Let $h:\mathbb{S}^2 \to \mathbb{R} $ be a smooth function satisfying:

$h(-x)=-h(x)$

For every hemisphere $A \subseteq \mathbb{S}^2$, $\int_{A}h\text{Vol}_{\mathbb{S}^2}=0$, where $\text{Vol}_\mathbb{S}^2$ is the standard volume form on the sphere.

**I am interested in finding an elementary proof that $h=0$.** (Without relying on the invertibility of the Funk transform, see details below on the connection of this to the problem above).

**Edit:** As commented by Alex Degtyarev, if we assume $h$ is everywhere non zero, then it's trivial. (In that case $h$ has constant sign, so assumption $1$ alone immediately implies $h=0$).

*Motivation:*

Let $\omega$ be a 2-form on $\mathbb{S}^2$ with the property that the induced area of all the hemispheres is the same.

I want to find an elementary proof that $\omega$ is invariant under the antipodal map, i.e $f^*\omega=\omega$, where $f(x)=-x$.

Denote $$V=\{ \omega \in \Omega^2(\mathbb{S}^2) \, | \, \int_{A}\omega=\int_{A}f^*\omega \, \text{ for every hemisphere $A \subseteq \mathbb{S}^2$} \},$$

$$W=\{ \omega \in \Omega^2(\mathbb{S}^2) \, | \, \omega=f^*\omega \}.$$

We want to prove $V \subseteq W$. Let $\omega \in V$, and define $\tilde \omega:=\omega-f^*\omega$. Since $V$ is a vector space, closed under the operation $\omega \to f^*\omega$, we have $\tilde \omega \in V$. Note that $f^*\tilde \omega=-\tilde \omega$, and that we need to show $\tilde \omega=0$.

Thus, the problem is equivalent to the following:

Let $\omega \in V$, satisfy $f^*\omega=-\omega$. Then $\omega=0$.

The assumptions imply $\int_A \omega=0$ for every hemisphere $A$. Writing $\omega=h\text{Vol}_{\mathbb{S}^2}$, we obtain the formulation of the question as stated in the beginning.

**Edit:** If we assume $\omega$ is non-degenerate (i.e everywhere non-zero), then the question becomes trivial: In that case $h$ has a constant sign, hence must be zero due to the property $h(-x)=-h(x)$.

It turns out that using flows by Killing fields, one can reduce this problem to the invertibility of the Funk transform, but this is a non-elementary result which I prefer to avoid.

(Essentially the idea is that if $\int_A\omega=0$ on any hemisphere, then $\int_{A}L_X\omega=0$ for every Killing field $X$). For details see here and here.

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