Let $P(z)$ be a non-null complex polynomial in $\nu$ variables $z=(z_1,\dots,z_n)$ of degree $\mu$: \begin{equation} P(z)=\sum_{|\alpha| \leq \mu} c_{\alpha} z^{\alpha}, \end{equation} where as usual for every $\alpha=(\alpha_1,\dots,\alpha_\nu) \in \mathbb{N}^{\nu}$ (here and in the following $\mathbb{N}$ denotes the set of all non-negative integers) we set $|\alpha|=\alpha_1+\dots+\alpha_\nu$, and $z^{\alpha}=z_1^{\alpha_1}\dots z_{\nu}^{\alpha_\nu}$. Consider $P$ as a polynomial function from $\mathbb{R}^\nu$ into $\mathbb{C}$: \begin{equation} P(x)=\sum_{|\alpha| \leq \mu} c_{\alpha} x^{\alpha} \quad (x \in \mathbb{R}^{\nu}). \end{equation} For any $m \in \mathbb{N}$, any $S \subseteq \mathbb{R}^\nu$, and any $\phi \in \mathcal{D}(\mathbb{R}^\nu)$ set: \begin{equation} ||\phi||_{m,S} = \sup_{\substack{x \in S \\ |\alpha| \leq m}} |(D^{\alpha} \phi)(x)|. \end{equation} Let $M > L > 0$ and put $Q=[-M,M]^\nu$ and $E=Q \backslash (-L,L)^\nu$. I am trying to prove that for any $m \in \mathbb{N}$, there exist $K > 0$ and $m' \in \mathbb{N}$ such that we have \begin{equation} ||\phi||_{m,E} \leq K ||P\phi||_{m',E} \quad \forall \phi \in \mathcal{D}_{Q} \tag{I}, \end{equation} where as usual $\mathcal{D}_{Q}$ is the set of all complex-valued functions $\phi \in C^{\infty}(\mathbb{R}^\nu)$ with support contained in $Q$. See the notes below for an explanation of the origin and relevance of this question.

Thank you very much in advance for your attention.

NOTE (1). If we take $L=0$, so that $E=Q$, then (I) is an immediate corollary of a remarkable result proved by Lars Hörmander in his wonderful work On the Division of Distributions by Polynomials. Indeed, inequality (4.3) of this work (taken with $k=0$) implies that for any $n, m \in \mathbb{N}$, there exist $K > 0$ and $n', m' \in \mathbb{N}$ such that \begin{equation} \sup_{\substack{x \in \mathbb{R}^\nu \\ |\alpha| \leq m}} (1+|x|)^n |(D^{\alpha} \phi) (x)| \leq K \sup_{\substack{x \in \mathbb{R}^\nu \\ |\alpha| \leq m'}} (1+|x|)^{n'} |(D^{\alpha} (P\phi)) (x)| \quad \forall \phi \in \mathcal{S}(\mathbb{R}^\nu) \tag{II}. \end{equation} We can state (II) in another way. Define the linear subspace $\mathcal{M}_{P}$ of $\mathcal{S}(\mathbb{R}^\nu)$: \begin{equation} \mathcal{M}_{P}=\{\psi \in \mathcal{S}(\mathbb{R}^\nu): \psi=P \phi, \phi \in \mathcal{S}(\mathbb{R}^\nu) \}, \end{equation} and consider the multiplication map $M_{P}:\mathcal{S}(\mathbb{R}^\nu) \rightarrow \mathcal{M}_{P}$ defined by \begin{equation} M_{P}(\phi)=P\phi \quad (\phi \in \mathcal{S}(\mathbb{R}^\nu)), \end{equation} Then (II) is equivalent to say that $M_{P}$ has a continuous inverse (this statement is Theorem (1) in Hörmander's work).

NOTE (2). Inequality (I) was stated without proof by ifw in his answer to the post Division of Distributions by Polynomials (see also my own answer for a comment). If true, (I) would allow to give a direct proof of Theorem (4) in Hörmander's paper, which states that every distribution can be divided by a non-null polynomial. In one of his comments, ifw said that (I) could be proved by localizing (II) or by modifying properly Hörmander's original proof. Even though I studied very carefully Hörmander's original proof (which can also be found in Trèves, Linear Partial Differential Equations with Constant Coefficients, $\S$ 5.5), I could not modify it in order to obtain (I) nor I could get (I) by localizing (II).


