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

### The convex and monotone functions associated with second-order cone

^{1}

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) = t*^{2}*, t*^{3}*, 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 IR* ^{n}*, also called Lorentz cone, is defined by

*K*^{n}*= {(x*_{1}*, x*_{2}*) ∈ IR × IR*^{n−1}*| kx*_{2}*k ≤ x*_{1}*},* (1)
*where k · k denotes the Euclidean norm. If n = 1, let K** ^{n}* denote the set of nonnegative
reals IR

_{+}

*. For any x, y in IR*

^{n}*, we write x º*

_{Kn}*y if x − y ∈ K*

^{n}*; 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(K*^{n}*). In other words, we have x º*_{Kn}*0 if and only if x ∈ K*^{n}*and x Â** _{Kn}* 0 if and

*only if x ∈ int(K*

^{n}

*). The relation º*

_{Kn}*is a partial ordering, but not a linear ordering in K*

*,*

^{n}*i.e., there exist x, y ∈ K*

^{n}*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) /∈ K*

^{n}*, y − x = (0, −1) /∈ K*

*.*

^{n}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 = (x*_{1}*, x*_{2}*) ∈ IR × IR*^{n−1}*, x can be decomposed as*

*x = λ*_{1}*u*^{(1)}*+ λ*_{2}*u*^{(2)}*,* (2)

*where λ*1*, λ*2 *and u*^{(1)}*, u*^{(2)} *are the spectral values and the associated spectral vectors of x*
given by

*λ*_{i}*= x*_{1}*+ (−1)*^{i}*kx*_{2}*k,* (3)

*u** ^{(i)}* =

1 2

µ

*1 , (−1)*^{i}*x*2

*kx*_{2}*k*

¶

*, if x*2 *6= 0,*

1 2

µ

*1 , (−1)*^{i}*w*

¶

*,* *if x*_{2} *= 0,*

(4)

*for i = 1, 2 with w being any vector in IR*^{n−1}*satisfying kwk = 1. If x*_{2} *6= 0, the decomposi-*
tion is unique.

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

*f*^{soc}*(x) = f (λ*1*)u*^{(1)}*+ f (λ*2*)u*^{(2)}*,* *∀x = (x*1*, x*2*) ∈ IR × IR*^{n−1}*.* (5)
*If f is defined only on a subset of IR, then f*^{soc} is defined on the corresponding subset
of IR^{n}*. The definition (5) is unambiguous whether x*_{2} *6= 0 or x*_{2} = 0. The cases of
*f*^{soc}*(x) = x*^{1/2}*, x*^{2}*, 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) =*
*t*^{2}*, t*^{3}*, 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 : IR*^{n}*→ IR, ∇f (x)*
*denotes the gradient of f at x. For any differentiable mapping F = (F*_{1}*, F*_{2}*, · · · , F** _{m}*)

*: IR*

^{T}

^{n}*→ IR*

^{m}*, ∇F (x) = [∇F*

_{1}

*(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 = (x*_{1}*, x*_{2}*) ∈ IR × IR*^{n−1}*and y = (y*_{1}*, y*_{2}*) ∈ IR × IR*^{n−1}*, we define their Jordan*
*product as*

*x ◦ y = (x*^{T}*y , y*_{1}*x*_{2}*+ x*_{1}*y*_{2}*).* (6)
*We write x*^{2} *to mean x ◦ x and write x + y to mean the usual componentwise addition of*
*vectors. Then ◦, +, together with e = (1, 0, . . . , 0)*^{T}*∈ IR** ^{n}* have the following basic proper-

*ties (see [9, 10]): (1) e ◦ x = x, for all x ∈ IR*

^{n}*. (2) x ◦ y = y ◦ x, for all x, y ∈ IR*

*. (3)*

^{n}*x ◦ (x*

^{2}

*◦ y) = x*

^{2}

*◦ (x ◦ y), for all x, y ∈ IR*

^{n}*. (4) (x + y) ◦ z = x ◦ z + y ◦ z, for all x, y, z ∈ IR*

*.*

^{n}*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 ∈ IR*

*. Thus, we may, without fear*

^{n}*of ambiguity, write x*

^{m}*for the product of m copies of x and x*

^{m+n}*= x*

^{m}*◦ x*

*for all positive*

^{n}*integers m and n. We define x*

^{0}

*= e. Besides, K*

*is not closed under Jordan product. For*

^{n}*example, x = (√*

*2, 1, 1) ∈ K*^{3}*, y = (√*

*2, 1, −1) ∈ K*^{3}*, but x ◦ y = (2, 2√*

*2, 0) /∈ K*^{3}.

