Positivity certificates and polynomial optimization on non-compact semialgebraic sets

Abstract

In a first contribution, we revisit two certificates of positivity on (possibly non-compact) basic semialgebraic sets due to Putinar and Vasilescu (C R Acad Sci Ser I Math 328(6):495–499, 1999). We use Jacobi’s technique from (Math Z 237(2):259–273, 2001) to provide an alternative proof with an effective degree bound on the sums of squares weights in such certificates. As a consequence, it allows one to define a hierarchy of semidefinite relaxations for a general polynomial optimization problem. Convergence of this hierarchy to a neighborhood of the optimal value as well as strong duality and analysis are guaranteed. In a second contribution, we introduce a new numerical method for solving systems of polynomial inequalities and equalities with possibly uncountably many solutions. As a bonus, one can apply this method to obtain approximate global optimizers in polynomial optimization.

Appendix

1.1 Proof of Theorem 5

Jacobi [18, Theorem 7] proves another result of Putinar and Vasilescu [47, Theorem 4.2], which states that if $f,g_1,\dots ,g_m$ are polynomials of even degree and $\tilde{f}>0$ on $S(\{\tilde{g}_1,\dots ,\tilde{g}_m\})\backslash \{0\}$, then $\theta ^kf\in Q(g)$ for k large enough, where $\tilde{h}$ is the homogenization of given polynomial h. The idea of Jacobi is to apply Putinar’s Positivstellensatz for the fact that $\tilde{f}>0$ on the intersection of $S(\{\tilde{g}_1,\dots ,\tilde{g}_m\})$ with the unit sphere. Our proof for Theorem 5 replaces $\tilde{f}$ here by the perturbation of $\tilde{f}$ and replaces the unit sphere by a sphere with an arbitrary radius $L^{1/2}$. This will have a direct implication on the complexity analysis derived in Proposition 1.

Proof

1. Let us prove the conclusion under condition (i). For every $h \in {\mathbb {R}}[ x ]$, we define by $\hat{h}$ the degree $2d_1 ( h )$ homogenization of h, i.e.,

$$\begin{aligned} \hat{h}\left( {x,{x_{n + 1}}} \right) = x_{n + 1}^{2d_1 ( h )}h\left( {\displaystyle {x}/{{{x_{n + 1}}}}} \right) . \end{aligned}$$

(1.34)

where ${d_1}( h ): = {1 + \lfloor {{{\deg ( h )}}/{2}} \rfloor }$. Let $\varepsilon > 0$ be fixed. Let $\tilde{f}$ be the degree 2d homogenization of f. Then $\tilde{f} + \varepsilon {\Vert {( {x,{x_{n + 1}}} )} \Vert _2^{2d}}$ is a homogeneous polynomial of degree 2d. Let $\hat{g} := \{ {{\hat{g}_1},\dots ,{\hat{g}_m}} \}$ and $L>0$. We will show that

$$\begin{aligned} \tilde{f} + \varepsilon {\Vert {\left( {x,{x_{n + 1}}} \right) }\Vert _2^{2d}} \ge \varepsilon L^{d}\text { on }S\left( \hat{g}, \left\{ L-\Vert {\left( {x,{x_{n + 1}}} \right) }\Vert _2^2\right\} \right) . \end{aligned}$$

(1.35)

Let $( {y,{y_{n + 1}}}) \in S( \hat{g}, \{L-\Vert ( x,x_{n + 1} \Vert _2^2\})$ be fixed. By (1.34), one has

$$\begin{aligned} y_{n + 1}^{2d_1 \left( g_j \right) }{g_j}\left( {\displaystyle {y}/{{{y_{n + 1}}}}} \right) ={{\hat{g}}_j}\left( {y,{y_{n + 1}}} \right) \ge 0,\,j = 1,\dots ,m. \end{aligned}$$

(1.36)

We consider the following two cases:

• Case 1: ${y_{n + 1}} \ne 0$. For $j=1,\dots ,m$, by (1.36) and since $y_{n + 1}^{2d_1 \left( g_j \right) }>0$, ${g_j}( {\displaystyle {y}/{{{y_{n + 1}}}}} ) \ge 0$. It implies that $\displaystyle {y}/{{{y_{n + 1}}}} \in S\left( g\right)$. By assumption, $f( \displaystyle {y}/{{{y_{n + 1}}}}) \ge 0$. From this, $\tilde{f}( {y,{y_{n + 1}}}) = y_{n + 1}^{2d}f( {{y}/{{{y_{n + 1}}}}} ) \ge 0$. From this and since $\Vert ( {y,{y_{n + 1}}} )\Vert _2^2=L$, $\tilde{f}( {y,{y_{n + 1}}}) + \varepsilon {\Vert {( {y,{y_{n + 1}}} )} \Vert _2^{2d}} \ge \varepsilon L^{d}$.

