The decompositions with respect to two core non-symmetric cones

Abstract

It is known that the analysis to tackle with non-symmetric cone optimization is quite different from the way to deal with symmetric cone optimization due to the discrepancy between these types of cones. However, there are still common concepts for both optimization problems, for example, the decomposition with respect to the given cone, smooth and nonsmooth analysis for the associated conic function, conic-convexity, conic-monotonicity and etc. In this paper, motivated by Chares’s thesis (Cones and interior-point algorithms for structured convex optimization involving powers and exponentials, 2009), we consider the decomposition issue of two core non-symmetric cones, in which two types of decomposition formulae will be proposed, one is adapted from the well-known Moreau decomposition theorem and the other follows from geometry properties of the given cones. As a byproduct, we also establish the conic functions of these cones and generalize the power cone case to its high-dimensional counterpart.

1 Introduction

Consider the following two core non-symmetric cones

$$\begin{aligned} {\mathcal {K}}_\alpha:= & {} \left\{ (x_1,{\bar{x}})\in {\mathbb {R}} \times {\mathbb {R}}^2\,\bigg |\, |x_1|\le {\bar{x}}^{\alpha _1}_1{\bar{x}}^{\alpha _2}_2,\ {\bar{x}}_1\ge 0,\ {\bar{x}}_2\ge 0\right\} , \end{aligned}$$

(1)

$$\begin{aligned} {\mathcal {K}}_{\exp }:= & {} \hbox {cl}\left\{ (x_1,{\bar{x}}) \in {\mathbb {R}}\times {\mathbb {R}}^2\,\bigg |\, x_1\ge {\bar{x}}_2\cdot \hbox {exp}\left( \frac{{\bar{x}}_1}{{\bar{x}}_2}\right) ,\ {\bar{x}}_2>0,\ x_1\ge 0\right\} , \end{aligned}$$

(2)

where ${\bar{x}}:=({\bar{x}}_1,{\bar{x}}_2)^T\in {\mathbb {R}}^2$, $\alpha :=(\alpha _1,\alpha _2)^T\in {\mathbb {R}}^2$, $\alpha _1, \alpha _2\in (0,1)$, $\alpha _1+\alpha _2=1$ and $\hbox {cl}(\varOmega )$ is the closure of $\varOmega$. We call ${\mathcal {K}}_{\alpha }$ the power cone and ${\mathcal {K}}_{\exp }$ the exponential cone¹, whose graphs are depicted in Fig. 1.

1.1 Motivations and literatures

Why do we pay attention to these two core non-symmetric cones? There are two main reasons. In theory, Chares [5] proposes two important concepts (i.e., $\alpha$-representable and extended $\alpha$-representable, see “Appendix 6.1”) involving powers and exponentials and plenty of famous cones can be generated from these two cones such as second-order cones [1, 8, 10, 15, 23, 24], p-order cones [2, 27, 44], geometric cones [4, 16, 17, 26], $L_p$ cones [18] and etc., one can refer to [5, chapter 4] for more examples. In applications, many practical problems can be cast into optimization models involving the power cone constraints and the exponential cone constraints, such as location problems [5, 21] and geometric programming problems [4, 31, 34]. Therefore, it becomes quite obvious that there is a great demand for providing systematic studies for these cones.

Location problem [5, 21]: The generalized location problem is to find a point $x\in {\mathbb {R}}^n$ whose sum of weight distances from a given set of locations $L_1,\ldots ,L_m$ is minimized, which has the following form

$$\begin{aligned} (P)\quad \begin{array}{ll} \min \nolimits _{x\in {\mathbb {R}}^n}&\sum \nolimits _{i=1}^m w_i\Vert x-L_i\Vert _{p_i} \end{array} \end{aligned}$$

where $\Vert \cdot \Vert _{p_i}(p_i\ge 1$) denotes the $p_i$-norm defined on ${\mathbb {R}}^n$. If $p_i$ is equal to 2, then the above problem reduces to the classical Weber-Point problem. Denote by $x:=(x_1,\ldots ,x_n)^T\in {\mathbb {R}}^n$ and $a:=(a_1,\ldots ,a_n)^T\in {\mathbb {R}}^n$, Problem (P) can be rewritten as

$$\begin{aligned} \begin{array}{ll} \min \nolimits _{x,a,y_i}&{}\sum \nolimits _{i=1}^m w_ia_i\\ s.t.&{}(y_{i,j},a_i,x_j-L_{i,j})\in {\mathcal {K}}_{\frac{1}{p_i}},\ i=1,\ldots ,m,\ j=1,\ldots ,n,\\ \ &{}\sum \nolimits _{j=1}^n y_{i,j}=a_i,\ i=1,\ldots ,m, \end{array} \end{aligned}$$

where $L_{i,j}$ and $y_{i,j}$ stand for the j-th component of $L_i\in {\mathbb {R}}^n$ and $y_i\in {\mathbb {R}}^n$, respectively.

Fig. 1

The power cone ${\mathcal {K}}_{\alpha }$ (left) and the exponential cone ${\mathcal {K}}_{\exp }$ (right)

Geometric programming [4, 31, 34]: Let $x:=(x_1,\ldots ,x_n)^T\in {\mathbb {R}}^n$ be a vector with real positive components $x_i$ . A real valued function m, of the form $m(x):=c\prod ^n_{i=1}x^{\alpha _i}_i$, is called a monomial function, where $c>0$ and $\alpha _i$ are its coefficient and exponents, respectively. A sum of one or more monomials, i.e., a function that looks like $f(x):=\sum _{k=1}^K m_k(x)$, is called a posynomial function, where $m_k(x):=c_k\prod ^n_{i=1}x^{\alpha _{i,k}}_i$. A geometric program is composed of a posynomial objective with posynomial inequality constraints and monomial equality constraints, which can be described as

$$\begin{aligned} (GP)\quad \begin{array}{ll} \min \nolimits _x&{}f_0(x)\\ s.t.&{}f_s(x)\le 1,\quad s=1,\ldots ,p,\\ \ &{}g_t(x)=1,\quad t=1,\ldots ,q, \end{array} \end{aligned}$$

where $f_s:=\sum _{k=1}^K c_{k,s}\prod ^n_{i=1}x^{\alpha _{i,k,s}}_i,\ s\in \{0,1,\ldots ,p\}$ and $g_t(x):=c_t\prod ^n_{i=1}x^{\alpha _{i,t}}_i, \ t\in \{1,\ldots ,q\}$. Using the following change of variables as $x_i:=\exp (u_i), c_{k,i}:=\exp (d_{k,i}), c_t:=\exp (d_t)$ and adding some additional variables, Problem (GP) can be rewritten as

$$\begin{aligned} \begin{array}{ll} \min \nolimits _{u_i,w,\xi _{k,0},\eta _{k,s}}&{}w\\ s.t.&{}(d_{k,0}+\sum \nolimits _{i=1}^n u_i\cdot \alpha _{i,k,0},\xi _{k,0},1) \in {\mathcal {K}}_{\exp },\ \sum \nolimits _{k=1}^K \xi _{k,0}=w,\\ \ &{}(d_{k,s}+\sum \nolimits _{i=1}^n u_i\cdot \alpha _{i,k,s},\eta _{k,s},1) \in {\mathcal {K}}_{\exp },\ \sum \nolimits _{k=1}^K \eta _{k,s}=1,\quad s=1,\ldots ,p,\\ \ &{}d_t+\sum \nolimits _{i=1}^n u_i\cdot \alpha _{i,t}=0,\quad t=1,\ldots ,q. \end{array} \end{aligned}$$

In the past three decades, a great deal of mathematical effort in conic programming has been devoted to the study of symmetric cones and it has been made extensive progress [9, 14, 29, 30, 33, 38], particularly for the second-order cone (SOC) [1, 8, 10, 15, 23, 24] and the positive semidefinite cone [35, 37, 39,40,41]. For example, consider the second-order cone

$$\begin{aligned} {\mathbb {L}}^n:=\{(x_1,{\bar{x}})\in {\mathbb {R}}\times {\mathbb {R}}^{n-1} \, | \, x_1\ge \Vert {\bar{x}}\Vert \}. \end{aligned}$$

For any given $z=(z_1,{\bar{z}})\in {\mathbb {R}}\times {\mathbb {R}}^{n-1}$, its decomposition with respect to ${\mathbb {L}}^n$ has the form

$$\begin{aligned} z=\lambda _1(z)\cdot u^{(1)}_z+\lambda _2(z)\cdot u^{(2)}_z, \end{aligned}$$

(3)

where $\lambda _i(z):=z_1+(-1)^i\Vert {\bar{z}}\Vert$ and $u^{(i)}_z$ is equal to $\frac{1}{2}\left( 1,(-1)^i\frac{{\bar{z}}}{\Vert {\bar{z}}\Vert }\right)$, if ${\bar{z}}\ne 0$; $\frac{1}{2}\left( 1,(-1)^iw\right)$, otherwise, which is applicable for $i=1,2$ with $w\in {\mathbb {R}}^{n-1}$ being any unit vector. For any scalar function $f:{\mathbb {R}}\rightarrow {\mathbb {R}}$, the associated conic function $f^{soc}(z)$ (called the SOC function) is given by

$$\begin{aligned} f^{soc}(z)=f(\lambda _1(z))\cdot u^{(1)}_z+f(\lambda _2(z))\cdot u^{(2)}_z. \end{aligned}$$

(4)

In light of the decomposition formula and its conic function, one can further establish their analytic properties (i.e., projection mapping, cone-convexity, conic-monotonicity) and design numerical algorithms (i.e., proximal-like algorithms and interior-point algorithms), see Fig. 2 for their relations and refer to the monograph [11] for more details. Similar results have also been established for the positive semidefinite cone [40, 43] and symmetric cones [14, 38]. Therefore, the past experience [11, 14, 43] indicate that how to derive the associated decomposition expression with respect to a given cone as the form (3) at a low cost becomes an important issue in the whole picture of researches.

Fig. 2

The relations between the decomposition with respect to SOC and other topics

As a fundamental tool in optimization, Moreau decomposition theorem [25] characterizes the key relationship between the decomposition with respect to a closed convex cone and its projection mappings. More concretely, for any given $z\in {\mathbb {R}}^n$, it can be uniquely decomposed into

$$\begin{aligned} z=\varPi _{{\mathcal {K}}}(z)+\varPi _{{\mathcal {K}}^\circ }(z) =\varPi _{{\mathcal {K}}}(z)-\varPi _{{\mathcal {K}}^*}(-z), \end{aligned}$$

(5)

where $\varPi _{{\mathcal {K}}}(z)$ is the projection mapping of $z\in {\mathbb {R}}^n$ onto ${\mathcal {K}}$ and ${\mathcal {K}}^\circ$ is the polar cone of ${\mathcal {K}}$, i.e.,

$$\begin{aligned} {\mathcal {K}}^\circ :=\{y\in {\mathbb {R}}^n\,|\,\ x^Ty\le 0,\ \forall x\in {\mathcal {K}}\}. \end{aligned}$$

In addition, ${\mathcal {K}}^*$ is the dual cone of ${\mathcal {K}}$ and satisfies the relation ${\mathcal {K}}^*=-{\mathcal {K}}^\circ$. It follows from (5) that if these projection mappings have closed-form expressions, the decomposition issue can be simply solved by this classical theorem. However, for most non-symmetric cones (except for the circular cone [7, 45], see “Appendix 6.2”), their projection mappings are usually not explicit, such as the power cone ${\mathcal {K}}_{\alpha }$ [21, section 2] and the exponential cone ${\mathcal {K}}_{\exp }$ [26, section 6]. Thus, one cannot employ the Moreau decomposition theorem directly and continue subsequent studies on optimization problems involving with these non-symmetric cones. This is a big hurdle for non-symmetric cone optimization problems.

In reality, there are plenty of non-symmetric cones in the literatures, such as homogeneous cones [6, 20, 42], matrix norm cones [12], p-order cones [2, 18, 27, 44], hyperbolicity cones [3, 19, 32], circular cones [7, 45] and copositive cones [13], etc. Unlike the symmetric cone optimization, there seems no systematic study due to the various features and very few algorithms are proposed to solve optimization problems with these non-symmetric cones constraints, except for some interior-point type methods [6, 22, 28, 36, 44]. For example, Xue and Ye [44] study an optimization problem of minimizing a sum of p-norms, in which two new barrier functions are introduced for p-order cones and a primal-dual potential reduction algorithm is presented. Chua [6] combines the T-algebra with the primal-dual interior-point algorithm to solve the homogeneous conic programming problems. Based on the concept of self-concordant barriers and the efficient computational experience of the long path-following steps, Nesterov [28] proposes a new predictor-corrector path-following method with an additional primal-dual lifting process (called Phase I). Skajaa and Ye [36] present a homogeneous interior-point algorithm for non-symmetric convex conic optimization, in which no Phase I method is needed. Recently, Karimi and Tuncel [22] present a primal-dual interior-point methods for convex optimization problems, in which a new concept called Domain-Driven Setup plays a crucial role in their theoretical analysis.

In contrast to these interior-point type methods, we pay more attention to the decomposition issue of the given cones. It is worth noting that the decompositions with respect to the second-order cone ${\mathbb {L}}^n$ and the circular cone ${\mathcal {L}}_\theta$ [see Eqs. (3) and (51)] show that any given point can be divided into two parts, one lies in the boundary of the given cone (i.e., $u^{(1)}_z\in \partial {\mathbb {L}}^n$, ${\tilde{u}}^{(1)}_z\in \partial {\mathcal {L}}_\theta$, where $\partial \varOmega$ is the boundary of $\varOmega$) and the other comes from the boundary of the given cone (i.e., $u^{(2)}_z\in \partial {\mathbb {L}}^n$) or its polar (i.e., ${\tilde{u}}^{(2)}_z\in \partial {\mathcal {L}}^\circ _\theta$). One can easily verify these results by the Moreau decomposition theorem in some cases (for example, the given point lies out the union of the given cone and its polar), but it is amazing that these decompositions are satisfied in all cases! These observations motivate us to study the boundary structures of the given cones more carefully.

1.2 Contributions and contents

In this paper, we successfully explore two new types of decompositions with respect to the power cone ${\mathcal {K}}_{\alpha }$ and the exponential cone ${\mathcal {K}}_{\exp }$, one is adapted from the well-known Moreau decomposition theorem, which looks like

$$\begin{aligned} z=s_x\cdot x+s_y\cdot y,\quad x\in \partial {\mathcal {K}},\quad y \in \partial {\mathcal {K}}^\circ ,\quad (s_x,s_y)\ne (0,0) \end{aligned}$$

(6)

and the other follows from geometric structures of the given cone, i.e.,

$$\begin{aligned} z=s_x\cdot x+s_y\cdot y,\quad x\in \partial {\mathcal {K}},\quad y \in \partial {\mathcal {K}}, \quad (s_x,s_y)\ne (0,0), \end{aligned}$$

(7)

where $z\in {\mathbb {R}}^n$, $s_x, s_y\in {\mathbb {R}}$, $x,y\in {\mathbb {R}}^n$, ${\mathcal {K}}$ has two choices, namely ${\mathcal {K}}_{\alpha }$ or ${\mathcal {K}}_{\exp }$, as defined in (1) and (2). In the sequel, we call (6) the Type I decomposition and (7) the Type II decomposition, respectively. To our best knowledge, no results about the decompositions with respect to these two non-symmetric cones have been reported. Hence, the purpose of this paper aims to fill this gap and the contributions of our research can be summarized as follows.

• (a) We propose a more compact description of the boundary for these two cones.

• (b) Two types of decompositions with respect to ${\mathcal {K}}_{\alpha }, {\mathcal {K}}_{\exp }$ are presented, which are do-able and computable.

  As a byproduct, the decomposition expressions with respect to the high-dimensional power cone are also derived.

• (c) We establish the conic functions of the power cone ${\mathcal {K}}_{\alpha }$ and the exponential cone ${\mathcal {K}}_{\exp }$ based on their decomposition formulae.

The remainder of this paper is organized as follows. In Sects. 2 and 3, we present the decomposition formulae with respect to the power cone ${\mathcal {K}}_{\alpha }$ and the exponential cone ${\mathcal {K}}_{\exp }$, respectively. In Sect. 4, we discuss some applications of these decompositions. Finally, we draw some concluding remarks in Sect. 5.

2 The decompositions with respect to the power cone ${\mathcal {K}}_\alpha$

In this section, we present two types of decompositions with respect to the power cone ${\mathcal {K}}_\alpha$. Before that, we present some analytic properties of ${\mathcal {K}}_\alpha$ in the following lemmas.

Lemma 1

${\mathcal {K}}_\alpha$ is a closed convex cone.

Proof

It can be easily verified by definition, see “Appendix 6.3” for more details. $\square$

Lemma 2

The dual cone ${\mathcal {K}}^*_\alpha$ can be described as

$$\begin{aligned} {\mathcal {K}}^*_{\alpha }=\left\{ (x_1,{\bar{x}})\in {\mathbb {R}}\times {\mathbb {R}}^2\, \bigg |\, |x_1|\le \left( \frac{{\bar{x}}_1}{\alpha _1}\right) ^{\alpha _1} \left( \frac{{\bar{x}}_2}{\alpha _2}\right) ^{\alpha _2},\ {\bar{x}}_1\ge 0,\ {\bar{x}}_2\ge 0\right\} , \end{aligned}$$

where ${\bar{x}}:=({\bar{x}}_1,{\bar{x}}_2)^T\in {\mathbb {R}}^2$, $\alpha :=(\alpha _1,\alpha _2)^T\in {\mathbb {R}}^2$, $\alpha _1, \alpha _2\in (0,1)$, $\alpha _1+\alpha _2=1$.

Proof

We refer the readers to [5, Theorem 4.3.1] for its verification.$\square$

From the relation ${\mathcal {K}}^\circ _{\alpha }=-{\mathcal {K}}^*_{\alpha }$ and Lemma 2, the polar cone ${\mathcal {K}}^\circ _\alpha$ has the following closed-form expression.

Corollary 1

The polar cone ${\mathcal {K}}^\circ _{\alpha }$ is given by

$$\begin{aligned} {\mathcal {K}}^\circ _{\alpha }=\left\{ (x_1,{\bar{x}})\in {\mathbb {R}}\times {\mathbb {R}}^2\, \bigg |\, |x_1|\le \left( \frac{-{\bar{x}}_1}{\alpha _1}\right) ^{\alpha _1} \left( \frac{-{\bar{x}}_2}{\alpha _2}\right) ^{\alpha _2},\ {\bar{x}}_1\le 0,\ {\bar{x}}_2\le 0\right\} . \end{aligned}$$

We now proceed to identify the structures of the power cone ${\mathcal {K}}_\alpha$, its dual ${\mathcal {K}}^*_\alpha$ and its polar ${\mathcal {K}}^\circ _\alpha$ more clearly, particularly for their interiors and boundaries.

