Prove that |f| is bounded by a quotient

[docnet]

Homework Statement

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Relevant Equations

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I just spent 15 minutes to re-type all the Latex again because I lost everything while editing. why does this happen?? This is a huge waste of time.

$f$ is entire so $f$ is holomorphic on $\mathbb{D}∪C$. Also, $\mathbb{D}∪C$ is a connected set. By the maximum principle, $f$ restricted to $\mathbb{D}∪C$ attains its maximum on $C$. With this, $|f|≤1$ on $C$ leads to $|f|≤1$ in $\mathbb{D}∪C$

We consider $\frac{z−a}{−\bar{a}z+1}$ The corresponding matrix is $$\begin{bmatrix}1&−a\\−\bar{a}&1\end{bmatrix}$$ The determinant of the matrix is $$1−|a|≠0$$ nonzero because $|a|<1$. Which means $\frac{z−a}{−\bar{a}z+1}$ is a Möbius transformation.

We decompose the transfomation $\frac{z−a}{−\bar{a}z+1}$

into the compositions of simple transformations $f_4(f_3(f_2(f_1(z))))=−\frac{1}{\bar{a}}+\frac{e}{z-\frac{1}{\bar{a}}},e=\frac{|a|−1}{\bar{a}^2}$ where

$f_1=z−\frac{1}{\bar{a}}$ or translation by $-\frac{1}{\bar{a}}$

$f_2=\frac{1}{z}$ or inversion and reflection with respect to the real axis

$f_3=ez=\frac{|a|−1}{\bar{a}^2}z$ or homothety and rotation

$f_4=z-\frac{1}{\bar{a}}$ or translation by $-\frac{1}{\bar{a}}$

questions for everyone:

1. is this is a correct way to prove the original problem?

2. Is the composition supposed to take the unit disk into itself? if so, how do we prove this, and how do we use this to show that $|\frac{z−a}{−\bar{a}z+1}|\geq 1$? (if this is a correct way to prove the question)

3. It isn't immedietly obvious that it should map the unit disc onto itself, but if so, $f_3$ must un-do the inversion done by $f_2$, and the translations by $f_1$ and $f_4$ must cancel out. does anyone have any smart explanations for the transformations in common everyday lingo? thank you.

 

[Infrared]

2. Is the composition supposed to take the unit disk into itself? if so, how do we prove this, and how do we use this to show that $|\frac{z−a}{−\bar{a}z+1}|\geq 1$? (if this is a correct way to prove the question)

I think you mean for the inequality to go the other way around (what if $z=\alpha$), but yes it's otherwise correct. You can compute $|z-\alpha|^2=|z|^2-2\text{Re}(\overline{\alpha}z)+|\alpha|^2$ and $|1-\bar{\alpha}|^2=1-2\text{Re}(\overline{\alpha}z)+|\alpha z|^2.$ Do you see why the first is bounded by the second?

You still have a little bit of work to do after checking this.

 

[docnet]

the first is strictly bounded by the second because
$$1+x^2y^2-x^2-y^2>0, \forall x\in \mathbb{D}$$
we know this is true because if we let $f=1+x^2y^2$ and $g=x^2+y^2$, there are no solutions for $f=g$ restricted to $\mathbb{D}$.

I'm confused by the logic though, by showing that $a\leq 1$ and $b\leq 1$, this doesn't lead to $a \leq b$ without doing more work. that's why I was trying to show $b\geq 1$ leading to $a \leq b$. but I guess this would not be true since it maps the unit disc onto itself??

 

[docnet]

@Infrared could you please drop another hint? I am not making any progress.

 

[Infrared]

Are you familiar with the Schwarz lemma?