# Fourier transform of function composition

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.

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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: $$(\widehat{f\circ A})(\xi) = |\det A^{-1}|\widehat{f}(A^{-T}\xi).$$
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 $$P(k,l) = \int_{x \in \mathcal{R}} e^{i2\pi(lg(x) - kx)}\,dx.$$ The FT of the composite function is $$\text{FT}\left[f\left(g(x)\right)\right](k) = \int_{l\in\mathcal{R}} \hat{f}(l)P(k,l) \,dl,$$ 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.
Nice answer, thanks. I need this for an engineering problem that I'm working on, have you seen a multi-dimensional version? That is, $f\circ g(\mathbf{x})$. – daaxix Sep 13 '15 at 0:13