8
$\begingroup$

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?

$\endgroup$
5
  • $\begingroup$ 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 '16 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 '16 at 1:37
  • $\begingroup$ So that's the maximum value. But no substitution please. $\endgroup$
    – Aditya Dev
    Feb 8 '16 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 '16 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 '16 at 19:38
11
$\begingroup$

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 $0$.

$\endgroup$
1
  • 2
    $\begingroup$ Very nice and simple approach. +1 $\endgroup$
    – Paramanand Singh
    Feb 9 '16 at 11:11
3
$\begingroup$

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 $f(5)=0$.

$\endgroup$
3
  • $\begingroup$ 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 '16 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 '16 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 '16 at 19:49

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.