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Inverse Function Theorem. Let $f: \mathbb{R}^{n} \to \mathbb{R}^{n}$ be a $C^{1}$ function. If $\det Df_{a} \neq 0$, there is open sets $U, V$ such that $f: U \to V$ is a diffeomorphism $C^{1}$ ($a \in U$ and $f(a) \in V$).
Why is the continuously differentiability necessary for this version? I'm trying to find a example with $f$ just differentiable, $\det Df_{a} \neq 0$ but $f$ is not a local diffeomorphism (at the point $a$)
Thanks for the advance!
Not a particularly flashy example, but consider the function $f: \mathbb{R}\to \mathbb{R}$ given by $$ f(x) = \begin{cases} x+2x^2\sin(\frac{1}{x})&\text{if }x\neq 0\\ 0&\text{if }x=0 \end{cases}. $$ You can show that $f$ is differentiable on all of $\mathbb{R}$ with $f'(0) = 1$, but $f'$ is discontinuous at 0. To see that $f$ is not a local diffeomorphism at the origin, you can find a sequence of intervals that approach the origin on which $f$ is decreasing, but since $f'(0)>0$ we can find a point $a$ with $0<a$ with $f(0)<f(a)$, so $f$ can't be a one-to-one mapping on any neighborhood of zero.
In detail, we have $$ f'(x) = \begin{cases} 1 + 4x\sin(\frac{1}{x})-2\cos(\frac{1}{x})&\text{if }x\neq 0\\ 1&\text{if }x=0 \end{cases}. $$ As $x$ approaches 0 the $4x\sin(\frac{1}{x})$ term can be made arbitrarily small while the $2\cos(\frac{1}{x})$ term oscillates between $-2$ and $2$. Consequently, we can find a sequence of points $x_n\to 0$ with $0<x_{n+1}<x_n$ such that $f'(x_n) = -1$ say. Since $f'$ is continuous away from the origin, $f$ is decreasing on a small neighborhood of each $x_n$ by the mean value theorem.
By the definition of the derivative we have $$ f'(0) = \lim_{x\to 0}\frac{f(x)-f(0)}{x-0} = \lim_{x\to 0}\frac{f(x)}{x} = 1>0. $$ In particular, for $a$ sufficiently close to zero and positive we have $f(a)>0$. Now if we take one of our $x_n$'s such that $0<x_n<a$ then $f$ is decreasing on some interval contained in $(0, a)$. But if $f(0)=0$, $f(a)>0$, and $f$ is decreasing on some interval between $0$ and $a$ then by continuity, $f$ can't be one-to-one on $[0,a]$.
More intuition than rigor: First verify that if $x\le f(x) \le x+x^2$ on $(-1,1),$ then $f'(0)=1.$
Now consider this situation: $1>a_1>1/2 >a_2>1/3 >a_3>1/4\cdots.$ Choose the $a_n$ so close to $1/n$ that the line segments $[a_n,1/n]\times \{1/n\}$ lie between the graphs of $x$ and $x+x^2.$ We can then connect these line segments together, smoothly, so that the result is the graph of a smooth function on $(0,1)$ that stays between the graphs of $x,x+x^2.$
Call this function $f$ and then extend it to $(-1,0]$ by setting $f(x)=x,x\le 0.$ Then $f$ is differentiable on $(-1,1)$ and $f'(0)=1.$ Yet $f$ is constant on each of the intervals $[a_n,1/n].$ This shows $f$ is not $1-1$ on any interval containing $0.$
