## SOC Functions and Their Applications

### by

### Jein-Shan Chen

Department of Mathematics National Taiwan Normal University

### January 07, 2019

### Preface

The second-order cone programs (SOCP) have been an attraction due to plenty of ap- plications in engineering, data science, and finance. To deal with this special type of optimization problems involving second-order cone (SOC). We believe that the following items are crucial concepts: (i) spectral decomposition associated with SOC, (ii) analy- sis of SOC functions, (iii) SOC-convexity and SOC-monotonicity. In this book, we go through all these concepts and try to provide the readers a whole picture regarding SOC functions and their applications.

As introduced in Chapter 1, the SOC functions are indeed vector-valued functions associated with SOC, which are accompanied by Jordan product. However, unlike the matrix multiplication, the Jordan product associated with SOC is not associative which is the main source of difficulty when we do the analysis. Therefore, the ideas for proofs are usually quite different from those for matrix-valued functions. In other words, al- though SOC and positive semidefinite cone both belong to symmetric cones, the analysis for them are different. In general, the arguments are more tedious and need subtle ar- rangements in the SOC setting. This is due to the feature of SOC.

To deal with second-order cone programs (SOCPs) and second-order cone complemen- tarity problems (SOCCPs), many methods rely on some SOC complementarity functions or merit functions to reformulate the KKT optimality conditions as a nonsmooth (or smoothing) system of equations or an unconstrained minimization problem. In fact, such SOC complementarity or merit functions are connected to SOC functions. In other words, 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 help- ful for developing and analyzing more solutions methods.

For SOCP, there are still many approaches without using SOC complementarity func- tions. In this case, the concepts of SOC-convexity and SOC-monotonicity introduced in Chapter 2 play a key to those solution methods. In Chapter 3, we present proximal-type algorithms in which SOC-convexity and SOC-monotonicity are needed in designing so- lution methods and proving convergence analysis.

In Chapter 4, we pay attention to some other types of applications of SOC-functions, SOC-convexity, and SOC-monotonicity introduced in this monograph. These include so-called SOC means, SOC weighted means, and a few SOC trace versions of Young, H¨older, Minkowski inequalities, and Powers-Størmer’s inequality. All these materials are newly discovered and we believe that they will be helpful in convergence analysis of var- ious optimizations involving SOC. Chapter 5 offers a direction for future investigation, although it is not very consummate yet.

This book is based on my series of study regarding second-order cone, SOCP, SOCCP, SOC-functions, etc. during the past fifteen years. It is dedicated to the memory of my supervisor, Prof. Paul Tseng, who guided me into optimization research, especially to second-order cone optimization. Without his encouragement, it is not possible to achieve the whole picture of SOC-functions, which is the main role of this monograph. His attitude towards doing research always remains in my heart, albeit he got missing in 2009. I would like to thank all my co-authors of the materials that appear in this book, including Prof. Shaohua Pan, Prof. Xin Chen, Prof. Jiawei Zhang, Prof. Yu-Lin Chang, Dr. Chien-Hao Huang, etc.. The collaborations with them are wonderful and enjoyable experiences. I also thank Dr. Chien-Hao Huang, Dr. Yue Lu, Dr. Liguo Jiao, Prof.

Xinhe Miao, and Prof. Chu-Chin Hu for their help on proofreading. Final gratitude goes to my family, Vivian, Benjamin, and Ian, who offer me support and stimulate endless strength in pursuing my exceptional academic career.

January 07, 2019 Taipei, Taiwan

### Notations

• Throughout this book, an n-dimensional vector x = (x_{1}, x_{2}, · · · , x_{n}) ∈ IR^{n} means
a column vector, i.e.,

x =

x1

x_{2}
...
xn

.

In other words, without ambiguity, we also write the column vector as x = (x_{1}, x_{2}, · · · , x_{n}).

• IR^{n}_{+} means {x = (x_{1}, x_{2}, . . . , x_{n}) | x_{i} ≥ 0, for all i = 1, 2, . . . , n}, whereas IR^{n}_{++} de-
notes {x = (x_{1}, x_{2}, . . . , x_{n}) | x_{i} > 0, ∀i = 1, 2, . . . , n}.

• h·, ·i denotes the Euclidean inner product.

• ^{T} means transpose of a vector or a matrix.

• B(x, δ) denotes the neighborhood of x with radius δ > 0.

• IR^{n×n} denotes the space of n × n real matrices.

• I represents an identity matrix of suitable dimension.

• For any symmetric matrices A, B ∈ IR^{n×n}, we write A B (respectively, A B)
to mean A − B is positive semidefinite (respectively, positive definite).

• S^{n}denotes the space of n × n symmetric matrices; and S_{+}^{n} means the space of n × n
symmetric positive semidefinite matrices.

• O denotes the set of P ∈ IR^{n×n} that are orthogonal, i.e., P^{T} = P^{−1}.

• k · k is the Euclidean norm.

• Given a set S, we denote ¯S, int(S) and bd(S) by the closure, the interior and the boundary of S, respectively.

• A function f : IR^{n} → (−∞, ∞] is said to be proper if f (ζ) < ∞ for at least one
ζ ∈ IR^{n} and f (ζ) > −∞ for all ζ ∈ IR^{n}.

• For a mapping f : IR^{n}→ IR, ∇f (x) denotes the gradient of f at x.

• For a closed proper convex function f : IR^{n} → (−∞, ∞], we denote its domain by
domf := { ζ ∈ IR^{n}| f (ζ) < ∞}.

• For a closed proper convex function f : IR^{n} → (−∞, ∞], we denote the subdiffer-
ential of f at bζ by

∂f (bζ) :=n

w ∈ IR^{n}| f (ζ) ≥ f (bζ) + hw, ζ − bζi, ∀ζ ∈ IR^{n}o
.

