Sum-of-squares relaxations for polynomial min–max problems over simple sets

Abstract

We consider min–max optimization problems for polynomial functions, where a multivariate polynomial is maximized with respect to a subset of variables, and the resulting maximal value is minimized with respect to the remaining variables. When the variables belong to simple sets (e.g., a hypercube, the Euclidean hypersphere, or a ball), we derive a sum-of-squares formulation based on a primal-dual approach. In the simplest setting, we provide a convergence proof when the degree of the relaxation tends to infinity and observe empirically that it can be finitely convergent in several situations. Moreover, our formulation leads to an interesting link with feasibility certificates for polynomial inequalities based on Putinar’s Positivstellensatz.

Appendices

Appendix A Convergence rates of matrix-valued SOS

We extend the proof of [11, Theorem 1] to matrix-valued polynomials, using the same technique as [21], and following the notations of [11] closely.

Proposition 3

Let $r>0$ and $s \geqslant 3r$, and $\varepsilon (s) = \big [ \big ( 1 - \frac{6r^2}{s^2} \big )^{-d} - 1 \big ] \sim _{s \rightarrow +\infty } \frac{ 6 r^2 d }{s^2}$. For any multivariate matrix-valued trigonometric polynomial f of degree less than 2r, written $f(x) = \sum _{\Vert \omega \Vert _\infty \leqslant 2r} \hat{f}(\omega ) e^{2i\pi \omega ^\top x}$,

$$\begin{aligned}{} & {} \forall x \in [0,1]^d, f(x) \succcurlyeq \varepsilon (s) \sum _{\Vert \omega \Vert _\infty \leqslant 2r, \ \omega \ne 0}\!\! \Vert \hat{f}(\omega ) \Vert _{\textrm{op}} \ \\{} & {} \quad \Rightarrow \ f \ \text{ is } \text{ a } \text{ sum } \text{ of } \text{ squares } \text{ of } \text{ polynomials } \text{ of } \text{ degree } s. \end{aligned}$$

Proof

We consider the following integral operator on 1-periodic matrix-valued functions on $[0,1]^d$, defined as

$$\begin{aligned} Th(x) = \int _{[0, 1]^d} |q(x-y)|^2 h(y) dy, \end{aligned}$$

(A1)

for a well-chosen 1-periodic function q which is a trigonometric polynomial of degree s. The function $x \mapsto |q(x-y)|^2$ is an element of the finite-dimensional cone of SOS polynomials of degree s, thus, by design, if h has positive semi-definite values, then Th is a sum of squares of matrix polynomials of degree less than s. We will find h such that $Th = f$.

In the Fourier domain, since convolutions lead to pointwise multiplication and vice-versa, we have for all $\omega \in \mathbb {Z}^d$, where $\hat{q} *\hat{q}(\omega )$ is a shorthand for $(\hat{q} *\hat{q})(\omega )$:

$$\begin{aligned} {\widehat{Th}}(\omega ) = \hat{q} *\hat{q}(\omega ) \cdot \hat{h}(\omega ), \end{aligned}$$

and thus, the candidate h is defined by its Fourier series, which is equal to zero for $\Vert \omega \Vert _\infty > 2r$, and to

$$\begin{aligned} \frac{\hat{f}(\omega ) }{ \hat{q} *\hat{q}(\omega )}\end{aligned}$$

otherwise. If we impose that $\hat{q} *\hat{q}(0)=1$, we then have

$$\begin{aligned} f - h= & {} \sum _{\omega \in \mathbb {Z}^d} \hat{f}(\omega ) \Big ( 1 - \frac{1}{\hat{q} *\hat{q}(\omega )} \Big ) \exp ( 2i \pi \omega ^\top \cdot ) \\ {}= & {} \sum _{\omega \ne 0 } \hat{f}(\omega ) \Big ( 1 - \frac{1}{\hat{q} *\hat{q}(\omega )} \Big ) \exp ( 2i \pi \omega ^\top \cdot ). \end{aligned}$$

We then get:

$$\begin{aligned} \sup _{x \in [0,1]^d} \Vert f(x) - h(x)\Vert _{\textrm{op}} \leqslant \sum _{\omega \ne 0} \big \Vert \hat{f}(\omega ) \big \Vert _{\textrm{op}} \cdot \max _{ \Vert \omega \Vert _\infty \leqslant 2r} \Big | \frac{1}{\hat{q} *\hat{q}(\omega )} \, - 1\Big |. \end{aligned}$$

(A2)

With the choice $\hat{q}(\omega ) = a \prod \nolimits _{i=1}^d \Big ( 1 - \frac{|\omega _i|}{s} \Big )_+,$ with a a normalizing constant, we get $\hat{q}*\hat{q}(0)=1$ and $\max _{ \Vert \omega \Vert _\infty \leqslant 2r} \big | \frac{1}{\hat{q} *\hat{q}(\omega )} \, - 1\big | \leqslant \varepsilon (s)$ (see [11] for details). Thus, for all $x \in [0,1]^d$, using Eq. (A2) and the assumption on f:

$$\begin{aligned} h(x) = f(x) - ( f(x) - h(x)) \succcurlyeq \varepsilon (s) \sum _{ \omega \ne 0} \Vert \hat{f}(\omega ) \Vert _{\textrm{op}} - \varepsilon (s) \sum _{ \omega \ne 0} \Vert \hat{f}(\omega ) \Vert _{\textrm{op}} = 0, \end{aligned}$$

which leads to the desired result. $\square$

Appendix B Alternating optimization for the two-stage approach

In this section, we explore briefly the possibility evoked in Sect. 4.1 of trying to minimize Eq. () with respect to $\Sigma$ as well. This is a non-convex problem, and alternating optimization has a particularly simple formulation. Indeed, in the kernelized version in Eq. (18), this corresponds to replacing $\mu$ by the previous value of $\alpha$ and iterating. Since the first upper-bound is minimized exactly, at the second iteration and all later ones, the matrix $\Sigma$ corresponds to a Dirac measure, and the upper-bounding polynomial is so that its value at this point is minimized. This is shown empirically in Fig. 5: even in the good attraction basin, the alternating optimization does not lead to the global optimum.

Fig. 5

Two-stage approach for trigonometric polynomials in one dimension, with alternating optimization and $r=2$, with 6 iterations