• Case 2: ${y_{n + 1}} = 0$. By definition of $d_1(f)$, $x_{n+1}$ divides $\tilde{f}$ so $\tilde{f}( {y,{y_{n + 1}}}) = 0$. From this and since $\Vert ( {y,{y_{n + 1}}})\Vert _2^2=L$,

  $$\begin{aligned} \tilde{f}\left( {y,{y_{n + 1}}}\right) + \varepsilon {\Vert {\left( {y,{y_{n + 1}}} \right) } \Vert _2^{2d}} = \varepsilon L^{d}. \end{aligned}$$

Thus, (1.35) holds. It is not hard to show that $S( \hat{g} \cup \{L-\Vert {( {x,{x_{n + 1}}} )}\Vert _2^2\})$ satisfies the Archimedean condition. From this and by applying Theorem 2 (i),

$$\begin{aligned} \tilde{f} + \varepsilon {\Vert {\left( {x,{x_{n + 1}}} \right) } \Vert ^{2d}} \in Q\left( \hat{g},{\left\{ { L - {{\Vert {\left( {x,{x_{n + 1}}} \right) } \Vert }^2_2}} \right\} } \right) . \end{aligned}$$

Then there exist $\psi _{j} \in \varSigma [ {x,{x_{n + 1}}} ],\ j = 0,1,\dots ,m$ and $\varphi \in {\mathbb {R}}[ {x,{x_{n + 1}}} ]$ such that

$$\begin{aligned} \tilde{f} + \varepsilon {\Vert {\left( {x,{x_{n + 1}}} \right) } \Vert _2^{2d}} = \psi _0 + \sum \limits _{j = 1}^m {\psi _j{{\hat{g}}_j}} + \left( {L - {{\Vert {\left( {x,{x_{n + 1}}} \right) } \Vert }_2^2}} \right) \varphi . \end{aligned}$$

Let $( {z,{z_{n + 1}}}) \in {{\mathbb {R}}^{n + 1}}\backslash \{ 0\}$. By replacing ${( {x,{x_{n + 1}}} )}$ in the last equality by $L^{1/2}\frac{{( {z,{z_{n + 1}}} )}}{{\Vert {( {z,{z_{n + 1}}} )} \Vert _2}}$ and the fact that ${\tilde{f} + \varepsilon {\Vert {( {x,{x_{n + 1}}})} \Vert _2^{2d}}}, \ \hat{g}_1,\dots ,\hat{g}_m$ are homogeneous polynomials of degree $2d,\ 2d_1 \left( g_1 \right) ,\dots ,2d_1 \left( g_m\right)$ respectively, one has

$$\begin{aligned}&\left( {\tilde{f}\left( {z,{z_{n + 1}}} \right) + \varepsilon {{\Vert {\left( {z,{z_{n+ 1}}} \right) } \Vert }_2^{2d}}} \right) L^d{\Vert {\left( {z,{z_{n + 1}}} \right) } \Vert _2^{ -2d}}\\&\quad = \psi _0 \left( L^{1/2}\frac{\left( z,z_{n + 1}\right) }{\Vert \left( z,z_{n + 1} \right) \Vert _2} \right) \\&\qquad + \sum \limits _{j = 1}^m {\psi _j \left( L^{1/2}\frac{\left( z,z_{n + 1} \right) }{\Vert \left( z,z_{n + 1}\right) \Vert } \right) \hat{g}_j\left( z,z_{n + 1} \right) L^{d_1(g_j)}\Vert \left( z,z_{n + 1} \right) \Vert _2^{ - 2d_1 ( g_j )}} . \end{aligned}$$

Set

$$\begin{aligned} K: = \max \left\{ {2d,\deg (\psi _0 ),2d_1\left( g_1\right) + \deg \left( \psi _1\right) ,\dots ,2d_1\left( g_m\right) + \deg \left( \psi _m\right) ,2+2\lceil \deg (\varphi )/2\rceil }\right\} . \end{aligned}$$

Obviously, K is even. After multiplying the two sides of the last equality with $\Vert ( z,z_{n + 1}) \Vert ^K$, one has

$$\begin{aligned}&\left( {\tilde{f}\left( {z,{z_{n + 1}}} \right) + \varepsilon {{\Vert {\left( {z,{z_{n + 1}}} \right) } \Vert }_2^{2d}}} \right) L^d{\Vert {\left( {z,{z_{n + 1}}} \right) } \Vert _2^{K -2d}}\\&\qquad = \bar{\psi }_0\left( z,z_{n + 1} \right) + \sum \limits _{j = 1}^m {L^{d_1\left( g_j\right) } \bar{\psi }_j \left( z,z_{n + 1} \right) \hat{g}_j\left( z,z_{n + 1} \right) } , \end{aligned}$$

where

$$\begin{aligned} \bar{\psi }_0&:=\psi _0 \left( L^{1/2}\frac{( x,x_{n + 1})}{\Vert ( x,x_{n + 1} ) \Vert _2} \right) \Vert ( x,x_{n + 1} ) \Vert _2^{ K}\, ,\\ \bar{\psi }_j&:=\psi _j \left( L^{1/2}\frac{( x,x_{n + 1} )}{\Vert ( x,x_{n + 1}) \Vert _2} \right) \Vert ( x,x_{n + 1} ) \Vert _2^{ K- 2d_1 ( g_j )},\,j=1,\dots ,m. \end{aligned}$$

