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Sorry for the not-perfect question. I am asking for a reference for the following relation:

$$\int f . g. h ...= \int_{\xi_1 +\xi_2 +...=0} \hat{f}(\xi_1) \hat{g}(\xi_2)... d\xi_1 d\xi_2...$$

Could you please show me where to find details, statements and proofs about this relationship? Thanks in Advance.

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    $\begingroup$ I guess this is off-topic here, but it's a one-line proof if you substitute for each $f,g,h...$ the Fourier integral $f(x)=\int \hat{f}(\xi)e^{i\xi x}d\xi$, and then integrate over $x$, using that $\int_{-\infty}^\infty e^{ix(\xi_1+\xi_2\cdots)}dx=2\pi\delta(\xi_1+\xi_2\cdots)$. $\endgroup$
    – Carlo Beenakker
    Commented Jun 30, 2022 at 6:49
  • $\begingroup$ @CarloBeenakker why did the integration domain change in the phase space? $\endgroup$
    – Mr. Proof
    Commented Jun 30, 2022 at 7:29
  • $\begingroup$ the integrand contains the delta function, which vanishes when $\xi_1+\xi_2+\cdots=0$. $\endgroup$
    – Carlo Beenakker
    Commented Jun 30, 2022 at 9:03
  • $\begingroup$ Better typesetting: $$ \begin{align} & \int f . g. h ...= \int_{\xi_1 +\xi_2 +...=0} \hat{f}(\xi_1) \hat{g}(\xi_2)... d\xi_1 d\xi_2... \\ & \qquad \text{versus} \\ & \int f \cdot g\cdot h \cdots = \int_{\xi_1 +\xi_2 +\cdots=0} \hat{f}(\xi_1) \hat{g}(\xi_2)\cdots \, d\xi_1 \, d\xi_2\cdots \end{align} $$ $\endgroup$
    – Michael Hardy
    Commented Jun 30, 2022 at 19:11

1 Answer 1

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If $f_1,\ldots,f_n$ belong to the Schwarz space, Fourier inversion formula and Fourier-convolution properties yield \begin{eqnarray*} \int_{\xi_1+\cdots+\xi_n=0} \hat{f_1}(\xi_1) \cdots \hat{f_n}(\xi_n) d\xi_1 \cdots d\xi_n &=& (\hat{f_1} * \cdots * \hat{f_n})(0) \\ &=& (2\pi)^{-1} \mathcal{F}(\overline{\mathcal{F}}(\hat{f_1} * \cdots * \hat{f_n}))(0) \\ &=& \mathcal{F}(f_1 \cdots f_n)(0) \\ &=& \int_\mathbb{R} f_1 \cdots f_n(x) dx. \end{eqnarray*}

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