Sparse polynomial interpolation: sparse recovery, super-resolution, or Prony?

Abstract

We show that the sparse polynomial interpolation problem reduces to a discrete super-resolution problem on the n-dimensional torus. Therefore, the semidefinite programming approach initiated by Candès and Fernandez-Granda (Commun. Pure Appl. Math. 67(6) 906–956, 2014) in the univariate case can be applied. We extend their result to the multivariate case, i.e., we show that exact recovery is guaranteed provided that a geometric spacing condition on the supports holds and evaluations are sufficiently many (but not many). It also turns out that the sparse recovery LP-formulation of ℓ₁-norm minimization is also guaranteed to provide exact recovery provided that the evaluations are made in a certain manner and even though the restricted isometry property for exact recovery is not satisfied. (A naive sparse recovery LP approach does not offer such a guarantee.) Finally, we also describe the algebraic Prony method for sparse interpolation, which also recovers the exact decomposition but from less point evaluations and with no geometric spacing condition. We provide two sets of numerical experiments, one in which the super-resolution technique and Prony’s method seem to cope equally well with noise, and another in which the super-resolution technique seems to cope with noise better than Prony’s method, at the cost of an extra computational burden (i.e., a semidefinite optimization).

Appendix

Lemma 1

Ifz⁽¹⁾,…, z^(d)are distinct points of$\mathbb {C}^{n}$,thenv_d− 1(z⁽¹⁾),…, v_d− 1(z^(d)) are linearly independent vectors, wherev_d(z) := (z^α)_|α≤d.

Proof

Consider some complex numbers c₁,…, c_d such as the following:

$$ \sum\limits_{k = 1}^{d} c_{k} (z^{(k)})^{\alpha} = 0 ~,~~~ \forall |\alpha|\leqslant d-1. $$

(6.1)

Given 1 ≤ l ≤ d, define the Lagrange interpolation polynomial as follows:

$$ L^{(l)}(z) := \prod\limits_{\scriptsize \begin{array}{c} 1 \leqslant k \leqslant d \\ k \neq l \end{array}} \frac{ z_{i(k)} - z^{(k)}_{i(k)} }{ z^{(l)}_{i(k)} - z^{(k)}_{i(k)} } $$

(6.2)

where i(k) ∈{1,…, n} is an index such that $z^{(k)}_{i(k)} \neq z^{(l)}_{i(k)}$. It satisfies L^(l)(z^(k)) = 1 if k = l and L^(l)(z^(k)) = 0 if k≠l. The degree of $L^{(l)}(z) =: {\sum }_{\alpha } L^{(l)}_{\alpha } z^{\alpha }$ is equal to d − 1. Thus, we may multiply the equation in (6.1) by $L^{(l)}_{\alpha }$ to obtain the following:

$$ \sum\limits_{k = 1}^{d} c_{k} ~ L^{(l)}_{\alpha} (z^{(k)})^{\alpha} = 0 ~,~~~ \forall |\alpha|\leqslant d-1. $$

(6.3)

Summing over all |α|≤ d − 1 yields ${\sum }_{k = 1}^{d} c_{k} ~ L^{(l)}(z^{(k)}) = c_{l} = 0$. □

Lemma 2

If $u_{1},\hdots ,u_{d} \in \mathbb {C}^{n}$ are linearly independent, and $c_{1},\hdots ,c_{d} \in \mathbb {C}\setminus \{0\}$ , then, $\mathcal {R}({\sum }_{i = 1}^{d} c_{i} u_{i} {u_{i}^{T}}) = \mathcal {R}({\sum }_{i = 1}^{d} c_{i} u_{i} u_{i}^{*}) = \text {span} \{u_{1},\hdots ,u_{d}\}$ where $\mathcal {R}$ denotes the range.

Proof

If $z \in \mathbb {C}^{n}$, then, $({\sum }_{i = 1}^{d} c_{i} u_{i} {u_{i}^{T}})z = {\sum }_{i = 1}^{d} (c_{i} {u_{i}^{T}} z) u_{i} \in \text {span} \{u_{1},\hdots ,u_{d}\}$ and $({\sum }_{i = 1}^{d} c_{i} u_{i} u_{i}^{*})z = {\sum }_{i = 1}^{d} (c_{i} u_{i}^{*} z) u_{i} \in \text {span} \{u_{1},\hdots ,u_{d}\}$. Conversely, an element of the span ${\sum }_{i = 1}^{d} \lambda _{i} u_{i}$ with $\lambda _{1},\hdots ,\lambda _{n} \in \mathbb {C}$ belongs to the range of ${\sum }_{i = 1}^{d} c_{i} u_{i} {u_{i}^{T}}$ if there exists $z\in \mathbb {C}^{n}$ such as the following:

$$ \sum\limits_{i = 1}^{d} \lambda_{i} u_{i} = \left( \sum\limits_{i = 1}^{d} c_{i} u_{i} {u_{i}^{T}} \right) z $$

which is equivalent to each of the next three lines:

$$ \sum\limits_{i = 1}^{d} [ \lambda_{i} - (c_{i} {u_{i}^{T}} z) ] u_{i} = 0, $$

(6.4)

$$ \lambda_{i} = (c_{i} u_{i})^{T} z ~,~ i = 1,\hdots,d, $$

(6.5)

$$ \lambda = (c_{1} u_{1} \hdots c_{d} u_{d} )^{T} z. $$

(6.6)

Since $(c_{1} u_{1} \hdots c_{d} u_{d} ) \in \mathbb {C}^{n \times d}$ has rank d, its transpose has rank d. Thus, there exists a desired $z \in \mathbb {C}^{n}$. Likewise, ${\sum }_{i = 1}^{d} \lambda _{i} u_{i}$ belongs to the range of ${\sum }_{i = 1}^{d} c_{i} u_{i} u_{i}^{*}$ if there exists $z\in \mathbb {C}^{n}$ such as the following:

$$ \lambda_{i} = (c_{i} u_{i})^{*} z ~,~ i = 1,\hdots,d. $$

Since $(c_{1} u_{1} \hdots c_{d} u_{d} ) \in \mathbb {C}^{n \times p}$ has rank d, its conjugate transpose has rank d. Thus, there exists a desired $z \in \mathbb {C}^{n}$. □