Let $f:[0,1] \to \mathbb{R}$ be continuous, differentiable on $(0,1)$ and $f(0) =0$ with $|f'(x)|\leq c|f(x)|$ ($c>0$)for each $x \in (0,1).$ Show that $f=0.$
I am aware of the several solutions to this problem on this site. My question is: Is the following attempt correct?
Let $\varepsilon>0$ be given. By uniform continuity of $f$ on $[0, 1]$ for $x, y \in [0,1],$ there exists $\delta >0$ such that $|x-y|<\delta$ implies $|f(x)-f(y)|<\frac{\varepsilon}{c}.$ Then for $x\in (0, \delta)$ we have: $|f'(x)|\leq c|f(x)|< \varepsilon.$ Since $\varepsilon$ was arbitrary, $f'(x)=0$ for $x \in (0, \delta).$ This implies that for $x\in (0, \delta)$, $f(x)= k$, for some constant $k$ but since $f(0)=0,$ by continuity we must have $k=0.$ By continuity at $x=\delta$ we also have $f(\delta)=0.$
Now choose $x\in [\delta, 2\delta).$ Since $|x-\delta|<\delta,$ once again we have $|f'(x)|\leq c|f(x)|<\varepsilon$ so by the same argument above $f(x)=0$ for $x\in [\delta, 2\delta].$ Continuing like this for $n$ steps, where $n =\lceil \frac{1}{\delta}\rceil$ we can show that $f(x)=0$ for $x \in [(i-1)\delta, i\delta]$ for $i=1,\ldots, n.$ and hence $f(x)=0$ for every $x \in [0,1].$
Is there a problem with this argument? Thanks in advance.
Yes, there is a problem with the argument. You've chosen your $\delta$ given a value of $\varepsilon > 0$. This value of $\delta$ will (in general) change depending on the value of $\varepsilon$. So, while you can always find some $\delta_\varepsilon$ such that $$x \in (0, \delta_\varepsilon) \implies |f'(x)| \le c|f(x)| < \varepsilon,$$ the interval $(0, \delta_\varepsilon)$ will shrivel up as you make $\varepsilon$ smaller. There may be no one point $x > 0$ such that $|f'(x)| = 0$, using this argument.