Lemma 3

The interior of the power cone ${\mathcal {K}}_\alpha$, its dual ${\mathcal {K}}^*_\alpha$ and its polar ${\mathcal {K}}^\circ _\alpha$ , denoted by $\hbox {int}\,{\mathcal {K}}_\alpha$, $\hbox {int}\,{\mathcal {K}}^*_\alpha$ and $\hbox {int}\,{\mathcal {K}}^\circ _\alpha$, are respectively given by

$$\begin{aligned} \hbox {int}\,{\mathcal {K}}_\alpha= & {} \left\{ (x_1,{\bar{x}})\in {\mathbb {R}} \times {\mathbb {R}}^2\,\bigg |\, |x_1|<\sigma _{\alpha }({\bar{x}}),\ {\bar{x}}_1>0,\ {\bar{x}}_2>0\right\} , \end{aligned}$$

(8)

$$\begin{aligned} \hbox {int}\,{\mathcal {K}}^*_\alpha= & {} \left\{ (x_1,{\bar{x}})\in {\mathbb {R}} \times {\mathbb {R}}^2\,\bigg |\, |x_1|<\eta _{\alpha }({\bar{x}}),\ {\bar{x}}_1>0,\ {\bar{x}}_2>0\right\} , \end{aligned}$$

(9)

$$\begin{aligned} \hbox {int}\,{\mathcal {K}}^\circ _\alpha= & {} \left\{ (x_1,{\bar{x}})\in {\mathbb {R}} \times {\mathbb {R}}^2\,\bigg |\, |x_1|<\eta _{\alpha }(-{\bar{x}}),\ {\bar{x}}_1<0,\ {\bar{x}}_2<0\right\} , \end{aligned}$$

(10)

where

$$\begin{aligned} \sigma _{\alpha }({\bar{x}}):={\bar{x}}^{\alpha _1}_1{\bar{x}}^{\alpha _2}_2,\quad \eta _{\alpha }({\bar{x}}):=\left( \frac{{\bar{x}}_1}{\alpha _1}\right) ^{\alpha _1} \left( \frac{{\bar{x}}_2}{\alpha _2}\right) ^{\alpha _2}. \end{aligned}$$

(11)

Proof

By definition, $(x_1,{\bar{x}})$ is an element of $\hbox {int}\,{\mathcal {K}}_\alpha$ if and only if there exists an open neighborhood of $(x_1,{\bar{x}})\in {\mathbb {R}}\times {\mathbb {R}}^2$ entirely included in ${\mathcal {K}}_\alpha$. Let us take $(x_1,{\bar{x}})\in {\mathcal {K}}_\alpha$. For any given strict positive scalars ${\bar{x}}_1,{\bar{x}}_2\in {\mathbb {R}}$, it is easy to see that (0, 0, 0), $(0,{\bar{x}}_1,0)$ and $(0,0,{\bar{x}}_2)$ are all outside of $\hbox {int}\,{\mathcal {K}}_\alpha$, due to the observation that every open neighborhood with respect to each of these points contains a point with the negative ${\bar{x}}_1$ or ${\bar{x}}_2$ component. For a point $(x_1,{\bar{x}}_1,{\bar{x}}_2)\in {\mathbb {R}}\times {\mathbb {R}}^2$ such that $\sigma _{\alpha }({\bar{x}})=|x_1|$ with ${\bar{x}}_1,{\bar{x}}_2>0$, where $\sigma _{\alpha }({\bar{x}})$ is defined as in (11). In this case, we can take a point $(x^\prime _1,{\bar{x}}^\prime _1,{\bar{x}}^\prime _2)$ with $0<{\bar{x}}^\prime _1<{\bar{x}}_1$, $0<{\bar{x}}^\prime _2<{\bar{x}}_2$, $|x^\prime _1|>|x_1|$ in every open neighborhood of $(x_1,{\bar{x}}_1,{\bar{x}}_2)\in {\mathbb {R}}\times {\mathbb {R}}^2$, which implies that $|x^\prime _1|>|x_1|= \sigma _{\alpha }({\bar{x}})> \sigma _{\alpha }({\bar{x}}^\prime )$, i.e., the point $(x^\prime _1,{\bar{x}}^\prime _1,{\bar{x}}^\prime _2)$ can not belong to ${\mathcal {K}}_\alpha$ and hence $(x_1,{\bar{x}}_1,{\bar{x}}_2)\notin \hbox {int}\,{\mathcal {K}}_\alpha$.

Next, we turn to show that all the remaining points that do not satisfy the above two cases, i.e., the points in the right-hand side of (8), belong to the interior of ${\mathcal {K}}_\alpha$. For sufficiently small scalar $\epsilon \in (0,\min \{{\bar{x}}_1,{\bar{x}}_2\})$, let ${\mathcal {B}}^\epsilon _{(x_1,{\bar{x}})}$ be a neighborhood of $(x_1,{\bar{x}})$ with the form

$$\begin{aligned} {\mathcal {B}}^\epsilon _{(x_1,{\bar{x}})}:= & {} \left\{ (x^\prime _1, {\bar{x}}^\prime ) \in {\mathbb {R}}\times {\mathbb {R}}^2\, \bigg | \, 0\le |x_1|-\epsilon \le |x^\prime _1|\le |x_1|+\epsilon ,\right. \\&\qquad \left. 0<{\bar{x}}_i-\epsilon \le {\bar{x}}^\prime _i\le {\bar{x}}_i+\epsilon ,\ i=1,2\right\} . \end{aligned}$$

Taking $(x_1,{\bar{x}})\in {\mathbb {R}}\times {\mathbb {R}}^2$ from the right-hand side of (8), i.e., $\sigma _{\alpha }({\bar{x}})>|x_1|$, ${\bar{x}}_i>0$, $i=1,2$. For all elements $(x^\prime _1,{\bar{x}}^\prime )\in {\mathcal {B}}^\epsilon _{(x_1,{\bar{x}})}$, we have

$$\begin{aligned} |x^\prime _1|-\sigma _\alpha ({\bar{x}}^\prime )\le |{\bar{x}}_1| +\epsilon -({\bar{x}}^\prime _1)^{\alpha _1}({\bar{x}}^\prime _2)^{\alpha _2} \le |{\bar{x}}_1|+\epsilon -({\bar{x}}_1-\epsilon )^{\alpha _1}({\bar{x}}_2-\epsilon )^{\alpha _2}. \end{aligned}$$

(12)

In addition, letting $\epsilon \rightarrow 0$, we obtain

$$\begin{aligned} \lim _{\epsilon \rightarrow 0}\left( |{\bar{x}}_1|+\epsilon -({\bar{x}}_1-\epsilon )^{\alpha _1} ({\bar{x}}_2-\epsilon )^{\alpha _2}\right) =|{\bar{x}}_1|-\sigma _\alpha ({\bar{x}})<0. \end{aligned}$$

Therefore, there exists a scalar $\epsilon ^*$ such that $|{\bar{x}}_1|+\epsilon ^*-({\bar{x}}_1-\epsilon ^*)^{\alpha _1} ({\bar{x}}_2-\epsilon ^*)^{\alpha _2}<0$. This together with (12) imply that

$$\begin{aligned} |x^\prime _1|-\sigma _\alpha ({\bar{x}}^\prime )<0,\ \forall (x^\prime _1,{\bar{x}}^\prime )\in {\mathcal {B}}^\epsilon _{(x_1,{\bar{x}})}, \end{aligned}$$

which is sufficient to show that ${\mathcal {B}}^\epsilon _{(x_1,{\bar{x}})}$ is entirely included in ${\mathcal {K}}_\alpha$ and hence $(x_1,{\bar{x}})\in \hbox {int}\,{\mathcal {K}}_\alpha$.

Applying a similar way to ${\mathcal {K}}^*_\alpha$ and ${\mathcal {K}}^\circ _\alpha$, their interiors can also be verified as the right-hand side of (9) and (10). $\square$

From the proof of Lemma 3, we further define the following sets

$$\begin{aligned} \begin{array}{l} S_1:=\left\{ (x_1,{\bar{x}})\in {\mathbb {R}}\times {\mathbb {R}}^2\,|\,\ x_1=0,\ {\bar{x}}_1>0,\ {\bar{x}}_2=0\right\} ,\\ S_2:=\left\{ (x_1,{\bar{x}})\in {\mathbb {R}}\times {\mathbb {R}}^2\,|\,\ x_1=0,\ {\bar{x}}_1=0,\ {\bar{x}}_2>0\right\} ,\\ S_3:=\left\{ (x_1,{\bar{x}})\in {\mathbb {R}}\times {\mathbb {R}}^2\,|\,\ |x_1|=\sigma _{\alpha }({\bar{x}}),\ {\bar{x}}_1>0,\ {\bar{x}}_2>0\right\} ,\\ S_4:=\left\{ (x_1,{\bar{x}})\in {\mathbb {R}}\times {\mathbb {R}}^2\,|\,\ |x_1|=\eta _{\alpha }({\bar{x}}),\ {\bar{x}}_1>0,\ {\bar{x}}_2>0\right\} ,\\ T_1:=\left\{ (x_1,{\bar{x}})\in {\mathbb {R}}\times {\mathbb {R}}^2\,|\,\ x_1=0,\ {\bar{x}}_1<0,\ {\bar{x}}_2=0\right\} =-S_1, \\ T_2:=\left\{ (x_1,{\bar{x}})\in {\mathbb {R}}\times {\mathbb {R}}^2\,|\,\ x_1=0,\ {\bar{x}}_1=0,\ {\bar{x}}_2<0\right\} =-S_2, \\ T_3:=\left\{ (x_1,{\bar{x}})\in {\mathbb {R}}\times {\mathbb {R}}^2\,|\,\ |x_1| =\eta _{\alpha }(-{\bar{x}}),\ {\bar{x}}_1<0,\ {\bar{x}}_2<0\right\} =-S_4. \end{array} \end{aligned}$$

(13)

Then, the boundary of ${\mathcal {K}}_\alpha , {\mathcal {K}}^*_{\alpha }$ and ${\mathcal {K}}^\circ _{\alpha }$ can be stated in a more compact form.

Lemma 4

The boundary of ${\mathcal {K}}_\alpha$ and ${\mathcal {K}}^*_{\alpha }$, denoted by $\partial {\mathcal {K}}_\alpha$ and $\partial {\mathcal {K}}^*_{\alpha }$, are respectively given by

$$\begin{aligned} \partial {\mathcal {K}}_{\alpha }:=S_1\cup S_2\cup S_3\cup \{0\},\quad \partial {\mathcal {K}}^*_{\alpha }:=S_1\cup S_2\cup S_4\cup \{0\}. \end{aligned}$$

Similarly, the boundary of ${\mathcal {K}}^\circ _\alpha$, denoted by $\partial {\mathcal {K}}^\circ _\alpha$, can be formulated as

$$\begin{aligned} \partial {\mathcal {K}}^\circ _{\alpha }:=T_1\cup T_2\cup T_3\cup \{0\}. \end{aligned}$$

Remark 1

It follows that the union set ${\mathcal {K}}_\alpha \cup {\mathcal {K}}^\circ _\alpha$ can be divided into seven parts

$$\begin{aligned} {\mathcal {K}}_\alpha \cup {\mathcal {K}}^\circ _\alpha =S_1\cup S_2 \cup T_1\cup T_2\cup P_1\cup P_2\cup \{0\}, \end{aligned}$$

where

$$\begin{aligned} \begin{array}{l} P_1:=\left\{ (x_1,{\bar{x}})\in {\mathbb {R}}\times {\mathbb {R}}^2\,|\,\ |x_1| \le \sigma _{\alpha }({\bar{x}}),\ {\bar{x}}_1>0,\ {\bar{x}}_2>0\right\} ,\\ P_2:=\left\{ (x_1,{\bar{x}})\in {\mathbb {R}}\times {\mathbb {R}}^2\,|\,\ |x_1| \le \eta _{\alpha }(-{\bar{x}}),\ {\bar{x}}_1<0,\ {\bar{x}}_2<0\right\} . \end{array} \end{aligned}$$

In addition, the boundary of ${\mathcal {K}}_\alpha$ and its polar ${\mathcal {K}}^\circ _{\alpha }$ are depicted in Fig. 3.

Fig. 3

The different parts of $\partial {\mathcal {K}}_{\alpha }$ (left) and $\partial {\mathcal {K}}^\circ _{\alpha }$ (right)

In order to make the classifications clear and neat, we adapt some notations as follows:

$$\begin{aligned} \begin{array}{c} z:=(z_1,{\bar{z}})\in {\mathbb {R}}\times {\mathbb {R}}^2,\ {\bar{z}}:=({\bar{z}}_1,{\bar{z}}_2)^T\in {\mathbb {R}}^2,\ {\bar{z}}_{\min }:=\min \{{\bar{z}}_1,{\bar{z}}_2\},\ {\bar{z}}_{\max }:=\max \{{\bar{z}}_1,{\bar{z}}_2\}. \end{array} \end{aligned}$$

(14)

Consequently, we divide the space ${\mathbb {R}}\times {\mathbb {R}}^2$ into the following four blocks

$$\begin{aligned} \begin{array}{ll} {\hbox {Block I}:}&{}\ B_1:=\left\{ (z_1,{\bar{z}})\in {\mathbb {R}}\times {\mathbb {R}}^2\,|\,{\bar{z}}_{\min } \cdot {\bar{z}}_{\max }>0\ \hbox {or}\ (z_1\ne 0\ \hbox {and}\ {\bar{z}}_{\min }={\bar{z}}_{\max }=0)\right\} .\\ {\hbox {Block II}:}&{}\ B_2:=\left\{ (z_1,{\bar{z}})\in {\mathbb {R}}\times {\mathbb {R}}^2\,|\,{\bar{z}}_{\min } \cdot {\bar{z}}_{\max }=0\ \hbox {and}\ {\bar{z}}_{\min }+{\bar{z}}_{\max }\ne 0\right\} .\\ {\hbox {Block III}:}&{}\ B_3:=\left\{ (z_1,{\bar{z}})\in {\mathbb {R}}\times {\mathbb {R}}^2\,|\,{\bar{z}}_{\min } \cdot {\bar{z}}_{\max }<0\right\} .\\ {\hbox {Block IV}:}&{}\ B_4:=\left\{ (z_1,{\bar{z}})\in {\mathbb {R}}\times {\mathbb {R}}^2\,|\,z_1=0\ \hbox {and}\ {\bar{z}}_{\min } ={\bar{z}}_{\max }=0\right\} . \end{array} \end{aligned}$$

(15)

The subcases of these blocks with respect to ${\mathcal {K}}_\alpha$ can be found in Table 1.

Table 1 The subcases of each block in (15) with respect to ${\mathcal {K}}_\alpha$

2.1 The Type I decomposition with respect to the power cone ${\mathcal {K}}_{\alpha }$

In this subsection, we present the Type I decomposition with respect to the power cone ${\mathcal {K}}_{\alpha }$. To proceed, we discuss four cases, in which the sets $S_i\subset {\mathcal {K}}\ (i=1,2,3,4)$ and $T_j\subset \partial {\mathcal {K}}^\circ \ (j=1,2,3)$ are defined as in (13).

Case 1:$(z_1,{\bar{z}})\in B_1$.

• (a) ${\bar{z}}_{\min }>0$. In this subcase, $(z_1,{\bar{z}})\in B_{11}$, i.e., ${\bar{z}}_1>0, {\bar{z}}_2>0$, which implies $\sigma _{\alpha }({\bar{z}})>0$ and $\eta _{\alpha }({\bar{z}})>0$. Then, we take $x={\dot{x}}^{(B_1,a)}$, $y={\dot{y}}^{(B_1,a)}$ and $s_x={\dot{s}}_x^{(B_1,a)}$, $s_y={\dot{s}}_y^{(B_1,a)}$, where

  $$\begin{aligned} {\dot{x}}^{(B_1,a)}:= & {} \left[ \begin{array}{c} 1 \\ \frac{{\bar{z}}}{\sigma _{\alpha }({\bar{z}})} \end{array} \right] \in S_3,\ {\dot{y}}^{(B_1,a)}:=\left[ \begin{array}{c} 1 \\ -\frac{{\bar{z}}}{\eta _{\alpha }({\bar{z}})} \end{array} \right] \in T_3. \end{aligned}$$

  (16)
  $$\begin{aligned} {\dot{s}}_x^{(B_1,a)}:= & {} \frac{z_1+\eta _{\alpha }({\bar{z}})}{\sigma _{\alpha }({\bar{z}})+\eta _{\alpha }({\bar{z}})}\cdot \sigma _{\alpha }({\bar{z}}),\ {\dot{s}}_y^{(B_1,a)}:=\frac{z_1-\sigma _{\alpha }({\bar{z}})}{\sigma _{\alpha }({\bar{z}})+\eta _{\alpha }({\bar{z}})}\cdot \eta _{\alpha }({\bar{z}}). \end{aligned}$$

  (17)

  It is easy to show that the above setting satisfies the decomposition formula (6).

