Convex Optimization · Problem 2.37

Convex functions — Boyd & Vandenberghe

Mathematics — Optimization · nonnegative-polynomials-and-hankel-lmis

Problem

Nonnegative polynomials and Hankel LMIs. Let KpolK_{\mathrm{pol}} be the set of (coefficients of) nonneg\-ative polynomials of degree 2k2k on R\mathbf{R}: Kpol={x∈R2k+1∣x1+x2t+x3t2+⋯+x2k+1t2k≥0 for all t∈R}.K_{\mathrm{pol}} = \{x \in \mathbf{R}^{2k+1} \mid x_1 + x_2 t + x_3 t^2 + \cdots + x_{2k+1} t^{2k} \geq 0 \text{ for all } t \in \mathbf{R}\}. ((a)) Show that KpolK_{\mathrm{pol}} is a proper cone. ((b)) A basic result states that a polynomial of degree 2k2k is nonneg\-ative on R\mathbf{R} if and only if it can be expressed as the sum of squares of two polynomials of degree kk or less. In other words, x∈Kpolx \in K_{\mathrm{pol}} if and only if the polynomial p(t)=x1+x2t+x3t2+⋯+x2k+1t2kp(t) = x_1 + x_2 t + x_3 t^2 + \cdots + x_{2k+1} t^{2k} can be expressed as p(t)=r(t)2+s(t)2p(t) = r(t)^2 + s(t)^2, where rr and ss are polynomials of degree kk. Use this result to show that Kpol={x∈R2k+1 | xi=∑m+n=i+1Ymn for some Y∈S+k+1}.K_{\mathrm{pol}} = \left\{ x \in \mathbf{R}^{2k+1} \,\middle|\, x_i = \sum_{m+n=i+1} Y_{mn} \text{ for some } Y \in \mathbf{S}^{k+1}_+ \right\}. In other words, p(t)=x1+x2t+x3t2+⋯+x2k+1t2kp(t) = x_1 + x_2 t + x_3 t^2 + \cdots + x_{2k+1} t^{2k} is nonneg\-ative if and only if there exists a matrix Y∈S+k+1Y \in \mathbf{S}^{k+1}_+ such that x1=Y11x2=Y12+Y21x3=Y13+Y22+Y31⋮x2k+1=Yk+1,k+1.\begin{aligned} x_1 &= Y_{11} \\ x_2 &= Y_{12} + Y_{21} \\ x_3 &= Y_{13} + Y_{22} + Y_{31} \\ &\vdots \\ x_{2k+1} &= Y_{k+1,k+1}. \end{aligned} ((c)) Show that Kpol∗=KhanK^*_{\mathrm{pol}} = K_{\mathrm{han}} where Khan={z∈R2k+1∣H(z)⪰0}K_{\mathrm{han}} = \{z \in \mathbf{R}^{2k+1} \mid H(z) \succeq 0\} and H(z)=[z1z2z3⋯zkzk+1z2z3z4⋯zk+1zk+2z3z4z5⋯zk+2zk+4⋮⋮⋮⋱⋮⋮zkzk+1zk+2⋯z2k−1z2kzk+1zk+2zk+3⋯z2kz2k+1].H(z) = \begin{bmatrix} z_1 & z_2 & z_3 & \cdots & z_k & z_{k+1} \\ z_2 & z_3 & z_4 & \cdots & z_{k+1} & z_{k+2} \\ z_3 & z_4 & z_5 & \cdots & z_{k+2} & z_{k+4} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ z_k & z_{k+1} & z_{k+2} & \cdots & z_{2k-1} & z_{2k} \\ z_{k+1} & z_{k+2} & z_{k+3} & \cdots & z_{2k} & z_{2k+1} \end{bmatrix}. (This is the Hankel matrix with coefficients z1,…,z2k+1z_1, \ldots, z_{2k+1}.) ((d)) Let KmomK_{\mathrm{mom}} be the conic hull of the set of all vectors of the form (1,t,t2,…,t2k)(1, t, t^2, \ldots, t^{2k}), where t∈Rt \in \mathbf{R}. Show that y∈Kmomy \in K_{\mathrm{mom}} if and only if y1≥0y_1 \geq 0 and y=y1(1,E u,E u2,…,E u2k)y = y_1(1, \mathbf{E}\, u, \mathbf{E}\, u^2, \ldots, \mathbf{E}\, u^{2k}) for some random variable uu. In other words, the elements of KmomK_{\mathrm{mom}} are nonneg\-ative multiples of the moment vectors of all possible distributions on R\mathbf{R}. Show that Kpol=Kmom∗K_{\mathrm{pol}} = K^*_{\mathrm{mom}}. ((e)) Combining the results of (c) and (d), conclude that Khan=cl KmomK_{\mathrm{han}} = \mathbf{cl}\, K_{\mathrm{mom}}. As an example illustrating the relation between KmomK_{\mathrm{mom}} and KhanK_{\mathrm{han}}, take k=2k = 2 and z=(1,0,0,0,1)z = (1, 0, 0, 0, 1). Show that z∈Khanz \in K_{\mathrm{han}}, z∉Kmomz \notin K_{\mathrm{mom}}. Find an explicit sequence of points in KmomK_{\mathrm{mom}} which converge to zz.

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