to appear in Journal of Inequalities and Applications, 2014

### Circular cone convexity and some inequalities associated with circular cones

Jinchuan Zhou ^{1}
Department of Mathematics

School of Science

Shandong University of Technology Zibo 255049, P.R. China E-mail: jinchuanzhou@163.com

Jein-Shan Chen ^{2}
Department of Mathematics
National Taiwan Normal University

Taipei 11677, Taiwan E-mail: jschen@math.ntnu.edu.tw

Hao-Feng Hung Department of Mathematics National Taiwan Normal University

Taipei 11677, Taiwan E-mail: 60040016s@ntnu.edu.tw

July 17, 2013

(revised on November 12, 2013)

Abstract. The study of this paper consists of two aspects. One is characterizing the so-
called circular cone convexity of f by exploiting the second order differentiability of f^{L}^{θ};
the other is introducing the concepts of determinant and trace associated with circular
cone and establishing their basic inequalities. These results show the essential role played
by the angle θ, which gives us new insight when looking into properties about circular
cone.

1The author’s work is supported by National Natural Science Foundation of China (11101248, 11271233) and Shandong Province Natural Science Foundation (ZR2010AQ026, ZR2012AM016).

2Corresponding author. Member of Mathematics Division, National Center for Theoretical Sciences, Taipei Office. The author’s work is supported by National Science Council of Taiwan.

Keywords. Circular cone, convexity, determinant, trace.

AMS subject classifications. 26A27, 26B05, 26B35, 49J52, 90C33, 65K05.

### 1 Introduction

Recently, much attention has been paid to the nonsymmetric cone optimization problems,
see [13, 14, 16, 17] and references therein. Unlike symmetric cones [11], there is no
unified structure for nonsymmetric cones. Hence, how to tackle with nonsymmetric cone
optimization is still an issue. For symmetric cone optimization, the algebraic structure
associated with symmetric cones, including second-order cone and positive semi-definite
matrix cones, allow us to study them via exploiting the unified Euclidean Jordan algebra
[11]. In general, the way to deal with nonsymmetric cone optimization depends on
the feature of the associated nonsymmetric cone. In this paper, we focus on a special
nonsymmetric cone, circular cone L_{θ}. The circular cone [10, 18, 19, 20] is a pointed
closed convex cone having hyperspherical sections orthogonal to its axis of revolution
about which the cone is invariant to rotation. Let its half-aperture angle be θ with
θ ∈ (0, 90^{◦}). Then, it is mathematically expressed as

L_{θ} := {x = (x_{1}, x_{2})^{T} ∈ IR × IR^{n−1}| x_{1} ≥ kxk cos θ}

= {x = (x_{1}, x_{2})^{T} ∈ IR × IR^{n−1}| x_{1} ≥ kx_{2}k cot θ}.

Real applications of circular cone lie in some engineering problems, for example, in the formulation for optimal grasping manipulation for multi-fingered robots, the grasping force of i-th finger is subject to a circular cone constraint, see [4, 12] and references for more details.

Although Lθ is a nonsymmetric cone, we can, due to its special structure, establish the explicit form of orthogonal decomposition (or spectral decomposition) [18] as

x = λ_{1}(x) · u^{(1)}_{x} + λ_{2}(x) · u^{(2)}_{x} (1)
where

λ_{1}(x) = x_{1}− kx_{2}k cot θ
λ2(x) = x1+ kx2k tan θ
and

u^{(1)}x = 1
1 + cot^{2}θ

1 0

0 cot θI_{n−1}

1

−¯x_{2}

=

sin^{2}θ

−(sin θ cos θ)¯x_{2}

u^{(2)}x = 1
1 + tan^{2}θ

1 0

0 tan θI_{n−1}

1

¯
x_{2}

=

cos^{2}θ
(sin θ cos θ)¯x_{2}

with ¯x_{2} = x_{2}/kx_{2}k if x_{2} 6= 0, and ¯x_{2} being any vector w in IR^{n−1} satisfying kwk = 1 if
x_{2} = 0. Clearly, x ∈ L_{θ} if and only if λ_{1}(x) ≥ 0.

The formula (1) allows us to define the following vector-valued function

f^{L}^{θ}(x) := f (λ1(x)) u^{(1)}_{x} + f (λ2(x)) u^{(2)}_{x} , (2)
where f is a real-valued function from J to IR with J being a subset in IR. Let S
be the set of all x ∈ IR^{n} whose spectral values λ_{i}(x) for i = 1, 2 belong to J , i.e.,
S := {x ∈ IR^{n}| λ_{i}(x) ∈ J, i = 1, 2}. According to [15], we know S is open if and only if
J is open. In addition, as J is an interval, then S is convex because

min{λ_{1}(x), λ_{1}(y)} ≤ λ_{1}(βx+(1−β)y) ≤ λ_{2}(βx+(1−β)y) ≤ max{λ_{2}(x), λ_{2}(y)}, ∀β ∈ [0, 1].

Throughout this paper, we always assume that J is an interval in IR. Clearly, as θ = 45^{◦},
L_{45}^{◦} reduces to the second-order cone and the above expressions (1) and (2) correspond
to the spectral decomposition and SOC-function associated with second-order cone, re-
spectively (see [6, 9] for more information regarding f^{soc}).

It is well-known that, to deal with symmetric cone optimization problems, such as
second-order cone optimization problems and positive semi-definite optimization prob-
lems, this type of vector-valued functions plays an essential role. Inspired by this, we
study the properties of f^{L}^{θ}, which is crucial for circular cone optimization problems. In
our previous works, we have studied the smooth and nonsmooth analysis of f^{L}^{θ} [4, 19];

and the circular cone monotonicity and second-order differentiability of f^{L}^{θ} [20]. Form
the aforementioned research, there has an interesting observation: some properties com-
monly shared by f^{soc} and f^{L}^{θ} are independent of the angle θ; for example, f^{L}^{θ} is direc-
tionally differentiable, Fr´echet differentiable, semismooth if and only if f is directionally
differentiable, Fr´echet differentiable, semismooth; while some properties are dependent
of the angle θ; for example, f^{L}^{θ} with f (t) = −1/t for t > 0 is circular cone monotone as
θ ∈ [45^{◦}, 90^{◦}), but not circular cone monotone as θ ∈ (0, 45^{◦}).

