UMD Analysis Qualifying Exam/Jan09 Complex

Problem 4

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

Solution 4

Define new function h(z)

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

h is continuous on the closure of D

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

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

h is analytic on D

This follows from above

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

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

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

$|h(z)|\le 1$ on $\bar{D}$.

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

$|f(0)|\le |g(0)|$

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 $\alpha \in ℂ$

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

$|f(z)|=|\alpha ||g(z)|\le |g(z)|$

which implies

$|\alpha |\le 1$

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

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

i.e.

$|f(z)|\le |g(z)|$

Since $0\in D$, we also have

$|f(0)|\le |g(0)|$