# Prove that $\left(\forall x\in \mathbb R\left(f'(x)=f(x)^2\right)\land f(0)=0\right)\implies f=\bf 0$

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.

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

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?
• 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 $(0,a)$? – Clement C. Aug 27 '16 at 21:34
• 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). – zhw. Aug 27 '16 at 21:39