Projection-slice theorem

Remarks

The proof is one application of Fubini's theorem: write $x = s\theta + y$ with $s \in \R$ and $y \in \theta^\perp$, so that $x \cdot \sigma\theta = s\sigma$ and $dx = ds\, dy$.

Example: for $f(x) = e^{-\abs{x}^2}$ one has $R_\theta f(s) = \pi^{(n-1)/2} e^{-s^2}$, whose one-dimensional Fourier transform is $\pi^{n/2} e^{-\sigma^2/4}$. This is $\hat f(\sigma\theta) = \pi^{n/2} e^{-\abs{\sigma\theta}^2/4}$, as the theorem says.

In words: the one-dimensional Fourier transform of a projection is the restriction of the $n$-dimensional Fourier transform to a line through the origin. Since these lines cover $\R^n$, the theorem shows at once that $R f$ determines $f$ for $f \in L^1(\R^n)$, by injectivity of the Fourier transform on $L^1$.

Going deeper

For a student

The theorem is the bridge between the Radon transform and Fourier analysis. Combined with Fourier inversion in polar coordinates, it gives the Radon inversion formula and the ramp filter $\abs{\sigma}^{n-1}$ of filtered back-projection. It also shows that the Radon transform is injective on $L^1(\R^n)$ and explains the moment conditions in the range conditions for the Radon transform. In practice it suggests Fourier reconstruction: compute the Fourier transforms of the projections, interpolate them onto a grid, and invert one $n$-dimensional Fourier transform.