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Asia-Pacific Journal of Operational Research, vol.24, pp. 401-420, 2007

On some NCP-functions based on the generalized Fischer-Burmeister function

Jein-Shan Chen 1 Department of Mathematics National Taiwan Normal University

Taipei, Taiwan 11677

June 2, 2005

(revised November 18, 2005) (second revised January 11, 2006)

Abstract In this paper, we study several NCP-functions for the nonlinear comple- mentarity problem (NCP) which are indeed based on the generalized Fischer-Burmeister function, φp(a, b) = k(a, b)kp−(a+b). It is well known that the NCP can be reformulated as an equivalent unconstrained minimization by means of merit functions involving NCP- functions. Thus, we aim to investigate some important properties of these NCP-functions that will be used in solving and analyzing the reformulation of the NCP.

Key words. NCP-function, complementarity, merit function, bounded level sets, sta- tionary point.

1 Introduction

The nonlinear complementarity problem (NCP) [13, 20] is to find a point x ∈ IRn such that

x ≥ 0, F (x) ≥ 0, hx, F (x)i = 0, (1)

where h·, ·i is the Euclidean inner product and F = (F1, F2, · · · , Fn)T maps from IRn to IRn. We assume that F is continuously differentiable throughout this paper. The NCP has attracted much attention due to its various applications in operations research, eco- nomics, and engineering [9, 13, 20].

There have been many methods proposed for solving the NCP [13, 20]. Among which, one of the most popular and powerful approaches that has been studied inten- sively recently is to reformulate the NCP as a system of nonlinear equations [17] or as an

1E-mail: jschen@math.ntnu.edu.tw, TEL: 886-2-29325417, FAX: 886-2-29332342.

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unconstrained minimization problem [8, 10, 14]. Such a function that can constitute an equivalent unconstrained minimization problem for the NCP is called a merit function.

In other words, a merit function is a function whose global minima are coincident with the solutions of the original NCP. For constructing a merit function, the class of func- tions, so-called NCP-functions and defined as below, serves an important role.

A function φ : IR2 → IR is called an NCP-function if it satisfies

φ(a, b) = 0 ⇐⇒ a ≥ 0, b ≥ 0, ab = 0. (2)

Many NCP-functions and merit functions have been explored during the past two decades [6, 16, 25, 26]. Among which, a popular NCP-function intensively studied recently is the well-known Fischer-Burmeister NCP-function [10, 11] defined as

φFB(a, b) =√

a2+ b2− (a + b). (3)

With the above characterization of φFB, the NCP is equivalent to a system of nonsmooth equations:

ΦFB(x) =

φFB(x1 , F1(x))

·

·

·

φFB(xn , Fn(x))

= 0. (4)

For each NCP-function, there is a natural merit function, ΨFB : IRn → IR+ given by ΨFB(x) := 1

2FB(x)k2 = 1 2

Xn

i=1

φFB(xi , Fi(x))2, (5) from which the NCP can be recast as an unconstrained minimization:

x∈IRminnΨFB(x). (6)

In this paper, we are particularly interested in the generalized Fischer-Burmeister function, i.e., φp : IR2 → IR given by

φp(a, b) := k(a, b)kp− (a + b), (7) where p is a positive integer greater than one and k(a, b)kp = qp|a|p+ |b|p means the p- norm of (a, b). Notice that φp reduces to the well known Fischer-Burmeister function φFB when p = 2 and its related properties were recently presented in [3, 4]. Corresponding to φp, we define ψp : IR2 → IR+ by

ψp(a, b) := 1

2p(a, b)|2. (8)

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Then both φp and ψp are NCP-functions and yield a merit function Ψp(x) :=

Xn

i=1

ψp(xi , Fi(x)) = 1 2

Xn

i=1

φp(xi , Fi(x))2, (9) from which the NCP can be reformulated as an unconstrained minimization:

x∈IRminnΨp(x). (10)

However, there has some limitations for the (generalized) Fischer-Burmeister functions and some of its variants when dealing with monotone complementarity problem. In particular, its natural merit function Ψp does not guarantee bounded level sets for this class of problem which is an important class (see page 4 of [1]). Some modifications to the Fischer-Burmeister have been proposed to conquer the above problem, see [16, 25]. In this paper, we extend these modifications to the generalized Fischer-Burmeister function φp. More specifically, we study the following NCP-functions:

φ1(a, b) := φp(a, b) − αa+b+, α > 0, φ2(a, b) := φp(a, b) − α(ab)+, α > 0, φ3(a, b) := qp(a, b)]2+ α(a+b+)2, α > 0, φ4(a, b) := qp(a, b)]2+ α[(ab)+]2, α > 0,

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The function φ1 is called penalized Fischer-Burmeister function when p = 2 and was studied in [1]. The functions φ2, φ3, φ4 generalize the merit functions of p = 2, which were discussed in [25, 27]. Note that for i = 1, 2, 3, 4, we have

φi(a, b) ≡ φp(a, b) (12)

for all (a, b) ∈ N (this notation is used in [25]) where

N := {(a, b)| ab ≤ 0}. (13)

Thus, φi where i = 1, 2, 3, 4 are only different in the first or third quadrant.

