Polynomials and Integrability

Question

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 \begin{equation} P(x,y)= \sum_{n=1}^{N} \sum_{m=1}^{M} c_{n,m} x^{n} y^{m}. \end{equation} Now define, whenever $(x,y) \in \mathbb{R}^2$ is such that $P(x,y) \neq 0$: \begin{equation} F(x,y)=\frac{1}{P(x,y)}. \end{equation} 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 $\mathbb{R}^2$.

Answer 1

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 \begin{equation} P(r \cos \theta, r \sin \theta) = \sum_{n,m} c_{n,m} r^{n+m} \cos^{n}(\theta) \sin^{m}(\theta), \end{equation} so that by putting \begin{equation} S_k(\theta)= \sum_{n+m=k} c_{n,m} \cos^{n}(\theta) \sin^{m}(\theta), \end{equation} we have \begin{equation} P(r \cos \theta, r \sin \theta) = \sum_{k=2}^{d} r^{k} S_k(\theta), \end{equation} where $d$ is the degree of $P(x,y)$. For $r \in (0,1)$, we then have \begin{equation} \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)|}. \end{equation} 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 \begin{equation} S_k(\theta)= \cos^{k} (\theta) \left[ \sum_{n+m=k} c_{n,m} \tan^{m}(\theta) \right], \end{equation} we have $S_k(\theta) \neq 0$, except for a finite number of values of $\theta$. Then we conclude that \begin{equation} \int_{[0,1] \times [0,2 \pi]} \frac{1}{r} \frac{1}{ \sum_{k=r}^{d} |S_k(\theta)|} dr d \theta = \infty, \end{equation} 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 \begin{equation} \int_{[-\sigma,\sigma]} |F(x,y)| dy = \infty, \end{equation} and we conlude that \begin{equation} \int_{[-\delta, \delta]} \int_{[-\sigma,\sigma]} |F(x,y)| dy dx = \infty. \end{equation} QED

Answer 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

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

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 $(0,0).$