# 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 $R$.

## A Warm-up: The Range of the Fourier Transform

For the Fourier Transform we know

• $\mathfrak{F}(L^2) = L^2$
• $\mathfrak{F}( \mathcal{S}) = \mathcal{S}$
• $\mathfrak{F}( \mathcal{S^\prime}) = \mathcal{S^\prime}$

But what is $\mathfrak{F}(C_0^{\infty})$? For $u \in C_0^\infty(\mathbb{R}^n)$, $\hat{u}$ is analytic (we can write down its power series by expanding the series of the exponential in the definition of the Fourier transform).  But an analytic function with compact support must be zero everywhere, so if $\hat{u}$ has compact support, it (and $u$) must be identically zero.

So $\mathfrak{F}(C_0^{\infty}(\mathbb{R}^n)) \neq C_0^{\infty}(\mathbb{R}^n)$.  We will talk about this more later when we discuss the Paley-Wiener theorem.

## The Range of $R$ Acting on $L^2(\mathbb{R}^n)$

Now let’s look at the Radon transform, $R$.  We know that the Radon transform of a function must be even, i.e. $Rf(s,\omega) = Rf(-s,-\omega)$, so the range of $r$ must be somewhat restricted.  Let’s start with $L^2$.

Proposition The range of $T = c_n|\partial_s|^{\frac{n-1}{2}}R$ acting on $L^2(\mathbb{R}^n)$ is $L^2(\mathbb{R}\times S^{n-1}) \cap \{ u | u(-s,-\omega) = u(s,\omega) \}$

In other words, the range of $L^2$ is everything it can be.

Proof We know that $T(L^2(\mathbb{R}^n) \subset L^2(\mathbb{R}\times S^{n-1})$.  First we will show that $Tu$ must be even (even though we’ve already seen this for $R$).

$Tu(-s,-\omega)= \int_{-\infty}^\infty e^{-is\rho}\mathfrak{F}_sTu(\rho,-\omega)d\rho$

$= c\int_{-\infty}^\infty e^{-is\rho}|\rho|^{\frac{n-1}{2}}\hat{u}(-\rho\omega)d\rho$

Now changing $\rho$ to $-\rho$

$= c\int_{-\infty}^\infty e^{is\rho}|\rho|^{\frac{n-1}{2}}\hat{u}(\rho\omega)d\rho = Tu(\rho,\omega)$

So the range of $R$ is a subset of the even functions in $L^2(\mathbb{R}\times S^{n-1})$.  Now we will go the other way and show that every even function in $L^2(\mathbb{R}\times S^{n-1})$ is in the range.

Let $g$ be an even function in $L^2(\mathbb{R}\times S^{n-1})$.  We need to find an $f \in L^2(\mathbb{R}^n)$ such that $Tf = g$.

Note that if $f$ exists, then

$\widehat{Tf} = |\rho|^{\frac{n-1}{2}}\hat{f}(\rho\omega) = \mathfrak{F}_sg(\rho,\omega)$

so we define $f$ with the Fourier transform

$\hat{f}(\rho\omega) \equiv \frac{\mathfrak{F}_sg(\rho,\omega)}{|\rho|^{\frac{n-1}{2}}}$

We need top be careful, this is a polar change of coordinates which is singular.

We need to check something first: $\rho\omega = (-\rho)(-\omega)$, so for $\hat{f}$ to be well defined we must have

$\mathfrak{F}_sg(\rho,\omega) = \mathfrak{F}_sg(-\rho,-\omega)$

To see this, note that

$\mathfrak{F}_s g(-\rho,-\omega) = \int_{\mathbb{R}} e^{is\rho}g(s,-\omega) ds$

$= \int_{\mathbb{R}} e^{is\rho}g(-s,\omega)ds$

$= \int_{\mathbb{R}} e^{-i\tilde{s}\rho}g(\tilde{s}, \omega)d\tilde{s} = \mathfrak{F}_sg(\rho,\omega)$

by applying the change of variables $\tilde{s} = -s$. Next we need to check that $\hat{f} \in L^2(\mathbb{R}^n)$.

$\|f\|_{L^2(\mathbb{R}^n)} = \int_{\mathbb{R}^n)} |\hat{f}(\xi)|^2 d\xi = \int_{S^{n-1}}\int_0^{\infty}\frac{|\mathfrak{F}_sg(\rho,-\omega)|^2}{|\rho|^{n-1}} |\rho|^{n-1} d\rho d\omega$

$= \int_{S^{n-1}}\int_0^\infty |\mathfrak{F}_sg(\rho,-\omega)|^2 d\rho d\omega = \|g\|^2_{L^2(\mathbb{R}\times S^{n-1})}$

So $f$ is in the domain of $T$, and $T$ is an isometric isomorphism from $L^2(\mathbb{R}^n)$ to $L^2_{\text{even}}(\mathbb{R}\times S^{n-1})$.

## The Range of $R$ on the Schwartz Class

Now let’s look at how $R$ acts on another space, $\mathcal{S}(\mathbb{R}^n)$.  We have already seen that $R: \mathcal{S}(\mathbb{R}^n) \rightarrow \mathcal{S}(\mathbb{R}\times S^{n-1})$ is linear and continuous.

So what is $R(\mathcal{S}(\mathbb{R}^n))$?  Is it the set of all even Schwartz functions?

