3 SOC-convex function and SOC-monotone function

Optimization, vol. 55, pp. 363-385, 2006

The convex and monotone functions associated with second-order cone

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

Taipei 11677, Taiwan

November 18, 2004 (revised April 2, 2006)

Abstract Like the matrix-valued functions used in solutions methods for semidefinite pro- gram (SDP) and semidefinite complementarity problem (SDCP), the vector-valued func- tions associated with second-order cone are defined analogously and also used in solutions methods for second-order cone program (SOCP) and second-order cone complementarity problem (SOCCP). In this paper, we study further about these vector-valued functions as- sociated with second-order cone. In particular, we define so-called SOC-convex and SOC- monotone functions for any given function f : IR → IR. We discuss the SOC-convexity and SOC-monotonicity for some simple functions, e.g., f (t) = t², t³, 1/t, t^1/2, |t|, and [t]+. 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.

Key words. Second-order cone, convex function, monotone function, complementarity, spectral decomposition

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

1 Introduction

The second-order cone (SOC) in IRⁿ, also called Lorentz cone, is defined by

Kⁿ = {(x₁, x₂) ∈ IR × IRⁿ⁻¹ | kx₂k ≤ x₁}, (1) where k · k denotes the Euclidean norm. If n = 1, let Kⁿ denote the set of nonnegative reals IR₊. For any x, y in IRⁿ, we write x º_Kn y if x − y ∈ Kⁿ; and write x Â_Kn y if

1This work is supported by National Science Council of Taiwan.

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

x − y ∈ int(Kⁿ). In other words, we have x º_Kn 0 if and only if x ∈ Kⁿ and x Â_Kn 0 if and only if x ∈ int(Kⁿ). The relation º_Kn is a partial ordering, but not a linear ordering in Kⁿ, i.e., there exist x, y ∈ Kⁿ such that neither x º_Kn y nor y º_Kn x. To see this, for n = 2, let x = (1, 1) , y = (1, 0). Then we have x − y = (0, 1) /∈ Kⁿ , y − x = (0, −1) /∈ Kⁿ.

Recently, the second-order cone has received much attention in optimization, particu- larly in the context of applications and solutions methods for second-order cone program (SOCP) [14] and second-order cone complementarity problem (SOCCP), [5, 6, 7, 8]. For those solutions methods, there needs spectral decomposition associated with SOC. The basic concepts are as below. For any x = (x₁, x₂) ∈ IR × IRⁿ⁻¹, x can be decomposed as

x = λ₁u⁽¹⁾+ λ₂u⁽²⁾, (2)

where λ1, λ2 and u⁽¹⁾, u⁽²⁾ are the spectral values and the associated spectral vectors of x given by

λ_i = x₁+ (−1)ⁱkx₂k, (3)

u⁽ⁱ⁾ =









1 2

µ

1 , (−1)ⁱ x2

kx₂k

¶

, if x2 6= 0,

1 2

µ

1 , (−1)ⁱw

¶

, if x₂ = 0,

(4)

for i = 1, 2 with w being any vector in IRⁿ⁻¹ satisfying kwk = 1. If x₂ 6= 0, the decomposi- tion is unique.

For any function f : IR → IR, the following vector-valued function associated with Kⁿ (n ≥ 1) was considered [8, 10]:

f^soc(x) = f (λ1)u⁽¹⁾+ f (λ2)u⁽²⁾, ∀x = (x1, x2) ∈ IR × IRⁿ⁻¹. (5) If f is defined only on a subset of IR, then f^soc is defined on the corresponding subset of IRⁿ. The definition (5) is unambiguous whether x₂ 6= 0 or x₂ = 0. The cases of f^soc(x) = x^1/2, x², exp(x) are discussed in the book of [9]. In fact, the above definition (5) is analogous to one associated with the semidefinite cone S₊ⁿ, see [19, 21].

In this paper, we further define so-called SOC-convex and SOC-monotone functions (see Sec. 3) which are parallel to matrix-convex and matrix-monotone functions (see [2, 11]).

