An SDP method for fractional semi-infinite programming problems with SOS-convex polynomials

Abstract

In this paper, we study a class of fractional semi-infinite polynomial programming problems involving sos-convex polynomial functions. For such a problem, by a conic reformulation proposed in our previous work and the quadratic modules associated with the index set, a hierarchy of semidefinite programming (SDP) relaxations can be constructed and convergent upper bounds of the optimum can be obtained. In this paper, by introducing Lasserre’s measure-based representation of nonnegative polynomials on the index set to the conic reformulation, we present a new SDP relaxation method for the considered problem. This method enables us to compute convergent lower bounds of the optimum and extract approximate minimizers. Moreover, for a set defined by infinitely many sos-convex polynomial inequalities, we obtain a procedure to construct a convergent sequence of outer approximations which have semidefinite representations (SDr). The convergence rate of the lower bounds and outer SDr approximations are also discussed.

Appendix A

We first recall some definitions and properties about the so-called needle polynomials required in the proof of Proposition 4.2.

Definition A.1

For $k\in \mathbb {N}$, the Chebyshev polynomial $T_k(t)\in {\mathbb R}[t]_k$ is defined by

$$\begin{aligned} T_k(t)=\left\{ \begin{aligned}&\cos (k \arccos t)&\text {for}\ |t|\le 1,\\&\frac{1}{2}(t+\sqrt{t^2-1})^k+\frac{1}{2}(t-\sqrt{t^2-1})^k&\text {for}\ |t|\ge 1. \end{aligned} \right. \end{aligned}$$

Definition A.2

[23] For $k\in \mathbb {N}$, $h\in (0, 1)$, the needle polynomial $v_k^h(t)\in {\mathbb R}[t]_{4k}$ is defined by

$$\begin{aligned} v_k^h(t)=\frac{T_k^2(1+h^2-t^2)}{T_k^2(1+h^2)}. \end{aligned}$$

Theorem A.1

[8, 22, 23] For $k\in \mathbb {N}$, $h\in (0, 1)$, the following properties hold for $v_k^h(t)$:

$$\begin{aligned} \begin{aligned}&v_k^h(0)=1,&\\&0\le v_k^h(t) \le 1&\text {for}\ t\in [-1, 1], \\&v_k^h(t) \le 4e^{-\frac{1}{2}kh}&\text {for}\ t\in [-1, 1]\ \text {with}\ |t|\ge h. \end{aligned} \end{aligned}$$

The following result gives a lower estimator which is used in the proof of Proposition 4.2 to lower bound the integral of the needle polynomial.

Proposition A.1

[44, Lemma 13] Let $\phi (t)\in {\mathbb R}[t]_k$ be a polynomial of degree up to $k\in \mathbb {N}$, which is nonnegative over ${\mathbb R}_{\ge 0}$ and satisfies $\phi (0)=1, \phi (t)\le 1$ for all $t\in [0,1]$. Let $\Phi _k : {\mathbb R}_{\ge 0} \rightarrow {\mathbb R}_{\ge 0}$ be defined by

$$\begin{aligned} \Phi _k(t)=\left\{ \begin{aligned}&1-2k^2t&\text {if}\ t\le \frac{1}{2k^2},\\&0&\text {otherwise}. \end{aligned} \right. \end{aligned}$$

Then $\Phi _k(t) \le p(t)$ for all $t\in {\mathbb R}_{\ge 0}$.

Proof of Proposition 4.2

For any $k\in \mathbb {N}$, let $\rho (k)=\frac{1}{16k^2}$ and $h(k):=(4n+2)\log k/\lfloor k/2\rfloor$. Then, there exists a $k'\in \mathbb {N}$ such that $\rho (k)\le h(k)<\min \{\epsilon _{\textbf{Y}}, 1\}$ for any $k\ge k'$. Fix a $k\ge k'$, a linear functional $\mathscr {L}_k\in ({\mathbb R}[x]_{2\textbf{d}})^*$ satisfying the conditions in (9), and a minimizer $y^{\star }$ of $\min _{y\in \textbf{Y}} -\mathscr {L}_k(p(x,y))$. Using the needle polynomial $v_k^h(t)\in {\mathbb R}[t]$, define $\sigma _k(y):=v_{\lfloor k/2\rfloor }^{h(k)}(\Vert y-y^{\star }\Vert /(2\sqrt{n}))$. Then, the polynomial $\tilde{\sigma }:=\sigma _k/\int _{\textbf{Y}}\sigma _k\textrm{d}y\in \Sigma _k^2[y]$ and feasible to (15). Hence, by Taylor’s theorem,

