Vrftsjtryk

Plancherel's theorem is stated as (e.g. in Rudin's Real and Complex Analysis)

If $fin L^1 cap L^2$ then
$$ |f|_2 = |hat f|_2 $$

where $hat f$ is the Fourier transform of $f$. On the other hand, Parseval's formula

$$ int f,overline{g}, d x = int hat{f}~overline{hat{g}}, d x$$

should hold whenever $f,hat f, gin L^1$.

My question is: is the requirement $fin L^2$ in Plancherel's theorem needed just to have the two norms to be finite or is there some (more or less hidden) detail that I'm missing and that makes the statement to actually be false if $fnotin L^2$?

If $f,g in L^{1}$ there is no reason $int f overline {g} , dx$ exists. Ex: $f(x)=g(x)=x^{-1/2}$ for $0<|x|<1$ and $0$ otherwise.

If $f,gin L^1$, then
$$
int f hat{g}dx = int ghat{f}dx.
$$

If $f,hat{f} in L^1$, then $f= (hat{f})^{vee}$, which allows you to swap $f$ and $hat{f}$ in the above in order to obtain the identity that you need. Then you need
$$
f,hat{f},g in L^1.
$$

However, $|f|_2 = |hat{f}|_2$ makes sense for any $fin L^2$ because $hat{f}$ can be defined as the $L^2$ limit of the truncated Fourier transform integral:
$$
hat{f} = L^2-lim_{Rrightarrowinfty}widehat{fchi_{[-R,R]}}
$$
and $fchi_{[-R,R]}in L^1cap L^2$ for $fin L^2$.