2
$\begingroup$

(UPDATED for rapid decay considerations + new question)

In dimension 2, the Radon transform range theorem states that a rapidly decaying (Schwartz) function $g(t,\theta)$ can be represented as a Radon transform of some function $f(x,y)$ (i.e. $g=R[f]$) if and only if, for all integers $n\geq0$ $$P_n(\theta) := \int\limits_{-\infty}^{\infty} t^n g(t,\theta) dt$$ is a homogeneous polynomial of degree $n$ in $\cos\theta$ and $\sin\theta$. This is often referred to as the moment or Cavalieri conditions. See e.g. Helgason's book p.5, Lemma 2.2 (2011 ed.) for the property that $P_n$ must be a homogeneous polynomial of degree $n$.

Question 1: If $f$ is a radial function then its Radon transform $g=R[f]$ is known to be independent from $\theta$. Therefore all moments $P_n(\theta)$ are also independent from $\theta$, in apparent violation of the property that $P_n$ is a homogeneous polynomial in $\cos\theta, \sin\theta$ of degree $n$ when $n\geq1$. What am I missing?

For example, consider $f(x,y) = e^{-x^2-y^2} / \sqrt\pi$. Then $g(t,\theta) = e^{-t^2}$ which is independent from $\theta$ as expected of radial functions. The first moment is 0 which is NOT a homogeneous polynomial of degree 1, the second moment is $\sqrt\pi/2$ which is NOT a homogeneous polynomial of degree 2, and so forth.

Question 2: What happens when the above integral does not converge? Usually this happens when there is no solution to the Radon transform inverse problem, but consider $g(t,\theta) = (1-e^{-1/t^2})/|t|$ which is independent from $\theta$. After calculations, the inversion formula for radial function gives $$f(x,y) = f_0\!\left(\sqrt{x^2+y^2}\right), \qquad f_0(r) = \frac2{\pi r^3}\mathfrak D\!\left(\frac 1r\right)$$ where $\mathfrak D(x) := e^{-x^2}\int_0^x e^{t^2} dt$ is Dawson's function. So there exists $f$ such that $g=R[f]$, and yet $$ P_n(\theta) = \int\limits_{-\infty}^{\infty} t^n \frac{1-e^{-1/t^2}}{|t|} dt $$ does not converge for $n\geq 2$.

My partial answer to Q2: This specific example is not a Schwartz function. Any references to range theorems for non-Schwartz functions appreciated. I found "A Range Theorem for the Radon Transform" (Madych and Solmon, 1988), other suggestions very appreciated.

Thanks! p.

$\endgroup$

1 Answer 1

2
$\begingroup$

To answer Q1: There are trig identities at play. First, 0 is usually accepted under the definition of "homogeneous polynomial" (i.e., it's a polynomial whose coefficients are all zero) so there is no contradiction there. But for the case of the second moment being constant, we have the identity $\sin^2(\theta) + \cos^2(\theta) = 1$, so actually a constant is representable by a homogeneous trig polynomial of degree 2 polynomial, so again, no contradiction.

Edit: for Q2, The moment conditions can indeed be violated when the function is not Schwartz. A good reference for this is the paper:

Solmon, D. C. (1987). Asymptotic formulas for the dual Radon transform and applications. Mathematische Zeitschrift, 195(3), 321-343.

The main theorem of this paper shows that the inverse Radon transform maps any even Schwartz function $\phi$ over $\mathbb{S}^{d-1}\times \mathbb{R}$ to a $C^\infty$-smooth function on $\mathbb{R}^d$ that decays like $O(\|x\|^{-d})$ as $\|x\|\rightarrow \infty$, i.e., absolutely integrable along hyperplanes, but not necessarily absolutely integrable over all of $\mathbb{R}^d$. Here $\phi$ does not have to satisfy any of the moment conditions, even though the defining integrals are convergent since $\phi$ is Schwartz.

$\endgroup$
7
  • $\begingroup$ Thanks for your answer. I realized this later. However: the degree of the zero polynomial is $-\infty$, not 1. In addition, if Pn is considered as a "spherical" homogeneous polynomial subject to trigonometric identity simplifications, how would you define its degree? $\endgroup$
    – phaedo
    Jan 21 at 11:25
  • 1
    $\begingroup$ I don't think there is a uniquely defined degree in this case. But I don't think that matters in defining the moment conditions. Simply take $H_n$ to be the space of all functions realizable as a homogeneous polynomial of degree $n$ evaluated on the sphere. Then the moment conditions say that $P_n$ must belong to $H_n$. $\endgroup$
    – Greg O.
    Jan 23 at 16:13
  • 1
    $\begingroup$ To be more clear, the proof of the moment conditions amounts to the identity: $\int Rf(w,t) t^n dt = \int_{\mathbb{R}^d} f(x) \langle w,x\rangle^n dx$ for all $w = (w_1,...,w_n) \in \mathbb{S}^{d-1}$. The right-hand side may be formally expanded into a homogeneous polynomial of degree $n$ in the variable $w$, and this is what is meant by a "degree n" spherical polynomial. $\endgroup$
    – Greg O.
    Jan 23 at 16:22
  • 1
    $\begingroup$ That's not true, though. For example, $H_1$ as I've defined it does not contain constant functions, since there is no linear combination of $\sin(\theta)$ and $\cos(\theta)$ that gives a constant. Likewise,$H_2$ does not contain $\sin(\theta)$ or $\cos(\theta)$, since there is no linear combination of $\sin^2(\theta)$, $\cos(\theta)\sin(\theta)$, $\cos^2(\theta)$ to give you those functions, etc. $\endgroup$
    – Greg O.
    Jan 25 at 19:51
  • 1
    $\begingroup$ I think that's essentially correct. But, actually, there is a cleaner way to characterize the $H_n$ using spherical harmonics: $H_n$ is the space of harmonic polynomials up to degree $n$, where a harmonic polynomial is any polynomial whose Laplacian vanishes. See Theorem 1.1.3. of Bai and Xu's "Approximation Theory and Harmonic Analysis on Spheres and Balls": link.springer.com/content/pdf/10.1007/978-1-4614-6660-4.pdf $\endgroup$
    – Greg O.
    Jan 29 at 21:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.