real analysis - Show a differentiable function has value of zero

Suppose $g$ is differentiable over $[a,b], g(a)=0$, $|g'(p)|\le Q|g(p)|$ over $[a,b]$ for some constant $Q$, show that $g=0$ over $[a,b]$.

I was thinking about using the properties in the following link, Show a function whose derivative is bounded is also bounded in an interval
but not sure how to proceed. Thanks.

Answer

Fix $x_0\in [a,b]$ let $M_0=\sup |f(x)|$, $M_1=\sup |f'(x)|$ for $[a,x_0]$. For any such $x$, $$|f(x)|\leqslant M_1(x_0-a)\leqslant A(x_0-a)M_0$$ Hence $M_0=0$ if $A(x_0-a)<1.$ That is, $f=0$ on $[a,x_0]$. Proceed this process.