This is in answer to Jessica and Zen. Yemon supposes that $\mathcal F$ restricts to a linear map $C_0(-1,1)\rightarrow L^1(\mathbb R)$. We want this to have a closed graph-- so if $(f_n)\subseteq C_0(-1,1)$ with $f_n\rightarrow 0$ and $\mathcal F(f_n)\rightarrow g\in L^1(\mathbb R)$, why is $g=0$?
Well, I'd be tempted to use that $\mathcal F$ extends to a unitary $L^2(\mathbb R)\rightarrow L^2(\mathbb R)$ (which can be proved in a quite elementary way). In particular, as $\|f_n\|_2\rightarrow 0$ we know that $\|\mathcal F(f_n)\|_2 \rightarrow 0$. So if $h$ is a compactly support continuous function, then use the embedding $L^1(\mathbb R)\rightarrow
C_0(\mathbb R)^*$ to see that
\[ \int_{\mathbb R} g(s) h(s) \ ds = \langle g, h\rangle_{(L^1(\mathbb R),C_0(\mathbb R))} = \lim_n \langle \mathcal F(f_n), h\rangle_{(L^1(\mathbb R),C_0(\mathbb R))} = \lim_n (\mathcal F(f_n)|\overline{h})_{L^2(\mathbb R)} = 0. \]$$\int_{\mathbb R} g(s) h(s) \ ds = \langle g, h\rangle_{(L^1(\mathbb R),C_0(\mathbb R))}
= \lim_n \langle \mathcal F(f_n), h\rangle_{(L^1(\mathbb R),C_0(\mathbb R))}
= \lim_n (\mathcal F(f_n)|\overline{h})_{L^2(\mathbb R)} = 0.$$
This shows that $g=0$ in $L^1(\mathbb R)$.
I'm not sure if that's what you're after, but it's the sort of "soft analysis" proof I'd use (that is, uses measure theory, Hilbert space theory, but no distributions etc.)