• C^{(i)}(J ) denotes the family of functions which are defined on J ⊆ IR^{n} to IR and have
continuous i-th derivative.

• For any differentiable mapping F = (F_{1}, F_{2}, · · · , F_{m}) : IR^{n} → IR^{m}, ∇F (x) =
[∇F_{1}(x) · · · ∇F_{m}(x)] is a n × m matrix which denotes the transpose Jacobian of F
at x.

• For any x, y ∈ IR^{n}, we write x _{Kn} y if x − y ∈ K^{n}; and write x _{Kn} y if
x − y ∈ int(K^{n}).

• For a real valued function 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 a mapping F : S ⊆ IR^{n} → IR^{m}, ∂F (x) denotes the subdifferential of F at x,
while ∂_{B}F (x) denotes the B-subdifferential of F at x.

## Chapter 1

## SOC Functions

During the past two decades, there have been active research for second-order cone pro- grams (SOCPs) and second-order cone complementarity problems (SOCCPs). Various methods had been proposed which include the interior-point methods [1, 102, 109, 123, 146], the smoothing Newton methods [51, 63, 71], the semismooth Newton methods [86, 120], and the merit function methods [43, 48]. All of these methods are proposed by using some SOC complementarity function or merit function to reformulate the KKT optimality conditions as a nonsmooth (or smoothing) system of equations or an uncon- strained minimization problem. In fact, such SOC complementarity functions or merit functions are closely connected to so-called SOC functions. In other words, studying SOC functions is crucial to dealing with SOCP and SOCCP, which is the main target of this chapter.

### 1.1 On the second-order cone

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} , (1.1)
where k · k denotes the Euclidean norm. If n = 1, let K^{n} denote the set of nonnegative
reals IR_{+}. For n = 2 and n = 3, the pictures of K^{n} are depicted in Figure 1.1(a) and
Figure 1.1(b), respectively. It is known that K^{n} is a pointed closed convex cone so that a
partial ordering can be deduced. More specifically, for any x, y in IR^{n}, we write x _{Kn} y if
x − y ∈ K^{n}; and write x _{Kn} y if x − y ∈ int(K^{n}). In other words, we have x _{Kn} 0 if and
only if x ∈ K^{n}; whereas x _{Kn} 0 if and only if x ∈ int(K^{n}). The relation _{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 _{Kn} y nor y _{Kn} x. To see this, for n = 2, let x = (1, 1) ∈ K^{2} and y = (1, 0) ∈ K^{2}.
Then, we have x − y = (0, 1) /∈ K^{2} and y − x = (0, −1) /∈ K^{2}.

1

(a) 2-dimensional SOC (b) 3-dimensional SOC

Figure 1.1: The graphs of SOC

The second-order cone has received much attention in optimization, particularly in the
context of applications and solutions methods for second-order cone program (SOCP) [1,
47, 48, 102, 115, 116, 118] and second-order cone complementarity problem (SOCCP), [42,
43, 45, 48, 63, 71, 117]. For those solutions methods, there needs spectral decomposition
associated with SOC whose basic concept is described below. For any x = (x_{1}, x_{2}) ∈
IR × IR^{n−1}, x can be decomposed as

x = λ_{1}(x)u^{(1)}_{x} + λ_{2}(x)u^{(2)}_{x} , (1.2)
where λ_{1}(x), λ_{2}(x) and u^{(1)}x , u^{(2)}x are the spectral values and the associated spectral
vectors of x given by

λ_{i}(x) = x_{1}+ (−1)^{i}kx_{2}k, (1.3)

u^{(i)}_{x} =

1 2

1, (−1)^{i} x2

kx_{2}k

, if x2 6= 0,

1

2(1, (−1)^{i}w) , if x2 = 0,

(1.4)

for i = 1, 2 with w being any vector in IR^{n−1} satisfying kwk = 1. If x2 6= 0, the decom-
position is unique.

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 = (hx, yi, y_{1}x_{2}+ x_{1}y_{2}) ∈ IR × IR^{n−1}. (1.5)
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^{n}. Thus, without fear
of ambiguity, we may write x^{m} for the product of m copies of x and x^{m+n}= x^{m}◦ x^{n} for
all positive integers m and n. The vector e = (1, 0, . . . , 0) is the unique identity element
for the Jordan product, and we define x^{0} = e for convenience. In addition, K^{n} is not
closed under Jordan product. For example, x = (√

2, 1, 1) ∈ K^{3}, y = (√

2, 1, −1) ∈ K^{3},

but x ◦ y = (2, 2√

2, 0) /∈ K^{3}. We point out that lacking associative property of Jordan
product and closedness of SOC are the main sources of difficulty when dealing with SOC.

We write x^{2} to denote x ◦ x and write x + y to mean the usual componentwise addition
of vectors. Then, “◦, +” together with e = (1, 0, . . . , 0) ∈ IR^{n} have the following basic
properties (see [61, 63]):

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

(3) 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}.

For each x = (x1, x2) ∈ IR × IR^{n−1}, the determinant and the trace of x are defined by
det(x) = x^{2}_{1}− kx2k^{2}, tr(x) = 2x1.

In view of the definition of spectral values (1.3), it is clear that the determinant, the
trace and the Euclidean norm of x can all be represented in terms of λ_{1}(x) and λ_{2}(x):

det(x) = λ_{1}(x)λ_{2}(x), tr(x) = λ_{1}(x) + λ_{2}(x), kxk^{2} = 1

2 λ_{1}(x)^{2}+ λ_{2}(x)^{2} . (1.6)
As below, we elaborate more about the determinant and trace by showing some
properties.

Proposition 1.1. For any x _{Kn} 0 and y _{Kn} 0, the following results hold.

(a) If x _{Kn} y, then det(x) ≥ det(y) and tr(x) ≥ tr(y).