*For each x = (x*1*, x*2*) ∈ IR × IR*^{n−1}*, the determinant and the trace of x are defined by*
*det(x) = x*^{2}_{1}*− kx*_{2}*k*^{2} *,* *tr(x) = 2x*_{1}*.*

*In general, det(x ◦ y) 6= det(x)det(y) unless x*_{2} *= y*_{2}*. A vector x = (x*_{1}*, x*_{2}*) ∈ IR × IR** ^{n−1}* is

*said to be invertible if det(x) 6= 0. If x is invertible, then there exists a unique y = (y*1

*, y*2

*) ∈*

*IR × IR*

^{n−1}*satisfying x ◦ y = y ◦ x = e. We call this y the inverse of x and denote it by x*

*. In fact, we have*

^{−1}*x** ^{−1}* = 1

*x*^{2}_{1}*− kx*_{2}*k*^{2}*(x*_{1} *, −x*_{2}) = 1
*det(x)*

µ

*tr(x)e − x*

¶

*.*

*Therefore, x ∈ int(K*^{n}*) if and only if x*^{−1}*∈ int(K*^{n}*). Moreover, if x ∈ int(K*^{n}), then
*x*^{−k}*= (x** ^{k}*)

^{−1}*is also well-defined. For any x ∈ K*

*, it is known that there exists a unique*

^{n}*vector in K*

^{n}*denoted by x*

^{1/2}*such that (x*

*)*

^{1/2}^{2}

*= x*

^{1/2}*◦ x*

^{1/2}*= x. Indeed,*

*x** ^{1/2}*=

µ

*s ,* *x*_{2}
*2s*

¶

*,* *where s =*

s1 2

µ

*x*_{1}+^{q}*x*^{2}_{1}*− kx*_{2}*k*^{2}

¶

*.*

*In the above formula, the term x*_{2}*/s is defined to be the zero vector if x*_{2} *= 0 and s = 0,*
*i.e., x = 0 .*

*For any x ∈ IR*^{n}*, we always have x*^{2} *∈ K*^{n}*(i.e., x*^{2} *º** _{Kn}* 0). Hence, there exists a unique

*vector (x*

^{2})

^{1/2}*∈ K*

^{n}*denoted by |x|. It is easy to verify that |x| º*

_{Kn}*0 and x*

^{2}

*= |x|*

^{2}for

*any x ∈ IR*

^{n}*. It is also known that |x| º*

_{Kn}*x. For any x ∈ IR*

^{n}*, we define [x]*+ to be the nearest point (in Euclidean norm, since Jordan product does not induce a norm) projection

*of x onto K*

*, which is the same definition as in IR*

^{n}

^{n}_{+}

*. In other words, [x]*

_{+}is the optimal

*solution of the parametric SOCP: [x]*

_{+}

*= argmin{ kx − yk | y ∈ K*

^{n}*}. It is well-known that*

*[x]*+=

^{1}

_{2}

*(x + |x|), see Property 2.2(f).*

*Next, for any x = (x*_{1}*, x*_{2}*) ∈ IR × IR** ^{n−1}*, we define a linear mapping from IR

*to IR*

^{n}*as*

^{n}*L*

*: IR*

_{x}

^{n}*−→ IR*

^{n}*y* *−→ L*_{x}*y :=*

"

*x*1 *x*^{T}_{2}
*x*_{2} *x*_{1}*I*

#

*y .*

*It can be easily verified that x ◦ y = L*_{x}*y, ∀y ∈ IR*^{n}*, and L** _{x}* is positive definite (and hence

