Error bounds for monomial convexification in polynomial optimization

Abstract

Convex hulls of monomials have been widely studied in the literature, and monomial convexifications are implemented in global optimization software for relaxing polynomials. However, there has been no study of the error in the global optimum from such approaches. We give bounds on the worst-case error for convexifying a monomial over subsets of . This implies additive error bounds for relaxing a polynomial optimization problem by convexifying each monomial separately. Our main error bounds depend primarily on the degree of the monomial, making them easy to compute. Since monomial convexification studies depend on the bounds on the associated variables, in the second part, we conduct an error analysis for a multilinear monomial over two different types of box constraints. As part of this analysis, we also derive the convex hull of a multilinear monomial over .

Appendix: Missing proofs

Proof of Proposition 3.4

Since and make for all , we have . The lower bound of 0 comes from

If , then implies that and so by (7c), we have . For the fourth claim we have . Denote this simplex by . The assumption means that for all . Substituting this point into (7c) gives us for all . This leads to . Since , . Note that . The positivity of ${\beta }$ then makes it clear that . Hence , where . Now,

Since we have already argued , it follows that . $\square$

Proof of Lemma 3.4

For nontriviality, assume .

(1) The first derivative is . If , then and for all and hence is strictly increasing over and for all .

(2 & 3) Now assume . Set and realize that and . Then we have for . Therefore is decreasing on , which implies for . Hence . The construction of also implies , and hence is increasing, for . Since , it follows that there is a unique real number in such that . Thus we have for and for . If is odd, the other root is obtained by applying Descartes’ rule of signs as in the first claim.

(4) Take and define . If , then the first claim in this lemma, with replaced by , gives us . Now assume . Applying the second claim in this lemma, after replacing with , tells us there is a unique real that is a root of in . Now because . Then the third claim in this lemma, with replaced by , gives us and consequently, the proposed fourth claim.

For the final part, note that the roots of and its complemented polynomial are in bijection under the relation . Descartes’ rule of signs tells us that has exactly one positive root besides . When , this root must be in because otherwise we would get a contradiction to not having any roots in . Descartes’ rule also tells us there is exactly one negative root when is odd. This translates to having a root in if and only if is odd. $\square$

Proof of of Proposition 4.2

Note that . We first claim that is strictly increasing on ${(0,t^{*})}$. In fact, we argue the stronger claim that for all . This claim is equivalent to showing that , which is equivalent to . The function is convex and is zero-valued at and . Therefore, by convexity, for all , and hence, we have for all .

Since , is strictly increasing on ${(0,t^{*})}$, and , the condition implies that yields the maximum value in the formula for ${\mathscr {D}_{r,n}}$. Now suppose . Since is a stationary point, . Now,

$$\begin{aligned} 0< & {} \mathscr {D}_{r,n}\le (1+{t^{**}}(r-1))^n-r^{n{t^{**}}} = r^{\frac{n^2{t^{**}}}{n-1}}\left( \frac{\ln r}{r-1}\right) ^{\frac{n}{n-1}} -r^{n{t^{**}}} \\= & {} r^{n{t^{**}}}\left( r^{\frac{n{t^{**}}}{n-1}}\left( \frac{\ln r}{r-1}\right) ^{\frac{n}{n-1}}-1\right) \\\le & {} r^{n-1}\left( r \left( \frac{\ln r}{r-1}\right) ^{\frac{n}{n-1}}-1\right) , \end{aligned}$$

where the last inequality uses and . Finally, if ${t^{*}< (n-1)/n < t^{**}}$, since can be arbitrarily close to , we can only bound and in above by and , respectively, to obtain the last proposed bound on ${\mathscr {D}_{r,n}}$. $\square$