0) ∈ Rⁿ× Rⁿwith x²+y²∈int(Kⁿ). Moreover, k∇ F(x, y)k, k∇G(x, y)k are uniformly bounded at such points.

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Proof. (a) That the function z is differentiable is an immediate consequence of [6]. See also [1, Prop. 4]. Since, z²(x, y) = x²+y², applyingLemma 3.1yields

2∇_xz(x, y)Lz(x,y)=2L_x, 2∇_yz(x, y)Lz(x,y)=2L_y. Hence, the desired results follow.

(b) For symmetry, it is enough to show that F is differentiable at(x, y) 6= (0, 0) with x²+y²∈int(Kⁿ) and that k∇xF(x, y)k, k∇_yF(x, y)k are uniformly bounded. It is clear that F is differentiable at such points. The key part is to show the uniform boundedness of k∇xF(x, y)k, k∇yF(x, y)k. Let λ1, λ2be the spectral values of x²+y², then

λ1:= kx k²+ kyk²−2kx1x2+y1y2k, λ2:= kx k²+ kyk²+2kx₁x₂+y₁y₂k.

Thus, z(x, y) := (x²+y²)¹^/2has the spectral values√ λ1,√

λ2and

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

√ λ1+

√λ2

2 ,

√λ2−

√λ1

2 w2

, (17)

wherew2:= ^x¹^x²^+y¹^y²

kx₁x₂+y₁y₂kif x₁x₂+y₁y₂6=0 and otherwisew2is any vector in Rⁿ⁻¹satisfying kw2k =1.

Now, let u := L⁻¹_z(x,y)(x + y). By applyingProperty 2.1, we compute u as below.

u = L⁻¹_z(x,y)(x + y)

= 1

det(z(x, y))





z1 −z^T₂

−z2

det(z(x, y)) z1

I + 1 z1

z2z^T₂





x₁+y₁ x₂+y₂

= 1

det(z(x, y))





(x1+y₁)z1−(x2+y₂)^Tz₂

−(x1+y1)z2+det(z)

z1 (x2+y2) +(x2+y₂)^Tz₂ z1

z2





:= u₁ u₂

.

We notice that F(x, y) = LxL⁻¹_z_(x,y)(x + y) = Lxu = x ◦ u. Then by applyingLemma 3.1, we obtain

∇_xF(x, y) = Lu+ ∇_xu(x, y)Lx and ∇_yF(x, y) = ∇yu(x, y)Lx. (18) To show that k∇_xF(x, y)k is uniformly bounded, we shall verify that both kLukand k∇_xu(x, y)Lxkare uniformly bounded. We prove them as follows.

(i) To see kLukis uniformly bounded, it is sufficient to argue that |u1|, ku2kare both uniformly bounded. First, we argue that

|u1|is uniformly bounded. From the above expression of u, we have

u₁= 1

det(z(x, y))(x1z₁−x₂^Tz₂) + 1

det(z(x, y))(y1z₁−y₂^Tz₂).

Following the similar arguments as in [2, Lemma 4] yields

u1 = 1

det(z(x, y))(x1z1−x₂^Tz2) + 1

det(z(x, y))(y1z1−y₂^Tz2)

=

"

O(1) +(x1−x₂^Tw2) 2√

λ1

# +

"

O(1) +(y1−y₂^Tw2) 2√

λ1

# ,

where O(1) denotes terms that are uniformly bounded with bound independent of (x, y). Moreover, by [2, Lemma 3], if x1x2+y1y26=0 then |x1−x₂^Tw2| ≤ kx2−x1w2k ≤√

λ1. If x1x2+y1y2=0 thenλ1= kx k²+ kyk²so that by choosingw2to further satisfy x^T₂w2=0 we obtain |x₁−x₂^Tw2| ≤ kx2−x1w2k ≤ kx k ≤

√λ1. Similarly, it can be verified that |y₁−y₂^Tw2| ≤

√λ1. Thus, |u₁|is uniformly bounded.

Secondly, we argue that ku2kis also uniformly bounded. Again, using the expression of u and following the similar arguments as in [2, Lemma 4], we obtain

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u2 = 1 det(z(x, y))

"

−x1z2+det(z(x, y))

z₁ x2+x₂^Tz2

z₁ z2

#

+ 1

det(z(x, y))

"

−y1z2+det(z(x, y))

z₁ y2+ y₂^Tz₂ z₁ z2

#

=



 O(1) − x1w2

2

√λ1

+

√λ2

√λ1(x₂^Tw2) 2(√

λ1+

√λ2)w2



 +



 O(1) − y1w2

2

√λ1

+

√λ2

√λ1(y₂^Tw2) 2(√

λ1+

√λ2)w2





=

"

O(1) − x₁w2

2(√ λ1+√

λ2)−

√λ2(x1−x₂^Tw2) 2(√

λ1+√ λ2)√

λ1

w2

#

+

"

O(1) − y1w2

2(√ λ1+

√λ2)−

√λ2(y1−y^T₂w2) 2(√

λ1+

√λ2)√ λ1

w2

# .

Using the same explanations as above for u₁yields that each term is uniformly bounded. Thus, ku₂kis uniformly bounded. This together with |u1|being uniformly bounded implies that k∇_xF(x, y)k = kLuk =

hu₁ u^T₂ u₂ u₁I

i

is also uniformly bounded.