*invertible) if and only if x ∈ int(K*

^{n}

*). However, L*

^{−1}

_{x}*y 6= x*

^{−1}*◦ y, for some x ∈ int(K*

^{n}) and

*y ∈ IR*

^{n}*, 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 = (x*_{1}*, x*_{2}*) ∈ IR×IR*^{n−1}*with the spectral values λ*_{1}*, λ*_{2} *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) λ*_{1}*, λ*_{2} *are nonnegative (positive) if and only if x ∈ K*^{n}*(x ∈ int(K*^{n}*)) , i.e.,*
*λ*_{i}*≥ 0 , ∀i = 1, 2 ⇐⇒ x º*_{Kn}*0 .*

*λ**i* *> 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 λ*_{1}*, λ*_{2} *:*

*det(x) = λ*_{1}*λ*_{2} *, tr(x) = λ*_{1}*+ λ*_{2} *, kxk*^{2} = 1

2*(λ*^{2}_{1}*+ λ*^{2}_{2}*) .*

*Property 2.2 For any x = (x*_{1}*, x*_{2}*) ∈ IR×IR*^{n−1}*with the spectral values λ*_{1}*, λ*_{2} *and spectral*
*vectors u*^{(1)}*, u*^{(2)} *given as in (3)-(4), we have*

*(a) x*^{2} *= λ*^{2}_{1}*u*^{(1)}*+ λ*^{2}_{2}*u*^{(2)} *.*
*(b) If x ∈ K*^{n}*, then x** ^{1/2}* =

*√*

*λ*1 *u*^{(1)}+*√*

*λ*2 *u*^{(2)} *.*
*(c) |x| = |λ*_{1}*|u*^{(1)}*+ |λ*_{2}*|u*^{(2)} *.*

*(d) [x]*_{+} *= [λ*_{1}]_{+}*u*^{(1)}*+ [λ*_{2}]_{+}*u*^{(2)} *, [x]*_{−}*= [λ*_{1}]_{−}*u*^{(1)}*+ [λ*_{2}]_{−}*u*^{(2)} *.*
*(e) |x| = [x]*_{+}*+ [−x]*_{+} *= [x]*_{+}*− [x]*_{−}*.*

*(f) [x]*+ = ^{1}_{2}*(x + |x|) , [x]**−* = ^{1}_{2}*(x − |x|) .*

*Property 2.3 (a) Any x ∈ IR*^{n}*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 ∈ IR*^{n}*, if x*^{2} *º*_{Kn}*y*^{2}*, then |x| º*_{Kn}*|y|.*

*(d) For any x ∈ IR*^{n}*, x º*_{Kn}*0 ⇐⇒ hx, yi ≥ 0,* *∀y º*_{Kn}*0.*

*(e) For any x º*_{Kn}*0 and y ∈ IR*^{n}*, x*^{2} *º*_{Kn}*y*^{2} *=⇒ 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 K*^{n}*and S** ^{n}* 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) = x*^{2}_{1}*− kx*2*k*^{2} *,* *tr(x) = 2x*1*,*
*det(y) = y*_{1}^{2}*− ky*_{2}*k*^{2} *,* *tr(y) = 2y*_{1}*.*

*Since x − y = (x*1*− y*1*, x*2 *− y*2*) º*_{Kn}*0, we have kx*2*− y*2*k ≤ x*1 *− y*1*. Thus, x*1 *≥ y*1, and
*then tr(x) ≥ tr(y). Besides, the assumption on x and y gives*

*x*1*− y*1 *≥ kx*2*− y*2*k ≥*

¯¯

¯¯*kx*2*k − ky*2*k*

¯¯

¯¯*,* (7)

*which is equivalent to x*_{1}*− kx*_{2}*k ≥ y*_{1}*− ky*_{2}*k > 0 and x*_{1}*+ kx*_{2}*k ≥ y*_{1} *+ ky*_{2}*k > 0. Hence,*
*det(x) = x*^{2}_{1}*− kx*2*k*^{2} *= (x*1*+ kx*2*k)(x*1*− kx*2*k) ≥ (y*1*+ ky*2*k)(y*1*− ky*2*k) = det(y).*

