DOI 10.1007/s10898-008-9373-z
Some characterizations for SOC-monotone and SOC-convex functions
Jein-Shan Chen · Xin Chen · Shaohua Pan · Jiawei Zhang
Received: 29 June 2007 / Accepted: 21 October 2008 / Published online: 7 November 2008
© Springer Science+Business Media, LLC. 2008
Abstract We provide some characterizations for SOC-monotone and SOC-convex functions by using differential analysis. From these characterizations, we particularly obtain that a continuously differentiable function defined in an open interval is SOC-monotone (SOC-convex) of order n ≥ 3 if and only if it is 2-matrix monotone (matrix convex), and furthermore, such a function is also SOC-monotone (SOC-convex) of order n ≤ 2 if it is 2-matrix monotone (matrix convex). In addition, we also prove that Conjecture 4.2 proposed in Chen (Optimization 55:363–385, 2006) does not hold in general. Some examples are included to illustrate that these characterizations open convenient ways to verify the SOC- monotonicity and the SOC-convexity of a continuously differentiable function defined on an open interval, which are often involved in the solution methods of the convex second-order cone optimization.
Keywords Second-order cone· SOC-monotone function · SOC-convex function Mathematics Subject Classification (2000) 26A48· 26A51 · 26B05 · 90C25
J.-S. Chen (
B
)Department of Mathematics, National Taiwan Normal University, Taipei 11677, Taiwan e-mail: jschen@math.ntnu.edu.tw
X. Chen
Department of Industrial and Enterprise System Engineering, University of Illinois at Urbana–Champaign, Urbana 61801, IL, USA
e-mail: xinchen@uiuc.edu S. Pan
School of Mathematical Sciences, South China University of Technology, Guangzhou 510640, China e-mail: shhpan@scut.edu.cn
J. Zhang
Department of Information, Operations and Management Sciences, New York University, New York 10012-1126, NY, USA
e-mail: jzhang@stern.nyu.edu
1 Introduction
The second-order cone (SOC) in IRn, also called the Lorentz cone, is a set defined by Kn:=
(x1, x2) ∈ IR × IRn−1| x2 ≤ x1
, (1)
where · denotes the Euclidean norm, andK1denotes the set of nonnegative reals IR+. It is known thatKnis a closed convex self-dual cone with nonempty interior int(Kn). For any x, y ∈ IRn, we write xKn y if x− y ∈Kn; and write xKn y if x− y ∈ int(Kn). In other words, we have xKn 0 if and only if x∈Knand xKn 0 if and only if x∈ int(Kn). The relationKn is a partial ordering, but not a linear ordering inKn, i.e., there exist x, y ∈Kn such that neither xKn y nor yKn x. To see this, let x = (1, 1), y = (1, 0), and then we have x− y = (0, 1) /∈K2, y − x = (0, −1) /∈K2.
For any x= (x1, x2), y = (y1, y2) ∈ IR × IRn−1, we define their Jordan product as x◦ y = (x, y , y1x2+ x1y2). (2) we write x2to mean x◦x and write x+y to mean the usual componentwise addition of vectors.
Then◦, +, and e = (1, 0, . . . , 0)T ∈ IRnhave the following basic properties (see [7,8]): (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. Note that the Jordan product is not associative. Besides,Knis not closed under Jordan product.
We recall from [7,8] that each x= (x1, x2) ∈ IR × IRn−1admits a spectral factorization, associated withKn, of the form
x= λ1(x) · u(1)x + λ2(x) · u(2)x , (3) whereλ1(x), λ2(x) and u(1)x , u(2)x are the spectral values and the associated spectral vectors of x given by
λi(x) = x1+ (−1)ix2, u(i)x = 1 2
1, (−1)i¯x2
for i= 1, 2, (4) with ¯x2= xx22if x2= 0 and otherwise ¯x2being any vector in IRn−1such that ¯x2 = 1. If x2 = 0, the factorization is unique. By the spectral factorization, for any f : IR → IR, we can define a vector-valued function associated withKn(n≥ 1) by
fsoc(x) = f (λ1(x))u(1)x + f (λ2(x))u(2)x , ∀x = (x1, x2) ∈ IR × IRn−1, (5) and call it the SOC-function induced by f . If f is defined only on a subset of IR, then fsocis defined on the corresponding subset of IRn. The definition is unambiguous whether x2= 0 or x2 = 0. The cases of fsoc(x) = x1/2, x2, exp(x) were discussed in [7]. In fact, the above definition (5) is analogous to one associated with the semidefinite cone; see [19,20].
