The convex and monotone functions associated with second-order cone

Vol. 55, No. 4, August 2006, 363–385

The convex and monotone functions associated with second-order cone

JEIN-SHAN CHEN*

Department of Mathematics, National Taiwan Normal University, Taipei 11677, Taiwan

(Received 2 December 2004; in final form 21 April 2006)

Like the matrix-valued functions used in solutions methods for semidefinite programs (SDPs) and semidefinite complementarity problems (SDCPs), the vector-valued functions associated with second-order cones are defined analogously and also used in solutions methods for second-order-cone programs (SOCPs) and second-order-cone complementarity problems (SOCCPs). In this article, we study further about these vector-valued functions associated with second-order cones (SOCs). In particular, we define the so-called SOC-convex and SOC-monotone functions for any given functionf:R!R. We discuss the SOC-convexity and SOC-monotonicity for some simple functions, e.g., fðtÞ ¼t², t³, 1=t, t¹⁼², jtj, and [t]_þ. Some characterizations of SOC-convex and SOC-monotone functions are studied, and some conjectures about the relationship between SOC-convex and SOC-monotone functions are proposed.

Keywords: Second-order cone; Convex function; Monotone function; Complementarity;

Spectral decomposition

Mathematics Subject Classifications 2000: 26A27; 26B05; 26B35; 49J52; 90C33

1. Introduction

The second-order cone (SOC) inRⁿ, also called the Lorentz cone, is defined by Kⁿ¼ ðx ₁,x₂Þ 2RRⁿ¹j kx₂k x₁

, ð1Þ

wherekk denotes the Euclidean norm. Ifn¼1, let Kⁿ denote the set of nonnegative reals R_þ. For any x, y in Rⁿ, we write xKⁿy if xy2 Kⁿ; and write xKⁿ y if xy2intðKⁿÞ. In other words, we have xKⁿ 0 if and only if x2 Kⁿ, and xKⁿ0 if and only ifx2intðKⁿÞ. The relationKⁿis a partial ordering but not a linear ordering

*Tel. 886-2-29325417. Fax: 886-2-29332342. Email: jschen@math.ntnu.edu.tw

Optimization

ISSN 0233-1934 print/ISSN 1029-4945 onlineß2006 Taylor & Francis http://www.tandf.co.uk/journals

DOI: 10.1080/02331930600819514

inKⁿ, i.e., there existx,y2 Kⁿ such that neitherx_Kⁿy nory_Kⁿx. To see this, for n¼2, letx¼(1, 1),y¼(1, 0). Then, we have xy¼ ð0, 1Þ2 K= ⁿ,yx¼ ð0,1Þ2 K= ⁿ.

Recently, the SOC has received much attention in optimization, particularly in the context of applications and solutions methods for the second-order-cone program (SOCP) [14] and second-order-cone complementarity problem (SOCCP) [5–8].

For those solutions methods, spectral decompositionassociated with SOC is required.

The basic concepts are as follows. For any x¼ ðx₁,x₂Þ 2RRⁿ¹, x can be decomposed as

x¼1u^ð1Þþ2u^ð2Þ, ð2Þ

where1,2andu⁽¹⁾,u⁽²⁾are the spectral values and the associated spectral vectors ofx given by

_i¼x₁þ ð1Þⁱkx₂k, ð3Þ

u^ðiÞ¼ 1

2 1,ð1Þⁱ x2

kx2k

, ifx2¼= 0, 1

21,ð1Þⁱw

, ifx2¼0, 8>

<

>: ð4Þ

for i¼1, 2 with w being any vector in Rⁿ¹ satisfying kwk ¼1. If x2¼= 0, the decomposition is unique.

For any functionf:R!R, the following vector-valued function associated withKⁿ (n1) was considered [8,10]:

f^socðxÞ ¼fð1Þu^ð1Þþfð2Þu^ð2Þ, 8x¼ ðx1,x2Þ 2RRⁿ¹: ð5Þ Iffis defined only on a subset ofR, thenf^socis defined on the corresponding subset of Rⁿ. The definition (5) is unambiguous whether x2¼= 0 or x2¼0. The cases of f^soc(x)¼x^1/2,x², exp(x) are discussed in [9]. In fact, the equation (5) is analogous to one associated with the semidefinite coneSⁿ_þ [19,21].

In this article, we further define the so-called SOC-convex and SOC-monotone functions (section 3), which are parallel to matrix-convex and matrix-monotone functions [2,11]. We study the SOC-convexity and SOC-monotinicity for some simple functions, e.g.,f(t)¼t²,t³, 1/t,t^1/2,jtj, and [t]þ. Then, we explore the characterizations of SOC-convex and SOC-monotone functions. In addition, we state some conjectures about the relationship between SOC-convex and SOC-monotone functions. It is our intention to extend the existing properties of matrix-convex and matrix-monotone functions shown as in [2,11]. As will be seen in section 3, the vector-valued functions associated with SOC are accompanied by the Jordan product (will be defined in section 2). However, unlike matrix multiplication, the Jordan product associated with SOC is not associative, which is the main source of difficulty when we do the extension.

Therefore, the ideas for proofs are usually quite different from those for matrix-valued functions. The vector-valued functions associated with SOC are heavily used in the solutions methods for SOCP and SOCCP. Therefore, further study on these functions will be helpful for developing and analyzing more solutions methods. That is one of the main motivations for this article.

