• 沒有找到結果。

2 Circular cone convexity

N/A
N/A
Protected

Academic year: 2022

Share "2 Circular cone convexity"

Copied!
20
0
0

加載中.... (立即查看全文)

全文

(1)

to appear in Journal of Inequalities and Applications, 2014

Circular cone convexity and some inequalities associated with circular cones

Jinchuan Zhou 1 Department of Mathematics

School of Science

Shandong University of Technology Zibo 255049, P.R. China E-mail: jinchuanzhou@163.com

Jein-Shan Chen 2 Department of Mathematics National Taiwan Normal University

Taipei 11677, Taiwan E-mail: jschen@math.ntnu.edu.tw

Hao-Feng Hung Department of Mathematics National Taiwan Normal University

Taipei 11677, Taiwan E-mail: 60040016s@ntnu.edu.tw

July 17, 2013

(revised on November 12, 2013)

Abstract. The study of this paper consists of two aspects. One is characterizing the so- called circular cone convexity of f by exploiting the second order differentiability of fLθ; the other is introducing the concepts of determinant and trace associated with circular cone and establishing their basic inequalities. These results show the essential role played by the angle θ, which gives us new insight when looking into properties about circular cone.

1The author’s work is supported by National Natural Science Foundation of China (11101248, 11271233) and Shandong Province Natural Science Foundation (ZR2010AQ026, ZR2012AM016).

2Corresponding author. Member of Mathematics Division, National Center for Theoretical Sciences, Taipei Office. The author’s work is supported by National Science Council of Taiwan.

(2)

Keywords. Circular cone, convexity, determinant, trace.

AMS subject classifications. 26A27, 26B05, 26B35, 49J52, 90C33, 65K05.

1 Introduction

Recently, much attention has been paid to the nonsymmetric cone optimization problems, see [13, 14, 16, 17] and references therein. Unlike symmetric cones [11], there is no unified structure for nonsymmetric cones. Hence, how to tackle with nonsymmetric cone optimization is still an issue. For symmetric cone optimization, the algebraic structure associated with symmetric cones, including second-order cone and positive semi-definite matrix cones, allow us to study them via exploiting the unified Euclidean Jordan algebra [11]. In general, the way to deal with nonsymmetric cone optimization depends on the feature of the associated nonsymmetric cone. In this paper, we focus on a special nonsymmetric cone, circular cone Lθ. The circular cone [10, 18, 19, 20] is a pointed closed convex cone having hyperspherical sections orthogonal to its axis of revolution about which the cone is invariant to rotation. Let its half-aperture angle be θ with θ ∈ (0, 90). Then, it is mathematically expressed as

Lθ := {x = (x1, x2)T ∈ IR × IRn−1| x1 ≥ kxk cos θ}

= {x = (x1, x2)T ∈ IR × IRn−1| x1 ≥ kx2k cot θ}.

Real applications of circular cone lie in some engineering problems, for example, in the formulation for optimal grasping manipulation for multi-fingered robots, the grasping force of i-th finger is subject to a circular cone constraint, see [4, 12] and references for more details.

Although Lθ is a nonsymmetric cone, we can, due to its special structure, establish the explicit form of orthogonal decomposition (or spectral decomposition) [18] as

x = λ1(x) · u(1)x + λ2(x) · u(2)x (1) where

 λ1(x) = x1− kx2k cot θ λ2(x) = x1+ kx2k tan θ and





u(1)x = 1 1 + cot2θ

1 0

0 cot θIn−1

  1

−¯x2



=

 sin2θ

−(sin θ cos θ)¯x2



u(2)x = 1 1 + tan2θ

1 0

0 tan θIn−1

  1

¯ x2



=

 cos2θ (sin θ cos θ)¯x2



with ¯x2 = x2/kx2k if x2 6= 0, and ¯x2 being any vector w in IRn−1 satisfying kwk = 1 if x2 = 0. Clearly, x ∈ Lθ if and only if λ1(x) ≥ 0.

(3)

The formula (1) allows us to define the following vector-valued function

fLθ(x) := f (λ1(x)) u(1)x + f (λ2(x)) u(2)x , (2) where f is a real-valued function from J to IR with J being a subset in IR. Let S be the set of all x ∈ IRn whose spectral values λi(x) for i = 1, 2 belong to J , i.e., S := {x ∈ IRn| λi(x) ∈ J, i = 1, 2}. According to [15], we know S is open if and only if J is open. In addition, as J is an interval, then S is convex because

min{λ1(x), λ1(y)} ≤ λ1(βx+(1−β)y) ≤ λ2(βx+(1−β)y) ≤ max{λ2(x), λ2(y)}, ∀β ∈ [0, 1].

Throughout this paper, we always assume that J is an interval in IR. Clearly, as θ = 45, L45 reduces to the second-order cone and the above expressions (1) and (2) correspond to the spectral decomposition and SOC-function associated with second-order cone, re- spectively (see [6, 9] for more information regarding fsoc).

