derivative bounded by a constant multiple of the function $f$ is differentiable on $[a,b]$, $f'(x) \leq A|f(x)|$ where $A$ is a non-negative constant.
If $f(a)=0$ show $f(x)=0, \forall x\in [a,b]$
I imagine the proof uses the Mean Value Theorem but I have not been able to get it to work.
I know $|f(x)|=|f'(c)|(x-a)$ where $c \in [a,x]$, so $|f(x)| \leq A\  |f(c)|(x-a)$ where $c\leq x$ And I guess I could sort of iterate this to keep getting a smaller $c$ but I don't see why it must go all the way to zero.
 A: \begin{align}
|f(x)| &= |f(x) - f(a)| = |f'(c)||x-a| \le A|f(c)||x-a| \\\
& = A|f(c) - f(a)||x-a| = A|f'(d)||d-a||x-a| \le A|f'(d)||x-a|^2 \\\
& \le |f(d)||A(x-a)|^2.
\end{align}
Now suppose there exists $c \in ]a,x[$ such that $|f(x)| \le |f(c)| |A(x-a)|^n$, by induction on $n$. Then
\begin{align}
|f(x)| & \le |f(c)||A(x-a)|^n = |f(c) - f(a)||A(x-a)|^n \\\
& = |f'(d)||d-a||A(x-a)|^n \le A|f(d)||d-a||A(x-a)|^n \\\
& \le |f(d)||A(x-a)|^{n+1}. 
\end{align}
Hence for every integer $n$ there exists $c_n$ such that $|f(x)| \le |f(c_n)||A(x-a)|^{n+1}$. Since $[a,b]$ is compact and $f$ is differentiable, $f$ is continuous, hence $|f|$ is continuous and attains a maximum $M$ over $[a,b]$, so that $|f(x)| \le |f(c_n)| |A(x-a)|^n \le M|A(x-a)|^n$. This means $f(x) = 0$ as long as $|x-a| < \frac 1A$, since the bound only depends on $n$, hence we can let it go to infinity. By continuity of $f$ you also get $f(x) = 0$ if $|x-a| \le \frac 1A$.
Since you can split the interval $[a,b]$ as $[a,a+\frac 1A]$, $[a+\frac 1A, a + \frac2A]$, $\dots$, $[a+\frac mA, b]$, you can repeat this process finitely many times and show that $f$ is zero over $[a,b]$. 
Hope that helps,