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

*λ*_{1}*(x) = x*_{1}*− kx*_{2}*k , λ*_{2}*(x) = x*_{1} *+ kx*_{2}*k and λ*_{1}*(y) = y*_{1}*− ky*_{2}*k , λ*_{2}*(y) = y*_{1}*+ ky*_{2}*k .*
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*

¶

*≥ α*^{2}*det(x) + (1 − α)*^{2}*det(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 kx*_{2}*k ≤ x*_{1} *and ky*_{2}*k ≤ y*_{1}, which implies

*|hx*2*, y*2*i| ≤ kx*2*k · ky*2*k ≤ x*1*y*1*.*

Hence, we obtain

*det(x + y)*

*= (x*1*+ y*1)^{2}*− kx*2*+ y*2*k*^{2}

=

µ

*x*^{2}_{1}*− kx*_{2}*k*^{2}

¶

+

µ

*y*_{1}^{2}*− ky*_{2}*k*^{2}

¶

+ 2

µ

*x*_{1}*y*_{1}*− hx*_{2}*, y*_{2}*i*

¶

*≥*

µ

*x*^{2}_{1}*− kx*_{2}*k*^{2}

¶

+

µ

*y*_{1}^{2}*− ky*_{2}*k*^{2}

¶

*= det(x) + det(y).*

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

*= hx, yi*^{2}*− kx*_{1}*y*_{2}*+ y*_{1}*x*_{2}*k*^{2}

=

µ

*x*_{1}*y*_{1} *+ hx*_{2}*, y*_{2}*i*

¶_{2}

*−*

µ

*x*^{2}_{1}*ky*_{2}*k*^{2} *+ 2x*_{1}*y*_{1}*hx*_{2}*, y*_{2}*i + y*_{1}^{2}*kx*_{2}*k*^{2}

¶

*= x*^{2}_{1}*y*_{1}^{2}*+ hx*_{2}*, y*_{2}*i*^{2}*− x*^{2}_{1}*ky*_{2}*k*^{2}*− y*^{2}_{1}*kx*_{2}*k*^{2}

*≤ x*^{2}_{1}*y*_{1}^{2}*+ kx*2*k*^{2}*· ky*2*k*^{2} *− x*^{2}_{1}*ky*2*k*^{2}*− y*^{2}_{1}*kx*2*k*^{2}

=

µ

*x*^{2}_{1}*− kx*2*k*^{2}

¶

*·*

µ

*y*_{1}^{2}*− ky*2*k*^{2}

¶

*= 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) = α*

^{2}

*det(x), for all α > 0. Hence,*

*det*

µ

*αx + (1 − α)y*

¶

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

*(d) For any x º*_{Kn}*0, we know det(x) = λ*_{1}*λ*_{2} *≥ 0, where λ*_{i}*are the spectral values of x.*

*Hence, det(e + x) = (1 + λ*_{1}*)(1 + λ*_{2}*) ≥ (1 +√*

*λ*_{1}*λ*_{2})^{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

*x*1*− kx*2*k ≥ 0,*
*y*_{1}*− ky*_{2}*k ≥ 0,*

*|hx*_{2}*, y*_{2}*i| ≤ kx*_{2}*k · ky*_{2}*k ≤ x*_{1}*y*_{1}*.*

(8)

*Also, we know det(e + x + y) = (1 + x*_{1}*+ y*_{1})^{2}*− kx*_{2}*+ y*_{2}*k*^{2} *, det(e + x) = (1 + x*_{1})^{2}*− kx*_{2}*k*^{2}
*and det(e + y) = (1 + y*_{1})^{2}*− ky*_{2}*k*^{2}. Hence,

