A Semidefinite Approach for Truncated K-Moment Problems

Abstract

A truncated moment sequence (tms) in n variables and of degree d is a finite sequence y=(y _α ) indexed by nonnegative integer vectors α:=(α ₁,…,α _n ) such that α ₁+⋯+α _n ≤d. Let K⊆ℝⁿ be a semialgebraic set. The truncated K-moment problem (TKMP) is: How can one check if a tms y admits a K-measure μ (a nonnegative Borel measure supported in K) such that $y_{\alpha}= \int_{K} x_{1}^{\alpha_{1}}\cdots x_{n}^{\alpha_{n}}\,\mathrm{d}\mu$ for every α? This paper proposes a semidefinite programming (SDP) approach for solving TKMP. When K is compact, we get the following results: whether a tms admits a K-measure or not can be checked via solving a sequence of SDP problems; when y admits no K-measure, a certificate for the nonexistence can be found; when y admits one, a representing measure for y can be obtained from solving the SDP problems under some necessary and some sufficient conditions. Moreover, we also propose a practical SDP method for finding flat extensions, which in our numerical experiments always found a finitely atomic representing measure when it exists.

Appendix: Optimization with Measures

We prove that for a dense set of optimization problems (4.3), the optimal solutions are measures having finite supports. This justifies condition (4.5) in Sect. 4.2.

Lemma A.1

Let y be a tms in and K be a subset of ℝⁿ. Assume the set meas(y,K) is nonempty. If a measure μ is an extreme point of meas(y,K), then μ is finitely atomic and $|\mathrm{supp}(\mu)|\leq\binom{n+d}{d}$.

Proof

We prove by contradiction. Suppose μ is extreme and $|\mbox {supp}(\mu )| > N:=\binom{n+d}{d}$. Then we can decompose supp(μ) in a way such that

$$\mathrm{supp}(\mu) = \bigcup_{j=1}^{N+1} S_j, \quad S_i \cap S_j = \emptyset, \ \mu(S_j) >0, \ \forall i\ne j. $$

For each j, let μ _j be the restriction of μ on S _j , i.e., $\mu_{j} = \mu|_{S_{j}}$. Note that dimℝ[x] _d =N. So there exists (a ₁,…,a _N+1)≠0 satisfying

$$\sum_{j=1}^{N+1} a_j \int _K x^\alpha \,\mathrm {d}\mu_j = 0, \quad\forall |\alpha| \leq d. $$

Define a measure ν (possibly non-positive) such that

$$\nu= \nu|_{S_1} + \cdots+ \nu|_{S_{N+1}}, \quad \nu|_{S_j} = a_j \mu_j, \ j=1,\ldots, N+1. $$

Note that supp(ν)⊆supp(μ). Then, for ϵ>0 sufficiently small, μ±ϵν are nonnegative Borel measures supported on K, representing the same tms y and

$$\mu= (\mu+\epsilon\nu)/2 + (\mu- \epsilon\nu)/2. $$

This contradicts the extremality of μ in meas(y,K). □

Theorem A.2

Let y be a tms in . For any real analytic function f on a compact set K and ϵ>0, there exists an real analytic function $\tilde{f}$ on K such that $\|\tilde{f} - f \|_{\infty} \leq\epsilon$ and every minimizer of

$$ \min_\mu\quad\int_K \tilde{f} \,\mathrm {d}\mu \quad\mathrm{s.t.} \quad\mu\in \mathrm{meas}(y,K) $$

(A.1)

has cardinality at most $\binom{n+d}{d}$. Let Sol be the set of all optimizers above. Then

$$ \bigg|\bigcup_{ \mu\in\texttt{Sol} } \bigl\{ \mathrm{supp}(\mu) \bigr\} \bigg| \leq \binom{n+d}{d}. $$

(A.2)

Proof

For a given real analytic function f on K, consider the optimization problem

$$ \min_\mu\quad\int_K f\,\mathrm {d}\mu \quad \mathrm{s.t.} \quad\mu\in \mathrm{meas}(y,K). $$

(A.3)

The set of positive measures on K whose total masses equal y ₀ is compact in the weak-∗ topology (denote this topology by $\mathcal{T}$) by the Alaoglu Theorem (cf. [2, Theorem V.3.1]). Recall that the weak-∗ topology is the topology on the measures regarded as the dual space of the Banach space . It is the weakest topology for which the convergence of a sequence of measures μ _n →μ implies that for every

$$ \int_K h \,\mathrm {d}\mu_n \to\int_K h\,\mathrm {d}\mu. $$

(A.4)

This implies that every moment of μ _n converges to the corresponding one of μ. Hence, meas(y,K) is $\mathcal{T}$-closed inside a compact set, and it is also $\mathcal{T}$-compact.

Thus, by compactness of its feasible set, the optimization problem (A.3) has a minimizer, which is a generalized convex combination of certain extreme points, by the Choquet–Bishop–de Leeuw Theorem (cf. [22]). To be more precise, this means that for every $\hat{\mu}\in \mathrm{meas}(y,K)$, there exists a probability measure Γ on the set $\mathcal{E}$ of extreme points of meas(y,K) such that

$$\ell(\hat{\mu})= \int_\mathcal{E}\ell( \mu)\,\mathrm {d}\varGamma(\mu) $$

for every affine function ℓ on meas(y,K).