We study the SOC-convexity and SOC-monotonicity for some simple functions, e.g., f (t) = t², t³, 1/t, t^1/2, |t|, and [t]₊. Then, we explore 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 intension to extend the existing prop- erties of matrix-convex and matrix-monotone functions shown as in [2, 11]. As will be seen in Sec. 3, the vector-valued functions associated with SOC are accompanied by Jordan product (will be defined in Sec. 2). However, unlike the matrix multiplication, the Jordan

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 paper.

In what follows and throughout the paper, h·, ·i denotes the Euclidean inner product and k·k is the Euclidean norm. The notation ”:=” means ”define”. For any f : IRⁿ→ IR, ∇f (x) denotes the gradient of f at x. For any differentiable mapping F = (F₁, F₂, · · · , F_m)^T : IRⁿ → IR^m, ∇F (x) = [∇F₁(x) · · · ∇F_m(x)] is a n by m matrix which denotes the transpose Jacobian of F at x. For any symmetric matrices A, B ∈ IR^n×n, we write A º B (respec- tively, A Â B) to mean A − B is positive semidefinite (respectively, positive definite).

2 Jordan product and related properties

For any x = (x₁, x₂) ∈ IR × IRⁿ⁻¹ and y = (y₁, y₂) ∈ IR × IRⁿ⁻¹, we define their Jordan product as

x ◦ y = (x^Ty , y₁x₂+ x₁y₂). (6) We write x² to mean x ◦ x and write x + y to mean the usual componentwise addition of vectors. Then ◦, +, together with e = (1, 0, . . . , 0)^T ∈ IRⁿ have the following basic proper- ties (see [9, 10]): (1) e ◦ x = x, for all x ∈ IRⁿ. (2) x ◦ y = y ◦ x, for all x, y ∈ IRⁿ. (3) x ◦ (x²◦ y) = x²◦ (x ◦ y), for all x, y ∈ IRⁿ. (4) (x + y) ◦ z = x ◦ z + y ◦ z, for all x, y, z ∈ IRⁿ. The Jordan product is not associative. For example, for n = 3, let x = (1, −1, 1) and y = z = (1, 0, 1) , then we have (x ◦ y) ◦ z = (4, −1, 4) 6= x ◦ (y ◦ z) = (4, −2, 4) . However, it is power associative, i.e., x◦(x◦x) = (x◦x)◦x , for all x ∈ IRⁿ. Thus, we may, without fear of ambiguity, write x^m for the product of m copies of x and x^m+n = x^m◦ xⁿ for all positive integers m and n. We define x⁰ = e. Besides, Kⁿ is not closed under Jordan product. For example, x = (√

2, 1, 1) ∈ K³, y = (√

2, 1, −1) ∈ K³, but x ◦ y = (2, 2√

2, 0) /∈ K³.

For each x = (x1, x2) ∈ IR × IRⁿ⁻¹, the determinant and the trace of x are defined by det(x) = x²₁− kx₂k² , tr(x) = 2x₁.

In general, det(x ◦ y) 6= det(x)det(y) unless x₂ = y₂. A vector x = (x₁, x₂) ∈ IR × IRⁿ⁻¹ is said to be invertible if det(x) 6= 0. If x is invertible, then there exists a unique y = (y1, y2) ∈ IR × IRⁿ⁻¹ satisfying x ◦ y = y ◦ x = e. We call this y the inverse of x and denote it by x⁻¹. In fact, we have

x⁻¹ = 1

x²₁− kx₂k²(x₁ , −x₂) = 1 det(x)

µ

tr(x)e − x

¶

.

Therefore, x ∈ int(Kⁿ) if and only if x⁻¹ ∈ int(Kⁿ). Moreover, if x ∈ int(Kⁿ), then x^−k = (x^k)⁻¹ is also well-defined. For any x ∈ Kⁿ, it is known that there exists a unique vector in Kⁿ denoted by x^1/2 such that (x^1/2)² = x^1/2◦ x^1/2 = x. Indeed,

x^1/2=

µ

s , x₂ 2s

¶

, where s =

s1 2

µ

x₁+^qx²₁− kx₂k²

¶

.

