Circular cone convexity and some inequalities associated with circular cones

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 $f^{{\mathcal{L}}_{\theta }}$; 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 a new insight when looking into properties about circular cone.

MSC:26A27, 26B05, 26B35, 49J52, 90C33, 65K05.

1 Introduction

Recently, much attention has been paid to the nonsymmetric cone optimization problems, see [1–4] and the references therein. Unlike symmetric cones [5], there is no unified structure for nonsymmetric cones. Hence, how to tackle 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, allows us to study them via exploiting the unified Euclidean Jordan algebra [5]. 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 ${\mathcal{L}}_{\theta }$. The circular cone [6–9] 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 $\theta \in (0,{90}^{\circ })$. Then, it is mathematically expressed as

$$\begin{array}{rcl}{\mathcal{L}}_{\theta }& :=& \{x={(x_1,x_2)}^T\in \mathbb{R}\times {\mathbb{R}}^{n-1}\mid x_1\ge \parallel x\parallel cos\theta \}\\ =& \{x={(x_1,x_2)}^T\in \mathbb{R}\times {\mathbb{R}}^{n-1}\mid x_1\ge \parallel x_2\parallel cot\theta \}.\end{array}$$

Real applications of a 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 [10, 11] and references for more details.

Although ${\mathcal{L}}_{\theta }$ is a nonsymmetric cone, we can, due to its special structure, establish the explicit form of orthogonal decomposition (or spectral decomposition) [7] as

$$x={\lambda }_1(x)\cdot u_x^{(1)}+{\lambda }_2(x)\cdot u_x^{(2)},$$

(1)

where

$$\{\begin{array}{c}{\lambda }_1(x)=x_1-\parallel x_2\parallel cot\theta ,\hfill \\ {\lambda }_2(x)=x_1+\parallel x_2\parallel tan\theta \hfill \end{array}$$

and

$$\{\begin{array}{c}{u}_x^{(1)}=\frac{1}{1+{cot}^2\theta }\left[\begin{array}{cc}1& 0\\ 0& cot\theta I_{n-1}\end{array}\right]\left[\begin{array}{c}1\\ -{\overline{x}}_2\end{array}\right]=\left[\begin{array}{c}{sin}^2\theta \\ -(sin\theta cos\theta ){\overline{x}}_2\end{array}\right],\hfill \\ u_x^{(2)}=\frac{1}{1+{tan}^2\theta }\left[\begin{array}{cc}1& 0\\ 0& tan\theta I_{n-1}\end{array}\right]\left[\begin{array}{c}1\\ {\overline{x}}_2\end{array}\right]=\left[\begin{array}{c}{cos}^2\theta \\ (sin\theta cos\theta ){\overline{x}}_2\end{array}\right]\hfill \end{array}$$

with ${\overline{x}}_2=x_2/\parallel x_2\parallel$ if $x_2\ne 0$, and ${\overline{x}}_2$ being any vector w in ${\mathbb{R}}^{n-1}$ satisfying $\parallel w\parallel =1$ if $x_2=0$. Clearly, $x\in {\mathcal{L}}_{\theta }$ if and only if ${\lambda }_1(x)\ge 0$.

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

$$f^{{\mathcal{L}}_{\theta }}(x):=f({\lambda }_1(x))u_x^{(1)}+f({\lambda }_2(x))u_x^{(2)},$$

(2)

where f is a real-valued function from J to ℝ with J being a subset in ℝ. Let S be the set of all $x\in {\mathbb{R}}^n$ whose spectral values ${\lambda }_i(x)$ for $i=1,2$ belong to J, i.e., $S:=\{x\in {\mathbb{R}}^n\mid {\lambda }_i(x)\in J,i=1,2\}$. According to [12], we know that S is open if and only if J is open. In addition, as J is an interval, then S is convex because

$$\begin{array}{rcl}min\{{\lambda }_1(x),{\lambda }_1(y)\}& \le & {\lambda }_1(\beta x+(1-\beta )y)\le {\lambda }_2(\beta x+(1-\beta )y)\\ \le & max\{{\lambda }_2(x),{\lambda }_2(y)\},\mathrm{\forall }\beta \in [0,1].\end{array}$$

Throughout this paper, we always assume that J is an interval in ℝ. Clearly, as $\theta ={45}^{\circ }$, ${\mathcal{L}}_{{45}^{\circ }}$ reduces to the second-order cone and the above expressions (1) and (2) correspond to the spectral decomposition and the SOC-function associated with the second-order cone, respectively (see [13, 14] for more information regarding $f^{\mathrm{soc}}$).

It is well known that in dealing with symmetric cone optimization problems, such as second-order cone optimization problems and positive semi-definite optimization problems, this type of vector-valued functions plays an essential role. Inspired by this, we study the properties of $f^{{\mathcal{L}}_{\theta }}$, which is crucial for circular cone optimization problems. In our previous works, we have studied the smooth and nonsmooth analysis of $f^{{\mathcal{L}}_{\theta }}$ [8, 10]; and the circular cone monotonicity and second-order differentiability of $f^{{\mathcal{L}}_{\theta }}$ [9]. From the aforementioned research, there is an interesting observation: some properties commonly shared by $f^{\mathrm{soc}}$ and $f^{{\mathcal{L}}_{\theta }}$ are independent of the angle θ; for example, $f^{{\mathcal{L}}_{\theta }}$ is directionally differentiable, Fréchet differentiable, semi-smooth if and only if f is directionally differentiable, Fréchet differentiable, semi-smooth; while some properties are dependent on the angle θ; for example, $f^{{\mathcal{L}}_{\theta }}$ with $f(t)=-1/t$ for $t>0$ is circular cone monotone as $\theta \in [{45}^{\circ },{90}^{\circ })$, but not circular cone monotone as $\theta \in (0,{45}^{\circ })$.

In this paper, we further study the circular cone convexity of f. More precisely, a real-valued function $f:J\to \mathbb{R}$ is said to be ${\mathcal{L}}_{\theta }$-convex of order n on S if for any $x,y\in S$,

$$f^{{\mathcal{L}}_{\theta }}(\beta x+(1-\beta )y){⪯}_{{\mathcal{L}}_{\theta }}\beta f^{{\mathcal{L}}_{\theta }}(x)+(1-\beta )f^{{\mathcal{L}}_{\theta }}(y),\mathrm{\forall }\beta \in [0,1].$$

The characterization of ${\mathcal{L}}_{\theta }$-convexity is based on the observation that f is ${\mathcal{L}}_{\theta }$-convex if and only if ${(f^{{\mathcal{L}}_{\theta }})}^{″}(x)(h,h)\in {\mathcal{L}}_{\theta }$ for all $h\in {\mathbb{R}}^n$. Our result shows that the circular cone convexity requires that the angle θ belongs in $[{45}^{\circ },{90}^{\circ })$. In particular, we show that f is ${\mathcal{L}}_{\theta }$-convex of order 2 if and only if $\theta \in [{45}^{\circ },{90}^{\circ })$ 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

$$det(x):={\lambda }_1(x){\lambda }_2(x)\text{and}tr(x):={\lambda }_1(x)+{\lambda }_2(x),$$

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 [15–17]. 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 on the angle, for example, for $x,y\in {\mathcal{L}}_{\theta }$, the inequality

$$det(e+x+y)\le det(e+x)det(e+y),$$

where $e=(1,0,\dots ,0)\in {\mathbb{R}}^n$, fails as $\theta \in (0,{45}^{\circ })$ but holds as $\theta \in [{45}^{\circ },{90}^{\circ })$; (iii) the third class always fails no matter what value of θ is chosen. These results give us a new insight into a circular cone and make us focus more on the role played by the angle θ.

The notation used in this paper is standard. For example, denote by ${\mathbb{R}}^n$ the n-dimensional Euclidean space and by ${\mathbb{R}}_{+}$ the set of all nonnegative real scalars, i.e., ${\mathbb{R}}_{+}=\{t\in \mathbb{R}\mid t\ge 0\}$. For $x,y\in {\mathbb{R}}^n$, the inner product is denoted by $x^{T}y$. Let ${\mathbb{S}}^n$ mean the spaces of all real symmetric matrices in ${\mathbb{R}}^{n\times n}$, and let ${\mathbb{S}}_{+}^n$ denote the cone of positive semi-definite matrices. We write $x{⪰}_{{\mathcal{L}}_{\theta }}y$ to stand for $x-y\in {\mathcal{L}}_{\theta }$. Finally, we define $\frac{0}{0}:=0$ for convenience.

2 Circular cone convexity

The main purpose of this section is to provide characterizations of ${\mathcal{L}}_{\theta }$-convex functions. First, we need the following technical lemma.

