Is the extrinsic metric and intrinsic metric equivalent if second fundamental form is bounded

Question

Let $M$ be a hypersurface in $\mathbb{R}^{n+1}$ with bounded second fundamental form $|A|\leq C$. Does intrinsic distance satisfy $d_g(p,q)\leq C'|p-q|$, where $C'$ only depends on $C$. Here $d_g(p,q)$ is the intrinsic metric on $M$, and $|p-q|$ is the extrinsic metric in $\mathbb{R}^{n+1}$.

Answer 1

No, this is not true.

For $L>0$, consider two lines in $\mathbb{R}^2$:

$\{(x,0) : 0 \leq x \leq L\}$

and

$\{(x,2) : 0 \leq x \leq L\}$.

These can be "capped" at either end with half-circles of radius $1$. This produces a closed curve in $\mathbb{R}^2$. If you want, you can smooth things to make the curve $C^\infty$ or even real analytic.

Notice that we can smooth the curves so that their curvature satisfies $|\kappa| \leq 2$, say. Importantly, we can do this independently of $L$.

Now, the points $(L/2,0)$ and $(L/2,2)$ are a distance $2$ apart extrinsically, but as $L\to\infty$, the intrinsic distance goes to infinity,