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, $g= R[ R^{-1}g ]$.

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

-
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 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 $|x|^{-2}$. – 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.
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