Recently, the concepts of SOC-monotone and SOC-convex functions are introduced in [5]. Especially, a function f: J → IR with J ⊆ IR is said to be SOC-monotone of order n if
x Kn y ⇒ fsoc(x) Kn fsoc(y) (6)
for any x, y ∈ dom fsoc ⊆ IRn, where dom fsocdenotes the domain of the function fsoc; and f is said to be SOC-convex of order n if, for any x, y ∈ dom fsoc,
fsoc(λx + (1 − λ)y) Kn λfsoc(x) + (1 − λ) fsoc(y) λ ∈ [0, 1]. (7)
The function f is said to be SOC-monotone (respectively, SOC-convex) if it is SOC- monotone of all order n (respectively, SOC-convex of all order n), and f is SOC-convex on J if and only if− f is SOC-concave on J. The concepts of SOC-monotone and SOC-convex functions are analogous to matrix monotone and matrix convex functions [2,10,11,14], and are special cases of operator monotone and operator convex functions [1,3,12]. For example, the function f is said to be n-matrix convex on J if
f(λA + (1 − λ)B) λf (A) + (1 − λ) f (B) λ ∈ [0, 1]
for arbitrary Hermitian n× n matrices A and B with spectra in J. It is clear that the set of SOC-monotone functions and the set of SOC-convex functions are closed under positive linear combinations and pointwise limits.
There has been systematic study on matrix monotone and matrix convex functions, and moreover, characterizations for such functions have been explored; see [4,10,11,13,14] and the references therein. To the contrast, the study on SOC-monotone and SOC-convex func- tions just starts its first step. One reason is that they were viewed as special cases of operator monotone and operator convex functions. However, we recently observed that SOC-monotone and SOC-convex functions play an important role in the design of solutions methods for con- vex second-order cone programs (SOCPs); for example, the proximal-like methods in [15]
and the augmented Lagrangian method introduced in Sect.5. On the other hand, we all know that the developments of matrix-valued functions have major contributions in the solution of optimization problems. Thus, we hope similar systematic study on SOC-functions can be exploited so that it can be readily adopted to optimization field. This is the main motivation of the paper.
Although some work was done in [5] for SOC-monotone and SOC-convex functions, the focus there is to provide some specific examples by the definition and it seems difficult to exploit the characterizations there to verify whether a given function is SOC-convex or not. In this paper, we employ differential analysis to establish some useful characterizations which will open convenient ways to verify the SOC-monotonicity and the SOC-convexity of a function defined on an open interval. Particularly, from these characterizations, we obtain that a continuously differentiable function defined on an open interval is SOC-monotone (SOC- convex) of order n ≥ 3 if and only if it is 2-matrix monotone (matrix convex), and such a function is also SOC-monotone (SOC-convex) of order n ≤ 2 if it is 2-matrix monotone (matrix convex). Thus, if such functions are 2-matrix monotone (matrix convex), then it must be SOC-monotone (SOC-convex). It should be pointed out that the analysis of this paper can not be obtained from those for matrix-valued functions. One of the reasons is that the matrix multiplication is associative whereas the Jordan product is not.
