Approximation methods for complex polynomial optimization

Abstract

Complex polynomial optimization problems arise from real-life applications including radar code design, MIMO beamforming, and quantum mechanics. In this paper, we study complex polynomial optimization models where the objective function takes one of the following three forms: (1) multilinear; (2) homogeneous polynomial; (3) symmetric conjugate form. On the constraint side, the decision variables belong to one of the following three sets: (1) the $m$-th roots of complex unity; (2) the complex unity; (3) the Euclidean sphere. We first discuss the multilinear objective function. Polynomial-time approximation algorithms are proposed for such problems with assured worst-case performance ratios, which depend only on the dimensions of the model. Then we introduce complex homogenous polynomial functions and establish key linkages between complex multilinear forms and the complex polynomial functions. Approximation algorithms for the above-mentioned complex polynomial optimization models with worst-case performance ratios are presented. Numerical results are reported to illustrate the effectiveness of the proposed approximation algorithms.

Appendix: Proofs of the lemmas

Lemma 3.2 Define $F_m:\mathbb {C}\mapsto \mathbb {C}$ with $F_m(x):=\frac{m(2 - \omega _m - \omega _m^{-1})}{8\pi ^2} \sum _{\ell =0}^{m-1} \omega _m^\ell \left( \arccos \left( -\hbox {Re}\,\omega _m^{-\ell } x\right) \right) ^2$.

• (1) If $a \in \mathbb {C}$ and $b \in \mathbf{\Omega }_m$, then $F_m (ab)=bF_m(a)$.

• (2) If $a \in \mathbb {R}$, then $F_m(a)\in \mathbb {R}$.

Proof

• (1) If $b \in \mathbf{\Omega }_m$, let $b = \omega _m^k$ for some $k\in \mathbb {Z}$. It holds that

  $$\begin{aligned} F_m(ab)=F_m(\omega _m^k a)&= \frac{m(2 - \omega _m - \omega _m^{-1})}{8\pi ^2} \sum _{\ell =0}^{m-1} \omega _m^\ell \left( \arccos \left( -\hbox {Re}\,\omega _m^{-\ell }\omega _m^k a \right) \right) ^2\\&= \omega _m^k\frac{m(2\!-\!\omega _m \!-\!\omega _m^{-1})}{8\pi ^2} \sum _{\ell =0}^{m-1} \omega _m^{\ell -k} \left( \arccos \left( \!-\!\hbox {Re}\,\omega _m^{-(\ell -k)} a \right) \right) ^2\\&= b \frac{m(2-\omega _m -\omega _m^{-1})}{8\pi ^2} \sum _{j=-k}^{m-1-k} \omega _m^{j} \left( \arccos \left( -\hbox {Re}\,\omega _m^{-j} a \right) \right) ^2\\&= b F_m(a). \end{aligned}$$

• (2) If $a \in \mathbb {R}$, then $\hbox {Re}\,\omega _m^{-k}a=a \hbox {Re}\,\omega _m^{-k}=a \hbox {Re}\,\omega _m^{k}= \hbox {Re}\,\omega _m^{k}a$ for any $k\in \mathbb {Z}$. Therefore,

  $$\begin{aligned} \overline{F_m(a)}&= \frac{m(2 - \omega _m^{-1} - \omega _m)}{8\pi ^2} \sum _{\ell =0}^{m-1} \omega _m^{-\ell } \left( \arccos \left( -\hbox {Re}\,\omega _m^{-\ell }a \right) \right) ^2 \\&= \frac{m(2 - \omega _m- \omega _m^{-1})}{8\pi ^2} \sum _{\ell =0}^{m-1} \omega _m^{-\ell } \left( \arccos \left( -\hbox {Re}\,\omega _m^{\ell }a \right) \right) ^2 \\&= \frac{m(2 - \omega _m- \omega _m^{-1})}{8\pi ^2} \sum _{j=1-m}^{0} \omega _m^{j} \left( \arccos \left( -\hbox {Re}\,\omega _m^{-j}a \right) \right) ^2\\&= F_m(a), \end{aligned}$$

implying that $F_m(a)\in \mathbb {R}$. $\square$

Lemma 4.1 Let $m\in \{3,4,\dots ,\infty \}$. Suppose $x^1,x^2,\dots ,x^d \in \mathbb {C}^{n}$, and $\mathcal {F}\in \mathbb {C}^{n^d}$ is a super-symmetric complex tensor with its associated multilinear form $L$ and homogeneous polynomial $H$. If $\xi _1,\xi _2,\dots ,\xi _d$ are i.i.d. uniform distribution on $\mathbf{\Omega }_m$, then

$$\begin{aligned} \mathbf{\mathsf E}\left[ \prod _{i=1}^d \overline{\xi _i} H\left( \sum _{k=1}^d \xi _k x^k\right) \right] = d!L(x^1,x^2\dots , x^d) \text{ and } \mathbf{\mathsf E}\left[ \prod _{i=1}^d \xi _i H\left( \sum _{k=1}^d \xi _k x^k\right) \right] = 0. \end{aligned}$$