Since $( {z,{z_{n + 1}}}) \in {{\mathbb {R}}^{n + 1}}\backslash \{ 0\}$ is arbitrary,

$$\begin{aligned} \left( {\tilde{f} + \varepsilon {{\Vert {( {x,{x_{n + 1}}} )} \Vert }_2^{2d}}} \right) L^d{\Vert {( {x,{x_{n + 1}}} )} \Vert _2^{K - 2d}}= \bar{\psi }_0 + \sum \limits _{j = 1}^m {L^{d_1(g_j)}\bar{\psi }_j \hat{g}_j} . \end{aligned}$$

(1.37)

Let $j\in \{0,\dots ,m\}$ be fixed and set $g_0:=1$ and $d_1(g_0):=0$. We will show that

$$\begin{aligned} \bar{\psi }_j=s_j+\xi _j \Vert (x,x_{n+1})\Vert _2, \end{aligned}$$

(1.38)

for some $s_j\in \varSigma [x,x_{n+1}]_{K/2-d_1(g_j)}$ and $\xi _j\in {\mathbb {R}}[x,x_{n+1}]$. By definition of K ($K- 2d_1 ( g_j )\ge \deg (\psi _j)$, $j=1,\dots ,m$) and since $\psi _j$ is SOS, $\bar{\psi }_j=\sum \nolimits _{t = 1}^{l} {\phi _{t}^2}$ with $\phi _{t}:=h_t+p_t\Vert (x,x_{n+1})\Vert _2$ for some $h_t\in {\mathbb {R}}[x,x_n]_{K/2-d_1(g_j)}$ and $p_t\in {\mathbb {R}}[x,x_n]_{K/2-d_1(g_j)-1}$. Then

$$\begin{aligned} \bar{\psi }_j=\sum \limits _{t = 1}^{l} {\left( h_t^2+p_t^2\Vert \left( x,x_{n+1}\right) \Vert _2^2\right) }+2\Vert \left( x,x_{n+1}\right) \Vert _2\sum \limits _{t = 1}^{l} {h_tp_t}. \end{aligned}$$

By setting $s_j:=\sum \nolimits _{t = 1}^{l} {(h_t^2+p_t^2\Vert (x,x_{n+1})\Vert _2^2)}$ and $\xi _j:=2\sum \nolimits _{t = 1}^{l} {h_tp_t}$, (1.38) follows. By (1.37) and (1.38),

$$\begin{aligned}&\left( {\tilde{f} + \varepsilon {{\Vert {\left( {x,{x_{n + 1}}} \right) } \Vert }_2^{2d}}} \right) L^d{\Vert {\left( {x,{x_{n + 1}}} \right) } \Vert _2^{K - 2d}}\\&\quad = s_0 + \sum \limits _{j = 1}^m {L^{d_1\left( g_j\right) }s_j \hat{g}_j}+\Vert \left( x,x_{n + 1} \right) \Vert _2\sum \limits _{j = 1}^m {L^{d_1\left( g_j\right) }\xi _j \hat{g}_j} . \end{aligned}$$

Note that $\Vert {( {x,{x_{n + 1}}} )} \Vert _2$ is not a polynomial. From the last equality and since the right hand side is polynomial, the left hand side must be a polynomial, so $\sum \nolimits _{j = 1}^m {L^{d_1(g_j)}\xi _j \hat{g}_j}=0$. Thus,

$$\begin{aligned} \left( {\tilde{f} + \varepsilon {{\Vert {\left( {x,{x_{n + 1}}} \right) } \Vert }_2^{2d}}} \right) L^d{\Vert {\left( {x,{x_{n + 1}}} \right) } \Vert _2^{K - 2d}}= s_0 + \sum \limits _{j = 1}^m {L^{d_1(g_j)}s_j \hat{g}_j} . \end{aligned}$$

By setting $k_\varepsilon :=K/2-d$, ${s_0} \in \varSigma {[ x,x_{n+1} ]_{{k_\varepsilon } + d}}$, ${s _j} \in \varSigma {[ x,x_{n+1} ]_{{k_\varepsilon } + d - d_1( {{g_j}})}},\,j = 1,\dots ,m$ and $L^d{\Vert {( {x,{x_{n + 1}}})} \Vert _2^{2{k_\varepsilon }}}( {\tilde{f} + \varepsilon {{\Vert {( {x,{x_{n + 1}}} )} \Vert }_2^{2d}}} ) = {s _0} + \sum \nolimits _{j = 1}^m {L^{d_1(g_j)}{s_j}{{\hat{g}}_j}}$. By replacing $x_{n+1}$ by 1, $L^d{\theta ^{{k_\varepsilon }}}( {f + \varepsilon {\theta ^{d}}} ) = {s _0}( {x,1} ) + \sum \nolimits _{j = 1}^m {L^{d_1(g_j)}{s_j}( {x,1} ){g_j}}$. Since ${s_0} \in \varSigma {[ x,x_{n+1}]_{{k_\varepsilon } + d}}$, ${s_j} \in \varSigma {[ x,x_{n+1} ]_{{k_\varepsilon } + d - d_1( {{g_j}} )}},\,j = 1,\dots ,m$, ${s _0}( {x,1} ) \in \varSigma {[ x ]_{{k_\varepsilon } + d}}$, ${s _j}( {x,1}) \in \varSigma {[ x]_{{k_\varepsilon } + d - d_1( {{g_j}})}},\,j = 1,\dots ,m$. Note that $d_1(g_j)\ge d_2(g_j)=u_j$, $j=1,\dots ,m$. Hence, ${\theta ^{{k_\varepsilon }}}( {f + \varepsilon {\theta ^{d}}} ) \in {Q_{{k_\varepsilon } + d}}( g )$ by the definition of truncated quadratic module.