Here is our first surprise.  The range of $R$ is quite a bit more restricted.

In the 1960’s Helgason and Ludwig found the following.  Let $f \in \mathcal{S}(\mathbb{R}^n)$ and consider the moments of $Rf$ in $s$:

$\mu_p(Rf)(\omega) = \int_{\mathbb{R}} s^p Rf(s,\omega) ds$

Then observe that

$\mu_p(Rf)(\omega) = \int_{\mathbb{R}} s^p Rf(s,\omega) ds$

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

But for fixed $x$, $(x\cdot\omega)^p$ is a homogeneous polynomial of degree $p$ in $\omega$.  Our integral is just summing coefficients of homogeneous polynomials, and must be homogeneous of degree $p$ as well.

So if $g$ is in the range or $R$, then $\mu_p(g)(\omega)$ must be a homogeneous polynomial of degree $p$!  Functions in $L^2$ don’t generally have finite moments, so we did not see this restriction there.

It turns out that this is the only real restriction on the range

Theorem [Helgason, Ludwig]

$R(\mathcal{S}(\mathbb{R}^n))$ is

$\{g\in\mathcal{S} | g(-s,-\omega) = g(s,\omega), \mu_p(g) \text{ hom. poly of deg. d in } \omega \}$

We have already seen the necessity of these moment conditions. We’ll get an idea of the sufficiency proof before treating it rigorously.

Say $g \in \mathcal{S}_{\text{even}}(\mathbb{R}\times S^{n-1})$ (We will add the moment conditions later).  We want to find an $f$with $Rf = g$, so with the Fourier Slice Theorem we write

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

It is easy to check that this function is in $\mathcal{S}(\mathbb{R}^n - \{0\})$.  We must check that $\hat{f}$ is smooth at zero.

What is $\hat{f}(0)$?

It is $\mathfrak{F}_sg(0,\omega)$.  But is this independent of $\omega$?

$\mathfrak{F}_sg(0,\omega) = \int g(s, \omega) ds = \int s^0g(s,\omega)ds = \mu_0g(\omega)$

But by our assumption, the 0-th moment is a homogeneous polynomial of degree zero in $\omega$.  This is a fancy way of saying it is a constant.  So by our definition of $f$, $\hat{f}(0)$ is well defined.

So what about $\partial^\alpha \hat{f}(0)$?

First we should ask “what should it be?”

$\hat{f}$ is differentiable at zero if and only if we can approximate it with a linear function:

$\hat{f}(\xi) = \hat{f}(0) + \mathbf{l}\cdot\xi + o(|\xi|)$

Writing this with $\xi = \rho\omega$ we see

$\hat{f}(\rho\omega) = \hat{f}(0) + \rho\mathbf{l}\cdot\omega + o(|\rho|)$

If we differentiate this with respect to $\rho$, we would see that

$\partial_\rho \hat{f}(\rho\omega)\Big |_{\rho = 0} = \mathbf{l}\cdot \omega$

In other words, $\partial_\rho\hat{f}(\rho\omega)\Big |_{\rho = 0}$ should be a homogeneous polynomial of degree one in $\omega$.

Now what is $\partial_\rho \hat{f}(\rho,\omega)$?

$\partial_\rho \hat{f}(\rho\omega) = \int_{\mathbb{R}}\partial_\rho e^{-is\rho}g(s,\omega)ds$

$= \int_{\mathbb{R}}e^{-is\rho}(-is)g(s,\omega)ds$

So

$(\partial_\rho \hat{f})(0) = \int_{\mathbb{R}} (-is)g(s,\omega)ds = -i\mu_1(g)(\omega)$

Which, by our moment condition, is a homogeneous polynomial of degree 1 in $\omega$, exactly what we were looking for.

We can continue like this, expanding the power series of $e^{-i\rho s}$ in the Fourier transform:

$\hat{f}(\rho\omega) = \sum_{n=0}^{\infty}\int_{-\infty}^{\infty}\frac{(-i\rho s)^n}{n!}g(s,\omega)ds$

$= \sum_{n=0}^{N}\rho^nP_n(\omega) + \int_{-\infty}^{\infty}\mathbf{e}_{N+1}(-i\rho s)g(s,\omega) ds$

Where $P_n$ is a homogenous polynomial in $\omega$ of degree $n$ and $\mathbf{e}_{N+1}(x) = \sum_{n = N+1}^{\infty} \frac{x^n}{n!}$ is the exponential function minus the first $N$ terms of its Taylor series. .We have used the moment conditions to get this expansion.

Now

$|\int_{-\infty}^{\infty}\mathbf{e}_{N+1}(-i\rho s)g(s,\omega) ds|$

$\leq \int_{-\infty}^{\infty}|\rho|^{N+1}|s|^{N+1}g(s,\omega) ds = C_{g, N+1}|\rho|^{N+1}$

So using expression we derived above for $\hat{f}(\rho\omega)$ we can write a Taylor series for $\hat{f}$:

$\hat{f}(\rho\omega) = \sum_{n=0}^{N}\rho^nP_n(\omega) + O(|\rho|^{N+1})$

This is not a proof of the theorem yet, and we need more information to make it rigorous (e.g. if we knew $f$ had compact support, we coul;d show that the power series converged and $\hat{f}$ was analytic) .  But it does show how the moment conditions appear naturally in this setting.

We will pursue a more technical, but straightforward proof in the next post.