I am considering holomorphic functions $g(z)$ on the unit disk such that $|g(z)| \leq M$ for some $M>0$ and also for some real $a < 1$ we have $$\int_0^a |g(z)|^2 \frac{dz}{1 - z^2} = 1$$ I want to show that for all $d > 0$, there is some $C>0$ such that $$\int_{z_1}^{z_2} |g(z)|^2 \frac{dz}{1 - z^2} \geq C$$ for any interval $[z_1, z_2] \subseteq [0,a]$ with $z_2 - z_1 \geq d$.
I think I should be using a compactness argument for this. But more importantly, I need a lower bound on $C$ in terms of the other parameters, which might be a lot more difficult.
Thanks :)