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Let $f:\Bbb{R}\to\Bbb{R}$ differentiable such that $f(0)=0$ and for all $x\in\Bbb{R}$ we have $f'(x)=(f(x))^2$. Show that $f(x)=0$ for all $x\in\Bbb{R}$

Just a Hint!

What can i use to solve this problem?

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    I would keep derivating – 2017-02-19
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    @RobArthan You can, because $f'(x)=f(x)^2$ implies that $f'(x)$ is differentiable by chain rule – 2017-02-19

3 Answers 3

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Let $U = \{ x \in \Bbb{R} : f(x) = 0 \}$ be the set of zeros of $f$. Clearly $U$ is closed and non-empty. We claim that $U$ is also open. Since $\Bbb{R}$ is connected, this is possible only when $U = \Bbb{R}$.

To this end, let $a \in U$ and $\delta > 0$ be arbitrary and let $M = \sup_{|x-a| < \delta} |f(x)| $. Then by the mean value theorem, for $x \in (a-\delta,a+\delta)$ we have

$$ |f(x)| = |f(x) - f(a)| \leq M^2|x - a| \leq M^2 \delta. $$

From this we have $M \leq M^2\delta$. If $\delta$ is sufficiently small, then we have $M\delta < 1$ and the previous inequality yields $M = 0$. This shows that $(a-\delta, a+\delta) \subseteq U$ and hence the claim follows.

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    Doesn't $M$ depend on $\delta$ ? So why should it hold that $M\delta<1$ when $\delta$ is small ? – 2017-12-06
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    @GabrielRomon, Although $M$ depends on $\delta$, the dependence is nice in the sense that $\delta \mapsto M(\delta)$ is increasing. So making $\delta$ small makes $M(\delta)$ small as well. – 2017-12-06
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It's immediate from the existence and uniqueness theorem for ordinary differential equations.

If you don't have that, you might note that (if $f(x) \ne 0$) $\dfrac{d}{dx} \dfrac{1}{f(x)} = -1$.

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Hint: For $x \ge 0$ $$f(x) = \int_0^x f(t)^2 dt$$ from the fundamental theorem of calculus and the fact that $f(0)=0$. Note also that $f$ is increasing and positive for $x \ge 0$ (why?). Hence for all $x \ge 0$,

$$f(x) \le xf(x)^2.$$

If $f(x)$ is not $0$ for some $0 \le y \le 1$, we can divide the above inequality by $f(y)$ to see that $f(y) \ge 1$. This is a contradiction (why? Hint: IVT). We can deduce that $f(x) = 0$ for all $x \in [0,1].$

From this we can deduce the following inequality:

$$f(x) \le (x-1)f(x)^2.$$

Applying the above logic inductively we can prove that $f(x)=0$ for all $x \ge 0$. (i.e suppose there is $1 \le y \le 2$ such that $f(y)$ is not $0$.....etc).

A similar argument works for $x<0$. This shows $f$ must be everywhere $0$.

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    You might want to write $f(x)=\int^x_0 f(t)^2\,dt$ – 2017-02-19
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    Yup - cheers! :) – 2017-02-19