Similarly, for each φi there is an associated ψi : IR2 → IR+ given by ψi(a, b) := 1

2i(a, b)|2 i = 1, 2, 3, 4, (14) which is also an NCP-function for every i. Moreover, for φ ∈ {φ1, φ2, φ3, φ4}, we can define

Φ(x) =

φ(x1 , F1(x))

·

·

· φ(xn , Fn(x))

, (15)

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from which the NCP is equivalent to the unconstrained minimization:

x∈IRminnΨ(x) (16)

where

Ψ(x) := 1

2kΦ(x)k2 = 1 2

Xn

i=1

φ(xi , Fi(x))2 (17)

is the natural merit function corresponding to φ ∈ {φ1, φ2, φ3, φ4}.

The paper is organized as follows. In Sec. 2, we review some background definitions including monotonicity, P0-function, semismoothness, e.t.c. and known results about Ψp and its related properties. In Sec. 3, we show that all (φi)2, i ∈ {1, 2, 3, 4} are continu- ously differentiable and investigate properties of the merit function Ψ constructed via φi

with i ∈ {1, 2, 3, 4}. In particular, it provides bounded level sets for a monotone NCP with a strictly feasible point. In addition, we give conditions under which a stationary point of Ψ is a solution of the NCP. In general, the analytic techniques used in this paper are similar to those in [1, 8, 25] since the work is somewhat considered the extensions of NCP-functions studied in those literatures.

Throughout this paper, IRn denotes the space of n-dimensional real column vectors and T denotes transpose. For any differentiable function f : IRn → IR, ∇f (x) denotes the gradient of f at x. For any differentiable mapping F = (F1, ..., Fm)T : IRn → IRm,

∇F (x) = [∇F1(x) · · · ∇Fm(x)] denotes the transpose Jacobian of F at x. We denote by kxkp the p-norm of x and by kxk the Euclidean norm of x. In this whole paper, we assume p is a positive integer greater than one.

2 Preliminaries

In this section, we recall some background concepts and materials which will play an important role in the subsequent analysis. Let F : IRn → IRn. Then, (1) F is mono- tone if hx − y, F (x) − F (y)i ≥ 0, for all x, y ∈ IRn. (2) F is strictly monotone if hx − y, F (x) − F (y)i > 0, for all x, y ∈ IRn and x 6= y. (3) F is strongly monotone with modulus µ > 0 if hx − y, F (x) − F (y)i ≥ µkx − yk2, for all x, y ∈ IRn. (4) F is a P0-function if max

1≤i≤n xi6=yi

(xi− yi)(Fi(x) − Fi(y)) ≥ 0, for all x, y ∈ IRn and x 6= y. (5) F is a P -function if max

1≤i≤n(xi− yi)(Fi(x) − Fi(y)) > 0, for all x, y ∈ IRn and x 6= y. (6) F is a uniform P -function with modulus µ > 0 if max

1≤i≤n(xi− yi)(Fi(x) − Fi(y)) ≥ µkx − yk2, for all x, y ∈ IRn. (7) F is a R0-function if for every sequence {xk} satisfying {kxkk} → ∞, lim inf

k→∞

minixki

kxkk ≥ 0, and lim inf

k→∞

miniFi(xk)

kxkk ≥ 0, there exists an index j such that

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{xkj} → ∞ and {Fj(xk)} → ∞.

It is clear that strongly monotone functions are strictly monotone, and strictly mono- tone functions are monotone. Moreover, F is a P0-function if F is monotone and F is a uniform P -function with modulus µ > 0 if F is strongly monotone with modulus µ > 0. In addition, when F is continuously differentiable, we have the following: (i) F is monotone if and only if ∇F (x) is positive semi-definite for all x ∈ IRn. (ii) F is strictly monotone if ∇F (x) is positive definite for all x ∈ IRn. (iii) F is strongly monotone if and only if ∇F (x) is uniformly positive definite. An R0-function can be viewed as a generalization of a uniform P -function since every uniform P -function is an R0-function (see [2, Prop. 3.11]).