Let γ ^∗ be the optimal value of (A.3) and $\tilde{\mu}$ be a minimizer, and let $\widetilde{\varGamma}$ denote the probability measure on $\mathcal{E}$ representing $\tilde{\mu}$ and let $\widetilde{\mathcal{E}}$ denote the support of $\widetilde{\varGamma}$. Then

$$\gamma^* = \int_K f \,\mathrm {d}\tilde{\mu} = \int _{ \widetilde{\mathcal{E}} } \biggl(\int_K f \,\mathrm {d}\mu \biggr)\,\mathrm {d}\widetilde{\varGamma} (\mu). $$

By optimality of γ ^∗, ∫ _K f dμ≥γ ^∗ for all $\mu\in\widetilde{\mathcal{E}}$. Indeed, ∫ _K f dμ=γ ^∗ for all $\mu\in\widetilde {\mathcal{E}}$. Otherwise, suppose ∫ _K f dμ>γ ^∗ on a set of μ having positive $\widetilde{\varGamma}$ measure. Then

$$\gamma^* = \int_{ \widetilde{\mathcal{E}} } \biggl(\int_K f \,\mathrm {d}\mu \biggr) \,\mathrm {d}\widetilde{\varGamma}(\mu) > \gamma^*, $$

which yields a contradiction. This implies ∫ _K f dμ=γ ^∗ on $\widetilde{\mathcal{E}}$. Choose a measure $\mu^{*} \in\widetilde{\mathcal{E}}$. It is extreme and the optimum of (A.3) is attained at μ ^∗.

By Lemma A.1, we have $|\mathrm{supp}(\mu^{*})|\leq N:=\binom{n+d}{d}$. Without loss of generality, we can normalize y ₀=∫ _K  dμ ^∗=1 and denote supp(μ ^∗)={u ₁,…,u _r }. Let e(x) be the exponential function defined as

$$e(x) = \epsilon\cdot\exp\bigl(-\|x-u_1\|_2^2 \cdots\|x-u_r\|_2^2\bigr). $$

Clearly, e(x)=ϵ for all x∈supp(μ ^∗) and 0<e(x)<ϵ for all $x \not\in\mathrm{supp}(\mu^{*})$. This implies that

$$\max_{\mu\in \mathrm{meas}(y,K) } \int_K e \,\mathrm {d}\mu= \int _K e \,\mathrm {d}\mu^* =\epsilon. $$

Set $\tilde{f}:= f- e$ and

$$ \beta:= \min_{\mu\in \mathrm{meas}(y,K) } \int_K \tilde{f}\,\mathrm {d}\mu. $$

(A.5)

Then

$$ \beta \geq \min_{\mu\in \mathrm{meas}(y,K) } \int_K f \,\mathrm {d}\mu- \max_{\mu\in \mathrm{meas}(y,K) } \int_K e \,\mathrm {d}\mu= \int _K f \,\mathrm {d}\mu^* - \epsilon. $$

(A.6)

On the other hand

$$\int_K f \,\mathrm {d}\mu^* - \epsilon= \int_K \tilde{f}\,\mathrm {d}\mu^* \geq \min_{\mu\in \mathrm{meas}(y,K) } \int_K \tilde{f}\,\mathrm {d}\mu=\beta $$

because μ ^∗ belongs to meas(y,K). Thus

$$\beta= \int_K f \,\mathrm {d}\mu^* - \epsilon = \int _K \tilde{f}\,\mathrm {d}\mu^* $$

and μ ^∗ is a minimizer of (A.5) as well as (A.1).

Suppose $\tilde{\mu}^{*}$ is another minimizer of (A.5). We wish to show that

$$ \mathrm{supp}\bigl({\tilde{\mu}^*}\bigr) \subseteq\{u_1, \ldots, u_r\}. $$

(A.7)

If this is not true, then $\int e \,\mathrm {d}{\tilde{\mu}^{*}} <\epsilon$ and

$$\beta= \int_K f \, \mathrm {d}\mu^*- \epsilon < \min_{\mu\in \mathrm{meas}(y,K) } \int_K f \,\mathrm {d}\mu- \int _K e \,\mathrm {d}\tilde{\mu}^*. $$

This implies

$$\beta< \int_K f \,\mathrm {d}\tilde{\mu}^*- \int _K e \,\mathrm {d}\tilde{\mu}^* = \int_K \tilde{f}\,\mathrm {d}{\tilde{\mu}}^* =\beta, $$

which is a contradiction.

The inequality (A.2) follows from the first part, because the average of all the minimizers is still a minimizer. □

Remark

As we can see in the above proof, if μ ^∗ is a unique minimizer of (4.3), then μ ^∗ must have finite support and $|\mathrm{supp}(\mu^{*})|\leq \binom {n+d}{d}$.

