Inverse Problems Course Notes — The Range 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 we are prepared to study the range of the X-Ray Transform. We will will look at the image of ${\mathcal{S}({\mathbb R}^n)}$ in detail — once we understand this we can easily use our understanding of the Radon transform to characterize the image of ${X}$ for smooth, compactly supported functions.

So let ${f \in \mathcal{S}({\mathbb R}^n)}$ . Recall that

$\displaystyle Xf(x,\theta) = \int_{\mathbb R} f(x + t\theta) dt$

It will be convenient to extend ${Xf}$ to all of ${{\mathbb R}^n\times({\mathbb R}^n - \{0\})}$ as follows

$\displaystyle \begin{array}{rcl} Xf(x,\xi) &=& \int_{\mathbb R} f(x + t\xi) dt, \quad \xi \in {\mathbb R}^n - \{0\}) \\ \\ &=& \frac{1}{|\xi|}\int_{{\mathbb R}}f(x + t\frac{\xi}{|\xi}) dt \\ \\ &=& \frac{1}{|\xi|} \int_{\mathbb R} f(x - \langle x, \frac{\xi}{|\xi|}\rangle\frac{\xi}{|\xi|} + t\frac{\xi}{|\xi|}) dt \\ \\ &=& \frac{1}{|\xi|}Xf(x-\langle x, \frac{\xi}{|\xi|}\rangle\frac{\xi}{|\xi|}, \frac{\xi}{|\xi|}) \end{array}$

So ${Xf}$ is said to be positive homogeneous of degree -1 in ${\xi}$ .

Claim 1 For ${x \in {\mathbb R}^n}$ , ${\xi \in {\mathbb R}^n - \{0\}}$ The partial differential equations
$\displaystyle \frac{\partial^2}{\partial x_i\partial \xi_j}Xf - \frac{\partial^2}{\partial x_j\partial \xi_i}Xf = 0$

hold for all ${1 \leq i, j \leq n}$ .

To show this, check the following

$\displaystyle \begin{array}{rcl} \frac{\partial^2}{\partial x_i\partial \xi_j}Xf &=& \int_{\mathbb R} \frac{\partial^2}{\partial x_i\partial \xi_j} f(x + t\xi) dt \\ &=& t\int_{\mathbb R} \partial^2_{ij}f(x + t\xi) dt \end{array}$

which is symmetric in ${i}$ and ${j}$ .

These equations are called John’s Equations, named for Fritz John who was the first to point this fact out. The main result is that these are the only conditions we need to characterize the range of ${X}$

Theorem 1 (Fritz John) If ${\varphi \in \mathcal{S}(T)}$ , define its extension to ${{\mathbb R}^n\times({\mathbb R}^n - \{0\})}$ as
$\displaystyle \psi(x,\xi) = \frac{1}{|\xi|}\varphi(x-\langle x, \frac{\xi}{|\xi|}\rangle\frac{\xi}{|\xi|}, \frac{\xi}{|\xi|})$

Then ${\varphi}$ is in the range of ${X}$ if and only if ${\varphi(x, \theta) = \varphi(x, -\theta)}$ and
$\displaystyle \frac{\partial^2}{\partial x_i\partial \xi_j}\psi - \frac{\partial^2}{\partial x_j\partial \xi_i}\psi = 0$

for all ${1 \leq i,j \leq n}$

This is nice: we characterize the range with only a finite number of conditions. Contrast this with the range characterization of the Radon Transform, which required an infinite number of conditions and could not be effectively checked.

Proof: We want to solve

$\displaystyle Xf = \varphi$

By the Fourier Slice Theorem for the X-Ray transform, this is solved if we take ${f}$ such that

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

But for any fixed ${\eta}$ , there are many ${\theta}$ such that ${\eta \in {\theta^\perp}}$ , so at first glance this definition looks problematic. To make it work we must show that, for fixed ${\eta}$ , ${\mathfrak{F}_{{\theta^\perp}}\varphi(\eta, \theta_1) = \mathfrak{F}_{{\theta^\perp}}\varphi(\eta, \theta_2)}$ for all ${\theta_1, \theta_2}$ orthogonal to ${\eta}$ .

We will do this by showing that the directional derivatives of ${\mathfrak{F}_{{\theta^\perp}}\varphi(\eta, \theta)}$ are zero in all directions orthogonal to ${\eta}$ . It is easier to work on ${{\mathbb R}^{2n}}$ than on the manifold ${T}$ , so we will study an extension of ${\mathfrak{F}_{{\theta^\perp}}\varphi}$ to ${{\mathbb R}^n\times{\mathbb R}^n}$ and show that the ${\xi}$ -gradient of this extension is parallel to ${\eta}$ , thus the directional derivatives in all directions orthogonal to ${\eta}$ will be zero.

Defining this extension, we need to make sure that it does not vary as we scale ${\xi}$ , or the ${\xi}$ -gradient will have a component in the ${\xi}$ direction and will not be parallel to ${\eta}$ . With that in mind, define the extension ${g}$ as follows

$\displaystyle \begin{array}{rcl} g(\eta, \xi) &=& \mathfrak{F}_{{\theta^\perp}}\varphi(\eta - \langle\eta, \frac{\xi}{|\xi|}\rangle\frac{\xi}{|\xi|}, \frac{\xi}{|\xi|}) \\ &=& \int_{x \in \xi^\perp}e^{-i\langle \eta - \langle\eta, \frac{\xi}{|\xi|}\rangle\frac{\xi}{|\xi|}, x\rangle}\varphi(x, \frac{\xi}{|\xi|})dx \\ &=& \int_{x \in \xi^\perp}e^{-i\langle\eta,x\rangle}\varphi(x, \frac{\xi}{|\xi|})dx \\ &=& |\xi|\int_{x \in \xi^\perp}e^{-i\langle\eta,x\rangle}\psi(x, \xi)dx \end{array}$

using the homogeneity in the definition of ${\psi}$ . Applying the inverse Fourier Transform gives