A matrix M ∈ IRn×n is a P0-matrix if every of its principal minors is nonnegative, and it is a P -matrix if every of its principal minors is positive. In addition, it is said to be a R0-matrix if the following system has only zero solution:

x ≥ 0,

Mix = 0 if xi > 0, Mix ≥ 0 if xi = 0,

It is obvious that every P -matrix is also a P0-matrix and it is known that the Jacobian of every continuously differentiable P0-function is a P0-matrix. For more properties about P -matrix and P0-matrix, please refer to [7]. It is also known that F is an R0-function if and only if M is an R0-matrix when F is an affine function, see [2, Prop. 3.10].

Next, we recall the definition of semismoothness. First, we introduce that F is strictly continuous (also called ‘locally Lipschitz continuous’) at x ∈ IRn [24, Chap. 9] if there exist scalars κ > 0 and δ > 0 such that

kF (y) − F (z)k ≤ κky − zk ∀y, z ∈ IRn with ky − xk ≤ δ, kz − xk ≤ δ;

and F is strictly continuous if F is strictly continuous at every x ∈ IRn. If δ can be taken to be ∞, then F is Lipschitz continuous with Lipschitz constant κ. We say F is directionally differentiable at x ∈ IRn if

F0(x; h) := lim

t→0+

F (x + th) − F (x)

t exists ∀h ∈ IRn;

and F is directionally differentiable if F is directionally differentiable at every x ∈ IRn. Assume F : IRn → IRm is strictly continuous. We say F is semismooth at x if F is directionally differentiable at x and, for any V ∈ ∂F (x + h) (the generalized Jacobian), we have

F (x + h) − F (x) − V h = o(khk).

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We say F is ρ-order semismooth at x (0 < ρ < ∞) if F is semismooth at x and, for any V ∈ ∂F (x + h), we have

F (x + h) − F (x) − V h = O(khk1+ρ).

We say F is semismooth (respectively, ρ-order semismooth) if F is semismooth (respec- tively, ρ-order semismooth) at every x ∈ IRk. We say F is strongly semismooth if it is 1-order semismooth. Convex functions and piecewise continuously differentiable func- tions are examples of semismooth functions. Examples of strongly semismooth functions include piecewise linear functions and LC1 functions meaning smooth functions with gradients being locally Lipschitz continuous (strictly continuous) [7, 22]. The composi- tion of two (respectively, ρ-order) semismooth functions is also a (respectively, ρ-order) semismooth function. The property of semismoothness plays an important role in nons- mooth Newton methods [21, 23] as well as in some smoothing methods mentioned in the previous section. For extensive discussions of semismooth functions, see [11, 18, 23].

To end this section, we collect some useful properties of φp, ψp defined as in (7) and (8), respectively, that will be used in the subsequent analysis. All the proofs can be found in [3].

Property 2.1 ([3, Prop. 3.1, Lem. 3.1]) Let φp : IR2 → IR be defined as (7). Then (a) φp is an NCP-function, i.e., it satisfies (2).

(b) φp is sub-additive, i.e., φp(w + w0) ≤ φp(w) + φ(w0) for all w, w0 ∈ IR2. (c) φp is positive homogeneous, i.e., φp(αw) = αφp(w) for all w ∈ IR2 and α ≥ 0.

(d) φp is convex, i.e., φp(αw + (1 − α)w0) ≤ αφp(w) + (1 − α)φp(w0) for all w, w0 ∈ IR2 and α ≥ 0.

(e) φp is Lipschitz continuous with L1 = 1 +

2, i.e., |φp(w) − φp(w0)| ≤ L1kw − w0k;

or with L2 = 1 + 2(1−1/p), i.e., |φp(w) − φp(w0)| ≤ L2kw − w0kp for all w, w0 ∈ IR2. (f) φp is semismooth.

(g) If {(ak, bk)} ⊆ IR2 with (ak → −∞) or (bk→ −∞) or (ak → ∞ and bk → ∞), then we have |φp(ak, bk)| → ∞ for k → ∞.

Property 2.2 ([3, Prop. 3.2]) Let φp, ψp be defined as (7) and (8), respectively. Then (a) ψp is an NCP-function, i.e., it satisfies (2).

(b) ψp(a, b) ≥ 0 for all (a, b) ∈ IR2.

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(c) ψp is continuously differentiable everywhere.

(d) ∇aψp(a, b) · ∇bψp(a, b) ≥ 0 for all (a, b) ∈ IR2. The equality holds ⇐⇒ φp(a, b) = 0.

(e) ∇aψp(a, b) = 0 ⇐⇒ ∇bψp(a, b) = 0 ⇐⇒ φp(a, b) = 0.

