UMD Analysis Qualifying Exam/Jan09 Complex

Problem 4

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

Solution 4

Define new function h(z)

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

h is continuous on the closure of D

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

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

h is analytic on D

This follows from above

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

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

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

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

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

$|f(0)|\leq |g(0)|\!\,$

Case 2: h(z) constant on D

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

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

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

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

which implies

$|\alpha |\leq 1\!\,$

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

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

i.e.

$|f(z)|\leq |g(z)|\!\,$

Since $0\in D\!\,$, we also have

$|f(0)|\leq |g(0)|\!\,$