Optimality conditions and finite convergence of Lasserre’s hierarchy

Abstract

Lasserre’s hierarchy is a sequence of semidefinite relaxations for solving polynomial optimization problems globally. This paper studies the relationship between optimality conditions in nonlinear programming theory and finite convergence of Lasserre’s hierarchy. Our main results are: (i) Lasserre’s hierarchy has finite convergence when the constraint qualification, strict complementarity and second order sufficiency conditions hold at every global minimizer, under the standard archimedean condition; the proof uses a result of Marshall on boundary hessian conditions. (ii) These optimality conditions are all satisfied at every local minimizer if a finite set of polynomials, which are in the coefficients of input polynomials, do not vanish at the input data (i.e., they hold in a Zariski open set). This implies that, under archimedeanness, Lasserre’s hierarchy has finite convergence generically.

Appendix: Non-identically vanishing of \(\fancyscript{R}\) and \(\fancyscript{D}\)

Given any positive degrees $d_0, d_1, \ldots , d_k, d_{k+1}$, we show that the polynomial $\fancyscript{R}(p_0,\ldots ,p_k;q)$ defined in (4.6) and the polynomial $\fancyscript{D}(p_0,p_1,\ldots ,p_k)$ defined in (4.11) do not vanish identically in $p_i \in \mathbb R [x]_{d_i}\,(i=0,\ldots ,k)$ and $q \in \mathbb R [x]_{d_{k+1}}$. Without loss of generality, we can assume all $d_1, \ldots , d_k>1$ because linear constraints in (4.2) can be removed by eliminating variables.

First, we prove that the polynomial $\fancyscript{R}$ defined in (4.6) does not vanish identically in the space $\mathbb R [x]_{d_0} \times \cdots \times \mathbb R [x]_{d_k} \times \mathbb R [x]_{d_{k+1}}$. We only consider the case $k<n$, because if $k=n$ then $\widetilde{p_1}(\tilde{x})=\cdots =\widetilde{p_k}(\tilde{x}) =\widetilde{q}(\tilde{x})$ has no nonzero complex solution in the generic case. By Proposition 4.1, it is enough to show that the homogeneous polynomial system (4.5) does not have a complex solution $\tilde{x} \ne 0$ for generic $p_0, p_1,\ldots , p_k, q$. We prove this in two cases:

• ($\mathbf x_0 \ne 0$)  We can scale as $x_0=1$, and the system (4.5) is then reduced to (4.4). When $p_0, p_1,\ldots , p_k$ are generic, the set $\mathcal{K }(p)$ defined in (4.3) is finite (cf. [17, Prop. 2.1]). Thus, when $q$ is also generic, (4.4) does not have a solution in $\mathbb C ^n$.

• ($\mathbf x_0 = 0$)  The system (4.5) is then reduced to

  $$\begin{aligned} \left\{ \begin{array}{c} \text{ rank }\left[ \nabla _x p_0^h(x) \nabla _x p_1^h (x) \cdots \nabla _x p_k^h(x)\right] \le k, \\ \quad p_1^h(x) = \cdots = p_k^h(x) = q^h(x) = 0. \end{array} \right. \end{aligned}$$

  (6.1)

  (Here, $f^h$ denotes the homogeneous part of the highest degree for a polynomial $f$.) When $p_1,\ldots , p_k$ are generic, we have $\varDelta (p_1^h,\ldots , p_k^h) \ne 0$. By definition of $\varDelta$ (cf. §2.4), if $p_1^h(x) = \cdots = p_k^h(x) =0$ and $x \ne 0$, then

  $$\begin{aligned} \text{ rank }\left[ \nabla _x p_1^h (x) \cdots \nabla _x p_k^h(x) \right] = k. \end{aligned}$$

  So, if $x$ satisfies (6.1), there must exist scalars $c_1,\ldots , c_k$ such that

  $$\begin{aligned} \nabla _x p_0^h(x) = c_1 \nabla _x p_1^h (x) + \cdots + c_k \nabla _x p_k^h(x). \end{aligned}$$

