I wonder if somebody could help me understand the following solution:
$\textbf{Problem:}$ Let $F=\{ f \ast f: f \in \mathcal{E}'(\mathbb{R}^n) \} \subset \mathcal{E}'(\mathbb{R}^n)$. Show that $F \neq \mathcal{E}'(\mathbb{R}^n)$ ($f\ast f$ is the convolution of $f$ with it self and $\mathcal{E'} $ is the space of distributions with compact support in $\mathbb{R}^n$).
$\textbf{Solution:}$ Assume the contrary. Then $\delta_1 -\delta_0 \in F$ so there is an $f \in \mathcal{E'}$ such that $f \ast f= \delta_1 -\delta_0$ Since $f \in \mathcal{E'}$ such the Fourier transform of $f$, $\hat{f}$, is complex analytic. Then $ (\hat{f}(\xi))^2= e^{-i\xi}-1 \in \mathcal{O}({\xi}) \implies \hat{f}(0)=0$.
I fail to see why this is a contradiction (also I don't get the last implication). Does anybody understand this?