# Minimizers of polynomial functions?

Let $p: \mathbb{R}^2 \to \mathbb{R}$ be a polynomial function of two real variables. Suppose that

(P) $\; \; \; \; \; \; p(x,y) \ge 0, \; \; \forall \; (x,y) \in \mathbb{R}^2$

1: Prove that if $p$ is of degree two, it has a minimizer over $\mathbb{R}$.

2: Prove that if a polynomial function $p$ verifies property (P) its degree is necessarily even.

3: Suppose that degree $p=2n$ with $n \gt 1$. Does property (P) imply the existence of a minimizer for $p$ over $\mathbb{R}$? $$\\ \\$$

My ideas:

Since $f(x,y)$ is a degree two polynomial, it'd be of the form $$f(x,y) = ax^2 +by^2 + cx + dy + kxy + l$$

Taking all the second derivatives to compute the Hessian gives:

$$\left\lbrack \matrix{2a & k\cr k & 2b} \right\rbrack$$

My hunch is that since the function has to be equal or greater than zero, $a, b$ and $k$ have to be non-negative, which would mean the Hessian is always non-negative, which I think would mean that there has to be a minimizer.

I'm not really sure about the other parts, but I think that the answer to part "3" is YES. At least a local minimizer I think

• By minimizer do you mean a point which achieves an absolute minimum? Commented Sep 8, 2016 at 4:54
• @NickR Yes, an absolute minimum. Commented Sep 10, 2016 at 22:10
• note that the exercise assumes a degree $2n$ with $n>1$ and not $2$, which leads to a $2n \times 2n$ hessian. Yes, a positive hessian is a sufficient condition for a local extremum which is a minimum here ...
– user354674
Commented Sep 11, 2016 at 5:47

(1) and (2) are true, but (3) is false, and no Hessians are needed.

(1) is true : specializing (P) with $x=0$, we see that $by^2+dy+l\geq 0$ for any $y$. So either $b>0,d^2-4bl \leq 0$ or $b=d=l=0$. In the latter case, we have $f(x,y)=ax^2+cx+kxy=x(ax+c+ky)$, so $f(1,y)=a+c+ky$ and (P) forces $k=c=0,a\geq 0$ ; then $0$ is a minimizer.

So we can assume without loss that $b>0,d^2-4bl \leq 0$. Similarly, interchanging $x$ and $y$, we can further assume $a>0,c^2-4al\geq 0$.

Next, we use the classic technique of "completing the square" ; i.e. we write

$$f(x,y)=ax^2+cx+b\bigg(\big(y+\frac{d+kx}{b}\big)^2-\big(\frac{d+kx}{b}\big)^2\bigg)+l= g(x,Y)$$ where $Y=y+\frac{d+kx}{b}$ and

$$g(x,Y)=ax^2+cx-b\big(\frac{d+kx}{b}\big)^2+bY^2+l$$

We see that $g$ satisfies $(P)$ just like $f$. So replacing $f$ with $g$, we may assume without loss that $d=k=0$. Similarly completing the square in $x$, we can further assume that $c=0$. Then $f(x,y)=ax^2+by^2+l$ with $a,b$ nonnegative, and $l$ is a minimizer.

(2) is clear by specializing $(P)$ to $f(t,t)$, which is a univariate polynomial with a lower bound, so it must be of even degree.

(3) Counter-example : $f(x,y)=x^2+(xy-1)^2$. Note that

$f(\frac{1}{n},n)=\frac{1}{n^2}$, so that $0$ is a lower bound, but it is never attained.