  Since each $p_i^h$ is a form, by Euler’s formula for homogeneous polynomials (cf. [18, §2]), we can get

  $$\begin{aligned} d_0 p_0^h(x) = x^T \nabla _x p_0^h(x) =\sum \limits _{i=1}^k c_i x^T \nabla _x p_k^h(x) =\sum \limits _{i=1}^k c_i d_k p_k^h(x) = 0. \end{aligned}$$

  This means that (6.1) implies

  $$\begin{aligned} p_0^h(x) = p_1^h(x) = \cdots = p_k^h(x) =0,\\ \text{ rank }\left[ \nabla _x p_0^h (x) \nabla _x p_1^h (x) \cdots \nabla _x p_k^h(x) \right] = k. \end{aligned}$$

  Any $x$ satisfying the above must be zero if $\varDelta (p_0^h, p_1^h,\ldots ,p_k^h) \ne 0$.

Combining the above two cases, we know the polynomial system (4.5) has no complex solution $\tilde{x} \ne 0$ when $p_0, p_1,\ldots , p_k,q$ are generic.

Second, we show that the polynomial $\fancyscript{D}(p_0,p_1,\ldots ,p_k)$ defined in (4.11) does not identically vanish in the space $\mathbb R [x]_{d_0} \times \mathbb R [x]_{d_1} \times \cdots \times \mathbb R [x]_{d_k}$. By Proposition 4.2, it is enough to prove that there exist $p_i \in \mathbb R [x]_{d_i} \,(i=0,\ldots ,k)$ such that (4.10) has no complex solution $(\tilde{x},y)$ with $\tilde{x}\ne 0, y\ne 0$. We prove this in two cases.

• ($\mathbf x_0 \ne 0$)  We scale as $x_0 = 1$, and (4.10) is then reduced to (4.9). Choose polynomials $\hat{p}_i$ as follows:

  $$\begin{aligned} \hat{p}_0:= f_0 \in \mathbb R [x_{k+1},\ldots ,x_n]_{d_0}, \, \hat{p}_1:=x_1^{d_1}-1, \ldots , \hat{p}_k:=x_k^{d_k}-1. \end{aligned}$$

  Clearly, on the variety $V(\hat{p}_1,\ldots ,\hat{p}_k)$, the gradients $\nabla _x \hat{p}_1,\ldots , \nabla _x \hat{p}_k$ are linearly independent, and so are

  $$\begin{aligned} \left[ \begin{array}{l} \nabla _x \hat{p}_i \\ (\nabla _x^2 \hat{p}_i) y \\ \end{array}\right] , \left[ \begin{array}{l} 0 \\ \nabla _x \hat{p}_i \\ \end{array}\right] \quad (i=1,\ldots ,k). \end{aligned}$$

  Thus, (4.9) is equivalent to (4.8). If $(x,\lambda )$ is a critical pair, then $\lambda _1 = \cdots = \lambda _k =0$ and $D:=\text{ diag }(d_1x_1^{d_1-1}, \ldots ,d_kx_k^{d_k-1})$ is invertible. Denote $x_{I}:=(x_1,\ldots ,x_k,)$ and $x_{II}:=(x_{k+1},\ldots ,x_n)$. Note that ($\hat{p}:=(\hat{p}_0,\ldots ,\hat{p}_k)$)

  $$\begin{aligned} H_{\hat{p}}(x,0) = \left[ \begin{array}{c|c|c} 0 &{} 0 &{} D\\ \hline 0 &{} \nabla _{x_{II} }^2 f_0 &{} 0 \\ \hline D &{} 0 &{} 0 \end{array} \right] . \end{aligned}$$