In what follows and throughout the article,h,idenotes the Euclidean inner product and kk the Euclidean norm. The notation ‘‘:¼’’ means ‘‘define’’. For any

f:Rⁿ!R, rfðxÞ, denotes the gradient of f and x. For any differentiable mapping F¼ ðF1,F2,. . .,FmÞ^T:Rⁿ!R^m, rFðxÞ ¼ ½rF1ðxÞ rFmðxÞ, is a nm matrix denotes the transposed Jacobian ofF at x. For any symmetric matrices A,B2Rⁿⁿ, we write AB (respectively, AB) to mean AB is positive semidefinite (respectively, positive definite). We also use p.s.d. (respectively, p.d.) to represent the abbreviation of positive semidefinite (respectively, positive definite).

2. Jordan product and related properties

For any x¼ ðx1,x2Þ 2RRⁿ¹ and y¼ ðy1,y2Þ 2RRⁿ¹, we define their Jordan productas

x y¼ ðx^Ty, y1x2þx1y2Þ: ð6Þ We writex²to meanx xand writexþyto mean the usual componentwise addition of vectors. Then , þ, together with e¼ ð1, 0,. . ., 0Þ^T2Rⁿ have the following basic properties [9,10]: (1) e x¼x, for all x2Rⁿ, (2) x y¼y x, for all x,y2Rⁿ, (3) x ðx² yÞ ¼x² ðx yÞ, for allx,y2Rⁿ and (4)ðxþyÞ z¼x zþy z, for all x,y,z2Rⁿ. The Jordan product is not associative. For example, for n¼3, let x¼(1,1, 1) and y¼z¼(1, 0, 1), then we have ðx yÞ z¼ ð4, 1, 4Þ¼= x ðy zÞ ¼ ð4, 2, 4Þ. However, it is power associative, i.e., x ðx xÞ ¼ ðx xÞ x for all x2Rⁿ. Thus, we may, without fear of ambiguity, writex^mfor the product of mcopies of xand x^mþn¼x^m xⁿ for all positive integersmand n. We define x⁰¼e.

Besides, Kⁿ is not closed under Jordan product. For example, x¼ ð ffiffiffi p2

, 1, 1Þ 2 K³, y¼ ð ffiffiffi

p2

, 1, 1Þ 2 K³, butx y¼ ð2, 2 ffiffiffi p2

, 0Þ2 K= ³.

For eachx¼ ðx1,x2Þ 2RRⁿ¹, thedeterminantand thetraceof xare defined by detðxÞ ¼x²₁ kx2k², trðxÞ ¼2x1:

In general, detðx yÞ¼= detðxÞdetðyÞunlessx2¼y2. A vectorx¼ ðx1,x2Þ 2RRⁿ¹is said to be invertible if det(x)¼= 0. If x is invertible, then there exists a unique y¼ ðy1,y2Þ 2RRⁿ¹ satisfyingx y¼y x¼e. We call thisy the inverse ofxand denote it byx¹. In fact, we have

x¹¼ 1

x²₁ kx₂k²ðx1, x2Þ ¼ 1

detðxÞðtrðxÞexÞ:

Therefore, x2intðKⁿÞ if and only if x¹2intðKⁿÞ. Moreover, if x2intðKⁿÞ, then x^k¼(x^k)¹is also well-defined. For anyx2 Kⁿ, it is known that there exists a unique vector inKⁿ denoted byx^1/2such thatðx¹⁼²Þ²¼x¹⁼² x¹⁼² ¼x. Indeed,

x¹⁼²¼ s,x2

2s

, wheres¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1

2 x₁þ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x²₁þ kx₂k²

q

s

:

In the preceding formula, the term x2/s is defined to be the zero vector if x2¼0 ands¼0, i.e.,x¼0.

For anyx2Rⁿ, we always havex²2 Kⁿ (i.e.,x²_Kⁿ0). Hence, there exist a unique vectorðx²Þ¹⁼² 2 Kⁿdenoted byjxj. It is easy to verify thatjxj _Kⁿ0 andx²¼ jxj²for any x2Rⁿ. It is also known thatjxj _Kⁿ x. For anyx2Rⁿ, we define [x]þto be the nearest point (in Euclidean norm, since Jordan product does not induce a norm) projection ofx

onto Kⁿ, which is the same definition as in Rⁿ

þ. In other words, [x]þis the optimal solution of the parametric SOCP:½x_þ¼arg minfkxyk jy2 Kⁿg. It is well-known that

½x_þ¼1=2ðxþ jxjÞ; see Property 2.2(f).

Next, for anyx¼ ðx1,x2Þ 2RRⁿ¹, we define a linear mapping fromRⁿ toRⁿ as L_x:Rⁿ!Rⁿ

y!Lxy:¼ x1 x^T₂ x₂ x₁I

y:

It can be easily verified thatx y¼Lxy,8y2Rⁿ, andLxis positive definite (and hence invertible) if and only if x2intðKⁿÞ. However, L¹_x y¼= x¹ y, for some x2intðKⁿÞ andy2Rⁿ, i.e.,L¹_x ¼= L_x¹.