(ii) Now, it comes to show that k∇_xu(x, y)Lxkis uniformly bounded. From the definition of u := L⁻¹_z_(x,y)(x + y), we know that z(x, y) ◦ u = x + y. ApplyingLemma 3.1gives

∇_xz(x, y)Lu+ ∇_xu(x, y)Lz(x,y)=I, which leads to

∇_xu(x, y)Lz(x,y)=I − ∇_xz(x, y)Lu=I −(LxL⁻¹_z(x,y))Lu

⇒ ∇_xu(x, y) =

I − LxL⁻¹_z_(x,y)Lu

L⁻¹_z_(x,y)

⇒ ∇_xu(x, y)Lx =

I − LxL⁻¹_z_(x,y)Lu

L⁻¹_z_(x,y)Lx

⇒ ∇_xu(x, y)Lx =L⁻¹_z_(x,y)L_x−L_xL⁻¹_z_(x,y)L_uL⁻¹_z_(x,y)L_x

⇒ ∇_xu(x, y)Lx =(LxL⁻¹_z_(x,y))^T−(LxL⁻¹_z_(x,y))Lu(LxL⁻¹_z_(x,y))^T. Therefore,

k∇_xu(x, y)Lxk ≤ k(LxL⁻¹_z_(x,y))^Tk + kL_xL⁻¹_z_(x,y)k · kL_uk · k(LxL⁻¹_z_(x,y))^Tk.

By [2, Lemma 4], kLxL⁻¹_z_(x,y)k is uniformly bounded, so is k(LxL⁻¹_z_(x,y))^Tk. This together with kLuk being uniformly bounded shown as above yields that k∇xu(x, y)Lxkis uniformly bounded.

From (i) and (ii), we conclude that k∇xF(x, y)k is uniformly bounded. Similar arguments apply to k∇yF(x, y)k; and hence, k∇F(x, y)k is uniformly bounded. Thus, we complete the proof.

Lemma 3.3. LetψFBbe defined as(7). Then, ∇ψFBis continuously differentiable everywhere except for(x, y) = (0, 0). Moreover, k∇²ψFB(x, y)k is uniformly bounded for all (x, y) 6= (0, 0).

Proof. For any(x, y) ∈ Rⁿ× Rⁿ, let z :=(x²+y²)¹^/2. We prove this lemma by considering the following two cases.

(i) Consider all points(x, y) 6= (0, 0) with x²+y²∈int(Kⁿ). Since

∇_xψFB(x, y) =

L_xL⁻¹_z −I φFB(x, y) = x − LxL⁻¹_z (x + y) − φFB(x, y),

∇_yψFB(x, y) =

L_yL⁻¹_z −I φFB(x, y) = y − LyL⁻¹_z (x + y) − φFB(x, y), we can compute ∇²ψFB(x, y) as follows:

∇_{x x}² ψFB(x, y) = I − ∇x

L_xL⁻¹_z (x + y)

−

L_xL⁻¹_z −I , (19)

∇_{x y}² ψFB(x, y) = −∇y

LxL⁻¹_z (x + y)

−

LyL⁻¹_z −I ,

∇²_yxψFB(x, y) = −∇x

LyL⁻¹_z (x + y)

−

LxL⁻¹_z −I ,

∇²_yyψFB(x, y) = I − ∇y

LyL⁻¹_z (x + y)

−

LyL⁻¹_z −I .

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The continuity of ∇²ψFB at (x, y) thus follows. It is easy to see that kLxL⁻¹_z k, kL_yL⁻¹_z k are uniformly bounded by [2, Lemma 4] (k · k and k · k_F are equivalent in R^n×n). Let F(x, y) := LxL⁻¹_z (x + y) and G(x, y) := LyL⁻¹_z (x + y).

By Lemma 3.2, we know that

∇_x LxL⁻¹_z (x + y) = k∇_xF(x, y)k is uniformly bounded. Likewise, we have that

∇_y LxL⁻¹_z (x + y) ,

∇_x LyL⁻¹_z (x + y) ,

∇_y LyL⁻¹_z (x + y) are all uniformly bounded. Thus, we can conclude that k∇_{x x}² ψFB(x, y)k, k∇_{x y}² ψFB(x, y)k, k∇²_yxψFB(x, y)k, k∇²_yyψFB(x, y)k are all uniformly bounded which implies that k∇²ψFB(x, y)k is also uniformly bounded.

