Suppose that $f$ is a function with a Fourier transform, and that $g:\mathbb{R}\rightarrow \mathbb{R}$ is a smooth function such that $g\circ f$ has a Fourier transform also.

Is there an expression for the Fourier transform of $g\circ f$ in terms of that of $f$?

  • 2
    $\begingroup$ I presume $(g\circ f)(x)=g[f(x)]$? formally you could expand $g[f]$ as a power series in $f$ and then the Fourier transform $F(g)$ of $g$ would be a series of repeated convolutions $F(f)\ast F(f)\ast F(f)\cdots$. $\endgroup$ Dec 16, 2018 at 11:48
  • $\begingroup$ Ya, but I was hoping if there was something more elegant... My intuition says maybe there is a relationship with Osculatory integral transforms? $\endgroup$
    – ABIM
    Dec 16, 2018 at 12:24
  • 1
    $\begingroup$ Let $G$ be an antiderivative of $f$. As $(G \circ f)'=(g \circ f).f'$, applying Fourier transform, one gets : $$(-2i \pi \nu) {\frak F} (G \circ f)(\nu)={\frak F} (g \circ f)(\nu) * (-2 i \pi \nu){\frak F}(f)(\nu) \ \ \ \ (1) $$ (with * for convolution operator). Then take back the inverse Fourier transform of (1)... But has $G$ a Fourier Transform ? $\endgroup$ Dec 16, 2018 at 15:31

1 Answer 1


Too long for a comment. I find your question interesting for a reason linked to the proof of the Faà de Bruno formula. Let $f,g$ be two functions from $\mathbb R$ into itself. Let me assume that $f(0)=0$ and let me give a formal expression for $(g\circ f)(x)$ using the Fourier transform of $g$: $$ g(f(x))=\int \hat g(\eta) e^{2iπ \eta f(x)} d\eta=\sum_{k\ge 0}\int \hat g(\eta)\frac{(2iπ \eta f(x))^k}{k!}d\eta, $$ so that $$ g(f(x))=\sum_{k\ge 0}\int \hat g(\eta)\frac{(2iπ \eta)^k}{k!}\prod_{1\le j\le k}\hat f(\xi_j) e^{2iπ x\xi_j}d\xi_j d\eta \\= \sum_{k\ge 0}\int \hat g(\eta)\frac{(2iπ \eta)^k}{k!}\prod_{1\le j\le k}\hat f(\xi_j) \sum_{m\ge 0}\frac{(x2iπ \sum \xi_j)^m}{m!} d\xi d\eta, $$ and after this iteration of Fourier transformations this yields for $m\ge 1$ the Faà de Bruno Formula, $$ \frac{(g\circ f)^{(m)}(0)}{m!}=\sum_{k\ge 1\atop \sum_{1\le j\le k} \alpha_j=m} \frac{g^{(k)}(f(0))}{k!}\prod_{1\le j\le k}\frac{f^{(\alpha_j)}(0)}{\alpha_j!}. $$


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