Journal of Nonlinear and Convex Analysis, vol. 13, no. 3, pp. 421-431, 2012
Using Schur Complement Theorem to prove convexity of some
SOC-functions
Jein-Shan Chen 1
Department of Mathematics
National Taiwan Normal University
Taipei 11677, Taiwan
E-mail: [email protected]
Tsun-Ko Liao
Department of Mathematics
National Taiwan Normal University
Taipei 11677, Taiwan
E-mail: [email protected]
Shaohua Pan 2
Department of Mathematics
South China University of Technology
Guangzhou 510640, China
March 10, 2012
Abstract. In this paper, we provide an important application of the Schur Complement
Theorem in establishing convexity of some functions associated with second-order cones
(SOCs), called SOC-trace functions. As illustrated in the paper, these functions play a
key role in the development of penalty and barrier functions methods for second-order
cone programs, and establishment of some important inequalities associated with SOCs.
Key words. Second-order cone, SOC-functions, convexity, positive semidefiniteness
AMS subject classifications. 26A27, 26B05, 26B35, 49J52, 90C33.
1
Member of Mathematics Division, National Center for Theoretical Sciences, Taipei Office. The
author’s work is supported by National Science Council of Taiwan.
2
The author’s work is supported by National Young Natural Science Foundation (No. 10901058) and
Guangdong Natural Science Foundation (No. 9251802902000001). E-mail: [email protected].
1
1
Introduction
The second-order cone (SOC) in IRn , also called Lorentz cone, is a set defined as
n
o
Kn := (x1 , x2 ) ∈ IR × IRn−1 | x1 ≥ kx2 k ,
(1)
where k·k denotes the Euclidean norm. When n = 1, Kn reduces to the set of nonnegative
real numbers IR+ . As shown in [12], Kn is also a set composed of the squared elements
from Jordan algebra (IRn , ◦), where the Jordan product “◦” is a binary operation defined
by
x ◦ y := (hx, yi, x1 y2 + y1 x2 )
(2)
for any x = (x1 , x2 ), y = (y1 , y2 ) ∈ IR × IRn−1 . Unless otherwise stated, in the rest of
this note, we use e = (1, 0, . . . , 0)T ∈ IRn to denote the unit element of Jordan algebra
(IRn , ◦), i.e., e ◦ x = x for any x ∈ IRn , and for any x ∈ IRn , use x1 to denote the first
component of x, and x2 to denote the vector consisting of the rest n − 1 components.
From [11, 12], we recall that each x ∈ IRn admits a spectral factorization associated
with Kn , of the following form
(2)
x = λ1 (x)u(1)
x + λ2 (x)ux ,
(3)
(i)
where λi (x) and ux for i = 1, 2 are the spectral values and the associated spectral vectors
of x, respectively, defined by
1
i
i
(i)
(4)
1, (−1) x̄2 ,
λi (x) = x1 + (−1) kx2 k, ux =
2
with x̄2 = kxx22 k if x2 6= 0, and otherwise x̄2 being any vector in IRn−1 such that kx̄2 k = 1.
When x2 6= 0, the spectral factorization is unique. The determinant and trace of x are
defined as det(x) := λ1 (x)λ2 (x) and tr(x) := λ1 (x) + λ2 (x), respectively.
With the spectral factorization above, for any given scalar function f : J ⊆ IR → IR,
we may define a vector-valued function f soc : S ⊆ IRn → IRn by
(2)
f soc (x) := f (λ1 (x))u(1)
x + f (λ2 (x))ux
(5)
where J is an interval (finite or infinite, open or closed) of IR, and S is the domain of f soc
determined by f . Obviously, f soc is well-defined whether x2 = 0 or not. Assume that f
is differentiable on intJ. Then, by Lemma 2.2 in Section 2, f soc is differentiable on intS,
and moreover, for any x ∈ intS, ∇f soc (x) has a structure which entails the application of
the Schur Complement Theorem in establishing its positive semidefiniteness or positive
definiteness. On the other hand, Lemma 2.2 shows that the SOC-trace function
f tr (x) := f (λ1 (x)) + f (λ2 (x)) = tr(f soc (x)) ∀x ∈ S
2
(6)
is differentiable on intS with ∇f tr (x) = (f 0 )soc (x) for any x ∈ intS, which means that
∇2 f tr (x) = ∇(f 0 )soc (x)
∀x ∈ intS.