$$\begin{aligned} \begin{aligned} E(\mathscr {L}_k)&\le \frac{1}{\int _{\textbf{Y}}\sigma _k(y)\textrm{d}y}\int _{\textbf{Y}}-\mathscr {L}_k({p(x,y)})\sigma _k(y)\textrm{d}y - p^{\star }_k\\&= \frac{1}{\int _{\textbf{Y}}\sigma _k(y)\textrm{d}y}\int _{\textbf{Y}}\sigma _k(y)(-\mathscr {L}_k({p(x,y)} - p^{\star }_k) \textrm{d}y \\&\le \frac{B_1}{\int _{\textbf{Y}}\sigma _k(y)\textrm{d}y}\int _{\textbf{Y}}\sigma _k(y)\Vert y-y^{\star }\Vert \textrm{d}y \end{aligned} \end{aligned}$$

(25)

Define two sets

$$\begin{aligned} \textbf{Y}_1:=\textbf{B}_{2\sqrt{n}h(k)}^n(y^{\star })\cap \textbf{Y}\quad \text {and}\quad \textbf{Y}_2:=\textbf{B}_{2\sqrt{n}\rho (k)}^n(y^{\star })\cap \textbf{Y}\subseteq \textbf{Y}_1. \end{aligned}$$

Then,

$$\begin{aligned} \int _{\textbf{Y}}\sigma _k(y)\textrm{d}y \ge \int _{\textbf{Y}_1}\sigma _k(y)\textrm{d}y \ge \int _{\textbf{Y}_2}\sigma _k(y)\textrm{d}y. \end{aligned}$$

(26)

As $\textbf{Y}\subseteq \textbf{H}^n$,

$$\begin{aligned} \begin{aligned} \int _{\textbf{Y}}\sigma _k(y)\Vert y-y^{\star }\Vert \textrm{d}y&= \int _{\textbf{Y}_1}\sigma _k(y)\Vert y-y^{\star }\Vert \textrm{d}y + \int _{\textbf{Y}\setminus \textbf{Y}_1}\sigma _k(y)\Vert y-y^{\star }\Vert \textrm{d}y\\&\le 2\sqrt{n}h(k)\int _{\textbf{Y}_1}\sigma _k(y)\textrm{d}y + 2\sqrt{n} \int _{\textbf{Y}\setminus \textbf{Y}_1}\sigma _k(y)\textrm{d}y. \end{aligned} \end{aligned}$$

(27)

By Theorem A.1, we have $\sigma _k(y)\le 4e^{-\frac{1}{2}h(k)\lfloor k/2\rfloor }$ for any $y\in \textbf{Y}\setminus \textbf{Y}_1$ and hence

$$\begin{aligned} \int _{\textbf{Y}\setminus \textbf{Y}_1}\sigma _k(y)\textrm{d}y \le 4e^{-\frac{1}{2}h(k)\lfloor k/2\rfloor }\cdot \text { vol}(\textbf{Y}\setminus \textbf{Y}_1)\le 4e^{-\frac{1}{2}h(k)\lfloor k/2\rfloor }\cdot \text { vol}(\textbf{H}^n). \end{aligned}$$

Moreover, by Proposition A.1, we have

$$\begin{aligned} \sigma _k(y)\ge \Phi _{2k}(\Vert y-y^{\star }\Vert /2\sqrt{n})=1-8k^2(\Vert y-y^{\star }\Vert /2\sqrt{n})\ge \frac{1}{2}, \end{aligned}$$

for all $y\in \textbf{Y}_2$. Therefore, A5 implies that

$$\begin{aligned} \int _{\textbf{Y}_2}\sigma _k(y)\textrm{d}y \ge \frac{1}{2}\text { vol}(\textbf{Y}_2)\ge \frac{1}{2}\eta _{\textbf{Y}} 2^n n^{n/2} \rho (k)^n\text { vol}(\textbf{B}^n) =\frac{\eta _{\textbf{Y}} n^{n/2}\text { vol}(\textbf{B}^n)}{2^{3n+1}k^{2n}}. \end{aligned}$$

(28)

Combining (25)-(28), we obtain

$$\begin{aligned} \begin{aligned} E(\mathscr {L}_k)&\le 2\sqrt{n}B_1\left( h(k) +4e^{-\frac{1}{2}h(k)\lfloor k/2\rfloor } \frac{2^{3n+1}k^{2n}\text { vol}(\textbf{H}^n)}{\eta _{\textbf{Y}} n^{n/2}\text { vol}(\textbf{B}^n)}\right) \\&=2\sqrt{n}B_1(h(k)+Ce^{-\frac{1}{2}h(k)\lfloor k/2\rfloor } k^{2n}) \end{aligned} \end{aligned}$$

(29)

The conclusion follows by substituting $h(k)=(4n+2)\log k/\lfloor k/2\rfloor$ in (29). $\square$