# Inverse Problems Course Notes — Motivating the Range Characterization for the Radon Transform

With the proofs of the range characterization and support theorems for the radon transform complete, let’s take a step back and look at the intuition behind these results again.  In particular, let’s try to understand the moment conditions (It might be good to read this post before reading the full proofs).  It was easy to directly verify that the moment conditions were necessary, so

$R(\mathcal{S}(\mathbb{R}^n) \subset \mathcal{S}_H(\mathbb{R}\times S^{n-1})$

We need to show it is sufficient — that $\mathcal{S}_H(\mathbb{R}\times S^{n-1}) \subset R(\mathcal{S}(\mathbb{R}^n)$.  Take $g \ in \mathcal{S}_H(\mathbb{R}\times S^{n-1})$, we need to find $f \in \mathcal{S}(\mathbb{R}^n)$ with $Rf = g$ Continue reading

# 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

# Inverse Problems Course Notes — The Support Theorem for the Radon Transform

In the last two posts we proved that when the Radon transform acts on Schwartz functions, its range is a subspace of Schwartz functions characterized by a set of moment conditions.  Now we will look at $R$ on an even more restricted domain: functions with compact support.

Clearly if $f$ has compact support, so does $Rf$, so  $R: C_0^{\infty}(\mathbb{R}^n) \rightarrow \mathcal{S}_H(\mathbb{R}\times S^{n-1}) \cap C_0(\mathbb{R}\times S^{n-1})$ is injective.  We want to show that there are no other restrictions on the range, i.e. this map is onto.

Thanks to our work on $\mathcal{S}$ we already know that for any $g \in \mathcal{S}_H(\mathbb{R}\times S^{n-1}) \cap C_0(\mathbb{R}\times S^{n-1})$ there exists some $f \in \mathcal{S}(\mathbb{R}^n)$ with $Rf = g$.  If we can prove that this $f$ must have compact support we will have our result.

Actually, we’ll prove an apparently stronger statement.

Theorem [The Support Theorem] If $f \in C(\mathbb{R}^n)$ satisfies

(i) [Rapid decrease] $|x|^k|f(x)|$ is bounded for all $k > 0$

(ii) [Compact support] $\exists A > 0$ such that $|\rho| > A \implies Rf(\rho,\omega) = 0$

Then $f(x) = 0$ for all $|x| > A$.

The first condition seems strange — is this really needed?

In fact, there are counterexamples. 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

# Inverse Problems Course Notes — Proof of Helgason-Ludwig’s Range Characterization

In the last post we stated and motivated a theorem characterizing the image of the Schwartz space under the Radon transform:

Theorem [Helgason-Ludwig] The Radon transform is a bijective linear map from $\mathcal{S}(\mathbb{R}^n)$ onto $\mathcal{S}_H(\mathbb{R}\times S^{n-1})$, where $\phi \in \mathcal{S}_H(\mathbb{R}\times S^{n-1})$ if and only if

1. $\phi \in \mathcal{S}(\mathbb{R}\times S^{n-1})$ , i.e. it is smooth and all of its derivatives decay faster than any polynomial.
2. It is even: $\phi(-s, -\omega) = \phi(s,\omega)$
3. It satisfies the “moment conditions”: $\mu_k\phi(\omega) = \int_{\mathbb{R}}\phi(s, \omega)s^kds$ is a homogeneous polynomial of degree $k$ in $\omega$.

Proof We need to prove the following:

1. $R$ is linear.  This is trivial
2. $R$ is injective.  This is a direct corollary of the Fourier Slice Theorem.
3. $f \in \mathcal{S}(\mathbb{R}^n)$ \implies Rf \in \mathcal{S}_H(\mathbb{R}\times S^{n-1})\$.  This was largely done in earlier posts, but we will consolidate the argument below.
4. $g \in \mathcal{S}_H(\mathbb{R}\times S^{n-1}) \implies \exists f \in \mathcal{S}(\mathbb{R}^n), Rf = g$, i.e. the map is surjective. Continue reading

# Inverse Problems Course Notes — The Range of the Radon Transform

Now we want to study the range of the Radon transform.  This is important for our inverse problem: when we want to reconstruct a function from its Radon transform, all we really have to work with is a finite number of error-prone samples of the Radon transform.  We need to analyze this statistically to find the function “most likely” to have produced the observed data.  We can only do this if we know what sort of functions can be Radon transforms of other functions.