(7)
The two sides show that we may establish the convexity of f tr with the help of the Schur
Complement Theorem. In fact, the Schur Complement Theorem is frequently used in the
topics related to semidefinite programming (SDP), but there is no paper to introduce its
application involving SOCs, to the best of our knowledge.
Motivated by this, in this paper we provide such an application of the Schur Complement Theorem in establishing the convexity of SOC-trace functions and the compounds
of SOC-trace functions and real-valued functions. As illustrated in the next section, some
of these functions are the key of penalty and barrier function methods for second-order
cone programs (SOCPs), as well as the establishment of some important inequalities
associated with SOCs, for which the proof of convexity of these functions is a necessity. But, this requires computation of the first and second-order derivatives, which is
technically much more demanding than in the linear and semidefinite cases. As will be
seen, Theorem 2.1 gives a simple way to achieve this objective via the Schur Complement
Theorem, by which one only needs to check the sign of the second-order derivative of a
scalar function.
Throughout this note, for any x, y ∈ IRn , we write x Kn y if x − y ∈ Kn ; and write
x Kn y if x − y ∈ intKn . For a real symmetric matrix A, we write A 0 (respectively,
A 0) if A is positive semidefinite (respectively, positive definite). For any f : J → IR,
f 0 (t) and f 00 (t) denote the first derivative and second-order derivative of f at the differentiable point t ∈ J, respectively; for any F : S ⊆ IRn → IR, ∇F (x) and ∇2 F (x) denote
the gradient and the Hessian matrix of F at the differentiable point x ∈ S, respectively.
2
Main results
The Schur Complement Theorem gives a characterization for the positive semidefiniteness
(definiteness) of a matrix via the positive semidefiniteness (definiteness) of the Schurcomplement with respect to a block partitioning of the matrix, which is stated as below.
Lemma 2.1 [13] (Schur Complement Theorem) Let A ∈ IRm×m be a symmetric positive
definite matrix, C ∈ IRn×n be a symmetric matrix, and B ∈ IRm×n . Then,
A B
0 ⇐⇒ C − B T A−1 B 0
(8)
BT C
and
A B
BT C
0
C − B T A−1 B 0.
⇐⇒
3
(9)
In this section, we focus on on the application of the Schur Complement Theorem in
establishing convexity of SOC-trace functions. To the end, we need the following lemma.
Lemma 2.2 For any given f : J ⊆ IR → IR, let f soc : S → IRn and f tr : S → IR be given
by (5) and (6), respectively. Assume that J is open. Then, the following results hold.
(a) The domain S of f soc and f tr is also open.
(b) If f is (continuously) differentiable on J, then f soc is (continuously) differentiable
on S. Moreover, for any x ∈ S, ∇f soc (x) = f 0 (x1 )I if x2 = 0, and otherwise


xT2
c(x)
 b(x)

kx2 k
soc


∇f (x) = 
(10)
T ,
x2 x2
x2
a(x)I + (b(x) − a(x))
c(x)
kx2 k
kx2 k2
where
b(x) =
f 0 (λ2 (x))− f 0 (λ1 (x))
f (λ2 (x))− f (λ1 (x))
f 0 (λ2 (x))+ f 0 (λ1 (x))
, c(x) =
, a(x) =
.
2
2
λ2 (x) − λ1 (x)
(c) If f is (continuously) differentiable, then f tr is (continuously) differentiable on S
with ∇f tr (x) = (f 0 )soc (x); if f is twice (continuously) differentiable, then f tr is
twice (continuously) differentiable on S with ∇2 f tr (x) = ∇(f 0 )soc (x).
Proof. (a) Fix any x ∈ S. Then λ1 (x), λ2 (x) ∈ J. Since J is an open subset of IR,
there exist δ1 , δ2 > 0 such that {t√∈ IR| |t − λ1 (x)| < δ1 } ⊆ J and {t ∈ IR| |t − λ2 (x)| <
δ2 } ⊆ J. Let δ := min{δ1 , δ2 }/ 2. Then, for any y satisfying ky − xk < δ, we have
|λ1 (y) − λ1 (x)| < δ1 and |λ2 (y) − λ2 (x)| < δ2 by noting that
(λ1 (x) − λ1 (y))2 + (λ2 (x) − λ2 (y))2
= 2(x21 + kx2 k2 ) + 2(y12 + ky2 k2 ) − 4(x1 y1 + kx2 kky2 k)
≤ 2(x21 + kx2 k2 ) + 2(y12 + ky2 k2 ) − 4(x1 y1 + hx2 , y2 i)
= 2 kxk2 + kyk2 − 2hx, yi = 2kx − yk2 ,
and consequently λ1 (y) ∈ J and λ2 (y) ∈ J. Since f is a function from J to IR, this means
that {y ∈ IRn | ky − xk < δ} ⊆ S, and therefore the set S is open.
