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

$\phi \in \mathcal{S}(\mathbb{R}\times S^{n-1})$ , i.e. it is smooth and all of its derivatives decay faster than any polynomial.
It is even: $\phi(-s, -\omega) = \phi(s,\omega)$
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:

$R$ is linear. This is trivial
$R$ is injective. This is a direct corollary of the Fourier Slice Theorem.
$f \in \mathcal{S}(\mathbb{R}^n)$ \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.
$g \in \mathcal{S}_H(\mathbb{R}\times S^{n-1}) \implies \exists f \in \mathcal{S}(\mathbb{R}^n), Rf = g$ , i.e. the map is surjective.

Items (1) and (2) have been proven. Let’s review the proof of (3). To see that $Rf \in \mathcal{S}_H$ we must show that it is a Schwartz function, it is even, and it satisfies the moment conditions.

The moment conditions follow from the definition of the Radon transform and Fubini’s theorem:

$\mu_pRf(\omega) = \int_{\mathbb{R}}s^pRf(s,\omega)ds$

$= \int_{\mathbb{R}}s^p\int_{x\cdot\omega = s}f(x)dHds = \int_{\mathbb{R}^n}(x\cdot\omega)^pf(x)dx$

Which is a homogeneous polynomial of degree $p$ in $\omega$ , so the moment conditions are satisfied. $Rf$ is even because

$Rf(-s,-\omega) = \int_{-x\cdot\omega = -s}f(x)dH = \int_{x\cdot\omega = s}f(x)dH = Rf(s,\omega)$

Finally we need to show that $Rf \in \mathcal{S}(\mathbb{R}\times S^{n-1})$ . This is the equivalent to showing that $\mathfrak{F}_s Rf \in \mathcal{S}(\mathbb{R}\times S^{n-1})$ . Since $f \in \mathcal{S}(\mathbb{R}^n)$ , we know that $\hat{f} \in \mathcal{S}(\mathbb{R}^n)$ , so it is smooth and with its derivatives decays faster than any polynomial (not exactly trivial, check smoothness at $s = 0$ ). Immediately this tells us that

$\mathfrak{F}_sRf(\rho,\omega) = \hat{f}(\rho\omega)$

is smooth, and it decays faster than any polynomial is $\rho$ . We need to verify the decay estimates for arbitrary derivatives of $\mathfrak{F}_sRf$ . For all $k, \ell \in \mathbb{N}$ and $D$ differential operators on $S^{n-1}$ we can make the following bound:

$|(1+ |s|^k)\partial_s^{\ell}DRf(s,\omega)|$

$= \frac{1}{2\pi}|\int_{-\infty}^{\infty}(1 + |s|^k)\rho^{|D^\prime|}(D^{\prime}\hat{f}(\rho\omega))(i\rho)^{\ell}e^{i\rho s}d\rho|$

thanks to the Fourier Slice Theorem. Here $D^\prime$ is differential operator on $\mathbb{R}^n$ given by applying $D$ to the $\omega$ variables in polar coordinates. $|D^\prime|$ is the degree of $D$ . But we know that $\hat{f} \in \mathcal{S}(\mathbb{R}^n)$ , so we can simplify this a bit and say there is a $g \in \mathcal{S}(\mathbb{R}^n)$ such that

$|(1+ |s|^k)\partial_s^{\ell}DRf(s,\omega)| \leq |\frac{1}{2\pi} \int(1 + |s|^k)e^{i\rho s}g(\rho\omega)d\rho|$

$= \lvert \frac{1}{2\pi}\int [\frac{(1 + |s|^k)}{(1 + |s|^{2k+2})}(1 - \frac{d^{2k+2}}{d\rho^{2k+2}})e^{i\rho s}] g(\rho\omega)d\rho \rvert$

$\leq c_1 + c_2 \lvert \int [\frac{d^{2k+2}}{d\rho^{2k+2}}e^{i\rho s}] g(\rho\omega)d\rho \rvert \leq C_1 < \infty$

For $s$ bounded away from zero because $g$ is in Schwartz and we can integrate the expression by parts:

$\int [\frac{d^{2k+2}}{d\rho^{2k+2}}e^{i\rho s}] g(\rho\omega)d\rho = \frac{1}{(is)^{2k+2}}\int e^{i\rho s}\frac{d^{2k+2}}{d\rho^{2k+2}}g(\rho\omega)d\rho$