It is well-known that, to deal with symmetric cone optimization problems, such as second-order cone optimization problems and positive semi-definite optimization prob- lems, this type of vector-valued functions plays an essential role. Inspired by this, we study the properties of fLθ, which is crucial for circular cone optimization problems. In our previous works, we have studied the smooth and nonsmooth analysis of fLθ [4, 19];

and the circular cone monotonicity and second-order differentiability of fLθ [20]. Form the aforementioned research, there has an interesting observation: some properties com- monly shared by fsoc and fLθ are independent of the angle θ; for example, fLθ is direc- tionally differentiable, Fr´echet differentiable, semismooth if and only if f is directionally differentiable, Fr´echet differentiable, semismooth; while some properties are dependent of the angle θ; for example, fLθ with f (t) = −1/t for t > 0 is circular cone monotone as θ ∈ [45, 90), but not circular cone monotone as θ ∈ (0, 45).

In this paper, we further study the circular cone convexity of f . More precisely, a real-valued function f : J → IR is said to be Lθ-convex of order n on S, if for any x, y ∈ S,

fLθ(βx + (1 − β)y) Lθ βfLθ(x) + (1 − β)fLθ(y), ∀β ∈ [0, 1].

The characterization of Lθ-convexity is based on the observation that f is Lθ-convex if and only if (fLθ)00(x)(h, h) ∈ Lθ for all h ∈ IRn. Our result shows that the circular cone convexity requires that the angle θ belongs in [45, 90). In particular, we show that f is Lθ-convex of order 2 if and only if θ ∈ [45, 90) and f is convex.

On the other hand, using the spectral decomposition (1), we define the determinant and trace of x in the framework of circular cone as

det(x) := λ1(x)λ2(x) and tr(x) := λ1(x) + λ2(x),

(4)

respectively. In the symmetric cone setting, the concepts of determinant and trace are the key ingredients of barrier and penalty functions which are used in barrier and penalty methods (including interior point methods) for symmetric cone optimization, see [1, 2, 3].

Here we further study some basic inequalities of det(x) and tr(x) in the framework of circular cone. As seen in Section 3, the obtained inequalities are classified into three categories: (i) the first class is independent of the angle (i.e., still holds in the framework of circular cone); (ii) the second class is dependent of the angle, for example, for x, y ∈ Lθ, the inequality

det(e + x + y) ≤ det(e + x)det(e + y),

where e = (1, 0, . . . , 0) ∈ IRn, fails as θ ∈ (0, 45), but holds as θ ∈ [45, 90); (iii) the third class always fails no matter what value of θ is chosen. These results give us new insight on circular cone and attract us more attention on the role played by the angle θ.

The notation used in this paper is standard. For example, denote by IRn the n- dimensional Euclidean space and by IR+ the set of all nonnegative real scalars, i.e., IR+ = {t ∈ IR| t ≥ 0}. For x, y ∈ IRn, the inner product is denoted by xTy. Let Sn mean the spaces of all real symmetric matrices in IRn×n and by Sn+ the cone of positive semi-definite matrices. We write x Lθ y to stand for x − y ∈ Lθ. Finally, we define

0

0 := 0 for convenience.

2 Circular cone convexity

The main purpose of this section is to provide characterizations of Lθ-convex functions.

First, we need the following technical lemma.

Lemma 2.1. Given αi ∈ IR for i = 1, . . . , 6 and βi ∈ IR for i = 1, 2, 3, we define

F (β1, β2, β3) := α1β14+ α2β34+ α3β12β32+ α4β22β32+ α5β12β22+ α6β1β2β32. (3) If F (β1, β2, β3) ≥ 0 for all (β1, β2, β3) ∈ IR3, then

α1 ≥ 0, α2 ≥ 0, α4 ≥ 0, α5 ≥ 0, α3 ≥ −2√ α1α2. Furthermore, if

α26 ≤ 4α2α5 for α3 ≥ 0 4[α2− (α23/4α1)]α5 for α3 ∈ [−2√

α1α2, 0) (4) then F (β1, β2, β3) ≥ 0 for all (β1, β2, β3) ∈ IR3.

Proof. If β1 = 0, then F (β1, β2, β3) = β322β32+ α4β22]. From F (β1, β2, β3) ≥ 0, we have α2β32+ α4β22 ≥ 0. Thus, α2 ≥ 0 by letting β2 → 0 and α4 ≥ 0 by letting β3 → 0.

(5)

If β3 = 0, then F (β1, β2, β3) = β121β12+ α5β22]. From F (β1, β2, β3) ≥ 0, we obtain α1 ≥ 0 and α5 ≥ 0.

