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Let $H:\mathbb{R} \rightarrow \mathbb{C}$ be a function, with $H \in C^{\infty}(\mathbb{R})$, and $S \in \mathscr{D}'(R)$ a distribution such that there exists $x_0 \in \mathbb{R}$ which belongs to both the support of $H$ and the support of $S$. Assume that $H$ and all its derivatives vanish in $x_0$: \begin{equation} (D^{n}H)(x_0)=0 \qquad (n=0,1,2,\dots). \end{equation} I am trying to prove that in these hypotheses, the division of $S$ by $H$ is not possible. This means that there exists no distribution $T \in \mathscr{D}'(R)$ such that \begin{equation} H \cdot T = S. \end{equation} I could prove this statement only in a particular case (see Note (2) below), but I am quite convinced that it is true. Any help is welcome.

Thank you very much in advance for your attention.

NOTE (1). The statement above is a conjecture of mine, inspired by Schwartz, Théories des Distributions, Chapitre V, $\S 4$, p.126. He states that if $H$ is a function as described above, that is $H \in C^{\infty}(\mathbb{R})$ and $H$ vanishes with all its derivatives in a point $x_0$, then the division of a distribution $S$ by $H$ is not possible. Obviously, this is not true for a generic distribution $S$. To see this, assume that $H$ has as only zero $x_0$, and that $x_0$ does not lie in the support of $S$. For any non-empty open subset $\Omega$ of $\mathbb{R}$, define $S_{\Omega} \in \mathscr{D'}(\Omega)$ as \begin{equation} S_{\Omega}(\psi)=S(\psi) \quad (\psi \in \mathscr{D}(\Omega)). \end{equation}

Then if $\Omega_1$ is the complement of the support of $S$, the zero distribution $T_1=0 \in \mathscr{D'}(\Omega_1)$ satisfies \begin{equation} H \cdot T_1 = S_{\Omega_1}, \end{equation} since we have \begin{equation} T_1(H\psi) = S(\psi)=0 \quad \forall \psi \in \mathscr{D}(\Omega_1). \end{equation} Now put $\Omega_2=\mathbb{R} \backslash \{x_0 \}$. Since $1/H \in C^{\infty}(\Omega_2)$, $T_2=\frac{1}{H} \cdot S_{\Omega_2}$ is a well defined element of $\mathscr{D'}(\Omega_2)$ and we have \begin{equation} H \cdot T_2 = S_{\Omega_2}, \end{equation} that is \begin{equation} T_2(H \psi) = S (\psi) \quad \forall \psi \in \mathscr{D}(\Omega_2). \end{equation} Now let $\xi_1 \in \mathscr{D}(\Omega_1)$ be such that $0 \leq \xi \leq 1$, and $\xi=1$ on an open set $V$ containing $x_0$, and define $\xi_2 = 1 - \xi_1$. Define \begin{equation} T(\phi)=T_1(\xi_1 \phi)+T_2(\xi_2 \phi) \quad (\phi \in \mathscr{D}(\mathbb{R})). \end{equation} It is immediate to see that $T \in \mathscr{D'}(\mathbb{R})$ and that \begin{equation} H \cdot T = S. \end{equation}

NOTE (2). I could prove the statement above in the specific case in which $S$ is the distribution defined by the constant function equal to one. In this case, take $\psi \in \mathscr{D}(R)$ such that \begin{equation} \int_{\mathbb{R}} \psi(x) dx =1, \end{equation} and define the sequence of test functions \begin{equation} \psi_{m}(x)=m \psi(m(x-x_0)+x_0) \quad (x \in \mathbb{R}, m=1,2,3,\dots). \end{equation} It is easy to see that $S(\psi_m) =1$ for all $m$. Now set \begin{equation} \phi_m(x) = H(x) \psi_m(x) \quad (x \in \mathbb{R}, m=1,2,3,\dots). \end{equation} Define for any non-negative integers $n, m$ the function \begin{equation} F_{n,m}(x) = \begin{cases} \frac{(D^n H)(x)}{(x-x_0)^m} & x \neq x_0, \\ 0 & x = x_0, \end{cases} \end{equation} We have $F_{n,m} \in C^{\infty}(\mathbb{R})$: in particular $F_{n,m}$ is continuous at $x_0$. From this observation and Leibniz formula we easily get that $\phi_m \rightarrow 0$ in $\mathscr{D}(\mathbb{R})$. So if there existed $T \in \mathscr{D'}(\mathbb{R})$ such that \begin{equation} H \cdot T = S, \end{equation} we should have $T(\phi_m) \rightarrow 0$. But we have $T(\phi_m)=T(H\psi_m)=S(\psi_m)=1$ for all $m$, a contradiction.

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  • $\begingroup$ not sure if I got your question right, but if $S=H$ then such a $T$ exists, namely $T=1$ $\endgroup$
    – user8268
    Jan 22, 2017 at 16:29
  • $\begingroup$ @user8268: you're perfectly right! How couldn't I think of this example? If you post it as an answer I will accept it. $\endgroup$ Jan 23, 2017 at 9:45

1 Answer 1

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My conjecture is false, as user8268 pointed out in his comment above. This is his counterexample.

Assume $H$ is as stated in the question, that is $H \in C^{\infty}(\mathbb{R})$ and $H$ vanishes with all its derivatives in a point $x_0$ belonging to the support of $H$. Let $S$ be the distribution defined by $H$. Then the support of $S$ is equal to the support of $H$, so that $x_0$ belongs to both, but clearly the distribution $T$ defined by the constant function equal to one satisfies \begin{equation} H \cdot T = S. \end{equation} Taking into account this counterexample and Note (2) in my post, we conclude that Schwartz's statement has to be meant in the following sense: if $H \in C^{\infty}(\mathbb{R})$ and $H$ vanishes with all its derivatives in $x_0$, then there exists some $S \in \mathscr{D'}(\mathbb{R})$ such that the division of $S$ by $H$ is not possible.

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