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

If $f\in L^1 \cap L^2$ 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 $f,\hat f, g\in L^1$.

My question is: is the requirement $f\in L^2$ 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 $f\notin L^2$?

Manlio
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  • If $p>2$, the this link says that there exists $f\in L^p$ whose Fourier transform, as in distribution sense, is not a proper function. So it seems that we do need certain restrictions on the regularity of $f$. – Sangchul Lee Nov 19 '18 at 22:48

2 Answers2

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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.

Kavi Rama Murthy
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  • Doesn't $\int f,\bar g dx$ simply diverges in your example? The same happens to the right-hand side of Parseval's equality, so you would still have $\infty = \infty$. Sorry, I'm probably missing something very basic. – Manlio Nov 19 '18 at 09:30
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If $f,g\in L^1$, then $$ \int f \hat{g}dx = \int g\hat{f}dx. $$

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 $f\in L^2$ because $\hat{f}$ can be defined as the $L^2$ limit of the truncated Fourier transform integral: $$ \hat{f} = L^2-\lim_{R\rightarrow\infty}\widehat{f\chi_{[-R,R]}} $$ and $f\chi_{[-R,R]}\in L^1\cap L^2$ for $f\in L^2$.

Disintegrating By Parts
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