The Lipschitz constant of convex sphere in $\mathbb{R}^3$

Question

Is every convex sphere (in the sense of Alexandorff, which is the boundary of some convex body in $\mathbb{R}^3$) with Alexandorff curvature $\geq 1$, admitting a bijective map to the unit round sphere in $\mathbb{R}^3$ with Lipschitz constant $\leq 1$?

Answer 1

The answer is positive, without the assumption on curvature (of course we do have curvature $\ge 0$, for any convex). Let $o$ be an interior point of the body $B$ whose boundary $\partial B$ is your convex sphere, and let $r$ be small enough that $S(o,r) \subset \operatorname{int}(B)$. Consider the radial projection $\mathbb{R}^3\to B(o,r)$, which is $1$-Lipschitz. All we have to prove is that its restriction $\pi$ from $\partial B$ to $S(o,r)$ is invertible and has Lipschitz inverse.

Every ray issued from $o$ interects $\partial B$, so that $\pi$ is onto. If such a ray did intersect $\partial B$ in several point, then $o$ would have to be a boundary point, which is excluded, so that $\pi$ is one-to-one.

Now, that $\pi^{-1}$ is not Lipschitz is equivalent to say that we can find rays $\gamma$ issued from o and support hyperplanes $H$ at the intersection point $\gamma\cap \partial B$ making arbitrarily small angle. If this where the case, taking a sequence $(\gamma_n,H_n)_n$ with angle going to $0$, we could extract a convergent subsequence. This would give in the limit a ray $\gamma_\infty$ and a support hyperplane $H_\infty$ such that $\gamma_\infty \subset H_\infty$. Then $o\in \gamma_\infty\subset H_\infty$ in contradiction with $o$ being an interior point.