Injectivity of the Funk transform for nonsmooth functions

Question

Let $S^{n-1}$ be the unit sphere in $\mathbb R^n$ and $\Gamma_n$ the collection of great circles on it. Assume $n\geq3$. The Funk transform of a function $f:S^{n-1}\to\mathbb R$ is a map $Ff:\Gamma_n\to\mathbb R$ defined by $$ Ff(\gamma)=\int_\gamma f \,ds. $$ It is clear that $F$ annihilates all odd functions. If we restrict our attention to smooth functions $f$, then this is the whole kernel of $F$.

It is a classical result (due to Funk) that if $n=3$, the Funk transform $Ff$ indeed determines the even part of $f\in C^\infty(S^2)$. If $n\geq3$, we can reduce the problem to this result by looking at $S^2$ — the sphere $S^{n-1}$ is a union of totally geodesic copies of $S^2$ in an obvious way, and we can recover the even part separately on each of these $S^2$s.

Assuming smoothness looks like a technical convenience rather than a necessary assumption, and I believe the result to remain true with much less regularity. Where could I find a (citable) proof of injectivity of the Funk transform for even $L^1$ functions? If $L^1$ is to much, other spaces are welcome, too, but I would like the space to include piecewise continuous functions.

If $f\in L^1$, then $Ff(\gamma)$ is not pointwise well defined; changing $f$ in a null set can change $Ff(\gamma)$. Therefore the Funk transform has to understood in a weaker sense. Changing $f$ in a null set should only affect $Ff(\gamma)$ for a null set of $\gamma$s (for a suitable measure on $\Gamma_n$), so that $Ff$ is well defined almost everywhere. The Radon and X-ray transforms (where one integrates over hyperplanes or lines in $\mathbb R^n$) have been studied for very low regularity (see eg. these lecture notes), so one should be able to study the Funk transform of $L^1$ functions as well.

It is possible that there is a nice duality argument showing that if $F$ is injective on even smooth functions, then it is injective on even distributions, but I don't recall seeing one. I know such an argument for Lie groups, but $S^{n-1}$ is only a symmetric space (for most $n$).

Answer 1

One can define the Radon transform $F$ on the space of generalized functions $F\colon C^{-\infty}(S^{n-1})\to C^{-\infty}(Gr_{n,2})$ by duality to the space of smooth functions, where $Gr_{n,2}$ denotes the Grassmannian of 2-planes in $\mathbb{R}^n$. (In particular $F$ is defined on $L^1$.) (Here I am using the identity $$\int_{S^{n-1}}fG(g)=\int_{Gr_{n,2}}F(f)g,$$ where $G$ is the Radon transform from $Gr_{n,2}$ to $S^{n-1}$ and $f,g$ are infinitely smooth functions.)

The claim is that the kernel of $F$ on generalized functions (resp. $L^1$ functions) is equal precisely to odd generalized functions (resp. odd $L^1$ functions).

To prove this, notice that the group $SO(n)$ intertwines $F$. Hence the kernel of $F$ of $SO(n)$-invariant closed subspace (in both weak* and strong topologies). It is well known that

1)for such a subspace the space of $SO(n)$-finite functions is dense (a vector in a representation is called $SO(n)$-finite if it is contained in a finite dimensional $SO(n)$-invariant subspace). (see Helgason's book "Groups and geometric analysis", Ch. IV, Lemma 1.9.)

2) All $SO(n)$-finite functions in $C^{-\infty}(S^{n-1})$ are smooth. (This follows e.g. from Ex. 1(i) on p. 481 in the above book by Helgason.)

(Actually both statements (1) and (2) are true for any homogeneous space of a compact Lie group).

But we know that smooth functions from the kernel of $F$ are odd. Hence their closure in the generalized functions coincides with the odd functions.