Prove $f(x) = 0 $for all $x \in [0, \infty)$ when $|f'(x)| \leq |f(x)|$ mathematicians! I want to ask to all wise people about a problem I met at the quiz to obtain some ideas. The problem is following. It may not be accurate since the problem is dependent on my memory 
Let $f : [0, \infty) \rightarrow \mathbb{R}$ and $f$ is differentiable. $|f'(x) |\; \leq\  |f(x)|$ for all $x \in [0, \infty)$. $f(0) = 0$. Then prove that $f(x) = 0$ for all $x \in [0,\infty)$.
At the quiz, I applied the Mean Value Theorem and the Cauchy-Schwarz inequality. 
For any $x$, $|f(x)| = |\{f(x) - f(0)\}(x-0)| = |f'(c_{1})(x)|$ for some $c_{1} \in (0,x)$, and then 
$$|f'(c_{1})(x)| \leq |f'(c_{1})||x| \leq |f(c_{1})||x|$$ by Cauchy-Schwarz and given assumption. By doing so consecutively, we can obtain the inequality that 
$$
|f(x)| \leq |f(c_{n})||x||c_{1}|\ldots |c_{n-1}|
$$
for any $x \in [0,\infty)$. As $\{c_{i}\}, \; i\in \mathbb{N}$ is decreasing sequence and $f$ is continuous, $|f(c_{n})|$ can be arbitrary close to $0$ when $c_{n} \rightarrow 0$. Thus the inequality above can be bounded by $\epsilon$.
I think my answer is right, but a bit uncertain. Could you give me a certainty? Thank you very much. 
 A: The function $|f|$ is continuous on $[0,\frac{1}{2}]$ , so it attains
a maximum value on it -call it $M$. Then $|f’(x)| \leq |f(x)| \leq  M$ for 
$x\in [0,\frac{1}{2}]$. Integrating, $|f(x)-f(0)| \leq M|x-0|$, i.e. $|f(x)| \leq Mx$
for $x\in [0,\frac{1}{2}]$. But then $|f(x)| \leq \frac{M}{2}$, on that interval, so
$M\leq \frac{M}{2}$ since $M$ is attained somewhere.
  Hence $M=0$.
We have shown $f=0$ on $[0,\frac{1}{2}]$. It is
  easy now to see that $f$ will be zero on each $[\frac{n}{2},\frac{n+1}{2}]$ by induction on 
  the integer $n$. So $f=0$ everywhere.
A: Your answer looks about right. 
Basically, the intuition is that for $f(x)$ to be anything above $0$, the derivative must be non-zero (so the function can go move up or down). But since the absolute value derivative is less than the absolute value of the function at all times and the function starts at $0$, the function is never given a reason to go up, cuz the derivative is stuck at $0$. So the function is constantly $0$.
A: If $f$ is not identically zero, let's say $f$ is positive for some $x$,
define $x_0=\inf\{x\in [0,\infty)|f(x) >0\}$. As
$$f'(x_0)=\lim_{h\to0}\frac{f(x_0+h)-f(x_0)}{h}
=\lim_{h\to0}\frac{f(x_0+h)}{h}
\leq f(x_0)=0
$$
and $f$ is continuous, there exist an interval
$(0,s)$, $s>0$, such for all $h<s$ we have
$f(x_0+h)/h\leq0$, that is $$f(x_0+h)\leq0.$$
OTOH there exist $t>0$ s.t. for all $h\in(0,t)$ we have $$f(x_0+h)>0.$$
Choose some $0<h<\min\{s,t\}$ to achieve a contradiction. 
In this prove there's no need to assume the integrability of $f'$.
