Volume 83, Number 287, May 2014, Pages 1143–1171 S 0025-5718(2013)02742-1
Article electronically published on July 15, 2013
ON THE GENERALIZED FISCHER-BURMEISTER MERIT FUNCTION FOR THE SECOND-ORDER CONE
COMPLEMENTARITY PROBLEM
SHAOHUA PAN, SANGHO KUM, YONGDO LIM, AND JEIN-SHAN CHEN
Abstract. It has been an open question whether the family of merit functions ψp(p > 1), the generalized Fischer-Burmeister (FB) merit function, associated to the second-order cone is smooth or not. In this paper we answer it partly, and show that ψp is smooth for p∈ (1, 4), and we provide the condition for its coerciveness. Numerical results are reported to illustrate the influence of p on the performance of the merit function method based on ψp.
1. Introduction
Given two continuously differentiable mappings F, G :Rn→ Rn, we consider the second-order cone complementarity problem (SOCCP): to seek a ζ ∈ Rn such that (1) F (ζ)∈ K, G(ζ) ∈ K, F (ζ), G(ζ) = 0,
where·, · denotes the Euclidean inner product that induces the norm · , and K is the Cartesian product of a group of second-order cones (SOCs). In other words, (2) K = Kn1× Kn2× · · · × Knm,
where n1, . . . , nm≥ 1, n1+· · · + nm= n, andKni is the SOC inRni defined by Kni :=
(xi1, xi2)∈ R × Rni−1 | xi1≥ xi2 .
As an extension of the nonlinear complementarity problem (NCP) over the non- negative orthant cone Rn+ (see [13]), the SOCCP has important applications in engineering problems [21] and robust Nash equilibria [19]. In particular, it also
Received by the editor August 15, 2010, and in revised form, April 18, 2011 and August 7, 2012.
2010 Mathematics Subject Classification. Primary 90C33, 90C25.
Key words and phrases. Second-order cones, complementarity problem, generalized FB merit function.
The first author’s work was supported by National Young Natural Science Foundation (No.
10901058) and the Fundamental Research Funds for the Central Universities (SCUT).
The second author’s work was supported by Basic Science Research Program through NRF Grant No. 2012-0001740.
The third author’s work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MEST) (No.2012-005191).
Corresponding author: The fourth author is a member of the Mathematics Division, National Center for Theoretical Sciences, Taipei Office. The fourth author’s work was supported by National Science Council of Taiwan.
2013 American Mathematical Societyc 1143
arises from the suitable reformulation for the Karush-Kuhn-Tucker (KKT) opti- mality conditions of the nonlinear second-order cone programming (SOCP):
(3) minimize f (x)
subject to Ax = b, x∈ K,
where f : Rn → R is a twice continuously differentiable function, A is an m × n real matrix with full row rank, and b∈ Rm. It is well known that the SOCP has very wide applications in engineering design, control, management science, and so on; see [1, 25] and the references therein.
In the past several years, there have various methods proposed for SOCPs and SOCCPs. They include the interior-point methods [2, 26, 28, 31, 33], the smoothing Newton methods [11, 15, 18], the semismooth Newton methods [22, 34], and the merit function methods [4, 12]. The merit function method aims to seek a smooth (continuously differentiable) function ψ :Rn× Rn→ R+ satisfying
(4) ψ(x, y) = 0 ⇐⇒ x ∈ K, y ∈ K, x, y = 0,
so that the SOCCP can be reformulated as an unconstrained minimization problem
(5) min
ζ∈RnΨ(ζ) := ψ(F (ζ), G(ζ))
in the sense that ζ∗ is a solution to (1) if and only if it solves (5) with zero op- timal value. We call such ψ a merit function associated with K. Note that the smooth merit functions also play a key role in the globalization of semismooth and smoothing Newton methods.