If β2 = 0, then

F (β1, β2, β3) = α1β14+ α2β34+ α3β12β32 = β12β32

 α11

β3)2+ α3+ α23 β1)2



(5) whenever β1 6= 0 and β3 6= 0. Let t = β13. From F (β1, β2, β3) ≥ 0, the equation (5) implies

α3 ≥ −α1t2− α2(1/t2), ∀t 6= 0, i.e.,

α3 ≥ max

t6=0 [−α1t2− α2(1/t2)] = − min

t6=01t2+ α2(1/t2)] = −2√ α1α2. Furthermore, if α3 ≥ 0, then

F (β1, β2, β3) ≥ α2β34 + α5β12β22+ α6β1β2β32 = [β32 β1β2]

 α2 α6/2 α6/2 α5

  β32 β1β2



≥ 0 where the last step is due to

 α2 α6/2 α6/2 α5



S2

+ O

which is ensured by the condition (4). Similarly, if α3 ∈ [−2√

α1α2, 0) (implying α1 6= 0 in this case), then

F (β1, β2, β3)

= √

α1β12+ α3 2√

α1β32

2

+



α2− α231



β34+ α4β22β32+ α5β12β22+ α6β1β2β32



α2− α231



β34+ α5β12β22+ α6β1β2β32

= [β32 β1β2]α2− (α23/4α1) α6/2 α6/2 α5

  β32 β1β2



≥ 0 where the last step is due to

2− (α23/4α1) α6/2 α6/2 α5



S2

+ O

which is ensured by the condition (4) and the fact α2− (α23/4α1) ≥ 0 since −2√

α1α2 ≤ α3 < 0. This completes the proof. 2

Lemma 2.2. [20, Theorem 3.1] Let f : J → IR and fLθ be defined as in (2). Then, fLθ is second order differentiable at x ∈ S if and only if f is second order differentiable at λi(x) ∈ J for i = 1, 2. Moreover, for u, v ∈ IRn, if x2 = 0, then

(fLθ)00(x)(u, v)

(6)

=









f00(x1)

 uTv u1v2+ v1u2



, either u2 = 0 or v2 = 0.

"

f00(x1)uTv

f00(x1)(v1u2+ u1v2) + 12f00(x1)(tan θ − cot θ)

ku2kv2+ ¯uT22kv2ku2

#

, otherwise.

If x2 6= 0, then

(fLθ)00(x)(u, v) = I1 I2



where

I1 := v1u1ξ + ˜˜ %

u1T2v2+ v1T2u2

+ ˜av2Tu2 +

˜ η − ˜a

¯

xT2v2T2u2 I2 := h

˜

η − ˜au1T2v2+ $ − 3 ˜d ¯xT2v2T2u2+ ˜%v1u1+ ˜η − ˜av1T2u2i

¯ x2 + ˜dh

¯

xT2u2v2+ v2Tu22+ ¯xT2v2u2i

+ ˜a u1v2 + v1u2 with

˜

a = f02(x)) − f01(x)) λ2(x) − λ1(x) , ξ =˜ f001(x))

1 + cot2θ + f002(x)) 1 + tan2θ,

˜

% = − cot θ

1 + cot2θf001(x)) + tan θ

1 + tan2θf002(x)),

˜

η = cot2θ

1 + cot2θf001(x)) + tan2θ

1 + tan2θf002(x)), d =˜ 1

kx2k

 cot2θ

1 + cot2θf01(x)) + tan2θ

1 + tan2θf02(x)) − f (λ2(x)) − f (λ1(x)) λ2(x) − λ1(x)

 ,

$ = − cot3θ

1 + cot2θf001(x)) + tan3θ

1 + tan2θf002(x)).

The characterization of Lθ-convexity is established below, which can be regarded as the extension of some results given in [5, 7, 8, 15] from second-order cone setting to circular cone setting.

Theorem 2.1. Suppose that f : J → IR is second order continuously differentiable. If f is Lθ-convex of order n on S, then tan θ ≥ 1, f is convex on J , and for all τ1, τ2 ∈ J with τ1 ≤ τ2,

f002)δ(τ2, τ1) ≥ 2

2− τ1)2δ(τ1, τ2)2 (6) and

h

tan2θδ(τ1, τ2) + (tan2θ − 1)δ(τ2, τ1)i

f001) − 2

2− τ1)2δ(τ2, τ1)2

(7)

≥ −f001) r

(tan2θ − 1)δ(τ2, τ1)h

2 tan2θδ(τ1, τ2) + (tan2θ − 1)δ(τ2, τ1)i

. (7) Furthermore, if

h

tan2θδ(τ1, τ2) + (tan2θ − 1)δ(τ2, τ1)i

f001) ≥ 2

2− τ1)2δ(τ2, τ1)2 (8) and

8 δ(τ2, τ1)δ(τ1, τ2)2 ≤h

2 tan2θδ(τ1, τ2) + (tan2θ − 1)δ(τ2, τ1)i

f001)f002)(τ2− τ1)4 (9) or if

h

tan2θδ(τ1, τ2) + (tan2θ − 1)δ(τ2, τ1) i

f001) < 2

2 − τ1)2δ(τ2, τ1)2 and

8δ(τ1, τ2)2δ(τ2, τ1)2(tan2θ − 1)f001)