In the above formula, the term x₂/s is defined to be the zero vector if x₂ = 0 and s = 0, i.e., x = 0 .

For any x ∈ IRⁿ, we always have x² ∈ Kⁿ (i.e., x² º_Kn 0). Hence, there exists a unique vector (x²)^1/2 ∈ Kⁿ denoted by |x|. It is easy to verify that |x| º_Kn 0 and x² = |x|² for any x ∈ IRⁿ. It is also known that |x| º_Kn x. For any x ∈ IRⁿ, we define [x]+ to be the nearest point (in Euclidean norm, since Jordan product does not induce a norm) projection of x onto Kⁿ, which is the same definition as in IRⁿ₊. In other words, [x]₊ is the optimal solution of the parametric SOCP: [x]₊ = argmin{ kx − yk | y ∈ Kⁿ}. It is well-known that [x]+= ¹₂(x + |x|), see Property 2.2(f).

Next, for any x = (x₁, x₂) ∈ IR × IRⁿ⁻¹, we define a linear mapping from IRⁿ to IRⁿ as L_x : IRⁿ −→ IRⁿ

y −→ L_xy :=

"

x1 x^T₂ x₂ x₁I

#

y .

It can be easily verified that x ◦ y = L_xy, ∀y ∈ IRⁿ, and L_x is positive definite (and hence invertible) if and only if x ∈ int(Kⁿ). However, L⁻¹_x y 6= x⁻¹◦ y, for some x ∈ int(Kⁿ) and y ∈ IRⁿ, i.e., L⁻¹_x 6= L_x⁻¹.

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

Property 2.1 For any x = (x₁, x₂) ∈ IR×IRⁿ⁻¹ with the spectral values λ₁, λ₂ and spectral vectors u⁽¹⁾, u⁽²⁾ given as in (3)-(4), we have

(a) u⁽¹⁾ and u⁽²⁾ are orthogonal under Jordan product and have length 1/√

2 , i.e., u⁽¹⁾◦ u⁽²⁾ = 0 , ku⁽¹⁾k = ku⁽²⁾k = 1

√2 .

(b) u⁽¹⁾ and u⁽²⁾ are idempotent under Jordan product, i.e., u⁽ⁱ⁾◦ u⁽ⁱ⁾ = u⁽ⁱ⁾ , i = 1, 2 .

(c) λ₁, λ₂ are nonnegative (positive) if and only if x ∈ Kⁿ (x ∈ int(Kⁿ)) , i.e., λ_i ≥ 0 , ∀i = 1, 2 ⇐⇒ x º_Kn 0 .

λi > 0 , ∀i = 1, 2 ⇐⇒ x Â_Kn 0 .

(d) The determinant, the trace and the Euclidean norm of x can all be represented in terms of λ₁, λ₂ :

det(x) = λ₁λ₂ , tr(x) = λ₁+ λ₂ , kxk² = 1

2(λ²₁+ λ²₂) .

Property 2.2 For any x = (x₁, x₂) ∈ IR×IRⁿ⁻¹ with the spectral values λ₁, λ₂ and spectral vectors u⁽¹⁾, u⁽²⁾ given as in (3)-(4), we have

(a) x² = λ²₁u⁽¹⁾+ λ²₂u⁽²⁾ . (b) If x ∈ Kⁿ , then x^1/2 =√

λ1 u⁽¹⁾+√

λ2 u⁽²⁾ . (c) |x| = |λ₁|u⁽¹⁾+ |λ₂|u⁽²⁾ .

(d) [x]₊ = [λ₁]₊u⁽¹⁾+ [λ₂]₊u⁽²⁾ , [x]₋ = [λ₁]₋u⁽¹⁾+ [λ₂]₋u⁽²⁾ . (e) |x| = [x]₊+ [−x]₊ = [x]₊− [x]₋ .

