0
$\begingroup$

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 $f$?

$\endgroup$
3
  • 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

1
$\begingroup$

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, $$ \frac{(g\circ f)^{(m)}(0)}{m!}=\sum_{k\ge 1\atop \sum_{1\le j\le k} \alpha_j=m} \frac{g^{(k)}(f(0))}{k!}\prod_{1\le j\le k}\frac{f^{(\alpha_j)}(0)}{\alpha_j!}. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.