On \(\mathbb{R}^d\), we define the following map to be the Fourier transform. \[\begin{equation} \mathcal{F}: \mathcal{S}(\mathbb{R}^{d}) \mapsto \mathcal{S}(\mathbb{R}^{d}) \end{equation}\]
\[\begin{equation} f \mapsto \mathcal{F}(f) = \hat{f} \end{equation}\] \[\begin{equation} \hat{f}(\xi) = \int_{\mathbb{R}^{d}} f(x)e^{-2\pi i \langle x,\xi \rangle}dx \end{equation}\]
Let \(g_{z}=e^{ \pi i z \| x \|^{2}}\). Then \(\mathcal{F}(g_{z})= \left( \frac{i}{z}\right)^{\frac{d}{2}} g_{-\frac{1}{z}}\). This is example of a Gaussian.
Let \(f:\mathbb{R}^{d} \rightarrow \mathbb{R}\) be an admissible function for the Fourier transform \(\mathcal{F}\).
Let \(f_{r}(x) = f(rx)\) be the \(r\)-dilate of \(f:\mathbb{R}^{d} \rightarrow \mathbb{R}\) for some \(r \in \mathbb{R}\). Then, we have \(\mathcal{F}{(f_{r})} = r^{d} \mathcal{F}(f)_{1/r}\).
Let \(v \in \mathbb{R}^{d}\) and let \(T_{v}(f)(x) = f(x+v)\). Then \[\begin{align} \mathcal{F} \circ T_{v} (f) (\xi) = e^{2\pi i \langle \xi,v\rangle} \hat{f}(\xi) \end{align}\]
See Fourier transform of radial functions
See here
You can also do a Fourier transform on a Hamming space
This page was updated on October 6, 2025.
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