(b) If x _{Kn} y, then λ_{i}(x) ≥ λ_{i}(y) for 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}) _{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, using the assumption on x and y gives

x_{1}− y_{1} ≥ kx_{2}− y_{2}k ≥

kx_{2}k − ky_{2}k

, (1.7)

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 (1.7) in the proof of part(a), the results follow immediately.
We point out that there may have other simpler ways to prove Proposition 1.1. The
approach here is straightforward and intuitive by checking definitions. The converse of
Proposition 1.1 does not hold, a counterexample occurs when taking x = (5, 3) ∈ K^{2} and
y = (3, −1) ∈ K^{2}. In fact, if (x_{1}, x_{2}) ∈ IR × IR^{n−1} serves as a counterexample for K^{n},
then (x_{1}, x_{2}, 0, . . . , 0) ∈ IR × IR^{m−1} is automatically a counterexample for K^{m} whenever
m ≥ n. Moreover, for any x _{Kn} y, there always have λ_{i}(x) ≥ λ_{i}(y) and tr(x) ≥ tr(y)
for i = 1, 2. There is no need to restrict x _{Kn} 0 and y _{Kn} 0 as in Proposition 1.1.

Proposition 1.2. Let x _{Kn} 0, y _{Kn} 0 and e = (1, 0, · · · , 0). Then, the following hold.

(a) det(x + y) ≥ det(x) + det(y).

(b) det(x ◦ y) ≤ det(x) det(y).

(c) det αx + (1 − α)y ≥ α^{2}det(x) + (1 − α)^{2}det(y) for all 0 < α < 1.

(d) det(e + x)1/2

≥ 1 + det(x)^{1/2}.
(e) det(e + x + y) ≤ det(e + x) det(e + y).

Proof. (a) For any x _{Kn} 0 and y _{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}

= x^{2}_{1}− kx_{2}k^{2} + y^{2}_{1}− ky_{2}k^{2} + 2 x_{1}y_{1}− hx_{2}, y_{2}i

≥ x^{2}_{1}− kx_{2}k^{2} + y^{2}_{1}− ky_{2}k^{2}

= det(x) + det(y).

(b) Applying the Cauchy inequality gives
det(x ◦ y) = hx, yi^{2}− kx1y2+ y1x2k^{2}

= x_{1}y_{1}+ hx_{2}, y_{2}i2

− x^{2}_{1}ky_{2}k^{2}+ 2x_{1}y_{1}hx_{2}, y_{2}i + y^{2}_{1}kx_{2}k^{2}

= x^{2}_{1}y_{1}^{2}+ hx_{2}, y_{2}i^{2}− x^{2}_{1}ky_{2}k^{2}− y_{1}^{2}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_{1}^{2}kx_{2}k^{2}

= x^{2}_{1}− kx2k^{2}

y^{2}_{1}− ky2k^{2}

= det(x) det(y).

(c) For any x _{Kn} 0 and y _{Kn} 0, it is clear that αx _{Kn} 0 and (1 − α)y _{Kn} 0 for every
0 < α < 1. In addition, we observe that det(αx) = α^{2}det(x). Hence,

det 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 _{Kn} 0, we know det(x) = λ_{1}(x)λ_{2}(x) ≥ 0, where λ_{i}(x) are the spectral
values of x. Hence,

det(e + x) = (1 + λ_{1}(x))(1 + λ_{2}(x)) ≥
1 +p

λ_{1}(x)λ_{2}(x)2

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

(e) Again, For any x _{Kn} 0 and y _{Kn} 0, we have the following inequalities

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}. (1.8)
Moreover, 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)

= (1 + x1)^{2}− kx2k^{2}

(1 + y1)^{2}− ky2k^{2} − (1 + x1 + y1)^{2}− kx2+ y2k^{2}

= 2x_{1}y_{1}+ 2hx_{2}, y_{2}i + 2x_{1}y_{1}^{2}+ 2x^{2}_{1}y_{1}− 2y_{1}kx_{2}k^{2}− 2x_{1}ky_{2}k^{2}
+x^{2}_{1}y^{2}_{1} − y_{1}^{2}kx_{2}k^{2}− x^{2}_{1}ky_{2}k^{2} + kx_{2}k^{2}ky_{2}k^{2}

= 2 x_{1}y_{1}+ hx_{2}, y_{2}i + 2x_{1} y_{1}^{2}− ky_{2}k^{2} + 2y_{1} x^{2}_{1}− kx_{2}k^{2}
+ x^{2}_{1}− kx2k^{2}

y^{2}_{1}− ky2k^{2}

≥ 0,

where we multiply out all the expansions to obtain the second equality and the last inequality holds by (1.8).

Proposition 1.2(c) can be extended to a more general case:

det αx + βy ≥ α^{2}det(x) + β^{2}det(y) ∀α ≥ 0, β ≥ 0.

Note that together with Cauchy-Schwartz inequality and properties of determinant, one may achieve other way to verify Proposition 1.2. Again, the approach here is only one choice of proof which is straightforward and intuitive. There are more inequalities about determinant, see Proposition 1.8 and Proposition 2.32, which are established by using the concept of SOC-convexity that will be introduced in Chapter 2. Next, we move to the inequalities about trace.

Proposition 1.3. For any x, y ∈ IR^{n}, we have

(a) tr(x + y) = tr(x) + tr(y) and tr(αx) = α tr(x) for any α ∈ IR. In other words, tr(·)
is a linear function on IR^{n}.

(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).

Proof. Part(a) is trivial and it remains to verify part(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(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)

= λ_{1}(x)λ_{1}(y) + λ_{2}(x)λ_{2}(y),
which completes the proof.

