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?