(f) [x]+ = ¹₂(x + |x|) , [x]− = ¹₂(x − |x|) .

Property 2.3 (a) Any x ∈ IRⁿ satisfies |x| º_Kn x.

(b) For any x, y º_Kn 0, if x º_Kn y, then x^1/2 º_Kn y^1/2. (c) For any x, y ∈ IRⁿ, if x² º_Kn y², then |x| º_Kn |y|.

(d) For any x ∈ IRⁿ, x º_Kn 0 ⇐⇒ hx, yi ≥ 0, ∀y º_Kn 0.

(e) For any x º_Kn 0 and y ∈ IRⁿ, x² º_Kn y² =⇒ x º_Kn y.

In the following propositions, we study and explore more characterizations about spec- tral values, determinant and trace of x as well as the partial order º_Kn. In fact, Prop.

2.1-2.4 are parallel results analogous to those associated with positive semidefinite cone, see [11]. Even though both Kⁿ and Sⁿ belong to self-dual cones and share similar prop- erties, 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 Â_Kn 0 and y Â_Kn 0, the following results hold.

(a) If x º_Kn y, then det(x) ≥ det(y) , tr(x) ≥ tr(y).

(b) If x º_Kn y, then λ_i(x) ≥ λ_i(y) , ∀i = 1, 2.

Proof. (a) From definition, we know that

det(x) = x²₁− kx2k² , tr(x) = 2x1, det(y) = y₁²− ky₂k² , tr(y) = 2y₁.

Since x − y = (x1− y1, x2 − y2) º_Kn 0, we have kx2− y2k ≤ x1 − y1. Thus, x1 ≥ y1, and then tr(x) ≥ tr(y). Besides, the assumption on x and y gives

x1− y1 ≥ kx2− y2k ≥

¯¯

¯¯kx2k − ky2k

¯¯

¯¯, (7)

which is equivalent to x₁− kx₂k ≥ y₁− ky₂k > 0 and x₁+ kx₂k ≥ y₁ + ky₂k > 0. Hence, det(x) = x²₁− kx2k² = (x1+ kx2k)(x1− kx2k) ≥ (y1+ ky2k)(y1− ky2k) = det(y).

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

λ₁(x) = x₁− kx₂k , λ₂(x) = x₁ + kx₂k and λ₁(y) = y₁− ky₂k , λ₂(y) = y₁+ ky₂k . Then, by the inequality (7) in the proof of part (a), the results follow immediately. 2

Proposition 2.2 For any x º_Kn 0 and y º_Kn 0, we have (a) det(x + y) ≥ det(x) + det(y).

(b) det(x ◦ y) ≤ det(x) · det(y).

(c) det

µ

αx + (1 − α)y

¶

≥ α²det(x) + (1 − α)²det(y), ∀ 0 < α < 1.

(d)

µ

det(e + x)

¶_1/2

≥ 1 + det(x)^1/2, ∀x º_Kn 0.

(e) det(e + x + y) ≤ det(e + x) · det(e + y).

Proof. (a) For any x º_Kn 0 and y º_Kn 0, we know kx₂k ≤ x₁ and ky₂k ≤ y₁, which implies

|hx2, y2i| ≤ kx2k · ky2k ≤ x1y1.

Hence, we obtain

det(x + y)

= (x1+ y1)²− kx2+ y2k²

=

µ

x²₁− kx₂k²

¶

+

µ

y₁²− ky₂k²

¶

+ 2

µ

x₁y₁− hx₂, y₂i

¶

≥

µ

x²₁− kx₂k²

¶

+

µ

y₁²− ky₂k²

¶

= det(x) + det(y).