Throughout the paper,·, · denotes the Euclidean inner product, IRn denotes the space of n-dimensional real column vectors, and IRn1× · · · × IRnm is identified with IRn1+···+nm. Thus,(x1, . . . , xm) ∈ IRn1× · · · × IRnm is viewed as a column vector in IRn1+···+nm. Also, I represents an identity matrix of suitable dimension; J is a subset of IR; and 0 is a zero matrix or vector of suitable dimension. The notationT means transpose and C(i)(J) denotes the family of functions which are defined on J ⊆ IR to IR and have the i-th continuous derivative. For a function f : IR → IR, f(i)(x) represents the i-th order derivative of f at x ∈ IR, and the first-order and the second-order derivative of f are also written as fand f, respectively. For any f : IRn → IR, ∇ f (x) denotes the gradient of f at x ∈ IRn and dom f denotes the domain of f . For any differentiable mapping F= (F1, . . . , Fm)T : IRn → IRm,
∇ F(x) = [∇ F1(x) · · · ∇ Fm(x)] is an n × m matrix which denotes the transposed Jacobian of F at x. For any symmetric matrices A, B ∈ IRn×n, we write A B (respectively, A B) to mean A− B is positive semidefinite (respectively, positive definite).
2 Preliminaries
In this section, we develop the second-order Taylor’s expansion for the vector-valued SOC- function fsoc defined as in (5) which is crucial in our subsequent analysis. To the end, we assume that f ∈ C(2)(J) with J being an open interval in IR and dom fsoc ⊆ IRn.
Given any x ∈ dom fsoc and h = (h1, h2) ∈ IR × IRn−1, we have x+ th ∈ dom fsoc for any sufficiently small t> 0. We wish to calculate the Taylor’s expansion of the function fsoc(x + th) at x for any sufficiently small t > 0. In particular, we are interested in finding matrices∇ fsoc(x) and Ai(x) for i = 1, 2, . . . , n such that
fsoc(x + th) = fsoc(x) + t∇ fsoc(x)h +1 2t2
⎡
⎢⎢
⎢⎣
hTA1(x)h hTA2(x)h
...
hTAn(x)h
⎤
⎥⎥
⎥⎦+ o(t2). (8)
For convenience, we omit the variable notion x inλi(x) for i = 1, 2 in the discussions below.
It is known that fsocis differentiable (respectively, smooth) if and only if f is differentiable (respectively, smooth); see [6,8]. Moreover, there holds that
∇ fsoc(x) =
⎡
⎢⎢
⎣
b(1) c(1) x2T
x2 c(1) x2
x2 a(0)I+ (b(1)− a(0))x2x2T
x22
⎤
⎥⎥
⎦ (9)
if x2= 0, and otherwise
∇ fsoc(x) = f(x1)I (10)
where
a(0)= f(λ2) − f (λ1)
λ2− λ1 , b(1)= f(λ2) + f(λ1)
2 , c(1) = f(λ2) − f(λ1)
2 . (11)
Therefore, we only need to derive the formula of Ai(x) for i = 1, 2, . . . , n in (8).
We first consider the case where x2 = 0 and x2+ th2= 0. By the definition (5),
fsoc(x + th) = 1
2f(x1+ th1− x2+ th2)
⎡
⎣ 1
− x2+ th2
x2+ th2
⎤
⎦
+1
2f(x1+ th1+ x2+ th2)
⎡
⎣ 1
x2+ th2
x2+ th2
⎤
⎦
=
⎡
⎢⎣
f(x1+ th1− x2+ th2) + f (x1+ th1+ x2+ th2) f(x1+ th1+ x2+ th2) − f (x21+ th1− x2+ th2)
2
x2+ th2
x2+ th2
⎤
⎥⎦
:=
1
2
. (12)
To derive the Taylor’s expansion of fsoc(x + th) at x with x2 = 0, we first write out and expandx2+ th2. Notice that
x2+ th2 =
x22+ 2tx2Th2+ t2h22= x2
1+ 2tx2Th2
x22 + t2h22
x22. Therefore, using the fact that
√1+ = 1 +1 2 − 1
82+ o(2), we may obtain
x2+ th2 = x2
1+ t α
x2+1 2t2 β
x22
+ o(t2), (13)
where
α = x2Th2
x2, β = h22−(x2Th2)2
x22 = h22− α2= hT2Mx2h2, with
Mx2 = I − x2x2T
x22.