  (In the above, the 0’s denote zero matrices of proper dimensions.) The matrix $H_{\hat{p}}(x,0)$ is nonsingular if and only if $\nabla _{x_{II} }^2 f_0$ is nonsingular. Therefore, (4.8) has a solution if and only if there exists $u \in \mathbb C ^{n-k}$ satisfying

  $$\begin{aligned} \nabla _{x_{II}} f_0(u) =0, \quad \det \, \nabla _{x_{II}}^2 f_0(u) = 0. \end{aligned}$$

  However, the above is possible only if

  $$\begin{aligned} \varDelta \left( \frac{\partial f_0}{\partial x_{k+1}}, \ldots , \frac{\partial f_0}{\partial x_{n}}\right) = 0. \end{aligned}$$

  So, if $f_0$ is generic, then $H_{\hat{p}}(x,0)$ is nonsingular for all $(x,\lambda )$ satisfying (4.2) corresponding to $\hat{p}_0, \hat{p}_1, \ldots , \hat{p}_k$. By continuity of roots of polynomials, $H_{p}(x,\lambda )$ is nonsingular for every pair $(x,\lambda )$ satisfying (4.2), if each $p_i$ is generic and close enough to $\hat{p_i}$.

• ($\mathbf x_0 = 0$)  The polynomial system (4.10) is then reduced to

  $$\begin{aligned} \left\{ \begin{array}{c} \text{ rank }\, Q(x,y) \le 2k, \quad p_1^h(x) = \cdots = p_k^h(x) = 0, \\ \big (\nabla _{x} p_1^h(x)\big )^Ty = \cdots = \big (\nabla _{x} p_1^h(x)\big )^Ty = 0. \end{array}\right. \end{aligned}$$

  (6.2)

  In the above, $Q(x,y)$ denotes the matrix

  $$\begin{aligned} \left[ \begin{array}{l} \nabla _{x} p_0^h \quad \cdots \quad \nabla _{x} p_k^h\\ \big (\nabla _{x}^2 p_0^h\big ) y \cdots \big (\nabla _{x}^2 p_k^h\big ) y \nabla _{x} p_1^h \quad \cdots \quad \nabla _{x} p_k^h \end{array}\right] . \end{aligned}$$

  We show that if $p_0,p_1,\ldots ,p_k$ are generic, then (6.2) has no complex solution $(x,y)$ with $x\ne 0, y\ne 0$. When all $p_i$ are generic, for every $x\ne 0$ satisfying

  $$\begin{aligned} p_1^h(x) = \cdots = p_k^h(x) = 0, \end{aligned}$$

  the gradients $\nabla _{x} p_1^h, \ldots , \nabla _{x} p_k^h$ are linearly independent. When $\text{ rank }\, Q(x,y) \le 2k$, there exist scalars $c_1, \ldots , c_k$ such that

  $$\begin{aligned} \nabla _x p_0^h(x) - \sum \limits _{i=1}^k c_i \nabla _x p_i^h(x) = 0. \end{aligned}$$

  (6.3)

  By Euler’s formula for homogeneous polynomials, the above implies

  $$\begin{aligned} d_0 p_0^h(x) = x^T \nabla _x p_0^h(x) = \sum \limits _{i=1}^k \lambda _i x^T \nabla _x p_i^h(x) = \sum \limits _{i=1}^k \lambda _i d_i p_i^h(x) = 0. \end{aligned}$$

  This means that if some $x\ne 0$ satisfies (6.2) then the polynomial system

  $$\begin{aligned} p_0^h(x) = p_1^h(x) = \cdots = p_k^h(x) = 0 \end{aligned}$$

  is singular. But this is impossible unless $\varDelta (p_0^h, p_1^h,\ldots , p_k^h) = 0$.

Combining the above two cases, we know that there exist polynomials $p_i \in \mathbb R [x]_{d_i}\,(i=0,\ldots ,k)$ such that there are no complex $\tilde{x} \ne 0, y\ne 0$ satisfying (4.10). This shows that $\fancyscript{D}(p_0,\ldots ,p_k)$ does not identically vanish.