≤ ("

(tan2θ − 1)f001)2 h

2 tan2θδ(τ1, τ2) + (tan2θ − 1)δ(τ2, τ1) i

δ(τ2, τ1)

#

(10)

"



tan2θδ(τ1, τ2) + (tan2θ − 1)δ(τ2, τ1)

f001) − 2

2− τ1)2δ(τ2, τ1)2

#2)

f002)(τ2− τ1)4,

then f is Lθ-convex. Here δ(τ, τ0) := f (τ ) − f (τ0) − f00)(τ − τ0) for τ, τ0 ∈ J.

Proof. According to [20, Theorem 3.2], f is Lθ-convex if and only if (fLθ)00(x)(h, h) ∈ Lθ for all x ∈ S and h ∈ IRn. We proceed the proof by considering the following three cases.

Case 1. For x2 = 0 and h2 = 0, it follows from Lemma 2.2 that (fLθ)00(x)(h, h) = f00(x1) h21

0

 . Hence, (fLθ)00(x)(h, h) ∈ Lθ if and only if f00(x1) ≥ 0.

Case 2. For x2 = 0 and h2 6= 0, it follows from Lemma 2.2 that (fLθ)00(x)(h, h) =

 f00(x1)khk2

2f00(x1)h1h2+ f00(x1)(tan θ − cot θ)kh2kh2

 .

Hence, (fLθ)00(x)(h, h) ∈ Lθ if and only if f00(x1) ≥ 0 and tan θkhk2

2h1+ (tan θ − cot θ)kh2k kh2k, i.e.,

− tan θ(h21+ kh2k2) ≤h

2h1+ (tan θ − cot θ)kh2ki

kh2k ≤ tan θ(h21+ kh2k2).

(8)

Dividing by kh2k2 and letting t = h1/kh2k yields

− tan θ(t2+ 1) ≤ 2t + tan θ − cot θ ≤ tan θ(t2+ 1)

⇐⇒ max

t∈IR − tan θ(t2+ 1) − 2t ≤ tan θ − cot θ ≤ min

t∈IRtan θ(t2 + 1) − 2t

⇐⇒ cot θ − tan θ ≤ tan θ − cot θ ≤ tan θ − cot θ

⇐⇒ tan θ ≥ 1.

Case 3. For x2 6= 0, due to the simplification of notation, let us denote µ1 := h1− cot θ¯xT2h2, µ2 := h1+ tan θ¯xT2h2, µ3 :=

q

kh2k2− (¯xT2h2)2. (11) Then,

¯

xT2h2 = µ2− µ1

tan θ + cot θ and h1 = tan θµ1+ cot θµ2

tan θ + cot θ . (12)

Note that µ1, µ2, and µ3 can be taken any value in IR × IR × IR+by taking suitable value of h (because the vector h has n variables). It follows from Lemma 2.2 that

(fLθ)00(x)(h, h) =

ξh˜ 21+ 2 ˜%¯xT2h2h1+ ˜akh2k2+ ˜η − ˜a(¯xT2h2)2 h

$ − 3 ˜d(¯xT2h2)2+ 2 ˜η − ˜a ¯xT2h2h1i

¯ x2+h

˜

%h21+ ˜dkh2k2i

¯ x2 +2h

˜

ah1+ ˜d¯xT2h2i h2

=:

 Θ1

Θ22+ Θ3h2



where

Θ1 = ˜ξh21 + 2 ˜%¯xT2h2h1+ ˜akh2k2+ ˜η − ˜a(¯xT2h2)2

Θ2 = $ − 3 ˜d(¯xT2h2)2+ 2 ˜η − ˜a ¯xT2h2h1 + ˜%h21+ ˜dkh2k2 Θ3 = 2

h

˜

ah1 + ˜d¯xT2h2

i . Hence, (fLθ)00(x)(h, h) ∈ Lθ is equivalent to

Θ1 ≥ 0 and Θ21tan2θ ≥ kΘ22+ Θ3h2k2. Note that

Θ1 = 1

1 + cot2θf001(x))h

h21− 2(¯xT2h2)h1cot θ + (¯xT2h2)2cot2θi

+ 1

1 + tan2θf002(x))h

h21+ 2(¯xT2h2)h1tan θ + (¯xT2h2)2tan2θi + ˜ah

kh2k2− (¯xT2h2)2i

= 1

1 + cot2θf001(x))µ21+ 1

1 + tan2θf002(x))µ22+ ˜aµ23. (13)

(9)

We now claim that Θ1 ≥ 0 for all h ∈ IRn if and only if

f001(x)) ≥ 0, f002(x)) ≥ 0, and ˜a ≥ 0. (14) The sufficiency is clear. Let us show the necessity. In particular, choosing h = (− tan θ, ¯x2) yields µ2 = 0 and µ3 = 0. It then follows from Θ1 ≥ 0 that f001(x)) ≥ 0. If we choose h = (cot θ, ¯x2), then we have f002(x)) ≥ 0. Finally, choosing h = (1, kz2) with k ∈ IR, kz2k = 1 and z2T2 = 0 gives

Θ1 = f001(x))

1 + cot2θ + f002(x))

1 + tan2θ + ˜ak2 ≥ 0.