Proof

First we observe that

$$\begin{aligned} \mathbf{\mathsf E}\left[ \prod _{i=1}^d \overline{\xi _i} H\left( \sum _{k=1}^d\xi _kx^k\right) \right]&= \mathbf{\mathsf E}\left[ \prod _{i=1}^d \overline{\xi _i} \sum _{1\le k_1,k_2,\dots ,k_d\le d} L\left( \xi _{k_1}x^{k_1},\xi _{k_2}x^{k_2}, \dots , \xi _{k_d}x^{k_d}\right) \right] \\&= \sum _{1\le k_1,k_2,\dots ,k_d\le d} \mathbf{\mathsf E}\left[ \left( \prod _{i=1}^d \overline{\xi _i}\right) \!\left( \prod _{j=1}^d \xi _{k_j}\right) L\left( x^{k_1},x^{k_2},\dots , x^{k_d} \right) \right] . \end{aligned}$$

If $(k_1,k_2,\dots ,k_d)\in \Pi (1,2,\dots ,d)$, i.e., a permutation of $\{1,2,\dots ,d\}$, then

$$\begin{aligned} \mathbf{\mathsf E}\left[ \left( \prod _{i=1}^d \overline{\xi _i}\right) \!\left( \prod _{j=1}^d \xi _{k_j}\right) \right] = \mathbf{\mathsf E}\left[ \prod _{i=1}^d \overline{\xi _i}\xi _i\right] =1; \end{aligned}$$

otherwise, there is $k_0 (1\le k_0 \le d)$ such that and $k_0\ne k_j$ for all $j=1,2,\dots ,d$. In the latter case,

$$\begin{aligned} \mathbf{\mathsf E}\left[ \left( \prod _{i=1}^d \overline{\xi _i}\right) \!\left( \prod _{j=1}^d \xi _{k_j}\right) \right] = \mathbf{\mathsf E}\left[ \overline{\xi _{k_0}}\right] \mathbf{\mathsf E}\left[ \left( \prod _{1\le i\le d, i\ne k_0} \overline{\xi _i}\right) \!\left( \prod _{j=1}^d \xi _{k_j}\right) \right] =0. \end{aligned}$$

Since the number of different permutations of $\{1,2,\dots ,d\}$ is $d!$, by taking into account the super-symmetric property of $L$, the first identity follows.

For the second identity, similarly we have

$$\begin{aligned} \mathbf{\mathsf E}\left[ \prod _{i=1}^d \xi _i H\left( \sum _{k=1}^d\xi _kx^k\right) \right] = \sum _{1\le k_1,k_2,\dots ,k_d\le d} \mathbf{\mathsf E}\left[ \left( \prod _{i=1}^d \xi _i\right) \!\left( \prod _{j=1}^d \xi _{k_j}\right) L\left( x^{k_1},x^{k_2},\dots , x^{k_d} \right) \right] . \end{aligned}$$

There exists $k_0 (1 \!\le \! k_0 \!\le \! d)$ such that $\xi _{k_0}$ appears once or twice in $\big (\prod _{i=1}^d \xi _i\big )\!\big (\prod _{j=1}^d \xi _{k_j}\big )$. For $m\in \{3,4,\dots ,\infty \}$, we notice that $\mathbf{\mathsf E}[\xi _i]=0$ and $\mathbf{\mathsf E}[\xi _i^2]=0$ for $i=1,2,\dots ,d$. By independence of $\xi _i$’s, $\mathbf{\mathsf E}\left[ \big (\prod _{i=1}^d \xi _i\big ) \! \big (\prod _{j=1}^d \xi _{k_j}\big )\right]$ is always zero, leading to the second identity. $\square$

