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I am trying to prove the Inverse Function Theorem in a version different of the classical, for strongly differentiable functions, whose definition is:

Definition: A function $f:U \rightarrow \mathbb{R}^n$, $U$ is a open of $\mathbb{R^m}$, is said strongly differentiable in $a \in U$ if there exists a linear transformation $T: \mathbb{R}^m \rightarrow \mathbb{R}^n$ such that $$f(x) - f(y) = T \cdot (x-y) + r_a(x,y)|x-y|,$$ for all $x,y \in U$ and such that $\displaystyle \lim_{(x,y) \rightarrow (a,a)} r_a(x,y) = 0$.

The classic version of Theorem proves the differentiability of the homeomorphism inverse. In this context I need to prove the strong differentiability of the homeomorphism inverse, ie, the next lemma:

Lemma: Let $f:U \rightarrow V$ it is a homeomorphism, where $U$ and $V$ are open of the $\mathbb{R}^m$. If $f$ is strongly differentiable in $a \in U$ and $f'(a): \mathbb{R}^m \rightarrow \mathbb{R}^m$ is a isomorphism, then $f^{-1}$ is strongly differentiable in $b = f(a)$.

In my reference, the author presents a lemma to prove this result, but I think it is more direct, like the classic version, but I am unable to prove it.

What does this concept of "strongly differentiable"? Was it as if he were going to prove the theorem on a point?

Thank you for your help.

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    $\begingroup$ Hi, I think this is a good topic, but I'm not sure what your question is. Are you asking for a proof of the lemma? But you were saying you are following a reference - what is the reference? Does the reference give a proof of the lemma or not? Or do you already have the proof, but you're asking what is the purpose of formulating the inverse function theorem in terms of strong differentiability? $\endgroup$ – echinodermata Oct 3 '20 at 23:22
  • $\begingroup$ What I wanted was proof of the lemma. But taking advantage of the opportunity, what is the purpose of formulating the inverse function theorem in terms of strong differentiability? This is not so clear to me, it seems to be defining the theorem on time, something like that. $\endgroup$ – Croos Nov 5 '20 at 18:19
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Proof of the Lemma: We assume $f$ is strongly differentiable at $a.$ Let $T=f'(a).$ Because $T$ is an isomorphism, there exist constants $0<c<C$ such that $c|x|\le |Tx|\le C|x|$ for all $x\in \mathbb R^m.$ So

$$|f(x)-f(y)| \ge |T(x-y)|-|r_a(x,y)||x-y|$$ $$\ge (c/2)|x-y|$$

for $(x,y)$ near $(a,a).$

We want to show

$$\tag 1 f^{-1}(u)-f^{-1}(v) - T^{-1}(u-v) = r_b(u,v)|u-v|$$

for $u,v\in V,$ where $r_b:V\times V\to \mathbb R^m$ and $r_b(u,v)\to 0$ as $(u,v)\to (b,b).$

Now $f$ is a homeomorphism, so we can make the change of variables $u=f(x),v=f(y),$ where $x,y\in U.$ The left side of $(1)$ is then

$$ f^{-1}(f(x))-f^{-1}(f(y)) - T^{-1}(T(x-y) +r_a(x,y)) $$

$$= x-y-(x-y)- T^{-1}(r_a(x,y)|x-y|)$$ $$\tag 2= - |x-y|T^{-1}(r_a(x,y)).$$

Now if $x\ne y$ (and that's all we need to think about), then

$$\tag 3 |x-y|= |f(x)-f(y)|\frac{|x-y|}{|f(x)-f(y)|}\le |f(x)-f(y)|\cdot \frac{1}{c}.$$

So going back to the $u,v$ notation, we can say $(2)$ is bounded above in absolute value by

$$|u-v|\cdot \frac{1}{c}\cdot |T^{-1}(r_a(f^{-1}(u),f^{-1}(u))|.$$

That has the form $|u-v|\cdot o(1)$ as $(u,v)\to (b,b).$ This completes the proof.

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