In general, det(x ◦ y) 6= det(x) det(y) unless x_{2} = αy_{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 = (y1, y2) ∈ IR × IR^{n−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

x^{2}_{1} − kx_{2}k^{2}(x_{1}, −x_{2}) = 1

det(x) 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} = (x^{−1})^{k} is also well-defined. For any x ∈ K^{n}, it is known that there
exists a unique vector in K^{n} denoted by x^{1/2} (also denoted by √

x sometimes) such that
(x^{1/2})^{2} = x^{1/2}◦ x^{1/2} = x. Indeed,

x^{1/2}=
s, x_{2}

2s

, where s = s

1 2

x_{1} +

q

x^{2}_{1}− kx_{2}k^{2}

.

In the above formula, the term ^{x}_{2s}^{2} is defined to be the zero vector if s = 0 (and hence
x2 = 0), i.e., x = 0 .

For any x ∈ IR^{n}, we always have x^{2} ∈ K^{n}(i.e., x^{2} _{Kn} 0). Hence, there exists a unique
vector (x^{2})^{1/2} ∈ K^{n} denoted by |x|. It is easy to verify that |x| _{Kn} 0 and x^{2} = |x|^{2} for
any x ∈ IR^{n}. It is also known that |x| _{Kn} x. For any x ∈ IR^{n}, we define [x]+ to be the
projection point of x onto K^{n}, which is the same definition as in IR^{n}_{+}. In other words,
[x]_{+} is the optimal solution of the parametric SOCP:

[x]_{+} = argmin{kx − yk | y ∈ K^{n}}.

Here the norm is in Euclidean norm since Jordan product does not induce a norm. Like-
wise, [x]_{−} means the projection point of x onto −K^{n}, which implies [x]_{−}= −[−x]_{+}. It is
well known that [x]_{+} = ^{1}_{2}(x + |x|) and [x]− = ^{1}_{2}(x − |x|), see Property 1.2(f).

The spectral decomposition along with the Jordan algebra associated with SOC entails some basic properties as below. We omit the proofs since they can be found in [61, 63].

Property 1.1. For any x = (x_{1}, x_{2}) ∈ IR × IR^{n−1} with the spectral values λ_{1}(x), λ_{2}(x)
and spectral vectors u^{(1)}x , u^{(2)}x given as in (1.3)-(1.4), we have

(a) u^{(1)}x and u^{(2)}x are orthogonal under Jordan product and have length ^{√}^{1}

2, i.e.,
u^{(1)}_{x} ◦ u^{(2)}_{x} = 0, ku^{(1)}_{x} k = ku^{(2)}_{x} k = 1

√2.

(b) u^{(1)}x and u^{(2)}x are idempotent under Jordan product, i.e.,
u^{(i)}_{x} ◦ u^{(i)}_{x} = u^{(i)}_{x} , i = 1, 2.

(c) λ_{1}(x), λ_{2}(x) are nonnegative (positive) if and only if x ∈ K^{n} (x ∈ int(K^{n})), i.e.,
λ_{i}(x) ≥ 0 for i = 1, 2 ⇐⇒ x _{Kn} 0.

λi(x) > 0 for i = 1, 2 ⇐⇒ x _{Kn} 0.

Although the converse of Proposition 1.1(b) does not hold as mentioned earlier, Prop-
erty 1.1(c) is useful in verifying whether a point x belongs to K^{n} or not.

Property 1.2. For any x = (x_{1}, x_{2}) ∈ IR × IR^{n−1} with the spectral values λ_{1}(x), λ_{2}(x)
and spectral vectors u^{(1)}x , u^{(2)}x given as in (1.3)-(1.4), we have

(a) x^{2} = λ_{1}(x)^{2}u^{(1)}x + λ_{2}(x)^{2}u^{(2)}x and x^{−1} = λ^{−1}_{1} (x)u^{(1)}x + λ^{−1}_{2} (x)u^{(2)}x .
(b) If x ∈ K^{n}, then x^{1/2} =pλ1(x) u^{(1)}x +pλ2(x) u^{(2)}x .

(c) |x| = |λ_{1}(x)|u^{(1)}x + |λ_{2}(x)|u^{(2)}x .

(d) [x]_{+} = [λ_{1}(x)]_{+}u^{(1)}x + [λ_{2}(x)]_{+}u^{(2)}x and [x]− = [λ_{1}(x)]−u^{(1)}x + [λ_{2}(x)]−u^{(2)}x .
(e) |x| = [x]_{+}+ [−x]_{+} = [x]_{+}− [x]−.

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

Property 1.3. Let x = (x_{1}, x_{2}) ∈ IR × IR^{n−1} and y = (y_{1}, y_{2}) ∈ IR × IR^{n−1}. Then, the
following hold.

(a) Any x ∈ IR^{n} satisfies |x| _{Kn} x.

(b) For any x, y _{Kn} 0, if x _{Kn} y, then x^{1/2} _{Kn} y^{1/2}.
(c) For any x, y ∈ IR^{n}, if x^{2} _{Kn} y^{2}, then |x| _{Kn} |y|.

(d) For any x ∈ IR^{n}, x _{Kn} 0 if and only if hx, yi ≥ 0 for all y _{Kn} 0.

(e) For any x _{Kn} 0 and y ∈ IR^{n}, if x^{2} _{Kn} y^{2}, then x _{Kn} y.

Note that for any x, y _{Kn} 0, if x _{Kn} y, one can also conclude that x^{−1} _{Kn} y^{−1}.
However, the arguments are not trivial by direct verifications. We present it by other
approach, see Proposition 2.3(a).

Property 1.4. For any x = (x_{1}, x_{2}) ∈ IR × IR^{n−1} with spectral values λ_{1}(x), λ_{2}(x) and
any y = (y_{1}, y_{2}) ∈ IR × IR^{n−1} with spectral values λ_{1}(y), λ_{2}(y), we have

|λ_{i}(x) − λ_{i}(y)| ≤√

2kx − yk, i = 1, 2.