(b) Applying the Cauchy inequality gives det(x ◦ y)

= hx, yi²− kx₁y₂+ y₁x₂k²

=

µ

x₁y₁ + hx₂, y₂i

¶₂

−

µ

x²₁ky₂k² + 2x₁y₁hx₂, y₂i + y₁²kx₂k²

¶

= x²₁y₁²+ hx₂, y₂i²− x²₁ky₂k²− y²₁kx₂k²

≤ x²₁y₁²+ kx2k²· ky2k² − x²₁ky2k²− y²₁kx2k²

=

µ

x²₁− kx2k²

¶

·

µ

y₁²− ky2k²

¶

= det(x) · det(y).

(c) For any x Â_Kn 0 and y Â_Kn 0, it is clear that αx Â_Kn 0 and (1 − α)y Â_Kn 0 for every α > 0. In addition, we observe that det(αx) = α²det(x), for all α > 0. Hence,

det

µ

αx + (1 − α)y

¶

≥ det(αx) + det((1 − α)y) = α²det(x) + (1 − α)²det(y), where the inequality is from part (a).

(d) For any x º_Kn 0, we know det(x) = λ₁λ₂ ≥ 0, where λ_i are the spectral values of x.

Hence, det(e + x) = (1 + λ₁)(1 + λ₂) ≥ (1 +√

λ₁λ₂)² = (1 + det(x)^1/2)². Then, taking square root both sides yields the desired result.

(e) Again, For any x º_Kn 0 and y º_Kn 0, we have







x1− kx2k ≥ 0, y₁− ky₂k ≥ 0,

|hx₂, y₂i| ≤ kx₂k · ky₂k ≤ x₁y₁.

(8)

Also, we know det(e + x + y) = (1 + x₁+ y₁)²− kx₂+ y₂k² , det(e + x) = (1 + x₁)²− kx₂k² and det(e + y) = (1 + y₁)²− ky₂k². Hence,

det(e + x) · det(e + y) − det(e + x + y)

=

µ

(1 + x₁)²− kx₂k²

¶µ

(1 + y₁)²− ky₂k²

¶

−

µ

(1 + x₁+ y₁)²− kx₂+ y₂k²

¶

= 2x₁y₁+ 2hx₂, y₂i + 2x₁y²₁ + 2x²₁y₁− 2y₁kx₂k²− 2x₁ky₂k² +x²₁y²₁− y²₁kx₂k²− x²₁ky₂k²+ kx₂k²· ky₂k²

= 2

µ

x₁y₁+ hx₂, y₂i

¶

+ 2x₁

µ

y₁²− ky₂k²

¶

+ 2y₁

µ

x²₁ − kx₂k²

¶

+

µ

x²₁− kx₂k²

¶µ

y²₁− ky₂k²

¶

≥ 0,

where we multiply out all the expansions to obtain the second equality and the last in- equality holds by (8). 2

Proposition 2.3 For any x, y ∈ IRⁿ, we have (a) tr(x + y) = tr(x) + tr(y).

(b) λ1(x)λ2(y) + λ1(y)λ2(x) ≤ tr(x ◦ y) ≤ λ1(x)λ1(y) + λ2(x)λ2(y).

(c) tr(αx + (1 − α)y) = α · tr(x) + (1 − α) · tr(y), ∀α ∈ IR.

Proof. Part(a) and (c) are trivial. Thus, it remains to verify (b). Using the fact that tr(x ◦ y) = 2hx, yi, we obtain

λ₁(x)λ₂(y) + λ₁(y)λ₂(x)

= (x₁− kx₂k)(y₁+ ky₂k₊(x₁+ kx₂k)(y₁− ky₂k)

= 2(x1y1− kx2kky2k)

≤ 2(x₁y₁+ hx₂, y₂i

= 2hx, yi = tr(x ◦ y)

≤ 2(x₁y₁+ kx₂kky₂k)

= (x1− kx2k)(y1− ky2k) + (x1+ kx2k)(y1+ ky2k), which completes the proof. 2

The following two lemmas are well known results in matrix analysis and are key to proving Prop. 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 x ∈ IRⁿ, the matrix xx^T is positive semidefinite (p.s.d.) with only one nonzero eigenvalue kxk².

