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 $f$?



Transformation of Fourier Transform
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,
$$
\frac{(g\circ f)^{(m)}(0)}{m!}=\sum\_{k\ge 1\atop \sum\_{1\le j\le k} \alpha\_j=m}
\frac{g^{(k)}(f(0))}{k!}\prod\_{1\le j\le k}\frac{f^{(\alpha\_j)}(0)}{\alpha\_j!}.
$$