Furthermore, from (13) and the fact that(1 + )−1= 1 − + 2+ o(2), it follows that
x2+ th2−1= x2−1
1− t α
x2+1 2t2
2 α2
x22 − β
x22
+ o(t2)
. (14) Combining Eqs. (13) and (14) then yields that
x2+ th2
x2+ th2 = x2
x2+ t
h2
x2− α
x2 x2
x2
+1 2t2
2 α2
x22 − β
x22
x2
x2− 2 h2
x2 α
x2
+ o(t2)
= x2
x2+ t Mx2
h2
x2 +1
2t2
3hT2x2x2Th2
x24 x2
x2−h22
x22 x2
x2− 2h2h2T
x22 x2
x2
+ o(t2). (15)
In addition, from (13), we have the following equalities f(x1+ th1− x2+ th2)
= f
x1+ th1−
x2
1+ t α
x2+1 2t2 β
x22
+ o(t2)
= f
λ1+ t(h1− α) −1 2t2 β
x2+ o(t2)
= f (λ1) + t f(λ1)(h1− α) + 1 2t2
− f(λ1) β
x2+ f(λ1)(h1− α)2
+ o(t2) (16)
and
f(x1+ th1+ x2+ th2)
= f
λ2+ t(h1+ α) + 1 2t2 β
x2+ o(t2)
= f (λ2) + t f(λ2)(h1+ α) +1 2t2
f(λ2) β
x2+ f(λ2)(h1+ α)2
+ o(t2). (17) For i= 0, 1, 2, we define
a(i)= f(i)(λ2) − f(i)(λ1)
λ2− λ1 , b(i)= f(i)(λ2) + f(i)(λ1)
2 , c(i)= f(i)(λ2) − f(i)(λ1)
2 ,
(18) where f(i)means the i -th derivative of f and f(0)is the same as the original f . Then, by the Eqs. (16)–(18), it can be verified that
1= 1 2
f(x1+ th1+ x2+ th2) + f (x1+ th1− x2+ th2)
= b(0)+ t
b(1)h1+ c(1)α +1
2t2
a(1)β + b(2)(h21+ α2) + 2c(2)h1α + o(t2)
= b(0)+ t
b(1)h1+ c(1)hT2 x2
x2
+1
2t2hTA1(x)h + o(t2), where
A1(x) =
⎡
⎢⎢
⎣
b(2) c(2) x2T
x2 c(2) x2
x2 a(1)I+
b(2)− a(1) x2x2T
x22
⎤
⎥⎥
⎦ . (19)
Note that in the above expression for1, b(0)is exactly the first component of fsoc(x) and
b(1)h1+ c(1)hT2 xx2
2
is the first component of∇ fsoc(x)h. Using the same techniques again, 1
2
f(x1+ th1+ x2+ th2) − f (x1+ th1− x2+ th2)
= c(0)+ t
c(1)h1+ b(1)α +1
2t2
b(1) β
x2+ c(2)(h21+ α2) + 2b(2)h1α
+ o(t2)
= c(0)+ t
c(1)h1+ b(1)α +1
2t2hTB(x)h + o(t2), (20)
where
B(x) =
⎡
⎢⎢
⎢⎣
c(2) b(2) x2T
x2 b(2) x2
x2 c(2)I+
b(1)
x2− c(2)
Mx2
⎤
⎥⎥
⎥⎦. (21)
Using Eqs. (15) and (20), we obtain that
2 = 1 2
f(x1+ th1+ x2+ th2) − f (x1+ th1− x2+ th2) x2+ th2
x2+ th2
= c(0) x2
x2+ t
x2
x2(c(1)h1+ b(1)α) + c(0)Mx2 h2
x2
+1
2t2W+ o(t2),
where
W = x2
x2hTB(x)h + 2Mx2
h2
x2
c(1)h1+ b(1)α +c(0)
3h2Tx2x2Th2
x24 x2
x2−h22
x22 x2
x2− 2h2hT2
x22 x2
x2
.
