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

1. $h(-x)=-h(x)$

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 $A$. 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.

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

1. $h(-x)=-h(x)$

2. 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.

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

1. $h(-x)=-h(x)$

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).

Motivation:

Let $\omega$ be a volume 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 $A$. Writing $\omega=h\text{Vol}_{\mathbb{S}^2}$, we obtain the formulation of the question as stated in the beginning.

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.

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

1. $h(-x)=-h(x)$

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 $A$. 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.