Inequality between function and its derivative

Recently, I stumbled across this problem in a real analysis textbook of mine

Let $$f:[0,1] \to \mathbb{R}$$ be a continuous function differentiable in $$(0,1)$$ such that $$f(0)=0$$ and also $$0 \leq f'(x) \leq 2f(x)$$. We are asked to show that $$f \equiv 0$$.

Since we don't know $$f$$ to be absolutely continuous, we cannot simply integrate the inequality with the derivative. I tried manipulating the inequality but that didn't work and also using the mean value theorem or fundamental theorem of calculus but nothing useful came up. I would certainly appreciate help on this.

Observe that: $$\left(\color{blue}{e^{-2x}f(x)}\right)'=e^{-2x}f'(x)-2e^{-2x}f(x)=\underbrace{\left(f'(x)-2f(x)\right)}_{\le \, 0}e^{-2x}$$ This means that $$\color{blue}{e^{-2x}f(x)}$$ is non-increasing, but:
• it is $$0$$ in $$x=0$$ (due to $$f(0)=0$$);
• $$e^{-2x}> 0$$ for all $$x$$;
• it is given that $$f(x) \ge 0$$.