In this paper, we further study the circular cone convexity of f . More precisely, a
real-valued function f : J → IR is said to be L_{θ}-convex of order n on S, if for any x, y ∈ S,

f^{L}^{θ}(βx + (1 − β)y) L_{θ} βf^{L}^{θ}(x) + (1 − β)f^{L}^{θ}(y), ∀β ∈ [0, 1].

The characterization of L_{θ}-convexity is based on the observation that f is L_{θ}-convex if
and only if (f^{L}^{θ})^{00}(x)(h, h) ∈ L_{θ} for all h ∈ IR^{n}. Our result shows that the circular cone
convexity requires that the angle θ belongs in [45^{◦}, 90^{◦}). In particular, we show that f is
L_{θ}-convex of order 2 if and only if θ ∈ [45^{◦}, 90^{◦}) and f is convex.

On the other hand, using the spectral decomposition (1), we define the determinant and trace of x in the framework of circular cone as

det(x) := λ_{1}(x)λ_{2}(x) and tr(x) := λ_{1}(x) + λ_{2}(x),

respectively. In the symmetric cone setting, the concepts of determinant and trace are the key ingredients of barrier and penalty functions which are used in barrier and penalty methods (including interior point methods) for symmetric cone optimization, see [1, 2, 3].

Here we further study some basic inequalities of det(x) and tr(x) in the framework of
circular cone. As seen in Section 3, the obtained inequalities are classified into three
categories: (i) the first class is independent of the angle (i.e., still holds in the framework
of circular cone); (ii) the second class is dependent of the angle, for example, for x, y ∈ L_{θ},
the inequality

det(e + x + y) ≤ det(e + x)det(e + y),

where e = (1, 0, . . . , 0) ∈ IR^{n}, fails as θ ∈ (0, 45^{◦}), but holds as θ ∈ [45^{◦}, 90^{◦}); (iii) the
third class always fails no matter what value of θ is chosen. These results give us new
insight on circular cone and attract us more attention on the role played by the angle θ.

The notation used in this paper is standard. For example, denote by IR^{n} the n-
dimensional Euclidean space and by IR_{+} the set of all nonnegative real scalars, i.e.,
IR_{+} = {t ∈ IR| t ≥ 0}. For x, y ∈ IR^{n}, the inner product is denoted by x^{T}y. Let S^{n}
mean the spaces of all real symmetric matrices in IR^{n×n} and by S^{n}+ the cone of positive
semi-definite matrices. We write x L_{θ} y to stand for x − y ∈ L_{θ}. Finally, we define

0

0 := 0 for convenience.

### 2 Circular cone convexity

The main purpose of this section is to provide characterizations of L_{θ}-convex functions.

First, we need the following technical lemma.

Lemma 2.1. Given αi ∈ IR for i = 1, . . . , 6 and βi ∈ IR for i = 1, 2, 3, we define

F (β_{1}, β_{2}, β_{3}) := α_{1}β_{1}^{4}+ α_{2}β_{3}^{4}+ α_{3}β_{1}^{2}β_{3}^{2}+ α_{4}β_{2}^{2}β_{3}^{2}+ α_{5}β_{1}^{2}β_{2}^{2}+ α_{6}β_{1}β_{2}β_{3}^{2}. (3)
If F (β_{1}, β_{2}, β_{3}) ≥ 0 for all (β_{1}, β_{2}, β_{3}) ∈ IR^{3}, then

α_{1} ≥ 0, α_{2} ≥ 0, α_{4} ≥ 0, α_{5} ≥ 0, α_{3} ≥ −2√
α_{1}α_{2}.
Furthermore, if

α^{2}_{6} ≤ 4α_{2}α_{5} for α_{3} ≥ 0
4[α2− (α^{2}_{3}/4α1)]α5 for α3 ∈ [−2√

α1α2, 0) (4)
then F (β_{1}, β_{2}, β_{3}) ≥ 0 for all (β_{1}, β_{2}, β_{3}) ∈ IR^{3}.

Proof. If β_{1} = 0, then F (β_{1}, β_{2}, β_{3}) = β_{3}^{2}[α_{2}β_{3}^{2}+ α_{4}β_{2}^{2}]. From F (β_{1}, β_{2}, β_{3}) ≥ 0, we have
α_{2}β_{3}^{2}+ α_{4}β_{2}^{2} ≥ 0. Thus, α_{2} ≥ 0 by letting β_{2} → 0 and α_{4} ≥ 0 by letting β_{3} → 0.

If β_{3} = 0, then F (β_{1}, β_{2}, β_{3}) = β_{1}^{2}[α_{1}β_{1}^{2}+ α_{5}β_{2}^{2}]. From F (β_{1}, β_{2}, β_{3}) ≥ 0, we obtain α_{1} ≥ 0
and α5 ≥ 0.

If β_{2} = 0, then

F (β_{1}, β_{2}, β_{3}) = α_{1}β_{1}^{4}+ α_{2}β_{3}^{4}+ α_{3}β_{1}^{2}β_{3}^{2} = β_{1}^{2}β_{3}^{2}

α_{1}(β_{1}

β_{3})^{2}+ α_{3}+ α_{2}(β_{3}
β_{1})^{2}

(5)
whenever β_{1} 6= 0 and β_{3} 6= 0. Let t = β_{1}/β_{3}. From F (β_{1}, β_{2}, β_{3}) ≥ 0, the equation (5)
implies

α_{3} ≥ −α_{1}t^{2}− α_{2}(1/t^{2}), ∀t 6= 0,
i.e.,

α_{3} ≥ max

t6=0 [−α_{1}t^{2}− α_{2}(1/t^{2})] = − min

t6=0[α_{1}t^{2}+ α_{2}(1/t^{2})] = −2√
α_{1}α_{2}.
Furthermore, if α_{3} ≥ 0, then

F (β_{1}, β_{2}, β_{3}) ≥ α_{2}β_{3}^{4} + α_{5}β_{1}^{2}β_{2}^{2}+ α_{6}β_{1}β_{2}β_{3}^{2} = [β_{3}^{2} β_{1}β_{2}]

α2 α6/2
α_{6}/2 α_{5}

β_{3}^{2}
β_{1}β_{2}

≥ 0 where the last step is due to

α2 α6/2
α_{6}/2 α_{5}

_{S}^{2}

+ O

which is ensured by the condition (4). Similarly, if α_{3} ∈ [−2√

α_{1}α_{2}, 0) (implying α_{1} 6= 0
in this case), then

F (β_{1}, β_{2}, β_{3})

