Chapter 4
SOC means and SOC inequalities
In this chapter, we present some other types of applications of the aforementioned SOC-functions, SOC-convexity, and SOC-monotonicity. These include so-called SOC means, SOC weighted means, and a few SOC trace versions of Young, H¨older, Minkowski in-equalities, and Powers-Størmer’s inequality. We believe that these results will be helpful in convergence analysis of optimizations involved with SOC. Many materials of this chap-ter are extracted from [36, 77, 78], the readers can look into them for more details.
(f) m(a, b) is continuous in a, b.
Many types of means have been investigated in the literature, to name a few, the arithmetic mean, geometric mean, harmonic mean, logarithmic mean, identric mean, contra-harmonic mean, quadratic (or root-square) mean, first Seiffert mean, second Seif-fert mean, and Neuman-Sandor mean, etc.. In addition, many inequalities describing the relationship among different means have been established. For instance, for any two positive real number a, b, it is well-known that
min{a, b} ≤ H(a, b) ≤ G(a, b) ≤ L(a, b) ≤ A(a, b) ≤ max{a, b}, (4.1) where
H(a, b) = 2ab a + b, G(a, b) = √
ab, L(a, b) =
a − b
ln a − ln b if a 6= b,
a if a = b,
A(a, b) = a + b 2 ,
represents the harmonic mean, geometric mean, logarithmic mean, and arithmetic mean, respectively. For more details regarding various means and their inequalities, please refer to [31, 65].
Recently, the matrix version of means have been generalized from the classical means, see [22, 24–26]. In particular, the matrix version of Arithmetic Geometric Mean Inequal-ity (AGM) is proved in [22, 23], and has attracted much attention. Indeed, let A and B be two n × n positive definite matrices, the following inequalities hold under the partial order induced by positive semidefinite matrices cone S+n:
(A : B) A#B 1
2(A + B), (4.2)
where
A : B = 2 A−1+ B−1−1 , A#B = A1/2 A−1/2BA−1/21/2
A1/2,
denote the matrix harmonic mean and the matrix geometric mean, respectively. For more details about matrix means and their related inequalities, please see [22, 24–26, 88]
and references therein.
Note that the nonnegative orthant, the cone of positive semidefinite matrices, and the second-order cone all belong to the class of symmetric cones [61]. This motivates us to consider further extension of means, that is, the means associated with SOC. More specifically, in this section, we generalize some well-known means to the SOC setting and build up some inequalities under the partial order induced by Kn. One trace inequality is established as well. For achieving these results, the SOC-monotonicity contributes a lot in the analysis. That is the application aspect of SOC-monotone function that we want to illustrate.
The relation Kn is not a linear ordering. Hence, it is not possible to compare any two vectors (elements) via Kn. Nonetheless, we note that for any a, b ∈ IR
max{a, b} = b + [a − b]+ = 1
2(a + b + |a − b|), min{a, b} = a − [a − b]+ = 1
2(a + b − |a − b|).
This motivates us to define the supremum and infimum of {x, y}, denoted by x ∨ y and x ∧ y respectively, in the SOC setting as follows. For any x, y ∈ IRn, we let
x ∨ y := y + [x − y]+= 1
2(x + y + |x − y|),
x ∧ y := x − [x − y]+= 12(x + y − |x − y|), if x + y Kn |x − y|;
0, otherwise.
In view of the above expressions, we define the SOC means in a similar way.
Definition 4.1. A binary operation (x, y) 7→ M (x, y) defined on int(Kn) × int(Kn) is called an SOC mean if the following conditions are satisfied:
(i) M (x, y) Kn 0;
(ii) x ∧ y Kn M (x, y) Kn x ∨ y;
(iiii) M (x, y) is monotone in x, y;
(iv) M (αx, αy) = αM (x, y), α > 0;
(v) M (x, y) is continuous in x, y.