$\displaystyle \psi(x,\xi) = \frac{1}{2\pi|\xi|}\int_{\eta \in \xi^\perp}e^{i\langle\eta, x\rangle}g(\eta, \xi) d\eta$

When ${x \in \xi^{\perp}}$ . For general ${x, \xi}$ we have

$\displaystyle \psi(x,\xi) = \frac{1}{2\pi|\xi|}\int_{\eta \in \xi^\perp}e^{i\langle\eta, x - \langle x, \frac{\xi}{|\xi|}\rangle \frac{\xi}{|\xi|}\rangle}g(\eta, \xi) d\eta$

Because ${\varphi}$ , and hence ${g}$ , is well behaved, we can differentiate this expression under the integral sign.

$\displaystyle \begin{array}{rcl} \frac{\partial^2\psi}{\partial x_i\partial\xi_j} &=& \frac{\partial}{\partial \xi_j} \int_{\eta \in \xi^\perp} \frac{1}{|\xi|}\left[\eta_i - \frac{\xi_i\langle\eta,\xi\rangle}{|\xi|^2} \right]e^{i\langle\eta, x - \langle x, \frac{\xi}{|\xi|}\rangle \frac{\xi}{|\xi|}\rangle}g(\eta, \xi) d\eta \\ &=& \int_{\eta \in \xi^\perp} \left[\frac{1}{|\xi|} \eta_i\frac{\partial}{\partial \xi_j} - \frac{\xi_j\eta_i}{|\xi|^3} -\frac{\xi_i\eta_j}{|\xi|^3} \right] e^{i\langle\eta, x - \langle x, \frac{\xi}{|\xi|}\rangle \frac{\xi}{|\xi|}\rangle}g(\eta, \xi) d\eta \end{array}$

at points where ${\langle x, \xi\rangle = \langle \eta, \xi\rangle = 0}$ . The general expression is quite complicated, but many of the terms are multiples of ${\langle x, \xi\rangle}$ and ${\langle \eta, \xi\rangle}$ . It is a worthwhile exercise to verify this expression.

Now we can use John’s equation to say that, when ${x \in \xi^\perp}$

$\displaystyle \begin{array}{rcl} 0 &=& \frac{\partial^2}{\partial x_i\partial \xi_j}\psi(x,\xi) - \frac{\partial^2}{\partial x_j\partial \xi_i}\psi(x, \xi) \\ &=& \frac{1}{|\xi|} \int_{\eta \in \xi^\perp}\left[\eta_i\frac{\partial}{\partial \xi_j} - \eta_j\frac{\partial}{\partial \xi_i} \right]g(\eta,\xi)e^{i\langle x, \eta\rangle} d\eta \end{array}$

But this is an (inverse) Fourier Transform of a function and can only vanish if the transformed function vanishes ( ${\mathfrak{F}^{-1}_{\theta^\perp}}$ is injective). So

$\displaystyle \left[\eta_i\frac{\partial}{\partial \xi_j} - \eta_j\frac{\partial}{\partial \xi_i} \right]g(\eta,\xi) = 0$

The expression on the left hand side is nothing but the ${i,j}$ coordinate of the wedge product ${\eta \wedge \nabla_\xi g(\eta, \xi)}$ . So this wedge product vanishes. That can only happen when the two (co)vectors are parallel, so there is a scalar valued function ${c(\eta, \xi)}$ such that

$\displaystyle \nabla_\xi g(\eta, \xi) = c(\eta, \xi)\eta$

If you aren’t comfortable with wedge products, it is just a short hand way of making the following calcuation.

$\displaystyle \begin{array}{rcl} 0 &=& \sum_{i,j} \eta_i\times 0 \times \frac{\partial g}{\partial\xi_j} \\ &=& \sum_{i,j} \eta_i \left[\eta_i\frac{\partial g}{\partial \xi_j} - \eta_j\frac{\partial g}{\partial \xi_i} \right]\frac{\partial g}{\partial\xi_j} \\ &=& \|\eta\|^2\|\nabla_\xi g\|^2 - \langle \eta, \nabla_\xi g\rangle^2 \end{array}$

which clearly implies that ${\eta}$ and ${\nabla_\xi g}$ are parallel.

This is exactly what we were looking for: ${g}$ is an extension of ${\mathfrak{F}_{{\theta^\perp}}\varphi}$ , and its derivatives vanish in all directions orthogonal to ${\eta}$ . So when we look at a small change in ${\theta}$ , ${\mathfrak{F}_{{\theta^\perp}}\varphi}$ as long as that change remains perpendicular to ${\eta}$ :

$\displaystyle \mathfrak{F}_{{\theta^\perp}}\varphi(\eta, \theta + h\hat{\theta}) - \mathfrak{F}_{{\theta^\perp}}\varphi(\eta, \theta) \approx h\hat{\theta}\cdot \nabla_\xi g(\eta, \xi) = 0$

So for fixed ${\eta}$ , ${\mathfrak{F}_{{\theta^\perp}}\varphi}$ is locally constant on ${\eta^\perp}$ . This set is connected, so ${\mathfrak{F}_{{\theta^\perp}}\varphi}$ is constant on ${\eta^\perp}$ and our definition of ${f}$ is sound.

The rest of the proof is straightforward. $\Box$

(A pdf version of these notes is available here.)

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Inverse Problems Course Notes — The Range of the X-Ray Transform

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