• (b) ${\bar{z}}_{\max }<0$. Similar to the argument in Case 1 (a), $(z_1,{\bar{z}})\in B_{12}$, i.e., ${\bar{z}}_1<0$, ${\bar{z}}_2<0$, which implies $\sigma _{\alpha }(-{\bar{z}})>0$ and $\eta _{\alpha }(-{\bar{z}})>0$. In this subcase, we set $x={\dot{x}}^{(B_1,b)}$, $y={\dot{y}}^{(B_1,b)}$ and $s_x={\dot{s}}_x^{(B_1,b)}$, $s_y={\dot{s}}_y^{(B_1,b)}$, where

  $$\begin{aligned} {\dot{x}}^{(B_1,b)}:= & {} \left[ \begin{array}{c} 1 \\ \frac{-{\bar{z}}}{\sigma _{\alpha }(-{\bar{z}})} \end{array} \right] \in S_3,\ {\dot{y}}^{(B_1,b)}:=\left[ \begin{array}{c} 1 \\ \frac{{\bar{z}}}{\eta _{\alpha }(-{\bar{z}})} \end{array} \right] \in T_3. \end{aligned}$$

  (18)
  $$\begin{aligned} {\dot{s}}_x^{(B_1,b)}:= & {} \frac{z_1-\eta _{\alpha }(-{\bar{z}})}{\sigma _{\alpha }(-{\bar{z}})+\eta _{\alpha }(-{\bar{z}})} \cdot \sigma _{\alpha }(-{\bar{z}}),\nonumber \\ {\dot{s}}_y^{(B_1,b)}:= & {} \frac{z_1+\sigma _{\alpha }(-{\bar{z}})}{\sigma _{\alpha }(-{\bar{z}})+\eta _{\alpha }(-{\bar{z}})} \cdot \eta _{\alpha }(-{\bar{z}}). \end{aligned}$$

  (19)

• (c) $z_1\ne 0$ and ${\bar{z}}_{\min }={\bar{z}}_{\max }=0$. In this subcase, $(z_1,{\bar{z}})\in B_{13}$, which implies $\sigma _{\alpha }({\bar{z}})=0$ and $\eta _{\alpha }({\bar{z}})=0$. Therefore, we set $x={\dot{x}}^{(B_1,c)}$, $y={\dot{y}}^{(B_1,c)}$ and $s_x={\dot{s}}_x^{(B_1,c)}$, $s_y={\dot{s}}_y^{(B_1,c)}$, where

  $$\begin{aligned} {\dot{x}}^{(B_1,c)}:= & {} \left[ \begin{array}{c} 1 \\ \frac{{\mathbf {1}}}{\sigma _{\alpha }({\mathbf {1}})} \end{array} \right] \in S_3,\ {\dot{y}}^{(B_1,c)}:=\left[ \begin{array}{c} 1 \\ -\frac{{\mathbf {1}}}{\eta _{\alpha }({\mathbf {1}})} \end{array} \right] \in T_3, \end{aligned}$$

  (20)
  $$\begin{aligned} {\dot{s}}^{(B_1,c)}_x:= & {} \frac{z_1}{\sigma _{\alpha }({\mathbf {1}}) +\eta _{\alpha }({\mathbf {1}})}\cdot \sigma _{\alpha }({\mathbf {1}}),\ {\dot{s}}^{(B_1,c)}_y:=\frac{z_1}{\sigma _{\alpha }({\mathbf {1}}) +\eta _{\alpha }({\mathbf {1}})}\cdot \eta _{\alpha }({\mathbf {1}}) \end{aligned}$$

  (21)

  with ${\mathbf {1}}:=(1,1)^T\in {\mathbb {R}}^2$.

Case 2:$(z_1,{\bar{z}})\in B_2$.

• (a) ${\bar{z}}_{\min }=0,\ {\bar{z}}_{\max }>0$. In this subcase, $(z_1,{\bar{z}})\in B_{21}$ or $(z_1,{\bar{z}})\in B_{22}$. Therefore, we set $x={\dot{x}}^{(B_2,a)}$, $y={\dot{y}}^{(B_2,a)}$ and $s_x=1$, $s_y=1$, where ${\dot{x}}^{(B_2,a)}=({\dot{x}}^{(B_2,a)}_1,\dot{{\bar{x}}}^{(B_2,a)})$ and ${\dot{y}}^{(B_2,a)}=({\dot{y}}^{(B_2,a)}_1,\dot{{\bar{y}}}^{(B_2,a)})$ with

  $$\begin{aligned} {\dot{x}}^{(B_2,a)}_1:=z_1,\dot{{\bar{x}}}^{(B_2,a)}:= & {} \left\{ \begin{array}{ll} \left[ \begin{array}{c} \left( \frac{|z_1|}{{\bar{z}}^{\alpha _2}_2}\right) ^{\frac{1}{\alpha _1}}\\ {\bar{z}}_2 \end{array} \right] &{}\hbox {if}\ (z_1,{\bar{z}})\in B_{21},\\ \left[ \begin{array}{c} {\bar{z}}_1\\ \left( \frac{|z_1|}{{\bar{z}}^{\alpha _1}_1}\right) ^{\frac{1}{\alpha _2}}\\ \end{array} \right] &{}\hbox {if}\ (z_1,{\bar{z}})\in B_{22},\\ \end{array} \right. \end{aligned}$$

  (22)
  $$\begin{aligned} {\dot{y}}^{(B_2,a)}_1:=0,\dot{{\bar{y}}}^{(B_2,a)}:= & {} \left\{ \begin{array}{ll} \left[ \begin{array}{c} -\left( \frac{|z_1|}{{\bar{z}}^{\alpha _2}_2}\right) ^{\frac{1}{\alpha _1}}\\ 0\end{array} \right] &{}\hbox {if}\ (z_1,{\bar{z}})\in B_{21},\\ \left[ \begin{array}{c} 0\\ -\left( \frac{|z_1|}{{\bar{z}}^{\alpha _1}_1}\right) ^{\frac{1}{\alpha _2}}\\ \end{array} \right] &{}\hbox {if}\ (z_1,{\bar{z}})\in B_{22}.\\ \end{array} \right. \end{aligned}$$

  (23)

  It is easy to see that

  $$\begin{aligned} \begin{array}{l} \hbox {(a)}\ (z_1,{\bar{z}})\in B_{21}, z_1=0 \Rightarrow x\in S_2, y=0;\quad \hbox {(b)}\ (z_1,{\bar{z}})\in B_{21}, z_1\ne 0 \Rightarrow x\in S_3, y\in T_1;\\ \hbox {(c)}\ (z_1,{\bar{z}})\in B_{22}, z_1=0 \Rightarrow x\in S_1, y=0;\quad \hbox {(d)}\ (z_1,{\bar{z}})\in B_{22}, z_1\ne 0 \Rightarrow x\in S_3, y\in T_2. \end{array} \end{aligned}$$

• (b) ${\bar{z}}_{\min }<0,\ {\bar{z}}_{\max }=0$. In this subcase, $(z_1,{\bar{z}})\in B_{23}$ or $(z_1,{\bar{z}})\in B_{24}$. We set $x={\dot{x}}^{(B_2,b)}$, $y={\dot{y}}^{(B_2,b)}$ and $s_x=-1$, $s_y=-1$, where ${\dot{x}}^{(B_2,b)}=({\dot{x}}^{(B_2,b)}_1,\dot{{\bar{x}}}^{(B_2,b)})$ and ${\dot{y}}^{(B_2,b)}=({\dot{y}}^{(B_2,b)}_1,\dot{{\bar{y}}}^{(B_2,b)})$ with

  $$\begin{aligned} {\dot{x}}^{(B_2,b)}_1:=-z_1,\dot{{\bar{x}}}^{(B_2,b)}:= & {} \left\{ \begin{array}{ll} \left[ \begin{array}{c} \left( \frac{|z_1|}{(-{\bar{z}}_2)^{\alpha _2}}\right) ^{\frac{1}{\alpha _1}}\\ -{\bar{z}}_2 \end{array} \right] &{}\hbox {if}\ (z_1,{\bar{z}})\in B_{23},\\ \left[ \begin{array}{c} -{\bar{z}}_1\\ \left( \frac{|z_1|}{(-{\bar{z}}_1)^{\alpha _1}}\right) ^{\frac{1}{\alpha _2}}\\ \end{array} \right] &{}\hbox {if}\ (z_1,{\bar{z}})\in B_{24},\\ \end{array} \right. \end{aligned}$$

  (24)
  $$\begin{aligned} {\dot{y}}^{(B_2,b)}_1:=0,\dot{{\bar{y}}}^{(B_2,b)}:= & {} \left\{ \begin{array}{ll} \left[ \begin{array}{c} -\left( \frac{|z_1|}{(-{\bar{z}}_2)^{\alpha _2}}\right) ^{\frac{1}{\alpha _1}}\\ 0 \end{array} \right] &{}\hbox {if}\ (z_1,{\bar{z}})\in B_{23},\\ \left[ \begin{array}{c} 0\\ -\left( \frac{|z_1|}{(-{\bar{z}}_1)^{\alpha _1}}\right) ^{\frac{1}{\alpha _2}}\\ \end{array} \right] &{}\hbox {if}\ (z_1,{\bar{z}})\in B_{24}.\\ \end{array} \right. \end{aligned}$$

  (25)

  Similar to the arguments in Case 2 (a), we obtain

  $$\begin{aligned} \begin{array}{l} \hbox {(a)}\ (z_1,{\bar{z}})\in B_{23}, z_1=0 \Rightarrow x\in S_2, y=0;\quad \hbox {(b)}\ (z_1,{\bar{z}})\in B_{23}, z_1\ne 0 \Rightarrow x\in S_3, y\in T_1;\\ \hbox {(c)}\ (z_1,{\bar{z}})\in B_{24}, z_1=0 \Rightarrow x\in S_1, y=0;\quad \hbox {(d)}\ (z_1,{\bar{z}})\in B_{24}, z_1\ne 0 \Rightarrow x\in S_3, y\in T_2. \end{array} \end{aligned}$$

Case 3:$(z_1,{\bar{z}})\in B_3$. In this subcase, $(z_1,{\bar{z}})\in B_{31}$ or $(z_1,{\bar{z}})\in B_{32}$. We set $x={\dot{x}}^{(B_3)}\in \partial {\mathcal {K}}_{\alpha }$, $y={\dot{y}}^{(B_3)}\in \partial {\mathcal {K}}^\circ _{\alpha }$ and $s_x=1$, $s_y=1$, where ${\dot{x}}^{(B_3)}=({\dot{x}}^{(B_3)}_1,\dot{{\bar{x}}}^{(B_3)})$ and ${\dot{y}}^{(B_3)}=({\dot{y}}^{(B_3)}_1,\dot{{\bar{y}}}^{(B_3)})$ with

$$\begin{aligned} {\dot{x}}^{(B_3)}_1:=z_1,\dot{{\bar{x}}}^{(B_3)}:= & {} \left\{ \begin{array}{ll} \left[ \begin{array}{c} \left( \frac{|z_1|}{{\bar{z}}^{\alpha _2}_2}\right) ^{\frac{1}{\alpha _1}}\\ {\bar{z}}_2 \end{array} \right] &{}\hbox {if}\ z\in B_{31},\\ \left[ \begin{array}{c} {\bar{z}}_1\\ \left( \frac{|z_1|}{{\bar{z}}^{\alpha _1}_1}\right) ^{\frac{1}{\alpha _2}}\\ \end{array} \right] &{}\hbox {if}\ z\in B_{32},\\ \end{array} \right. \end{aligned}$$

(26)

$$\begin{aligned} {\dot{y}}^{(B_3)}_1:=0,\dot{{\bar{y}}}^{(B_3)}:= & {} \left\{ \begin{array}{ll} \left[ \begin{array}{c} {\bar{z}}_1-\left( \frac{|z_1|}{{\bar{z}}^{\alpha _2}_2}\right) ^{\frac{1}{\alpha _1}}\\ 0\end{array} \right] &{}\hbox {if}\ z\in B_{31},\\ \left[ \begin{array}{c} 0\\ {\bar{z}}_2-\left( \frac{|z_1|}{{\bar{z}}^{\alpha _1}_1} \right) ^{\frac{1}{\alpha _2}}\\ \end{array} \right] &{}\hbox {if}\ z\in B_{32}.\\ \end{array} \right. \end{aligned}$$

(27)

More concretely, we obtain

$$\begin{aligned} \begin{array}{l} \hbox {(a)}\ (z_1,{\bar{z}})\in B_{31}, z_1=0 \Rightarrow x\in S_2, y\in T_1;\quad \hbox {(b)}\ (z_1,{\bar{z}})\in B_{31}, z_1\ne 0 \Rightarrow x\in S_3, y\in T_1;\\ \hbox {(c)}\ (z_1,{\bar{z}})\in B_{32}, z_1=0 \Rightarrow x\in S_1, y\in T_2;\quad \hbox {(d)}\ (z_1,{\bar{z}})\in B_{32}, z_1\ne 0 \Rightarrow x\in S_3, y\in T_2. \end{array} \end{aligned}$$

Case 4:$(z_1,{\bar{z}})\in B_4$. In this subcase, we set $x={\dot{x}}^{(B_4)}$, $y={\dot{y}}^{(B_4)}$ and $s_x=1$, $s_y=1$, where

$$\begin{aligned} \begin{array}{ll} {\dot{x}}^{(B_4)}:=\left[ \begin{array}{c} 0\\ 1\\ 0\end{array} \right] \in S_1,\quad &{}{\dot{y}}^{(B_4)}:=\left[ \begin{array}{c} 0\\ -1\\ 0\end{array} \right] \in T_1, \end{array} \end{aligned}$$

(28)

or

$$\begin{aligned} \begin{array}{ll} {\dot{x}}^{(B_4)}:=\left[ \begin{array}{c} 0\\ 0\\ 1\end{array} \right] \in S_2,\quad &{}{\dot{y}}^{(B_4)}:=\left[ \begin{array}{c} 0\\ 0\\ -1\end{array} \right] \in T_2. \end{array} \end{aligned}$$

(29)

To sum up these discussions, we present the Type I decomposition with respect to the power cone ${\mathcal {K}}_{\alpha }$ in the following theorem.

Theorem 1

For any given $z=(z_1,{\bar{z}})\in {\mathbb {R}}\times {\mathbb {R}}^2$, its Type I decomposition with respect to ${\mathcal {K}}_{\alpha }$ is given by

• (a) If $z\in B_1$, then

  $$\begin{aligned} z=\left\{ \begin{array}{ll} {\dot{s}}_x^{(B_1,a)}\cdot {\dot{x}}^{(B_1,a)}+{\dot{s}}_y^{(B_1,a)}\cdot {\dot{y}}^{(B_1,a)},&{}\hbox {if}\ z\in B_{11},\\ {\dot{s}}_x^{(B_1,b)}\cdot {\dot{x}}^{(B_1,b)}+{\dot{s}}_y^{(B_1,b)}\cdot {\dot{y}}^{(B_1,b)},&{}\hbox {if}\ z\in B_{12},\\ {\dot{s}}_x^{(B_1,c)}\cdot {\dot{x}}^{(B_1,c)}+{\dot{s}}_y^{(B_1,c)}\cdot {\dot{y}}^{(B_1,c)},&{}\hbox {if}\ z\in B_{13}, \end{array} \right. \end{aligned}$$

  where ${\dot{x}}^{(B_1,a)}$, ${\dot{y}}^{(B_1,a)}$, ${\dot{s}}_x^{(B_1,a)}$, ${\dot{s}}_y^{(B_1,a)}$ are defined as in (16)–(17), ${\dot{x}}^{(B_1,b)}$, ${\dot{y}}^{(B_1,b)}$, ${\dot{s}}_x^{(B_1,b)}$, ${\dot{s}}_y^{(B_1,b)}$ are defined as in (18)–(19) and ${\dot{x}}^{(B_1,c)}$, ${\dot{y}}^{(B_1,c)}$, ${\dot{s}}_x^{(B_1,c)}$, ${\dot{s}}_y^{(B_1,c)}$ are defined as in (20)–(21).

• (b) If $z\in B_2$, then

  $$\begin{aligned} z=\left\{ \begin{array}{ll} {\dot{x}}^{(B_2,a)}+{\dot{y}}^{(B_2,a)},&{}\hbox {if}\ z\in B_{21}\ \hbox {or}\ z\in B_{22},\\ (-1)\cdot {\dot{x}}^{(B_2,b)}+(-1)\cdot {\dot{y}}^{(B_2,b)},&{}\hbox {if}\ z\in B_{23}\ \hbox {or}\ z\in B_{24}, \end{array} \right. \end{aligned}$$

  where ${\dot{x}}^{(B_2,a)}$, ${\dot{y}}^{(B_2,a)}$ are defined as in (22)–(23), ${\dot{x}}^{(B_2,b)}$, ${\dot{y}}^{(B_2,b)}$ are defined as in (24)–(25).

• (c) If $z\in B_3$, then $z={\dot{x}}^{(B_3)}+{\dot{y}}^{(B_3)}$, where ${\dot{x}}^{(B_3)}$, ${\dot{y}}^{(B_3)}$ are defined as in (26)–(27).

• (d) If $z\in B_4$, then $z={\dot{x}}^{(B_4)}+{\dot{y}}^{(B_4)}$, where ${\dot{x}}^{(B_4)}$ and ${\dot{y}}^{(B_4)}$ are defined as in (28) or (29).

In addition, the locations of the x-part and y-part in each case are shown in Table 2, where $S_i,T_i\ (i=1,2,3,4)$ are defined as in (13) and $x_{loc}, y_{loc}$ denote the locations of x and y, respectively.

Table 2 The locations of the x-part and y-part in the Type I decomposition with respect to ${\mathcal {K}}_{\alpha }$

2.2 The Type II decomposition with respect to the power cone ${\mathcal {K}}_{\alpha }$

In this subsection, we present the Type II decomposition with respect to the power cone ${\mathcal {K}}_{\alpha }$. Similarly, we consider the following four cases.

Case 1:$(z_1,{\bar{z}})\in B_1$.

• (a) ${\bar{z}}_{\min }>0$. In this subcase, $(z_1,{\bar{z}})\in B_{11}$ and $\sigma _{\alpha }({\bar{z}})>0$. Then, we take $x=\ddot{x}^{(B_1,a)}$, $y=\ddot{y}^{(B_1,a)}$ and $s_x=\ddot{s}_x^{(B_1,a)}$, $s_y=\ddot{s}_y^{(B_1,a)}$, where

  $$\begin{aligned} \ddot{x}^{(B_1,a)}:= & {} \left[ \begin{array}{c} 1 \\ \frac{{\bar{z}}}{\sigma _{\alpha }({\bar{z}})} \end{array} \right] \in S_3,\ \ddot{y}^{(B_1,a)}:=\left[ \begin{array}{c} -1 \\ \frac{{\bar{z}}}{\sigma _{\alpha }({\bar{z}})} \end{array} \right] \in S_3. \end{aligned}$$

  (30)
  $$\begin{aligned} \ddot{s}_x^{(B_1,a)}:= & {} \frac{z_1+\sigma _{\alpha }({\bar{z}})}{2},\ \ddot{s}_y^{(B_1,a)}:=\frac{\sigma _{\alpha }({\bar{z}})-z_1}{2}. \end{aligned}$$

  (31)

  Similarly, we can show that the above setting satisfies the decomposition formula (7).

