Let $F(z) = e^{−z^2}$ . I am trying to prove that there is no $f \in L^1[−1, 1]$ such that $F(z) =\int_1^{-1} f(t)e^{itz}dt$.
I first tried to show that $\int_1^{-1} f(t)e^{itz}dt$ is not an entire function of $z$ . But what I actually got is that $\int_1^{-1} f(t)e^{itz}dt$ is entire (by Morera's theorem). So I think I need a different approach, but I can't find such one. Any hints?
It is a standard calculation (complete the square then use Cauchy's theorem to reduce to $z=0$) that $$F(z)=\int_{\mathbb R} \frac1{2\sqrt \pi}e^{-t^2/4} e^{-itz} dt.$$ Since the Fourier transform is injective, (WLOG switching $f(-t)$ for $f(t)$), $$ F(z) = \int_{\mathbb R} f(t) \mathbb 1_{[-1,1]}(t) e^{-itz} dt$$ would imply that almost everywhere in $\mathbb R$, $$\frac1{2\sqrt \pi}e^{-t^2/4}=f(t) \mathbb 1_{[-1,1]}(t),$$ which is immediately absurd for any $|t|\ge1$.
