5

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!

Nen
  • 517

1 Answers1

4

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 $H$.

Antoine
  • 1,665
  • So can I interpret your answer as, we were initially performing Fourier transform on a smooth function that is defined on a $(n-1)$ dimensional submanifold of $\mathbb{R}^n$? I have only learned about Fourier analysis for functions defined on $\mathbb{R}^n$ and not on manifolds. Could you point me to a reference so that I can learn more about Fourier analysis on manifolds (e.g. can we even define a Schwartz space on a linear space? what about general manifolds?) – Nen Sep 08 '18 at 19:21
  • 1
    Yes you're right, technically you are performing a Fourier transform on a submanifold of $\mathbb{R}^n$, but this is a very special one : it is a linear subspace. This means that it is isomorphic to $\mathbb{R}^p$ for some $p <n$ (in your case, $p=n-1$). Therefore you can forget that it is some submanifold, and use all the theory of Fourier transform in $\mathbb{R}^p$. – Antoine Sep 09 '18 at 10:52
  • 1
    If you want to know what happens on a generic manifold, then the theory is much richer. For instance, take a two-dimensional sphere $S^2$ inside $\mathbb{R}^3$. Then Fourier analysis exists, but is no longer uses the complex exponentials -- simply because these are not even defined on a sphere! Rather, you need to use the spherical harmonics $Y_{l,m} (\theta , \varphi)$. So Fourier theory on manifolds goes under the name of harmonic analysis (you will find plenty of books and articles dealing with that). – Antoine Sep 09 '18 at 10:56
  • 1
    A last comment : there is certainly one manifold on which you know the harmonic analysis, although maybe you are not aware of it : it is the circle $S^1$. There, the functions you use are the periodic exponentials, $x \mapsto e^{2 \pi i n x}$ with $n \in \mathbb{Z}$. Indeed, a function on a circle is nothing but a periodic function on $\mathbb{R}$. Hence the theory of Fourier series is harmonic analysis on a circle, in the same way as Fourier transforms correspond to harmonic analysis on the real line. – Antoine Sep 09 '18 at 10:58
  • Thanks a lot for your answers! One final question, I'm not sure if I follow your "Therefore you can forget that it is some submanifold". By the linear subspace "isomorphic" to $\mathbb{R}^p$ you mean we can find isometries that map the linear subspace to $\mathbb{R}^p$? Why does that justify Fourier transform on these manifolds? Is there a solid definition/justification for Fourier transform on linear subspaces? I'm sorry if my questions are too shallow, it has been quite some time since I have done serious math... – Nen Sep 09 '18 at 14:49
  • 1
    The reason why you can do Fourier analysis on a linear subspace is that indeed you can find an isometry that maps your subspace to $\mathbb{R}^p$. Equivalently, choosing a basis on the subspace, the coordinates of any vector in that basis is represented by an element of $\mathbb{R}^p$, and operators like the Laplacian have their standard expression in these coordinates. Given that, there is no need for a theory of Fourier transform on a linear subspace, it is really just the usual Fourier transform. That's why you probably won't see that explicitly explained in many books. – Antoine Sep 10 '18 at 12:12