Supremum of a linear fractional function

Question

Let $(x_1,x_2,x_3)\in \mathbb{C}^3$. Let $\mathbb{D}$ be the open unit disc in $\mathbb{C}$. Define a linear fractional function as $\phi(z)=\frac{x_3z-x_1}{x_2z-1}$, for $z\in \mathbb{C}$. Can you tell why if $|x_2|<1,$ then $\sup_{z\in\mathbb{D}}|\phi(z)|=\frac{|x_1-\bar{x_2}x_3|+|x_1x_2-x_3|}{1-|x_2|^2}$?

Answer 1

Since dealing with indices can obscure the argument, let's call $x_3=a,x_1=b, x_2=c$ and notice that $\frac{az-b}{cz-1}-\frac{a\bar c-b}{c\bar c-1}=\frac{bc-a}{c\bar c-1}\frac{z-\bar c}{cz-1}$.

But now if $|c|<1$ we know that $\frac{z-\bar c}{cz-1}$ is a Mobius automorphism of the unit disc onto itself so it sends the unit circle onto itself also, hence $|\frac{z-\bar c}{cz-1}|\le 1$ in the closed unit disc with equality everywhere on the unit circle.

By the triangle inequality, this means:

$|\frac{az-b}{cz-1}| \le |\frac{a\bar c-b}{c\bar c-1}|+|\frac{bc-a}{c\bar c-1}|$ on the closed unit disc and we actually have equality for the generally unique $z=e^{i\theta}$ for which $\frac{z-\bar c}{cz-1}=e^{i \psi}$ makes $\frac{a\bar c-b}{c\bar c-1}$ and $e^{i \psi}\frac{bc-a}{c\bar c-1}$ have the same argument (obviously any $z$ on the unit circle would work if $a\bar c=b$ or $bc=a$, the first case making the original function a scaled Mobius transform, the second making it constant).

So we get that our supremum on the open unit disc is a maximum on the closed disc attained as above and with value $\frac{|a\bar c-b|+|bc-a|}{|c\bar c-1|}$, which is the required answer once we notice that $|c\bar c-1|=1-|c|^2$ since $|c|<1$ and then go back to the original variables with indices!