Lemma 2.1 Given ${\alpha }_i\in \mathbb{R}$ for $i=1,\dots ,6$ and ${\beta }_i\in \mathbb{R}$ for $i=1,2,3$, we define

$$\mathcal{F}({\beta }_1,{\beta }_2,{\beta }_3):={\alpha }_1{\beta }_1^4+{\alpha }_2{\beta }_3^4+{\alpha }_3{\beta }_1^2{\beta }_3^2+{\alpha }_4{\beta }_2^2{\beta }_3^2+{\alpha }_5{\beta }_1^2{\beta }_2^2+{\alpha }_6{\beta }_1{\beta }_2{\beta }_3^2.$$

(3)

If $\mathcal{F}({\beta }_1,{\beta }_2,{\beta }_3)\ge 0$ for all $({\beta }_1,{\beta }_2,{\beta }_3)\in {\mathbb{R}}^3$, then

$${\alpha }_1\ge 0,{\alpha }_2\ge 0,{\alpha }_4\ge 0,{\alpha }_5\ge 0,{\alpha }_3\ge -2\sqrt{{\alpha }_1{\alpha }_2}.$$

Furthermore, if

$${\alpha }_6^2\le \{\begin{array}{cc}4{\alpha }_2{\alpha }_5\hfill & \mathit{\text{for }}{\alpha }_3\ge 0,\hfill \\ 4[{\alpha }_2-({\alpha }_3^2/4{\alpha }_1)]{\alpha }_5\hfill & \mathit{\text{for }}{\alpha }_3\in [-2\sqrt{{\alpha }_1{\alpha }_2},0),\hfill \end{array}$$

(4)

then $\mathcal{F}({\beta }_1,{\beta }_2,{\beta }_3)\ge 0$ for all $({\beta }_1,{\beta }_2,{\beta }_3)\in {\mathbb{R}}^3$.

Proof If ${\beta }_1=0$, then $\mathcal{F}({\beta }_1,{\beta }_2,{\beta }_3)={\beta }_3^2[{\alpha }_2{\beta }_3^2+{\alpha }_4{\beta }_2^2]$. From $\mathcal{F}({\beta }_1,{\beta }_2,{\beta }_3)\ge 0$, we have ${\alpha }_2{\beta }_3^2+{\alpha }_4{\beta }_2^2\ge 0$. Thus, ${\alpha }_2\ge 0$ by letting ${\beta }_2\to 0$ and ${\alpha }_4\ge 0$ by letting ${\beta }_3\to 0$.

If ${\beta }_3=0$, then $\mathcal{F}({\beta }_1,{\beta }_2,{\beta }_3)={\beta }_1^2[{\alpha }_1{\beta }_1^2+{\alpha }_5{\beta }_2^2]$. From $\mathcal{F}({\beta }_1,{\beta }_2,{\beta }_3)\ge 0$, we obtain ${\alpha }_1\ge 0$ and ${\alpha }_5\ge 0$.

If ${\beta }_2=0$, then

$$\mathcal{F}({\beta }_1,{\beta }_2,{\beta }_3)={\alpha }_1{\beta }_1^4+{\alpha }_2{\beta }_3^4+{\alpha }_3{\beta }_1^2{\beta }_3^2={\beta }_1^2{\beta }_3^2[{\alpha }_1{\left(\frac{{\beta }_1}{{\beta }_3}\right)}^2+{\alpha }_3+{\alpha }_2{\left(\frac{{\beta }_3}{{\beta }_1}\right)}^2]$$

(5)

whenever ${\beta }_1\ne 0$ and ${\beta }_3\ne 0$. Let $t={\beta }_1/{\beta }_3$. From $\mathcal{F}({\beta }_1,{\beta }_2,{\beta }_3)\ge 0$, equation (5) implies

$${\alpha }_3\ge -{\alpha }_1{t}^2-{\alpha }_2(1/t^2),\mathrm{\forall }t\ne 0,$$

i.e.,

$${\alpha }_3\ge \underset{t\ne 0}{max}[-{\alpha }_1{t}^2-{\alpha }_2(1/t^2)]=-\underset{t\ne 0}{min}[{\alpha }_1{t}^2+{\alpha }_2(1/t^2)]=-2\sqrt{{\alpha }_1{\alpha }_2}.$$

Furthermore, if ${\alpha }_3\ge 0$, then

$$\mathcal{F}({\beta }_1,{\beta }_2,{\beta }_3)\ge {\alpha }_2{\beta }_3^4+{\alpha }_5{\beta }_1^2{\beta }_2^2+{\alpha }_6{\beta }_1{\beta }_2{\beta }_3^2=\left[\begin{array}{cc}{\beta }_3^2& {\beta }_1{\beta }_2\end{array}\right]\left[\begin{array}{cc}{\alpha }_2& {\alpha }_6/2\\ {\alpha }_6/2& {\alpha }_5\end{array}\right]\left[\begin{array}{c}{\beta }_3^2\\ {\beta }_1{\beta }_2\end{array}\right]\ge 0,$$

where the last step is due to

$$\left[\begin{array}{cc}{\alpha }_2& {\alpha }_6/2\\ {\alpha }_6/2& {\alpha }_5\end{array}\right]{⪰}_{{\mathbb{S}}_{+}^2}O,$$

which is ensured by condition (4). Similarly, if ${\alpha }_3\in [-2\sqrt{{\alpha }_1{\alpha }_2},0)$ (implying ${\alpha }_1\ne 0$ in this case), then

$$\begin{array}{c}\mathcal{F}({\beta }_1,{\beta }_2,{\beta }_3)\hfill \\ ={(\sqrt{{\alpha }_1}{\beta }_1^2+\frac{{\alpha }_3}{2\sqrt{{\alpha }_1}}{\beta }_3^2)}^2+({\alpha }_2-\frac{{\alpha }_3^2}{4{\alpha }_1}){\beta }_3^4+{\alpha }_4{\beta }_2^2{\beta }_3^2+{\alpha }_5{\beta }_1^2{\beta }_2^2+{\alpha }_6{\beta }_1{\beta }_2{\beta }_3^2\hfill \\ \ge ({\alpha }_2-\frac{{\alpha }_3^2}{4{\alpha }_1}){\beta }_3^4+{\alpha }_5{\beta }_1^2{\beta }_2^2+{\alpha }_6{\beta }_1{\beta }_2{\beta }_3^2\hfill \\ =\left[\begin{array}{cc}{\beta }_3^2& {\beta }_1{\beta }_2\end{array}\right]\left[\begin{array}{cc}{\alpha }_2-({\alpha }_3^2/4{\alpha }_1)& {\alpha }_6/2\\ {\alpha }_6/2& {\alpha }_5\end{array}\right]\left[\begin{array}{c}{\beta }_3^2\\ {\beta }_1{\beta }_2\end{array}\right]\ge 0,\hfill \end{array}$$

where the last step is due to

$$\left[\begin{array}{cc}{\alpha }_2-({\alpha }_3^2/4{\alpha }_1)& {\alpha }_6/2\\ {\alpha }_6/2& {\alpha }_5\end{array}\right]{⪰}_{{\mathbb{S}}_{+}^2}O,$$

which is ensured by condition (4) and the fact ${\alpha }_2-({\alpha }_3^2/4{\alpha }_1)\ge 0$ since $-2\sqrt{{\alpha }_1{\alpha }_2}\le {\alpha }_3<0$. This completes the proof. □

Lemma 2.2 [[9], Theorem 3.1]

Let $f:J\to \mathbb{R}$ and $f^{{\mathcal{L}}_{\theta }}$ be defined as in (2). Then $f^{{\mathcal{L}}_{\theta }}$ is second-order differentiable at $x\in S$ if and only if f is second-order differentiable at ${\lambda }_i(x)\in J$ for $i=1,2$. Moreover, for $u,v\in {\mathbb{R}}^n$, if $x_2=0$, then

$$\begin{array}{c}{\left(f^{{\mathcal{L}}_{\theta }}\right)}^{″}(x)(u,v)\hfill \\ =\{\begin{array}{cc}{f}^{″}(x_1)\left[\begin{array}{c}{u}^{T}v\\ u_1{v}_2+v_1{u}_2\end{array}\right],\hfill & \mathit{\text{either }}{u}_2=0\mathit{\text{ or }}{v}_2=0,\hfill \\ \left[\begin{array}{c}{f}^{″}(x_1)u^{T}v\\ f^{″}(x_1)(v_1{u}_2+u_1{v}_2)+\frac{1}{2}{f}^{″}(x_1)(tan\theta -cot\theta )(\parallel u_2\parallel v_2+{\overline{u}}_2^T{\overline{v}}_2\parallel v_2\parallel u_2)\end{array}\right],\hfill & \mathit{\text{otherwise}}.\hfill \end{array}\hfill \end{array}$$

