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. In $\mathbb{R}^2$ consider the function $f(x,y) = \frac{1}{(x + iy)^5}$ for $x,y$ outside some small ball around the origin. Paste it together in that ball to make it continuous. Then the hyperplane integrals in the Radon transform are just line integrals of the complex function $\frac{1}{z}$ . A contour integration argument shows that

$\int_{\ell} \frac{1}{z^5} dz = 0$

So we will certainly need condition (i).

Proof

Step 1: reduction to smooth $f$

First we will argue that it is sufficient to assume that $f$ is smooth. To do this, we mollify $f$ and show that the mollified function still satifies the hypotheses of the theorem, replacing $A$ by $A + \epsilon$ . But if the we prove that mollified version of $f$ has compact support, then $f$ must have compact support as well.

So pick a positive normal bump function $\phi > 0$ supported in $B_{\epsilon}(0)$ . Then $\phi*f$ satisfies the hypotheses of the theorem with $A$ replaced by $A + \epsilon$ :

(i) We write

$|\phi*f(x)| \leq \int|\phi(y)|\sup_{\xi \in B_{\epsilon}(x)}|f(\xi)| dy \leq \sup_{\xi \in B_{\epsilon}(x)} |f(x)|$

which satisfies the bounds of condition (i).

(ii) Compact support:

$R(\phi*f)(\rho, \omega) = \int_{\mathbb{R}^n} \phi(y)Rf(\omega, \rho - y\cdot\omega) dy$

and if $\rho > A+\epsilon$ , this is zero.

Step 2: reduction to radial $f$

Now we will make one more reduction, and show that we only need to prove the theorem for radial functions. So let’s assume that the theorem is true for radial functions.

Let $f \in C^{\infty}(\mathbb{R}^n)$ satisfy the hypotheses of the theorem. Let $dk$ be the left invariant Haar measure on $altex O(n)$ normalized so that $\int_{O(n)}dk = 1$ .

Define

$g_x(y) = \int_{O(n)} f(x + k\cdot y) dk$

This is the average of $f$ over the sphere of radius $|y|$ centered at $x$ .

Then — and this is a key point — $g_x$ is radial.

Now

$Rg_x(\omega, \rho) = \int_{y\cdot \omega = \rho}\int_{O(n)} f(x + k\cdot y) dk dH(y)$

$= \int_{O(n)} Rf_x(k\cdot\omega, \rho) dk$

Which is the average of $Rf_x$ over all hyperplanes distance $\rho$ from $x$ . Here $f_x(y) \equiv f(x+y)$ . If $\rho > A+|x|$ then $B_0(A) \subset B_\rho(x)$ , then $Rg_x(\omega, \rho) = 0$ , because $f_x$ will be identically zero on all hyperplanes at distance $\rho$ .

If our theorem is true for radial functions (to be proven later), then this allows us to deduce that

$g_x(y) = 0$ for $|y| > A + |x|$

This means that for $|y| > A+ |x|$

$\int_{S_r(x)}f(\omega) d\omega = 0$ if $r > A + |x|$

We will show that this implies that $f\Big |_{S_r(x)} = 0$ . Here is the idea: we will perturb the identity to get more identities, then put them together to see that $f$ must vanish on that sphere. This is where the rapid decay condition comes in.

Since $f$ has rapid decay, we know it is at least absolutely integrable, so we can safely write

$\int_{B_r(x)} f(y) dy = int_{B_r(x)} f(y) dy + \int_{\rho > r}\int_{s_\rho(x)} f(\rho\omega) \rho^{n-1}d\omega d\rho = \int_{\mathbb{R}^n} f(y) dy < \infty$

Now differentiate both sides with respect to $x$ :

$0 = \frac{\partial}{\partial x_i}(\int_{B_r(0)}f(x+y) dy)$

$= \int_{B_r(0)}\frac{\partial f}{\partial x_i}(x+y) dy$

$= \int_{B_r(0)}\text{div}[f(x+y)\frac{\partial}{\partial y_i} dy$

$= \int_{S_r(0)}\frac{y_i}{r}f(x+y) d\omega(y)$

where the last step used the divergence theorem (sort of a standard trick). So we see that

$\int_{S_r(0)} y_if(x+y) dy = 0$

We can repeat this sort of argument — using the rapid decay of $f$ – to show that for all polynomials $p$

$\int_{S_r(0)}p(\omega)f(\omega) d\omega = 0$

So $f$ is orthogonal to all polynomials on the sphere (including, e.g. the spherical harmonics), and it must vanish. So $f = 0$ on $S_r(x) \forall x \in \mathbb{R}^n, r > A + |x|$ . Take $x = 0$ to see that $f$ must vanish outside of $B_A(0)$ , proving our theorem.

