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 $u \in U$, $|g(u)| < M$".
Sufficient conditions for a function $g: A \subset \mathbb{R}^n \to \mathbb{R}^n$ to be locally bounded
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analysis
multivariable-calculus
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0I wanted to say Inverse Function Theorem. – 2012-07-19
1 Answers
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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 $x$.