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Suppose that $f: \mathbb{R}^d \to \mathbb{R}$ denotes the characteristic function of the unit open ball in $\mathbb{R}^d$. Clearly, this belongs to $L^p(\mathbb{R}^d)$ for all $p \in \mathbb{N}$. But then $f$ has Fourier transform (up to a multiplicative constant) given by

$$\displaystyle \mathcal{F}(f)(\xi) = \frac{J_{d/2}(|\xi|)}{|\xi|^{d/2}},$$

where $|\xi|$ is the Euclidean norm of $\xi$ and $J_{\nu}$ denotes the Bessel function of the first kind. However, this function is not in $L^{1}(\mathbb{R}^d)$, which can easily be seen using the asymptotics of the Bessel function and switching to polar coordinates, say. Why is this? I always thought that the Fourier transform of an $L^1$ function was always in $L^1$.

Moreover, does this mean we can't use the usual "parity rule" which says that $(\mathcal{F} \circ \mathcal{F})(f)(\xi) = f(-\xi)$?

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    No, if $f, \hat f\in L^1$ then $f= F^{-1}(\hat f),$ which implies $f$ is continuous. – 2017-02-19
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    The Fourier transform of an $L^2$ function is always in $L^2$ (by Plancherel's formula), but the Fourier transform of an $L^1$ function is not necessarily in $L^1$ (...as your example shows). The Fourier transform of an $L^1$ function is continuous and bounded. – 2017-02-19

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