# Why is continuous differentiability necessary for Inverse Function Theorem?

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 $$a$$)

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 $$0 with $$f(0), 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 $$0 such that $$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 $$0 then $$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]$$.

• It might be easier to mention that $f'$ takes positive and negative values in any neighborhood of $0$ ($f'(1/(n\pi))=3$ for $n$ odd and $f'(1/(n\pi))=-1$ for $n$ even). Since a continuous bijection on an interval is necessarily monotonic, we're done. – zhw. Mar 30 '19 at 16:14

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 $$0.$$