Does $f(0)=0$ and $\left|f^\prime(x)\right|\leq\left|f(x)\right|$ imply $f(x)=0$? Let $f:\mathbb{R}\to\mathbb{R}$ be a function such that $f(0)=0$ for all real numbers $x$, $\left|f^\prime(x)\right|\leq\left|f(x)\right|$. Can $f$ be a function other than the constant zero function? 
I coudn't find any other function satisfying the property. The bound on $f^\prime(x)$ may mean that $f(x)$ may not change too much but does it mean that $f$ is constant?
I thought for a while and found that $f^\prime(0)=0$ and by using mean value theorem, if $x\neq0$ then there's a real number $y$ between $0$ and $x$ such that $\left|f(x)\right|=\left|xf^\prime(y)\right|\leq\left|xf(y)\right|$. Anything further?

Continuing the idea that I mentioned after proposing the question:
Let's define  $y_1:=y$. By the mean value theorem there's a real number $y_2$  between $0$ and $y_1$ such that $\left|f(y_1)\right|\leq\left|y_1f^\prime(y_2)\right|\leq\left|y_1f(y_2)\right|$ so $\left|f(x)\right|\leq\left|xy_1f(y_2)\right|\leq\left|x^2f(y_2)\right|$. Continuing this way, we inductively conclude that for every positive integer $n$, there's a real number $y_n$ between $0$ and $x$ such that $\left|f(x)\right|\leq\left|x^nf(y_n)\right|$. So if $0<x<1$, using the fact that $f$ is bounded on a bounded interval, we take the limit of the right-hand side of the last equation as $n$ tends to infinity and conclude that $f(x)=0$. Since $f$ is continuous, we have $f(x)=0$ for $0\leq x\leq1$. Now, if $m$ is a positive integer, the function $g(x)=f(x+m)$ has the property $\left|g^\prime(x)\right|\leq\left|g(x)\right|$. This lets us to prove that $f(x)=0$ for $m\leq x\leq m+1$ inductively. The function $h(x)=f(-m+1-x)$ can be treated in the same manner and that allows us to prove $f(x)=0$ for every real number $x$.
