Fourier transform of an $L^p$ function need not be a function

Question

Fourier transforms can be defined for any $L^p$ function for $1\leq p \leq \infty$ by treating them as a tempered distribution. I am looking for an explicit example of an $L^p$ function whose Fourier transform is a tempered distribution which is not generated by any function. Explicit counter example, with a justification as to why its Fourier transform is not of function type will be greatly appreciated. Thanks in advance.

Answer 1

The constant function $x\mapsto 1$ belongs to $L^\infty(\mathbb R)$ and has Fourier transform $2\pi\delta,$ which is not a function.

Answer 2

We can replace "Fourier transform" in your question with "inverse Fourier transform", by Fourier inversion, in the sense of distributions. Recall the $\text{pv}(\frac{1}{x})$ distribution defined by

$$ \text{pv}(\frac{1}{x})(\varphi) = \lim_{\epsilon \to 0^+}\int_{|x|\ge\epsilon} \frac{\varphi(x)}{x}\ \mathrm{d}x $$ for Schwartz $\varphi$. The Fourier transform of this distribution is $-i\pi \text{sgn}(\xi)$ which belongs to $L^{\infty}(\mathbb{R})$, yet $\text{pv}(\frac{1}{x})$ is not a function. If it were, then it would agree with $1/x$ away from the origin, and in particular almost everywhere. However $1/x$ is not locally integrable, so it is not a distribution on $\mathbb{R}$.