Fourier transform of a tempered distribution zero has supported at the origin?

Question

Hello. I am reviewing the book Classical Analysis by Grafakos and I cannot fully understand this proof.

Why $\widehat{u}$ is supported at the origin?

pd:

Answer 1

The product between tempered distributions and smooth functions with polinomially growing derivatives is well-defined, so the product $|\xi|^2\widehat u$ is well defined. From the proof we know that $|\xi|^2\widehat u=0$ as a distribution, and since the function $\xi\mapsto |\xi|^2$ is non-zero except at $\xi=0$, it follows that the support of $\widehat u$ is at the origin. Conceptually speaking, it’s because we can divide the equation by $|\xi|^2$ outside the origin.

To prove that step rigorously (just a sketch), consider a test function $\varphi$ whose support does not contain the origin. Then $|\xi|^{-2} \phi(\xi)$ is well-defined and it’s a test function as well, and since $|\xi|^2\widehat u=0$, we have that $$ 0=|\xi|^2\widehat u(|\xi|^{-2} \phi(\xi))=\widehat u(\phi)=0. $$ So, $\widehat u$ yelds zero when tested against any test function whose support does not contain the origin; thus, by definition, the support of $\widehat u$ is contained in the origin.