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I have the following problem:

\begin{align*} \sup_y&\quad \big | \langle u,y \rangle\big |\\ \mbox{s.t.}&\quad \frac{1}{2}\langle y,y \rangle\ + \langle b,y \rangle\ \geq \gamma. \end{align*}
where $u$ is a constant vector.

I am confused about the following:

  1. $\big | \langle u,y \rangle\big |$ is a convex function, then it seems that this problem is infeasible.

To solve it, I try to find its dual. So the Lagrangian. The first step is to rewrite it in a familiar form:

\begin{align*} -\inf_y&\quad -\big | \langle u,y \rangle\ \big |\\ \mbox{s.t.}&\quad \gamma - \frac{1}{2}\langle y,y \rangle\ - \langle b,y \rangle\ \leq 0. \end{align*}

$$L(x,y) = -\big | \langle u,y \rangle\ \big| +x\big(\ \gamma - \frac{1}{2}\langle y,y \rangle\ - \langle b,y \rangle\big)$$

Then how to solve it? How to deal with the absolute value?

  • 2
    This is not convex. $|\langle u, y\rangle|$ is convex, but in a convex optimization model, you can only _minimize_ convex functions. – 2017-02-12

2 Answers 2

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Here is one approach that involves modifying the original program.

I claim that your problem

\begin{equation*} \begin{array}{rcl} \sup_y &&|\langle u,y\rangle|\hspace{1.5 in}(1)\\ s.t. &&\frac{1}{2}\langle y,y\rangle+\langle b,y\rangle \geq \gamma \end{array} \end{equation*}

is equivalent to

\begin{equation*} \begin{array}{rcl} \sup_{y,w} &&w\hspace{1.75 in}(2)\\ s.t. &&\frac{1}{2}\langle y,y\rangle+\langle b,y\rangle \geq \gamma\\ && w^2\leq \langle u,y\rangle^2\\ &&w \geq 0. \end{array} \end{equation*}

In order to see the equivalence of (1) and (2), let $y^*$ be an optimal solution to (1) and $(\bar{y}, \bar{w})$ be an optimal solution to (2). The constraints $w^2\leq \langle u,y\rangle^2$ and $w \geq 0$ in (2) imply that \begin{equation} 0\leq \bar{w}\leq |\langle u,\bar{y}\rangle|\leq |\langle u,y^*\rangle|. \hspace{.5 in}(*) \end{equation} Note that $(y^*, w^*)$ for $w^* = |\langle u, y^*\rangle|$ is a feasible solution for (2) and so $$|\langle u, y^*\rangle| = w^* \leq\bar{w} \leq |\langle u,\bar{y}\rangle|.\hspace{.5 in} (**)$$

From $(*)$ and $(**)$, we get the desired inclusion.

Now (2) does not have any absolute values, but it does have some quadratic terms. However, your initial problem (1) also had quadratic terms.

  • 2
    Note that this reformulation, while equivalent, is not convex either. – 2017-02-12
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Given $\mathrm c, \mathrm y \in \mathbb R^n$ and $\rho > 0$,

$$\begin{array}{ll} \text{supremize} & | \mathrm c^{\top} \mathrm x |\\ \text{subject to} & \| \mathrm x - \mathrm y \|_2 \geq \rho\end{array}$$

Since $-\| \mathrm c \|_2 \| \mathrm x \|_2 \leq \mathrm c^{\top} \mathrm x \leq \| \mathrm c \|_2 \| \mathrm x \|_2$, we drop the absolute value. Hence, we have the following non-convex quadratically constrained linear program (QCLP)

$$\begin{array}{ll} \text{supremize} & \mathrm c^{\top} \mathrm x\\ \text{subject to} & \| \mathrm x - \mathrm y \|_2 \geq \rho\end{array}$$

The feasible region is $\mathbb R^n \setminus \mathbb B_{\rho} (\mathrm y)$, where $\mathbb B_{\rho} (\mathrm y)$ is the open Euclidean ball of radius $\rho$ centered at $\mathrm y$.

  • If $\mathrm c = 0_n$, then the maximum is zero and every feasible point is a maximizer.

  • If $\mathrm c \neq 0_n$, then we can make $\mathrm c^{\top} \mathrm x$ arbitrarily large because the feasible region is isotropically unbounded. In this case, the supremum is $\infty$.

  • 0
    Note that $$\mathrm b := - \mathrm y$$ and $$\gamma := \frac 12 \left( \rho^2 - \| \mathrm y \|_2^2 \right)$$ – 2017-02-13