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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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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