Now we denote
d := b(1)− a(0)
x2 = 2(b(1)− a(0)) λ2− λ1
, U := hTC(x)h
V := 2c(1)h1+ b(1)α
x2 − c(0)2x2Th2
x23 = 2a(1)h1+ 2dx2Th2
x2, where
C(x) :=
⎡
⎢⎢
⎣
c(2) (b(2)− a(1)) x2T
x2 (b(2)− a(1)) x2
x2 dI+
c(2)− 3d x2x2T
x22
⎤
⎥⎥
⎦ . (22)
Then U can be further recast as
U = hTB(x)h + c(0)3h2Tx2x2Th2
x24 − c(0)h22
x22 − 2x2Th2
x22(c(1)h1+ b(1)α).
Consequently,
W = x2
x2U+ h2V.
We next consider the case where x2= 0 and x2+ th2= 0. By definition (5),
fsoc(x + th) = f(x1+ t(h1− h2)) 2
⎡
⎣ 1
− h2
h2
⎤
⎦ + f(x1+ t(h1+ h2)) 2
⎡
⎣ 1 h2
h2
⎤
⎦
=
⎡
⎢⎣
f(x1+ t(h1− h2)) + f (x1+ t(h1+ h2)) f(x1+ t(h1+ h2)) − f (x2 1+ t(h1− h2))
2
h2
h2
⎤
⎥⎦ . (23)
Using the Taylor expansion of f at x1, we can obtain that 1
2
f(x1+ t(h1− h2)) + f (x1+ t(h1+ h2))
= f (x1) + t f(1)(x1)h1+1
2t2f(2)(x1)hTh+ o(t2), 1
2
f(x1+ t(h1− h2)) − f (x1+ t(h1+ h2))
= t f(1)(x1)h2+1
2t2f(2)(x1)2h1h2+ o(t2).
Therefore,
fsoc(x + th) = fsoc(x) + t f(1)(x1)h + 1
2t2f(2)(x1) hTh
2h1h2
. (24)
Thus, under this case, we have that
A1(x) = f(2)(x1)I, Ai(x) = f(2)(x1)
0 ¯ei−1T
¯ei−1 0
i= 2, . . . , n, (25) where ¯ej ∈ IRn−1is the vector whose j th component is 1 and the others are 0.
Summing up the above discussions, we may obtain the following conclusion.
Proposition 2.1 Let f ∈ C(2)(J) with J being an open interval in IR and dom fsoc ⊆ IRn. Then, for given x∈ dom fsoc, h∈ IRnand any sufficiently small t> 0,
fsoc(x + th) = fsoc(x) + t∇ fsoc(x)h +1 2t2
⎡
⎢⎢
⎢⎣
hTA1(x)h hTA2(x)h
...
hTAn(x)h
⎤
⎥⎥
⎥⎦+ o(t2),
where∇ fsoc(x) and Ai(x), i = 1, 2, . . . , n are given by (10) and (25) if x2 = 0; and otherwise∇ fsoc(x) and A1(x) are given by (9) and (19), respectively, and for i ≥ 2,
Ai(x) = C(x) x2i
x2+ Bi(x) where
Bi(x) = veTi + eivT, v =
a(1) d x2T
x2
T .
From Proposition 4.3 of [5] and Proposition2.1, we readily have the following result.
Proposition 2.2 Let f ∈ C(2)(J) with J being an open interval in IR and dom fsoc ⊆ IRn. Then, f is SOC-convex if and only if for any x∈ dom fsocand h∈ IRn, the vector
⎡
⎢⎢
⎢⎣
hTA1(x)h hTA2(x)h
...
hTAn(x)h
⎤
⎥⎥
⎥⎦∈Kn.
3 Characterizations of SOC-monotone functions
Now we are ready to show our main result concerning the characterization of SOC-monotone functions. We need the following technical lemmas for the proof. The first one is so-called S-Lemma whose proof can be found in [16,18].