(b) The proof is direct by using the same arguments as those of [9, Props. 4 and 5].
(c) If f is (continuously) differentiable, then from part (b) and f tr (x) = eT f soc (x) it
follows that f tr is (continuously) differentiable. In addition, a simple computation yields
that ∇f tr (x) = ∇f soc (x)e = (f 0 )soc (x). Similarly, by part (b), the second part follows.
2
4
Theorem 2.1 For any given f : J → IR, let f soc : S → IRn and f tr : S → IR be given by
(5) and (6), respectively. Assume that J is open. If f is twice differentiable on J, then
(a) f 00 (t) ≥ 0 for any t ∈ J ⇐⇒ ∇(f 0 )soc (x) 0 for any x ∈ S ⇐⇒ f tr is convex in S.
(b) f 00 (t) > 0 for any t ∈ J ⇐⇒ ∇(f 0 )soc (x) 0 ∀x ∈ S =⇒ f tr is strictly convex in S.
Proof. (a) By Lemma 2.2(c), ∇2 f tr (x) = ∇(f 0 )soc (x) for any x ∈ S, and the second
equivalence follows by Prop. B.4(a) and (c) of [4]. We next come to the first equivalence.
By Lemma 2.2(b), for any fixed x ∈ S, ∇(f 0 )soc (x) = f 00 (x1 )I if x2 = 0, and otherwise
∇(f 0 )soc (x) has the same expression as in (10) except that
b(x) =
f 00 (λ2 (x))+ f 00 (λ1 (x))
f 00 (λ2 (x))− f 00 (λ1 (x))
f 0 (λ2 (x))− f 0 (λ1 (x))
, c(x) =
, a(x) =
.
2
2
λ2 (x) − λ1 (x)
Assume that ∇(f 0 )soc (x) 0 for any x ∈ S. Then, we readily have b(x) ≥ 0 for any
x ∈ S. Noting that b(x) = f 00 (x1 ) when x2 = 0, we particularly have f 00 (x1 ) ≥ 0 for all
x1 ∈ J, and consequently f 00 (t) ≥ 0 for all t ∈ J. Assume that f 00 (t) ≥ 0 for all t ∈ J. Fix
any x ∈ S. Clearly, b(x) ≥ 0 and a(x) ≥ 0. If b(x) = 0, then f 00 (λ1 (x)) = f 00 (λ2 (x)) = 0,
and consequently c(x) = 0, which in turn implies that
"
#
0
0
0.
∇(f 0 )soc (x) =
(11)
x xT
0 a(x) I − kx22 k22
If b(x) > 0, then by the first equivalence of Lemma 2.1 and the expression of ∇(f 0 )soc (x)
it suffices to argue that the following matrix
a(x)I + (b(x) − a(x))
c2 (x) x2 xT2
x2 xT2
−
kx2 k2
b(x) kx2 k2
(12)
is positive semidefinite. Since the rank-one matrix x2 xT2 has only one nonzero eigenvalue
kx2 k2 , the matrix in (12) has one eigenvalue a(x) of multiplicity n − 1 and one eigenvalue
2 −c(x)2
b(x)2 −c(x)2
of multiplicity 1. Since a(x) ≥ 0 and b(x)b(x)
= f 00 (λ1 (x))f 00 (λ2 (x)) ≥ 0, the
b(x)
matrix in (12) is positive semidefinite. By the arbitrary of x, we have that ∇(f 0 )soc (x) 0
for all x ∈ S.
(b) The first equivalence is direct by using (9) of Lemma (2.1), noting ∇(f 0 )soc (x) 0
implies a(x) > 0 when x2 6= 0, and following the same arguments as part (a). The second
part is due to [4, Prop. B.4(b)].
