Holomorphic function on an annulus
up vote
3
down vote
favorite
Let $f$ be a holomorphic function on the set $U={z in mathbb{C}: 1 leq |z| leq pi }$ . Assume that $max_{|z|=1}|f(z)| leq 1$ and $max_{|z|=pi}|f(z)| leq pi^{pi}$ . How to prove that $max_{|z|=e}|f(z)| leq e^{pi}$ ? Usually in such exercises one considers $g(z)$ of the form $alpha z^nf(z)$ where $alpha$ and $n$ are such that our assumptions on $f$ gives $|g(z)| leq 1$ on both circles being the boundary of $U$ . However it is not guarantedd that $n$ would be integer thus in general we don't get a holomorphic function $g$ for which we could apply the maximum principle. If $n$ happens to be rational one can overcome this difficulty by considering $(alpha z^n f(z))^q$ where $n=frac{p}{q}$ . However in our example $n$ turns out to be $pi$ and by considering $g(z)=f(z)/z^{pi}$ we don...