The spectral decomposition along with the Jordan algebra associated with SOC entail some basic properties as listed in the following text. We omit the proofs since they can be found in [9,10].

Property 2.1 For any x¼ ðx₁,x₂Þ 2RRⁿ¹ with the spectral values 1, 2 and spectral vectorsu⁽¹⁾,u⁽²⁾given as in equations (3)–(4), we have

(a) u⁽¹⁾andu⁽²⁾are orthogonal under Jordan product and have length 1= ffiffiffi p2

, i.e., u^ð1Þ u^ð2Þ¼0, ku^ð1Þk ¼ ku^ð2Þk ¼ 1

ffiffiffi2 p : (b) u⁽¹⁾andu⁽²⁾are idempotent under Jordan product, i.e.,

u^ðiÞ u^ðiÞ ¼u^ðiÞ, i¼1, 2:

(c) 1,2are nonnegative (positive) if and only ifx2 Kⁿðx2intðKⁿÞÞ, i.e., _i0, 8i¼1, 2()xKⁿ 0:

i>0, 8i¼1, 2()x_Kⁿ 0:

(d) The determinant, the trace and the Euclidean norm of x can all be represented in terms of1,2:

detðxÞ ¼12, trðxÞ ¼1þ2, kxk² ¼1

2²₁þ²₂ :

Property 2.2 For any x¼ ðx1,x2Þ 2RRⁿ¹ with the spectral values 1, 2 and spectral vectorsu⁽¹⁾,u⁽²⁾given as in equations (3)–(4), we have

(a) x²¼²₁u^ð1Þþ²₂u^ð2Þ. (b) Ifx2 Kⁿ, thenx¹⁼²¼ ffiffiffiffiffi

1

p u^ð1Þþ ffiffiffiffiffi 2

p u^ð2Þ. (c) jxj ¼ j1ju^ð1Þþ j2ju^ð2Þ.

(d) ½x_þ¼ ½1_þu^ð1Þþ ½2_þu^ð2Þ, ½x¼ ½1u^ð1Þþ ½2u^ð2Þ. (e) jxj ¼ ½x_þþ ½x_þ¼ ½x_þ ½x.

(f) ½x_þ¼1=2ðxþ jxjÞ, ½x¼1=2ðx jxjÞ.

Property 2.3

(a) Any x2Rⁿ satisfiesjxj _Kⁿx.

(b) For anyx,y_Kⁿ0, ifx_Kⁿy, thenx¹⁼²_Kⁿy¹⁼².

(c) For anyx,y2Rⁿ, ifx²_Kⁿy², thenjxj Kⁿjyj.

(d) For anyx2Rⁿ,x_Kⁿ 0, hx,yi 0,8y_Kⁿ0.

(e) For anyx_Kⁿ0 and y2Rⁿ,x²_Kⁿy²)x_Kⁿy.

In the following propositions, we study and explore more characterizations about spectral values, determinant, and trace ofx as well as the partial order _Kⁿ. In fact, Propositions 2.1–2.4 are parallel results analogous to those associated with positive semidefinite cone [11]. Even though bothKⁿandSⁿbelong to self-dual cones and share similar properties, as we will see, the ideas for proving these results are quite different.

One reason is that the Jordan product is not associative as mentioned earlier.

PROPOSITION 2.1 For any x_Kⁿ0and y_Kⁿ0,the following results hold.

(a) If x_Kⁿy, thendetðxÞ detðyÞ, trðxÞ trðyÞ.

(b) If x_Kⁿy, theniðxÞ iðyÞ,8i¼1, 2.

Proof (a) From definition, we know that

detðxÞ ¼x²₁ kx2k², trðxÞ ¼2x₁, detðyÞ ¼y²₁ ky2k², trðyÞ ¼2y1:

Since xy¼ ðx1y1,x2y2Þ Kⁿ0, we have kx2y2k x1y1. Thus, x1y1, and then tr(x)tr(y). Besides, the assumption on xandygives

x1y1 kx2y2k kx 2k ky2k, ð7Þ which is equivalent tox₁ kx₂k y₁ ky₂k>0 andx₁þ kx₂k y₁þ ky₂k>0. Hence,

detðxÞ ¼x²₁ kx2k² ¼ ðx1þ kx2kÞðx1 kx2kÞ ðy1þ ky2kÞðy1 ky2kÞ ¼detðyÞ:

(b) From definition of spectral values, we know that

1ðxÞ ¼x1 kx2k, 2ðxÞ ¼x1þ kx2k and 1ðyÞ ¼y1 ky2k, 2ðyÞ ¼y1þ ky2k:

Then, by the inequality (7) in the proof of part (a), the results follow immediately. g PROPOSITION 2.2 For any x_Kⁿ0and y_Kⁿ0,we have

(a) detðxþyÞ detðxÞ þdetðyÞ.

(b) detðx yÞ detðxÞ detðyÞ.

(c) detðxþ ð1ÞyÞ ²detðxÞ þ ð1Þ²detðyÞ, 80< <1:

(d) ðdetðeþxÞÞ¹⁼²1þdetðxÞ¹⁼², 8xKⁿ0:

(e) detðeþxþyÞ detðeþxÞ detðeþyÞ.