Dividing by k2 on both sides and taking the limits as k → ∞, we obtain ˜a ≥ 0. Since λi(x) can be taken arbitrary value in J , it is clear that (14) is equivalent to saying that f00(τ ) ≥ 0 for all τ ∈ J , i.e., f is convex on J . Indeed, the condition ˜a ≥ 0 is ensured by the fact that ˜a = f0λ2(x))−f01(x))

2(x)−λ1(x) = f00(t0) ≥ 0 for some t0 ∈ (λ1(x), λ2(x)).

Now we calculate the value of Θ2 and Θ3, respectively.

Θ2

= − cot θ

1 + cot2θf001(x))µ21 + tan θ

1 + tan2θf002(x))µ22+ ˜dµ23− 2( ˜d¯xT2h2+ ˜ah1)(¯xT2h2)

= − cot θ

1 + cot2θf001(x))µ21 + tan θ

1 + tan2θf002(x))µ22+ ˜dµ23− (¯xT2h23. (15) Meanwhile, it follows from (12) that

Θ3 = 2



˜

atan θµ1+ cot θµ2

tan θ + cot θ + ˜d µ2 − µ1 tan θ + cot θ



= 2

tan θ + cot θ h

µ1



˜

a tan θ − ˜d

 + µ2



˜

a cot θ + ˜d

i

. (16)

Note that

22+ Θ3h2k2 = Θ22+ 2Θ2Θ3T2h2+ Θ23kh2k2

= Θ22+ 2Θ2Θ3T2h2+ Θ23 h

µ23+ (¯xT2h2)2 i

= 

Θ2+ Θ3T2h22

+ Θ23µ23. (17)

Putting (13) and (15)-(17) together, the condition Θ21tan2θ ≥ kΘ22 + Θ3h2k2 can be rewritten equivalently as

tan2θ f001(x))

1 + cot2θµ21+ f002(x))

1 + tan2θµ22+ ˜aµ23

2

(10)



− cot θ

1 + cot2θf001(x))µ21+ tan θ

1 + tan2θf002(x))µ22+ ˜dµ23

2

+ 4

(tan θ + cot θ)2 h

µ1



˜

a tan θ − ˜d

 + µ2



˜

a cot θ + ˜d

i2

µ23 i.e.,



tan4θ − 1

f001(x))2µ41+

tan θ + cot θ2

˜

a2tan2θ − ˜d2 µ43 +2h

tan θ + cot θ

˜

a tan3θ + ˜d

f001(x)) − 2

˜

a tan θ − ˜d2i

µ21µ23 (18) +2h

tan θ + cot θ

˜

a tan θ − ˜d

f002(x)) − 2

˜

a cot θ + ˜d2i µ22µ23 +2



tan2θ + 1



f001(x))f002(x))µ21µ22− 8

˜

a tan θ − ˜d



˜

a cot θ + ˜d



µ1µ2µ23 ≥ 0.

To apply Lemma 2.1, we need to compute each coefficient in (18). By calculation, we have

˜

a tan θ − ˜d

= f02(x)) − f01(x))

λ2(x) − λ1(x) tan θ − 1 kx2k

 cot2θ

1 + cot2θf01(x)) + tan2θ

1 + tan2θf02(x))



+ 1 kx2k

f (λ2(x)) − f (λ1(x)) λ2(x) − λ1(x)

= f02(x)) − f01(x))

λ2(x) − λ1(x) tan θ − 1 kx2k

 cot θ

tan θ + cot θf01(x)) + tan θ

tan θ + ctanθf02(x))



+ 1 kx2k

f (λ2(x)) − f (λ1(x)) λ2(x) − λ1(x)

= −tan θ + ctanθ

λ2(x) − λ1(x)f01(x)) + tan θ + ctanθ [λ2(x) − λ1(x)]2

h

f (λ2(x)) − f (λ1(x))i

=

(tan θ + cot θ)h

f (λ2(x)) − f (λ1(x)) − f01(x))(λ2(x) − λ1(x))i [λ2(x) − λ1(x)]2

= tan θ + cot θ

2(x) − λ1(x)]2δ λ2(x), λ1(x),

where the third equation follows from the fact λ2(x) − λ1(x) = (tan θ + ctanθ)kx2k.

