Given two functions $f$ and $g$, is there a formula for the Fourier transform of $f \circ g$ in terms of the Fourier transforms of $f$ and $g$ individually?

I know you can do this for the sum, the product and the convolution of two functions. But I haven't seen a formula for the composition of two functions.


There is no such rule in general. The key here is variable substitution: If $g$ is a bijection and smooth enough then, if all integrals exist: $$ (\widehat{f\circ g})(\xi) = \int f(g(x))\exp(ix\xi)dx = \int f(y)\exp(ig^{-1}(y)\xi)|\det g'(y)|^{-1}dy.$$ This does only rarely lead to something interesting, e.g. in the case of scaling (i.e. linear transformation of the variable): Working in $\mathbb{R}^d$ with $A\in\mathbb{R}^{d\times d}$ invertible: $$ (\widehat{f\circ A})(\xi) = |\det A^{-1}|\widehat{f}(A^{-T}\xi). $$

  • $\begingroup$ "There is no such rule in general." ... The answer by Lee gives a rule. $\endgroup$ – becko Jul 26 '17 at 22:50
  • $\begingroup$ @becko There is no "useful" rule is the more accurate statement (in the same manner as it's no "useful" analytical formula for the $n$th prime). The expression you link to you gives the answer in terms of an infinite number of integrals that needs to be convolved. Compare this with the problem we are trying to solve which is just one a integral. Such an expression might be useful for some purposes, but it's completely useless for the wast majority of applications. $\endgroup$ – Winther Jul 26 '17 at 23:06
  • $\begingroup$ @Winther Why do you say "infinite number of integrals that need to be convolved"? At least in the 1-dimensional case, there is only one integral, of the Fourier transform times a function $P(k,l)$ that depends on $g$. I haven't applied my self to anything, but it does look usable. $\endgroup$ – becko Jul 26 '17 at 23:19
  • $\begingroup$ @becko I was thinking about each $\ell$ as one integral (analytical solutions are likely rare). Anyway, try it for any non-trivial case and see if it works. I suspect you will find it's much harder than the original problem you are trying to solve and/or doing exactly the same in a more complicated manner. If it turns out to be useful then that would be great, I'm just not seeing it. Would be happy to be proven wrong here. $\endgroup$ – Winther Jul 26 '17 at 23:47
  • $\begingroup$ @Winther See for example: math.stackexchange.com/q/2372929/10063. The analogue in Laplacian transform. In the particular example there, the analogue of $P(k,l)$ turns out to be analytic. $\endgroup$ – becko Jul 26 '17 at 23:55

Have a look at Bergner et al. 2006 "A Spectral Analysis of Function Composition and Its Implications for Sampling in Direct Volume Visualization" IEEE TRANSACTIONS ON VISUALIZATION AND COMPUTER GRAPHICS, VOL. 12, NO. 5. Sec. 3.

Unfortunately it's not quite as simple as the transform of a convolution or even the derivative. The short answer is, in one dimension let \begin{equation} P(k,l) = \int_{x \in \mathcal{R}} e^{i2\pi(lg(x) - kx)}\,dx. \end{equation} The FT of the composite function is \begin{equation} \text{FT}\left[f\left(g(x)\right)\right](k) = \int_{l\in\mathcal{R}} \hat{f}(l)P(k,l) \,dl, \end{equation} where $\hat{f}(l)$ is the Fourier transform of $f(x)$. As you can see the transformation involves the inner product of $\hat{f}(l)$ with a slightly awkward two dimensional function. In the discrete case this would be implemented as a matrix multiplication.

  • 1
    $\begingroup$ Nice answer, thanks. I need this for an engineering problem that I'm working on, have you seen a multi-dimensional version? That is, $f\circ g(\mathbf{x})$. $\endgroup$ – daaxix Sep 13 '15 at 0:13
  • $\begingroup$ +1 I would like to understand the multidimensional implications of this. $\endgroup$ – MadcowD Jul 12 '16 at 1:13

This is the Lee that gave the answer in 2014, now posting from a registered account. That answer is a simplified version of the multidimensional case. If you want to understand the multidimensional case I suggest a Google search for the paper I referenced.

Best of luck.


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