Help to show that f is not differentiable. Let $ 0 < \alpha < 1 $. If $\vert f(x)\vert \geq \vert x\vert^\alpha$ for all $x$ and $f(0) =0$, then $f$ is not differentiable at $x=0$.
I would appreciate some help to prove this.
Edit : I supposed that $f$ is differentiable at 0 and used the definition of derivative as a limit. Since $f(0)=0$ , I get $ \lim_{h \to 0} \frac{f(h)}{h} \geq  \lim_{h \to 0} \frac{1}{h^{1- \alpha}} $ by hypothesis and the fact that limit preserves inequality. Since the second limit goes to infinity, then the first limit goes to infinity too, and this is a contradiction with the fact that $f’(0)$ exist. Its that right?

Suppose it is, then $\left|\dfrac{f(x) - f(0)}{x-0}\right|\ge \dfrac{1}{x^{1-\alpha}}\implies f'(0) = \infty$, contradiction to the fact that $f$ being differentiable at $x = 0$, hence $f'(0)$ must be finite.
