Upper bound on $\inf_{|z|=1}|f(z)|$ of holomorphic function $U\supset\overline{\Bbb D}\to\Bbb C$

Question

I am studying for an analysis final, and this is an old exam problem.

Let $\Bbb D$ be the open unit disc. Suppose that $f(z)=\sum_{n=0}^\infty a_nz^n$ is a holomorphic function defined on a neighborhood of the closed unit disc $\overline{\Bbb{D}}$. Show that if $f(z)$ has exactly $m$ zeros (counted with multiplicity) inside $\Bbb D$ then $$ \inf_{|z|=1}|f(z)| \le |a_0| + |a_1| + \dotsb + |a_m|. $$ I have given this problem some thought, but I am having trouble getting a foot-hold. Concepts that come to mind are the maximum modulus principle, Cauchy integral formula and estimates as well as the argument principle and the residue theorem. The $\inf$ is throwing me a bit. Any hints and suggestions are welcome, but please no full solutions.

Answer 1

Write $f=P+g$ where $P(z)=a_0+\dots+a_mz^m$. Proving by contradiction, suppose your inequality does not hold. Then apply Rouche's theorem to arrive at a contradiction.

Answer 2

Based on Professor Eremenko's suggestion, I believe I have a solution.

We have two cases. If $f(z) = a_0 + a_1z + \dotsb + a_mz^m$, then the conclusion follows with little work. Otherwise, there are terms in the power series of $f$ that involve powers of $z^k$ with $k > m$. Let $k_0 >m$ be the first integer such that $a_{k_0} \ne 0$. Hence, write $f = P + g$ with $P(z) = a_0 + a_1z + \dotsb + a_mz^m$ and suppose for the sake of contradiction that $\inf_{|z|=1}|f(z)| > |a_0| + |a_1| + \dotsb + |a_m|$.

Applying Rouché's theorem to the functions $f$ and $g$, we see that for $|z| = 1$, $$ |f(z) - g(z)| = |P(z)| \le |a_0| + |a_1| + \dotsb + |a_m| < |f(z)|. $$ Hence $f$ and $g$ have the same number of zeros in $\Bbb D$. However, $g(z) = z^{k_0}\cdot\sum_{n\ge0}a_{k_0+n}z^n$, hence $g$ has at least $k_0>m$ zeros in $\Bbb D$, a contradiction since we assumed that $f$ had only $m$ zeros in $\Bbb D$.