2. Let us show the conclusion under condition (ii). We do a similar process as part 1 (under condition (i)). The difference is that $\hat{h}$ here is defined as the degree $2d_2 ( h )$ (instead of $2d_1 ( h )$) homogenization of $h \in {\mathbb {R}}[ x ]$ and the proof for (1.35). To show (1.35) in Case 2: $y_{n+1}=0$ here, we rely on the constraint $g_m=f+\lambda$ for some $\lambda \ge 0$ (instead of $d \ge d_1(f)$ and that $x_{n+1}$ divides $\tilde{f}$). More explicitly, we assume by contradiction that

$$\begin{aligned} \tilde{f}({y,{y_{n + 1}}}) + \varepsilon {\Vert {( {y,{y_{n + 1}}} )}\Vert _2^{2d}} <\varepsilon L^{d}. \end{aligned}$$

From this and since $0 \le {\hat{g}_m}(y,{y_{n + 1}}) = \hat{f}(y,0)$,

$$\begin{aligned} \begin{array}{rl} \varepsilon \Vert ({y,{y_{n + 1}}})\Vert _2^{2d} &{}\le 0^{2d-2d_2(f)}\hat{f}(y,0) + \varepsilon \Vert ({y,{y_{n + 1}}})\Vert _2^{2d}\\ &{}= y_{n+1}^{2d-2d_2(f)}\hat{f}(y,y_{n+1}) + \varepsilon \Vert ({y,{y_{n + 1}}})\Vert _2^{2d} \\ &{}= \tilde{f}({y,{y_{n + 1}}}) + \varepsilon {\Vert {( {y,{y_{n + 1}}} )}\Vert _2^{2d}} <\varepsilon L^{d}. \end{array} \end{aligned}$$

It follows that $\Vert ({y,{y_{n + 1}}})\Vert _2^2<L=\Vert ({y,{y_{n + 1}}})\Vert _2^2$. It is impossible. Thus, $\tilde{f}({y,{y_{n + 1}}}) + \varepsilon {\Vert {( {y,{y_{n + 1}}} )}\Vert _2^{2d}} \ge \varepsilon L^{d}$. $\square$

1.2 Proof of Proposition 1

Proof

We keep all the notation from the proof of Theorem 5. Without loss of generality, let us assume that $L=1/4$. By (1.35), $S( \hat{g},{\{ { L - {{\Vert {( {x,{x_{n + 1}}} )} \Vert }^2_2}} \}} )\subset (-1,1)^{n+1}$ and

$$\begin{aligned} \min \{\tilde{f} + \varepsilon {\Vert {( {x,{x_{n + 1}}} )} \Vert _2^{2d}}\,:( {x,{x_{n + 1}}} )\in S( \hat{g},{\{ { L - {{\Vert {( {x,{x_{n + 1}}} )} \Vert }^2_2}} \}} )\}\ge 2^{-2d}\varepsilon .\qquad \end{aligned}$$

(1.39)

By definition of $\hat{g}$ and since $0_{{\mathbb {R}}^n}\in S(g)$, $(0_{{\mathbb {R}}^n},L^{1/2})\in S( \hat{g},{\{ { L - {{\Vert {( {x,{x_{n + 1}}} )} \Vert }^2_2}} \}} )$. Thus, $S( \hat{g},{\{ { L - {{\Vert {( {x,{x_{n + 1}}} )} \Vert }^2_2}} \}} )$ is nonempty. By definition of K,

$$\begin{aligned} \tilde{f} + \varepsilon {\Vert {( {x,{x_{n + 1}}} )} \Vert _2^{2d}} \in Q_{K/2}( \hat{g},{\{ { L - {{\Vert {( {x,{x_{n + 1}}} )} \Vert }^2_2}} \}} ). \end{aligned}$$

Since $\Vert {(x,{x_{n + 1}})} \Vert _2^{2d} = \sum \limits _{\bar{\alpha }\in {\mathbb {N}}_d^{n + 1}} {c_{n+1}(\bar{\alpha })(x,{x_{n + 1}})^{2\bar{\alpha }}}$, one has

$$\begin{aligned} \begin{array}{rl} \Vert {\tilde{f} + \varepsilon \Vert {(x,{x_{n + 1}})} \Vert _2^{2d}} \Vert _{\max }&{}\le \Vert {\tilde{f} } \Vert _{\max } +\varepsilon \Vert {\Vert {(x,{x_{n + 1}})} \Vert _2^{2d}} \Vert _{\max }\\ &{}=\Vert {\tilde{f} } \Vert _{\max } +\varepsilon \max \{ \displaystyle {c_{n+1}(\bar{\alpha })}/{c_{n+1}(2\bar{\alpha })}\,:{\bar{\alpha }\in {\mathbb {N}}^{n+1}_d}\}. \end{array} \end{aligned}$$