(ii) Consider all points(x, y) 6= (0, 0) with x²+y²6∈int(Kⁿ). Since

∇_xψFB(x, y) =



 x1

q x²₁+y₁²

−1



φFB(x, y) = x − x1

q x₁²+y₁²

(x + y) − φFB(x, y),

∇_yψFB(x, y) =



 y1

q x²₁+y₁²

−1



φFB(x, y) = y − y1

q x₁²+y₁²

(x + y) − φFB(x, y),

we can compute ∇²ψFB(x, y) as follows:

∇²_{x x}ψFB(x, y) = I −



 x₁ q

x₁²+y₁²

I + x1y₁²+y₁³ (x²₁+y₁²)^3/2

1 0 0 0



 −



 x₁ q

x₁²+y₁²

−1



 I, (20)

∇²

x yψFB(x, y) = −



 x1

q x²₁+y₁²

I − x₁²y₁+x₁y₁² (x₁²+y₁²)³^/2

1 0 0 0



 −



 y1

q x₁²+y₁²

−1



 I,

∇²_yxψFB(x, y) = −



 y1

q x₁²+y₁²

I − x₁²y1+x1y₁² (x₁²+y₁²)³^/2

1 0 0 0



 −



 x1

q x₁²+y₁²

−1



 I,

∇²_yyψFB(x, y) = I −



 y₁ q

x₁²+y₁²

I + x₁³+x₁²y1

(x₁²+y₁²)³^/2

1 0 0 0



 −



 y₁ q

x₁²+y₁²

−1



 I, where 0 denotes the(n − 1) × (n − 1) zero matrix.

Now we provide a sketch proof to verify that ∇_{x x}ψFB is continuous. Let(a, b) 6= (0, 0) and a²+b² 6∈ int(Kⁿ). We want to prove that

∇_{x x}ψFB(x, y) → ∇x xψFB(a, b), as (x, y) → (a, b). (21)

Due to the neighborhood of such(a, b), we have to consider two subcases: (1) (x, y) 6= (0, 0) with x²+y² ∈ int(Kⁿ) and (2) (x, y) 6= (0, 0) with x²+y²6∈int(Kⁿ). It is clear that(21)holds in subcase (2) because the formula given in(20)is continuous. In subcase (1), we have

∇_{x x}ψFB(x, y) = I − ∇x

L_xL⁻¹_z (x + y)

−

L_xL⁻¹_z −I

= I −

Lu+

LxL⁻¹_z T

−

LxL⁻¹_z (Lu)

LxL⁻¹_z T

−

LxL⁻¹_z −I . (22)

In view of(19),(20)and(22), it suffices to show the following three statements for(21)to be held in this subcase (1):

(a) L_xL⁻¹_z → _q ^a¹

a₁²+b²₁

I, as(x, y) → (a, b).

(b) L_u→ _q^a¹^+b¹

a²₁+b²₁

I, as(x, y) → (a, b).

(c) L_u−(LxL⁻¹_z )(Lu)(LxL⁻¹_z )^T→ ^a

2 1(a1+b1)

(a₁²+b₁²)^3/2 I, as(x, y) → (a, b).

First, we know from [2, Prop. 2] that there holds LxL⁻¹_z (x + y) → a₁

q a²₁+b₁²

(a + b) as (x, y) → (a, b),

which implies that L_xL⁻¹_z → _q ^a¹

a²₁+b²₁

I, as(x, y) → (a, b) since both (x + y) and LxL⁻¹_z are continuous and(x + y) → (a + b) when(x, y) → (a, b). Secondly, if we look into the entries of Luand compare them with the entries of LxL⁻¹_z (see [2, eq. (27)]),

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then it is clear that L_u → q^a¹^+b¹ a₁²+b²₁

I, as (x, y) → (a, b). Finally, part(c) follows immediately from part (a) and (b). Thus, we complete the verifications of(21). The other cases can be argued similarly for ∇_{x y}ψFB, ∇_yxψFB, and ∇_yyψFB. In addition, it is also clear that each term in the above expressions(20)is uniformly bounded. Thus, we obtain that ∇²ψFBis continuously differentiable near(x, y) and k∇²ψFB(x, y)k is uniformly bounded.

