Supremum of supremum... I have $|f(z)|≤|f(0)|≤M_r$ where $M_r$ is the supremum of $|f(z)| : |z|=r$ Can I take the supremum of the inequality and conclude $M_r≤|f(0)|≤M_r$ $\sup\sup(|f(z)|)= \sup M

Supremum of supremum... I have $|f(z)|≤|f(0)|≤M_r$ where $M_r$ is the supremum of $|f(z)| : |z|=r$ Can I take the supremum of the inequality and conclude $M_r≤|f(0)|≤M_r$ $\sup\sup(|f(z)|)= \sup M_r=\text{?}$

I know what assignment this is. To be pedantic, you need to take a supremum over a set. Yes, you may conclude $M_r = |f(0)|$.

I don't see how this is relevant to a double sup? The sup of a constant is itself, maybe thats what you are trying to ask? If yes, then maybe your notation could use some cleaning up (and you'd get more for clarity) if you wrote $\sup_{z:|z|=r}$ instead of simply $\sup$.