Note that $\tilde{f} = \sum \nolimits _\alpha {f_\alpha x^\alpha x_{n+1}^{2d-|\alpha |}}$, so $\Vert \tilde{f} \Vert _{\max }=\Vert f \Vert _{\max ,d}$. Thus,

$$\begin{aligned} \Vert {\tilde{f} + \varepsilon \Vert {(x,{x_{n + 1}})} \Vert _2^{2d}} \Vert _{\max }\le \Vert { f } \Vert _{\max , d} +\varepsilon \max \{ \displaystyle {c_{n+1}(\bar{\alpha })}/{c_{n+1}(2\bar{\alpha })}\,:{\bar{\alpha }\in {\mathbb {N}}^{n+1}_d}\}. \end{aligned}$$

(1.40)

From these and using Theorem 3, one can choose

$$\begin{aligned} \frac{K}{2}\ge C\exp \left( \left( 4^{d+1}d^2(n + 1)^{2d}\left( \varepsilon ^{-1} \Vert f \Vert _{\max , d}+ \max \left\{ \displaystyle \frac{c_{n+1}(\bar{\alpha })}{c_{n+1}(2\bar{\alpha })}\,:{\bar{\alpha }\in {\mathbb {N}}^{n+1}_d}\right\} \right) \right) ^C \right) , \end{aligned}$$

for some $C>0$. The right hand side of this inequality comes from (1.39), (1.40) and the fact that the function $t \mapsto c\exp \left( {b{t^c}} \right)$ with positive constants b and c is increasing on $[0,\infty )$. By setting $k_\varepsilon =K/2-d$, the conclusion follows. $\square$

1.3 Proof of Proposition 3

Proof

We denote by P and $P^\star$ feasible set and optimal solutions set for the moment SDP (4.22), respectively. We claim that P is nonempty. Indeed, with z being the moment sequence of the 1-atomic measure $\theta (\bar{x})^{-1}\delta _{\bar{x}}$ for some $\bar{x}\in S(g)$, it is not hard to check that the truncation of z is a feasible solution of (4.22). By setting $C: = {L_z}({\theta ^k}(f + \varepsilon {\theta ^d}))$, $P^\star$ is also the set of optimal solutions for

$$\begin{aligned} \begin{array}{rl} {\tau _k^2}(\varepsilon ) = \inf &{}{L_y}( {{\theta ^k}( {f + \varepsilon {\theta ^d}} )} )\\ \text {s.t.}&{}y = {(y_\alpha )_{\alpha \in {\mathbb {N}}^n_{2( {d + k})}}} \subset {\mathbb {R}},\\ &{}{L_y}( {{\theta ^k}( {f + \varepsilon {\theta ^d}} )} )\le C,\\ &{}{M_{k + d - {u_j}}}( {{g_j}y}) \succeq 0 ,\;j = 0,\dots ,m,\\ &{}{L_y}( {{\theta ^k}}) = 1, \end{array} \end{aligned}$$

(1.41)

with $g_0=1$. Denote by $\bar{P}$ the feasible set of (1.41). Note that $\bar{P}$ is nonempty since the truncation of z is also a feasible solution of (1.41). We will prove that $\bar{P}$ is bounded. By definition of $\theta$, ${\theta ^r} = \sum \nolimits _{\alpha \in {\mathbb {N}}_r^n} {{C_\alpha ^r }{x^{2\alpha }}}$ for all $r\in {\mathbb {N}}$, where $C_\alpha ^r\ge 1$ for all $\alpha \in {\mathbb {N}}_r^n$. Since $f-f^\star \ge 0$ on S(g), by Theorem 5, there exists $K\in {\mathbb {N}}$ such that

$$\begin{aligned} {\theta ^K}\left( {f - f^\star + \displaystyle \frac{\varepsilon }{2} {\theta ^d}} \right) \in Q_{K + d}{( g )}. \end{aligned}$$

Note that K depends only on f, g and $\varepsilon$. Assume that $k\ge K$. Since $\theta ^{k-K} \in \varSigma {[ x ]_{k-K}}$, one has

$$\begin{aligned} {\theta ^k}\left( {f - {f^\star }+ \displaystyle \frac{\varepsilon }{2} {\theta ^d} }\right) = \theta ^{k - K} {\theta ^{K}} \left( {f - {f^\star }+ \displaystyle \frac{\varepsilon }{2} {\theta ^d}} \right) \in Q _{k + d}{( g)}. \end{aligned}$$

Then there exist $G_j\succeq 0$, $j=1,\dots ,m$ such that

$$\begin{aligned} {\theta ^k}\left( f - {f^ \star } + \frac{\varepsilon }{2}{\theta ^d}\right) = \sum \limits _{j = 0}^m {v_{k + d - {u_j}}^T{G_j}{v_{k + d - {u_j}}}{g_j}} = \sum \limits _{j = 0}^m {{\hbox {trace}}({G_j}{v_{d + k - {u_j}}}v_{d + k - {u_j}}^Tg_j)} . \end{aligned}$$

Let $y\in \bar{P}$. From these and since ${M_{k + d - {u_j}}}( {{g_j}y})\succeq 0$,

$$\begin{aligned} {L_y}\left( {{\theta ^k}\left( {f - {f^ \star } + \frac{\varepsilon }{2}{\theta ^d}} \right) } \right) = \sum \limits _{j = 0}^m {{\hbox {trace}}({G_j}{M_{d + k - {u_j}}}({g_j}y))} \ge 0. \end{aligned}$$

