MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $p=s+qt$.


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.

share|cite|improve this question

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


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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.