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I have a basic and not very deep understanding of continuous and discrete Fourier transforms in one dimension.

In a multidimensional Fourier transform, the exponent of $e$ includes the inner product of the arguments of the original function and the transformed function. I'm having trouble finding an explanation of why this is the right way to extend the idea of a Fourier transform to more than one dimension. What is the intuition, or perhaps a more formal reason for this idea? In one dimension, the original function's and transformed function's arguments are multiplied, and using an inner product in a multidimensional case is one natural extension of that idea, but that's not enough to justify it.

I see that using the inner product in the exponential means that the integral or sum (for a discrete Fourier transform) is over a product of $\cos x_i + i \sin x_i$ sums, but I am not sure why that makes sense, or even whether that's a useful way to think about it. (I can picture waves in two real dimensions, and multiplying one-dimensional wave equations kind of feels like a good way to represent that, but I still don't have a clear understanding.)

Pointers to texts as well as explanations here would be welcome.

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    $\begingroup$ Since $\exp(ix\cdot \xi) = \exp(\sum ix_j\xi_j) = \prod \exp(ix_j\xi_j)$ its just fourier transforming repeatedly in each dimension $\endgroup$ – Calvin Khor Jun 1 at 19:32
  • $\begingroup$ It is exponential law. Multiplication of two exp-expressions adds the exponents. $\endgroup$ – mathreadler Jun 1 at 19:40
  • $\begingroup$ @CalvinKhor, thanks--but why is multiplying the way to do that? (Contrast: You can't use multiplication to analyze a multidimensional probability distribution unless the marginal r.v.s are independent. I recognize that this is a different case, but the general lesson, for me, is that you have to have a reason that multiplication is appropriate.) $\endgroup$ – Mars Jun 2 at 4:49
  • $\begingroup$ I am not claiming that the integrals split into a product of integrals. You should explicitly write out the repeated Fourier transform $\mathcal F_x \mathcal F_y [u]$ ie once in $y$ then once in $x$ and use what I said. $\endgroup$ – Calvin Khor Jun 2 at 6:48
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    $\begingroup$ @Mars Yeah I think the note is fine. You're welcome. $\endgroup$ – Calvin Khor Jun 6 at 22:56
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Following @CalvinKhor's suggestions in comments, including the suggestion that I post an answer:

First treat $f(x,y)$ as a function of $x$, and apply a Fourier transform to define $H(\omega,y)$ as a function of $\omega$ for fixed $y$:

$$H(\omega,y)=\int_{-\infty}^{+\infty}f(x,y)\,e^{-i\omega x}dx$$

Now apply a Fourier transform to the result with respect to $y$, to define a function $F$ as a function of $\xi$:

$$F(\omega,\xi)=\int_{-\infty}^{+\infty}H(\omega,y)\,e^{-i\xi y}dy = \int_{-\infty}^{+\infty}\left(\int_{-\infty}^{+\infty}f(x,y)\,e^{-i\omega x}dx\right)e^{-i\xi y}dy =$$

$$\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty}f(x,y)\,e^{-i(\omega x + \xi y)} dx \, dy$$

This shows that the inner product of the two vectors inside the transform is simply the result of iteratively applying the Fourier transform to different component variables. One can repeat the process for more dimensions.

You can combine the integrals if you want to treat them as a single integral applying to vectors as such.

(I think it's pretty clear that the same strategy will work with a discrete Fourier transform. There are some nuances for DFT that have to be specified, but in the end you are doing the same sort of thing with sums instead of integrals.)

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