Step 3: Proof of the result for radial functions

This section of the proof is elementary but a bit technical. I have proposed an alternative proof that I find more satisfying here. It leans on the Paley-Wiener theorem and some basic facts about radial functions. But now let’s proceed with the proof from class.

Assume that $f(x) = F(|x|)$ is a radial function on $\mathbb{R}^n$ . We can write

$Rf(\rho, \omega) = \int_{\langle x,\omega\rangle = \rho} F(|x|) dH = c_n\int_0^{\infty} F(\sqrt{\rho^2 + t^2}) t^{n-2} dt$

Where we are using polar coordinates on the plane $\langle x,\omega\rangle = \rho$ : $\rho$ is the distance from the hyperplane to the origin, and $t$ is the distance of a point in this hyperplane to the “center” of the hyperplane (the point closest to the origin).

Substitute $s = \sqrt{\rho^2 + t^2}$ , which gives $s^3ds = -t dt$ and $t = \sqrt{\frac{1}{s^2} - \rho^2}$ . This gives us the rewriting

$Rf(\rho, \omega) = c_n \int_0^{\frac{1}{\rho}} F(\frac{1}{s})s^3(\frac{1}{s^2} - \rho^2)^{\frac{n-3}{2}} ds$

Set $u = \rho^{-1}$ and get\

$Rf(u^{-1},\omega) = \int_0^u F(\frac{1}{s})s^3(\frac{1}{s^2} - \frac{1}{u^2})^{\frac{n-3}{2}} ds$

$u^{n-3}Rf(u^{-1}, \omega) = \int_0^u F(\frac{1}{s}) s^{-n}(u^2 - s^2)^{\frac{n-3}{2}} ds$

Simplify notation by letting $h(u) = u^{n-3}Rf(u^{-1}, \omega)$ and $g(u) \equiv F(\frac{1}{s})s^{-n}$ . Then

$h(u) = \int_o^u g(s)(u^2 - s^2)^{\frac{n-3}{2}} ds$

Which is similar to the Abel integral equation.

Multiply both sides by something, then integrate:

$\int_0^t h(u)(t^2 - u^2)^{\frac{n-3}{2}} u du = \int_0^t \int _0^u g(s)[(u^2 - s^2)(t^2 - u^2)]^{\frac{n-3}{2}} u ds du$

$= \int_0^tg(s)\int _s^t [(u^2 - s^2)(t^2 - u^2)]^{\frac{n-3}{2}} u du ds$

Now substitute $altex (t^2 – s^2)v = (t^2 + s^2) 2u^2$ to get

$= \int_0^tg(s)(t^2 - s^2)^{n-2}\int_{-1}^{1}(1-v^2)^{\frac{n-3}{2}} dv ds$

So we have pulled out the dependence on $u$ and can write

$\int_0^t h(t)(t^2 - u^2)^{\frac{n-3}{2}} du = c_n \int_0^t g(s)(t^2 - s^2)^{n-2} ds$

Differentiate both sides with respect to $t^2$ , $latex \frac{\partial}{\partial(t^2)} = \frac{1}{2t}\frac{\partial}{\partial t}$. Apply $n-1$ times to both sides to get

$(\frac{\partial}{\partial(t^2)})^{n-1}\int_0^th(u)(t^2 - u^2)^{\frac{n-3}{2}} u du = c_n^{\prime}(n-2)!\frac{g(t)}{2t}$

$g(t) = Ct(\frac{\partial}{\partial (t^2)})^{n-1}\int_0^t h(t)*t^2 - u^2) d du$

Remembering the definitions of $g, h$ this is

$F(t^{-1}) = a(t)\int Rf(u^{-1}, \omega)b(u) du$

So $t^{-1} > A, t < A^{-1}$ implies (since $u < t$ in the integral) $u^{-1} > A \implies Rf(u^{-1}, \omega) = 0$ . So $F(t^{-1}) = 0$ , completing the proof.

Now let’s see a quick corollary.

Corollary Let $B$ be any compact convex set, then if $f$ satisfies condition (i) of the theorem and the integral of $f$ vanishes over any hyperplane outside of $B$ , then $f$ vanishes outside of $B$

Proof The set $B$ is the intersection of all balls containing $B$ . Apply the theorem to see that $altex f$ must vanish outside each of those balls, hence it must vanish outside $B$ .

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This entry was posted on November 3, 2009 at 1:25 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.

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