Radial distributions

Question

here is a theorem I am not able to solve, it is about distributions (Schwartz).

Let $\Omega = \mathbb{R}^2 \setminus \{0\}$. Show that for all $S \in \mathcal{D}'(\mathbb{R}_+^*)$ there exists a unique $T \in \mathcal{D}'(\Omega)$ such that : $$\langle S, \phi \rangle = \langle T, \phi \circ N \rangle$$ and satisfies $(x \partial _y - y \partial _x) T=0$ in $\mathcal{D}'(\Omega)$ and where $N(x,y)=x^2+y^2$

What I've done :

1. If $T$ is a $\mathcal{C}^1$ function then by setting $x=r \cos \theta$ and $y= r \sin \theta$ we have $\partial _{\theta} T=0$ so that $T(r, \theta)=T_0(r)$ and that makes the theorem I want to prove a generalisation of this to distributions.

2. For $T \in \mathcal{D}'(\Omega)$ such that $(x \partial _y - y \partial _x) T=0$ in $\mathcal{D}'(\Omega)$ I showed that $T \circ R_{\theta}=T$ where $R_{\theta}$ is the rotation of the plan of angle $\theta \in \mathbb{R}$.

Can you provide me a hint to prove it ?

Thanks in advance.

Answer 1

For unicity it is sufficient to show that for $S=0$ the only possible $T$ is equal to $0$.

The condition $S=0$ implies that $T$ is zero on all test functions $\phi$ which depend only on $r=\sqrt{x^2+y^2}$.

The second condition, as you've shown, implies that $T$ is invariant under rotations $R_\theta$.

edit

Define $R_\theta$ as a rotation by angle $\theta$. Then any point $(x,y)$ with $r=\sqrt{x^2+y^2}=\sqrt{x_0^2+y_0^2}>0$ satisfies $$(x,y)=R_\theta[(x_0,y_0)]$$ for some angle $\theta\in[0,2\pi)$.

Then you write $$\psi (x_0,y_0) =\frac{1}{|C|}\int_{\sqrt{x^2+y^2}=\sqrt{x_0^2+y_0^2}}\phi(x,y)ds.$$ The equation $\sqrt{x^2+y^2}=\sqrt{x_0^2+y_0^2}$ gives you a manifold (a circle $C$), $ds$ is a measure on this manifold, and $|C|$ is the total measure of this manifold.

Upon changing variables you can say that

$$\psi (x_0,y_0) =\frac{1}{2\pi}\int_{\theta\in[0,2\pi)}\phi(R_\theta [(x_0,y_0)])d\theta.$$

This will be a test function depending only on $r=\sqrt{x_0^2+y_0^2}$, hence $\langle T,\psi\rangle=0$. On the other hand,

$$\langle T,\psi\rangle= \frac{1}{2\pi}\left\langle T,\int_{[0,2\pi]}\phi(R_\theta [(x_0,y_0)])d\theta \right\rangle = \frac{1}{2\pi} \int_{[0,2\pi]}\left\langle T,\phi(R_\theta [(x_0,y_0)])\right\rangle d\theta =\frac{1}{2\pi} \int_{[0,2\pi]} \langle T\circ R_{-\theta},\phi \rangle d\theta= \frac{1}{2\pi} \int_{[0,2\pi]} \langle T ,\phi \rangle d\theta = \langle T ,\phi \rangle . $$ We can conclude that $T=0$ and thus the unicity holds.

end edit

Answer 2

After long thinking I came to a partial solution :

For the existence just set : $$\langle T, \phi \rangle = \frac{1}{2 \pi} \left\langle S , \int_{\theta \in [0, 2\pi]} \phi(\cdot , \theta) \mathrm{d} \theta \right \rangle $$ This defines a distribution and it is easily seen that $\partial_{\theta} T=0$ i.e $(x \partial _y - y \partial _x)T=0$

What about the uniqueness, I know it comes from the fact that $T \circ R_{\theta} = T$ but impossible to prove it in a clean way ..