2
Remark 2.1 Note that the strict convexity of f tr does not necessarily imply the positive
definiteness of ∇2 f tr (x). Consider f (t) = t4 for t ∈ IR. We next show that f tr is strictly
5
convex. Indeed, f tr is convex in IRn by Theorem 2.1(a) since f 00 (t) = 12t2 ≥ 0. Taking
into account that f tr is continuous, it remains to prove that
f tr (x) + f tr (y)
x+y
tr
f
=
=⇒ x = y.
(13)
2
2
Since h(t) = (t0 + t)4 + (t0 − t)4 for some t0 ∈ IR is increasing on [0, +∞), and the
function f (t) = t4 is strictly convex in IR, we have that
f
tr
x+y
2
4
x+y
λ1
+ λ2
2
4 4
x1 + y1 − kx2 + y2 k
x1 + y1 + kx2 + y2 k
+
2
2
4 4
x1 + y1 − kx2 k − ky2 k
x1 + y1 + kx2 k + ky2 k
+
2
2
4 4
λ1 (x) + λ1 (y)
λ2 (x) + λ2 (y)
+
2
2
4
4
4
(λ1 (x)) + (λ1 (y)) + (λ2 (x)) + (λ2 (y))4
2
f tr (x) + f tr (y)
,
2
=
=
≤
=
≤
=
x+y
2
4
and moreover, the above inequalities become the equalities if and only if
kx2 + y2 k = kx2 k + ky2 k, λ1 (x) = λ1 (y), λ2 (x) = λ2 (y).
It is easy to verify that the three equalities hold if and only if x = y. So, the implication
in (13) holds, i.e., f tr is strictly convex. However, by Theorem 2.1(b), ∇(f 0 )soc (x) 0
does not hold for all x ∈ IRn since f 00 (t) > 0 does not hold for all t ∈ IR.
It should be mentioned that the fact that the strict convexity of f implies the strict
convexity of f tr was proved in [2, 8] via the definition of convex function, but here we use
the Shcur Complement Theorem and the relation between ∇(f 0 )soc and ∇2 f tr to establish
the convexity of SOC-trace functions. In addition, we also note that the necessity involved
in the first equivalence of Theorem 2.1(a) was given in [11] via a different way. Next, we
illustrate the application of Theorem 2.1 with some SOC-trace functions.
Proposition 2.1 The following functions associated with Kn are all strictly convex.
(a) F1 (x) = − ln(det(x)) for x ∈ intKn .
(b) F2 (x) = tr(x−1 ) for x ∈ intKn .
6
(c) F3 (x) = tr(φ(x)) for x ∈ intKn , where
( p+1
1−q
x
−e
+ x q−1−e if p ∈ [0, 1], q > 1;
p+1
φ(x) =
xp+1 −e
− ln x if p ∈ [0, 1], q = 1.
p+1
(d) F4 (x) = − ln(det(e − x)) for x ≺Kn e.
(e) F5 (x) = tr((e − x)−1 ◦ x) for x ≺Kn e.
(f ) F6 (x) = tr(exp(x)) for x ∈ IRn .
(g) F7 (x) = ln(det(e + exp(x))) for x ∈ IRn .
x + (x2 + 4e)1/2
for x ∈ IRn .
(h) F8 (x) = tr
2
Proof. Note that F1 (x), F2 (x) and F3 (x) are the SOC-trace functions associated with
f1 (t) = − ln t (t > 0), f2 (t) = t−1 (t > 0) and f3 (t) (t > 0), respectively, where
( p+1
1−q
t
−1
+ t q−1−1 if p ∈ [0, 1], q > 1,
p+1
f3 (t) =
tp+1 −1
− ln t if p ∈ [0, 1], q = 1;
p+1
F4 (x) is the SOC-trace function associated with f4 (t) = − ln(1 − t) (t < 1), F5 (x) is the
t
SOC-trace function associated with f5 (t) = 1−t
(t < 1) by noting that
(e − x)−1 ◦ x =
λ2 (x) (2)
λ1 (x) (1)
ux +
u ;
λ1 (e − x)
λ2 (e − x) x
F6 (x) and F7 (x) are the SOC-trace functions associated with f6 (t) = exp(t) (t ∈ IR)
and f7 (t) = ln(1 + exp(t)) (t ∈ √IR), respectively,
and F8 (x) is the SOC-trace function
associated with f7 (t) = 2−1 t + t2 + 4 (t ∈ IR). It is easy to verify that the functions
f1 -f8 have positive second-order derivatives in their respective domain, and therefore
F1 -F8 are strictly convex functions by Theorem 2.1(b).
