# Inverse Problems Course Notes — An Alternative Development of the X-Ray Transform

These notes are based on Gunther Uhlmann’s lectures for MATH 581 taught at the University of Washington in Autumn 2009.

An index to all of the notes is available here.

Now that we’ve sketched the basic facts about the X-Ray transform, let’s look at it from a different perspective. When we developed the theory of the Radon transform, we made extensive use of the Fourier Slice Theorem — something we did not mention at all for the X-Ray transform. We also notice that our inversion formula looked different from the one we developed for ${R}$.

In this post we will see that these differences are superficial. We will find the analog for the Fourier Slice Theorem and use it to redevelop our X-Ray transform results, mimicking our theory of the Radon transform.

1. The Fourier Slice Theorem

When analyzing the Radon Transform, we noticed that the “wave functions” used to define the Fourier Transform are constant on hyperplanes, so the Fourier Transform of a function could be represented as a phase-weighted sum of integrals over these constant hyperplanes, i.e. as a weighted integral of the Radon Transform of the function. Specifically

$\displaystyle \hat{f}(\rho\omega) = \int_{-\infty}^{\infty}e^{-i\rho s}Rf(s, \omega) ds = \mathfrak{F}_sRf(\rho, \omega)$

This was the Fourier Slice Theorem. If the waves are constant over hyperplanes, they are certainly constant over lines, so we can find a similar formula for the X-Ray Transform. There are many ways we can foliate a hyperplane into a family of parallel lines, so we should expect to get many formulas. This is exactly what happens. (I try to give a bit more intuition in a separate post.)

Given ${\eta}$, select an arbitrary orthogonal unit vector ${\theta \in S^{n-1}, \eta\cdot\theta = 0}$. Then

$\displaystyle \begin{array}{rcl} \mathfrak{F}_{\theta^{\perp}}Xf(\eta, \theta) &=& \int_{\theta^\perp}e^{-y\cdot\eta}Xf(y,\theta)dH_{\theta^\perp} \\ \\ &=& \int_{\theta^\perp}e^{-iy\cdot\eta}\int_{-\infty}^\infty f(y + t\theta)dt dH_{\theta^\perp} \end{array}$

As ${y}$ varies over ${\theta^\perp}$ and ${t}$ varies over ${\mathbb{R}}$, ${x = y + t\theta}$ varies over ${\mathbb{R}^n}$ and ${dx = dH_{\theta^\perp}dt}$ is standard Lebesgue measure. In this case, notice that ${y\cdot\eta = (x - t\theta)\cdot\eta = x\cdot\theta}$ because ${\eta\cdot\theta = 0}$ so

$\displaystyle \mathfrak{F}_{\theta^{\perp}}Xf(\eta, \theta) = \int_{\mathbb{R}^n}e^{-ix\cdot\eta}f(x) dx = \hat{f}(\eta)$

As promised, this gives us a different formula for each choice of ${\theta}$. Let’s summarize the result in a theorem

Theorem 1 (Fourier Slice Theorem for the X-Ray Transform) For all ${\eta \in \mathbb{R}^n}$ and ${\theta}$ with ${\eta \in \theta^\perp}$

$\displaystyle \hat{f}(\eta) = \mathfrak{F}_{\theta^\perp}Xf(\eta, \theta)$

For the Radon transform, the slice theorem immediately allowed us to prove that the Radon Transform intertwines (powers of) the Laplacian on ${\mathbb{R}^n}$ with ${\partial_s^2}$ on ${\mathbb{R}\times S^{n-1}}$. We can do the same thing for the X-Ray transform.