Similarly, we have

˜

a tan θ + ˜d

=

(tan θ + cot θ)h

f (λ1(x)) − f (λ2(x)) +

2 tan θ

tan θ+cot θf02(x)) + cot θ−tan θ

tan θ+cot θf01(x))

λ2(x) − λ1(x)i [λ2(x) − λ1(x)]2

= tan θ + cot θ [λ2(x) − λ1(x)]2

 2 tan2θ

tan2+1δ λ1(x), λ2(x) + tan2θ − 1

tan2θ + 1δ λ2(x), λ1(x)

 .

(11)

˜

a cot θ + ˜d =

(tan θ + cot θ)h

f (λ1(x)) − f (λ2(x)) − f02(x))(λ1(x) − λ2(x))i [λ2(x) − λ1(x)]2

= tan θ + cot θ

2(x) − λ1(x)]2δ λ1(x), λ2(x).

˜

a tan3θ + ˜d

=

(tan θ + cot θ)h

f (λ1(x)) − f (λ2(x)) − [tan2θf02(x)) + (1 − tan2θ)f01(x))](λ1(x) − λ2(x))i [λ2(x) − λ1(x)]2

= tan θ + cot θ [λ2(x) − λ1(x)]2

h

tan2θδ λ1(x), λ2(x) + (tan2θ − 1)δ λ2(x), λ1(x)i .

Corresponding each coefficient in (18) to (3), we know









































α1 = (tan4θ − 1)f001(x))2 α2 = (tan θ+cot θ)4

2(x)−λ1(x)]4δ λ2(x), λ1(x)h

2 tan2θ

tan2θ+1δ λ1(x), λ2(x) +tantan22θ−1θ+1δ λ2(x), λ1(x)i α3 = 2(tan θ+cot θ)2

2(x)−λ1(x)]2

( h

tan2θδ λ1(x), λ2(x) + (tan2θ − 1)δ λ2(x), λ1(x)i

f001(x))

−2δ λ2(x),λ1(x)

2

2(x)−λ1(x)]2

)

α4 = 2(tan θ+cot θ)2 2(x)−λ1(x)]2

"

δ λ2(x), λ1(x)f002(x)) − 2δ λ1(x),λ2(x)

2

2(x)−λ1(x)]2

#

α5 = 2(tan2θ + 1)f001(x))f002(x)) α6 = −8(tan θ+cot θ)2

2(x)−λ1(x)]4δ λ1(x), λ2(x)δ λ2(x), λ1(x).

In view of Lemma 2.1, the condition α1 ≥ 0 means tan θ ≥ 1, α2, α5 ≥ 0 is ensured by the convexity of f (see (14)), α4 ≥ 0 corresponds to (6), and α3 ≥ −2√

α1α2 corresponds to (7). In addition, the condition (4) takes the special form (9) and (10), respectively.

2

Theorem 2.2. Suppose that f : J → IR is second order continuously differentiable.

Then, f is Lθ-convex of order 2 on S if and only if tan θ ≥ 1 and f is convex on J . Proof. The necessity is clear from Theorem 2.1. For sufficiency, note that in (11) µ3 = 0 since ¯x2 = ±1 in this case. Hence, (18) takes the form of

(tan4θ − 1)f001(x))2µ41+ 2(tan2θ + 1)f001(x))f002(x))µ21µ22 ≥ 0

(12)

for all µ1 and µ2, which is equivalent to verifying

tan θ ≥ 1 and f001(x))f002(x)) ≥ 0.

This is ensued by the conditions that tan θ ≥ 1 and f is convex on J . Thus, the proof is complete. 2

If, in particular, θ = 45o, then (6) and (7) reduces to [15, (21) in Proposition 4.2]; (9) reduces to [15, (22) in Proposition 4.2]. In addition, due to (7), (8) holds automatically in this case. The above results indicate that the Lθ-convexity is dependent of the properties of f and the angle θ together.

3 Inequalities associated with circular cone

In this section, we establish some inequalities associated with circular cone, which we believe that they will be useful for further analyzing properties of fLθ and proving the convergence of interior point methods for optimization problems involved in circular cones.

In [5], the author establishes the following results in the framework of second-order cone. More specifically, for x L45◦ 0 and y L45◦ 0, then

(a) det(e + x)1/2≥ 1 + det(x)1/2 (b) det(x + y) ≥ det(x) + det(y)

(c) det(αx + (1 − α)y) ≥ α2det(x) + (1 − α)2det(y), ∀α ∈ [0, 1]

(d) det(e + x + y) ≤ det(e + x)det(e + y)

(e) If x L45◦ y L45◦ 0, then det(x) ≥ det(y), tr(x) ≥ tr(y), and λi(x) ≥ λi(y) for i = 1, 2

(f) tr(x + y) = tr(x) + tr(y) and det(γx) = γ2det(x) for all γ ∈ IR.

