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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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    $\begingroup$ If $H$ is locally integrable things are simple because then the only possible problem would be at infinity, and thre is no problem there because $P$ is polynomial. What you do in non-integrable cases is the trickier situation. Try for example to define $1/x$ as a tempered distribution on $\mathbb{R}$. There is no unique answer. Any two answers differ by a distribution concentrated at $0$. $\endgroup$
    – Liviu Nicolaescu
    Commented Feb 25, 2017 at 10:27
  • $\begingroup$ 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 and idea of the fact that the inequality proved by Hörmander is not trivial at all! $\endgroup$
    – Maurizio Barbato
    Commented Feb 26, 2017 at 9:04
  • $\begingroup$ Let me also notice another simple fact: clearly not every non-null polynomial $P$ defines a locally integrable $H$. For example, if $n=1$ and $P(x)$ has a real zero, then for sure $H$ is not locally integrable. For $n=2$, no real polynomial (that is with real coefficients) having a real zero $(x,y)$ defines a locally integrable function. The proof of this fact can be found in my post Polynomials and Integrability. $\endgroup$
    – Maurizio Barbato
    Commented Feb 26, 2017 at 9:07
  • $\begingroup$ I made several points: when $1/P$ is locally integrable on $\mathbb{R}^n$ then it defines a distribution; even when the local integrability fails one can still define $1/P$; the answer need not be unique. E.g., $1/x$ admits many extensions as a distribution on $\mathbb{R}$. $1/(x\pm 0 \cdot i)$, $i=\sqrt{-1}$ are two such possible extensions. $\endgroup$
    – Liviu Nicolaescu
    Commented Feb 26, 2017 at 16:04
  • $\begingroup$ @LiviuNicolaescu Dear Liviu, I know all that: these are standard results that can be found e.g. in the wonderful treatise of Laurent Schwartz. But this is of no help for the question I asked. $\endgroup$
    – Maurizio Barbato
    Commented Feb 26, 2017 at 16:06

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