cambrassk3

2022-07-08

Let $f(z)$ be a rational function in the complex plane such that $f$ does not have any poles in $\{z:\mathrm{\Im}z\ge 0\}$.

Prove that $sup\{|f(z)|:\mathrm{\Im}z\ge 0\}=sup\{|f(z)|:\mathrm{\Im}z=0\}$.

Let ${\mathrm{\Gamma}}_{r}$ be a half circle counter such that ${\mathrm{\Gamma}}_{r}={\mathrm{\Gamma}}_{{r}_{1}}+{\mathrm{\Gamma}}_{{r}_{2}}$ when ${\mathrm{\Gamma}}_{{r}_{1}}=\{z:\mathrm{\Im}z=0,|z|=r\}$, ${\mathrm{\Gamma}}_{{r}_{2}}=\{z:\mathrm{\Im}z>0,|z|=r\}$. Using the Maximum modulus principle on the insides of ${\mathrm{\Gamma}}_{r}$, $|f|$ Gets is maximum value on ${\mathrm{\Gamma}}_{r}$. As r gets bigger if |f| got it's maximum on ${\mathrm{\Gamma}}_{{r}_{2}}$ than it's still smaller the the value on ${\mathrm{\Gamma}}_{r+1}$ which does not contain ${\mathrm{\Gamma}}_{{r}_{2}}$. I would like a hint on how to proceed.

Prove that $sup\{|f(z)|:\mathrm{\Im}z\ge 0\}=sup\{|f(z)|:\mathrm{\Im}z=0\}$.

Let ${\mathrm{\Gamma}}_{r}$ be a half circle counter such that ${\mathrm{\Gamma}}_{r}={\mathrm{\Gamma}}_{{r}_{1}}+{\mathrm{\Gamma}}_{{r}_{2}}$ when ${\mathrm{\Gamma}}_{{r}_{1}}=\{z:\mathrm{\Im}z=0,|z|=r\}$, ${\mathrm{\Gamma}}_{{r}_{2}}=\{z:\mathrm{\Im}z>0,|z|=r\}$. Using the Maximum modulus principle on the insides of ${\mathrm{\Gamma}}_{r}$, $|f|$ Gets is maximum value on ${\mathrm{\Gamma}}_{r}$. As r gets bigger if |f| got it's maximum on ${\mathrm{\Gamma}}_{{r}_{2}}$ than it's still smaller the the value on ${\mathrm{\Gamma}}_{r+1}$ which does not contain ${\mathrm{\Gamma}}_{{r}_{2}}$. I would like a hint on how to proceed.

potamanixv

Beginner2022-07-09Added 15 answers

To develop the comment of Daniel Fischer. Let $T(z)={\displaystyle \frac{i-z}{1-iz}}$. Then T is a conformal mapping of D(0,1) onto the upper half plane $\{z:\mathrm{\Im}z>0\}$. $f\circ T$ is a rational function which is analytic in D(0,1) and continuous on $\overline{D(0,1)}$, according to the assumption. So the maximum of $|f\circ T|$ on $\overline{D(0,1)}$ is attained on the boundary of the disk that is

$\underset{D(0,1)}{sup}|f\circ T|=\underset{C(0,1)}{sup}|f\circ T|$

or equivalently

$\underset{\{z:\mathrm{\Im}z>0\}}{sup}|f(z)|=\underset{\{z:\mathrm{\Im}z=0\}}{sup}|f(z)|.$

Because $T(D(0,1))=\{z:\mathrm{\Im}z>0\}$, and $T(C(0,1)\setminus \{-i\})=\{z:\mathrm{\Im}z=0\}$.

$\underset{D(0,1)}{sup}|f\circ T|=\underset{C(0,1)}{sup}|f\circ T|$

or equivalently

$\underset{\{z:\mathrm{\Im}z>0\}}{sup}|f(z)|=\underset{\{z:\mathrm{\Im}z=0\}}{sup}|f(z)|.$

Because $T(D(0,1))=\{z:\mathrm{\Im}z>0\}$, and $T(C(0,1)\setminus \{-i\})=\{z:\mathrm{\Im}z=0\}$.

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