This paper is concerned with the generalized Fischer-Burmeister (FB) merit function
(6) ψp(x, y) := 1
2φp(x, y)2,
where p is a fixed real number from (1, +∞), and φp:Rn× Rn→ Rn is defined by
(7) φp(x, y) :=p
|x|p+|y|p− (x + y)
with |x|p being the vector-valued function (or L¨owner function) associated with
|t|p (t∈ R) (see Section 2 for the definition). Clearly, when p = 2, ψp reduces to the FB merit function
ψFB(x, y) := 1
2φFB(x, y)2,
where φFB :Rn× Rn→ Rn is the FB SOC complementarity function defined by φFB(x, y) :=
x2+ y2− (x + y),
with x2 = x◦ x being the Jordan product of x with itself, and √
x with x ∈ K being the unique vector such that√
x◦√
x = x. The function ψFB is shown to be a smooth merit function with globally Lipschitz continuous derivative [10, 12]. Such a desirable property is also proved for the FB matrix-valued merit function [30, 32].
In this paper, we study the favorable properties of ψp. The motivations for us to study this family of merit functions are as follows. In the setting of NCPs, ψpis shown to share all favorable properties as the FB merit function holds (see [9, 8]), and the performance profile in [5] indicates that the semismooth Newton method based on φp with a smaller p has better performance than a larger p. Thus, it is very natural to ask whether ψphas the desirable properties of the FB merit function or not in the setting of SOCCPs, and what performance the merit function method
and the Newton-type methods based on φp display with respect to p. This work is the first step in resolving these questions. Although there are some papers [4, 6, 12]
to study the smoothness of merit functions for the SOCCPs, the analysis techniques therein are not applicable for the general function ψp. We wish that the analysis technique of this paper would be helpful in handling general L¨owner operators.
The main contribution of this paper is to show that ψpwith p∈ (1, 4) is a smooth merit function associated with K, and to establish the coerciveness of Ψp(ζ) :=
ψp(F (ζ), G(ζ)) under the uniform Jordan P -property and the linear growth of F . Throughout this paper, we will focus on the case ofK = Kn, and all the analysis can be carried over to the general case whereK is the Cartesian product of Kni. To this end, for any given x∈ Rnwith n > 1, we write x = (x1, x2) where x1is the first component of x, and x2 is the column vector consisting of the rest components of x; and let x2= xx2
2 whenever x2 = 0, and otherwise let x2be an arbitrary vector in Rn−1 with x2 = 1. We denote intKn, bdKn and bd+Kn by the interior, the boundary, and the boundary excluding the origin, respectively, ofKn. For any x, y∈ Rn, xKn y means x− y ∈ Kn; and xKn y means x− y ∈ intKn. For a real symmetric matrix A, we write A 0 (respectively, A 0) to mean that A is positive semidefinite (respectively, positive definite). For a differentiable mapping F :Rn → Rm,∇F (x) denotes the transposed Jacobian of F at x. For nonnegative α and β, α = O(β) means α ≤ Cβ for some C > 0 independent of α and β. The notation I always represents an identity matrix of appropriate dimension.
2. Preliminaries
The Jordan product of any two vectors x and y associated with Kn (see [14]) is defined as
x◦ y := (x, y, y1x2+ x1y2).
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 product is e = (1, 0, . . . , 0)T ∈ Rn. For any given x ∈ Rn, define Lx:Rn → Rn by
Lxy :=
x1 xT2 x2 x1I
y = x◦ y ∀y ∈ Rn.
Recall from [14] that each x∈ Rn has a spectral factorization associated withKn: x = λ1(x)u(1)x + λ2(x)u(2)x ,
(8)
where λi(x) and u(i)x for i = 1, 2 are the spectral values of x and the corresponding spectral vectors, respectively, defined by
λi(x) := x1+ (−1)ix2 and u(i)x :=1 2
1, (−1)ix2
. (9)
The factorization is unique when x2 = 0. The following lemma states the relation between the spectral factorization of x and the eigenvalue decomposition of Lx. Lemma 2.1 ([14, 15]). For any given x∈ Rn, let λ1(x), λ2(x) be the spectral values of x, and let u(1)x , u(2)x be the corresponding spectral vectors. Then, we have
Lx= Uxdiag (λ2(x), x1, . . . , x1, λ1(x)) UxT
with Ux = [√
2u(2)x u(3) · · · u(n) √
2u(1)x ] ∈ Rn×n beings an orthogonal matrix, where u(i)= (0, ui) for i = 3, . . . , n with u3, . . . , un being any unit vectors to span the linear subspace orthogonal to x2.