We start with the simple SOC arithmetic mean A(x, y) : int(Kn)×int(Kn) → int(Kn), which is defined by
A(x, y) = x + y
2 . (4.3)
It is clear that A(x, y) satisfies all the above conditions. Besides, it is not hard to verify that the SOC harmonic mean of x and y, H(x, y) : int(Kn) × int(Kn) → int(Kn), can be defined as
H(x, y) = x−1+ y−1 2
−1
. (4.4)
The relation between A(x, y) and H(x, y) is described as below.
Proposition 4.1. Let A(x, y), H(x, y) be defined as in (4.3) and (4.4), respectively. For any x Kn 0, y Kn 0, there holds
x ∧ y Kn H(x, y) Kn A(x, y) Kn x ∨ y.
Proof. (i) To verify the first inequality, if 12(x + y − |x − y|) /∈ Kn, the inequality holds clearly. Suppose 12(x + y − |x − y|) Kn 0, we note that 12(x + y − |x − y|) Kn x and
1
2(x + y − |x − y|) Kn y. Then, using the SOC-monotonicity of f (t) = −t−1 shown in Proposition 2.3, we obtain
x−1 Kn x + y − |x − y|
2
−1
and y−1 Kn x + y − |x − y|
2
−1
, which imply
x−1+ y−1
2 Kn x + y − |x − y|
2
−1
. Next, applying the SOC-monotonicity again, we conclude that
x + y − |x − y|
2 Kn x−1+ y−1 2
−1 .
(ii) To see the second inequality, we first observe that
x−1+ y−1 2
−1
Kn 1
2(x−1)−1+ 1
2(y−1)−1 = x + y 2 , where the inequality comes from the SOC-convexity of f (t) = t−1. (iii) To check the last inequality, we observe that
x + y
2 Kn x + y + |x − y|
2 ⇐⇒ 0 Kn |x − y|
2 ,
where it is clear |x − y| Kn 0 always holds for any element x, y. Then, the desired result follows.
Now, we consider the SOC geometric mean, denoted by G(x, y), which can be bor-rowed from the geometric mean of symmetric cone, see [101]. More specifically, let V
be a Euclidean Jordan algebra, K be the set of all square elements of V (the associated symmetric cone), and Ω := intK (the interior symmetric cone). For x ∈ V , let L(x) denote the linear operator given by L(x)y := x ◦ y, and let
P (x) := 2L(x)2− L(x2). (4.5)
The mapping P is called the quadratic representation of V . If x is invertible, then we have
P (x)K = K and P (x)Ω = Ω.
Suppose that x, y ∈ Ω, the geometric mean of x and y, denoted by x#y, is x#y := P (x12)(P (x−12)y)12.
On the other hand, it turns out that the cone Ω admits a G(Ω)-invariant Riemannian metric [61]. The unique geodesic curve joining x and y is
t 7→ x#ty := P (x12)
P (x−12)yt
,
and the geometric mean x#y is the midpoint of the geodesic curve. In addition, Lim establishes the arithmetic-geometric-harmonic means inequalities [101, Theorem 2.8],
x−1+ y−1 2
−1
K x#y K
x + y
2 , (4.6)
where K is the partial order induced by the closed convex cone K. The inequality (4.6) includes the inequality (4.2) as a special case. For more details, please refer to [101]. As an example of Euclidean Jordan algebra, for any x and y in int(Kn), we therefore adopt the geometric mean G(x, y) as
G(x, y) := P (x12)
P (x−12)y
12
. (4.7)
Then, we immediately have the following parallel properties of SOC geometric mean.
Proposition 4.2. Let A(x, y), H(x, y), G(x, y) be defined as in (4.3), (4.4) and (4.7), respectively. Then, for any x Kn 0 and y Kn 0, we have
(a) G(x, y) = G(y, x).
(b) G(x, y)−1 = G(x−1, y−1).
(c) H(x, y) Kn G(x, y) Kn A(x, y).
Next, we look into another type of SOC mean, the SOC logarithmic mean L(x, y).
First, for any two positive real numbers a, b, Carlson [32] has set up the integral repre-sentation:
L(a, b) =
Z 1 0
dt ta + (1 − t)b
−1 ,
whereas Neuman [112] has also provided an alternative integral representation:
L(a, b) = Z 1
0
a1−tbtdt.