In the following, we show that, in the framework of circular cone, the above inequali- ties can be classified into three categories. The first class holds independent of the angle, e.g., (a); the second class holds dependent of the angle, e.g., (b)-(e); the third class fails no matter what value of the angle is chosen, e.g., (f).

Theorem 3.1. Let x = (x1, x2) ∈ IR × IRn−1 possess spectral factorization associated with circular cone given as in (1). Then,

(a) [det(e + x)]1/2 ≥ 1 + det(x)1/2 for all x ∈ Lθ

(13)

(b) If x Lθ y, then λ1(x) ≥ λ1(y).

Proof. (a) Note that det(x) ≥ 0 and det(e + x) ≥ 0 since x, x + e ∈ Lθ. Therefore, [det(e + x)]1/2 ≥ 1 + det(x)1/2

⇐⇒ det(e + x) ≥ 1 + 2det(x)1/2+ det(x)

⇐⇒ λ1(e + x)λ2(e + x) ≥ 1 + 2p

λ1(x)λ2(x) + λ1(x)λ2(x)

⇐⇒ (x1 + 1 − kx2k cot θ)(x1+ 1 + kx2k tan θ) ≥ 1 + 2p

λ1(x)λ2(x) + λ1(x)λ2(x)

⇐⇒ (λ1(x) + 1)(λ2(x) + 1) ≥ 1 + 2p

λ1(x)λ2(x) + λ1(x)λ2(x)

⇐⇒ λ1(x)λ2(x) + λ1(x) + λ2(x) + 1 ≥ 1 + 2p

λ1(x)λ2(x) + λ1(x)λ2(x)

⇐⇒ λ1(x) + λ2(x) ≥ 2p

λ1(x)λ2(x)

⇐⇒ λ1(x) + λ2(x)

2 ≥p

λ1(x)λ2(x).

Hence, to prove the desired result, it suffices to show that λ1(x) + λ2(x)

2 ≥p

λ1(x)λ2(x)

which is clearly true by the arithmetic mean-geometric mean (AM-GM) inequality.

(b) Since x − y ∈ Lθ, we know

x1− y1 ≥ kx2− y2k cot θ ≥h

kx2k − ky2ki cot θ, i.e., λ1(x) = x1− kx2k cot θ ≥ y1− ky2k cot θ = λ1(y). 2

Theorem 3.2. Let x = (x1, x2) ∈ IR × IRn−1 possess spectral factorization associated with circular cone given as in (1). Then, the following hold.

(a) For all x, y ∈ Lθ,

det(x + y) ≥ det(x) + det(y) + (kx2k2+ ky2k2) csc2θ − (x21+ y12) sec2θ.

In particular, when θ ∈ (0, 45], we have

det(x + y) ≥ det(x) + det(y). (19)

(b) For all x, y ∈ Lθ and α ∈ [0, 1], det(αx + (1 − α)y)

≥ α2det(x) + (1 − α)2det(y) + (α2kx2k2+ (1 − α)2ky2k2) csc2θ − (α2x21+ (1 − α)2y12) sec2θ.

In particular, when θ ∈ (0, 45], we have

det(αx + (1 − α)y) ≥ α2det(x) + (1 − α)2det(y).

(14)

(c) If x, y ∈ Lθ and θ ∈ [45, 90), then

det(e + x + y) ≤ det(e + x)det(e + y). (20) (d) If x Lθ y Lθ 0 and θ ∈ (0, 45], then

λ2(x) ≥ λ2(y), det(x) ≥ det(y), and tr(x) ≥ tr(y). (21) Proof. (a) Notice that

det(x + y)

= λ1(x + y) · λ2(x + y)

= (x1+ y1− kx2+ y2k cot θ) (x1+ y1+ kx2+ y2k tan θ)

= (x1+ y1)2+ (x1 + y1)kx2+ y2k tan θ − (x1+ y1)kx2+ y2k cot θ − kx2+ y2k2 and

det(x) + det(y)

= λ1(x)λ2(x) + λ1(y)λ2(y)

= (x1− kx2k cot θ)(x1+ kx2k tan θ) + (y1− ky2k cot θ)(y1+ ky2k tan θ)

= x21+ x1kx2k tan θ − x1kx2k cot θ − kx2k2+ y21+ y1ky2k tan θ − y1ky2k cot θ − ky2k2

= x21+ y21 + x1kx2k tan θ + y1ky2k tan θ − x1kx2k cot θ − y1ky2k cot θ − kx2k2− ky2k2. Then, we have

det(x + y) − det(x) − det(y)

= 2x1y1− 2xT2y2+ (x1kx2+ y2k + y1kx2 + y2k − x1kx2k − y1ky2k) tan θ

−(x1kx2 + y2k + y1kx2+ y2k − x1kx2k − y1ky2k) cot θ.

Using x, y ∈ Lθ (and hence x + y ∈ Lθ) gives

x1tan θ ≥ kx2k, −x1tan θ ≤ −kx2k, x1 ≥ kx2k cot θ, −x1 ≤ −kx2k cot θ,

−(x1+ y1) ≤ −kx2+ y2k cot θ.

