If $g$ is holomorphic on $H = \{z:|z|\le1\}$ and $|\alpha|<1$ show that:
$$(1-|\alpha|^2)|g(\alpha)|\le\frac{1}{2\pi}\int_0^{2\pi}|g(e^{i\theta})|d\theta$$
I think I have solved it however, I am not sure if it is entirely correct:
My solution:
Let us consider the integral:
$$\frac{1}{2\pi i}\oint_{|z|=1}g(z)\frac{1-\overline{\alpha}z}{z-\alpha}dz=\frac{1}{2\pi i}\oint_{|z|=1}\frac{g(z)}{z-\alpha}dz-\frac{1}{2\pi i}\oint_{|z|=1}\frac{g(z)\overline{\alpha}z}{z-\alpha}dz=g(\alpha)(1-|\alpha|^2)$$ $$|g(\alpha)(1-|\alpha|^2)|=\bigg|\frac{1}{2\pi i}\oint_{|z|=1}g(z)\frac{1-\overline{\alpha}z}{z-\alpha}dz\bigg|\le\frac{1}{2\pi}\oint_{|z|=1}|g(z)|\bigg|\frac{1-\overline{\alpha}z}{z-\alpha}\bigg||dz|=\frac{1}{2\pi}\int_0^{2\pi}|g(e^{i\theta})|d\theta$$
This is what I got, but I am not entirely sure if it is correct. Could anyone see if the solution is fine? Thank you.