A function where its Fourier transform satisfies $\operatorname{Supp} \hat{f}=\{0\}$

Question

I did not understand why, if $\operatorname{Supp} \,\hat{f} = \{0\} $ for a given function $f$ in $L^p(\mathbb R^n)$; $p<\infty$, then $f=0$ ?

Thank you in advance

Answer 1

If $f \in L^p$ then $f$ generates a tempered distribution $T_f$ through $\langle T_f, \phi \rangle = \int f(x) \, \phi(x) \, dx$, and also $\hat f$ is a tempered distribution.

In the space of (tempered) distributions, if $\operatorname{supp} \hat f = \{0\}$ then $\hat f = \sum_\alpha c_\alpha \partial^\alpha \delta$ for some constants $\alpha$. Here $\alpha$ is a multi-index. This means that $f(x) = \sum_{|\alpha|\leq N} \bar c_\alpha x^\alpha$ for some $N \in \mathbb N$, where $\bar c_\alpha$ is closely related to $c_\alpha$ (by a factor $(2\pi)^n$ or something like that).

But $\sum_{|\alpha|\leq N} \bar c_\alpha x^\alpha \not\in L^p$ unless $\bar c_\alpha = 0$ for all $\alpha$, i.e. unless $f = 0$.

Thus, $f = 0$.