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Matthew Daws
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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 \[ \int_{\mathbb R} g(s) h(s) \ ds = \langle g, h\rangle_{(L^1(\mathbb R),C_0(\mathbb R))} = \lim_n \langle \mathcal F(f_n), h\rangle_{(L^1(\mathbb R),C_0(\mathbb R))} = \lim_n (\mathcal F(f_n)|\overline{h})_{L^2(\mathbb R)} = 0. \] 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.)

Matthew Daws
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