= √

α_{1}β_{1}^{2}+ α_{3}
2√

α_{1}β_{3}^{2}

2

+

α_{2}− α^{2}_{3}
4α_{1}

β_{3}^{4}+ α_{4}β_{2}^{2}β_{3}^{2}+ α_{5}β_{1}^{2}β_{2}^{2}+ α_{6}β_{1}β_{2}β_{3}^{2}

≥

α_{2}− α^{2}_{3}
4α_{1}

β_{3}^{4}+ α_{5}β_{1}^{2}β_{2}^{2}+ α_{6}β_{1}β_{2}β_{3}^{2}

= [β_{3}^{2} β_{1}β_{2}]α_{2}− (α^{2}_{3}/4α_{1}) α_{6}/2
α_{6}/2 α_{5}

β_{3}^{2}
β_{1}β_{2}

≥ 0 where the last step is due to

α_{2}− (α^{2}_{3}/4α1) α6/2
α_{6}/2 α_{5}

_{S}^{2}

+ O

which is ensured by the condition (4) and the fact α_{2}− (α^{2}_{3}/4α_{1}) ≥ 0 since −2√

α_{1}α_{2} ≤
α_{3} < 0. This completes the proof. 2

Lemma 2.2. [20, Theorem 3.1] Let f : J → IR and f^{L}^{θ} be defined as in (2). Then, f^{L}^{θ}
is second order differentiable at x ∈ S if and only if f is second order differentiable at
λi(x) ∈ J for i = 1, 2. Moreover, for u, v ∈ IR^{n}, if x2 = 0, then

(f^{L}^{θ})^{00}(x)(u, v)

=

f^{00}(x_{1})

u^{T}v
u1v2+ v1u2

, either u_{2} = 0 or v_{2} = 0.

"

f^{00}(x_{1})u^{T}v

f^{00}(x_{1})(v_{1}u_{2}+ u_{1}v_{2}) + ^{1}_{2}f^{00}(x_{1})(tan θ − cot θ)

ku_{2}kv_{2}+ ¯u^{T}_{2}v¯_{2}kv_{2}ku_{2}

#

, otherwise.

If x_{2} 6= 0, then

(f^{L}^{θ})^{00}(x)(u, v) = I_{1}
I_{2}

where

I_{1} := v_{1}u_{1}ξ + ˜˜ %

u_{1}x¯^{T}_{2}v_{2}+ v_{1}x¯^{T}_{2}u_{2}

+ ˜av_{2}^{T}u_{2} +

˜ η − ˜a

¯

x^{T}_{2}v_{2}x¯^{T}_{2}u_{2}
I_{2} := h

˜

η − ˜au_{1}x¯^{T}_{2}v_{2}+ $ − 3 ˜d ¯x^{T}_{2}v_{2}x¯^{T}_{2}u_{2}+ ˜%v_{1}u_{1}+ ˜η − ˜av_{1}x¯^{T}_{2}u_{2}i

¯
x_{2}
+ ˜dh

¯

x^{T}_{2}u_{2}v_{2}+ v_{2}^{T}u_{2}x¯_{2}+ ¯x^{T}_{2}v_{2}u_{2}i

+ ˜a u_{1}v_{2} + v_{1}u_{2}
with

˜

a = f^{0}(λ2(x)) − f^{0}(λ1(x))
λ_{2}(x) − λ_{1}(x) ,
ξ =˜ f^{00}(λ_{1}(x))

1 + cot^{2}θ + f^{00}(λ_{2}(x))
1 + tan^{2}θ,

˜

% = − cot θ

1 + cot^{2}θf^{00}(λ1(x)) + tan θ

1 + tan^{2}θf^{00}(λ2(x)),

˜

η = cot^{2}θ

1 + cot^{2}θf^{00}(λ1(x)) + tan^{2}θ

1 + tan^{2}θf^{00}(λ2(x)),
d =˜ 1

kx_{2}k

cot^{2}θ

1 + cot^{2}θf^{0}(λ1(x)) + tan^{2}θ

1 + tan^{2}θf^{0}(λ2(x)) − f (λ_{2}(x)) − f (λ_{1}(x))
λ_{2}(x) − λ_{1}(x)

,

$ = − cot^{3}θ

1 + cot^{2}θf^{00}(λ_{1}(x)) + tan^{3}θ

1 + tan^{2}θf^{00}(λ_{2}(x)).

The characterization of L_{θ}-convexity is established below, which can be regarded as
the extension of some results given in [5, 7, 8, 15] from second-order cone setting to
circular cone setting.

Theorem 2.1. Suppose that f : J → IR is second order continuously differentiable. If
f is L_{θ}-convex of order n on S, then tan θ ≥ 1, f is convex on J , and for all τ_{1}, τ_{2} ∈ J
with τ1 ≤ τ2,

f^{00}(τ_{2})δ(τ_{2}, τ_{1}) ≥ 2

(τ_{2}− τ_{1})^{2}δ(τ_{1}, τ_{2})^{2} (6)
and

h

tan^{2}θδ(τ_{1}, τ_{2}) + (tan^{2}θ − 1)δ(τ_{2}, τ_{1})i

f^{00}(τ_{1}) − 2

(τ_{2}− τ_{1})^{2}δ(τ_{2}, τ_{1})^{2}

≥ −f^{00}(τ_{1})
r

(tan^{2}θ − 1)δ(τ_{2}, τ_{1})h

2 tan^{2}θδ(τ_{1}, τ_{2}) + (tan^{2}θ − 1)δ(τ_{2}, τ_{1})i

. (7) Furthermore, if

h

tan^{2}θδ(τ_{1}, τ_{2}) + (tan^{2}θ − 1)δ(τ_{2}, τ_{1})i

f^{00}(τ_{1}) ≥ 2

(τ_{2}− τ_{1})^{2}δ(τ_{2}, τ_{1})^{2} (8)
and

8 δ(τ_{2}, τ_{1})δ(τ_{1}, τ_{2})^{2} ≤h

2 tan^{2}θδ(τ_{1}, τ_{2}) + (tan^{2}θ − 1)δ(τ_{2}, τ_{1})i

f^{00}(τ_{1})f^{00}(τ_{2})(τ_{2}− τ_{1})^{4} (9)
or if

h

tan^{2}θδ(τ1, τ2) + (tan^{2}θ − 1)δ(τ2, τ1)
i

f^{00}(τ1) < 2

(τ_{2} − τ_{1})^{2}δ(τ2, τ1)^{2}
and

8δ(τ_{1}, τ_{2})^{2}δ(τ_{2}, τ_{1})^{2}(tan^{2}θ − 1)f^{00}(τ_{1})

