Prove: $h \in T_xM \iff \operatorname{dist}(x+\epsilon h,M) = o(\epsilon)$ For $M$ a smooth manifold in $\mathbb{R}^n$, $h\in\mathbb{R}^n$, $x \in M$. I know that $T_xM = \operatorname{Im}(Df) = \ker(Dg) $ where $f$, $g$ are a chart and co-chart at point $x$, respectively. I just don't see how I could relate the definition to what I need to prove.
Any help would be appreciated!
Here you go:
$ " \Rightarrow " $ Let $\gamma: \mathbb{R} \to M$ be a curve s.t. $\gamma(0)=x$ and $\gamma'(0)=h$. You can always find such a curve. Now you have:
$$ \operatorname{dist}(x+\epsilon h,M) \leq \operatorname{dist}(x+\epsilon h,\gamma(\epsilon)) $$
So what we want to show is that
$$ \lim_{\epsilon \to 0} \frac{\operatorname{dist}(x+\epsilon h,\gamma(\epsilon)) }{\epsilon}=0 $$
We can rewrite
\begin{eqnarray} \frac{\operatorname{dist}(x+\epsilon h,\gamma(\epsilon)) }{\epsilon}&=&\frac{\Vert\gamma(\epsilon)-\epsilon\gamma'(0)-\gamma(0) \Vert }{\epsilon}\\ &\leq & \left\Vert\frac{\gamma(\epsilon)-\gamma(0) }{\epsilon}- \gamma'(0) \right\Vert \end{eqnarray} Now letting $\epsilon \to 0$ we obtain that
$$ \lim_{\epsilon \to 0}\frac{\gamma(\epsilon)-\gamma(0) }{\epsilon}= \gamma'(0) $$
This shows our claim.
"$\Leftarrow$" We want to show that $h \in \ker(Dg_x)$. Choose a domain of definition $V \subset U$ for $g$ s.t. $\overline{V}\subset U$ and $\overline{V}$ is compact in $U$. By this we can assume that $g$ is Lipschitz continous with Lipschitz constant $L$.
Now let $\gamma(\epsilon):=x+\epsilon h$, notice that $\gamma(0)=x$ and $\gamma'(0)=h$. Furthermore let $\phi:\mathbb{R}\to M$ be the function that associates to $\epsilon$ one of the $y \in \overline{V} \cap M$ with $\operatorname{dist}(\gamma(\epsilon),M)=\operatorname{dist}(\gamma(\epsilon),y)$. Notice that as long as we restrict ourselfs to $\epsilon$ that are small enough this is well defined (You maybe want your $\epsilon$ so small that there is a ball $B$ around $\gamma(\epsilon)$ which contains $x$ and s.t. $B\subset V$ .). Also note that $\phi$ is not continous but just a function.
Now notice that $g(\gamma(0))=g(\phi(\epsilon))=0$. By the chain rule one gets:
\begin{eqnarray} \Vert Dg_x(h) \Vert &=& \Vert Dg_x(\gamma'(0))\Vert\\ &=& \left\Vert \frac{d}{d\epsilon}g(\gamma(\epsilon))_{\vert\epsilon=0}\right\Vert\\ &=& \lim_{\epsilon\to 0}\left\Vert \frac{g(\gamma(\epsilon))-g(\gamma(0))}{\epsilon} \right\Vert\\ &=&\lim_{\epsilon\to 0}\frac{1}{\epsilon}\left\Vert g(\gamma(\epsilon))-g(\phi(\epsilon)) \right\Vert\\ &\leq&\lim_{\epsilon\to 0}\frac{1}{\epsilon} L \left\Vert \gamma(\epsilon)-\phi(\epsilon) \right\Vert\\ &=& L\lim_{\epsilon\to 0}\frac{\operatorname{dist}(\gamma(\epsilon),M)}{\epsilon} \end{eqnarray} But this is zero by assumption. So all in all we get $Dg_x(h)=0$ which finishes the proof.