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

=

µ

*(1 + x*_{1})^{2}*− kx*_{2}*k*^{2}

¶µ

*(1 + y*_{1})^{2}*− ky*_{2}*k*^{2}

¶

*−*

µ

*(1 + x*_{1}*+ y*_{1})^{2}*− kx*_{2}*+ y*_{2}*k*^{2}

¶

*= 2x*_{1}*y*_{1}*+ 2hx*_{2}*, y*_{2}*i + 2x*_{1}*y*^{2}_{1} *+ 2x*^{2}_{1}*y*_{1}*− 2y*_{1}*kx*_{2}*k*^{2}*− 2x*_{1}*ky*_{2}*k*^{2}
*+x*^{2}_{1}*y*^{2}_{1}*− y*^{2}_{1}*kx*_{2}*k*^{2}*− x*^{2}_{1}*ky*_{2}*k*^{2}*+ kx*_{2}*k*^{2}*· ky*_{2}*k*^{2}

= 2

µ

*x*_{1}*y*_{1}*+ hx*_{2}*, y*_{2}*i*

¶

*+ 2x*_{1}

µ

*y*_{1}^{2}*− ky*_{2}*k*^{2}

¶

*+ 2y*_{1}

µ

*x*^{2}_{1} *− kx*_{2}*k*^{2}

¶

+

µ

*x*^{2}_{1}*− kx*_{2}*k*^{2}

¶µ

*y*^{2}_{1}*− ky*_{2}*k*^{2}

¶

*≥ 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 ∈ IR*^{n}*, we have*
*(a) tr(x + y) = tr(x) + tr(y).*

*(b) λ*1*(x)λ*2*(y) + λ*1*(y)λ*2*(x) ≤ tr(x ◦ y) ≤ λ*1*(x)λ*1*(y) + λ*2*(x)λ*2*(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*

*λ*_{1}*(x)λ*_{2}*(y) + λ*_{1}*(y)λ*_{2}*(x)*

*= (x*_{1}*− kx*_{2}*k)(y*_{1}*+ ky*_{2}*k*_{+}*(x*_{1}*+ kx*_{2}*k)(y*_{1}*− ky*_{2}*k)*

*= 2(x*1*y*1*− kx*2*kky*2*k)*

*≤ 2(x*_{1}*y*_{1}*+ hx*_{2}*, y*_{2}*i*

*= 2hx, yi = tr(x ◦ y)*

*≤ 2(x*_{1}*y*_{1}*+ kx*_{2}*kky*_{2}*k)*

*= (x*1*− kx*2*k)(y*1*− ky*2*k) + (x*1*+ kx*2*k)(y*1*+ ky*2*k),*
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 ∈ IR*^{n}*, the matrix xx*^{T}*is positive semidefinite*
*(p.s.d.) with only one nonzero eigenvalue kxk*^{2}*.*

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*^{T}*A*^{−1}*B.*

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(K*^{n}*).*

*(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(K*^{n}*).*

*(d) the real-valued function f (x) = tr(x*^{−1}*) is convex on int(K*^{n}*).*

*Proof. (a) Since int(K*^{n}*) is a convex set, it is enough to show that ∇*^{2}*f (x) is negative*
semidefinite. From direct computation, we know

*∇f (x) =*

µ *2x*_{1}

*x*^{2}_{1}*− kx*2*k*^{2}*,* *−2x*_{2}
*x*^{2}_{1}*− kx*2*k*^{2}

¶

*= 2x*^{−1}*,*
and

*∇*^{2}*f (x) =*

*−2x*^{2}_{1}*−2kx*2*k*^{2}
*(x*^{2}_{1}*−kx*2*k*^{2})^{2}

*4x*1*x*^{T}_{2}
*(x*^{2}_{1}*−kx*2*k*^{2})^{2}
*4x*1*x*2

*(x*^{2}_{1}*−kx*2*k*^{2})^{2}

*−2(x*^{2}_{1}*−kx*2*k*^{2}*)I−4x*2*x*^{T}_{2}
*(x*^{2}_{1}*−kx*2*k*^{2})^{2}

= * _{(x}*2

*1*

^{−2}*−kx*2

*k*

^{2})

^{2}

"

*(x*^{2}_{1}*+ kx*_{2}*k*^{2})^{2} *−2x*_{1}*x*^{T}_{2}

*−2x*_{1}*x*_{2} *(x*^{2}_{1}*− kx*_{2}*k*^{2}*)I + 2x*_{2}*x*^{T}_{2}

#

*.*

*Let ∇*^{2}*f (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*^{T}*B*

*= (x*^{2}_{1}*+ kx*_{2}*k*^{2})

