How to show $f\equiv 0$?

Question

$f(x)$ is differentiable, and for any $x\in \mathbb R$, $|f'(x)|\le \lambda |x|$, then how to show $f\equiv 0$ ?

This is a question my student ask me, but I don't know how to deal it. So ask help here. Thanks for any hint or answer.

Answer 1

This is not correct: $f(x)=x^{2}$, $|f'(x)|=2|x|$.

Even that $\lambda$ is required to be $\lambda\in(0,1)$ is still not correct: $f(x)=\lambda^{2}\log(3\lambda+x^{2})$, then $|f'(x)|=\dfrac{2\lambda^{2}|x|}{3\lambda+x^{2}}\leq\dfrac{2}{3}\cdot\lambda|x|$.