Problem: Let $P_{\theta}f(x) = \int_{\mathbb{R}}f(x + s \theta) ds$ be defined as the x-ray transform, where $\theta \in S^{n-1}$, and $x$ belongs to $\Theta^{\perp}$, the hyperplane that passes through the origin and is orthogonal to $\theta$. Now, why is it that $\widehat{P_{\theta}f}(\xi) = 0$ for $\xi \in \Theta^{\perp}$ implies that $P_{\theta}f = 0$ ($\hat{f}$ means Fourier transform of $f$)? This result occured in the paper "Practical and Mathematical Aspects of the Problem of Reconstructing Objects from Radiographs" (Theorem 4.2 proof), the author did not give any explanation for it.
Attempt at an explanation: I was originally thinking maybe the injectivity of the Fourier transform was playing a role here, but I can't see how it can actually be true on a hyperplane instead of $\mathbb{R}^n$. Using the Fourier inversion formula also doesn't seem to give me $P_{\theta}f = 0$, since $\widehat{P_{\theta}f}(\xi)$ only vanishes on $\Theta^{\perp}$ instead of on all of $\mathbb{R}^n$. Am I missing something very obvious here? Any help is appreciated!
It is a general fact that if a function (say from $\mathbb{R}^N$ to $\mathbb{C}$) vanishes, then its Fourier transform vanishes as well.
Let $\theta \in S^{n-1}$ be a fixed angle, and let $f : \mathbb{R}^n \rightarrow \mathbb{C}$. Then $P_{\theta} f$ is a function from the hyperplane $H$ orthgonal to $\theta$ to $\mathbb{C}$. This hyperplane can be seen (by a change of basis for instance) as $\mathbb{R}^{n-1}$. So you have a function $P_{\theta} f : H = \mathbb{R}^{n-1} \rightarrow \mathbb{C}$ whose Fourier transform vanishes, and applying the above general fact, you deduce that $P_{\theta} f$ vanishes everywhere on $H$. This is what is claimed in the paper.
I think your confusion comes from the fact that you thought $P_{\theta} f$ was defined on all $\mathbb{R}^n$, but that's not the case, it's only defined on $H$.