For $s$ near zero the bound is easier, since

$|(1+ |s|^k)\partial_s^{\ell}DRf(s,\omega)| \leq |\frac{1}{2\pi} \int(1 + |s|^k)e^{i\rho s}g(\rho\omega)d\rho|$

$\leq c\lvert \int e^{i\rho s}g(\rho\omega) d\rho \rvert \leq \int |g(\rho,\omega)| d\rho \leq C_2 < < \infty$

Proving that $R$ is surjective

Now we prove (4), for every $\phi \in \mathcal{S}_H(\mathbb{R}\times S^{n-1})$ there is an $f \in \mathcal{S}(\mathbb{R}^n)$ with $Rf = \phi$ . As in the last post, we define the Fourier transform of $f$

$\hat{f}(\rho\omega) =\int_{-\infty}^{\infty}\phi(\omega,s)e^{-i\rho s}ds$

Because $\phi$ is even, $\hat{f}$ is well defined. If we can show that $\hat{f} \in \mathcal{S}(\mathbb{R}^n)$ then we are done.

We will need to show that (1) $\hat{f} \in \mathcal{S}(\mathbb{R}^n - \{0\})$ and (2) show that $\hat{f}$ is smooth at zero. (1) is technical, and (2) is where we will use the moment conditions.

Let’s start by showing that $\hat{f}$ is well behaved away from the origin. We will do this by changing between polar and rectangular coordinates. To simplify the notation, define $F^*$ on $\mathbb{R}\times S^{n-1}$ by

$F^*(\rho, \omega) \equiv \hat{f}(\rho\omega)$

Now we will specify a coordinate change on part of $\mathbb{R}^n$ . In the region where $x_n > 0$ define

$\xi_1 = \rho\omega_1$

$\xi_2 = \rho\omega_2$

…

$\xi_n = \rho\sqrt{1 = \sum_{i=0}^{n-1}\omega_i^2}$

Then we can use the chain rule to write

$\frac{\partial \hat{f}}{\partial \xi_j} = \frac{\partial F^*}{\partial \rho}\frac{\partial \rho}{\partial \xi_j} + \sum_{k=1}^{n-1}\frac{\partial F^*}{\partial\omega_k}\frac{\partial \omega_k}{\partial x_j}$

$= \omega_j\frac{\partial F^*}{\partial\rho} + \frac{1}{\rho}\sum_{k=1}^{n-1}(\delta_{jk} - \omega_j\omega_k)\frac{\partial F^*}{\partial\omega_k}$

for $j = 1, ... n-1$ . A different expression is needed for $x_n$ . With this in hand we can make Schwartz estimates away from the origin:

$|(1 + |\xi|^k)\frac{\partial \hat{f}}{\partial x_j}(\xi)| \leq |(1+|\rho|^k)(\omega_j\frac{\partial F^*}{\partial\rho} + \frac{1}{\rho}\sum_{k=1}^{n-1}(\delta_{jk} - \omega_j\omega_k)\frac{\partial F^*}{\partial\omega_k})|$

$\leq |(1 + |\rho|^k)\int_{-\infty}^{\infty} (\omega_j(-is) + \frac{1}{\rho}D)\phi(\omega, s) e^{-i\rho s} ds|$

using the definition of $F^*$ . We can show that this is bounded using the integration by parts trick we used earlier. This shows that $\hat{f} \in \mathcal{S}(\mathbb{R}^n - 0)$ .

Now we will complete the proof by showing that all of the derivatives of $\hat{f}$ are bounded near the origin.

Exercise Show that if a function and all of its derivatives are bounded in $B_1(0) - B_{\epsilon}(0)$ for all $\epsilon > 0$ , with bounds independent of $\epsilon$ , then the function is smooth at zero.

We will start with our expression

$\frac{\partial \hat{f}}{\partial \xi_j} = \omega_j\frac{\partial F^*}{\partial\rho} + \frac{1}{\rho}\sum_{k=1}^{n-1}(\delta_{jk} - \omega_j\omega_k)\frac{\partial F^*}{\partial\omega_k}$

and bound the terms $\omega_j\frac{\partial F^*}{\partial\rho}$ and $\frac{1}{\rho}\frac{\partial F^*}{\partial\omega_k}$ separately. First note that

$F^*(s,\omega) = \int_{-\infty}^{\infty}\phi(\omega,\rho)(e^{-i\rho s} - 1)d\rho + \int_{-\infty}^{\infty}\phi(\omega, \rho)d\rho$

But thanks to the moment condition, the second integral is a homogeneous polynomial of degree 0 (i.e. a constant), so it will disappear when we apply a differential operator. Let’s make a bound.