In other words, we can only do this if we know the Range of $R$. Continue reading

# Inverse Problems Course Notes — More About Stability

We have seen that Radon inversion is stable — if $Rf_1 \approx Rf_2$ then $f_1 \approx f_2$.  Now we want to look at the other direction, and study the stability of the “forward” Radon transform.  That is, we want to know: if $F_1 \approx f_2$, is $Rf_1 \approx Rf_2$?

Definition Let $K \Subset \mathbb{R}^n$ ($K$ is compactly supported in $\mathbb{R}^n$).  Then define

$H^2(K) \equiv \{ u \in H^s(\mathbb{R}^n) | \text{supp} u \subset K \}$

This lets us state

Proposition $R: H^s(K) \rightarrow H^{s + \frac{n-1}{2}}(\mathbb{R}\times S^{n-1})$ and is bounded $\forall s, K \Subset \mathbb{R}^n$, with a norm depending on $K$.  In other words

$\|Ru\|_{H^{s + \frac{n-1}{2}}} \leq C(K)\|u\|_{H^s(K)}$ for some $C(K) > 0$.

Unfortunately this is sharp — there is no estimate independent of $K$. Continue reading

# Inverse Problems Course Notes — Stability of Radon Inversion

With the tools of Sobolev spaces at our disposal, we are ready to go back to one of our basic questions: is our solution to the Radon transform inverse problem stable?  In other words if\

$Rf_1 \approx Rf_2$, do we know that $f_1 \approx f_2$.

If we don’t know this, we can’t trust our reconstruction — even tiny rounding errors could be disastrous.

## A Formal Look

Let’s consider the case when $n$ is odd (so that $n-1$ is even).  If $u \in C_0^{\infty}(\mathbb{R}^n)$,  we know that $u = c_nR^t\partial_s^{n-1}Ru$, so

$\langle u, u\rangle = \langle c_nR^t\partial_s^{n-1}Ru, u\rangle = c_n\langle \partial_s^{n-1}Ru, Ru\rangle$

$= c\langle \partial_s^{\frac{n-1}{2}}Ru, \partial_s^{\frac{n-1}{2}}Ru \rangle = \lVert \partial_s^{\frac{n-1}{2}}Ru \rVert_{L^2(\mathbb{R}\times S^{n-1})}$

This suggests that $Ru \in H^{\frac{n-1}{2}}(\mathbb{R}\times S^{n-1})$.  To prove it, we need to define that space. Continue reading

# Inverse Problems Course Notes — Sobolev Spaces

We will define Sobolev spaces using the Fourier transform and $L^2$.

Definition For $s \in \mathbb{R}$ let $H^s(\mathbb{R}^n)$ be $\{u \in \mathcal{S}^\prime(\mathbb{R}^n) | \hat{u}(\xi) \text{ is a function and } \int (1 + |\xi|^2)^s|\hat{u}(\xi)|^2d\xi < \infty \}$

So $H^0(\mathbb{R}^n) = L^2(\mathbb{R}^n)$.

Claim $H^1(\mathbb{R}^n) = \{u \in \mathcal{S}^\prime(\mathbb{R}^n) | \partial^\alpha u \in L^2(\mathbb{R}^n) |\alpha| \leq 1\}$ Continue reading

# Inverse Problems Course Notes — Extending the Domain of R to Distributions

The Radon Inversion formula developed in the last section is nice, but there is a problem.  It only applies to smooth functions, and the functions we are interested in are not smooth.  Heart tissue does not smoothly change into lung tissue; body parts have sharp boundaries.  If we want to use this for applications like medical imaging we must extend the results to a larger domain. Continue reading