2
The functions F1 , F2 and F3 are the popular barrier functions which play a key role in
the development of interior point methods for SOCPs, see, e.g., [6, 7, 14, 15, 17], where
F3 covers a wide range of barrier functions, including the classical logarithmic barrier
function, the self-regular functions and the non-self-regular functions; see [7] for details.
The functions F4 and F5 are the popular shifted barrier functions [1, 2, 3] for SOCPs, and
F6 -F8 can be used as penalty functions for second-order cone programs (SOCPs), and
these functions are added to the objective of SOCPs for forcing the solution to be feasible.
Besides the application in establishing convexity for SOC-trace functions, the Schur
complement theorem can be employed to establish convexity of some compound functions
of SOC-trace functions and scalar-valued functions, which is usually difficult to achieve by
the definition of convex function. The following proposition presents such an application.
7
Proposition 2.2 For any x ∈ Kn , let F9 (x) := −[det(x)]1/p with p > 1. Then,
(a) F9 is twice continuously differentiable in intKn .
(b) F9 is convex when p ≥ 2, and moreover, it is strictly convex when p > 2.
Proof. (a) Note that −F9 (x) = exp (p−1 ln(det(x))) for any x ∈ intKn , and ln(det(x)) =
f tr (x) with f (t) = ln(t) for t ∈ IR++ . By Lemma 2.2(c), ln(det(x)) is twice continuously
differentiable in intKn . Hence −F9 (x) is twice continuously differentiable in intKn . The
result then follows.
(b) In view of the continuity of F9 , we only need to prove its convexity over intKn . By
part (a), we next achieve this goal by proving that the Hessian matrix ∇2 F9 (x) for any
x ∈ intKn is positive semidefinite when p ≥ 2, and positive definite when p > 2. Fix any
x ∈ intKn . From direct computations, we obtain
"
1 −1 #
1
(2x1 ) x21 − kx2 k2 p
∇F9 (x) = −
1
p (−2x2 ) x21 − kx2 k2 p −1
and

∇2 F9 (x) =
1
p−1
(det(x)) p −2 
2
p
4x21
−
2p(x21 −kx2 k2 )
p−1
−4x1 x2
−4x1 xT2
2p(x21 −kx2 k2 )
4x2 xT2 +
I
p−1

.
(14)
Since x ∈ intKn , we have x1 > 0 and det(x) = x21 − kx2 k2 > 0, and therefore
2p (x21 − kx2 k2 )
2p
2p
2
a1 (x) := 4x1 −
= 4−
x21 +
kx2 k2 .
p−1
p−1
p−1
We next proceed the arguments by the following two cases: (1) a1 (x) = 0; (2) a1 (x) > 0.
Case 1: a1 (x) = 0. Since p ≥ 2, under this case we must have x2 = 0, and consequently,
2
0
0
p−1
−4
2
∇ F9 (x) =
(x1 ) p
0.
2p 2
0 p−1
x1 I
p2
Case 2: a1 (x) > 0. Under this case, we calculate that
2p (x21 − kx2 k2 )
2p (x21 − kx2 k2 )
2
T
4x1 −
4x2 x2 +
I − 16x21 x2 xT2
p−1
p−1
2
2
4p (x1 − kx2 k ) p − 2 2
p
2
T
=
x I+
kx2 k I − 2x2 x2 .
p−1
p−1 1
p−1
(15)
Since the rank-one matrix 2x2 xT2 has only one nonzero eigenvalue 2kx2 k2 , the matrix in
the bracket of the right hand side of (15) has one eigenvalue of multiplicity 1 given by
p−2 2
p
p−2 2
x1 +
kx2 k2 − 2kx2 k2 =
x1 − kx2 k2 ≥ 0,
p−1
p−1
p−1
8
p
and one eigenvalue of multiplicity n − 1 given by p−2
x2 + p−1
kx2 k2 ≥ 0. Furthermore, we
p−1 1
see that these eigenvalues must be positive when p > 2 since x21 > 0 and x21 − kx2 k2 >
0. This means that the matrix on the right hand side of (15) is positive semidefinite,
and moreover, it is positive definite when p > 2. Applying Lemma 2.1, we have that
∇2 F9 (x) 0, and furthermore ∇2 F9 (x) 0 when p > 2.