By Lemma 2.1, clearly, Lx 0 iff (if and only if) x Kn 0, Lx 0 iff x Kn 0, and Lxis invertible iff x1 = 0 and det(x) := x21−x22 = 0. Also, if Lxis invertible,
L−1x = 1 det(x)
x1 −xT2
−x2 det(x) x1 I +x1
1x2xT2
. (10)
Given a scalar function g :R → R, define a vector function gsoc:Rn→ Rn by gsoc(x) := g(λ1(x))u(1)x + g(λ2(x))u(2)x .
(11)
If g is defined on a subset of R, then gsoc is defined on the corresponding subset of Rn. The definition of gsoc is unambiguous whether x2 = 0 or x2= 0. In this paper, we often use the vector-valued functions associated with |t|p (t ∈ R) and
√p
t (t≥ 0), respectively, written as
|x|p:=|λ1(x)|p u(1)x +|λ2(x)|p u(2)x ∀ x ∈ Rn,
√p
x := p
λ1(x) u(1)x +p
λ2(x) u(2)x ∀ x ∈ Kn. The two functions show that φp in (7) is well defined for any x, y∈ Rn.
We next present four lemmas that will often be used in the subsequent analysis.
Lemma 2.2 ([23, 24]). For any given 0≤ ρ ≤ 1, ξρKn ηρ when ξKnηKn0.
Lemma 2.3. For any nonnegative real numbers a and b, the following results hold:
(a): (a + b)ρ ≥ aρ+ bρ if ρ > 1, and the equality holds iff ab = 0;
(b): (a + b)ρ≤ aρ+ bρ if 0 < ρ < 1, and the equality holds iff ab = 0.
Proof. Without loss of generality, we assume that a ≤ b and b > 0. Consider the function h(t) = (t + 1)ρ− (tρ+ 1) (t≥ 0). It is easy to verify that h is increasing on [0, +∞) when ρ > 1. Hence, h(a/b) ≥ h(0) = 0, i.e., (a + b)ρ ≥ aρ+ bρ. Also, h(a/b) = h(0) if and only if a/b = 0. That is, (a + b)ρ = aρ+ bρ if and only if ab = 0. This proves part (a). Note that h is decreasing on [0, +∞) when 0 < ρ < 1,
and a similar argument leads to part (b).
Lemma 2.4. For any ξ, η∈ Kn, if ξ + η∈ bdKn, then one of the following cases must hold: (i) ξ = 0, η ∈ bdKn; (ii) ξ∈ bdKn, η = 0; (iii) ξ = γη for some γ > 0 with η∈ bd+Kn.
Proof. From ξ, η∈ Kn and ξ + η∈ bdKn, we immediately obtain that
ξ2 + η2 ≥ ξ2+ η2 = ξ1+ η1≥ ξ2 + η2.
This shows that ξ2 = 0, or η2= 0, or ξ2= γη2 = 0 for some γ > 0. Substituting ξ2= 0, or η2= 0, or ξ2= γη2 into ξ2+ η2 = ξ1+ η1 yields the result. To close this section, we show that φpin (7) is an SOC complementarity function, and then its squared norm ψp is a merit function associated withKn.
Lemma 2.5. Let φp be defined by (7). Then, for any x, y∈Rn, it holds that φp(x, y) = 0 ⇐⇒ x ∈ Kn, y∈ Kn, x, y = 0.
Proof. “⇐”. From [16, Proposition 6], there exists a Jordan frame
u(1), u(2) such that x = λ1u(1)+ λ2u(2) and y = μ1u(1)+ μ2u(2)with λi, μi≥ 0 for i = 1, 2. Then,
(x + y)p = (λ1+ μ1)pu(1)+ (λ2+ μ2)pu(2), xp+ yp = (λp1+ μp1)u(1)+ (λp2+ μp2)u(2).
