- #1

joypav

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$$\vert{f(z)}\vert\le{M\vert{\frac{z-\alpha}{1-\overline{\alpha}z}}\vert}$$

What I have:

Let $g(z)=f((1-\overline{\alpha}z)z+\alpha)$. Then g is analytic on the disc $\vert{z}\vert<1$ and $g(0)=f(\alpha)=0$.

If $\vert{f(z)}\vert\le{M}$ if $\vert{z}\vert<1$, then $\vert{g(z)}\vert\le{M}$ if $\vert{z}\vert<1$

1. $\vert{f((1-\overline{\alpha}z)z+\alpha)}\vert=\vert{g(z)}\vert\le{M\vert{z}\vert} \rightarrow \vert{f(z)}\vert\le{M\vert{\frac{z-\alpha}{1-\overline{\alpha}z}}}\vert$

(By substituting $z=\frac{z-\alpha}{1-\overline{\alpha}z}$)

2. Show that $g$ does satisfy $\vert{g(z)}\vert\le{M\vert{z}\vert}$.

Let $h(z)=\frac{g(z)}{z}$. Then h(z) is analytic on $\vert{z}\vert<1$, (removable singularity at z=0). Then by the Max Modulus Theorem, $\vert{h(z)}\vert$ attains its max somewhere on the boundary of $\vert{z}\vert<1$. Call this max M. Then,

$$\forall{z}, \vert{h(z)}\vert\le{M} \rightarrow \frac{\vert{g(z)}\vert}{\vert{z}\vert}\le{M} \rightarrow \vert{g(z)}\vert\le{M\vert{z}\vert}$$

Then, show that

$$M\vert{\frac{f(z)-f(z_0)}{M^2-\overline{f(z_0)}f(z)}}\vert \le{\vert{\frac{z-z_0}{1-\overline{z_0}z}}}\vert$$

for all $z, z_0$ in $w:\vert{w}\vert<1$.I'm just wondering if my first part is correct. If so, I assume I'll utilize the first to prove the second?