# 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

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