If the derivative of a function is square of it then it is constant [duplicate]

Let $$f:\mathbb{R}\to\mathbb{R}$$ is differentiable and $$f(0)=0$$. Also $$\forall x\in \mathbb{R}$$ we have $$f'(x)=f^2(x)$$. Prove that $$f(x)=0$$, for every $$x$$.

I tried to use MVT for both derivative and integral. But I got nowhere.

I just found out that

$$f$$ is increasing.

for positive values $$f$$ is non negative.

$$\forall x>0$$, there exists some $$c\in (0,x)$$ s.t. $$f(x)=xf^2(c).$$

Intuitively, it seems one can start by a small interval around zero and show that $$f=0$$ and so on.

Any comment!

• Well, what hindered you from using your idea to prove the result? You pick an $x_0<1$ with $f(x_0)<1$, then use that $f$ is continuous and monotone, so if $f(x_0)>0$ we'd have $f(x_0)>x_0 \cdot f^2(c)$, and therefore get that $f(x)=0, x\le x_0$. Then all that's left to do is argue, that by inductively using this method, we get for any $x\in\mathbb R$ that $f(x)=0$. If that version of induction is to vague for you, you can also do a formal one: math.uga.edu/~pete/realinduction.pdf Jul 14, 2019 at 5:45
• Jul 14, 2019 at 6:17

Suppose for some $$x_0$$ you have $$f(x_0)\neq0$$. Then you can solve the differential equation $$\frac{f'(x)}{f^2(x)}=1$$ with the initial condition $$f(x_0)$$, which gives $$\frac{-1}{f(x)}+\frac{1}{f(x_0)}=x-x_0\iff f(x)=\frac{1}{c-x}$$ where $$c$$ is a constant, and this holds for every $$x$$ that is in the same component of $$\mathbb R\backslash\{c\}$$ with $$x_0$$, say $$I=(-\infty,c)$$. This goes to $$\infty$$ as $$x$$ approaches $$c$$, hence it cannot be the restriction of a function that is differentiable over $$\mathbb R$$ to $$I$$.
On the set where $$f(x) \neq 0$$ the derivative of $$-1/f$$ is $$1$$ so we get $$f(x) (x+c)=-1$$ for some constant $$c$$. It follows that the continuous function $$f(x) (x+c)$$ takes only two values $$0$$ and$$-1$$. Hence it is a constant. But $$f(0)=0$$ so $$f$$ must vanish identically. [We get $$f(x)=0$$ for $$x \neq -c$$ but $$f(-c)$$ is also $$0$$ by continuity].
Some additional details: The set where $$f \neq 0$$ is an open set, so it is a countable disjoint union of open intervals. If $$(a,b)$$ is one of these intervals then there exits $$c$$ such that $$f(x)(x+c)=-1$$ in $$(a,b)$$ and it is $$0$$ at the end points. This contradicts continuity of $$f$$. Conclusion: there is no point $$x$$ with $$f(x) \neq 0$$.
• I can't follow the details of your A, but the first line is the key idea. If $x>0$ and $f(x)\ne 0$ let $x_1=\max ([0,x]\cap f^{-1}\{0\}).$ Then $x_1<x$ and $f(y)=-1/(y+c)$ for all $y\in (x_1,x).$ But then $0=f(x_1)=\lim_{y\to x_1^+ }(-1/(y+c))=-1/(x_1+c)$ which is absurd...And similarly if $x<0$ and $f(x)\ne 0.$ Jul 14, 2019 at 5:10