Suppose I have a polynomial $p\in R[x_1,\ldots,x_n]$ and I look at the set $S:=\{ x\in R^n : p(x)\geq 0\}$. Are there algebraic certificates on $p$ that will certify that $S$ is compact?
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$\begingroup$ Surely it is necessary and sufficient for $\limsup_{x\to\infty} p(x) < 0$. This should be detected by the top order terms. So something like: take the top order terms of $p(x)$ and restrict them to the unit sphere in $\mathbb{R}^n$. It is both necessary and sufficient that this function is bounded above by $-c$ for some $c>0$. $\endgroup$– Otis ChodoshCommented Sep 16, 2013 at 20:09
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Let $p \in \mathbb{R}[x_1, \ldots, x_n]$. The set $S=\{x \in \mathbb{R}^n: p(x) \geq 0\}$ is compact if and only if there is a natural number $N$ and polynomials $g_i, h_i \in \mathbb{R}[x_1, \ldots, x_n]$, such that $N-\sum_{i=1}^n x_i^2 = \sum_{i=1}^rg_i^2+p \sum_{i=1}^s h_i^2$. It is easy to see that this criterion is sufficient. Schmüdgen's Positivstellensatz says that it is also necessary.
