Let $\Omega$ be an open set of $\mathbb{R}^n$ and let $f:\Omega\to\mathbb{R}^n$ a function of $\mathscr{C}^r(r\in\mathbb{N}\cup\{\infty\})$. Let $a\in\Omega$ such that $f´(a):\mathbb{R}^n\to\mathbb{R}^n$ be a linear isomorphism and let $b=f(a)$. So there is an open neighbourhood $U$ of $a$ in $\Omega$, and there is an open neighbourhood $V$ of $b$ in $\mathbb{R}^n$ such that $f(U)=V$ and $f:U\to V$ is a diffeomorphism of $\mathscr{C}^r class$.

Proof $f$ is locally injective

Let $A=f´(a)$ and let $\lambda=\frac{1}{2||A^{-1}||}$

As $f´$ is continuous in $a$ there exists an open connected $U$ that contains $A$ and is contained in $\Omega$ such that $||f(x)-A||<\lambda\:\:\forall x\in U$ Given $y\in\mathbb{R}^n$ consider the function $\psi=\psi_y$ defined as:

$\psi_y:U\to\mathbb{R}^n\\x\to x+A^{-1}(y-f(x))$

Note $f(x)=y$ iff $\psi(x)=x$. For each $x\in\mathbb{R}^n$

$\psi´(x)=A^{-1}(A-f´(x))$ so that $||\psi´(x)||<\frac{1}{2}\:\:\forall x\in U$

By the intermdeite value theorem:

$||\psi(x_1)-\psi(x_2)||\leqslant\frac{1}{2}||x_1-x_2||\:\:\forall x_1,x_2\in U$.

By Banach fixed point Theorem we have $x\in U$ such that $\psi(x)=x$.


1)I guess $\psi´(x)=A^{-1}(A-f´(x))$ is wrong once $(A^{-1}f(x))´=-A^{-2}f´(x)$, right?

2)Why does the author goes for this method to prove the function is locally injective? Why does not the author take the derivative as in the inverse theorem in one dimension? Why complicate so much?

Thanks in advance!

  • 1
    $\begingroup$ 1) The number $y$ is fixed and the matrix $A$ is constant, so $\psi’(x)=(x+A^{-1}(y-f(x)))’=I-A^{-1}f’(x)=A^{-1}(A-f´(x))$, right? $\endgroup$ – Alex Ravsky Nov 8 '17 at 4:08
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    $\begingroup$ What is it about taking derivatives do you think implies local injectivity? In one dimension, if the derivative is strictly positive (resp. negative) on some open interval containing $a$, then the function is strictly increasing (resp. decreasing) on that interval; no such simple rule exists in multiple dimensions. (Indeed, this is why the inverse function thereom is interesting.) $\endgroup$ – Jason Nov 8 '17 at 4:08

Note that even in one dimension one has to be careful. Even if $f:{\Bbb R}\rightarrow {\Bbb R}$ is differentiable at every point and $f'(0)= 1$ then $f$ need not be locally injective at zero. A counter-example: $$ f(x) = x + x^2 \sin\left(\frac{1}{x^2}\right) .$$ This function is not injective on any neighborhood of zero.

To get a correct result you need continuity of the derivative at zero (and a proof may use a Banach fixed point argument as above) or alternative knowledge of a definite sign of the derivative in a neighborhood of the selected point.

The last condition may be converted into a fairly elementary proof using the intermediate value theorem. However, it uses strongly the ordering of the reals. In dimension larger than 2 this ordering is not a priori available, whence the more analytic proof that you cite.


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