From these properties, it was proved in [3] that Ψp(x) ≥ 0 for all x ∈ IRn and Ψp(x) = 0 if and only if x solves the NCP (1), where Ψp : IRn → IR is defined as (9).

Moreover, suppose that the NCP has at least one solution. Then x is a global minimizer of Ψp if and only if x solves the NCP. In addition, it was also shown in [3] that if F is either monotone or P0-function, then every stationary point of Ψp is a global minima of (10); and therefore solves the original NCP. We will investigate the analogous results for the merit function Ψ which is based on φi studied in this paper. On the other hand, as mentioned the natural merit function induced from the generalized Fischer-Burmeister (which behaves like the Fischer-Burmeister function) does not guarantee bounded level sets under the assumption of F being monotone. Instead, there needs that F is strongly monotone or uniform P -function to ensure that the property is held. Another main purpose of this work is to obtain same results for the merit function Ψ studied in this paper under the weaker assumption that F is monotone only (see Sec. 4).

3 Properties of φ and ψ

In this section, we investigate properties of φ ∈ {φ1, φ2, φ3, φ4} and ψ ∈ {ψ1, ψ2, ψ3, ψ4} defined as in (11) and (14), respectively. These include strong semismoothness of φ and continuous differentiability of ψ. First, we denote

Nφ:= {(a, b)| a ≥ 0, b ≥ 0, ab = 0}. (18) This notation is adopted from [1] and it is easy to see that (a, b) ∈ Nφif and only if (a, b) satisfies (2). Now we are ready to show the favorable properties of φ and ψ.

Proposition 3.1 Let φ ∈ {φ1, φ2, φ3, φ4} be defined as in (11). Then (a) φ(a, b) = 0 ⇐⇒ (a, b) ∈ Nφ ⇐⇒ (a, b) satisfies (2).

(b) φ is strongly semismooth.

(c) Let {ak}, {bk} ⊆ IR be any two sequences such that either ak+bk+ → ∞ or ak → −∞

or bk → −∞. Then |φ(ak, bk)| → ∞ for k → ∞.

Proof. (a) It is enough to prove the first equivalence. Suppose φ(a, b) = 0, for i = 2, 3, 4, φi(a, b) = 0 yields φp(a, b) = 0 which says (a, b) ∈ Nφ by Property 2.1(a). For i = 1, φ1(a, b) = 0 implies φp(a, b) = αa+b+. Since α could be any arbitrary positive number,

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the above leads to φp(a, b) = a+b+ = 0 which which says (a, b) ∈ Nφ by Property 2.1(a) again.

On the other hand, suppose (a, b) ∈ Nφ then φp(a, b) = 0 by by Property 2.1(a). Since a ≥ 0, b ≥ 0, we obtain a+b+ = ab = 0. Hence we see that all φi(a, b) = 0, i = 1, 2, 3, 4.

(b) The verification of strong semismoothness of φ is a routine work which can be done as in [25, Lemma 1]. We omit it.

(c) This follows from Property 2.1(g) and definition of (·)+. 2

Proposition 3.2 Let Φ be defined as in (15) with φ ∈ {φ1, φ2, φ3, φ4}. Then (a) Φ is semismooth.

(b) Φ is strongly semismooth if every Fi is LC1 function.

Proof. By using Prop. 3.1(b) and the fact that every LC1 function is strongly semis- mooth, the results follow. 2

The following is a technical lemma which describes the generalized gradients of all φi, i = 1, 2, 3, 4 defined as in (11). It will be used for proving Prop. 3.3.

Lemma 3.1 Let φ1, φ2, φ3, φ4 be defined as (11).

(a) The generalized gradient ∂φ1(a, b) of φ1 at a point (a, b) is equal to the set of all (va, vb) such that

(va, vb) =

µ

ap−1

k(a,b)kp−1p − 1, k(a,b)kbp−1p−1

p − 1

− α

µ

b+∂a+, a+∂b+

, if (a, b) 6= (0, 0) and p is even,

µ

sgn(a)·ap−1

k(a,b)kp−1p − 1, sgn(b)·bk(a,b)kp−1p−1

p − 1

− α

µ

b+∂a+, a+∂b+

, if (a, b) 6= (0, 0) and p is odd, (ξ − 1, ζ − 1), if (a, b) = (0, 0),

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where (ξ, ζ) is any vector satisfying |ξ|p−1p + |ζ|p−1p ≤ 1 and

∂z+ =

1, if z > 0, [0, 1], if z = 0, 0, if z < 0.

