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