• (b) ${\bar{z}}_{\max }<0$. Similar to the argument in Case 1 (a), $(z_1,{\bar{z}})\in B_{12}$ and $\sigma _{\alpha }(-{\bar{z}})>0$. In this subcase, we set $x=\ddot{x}^{(B_1,b)}$, $y=\ddot{y}^{(B_1,b)}$ and $s_x=\ddot{s}_x^{(B_1,b)}$, $s_y=\ddot{s}_y^{(B_1,b)}$, where

  $$\begin{aligned} \ddot{x}^{(B_1,b)}:= & {} \left[ \begin{array}{c} 1 \\ \frac{-{\bar{z}}}{\sigma _{\alpha }(-{\bar{z}})} \end{array} \right] \in S_3,\ \ddot{y}^{(B_1,b)}:=\left[ \begin{array}{c} -1 \\ \frac{-{\bar{z}}}{\sigma _{\alpha }(-{\bar{z}})} \end{array} \right] \in S_3. \end{aligned}$$

  (32)
  $$\begin{aligned} \ddot{s}_x^{(B_1,b)}:= & {} \frac{z_1-\sigma _{\alpha }(-{\bar{z}})}{2},\ \ddot{s}_y^{(B_1,b)}:=\frac{-\sigma _{\alpha }(-{\bar{z}})-z_1}{2}. \end{aligned}$$

  (33)

• (c) $z_1\ne 0$ and ${\bar{z}}_{\min }={\bar{z}}_{\max }=0$. In this subcase, $(z_1,{\bar{z}})\in B_{13}$ and $\sigma _{\alpha }({\bar{z}})=0$. Thus, we set $x=\ddot{x}^{(B_1,c)}$, $y=\ddot{y}^{(B_1,c)}$ and $s_x=\ddot{s}_x^{(B_1,c)}$, $s_y=\ddot{s}_y^{(B_1,c)}$, where

  $$\begin{aligned} \ddot{x}^{(B_1,c)}:= & {} \left[ \begin{array}{c} 1 \\ \frac{{\mathbf {1}}}{\sigma _{\alpha }({\mathbf {1}})} \end{array} \right] \in S_3,\ \ddot{y}^{(B_1,c)}:=\left[ \begin{array}{c} -1 \\ \frac{{\mathbf {1}}}{\sigma _{\alpha }({\mathbf {1}})} \end{array} \right] \in S_3, \end{aligned}$$

  (34)
  $$\begin{aligned} \ddot{s}_x^{(B_1,c)}= & {} \frac{z_1}{2},\ \ddot{s}_y^{(B_1,c)}=-\frac{z_1}{2}. \end{aligned}$$

  (35)

Case 2:$(z_1,{\bar{z}})\in B_2$.

• (a) ${\bar{z}}_{\min }=0,\ {\bar{z}}_{\max }>0$. In this subcase, $(z_1,{\bar{z}})\in B_{21}$ or $(z_1,{\bar{z}})\in B_{22}$, we set $x=\ddot{x}^{(B_2,a)}$, $y=\ddot{y}^{(B_2,a)}$ and $s_x=1$, $s_y=-1$, where $\ddot{x}^{(B_2,a)}=(\ddot{x}^{(B_2,a)}_1,\ddot{{\bar{x}}}^{(B_2,a)})$ and $\ddot{y}^{(B_2,a)}=(\ddot{y}^{(B_2,a)}_1,\ddot{{\bar{y}}}^{(B_2,a)})$ with

  $$\begin{aligned} \ddot{x}^{(B_2,a)}_1:=z_1,\ddot{{\bar{x}}}^{(B_2,a)}:= & {} \left\{ \begin{array}{ll} \left[ \begin{array}{c} \left( \frac{|z_1|}{{\bar{z}}^{\alpha _2}_2}\right) ^{\frac{1}{\alpha _1}}\\ {\bar{z}}_2 \end{array} \right] &{}\hbox {if}\ (z_1,{\bar{z}})\in B_{21},\\ \left[ \begin{array}{c} {\bar{z}}_1\\ \left( \frac{|z_1|}{{\bar{z}}^{\alpha _1}_1}\right) ^{\frac{1}{\alpha _2}}\\ \end{array} \right] &{}\hbox {if}\ (z_1,{\bar{z}})\in B_{22},\\ \end{array} \right. \end{aligned}$$

  (36)
  $$\begin{aligned} \ddot{y}^{(B_2,a)}_1:=0,\ddot{{\bar{y}}}^{(B_2,a)}:= & {} \left\{ \begin{array}{ll} \left[ \begin{array}{c} \left( \frac{|z_1|}{{\bar{z}}^{\alpha _2}_2}\right) ^{\frac{1}{\alpha _1}}\\ 0 \end{array} \right] &{}\hbox {if}\ (z_1,{\bar{z}})\in B_{21},\\ \left[ \begin{array}{c} 0\\ \left( \frac{|z_1|}{{\bar{z}}^{\alpha _1}_1}\right) ^{\frac{1}{\alpha _2}}\\ \end{array} \right] &{}\hbox {if}\ (z_1,{\bar{z}})\in B_{22}.\\ \end{array} \right. \end{aligned}$$

  (37)

  It is easy to see that

  $$\begin{aligned} \begin{array}{l} \hbox {(a)}\ (z_1,{\bar{z}})\in B_{21}, z_1=0 \Rightarrow x\in S_2, y=0;\quad \hbox {(b)}\ (z_1,{\bar{z}})\in B_{21}, z_1\ne 0 \Rightarrow x\in S_3, y\in S_1;\\ \hbox {(c)}\ (z_1,{\bar{z}})\in B_{22}, z_1=0 \Rightarrow x\in S_1, y=0;\quad \hbox {(d)}\ (z_1,{\bar{z}})\in B_{22}, z_1\ne 0 \Rightarrow x\in S_3, y\in S_2. \end{array} \end{aligned}$$

• (b) ${\bar{z}}_{\min }<0,\ {\bar{z}}_{\max }=0$. In this subcase, $(z_1,{\bar{z}})\in B_{23}$ or $(z_1,{\bar{z}})\in B_{24}$. We set $x=\ddot{x}^{(B_2,b)}$, $y=\ddot{y}^{(B_2,b)}$ and $s_x=-1$, $s_y=1$, where $\ddot{x}^{(B_2,b)}=(\ddot{x}^{(B_2,b)}_1,\ddot{{\bar{x}}}^{(B_2,b)})$ and $\ddot{y}^{(B_2,b)}=(\ddot{y}^{(B_2,b)}_1,\ddot{{\bar{y}}}^{(B_2,b)})$ with

  $$\begin{aligned} \ddot{x}^{(B_2,b)}_1:=-z_1,\ddot{{\bar{x}}}^{(B_2,b)}:= & {} \left\{ \begin{array}{ll} \left[ \begin{array}{c} \left( \frac{|z_1|}{(-{\bar{z}}_2)^{\alpha _2}}\right) ^{\frac{1}{\alpha _1}}\\ -{\bar{z}}_2 \end{array} \right] &{}\hbox {if}\ (z_1,{\bar{z}})\in B_{23},\\ \left[ \begin{array}{c} -{\bar{z}}_1\\ \left( \frac{|z_1|}{(-{\bar{z}}_1)^{\alpha _1}}\right) ^{\frac{1}{\alpha _2}}\\ \end{array} \right] &{}\hbox {if}\ (z_1,{\bar{z}})\in B_{24},\\ \end{array} \right. \end{aligned}$$

  (38)
  $$\begin{aligned} \ddot{y}^{(B_2,b)}_1:=0,\ddot{{\bar{y}}}^{(B_2,b)}:= & {} \left\{ \begin{array}{ll} \left[ \begin{array}{c} \left( \frac{|z_1|}{(-{\bar{z}}_2)^{\alpha _2}}\right) ^{\frac{1}{\alpha _1}}\\ 0\end{array} \right] &{}\hbox {if}\ (z_1,{\bar{z}})\in B_{23},\\ \left[ \begin{array}{c} 0\\ \left( \frac{|z_1|}{(-{\bar{z}}_1)^{\alpha _1}}\right) ^{\frac{1}{\alpha _2}}\\ \end{array} \right] &{}\hbox {if}\ (z_1,{\bar{z}})\in B_{24}.\\ \end{array} \right. \end{aligned}$$

  (39)

  Similar to the arguments in Case 2 (a), we obtain

  $$\begin{aligned} \begin{array}{l} \hbox {(a)}\ (z_1,{\bar{z}})\in B_{23}, z_1=0 \Rightarrow x\in S_2, y=0;\quad \hbox {(b)}\ (z_1,{\bar{z}})\in B_{23}, z_1\ne 0 \Rightarrow x\in S_3, y\in S_1;\\ \hbox {(c)}\ (z_1,{\bar{z}})\in B_{24}, z_1=0 \Rightarrow x\in S_1, y=0;\quad \hbox {(d)}\ (z_1,{\bar{z}})\in B_{24}, z_1\ne 0 \Rightarrow x\in S_3, y\in S_2. \end{array} \end{aligned}$$

Case 3:$(z_1,{\bar{z}})\in B_3$. In this subcase, $(z_1,{\bar{z}})\in B_{31}$ or $(z_1,{\bar{z}})\in B_{32}$. We set $x=\ddot{x}^{(B_3)}$, $y=\ddot{y}^{(B_3)}$ and $s_x=1$, $s_y=-1$, where $\ddot{x}^{(B_3)}=(\ddot{x}^{(B_3)}_1,\ddot{{\bar{x}}}^{(B_3)})$ and $\ddot{y}^{(B_3)}=(\ddot{y}^{(B_3)}_1,\ddot{{\bar{y}}}^{(B_3)})$ with

$$\begin{aligned} \ddot{x}^{(B_3)}_1:=z_1,\ddot{{\bar{x}}}^{(B_3)}:= & {} \left\{ \begin{array}{ll} \left[ \begin{array}{c} \left( \frac{|z_1|}{{\bar{z}}^{\alpha _2}_2}\right) ^{\frac{1}{\alpha _1}}\\ {\bar{z}}_2 \end{array} \right] &{}\hbox {if}\ z\in B_{31},\\ \left[ \begin{array}{c} {\bar{z}}_1\\ \left( \frac{|z_1|}{{\bar{z}}^{\alpha _1}_1}\right) ^{\frac{1}{\alpha _2}}\\ \end{array} \right] &{}\hbox {if}\ z\in B_{32},\\ \end{array} \right. \end{aligned}$$

(40)

$$\begin{aligned} \ddot{y}^{(B_3)}_1:=0,\ddot{{\bar{y}}}^{(B_3)}_j:= & {} \left\{ \begin{array}{ll} \left[ \begin{array}{c} -{\bar{z}}_1+\left( \frac{|z_1|}{{\bar{z}}^{\alpha _2}_2}\right) ^{\frac{1}{\alpha _1}}\\ 0 \end{array} \right] &{}\hbox {if}\ z\in B_{31},\\ \left[ \begin{array}{c} 0\\ -{\bar{z}}_2+\left( \frac{|z_1|}{{\bar{z}}^{\alpha _1}_1} \right) ^{\frac{1}{\alpha _2}}\\ \end{array} \right] &{}\hbox {if}\ z\in B_{32}.\\ \end{array} \right. \end{aligned}$$

(41)

More concretely, we obtain

$$\begin{aligned} \begin{array}{l} \hbox {(a)}\ (z_1,{\bar{z}})\in B_{31}, z_1=0 \Rightarrow x\in S_2, y\in S_1;\quad \hbox {(b)}\ (z_1,{\bar{z}})\in B_{31}, z_1\ne 0 \Rightarrow x\in S_3, y\in S_1;\\ \hbox {(c)}\ (z_1,{\bar{z}})\in B_{32}, z_1=0 \Rightarrow x\in S_1, y\in S_2;\quad \hbox {(d)}\ (z_1,{\bar{z}})\in B_{32}, z_1\ne 0 \Rightarrow x\in S_3, y\in S_2. \end{array} \end{aligned}$$

Case 4:$(z_1,{\bar{z}})\in B_4$. In this subcase, we set $x=\ddot{x}^{(B_4)}\in \partial {\mathcal {K}}_{\alpha }$, $y=\ddot{y}^{(B_4)}\in \partial {\mathcal {K}}_{\alpha }$ and $s_x=1$, $s_y=-1$, where $\ddot{x}^{(B_4)}=(\ddot{x}^{(B_4)}_1,\ddot{{\bar{x}}}^{(B_4)})$ and $\ddot{y}^{(B_4)}=(\ddot{y}^{(B_4)}_1,\ddot{{\bar{y}}}^{(B_4)})$ with

$$\begin{aligned} \begin{array}{ll} \ddot{x}^{(B_4)}:=\left[ \begin{array}{c} 0\\ 1\\ 0\end{array} \right] \in S_1,\quad &{}\ddot{y}^{(B_4)}:=\left[ \begin{array}{c} 0\\ 1\\ 0\end{array} \right] \in S_1, \end{array} \end{aligned}$$

(42)

or

$$\begin{aligned} \begin{array}{ll} \ddot{x}^{(B_4)}:=\left[ \begin{array}{c} 0\\ 0\\ 1\end{array} \right] \in S_2,\quad &{}\ddot{y}^{(B_4)}:=\left[ \begin{array}{c} 0\\ 0\\ 1\end{array} \right] \in S_2. \end{array} \end{aligned}$$

(43)

As mentioned above, the next theorem presents the Type II decomposition with respect to the power cone ${\mathcal {K}}_{\alpha }$.

Theorem 2

For any given $z=(z_1,{\bar{z}})\in {\mathbb {R}}\times {\mathbb {R}}^2$, its Type II decomposition with respect to ${\mathcal {K}}_{\alpha }$ is given by

• (a) If $z\in B_1$, then

  $$\begin{aligned} z=\left\{ \begin{array}{ll} \ddot{s}_x^{(B_1,a)}\cdot \ddot{x}^{(B_1,a)}+\ddot{s}_y^{(B_1,a)}\cdot \ddot{y}^{(B_1,a)},&{}\hbox {if}\ z\in B_{11},\\ \ddot{s}_x^{(B_1,b)}\cdot \ddot{x}^{(B_1,b)}+\ddot{s}_y^{(B_1,b)}\cdot \ddot{y}^{(B_1,b)},&{}\hbox {if}\ z\in B_{12},\\ \ddot{s}_x^{(B_1,c)}\cdot \ddot{x}^{(B_1,c)}+\ddot{s}_y^{(B_1,c)}\cdot \ddot{y}^{(B_1,c)},&{}\hbox {if}\ z\in B_{13}, \end{array} \right. \end{aligned}$$

  where $\ddot{x}^{(B_1,a)}$, $\ddot{y}^{(B_1,a)}$, $\ddot{s}_x^{(B_1,a)}$, $\ddot{s}_y^{(B_1,a)}$ are defined as in (30)–(31), $\ddot{x}^{(B_1,b)}$, $\ddot{y}^{(B_1,b)}$, $\ddot{s}_x^{(B_1,b)}$, $\ddot{s}_y^{(B_1,b)}$ are defined as in (32)–(33) and $\ddot{x}^{(B_1,c)}$, $\ddot{y}^{(B_1,c)}$, $\ddot{s}_x^{(B_1,c)}$, $\ddot{s}_y^{(B_1,c)}$ are defined as in (34)–(35).

• (b) If $z\in B_2$, then

  $$\begin{aligned} z=\left\{ \begin{array}{ll} \ddot{x}^{(B_2,a)}+(-1)\cdot \ddot{y}^{(B_2,a)},&{}\hbox {if}\ z\in B_{21}\ \hbox {or}\ z\in B_{22},\\ (-1)\cdot \ddot{x}^{(B_2,b)}+\ddot{y}^{(B_2,b)},&{}\hbox {if}\ z\in B_{23}\ \hbox {or}\ z\in B_{24},, \end{array} \right. \end{aligned}$$

  where $\ddot{x}^{(B_2,a)}$, $\ddot{y}^{(B_2,a)}$ are defined as (36)–(37), $\ddot{x}^{(B_2,b)}$, $\ddot{y}^{(B_2,b)}$ are defined as in (38)–(39).

• (c) If $z\in B_3$, then $z=\ddot{x}^{(B_3)}+(-1)\cdot \ddot{y}^{(B_3)}$, where $\ddot{x}^{(B_3)}$, $\ddot{y}^{(B_3)}$ are defined as in (40)–(41).

• (d) If $z\in B_4$, then $z=\ddot{x}^{(B_4)}+(-1)\cdot \ddot{y}^{(B_4)}$, where $\ddot{x}^{(B_4)}$ and $\ddot{y}^{(B_4)}$ are defined as in (42) or (43).

In addition, the locations of the x-part and y-part in each case are summarized in Table 3.

Table 3 The locations of the x-part and y-part in the Type II decomposition with respect to ${\mathcal {K}}_{\alpha }$

2.3 Manipulation of a real example

In this subsection, we elaborate more about how to implement the Type I and Type II decomposition with respect to the power cone ${\mathcal {K}}_\alpha$ explicitly by manipulating an example. Without loss of generality, we set the parameters $\alpha _1=\alpha _2=\frac{1}{2}$.

