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If $f$ is entire and $|f|=1$ on $|z|=1$,then $f(z)=cz^n$ for some $c$.

First consider $g(z)=f(z)/\prod(z-a_i)/(1-\overline{a_i}z)$,where $a_i$ are zeros of $f(z)$.

Then I want to apply the maximum and minimum modulus theorem to argue that all $a_i$'s are zero. But what am I supposed to do? Do I need to first show that $g(z)$ is constant?

user45955
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1 Answers1

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  • Show that $|g(z)|=1$ if $|z|=1$, and that $g$ never vanishes on the unit disk.
  • Applying maximum modulus principle to $g$ and $1/g$ we get that $g(z)=c$ on the unit disk, where $|c|=1$.
  • So we can write $f(z)=c\prod_{j=1}^n\frac{z-a_j}{1-\bar a_jz}$, and we are reduced to show that $a_j=0$ for all $j$. As $f$ is assumed entire, we need to remove the singularities in $\frac 1{\bar{a_j}}$, which has modulus $>1$. These one can't be canceled unless $a_j=0$.

Note that we need the assumption entire, otherwise, $f$ could be of the form $f(z)=c\prod_{j=1}^n\frac{z-a_j}{1-\bar a_jz}$ where $|a_j|<1$, but not necessarily equal to $0$.

Davide Giraudo
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  • How can you show that $g$ never vanishes? And why does $1/\overline{a_j}$ have modulus>1? – user45955 Nov 04 '12 at 19:18
  • I meant on the unit ball. To see that, use the fact that the zeros are counted with their multiplicity. As $|\bar{a_j}|<1$, we have an upper bound for $\frac 1{|a_j|}$. – Davide Giraudo Nov 04 '12 at 20:26
  • There is at least one unremovable singularity if $a_j\neq 0$ for some $j$. Indeed, the expression of $f$ should be valid for $|z|>1$, and the limit $\lim_{z\to\frac{\prod_j{z-a_j}}{\bar{a_j}}$ wouldn't exist. – Davide Giraudo Apr 10 '13 at 14:51
  • Can you fix the TeX? I'm having trouble understanding what you meant. Thanks! – Max Apr 10 '13 at 14:57
  • Ok: $\lim_{z\to\frac 1{\bar{a_j}}}\frac{\prod_j(z-a_j)}{1-z\bar{a_j}}$ – Davide Giraudo Apr 10 '13 at 15:05