counter example of inverse function theorem

Question

Prove that in the inverse function theorem, the hypothesis that $f$ is $C^1$ cannot be weakened to the hypothesis that $f$ is differentiable. I read an example of my teacher, but I can't have any analysis argument for the fact that $f$ is not one to one in any neighborhood of $0$. Here it is,

$$f(0)=0, \qquad f(x)=x + 2x^2\sin \frac{1}{x}, x\neq 0.$$

Answer 1

Except at $0$, the function is $C^1$, so you can analyse it by examining its derivate on the interval $(0,\varepsilon)$. If $f$ was injective on $(0,\varepsilon)$, then it would have to be monotone and hence its derivative could not have a change of sign there. Show that it does.

Note: I had originally and erroneously looked at the function $f(x)=x + x^2\sin \frac{1}{x}$, which is harder to analyse, and I analysed it sloppily and wrongly.

Answer 2

The upper and lower bounds for the function tell us that $0<y=f(x)$ may have solutions between the positive solutions of

$y=x\pm2x^2=2(x\pm\tfrac14)^2-\tfrac18$ or $x=\tfrac14(\sqrt{1+8y}\mp1)$.

This segment of length $\tfrac12$ at height $y$ is long enough for multiple oscillations of the $\sin(\frac1x)$ factor, especially if $y$ and thus the lower bound for $x$ is close to zero. So however close to $0$ one restricts the interval under consideration, there is always a value of $y$ small enough to get the left bound close enough to zero to have multiple oscillations cross over the line at height $y$.