Let $f: \mathbb R^2 \rightarrow \mathbb R$. The Radon transform of $f$ is given by the set of integrals along lines: $$ \mathcal Rf(t,\theta)=\int_C f(x,y) \; ds $$ where the path of integration lies on the line $C = \{(x,y): x\cos\theta+y\sin\theta = t\}$ and $ds$ is the length measure supported on this line.

The "inversion" formula is often derived with the help of a function $h(t)$ whose Fourier transform is $|\omega|$ and the projection slice property of the Radon transform (eg. Theorem 2): $$ f(x,y)=\frac{1}{(2\pi)^2} \int_0^\pi [\mathcal R f(\cdot,\theta)*h(\cdot)](x\cos\theta+y\sin\theta)\;d\theta. $$ The problem is that I can't think of any nice function $h(t)$ in $L^1$ or in $L^2$ which has a Fourier transform $|\omega|$. How do we deal with this inversion formula in a mathematically rigorous way, using functions that actually exist? How do I formalize the notion of "niceness" for $f$ and $h$ here?

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    $\begingroup$ Your "inversion formula" is filtered backprojection, which is a heuristic method that works reasonably well in practice, but cannot be called the inverse Radon transform. The filter is typically applied in the Fourier domain, I believe. $\endgroup$ Commented Apr 9, 2018 at 0:24
  • $\begingroup$ I agree. That begs the question---how do I deal with the inversion formula for the Radon transform in a mathematically rigorous way? $\endgroup$
    – Atul Ingle
    Commented Apr 9, 2018 at 3:02

1 Answer 1



  • To formalize the niceness of $f$, one can restrict it to a space of functions that are smooth and decay rapidly away from the origin.
  • $h$ is not a function in the usual sense, but a distribution (generalized function) $h(t)=\frac{d}{dt}\mathrm{p.v.}\left(\frac{1}{t}\right).$


A nice treatment of these questions is presented in [Deans1983]. A rigorous way to define the Radon transform of a function $f:\mathbb{R}^2\rightarrow\mathbb{R}$ is to restrict $f \in \mathcal{S}(\mathbb R ^2)$, the Schwartz space of functions that are infinitely differentiable with derivatives that decay rapidly. Then the Radon transform $\mathcal R f \in \mathcal{S}(\mathbb R \times S^1)$ where $S^1$ is the unit circle in $\mathbb R^2$. One can then derive the following inverse mapping: $$ f(x,y)=\frac{-1}{2\pi^2}\int_0^\pi d\theta\int_{-\infty}^\infty dt \frac{\frac{\partial}{\partial t}\mathcal Rf(t,\theta)}{t-x\cos\theta-y\sin\theta} $$ where the inner integral must be understood as a Cauchy principal value.

For the inner integral we note that $$ \int_{-\infty}^\infty dt \frac{\frac{\partial}{\partial t}\mathcal Rf(t,\theta)}{t-p} = \frac{-1}{\pi} \mathcal Rf(p,\theta)* \frac{-1}{p ^2} $$ where, once again, the Cauchy principal value is understood for the convolution integral, and we have the following generalized function representation for $h$ (see also this math.SE question): $$ h(t)=\frac{d}{dt}\mathrm{p.v.}\left(\frac{1}{t}\right). $$

Putting this all together, we have $$ f(x,y)=\frac{1}{2\pi} \int_0^\pi d\theta \;\;\left. \mathcal Rf(t,\theta)*\frac{d}{dt}\mathrm{p.v.}\left(\frac{1}{t}\right) \right|_{t=x\cos\theta+y\sin\theta} $$ Note a slight discrepancy in the constant $2 \pi$ because [Deans1983] does not use a $1/2\pi$ in the definition of inverse Fourier transform.


[Deans1983] S. Deans, The Radon Transform and Some of Its Applications (1983)


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