Suppose $M \subset \mathbb R^d$ is a $C^2$ $(d-1)$-manifold. In particular I am interested in when $M$ is the set $\{x \in \mathbb R^d: f(x) \le 1\}$ for some $C^2$ function $f:\mathbb R^d \to \mathbb R$. Suppose also the curvature at each point at least $m \ge 0$. There are loads of equivalent definitions of this. For example
For every twice-differentiable path $\gamma:(0,1) \to M$ with all $\|\gamma'(t)\| =1 $ we have $\|\gamma''(t)\| \ge m$ for all and $t \in (0,1)$. Here $\|\cdot\| $ is the Euclidean norm.
For each point $x \in M$ if we locally represent the manifold as the graph of a function $F:\mathbb R^{d-1} \to \mathbb R$ that is minimised at $x$ then the Hessian $H$ of $F$ at $x$ has $v^T H v \ge m \|v\|^2$ for all $v$ in the tangent space at $x$.
The curvature at $x$ describes how quickly the unit normal $N_y$ changes as we vary the basepoint $y$ in a small region about $x$. Thus one would expect we could bound $\|N_a-N_b\|$ from below by integrating the infinitesimal change along a path from $a$ to $b$.
Suppose $M \subset \mathbb R^d$ is a $C^2$ $(d-1)$-manifold with curvature at each point at least $m \ge 0$. Is there $C \ge 0$ such that $\|N_x-N_y\| \ge Cm \|x-y\|$ for all $x,y \in M$ with unit normals $N_x,N_y$?
I suspect if there is no uniform bound on the curvature this is impossible, as $N_x$ might vary in one direction and then back the other direction, resulting in $N_a =N_b$. But if we have $m \ge 0$ then $M$ is the boundary of a convex set and it's hard to imagine this happening.
For the special case when $M =\{x \in \mathbb R^d: f(x) \le 1\}$ with $\|\nabla f(x)\|=1$ for $x \in M$ then we have $N_x = \nabla f(x)$ and I think a proof might look something like this:
For any $x,y$ let $\gamma:[0,1]$ be a geodesic from $y$ to $x$ with constant speed $s$ equal to the geodesic distance between the points. Since $\nabla f(x)$ has derivative $\nabla ^2f(x)$ some version of the fundamental theorem of algebra gives
$$\nabla f(x) - \nabla f(x) = \int_{0}^1 \nabla^2 f(\gamma(t)) \gamma'(t) dt. $$ Take squares to get $ \|\nabla f(x) - \nabla f(x)\|^2 $ equal to $$ \left(\int_{0}^1 \nabla^2 f(\gamma(t)) \gamma'(t) dt \right) \cdot \left( \int_{0}^1 \nabla^2 f(\gamma(t)) \gamma'(t) dt \right) $$
what I would like to do now is say the right-hand side is at least $$ \int_{0}^1 \Big(\nabla^2 f(\gamma(t)) \gamma'(t) \Big) \cdot \Big (\nabla^2 f(\gamma(t)) \gamma'(t) \Big)dt $$ $$ = \int_{0}^1 \gamma'(t) \nabla^2 f(\gamma(t)) \nabla^2 f(\gamma(t)) \gamma'(t) dt.$$
By the Hessian condition the integrand should be at least $m^2\|\gamma'(t)\|^2 = m^2 s^2 \ge m^2 \|x-y\|^2$. Now take square roots to get $\|\nabla f(x)-\nabla f(y)\| \ge m \|x-y\|$.
This only works in a very special case, and the inequality step is just wishful thinking. But I have been staring at this for far too long and am making no progress. Would someone with more geometry experience mind taking a look?