Proof. First, we compute that

|λ_{1}(x) − λ_{1}(y)| = |x_{1} − kx_{2}k − y_{1}+ ky_{2}k|

≤ |x_{1} − y_{1}| + |kx_{2}k − ky_{2}k|

≤ |x_{1} − y_{1}| + kx_{2}− y_{2}k

≤ √

2 |x_{1}− y_{1}|^{2}+ kx_{2}− y_{2}k^{2}1/2

= √

2kx − yk,

where the second inequality uses kx_{2}k ≤ kx_{2}−y_{2}k+ky_{2}k and ky_{2}k ≤ kx_{2}−y_{2}k+kx_{2}k; the
last inequality uses the relation between the 1-norm and the 2-norm. A similar argument
applies to |λ2(x) − λ2(y)|.

In fact, Property 1.1-1.3 are parallel results analogous to those associated with positive
semidefinite cone S_{+}^{n}, see [74]. Even though both K^{n} and S_{+}^{n} belong to the family of
symmetric cones [61] and share similar properties, as we will see, the ideas and techniques
for proving these results are quite different. One reason is that the Jordan product is not
associative as mentioned earlier.

### 1.2 SOC function and SOC trace function

In this section, we introduce two types of functions, SOC function and SOC trace func- tion, which are very useful in dealing with optimization involved with SOC. Some in- equalities are established in light of these functions.

Let x = (x_{1}, x_{2}) ∈ IR × IR^{n−1} with spectral values λ_{1}(x), λ_{2}(x) given as in (1.3) and
spectral vectors u^{(1)}x , u^{(2)}x given as in (1.4). We first define its corresponding SOC function
as below. For any real-valued function f : IR → IR, the following vector-valued function
associated with K^{n} (n ≥ 1) was considered [45, 63]:

f^{soc}(x) := f (λ_{1}(x))u^{(1)}_{x} + f (λ_{2}(x))u^{(2)}_{x} , ∀x = (x_{1}, x_{2}) ∈ IR × IR^{n−1}. (1.9)
The definition (1.9) is unambiguous whether x_{2} 6= 0 or x_{2} = 0. The cases of f^{soc}(x) = x^{1/2},
x^{2}, exp(x), which correspond to f (t) = t^{1/2}, t^{2}, e^{t}, are already discussed in the book [61].

Indeed, the above definition (1.9) is analogous to one associated with the semidefinite
cone S_{+}^{n}, see [140, 145]. For subsequent analysis, we also need the concept of SOC trace
function [46] defined by

f^{tr}(x) := f (λ1(x)) + f (λ2(x)) = tr(f^{soc}(x)). (1.10)
If f is defined only on a subset of IR, then f^{soc} and f^{tr}are defined on the corresponding
subset of IR^{n}. More specifically, from Proposition 1.4 shown as below, we see that the
corresponding subset for f^{soc} and f^{tr} is

S = {x ∈ IR^{n}| λ_{i}(x) ∈ J, i = 1, 2.} (1.11)
provided f is defined on a subset of J ⊆ IR. In addition, S is open in IR^{n} whenever J is
open in IR. To see this assertion, we need the following technical lemma.

Lemma 1.1. Let A ∈ IR^{m×m} be a symmetric positive definite matrix, C ∈ IR^{n×n} be a
symmetric matrix, and B ∈ IR^{m×n}. Then,

A B

B^{T} C

O ⇐⇒ C − B^{T}A^{−1}B O (1.12)

and

A B

B^{T} C

O ⇐⇒ C − B^{T}A^{−1}B O. (1.13)

Proof. This is indeed the Schur Complement Theorem, please see [21, 22, 74] for a proof.

Proposition 1.4. For any given f : J ⊆ IR → IR, let f^{soc} : S → IR^{n} and f^{tr} : S → IR
be given by (1.9) and (1.10), 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}(x_{1})I if x_{2} = 0, and otherwise

∇f^{soc}(x) =

b(x) c(x) x^{T}_{2}

kx_{2}k
c(x) x_{2}

kx_{2}k a(x)I + (b(x) − a(x))x_{2}x^{T}_{2}
kx_{2}k^{2}

, (1.14)

where

a(x) = f (λ_{2}(x)) − f (λ_{1}(x))
λ_{2}(x) − λ_{1}(x) ,
b(x) = f^{0}(λ_{2}(x)) + f^{0}(λ_{1}(x))

2 ,

c(x) = f^{0}(λ_{2}(x)) − f^{0}(λ_{1}(x))

2 .

(c) If f is (continuously) differentiable, then f^{tr} is (continuously) differentiable on S
with ∇f^{tr}(x) = 2(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 λ_{1}(x), λ_{2}(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)| <

δ_{2}} ⊆ J. Let δ := min{δ_{1}, δ_{2}}/√

2. Then, for any y satisfying ky − xk < δ, we have

|λ_{1}(y) − λ_{1}(x)| < δ_{1} and |λ_{2}(y) − λ_{2}(x)| < δ_{2} by noting that
(λ1(x) − λ1(y))^{2}+ (λ2(x) − λ2(y))^{2}

= 2(x^{2}_{1}+ kx_{2}k^{2}) + 2(y^{2}_{1} + ky_{2}k^{2}) − 4(x_{1}y_{1}+ kx_{2}kky_{2}k)

≤ 2(x^{2}_{1}+ kx_{2}k^{2}) + 2(y^{2}_{1} + ky_{2}k^{2}) − 4(x_{1}y_{1}+ hx_{2}, y_{2}i)

= 2 kxk^{2}+ kyk^{2}− 2hx, yi

= 2kx − yk^{2},

and consequently λ_{1}(y) ∈ J and λ_{2}(y) ∈ J . Since f is a function from J to IR, this means
that {y ∈ IR^{n}| ky − xk < δ} ⊆ S, and therefore the set S is open. In addition, from the
above, we see that S is characterized as in (1.11).

(b) The arguments are similar to Proposition 1.13 and Proposition 1.14 in Section 1.3.

