Lipschitz maps between Riemannian manifolds

Let $U \subset \mathbb R^n$ and $V \subset \mathbb R^m$ be two open subsets with $U$ convex and $f: U \to V$ a $C^1$-map. Then, using the fundamental theorem of calculus among other things, one can show that the following are equivalent:

1) there exists $M \in \mathbb R$ such that $\forall x,y \in U$: $|f(y)-f(x)| \leq M|y-x|$.

2) $\forall x \in U: ||Df(x)|| \leq M$.

Now let $(U,g)$ and $(V,h)$ be two Riemannian manifolds and $f: U \to V$ a $C^1$-map. I want to know in how far the above equivalence can be extended in this situation. It seems that I cannot simply do the same proof. Any hint would be appreciated. Thanks!

Solution 1:

The result can be generalized to the following: Let $(M, g), (N, h)$ be Riemannian manifolds, $M$ connnected, and $f: M\to N$ a $C^1$ maps. Then $d_N( f(x), f(y)) \le C d_M(x, y)$ for all $x, y\in M$ if and only if $\| Df (x)\|\le C$ for all $x\in M$.

We assume $C>0$, the case $C=0$ is easy.

Proof$: (\Leftarrow)$ Let $x, y\in M$. Let $\epsilon >0$. Then by definition of $d_M$, there is a differentiable path $\gamma : [0, 1]\to M$ so that $\gamma(0) = x, \gamma(1) = y$ and $\int_0^1 |\dot\gamma (t) |dt \le d_M(x, y) + \epsilon$. Then the curve $f\circ \gamma$ connects $f(x)$ and $f(y)$. Thus

$$d_N(f(x), f(y)) \le \int_0^1 \left| \frac{\partial}{\partial t}(f\circ \gamma)\right|dt = \int_0^1 |Df (\dot \gamma)| \le C(d_M(x, y) +\epsilon).$$

Take $\epsilon \to 0$ and we are done.

$(\Rightarrow)$. Assume the contrary that $\|D f\| > C$ at some point $x\in M$. That is, there is $v\in T_xM$ so that $\|v\|=1$ and $\| Df(x) (v)\| >C$. Let $\delta >0$ be small so that $x(t) = \exp_x(tv)$ is defined for $t\in [0, \delta]$, as $\|Df\|$ is continuous, we have

$$\|Df(x(t)) (\dot x(t))\| > C + \epsilon_0$$

for all $t \in [0, \delta]$ and for some fixed $\epsilon_0$.

Thus we have for all $s\in [0, \delta]$,

$$ \int_0^s \|\dot\gamma (t)\| dt> (C+\epsilon_0) s,$$

where $\gamma(t) = f(x(t))$. Note that when $s$ is small enough, $s = d_M(x, x(s))$. So

$$ (*)\ \ \ \ \int_0^s \|\dot\gamma (t)\| dt> (C+\epsilon_0) d_M(x, x(s)).$$ Now we use the following fact

Let $\gamma(t)$ be a $C^1$-curve starting at $y$ and $\|\dot\gamma(0)\|\neq 0$. Then $$\lim_{t\to 0} \frac{d(y, \gamma(t))}{\int_0^t |\dot\gamma|} = 1.$$

Using this and $(*)$, we can find $s_0$ small so that

$$d_N(f(x), f(\eta(s_0))> (C+\epsilon_0/2) d_M(x, \eta(s_0))$$

and hence in required inequality is not satisfied.