Since 0 = 2x, y = λ1μ1+ λ2μ2 implies λ1μ1 = λ2μ2 = 0, from the last two equalities and Lemma 2.3(a) we obtain (x + y)p= xp+ yp, and then φp(x, y) = 0.
“⇒”. Since φp(x, y) = 0, we have x =
|x|p+|y|p− y Kn |y| − y ∈ Kn, where the inequality is due to Lemma 2.2. Similarly, we have y = p
|x|p+|y|p− x Kn
|x| − x ∈ Kn. Now from φp(x, y) = 0, we have (x + y)p= xp+ yp, and then (λ1(x + y))p+ (λ2(x + y))p= (λ1(x))p+ (λ2(x))p+ (λ1(y))p+ (λ2(y))p. Noting that h(t) = (t0+ t)p+ (t0− t)pfor a fixed t0≥ 0 is increasing on [0, t0], we also have
[λ1(x + y)]p+ [λ2(x + y)]p≥ (x1+ y1− x2 + y2)p+ (x1+ y1+x2 − y2)p
= (λ1(x) + λ2(y))p+ (λ2(x) + λ1(y))p
≥ (λ1(x))p+ (λ2(y))p+ (λ2(x))p+ (λ1(y))p, (12)
where the last inequality is due to Lemma 2.3(a) and x, y ∈ Kn. The last two equations imply that all the inequalities on the right-hand side of (12) become equalities. Therefore,
(13) x2+ y2 = x2 − y2, λ1(x)λ2(y) = 0, λ2(x)λ1(y) = 0.
Assume that x2 = 0 and y2 = 0. Since x, y ∈ Kn, from the equalities in (13), we get x1=x2, y1=y2, and x2= γy2for some γ < 0, which implies x, y = 0. When x2= 0 or y2= 0, using the continuity of the inner product yieldsx, y = 0.
3. Differentiability of ψp
Unless otherwise stated, in the rest of this paper, we assume that p > 1 with q = (1−p−1)−1, and gsoc is the vector-valued function associated with|t|p (t∈ R), i.e., gsoc(x) =|x|p. For any x, y∈ Rn, we define
w = w(x, y) :=|x|p+|y|p and z = z(x, y) :=p
|x|p+|y|p. (14)
By definitions of|x|p and|y|p, clearly,
w1:= w1(x, y) = |λ2(x)|p+|λ1(x)|p
2 +|λ2(y)|p+|λ1(y)|p
2 ,
w2:= w2(x, y) = |λ2(x)|p− |λ1(x)|p
2 x2+|λ2(y)|p− |λ1(y)|p
2 y2,
(15)
where x2 = xx22 if x2 = 0, and otherwise x2 is an arbitrary vector inRn−1 with
x2 = 1, and y2 has a similar definition.
Noting that z(x, y) = p
w(x, y), we have z1= z1(x, y) =
p
λ2(w) +p λ1(w)
2 ,
z2= z2(x, y) =
p
λ2(w)−p λ1(w)
2 w2,
(16)
where w2= ww2
2 if w2 = 0, and otherwise w2 is an arbitrary vector inRn−1 with
w2 = 1.
To study the differentiability of ψp, we need the following two crucial lemmas.
The first one gives the properties of the points (x, y) satisfying w(x, y) ∈ bdKn, and the second one provides a sufficient characterization for the continuously dif- ferentiable points of z(x, y).
Lemma 3.1. For any (x, y) with w(x, y)∈ bdKn, we have the following equalities:
w1(x, y) =w2(x, y) = 2p−1(|x1|p+|y1|p), x21=x22, y21=y22, x1y1= xT2y2, x1y2= y1x2. (17)
If, in addition, w2(x, y) = 0, the following equalities hold with w2(x, y) = ww2(x,y)
2(x,y): (18) xT2w2(x, y) = x1, x1w2(x, y) = x2, yT2w2(x, y) = y1, y1w2(x, y) = y2. Proof. Fix any (x, y) with w(x, y)∈ bdKn. Since|x|p,|y|p∈ Kn, applying Lemma 2.4 with ξ =|x|pand η =|y|p, we have|x|p ∈ bdKn and|y|p∈ bdKn. This means that|λ2(x)|p·|λ1(x)|p= 0 and|λ2(y)|p·|λ1(y)|p= 0. So, x21=x22and y12=y22. Substituting this into w1(x, y), we readily obtain w1(x, y) = 2p−1(|x1|p+|y1|p).
