2
$\begingroup$

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?

Thanks!

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 $g\neq R[R^{-1}g]$.

$\endgroup$
13
  • $\begingroup$ 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 $\endgroup$
    – Yemon Choi
    Oct 31, 2010 at 19:41
  • $\begingroup$ 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 $\mathbb{R}\times [0;\pi]$. $\endgroup$
    – Oleg
    Oct 31, 2010 at 20:07
  • 1
    $\begingroup$ 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 www-math.mit.edu/~helgason/Radonbook.pdf 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 $|x|^{-2}$. $\endgroup$ Nov 1, 2010 at 19:42
  • 1
    $\begingroup$ 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. $\endgroup$ Nov 1, 2010 at 23:36
  • 1
    $\begingroup$ The last integral does not converge. $\endgroup$ Nov 2, 2010 at 21:44

1 Answer 1

3
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ 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! $\endgroup$
    – Oleg
    Nov 2, 2010 at 20:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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