On Convex Hulls of Epigraphs of QCQPs

Abstract

Quadratically constrained quadratic programs (QCQPs) are a fundamental class of optimization problems well-known to be NP-hard in general. In this paper we study sufficient conditions for a convex hull result that immediately implies that the standard semidefinite program (SDP) relaxation of a QCQP is tight. We begin by outlining a general framework for proving such sufficient conditions. Then using this framework, we show that the convex hull result holds whenever the quadratic eigenvalue multiplicity, a parameter capturing the amount of symmetry present in a given problem, is large enough. Our results also imply new sufficient conditions for the tightness (as well as convex hull exactness) of a second order cone program relaxation of simultaneously diagonalizable QCQPs.

A Proof Sketch of Lemma 4

For simplicity, we will assume that $\varGamma$ is a polytope in this proof sketch. Let $(\hat{x},\hat{t})$ satisfy the assumptions of Lemma 4. Without loss of generality, we may assume that $\sup _{\gamma \in \varGamma } q(\gamma ,\hat{x}) = 2\hat{t}$.

We claim that the following system in variables v and s

$$\begin{aligned} {\left\{ \begin{array}{ll} \left\langle b(\gamma ), v \right\rangle = s,\,\forall \gamma \in \mathcal F\\ v\in \mathcal V(\mathcal F),\,s\in {\mathbb {R}}\end{array}\right. } \end{aligned}$$

has a nonzero solution. Indeed, we may replace the first constraint with at most

$$\begin{aligned} \mathrm {aff}\dim (\left\{ b(\gamma ):\, \gamma \in \mathcal F\right\} ) + 1 \le \dim (\mathcal V(\mathcal F)) \end{aligned}$$

homogeneous linear equalities in the variables v and s. The claim then follows by noting that the equivalent system is an under-constrained homogeneous system of linear equalities and thus has a nonzero solution (v, s). It is easy to verify that $v\ne 0$ and hence, by scaling, we may take $v\in \mathbf{S}^{N-1}$.

We will modify $(\hat{x},\hat{t})$ in the (v, s) direction. For $\alpha \in {\mathbb {R}}$, define

$$\begin{aligned} (x_\alpha ,t_\alpha ):= \left( \hat{x} + \alpha v,\, \hat{t} + \alpha s\right) . \end{aligned}$$

We will sketch the existence of an $\alpha >0$ such that $(x_\alpha ,t_\alpha )$ satisfies the conclusions of Lemma 4. A similar line of reasoning will produce an analogous $\alpha <0$. This will complete the proof sketch.

Suppose $\gamma \in \mathcal F$. Then, by our choice of v and s, the function $\alpha \mapsto q(\gamma ,x_\alpha )-2t_\alpha = q(\gamma ,\hat{x}) -2t = 0$ is identically zero. Now suppose $\gamma \in \varGamma \setminus \mathcal F$. Then, the function $\alpha \mapsto q(\gamma ,x_\alpha )-2t_\alpha$ is a convex quadratic function which is negative at $\alpha = 0$.

We conclude that the following set

$$\begin{aligned} \mathcal Q := \left\{ \alpha \mapsto q(\gamma ,x_\alpha ) - 2 t_\alpha :\, \gamma \in {\mathop {\mathrm {extr}}}(\varGamma )\right\} \setminus \left\{ 0\right\} , \end{aligned}$$

consists of convex quadratic functions which are negative at $\alpha = 0$. The finiteness of this set follows from the assumption that $\varGamma$ is polyhedral.

Assumption 1 implies that at least one of the functions in $\mathcal Q$ is strictly convex. Then as $\mathcal Q$ is a finite set, there exists an $\alpha _+ >0$ such that $q(\alpha _+)\le 0$ for all $q\in \mathcal Q$ with at least one equality. We emphasize that this is the step where Assumption 2 cannot be dropped.

Finally, it is easy to check that $(x_{\alpha _+}, t_{\alpha _+})$ satisfies the conclusions of Lemma 4.