Please check them for details.

(c) If f is (continuously) differentiable, then from part(b) and f^{tr}(x) = 2e, f^{soc}(x) it
follows that f^{tr} is (continuously) differentiable. In addition, a simple computation yields
that ∇f^{tr}(x) = 2∇f^{soc}(x)e = 2(f^{0})^{soc}(x). Similarly, by part(b), the second part follows.

Proposition 1.5. For any f : J → IR, let f^{soc} : S → IR^{n} and f^{tr} : S → IR be given by
(1.9) and (1.10), 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) O for any x ∈ S ⇐⇒ f^{tr} is convex in S.

(b) f^{00}(t) > 0 for any t ∈ J ⇐⇒ ∇(f^{0})^{soc}(x) O for any x ∈ S =⇒ f^{tr} is strictly
convex in S.

Proof. (a) By Proposition 1.4(c), ∇^{2}f^{tr}(x) = 2∇(f^{0})^{soc}(x) for any x ∈ S, and the second
equivalence follows by [20, Prop. B.4(a) and (c)]. We next come to the first equivalence.

By Proposition 1.4(b), for any fixed x ∈ S, ∇(f^{0})^{soc}(x) = f^{00}(x_{1})I if x_{2} = 0, and otherwise

∇(f^{0})^{soc}(x) has the same expression as in (1.14) except that
a(x) = f^{0}(λ_{2}(x)) − f^{0}(λ_{1}(x))

λ_{2}(x) − λ_{1}(x) ,
b(x) = f^{00}(λ_{2}(x)) + f^{00}(λ_{1}(x))

2 ,

c(x) = f^{00}(λ_{2}(x)) − f^{00}(λ_{1}(x))

2 .

Assume that ∇(f^{0})^{soc}(x) O for any x ∈ S. Then, we readily have b(x) ≥ 0 for any
x ∈ S. Noting that b(x) = f^{00}(x_{1}) when x_{2} = 0, we particularly have f^{00}(x_{1}) ≥ 0 for all
x_{1} ∈ 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

∇(f^{0})^{soc}(x) = " 0 0
0 a(x)

I − _{kx}^{x}^{2}^{x}^{T}^{2}

2k^{2}

#

O. (1.15)

If b(x) > 0, then by the first equivalence of Lemma 1.1 and the expression of ∇(f^{0})^{soc}(x)
it suffices to argue that the following matrix

a(x)I + (b(x) − a(x)) x_{2}x^{T}_{2}

kx_{2}k^{2} −c^{2}(x)
b(x)

x_{2}x^{T}_{2}

kx_{2}k^{2} (1.16)

is positive semidefinite. Since the rank-one matrix x2x^{T}_{2} has only one nonzero eigenvalue
kx_{2}k^{2}, the matrix in (1.16) has one eigenvalue a(x) of multiplicity n−1 and one eigenvalue

b(x)^{2}−c(x)^{2}

b(x) of multiplicity 1. Since a(x) ≥ 0 and ^{b(x)}^{2}_{b(x)}^{−c(x)}^{2} = f^{00}(λ_{1}(x))f^{00}(λ_{2}(x)) ≥ 0, the
matrix in (1.16) is positive semidefinite. By the arbitrary of x, we have that ∇(f^{0})^{soc}(x)
O for all x ∈ S.

(b) The first equivalence is direct by using (1.13) of Lemma 1.1, noting ∇(f^{0})^{soc}(x) O
implies a(x) > 0 when x_{2} 6= 0, and following the same arguments as part(a). The second
part is due to [20, Prop. B.4(b)].

Remark 1.1. Note that the strict convexity of f^{tr} does not necessarily imply the positive
definiteness of ∇^{2}f^{tr}(x). Consider f (t) = t^{4} for t ∈ IR. We next show that f^{tr} is strictly
convex. Indeed, f^{tr} is convex in IR^{n} by Proposition 1.5(a) since f^{00}(t) = 12t^{2} ≥ 0. Taking
into account that f^{tr} is continuous, it remains to prove that

f^{tr} x + y
2

= f^{tr}(x) + f^{tr}(y)

2 =⇒ x = y. (1.17)

Since h(t) = (t_{0} + t)^{4} + (t_{0} − t)^{4} for some t_{0} ∈ IR is increasing on [0, +∞), and the
function f (t) = t^{4} is strictly convex in IR, we have that

f^{tr} x + y
2

=

λ_{1} x + y
2

4

+

λ_{2} x + y
2

4

= x_{1}+ y_{1}− kx_{2}+ y_{2}k
2

4

+ x_{1}+ y_{1}+ kx_{2}+ y_{2}k
2

4

≤ x_{1}+ y_{1}− kx_{2}k − ky_{2}k
2

4

+ x_{1} + y_{1}+ kx_{2}k + ky_{2}k
2

4

= λ_{1}(x) + λ_{1}(y)
2

4

+ λ_{2}(x) + λ_{2}(y)
2

4

≤ (λ_{1}(x))^{4} + (λ_{1}(y))^{4}+ (λ_{2}(x))^{4}+ (λ_{2}(y))^{4}
2

= f^{tr}(x) + f^{tr}(y)

2 ,

and moreover, the above inequalities become the equalities if and only if
kx_{2}+ y_{2}k = kx_{2}k + ky_{2}k, λ_{1}(x) = λ_{1}(y), λ_{2}(x) = λ_{2}(y).

It is easy to verify that the three equalities hold if and only if x = y. Thus, the implication
in (1.17) holds, i.e., f^{tr} is strictly convex. However, by Proposition 1.5(b), ∇(f^{0})^{soc}(x)
O does not hold for all x ∈ IR^{n} since f^{00}(t) > 0 does not hold for all t ∈ IR.

