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

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