# Decaying Fourier transform and smoothness

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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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.