Proof. The proof is routine, we omit it. 2

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 C Â B^TA⁻¹B.

Proof. This is Theorem 7.7.6 in [11]. 2

Proposition 2.4 For any x Â_Kn 0 and y Â_Kn 0, we have

(a) the real-valued function f (x) = ln(det(x)) is concave on int(Kⁿ).

(b) det(αx + (1 − α)y) ≥ (det(x))^α(det(y))^1−α, ∀ 0 < α < 1.

(c) the function real-valued f (x) = ln(det(x⁻¹)) is convex on int(Kⁿ).

(d) the real-valued function f (x) = tr(x⁻¹) is convex on int(Kⁿ).

Proof. (a) Since int(Kⁿ) is a convex set, it is enough to show that ∇²f (x) is negative semidefinite. From direct computation, we know

∇f (x) =

µ 2x₁

x²₁− kx2k², −2x₂ x²₁− kx2k²

¶

= 2x⁻¹, and

∇²f (x) =





−2x²₁−2kx2k² (x²₁−kx2k²)²

4x1x^T₂ (x²₁−kx2k²)² 4x1x2

(x²₁−kx2k²)²

−2(x²₁−kx2k²)I−4x2x^T₂ (x²₁−kx2k²)²





= _(x2 ⁻² 1−kx2k²)²

"

(x²₁+ kx₂k²)² −2x₁x^T₂

−2x₁x₂ (x²₁− kx₂k²)I + 2x₂x^T₂

#

.

Let ∇²f (x) be denoted by the matrix

"

A B

B^T C

#

given as in Lemma 2.2 (here A is a scalar). Then, we have

AC − B^TB

= (x²₁+ kx₂k²)

µ

(x²₁− kx₂k²)I + 2x₂x^T₂

¶

− 4x²₁x₂x^T₂

= (x⁴₁− kx₂k⁴)I − 2(x²₁− kx₂k²)x₂x^T₂

= (x²₁− kx₂k²)

µ

(x²₁+ kx₂k²)I − 2x₂x^T₂

¶

= (x²₁− kx₂k²) · M,

where we denote the whole matrix in the big parenthesis of the last second equality by M.

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 (x²₁ + kx₂k²) − 2kx₂k² = x²₁ − kx₂k² and

x²₁+ kx₂k² with multiplicity of n − 2, which are all positive. Thus, M is positive definite which implies that ∇²f (x) is negative definite and hence negative semidefinite.

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

µ

det(αx + (1 − α)y)

¶

≥ α ln(det(x)) + (1 − α) ln(det(y))

= ln

µ

det(x)

¶_α

+ ln

µ

det(y)

¶_1−α

= ln

µ

det(x)

¶_αµ

det(y)

¶_1−α

.

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

(c) We observe that det(x⁻¹) = 1/det(x), for all x ∈ int(Kⁿ). Therefore, ln det(x⁻¹) =

− ln det(x) is a convex function by part(a).

(d) The idea for proving this is the same as the one for part(a). Since int(Kⁿ) is a convex set, it is enough to show that ∇²f is positive semidefinie. Note that f (x) = tr(x⁻¹) =

2x₁

x²₁− kx2k². Thus, from direct computations, we have

∇²f (x) = 2 (x²₁− kx2k²)³



 2x³₁+ 6x₁kx₂k², −(6x²₁+ 2kx₂k²)x^T₂

−(6x²₁+ 2kx2k²)x2, 2x1

µ

(x²₁− kx2k²)I + 4x2x^T₂

¶



.

