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I'm trying to express the integral

$$I = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} f(x_1, x_2) \; g(x_1, x_2) \; \mathrm{d}x_1 \mathrm{d}x_2$$

of two real valued functions $f(x_1,x_2)$ and $g(x_1,x_2)$ in terms of their Fourier transforms $\tilde{f}(\omega_1, \omega_2)$ and $\tilde{g}(\omega_1, \omega_2)$. However, as $f(x_1,x_2)$ and $g(x_1,x_2)$ are real-valued functions and $I$ is not defined as the the product of $f(x_1,x_2)$ and $g(x_1,x_2)^\ast$ I'm not working with the complex conjugate).

$$f(x_1,x_2) = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \tilde{f}(\omega_1, \omega_2) \; e^{2 \pi i ( \;\omega_1 x_1 \;+ \;\omega_2 x_2 \;) } \; \mathrm{d}\omega_1 \mathrm{d}\omega_2$$


$$g(x_1,x_2) = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \tilde{g}(\omega_1, \omega_2) \; e^{2 \pi i ( \;\omega_1 x_1 \;+ \;\omega_2 x_2 \;) } \; \mathrm{d}\omega_1 \mathrm{d}\omega_2$$

The integral becomes

$$I = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \; \left[ \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \tilde{f}(\omega_1, \omega_2) \; e^{2 \pi i ( \;\omega_1 x_1 \;+ \;\omega_2 x_2 \;) } \; \mathrm{d}\omega_1 \mathrm{d}\omega_2 \right] \; \left[ \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \tilde{g}(\omega_1, \omega_2) \; e^{2 \pi i ( \;\omega_1 x_1 \;+ \;\omega_2 x_2 \;) } \; \mathrm{d}\omega_1 \mathrm{d}\omega_2 \right] \mathrm{d}x_1 \mathrm{d}x_2$$

This is where it starts to get confusing. I'm not sure if I should write:

$$I = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \mathrm{d}\omega_1 \; \tilde{f}(\omega_1, \omega_2) \; \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \mathrm{d}\omega_2 \tilde{g}(\omega_1, \omega_2) \; \; \int_{-\infty}^{+\infty} \; e^{2 \pi i x_1 ( \; \omega_1 \;+ \;\omega_2 \;) } \; \mathrm{d}x_1 \; \int_{-\infty}^{+\infty} \; e^{2 \pi i x_2 ( \; \omega_1 \;+ \;\omega_2 \;) } \; \mathrm{d}x_2$$

and extract two one-dimensional dirac deltas, or write:

$$I = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \mathrm{d}\omega_1 \; \tilde{f}(\omega_1, \omega_2) \; \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} d\omega_2 \tilde{g}(\omega_1, \omega_2) \; \;\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} e^{2 \pi i x_1 ( \; \omega_1 \;+ \;\omega_2 \;) + x_2 ( \; \omega_1 \;+ \;\omega_2 \;) } \; \mathrm{d}x_1 \; \mathrm{d}x_2$$

with the hope of extracting the two-dimensional dirac delta. (I'm not even sure what the inverse Fourier transform of the two dimensional dirac delta is).

I'd appreciate advise on how to proceed. (Thanks.)

share|improve this question
This is Plancherel's theorem (not Parseval's). The one-dimensional proof carries over essentially verbatim to $n$ dimensions. See www.ma.utexas.edu/users/odiaz/notas/fourier.pdf for an account by Oliver Diaz that eschews hand-waving such as the use of the "Dirac function". –  Robin Chapman Sep 8 '10 at 12:17
There is no such thing as the "Dirac function" — the Dirac delta is a distribution (i.e. generalized function, not probability distribution). –  You Sep 9 '10 at 20:30
Thanks Robin. I finally understand the proof. I would have given you credit for the answer if you'd posted a full reply instead of a comment. –  Olumide Sep 12 '10 at 7:18

1 Answer 1

I don't see what you're doing. Are you trying to show that:

$$\int_{\mathbb R^2} f(x) g(x) \, dx = \int_{\mathbb R^2} \widehat{f(\xi)} \overline{\widehat{g(\xi)}} \, d\xi$$? This is usually called Parseval's identity.

You can prove it as follows (even in $\mathbb R^n$), first prove by Fubini (very simple) that:

$$\int_{\mathbb R^2} f(x) \widehat{g(x)} \, dx = \int_{\mathbb R^2} \widehat{g(x)} f(x) \, dx$$

Then let $g = \overline{\hat{h}}$ and note then that $\hat{g} = \overline{h}$ (by Fourier inversion), then we get

$$\int_{\mathbb R^2} f(x) h(x) \, dx = \int_{\mathbb R^2} \widehat{g(x)} \overline{\widehat{h(x)}} \, dx$$ (assuming $h$ is real, otherwise we get the conjugate).

share|improve this answer
Yes I'm trying to prove that $\int_{\mathbb R^2} f(x) g(x) \, dx = \int_{\mathbb R^2} \hat{f(\xi)} \overline{\hat{g(\xi)}} \, d\xi$ . I understand that the hat means a Fourier transform. Why is the hat in the first equation small? Is this a typo? –  Olumide Sep 9 '10 at 1:26
BTW, the integral on the LHS of the first equation should be something like $\int_{\mathbb R^2} f(x_1,x_2) g(x_1,x_2) \, dx_1 dx_2$ –  Olumide Sep 9 '10 at 13:06
Yes, that is a typo hat instead of widehat, it is the same, $x = (x_1, x_2)$, it is just a shorthand. –  Jonas Teuwen Sep 9 '10 at 15:45

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