Thus,

det(x + y) − det(x) − det(y)

≥ 2x1y1− 2xT2y2+ kx2kkx2+ y2k + ky2kkx2+ y2k − x21tan2θ − y21tan2θ

−x1(x1+ y1) − y1(x1+ y1) + kx2k2cot2θ + ky2k2cot2θ

= 2x1y1− 2xT2y2+ kx2+ y2k(kx2k + ky2k) − (x21+ y12) tan2θ

−(x1 + y1)2+ (kx2k2+ ky2k2) cot2θ

(15)

≥ kx2+ y2k2− (x21+ y12) tan2θ − x21− y12− 2xT2y2+ (kx2k2+ ky2k2) cot2θ

= kx2k2+ ky2k2− (x21+ y12) tan2θ − x21− y12+ (kx2k2+ ky2k2) cot2θ

= (kx2k2+ ky2k2)(1 + cot2θ) − (x21+ y12)(1 + tan2θ)

= (kx2k2+ ky2k2) csc2θ − (x21+ y12) sec2θ which is the desired result.

When θ ∈ (0, 45], we know tan θ ≤ cot θ. Since x, y ∈ Lθ, i.e., x1 ≥ kx2k cot θ and y1 ≥ ky2k cot θ, there exist a, b ≥ 0 such that x1 = kx2k cot θ + a and y1 = ky2k cot θ + b.

Hence,

det(x + y) − det(x) − det(y)

= 2x1y1 − 2xT2y2+ (x1kx2+ y2k + y1kx2+ y2k − x1kx2k − y1ky2k) tan θ

−(x1kx2+ y2k + y1kx2 + y2k − x1kx2k − y1ky2k) cot θ

= 

kx2k + ky2kh

kx2k + ky2k − kx2+ y2ki cot2θ +kx2+ y2k

kx2k + ky2k − kx2+ y2k + 2ab +a cot θ

ky2k + kx2k − kx2+ y2k

+ a tan θ

ky2k cot2θ + kx2+ y2k − kx2k +b cot θ

ky2k + kx2k − kx2+ y2k

+ b tan θ

kx2k cot2θ + kx2+ y2k − ky2k

≥ 0,

where the last step is due to kx2k + ky2k ≥ kx2+ y2k, kx2k cot2θ + kx2+ y2k − ky2k ≥ kx2k+kx2+y2k−ky2k ≥ 0, and ky2k cot2θ+kx2+y2k−kx2k ≥ ky2k+kx2+y2k−kx2k ≥ 0 since cot θ ≥ 1, due to θ ∈ (0, 45].

(b) The result follows from the fact that det(γx) = γ2det(x) for all γ ≥ 0.

(c) Since θ ∈ [45, 90), cot θ ≤ 1. For x, y ∈ Lθ, there exists two nonnegative scalars a, b ≥ 0 such that x1 = kx2k cot θ + a and y1 = ky2k cot θ + b. This implies

det(e+x) = (x1+1−kx2k cot θ)(x1+1+kx2k tan θ) = (a+1)(cot θ +tan θ)kx2k+(a+1)2, det(e + y) = (y1+ 1 − ky2k cot θ)(y1+ 1 + ky2k tan θ) = (b + 1)(cot θ + tan θ)ky2k + (b + 1)2. Thus, we obtain

det(e + x)det(e + y)

= (a + 1)(b + 1)(cot θ + tan θ)2kx2kky2k + (a + 1)(b + 1)2(cot θ + tan θ)kx2k +(a + 1)2(b + 1)(cot θ + tan θ)ky2k + (a + 1)2(b + 1)2. (22) On the other hand,

det(e + x + y)

參考文獻

相關文件

It is well known that second-order cone programming can be regarded as a special case of positive semidefinite programming by using the arrow matrix.. This paper further studies

In section 4, based on the cases of circular cone eigenvalue optimization problems, we study the corresponding properties of the solutions for p-order cone eigenvalue

Chen, Conditions for error bounds and bounded level sets of some merit func- tions for the second-order cone complementarity problem, Journal of Optimization Theory and

Chen, Properties of circular cone and spectral factorization associated with circular cone, to appear in Journal of Nonlinear and Convex Analysis, 2013.

According to the authors’ earlier experience on symmetric cone optimization, we believe that spectral decomposition associated with cones, nonsmooth analysis regarding cone-

For circular cone, a special non-symmetric cone, and circular cone optimization, like when dealing with SOCP and SOCCP, the following studies are cru- cial: (i) spectral

Taking second-order cone optimization and complementarity problems for example, there have proposed many ef- fective solution methods, including the interior point methods [1, 2, 3,

For finite-dimensional second-order cone optimization and complementarity problems, there have proposed various methods, including the interior point methods [1, 15, 18], the