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?

  • $\begingroup$ You have have shown that $f$ is injective on $B(\xi,\delta)$, the ball with radius $\delta$ centered in $\xi$, right? $\endgroup$
    – Filippo
    Nov 24 at 7:10
  • 1
    $\begingroup$ 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. $\endgroup$
    – copper.hat
    Nov 25 at 4:33

This is not true.

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 $x=0$.

This is slightly counterintuitive, maybe because we imagine derivatives to be continuous.

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.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.