Is the function $f(x)=\begin{cases} x+0.2x^2 \sin (1/x) & x \ne 0\\ 0 & x=0\end{cases}$ invertible in a neighborhood of origin?

Is the function
$$f(x)=\begin{cases} x+0.2x^2 \sin (1/x) & x \ne 0\\ 0 & x=0 \end{cases}$$invertible in a neighborhood of origin?

I just know that $f$ is continuous on $\mathbb{R}$ and $$f'(x)=\begin{cases} 1+0.4x\sin\left(\frac{1}{x}\right)-0.2\cos\left(\frac{1}{x}\right)& x \ne 0\\ 1 & x=0, \end{cases}$$

$f'$ is not continuous at $0$, it's not satisfies conditions in Inverse Theorem.

But when I've tried to prove like this:

Choose $B=(-0.1,0.1)\subset \mathbb{R}$, $f(x) \ne 0,\forall x \in B$. Assume $f'(x) \ne 0$, we have $$\|f'(x)-I\|=\left|0.4xsin\left(\frac{1}{x}\right)-0.2cos\left(\frac{1}{x}\right)\right|\leq 0.4r+0.2=0.24<\frac{1}{2}.$$ Let $g(x)=x-f(x)$ so $\|g'(x)\|<\frac{1}{2}$. For all $x_1,x_2\in B$ exits $c\in(x_1,x_2)$ such that $$g(x_2)-g(x_1)=g'(c)(x_2-x_1).$$ Then $$|g(x_2)-g(x_1)|<\frac{1}{2}$$ For all $y \in (-0.05,-0.05)$ let $g_y(x)=g(x)+y, x \in B$. As $g(0)=0$ so $$|g_y(x)|<|g(x)+y|<|g(x)|+|y|<0.1,\forall x \in B.$$ So $g_y(B)\subset B$. Moreover, $$|g_y(x_2)-g_y(x_1)|=|g(x_2)-g(x_1)|<\frac{1}{2}|x_2-x_1|.$$ So $g_y$ satisfies Lipschitz condition in x, then exits unique $x_y \in B$ such that $x_y=g_y(x_y)$ or $f(x_y)=y$. Let $x_y:=f^{-1}(y)$. So $f^{-1}(y)$ exist for all $y \in (-0.05,0.05)$. And I can check $f^{-1}(y)$ is continuos. I'm so confused about this. Help me, thank you so much.

Hint: we have $f'(x)> 0.4>0$ for all $x$.

Conclusion ?

Your computations of $f'$ tell you that in the set $[-1;0) \cup (0;1]$ you have $$f'(x) \ge 1-0.2-0.4=0.4$$

Note that $f(x)=x(1+x \sin (1/x))$ is equal to $x$ times a strictly positive quantity. Thus

1. $f(x) <0$ if and only if $x<0$;
2. $f(x) >0$ if and only if $x>0$;
3. $f(x) =0$ if and only if $x=0$

Now, suppose by contradiction that $f(x)$ is not invertible on $[-1;1]$. This means that there exist two distinct numbers $a < b$ such that $f(a)=f(b)$. The observation on the sign made above tells us that $a,b \neq 0$ and they have the same sign.

Now apply Rolle's theorem on the interval $[a;b]$ (which does not contain $0$!): there exists some $c$ such that $$f'(c)=0$$ and this is a contradiction, since on $[a;b]$ you have $$f'(c) \ge 0.4$$

• Thanks, I've another question is that if $f^{-1}$ is differentiable? – Thu Le Jul 2 '18 at 14:45