Minkowski's Conjecture

Problem

Can one always write a nonnegative polynomial as a sum of squares? That is, if the real polynomial $P(x_1,x_2,\dots ,x_n)$ satisfies $P(x_1,x_2,\dots ,x_n)\ge 0\text{for all }(x_1,x_2,\dots ,x_n)\in {\mathbb{R}}^n,$ can one find a set of $s$ real polynomials $Q_k(x_1,x_2,\dots ,x_n)$, $1\le k\le s$, such that $P(x_1,x_2,\dots ,x_n)=Q_1^2+Q_2^2+\cdots +Q_s^2?$

If we restricted $Q_k$ to rational functions, it becomes Hilbert 17th problem:

• Given a multivariate polynomial that takes only non-negative values over the reals, can it be represented as a sum of squares of rational functions?

For one variable case, the answer to the problem is true. The proof is inspired a special case of Lagrange identity $(a_1^2+a_2^2)(b_1^2+b_2^2)=(a_1{b}_1+a_2{b}_2{)}^2+(a_1{b}_2-a_2{b}_1{)}^2.$ This identity implies if $P(x)=Q(x)R(x)$ where $Q(x)$ and $R(x)$ have the representations $Q(x)=Q_1^2(x)+Q_2^2(x)$ and $R(x)=R_1^2(x)+R_2^2(x)$, then $P(x)$ also has a representation as a sum of two squares. So we consider induction. For $P(x)$ has a real root $P(x)=(x-r{)}^{mR(x)}\text{where}R(r)=0,$ so, if we set $x=r+ϵ$, then we have $P(r+ϵ)={ϵ}^{m}R(r+ϵ)$. Also, by the continuity of $R$, there is a $\delta$ such that $R(r+ϵ)$ has the same sign for all $ϵ$ with $|ϵ|\le \delta$. Since $P(x)$ is always nonnegative, we then see that ${ϵ}^m$ has the same sign for all $|ϵ|\le \delta$, so $m$ must be even. If we set $m=2k$, we see that

$$P(x)=Q^2(x)R(x)\text{where }Q(x)=(x-r{)}^k,$$

For $P(x)$ has no real roots.

$$P(x)=(x-r)(x-\bar{r})R(x)=Q(x)R(x).$$

Minkowski’s conjecture

For polynomials in two variables, there exist nonnegative polynomials of two variables that cannot be written as the sum of squares of real polynomials.

From $(a_1{a}_2\cdots a_n{)}^{\frac{1}{n}}\le \frac{a_1+a_2\cdots +a_n}{n},$ consider $a_1=x^2$, $a_2=y^2$ and $a_3=1/x^2{y}^2$, we get $1\le \frac{1}{3}(x^2+y^2+\frac{1}{x^2{y}^2}),$ which means the polynomial $P(x,y)=x^4{y}^2+x^2{y}^4-3x^2{y}^2+1$ is non-negative.

Observe that $Q_k(x,y)$ has degree no greater than 3, we see that they must be of the form $Q_k(x,y)=a_k+b_{k}xy+c_k{x}^{2}y+d_{k}x{y}^2,$ then from coefficient of $x^2{y}^2$ in ${\sum }_{k=1}^s{Q}_k(x,y)$ would be $b_1^2+\cdots b_s^2$ which can’t be -3, giving us $P(x,y)$ is the desired counterexample.