Proving an estimate on a disk 
Suppose that $f$ is analytic on this unit disc $D$, and $$f(0)=f'(0)=0,\,\,\,\text{and}\,\,\, \lvert \,f'(z)\rvert\le 1\,\,\,\text{for all}\,\,\,z \in D.$$  Prove that 
  $$
\lvert \,f'(z)\rvert\le\frac{\lvert z\rvert^2}{2}\quad \text{for every}\,\,\, z \in D.
$$
  For which functions $f$ can equality hold in this estimate at some $z\neq 0$.

Since $f$ is analytic on the disk we know that $f(z)-f(0)=\int_{0}^zf'(\zeta)\,d\zeta$.  Since $f(0)=0$, we have that $f(z)=\int_{0}^{z}f'(\zeta)\,d\zeta$.  
Let $\gamma(t)=zt$. Now we have $f(z)=\int_\gamma f'(\zeta)\,d\zeta=\int_{0}^{1}f'(\gamma(t))\gamma'(t)dt=\int_{0}^{1}f'(zt)z\,dt.$
Then $|f(z)|\leq \left|\int_{0}^{1}f(zt)z\,dt \right|\leq \int_{0}^{1}|f(zt)||z|\,dt$.  Since $|f'(zt)|\leq 1$, then $|f'(zt)||z|\leq |z|$.  So now we have $|f(z)|\leq \left|\int_{0}^{1}f(zt)z\,dt \right|\leq \int_{0}^{1}|z|^2\,dt$
As you can see...I'm not exactly getting the estimate of $|z|^2/2$.  I'm not sure why.

Hint. Since $|f'(z)|\le 1$, using Schwarz's Lemma we obtain that $|f'(z)|\le |z|$, and finally
$$
f(z)=\int_0^1\frac{df(zt)}{dt}dt=\int_0^1zf'(zt)\,dt,
$$
and thus
$$
|f(z)|\le \int_0^1|z|\, |f'(zt)|\,dt\le \int_0^1|z|\, |z t|\,dt=|z|^2
\int_0^1 t\,dt=\frac{|z|^2}{2}.
$$
fix the gaps.