Moreover, Bhatia [22, page 229] proposes the matrix logarithmic mean of two positive definite matrices A and B as
L(A, B) = A1/2 Z 1
0
A−1/2BA−1/2t
dt A1/2. In other words,
L(A, B) = Z 1
0
A#tB dt, where A#tB =: A1/2 A−1/2BA−1/2t
A1/2 = P (A1/2)(P (A−1/2)B)tis called the t-weighted geometric mean. We remark that A#tB = A1−tBt for AB = BA, and the definition of logarithmic mean coincides with the one of real numbers. This integral representation motivates us to define the SOC logarithmic mean on int(Kn) × int(Kn) as
L(x, y) = Z 1
0
x#ty dt. (4.8)
To verify it is an SOC mean, we need the following technical lemmas. The first lemma is the symmetric cone version of Bernoulli inequality.
Lemma 4.1. Let V be a Euclidean Jordan algebra, K be the associated symmetric cone, and e be the Jordan identity. Then,
(e + s)tK e + ts,
where 0 ≤ t ≤ 1, s K −e, and the partial order is induced by the closed convex cone K.
Proof. For any s ∈ V , we denote the spectral decomposition of s as
r
X
i=1
λici. Since s K −e, we obtain that each eigenvalue λi ≥ −1. Then, we have
(e + s)t = (1 + λ1)tc1+ (1 + λ2)tc2+ · · · + (1 + λr)tcr
K (1 + tλ1)c1+ (1 + tλ2)c2+ · · · + (1 + tλr)cr
= e + ts,
where the inequality holds by the real number version of Bernoulli inequality. Lemma 4.1 is the Bernoulli Inequality associated with symmetric cone although we will use it only in the SOC setting.
Lemma 4.2. Suppose that u(t) : IR → IRn is integrable on [a, b].
(a) If u(t) Kn 0 for any t ∈ [a, b], then Rb
au(t)dt Kn 0.
(b) If u(t) Kn 0 for any t ∈ [a, b], thenRb
a u(t)dt Kn 0.
Proof. (a) Consider the partition P = {t0, t1, . . . , tn} of [a, b] with tk = a + k(b − a)/n and some ¯tk ∈ [tk−1, tk], we have
Z b a
u(t)dt = lim
n→∞
n
X
k=1
u(¯tk)b − a
n Kn 0 because u(t) Kn 0 and Kn is closed.
(b) For convenience, we write u(t) = (u1(t), u2(t)) ∈ IR × IRn−1, and let
¯
u(t) = (ku2(t)k, u2(t)) ,
˜
u(t) = (u1(t) − ku2(t)k, 0) . Then, we have
u(t) = ¯u(t) + ˜u(t) and
u(t) ¯ Kn 0,
u1(t) − ku2(t)k > 0.
Note that Rb
au(t)dt = (˜ Rb
a(u1(t) − ku2(t)k)dt, 0) Kn 0 since u1(t) − ku2(t)k > 0. This together with Rb
au(t)dt ¯ Kn 0 yields that Z b
a
u(t)dt = Z b
a
¯
u(t)dt + Z b
a
˜
u(t)dt Kn 0.
Thus, the proof is complete.
Proposition 4.3. Suppose that u(t) : IR → IRn and v(t) : IR → IRn are integrable on [a, b].
(a) If u(t) Kn v(t) for any t ∈ [a, b], then Rb
a u(t)dt Kn Rb
a v(t)dt.
(b) If u(t) Kn v(t) for any t ∈ [a, b], then Rb
a u(t)dt Kn Rb
a v(t)dt.
Proof. It is an immediate consequence of Lemma 4.2.
Proposition 4.4. Let A(x, y), G(x, y), and L(x, y) be defined as in (4.3), (4.7), and (4.8), respectively. For any x Kn 0, y Kn 0, there holds
G(x, y) Kn L(x, y) Kn A(x, y), and hence L(x, y) is an SOC mean.
Proof. (i) To verify the first inequality, we first note that G(x, y) = P (x12)(P (x−12)y)12 =
Z 1 0
P (x12)(P (x−12)y)12dt.
