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Let $g: A \subset \mathbb{R}^n \to \mathbb{R}^n$ be an injective continuously differentiable function such that $\forall x \in A, \det g'(x) \neq 0$. Can I say that $g$ is locally bounded? By "$g$ is locally bounded", I means "for all $x \in A$, there is a neighborhood $U$ of $x$ in $A$ and a $M > 0$ such that for all $u \in U$, $|g(u)| < M$".

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What hypotheses do we have on $A$? In particular, is $A$ closed? –  Alex Becker Jul 19 '12 at 2:08
    
$A$ is open. It is in general an implicit assumption, since $g$ is differentiable on $A$. –  João Júnior Jul 19 '12 at 2:10
    
I can't think of a counterexample. Maybe it can be proved by Implict Function Theorem, and/or Mean Value Theorem... –  João Júnior Jul 19 '12 at 2:16
    
I wanted to say Inverse Function Theorem. –  João Júnior Jul 19 '12 at 3:32

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up vote 2 down vote accepted

Any continuous function $g:A\to\mathbb R^n$ is locally bounded. This is because for any $x\in A$, we have some open ball $B_r(x)\subset A$ and thus the closed ball $C:=\overline{B_{r/2}(x)}\subset A$ as well. Since any closed ball around a point in $\mathbb R^n$ is compact, we have that $g|_C$ is uniformly continuous, as for any $\epsilon>0$ we have open balls $B_i$ such that $x,y\in B_i\implies \|g|_C(x)-g|_C(y)\|<\epsilon$ which form an open cover of $C$ hence there exists a finite subcover, so if we let $\delta$ be the minimum of their radii we have $\|x-y\|<\delta\implies \|g|_C(x)-g|_C(y)\|<\epsilon$. Letting $\epsilon=1$ we see that $g|_C$ varies by at most $r/\delta$ over $B_{r/2}(x)$ thus $g$ is bounded on a neighborhood of $x$.

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Thank you, Alex. –  João Júnior Jul 19 '12 at 2:27

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