Proof

(a) For anyx_Kⁿ0 and y_Kⁿ0, we knowkx2k x1andky2k y1, which implies jhx2,y2ij kx2k ky2k x1y1:

Hence, we obtain

detðxþyÞ ¼ ðx1þy1Þ² kx2þy2k²

¼ ðx²₁ kx₂k²Þ þ ðy²₁ ky₂k²Þ þ2ðx₁y₁ hx₂,y₂iÞ ðx²₁ kx2k²Þ þ ðy²₁ ky2k²Þ

¼detðxÞ þdetðyÞ:

(b) Applying the Cauchy inequality gives detðx yÞ ¼ hx,yi² kx1y2þy1x2k²

¼ðx1y1þ hx2,y2iÞ²x²₁ky2k²þ2x₁y1hx2,y2i þy²₁kx2k²

¼x²₁y²₁þ hx2,y2i²x²₁ky2k²y²₁kx2k² x²₁y²₁þ kx2k² ky2k²x²₁ky2k²y²₁kx2k²

¼ ðx²₁ kx₂k²Þ ðy²₁ ky₂k²Þ

¼detðxÞ detðyÞ:

(c) For anyx_Kⁿ0 andy_Kⁿ 0, it is clear thatx_Kⁿ 0 andð1Þy_Kⁿ0 for every 0 << 1. In addition, we observe that detðxÞ ¼²detðxÞ, for all> 0. Hence,

detðxþ ð1ÞyÞ detðxÞ þdetðð1ÞyÞ ¼²detðxÞ þ ð1Þ²detðyÞ, where the inequality is from part (a).

(d) For anyx_Kⁿ0, we know detðxÞ ¼120, whereiare the spectral values ofx.

Hence, detðeþxÞ ¼ ð1þ1Þð1þ2Þ ð1þ ffiffiffiffiffiffiffiffiffiffi 12

p Þ²¼ ð1þdetðxÞ¹⁼²Þ². Then, taking square root both sides yields the desired result.

(e) Again, For anyxKⁿ0 and yKⁿ 0, we have x1 kx2k 0, y₁ ky₂k 0,

jhx2,y2ij kx2k ky2k x1y1: 8>

<

>: ð8Þ

Also, we know detðeþxþyÞ ¼ ð1þx1þy1Þ² kx2þy2k², detðeþxÞ ¼ ð1þx1Þ² kx2k² and detðeþyÞ ¼ ð1þy1Þ² ky2k². Hence,

detðeþxÞ detðeþyÞ detðeþxþyÞ

¼ ð1 þx1Þ² kx2k²

ð1þy1Þ² ky2k²

ð1 þx₁þy₁Þ² kx₂þy₂k²

¼2x₁y₁þ2hx₂,y₂i þ2x₁y²₁þ2x²₁y₁2y₁kx₂k²2x₁ky₂k² þx²₁y²₁y²₁kx₂k²x²₁ky₂k²þ kx₂k² ky₂k²

¼2ðx1y1þ hx2,y2iÞ þ2x₁ðy²₁ ky2k²Þ þ2y₁ðx²₁ kx2k²Þ þ ðx²₁ kx2k²Þðy²₁ ky2k²Þ

0,

where we multiply out all the expansions to obtain the second equality and the last

inequality holds by (8). g

PROPOSITION 2.3 For any x,y2Rⁿ,we have (a) trðxþyÞ ¼trðxÞ þtrðyÞ.

(b) ₁ðxÞ₂ðyÞ þ₁ðyÞ₂ðxÞ trðx yÞ ₁ðxÞ₁ðyÞ þ₂ðxÞ₂ðyÞ.

(c) trðxþ ð1ÞyÞ ¼trðxÞ þ ð1Þ trðyÞ,82R.

Proof Parts (a) and (c) are trival. Thus, it remains to verify (b). Using the fact that trðx yÞ ¼2hx,yi, we obtain

₁ðxÞ₂ðyÞ þ₁ðyÞ₂ðxÞ ¼ ðx₁ kx₂kÞðy₁þ ky₂k_þðx₁þ kx₂kÞðy₁ ky₂kÞ

¼2ðx1y1 kx2kky2kÞ 2ðx1y1þ hx2,y2i

¼2hx,yi ¼trðx yÞ 2ðx1y1þ kx2kky2kÞ

¼ ðx1 kx2kÞðy1 ky2kÞ þ ðx1þ kx2kÞðy1þ ky2kÞ,

which completes the proof. g

The following two lemmas are well-known results in matrix analysis and are key to proving Proposition 2.4, which is an important extension about the function ln det() from positive semidefinite cone to SOC.

LEMMA 2.1 For any nonzero vector x2Rⁿ, the matrix xx^T is positive semidefinite (p.s.d.)with only one nonzero eigenvaluekxk².

Proof The proof is routine, hence we omit it. g

LEMMA 2.2 Suppose that a symmetric matrix is partitioned as

A B

B^T C

,

where A and C are square. Then this matrix is positive definite (p.d.)if and only if A is positive definite and CB^TA¹B.

Proof See Theorem 7.7.6 in [11]. g

PROPOSITION 2.4 For any x_Kⁿ0and y_Kⁿ0,we have

(a) The real-valued function fðxÞ ¼lnðdetðxÞÞis concave on intðKⁿÞ.

