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Might the following be true:

Let $p,q\in\mathbb{R}[x,y,z]$ be homogeneous polynomials, with $\deg(p)\leq 4$ and $\deg(q)= 2$. Suppose $q(x,y,z)>0$ for some $x,y,z\in\mathbb{R}$. Then the following are equivalent:

1) $q(x,y,z)\geq 0\Rightarrow p(x,y,z)\geq 0$ for all $x,y,z\in \mathbb{R}$

2) there are nonnegative homogeneous polynomials $s,t$ with $\deg(s)\leq 4$ and $\deg(t)\leq 2$ such that $p=s+qt$.

Background:

A) The case of the above conjecture that $\deg(p)=2$ is exactly the S-lemma for polynomials $p,q$ in three variables.

B) The analogous statement for homogeneous polynomials $p,q$ in two variables is also known to be true.

C) By a theorem of Hilbert, a homogeneous polynomial $p\in\mathbb{R}[x,y,z]$ of degree 4 is nonnegative if and only if it is a sum of three squares of quadratic polynomials.

If the above conjecture holds up then $$\min\{p(x,y)\mid q(x,y)\geq 0\}$$ for inhomogeneous polynomials $p,q$ with $\deg(p)=4$ and $\deg(q)=2$ can be cast as a semidefinite optimization problem.

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1 Answer 1

There is the following counterexample:

Set $f=\mathrm{x}_1^3\mathrm{x}_3+\mathrm{x}_1^3\mathrm{x}_2+\mathrm{x}_2^2\mathrm{x}_3^2$ and $g=\mathrm{x}_1\mathrm{x}_3+\mathrm{x}_2\mathrm{x}_3+\mathrm{x}_1\mathrm{x}_2$. The two homogeneous polynomials satisfy condition (1): The easiest way to verify this, is to type in the following command

Reduce[ForAll[{x1,x2,x3},Implies[x1*x3+x2*x3+x1*x2>=0,x1^3*x3+x1^3*x2+x2^2*x3^2>=0]]]

in Mathematica.

But condition (2) is violated: Suppose we could find a polynomial $t=a_1\mathrm{x}_1^2+a_2\mathrm{x}_2^2+a_3\mathrm{x}_1\mathrm{x}_2+a_4\mathrm{x}_1\mathrm{x}_3+a_5\mathrm{x}_2\mathrm{x}_3+a_6\mathrm{x}_3^2$ ($a_1,\ldots,a_6\in \mathbb{R}$) such that $f(y)-t(y)g(y)\geq 0$ for all $y\in \mathbb{R}^3$. A simple calculation shows that $f-tg=\mathrm{x}_1^3\mathrm{x}_2-a_1\mathrm{x}_1^3\mathrm{x}_2-a_3\mathrm{x}_1^2\mathrm{x}_2^2-a_2\mathrm{x}_1\mathrm{x}_2^3+\mathrm{x}_1^3\mathrm{x}_3-a_1\mathrm{x}_1^3\mathrm{x}_3-a_1\mathrm{x}_1^2\mathrm{x}_2\mathrm{x}_3-a_3\mathrm{x}_1^2\mathrm{x}_2\mathrm{x}_3-a_4\mathrm{x}_1^2\mathrm{x}_2\mathrm{x}_3-a_2\mathrm{x}_1\mathrm{x}_2^2\mathrm{x}_3-a_3\mathrm{x}_1\mathrm{x}_2^2\mathrm{x}_3-a_5\mathrm{x}_1\mathrm{x}_2^2\mathrm{x}_3-a_2\mathrm{x}_2^3\mathrm{x}_3-a_4\mathrm{x}_1^2\mathrm{x}_3^2-a_4\mathrm{x}_1\mathrm{x}_2\mathrm{x}_3^2-a_5\mathrm{x}_1\mathrm{x}_2\mathrm{x}_3^2-a_6\mathrm{x}_1\mathrm{x}_2\mathrm{x}_3^2+\mathrm{x}_2^2\mathrm{x}_3^2-a_5\mathrm{x}_2^2\mathrm{x}_3^2-a_6\mathrm{x}_1\mathrm{x}_3^3-a_6\mathrm{x}_2\mathrm{x}_3^3.$ Thus $f(\mathrm{x}_1,0,1)-t(\mathrm{x}_1,0,1)g(\mathrm{x}_1,0,1)=\mathrm{x}_1^3-a_1\mathrm{x}_1^3-a_4\mathrm{x}_1^2-a_6\mathrm{x}_1$. We know that $f-tg$ is a non-negative polynomial. This is only possible if $a_1=1$, $a_6=0$, and $a_4\leq 0$. Since $t$ is non-negative and $a_6=0$, the two coefficients $a_4$, $a_5$ must also vanish. The leading term of $f(0,\mathrm{x}_2,1)-t(0,\mathrm{x}_2,1)g(0,\mathrm{x}_2,1)$ is $-a_2\mathrm{x}_2^3$, and therefore $a_2=0$. Now only $a_3$ is not determined. But it is easy to see that $a_3$ must also vanish. Thus $f-tg$ reduces to $f-tg=-\mathrm{x}_1^2\mathrm{x}_2\mathrm{x}_3+\mathrm{x}_2^2\mathrm{x}_3^2$, which is obviously not non-negative.

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