# Polynomials and Integrability

Let $c_{n, m}$ be complex numbers ($n=0,1,\dots,N$ and $m=0,1,\dots,M$) not all equal to zero, and consider the polynomial function $P:\mathbb{R}^2 \rightarrow \mathbb{C}$ defined for each $(x,y) \in \mathbb{R}^2$ by $$P(x,y)= \sum_{n=1}^{N} \sum_{m=1}^{M} c_{n,m} x^{n} y^{m}.$$ Now define, whenever $(x,y) \in \mathbb{R}^2$ is such that $P(x,y) \neq 0$: $$F(x,y)=\frac{1}{P(x,y)}.$$ Let $Q(x,y)$ and $R(x,y)$ be respectvely the real and imaginary part of $P(x,y)$. The zero set of $P(x,y)$ is the intersection of the zero sets of the two real polynomials $Q(x,y)$ and $R(x,y)$. Since the zero set of a non null real polynomial has zero Lebesgue measure (for a very simple proof see Daniel Fischer's answer in Zero Set of a Polynomial), we conclude that $F$ is defined a.e. Assume that $F$ is locally integrable. Can we conclude that $P$ has no zero on $\mathbb{R}^2$?

I think the answer is yes, but I could find no proof of this fact.

NOTE (1). Let us not that the analogous question for polynomials in one variable has clearly a positive answer. Indeed for the Fundamental Theorem of Algebra we have in this case $P(x)=(x-z_1)^{n_1}\dots(x-z_k)^{n_k}$, and if some of the $z_1,\dots,z_k$ is real, clearly $F$ is not integrable.

NOTE (2).Let us also note that the analogous question on $\mathbb{R}^n$, for $n \geq 3$, has a negative answer, as the polynomial $P(x_1,\dots,x_n)=x_1^2+\dots+x_n^2$ shows. So the only critical case is that of $\mathbb{R}^2$.

Here's another solution. I'll assume all polynomials are real. I'll start with an easy result:

Lemma: Suppose $P$ is a polynomial in one variable and on the interval $[a,b]$ we have $P(a) < 0, P(b)>0.$ Then

$$\tag 1\int_a^b \frac{1}{|P(t)|}\, dt =\infty.$$

Proof: By the intermediate value theorem, $P(c) = 0$ for some $c\in (a,b).$ Thus there is a positive $C$ such that $|P(t)|\le C|t-c|$ for $t$ near $c.$ This implies $1/|P(t)|\ge 1/(C|t-c|),$ which gives $(1).$

Suppose now $P$ is a nonconstant polynomial in two variables, with $P(0,0) = 0.$

Case 1: $\nabla P(0,0) = (0,1).$ Then $P(x,y) = y + Q(x,y),$ where $Q(0,0) = 0$ and $\nabla Q(0,0) = (0,0).$ Every term in the polynomial $Q$ is then $O(x^2+y^2).$ It follows that there is a constant $C$ such that

$$\tag 2 |Q(x,y)|\le C(x^2+y^2) \text { for } (x,y) \text { close to } (0,0).$$

Choose $h_0$ such $(x,y) \in [-h_0,h_0]^2$ implies $(2).$ By going to a smaller $h_0,$ if necessary, we can assume $2Ch^2<h$ for $0<h<h_0.$ Then for any such $h,$

$$\tag 3 \int_{-h}^h \int_{-h}^h \frac{1}{|y+Q(x,y)|}\, dy\,dx = \int_{-h}^h \infty \,dx = \infty.$$

Why? For any fixed $x \in [-h,h],$ the polynomial $y\to y+Q(x,y)$ is positive at $h$ and negative at $-h.$ By the lemma, the integral with respect to $y$ is $\infty$ for this $x.$ Fubini then gives $(3).$ Conclusion: $1/|P|$ is not locally integrable at $(0,0).$