≤ ("

(tan^{2}θ − 1)f^{00}(τ1)^{2}
h

2 tan^{2}θδ(τ1, τ2) + (tan^{2}θ − 1)δ(τ2, τ1)
i

δ(τ2, τ1)

#

(10)

−

"

tan^{2}θδ(τ_{1}, τ_{2}) + (tan^{2}θ − 1)δ(τ_{2}, τ_{1})

f^{00}(τ_{1}) − 2

(τ2− τ1)^{2}δ(τ_{2}, τ_{1})^{2}

#2)

f^{00}(τ_{2})(τ_{2}− τ_{1})^{4},

then f is L_{θ}-convex. Here δ(τ, τ^{0}) := f (τ ) − f (τ^{0}) − f^{0}(τ^{0})(τ − τ^{0}) for τ, τ^{0} ∈ J.

Proof. According to [20, Theorem 3.2], f is L_{θ}-convex if and only if (f^{L}^{θ})^{00}(x)(h, h) ∈ L_{θ}
for all x ∈ S and h ∈ IR^{n}. We proceed the proof by considering the following three cases.

Case 1. For x_{2} = 0 and h_{2} = 0, it follows from Lemma 2.2 that
(f^{L}^{θ})^{00}(x)(h, h) = f^{00}(x_{1}) h^{2}_{1}

0

.
Hence, (f^{L}^{θ})^{00}(x)(h, h) ∈ L_{θ} if and only if f^{00}(x_{1}) ≥ 0.

Case 2. For x2 = 0 and h2 6= 0, it follows from Lemma 2.2 that
(f^{L}^{θ})^{00}(x)(h, h) =

f^{00}(x1)khk^{2}

2f^{00}(x_{1})h_{1}h_{2}+ f^{00}(x_{1})(tan θ − cot θ)kh_{2}kh_{2}

.

Hence, (f^{L}^{θ})^{00}(x)(h, h) ∈ Lθ if and only if f^{00}(x1) ≥ 0 and
tan θkhk^{2} ≥

2h_{1}+ (tan θ − cot θ)kh_{2}k
kh_{2}k,
i.e.,

− tan θ(h^{2}_{1}+ kh_{2}k^{2}) ≤h

2h_{1}+ (tan θ − cot θ)kh_{2}ki

kh_{2}k ≤ tan θ(h^{2}_{1}+ kh_{2}k^{2}).

Dividing by kh_{2}k^{2} and letting t = h_{1}/kh_{2}k yields

− tan θ(t^{2}+ 1) ≤ 2t + tan θ − cot θ ≤ tan θ(t^{2}+ 1)

⇐⇒ max

t∈IR − tan θ(t^{2}+ 1) − 2t ≤ tan θ − cot θ ≤ min

t∈IRtan θ(t^{2} + 1) − 2t

⇐⇒ cot θ − tan θ ≤ tan θ − cot θ ≤ tan θ − cot θ

⇐⇒ tan θ ≥ 1.

Case 3. For x_{2} 6= 0, due to the simplification of notation, let us denote
µ_{1} := h_{1}− cot θ¯x^{T}_{2}h_{2}, µ_{2} := h_{1}+ tan θ¯x^{T}_{2}h_{2}, µ_{3} :=

q

kh_{2}k^{2}− (¯x^{T}_{2}h_{2})^{2}. (11)
Then,

¯

x^{T}_{2}h_{2} = µ_{2}− µ_{1}

tan θ + cot θ and h_{1} = tan θµ_{1}+ cot θµ_{2}

tan θ + cot θ . (12)

Note that µ_{1}, µ_{2}, and µ_{3} can be taken any value in IR × IR × IR_{+}by taking suitable value
of h (because the vector h has n variables). It follows from Lemma 2.2 that

(f^{L}^{θ})^{00}(x)(h, h) =

ξh˜ ^{2}_{1}+ 2 ˜%¯x^{T}_{2}h_{2}h_{1}+ ˜akh_{2}k^{2}+ ˜η − ˜a(¯x^{T}_{2}h_{2})^{2}
h

$ − 3 ˜d(¯x^{T}_{2}h_{2})^{2}+ 2 ˜η − ˜a ¯x^{T}_{2}h_{2}h_{1}i

¯
x_{2}+h

˜

%h^{2}_{1}+ ˜dkh_{2}k^{2}i

¯
x_{2}
+2h

˜

ah_{1}+ ˜d¯x^{T}_{2}h_{2}i
h_{2}

=:

Θ_{1}

Θ2x¯2+ Θ3h2

where

Θ_{1} = ˜ξh^{2}_{1} + 2 ˜%¯x^{T}_{2}h_{2}h_{1}+ ˜akh_{2}k^{2}+ ˜η − ˜a(¯x^{T}_{2}h_{2})^{2}

Θ2 = $ − 3 ˜d(¯x^{T}_{2}h2)^{2}+ 2 ˜η − ˜a ¯x^{T}_{2}h2h1 + ˜%h^{2}_{1}+ ˜dkh2k^{2}
Θ3 = 2

h

˜

ah1 + ˜d¯x^{T}_{2}h2

i
.
Hence, (f^{L}^{θ})^{00}(x)(h, h) ∈ L_{θ} is equivalent to

Θ_{1} ≥ 0 and Θ^{2}_{1}tan^{2}θ ≥ kΘ_{2}x¯_{2}+ Θ_{3}h_{2}k^{2}.
Note that

Θ_{1} = 1

1 + cot^{2}θf^{00}(λ_{1}(x))h

h^{2}_{1}− 2(¯x^{T}_{2}h_{2})h_{1}cot θ + (¯x^{T}_{2}h_{2})^{2}cot^{2}θi

+ 1

1 + tan^{2}θf^{00}(λ_{2}(x))h

h^{2}_{1}+ 2(¯x^{T}_{2}h_{2})h_{1}tan θ + (¯x^{T}_{2}h_{2})^{2}tan^{2}θi
+ ˜ah

kh_{2}k^{2}− (¯x^{T}_{2}h_{2})^{2}i

= 1

1 + cot^{2}θf^{00}(λ_{1}(x))µ^{2}_{1}+ 1

1 + tan^{2}θf^{00}(λ_{2}(x))µ^{2}_{2}+ ˜aµ^{2}_{3}. (13)