(b) detðxþ ð1ÞyÞ ðdetðxÞÞðdetðyÞÞ¹, 80< <1:

(c) The function real-valued f(x)¼ln(det(x¹)is convex on intðKⁿÞ.

(d) The real-valued function f(x)¼tr(x1) is convex onintðKⁿÞ.

Proof

(a) Since intðKⁿÞ is a convex set, it is enough to show that r²fðxÞ is negative semidefinite. From direct computation, we know

rfðxÞ ¼ 2x1

x²₁ kx₂k², 2x2

x²₁ kx₂k²

!

¼2x¹,

and

r²fðxÞ ¼

2x²₁2kx₂k² ðx²₁ kx2k²Þ²

4x₁x^T₂ ðx²₁ kx2k²Þ² 4x₁x₂

ðx²₁ kx₂k²Þ²

2ðx²₁ kx₂k²ÞI4x₂x^T₂ ðx²₁ kx₂k²Þ² 2

66 64

3 77 75

¼ 2

ðx²₁ kx2k²Þ²

ðx²₁þ kx2k²Þ² 2x1x^T₂

2x₁x₂ ðx²₁ kx₂k²ÞIþ2x₂x^T₂

" #

:

Letr²fðxÞbe denoted by the matrix

A B

B^T C

given as in Lemma 2.2 (hereAis a scalar). Then, we have ACB^TB¼ ðx²₁þ kx2k²Þ ðx ²₁ kx2k²ÞIþ2x2x^T₂

4x²₁x2x^T₂

¼ ðx⁴₁ kx₂k⁴ÞI2ðx²₁ kx₂k²Þx₂x^T₂

¼ ðx²₁ kx2k²Þ ðx ²₁þ kx2k²ÞI2x2x^T₂

¼ ðx²₁ kx2k²Þ M,

were we denote the whole matrix in the big parenthesis of the last second equality byM. From Lemma 2.1, we know that x₂x^T₂ is a p.s.d. with only one nonzero eigenvalue kx2k². Hence, all the eigenvalues of the matrix M are ðx²₁þ kx2k²Þ 2kx2k²¼ x²₁ kx2k² and x²₁þ kx2k² with multiplicity of n2, which are all positive. Thus,M is positive definite which implies that r²fðxÞ is negative definite and hence negative semidefinite.

(b) From part (a), for all 0 << 1, we have

ln detðxð þ ð1ÞyÞÞ ln detðxÞð Þ þ ð1Þln detðyÞð Þ

¼ln detðxÞð Þþln detðyÞð Þ¹

¼ln detðxÞð ÞðdetðyÞÞ¹:

Since natural logarithm is an increasing function, the desired result follows.

(c) We observe that det(x¹)¼1/det(x), for all x2intðKⁿÞ. Therefore, ln detðx¹Þ ¼ ln detðxÞis a convex function by part (a).

(d) The idea for proving this is the same as the one for part (a). Since intðKⁿÞ is a convex set, it is enough to show that r²f is positive semidefinite. Note that fðxÞ ¼trðx¹Þ ¼2x1=ðx²₁ kx2k²Þ. Thus, from direct computations, we have

r²fðxÞ ¼ 2 ðx²₁ kx2k²Þ³

2x³₁þ6x1kx2k², ð6x²₁þ2kx2k²Þx^T₂ ð6x²₁þ2kx₂k²Þx₂, 2x₁ðx²₁ kx₂k²ÞIþ4x₂x^T₂

" #

:

Again, letr²fðxÞbe denoted by the matrix

A B

B^T C

given as in Lemma 2.2 (hereAis a scalar). Then, we have ACB^TB¼2x12x³₁þ6x1kx2k²

ðx²₁ kx2k²ÞIþ4x2x^T₂

6x²₁þ2kx2k²2

x2x^T₂

¼4x⁴₁þ12x²₁kx₂k²

x²₁ kx₂k²

I20x⁴₁24x²₁kx₂k²þ4kx₂k⁴ x₂x^T₂

¼4x⁴₁þ12x²₁kx2k²

x²₁ kx2k²

I4 5x ²₁ kx2k²

x²₁ kx2k²

x2x^T₂

¼x²₁ kx₂k²

4x⁴₁þ12x²₁kx₂k²

I4 5x ²₁ kx₂k² x₂x^T₂

¼x²₁ kx2k² M,

where we denote the whole matrix in the big parenthesis of the last second equality byM. From Lemma 2.1, we know that x2x^T₂ is a p.s.d. with only one nonzero eigenvalue kx2k². Hence, all the eigenvalues of the matrix M are ð4x⁴₁þ 12x²₁kxk²20x²₁kx₂k²þ4kx₂k⁴Þ and 4x⁴₁þ12x²₁kx₂k² with multiplicity of n2, which are all positive since

4x⁴₁þ12x²₁kx2k²20x²₁kx2k²þ4kx2k⁴¼4x⁴₁8x²₁kx2k²þ4kx2k⁴

¼4x²₁ kx₂k²

>0:

Thus, by Lemma 2.2, we obtain that r²fðxÞ is positive definite and hence is positive semidefinite. Therefore,fis convex on int(Kⁿ). g

3. SOC-convex function and SOC-monotone function

In this section, we define the SOC-convexity and SOC-monotonicity and the study some examples of such functions.

