Why is the Fourier transform for $L^p$ not defined?

Question

I have a conceptual doubt. I know that if $f\in S$ with $S$ Schwartz space then $\mathcal{F}(f)\in S$. If $f\in L^p$ and $S$ dense in $L^p$ then $\mathcal{F}(f)$ makes sense?

Answer 1

The key to extension by density results is an inequality like $\|Ff\|_X \le C\|f\|_Y$. Then if $f_n$ are from a nice subset of $Y$ where $f_n\to f$ in $Y$, $Ff_n$ forms a Cauchy sequence in $X$, so a limit exists.

This works for $p\in[1,2]$ by the Hausdorff-Young inequality $\| Ff\|_{L^{p'}} \le C\|f\|_{L^p}$ but no such inequality is possible for $p>2$. This can be shown using with a family of counterexamples, or a 'random counterexample' with Khintchine's inequality, as in the notes of Prof Terence Tao here.

However that is not to say you cannot define the Fourier transform of such functions; they just end up being distributional Fourier transforms. For instance the Fourier transform of $1\in L^\infty$ is the "dirac delta" distribution.