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If $f \in L^1$ then its Fourier transform is easy to define: $\widehat{f}(\xi) = \int f(x)e^{-2 \pi i \xi x}dx$. If $\widehat{f} \in L^1$ then we recover $f(x) = \int \widehat{f}(\xi)e^{2 \pi i \xi x} d \xi$, a nice formula; if not, there's the not-so-nice formula $\|f \ast K_{\delta} - f\|_1 \to 0$ as $\delta \to 0$ (where $K_{\delta}$ is the Gauss kernel), and $f \ast K_{\delta}$ can be written in terms of $\widehat{f}$. My question is: is there a nice characterization of when I can apply the first formula, that is, which $L^1$ functions have $L^1$ Fourier transforms? If not, what sorts of classes of functions satisfy this? The Schwartz class would be such an example.

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  • $\begingroup$ There is none AFAIK. The space of functions with the property you mention is sometimes called inversion space and denoted by $\mathcal{I}(\mathbb{R}^n)$. $\endgroup$ – Giuseppe Negro Mar 12 '14 at 21:58
  • $\begingroup$ @GiuseppeNegro That's a bummer. I edited my question to request examples of subspaces of $\mathcal{I}(\mathbb{R}^n)$, then. $\endgroup$ – Julien Clancy Mar 13 '14 at 0:29
  • $\begingroup$ You could probably expand even further away from Schwartz to functions with exponential decay. In this case, the Fourier transform is definitely not an automorphism but the functions should be integrable. $\endgroup$ – Cameron Williams Mar 13 '14 at 0:31
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As commenters said, there is no explicit characterization of functions with an integrable Fourier transforms. A natural and useful class of functions that have integrable Fourier transforms is the Sobolev space $H^s(\mathbb R^n)$ with $s>n/2$: $$f\in H^s \iff f\in L^2 \text{ and } |\xi|^s \hat f(\xi) \in L^2 $$ Indeed, for such functions $$\|\hat f\|_{L^1} = \int_{\mathbb R^n} \Big((1+|\xi|^s) \hat f(\xi) \Big) \cdot \frac{1}{1+|\xi|^s}\,d\xi <\infty $$ because both factors in the last integral are in $L^2$.

In classical terms: if $f$ is square integrable and so are its $k$th order derivatives for some $k>n/2$, then the above applies.

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