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Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function such that $f'(x)$ is continuous and $|f'(x)|\le|f(x)|$ for all $x\in\mathbb{R}$. If $f(0)=0$, find the maximum value of $f(5)$.

$f'(x)=f(x)$ is true when $f(x)=ke^x$.

$f(x)=0$ satisfies the condition. So $f(5)=0$ which is also the correct answer. But is there any method other than substitution?

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  • $\begingroup$ You have shown it can be $0$. Now you have to prove that is the maximum. I'm not sure how your first observation plays into this. $\endgroup$ Feb 8, 2016 at 1:26
  • $\begingroup$ I am not clear what it means that find the maximum value of f(5)? The function who meets the condition is zero function. $\endgroup$
    – runaround
    Feb 8, 2016 at 1:37
  • $\begingroup$ So that's the maximum value. But no substitution please. $\endgroup$
    – Aditya Dev
    Feb 8, 2016 at 2:15
  • $\begingroup$ I think maybe you can use Gronwal's inequality to show that only the trivial case holds. $\endgroup$
    – Vim
    Feb 8, 2016 at 15:13
  • $\begingroup$ If $f$ satisfies the given conditions, then so does $\alpha f$ for any $\alpha \in \mathbb R$. Hence, if $f(5)$ has a maximum possible value, that value must be $0$. $\endgroup$
    – TonyK
    Feb 8, 2016 at 19:38

2 Answers 2

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Chose $N\in\mathbb N$ arbitrarily. Then $f$ is continuous in $[0,N]$ and therefore attains its maximum and minimum, so $\exists M>0:|f(x)|\leq M,\,\forall x\in [0,N]$.

For each $x\in[0,N]$ you have $$f(x)=\int\limits_{0}^{x}{f'(t)dt}\Rightarrow |f(x)|\leq \int\limits_{0}^{x}{|f(t)|dt}\leq \int\limits_{0}^{x}{Mdt}=Mx$$ Take the inequality $|f(t)|\leq Mt,\forall t\in [0,N]$ and integrate it again in $[0,x]$ for arbitrary $x\in [0,N]$ to get $$|f(x)|\leq \int\limits_{0}^{x}{|f(t)|dt}\leq \int\limits_{0}^{x}{Mtdt}=\frac{Mx^2}{2}$$ and continue the same way. You get that $|f(x)|\leq \frac{Mx^n}{n!},\forall n\in\mathbb N$ which means that $|f(x)|\equiv 0,\,\forall x\in [0,N]$ because $\lim\limits_{n\to\infty}\frac{x^n}{n!}\to 0,\,\forall x\in\mathbb R$. Because $N$ was arbitrarily big it follows that $f(x)\equiv 0,\,\forall x\in \mathbb R$ and this is the only function satisfying the conditions in the question $\Rightarrow $ the maximum value at $x=5$ is the only possible value $0$.

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    $\begingroup$ Very nice and simple approach. +1 $\endgroup$
    – Paramanand Singh
    Feb 9, 2016 at 11:11
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The inequality indicates $$ \left|\frac{\mathrm{d}}{\mathrm{d}x}\log\left|f(x)\right|\,\right|\le1 $$ If $\log\left|f(5)\right|=a$ is finite, then $\log\left|f(0)\right|\ge a-5$. Since $f(0)=0$, we must have $f(5)=0$.

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  • $\begingroup$ I'm unclear how you get this -- what is the formal derivation of the inequality? (given that $f(0)=0$, division by $f(x)$ seems risky -- isn't it?) $\endgroup$
    – Clement C.
    Feb 8, 2016 at 19:35
  • $\begingroup$ If we assume that $\log\left|f(5)\right|=a$ is finite, there is no problem. We arrive at a contradiction that $\log\left|f(0)\right|\ge a-5$ is finite. Therefore, we conclude that $f(5)=0$. $\endgroup$
    – robjohn
    Feb 9, 2016 at 19:48
  • $\begingroup$ Ack -- I was missing that part of the proof (essentially, that it's by contradiction). Thanks! $\endgroup$
    – Clement C.
    Feb 9, 2016 at 19:49

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