Find all solutions $T$ of $x^{2006} T = 0$ in the space of tempered distributions, $\mathcal{S}'(\mathbb{R})$

Question

I'm modeling my solution after this answer to a similar question. This is as far as I've gotten:

Every $\phi \in \mathcal{S}(\mathbb{R})$ that vanishes at $0$ can be expressed as $\phi(x) = x \psi (x)$. Then, $T\phi = xT(\psi) = 0$ by assumption.

Fix $\chi \in \mathcal{S}(\mathbb{R})$ such that $\chi(0) = 1$. Let $T\chi = a$. Then, for any $\phi \in \mathcal{S}(\mathbb{R})$, $$T\phi = T(\phi - \phi(0) \chi + \phi(0) \chi) = T(\phi - \phi(0) \chi) + T(\phi(0) \chi).$$

This is where I've gotten stuf. I'm not sure how, in the linked solution, the answer reduces from $T(\phi - \phi(0) \chi) + T(\phi(0) \chi)$ to $0 + a \phi(0)$ (my primary confusion is $T(\phi - \phi(0) \chi) = 0$) nor how to adapt that for my own question.

Any suggestions?

Answer 1

Let $\psi \in C^\infty_c([-1,1]), \int_\Bbb{R} \psi(x)dx = 1$ then $ \Psi = \psi \ast 1_{x \in [-3,3]} \in C^\infty_c(\Bbb{R})$ is constant $=1$ on $[-1,1]$.

For $\phi \in S(\Bbb{R})$ let $$A(\phi) = \phi- \sum_{n=0}^N \frac{\phi^{(n)}(0)}{n!} x^n \Psi \in S(\Bbb{R})$$

Because the derivatives at $0$ vanish then $\frac{A(\phi)}{x^N} \in S(\Bbb{R})$.

Let $T$ be a tempered distribution such that $x^N T = 0$ that is $$\forall \varphi \in S(\Bbb{R}),\qquad \langle T,x^N\varphi \rangle = 0$$ Then $$\langle T,\phi\rangle = \langle x^N T,\frac{A(\phi)}{x^N}\rangle+ \sum_{n=0}^N \frac{\phi^{(n)}(0)}{n!} \langle T,x^n \Psi \rangle = \langle \sum_{n=0}^N c_n\delta^{(n)}, \phi \rangle \\c_n = \frac{(-1)^n}{n!} \langle T,x^n \Psi \rangle$$

Whence $T = \sum_{n=0}^N c_n\delta^{(n)}$