If all partial derivatives 0f $f:\mathbb{R^2}\rightarrow \mathbb{R}$ equal $0$ at $(x,y)$ then $f$ is not injective in neighborhood of $(x,y)$

Question

If you have a map $f:\mathbb{R^2}\rightarrow \mathbb{R}$ and know that a point $(x,y)$ is a maximum, then $D_1f(x,y)=0$ and $D_2f(x,y)=0$. It's clear that the converse doesn't always hold, but is it true that in some neighbhourood of $(x,y)$ that $D_1f(x,y)=0$ and $D_2f(x,y)=0$ imply that $f$ is not injective n a neighborhood of $(x,y)$?

Answer 1

There is actually no continuous function $f:\mathbb{R}^2\to\mathbb{R}$ that is injective in any neighborhood of any point. Intuitively, the idea is that you can't squeeze a plane injectively into a line, since a line has only one dimension. More rigorously, you can prove this by restricting $f$ to a small circle in your neighborhood and considering the path $f$ takes along that circle. For $f$ to be injective and continuous, it would need to be strictly monotone as you go around the circle. But that is impossible, since it must come back to where it started after looping all the way around the circle.

If you don't assume that $f$ is continuous everywhere and merely assume that $D_1f$ and $D_2f$ exist and are $0$ at your single point $(x,y)$, then there are counterexamples. As a sketch, you can take $(x,y)=(0,0)$ and $f(0,0)=0$ and let $f$ map annuli around the origin bijectively to intervals approaching $0$, with the intervals getting to $0$ fast enough that $D_1f(0,0)$ and $D_2f(0,0)$ are $0$.