Thus, for every $\beta \in {\mathbb {N}}_{k + d}^n$,

$$\begin{aligned} \begin{array}{rl} \displaystyle \frac{\varepsilon }{2} {y_{2\beta }} &{}\le \displaystyle \frac{\varepsilon }{2} C_\beta ^{k + d}{y_{2\beta }} \le \displaystyle \frac{\varepsilon }{2} \sum \limits _{\alpha \in {\mathbb {N}}_{k + d}^n} {C_\alpha ^{k + d}{y_{2\alpha }}} = \displaystyle \frac{\varepsilon }{2} \sum \limits _{\alpha \in {\mathbb {N}}_{k + d}^n} {C_\alpha ^{k + d}{L_y}({x^{2\alpha }})} \\ &{} = \displaystyle \frac{\varepsilon }{2} {L_y}({\theta ^{k + d}}) \le \displaystyle \frac{\varepsilon }{2} {L_y}({\theta ^{k + d}}) + {L_y}\left( {\theta ^k}\left( f - f^\star +\displaystyle \frac{\varepsilon }{2} \theta ^d\right) \right) \\ &{}= {L_y}({\theta ^k}(f + \varepsilon {\theta ^d})) - f^\star {L_y}({\theta ^k})\le C - f^\star , \end{array}\ \end{aligned}$$

since every element $y_{2\alpha }$ on the diagonal of the positive semidefinite matrix ${M_{k + d}}( {y})$ is nonnegative. Thus, ${y_{2\beta }}\le 2(C - f^\star )\varepsilon ^{-1}$ for every $\beta \in {\mathbb {N}}_{k + d}^n$. Since ${M_{k + d }}( {y})\succeq 0$, $| {{y_{\alpha + \beta }}} | \le {y_{2\beta }} \le 2(C - \underline{f})\varepsilon ^{-1}$ for all $\alpha ,\beta \in {\mathbb {N}}_{k + d}^n$. This implies that $\Vert y\Vert _2$ is bounded by $2(C - \underline{f})\varepsilon ^{-1}\sqrt{s(2(d + k))}$. Since the objective function of (1.41) is linear and the feasible set of (1.41) is closed and bounded, the set $P^\star$ of optimal solutions of (1.41) is nonempty and bounded. By using Trnovska’s result [59, Corrollary 1], $\rho _k^2(\varepsilon )=\tau _k^2(\varepsilon )$, yielding the desired conclusion. $\square$

1.4 Proof of Proposition 4

Proof

We do a similar process as the proof of Proposition 3. The difference is the bound of y here obtained by the inequality constraint $g_m=f-\underline{f}\ge 0$ in stead of Positivstellensatz. Thus, we do not need k sufficient large here. More explicitly, one has $\Vert y\Vert _2 \le (C - \underline{f})\varepsilon ^{-1}\sqrt{s(2(d + k))}$, since for every $\beta \in {\mathbb {N}}_{k + d}^n$,

$$\begin{aligned} \begin{array}{rl} \varepsilon {y_{2\beta }} &{}\le \varepsilon C_\beta ^{k + d}{y_{2\beta }} \le \varepsilon \sum \limits _{\alpha \in {\mathbb {N}}_{k + d}^n} {C_\alpha ^{k + d}{y_{2\alpha }}} = \varepsilon \sum \limits _{\alpha \in {\mathbb {N}}_{k + d}^n} {C_\alpha ^{k + d}{L_y}({x^{2\alpha }})} \\ &{}\le \varepsilon \sum \limits _{\alpha \in {\mathbb {N}}_{k + d}^n} {C_\alpha ^{k + d}{L_y}({x^{2\alpha }})} + \sum \limits _{\alpha \in {\mathbb {N}}_k^n} {C_\alpha ^k{L_y}({x^{2\alpha }}{g_m})} \\ &{} = \varepsilon {L_y}({\theta ^{k + d}}) + {L_y}({\theta ^k}(f - \underline{f} )) = {L_y}({\theta ^k}(f + \varepsilon {\theta ^d})) - \underline{f} {L_y}({\theta ^k})\\ &{}\le C - \underline{f}. \end{array} \end{aligned}$$

1.5 Proof of Lemma 4

Proof

Let us show that $(\xi _t)_{t=0,\dots ,n}$ is real sequence. Obviously, $\xi _t\ge 0$, $t=0,\dots ,n$. Set $r_t=\xi _t^{1/2}$, $t=0,\dots ,n$. Then

$$\begin{aligned} \left\{ \begin{array}{l} {r_0} = d( {{a_0},S( g,h )} )\ge 0,\\ {r_t} = d( {{a_t},S( g,h ) \cap \partial {B( {{a_0},{r_0}} )} \cap \dots \cap \partial {B( {{a_{t - 1}},{r_{t - 1}}} )} } )\ge 0,\, t = 1,\dots ,n, \end{array} \right. \end{aligned}$$