If $x_2\ne 0$, then

$${\left(f^{{\mathcal{L}}_{\theta }}\right)}^{″}(x)(u,v)=\left[\begin{array}{c}{I}_1\\ I_2\end{array}\right],$$

where

$$\begin{array}{c}{I}_1:=v_1{u}_1\tilde{\xi }+\tilde{\varrho }(u_1{\overline{x}}_2^T{v}_2+v_1{\overline{x}}_2^T{u}_2)+\tilde{a}{v}_2^T{u}_2+(\tilde{\eta }-\tilde{a}){\overline{x}}_2^T{v}_2{\overline{x}}_2^T{u}_2,\hfill \\ I_2:=[(\tilde{\eta }-\tilde{a})u_1{\overline{x}}_2^T{v}_2+(\varpi -3\tilde{d}){\overline{x}}_2^T{v}_2{\overline{x}}_2^T{u}_2+\tilde{\varrho }{v}_1{u}_1+(\tilde{\eta }-\tilde{a})v_1{\overline{x}}_2^T{u}_2]{\overline{x}}_2\hfill \\ \phantom{I_2:=}+\tilde{d}[{\overline{x}}_2^T{u}_2{v}_2+v_2^T{u}_2{\overline{x}}_2+{\overline{x}}_2^T{v}_2{u}_2]+\tilde{a}(u_1{v}_2+v_1{u}_2)\hfill \end{array}$$

with

$$\begin{array}{c}\tilde{a}=\frac{f^{\prime }({\lambda }_2(x))-f^{\prime }({\lambda }_1(x))}{{\lambda }_2(x)-{\lambda }_1(x)},\hfill \\ \tilde{\xi }=\frac{f^{″}({\lambda }_1(x))}{1+{cot}^2\theta }+\frac{f^{″}({\lambda }_2(x))}{1+{tan}^2\theta },\hfill \\ \tilde{\varrho }=-\frac{cot\theta }{1+{cot}^2\theta }{f}^{″}({\lambda }_1(x))+\frac{tan\theta }{1+{tan}^2\theta }{f}^{″}({\lambda }_2(x)),\hfill \\ \tilde{\eta }=\frac{{cot}^2\theta }{1+{cot}^2\theta }{f}^{″}({\lambda }_1(x))+\frac{{tan}^2\theta }{1+{tan}^2\theta }{f}^{″}({\lambda }_2(x)),\hfill \\ \tilde{d}=\frac{1}{\parallel x_2\parallel }[\frac{{cot}^2\theta }{1+{cot}^2\theta }{f}^{\prime }({\lambda }_1(x))+\frac{{tan}^2\theta }{1+{tan}^2\theta }{f}^{\prime }({\lambda }_2(x))-\frac{f({\lambda }_2(x))-f({\lambda }_1(x))}{{\lambda }_2(x)-{\lambda }_1(x)}],\hfill \\ \varpi =-\frac{{cot}^3\theta }{1+{cot}^2\theta }{f}^{″}({\lambda }_1(x))+\frac{{tan}^3\theta }{1+{tan}^2\theta }{f}^{″}({\lambda }_2(x)).\hfill \end{array}$$

The characterization of ${\mathcal{L}}_{\theta }$-convexity is established below, which can be regarded as the extension of some results given in [12, 18–20] from the second-order cone setting to the circular cone setting.

Theorem 2.1 Suppose that $f:J\to \mathbb{R}$ is second-order continuously differentiable. If f is ${\mathcal{L}}_{\theta }$-convex of order n on S, then $tan\theta \ge 1$, f is convex on J, and for all ${\tau }_1,{\tau }_2\in J$ with ${\tau }_1\le {\tau }_2$,

$$f^{″}({\tau }_2)\delta ({\tau }_2,{\tau }_1)\ge \frac{2}{{({\tau }_2-{\tau }_1)}^2}\delta {({\tau }_1,{\tau }_2)}^2$$

(6)

and

$$\begin{array}{c}[{tan}^2\theta \delta ({\tau }_1,{\tau }_2)+({tan}^2\theta -1)\delta ({\tau }_2,{\tau }_1)]f^{″}({\tau }_1)-\frac{2}{{({\tau }_2-{\tau }_1)}^2}\delta {({\tau }_2,{\tau }_1)}^2\hfill \\ \ge -f^{″}({\tau }_1)\sqrt{({tan}^2\theta -1)\delta ({\tau }_2,{\tau }_1)[2{tan}^2\theta \delta ({\tau }_1,{\tau }_2)+({tan}^2\theta -1)\delta ({\tau }_2,{\tau }_1)]}.\hfill \end{array}$$

(7)

Furthermore, if

$$[{tan}^2\theta \delta ({\tau }_1,{\tau }_2)+({tan}^2\theta -1)\delta ({\tau }_2,{\tau }_1)]f^{″}({\tau }_1)\ge \frac{2}{{({\tau }_2-{\tau }_1)}^2}\delta {({\tau }_2,{\tau }_1)}^2$$

(8)

and

$$8\delta ({\tau }_2,{\tau }_1)\delta {({\tau }_1,{\tau }_2)}^2\le [2{tan}^2\theta \delta ({\tau }_1,{\tau }_2)+({tan}^2\theta -1)\delta ({\tau }_2,{\tau }_1)]f^{″}({\tau }_1)f^{″}({\tau }_2){({\tau }_2-{\tau }_1)}^4,$$

(9)

or if

$$[{tan}^2\theta \delta ({\tau }_1,{\tau }_2)+({tan}^2\theta -1)\delta ({\tau }_2,{\tau }_1)]f^{″}({\tau }_1)<\frac{2}{{({\tau }_2-{\tau }_1)}^2}\delta {({\tau }_2,{\tau }_1)}^2$$

and

$$\begin{array}{c}8\delta {({\tau }_1,{\tau }_2)}^2\delta {({\tau }_2,{\tau }_1)}^2({tan}^2\theta -1)f^{″}({\tau }_1)\hfill \\ \le \{[({tan}^2\theta -1)f^{″}{({\tau }_1)}^2[2{tan}^2\theta \delta ({\tau }_1,{\tau }_2)+({tan}^2\theta -1)\delta ({\tau }_2,{\tau }_1)]\delta ({\tau }_2,{\tau }_1)]\hfill \\ -{[({tan}^2\theta \delta ({\tau }_1,{\tau }_2)+({tan}^2\theta -1)\delta ({\tau }_2,{\tau }_1))f^{″}({\tau }_1)-\frac{2}{{({\tau }_2-{\tau }_1)}^2}\delta {({\tau }_2,{\tau }_1)}^2]}^2\}\hfill \\ \times f^{″}({\tau }_2){({\tau }_2-{\tau }_1)}^4,\hfill \end{array}$$

(10)

then f is ${\mathcal{L}}_{\theta }$-convex. Here $\delta (\tau ,{\tau }^{\prime }):=f(\tau )-f({\tau }^{\prime })-f^{\prime }({\tau }^{\prime })(\tau -{\tau }^{\prime })$ for $\tau ,{\tau }^{\prime }\in J$.

Proof According to [[9], Theorem 3.2], f is ${\mathcal{L}}_{\theta }$-convex if and only if ${(f^{{\mathcal{L}}_{\theta }})}^{″}(x)(h,h)\in {\mathcal{L}}_{\theta }$ for all $x\in S$ and $h\in {\mathbb{R}}^n$. 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

$${\left(f^{{\mathcal{L}}_{\theta }}\right)}^{″}(x)(h,h)=f^{″}(x_1)\left[\begin{array}{c}{h}_1^2\\ 0\end{array}\right].$$

Hence, ${(f^{{\mathcal{L}}_{\theta }})}^{″}(x)(h,h)\in {\mathcal{L}}_{\theta }$ if and only if $f^{″}(x_1)\ge 0$.

Case 2. For $x_2=0$ and $h_2\ne 0$, it follows from Lemma 2.2 that

$${\left(f^{{\mathcal{L}}_{\theta }}\right)}^{″}(x)(h,h)=\left[\begin{array}{c}{f}^{″}(x_1){\parallel h\parallel }^2\\ 2f^{″}(x_1)h_1{h}_2+f^{″}(x_1)(tan\theta -cot\theta )\parallel h_2\parallel h_2\end{array}\right].$$

Hence, ${(f^{{\mathcal{L}}_{\theta }})}^{″}(x)(h,h)\in {\mathcal{L}}_{\theta }$ if and only if $f^{″}(x_1)\ge 0$ and

$$tan\theta {\parallel h\parallel }^2\ge |2h_1+(tan\theta -cot\theta )\parallel h_2\parallel |\parallel h_2\parallel ,$$

i.e.,

$$-tan\theta (h_1^2+{\parallel h_2\parallel }^2)\le [2h_1+(tan\theta -cot\theta )\parallel h_2\parallel ]\parallel h_2\parallel \le tan\theta (h_1^2+{\parallel h_2\parallel }^2).$$

Dividing by ${\parallel h_2\parallel }^2$ and letting $t=h_1/\parallel h_2\parallel$ yields

$$\begin{array}{c}-tan\theta (t^2+1)\le 2t+tan\theta -cot\theta \le tan\theta (t^2+1)\hfill \\ ⟺\underset{t\in \mathbb{R}}{max}-tan\theta (t^2+1)-2t\le tan\theta -cot\theta \le \underset{t\in \mathbb{R}}{min}tan\theta (t^2+1)-2t\hfill \\ ⟺cot\theta -tan\theta \le tan\theta -cot\theta \le tan\theta -cot\theta \hfill \\ ⟺tan\theta \ge 1.\hfill \end{array}$$

Case 3. For $x_2\ne 0$, due to the simplification of notation, let us denote

$${\mu }_1:=h_1-cot\theta {\overline{x}}_2^T{h}_2,{\mu }_2:=h_1+tan\theta {\overline{x}}_2^T{h}_2,{\mu }_3:=\sqrt{{\parallel h_2\parallel }^2-{\left({\overline{x}}_2^T{h}_2\right)}^2}.$$

(11)