We now claim that Θ_{1} ≥ 0 for all h ∈ IR^{n} if and only if

f^{00}(λ_{1}(x)) ≥ 0, f^{00}(λ_{2}(x)) ≥ 0, and ˜a ≥ 0. (14)
The sufficiency is clear. Let us show the necessity. In particular, choosing h = (− tan θ, ¯x_{2})
yields µ_{2} = 0 and µ_{3} = 0. It then follows from Θ_{1} ≥ 0 that f^{00}(λ_{1}(x)) ≥ 0. If we choose
h = (cot θ, ¯x_{2}), then we have f^{00}(λ_{2}(x)) ≥ 0. Finally, choosing h = (1, kz_{2}) with k ∈ IR,
kz_{2}k = 1 and z_{2}^{T}x¯_{2} = 0 gives

Θ_{1} = f^{00}(λ_{1}(x))

1 + cot^{2}θ + f^{00}(λ_{2}(x))

1 + tan^{2}θ + ˜ak^{2} ≥ 0.

Dividing by k^{2} on both sides and taking the limits as k → ∞, we obtain ˜a ≥ 0. Since
λ_{i}(x) can be taken arbitrary value in J , it is clear that (14) is equivalent to saying that
f^{00}(τ ) ≥ 0 for all τ ∈ J , i.e., f is convex on J . Indeed, the condition ˜a ≥ 0 is ensured by
the fact that ˜a = ^{f}^{0}^{(λ}_{λ}^{2}^{(x))−f}^{0}^{(λ}^{1}^{(x))}

2(x)−λ1(x) = f^{00}(t0) ≥ 0 for some t0 ∈ (λ1(x), λ2(x)).

Now we calculate the value of Θ_{2} and Θ_{3}, respectively.

Θ_{2}

= − cot θ

1 + cot^{2}θf^{00}(λ_{1}(x))µ^{2}_{1} + tan θ

1 + tan^{2}θf^{00}(λ_{2}(x))µ^{2}_{2}+ ˜dµ^{2}_{3}− 2( ˜d¯x^{T}_{2}h_{2}+ ˜ah_{1})(¯x^{T}_{2}h_{2})

= − cot θ

1 + cot^{2}θf^{00}(λ_{1}(x))µ^{2}_{1} + tan θ

1 + tan^{2}θf^{00}(λ_{2}(x))µ^{2}_{2}+ ˜dµ^{2}_{3}− (¯x^{T}_{2}h_{2})Θ_{3}. (15)
Meanwhile, it follows from (12) that

Θ_{3} = 2

˜

atan θµ_{1}+ cot θµ_{2}

tan θ + cot θ + ˜d µ_{2} − µ_{1}
tan θ + cot θ

= 2

tan θ + cot θ h

µ1

˜

a tan θ − ˜d

+ µ2

˜

a cot θ + ˜d

i

. (16)

Note that

kΘ_{2}x¯_{2}+ Θ_{3}h_{2}k^{2} = Θ^{2}_{2}+ 2Θ_{2}Θ_{3}x¯^{T}_{2}h_{2}+ Θ^{2}_{3}kh_{2}k^{2}

= Θ^{2}_{2}+ 2Θ2Θ3x¯^{T}_{2}h2+ Θ^{2}_{3}
h

µ^{2}_{3}+ (¯x^{T}_{2}h2)^{2}
i

=

Θ_{2}+ Θ_{3}x¯^{T}_{2}h_{2}2

+ Θ^{2}_{3}µ^{2}_{3}. (17)

Putting (13) and (15)-(17) together, the condition Θ^{2}_{1}tan^{2}θ ≥ kΘ2x¯2 + Θ3h2k^{2} can be
rewritten equivalently as

tan^{2}θ f^{00}(λ1(x))

1 + cot^{2}θµ^{2}_{1}+ f^{00}(λ2(x))

1 + tan^{2}θµ^{2}_{2}+ ˜aµ^{2}_{3}

^{2}

≥

− cot θ

1 + cot^{2}θf^{00}(λ_{1}(x))µ^{2}_{1}+ tan θ

1 + tan^{2}θf^{00}(λ_{2}(x))µ^{2}_{2}+ ˜dµ^{2}_{3}

2

+ 4

(tan θ + cot θ)^{2}
h

µ1

˜

a tan θ − ˜d

+ µ2

˜

a cot θ + ˜d

i2

µ^{2}_{3}
i.e.,

tan^{4}θ − 1

f^{00}(λ_{1}(x))^{2}µ^{4}_{1}+

tan θ + cot θ2

˜

a^{2}tan^{2}θ − ˜d^{2}
µ^{4}_{3}
+2h

tan θ + cot θ

˜

a tan^{3}θ + ˜d

f^{00}(λ_{1}(x)) − 2

˜

a tan θ − ˜d2i

µ^{2}_{1}µ^{2}_{3} (18)
+2h

tan θ + cot θ

˜

a tan θ − ˜d

f^{00}(λ_{2}(x)) − 2

˜

a cot θ + ˜d2i
µ^{2}_{2}µ^{2}_{3}
+2

tan^{2}θ + 1

f^{00}(λ_{1}(x))f^{00}(λ_{2}(x))µ^{2}_{1}µ^{2}_{2}− 8

˜

a tan θ − ˜d

˜

a cot θ + ˜d

µ_{1}µ_{2}µ^{2}_{3} ≥ 0.

