Showing that a derivable function $f$ (satisfying some conditions) is null. Enunciate:


*

*Let $f:\mathbb{R}\to\mathbb{R}$ derivable, such that $f(0)=0$ and, for all $x\in\mathbb{R}$, satisfy $f'(x)=(f(x))^2$. Prove that $f\equiv 0$.


How can I prove it? I'm trying to integrate something in someway, but nothing that I tried works.
 A: Suppose $f\not \equiv 0.$ Then the set where $f \ne 0$ is open (by continuity) and nonempty, hence is a pairwise disjoint union of open intervals. Suppose one of these intervals has the form $(a,b),$ where $a\in \mathbb R, b \in (a,\infty].$ Then in $(a,b)$ we have
$$-\frac{1}{f(x)} = x + c$$
for some constant $c.$ Why? Because the derivative of the left side equals the derivative of the right side on this interval. This implies the limit of the left side is finite as $x\to a^+.$ That is a contradiction, since $f(a) = 0.$
The other possibility, where $a \in [-\infty,b), b\in \mathbb R,$ is handled similarly.
A: Using only “basic” tools: MVT and continuity. No integrals and no differential equations. No $\varepsilon-\delta$.
First observe that since $f'=f^2\geq0$, the function $f$ is non-decreasing. Now:


*

*Let $x>0$. Then, by the Mean Value Theorem, there exists $c\in(0,x)$ such that
$$f(x)=xf'(c)=xf(c)^2\leq xf(x)^2$$
(since $f(0)=0$, $f$ is non-decreasing and $0<c<x$).
If $f(x)\neq0$, then $f(x)>0$ (since $x>0$, $f(0)=0$ and $f$ is non-decreasing) and hence
$$1\leq xf(x).$$

*Let $x<0$, there exists $c\in(x,0)$ such that
$$f(x)=xf'(c)=xf(c)^2\geq xf(x)^2$$
(since $f$ is non-decreasing, $x<c<0$). If $f(x)\neq0$ then
$f(x)<0$ (since $x<0$, $f(0)=0$ and $f$ is non-decreasing), and hence
$$1\leq xf(x).$$


Conclusion: for all $x\in\mathbb{R}$,
$$f(x)=0\qquad\text{or}\qquad 1\leq xf(x).$$
Since the function $x\mapsto xf(x)$ is continuous on $\mathbb{R}$ (and its value is $0$ at $0$), we conclude (a full justification could be done using the Intermediate Value Theorem) that
$$\forall x\in\mathbb{R},\ xf(x)=0,$$
and hence (by continuity of $f$ at $0$), $f$ is nil.
A: You can use uniform continuity. Let $a>0$ be fixed, and let's show that $f$ is zero on $[0,a]$. Note that since $f'=f^2\geq 0$, $f$ is non-decreasing. Chose $\delta>0$ such that $|x-y|<\delta$ in $[0,a]$ implies $|f(x)-f(y)|<1$. We can assume $\delta<1$. Let $0=x_0<x_1<\ldots<x_n=a$ be a partition of $[0,a]$ with $|x_i-x_{i+1}|<\delta$. Let's show that $f(x_i)=0$ for all $i$ by induction. The case $i=0$ is from the hypothesis. If $f(x_i)=0$, then $f(x_{i+1})<1$, so
$$f(x_{i+1})=\int_{x_i}^{x_{i+1}} f'(t)dt=\int_{x_i}^{x_{i+1}}f(t)^2dt\leq(x_{i+1}-x_i)f(x_{i+1})^2$$
If $f(x_{i+1})$ were not zero, this would imply $1<\delta$, a contradiction.
Therefore, $f(x_i)=0$ for all $i$. In particular, $f(a)=f(x_n)=0$. Since $a>0$ is arbitrary, we see that $f(x)=0$ for $x\geq 0$.
Considering the function $g(x)=f(-x)$, $g$ also satisfies the hypothesis of the question, so the previous argument implies $g(x)=0$ for $x\geq 0$, that is, $f(x)=0$ for $x<0$.