Let s = P (x−12)y = λ1u(1)s + λ2u(2)s . Then, we have L(x, y) − G(x, y)
= Z 1
0
P (x12)(P (x−12)y)t dt − P (x12)(P (x−12)y)12
= Z 1
0
P (x12) λt1u(1)s + λt2u(2)s
dt − P (x12)p
λ1u(1)s +p
λ2u(2)s
=
Z 1 0
λt1dt
P (x12)u(1)s +
Z 1 0
λt2dt
P (x12)u(2)s − P (x12)p
λ1u(1)s +p
λ2u(2)s
=
λ1− 1
ln λ1− ln 1 −p λ1
P (x12)u(1)s +
λ2− 1
ln λ2− ln 1 −p λ2
P (x12)u(2)s
= [L(λ1, 1) − G(λ1, 1)] P (x12)u(1)s + [L(λ2, 1) − G(λ2, 1)] P (x12)u(2)s
Kn 0,
where last inequality holds by (4.1) and P (x12)u(i)s ∈ Kn. Thus, we obtain the first inequality.
(ii) To see the second inequality, we let s = P (x−12)y − e. Then, we have s Kn −e, and applying Lemma 4.1 gives
e + P (x−12)y − et
Kn e + th
P (x−12)y − ei , which is equivalent to
0 Kn (1 − t)e + th
P (x−12)yi
−
P (x−12)yt
. Since P (x12) is invariant on Kn, we have
0 Kn P (x12)
(1 − t)e + th
P (x−12)yi
−
P (x−12)yt
= (1 − t)x + ty − x#ty.
Hence, by Proposition 4.3, we obtain L(x, y) =
Z 1 0
x#ty dt Kn Z 1
0
[(1 − t)x + ty] dt = A(x, y).
The proof is complete.
Finally, for SOC quadratic mean, it is natural to consider the following Q(x, y) := x2+ y2
2
1/2
.
It is easy to verify A(x, y) Kn Q(x, y). However, Q(x, y) does not satisfy the property(ii) mentioned in the definition of SOC mean. Indeed, taking x =
31 10
−20
∈ Kn and y =
10
9 0
∈ Kn, it is obvious that x Kn y. In addition, by simple calculation, we have
x2+ y2 2
1/2
=
s
400
−6202s 2s
≈
24.30
8.23
−12.76
,
where s = r
1 2
821 +p8212 − (4002+ 6202)
≈ 24.30. However,
x ∨ y − x2+ y2 2
1/2
≈
6.7 1.77
−7.24
is not in Kn. Hence, this definition of Q(x, y) cannot officially serve as an SOC mean.
To sum up, we already have the following inequalities
x ∧ y Kn H(x, y) Kn G(x, y) Kn L(x, y) Kn A(x, y) Kn x ∨ y,
but we do not have SOC quadratic mean. Nevertheless, we still can generalize all the means inequalities as in (4.1) to SOC setting when the dimension is 2. To see this, the Jordan product on second-order cone of order 2 satisfies the associative law and closedness such that the geometric mean
G(x, y) = x1/2◦ y1/2 and the logarithmic mean
L(x, y) = Z 1
0
x1−t◦ yt dt
are well-defined (note this is true only when n = 2) and coincide with the definition (4.7), (4.8). Then, the following inequalities
x ∧ y
K2 H(x, y)
K2 G(x, y)
K2 L(x, y)
K2 A(x, y)
K2 Q(x, y)
K2 x ∨ y hold as well.
By applying Proposition 1.1(a), we immediately obtain one trace inequality for SOC mean.
Proposition 4.5. Let A(x, y), H(x, y), G(x, y) and L(x, y) be defined as in (4.3)-(4.4), (4.7)-(4.8), respectively. For any x Kn 0, y Kn 0, there holds
tr(x ∧ y) ≤ tr(H(x, y)) ≤ tr(G(x, y)) ≤ tr(L(x, y)) ≤ tr(A(x, y)) ≤ tr(x ∨ y).