µ

*(x*^{2}_{1}*− kx*_{2}*k*^{2}*)I + 2x*_{2}*x*^{T}_{2}

¶

*− 4x*^{2}_{1}*x*_{2}*x*^{T}_{2}

*= (x*^{4}_{1}*− kx*_{2}*k*^{4}*)I − 2(x*^{2}_{1}*− kx*_{2}*k*^{2}*)x*_{2}*x*^{T}_{2}

*= (x*^{2}_{1}*− kx*_{2}*k*^{2})

µ

*(x*^{2}_{1}*+ kx*_{2}*k*^{2}*)I − 2x*_{2}*x*^{T}_{2}

¶

*= (x*^{2}_{1}*− kx*_{2}*k*^{2}*) · 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 x*2*x*^{T}_{2} *is a p.s.d. with only one nonzero eigenvalue kx*2*k*^{2}.
*Hence, all the eigenvalues of the matrix M are (x*^{2}_{1} *+ kx*_{2}*k*^{2}*) − 2kx*_{2}*k*^{2} *= x*^{2}_{1} *− kx*_{2}*k*^{2} and

*x*^{2}_{1}*+ kx*_{2}*k*^{2} *with multiplicity of n − 2, which are all positive. Thus, M is positive definite*
*which implies that ∇*^{2}*f (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)*

¶* _{α}*µ

*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(K*^{n}*). 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(K*^{n}) is a convex
*set, it is enough to show that ∇*^{2}*f is positive semidefinie. Note that f (x) = tr(x** ^{−1}*) =

*2x*_{1}

*x*^{2}_{1}*− kx*2*k*^{2}. Thus, from direct computations, we have

*∇*^{2}*f (x) =* 2
*(x*^{2}_{1}*− kx*2*k*^{2})^{3}

*2x*^{3}_{1}*+ 6x*_{1}*kx*_{2}*k*^{2}*,* *−(6x*^{2}_{1}*+ 2kx*_{2}*k*^{2}*)x*^{T}_{2}

*−(6x*^{2}_{1}*+ 2kx*2*k*^{2}*)x*2*, 2x*1

µ

*(x*^{2}_{1}*− kx*2*k*^{2}*)I + 4x*2*x*^{T}_{2}

¶

*.*

*Again, let ∇*^{2}*f (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*^{T}*B*

*= 2x*1

µ

*2x*^{3}_{1} *+ 6x*1*kx*2*k*^{2}

¶µ

*(x*^{2}_{1}*− kx*2*k*^{2}*)I + 4x*2*x*^{T}_{2}

¶

*−*

µ

*6x*^{2}_{1}*+ 2kx*2*k*^{2}

¶_{2}

*x*2*x*^{T}_{2}

=

µ

*4x*^{4}_{1}*+ 12x*^{2}_{1}*kx*_{2}*k*^{2}

¶µ

*x*^{2}_{1}*− kx*_{2}*k*^{2}

¶

*I−*

µ

*20x*^{4}_{1}*− 24x*^{2}_{1}*kx*_{2}*k*^{2} *+ 4kx*_{2}*k*^{4}

¶

*x*_{2}*x*^{T}_{2}

=

µ

*4x*^{4}_{1}*+ 12x*^{2}_{1}*kx*_{2}*k*^{2}

¶µ

*x*^{2}_{1}*− kx*_{2}*k*^{2}

¶

*I − 4*

µ

*5x*^{2}_{1}*− kx*_{2}*k*^{2}

¶µ

*x*^{2}_{1}*− kx*_{2}*k*^{2}

¶

*x*_{2}*x*^{T}_{2}

=

µ

*x*^{2}_{1}*− kx*_{2}*k*^{2}

¶ ·µ

*4x*^{4}_{1}*+ 12x*^{2}_{1}*kx*_{2}*k*^{2}

¶

*I − 4*

µ

*5x*^{2}_{1}*− kx*_{2}*k*^{2}

¶

*x*_{2}*x*^{T}_{2}

¸

=

µ

*x*^{2}_{1}*− kx*_{2}*k*^{2}

¶

*· 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 x*_{2}*x*^{T}_{2} *is a p.s.d. with only one nonzero eigenvalue kx*_{2}*k*^{2}.
*Hence, all the eigenvalues of the matrix M are (4x*^{4}_{1}*+ 12x*^{2}_{1}*kx*^{2}_{k}*− 20x*^{2}_{1}*kx*_{2}*k*^{2}*+ 4kx*_{2}*k*^{4}) and