Case 2: $\nabla P(0,0)\ne (0,0).$ Then there is a nonsingular linear transformation $T:\mathbb R^2 \to \mathbb R^2$ such that $\nabla (P\circ T)(0,0) = (0,1).$ By case 1, $1/|P\circ T|$ is not locally integrable at $(0,0).$ The linear change of variables $(x,y) \to T^{-1}(x,y)$ then shows $1/|P|$ is not locally integrable at $(0,0).$

Case 3: $\nabla P(0,0)= (0,0).$ This is the easy case. We have $|P(x,y)| \le C(x^2+y^2)$ near $(0,0).$ Hence $1/|P(x,y)| \ge 1/[C(x^2+y^2)]$ for these $(x,y).$ In polar coordinates this implies a singularity on the order of $1/r^2$ at the origin. Integrating in polar coordinates then shows $1/|P|$ is not locally integrable at $(0,0).$

• Thank you for having answered my question. The proof is essentially the same, but it has the merit of avoiding the use of the implicit function theorem, which is of course a big result! – Maurizio Barbato Jan 21 '17 at 12:30

Finally, I realized that the answer to my question is generally negative if we assume that the coefficients $c_{n,m}$ are complex, as the simple example $P(x,y)=x+iy$ shows.

But if we make the assumption that all the coefficients $c_{n,m}$ are real, then the answer is in the affirmative, and this is the proof.

Assume that $(x_0,y_0) \in \mathbb{R}^2$ is such that $P(x_0,y_0)=0$. By considering eventually the polynomial $\tilde{P}(x,y)=P(x+x_0,y+y_0)$, we can assume without loss of generality that $(x_0,y_0)=(0,0)$. So $P$ has no constant term, that is $c_{0,0}=0$.

First consider the case $c_{1,0}=c_{0,1}=0$. By considering polar coordinates, we then have $$P(r \cos \theta, r \sin \theta) = \sum_{n,m} c_{n,m} r^{n+m} \cos^{n}(\theta) \sin^{m}(\theta),$$ so that by putting $$S_k(\theta)= \sum_{n+m=k} c_{n,m} \cos^{n}(\theta) \sin^{m}(\theta),$$ we have $$P(r \cos \theta, r \sin \theta) = \sum_{k=2}^{d} r^{k} S_k(\theta),$$ where $d$ is the degree of $P(x,y)$. For $r \in (0,1)$, we then have $$\frac{r}{|F(r \cos \theta, r \sin \theta)|} \geq \frac{1}{\sum_{k=2}^{d} r^{k-1} |S_k(\theta)|} \geq \frac{1}{r \sum_{k=2}^{d} |S_k(\theta)|}.$$ Let us make the simple observation that if we have $c_{n,m} \neq 0$ for some $n, m$ such that $n+m=k$, then, since $$S_k(\theta)= \cos^{k} (\theta) \left[ \sum_{n+m=k} c_{n,m} \tan^{m}(\theta) \right],$$ we have $S_k(\theta) \neq 0$, except for a finite number of values of $\theta$. Then we conclude that $$\int_{[0,1] \times [0,2 \pi]} \frac{1}{r} \frac{1}{ \sum_{k=r}^{d} |S_k(\theta)|} dr d \theta = \infty,$$ so $F$ is not locally integrable.

Consider now the case in which at least one of $c_{0,1}$ and $c_{1,0}$ is distinct from zero, e.g. $c_{0,1} \neq 0$. By the Implicit Function Theorem, there is $\delta > 0$ and $\sigma > 0$ such that for every $x \in (-\delta, \delta)$ there exists a unique $y \in (-\sigma,\sigma)$ such that $P(x,y)=0$. We then have for every fixed $x \in (-\delta, \delta)$ that the polynomial $Q(y)=P(x,y)$ has (exactly) one zero on the interval $(-\sigma,\sigma)$, so that by the result in NOTE (1) we have $$\int_{[-\sigma,\sigma]} |F(x,y)| dy = \infty,$$ and we conlude that $$\int_{[-\delta, \delta]} \int_{[-\sigma,\sigma]} |F(x,y)| dy dx = \infty.$$ QED