(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.

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.

• 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? Jan 21 at 11:25
• 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 $H_n$. Jan 23 at 16:13
• 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. Jan 23 at 16:22
• 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. Jan 25 at 19:51
• 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 Jan 29 at 21:10