Then

$${\overline{x}}_2^T{h}_2=\frac{{\mu }_2-{\mu }_1}{tan\theta +cot\theta }\text{and}{h}_1=\frac{tan\theta {\mu }_1+cot\theta {\mu }_2}{tan\theta +cot\theta }.$$

(12)

Note that ${\mu }_1$, ${\mu }_2$, and ${\mu }_3$ can take any value in $\mathbb{R}\times \mathbb{R}\times {\mathbb{R}}_{+}$ by taking a suitable value of h (because the vector h has n variables). It follows from Lemma 2.2 that

$$\begin{array}{c}{\left(f^{{\mathcal{L}}_{\theta }}\right)}^{″}(x)(h,h)\hfill \\ =\left[\begin{array}{c}\tilde{\xi }{h}_1^2+2\tilde{\varrho }{\overline{x}}_2^T{h}_2{h}_1+\tilde{a}{\parallel h_2\parallel }^2+(\tilde{\eta }-\tilde{a}){({\overline{x}}_2^T{h}_2)}^2\\ [(\varpi -3\tilde{d}){({\overline{x}}_2^T{h}_2)}^2+2(\tilde{\eta }-\tilde{a}){\overline{x}}_2^T{h}_2{h}_1]{\overline{x}}_2+[\tilde{\varrho }{h}_1^2+\tilde{d}{\parallel h_2\parallel }^2]{\overline{x}}_2+2[\tilde{a}{h}_1+\tilde{d}{\overline{x}}_2^T{h}_2]h_2\end{array}\right]\hfill \\ =:\left[\begin{array}{c}{\mathrm{\Theta }}_1\\ {\mathrm{\Theta }}_2{\overline{x}}_2+{\mathrm{\Theta }}_3{h}_2\end{array}\right],\hfill \end{array}$$

where

$$\begin{array}{c}{\mathrm{\Theta }}_1=\tilde{\xi }{h}_1^2+2\tilde{\varrho }{\overline{x}}_2^T{h}_2{h}_1+\tilde{a}{\parallel h_2\parallel }^2+(\tilde{\eta }-\tilde{a}){\left({\overline{x}}_2^T{h}_2\right)}^2,\hfill \\ {\mathrm{\Theta }}_2=(\varpi -3\tilde{d}){\left({\overline{x}}_2^T{h}_2\right)}^2+2(\tilde{\eta }-\tilde{a}){\overline{x}}_2^T{h}_2{h}_1+\tilde{\varrho }{h}_1^2+\tilde{d}{\parallel h_2\parallel }^2,\hfill \\ {\mathrm{\Theta }}_3=2[\tilde{a}{h}_1+\tilde{d}{\overline{x}}_2^T{h}_2].\hfill \end{array}$$

Hence, ${(f^{{\mathcal{L}}_{\theta }})}^{″}(x)(h,h)\in {\mathcal{L}}_{\theta }$ is equivalent to

$${\mathrm{\Theta }}_1\ge 0\text{and}{\mathrm{\Theta }}_1^2{tan}^2\theta \ge {\parallel {\mathrm{\Theta }}_2{\overline{x}}_2+{\mathrm{\Theta }}_3{h}_2\parallel }^2.$$

Note that

$$\begin{array}{rcl}{\mathrm{\Theta }}_1& =& \frac{1}{1+{cot}^2\theta }{f}^{″}({\lambda }_1(x))[h_1^2-2\left({\overline{x}}_2^T{h}_2\right)h_{1}cot\theta +{\left({\overline{x}}_2^T{h}_2\right)}^2{cot}^2\theta ]\\ +\frac{1}{1+{tan}^2\theta }{f}^{″}({\lambda }_2(x))[h_1^2+2\left({\overline{x}}_2^T{h}_2\right)h_{1}tan\theta +{\left({\overline{x}}_2^T{h}_2\right)}^2{tan}^2\theta ]\\ +\tilde{a}[{\parallel h_2\parallel }^2-{\left({\overline{x}}_2^T{h}_2\right)}^2]\\ =& \frac{1}{1+{cot}^2\theta }{f}^{″}({\lambda }_1(x)){\mu }_1^2+\frac{1}{1+{tan}^2\theta }{f}^{″}({\lambda }_2(x)){\mu }_2^2+\tilde{a}{\mu }_3^2.\end{array}$$

(13)

We now claim that ${\mathrm{\Theta }}_1\ge 0$ for all $h\in {\mathbb{R}}^n$ if and only if

$$f^{″}({\lambda }_1(x))\ge 0,f^{″}({\lambda }_2(x))\ge 0,\text{and}\tilde{a}\ge 0.$$

(14)

The sufficiency is clear. Let us show the necessity. In particular, choosing $h=(-tan\theta ,{\overline{x}}_2)$ yields ${\mu }_2=0$ and ${\mu }_3=0$. It then follows from ${\mathrm{\Theta }}_1\ge 0$ that $f^{″}({\lambda }_1(x))\ge 0$. If we choose $h=(cot\theta ,{\overline{x}}_2)$, then we have $f^{″}({\lambda }_2(x))\ge 0$. Finally, choosing $h=(1,k{z}_2)$ with $k\in \mathbb{R}$, $\parallel z_2\parallel =1$ and $z_2^T{\overline{x}}_2=0$ gives

$${\mathrm{\Theta }}_1=\frac{f^{″}({\lambda }_1(x))}{1+{cot}^2\theta }+\frac{f^{″}({\lambda }_2(x))}{1+{tan}^2\theta }+\tilde{a}{k}^2\ge 0.$$

Dividing by $k^2$ both sides and taking the limits as $k\to \mathrm{\infty }$, we obtain $\tilde{a}\ge 0$. Since ${\lambda }_i(x)$ can take an arbitrary value in J, it is clear that (14) is equivalent to saying that $f^{″}(\tau )\ge 0$ for all $\tau \in J$, i.e., f is convex on J. Indeed, the condition $\tilde{a}\ge 0$ is ensured by the fact that $\tilde{a}=\frac{f^{\prime }({\lambda }_2(x))-f^{\prime }({\lambda }_1(x))}{{\lambda }_2(x)-{\lambda }_1(x)}=f^{″}(t_0)\ge 0$ for some $t_0\in ({\lambda }_1(x),{\lambda }_2(x))$.

Now we calculate the values of ${\mathrm{\Theta }}_2$ and ${\mathrm{\Theta }}_3$, respectively.

$$\begin{array}{rcl}{\mathrm{\Theta }}_2& =& -\frac{cot\theta }{1+{cot}^2\theta }{f}^{″}({\lambda }_1(x)){\mu }_1^2+\frac{tan\theta }{1+{tan}^2\theta }{f}^{″}({\lambda }_2(x)){\mu }_2^2\\ +\tilde{d}{\mu }_3^2-2(\tilde{d}{\overline{x}}_2^T{h}_2+\tilde{a}{h}_1)\left({\overline{x}}_2^T{h}_2\right)\\ =& -\frac{cot\theta }{1+{cot}^2\theta }{f}^{″}({\lambda }_1(x)){\mu }_1^2+\frac{tan\theta }{1+{tan}^2\theta }{f}^{″}({\lambda }_2(x)){\mu }_2^2+\tilde{d}{\mu }_3^2-\left({\overline{x}}_2^T{h}_2\right){\mathrm{\Theta }}_3.\end{array}$$

(15)

Meanwhile, it follows from (12) that

$$\begin{array}{rcl}{\mathrm{\Theta }}_3& =& 2[\tilde{a}\frac{tan\theta {\mu }_1+cot\theta {\mu }_2}{tan\theta +cot\theta }+\tilde{d}\frac{{\mu }_2-{\mu }_1}{tan\theta +cot\theta }]\\ =& \frac{2}{tan\theta +cot\theta }[{\mu }_1(\tilde{a}tan\theta -\tilde{d})+{\mu }_2(\tilde{a}cot\theta +\tilde{d})].\end{array}$$

(16)

Note that

$$\begin{array}{rcl}{\parallel {\mathrm{\Theta }}_2{\overline{x}}_2+{\mathrm{\Theta }}_3{h}_2\parallel }^2& =& {\mathrm{\Theta }}_2^2+2{\mathrm{\Theta }}_2{\mathrm{\Theta }}_3{\overline{x}}_2^T{h}_2+{\mathrm{\Theta }}_3^2{\parallel h_2\parallel }^2\\ =& {\mathrm{\Theta }}_2^2+2{\mathrm{\Theta }}_2{\mathrm{\Theta }}_3{\overline{x}}_2^T{h}_2+{\mathrm{\Theta }}_3^2[{\mu }_3^2+{\left({\overline{x}}_2^T{h}_2\right)}^2]\\ =& {({\mathrm{\Theta }}_2+{\mathrm{\Theta }}_3{\overline{x}}_2^T{h}_2)}^2+{\mathrm{\Theta }}_3^2{\mu }_3^2.\end{array}$$

(17)

