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 $f(x) \ge 0$.


| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.