I have certain doubts about a classical Fourier inversion theorem. According to it (this is a theorem from "Panorama of Harmonic Analysis" by Krantz), if $f$ and $\hat{f}$ are both in $L_1(R)$ and continuous, then almost everywhere f is equal to the inverse Fourier transform of $\hat f$.

It seems to me that some conditions may be missing here. Indeed, if the function $f$ is the inverse Fourier transform of $\hat f$ then, as a Fourier transform, $f$ must be bounded, uniformly continuous on $R$ and satisfy the conditions $ f(y) \to 0$ as $y \to \infty$ and as $y \to -\infty$. Do these conditions really hold automatically or no if $f$ and $\hat{f}$ are both in $L_1(R)$ and continuous?

Are there any easy-to-check sufficient conditions for a function to be a Fourier transform of a function from $L_1(R)$?

Thanks in advance.

  • $\begingroup$ Everything you are asking for can be found in Katznelson book on Harmonic analysis, which is I believe available online. $\endgroup$ – Alvin Mar 5 '14 at 18:25

(Too long for a comment, sorry). If $f$ is in $L^1$ then the integral defining the Fourier transform is defined at every point and defines a continuous function vanishing at infinity. Now if the transform also happens to be in $L^1$, the inverse Fourier transform is also a continuous function vanishing at infinity, which coincides a.e. with $f$. Does this answer your question? as to a characterization of the so called Wiener algebra (Fourier transforms of $L^1$ functions) I do not think there is any useful characterization.


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