Putting (13) and (15)-(17) together, the condition ${\mathrm{\Theta }}_1^2{tan}^2\theta \ge {\parallel {\mathrm{\Theta }}_2{\overline{x}}_2+{\mathrm{\Theta }}_3{h}_2\parallel }^2$ can be rewritten equivalently as

$$\begin{array}{c}{tan}^2\theta {[\frac{f^{″}({\lambda }_1(x))}{1+{cot}^2\theta }{\mu }_1^2+\frac{f^{″}({\lambda }_2(x))}{1+{tan}^2\theta }{\mu }_2^2+\tilde{a}{\mu }_3^2]}^2\hfill \\ \ge {[-\frac{cot\theta }{1+{cot}^2\theta }{f}^{″}({\lambda }_1(x)){\mu }_1^2+\frac{tan\theta }{1+{tan}^2\theta }{f}^{″}({\lambda }_2(x)){\mu }_2^2+\tilde{d}{\mu }_3^2]}^2\hfill \\ +\frac{4}{{(tan\theta +cot\theta )}^2}{[{\mu }_1(\tilde{a}tan\theta -\tilde{d})+{\mu }_2(\tilde{a}cot\theta +\tilde{d})]}^2{\mu }_3^2,\hfill \end{array}$$

i.e.,

$$\begin{array}{c}({tan}^4\theta -1)f^{″}{({\lambda }_1(x))}^2{\mu }_1^4+{(tan\theta +cot\theta )}^2({\tilde{a}}^2{tan}^2\theta -{\tilde{d}}^2){\mu }_3^4\hfill \\ +2[(tan\theta +cot\theta )(\tilde{a}{tan}^3\theta +\tilde{d})f^{″}({\lambda }_1(x))-2{(\tilde{a}tan\theta -\tilde{d})}^2]{\mu }_1^2{\mu }_3^2\hfill \\ +2[(tan\theta +cot\theta )(\tilde{a}tan\theta -\tilde{d})f^{″}({\lambda }_2(x))-2{(\tilde{a}cot\theta +\tilde{d})}^2]{\mu }_2^2{\mu }_3^2\hfill \\ +2({tan}^2\theta +1)f^{″}({\lambda }_1(x))f^{″}({\lambda }_2(x)){\mu }_1^2{\mu }_2^2\hfill \\ -8(\tilde{a}tan\theta -\tilde{d})(\tilde{a}cot\theta +\tilde{d}){\mu }_1{\mu }_2{\mu }_3^2\ge 0.\hfill \end{array}$$

(18)

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

$$\begin{array}{c}\tilde{a}tan\theta -\tilde{d}\hfill \\ =\frac{f^{\prime }({\lambda }_2(x))-f^{\prime }({\lambda }_1(x))}{{\lambda }_2(x)-{\lambda }_1(x)}tan\theta -\frac{1}{\parallel x_2\parallel }[\frac{{cot}^2\theta }{1+{cot}^2\theta }{f}^{\prime }({\lambda }_1(x))+\frac{{tan}^2\theta }{1+{tan}^2\theta }{f}^{\prime }({\lambda }_2(x))]\hfill \\ +\frac{1}{\parallel x_2\parallel }\frac{f({\lambda }_2(x))-f({\lambda }_1(x))}{{\lambda }_2(x)-{\lambda }_1(x)}\hfill \\ =\frac{f^{\prime }({\lambda }_2(x))-f^{\prime }({\lambda }_1(x))}{{\lambda }_2(x)-{\lambda }_1(x)}tan\theta -\frac{1}{\parallel x_2\parallel }[\frac{cot\theta }{tan\theta +cot\theta }{f}^{\prime }({\lambda }_1(x))+\frac{tan\theta }{tan\theta +ctan\theta }{f}^{\prime }({\lambda }_2(x))]\hfill \\ +\frac{1}{\parallel x_2\parallel }\frac{f({\lambda }_2(x))-f({\lambda }_1(x))}{{\lambda }_2(x)-{\lambda }_1(x)}\hfill \\ =-\frac{tan\theta +ctan\theta }{{\lambda }_2(x)-{\lambda }_1(x)}{f}^{\prime }({\lambda }_1(x))+\frac{tan\theta +ctan\theta }{{[{\lambda }_2(x)-{\lambda }_1(x)]}^2}[f({\lambda }_2(x))-f({\lambda }_1(x))]\hfill \\ =\frac{(tan\theta +cot\theta )[f({\lambda }_2(x))-f({\lambda }_1(x))-f^{\prime }({\lambda }_1(x))({\lambda }_2(x)-{\lambda }_1(x))]}{{[{\lambda }_2(x)-{\lambda }_1(x)]}^2}\hfill \\ =\frac{tan\theta +cot\theta }{{[{\lambda }_2(x)-{\lambda }_1(x)]}^2}\delta ({\lambda }_2(x),{\lambda }_1(x)),\hfill \end{array}$$

where the third equation follows from the fact ${\lambda }_2(x)-{\lambda }_1(x)=(tan\theta +ctan\theta )\parallel x_2\parallel$. Similarly, we have

$$\begin{array}{c}\tilde{a}tan\theta +\tilde{d}\hfill \\ =\frac{(tan\theta +cot\theta )[f({\lambda }_1(x))-f({\lambda }_2(x))+(\frac{2tan\theta }{tan\theta +cot\theta }{f}^{\prime }({\lambda }_2(x))+\frac{cot\theta -tan\theta }{tan\theta +cot\theta }{f}^{\prime }({\lambda }_1(x)))({\lambda }_2(x)-{\lambda }_1(x))]}{{[{\lambda }_2(x)-{\lambda }_1(x)]}^2}\hfill \\ =\frac{tan\theta +cot\theta }{{[{\lambda }_2(x)-{\lambda }_1(x)]}^2}[\frac{2{tan}^2\theta }{{tan}^2+1}\delta ({\lambda }_1(x),{\lambda }_2(x))+\frac{{tan}^2\theta -1}{{tan}^2\theta +1}\delta ({\lambda }_2(x),{\lambda }_1(x))],\hfill \\ \tilde{a}cot\theta +\tilde{d}=\frac{(tan\theta +cot\theta )[f({\lambda }_1(x))-f({\lambda }_2(x))-f^{\prime }({\lambda }_2(x))({\lambda }_1(x)-{\lambda }_2(x))]}{{[{\lambda }_2(x)-{\lambda }_1(x)]}^2}\hfill \\ \phantom{\tilde{a}cot\theta +\tilde{d}}=\frac{tan\theta +cot\theta }{{[{\lambda }_2(x)-{\lambda }_1(x)]}^2}\delta ({\lambda }_1(x),{\lambda }_2(x)),\hfill \\ \tilde{a}{tan}^3\theta +\tilde{d}\hfill \\ =\frac{(tan\theta +cot\theta )[f({\lambda }_1(x))-f({\lambda }_2(x))-[{tan}^2\theta f^{\prime }({\lambda }_2(x))+(1-{tan}^2\theta )f^{\prime }({\lambda }_1(x))]({\lambda }_1(x)-{\lambda }_2(x))]}{{[{\lambda }_2(x)-{\lambda }_1(x)]}^2}\hfill \\ =\frac{tan\theta +cot\theta }{{[{\lambda }_2(x)-{\lambda }_1(x)]}^2}[{tan}^2\theta \delta ({\lambda }_1(x),{\lambda }_2(x))+({tan}^2\theta -1)\delta ({\lambda }_2(x),{\lambda }_1(x))].\hfill \end{array}$$

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

$$\{\begin{array}{c}{\alpha }_1=({tan}^4\theta -1)f^{″}{({\lambda }_1(x))}^2,\hfill \\ {\alpha }_2=\frac{{(tan\theta +cot\theta )}^4}{{[{\lambda }_2(x)-{\lambda }_1(x)]}^4}\delta ({\lambda }_2(x),{\lambda }_1(x))[\frac{2{tan}^2\theta }{{tan}^2\theta +1}\delta ({\lambda }_1(x),{\lambda }_2(x))+\frac{{tan}^2\theta -1}{{tan}^2\theta +1}\delta ({\lambda }_2(x),{\lambda }_1(x))],\hfill \\ {\alpha }_3=2\frac{{(tan\theta +cot\theta )}^2}{{[{\lambda }_2(x)-{\lambda }_1(x)]}^2}\{[{tan}^2\theta \delta ({\lambda }_1(x),{\lambda }_2(x))+({tan}^2\theta -1)\delta ({\lambda }_2(x),{\lambda }_1(x))]f^{″}({\lambda }_1(x)),\hfill \\ \phantom{{\alpha }_3=}-2\frac{\delta {({\lambda }_2(x),{\lambda }_1(x))}^2}{{[{\lambda }_2(x)-{\lambda }_1(x)]}^2}\},\hfill \\ {\alpha }_4=2\frac{{(tan\theta +cot\theta )}^2}{{[{\lambda }_2(x)-{\lambda }_1(x)]}^2}[\delta ({\lambda }_2(x),{\lambda }_1(x))f^{″}({\lambda }_2(x))-2\frac{\delta {({\lambda }_1(x),{\lambda }_2(x))}^2}{{[{\lambda }_2(x)-{\lambda }_1(x)]}^2}],\hfill \\ {\alpha }_5=2({tan}^2\theta +1)f^{″}({\lambda }_1(x))f^{″}({\lambda }_2(x)),\hfill \\ {\alpha }_6=-8\frac{{(tan\theta +cot\theta )}^2}{{[{\lambda }_2(x)-{\lambda }_1(x)]}^4}\delta ({\lambda }_1(x),{\lambda }_2(x))\delta ({\lambda }_2(x),{\lambda }_1(x)).\hfill \end{array}$$

In view of Lemma 2.1, the condition ${\alpha }_1\ge 0$ means $tan\theta \ge 1$, ${\alpha }_2,{\alpha }_5\ge 0$ is ensured by the convexity of f (see (14)), ${\alpha }_4\ge 0$ corresponds to (6), and ${\alpha }_3\ge -2\sqrt{{\alpha }_1{\alpha }_2}$ corresponds to (7). In addition, condition (4) takes the special form (9) and (10), respectively. □

Theorem 2.2 Suppose that $f:J\to \mathbb{R}$ is second-order continuously differentiable. Then f is ${\mathcal{L}}_{\theta }$-convex of order 2 on S if and only if $tan\theta \ge 1$ and f is convex on J.

