Radon inversion formula

Remarks

The proof starts from the Fourier inversion formula $f(x) = (2\pi)^{-n} \int_{\R^n} \hat f(\xi)\, e^{i x \cdot \xi}\, d\xi$ in polar coordinates $\xi = \sigma\theta$. Writing the integral over $\sigma \in \R$ and $\theta \in S^{n-1}$ counts each $\xi$ twice, which gives the factor $\frac12$ and the weight $\abs{\sigma}^{n-1}$. The projection-slice theorem replaces $\hat f(\sigma\theta)$ by the Fourier transform of $R_\theta f$, and the remaining $\theta$-integral is $R^*$.

Example: for $n = 3$ the formula is $f(x) = -\frac{1}{8\pi^2} \int_{S^2} (\partial_s^2 R f)(\theta, x \cdot \theta)\, d\theta$, so $f(x)$ depends only on the integrals of $f$ over planes close to $x$.

For even $n$ the multiplier $\abs{\sigma}^{n-1}$ is not a polynomial, $\Lambda$ involves the Hilbert transform in $s$, and the formula is not local: $f(x)$ depends on integrals over all hyperplanes.

Going deeper

For a student

The formula is the continuous model of filtered back-projection, the standard reconstruction algorithm of computed tomography. The filter $\abs{\sigma}^{n-1}$ grows at high frequencies, which shows why inversion amplifies noise and why practical filters cut it off. In odd dimensions the inversion is local, a fact used in three-dimensional imaging; in even dimensions, such as the plane, it is not. The same formula, read with the Riesz potential, extends to wider classes of functions and to distributions.