Since a1 (x) > 0 must hold when p > 2, the arguments above show that F9 (x) is
convex over intKn when p ≥ 2, and strictly convex over intKn when p > 2.
2
It is worthwhile to point out that det(x) is neither convex nor concave on Kn , and it
is difficult to argue the convexity of those compound functions involving det(x) by the
definition of convex function. But, as shown in Proposition 2.2, the Schur Complement
Theorem offers a simple way to prove their convexity.
To close section, we take a look at the application of some of convex functions above
in establishing inequalities associated with SOCs. Some of these inequalities have been
used to analyze the properties of SOC-function f soc [10] and the convergence of interior
point methods for SOCPs [2].
Proposition 2.3 For any x Kn 0 and y Kn 0, the following inequalities hold.
(a) det(αx + (1 − α)y) ≥ (det(x))α (det(y))1−α for any 0 < α < 1.
2
(b) det(x + y)1/p ≥ 2 p −1 det(x)1/p + det(y)1/p for any p ≥ 2.
(c) det(x + y) ≥ det(x) + det(y).
(d) det(αx + (1 − α)y) ≥ α2 det(x) + (1 − α)2 det(y) for any 0 < α < 1.
(e) [det(e + x)]1/2 ≥ 1 + det(x)1/2 .
(f ) If x Kn y, then det(x) ≥ det(y).
1
1/2
tr(x ◦ y) : det(y) = 1, y Kn 0 . Furthermore, when x Kn 0,
(g) det(x)
= inf
2
the same relation holds with inf replaced by min.
(h) tr(x ◦ y) ≥ 2det(x)1/2 det(y)1/2 .
Proof. (a) From Prop. 2.1(a), we know that ln(det(x)) is strictly concave in intKn .
Thus,
ln(det(αx + (1 − α)y)) ≥ α ln(det(x)) + (1 − α)det(y)
= ln(det(x)α ) + ln(det(x)1−α )
9
for any 0 < α < 1 and x, y ∈ intKn . This, together with the increasing of ln t (t > 0)
and the continuity of det(x), implies the desired result.
(b) By Prop. 2.2(b), det(x)1/p is concave over Kn . So, for any x, y ∈ Kn , we have that
1/p
x+y
1
det
≥
det(x)1/p + det(y)1/p
2
2
"
#1/p
2 x2 + y2 2
1/p
1/p
x1 + y 1
≥ x21 − kx2 k2
+ y12 − ky2 k2
⇐⇒ 2
−
2
2
1
1/p i
1/p
4p h 2
2
2 1/p
+ y12 − ky2 k2
≥
⇐⇒ (x1 + y1 ) − kx2 + y2 k
x1 − kx2 k2
2
2
−1
1/p
1/p
p
⇐⇒ det(x + y) ≥ 2
det(x) + det(y)1/p ,
which is the desired result.
(c) Using the inequality in part (b) with p = 2, we have
det(x + y)1/2 ≥ det(x)1/2 + det(y)1/2 .
Squaring both sides yields
det(x + y) ≥ det(x) + det(y) + 2det(x)1/2 det(y)1/2 ≥ det(x) + det(y),
where the last inequality is by the nonnegativity of det(x) and det(y) since x, y ∈ Kn .
(d) The inequality is direct by part (c) and the fact det(αx) = α2 det(x).
(e) The inequality follows from part (b) with p = 2 and the fact that det(e) = 1.
(f) Using part (c) and noting that x Kn y, it is easy to verify that
det(x) = det(y + x − y) ≥ det(y) + det(x − y) ≥ det(y).
(g) Using the Cauchy-Schwartz inequality, it is easy to verify that
tr(x ◦ y) ≥ λ1 (x)λ2 (y) + λ1 (y)λ2 (x)
∀x, y ∈ IRn .
For any x, y ∈ Kn , this along with the arithmetic-geometric mean inequality implies that
λ1 (x)λ2 (y) + λ1 (y)λ2 (x)
tr(x ◦ y)
≥
2
2
p
≥
λ1 (x)λ2 (y)λ1 (y)λ2 (x)
= det(x)1/2 det(y)1/2 ,
1
tr(x ◦ y) : det(y) = 1, y Kn 0 = det(x)1/2 for a fixed x ∈
which means that inf
2
−1
Kn . If x Kn 0, then we can verify that the feasible point y ∗ = √x
is such that
det(x)
1
tr(x ◦ y ∗ ) = det(x)1/2 , and the second part follows.
