MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let us consider the Radon transform in two dimensions:

$$\tag{1}Rf(r,\theta):=\int\limits_{-\infty}^{\infty} f(r\cos\theta-t\sin\theta,r\sin\theta+t\cos\theta) dt,$$

where $r\in\mathbb{R}$ and $0\leq\theta\leq \pi$. There is a well known theorem about the range of the transform.

Theorem. A function $g(r,\theta)$ can be represented as a Radon transform of some function $f(x,y)$ (i.e. $g=R[f]$) if and only if for all integers $n\geq0$ $$\int\limits_{-\infty}^{\infty} r^ng(r,\theta) dr$$ is a homogeneous polynomial of $\cos\theta$ and $\sin\theta$.

Obviously, if $g(r,\theta)$ belongs to the range of the Radon transform then the inverse Radon transform of the function $g(r,\theta)$ is $f(x,y)$.

Now let us consider a function which DOES NOT belong to the range of the transform.

QUESTION: What we would receive if we apply the inverse Radon transform to a function not from the range of the transform?

For example, consider function $g(r,\theta):= e^{-r^2}$ if $0\leq\theta\leq \pi/2$ and $g(r,\theta):= e^{-r^2(1-\cos\theta\sin\theta)}$ if $\pi/2\leq\theta\leq\pi$. This function does not belong to the range of the Radon transform. Then, on the one hand, there is no function $f$ such that $g=R[f]$. On the other hand, $g= R[ R^{-1}g ]$.

What's wrong with this paradox?


UPDATE: Let us notice that $R[R^{-1}g]$ is defined correctly, but it is not equal to $g$.

Indeed, if $g=R[R^{-1}g]$, then

$$\int\limits_{-\infty}^{\infty} r^ng(r,\theta) dr = \int\limits_{-\infty}^{\infty} r^n R[R^{-1}g] (r,\theta) dr=$$

$$=\int\int r^n [R^{-1}g] (r\cos\theta−t\sin\theta,r\sin\theta+t\cos\theta)drdt=$$

$$=\int\int (u\cos\theta+v\sin\theta)^n [R^{-1}g] (u,v)dudv,$$

which is a homogeneous polynomial of $\cos\theta$ and $\sin\theta$ (we just have to expand the brackets). On the other hand it is NOT a homogeneous polynomial (by assumption). Therefore $g\neq R[R^{-1}g]$.

share|cite|improve this question
Without specifying the domain of the Radon transform, it does not make sense to talk about its range. My feeling is that this is the root of your supposed paradox – Yemon Choi Oct 31 '10 at 19:41
Yemon, I don't quite understand your comment. f(x,y) is defined on $\mathbb{R}^2$ and Radon transform, $R[f](r,\theta)$ is defined on $\mathbb{R}\times [0;\pi]$. – Oleg Oct 31 '10 at 20:07
Actually, no. If you define the domain to be all functions for which the integral (1) converges, your "well known theorem" is false. The Radon and inverse Radon transforms establishes a bijection between Schwarz functions on R^2 and with Schwarz functions on $S^1 \times R$ that satisfies the homogeneous polynomial condition its moment. See chapter 1 of Helgason's book See also Dirk's answer below. So most likely when you take the inverse transform of you function, you get something that decays only slightly faster than $|x|^{-2}$. – Willie Wong Nov 1 '10 at 19:42
No, if $g$ is such that $R^{-1}$ is well defined and $R^{-1}g$ is such that $RR^{-1}g$ is well defined, $RR^{-1}g = g$ by definition. The problem is that $R: \mathcal{S}(\mathbb{R}^2) \to \mathcal{S}(\mathbb{P})$ with the image being functions satisfying the homogeneous polynomial condition, while $R$ may still send a bigger space (say, a space of functions for which the trace on all lines are defined and absolutely integrable) to something else. For comparison, think of the Fourier transform. It is a bijection of Schwarz functions, but it also is a bijection of $L^2$ function with itself. – Willie Wong Nov 1 '10 at 23:36
The last integral does not converge. – Willie Wong Nov 2 '10 at 21:44

Probably you refer to some theorem in "Mathematics of Computerized Tomography" by Frank Natterer (e.g. Theorem 4.2)? Then you are assuming that the domain in $\mathcal{S}$ and if I remember correctly, in that book this denotes the Schwartz space of rapidly decaying $C^\infty$-functions. Hence you paradox is resolved by the fact that $R^{-1} g$ is not a Schwartz function.

share|cite|improve this answer
Hello Dirk! Thanks for this explanation. But anyway, it is easy to see, that one can apply a Radon transform to function $R^{-1}g$ and $RR^{-1}g$ is defined correctly. What is your opinion, is $RR^{-1}g$ equal to $g$ (see update of the question)? Thanks! – Oleg Nov 2 '10 at 20:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.