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(b) The generalized gradient ∂φ2(a, b) of φ2 at a point (a, b) is equal to the set of all (va, vb) such that

(va, vb) =

µ ap−1 k(a, b)kp−1p

− 1, bp−1 k(a, b)kp−1p

− 1

− α(b, a), if (a, b) 6= (0, 0), ab > 0 and p is even, (va, vb) =

µ ap−1 k(a, b)kp−1p

− 1, bp−1 k(a, b)kp−1p

− 1

− α(b, a) · [0, 1], if (a, b) 6= (0, 0), ab = 0 and p is even,

(va, vb) =

µ ap−1 k(a, b)kp−1p

− 1, bp−1 k(a, b)kp−1p

− 1

, if (a, b) 6= (0, 0), ab < 0 and p is even, (va, vb) =

µsgn(a) · ap−1 k(a, b)kp−1p

− 1, sgn(b) · bp−1 k(a, b)kp−1p

− 1

− α(b, a),

if (a, b) 6= (0, 0), ab > 0 and p is odd, (20) (va, vb) =

µsgn(a) · ap−1 k(a, b)kp−1p

− 1, sgn(b) · bp−1 k(a, b)kp−1p

− 1

− α(b, a) · [0, 1], if (a, b) 6= (0, 0), ab = 0 and p is odd,

(va, vb) =

µsgn(a) · ap−1 k(a, b)kp−1p

− 1, sgn(b) · bp−1 k(a, b)kp−1p

− 1

, if (a, b) 6= (0, 0), ab < 0 and p is odd, (va, vb) = (ξ − 1, ζ − 1) − α(b, a) · [0, 1],

if (a, b) = (0, 0),

where (ξ, ζ) is any vector satisfying |ξ|p−1p + |ζ|p−1p ≤ 1.

(c) φ3 is continuously differentiable everywhere except at (0, 0) with

aφ3(a, b) =

φp(a,b)·

·

ap−1 k(a,b)kp−1

p

−1

¸

+α(a+)(b+)2

φ3(a,b) ,

if (a, b) 6= (0, 0), and p is even,

φp(a,b)·

·

sgn(a)·ap−1 k(a,b)kp−1

p

−1

¸

+α(a+)(b+)2

φ3(a,b) ,

if (a, b) 6= (0, 0), and p is odd,

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bφ3(a, b) =

φp(a,b)·

·

bp−1 k(a,b)kp−1

p

−1

¸

+α(a+)2(b+)

φ3(a,b) ,

if (a, b) 6= (0, 0), and p is even,

φp(a,b)·

·

sgn(b)·bp−1 k(a,b)kp−1

p

−1

¸

+α(a+)2(b+)

φ3(a,b) ,

if (a, b) 6= (0, 0), and p is odd,

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(10)

and ∂φ3(0, 0) = (va, vb) where (va, vb) ∈ (−∞, ∞).

(d) φ4 is continuously differentiable everywhere except at (0, 0) with

aφ4(a, b) =

φp(a,b)·

·

ap−1 k(a,b)kp−1

p

−1

¸

+α(ab)+·b

φ4(a,b) ,

if (a, b) 6= (0, 0), and p is even,

φp(a,b)·

·

sgn(a)·ap−1 k(a,b)kp−1

p

−1

¸

+α(ab)+·b

φ4(a,b) ,

if (a, b) 6= (0, 0), and p is odd,

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bφ4(a, b) =

φp(a,b)·

·

bp−1 k(a,b)kp−1

p

−1

¸

+α(ab)+·a

φ4(a,b) ,

if (a, b) 6= (0, 0), and p is even,

φp(a,b)·

·

sgn(b)·bp−1 k(a,b)kp−1

p

−1

¸

+α(ab)+·a

φ4(a,b) ,

if (a, b) 6= (0, 0), and p is odd,

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and ∂φ4(0, 0) = (va, vb) where (va, vb) ∈ (−∞, ∞).

Proof. (a) First, we note that φp is continuously differentiable everywhere except at (0, 0) (see [3]). Hence, by the Corollary to Prop. 2.2.1 in [5], φp is strictly differentiable everywhere except at the origin. Let φ+(a, b) := a+b+. Then φ+ is strictly differentiable at the origin as proved in [1, Prop. 2.1]. Both φ1 and φ+ are strongly semismooth functions, we know that they are locally Lipschitz (strictly continuous) functions. Thus, the Corollary 2 to Prop. 2.3.3 in [5] yields

∂φ1(a, b) = ∂φp(a, b) − α · ∂φ+(a, b).

