$f:[0,1] \to \mathbb{R}$ is differentiable and $|f'(x)|\le|f(x)|$ $\forall$ $x \in [0,1]$,$f(0)=0$.Show that $f(x)=0$ $\forall$ $x \in [0,1]$ $f:[0,1] \to \mathbb{R}$ is differentiable and $|f'(x)|\le|f(x)|$ $\forall$ $x \in [0,1]$,$f(0)=0$.Show that $f(x)=0$ $\forall$ $x \in [0,1]$
I used the definition of derivative:
$f'(x)=\left|\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\right| \le |f(x)|$
Now, checking differentiability at $x=0$,
$\left|\lim_{h \rightarrow 0} \frac{f(h)}{h}\right| \le 0$ which should give a contradiction as modulus of something is negative So, $|f(h)| \le 0 \implies f(x)=0$ But I am not confident and believe that something is wrong..please clarify
 A: Since $f$ is continuous on the closed and bounded interval $[0,1]$ there is a constant $c$ so $$|f(x)|\le c\quad(x\in[0,1]).$$
So $|f'(x)|\le c$. Since $f(0)=0$ this implies $|f(x)|\le cx$. 
So $|f'(x)|\le cx$, hence $|f(x)|\le cx^2/2$.
And so on; by induction you see $$|f(x)|\le cx^n/n!.$$
This implies $|f(x)|\le 1/n!$ for all $n$, so $f(x)=0$.
A: Here's another approach, using the Mean Value Theorem. Fix a small $\epsilon\in(0,1)$ and define $M(\epsilon):=\sup\{|f(x)|: 0\le x\le 1-\epsilon\}$. Suppose that $M(\epsilon)>0$, and let $x_0$ be the smallest point $x\in (0,1-\epsilon]$ at which $|f(x)|$ attains the value $M(\epsilon)$. By the MVT there exists $\xi\in(0,x_0)$ such that $f(x_0)=f(x_0)-f(0)=x_0\cdot f'(\xi)$. We then have 
$$
M(\epsilon)=|f(x_0)|=x_0|f'(\xi)|\le x_0|f(\xi)|\le x_0\cdot M(\epsilon)\le(1-\epsilon)M(\epsilon).
$$
 This contradicts the assumption that $M(\epsilon)>0$. Thus $M(\epsilon)=0$ for all $\epsilon\in(0,1)$.
A: Let $f(x_0)=M$, $f(x_1)=m$, a maximum and minimum respectively, and $c=max\{|m|,|M|\}$. Then, $|f(x_i)|= c$ for some $i \in \{0,1\}$. So, by MVT, $|f(x_i)-f(0)|= x_i|f'(d)|$ for some $d \in (0,x_i)$. So, $|f(x_i)|\leq x_i|f(d)|\leq x_i|f(x_i)|$. This implies $|f(x_i)|=0$, that is $f(x)=0 \forall x\in [0,1]$.
Is this correct? Sorry for bad english :(
