Given the polynomials $p_1(x_1, \ldots, x_n), \ldots,p_k(x_1, \ldots, x_n)$, let $\Lambda$ be the set of $x \in R^n$ where all of these polynomials are nonnegative. Does there exist a polynomial $q(x_1, \ldots, x_n)$ such that $\Lambda$ is the set of points where $q(x_1, \ldots, x_n)$ is nonnegative, i.e., $\Lambda = \{ x ~|~ q(x) \geq 0\}$?
About the set of points where a polynomial is nonnegative
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polynomials
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1Let $n=2$, let $p_1=x_1,p_2=x_2$. Is there a polynomial that is non-negative precisely on the (closed) first quadrant? – 2011-09-08
1 Answers
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Not necessarily.
Consider the special case $p_1(x,y)=x$, $p_2(x,y)=y$. Then $q(x,y)$ should be nonnegative exactly when $x\ge 0$ and $y\ge 0$. The positive x axis is on the boundary of this set, so by continuity $q(x,0)=0$ for $x\ge 0$.
Now consider the function $x\mapsto q(x,0)$. This is a polynomial in one variable (it arises by dropping all terms with positive exponent in $y$). We've concluded that it must be identically zero for $x\ge 0$. But that must mean that it is the zero polynomial, because a nonzero polynomial in one variable cannot have more zeroes than its degree.
So in particular, for example, $q(-1,0)=0$. But that contradicts the assumption that $q(x,y)\ge0$ only for $x\ge 0$.
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1My comment would be relevant if you were talking about \{(x,y): q(x,y) > 0\}. – 2011-09-08