Example 1

The power cone ${\mathcal {K}}_{\frac{1}{2}}$ and its polar cone ${\mathcal {K}}^\circ _{\frac{1}{2}}$ are respectively given by

$$\begin{aligned} \begin{array}{l} {\mathcal {K}}_{\frac{1}{2}}=\left\{ (x_1,{\bar{x}})\in {\mathbb {R}}\times {\mathbb {R}}^2\, \bigg |\,|x_1|\le {\bar{x}}^{\frac{1}{2}}_1{\bar{x}}^{\frac{1}{2}}_2,\ {\bar{x}}_1\ge 0,\ {\bar{x}}_2\ge 0\right\} ,\\ {\mathcal {K}}^\circ _{\frac{1}{2}}=\left\{ (x_1,{\bar{x}})\in {\mathbb {R}}\times {\mathbb {R}}^2\, \bigg |\,|x_1|\le (-2{\bar{x}}_1)^{\frac{1}{2}}(-2{\bar{x}}_2)^{\frac{1}{2}},\ {\bar{x}}_1\le 0,\ {\bar{x}}_2\le 0\right\} . \end{array} \end{aligned}$$

According to the four blocks defined as in (15), we pick different points to figure out their decompositions with respect to ${\mathcal {K}}_{\frac{1}{2}}$. For example, we take $z=(1,-1,-2)^T\in B_{12}$. In this case, $z_1=1, {\bar{z}}=(-1,-2)^T$, $\sigma _{\frac{1}{2}}(-{\bar{z}})=\sqrt{2},\ \eta _{\frac{1}{2}}(-{\bar{z}})=2\sqrt{2}$. From the relations (18)–(19) and (32)–(33), we obtain

$$\begin{aligned} {\dot{x}}^{(B_1,b)}:= & {} \left[ \begin{array}{c} 1 \\ \frac{-{\bar{z}}}{\sigma _{\frac{1}{2}}(-{\bar{z}})} \end{array} \right] =\left[ \begin{array}{c}1\\ \frac{1}{\sqrt{2}}\\ \frac{2}{\sqrt{2}}\end{array}\right] \in S_3,\quad {\dot{y}}^{(B_1,b)}:=\left[ \begin{array}{c} 1 \\ \frac{{\bar{z}}}{\eta _{\frac{1}{2}}(-{\bar{z}})} \end{array} \right] =\left[ \begin{array}{c}1\\ -\frac{1}{2\sqrt{2}}\\ -\frac{1}{\sqrt{2}}\end{array}\right] \in T_3,\\ {\dot{s}}_x^{(B_1,b)}:= & {} \frac{z_1-\eta _{\frac{1}{2}}(-{\bar{z}})}{\sigma _{\frac{1}{2}}(-{\bar{z}})+\eta _{\frac{1}{2}}(-{\bar{z}})} \cdot \sigma _{\frac{1}{2}}(-{\bar{z}})=\frac{1-2\sqrt{2}}{\sqrt{2}+2\sqrt{2}}\cdot \sqrt{2}=\frac{1-2\sqrt{2}}{3},\\ {\dot{s}}_y^{(B_1,b)}:= & {} \frac{z_1+\sigma _{\frac{1}{2}}(-{\bar{z}})}{\sigma _{\frac{1}{2}}(-{\bar{z}})+\eta _{\frac{1}{2}}(-{\bar{z}})} \cdot \eta _{\frac{1}{2}}(-{\bar{z}})=\frac{1+\sqrt{2}}{\sqrt{2}+2\sqrt{2}}\cdot 2\sqrt{2}=\frac{2(1+\sqrt{2})}{3}.\\ \ddot{x}^{(B_1,b)}:= & {} \left[ \begin{array}{c} 1 \\ \frac{-{\bar{z}}}{\sigma _{\frac{1}{2}}(-{\bar{z}})} \end{array} \right] =\left[ \begin{array}{c}1\\ \frac{1}{\sqrt{2}}\\ \frac{2}{\sqrt{2}}\end{array}\right] \in S_3,\quad \ddot{y}^{(B_1,b)}:=\left[ \begin{array}{c} -1 \\ \frac{-{\bar{z}}}{\sigma _{\frac{1}{2}}(-{\bar{z}})} \end{array} \right] =\left[ \begin{array}{c}-1\\ \frac{1}{\sqrt{2}}\\ \frac{2}{\sqrt{2}}\end{array}\right] \in S_3,\\ \ddot{s}_x^{(B_1,b)}:= & {} \frac{z_1-\sigma _{\frac{1}{2}}(-{\bar{z}})}{2} =\frac{1-\sqrt{2}}{2},\quad \ddot{s}_y^{(B_1,b)}:=\frac{-\sigma _{\frac{1}{2}}(-{\bar{z}})-z_1}{2} =\frac{-\sqrt{2}-1}{2}. \end{aligned}$$

Therefore, the corresponding two types of decompositions with respect to ${\mathcal {K}}_{\frac{1}{2}}$ are respectively given by

$$\begin{aligned} \begin{array}{ll} \hbox {Type I:} \qquad &{}\left[ \begin{array}{c}1\\ -1\\ -2\end{array}\right] =\frac{1-2\sqrt{2}}{3}\left[ \begin{array}{c}1\\ \frac{1}{\sqrt{2}}\\ \frac{2}{\sqrt{2}}\end{array}\right] +\frac{2(1+\sqrt{2})}{3} \cdot \left[ \begin{array}{c}1\\ -\frac{1}{2\sqrt{2}}\\ -\frac{1}{\sqrt{2}}\end{array}\right] ,\\ \\ \hbox {Type II:}\qquad &{}\left[ \begin{array}{c}1\\ -1\\ -2\end{array}\right] =\frac{1-\sqrt{2}}{2}\left[ \begin{array}{c}1\\ \frac{1}{\sqrt{2}}\\ \frac{2}{\sqrt{2}}\end{array}\right] +\frac{-\sqrt{2}-1}{2} \cdot \left[ \begin{array}{c}-1\\ \frac{1}{\sqrt{2}}\\ \frac{2}{\sqrt{2}}\end{array}\right] , \end{array} \end{aligned}$$

whose graphs are depicted in Figs. 4 and 5 , respectively. The other cases for testing the decompositions with respect to ${\mathcal {K}}_{\frac{1}{2}}$ can be seen in Table 4, in which $x_{loc}, y_{loc}, z_{loc}$ denote the locations of x, y, z, respectively.

Fig. 4

The Type I decomposition for Example 1

Fig. 5

The Type II decomposition for Example 1

Table 4 Examples of two types of decompositions with respect to ${\mathcal {K}}_{\frac{1}{2}}$

Remark 2

As shown in Example 1, these two types of decompositions for any given nonzero vectors with respect to the power cone ${\mathcal {K}}_\alpha$ are easy to implement, which is a new feature to the progress of this core non-symmetric cone and plays a crucial role in continuing subsequent study on this topic, for instance generating conic functions like the SOC-function [8] and Löwner’s operator for the positive semidefinite cone [37, 38] as mentioned above. Moreover, through comparing the above two types of decompositions established in Theorem 1 and 2 , we rewrite them as follows:

$$\begin{aligned} \begin{array}{ll} \hbox {Type I:} \qquad &{}z=s^I_x\cdot x^I+s^I_y\cdot y^I.\\ \hbox {Type II:} \qquad &{}z=s^{II}_x\cdot x^{II}+s^{II}_y\cdot y^{II}. \end{array} \end{aligned}$$

It is easy to see that if $\eta _{\alpha }(z)=\sigma _{\alpha }(z)$, then $x^I=x^{II}$, $y^I=-y^{II}$, $s^I_x=s^{II}_x$ and $s^I_y=-s^{II}_y$, where $\eta _{\alpha },\sigma _{\alpha }$ are defined as in (11). On the other hand, we also find that the $s_x$-part and $s_y$-part of the Type I decomposition are more complicated than the Type II counterpart in general. Therefore, we prefer the Type II decomposition with respect to ${\mathcal {K}}_\alpha$ for further studies, see Sect. 4 for more details.

3 The decompositions with respect to the exponential cone ${\mathcal {K}}_{\exp }$

In this section, we present two types of decompositions with respect to the exponential cone ${\mathcal {K}}_{\exp }$. Again, we also present its analytic properties. Due to similar procedures as Sect. 2, we omit their proofs and only list some results. For the dual of the exponential cone ${\mathcal {K}}_{\exp }$, we refer the readers to [5, Theorem 4.3.3] for its verification.

Lemma 5

${\mathcal {K}}_{\exp }$ is a closed convex cone.

Lemma 6

The dual cone ${\mathcal {K}}^*_{\exp }$ can be described as

$$\begin{aligned} {\mathcal {K}}^*_{\exp }:=\hbox {cl}\left\{ (x_1,{\bar{x}}) \in {\mathbb {R}}\times {\mathbb {R}}^2\, \bigg |\,x_1\ge -\frac{{\bar{x}}_1}{e}\cdot \hbox {exp}\left( \frac{{\bar{x}}_2}{{\bar{x}}_1}\right) ,\ {\bar{x}}_1<0,\ x_1\ge 0\right\} . \end{aligned}$$

Correspondingly, the polar ${\mathcal {K}}^\circ _{\exp }$ is given by

$$\begin{aligned} {\mathcal {K}}^\circ _{\exp }:=\hbox {cl}\left\{ (x_1,{\bar{x}}) \in {\mathbb {R}}\times {\mathbb {R}}^2\, \bigg |\,x_1\le -\frac{{\bar{x}}_1}{e}\cdot \hbox {exp} \left( \frac{{\bar{x}}_2}{{\bar{x}}_1}\right) ,\ {\bar{x}}_1>0,\ x_1\le 0\right\} . \end{aligned}$$

Similar to Lemma 4, we also define the following sets

$$\begin{aligned} \begin{array}{l} {\hat{S}}_1:=\left\{ (x_1,{\bar{x}})\in {\mathbb {R}}\times {\mathbb {R}}^2\,|\,\ x_1>0,\ {\bar{x}}_1<0,\ {\bar{x}}_2=0\right\} ,\\ {\hat{S}}_2:=\left\{ (x_1,{\bar{x}})\in {\mathbb {R}}\times {\mathbb {R}}^2\,|\,\ x_1=0,\ {\bar{x}}_1<0,\ {\bar{x}}_2=0\right\} ,\\ {\hat{S}}_3:=\left\{ (x_1,{\bar{x}})\in {\mathbb {R}}\times {\mathbb {R}}^2\,|\,\ x_1>0,\ {\bar{x}}_1=0,\ {\bar{x}}_2=0\right\} ,\\ {\hat{S}}_4:=\left\{ (x_1,{\bar{x}})\in {\mathbb {R}}\times {\mathbb {R}}^2\,|\,\ x_1\ge 0,\ \sigma _{\exp }({\bar{x}})=x_1,\ {\bar{x}}_2>0\right\} ,\\ {\hat{S}}_5:=\left\{ (x_1,{\bar{x}})\in {\mathbb {R}}\times {\mathbb {R}}^2\,|\,\ x_1>0,\ {\bar{x}}_1=0,\ {\bar{x}}_2>0\right\} ,\\ {\hat{S}}_6:=\left\{ (x_1,{\bar{x}})\in {\mathbb {R}}\times {\mathbb {R}}^2\,|\,\ x_1=0,\ {\bar{x}}_1=0,\ {\bar{x}}_2>0\right\} ,\\ {\hat{S}}_7:=\left\{ (x_1,{\bar{x}})\in {\mathbb {R}}\times {\mathbb {R}}^2\,|\,\ x_1>0,\ {\bar{x}}_1=0,\ {\bar{x}}_2=0\right\} ,\\ {\hat{S}}_8:=\left\{ (x_1,{\bar{x}})\in {\mathbb {R}}\times {\mathbb {R}}^2\,|\,\ x_1\ge 0,\ {\bar{x}}_1<0,\ -\eta _{\exp }({\bar{x}})=x_1\right\} ,\\ {\hat{T}}_1:=\left\{ (x_1,{\bar{x}})\in {\mathbb {R}}\times {\mathbb {R}}^2\,|\,\ x_1<0,\ {\bar{x}}_1=0,\ {\bar{x}}_2<0\right\} ,\\ {\hat{T}}_2:=\left\{ (x_1,{\bar{x}})\in {\mathbb {R}}\times {\mathbb {R}}^2\,|\,\ x_1=0,\ {\bar{x}}_1=0,\ {\bar{x}}_2<0\right\} ,\\ {\hat{T}}_3:=\left\{ (x_1,{\bar{x}})\in {\mathbb {R}}\times {\mathbb {R}}^2\,|\,\ x_1<0,\ {\bar{x}}_1=0,\ {\bar{x}}_2=0\right\} ,\\ {\hat{T}}_4:=\left\{ (x_1,{\bar{x}})\in {\mathbb {R}}\times {\mathbb {R}}^2\,|\,\ x_1\le 0,\ {\bar{x}}_1>0,\ -\eta _{\exp }({\bar{x}})=x_1\right\} . \end{array} \end{aligned}$$

(44)

Consequently, the boundary of ${\mathcal {K}}_{\exp }$ and ${\mathcal {K}}^\circ _{\exp }$ can be described in a more compact form.

Lemma 7

The boundary of ${\mathcal {K}}_{\exp }$ and ${\mathcal {K}}^*_{\exp }$, denoted by $\partial {\mathcal {K}}_{\exp }$ and $\partial {\mathcal {K}}^*_{\exp }$, are respectively given by

$$\begin{aligned} \partial {\mathcal {K}}_{\exp }:={\hat{S}}_1\cup {\hat{S}}_2\cup {\hat{S}}_3\cup {\hat{S}}_4\cup \{0\},\quad \partial {\mathcal {K}}^*_{\exp }:={\hat{S}}_5\cup {\hat{S}}_6\cup {\hat{S}}_7\cup {\hat{S}}_8\cup \{0\}, \end{aligned}$$

where

$$\begin{aligned} \sigma _{\exp }({\bar{x}}):={\bar{x}}_2\cdot \exp \left( \frac{{\bar{x}}_1}{{\bar{x}}_2}\right) ,\quad \eta _{\exp }({\bar{x}}):=\frac{{\bar{x}}_1}{e}\cdot \exp \left( \frac{{\bar{x}}_2}{{\bar{x}}_1}\right) . \end{aligned}$$

(45)

Similarly, the boundary of ${\mathcal {K}}^\circ _{\exp }$ can be formulated as

$$\begin{aligned} \partial {\mathcal {K}}^\circ _{\exp }:={\hat{T}}_1\cup {\hat{T}}_2\cup {\hat{T}}_3\cup {\hat{T}}_4\cup \{0\}. \end{aligned}$$

Remark 3

Similar to Remark 1, the set ${\mathcal {K}}_{\exp }\cup {\mathcal {K}}^\circ _{\exp }$ can also be divided into the following nine parts

$$\begin{aligned} {\mathcal {K}}_{\exp }\cup {\mathcal {K}}^\circ _{\exp }={\hat{S}}_1\cup {\hat{S}}_2\cup {\hat{S}}_3\cup {\hat{T}}_1 \cup {\hat{T}}_2\cup {\hat{T}}_3\cup {\hat{P}}_1\cup {\hat{P}}_2\cup \{0\}, \end{aligned}$$

where

$$\begin{aligned} \begin{array}{l} {\hat{P}}_1:=\left\{ (x_1,{\bar{x}})\in {\mathbb {R}}\times {\mathbb {R}}^2\,|\,\ x_1 \ge 0,\ \sigma _{\exp }({\bar{x}})\le x_1,\ {\bar{x}}_2>0\right\} ,\\ {\hat{P}}_2:=\left\{ (x_1,{\bar{x}})\in {\mathbb {R}}\times {\mathbb {R}}^2\,|\,\ x_1 \le 0,\ {\bar{x}}_1>0,\ -\eta _{\exp }({\bar{x}})\ge x_1\right\} . \end{array} \end{aligned}$$

Fig. 6

The different parts of $\partial {\mathcal {K}}_{\exp }$ (left) and $\partial {\mathcal {K}}^\circ _{\exp }$ (right)

In addition, the boundary of ${\mathcal {K}}_{\exp }$ and its polar ${\mathcal {K}}^\circ _{\exp }$ are depicted in Fig. 6.

3.1 The Type I decomposition with respect to the power cone ${\mathcal {K}}_{\exp }$

In this subsection, we present the Type I decomposition with respect to the exponential cone ${\mathcal {K}}_{\exp }$, in which we divide the space ${\mathbb {R}}\times {\mathbb {R}}^2$ into the following four blocks:

$$\begin{aligned} \begin{array}{ll} {\hbox {Block I}:}&{}\ {\tilde{B}}_1:=\left\{ (z_1,{\bar{z}})\in {\mathbb {R}}\times {\mathbb {R}}^2 \, | \,{\bar{z}}_1\cdot {\bar{z}}_2>0\ \hbox {or}\ (z_1\ne 0\ \hbox {and}\ {\bar{z}}=0)\right\} .\\ {\hbox {Block II}:}&{}\ {\tilde{B}}_2:=\left\{ (z_1,{\bar{z}})\in {\mathbb {R}}\times {\mathbb {R}}^2 \, | \,({\bar{z}}_1=0,\ {\bar{z}}_2\ne 0)\ \hbox {or}\ ({\bar{z}}_1<0,\ {\bar{z}}_2> 0)\right\} .\\ {\hbox {Block III}:}&{}\ {\tilde{B}}_3:=\left\{ (z_1,{\bar{z}})\in {\mathbb {R}}\times {\mathbb {R}}^2 \, | \,({\bar{z}}_1\ne 0,\ {\bar{z}}_2=0)\ \ \hbox {or}\ ({\bar{z}}_1>0,\ {\bar{z}}_2< 0)\right\} .\\ {\hbox {Block IV}:}&{}\ {\tilde{B}}_4:=\left\{ (z_1,{\bar{z}})\in {\mathbb {R}}\times {\mathbb {R}}^2 \, | \, z_1=0\ \hbox {and}\ {\bar{z}}=0\right\} . \end{array} \end{aligned}$$

(46)

Their subcases with respect to ${\mathcal {K}}_{\exp }$ for the Type I decomposition can be found in Table 5.

Similar to Theorem 1, we now present the Type I decomposition with respect to ${\mathcal {K}}_{\exp }$.

Theorem 3

For any given $z=(z_1,{\bar{z}})\in {\mathbb {R}}\times {\mathbb {R}}^2$, its Type I decomposition with respect to ${\mathcal {K}}_{\exp }$ is given by

• (a) If $z\in {\tilde{B}}_1$, then

  $$\begin{aligned} z=\left\{ \begin{array}{ll} \displaystyle \frac{(z_1 +\eta _{\exp }({\bar{z}}))\cdot \sigma _{\exp }({\bar{z}})}{\sigma _{\exp }({\bar{z}}) +\eta _{\exp }({\bar{z}})}\cdot \left[ \begin{array}{c} 1 \\ \frac{{\bar{z}}}{\sigma _{\exp }({\bar{z}})} \end{array} \right] \\ \quad +\frac{(\sigma _{\exp }({\bar{z}})-z_1)\cdot \eta _{\exp } ({\bar{z}})}{\sigma _{\exp }({\bar{z}})+\eta _{\exp }({\bar{z}})}\cdot \left[ \begin{array}{c} -1 \\ \frac{{\bar{z}}}{\eta _{\exp }({\bar{z}})} \end{array} \right] ,\quad \hbox {if}\ z\in {\tilde{B}}_{11}\ \hbox {or}\ z\in {\tilde{B}}_{12}, \\ \\ \displaystyle \frac{z_1\cdot \sigma _{\exp }({\mathbf {1}})}{\sigma _{\exp }({\mathbf {1}}) +\eta _{\exp }({\mathbf {1}})}\cdot \left[ \begin{array}{c} 1 \\ \frac{{\mathbf {1}}}{\sigma _{\exp }({\mathbf {1}})} \end{array} \right] +\frac{-z_1\cdot \eta _{\exp }({\mathbf {1}})}{\sigma _{\exp }({\mathbf {1}}) +\eta _{\exp }({\mathbf {1}})}\cdot \left[ \begin{array}{c} -1 \\ \frac{{\mathbf {1}}}{\eta _{\exp }({\mathbf {1}})} \end{array} \right] , \quad \hbox {if}\ z\in {\tilde{B}}_{13}, \end{array} \right. \end{aligned}$$

  where ${\mathbf {1}}:=(1,1)^T\in {\mathbb {R}}^2$ and $\sigma _{\exp }({\bar{z}}),\eta _{\exp }({\bar{z}})$ are defined as in (45).

• (b) If $z\in {\tilde{B}}_2$, then

  $$\begin{aligned} z=\sigma _{\exp }({\bar{z}})\cdot \left[ \begin{array}{c} 1 \\ \frac{{\bar{z}}}{\sigma _{\exp }({\bar{z}})} \end{array} \right] +\hbox {sgn}(\sigma _{\exp }({\bar{x}})-z_1)\cdot \left[ \begin{array}{c} -|z_1-\sigma _{\exp }({\bar{z}})| \\ 0 \end{array} \right] , \end{aligned}$$

  where $\hbox {sgn}(t)$ denotes the sign of the variable $t\in {\mathbb {R}}$.