To prove other equalities in (17) and (18), we first consider the case where x1+x2 = 0 and y1− y2 = 0 with x2 = 0 and y2 = 0. Under this case,
w1= |λ1(x)|p+|λ2(y)|p
2 = |λ1(x)|p 2
x2
x2−|λ2(y)|p 2
y2
y2 2,
which implies that xT2y2 = −x2y2 = x1y1. Together with x21 = x22 and y21=y22, we have that x1y2= y1x2. From the definition of w2, it follows that
xT2w2=−|λ1(x)|p
2 x2 + |λ2(y)|p 2
x1y1
y2 = 2p−1(|x1|p+|y1|p) x1=w2x1, x1w2=−|λ1(x)|p
2
x1x2
x2 +|λ2(y)|p 2
y1x2
y2 = 2p−1(|x1|p+|y1|p) x2=w2x2. Similarly, we also have y2Tw2=w2y1 and yT1w2=w2y2. The above arguments show that equations (17) and (18) hold under the case where x1 = −x2, y1 =
y2. Using the same arguments, we can prove that (17) and (18) hold under any one of the following cases: x1 = x2, y1 = y2; or x1 = −x2, y1 = y2; or
x1=−x2, y1=−y2.
Lemma 3.2. z(x, y) is continuously differentiable at (x, y) with w(x, y)∈ intKn, and
∇xz(x, y) =∇gsoc(x)∇gsoc(z)−1 and ∇yz(x, y) =∇gsoc(y)∇gsoc(z)−1, where∇gsoc(z)−1= (p√q
w1)−1I if w2= 0, and otherwise
∇gsoc(z)−1 = 1 2p
⎡
⎢⎣
1
√q
λ2(w) + √q 1
λ1(w)
wT2
√q
λ2(w) − √qwT2
λ1(w) w2
√q
λ2(w) −√q w2
λ1(w)
2p(I−w2wT2)
a(z) +√qw2wT2
λ2(w)+√qw2wT2
λ1(w)
⎤
⎥⎦ .
Proof. Since |t|p (t ∈ R) and √p
t (t > 0) are continuously differentiable, by [15, Proposition 5.2] or [7, Proposition 5], the functions gsoc(x) and√
x are continuously differentiable in Rn and intKn, respectively. This implies the first part of this
lemma. A simple calculation gives the expression of ∇z(x, y). By the formula in [15, Proposition 5.2],
∇gsoc(x) =
⎧⎨
⎩
p sign(x1)|x1|p−1I if x2= 0;
b(x) c(x)xT2
c(x)x2 a(x)I + (b(x)− a(x))x2xT2
if x2 = 0, (19)
where
x2= x2
x2, a(x) =|λ2(x)|p− |λ1(x)|p λ2(x)− λ1(x) , b(x) = p
2
sign(λ2(x))|λ2(x)|p−1+ sign(λ1(x))|λ1(x)|p−1 , c(x) = p
2
sign(λ2(x))|λ2(x)|p−1− sign(λ1(x))|λ1(x)|p−1 . (20)
We next derive the formula of∇gsoc(z)−1. When w2= 0, we have λ1(w) = λ2(w) = w1 > 0, which by (16) implies z1 = √p
w1 and z2= 0. From formula (19), it then follows that ∇gsoc(z) = p|z1|p−1I = p√q
w1I. Consequently, ∇gsoc(z)−1 = p√q1w1I.
When w2 = 0, since p
λ2(w) > p
λ1(w), we have z2 = 0 and z2 = zz2
2 = w2
by (16). Using the expression of ∇gsoc(z), it is easy to verify that b(z) + c(z) and b(z)− c(z) are the eigenvalues of ∇gsoc(z) with (1, w2) and (1,−w2) being the corresponding eigenvectors, and a(z) is the eigenvalue of multiplicity n− 2 with corresponding eigenvectors of the form (0, vi), where v1, . . . , vn−2are any unit vectors inRn−1 that span the subspace orthogonal to w2. Hence,
∇gsoc(z) = U diag (b(z)− c(z), a(z), . . . , a(z), b(z) + c(z)) UT, where U = [u1 v1 · · · vn−2 u2]∈ Rn×n is an orthogonal matrix with
u1=
1
−w2
, u2=
1 w2
, vi=
0 vi
for i = 1, . . . , n− 2.
