Problem is from a course in Real Analysis from the chapter on differentiation.
Problem is as follows:
Let $f:[0,1]\mapsto \mathbb{R}$ be differentiable and $f(0)=0$
Suppose $|f'(x)|\leq|f(x)|$ for all $x\in[0,1]$
Show that for all $x\in[0,1]$, $f(x)=0$.
My proof is as follows:
Suppose $\exists a\in[0,1]$ such that $f(a)>0$. Since [0,1] is compact, by the Extreme Value Theorem, we know there exists some $b\in[0,1]$ such that $|f(b)| \geq |f(x)|$ for all $x\in[0,1]$. Now consider the interval $[0,b]$. By the Mean Value Theorem, we know $\exists c\in(0,b)$ such that $|f'(c)| = |\frac{f(b)}{b}|\geq|f(b)|\geq|f(c)|$ since $b<=1$. Thus we have a contradiction. We can repeat a similar proof if $\exists a\in[0,1]$ such that $f(a)<0$. Thus for all $x\in[0,1]$, $f(x)=0$.
Can I check, if there are any errors in my proof, or did I miss out on anything?