We point out that the fact that the strict convexity of f implies the strict convexity
of f^{tr} was proved in [7, 15] via the definition of convex function, but here we use the
Schur Complement Theorem and the relation between ∇(f^{0})^{soc} and ∇^{2}f^{tr} to establish
the convexity of SOC trace functions. Next, we illustrate the application of Proposition
1.5 with some SOC trace functions.

Proposition 1.6. The following functions associated with K^{n} are all strictly convex.

(a) F_{1}(x) = − ln(det(x)) for x ∈ int(K^{n}).

(b) F_{2}(x) = tr(x^{−1}) for x ∈ int(K^{n}).

(c) F_{3}(x) = tr(φ(x)) for x ∈ int(K^{n}), where
φ(x) =

( _{x}p+1−e

p+1 +^{x}^{1−q}_{q−1}^{−e} if p ∈ [0, 1], q > 1,

x^{p+1}−e

p+1 − ln x if p ∈ [0, 1], q = 1.

(d) F_{4}(x) = − ln(det(e − x)) for x ≺_{Kn} e.

(e) F5(x) = tr((e − x)^{−1}◦ x) for x ≺_{Kn} e.

(f ) F_{6}(x) = tr(exp(x)) for x ∈ IR^{n}.

(g) F_{7}(x) = ln(det(e + exp(x))) for x ∈ IR^{n}.
(h) F_{8}(x) = tr x + (x^{2} + 4e)^{1/2}

2

for x ∈ IR^{n}.

Proof. Note that F_{1}(x), F_{2}(x) and F_{3}(x) are the SOC trace functions associated with
f_{1}(t) = − ln t (t > 0), f_{2}(t) = t^{−1} (t > 0) and f_{3}(t) (t > 0), respectively, where

f3(t) =

( _{t}p+1−1

p+1 +^{t}^{1−q}_{q−1}^{−1} if p ∈ [0, 1], q > 1,

t^{p+1}−1

p+1 − ln t if p ∈ [0, 1], q = 1;

Next, F_{4}(x) is the SOC trace function associated with f_{4}(t) = − ln(1 − t) (t < 1), F_{5}(x)
is the SOC trace function associated with f_{5}(t) = _{1−t}^{t} (t < 1) by noting that

(e − x)^{−1}◦ x = λ_{1}(x)

λ_{1}(e − x)u^{(1)}_{x} + λ_{2}(x)
λ_{2}(e − x)u^{(2)}_{x} ;

In addition, F_{6}(x) and F_{7}(x) are the SOC trace functions associated with f_{6}(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 f_{8}(t) = ^{1}_{2} t +√

t^{2}+ 4 (t ∈ IR). It is easy to verify that all
the functions f_{1}-f_{8} have positive second-order derivatives in their respective domain, and
therefore F_{1}-F_{8} are strictly convex functions by Proposition 1.5(b).

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., [14, 19, 109, 123, 146],
where F_{3} covers a wide range of barrier functions, including the classical logarithmic
barrier function, the self-regular functions and the non-self-regular functions; see [14]

for details. The functions F4 and F5 are the popular shifted barrier functions [6, 7, 9]

for SOCPs, and F_{6}-F_{8} 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 solu-
tion 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 func- tions of SOC trace functions and scalar-valued functions, which are usually difficult

to achieve by checking the definition of convexity directly. The following proposition presents such an application.

Proposition 1.7. For any x ∈ K^{n}, let F9(x) := −[det(x)]^{1/p} with p > 1. Then,
(a) F_{9} is twice continuously differentiable in int(K^{n}).

(b) F9 is convex when p ≥ 2, and moreover, it is strictly convex when p > 2.

Proof. (a) Note that −F_{9}(x) = exp (p^{−1}ln(det(x))) for any x ∈ int(K^{n}), and ln(det(x)) =
f^{tr}(x) with f (t) = ln(t) for t ∈ IR++. By Proposition 1.4(c), ln(det(x)) is twice contin-
uously differentiable in int(K^{n}). Hence −F_{9}(x) is twice continuously differentiable in
int(K^{n}). The result then follows.

(b) In view of the continuity of F_{9}, we only need to prove its convexity over int(K^{n}). By
part(a), we next achieve this goal by proving that the Hessian matrix ∇^{2}F_{9}(x) for any
x ∈ int(K^{n}) is positive semidefinite when p ≥ 2, and positive definite when p > 2. Fix
any x ∈ int(K^{n}). From direct computations, we obtain

∇F_{9}(x) = −1
p

"

(2x1) x^{2}_{1}− kx2k^{2}^{1}_{p}−1

(−2x2) x^{2}_{1}− kx2k^{2}^{1}_{p}−1

#

and

∇^{2}F_{9}(x) = p − 1

p^{2} (det(x))^{1}^{p}^{−2}

4x^{2}_{1}−^{2p}(^{x}^{2}1−kx2k^{2})

p−1 −4x_{1}x^{T}_{2}

−4x_{1}x_{2} 4x_{2}x^{T}_{2} +^{2p}(^{x}^{2}1−kx2k^{2})

p−1 I

.

Since x ∈ int(K^{n}), we have x_{1} > 0 and det(x) = x^{2}_{1}− kx_{2}k^{2} > 0, and therefore
a1(x) := 4x^{2}_{1}− 2p (x^{2}_{1}− kx_{2}k^{2})

p − 1 =

4 − 2p p − 1

x^{2}_{1}+ 2p

p − 1kx_{2}k^{2}.

We next proceed the arguments by the following two cases: a_{1}(x) = 0 or a_{1}(x) > 0.

Case 1: a1(x) = 0. Since p ≥ 2, under this case we must have x2 = 0, and consequently,

∇^{2}F_{9}(x) = p − 1

p^{2} (x_{1})^{2}^{p}^{−4} 0 0
0 _{p−1}^{2p} x^{2}_{1}I

O.

