# Continuous, integrable fourier transform of an $L^{2}(\mathbb{R})$ function, integrable.

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?

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## 1 Answer

Did you mean "how the smoothness of a Fourier transform lead to a faster decay in the function" [not of its transform]?

First of all, $f$ is the inverse Fourier transform of $\hat f$, and the inverse FT is the same as FT up to an inconsequential sign. Since $\hat f\in L^1$, we have $f$ bounded (and continuous) automatically; this hypothesis is redundant. Your question can be stated as: does the (inverse) Fourier transform map $L^1\cap C^0$ into $L^1$?

The answer is no: see this mathoverflow post by Yemon Choi. (Begin reading with the last paragraph.)

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