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.

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). $\square$