On the other hand, the generalized gradient of φp can be verified as below (see [3]):

∂φp(a, b) =

µ ap−1 k(a, b)kp−1p

− 1, bp−1 k(a, b)kp−1p

− 1

, if (a, b) 6= (0, 0) and p is even,

∂φp(a, b) =

µsgn(a) · ap−1 k(a, b)kp−1p

− 1, sgn(b) · bp−1 k(a, b)kp−1p

− 1

, if (a, b) 6= (0, 0) and p is odd,

∂φp(a, b) = (ξ − 1, ζ − 1), if (a, b) = (0, 0), (25)

where (ξ, ζ) is any vector satisfying |ξ|p−1p + |ζ|p−1p ≤ 1. In addition, it was already shown in [1, Prop. 2.1] that

∂φ+(a, b) = (b+∂a+, a+∂b+).

Thus, the desired results follow.

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(b) Following the same arguments as in part(a) and using the fact that

∂(ab)+ =

(b, a), if ab > 0, (0, 0), if ab < 0, (b, a) · [0, 1], if ab = 0, the desired results hold.

(c) It is known that (φp)2 and (a+b+)2 are continuously differentiable. Then the desired results follow by direct computations using the chain rule and the fact that

∂(√ z) =

1 2

z, if z > 0, (−∞, ∞), if z = 0.

(d) Same arguments as part(c). 2

Proposition 3.3 Let ψ ∈ {ψ1, ψ2, ψ3, ψ4} be defined as in (14). Then (a) ψ(a, b) = 0 ⇐⇒ (a, b) ∈ Nφ⇐⇒ (a, b) satisfies (2).

(b) ψ is continuously differentiable on IR2. (c) ∇aψ(a, b) · ∇bψ(a, b) ≥ 0 for all (a, b) ∈ IR2.

(d) ψ(a, b) = 0 ⇐⇒ ∇ψ(a, b) = 0 ⇐⇒ ∇aψ(a, b) = 0 ⇐⇒ ∇bψ(a, b) = 0.

Proof. (a) The proof is straightforward by the same arguments as in Prop. 3.1(a).

(b) The ideas for the proof are indeed borrowed from [8, Prop. 3.4].

For i = 1 and p is even, ψ1(a, b) = 121(a, b))2. By the chain rule (see [5, Theorem 2.2.4]), we obtain ∂ψ1(a, b) = ∂φ1(a, b)Tφ1(a, b). We will show that ∂φ1(a, b)Tφ1(a, b) is single-valued for all (a, b) ∈ IR2 because the zero of φ1 cancels the multi-valued portion of ∂φ1(a, b)T. To see this, we discuss several cases as below.

(i) If a > 0, b > 0, then (b+∂a+, a+∂b+) = (b, a) which is single-valued. Hence, by (19), it is easy to see that ∂φ1(a, b)Tφ1(a, b) is single-valued.

(ii) If a > 0, b < 0, then (b+∂a+, a+∂b+) = (0, a) which is single-valued. Hence, by (19),

∂φ1(a, b)Tφ1(a, b) is single-valued.

(iii) If a > 0, b = 0, then (b+∂a+, a+∂b+) = (0, a · [0, 1]) which is multi-valued. However, under this case, we observe that φ1(a, b) = k(a, b)kp − (a + b) − αa+b+ = 0. Hence,

∂φ1(a, b)Tφ1(a, b) is still single-valued.

(iv) If a < 0, b > 0 or a < 0, b < 0, or a < 0, b = 0, then (b+∂a+, a+∂b+) all equals (0, 0) which is single-valued. Hence, by (19), ∂φ1(a, b)Tφ1(a, b) is single-valued.

(v) If a = 0, b > 0, then (b+∂a+, a+∂b+) = (b · [0, 1], 0) which is multi-valued. However,

(12)

under this case, we observe that φ1(a, b) = k(a, b)kp − (a + b) − αa+b+ = 0. Hence,

∂φ1(a, b)Tφ1(a, b) is still single-valued.

(vi) If a = 0, b < 0, then then (b+∂a+, a+∂b+) = (0, 0) which is single-valued. Hence, by (19), ∂φ1(a, b)Tφ1(a, b) is single-valued.

(vii) If a = 0, b = 0 then φ1(a, b) = 0. Hence, ∂φ1(a, b)Tφ1(a, b) is single-valued.

Thus, by applying the Corollary to Theorem 2.2.4 in [5], the above facts yield that ψ1

is continuously differentiable everywhere. For p is odd, going over the same cases, the proof follows.

For i = 2, ψ2(a, b) = 122(a, b))2. We discuss the following cases : (i) (a, b) 6= (0, 0) and ab > 0, (ii) (a, b) 6= (0, 0) and ab = 0, (iii) (a, b) 6= (0, 0) and ab < 0, (iv) (a, b) = (0, 0).