Proof The necessity is clear from Theorem 2.1. For sufficiency, note that in (11) ${\mu }_3=0$ since ${\overline{x}}_2=\pm 1$ in this case. Hence, (18) takes the form of

$$({tan}^4\theta -1)f^{″}{({\lambda }_1(x))}^2{\mu }_1^4+2({tan}^2\theta +1)f^{″}({\lambda }_1(x))f^{″}({\lambda }_2(x)){\mu }_1^2{\mu }_2^2\ge 0$$

for all ${\mu }_1$ and ${\mu }_2$, which is equivalent to verifying

$$tan\theta \ge 1\text{and}{f}^{″}({\lambda }_1(x))f^{″}({\lambda }_2(x))\ge 0.$$

This is ensued by the conditions that $tan\theta \ge 1$ and f is convex on J. Thus, the proof is complete. □

If, in particular, $\theta ={45}^{\circ }$, then (6) and (7) reduce to [[12], (21) in Proposition 4.2]; (9) reduces to [[12], (22) in Proposition 4.2]. In addition, due to (7), (8) holds automatically in this case. The above results indicate that the ${\mathcal{L}}_{\theta }$-convexity is dependent on 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 will be useful for further analyzing the properties of $f^{{\mathcal{L}}_{\theta }}$ and proving the convergence of interior point methods for optimization problems involved in circular cones.

In [18], the author establishes the following results in the framework of second-order cone. More specifically, for $x{⪰}_{{\mathcal{L}}_{{45}^{\circ }}}0$ and $y{⪰}_{{\mathcal{L}}_{{45}^{\circ }}}0$, then

• (a) $det{(e+x)}^{1/2}\ge 1+det{(x)}^{1/2}$,

• (b) $det(x+y)\ge det(x)+det(y)$,

• (c) $det(\alpha x+(1-\alpha )y)\ge {\alpha }^{2}det(x)+{(1-\alpha )}^{2}det(y)$, $\mathrm{\forall }\alpha \in [0,1]$,

• (d) $det(e+x+y)\le det(e+x)det(e+y)$,

• (e) If $x{⪰}_{{\mathcal{L}}_{{45}^{\circ }}}y{⪰}_{{\mathcal{L}}_{{45}^{\circ }}}0$, then $det(x)\ge det(y)$, $tr(x)\ge tr(y)$, and ${\lambda }_i(x)\ge {\lambda }_i(y)$ for $i=1,2$,

• (f) $tr(x+y)=tr(x)+tr(y)$ and $det(\gamma x)={\gamma }^{2}det(x)$ for all $\gamma \in \mathbb{R}$.

In the following, we show that, in the framework of circular cone, the above inequalities can be classified into three categories. The first class holds independent of the angle, e.g., (a); the second class holds dependent on 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 $x=(x_1,x_2)\in \mathbb{R}\times {\mathbb{R}}^{n-1}$ possess spectral factorization associated with circular cone given as in (1). Then

• (a) ${[det(e+x)]}^{1/2}\ge 1+det{(x)}^{1/2}$ for all $x\in {\mathcal{L}}_{\theta }$;

• (b) If $x{⪰}_{{\mathcal{L}}_{\theta }}y$, then ${\lambda }_1(x)\ge {\lambda }_1(y)$.

Proof (a) Note that $det(x)\ge 0$ and $det(e+x)\ge 0$ since $x,x+e\in {\mathcal{L}}_{\theta }$. Therefore,

$$\begin{array}{c}{[det(e+x)]}^{1/2}\ge 1+det{(x)}^{1/2}\hfill \\ ⟺det(e+x)\ge 1+2det{(x)}^{1/2}+det(x)\hfill \\ ⟺{\lambda }_1(e+x){\lambda }_2(e+x)\ge 1+2\sqrt{{\lambda }_1(x){\lambda }_2(x)}+{\lambda }_1(x){\lambda }_2(x)\hfill \\ ⟺(x_1+1-\parallel x_2\parallel cot\theta )(x_1+1+\parallel x_2\parallel tan\theta )\ge 1+2\sqrt{{\lambda }_1(x){\lambda }_2(x)}+{\lambda }_1(x){\lambda }_2(x)\hfill \\ ⟺({\lambda }_1(x)+1)({\lambda }_2(x)+1)\ge 1+2\sqrt{{\lambda }_1(x){\lambda }_2(x)}+{\lambda }_1(x){\lambda }_2(x)\hfill \\ ⟺{\lambda }_1(x){\lambda }_2(x)+{\lambda }_1(x)+{\lambda }_2(x)+1\ge 1+2\sqrt{{\lambda }_1(x){\lambda }_2(x)}+{\lambda }_1(x){\lambda }_2(x)\hfill \\ ⟺{\lambda }_1(x)+{\lambda }_2(x)\ge 2\sqrt{{\lambda }_1(x){\lambda }_2(x)}\hfill \\ ⟺\frac{{\lambda }_1(x)+{\lambda }_2(x)}{2}\ge \sqrt{{\lambda }_1(x){\lambda }_2(x)}.\hfill \end{array}$$

Hence, to prove the desired result, it suffices to show that

$$\frac{{\lambda }_1(x)+{\lambda }_2(x)}{2}\ge \sqrt{{\lambda }_1(x){\lambda }_2(x)},$$

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

• (b) Since $x-y\in {\mathcal{L}}_{\theta }$, we know

  $$x_1-y_1\ge \parallel x_2-y_2\parallel cot\theta \ge [\parallel x_2\parallel -\parallel y_2\parallel ]cot\theta ,$$

i.e., ${\lambda }_1(x)=x_1-\parallel x_2\parallel cot\theta \ge y_1-\parallel y_2\parallel cot\theta ={\lambda }_1(y)$. □

Theorem 3.2 Let $x=(x_1,x_2)\in \mathbb{R}\times {\mathbb{R}}^{n-1}$ possess spectral factorization associated with circular cone given as in (1). Then the following hold.

• (a) For all $x,y\in {\mathcal{L}}_{\theta }$,

  $$det(x+y)\ge det(x)+det(y)+({\parallel x_2\parallel }^2+{\parallel y_2\parallel }^2){csc}^2\theta -(x_1^2+y_1^2){sec}^2\theta .$$

In particular, when $\theta \in (0,{45}^{\circ }]$, we have

$$det(x+y)\ge det(x)+det(y).$$

(19)

• (b) For all $x,y\in {\mathcal{L}}_{\theta }$ and $\alpha \in [0,1]$,

  $$\begin{array}{c}det(\alpha x+(1-\alpha )y)\hfill \\ \ge {\alpha }^{2}det(x)+{(1-\alpha )}^{2}det(y)+({\alpha }^2{\parallel x_2\parallel }^2+{(1-\alpha )}^2{\parallel y_2\parallel }^2){csc}^2\theta \hfill \\ -({\alpha }^2{x}_1^2+{(1-\alpha )}^2{y}_1^2){sec}^2\theta .\hfill \end{array}$$

In particular, when $\theta \in (0,{45}^{\circ }]$, we have

$$det(\alpha x+(1-\alpha )y)\ge {\alpha }^{2}det(x)+{(1-\alpha )}^{2}det(y).$$

• (c) If $x,y\in {\mathcal{L}}_{\theta }$ and $\theta \in [{45}^{\circ },{90}^{\circ })$, then

  $$det(e+x+y)\le det(e+x)det(e+y).$$

  (20)

• (d) If $x{⪰}_{{\mathcal{L}}_{\theta }}y{⪰}_{{\mathcal{L}}_{\theta }}0$ and $\theta \in (0,{45}^{\circ }]$, then

  $${\lambda }_2(x)\ge {\lambda }_2(y),det(x)\ge det(y),\mathit{\text{and}}tr(x)\ge tr(y).$$

  (21)