Again, let ∇²f (x) be denoted by the matrix

"

A B

B^T C

#

given as in Lemma 2.2 (here A is a scalar). Then, we have

AC − B^TB

= 2x1

µ

2x³₁ + 6x1kx2k²

¶µ

(x²₁− kx2k²)I + 4x2x^T₂

¶

−

µ

6x²₁+ 2kx2k²

¶₂

x2x^T₂

=

µ

4x⁴₁+ 12x²₁kx₂k²

¶µ

x²₁− kx₂k²

¶

I−

µ

20x⁴₁− 24x²₁kx₂k² + 4kx₂k⁴

¶

x₂x^T₂

=

µ

4x⁴₁+ 12x²₁kx₂k²

¶µ

x²₁− kx₂k²

¶

I − 4

µ

5x²₁− kx₂k²

¶µ

x²₁− kx₂k²

¶

x₂x^T₂

=

µ

x²₁− kx₂k²

¶ ·µ

4x⁴₁+ 12x²₁kx₂k²

¶

I − 4

µ

5x²₁− kx₂k²

¶

x₂x^T₂

¸

=

µ

x²₁− kx₂k²

¶

· M,

where we denote the whole matrix in the big parenthesis of the last second equality by M.

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 (4x⁴₁+ 12x²₁kx²_k− 20x²₁kx₂k²+ 4kx₂k⁴) and

4x⁴₁+ 12x²₁kx₂k² with multiplicity of n − 2, which are all positive since 4x⁴₁ + 12x²₁kx₂k²− 20x²₁kx₂k² + 4kx₂k⁴

= 4x⁴₁ − 8x²₁kx₂k²+ 4kx₂k⁴

= 4

µ

x²₁ − kx₂k²

¶

> 0.

Thus, by Lemma 2.2, we obtain that ∇²f (x) is positive definite and hence is positive semidefinite. Therefore, f is convex on int(Kⁿ). 2

3 SOC-convex function and SOC-monotone function

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

Definition 3.1 Let f : IR → IR. Then

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

x º_Kn y =⇒ f^soc(x) º_Kn f^soc(y).

(b) f is said to be SOC-convex of order n if the corresponding vector-valued function f^soc satisfies the following:

f^soc

µ

(1 − λ)x + λy

¶

¹_Kn (1 − λ)f^soc(x) + λf^soc(y) , (9) for all x, y ∈ IRⁿ and 0 ≤ λ ≤ 1.

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

f^soc

µx + y 2

¶

¹_Kn 1 2

µ

f^soc(x) + f^soc(y)

¶

. (10)

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 point-wise limits.

Proposition 3.1 Let f : IR → IR be f (t) = α + βt, then

(a) f is SOC-monotone on IR for every α ∈ IR and β ≥ 0.

(b) f is SOC-convex on IR for all α, β ∈ IR.

Proof. The proof is straightforward by checking that definition 3.1 is satisfied. 2

Proposition 3.2 (a) Let f : IR → IR be f (t) = t², then f is SOC-convex on IR.

(b) Hence, the function g(t) = α + βt + γt² is SOC-convex on IR for all α, β ∈ IR and γ ≥ 0.

Proof. (a) For any x, y ∈ IRⁿ, we have 1

2

µ

f (x) + f (y)

¶

− f

µx + y 2

¶

= x²+ y²

2 −

µx + y 2

¶₂

= 1

4(x − y)² º_Kn 0.

Since f is continuous, the above implies that f is SOC-convex.

(b) It’s an immediate consequence of (a). 2

Example 3.1 The function f (t) = t² is not SOC-monotone on IR.

To see this, let x = (1, 0), y = (−2, 0), then x − y = (3, 0) º_Kn 0. But, x² − y² = (1, 0) − (4, 0) = (−3, 0) 6º_Kn 0. 2

It is clear that f (t) = t² is also SOC-convex on the smaller interval [0, ∞) by Prop.

3.2(a). We may ask a natural question: Is f (t) = t² SOC-monotone on the interval [0, ∞)?

The answer is: it’s true only for n = 2, however, false for general n ≥ 3. We show this in the next example.

Example 3.2 (a) The function f (t) = t² is SOC-monotone on [0, ∞) for n = 2.

(b) However, f (t) = t² is not SOC-monotone on [0, ∞) for n ≥ 3.

