Characterizing polynomials whose associated semi-algebraic set is bounded? Hello,
I'm a theoretical computer scientist, and would be interested by the following problem: given a polynomial p, over R, multi-variate, is there a sufficient condition on its parameters such that {$x~ | ~p(x)\leq 0$} is bounded?
The idea would be to somewhat generalize the fact that given a semi-definite matrix positive P, the set {$x ~|~ (1,~ x^t~ ;~ x~ P)\text{ is sdp}$} describes an ellipsoid, potentially flat in some dimensions.
The problem seems to be too easily formulated for having not been deeply considered by mathematicians, but I did not manage to find any reference.
Any help and/or pointers would be appreciated.
Best,
Yannick.
 A: The question is a bit open-ended because I could give you sufficient and very restrictive conditions for this to happen. Here are a couple of elementary remarks that might help.
Let $[p]$ denote the top degree  part of $p$.   This is a homogeneous polynomial of  degree $k$. If you require that $[p](u)>0$ for any $u\in \mathbb{R}^n$ of length $1$ then you get the desired conclusion. Note that this forces the degree  $k$ to be even.  This reduces   the problem to characterizing  even degree homogeneous polynomial which are positive definite. There are many of those, but I don't know how to characterize them. Here is a large  supply. Suppose that $Q(y_1,\dotsc, y_m)$ is a nontrivial  homogeneous polynomial with  nonnegative coefficients, and $q_1,\dotsc, q_m$ are positive definite  quadratic forms in $n$ variables $x_1,\dotsc, x_n$. Then  $Q(q_1,\dotsc, q_m)$ is a positive definite  homogeneous polynomial in the variables $x_1,\dotsc, x_n$.
The minimum value  of a polynomial $P$ of  even degree $k$ on the unit sphere is obtained via Lagrange multipliers. Look at the solutions of the system
$$ \lambda\in\mathbb{R},\;\;u\in\mathbb{R}^n,\;\;|u|=1,\;\;  \nabla P(u)=\lambda u. \tag{1}$$
The Euler identity for   homogeneous functions implies
$$ k P(u)= u\cdot \nabla P(u) =\lambda |u|^2 =\lambda. $$
The polynomial  $P$ will be positive definite  if the system (1)  has no solutions $(\lambda, u)$  with $\lambda\leq 0$. 