Definition 3.1 Let f:R!R. Then

(a) f is said to be SOC-monotone of order n if the corresponding vector-valued functionf^socsatisfies the following:

xKⁿy)f^socðxÞ Kⁿ f^socðyÞ:

(b) fis said to be SOC-convex of ordernif the corresponding vector-valued function f^socsatisfies the following:

f^socðð1ÞxþyÞ_Kⁿð1Þf^socðxÞ þf^socðyÞ, ð9Þ for allx,y2Rⁿ and 01.

We say f is SOC-monotone (respectively, SOC-convex) if f is SOC-monotone of all ordern(respectively, SOC-convex of all ordern). Iffis continuous, then the condition in equation (9) can be replaced by the more special condition:

f^soc xþy 2

_Kⁿ1

2ðf^socðxÞ þf^socðyÞÞ: ð10Þ It is clear that the set of SOC-monotone functions and the set of SOC-convex functions are both closed under positive linear combinations and under pointwise limits.

PROPOSITION 3.1 Let f:R!Rbe f(t)¼þt,then (a) f is SOC-monotone on Rfor every2Rand0.

(b) f is SOC-convex onRfor all,2R.

Proof The proof is straightforward by checking that Definition 3.1 is satisfied. g PROPOSITION 3.2

(a) Let f:R!Rbe f(t)¼t²,then f is SOC-convex onR.

(b) Hence, the function gðtÞ ¼þtþt² is SOC-convex on R for all ,2R and 0.

Proof

(a) For anyx,y2Rⁿ, we have

1

2ðfðxÞ þfðxÞÞ f xþy 2

¼x²þy²

2 xþy

2

²

¼1

4ðxyÞ²Kⁿ0:

Sincefis continuous, the above implies thatfis SOC-convex.

(b) This is an immediate consequence of (a). g

Example 3.1 The functionf(t)¼t²is not SOC-monotone on R.

To see this, let x¼(1, 0), y¼(2, 0), then xy¼ ð3, 0Þ Kⁿ0. But, x²y²¼ ð1, 0Þ ð4, 0Þ ¼ ð3, 0Þ 6_Kⁿ0.

It is clear that f(t)¼t² is also SOC-convex on the smaller interval [0,1) by Proposition 3.2(a). We may ask a natural question: Isf(t)¼t²SOC-monotone on the interval [0,1)? The answer is: it is true only for n¼2 and is, false for general n3.

We show this in the following example.

Example 3.2

(a) The function f(t)¼t²is SOC-monotone on [0,1) forn¼2.

(b) However,f(t)¼t²is not SOC-monotone on [0,1) forn3.

(a) Let x¼ ðx1,x2Þ _K2 y¼ ðy1,y2Þ _K2 0. Then we have the following inequalities:

jx2j x1, jy2j y1, jx2y2j x1y1,

which implies

x1x2y1y20, x₁þx₂y₁þy₂0:

8<

: ð11Þ

We want to prove that fðxÞ fðyÞ ¼ ðx²₁þx²₂y²₁y²₂, 2x₁x₂2y₁y₂Þ _K20, which is enough to verify thatx²₁þx²₂y²₁y²₂ j2x₁x₂2y₁y₂j. This can been

seen by

x²₁þx²₂y²₁y²₂ j2x1x22y1y2j

¼ x²₁þx²₂y²₁y²₂ ð2x1x22y1y2Þ if x1x2y1y20 x²₁þx²₂y²₁y²₂ ð2y1y22x1x2Þ if x1x2y1y20 (

¼ ðx1x2Þ² ðy1y2Þ² ifx1x2y1y20 ðx₁þx₂Þ² ðy₁þy₂Þ² ifx₁x₂y₂y₂0 (

0,

where the inequalities are true due to the inequalities (11).

(b) Forn3, we give a counterexample to show thatf(t)¼t²is not SOC-monotone on the interval [0,1). Let x¼ ð3, 1,2Þ 2 K³ and y¼ ð1, 1, 0Þ 2 K³. It is clear that xy¼ ð2, 0,2Þ _K³0. But x²y²¼ ð14, 6, 12Þ ð2, 2, 0Þ ¼ ð12, 4, 12Þ 6_K3 0.

Now we look at the function f(t)¼t³. As expected,f(t)¼t³is not SOC-convex.

However, it is true thatf(t)¼t³is SOC-convex on [0, 1) forn¼2, whereas it is false forn3. Besides, we will sef(t)¼t³is neither SOC-monotone onRnor SOC-monotone on the interval [0,1) in general. Nonetheless, it is true that it is SOC-monotone on the interval [0,1) forn¼2. The following two examples show what we have just said.

Example 3.3

(a) The function f(t)¼t³is not SOC-convex onR.

(b) Moreover,f(t)¼t³is not SOC-convex on [0,1) forn3.

(c) However,f(t)¼t³is SOC-convex on [0,1) forn¼2.

To see (a), let x¼(0,2), y¼(1, 0). It can be verified that 1=2ðfðxÞ þfðyÞÞ fððxþyÞ=2Þ ¼ 9=8,ð 9=4Þ 6_K20, which says f(t)¼t³ is not SOC-convex onR.

