I have read in a few places that the formula $$ \int_\mathbb{R} x(t)^3 \, dt \ = \ \int_{\mathbb{R}^2} \hat{x}(f_1)\hat{x}(f_2)\hat{x}(-f_1-f_2) \, d(f_1,f_2) $$ holds (where $\hat{x}$ denotes the Fourier transform of $x$), but without any description of the class of functions that $x$ belongs to.

Straightforward iterated application of the convolution theorem shows that this formula is true at least if $x \in \mathcal{S}(\mathbb{R})$. Being a bit more precise, it will (I think) be true at least for all $x \in W^{2,1}(\mathbb{R})$. But more generally:

Is it the case that for all $x \in L^2(\mathbb{R}) \cap L^3(\mathbb{R})$, the map $\,(f_1,f_2) \mapsto \hat{x}(f_1)\hat{x}(f_2)\hat{x}(-f_1-f_2)\,$ [known as the bispectrum of $x$] is absolutely integrable on $\mathbb{R}^2$ and the above formula holds?

(And if so, is there a reference that contains this fact?)


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