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This question already has an answer here:

I was given the following problem:

Let $f:[0,1] \to \Bbb R $ be a differentiable function on $[0,1]$ such that $f(0)=0$, and $\forall x \in[0,1]$ : $\lvert f'(x)\rvert \le \lvert f(x)\rvert$. Prove that $f(x)=0$ $\forall x \in [0,1]$.

I tried to solve it using mean value theorem iteratively, and got to a point where I have a series that holds $f(x_1) \ge f(x_2) \ge f(x_3) \ge \dots $ pretty much stuck from here, is it a good start? any advice on how to move on / solve it differently? Thanks

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marked as duplicate by Paramanand Singh calculus Jan 19 '17 at 19:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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It suffices to show that $f \equiv 0$ on $[0,1)$, for then $f(1) = \lim\limits_{n\to \infty} f(1-1/n) = \lim\limits_{n\to \infty} 0 = 0$.

Let $0 < a < 1$. We want to show that $f(t) = 0$ for all $t\in [0,a]$. By the extreme value theorem, there is a point $b\in [0,a]$ such that $\lvert f(b)\rvert = \max_{0 \le t \le a} \lvert f(t)\rvert$. If $b = 0$, then claim follows as $f(0) = 0$. If $b \neq 0$, then since $f(0) = 0$, the mean value theorem gives $f(b) = f'(c)b$ for some $c\in (0,b)$. Then $$\lvert f(b)\rvert = \lvert f'(c)\rvert b \le \lvert f(c)\rvert b \le \lvert f(b)\rvert b$$ As $b < 1$, the inequality forces $f(b) = 0$. So again, $f(t) = 0$ for all $t\in [0,a]$.

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  • $\begingroup$ you beat me to it. +1 $\endgroup$ – Paramanand Singh Jan 19 '17 at 19:08
  • $\begingroup$ Aptly done. (+1) $\endgroup$ – Mark Viola Jan 19 '17 at 19:27
  • $\begingroup$ I'm missing something, doesnt it only show that the maximum value in $[0,a]$ is 0? $\endgroup$ – user401516 Jan 19 '17 at 21:15
  • $\begingroup$ @user401516 if the max value, $M$, of $\lvert f\rvert$ on $[0,a]$ is $0$, then for all $t\in [0,a]$, $\lvert f(t)\rvert \le M = 0$. That implies $f(t) = 0$ for all $t\in [0,a]$. Since $a$ was arbitrary, $f(t) = 0$ for all $t\in [0,1)$. $\endgroup$ – kobe Jan 19 '17 at 21:21
  • $\begingroup$ Thanks, I missed the absolute value. Great proof! $\endgroup$ – user401516 Jan 19 '17 at 21:27

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