I would like to prove that:
Lemma. Let $M\subset\mathbb R^{n+1}$ be a connected and compact $n$-dimensional smooth submanifold (i.e. a hypersurface). Suppose that for every direction, there exists a hyperplane such that $M$ reflected at this hyperplane equals $M$. Or put more formally, for every $v\in\mathbb R^{n+1}$ with $|v|=1$, there exists a constant $c(v)\in\mathbb R$ such that $M$ reflected at $$\{x\in\mathbb R^{n+1}\mid \langle x,v\rangle = c(v)\}$$ is equal to $M$. Then $M$ is a sphere.
It is said that this Lemma is due to Hopf, but I don't know how to prove it.
I found the following paper: ALEKSANDROV’S THEOREM: CLOSED SURFACES WITHCONSTANT MEAN CURVATURE in which Lemma 1.4 is almost identical to my Lemma. But I can't understand the proof. Also, the Lemma 1.4 has the additional assumption that $M$ is a closed hypersurface.
How can I prove this Lemma?
The argument in the article is as follows: we can translate $M$ so that its center $G$ of mass is $0$. So, for any affine hyperplane $H$, if $M$ reflected through $H$ is $M$, then $G$ reflected through $H$ is $G$, so $0=G \in H$, hence $H$ is a “linear” hyperplane.
Let $H$ be any linear hyperplane. There is an affine hyperplane $H’$ parallel to $H$ such that $M$ reflected through $H’$ is $M$. Now, by the above, $0 \in H’$, so $H’=H$, hence $M$ reflected through $H$ is $M$.
Let $x \in M$, $y$ with the same norm. There is a linear hyperplane $H$ such that $x$ reflects to $y$ through $H$. Therefore, $y \in M$. We thus have:
$\forall r \geq 0,\, M \supset rS^n \Leftrightarrow M \cap rS^n \neq \emptyset$.
Let $U=\{|x|,\, x \in M\}$. $U$ is an interval of $[0,\infty)$ and $M=US^n$. Since $M$ has dimension $n$, it has empty interior, so $US^n$ has empty interior. It follows that $U$ must be a point, so $M$ is a sphere.