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Liouville's Theorem states that if a tempered distribution is harmonic, $\Delta{u}=0$, then $u$ is given by a polynomial. For the argument, we take Fourier transform of $\Delta{u}=0$ to obtain $4\pi^2|\xi|^2\hat{u}=0$. I do not understand why this should imply that $\hat{u}$ has support in $\{0\}$. I should be able to conclude that if a Schwartz class function $\phi$ is supported away from $0$, then $\hat{u}(\phi)=0$. But all we have is $\hat{u}(|\xi|^2\phi)=0$. In fact, any polynomial which is non-zero other than at zero would do the job and not only $|\xi|^2$.

I am surely missing something tiny.

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Here is a proof:… – Siminore Aug 29 '12 at 13:39

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