Lemma 3.1 Let A, B be symmetric matrices and yTAy> 0 for some y. Then, the implica- tion
zTAz≥ 0 ⇒ zTBz≥ 0
is valid if and only if B λA for some λ ≥ 0.
Lemma 3.2 Givenθ ∈ IR, a ∈ IRn−1, and a symmetric matrix A ∈ IRn×n. LetBn−1 :=
{z ∈ IRn−1| z ≤ 1}. Then, the following results hold:
(a) For any h∈Kn, Ah∈Knis equivalent to A 1
z
∈Knfor any z∈Bn−1. (b) For any z∈Bn−1,θ + aTz≥ 0 is equivalent to θ ≥ a.
(c) If A = θ aT
a H
with H being an(n − 1) × (n − 1) symmetric matrix, then for any h∈Kn, Ah∈Knis equivalent toθ ≥ a and there exists λ ≥ 0 such that the matrix
θ2− a2− λ θaT − aTH θa − HTa aaT − HTH+ λI
O.
Proof (a) For any h∈Kn, suppose that Ah∈Kn. Let h= 1
z
where z∈Bn−1. Then h∈Knand the desired result follows. For the other direction, if h= 0, the conclusion is obvious. Now let h := (h1, h2) be any nonzero vector inKn. Then, h1 > 0 and
h2 ≤ h1. Consequently, h2
h1 ∈Bn−1and A
⎡
⎣1 h2 h1
⎤
⎦ ∈Kn. SinceKn is a cone, we have
h1A
⎡
⎣1 h2
h1
⎤
⎦ = Ah ∈Kn.
(b) For z∈Bn−1, supposeθ + aTz≥ 0. If a = 0, then the result is clear since θ ≥ 0. If a = 0, let z := −a/a. Clearly, z ∈Bn−1and henceθ + −aTa
a ≥ 0 which gives θ − a ≥ 0. For the other direction, the result follows from the Cauchy Schwarz inequality:
θ + aTz≥ θ − a · z ≥ θ − a ≥ 0.
(c) From part (a), Ah∈Knfor any h∈Knis equivalent to A 1
z
∈Knfor any z∈Bn−1. Notice that
A 1
z
= θ aT
a H 1
z
=
θ + aTz a+ Hz
. Then, Ah∈Knfor any h∈Kn is equivalent to the following two things:
θ + aTz≥ 0, for any z ∈Bn−1 (26)
and
(a + Hz)T(a + Hz) ≤ (θ + aTz)2, for any z ∈Bn−1. (27) By part (b), (26) is equivalent toθ ≥ a. Now, we write the expression of (27) as below:
zT(aaT− HTH)z + 2(θaT − aTH)z + θ2− aTa≥ 0, for any z ∈Bn−1, which can be further simplified as
1 zT θ2− a2 θaT − aTH θa − HTa aaT− HTH
1 z
≥ 0, for any z ∈Bn−1.
Observe that z∈Bn−1is the same as
1 zT 1 0 0−I
1 z
≥ 0.
Thus, by applying the S-Lemma (Lemma3.1), there existsλ ≥ 0 such that θ2− a2 θaT− aTH
θa − HTa aaT − HTH
− λ 1 0
0−I
O
This completes the proof of part (c).
Theorem 3.1 Let f ∈ C(1)(J) with J being an open interval and dom fsoc⊆ IRn. Then, (i) when n= 2, f is SOC-monotone if and only if f(τ) ≥ 0 for any τ ∈ J;
(ii) when n≥ 3, f is SOC-monotone if and only if the 2 × 2 matrix
⎡
⎢⎣
f(1)(t1) f(t2) − f (t1) t2− t1
f(t2) − f (t1) t2− t1
f(1)(t2)
⎤
⎥⎦ O for all t1, t2∈ J.