To apply Lemma 2.1, we need to compute each coefficient in (18). By calculation, we have

˜

a tan θ − ˜d

= f^{0}(λ_{2}(x)) − f^{0}(λ_{1}(x))

λ_{2}(x) − λ_{1}(x) tan θ − 1
kx_{2}k

cot^{2}θ

1 + cot^{2}θf^{0}(λ_{1}(x)) + tan^{2}θ

1 + tan^{2}θf^{0}(λ_{2}(x))

+ 1
kx_{2}k

f (λ_{2}(x)) − f (λ_{1}(x))
λ_{2}(x) − λ_{1}(x)

= f^{0}(λ_{2}(x)) − f^{0}(λ_{1}(x))

λ_{2}(x) − λ_{1}(x) tan θ − 1
kx_{2}k

cot θ

tan θ + cot θf^{0}(λ_{1}(x)) + tan θ

tan θ + ctanθf^{0}(λ_{2}(x))

+ 1
kx_{2}k

f (λ_{2}(x)) − f (λ_{1}(x))
λ_{2}(x) − λ_{1}(x)

= −tan θ + ctanθ

λ_{2}(x) − λ_{1}(x)f^{0}(λ_{1}(x)) + tan θ + ctanθ
[λ_{2}(x) − λ_{1}(x)]^{2}

h

f (λ_{2}(x)) − f (λ_{1}(x))i

=

(tan θ + cot θ)h

f (λ_{2}(x)) − f (λ_{1}(x)) − f^{0}(λ_{1}(x))(λ_{2}(x) − λ_{1}(x))i
[λ_{2}(x) − λ_{1}(x)]^{2}

= tan θ + cot θ

[λ_{2}(x) − λ_{1}(x)]^{2}δ λ_{2}(x), λ_{1}(x),

where the third equation follows from the fact λ_{2}(x) − λ_{1}(x) = (tan θ + ctanθ)kx_{2}k.

Similarly, we have

˜

a tan θ + ˜d

=

(tan θ + cot θ)h

f (λ_{1}(x)) − f (λ_{2}(x)) +

2 tan θ

tan θ+cot θf^{0}(λ_{2}(x)) + cot θ−tan θ

tan θ+cot θf^{0}(λ_{1}(x))

λ_{2}(x) − λ_{1}(x)i
[λ_{2}(x) − λ_{1}(x)]^{2}

= tan θ + cot θ
[λ_{2}(x) − λ_{1}(x)]^{2}

2 tan^{2}θ

tan^{2}+1δ λ_{1}(x), λ_{2}(x) + tan^{2}θ − 1

tan^{2}θ + 1δ λ_{2}(x), λ_{1}(x)

.

˜

a cot θ + ˜d =

(tan θ + cot θ)h

f (λ_{1}(x)) − f (λ_{2}(x)) − f^{0}(λ_{2}(x))(λ_{1}(x) − λ_{2}(x))i
[λ2(x) − λ1(x)]^{2}

= tan θ + cot θ

[λ_{2}(x) − λ_{1}(x)]^{2}δ λ_{1}(x), λ_{2}(x).

˜

a tan^{3}θ + ˜d

=

(tan θ + cot θ)h

f (λ_{1}(x)) − f (λ_{2}(x)) − [tan^{2}θf^{0}(λ_{2}(x)) + (1 − tan^{2}θ)f^{0}(λ_{1}(x))](λ_{1}(x) − λ_{2}(x))i
[λ_{2}(x) − λ_{1}(x)]^{2}

= tan θ + cot θ
[λ_{2}(x) − λ_{1}(x)]^{2}

h

tan^{2}θδ λ_{1}(x), λ_{2}(x) + (tan^{2}θ − 1)δ λ_{2}(x), λ_{1}(x)i
.

Corresponding each coefficient in (18) to (3), we know

α_{1} = (tan^{4}θ − 1)f^{00}(λ_{1}(x))^{2}
α_{2} = (tan θ+cot θ)^{4}

[λ2(x)−λ1(x)]^{4}δ λ_{2}(x), λ_{1}(x)h

2 tan^{2}θ

tan^{2}θ+1δ λ_{1}(x), λ_{2}(x) +^{tan}_{tan}^{2}2^{θ−1}θ+1δ λ_{2}(x), λ_{1}(x)i
α_{3} = 2(tan θ+cot θ)^{2}

[λ2(x)−λ1(x)]^{2}

( h

tan^{2}θδ λ_{1}(x), λ_{2}(x) + (tan^{2}θ − 1)δ λ_{2}(x), λ_{1}(x)i

f^{00}(λ_{1}(x))

−2^{δ λ}^{2}^{(x),λ}^{1}^{(x)}

^{2}

[λ2(x)−λ1(x)]^{2}

)

α_{4} = 2(tan θ+cot θ)^{2}
[λ2(x)−λ1(x)]^{2}

"

δ λ_{2}(x), λ_{1}(x)f^{00}(λ_{2}(x)) − 2^{δ λ}^{1}^{(x),λ}^{2}^{(x)}

^{2}

[λ2(x)−λ1(x)]^{2}

#

α_{5} = 2(tan^{2}θ + 1)f^{00}(λ_{1}(x))f^{00}(λ_{2}(x))
α_{6} = −8(tan θ+cot θ)^{2}

[λ2(x)−λ1(x)]^{4}δ λ_{1}(x), λ_{2}(x)δ λ_{2}(x), λ_{1}(x).

In view of Lemma 2.1, the condition α_{1} ≥ 0 means tan θ ≥ 1, α_{2}, α_{5} ≥ 0 is ensured by
the convexity of f (see (14)), α_{4} ≥ 0 corresponds to (6), and α_{3} ≥ −2√

α_{1}α_{2} corresponds
to (7). In addition, the condition (4) takes the special form (9) and (10), respectively.

2

Theorem 2.2. Suppose that f : J → IR is second order continuously differentiable.

Then, f is Lθ-convex of order 2 on S if and only if tan θ ≥ 1 and f is convex on J .
Proof. The necessity is clear from Theorem 2.1. For sufficiency, note that in (11) µ_{3} = 0
since ¯x_{2} = ±1 in this case. Hence, (18) takes the form of

(tan^{4}θ − 1)f^{00}(λ_{1}(x))^{2}µ^{4}_{1}+ 2(tan^{2}θ + 1)f^{00}(λ_{1}(x))f^{00}(λ_{2}(x))µ^{2}_{1}µ^{2}_{2} ≥ 0

for all µ_{1} and µ_{2}, which is equivalent to verifying

tan θ ≥ 1 and f^{00}(λ_{1}(x))f^{00}(λ_{2}(x)) ≥ 0.

This is ensued by the conditions that tan θ ≥ 1 and f is convex on J . Thus, the proof is complete. 2

If, in particular, θ = 45^{o}, then (6) and (7) reduces to [15, (21) in Proposition 4.2]; (9)
reduces to [15, (22) in Proposition 4.2]. In addition, due to (7), (8) holds automatically in
this case. The above results indicate that the L_{θ}-convexity is dependent of the properties
of f and the angle θ together.