1 Answer 1


Finally, I realized how to modify Hörmander's proof in order to prove (I). I will describe here the necessary changes we have to do to Hörmander's proof. Clearly, all the notation is that of Hörmander's paper.

Let $C$ be a closed, convex set of $\mathbb{R}^\nu$. Then make the following changes in Section (4) of Hörmander's paper On the Division of Distributions by Polynomials:

$\bullet$ take the suprema comparing in all the relations over $C$;

$\bullet$ after relation (4.2), replace everywhere $N^k$ with $N^k \cap C$ and $N^{k+1}$ with $N^{k+1} \cap C$;

$\bullet$ replace $\mathbb{R}^\nu$ with $C$ in (4.13);

$\bullet$ replace in every relation $S(\xi)$ with $S(\xi) \cap C$ and $S(\eta)$ with $S(\eta) \cap C$;

$\bullet$ replace in every relation $S(\xi,\eta)$ with $S(\xi,\eta) \cap C$ and $S_{1}(\xi,\eta)$ with $S_{1}(\xi,\eta) \cap C$.

With these changes, Hörmander's proof shows that to all $n, m \in \mathbb{N}$ and $k \leq \mu$ there are $n', m' \in \mathbb{N}$ and a constant $K > 0$ such that \begin{equation} \sup_{\xi \in C} (1+|\xi|)^n |f|_{m, (N^k \cap C)_{\xi}} \leq K \sup_{\xi \in C} (1+|\xi|)^{n'} |Pf|_{m',\xi}, \quad (f \in C^{m'}(\mathbb{R}^\nu)). \end{equation} So, if $C$ is a compact convex set, the previous inequality, taken with $k=0$, implies that to all $m \in \mathbb{N}$ there are $ m' \in \mathbb{N}$ and a constant $K > 0$ such that \begin{equation} ||f||_{m,C} \leq K ||Pf||_{m',C} \quad (f \in C^{m'}(\mathbb{R}^\nu)) \tag{III}. \end{equation} Now, assume that $E$ is a finite union of compact convex subsets $C_1,\dots, C_p$ of $\mathbb{R}^\nu$. By (III), for each $i=1,\dots,p$, there exist $m'_{i} \in \mathbb{N}$ and $K_i > 0$ such that \begin{equation} ||f||_{m,C_i} \leq K_i ||Pf||_{m'_{i},C_i} \quad (f \in C^{m'_{i}}(\mathbb{R}^\nu)) \tag{III}. \end{equation} By taking $K=\max\{K_1,\dots,K_p\}$ and $m'=\max\{m'_{1},\dots,m'_{p}\}$, we then get \begin{equation} ||f||_{m,E} \leq K ||Pf||_{m',E} \quad (f \in C^{m'}(\mathbb{R}^\nu)) \tag{IV}. \end{equation} Clearly (I) is a particular case of (IV), so we are done. QED

REMARK. One could think that (IV) still holds if we take $E$ to be any compact set, but this is not the case, as the following counterexample shows. Let $\nu=1$, and set \begin{equation} E=\{ 0 \} \cup \left\{ \frac{1}{n}: n=1,2,3,\dots \right\}. \end{equation} Take $P(x)=x$ and let $K > 0$. Choose $N > K$. By the celebrated Whitney's Extension Theorem (which is Theorem (I) in Analytic Extensions of Differentiable Functions Defined in Closed Sets), there exists $f \in C^{\infty}(\mathbb{R})$, with support contained in $\left[ \frac{1}{N+1},2 \right]$, such that for $n=1,2,3,\dots,N$ \begin{equation} f^{(m)}\left( \frac{1}{n} \right) = (-1)^m m! n^{m+1} \quad (m \in \mathbb{N}). \end{equation} We have $D^{m}(Pf)(x)=0$ for all $x \in E$ and all $m=1,2,\dots$, so that $||Pf||_{m',E}=1$ for all $m' \in \mathbb{N}$. But $||f||_{0,E}=N > K$, and we conclude that (IV) cannot hold.


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