Proof By the definition of SOC-monotonicity, f is SOC-monotone if and only if
fsoc(x + h) − fsoc(x) ∈Kn (28)
for any x ∈ dom fsoc and h ∈ Kn such that x+ h ∈ dom fsoc. By the first-order Taylor expansion of fsoc, i.e.,
fsoc(x + h) = fsoc(x) + ∇ fsoc(x + th)h for some t ∈ (0, 1),
it is clear that (28) is equivalent to∇ fsoc(x + th)h ∈Knfor any x ∈ dom fsocand h ∈Kn such that x+ h ∈ dom fsoc, and some t ∈ (0, 1). Let y := x + th = µ1v(1)+ µ2v(2)for such x, h and t. We next proceed the arguments by the two cases of y2= 0 and y2= 0.
Case (1): y2= 0. Under this case, we notice that
∇ fsoc(y) = θ aT
a H
,
where
θ = ˜b(1), a = ˜c(1) y2
y2, and H = ˜a(0)I+ ( ˜b(1)− ˜a(0))y2y2T
y22, with
˜a(0)= f(µ2) − f (µ1)
µ2− µ1 , ˜b(1)= f(µ2) + f(µ1)
2 , ˜c(1) = f(µ2) − f(µ1)
2 . (29)
In addition, we also observe that
θ2− a2= ( ˜b(1))2− (˜c(1))2, θaT − aTH= 0 and
aaT − HTH= −(˜a(0))2I+
(˜c(1))2− ( ˜b(1))2+ (˜a(0))2 y2y2T
y22. Thus, by Lemma3.2, f is SOC-monotone if and only if
(a) ˜b(1)≥ |˜c(1)|;
(b) and there existsλ ≥ 0 such that the matrix
⎡
⎣( ˜b(1))2− (˜c(1))2− λ 0 0 (λ − (˜a(0))2)I +
(˜c(1))2− ( ˜b(1))2+ (˜a(0))2 y2y2T
y22
⎤
⎦ O.
When n= 2, (a) together with (b) is equivalent to saying that f(µ1) ≥ 0 and f(µ2) ≥ 0.
Then we conclude that f is SOC-monotone if and only if f(τ) ≥ 0 for any τ ∈ J.
When n≥ 3, (b) is equivalent to saying that ( ˜b(1))2−(˜c(1))2= λ ≥ 0 and λ−(˜a(0))2≥ 0, i.e.,( ˜b(1))2− (˜c(1))2≥ (˜a(0))2. Therefore, (a) together with (b) is equivalent to
⎡
⎢⎣
f(1)(µ1) f(µ2) − f (µ1) µ2− µ1
f(µ2) − f (µ1) µ2− µ1
f(1)(µ2)
⎤
⎥⎦ O
for any x∈ IRn, h ∈Knsuch that x+ h ∈ dom fsoc, and some t∈ (0, 1). Thus, we conclude that f is SOC-monotone if and only if
⎡
⎢⎣
f(1)(t1) f(t2) − f (t1) t2− t1
f(t2) − f (t1) t2− t1
f(1)(t2)
⎤
⎥⎦ O for all t1, t2∈ J.
Case (2): y2 = 0. Now we have µ1 = µ2 and∇ fsoc(y) = f(1)(µ1)I = f(1)(µ2)I . Hence, f is SOC-monotone is equivalent to f(1)(µ1) ≥ 0, which is also equivalent to
⎡
⎢⎣
f(1)(µ1) f(µ2) − f (µ1) µ2− µ1
f(µ2) − f (µ1) µ2− µ1
f(1)(µ2)
⎤
⎥⎦ O
since f(1)(µ1) = f(1)(µ2) and f(µµ22)− f (µ−µ1 1) = f(1)(µ1) = f(1)(µ2) by the Taylor formula andµ1= µ2. Thus, similar to Case (1), the conclusion also holds under this case.
From Theorem3.1and [11, Theorem 6.6.36], we immediately have the following results.
Corollary 3.1 Let f ∈ C(1)(J) with J being an open interval in IR. Then,
(a) f is SOC-monotone of order n ≥ 3 if and only if it is 2-matrix monotone, and f is SOC-monotone of order n≤ 2 if it is 2-matrix monotone.
