# Injectivity of the Funk transform for nonsmooth functions

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 $n$).

• Is it even clear $Ff$ is well defined for $f\in L^1$? (What happens if I change $f$ on a null set?) – Christian Remling Jul 8 '15 at 22:29
• @ChristianRemling, it should be enough to have $Ff(\gamma)$ well defined for almost every $\gamma$. Changing $f$ in a null set may change some values of $Ff$, but only in a null set of great circles. Analyzing the Radon or X-ray transform (where one integrates over lines or hyperplanes of $\mathbb R^n$) of $L^p$ functions or even distributions has been done. Of course, the proof should come with a suitable weak definition of the Funk transform, but I don't think this is a big issue. – Joonas Ilmavirta Jul 9 '15 at 8:55
• @JoonasIlmavirta Do you have a recommended reference for the injectivity of the Funk transform on the space of smooth even functions where $n=3$? (There is always the original paper I guess, but perhaps you know a better presentation). Thanks. – Asaf Shachar Oct 31 '17 at 6:42
• @AsafShachar See Helgason's The Radon Transform, Theorem 1.7, Section 1B, Chapter III. I don't know if anyone has published a more readable proof. The transform is also discussed in some detail in section 4 of this paper (also on arXiv). – Joonas Ilmavirta Oct 31 '17 at 16:37

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.

• If I understood correctly, for smooth functions $\int_{S^{n-1}}fFg=\int_{S^{n-1}}gFf$ and it is this identity that gives an easy definition by duality, and $F$ is continuous for smooth functions and distributions. Then a distribution in the kernel can be approximated by smooth functions in the kernel, whence the distribution has to be odd. Do you have a reference for the tools you used (about $SO(n)$-finite functions)? (Side note: Your $F$ agrees with my $F$ only for $n=3$, but your result is enough to give what I want for $L^1$ functions, so that's fine.) – Joonas Ilmavirta Jul 10 '15 at 16:51
• 1) Your equality is correct. 2)I will try to find a reference. 3) I think your $F$ and my $F$ agree for any $n$. – MKO Jul 10 '15 at 19:11
• It seems to me that you are integrating over a sphere of codimension one but I'm integrating over a sphere of dimension one (a circle). That is, your $F$ is the Radon transform and mine is the X-ray transform. If you are using circles instead, how do you parametrize them with $S^{n-1}$? – Joonas Ilmavirta Jul 10 '15 at 19:20
• You are right, sorry. I have made appropriate corrections in the answer. In fact the whole argument works even in greater generality if you integrate over great spheres of any dimension $0<k<n-1$. – MKO Jul 11 '15 at 6:44
• Excellent, thank you! I will check the book as soon as I can lay my hands on it. – Joonas Ilmavirta Jul 11 '15 at 12:53