Duality of nonconvex optimization with positively homogeneous functions

Abstract

We consider an optimization problem with positively homogeneous functions in its objective and constraint functions. Examples of such positively homogeneous functions include the absolute value function and the p-norm function, where p is a positive real number. The problem, which is not necessarily convex, extends the absolute value optimization proposed in Mangasarian (Comput Optim Appl 36:43–53, 2007). In this work, we propose a dual formulation that, differently from the Lagrangian dual approach, has a closed-form and some interesting properties. In particular, we discuss the relation between the Lagrangian duality and the one proposed here, and give some sufficient conditions under which these dual problems coincide. Finally, we show that some well-known problems, e.g., sum of norms optimization and the group Lasso-type optimization problems, can be reformulated as positively homogeneous optimization problems.

Appendix

The following proposition shows that the dual of the p-norm function is the $\infty$-norm even when p is less than 1.

Proposition A.1

Suppose that $p \in (0, 1)$. Then, the dual of the p-norm function is equal to the $\infty$-norm.

Proof

Let $y \in \mathbb {R}^n$ be an arbitrary vector. If $y = 0$, this proposition clearly holds. If $y \ne 0$, from Definition 2.3, we obtain

$$\begin{aligned} \Vert y \Vert _p^*= & {} \sup \{ x^T y \, | \, \Vert x \Vert _p \le 1 \} \\\le & {} \sup \{ | x^T y | \, | \, \Vert x \Vert _p \le 1 \} \\\le & {} \sup \biggl \{ \sum _{i=1}^n | x_i | | y_i | \, | \, \Vert x \Vert _p \le 1 \biggr \} \\\le & {} \max _j | y_j | \left( \sup \biggl \{ \sum _{i=1}^n | x_i | \, | \, \Vert x \Vert _p \le 1 \biggr \} \right) \\= & {} \max _j | y_j | \biggl ( \sup \{ \Vert x \Vert _1 \, | \, \Vert x \Vert _p \le 1 \} \biggr ). \end{aligned}$$

Since $p \in (0, 1)$, we note that $\Vert x \Vert _1 \le \Vert x \Vert _p$ holds [11]. Then, we have

$$\begin{aligned} \Vert y \Vert _p^* \le \max _j | y_j | \biggl ( \sup \{ \Vert x \Vert _p \, | \, \Vert x \Vert _p \le 1 \} \biggr ) = \max _j | y_j | = \Vert y \Vert _\infty . \end{aligned}$$

Now, take an arbitrary $i_0 \in \mathop {\mathrm{argmax}}\limits _i | y_i |$, and define $\bar{x}_i$ as follows:

$$\begin{aligned} \bar{x}_i = \left\{ \begin{array}{ll} {\text {sign}}(y_i), &{} \quad \mathrm {if} \quad i = i_0, \\ 0, &{} \quad \mathrm {otherwise}, \end{array} \right. \end{aligned}$$

where

$$\begin{aligned} {\text {sign}}(y_i) = \left\{ \begin{array}{ll} 1, &{} \quad \mathrm {if} \quad y_i > 0, \\ 0, &{} \quad \mathrm {if} \quad y_i = 0, \\ -\,1, &{} \quad \mathrm {if} \quad y_i < 0. \end{array} \right. \end{aligned}$$

Then, $\Vert \bar{x} \Vert _p = 1$ and we have

$$\begin{aligned} \Vert y \Vert _p^* = \sup \{ x^T y \, | \, \Vert x \Vert _p \le 1 \} \ge \bar{x}^T y = \max _i |y_i | = \Vert y \Vert _\infty , \end{aligned}$$

which completes the proof. $\square$