If $f(0)=0$ and $0\le f'(x)\le2f(x)$ for all $x\in(0,1)$, prove that $f\equiv0$. Let $f: [0,1]\to \Bbb R$ be a continuous function such that $f(0)=0$ and it's differentiable on $(0,1)$. In addition, we know that $0\le f'(x)\le2f(x)$. Prove that $f=0$.
I know that if the function $f$ is continuous on $[a,b]$, differentiable on $(a,b)$ and $f'(x) = 0$ on $(a,b)$, then $f$ must be a constant function on $[a,b]$.
But I do not quite know how to use the inequality.
 A: Since $f(0)=0$ and $f'(x)\ge 0$ on $(0,1)$, $f(x)\ge 0$ on $[0,1]$. Let $g(x)=f(x)e^{-2x}$ on $[0,1]$. Then $g$ is continuous on $[0,1]$ and differentiable on  $(0,1)$. On the one hand, $g(x)\ge 0$ on $[0,1]$. On the other hand, since $g(0)=0$ and $g'(x)=e^{-2x}(f'(x)-2f(x))\le 0$ on $(0,1)$, $g(x)\le 0$ on $[0,1]$. The conclusion follows.
A: It is true even if you ignore the assumption of being constant (which surely must be a typo.)
Note that since $f'(x) \geq 0$, $f$ is non-decreasing. Hence $f(x) \geq 0$ for $x \in [a,b]$. Also, since $f$ is differentiable, it is continuous.
Let $0 \leq a < b \leq 1$, then $f(b)-f(a) \leq \sup_{\xi \in (a,b)} f'(\xi) (b-a) \leq 2 f(b)(b-a)$. Now suppose $f(a) = 0$ and $b-a < \frac{1}{2}$. Then the inequality gives $f(b) \leq 2 f(b) (b-a)$, or $0 \leq f(b)(2(b-a)-1)$. Hence $f(b) = 0$.
Take $a=0,b<\frac{1}{2}$. Then the above shows that $f(b) = 0$ for $b \in (0,\frac{1}{2})$, and continuity shows $f(\frac{1}{2}) = 0$. Now apply the same argument with $a=\frac{1}{2}, b<1$, to get $f(1) = 0$, from which the desired conclusion follows.