The following lemmas are used in the proof of Theorem 4.3.

Lemma A.3

Suppose K⊆B(0,R) with R<1 and c is in . If a tms w belongs to E _k (y)∪E _∞(y), then

$$\|w\|_2\leq y_0/\bigl(1-R^2\bigr), $$

and for any integer t>0 we have

$$ \bigg| \sum_{ |\alpha| >2t } c_\alpha w_\alpha \bigg| \leq\|c\|_2 \cdot y_0 \cdot R^{2t+2}/\bigl(1-R^2\bigr). $$

(A.8)

Proof

First, we consider the case that w∈E _k (y). The condition M _t−1(ρ∗w)⪰0 in (4.1) implies that for every t=1,…,k

A repeated application of the above gives

Since M _k (w) is positive semidefinite, we can see that

$$\|w\|_2 \leq\big\|M_k(w)\big\|_F \leq \mathrm{Trace} \bigl(M_k(w)\bigr) = \sum_{t=0}^k \sum_{|\alpha| = t}\, w_{2\alpha}, $$

Thus, we have

$$\|w\|_2 \leq y_0\bigl(1+R^2+ \cdots+R^{2k}\bigr) \leq y_0/\bigl(1-R^2\bigr). $$

When k>t, we have

$$\bigg| \sum_{ 2t< |\alpha| \leq2k } \, c_\alpha w_\alpha \bigg|^2 \leq \biggl(\sum_{ 2t< |\alpha| \leq2k } \, c_\alpha^2 \biggr) \biggl(\sum_{ 2t< |\alpha| \leq2k } \, w_\alpha^2 \biggr). $$

Let M _t,k be the principal submatrix of M _k (w) whose row and column indices β satisfy t<|β|≤k. Clearly, M _t,k ⪰0 and we similarly have

Combining all of the above, we get

$$\bigg| \sum_{ 2t< |\alpha| \leq2k } c_\alpha w_\alpha \bigg| \leq \|c\|_2 \cdot y_0 \cdot\bigl(R^{2t+2}+ \cdots+R^{2k} \bigr) \leq\|c\|_2 \cdot y_0 \cdot R^{2t+2}/\bigl(1-R^2\bigr). $$

Note that w _α =0 if |α|>2k. So, (A.8) is true.

Second, we consider the case that w∈E _∞(y). The sequence w admits a K-measure μ such that w _α =∫ _K x ^α dμ for every α. For every k, consider the truncation z=∫ _K [x]_2k  dμ. Clearly, the tms z is feasible for (4.1). By part (i), we have

$$\|z\|_2\leq y_0/\bigl(1-R^2\bigr), \qquad \bigg| \sum_{ 2t< |\alpha| \leq2k } c_\alpha z_\alpha \bigg| \leq\|c\|_2 \cdot y_0 \cdot R^{2t+2}/ \bigl(1-R^2\bigr). $$

The above is true for all k, and implies (A.8) by letting k→∞. □

Lemma A.4

Assume K⊆B(0,R) with R<1, and c is in . Let γ _k ,γ be the optimal values of (4.1) and (4.3), respectively. Then, we have the convergence γ _k →γ as k→∞.

Proof

By Lemma A.3, for an arbitrary ϵ>0, there exists t such that

$$ \bigg| \sum_{ |\alpha| > t } c_\alpha w_\alpha \bigg| < \epsilon, \quad\forall w \in E_k(y) \cup E_\infty(y). $$

(A.9)

This implies that for every w∈E _k (y)∪E _∞(y) with k≥t, we have

$$ \big|\langle c, w \rangle- c_t^{\mathrm{T}}w|_{2t}\big| < \epsilon. $$

(A.10)

Now we consider the truncated optimization problems

(A.11)

and

$$ \min_{ w } \quad c_t^{\mathrm{T}}w|_{2t} \quad \mathrm{s.t.} \quad w \in E_\infty(y). $$

(A.12)

Let τ _k and τ be the optimal values of (A.11) and (A.12) respectively. Note that (A.11) is a semidefinite relaxation of (A.12). Thus, we have τ _k →τ, by Theorem 1 of Lasserre [16]. The inequality (A.10) implies that for all k≥t

$$\underset{ w\in E_k(y) }{\min} c_t^{\mathrm{T}} w|_{2t} - \epsilon \leq\underset{ w\in E_k(y) }{\min} c_k^{\mathrm{T}}w \leq \underset{ w\in E_k(y) }{\min} c_t^{\mathrm{T}} w|_{2t} + \epsilon. $$

This shows that |γ _k −τ _k |≤ϵ for k≥t. Applying a similar argument to (A.12), we can get |γ−τ|≤ϵ. Note that

$$| \gamma_k - \gamma| \leq| \gamma_k - \tau_k | + | \tau_k - \tau| + | \tau- \gamma|. $$

Because τ _k →τ, one gets

$$0 \leq\limsup_{k\to\infty} | \gamma_k - \gamma| \leq2\epsilon. $$

Since ϵ>0 is arbitrary, we must have γ _k →γ as k→∞. □