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Let $f:\mathbb{R}\to\mathbb{R}$ differentiable, $f(0)=0$ and $f'(x)=f(x)^2\; \forall x\in\mathbb{R}$. Show that $f(x) = 0\; \forall x\in\mathbb{R}$.

Some thoughts:

It can be shown that $f^{(n)}(x) = n!f(x)^{n+1}$, therefore $f^{(n)}(0)=0$. I thought of using Taylor series but that is only useful if the function is analytical. I also tried something with $f(x) = \int_0^x f(t)^2$ but no luck.

Solution

Since $f'(x)=f(x)^2$, $f$ is increasing. Then $f(x) \geq 0$ for $x>0$ and $f(x) \leq 0$ for $x<0$. Suppose that for some $a>0$ we have $f(a) > 0$, then $\forall x\in(a,\infty),\; f(x)>0$. Then in this interval, we can proceed similarly to the answer provided by @zhw below, to get $f(x) = \frac{-1}{x+c}$, but for $x$ sufficiently large, $f(x)$ would be negative, contradiction.

The case for the negative part is similar.

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  • $\begingroup$ the question is: show that for every real number $x$ with $x\neq 0, f(x)=0$, right? $\endgroup$ – Domates Aug 27 '16 at 20:55
  • $\begingroup$ Yes, sorry about that. I will edit it. $\endgroup$ – Michael Aug 27 '16 at 21:00
  • $\begingroup$ If $ f(x) $ is not equal to zero, then what is the derivative of $ 1/f(x)$ ? $\endgroup$ – Ashar Tafhim Aug 27 '16 at 21:17
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Just to get you started: Suppose $f>0$ in $(0,a)$ for some $a>0.$ Then

$$\frac{f'(x)}{f(x)^2} = 1,\,\, x\in (0,a).$$

This implies $(-1/f)' = 1$ on $(0,a).$ Therefore $-1/f(x) = x+c$ on $(0,a).$ Is that possible?

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  • $\begingroup$ "Just to get you started and almost finished". Up voted anyway. $\endgroup$ – Git Gud Aug 27 '16 at 21:22
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    $\begingroup$ Just to check: this shows that there is no interval $(0,a)$ on which $f$ does not cancel. (And by a similar argument, no $(a,0)$ either.) From there, how to conclude that $f$ is identically $0$ on every $(0,a)$? $\endgroup$ – Clement C. Aug 27 '16 at 21:34
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    $\begingroup$ The set where $f=0$ is closed. Its complement is thus a pwdj union of intervals $ (a_n,b_n)$ On each of those we can do the above (essentially). $\endgroup$ – zhw. Aug 27 '16 at 21:39
  • $\begingroup$ @GitGud I wouldn't say it's almost finished, but it is the main part, yes. $\endgroup$ – Michael Aug 27 '16 at 23:24

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