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Let $\{g_{n}:B_{1}(0) \rightarrow \mathbb{R}^{2}\}_{n\in \mathbb{N}}$ be a sequence of smooth maps where $B_{1}(0)=\{x\in \mathbb{R}^{2} \mid |x|<1\}$ is the unit ball in $\mathbb{R}^{2}$. Assume that $g_{n} \rightarrow \text{id}$ in $C^{\infty}_{\text{loc}}(B_{1}(0))$ for $n\rightarrow \infty$, where $\text{id}:B_{1}(0)\rightarrow \mathbb{R}^{2},x\mapsto x$ is the identity map. Also assume that $g_{n}(0)=0,\forall n$.

What I am trying to show is: $\exists \epsilon > 0, \exists n_{\epsilon}\in \mathbb{N}$ such that $\forall n \geq n_{\epsilon}$: $g_{n}:B_{\epsilon}(0)\rightarrow g_{n}(B_{\epsilon}(0))$ is a bijection.

I know that in dimension $1$ this is true, since monotone function on an interval is bijective. But I don't know if it is also true in dimension $2$. Is this true or is there some counterexample, what is your idea? Hope you can help me out.

Martin

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  • $\begingroup$ Forgot to mention: I also assume that $g_{n}(0)=0,\forall n$. $\endgroup$
    – Martin
    Commented Jan 24, 2015 at 15:27
  • $\begingroup$ What is the definition of $C^\infty_{\mathrm{loc}}$ convergence? $\endgroup$
    – Gabriel C. Drummond-Cole
    Commented Jan 24, 2015 at 15:47
  • $\begingroup$ $g_{n} \rightarrow \text{id}$ in $C_{\text{loc}}^{\infty}(B_{1}(0))$ if: $\forall K \subset B_{1}(0)$ compact and $\forall k \in \mathbb{N}_{0}$ we have $\left \Vert g_{n} - \text{id} \right \Vert_{C^{k}(K)}\rightarrow 0$ as $n \rightarrow \infty$. $\endgroup$
    – Martin
    Commented Jan 24, 2015 at 15:52
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    $\begingroup$ If you look at a proof of the inverse mapping theorem, for instance in Terry Tao's Analysis, you find that the radius of a ball, in which a local inverse exists, depends on the norm of the difference of the differentials, which in your case converges to zero. so your claim follows. $\endgroup$
    – user1688
    Commented Jan 24, 2015 at 16:05
  • $\begingroup$ Can you tell me exactly where to find this in terry tao's book, please? $\endgroup$
    – Martin
    Commented Jan 24, 2015 at 16:11

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First of all in the statement of what you want to show, there is probably a mistake: you want $\forall\epsilon<1$ in the beginning. Anyway, this is true and this is a stronger statement.

Sketch of the proof. What you want to prove is that $g_n$ is injective in $|z|<r$, for a fixed $r$. You have that $g_n(z)=z+h_n(z)$ where $h_n$ is small, together with all derivatives. Write $$|g_n(z_1)-g_n(z_2)|\geq |z_1-z_2|-|h_n(z_1)-h_n(z_2)|.$$ But $h_n$ has small derivative in $|z|<r$, say less than $\delta<1$. So the RHS is greater than $|z_1-z_2|(1-\delta)\neq 0$. This proves your statement.

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