(1.42)

where $d(a,A):=\inf \{\Vert x-a\Vert _2:\, x\in A\}$ for $a\in {\mathbb {R}}^n$ and $A\subset {\mathbb {R}}^n$. It is sufficient to prove that $r_t$ is real, $t=0,\dots ,n$. It is easy to see that S(g, h) is closed and S(g, h) is nonempty by assumption. From this, ${r_0}$ is nonnegative real and $S( g,h ) \cap \partial {B( {{a_0},{r_0}} )}$ is also closed and nonempty. It implies that $r_1$ is nonnegative real and

$$\begin{aligned} {S( g,h) \cap \partial {B( {{a_0},{r_0}} )} \cap \partial {B( {{a_{1}},{r_{1}}} )} } \end{aligned}$$

is also closed and nonempty. By induction, for $t \in \{ {0,\dots ,n} \}$, $r_t$ is nonnegative real and

$$\begin{aligned} {S( g,h ) \cap \partial {B( {{a_0},{r_0}} )} \cap \dots \cap \partial {B( {{a_{t}},{r_{t}}} )} } \end{aligned}$$

is closed and nonempty. Thus,

$$\begin{aligned}&S\left( {g,h \cup \{ {{\xi _t} - {{ \Vert {x - {a_t}} \Vert }_2^2}:\, t = 0,\dots ,n} \}} \right) \\&\qquad =S( g,h ) \cap \partial {B( {{a_0},{r_0}})} \cap \dots \cap \partial {B\left( {{a_n},{r_n}} \right) } \ne \emptyset . \end{aligned}$$

Let ${x^\star } \in S( {g,h \cup \{ {{\xi _t} - {{ \Vert {x - {a_t}} \Vert }_2^2}:\, t = 0,\dots ,n} \}} )$. Then ${x^\star } \in \partial B\left( {{a_0},{r_0}} \right) \cap \dots \cap \partial B( {{a_n},{r_n}} )$. It follows that ${\Vert {{x^\star } - {a_t}} \Vert _2^2} = r_t^2=\xi _t,\, t = 0,\dots ,n$. For $t = 1,\dots ,n$,

$$\begin{aligned} \xi _t - \xi _0 = {\Vert {{x^\star } - {a_t}} \Vert _2^2} - {\Vert {{x^\star } - {a_0}} \Vert _2^2} = -2{( {{a_t} - {a_0}} )^T}{x^\star } + {\Vert {{a_t}} \Vert _2^2} - {\Vert {{a_0}} \Vert _2^2}. \end{aligned}$$

It implies (4.31). Denote

$$\begin{aligned} A = \left( {\begin{array}{*{20}{c}} {{{( {{a_1} - {a_0}} )}^T}}\\ {\dots }\\ {{{( {{a_n} - {a_0}} )}^T}} \end{array}} \right) \qquad \text { and }\qquad b = -\frac{1}{2}\left( {\begin{array}{*{20}{c}} {\xi _1 - \xi _0 - {{\Vert {{a_1}} \Vert }_2^2} + {{\Vert {{a_0}} \Vert }_2^2}}\\ {\dots }\\ {\xi _n - \xi _0 - {{\Vert {{a_n}} \Vert }_2^2} + {{\Vert {{a_0}} \Vert }_2^2}} \end{array}} \right) . \end{aligned}$$

The system (4.31) can be rewritten as $Ax^\star =b$. Since ${a_j} - {a_0},\, j = 1,\dots ,n$ are linearly independent in ${\mathbb {R}}^n$, A is invertible. Hence, $x^\star$ is determined uniquely by $x^\star =A^{-1}b$. $\square$

1.6 Proof of Theorem 8

Proof

We will prove by induction that for $t \in \left\{ {0,\dots ,n} \right\}$,

$$\begin{aligned} \exists {K_t} \in {\mathbb {N}}:\,\forall k \ge {K_t},\,\forall j \in \{ {0,\dots ,t} \},\, \eta _k^j = {\xi _j}. \end{aligned}$$

(1.43)

For $t=0$, (4.33) is the SDP relaxation of order $k+w$ of

$$\begin{aligned} {\xi _0} = \min \left\{ {{{\Vert {x - {a_0}} \Vert }_2^2}:\, x \in S\left( {g\cup \{ {L - {{\Vert {x } \Vert }_2^2}} \},h} \right) } \right\} . \end{aligned}$$

By assumption, ${( {\eta _k^0} )_{k \in {\mathbb {N}}}}$ finitely converges to $\xi _0$, i.e. there exist $K_0\in {\mathbb {N}}$ such that $\eta _k^0 = {\xi _0}$ for all $k\ge K_0$. It follows that (1.43) is true for $t=0$. Assume that (1.43) is true for $t=T$, i.e.

$$\begin{aligned} \exists {K_T} \in {\mathbb {N}}:\,\forall k \ge {K_T},\,\forall j \in \{ {0,\dots ,T} \},\, \eta _k^j = {\xi _j}. \end{aligned}$$

(1.44)

