Let $f(x)$ be a differential real function defined on the real line. If $f(0)=0$ and $f'(x)=[f(x)]^2$, then $f(x)=0$ foi any $x$. Again, $f:\mathbb{R}\to\mathbb{R}$ is differentiable, $f(0)=0$, and $f'(x)=[f(x)]^2$ for every $x$. A friend suggested the following argument:
If exists $c$ such that $f(c)\neq0$, there exists an interval $I$ around $c$ such that $f(x)\neq0$ if $x\in I$ (because $f$ is continuous since it is differentiable). In that interval, we could define $g(x)=x+\frac{1}{f(x)}$. This function $g$ would be differentiable and $g'(x)=0$. Then $g(x)$ is constant, for example, $k$. Then, $f(x)=\frac{1}{k-x}$ for $x\in I$
But I don't know where to find an absurd. What should I do next?
I think I should use the fundamental theorem of calculus and try to find an absurd with $f(x)=\int_0^x f'(t)dt=\int_0^x [f(t)]^2 dt$, but I also didn't got anywhere.
Thanks in advance.
 A: Suppose $f(x)$ exists on some closed interval $[0,\delta]$.  Let $F=F(\delta)$ be its maximum value on this interval.  If $x\in(0,\delta)$, then your integral equation $f=\int f^2$ yields
$$
|f(x)|\le \int_0^xf(t)^2dt\le x F^2.
$$
Hence, 
$$
F(\delta)\le \delta F(\delta)^2.
$$
If $F(\delta)\neq 0$, then
$$
1/\delta\le F(\delta).
$$
This contradicts continuity of $f(x)$ at $x=0$, since we can choose $\delta$ arbitrarily small.
A: Your friend's argument is actually very close to a complete proof.  It proves that for any interval $I$ on which $f$ is nonzero, there exists a constant $k$ such that $f(x)=\frac{1}{k-x}$ for all $x\in I$.  So take a maximal such interval $I$; in other words, a connected component of the open set $\{x\in\mathbb{R}:f(x)\neq 0\}$.  Then $0\not\in I$, so $I=(a,b)$ is an open interval that is not all of $\mathbb{R}$.  That is, either $a$ or $b$ is finite; let us suppose $a$ is finite (the other case is similar).  Now simply observe that $f(a)=0$ (otherwise we could enlarge $I$ to contain $a$), but $f(x)=\frac{1}{k-x}$ does not approach $0$ as $x$ approaches $a$ from above, so $f$ is discontinuous at $a$.  This is a contradiction.
A: Let $a < b \in \mathbb{R}$. Since $f: [a,b] \longrightarrow \mathbb{R}$ is continuous, theres exist $F: [a,b] \longrightarrow \mathbb{R}$ such that $F' = f$. Let $g: [a,b] \longrightarrow \mathbb{R}$ defined by $g(x) = f(x)e^{-F(x)}$. Note that $g$ is differentiable, then
$$g'(x) = f'(x)e^{-F(x)} + f(x)e^{-F(x)}(-f(x)) = (f(x))^{2}e^{-F(x)} - (f(x))^{2}e^{-F(x)} = 0.$$
Therefore, $g$ is constant. Since $g(0) = 0$, $g(x) = 0$ for all $x \in [a,b]$, so $f(x) = 0$ for all $x \in [a,b]$. Since $a, b$ were chosen arbitrarily, the result follows.
A: Say $f_1$ is a solution of the differential equation. Let $J \supseteq \{0\}$ be an interval of maximal length with $f_1|_{J}=0$. By continuity, it is closed. 
Assume $b=\sup J < \infty$. Then, by Picard-Lindelöf there exists an open interval containing $b$, say $\tilde J$, such that $f_1$ agrees with the 0-function on $\tilde J$, which contradicts that $J$ is maximal. That is, the assumption $b<\infty$ is wrong.
A: By contradiction.  First, suppose $f(x)\ne 0$ for some $x>0.$ For $x\geq 0,$ for brevity let $F(x)=\sup_{0\leq y\leq x}|f(y)|=\sup_{0\leq y\leq x}|f(y)-f(0)|.$
Let $r=\sup \{x\geq 0: F(x)=0\}=\sup \{x\geq 0:\forall y\in [0,x]\;(f(y)=0)\}.$
For $x> r:$  Since $f(t)=0$ for $t\in [0,r],$  we have $F(x)=\sup_{r\leq y\leq x}|f(y)|=\sup_{r\leq y\leq x}|f(y)-f(r)|.$ Therefore $$F(x)=   \sup_{r\leq y\leq x}|\int_r^yf(t)^2dt|\leq$$ $$\leq    \sup_{r\leq y\leq x}\int_r^y F(x)^2dt=$$ $$=   \sup_{r\leq y\leq x} (y-r)F(x)^2=$$ $$=(x-r)F(x)^2.$$ So $x>r\implies F(x)\leq (x-r)F(x)^2.$  
Now by def'n of $r,$ for every $x>r$ there exists $x'\in (r,x)$ with $f(x')\ne 0,$ so if $x>r$ then $F(x)\geq |f(x')|>0.$
So for $X>r$ we have  $0<F(x)\leq (x-r)F(x)^2,$ implying $\frac {1}{x-r}\leq F(x).$ 
So $\lim_{x\to r^+}F(x)=\infty.$ This is impossible  because $\sup \{F(x): r< x\leq r+1\}\leq F(r+1)<\infty. $
Second, the case where $f(x)\ne 0$ for some $x<0$ is handled similarly.  
A: Consider the differential equation (y=f(x)) $y'=y^2$.  The solution is $y=\frac{1}{c-x}$ where c is an arbitrary constant.  If f(0)=0, then c must be infinite, so that f(x) is identically 0.