### 3 Inequalities associated with circular cone

In this section, we establish some inequalities associated with circular cone, which we
believe that they will be useful for further analyzing properties of f^{L}^{θ} and proving the
convergence of interior point methods for optimization problems involved in circular
cones.

In [5], the author establishes the following results in the framework of second-order
cone. More specifically, for x _{L}_{45◦} 0 and y _{L}_{45◦} 0, then

(a) det(e + x)^{1/2}≥ 1 + det(x)^{1/2}
(b) det(x + y) ≥ det(x) + det(y)

(c) det(αx + (1 − α)y) ≥ α^{2}det(x) + (1 − α)^{2}det(y), ∀α ∈ [0, 1]

(d) det(e + x + y) ≤ det(e + x)det(e + y)

(e) If x _{L}_{45◦} y _{L}_{45◦} 0, then det(x) ≥ det(y), tr(x) ≥ tr(y), and λ_{i}(x) ≥ λ_{i}(y) for
i = 1, 2

(f) tr(x + y) = tr(x) + tr(y) and det(γx) = γ^{2}det(x) for all γ ∈ IR.

In the following, we show that, in the framework of circular cone, the above inequali- ties can be classified into three categories. The first class holds independent of the angle, e.g., (a); the second class holds dependent of the angle, e.g., (b)-(e); the third class fails no matter what value of the angle is chosen, e.g., (f).

Theorem 3.1. Let x = (x_{1}, x_{2}) ∈ IR × IR^{n−1} possess spectral factorization associated
with circular cone given as in (1). Then,

(a) [det(e + x)]^{1/2} ≥ 1 + det(x)^{1/2} for all x ∈ L_{θ}

(b) If x L_{θ} y, then λ_{1}(x) ≥ λ_{1}(y).

Proof. (a) Note that det(x) ≥ 0 and det(e + x) ≥ 0 since x, x + e ∈ L_{θ}. Therefore,
[det(e + x)]^{1/2} ≥ 1 + det(x)^{1/2}

⇐⇒ det(e + x) ≥ 1 + 2det(x)^{1/2}+ det(x)

⇐⇒ λ1(e + x)λ2(e + x) ≥ 1 + 2p

λ1(x)λ2(x) + λ1(x)λ2(x)

⇐⇒ (x_{1} + 1 − kx_{2}k cot θ)(x_{1}+ 1 + kx_{2}k tan θ) ≥ 1 + 2p

λ_{1}(x)λ_{2}(x) + λ_{1}(x)λ_{2}(x)

⇐⇒ (λ_{1}(x) + 1)(λ_{2}(x) + 1) ≥ 1 + 2p

λ_{1}(x)λ_{2}(x) + λ_{1}(x)λ_{2}(x)

⇐⇒ λ_{1}(x)λ_{2}(x) + λ_{1}(x) + λ_{2}(x) + 1 ≥ 1 + 2p

λ_{1}(x)λ_{2}(x) + λ_{1}(x)λ_{2}(x)

⇐⇒ λ1(x) + λ2(x) ≥ 2p

λ1(x)λ2(x)

⇐⇒ λ1(x) + λ2(x)

2 ≥p

λ_{1}(x)λ_{2}(x).

Hence, to prove the desired result, it suffices to show that λ1(x) + λ2(x)

2 ≥p

λ_{1}(x)λ_{2}(x)

which is clearly true by the arithmetic mean-geometric mean (AM-GM) inequality.

(b) Since x − y ∈ L_{θ}, we know

x_{1}− y_{1} ≥ kx_{2}− y_{2}k cot θ ≥h

kx_{2}k − ky_{2}ki
cot θ,
i.e., λ_{1}(x) = x_{1}− kx_{2}k cot θ ≥ y_{1}− ky_{2}k cot θ = λ_{1}(y). 2

Theorem 3.2. Let x = (x_{1}, x_{2}) ∈ IR × IR^{n−1} possess spectral factorization associated
with circular cone given as in (1). Then, the following hold.

(a) For all x, y ∈ L_{θ},

det(x + y) ≥ det(x) + det(y) + (kx_{2}k^{2}+ ky_{2}k^{2}) csc^{2}θ − (x^{2}_{1}+ y_{1}^{2}) sec^{2}θ.

In particular, when θ ∈ (0, 45^{◦}], we have

det(x + y) ≥ det(x) + det(y). (19)

(b) For all x, y ∈ L_{θ} and α ∈ [0, 1],
det(αx + (1 − α)y)

≥ α^{2}det(x) + (1 − α)^{2}det(y) + (α^{2}kx2k^{2}+ (1 − α)^{2}ky2k^{2}) csc^{2}θ − (α^{2}x^{2}_{1}+ (1 − α)^{2}y_{1}^{2}) sec^{2}θ.

In particular, when θ ∈ (0, 45^{◦}], we have

det(αx + (1 − α)y) ≥ α^{2}det(x) + (1 − α)^{2}det(y).

(c) If x, y ∈ L_{θ} and θ ∈ [45^{◦}, 90^{◦}), then

det(e + x + y) ≤ det(e + x)det(e + y). (20)
(d) If x L_{θ} y L_{θ} 0 and θ ∈ (0, 45^{◦}], then

λ2(x) ≥ λ2(y), det(x) ≥ det(y), and tr(x) ≥ tr(y). (21) Proof. (a) Notice that

det(x + y)

= λ_{1}(x + y) · λ_{2}(x + y)

= (x_{1}+ y_{1}− kx_{2}+ y_{2}k cot θ) (x_{1}+ y_{1}+ kx_{2}+ y_{2}k tan θ)

= (x1+ y1)^{2}+ (x1 + y1)kx2+ y2k tan θ − (x1+ y1)kx2+ y2k cot θ − kx2+ y2k^{2}
and

det(x) + det(y)

= λ_{1}(x)λ_{2}(x) + λ_{1}(y)λ_{2}(y)

= (x_{1}− kx_{2}k cot θ)(x_{1}+ kx_{2}k tan θ) + (y_{1}− ky_{2}k cot θ)(y_{1}+ ky_{2}k tan θ)

= x^{2}_{1}+ x_{1}kx_{2}k tan θ − x_{1}kx_{2}k cot θ − kx_{2}k^{2}+ y^{2}_{1}+ y_{1}ky_{2}k tan θ − y_{1}ky_{2}k cot θ − ky_{2}k^{2}

= x^{2}_{1}+ y^{2}_{1} + x_{1}kx_{2}k tan θ + y_{1}ky_{2}k tan θ − x_{1}kx_{2}k cot θ − y_{1}ky_{2}k cot θ − kx_{2}k^{2}− ky_{2}k^{2}.
Then, we have

det(x + y) − det(x) − det(y)

= 2x1y1− 2x^{T}_{2}y2+ (x1kx2+ y2k + y1kx2 + y2k − x1kx2k − y1ky2k) tan θ

−(x_{1}kx_{2} + y_{2}k + y_{1}kx_{2}+ y_{2}k − x_{1}kx_{2}k − y_{1}ky_{2}k) cot θ.