Theorem 2 (Intertwining) For ${f \in \mathcal{S}(\mathbb{R}^n)}$ and ${\alpha > -(n-1)}$

$\displaystyle X(-\bigtriangleup)^{\frac{\alpha}{2}}f = (-\bigtriangleup_{\theta^\perp})^{\frac{\alpha}{2}}Xf$

Where ${(-\bigtriangleup)^{\frac{\alpha}{2}}}$ is defined by ${\mathfrak{F}_{\theta^\perp}((-\bigtriangleup)^{\frac{\alpha}{2}}g)(\eta,\theta) = |\eta|^{\alpha}\mathfrak{F}_{\theta^\perp}g(\eta,\theta)}$.

The condition that ${\alpha > -(n-1)}$ insures that the operator ${(-\bigtriangleup)^{\frac{\alpha}{2}}}$ is defined by an absolutely convergent integral on ${\theta^{\perp}}$.

Proof: The proof is exactly the same as it was for the Radon Transform. We use the slice Theorem for ${\eta \in \theta^\perp}$:

$\displaystyle \begin{array}{rcl} \mathfrak{F}_{\theta^\perp}X(-\bigtriangleup)^{\frac{\alpha}{2}} f(\eta, \theta) &=& \widehat{(-\bigtriangleup)^{\frac{\alpha}{2}} f}(\eta) \\ &=& |\eta|^\alpha\hat{f}(\eta) \\ &=& |\eta|^\alpha\widehat{Xf}(\eta, \theta) \\ &=& \mathfrak{F}_{\theta^\perp}(-\bigtriangleup_{\theta^\perp})^{\frac{\alpha}{2}}Xf(\eta, \theta) \end{array}$

$\Box$

We can do the same thing for the transpose.

Theorem 3 (Intertwining) For ${g \in \mathcal{S}(T)}$ and ${\alpha > -(n-1)}$

$\displaystyle X^t(-\bigtriangleup_{\theta^\perp})^{\frac{\alpha}{2}}g = (-\bigtriangleup)^{\frac{\alpha}{2}}X^tg$

Proof: As often happens with dual results, the proof is just a formal manuipulation. First note that ${(-\bigtriangleup)^{\frac{\alpha}{2}}}$ is its own transpose:

$\displaystyle \begin{array}{rcl} \langle(-\bigtriangleup)^{\frac{\alpha}{2}}f, g\rangle_{L^2(\mathbb{R}^n)} &=& \langle|\eta|^\alpha\hat{f}, \hat{g}\rangle_{L^2(\mathbb{R}^n)} \\ &=& \langle \hat{f}, |\eta|^\alpha\hat{g}\rangle_{L^2(\mathbb{R}^n)} \\ &=& \langle f, (-\bigtriangleup_{\theta^\perp})^{\frac{\alpha}{2}}g\rangle_{L^2(\mathbb{R}^n)} \end{array}$

A similar argument works for ${\bigtriangleup_{\theta^\perp}}$ on ${T}$, so we can write

$\displaystyle \begin{array}{rcl} \langle(-\bigtriangleup)^{\frac{\alpha}{2}}X^tg, h\rangle_{L^2(\mathbb{R}^n)} &=& \langle X^tg, (-\bigtriangleup)^{\frac{\alpha}{2}}h\rangle_{L^2(\mathbb{R}^n)} \\ &=& \langle g, X((-\bigtriangleup)^{\frac{\alpha}{2}}h \rangle_{L^2(T)} \\ &=& \langle g, (-\bigtriangleup_{\theta^\perp})^{\frac{\alpha}{2}}Xh \rangle_{L^2(T)} \\ &=& \langle X^t(-\bigtriangleup_{\theta^\perp})^{\frac{\alpha}{2}}g, h\rangle_{L^2(\mathbb{R}^n)} \end{array}$

$\Box$

This result immediately lets us rewrite the inversion formula in a style that looks more like our original Radon inversion formula

Theorem 4 (X-Ray Inversion II) For all ${f \in C_0^\infty(\mathbb{R}^n)}$

$\displaystyle X^t(-\bigtriangleup_{\theta^\perp})^{\frac{1}{2}}Xf = c_nf$