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Suppose that $f\in L^1 (\mathbb{R})$ and that for any $n\in \mathbb{N}$ there is $C_n > 0$ such that its Fourier transform satisfies $$ |\hat{f}(\xi )| \le C_n(1+|\xi |^2)^{-n}. $$ I want to show that $f\in C^\infty (\mathbb{R})$.

By the inversion formula we have $$ f(x) = \frac{1}{2\pi } \int \limits _\mathbb{R} \hat{f}(\xi ) e^{ix\xi } \, d\xi , $$
but as far as I can see this is only valid almost everywhere. Clearly the right hand side is continuous since $\hat{f}\in L^1$ (and even smooth), but my problem is that the formula is just an identity in $L^1(\mathbb{R})$.

Why is $f$ necessarily continuous?

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up vote 3 down vote accepted

Actually, what we have to show is that in the equivalence class of $f$ for equality almost everywhere, there is a smooth function. Inversion formula gives that $g\colon x\mapsto \frac 1{2\pi}\int_{\Bbb R}\widehat f(\xi)e^{i\xi x}d\xi$ is in this equivalence class, so what we have to show that is that $g$ is smooth.

The decay condition used with dominated convergence theorem allows us to show that $$g^{(d)}(x)=\frac 1{2\pi}i^d\int_\Bbb R\widehat f(\xi)e^{i\xi x}\xi^dd\xi,$$ which is a differentiable function.

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Yes, perhaps that is how I should think of the problem, not like I am given a fixed representative to begin with. Thank you. –  flavio Dec 5 '12 at 13:47

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