Immersion on a compact set Let $K\subset \mathbb R^n$ be compact and convex. Let $U$ be an open neighborhood of $K$ and $f : U\rightarrow \mathbb R^m$ an injective immersion. Must there exist constants $a,b >0$ such that if $x,y\in K$ and $|x-y| < a$, then $|f(x) - f(y)| \geq b |x-y|$? This seems intuitively like it should be true, since the differential of $f$ is bounded below and $K$ is compact. I couldn't find a way to make it precise though. Does anyone have a proof (or counterexample)?
 A: Argue by contradiction. Suppose there are sequences $x_n$ and $y_n$ in $K$ such that $\displaystyle \lim_{n\to\infty } \frac{|f(x_n)-f(y_n)|}{|x_n-y_n|} =0$. By compactness of $K$ we can extract convergent subsequences; after relabeling, $x_n\to x $ and $y_n\to y$. There are two cases: 


*

*$x\ne y$. Then $f(x)-f(y)=0$, contradicting the injectivity of $f$. 

*$x=y$. Since $Df(x)$ is an injective linear map, the constant rank theorem implies that in a neighborhood of $x$ the map $f$ is a composition of an $n$-dimensional diffeomorphism, injective linear map, and an $m$-dimensional diffeomorphism. All three of these satisfy the desired lower bound, and so does their composition. 

A: I don't think a lower bound on the derivative can be used to get such an estimate, because of the following example. Let $U=(0,2\pi)$ and let $f:U\to \mathbb{R}^2$ be a unit speed parametrisation of the circle minus a point. If $K=[\epsilon, 2\pi-\epsilon]$ for infinitesimal $\epsilon>0$, the endpoints of $K$ are mapped to points that are infinitely close. This will force $b$ to be infinitesimal.
