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!

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    $\begingroup$ 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 $\endgroup$
    – Sudix
    Jul 14, 2019 at 5:45
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    $\begingroup$ Also: math.stackexchange.com/q/2683895/42969. $\endgroup$
    – Martin R
    Jul 14, 2019 at 6:17

2 Answers 2


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 $f(x) \neq 0$.

  • $\begingroup$ 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 $f(x)\ne 0.$ $\endgroup$ Jul 14, 2019 at 5:10
  • $\begingroup$ @DanielWainfleet I have edited my answer with more details. I hope the proof is clear now. $\endgroup$ Jul 14, 2019 at 5:18
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    $\begingroup$ Quite clear now. ......+1 $\endgroup$ Jul 14, 2019 at 5:43

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