From (20), we know that ∂φ2(a, b) becomes multi-valued when ab = 0 or (a, b) = (0, 0).

However, φ2(a, b) = 0 under these two cases which implies that ∂φ2(a, b)Tφ2(a, b) is still single-valued. Hence, ψ2 is continuously differentiable everywhere by the Corollary to Theorem 2.2.4 in [5] again.

For i = 3, 4, from (21)-(24), it is trivial that ∂φ3(a, b), ∂φ4(a, b) are single-valued when (a, b) 6= (0, 0). When (a, b) = (0, 0), we observe that φ3(a, b) = φ4(a, b) = 0. Hence,

∂φ3(a, b)Tφ3(a, b) and ∂φ4(a, b)Tφ4(a, b) are still single-valued, which yield that ψ3, ψ4

are continuously differentiable everywhere by the same reason as above.

(c) For i = 1, ψ1 = 121)2, we employ and go over the cases discussed as in part (b).

(i) If a > 0, b > 0, then (b+∂a+, a+∂b+) = (b, a). Hence, from (19), we obtain that

aψ1(a, b) =

à ap−1 k(a, b)kp−1p

− 1 − αb

!

φ1(a, b), ∇bψ1(a, b) =

à bp−1 k(a, b)kp−1p

− 1 − αa

!

φ1(a, b), for both p are even and odd. Then, ∇aψ1(a, b) · ∇bψ1(a, b) equals

µ ap−1 k(a, b)kp−1p

− 1 − αb

¶µ bp−1 k(a, b)kp−1p

− 1 − αa

φ21(a, b).

Since,

¯¯

¯¯ a

p−1

k(a,b)kp−1p

¯¯

¯¯≤ 1,

¯¯

¯¯ b

p−1

k(a,b)kp−1p

¯¯

¯¯≤ 1, and αa > 0, αb > 0, we know

µ ap−1 k(a, b)kp−1p

− 1 − αb

< 0 and

µ bp−1 k(a, b)kp−1p

− 1 − αa

< 0, which implies that ∇aψ1(a, b) · ∇bψ1(a, b) ≥ 0.

(ii) If a > 0, b < 0, then (b+∂a+, a+∂b+) = (0, a). Hence, from (19), we have

aψ1(a, b) =

à ap−1 k(a, b)kp−1p

− 1

!

φ1(a, b), ∇bψ1(a, b) =

à bp−1 k(a, b)kp−1p

− 1 − αa

!

φ1(a, b), for p is even; and

aψ1(a, b) =

à ap−1 k(a, b)kp−1p

− 1

!

φ1(a, b), ∇bψ1(a, b) =

à −bp−1 k(a, b)kp−1p

− 1 − αa

!

φ1(a, b),

(13)

for p is odd. Again, since

¯¯

¯¯ a

p−1

k(a,b)kp−1p

¯¯

¯¯ ≤ 1,

¯¯

¯¯ b

p−1

k(a,b)kp−1p

¯¯

¯¯ ≤ 1, and αa > 0, it can be easily verified that ∇aψ1(a, b) · ∇bψ1(a, b) ≥ 0.

(iii) If a > 0, b = 0, then φ1(a, b) = 0 which says ∇aψ1(a, b) = 0 = ∇bψ1(a, b). Hence,

aψ1(a, b) · ∇bψ1(a, b) = 0.

(iv) If a < 0, b > 0 or a < 0, b < 0, or a < 0, b = 0, then (b+∂a+, a+∂b+) = (0, 0). Hence, from (19), we have

aψ1(a, b) =

à ap−1 k(a, b)kp−1p

− 1

!

φ1(a, b), ∇bψ1(a, b) =

à bp−1 k(a, b)kp−1p

− 1

!

φ1(a, b), for p is even; and

aψ1(a, b) =

Ãsgn(a) · ap−1 k(a, b)kp−1p

− 1

!

φ1(a, b), ∇bψ1(a, b) =

Ãsgn(b) · bp−1 k(a, b)kp−1p

− 1

!

φ1(a, b),

for p is odd. Again, by

¯¯

¯¯ a

p−1

k(a,b)kp−1p

¯¯

¯¯≤ 1, and

¯¯

¯¯ b

p−1

k(a,b)kp−1p

¯¯

¯¯≤ 1, the desired inequality holds.

(v) If a = 0, b > 0, then φ1(a, b) = 0 which says ∇aψ1(a, b) = 0 = ∇bψ1(a, b). Hence,

aψ1(a, b) · ∇bψ1(a, b) = 0.