*4x*^{4}_{1}*+ 12x*^{2}_{1}*kx*_{2}*k*^{2} *with multiplicity of n − 2, which are all positive since*
*4x*^{4}_{1} *+ 12x*^{2}_{1}*kx*_{2}*k*^{2}*− 20x*^{2}_{1}*kx*_{2}*k*^{2} *+ 4kx*_{2}*k*^{4}

*= 4x*^{4}_{1} *− 8x*^{2}_{1}*kx*_{2}*k*^{2}*+ 4kx*_{2}*k*^{4}

= 4

µ

*x*^{2}_{1} *− kx*_{2}*k*^{2}

¶

*> 0.*

*Thus, by Lemma 2.2, we obtain that ∇*^{2}*f (x) is positive definite and hence is positive*
*semidefinite. Therefore, f is convex on int(K*^{n}). *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 ∈ IR*^{n}*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}

µ*x + 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) = t*^{2}*, then f is SOC-convex on IR.*

*(b) Hence, the function g(t) = α + βt + γt*^{2} *is SOC-convex on IR for all α, β ∈ IR and*
*γ ≥ 0.*

*Proof. (a) For any x, y ∈ IR** ^{n}*, we have
1

2

µ

*f (x) + f (y)*

¶

*− f*

µ*x + y*
2

¶

= *x*^{2}*+ y*^{2}

2 *−*

µ*x + 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) = t*^{2} *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, x*^{2} *− y*^{2} =
*(1, 0) − (4, 0) = (−3, 0) 6º** _{Kn}* 0.

*2*

*It is clear that f (t) = t*^{2} *is also SOC-convex on the smaller interval [0, ∞) by Prop.*

*3.2(a). We may ask a natural question: Is f (t) = t*^{2} *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) = t*^{2} *is SOC-monotone on [0, ∞) for n = 2.*

*(b) However, f (t) = t*^{2} *is not SOC-monotone on [0, ∞) for n ≥ 3.*

*(a) Let x = (x*_{1}*, x*_{2}*) º*_{K}^{2} *y = (y*_{1}*, y*_{2}*) º*_{K}^{2} 0. Then we have the following inequalities:

*|x*2*| ≤ x*1 *, |y*2*| ≤ y*1 *, |x*2*− y*2*| ≤ x*1 *− y*1*,*

which implies _{(}

*x*_{1}*− x*_{2} *≥ y*_{1}*− y*_{2} *≥ 0,*

*x*_{1}*+ x*_{2} *≥ y*_{1}*+ y*_{2} *≥ 0.* (11)

*We want to prove that f (x) − f (y) = (x*^{2}_{1}*+ x*^{2}_{2} *− y*_{1}^{2}*− y*_{2}^{2} *, 2x*_{1}*x*_{2} *− 2y*_{1}*y*_{2}*) º** _{K2}* 0, which is

*enough to verify that x*

^{2}

_{1}

*+ x*

^{2}

_{2}

*− y*

_{1}

^{2}

*− y*

_{2}

^{2}

*≥ |2x*

_{1}

*x*

_{2}

*− 2y*

_{1}

*y*

_{2}

*|. This can been seen by*

*x*^{2}_{1}*+ x*^{2}_{2}*− y*_{1}^{2}*− y*^{2}_{2}*−*

¯¯

¯¯*2x*_{1}*x*_{2}*− 2y*_{1}*y*_{2}

¯¯

¯¯

=

( *x*^{2}_{1}*+ x*^{2}_{2}*− y*_{1}^{2}*− y*_{2}^{2}*− (2x*1*x*2 *− 2y*1*y*2*), if x*1*x*2*− y*1*y*2 *≥ 0*
*x*^{2}_{1}*+ x*^{2}_{2}*− y*_{1}^{2}*− y*_{2}^{2}*− (2y*_{1}*y*_{2}*− 2x*_{1}*x*_{2}*), if x*_{1}*x*_{2}*− y*_{1}*y*_{2} *≤ 0*