Proof (a) Notice that

$$\begin{array}{c}det(x+y)\hfill \\ ={\lambda }_1(x+y)\cdot {\lambda }_2(x+y)\hfill \\ =(x_1+y_1-\parallel x_2+y_2\parallel cot\theta )(x_1+y_1+\parallel x_2+y_2\parallel tan\theta )\hfill \\ ={(x_1+y_1)}^2+(x_1+y_1)\parallel x_2+y_2\parallel tan\theta -(x_1+y_1)\parallel x_2+y_2\parallel cot\theta -{\parallel x_2+y_2\parallel }^2\hfill \end{array}$$

and

$$\begin{array}{c}det(x)+det(y)\hfill \\ ={\lambda }_1(x){\lambda }_2(x)+{\lambda }_1(y){\lambda }_2(y)\hfill \\ =(x_1-\parallel x_2\parallel cot\theta )(x_1+\parallel x_2\parallel tan\theta )+(y_1-\parallel y_2\parallel cot\theta )(y_1+\parallel y_2\parallel tan\theta )\hfill \\ =x_1^2+x_1\parallel x_2\parallel tan\theta -x_1\parallel x_2\parallel cot\theta -{\parallel x_2\parallel }^2+y_1^2+y_1\parallel y_2\parallel tan\theta -y_1\parallel y_2\parallel cot\theta -{\parallel y_2\parallel }^2\hfill \\ =x_1^2+y_1^2+x_1\parallel x_2\parallel tan\theta +y_1\parallel y_2\parallel tan\theta -x_1\parallel x_2\parallel cot\theta -y_1\parallel y_2\parallel cot\theta -{\parallel x_2\parallel }^2-{\parallel y_2\parallel }^2.\hfill \end{array}$$

Then we have

$$\begin{array}{c}det(x+y)-det(x)-det(y)\hfill \\ =2x_1{y}_1-2x_2^T{y}_2+(x_1\parallel x_2+y_2\parallel +y_1\parallel x_2+y_2\parallel -x_1\parallel x_2\parallel -y_1\parallel y_2\parallel )tan\theta \hfill \\ -(x_1\parallel x_2+y_2\parallel +y_1\parallel x_2+y_2\parallel -x_1\parallel x_2\parallel -y_1\parallel y_2\parallel )cot\theta .\hfill \end{array}$$

Using $x,y\in {\mathcal{L}}_{\theta }$ (and hence $x+y\in {\mathcal{L}}_{\theta }$) gives

$$\begin{array}{c}{x}_{1}tan\theta \ge \parallel x_2\parallel ,-x_{1}tan\theta \le -\parallel x_2\parallel ,x_1\ge \parallel x_2\parallel cot\theta ,-x_1\le -\parallel x_2\parallel cot\theta ,\hfill \\ -(x_1+y_1)\le -\parallel x_2+y_2\parallel cot\theta .\hfill \end{array}$$

Thus,

$$\begin{array}{c}det(x+y)-det(x)-det(y)\hfill \\ \ge 2x_1{y}_1-2x_2^T{y}_2+\parallel x_2\parallel \parallel x_2+y_2\parallel +\parallel y_2\parallel \parallel x_2+y_2\parallel -x_1^2{tan}^2\theta -y_1^2{tan}^2\theta \hfill \\ -x_1(x_1+y_1)-y_1(x_1+y_1)+{\parallel x_2\parallel }^2{cot}^2\theta +{\parallel y_2\parallel }^2{cot}^2\theta \hfill \\ =2x_1{y}_1-2x_2^T{y}_2+\parallel x_2+y_2\parallel (\parallel x_2\parallel +\parallel y_2\parallel )-(x_1^2+y_1^2){tan}^2\theta \hfill \\ -{(x_1+y_1)}^2+({\parallel x_2\parallel }^2+{\parallel y_2\parallel }^2){cot}^2\theta \hfill \\ \ge {\parallel x_2+y_2\parallel }^2-(x_1^2+y_1^2){tan}^2\theta -x_1^2-y_1^2-2x_2^T{y}_2+({\parallel x_2\parallel }^2+{\parallel y_2\parallel }^2){cot}^2\theta \hfill \\ ={\parallel x_2\parallel }^2+{\parallel y_2\parallel }^2-(x_1^2+y_1^2){tan}^2\theta -x_1^2-y_1^2+({\parallel x_2\parallel }^2+{\parallel y_2\parallel }^2){cot}^2\theta \hfill \\ =({\parallel x_2\parallel }^2+{\parallel y_2\parallel }^2)(1+{cot}^2\theta )-(x_1^2+y_1^2)(1+{tan}^2\theta )\hfill \\ =({\parallel x_2\parallel }^2+{\parallel y_2\parallel }^2){csc}^2\theta -(x_1^2+y_1^2){sec}^2\theta ,\hfill \end{array}$$

which is the desired result.

When $\theta \in (0,{45}^{\circ }]$, we know $tan\theta \le cot\theta$. Since $x,y\in {\mathcal{L}}_{\theta }$, i.e., $x_1\ge \parallel x_2\parallel cot\theta$ and $y_1\ge \parallel y_2\parallel cot\theta$, there exist $a,b\ge 0$ such that $x_1=\parallel x_2\parallel cot\theta +a$ and $y_1=\parallel y_2\parallel cot\theta +b$. Hence,

$$\begin{array}{c}det(x+y)-det(x)-det(y)\hfill \\ =2x_1{y}_1-2x_2^T{y}_2+(x_1\parallel x_2+y_2\parallel +y_1\parallel x_2+y_2\parallel -x_1\parallel x_2\parallel -y_1\parallel y_2\parallel )tan\theta \hfill \\ -(x_1\parallel x_2+y_2\parallel +y_1\parallel x_2+y_2\parallel -x_1\parallel x_2\parallel -y_1\parallel y_2\parallel )cot\theta \hfill \\ =(\parallel x_2\parallel +\parallel y_2\parallel )[\parallel x_2\parallel +\parallel y_2\parallel -\parallel x_2+y_2\parallel ]{cot}^2\theta \hfill \\ +\parallel x_2+y_2\parallel (\parallel x_2\parallel +\parallel y_2\parallel -\parallel x_2+y_2\parallel )+2ab\hfill \\ +acot\theta (\parallel y_2\parallel +\parallel x_2\parallel -\parallel x_2+y_2\parallel )+atan\theta (\parallel y_2\parallel {cot}^2\theta +\parallel x_2+y_2\parallel -\parallel x_2\parallel )\hfill \\ +bcot\theta (\parallel y_2\parallel +\parallel x_2\parallel -\parallel x_2+y_2\parallel )+btan\theta (\parallel x_2\parallel {cot}^2\theta +\parallel x_2+y_2\parallel -\parallel y_2\parallel )\hfill \\ \ge 0,\hfill \end{array}$$

where the last step is due to $\parallel x_2\parallel +\parallel y_2\parallel \ge \parallel x_2+y_2\parallel$, $\parallel x_2\parallel {cot}^2\theta +\parallel x_2+y_2\parallel -\parallel y_2\parallel \ge \parallel x_2\parallel +\parallel x_2+y_2\parallel -\parallel y_2\parallel \ge 0$, and $\parallel y_2\parallel {cot}^2\theta +\parallel x_2+y_2\parallel -\parallel x_2\parallel \ge \parallel y_2\parallel +\parallel x_2+y_2\parallel -\parallel x_2\parallel \ge 0$ since $cot\theta \ge 1$, due to $\theta \in (0,{45}^{\circ }]$.

• (b) The result follows from the fact that $det(\gamma x)={\gamma }^{2}det(x)$ for all $\gamma \ge 0$.

• (c) Since $\theta \in [{45}^{\circ },{90}^{\circ })$, $cot\theta \le 1$. For $x,y\in {\mathcal{L}}_{\theta }$, there exist two nonnegative scalars $a,b\ge 0$ such that $x_1=\parallel x_2\parallel cot\theta +a$ and $y_1=\parallel y_2\parallel cot\theta +b$. This implies

  $$\begin{array}{c}det(e+x)=(x_1+1-\parallel x_2\parallel cot\theta )(x_1+1+\parallel x_2\parallel tan\theta )\hfill \\ \phantom{det(e+x)}=(a+1)(cot\theta +tan\theta )\parallel x_2\parallel +{(a+1)}^2,\hfill \\ det(e+y)=(y_1+1-\parallel y_2\parallel cot\theta )(y_1+1+\parallel y_2\parallel tan\theta )\hfill \\ \phantom{det(e+y)}=(b+1)(cot\theta +tan\theta )\parallel y_2\parallel +{(b+1)}^2.\hfill \end{array}$$

Thus, we obtain

$$\begin{array}{c}det(e+x)det(e+y)\hfill \\ =(a+1)(b+1){(cot\theta +tan\theta )}^2\parallel x_2\parallel \parallel y_2\parallel +(a+1){(b+1)}^2(cot\theta +tan\theta )\parallel x_2\parallel \hfill \\ +{(a+1)}^2(b+1)(cot\theta +tan\theta )\parallel y_2\parallel +{(a+1)}^2{(b+1)}^2.\hfill \end{array}$$

(22)

