Inverse Problems Coruse Notes — Stability Estimates for 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.

With an inversion formula in hand, we are ready to state and prove the basic stability estimates for the X-Ray Transform. They are almost identical to the estimates for the Radon transform, the only difference coming from the fact that {X} smoths by {1/2} of a derivative, where {R} smooths by {n/2} derivatives.

Theorem 1 For all {K \subset {\mathbb R}^n}, {K} compactly supported, and {f \in H^s(K)}, the following estimates hold

\displaystyle  \|f\|^2_{H^s({\mathbb R}^n)} \leq C_s\|Xf\|^2_{H^{s + 1/2}(T)} \leq C_s(K)\|f\|^2_{H^s({\mathbb R}^n)}

The first inequality holds even when {f} does not have compact support.

In other words X-Ray inversion is stable, and so is the X-Ray transform itself, as long as you measure errors with the right norms.

Proof: Step 1. {\|f\|_{H^s} \leq c\|Xf\|_{H^{s + 1/2}}}Let’s start by proving the first inequality — stability, or boundedness, of X-Ray inversion — in the {L^2} case ({s = 0}). The argument goes as follows: use the inversion formula to rewrite the formula for the norm of {f}, then use Plancherel’s Theorem and the Fourier Slice Theorem to move everything to the frequency domain. Here are the details.

\displaystyle  \begin{array}{rcl}  \|f\|^2_{L^2({\mathbb R}^n)} = \langle f,f\rangle_{L^2({\mathbb R}^n)} &=& \langle({(-\bigtriangleup)^{1/2}} X^tXf,f\rangle_{L^2({\mathbb R}^n)} \\ &=& \langle Xf, X{(-\bigtriangleup)^{1/2}} f\rangle_{L^2(T)} \\ &=& \langle \mathfrak{F}_{{\theta^\perp}}Xf, \mathfrak{F}_{{\theta^\perp}}X{(-\bigtriangleup)^{1/2}} f\rangle_{L^2(T)} \\ &=& \langle \hat{f}, |\eta|\hat{f}\rangle_{L^2({\mathbb R}^n)} \\ &=& \int_{{\mathbb R}^n}|\eta||\hat{f}(\eta)|^2 d\eta \\ &\leq& \int_{{\mathbb R}^n}(1 + |\eta|^2)^{1/2}|\hat{f}(\eta)|^2 d\eta \\ &=& \|f\|_{H^{1/2}({\mathbb R}^n)} \end{array}

Now we can extend this to general {s} by replacing {f} with {(I-\bigtriangleup)^{s/2}}. Then

\displaystyle  \begin{array}{rcl}  \|f\|_{H^s} = \|(I-\bigtriangleup)^{s/2}f\|_{L^2} &\leq& c_n\|X(I-\bigtriangleup)^{s/2}f\|_{H^{1/2}} \\ &=& c_n\|(I-\bigtriangleup_{{\theta^\perp}})^{s/2}Xf\|_{H^{1/2}} = c_n\|Xf\|_{H^{s+1/2}} \end{array}

Proving the first inequality. Here we relied on the following result, an easy generalization of the intertwining results for {X}, {\bigtriangleup} and {\mathfrak{F}} we used before.

Claim 1

\displaystyle  X(I-\bigtriangleup)^{s/2}f = (I-\bigtriangleup_{{\theta^\perp}})^{s/2}Xf

The proof is left as an exercise.

Step 2. {\|Xf\|_{H^{s + 1/2}} \leq C_s(K)\|f\|_{H^s}} Again, we will start with the {L^2} case.

\displaystyle  \begin{array}{rcl}  \|Xf\|_{H^{1/2}(T)} &=& \int_{S^{n-1}}\int_{{\theta^\perp}} |\mathfrak{F}_{{\theta^\perp}}Xf|^2(\eta,\theta)(1 + |\eta|^2)^{1/2} dH_{{\theta^\perp}}d\theta \\ &=& \int_{S^{n-1}}\int_{{\theta^\perp}} |\hat{f}(\eta)|^2(1 + |\eta|^2)^{1/2} dH_{{\theta^\perp}}d\theta \end{array}

To go forward we need a result that relates the measure {dH_{{\theta^\perp}}d\theta} to standard Lebesgue measure. As {\theta} varies over {S^{n-1}}, {H_{{\theta^\perp}}} clearly covers all of {{\mathbb R}^n}, but it covers it more than once and points close to the origin are covered “more densely” than points far away. The following claim makes this intuition precise, and the proof can be found in the appendix of Natterer’s book.

Lemma 2

\displaystyle  \begin{array}{rcl}  \int_{S^{n-1}}\int_{{\theta^\perp}} g(\eta) dH_{{\theta^\perp}}d\theta &=& \frac{1}{\text{Vol}(S^{n-2})}\int_{{\mathbb R}^n}\frac{g(y)}{|y|}dy \end{array}

This lets us continue, writing

\displaystyle  \|Xf\|_{H^{1/2}(T)} = c_n\int_{{\mathbb R}^n}\frac{|\hat{f}(\eta)|^2}{|\eta|}(1 + |\eta|^{2})^{1/2} d\eta

We have not used the compact support of {f} yet, but now it appears in a move that should seem familiar. We will split the integral into a sum of two integrals, one over the low fequencies, and the other over the high frequencies. Define

\displaystyle  \begin{array}{rcl}  \text{I} &=& c_n\int_{|\eta| \geq 1}\frac{|\hat{f}(\eta)|^2}{|\eta|}(1 + |\eta|^2)^{1/2} d\eta \\ \text{II} &=& c_n\int_{|\eta| \leq 1}\frac{|\hat{f}(\eta)|^2}{|\eta|}(1 + |\eta|^2)^{1/2} d\eta \end{array}

Then

\displaystyle  \|Xf\|_{H^{1/2}(T)} = {\text{I}} + {\text{II}}

But

\displaystyle  \text{I}\quad \leq \quad c\int_{|\eta| \geq 1} |\hat{f}(\eta)|^2d\eta \leq c\|f\|^2_{L^2}

And

\displaystyle  \text{II}\quad \leq \quad \sup_{|\eta| \leq 1}|\hat{f}(\eta)|^2 \cdot \int_{|\eta| \leq 1}\frac{(1 + |\eta|^2)^{1/2}}{|\eta|} d\eta \leq C(\text{Vol } K)^{1/2}\|f\|_{L^2}

We can repeat this argument for any {s} with the usual modifications (pick smooth compactly supported {\varphi \equiv 1} on {K} and use {\langle \varphi, f\rangle_{L^2} \leq \|\varphi\|_{H^{-s}}\|f\|_{H^s}}). \Box

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

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s