Radon transform range theorem and radial functions

Question

(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.

Answer 1

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.