To see (b), let x¼(2, 1,1), y¼ ð1, 1, 0Þ _K30, then we have 1=2ðfðxÞ þ fðyÞÞ fðxþy=2Þ ¼ ð3, 1,3Þ 6_K30, which implies f(t)¼t³ is not even SOC-convex on the interval [0,1).

To see (c), it is enough to show that fððxþyÞ=2Þ _K²1=2ðfðxÞ þfðyÞÞ for any x,yinK² 0. Letx¼ ðx₁,x₂Þ _K² 0 andy¼ ðy₁,y₂Þ _K²0, then we have

x³ ¼ ðx³₁þ3x1x²₂, 3x²₁x2þx³₂Þ, y³¼ ðy³₁þ3y1y²₂, 3y²₁y2þy³₂Þ, (

which yields f xþy

2

¼1

8ðx1þy2Þ³þ3ðx1þy1Þðx2þy2Þ², 3ðx1þy1Þ²ðx2þy2Þ þ ðx2þy2Þ³ , 1

2ðfðxÞ þfðyÞÞ ¼1

2 x³₁þy³₁þ3x₁x²₂þ3y₁y²₂,x³₂þy³₂þ3x²₁x2þ3y²₁y2

: 8>

<

>:

After simplifications, we denote 1=2ðfðxÞ þfðyÞÞ fððxþyÞ=2Þ:¼1=8ð₁,₂Þ, where ₁¼4x³₁þ4y³₁þ12x₁x²₂þ12y₁y³₂ ðx₁þy₁Þ³3ðx₁þy₁Þðx₂þy₂Þ², ₂¼4x³₂þ4y³₂þ12x²₁x₂þ12y²₁y₂ ðx₂þy₂Þ³3ðx₁þy₁Þ²ðx₂þy₂Þ:

(

To show that₁ j₂jwe discuss two cases. First, if ₂0, then

1 j2j ¼ ð4x³₁þ12x1x²₂12x²₁x24x³₂Þ þ ð4y³₁þ12y1y²₂12y²₁y24y³₂Þ ðx 1þy1Þ³þ3ðx₁þy1Þðx2þy2Þ²3ðx₁þy1Þ²ðx2þy2Þ ðx2þy2Þ³

¼4ðx₁x₂Þ³þ4ðy₁y₂Þ³ ðxð ₁þy₁Þ ðx₂þy₂ÞÞ³

¼4ðx₁x₂Þ³þ4ðy₁y₂Þ³ ðxð ₁x₂Þ þ ðy₁y₂ÞÞ³

¼3ðx₁x₂Þ³þ3ðy₁y₂Þ³3ðx₁x₂Þ²ðy₁y₂Þ 3ðx₁x₂Þðy₁y₂Þ²

¼3ððx₁x₂Þ þ ðy₁y₂ÞÞ ðx ₁x₂Þ² ðx₁x₂Þðy₁y₂Þ þ ðy₁y₂Þ² 3ðx1x2Þðy1y2Þ ðxð 1x2Þ þ ðy1y2ÞÞ

¼3ððx1x2Þ þ ðy1y2ÞÞ ðxð 1x2Þ ðy1y2ÞÞ² 0,

where the inequality is true sincex,y2 K². Similarly, if20, we also have, 1 j2j ¼ ð4x³₁þ12x1x²₂þ12x²₁x2þ4x³₂Þ þ ð4y³₁þ12y1y²₂þ12y²₁y2þ4y³₂Þ

ðx 1þy1Þ³þ3ðx1þy1Þðx2þy2Þ²þ3ðx1þy1Þ²ðx2þy2Þ þ ðx2þy2Þ³

¼4ðx1þx2Þ³þ4ðy1þy2Þ³ ðxð 1þy1Þ þ ðx2þy2ÞÞ³

¼4ðx1þx2Þ³þ4ðy1þy2Þ³ ðxð 1þx2Þ þ ðy1þy2ÞÞ³

¼3ðx1þx2Þ³þ3ðy1þy2Þ³3ðx1þx2Þ²ðy1þy2Þ 3ðx1þx2Þðy1þy2Þ²

¼3ððx1þx2Þ þ ðy1þy2ÞÞ ðx 1þx2Þ² ðx1þx2Þðy1þy2Þ þ ðy1þy2Þ² 3ðx1þx2Þðy1þy2Þ ðxð 1þx2Þ þ ðy1þy2ÞÞ

¼3ððx1þx2Þ þ ðy1þy2ÞÞ ðxð 1þx2Þ ðy1þy2ÞÞ² 0,

where the inequality is true since x,y2 K². Thus, we have verified that f(t)¼t³ is SOC-convex on [0,1) forn¼2.

Example 3.4

(a) The function f(t)¼t³is not SOC-monotone onR.

(b) Moreover,f(t)¼t³is not SOC-monotone on [0,1) forn3.

(c) However,f(t)¼t³is SOC-monotone on [0,1) forn¼2.

To see (a) and (b), let x¼ ð2, 1,1Þ _K³0 and y¼ ð1, 1, 0Þ _K³ 0. It is clear that x_K3y. But, we have f(x)¼x³¼(20, 14,14) and f(y)¼y³¼(4, 4, 0), which

gives fðxÞ fðyÞ ¼ ð16, 10, 14Þ 6_K30. Thus, we show that f(t)¼t³ is not even SOC-monotone on the interval [0,1).

