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Consider a complex polynomial in $n$ variables $z=(z_1,\dots,z_n)$: \begin{equation} P(z)=\sum_{|\alpha| \leq N} c_{\alpha} z^{\alpha}, \end{equation} where as usual for every $\alpha=(\alpha_1,\dots,\alpha_n) \in \mathbb{N}^{n}$ we set $|\alpha|=\alpha_1+\dots+\alpha_n$, and $z^{\alpha}=z_1^{\alpha_1}\dots z_n^{\alpha_n}$. Let $Z= \{ x \in \mathbb{R}^n : P(x) = 0 \}$ and define $H:\mathbb{R}^n \rightarrow \mathbb{C}$ as \begin{equation} H(x)= \begin{cases} \frac{1}{P(x)} & \textit{if } x \in \mathbb{R}^n \backslash Z,\\ 0 & \textit{if }x \in Z. \end{cases} \end{equation} Assume that $H \in L_{loc}^{1}(\mathbb{R}^n)$. Is it then true that $H$ defines a tempered distribution? More explicitly, if we set \begin{equation} T(\phi)=\int_{\mathbb{R}^n} H(x) \phi(x) dx, \quad (\phi \in \mathscr{D}(\mathbb{R}^n)), \end{equation} does $T$ extends to a continuous linear functional on $\mathscr{S}(\mathbb{R}^n)$?

I guess the answer is positive, but I have no idea of a possible proof.

NOTE (1). Let $x \in \mathbb{R}^n$, and let $Q(x)$ and $R(x)$ be respectvely the real and imaginary part of $P(x)$. $Z$ is the intersection of the zero sets of the two real polynomials $Q(x)$ and $R(x)$. 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 $Z$ has zero Lebesgue measure, and for our question is totally irrelevant how we define $H$ on $Z$.

NOTE (2). Let us note that $H$ can be locally integrable even if $Z$ is not empty when $n \geq 2$. Take e.g. for $n=2$ the polynomial $P(z_1,z_2)=z_1+ i z_2$, or for $n=3$ consider $P(z_1,z_2,z_3)=z_1^{2}+z_2^{2}+z_3^{2}$ or even $P(z_1,z_2,z_3)=z_1+ i z_2$.

NOTE (3). If $Z= \emptyset$, then clearly $H \in L_{loc}^{1}(\mathbb{R}^n)$. In this case $H$ defines a tempered distribution. Indeed, by a remarkable result of Hörmander (see Lemma (2) in On the Division of Distributions by Polynomials) there exist $C > 0$ and $\mu > 0$ such that \begin{equation} |P(x)| \geq C (1+|x|^2)^{-\mu} \quad \forall x \in \mathbb{R}^n. \end{equation} So for $M > 0$ big enough we have in this case \begin{equation} \int_{\mathbb{R}^n} (1+|x|^2)^{-M} |H(x)| dx < \infty, \end{equation} and we conclude that $H$ defines a temepered distribution.

When $Z \neq \emptyset$, then Hörmander proves that there exists positive constants $C, \mu, \nu$ such that \begin{equation} |P(x)| \geq C (1+|x|^2)^{-\mu} [d(x,Z)]^{\nu} \quad \forall x \in \mathbb{R}^n, \end{equation} where \begin{equation} d(x,Z)=\inf_{y \in Z} |x - y| \quad (x \in \mathbb{R}^n). \end{equation} I don't know if this remarkable inequality together with the assumption that $H \in L_{loc}^{1}(\mathbb{R}^n)$ implies that $H$ defines a tempered distribution.

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    Out of curiosity - could you give an example when $P$ is not a constant polynomial, yet $Z=\emptyset$? – 2017-01-20
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    @TZakrevskiy For example, $x^2+1$ (the set $Z$ is restricted to the real part of the space) – 2017-01-21
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    Thank you. I completely misread the definition of $Z$, it seems. – 2017-01-21
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    Let me stress one thing. Even in the case in which $P$ is assumed to be a real polynomial (that is $P$ has real coefficients) and $P(x) > 0$ for all $x \in \mathbb{R}^n$ it is not true that $P$ is bounded from below on $\mathbb{R}^n$ by some positive constant $C$. Take e.g. the polynomial in two variables: $P(x,y)=x^2+(xy-1)^2$. This gives you an idea of the fact that the inequality proved by Hörmander is not trivial at all! – 2017-01-21
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    Let me also notice another simple fact: clearly not every non-null polynomial $P$ defines a locally integrable $H$. For example, for $n=1$, if $P(x)$ has a real zero, then for sure $H$ is not locally integrable. For $n=2$, no real polynomial (that is with real coefficients) $P(x,y)$ having some real zero $(x,y)$ defines a locally integrable function. The proof of this result can be found in my post [Polynomials and Integrability](http://math.stackexchange.com/questions/2096573/polynomials-and-integrability). – 2017-01-24
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    @zaq: I believe the statement to be false if $P$ has real zeros. Consider $P(x)=\|x\|^{2n}$. Let $\phi$ be a *positive* function with compact support (therefore rapidly decreasing), with $\phi (0) = 1$. Then $\int \frac 1 P \phi$ essentially behaves like $\int \frac 1 {r^{n+1}} \Bbb d r$ which diverges, therefore $\frac 1 P$ is not even a *linear functional* on $\mathscr S (\Bbb R^n \setminus \{0\})$, let alone a tempered distribution! The problem, in general, is that the zeros of $P$ are not isolated (they form a smooth submanifold with singularities), and I don't know how to tackle this. – 2017-01-29
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    @AlexM. Perhaps you missed the assumption that $1/P$ is locally integrable. – 2017-01-29
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    @zaq: Indeed, I totally missed that. But wait, is it possible for $H$ to be locally-integrable? If $P(x)=0$ and $K$ is a compact neighbourhood of $x$, then is it possible for $H$ to be integrable on $K$? Isn't that assumption vacuous for polynomials with zeros? – 2017-01-29
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    @AlexM. Apparently, you missed Note (2) in the post as well. To avoid further misunderstanding, I'll also emphasize that the question is about _tempered_ distributions. The assumption of local integrability guarantees that $1/P$ defines a distribution, but the "tempered" part requires an analysis of its behavior at infinity. – 2017-01-29
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    @zaq: True, again. – 2017-01-29

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