I've come across a number of sources claiming a smoothness-decay duality between a function and its Fourier transform. But most seem to give results about how the smoothness of a function leads to faster decay in the Fourier transform. I'm more interested in results about how the smoothness of a Fourier transform lead to a faster decay in the Fourier transform.
In particular, I'd like to know: Given a bounded function $f \in L^{2}(\mathbb{R})$ with fourier transform $\hat{f}$ integrable and continuous, do I have that $f$ itself is integrable?