• (c) If $z\in {\tilde{B}}_3$, then

  $$\begin{aligned} z=\hbox {sgn}(z_1+\eta _{\exp }({\bar{z}}))\cdot \left[ \begin{array}{c} |z_1+\eta _{\exp }({\bar{z}})| \\ 0 \end{array} \right] +\eta _{\exp }({\bar{z}})\cdot \left[ \begin{array}{c} -1 \\ \frac{{\bar{z}}}{\eta _{\exp }({\bar{z}})} \end{array} \right] . \end{aligned}$$

• (d) If $z\in {\tilde{B}}_4$, then

  $$\begin{aligned} z=1\cdot \left[ \begin{array}{c} \max \{0,w\} \\ 0 \end{array} \right] +1\cdot \left[ \begin{array}{c} \min \{0,-w\} \\ 0 \end{array} \right] , \end{aligned}$$

  where w is any scalar in ${\mathbb {R}}$.

Table 5 The subcases of each block in (46) with respect to ${\mathcal {K}}_{\exp }$ for the Type I decomposition

In addition, the locations of the x-part and y-part in each case are summarized in Table 6, where ${\tilde{S}}_i,{\tilde{T}}_i\ (i=1,2,3,4)$ are defined as in (44).

Table 6 The locations of the x-part and y-part in the Type I decomposition with respect to ${\mathcal {K}}_{\exp }$

3.2 The Type II decomposition with respect to the power cone ${\mathcal {K}}_{\exp }$

In this subsection, we present the Type II decomposition of the power cone ${\mathcal {K}}_{\exp }$. By contrast with the Type I case, we present a new space division for ${\mathbb {R}}\times {\mathbb {R}}^2$ as follows:

$$\begin{aligned} \begin{array}{ll} {\hbox {Block I}:}&{}\ {\bar{B}}_1:=\left\{ (z_1,{\bar{z}})\in {\mathbb {R}}\times {\mathbb {R}}^2 \, | \,{\bar{z}}_2\ne 0\right\} .\\ {\hbox {Block II}:}&{}\ {\bar{B}}_2:=\left\{ (z_1,{\bar{z}})\in {\mathbb {R}}\times {\mathbb {R}}^2 \, | \,(z_1\ne 0, {\bar{z}}=0)\ \hbox {or}\ ({\bar{z}}_1<0,\ {\bar{z}}_2=0)\right\} .\\ {\hbox {Block III}:}&{}\ {\bar{B}}_3:=\left\{ (z_1,{\bar{z}})\in {\mathbb {R}}\times {\mathbb {R}}^2 \, | \, {\bar{z}}_1>0,\ {\bar{z}}_2=0\right\} .\\ {\hbox {Block IV}:}&{}\ {\bar{B}}_4:=\left\{ (z_1,{\bar{z}})\in {\mathbb {R}}\times {\mathbb {R}}^2 \, | \, z_1=0\ \hbox {and}\ {\bar{z}}=0\right\} . \end{array} \end{aligned}$$

(47)

Table 7 indicates their subcases of these blocks with respect to ${\mathcal {K}}_{\exp }$ for the Type II decomposition.

Table 7 The subcases of each block in (47) with respect to ${\mathcal {K}}_{\exp }$ for the Type II decomposition

Similar to Theorem 2, the next theorem presents the Type II decomposition with respect to ${\mathcal {K}}_{\exp }$.

Theorem 4

For any given $z=(z_1,{\bar{z}})\in {\mathbb {R}}\times {\mathbb {R}}^2$, its Type II decomposition with respect to ${\mathcal {K}}_{\exp }$ is given by

• (a) If $z\in {\bar{B}}_1$, then

  $$\begin{aligned} z=\sigma _{\exp }({\bar{z}})\cdot \left[ \begin{array}{c} 1 \\ \frac{{\bar{z}}}{\sigma _{\exp }({\bar{z}})} \end{array} \right] +\hbox {sgn}(z_1-\sigma _{\exp }({\bar{z}}))\cdot \left[ \begin{array}{c} |z_1-\sigma _{\exp }({\bar{z}})| \\ 0 \end{array} \right] , \end{aligned}$$

  where $\hbox {sgn}(t)$ denotes the sign of the variable $t\in {\mathbb {R}}$.

• (b) If $z\in {\bar{B}}_2$, then

  $$\begin{aligned} z=1\cdot \left[ \begin{array}{c} \max \{0,z_1\} \\ {\bar{z}}\end{array} \right] +(-1)\cdot \left[ \begin{array}{c} -\min \{0,z_1\} \\ 0 \end{array} \right] . \end{aligned}$$

• (c) If $z\in {\bar{B}}_3$, then

  $$\begin{aligned} z=1\cdot \left[ \begin{array}{c} \max \{0,z_1\} \\ 0 \end{array} \right] +(-1)\cdot \left[ \begin{array}{c} -\min \{0,z_1\} \\ -{\bar{z}} \end{array} \right] . \end{aligned}$$

• (d) If $z\in {\bar{B}}_4$, then

  $$\begin{aligned} z=1\cdot \left[ \begin{array}{c} \max \{0,w\} \\ 0 \end{array} \right] +(-1)\cdot \left[ \begin{array}{c} -\min \{0,-w\} \\ 0 \end{array} \right] , \end{aligned}$$

  where w is any scalar in ${\mathbb {R}}$.

In addition, the locations of the x-part and y-part in each case are summarized in Table 8.

Table 8 The locations of the x-part and y-part in the Type II decomposition with respect to ${\mathcal {K}}_{\exp }$

Remark 4

Similar to the power cone ${\mathcal {K}}_\alpha$ case discussed in Sect. 2.3, Theorems 3 and 4 also show that our decompositions with respect to the exponential cone ${\mathcal {K}}_{\exp }$ are easy to calculate. Implementing a real example is routine, we do not repeat it again there. On the other hand, different from the power cone case, the $s_x$-part and $s_y$-part of the Type I decomposition with respect to ${\mathcal {K}}_{\exp }$ seems to be more regular than the Type II counterpart in general, due to the appearance of the “wall” part in Fig. 6 (see ${\hat{S}}_1\cup {\hat{S}}_2\cup {\hat{S}}_3\cup \{0\}$). Therefore, we suggest to use the Type I decomposition with respect to ${\mathcal {K}}_{\exp }$ in the sequential studies.

4 Applications

In this section, we discuss some applications of these decompositions with respect to the power cone ${\mathcal {K}}_{\alpha }$ and the exponential cone ${\mathcal {K}}_{\exp }$.

4.1 Conic functions

As mentioned before, an important application of the decomposition with respect to the given cone is to establish its associated conic function. In this subsection, we focus on the conic functions for the power cone ${\mathcal {K}}_{\alpha }$ and the exponential cone ${\mathcal {K}}_{\exp }$.

According to Theorem 2 and Remark 3, the conic function with respect to the power cone ${\mathcal {K}}_{\alpha }$ is defined in the following form.

Definition 1

For any given $z=(z_1,{\bar{z}})\in {\mathbb {R}}\times {\mathbb {R}}^2$, let f be a scalar function defined in ${\mathbb {R}}$ and $f^{power}$ be the conic function with respect to the power cone ${\mathcal {K}}_{\alpha }$. Denote ${\mathbf {1}}:=(1,1)^T\in {\mathbb {R}}^2$ and the space division of ${\mathbb {R}}\times {\mathbb {R}}^2$ is defined as in (15). Then, we have

• (a) If $z\in B_1$, then

  $$\begin{aligned} f^{power}(z):=\left\{ \begin{array}{ll} f\left( \frac{z_1+\sigma _{\alpha }({\bar{z}})}{2}\right) \cdot \left[ \begin{array}{c} 1\\ \frac{{\bar{z}}_1}{\sigma _{\alpha }({\bar{z}})}\\ \frac{{\bar{z}}_2}{\sigma _{\alpha }({\bar{z}})}\end{array} \right] +f\left( \frac{\sigma _{\alpha }({\bar{z}})-z_1}{2} \right) \cdot \left[ \begin{array}{c} -1\\ \frac{{\bar{z}}_1}{\sigma _{\alpha }({\bar{z}})}\\ \frac{{\bar{z}}_2}{\sigma _{\alpha }({\bar{z}})}\end{array}\right] &{}\quad \hbox {if}\ z\in B_{11},\\ \\ f\left( \frac{z_1-\sigma _{\alpha }(-{\bar{z}})}{2}\right) \cdot \left[ \begin{array}{c} 1\\ \frac{-{\bar{z}}_1}{\sigma _{\alpha }(-{\bar{z}})}\\ \frac{-{\bar{z}}_2}{\sigma _{\alpha }(-{\bar{z}})}\end{array} \right] +f\left( \frac{-\sigma _{\alpha }({\bar{z}})-z_1}{2} \right) \cdot \left[ \begin{array}{c} -1\\ \frac{-{\bar{z}}_1}{\sigma _{\alpha }({\bar{z}})}\\ \frac{-{\bar{z}}_2}{\sigma _{\alpha }({\bar{z}})}\end{array} \right] &{}\quad \hbox {if}\ z\in B_{12},\\ \\ f\left( \frac{z_1}{2}\right) \cdot \left[ \begin{array}{c} 1\\ \frac{1}{\sigma _{\alpha }({\mathbf {1}})}\\ \frac{1}{\sigma _{\alpha }({\mathbf {1}})}\end{array} \right] +f\left( \frac{-z_1}{2}\right) \cdot \left[ \begin{array}{c} -1\\ \frac{1}{\sigma _{\alpha }({\mathbf {1}})}\\ \frac{1}{\sigma _{\alpha }({\mathbf {1}})}\end{array}\right]&\quad \hbox {if}\ z\in B_{13}, \end{array} \right. \end{aligned}$$

  where $\sigma _{\alpha }({\bar{z}})$ is defined as in (11).

• (b) If $z\in B_2$, then

  $$\begin{aligned} f^{power}(z):=\left\{ \begin{array}{ll} f(1)\cdot \left[ \begin{array}{c} z_1\\ \left( \frac{|z_1|}{{\bar{z}}^{\alpha _2}_2}\right) ^{\frac{1}{\alpha _1}}\\ {\bar{z}}_2\end{array}\right] +f(-1)\cdot \left[ \begin{array}{c} 0\\ \left( \frac{|z_1|}{{\bar{z}}^{\alpha _2}_2}\right) ^{\frac{1}{\alpha _1}}\\ 0\end{array}\right] &{}\quad \hbox {if}\ z\in B_{21},\\ \\ f(1)\cdot \left[ \begin{array}{c} z_1\\ {\bar{z}}_1\\ \left( \frac{|z_1|}{{\bar{z}}^{\alpha _1}_1}\right) ^{\frac{1}{\alpha _2}} \end{array}\right] +f(-1)\cdot \left[ \begin{array}{c} 0\\ 0\\ \left( \frac{|z_1|}{{\bar{z}}^{\alpha _1}_1}\right) ^{\frac{1}{\alpha _2}}\end{array}\right] &{}\quad \hbox {if}\ z\in B_{22},\\ \\ f(-1)\cdot \left[ \begin{array}{c} -z_1\\ \left( \frac{|z_1|}{(-{\bar{z}}_2)^{\alpha _2}}\right) ^{\frac{1}{\alpha _1}}\\ -{\bar{z}}_2\end{array}\right] +f(-1)\cdot \left[ \begin{array}{c} 0\\ \left( \frac{|z_1|}{(-{\bar{z}}_2)^{\alpha _2}}\right) ^{\frac{1}{\alpha _1}}\\ 0\end{array}\right] &{}\quad \hbox {if}\ z\in B_{23},\\ \\ f(-1)\cdot \left[ \begin{array}{c} -z_1\\ -{\bar{z}}_1\\ \left( \frac{|z_1|}{(-{\bar{z}}_1)^{\alpha _1}}\right) ^{\frac{1}{\alpha _2}} \end{array}\right] +f(-1)\cdot \left[ \begin{array}{c} 0\\ 0\\ \left( \frac{|z_1|}{(-{\bar{z}}_1)^{\alpha _1}}\right) ^{\frac{1}{\alpha _2}}\end{array}\right]&\quad \hbox {if}\ z\in B_{24}. \end{array} \right. \end{aligned}$$

• (c) If $z\in B_3$, then

  $$\begin{aligned} f^{power}(z):=\left\{ \begin{array}{ll} f(1)\cdot \left[ \begin{array}{c} z_1\\ \left( \frac{|z_1|}{{\bar{z}}^{\alpha _2}_2}\right) ^{\frac{1}{\alpha _1}}\\ {\bar{z}}_2\end{array}\right] +f(-1)\cdot \left[ \begin{array}{c} 0\\ -{\bar{z}}_1+\left( \frac{|z_1|}{{\bar{z}}^{\alpha _2}_2}\right) ^{\frac{1}{\alpha _1}}\\ 0\end{array}\right] &{}\quad \hbox {if}\ z\in B_{31},\\ \\ f(1)\cdot \left[ \begin{array}{c} z_1\\ {\bar{z}}_1\\ \left( \frac{|z_1|}{{\bar{z}}^{\alpha _1}_1}\right) ^{\frac{1}{\alpha _2}}\end{array}\right] +f(-1)\cdot \left[ \begin{array}{c} 0\\ 0\\ -{\bar{z}}_2+\left( \frac{|z_1|}{{\bar{z}}^{\alpha _1}_1} \right) ^{\frac{1}{\alpha _2}}\end{array}\right]&\quad \hbox {if}\ z\in B_{32}. \end{array} \right. \end{aligned}$$

• (d) If $z\in B_4$, then

  $$\begin{aligned}&f^{power}(z):=f(1)\cdot \left[ \begin{array}{c} 0\\ 1\\ 0\end{array}\right] +f(-1)\cdot \left[ \begin{array}{c} 0\\ -1\\ 0\end{array}\right] \quad \hbox {or}\\&f^{power}(z):=f(1)\cdot \left[ \begin{array}{c} 0\\ 0\\ 1\end{array}\right] +f(-1)\cdot \left[ \begin{array}{c} 0\\ 0\\ -1\end{array}\right] . \end{aligned}$$

Similarly, according to Theorem 3 and Remark 4, the conic function with respect to the exponential cone ${\mathcal {K}}_{\exp }$ has the following explicit description.

Definition 2

For any given $z=(z_1,{\bar{z}})\in {\mathbb {R}}\times {\mathbb {R}}^2$, let f be a scalar function defined in ${\mathbb {R}}$ and $f^{exp}$ be the conic function with respect to the exponential cone ${\mathcal {K}}_{\exp }$, the space division of ${\mathbb {R}}\times {\mathbb {R}}^2$ is defined as in (46). Then, we have

• (a) If $z\in {\tilde{B}}_1$, then

  $$\begin{aligned}&f^{exp}(z)\\&\quad :=\left\{ \begin{array}{ll} f\left( \frac{(z_1 +\eta _{\exp }({\bar{z}}))\cdot \sigma _{\exp }({\bar{z}})}{\sigma _{\exp }({\bar{z}}) +\eta _{\exp }({\bar{z}})}\right) \cdot \left[ \begin{array}{c} 1 \\ \frac{{\bar{z}}}{\sigma _{\exp }({\bar{z}})} \end{array} \right] +f\left( \frac{(\sigma _{\exp }({\bar{z}})-z_1)\cdot \eta _{\exp } ({\bar{z}})}{\sigma _{\exp }({\bar{z}})+\eta _{\exp }({\bar{z}})}\right) \cdot \left[ \begin{array}{c} -1 \\ \frac{{\bar{z}}}{\eta _{\exp }({\bar{z}})} \end{array} \right] , &{}\hbox {if}\ z\in {\tilde{B}}_{11}\cup {\tilde{B}}_{12}, \\ \\ f\left( \frac{z_1\cdot \sigma _{\exp }({\mathbf {1}})}{\sigma _{\exp }({\mathbf {1}}) +\eta _{\exp }({\mathbf {1}})}\right) \cdot \left[ \begin{array}{c} 1 \\ \frac{{\mathbf {1}}}{\sigma _{\exp }({\mathbf {1}})} \end{array} \right] +f\left( \frac{-z_1\cdot \eta _{\exp }({\mathbf {1}})}{\sigma _{\exp }({\mathbf {1}}) +\eta _{\exp }({\mathbf {1}})}\right) \cdot \left[ \begin{array}{c} -1 \\ \frac{{\mathbf {1}}}{\eta _{\exp }({\mathbf {1}})} \end{array} \right] ,&\hbox {if}\ z\in {\tilde{B}}_{13}, \end{array} \right. \end{aligned}$$

  where $\sigma _{\exp }({\bar{z}}),\eta _{\exp }({\bar{z}})$ are defined as in (45).

• (b) If $z\in {\tilde{B}}_2$, then

  $$\begin{aligned} f^{exp}(z):=f(\sigma _{\exp }({\bar{z}}))\cdot \left[ \begin{array}{c} 1 \\ \frac{{\bar{z}}}{\sigma _{\exp }({\bar{z}})} \end{array} \right] +f(\hbox {sgn}(\sigma _{\exp }({\bar{x}})-z_1))\cdot \left[ \begin{array}{c} -|z_1-\sigma _{\exp }({\bar{z}})| \\ 0 \end{array} \right] , \end{aligned}$$

  where $\hbox {sgn}(t)$ denotes the sign of the variable $t\in {\mathbb {R}}$.

• (c) If $z\in {\tilde{B}}_3$, then

  $$\begin{aligned} f^{exp}(z):=f(\hbox {sgn}(z_1+\eta _{\exp }({\bar{z}})))\cdot \left[ \begin{array}{c} |z_1+\eta _{\exp }({\bar{z}})| \\ 0 \end{array} \right] +f(\eta _{\exp }({\bar{z}}))\cdot \left[ \begin{array}{c} -1 \\ \frac{{\bar{z}}}{\eta _{\exp }({\bar{z}})} \end{array} \right] . \end{aligned}$$

• (d) If $z\in {\tilde{B}}_4$, then

  $$\begin{aligned} f^{exp}(z):=f(1)\cdot \left[ \begin{array}{c} \max \{0,w\} \\ 0 \end{array} \right] +f(1)\cdot \left[ \begin{array}{c} \min \{0,-w\} \\ 0 \end{array} \right] , \end{aligned}$$

  where w is any scalar in ${\mathbb {R}}$.

