Schmüdgen's Positivstellensatz requires the polynomial to be strictly positive on a semialgebraic set. While trying to understand it, I am just wondering if the strictly positive condition can be weakened to non-negative, and if not, what is an example on which the Positivstellensatz does not hold? Thanks.
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$\begingroup$ I gave the following answer, and Emil Jerabek pointed out my error: Consider the set $[-1,1]$ defined by $1+x\ge0$ and $1-x\ge0$. The polynomial $x^2$ is non-negative on the whole set. Any polynomial in $1+x$ and $1-x$ with positive coefficients will have a positive constant term. So there is no way to write $x^2$ in that polynomial form. But, as Emil pointed out, the coefficients only need to be sum-of-squares in the polynomial ring, and not in the base field, so the conclusion of the Positivstellensatz is satisfied here by $$x^2=\left(\frac{x}{\sqrt{2}}\right)^2((1+x)+(1-x))$$ $\endgroup$– user44143Commented Mar 16, 2020 at 16:02
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$\begingroup$ I recommend reading the article C. Scheiderer: Positivity and sums of squares: a guide to recent results. In Emerging applications of algebraic geometry, Volume 149 of IMA Vol. Math. Appl., pages 271–324. Springer, New York, 2009. Section 3 seems to contain results and counter-examples you are looking for. $\endgroup$– Johannes HuismanCommented Mar 16, 2020 at 21:25
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$\begingroup$ The paper above is at math.uni-konstanz.de/~scheider/preprints/GUIDE.pdf $\endgroup$– user44143Commented Mar 17, 2020 at 11:13
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$\begingroup$ @JohannesHuisman Thank you for the article, but what is an example on which the Positivstellensatz does not hold? Thanks a lot. $\endgroup$– SophiaACommented Mar 21, 2020 at 10:59
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This answer may be a bit late, but every homogeneous polynomial which is not a sum of squares of homogeneous polynomials (like $Z^6 + X^4 Y^2 + X^2 Z^4 - 3(XYZ)^2$), when restricted to any ball of finite positive radius, gives such a counterexample.
