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The statement is:

1. Let $x_0,y_0 \in \mathbb R^n$ and let $J_0, I_0$ be their respective neighborhoods. Let then $\mathbf f : J_0 \to I_0$ bijective and differentiable at $x_0$ such that $y_0 = \mathbf f(x_0)$, and let $d\mathbf f_{x_0} : \mathbb R^n \to \mathbb R^n$ denote the unique linear application such that, for $\mathbf h \in \mathbb R^n$ small enough, we may write $$\mathbf f(x_0 + \mathbf h) = \mathbf f(x_0) + d\mathbf f_{x_0}(\mathbf h) + |\mathbf h|\mathbf q(\mathbf h) \tag 1$$ for some $\mathbf q : \mathbb R^n \to \mathbb R^n$ such that $\lim_{\mathbf h\to \mathbf 0}\mathbf q(\mathbf h) = \mathbf 0$. Then, if $d\mathbf f_{x_0}$ is an isomorphism, $\mathbf f^{-1}$ is differentiable at $y_0$, and we have $$\tag 2d(\mathbf f^{-1})_{y_0} = (d\mathbf f_{x_0})^{-1} $$

The converse, that is,

2. If $\mathbf f^{-1}$ is differentiable at $y_0$, then $(2)$ holds and $d\mathbf f_{x_0}$ is an isomorphism

is true and can be proved using the general result about differentiability of composite functions.

My real analysis textbook does provide an ingenious counterexample that shows the falsity of statement 1., and therefore demonstrates the need for further hypotheses (specifically, continuity of $\mathbf f^{-1}$ at $y_0$) to achieve an invertible result. However, it is very elaborate, and even the author of the textbook admits that it took him a while to find it.

Are there more direct examples that show that 1. is false?

1 Answers 1

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Here is the counterexample.

Defining the bijective function. Consider the sets \begin{equation} \begin{split} A &\doteq \{x \in \mathbf R\ |\ \exists n \in \mathbf N, n\geq 2,\ \mathrm{s.t.}\ x = 1/n\} \\ B&\doteq \{n \in \mathbf N\ |\ n \geq 2\} \end{split} \end{equation} Let us define the value of $f(x)$ for $x\in A$ as follows: $$x \in A \Rightarrow f(x) = \frac 1 {n+m+1}$$ if $x = \frac 1 n$ for some $n\geq 2$, where $m$ is the only non-negative integer such that $2^m < n \leq 2^{m+1}$. It is clear that the image of the restriction of $f$ to values in $A$ is a subset of $A$ itself, and the complement set is countable – precisely, it the image of the following sequence (check to make sure!)

$$y_1 = \frac 1 2, \quad y_2 = \frac 1 4, \quad y_3 = \frac 1 7, \quad y_4 = \frac 1 {12}, \quad y_5 = \frac 1 {21}, \quad y_6 = \frac {1}{38}, \quad \dots $$

So let's define the value of $f(x)$ for $x\in B$ as follows: $$x \in B \Rightarrow f(x) = \begin{cases} y_m & x = 2m\ \mathrm{even} \\ m+1 & x = 2m+1\ \mathrm{odd} \end{cases} $$ These two definitions, taken together, lead to a function from $A\cup B$ to itself that is invertible. Then it is clear that requiring $$x \in \mathbf R \setminus A\cup B \Rightarrow f(x) = x $$ preserves the bijectivity of the function.

Discontinuity of the inverse. Clearly, the inverse function $f^{-1}$ is discontinuous in $f(0) = 0$, since $\{y_m\}$ tends to $0$, while $f^{-1}(y_m) = 2m$ for $m = 1,2,\dots$

Differentiability. Now let's check that $f$ be differentiable at $0$, and that $f'(0) = 1$. For $x = \frac 1 n \in A$ for some $n \geq 2$, having determined the integer $m \geq 0$ such that $2^m < x \leq 2^{m+1}$, we get $$x - f(x) = \frac 1 n - \frac 1 {n + m +1} = \frac 1 n \cdot \frac {1+m}{n + (m+1)} \leq \frac 1 n \cdot \frac {1 + \log_2 n}{n + \log_2 n}$$ We have then $$0 \leq \frac{x - f(x)}{x} \leq \frac{1 +\log_2 n}{n + \log_2 n} $$ But the last element of the chain has limit $0$ as $n\to\infty$, so the restriction to $A$ of $x \mapsto \frac{f(x) - x}{x}$ must tend to $0$ as $x \to 0$. Now, since $f(x) = x$ if $x \in (-1,1) \setminus A$, we conclude $$\frac{f(x) - x}{x} \to 0\quad \mathrm{as}\quad x\to 0. $$ This is equivalent to saying $f(x) = x + o(x)$ as $x \to 0$, therefore we have proved the result we needed.