=

( *(x*_{1} *− x*_{2})^{2}*− (y*_{1}*− y*_{2})^{2}*, if x*_{1}*x*_{2}*− y*_{1}*y*_{2} *≥ 0*
*(x*_{1} *+ x*_{2})^{2}*− (y*_{1} *+ y*_{2})^{2}*, if x*_{1}*x*_{2}*− y*_{1}*y*_{2} *≤ 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) = t*^{2} is not SOC-monotone
*on the interval [0, ∞). Let x = (3, 1, −2) ∈ K*^{3} *and y = (1, 1, 0) ∈ K*^{3}. It is clear that
*x − y = (2, 0, −2) º*_{K3}*0. But, x*^{2}*− y*^{2} *= (14, 6, −12) − (2, 2, 0) = (12, 4, −12) 6º** _{K3}* 0.

*2*

*Now we look at the function f (t) = t*^{3}*. As expected, f (t) = t*^{3} is not SOC-convex.

*However, it is true that f (t) = t*^{3} *is SOC-convex on [0, ∞) for n = 2, whereas false for*
*n ≥ 3. Besides, we will see f (t) = t*^{3} 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) = t*^{3} *is not SOC-convex on IR.*

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

*(c) However, f (t) = t*^{3} *is SOC-convex on [0, ∞) for n = 2.*

*To see (a), let x = (0, −2), y = (1, 0). It can be verified that* ^{1}_{2}

µ

*f (x) + f (y)*

¶

*− f*

µ

*x+y*
2

¶ µ =

*−* ^{9}_{8}*, −*^{9}_{4}

¶

*6º*_{K2}*0, which says f (t) = t*^{3} is not SOC-convex on IR.

*To see (b), let x = (2, 1, −1), y = (1, 1, 0) º** _{K3}* 0, then we have

^{1}

_{2}

µ

*f (x) + f (y)*

¶

*− f*

µ

*x+y*
2

¶

=
*(3, 1, −3) 6º*_{K3}*0, which implies f (t) = t*^{3} *is not even SOC-convex on the interval [0, ∞).*

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

µ*x + y*
2

¶

*¹** _{K2}* 1
2

µ

*f (x) + f (y)*

¶

*, for any x, y º** _{K2}* 0.

*Let x = (x*_{1}*, x*_{2}*) º*_{K2}*0 and y = (y*_{1}*, y*_{2}*) º** _{K2}* 0, then we have

( *x*^{3} *= (x*^{3}_{1}*+ 3x*_{1}*x*^{2}_{2} *, 3x*^{2}_{1}*x*_{2}*+ x*^{3}_{2}*),*
*y*^{3} *= (y*^{3}_{1}*+ 3y*_{1}*y*_{2}^{2} *, 3y*_{1}^{2}*y*_{2}*+ y*_{2}^{3}*),*
which yields

*f (*^{x+y}_{2} ) = ^{1}_{8}

µ

*(x*_{1}*+ y*_{1})^{3}*+ 3(x*_{1}*+ y*_{1}*)(x*_{2}*+ y*_{2})^{2} *, 3(x*_{1}*+ y*_{1})^{2}*(x*_{2}*+ y*_{2}*) + (x*_{2}*+ y*_{2})^{3}

¶

*,*

1 2

µ

*f (x) + f (y)*

¶

= ^{1}_{2}

µ

*x*^{3}_{1}*+ y*^{3}_{1} *+ 3x*_{1}*x*^{2}_{2}*+ 3y*_{1}*y*^{2}_{2} *, x*^{3}_{2}*+ y*^{3}_{2}*+ 3x*^{2}_{1}*x*_{2}*+ 3y*^{2}_{1}*y*_{2}

¶

*.*