Problem: Show that if $\vert x\vert^2u=0$ for $u\in\mathcal D'(\mathbb R^d)$, then for all $\psi\in\mathcal C_c^\infty(\mathbb R^d)$ such that $\operatorname{supp}(\psi)\cap B(0,10)=\varnothing$ we have $u(\psi)=0.$
My Thoughts: I think that the problem statement is not clear, as there seems to be nothing special about $B(0,10)$. For example, fix $\varepsilon>0$ and consider $B(0,\varepsilon)$. Note that $x\mapsto \vert x\vert^2$ is smooth on $\mathbb R^d\setminus B(0,\varepsilon)$. Therefore, if $\psi\in\mathcal D(\mathbb R^d)$ has $\operatorname{supp}(\psi)\cap B(0,\varepsilon)=\varnothing$, then
$$u(\psi)=\vert x\vert^2u\left(\frac{\psi}{\vert x\vert^2}\right)=0.$$
I justify the above calculation by noticing that $\psi/\vert x\vert^2\in\mathcal D(\mathbb R^d\setminus B(0,\varepsilon))$ and $\operatorname{supp}(\psi/\vert x\vert^2)=\operatorname{supp}(\psi)$. Also if $u\in\mathcal D'(\mathbb R^d)$, then $u\vert_{\mathbb R^d\setminus B(0,\varepsilon)}\in\mathcal D'(\mathbb R^d\setminus B(0,\varepsilon)).$
Therefore, it seems that the result holds for any ball centered at the origin.
Could anyone tell me if I am on the right track or if I have made a horrible mistake at the beginning in my understanding of the definitions?
Thank you for your time and appreciate the valuable feedback.
You are correct that there is nothing special about radius $10$. You can as well take radius $R>0$ or $\epsilon>0.$ That doesn't make the problem statement non-clear, though. But I understand that it might be confusing.
I was about to write that I do not understand how you justify $u(\psi)=|x|^2u(\psi/|x|^2)$ as $|x|$ is not a constant. But then I realized that you do not mean $|x|^2$ multiplied with $u(\psi/|x|^2)$ but the distribution $|x|^2 u$ applied to $\psi/|x|^2$. In the "bracket" notation that I'm used to it would be $\langle u, \psi \rangle = \langle |x|^2 u, |x|^{-2}\psi \rangle.$ So, yes, you are on the right track; $|x|^{-2}$ is smooth outside of the ball.