A Tensor Analogy of Yuan’s Theorem of the Alternative and Polynomial Optimization with Sign structure

Abstract

Yuan’s theorem of the alternative is an important theoretical tool in optimization, which provides a checkable certificate for the infeasibility of a strict inequality system involving two homogeneous quadratic functions. In this paper, we provide a tractable extension of Yuan’s theorem of the alternative to the symmetric tensor setting. As an application, we establish that the optimal value of a class of nonconvex polynomial optimization problems with suitable sign structure (or more explicitly, with essentially nonpositive coefficients) can be computed by a related convex conic programming problem, and the optimal solution of these nonconvex polynomial optimization problems can be recovered from the corresponding solution of the convex conic programming problem. Moreover, we obtain that this class of nonconvex polynomial optimization problems enjoy exact sum-of-squares relaxation, and so, can be solved via a single semidefinite programming problem.

Appendix

Proof of Proposition 2.1

Proof

As any SOS polynomial takes non-negative value, $\mathrm{SOS}_{m,n} \cap E_{m,n} \subseteq \mathrm{PSD}_{m,n} \cap E_{m,n}$ always holds. We only need to show the converse inclusion. To establish this, let $\mathcal {A} \in \mathrm{PSD}_{m,n} \cap E_{m,n}$, and let us consider the associated homogeneous polynomial:

$$\begin{aligned} f(x)=\langle \mathcal {A}, x^{\otimes m}\rangle =\sum _{i_1,\ldots ,i_m=1}^{n} \mathcal {A}_{i_1\cdots i_m}x_{i_1}\cdots x_{i_m}. \end{aligned}$$

Then, $f$ is a polynomial which takes non-negative value. Note that

$$\begin{aligned} f(x)=\sum _{i_1,\ldots ,i_m=1}^{n} \mathcal {A}_{i_1\cdots i_m}x_{i_1}\cdots x_{i_m}=\sum _{i=1}^n(\mathcal {A}_{ii\cdots i})x_i^{m}+\sum _{(i_1,\ldots ,i_m) \notin I}(\mathcal {A}_{i_1\cdots i_m})x_{i_1} \cdots x_{i_m}, \end{aligned}$$

where $I:=\{(i,i,\ldots ,i) \in \mathbb {N}^m: 1 \le i \le n\}.$ As $\mathcal {A}$ is essentially nonpositive, $\mathcal {A}_{i_1i_2\cdots i_m} \le 0$ for all $(i_1,\ldots ,i_m) \notin I$. Now, let $f(x)=\sum _{i=1}^n f_{m,i} x_i^{m}+\sum _{\alpha \in \Omega _f}f_{\alpha }x^{\alpha }$. Then, $f_{m,i}=\mathcal {A}_{ii\cdots i}$ and $f_{\alpha } < 0$ for all $\alpha \in \Omega _f$ where $\Omega _f=\{\alpha =(\alpha _1,\ldots ,\alpha _n) \in (\mathbb {N}\cup \{0\})^n: f_{\alpha } \ne 0 \text { and } \alpha \ne m e_i, \ i=1,\ldots ,n\},$ and $e_i$ is the vector where its $i$th component is one and all the other components are zero. Recall that $\Delta _f = \{\alpha =(\alpha _1,\ldots ,\alpha _n) \in \Omega _f: f_{\alpha } < 0 \text { or } \alpha \notin (2\mathbb {N}\cup \{0\})^n\}.$ Note that $f_{\alpha }<0$ for all $\alpha \in \Omega _f$, and so, $\Delta _f=\Omega _f$. It follows that

$$\begin{aligned} \hat{f}(x)&:= \sum _{i=1}^n f_{m,i} x_i^{m}-\sum _{\alpha \in \Delta _f}|f_{\alpha }|x^{\alpha }\\&= \sum _{i=1}^n f_{m,i} x_i^{m}+\sum _{\alpha \in \Delta _f}f_{\alpha }x^{\alpha } \\&= \sum _{i=1}^n f_{m,i} x_i^{m}+\sum _{\alpha \in \Omega _f}f_{\alpha }x^{\alpha } = f(x). \end{aligned}$$

So, $\hat{f}$ is also a polynomial which takes non-negative value. Thus, the conclusion follows by Lemma 2.1. $\square$