4.2 The generalization to the high-dimensional power cone

In this subsection, we extend the discussion for the power cone ${\mathcal {K}}_{\alpha }$ to its high-dimensional version

$$\begin{aligned} {\mathcal {K}}^{(n)}_\alpha :=\left\{ (x_1,{\bar{x}})\in {\mathbb {R}}\times {\mathbb {R}}^n\, \bigg |\, |x_1|\le \prod ^n_{i=1}{\bar{x}}^{\alpha _i}_i,\ {\bar{x}}_i\ge 0,\ i=1,2,\ldots ,n\right\} , \end{aligned}$$

(48)

where ${\bar{x}}:=({\bar{x}}_1,{\bar{x}}_2,\ldots ,{\bar{x}}_n)^T\in {\mathbb {R}}^n$. In order to make the classifications clear and neat, we similarly adapt some notations as follows:

$$\begin{aligned} \begin{array}{c} {\bar{z}}:=({\bar{z}}_1,{\bar{z}}_2,\ldots ,{\bar{z}}_n)^T\in {\mathbb {R}}^n,\ {\bar{z}}_{\min } :=\min \{{\bar{z}}_1,{\bar{z}}_2,\ldots ,{\bar{z}}_n\},\ {\bar{z}}_{\max } :=\max \{{\bar{z}}_1,{\bar{z}}_2,\ldots ,{\bar{z}}_n\},\\ {\mathbf {1}}^{(n)}:=(1,1,\ldots ,1)^T\in {\mathbb {R}}^n,\ {\mathbf {1}}_k :=(0,\ldots ,1,\cdots ,0)^T\in {\mathbb {R}}^n,\ [{\mathbf {n}}] :=\{1,2,\ldots ,n\},\\ I_-:=\{i\in [{\mathbf {n}}]\,|\,{\bar{z}}_i<0\},\ I_0 :=\{i\in [{\mathbf {n}}]\,|\,{\bar{z}}_i=0\},\ I_+ :=\{i\in [{\mathbf {n}}]\,|\,{\bar{z}}_i>0\},\\ \sigma ^{(n)}_{\alpha }({\bar{z}}):=\prod \nolimits _{i=1}^n{\bar{z}}^{\alpha _i}_i,\quad \eta ^{(n)}_{\alpha }({\bar{z}}):=\prod \nolimits _{i=1}^n\left( \frac{{\bar{z}}_i}{\alpha _i}\right) ^{\alpha _i}, \end{array} \end{aligned}$$

(49)

where ${\mathbf {1}}_k\ (k=1,2,\ldots ,n)$ is the kth column of the identity matrix $I_n\in {\mathbb {R}}^{n\times n}$. Now, the space ${\mathbb {R}}\times {\mathbb {R}}^n$ can be divided into the following four blocks

$$\begin{aligned} \begin{array}{ll} {\hbox {Block I}:}&{}\ B^{(n)}_1:=\left\{ (z_1,{\bar{z}})\in {\mathbb {R}}\times {\mathbb {R}}^n\,|\,{\bar{z}}_{\min } \cdot {\bar{z}}_{\max }>0\ \hbox {or}\ ({\bar{z}}_{\min }={\bar{z}}_{\max }=0\ \hbox {and}\ z_1\ne 0)\right\} .\\ {\hbox {Block II}:}&{}\ B^{(n)}_2:=\left\{ (z_1,{\bar{z}})\in {\mathbb {R}}\times {\mathbb {R}}^n\,|\,{\bar{z}}_{\min } \cdot {\bar{z}}_{\max }=0\ \hbox {and}\ {\bar{z}}_{\min }+{\bar{z}}_{\max }\ne 0\right\} .\\ {\hbox {Block III}:}&{}\ B^{(n)}_3:=\left\{ (z_1,{\bar{z}})\in {\mathbb {R}}\times {\mathbb {R}}^n\,|\,{\bar{z}}_{\min } \cdot {\bar{z}}_{\max }<0\right\} .\\ {\hbox {Block IV}:}&{}\ B^{(n)}_4:=\left\{ (z_1,{\bar{z}})\in {\mathbb {R}}\times {\mathbb {R}}^n\,|\,{\bar{z}}_{\min } ={\bar{z}}_{\max }=0\ \hbox {and}\ z_1=0\right\} . \end{array} \end{aligned}$$

(50)

We now establish two types of decompositions with respect to ${\mathcal {K}}^{(n)}_\alpha$ defined as in (48) in the following theorems. The proofs are adapted from Theorem 1 and 2 , we omit their details and only list the results.

Theorem 5

For any given $z=(z_1,{\bar{z}})\in {\mathbb {R}}\times {\mathbb {R}}^n$, its Type I decomposition with respect to ${\mathcal {K}}^{(n)}_\alpha$ is given by

• (a) If $z\in {B}^{(n)}_1$, then

  $$\begin{aligned} z=\left\{ \begin{array}{ll} \frac{(z_1+\eta ^{(n)}_{\alpha }({\bar{z}}))\cdot \sigma ^{(n)}_{\alpha }({\bar{z}})}{\sigma ^{(n)}_{\alpha }({\bar{z}})+\eta ^{(n)}_{\alpha }({\bar{z}})}\cdot \left[ \begin{array}{c} 1 \\ \frac{{\bar{z}}}{\sigma ^{(n)}_{\alpha }({\bar{z}})} \end{array} \right] +\frac{(z_1-\sigma ^{(n)}_{\alpha }({\bar{z}}))\cdot \eta ^{(n)}_{\alpha }({\bar{z}})}{\sigma ^{(n)}_{\alpha }({\bar{z}})+\eta ^{(n)}_{\alpha }({\bar{z}})}\cdot \left[ \begin{array}{c} 1 \\ -\frac{{\bar{z}}}{\eta ^{(n)}_{\alpha }({\bar{z}})} \end{array} \right] ,&{}\hbox {if}\ |I_+|=n,\\ \\ \frac{(z_1-\eta ^{(n)}_{\alpha }(-{\bar{z}}))\cdot \sigma ^{(n)}_{\alpha }(-{\bar{z}})}{\sigma ^{(n)}_{\alpha }(-{\bar{z}})+\eta ^{(n)}_{\alpha }(-{\bar{z}})}\cdot \left[ \begin{array}{c} 1 \\ \frac{-{\bar{z}}}{\sigma ^{(n)}_{\alpha }(-{\bar{z}})} \end{array} \right] +\frac{(z_1+\sigma ^{(n)}_{\alpha }(-{\bar{z}})) \cdot \eta ^{(n)}_{\alpha }(-{\bar{z}})}{\sigma ^{(n)}_{\alpha }(-{\bar{z}})+\eta ^{(n)}_{\alpha }(-{\bar{z}})}\cdot \left[ \begin{array}{c} 1 \\ \frac{{\bar{z}}}{\eta ^{(n)}_{\alpha }(-{\bar{z}})} \end{array} \right] ,&{}\hbox {if}\ |I_-|=n,\\ \\ \frac{z_1\cdot \sigma ^{(n)}_{\alpha }({\mathbf {1}}^{(n)})}{\sigma ^{(n)}_{\alpha }({\mathbf {1}}^{(n)})+\eta ^{(n)}_{\alpha } ({\mathbf {1}}^{(n)})}\cdot \left[ \begin{array}{c} 1 \\ \frac{{\mathbf {1}}^{(n)}}{\sigma ^{(n)}_{\alpha }({\mathbf {1}}^{(n)})} \end{array} \right] +\frac{z_1\cdot \eta ^{(n)}_{\alpha }({\mathbf {1}}^{(n)})}{\sigma ^{(n)}_{\alpha }({\mathbf {1}}^{(n)})+\eta ^{(n)}_{\alpha } ({\mathbf {1}}^{(n)})}\cdot \left[ \begin{array}{c} 1 \\ -\frac{{\mathbf {1}}^{(n)}}{\eta ^{(n)}_{\alpha }({\mathbf {1}}^{(n)})} \end{array} \right] ,&\hbox {if}\ |I_0|=n, \end{array} \right. \end{aligned}$$

  where ${\mathbf {1}}^{(n)}, \sigma ^{(n)}_{\alpha }({\bar{x}}), \eta ^{(n)}_{\alpha }({\bar{x}})$ are defined as in (49) and |I| denotes the cardinality of I.

• (b) If $z\in B^{(n)}_2$, then

  $$\begin{aligned} z=\left\{ \begin{array}{ll} 1\cdot \left[ \begin{array}{c} z_1 \\ \dot{{\bar{x}}}^{(B^{(n)}_2,a)} \end{array} \right] +1\cdot \left[ \begin{array}{c} 0 \\ \dot{{\bar{y}}}^{(B^{(n)}_2,a)} \end{array} \right] ,&{}\hbox {if}\ |I_-|=0,\\ \\ (-1)\cdot \left[ \begin{array}{c} -z_1 \\ \dot{{\bar{x}}}^{(B^{(n)}_2,b)} \end{array} \right] +(-1)\cdot \left[ \begin{array}{c} 0 \\ \dot{{\bar{y}}}^{(B^{(n)}_2,b)} \end{array} \right] ,&\hbox {if}\ |I_+|=0, \end{array} \right. \end{aligned}$$

  where k is the smallest index in $I_0$ and $\dot{{\bar{x}}}^{(B^{(n)}_2,a)},\dot{{\bar{y}}}^{(B^{(n)}_2,a)} ,\dot{{\bar{x}}}^{(B^{(n)}_2,b)},\dot{{\bar{y}}}^{(B^{(n)}_2,b)}$ are respectively defined as follows:

  $$\begin{aligned}&\left( \dot{{\bar{x}}}^{(B^{(n)}_2,a)}_j,\dot{{\bar{y}}}^{(B^ {(n)}_2,a)}_j\right) \\&\quad :=\left\{ \begin{array}{ll} ({\bar{z}}_j,0)&{}\hbox {if}\ j\in I_+,\\ (1,-1)&{}\hbox {if}\ j\in I_0\ \hbox {and}\ j\ne k,\\ \left( \left( \frac{|z_1|}{\prod _{i\ne k}\left( \dot{{\bar{x}}}^{(B^{(n)}_2,a)}_i\right) ^{\alpha _i}} \right) ^{\frac{1}{\alpha _k}},-\left( \frac{|z_1|}{\prod _{i\ne k}\left( \dot{{\bar{x}}}^{(B^{(n)}_2,a)}_i\right) ^{\alpha _i}} \right) ^{\frac{1}{\alpha _k}}\right) &{}\hbox {if}\ j=k. \end{array} \right. \\&\left( \dot{{\bar{x}}}^{(B^{(n)}_2,b)}_j,\dot{{\bar{y}}}^{(B^{ (n)}_2,b)}_j\right) \\&\quad :=\left\{ \begin{array}{ll} (-{\bar{z}}_j,0)&{}\hbox {if}\ j\in I_-,\\ (1,-1)&{}\hbox {if}\ j\in I_0\ \hbox {and}\ j\ne k,\\ \left( \left( \frac{|z_1|}{\prod _{i\ne k}\left( \dot{{\bar{x}}}^{(B^{(n)}_2,b)}_i\right) ^{\alpha _i}} \right) ^{\frac{1}{\alpha _k}},-\left( \frac{|z_1|}{\prod _{i\ne k}\left( \dot{{\bar{x}}}^{(B^{(n)}_2,b)}_i\right) ^{\alpha _i}} \right) ^{\frac{1}{\alpha _k}}\right) &{}\hbox {if}\ j=k. \end{array} \right. \end{aligned}$$

• (c) If $z\in B^{(n)}_3$, then

  $$\begin{aligned} z=\left\{ \begin{array}{ll} 1\cdot \left[ \begin{array}{c} z_1 \\ \dot{{\bar{x}}}^{(B^{(n)}_3,a)} \end{array} \right] +1\cdot \left[ \begin{array}{c} 0 \\ \dot{{\bar{y}}}^{(B^{(n)}_3,a)} \end{array} \right] ,&{}\hbox {if}\ |I_0|=0,\\ \\ 1\cdot \left[ \begin{array}{c} z_1 \\ \dot{{\bar{x}}}^{(B^{(n)}_3,b)} \end{array} \right] +1\cdot \left[ \begin{array}{c} 0 \\ \dot{{\bar{y}}}^{(B^{(n)}_3,b)} \end{array} \right] ,&\hbox {if}\ |I_0|\ne 0, \end{array} \right. \end{aligned}$$

  where t is the smallest index in $I_-$ and $\dot{{\bar{x}}}^{(B^{(n)}_3,a)},\dot{{\bar{y}}}^{(B^{(n)}_3,a)}$ are respectively defined as follows:

  $$\begin{aligned}&\left( \dot{{\bar{x}}}^{(B^{(n)}_3,a)}_j,\dot{{\bar{y}}}^{ (B^{(n)}_3,a)}_j\right) \\&\quad :=\left\{ \begin{array}{l@{\quad }l} ({\bar{z}}_j,0)&{}\hbox {if}\ j\in I_+,\\ (-{\bar{z}}_j,2{\bar{z}}_j)&{}\hbox {if}\ j\in I_-\ \hbox {and}\ j\ne t,\\ \left( \left( \frac{|z_1|}{\prod _{i\ne t}\left( \dot{{\bar{x}}}^{(B^{(n)}_3,a)}_i\right) ^{\alpha _i}} \right) ^{\frac{1}{\alpha _t}},{\bar{z}}_t-\left( \frac{|z_1|}{\prod _{i\ne t}\left( \dot{{\bar{x}}}^{(B^{(n)}_3,a)}_i\right) ^{\alpha _i}} \right) ^{\frac{1}{\alpha _t}}\right) &{}\hbox {if}\ j=t. \end{array} \right. \end{aligned}$$

  Similarly, q is the smallest index in $I_0$ and $\dot{{\bar{x}}}^{(B^{(n)}_3,b)},\dot{{\bar{y}}}^{(B^{(n)}_3,b)}$ are respectively defined as follows:

  $$\begin{aligned}&\left( \dot{{\bar{x}}}^{(B^{(n)}_3,b)}_j,\dot{{\bar{y}}}^{ (B^{(n)}_3,b)}_j\right) \\&\quad :=\left\{ \begin{array}{l@{\quad }l} ({\bar{z}}_j,0)&{}\hbox {if}\ j\in I_+,\\ (-{\bar{z}}_j,2{\bar{z}}_j)&{}\hbox {if}\ j\in I_-,\\ (1,-1)&{}\hbox {if}\ j\in I_0\ \hbox {and}\ j\ne q,\\ \left( \left( \frac{|z_1|}{\prod _{i\ne q}\left( \dot{{\bar{x}}}^{(B^{(n)}_3,b)}_i\right) ^{\alpha _i}} \right) ^{\frac{1}{\alpha _q}},-\left( \frac{|z_1|}{\prod _{i\ne q}\left( \dot{{\bar{x}}}^{(B^{(n)}_3,b)}_i\right) ^{\alpha _i}} \right) ^{\frac{1}{\alpha _q}}\right) &{}\hbox {if}\ j=q. \end{array} \right. \end{aligned}$$

• (d) If $z\in B^{(n)}_4$, then

  $$\begin{aligned} z=1\cdot \left[ \begin{array}{c} 0\\ {\mathbf {1}}^{(n)}-{\mathbf {1}}_k \end{array} \right] +1\cdot \left[ \begin{array}{c} 0\\ {\mathbf {1}}_k-{\mathbf {1}}^{(n)}\end{array} \right] , \end{aligned}$$

  where ${\mathbf {1}}_k\ (k=1,2,\ldots ,n)$ is the kth column of the identity matrix $I_{n}$.

Theorem 6

For any given $z=(z_1,{\bar{z}})\in {\mathbb {R}}\times {\mathbb {R}}^n$, its Type II decomposition with respect to ${\mathcal {K}}^{(n)}_\alpha$ is given by

• (a) If $z\in {B}^{(n)}_1$, then

  $$\begin{aligned} z=\left\{ \begin{array}{l@{\quad }l} \frac{z_1+\sigma ^{(n)}_{\alpha }({\bar{z}})}{2}\cdot \left[ \begin{array}{c} 1 \\ \frac{{\bar{z}}}{\sigma ^{(n)}_{\alpha }({\bar{z}})} \end{array} \right] +\frac{\sigma ^{(n)}_{\alpha }({\bar{z}})-z_1}{2}\cdot \left[ \begin{array}{c} -1 \\ \frac{{\bar{z}}}{\sigma ^{(n)}_{\alpha }({\bar{z}})} \end{array} \right] ,&{}\hbox {if}\ |I_+|=n,\\ \\ \frac{z_1-\sigma ^{(n)}_{\alpha }(-{\bar{z}})}{2}\cdot \left[ \begin{array}{c} 1 \\ \frac{-{\bar{z}}}{\sigma ^{(n)}_{\alpha }(-{\bar{z}})} \end{array} \right] +\frac{-\sigma ^{(n)}_{\alpha }({\bar{z}})-z_1}{2}\cdot \left[ \begin{array}{c} -1 \\ \frac{-{\bar{z}}}{\sigma ^{(n)}_{\alpha }(-{\bar{z}})} \end{array} \right] ,&{}\hbox {if}\ |I_-|=n,\\ \\ \frac{z_1}{2}\cdot \left[ \begin{array}{c} 1 \\ \frac{{\mathbf {1}}^{(n)}}{\sigma ^{(n)}_{\alpha }({\mathbf {1}}^{(n)})} \end{array} \right] +\frac{-z_1}{2}\cdot \left[ \begin{array}{c} -1 \\ \frac{{\mathbf {1}}^{(n)}}{\sigma ^{(n)}_{\alpha }({\mathbf {1}}^{(n)})} \end{array} \right] ,&\hbox {if}\ |I_0|=n. \end{array} \right. \end{aligned}$$

