# If $f$ has an invertible derivative at some point $x$, then $f$ is 1-1 on some neighborhood of $x$.

Let $$(V,\lvert\cdot\rvert)$$ be a normed vector space, $$U$$ an open subset of $$V$$, and $$f:U\rightarrow V$$. Suppose $$f$$ is differentiable at $$\xi$$ and $$Df(\xi)$$ is invertible. Show there exists a neighborhood $$N$$ of $$\xi$$ on which $$f$$ is injective.

Proof$$\quad$$ Put $$T=Df(\xi)$$. Since $$T$$ is invertible, $$\lVert T^{-1}\rVert\not=0$$, and $$\lvert x-y\rvert=\lvert T^{-1}(Tx-Ty)\rvert\leq\lVert T^{-1}\rVert\lvert Tx-Ty\rvert\mbox{,}$$ so that $$\frac{\lvert x-y\rvert}{\lVert T^{-1}\rVert}\leq\lvert Tx-Ty\rvert$$ for any $$x,y\in V$$.

$$\qquad$$ Since $$f$$ is differentiable at $$\xi$$, we have $$f(x)=f(\xi)+T(x-\xi)+\lvert x-\xi\rvert R(x-\xi)\mbox{,}$$ where $$R(x-\xi)\rightarrow 0$$ as $$x\rightarrow \xi$$. Choose $$\delta>0$$ so that $$\vert x-\xi\rvert<\delta$$ implies $$\lvert R(x-\xi)\rvert<\frac{1}{2\lVert T^{-1}\rVert}$$. It follows then that for $$\lvert x-\xi\rvert<\delta$$, $$\lvert y-\xi\rvert<\delta$$, \begin{aligned} \lvert f(x)-f(y)\rvert &= \lvert Tx-Ty+\lvert x-\xi\rvert R(x-\xi)+\lvert y-\xi\rvert R(y-\xi)\rvert\\ &\geq \lvert Tx-Ty\rvert-\lvert x-\xi\rvert\lvert R(x-\xi)\rvert-\lvert y-\xi\rvert\lvert R(y-\xi)\rvert \\ &> \frac{\vert x-y\rvert}{\lVert T^{-1}\rVert}-(\lvert x-\xi\rvert+\lvert y-\xi\rvert)\frac{1}{2\lVert T^{-1}\rVert}\\ &> \frac{\lvert x-y\rvert}{2\lVert T^{-1}\rVert}\mbox{.} \end{aligned}

Is this correct?

• You have have shown that $f$ is injective on $B(\xi,\delta)$, the ball with radius $\delta$ centered in $\xi$, right? Nov 24 at 7:10
• How did you go from the second last line to the last line in the equation above? I think you have a mistake in your estimate, we have $|x -\xi|+|y-\xi| \ge |x-y|$, not the other way around. Nov 25 at 4:33

Let $$f(x) = {x \over 2} + x^2 \sin {1 \over x}$$ for $$x \neq 0$$ and $$f(0) = 0$$. Note that $$f$$ is smooth for $$x \neq 0$$. We see that $$f'(0) = {1 \over 2} >0$$ and for $$x \neq 0$$ we have $$f'(x) = {1 \over 2} + 2x \sin {1 \over x} - \cos {1 \over x}$$.
We see that there are points $$x_+,x_- \neq 0$$ arbitrarily close to $$0$$ such that $$f'(x_+)>0$$ and $$f'(x_-)<0$$. Since $$f'$$ is continuous, this means that $$f$$ is increasing locally around $$x_+$$ and decreasing locally around $$x_-$$. In particular, $$f$$ cannot be locally injective at $$x=0$$.
A related result is found in Tao's answer which is an answer to his question on MO here. The difference is (apart from $$V=\mathbb{R}^n$$) that $$f$$ is assumed to have an invertible derivative everywhere. (And frankly, that this results in a positive answer is very surprising to me and highlights my lack of understanding things global. Tao's answer is fairly involved.)