(vi) If a = 0, b < 0, then then (b+∂a+, a+∂b+) = (0, 0). Hence, from (19), we have

aψ1(a, b) = −φ1(a, b), ∇bψ1(a, b) =

à bp−1 k(a, b)kp−1p

− 1

!

φ1(a, b), for p is even; and

aψ1(a, b) = −φ1(a, b), ∇bψ1(a, b) =

à −bp−1 k(a, b)kp−1p

− 1

!

φ1(a, b),

for p is odd. By the same reasons as in previous discussions, we obtain that ∇aψ1(a, b) ·

bψ1(a, b) ≥ 0.

(vii) If a = 0, b = 0, then φ1(a, b) = 0. Hence, ∇aψ1(a, b) = 0 = ∇bψ1(a, b) and

aψ1(a, b) · ∇bψ1(a, b) = 0.

For i = 2, ψ2 = 122)2, we discuss discuss four cases as in part (b).

(i) If (a, b) 6= (0, 0) and ab > 0, from (20), we have

aψ2(a, b) =

à ap−1 k(a, b)kp−1p

− 1 − αb

!

φ2(a, b), ∇bψ2(a, b) =

à bp−1 k(a, b)kp−1p

− 1 − αa

!

φ2(a, b), for p is even; and

aψ2(a, b) =

Ãsgn(a) · ap−1 k(a, b)kp−1p

− 1 − αb

!

φ2(a, b), ∇bψ2(a, b) =

Ãsgn(b) · bp−1 k(a, b)kp−1p

− 1 − αa

!

φ2(a, b), for p is odd. By the same reasons as in previous discussions, it can be easily verified that

aψ1(a, b) · ∇bψ1(a, b) ≥ 0.

(14)

(ii) If (a, b) 6= (0, 0) and ab = 0, then φ2(a, b) = 0. Hence, ∇aψ2(a, b) = 0 = ∇bψ2(a, b) and ∇aψ2(a, b) · ∇bψ2(a, b) = 0.

(iii) If (a, b) 6= (0, 0) and ab < 0, the arguments are the same as case (iv) for i = 1 except that φ1 is replaced by φ2.

(iv) If (a, b) = (0, 0), then φ2(a, b) = 0. Hence, ∇aψ2(a, b) = 0 = ∇bψ2(a, b) and

aψ2(a, b) · ∇bψ2(a, b) = 0.

For i = 3, ψ3 = 123)2, we have two cases as below.

(i) If (a, b) 6= (0, 0), from (21)-(22), we have

aψ3(a, b) = φp(a, b)

µ ap−1 k(a, b)kp−1p

− 1

+ α(a+)(b+)2,

bψ3(a, b) = φp(a, b)

µ bp−1 k(a, b)kp−1p

− 1

+ α(a+)2(b+), for p is even; and

aψ3(a, b) = φp(a, b)

µsgn(a) · ap−1 k(a, b)kp−1p

− 1

+ α(a+)(b+)2,

bψ3(a, b) = φp(a, b)

µsgn(b) · bp−1 k(a, b)kp−1p

− 1

+ α(a+)2(b+),

for p is odd. Thus, ∇aψ3(a, b) · ∇bψ3(a, b) equals φ2p(a, b)

µ

ap−1 k(a,b)kp−1p − 1

¶µ

bp−1 k(a,b)kp−1p − 1

+ α2(a+)3(b+)3 + φp(a, b)

µ

ap−1 k(a,b)kp−1p − 1

α(a+)2(b+) + φp(a, b)

µ

bp−1 k(a,b)kp−1p − 1

α(a+)(b+)2 or

φ2p(a, b)

µ

sgn(a)·ap−1 k(a,b)kp−1p − 1

¶µ

sgn(b)·bp−1 k(a,b)kp−1p − 1

+ α2(a+)3(b+)3 + φp(a, b)

µ

sgn(a)·ap−1 k(a,b)kp−1p − 1

α(a+)2(b+) + φp(a, b)

µ

sgn(b)·bp−1 k(a,b)kp−1p − 1

α(a+)(b+)2.

Note that in the above expressions, it is trivial that the first and second terms are nonnegative. We also notice that

(a+)(b+) =

( ab, if a > 0, b > 0 0, else.

Therefore, we only need to consider the subcase of a > 0, b > 0 for the third and fourth terms. In fact, summing up the third and fourth term under this subcase gives

αab · φp(a, b)

ap−1 k(a, b)kp−1p

− 1

a +

µ bp−1 k(a, b)kp−1p

− 1

b

#

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