• (b) If $z\in B^{(n)}_2$, then

  $$\begin{aligned} z=\left\{ \begin{array}{l@{\quad }l} 1\cdot \left[ \begin{array}{c} z_1 \\ \ddot{{\bar{x}}}^{(B^{(n)}_2,a)} \end{array} \right] +(-1)\cdot \left[ \begin{array}{c} 0 \\ \ddot{{\bar{y}}}^{(B^{(n)}_2,a)} \end{array} \right] ,&{}\hbox {if}\ |I_-|=0,\\ \\ (-1)\cdot \left[ \begin{array}{c} -z_1 \\ \ddot{{\bar{x}}}^{(B^{(n)}_2,b)} \end{array} \right] +1\cdot \left[ \begin{array}{c} 0 \\ \ddot{{\bar{y}}}^{(B^{(n)}_2,b)} \end{array} \right] ,&\hbox {if}\ |I_+|=0, \end{array} \right. \end{aligned}$$

  where k is the smallest index in $I_0$ and $\ddot{{\bar{x}}}^{(B^{(n)}_2,a)},\ddot{{\bar{y}}}^{(B^{ (n)}_2,a)},\ddot{{\bar{x}}}^{(B^{(n)}_2,b)}, \ddot{{\bar{y}}}^{(B^{(n)}_2,b)}$ are respectively defined as follows:

  $$\begin{aligned}&\left( \ddot{{\bar{x}}}^{(B^{(n)}_2,a)}_j,\ddot{{\bar{y}}}^{ (B^{(n)}_2,a)}_j\right) \\&\quad :=\left\{ \begin{array}{l@{\quad }l} ({\bar{z}}_j,0)&{}\hbox {if}\ j\in I_+,\\ (1,1)&{}\hbox {if}\ j\in I_0\ \hbox {and}\ j\ne k,\\ \left( \left( \frac{|z_1|}{\prod _{i\ne k}\left( \ddot{{\bar{x}}}^{(B^{(n)}_2,a)}_i\right) ^{\alpha _i}} \right) ^{\frac{1}{\alpha _k}},\left( \frac{|z_1|}{\prod _{i\ne k}\left( \ddot{{\bar{x}}}^{(B^{(n)}_2,a)}_i\right) ^{\alpha _i}} \right) ^{\frac{1}{\alpha _k}}\right) &{}\hbox {if}\ j=k. \end{array} \right. \\&\left( \ddot{{\bar{x}}}^{(B^{(n)}_2,b)}_j,\ddot{{\bar{y}}}^{ (B^{(n)}_2,b)}_j\right) \\&\quad :=\left\{ \begin{array}{l@{\quad }l} (-{\bar{z}}_j,0)&{}\hbox {if}\ j\in I_-,\\ (1,-1)&{}\hbox {if}\ j\in I_0\ \hbox {and}\ j\ne k,\\ \left( \left( \frac{|z_1|}{\prod _{i\ne k}\left( \ddot{{\bar{x}}}^ {(B^{(n)}_2,b)}_i\right) ^{\alpha _i}}\right) ^{\frac{1}{\alpha _k}} ,\left( \frac{|z_1|}{\prod _{i\ne k}\left( \ddot{{\bar{x}}}^{(B^{(n)}_2,b)}_i\right) ^{\alpha _i}} \right) ^{\frac{1}{\alpha _k}}\right) &{}\hbox {if}\ j=k. \end{array} \right. \end{aligned}$$

• (c) If $z\in B^{(n)}_3$, then

  $$\begin{aligned} z=\left\{ \begin{array}{l@{\quad }l} 1\cdot \left[ \begin{array}{c} z_1 \\ \ddot{{\bar{x}}}^{(B^{(n)}_3,a)} \end{array} \right] +(-1)\cdot \left[ \begin{array}{c} 0 \\ \ddot{{\bar{y}}}^{(B^{(n)}_3,a)} \end{array} \right] ,&{}\hbox {if}\ |I_0|=0,\\ \\ 1\cdot \left[ \begin{array}{c} z_1 \\ \ddot{{\bar{x}}}^{(B^{(n)}_3,b)} \end{array} \right] +(-1)\cdot \left[ \begin{array}{c} 0 \\ \ddot{{\bar{y}}}^{(B^{(n)}_3,b)} \end{array} \right] ,&\hbox {if}\ |I_0|\ne 0, \end{array} \right. \end{aligned}$$

  where t is the smallest index in $I_-$ and $\ddot{{\bar{x}}}^{(B^{(n)}_3,a)},\ddot{{\bar{y}}}^{(B^{(n)}_3,a)}$ are respectively defined as follows:

  $$\begin{aligned}&\left( \ddot{{\bar{x}}}^{(B^{(n)}_3,a)}_j,\ddot{{\bar{y}}}^{(B^{(n)}_3, a)}_j\right) \\&\quad :=\left\{ \begin{array}{l@{\quad }l} ({\bar{z}}_j,0)&{}\hbox {if}\ j\in I_+,\\ (-{\bar{z}}_j,-2{\bar{z}}_j)&{}\hbox {if}\ j\in I_-\ \hbox {and}\ j\ne t,\\ \left( \left( \frac{|z_1|}{\prod _{i\ne t}\left( \ddot{{\bar{x}}}^{(B^{(n)}_3,a)}_i\right) ^{\alpha _i}} \right) ^{\frac{1}{\alpha _t}},-{\bar{z}}_t+\left( \frac{|z_1|}{\prod _{i\ne t}\left( \ddot{{\bar{x}}}^{(B^{(n)}_3,a)}_i\right) ^{\alpha _i}} \right) ^{\frac{1}{\alpha _t}}\right) &{}\hbox {if}\ j=t. \end{array} \right. \end{aligned}$$

  Similarly, q is the smallest index in $I_0$ and $\ddot{{\bar{x}}}^{(B^{(n)}_3,b)},\ddot{{\bar{y}}}^{(B^{(n)}_3,b)}$ are respectively defined as follows:

  $$\begin{aligned}&\left( \ddot{{\bar{x}}}^{(B^{(n)}_3,b)}_j,\ddot{{\bar{y}}} ^{(B^{(n)}_3,b)}_j\right) \\&\quad :=\left\{ \begin{array}{l@{\quad }l} ({\bar{z}}_j,0)&{}\hbox {if}\ j\in I_+,\\ (-{\bar{z}}_j,-2{\bar{z}}_j)&{}\hbox {if}\ j\in I_-,\\ (1,1)&{}\hbox {if}\ j\in I_0\ \hbox {and}\ j\ne q,\\ \left( \left( \frac{|z_1|}{\prod _{i\ne q}\left( \ddot{{\bar{x}}}^{(B^{(n)}_3,b)}_i\right) ^{\alpha _i}} \right) ^{\frac{1}{\alpha _q}},\left( \frac{|z_1|}{\prod _{i\ne q}\left( \ddot{{\bar{x}}}^{(B^{(n)}_3,b)}_i\right) ^{\alpha _i}} \right) ^{\frac{1}{\alpha _q}}\right) &{}\hbox {if}\ j=q. \end{array} \right. \end{aligned}$$

• (d) If $z\in B^{(n)}_4$, then

  $$\begin{aligned} z=1\cdot \left[ \begin{array}{c} 0\\ {\mathbf {1}}^{(n)}-{\mathbf {1}}_k \end{array} \right] +(-1)\cdot \left[ \begin{array}{c} 0\\ {\mathbf {1}}^{(n)}-{\mathbf {1}}_k \end{array} \right] . \end{aligned}$$

5 Concluding remarks

In this paper, we propose two types of decomposition approaches for the power cone ${\mathcal {K}}_{\alpha }$ and the exponential cone ${\mathcal {K}}_{\exp }$, which are the generators of many well-known nonsymmetric cones. In particular, the corresponding explicit decomposition formulas are established based on different classifications for the reference points with respect to the given cones and the decomposition types. In contrast to the setting of ${\mathcal {K}}_{\exp }$, the power cone ${\mathcal {K}}_{\alpha }$ seems to be more regular, because its two types of decompositions share the same space division. At the same time, we also define their conic functions, namely $f^{power}$ and $f^{exp}$ as Definition 1 and 2 . As a byproduct, we can extend the decomposition results of the power cone ${\mathcal {K}}_{\alpha }$ to its high-dimensional case ${\mathcal {K}}^{(n)}_{\alpha }$ by slight modifications.

Although the results are not quite consummate due to the difficulty of handling nonsymmetric cones, they are very crucial to subsequent study towards nonsymmetric cone optimization. Further investigations are definitely desirable. We summarize and list out some future topics as below.

• 1. Exploring more structures and properties for the power cone and the exponential cone, such as their variational geometries including normal cones, tangent cones, second-order tangent sets, critical cone and “sigma” terms.

• 2. Similar to the second order cone setting and its generalization like circular cone, can the properties of continuity, strict continuity, Lipschitz continuity, directional differentiability, differentiability, continuous differentiability, and semismoothness be each inherited by $f^{power}$ and $f^{exp}$ from f?

• 3. Designing new algorithms for these nonsymmetric cones based on the non-interior-point framework, such as augmented Lagrangian method, proximal point method and their variants.

On the other hand, there are so many non-symmetric cones in real world. Can we figure out a way to clarify them? This is another important direction for our future study.

Appendix

1.1 The concepts of $\alpha$-representable and extended $\alpha$-representable sets

For a given convex set ${\mathcal {K}}$, it is $\alpha$-representable [5, p. 110] if there exist a finite integer M, scalars $\alpha _i\in [0,1]$, $i=1,2,\ldots ,M$, vectors $c_1,c_2,\ldots ,c_M\in {\mathbb {R}}^3$, matrices $A_1,A_2,\ldots ,A_M$ with three columns and an appropriate number of rows, a matrix $A_f$ and a vector $c_f$ such that

$$\begin{aligned} u\in {\mathcal {K}}\ \Leftrightarrow \ c_i-A^T_i\left[ \begin{array}{c} u\\ v \end{array}\right] \in {\mathcal {K}}_{\alpha _i}\ (i=1,2,\ldots ,M),\ A^T_f\left[ \begin{array}{c} u\\ v \end{array}\right] =c_f \end{aligned}$$

for some artificial variables or modelling variables v. Similarly, the set ${\mathcal {K}}$ is extended $\alpha$-representable [5, p. 122] if there exist finite integers $M_1, M_2$, matrices $A_\alpha ,A_{\exp },A_f$ and vectors $c_\alpha , c_{\exp }, c_f$ of appropriate sizes such that

$$\begin{aligned} u\in {\mathcal {K}}\ \Leftrightarrow \ c_\alpha -A^T_\alpha \left[ \begin{array}{c} u\\ v \end{array}\right] \in \prod ^{M_1}_{i=1}{\mathcal {K}}_{\alpha _i},\ c_{\exp }-A^T_{\exp }\left[ \begin{array}{c} u\\ v \end{array}\right] \in \prod ^{M_2}_{i=1}{\mathcal {K}}_{\exp },\ A^T_f\left[ \begin{array}{c} u\\ v \end{array}\right] =c_f. \end{aligned}$$

1.2 The decomposition with respect to the circular cone

Consider the circular cone

$$\begin{aligned} {\mathcal {L}}_\theta := \{(x_1,{\bar{x}})\in {\mathbb {R}}\times {\mathbb {R}}^{n-1}\,| \,x_1\tan \theta \ge \Vert {\bar{x}}\Vert \}. \end{aligned}$$

For any given $z=(z_1,{\bar{z}})\in {\mathbb {R}}\times {\mathbb {R}}^{n-1}$, the projection mappings $\varPi _{{\mathcal {L}}_\theta }(z), \varPi _{{\mathcal {L}}^\circ _\theta }(z)$ are respectively given by

$$\begin{aligned} \varPi _{{\mathcal {L}}_\theta }(z):=\left\{ \begin{array}{ll} z,&{}\hbox {if}\ z\in {\mathcal {L}}_\theta ,\\ 0,&{}\hbox {if}\ z\in {\mathcal {L}}^\circ _\theta ,\\ u,&{}\hbox {otherwise}, \end{array} \right. \qquad \varPi _{{\mathcal {L}}^\circ _\theta }(z):=\left\{ \begin{array}{ll} 0,&{}\hbox {if}\ z\in {\mathcal {L}}_\theta ,\\ z,&{}\hbox {if}\ z\in {\mathcal {L}}^\circ _\theta ,\\ v,&{}\hbox {otherwise}, \end{array} \right. \end{aligned}$$

where

$$\begin{aligned} u=\left[ \begin{array}{c} \displaystyle \frac{z_1+\Vert z_2\Vert \tan \theta }{1+\tan ^2\theta }\\ \displaystyle \left( \frac{z_1+\Vert z_2\Vert \tan \theta }{1 +\tan ^2\theta }\tan \theta \right) \frac{z_2}{\Vert z_2\Vert } \end{array} \right] ,\qquad v=\left[ \begin{array}{c} \displaystyle \frac{z_1-\Vert z_2\Vert \cot \theta }{1+\cot ^2\theta }\\ \displaystyle \left( \frac{z_1-\Vert z_2\Vert \cot \theta }{1+\cot ^2\theta } \cot \theta \right) \frac{-z_2}{\Vert z_2\Vert } \end{array} \right] . \end{aligned}$$

Combining these results with the Moreau decomposition theorem, the decomposition with respect to ${\mathcal {L}}_\theta$ is

$$\begin{aligned} z={\tilde{\lambda }}_1(z)\cdot {\tilde{u}}^{(1)}_z+{\tilde{\lambda }}_2(z)\cdot {\tilde{u}}^{(2)}_z, \end{aligned}$$

(51)

where

$$\begin{aligned} {\tilde{\lambda }}_1(z):= & {} z_1-\Vert {\bar{z}}\Vert \cot \theta , \qquad \qquad {\tilde{\lambda }}_2(z) :=z_1+\Vert {\bar{z}}\Vert \tan \theta ,\\ {\tilde{u}}^{(1)}_z:= & {} \displaystyle \frac{1}{1+\cot ^2\theta }\left[ \begin{array}{cc} 1 &{}\ 0 \\ 0 &{}\ \cot \theta \cdot I_{n-1} \end{array} \right] \left[ \begin{array}{c} 1 \\ -w \end{array} \right] ,\\ {\tilde{u}}^{(2)}_z:= & {} \displaystyle \frac{1}{1+\tan ^2\theta }\left[ \begin{array}{cc} 1 &{}\ 0 \\ 0 &{}\ \tan \theta \cdot I_{n-1} \end{array} \right] \left[ \begin{array}{c} 1 \\ w \end{array} \right] \end{aligned}$$

with $w=\frac{{\bar{z}}}{\Vert {\bar{z}}\Vert }$ if ${\bar{x}}\ne 0$ and w is any unit vector in ${\mathbb {R}}^{n-1}$ if ${\bar{x}}=0$ and $I_{n-1}$ is the identity matrix of order $n-1$. It is easy to see that

$$\begin{aligned} \varPi _{{\mathcal {L}}_\theta }(z)=\max \{0,{\tilde{\lambda }}_1(z)\}\cdot {\tilde{u}}^{(1)}_z+\max \{0,{\tilde{\lambda }}_2(z)\}\cdot {\tilde{u}}^{(2)}_z. \end{aligned}$$

More properties of the circular cone can be found in [45, Section 3].

1.3 Proof of Lemma 1

By definition, ${\mathcal {K}}_\alpha$ is closed, since the functions ${\bar{x}}^{\alpha _1}_1{\bar{x}}^{\alpha _2}_2$ and $|x_1|$ are continuous on ${\mathbb {R}}^2_+$ and ${\mathbb {R}}$, respectively. To proof that ${\mathcal {K}}_\alpha$ is a convex cone, we only need to verify that it is closed under the addition and the nonnegative multiplication. For any given $(x_1,{\bar{x}})\in {\mathcal {K}}_\alpha$ and $\beta \ge 0$, one can obtain that

$$\begin{aligned} (\beta {\bar{x}}_1)^{\alpha _1}(\beta {\bar{x}}_2)^{\alpha _2} =\beta {\bar{x}}_1^{\alpha _1}{\bar{x}}_2^{\alpha _2}\ge \beta |x_1|=|\beta x_1|,\ \beta {\bar{x}}_1\ge 0,\ \beta {\bar{x}}_2\ge 0, \end{aligned}$$

where the first equation uses the fact $\alpha _1+\alpha _2=1$. Therefore, we have $\beta (x_1,{\bar{x}})\in {\mathcal {K}}_\alpha$. For any given $(x_1,{\bar{x}}),(y_1,{\bar{y}})\in {\mathcal {K}}_\alpha$, we know

$$\begin{aligned} \begin{array}{lll} |x_1|\le {\bar{x}}^{\alpha _1}_1{\bar{x}}^{\alpha _2}_2,&{}{\bar{x}}_1\ge 0,&{}{\bar{x}}_2\ge 0,\\ |y_1|\le {\bar{y}}^{\alpha _1}_1{\bar{y}}^{\alpha _2}_2,&{}{\bar{y}}_1\ge 0,&{}{\bar{y}}_2\ge 0. \end{array} \end{aligned}$$

It is easy to see that ${\bar{x}}_1+{\bar{y}}_1\ge 0$, ${\bar{x}}_2+{\bar{y}}_2\ge 0$ and $|x_1+y_1|\le |x_1|+|y_1|\le {\bar{x}}^{\alpha _1}_1{\bar{x}}^{\alpha _2}_2 +{\bar{y}}^{\alpha _1}_1{\bar{y}}^{\alpha _2}_2$. In order to finish our proof, it suffices to show that

$$\begin{aligned} {\bar{x}}^{\alpha _1}_1{\bar{x}}^{\alpha _2}_2+{\bar{y}}^ {\alpha _1}_1{\bar{y}}^{\alpha _2}_2\le ({\bar{x}}_1+{\bar{y}}_1)^{\alpha _1}({\bar{x}}_2+{\bar{y}}_2)^{\alpha _2},\ \forall (x_1,{\bar{x}}),(y_1,{\bar{y}})\in {\mathcal {K}}_\alpha . \end{aligned}$$

(52)

We divide it into the following two cases. Suppose that there exists an index $i\in \{1,2\}$ such that ${\bar{x}}_i=0$ or ${\bar{y}}_i=0$, it is trivial to show (52). Otherwise, we obtain ${\bar{x}},{\bar{y}}\in {\mathbb {R}}^2_{++}$. Consider the function $f:{\mathbb {R}}^2_{++}\rightarrow {\mathbb {R}}$:

$$\begin{aligned} f({\bar{x}})={\bar{x}}^{\alpha _1}_1{\bar{x}}^{\alpha _2}_2, \end{aligned}$$

where ${\bar{x}}:=({\bar{x}}_1,{\bar{x}}_2)^T\in {\mathbb {R}}^2$ and ${\bar{x}}_1,{\bar{x}}_2>0$. By calculation, we obtain

$$\begin{aligned} \nabla ^2f({\bar{x}})=\left[ \begin{array}{cc} \alpha _1(\alpha _1-1){\bar{x}}^{\alpha _1-2}_1&{}0\\ 0&{}\alpha _2(\alpha _2-1){\bar{x}}^{\alpha _2-2}_2 \end{array} \right] \end{aligned}$$

Since $\alpha _i\in (0,1)$ and ${\bar{x}}_i$ is strictly positive, the Hessian matrix $\nabla ^2f({\bar{x}})$ is negative definite, which shows that f is concave defined on ${\mathbb {R}}^2_{++}$. Therefore, we have

$$\begin{aligned} f\left( \frac{{\bar{x}}+{\bar{y}}}{2}\right) \ge \frac{1}{2}\left( f({\bar{x}})+f({\bar{y}})\right) , \end{aligned}$$

which is equivalent to the above inequality (52). $\square$