Strong differentiability of the homeomorphism inverse. Inverse Function Theorem.

Question

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.

Answer 1

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.