Say we a function $u$ and a bounded region $\Omega \subset \mathbb{R}^2$, such that $(\Delta+\lambda)u = 0$ everywhere, and $u=0$ on the boundary. We extend it to the entire plane by defining $u=0$ everywhere outside of $\Omega$. Wikipedia states that, for the Radon transform, $R\Delta = \partial^2 / \partial s^2 R$, hence we can derive the ODE
$ \left( \frac{\partial^2}{\partial s^2} + \lambda \right) Ru(\theta,s) = 0,$
therefore we can reason that
$ Ru(\theta,s) = a(\theta) e^{i \sqrt{\lambda} s} + b(\theta) e^{-i \sqrt{\lambda} s}.$
But this is periodic in $s$ with period $2\pi/\sqrt{\lambda}$, which would imply either that the above vanishes everywhere, or that it's possible to integrate $u$ over lines arbitrarily far away from $\Omega$ and still get a positive value, despite $u$ vanishing identically on said line. Contradiction.
So does this mean:
- I can't use $R\Delta = \partial^2 / \partial s^2 R$ for functions that aren't sufficiently smooth, or
- The Radon transform is discontinuous at any line tangent to $\partial \Omega$?