Prove $f $ is identically zero if $f(0)=0$ and $|f'(x)|\le|f(x)|$ $f:\Bbb R \to\Bbb R $ is differentiable, $f(0)=0$ and $|f'(x)|\le|f(x)|$ for all $x$ then prove $f$ is identically zero.
I tried to use mean value theorem and end up in $|f(x)|\le |x||f(c)|$ for some constant $c$ which depends on $x$. Can you help me little bit...
 A: First, I will show that $f$ is identically zero on $[0,1]$.
Suppose that $|f|$ attains a maximum at a point $m$, which is guaranteed by the Extreme Value Theorem. We can assume that $m < 1$, since 
$|f(1)| \le |f(c)|$ for some $c \in (0,1)$.
It follows that 
$|f(m)| \le m |f(c_m)|$, where $0<c_m < m$.
If $|f(c_m)|\ne 0$, then we have $|f(m)| < |f(c_m)|$, which contradicts the maximality of $|f(m)|$. Therefore, $f(c_m) = 0$, and $f(m) = 0$, and $f$ must be identically zero on $[0,1]$.
Now, you can apply the mean value theorem to obtain $|f(x)| < |x-1| |f(c)|$ for $c$ in $(1,x)$, and apply the same argument to $[1,2]$, then to $[2,3]$, etc.
Note that same argument can be used to $[-1,0]$, $[-2,-1]$, then to $[-3,-2]$, etc.
A: Iterate the estimate. Fix $x = x_0$ with $\lvert x\rvert < 1$, then for every $n \geqslant 1$, there is an $x_n$ (with $0 < \lvert x_n\rvert < \lvert x_{n-1}\rvert$) such that $$\lvert f(x)\rvert \leqslant \lvert x\rvert^n\cdot \lvert f(x_n)\rvert.$$ Deduce $f \equiv 0$ on $(-1,1)$. Use a similar argument to deduce $f \equiv 0$ on $\mathbb{R}$.