Using x, y ∈ L_{θ} (and hence x + y ∈ L_{θ}) gives

x1tan θ ≥ kx2k, −x1tan θ ≤ −kx2k, x1 ≥ kx2k cot θ, −x1 ≤ −kx2k cot θ,

−(x_{1}+ y_{1}) ≤ −kx_{2}+ y_{2}k cot θ.

Thus,

det(x + y) − det(x) − det(y)

≥ 2x_{1}y_{1}− 2x^{T}_{2}y_{2}+ kx_{2}kkx_{2}+ y_{2}k + ky_{2}kkx_{2}+ y_{2}k − x^{2}_{1}tan^{2}θ − y^{2}_{1}tan^{2}θ

−x_{1}(x_{1}+ y_{1}) − y_{1}(x_{1}+ y_{1}) + kx_{2}k^{2}cot^{2}θ + ky_{2}k^{2}cot^{2}θ

= 2x1y1− 2x^{T}_{2}y2+ kx2+ y2k(kx2k + ky2k) − (x^{2}_{1}+ y_{1}^{2}) tan^{2}θ

−(x_{1} + y_{1})^{2}+ (kx_{2}k^{2}+ ky_{2}k^{2}) cot^{2}θ

≥ kx_{2}+ y_{2}k^{2}− (x^{2}_{1}+ y_{1}^{2}) tan^{2}θ − x^{2}_{1}− y_{1}^{2}− 2x^{T}_{2}y_{2}+ (kx_{2}k^{2}+ ky_{2}k^{2}) cot^{2}θ

= kx_{2}k^{2}+ ky_{2}k^{2}− (x^{2}_{1}+ y_{1}^{2}) tan^{2}θ − x^{2}_{1}− y_{1}^{2}+ (kx_{2}k^{2}+ ky_{2}k^{2}) cot^{2}θ

= (kx_{2}k^{2}+ ky_{2}k^{2})(1 + cot^{2}θ) − (x^{2}_{1}+ y_{1}^{2})(1 + tan^{2}θ)

= (kx_{2}k^{2}+ ky_{2}k^{2}) csc^{2}θ − (x^{2}_{1}+ y_{1}^{2}) sec^{2}θ
which is the desired result.

When θ ∈ (0, 45^{◦}], we know tan θ ≤ cot θ. Since x, y ∈ L_{θ}, i.e., x_{1} ≥ kx_{2}k cot θ and
y_{1} ≥ ky_{2}k cot θ, there exist a, b ≥ 0 such that x_{1} = kx_{2}k cot θ + a and y_{1} = ky_{2}k cot θ + b.

Hence,

det(x + y) − det(x) − det(y)

= 2x_{1}y_{1} − 2x^{T}_{2}y_{2}+ (x_{1}kx_{2}+ y_{2}k + y_{1}kx_{2}+ y_{2}k − x_{1}kx_{2}k − y_{1}ky_{2}k) tan θ

−(x_{1}kx_{2}+ y_{2}k + y_{1}kx_{2} + y_{2}k − x_{1}kx_{2}k − y_{1}ky_{2}k) cot θ

=

kx_{2}k + ky_{2}kh

kx_{2}k + ky_{2}k − kx_{2}+ y_{2}ki
cot^{2}θ
+kx2+ y2k

kx2k + ky2k − kx2+ y2k + 2ab +a cot θ

ky_{2}k + kx_{2}k − kx_{2}+ y_{2}k

+ a tan θ

ky_{2}k cot^{2}θ + kx_{2}+ y_{2}k − kx_{2}k
+b cot θ

ky_{2}k + kx_{2}k − kx_{2}+ y_{2}k

+ b tan θ

kx_{2}k cot^{2}θ + kx_{2}+ y_{2}k − ky_{2}k

≥ 0,

where the last step is due to kx_{2}k + ky_{2}k ≥ kx_{2}+ y_{2}k, kx_{2}k cot^{2}θ + kx_{2}+ y_{2}k − ky_{2}k ≥
kx_{2}k+kx_{2}+y_{2}k−ky_{2}k ≥ 0, and ky_{2}k cot^{2}θ+kx_{2}+y_{2}k−kx_{2}k ≥ ky_{2}k+kx_{2}+y_{2}k−kx_{2}k ≥ 0
since cot θ ≥ 1, due to θ ∈ (0, 45^{◦}].

(b) The result follows from the fact that det(γx) = γ^{2}det(x) for all γ ≥ 0.

(c) Since θ ∈ [45^{◦}, 90^{◦}), cot θ ≤ 1. For x, y ∈ L_{θ}, there exists two nonnegative scalars
a, b ≥ 0 such that x_{1} = kx_{2}k cot θ + a and y_{1} = ky_{2}k cot θ + b. This implies

det(e+x) = (x_{1}+1−kx_{2}k cot θ)(x_{1}+1+kx_{2}k tan θ) = (a+1)(cot θ +tan θ)kx_{2}k+(a+1)^{2},
det(e + y) = (y_{1}+ 1 − ky_{2}k cot θ)(y_{1}+ 1 + ky_{2}k tan θ) = (b + 1)(cot θ + tan θ)ky_{2}k + (b + 1)^{2}.
Thus, we obtain

det(e + x)det(e + y)

= (a + 1)(b + 1)(cot θ + tan θ)^{2}kx_{2}kky_{2}k + (a + 1)(b + 1)^{2}(cot θ + tan θ)kx_{2}k
+(a + 1)^{2}(b + 1)(cot θ + tan θ)ky2k + (a + 1)^{2}(b + 1)^{2}. (22)
On the other hand,

det(e + x + y)