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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?

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eraldcoil
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  • What does the Corollary $2.4.2$ say? – Gonçalo Nov 12 '23 at 01:21
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    it's easy to verify directly that $\langle \hat{u}, \phi \rangle = 0$ for $\phi \in C_c^{\infty}(\mathbb{R}^n \setminus 0)$. – Kakashi Nov 12 '23 at 01:24
  • But, if $\xi\neq 0$ then $|\xi|^2\widehat{u}=0$ implies that $\widehat{u}=0$, and therefore $\langle \widehat{u},\phi\rangle =0$ for all $\phi$? – eraldcoil Nov 12 '23 at 01:32
  • @eraldcoil Just $$\langle \hat{u}, \phi \rangle = \langle |\xi|^2\hat{u}, \frac{1}{|\xi|^2}\phi \rangle = 0.$$ – Kakashi Nov 12 '23 at 02:20

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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.

Lorenzo Pompili
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