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I'm having difficulty with the following question which was given to me following studying the inverse mapping theorem.

Let $U\subseteq\mathbb{R}^{n}$ be an open set and let $f:U\to\mathbb{R}^{n}$ be injective and differentiable in $U$ , assume also $f\left(U\right)$ is an open set and let $g:f\left(U\right)\to U$ be the inverse of $f$ . Prove that if $g$ is Lipschitzian then it is differentiable.

I'm assuming the main thing I'm missing is how to use the condition that $f(U)$ is open.

Help would be most appreciated!

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Please fix the typo in your title in case anyone is searching. –  Amzoti Dec 22 '12 at 13:14
    
Done, thanks for pointing it out to me :) –  Serpahimz Dec 22 '12 at 13:17

1 Answer 1

up vote 3 down vote accepted

In fact, due to invariance of domain, if $U\subset\mathbb{R}^n$ is open and if $f:U\to\mathbb{R}^n$ is injective and continuous, then $f(U)$ is open.

Given $x\in U$, let us show that $g$ is differentiable at $y=f(x)$. Since $f$ is differentiable, there exist $L_x>0$ and $r_x>0$, such that if $\|x'-x\|\le r_x$, then $x'\in U$ and $\|f(x')-y\|\le L_x\|x'-x\|$. Similarly, since $g$ is Lipschitz, there exist $L_y>0$ and $r_y>0$, such that if $\|y'-y\|\le r_y$, then $y'\in f(U)$ and $\|g(y')-x\|\le L\|y'-y\|$.

Let $r=\min(r_x, L_x^{-1}r_y)$. If $\|x'-x\|\le r$, then for $y'=f(x')$, $\|y'-y\|\le r_y$, and hence $$\|x'-x\|=\|g(y')-x\|\le L_y\|y'-y\|=L_y\|f(x')-y\|.$$ It implies that $f'(x)$, the Jacobi matrix of $f$ at $x$, is invertible, i.e. $\det f'(x)\ne 0$. Then it is easy to check that $$\limsup_{y'\to y}\frac{\|g(y')-x- f'(x)^{-1}(y'-y)\|}{\|y'-y\|}\le\|f'(x)^{-1}\|\cdot\lim_{x'\to x}\frac{\|f(x)-y- f'(x)(x'-x)\|}{\|x'-x\|}=0,$$ i.e. $g$ is differentiable at $y$ and $g'(y)=f'(x)^{-1}$.

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There's one part I don't get, you said that "Since $f$ is differentiable, there exist $Lx>0$ and $rx>0$, such that if $∥x′−x∥≤rx$, then $x′∈U$ and $∥f(x′)−y∥≤Lx∥x′−x∥$. I'm not sure how to derive that from the definition of differntiability that I'm familiar with which is that $f$ is differentiable in x if there's a linear operator such that: $\lim_{h\to0}\frac{f\left(x\right)-f\left(x^{'}\right)-D_{f}\left(x\right)\left(‌​x-x^{'}\right)}{\left\Vert h\right\Vert }=0$ –  Serpahimz Dec 22 '12 at 14:33
    
That limit should be: $$\lim_{x^{'}\to x}\frac{f\left(x\right)-f\left(x^{'}\right)-D_{f}\left(x\right)\left(x-x^{'}\rig‌​ht)}{\left\Vert x-x^{'}\right\Vert }=0$$ –  Serpahimz Dec 22 '12 at 14:42
    
For some reason I can't get the LaTeX to display properly there.. –  Serpahimz Dec 22 '12 at 14:47
    
@Serpahimz: If you rephrase your limit in the definition of differentiablity in $\epsilon-\delta$ language, cannot you see the existence of $r_x$ and $L_x$ ? Note that both of them depend on $x$. –  23rd Dec 22 '12 at 15:54
    
Hmm, I'll give it a try. Just one final question, after some additional thought I think I don't actually understand why the inequality $∥x′−x∥=∥g(y′)−x∥≤Ly∥y′−y∥=Ly∥f(x′)−y∥$ Shows that the Jacobi Matrix is invertible at $x$ Thanks a lot for your help Richard! –  Serpahimz Dec 22 '12 at 16:00

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