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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.

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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 $f(x) \ge 0$.

So...

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