By this, we know that∇gsoc(z)−1 has the expression given as in the lemma. Now we are in a position to prove the following main result of this section.
Proposition 3.1. The function ψp for p∈ (1, 4) is differentiable everywhere. Also, for any given x, y ∈ Rn, if w(x, y) = 0, then ∇xψp(x, y) = ∇yψp(x, y) = 0; if w(x, y)∈ intKn, then
∇xψp(x, y) =
∇gsoc(x)∇gsoc(z)−1− I
φp(x, y),
∇yψp(x, y) =
∇gsoc(y)∇gsoc(z)−1− I
φp(x, y);
(21)
and if w(x, y)∈ bd+Kn, then
∇xψp(x, y) =
sign(x1)|x1|p−1
q
|x1|p+|y1|p − 1
φp(x, y),
∇yψp(x, y) =
sign(y1)|y1|p−1
q
|x1|p+|y1|p − 1
φp(x, y).
(22)
Proof. Fix any (x, y)∈ Rn×Rn. If w(x, y)∈ intKn, the result is implied by Lemma 3.2 since φp(x, y) = z(x, y)− (x + y). In fact, in this case, ψp is continuously differentiable at (x, y). Hence, it suffices to consider the cases w(x, y) = 0 and w(x, y) ∈ bd+Kn. In the following arguments, x and y are arbitrary vectors in
Rn, and μ1(x, y), μ2(x, y) are the spectral values of w(x, y) with ξ(1), ξ(2)∈ Rn being the corresponding spectral vectors.
Case 1. w(x, y) = 0. Note that (x, y) = (0, 0) in this case. Hence, we only need to prove, for any x, y∈ Rn,
ψp(x, y)− ψp(0, 0) = 1
2z(x, y)− (x+ y)2= O((x, y)), (23)
which shows that ψp is differentiable at (0, 0) with ∇xψp(0, 0) = ∇yψp(0, 0) = 0.
Indeed,
z(x, y)− (x+ y) =p
μ1(x, y) ξ(1)+p
μ2(x, y) ξ(2)− (x+ y)
≤√ 2p
μ2(x, y) +x + y.
(24)
From the definition of w1(x, y) and w2(x, y), it is easy to obtain that
μ2(x, y) = w1(x, y) + w2(x, y)≤ |λ2(x)|p+|λ1(x)|p+|λ2(y)|p+|λ1(y)|p. Using the nondecreasing property of √p
t and Lemma 2.3(b), it then follows that
p
μ2(x, y)≤ (|λ2(x)|p+|λ1(x)|p+|λ2(y)|p+|λ1(y)|p)1/p
≤ |λ2(x)| + |λ1(x)| + |λ2(y)| + |λ1(y)| ≤ 2(x + y).
This, together with (24), implies that equation (23) holds.
Case 2. w(x, y)∈ bd+Kn. Now w1(x, y) =w2(x, y) = 0, and one of x2and y2is nonzero by (18). We proceed with the arguments in three steps, as shown below.
Step 1. We prove that w1(x, y) and w2(x, y) are p times differentiable at (x, y) = (x, y), wherep denotes the maximum integer not greater than p. Since one of x2 and y2 is nonzero, we prove this result by considering three possible cases: (i) x2 = 0, y2 = 0; (ii) x2= 0, y2 = 0; and (iii) x2 = 0, y2= 0. For case (i), since xx2
2, yy2
2, λ2(x), λ1(x), λ2(y), and λ1(y) are infinite times differentiable at (x, y), and|t|pisp times continuously differentiable in R, it follows that w1(x, y) and w2(x, y) are p times differentiable at (x, y). Now assume that case (ii) is satisfied. From the arguments in case (i), we know that
|λ2(y)|p+|λ1(y)|p
2 and |λ2(y)|p− |λ1(y)|p 2
y2
y2
are p times differentiable at (x, y). In addition, since |λi(x)|p ≤ 2p2xp for i = 1, 2, and x = 0 in this case, we have that |λ2(x)|p +|λ1(x)|p and
1
2(|λ2(x)|p− |λ1(x)|p)x2 are p times differentiable at x with the first p − 1 order derivatives being zero. Thus, w1(x, y) and w2(x, y) arep times differen- tiable at (x, y). By the symmetry of x, y in w(x, y) and the arguments in case (ii), the result also holds for case (iii).