Lemma 5.1 Let $m\in \{3,4,\dots ,\infty \}$. Suppose $x^1,x^2,\dots ,x^{2d} \in \mathbb {C}^{n}$, and $\mathcal {F}\in \mathbb {C}^{n^{2d}}$ is a conjugate partial-symmetric tensor with its associated multilinear form $L$ and symmetric conjugate form $C$. If $\xi _1,\xi _2,\dots ,\xi _{2d}$ are i.i.d. uniform distribution on $\mathbf{\Omega }_m$, then

$$\begin{aligned}&\mathbf{\mathsf E}\left[ \left( \prod _{i=1}^d \xi _i\right) \! \left( \prod _{i=d+1}^{2d} \overline{\xi _i}\right) C\left( \sum _{k=1}^d \overline{\xi _k} x^k + \!\sum _{k=d+1}^{2d} \overline{\xi _k x^k},\sum _{k=1}^d \xi _k\overline{x^k} + \! \sum _{k=d+1}^{2d} \xi _k x^k\right) \right] \\&= (d!)^2 L(x^1, x^2,\dots , x^{2d}). \end{aligned}$$

Proof

We first consider the following

$$\begin{aligned}&\mathbf{\mathsf E}\left[ \left( \prod _{i=1}^d \xi _i\right) \! \left( \prod _{i=d + 1}^{2d} \overline{\xi _i}\right) C\left( \sum _{k=1}^{2d} \overline{\xi _k x^k},\sum _{k=1}^{2d} \xi _k x^k\right) \right] \\&= \mathbf{\mathsf E}\left[ \left( \prod _{i=1}^d \xi _i\right) \! \left( \prod _{i=d + 1}^{2d}\overline{\xi _i}\right) \sum _{1\le k_1,\dots ,k_{2d}\le 2d} L\left( \overline{\xi _{k_1}x^{k_1}},\dots ,\overline{\xi _{k_d}x^{k_d}},\xi _{k_{d+1}}x^{k_{d+1}},\dots , \xi _{k_{2d}}x^{k_{2d}}\right) \right] \\&=\sum _{1\le k_1,\dots ,k_{2d}\le 2d}\mathbf{\mathsf E}\left[ \left( \prod _{i=1}^d \xi _i\right) \! \left( \prod _{i=d + 1}^{2d}\overline{\xi _i}\right) \! \left( \prod _{j=1}^d \overline{\xi _{k_j}}\right) \! \left( \prod _{j=d + 1}^{2d}\xi _{k_j}\right) \right] \\&\quad \cdot L\left( \overline{x^{k_1}},\dots ,\overline{x^{k_d}}, x^{k_{d+1}},\dots ,x^{k_{2d}}\right) . \end{aligned}$$

For $m\in \{3,4,\dots ,\infty \}$, we observe that $\mathbf{\mathsf E}[\xi _i]=0$ and $\mathbf{\mathsf E}[\xi _i^2]=0$ for $i=1,2,\dots ,2d$. Using a similar argument in the proof of Lemma 4.1, we have

$$\begin{aligned}&\mathbf{\mathsf E}\left[ \left( \prod _{i=1}^d \xi _i\right) \! \left( \prod _{i=d + 1}^{2d}\overline{\xi _i}\right) \! \left( \prod _{j=1}^d \overline{\xi _{k_j}}\right) \! \left( \prod _{j=d + 1}^{2d}\xi _{k_j}\right) \right] \\&\quad =\left\{ \begin{array}{l@{\quad }l}1 &{} (k_1,\dots ,k_d)\in \Pi (1,\dots ,d) \text{ and } (k_{d+1},\dots ,k_{2d})\in \Pi (d+1,\dots ,2d); \\ 0 &{}\text{ otherwise }. \end{array}\right. \end{aligned}$$

By noticing that $\mathcal {F}$ is conjugate partial-symmetric (see Definition 2.2), and considering numbers of permutations, it follows that

$$\begin{aligned} \mathbf{\mathsf E}\left[ \left( \prod _{i=1}^d \xi _i\right) \! \left( \prod _{i=d + 1}^{2d} \overline{\xi _i}\right) C\left( \sum _{k=1}^{2d} \overline{\xi _k x^k},\sum _{k=1}^{2d} \xi _k x^k\right) \right] \!=\! (d!)^2 L\left( \overline{x^1}, \dots , \overline{x^d},x^{d+1},\dots ,x^{2d}\right) . \end{aligned}$$

Finally, replacing $\overline{x^k}$ by $x^k$ for $k=1,2,\dots ,d$ in the above identity leads to the desired result. $\square$