$\lvert \frac{1}{s}\frac{\partial F^*}{\partial \omega_j} \rvert \leq \lvert \int_{-\infty}^{\infty}D\phi(\frac{e^{-i\rho s - 1}}{s})d\rho \rvert$

For a first order differential operator $D$ The term $D\phi$ is rapidly decreasing because $\phi$ is Schwartz. The term $\frac{e^{-i\rho s - 1}}{s}$ is bounded by $|\rho|$ . Thus we can, for example, say that

$\lvert \frac{1}{s}\frac{\partial F^*}{\partial \omega_j} \rvert \leq C \int_{-\infty}^{\infty} \frac{|\rho|}{1 + |\rho|^4} d\rho < C^{\prime}$

everywhere. Other first order derivatives can be bounded similarly. To bound higher derivatives we need the following.

Claim

$\frac{\partial^\alpha \hat{f}}{\partial \xi_{i_1}\partial \xi_{i_2}...\partial \xi_{i_q}}$

$= \sum_{1 \leq j + \ell \leq q, 1 \leq k_1, k_2, ... k_{\ell} \leq n-1}s^{j-q}a_{k_1 ...k_{\ell}}(\omega) \frac{\partial^{j + \ell}F^*}{\partial\omega_{k_1}...\partial\omega_{k_\ell}\partial s^j}$

The proof is by induction, and is left as an exercise.

Now we use the same sort of trick.

$F^*(s,\omega) = \int_{-\infty}^{\infty}\phi(\omega, \rho)[e^{-i\rho s} - \sum_{m=0}^{q-1} \frac{(-i\rho s)^m}{m!} ] d\rho + \int_{-\infty}^{\infty} \phi(\omega, \rho) sum_{m=0}^{q-1} \frac{(-i\rho s)^m}{m!} d\rho$

But by the moment condition, the last integrand is a polynomial of degree $m-1$ and will vanish when we apply a differential operator of order $m$ . Thus we have

$|\frac{\partial^{\alpha} \hat{f}}{\partial \xi_{i_1}...\partial \xi_{i_q}} |\leq |s^{j-q}a(\omega)\frac{\partial^{j+\ell}}{\partial \omega_{k_1}...\partial\omega_{k_\ell}\partial s^j}\int_{-\infty}^{\infty} \phi(\omega, \rho) \mathbf{e}_q(-i\rho s) d\rho |$

$\leq \int_{-\infty}^{\infty} a(\omega)|D\phi(\omega, \rho)| |s^{j-q}| |\frac{\partial \mathbf{e}_q(-i\rho s)}{\partial s^j}| d\rho$

$\leq \int_{-\infty}^{\infty} a(\omega)|D\phi(\omega, \rho)| |s^{j-q}(-i\rho)^j \mathbf{e}_{q-j}(-i\rho s)| d\rho$

Now we notice that $D\phi$ is Schwartz so we can say, for example, that $|D\phi(\omega, \rho)| < C(1+|\rho|^{2q})$ . Also,

$|s^{j-q}(-i\rho)^j \mathbf{e}_{q-j}(-i\rho s)| = |(-i\rho s)^{j-q} \mathbf{e}_{q-j}(-i\rho s) (-i\rho)^q| \leq |\rho|^q$

Putting these together we see that

$|\frac{\partial^{\alpha} \hat{f}}{\partial \xi_{i_1}...\partial \xi_{i_q}} |\leq C\int_{-\infty}^{\infty} \frac{|\rho|^q}{1 + |\rho|^{2q}} d\rho < C^{\prime}$

And we have proven that all derivatives of $\hat{f}$ are bounded, completing our proof of the theorem.

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This entry was posted on October 30, 2009 at 6:18 am and is filed under Inverse Problems. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

2 Responses to “Inverse Problems Course Notes — Proof of Helgason-Ludwig’s Range Characterization”

Inverse Problems Course Notes — The Support Theorem for the Radon Transform « Rolfe's Lecture Notes Says:
November 3, 2009 at 1:25 am | Reply
[...] Notes — The Support Theorem for the Radon Transform By Rolfe Schmidt In the last two posts we proved that when the Radon transform acts on Schwartz functions, its range is a subspace of [...]
Inverse Problems Course Notes — Motivating the Range Characterization for the Radon Transform « Rolfe's Lecture Notes Says:
November 12, 2009 at 11:18 pm | Reply
[...] the Range Characterization for the Radon Transform By Rolfe Schmidt 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 [...]

Rolfe's Lecture Notes