Theorem 3.1. LetψFBbe defined as(7). Then, ∇ψFBis globally Lipschitz continuous, i.e., there exists a constant C such that for all(x, y), (a, b) ∈ Rⁿ× Rⁿ,

k∇_xψFB(x, y) − ∇xψFB(a, b)k ≤ Ck(x, y) − (a, b)k, (23)

k∇_yψFB(x, y) − ∇yψFB(a, b)k ≤ Ck(x, y) − (a, b)k and is semismooth everywhere.

Proof. Owing to symmetry, we only need to show that the first part of(23)holds. For any(x, y) ∈ Rⁿ× Rⁿ, let z :=(x²+y²)¹^/2. (i) First, we prove that ∇_xψFBis Lipschitz continuous at(0, 0). We have to discuss three subcases for completing the proof of this part.

If(x, y) = (0, 0), it is obvious that(23)is satisfied.

If(x, y) 6= (0, 0) with x²+y²∈int(Kⁿ), then

k∇_xψFB(x, y) − ∇xψFB(0, 0)k = k∇xψFB(x, y)k = kx − LxL⁻¹_z (x + y) − φFB(x, y)k.

It is already known that x andφFB(x, y) are Lipschitz continuous (see [10, Cor. 3.3]). In addition, Theorem 3.2.4 of [7, pp. 70] says that the uniform boundedness of ∇ L_xL⁻¹_z (x + y) (byLemma 3.2) yields the Lipschitz continuity. Thus,(23)is satisfied for this subcase. If(x, y) 6= (0, 0) with x²+y²6∈int(Kⁿ), then

k∇_xψFB(x, y) − ∇xψFB(0, 0)k = k∇xψFB(x, y)k =

x − x₁ q

x₁²+y₁²

(x + y) − φFB(x, y) .

Since

x₁ q

x₁²+y₁²

≤1 and both(x + y), φFB(x, y) are known Lipschitz continuous, the desired result follows.

(ii) Secondly, we prove that ∇_xψFBis Lipschitz continuous at(a, b) 6= (0, 0). Let (x, y) ∈ Rⁿ× Rⁿ, we wish to show that(23) is satisfied. In fact, if the line segment [(a, b), (x, y)] does not contain the origin, then we can write

k∇_xψFB(x, y) − ∇xψFB(a, b)k ≤

Z 1 0

∇²ψFB[(a, b) + t((x, y) − (a, b))]dt

≤C k(x, y) − (a, b)k,

where the first inequality is from the Mean Value Theorem (see [7, Theorem 3.2.3]), and the second inequality is byLemma 3.3. On the other hand, if the line segment [(a, b), (x, y)] contains the origin, we can construct a sequence {(x^k, y^k)} converging to (x, y) but for each k, the line segment [(a, b), (x^k, y^k)] does not contain the origin and apply the above inequalities to get

k∇_xψFB(x^k, y^k) − ∇xψFB(a, b)k ≤ Ck(x^k, y^k) − (a, b)k, which, by the continuity, implies

k∇_xψFB(x, y) − ∇xψFB(a, b)k ≤ Ck(x, y) − (a, b)k.

Thus,(23)is satisfied.

To complete the proof of this theorem, we only need to show that ∇ψFBis semismooth at the origin as, byLemma 3.3, ∇ψFB

is continuously differentiable near any (0, 0) 6= (x, y) ∈ Rⁿ × Rⁿ. From (14)and (15), we know that for any t ∈ R+ and (x, y) ∈ Rⁿ× Rⁿwe have

∇ψFB(tx, ty) = t∇ψFB(x, y).

Thus, ∇ψFBis directionally differentiable at the origin and for any(0, 0) 6= (x, y) ∈ Rⁿ× Rⁿ

∇²ψFB(x, y)(x, y) = (∇ψFB)⁰((x, y); (x, y)) = ∇ψFB(x, y).

This means that for any(0, 0) 6= (x, y) ∈ Rⁿ× Rⁿconverging to(0, 0),

∇ψFB(x, y) − ∇ψFB(0, 0) − ∇²ψFB(x, y)(x, y) = ∇ψFB(x, y) − 0 − ∇ψFB(x, y) = 0 ,

which, together with the Lipschitz continuity of ∇ψFBand the directional differentiability of ∇ψFBat the origin (∇ψFBis, however, not differentiable at the origin), shows that ∇ψFB(x, y) is (strongly) semismooth at the origin. The proof is completed.

FromTheorem 3.1, we immediately obtain that the functionψFBdefined as in(7)is an SC¹function as well as an LC¹function.

(8)

Acknowledgements

Jein-Shan Chen is a member of Mathematics Division, National Center for Theoretical Sciences, Taipei Office and his research is partially supported by the National Science Council of Taiwan. The research of Defeng Sun is partially supported by the Academic Research Fund under Grant No. R-146-000-104-112 of the National University of Singapore. The research of Jie Sun is partially supported by Singapore-MIT Alliance.

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