To see (c), let x¼ ðx1,x2Þ _K2y¼ ðy1,y2Þ _K2 0. Again, we have the following inequalities:

jx2j x1, jy2j y1, jx2y2j x1y1, which leads to the inequalities (11). In addition, we know

fðxÞ ¼x³ ¼x³₁þ3x₁x²₂, 3x²₁x₂þx³₂ , fðyÞ ¼y³¼y³₁þ3y1y²₂, 3y²₁y2þy³₂

:

We denotefðxÞ fðyÞ:¼ ð1,2Þ, where

₁¼x³₁y³₁þ3x₁x²₂3y₁y²₂, 2¼x³₂y³₂þ3x²₁x23y²₁y2: (

We wish to prove thatfðxÞ fðyÞ ¼x³y³_K2 0, which is enough to show1|2|.

This is true because

x³₁y³₁þ3x1x²₂3y1y²₂ x³₂y³₂þ3x²₁x23y²₁y2

¼ x³₁y³₁þ3x₁x²₂3y₁y²₂ ðx³₂y³₂þ3x²₁x₂3y²₁y₂Þ if₂0 x³₁y³₁þ3x1x²₂3y1y²₂þ ðx³₂y³₂þ3x²₁x23y²₁y2Þ if20 (

¼ ðx1x2Þ³ ðy1y2Þ³ if20 ðx₁þx₂Þ³ ðy₁þy₂Þ³ if₂0 (

0,

where the inequalities are true by the inequalities (11).

Hence, we complete the verification.

Now, we move to another simple functionf(t)¼1/t. We will prove that1/tis SOC- monotone on the interval (0,1) and 1/tis SOC-convex on the interval (0,1) as well.

For the proof, we need the following technical lemmas.

LEMMA 3.1 For any ab> 0and cd> 0,we always have a

b

c d

aþc

bþd ð12Þ

Proof The proof is followed byacðbþdÞ bdðaþcÞ ¼abðcdÞ þcdðabÞ 0. g LEMMA 3.2 For any x¼ ðx₁,x₂Þ, y¼ ðy₁,y₂Þ 2 Kⁿ,we have

(a) ðx₁þy₁Þ² ky₂k²4x₁ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi y²₁ ky₂k²

q .

(b) ðx₁þy₁ ky₂kÞ² 4x₁ðy₁ ky₂kÞ.

(c) ðx1þy1þ ky2kÞ² 4x1ðy1þ ky2kÞ:

(d) x1y1 hx2,y2i

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x²₁ kx₂k² q

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi y²₁ ky₂k² q

. (e) ðx₁þy₁Þ² kx₂þy₂k²4 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

x²₁ kx₂k² q

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi y²₁ ky₂k² q

.

Proof

(a) The proof follows from

ðx1þy1Þ² ky2k² ¼x²₁þ ðy²₁ ky2k²Þ þ2x1y1

2x1

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi y²₁ ky2k² q

þ2x1y1

2x1

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi y²₁ ky2k² q

þ2x1

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi y²₁ ky2k² q

¼4x₁

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi y²₁ ky₂k² q

,

where the first inequality is true due to the fact thataþb2 ffiffiffiffiffi pab

for any positive numbersaandb.

(b) The proof follows from x₁þy₁ ky₂k

ð Þ²4x₁ðy₁ ky₂kÞ ¼x²₁þy²₁þ ky₂k²2x₁y₁2y₁ky₂k þ2x₁ky₂k

¼ðx₁y₁þ ky₂kÞ²0:

(c) Similarly, the proof follows from x1þy1þ ky2k

ð Þ²4x1ðy1þ ky2kÞ ¼x²₁þy²₁þ ky2k²2x1y1þ2y1ky2k 2x1ky2k

¼ðx1y1 ky2kÞ²0:

(d) We know thatx1y1 hx2,y2i x1y1 kx2k ky2k 0, and x1y1 kx2k ky2k

ð Þ²x²₁ kx2k²

y²₁ ky2k²

¼x²₁ky2k²þy²₁kx2k²2x1y1kx2k ky2k

¼ðx1ky2k y1kx2kÞ²0:

Hence, we obtainx₁y₁ hx₂,y₂i x₁y₁ kx₂k ky₂k ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x²₁ kx₂k² q

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi y²₁ ky₂k²

q .

(e) The proof follows from

ðx₁þy₁Þ² kx₂þy₂k²¼x²₁ kx₂k²

þy²₁ ky₂k²

þ2ðx₁y₁ hx₂,y₂iÞ 2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðx²₁ kx₂k²Þðy²₁ ky₂k²Þ q

þ2ðx₁y₁ hx₂,y₂iÞ 2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðx²₁ kx2k²Þðy²₁ ky2k²Þ q

þ2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðx²₁ kx2k²Þðy²₁ ky2k²Þ q

¼4

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðx²₁ kx2k²Þðy²₁ ky2k²Þ q

, where the first inequality is true sinceaþb2 ffiffiffiffiffi

ab

p for all positive a,b and the

second inequality is from part(d). g

PROPOSITION 3.3 Let f:ð0,1Þ ! ð0,1Þbe f(t)¼1/t.Then (a) f is SOC-monotone on(0,1).

(b) f is SOC-convex on(0,1).