# UMD Analysis Qualifying Exam/Jan09 Complex

## Problem 4

 Suppose that ${\displaystyle f,g\!\,}$ are analytic on ${\displaystyle \{|z|\leq 1\}\!\,}$ with ${\displaystyle g\neq 0\!\,}$ on ${\displaystyle \{|z|<1\}\!\,}$. Prove that ${\displaystyle |f(z)|\leq |g(z)|\!\,}$ for all ${\displaystyle z\in \{|z|=1\}\!\,}$ implies ${\displaystyle |f(0)|\leq |g(0)|\!\,}$

## Solution 4

### Define new function h(z)

Define ${\displaystyle h(z)={\frac {f(z)}{g(z)}}\!\,}$.

#### h is continuous on the closure of D

Since ${\displaystyle g\neq 0\!\,}$ on ${\displaystyle D\!\,}$, then by the Maximum Modulus Principle, ${\displaystyle g\!\,}$ is not zero in ${\displaystyle {\overline {D}}\!\,}$.

Hence, since ${\displaystyle f\!\,}$ and ${\displaystyle g\!\,}$ are analytic on ${\displaystyle {\overline {D}}\!\,}$ and ${\displaystyle g\neq 0\!\,}$ on ${\displaystyle {\overline {D}}\!\,}$, then ${\displaystyle h\!\,}$ is analytic on ${\displaystyle {\overline {D}}\!\,}$ which implies ${\displaystyle h\!\,}$ is continuous on ${\displaystyle {\overline {D}}\!\,}$

#### h is analytic on D

This follows from above

### Case 1: h(z) non-constant on D

If ${\displaystyle h\!\,}$ is not constant on ${\displaystyle {\overline {D}}\!\,}$, then by Maximum Modulus Principle, ${\displaystyle |h|\!\,}$ achieves its maximum value on the boundary of ${\displaystyle D\!\,}$.

But since ${\displaystyle |h(z)|\leq 1\!\,}$ on ${\displaystyle \partial D\!\,}$ (by the hypothesis), then

${\displaystyle |h(z)|\leq 1\!\,}$ on ${\displaystyle {\overline {D}}\!\,}$.

In particular ${\displaystyle |h(0)|\leq 1\!\,}$, or equivalently

${\displaystyle |f(0)|\leq |g(0)|\!\,}$

### Case 2: h(z) constant on D

Suppose that ${\displaystyle h(z)\!\,}$ is constant. Then

${\displaystyle |h(z)|=\left|{\frac {f(z)}{g(z)}}\right|=|\alpha |\!\,}$ where ${\displaystyle \alpha \in \mathbb {C} \!\,}$

Then from hypothesis we have for all ${\displaystyle z\in \{|z|=1\}\!\,}$,

${\displaystyle |f(z)|=|\alpha ||g(z)|\leq |g(z)|\ \!\,}$

which implies

${\displaystyle |\alpha |\leq 1\!\,}$

Hence, by maximum modulus principle, for all ${\displaystyle z\in D\!\,}$

${\displaystyle \left|{\frac {f(z)}{g(z)}}\right|=|\alpha |\leq 1\!\,}$

i.e.

${\displaystyle |f(z)|\leq |g(z)|\!\,}$

Since ${\displaystyle 0\in D\!\,}$, we also have

${\displaystyle |f(0)|\leq |g(0)|\!\,}$