Let $f$ be such that it satisfies $f'(t) = h(t) \cdot f(t)$ for all $t$, where $h(t)$ is an analytic function in a neighborhood of $0$. I would like to show that $f(0) = 0 \implies f(t) = 0$ for all $t$.
Initially, I thought I could show this by showing that all the $n$-th derivatives $f^{(n)}(0)$ are $0$ (because they are multiples of $f(0)$). However, then I came to know of "flat functions", which would make the above argument insufficient.
My hunch is that the statement should be true, but I am not sure. Any help here would be appreciated.
In a neighborhood of $0$, the function $h$ has an anti-derivative $H$. Consider $f(t)e^{-H(t)}$. Check that its derivative is $0$. So it a constant $c$ and we get $f(t)=ce^{H(t)}$. Put $t=0$ to see that $c=0$. Hence, $f(t)=0$ in a neighborhood of $0$.