# The Geometry of the Slice Theorems

﻿﻿﻿When studying the Radon transform, we saw that we could reconstruct the Fourier transform of a function from its Radon Transform:

• The Fourier Transform integrates the product of a function with “waves” that are constant on hyperplanes.
• The Radon Transform computes the integral of a function over these hyperplanes.
• So the Fourier Transform of a function is a sort of phase-weighted sum of integrals over hyperplanes, i.e. a phase-weighted sum of the Radon transform.

When evaluating the Fourier Transform of a function $f$ at a point $\rho\omega$, the hyperplane $x\cdot\omega = s$ appears with “weight” $e^{-i\rho x\cdot\omega} = e^{-i\rho s}$. So we can write the Fourier transform of $f$ in terms of the Radon Transform as follows: $\hat{f}(\rho\omega) = \int_{-\infty}^{\infty}e^{-i\rho s}Rf(s, \omega) ds = \mathfrak{F}_sRf(\rho, \omega)$

The same arguments apply to the X-Ray transform.  In fact, we can compute the Radon transform from the X-Ray transform at a hyperplane by integrating $Xf$ over a set of parallel lines that covers the hyperplane (there is more than one way to do this!).

To compute the Fourier Transform directly from the X-Ray transform, consider the following.  Say we want to compute the Fourier Transform of $f$ at a frequency vector $\eta$.  We can pick any orthogonal $\theta$, and have $\eta \in \theta^\perp$.  Now the “wave” with frequency $\eta$ will be constant in direction $\theta$, so to compute the Fourier Transform of $f$ at $\eta$ we just need to add up the integrals of $f$ along the lines in direction $\theta$ $Xf(y, \theta)$ — weighted it by the value of the wave on these lines.  In other words $\hat{f}(\eta) = \int_{\theta^\perp}e^{-iy\cdot\eta}Xf(y,\theta)dH_{\theta^\perp}(y) = \mathfrak{F}_{\theta^\perp}Xf(\eta, \theta)$

for $\eta \in \theta^\perp$.  Notice that our choice of $\theta$ was arbitrary, so this really gives us a continuum of formulas, one for each $\theta \in \eta^\perp$.

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# Expressing Exponential-Type Conditions in Terms of Derivative Growth Rates

When I talked about my alternative proof of Helgason’s support theorem in class, I  asserted that our function $\hat{f}$ had compact support in $B_A(0)$ if and only if its partial derivatives satisfied $|\partial^{\alpha}_\xi\hat{f}(0)| \leq CA^{|\alpha|}$ for any multi-index $\alpha$.

In other words, a function is of exponential type if its partial derivatives grow like those of an exponential function.

This is true, but the claims I stated on the board were not correct — I did not mention the fact that we needed to use extra information about $f$, in particular the fact that $f \in \mathcal{S}(\mathbb{R}^n)$ is all we need to complete our proof.

It is possible to state much more general results, but to get the idea across let me state the following result  in one dimension:

Claim Assume $f \in \mathcal{S}(\mathbb{R})$  then $\text{supp} f \subset [-A, A]$ if and only if $\hat{f}$ is entire analytic and for all $n$ $\big |\frac{d^n\hat{f}}{dz^n}(0) \big | \leq C A^n$

for some constant $C$. Continue reading

# An Alternative Proof of the Radon Transform Support Theorem for Radial Functions

A key technical step in the proof of the Radon Transform support theorem is proving that the result holds for radial functions.  The proof presented in class was elementary but technical and, in my opinion, bewildering.  Not the sort of thing I’d come up with.

Yes, I know, I need to work on my skills.

Instead of doing that, I came up with another proof that I find more satisfying.  It is fairly simple, and uses nothing more than the Paley-Wiener theorem and a bit of Calculus.

Theorem if $Rf = g$, $g \in \mathcal{S}_h(\mathbb{R}\times S^{n-1}) \cap C_0(\mathbb{R}\times S^{n-1})$, $s > A \implies g(s,\omega) = 0$, and $f(x) = F(|x|^2)$ is radial, then $f$ has compact support and $|x| > A \implies f(x) = 0$.

The rest of this post will be the proof.  Here is a quick sketch:

1. $\mathfrak{F}_s(\rho, \omega)$ is analytic of exponential type in $\rho$, by the Paley-Wiener Theorem.
2. $\frac{\partial_n\hat{f}}{\partial\xi_i^n} \Big |_{\xi = 0} = \frac{\partial^n}{\partial\rho^n}\mathfrak{F}_s(\rho, \omega)\Big |_{\rho = 0}$, so the exponential type bounds we got in for $\mathfrak{F}_sg$ in (1) transfer to bounds on the pure partial derivatives of $\hat{f}$
3. If $f(x) = F(|x|^2)$ is radial, then at the origin $\partial^{2\alpha}_xf = F^{|\alpha|}(0)\prod \frac{(2\alpha_i - 1)!}{\alpha_i!}$
for all multi-indices $\alpha$
4. This shows that the mixed partials of $\hat{f}$ are dominated by the pure partials at the origin, and the exponential type estimates in (2) hold for all partials
5. So $\hat{f}$ is analytic of exponential type. Thus $f$ has compact support. Continue reading