2
10
(h) Using part (g), for any x ∈ Kn and y ∈ int(Kn ), we have that
!
p
tr(x ◦ y)
y
p
= tr x ◦ p
≥ det(x),
2 det(y)
det(y)
which together with the continuity of det(x) and tr(x) implies that
tr(x ◦ y) ≥ 2det(x)1/2 det(y)1/2
Thus, we complete the proof.
∀x, y ∈ Kn .
2
Note that some of the inequalities in Prop. 2.3 were ever established with the help of
the Schwartz-inequality [10], but here we achieve the goal easily by using the convexity
of SOC-functions. These inequalities all have the corresponding counterparts for matrix
inequalities [5, 13, 16]. For example, Prop. 2.3(b) with p = 2, i.e., p equal to the rank of
Jordan algebra (IRn , ◦), corresponds to the Minkowski inequality of matrix case:
det(A + B)1/n ≥ det(A)1/n + det(B)1/n
for any n × n positive semidefinite matrices A and B.
3
Conclusions
We studied an application of the Schur complement theorem in establishing convexity
of SOC-functions, especially for SOC-trace functions, which are the key of penalty and
barrier function methods for SOCPs and some important inequalities associated with
SOCs. One possible future direction is proving self-concordancy for such barrier/penalty
functions associated with SOCs. We also believe that the results in this paper will be
helpful towards establishing further properties of other SOC-functions.
References
[1] A. Auslender, Penalty and barrier methods: a unified framework, SIAM Journal
on Optimization, vol. 10, pp. 211-230, 1999.
[2] A. Auslender, Variational inequalities over the cone of semidefinite positive symmetric matrices and over the Lorentz cone, Optimization Methods and Software, vol.
18, pp. 359-376, 2003.
[3] A. Auslender and H. Ramirez, Penalty and barrier methods for convex semidefinite programming, Mathematical Methods of Operations Research, vol. 63, pp. 195219, 2006.
11
[4] D. P. Bertsekas, Nonlinear Programming, 2nd edition, Athena Scientific, 1999.
[5] R. Bhatia, Matrix Analysis, Springer-Verlag, New York, 1997.
[6] A. Ben-Tal and A. Nemirovski, Lectures on Modern Convex Optimization: Analysis, Algorithms and Engineering Applications, MPS-SIAM Series on Optimization.
SIAM, Philadelphia, USA, 2001.
[7] Y.-Q. Bai and G. Q. Wang, Primal-dual interior-point algorithms for second-order
cone optimization based on a new parametric kernel function, Acta Mathematica
Sinica, vol. 23, pp. 2027-2042, 2007.
[8] H. Bauschke, O. Güler, A. S. Lewis and S. Sendow, Hyperbolic polynomial
and convex analysis, Canadian Journal of Mathematics, vol. 53, pp. 470–488, 2001.
[9] J.-S. Chen, X. Chen and P. Tseng, Analysis of nonsmooth vector-valued functions associated with second-oredr cone, Mathmatical Programming, vol. 101, pp.
95-117, 2004.
[10] J.-S. Chen, The convex and monotone functions associated with second-order cone,
Optimization, vol. 55, pp. 363-385, 2006.
[11] M. Fukushima, Z.-Q Luo and P. Tseng, Smoothing functions for second-order
cone complementarity problems, SIAM Journal on Optimazation, vol. 12, pp. 436-460,
2002.
[12] J. Faraut and A. Korányi, Analysis on symmetric Cones, Oxford Mathematical
Monographs, Oxford University Press, New York, 1994.
[13] R. A. Horn and C. R. Johnson, Matrix Analysis, Cambridge University Press,
Cambridge, 1986.
[14] R. D. C. Monteiro and T. Tsuchiya, Polynomial convergence of primal-dual
algorithms for the second-order cone programs based on the MZ-family of directions,
Mathematical Programming, vol. 88, pp. 61–83, 2000.
[15] J. Peng, C. Roos and T. Terlaky, Self-Regularity, A New Paradigm for PrimalDual Interior-Point Algorithms, Princeton University Press, 2002.
[16] J. Schott, Matrix Analysis for Statistics, John Wiley and Sons, 2nd edition, 2005.
[17] T. Tsuchiya, A convergence analysis of the scaling-invariant primal-dual pathfollowing algorithms for second-order cone programming, Optimization Methods and
Software, vol. 11, pp. 141–182, 1999.
12