On the other hand,

$$\begin{array}{c}det(e+x+y)\hfill \\ =(x_1+y_1+1-\parallel x_2+y_2\parallel cot\theta )(x_1+y_1+1+\parallel x_2+y_2\parallel tan\theta )\hfill \\ =([\parallel x_2\parallel +\parallel y_2\parallel -\parallel x_2+y_2\parallel ]cot\theta +(a+b+1))\hfill \\ \times ([\parallel x_2\parallel +\parallel y_2\parallel ]cot\theta +\parallel x_2+y_2\parallel tan\theta +(a+b+1))\hfill \\ =(\parallel x_2\parallel +\parallel y_2\parallel -\parallel x_2+y_2\parallel )(\parallel x_2\parallel +\parallel y_2\parallel ){cot}^2\theta \hfill \\ +(\parallel x_2\parallel +\parallel y_2\parallel -\parallel x_2+y_2\parallel )\parallel x_2+y_2\parallel +(a+b+1)(\parallel x_2\parallel +\parallel y_2\parallel -\parallel x_2+y_2\parallel )cot\theta \hfill \\ +(a+b+1)(\parallel x_2\parallel +\parallel y_2\parallel )cot\theta +(a+b+1)\parallel x_2+y_2\parallel tan\theta +{(a+b+1)}^2\hfill \\ =2{cot}^2\theta \parallel x_2\parallel \parallel y_2\parallel +2(a+b+1)cot\theta \parallel x_2\parallel +2(a+b+1)cot\theta \parallel y_2\parallel +{(a+b+1)}^2\hfill \\ +[{\parallel x_2\parallel }^2+{\parallel y_2\parallel }^2]{cot}^2\theta +(1-{cot}^2\theta )\parallel x_2+y_2\parallel (\parallel x_2\parallel +\parallel y_2\parallel )\hfill \\ +(a+b+1)(tan\theta -cot\theta )\parallel x_2+y_2\parallel -{\parallel x_2+y_2\parallel }^2\hfill \\ \le 2{cot}^2\theta \parallel x_2\parallel \parallel y_2\parallel +2(a+b+1)cot\theta \parallel x_2\parallel +2(a+b+1)cot\theta \parallel y_2\parallel +{(a+b+1)}^2\hfill \\ +[{\parallel x_2\parallel }^2+{\parallel y_2\parallel }^2]{cot}^2\theta +(1-{cot}^2\theta ){(\parallel x_2\parallel +\parallel y_2\parallel )}^2\hfill \\ +(a+b+1)(tan\theta -cot\theta )(\parallel x_2\parallel +\parallel y_2\parallel )-({\parallel x_2\parallel }^2+{\parallel y_2\parallel }^2)+2\parallel x_2\parallel \parallel y_2\parallel \hfill \\ =2{cot}^2\theta \parallel x_2\parallel \parallel y_2\parallel +(a+b+1)(cot\theta +tan\theta )\parallel x_2\parallel +(a+b+1)(cot\theta +tan\theta )\parallel y_2\parallel \hfill \\ +{(a+b+1)}^2+(1-{cot}^2\theta )[{(\parallel x_2\parallel +\parallel y_2\parallel )}^2-({\parallel x_2\parallel }^2+{\parallel y_2\parallel }^2)]+2\parallel x_2\parallel \parallel y_2\parallel \hfill \\ =4\parallel x_2\parallel \parallel y_2\parallel +(a+b+1)(cot\theta +tan\theta )\parallel x_2\parallel +(a+b+1)(ctan\theta +tan\theta )\parallel y_2\parallel \hfill \\ +{(a+b+1)}^2.\hfill \end{array}$$

(23)

Note that $(a+1)(b+1){(cot\theta +tan\theta )}^2\ge {(cot\theta +tan\theta )}^2\ge 4$ and

$$\begin{array}{c}(a+1){(b+1)}^2\ge a+b+1,\hfill \\ {(a+1)}^2(b+1)\ge a+b+1,\hfill \\ {(a+1)}^2{(b+1)}^2\ge {(a+b+1)}^2.\hfill \end{array}$$

Hence, comparing (22) and (23) yields

$$det(e+x+y)\le det(e+x)det(e+y).$$

• (d) For $\theta \in (0,{45}^{\circ }]$, since $cot\theta \ge tan\theta$ and $x-y\in {\mathcal{L}}_{\theta }$, we know

  $$x_1-y_1\ge \parallel x_2-y_2\parallel cot\theta \ge \parallel x_2-y_2\parallel tan\theta \ge [\parallel y_2\parallel -\parallel x_2\parallel ]tan\theta ,$$

which means

$${\lambda }_2(x)=x_1+\parallel x_2\parallel tan\theta \ge y_1+\parallel y_2\parallel tan\theta ={\lambda }_2(y).$$

This together with the fact ${\lambda }_1(x)\ge {\lambda }_1(y)$ by Part (b) in Theorem 3.1 and ${\lambda }_i(x),{\lambda }_i(y)\ge 0$ for $i=1,2$ (due to $x,y\in {\mathcal{L}}_{\theta }$) further yields

$$det(x)={\lambda }_1(x){\lambda }_2(x)\ge {\lambda }_1(y){\lambda }_2(y)=det(y).$$

Meanwhile, we obtain

$$tr(x)={\lambda }_1(x)+{\lambda }_2(x)\ge {\lambda }_1(y)+{\lambda }_2(y)=tr(y).$$

 □

Here are some remarks for Theorem 3.2.

• (i) Inequality (19) fails when $\theta \in ({45}^{\circ },{90}^{\circ })$. For example, let $x=(1,3,4)$, $y=(1,-3,-4)$, and $cot\theta =0.1$. Then $det(x)=det(y)=51/2$ and $det(x+y)=4$, which says $det(x+y)=4<51=det(x)+det(y)$.

• (ii) Inequality (20) fails when $\theta \in (0,{45}^{\circ })$. For example, let $x=(3/10,1/10)$, $y=(3/10,-1/10)$, and $cot\theta =2$. Then $det(e+x+y)={(1.6)}^2>{(1.485)}^2=det(e+x)det(e+y)$.

• (iii) Inequality (21) fails when $\theta \in ({45}^{\circ },{90}^{\circ })$. For example, for $x=(1.1,1)$, $y=(1,2)$, and $cot\theta =0.1$. Then $x{⪰}_{{\mathcal{L}}_{\theta }}y$, ${\lambda }_2(x)=11.1<21={\lambda }_2(y)$, $det(x)=(1.1-0.1)(1.1+10)=11.1<16.8=(1-0.2)(1+20)=det(y)$, and $tr(x)=12.1<21.8=tr(y)$.

Next, let us move from inequalities to equalities. In particular, we focus on two identities in the framework of second-order cone as below

$$tr(x+y)=tr(x)+tr(y)\text{and}det(\gamma x)={\gamma }^{2}det(x),\mathrm{\forall }\gamma \in \mathbb{R}.$$

(24)

But these two identities fail to hold in the circular cone setting no matter what value of the angle is chosen. In fact, in the second-order cone case,

$$tr(x)=2x_1\text{and}det(x)=x_1^2-{\parallel x_2\parallel }^2.$$

Hence, (24) holds trivially. For the circular cone setting, we have

$$tr(x)=2x_1+\parallel x_2\parallel (tan\theta -cot\theta )\text{and}det(x)=(x_1-\parallel x_2\parallel cot\theta )(x_1+\parallel x_2\parallel tan\theta ).$$

Thus, $tr(x)$ is not linear any more, i.e., $tr(x+y)\ne tr(x)+tr(y)$; e.g., for $x=(1,2)$ and $y=(1,-2)$, and $cot\theta =1/2$ (or $cot\theta =2$). Then

$$tr(x+y)=4\ne 10=tr(x)+tr(y)(\text{or }tr(x+y)=4\ne -2=tr(x)+tr(y)).$$

In addition, $det(\gamma x)={\gamma }^{2}det(x)$ holds as $\gamma \ge 0$ but not true as $\gamma <0$; e.g., for $x=(3,4)$, $\gamma =-2$, and $cot\theta =1/2$ (or $cot\theta =2$), then

$$det(-2x)=-100\ne 44={(-2)}^{2}det(x)(\text{or }det(-2x)=44\ne -100={(-2)}^{2}det(x)).$$

The precise relationship between $tr(x+y)$ and $tr(x)+tr(y)$ is provided as below.

Theorem 3.3

$$tr(x+y)\{\begin{array}{cc}\ge tr(x)+tr(y)\hfill & \mathit{\text{as }}\theta \in (0,{45}^{\circ }],\hfill \\ \le tr(x)+tr(y)\hfill & \mathit{\text{as }}\theta \in [{45}^{\circ },{90}^{\circ }).\hfill \end{array}$$

Proof The result follows from the fact that

$$tr(x+y)-tr(x)-tr(y)=[\parallel x_2\parallel +\parallel y_2\parallel -\parallel x_2+y_2\parallel ](cot\theta -tan\theta ).$$

 □

Note that $tr(x)$ is positively homogeneous, i.e., $tr(\gamma x)=\gamma tr(x)$ for all $\gamma \ge 0$. This together with Theorem 3.3 yields the following result.

Corollary 3.1 The trace $tr(x)$ is concave as $\theta \in (0,{45}^{\circ }]$ and is convex as $\theta \in [{45}^{\circ },{90}^{\circ })$.

These results further indicate that the angle plays an essential role for a circular cone. As in symmetric cone optimization, we believe that these inequalities about $det(x)$ and $tr(x)$ are key ingredients in penalty and barrier functions which can be adapted in designing barrier and penalty algorithms (including interior point algorithm) for circular cone optimization. This merits our further research.

Authors’ information

The second author is a member of Mathematics Division, National Center for Theoretical Sciences, Taipei Office.