We will show that (1.43) is true for $t=T+1$. By (4.33) and (1.44), for all $k\ge K_T$,

$$\begin{aligned} \begin{array}{rl} \eta _k^{T+1} = \mathop {\inf }\limits _{y \in {{\mathbb {R}}^{s\left( {2(k+w)} \right) }}}&{} {L_y}\left( {{{\Vert {x - {a_{T+1}}} \Vert }_2^2}} \right) \\ \text {s.t. }&{}{M_{k+w - {u_j}}}\left( {{g_j}y} \right) \succeq 0,\\ &{}{M_{k+w - {w_q}}}\left( {h_qy} \right) = 0,\\ &{}{M_{k+w - 1}}\left( {\left( {{\xi _ j} - {{\Vert {x - {a_j}} \Vert }_2^2}}\right) y}\right) = 0,\, j = 0,\dots ,T,\\ &{}{y_0} = 1 \end{array} \end{aligned}$$

is the SDP relaxation of order $k+w$ of problem

$$\begin{aligned} {\xi _{T+1}} = \min \left\{ {{{\Vert {x - {a_t}} \Vert }_2^2}:\, x \in S\left( {g,h \cup \left\{ {{\xi _j} - {{\Vert {x- {a_j}} \Vert }_2^2},\, j = 0,\dots ,T} \right\} } \right) } \right\} . \end{aligned}$$

By assumption, ${( {\eta _k^{T+1}})_{k \ge K_T}}$ finitely converges to $\xi _{T+1}$, i.e. there exist $K_{T+1}\ge K_T$ such that $\eta _k^{T+1} = {\xi _{T+1}}$ for all $k\ge K_{T+1}$. It follows that (1.44) is true for $t=T+1$. Thus, (1.44) is true for $t=0,\dots ,n$. For $t=n$, there exists $K=K_n\in {\mathbb {N}}$ such that for all $k\ge K$, $\eta _k^t = {\xi _t}$, $t = 0,\dots ,n$. Let $k\ge K$ be fixed. Let y be the solution of problem

$$\begin{aligned} \begin{array}{rl} \eta _k^n: = \mathop {\inf }\limits _{y \in {{\mathbb {R}}^{s\left( {2(k+w)}\right) }}}&{} {L_y}\left( {{{\Vert {x - {a_n}} \Vert }_2^2}} \right) \\ \text {s.t. }&{}{M_{k+w - {u_j}}}\left( {{g_j}y} \right) \succeq 0,\\ &{}{M_{k+w - {w_q}}}\left( {h_qy}\right) = 0,\\ &{}{M_{k+w - 1}}\left( {\left( {{\xi _ j} - {{\Vert {x - {a_j}} \Vert }_2^2}} \right) y} \right) \succeq 0,\, j = 0,\dots ,n-1,\\ &{}{y_0} = 1, \end{array} \end{aligned}$$

which is the SDP hierarchy relaxation of order $k+w$ of problem

$$\begin{aligned} {\xi _{n}} = \min \left\{ {{{\Vert {x - {a_n}} \Vert }_2^2}:\, x \in S\left( {g,h \cup \left\{ {{\xi _j} - {{\Vert {x - {a_j}} \Vert }_2^2}:\, j = 0,\dots ,n-1} \right\} } \right) } \right\} . \end{aligned}$$

We will prove that this latter problem has a unique minimizer $x^\star$. Set $\hat{h}:=h\cup \{ \xi _0-\Vert x-a_0\Vert _2^2\}$. Then

$$\begin{aligned} \left\{ \begin{array}{l} {\xi _0} = \min \{ {{{\Vert {x - {a_0}} \Vert }_2^2}:\, x \in S( g,\hat{h})} \},\\ {\xi _t} = \min \{ {{{\Vert {x - {a_t}} \Vert }_2^2}:\, x \in S( { g,\hat{h} \cup \{ {{\xi _j} - {{\Vert {x - {a_j}} \Vert }_2^2}:\, j = 0,\dots ,t - 1} \}} )} \},\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ \ \ \ \ t = 1,\dots ,n. \end{array} \right. \ \end{aligned}$$

By Lemma 4 (with $S(g,h):=S(g,\hat{h})$), there exists $x^\star$ such that

$$\begin{aligned}&S\left( {g,h \cup \left\{ {{\xi _j} - {{\Vert {x - {a_j}} \Vert }_2^2}:\,j = 0,\dots ,n } \right\} } \right) \nonumber \\&\quad = S\left( { g, \hat{h} \cup \left\{ {{\xi _j} - {{\Vert {x - {a_j}} \Vert }_2^2}:\,j = 0,\dots ,n }\right\} } \right) = \left\{ {{x^\star }} \right\} . \end{aligned}$$

(1.45)

Let a be a minimizer of the above POP with value $\xi _n$. Then

$$\begin{aligned} a\in S( {g,h \cup \{ {{\xi _j} - {{\Vert {x - {a_j}} \Vert }_2^2}:\, j = 0,\dots ,n-1} \}} ), \end{aligned}$$

and ${\Vert {a - {a_n}} \Vert }_2^2=\xi _n$. It follows that

$$\begin{aligned} a\in S( {g,h \cup \{ {{\xi _j} - {{\Vert {x - {a_j}} \Vert }_2^2}:\,j = 0,\dots ,n } \}} ). \end{aligned}$$

From this and by (1.45), $a=x^\star$. Since the above POP with optimal value $\xi _n$ has a unique minimizer $x^\star$ and its Lasserre’s hierarchy has finite convergence, the solution y of the SDP with optimal value $\eta _k^n$ must have a representing 1-atomic measure $\mu$ supported on $x^\star$. Then y satisfies the flat extension condition when k is large enough and ${\mathop \mathrm{supp}\nolimits }( \mu ) = \{ {{x^\star }}\}\subset S(g,h)$. The conclusion follows. $\square$