$ f(x):[0,1] \to R $ is continuous on $[0,1]$ & differentiable on $(0,1)$ such that for all $x\in(0,1)$
$$|f'(x)|\le c |f(x)|$$ for some $c\in (0,1). $ If $f(0)= 0$, prove that $f(x)=0 $ on whole $[0,1]$
Here what I have done till now. Using Mean Value Theorem, we get $|f(x)-f(y)|= |x-y||f'(\Omega)|\le c|x-y||f(\Omega)|$ for some $\Omega\in(x,y)\subset[0,1]$.
As f is defined on compact set, so it has maxima and minima in that set. Let $M$ be maximum value of f(x) at $x_o$.
So, our inequality gives us $f(x_o) =M < f(x) + \delta |f(\mu)|$ where we choose $|x-x_o|<\delta/c$ where $\mu\in(x,x_o)$ or $(x_o,x)$ whatever suitable.
I am stuck here and not been able to proceed further. I feel like an argument like this may work : - we can choose $\delta$ as small as possible and so equality will hold only if function must have value $M$ in an interval. As $f(0)=0$ so $M=0$
I will be grateful, if someone can provide different solutions.
As you started: $[0,1]$ is compact and $|f|$ is continuous, so let $M=\max_{x\in[0,1]}|f(x)|$ and $x_0\in[0,1]$ with $|f(x_0)|=M$. By the Mean Value Theorem, there exists $x_1\in(0,x_0)$ such that $f(x_0)-f(0)=(x_0-0)f'(x_1)$. As $f(0)=0$ and $|x_0|<1$, we get $$ M=|f(x_0)-f(0)|=|x_0|\cdot |f'(x_1)|\le |x_0|\cdot c\cdot |f(x_1)|\le |x_0|\cdot c\cdot M.$$ If $M>0$, this implies $|x_0|\cdot c\ge 1$, but clearly $0<x_0<1$ and we are given that $0<c<1$. We conclude that $M=0$.