(b) Suppose that n≥ 3 and f is SOC-monotone of order n. Then, f(t0) = 0 for some t0∈ J if and only if f(·) is a constant function on J.
Note that the SOC-monotonicity of order 2 does not imply the 2-matrix monotonicity.
For example, f(t) = t2is SOC-monotone of order 2 on(0, +∞) by Example 3.2 (a) in [5], but by [11, Theorem 6.6.36] we can verify that it is not 2-matrix monotone. Corollary3.1 (a) implies that a continuously differentiable function defined on an open interval must be SOC-monotone if it is 2-matrix monotone. In addition, from the following proposition, we also have that the compound of two simple SOC-monotone functions is SOC-monotone.
Proposition 3.1 If f : J1→ J and g : J → IR with J1, J ⊆ IR are SOC-monotone on J1
and J , respectively, then the function g◦ f : J1→ IR is SOC-monotone on J.
Proof It is easy to verify that for all x, y ∈ IRn, xKn y if and only ifλi(x) ≥ λi(y) with i= 1, 2. In addition, g is monotone on J since it is SOC-monotone. From the two facts, we
immediately obtain the result.
4 Characterizations of SOC-convex functions
In this section, we exploit Peirce decomposition to derive some characterizations for SOC- convex functions. Let f ∈ C(2)(J) with J being an open interval in IR and dom fsoc⊆ IRn.
For any x∈ dom fsoc and h∈ IRn, if x2= 0, then from Proposition2.1we have that
⎡
⎢⎢
⎢⎣
hTA1(x)h hTA2(x)h
...
hTAn(x)h
⎤
⎥⎥
⎥⎦= f(2)(x1) hTh
2h1h2
.
Since(hTh, 2h1h2) ∈Kn, from Proposition2.2it follows that f is SOC-convex if and only if f(2)(x1) ≥ 0. By the arbitrariness of x1, f is SOC-convex if and only if f is convex on J . In what follows, we assume that x2 = 0. Let x = λ1(x)u(1)x + λ2(x)u(2)x , where u(1)x and u(2)x are given by (4) with ¯x2 = xx22. Let u(i)x = (0, υ2(i)) for i = 3, . . . , n, where υ2(3), . . . , υ2(n)is any orthonormal set of vectors that span the subspace of IRn−2orthogonal to x2. It is easy to verify that the vectors u(1)x , u(2)x , u(3)x , . . . , u(n)x are linearly independent.
Hence, for any given h= (h1, h2) ∈ IR × IRn−1, there existsµi, i = 1, 2, . . . , n such that h = µ1
√2u(1)x + µ2
√2u(2)x +
n i=3
µiu(i)x .
From (19), we can verify that b(2)+ c(2)and b(2)− c(2)are the eigenvalues of A1(x) with u(2)x and u(1)x being the corresponding eigenvectors, and a(1)is the eigenvalue of multiplicity n−2 with u(i)x = (0, υ2(i)) for i = 3, . . . , n being the corresponding eigenvectors. Therefore,
hTA1(x)h = µ21(b(2)− c(2)) + µ22(b(2)+ c(2)) + a(1)
n i=3
µ2i
= f(2)(λ1)µ21+ f(2)(λ2)µ22+ a(1)µ2, (30) where
µ2=n
i=3µ2i.
Similarly, we can verify that c(2)+ b(2)− a(1)and c(2)− b(2)+ a(1)are the eigenvalues of
⎡
⎢⎢
⎣
c(2) (b(2)− a(1)) x2T
x2 (b(2)− a(1)) x2
x2 dI+
c(2)− d x2x2T
x22
⎤
⎥⎥
⎦
with u(2)x and u(1)x being the corresponding eigenvectors, and d is the eigenvalue of multiplicity n− 2 with u(i)x = (0, υ2(i)) for i = 3, . . . , n being the corresponding eigenvectors. Notice that C(x) in (22) can be decomposed the sum of the above matrix and
⎡
⎣0 0
0−2dx2x2T
x22
⎤
⎦ .