Division of distributions by polynomials. I'm trying to solve the equation
$(1-|x|^2)T = 0$,
where $T$ is a tempered distribution. I know how to do this (it is a common exercise) in dimension $1$. How can I solve it in higher dimensions?
Thank you very much.
 A: The solution is the direct product of $\delta(|x|-1)$ and any distribution on the sphere.
A: This is not a different answer. I only give here the details of Michael Renardy's solution. 
Theorem. A tempered distribution $T$ on $\mathbb{R}^n$ solves $(1-|x|^2)T=0$ if and only if for some distribution $V$ on the unit sphere $S^{n-1}=\{x \in \mathbb{R}^n : |x|=1 \}$, we have $T(\phi)=V(\phi_{|S^{n-1}})$, for every $\phi \in \mathcal{S}(\mathbb{R}^n)$.
Proof. If for some distribution $V$ on $S^{n-1}$, we have $T(\phi)=V(\phi_{|S^{n-1}})$, for every $\phi \in \mathcal{D}(\mathbb{R}^n)$, then clearly $T$ is tempered, that is it can be extended to a continuous linear functional on $\mathcal{S}(\mathbb{R}^n)$ by putting $T(\phi)=V(\phi_{|S^{n-1}})$, for every $\phi \in \mathcal{S}(\mathbb{R}^n)$. Moreover, we clearly have $T((1-|x|^2) \phi(x))=V(0)=0$ for every $\phi \in \mathcal{S}(\mathbb{R}^n)$. 
Now, let $T$ any distribution (tempered or not) such that $(1-|x|^2)T=0$. Then the support of $T$ is contained in $S^{n-1}$. Let $\epsilon \in \left(0,\frac{1}{2} \right)$ and $\xi \in \mathcal{D}(\mathbb{R}^n)$ a function such that $\xi=1$ on $\{x \in \mathbb{R}^n: 1 - \epsilon < |x| < 1 + \epsilon \}$, and such that the support of $\xi$ is constrained in $\{x \in \mathbb{R}^n: 1 - 2 \epsilon < |x| < 1 + 2 \epsilon \}$. Then we have $T(\xi \phi)=T(\phi)$ for every $\phi \in \mathcal{D}(\mathbb{R}^n)$.
Now let $\phi \in \mathcal{D}(\mathbb{R}^n)$, and define
\begin{equation}
\alpha(x)= \xi(x) \phi \left( \frac{x}{|x|} \right),
\end{equation}
\begin{equation}
\beta(x)= \xi(x) \left[\phi(x) - \phi \left( \frac{x}{|x|} \right)\right].
\end{equation}
Then $\alpha$ and $\beta$ clearly belong to $\mathcal{D}(\mathbb{R}^n)$ and we have $\xi \phi= \alpha + \beta$. Set
\begin{equation}
\chi(x) = \begin{cases} \frac{\beta(x)}{1-|x|^2} & \textit{if } |x| \neq 1,\\
- \frac{1}{2} (\nabla \phi)(x) \cdot \frac{x}{|x|} & \textit{if } |x|=1,\end{cases}
\end{equation}
It is easy to see by induction that $\chi$ and all its partial derivatives are  differentiable, so that $\chi \in \mathcal{D}(\mathbb{R}^n)$. From the equation
\begin{equation}
\xi(x) \phi(x)= \alpha(x) + (1-|x|^2) \chi(x),
\end{equation}
we conclude that $T(\phi)=T(\alpha(x))$. So, if we define for any smooth function $\gamma$ on $S^{n-1}$
\begin{equation}
V(\gamma)=T\left( \xi(x) \gamma\left( \frac{x}{|x|} \right) \right),
\end{equation}
clearly $V$ is a distribution on $S^{n-1}$, and we have $T(\phi)=V(\phi_{|S^{n-1}})$ for any $\phi \in \mathcal{D}(\mathbb{R}^n)$.
QED