Case 2: a1(x) > 0. Under this case, we calculate that

4x^{2}_{1}− 2p (x^{2}_{1} − kx_{2}k^{2})
p − 1

4x_{2}x^{T}_{2} +2p (x^{2}_{1}− kx_{2}k^{2})
p − 1 I

− 16x^{2}_{1}x_{2}x^{T}_{2}

= 4p (x^{2}_{1}− kx2k^{2})
p − 1

p − 2

p − 1x^{2}_{1}I + p

p − 1kx_{2}k^{2}I − 2x_{2}x^{T}_{2}

. (1.18)

Since the rank-one matrix 2x_{2}x^{T}_{2} has only one nonzero eigenvalue 2kx_{2}k^{2}, the matrix in
the bracket of the right hand side of (1.18) has one eigenvalue of multiplicity 1 given by

p − 2

p − 1x^{2}_{1}+ p

p − 1kx_{2}k^{2}− 2kx_{2}k^{2} = p − 2

p − 1 x^{2}_{1}− kx_{2}k^{2} ≥ 0,

and one eigenvalue of multiplicity n − 1 given by ^{p−2}_{p−1}x^{2}_{1}+_{p−1}^{p} kx2k^{2} ≥ 0. Furthermore, we
see that these eigenvalues must be positive when p > 2 since x^{2}_{1} > 0 and x^{2}_{1}− kx_{2}k^{2} > 0.

This means that the matrix on the right hand side of (1.18) is positive semidefinite, and moreover, it is positive definite when p > 2. Applying Lemma 1.1, we have that

∇^{2}F_{9}(x) O, and furthermore ∇^{2}F_{9}(x) O when p > 2.

Since a_{1}(x) > 0 must hold when p > 2, the arguments above show that F_{9}(x) is
convex over int(K^{n}) when p ≥ 2, and strictly convex over int(K^{n}) when p > 2.

It is worthwhile to point out that det(x) is neither convex nor concave on K^{n}, and
it is difficult to argue the convexity of those compound functions involving det(x) by
the definition of convex function. But, our SOC trace function offers a simple way to
prove their convexity. Moreover, it helps on establishing more inequalities associated
with SOC. Some of these inequalities have been used to analyze the properties of SOC
function f^{soc} [41] and the convergence of interior point methods for SOCPs [7].

Proposition 1.8. 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.

(b) det(x + y)^{1/p}≥ 2^{2}^{p}^{−1} det(x)^{1/p}+ det(y)^{1/p} for any p ≥ 2.

(c) det(αx + (1 − α)y) ≥ α^{2}det(x) + (1 − α)^{2}det(y) for any 0 < α < 1.

(d) [det(e + x)]^{1/2} ≥ 1 + det(x)^{1/2}.
(e) det(x)^{1/2} = inf 1

2tr(x ◦ y)

det(y) = 1, y _{Kn} 0

. Furthermore, when x _{Kn} 0,
the same relation holds with inf replaced by min.

(f ) tr(x ◦ y) ≥ 2 det(x)^{1/2}det(y)^{1/2}.

Proof. (a) From Proposition 1.6(a), we know that ln(det(x)) is strictly concave in
int(K^{n}). With this, we have

ln(det(αx + (1 − α)y)) ≥ α ln(det(x)) + (1 − α) ln(det(y))

= ln(det(x)^{α}) + ln(det(x)^{1−α})

for any 0 < α < 1 and x, y ∈ int(K^{n}). This, together with the increasing of ln t (t > 0)
and the continuity of det(x), implies the desired result.

(b) By Proposition 1.7(b), det(x)^{1/p} is concave over K^{n}. Then, for any x, y ∈ K^{n}, we
have

det x + y 2

1/p

≥ 1

2det(x)^{1/p}+ det(y)^{1/p}

⇐⇒ 2

"

x_{1}+ y_{1}
2

2

−

x_{2}+ y_{2}
2

2#1/p

≥ x^{2}_{1}− kx2k^{2}1/p

+ y_{1}^{2}− ky2k^{2}1/p

⇐⇒ (x_{1}+ y_{1})^{2}− kx_{2}+ y_{2}k^{2}1/p

≥ 4^{1}^{p}
2

h

x^{2}_{1}− kx_{2}k^{2}1/p

+ y_{1}^{2}− ky_{2}k^{2}1/pi

⇐⇒ det(x + y)^{1/p} ≥ 2^{2}^{p}^{−1} det(x)^{1/p}+ 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) + 2 det(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 ∈ K^{n}.
This together with the fact det(αx) = α^{2}det(x) leads to the desired result.

(d) This inequality is presented in Proposition 1.2(d). Nonetheless, we provide a different approach by applying part(b) with p = 2 and the fact that det(e) = 1.

(e) From Proposition 1.3(b), we have

tr(x ◦ y) ≥ λ_{1}(x)λ_{2}(y) + λ_{1}(y)λ_{2}(x), ∀x, y ∈ IR^{n}.

For any x, y ∈ K^{n}, this along with the arithmetic-geometric mean inequality implies that
tr(x ◦ y)

2 ≥ λ_{1}(x)λ_{2}(y) + λ_{1}(y)λ_{2}(x)
2

≥ p

λ_{1}(x)λ_{2}(y)λ_{1}(y)λ_{2}(x)

= det(x)^{1/2}det(y)^{1/2},
which means that inf 1

2tr(x ◦ y)

det(y) = 1, y _{Kn} 0

= det(x)^{1/2} for a fixed x ∈ K^{n}.
If x _{Kn} 0, then we can verify that the feasible point y^{∗} = √^{x}^{−1}

det(x) is such that 1

2tr(x◦y^{∗}) =
det(x)^{1/2}, and the second part follows.

(f) Using part(e), for any x ∈ K^{n} and y ∈ int(K^{n}), we have
tr(x ◦ y)

2pdet(y) = 1

2tr x ◦ y pdet(y)

!

≥p

det(x),