(a) Let x = (x₁, x₂) º_K² y = (y₁, y₂) º_K² 0. Then we have the following inequalities:

|x2| ≤ x1 , |y2| ≤ y1 , |x2− y2| ≤ x1 − y1,

which implies ₍

x₁− x₂ ≥ y₁− y₂ ≥ 0,

x₁+ x₂ ≥ y₁+ y₂ ≥ 0. (11)

We want to prove that f (x) − f (y) = (x²₁+ x²₂ − y₁²− y₂² , 2x₁x₂ − 2y₁y₂) º_K2 0, which is enough to verify that x²₁+ x²₂ − y₁²− y₂² ≥ |2x₁x₂− 2y₁y₂|. This can been seen by

x²₁+ x²₂− y₁²− y²₂−

¯¯

¯¯2x₁x₂− 2y₁y₂

¯¯

¯¯

=

( x²₁+ x²₂− y₁²− y₂²− (2x1x2 − 2y1y2), if x1x2− y1y2 ≥ 0 x²₁+ x²₂− y₁²− y₂²− (2y₁y₂− 2x₁x₂), if x₁x₂− y₁y₂ ≤ 0

=

( (x₁ − x₂)²− (y₁− y₂)², if x₁x₂− y₁y₂ ≥ 0 (x₁ + x₂)²− (y₁ + y₂)², if x₁x₂− y₁y₂ ≤ 0

≥ 0 ,

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

(b) For n ≥ 3, we give a counterexample to show that f (t) = t² is not SOC-monotone on the interval [0, ∞). Let x = (3, 1, −2) ∈ K³ and y = (1, 1, 0) ∈ K³. It is clear that x − y = (2, 0, −2) º_K3 0. But, x²− y² = (14, 6, −12) − (2, 2, 0) = (12, 4, −12) 6º_K3 0. 2

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

However, it is true that f (t) = t³ is SOC-convex on [0, ∞) for n = 2, whereas false for n ≥ 3. Besides, we will see f (t) = t³ is neither SOC-monotone on IR nor SOC-monotone on the interval [0, ∞) in general. Nonetheless, it is true that it is SOC-monotone on the interval [0, ∞), for n = 2. The following two examples show what we have just said.

Example 3.3 (a) The function f (t) = t³ is not SOC-convex on IR.

(b) Moreover, f (t) = t³ is not SOC-convex on [0, ∞) for n ≥ 3.

(c) However, f (t) = t³ is SOC-convex on [0, ∞) for n = 2.

To see (a), let x = (0, −2), y = (1, 0). It can be verified that ¹₂

µ

f (x) + f (y)

¶

− f

µ

x+y 2

¶ µ =

− ⁹₈, −⁹₄

¶

6º_K2 0, which says f (t) = t³ is not SOC-convex on IR.

To see (b), let x = (2, 1, −1), y = (1, 1, 0) º_K3 0, then we have ¹₂

µ

f (x) + f (y)

¶

− f

µ

x+y 2

¶

= (3, 1, −3) 6º_K3 0, which implies f (t) = t³ is not even SOC-convex on the interval [0, ∞).

To see (c), it is enough to show that f

µx + y 2

¶

¹_K2 1 2

µ

f (x) + f (y)

¶

, for any x, y º_K2 0.

Let x = (x₁, x₂) º_K2 0 and y = (y₁, y₂) º_K2 0, then we have

( x³ = (x³₁+ 3x₁x²₂ , 3x²₁x₂+ x³₂), y³ = (y³₁+ 3y₁y₂² , 3y₁²y₂+ y₂³), which yields









f (^x+y₂ ) = ¹₈

µ

(x₁+ y₁)³+ 3(x₁+ y₁)(x₂+ y₂)² , 3(x₁+ y₁)²(x₂+ y₂) + (x₂+ y₂)³

¶

,

1 2

µ

f (x) + f (y)

¶

= ¹₂

µ

x³₁+ y³₁ + 3x₁x²₂+ 3y₁y²₂ , x³₂+ y³₂+ 3x²₁x₂+ 3y²₁y₂

¶

.