Step 2. We show that ψp is differentiable at (x, y). By the definition of ψp, we have 2ψp(x, y) =x+ y2+z(x, y)2− 2z(x, y), x+ y.
Sincex+ y2is differentiable, it suffices to argue that the last two terms on the right-hand side are differentiable at (x, y). By formulas (8)-(9), it is not hard to
calculate that
2z(x, y)2= (μ2(x, y))2p+ (μ1(x, y))2p, (25)
2z(x, y), x+ y =p
μ2(x, y)
x1+ y1+(w2(x, y))T(x2+ y2)
w2(x, y)
+p
μ1(x, y)
x1+ y1 −(w2(x, y))T(x2+ y2)
w2(x, y)
. (26)
Since w2(x, y) = 0, μ2(x, y) = λ2(w) > 0, and w1(x, y) and w2(x, y) are differ- entiable at (x, y), by Step 1 we have that (μ2(x, y))2p and the first term on the right-hand side of (26) is differentiable at (x, y). Thus, it suffices to prove that (μ1(x, y))2p and the last term on the right-hand side of (26) are differentiable at (x, y).
We first argue that (μ1(x, y))2p is differentiable at (x, y). Since w2(x, y) = 0, and w1(x, y) and w2(x, y) are p times differentiable at (x, y) by Step 1, the function μ1(x, y) is p times differentiable at (x, y). When p < 2, by the mean-value theorem and μ1(x, y) = λ1(w) = 0, it follows that μ1(x, y) = O(x − x + y − y) for any (x, y) sufficiently close to (x, y), and therefore (μ1(x, y))2p = O[(x− x + y− y)2p]. This shows that (μ1(x, y))2p is differ- entiable at (x, y) with zero derivative. When p ≥ 2, μ1(x, y) is infinite times differentiable at (x, y), and its first derivative equals zero by the result in the Ap- pendix. From the second-order Taylor expansion of μ1(x, y) at (x, y), it follows that (μ1(x, y))2p = O[(x− x + y − y)p4]. This implies that (μ1(x, y))2p is differentiable at (x, y) with zero gradient when 2 ≤ p < 4. Thus, we prove that (μ1(x, y))2p is differentiable at (x, y) with zero gradient when p∈ (1, 4).
We next consider the last term on the right-hand side of (26). Observe that
x1+ y1 −(w2(x, y))T(x2+ y2)
w2(x, y)
is differentiable at (x, y), and its function value at (x, y) equals zero by (18).
Hence, this term is O(x − x + y − y), which, along with μ1(x, y) = O(x − x + y − y), means that the last term of (26) is O((x − x+
y− y)1+1p) = o(x − x + y − y). This shows that the last term of (26) is differentiable at (x, y) with zero derivative.
Step 3. We derive the formula of∇xψp(x, y). From Step 2, we see that 2∇ψp(x, y) equals the difference between the gradient of 12(μ2(x, y))2p+x+ y2and that of the first term on the right-hand side of (26), evaluated at (x, y). By the Appendix, the gradients of (μ2(x, y))1/p and (μ2(x, y))2/p with respect to x, evaluated at (x, y) = (x, y), are
∇x(μ2(x, y))1/p|(x,y)=(x,y)= (λ2(w))1p−12p−1 sign(x1)|x1|p−1
1 w2
, (27)
∇x(μ2(x, y))2/p|(x,y)=(x,y)= (λ2(w))2p−12p sign(x1)|x1|p−1
1 w2
. (28)
