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Let $S\subseteq\mathbb{R}^3$ a closed surface and let $X\in\mathfrak{X} (S)$ a vector field on $S$ such that $\mid\mid X_p\mid\mid \le M$ $\forall p\in S$ for some constant $M>0$. Prove that $X$ is complete.

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For 'closed' surface you mean a compact surface without boundary or a surface (with or without boundary) closed as a subset of $ R^3 $? Note the first case is a particular case of the second one. – user55449 Feb 16 '13 at 11:55
probably you mean the second option since you are assuming boundness of the norm of $ X $, that in the first case would be guarantied by compactness – user55449 Feb 16 '13 at 11:58
Yes, I mean that $S$ is a surface (without boundary) closed as a subset of $\mathbb{R}^3$. – Frankenstein Feb 16 '13 at 12:11
up vote 1 down vote accepted

Let $p$ be a point in $S$.

By Cauchy-Lipschitz theorem, let $c:]a,b[\rightarrow S$ be the unique maximal solution curve to $X$ with initial condition $c(0)=p$.

You want to prove that $]a,b[=\mathbb R$. So assume that $b\neq +\infty$.

First, you can see that the derivative of $c$ is bounded by $M$ and by the mean value inequality you get that $c$ is $M-$Lipschtizian.

Next, show that if $(t_n)$ is a sequence in $]a,b[$ converging to $b$ then the sequence $(c(t_n))$ is a Cauchy sequence in $\mathbb R^3$ and then it converges in $\mathbb R^3$.

Prove that this limit does not depend on the sequence $(t_n)$. Then if we denote this limit by $\alpha$, prove that $\alpha$ is still in $S$. (Here you will need that $S$ is closed in $\mathbb R^3$).

It finally means that you can (smoothly) extend the curve $c$ to $]a,b]$ with $c(b)=\alpha$: a contradiction.

One can do the same if $a\neq -\infty$.

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